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  <row Id="1" PostTypeId="1" AcceptedAnswerId="2" CreationDate="2015-01-20T18:35:59.493" Score="34" ViewCount="8537" Body="&lt;p&gt;Is it possible to &quot;pump&quot; a powder the same way liquids can be pumped?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what are the challenges? If not, what are some alternatives?&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="148" LastEditDate="2015-01-21T08:28:32.950" LastActivityDate="2018-11-14T01:49:37.590" Title="Is it possible to &quot;pump&quot; a powder?" Tags="|pumps|liquid|fluid-mechanics|" AnswerCount="8" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2" PostTypeId="2" ParentId="1" CreationDate="2015-01-20T18:41:00.530" Score="29" Body="&lt;p&gt;I believe it's termed &quot;injecting&quot; not &quot;pumping&quot; but yes, it's certainly possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coal fired electric power plants grind the coal to a very fine dust and then blow the dust into the central burner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.netl.doe.gov/File%20Library/Research/Coal/major%20demonstrations/cctdp/Round3/bethstl.pdf&quot;&gt;This particular article&lt;/a&gt; is a summary of some DOE research for a Blast Furnace Granulated Coal Injection System Demonstration Project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;a href=&quot;http://www.nfpa.org/~/media/Files/Research/Research%20Foundation/foundation%20proceedings/david_snell.pdf&quot;&gt;this article&lt;/a&gt; discusses some of the hazards from a National Fire Protection Association perspective.  Coal dust having the particular dangerous property of being explosive and all that...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as specifics regarding danger, I suspect it will vary depending upon the material(s) involved in the powder.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-01-20T18:41:00.530" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3" PostTypeId="1" CreationDate="2015-01-20T18:41:20.310" Score="11" ViewCount="3007" Body="&lt;p&gt;What would be the best type of steel to specify? Would a heavy galvanization suffice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there specialty paints/coatings available for this particular application? If so, how long do they last?&lt;/p&gt;&#xA;" OwnerUserId="3" LastEditorUserId="33" LastEditDate="2015-10-18T18:40:48.110" LastActivityDate="2015-10-18T18:40:48.110" Title="What kind of protection should I give structural steel in a natatorium (swiming pool) environment?" Tags="|steel|structures|corrosion|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4" PostTypeId="1" AcceptedAnswerId="248" CreationDate="2015-01-20T18:41:51.527" Score="6" ViewCount="172" Body="&lt;p&gt;What is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Barometric_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;barometric height formula&lt;/a&gt; for a mixture of gases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem I can see, the curves describing the minimal energy (kinetic + gravitational/centrifugal) won't be any more a simple derivative, but a functional equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, this or similar equations could be useful in very separate areas (isotope separation, estimating high atmosphere components of planets, etc.) Despite I couldn't find anything with google about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is: we have a high gas mixture in constant g gravity and T temperature. We know the &lt;span class=&quot;math-container&quot;&gt;$p_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$p_2$&lt;/span&gt;, ... partial pressures of the gases on the &lt;span class=&quot;math-container&quot;&gt;$h=0$&lt;/span&gt;. Of course we know every relevant constants of the gases, too. What is their minimal-energy, isothermic, height-dependent partial pressure function &lt;span class=&quot;math-container&quot;&gt;$p_n(h)$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditDate="2020-01-04T20:57:17.173" LastActivityDate="2020-01-04T20:57:17.173" Title="Multi-gas barometric height formula in isotope separation devices?" Tags="|pressure|gas|nuclear-technology|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="6" PostTypeId="1" AcceptedAnswerId="8" CreationDate="2015-01-20T19:49:53.683" Score="12" ViewCount="2429" Body="&lt;p&gt;The United States have different rules about exactly how one obtains an engineering license, but the general process is the FE exam, a few years of work, and the PE exam. Suppose one then wishes to practice engineering in Germany. What are the legal requirements for doing so?&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2015-01-21T18:46:54.050" Title="What is the process of transferring an engineering license from the United States to Germany?" Tags="|licensure|international|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8" PostTypeId="2" ParentId="6" CreationDate="2015-01-20T20:27:48.367" Score="8" Body="&lt;p&gt;Unlike within the &lt;a href=&quot;http://en.wikipedia.org/wiki/Regulation_and_licensure_in_engineering&quot; rel=&quot;nofollow&quot;&gt;United States and the PE license&lt;/a&gt;, it doesn't appear that it's possible to receive an &lt;a href=&quot;http://en.wikipedia.org/wiki/European_Engineer&quot; rel=&quot;nofollow&quot;&gt;EUR ING&lt;/a&gt; license by &lt;a href=&quot;http://www.nspe.org/resources/licensure/resources/licensure-comity&quot; rel=&quot;nofollow&quot;&gt;direct comity&lt;/a&gt;.  EU Engineering licensing is handled by &lt;a href=&quot;http://www.feani.org/site/index.php&quot; rel=&quot;nofollow&quot;&gt;FEANI&lt;/a&gt;, and on their &lt;a href=&quot;http://www.feani.org/site/index.php?id=261&quot; rel=&quot;nofollow&quot;&gt;EUR ING&lt;/a&gt; page they state: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Application is open only to individuals if they are members of an engineering association represented in FEANI through a National Member&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But it may be possible to join one of the &lt;a href=&quot;http://www.feani.org/site/index.php?id=57&quot; rel=&quot;nofollow&quot;&gt;German FEANI members&lt;/a&gt; and see if they can support a reciprocity or comity process.&lt;sup&gt;1&lt;/sup&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;There are other member organization in other countries, but your question specifically mentioned Germany.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Digging a bit deeper, it does appear that there may be limited cases where an EUR ING will be awarded based upon sufficient experience as a professional engineer.  Take a look through the &lt;a href=&quot;http://www.feani.org/site/index.php?eID=tx_nawsecuredl&amp;amp;u=0&amp;amp;file=fileadmin/PDF_Documents/EUR_ING_Tittle/Guide_to_the_Register_FINAL_approved_GA_2013.pdf&amp;amp;t=1421941885&amp;amp;hash=f78dd0dcf1b44b41a9cc45e17cc6500d1f436621&quot; rel=&quot;nofollow&quot;&gt;FEANI EUR ING guide&lt;/a&gt;, in particular Section 5 which details the minimum requirements.  Most of the cases cover qualifications based upon an education received within an FEANI member country.  Section 5.4b deals with special circumstances and provides the potential pathway that you could receive an EUR ING license.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nevertheless it is possible to consider such alternative routes. Very strict procedures, however, have then to be followed, (see 7.1), and the applicant must have at least 15 years of Professional Engineering Experience recognized by FEANI.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it does appear to be possible, but again, it's not as simple as the comity process within the United States.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-01-21T18:46:54.050" LastActivityDate="2015-01-21T18:46:54.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9" PostTypeId="1" CreationDate="2015-01-20T20:31:25.093" Score="13" ViewCount="2273" Body="&lt;p&gt;I am currently having an issue where features of my machined parts are warping. I am using Alloy 7050 Type I and Type III in this particular case. Using a feeler gauge and a certified, calibrated granite slab, I am getting out-of-flatness measurements up to .009, which is significantly out of tolerance relative to my requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have decided to construct a statistically-designed experiment (&lt;a href=&quot;https://en.wikipedia.org/wiki/Design_of_experiments#Experimental_designs_after_Fisher&quot; rel=&quot;noreferrer&quot;&gt;design of experiments&lt;/a&gt;) to find which specific factors are playing a role, but I want to ensure that I am including all of the factors which may contribute.  I don't want to lead the conversation too much by posting what I already listed, but I am looking for others' opinions on what might cause this kind of warping/deflection.&lt;/p&gt;&#xA;" OwnerUserId="35" LastEditorUserId="368" LastEditDate="2015-04-27T16:25:03.123" LastActivityDate="2018-06-24T10:59:19.160" Title="In aluminum machining, what are the most common factors that might cause a machined component to have warping issues?" Tags="|mechanical-engineering|machining|aluminum|quality-engineering|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10" PostTypeId="1" CreationDate="2015-01-20T21:19:03.447" Score="3" ViewCount="281" Body="&lt;p&gt;I am currently considering turning my fascination with magnets into a diy activity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How feasible is it for a hobbyist to construct a working model of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Maglev&quot; rel=&quot;nofollow&quot;&gt;maglev&lt;/a&gt; device?&#xA;I'm looking to build something  really small for now,just a demonstrative prototype that can hover over a surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a rough idea of cost, time and materials. Is it something that can be attempted while working a full time job?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of any good starting points in terms of literature? &lt;/p&gt;&#xA;" OwnerUserId="42" LastEditorUserId="13974" LastEditDate="2017-12-02T15:58:39.070" LastActivityDate="2017-12-02T15:58:39.070" Title="Magnetic levitation" Tags="|electrical-engineering|magnets|electromagnetism|propulsion|linear-motors|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ClosedDate="2015-01-21T23:07:02.393" ContentLicense="CC BY-SA 3.0" />
  <row Id="12" PostTypeId="1" CreationDate="2015-01-20T21:45:26.513" Score="14" ViewCount="456" Body="&lt;p&gt;I am examining a watershed and determining runoff using the RTK method. I basically use a 'guess and check the hydrograph' method right now. &lt;a href=&quot;http://www.innovyze.com/products/infoswmm/calibrator.aspx&quot;&gt;I've seen calibrators&lt;/a&gt; that use a mutation and convergence criteria using root mean square but I've had limited success using them. How can I perform this analysis more efficiently than using my 'guess and check' method?&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="368" LastEditDate="2015-02-26T16:43:41.073" LastActivityDate="2022-08-05T13:00:30.060" Title="Determining RTK parameters for runoff analysis" Tags="|hydrology|water-resources|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13" PostTypeId="2" ParentId="1" CreationDate="2015-01-20T21:49:29.263" Score="11" Body="&lt;p&gt;You may also look into pneumatic conveying.  It involves &lt;a href=&quot;http://en.wikipedia.org/wiki/Fluidization&quot;&gt;fluidization of the particles&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simply uses pressure differential to move any bulk material. It's used a lot in industrial applications.  You can read a &lt;a href=&quot;http://www.nol-tec.com/answers-to-eight-common-questions.html&quot;&gt;summary of those systems, here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="104" LastEditDate="2015-01-21T02:48:11.267" LastActivityDate="2015-01-21T02:48:11.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14" PostTypeId="1" AcceptedAnswerId="142" CreationDate="2015-01-20T22:03:47.020" Score="18" ViewCount="2872" Body="&lt;p&gt;I think, here is the main problem the difference between the internal and the external temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in Saudi Arabia, in 50 C, a passive house needed probably much sophisticated planning as in Paris.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to the traditional cooling systems, in the second case is enough only to get a cooling system with bigger power. I think, they are much more scalable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it anyways possible?&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="46" LastEditDate="2015-01-24T19:49:44.740" LastActivityDate="2018-04-02T04:41:42.393" Title="How are passive houses made in very hot regions (like Saudi Arabia)?" Tags="|civil-engineering|architecture|cooling|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16" PostTypeId="1" AcceptedAnswerId="19" CreationDate="2015-01-20T22:16:03.560" Score="8" ViewCount="2951" Body="&lt;p&gt;Liquid hydrogen cooling could be some intermediate solution between helium and liquid nitrogen based cooling.  For example, some type of superconductors could be cooled by liquid H&lt;sub&gt;2&lt;/sub&gt;, and this could be much cheaper than the liquid He.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are their most characteristic applications in the industry?&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="63" LastEditDate="2015-01-20T22:58:57.157" LastActivityDate="2015-02-03T12:54:06.913" Title="Do liquid hydrogen-based cooling devices exist and what are they used for?" Tags="|superconduction|cryogenics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17" PostTypeId="2" ParentId="16" CreationDate="2015-01-20T22:31:40.093" Score="6" Body="&lt;p&gt;Gaseous hydrogen is often used to cool the large generators in power plants. &quot;Hydrogen’s low gas density, high specific heat, and high thermal conductivity&quot; are listed in the GE link as the major drivers behind its use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen-cooled_turbo_generator&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Hydrogen-cooled_turbo_generator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.control.com/thread/1267097548&quot; rel=&quot;nofollow&quot;&gt;http://www.control.com/thread/1267097548&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://powergen.gepower.com/plan-build/products/generators/hydrogen-cooled.html&quot; rel=&quot;nofollow&quot;&gt;https://powergen.gepower.com/plan-build/products/generators/hydrogen-cooled.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the matter of explosions - hydrogen is not particularly explosive without oxygen present. The goal, then, is to provide a system that keeps the hydrogen free from impurities (air) by sealing it off from the outside air. The purity of the hydrogen in the system is constantly monitored, and in an emergency the gas is evacuated by CO2. It's also worth noting that while helium is not explosive and very similar to H2 physically, it is too expensive to use as a cooling medium.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="45" LastEditDate="2015-02-03T12:54:06.913" LastActivityDate="2015-02-03T12:54:06.913" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18" PostTypeId="2" ParentId="3" CreationDate="2015-01-20T22:34:59.617" Score="4" Body="&lt;p&gt;Safety issues in a &quot;natatorium&quot; (swimming pool) environment include the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Chlorine and other high acid gases that can corrode the steel. Both galvanization and paints/coatings can help, until the weather wears the protection away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Water freezing and thawing. Basically, you need to keep water away from the pool at times when in it is likely to expand or contract through freezing and thawing, so that this doesn't happen. Mostly a problem for northern climes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Algae, molds, etc. You need to keep away organic products that will encourage them to grow, and create destructive acids or other compounds.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditDate="2015-01-20T23:36:38.757" LastActivityDate="2015-01-20T23:36:38.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19" PostTypeId="2" ParentId="16" CreationDate="2015-01-20T22:36:55.360" Score="13" Body="&lt;p&gt;The combustion chamber and nozzle of the Space Shuttle main engine were cooled with liquid hydrogen.  Liquid hydrogen was also the fuel.  It was used as a coolant before it was burned (&lt;a href=&quot;https://en.wikipedia.org/wiki/Regenerative_cooling_%28rocket%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;regenerative cooling&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/b/bc/Ssme_schematic.svg&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the &lt;a href=&quot;http://en.wikipedia.org/wiki/SABRE_%28rocket_engine%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;air breathing rocket engine SABRE&lt;/a&gt;.  Here too liquid hydrogen is used as fuel.  It's also used for liquefying the atmospheric oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PqrXj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the industry, one advantage that liquid helium and nitrogen have over hydrogen is that helium and nitrogen are inert and non-combustible.  Probably less corrosive too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;p.s.&lt;/em&gt; The question has a &lt;a href=&quot;/questions/tagged/superconduction&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;superconduction&amp;#39;&quot; rel=&quot;tag&quot;&gt;superconduction&lt;/a&gt; tag.  So, I don't know if this is the sort of application of liquid hydrogen cooling that the O.P. is looking for.  &lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:49.870" LastActivityDate="2015-01-31T06:30:22.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20" PostTypeId="1" AcceptedAnswerId="91" CreationDate="2015-01-20T22:41:57.420" Score="8" ViewCount="270" Body="&lt;p&gt;I'm making a robot for the Sparkfun AVC. I was curious if I could use only the knowledge about&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How the car is being steered at every time interval,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How many times the wheels have rotated,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;to get a general sense of where the car is. I would use computer vision to avoid immediate dangers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest problem is any slipping causing false counts.&lt;/p&gt;&#xA;" OwnerUserId="74" LastEditorDisplayName="user16" LastEditDate="2015-01-24T02:45:20.707" LastActivityDate="2015-03-23T23:47:36.393" Title="Knowing how many times the rear wheels have rotated, and the angle of the front wheels at every time interval, how do I find the location of a car?" Tags="|control-engineering|robotics|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21" PostTypeId="2" ParentId="1" CreationDate="2015-01-20T22:49:09.513" Score="2" Body="&lt;p&gt;You can pump both liquids and powders into a &quot;mold&quot; using &lt;a href=&quot;http://en.wikipedia.org/wiki/Injection_moulding&quot; rel=&quot;nofollow&quot;&gt;injection molding&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works more easily with liquids, but it can also be done with small particles such as powders that are injected into heated molds with sufficient force.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2015-01-20T22:49:09.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="22" PostTypeId="1" AcceptedAnswerId="83" CreationDate="2015-01-20T22:53:44.837" Score="7" ViewCount="10812" Body="&lt;p&gt;I recently had a project where I made an aluminum linkage (similar to a scissor lift) operated by linear actuators. Aluminum pieces were fabricated from 6061 &quot;standard&quot; aluminum stock, 6mm (0.25 inch) thick. The bars are at most 750x75mm (30x3&quot;). The application was mostly cosmetic (for a special effect) rather than lifting or exerting much force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used shoulder bolts with a nylon washer between the aluminum pieces to reduce friction, and standard zinc-plated steel washers on the exterior sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I encountered was tightening the bolt too much would result in too much friction, whereas too loose would result in one of the aluminum pieces coming slightly off the shoulder of the bolt, skewing, and causing similar problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be a better way to create a reliable pivot point that doesn't require constant adjustment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put together this illustration as an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m1Psy.png&quot; alt=&quot;Linkage Description&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="57" LastEditorUserId="33" LastEditDate="2015-10-18T18:56:36.563" LastActivityDate="2024-03-02T14:59:46.317" Title="What method can I use to make a pivot point in an aluminum linkage?" Tags="|bolting|aluminum|linkage|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="23" PostTypeId="1" CreationDate="2015-01-20T22:58:36.060" Score="10" ViewCount="174" Body="&lt;p&gt;I'm investigating comminution processes at work, and from my studies I know that there are different breakage models for different modes (Compression, attrition, fracture etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have only seen population models for ball mills in use. These models use time, size fraction and axial distance in the mill to specify the mass fraction of material at any given location in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if similar models exist for other comminution devices (Gyratory or Jaw crushers) and if so  what do they look like? &lt;/p&gt;&#xA;" OwnerUserId="79" LastEditorDisplayName="user6335" LastEditDate="2017-09-29T13:54:37.670" LastActivityDate="2018-02-26T17:34:59.047" Title="Recommended models for mineral liberation in crushers" Tags="|process-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="24" PostTypeId="2" ParentId="3" CreationDate="2015-01-20T22:58:46.440" Score="3" Body="&lt;p&gt;Most stainless steels contain nickel around 10%. It is combined normally with chrome and molybdenum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem, that a natatorium (i.e. swimming pool) is a highly corrosive environment, and most &quot;stainless&quot; steel alloy is a little bit, slowly, corroding here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ourHW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even aluminium gets a greyish-white surface layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of constructive materials, the situation is a little bit better. First, because they are mostly in the wall, and the concrete around them partially protects them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of freely available surfaces, plastic or painting is yet used.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2015-01-20T22:58:46.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="25" PostTypeId="2" ParentId="9" CreationDate="2015-01-20T23:07:59.383" Score="1" Body="&lt;p&gt;Compared to most other metals, aluminum is a relatively low performance material. It's useful mainly for low speed, low volume, low temperature variation, and &quot;coarse&quot; applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It tends to &quot;warp,&quot; or otherwise break down in high speed, high volume applications requirement &quot;fine&quot; performance from metals such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Injection_moulding&quot; rel=&quot;nofollow&quot;&gt;injection molding&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2015-01-20T23:07:59.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="26" PostTypeId="1" AcceptedAnswerId="112" CreationDate="2015-01-20T23:14:48.790" Score="7" ViewCount="859" Body="&lt;p&gt;As far I know, the depleted fuel cells are crushed, and solved in nitric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is coming after that? This nitric acid should probably contain a very wide spectrum of different salts (practically, all of the elements between 35-65, and yet a lot of transurans, and a lot of uranium (both of 235 ad 238), and plutonium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reach an efficient reprocessing, the uranium (or, at least, the actinid elements) should be somehow separated from this solution. But, AFAIK, they have very different chemical properties. How is it possible to separate only the transuranic materials?&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2015-01-21T19:34:43.540" Title="How does nuclear fuel reprocessing work?" Tags="|nuclear-technology|nuclear-reprocessing|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="27" PostTypeId="2" ParentId="22" CreationDate="2015-01-20T23:24:03.133" Score="5" Body="&lt;p&gt;A shoulder bolt is a good approach in applications with light loading, but it needs to be specified such that it can be fully tightened while still allowing the necessary clearance for movement without binding. Use shims where needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can reduce some of the play in the joint by having the bolt threaded into one of the aluminum pieces (but still backed up with a nut for strength). In other words, only one of the aluminum bars needs to turn freely on the bolt.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2015-01-21T03:32:44.023" LastActivityDate="2015-01-21T03:32:44.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="28" PostTypeId="1" CreationDate="2015-01-20T23:24:34.500" Score="12" ViewCount="910" Body="&lt;p&gt;I am wondering if there is a spring strong enough and light enough to inflate an airtight membrane so that it floats at Standard Temp &amp;amp; Pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, a 10cm cylinder would displace just under 1 gram of air, but a cylindrical spring would have to oppose 162kg of atmospheric force in that configuration.  Is there such a thing as a 162kg 10cm cylindric spring that weighs less than 1 gram?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's just one spring design.  More generally, what spring designs provide the highest rate/weight ratio?  What are the current practical limits on spring rate vs spring mass, and using what materials?&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="77" LastEditDate="2015-01-23T14:21:03.330" LastActivityDate="2016-02-04T18:28:19.247" Title="What are the lightest springs?" Tags="|springs|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="29" PostTypeId="2" ParentId="20" CreationDate="2015-01-20T23:35:57.150" Score="3" Body="&lt;p&gt;Knowing the diameter of the wheels and the number of times the wheels turn over in a time interval will give both distance travelled and average speed for that time interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have to have a continual monitoring system which would record time, wheel revolutions and the angle of the wheels used for steering, most likely relative to the central longitudinal axis of the vehicle. You will need to vectorize the steering data and add the vectors to give the position relative to the  vehicle's starting location.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-01-20T23:35:57.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="30" PostTypeId="2" ParentId="9" CreationDate="2015-01-20T23:39:55.953" Score="10" Body="&lt;p&gt;The work history and temperature history of any metal can make a big difference in the final shape after machining. If the material was cold worked (e.g. rolled) there may be significant residual stresses within the material.  When you start selectively removing material, those stresses may cause the part to warp into a new shape. Cast metals and ones that have been given surface treatments can have the same problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to address the problem by roughing the part to near-final shape and then running a finishing pass to remove that last little bit of material. Otherwise take a look at the specifications of the material. You may need to heat treat the part to relieve those stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a problem that can exist with just about any metal, not just Al.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-01-20T23:39:55.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="31" PostTypeId="1" CreationDate="2015-01-20T23:41:13.397" Score="6" ViewCount="211" Body="&lt;p&gt;I have some height-adjustment nuts screwed onto vertical threads to support my print-bed for a 3D printer. They are domed nuts and the aluminium build-plate sits on top of the domes, three holes in the plate align with the domes to hold the plate in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I protect these nuts from vibration loosening without using thread-locker or similar, so that they remain adjustable?&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-07-22T15:24:39.187" Title="How to prevent vibration loosening whilst maintaining adjustability?" Tags="|vibration|3d-printing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="32" PostTypeId="1" AcceptedAnswerId="205" CreationDate="2015-01-20T23:41:29.103" Score="6" ViewCount="1246" Body="&lt;p&gt;I've found myself tapping and threading a lot lately, and doing so with hand taps and dies.  What would be the next step up from using a drill-press to cut holes for tapping, and a lathe for turning parts to diameter for threading?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are options for making this easier and more precise short of a 6-figure screw or CNC machine?&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2015-01-24T03:27:27.790" Title="What machinery should I invest in for tapping and threading?" Tags="|machining|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="33" PostTypeId="1" AcceptedAnswerId="75" CreationDate="2015-01-20T23:46:47.343" Score="11" ViewCount="1381" Body="&lt;p&gt;It's true that most bridges are &quot;two directional.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;a href=&quot;http://en.wikipedia.org/wiki/Three-way_bridge&quot;&gt;three way bridges&lt;/a&gt; are pretty rare, globally. I can understand why there wouldn't be many for rivers, but if bridges are designed based on the lie of the surrounding ground, why wouldn't there be a large number of non-river sites that would support such bridges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, three out of the world's bridges exist in Michigan (and only ten or so elsewhere in the United States). What is it about the land, topography, or other features of Michigan that cause it to have a disproportionate number of the country's and world's three way bridges.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditDate="2015-01-21T21:31:55.910" LastActivityDate="2015-04-03T03:09:40.567" Title="Why are three way bridges rare?" Tags="|structures|design|civil-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="34" PostTypeId="1" AcceptedAnswerId="41" CreationDate="2015-01-20T23:50:02.410" Score="12" ViewCount="2255" Body="&lt;p&gt;My design uses a round steel bar inserted through a hole in an aluminium component with a tight fit (no movement). The operating environment is dry, 20-40°C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to worry about galvanic corrosion, and if so how long will this last before starting to corrode? Also, what storage conditions should I avoid that would accelerate the corrosion?&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-25T20:46:59.583" Title="Do I need to worry about galvanic corrosion when inserting a steel bar through aluminium?" Tags="|steel|corrosion|materials|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="35" PostTypeId="1" AcceptedAnswerId="99" CreationDate="2015-01-20T23:51:19.933" Score="10" ViewCount="10471" Body="&lt;p&gt;If I have structural or tool steel that has been treated to some standard (ASTM, SAE, ISO -- e.g., for hardness) but I don't know the details of the treatment, is there a &quot;safe&quot; temperature below which I can work the steel without affecting its performance characteristics?&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="77" LastEditDate="2015-01-22T16:46:52.107" LastActivityDate="2023-11-16T11:47:06.237" Title="At what temperature do I risk altering the structure of steel?" Tags="|steel|machining|metallurgy|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="36" PostTypeId="2" ParentId="20" CreationDate="2015-01-20T23:57:21.430" Score="6" Body="&lt;p&gt;The main problem which I can see in your idea, that your system will have a cumulative error. Only calculating this won't be enough, you will have to find alternate solutions, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In similar (but maybe bigger) scenarios, for example for drones, there is a similar problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is using the wheel rotation counters to get a fast, real-time, but buggy input data (which contains the cumulative error as well). In case of flying drones, this data is coming from gyros and accelerometers, your task is much simpler compared to their.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But&lt;/em&gt;, you should get an alternate information source, too! (In case of the drones it is normally the GPS). This can be GPS, or some other thing - there is a wide spectrum of possibilities. Visual image processing? Pre-calibrated ultrasound markers? Marker paintings on the floor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I suspect correctly the size of your experiment, maybe the last would be most promising to you.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="282" LastEditDate="2015-03-23T23:47:36.393" LastActivityDate="2015-03-23T23:47:36.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="37" PostTypeId="1" AcceptedAnswerId="53" CreationDate="2015-01-20T23:59:10.837" Score="10" ViewCount="5458" Body="&lt;p&gt;I am starting to use steel in some of my designs, and while trying to learn more about this material and its properties I have realized that there is so much more to it than I thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the grading system used for steel work? For instance, what does &quot;Grade 11SMn30&quot; mean, and how does this affect the properties of the material?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answers/comments indicate there are multiple standards, which I did not know, but I've found out that this is an EN grading.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="120" LastEditDate="2015-01-21T03:01:00.333" LastActivityDate="2015-01-21T03:01:00.333" Title="What does the steel grade &quot;11SMn30&quot; mean?" Tags="|steel|materials|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="38" PostTypeId="1" AcceptedAnswerId="1945" CreationDate="2015-01-21T00:03:53.533" Score="7" ViewCount="164" Body="&lt;p&gt;The LTE specification specifies SC-FDMA as the encoding scheme for the reverse link (mobile to base station) when the forward link uses OFDMA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for the difference?&lt;/p&gt;&#xA;" OwnerUserId="54" LastEditorUserId="7083" LastEditDate="2019-08-07T20:36:24.387" LastActivityDate="2019-08-07T20:36:24.387" Title="Why does the LTE Specification define SC-FDMA as the modulation scheme for the reverse link?" Tags="|electrical-engineering|telecommunication|lte|modulation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="39" PostTypeId="2" ParentId="10" CreationDate="2015-01-21T00:12:55.177" Score="4" Body="&lt;p&gt;Paradoxically, maybe the home-sized magnetic levitation could be a narrow area, where superconducting solutions could be cheap. I explain, why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although room-temperature superconductivity is a dream, room superconductivity is not. :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are soon superconducting materials which can be cooled by liquid air, and they aren't even costly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid air is also relatively cheap. Maybe a liquid air processing machine were costly, but you don't need to buy that. Only to buy the superconducting alloy, and the liquid air (nitrogen). For the first, I tried ebay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, it depends what you want to reach. I once played chess with figures levitated by Meissner-effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result could be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hxFn9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2015-01-21T00:12:55.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="40" PostTypeId="2" ParentId="31" CreationDate="2015-01-21T00:13:45.360" Score="5" Body="&lt;p&gt;Double nutting can be an easy option for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was about to describe it myself, but &lt;a href=&quot;http://www.boltscience.com/pages/twonuts.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;these folks&lt;/a&gt; do a great job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image, &lt;a href=&quot;http://www.boltscience.com/pages/twonuts.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;from the article&lt;/a&gt;, models the forces that do the job.  Take note of the contact surfaces between the threading of the bolt, in the top vs. bottom nut:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/htgru.gif&quot; alt=&quot;force model for double nutting&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="368" LastEditDate="2015-07-21T23:26:51.477" LastActivityDate="2015-07-21T23:26:51.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="41" PostTypeId="2" ParentId="34" CreationDate="2015-01-21T00:29:30.307" Score="7" Body="&lt;p&gt;Very probably, no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Galvanic corrosion required the existence of a reactive &lt;em&gt;and&lt;/em&gt; conductive medium, which you don't have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GraHA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is some (for example, at least moist air, or similar) best you can do to paint &lt;em&gt;both&lt;/em&gt; of the metals, if you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, if none of the materials would corrode in an environment, galvanic corrosion is a non-issue (in that environment).&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2015-01-21T00:29:30.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="42" PostTypeId="2" ParentId="37" CreationDate="2015-01-21T00:30:44.500" Score="3" Body="&lt;p&gt;The numbers from 1 to 9 indicate the type of steel. The first 1 is for carbon steel, and the second 1 and the S mean that sulfur has been added to it. The Mn stands for &quot;manganese&quot; and 30 is the amount of carbon in the steel, expressed as a fraction of 1%. It really means, .30% carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information can be found in &lt;a href=&quot;http://en.wikipedia.org/wiki/SAE_steel_grades&quot; rel=&quot;nofollow&quot;&gt;sources&lt;/a&gt; such as this one. The above answer would be responsive assuming that it is an American designation, but a commenter pointed out that it could be an English code, which would mean something different. In any event, it is the responsibility of the questioner to define these things.&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditDate="2015-01-21T00:36:46.960" LastActivityDate="2015-01-21T00:36:46.960" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="43" PostTypeId="2" ParentId="26" CreationDate="2015-01-21T00:32:02.570" Score="6" Body="&lt;p&gt;You could try &lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_reprocessing&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, but they note that the article may not meet there quality standards, so I'll stay away from there as a source. The &lt;a href=&quot;http://world-nuclear.org/info/Nuclear-Fuel-Cycle/Fuel-Recycling/Processing-of-Used-Nuclear-Fuel/&quot; rel=&quot;noreferrer&quot;&gt;World Nuclear Association&lt;/a&gt; seems more credible, anyway. Their page discusses the process. Quoting the section entitled &amp;quot;Reprocessing today – PUREX&amp;quot; (emphasis mine):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;All commercial reprocessing plants use the well-proven hydrometallurgical PUREX (plutonium uranium extraction) process. This involves dissolving the fuel elements in concentrated nitric acid. C&lt;strong&gt;hemical separation of uranium and plutonium is then undertaken by solvent extraction steps (neptunium – which may be used for producing Pu-238 for thermo-electric generators for spacecraft – can also be recovered if required).&lt;/strong&gt; The Pu and U can be returned to the input side of the fuel cycle – the uranium to the conversion plant prior to re-enrichment and the plutonium straight to MOX fuel fabrication.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Block of text &lt;span class=&quot;math-container&quot;&gt;$\to$&lt;/span&gt; coherent summary: Chemical solvents are used.&lt;/p&gt;&#xA;&lt;p&gt;Further below, the WNA notes&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The used fuel is chopped up and dissolved in hot concentrated nitric acid. The first stage separates the uranium and plutonium in the aqueous nitric acid stream from the fission products and minor actinides by a countercurrent solvent extraction process, using tributyl phosphate dissolved in kerosene or dodecane. In a pulsed column uranium and plutonium enter the organic phase while the fission products and other elements remain in the aqueous raffinate.&lt;/p&gt;&#xA;&lt;p&gt;In a second pulsed column uranium is separated from plutonium by reduction with excess U4+ added to the aqueous stream. Plutonium is then transferred to the aqueous phase while the mixture of U4+ and U6+ remains in the organic phase. It is then stripped from the organic solvent with dilute nitric acid.&lt;/p&gt;&#xA;&lt;p&gt;The plutonium nitrate is concentrated by evaporation then subject to an oxalate precipitation process followed by calcination to produce PuO2 in powder form. The uranium nitrate is concentrated by evaporation and calcined to produce UO3 in powder form. It is then converted to UO2 product by reduction in hydrogen.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Compounds of importance: tributyl phosphate, kerosene and dodecane, and aqueous affinate are used as solvents after the nitric acid. Then there is a reaction with &lt;span class=&quot;math-container&quot;&gt;$U^{4+}$&lt;/span&gt;, and uranium and plutonium are separated.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;That's the chemical part. I haven't been able to find much information on the specific mechanical methods (i.e. the pure engineering you were looking for). In fact, &lt;a href=&quot;http://worldwide.espacenet.com/publicationDetails/originalDocument;jsessionid=VagyIBjpGc8AsrFdt8frL6RZ.espacenet_levelx_prod_2?CC=US&amp;amp;NR=2924506A&amp;amp;KC=A&amp;amp;FT=D&amp;amp;ND=&amp;amp;date=19600209&amp;amp;DB=&amp;amp;locale=en_EP&quot; rel=&quot;noreferrer&quot;&gt;the original patent&lt;/a&gt; for PUREX does not cover the mechanical processes, leading me to believe that the exact setup can vary.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.inl.gov/technicalpublications/Documents/4460757.pdf&quot; rel=&quot;noreferrer&quot;&gt;This pdf&lt;/a&gt; is somewhat helpful. It says that the input materials are first chopped up. Perforated baskets are then used to separate this machinery from the rest of the solution; a &amp;quot;wheel dissolver&amp;quot; may be used.&lt;/p&gt;&#xA;&lt;p&gt;I do have some information, albeit for a non-PUREX method. For aqueous reprocessing (go to page 12), mixer settlers (mixing chambers with settling chambers), columns and centrifuges are the main pieces of equipment.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-01-21T01:16:33.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="44" PostTypeId="1" CreationDate="2015-01-21T00:33:25.187" Score="10" ViewCount="1212" Body="&lt;p&gt;I am working with 1&quot; medium pressure steam mains (45 psig) for a batch process. Time between batches is usually short (to the order of minutes) but depending on the day the delay can be hours, and even longer over weekends or holidays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We recently installed a full set of orifice (venturi) type steam traps. When we start up, the orifices can't handle the condensate load and the back up condensate is causing water hammer in some sections of the piping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are a little tight on budget but will look to installing additional (non-orifice) traps in the problem portions of the piping.  In the meantime, we are resorting to manually opening/closing the bypass valves to vent the initial condensate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better way to prevent condensate backup at startup on a batch process when using orifice/venturi steam traps?&lt;/p&gt;&#xA;" OwnerUserId="112" LastEditorUserId="110" LastEditDate="2015-02-10T11:45:32.497" LastActivityDate="2019-05-01T22:02:20.437" Title="Preventing condensate backup at startup with orifice (venturi) steam traps on medium pressure batch process mains?" Tags="|mechanical-engineering|steam|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="45" PostTypeId="5" CreationDate="2015-01-21T00:34:49.273" Score="0" Body="&lt;p&gt;LTE, originally name Long Term Evolution, is a wireless telecommunications standard evolved from the 3GPP family (which includes GSM and UMTS).  The official standard called LTE began with &lt;a href=&quot;http://www.3gpp.org/specifications/releases/72-release-8&quot; rel=&quot;nofollow&quot;&gt;Release 8&lt;/a&gt; of the 3GPP specification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While commonly referred as a 4G technology, 3GPP Releases 8 and 9 do not officially meet the requirements of 4G but the &lt;a href=&quot;http://en.wikipedia.org/wiki/ITU-R&quot; rel=&quot;nofollow&quot;&gt;ITU&lt;/a&gt;.  It is not until 3GPP Release 10 (Commonly known as LTE Advanced, does it meet the officially meet the &lt;a href=&quot;http://en.wikipedia.org/wiki/IMT_Advanced&quot; rel=&quot;nofollow&quot;&gt;4G requirements&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="54" LastEditorUserId="54" LastEditDate="2015-01-21T02:48:22.747" LastActivityDate="2015-01-21T02:48:22.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="46" PostTypeId="4" CreationDate="2015-01-21T00:34:49.273" Score="0" Body="Use this tag for questions about the Long Term Evolution (LTE) wireless telecommunications standard." OwnerUserId="54" LastEditorUserId="368" LastEditDate="2015-05-06T16:16:38.433" LastActivityDate="2015-05-06T16:16:38.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="47" PostTypeId="5" CreationDate="2015-01-21T00:39:11.520" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-21T00:39:11.520" LastActivityDate="2015-01-21T00:39:11.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="48" PostTypeId="4" CreationDate="2015-01-21T00:39:11.520" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-21T00:39:11.520" LastActivityDate="2015-01-21T00:39:11.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="49" PostTypeId="1" CreationDate="2015-01-21T00:46:00.803" Score="5" ViewCount="7552" Body="&lt;p&gt;This is similar to my Physics.SE question &lt;a href=&quot;https://physics.stackexchange.com/questions/155666/microtops-sunphotometer-sun-target-optics&quot;&gt;Microtops sunphotometer sun target optics&lt;/a&gt; - there I was asking about the optics principles (the answer is no good for my understanding, but I have been able to figure it out myself).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose behind a sunspotter is to show when a device is lined up in direct view of the sun - vital for fields such as sunphotometry etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is given that I now understand the optics principles involved (similar to a magnifying glass, but smaller and does not burn when focused) - &lt;strong&gt;what method can be used to make a rudimentary sunspotter?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I am interested in:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a suitable material that would be used for the sun target (as per the Microtops picture on the linked page).&lt;/li&gt;&#xA;&lt;li&gt;the same orientation of the sun target being perpendicular to the incident light.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I am not interested in:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;any device how to make a device that allows viewing of things like eclipses.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user52" LastEditorUserId="-1" LastEditorDisplayName="user52" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2017-02-12T00:45:19.470" Title="How can I make a rudimentary sunspotter?" Tags="|materials|optics|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="50" PostTypeId="1" AcceptedAnswerId="96" CreationDate="2015-01-21T00:46:44.707" Score="3" ViewCount="1588" Body="&lt;p&gt;Recently, both Zigbee and Z-wave tend to be the most commonly used protocol to communicate between Internet of Things (IoT) devices. Most of these IoT devices tend to have a combination of mechanical or optical related transducer or actuator device component coupled with a micro-controller and respective software stack. What are the major differences between Zigbee 3.0 and Z-Wave in relation to communication between IoT devices?&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-05-18T09:42:39.153" LastActivityDate="2016-07-21T00:31:00.460" Title="What are the pros and cons of between Zigbee and Z-Wave communication protocols used in IoT device space?" Tags="|electrical-engineering|telecommunication|iot|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="51" PostTypeId="2" ParentId="34" CreationDate="2015-01-21T00:51:59.367" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_corrosion&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; defines galvanic corrosion as: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Galvanic corrosion is an electrochemical process in which one metal corrodes preferentially to another when both metals are in electrical contact, in the presence of an electrolyte. This same galvanic reaction is exploited in primary batteries to generate an electrical voltage.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't see any electrolytes in your set-up to generate the required electric voltage. As long as you keep your system free of liquids that could carry such electrolytes, you're fine.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2015-01-21T00:51:59.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="52" PostTypeId="1" CreationDate="2015-01-21T01:02:00.727" Score="4" ViewCount="479" Body="&lt;p&gt;I am looking for multi-position bi-stable solenoids. So far, I have only found one vendor&lt;sup&gt;1&lt;/sup&gt; with a 3 position bi-stable solenoid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there multi-position bi-stable solenoids with more than three positions? How can a 3 position bi-stable solenoid be mechanically amplified to multiple positions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are these such a rare commodity? What are the alternatives other than stepper motors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;Note, I'm not looking for a vendor recommendation, I'm trying to find out if I'm looking for the right thing. &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="282" LastEditDate="2015-03-04T00:19:10.097" LastActivityDate="2015-03-04T13:09:07.073" Title="Multi-position bi-stable solenoids in a small form factor" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="53" PostTypeId="2" ParentId="37" CreationDate="2015-01-21T01:04:32.327" Score="12" Body="&lt;p&gt;Based on it being an EN steel grade:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first number is 100x the carbon content percentage (so 0.11%), the letters are added elements (sulphur and maganese), and the last number is the sulfur content (0.30%).  &lt;a href=&quot;http://www.steelnumber.com/en/steel_composition_eu.php?name_id=155&quot;&gt;You can see the full details here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The full format seems to be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[X][% carbon][added elements][% of added elements, hyphenated]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that the X is only present for stainless steels.  &lt;a href=&quot;http://mobile.euro-inox.org/map/EN_name/EN_en_name.php&quot;&gt;Here is a good example&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that this scheme is somewhat ambiguous.  The percentages are only an approximation, and the example you gave is interesting because it lists Sulfur before Maganese, despite the naming convention stating that they should be listed in order of content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's alright for getting quick basic info about the steel, but for anything else you may want to use the EN number, 1.0715, rather than the name.  &lt;a href=&quot;http://en.wikipedia.org/wiki/Steel_grades#European_standard_steel_number&quot;&gt;Wikipedia has details on the format&lt;/a&gt;.  Given this classification you can find out much more about the steel's properties and see what general category it fits into.  The site I linked first says this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EN 10277-3: 2008 Bright steel products&lt;br&gt;&#xA;  Technical delivery conditions. Free-cutting steels  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;EN 10087: 1999 Free cutting steels&lt;br&gt;&#xA;  Technical delivery conditions for semi-finished products, hot rolled bars and rods&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="53" LastActivityDate="2015-01-21T01:04:32.327" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="54" PostTypeId="1" AcceptedAnswerId="116" CreationDate="2015-01-21T01:19:55.740" Score="7" ViewCount="204" Body="&lt;p&gt;I am thinking mainly in $\frac{kWh}{m^3}$ and $\frac{\$}{m^3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past some decades a wide range of surprisingly efficient water desalination plants were built, mainly in desert regions (Middle East). These plants use reverse osmosis via a system of multiple pressered membranes. This solution seems to be very effective in the sense of energy usage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is not enough. Comparing the desalination prices (coming mainly from energy costs) to the alternatives, a further 60-90% reduction is yet needed.  Comparing them, what development potential is there in water desalination?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think water desalination has probably a theoretical energy limit, which could be maybe calculated from entropy and binding energy formulas.  How near are we to this theoretical limit?&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="368" LastEditDate="2015-07-21T23:03:43.630" LastActivityDate="2017-03-28T06:53:42.807" Title="What development possibilities could yet exist in water desalination?" Tags="|water-resources|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="55" PostTypeId="1" AcceptedAnswerId="58" CreationDate="2015-01-21T01:23:13.033" Score="10" ViewCount="3389" Body="&lt;p&gt;I have been investigating recapturing energy from house lighting using solar panels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there solar panels specifically designed to capture indoor lighting with a higher efficiency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what are the differences between the outdoor and indoor solar panels designed for this purpose?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://solarcellcentral.com/limits_page.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Efficiency Limits&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://classroom.synonym.com/can-solar-batteries-charge-using-incandescent-light-7189.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Can Solar Batteries Charge Using Incandescent Light?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.linkedin.com/groups/Can-solar-panels-charge-from-96493.S.209515615&quot; rel=&quot;nofollow noreferrer&quot;&gt;Can solar panels charge from street lights, during the night? As it does charge from the sun during the day!&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-08-22T23:22:38.750" LastActivityDate="2023-01-04T20:34:38.953" Title="Are there different solar panels for recapturing energy from household lighting?" Tags="|electrical-engineering|photovoltaics|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="56" PostTypeId="5" CreationDate="2015-01-21T01:28:43.910" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-21T01:28:43.910" LastActivityDate="2015-01-21T01:28:43.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="57" PostTypeId="4" CreationDate="2015-01-21T01:28:43.910" Score="0" Body="Questions related to Design of Experiments, as a methodological practice. " OwnerUserId="104" LastEditorUserId="160" LastEditDate="2015-01-22T18:50:14.550" LastActivityDate="2015-01-22T18:50:14.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="58" PostTypeId="2" ParentId="55" CreationDate="2015-01-21T01:42:10.390" Score="14" Body="&lt;p&gt;I don't think they existed, and it has its reason.&lt;/p&gt;&#xA;&lt;p&gt;First: a solar panel can be characterized mainly by its efficiency spectrum: on which wavelength, which ratio of light energy can it convert to electric power.&lt;/p&gt;&#xA;&lt;p&gt;This needs to have its maximum around the visible light, because the Sun gives most of its energy in this wavelength interval.&lt;/p&gt;&#xA;&lt;p&gt;This is because our eyes can see best in this spectrum. We simply evolved to the sunlight.&lt;/p&gt;&#xA;&lt;p&gt;And this is why home light is also in this wavelength: this is what we, human, like the most.&lt;/p&gt;&#xA;&lt;p&gt;There was no need for different solar panels.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;But the power of the sunlight is around some hundred &lt;span class=&quot;math-container&quot;&gt;$\frac{W}{m^2}$&lt;/span&gt;, although it varies very heavily:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BGrY1.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The light power of a house bulb is around sometimes 10W - not for a &lt;span class=&quot;math-container&quot;&gt;$m^2$&lt;/span&gt;, but for a whole room! Maybe, we see in a well lighted room just as good as in sunlight, but it is only because our eye is very adaptive. The actual light power density is a tenth, or even hundredth smaller, compared to sunlight.&lt;/p&gt;&#xA;&lt;p&gt;And the efficiency of most solar panels is around 10-20%. There are experimental, very costly versions reaching 40%. A solar panel in a room couldn't produce valuable energy, at most some watts - at the cost of the price of a solar panel on the roof. And the cost is their main problem even with the many times bigger solar power.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="2824" LastEditDate="2023-01-04T20:34:38.953" LastActivityDate="2023-01-04T20:34:38.953" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="59" PostTypeId="2" ParentId="55" CreationDate="2015-01-21T01:46:29.707" Score="11" Body="&lt;p&gt;My gut feel is that you would be able to save more energy than you could re-capture by eliminating light wastage and lowering the energy usage of the light source accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me use a thought experiment to explain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were in a white cube-shaped room with a light in the roof and a large window in one wall with black curtains. At night when there's no light from the window if your curtains are open, you lose all that light, if you draw the curtains it's no better as they are black so they absorb it and the room is no lighter. However, if you make the curtains white the room will be a bit brighter and you can lower the power consumption of your light to get the same brightness in the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you put a solar panel all over one of the white walls of your room, you now have something &quot;re-capturing&quot; the energy, but the room is darker because that light is no longer adding to the brightness of the room by being reflected, so you need to increase the energy of your light to get the same brightness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to inefficiencies of energy conversion from electricity to light and in re-capturing that and converting it back to electricity, you will always lose more than the extra you have to put in in order to maintain the same room-brightness. Therefore it's more efficient to lower the light usage and optimise the room for reflecting more of the light back in (white surfaces).&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-21T01:46:29.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="60" PostTypeId="2" ParentId="20" CreationDate="2015-01-21T02:01:10.593" Score="3" Body="&lt;p&gt;The basic equation would be&#xA;$$\text{Distance traveled}=2 \pi r \times n$$&#xA;where $r$ is the radius of the wheel and $n$ is the number of revolutions of the wheel. This assumes, though, that there's no slipping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/158728/wheel-slip-in-vehicle-dynamics&quot;&gt;This question&lt;/a&gt; says that for the car to move,&#xA;$$F_k&amp;gt;f_cG_k$$&#xA;where $F_k$ is the force applied that's moving the far, $f_c$ is the coefficient of cohesion, and $G_k$ is the weight of the wheel plus the weight of the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total force is&#xA;$$\sum F=F_k-f_cG_k$$&#xA;This produces torque, $\tau$:&#xA;$$\tau=Fr$$&#xA;So&#xA;$$\tau=(f_k-f_cG_k)r=I \alpha$$&#xA;where $I$ is the moment of inertia of the wheel and $\alpha$ is the angular acceleration. So&#xA;$$\alpha=\frac{(f_k-f_cG_k)r}{I}$$&#xA;You have to measure $f_c$ beforehand, and you'll have to figure out the force applied by the eninge ($F_c$). But then you can use kinematics to figure out how many rotations the car's wheel will actually undertake:&#xA;$$\theta=\omega_0 t+\frac{1}{2}\alpha t^2$$&#xA;where $\theta$ is the number of revolutions, $\omega_0$ is the initial velocity, and $t$ is the time. So, with $\text{Distance traveled}=r \theta$, your equation should be&#xA;$$\text{Distance traveled}=r \left( \omega_0t+\frac{1}{2} \left(\frac{(f_k-f_cG_k)r}{I} \right)t^2 \right)$$&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2015-01-21T02:01:10.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="61" PostTypeId="2" ParentId="49" CreationDate="2015-01-21T02:27:40.823" Score="-1" Body="&lt;p&gt;Try this experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carefully stab a cocktail stick into a grape at one end (longways so that the grape looks round when you look at it end-on and the cocktail stick is central). Stab the other end of the stick into a cube of cheese (slightly bigger than the grape). Now balance the stick on the cheese so that the grape is vertically above the cheese.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Procedure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wait until night or otherwise darken the room. Grab an LED torch and turn off the lights, hold the torch high above the grape pointing downwards. Look at the circular shadow of the grape on the cheese, when it is central around the stick it means the light is directly overhead (or at least aligned with the stick - if the stick isn't completely vertical).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine where light sensors would need to be placed on the surface of the cheese and around the outside (beyond the shadowed area) in order to know when the equipment is lined up with the light source. Once you get it, reward yourself by eating the grape and cheese.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-21T02:27:40.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="62" PostTypeId="1" AcceptedAnswerId="64" CreationDate="2015-01-21T02:29:22.187" Score="3" ViewCount="20101" Body="&lt;p&gt;I would like be able to measure LED brightness using a cost effective instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the available instruments to measure LED brightness and respective cost effective techniques?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: I'm interested in converting what the customer thinks is acceptable brightness to actual LED brightness specification?.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/150511/how-can-i-measure-the-brightness-of-a-led&quot;&gt;How can I measure the brightness of a LED?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2016-01-26T12:14:07.420" Title="How can I measure the brightness of a LED?" Tags="|electrical-engineering|optics|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="63" PostTypeId="1" AcceptedAnswerId="76" CreationDate="2015-01-21T02:42:04.263" Score="35" ViewCount="45964" Body="&lt;p&gt;The debate of traffic circles (also called roundabouts or rotaries) versus traffic light intersections has been in progress for a while. Those in favor of traffic circles &lt;a href=&quot;http://www.wsdot.wa.gov/Safety/roundabouts/benefits.htm&quot; rel=&quot;noreferrer&quot;&gt;say that&lt;/a&gt;, among other things, that they are safer than traffic light intersections. This claim has been scientifically proven. On the other hand, traffic light intersections are more space-inefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even &lt;a href=&quot;http://www.treehugger.com/cars/mythbusters-roundabouts-vs-4-way-stop-intersection-which-more-efficient-video.html&quot; rel=&quot;noreferrer&quot;&gt;Mythbusters&lt;/a&gt; has joined the fun, testing the efficiency (which is one of the main arguments both sides seem to concern themselves with) of each method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, here's a quick picture of a traffic circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5wI9.jpg&quot; alt=&quot;Traffic circle&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of a four-way traffic light intersection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QL3y9.jpg&quot; alt=&quot;Intersection&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;strong&gt;what are the pros and cons of a traffic circle versus a traffic light intersection?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="106" LastEditDate="2015-01-28T00:17:58.240" LastActivityDate="2019-04-21T04:11:16.163" Title="What are the pros and cons of a traffic circle versus a traffic light intersection?" Tags="|traffic-light|highway-engineering|" AnswerCount="12" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="64" PostTypeId="2" ParentId="62" CreationDate="2015-01-21T02:54:56.297" Score="6" Body="&lt;p&gt;Customer specified &quot;acceptable brightness&quot; is going to be a subjective affair unless you have something to calibrate their responses against.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loosely, you'll need to measure the available light within a particular area and then correlate that measurement with customer surveys.  With a large enough sample of responses, you should be able to determine what constitutes an acceptable level of lighting for your customers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect that customer responses will fall along a curve of some sort and you'll eventually be able to determine what the standard deviations are for &quot;acceptable.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One factor to be concerned about is that differing rooms will have differing levels of reflection.  So even though two rooms may be of similar size and furnishings, the more darkly painted room will reflect less light and will be viewed as &quot;more dim&quot; on a subjective basis.  To workaround that, you might create a dedicated &quot;test room&quot; and have potential customers come through and provide their feedback.  While this is a more expensive approach, it's a very good way to isolate out variables that you can't otherwise control against.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To get you started with measuring available light, you'll likely need a light meter of some sort.  &lt;a href=&quot;http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_131764_-1&quot; rel=&quot;noreferrer&quot;&gt;This&lt;/a&gt; and &lt;a href=&quot;http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_2140251_-1&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; are just two examples of what I'm sure you'll be able to readily find with some searching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have a lux measurement for the area, you can convert that to lumens using a standard calculator.  Searching will turn up a number of sites, such as &lt;a href=&quot;http://www.rapidtables.com/calc/light/how-lux-to-lumen.htm&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt;. I didn't put the equations here as there are a number of steps involved, including determining the area of the room.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-01-21T03:25:53.973" LastActivityDate="2015-01-21T03:25:53.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="65" PostTypeId="2" ParentId="63" CreationDate="2015-01-21T02:55:19.950" Score="3" Body="&lt;p&gt;This answer is based on personal observation and commonsense. There is little science but some engineering intuition. I don't claim any civil or traffic engineering expertise. Like most engineering solution no one size fits all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Traffic Circle (Round about)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traffic Circles are good for areas with low traffic volume, warm climate and with large population of discipline. law abiding, courteous drivers who rarely use cell phone while driving. This will enable the traffic to flow smoothly and everyone will be happy drivers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traffic circles are not so great for areas prone to heavy snow or back ice. During heavy snow it is extreme difficult for snow plows to clean the roads properly, thus leaves snow which eventually turn to ice and cause accidents. Using extra salt is and option but then in return salt eventually pollutes water sources.  Therefore low cost stop signs are a great alternative, as well as more expensive traffic lights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Traffic Lights&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from the power requirements, overzealous drivers trying to beat the red light, and driver under influence who fail to recognize traffic light, I believe traffic light have good benefit. Traffic light give the ability automate the follow of traffic, allow an override option for pedestrians as well keep those unlawful, rude drivers in check. &#xA;In summary a good engineer will pick the best solution for the particular application based on data, analysis and skill. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to all those Civil and Traffic engineers, do you think you can find a solution to this situation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happen about a week ago, not too far away from my residence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=O7ApxVyskuI&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=O7ApxVyskuI&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-01-23T23:01:15.303" LastActivityDate="2015-01-23T23:01:15.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="66" PostTypeId="1" CreationDate="2015-01-21T03:00:07.033" Score="6" ViewCount="203" Body="&lt;p&gt;I know that my input signal contains a main sinusoidal signal, and I know its frequency but not its amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until now, I have been successfully using the least-squares method to generate a sinusoidal signal with the least difference from my input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this method breaks down when faced with slightly more complicated input signals, where the noise is big enough to disturb the list square. As a consequence, it's not efficient anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some other ways I can accomplish this filtering that don't have the same issue?&lt;/p&gt;&#xA;" OwnerUserId="123" LastEditorUserId="368" LastEditDate="2015-11-20T01:09:57.333" LastActivityDate="2015-11-20T01:09:57.333" Title="How can I filter out a signal while extracting only the desired one?" Tags="|electrical-engineering|signal-processing|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="67" PostTypeId="2" ParentId="63" CreationDate="2015-01-21T03:04:56.383" Score="11" Body="&lt;p&gt;&lt;strong&gt;Roundabout:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pros&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As you've mentioned,  the Mythbusters' test results showed that roundabouts were about 20% more efficient for cars, good for high traffic&lt;/li&gt;&#xA;&lt;li&gt;Allows more cars to cross at a time.&lt;/li&gt;&#xA;&lt;li&gt;Doesn't need traffic lights.&lt;/li&gt;&#xA;&lt;li&gt;Safer&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Cons&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;As you've also mentioned, it takes up more space.&lt;/li&gt;&#xA;&lt;li&gt;Uses more material to make.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4-way intersection&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pros&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Easier for pedestrians to cross street&lt;/li&gt;&#xA;&lt;li&gt;Space efficient&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Cons&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;More crashes&lt;/li&gt;&#xA;&lt;li&gt;requires traffic lights&lt;/li&gt;&#xA;&lt;li&gt;Slower than Roundabout&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Overall, the roundabout has more Pros, so it is a better choice. But when you don't have space, use the classic 4 way intersection.&lt;/p&gt;&#xA;" OwnerUserId="116" LastActivityDate="2015-01-21T03:04:56.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="68" PostTypeId="1" AcceptedAnswerId="69" CreationDate="2015-01-21T03:15:17.020" Score="21" ViewCount="37683" Body="&lt;p&gt;I have a 2 mm thick steel plate which is 300 mm long and 30 mm wide, supported at either end. It supports a weight-bearing wheel that can roll along the plate. It currently supports the maximum weight that I expect it to support when the wheel is in the middle, but it flexes a little bit too much. Would making it wider help to support the weight and increase its stiffness, or do I need to make it thicker?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also is there a way to calculate how the stiffness will change with the thickness (or width if that would affect it)?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="368" LastEditDate="2015-04-27T15:56:03.283" LastActivityDate="2021-08-03T12:36:46.440" Title="How does width and thickness affect the stiffness of steel plate?" Tags="|mechanical-engineering|steel|stiffness|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="69" PostTypeId="2" ParentId="68" CreationDate="2015-01-21T03:20:03.017" Score="21" Body="&lt;p&gt;&lt;em&gt;Short answer&lt;/em&gt;: make it thicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Long answer&lt;/em&gt;: The moment of inertia affects the beam's ability to resist flexing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use one of the many, free, online moment of inertia calculators (like &lt;a href=&quot;http://bendingmomentdiagram.com/free-calculator/moment-of-inertia-calculator/&quot;&gt;this one&lt;/a&gt;) to see how increasing the height of the beam will have an exponential effect on increasing the stiffness of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;a href=&quot;http://www.engineeringcalculator.net/beam_calculator.html&quot;&gt;this site&lt;/a&gt; helps provide a pictorial view of the load(s) upon a beam depending upon differing configurations, such as where the supports are and where the load is applied.  It also provides a calculator to determine the forces involved.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a decent &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_area_moments_of_inertia&quot;&gt;article for area moments of inertia.&lt;/a&gt;  In your particular case, you're asking about a &lt;code&gt;filled rectangular area&lt;/code&gt; and I&lt;sub&gt;x&lt;/sub&gt; = &lt;em&gt;bh&lt;sup&gt;3&lt;/sup&gt;&lt;/em&gt;/12.  The height has an exponential factor of 3 whereas increasing the base does not have an exponential factor.  So for the same amount of material, increasing the height stiffens the beam better.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To be clear, you &lt;em&gt;can&lt;/em&gt; make the beam sag less by increasing the width of the plate.  It's just more effective to make the plate thicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current moment: I&lt;sub&gt;x&lt;/sub&gt; = 30 * 2&lt;sup&gt;3&lt;/sup&gt; / 12 = 20 mm&lt;sup&gt;4&lt;/sup&gt;&lt;br&gt;&#xA;Increase width by 1mm: I&lt;sub&gt;x&lt;/sub&gt; = 31 * 2&lt;sup&gt;3&lt;/sup&gt; / 12 = 20.6 mm&lt;sup&gt;4&lt;/sup&gt;&lt;br&gt;&#xA;Increase height by 1mm: I&lt;sub&gt;x&lt;/sub&gt; = 30 * 3&lt;sup&gt;3&lt;/sup&gt; / 12 = 67.5 mm&lt;sup&gt;4&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And if for some reason you can't easily increase the thickness of the plate, you can consider a different beam structure.  Currently, your beam is a simple rectangle.  You can easily use a &lt;a href=&quot;http://www.amesweb.info/SectionalPropertiesTabs/SectionalPropertiesTbeam.aspx#.VL_-Ti5WS7k&quot;&gt;T-beam&lt;/a&gt; or an &lt;a href=&quot;http://www.amesweb.info/SectionalPropertiesTabs/SectionalPropertiesIbeam.aspx#.VL_-Zi5WS7k&quot;&gt;I-beam&lt;/a&gt; in order to stiffen the plate instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Again, while I've provided some suggested links to online calculators feel free to search for and use others that you may prefer.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-01-21T19:42:23.003" LastActivityDate="2015-01-21T19:42:23.003" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="70" PostTypeId="2" ParentId="66" CreationDate="2015-01-21T03:27:09.800" Score="5" Body="&lt;p&gt;It sounds like what you're looking for is a bandpass filter. It will filter out any signal outside of a given frequency range. If the noise isn't too strong, the main signal should come through fairly clearly. The actual design of the bandpass filter is another matter, depending on what frequency range you need and how prevalent the noise is. It may not spit out a perfect sinusoid, but it might get you close enough that you could use your least-squares method on the bandpass output to generate the sinusoid. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-01-21T03:27:09.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="71" PostTypeId="2" ParentId="68" CreationDate="2015-01-21T03:36:32.740" Score="15" Body="&lt;p&gt;The stiffness of a rectangular cross section, be it steel, concrete, wood, or any other material, is related almost entirely to it's modulus of elasticity, $E$, and it's moment of inertia about the axis of bending, $I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you already have your material set, steel, you cannot change $ E $. What you can change is your $ I $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment of inertia for a rectangular cross section about its neutral axis is $ \frac{b \cdot d^3}{12} $. You will increase your stiffness exponentially by increasing the depth of your plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to note that you can increase your I in other ways as well. For example, if you welded another plate to your existing plate to make a T shape in cross section, you will significantly increase your $I$ and greatly stiffen your member.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2016-03-07T18:51:03.853" LastActivityDate="2016-03-07T18:51:03.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="73" PostTypeId="1" CreationDate="2015-01-21T03:49:49.600" Score="14" ViewCount="739" Body="&lt;p&gt;A highly desirable property of metals like gallium and indium is that they are able to wet glass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, as metals, they wouldn't be able to form an airtight seal when in contact with glass.  But they are able to do so, and they can be used in applications where an airtight seal is required (e.g. vacuum gaskets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, indium and gallium are somewhat expensive (at press time, around 700 USD/kg and 250 USD/kg, respectively).  Industrial compounds that also have similar properties, such as Galinstan (even though the glass still needs a wetting with gallium oxide first), aren't much better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some metal substances that are both cost-effective and able to wet glass with an airtight seal that can be used in industrial applications?&lt;/p&gt;&#xA;" OwnerUserId="120" LastEditorUserId="368" LastEditDate="2015-07-21T22:54:46.297" LastActivityDate="2015-07-21T22:54:46.297" Title="What are some cost-effective metal substances that can wet glass?" Tags="|materials|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="74" PostTypeId="1" AcceptedAnswerId="314" CreationDate="2015-01-21T04:05:04.740" Score="4" ViewCount="244" Body="&lt;p&gt;One of the valves on my 3 bar line just failed. Without it, our production has ground to a stop and we can't operate until we get a replacement (which we have already ordered). In order to make a deadline, I am trying to find a way to use a (properly rated for pressure) DN 16 flanged valve to fit with the existing DN 10 piping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to (safely) connect these two flanges together temporarily?&lt;/p&gt;&#xA;" OwnerUserId="112" LastActivityDate="2015-01-27T22:29:01.513" Title="How can I fit a DN16 flange to a DN10 flanged pipe?" Tags="|steam|piping|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="75" PostTypeId="2" ParentId="33" CreationDate="2015-01-21T04:42:38.530" Score="23" Body="&lt;p&gt;Most bridges (and overpasses) are built to cross over something. With a few notable exceptions, most of these &quot;somethings&quot; are relatively long perpendicular to the desired crossing direction and fairly narrow parallel to it. Therefore a simple &quot;two directional&quot; bridge best meets the needs of the engineering problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers always try to solve a problem in the simplest possible way to prevent introducing more problems than they have solved. If a simple two-way bridge solves the problem, then there is no reason to complicate the solution. Along those lines, vehicular traffic flow on a three-way bridge can be fairly complicated and may not easily facilitate high traffic volume. I believe that is why you will find more pedestrian bridges created in this style.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for Michigan, I don't know. It's entirely possible that a structural engineer or engineering company in the area had a fondness for that type of structure and bid on those projects with three-way designs.&lt;/p&gt;&#xA;" OwnerUserId="131" LastActivityDate="2015-01-21T04:42:38.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="76" PostTypeId="2" ParentId="63" CreationDate="2015-01-21T04:50:55.950" Score="30" Body="&lt;p&gt;I think you are talking about roundabouts, not traffic circles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is baffling to those of us in the UK that Americans think roundabouts are a new idea. In the UK we have so many variants, from mini-roundabouts all the way up to full motorway junctions (a giant roundabout above or below the motorway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ULmaM.jpg&quot; alt=&quot;Motorway Roundabout&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So do roundabouts take up more space? Not necessarily, this is a mini roundabout:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UlBYR.jpg&quot; alt=&quot;Mini Roundabout&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's nothing more than a slightly domed area of paint on the road, no lights are necessary, you can actually drive straight over the top of it rather than around it, its main  purpose is simply to dictate who has &lt;em&gt;right of way&lt;/em&gt; so that everyone knows who should yield and who should go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a busy town or city environment, roundabouts do not work well because excessive traffic from one direction with right of way can completely stop all other traffic causing congestion in other directions. Some roundabouts have lights or peak-time signals to prevent this. One great thing is that they're easy to modify (adding lights, making it mini (drive-overable), adding another entry-point, etc. Everywhere other than in busy grid-based towns/cities they are ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from a highway engineering perspective...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main pros are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cheap to build&lt;/li&gt;&#xA;&lt;li&gt;Agile (Flexible / extensible)&lt;/li&gt;&#xA;&lt;li&gt;Scalable to suit any junction size&lt;/li&gt;&#xA;&lt;li&gt;Mutable (add peak signals, bypass lanes, extra incoming roads)&lt;/li&gt;&#xA;&lt;li&gt;Modular (google image search for &quot;double roundabout&quot;, &quot;magic roundabout&quot;)&lt;/li&gt;&#xA;&lt;li&gt;Easy and safe for drivers to use (rules don't change in any configuration)&lt;/li&gt;&#xA;&lt;li&gt;Aids navigation of complex junctions from simple road-signs (just count the exits)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cons are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Annoys drivers on country roads when you'd like to just bypass&lt;/li&gt;&#xA;&lt;li&gt;Not suitable for busy city grids&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-21T04:50:55.950" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="77" PostTypeId="2" ParentId="32" CreationDate="2015-01-21T05:10:49.583" Score="2" Body="&lt;p&gt;LIMITED SCOPE: this answer is limited to the tap aspect ... it's a way to mitigate the tap-hole alignment concern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you already have the drill press and piece, you can put the tap in the drill press somewhat loosely in order to get the tap in the first 1/4&quot; or more (presuming a countersunk hole), from which you can do the rest by hand.  This reduces the lining-up-of-the-tap-with-hole aspect which often requires special care &amp;amp; time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Video demo: &lt;a href=&quot;https://www.youtube.com/watch?v=KmtbzL8YlxM&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=KmtbzL8YlxM&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="92" LastActivityDate="2015-01-21T05:10:49.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="78" PostTypeId="2" ParentId="35" CreationDate="2015-01-21T05:43:13.587" Score="5" Body="&lt;p&gt;The answer to your question depends a lot based on what kind of steel and what kind of heat treatment you're thinking of. For one point of reference, if you were working on a steel structure in the United States, AWS D1.1 would limit the maximum heat in quenched and tempered steels to 1100 deg F. This temperature is compatible with preheating for welds or heat straightening.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-01-21T05:43:13.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="80" PostTypeId="1" CreationDate="2015-01-21T06:21:26.753" Score="5" ViewCount="667" Body="&lt;p&gt;In an industrial jacketed reactor, we have two temperature probes: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;one on a baffle the extends down about to the 50% fill line&lt;/li&gt;&#xA;&lt;li&gt;one built into reactor, very close to the bottom valve. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When is it more reliable and practical to use one location over the other when monitoring and controlling the temperature in a chemical process? Please ignore  obvious situations, such as the liquid level is too low for the baffle so the bottom must be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard some reasons for each location. For example, baffle temperature is best when reacting at a constant temperature, or when a slurry presents possible solids buildup on the bottom probe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any solid industry best practices relating to this?&lt;/p&gt;&#xA;" OwnerUserId="140" LastEditorUserId="368" LastEditDate="2015-06-05T18:49:17.973" LastActivityDate="2015-08-07T22:06:46.487" Title="Baffle vs Bottom Temperature Measurement/Control" Tags="|chemical-engineering|control-engineering|process-engineering|temperature|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="81" PostTypeId="2" ParentId="6" CreationDate="2015-01-21T07:01:01.480" Score="4" Body="&lt;p&gt;Take all this with a grain of salt. I work in Germany as an engineer and don't have any special license sides my degree, but I'm employed and your situation may be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, engineer is a free profession in Germany. This means, if you are an engineer (have an engineering degree from a university or a university of applied science), you can freelance as an engineer. Each state has an &lt;em&gt;Ingenieurgesetz&lt;/em&gt; that states how to handle foreign degrees. Mostly, if the degree is equivalent (Master/BA of engineering) you can request the right to call yourself an engineer (But I don't know yet where).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, I &lt;em&gt;think&lt;/em&gt; for certain tasks in civil engineering you have to be part of an engineering association.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, if you work with a company, you generally don't need a special license (for example to sign of plans)&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-01-21T07:01:01.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="82" PostTypeId="2" ParentId="63" CreationDate="2015-01-21T07:50:04.870" Score="8" Body="&lt;p&gt;Like the British, Australia has had roundabouts for at least five decades in some shape or form and there prevalence has increased over the last thirty years or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the city were I live there is a five-way intersection and it is controlled by a single roundabout, nothing else. It's been operating for at least forty years without any issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easier and simpler to install a roundabout for such an intersection than to install a system of traffic lights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending to the zealotry of the relevant traffic authority, if roundabouts are placed at every intersection on a major road for a considerable length of road, it can be very frustrating for drivers on the major road. In this situation the roundabout can slow traffic. Prior to the installation of the roundabout at those intersection the minor road intersecting the major road were controlled by Stop or Give Way signs, no traffic lights, and at when the major road was busy vehicles trying to enter the major road had very long waiting times. Hence, the roundabouts made the traffic flow easier for the vehicles on the minor roads faster and fairer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As jhabbott states in his/her answer about, roundabouts can be small and cheap. A small painted circle and on the road and signs on the roads at the intersection informing drivers that the intersection is a roundabout. Even very large roundabouts can be virtually level with the rest of the road, allowing large heavy vehicles like buses and semi-trailers to drive straight through providing it is safe to do so.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-01-21T07:50:04.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="83" PostTypeId="2" ParentId="22" CreationDate="2015-01-21T07:56:11.967" Score="7" Body="&lt;p&gt;You can use a sort of &amp;quot;double diameter washer&amp;quot;. I've made that name up, but I am pretty sure such a device exists:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mEKcj.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;That's a cross section of one of your linkages. The white things are the aluminum pieces, the grey one is the screw, the yellow is the nut while the red is the &amp;quot;double diameter washer&amp;quot;. If you can found some washer that is slightly longer than your aluminum bars thikness you are good to go. Of course if it's too long the linkage may become lousy, but with a good cutter you can trim them down, or you can also add some regular washers between the red one and the nut.&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-01-21T07:56:11.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="84" PostTypeId="1" CreationDate="2015-01-21T08:18:50.177" Score="7" ViewCount="264" Body="&lt;p&gt;I need to pump a thick slurry from a tank. The slurry can be pumped, but won't flow well enough that I can use a suction pump from outside the tank. Normally, I'd use a centrifugal immersion pump that can be drawn from the tank for service. But with the thick slurry I need to use a PD pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One option that I'm looking at is a &lt;a href=&quot;http://www.netzsch-pumpen.de/de/produkte-loesungen/nemo-exzenterschneckenpumpe/nemo-bt-eintauchpumpe.html&quot; rel=&quot;nofollow&quot;&gt;monopump housed in a long pipe&lt;/a&gt;, so the rotor/stator packet is submerged and operated by a long shaft in the pipe. But this will probably get expensive and I'm open for other options. Another consideration is ease of service, the immersion pump linked above would have to be lifted with a crane out of the tank for service. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The application is agricultural / bio-energy; tank depth is 4m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm looking for a PD pump that is ...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;capable of pumping thick slurries&lt;/li&gt;&#xA;&lt;li&gt;can be put into my sump pump and retrieved for service with low effort (which probably rules out long driveshafts)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm mostly hunting for ideas, no need for a concrete product recommendation.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorDisplayName="user16" LastEditDate="2015-02-04T10:24:16.870" LastActivityDate="2015-05-27T18:29:33.137" Title="What is a PD pump that I can use as an immersion pump for a slurry?" Tags="|process-engineering|pumps|environmental-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="85" PostTypeId="2" ParentId="66" CreationDate="2015-01-21T08:45:48.627" Score="6" Body="&lt;p&gt;First, I would recommend that you create a frequency spectrum of your signal. From your question I interpret that you already know the frequency of your &lt;em&gt;sinusoidal signal&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common way for the computation of a frequency spectrum is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fast_Fourier_transform&quot; rel=&quot;noreferrer&quot;&gt;Fast Fourier Transform&lt;/a&gt;. Most software packages have a ready to use algorithm for this implemented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a look at your frequency spectrum, there are different types of noise, most common are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/White_noise&quot; rel=&quot;noreferrer&quot;&gt;White Noise&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pink_noise&quot; rel=&quot;noreferrer&quot;&gt;Pink Noise&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Brownian_noise&quot; rel=&quot;noreferrer&quot;&gt;Red (or Brownian) Noise&lt;/a&gt; and&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Grey_noise&quot; rel=&quot;noreferrer&quot;&gt;Grey Noise&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Know comes the tricky part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;The filtering of noise&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your frequency range, in which you want to acquire a signal, and the noise you record as well, there are three different filter types you can use&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Low-pass_filter&quot; rel=&quot;noreferrer&quot;&gt;Low-Pass&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Band-pass_filter&quot; rel=&quot;noreferrer&quot;&gt;Band-Pass&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/High-pass_filter&quot; rel=&quot;noreferrer&quot;&gt;High-Pass&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attention&lt;/strong&gt;: Filtering of the signal &lt;em&gt;after&lt;/em&gt; it was converted into a digital signal only makes sense if you can make shure that you do not get trapped with &lt;a href=&quot;https://en.wikipedia.org/wiki/Aliasing&quot; rel=&quot;noreferrer&quot;&gt;aliasing&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="148" LastActivityDate="2015-01-21T08:45:48.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="86" PostTypeId="1" AcceptedAnswerId="90" CreationDate="2015-01-21T09:16:09.513" Score="8" ViewCount="796" Body="&lt;p&gt;For one-off, bespoke electronic products, or ventures on low budgets, what is the cheapest way of ensuring the product conforms to EMI/EMC requirements of CE marking? This can be at design time and / or  throughout the manufacturing process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A small-run, electronic microprocessor based controller (using a CE-Marked zero-cross solid state relay) intended to switch on or off a 2 kW load such as a consumer grade 240 V, 2 kW blower heater. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A one-off CNC machine that utilizes CE-marked, low voltage motors and servos, and is controlled by a microcontroller and miscellaneous circuitry&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A new venture into producing LED lighting with controllers - typically switching 10 W loads; using PWM to control brightness; controlled by a micro-controller; using SMD LEDs, and a CE-marked switched-mode 24 V, 3 A PSU.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Obviously there must be logical principles which one can apply to allow one to bypass other expensive conforming methods.  For example,  connecting two CE-marked appliances to each other where they are intended to do so would most likely not need any further testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that for more complicated scenarios getting products approved by external companies that provide CE-marking assessments can be incredibly expensive. Also, I have not found any particularly affordable equipment/methods to carry out EMI/EMC testing - and not being an expert in EMI/EMC means I would not be confident about using such equipment effectively. &lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="368" LastEditDate="2015-07-21T22:49:01.270" LastActivityDate="2015-07-21T22:49:01.270" Title="CE Marking on low budgets: conforming to EMI / EMC requirements in ELV and mains circuits" Tags="|electrical-engineering|emc|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="87" PostTypeId="1" AcceptedAnswerId="114" CreationDate="2015-01-21T10:09:38.163" Score="7" ViewCount="698" Body="&lt;p&gt;There has recently been a series of &lt;a href=&quot;http://www.theguardian.com/business/2015/jan/14/cheesegrater-leadenhall-building-bolt-city-london&quot; rel=&quot;nofollow&quot;&gt;high profile bolt failures&lt;/a&gt; on &lt;a href=&quot;https://en.wikipedia.org/wiki/122_Leadenhall_Street&quot; rel=&quot;nofollow&quot;&gt;'The Cheesegrater'&lt;/a&gt; in London, which have been attributed to &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_embrittlement&quot; rel=&quot;nofollow&quot;&gt;hydrogen embrittlement&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can engineers do to ensure that hydrogen embrittlement doesn't effect their structures?&lt;/p&gt;&#xA;" OwnerUserId="151" LastEditorUserId="272" LastEditDate="2015-09-29T16:01:25.927" LastActivityDate="2019-08-08T16:51:45.777" Title="How can we prevent/detect hydrogen embrittlement?" Tags="|mechanical-engineering|civil-engineering|structures|metallurgy|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="88" PostTypeId="1" AcceptedAnswerId="93" CreationDate="2015-01-21T10:24:26.413" Score="6" ViewCount="11276" Body="&lt;p&gt;When designing a steel beam, the resistance to buckling is related to M&lt;sub&gt;cr&lt;/sub&gt;; the elastic critical moment for lateral-torsional buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the Eurocodes give no advice about how to calculate this parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you calculate it?&lt;/p&gt;&#xA;" OwnerUserId="151" LastEditorUserId="368" LastEditDate="2015-02-13T23:57:15.467" LastActivityDate="2019-05-25T16:00:53.007" Title="How do you calculate Mcr (critical buckling moment)" Tags="|civil-engineering|steel|beam|eurocodes|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="89" PostTypeId="1" AcceptedAnswerId="115" CreationDate="2015-01-21T10:34:49.407" Score="13" ViewCount="417" Body="&lt;p&gt;It is common to lift bridges to replace bearings, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an ideal world the lifting capacity required of the jacks would be the self-weight of the bridge divided by the number of jacks (+ allowances for wind/snow, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my (limited) experience, however, bridges begin to 'stick' to their bearings, and an additional allowance for over-coming this has to be provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any guidance about how to determine this figure?&lt;/p&gt;&#xA;" OwnerUserId="151" LastEditorUserId="151" LastEditDate="2015-01-22T17:21:19.793" LastActivityDate="2015-01-22T17:21:19.793" Title="Additional jacking force to overcome stiction/dry friction" Tags="|civil-engineering|structures|bridges|maintenance|friction|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="90" PostTypeId="2" ParentId="86" CreationDate="2015-01-21T10:41:59.097" Score="12" Body="&lt;p&gt;First of all, the main documents you need to know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;DIRECTIVE 2004/108/EC relating to electromagnetic compatibility: &#xA;&lt;a href=&quot;http://eur-lex.europa.eu/LexUriServ/LexUriServ.do?uri=OJ:L:2004:390:0024:0037:en:PDF&quot; rel=&quot;nofollow&quot;&gt;http://eur-lex.europa.eu/LexUriServ/LexUriServ.do?uri=OJ:L:2004:390:0024:0037:en:PDF&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;DIRECTIVE 2006/95/EC on low voltage devices: &lt;a href=&quot;http://eur-lex.europa.eu/LexUriServ/LexUriServ.do?uri=OJ:L:2006:374:0010:0019:en:PDF&quot; rel=&quot;nofollow&quot;&gt;http://eur-lex.europa.eu/LexUriServ/LexUriServ.do?uri=OJ:L:2006:374:0010:0019:en:PDF&lt;/a&gt; (and the version entering into force in April 2016: &lt;a href=&quot;http://eur-lex.europa.eu/legal-content/EN/TXT/PDF/?uri=CELEX:32014L0035&amp;amp;from=EN&quot; rel=&quot;nofollow&quot;&gt;http://eur-lex.europa.eu/legal-content/EN/TXT/PDF/?uri=CELEX:32014L0035&amp;amp;from=EN&lt;/a&gt; )&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Those documents form the basis of your work. Also, they define if your device is subject to the directive or not. Depending on the case, you could also be subject to the R&amp;amp;TTE Directive and RoHs Directive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Annex I of both documents informs us on the essentials requirements to meet. Mainly, it says that your product must be safe for humans, pets and property and that it should not be impacted by the electromagnetic emissions of other nearby devices nor it should impact them with its own emissions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's really interesting lies in the Annex II of the first document and in Annex IV of the second document. These parts defines the conformity assessment procedure you'll have to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 8 procedures (called modules) of conformity assessment: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;internal production control (module A);&lt;/li&gt;&#xA;&lt;li&gt;CE type-examination (module B);&lt;/li&gt;&#xA;&lt;li&gt;conformity to type (module C);&lt;/li&gt;&#xA;&lt;li&gt;production quality assurance (module D);&lt;/li&gt;&#xA;&lt;li&gt;product quality assurance (module E);&lt;/li&gt;&#xA;&lt;li&gt;product verification (module F);&lt;/li&gt;&#xA;&lt;li&gt;unit verification (module G);&lt;/li&gt;&#xA;&lt;li&gt;full quality assurance (module H).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In our case, the one to follow is called Internal Production Control. Basically, what you have to do is write the technical documentation. This document should be made of those parts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;a general description of the electrical equipment,&lt;/li&gt;&#xA;  &lt;li&gt;conceptual design and manufacturing drawings and schemes of components, sub-assemblies, circuits, etc.,&lt;/li&gt;&#xA;  &lt;li&gt;descriptions and explanations necessary for the understanding of said drawings and schemes and the operation of the electrical equipment,&lt;/li&gt;&#xA;  &lt;li&gt;a list of the standards applied in full or in part, and descriptions of the solutions adopted to satisfy the safety aspects of this Directive where standards have not been applied,&lt;/li&gt;&#xA;  &lt;li&gt;results of design calculations made, examinations carried out, etc.,&lt;/li&gt;&#xA;  &lt;li&gt;test reports.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We learn what we have to do in order to respect the directives. The main constraint here will be showing that you have indeed accessed the harmonised standards. This can cost quite a hefty sum of money. However Afnor (the French certification organism) provides subscription to their database, which can be cost effective compared to buying individual standards). Your local certification organism should have a similar offer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The harmonised standards list to be followed is available here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;for the EMC directive: &lt;a href=&quot;http://ec.europa.eu/enterprise/policies/european-standards/harmonised-standards/electromagnetic-compatibility/index_en.htm&quot; rel=&quot;nofollow&quot;&gt;http://ec.europa.eu/enterprise/policies/european-standards/harmonised-standards/electromagnetic-compatibility/index_en.htm&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;for the LVD directive: &lt;a href=&quot;http://ec.europa.eu/enterprise/policies/european-standards/harmonised-standards/low-voltage/index_en.htm&quot; rel=&quot;nofollow&quot;&gt;http://ec.europa.eu/enterprise/policies/european-standards/harmonised-standards/low-voltage/index_en.htm&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This document will then have to be kept for up to ten years after the last product manufactured was sold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing this will allow you to affix the CE marking and provide a certificate of conformity to your customer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if this is a lighter method for CE certification, it still involves a bit of work. However, this is something that should be done anyway (if just to make sure your product will perform well and as expected in all conditions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT :&lt;/strong&gt;&#xA;There is a European Union website that shows all the currents harmonized standards and all the different Directives your product may be subject to:&#xA;&lt;a href=&quot;http://www.newapproach.org/Directives/&quot; rel=&quot;nofollow&quot;&gt;http://www.newapproach.org/Directives/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; At least two directives applies, one related to EMC, the other to Low Voltage Devices. The procedure to follow is called Internal Production Control. To comply, one has to draw up a technical documentation whose main part is the harmonised standard list followed is added.&lt;/p&gt;&#xA;" OwnerUserId="154" LastEditorUserId="154" LastEditDate="2015-03-15T11:38:11.247" LastActivityDate="2015-03-15T11:38:11.247" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="91" PostTypeId="2" ParentId="20" CreationDate="2015-01-21T10:57:29.683" Score="6" Body="&lt;p&gt;Taking this in stages, initially lets consider only a single front wheel, with no slip in any direction, where we have an accurate continuous measure of both rotational position and absolute angle. In this case calculating current position (relative to our starting point) is a relatively straightforward trigonometry and calculus problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately we don't know our absolute front wheel angle (unless we are free to equip it with a magnetometer or similar), instead we know our front wheel angle relative to the body of our vehicle. We can effectively define this as a bicycle type configuration, with a pivoting front wheel and a fixed rear wheel. The absolute angle of the front wheel can now be calculated from the absolute angle of the body (equal to the angle of the rear wheel) plus the relative angle of the front wheel to the body. An additional calculation is therefore required to determine the absolute angle of the rear wheel, based on our measurements from the front wheel. This will depend upon the distance between the front and rear wheels (consider the difference in turning circles between a single bike and a tandem to picture this). Again this is a trigonometry and calculus problem, this time in the reference frame of the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extending this to a four wheeled vehicle causes complications, as some of the wheels now have to slip. My mental image for this is two bicycles side-by-side, with bars connecting the frames together, and some mechanism ensuring that their steering is synchronised. If the bars connecting them are short then there is very little slip required. If they are very long (making the bikes far apart) then one or both of the front tyres will need to slip sideways to make tight turns. The distances travelled will also differ between the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further physical analysis from this point is likely to get very involved, and will depend upon frictions and masses on each wheel. A practical approach might be to take measurements from the rotation of the two front wheels separately and then average at some point in the equations discussed above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further complications arise when rotational slip in any of the wheels on the ground is considered, or if any of our measures are subject to error. To incorporate these would require a very detailed model of your vehicle and a better approach would probably be to fuse an estimate from our simplified analysis above with information from other sensors using a Kalman filter or similar. In this case it may be worth considering what states are estimated in your filter, as including absolute orientation as an explicit state and using it within your calculations may give you a better overall estimate of position. A clever filter might also include an estimate of slip as part of its measurement uncertainty.&lt;/p&gt;&#xA;" OwnerUserId="157" LastActivityDate="2015-01-21T10:57:29.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="92" PostTypeId="2" ParentId="49" CreationDate="2015-01-21T11:03:48.480" Score="4" Body="&lt;p&gt;I you look up sunspotter on the internet you get many things that look like &lt;a href=&quot;http://www.scientificsonline.com/product/sunspotter&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zmNed.gif&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this appears to do precisely what you say you don't want, i.e. imaging the sun whith a bit of calibration it could also be used to determine alignment. Effectively this is just a folded up telescope. You could align it by firing a light source perpendicular to the lens, a laser pointer would probably work well here, and marking a cross where it forms an image. If the sun image is centred on the same point then it is also aligned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could achieve the same thing with a standard telescope or one side of a pair of binoculars and some rig to hold the viewing screen in the correct position as you rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your other question the sun target in the microtops picture is just a lens to collect the light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also do it something like this. Just use a long stick or pole and find where the shadow disappears. Then use a plumb line and protractor to get the angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Fq8H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-01-21T11:03:48.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="93" PostTypeId="2" ParentId="88" CreationDate="2015-01-21T11:46:37.793" Score="6" Body="&lt;p&gt;In case Eurocodes do not provide enough information, some sources exist. In the case of elastic critical moment for lateral-torsional buckling, an NCCI (Non-contradictory, complementary information) document exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The document code is SN 003, and one version (maybe not the latest) can be accessed &lt;a href=&quot;http://eurocodes.jrc.ec.europa.eu/doc/WS2008/SN003a-EN-EU.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Hopefully, this will cover your current needs. In addition, the French technical centre for steel construction has developped a couple pieces of software that can run the calculation for you:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.cticm.com/content/logiciels&quot; rel=&quot;nofollow&quot;&gt;LTbeam&lt;/a&gt;: for beams under bending sollicitations&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.cticm.com/content/logiciels&quot; rel=&quot;nofollow&quot;&gt;LTbeamN&lt;/a&gt;: for beams under bending and compression sollicitations&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="137" LastEditorUserId="1832" LastEditDate="2016-01-05T10:51:15.697" LastActivityDate="2016-01-05T10:51:15.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="94" PostTypeId="5" CreationDate="2015-01-21T12:02:52.353" Score="0" Body="&lt;p&gt;The civil engineering tag covers the design, construction and maintenance of the built environment; including bridges, infrastructure and tall structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Civil engineering is typically involved when projects are large in scale or affect a large area of land.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use this tag when someone who has trained as a civil engineer is likely to be able to answer/ask your question. As with most disciplines, civil engineering encompasses a lot of specialisms and it may be necessary to further sub-categorize your question using additional tags.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Civil Engineering Sub-disciplines&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Civil engineering is a very broad field. If your question is specific to any of the following disciplines, consider using the more specific tag instead of or in addition to this one:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Geotechnical Engineering&lt;/li&gt;&#xA;&lt;li&gt;Structural Engineering&lt;/li&gt;&#xA;&lt;li&gt;Transportation Engineering&lt;/li&gt;&#xA;&lt;li&gt;Environmental Engineering&lt;/li&gt;&#xA;&lt;li&gt;Water Resources Engineering&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Engineering projects are often multi- and inter-disciplinary and the above list is not exhaustive.&lt;/p&gt;&#xA;" OwnerUserId="151" LastEditorUserId="33" LastEditDate="2016-02-23T02:41:13.763" LastActivityDate="2016-02-23T02:41:13.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="95" PostTypeId="4" CreationDate="2015-01-21T12:02:52.353" Score="0" Body="Questions that fall into the problem domain of civil engineering or one of its sub-disciplines. Questions about buildings or other structures could usually use the &quot;structural-engineering&quot; tag also." OwnerUserId="151" LastEditorUserId="33" LastEditDate="2016-02-23T02:41:13.763" LastActivityDate="2016-02-23T02:41:13.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="96" PostTypeId="2" ParentId="50" CreationDate="2015-01-21T12:58:52.213" Score="7" Body="&lt;p&gt;I will leave the main part of the previous answer below, as it contains more information about the differences between several radio protocols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the differences between Zigbee 3.0 and Z-Wave, let's do a side by side comparison:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Frequency: Zigbee uses the 2.4GHz band, while Z-wave uses the 868MHz band in Europe and the 900MHz ISM band in the US (see the frequency coverage &lt;a href=&quot;http://z-wave.sigmadesigns.com/docs/Z-Wave_Frequency_Coverage.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;OSI Layers: Zigbee is based on the IEEE 802.15.4 spec for the PHY/MAC Layers, where Z-Wave is based on ITU-T G.9959 rPHY/MAC. The Z-Wave protocol stack was developed by Sigma Designs (and the standard is maintained by the Z-Wave Alliance), whereas Zigbee is developed by the Zigbee Alliance. Both networks have mesh capabilities&lt;/li&gt;&#xA;&lt;li&gt;Ecosystem: Z-Wave claims to have &quot;more than 1200 products&quot; on the market, Zigbee claims a similar figure.&lt;/li&gt;&#xA;&lt;li&gt;Documentation: Zigbee 3.0 spec is (for now) only available to Alliance members (the other Zigbee standards can be downloaded after entering some information (which can be bogus if you somehow want to stay anonymous)). Z-Wave requires you to buy the development kit or join the Alliance to access the spec.&lt;/li&gt;&#xA;&lt;li&gt;Certification: Both protocols will require a certification to be fully certified and to be able to use the protocol logo.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;More generally, here is what you need to take into account when choosing a protocol:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Frequency: The protocols you named all uses the 2.4GHz band, but a lot of other protocols uses the 868MHz, the 915MHz, the 433MHz band. The idea is the lower the frequency, the lower the energy use and greater the range. At least in theory. One also have to consider the physical size of the antenna (the lower the frequency, the bigger the antenna).&lt;/li&gt;&#xA;&lt;li&gt;OSI Layers used: Some of the &quot;protocols&quot; out there only implements one layer of the OSI model, and you will need to have another layer implemented to be able to communicate&lt;/li&gt;&#xA;&lt;li&gt;Ease of use of protocol: some protocols are very easy to use (for example EnOcean), some are more complicated (Zigbee, BLE). They differ in their capabilities and the freedom you have to send the information&lt;/li&gt;&#xA;&lt;li&gt;Bitrate, frame size, useful payload: How much information you can send in a given timeframe. Depending on your use, you may need to send a lot of data, where different protocols may or may not behave well with this.&lt;/li&gt;&#xA;&lt;li&gt;Collision and packet collision avoidance: how does the protocol manages the band use and prevent packet collision (two objects sending at the same time).&lt;/li&gt;&#xA;&lt;li&gt;Ecosystem: Are there a lot of other devices out there using this protocol? What do you want to connect your object to (smartphone directly? Internet directly? Something else)? Do you need a local gateway to access internet?&lt;/li&gt;&#xA;&lt;li&gt;Documentation availability: Is all the information necessary freely available? Is it locked (does it require membership to an association to be read)? Is it behind a paywall?&lt;/li&gt;&#xA;&lt;li&gt;Last but not least: Cost of use. To be implemented and sold in your products, some protocols will require you to join the Alliance so you can certify your product compliance with the protocol. This often cost a lot of money (depends on the protocol).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I thinks this pretty much sums it all, and should give good pointers to the right direction to look for more information.&lt;/p&gt;&#xA;" OwnerUserId="154" LastEditorUserId="7083" LastEditDate="2016-07-21T00:31:00.460" LastActivityDate="2016-07-21T00:31:00.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="97" PostTypeId="2" ParentId="14" CreationDate="2015-01-21T13:04:29.280" Score="4" Body="&lt;p&gt;Farther from the equator, a passive house is mainly a very well insulated house with good exposure to solar lighting for passive solar heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in hot climate, one has to rethink completely the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here what matters more is the way you can use the night &quot;cool&quot; air to cool the house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you have to take great care to shading as best as you can all the exposed surface to the Sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, active systems (like air conditionning) can help you reach the desired level of comfort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something else is equally important: you do not need to cool down the house as much as you think. A temperature difference around 10-15°C with the outside is already a lot.&lt;/p&gt;&#xA;" OwnerUserId="154" LastActivityDate="2015-01-21T13:04:29.280" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="98" PostTypeId="2" ParentId="88" CreationDate="2015-01-21T13:55:56.583" Score="2" Body="&lt;p&gt;There are three options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Using paragraph 6.3.2.4 &quot;Simplified assessment methods form beams with restraints in buildings&quot; Which gives a conservative way to design a beam resistant to lateral torsion buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Looking in the Apendices and Supplementary informations of the National Annex for the country in question. In the Belgian annex, for example, there is a lengthy method for calculating Mcr by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Using a FE-program, capable of doing a stability analysis.&lt;/p&gt;&#xA;" OwnerUserId="80" LastEditorUserId="80" LastEditDate="2015-01-21T15:16:40.497" LastActivityDate="2015-01-21T15:16:40.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="99" PostTypeId="2" ParentId="35" CreationDate="2015-01-21T14:05:49.723" Score="14" Body="&lt;p&gt;In general, you want to stay below the recrystallization temperature. Steel is composed of grains, and different types of steel have different grain sizes. The size of these grains affects the steels behavior once it gets past the yield point. At the recrystallization temperature, new grains will nucleate and grow, which undoes any sort of hardening that the steel may have previously gone through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this temperature will vary depending on the alloying elements in the steel, so if you don't know the grade, it'll be hard to know the recrystallization temperature. Unless you're going to be working it around 900°F (500°C) I don't think that should be an issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This chart shows the temperatures at which different heat treatments are done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wl3VO.jpg&quot; alt=&quot;http://practicalmaintenance.net/?p=1329&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://practicalmaintenance.net/?p=1329&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="41" LastEditDate="2015-01-22T04:12:34.683" LastActivityDate="2015-01-22T04:12:34.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="100" PostTypeId="1" CreationDate="2015-01-21T15:10:36.230" Score="38" ViewCount="43659" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Prince_Rupert&amp;#39;s_Drop&quot;&gt;Prince Rupert's Drops&lt;/a&gt; are glass objects created by dripping molten glass into cold water. While the outside of the drop quickly cools, the inside remains hot for a longer time. When it eventually cools, it shrinks, setting up very large compressive stresses on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DxUTZ.jpg&quot; alt=&quot;courtesy of wikipedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is a sort of toughened glass: you can hammer the drop head without damaging it, but a scratch on the tail leads to an explosive disintegration. Check out &lt;a href=&quot;https://www.youtube.com/watch?v=xe-f4gokRBs#t=83&quot;&gt;this video.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is it possible to build spherical Prince Rupert's drops? And if so, how? &#xA;One example of an application is as a replacement for traditional ball bearing spheres. There will be improvements in wear resistance and maximum loads tolerable, and a glass sphere would cost less anyway.&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="421" LastEditDate="2015-09-29T01:17:12.017" LastActivityDate="2020-08-23T03:54:17.020" Title="Is it possible to build a perfectly spherical Prince Rupert's drop?" Tags="|materials|" AnswerCount="10" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="101" PostTypeId="1" CreationDate="2015-01-21T15:27:42.147" Score="8" ViewCount="1480" Body="&lt;p&gt;Sometimes underground pipelines are exposed due to vertical scour or lateral migration of streams (or other erosion). One option is to lower the (in-service) pipeline to a new depth that provides sufficient ground cover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine the mechanical stresses on the pipe, and determine how much the line can be lowered? Also required is determining what length of trench will be required to accommodate the lowering. What are the relevant codes or standards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is in relation to steel pipelines for both gas and liquids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1JXrL.png&quot; alt=&quot;Pipeline Lowering Profile&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="167" LastActivityDate="2015-01-21T17:27:54.007" Title="How can I calculate the stresses and limits for the physical lowering of in-service pipelines?" Tags="|mechanical-engineering|pipelines|stresses|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="102" PostTypeId="2" ParentId="88" CreationDate="2015-01-21T15:44:32.220" Score="3" Body="&lt;p&gt;I know it's tagged Eurocodes, but another answer for the Americans: specification section F of the AISC manual (pg. 16.1-44 of the 13th edition) delineates the equations applicable to lateral-torsional buckling. For the usual case of doubly symmetric I shapes bent about the major axis (and normal, &quot;compact&quot; flanges), it's section F2 (pg. 16.1-47). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, unless you're creating a program to do your designs automatically (like in Mathcad or Excel or something), it is so much easier to just use the design tables (see section 3-5). &lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2015-01-21T15:44:32.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="103" PostTypeId="2" ParentId="63" CreationDate="2015-01-21T17:01:36.060" Score="15" Body="&lt;p&gt;There's a hierarchy of junctions, that you'll find in most standard highway design guides, such as the UK's immense &lt;a href=&quot;http://www.standardsforhighways.co.uk/dmrb/&quot;&gt;Design Manual for Roads and Bridges&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different junction designs have different motor-vehicle capacities, and operate best at different speeds. Within towns and cities, junction capacity is critical: it is that that determines the network's capacity. It's different for motorways, where it's link capacity that determines network capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Junction design can take a whole bunch of different factors into effect: available space; movement of non-motor vehicles, pedestrian connectivity, and so on. Nevertheless, for a few unfortunate decades in the late 20th Century, and even today in less enlightened places, junction capacity was/is used as &lt;em&gt;the&lt;/em&gt; determining factor to select which form of junction to use, within spatial constraints. In other places, it is just one consideration amongst many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest-capacity junctions are grade-separated. The next highest are signalised roundabouts, then unsignalised roundabouts, then signalised cross-roads, then unsignalised cross-roads have the lowest capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the biggest criterion that separates roundabouts from cross-roads. All other factors are a question of design, regulation, and patterns of behaviour: safety of all, comfort for pedestrians, ease of use for cyclists, landscape impact, cost - any of these can be better for roundabouts, or worse for roundabouts, depending on the particular design.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2015-01-21T17:01:36.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="104" PostTypeId="2" ParentId="101" CreationDate="2015-01-21T17:27:54.007" Score="6" Body="&lt;p&gt;This is something that I've looked at with tunnels, rather than pipes, and arguably with smaller deflections. Hopefully it'll be some help, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can satisfy yourself that the rate of curvature is relatively small- then you can approximate the stress generated by the lowering using &lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Beams/Beam_theory.html&quot;&gt;simple beam theory&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma=\frac{Ey}{R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;E is the Youngs Modulus&lt;/li&gt;&#xA;&lt;li&gt;y is the distance from the neutral axis you are measure the stress at (i.e. the radius of the pipe itself)&lt;/li&gt;&#xA;&lt;li&gt;R is the &lt;a href=&quot;http://www.intmath.com/applications-differentiation/8-radius-curvature.php&quot;&gt;radius of curvature&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Mechanically, you'll need to check that the stresses in the pipe along the parabaola that defines your curve never exceed the limiting stress of the pipe material, and (probably more critcally) the capacity of the connections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a bit of a simplification, and depending on the type of pipe, the joints themselves will have some play.&lt;/p&gt;&#xA;" OwnerUserId="151" LastActivityDate="2015-01-21T17:27:54.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="105" PostTypeId="2" ParentId="49" CreationDate="2015-01-21T17:33:39.910" Score="3" Body="&lt;p&gt;Two points make a line, so a long tube with a photodiode on the bottom should be able to find the sun... (assuming no clouds.)   In regards to Solar Photometry I think &lt;a href=&quot;http://www.forrestmims.org/scientificresearch.html&quot; rel=&quot;nofollow&quot;&gt;Forrest Mims&lt;/a&gt; is making photometers from the wavelength response of different LEDS.. using LEDs as detectors.  &lt;/p&gt;&#xA;" OwnerUserId="168" LastActivityDate="2015-01-21T17:33:39.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="106" PostTypeId="2" ParentId="66" CreationDate="2015-01-21T17:36:30.193" Score="2" Body="&lt;p&gt;(How do you know the signal frequency?)  OK this may be over the top for your application, but if you have access to the drive (reference) signal, then you can make a lock-in or synchronous detector.  &lt;/p&gt;&#xA;" OwnerUserId="168" LastActivityDate="2015-01-21T17:36:30.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="107" PostTypeId="2" ParentId="100" CreationDate="2015-01-21T17:50:59.957" Score="1" Body="&lt;p&gt;Well forget the &quot;perfect&quot; sphere, but I don't see why it couldn't be made in any shape.  You just have to cool the outside fast.  I seem to recall that pyrex is made this way, with built in stresses.. but I couldn't find a link.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Toughened_glass&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; may be helpful.  &lt;/p&gt;&#xA;" OwnerUserId="168" LastActivityDate="2015-01-21T17:50:59.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="108" PostTypeId="2" ParentId="62" CreationDate="2015-01-21T18:24:44.647" Score="5" Body="&lt;p&gt;If you have a DMM with a uA current range, then a photodiode is the simplest detector.  I get about 15 uA of current in my lab (PD area is 44 mm^2)  An LED held up close can give more than 1 mA of current.    &lt;/p&gt;&#xA;" OwnerUserId="168" LastActivityDate="2015-01-21T18:24:44.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="109" PostTypeId="5" CreationDate="2015-01-21T18:37:15.673" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-21T18:37:15.673" LastActivityDate="2015-01-21T18:37:15.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="110" PostTypeId="4" CreationDate="2015-01-21T18:37:15.673" Score="0" Body="Use for questions about designing, manufacturing or synthesizing engineered materials or about the properties of specific materials. Questions concerning common construction materials, such as wood, steel or masonry do not need to use this tag; use the more specific tag instead." OwnerUserId="95" LastEditorUserId="368" LastEditDate="2015-03-26T18:37:45.563" LastActivityDate="2015-03-26T18:37:45.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="111" PostTypeId="2" ParentId="28" CreationDate="2015-01-21T19:21:53.060" Score="4" Body="&lt;p&gt;I know your question is about springs, but I don't think that springs are the answer to your desired application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mentioned, the inward force from air pressure is very large so not only do you need a strong spring, but you also need a material for the paneling which is light and can take those kind of forces. Additionally, you would need to evacuate the air that seeps in through the material at a molecular level. There are good reasons lighter-than-air aircraft use a low-density gas inside the balloon:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is no requirement for exceptionally strong and light materials.&lt;/li&gt;&#xA;&lt;li&gt;There is no (or very low) pressure differential between the inside and outside, and therefore no tendency for the gas to escape or the atmosphere to seep in.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You may be interested to read the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum_airship&quot; rel=&quot;nofollow noreferrer&quot;&gt;vacuum airships&lt;/a&gt;, in particular the section on &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum_airship#Material_constraints&quot; rel=&quot;nofollow noreferrer&quot;&gt;material constraints&lt;/a&gt;. This shows the proof that even using a strong shape (hollow sphere) made of diamond, the pressures are too great for the structure to withstand once you thin it down to be light enough to become buoyant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7yfAv.png&quot; alt=&quot;Flying Boat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that in these ideal conditions it is not possible, it seems unlikely that a spring in a balloon would be possible. However, the Wikipedia article does leave open the possibility that if the wall of the sphere was not solid, for example if it was made of a honeycomb structure, then it may still be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: Apparently the &lt;a href=&quot;http://worldwide.espacenet.com/textdoc?DB=EPODOC&amp;amp;IDX=US2007001053&quot; rel=&quot;nofollow noreferrer&quot;&gt;maths has been done&lt;/a&gt; to show this is possible in theory, but it's not clear if it has been successfully implemented.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="95" LastEditDate="2015-04-30T12:05:19.350" LastActivityDate="2015-04-30T12:05:19.350" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="112" PostTypeId="2" ParentId="26" CreationDate="2015-01-21T19:34:43.540" Score="6" Body="&lt;p&gt;&lt;strong&gt;Fission Products Extraction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seperation of plutionium and uranium from the other fission products is done with the organic molecule &lt;a href=&quot;http://en.wikipedia.org/wiki/Tributyl_phosphate&quot; rel=&quot;noreferrer&quot;&gt;tributyl phosphate&lt;/a&gt; by &lt;a href=&quot;http://en.wikipedia.org/wiki/Liquid%E2%80%93liquid_extraction&quot; rel=&quot;noreferrer&quot;&gt;liquid-liquid-extraction&lt;/a&gt;. You have two phases, one organic and one aqueous. The fission products will solve in the aqueous phase and uranium/plutonium will solve in the organic phase with the tributyl phosphate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Uranium/Plutonium Separation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To seperate uranium from plutonium you have to reduce the plutonium with uranium nitrate. Again, you have two streams: One with the uranium/plutonium from before and one aqueous stream with uranium nitrate (U4+). Plutonium will be reduced and solve in the aqueous solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To accomplish the aforementioned chemical processes, the liquid-liquid extraction you can use the following desgins:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mixer-Settler&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lOtdo.png&quot; alt=&quot;Mixer-Settler&quot;&gt; (&lt;a href=&quot;https://scholar.google.de/scholar?cluster=16705612692502180121&amp;amp;hl=de&amp;amp;as_sdt=0,5&amp;amp;sciodt=0,5&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pulse Column&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AbjZf.png&quot; alt=&quot;Column&quot;&gt; (&lt;a href=&quot;https://scholar.google.de/scholar?cluster=16705612692502180121&amp;amp;hl=de&amp;amp;as_sdt=0,5&amp;amp;sciodt=0,5&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle is always: Organic phase is lighter than the aqueous phase. Both phases are first separated, than mixed, and then again separated. During the mixing the chemical reactions occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see the mentioned &lt;a href=&quot;https://scholar.google.de/scholar?cluster=16705612692502180121&amp;amp;hl=de&amp;amp;as_sdt=0,5&amp;amp;sciodt=0,5&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt; for more information. &lt;/p&gt;&#xA;" OwnerUserId="46" LastActivityDate="2015-01-21T19:34:43.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="113" PostTypeId="2" ParentId="62" CreationDate="2015-01-21T19:43:24.740" Score="3" Body="&lt;p&gt;The brightness of a light emitting diode (or anything else) is measured by &lt;a href=&quot;http://en.wikipedia.org/wiki/Luminous_flux&quot; rel=&quot;nofollow&quot;&gt;luminous flux&lt;/a&gt;, which is the impact of the light on an eye, adjusted for different wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luminous flux is measured by lumens, which corresponds to candelas of light, or luminous intensity, emitted over a solid angle.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2015-01-21T19:43:24.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="114" PostTypeId="2" ParentId="87" CreationDate="2015-01-21T20:15:46.797" Score="4" Body="&lt;p&gt;There are a number of standards to address hydrogen embrittlement in the manufacture and coating of fasteners. Here in the US, ASTM F1940 and ASTM F519 would apply. Hydrogen embrittlement is also a concern for structural welds, but should be addressed by your welding code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem doesn't seem to be in the spec as often as in the quality control at the point of manufacture. In situations where this is a big risk, requiring third party inspections and testing are likely the most effective step the engineer can take.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-01-21T20:15:46.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="115" PostTypeId="2" ParentId="89" CreationDate="2015-01-21T20:19:35.730" Score="11" Body="&lt;p&gt;So there's an incorrect assumption underlying your question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In an ideal world the lifting capacity required of the jacks would be the self-weight of the bridge divided by the number of jacks (+ allowances for wind/snow, etc.).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the assumption there is that the lifting capacity is equivalent only to the weight of the bridge.  The problem is that if anything goes wrong, you're likely to see a catastrophic failure of some sort which could lead to irreparable damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real world lifts don't operate in that &quot;ideal&quot; manner, and instead rely upon a &lt;a href=&quot;http://en.wikipedia.org/wiki/Factor_of_safety&quot;&gt;safety factor&lt;/a&gt; in order to make sure that the lifted weight is well within the limitations of the equipment.  And in some cases, the safe working limit (SWL) may be derated further from the working lift limit (WLL) if there are any extenuating circumstances such as worn equipment or hazardous weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the ideal lift capacity is one that is significantly larger than load to be lifted.  The actual lift capacity used is tempered by the fact that you generally pay for that lift capacity whether you need it or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Factor_of_safety&quot;&gt;safety factors&lt;/a&gt;, a factor of 2 is common with building materials and 3 is common for automobiles.  You need to weigh the risk to human health or safety within the lift you're considering and use an appropriate safety factor.  A conservative approach would be to use a higher safety factor of 3, so you need at least 3 times the bridge weight for the lifting capacity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Assuming you're staying within the SWL and WLL of the lift equipment you're using, that still doesn't necessarily account for the binding forces caused by corrosion between the bridge and the bearings supporting it.  Static friction can also come into play if the bridge itself has to be slid out of the supporting structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, it's hard to determine what that binding force is going to add up to without a lot more detail.  At a minimum, you would need to know the materials involved and the cross sectional area of contact between them.  You would also want to approximate how long they've been exposed to the elements and what sort of conditions the elements have brought - such as salt water exposure vs mountain air.&lt;br&gt;&#xA;&lt;sub&gt;This is where I'm going to wave my hands in an airy fashion and not attempt to swag the binding force created by corrosion.&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Friction&quot;&gt;Static frictional force&lt;/a&gt; can be guessed at a little more easily than the corrosion binding.  $\mu_{static}$ ranges from 0.6 to 0.8 for various materials in contact with steel.  And while the general formula for calculating the &lt;code&gt;Force&lt;/code&gt; from friction is $F_{static} = \mu_{static} * F_{normal}$, that equation also assumes horizontal movement.  As you're likely lifting vertically, not sliding horizontally, the static frictional forces will be less since the equivalent $F_{normal}$ for vertical motion will be less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based upon that, I'd use a conservative guess of $\frac 23$ or $\frac 34$ of the weight of the bridge to estimate the static friction forces involved in the lift.  Experience and the particulars of the lift will guide you in adjusting that guess up or down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So depending upon your safety factor, it could very well be that the SWL of the equipment will provide sufficient lift to overcome any static friction or binding caused by corrosion.  Or it could be that you need to increase the lift capacity requirements to overcome that effect.  And it's worth pointing out that equipment can exceed those limits, so a lower safety factor may be &quot;good enough&quot; to get past the effects of static friction at the start of the lift.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-01-22T12:58:23.793" LastActivityDate="2015-01-22T12:58:23.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="116" PostTypeId="2" ParentId="54" CreationDate="2015-01-21T21:20:10.177" Score="3" Body="&lt;p&gt;Considering that reverse osmosis is not the only way to desalinate water, I think that yes, there is a lot of development potential in desalination, but that potential might not lie in improvements to existing techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To justify this conclusion and illustrate some areas where there could be a lot of development potential, I present to you my idea for a combined wave, wind, and solar desalination and power plant. I haven't done any maths on this to calculate the area of land needed, or costs, or output, so it might not be viable as-is. But I think the concepts described below (and remember this is just one idea) demonstrate that there is development potential in the following areas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Using renewable energy sources on-site to power the plant&lt;/li&gt;&#xA;&lt;li&gt;Using direct-drive energy instead of electrically transmitted energy&lt;/li&gt;&#xA;&lt;li&gt;Directing and amplifying natural processes of desalination&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Combined wave, wind, and solar desalination and power plant&lt;/h2&gt;&#xA;&#xA;&lt;h3&gt;Inputs&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No external energy input&lt;/li&gt;&#xA;&lt;li&gt;Cleverly harnessed wave, wind, and solar&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Outputs&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Energy (electricity)&lt;/li&gt;&#xA;&lt;li&gt;Fresh water&lt;/li&gt;&#xA;&lt;li&gt;Cool air&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Location&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This plant requires a hot location with large area of cheap land by the ocean and a relatively consistent wind.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Stage 1 - Wave Pump&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A wave-powered pump raises sea water into a large lake on land. Here is an example of a direct &lt;a href=&quot;http://machinedesign.com/news/wave-driven-water-pump-has-no-moving-parts&quot; rel=&quot;nofollow&quot;&gt;wave-powered pump&lt;/a&gt;, other types of wave power harnessing typically convert mechanical motion into electricity. However, that motion can be easily used to directly drive a pump.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Stage 2 - Evaporation Lake&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The evaporation lake is a large shallow area, covered in a greenhouse-like way to aid evaporation. The sea-water flows away from the ocean along channels in the lake-bed then back again towards the ocean in the next adjacent channel where it drains back into the sea. This prevents the build-up of deposits as the returning sea-water will take them with it and return to the sea more concentrated. The roof may contain Fresnel lenses or other solar concentrators to help evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wind-catching tower blows air across the lake to lower air pressure and aid in evaporation. This tower could be like &lt;a href=&quot;http://www.triplepundit.com/2011/01/masdar-wind-tower-abu-dhabi/&quot; rel=&quot;nofollow&quot;&gt;those used in Masdar City&lt;/a&gt;, or a standard wind turbine tower with either electrical or direct transmission to a series of fans. The result is a continuous airflow across the lake which carries the water vapor to the far side where it is channeled up a wide column into the next stage.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Stage 3 - Condensing Tower&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The water vapor is channeled up a large column to a condensing chamber high on the tower. Here, a series of fins are cooled by a heat-pump driven directly by a wind-turbine on top of the tower. the water condenses on the fins and drains into a fresh-water tank near the top of the tower.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Stage 4 - Power Generation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The water from the condensing tower is lowered to a height suitable for a standard water-tower through one or more water turbines to generate power.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Stage 5 - Filtering and treatment&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The salty sea-air will also condense on the fins, and there may be small airborne particles and particles from wear on stages within this process that get into the water, so it will probably need further filtering and treatment to make it drinkable. Some of the power from the water turbine may be used for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There you have it, you have clean water, above ground level so pressure is already available, and hopefully some excess electricity and cool dry air as by-products.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="95" LastEditDate="2015-01-22T00:19:35.460" LastActivityDate="2015-01-22T00:19:35.460" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="117" PostTypeId="2" ParentId="87" CreationDate="2015-01-21T21:27:02.513" Score="5" Body="&lt;p&gt;Like any number of engineering problems, solutions for hydrogen &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrogen_embrittlement&quot; rel=&quot;noreferrer&quot;&gt;embrittlement&lt;/a&gt; fall into two categories: 1) prevention and 2) cure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Embrittlement,&quot; in this context, arises from the diffusion of hydrogen into metal. If it is not too severe, a possible cure is &quot;reverse diffusion&quot; that takes hydrogen away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cure doesn't always work, if the hydrogen finds its way into small pockets of metal and combines with other gases, e.g. carbon, to produce methane, a corrosive. Then, the only real solution is prevention. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One form of prevention is coating the metals to prevent contact with the hydrogen. In welding applications, it is also possible to separate the hydrogen from the metal by heating the metal.&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2015-01-21T21:27:02.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="118" PostTypeId="2" ParentId="84" CreationDate="2015-01-21T21:36:41.663" Score="0" Body="&lt;p&gt;I'm not a pump expert but I would have thought that an Archimedes screw pump would be ideal in such a situation. &lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-21T21:36:41.663" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="119" PostTypeId="1" CreationDate="2015-01-21T21:59:10.387" Score="12" ViewCount="425" Body="&lt;p&gt;I have recently been told that my &quot;salesmanship&quot; is lacking when it comes to getting my ideas implemented.  I receive the feedback that my ideas are sound and the data to implement is compelling, and even that the ideas are presented in a way that management can understand, but I am not doing a good job of &quot;selling&quot; the idea when it comes time to implement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel a little odd about this because on the one hand I can understand that, as an engineer, I haven't honed this skill.  However, on the other hand I feel almost blindsided that this is a skill I am expected to have; the decision-makers have already said, &quot;go implement&quot;, but when I go to implement, the people with whom I am supposed to partner put up barriers to progress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering how others deal with this.&lt;/p&gt;&#xA;" OwnerUserId="35" LastActivityDate="2015-02-09T10:17:37.293" Title="In Engineering fields, how often do you find yourself being asked to be a &quot;salesperson&quot;, and what is your reaction?" Tags="|communication|" AnswerCount="6" CommentCount="0" ClosedDate="2015-12-21T15:34:08.683" ContentLicense="CC BY-SA 3.0" />
  <row Id="120" PostTypeId="1" AcceptedAnswerId="123" CreationDate="2015-01-21T22:09:23.393" Score="5" ViewCount="757" Body="&lt;p&gt;I've looked around for different types of &lt;a href=&quot;http://en.wikipedia.org/wiki/Concrete_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;concrete pump&lt;/a&gt;s and simply don't know if there are many different types, what their mechanisms are and which ones are the most predominant ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've found:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This &lt;a href=&quot;https://www.youtube.com/watch?v=dgmRN3BDE4E&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; explains how a pump with an S-shaped tube operates.&lt;/li&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Concrete_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;relevant wikipedia page&lt;/a&gt; explains another type, which does differ in the way how the concrete is released&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l9sDNm.jpg&quot; alt=&quot;An example of a concrete pump&quot;&gt; (&lt;em&gt;a concrete pump&lt;/em&gt;)&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="46" LastEditDate="2015-01-26T11:34:36.777" LastActivityDate="2017-04-02T12:49:39.267" Title="What are the predominant types of concrete pumps and what are their working mechanisms?" Tags="|mechanical-engineering|pumps|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="121" PostTypeId="1" AcceptedAnswerId="134" CreationDate="2015-01-21T22:18:05.123" Score="12" ViewCount="2899" Body="&lt;p&gt;How are spindle speeds calculated for a drill bit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen dozens of charts that highlight the rpm that should be used for specific drill bit types, bit diameter, and material. However, what if my chart doesn't have the particular type of material or bit that I am using? I'd also like to have some intuition to know if the chart looks right or wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon some quick research, it appears that the &quot;cutting speed&quot; is what is ultimately needed for a particular material. I assume the cutting speed for each material must be looked up? Is there a standard or &quot;go-to&quot; place to find these? Then, information about the drill bit can be used to determine spindle speed. Again, what if I'm using a big hole saw or circle cutter, and it's not listed? How do I model the bit to use the cutting speed to determine rpm (for a given material of course)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to know to calculate feed speeds for a drill or mill, but there are presumably more variables. It is probably better answered in another question.&lt;/p&gt;&#xA;" OwnerUserId="161" LastActivityDate="2021-03-02T21:47:12.207" Title="Determining Drill Speed" Tags="|machining|mechanical-engineering|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="122" PostTypeId="1" CreationDate="2015-01-21T22:20:09.830" Score="3" ViewCount="27140" Body="&lt;p&gt;All of these theories are in use today, and anyone can look up the definition of each theory in a textbook. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what are the practical applications for each theory, or said another way, in what situations should one theory be preferred to the others? Compilation of a general list would be most useful. &lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2015-01-22T12:25:53.843" Title="Applications for Rankine, Coulomb, and Terzaghi horizontal earth pressure theories" Tags="|civil-engineering|geotechnical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="123" PostTypeId="2" ParentId="120" CreationDate="2015-01-21T22:20:42.393" Score="3" Body="&lt;p&gt;I've found &lt;a href=&quot;http://www.ritchiewiki.com/wiki/index.php/Concrete_Pump&quot; rel=&quot;nofollow&quot;&gt;a fairly comprehensive source&lt;/a&gt;. There are five main types:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mechanical pumps:&lt;/strong&gt; A piston and a rotary valve are used the push the concrete through.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hydraulic pumps:&lt;/strong&gt; Hydraulics are used to pump the concrete.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Schwing pumps:&lt;/strong&gt; Two cylinders are used: One to receive concrete from the input hopper and one to release it into the output pipe. The two cylinders are connected by a flat gate valve. Apparently, Fredrich Schwing started &lt;a href=&quot;http://schwing.com/about-us/&quot; rel=&quot;nofollow&quot;&gt;a company&lt;/a&gt; to manufacture concrete pumps. Unfortunately, their website does not have much information.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Thomsen pumps:&lt;/strong&gt; Similarly to the Schwing pump, two cylinders are used with internal pistons. The difference is that a flapper valve (a type of &lt;a href=&quot;https://en.wikipedia.org/wiki/Check_valve&quot; rel=&quot;nofollow&quot;&gt;check valve&lt;/a&gt;) is used.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Squeeze-crete pumps:&lt;/strong&gt; A rotating cylinder is used to squeeze the concrete through.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-01-25T01:03:18.137" LastActivityDate="2015-01-25T01:03:18.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="124" PostTypeId="2" ParentId="119" CreationDate="2015-01-21T22:49:02.770" Score="11" Body="&lt;p&gt;I think I know what you mean. The terms &quot;selling&quot; and &quot;salesmanship&quot; are somewhat confusing in this context. Really you just need to be able to explain the situation and the reasons behind it in a context that the other side can understand. This differs when talking to management or engineers. I would call this more justification than selling.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;1. Justifying to management&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If you are talking to management to get a green-light for an idea, you need to not only explain why it is technically viable, but also what the business benefits are of this idea above other ideas. For example: it costs less to initially implement; it is easier to maintain (costs less); it is harder to maintain (creates lucrative service contracts)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is 'good' really depends on the business strategy of the company as highlighted by the last two examples which are opposite from a technical perspective, but both might be considered beneficial under different business strategies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't like or disagree with the business strategy of your company, leave and start your own business and treat your customers better. Or at least you can begin to understand why your ideas are not adopted despite being technically fantastic.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;2. Justifying to engineers&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;It sounds like you've already got the go ahead from management and now you're struggling to convince other engineers why this way is actually better. Well, the reason you're struggling is because those other engineers don't understand the business strategy and how your idea fits in with it. So now you find yourself in the odd position of having to explain the business strategy to engineers. You don't have to justify the business strategy itself, only why your technical solution fits in better with the business strategy dictated to you from management. You need to do this in terms that non-business oriented engineers will understand, which can be tricky because quite often the best technical solution is at odds with the business strategy.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I'll use a real-world example that I had to deal with recently to explain. This is from a software company, but applicable to any engineering situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The company wanted to have two separate work streams (two teams) doing essentially the same thing, creating two different products (that do the same thing) for two markets. Utter madness technically, especially when both products could share so many components and in fact could even be the same product with some minor configuration options to address the differences in these two markets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers were struggling to agree with this approach until I explained to them that 80% of the revenue was coming from one market, and the competition there was also much more fierce, so the business strategy was to focus one team solely on this market, without any consideration for supporting the other market so that they could move quickly and stay ahead of the competition. The secondary market was still worth exploiting and growing, so the second team would focus on that. This is the strategy given to us, this is the problem we need to solve, not that other problem of making the single-product.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Conclusion&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I don't think you need to learn sales, you just need to be able to identify when other engineers are trying to solve a different problem than you (normally caused by ridiculous non-technical things like customers, profit, economics, and all that malarkey). Then you need to be able to explain what the problem actually is and why, so that they jump on board with your (already accepted by management) idea.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-21T22:49:02.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="125" PostTypeId="2" ParentId="119" CreationDate="2015-01-21T23:14:38.160" Score="6" Body="&lt;p&gt;&quot;Salesmanship&quot; in this context, means &quot;being able to overcome resistance.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Resistance&quot; can take one of two forms. It could be &quot;reasonable&quot; resistance for e.g. economic reasons, whereby the proposed technical solution is not economically feasible. In this case, you might do well to find a cheaper solution, or at least an acceptable compromise. Here, you don't need classical &quot;selling&quot; skills but you do need some &quot;marketing&quot; skills to put together a package that will convince the bean counters that you won't break the bank. These people will at least be direct about their objections that &quot;the math doesn't work,&quot; or &quot;the numbers don't add up&quot; [to a profit].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;unreasonable&quot; kind of resistance takes place if you happen to be going against vested interests. Then you need to find out what those interests are, who the &quot;decision maker&quot; is, and whether your technical solution can be revamped so that those objections disappear. This is more of an internal, &quot;political&quot; situation than a straight &quot;selling&quot; situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not everyday engineering issues, but do occur from time to time. I'd field them as well as possible because these events could make or break an engineering career. This is particularly true if you have aspirations to leave engineering and go into &quot;management.&quot;&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditDate="2015-02-03T21:52:58.327" LastActivityDate="2015-02-03T21:52:58.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="126" PostTypeId="1" AcceptedAnswerId="408" CreationDate="2015-01-21T23:55:26.270" Score="4" ViewCount="3412" Body="&lt;p&gt;What is an effective method to re-engineer a current product to reduce engineering bill of material (EBOM) cost?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a continues improvement project for a current product in electronics space. Attached is an example of a PCB board. The product consists of many similar boards. The BOM includes active and passive electronic components such as capacitors, resistor, microprocessors, transistors, PCB boards, connectors etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is an effective method to tackle this engineering cost reduction problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UCzai.jpg&quot; alt=&quot;Beaglebone black&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="282" LastEditDate="2015-02-07T17:24:08.267" LastActivityDate="2015-05-19T17:48:41.350" Title="How to reduce Engineering Bill of Material Cost for electronics products?" Tags="|electrical-engineering|cost-engineering|" AnswerCount="6" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="127" PostTypeId="5" CreationDate="2015-01-22T00:23:15.273" Score="0" Body="&lt;p&gt;3D-printing is the process of creating a three-dimensional object using a the same principles as printing with ink. Layers of plastic or other materials are slowly built up inside a device typically the size of a large desktop printer. 3D printing is a somewhat recent invention, but it has revolutionized the maker movement by making prototypes and models easy to make.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="33" LastEditDate="2016-02-23T03:06:55.637" LastActivityDate="2016-02-23T03:06:55.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="128" PostTypeId="4" CreationDate="2015-01-22T00:23:15.273" Score="0" Body="The process of creating three-dimensional objects out of plastic or other materials using a device similar to a printer." OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-01-22T01:04:49.463" LastActivityDate="2015-01-22T01:04:49.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="129" PostTypeId="5" CreationDate="2015-01-22T00:25:17.873" Score="0" Body="&lt;p&gt;Questions about the properties of liquids or systems where the properties of liquids play a crucial role.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="33" LastEditDate="2016-02-23T03:19:53.713" LastActivityDate="2016-02-23T03:19:53.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="130" PostTypeId="4" CreationDate="2015-01-22T00:25:17.873" Score="0" Body="For questions about the specific properties of materials in their liquid states, and their engineering applications, such as in hydraulics, heating/cooling systems, and chemical reactions." OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-01-22T18:03:42.177" LastActivityDate="2015-01-22T18:03:42.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="131" PostTypeId="2" ParentId="126" CreationDate="2015-01-22T00:38:13.123" Score="8" Body="&lt;p&gt;This is quite a vague question, so only a very broad and vague answer can be given. Here are some ways to achieve this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Invest in marketing to increase sales, or hold more stock, so that you can buy parts in larger quantities and get better pricing.&lt;/li&gt;&#xA;&lt;li&gt;Negotiate with suppliers and make a commitment to buy a large quantity over time in order to get better pricing without a large initial investment.&lt;/li&gt;&#xA;&lt;li&gt;Look for areas of your design where some parts are potentially unnecessary. For example non load-bearing parts may be clipped on or glued in place rather than using bolts. Fewer or smaller parts may be used in some areas, etc.&lt;/li&gt;&#xA;&lt;li&gt;Split your product into a core product with optional add-ons to reduce the base-price for entry-level customers.&lt;/li&gt;&#xA;&lt;li&gt;Look for alternate materials and alternate parts that achieve the same function without compromising the product design goals.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-22T00:38:13.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="132" PostTypeId="2" ParentId="126" CreationDate="2015-01-22T00:52:27.317" Score="6" Body="&lt;p&gt;I agree with jhabbott that this is pretty vague. Here are some more ideas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Buy in bulk.&lt;/strong&gt; In most cases, it's much cheaper per item to buy ten items that it is to buy one. The companies selling you the materials will most likely give you a slightly cheaper price, which they'll sacrifice in return for the larger order. This rule may not apply for any materials that have to be used very quickly after they are purchased.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Buy more, and buy what you can sell.&lt;/strong&gt; This is a little bit unorthodox, but it could still work. Buy more than your plans indicate you'll need. This way, even if you've properly planned everything out, if a problem crops up, you'll have extra materials, and you won't have to put the project on hold to get more. The latter part of this bullet means that you should be prepared to resell additional materials that you don't use. Clearly, there will be a whole lot of things that can't be resold. But some surplus items may be marketable when the job is done. For example, if you were 3D printing an object, you could resell some of the unused plastic. Alternatively, save this materials for another, later project.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Get some of the above materials.&lt;/strong&gt; In many cases, it might be cheap to buy materials from a company following the above bullet. They might be happy to get some items off their hands, and you could get a good price for it. This reminds me of banks foreclosing (not a prefect analogy). They'd like to get rid of the house because while they possess it, their asset is simply sitting there. They can't profit from it. So they'd like someone to buy it.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Haggling with suppliers.&lt;/strong&gt; Curses, jhabbott beat me to this. Anyway, be aware that the purchase price isn't necessarily the price you have to pay. Again, some companies might be willing to offload some materials for a lower price. Don't be afraid to haggle if you have to, although large-scale project managers may not be too happy to see this technique used.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-01-22T00:52:27.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="133" PostTypeId="2" ParentId="121" CreationDate="2015-01-22T01:24:20.457" Score="3" Body="&lt;p&gt;I don't know that there's much materials science behind those charts.  I suspect they are amalgamations of collective experience and wisdom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general rule, the bigger the bit the slower the speed should be.  So there's an inverse relationship between acceptable maximum speed and the size of the hole that's being cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, you can cut a small hole at a small speed if you're patient.  But most folk would rather move on to the next step and are looking for the fastest speed that they can cut that particular size hole with.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So why is that general rule in place then?  As a bit grinds against a particular material, it's going to generate heat from the frictional forces involved.  Bigger bits mean bigger surface areas so you therefore have greater amounts of heat.  That heat generation is multiplied by the rotational speed of the bit as it reflects the amount of area ground per period of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by slowing the really big bit down, you're decreasing the amount of surface area covered in the same amount of time.  That keeps the amount of frictional heat down (since it can dissipate more easily) and reduces the chances that you'll destroy the temper on the bit.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-01-22T01:24:20.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="134" PostTypeId="2" ParentId="121" CreationDate="2015-01-22T01:56:37.213" Score="9" Body="&lt;p&gt;You are correct that the cutting speed of the material is what determines the rpm for your drill-bit. This actually makes the calculation very simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{Spindle speed (RPM)} = \frac{\text{Cutting speed}}{\text{Circumference}} = \frac{\text{Cutting speed}}{π \cdot \text{Diameter}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing you need to be careful of is the units of cutting speed and diameter. For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Metric:&lt;/strong&gt; If your cutting speed is in $m/min$ and your diameter is in $mm$ then you need to multiply your cutting speed by 1000 so that it is in $mm/min$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Imperial:&lt;/strong&gt; If your cutting speed is in $ft/min$ and your diameter is in $inches$ then you need to multiply your cutting speed by 12 so that it is in $inches/min$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For more information see &lt;a href=&quot;http://en.wikipedia.org/wiki/Speeds_and_feeds#Spindle_speed_calculations&quot; rel=&quot;noreferrer&quot;&gt;spindle speed calculations&lt;/a&gt; on Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="95" LastEditDate="2015-01-23T00:19:39.207" LastActivityDate="2015-01-23T00:19:39.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="135" PostTypeId="2" ParentId="14" CreationDate="2015-01-22T02:02:31.000" Score="9" Body="&lt;p&gt;Insulation, shade, the use of low thermal conductivity materials and utilising any cool breeze at any time of the day are going to be key to passively cooling buildings in a hot climate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different solutions will be more applicable to houses than non residential buildings. The other thing to consider is whether the climate hot and dry or hot and humid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In hot dry regions of Australia passive cooling measures used until the mid 20th Century included thick exterior walls, high ceilings, verandas, double roofs and where double roofs weren't used roof ventilators. The verandas and double roofs off shade. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the hot humid parts of Australia houses were built on stilts, approximately 2m to 3m high. This continued until the late 1970s in Darwin. Such houses were also built to encourage airflow inside by have louvred internal walls and sometimes short height internal walls so the rooms looked like cubicles in modern day offices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://melbourneblogger.blogspot.com.au/2010/01/traditional-queenslander-house.html&quot;&gt;http://melbourneblogger.blogspot.com.au/2010/01/traditional-queenslander-house.html&lt;/a&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Queenslander_%28architecture%29&quot;&gt;https://en.wikipedia.org/wiki/Queenslander_%28architecture%29&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.territorystories.nt.gov.au/handle/10070/19572&quot;&gt;http://www.territorystories.nt.gov.au/handle/10070/19572&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that is sometimes overlooked in the passive cooling of buildings is the used green foliage gardens to offer shade and to reduce the amount of radiated and reflected heat reaching buildings. &lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-01-22T02:02:31.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="137" PostTypeId="1" AcceptedAnswerId="145" CreationDate="2015-01-22T10:12:39.987" Score="18" ViewCount="4625" Body="&lt;p&gt;Given a tolerance within which your workpiece should be manufactured, say some length should be $10\pm1$mm. If you determine that your uncertainty in measuring this length is $0.2$mm (at 95%). How should a measurement of $9.1$mm be treated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly there is a significant probability that this value will actually be outside of tolerance. Do you need to decrease you tolerance range based on the uncertainty in your measurement?&lt;/p&gt;&#xA;" OwnerUserId="153" LastEditorUserId="153" LastEditDate="2015-01-23T09:47:40.967" LastActivityDate="2015-12-06T03:13:19.553" Title="How does measurement uncertainty combine with tolerances?" Tags="|tolerance|measurements|statistics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="138" PostTypeId="2" ParentId="137" CreationDate="2015-01-22T11:46:22.633" Score="6" Body="&lt;p&gt;There are two different aspects in your measurement. On the one hand, you are dealing with tolerances. On the other hand, you cover probabilities in measurement systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for a rough calculation:&#xA;The probability of the real length to be within 9.8mm and 10mm is 95%. The certainty of this measurement depends on the distribution of your probability. For example, assuming a &lt;a href=&quot;https://en.wikipedia.org/wiki/Normal_distribution&quot;&gt;Gaussian Distribution&lt;/a&gt; (or any other symmetric distribution), your certainty is higher than the 95%. If you're lucky, you can get the certainty range for 99% or 99.5% percent as well from the supplier. Another option is to do a lot of measurements and find the certainty ranges on your own.&lt;/p&gt;&#xA;" OwnerUserId="148" LastActivityDate="2015-01-22T11:46:22.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="139" PostTypeId="2" ParentId="122" CreationDate="2015-01-22T12:25:53.843" Score="11" Body="&lt;p&gt;The primary difference between Rankine and Coulomb earth pressure theories is that Coulomb's considers a frictional retaining wall. In other words, the interface between the soil and the retaining wall is not assumed frictionless (as it is in Rankine theory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, it is typically considered that &lt;strong&gt;Rankine underpredicts&lt;/strong&gt; the true orientation of the failure surface, whereas &lt;strong&gt;Coulomb overpredicts&lt;/strong&gt; the orientation. In that sense, you could use both methods, and use the two solutions to bound what will likely occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terzaghi (and Peck)'s method is largely empirical. It simply uses the soil's classification and the backfill slope, then they simply tabulated coefficients of lateral earth pressure. That being said, they aren't bad, it's just that any empirical solutions like this tend to be relatively site specific, so the solution needs to be taken with a grain of salt.&lt;/p&gt;&#xA;" OwnerUserId="34" LastActivityDate="2015-01-22T12:25:53.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="140" PostTypeId="5" CreationDate="2015-01-22T12:44:00.863" Score="0" Body="&lt;p&gt;Geotechnical engineering is the branch of civil engineering concerned with the engineering behavior of earth materials. It uses principles of soil mechanics and rock mechanics to investigate subsurface conditions and materials; determine the relevant physical/mechanical and chemical properties of these materials; evaluate stability of natural slopes and man-made soil deposits; assess risks posed by site conditions; design earthworks and structure foundations; and monitor site conditions, earthwork and foundation construction.&lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="34" LastEditDate="2015-01-22T18:00:40.550" LastActivityDate="2015-01-22T18:00:40.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="141" PostTypeId="4" CreationDate="2015-01-22T12:44:00.863" Score="0" Body="Geotechnical engineering is the branch of civil engineering concerned with the engineering behavior of earth materials." OwnerUserId="34" LastEditorUserId="34" LastEditDate="2015-01-22T18:00:54.663" LastActivityDate="2015-01-22T18:00:54.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="142" PostTypeId="2" ParentId="14" CreationDate="2015-01-22T13:26:04.173" Score="18" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Yakhchal&quot; rel=&quot;noreferrer&quot;&gt;yahkchal&lt;/a&gt; is an example of a type of passively cooled building in Iran&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QW8EA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They utilise a combination of passive evaporative cooling and thick thermally insulating walls in order to keep the interior temperatures low enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, wind is directed into underground aquifers known as &lt;em&gt;qanat&lt;/em&gt;. They are then cooled due to the low humidity desert air causing water to evaporate. The cooled air then flows through the interior of the yakhchal, cooling the interior. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z24CY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thick insulating walls (filled with earth and various insulating materials such as straw and feathers) help to insulate the cool interior from the hot exterior, therefore maintaining a low temperature inside the yakhchal. &lt;/p&gt;&#xA;" OwnerUserId="164" LastActivityDate="2015-01-22T13:26:04.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="143" PostTypeId="5" CreationDate="2015-01-22T15:50:56.003" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-22T15:50:56.003" LastActivityDate="2015-01-22T15:50:56.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="144" PostTypeId="4" CreationDate="2015-01-22T15:50:56.003" Score="0" Body="Principal Component Regression (PCR) is a statistical technique based upon Principal Component Analysis (PCA).  Questions involving PCR should have a clear focus on Engineering technique or analysis." OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-01-22T18:04:31.520" LastActivityDate="2015-01-22T18:04:31.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="145" PostTypeId="2" ParentId="137" CreationDate="2015-01-22T16:30:55.117" Score="13" Body="&lt;p&gt;You need to ensure that even in the worst case scenario, you still meet your measurement spec of $10 \pm 1\text{mm}$. If your tolerance is $0.2\text{mm}$ of your measurement, then a measurement of $11\text{mm}$, while may look like it meets spec, it doesn't because it could be $11.1\text{mm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the worst case that still meets your spec is a measurement of $10.9\text{mm}$, because then with a max tolerance of $0.2\text{mm}$, you still meet $11\text{mm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a $0.2\text{mm}$ tolerance, your $10 \pm 1\text{mm}$ spec becomes $10 \pm 0.9\text{mm}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How should a measurement of $9.9\text{mm}$ be treated?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So revised spec is between $9.1\text{mm}$ and $10.9\text{mm}$, so $9.9\text{mm}$ is within spec.&lt;/p&gt;&#xA;" OwnerDisplayName="user133" LastEditorUserId="2538" LastEditDate="2015-12-06T03:13:19.553" LastActivityDate="2015-12-06T03:13:19.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="146" PostTypeId="1" CreationDate="2015-01-22T16:41:41.347" Score="5" ViewCount="80" Body="&lt;p&gt;I want to quantify friction between a bullet and barrel to test different friction-proofing coatings and treatments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a controlled test I could do at low speeds in a lab between samples of treated copper and steel that would be theoretically applicable at velocities up to 1000 m/s?  I.e., is the dynamic coefficient of friction considered constant between two materials like copper and steel up to speeds of 1000 m/s, or is there a known relationship between speed and friction that would allow extrapolation from lab-measured coefficients?  And what lab tests would be suited to measuring that coefficient to high precision?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I realize that the most obvious way to approach this problem would be empirically: i.e., just shoot the different projectiles and compare the resulting velocities.  I have tried that and gotten confusing results that appear to be attributable to the sources of error in internal ballistics: in particular to the fact that different coatings are probably altering the diameter of the projectiles at the micron level, and dimensional changes can overwhelm the changes in friction, which render A-B comparisons invalid.)&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2015-01-22T17:01:53.853" Title="Quantifying and measuring friction under extreme conditions" Tags="|friction|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="147" PostTypeId="5" CreationDate="2015-01-22T16:47:26.883" Score="0" Body="&lt;p&gt;As steel are referred metallic alloys, whose main constituents are iron and carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This general, common since the early 20th century definition includes the term steel also at that time hardly produced wrought iron, which also had a carbon content of less than 2.06%. The term wrought iron does not describe the components of the alloy, but the method developed over many centuries for producing ductile iron parts, which were essentially based on different types of refining of iron blooms and subsequent forging. Although wrought iron has similar carbon contents as today's steel, but they are not identical due to a slightly different alloy with other substances and not completely removed slag parts.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="33" LastEditDate="2016-02-23T02:48:11.750" LastActivityDate="2016-02-23T02:48:11.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="148" PostTypeId="4" CreationDate="2015-01-22T16:47:26.883" Score="0" Body="For questions related to the alloy steel: Its production, material properties and applications. The carbon content of steel is between 0.002% and 2.1% by weight for plain iron-carbon alloys." OwnerUserId="77" LastEditorUserId="46" LastEditDate="2015-01-28T16:51:06.560" LastActivityDate="2015-01-28T16:51:06.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="149" PostTypeId="2" ParentId="146" CreationDate="2015-01-22T17:01:53.853" Score="3" Body="&lt;p&gt;I would think the friction regimes will be very different, especially for something like a gum barrel where the projectile is contained. At high speeds plastic deformation of the surface will be much more significant as well as possible chemical reactions due to the heat/energy. Therefore I wouldn't necessarily expect the coeffcients to be similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could possibly could set up a test rig using a high speed touching a flat plate to test the friction at high speed outside a barrel. However, I haven't thought about the details and this would probably be difficult\expensive\dangerous to set up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick google gives me &lt;a href=&quot;http://www.jstor.org/discover/10.2307/100521?sid=21105677463913&amp;amp;uid=4&amp;amp;uid=70&amp;amp;uid=3738032&amp;amp;uid=2129&amp;amp;uid=2&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; which supports my theory that friction would not behave consistently.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-01-22T17:01:53.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="150" PostTypeId="1" AcceptedAnswerId="153" CreationDate="2015-01-22T17:56:18.063" Score="14" ViewCount="860" Body="&lt;p&gt;As I heard, the life expectancy of photovoltaic cells is normally some decades, as so. This is very long, regarding their high cost, it has significant (negative) effect on the total costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do they age? I can see the only possibility which could damage their atomic structure, and this is the very high energetic ultraviolet (or even soft röntgen) photon spectrum of the sun (which is only a small part of its power).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, I think, these photons could be maybe easily filtered out by a transparent plastic layer over the silicon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, why do the solar cells age, and what happens in them on the atomic level?&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="368" LastEditDate="2015-08-10T16:13:18.137" LastActivityDate="2015-08-10T16:13:18.137" Title="Do solar cells age? Why, and how?" Tags="|photovoltaics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="151" PostTypeId="1" AcceptedAnswerId="192" CreationDate="2015-01-22T18:25:15.817" Score="3" ViewCount="706" Body="&lt;p&gt;I think, it could be a reactor utilizing californium-242 (or, at least, weapon-grade U-235) cooled and moderated by heavy water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, it were similar to an atomic bomb, but - of course - it would be optimized for stay around the equilibrial state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result were probably a very strong neutron source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, it could be used for various things, mainly in the space applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does any cost/size estimations about this ever created?&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditDate="2015-01-23T18:13:22.733" LastActivityDate="2020-03-10T16:10:35.613" Title="Smallest possible controlled chain reaction-based nuclear fission reactor?" Tags="|nuclear-technology|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="152" PostTypeId="2" ParentId="151" CreationDate="2015-01-22T18:59:31.153" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator&quot; rel=&quot;nofollow&quot;&gt;Radioisotope thermoelectric generators&lt;/a&gt; have been used in space for decades, both on the moon missions and on deep-space probes where solar cells would not be practical.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-01-22T18:59:31.153" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="153" PostTypeId="2" ParentId="150" CreationDate="2015-01-22T19:05:37.617" Score="12" Body="&lt;p&gt;The National Renewable Energy Lab (NREL) estimates something like 0.4%-1% degradation per year for solar panels. I don't know what the chemical action is here (the electrical result being an increase in series resistance of the PV cells), but the long-term degradation of the panels is correlated with UV exposure. Although damaging to the PV material, the UV rays do contain energy that we want to collect, so you won't be able to filter it out completely without defeating the purpose of the panel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other causes for degradation are mostly weather-related, but in the form of mechanical stresses to the system. Anything from wind to snowfall can damage the panels by bending the panels to produce minor cracks in the solar cells and embedded electrical connections. Other factors in a degraded PV panel include corrosion, or the lifetime of the supporting electrical devices needed to maintain a functioning system (these other devices may have shorter lifespans than the panel, and if they are expensive enough it would cause the system to be unsustainable financially compared to replacing the entire panel system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.engineering.com/ElectronicsDesign/ElectronicsDesignArticles/ArticleID/7475/What-Is-the-Lifespan-of-a-Solar-Panel.aspx&quot;&gt;http://www.engineering.com/ElectronicsDesign/ElectronicsDesignArticles/ArticleID/7475/What-Is-the-Lifespan-of-a-Solar-Panel.aspx&lt;/a&gt; (See the Read More link for the NREL report)&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2015-01-22T19:05:37.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="154" PostTypeId="5" CreationDate="2015-01-22T19:39:56.743" Score="0" Body="&lt;h1&gt;Suitable questions for this tag&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;As it is a basic engineering discipline this tag can cover many questions. However in order to use this tag the question should be about a &lt;strong&gt;mechanical system&lt;/strong&gt; or &lt;strong&gt;machinery&lt;/strong&gt;. Below is a brief outline about topics covered by mechanical engineering.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Overview of Mechanical Engineering&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Mechanical Engineering is engineering field that deals with: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;design &lt;/li&gt;&#xA;&lt;li&gt;analysis &lt;/li&gt;&#xA;&lt;li&gt;production&lt;/li&gt;&#xA;&lt;li&gt;operation and&lt;/li&gt;&#xA;&lt;li&gt;maintenance of &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;mechanical systems&lt;/strong&gt; and &lt;strong&gt;machinery&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of these mechanical systems and machinery are manufacturing machines, heating and cooling systems, aircraft, watercraft, weapons, medical devices and machinery, robotics, automobiles, rolling stock, fluid systems. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Scope of Mechanical Engineering&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Mechanical Engineering fields of study can be summarized as: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mechanics&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Statics&lt;/li&gt;&#xA;&lt;li&gt;Dynamics&lt;/li&gt;&#xA;&lt;li&gt;Mechanics of Materials&lt;/li&gt;&#xA;&lt;li&gt;Fluid Mechanics&lt;/li&gt;&#xA;&lt;li&gt;Kinematics&lt;/li&gt;&#xA;&lt;li&gt;Continuum Mechanics&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Mechatronics and robotics&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Micro Electromechanic Systems (MEMS) &lt;/li&gt;&#xA;&lt;li&gt;Control System Design&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Structural Analysis&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Failure Analysis&lt;/li&gt;&#xA;&lt;li&gt;Fatigue Analysis&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Thermodynamics and thermo-science&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Heat Transfer&lt;/li&gt;&#xA;&lt;li&gt;Thermofluids&lt;/li&gt;&#xA;&lt;li&gt;Energy Conversion&lt;/li&gt;&#xA;&lt;li&gt;Heating Ventilation and Air Conditioning (HVAC)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Design and Drafting&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Computer Aided Design (CAD)&lt;/li&gt;&#xA;&lt;li&gt;Computer Aided Manufacturing (CAM)&lt;/li&gt;&#xA;&lt;li&gt;Finite Element Analysis (FEA)&lt;/li&gt;&#xA;&lt;li&gt;Computational Fluid Dynamics (CFD)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Mechanical Engineering and other fields&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Because mechanical engineering covers a vast area of knowledge, it overlaps with many basic science and engineering disciplines with varying amounts. These are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Physics&lt;/li&gt;&#xA;&lt;li&gt;Chemistry&lt;/li&gt;&#xA;&lt;li&gt;Materials Science&lt;/li&gt;&#xA;&lt;li&gt;Electrical and Electronics Engineering&lt;/li&gt;&#xA;&lt;li&gt;Aerospace Engineering&lt;/li&gt;&#xA;&lt;li&gt;Chemical Engineering&lt;/li&gt;&#xA;&lt;li&gt;Metallurgical Engineering&lt;/li&gt;&#xA;&lt;li&gt;Civil Engineering&lt;/li&gt;&#xA;&lt;li&gt;Biomedical Engineering&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="879" LastEditorUserId="33" LastEditDate="2016-02-23T02:37:05.510" LastActivityDate="2016-02-23T02:37:05.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="155" PostTypeId="4" CreationDate="2015-01-22T19:39:56.743" Score="0" Body="Questions within the problem domain of mechanical engineering. Mechanical Engineering can be a broad field; consider choosing more specific tags if they apply." OwnerUserId="34" LastEditorUserId="33" LastEditDate="2016-02-22T22:49:00.347" LastActivityDate="2016-02-22T22:49:00.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="156" PostTypeId="2" ParentId="151" CreationDate="2015-01-22T19:49:41.420" Score="2" Body="&lt;p&gt;It largely depends on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Critical_mass&quot; rel=&quot;nofollow noreferrer&quot;&gt;critical mass&lt;/a&gt; of the fuel - that is, the smallest amount of fuel that is needed for a controlled, sustained reaction to occur. This is often on the order of ten or so kilograms, but can vary with the other aspects of the design (see, for example, &lt;a href=&quot;http://sti.srs.gov/fulltext/ms9900313/ms9900313.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; paper on $^{238}\text{Pu}$). The Wikipedia page also has a nice summary from various sources of the size of a sphere containing the critical mass (and the critical mass itself) of quite a few different isotopes. The diameters range from 6.9 centimeters to 30-35 centimeters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The DOE has &lt;a href=&quot;http://www.newton.dep.anl.gov/newton/askasci/1993/physics/PHY128.HTM&quot; rel=&quot;nofollow noreferrer&quot;&gt;a page&lt;/a&gt; on the size of fusion reactors; one of the expert responses mentions fission reactors and, while admitting that the critical mass can be very low,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A typical critical mass for a fission reaction would be several kilograms. This would be very portable, if it was not for the shielding needed to keep from frying everybody in the vicinity. Plus, you need cooling and heat exchangers to use the heat that is produced. Both fusion and fission reactors put out heat as their primary method of energy production.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://depletedcranium.com/why-you-cannot-build-a-nuclear-fission-reactor-at-home/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Depleted Cranium&lt;/a&gt; blog notes that the first nuclear reactor, &lt;a href=&quot;http://en.wikipedia.org/wiki/Chicago_Pile-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;CP-1&lt;/a&gt;, was really the smallest ever. It used 40 tons of uranium and &lt;em&gt;400 tons of graphite&lt;/em&gt; to be used as a nuclear moderator. This graphite was made into 45,000 blocks, which created &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/f/fe/Stagg_Field_reactor.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;a structure like this drawing&lt;/a&gt;:&#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/f/fe/Stagg_Field_reactor.jpg&quot; alt=&quot;CP-1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a material like graphite as a moderator, and you need it to be in substantial quantities. CP-1 used the smallest amount possible and produced only a small amount of energy. So you're not going to get under about 450 tons if you want to actually generate a substantial amount of energy.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:49.903" LastActivityDate="2015-01-23T16:00:19.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="157" PostTypeId="5" CreationDate="2015-01-22T19:53:54.007" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-22T19:53:54.007" LastActivityDate="2015-01-22T19:53:54.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="158" PostTypeId="4" CreationDate="2015-01-22T19:53:54.007" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-22T19:53:54.007" LastActivityDate="2015-01-22T19:53:54.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="159" PostTypeId="2" ParentId="150" CreationDate="2015-01-22T19:54:04.867" Score="12" Body="&lt;p&gt;Keep in mind that P-N junctions are created in the first place by diffusing impurities (dopants) into the silicon at elevated temperatures (1000 - 1500 K). The efficiency of the junction is related to how sharply the dopant concentration changes at the junction (its gradient).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At lower temperatures, such as those to which a solar array is exposed (say, 270 - 330 K), diffusion is much, much slower, but it does not completely stop. Given enough time, the dopant gradients become less steep, reducing the effectiveness of the P-N junction. Over the extreme long term (millennia), semiconductor devices will stop working as their junctions disappear altogether.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-01-22T19:54:04.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="160" PostTypeId="1" AcceptedAnswerId="161" CreationDate="2015-01-22T20:40:52.330" Score="10" ViewCount="2281" Body="&lt;p&gt;I punch metals before I start a bit in them, but with small-diameter bits I am finding that they wander inside the material.  E.g., drilling a 1/8&quot; hole in 1/2&quot; of aluminum the (titanium-coated HSS) bit will come out the other end a measurable distance to a side (orthogonal to the drill feed direction) from where it entered.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, the bit starts where I want it to since I punched the metal. But the bit appears to be flexing within the material, causing the resulting hole to be angled.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prevent this?  Is there something about the material that I need to understand in order to prevent the bit from wandering and causing the hole to be skewed?  I'd like to know the mechanics behind the materials that is causing these skewed holes.&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorDisplayName="user16" LastEditDate="2015-01-26T15:19:56.493" LastActivityDate="2015-01-26T15:19:56.493" Title="How do I keep my bit from wandering on a drill press?" Tags="|materials|machining|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="161" PostTypeId="2" ParentId="160" CreationDate="2015-01-22T20:54:03.830" Score="11" Body="&lt;p&gt;It's common to start with a shorter, stiffer tool such as a center drill or a spotting drill. In addition, using the shortest drill bit that drills the hole you need will increase stiffness. Because of the flutes in a drill, the stiffness goes down geometrically as the length increases. The other variable you have control over is how you are holding the tool. A collet will keep the tool in better alignment than a chuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're doing this often, it's worth making sure you have a good grasp of feeds and speeds, and that you're using the best type of drill for the job. Both of those could have some affect on hole perpendicularity.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-01-22T20:54:03.830" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="162" PostTypeId="1" AcceptedAnswerId="164" CreationDate="2015-01-22T20:59:09.643" Score="13" ViewCount="340" Body="&lt;p&gt;I want to design a small wind turbine that can be easily carried out of the classroom into the sports field in relatively low-wind conditions (5-15km/h) that can power a small ultra-bright 5V LED - just enough to show that it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the power I can get from different blade diameters in these wind conditions, and the power needed to drive a small DC motor/generator enough to light the LED?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="368" LastEditDate="2015-02-27T20:17:24.480" LastActivityDate="2021-07-21T21:18:57.517" Title="How to correctly size the blades for a small educational wind turbine?" Tags="|electrical-engineering|renewable-energy|power-engineering|wind-power|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="163" PostTypeId="2" ParentId="162" CreationDate="2015-01-22T21:32:25.100" Score="2" Body="&lt;p&gt;Try to cut a few from balsa wood to see what you get. Experimentation is probably the best method on such a small scale, and the math won't get you very far without an extensive knowledge of the materials (i.e. the cheaper the material, the less likely the blade will look like the big ones). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I worked summer activities at my university using materials from KidWind, and it's actually a ton of fun to build your own and experiment with it. We had always set up a wind tunnel and turbine stands/rotor that the kids would then design and attach blades to. Tons of fun, and you'll be amazed at what kinds of shapes actually work worth a darn. The activity even took less than an hour to complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://challenge.kidwind.org/events/building-a-turbine&quot; rel=&quot;nofollow&quot;&gt;http://challenge.kidwind.org/events/building-a-turbine&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://learn.kidwind.org/learn/science_fair_projects&quot; rel=&quot;nofollow&quot;&gt;http://learn.kidwind.org/learn/science_fair_projects&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A page with a photo of a tunnel (we modeled ours off of this): &lt;a href=&quot;http://challenge.kidwind.org/?/national/tunnel&quot; rel=&quot;nofollow&quot;&gt;http://challenge.kidwind.org/?/national/tunnel&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2015-01-22T21:32:25.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="164" PostTypeId="2" ParentId="162" CreationDate="2015-01-22T21:56:30.197" Score="9" Body="&lt;p&gt;As it happens, I just recently went through that calculation myself for a different site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the following facts from a quick web search, it isn't difficult to work out the numbers.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The maximum efficiency of a (large) windmill is about 40%.&lt;/li&gt;&#xA;&lt;li&gt;The density of air is 1.225 kg/m&lt;sup&gt;3&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;You need about 50 mW (10 mA at 5V) to light up an LED&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;First, we'll need about 50 mW / 0.40 = 125 mW of air power flowing through the windmill to create the electricity we need (ignoring other factors such as the actual efficiency of a small windmill and the efficiency of the generator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power of the air flowing through the windmill is 0.5mv&lt;sup&gt;2&lt;/sup&gt;, where m is the mass rate of the air flowing through the &quot;disk&quot; defined by the diameter of the blades. For example, suppose we have a disk of 0.03m&lt;sup&gt;2&lt;/sup&gt; (about 20cm in diameter). The mass rate of the air is the area of the disk multiplied by the air velocity, multiplied by the density of the air:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Mass rate} = 0.03 \text {m}^2 \cdot v \cdot 1.225 \text{ kg/m}^3 = v \cdot 0.03675 \text{ kg/m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power of that air is therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = 0.5 \cdot \text{Mass rate} \cdot v^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting and solving for $v$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \sqrt[3]{\frac{0.125 \text{ W}}{0.5 \cdot 0.03675 \text{ kg/m}}} = 1.9 \text{ m/s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... or about 7 km/h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking into account the efficiencies we ignored earlier, plus the losses in a gear train that might be needed to get the generator RPMs up to a usable level, I would probably shoot for about 4&amp;times; the area, or about 2&amp;times; the diameter (40-50 cm), in order to get reasonable results.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="106" LastEditDate="2015-01-22T22:46:58.063" LastActivityDate="2015-01-22T22:46:58.063" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="165" PostTypeId="1" AcceptedAnswerId="177" CreationDate="2015-01-22T23:16:05.397" Score="14" ViewCount="774" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pontoon_bridge&quot;&gt;Pontoon bridges&lt;/a&gt; differ from traditional bridges in that they are supported not by structures anchored to the floor of the body being spanned but by floating pontoons that are connected by a more rigid structure that supports a roadway. They're often used by militaries to provide a temporary crossing point, but they're also &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_pontoon_bridges&quot;&gt;used for permanent civilian crossings&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that they make it easier to cross larger bodies of water because there is less structure to be secured below the surface. In areas of deep water, support structures can become unfeasibly large. These larger spans could, though, make the pontoon bridges susceptible to damage from strong winds and currents. Are there plans to use pontoon bridges to cross long distances?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-02-14T20:48:56.860" LastActivityDate="2015-02-14T20:48:56.860" Title="Are pontoon bridges being considered to extend bridge span?" Tags="|civil-engineering|bridges|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="166" PostTypeId="1" AcceptedAnswerId="167" CreationDate="2015-01-22T23:16:44.680" Score="7" ViewCount="658" Body="&lt;p&gt;By &quot;modern computers,&quot; I mean electronic programmable computers such as those that were developed about the middle of the 20th century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that early computers such as Charles Babbage's &quot;calculating machines&quot; worked mainly on mechanical principles, like a modern &quot;abacus&quot;. Apparently later machines were more electronically based. I'm talking about computers that evolved from using vacuum tubes, to transistors to integrated circuits, and ultimately silicon chips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What engineering advances enabled the above transition to occur from physical means (e.g. vacuum tubes) to electronic means (e.g. transistors) when it did (middle 20th century)?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="106" LastEditDate="2015-01-25T15:24:30.873" LastActivityDate="2015-05-30T14:46:09.123" Title="What engineering advances paved the way for &quot;modern&quot; computers?" Tags="|computer|electrical-engineering|engineering-history|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="167" PostTypeId="2" ParentId="166" CreationDate="2015-01-23T00:01:28.223" Score="9" Body="&lt;p&gt;&lt;strong&gt;The transistor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Transistor&quot; rel=&quot;nofollow&quot;&gt;transistor&lt;/a&gt; was the revolutionary replacement of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Vacuum_tube&quot; rel=&quot;nofollow&quot;&gt;vacuum tube&lt;/a&gt;, which had been at the heart of computers for the first half of the 20th century. Vacuum tubes themselves had only two main problems: They were power-hungry and they were relatively big. Relative to their replacements, that is. They also had a tendency to burn out or leak during operation, which could prove disastrous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1947, &lt;a href=&quot;http://www.101science.com/transistor.htm&quot; rel=&quot;nofollow&quot;&gt;John Bardeen and Walter Brattain&lt;/a&gt;, along with William Shockley, their boss at Bell Labs, successfully amplified an electric current, using germanium. This &lt;a href=&quot;http://electronics.howstuffworks.com/transistor.htm&quot; rel=&quot;nofollow&quot;&gt;&quot;point-contact transistor&quot;&lt;/a&gt;, as it was called, was soon used to speed up computing and to make computers smaller and more efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example of the transistor and the vacuum tube is the construction of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Manchester_computers&quot; rel=&quot;nofollow&quot;&gt;Manchester Computers&lt;/a&gt;, developed at the University of Manchester. The first, the Small-Scale Experimental Machine (SSEM) (developed in 1947), was a state-of-the-art testbed for new innovations in computing, such as the Williams tube. But it still used vacuum tubes. It had 550 valves and took in 3500 watts of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SSEM's successor, the Manchester Mark 1 (developed in 1949), was much more powerful. It used 4050 valves and consumed 25000 watts of power. Yet the aptly-named Transistor Computer, built in 1955, used only 200 transistors and 1,300 diodes, and only used 150 watts. It wasn't the first computer to exclusively use vacuum tubes, but it was a huge step forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's tough to say exactly why the transistor was created when it was (I'm answering the last part of your question now), but it could be argued that the computational advances of World War II (such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Harvard_Mark_I&quot; rel=&quot;nofollow&quot;&gt;Harvard Mark I&lt;/a&gt;) ensured that many new advances would be made in computing; the transistor happened to be one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The integrated circuit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Integrated_circuit&quot; rel=&quot;nofollow&quot;&gt;integrated&lt;/a&gt; circuit, developed about a decade after the transistor, also had profound effects on computing. &lt;a href=&quot;http://www.nobelprize.org/educational/physics/integrated_circuit/history/&quot; rel=&quot;nofollow&quot;&gt;It was developed by Jack Kilby in 1958&lt;/a&gt; - though many others were involved along the way, and &lt;a href=&quot;http://www.pbs.org/transistor/background1/events/icinv.html&quot; rel=&quot;nofollow&quot;&gt;there are disputes&lt;/a&gt; as to who should get the credit for inventing it first - at Texas Instruments. He used &lt;a href=&quot;http://en.wikipedia.org/wiki/Semiconductor&quot; rel=&quot;nofollow&quot;&gt;semiconductors&lt;/a&gt; to create an entire computer chip - the integrated circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An integrated circuit can contain incredible amounts of transistors, and it is this complexity and compactness that make it so useful. Manufacturing was also much easier and quicker. Integrated circuits started a second computing revolution, which laid the groundwork for cheaper computers that could be available to the masses.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now that the question is focused on the transition between vacuum tubes and transistors, I'd like to add something about &lt;a href=&quot;http://en.wikipedia.org/wiki/Semiconductor_device&quot; rel=&quot;nofollow&quot;&gt;semiconductors&lt;/a&gt;, because they play a key role in transistors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Semiconductors allow for good conduction of electricity, but one of their really useful properties is that their conduction can be modified in a process called &lt;a href=&quot;http://en.wikipedia.org/wiki/Doping_(semiconductor)&quot; rel=&quot;nofollow&quot;&gt;semiconductor doping&lt;/a&gt;. This adds &quot;impurities&quot; to the semiconductor, adding either electrons or holes. &lt;a href=&quot;http://electronics.howstuffworks.com/diode1.htm&quot; rel=&quot;nofollow&quot;&gt;Semiconductors can be either n-type or p-type&lt;/a&gt; - n-type semiconductors have an excess of electrons, while p-type semiconductors have an excess of holes. These can be combined to form a diode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another relevant development was the creation of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Czochralski_process&quot; rel=&quot;nofollow&quot;&gt;Czochralski process&lt;/a&gt;, which allows crystals to be grown for semiconductors. This also involves semiconductor doping, and has allowed for semiconductors to be produced on a large scale, making it easier to build transistors.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Are there other technologies that have been crucial to computer development? Of course. I could cite the vacuum tube, cathode ray tube, solid-state drives, and a host of others as crucial to computer development. But the transistor and the integrated circuit were the two main players in the development of the &quot;modern computer&quot; in the relevant time period here - the mid-20th century. I suppose you could make cases for other technologies, but I'd certainly rank them at the top of the list.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-05-30T14:46:09.123" LastActivityDate="2015-05-30T14:46:09.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="168" PostTypeId="5" CreationDate="2015-01-23T00:52:09.490" Score="0" Body="" OwnerUserId="120" LastEditorUserId="13" LastEditDate="2015-01-23T17:11:34.963" LastActivityDate="2015-01-23T17:11:34.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="169" PostTypeId="4" CreationDate="2015-01-23T00:52:09.490" Score="0" Body="Study and applications of electricity, electronics, and electromagnetism. Use this tag when someone who has trained as an electrical engineer is likely to be able to answer your question." OwnerUserId="-1" LastEditorUserId="368" LastEditDate="2017-03-02T20:06:25.217" LastActivityDate="2017-03-02T20:06:25.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="170" PostTypeId="1" AcceptedAnswerId="173" CreationDate="2015-01-23T01:27:14.347" Score="7" ViewCount="763" Body="&lt;p&gt;I understand that you can have a device with angular measurements for rotation and elevation, and use trigonometry to calculate the distances... but only if you have some distances to start with. How did they accurately measure the first straight-line distance over any distance significantly long enough to give usable angles? And wouldn't the error in other distances calculated from this very quickly accumulate out of control?&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-11-12T11:36:29.397" Title="How was surveying for maps done before lasers?" Tags="|engineering-history|surveying|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="171" PostTypeId="1" CreationDate="2015-01-23T02:28:55.283" Score="3" ViewCount="278" Body="&lt;p&gt;Within most engineering development processes, product testing and validation is an integral part of the product development process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All prototype / engineering development samples are tested. And all failures are handed over to product engineering with failure reports, which includes test data documented in various forms (paper, word documents, excel tables, etc). The product development team then gets 1-2 weeks to perform root cause analysis and develop corrective action before the product goes into production. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current root cause analysis / corrective action process that I'm working with can be characterized at best as chaotic, brute force, or uncoordinated. Needless to say, some product comes back as warranty returns or field returns.  We'd like to drive the return rate down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The business conditions, competition, and resource levels don’t permit the luxury of additional time for root cause analysis and proper (the way I prefer) corrective action. We do understand there are areas in the engineering development process that can be improved, and plans are underway to implement these improvements.  Currently, we are following waterfall project management methodology. And it's worth noting that I am responsible for the electrical engineering component only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is an engineering product develop process strategy that will help bring order to the current chaotic, brute force, uncoordinated root cause analysis and corrective action processes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We currently have a few suggestions on the table, but would like to probe the community for additional insight.&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="33" LastEditDate="2015-10-31T17:47:54.590" LastActivityDate="2015-10-31T17:47:54.590" Title="How can I efficiently find the root cause of validation failures?" Tags="|product-testing|engineering-managment|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="172" PostTypeId="2" ParentId="170" CreationDate="2015-01-23T02:52:50.067" Score="2" Body="&lt;p&gt;The &quot;first straight-line distance&quot; in any survey is called the &quot;baseline&quot;, and surveyors over the centuries have worked out many ways to minimize the error in the baseline to begin with, and to account for error accumulation as they survey additional points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also many ways to cross-check measurements using independent methods that allows some errors to be mitigated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic instruments of surveying are the transit for measuring horizontal and vertical angles and the rod and the chain (and the tape, in modern times) for measuring distances. But the real art in surveying is not how well you can take the raw measurements, but your ability to account for the errors and minimize them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this was developed long before the Internet, and the best information on these techniques will be found in books, not online.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The key instrument for establishing a baseline distance is the steel tape. Under proper conditions &amp;mdash; paying attention to details such as temperature and tension &amp;mdash; a 100-foot tape can typically be read with an accuracy of 1/100 of a foot (about 1/8 inch), giving an overall accuracy of 1 part in 10,000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A land survey will generally start with a measured baseline at one end of the area, and then after surveying the intermediate points, a second baseline between two points at the far end of the area will be measured as well. This allows most of the systematic errors for all of the intermediate points to be cancelled out.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2015-01-23T15:55:13.333" LastActivityDate="2015-01-23T15:55:13.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="173" PostTypeId="2" ParentId="170" CreationDate="2015-01-23T03:50:03.507" Score="16" Body="&lt;p&gt;The device for taking horizontal and vertical angles that you mention is called a theodolite. Theodolites only started being phased out as the main surveying tool in the 1980s when total stations where introduced. Below is a Soviet theodolite from 1958, (ex Wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V1KXM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theodolites were analogue devices and the angles measured had to be written in a notebook. Total stations were electronic devices, essentially electronic theodolites, with electronic distance measuring devices, based on infrared signals. These devices could be connected to a portable electronic memory unit with a keypad to store the measurements. The surveyor still had to manually enter a point identifier for each reading, but didn't have to enter the measured angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When starting a survey, a reference marker from the nation system of surveying markers closest to the surveying region was chosen as this had a known/established northing, easting and elevation. A picture of US survey marker follows (from Wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8c4xc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theodolite would be set up and the first reading would be to the known marker peg to establish the baseline for the survey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For very accurate surveys a surveying target, on tripod, was placed over the survey marker; either a plate with a cross on it or a short pointed rod with the point upwards. A similar target would then be placed on a temporary marker and the horizontal angle between the two targets measured. The vertical angle from theodolite's horizontal plane (in the eye piece) to the first target would be measured as would the vertical angle to the second target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each theodolite has specifically marker dot on it, at eye piece (telescope) height. This is the reference marker for the theodolite from which lateral distances are measured. A measuring tape was place against the dot on the theodolite and the other end of the tape was place at the centre of each target cross or the tips of each pointed target rod, to measure the slope distances. The measuring tape had to have a certain tension applied and the readings would be recorded. Later, in the office, the measured slope distances would be corrected for tape sag. Additionally, the heights of the theodolite and the two targets, above the ground, would be measured with a tape measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having done all of that, another temporary marker would be established, the theodolite moved between the last two pegs and the process repeated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each set up, the heights of the theodolite and targets was needed as were the slope distances, vertical angles and horizontal angle. Using trigonometry on all this data one could determine the co-ordinates and elevation of each peg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method used of measuring was called stadia. This used a theodolite but instead of a cross targets or pointed rod targets being used to sight to at each of the survey pegs, surveying rods were used. See the picture below from &lt;a href=&quot;http://www.tigersupplies.com&quot; rel=&quot;noreferrer&quot;&gt;http://www.tigersupplies.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ohAUd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surveying rod would be placed on each peg and three height reading were taken from the surveying rod: the top cross hair, the central (main) cross hairs and the bottom cross. See the picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1PAlf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reading from the central cross hairs gives the height for the elevation. The difference between the upper and lower cross hair readings multiplied by an optical constant for the optics of the theodolite gave the distance between the surveying rod and the theodolite. Except for some Japanese theodolites, the optical constant was 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture above, the cross hair readings are 1.500, 1.422 and 1.344.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irrespective of which method was used. To make adjustments for surveying errors, a closed traverse was done whereby after everything that needed to be surveyed was measured, the last reading was back to the first peg surveyed. If the co-ordinates in 3D matched there were no errors. If they didn't each of the readings would need to be adjusted to close the traverse with &quot;no errors&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To minimize errors, the shorter the lateral distances the better, as there was less tape sag. For measurements that required high levels of accuracy, such as when assembling large equipment in hot climates the work would be done during the early mornings to minimize or eliminate heat shimmer. &lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-01-23T03:50:03.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="174" PostTypeId="2" ParentId="171" CreationDate="2015-01-23T04:17:50.497" Score="5" Body="&lt;p&gt;Generically speaking, the first step would be to determine if your failures have any factors in common. You could also examine your successes and see if &lt;strong&gt;they&lt;/strong&gt; have anything in common. These commonalities could be things such as the employee working on the product, location where the tests were performed, specific piece of equipment on which the tests were performed, time of day, etc. Once you have identified some commonalities you can attempt to determine if any of them are the proximate cause of the failures and work backwards until you can determine the ultimate cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the details of how to do this depend on the details of your situation.&lt;/p&gt;&#xA;" OwnerUserId="131" LastActivityDate="2015-01-23T04:17:50.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="175" PostTypeId="2" ParentId="165" CreationDate="2015-01-23T04:17:56.963" Score="10" Body="&lt;p&gt;The problem with long span pontoon bridges is not obvious at first. You would think that each section is perfectly capable of withstanding the relatively small amount of force from wind or currents applied to its adjacent section. However, when you think a bit more deeply, it becomes apparent that each section is pulling its adjacent section and that section is also affected by the same wind and currents, so it pulls a bit harder on its adjacent section. These forces accumulate until they are too great and the bridge would snap.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-23T04:17:56.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="176" PostTypeId="2" ParentId="20" CreationDate="2015-01-23T04:52:46.470" Score="4" Body="&lt;p&gt;The correct way to do this is by using what is known as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Particle_filter&quot; rel=&quot;nofollow&quot;&gt;particle filter&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maths for estimating your next position is quite simple and other answers have already provided that, but this is how you deal with the uncertainty. &lt;a href=&quot;https://www.youtube.com/watch?v=aUkBa1zMKv4&quot; rel=&quot;nofollow&quot;&gt;This video&lt;/a&gt; explains the basic principle rather well. You will notice that you need to take a measurement of some aspects relating to where you are, for example distance to known objects, compass reading, position of visible markings, etc. (Use your computer vision for that, plus anything else you can such as sonic range finders). It doesn't have to be perfect, the particle filter deals really well with 'noisy' measurements and 'noisy' (error-prone) predictions of where you are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also do a search for &quot;slam particle filter&quot; to get some further insight.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-23T04:52:46.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="177" PostTypeId="2" ParentId="165" CreationDate="2015-01-23T12:21:51.530" Score="16" Body="&lt;p&gt;Theoretically pontoon bridges with rope anchors keeping them to the bottom would work against wind and flow, overcoming the problem jhabbot mentioned in his answer (same as train length limit - stretching force).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice these come with more problems of their own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They drift on water surface and as result, rise and fall with water waves. The larger the body of water they span, the higher the waves; at certain point in stormy weather the bridge would just launch the vehicles into the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The anchoring isn't exactly simple if it's to withstand such forces. You could just as well go with pillars, these don't add much to complexity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are only a short way above the surface. Waves could roll over them washing vehicles off. Also, they stay level to the local surface - a wave wouldn't need to roll over it - it could just flip a vehicle sideways by twisting the bridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since segments need to be mobile relative to each over, their joints will be uneven, forcing a severe speed limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless they float freely, they'd be very limited with water level. If you anchor them firmly, water rising (even due to a storm) may submerge them. And yet again, as the segments need to be at least partially mobile, a longitudal force stretching one side of the bridge may lead to the other side to stack segments against each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have construction technologies that are extremely durable against longitudal forces (stretching, compressing) - reinforced concrete, steel ropes etc. But add lateral forces, and the design becomes much harder to keep strong; buckling, twisting, and loss of stability become very severe. With keeping the bridge suspended you keep lateral forces limited to wind. If the bridge is partially submerged, this goes out of the window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary upside of pontoon bridges is the simplicity - they can be deployed in matter of hours, and as such they play huge role in the military. But since they are vulnerable against weather and due to the slew of problems they create - especially with increasing span = size of the body of water = mechanical influence of water conditions, they make very poor permanent bridges, and so firmly supported bridges are simply superior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...also, where the ratio of vehicle traffic demand versus length of the span of water to cross is too low, the right solution is a ferry. Ferries can take many vehicles on board and cross the distance (and depths!) not viable for any bridge, and of course their cost is a tiny fraction of cost of the bridge covering that distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note pontoon bridges are okay as temporary solution (say, in place of a bridge washed off by flood, or for duration of construction of permanent bridge, or in locations where building a permanent bridge would be overly expensive or difficult), but they are always considered a poor man's substitute - and while they are okay for crossing a moderately sized river, the engineering problems scale up to insurmountable levels as the distance covered rises - they are really unsuitable for very long spans of water.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2015-01-30T21:03:19.057" LastActivityDate="2015-01-30T21:03:19.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="178" PostTypeId="2" ParentId="1" CreationDate="2015-01-23T12:56:39.927" Score="20" Body="&lt;p&gt;While pumping powders as a mix with swiftly moving gas is viable, one of frequently used alternatives for both powders and granular materials is &lt;a href=&quot;http://en.wikipedia.org/wiki/Archimedes%27_screw&quot; rel=&quot;noreferrer&quot;&gt;Archimedes' Screw&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KE01J.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, a helix in a diagonal pipe making the material &quot;pour/roll&quot; upwards the curve, allowing lifting it a long distance. This solution is frequently used in farming equipment for transporting grain or granulates. While it doesn't provide &quot;pressure&quot; for moving it through inert pipes, a &quot;pipeline&quot; of such screws can go a long way, and while it restricts transport angle, it provides a viable alternative especially for heavier materials where extreme air flows would be required to move them.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-01-23T12:56:39.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="179" PostTypeId="1" AcceptedAnswerId="388" CreationDate="2015-01-23T13:38:05.777" Score="13" ViewCount="3655" Body="&lt;p&gt;The Eurocodes gives the following equation for estimating a &quot;simply supported bridge subject to bending only&quot;*:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_0 = \frac{17.75}{\sqrt{\delta_0}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$n_0$ is the natural frequency &lt;em&gt;in hertz&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;$\delta_0$ is the deflection at mid-span under permanent actions &lt;em&gt;in mm&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The equation is seemly plucked from thin air, and there is no explanation as to where the constant 17.75 comes from. As an engineer I'm loath to use a formula I don't understand, but more than that it would be helpful to learn the fundamentals behind it so that I can see if it can be altered to work with other support conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone provide a derivation / fundamental origin to this relationship?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;*Full reference is: EN 1991-2:2003 6.4.4 [Note 8] (Equation 6.3), if that helps.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="151" LastEditorUserId="368" LastEditDate="2015-02-20T23:02:00.390" LastActivityDate="2015-09-14T13:04:30.707" Title="Derivation for bridge natural frequency estimate in Eurocodes" Tags="|bridges|dynamics|eurocodes|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="180" PostTypeId="1" CreationDate="2015-01-23T13:48:27.203" Score="3" ViewCount="336" Body="&lt;p&gt;I'm working on design of a 3D movement drive (for 3D printers/CNC mills etc) using DC motors and helical drive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A DC motor drives the leadscrew actuator which supports the working elements (head, table or head bridge). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current of the motor shaft -&gt; position of the driven element is read with optical encoder. (I can derive speed, acceleration, momentum etc from that.) I can read the current (voltage across a fixed low resistance in series with the motor), and the voltage across the motor terminals, resulting in measurement of the power -&gt; torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can drive the power supplied to the motor through PWM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For optimal drive, I should maintain movement speed as close as possible to given (right for the used machining parameters), and as I'm machining material, the resistance of the material against the tool will change in more or less unexpected pattern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is pretty simple with stepper motors where we just assume the output directly follows the input. In case of DC motor though, the control must be much more reactive, increasing power as torque rises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What control algorithm would allow me to maintain movement speed closest to preset - meeting the additional condition: restricting or rapidly minimizing exceeding the preset speed; rapidly reduced load as the head exits material could result in speed increase and drive the head into a neighbour element, damaging the work piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(please note, this is one of the questions that take long rather than short answers; control algorithms in automatics is a subject of good several semesters of study, and choosing and explaining the control algorithm meeting specified criteria is bound to require some work.)&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="368" LastEditDate="2015-05-12T16:28:01.920" LastActivityDate="2015-05-12T16:28:01.920" Title="Given drive rotation angle, controlling torque, how do I maintain movement speed with varied load?" Tags="|control-engineering|feedback-loop|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="181" PostTypeId="2" ParentId="180" CreationDate="2015-01-23T14:14:26.427" Score="3" Body="&lt;p&gt;This is typically done with a PID (Proportional, Integral, Derivative) control algorithm. There are heaps of literature about designing and optimizing PID controllers, so there's not much sense going into a more specific detail here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically you use the PID controller to regulate speed. Assuming your stopping point is also critical, there will be some trade-off between accuracy of position and accuracy of deceleration at the end point. With relatively low inertia systems and reasonable deceleration values this is not usually a very big issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a very in depth article on the design of PID controllers at &lt;a href=&quot;https://en.wikipedia.org/wiki/PID_controller&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/PID_controller&lt;/a&gt; . There are a number of sample and open source implementations available as well, depending on what platform you are working on. If you have already done this basic research and have a more specific question, please clarify.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-01-23T14:14:26.427" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="182" PostTypeId="2" ParentId="179" CreationDate="2015-01-23T16:24:08.883" Score="3" Body="&lt;p&gt;Here's a possible answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=bridge%20natural%20frequency%20estimate&amp;amp;source=web&amp;amp;cd=10&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CGMQFjAJ&amp;amp;url=http%3A%2F%2Fwww.eng.uwo.ca%2Fpeople%2Fesavory%2Fstudy%2520note%2520on%2520dynamics.doc&amp;amp;ei=vnDCVLCQF6PCsATguYGIDA&amp;amp;usg=AFQjCNHdBl_-R5k8O2FjTInpIe5gpxgYfA&amp;amp;sig2=itv9qJF31tDNA7aJn-NxYg&amp;amp;bvm=bv.84349003,d.cWc&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; document (not sure of the exact source), which contains a related derivation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a simple harmonic motion problem,&#xA;$$n_0=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$$&#xA;where $k$ is the elastic stiffness and $m$ is the mass undergoing vibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k=\frac{\text{load}}{\text{deflection}}=\frac{F}{\delta}$$&#xA;where $F$ is force and $\delta$ is the deflection. Thus,&#xA;$$n_0=\frac{1}{2 \pi}\sqrt{\frac{F}{m\delta}}=\frac{1}{2 \pi}\sqrt{\frac{ma}{m \delta}}=\frac{1}{2 \pi}\sqrt{\frac{a}{\delta}}$$&#xA;But the deflection in your example is in millimeters, while it's in meters here, so I get about&#xA;$$n_0=5.03 \sqrt{\frac{a}{\delta}}$$&#xA;If $a=12.4382$, we get your equation. But I'm not sure where this value comes from. It could be that another unit switch is needed, or it could be that this constant is only for a small subset of cases, where the acceleration is along those lines.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-01-23T16:45:55.183" LastActivityDate="2015-01-23T16:45:55.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="183" PostTypeId="1" CreationDate="2015-01-23T16:25:41.180" Score="11" ViewCount="2529" Body="&lt;p&gt;I am working on rub-induced vibrations in compressor stages of aircraft engines. The work is carried out in partnership with a company, therefore there is a limited amount of it that could be published openly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it feasible to release &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Open_source&quot; rel=&quot;nofollow&quot;&gt;open source&lt;/a&gt;&lt;/strong&gt; compressor or turbine blade designs (e.g., NACA airfoil profiles for wings), as is commonly done in the software industry? By which I mean, detailed dimensions—in particular for the cross-section profiles and stacking laws—and the associated material properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this has already been done, I'm not aware of it. Are there specific obstacles preventing the open source movement from spreading from the programming world into the engineering world? If so, are they technical or non-technical obstacles and how could they be overcome?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only would these specific designs be very useful for my own work, displaying relevant characteristics and realistic behaviors from an engineering perspective, while sharing all geometry and material property details, would allow other researchers to reproduce published results.&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="368" LastEditDate="2015-05-11T03:37:16.380" LastActivityDate="2015-05-11T03:37:42.460" Title="Detailed dimensions of an aircraft engine compressor blade" Tags="|mechanical-engineering|vibration|dynamics|aircraft-design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="184" PostTypeId="2" ParentId="183" CreationDate="2015-01-23T16:52:51.040" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;http://www.spaceflight.esa.int/impress/text/education/Circular%20Motion/Turbine_Blades.html&quot; rel=&quot;noreferrer&quot;&gt;ESA&lt;/a&gt; has a page on compressor blades. They give a good dimensioned diagram of an approximate shape; here are some basic dimensions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Length:&lt;/strong&gt; 300 mm&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Width:&lt;/strong&gt; 30 mm&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Height:&lt;/strong&gt; 70 mm&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Thickness:&lt;/strong&gt; 5 mm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can't find any &lt;em&gt;complete&lt;/em&gt; open source designs (i.e. high-quality, technical engineering drawings), but this is a good approximation.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-01-23T16:52:51.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="186" PostTypeId="1" AcceptedAnswerId="8521" CreationDate="2015-01-23T17:34:03.773" Score="7" ViewCount="3443" Body="&lt;p&gt;There are numerous steel alloys, containing mostly iron, carbon, and some other metals. Generally speaking, we can think of them as if they were some type of steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: do &quot;non-ferrous steels&quot; exist? I am thinking of pure, non-iron metals, containing a little carbon, just as is added to iron to turn it into steel.  Or asked another way, are there other metals besides iron that are doped with carbon in order to form an alloy like steel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, how does the addition of carbon affect the properties of these metals?&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorDisplayName="user16" LastEditDate="2015-01-24T15:43:43.923" LastActivityDate="2016-04-22T04:16:54.383" Title="Do &quot;non-iron steels&quot; exist?" Tags="|steel|alloys|metals|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="187" PostTypeId="5" CreationDate="2015-01-23T17:37:20.013" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-23T17:37:20.013" LastActivityDate="2015-01-23T17:37:20.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="188" PostTypeId="4" CreationDate="2015-01-23T17:37:20.013" Score="0" Body="Metallurgy is the science of the production and the processing of different metals and other metallurgic usable elements." OwnerUserId="9" LastEditorUserId="9" LastEditDate="2015-01-27T14:51:19.077" LastActivityDate="2015-01-27T14:51:19.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="189" PostTypeId="2" ParentId="186" CreationDate="2015-01-23T17:48:57.923" Score="18" Body="&lt;p&gt;Steel is defined as an alloy of iron and carbon; there is no such thing as a non-ferrous steel. If you alloy some other metal with carbon, it becomes something other than steel. Looking for a steel without iron in it would be like looking for brass or bronze without copper. You can alloy things other than copper with zinc, tin, or aluminum, but those would not be kinds of brass or bronze. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as other alloys that contain carbon, &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_alloys&quot;&gt;this Wikipedia article&lt;/a&gt; has a good list of various kinds of alloys (as you can see, there are a lot of them), and searching through it, you'll see that there aren't a lot of other things that are alloyed with carbon besides iron. As for why this is, I don't have a good answer. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-01-23T17:48:57.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="191" PostTypeId="1" AcceptedAnswerId="529" CreationDate="2015-01-23T18:03:09.120" Score="10" ViewCount="3174" Body="&lt;p&gt;As I know, a very considerable part of the currently processed steels (around half of it) is coming from recycling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But during the steels coming into the recycling process are coming normally from various sources, and thus they are containing very different alloying materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the output of the reprocessed steel must be steel containing alloys exactly in the specified ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do some type of &quot;separation&quot;, or &quot;removal&quot; of the previous alloys of the recycled steel happen? And if yes, how does it work?&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2019-02-18T20:43:13.337" Title="How can the alloying materials of recycled steels be separated?" Tags="|steel|metallurgy|alloys|recycling|" AnswerCount="4" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="192" PostTypeId="2" ParentId="151" CreationDate="2015-01-23T18:05:24.680" Score="9" Body="&lt;p&gt;The RM-1 Russian submarine reactor had a core of less than one cubic metre. It had about 100kg fuel load, which was 90% enriched (i.e. 90kg) Uranium 235. This was liquid-metal cooled [specifically a &amp;quot;eutectic lead-bismuth alloy (44.5 wt% lead, 55.5 wt% bismuth)&amp;quot; - source as below, p40], so didn't need a moderator.&lt;/p&gt;&#xA;&lt;p&gt;Submarine 901 had in its right-board reactor just 30.6 kg of Uranium 235; this was at 20% enrichment, so a total fuel load of 153 kg.&lt;/p&gt;&#xA;&lt;p&gt;These were controllable chain-reaction based reactors.&lt;/p&gt;&#xA;&lt;h3&gt;Source:&lt;/h3&gt;&#xA;&lt;p&gt;&lt;strong&gt;NKS-138 Russian Nuclear Power Plants for Marine Applications&lt;/strong&gt;&lt;br /&gt;&#xA;Ole Reistad, Norwegian Radiation Protection Authority, Norway&lt;br /&gt;&#xA;Povl L. Ølgaard, Risø National Laboratory, Denmark&lt;br /&gt;&#xA;Published by Nordic Nuclear Safety Research, April 2006&lt;br /&gt;&#xA;ISBN: 87-7893-200-9&lt;br /&gt;&#xA;&lt;a href=&quot;http://www.nks.org/scripts/getdocument.php?file=111010111120029&quot; rel=&quot;noreferrer&quot;&gt;http://www.nks.org/scripts/getdocument.php?file=111010111120029&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-01-23T18:21:13.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="193" PostTypeId="1" AcceptedAnswerId="194" CreationDate="2015-01-23T18:05:35.530" Score="16" ViewCount="7907" Body="&lt;p&gt;Consider a large auditorium, a church, or some other very large, essentially one roomed building with a high ceiling.  Suppose that the building has many entrances which enable cold air in/hot air out and traffic in and out of the building is unavoidably large.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine that any attempts to control the temperature of such a large building would be very inefficient in terms of energy and cost, particularly because warm air rises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that it's very cold outside and we're interested primarily in keeping the building warm at the ground level so that it is comfortable for humans to work and interact,  what is the best method to keep such a large, essentially one roomed building with a high ceiling warm when outdoor cold air exposure is frequent and unavoidable?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I say &quot;best,&quot; I'm interested in balancing energy, maintenance, and monetary costs over the life of the building.  &lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="368" LastEditDate="2015-05-01T19:31:41.237" LastActivityDate="2015-05-01T19:31:41.237" Title="What is the most efficient means of warming a building with a high ceiling?" Tags="|civil-engineering|building-physics|energy|hvac|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="194" PostTypeId="2" ParentId="193" CreationDate="2015-01-23T18:18:43.203" Score="15" Body="&lt;p&gt;Your question sort of has two parts: How to supply heat, and how to keep it in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large open rooms with a high ceilings are most efficiently warmed with radiant ceiling heat.  Warm air rises, which renders forced-air systems inefficient because the pumped heat ends up at the ceiling and the coldest part of the room is near the floor where you actually want the heat.  Radiant floor systems are limited to about 87F because they are in contact with occupants, and so their peak output may not be enough to keep the space comfortable.  They also lose more heat to convection than radiant ceilings. (See &lt;a href=&quot;http://www.tesmar.com/html/radiant_floors_vs_radiant_ceilings.html&quot;&gt;this ref&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for keeping heat in, besides solid insulation/barriers: air doors (a.k.a. air curtains) are the standard solution in high traffic passages.&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="77" LastEditDate="2015-01-23T22:41:36.817" LastActivityDate="2015-01-23T22:41:36.817" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="195" PostTypeId="5" CreationDate="2015-01-23T18:25:26.840" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-23T18:25:26.840" LastActivityDate="2015-01-23T18:25:26.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="196" PostTypeId="4" CreationDate="2015-01-23T18:25:26.840" Score="0" Body="Questions about air heating, ventilation, and conditioning, including indoor air quality (IAQ)." OwnerUserId="77" LastEditorUserId="77" LastEditDate="2015-01-27T14:51:38.230" LastActivityDate="2015-01-27T14:51:38.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="197" PostTypeId="2" ParentId="193" CreationDate="2015-01-23T18:34:43.897" Score="4" Body="&lt;p&gt;Here are some ways to keep the warm air from escaping:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Revolving doors&lt;/strong&gt; prevent a constant stream of air exchange between the inside and outside.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Air lock&lt;/strong&gt; areas - a double-set of sliding doors can also prevent continuous air-exchange.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Positive pressure&lt;/strong&gt; by the entrance, usually supplied by a fan heater above the inner entrance door ensures the main bulk of air stays inside and the air from the fan heater escapes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To keep the lower areas warm, the warm air at the top of the building can be blown down to floor-vents at very slow speed. Either directly or via a heat exchanger so that the heat from the stale air is transferred to a fresh stream of air from outside. This is also a relatively efficient way to heat the air in the first place. A heat-pump can also be used to heat the floor using under-floor heating pipes. This helps to prevent heat-loss from the warm air into the floor and gently keeps the physical mass in the lower area warm.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-23T18:34:43.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="198" PostTypeId="2" ParentId="191" CreationDate="2015-01-23T19:03:54.080" Score="7" Body="&lt;p&gt;To the best of my knowledge, such separation of components is not attempted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a friend who at one time worked for Lukens Steel in Coatesville, PA. His job was to write computer software that kept track of the composition of all of the scrap steel they had in their yards and to come up with the correct proportions of which kinds of scrap to use for any new melts. Obviously, this implies that they did a fairly comprehensive analysis of all incoming scrap and sorted similar alloys into separate piles.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-01-23T19:03:54.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="200" PostTypeId="1" AcceptedAnswerId="206" CreationDate="2015-01-23T20:21:06.483" Score="1" ViewCount="4785" Body="&lt;p&gt;For a &lt;a href=&quot;http://en.wikipedia.org/wiki/Piston_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;piston pump&lt;/a&gt; you can calculate/measure things like,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Average Volume flow rate (m³/s)&lt;/li&gt;&#xA;&lt;li&gt;Geometric Values (Surface)&lt;/li&gt;&#xA;&lt;li&gt;Pressure over time&lt;/li&gt;&#xA;&lt;li&gt;closing times of the valves&lt;/li&gt;&#xA;&lt;li&gt;energy conversion efficiency $\eta$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;besides those, what charateristic values are there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c6X9H.gif&quot; alt=&quot;enter image description here&quot;&gt; (&lt;em&gt;example of a type of piston pump&lt;/em&gt; &lt;a href=&quot;https://nationalvetcontent.edu.au/alfresco/d/d/workspace/SpacesStore/dc6a7f1f-e3d3-44d8-9bf1-984ea8cb3c01/204/pmaops201b/images/proc201_010100_g04.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="46" LastEditDate="2015-01-26T11:31:54.410" LastActivityDate="2021-08-14T06:46:00.820" Title="What are characteristic values for a piston pump?" Tags="|mechanical-engineering|pumps|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="201" PostTypeId="2" ParentId="73" CreationDate="2015-01-23T21:10:48.197" Score="7" Body="&lt;p&gt;My first thought was to use a small amount of chromium and then another metal (gold). This is how its done for electronics structures on glass. The chromium does react with the silicium-oxid and hence has strong bonding. I've done this only with layers of very small thickness (sub µm), and only via a &lt;a href=&quot;http://en.wikipedia.org/wiki/Physical_vapor_deposition&quot; rel=&quot;noreferrer&quot;&gt;PVD&lt;/a&gt;-technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt;, as you can read on &lt;a href=&quot;http://en.wikipedia.org/wiki/Glass-to-metal_seal&quot; rel=&quot;noreferrer&quot;&gt;this Wikipedia page&lt;/a&gt;, it can be done in this way for glass-metal seals. For example, with chromium and stainless steel. As such, I think it is rather cost effective, at least if you compare it with pure indium and gallium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your application has temperature gradients, be aware of the potentially very different temperature coefficients.&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="368" LastEditDate="2015-07-21T22:53:41.243" LastActivityDate="2015-07-21T22:53:41.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="202" PostTypeId="2" ParentId="166" CreationDate="2015-01-24T00:41:19.003" Score="8" Body="&lt;p&gt;I'm still confused about what you mean by &quot;engineering advances&quot;. To me, things like vacuum tubes, transistors and ICs are &lt;em&gt;technologies&lt;/em&gt;, and engineering is all about &lt;em&gt;applying&lt;/em&gt; those technologies to real-world problems, such as building computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers have typically taken each available technology to its practical limit &amp;mdash; to the point where other issues become predominant. Examples of this would include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ability to machine the individual parts of Babbage's mill to sufficient precision so that it worked accurately and reliably. This became the limit on how complex a mechanical calculator could be.&lt;/li&gt;&#xA;&lt;li&gt;Making vacuum tubes &lt;em&gt;reliable&lt;/em&gt; enough that you could operate tens of thousands of them for at least a few hours before one of them burned out. This became the limit on how complex a vacuum tube computer could be.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are these the sorts of &lt;em&gt;engineering&lt;/em&gt; issues you're looking for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With integrated circuits, the current mainstream technology, the engineering issues include things like minimum feature size, static and dynamic power dissipation, and defect densities. It's a complex problem to find the best combination of parameters that results in the most cost-effective (and profitable!) chips.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How did the discovery of vacuum tubes, transistors, and integrated circuits allow people to &quot;engineer&quot; computers in a way they hadn't done so before?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I said, the physical properties of mechanical parts and the ability to machine them precisely limited both the speed and the complexity of mechanical and electromechanical (relay-based) computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ability to mass-produce reliable vacuum tubes made it possible to build purely electronic circuits that matched (and eventually exceeded) the complexity of relay-based logic, while operating at much higher speeds: tens and hundredss of kHz, rather than tens of cycles per second or less. However, the reliability of the tube filaments eventually became the dominant issue, once computers got to the point of needing tens of thousands of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The development of the transistor allowed circuits to be shrunk by at least an order of magnitude, but more importantly, the MTBF of transistors exceeded that of tubes by several orders of magnitude. Both of these advantages enabled the development of more complex computers. But the physical size of the transistor in its package still presented a problem in terms of the size of the computer, and the large number of discrete connections still presented a reliability issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integrated circuit has solved both of those issues, with millions of transistors crammed into the volume that a single discrete transistor requires, and the connections being made by patterning metal directly on the IC surface. Both the complexity and the speed of computers has grown by leaps and bounds as we learned to make ever-finer patterns on the surface of the chips. We are only just beginning to see some fundamental limits on how far we can push this before we need a new fundamental technology.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2015-01-24T17:27:00.610" LastActivityDate="2015-01-24T17:27:00.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="203" PostTypeId="1" CreationDate="2015-01-24T01:02:20.867" Score="9" ViewCount="361" Body="&lt;p&gt;I'm trying to bond semiconductor samples (Si and Ge) area ~1-2cm^2 to &#xA;fiberglass printed circuit board (PCB)&#xA;We use West system marine epoxy for other things. 105 resin 209 hardener (long cure time.)&#xA;So that's what I used. (mixed at standard ratio)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted a controlled thickness for electrical isolation.&#xA;So I also tried adding some fillers to the epoxy.&#xA;Glass beads (9.8 mil .. kinda thick IMO, sprinkled on surface.)&#xA;Alumina oxide, 240 grit. (~1 part Al2O3 to 2 parts epoxy, by weight.)&#xA;All samples (but one Ge) were from an old piece of Si wafer.&#xA;Samples and pcb were cleaned in acetone, &#xA;and scrubbed with cotton tipped applicator.&#xA;Epoxy mixed. applied and samples pushed into place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And allowed to cure for 24 hours.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They where then dunked into liquid nitrogen (LN2).&#xA;After a few dunks, warming in room air, &#xA;the samples glued with Al2O3 filler had fallen off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After more dunking the samples held with raw epoxy and beads fell off.&#xA;A few more tortures that also included warming more rapidly &#xA;with a heat gun.  And I'd lost everything but the Ge sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final abuse the Ge sample was taken from LN2 and placed into&#xA;a cup of warm water, several times.  It remained attached. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the bonds failed at the Si interface, and the epoxy remained attached &#xA;to the PCB  (except for the glass beads, that failed everywhere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought was about the coefficients of thermal expansion (CTE) &lt;a href=&quot;http://www.cleanroom.byu.edu/CTE_materials.phtml&quot; rel=&quot;noreferrer&quot;&gt;Here's a link to some values&lt;/a&gt;, Si is very low.&lt;br&gt;&#xA;The pcb is ~ &lt;a href=&quot;http://en.wikipedia.org/wiki/FR-4#Specifications&quot; rel=&quot;noreferrer&quot;&gt;12-14 ppm.&lt;/a&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then thought about cleaning.  The old Si samples may have all sorts of hand grease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final difference is that the Si samples are polished on both sides,&#xA;while the Ge was only onthe top... the bottom was rough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;wow, that was a long question, (sorry)&lt;br&gt;&#xA;I made up a new batch of samples today to try and answer those questions.  They will cure over the weekend.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wonder if I need a different epoxy?  The West 105 stays somewhat pliable.&lt;br&gt;&#xA;I don't know if that's a good or bad thing.  &lt;/p&gt;&#xA;" OwnerUserId="168" LastEditorUserId="110" LastEditDate="2015-02-09T11:59:52.517" LastActivityDate="2017-10-26T02:04:18.977" Title="Bonding Semiconductors to PCB" Tags="|electrical-engineering|joining|epoxy|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="204" PostTypeId="2" ParentId="150" CreationDate="2015-01-24T03:15:42.027" Score="9" Body="&lt;p&gt;For a different angle, my experience with solar cells is space-based, rather than terrestrial. The biggest drivers for degradation of space based solar cells are particle radiation and orbital debris impacts. The particle radiation degrades by wearing down the junction, and the orbital debris degrades primarily by increasing the internal series resistance of the cells. Because there is no air to cool the cells, reverse bias heating can also be a serious problem if a solar array gets partially shadowed by other structure.&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2015-01-24T03:15:42.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="205" PostTypeId="2" ParentId="32" CreationDate="2015-01-24T03:27:27.790" Score="5" Body="&lt;p&gt;For tapping threads, there are a few steps between free hand tapping and fully automatic machines. Which one is right for you will depend on the material you're tapping, the size of the threads you're tapping, and how large your production run is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step would be to buy a guided tap wrench, which is a regular hand tap wrench with a bushing you can put on the top and chuck into a drill press (or mill) or lathe. This helps keep your tap square to the workpiece and also lets you apply a little pressure with the quill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get into power tapping (And need more precision than a cordless drill with the clutch turned down,) there are reversing tapping heads you can attach to a drill which reverses the tap out when threading is complete. These are available for drill presses or hand-held drills. They run around \$500-$1,000. here's another style of tap chuck called a tension compression chuck, which allows a little vertical give but I believe that is only of much use if your spindle is computer controlled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a more robust solution and have some money, there are tapping presses that start at around $3,000. It's not quite 6 figures, but they aren't cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what you are making, self threading screws may also be a valid option. There are a wide variety of types that work in different situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for external threads, after die cutting, the only cheaper method I'm aware of is roll threading. This requires a dedicated machine that is relatively expensive unless you're working on very small threads.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-01-24T03:27:27.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="206" PostTypeId="2" ParentId="200" CreationDate="2015-01-24T04:05:35.140" Score="3" Body="&lt;p&gt;Here are some other criteria:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cycle time&lt;/li&gt;&#xA;&lt;li&gt;Piston speed&lt;/li&gt;&#xA;&lt;li&gt;Piston friction&lt;/li&gt;&#xA;&lt;li&gt;Device lifespan (most likely a function of the number of cycles completed)&lt;/li&gt;&#xA;&lt;li&gt;Number of suction/discharge ports&lt;/li&gt;&#xA;&lt;li&gt;Type of valves used&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="131" LastActivityDate="2015-01-24T04:05:35.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="207" PostTypeId="1" AcceptedAnswerId="208" CreationDate="2015-01-24T05:36:09.180" Score="21" ViewCount="2678" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Disclaimer&lt;/em&gt;&lt;/strong&gt;&#xA;I'm an applied mathematician by training, not an engineer.  My work research primarily focuses on creating new &quot;methods&quot; to solve different PDE's related to solid deformation (elasticity) and fluid mechanics.  In this sense, i know how to solve a pde problem computationally.  From my perspective, engineers use my work as &quot;tools&quot; to accomplish their work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, due to my lack of education/experience in engineering, i admit i'm actually rather clueless on how numerical solutions to pde's are really used in an engineers actual practice.  &lt;strong&gt;&lt;em&gt;The primary source of my confusion is the following:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been told that engineers never (or should never) conduct numerical simulations (e.g. finite element analysis, CFD, etc...) without knowing or having a good idea ahead of time what the simulation &quot;should&quot; look like.  This helps engineers discriminate realistic results from questionable ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, i argue that  &lt;strong&gt;&lt;em&gt;if the engineer already knows what is supposed to happen in the simulation, then what's the point of simulation in the first place???&lt;/em&gt;&lt;/strong&gt; I've always assumed that simulations are needed for predictive purposes, which assumes ignorance of what is to come.  That is, &lt;strong&gt;&lt;em&gt;I think of a simulation as a stand-alone tool to predict the future when you don't know what to expect&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i'm looking for is a broader perspective into how/when/why engineers use numerical simulations like CFD and Finite Element Analysis, especially if good engineering practice dictates that you should already know what to expect when you're simulating?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="88" LastEditDate="2015-02-02T20:13:10.587" LastActivityDate="2017-12-01T13:42:18.210" Title="How do engineers really use numerical simulation?" Tags="|modeling|" AnswerCount="8" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="208" PostTypeId="2" ParentId="207" CreationDate="2015-01-24T07:35:36.857" Score="24" Body="&lt;p&gt;I have written mostly about CFD in this answer, however same points should also work for FEA or other simulation techniques. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;CFD is mostly used for design optimization and parametric study of the design. &#xA;Following are a few examples showing how engineers use simulations&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Selection of a design&lt;/strong&gt;:&#xA;Read: &lt;a href=&quot;http://soar.wichita.edu/bitstream/handle/10057/5048/t11063_Ghoddoussi.pdf?sequence=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;A conceptual study of airfoil performance enhancement using CFD.&lt;/a&gt; &#xA;This thesis shows use of CFD for selecting the best design out of a number of candidate designs. Engineers often go for simulations to &lt;strong&gt;select 'the one' out of many&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Shape optimization using CFD&lt;/strong&gt;:&#xA;&lt;a href=&quot;http://webdocs.cs.ualberta.ca/~mgarcia/vwt_public/papers/paper_EngOpt_SG.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; gives an example of wing shape optimization using CFD. And &lt;a href=&quot;https://www.youtube.com/watch?v=YZUNRmwoijw&quot; rel=&quot;nofollow noreferrer&quot;&gt;this amazing YouTube video&lt;/a&gt; is an excellent example of the way an engineer would use a CFD software (&lt;a href=&quot;http://www.openfoam.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;OpenFOAM&lt;/a&gt;) and genetic algorithm. CFD makes it possible to arrive at a better design without actually building a number of prototypes and testing (which is an expensive and long process). Actually design optimization is the most common way the CFD is used. According to &lt;a href=&quot;http://www.mentor.com/company/news/survey-highlights-potential-for-increased-use-of-cfd-by-mechanical&quot; rel=&quot;nofollow noreferrer&quot;&gt;this survey&lt;/a&gt;, mechanical design engineers make the use of CFD the most (note: I do not know the authenticity of the report). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Using simulations where experiments are difficult to carry out / might cost a lot of resources (or life)&lt;/strong&gt;:&#xA;Applications where experiments are not possible to carry out, such as the heat transfer in hypersonic re-entry vehicles (&lt;a href=&quot;http://www.hypersonic-cfd.com/research.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;examples here&lt;/a&gt;), or &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3173660/&quot; rel=&quot;nofollow noreferrer&quot;&gt;blood flow in human body&lt;/a&gt;, can be simulated with a computer and final design can be tested. Another example; CFD is used for placement of probes on a wind tunnel model. CFD gives, for example, the position of the stagnation point on a surface of the model, and there we can have the pressure probe placed and then test the model in actual wind tunnel. This presentation explains how &lt;a href=&quot;http://www.cd-adapco.com/sites/default/files/Presentation/Todd_Leighton.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;CFD and wind tunnel are complimentary to each other&lt;/a&gt;. Also CFD is used to predict the results where experimental results are not available (one can not have probes everywhere on the model). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Design and optimization of the experimentation facility itself&lt;/strong&gt;:&#xA;Simulations are commonly used for design of the facility itself. For example, &lt;a href=&quot;http://www.academia.edu/8136650/A_validated_design_methodology_for_a_closed-loop_subsonic_wind_tunnel&quot; rel=&quot;nofollow noreferrer&quot;&gt;this report&lt;/a&gt; describes how CFD is used for design of the wind tunnel. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;To develop a theoretical model&lt;/strong&gt;:&#xA;This is often seen in cosmology. Scientists carry out simulations based on a model and validate with the experimental data. This iterative process results in better understanding of the physics and working of the universe. &#xA;NASA astrophysics group have done some simulation of Supermassive Black Holes, &lt;a href=&quot;http://www.space.com/21573-stellar-mass-black-hole-simulated-by-supercomputer-video.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video talks more about it&lt;/a&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;In movies, art and animations&lt;/strong&gt;:&#xA;&lt;a href=&quot;https://scicomp.stackexchange.com/questions/7305/is-cfd-used-in-animations-movies-art/7306#7306&quot;&gt;This question and following answers&lt;/a&gt; on Scicomp.SE show, how much a role CFD has to play in movies and animations... (disclaimer: I have asked the question). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some other applications:&#xA;&lt;a href=&quot;http://waset.org/publications/9998402/cfd-modeling-of-insect-flight-at-low-reynolds-number&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aerodynamics of the insect flight&lt;/a&gt;, &lt;a href=&quot;http://www.cs.odu.edu/~mln/ltrs-pdfs/NASA-99-10tfaw-dpl.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Noise computation using CAA&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Computational_electromagnetics&quot; rel=&quot;nofollow noreferrer&quot;&gt;design of antennas and stealth technology using CEM&lt;/a&gt;, &lt;a href=&quot;http://ecaaser3.ecaa.ntu.edu.tw/weifang/eBook/greenhouse-search/Applications%20of%20CFD%20in%20the%20food%20industry%20-%20a%20review.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Applications of CFD in food industry&lt;/a&gt; etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The list will go on...&#xA;End of the day, CFD is a virtual wind tunnel, its a workbench where an engineer can test his idea without manufacturing / building anything. So if the results are validated against a known model / experiment, then one can rely on the CFD methodology for a slight change in geometry or shape. Also because of the CFD results, an engineer can have confidence in his/her experimental results. Thats why the term validation. A good resource for validation test cases &lt;a href=&quot;http://www.cfd-online.com/Wiki/Validation_and_test_cases&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!   &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="-1" LastEditDate="2017-04-13T12:53:56.163" LastActivityDate="2015-01-24T17:49:54.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="209" PostTypeId="2" ParentId="207" CreationDate="2015-01-24T08:08:50.283" Score="8" Body="&lt;p&gt;As my Fluids lecturer said many years ago, “if the mathematics doesn't agree with reality the mathematics is wrong”. You can easily substitute the words model, theory or simulation for the word mathematics..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers who use simulations should have a very good idea of what to expect for a solution, not necessarily know what the answer will be for a simulation. There is a difference. That's where the engineer's experience is critical and why inexperienced engineers should always be well supervised when doing simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers use simulations for a variety of reasons, depending on the field of engineering they work in and what they are doing. Some engineers use simulations to confirm their designs while others use simulations to look for potential weaknesses in designs or materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other aspect of simulations is they allow engineers to consider a number of “what-if scenarios” to ascertain what could happen when parameters are changed. This can be used  to look at upper and lower bound performance limits or it can lead to design changes and in some cases a total redesign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, depending on the field of engineering, simulations are also useful when considering when something need to be added to or increased in scale, such the affect on a water distribution system by adding a new development, or changes that need to be made to the ventilation system of an underground mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simulations can also be done to look at:&#xA; - the impact on the flow of materials and resources: oil or water in&#xA;   their respective piping networks, air in ventilation networks, ore&#xA;   from a mine  or several mines to a processing plant or a number of&#xA;   processing plants&#xA; - blending of mineral products extending public&#xA; - transport infrastructure like railways, road, electricity &amp;amp;&#xA;   communication networks&#xA; - traffic movement when changes are made to a traffic system: road&#xA;   blocked or widened, reorganized for one-way traffic, the introduction&#xA;   of clearways and the prohibiting of parking on the sides of roads the&#xA; - the design of underground spaces for civil applications such as&lt;br&gt;&#xA;   underground parking areas, train stations or tunnels and stopes in an&#xA;   underground mine.&#xA; - financial NPV evaluations for project economics and    investment&#xA;   purposes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's always cheaper and prudent to run a number of simulations than to construct something and have it fail catastrophically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another one of my university lecturer's also said, way back when, “Doctor's bury their mistakes, architects plan vines around their mistakes, engineers are killed by their mistakes”.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-01-24T08:08:50.283" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="210" PostTypeId="2" ParentId="207" CreationDate="2015-01-24T12:39:45.997" Score="13" Body="&lt;p&gt;To summarize the other answers: An engineer needs to know qualitatively how the simulation will go, but he still needs to run the simulation to get the quantitative answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, simulation allows the engineer to vary the parameters slightly (&lt;a href=&quot;http://en.wikipedia.org/wiki/Monte_Carlo_method&quot;&gt;Monte Carlo simulation&lt;/a&gt;) in order to evaluate the stability or margin of error of the solution. This is frequently done in electrical circuit simulation, for example, in order to evaluate a design's sensitivity to component value tolerances.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-01-24T12:39:45.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="211" PostTypeId="1" AcceptedAnswerId="222" CreationDate="2015-01-24T13:13:19.357" Score="12" ViewCount="923" Body="&lt;p&gt;These days most of modern electronics use rechargeable batteries as a power source. Also, these days most modern rechargeable batteries are &lt;strong&gt;Lithium Ion&lt;/strong&gt; or &lt;strong&gt;Lithium Polymer&lt;/strong&gt; based. Like any other devices, over time these rechargeable batteries lose the ability to recharge, retain and discharge energy thus users have to replace the devices or rechargeable batteries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my understanding that the rise in battery internal resistance is the primary cause the rechargeable battery aging. Is this accurate? If so what can be done to lower or eliminate the internal resistance in rechargeable batteries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my understanding is inaccurate what is the cause for rechargeable battery aging?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the causes for battery aging are understood, how can electronic engineers design charging and discharging circuits to extend the rechargeable battery life?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://batteryuniversity.com/&quot; rel=&quot;nofollow&quot;&gt;Battery University&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eetimes.com/author.asp?section_id=36&amp;amp;doc_id=1320292&quot; rel=&quot;nofollow&quot;&gt;All About Batteries, Part 1: Introduction&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://All%20About%20Batteries,%20Part%202:%20Specifications%20&amp;amp;%20Terminology&quot;&gt;All About Batteries, Part 2: Specifications &amp;amp; Terminology&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eetimes.com/author.asp?doc_id=1322276&quot; rel=&quot;nofollow&quot;&gt;All About Batteries, Part 7: Lithium Thionyl Chloride&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-05-11T10:23:34.773" LastActivityDate="2016-05-11T10:23:34.773" Title="What causes rechargeable batteries to age? What can be done to extend life of these batteries?" Tags="|energy|electrical-engineering|energy-storage|chemical-engineering|battery|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="212" PostTypeId="1" AcceptedAnswerId="235" CreationDate="2015-01-24T15:27:54.717" Score="-1" ViewCount="814" Body="&lt;p&gt;EMI/EMC issues are common in most electronic designs. Electronic and Hardware Engineers use many different strategies to mitigate these issues: These include (But not limited too)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Designing filter circuits &lt;/li&gt;&#xA;&lt;li&gt;Strategic selection of micro-controller clocking frequency &lt;/li&gt;&#xA;&lt;li&gt;PCB layout consideration including good grounding practices&lt;/li&gt;&#xA;&lt;li&gt;Mechanical Shielding of critical electrical circuits for immunity or suppression&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; This particular design in question, has a plastic enclosure, LCD display and USB type to connector for recharge a battery. What are some of the mechanical engineering best practices to help design product meet EMC/EMI requirement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ieee.li/pdf/viewgraphs/emc_design_fundamentals.pdf&quot; rel=&quot;nofollow&quot;&gt;EMC Design Fundamentals&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/sc/docs/apps/msp/intrface/usb/emitest.pdf&quot; rel=&quot;nofollow&quot;&gt;EMI Design Guideline for USB components&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cache.freescale.com/files/microcontrollers/doc/app_note/AN4438.pdf&quot; rel=&quot;nofollow&quot;&gt;EMC Design Considerations&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="110" LastEditorUserId="282" LastEditDate="2015-02-17T23:31:20.483" LastActivityDate="2015-02-17T23:31:20.483" Title="Can Mechanical Design help meet EMC/EMI requirements for electronic products?" Tags="|mechanical-engineering|electrical-engineering|emc|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="213" PostTypeId="1" CreationDate="2015-01-24T15:42:39.547" Score="7" ViewCount="584" Body="&lt;p&gt;The design for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cloud_chamber&quot; rel=&quot;noreferrer&quot;&gt;cloud chamber&lt;/a&gt; is rather simple, and many factors can be changed without producing different results. There is one curious point that I've stumbled upon, though: The type of alcohol vapor used. I've read about methanol, ethyl alcohol, and isospropyl alcohol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will any of these choices produce different particle tracks? The factors I'm considering include the length of time the track exists for before it fades and the clarity of the track (i.e. opacity/transparency relative to the rest of the vapor).&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="131" LastEditDate="2015-01-27T13:30:12.137" LastActivityDate="2017-11-02T21:12:36.117" Title="Can using different types of alcohol in a cloud chamber produce different particle tracks?" Tags="|gas|experimental-physics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="214" PostTypeId="1" AcceptedAnswerId="252" CreationDate="2015-01-24T16:22:45.257" Score="19" ViewCount="973" Body="&lt;p&gt;Austenite is &lt;strong&gt;non magnetic&lt;/strong&gt; while $\alpha$-ferrite and pearlite &lt;strong&gt;are magnetic&lt;/strong&gt;. (&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/030488539390454A&quot;&gt;Magnetic properties of pearlite vary as a function of carbon content&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a strong magnetic field is applied in a particular direction while the steel is being quenched (rather, austenite is being quenched!), would the grain structure change? Is it possible to get superior grain structure and hence tougher steel by application of cyclic magnetic field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My speculation is that, at the &lt;a href=&quot;http://www-g.eng.cam.ac.uk/mmg/teaching/typd/addenda/eutectoidreaction1.html&quot;&gt;eutectoid point&lt;/a&gt; while quenching the steel, as low carbon zone of pearlite has more permeability, that zone should align itself to the strong magnetic field by pushing the carbon in orthogonal direction, so grain boundaries should take a different shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it actually happen? &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="46" LastEditDate="2015-01-26T12:32:21.517" LastActivityDate="2016-07-26T08:55:57.330" Title="Can we change steel properties by application of magnetic field while quenching?" Tags="|steel|process-engineering|materials|metallurgy|magnets|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="215" PostTypeId="1" AcceptedAnswerId="317" CreationDate="2015-01-24T16:41:54.090" Score="10" ViewCount="222" Body="&lt;p&gt;Localities in the US have different road layout and setups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the &lt;a href=&quot;http://en.wikipedia.org/wiki/Michigan_left&quot;&gt;Michigan Left&lt;/a&gt;, &lt;a href=&quot;http://en.wikipedia.org/wiki/Jughandle&quot;&gt;Jersey Left/Jug Handle&lt;/a&gt;, vs a standard 4 way stop with left turns at the intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have any of them come out as clear winners? &lt;/p&gt;&#xA;" OwnerUserId="231" LastEditorUserId="231" LastEditDate="2015-01-27T22:09:16.800" LastActivityDate="2015-01-28T00:37:13.247" Title="Has a specific type of alternative intersection been shown to reduce congestion &amp; accidents?" Tags="|civil-engineering|highway-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="217" PostTypeId="1" CreationDate="2015-01-24T17:53:00.980" Score="7" ViewCount="10941" Body="&lt;p&gt;If I install a small air compressor/engine, generator, and tank in my garage, and I charge it up at night when electricity costs $\frac{5.4p}{kWh}$ and use it during the day to power my house when electricity costs $\frac{12.5p}{kWh}$ would this be able to save money on the electricity bill or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tank, engine, and generator should be sized to charge up over 7 hours between 01:00 and 08:00 and release the energy during the other 17 hours of the day. I'm assuming that regardless of when during the day the energy is actually released it is either used by me or goes back into the grid and runs the meter backwards to offset my day-usage.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2017-03-01T12:53:58.183" Title="Is compressed air energy storage economically viable in the home?" Tags="|energy|energy-storage|compressed-air|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="218" PostTypeId="1" AcceptedAnswerId="221" CreationDate="2015-01-24T19:17:31.397" Score="8" ViewCount="9026" Body="&lt;p&gt;I was looking at high pressure pipes and their pressure ratings. I'd like to know how these pressure ratings are determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that pipes are tested until failure, and the failure pressure is multiplied by some safety factor to determine its 'rated' pressure, but is there a formula to calculate what the failure pressure should be before testing it (based on the material, wall thickness, diameter, or other measurements of the pipe)?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="46" LastEditDate="2015-01-24T20:27:37.010" LastActivityDate="2016-12-07T17:00:55.267" Title="How to calculate the pressure a pipe can withstand?" Tags="|mechanical-engineering|pressure|mechanical-failure|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="219" PostTypeId="2" ParentId="217" CreationDate="2015-01-24T19:28:29.070" Score="2" Body="&lt;p&gt;I'm going to do some naïve math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while charging your setup at night, you take in $x$ kilowatt-hours per hour. Multiply this by 7 hours, and you've taken in $7x$ kilowatt-hours. At a cost of $\frac{5.4\text{p}}{\text{kWh}}$, you've paid $37.8x \text{ p}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that when the electricity is on, you use $y$ kilowatt-hours per hour. Multiply this by 17, and you've used $17y$ kilowatt-hours. At a cost of $\frac{12.5\text{p}}{\text{kWh}}$, you've paid $212.5y \text{ p}$ &lt;em&gt;if you were to take in energy during the day&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your setup to save money,&#xA;$$37.8x&amp;lt;212.5y \to x&amp;lt;5.62y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, you have to take in as much energy during the night as you need during the day:&#xA;$$7x \ge 17y \to x \ge 2.43y$$&#xA;so now we get our final equation (slightly simplified for convenience) of the ratio $\frac{x}{y}$:&#xA;$$2.43 \le \frac{x}{y} &amp;lt; 5.62$$&#xA;And that's your naïve energy usage equation. You can't take in too little, or you won't have enough energy to run your home during the day. You can't take in too much or you'll pay too much (unless some goes backwards through your meter, in which case there's no upper bound). It also doesn't take into account the fact that some energy will be lost during the storage, in which case the lower bound may need to be higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then have to figure out how much money you save:&#xA;$$\text{Energy saved}=212.5y-37.8x$$&#xA;and how many cycles it will take to pay off the cost of the system:&#xA;$$\text{Number of 24-hour cycles}=\frac{\text{Cost of the system}}{\text{Energy saved}}=\frac{\text{Cost of the system}}{212.5y-37.8x}$$&#xA;Again, this doesn't take energy loss into account, or running your meter backward.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That's the theoretical aspect. It's not incredibly helpful, because it doesn't take some factors into account, and it's general. The University of Dayton has a much, much more comprehensive analysis &lt;a href=&quot;http://academic.udayton.edu/kissock/http/Publications/Abels_Kissock_OptComAirStorageForEngyEff_kk.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. There are some minor problems with the analysis, namely that it doesn't cover the money saved in your scenario and that it doesn't cover specific, small-scale scenarios, as yours presumably is (it's also a bit boring unless you know exactly what you're looking for or you're a compressed air nut).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's an interesting diagram on page 3 of the paper, though, discussing efficiency. A quick inspection finds that of all the energy storage methods discussed, compressed air storage was second-lowest in efficiency (beaten out only by fuels cells, at 59%). Compressed air technologies have an efficiency of 70% (ouch!), meaning that the lower bounds of the equation need to be raised. In terms of efficiency, it's &lt;em&gt;not&lt;/em&gt; the best choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some points in its favor: low maintenance cost, environmentally friendly, and an extremely long lifespan (30 years!). So it's up there in the technologies to be considered, as it should. But in your scenario, it might not save you as much money as, say rechargeable batteries. You could choose to run it slightly longer during each cycle, or use slightly less electricity. But as it stands, you're only going to save a tiny amount. Although, let's face it: In an age when energy consumption is one of the biggest issues, and cheaper an environmentally-friendlier are better, everything counts. Most likely, the system won't hurt you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://www.comed.com/documents/business-savings/fact-sheets/compressed%20air_fs.pdf?FileTracked=true&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for some interesting information on large-scale compressed air systems.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-01-24T19:28:29.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="220" PostTypeId="2" ParentId="217" CreationDate="2015-01-24T19:42:38.997" Score="5" Body="&lt;p&gt;It isn't completely infeasible. Just to get a ROM (rough order of magnitude), let's assume that a typical household that isn't using electricity for heating uses about 1 kW on average, and that you'd like to be able to store a half day's energy, or 12 kWh, which is roughly 45 MJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commercial air compressors can easily achieve 15 bar or so (over 200 PSI). The energy in a tank of compressed air is equal to the pressure times the volume, so to store 45 MJ at 15 bar would require&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{45\ \text{MJ}}{15\ \text{bar}} \cdot \frac{1\ \text{bar}}{10^5\ \text{Pa}} = 30\ \text{m}^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... or about 8000 gallons. Can you get or build a tank that size that can hold the pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real key is what kind of thermodynamic efficiency you can achieve while making the conversion from electricity to pressure and back again. When you compress the air, it will get hot, and some of that heat will be lost to the environment. However, you can get some of that heat back if you run the expanding gas through a suitable heat exchanger (and get some &quot;free&quot; air conditioning in the process).&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="368" LastEditDate="2016-06-30T15:58:59.380" LastActivityDate="2016-06-30T15:58:59.380" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="221" PostTypeId="2" ParentId="218" CreationDate="2015-01-24T20:35:27.630" Score="13" Body="&lt;p&gt;According to the ASME Process Piping Code (B31.3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = \frac{2 * t * S * E}{D - 2 * t * Y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p$ = internal pressure&lt;br&gt;&#xA;$t$ = wall thickness&lt;br&gt;&#xA;$S$ = material's tensile strength&lt;br&gt;&#xA;$D$ = outer diameter&lt;br&gt;&#xA;$Y$ = wall thickness coefficient (B31.3-1999, Table 304.1.1)&lt;br&gt;&#xA;$E$ = material and pipe construction quality factor (B31.3-1999, Table A-1A)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this equation does indeed have a safety factor included.&lt;/p&gt;&#xA;" OwnerUserId="34" LastActivityDate="2015-01-24T20:35:27.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="222" PostTypeId="2" ParentId="211" CreationDate="2015-01-24T20:38:47.170" Score="7" Body="&lt;p&gt;One of the problems that plagued older rechargeable batteries (e.g. Nickel Cadmium ($\text{NiCad}$) and Nickel Metal Hydride ($\text{NiMH}$)) was the &lt;a href=&quot;https://en.wikipedia.org/wiki/Memory_effect&quot;&gt;memory effect&lt;/a&gt;. The memory effect occurs when a rechargeable battery is not fully discharged. It then &quot;forgets&quot; that it has a greater capacity than it thinks it has, and so in the future it discharges less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example is a water bottle. Initially, water bottles have a certain capacity for water. Let's say that I drink most of the water in a water bottle during one usage. If the memory effect affected water bottles, I would not be able to drink any water in the future occupying the space that had held the water that had not been drunken the last time. That extra space would be forever lost. Over time, this can wear down a rechargeable battery. Fortunately, &lt;a href=&quot;http://copquest.com/battery_care.htm&quot;&gt;this generally only affects $\text{NiCad}$ and $\text{NiMH}$ rechargeable batteries&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to find much about effects that influence only lithium ion batteries, but there are a lot of &lt;a href=&quot;http://www.mpoweruk.com/life.htm&quot;&gt;across-the-board factors&lt;/a&gt;. Here's a short list:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;ul&gt;&#xA;&lt;li&gt;Chemicals breaking down&lt;/li&gt;&#xA;&lt;li&gt;Passivation (which affects lithium ion batteries), which is when a layer of unwanted chemicals form on the battery cell. &lt;a href=&quot;http://pubs.acs.org/doi/pdf/10.1021/cr020730k&quot;&gt;This&lt;/a&gt; discusses a related phenomenon on page 4258:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unfortunately, on recharge, the lithium has a strong tendency to form mossy deposits and dendrites in the usual liquid organic solvents (cf. Figure 15B). This limits the cycle life to 100-150 cycles (considerably lower that the 300 cycles required for a commercial cell), as well as increasing the risk of a safety incident.&lt;/li&gt;&#xA;  &lt;li&gt;Mechanical stresses and leaking. Batteries can be damaged in a variety of ways, causing internal components to break and causing chemicals to leak out. This can be very dangerous to humans.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  There are other long-term factors that increase battery aging. The page I linked for the above list seems to be fond of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Arrhenius_equation&quot;&gt;Arrhenius equation&lt;/a&gt;:&#xA;  $$k=Ae^{-E_a/RT}$$&#xA;  which shows that the rate of chemical reations changes as temperature changes. High temperatures mean faster reactions but also possibly a shorter life; this can affect non-rechargeable batteries significantly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Finally, there's the phenomenon of &lt;a href=&quot;http://www.corrosion-doctors.org/Batteries/self-compare.htm&quot;&gt;self-discharge&lt;/a&gt;, which is when unwanted reactions in the battery &quot;eat away,&quot; so to speak, at the battery's capacity. &lt;a href=&quot;https://en.wikipedia.org/wiki/Self-discharge&quot;&gt;The process can differ&lt;/a&gt; based on the type of battery. &lt;a href=&quot;http://batteryuniversity.com/learn/article/elevating_self_discharge&quot;&gt;Battery University has a page on it&lt;/a&gt;, which you may have already seen. It reiterates that temperature can speed up this process. Scarily enough, lithium ion batteries may discharge as much as 5% within the first 24 hours, slowing down to 1-2% per month after that.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-01-25T22:46:30.857" LastActivityDate="2015-01-25T22:46:30.857" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="223" PostTypeId="1" AcceptedAnswerId="225" CreationDate="2015-01-24T20:39:22.510" Score="3" ViewCount="117" Body="&lt;p&gt;Weight lifters will use the term &quot;one rep maximum&quot; in order to refer to the maximum amount of weight that they can lift one time.  The one rep max is often used as a proxy for how strong someone is (or isn't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recreational lifters, other types of athletes, and the general population rarely bother with testing themselves in order to determine their maximum strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a human factors and usability point of view, it can be useful to know what the average maximum strength of a population is so that design can take those limitations into account.  Please see the footnotes &lt;sup&gt;1,2,3&lt;/sup&gt; for some practical applications of this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if it's possible to estimate a person's maximum strength for a given activity by seeing how many repetitions they can complete at a lower weight.  For example, if someone can bench-press 100 lbs 10 times in a row, can that be used to estimate the maximum amount they could bench-press once?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, does that estimation model:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;hold consistently across activity types?&lt;/li&gt;&#xA;&lt;li&gt;hold consistently across increasing amounts of weight?&lt;/li&gt;&#xA;&lt;li&gt;hold consistently across age and gender distributions?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;An entertaining example: amusement park / carnival attractions can be designed differently in order to accommodate the various distributions of strength.&lt;/sub&gt;&lt;br&gt;&#xA;&lt;sup&gt;2&lt;/sup&gt;&lt;sub&gt;A mobility example: aids for the disabled can be enhanced if it's found that their maximum strength is below the threshold for a device.&lt;/sub&gt;&lt;br&gt;&#xA;&lt;sup&gt;3&lt;/sup&gt;&lt;sub&gt;A diagnostics example: restraints can be designed just to the maximum of a particular demographic's abilities in order to provide a more comforting or friendly appearance.  This provides a humanizing touch while maintaining the accuracy of the diagnostic test.&lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-01-25T15:44:41.150" LastActivityDate="2015-01-25T15:44:41.150" Title="Estimating maximum lift or strength" Tags="|statistics|biomechanics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="224" PostTypeId="1" AcceptedAnswerId="228" CreationDate="2015-01-24T20:46:33.150" Score="3" ViewCount="549" Body="&lt;p&gt;&lt;a href=&quot;https://www.refrigerationbasics.com/RBIII/controls1.htm&quot; rel=&quot;nofollow&quot;&gt;Cut-in, cut-out&lt;/a&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Thermostat&quot; rel=&quot;nofollow&quot;&gt;thermostats&lt;/a&gt; are commonly used with refrigeration devices in order to hold a consistent temperature range.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their general operations is such that when the temperature by the probe reaches a particular temperature set as the upper limit, the thermostat cuts-in and the compressor is turned on providing cold air.&lt;br&gt;&#xA;And when the temperature by the probe reaches the lower temperature set point, the thermostat cuts-out and turns off the compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What materials are used in within an analog or mechanical cut-in, cut-out thermostat?  Likewise, how is the internal circuit designed such that one probe feeds both cut circuits and what control is put in place for the cut-out to override the cut-in signal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help scope this question: based upon the materials used in an analog cut-in, cut-out thermostat, would it be reasonable to attempt to test and adjust the set points of a thermostat by using jars of water at known temperatures and measuring the output off of the relay within the thermostat?&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-01-25T15:18:51.300" LastActivityDate="2015-01-25T15:18:51.300" Title="Thermostat design properties" Tags="|electrical-engineering|circuit-design|refrigeration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="225" PostTypeId="2" ParentId="223" CreationDate="2015-01-24T21:05:38.600" Score="2" Body="&lt;p&gt;This actually isn't as much of an engineering question as it is a physiology question. There are actually a number of widely used estimates to predict your one-rep maximum, aka &quot;1RM&quot;, if you know how many repetitions you can do at a lower weight. &lt;a href=&quot;http://www.weightrainer.net/training/coefficients.html&quot; rel=&quot;nofollow&quot;&gt;See here for more info&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the methods are based on empirical studies, and are basically look-up tables of coefficients.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/One-repetition_maximum#Brzycki&quot; rel=&quot;nofollow&quot;&gt;Brzycki&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/One-repetition_maximum#Epley_Formula&quot; rel=&quot;nofollow&quot;&gt;Epley&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/One-repetition_maximum#Lander&quot; rel=&quot;nofollow&quot;&gt;Lander&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/One-repetition_maximum#Lombardi&quot; rel=&quot;nofollow&quot;&gt;Lombardi&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/One-repetition_maximum#Mayhew_et_al.&quot; rel=&quot;nofollow&quot;&gt;Mayhew et al.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/One-repetition_maximum#O.27Conner_et_al.&quot; rel=&quot;nofollow&quot;&gt;Connor et al.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/One-repetition_maximum#Wathen&quot; rel=&quot;nofollow&quot;&gt;Wathen&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;National Strength And Conditioning Association (NSCA) Coefficients&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="34" LastActivityDate="2015-01-24T21:05:38.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="226" PostTypeId="1" AcceptedAnswerId="451" CreationDate="2015-01-24T21:51:11.753" Score="13" ViewCount="9670" Body="&lt;p&gt;Some dc motors can be used as generators as well by applying mechanical torque to the output shaft to induce a current.  However, even if a dc motor &lt;em&gt;can&lt;/em&gt; do this, I imagine they were not designed for this purpose and thus perform less efficiently when used as a generator rather than as a motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my admittedly naive understanding, dc generators and dc motors are essentially the same machinery, but with inputs and outputs reversed.   This leads me to believe that some other design considerations are used to make one direction more efficient than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;How differently are dc generators and dc motors designed to make one direction of input/output more efficient than the other?&lt;/em&gt;&lt;/strong&gt;  What can one do electrically or mechanically to improve the efficiency in either direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, I'm interested in converting a dc motor into a generator and want to know how I can improve its efficiency in converting mechanical energy into electrical energy.&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="110" LastEditDate="2015-01-24T22:54:03.263" LastActivityDate="2020-04-13T13:37:05.047" Title="How different are dc motor designs from dc generator designs?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="227" PostTypeId="1" CreationDate="2015-01-25T06:31:07.680" Score="20" ViewCount="526" Body="&lt;p&gt;I understand that valves in two-phase vapour / liquid were sometimes sized by considering the required vapour and liquid flow rates separately, and then adding the results together. But this is now considered an outdated practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also heard of the &lt;a href=&quot;http://wins.engr.wisc.edu/teaching/mpfBook/node12.html&quot;&gt;homogeneous equilibrium model (HEM)&lt;/a&gt;, but I am under the impression that there are a number of other accepted sizing techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My primary question is: what are the currently accepted practices / standards for sizing relief devices with two-phase vapour / liquid flow? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, if the older techniques are still valid, under what conditions or assumptions behind each technique are they still valid?&lt;/p&gt;&#xA;" OwnerUserId="136" LastEditorDisplayName="user16" LastEditDate="2015-02-20T12:46:31.347" LastActivityDate="2015-07-28T12:48:28.063" Title="Sizing a pressure safety valve for two-phase flow" Tags="|mechanical-engineering|process-engineering|pressure|valves|multiphase-flow|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="228" PostTypeId="2" ParentId="224" CreationDate="2015-01-25T08:55:43.323" Score="4" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Bimetallic_strip&quot; rel=&quot;nofollow noreferrer&quot;&gt;bimetallic strip&lt;/a&gt; is used. Two different metals like steel and copper expand at different rate and a strip of the two bound together as result bends with temperature changes. A contact placed at the end of the strip will close when the temperature is right - and the temperature can be tuned by turning a screw that pushes the strip closer or farther from the contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simpler systems the natural inertia of the system is used to create the hysteresis - as the refrigerator compressor works, it takes time for the temperature drop to reach the bimetallic strip, and switch it off, so the temperature is brought below the cut-off point by some hard to control factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more complex systems, a two-level switch (either through two strips or just mechanism that shorts two switches at two different temperature levels), in connection with an analog RS switch (usually based on a relay) creates a better-controllable hysteresis.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/x8eOO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As temperature is too high, both switches are engaged and so the cooling compressor starts. Temperature drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At cut-in temperature the &quot;On&quot; switch disengages, but power to the relay is still supplied through the &quot;Off&quot; switch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At cut-out temperature the &quot;Off&quot; switch disengages, and the cooling process stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With temperature rising, the &quot;Off&quot; switch engages, but since the relay is open, it doesn't supply power to the coil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further rise causes the &quot;On&quot; switch to power up the coil and the relay switches both the compressor and the &quot;Off&quot; switch circuit on. With temperature drop, the &quot;on&quot; switch will disconnect but the coil powered through the relay will &quot;remember&quot; the state until &quot;Off&quot; opens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(and I'm sorry, but I don't know about calibrating thermostats with jars of water. My vote would be &quot;against&quot; as they are rarely submersible.)&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-01-25T08:55:43.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="229" PostTypeId="1" AcceptedAnswerId="232" CreationDate="2015-01-25T12:13:28.800" Score="12" ViewCount="6343" Body="&lt;p&gt;Since the units are the same ($\frac{N}{m^2}$), what's the difference between pressure and stress?&lt;/p&gt;&#xA;" OwnerDisplayName="user133" LastEditorUserId="95" LastEditDate="2015-01-25T19:59:35.237" LastActivityDate="2016-03-01T13:53:51.210" Title="What is the difference between pressure and stress?" Tags="|pressure|stresses|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="231" PostTypeId="1" AcceptedAnswerId="288" CreationDate="2015-01-25T16:00:44.127" Score="14" ViewCount="2083" Body="&lt;p&gt;I want to do some very simplified drag calculations on a vessel. My hope was that calculating the skin-friction resistance would be enough to get a good estimate of the surge resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because wave-making resistance is very speed dependent, I assume you can neglect it when the vessel is below a certain speed. I also assume that I have to work with &lt;a href=&quot;https://en.wikipedia.org/wiki/Froude_number&quot;&gt;Froude numbers&lt;/a&gt; instead of velocities, to take vessel size into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen Froude numbers below Fn = 0.1 and Fn = 0.2 mentioned in books and on the Internet, but if you calculate the velocity for a vessel with a 100 m long waterline you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = 0.1 \cdot \sqrt{9.81\ \text{m/s}^2 \cdot 100\ \text{m}} \approx 3.13\ \text{m/s} \approx 6.08\ \text{knots}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = 0.2 \cdot \sqrt{9.81\ \text{m/s}^2 \cdot 100\ \text{m}} \approx 6.26\ \text{m/s} \approx 12.16\ \text{knots}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This values seem way too high in my opinion. 12.16 knots is almost service speed for some vessels and 6 knots is also quite high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are Fn = 0.1 and Fn = 0.2 reasonable numbers, and if not, below what Froude numbers should I stay to be able to neglect wave-making resistance? &lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="95" LastEditDate="2015-01-25T20:28:57.593" LastActivityDate="2015-09-30T00:59:36.667" Title="Below what Froude number can I neglect wave-making resistance on a vessel?" Tags="|drag|marine-engineering|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="232" PostTypeId="2" ParentId="229" CreationDate="2015-01-25T16:28:57.073" Score="5" Body="&lt;p&gt;Pressure is a force applied against the surface of the material in question. It is divided by area because it describes distributed forces (eg. force from a compressed gas or liquid, or stacked/piled solids.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stress is a force distributed through the thickness of the material in question. It is divided by area because force gets shared (though not always evenly) by the cross section of the material. For example if you have a solid block of material supporting a weight, the force from the weight, divided by the width and depth of that bock, gives you the stress.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-01-25T16:28:57.073" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="234" PostTypeId="2" ParentId="137" CreationDate="2015-01-25T16:44:08.863" Score="4" Body="&lt;p&gt;Measurement variations are very common and should be taken in to consideration when engineering systems. In most cases high precision equipment is available but might be cost prohibitive to justify purchasing for the project. Therefore, the goal of the engineer is to design the system to account for measurement variation. In this case the min and max limits are 9mm and 11mm with 10mm being nominal.  There are few strategies that can be used. They are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Define an upper and lower control measurement limit to account for maximum 0.2 mm variation. Therefore the LCL and UCL would be 9.2mm and 10.8mm. This will guarantee the work pieces are always $10\pm1$mm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another would be to perform a gauge R&amp;amp;R study to understand true measurement variation and include this data in the design. Make sure calibration is included in the preventive maintenance schedule.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using a Design for Six Sigma (DFSS) might be a better approach. Hopefully, the design is 6 sigma capable, after accounting for 0.2mm worst-case measurement variation. If so measurement variation might be insignificant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In most case a combination of the above plus other strategies will be required to achieve a good design&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.qualitydigest.com/inside/metrology-article/basics-gauge-rr.html&quot; rel=&quot;nofollow&quot;&gt;Basics of Gauge R&amp;amp;R&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Design_for_Six_Sigma&quot; rel=&quot;nofollow&quot;&gt;Design for Six Sigma&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-12-05T18:19:04.900" LastActivityDate="2015-12-05T18:19:04.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="235" PostTypeId="2" ParentId="212" CreationDate="2015-01-25T18:33:24.643" Score="3" Body="&lt;p&gt;Generally speaking meeting the EMC/EMI requirements is the EE job. The biggest part of radiated EM waves usually comes from a poor PCB, and of course there is little the ME can do about that. Providing more space, if possible, can relax some constraints for the EE that will have more space to properly route his/hers tracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case it appears to me that no EM problem should be present: the fastest signal lying around would pass through the USB connection, but you say it's used only for power. Your device is also battery power and that gives a huge help since there's no risk to inject some unwanted frequencies in the mains through the power supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's some high speed clock inside, and I guess there is, proper routing should be enough. A great help would come from a metallic enclosure, as you probably know a grounded metallic enclouser greatly helps to keep unwanted EM from escaping your device. Bonus points: it also keeps unwanted EM from messing with your circuit, that might be a bigger issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help with the routing, relaxing the position of the usb connector and/or the lcd can help, but really in this case it seems quite trivial to me to make a circuit that has no problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without additional informations on the specific device in question I'd say just relax, make a prototype and test it, you're probably getting away with it without any problem.&lt;/p&gt;&#xA;" OwnerUserId="48" LastActivityDate="2015-01-25T18:33:24.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="236" PostTypeId="2" ParentId="34" CreationDate="2015-01-25T20:46:59.583" Score="7" Body="&lt;p&gt;Aluminum and steel are very near to each other in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_series&quot; rel=&quot;noreferrer&quot;&gt;Galvanic Series&lt;/a&gt; for seawater. So even if your apparatus is exposed to a corrosive environment along those lines, it should not corrode very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aluminum will corrode first, if corrosion occurs. If the exposed surface area of the steel is small compared to that of the aluminum, the corrosion that occurs will be very slow. This is part of the principle behind galvanized steel pipe: even if the zinc is scratched through, the remainder of it will corrode slowly and galvanically protect the steel. Because the surface area of the exposed steel is so much less than that of the zinc, the corrosion of the zinc is negligible.&lt;/p&gt;&#xA;" OwnerUserId="244" LastActivityDate="2015-01-25T20:46:59.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="237" PostTypeId="2" ParentId="100" CreationDate="2015-01-25T20:59:45.203" Score="5" Body="&lt;p&gt;I think the tail forms as a result of how the glass is dropped. In the video, the molten glass separates from the rest of the lump and stretches - like Silly Putty or molten mozzarella cheese. I expect that you could at least shorten the tail by cutting the gooey glass - but there's a possibility that the result would explode on cooling, as suggested in nivag's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sufficiently spherical glass balls would be pretty difficult. Maybe it could be done using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Shot_tower&quot;&gt;shot tower&lt;/a&gt; concept, or some kind of molding method.&lt;/p&gt;&#xA;" OwnerUserId="244" LastActivityDate="2015-01-25T20:59:45.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="238" PostTypeId="2" ParentId="226" CreationDate="2015-01-26T02:41:22.213" Score="4" Body="&lt;p&gt;DC generators, or &lt;a href=&quot;http://en.wikipedia.org/wiki/Dynamo&quot; rel=&quot;nofollow&quot;&gt;dynamos&lt;/a&gt; of any significant size are rather rare these days. It's much more common to use an AC generator (alternator) with an external rectifier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have an authoritative answer to your question, but I can't really see any reason that they'd be different in any significant way. All of the issues relating to mechanical strength, magnetic field design and the electrical characteristics are pretty much the same, regardless of which way the power is actually flowing.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-01-26T02:41:22.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="239" PostTypeId="2" ParentId="100" CreationDate="2015-01-26T02:41:47.763" Score="4" Body="&lt;p&gt;It was stated previously that a &quot;perfect&quot; sphere cannot exist in terms of engineering or manufacturing, but ignoring trivialities, let's answer the question. A Prince Rupert's drop is such that molten glass is viscous enough to droop off your rod and into a bucket of water, causing the glass to cool rapidly enough to create high amounts of internal tension, which causes the famed effect of making an unbreakable teardrop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you were to spin the rod quickly so as to not have a long tail, some thin dragging would still exist and make a tail. It may be small, but it would still be there. If you were interested in making it more spherical, you might think to shave off the tail end, but as you know, a single nick or disturbance to the tail end results in a solid glass explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you spun the rod in a way (in a magical world) so that there was no tail. Then you wouldn't have a Prince Rupert's drop!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question is no, it is not possible to make a spherical Prince Rupert's drop because either the glass would explode, or you simple don't have the drop you were looking for.&lt;/p&gt;&#xA;" OwnerUserId="251" LastActivityDate="2015-01-26T02:41:47.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="241" PostTypeId="1" CreationDate="2015-01-26T03:36:32.143" Score="6" ViewCount="239" Body="&lt;p&gt;If the goal is to transmit power over a large distance, I can see the advantage of an alternator over a generator:  more power, less loss.  If the target DC load is small and very close by, would an alternator still be more advantageous in terms of power production?  Is there any case where DC power generation is more advantageous to AC power generation?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2015-01-26T14:17:48.710" LastActivityDate="2015-01-26T14:25:50.183" Title="When is DC power generation superior over AC power generation?" Tags="|electrical-engineering|power-electronics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="242" PostTypeId="2" ParentId="241" CreationDate="2015-01-26T07:10:39.007" Score="6" Body="&lt;p&gt;There are cases where DC power generation is superior to AC.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If it's going to be transmitted by DC, and you can generate at the transmission voltage, or you can step up the voltage to transmission levels without too much loss, DC is superior.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If your load is DC, and is close to your generator (in space and in voltage), DC is superior.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both of these assume you've got suitable DC breakers available: electrical systems are usual designed to be fault tolerant, to the degree that no single fault can be dangerous to humans.  This typically means ensuring the system fails to safe, and sometimes that requires breaking the current. With the development of DC breakers even for HVDC applications, this has improved considerably: previously, it was one area where AC had the advantage: in an AC circuit, current hits zero twice every cycle, making safe circuit-breaking much easier.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;And sometimes you just don't have a choice, and the technology (e.g. photovoltaics) will &lt;em&gt;only&lt;/em&gt; generate DC, and if you want AC, you're going to have to convert it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-01-26T14:25:50.183" LastActivityDate="2015-01-26T14:25:50.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="243" PostTypeId="2" ParentId="241" CreationDate="2015-01-26T08:36:45.497" Score="4" Body="&lt;p&gt;When chemical/photovoltaic power generation is readily available, versus electromagnetic (based on mechanical power).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If locally you have sources of mechanical power (wind turbines, steam turbines, diesel motor, hybrid car gasoline motor, etc) you use a generator and either rectify the AC to DC (with losses) or adapt your recipient devices to accept AC, or send AC into the grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a fuel cell, a battery bank, a solar cell, you'll prefer DC loads, and you'll need to use an inverter (lossy!) to provide AC current, or send it over long distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...then there are cases where you work with dual power, and then simplicity decides DC is preferred. Say, a normal car which must be able to run off batteries and off the alternator. In this case, simplicity trumps other factors; first, if you used a generator, you'd get variable frequency proportional to engine speed, then you'd need an inverter to provide AC from the batteries, and then a lot of low-voltage circuits that require DC would need rectifiers locally to adapt to AC supply. It's just much easier to rectify the supplied power and keep it DC everywhere in the car.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-01-26T08:36:45.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="244" PostTypeId="2" ParentId="226" CreationDate="2015-01-26T08:49:31.197" Score="3" Body="&lt;p&gt;Well, first, the dynamos do need either some kick-start power or permanent magnets, as they can't depend on electromagnets in the stator only; you don't generate any electricity by moving a wire without any current next to another wire without any current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A permanent magnet DC motor will act as a dynamo if you provide rotation power to its axis, so there's not much difference here. The commutator may be aligned a little bit differently, ignoring the need to cut off early, when the magnetic field would brake the motor; there would be no concern about not creating a 'dead zone' where the motor wouldn't start, not pulled by a neighbor magnet. If anything, the device would be considerably simpler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you want to engage electromagnets for the stator, there will be some differences... usually they would be powered from an external source, which could be then charged from the dynamo. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-01-26T08:49:31.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="245" PostTypeId="1" CreationDate="2015-01-26T09:01:49.870" Score="13" ViewCount="1299" Body="&lt;p&gt;I have a beam subject to twisting and/or bending forces as well as linear compression force along its main axis. It is modelled as an &lt;a href=&quot;http://en.wikipedia.org/wiki/Isotropy&quot; rel=&quot;nofollow&quot;&gt;isotropic&lt;/a&gt; beam, but if anisotropic is not too far off then that's okay too. The beam is capable of large deformations such that its maximum deformations are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;140 degrees in pure bending&lt;/li&gt;&#xA;&lt;li&gt;140 degrees in pure twisting&lt;/li&gt;&#xA;&lt;li&gt;70 degrees bending + 70 degrees twisting&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is an applicable nonlinear beam theory I can apply to this problem using equations rather than any software-based solutions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like using the basic undergrad &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow&quot;&gt;Euler-Bernoulli beam theory&lt;/a&gt;, but the assumptions make it invalid in this case and I'm looking for something that is in the same vein as far as calculations goes and does not require significantly more advanced mathematics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally a theory that reduces the problem to a set of equations that can be solved without requiring multiple pages of tensor calculations that are hard to follow.&lt;/p&gt;&#xA;" OwnerUserId="254" LastEditorUserId="272" LastEditDate="2015-05-27T18:14:13.467" LastActivityDate="2015-09-18T22:01:59.517" Title="Easy nonlinear model for large deformation of a beam" Tags="|mechanical-engineering|beam|applied-mechanics|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="246" PostTypeId="2" ParentId="229" CreationDate="2015-01-26T09:06:31.040" Score="5" Body="&lt;p&gt;Pressure and stress are both forces distributed on a surface, but are in essence two quite different concepts.&#xA;The main difference between them is that pressure is &lt;em&gt;external&lt;/em&gt; and stress is &lt;em&gt;internal&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have an object, &lt;strong&gt;pressure&lt;/strong&gt; is the surface-force perpendicular on the 'skin' of this object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To define &lt;strong&gt;stress&lt;/strong&gt; it is useful to imagine a solid object with a set of external forces (actions and reactions) working on its surface. Because of these forces the object gets deformed, until it is in a state of equilibrium. When you would make a cut through this object and remove a part of it, forces on the surface exposed by the cut would be needed to keep the object in the same deformed state and to keep it in equilibrum. These &lt;em&gt;internal&lt;/em&gt; surface-forces are called stresses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While pressure is defined to be perpendicular on the surface of the object, this restriction does not apply to stresses. Stresses can be applied in any direction on the internal surface. This is another difference between pressure and stress. Stresses &lt;em&gt;perpendicular&lt;/em&gt; to the internal surface are called 'normal stresses' (compression or tension). Stresses &lt;em&gt;parallel&lt;/em&gt; to the internal surface are called 'shear stresses'.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2015-01-26T09:06:31.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="247" PostTypeId="2" ParentId="52" CreationDate="2015-01-26T09:07:37.660" Score="3" Body="&lt;p&gt;I don't know about any ready-made multi-position bi-stable solenoids, but let me help with &quot;What are the alternatives?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what force you need to exert and distance to cover, a monostable solenoid driven by PWM might suffice, but that seriously limits the range (as the magnet power rises with square of distance, after certain threshold the shaft will &quot;drop&quot; into the solenoid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can stack two 3-position ones; needle of one moving the frame of another. For more than two you'll be better off using some servo/stepper actuators instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And last but not least, get a 2-position bi-stable solenoid and some position sensors - either a linear optical encoder or just some open-frame optocouplers along the way, at the stop points. Power the coils with PWM, adapting to current position of the shaft, modifying the pulse width ratio so that given optocoupler gives out half the normal level (covered halfway), stopping the shaft wherever you want.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-01-26T09:07:37.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="248" PostTypeId="2" ParentId="4" CreationDate="2015-01-26T09:20:50.127" Score="5" Body="&lt;p&gt;The barometric height formula is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p(z) = p(0)\exp(-Mgz/(RT))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a centrifuge the upper expression is transformed to:&#xA;$$ p(r) = p(0)\exp(M\omega^2r^2/(2RT))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see the $g$ is transformed to the radial acceleration $a = \omega^2 r$. The factor $2$ stems from the integration which you have to do during the derivation of the barometric height formula (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Barometric_formula#Derivation&quot; rel=&quot;noreferrer&quot;&gt;barometric height formula&lt;/a&gt;, derivation). And of course, the minus sign disappeared because the force points outward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both formulas hold also for a mixture of gases, I cite &quot;&lt;a href=&quot;https://scholar.google.de/scholar?cluster=8946199746943967206&amp;amp;hl=de&amp;amp;as_sdt=0,5&amp;amp;sciodt=0,5&quot; rel=&quot;noreferrer&quot;&gt;Kemp, R. Scott. &quot;Gas Centrifuge Theory and Development: A Review of US Programs.&quot; Science and Global Security 17 (2009): 1-19.&lt;/a&gt;&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the rotor contains a mixture of gases, the distribution holds independently for each species.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, as a first approximation, it is as I suggested in the comment section. The cited source gives the following equations for a two-isotope gas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p_A(r) = p_A(0)\exp(M_A\omega^2r^2/(2RT))$$&#xA;$$ p_B(r) = p_B(0)\exp(M_B\omega^2r^2/(2RT))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A separation factor can be calculated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \alpha_0 = \frac{p_A(0)}{p_B(0)}/\frac{p_A(r)}{p_B(r)} =\exp((M_B- M_A)\omega^2r^2/(2RT))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that all formulas here don't include convection of any kind. In earth's atmosphere, we have too much convection and temperature differences for those formulas to work. Furthermore, in centrifuges, to improve their performance, there is a current-flow introduced (see the &lt;a href=&quot;https://scholar.google.de/scholar?cluster=8946199746943967206&amp;amp;hl=de&amp;amp;as_sdt=0,5&amp;amp;sciodt=0,5&quot; rel=&quot;noreferrer&quot;&gt;cited source&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="46" LastActivityDate="2015-01-26T09:20:50.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="249" PostTypeId="2" ParentId="207" CreationDate="2015-01-26T09:32:02.007" Score="9" Body="&lt;p&gt;Engineers should have a general idea of the expected result (Balpark values, expected behaviour)when using a complicated computer model. Most of the time these conclusions are based on a (much) simpler model, that preferably can be checked by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biggest reason for this is to eliminate the possibility of &lt;strong&gt;human error&lt;/strong&gt; in constructing the model itself. Using modeling software as a black box is seriously frowned upon and considered very unprofessional and risky. When results are very different from the expected, the first question one should ask is 'is the model well constructed?, didn't I make a (stupid) mistake?'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second reason is to gain &lt;strong&gt;control&lt;/strong&gt; of the model by understanding it. The simpler model acts as a stepping stone in the understanding process. When a model is understood it is easier to know what to change to find the solution to the engineering problem. As such the model is a tool in the design proces.&lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2015-01-26T09:32:02.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="250" PostTypeId="2" ParentId="229" CreationDate="2015-01-26T09:32:53.513" Score="5" Body="&lt;p&gt;One could say they are closely related, but while pressure is more generic, omnidirectional (like in gas), stress is defined in a solid, and is a tensor - with factors responsible for displacement force in 3 dimensions plus twisting force in 3 axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With pressure, you take an imaginary piston in cyllinder with vacuum, with a dynamometer attached to the piston, and measure what force the medium exerts on that wall, dividing it by the piston surface. No matter how you turn it, the value is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take a bunch of &lt;a href=&quot;http://en.wikipedia.org/wiki/Strain_gauge&quot; rel=&quot;noreferrer&quot;&gt;strain guages&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JLv4Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and cover them with concrete, forming a concrete beam. At first they will all show the same, pressure of liquid concrete. But as the concrete solidifies, the readouts will change. Some will show negative values as the beam bends and strains along the outer side. Others will show lateral pressure of the beam exerting its own weight perpendicular to its length. If you compress the beam, you'll get quite extreme values length-wise, but tiny negatives outwards from the axis as the compressed material expands to the sides. If you try to bend the beam, you'll get some small negatives on the outer side of the bend, some small positives on the inner side, and then the beam will snap; it's way weaker against negative forces (pulling it apart) and these are exerted on the outer side of the bend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when using the 'stress' value, unless you give the full tensor, it's always essential to write which direction of the stress you're describing - just putting it down like pressure isn't all that helpful.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-01-26T09:32:53.513" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="251" PostTypeId="1" CreationDate="2015-01-26T09:48:59.113" Score="5" ViewCount="24246" Body="&lt;p&gt;When designing reinforced concrete slabs it is common to divide the slab into 'design strips', which are then used to rationalise the reinforcement design along (typically orthogonal) lengths of the slab (also known as the Hilleborg Strip Method).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have mostly seen this approach used in design software (for example, SAFE), however they offer little to no advice as to how and where these design strips should be located/sized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What rules and considerations do engineers apply when locating/sizing design strips?&lt;/p&gt;&#xA;" OwnerUserId="151" LastEditorUserId="151" LastEditDate="2015-01-27T13:29:28.320" LastActivityDate="2016-05-29T12:55:11.687" Title="How to determine reinforcement design strip widths and locations" Tags="|concrete|reinforcement|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="252" PostTypeId="2" ParentId="214" CreationDate="2015-01-26T09:54:06.883" Score="16" Body="&lt;p&gt;This will at least depend on the:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rate of Cooling&lt;/li&gt;&#xA;&lt;li&gt;Magnetic field strength&lt;/li&gt;&#xA;&lt;li&gt;Exact composition&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The magnetic field will alter the microstructure as you can read in, for example,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Yudong Zhang, Nathalie Gey, Changshu He, Xiang Zhao, Liang Zuo, Claude Esling, High temperature tempering behaviors in a structural steel under high magnetic field, Acta Materialia, Volume 52, Issue 12, 12 July 2004, Pages 3467-3474, ISSN 1359-6454, &lt;a href=&quot;http://dx.doi.org/10.1016/j.actamat.2004.03.044&quot;&gt;http://dx.doi.org/10.1016/j.actamat.2004.03.044&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;G.M. Ludtka, R.A. Jaramillo, R.A. Kisner, D.M. Nicholson, J.B. Wilgen, G. Mackiewicz-Ludtka, P.N. Kalu, In situ evidence of enhanced transformation kinetics in a medium carbon steel due to a high magnetic field, Scripta Materialia, Volume 51, Issue 2, July 2004, Pages 171-174, ISSN 1359-6462, &lt;a href=&quot;http://dx.doi.org/10.1016/j.scriptamat.2004.03.029&quot;&gt;http://dx.doi.org/10.1016/j.scriptamat.2004.03.029&lt;/a&gt;.&#xA;(&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1359646204001770&quot;&gt;http://www.sciencedirect.com/science/article/pii/S1359646204001770&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For me, it is behind a paywall. But as you can read in the abstract, 30 Tesla will result in, for example, more ferrite. The other paper reveals that for a hyper eutectoid steel, you will have particle like cementite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not aware of any models with which you can make predictions about tensile strength and so on. But, for the question &lt;strong&gt;Can we change steel properties by application of magnetic field while quenching?&lt;/strong&gt;, it is a clear &lt;strong&gt;yes&lt;/strong&gt;. A more complete lookup in the literature would be the next step for models and experiments for a more specific case.&lt;/p&gt;&#xA;" OwnerUserId="46" LastActivityDate="2015-01-26T09:54:06.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="253" PostTypeId="1" CreationDate="2015-01-26T11:01:16.890" Score="10" ViewCount="213" Body="&lt;p&gt;As an example, consider a P-T1-system with a PID-controller. First look only at the P-T1 system, set a $y_r$ and wait a long time - then we take a look on its output $x$ and see that it has still a disturbance $d$ which variates with time (see the plot, system output $= x$). In this model, the system output is, after you wait a long time, a constant plus $d(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GT04q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fbT3H.png&quot; alt=&quot;Example-Plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to introduce a PID-controller:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/O1fkM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this loop alone we could just use some experience-based technique like the Ziegler and Nichols procedure to adjust its parameters $K_p$, $K_i$ and $K_d$ optimally. If we switch to discrete control loop, because the controller is digital,  we will have one additional parameter: The $\Delta t$ at which the controller operates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What $\Delta t$ is required for the control loop to diminish the effects of $d$ on the system output? The trend will of course be the smaller $\Delta t$ the better, but is there a general rule for the maximum $\Delta t$?&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="46" LastEditDate="2015-01-31T14:26:56.573" LastActivityDate="2015-02-07T16:09:01.173" Title="In the context of a known disturbance $d(t)$ in a control loop, what is the $\Delta t$ at which the control loop has to be executed?" Tags="|process-engineering|control-engineering|" AnswerCount="1" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="254" PostTypeId="5" CreationDate="2015-01-26T11:13:48.907" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-26T11:13:48.907" LastActivityDate="2015-01-26T11:13:48.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="255" PostTypeId="4" CreationDate="2015-01-26T11:13:48.907" Score="0" Body="Questions about using applied mechanics to model engineering problems." OwnerUserId="46" LastEditorUserId="368" LastEditDate="2015-03-08T15:37:44.697" LastActivityDate="2015-03-08T15:37:44.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="256" PostTypeId="5" CreationDate="2015-01-26T11:22:21.743" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-26T11:22:21.743" LastActivityDate="2015-01-26T11:22:21.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="257" PostTypeId="4" CreationDate="2015-01-26T11:22:21.743" Score="0" Body="For questions about the material concrete, or its application in an engineering context. Concrete is a composite material composed mainly of water, aggregate, and cement." OwnerUserId="46" LastEditorUserId="46" LastEditDate="2015-01-27T14:51:44.913" LastActivityDate="2015-01-27T14:51:44.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="258" PostTypeId="5" CreationDate="2015-01-26T11:23:25.763" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-26T11:23:25.763" LastActivityDate="2015-01-26T11:23:25.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="259" PostTypeId="4" CreationDate="2015-01-26T11:23:25.763" Score="0" Body="Questions about the models and theories of control engineering and the application of control theory to design systems with desired behaviors." OwnerUserId="46" LastEditorUserId="368" LastEditDate="2015-06-15T16:17:33.093" LastActivityDate="2015-06-15T16:17:33.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="260" PostTypeId="5" CreationDate="2015-01-26T11:25:39.877" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-26T11:25:39.877" LastActivityDate="2015-01-26T11:25:39.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="261" PostTypeId="4" CreationDate="2015-01-26T11:25:39.877" Score="0" Body="Questions about solar applications that make use of the photovoltaic effect, especially solar panel and thin-film technologies." OwnerUserId="-1" LastEditorUserId="368" LastEditDate="2015-09-02T17:52:54.667" LastActivityDate="2015-09-02T17:52:54.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="262" PostTypeId="5" CreationDate="2015-01-26T11:27:28.110" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-26T11:27:28.110" LastActivityDate="2015-01-26T11:27:28.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="263" PostTypeId="4" CreationDate="2015-01-26T11:27:28.110" Score="0" Body="Process engineering focuses on the design, operation, control, and optimization of chemical, physical, and biological processes." OwnerUserId="46" LastEditorUserId="46" LastEditDate="2015-01-27T14:52:05.790" LastActivityDate="2015-01-27T14:52:05.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="264" PostTypeId="5" CreationDate="2015-01-26T11:29:52.570" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-26T11:29:52.570" LastActivityDate="2015-01-26T11:29:52.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="265" PostTypeId="4" CreationDate="2015-01-26T11:29:52.570" Score="0" Body="Use this tag for questions within the problem domain of chemical engineering." OwnerUserId="46" LastEditorUserId="368" LastEditDate="2015-02-18T18:17:18.107" LastActivityDate="2015-02-18T18:17:18.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="266" PostTypeId="5" CreationDate="2015-01-26T11:33:34.027" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-26T11:33:34.027" LastActivityDate="2015-01-26T11:33:34.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="267" PostTypeId="4" CreationDate="2015-01-26T11:33:34.027" Score="0" Body="For questions about the design, classification and working principle of pumps. A pump is a device that moves fluids (liquids or gases), or sometimes slurries, by mechanical action." OwnerUserId="46" LastEditorUserId="46" LastEditDate="2015-01-27T14:51:49.303" LastActivityDate="2015-01-27T14:51:49.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="268" PostTypeId="5" CreationDate="2015-01-26T11:34:35.300" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-26T11:34:35.300" LastActivityDate="2015-01-26T11:34:35.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="269" PostTypeId="4" CreationDate="2015-01-26T11:34:35.300" Score="0" Body="Questions relating to the historical development of an engineering practice or process. Also, questions about the people and places involved in significant engineering developments and achievements." OwnerUserId="95" LastEditorUserId="95" LastEditDate="2015-01-27T14:50:42.543" LastActivityDate="2015-01-27T14:50:42.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="270" PostTypeId="1" AcceptedAnswerId="276" CreationDate="2015-01-26T11:41:47.390" Score="28" ViewCount="30512" Body="&lt;p&gt;As a hydraulics layman thinking about hydraulic systems, it seems that the important factor is to have a liquid that doesn't compress much or at all. Doesn't water meet this requirement, and what other properties should the liquid have (if any) that water doesn't?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="368" LastEditDate="2015-04-14T22:04:09.203" LastActivityDate="2020-01-03T01:45:50.523" Title="Why do hydraulic systems use special fluid - what's wrong with water?" Tags="|mechanical-engineering|hydraulics|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="271" PostTypeId="1" AcceptedAnswerId="282" CreationDate="2015-01-26T12:03:23.730" Score="11" ViewCount="2200" Body="&lt;p&gt;As this will depend on Pressure drop $\Delta p$, assume that it does not leave the range from 0 to 100 bar. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation&quot; rel=&quot;noreferrer&quot;&gt;Hagen-Poiseuille equation&lt;/a&gt; for an incompressible fluid is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{V} = \frac{\pi R^4 \Delta p}{8\eta L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that it wont be applicable for very small (nm) diameters, so this question is in the context of microfluidics.  Fluids of interest in this case have a kinematic viscosity of 1 cSt to 10000 cSt.&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="46" LastEditDate="2015-01-27T10:03:40.070" LastActivityDate="2021-01-05T07:50:14.947" Title="Can you use the Hagen-Poiseuille equation for a pipe which radius is in the sub-millimeter region?" Tags="|fluid-mechanics|microfluidics|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="272" PostTypeId="5" CreationDate="2015-01-26T12:04:31.057" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-26T12:04:31.057" LastActivityDate="2015-01-26T12:04:31.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="273" PostTypeId="4" CreationDate="2015-01-26T12:04:31.057" Score="0" Body="For questions which are in the context of microfluidics. Microfluidics is a multidisciplinary field intersecting engineering, physics, chemistry, biochemistry, nanotechnology, and biotechnology, with practical applications to the design of systems in which small volumes of fluids will be handled. This is typically the sub-milimeter scale." OwnerUserId="46" LastEditorUserId="46" LastEditDate="2015-01-27T14:52:12.880" LastActivityDate="2015-01-27T14:52:12.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="274" PostTypeId="5" CreationDate="2015-01-26T12:17:40.567" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-26T12:17:40.567" LastActivityDate="2015-01-26T12:17:40.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="275" PostTypeId="4" CreationDate="2015-01-26T12:17:40.567" Score="0" Body="For questions about the design of circuits. This may cover everything from processing steps to models to high level circuit design issues." OwnerUserId="46" LastEditorUserId="46" LastEditDate="2015-01-27T14:50:53.787" LastActivityDate="2015-01-27T14:50:53.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="276" PostTypeId="2" ParentId="270" CreationDate="2015-01-26T12:19:34.310" Score="49" Body="&lt;p&gt;Water meets the low compressibility requirement, but there are many other considerations in the design of a hydraulic system:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Boiling point/vapor pressure:&lt;/strong&gt; If the system warms up during operation, the fluid may boil, which results in high compressibility and thus decreased effectiveness of the hydraulic system. Hydraulic fluid has a higher boiling point than water to help combat this. Related to this is the concept of vapor pressure. Hydraulic systems often involve small orifices, which can cause cavitation (localized boiling). This cavitation has the same effects as boiling and can cause pitting damage to the components near the cavitated region. Hydraulic fluid has a lower vapor pressure which helps here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Freezing point:&lt;/strong&gt; It would not be a good thing if your car's brake lines froze every time it got cold outside. Most hydraulic fluids have much lower freezing points to prevent this from happening under normal circumstances.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Oxidation/corrosion:&lt;/strong&gt; Water, being an electrolyte, will cause rust inside the lines as soon as air inevitably leaks into the system or the system isn't bled properly. Water will also exacerbate galvanic corrosion when dissimilar metals are used in the system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Lubrication:&lt;/strong&gt; Hydraulic components use seals and often involve sliding interfaces (cylinders and spools, for examples). Using an oil as the fluid means the working fluid can also function as a lubricant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Organic growth:&lt;/strong&gt; If perfectly distilled water and a closed system could be guaranteed, this would be a non-issue. But in practice, this is never the case. Oil-based hydraulic fluids are much less conducive to organic growth than water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Water is used in some systems where other considerations trump these (for example, some food-grade applications), but for a wide variety of applications, oil-based hydraulic fluids are the better choice because of the design considerations above.&lt;/p&gt;&#xA;" OwnerUserId="166" LastEditorUserId="166" LastEditDate="2015-01-26T12:59:01.743" LastActivityDate="2015-01-26T12:59:01.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="277" PostTypeId="1" CreationDate="2015-01-26T12:29:47.557" Score="6" ViewCount="1387" Body="&lt;p&gt;I have a 200 litre &lt;a href=&quot;http://en.wikipedia.org/wiki/Rainwater_tank&quot; rel=&quot;nofollow&quot;&gt;water butt&lt;/a&gt; and plenty of 22 mm plumbing pipe and fittings and I want to build an irrigation system for my vegetable patch so that I can open the tap after a few dry days and ensure all my plants get equal amounts of water from the butt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end of the patch is about 5 m from the water butt and at the same level, so the head is only the depth of the water-butt (1 m) plus about 10 cm as it sits on a small platform I built. I will bury a hose 20-30 cm under the ground to the end of the vegetable patch, where it will come back up to the surface and split into 4 rows of pipe about 15 m long each with holes drilled to release the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I should taper the holes so that the ones at the start will be smaller and the ones at the far end will be bigger, but given the low pressure I don't know how small these should be or what the size-difference in holes I should use from one end to the other to ensure that water gets all the way along and is evenly distributed?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="368" LastEditDate="2015-07-22T00:10:54.373" LastActivityDate="2016-11-14T01:42:25.923" Title="How to equalise the water distribution in a leaky pipe irrigation system?" Tags="|civil-engineering|pressure|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="278" PostTypeId="2" ParentId="277" CreationDate="2015-01-26T14:16:51.673" Score="5" Body="&lt;p&gt;Controlling the pressure with variously sized holes will be a fairly difficult task. It may be simpler to run hose with no perforations along the length of the planter, and then place the hoses with perforations as branches off of your main line. That still won't lead to a perfect distribution, but would help. Another common approach to this is to use so called 'drip emitters' which are very cheap regulated orifices. They are designed to put out a certain volume per time, relatively independent of the pressure. I don't know if they can be buried.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-01-26T14:16:51.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="279" PostTypeId="1" CreationDate="2015-01-26T16:58:06.593" Score="7" ViewCount="919" Body="&lt;p&gt;What are the benefits of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Diaphragm_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;diaphragm pump&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://animatedsoftware.com/pumpglos/diaphram.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zuThK.gif&quot; alt=&quot;diaphragm pump image&quot;&gt; &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;over a &lt;a href=&quot;http://en.wikipedia.org/wiki/Single-_and_double-acting_cylinders&quot; rel=&quot;nofollow noreferrer&quot;&gt;simplex pump&lt;/a&gt;?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://animatedsoftware.com/pumpglos/simplex.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NGMkj.gif&quot; alt=&quot;simplex pump image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any specific design criteria that should be used/stated in an engineering specification for a diaphragm pump?&lt;/p&gt;&#xA;" OwnerUserId="258" LastEditorUserId="106" LastEditDate="2015-01-30T03:20:32.947" LastActivityDate="2015-01-30T03:20:32.947" Title="What are the benefits of a diaphragm pump vs a simplex pump?" Tags="|mechanical-engineering|civil-engineering|pumps|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="280" PostTypeId="1" AcceptedAnswerId="294" CreationDate="2015-01-26T17:56:37.703" Score="22" ViewCount="1749" Body="&lt;p&gt;For a project I had built a &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Laval_nozzle&quot; rel=&quot;nofollow noreferrer&quot;&gt;convergent divergent nozzle&lt;/a&gt; designed for Mach number = 3. In that project, I could know the flow has gone supersonic by seeing the manometer fixed between the throat and the divergent section (drop in pressure, as the divergent section acts like a nozzle for the supersonic flow). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this got me thinking, If I am to build a nozzle for the propulsion purpose (or any practical purpose), it is not desirable to have holes in it for the manometer in order to maintain the uniform strength. My theoretical calculations tell me that the flow should go supersonic and no shock in the nozzle, but while building, the surface finish, geometric tolerances and supply pressure might not be what I expect. In that case, &lt;strong&gt;how do I know if the flow has gone supersonic?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about following ways. So far I haven't tried any of them. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Using a Pitot tube might not be useful since there will be a bow shock in front of the tube if in case the flow is indeed supersonic (as shown in the figure), &lt;img src=&quot;https://i.stack.imgur.com/oc1oR.jpg&quot; alt=&quot;enter image description here&quot;&gt; which will increase the total pressure. We can use &lt;a href=&quot;http://nptel.ac.in/courses/101103004/module7/lec6/3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reyleigh pitot tube formula&lt;/a&gt;, but how to compute static free stream pressure without affecting the flow / nozzle? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Schlieren_photography&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schlieren Photography&lt;/a&gt;: If we see oblique shocks / shock diamonds, then the inference will be: 'flow is supersonic'. This will work only when the shock features are super clear.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="223" LastEditorUserId="5560" LastEditDate="2019-06-21T23:41:20.287" LastActivityDate="2019-06-21T23:41:20.287" Title="How to know whether the flow is supersonic in a nozzle?" Tags="|fluid-mechanics|gas|pipelines|measurements|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="281" PostTypeId="2" ParentId="207" CreationDate="2015-01-26T19:35:41.877" Score="5" Body="&lt;p&gt;In my particular field (buried culvert design), we run &lt;a href=&quot;http://www.candeforculverts.com/home.html&quot; rel=&quot;nofollow&quot;&gt;finite element analyses&lt;/a&gt; constantly. We almost never change a design based on the results; we know going in (from a variety of factors, mostly prior experience and conservative assumptions) whether or not the design is good. We run the analyses to &lt;em&gt;demonstrate to others&lt;/em&gt; that our design is good. We may tweak something, but it is never changed substantially. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very often, building codes and regulatory agencies specify certain requirements for demonstration of design acceptability. Sometimes running the model is more-or-less jumping through these hoops, so that a person with less knowledge and time can quickly ascertain the relevant facts without getting bogged down in the minutia. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To summarize - and it's not my intention to be glib, but: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers use FEA/numerical simulation so we can have something to present in a courtroom other than the contents of our brain matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ADDENDUM:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our reports, we also like (and our insurance carriers REALLY REALLY like our) being able to say &lt;em&gt;&quot;The model says...&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2015-01-26T21:19:31.160" LastActivityDate="2015-01-26T21:19:31.160" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="282" PostTypeId="2" ParentId="271" CreationDate="2015-01-26T19:35:53.243" Score="8" Body="&lt;p&gt;Short Answer: &lt;strong&gt;YES&lt;/strong&gt; you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A) Limits of continuum mechanics:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The continuum model of fluid dynamics is valid only till the fluid behaves as a continuous medium. This is characterized by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Knudsen_number&quot; rel=&quot;noreferrer&quot;&gt;Knudsen number&lt;/a&gt;. The Knudsen number is given by $Kn = \frac{\lambda}{l_s}$, where $\lambda$ is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_free_path&quot; rel=&quot;noreferrer&quot;&gt;mean free path&lt;/a&gt; and $l_s$ is the characteristic dimension of the channel (diameter in the case of the circular pipe). Non equilibrium effects start to happen if $Kn &amp;gt; 10^{-3}$. Modified slip boundary conditions can be used for $10^{-3} &amp;lt; Kn &amp;lt; 10^{-1}$, and condinuum model completely breaks if $Kn &amp;gt; 1$. (&lt;strong&gt;Fun fact:&lt;/strong&gt; because the distance between two vehicles on a crowded road is much smaller than straight portion of the road itself (length scale in $1d$ flow), we can &lt;a href=&quot;http://www.math.nyu.edu/faculty/childres/traffic2.pdf&quot; rel=&quot;noreferrer&quot;&gt;model the traffic flow with a PDE&lt;/a&gt;! However it will not work if there is only one car on a long stretch of road)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming back to water, as the water molecules are not freely moving and are loosely bound, we consider the lattice spacing $\delta$ for computing $Kn$. For water $\delta$ is about $3 nm$. So continuum theory will hold good for a tube of diameter, $300 nm$ or larger $^*$. Now this is a good news!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$^*$ Reference: &lt;a href=&quot;http://microfluidics.stanford.edu/Publications/ElectrokineticFlows/Sharp%20Microfluidics%20Chapter%2010%20in%20CRC%20MEMS%20Handbook.pdf&quot; rel=&quot;noreferrer&quot;&gt;Liquid flows in microchannels&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B) Applicability of Hagen Poiseuille equation:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your tube is in sub-millimeters range, it is much larger than the minimum diameter required (sub-micrometer)for the continuity equation. However, &lt;strong&gt;depending on the shape of cross section of the tube, the results will differ&lt;/strong&gt; (&lt;a href=&quot;http://orbit.dtu.dk/fedora/objects/orbit:20967/datastreams/file_4757131/content&quot; rel=&quot;noreferrer&quot;&gt;Link to ref.&lt;/a&gt;). Liquid flows are much simpler to analyse since they are characterized by much smaller Reynold's number and velocities. Also density essentially remains constant. So there should not be a problem in considering the theory to be valid.&#xA;Now since the Hagen Poiseuille flow is derived from the Navier Stokes equations, it follows the assumption of continuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your flow is through a porous medium, you might have to consider effects like &lt;a href=&quot;http://en.wikipedia.org/wiki/Electrokinetic_phenomena&quot; rel=&quot;noreferrer&quot;&gt;electrokinetic effect&lt;/a&gt;. There might be other complications in straightforward application of H-P equations to microfluidic flows, but I am unable to comment since do not know much in this field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C) Some examples&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;a href=&quot;http://tesi.cab.unipd.it/42238/1/Tesi_Magistrale_Andrea_Biral.pdf&quot; rel=&quot;noreferrer&quot;&gt;report on &quot;microfluidics networking&quot;&lt;/a&gt;, Biral has used the continuum theory for modeling and simulation (in OpenFOAM) of the microfluidic flows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fillips discusses more about the Knudsen number in his paper- &lt;a href=&quot;http://arc.aiaa.org/doi/pdf/10.2514/1.44755&quot; rel=&quot;noreferrer&quot;&gt;Limits of continuum aerodynamics.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.elveflow.com/microfluidic-reviews-and-tutorials/flow-control-in-microfluidic-device&quot; rel=&quot;noreferrer&quot;&gt;This report&lt;/a&gt; clearly mentions that HP equation is applicable even to microfluidic flows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://viscometer.weebly.com/theory-behind-device.html&quot; rel=&quot;noreferrer&quot;&gt;This document on PDMS Viscometer&lt;/a&gt; gives derivation of HP equation for microfluidic flows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally &lt;a href=&quot;https://www.youtube.com/watch?v=7KW6Q0tYvJk&amp;amp;noredirect=1&quot; rel=&quot;noreferrer&quot;&gt;here is a YouTube video&lt;/a&gt; discussing about matrix formalism for solving the Hagen-Poiseuille law in microfluidic hydraulic circuits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on these references, it should be safe to assume that H-P equation can be applied to microfluidic flows. However, experts are welcome to enlighten us in this regards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="223" LastEditDate="2015-01-26T19:41:29.870" LastActivityDate="2015-01-26T19:41:29.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="283" PostTypeId="1" AcceptedAnswerId="286" CreationDate="2015-01-26T20:33:45.380" Score="6" ViewCount="290" Body="&lt;p&gt;A friend and I wrote a paper for a fluids class discussing the details (as they relate to fluid dynamics) of constructing a cannon that could shoot a steak fast enough to cook it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We quickly discovered (but not quickly enough to change our topic) that our paper was a bit too ambitious for two twenty year old undergrads taking an introductory course in fluid mechanics. Nonetheless, we still busted out a ballistics simulator, a cookbook, and a compressive heating calculator and did our best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the issues that stumped us was compressing the gas we used to launch the steak. We chose Helium because it was the least dense gas that probably won't burst into flames (like Hydrogen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a compressive heating calculator, we found the velocity we needed to shoot the steak, and were using Bernoulli's equation to find the pressure we needed to launch at our chosen velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue we ran into was density is dependent on pressure, but we needed the density to calculate the pressure needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one determine the pressure given the issue above? Is it simply several rounds of iteration until an acceptable answer is found? &lt;/p&gt;&#xA;" OwnerUserId="3" LastActivityDate="2018-08-17T19:10:55.007" Title="Determining pressures of a compressible gas" Tags="|fluid-mechanics|gas|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="284" PostTypeId="2" ParentId="207" CreationDate="2015-01-26T20:35:07.647" Score="3" Body="&lt;p&gt;I design electric motors and I use electromagnetic FEA as part of that design process.  Motor designers have a lot of good analytic techniques that get us very close to the actual performance of the motors for certain key parameters (torque, current draw, speed, etc.).  However, this requires that we make certain assumptions that may or may not be valid.  For example, I might assume that the flux through a certain path of steel is evenly distributed or I might assume a certain amount of flux leakage through a slot.  Those types of assumptions are often totally valid ones to make.  One reason I use FEA is to confirm that the assumptions I made were valid.  If they are valid, the FEA results will give me pretty much what I expected.  If they aren't valid, the FEA results will help me figure out what my bad assumptions were.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason I use it is that there are some motor parameters that can't be determined very well using analytic techniques.  For example, torque ripple (the amount of variation in torque as the rotor rotates) is difficult to do with analytic techniques.  I know certain motor types have worse ripple and I know certain combinations of poles to slots has better ripple than other combinations and other rules of thumb, but FEA can help you quantify that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other reason I use FEA is to really fine-tune a design.  If I have a design that pretty much does what I want it to, I can then try to get the efficiency up a bit or reduce the magnet thickness or whatever.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I use it to 1) check my assumptions, 2) solve problems that can't be done easily with analytic techniques and 3) fine-tune my designs to increase performance or decrease cost or just make it better.  All 3 of these require that I have a pretty good handle on the design before I start the FEA process.  That doesn't mean I'm never surprised by the results or don't learn things, but when those surprises happen, you can be sure I'll be going back and trying to figure out what went wrong.  &lt;/p&gt;&#xA;" OwnerUserId="198" LastActivityDate="2015-01-26T20:35:07.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="285" PostTypeId="2" ParentId="283" CreationDate="2015-01-26T21:40:06.977" Score="1" Body="&lt;p&gt;Without seeing more detail of your equations, it sounds like iteration is your answer, as that's not uncommon in fluid mechanics. Make an educated guess on a density for the helium (&lt;a href=&quot;http://www.engineeringtoolbox.com/gas-density-d_158.html&quot; rel=&quot;nofollow&quot;&gt;Engineering Toolbox&lt;/a&gt; lists the density of helium at STP as $\rho=0.1785 kg/m^3$ and also gives an NTP density of $\rho=0.1664kg/m^3$.) Use these values as a rough starting point, or you may be able to find a resource that allows you to make more accurate determinations of density for given pressure and temperature numbers. Plug in the density values, then solve for pressure and use those pressure numbers to solve for new density values, and see what kind of error you get. Hopefully after a couple of such iterations, you'll narrow it down to a small percentage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I hate to be the bearer of bad news, but you're &lt;a href=&quot;https://what-if.xkcd.com/28/&quot; rel=&quot;nofollow&quot;&gt;not the first&lt;/a&gt; to try and figure something like this out, and it looks like your issue is always going to be terminal velocity working against you, regardless of how much initial velocity you impart to the steak. And I have a feeling that, even though Randall doesn't go much into it, if you go sufficiently fast enough to actually cook it before it slows down, you'll break the steak apart into chunks for beef stew, which probably isn't your desired result, even though they'll cook faster that way. [source needed]&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-01-26T21:40:06.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="286" PostTypeId="2" ParentId="283" CreationDate="2015-01-26T22:08:59.693" Score="6" Body="&lt;p&gt;Seriously?!  :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are going to be two parts to the solution you are looking for,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Till the steak is in the cannon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) The steak leaves the cannon, is into the air and cooking starts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A) Internal ballistics:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are dealing with compressible flow. NEVER use simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bernaulli's equation&lt;/a&gt; beyond Mach no. 0.3. Make sure you are using correction terms till Mach number 0.7 and beyond that, use equations of &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas_dynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;gas dynamics&lt;/a&gt; (refer &lt;strong&gt;Modern Compressible Flow&lt;/strong&gt; by John Anderson). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, your case is same as an air rifle case. Instead of pellet, you are shooting steaks. So if you know the muzzle velocity, you can design your cannon as shown in &lt;a href=&quot;http://www.kiledjian.elac.org/phys%20001/Ballistics%20of%20Air%20Gun.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;. Now your question is how does one get $P_0$ mentioned in this paper, right? For that you will have to do reverse calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B) Steak leaves the cannon&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that you want your steak medium (as &lt;a href=&quot;https://cooking.stackexchange.com/questions/9/what-is-the-internal-temperature-a-steak-should-be-cooked-to-for-rare-medium-rar&quot;&gt;rare is not recommended apparently!&lt;/a&gt;), figure out the internal and surface temperatures for cooking. Also time required for cooking. At these temperature, your steak most probably will be flying at &lt;a href=&quot;http://en.wikipedia.org/wiki/Supersonic_speed&quot; rel=&quot;nofollow noreferrer&quot;&gt;supersonic speeds&lt;/a&gt;. Then there will be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bow_shock_(aerodynamics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;bow shock&lt;/a&gt; in front of the steak. You can safely approximate it as a normal shock and use &lt;a href=&quot;http://en.wikipedia.org/wiki/Normal_shock_tables&quot; rel=&quot;nofollow noreferrer&quot;&gt;normal shock relations&lt;/a&gt; to calculate total temperature ratio across the shock. Now $T_{01}$ becomes the atmospheric temperature and $T_{02}$ becomes the surface temperature on the steak (using &lt;a href=&quot;http://en.wikipedia.org/wiki/Total_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;total pressure&lt;/a&gt; ratio and gas dynamics relations). This will give you required shock strength and hence the flying Mach number. Assuming STP conditions at sea level, find &lt;a href=&quot;http://en.wikipedia.org/wiki/Speed_of_sound&quot; rel=&quot;nofollow noreferrer&quot;&gt;acoustic velocity&lt;/a&gt; and hence the steak velocity. Now this is average steak velocity. But there is going to be wave and pressure drag on the steak all the time. Use this &lt;a href=&quot;http://adg.stanford.edu/aa241/drag/ssdragcalc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stanford supersonic wing drag calculator&lt;/a&gt; to calculate this drag. In this take aspect ratio (AR) = 1, $C_L = 0$, put length of steak and its thickness / length as t/c. So compute the muzzle velocity by using newton's second and then first law. Now substitute this muzzle velocity in point A discussed above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That will give you your chamber pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I found &lt;a href=&quot;http://home2.fvcc.edu/~dhicketh/DiffEqns/spring07projects/StephenCompton/SpringAirModel.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;one report&lt;/a&gt; in which internal ballistics of spring loaded gun is considered. There is a matlab code as well. You can take author's permission to use it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue is, as you are going to use pre compressed pneumatic cylinder, the temperature is going to drop considerably when expansion happens. So flames is not a problem, however, during compressing of the gas in that cylinder, things are going to heat up, so using helium is smart move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way you can do this exercise is to write a small code in your favorite language and carry out those iterations you mentioned. However, don't use Bernaulli's equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the best for your paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speculation: If your hypothetical steak is flying supersonic for several minutes through the air, its most likely going to be eaten by some dog some hundred miles away from you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:38.773" LastActivityDate="2015-01-26T22:14:32.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="287" PostTypeId="2" ParentId="283" CreationDate="2015-01-27T03:07:59.197" Score="1" Body="&lt;p&gt;I'm not sure what you mean by compressible gas since all gasses are compressible.&lt;br&gt;&#xA;To answer the question, for an ideal gas (of which He is perhaps the idealist... closest to being ideal.) you can relate pressure and density by the ideal gas law.&lt;br&gt;&#xA;Which I write as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PV = NkT $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P$ is pressure&lt;br&gt;&#xA;$V$ is volume&lt;br&gt;&#xA;$N$ is the number of atoms&lt;br&gt;&#xA;$k$ is Boltzmann's constant&lt;br&gt;&#xA;And $T$ is the temperature  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get density you take the mass of a helium atom, multiple by $n$ and divide by the volume.&lt;/p&gt;&#xA;" OwnerUserId="168" LastEditorUserId="168" LastEditDate="2015-07-12T15:07:50.717" LastActivityDate="2015-07-12T15:07:50.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="288" PostTypeId="2" ParentId="231" CreationDate="2015-01-27T03:30:42.160" Score="4" Body="&lt;p&gt;Your intuition is correct, those are high. However, you would need to be moving &lt;strong&gt;very&lt;/strong&gt; slowly for wave-making resistance to be negligible. And since it is typically higher than skin-friction I don't think that you can realistically  expect to have a significant skin friction and a negligible wave-making resistance. Perhaps a better simplified approach would be to ignore the skin friction and focus only on the wave-making resistance. &lt;/p&gt;&#xA;" OwnerUserId="131" LastActivityDate="2015-01-27T03:30:42.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="289" PostTypeId="1" CreationDate="2015-01-27T03:34:07.273" Score="22" ViewCount="31581" Body="&lt;p&gt;Modern LED bulbs must convert the standard household supply (for example $240\text{V AC}$ in the UK) into a DC supply at a lower voltage (usually $12\text{V DC}$ I think) for the LED array. This is done on a per-bulb basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this waste much power in the conversion?&#xA;If all of your bulbs are LED bulbs, would it make sense instead to have a $12\text{V DC}$ circuit for lighting around the whole house and LED bulbs without the per-bulb conversion?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="3633" LastEditDate="2016-07-29T13:41:46.593" LastActivityDate="2017-08-01T18:59:43.563" Title="Would it make sense to have a 12V lighting circuit in a house?" Tags="|electrical-engineering|energy-efficiency|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="290" PostTypeId="2" ParentId="289" CreationDate="2015-01-27T04:25:06.770" Score="12" Body="&lt;p&gt;Yes:  Something like this has been commonly done in the U.S. for decades.  Both Incandescent undercabinet lights and exterior low-voltage &quot;landscape&quot; lighting are often run from a single 12V transformer over 12-gauge wires.  So there is code for it, and anyone who has built a computer recently knows that high-efficiency DC converters carry an upfront premium so one doubts the converter built into every LED bulb is as efficient ... not to mention that distancing the transformer from the bulb will increase its lifespan by reducing heat exposure.&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2015-01-27T04:25:06.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="291" PostTypeId="2" ParentId="289" CreationDate="2015-01-27T04:37:35.673" Score="14" Body="&lt;p&gt;Distributed DC power is actually used in some new construction. It's driven less by the efficiency of the transformer than by other logistics. Here in California, at least, we have a law (known as Title 24) which requires some fairly sophisticated controls of lighting, as well as fairly low power consumption per square foot. The control requirements include compensating for daylight by automatically dimming lights, occupancy sensing, and brownout usage reductions. This means that the controllers are fairly sophisticated and expensive, and that LED light fixtures are sometimes preferred in new construction. As a result, some systems are being sold with the controllers that output dimmed DC which directly powers the LED fixtures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;DC distribution reduces the number of wires that have to go to each fixture (DC+ and DC- instead of AC hot, neutral, and a separate control line) and saves some money on electronics. As far as I know though, there is still at least one controller per room. I'm not aware of any systems that distribute DC all throughout a building. I imagine this is because as lengths and currents increase, the advantages of high voltage in reducing wire size become more significant. &lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-01-27T04:37:35.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="292" PostTypeId="2" ParentId="279" CreationDate="2015-01-27T07:17:34.520" Score="4" Body="&lt;p&gt;I can't give a definite answer, since I hardly plan either type. So please read this answer as an incomplete list of questions you could ask the sales engineers of your pumps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both can be operated hydraulically or pneumatically (for example if you don't want to bring a motor into an ex-zone or there would be cooling problems). In the case of piston pumps you'd need extra air/oil pistons for that. The diaphragm can be operated directly by the fluid, so the pump can be simpler. So for a sump application, you could have a diaphragm pump in your sump powered by a compressor nearby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen small and huge piston pumps, but I've never seen huge diaphragm pumps (huge= flowrate &gt;10m³/h). I assume some economy of scale here that makes large piston pumps cheaper than comparable diaphragm pumps (But I could be totally wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diaphragm pumps have no seals that touch the medium. This is an advantage since you &lt;em&gt;probably&lt;/em&gt; need to service them less. This also means that they are better for aggressive media or applications where cleanliness is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With any pump, you need to know if there are any solids or abrasive stuff in your medium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your question what specifications should be stated, ask your vendor(s) &quot;What do you need to know to size and configure a pump?&quot;&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2015-01-27T07:30:14.343" LastActivityDate="2015-01-27T07:30:14.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="293" PostTypeId="2" ParentId="231" CreationDate="2015-01-27T11:25:24.550" Score="3" Body="&lt;p&gt;Whether these are sensible Froude numbers or not depends on the length of the vessel in question. For a 100 m ship these are probably high but for a 5 m dinghy these would be quite low numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed-to-length ratio is the critical factor that determines the importance of skin and wave friction. Skin friction scales with $V^2$, whereas wave drag increases much more rapidly. I haven't been able to find the exact formula in a brief search, I seen to remember being told a factor of $V^6$. The wave drag places a practical limit on the speed of a displacement (non-planing) vessel of $1.34L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small speed-to-length ratio skin friction will be dominant, whereas for large ratios wave friction is important. An &lt;a href=&quot;http://www.sailboat-cruising.com/hull-drag.html&quot; rel=&quot;nofollow&quot;&gt;example value&lt;/a&gt; I found was that skin friction is ~65% of total drag at speed/length=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general large ships will have a low speed/length and skin friction will dominate. On the other hand, small displacement boats like dinghies or kayaks will have skin friction dominated by wave drag.&lt;/p&gt;&#xA;" OwnerUserId="153" LastEditorUserId="368" LastEditDate="2015-06-10T20:26:50.067" LastActivityDate="2015-06-10T20:26:50.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="294" PostTypeId="2" ParentId="280" CreationDate="2015-01-27T12:03:21.420" Score="8" Body="&lt;p&gt;From my brief involvement in shocks, I think the most likely solution would be to image the exhaust, probably optically, but maybe using interferometry or something depending on what the exhaust is. The most obvious indication that you have supersonic flow is if you can see a &lt;a href=&quot;http://en.wikipedia.org/wiki/Shock_diamond&quot;&gt;shock diamond&lt;/a&gt;. I think you could probably also work it out from the length of the exhaust but I cannot remember how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you could also look at the thrust generated. You should be able to calculate the expected thrust. This is what they do when testing rockets/jet engines as they don't actually care if the flow is supersonic, just that it generates enough power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple way for pipes is to just measure the exit flow. Its a pipe so flow should be constant. However, in practice I suspect long pipes also have regular inspection hatches/areas where they measure the flow somehow to check for leaks/faults.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-01-27T12:03:21.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="295" PostTypeId="2" ParentId="289" CreationDate="2015-01-27T13:09:45.997" Score="3" Body="&lt;p&gt;RV, camping, boating, and solar power industry might have had a significant influence in developing LED blubs powered by 12V DC.  The automotive industry too might have had some influence. Some of the issues that make 12V DC lighting not be practical are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Structural: Majority of the home are power using 110/120/220/230/240 VAC. Converting these too 12V DC for lighting is costly. There is a signification structural cost to start using 12VDC for lighting lonely.&lt;/li&gt;&#xA;&lt;li&gt;All other appliances in a home in most case require 110/120/220/230/240 VAC depend on the region. Therefore a home will require a 110/120/220/230/240 VAC plus a 12V DC network. This is added cost to the consumer. &lt;/li&gt;&#xA;&lt;li&gt;Also it might be cost prohibitive to have a 12V DC network in larger office complexes, hospitals, hospitality, airports, stadiums etc.&lt;/li&gt;&#xA;&lt;li&gt;The DC power drop from point to point in these larger building might be higher too. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Converting AC power to 12V DC power locally for &lt;strong&gt;LED&lt;/strong&gt; (Light Emitted Diodes) bulb lighting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental building block of an LED is a Diode. A typical diode switches on at 0.7V forward voltage. Current/Voltage (I-V ) characteristic for a typical diode is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lotTq.png&quot; alt=&quot;I-V Diagram for a diode&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the LED color the forward voltage varies. The table below lists typical forward voltage for a different colored LED’s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I9Mn4.jpg&quot; alt=&quot;Typical Characteristics for colored LED&amp;#39;s&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forward voltage for a typical &lt;strong&gt;white LED&lt;/strong&gt; is between 3V to 4V. Below is I-V graph for typical coloured LED's &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EQR3t.gif&quot; alt=&quot;I-V curves for different colored LED&amp;#39;s&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical AC supply (120V in USA, 240V UK) need to be rectified to a lower DC voltage. Transmitting DC power over a long distance is inefficient. Thus AC supply is a far more efficient to transmit power. There is always an energy loss when you convert AC to DC, but current technology has overcome most of these efficiencies. There are small cost effective rectifier circuit modules available in the market. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an example of one such device. (RECOM Power RAC01-05SC)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T3Beh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6pZfZ.jpg&quot; alt=&quot;RECOM Power RAC01-05SC&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.recom-international.com/pdf/Powerline-AC-DC/RAC01_02-SC.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;RECOM Power RAC01-05SC&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although converting AC to DC is efficient, the other structural cost outweigh the benefits of a separate 12V DC line in homes. The growth in technology and markets it is very likely there will be universal AC LED bulbs where the conversion will occur seamlessly regardless of input AC voltage.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Light-emitting_diode&quot; rel=&quot;nofollow noreferrer&quot;&gt;Light-emitting diode&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.electronics-tutorials.ws/diode/diode_8.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Light Emitting Diode&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.recom-international.com/pdf/Powerline-AC-DC/RAC01_02-SC.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;RECOM Power RAC01-05SC&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/33036/least-expensive-way-to-plug-in-a-microcontroller-into-120v-without-wall-wart-or&quot;&gt;Least expensive way to plug in a micro-controller into 120V&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.edn.com/design/analog/4337310/Voltage-to-current-converter-drives-white-LEDs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voltage-to-current converter drives white LEDs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=YUXfWSxDUvc&quot; rel=&quot;nofollow noreferrer&quot;&gt;LED E27 12V Light Bulb&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2017-01-08T00:56:36.137" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="296" PostTypeId="2" ParentId="289" CreationDate="2015-01-27T13:40:00.000" Score="2" Body="&lt;p&gt;I'd say it depends on the size of the 12V network and the number of transformers and their size. The problem is two-fold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, we have to think about the conversion losses incurred. If (and that's a big If) the efficiency is the same on the bulbs than it would be on a central transformer, then the losses incurred by having a central transformer are on par with the one incurred by the distributed light bulbs. However, usually a transformer has an efficiency peak (for example, if the peak power of the transformer is 500W, the efficiency peak could be at 460W. When the system uses more or less power, the transformer is less efficient). This has to be taken into account when calculation potential losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, there's a wiring problem. When you route 220V or 110V, the cable section takes into account the maximal current that will have to be transported. In France, you can route lighting wiring on 1.5mm² cable (AWG15 or 16). Using LEDs will reduce the power used, but yet, as you reduced the voltage, the voltage drop may increase if you use a cable too thin. For example, if you use 12W bulbs, they will draw 1A of current at 12V instead of 0.05A at 220V. A cable too thin may incur a voltage drop too important that will prevent the LEDs from lighting up.&lt;/p&gt;&#xA;" OwnerUserId="154" LastActivityDate="2015-01-27T13:40:00.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="297" PostTypeId="2" ParentId="251" CreationDate="2015-01-27T13:51:31.847" Score="0" Body="&lt;p&gt;Usually, this is done through trial and error, and you optimise for the economic cost (weight of concrete and steel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do this by using a calculation sheet (in excel, or MathCad) where the end result is the weight or cost of both the concrete and the steel. And you try and find the best compromise between the distance between the strips, their size, and the resulting slab size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I answered you question perfectly though...&lt;/p&gt;&#xA;" OwnerUserId="154" LastActivityDate="2015-01-27T13:51:31.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="298" PostTypeId="1" AcceptedAnswerId="301" CreationDate="2015-01-27T14:31:24.163" Score="7" ViewCount="7406" Body="&lt;p&gt;This is inspired by a project I undertook for my Physics class last year; I'd like to apply it in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt; For the project, I and a few classmates were required to build a small car powered by any of a number of sources; we opted for a mousetrap that turned two rear wheels on a four-wheel car as it was released. Our group's goal was to make the car go as far as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem was that the power source was not constant. The lever arm of the trap lifted in an arc for about five seconds, then returned to its resting position. During this time, it propelled the car. The car then coasted for the rest of the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the powered stage, we wanted to increase traction. The wheels were CDs (because we had a $5.00 budget), which have a tendency to spin out. So we attached pieces of cloth to them to get better traction. On the second stage, however, we found that this cloth slowed down the car quite a bit because it dramatically increased friction (as we found out after several hours of testing different combinations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other groups used duct tape to cover the wheels, and some used records, which seemed to do slightly better than CDs (though that violated the size restrictions). Cloth seemed to have the best traction, though - we didn't have many spinouts. The tests were done on a typical classroom floor (I'm not sure what it's made of - linoleum? - but it's the same as in almost every school, at least in America).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a wheeled vehicle in general - obviously not just a small car powered by a mousetrap - how can I substantially improve traction on the wheels while the power source is on yet reduce friction while it coasts? Is it as simple as choosing certain tires, or is there a bigger and better engineering solution?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a final wrap-up: My thought would be to have a shifting center of mass for the vehicle, where the powered wheels have a lesser tendency to slip and the front wheels are nearly frictionless. During the powered stage, the center of mass would be near the back, while in the coasting stage, it would be near the front. This could help reduce the normal force on certain wheels and thereby produce or avoid extra friction.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="282" LastEditDate="2015-02-09T22:54:49.257" LastActivityDate="2016-04-11T23:19:32.460" Title="How can I improve traction on a wheeled vehicle yet reduce friction?" Tags="|mechanical-engineering|friction|applied-mechanics|" AnswerCount="5" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="299" PostTypeId="2" ParentId="298" CreationDate="2015-01-27T14:57:16.657" Score="4" Body="&lt;p&gt;Frictional forces are related to the surface type and the material type.  Traction is related to the amount of friction the wheel can generate as well as the weight applied to the wheel.  This &lt;a href=&quot;http://auto.howstuffworks.com/four-wheel-drive1.htm&quot; rel=&quot;nofollow&quot;&gt;how-stuff-works&lt;/a&gt; article does a pretty good job at covering the relationship between the two properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One popular misconception is that the surface area affects traction of the wheel.  And &lt;a href=&quot;http://en.wikipedia.org/wiki/Drag_racing&quot; rel=&quot;nofollow&quot;&gt;dragster style race cars&lt;/a&gt; are sometimes cited as an example of proving that point.  Those vehicles have special rear tires that are wide when at rest but become more narrow as their rotational speed increases.  The belief is that this has the effect of increasing starting traction, but reduced rolling resistance as the area of the wheel in contact with the ground decreases with rotational speed.  However, &lt;a href=&quot;http://stao.ca/VLresources/cruciblearticles/07Nov2Cdragsters.pdf&quot; rel=&quot;nofollow&quot;&gt;this writeup&lt;/a&gt; does a good job at explaining how resulting frictional forces remain the same regardless of the surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the &lt;a href=&quot;http://www.engineeringtoolbox.com/tractive-effort-d_1783.html&quot; rel=&quot;nofollow&quot;&gt;equation for traction&lt;/a&gt; $F_t = \mu_tmg$ is equivalent to the friction $F_f = \mu_fmg$&lt;sup&gt;1&lt;/sup&gt;, it's a balancing act between available torque; surface material; and available wheel materials.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;That's not to say that $\mu_t$ is exactly the same as $\mu_f$ but they are close enough for this case.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-01-27T15:39:19.460" LastActivityDate="2015-01-27T15:39:19.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="300" PostTypeId="2" ParentId="119" CreationDate="2015-01-27T15:14:43.500" Score="2" Body="&lt;p&gt;I find that often clients are resistant to change (internal or external clients), and even when they agree that technically something sounds/ appears to be a good solution, they do not want to let go of their existing process or product. (Why? Because that's what we've always done) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have all of this beautiful data - and an elegant solution to a clients problem. This is where the real work begins. This is also where it gets tricky, within the company I work for this has been handled by encouraging engineering staff to pursue MBA degrees to help develop their business sense, we also hired a dedicated marketing staff to help polish proposals and presentations, because the truth is not all engineers are great &quot;salespeople&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as helpful advice, I guess what I for you as a basic starting point would to find a good book on marketing, or persuasive speaking, as it doesn't sound like the problem lies within your technical solutions.   &lt;/p&gt;&#xA;" OwnerUserId="258" LastActivityDate="2015-01-27T15:14:43.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="301" PostTypeId="2" ParentId="298" CreationDate="2015-01-27T15:24:11.293" Score="6" Body="&lt;p&gt;You need to control how fast you apply the power. Some sort of damper mechanism that would allow power to be gradually applied and not cause it to spin out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try and wrap the 'power string' over different sized pulleys so that you changed the gear ratio. If you used a cone it would be similar to a CVT where starting out you would have a low gear ratio to get moving then as the string unwound down the cone it would propel the car faster rather than just spinning out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vTIDP.png&quot; alt=&quot;Something like this for your rear axle&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: You can either guess at the gear ratio or use some basic statics to figure out coefficient of friction and how to avoid slip while applying torque.&lt;/p&gt;&#xA;" OwnerUserId="231" LastEditorUserId="231" LastEditDate="2015-01-27T15:46:25.780" LastActivityDate="2015-01-27T15:46:25.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="302" PostTypeId="1" AcceptedAnswerId="328" CreationDate="2015-01-27T15:52:56.227" Score="2" ViewCount="340" Body="&lt;p&gt;It's been a long time since I went over heat exchangers and even then it was all theoretical. I'm looking for anyone with real world experience with different types of heat exchangers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking at building a external wood burning stove. &lt;a href=&quot;http://www.crownroyalstoves.com/Websites/crownroyalstoves/images/crs/waterjacket.jpg&quot; rel=&quot;nofollow&quot;&gt;Most of the designs are pretty much just a firebox surrounded by water&lt;/a&gt; which is a step up from just putting a pot of water over a stove. They a bit overpriced/simple for what they are and most are 'dumb' devices. Temp too cold, turn on fan, temp too hot, turn off fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of my combustion experience is with events that occur multiple times/second rather than in hours. I think there is a lot of room for improvement starting with a proper heat exchanger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rough initial estimates:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Air Temp: ~1500F | ~800C&lt;/li&gt;&#xA;&lt;li&gt;Water Temp: &amp;lt;212F | &amp;lt;100C (Not making a boiler due to safety).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="231" LastActivityDate="2015-01-28T07:59:25.730" Title="Most efficient heat exchanger for dirty air/water medium?" Tags="|mechanical-engineering|thermodynamics|heat-exchanger|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="303" PostTypeId="1" CreationDate="2015-01-27T16:23:35.050" Score="10" ViewCount="584" Body="&lt;p&gt;(This is closely linked with &lt;a href=&quot;https://engineering.stackexchange.com/questions/280/how-to-know-whether-the-flow-is-supersonic-in-a-nozzle&quot;&gt;measuring the Mach number inside a nozzle&lt;/a&gt; but it is not regarding the supersonic flow)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friction and heat transfer have effects on the Mach number of the compressible flow (&lt;a href=&quot;http://en.wikipedia.org/wiki/Fanno_flow&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fanno&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Rayleigh_flow&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rayleigh flow&lt;/a&gt;). As it is extremely important to have tight control over the flow properties, here are my questions:-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;how does one know the Mach number of the flow inside long pipelines carrying some sort of gas (e.g. &lt;a href=&quot;http://en.wikipedia.org/wiki/West%E2%80%93East_Gas_Pipeline&quot; rel=&quot;nofollow noreferrer&quot;&gt;WEPP&lt;/a&gt;)? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What Mach number is maintained through these pipelines? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Considering the cyclic temperature changes and friction inside the pipes, how is the Mach number maintained constant / within a range? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="223" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-01-28T16:22:55.330" Title="How is gas flow through (extremely long) pipelines monitored and controlled?" Tags="|fluid-mechanics|gas|pipelines|measurements|fluid-dynamics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="304" PostTypeId="2" ParentId="229" CreationDate="2015-01-27T16:51:40.417" Score="14" Body="&lt;p&gt;While some of these answers are close, they are (at the time this answer is written) all incorrect to some degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure and stress are very closely related -- in fact, one could argue that pressure is, in a sense, a subset of stress.  To be specific, the pressure in a material is the isotropic part of the total stress in a material.  Pressure is a scalar quantity -- the same in every direction, while stress is a tensor quantity that captures all deforming forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure and stress are related as follows:  if the components of the stress tensor are given by $\sigma_{ij}$, then the pressure is (using Einstein notation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = -\frac{1}{3}\sigma_{ii}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say, the pressure is the opposite of the average of the diagonal elements of the stress tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When speaking more specifically in terms of a boundary condition or an applied load for a structural analysis problem, it refers specifically to an applied normal stress over a given area.&lt;/p&gt;&#xA;" OwnerUserId="215" LastActivityDate="2015-01-27T16:51:40.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="305" PostTypeId="1" CreationDate="2015-01-27T17:56:40.497" Score="3" ViewCount="130" Body="&lt;p&gt;I'm not sure quite how many people are familiar with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stratolaunch_Systems&quot; rel=&quot;nofollow&quot;&gt;Stratolaunch project&lt;/a&gt;, so I'll give a quick summary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paul Allen and Burt Rutan (of Scale Composites) teamed up a few years ago to create a rocket that would be air-launched from &lt;a href=&quot;https://en.wikipedia.org/wiki/Scaled_Composites_Stratolaunch&quot; rel=&quot;nofollow&quot;&gt;an airplane&lt;/a&gt; - in fact, the aircraft with the largest wingspan ever. This airplane, affectionately dubbed &quot;the Roc,&quot; will, if things go well, have two fuselages, six jet engines and a 385-foot wingspan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The airplane took several parts from two 747-400s, including its engines. Wikipedia states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The aircraft will be powered by six 205–296 kN (46,000–66,500 lbf) thrust-range jet engines, that are planned to be sourced from two used 747-400s that will be cannibalized for engines, avionics, flight deck, landing gear and other proven systems that can be recycled to cut development costs. The two aircraft have been purchased and have arrived on site as of March 2012.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Stratolaunch has &lt;a href=&quot;http://www.stratolaunch.com/news_archives12/feb12.html&quot; rel=&quot;nofollow&quot;&gt;two&lt;/a&gt; &lt;a href=&quot;http://www.stratolaunch.com/news_archives12/mar12.html&quot; rel=&quot;nofollow&quot;&gt;pages&lt;/a&gt; on the purchases, but I can only find a small bit of relevant information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any specific plans for the unused parts of the 747s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I believe this is our first aerospace question (and our 100th question!).&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-01-27T18:24:53.007" LastActivityDate="2015-01-29T02:07:04.393" Title="Are there any specific plans for the other parts of the 747s scavenged to make the Stratolaunch carrier aircraft?" Tags="|aerospace-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="306" PostTypeId="1" CreationDate="2015-01-27T19:42:17.510" Score="14" ViewCount="855" Body="&lt;p&gt;Adaptive suspensions with adjustment rates on the order of milliseconds are going mainstream in consumer vehicles.  Air suspensions have been common on commercial trucks and some SUVs for some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are pneumatic tires necessary or helpful on vehicles equipped with such suspensions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand why a wearable tire is necessary: the friction coating on the wheels does need to be replaced.  Recent years have seen a proliferation of very low-profile tires for road vehicles, but to my knowledge they are still always pneumatic.  Why?  The downsides to inflatable tires are that they perform poorly when improperly inflated and are subject to failure, including catastrophic decompression.  What are their upsides?  For example, do they provide damping responses that can't be replicated with an air or adaptive suspension?&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="282" LastEditDate="2015-02-15T19:37:27.970" LastActivityDate="2016-04-28T18:27:32.060" Title="Are pneumatic tires obsolete?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="307" PostTypeId="2" ParentId="298" CreationDate="2015-01-27T19:47:07.643" Score="0" Body="&lt;p&gt;Traction, friction, terrain versatility, choose two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's to operate in slippery terrain, it will need high friction to improve traction. But if you can restrict the terrain (and keep reasonable acceleration; you can always trade it off for the remaining three), you can provide a very good traction with minimal friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, your wheels get cut-outs like a saw blade, angled backwards. You provide a &quot;starting strip&quot; of matching saw-like track with teeth directed forward. That way you prevent all slippage without sacrificing much of friction. The wheels can then either ride over flat terrain adding a little friction from the sawtooth surface, or you can use a dual system, round &quot;rims&quot; extending past the &quot;tracked&quot; sawtooth ones, so once the vehicle leaves the track, it moves on perfectly round wheels the rest of the way.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-01-27T19:47:07.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="308" PostTypeId="2" ParentId="303" CreationDate="2015-01-27T21:19:34.803" Score="4" Body="&lt;p&gt;For question 1, as far as I recall, fiscal metering of long hydrocarbon gas pipelines is done with coriolis effect mass flow meters. To calculate Mach number, all that would be required is the mass flow and the temperature from a thermocouple (along with some estimate of the mixture's heat capacity ratio.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer needs checking against reliable references. (CW as I didn't check any books.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions 2 and 3 are more difficult, and again may need to be separated.&lt;/p&gt;&#xA;" OwnerUserId="160" LastActivityDate="2015-01-27T21:19:34.803" CommentCount="0" CommunityOwnedDate="2015-01-27T21:19:34.803" ContentLicense="CC BY-SA 3.0" />
  <row Id="309" PostTypeId="1" AcceptedAnswerId="315" CreationDate="2015-01-27T21:37:38.480" Score="4" ViewCount="96" Body="&lt;p&gt;With the huge metal shredders that can shred an entire car or a bus, they can shred parts like the axle and engine which are large solid chunks of metal, just like the massive spinning shredder blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does the car get shredded and not the shredder? Are the blades made from harder/stronger metal, or is there something about their shape that makes them stronger (they just look like large plates with notches on)?&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-27T22:34:50.723" Title="Why don't massive industrial shredders shred themselves?" Tags="|crushing|shredding|waste-disposal|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="310" PostTypeId="2" ParentId="306" CreationDate="2015-01-27T22:05:52.987" Score="11" Body="&lt;p&gt;While I don't see them going obsolete any time soon, you make some great points that really made me think. I thought of some upsides to pneumatic tyres that overcome some edge-cases, but actually they all boil down to the same thing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Any protrusion will cause a focus point for the weight which is usually spread over the entire contact surface of the tyre.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pneumatic tyre will distort around such protrusions up to a point. If the entire weight usually supported by the tyre were focused onto a small area, this could cause damage to the road surface or the tyre. Here are some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Small rock in the road:&lt;/strong&gt; The rock may safely crush; or it may ping out at a high speed; or it could embed itself in the tyre causing damage to the tyre and road surface at multiple points further down the road; or it could crack the road surface and embed itself where water could get in and freeze causing more damage.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Exposed metalwork:&lt;/strong&gt; The suspension may ensure that the dynamic impact is not too severe, but the weight focused on a single sharp corner could significantly cut or dent the hard tyre.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Driving up a curb:&lt;/strong&gt; Repeated driving up a curb would wear the corner of the curb away much more quickly than a pneumatic tyre, which would distort and start to take weight on the high part of the curb before the weight was fully removed from the lower side.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Accidents:&lt;/strong&gt; A human leg (for example) being literally run over by a pneumatic tyre, while not pleasant at all, would suffer much less damage than a hard tyre. The hard tyre would probably crush/shatter bones into very small pieces and completely destroy muscle and blood vessels by flattening them into the bone fragments.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is one other upside that I can think of which doesn't fit into the above category:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pneumatic tyres are adjustable:&lt;/strong&gt; You can lower the air pressure to get more contact surface (and therefore grip) in slippery conditions and raise the pressure to get lower friction if grip is not an issue.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-27T22:05:52.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="311" PostTypeId="2" ParentId="306" CreationDate="2015-01-27T22:06:35.903" Score="14" Body="&lt;p&gt;A pneumatic tire provides the mechanical &quot;decoupling&quot; of the tiniest variations in a road surface (the highest freqeuncies), involving a small &quot;unsprung mass&quot; (the rubber of the tread) and a spring (the air pressure) working against the &quot;sprung mass&quot; of the wheel and the axle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vehicle's suspension (springs, shock absorbers, etc.), working between the axle and the frame of the vehicle, decouple the larger, slower (lower frequency) variations in the road surface. In this case, the entire tire + wheel + axle assembly is considered the &quot;unsprung mass&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While there are &lt;a href=&quot;https://www.youtube.com/watch?v=4jYcX_D09ig&quot;&gt;other tire designs&lt;/a&gt; that have similar qualities, they are generally heavier and much more complex to manufacture. At least for now, the pneumatic tire remains the most cost-effective way to get the desired overall performance. (Not to mention the fact that the entire vehicle service infrastructure is currently set up to deal with pneumatic tires.)&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2015-01-27T22:11:51.360" LastActivityDate="2015-01-27T22:11:51.360" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="312" PostTypeId="1" CreationDate="2015-01-27T22:16:53.640" Score="12" ViewCount="1035" Body="&lt;p&gt;Attach is a Pareto Diagram with &lt;strong&gt;Number of Defects&lt;/strong&gt; in Y-axis vs &lt;strong&gt;Causes&lt;/strong&gt; in X-axis for an engine overheating problem. How can this Pareto diagram help an engineer solve the engine overheating problem?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tWxlc.png&quot; alt=&quot;Cause of Engine Overheating&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pareto_analysis&quot; rel=&quot;noreferrer&quot;&gt;Pareto analysis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="282" LastEditDate="2015-03-04T00:11:26.617" LastActivityDate="2015-03-04T00:11:26.617" Title="How can Pareto Analysis help solve engineering problems?" Tags="|statistics|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="313" PostTypeId="1" AcceptedAnswerId="319" CreationDate="2015-01-27T22:28:43.523" Score="18" ViewCount="2385" Body="&lt;p&gt;I live in Louisiana these days, in an area that is known for its numerous antebellum plantation homes (circa early 1800s). While touring one of these homes it was clear that almost everything about the house was designed around keeping cool in the summer.  Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;4-meter-high ceilings to allow hot air to rise to the ceiling.&lt;/li&gt;&#xA;&lt;li&gt;Floor-to-ceiling windows to allow hot air at the top to escape and cool air to be drawn in at the bottom.  &lt;/li&gt;&#xA;&lt;li&gt;Porches on the sunny sides of the house to prevent sunlight from entering the windows.&lt;/li&gt;&#xA;&lt;li&gt;Large central staircases to allow hot air to rise to the second floor, drawing cool air in on the bottom floor.&lt;/li&gt;&#xA;&lt;li&gt;Some have a cupola, a central observation room at the top of the house, again to allow hot air to escape at the top of the house and draw air in from the bottom.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My question is: Given our modern understanding of thermodynamics, how could one design a home today to be cooled passively? Could we do any better than the plantation owners of the 1800s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's define cooling as making the house more comfortable for humans. This means that it is not only important to reduce the temperature, but also to block sunlight and maintain airflow. Also, if possible, it would be very beneficial to extract moisture from the air. &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="368" LastEditDate="2015-02-13T23:44:18.403" LastActivityDate="2015-02-13T23:44:18.403" Title="How to design a house to be cooled passively?" Tags="|civil-engineering|energy-efficiency|heat-exchange|cooling|" AnswerCount="1" CommentCount="1" ClosedDate="2015-05-12T14:24:58.247" ContentLicense="CC BY-SA 3.0" />
  <row Id="314" PostTypeId="2" ParentId="74" CreationDate="2015-01-27T22:29:01.513" Score="1" Body="&lt;p&gt;If you have access to a lathe you should be able to quickly machine a flanged reducer out of &lt;a href=&quot;http://en.wikipedia.org/wiki/Nylon_6&quot; rel=&quot;nofollow&quot;&gt;Nylon 6&lt;/a&gt;. Not sure what temperature you're operating at but make it a bit thick and it will easily be able to handle 3 bar.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-27T22:29:01.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="315" PostTypeId="2" ParentId="309" CreationDate="2015-01-27T22:34:50.723" Score="5" Body="&lt;p&gt;The blades are indeed made of, or tipped with a hard steel/carbide, but they also don't contact each other. The shredding action is created by the shear forces of the two corotating drums of &quot;teeth&quot;, and this non-contact between the hard points, high shear strength, and gearing advantages allow it to power through whatever you throw at it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=aVkTj9VrH4o&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=aVkTj9VrH4o&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="249" LastActivityDate="2015-01-27T22:34:50.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="317" PostTypeId="2" ParentId="215" CreationDate="2015-01-27T23:00:20.460" Score="9" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Michigan left&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.archive.org/web/20090510121808/http://www.tfhrc.gov/safety/pubs/04091/10.htm#1022&quot;&gt;This page&lt;/a&gt; (pdf &lt;a href=&quot;http://web.archive.org/web/20090510122331/http://www.tfhrc.gov/safety/pubs/04091/04091.pdf&quot;&gt;here&lt;/a&gt;) is very informative, though you really have to dig to get what you want. In a sub-section of 10.2.2 Median U-Turn Crossover, I found this (A &quot;Michigan left&quot; is referred to as a &quot;median U-turn crossover&quot;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A study on a Michigan corridor used simulation to compare median U-turn crossovers with two-way left-turn lanes (TWLTL). The study showed that during peak hours, the corridor with median U-turn crossovers had a lower travel time by 17 percent and a 25 percent higher average speed than the same corridor with a TWLTL. However, vehicles made more stops on the arterial with median U-turn crossovers. In nonpeak hours, the median U-turn crossovers had the same efficiency as the TWLTL, even though a higher delay for left-turning vehicles had been expected due to the higher travel distance a vehicle must cover to turn left using a median crossover.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So that's a yes for stopping congestion during peak hours. More information on that specific simulation can be found under found under footnote 149, which isn't too easy to find. Other simulations reportedly found similar results:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Simulation studies using a range of intersection configurations (number of through lanes on the major and minor street) and volumes from intersections in Virginia and North Carolina suggest a reduction in overall travel time for all movements through the intersection when compared to a conventional intersection: -21 to -2 percent during off-peak conditions, and -21 to +6 percent during peak conditions. The studies also show a general increase in the overall percent of stops when compared to a conventional intersection: -20 to +76 percent during off-peak conditions, and -2 to +30 percent during peak conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The rest of 10.2.2 has some more safety information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The collision rate is lowered slightly&lt;/li&gt;&#xA;&lt;li&gt;There are less &quot;conflict points&quot; (i.e locations where collisions are likely to happen)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Jersey jughandle&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Jersey jughandle (referred to as simply a &quot;Jughandle&quot;) does reduce conflict points, though not as much as a Michigan left. It, too, appears to increase efficiency:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Simulation studies using a range of intersection configurations (number of through lanes on the major and minor street) and volumes from intersections in Virginia and North Carolina suggest a reduction in overall travel time through the intersection when compared to a conventional intersection: -6 to +51 percent during off-peak conditions, and +4 to +45 percent during peak conditions. The studies also show a large increase in the overall percent of stops when compared to a conventional intersection: +15 to +193 percent during off-peak conditions, and +19 to +108 percent during peak conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is there a clear winner? Both clearly reduce travel time and congestion, so the answer to your question is a definite yes. The Michigan left has many less conflict points (16) than the Jersey Jughandle (26), which I consider quite the advantage (the standard four-way intersection has 32). It also has a lesser increase in stops. I'd give the edge here to the Michigan left, though both are probably improvements over your standard four-way intersection.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-01-28T00:37:13.247" LastActivityDate="2015-01-28T00:37:13.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="318" PostTypeId="2" ParentId="298" CreationDate="2015-01-27T23:05:47.110" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;how can I substantially improve traction on the wheels while the power&#xA;  source is on yet reduce friction while it coasts?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Considering that, for most purposes, traction is really a function of friction, I think addressing the two cases is better than addressing one in all cases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to dynamically inflate/deflate the tires?  If so, deflating the tires a little while the power is on, then inflating them back up when it's coasting might do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, what about keeping extra wheels retracted, then dropping them down like a bike's training wheels when you need more turning maneuverability?&lt;/p&gt;&#xA;" OwnerUserId="177" LastActivityDate="2015-01-27T23:05:47.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="319" PostTypeId="2" ParentId="313" CreationDate="2015-01-27T23:25:35.833" Score="10" Body="&lt;p&gt;In dry/arid areas a &lt;a href=&quot;http://en.wikipedia.org/wiki/Windcatcher&quot; rel=&quot;noreferrer&quot;&gt;wind catcher&lt;/a&gt; tower in conjunction with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Qanat&quot; rel=&quot;noreferrer&quot;&gt;qanat&lt;/a&gt; is a great way to keep buildings cool. The underground water stays cool and cools the air passing over it that is drawn in through the wind catcher.&lt;/p&gt;&#xA;&lt;p&gt;However, in a humid climate you would want to isolate the incoming warm/moist air from the cool underground water. You could pass pipes through a large underground water tank and draw the air through those, this may also condense some of the moisture out of the air and you'd need drainage outlets from the air-pipe that come outside of the water tank.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tPl5D.png&quot; alt=&quot;Passive house cooling system&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Obviously you'd use a much better heat exchange method than I drew in this rather crude picture, but hopefully it explains the basic idea.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-01-27T23:25:35.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="320" PostTypeId="2" ParentId="312" CreationDate="2015-01-28T00:12:15.770" Score="12" Body="&lt;p&gt;This gives you the starting point for a return on investment calculation. It tells you the value in fixing each issue in terms of how many defects would be prevented. The idea is that the curve shows the cumulative percentage of issues vs. the total issues. I've drawn a red cross-hair at where the curve crosses the 56.8 mark - 56.8 is 80% of the 71 issues. This means if you fix the &quot;Important&quot; issues on the left of that point, you will have solved 80% of the problems, so they are more valuable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VyJ2X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engineer should then estimate the effort/cost for each possible course of action for those important issues on the left. Estimates just need to be relative to each other, not in any absolute terms. You can then divide the value by the cost to get the relative return on investment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if I estimated to fix the radiator core issue is about 90 units of difficulty/cost, to fix the fans is 20, and to fix the thermostat is 15, my ROI calculations would be as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \text{Fix Radiator Core ROI} = \frac{31}{90} = 0.34444 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \text{Fix Faulty Fans ROI} = \frac{20}{20} = 1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \text{Fix Faulty Thermostat ROI} = \frac{8}{15} = 0.53333 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a good idea of where to start in order to maximise the value added as quickly as possible (you are maximising the area under the curve if you plotted total value added against time as you work on fixing these issues). As you can see, it's not as simple as starting with the first one because it will fix the most issues - this may take a long time due to the effort and in the meantime you could have fixed the simpler fans and thermostat issues. The ROI gives you the order to work on these issues, fixing the fans gives you the best return, then the thermostat, then the radiator core. You've then fixed 80% of the problems and you can do the same again to work on the other 20%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there may be other factors that add to the 'value' component - such as negative press around one issue. Likewise the effort estimate may have different weightings for time vs. financial costs, opportunity costs of using your resources on this as opposed to something else, etc. All of these factors should be considered in the 'value' and 'cost' components of this ROI calculation. It depends how detailed you want to get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you might think you could just work out the ROI for all issues including the less important ones, but in practice doing the estimates themselves may be quite a lot of work, especially if there are lots of issues (this example only has 6). So the Pareto analysis allows you to very quickly work out the most important issues so that you can estimate those first.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-28T00:12:15.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="321" PostTypeId="2" ParentId="302" CreationDate="2015-01-28T00:42:19.583" Score="3" Body="&lt;p&gt;I'm going to have to make this an answer.  If someone feels it's inappropriate I can delete it, what follows is the start of my comment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grin, I bought my house with an outdoor wood stove, we have a love/ hate relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I love:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It burns all my waste paper and meat products,  (Including the treats our cats bring us, and the scraps the dogs drag home during deer season.)&lt;/li&gt;&#xA;&lt;li&gt;It heats my garage and preheats hot water.  (It could heat the house, but... more wood.)&lt;/li&gt;&#xA;&lt;li&gt;Gathering and processing of wood, yes it's work, but it's also an excuse to be in my woods which I enjoy. I sit on my butt most of the day, so wood = exercise.&lt;/li&gt;&#xA;&lt;li&gt;That the ashes are all outside.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I hate:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When something breaks down, it's always on the coldest day of the year. &#xA;(of course that'a not true, but plumbing and outside, windy, below freezing is not fun.) &lt;/li&gt;&#xA;&lt;li&gt;It belches smoke because of poor burning.  &lt;/li&gt;&#xA;&lt;li&gt;If I leave my house during the winter, (say Xmas) then I either have to &#xA;drain it or have someone come over and feed it.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course some of that is open to engineering.&lt;br&gt;&#xA;I'm leary of overly complicated.&lt;br&gt;&#xA;and I burn a lot of wood on cold winter nights.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a pic, It's cold here tonight (7 F) and it's burning nicely.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/lJP1o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="168" LastActivityDate="2015-01-28T00:42:19.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="322" PostTypeId="5" CreationDate="2015-01-28T00:48:46.307" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-28T00:48:46.307" LastActivityDate="2015-01-28T00:48:46.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="323" PostTypeId="4" CreationDate="2015-01-28T00:48:46.307" Score="0" Body="Questions relating to fluids at rest (fluid statics) or the forces and effects of fluids in motion (fluid dynamics)." OwnerUserId="95" LastEditorUserId="368" LastEditDate="2015-05-08T17:58:24.853" LastActivityDate="2015-05-08T17:58:24.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="325" PostTypeId="1" CreationDate="2015-01-28T01:02:00.617" Score="7" ViewCount="794" Body="&lt;p&gt;LED brightness can be controlled using digitally controlled potentiometer such as a Microchip &lt;a href=&quot;http://ww1.microchip.com/downloads/en/DeviceDoc/22060b.pdf&quot; rel=&quot;nofollow&quot;&gt;MCP413X&lt;/a&gt; or a PWM signal. In both cases the signal will be controlled/ generated from a micro-controller. Which method is better using a digitally controlled potentiometer or PWM signal? Why?&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-05-04T02:43:47.677" LastActivityDate="2019-05-04T02:43:47.677" Title="What is the best way to control brightness of a LED?" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="326" PostTypeId="1" CreationDate="2015-01-28T06:12:26.843" Score="6" ViewCount="1015" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ground_effect_vehicle&quot;&gt;Wind In Ground (WIG)&lt;/a&gt;, is an old concept. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In short:&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Wingtip_vortices&quot;&gt;Wing-tip vortices&lt;/a&gt; create an additional drag on the aircraft known as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Lift-induced_drag&quot;&gt;induced drag&lt;/a&gt;. However, when the flying craft (or &lt;a href=&quot;http://www.loyno.edu/lucec/natural-history-writings/bird-flight-over-water&quot;&gt;bird&lt;/a&gt;) flies close to the ground or water, these vortices don't get enough space to develop which results in additional lift and reduced drag. This phenomenon has been effectively used in transportation vehicles to create ground effect or WIG crafts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although a paper titled '&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0376042106000637&quot;&gt;Wing-in-ground effect vehicles&lt;/a&gt;' declares that many technical difficulties have either been solved or can be solved, according to &lt;a href=&quot;http://en.wikipedia.org/wiki/Ground_effect_vehicle&quot;&gt;this wiki article&lt;/a&gt;, there are some practical issues in using these crafts for regular passenger traffic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;a href=&quot;http://en.wikipedia.org/wiki/Hovercraft&quot;&gt;hovercrafts&lt;/a&gt; also glide on air cushion, and yet they are not that rare, what makes WIG so uncommon? (Analogy wise, hovercraft is like helicopter while WIGcraft is like airplane) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it the engineering aspects that makes them uncommon? If yes, which aspects?&lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2018-12-26T06:27:15.850" Title="What are the engineering challenges in WIG craft development?" Tags="|aircraft-design|aerospace-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="327" PostTypeId="2" ParentId="325" CreationDate="2015-01-28T06:39:21.450" Score="4" Body="&lt;p&gt;This depends on your circuit. For efficiency, something that is often sought after with LED circuits, and especially things that run off batteries, the most efficient solution is to use PWM. Running a potentiometer works, but has more losses than the PWM setup, especially with a larger load. GreatScott! on YouTube provides a great explanation of this, but with a fan for an example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=90g6RpvXBYY&quot; rel=&quot;nofollow&quot;&gt;PWM VS Potentiometer! When to use which technique?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="249" LastEditorUserId="1586" LastEditDate="2016-02-18T19:15:09.663" LastActivityDate="2016-02-18T19:15:09.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="328" PostTypeId="2" ParentId="302" CreationDate="2015-01-28T07:59:25.730" Score="2" Body="&lt;p&gt;I can't speak from real (=hands on with my own hands) experience, but in industrial applications I see mostly &lt;strong&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Shell_and_tube_heat_exchanger&quot; rel=&quot;nofollow&quot;&gt;shell and tube&lt;/a&gt;&lt;/strong&gt; type HX. Exhaust is in the tubes, the shell has the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pros:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;lots of surface&lt;/li&gt;&#xA;&lt;li&gt;when you add flanges opposite the tube openings, cleaning is doable&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cons: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pressure loss in the small tubes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An alternative would be to &lt;strong&gt;jacket the (existing) chimney&lt;/strong&gt;, with water between chimney and jacket. Insulation would be addded on the outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pros:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;easier to retrofit&lt;/li&gt;&#xA;&lt;li&gt;little change to convection etc. characteristics of the stove&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Less surface, so less heat transfer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-01-28T07:59:25.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="329" PostTypeId="1" AcceptedAnswerId="2258" CreationDate="2015-01-28T08:12:50.767" Score="6" ViewCount="382" Body="&lt;p&gt;To size a heat exchanger, I need to know (among other things) the Reynolds number (Re) as an indication of the flow conditions. The Re number depends on the viscosity. In a &lt;a href=&quot;http://en.wikipedia.org/wiki/Shear_thinning&quot; rel=&quot;nofollow noreferrer&quot;&gt;shear thinning fluid&lt;/a&gt;, I can't assume a constant viscosity, instead it will depend on the shear rate that is not constant throughout the pipe. &lt;a href=&quot;https://physics.stackexchange.com/questions/86178/what-is-the-apparent-viscosity-in-shear-thinning-turbulent-flow-through-a-pipe&quot;&gt;I've asked on Physics SE about the shear rate in turbulent shear thinning fluid flow in a pipe&lt;/a&gt;, but received no helpful answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here at Engineering.SE, I am not looking for an in-depth examination of the beauty of the Navier-Stokes equation, but for a practical approach to sizing a heat exchange for such a fluid. How to go about it?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2015-03-31T18:50:35.783" Title="How to size a heat exchanger for a shear thinning fluid?" Tags="|mechanical-engineering|heat-exchanger|fluid-dynamics|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="330" PostTypeId="2" ParentId="325" CreationDate="2015-01-28T08:58:59.913" Score="10" Body="&lt;p&gt;The PWM solution is better under any aspect, plus the chip you linked has a quite high &quot;full scale&quot; resistance, $5\text{k}\Omega$, that will lead to use less steps than available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PWM is better because its efficiency is near unity and because it does not require any external IC if the micro can handle the led current, which is usually true. When you use high power leds their drivers usually have a PWM input anyway, and a resistor would be impractical anyway because of the amount of power it should dissipate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Digital potentiometers are used to attenuate an analog signal, such as a line level audio signal: you're trading some power dissipation for linearity. Such a signal is essentially a voltage signal, little to no current should flow so power dissipation in the resistors is not really a concern. When you want to power an LED instead you actually want power to flow into it, that's why you say &quot;power an LED&quot;, and a resistor dissipates power, that's inevitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd just add a series fixed resistor to protect the led or to use the whole PWM scale, if that's a concern. A standard red led wants some 2V @5-10mA to work properly, a 3V3 output is enough to burn it, adding a series small resistor, in the $200\Omega$ range, allows you to set the pwm up to 100% and protects the led in case of an hw/sw fault.&lt;/p&gt;&#xA;" OwnerUserId="48" LastActivityDate="2015-01-28T08:58:59.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="331" PostTypeId="2" ParentId="62" CreationDate="2015-01-28T13:35:18.607" Score="7" Body="&lt;p&gt;You can easily measure the brightness of any light source in the visible with a silicon (Si) or indium gallium arsenide (InGaAs) &lt;a href=&quot;http://en.wikipedia.org/wiki/Photodiode&quot; rel=&quot;noreferrer&quot;&gt;photodiode&lt;/a&gt;.  Both can be picked up at &lt;a href=&quot;http://www.digikey.com/product-search/en/sensors-transducers/optical-sensors-photodiodes/1967048&quot; rel=&quot;noreferrer&quot;&gt;relatively low cost&lt;/a&gt; especially if buying in bulk.  A photodiode works in the reverse manner to an LED.  With an LED running a current through the device causes it to emit light; with a photodiode light incident on the device causes it to become a current source. The responsivity of a Silicon detector is shown below.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Response_silicon_photodiode.svg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HSwXk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have your photodiode you need to build a simple circuit around it which converts current to voltage, and the voltage readout gives you a measure of the light intensity incident upon the device.  This type of circuit is typically called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Transimpedance_amplifier&quot; rel=&quot;noreferrer&quot;&gt;transimpedance amplifier&lt;/a&gt;.  A high SNR transimpedance amplifier can be built with a single op-amp so the cost again is relatively low.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This setup will make relative brightness measurements very precisely as long as you don't saturate the photodiode or circuitry with too much light.  Converting the measured voltage to an absolute intensity of the light is also possible though any errors in the parameters will turn up as errors in the inferred intensity.  To do so you use the circuit parameters to calculate the amount of current generated by the photodiode. The photodiode &lt;a href=&quot;http://www.rp-photonics.com/responsivity.html&quot; rel=&quot;noreferrer&quot;&gt;responsivity&lt;/a&gt; is then used to convert to incident intensity.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-01-28T13:35:18.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="332" PostTypeId="1" AcceptedAnswerId="335" CreationDate="2015-01-28T15:00:33.337" Score="3" ViewCount="995" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Helicopter_rotor#Flybar_.28stabilizer_bar.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;helicopter flybar&lt;/a&gt; is a kind of gyroscopic stabilizer which acts in feedback to the pitch of the blades.  The effect, in words, is that as the helicopter starts to pitch or roll in a particular direction more lift is applied on that side, keeping the helicopter level. These devices are commonly found on the inexpensive RC helicopters which are on sale at all major department stores.  An example of the flybar from one of these RC helicopters is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.modifyrcheli.com/what-is-flybar-shorter-or-longer-better&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tGFwG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it isn't obvious from the picture, there is an angle between the flybar and the rotors.  On my RC helicpoter at home the angle is roughly $35^\circ$.  I believe that this angle has the effect of introducing extra phase in the feedback loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the effect of changing the angle between the flybar and the rotor?&lt;/li&gt;&#xA;&lt;li&gt;How can I mathematically describe the effect of this angle?&lt;/li&gt;&#xA;&lt;li&gt;What would happen if the angle was changed to $90^\circ$? What about $0^\circ$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-01-28T17:00:08.687" Title="What is the effect of changing the angle of a helicopter flybar?" Tags="|aerospace-engineering|feedback-loop|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="333" PostTypeId="2" ParentId="245" CreationDate="2015-01-28T15:07:17.677" Score="2" Body="&lt;p&gt;This may not fully answer your question but hopefully it will be a good start. I thought a distributed mass model would be a good approach for this so I did some searching and found this paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://research.ncl.ac.uk/game/research/publications/Real-Time%20Deformable%20Soft-Body%20Simulation.pdf&quot; rel=&quot;nofollow&quot;&gt;Real-Time Deformable Soft-Body Simulation using Distributed Mass-Spring Approximations&lt;/a&gt; (PDF)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found this, which goes beyond what you need, including variable cross-sections and sheer stresses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hindawi.com/journals/isrn/2013/916581/&quot; rel=&quot;nofollow&quot;&gt;Bars under Torsional Loading: A Generalized Beam Theory Approach&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this second one is what you need, I included the first one because I can actually understand it whereas the second one is way beyond me. If you can simplify out the bits you don't need by substituting in suitable constants, it might be what you're looking for.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-28T15:07:17.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="334" PostTypeId="2" ParentId="303" CreationDate="2015-01-28T16:22:55.330" Score="14" Body="&lt;p&gt;It's important to remember that these pipelines don't exist as a single isolated line and will have a number of branches that tie into the main pipeline and split off for gas to be sold to different locations. This touches on a very broad subject of &lt;a href=&quot;https://en.wikipedia.org/wiki/Flow_assurance&quot;&gt;flow assurance&lt;/a&gt; and pipeline network modeling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pipeline will also be broken into smaller sections by &lt;a href=&quot;https://en.wikipedia.org/wiki/Compressor_station&quot;&gt;compressor stations&lt;/a&gt; because, as you mentioned, friction losses will require the gas to be re-compressed as it travels along the pipeline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, the gas flow rate will be regularly measured using a variety of &lt;a href=&quot;https://en.wikipedia.org/wiki/Flow_measurement&quot;&gt;flow meters&lt;/a&gt; such as an orifice meter, ultrasonic meter, Coriolis meter, turbine meter etc. This will occur at any point were gas flows into the main pipeline or is branched off the be sold. It also may occur at regular intervals (i.e. the compressor stations) along the pipeline. If you want to determine the mach number you could simply calculate it from your measured velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not aware of any target mach number or velocity in pipelines, it will certainly be subsonic, the rest will depend on the specifics of the section of pipe. For example, the elevation profile, the presence of liquids, corrosion, and the required pressure at the destination all will have an effect on the velocity you can transport the gas.&lt;/p&gt;&#xA;" OwnerUserId="136" LastActivityDate="2015-01-28T16:22:55.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="335" PostTypeId="2" ParentId="332" CreationDate="2015-01-28T17:00:08.687" Score="5" Body="&lt;p&gt;This angle is determined by the lift characteristics of the rotor and the rotational inertia of the whole helicopter. You are exactly correct that the angle changes the phase of the feedback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purpose of discussion let's assume the helicopter has pitched forwards slightly and needs to be corrected backwards, also let's assume a single rotor that spins anti-clockwise when viewed from above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we adjust the rotor when it is exactly at the front to give some more lift, the time it takes for the additional lift to overcome the rotational inertia of the whole helicopter would mean that the net effect of the lift was applied at some point to the front-left of the helicopter instead of exactly at the front. The next correction would then be slightly further around, and the next further round again. This would cause the helicopter to oscillate in the same manner that a coin does when you knock it over on the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you pretty much answered it yourself, the angle controls the phase offset between the detected error and where the lift &lt;em&gt;begins&lt;/em&gt; to be applied. Remembering that the net effect of the lift will be as if applied between where the rotor is adjusted and where the rotor is reset (after some time the helicopter has moved to correct the error).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the offset angle of the flybar will change the angle at which the correcting force is applied relative to the error. The effect of this will be an oscillation in one direction or the other. Up to 90 degrees from where it should be the flybar will give negative feedback out of phase, so should (at least mathematically) remain in a stable oscillation. Beyond 90 degrees from where it should be the flybar will start to give positive feedback instead of negative feedback, again out of phase until 180 degrees, causing the thing to exponentially spin out of control.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-28T17:00:08.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="336" PostTypeId="1" AcceptedAnswerId="344" CreationDate="2015-01-28T19:27:50.297" Score="12" ViewCount="5031" Body="&lt;p&gt;Many commercial and open-source CFD codes implement several closure methods for the non-linear convective acceleration term of the Reynolds-averaged Navier-Stokes (RANS) equations. Common methods (also known as &lt;em&gt;turbulence models&lt;/em&gt;) include&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Spalart%E2%80%93Allmaras_turbulence_model&quot;&gt;Spalart–Allmaras&lt;/a&gt; (S–A)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/K-epsilon_turbulence_model&quot;&gt;k–ε&lt;/a&gt; (k–epsilon)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/K-omega_turbulence_model&quot;&gt;k–ω&lt;/a&gt; (k–omega)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/SST_(Menter%E2%80%99s_Shear_Stress_Transport)&quot;&gt;SST&lt;/a&gt; (Menter’s Shear Stress Transport)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Reynolds_stress_equation_model&quot;&gt;Reynolds stress equation model&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which of these are suitable for CFD simulation of a streamlined vehicle body? The purpose of the simulations is to guide the refinement of the body shape to minimize aerodynamic drag forces. An exemplary answer would briefly outline the advantages and disadvantages of each method for this simulation application.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Potentially useful details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vehicle  is a small one-person vehicle with approximate dimensions&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;L = 2.5 m,&lt;/li&gt;&#xA;&lt;li&gt;W = 0.7 m, and&lt;/li&gt;&#xA;&lt;li&gt;H = 0.5 m.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It will be travelling at speeds ranging from 0 m/s to approximately 12 m/s. All three wheels are enclosed by the body envelope, and the vehicle has an approximate ground clearance of 15 cm except near the wheels, where the body shell extends down to within 1 cm of the road surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally aerodynamic forces at these speeds are very nearly negligible, but assume that this vehicle is being designed to compete in a &quot;Super Mileage&quot; competition on a smooth track, is very light-weight, and uses low friction drivetrain components throughout, so the aerodynamic forces have a significant effect on the achievable fuel consumption.&lt;/p&gt;&#xA;" OwnerUserId="166" LastEditorUserId="223" LastEditDate="2015-01-29T05:13:12.760" LastActivityDate="2015-04-18T09:41:38.220" Title="Which turbulence models are suitable for CFD analysis on a streamlined vehicle body?" Tags="|mechanical-engineering|fluid-mechanics|fluid-dynamics|simulation|modeling|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="337" PostTypeId="1" AcceptedAnswerId="341" CreationDate="2015-01-28T22:00:19.550" Score="5" ViewCount="221" Body="&lt;p&gt;The specific example I have in mind is a car tire with a small leak in it.  As the pressure increases, does the outflow of air increase linearly, i.e. $v\propto P$, or does it have some more interesting behavior?&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-01-29T06:25:04.050" Title="For a small valve in a fluid system, what is the relationship between pressure and flow rate?" Tags="|fluid-mechanics|pressure|fluid-dynamics|valves|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="338" PostTypeId="1" CreationDate="2015-01-28T22:35:37.500" Score="12" ViewCount="641" Body="&lt;p&gt;In reading about &lt;a href=&quot;http://en.wikipedia.org/wiki/Faraday_cage&quot;&gt;Faraday cages&lt;/a&gt;, I have found that most &quot;real world&quot; designs incorporate multiple layers of mesh.  For example, &lt;a href=&quot;https://nationalmaglab.org/about/around-the-lab/what-the/faraday-cage&quot;&gt;this article from the National High Magnetic Field Lab&lt;/a&gt; discusses several different shielded rooms built of varying layers.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the MagLab's High B/T Facility at the University of Florida's Microkelvin Laboratory in Gainesville, for example, all electromagnetic waves must be excluded. The magnet there is in a &quot;tempest&quot; quality shielded room, featuring walls made of layers of copper and welded steel that absorb the entire spectrum of EM radiation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My presumption is that each layer is tuned to or built for a particular range of frequencies. It's not clear to me if the tuning is due to thickness requirements for the layer and therefore a cost factor, or if there are other factors driving the design of each layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question then, is can (or should) a Faraday cage be treated as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Band-stop_filter&quot;&gt;notch filter&lt;/a&gt; as opposed to a &lt;a href=&quot;http://en.wikipedia.org/wiki/High-pass_filter&quot;&gt;high-pass filter&lt;/a&gt;?  If so, how do you determine the lower end of the notch?&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorUserId="368" LastEditDate="2015-11-20T01:15:56.450" LastActivityDate="2015-11-20T01:15:56.450" Title="Does a Faraday cage effectively act as a notch filter?" Tags="|electrical-engineering|signal-processing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="339" PostTypeId="2" ParentId="338" CreationDate="2015-01-29T01:36:01.727" Score="2" Body="&lt;p&gt;I would say a Faraday cage (at it's simplest) is more like a high pass filter.&lt;br&gt;&#xA;And more specifically one that blocks only Electric fields.&lt;br&gt;&#xA;The metal box we put our electronics in is a Faraday shield.&lt;br&gt;&#xA;Blocking magnetic fields is harder, it takes more layers, of special materials.&lt;br&gt;&#xA;(I didn't read NHMFL artcile, I'm at home w/ limited download, I'll read it tomorrow and edit my answer) &lt;/p&gt;&#xA;" OwnerUserId="168" LastActivityDate="2015-01-29T01:36:01.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="340" PostTypeId="2" ParentId="305" CreationDate="2015-01-29T02:07:04.393" Score="4" Body="&lt;p&gt;I suspect you won't find a definitive answer on the fate of those two particular 747's for the following reasons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The Stratolaunch project is likely still actively using the parts from those planes, so there's no need to dispose of them yet.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;747 parts are becoming quite commonplace and their value in the parts market continues to decline.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So that's a bit of a killjoy answer, for which I apologize.  Aeronautics is a wonderful field for generating compelling, romantic ideals that appeal to our dreams.  The thought of a plane (or two) being used for historic progress and then just being parted out is frankly somewhat depressing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when we look at the project management behind an engineering project like the Stratolaunch, the numbers become pretty telling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bloomberg.com/bw/articles/2012-06-21/boeings-747-400-a-faded-queen-of-the-skies&quot; rel=&quot;nofollow&quot;&gt;This Bloomberg article&lt;/a&gt; details how the 747-400 is increasingly being sold off by various airlines.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now, ten-year-old passenger 747-400s are worth a record-low $36 million, about 10 percent less than similarly aged planes last year, according to London-based aviation consultancy Ascend, as carriers seek more fuel-efficient models. There’s even little interest in converting the passenger jets into ﻿air freighters because of a slump in air cargo demand.&lt;br&gt;&#xA;  Some 48 of the humpbacked passenger 747-400s worldwide have also been placed in storage, according to Ascend. The onetime “Queen of the Skies” has been shunned in favor of Boeing’s smaller 777 widebody (which has two fewer engines sucking fuel) or Airbus’s mammoth A380 double-decker. “There’s not a lot of demand for the 747,” says Paul Sheridan, Ascend’s head of consultancy Asia. “They’re mostly being broken up for parts.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And &lt;a href=&quot;http://aviationweek.com/blog/photos-what-happens-all-those-airliners-when-they-retire-see-tear-down-747-400&quot; rel=&quot;nofollow&quot;&gt;this Aviation Week article&lt;/a&gt; highlights how 747-400s are increasingly being sold off for parts as it's more cost-effective to sell them than to continue operating them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;with airfield space often at a premium, airliners get withdrawn, re-cycled and chopped up in a much smaller timeframe. ...&lt;br&gt;&#xA;  The airline can recoup more money selling the aircraft than operating it for several more years before it might otherwise have retired. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Likewise, &lt;a href=&quot;http://www.npr.org/2014/03/29/296173820/boeings-iconic-747-may-be-flying-into-the-sunset&quot; rel=&quot;nofollow&quot;&gt;this NPR article&lt;/a&gt; piles on the bad news for the 747-400.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Boeing has scaled back production to only one and a half new 747s per month. Aviation consultant Ernest Arvai expects the company to keep the line running just long enough to replace Air Force One.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To draw that together:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a number of elements pointing us towards the fact that 747s and their associated parts are depreciating assets that are possibly depreciating faster than traditional valuations would have accounted for.  Depreciating assets receive proportionately less attention as their value continues to decline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stratolaunch is well funded and is focused on pushing the edges of space exploration technology.  While they may be concerned about how to wind down operations once they're done, I tend to doubt it.  Nor are they likely worried about maximizing resale value of the 747-400 parts due to their current funding levels.  I suspect they are more focused on how to solve the problems they set out to tackle than they are on &quot;afterwards.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have parts whose value is approaching that of salvage only and a company focused on broader objects without much need to optimize short-term economic factors.  From the projects I've been involved with, that translates into those two 747-400s becoming a footnote in the company's history; a footnote that the majority of us will never find out about how it ended.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-01-29T02:07:04.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="341" PostTypeId="2" ParentId="337" CreationDate="2015-01-29T03:55:45.973" Score="6" Body="&lt;p&gt;I think the best (and simplest) way to describe something like this is Bernoulli's equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P+\rho gh+\frac12 \rho v^2=constant$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use this, we're looking only at instantaneous velocity, because as the air leaks, the pressure will go down. We also should assume that the &quot;valve&quot; is really more of a small hole than anything that fluctuates too much with pressure variations, because that complicates it a bit more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant in the Bernoulli equation gets applied to any point in a continuous stream. So what we want to do is pick two points, one on either side of the hole, and relate those two points using the Bernoulli equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we'll get will look something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{tire}+\rho gh_0+\frac12 \rho v_0^2 =P_{atm} + \rho gh_1 +\frac12 \rho v_1^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, we'll say that any vertical movement of the air is small enough to neglect. Also, the velocity of the air inside the tire is negligible as well, if not in practice, than for the case of determining the relationship between pressure and velocity. Lastly, there's an important distinction between absolute pressure (which is in the equations above) and gauge pressure (which would be what we measure with a tire pressure gauge [go figure]). Gauge pressure is defined as $P_{gage}=P_{abs}-P_{atm}$. Bringing that all together we get the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{tire,gauge}=\frac12 \rho v_1^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other couple of important points to make with this is that it's valid for inviscid flow (no friction) with constant velocity. The first assumption is pretty valid, if the pressure differentials are significant, the second may not be, especially since that starts to drive up the fluid velocity, and constant density goes out the window when we get to compressible flows ($Ma&amp;gt;0.3$). Again though, for the simple case of examining the nature of the relationship, this evaluation should be fine. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-01-29T03:55:45.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="342" PostTypeId="2" ParentId="336" CreationDate="2015-01-29T06:02:21.307" Score="2" Body="&lt;p&gt;I can't say that this will be the ideal answer, but it should get you started. As will be apparent, I not a true expert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quality of these models will generally increase with their sophistication, which in this case basically tracks with the number of equations use.  So (S-A) would be least effective while k-$\epsilon$, k-$\omega$ and SST would be better. RSM would be the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the middle three, SST is (so I'm told) better at correctly predicting flow separation. The other two have a habit of not predicting separation when they should. Given that separation generally causes drag, these could result in a flawed design seeming good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While RSM would definitely be preferred if possible, it will be the most time consuming because it adds 7 equations on top of N-S. 10 years ago, you might have had to make a tough choice here.  Theses days you should be able to turn around RSM models of this kind of vehicle in reasonable amounts of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been been working on an FSAE (open wheeled single seat race car) aero design for the past couple of months and have found the use of RSM to be reasonable to run on a fairly high end laptop or any respectable gaming rig desktop. You can also find places where you can rent run time if you need to evaluate a large number of design iterations. I can add the name of a company that we used that was set up to run the software that we needed and helped us out with student prices (somebody please comment if that's appropriate for SE).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slight tangent:&#xA;I'd strongly recommend that you look for papers (ideally experimental) that you can use to validate your methods.  We made very sure that we could recreate (within reason) results from wind tunnel experiments before we proceeded to running our own designs.  It's also important to run mesh sensitivity analysis to make sure that your resolving the structure of the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, prism layers coming off of your surfaces (to better resolve boundary layers) are important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last: &lt;a href=&quot;http://www.scribd.com/doc/48010874/External-Aerodynamics#scribd&quot; rel=&quot;nofollow&quot;&gt;this document&lt;/a&gt; from the folks at Fluent is a bit old, but was still very helpful in getting us started. (sorry for the scribd link.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-01-29T06:02:21.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="343" PostTypeId="2" ParentId="337" CreationDate="2015-01-29T06:25:04.050" Score="5" Body="&lt;p&gt;Yep, the answer is a bit more interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass flow rate ($\dot m$) will vary as $C_d A \sqrt{2 \Delta P}$ (discharge coefficient, cross sectional area, and change in pressure across the valve). Barring compression of the fluid, velocity will behave the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The devil of the thing is in $C_d$. This is pretty much something that you have, measure experimentally or (try to) model numerically with computational fluid dynamics.  That little number captures all of the non-ideal aspects of the flow involved (viscosity, turbulence). It's always less than one (nothing is ever ideal), so a simple prediction from Bernoulli will always over-predict (just a question of how much).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, $C_d$ will change as you adjust the valve (as will area).  So the manufacturer will usually just give you a few values or a curve for flow as a function of valve position and pressure or a total flow coefficient $C_V$ as a function of valve position ($\sqrt{\Delta P }$ dependence assumed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the other answer, this all goes funny if the flow becomes compressible.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-01-29T06:25:04.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="344" PostTypeId="2" ParentId="336" CreationDate="2015-01-29T07:53:02.103" Score="12" Body="&lt;p&gt;&lt;a href=&quot;http://www.innovative-cfd.com/turbulence-model.html&quot;&gt;The turbulence model can make a big difference in your simulation&lt;/a&gt;. There are many turbulence models around. It becomes a tough job to select one out of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no perfect turbulence model. It all depends on several parameters like Reynold's number, whether the flow is separated, pressure gradients, boundary layer thikness and so on. In this answer, brief information about a few popular models is given along with pros and cons and potential applications. However, interested users can see &lt;a href=&quot;http://turbmodels.larc.nasa.gov/index.html&quot;&gt;this excellent NASA website&lt;/a&gt; and references therein to know more about turbulence modeling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A) ONE EQUATION MODEL:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Spalart-Allmaras&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model solves for one additional variable for Spalart-Allmaras viscosity. According to &lt;a href=&quot;http://turbmodels.larc.nasa.gov/spalart.html&quot;&gt;a NASA document&lt;/a&gt;, there are many modifications in this model targeted for specific purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pros&lt;/strong&gt;: Less memory intensive, Very robust, fast convergence &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cons&lt;/strong&gt;: Not suitable for separated flow, free shear layers, decaying turbulence, complex internal flows  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Uses&lt;/strong&gt;: Computations in boundary layers, entire flowfield if mild or no separation, aerospace and automobile applications, for initial computations before going to higher model, compressible flow computations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Applicability to your case&lt;/strong&gt;: a good candidate for reducing simulation time. You can predict the drag fairly well with this model. However, if you are interested in knowing the flow separation region, this model will not give highly accurate results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;________________________________________________________________________________&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B) TWO-EQUATIONS MODELS:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;$k$-$\epsilon$ turbulence model:&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;general purpose&lt;/strong&gt; model. This model solves for kinetic energy ($k$) and turbulent dissipation ($\epsilon$). The equations for this models can be found at &lt;a href=&quot;http://www.cfd-online.com/Wiki/Standard_k-epsilon_model&quot;&gt;this cfd-online page.&lt;/a&gt;  This model requires &lt;a href=&quot;http://www.cfd-online.com/Wiki/Wall_functions&quot;&gt;wall functions&lt;/a&gt; to be computed for the implementation. Suitable only for fully turbulent flows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pros&lt;/strong&gt;: simple to implement, fast convergence, predicts the flows in many practical cases, good for external aerodynamics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cons&lt;/strong&gt;: Not suitable for axi-symmetric jets, vortex flows and strong separation. Very low sensitivity for the adverse pressure gradients, difficult to start (need initialization with Spalart-Allmaras), not suitable for near wall applications &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Uses&lt;/strong&gt;: Suitable for initial iterations, good for external flows around complex geometries, good for shear layers and free non wall bounded flows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Applicability in your case&lt;/strong&gt;: Although this model is good for external bluff body computation, it is suitable only for turbulent flows. Since the velocities are low, flow is going to experience transition from laminar to turbulent (max $Re = 1.98*10^6$ using &lt;a href=&quot;http://airfoiltools.com/calculator/reynoldsnumber?MReNumForm%5Bvel%5D=12&amp;amp;MReNumForm%5Bchord%5D=2.5&amp;amp;MReNumForm%5Bkvisc%5D=1.5111E-5&amp;amp;yt0=Calculate&quot;&gt;this calculator&lt;/a&gt;). You might benefit better with a variant like realizable $k$-$\epsilon$ model. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. $k$-$\omega$ turbulence model&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solves for $k$ and turbulence frequency $\omega$. Gives better results for near wall flows. Predicts transition (although early sometimes). Quite sensitive to the initial guess and hence initial few iterations are performed with $k$-$\epsilon$ model. &lt;a href=&quot;http://www.cfd-online.com/Wiki/Near-wall_treatment_for_k-omega_models&quot;&gt;This article&lt;/a&gt; gives near wall treatment for this model.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pros&lt;/strong&gt;: Excellent for boundary layers, works in adverse pressure gradient, works for strong separated flows, jets and free shear layers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cons&lt;/strong&gt;: Time required for convergence is more, memory intensive, Requires mesh resolution near the wall, predicts early and excessive separation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Uses&lt;/strong&gt;: Internal flows, Pipe flows, Jet flows, vortices&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Applicability in your case&lt;/strong&gt;: Not completely suitable for your case since the boundary layer values depend strongly on free stream $\omega$. This requires a very fine grid to resolve and hence long computation time. Also it does not account for the transport of turbulent shear stress. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. $k$-$\omega$ SST&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best of both worlds! &lt;a href=&quot;http://www.cfd-online.com/Wiki/SST_k-omega_model&quot;&gt;This model&lt;/a&gt; has a blending function which uses $k$-$\omega$ near the wall and $k$-$\epsilon$ in the free stream. It does not use wall functions.&lt;br&gt;&#xA;All the variants of this model can be found at this &lt;a href=&quot;http://turbmodels.larc.nasa.gov/sst.html&quot;&gt;NASA page&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pros&lt;/strong&gt;: Accounts for turbulent shear stress while giving all the benefits of $k$-$\omega$ model, Highly accurate prediction of separation and transition, Very good free stream as well as boundary layer results&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cons&lt;/strong&gt;: Not suitable for free shear and vortex flows as much as standard $k$-$\omega$, Not suitable for jet flows, Requires fine mesh resolution near walls&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Uses&lt;/strong&gt;: External aerodynamics, separated flows, Boundary layers and adverse pressure gradients&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Applicability in your case&lt;/strong&gt;: Highly applicable. If you want better results, use a variant of sst model which uses $k$-$\epsilon$ RNG or realizable model away from the walls&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So which model is most appropriate?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess would be $k$-$\omega$ SST model. &#xA;Since it will give better transition, separation and works even under adverse pressure gradients, you will get better &lt;a href=&quot;http://en.wikipedia.org/wiki/Skin_friction_drag&quot;&gt;skin friction drag&lt;/a&gt;. At the same time, it works well away from the walls, which will give you good pressure drag and hence &lt;a href=&quot;http://en.wikipedia.org/wiki/Parasitic_drag#Form_drag&quot;&gt;parasitic drag&lt;/a&gt;. You will get better flow visualization. You can very well use the Spalart-Allmaras model, but if you see &lt;a href=&quot;http://www.innovative-cfd.com/turbulence-model-2.html&quot;&gt;this study&lt;/a&gt;, you will notice how much difference the SST model makes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And don't take my word for it. A report on '&lt;a href=&quot;http://dc.uwm.edu/cgi/viewcontent.cgi?article=1031&amp;amp;context=etd&quot;&gt;Aerodynamic Analysis and Drag Coefficient Evaluation of Time-Trial Bicycle Riders&lt;/a&gt; ' uses the SST model. &lt;a href=&quot;https://lirias.kuleuven.be/bitstream/123456789/297656/2/Defraeye+et+al._2010_Computational+Fluid+Dynamics+analysis+of+cyclist+aerodynamics+-+Performance+of+different+turbulence-modelling+and+boundary-layer+modelling+approaches.pdf&quot;&gt;This paper&lt;/a&gt; compares all the turbulence models results for cyclist aerodynamics and arrives at a conclusion that the SST model gives the best overall results. I am citing these results because Reynold's number wise and dimensions wise, a bicycle goes most closer to your case, for which tons of studies are available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if time is limited in your case, go for Spalart-Allmaras model. You can also go for RNG $k$-$\epsilon$ or realizable $k$-$\epsilon$ in that case. However, &lt;a href=&quot;http://www.altairhyperworks.com/html/en-us/rl/ACUSIM/papers/AIAA10_2010-1431_MNGodo.pdf&quot;&gt;this study of a bicycle wheel&lt;/a&gt; shows, S-A model gives better results than $k$-$\epsilon$ (this is very much geometry specific, different model might work for your geometry). If you have all the time in the world, conduct studies using SST and epsilon model and publish your comparison so others might also benefit from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have better computational resources, go for &lt;a href=&quot;http://en.wikipedia.org/wiki/Large_eddy_simulation&quot;&gt;LES&lt;/a&gt;. But I feel it is not called for in this case and it might not be appropriate. I do not have experience with LES, so can't comment.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some interesting resources:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://the-foam-house5.webnode.es/&quot;&gt;The FOAM house&lt;/a&gt;: If you want to learn OpenFOAM step by step &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0307904X14003448&quot;&gt;Recent advances on the numerical modeling of turbulent flows&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.turbulence-online.com/Publications/Lecture_Notes/Turbulence_Lille/TB_16January2013.pdf&quot;&gt;Lectures in Turbulence for the $21^{st}$ century&lt;/a&gt; - highly recommended reading if you wish to understand turbulence&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://people.nas.nasa.gov/~pulliam/Turbulence/Turbulence_Guide_v4.01.pdf&quot;&gt;Turbulence Models and Their Application to Complex Flows&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All the best!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="166" LastEditDate="2015-01-30T15:57:21.450" LastActivityDate="2015-01-30T15:57:21.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="346" PostTypeId="1" AcceptedAnswerId="349" CreationDate="2015-01-29T08:59:38.777" Score="7" ViewCount="218" Body="&lt;p&gt;Suppose I have a concrete tank, round, vertical, diameter in the range 16m-24m. One half the floor is filled with gravel or concrete to a height of maybe 2 meters. When the tank is empty, that will mean I have about 5 tons per m² more weight on this half, when the tank is full (with water or a slurry that's mostly water) it's still 3 tons (assuming 2.5 t/m³ density which is exact enough for my ballpark). Most of the times (&gt;90%) it will be full. The tank will be above ground, 8-10 m high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know if the tank will tilt during its lifetime, say 20 years. I'm not a civil engineer and I have no feeling for the numbers involved. My gut feeling is that my tank will tilt visibly in a matter of a few years and that my idea is not feasible as is. Can someone weigh in and comment on ...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Will I have tilt/uneven settling problems? At what magnitude over the tanks lifetime?&lt;/li&gt;&#xA;&lt;li&gt;What's the easiest (=cheapest) remedy, leaving the tank interior alone?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Clarifying points&lt;/em&gt;: The tank is not yet built or even planned. It's just an idea I'm thinking about that calls for half filling the tank to create a sort of funnel. I wonder if this idea is worth pursuing, and uneven loading/settling is one issue want to consider. I'm not in the &quot;call a structural engineer and let him calculate the static&quot; phase, I'm in the tossing around harebrained ideas in my skull phase. I'm sure such a tank can be built to last for 20 or 200 years, but at what price? &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="88" LastEditDate="2015-01-29T20:47:20.193" LastActivityDate="2015-01-29T21:26:39.580" Title="Settling of unevenly loaded storage tank" Tags="|civil-engineering|soil|geotechnical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="347" PostTypeId="2" ParentId="338" CreationDate="2015-01-29T14:30:40.553" Score="4" Body="&lt;p&gt;First, &lt;a href=&quot;https://physics.stackexchange.com/questions/89077/does-a-faraday-cage-block-all-magnetic-field-frequencies-lower-than-x&quot;&gt;here's a great answer to a similar question on the physics.SE&lt;/a&gt;. I will attempt to summarize it a bit and tune it to your question, but I think that all of your answers are already in that thread if you don't wish to read it from me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think that the mesh is tuned to particular frequencies for EM waves. It's just cheaper than building walls out of thick sheet metal. The thickness of the resulting mesh must be thick enough to attenuate the frequency (see Skin Effect), and the holes in the material must be smaller than the wavelength of signal. Overlapping layers of mesh provides the right thickness, and the randomness of one layer of mesh on top of another should eliminate the holes for most applicable wavelengths, making it a high-pass filter.&lt;/p&gt;&#xA;" OwnerUserId="45" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2015-01-29T14:30:40.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="349" PostTypeId="2" ParentId="346" CreationDate="2015-01-29T19:01:50.663" Score="5" Body="&lt;p&gt;Here's a really (and I mean really!) quick and dirty set of calculations that might give you an idea of the magnitudes of settlement you could be dealing with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The settlement potential of the tank location can be determined a number of ways, but probably the best thing to do would be a plate load bearing test. The test can be run to simulate the range (though not the duration) of loads you are expecting. A test like this will give you a spring constant $k$ that represents the &quot;modulus of subgrade reaction&quot; of the bearing soil (for the tested loading range). However, it's a short term test that doesn't take into account creep, so the long-term $k$ value will be lower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, a short-term $k$ will run from something like 80pci for a very soft clay to something like 250pci for a very dense sand (caveat: this is just from the top of my head without looking anything up). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's use the worst case scenario here, and to take into account creep, let's do what geotech engineers do best and slap a 2.5 safety factor on it. So we have about a 30pci modulus of subgrade reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's also assume that most of the differential settlement will occur as a result of the uneven loading of the empty tank, and that the emptying/filling of the tank is going to have a negligible contribution to the differential settlement. This isn't too terrible of an assumption, since the difference in applied surface pressure (which determines differential settlement) is much greater in the empty state, and it's also conservative because it will only be empty 10% of the time anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here we go (I'm American, so we're doing everything other than what you gave for dimensions in imperial-scum units first and then converting - sorry!): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k = 30 \frac{lbf}{in^3}$, $\gamma_{concrete}=150\frac{lbf}{ft^3}$, $H_{concrete}=2m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applied pressure under half the tank: $q_c=H_c\times\gamma_c=0.98ksf=5.3 \frac{tonf}{m^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Settlement under loaded half of tank: $S=\frac{q_c}{k} = 0.23in = 5.8mm$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume the other side of the tank does not settle at all, our differential settlement comes out to about 6mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this number assumes the loaded side of the tank is free to settle while the unloaded side remains static. This is not the case. Assuming the tank is nice and stiff, some of the applied pressure on the loaded side will be transferred to the unloaded side (which will reduce the settlement of the loaded side). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what the application is for this tank, but the above is probably a pretty conservative analysis of the situation you described. I would be surprised if differential settlement potential turns out to be a problem for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: One thing to note is that the tank will &quot;wiggle&quot; when it is being filled/drained. What I mean is, the entire thing will settle more when it is filled, but it will settle &lt;em&gt;more&lt;/em&gt; in the unloaded side (thereby undoing some of the differential settlement in the empty condition). Then when drained, the soil will rebound and the tank will return to the more-tilted empty condition when the unloaded side rebounds more than the loaded side (though it is likely neither side would rebound fully). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the 6mm of settlement from above, the deflection angle for the 24m diameter tank comes out to be $\arctan\frac{6mm}{24m}=0.014^{\circ}$. Pretty tiny. &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2015-01-29T21:26:39.580" LastActivityDate="2015-01-29T21:26:39.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="351" PostTypeId="2" ParentId="63" CreationDate="2015-01-29T20:26:43.500" Score="6" Body="&lt;p&gt;Roundabouts are only good for equal distribution of incoming traffic. As soon as one road feeds more traffic it will starve the other roads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small roundabouts are less efficient than large ones. So to get an efficient roundabout you need some space to lay it. Mini's are only good to specify equal right of way to the roads in low traffic areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double lane roundabouts are only 30% more efficient than single lane roundabouts. Compare that to traffic light where doubling the lanes will actually double the capacity.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Traffic lights however can be tuned to prioritize certain roads. For example a left turn at an intersection crossing opposing traffic can get a phase where they can go and the opposing direction has red and then don't have to worry about giving way to that traffic (a conflict-free left turn in countries that drive on the right). Or directions can only come up when there is traffic waiting for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traffic lights can also filter traffic to spread out the incoming flow of a subsequent intersection instead of letting it jam. Or coordinate them to get a &quot;green wave&quot; so traffic doesn't need to wait at the second intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things like public transport can also benefit as buses and trams can get be equipped with transponders to &quot;log in&quot; the intersection and better green times (earlier and/or longer). This is usually at the cost of other regular traffic though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A well designed traffic light will be so you only need to queue up at most once before you can get through, waiting at most 2 minutes. &lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="255" LastEditDate="2017-07-12T11:28:18.217" LastActivityDate="2017-07-12T11:28:18.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="352" PostTypeId="1" CreationDate="2015-01-29T20:44:11.430" Score="10" ViewCount="240" Body="&lt;p&gt;I've seen light fixtures with varying wattage rating. From 30W to 100W even though the actual socket is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the limiting factor, why can't I put say a 75W bulb in a fixture rated at 50W?&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorDisplayName="user16" LastEditDate="2015-01-29T21:09:44.173" LastActivityDate="2016-01-15T18:50:43.520" Title="How is the wattage of an incandescent light fixture determined?" Tags="|electrical-engineering|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="353" PostTypeId="2" ParentId="352" CreationDate="2015-01-29T21:03:41.680" Score="3" Body="&lt;p&gt;The rating on the fixture itself, as opposed to the bulb you put into it, is a safety rating. Above the fixture's rated wattage, the fixture will be unsafe. This will probably be due to heat issues that may breakdown the fixture's materials or even damage an installation's surrounding (i.e. it will set your house on fire if you try to install too big of a bulb in the socket). Other breakdown issues may include an arc hazard, but this is unlikely compared to the heat issue.&lt;/p&gt;&#xA;" OwnerUserId="45" LastActivityDate="2015-01-29T21:03:41.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="354" PostTypeId="2" ParentId="352" CreationDate="2015-01-29T21:04:44.000" Score="6" Body="&lt;p&gt;The rating of a lightbulb fixture is determined by the robustness of the materials used in making it.  Putting a lightbulb which uses significantly higher power than the fixture is rated for is dangerous.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, consider a cheap desktop lamp.  If you put a high wattage lightbulb into the socket, the amount of current draw will be larger than the wiring of the lamp is designed to handle.  Since some power is dissipated in the wiring of the lamp, this excess current will cause the lamp to heat up and potentially catch on fire.  A fuse is often included in devices such as desktop lamps to prevent this from happening though.&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-01-29T21:17:41.480" LastActivityDate="2015-01-29T21:17:41.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="355" PostTypeId="2" ParentId="352" CreationDate="2015-01-29T21:08:50.970" Score="10" Body="&lt;p&gt;This is related to the amount of current (Amps) the electrical parts of the fixture can take without overheating and causing damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Watts} = \text{Amps} * \text{Volts}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at 120v a 50W bulb will draw $\frac{50}{120}$ (0.416) Amps and a 75W bulb will draw $\frac{75}{120}$ (0.625) Amps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All electrical conductors have some resistance, this makes them warm up when you pass a current through them. The higher the current (more Amps) the faster it will get hot and the hotter it will get. There are many factors that determine how hot something can get before it will fail, for example how fast it can dissipate heat, what temperature the insulation can withstand. The rating is there so that you don't put too high a current through the electrical components of the fixture, which would cause it to get too hot and melt, catch fire, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How hot something will get can be predicted mathematically, but ultimately the rating value is determined by testing the fixture in a lab with light bulbs of different powers, and measuring how hot it gets. The rating is set such that it will always remain at a safe temperature, within a safety factor.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="95" LastEditDate="2015-01-29T21:15:32.720" LastActivityDate="2015-01-29T21:15:32.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="356" PostTypeId="1" AcceptedAnswerId="426" CreationDate="2015-01-29T21:18:31.507" Score="5" ViewCount="284" Body="&lt;p&gt;This is related to the question &lt;a href=&quot;https://diy.stackexchange.com/questions/58612/will-dark-gutters-stay-ice-free-better-than-light-colored-gutters&quot;&gt;Will dark gutters stay ice free better than light colored gutters?&lt;/a&gt; The current consensus seems to be that the mass of the gutters and limited light may make simply painting them black insufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my particular application the two major gutter runs are facing East and West so get decent amount of sun light. Physical location is Pittsburgh PA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an engineering perspective, these are some of the questions that would need to be addressed: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How much would I need to raise the temperature of the gutters to prevent ice build-up? &lt;/li&gt;&#xA;&lt;li&gt;How can I calculate the energy required to raise the temperature of the gutters?&lt;/li&gt;&#xA;&lt;li&gt;How much energy can I theoretically obtain from sunlight throughout the day for a given area during the winter at my latitude?&lt;/li&gt;&#xA;&lt;li&gt;How would the analysis change if I also wanted to melt ice that had built up during the night?&lt;/li&gt;&#xA;&lt;li&gt;What other factors should one consider when designing such a system?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="257" LastEditorUserId="-1" LastEditDate="2017-04-13T12:22:17.377" LastActivityDate="2015-08-10T16:11:18.013" Title="Is it possible to modify a home gutter to absorb sufficient solar energy to melt ice?" Tags="|solar-energy|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="357" PostTypeId="1" CreationDate="2015-01-30T00:06:32.067" Score="5" ViewCount="1617" Body="&lt;p&gt;There are many &lt;strong&gt;Lithium Ion&lt;/strong&gt; and &lt;strong&gt;Lithium Polymer&lt;/strong&gt; rechargeable battery related accidents. Are some of these accidents are battery design related? Are some of them misuse of rechargeable batteries. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the chemical composition of Lithium Ion and Lithium Polymer battery?&lt;/li&gt;&#xA;&lt;li&gt;Why do rechargeable Lithium Ion  &amp;amp; Lithium Polymer batteries explode or catch fire? &lt;/li&gt;&#xA;&lt;li&gt;Why cannot the rechargeable battery explosions and catching fire be prevented? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://computer.howstuffworks.com/dell-battery-fire.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;What causes laptop batteries to overheat?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.rcgroups.com/forums/showthread.php?t=209187&quot; rel=&quot;nofollow noreferrer&quot;&gt;Complete Guide to Lithium Polymer Batteries and LiPo Failure Reports&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.economist.com/blogs/economist-explains/2014/01/economist-explains-19&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why lithium batteries keep catching fire&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://globalnews.ca/news/1714748/why-lithium-ion-smartphone-batteries-keep-exploding/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why lithium-ion smartphone batteries keep exploding&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="106" LastEditDate="2015-01-30T00:45:22.680" LastActivityDate="2018-07-19T04:32:32.797" Title="Why do rechargeable Lithium Ion &amp; Lithium Polymer batteries explode or catch fire?" Tags="|electrical-engineering|chemical-engineering|battery|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="358" PostTypeId="2" ParentId="357" CreationDate="2015-01-30T00:43:32.957" Score="4" Body="&lt;p&gt;Disclaimer - I'm not an expert in batteries, but I do some related work with those particular types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium-ion batteries are a little different from &quot;regular&quot; lead acid batteries in that the Li-ion batteries can provide more power for short periods of time than what their nameplate rating would otherwise indicate.  The trade measurement term for this effect is called &lt;a href=&quot;http://batteryuniversity.com/learn/article/what_is_the_c_rate&quot; rel=&quot;nofollow&quot;&gt;C-rate&lt;/a&gt;.  The downside of exceeding a 1C (normal nameplate discharge) rate is that it creates more heat within the battery, wears out the battery faster, and can only safely be done for short periods of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the ability to exceed the regular 1C rate, and the fact that batteries tend to be exposed to a lot of environmental wear from being banged about leads to problems within the internal connections of the battery.  This &lt;a href=&quot;http://www.economist.com/blogs/economist-explains/2014/01/economist-explains-19&quot; rel=&quot;nofollow&quot;&gt;blogpost from The Economist&lt;/a&gt; does a good job explaining the thermal cascade that causes the thermal breakdown.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the battery is charged, lithium ions are driven from the electrolyte into a carbon anode. When the battery is discharged they flow back, creating a balancing flow of electrons in a circuit that powers the device. The trouble comes about if there is a small fault or damage is caused to the extremely thin separators that keep the elements of the battery apart. This can lead to an internal short-circuit and a subsequent build-up of heat. This can trigger what is known as a “thermal runaway” in which the battery overheats and can burst into flame.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-01-30T00:43:32.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="359" PostTypeId="5" CreationDate="2015-01-30T02:05:27.077" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-30T02:05:27.077" LastActivityDate="2015-01-30T02:05:27.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="360" PostTypeId="4" CreationDate="2015-01-30T02:05:27.077" Score="0" Body="Questions about the characteristics of fluids in motion." OwnerUserId="272" LastEditorUserId="368" LastEditDate="2015-06-23T15:44:57.267" LastActivityDate="2015-06-23T15:44:57.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="361" PostTypeId="5" CreationDate="2015-01-30T02:40:39.123" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-30T02:40:39.123" LastActivityDate="2015-01-30T02:40:39.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="362" PostTypeId="4" CreationDate="2015-01-30T02:40:39.123" Score="0" Body="Questions about the equipment and machinery used for crushing materials and objects. Also about the compression forces that result in crushing, either by deliberate mechanical means or by natural means such as pressure, gravity, build up of natural materials, etc. in order to engineer against them." OwnerUserId="95" LastEditorUserId="95" LastEditDate="2015-01-30T13:51:23.550" LastActivityDate="2015-01-30T13:51:23.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="363" PostTypeId="5" CreationDate="2015-01-30T02:42:12.717" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-30T02:42:12.717" LastActivityDate="2015-01-30T02:42:12.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="364" PostTypeId="4" CreationDate="2015-01-30T02:42:12.717" Score="0" Body="Questions about the equipment and machinery used for shredding materials and objects. Also about the shear forces that result in shredding either by deliberate mechanical means or by natural means in order to engineer against them." OwnerUserId="95" LastEditorUserId="95" LastEditDate="2015-01-30T13:51:14.460" LastActivityDate="2015-01-30T13:51:14.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="365" PostTypeId="5" CreationDate="2015-01-30T02:58:30.250" Score="0" Body="&lt;p&gt;&lt;strong&gt;Do not ask about individual academic guidance.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tag is mainly for questions about engineering &lt;a href=&quot;https://www.google.com/search?q=define%3A%20pedagogy&quot; rel=&quot;nofollow noreferrer&quot;&gt;pedagogy&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the method and practice of teaching, especially as an academic subject or theoretical concept&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, one of the first questions on the site asks about &lt;a href=&quot;https://engineering.stackexchange.com/q/162/368&quot;&gt;designing a small wind turbine for use in a classroom environment&lt;/a&gt;. Though it was tagged &lt;a href=&quot;/questions/tagged/education&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;education&amp;#39;&quot; rel=&quot;tag&quot;&gt;education&lt;/a&gt;, this question is mainly concerned with choosing a turbine blade diameter given simple, specific technical constraints. It would make just as much sense (and be as answerable) without knowing the context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question from the same user asks &lt;a href=&quot;https://engineering.stackexchange.com/q/367/368&quot;&gt;how to demonstrate an ancient method of measuring the radius of the Earth&lt;/a&gt;. In this case, although the question is technical, its constraints are much more closely tied to the educational context. Some knowledge of what is practical to perform with students, and how students learn best, could be required to write a good answer here, so the tag is appropriate.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Academic Guidance&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Questions asking for academic guidance are covered in &lt;a href=&quot;https://engineering.meta.stackexchange.com/q/261/368&quot;&gt;this discussion on Meta&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An &lt;strong&gt;academic guidance question&lt;/strong&gt; can be considered a question that inquires about the study of the various engineering fields as it relates to a specific person's academic and professional career. Prime examples are those that ask about a choice of institutions, degrees, courses, or topics of study.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Academic guidance questions are considered off-topic here because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They can and should be answered individually by someone whose job it is to advise students.&lt;/li&gt;&#xA;&lt;li&gt;Guidance will vary, and the decision of which guidance to follow is a personal one.&lt;/li&gt;&#xA;&lt;li&gt;Answering these questions on Engineering SE would be counter-productive for both the student and the site.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9" LastEditorUserId="6767" LastEditDate="2018-10-29T09:57:01.197" LastActivityDate="2018-10-29T09:57:01.197" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="366" PostTypeId="4" CreationDate="2015-01-30T02:58:30.250" Score="0" Body="Use this tag for questions about engineering education, including but not limited to curriculum, accreditation programs, mentoring programs and educational internships. Do not request individual academic guidance." OwnerUserId="95" LastEditorUserId="33" LastEditDate="2016-02-23T03:11:37.157" LastActivityDate="2016-02-23T03:11:37.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="367" PostTypeId="1" CreationDate="2015-01-30T04:31:35.177" Score="6" ViewCount="317" Body="&lt;p&gt;I want to design an experiment to show (in an educational context) how we can measure the radius of the Earth without using modern technology in order to demonstrate the engineering achievements of those who did it so long ago, and to show some practical uses for the maths involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The methods I know how to do this are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Place a stick vertically in the ground and another one many kilometers North of it and measure the shadow of both at exactly noon.&lt;/li&gt;&#xA;&lt;li&gt;Know the height of a mountain by a large lake or the sea and measure the distance you have to go until it disappears.&lt;/li&gt;&#xA;&lt;li&gt;Watch the sunset lying down and then stand up and watch it again, measuring the time difference between them and know your height.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first two are impractical to do with the kids, but the third one shows potential. However, is there a way to do this during the day using shadows cast by buildings, or must the shadow come from the horizon? How could I design some simple equipment to allow me to do this experiment and calculate the radius of the Earth during the day?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I am happy to use a digital stopwatch as the kids can easily see how people in the past might have used a mechanical clock to do the same thing (including measuring the length of a full day).&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-01-30T08:26:12.823" Title="Making a mechanical device to measure the radius of the Earth" Tags="|engineering-history|education|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="368" PostTypeId="2" ParentId="367" CreationDate="2015-01-30T08:11:05.140" Score="5" Body="&lt;p&gt;The experiment can be done during daylight but you need a very tall building overlooking the ocean &amp;amp; having an uninterrupted view of the horizon. This video show a group of US students doing the experiment &lt;a href=&quot;https://www.youtube.com/watch?v=dZ0l5ydpzW4&quot; rel=&quot;noreferrer&quot;&gt;You Tube - Measuring the Radius of the Earth&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an accurate measurement you will need:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;to be able to view the horizon unimpeded&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;have reasonably accurate way to measure vertical angle, a theodolite&#xA;would be ideal, but not practical with younger kids&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;know the height of the viewing platform&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The main reason why an unimpeded view of the horizon is required is for accuracy: the longer the target distance the more of the Earth's curvature is being considered. Also, the higher the building the further the viewable horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing is to decide which point on the horizon to measure to consistently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the problems will be weather and air clarity, particularly heat shimmer, haze absence of fog etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guys in the video had an error of 78% with their measurement due to air clarity, using a thick point marker and not having very accurate angle measuring equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third option you list could be done during a school camp by the sea, but I don't know that is practical or even possible in your situation.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-01-30T08:26:12.823" LastActivityDate="2015-01-30T08:26:12.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="370" PostTypeId="1" CreationDate="2015-01-30T09:20:43.520" Score="12" ViewCount="9876" Body="&lt;p&gt;I have a design for a desk, and I'd like to not just guess at how strong it'll be, but I can't find an explanation on how to figure out all the forces involved that doesn't assume I already know a lot about engineering already.&lt;/p&gt;&#xA;&lt;p&gt;So, if I were to apply 300lbf (1334 newtons) straight down on the front corner of the desk, how could I calculate the stress from the desktop to the upright beams, to the diagonal braces, to the ground?&lt;/p&gt;&#xA;&lt;p&gt;Assume A500 steel, 1x2x16ga.&lt;/p&gt;&#xA;&lt;h3&gt;Diagrams&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/UfNDS.jpg&quot; rel=&quot;noreferrer&quot;&gt;Overview&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/voK4S.jpg&quot; rel=&quot;noreferrer&quot;&gt;Legs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/qKO3H.jpg&quot; rel=&quot;noreferrer&quot;&gt;Desktop&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="285" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-07-15T18:03:31.007" Title="How do I calculate the forces on a desk and its legs?" Tags="|mechanical-engineering|beam|statics|" AnswerCount="4" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="371" PostTypeId="1" AcceptedAnswerId="374" CreationDate="2015-01-30T10:54:13.783" Score="10" ViewCount="8189" Body="&lt;p&gt;I recall learning about the bronze age in school. Why was bronze so important to the technological development of humans?  Why not some other metal?  &lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="368" LastEditDate="2015-02-20T23:04:28.253" LastActivityDate="2015-02-20T23:04:28.253" Title="What made bronze the first widely used (non-ornamental) metal in human civilization?" Tags="|materials|metallurgy|engineering-history|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="372" PostTypeId="1" CreationDate="2015-01-30T11:47:07.137" Score="10" ViewCount="1901" Body="&lt;p&gt;During CES2015 a well know Graphic Processor Unit (GPU) manufacturer introduced a new GPU. These GPUs need complex thermal management to keep the processor cool. Most GPU accelerator card manufacturers develop new graphics accelerator cards that use primitive thermal management technology compared to today's advanced technologies. Most of us know this technology as fans that manage this unwanted thermal energy, as seen in this image of a graphics accelerator card:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3KsTF.jpg&quot; alt=&quot;Inno3D-iChill-GeForce-GTX-980-HerculeZ-x4_1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What barriers do engineers need to overcome to convert this wasteful thermal energy to useful electrical energy?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a temperature profile of a GPU card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qgIdA.jpg&quot; alt=&quot;GPU Temperature Profile&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://spectrum.ieee.org/energywise/green-tech/conservation/a-thermoelectric-generator-that-runs-on-exhaust-fumes&quot; rel=&quot;noreferrer&quot;&gt;A Thermoelectric Generator That Runs on Exhaust Fumes&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.electrochem.org/dl/interface/fal/fal08/fal08_p54-56.pdf&quot; rel=&quot;noreferrer&quot;&gt;Small Thermoelectric Generators&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=CwQ8lHYebPU&quot; rel=&quot;noreferrer&quot;&gt;Thermoelectric Technology Overview&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.lifehacker.com.au/2011/12/extend-your-video-cards-life-with-custom-auto-fan-speed-settings/&quot; rel=&quot;noreferrer&quot;&gt;Extend Your Video Card's Life With Custom Auto Fan Speed Settings&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="368" LastEditDate="2015-02-13T19:57:13.260" LastActivityDate="2021-04-04T04:47:45.187" Title="Harvesting thermoelectric energy from computer graphics cards" Tags="|mechanical-engineering|electrical-engineering|thermodynamics|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="373" PostTypeId="2" ParentId="165" CreationDate="2015-01-30T13:21:48.807" Score="9" Body="&lt;p&gt;Yes, and they have been used successfully in some major applications.  One of the example in your list is &lt;a href=&quot;https://en.wikipedia.org/wiki/Evergreen_Point_Floating_Bridge&quot;&gt;Evergreen Point Floating Bridge&lt;/a&gt; &quot;Its 2,310 meters (7,580 ft) floating section is the longest floating bridge in the world.&quot;. It carries 4 lanes of traffic on SR 520.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having lived in the area and used the bridge, I can recall a few times that bridge has been closed, or had restricted traffic, I currently live in the Pittsburgh PA area where tunnels on major interstates are routinely closed.  There is not a significantly different traffic impact when compare a pontoon bridge to a tunnel. Different sources for the impact, but overall driver impact is similar. In my experience pontoon bridges are a viable solution in some cases.  I think probably the major limiter is tidal.  A pontoon bridge subject to tidal (ocean) impacts or seasonal variation (river), are going to have significantly higher stress issues.  &lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2015-01-30T13:21:48.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="374" PostTypeId="2" ParentId="371" CreationDate="2015-01-30T14:27:48.150" Score="19" Body="&lt;p&gt;The first thing to remember is that the naming of eras such as the Stone Age or the Bronze Age is never done by those living during the period. It was always done by others much later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To a certain degree, the reason why bronze was the first important &lt;strong&gt;alloy&lt;/strong&gt; was luck. For whatever reason, design or mistake, someone at some stage during antiquity mixed copper and tin  in a furnace and bronze was produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to the use of bronze, copper was used. Copper being a soft metal became blunt very quick when used in tools and needed to be sharpened at frequent intervals. Additionally, copper corrodes easy compared to bronze. When bronze proved superior to copper, copper was abandoned as the metal of choice for tools. During the time of the Pharaohs in Egypt, copper was for tool making. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Bronze Age was from about 3300 to about 800 BCE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first bronze made was arsenic bronze. When tin was discovered it replaced arsenic as the alloying metal. Tin bronze was better than arsenic bronze because the alloying process was more easily controlled and the resulting alloy was stronger and easier to cast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bronze became important because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It was a strong metal&lt;/li&gt;&#xA;&lt;li&gt;It was easy to cast&lt;/li&gt;&#xA;&lt;li&gt;It was easy to sharpen&lt;/li&gt;&#xA;&lt;li&gt;It maintained its sharp edge for a long time&lt;/li&gt;&#xA;&lt;li&gt;Weapons that maintained their sharp edges were very useful in&#xA;battles. Likewise for non military uses such as knives and chisels&lt;/li&gt;&#xA;&lt;li&gt;It is resistant to salt water corrosion, making it useful for&#xA;fittings in boats and ships&lt;/li&gt;&#xA;&lt;li&gt;It has a high resistant to corrosion and has fatigue endurance&lt;/li&gt;&#xA;&lt;li&gt;It does not oxidise beyond the surface&lt;/li&gt;&#xA;&lt;li&gt;It was useful as armour in ancient times&lt;/li&gt;&#xA;&lt;li&gt;It was fashioned into tiles for building construction&lt;/li&gt;&#xA;&lt;li&gt;It has a relatively low casting temperature&lt;/li&gt;&#xA;&lt;li&gt;When struck against a hard surface it does not generate sparks&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's heavy usage in ornamental work cannot be ignored. Statues and ornaments were important in ancient times and bronze was easy to produce and cast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, there was a need for metals and bronze was available. Iron only started being used when the tin trade was disrupted. Steel wasn't invented until much later and until it was, iron was very soft, it corroded easily and it was not as useful as bronze.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-01-30T14:27:48.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="375" PostTypeId="5" CreationDate="2015-01-30T14:47:55.050" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-30T14:47:55.050" LastActivityDate="2015-01-30T14:47:55.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="376" PostTypeId="4" CreationDate="2015-01-30T14:47:55.050" Score="0" Body="Questions pertaining to pressurized systems and their applications,  from containment to system losses to using as a benchmark for measurement." OwnerUserId="2015" LastEditorUserId="368" LastEditDate="2015-06-23T15:25:48.580" LastActivityDate="2015-06-23T15:25:48.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="377" PostTypeId="5" CreationDate="2015-01-30T14:49:25.847" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-01-30T14:49:25.847" LastActivityDate="2015-01-30T14:49:25.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="378" PostTypeId="4" CreationDate="2015-01-30T14:49:25.847" Score="0" Body="Aerospace engineering is the primary branch of engineering concerned with the research, design, development, construction, and testing of aircraft and spacecraft." OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-01-31T00:11:15.917" LastActivityDate="2015-01-31T00:11:15.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="379" PostTypeId="2" ParentId="372" CreationDate="2015-01-30T15:03:40.453" Score="15" Body="&lt;p&gt;There is heat that can be recaptured, but you won't get much of it away. As one of the commenters has mentioned, your absolute maximum is the Carnot efficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta_{Carnot}=1-\frac{T_c}{T_h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an idealized condition, you'll never reach this efficiency. But to find our limit, let's figure it out anyway. $T_c$ will just be room temperature, it might be slightly warmer inside the tower, but we'll give ourselves the benefit of the doubt and pick a nice round number at 20C (293K). $T_h$ will vary as the GPU works harder (this is one of the issues with this design in general; the power you get from the cooling system won't be consistent because GPU temperature varies depending on how much you're stressing the chip.) We don't want to run it too hot and damage the card, that defeats the purpose of a cooling system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some quick searching (Google &quot;GPU operating temperatures,&quot; you'll see a bunch of forum posts that give a lot of different numbers, none of which I think are strong enough to cite, but I'm pooling their data to make my own assumption) it looks like most cards have a strong upper limit of ~100C before you start to do serious damage. However, running that hot will still reduce the life of your card, and judging by the picture in the question, this is a nice card for which we've paid a pretty penny, and we want to keep it around as long as we can. 70C is a good place to shoot for, but 80C (353K) is still probably pretty safe, and we want our best possible case. With those numbers, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta_{Carnot}=1-\frac{293K}{353K}=0.17$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that, at the very maximum, the best we can do is to get 17% of the heat we're generating on the card back as electricity to power something in the tower. We can vary the card temperature, and as it goes between 60C and 100C, efficiency goes between 12% and 21%. Regardless, we're not getting a lot back. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the max efficiency though. &lt;a href=&quot;http://espressomilkcooler.com/teg-power-generator-thermoelectric-generator/&quot;&gt;This site&lt;/a&gt;, which sells thermoelectric generators, says that the top of the line TEGs will run at 8% efficiency. While this is better than the nothing that we'd have been getting before, the real issue here is cost and implementation. TEGs are not cheap, and cooling fans are. A basic cooling system is also much easier to install. Even if we can hook up a TEG to cool the card, we have to find something we can do with that electricity, and we don't want the variable power to be used for an critical components. Tower lights and extra fans are probably the extent of our usage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your actual question in there, I'm sure we can find all kinds of creative ways to get that heat converted into electrical or mechanical work. Making it &quot;useful&quot; is an entirely different story. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-01-30T15:03:40.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="380" PostTypeId="1" AcceptedAnswerId="381" CreationDate="2015-01-30T16:30:13.347" Score="6" ViewCount="158" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Otto_Lilienthal&quot;&gt;Otto Lilienthal&lt;/a&gt;, sometimes hailed as the first aviator, became famous for his many gliders, which pioneered new experiments in aeronautics. However, powered aircraft didn't take off - pun intended - until the Wright brothers successfully flew their powered airplane. Did Lilienthal ever experiment with non-human-powered flight? Wikipedia says he built a small engine, but doesn't say whether or not it was ever used in an aircraft.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-01-30T16:30:13.347" Title="Did Otto Lilienthal ever experiment with non-human-powered aircraft?" Tags="|engineering-history|aerospace-engineering|aircraft-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="381" PostTypeId="2" ParentId="380" CreationDate="2015-01-30T16:30:13.347" Score="5" Body="&lt;p&gt;I'm going to answer my own question because just before posting it I found an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lilienthal-museum.de/olma/e213.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Otto Lilienthal Museum&lt;/a&gt; has a comprehensive list of Lilienthal's designs. One is listed as the &quot;small wing-flapping machine.&quot; It didn't use a propeller (or jet engine, of course!) but instead used a small engine weighing about 22 pounds when fully fueled. Its wingspan was 22 feet, which was quite impressive. &lt;a href=&quot;http://en.wikipedia.org/wiki/Otto_Lilienthal&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; lists it as &quot;Small Ornithopter.&quot; &lt;a href=&quot;http://en.wikipedia.org/wiki/Ornithopter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another page&lt;/a&gt; calls it the &lt;em&gt;kleiner Schlagflügelapparat&lt;/em&gt;, which Google Translate translates to &quot;small flapping wing apparatus.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/d/d9/Otto_is_going_to_fly.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's&lt;/a&gt; a picture of Lilienthal with it, pre-flight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/d/d9/Otto_is_going_to_fly.jpg&quot; alt=&quot;Small ornithopter pre-flight&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:49.920" LastActivityDate="2015-01-30T16:30:13.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="382" PostTypeId="1" AcceptedAnswerId="385" CreationDate="2015-01-30T18:28:07.380" Score="5" ViewCount="647" Body="&lt;p&gt;The cabin noise of vehicles is &lt;a href=&quot;http://elevatingsound.com/the-top-30-quietest-cars-a-cabin-noise-test-by-auto-bild/&quot; rel=&quot;nofollow&quot;&gt;a large concern in the luxury market&lt;/a&gt;. Why haven't any automakers used the same techniques as are employed in &lt;a href=&quot;http://en.wikipedia.org/wiki/Noise-cancelling_headphones&quot; rel=&quot;nofollow&quot;&gt;active noise cancelling headphones&lt;/a&gt;.  In these devices a microphone is used as an in-loop sensor to suppress low frequency ($f\lt500\ \text{Hz}$) pressure fluctuations in the enclosed space near the ear.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why isn't similar technology used in vehicles?  What technical improvements are required over the standard vehicle sound system to implement it?  Given the increased volume of the system, what differences can be expected between vehicles and headphones?  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="282" LastEditDate="2015-02-09T22:58:42.840" LastActivityDate="2016-01-14T14:50:04.607" Title="Is active noise cancellation useful in vehicles?" Tags="|electrical-engineering|feedback-loop|audio-engineering|automotive-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="384" PostTypeId="2" ParentId="371" CreationDate="2015-01-30T19:03:46.670" Score="5" Body="&lt;p&gt;I'd like to add to what @Fred said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bronze wasn't the first. Before the Bronze Age, there was a comparatively brief &lt;a href=&quot;http://en.wikipedia.org/wiki/Copper_Age&quot; rel=&quot;nofollow&quot;&gt;Copper Age&lt;/a&gt; [also &lt;a href=&quot;http://en.wikipedia.org/wiki/Metallurgy_during_the_Copper_Age_in_Europe&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;].  Copper is comparatively abundant, and it sometimes naturally occurs in pure state (nuggets), as well as ores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some places, polymetallic ores were used for producing copper.  Early metalworkers noticed that the resulting &quot;copper&quot; had different properties.  I've used quotes, because it wasn't copper anymore: accidentally, it was bronze.  Subsequently, alloying materials were added on purpose.&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2015-01-30T20:48:19.683" LastActivityDate="2015-01-30T20:48:19.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="385" PostTypeId="2" ParentId="382" CreationDate="2015-01-31T00:28:05.860" Score="8" Body="&lt;p&gt;Some reasons why noise reduction in vehicle cabins is not a standard feature, yet:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As @Trevor Archibald states, safety is a very good reason. There is still a need to hear&#xA;some noises from outside the vehicle such as the sirens of emergency&#xA;vehicles: police, ambulance, fire fighters truck&lt;/li&gt;&#xA;&lt;li&gt;Hearing car horns from other drivers is still needed&lt;/li&gt;&#xA;&lt;li&gt;The sound of the engine lets people know if the engine is performing as&#xA;it should&lt;/li&gt;&#xA;&lt;li&gt;It's an added cost some people may not want to pay&lt;/li&gt;&#xA;&lt;li&gt;People haven't asked for it&lt;/li&gt;&#xA;&lt;li&gt;Most people don't object to hearing some noise, as long as it's not intrusive&lt;/li&gt;&#xA;&lt;li&gt;Insulating vehicle cabins against noise by using sound proofing&#xA;materials has suited most peoples needs until now&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It has been introduced in a small number of cars: &lt;a href=&quot;http://www.usatoday.com/story/money/cars/driveon/2012/09/26/cars-noise-cancellation-technology/1590703/&quot; rel=&quot;nofollow&quot;&gt;Auto Makers Shush Cars&lt;/a&gt;, but these are a bit more expensive than the average car. See also: &lt;a href=&quot;http://www.nydailynews.com/autos/ford-honda-noise-canceling-tech-new-cars-bmw-porsche-bring-noise-boosting-features-article-1.1168670&quot; rel=&quot;nofollow&quot;&gt;Cars Go Quiet&lt;/a&gt;, &lt;a href=&quot;http://www.autoblog.com/2013/12/22/bose-noise-cancelling-in-new-cars/&quot; rel=&quot;nofollow&quot;&gt;Bose Noise Cancelling in Cars&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, introducing electronic noise reduction technology in cars, could reduce the weight of cars by reducing the amount of sound proofing materials used, &lt;a href=&quot;http://news.harman.com/Press-Releases/Industry-first-HARMAN-Quiets-the-Car-by-Minimizing-Road-Noise-1ef.aspx&quot; rel=&quot;nofollow&quot;&gt;Harman Quiets the Car&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a different angle, in the 1980s electronic noise cancelling had been used to cancel the engine noise made by heavy vehicles used in inner urban development site to reduce the amount of noise heard by nearby residents.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="272" LastEditDate="2016-01-14T14:50:04.607" LastActivityDate="2016-01-14T14:50:04.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="386" PostTypeId="2" ParentId="372" CreationDate="2015-01-31T02:45:29.563" Score="13" Body="&lt;p&gt;Trevor Archibald has given you a really good answer, but I see from your comments a diffferent answer might be useful, as you still think this could be viable with the right economics.&lt;/p&gt;&#xA;&lt;p&gt;It wouldn't be. The issue is engineering, not economics. It's a bad idea from the economic perspective, certainly; but changing prices won't make it a good idea. It would still be a bad idea. Let me explain.&lt;/p&gt;&#xA;&lt;h3&gt;low-grade heat&lt;/h3&gt;&#xA;&lt;p&gt;Low-grade heat is heat that is a few Kelvin or tens of Kelvin above room temperature.&lt;/p&gt;&#xA;&lt;h3&gt;getting rid of the heat quickly is the name of the game&lt;/h3&gt;&#xA;&lt;p&gt;George Herold points you in a comment to one reason why energy harvesting on the card would be a bad idea: the thermal conductivity of the card is &lt;em&gt;designed&lt;/em&gt; to be high.&lt;/p&gt;&#xA;&lt;p&gt;Getting rid of heat quickly is particularly important in IT equipment, where the electrical efficiency of the equipment is really really staggeringly poor. And that means that of the electricity you put in, almost all of it is going to get turned straight into heat. There is a theoretical minimum amount of energy needed to flip a bit, regardless of the medium on which the bit is stored. All the rest of the energy put in above that minimum, is going to turn into heat straight away. In order to protect the equipment, you need to get rid of that heat as fast as possible.&lt;/p&gt;&#xA;&lt;p&gt;So the card is designed to get rid of heat as fast as possible. Anything you put in the way, such as your proposed energy harvesting device, will slow the rate at which heat leaves the card. That will raise the card's equilibrium temperature. And that will radically shorten the life of the card. That will happen regardless of the price of electricity.&lt;/p&gt;&#xA;&lt;h3&gt;it's not about the price of electricity&lt;/h3&gt;&#xA;&lt;p&gt;And this idea that if the price of electricity were high enough, it would make harvesting low-grade heat worthwhile, is simply wrong. If electricity is that valuable, then it's worth making the card more efficient in the first place, so that there was less waste heat: first, reduce the consumption of high-value energy, before trying to recycle low-value energy. And that brings me to ...&lt;/p&gt;&#xA;&lt;h3&gt;energy versus exergy&lt;/h3&gt;&#xA;&lt;p&gt;Heat is, in a large proportion of cases, a waste product. It's almost always the least useful form of energy. That's really what the Carnot efficiency limit is telling you: that to get any work out of low grade heat, you can only do so with very low efficiency; that is, almost all of the heat will stay as heat.&lt;/p&gt;&#xA;&lt;p&gt;When doing engineering with heat and other forms of energy, it's very useful to build up an intuition to distinguish between &lt;em&gt;energy&lt;/em&gt; (the thing measured in joules) and &lt;em&gt;exergy&lt;/em&gt; (the thing that gets work done). The form that energy is in, determines how much work it can do. Electricity can do huge amounts of work efficiently - it has very high exergy. Low-grade heat can do very little work - it has very low exergy.&lt;/p&gt;&#xA;&lt;p&gt;Once you've created low-grade heat, you're already at the end of the line for exergy (useful energy). Almost all uses of energy end up at low-grade heat. It's the final form for pretty much every chain of energy conversions. And, at the cosmic scale, it &lt;em&gt;is&lt;/em&gt; (as far as we can tell) the final form for every single joule, in the heat death of the universe.&lt;/p&gt;&#xA;&lt;p&gt;Low-grade heat is the end of the road. If you want more work out of those joules, then get that work done &lt;em&gt;before&lt;/em&gt; those joules are in the form of low-grade heat.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-01-31T03:59:10.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="387" PostTypeId="2" ParentId="370" CreationDate="2015-01-31T04:32:40.443" Score="4" Body="&lt;p&gt;To begin with I am assuming each of your horizontal surfaces: desktop and the three selves are each made the same material. To use a ridiculous and exaggerated case, the left half if the desktop is not heavy marble and the right side is not light weight balsa wood. The desktop is composed of one uniform material and each self is made of its own uniform material: wood, glass, metal, particle board and laminex, plywood, whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown, each of the shelves and the desktop are independently attached to the vertical supports that acts as the legs. Hence, the weight of each horizontal surface is directly transferred to the vertical supports. All horizontal surfaces are of uniform materials that have a uniform weight distribution. Consequently, each leg is carrying half of the combined weight of all the horizontal surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The section of each leg that experiences the full weight of what is above it is the short section between the two triangular braces: the one for the desktop and the other for the stand/foot of the desk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stress in each of these short sections of leg will be the weight carried by each leg divided the cross sectional area of the leg in the z-plane (breadth by width of the leg)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sloping part of the desktop brace will carry some of the weight of the desktop. Whereas, the short vertical part of the desktop brace carries all the weight of the vertical support above the desktop, the three shelves and some of the weight of the desktop. What the proportion of the desktop weight carried will be will depend on the perpendicular (normal) distance from the vertical support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise at the base of the leg, the triangular brace will redistribute the load in the foot according to your triangular configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a general overview of how to think about things relating to your design. As @Rick Teachey states, you really need to do a course in statics, get numbers for weights and cross-sectional dimensions of supports and plug it all into some formulae.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-01-31T04:32:40.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="388" PostTypeId="2" ParentId="179" CreationDate="2015-01-31T12:55:32.643" Score="10" Body="&lt;p&gt;If we simplify the whole bridge into 2D thin beam with a constant section size, no internal damping and subject only to small vertical deflections, then the natural frequency is determined by simple harmonic motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n_0 = \frac{1}{2 \pi} \sqrt{ \frac{ k } { m } } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $ n_0 $ is the natural frequency, $ k $ is the ratio between restorative force and deflection (the equivalent 'spring stiffness') and $m$ is the mass per unit length of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a beam the restorative force is the internal shear caused by the deflected shape. As the force exhibited by a beam is proportional to the rate of change of shear, which is related to the stiffness ($EI$) and the rate of change of moment &lt;a href=&quot;http://www.mae.ncsu.edu/eischen/courses/mae316/docs/FEBeamNotes.pdf&quot;&gt;it can be shown&lt;/a&gt; &lt;em&gt;(note: the deflection is proportional to the length of the beam)&lt;/em&gt; that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ k = \alpha \frac{ EI } { L^4 } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $E$ is the Young's Modulus of the beam material, $I$ is the Second Moment of Inertia of the beam section, $L$ is the length of the beam and $\alpha$ is a constant determined by the support conditions and mode number of the response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the literature I have seen expresses this in a way that  more convenient for the frequency equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ k =  \left( \frac{ K }{L^2} \right)^2 (EI) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting back in,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n_0 = \frac{ K }{ 2 \pi L^2 } \sqrt{ \frac{ EI } {m} } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the value of $K$ is quite involved, and there is an exact approach for simple solutions, and approximate methods including the free energy method and Raleigh Ritz. A few deviations for a &lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Vibrations/Natural_Vibrations_derivation.html&quot;&gt;simply supported beam can be found here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that this equation would have been enough, but as it requires a table for $K$ and the calculation of a value of $EI$ that represents the bridge as a homogenous beam, the authors of the Eurocode seem to have decided it would be better re-integrate the assumption that $k$ is constant along the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this they have used the following relationship:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta_0 = C \frac { w L^4 } { EI } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\delta_0$ is the maximum deflection, $C$ is a constant dictated by the support conditions, $w$ is a constant uniformly distributed load across the length of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under self-weight $w = gm$, where $g$ is acceleration due to gravity (9810 &lt;em&gt;mm/s&lt;/em&gt;&lt;sup&gt;2&lt;/sup&gt;; as deflection in this equation is given in &lt;em&gt;mm&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore (re-arranged:)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sqrt { \frac { EI } { m } } = L^2 \sqrt { 9810 } \frac { \sqrt { C } } { \sqrt { \delta_0 } } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ n_0 = \frac { 15.764 K \sqrt { C } } { \sqrt { \delta_0 } } $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General values for $K$ and $C$ can be found in structural tables- for example &lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Vibrations/Natural_Vibrations.html&quot;&gt;here&lt;/a&gt;, and &lt;a href=&quot;http://ruina.tam.cornell.edu/Courses/ME4730%20Fall%202013/Rand4770Vibrations/BeamFormulas.pdf&quot;&gt;here&lt;/a&gt;, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simply supported beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ K = \pi  ^ 2 \text{ and } C = \frac { 5 } { 384 } $$&#xA;$$ 15.764 K \sqrt { C } = 17.75 $$&#xA;$$ n_0 = \frac{ 17.75 } { \sqrt { \delta } } $$&lt;/p&gt;&#xA;" OwnerUserId="151" LastEditorUserId="368" LastEditDate="2015-07-22T00:02:17.890" LastActivityDate="2015-07-22T00:02:17.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="389" PostTypeId="1" AcceptedAnswerId="1734" CreationDate="2015-01-31T19:04:29.037" Score="12" ViewCount="1376" Body="&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&lt;br&gt;&#xA;In an automobile, only &lt;sup&gt;1&lt;/sup&gt;/&lt;sub&gt;3&lt;/sub&gt; of the potential energy in the fuel is converted into mechanical energy and significant portion of the energy is lost as heat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been previous attempts to recuperate this lost energy.  In the early 1990's, Porsche developed automotive thermoelectric generators (ATEG) which didn't go past prototyping stage. Currently, Porsche Motorsports is testing a thermal energy harvesting system in their LeMans series Race car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KbiMM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to Porsche's research, GM is in collaboration with Future Tech, LLC. to explore the idea of using themoelectric technology to harvest energy from internal combustion engines.  Other automotive manufactures, such as BMW, are also exploring this technology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently the power usage in a&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Small car is approximately 150 W&lt;/li&gt;&#xA;&lt;li&gt;Full size truck is approximately 500 W&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If this technology can successfully be implemented, then components such as the radiator, water pump, and alternator could effectively have reduced workload or removed from the system, thus reducing the load to the internal combustion engine.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;br&gt;&#xA;With the growing interest in green technology, are there technology barriers &lt;strong&gt;beside efficiency&lt;/strong&gt; that are preventing the implementation of energy harvesting from internal combustion engines using thermoelectric technology?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/197038/which-one-must-be-used-matched-output-voltage-or-open-circuit-voltage&quot;&gt;Which one must be used matched output voltage or open circuit voltage?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://energy.gov/sites/prod/files/2014/03/f13/stabler.pdf&quot; rel=&quot;noreferrer&quot;&gt;Benefits of Thermoelectric Technology for the Automobile&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.o-flexx.com/fileadmin/media/documents/chip_silver_edition_-_The_promise_and_problems_of_thermoelectric_generators.pdf&quot; rel=&quot;noreferrer&quot;&gt;The Promise and Problems of Thermoelectric Generators&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.academia.edu/3648232/Modeling_of_an_Automotive_ThermoElectric_Generator_ATEG_&quot; rel=&quot;noreferrer&quot;&gt;Modeling of an Automotive ThermoElectricGenerator (ATEG)&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.gizmag.com/thermoelectric-cars-improve-mpg/10928/&quot; rel=&quot;noreferrer&quot;&gt;Thermoelectrics to replace car alternators and improve MPG&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ijiset.com/v1s3/IJISET_V1_I3_47.pdf&quot; rel=&quot;noreferrer&quot;&gt;Thermo-Electric Generator in Turbocharged Diesel Engine&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.kettering.edu/news/recovering-wasted-energy&quot; rel=&quot;noreferrer&quot;&gt;Kettering University researchers are working with General Dynamics to convert the unused heat energy of their propulsion systems to useful and clean energy&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.thetruthaboutcars.com/2014/03/porsche-919-hybrid-lemans-racer-goes-after-the-two-thirds-of-gasolines-energy-thats-wasted-as-heat/&quot; rel=&quot;noreferrer&quot;&gt;Porsche 919 Hybrid LeMans Racer Goes After The Two Thirds of Gasoline’s Energy That’s Wasted As Heat&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ecomodder.com/forum/showthread.php/germans-trying-replace-alternator-thermoelectric-generators-tegs-3124.html&quot; rel=&quot;noreferrer&quot;&gt;Germans trying to replace Alternator with Thermoelectric Generators or TEGs&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.greencarcongress.com/2014/03/20140305-919.html&quot; rel=&quot;noreferrer&quot;&gt;Porsche 919 Hybrid LMP1 Le Mans prototype...&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Footnote&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://engineering.stackexchange.com/questions/372/energy-harvesting-using-thermoelectric-technology&quot;&gt;suggested duplicate&lt;/a&gt; is related, but still distinctly different.  The order of magnitude of energy available to recover from an internal combustion engine is significantly greater than within the GPU of a video card.  As such, the economies of scale are different and different solutions are therefore possible. &lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-12-04T02:29:49.500" Title="Thermoelectric Technology to Harvest Energy from Internal Combustions Engines" Tags="|mechanical-engineering|electrical-engineering|thermodynamics|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="390" PostTypeId="1" CreationDate="2015-02-01T00:04:12.877" Score="6" ViewCount="285" Body="&lt;p&gt;The power requirement for a sub system is 5V +/- 10% @ 1A of current. A &lt;a href=&quot;https://en.wikipedia.org/wiki/Request_for_quotation&quot; rel=&quot;nofollow&quot;&gt;RFQ&lt;/a&gt; was issued and a prospect vendor was selected. A first article of inspection (FAIR) has been submitted which includes 15 power supply samples. The samples have been tested with and without a 5 ohm resistor load using two separate calibrated DMM’s . Without the resistor the output voltage is 5V with little variation. Under a 5 ohm load the numbers are not too great. Below are the measurements (V):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.56, 4.71, 4.95, 5.12, 4.61, 4.83, 4.65, 4.69, 4.79, 4.81, 4.67, 4.74, 4.63, 4.74, 4.81&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should the vendor be approved as a supplier?&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="261" LastEditDate="2015-02-04T00:02:56.863" LastActivityDate="2015-06-30T21:45:29.817" Title="How can the vendor be approved/disapproved based on validation test data?" Tags="|electrical-engineering|statistics|quality-engineering|manufacturing-engineering|product-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="391" PostTypeId="2" ParentId="390" CreationDate="2015-02-01T02:11:39.377" Score="9" Body="&lt;p&gt;&lt;strong&gt;TL;DR: It depends, but probably not&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might argue that &quot;all 15 samples are within spec, so the supplier should be approved&quot;. Not so fast: depending on the parameters of your full production run, the 15 samples may or may not be statistically significant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There are numerous calculators online to do these calculations; I used &lt;a href=&quot;http://www.surveysystem.com/sscalc.htm&quot;&gt;this one&lt;/a&gt;. A good resource for the actual formulas is online at NIST's &lt;a href=&quot;http://www.itl.nist.gov/div898/handbook/prc/section2/prc242.htm&quot;&gt;Engineering Statistics Handbook&lt;/a&gt; or in any undergraduate Statistics textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the spec, we know that our confidence interval is 10% (two-sided). This means that any parts within 10% of nominal are acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we need to determine the confidence level. This is usually 95% but sometimes 99%, and represents &quot;how sure&quot; we can be of the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final piece of information we can give if it's known is the population size. In this scenario, this is the total number of parts to be ordered from the vendor over the lifetime of the product/process. Most calculators allow this to be left blank and, if missing, assume a large value, because the effect of the population size decreases as it grows in relation to the sample size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a 95% confidence level and our 10% confidence interval, with population left blank, we need a sample size of &lt;strong&gt;96 parts&lt;/strong&gt; in order to have a statistically significant result. Increasing to a confidence level of 99% requires a sample size of &lt;strong&gt;166 parts&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for industry-standard confidence levels, &lt;strong&gt;we cannot conclude that the vendor be approved based on the initial sample of 15 parts alone&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wait a minute—the entire FAIR run was within spec! Well yes, but what's to say the next 15 won't be out of spec? We don't know—that's why we have statistics! :-)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Well, under what circumstances &lt;em&gt;is&lt;/em&gt; our sample significant?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for illustration, I entered population values until I found a statistically-significant result at $n=15$: for 95% confidence level, $n=15$ samples would be significant only if the population (total production run) is 18 parts! For a confidence level of 99%, the situation is even worse: significance only if the production run is 16 parts!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other Notes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations above assume that the process follows a normal distribution and that the sample is representative of the population. In practice, both of these assumptions may be inaccurate.&lt;/p&gt;&#xA;" OwnerUserId="166" LastEditorUserId="166" LastEditDate="2015-02-01T05:07:37.777" LastActivityDate="2015-02-01T05:07:37.777" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="392" PostTypeId="1" AcceptedAnswerId="396" CreationDate="2015-02-01T07:04:55.137" Score="19" ViewCount="15587" Body="&lt;p&gt;Some traffic lights don't operate periodically but instead detect when a car is close by and then turns green. I have heard that they use a magnetic sensor embedded in the road to sense cars as they come near. Is this correct? Do they use other means as well?&lt;/p&gt;&#xA;" OwnerUserId="302" LastActivityDate="2016-04-17T09:37:05.180" Title="How does a traffic light sense the proximity of vehicles?" Tags="|electrical-engineering|civil-engineering|traffic-light|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="393" PostTypeId="2" ParentId="392" CreationDate="2015-02-01T07:31:00.230" Score="13" Body="&lt;p&gt;There are two techniques commonly in use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is induction loops cut into the road surface. The metal in a vehicle body induces a current in the loop, as the vehicle passes over it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Pyw8.jpg&quot; height=&quot;330&quot;&gt; (public domain pic from wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is cameras / radar (either Doppler microwave for moving vehicles, or infra-red for static vehicles at the stop line), typically mounted on top of the traffic light:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hnUhj.jpg&quot; alt=&quot;enter image description here&quot;&gt; (pic from the &lt;a href=&quot;http://webarchive.nationalarchives.gov.uk/20090104005813/http://www.dft.gov.uk/stellent/groups/dft_roads/documents/page/dft_roads_504814-01.hcsp&quot; rel=&quot;noreferrer&quot;&gt;archived UK DfT site&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other techniques, much rarer, that might include visible-light camera for special circumstances; and pneumatic loops for temporary lights. Oh, and someone somewhere has probably used piezo-electric sensors too.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-02-01T11:46:01.103" LastActivityDate="2015-02-01T11:46:01.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="394" PostTypeId="2" ParentId="392" CreationDate="2015-02-01T11:27:47.970" Score="5" Body="&lt;p&gt;The induction loop is the most reliable for switching on demand. Though it has the down side that motorcycles are more difficult to detect because they have less metal to detect. a similar issue exists for carbon fiber cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radar is typically only used to extend the green times as it can only detect moving vehicles reliably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some intersections use a camera and image recognition to trigger virtual detectors. These have the advantage that it can deal with a truck unloading right at the intersection (ignoring it and moving the spot to detect to next to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Public transport vehicles (trams and buses) can be equipped with transponders that log in special loops which allows the intersection controller to know which way the bus needs to go and give it green earlier and extend the green beyond the normal limit (or give it green at all).&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-02-01T11:27:47.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="396" PostTypeId="2" ParentId="392" CreationDate="2015-02-01T12:59:17.620" Score="30" Body="&lt;p&gt;As others stated before, induction loops are the primary - most reliable method: the coils (usually just several loops of wire) embedded in the road; fed given frequency from a generator, in presence of metal the frequency of the LC circuit changes and the sensor circuitry detects the change of frequency, producing a presence signal. In some cases these may fail to detect bicycles, but they are by far most common as they aren't affected by weather (or more precisely, the detection circuit tunes in to slow changes of frequency caused by weather) and are immune to accidental false positives. Note the loops can be localized (~2m size) or cover a lengthy part of a lane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Detection is performed by cards like these:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/jkprp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by induction loops made with wire laid in grooves like these:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/uBxKC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or placed in pipes under the road surface at construction time (in the photo is a loop for tram detection, but pre-built loops are similar)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nkN0j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Videodetection - cameras connecting to a specialized card with &quot;detection zones&quot; defined through specialized software detect the vehicles. They are vulnerable to bad weather and tend to produce false positives from glare of car headlights, shadows of vehicles on neighbor lane and such, but in certain cases - primarily where road surface makes installing detection loops impossible (gravel, or bad road surface) they are preferred. Additionally, the video detection cards are significantly more expensive than cards for detection loops. &lt;img src=&quot;https://i.stack.imgur.com/l8tW8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few lesser used techniques like geomagnetic (detecting changes in magnetic field; These largely depend on size of the vehicle, so a large truck can trigger a sensor in neighbor lane - but they are more durable), radar (detect only moving vehicles&lt;sup&gt;*&lt;/sup&gt; - but are frequently used to detect pedestrians as they rarely stay immobile), laser (measuring distance to road surface; vehicle in the way changes the distance measured. Quite reliable but only point-detection, no area detection).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pictured below is a geomagnetic sensor:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/klRtO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cs4UO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a radar sensors (short range for pedestrians and bicycles, and long range, for cars):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/IpbJd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/oKbmV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard of pneumatic and piezzoelectric, but I've never seen these in use for traffic control - probably problems of wear and durability; I know these are used for automated barriers for parking lots, but they obviously support an order of magnitude lower traffic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For city transport traffic, the vehicles are equipped with an on-board computer with a short-range radio (up to 500m) and GPS, and they broadcast messages about entering pre-defined &quot;checkpoints&quot; to the traffic system, alongside with data about intended turn direction, delay against schedule and some others, allowing the controller to prioritize. An alternative is a system that feeds vehicle position to a central unit, which then contacts controllers with messages about prioritizing these vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last but not least, cameras/sensors detecting strobe lights of specific frequency give immediate priority to oncoming emergency vehicles. (and take a photo of the vehicle in question, to prevent abuse.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Controllers can communicate with each other, and share their detector states, so two controllers can use each other's detectors, for example when they are a short way away from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two induction loops in a short distance (~1m) from each other are used to determine speed and length of vehicles, making adapting to longer or slower vehicles possible. Another application of pairs of detection loops near to each other is in directional detectors - basing on the order the neighbor loops are activated one can determine the direction the vehicle is moving. This is rarely used for cars but if a single rail line with trams (street cars) moving in both directions crosses a road, the same two pair of detectors can activate the green light for the vehicle and then register it finished crossing the street, regardless of its direction as the pairs can generate &quot;approaching / departing&quot; signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A special &quot;virtual&quot; detector composed of two loops in one lane in a considerable distanc measures the length of queue of cars, allowing prediction of time necessary to vacate the lane (and making &quot;time countdown displays&quot; viable.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another special type of detector is a &quot;blocking&quot; one, placed either in the middle of the crossing (camera) or behind it, on the &quot;departing&quot; lane (usually a detection loop); its purpose is to delay/block entry until the crossing is vacated, or prevent blocking the crossing if a traffic jam formed in the &quot;exit&quot; lane and new vehicles would be unable to depart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note this is the &quot;standard&quot; set, but since the controllers can accept a standarized 24V/'contact' signal, any generic source can be used, for example an infrared remote control to enable that one specific direction which is used in 0.1% cases, activated by the owner of the house with driveway right into the crossing, or by a manual trigger from a factory gate to enable a truck to enter/leave, or whatever need arises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a generic 16 inputs/16 outputs card. These are usually used for pedestrian buttons (and lamps) but they can provide signal from arbitrary sources and control arbitrary end-point devices.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/sxbgs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cities detectors work in &quot;pairs&quot; of two types; for example detection loops are very reliable for detecting vehicles, but mechanical stress from heavy transport can damage them, and repairing them is not a trivial matter. The card can detect a damaged loop (usually open circuit -&gt; no frequency or short circuit -&gt; very high frequency) and in such case the controller starts using a backup sensor, for example radar or laser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just a screenshot from one of the controllers showing the map with detectors displaying their state live (blue = active). Note that detector on the far right - it doesn't belong to this controller; it's composite data from a neighbor controller, so that the short road connecting the two doesn't get congested - as long as there are cars waiting in the potential congestion zone no more will be allowed into it from the other directions.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NjGXg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*&lt;/sup&gt;Note that while radar detectors can only detect cars in motion, that doesn't mean they can't be used as a standalone solution (&quot;just support&quot;). Sometimes the induction loops are placed at wrong locations as well (for various reasons, incompetence of the investor not the least of them), so cars stop behind/between them and don't trigger them during red light. This is still not a very big problem as any detector can be set as one with &quot;memory&quot;. Any vehicle even momentarily activating such a detector cause it to keep the active state until green light on the associated lane, then act as normal (&quot;forgetful&quot;) during the green light. Also note this is the default behavior for pedestrian pushbuttons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is not ideal, as a vehicle may get stuck right out of the detection zone exactly during the change from green to red, or (say, due to driver's fault) miss the whole green cycle altogether. Still, these are relatively rare cases, especially that another approaching vehicle will usually trigger the detector anyway.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2015-07-21T12:56:11.360" LastActivityDate="2015-07-21T12:56:11.360" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="397" PostTypeId="1" CreationDate="2015-02-02T00:29:14.023" Score="4" ViewCount="62" Body="&lt;p&gt;Various &lt;a href=&quot;http://en.wikipedia.org/wiki/Implant_(medicine)&quot; rel=&quot;nofollow&quot;&gt;materials are implanted&lt;/a&gt; within the human body as a result of of medical need.  Dental implants, stabilizing plates, arterial stents, and orthopedic replacements are all common examples where a material is implanted within the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical response of the immune system to any foreign body is to attack the implant.  If the immune system cannot eradicate the foreign body, then it attempts to encapsulate it instead.  This self-preserving attack from the immune system can lead to &lt;a href=&quot;http://en.wikipedia.org/wiki/Implant_failure&quot; rel=&quot;nofollow&quot;&gt;implant failure&lt;/a&gt;, and the likelihood of failure is, in part, dependent upon the bio-reactivity of the material.  For example, stainless steel has been evaluated to be less reactive than more base steel alloys, and titanium has been found to be less reactive than stainless steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is bio-reactivity characterized, and how is it measured over the conditions of a long term implantation?&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-02-02T03:45:37.840" Title="How is the long term reactivity of implant materials evaluated?" Tags="|biomechanics|biomedical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="398" PostTypeId="2" ParentId="397" CreationDate="2015-02-02T01:34:44.820" Score="4" Body="&lt;p&gt;Chronic animal experiment.  This is not the only stage of evaluation of the material, but it's a mandatory stage.  &lt;em&gt;Chronic&lt;/em&gt; means that the animal is survived after implantation for a period of time and observed.  Afterwards, the animal is sacrificed and autopsied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Humans have a strong immune system, compared to other animals.  So, you would want to use an animal that also has a strong immune system.  Pigs become likely candidates&lt;sup&gt;1&lt;/sup&gt;.  Dogs too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Indeed, since the immune response and repair functions in the body are so complicated it is not adequate to describe the biocompatibility of a single material in relation to a single cell type or tissue. Sometimes one hears of biocompatibility testing that is a large battery of &lt;em&gt;in vitro&lt;/em&gt; test that is used in accordance with &lt;a href=&quot;http://en.wikipedia.org/wiki/ISO_10993&quot; rel=&quot;nofollow&quot;&gt;ISO 10993&lt;/a&gt; (or other similar standards) to determine if a certain material (or rather biomedical product) is biocompatible. These tests do not determine the biocompatibility of a material, but they constitute an important step towards the animal testing and finally clinical trials that will determine the biocompatibility of the material in a given application, and thus medical devices such as implants or drug delivery devices.&lt;br&gt;&#xA;  [&lt;a href=&quot;http://en.wikipedia.org/wiki/Biocompatibility&quot; rel=&quot;nofollow&quot;&gt;exceprt from wiki&lt;/a&gt;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;Pig's also weigh roughly the same as humans.&lt;/sub&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; &lt;sub&gt;Nick worked on several projects developing electronics for surgical devices.  As a part of my duties, I assisted with a handful of animal experiments.  However, those devices weren't of an implanted sort.  Biocompatibility was not one of my direct responsibilities; the mechanical design team was responsible for it.&lt;br&gt;&#xA;An actual biocompatibility expert can provide this question deeper justice.  I'm writing this answer only as a fallback.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2015-02-02T03:45:37.840" LastActivityDate="2015-02-02T03:45:37.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="399" PostTypeId="2" ParentId="326" CreationDate="2015-02-02T04:34:10.970" Score="5" Body="&lt;p&gt;This is mostly an issue of safety trade-offs and not necessarily technical shortcomings. While a WIG vehicle, often referred to as a Ground Effect Vehicle (GEV), has improved efficiency it is also forced to fly &lt;strong&gt;very&lt;/strong&gt; low. Above around 50 feet (wingspan-dependent) you will not see much in the way of ground effect benefits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that, even over water, there are numerous obstacles that could be in a WIG vehicle's flight path. Commercial aircraft normally fly above 18,000 feet (5,500 meters) so that they stay in &lt;a href=&quot;http://en.wikipedia.org/wiki/Airspace_class&quot; rel=&quot;noreferrer&quot;&gt;Class A airspace&lt;/a&gt;. This allows them to remain &lt;strong&gt;far&lt;/strong&gt; above obstacles and it also allows air traffic control to deconflict air traffic and prevent dangerous proximity. There is no practical way to deconflict like this near the surface of the water where ubiquitous craft such as sailboats would pose deadly peril to a WIG vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it is very difficult to maneuver at such a low altitude. Aircraft can only safely pull upwards from a WIG vehicle flight altitude and that is the least-advantageous direction from an aircraft power perspective. Also, due to the low altitude, an obstacle may not be visible until very shortly before a possible impact; even on a clear day with good visibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for why hovercraft are more common: they are able to (and typically do) move at relatively slow speeds, and even stop on the water. They can slow down, stop, and make sharp turns more easily and more safely than a WIG vehicle. Due to their relatively slow speed, they also typically have more time to react to an emergency situation. Finally, hovercraft technology is much less complex and much less expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference: Aeronautical engineering experience and pilot experience&lt;/p&gt;&#xA;" OwnerUserId="131" LastActivityDate="2015-02-02T04:34:10.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="400" PostTypeId="1" CreationDate="2015-02-02T08:11:58.230" Score="9" ViewCount="975" Body="&lt;p&gt;Biogas slurry often contains sand and grit and organic matter that eventually sediments. Clearing thetank from sedimentation is expensive (plant downtime, additional equipment) and dangerous. So it would be interesting to know how much sediment there is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical tank heights are around 8m, so I think opening a hatch in the roof and poking with a long stick is not an option. What are other ways to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid level gauging is done with a pressure sensor mounted in the lower part of the wall, or a radar sensor on the roof.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-07-04T00:54:29.043" Title="How can I measure the thickness of a sedimentation layer in a closed tank?" Tags="|civil-engineering|process-engineering|sensors|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="401" PostTypeId="2" ParentId="400" CreationDate="2015-02-02T09:06:16.917" Score="4" Body="&lt;p&gt;You can put a rod with a blades on the bottom rotate it along the bottom. when the resistance is high then the sediment has reached above the blades and it is time to clean the tank.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-02-02T09:06:16.917" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="403" PostTypeId="2" ParentId="1" CreationDate="2015-02-02T10:55:49.627" Score="6" Body="&lt;p&gt;There is a key difference between liquids and powders:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Liquids have internal cohesion and can be sucked, but with a powder, you need a carrying medium.  &lt;/li&gt;&#xA;&lt;li&gt;The internal friction in a powder will usually be far higher than in a liquid.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In practice, this means that to pump a powder, you inject air or another gas as fluidizing agent. The pumped or blown air carries the powder, like a dust-storm within the pipe.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorDisplayName="user16" LastEditDate="2015-02-02T12:50:04.930" LastActivityDate="2015-02-02T12:50:04.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="404" PostTypeId="2" ParentId="400" CreationDate="2015-02-02T14:45:57.883" Score="4" Body="&lt;p&gt;If you want this automated, probably Rachet's solution is better; for a more low-tech approach for periodic manual testing, a wide, flat weight on a rope will work; lower it below water surface, let it rest on top of the sediment; the moment the rope goes loose, you've touched the sediment surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more hi-tech solution, an ultrasound distance sensor will be helpful, submerged under water surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're not interested in measuring the level so much as determining whether the sediment has crossed a certain critical level, a valve installed in the wall at given height will allow you to pick a small sample of whatever is at the valve level, helping you determine if it's covered by the sediment yet.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-02-02T14:45:57.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="405" PostTypeId="1" CreationDate="2015-02-02T14:52:52.810" Score="19" ViewCount="1850" Body="&lt;p&gt;What difference in education occurs between two graduates, one with an Engineering Technology degree (be it Civil, Mechanical, Electrical...) and the other with the equivalent Engineering degree &quot;proper?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming both degrees are 4 year degrees and both are able to become PEs, how do potential employers feel about the different degrees? Is there a difference in pay?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What options does the graduate with an ET degree have to advance his/her education? For example, can one earn a masters with an ET degree?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for advantages/disadvantages of either degree in terms of employability and career progress, from the point of view of an engineer who has hired people with these backgrounds.&lt;/p&gt;&#xA;" OwnerUserId="3" LastEditorUserId="368" LastEditDate="2015-03-19T21:04:25.460" LastActivityDate="2022-04-14T20:09:01.657" Title="What are the major differences between Engineering degrees and Engineering Technology degrees in terms of employability?" Tags="|education|licensure|" AnswerCount="4" CommentCount="0" ClosedDate="2022-04-15T01:40:26.317" ContentLicense="CC BY-SA 3.0" />
  <row Id="406" PostTypeId="1" AcceptedAnswerId="425" CreationDate="2015-02-02T16:13:17.497" Score="3" ViewCount="3175" Body="&lt;p&gt;It's not exactly my area of expertize and I've spent a couple hours ears deep in norms, and I'm not really sure if what I found is correct (the figure I found was to the order of 1$\mu$m).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When using standard (off-the-shelf, not specialized) threaded pin / bolt and a matching nut for a lead-screw actuator, what degree of imprecision (free lengthwise movement of the actuator without the shaft turning) can I expect coming from the allowed technical tolerances of the thread?  In other words, what amount of play can I expect between a threaded bolt and nut?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific,  I realize there will be movement from lateral and longitudinal flexibility of the shaft; from its mounting bearings; from its axial twist against the drive; and probably quite a few other sources, but I'd like to know this specific factor - how much of a &quot;give&quot; or play can there be in a thread, allowing movement of the nut along the bolt without either of them turning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that this changes with the shaft diameter, so a table or a method to calculate the amount of play for most common metric threads would be preferred for an answer.  If that's too difficult or doesn't exist, I'm most interested in the value for M10 thread.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorDisplayName="user16" LastEditDate="2015-02-04T00:02:59.730" LastActivityDate="2016-02-28T06:28:03.200" Title="What are the longitudal tolerances of thread on metric nuts and bolts?" Tags="|mechanical-engineering|tolerance|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="407" PostTypeId="2" ParentId="405" CreationDate="2015-02-02T17:16:47.967" Score="15" Body="&lt;p&gt;One is training to be an engineer and one is training to be an engineering tecnologist.&lt;/p&gt;&#xA;&lt;p&gt;The best way to see the differences are to look at the course curriculum for each:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.purdue.edu/ECE/Academics/Undergraduates/UGO/pdf/BSEE%20Sample%20Plan.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purdue BS EE Plan of Study&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://tech.purdue.edu/sites/default/files/ECET-fall-2011.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purdue BS EET Plan of Study&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.purdue.edu/ME/Academics/Undergraduate/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purdue BS ME Plan of Study&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://tech.purdue.edu/new-albany/degrees/mechanical-engineering-technology/plan-of-study&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purdue BS MET Plan of Study&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;both are able to become PEs&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This depends on your local rules for becoming a PE. Also note that being a PE isn't required for a lot of ME &amp;amp; EE engineering jobs &lt;em&gt;in the USA&lt;/em&gt; because they fall under industrial exemption. Canada and other countries where Engineer is a protected title have different rules (in Canada Engineering Technologists with a BTech degree can apply to get PEng license which gives them the engineer title mentioned above).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I was hoping to get answers from the point of view from an engineer who has hired people with these backgrounds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I haven't directly hired both but I've interviewed both and personally the question comes down to what is needed. To me it's like asking if you should go to a Dentist (DDS) or a Medical Doctor (MD), they both share the word &amp;quot;doctor&amp;quot; but they are different jobs with different training.&lt;/p&gt;&#xA;&lt;p&gt;If I need a technician I wouldn't hire an engineer. If I needed an engineer I wouldn't hire a technician.&lt;/p&gt;&#xA;&lt;p&gt;Like all jobs and majors the longer you are out the less it matters as knowledge learned on the job overshadows schooling. So 5-10 years down the road both could have the potential to end up in the same spot.&lt;/p&gt;&#xA;&lt;p&gt;Anecdotally engineers make more than technicians but it still comes down to company and location.&lt;/p&gt;&#xA;&lt;p&gt;Edit: I want to point out that I'm using technician/technologist interchangeably here. I've heard it both ways but I know some areas may be separate jobs. &lt;a href=&quot;http://en.wikipedia.org/wiki/Engineering_technologist&quot; rel=&quot;nofollow noreferrer&quot;&gt;The wiki on technologist&lt;/a&gt; is pretty good.&lt;/p&gt;&#xA;&lt;p&gt;And don't pick your answer based on which one makes more. It's irrelevant since I know a lot of people that have tried to force their way through one way or the other and ended up hating their education/jobs because of it. I would explore both and see which one you like better.&lt;/p&gt;&#xA;" OwnerUserId="231" LastEditorUserId="-1" LastEditDate="2022-04-13T10:45:16.403" LastActivityDate="2022-04-13T10:45:16.403" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="408" PostTypeId="2" ParentId="126" CreationDate="2015-02-02T18:05:50.310" Score="3" Body="&lt;p&gt;There are few generally accepted methods to use to reduce EBOM cost in electronic industry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Group EBOM components using one or both of the following criteria&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Type of component, (example: capacitors, microprocessor, inductors, resistors, transistors etc)&lt;/li&gt;&#xA;&lt;li&gt;Vendor or Manufacture  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then start looking for common components. These could be simple as looking for such 1uF value capacitors. Don’t forget to look for components in the adjacent in value. There might be an opportunity in the design to use a larger or small component without affecting the design. Then look for cheaper alternatives. This also helps to obtain volume based discounts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method is to group the components by price, start from most expensive components to the least expensive. The tabulate your results in a Pareto chart and focus on the top 20% of the components. Use the 80/20 rule. This concept commonly used in Supply Chain for insurance purposes which states 80% inventory cost is in 20% of the high value inventory. Therefore, cost reduction from the top 20% of component (cost base) will have 80% impact on the overall cost reduce effort. This is a concept that is hard to grasp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at component size is also a good idea. Consider changing larger 1206 or 0805 type SMT components to smaller 0402 or 0201 from factor components. Make sure the product design is not compromised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes the PCB board is a good candidate for too. If the current design is 12 layers then look at possibility of reducing the number of layers without impacting the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have module such as a blue tooth, WIFI or ZIGBEE module consider replacing the module by discrete components. In this case you might have perform test EMC testing to verify if the system meets regulatory requirements&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For cost reduction purpose outside of the EBOM&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the SMT and other manufacturing process such as Test. Using DFM &amp;amp; DFT strategies there is a significant opportunities to reduce product cost. SMT machines have a limited capacity, therefore strategic parts placement can gain higher throughput reducing product cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt; There is no golden rule, other than professional work experience and seeking help from peers.&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="282" LastEditDate="2015-02-07T15:37:31.060" LastActivityDate="2015-02-07T15:37:31.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="409" PostTypeId="2" ParentId="400" CreationDate="2015-02-02T18:35:20.427" Score="3" Body="&lt;p&gt;It sounds like the problematic solids level is/can be known ahead of time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming this is the case, you could install a vibratory apparatus at the bottom of the tank, and some sort of acoustic sensor at the level of interest (probably several of them positioned around the tank wall would work best). Since tranverse/secondary waves don't travel through liquid, you will know the sensor is covered by solids if it detects these kinds of waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;CHALLENGES: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensor would have to be sufficiently separated or damped from the tank itself so it does not detect waves propagating through the tank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensors may not work well when the sedimentation layer is very very loose (as it probably would be near the very top); the vibrating might help with this by &quot;compacting&quot; the sediment, however. &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2015-02-02T22:32:40.403" LastActivityDate="2015-02-02T22:32:40.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="410" PostTypeId="1" AcceptedAnswerId="2899" CreationDate="2015-02-02T19:03:47.180" Score="10" ViewCount="1188" Body="&lt;p&gt;I am trying to build an optical sensor for my old school, state property, standard french watt-o-meter. EDF ( Electricité de France ) has a gig which turns a wheel depending on watt consumption. The higher the wattage, the faster a tiny disk turns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the disk can be read by eye, so here is my question: why not use a hacked optical mouse to gather data ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contraints are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;can't open the gig&lt;/li&gt;&#xA;&lt;li&gt;the wheel is 1.5 cm away from a protective glass&lt;/li&gt;&#xA;&lt;li&gt;I have an optical mouse&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The CMOS has a half milimeter opening, and a 2mm lens from the standard casing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What optical system do I need to read half milimeters from 15 milimeter distance using a classic optical mouse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody help me calculate the correct lenses for this, or point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="313" LastEditorUserId="261" LastEditDate="2015-02-04T00:02:38.663" LastActivityDate="2015-05-21T15:06:45.503" Title="What is the focal distance of an optical mouse?" Tags="|optics|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="411" PostTypeId="2" ParentId="1" CreationDate="2015-02-02T21:58:23.987" Score="3" Body="&lt;p&gt;In food process you can &quot;pump&quot; powder with vacuum, sugar works very well, and some other products need sometimes some technical adjustment (pression or open/close cycles valves) .&#xA;But water in the pipes is ban  and last point it's only for small quantities and short distance 20meters max i think.&lt;/p&gt;&#xA;" OwnerUserId="322" LastActivityDate="2015-02-02T21:58:23.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="414" PostTypeId="1" CreationDate="2015-02-02T23:26:48.550" Score="8" ViewCount="306" Body="&lt;p&gt;Electro-mechanical systems typically used in robots have very low delays and are quite simple to build servo controllers for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydraulic systems are much less common in robotics and have delay due to the length of the hydraulic lines. This delay may vary considerably depending on the system conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What methods exist for dealing with this variable delay when designing a position or velocity servo controllers for hydraulic cylinders?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;I have spent some time checking academic literature and have found the following books:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Introduction to Time-Delay Systems - Fridman&lt;/li&gt;&#xA;&lt;li&gt;Nonlinear Control Under Nonconstant Delays - Bekiaris-Liberis and Krstic&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have not had a chance to read these enough to have a clear answer to my problem yet. Hopefully this can provide others with a starting point at least. Both authors have many papers on topics regarding control with time delay systems that are nonlinear or varying.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditDate="2015-03-20T07:18:41.070" LastActivityDate="2015-03-20T07:18:41.070" Title="Controlling hydraulic cylinders with variable delay" Tags="|mechanical-engineering|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="415" PostTypeId="2" ParentId="400" CreationDate="2015-02-03T02:02:02.337" Score="6" Body="&lt;p&gt;&lt;strong&gt;Buy a &lt;a href=&quot;http://www.amazon.com/s/?ie=UTF8&amp;amp;keywords=depth+finder&amp;amp;tag=googhydr-20&amp;amp;index=aps&amp;amp;hvadid=49843134385&amp;amp;hvpos=1t1&amp;amp;hvexid=&amp;amp;hvnetw=g&amp;amp;hvrand=1988141966228925601&amp;amp;hvpone=&amp;amp;hvptwo=&amp;amp;hvqmt=e&amp;amp;hvdev=c&amp;amp;ref=pd_sl_1zgna5kll0_e&quot;&gt;depth finder&lt;/a&gt;!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @SF pointed out, you can use an ultrasonic device to measure the depth of the water.  These &lt;a href=&quot;http://en.wikipedia.org/wiki/Fishfinder&quot;&gt;devices work&lt;/a&gt; by sending out an acoustic impulse and recording the return signals.  The acoustic impulse reflects off of any density boundaries in the medium.  By measuring the time for the signal to come back the distance from the transducer can be inferred by knowing the acoustic speed in the medium.  If the sediment is very dense then you will only get a signal from the sediment boundary, but if the sediment is not so dense (like muck on the bottom of a lake) then you will get a signal from both the sediment boundary and the bottom of the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might think that this sounds like an expensive device to design and build, but they are mass produced for boaters.  Almost every lake-going boat I've ever been on has one.  They range in price from &lt;a href=&quot;http://www.amazon.com/s/?ie=UTF8&amp;amp;keywords=depth+finder&amp;amp;tag=googhydr-20&amp;amp;index=aps&amp;amp;hvadid=49843134385&amp;amp;hvpos=1t1&amp;amp;hvexid=&amp;amp;hvnetw=g&amp;amp;hvrand=1988141966228925601&amp;amp;hvpone=&amp;amp;hvptwo=&amp;amp;hvqmt=e&amp;amp;hvdev=c&amp;amp;ref=pd_sl_1zgna5kll0_e&quot;&gt;\$50 to \$350 on Amazon&lt;/a&gt; depending on features and accuracy.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-02-03T02:02:02.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="416" PostTypeId="1" AcceptedAnswerId="419" CreationDate="2015-02-03T02:24:17.837" Score="3" ViewCount="142" Body="&lt;p&gt;On espresso machines, there is some kind of pipe joint which allows the steam wand to move while maintaining its seal. What is this component called?&lt;/p&gt;&#xA;" OwnerUserId="327" LastEditorUserId="282" LastEditDate="2015-02-15T19:38:11.373" LastActivityDate="2015-02-15T19:38:11.373" Title="What is the name of the part that allows a steam wand to have a moveable joint?" Tags="|mechanical-engineering|steam|piping|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="417" PostTypeId="1" CreationDate="2015-02-03T02:48:51.363" Score="20" ViewCount="16261" Body="&lt;p&gt;I am making a small (desktop) wind tunnel for educational purposes, I want to have 10 fairly thick smoke-streams about 3cm apart. I have experimented with incense but the stream is not thick enough and barely visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of using a pipe with holes drilled in it to get the 10 smoke-streams, the low pressure of the airflow drawing the smoke out through the pipe from a container with the burning substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I burn that will produce a nice thick white smoke and is non-toxic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a different/better way to get non-toxic 'smoke' without burning anything?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the pipe a good way to distribute the smoke or are there any other/better techniques to getting the multiple streams?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="368" LastEditDate="2015-02-27T20:16:14.543" LastActivityDate="2017-06-27T01:40:21.927" Title="How to make smoke for a small wind tunnel?" Tags="|aerospace-engineering|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="418" PostTypeId="2" ParentId="406" CreationDate="2015-02-03T02:53:25.580" Score="3" Body="&lt;p&gt;Standard threads are classified for accuracy by a &lt;em&gt;tolerance class&lt;/em&gt; (You can see a bit about the metric thread fit classes at &lt;a href=&quot;http://www.amesweb.info/Screws/IsoMetricScrewThread.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.amesweb.info/Screws/IsoMetricScrewThread.aspx&lt;/a&gt; .) The screws you find at your local hardware store will probably be a relatively rough tolerance class, meaning that the threads are designed to have some gap between them, and also that there will be a large variation in the dimensions of each individual bolt or nut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately for your question, the requirements of the tolerance classes are about the diameter of the crests and roots (and therefore truncation) of each thread, but not about the accuracy of the pitch (how many threads are in any given unit length.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there isn't a great theoretical way to predict the backlash for a thread that wasn't really designed for use as a leadscrew. I think your estimate of 1μm is pretty optimistic though. For point of reference, a well maintained manual milling machine might have as little as .001&quot; (~25μm) and that's using a specifically designed leadscrew with a low-backlash design. A less pristine machine could easily have as much as .010&quot; (~.25 mm) of backlash. If your design is using a smaller diameter leadscrew, there will be some improvement, but the tolerances designed in standard threadforms will make it worse. So your backlash would probably be on the order of magnitude of .1mm - 1mm for common leadscrew sizes (with 10mm being at the lower end of that range.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are strategies for dealing with backlash, including a 'split nut' where two separate nuts are coupled together and with either spring tension or fine adjustment forced into contact with opposite edges of the bolt thread. The other strategy is to always approach the target from the same side. If this is an option for you, it is the simplest way to solve the problem of backlash without any materials cost.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-02-03T14:13:12.677" LastActivityDate="2015-02-03T14:13:12.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="419" PostTypeId="2" ParentId="416" CreationDate="2015-02-03T03:21:16.313" Score="4" Body="&lt;p&gt;It is called a &quot;steam wand gasket&quot;. Technically, it's a ball-and-socket joint made out of two concave &lt;a href=&quot;http://en.wikipedia.org/wiki/Polytetrafluoroethylene&quot; rel=&quot;nofollow&quot;&gt;PTFE&lt;/a&gt; (Teflon) gaskets and an &lt;a href=&quot;http://en.wikipedia.org/wiki/EPDM_rubber&quot; rel=&quot;nofollow&quot;&gt;EDPM&lt;/a&gt; gasket for the seal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be other designs too; I learned this by watching &lt;em&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=fulyy9A9CYw&quot; rel=&quot;nofollow&quot;&gt;Saeco StarBucks Barista - How to Replace Steam Wand Seals and Gasket&lt;/a&gt;&lt;/em&gt; (YouTube video).&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="330" LastEditDate="2015-02-03T11:38:20.773" LastActivityDate="2015-02-03T11:38:20.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="420" PostTypeId="1" CreationDate="2015-02-03T03:32:43.320" Score="7" ViewCount="134" Body="&lt;p&gt;I'd like to build an exploratory robot that can fit within confined spaces&lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What traction mechanisms should I consider to propel a robot along the inside of a 4&quot; pipe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pipe to be traversed will likely have fluid (assume water) within it to varying degrees from partially full to completely full.  I'm initially planning on tethering the robot with coaxial cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I have considered a 3 and 4 point wheeled approach, such as with a &lt;code&gt;Y&lt;/code&gt; or &lt;code&gt;X&lt;/code&gt; configuration of wheels.  And I have considered treads instead of wheels.  The robot will need to be able to navigate 90&lt;sup&gt;o&lt;/sup&gt; turns on the way in, but can be hauled out via the coaxial cable if need be.  Preferably, it will be able to back its way out for the majority of the distance it can cover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The robot would not need to climb unassisted.  The initial use will be in a pipe that drops several feet from the access point, and then slopes down approximately 5 feet over a 60 foot run.  Retrieving the robot will be assisted by the coaxial cable or a tether line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;It's a rather boring application, actually.  I want to see how badly the roots are creeping into the sewer line from my house.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-02-03T16:15:35.487" LastActivityDate="2015-02-03T21:19:24.140" Title="Traction mechanisms for a robot in confined spaces" Tags="|mechanical-engineering|propulsion|systems-design|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="421" PostTypeId="2" ParentId="420" CreationDate="2015-02-03T04:06:38.827" Score="4" Body="&lt;p&gt;First, I'll say that from personal experience, locating and implementing treads on a robot that size can be VERY difficult. I was building a robot that was supposed to be 8&quot;x8&quot;x20&quot; (WxHxL), including the treads, and we had a lot of trouble finding usable treads that would be small enough but still hold up to some more intense use than Lego or K'nex robots. If you can find treads, you still need to be able to tension them and prevent them from traveling axially along the guide wheels (i.e. slipping off while the robot is driving.) So my first recommendation is to go with wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as getting traction I would consider some sort of rudimentary suspension system, or inflated or soft rubber tires that have some give in them. Basically, something that allows you to design the robot to have a resting diameter of 4.5&quot;, but then compress to 4&quot; to fit inside the pipe, put a higher normal force on each wheel, and increase the friction between the tire and the pipe. (Those numbers are examples, I'd do some more work to get an actual determination of ID, 4&quot; probably is the OD, and to determine how much flex you can have. And be careful, too much friction will prevent the robot from actually moving.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other thoughts: I like the Y configuration the best, and you may actually want the top wheel to be smaller, just to stabilize the robot, and not provide any driving power through it. Consider how you're going to get power to the wheels if they're not on the same axis. You may have to power each direction of wheels independently, or your gearing may get complex. And you may not be able to offset the power transmission primarily off the axis, there may be radial translation as well. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-02-03T04:06:38.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="422" PostTypeId="2" ParentId="417" CreationDate="2015-02-03T06:40:22.013" Score="16" Body="&lt;p&gt;The testing of ventilation systems and of building air-tightness is done using &lt;a href=&quot;https://www.bsria.co.uk/instrument/sales/leakage-testing/smoke-generators/&quot; rel=&quot;noreferrer&quot;&gt;smoke sticks or similar&lt;/a&gt; - non-toxic smoke generators that produce no ash residue. One of these may be suitable for your needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smoke generators typically use mixes of two or more of: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;alcohol &lt;/li&gt;&#xA;&lt;li&gt;glycol &lt;/li&gt;&#xA;&lt;li&gt;glycerol &lt;/li&gt;&#xA;&lt;li&gt;water  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I suppose strictly speaking they're fog generators rather than smoke generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a wide variety of parameters to chose from.  There are fast-dispersing mixes, and lingering ones. There are generators that run off batteries, and some that run off mains power. Some generate high flows, some low flows. Some start generating within seconds of being switched on, others take a few minutes before they start producing the fog (smoke).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kzPX4.jpg&quot; width=&quot;300&quot;&gt;&lt;br&gt;&#xA;&lt;sub&gt;(Drager Flow Check - source as above)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-02-03T17:20:40.413" LastActivityDate="2015-02-03T17:20:40.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="423" PostTypeId="1" CreationDate="2015-02-03T08:16:40.050" Score="2" ViewCount="751" Body="&lt;p&gt;I have a truck which weighs 190 kg and will travel 100m down a 38° slope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How I can find the trucks speed at the end of the 100m? &lt;/p&gt;&#xA;" OwnerUserId="331" LastEditorUserId="2819" LastEditDate="2016-03-15T05:43:40.870" LastActivityDate="2016-03-15T05:43:40.870" Title="Final speed of a truck on a slope?" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="424" PostTypeId="2" ParentId="423" CreationDate="2015-02-03T10:04:38.237" Score="9" Body="&lt;p&gt;This is a constant acceleration problem on an inclined plane which ignores friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force acting on the truck is the force of gravity. The acceleration the truck will experience will be solely due to gravity, $9.8\text{ m/s}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the 38 degrees for the slope you have to find the component of gravity/acceleration going down the slope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the truck will travel 100 m down the slope and the acceleration in the plane of the slope that you calculated, you have to rearrange the following equation to get the time required to travel the distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s=ut+0.5at^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial velocity, $u$, will be zero if the truck starts rolling from a standing start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the final velocity of the truck you use the following equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v^2=u^2+2as$$&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="106" LastEditDate="2015-02-04T00:02:33.427" LastActivityDate="2015-02-04T00:02:33.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="425" PostTypeId="2" ParentId="406" CreationDate="2015-02-03T10:40:07.740" Score="2" Body="&lt;p&gt;The link provided by Ethan48 added a lot of informations.&lt;/p&gt;&#xA;&lt;p&gt;Tolerance classes:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Note : * 6g is normally selected for commercial external (bolt) threads.&lt;/p&gt;&#xA;&lt;p&gt;Note : ** 6H is normally selected for commercial interrnal (nut) threads.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tJLVA.jpg&quot; alt=&quot;ISO metric screw external thread root contour according to ISO 965-1 for positions e, f, g&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The amount of movement allowed for external thread would be determined by axial counterpart to (radial) &lt;span class=&quot;math-container&quot;&gt;$T_{d2}/2 + es/2$&lt;/span&gt;. With thread angle of 60 degrees that would be 3/4 the value, and with movement in either direction, the value should be doubled, so the part of backslash from external thread would be &lt;span class=&quot;math-container&quot;&gt;${3 \over 4} (T_{d2} + es)$&lt;/span&gt;. The part for internal thread would be analogous, &lt;span class=&quot;math-container&quot;&gt;${3 \over 4} (T_{D2} + EI)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;As result, the &lt;em&gt;backslash&lt;/em&gt; = &lt;span class=&quot;math-container&quot;&gt;${3 \over 4} (T_{d2} + T_{D2} + es + EI)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let's take a common commercial M10x1 bolt with tolerance of 6g and a matching 6H nut. That's 1mm pitch.&lt;/p&gt;&#xA;&lt;p&gt;According to iso 965-1:&lt;/p&gt;&#xA;&lt;p&gt;es = 26 &lt;span class=&quot;math-container&quot;&gt;$\mu m$&lt;/span&gt;&lt;br&gt;&#xA;EI = 0&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T_{D2}$&lt;/span&gt; = 150 &lt;span class=&quot;math-container&quot;&gt;$\mu m$&lt;/span&gt;&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$T_{d2}$&lt;/span&gt; = 112 &lt;span class=&quot;math-container&quot;&gt;$\mu m$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${3 \over 4} \cdot (150+112+26+0) = 216 \mu m$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, for typical M10x1 commercial screws most pessimistic backslash is 0.216mm. In reality it should be significantly lower as the parts are always manufactured with some safety margin from the normative limits (manufacturers aiming at halfway the allowed range), and even small mismatch (imprecision) of thread pitch between the thread and the bolt further reduces the backslash. (and of course, over time, wear will make it worse.)&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-02-03T10:40:07.740" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="426" PostTypeId="2" ParentId="356" CreationDate="2015-02-03T11:16:04.187" Score="4" Body="&lt;p&gt;And so, as promised, I'll do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First assumption:&#xA;Let's work with 1m long strips of gutter. It'll be easier to calculate everything starting from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the gutter is already full of ice. (We'll work on the ice filling problem later on)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard gutter size (according to &lt;a href=&quot;http://stormmaster.com/gutter-sizing/&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;) is 5-inch K-Style, or 6-inch half round. If we use the half round version, we can &lt;a href=&quot;http://www.wolframalpha.com/input/?i=6%20inch%20diameter%201m%20long%20cylinder%20volume%20in%20Liters&quot; rel=&quot;nofollow&quot;&gt;learn&lt;/a&gt; it holds around 9L of ice. The &lt;a href=&quot;http://www.wolframalpha.com/input/?i=9Liter%20of%20ice&quot; rel=&quot;nofollow&quot;&gt;latent heat of fusion&lt;/a&gt; of 9L of ice is 3000kJ, or 833Wh. This means that if you wanted to melt this ice in an hour, you'll need 833 Watts of power. For each meter of gutter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what's the available solar energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second assumption: Let's say the sun shines all day, with the perfect orientation regarding the gutter, and the gutter absorb all the energy it receives from the sun. Let's assume it's the shortest day of the year, at the US mean latitude (around 38°N).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://pveducation.org/pvcdrom/properties-of-sunlight/calculation-of-solar-insolation&quot; rel=&quot;nofollow&quot;&gt;the third chart&lt;/a&gt;, the length of day on the shortest day of the year is around 9h30m. Let's round this up to 10h (I like round and easy numbers for my ballpark calculations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross section of our gutter &lt;a href=&quot;http://www.wolframalpha.com/input/?i=6%20inch%20by%201m%20long%20rectangle%20area%20in%20m%C2%B2&quot; rel=&quot;nofollow&quot;&gt;is around&lt;/a&gt; 0.15m². The sun irradiance reaching the ground stands at around 1000W/m². This means around 150W reaches our gutter. On a 10h day course, that would be 1500Wh. Hey, that would be enough! Well, yes, if you take your gutter and put in the perfect orientation regarding the sun. Which it's not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the energy received will be lower. Moreover, one also has to take into account the efficiency of the energy conversion. High quality solar thermal collector (which collect solar radiation and convert it to heat) typically have efficiency at around 60%. This means, in our case (where the efficiency will be lower, we'll receive at best 900Wh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take into account the fact that our gutter has a fixed orientation, the energy received will be even lower than that. Thus, we won't have enough energy to melt the ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this data, I'd say it's not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the ice filling the gutter. The problem is still the same. Making sure that the water flowing from the roof stays hot enough will still means providing enough energy to keep it above freezing. Also, usually, water melts on the roof, but doesn't flow alone, i.e. it brings down some snow and ice in the gutter, which you'll have to melt of you want to prevent ice buildup in the gutter.&lt;/p&gt;&#xA;" OwnerUserId="154" LastActivityDate="2015-02-03T11:16:04.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="427" PostTypeId="2" ParentId="405" CreationDate="2015-02-03T12:57:50.553" Score="4" Body="&lt;p&gt;IMHO Engineering Technology degrees focus on Application engineering whereas engineering degrees are very scientific based. A review of the web links from @JedF support this fact&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://web.archive.org/web/20150919163657/https://engineering.purdue.edu/ECE/Academics/Undergraduates/UGO/pdf/BSEE%20Sample%20Plan.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purdue BS EE Plan of Study&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://web.archive.org/web/20140813191504/https://tech.purdue.edu/degrees/ms-technology-eet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purdue BS EET Plan of Study&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;A quick compare of the above two curriculum's indicate the inclusion of course like &lt;strong&gt;Product and Program Management&lt;/strong&gt; in EET curriculum and an absence of such courses in its counterpart. Based on my interaction with peers following EET programs it was quite evident the EET curriculum focuses more on laboratory base hands on study approach whereas the contemporary EE programs were more scientific base curriculums.&lt;/p&gt;&#xA;&lt;p&gt;These support the claim that EET programs are aligned with application engineering where as the contemporary EE programs are aligned with industries requiring deep technical analysis. Both programs add value to the engineering discipline.&lt;/p&gt;&#xA;&lt;p&gt;From an employability stand point for the &lt;strong&gt;most part&lt;/strong&gt; EET degree candidates excel in hand on base work environment where as contemporary EE degree candidates excel in research oriented work environments. Good example would be that EET degree based would perform a required resistance calculation to an accuracy of 2 decimal places and use a 10% tolerance resistor where as a EE degree based engineer would do a same calculation to 10 decimal places and use a 1% tolerance resistor. Both get the job done but slightly different approaches.&lt;/p&gt;&#xA;&lt;p&gt;From an education standpoint there are many Masters in Electrical Engineering Technology programs. One such is the &lt;a href=&quot;https://web.archive.org/web/20140813191504/https://tech.purdue.edu/degrees/ms-technology-eet&quot; rel=&quot;nofollow noreferrer&quot;&gt;Masters in Electrical Engineering Technology Program at Purdue University&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Unfortunately, contemporary EE degree candidates attain a slightly higher starting salary but there are many EET degree engineers who have obtain MBA’s and have gone on to hold executive leadership positions in reputable organizations.&lt;/p&gt;&#xA;&lt;p&gt;At the end of the day what matters most is not the letter after your name, but what you do with those skills. There are many with and without letter after the name that has made a significant impact to the engineering discipline as well as society. Some of them are&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Steve_Jobs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Steve Jobs&lt;/a&gt; – Apple&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bill_Gates&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bill Gates&lt;/a&gt; – Microsoft&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Michael_Dell&quot; rel=&quot;nofollow noreferrer&quot;&gt;Michael Dell&lt;/a&gt; - Dell&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Richard_Branson&quot; rel=&quot;nofollow noreferrer&quot;&gt;Richard Branson&lt;/a&gt; – Virgin&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I don’t own any apple products, but when I learnt the passing away of Steve Jobs it gave me goosebumps. People left flowers in front of Apple stores. To me that was a confirmation of the impact Steve Jobs had made in the society.&lt;/p&gt;&#xA;&lt;p&gt;Besides obtaining an engineering degree it is also important to&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Develop soft skills&lt;/li&gt;&#xA;&lt;li&gt;Develop Leadership skills&lt;/li&gt;&#xA;&lt;li&gt;Develop Communication skills&lt;/li&gt;&#xA;&lt;li&gt;Have the ability to work in a team&lt;/li&gt;&#xA;&lt;li&gt;Be respectful and considerate toward others&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The above skills will have a significant impact on your career after you obtain your educational credentials.&lt;/p&gt;&#xA;&lt;p&gt;If you can spare 75 minutes, I suggest watching this youtube video &lt;a href=&quot;https://www.youtube.com/watch?v=ji5_MqicxSo&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Last Lecture&lt;/strong&gt; by Professor Randy Pausch&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="6767" LastEditDate="2022-02-25T22:16:01.607" LastActivityDate="2022-02-25T22:16:01.607" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="428" PostTypeId="2" ParentId="62" CreationDate="2015-02-03T13:56:24.067" Score="3" Body="&lt;p&gt;There are two aspects that you need to consider here: One is physical measurement of luminance, luminous flux, or illuminance (I'll come on to the differences in a moment), and the other is the much less clear-cut issue of human perception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's deal with the physics first. There are three concepts to explain, which are all related:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Luminous flux&lt;/strong&gt; is the total amount of light emitted by a light source, in all directions. Its SI unit is lumens. This is difficult to measure without expensive equipment (an &lt;a href=&quot;https://en.wikipedia.org/wiki/Integrating_sphere&quot; rel=&quot;nofollow&quot;&gt;integrating sphere&lt;/a&gt;). Luminous flux is relevant because it gives a measure of the total amount of light that a source is putting into a room, and will typically be given in the specifications of a light source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Illuminance&lt;/strong&gt; is the amount of light that falls on a surface of unit area. Its SI unit is lux, where 1 lux = 1 lm/m&lt;sup&gt;2&lt;/sup&gt;. It is easy to measure with commonly available &quot;light meters&quot; and photodiodes. Illuminance is the most important value when asking, for example, &quot;will enough light fall on this desk for somebody to work by&quot;. Building codes or standards often specify the illuminance required on floors, walls, or work surfaces, for specific tasks. This is a function of the overall design of lighting in the space rather than just of the light source - type of fixture, quantity and spacing, colour of the walls, etc., are all important -  but marketing materials for some LED light sources in particular will claim &quot;equivalent to xxx wattage incandescent&quot; by comparing the illuminance on a small area of work surface from an omnidirectional incandescent lamp vs a narrow-beam LED, and imply that they are equivalent - despite them having radically different effects everywhere else in the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Luminance&lt;/strong&gt; is a measure of the amount of light being &lt;em&gt;given off&lt;/em&gt; by a surface &lt;em&gt;in a specific direction&lt;/em&gt;, per unit of area. If that surface is the surface of a light source - e.g. if you look at a light bulb - then this is related to how bright the source appears. This measurement is relevant because if a customer looks directly at a range of light sources to compare &quot;how bright they are&quot;, it is differences in luminance that they will see. Luminance meters are much more expensive than illuminance meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Luminous flux is the total amount of light given off by a source&lt;/li&gt;&#xA;&lt;li&gt;Illuminance is how brightly a surface is lit&lt;/li&gt;&#xA;&lt;li&gt;Luminance is (part of) how bright the light source appears to the eye&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which one(s) you need to measure will depend on exactly what you are trying to achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Human perception&lt;/strong&gt; is the tricky part here. There is more to a feeling of &quot;brightness&quot; than either of the measures above. For example, people tend to perceive light of a higher colour temperature as being brighter than light of a lower colour temperature - a common issue when comparing LEDs to other sources. People are also strongly influenced by contrast; we mostly see relative rather than absolute brightness, so a patch of light on a small area of workspace in an otherwise dark room may seem brighter than the same illuminance throughout the space, especially once the viewer's eyes have adapted to the dark. So &lt;em&gt;be very careful&lt;/em&gt; if giving practical demonstrations, or you may find that the customer's perception does not match your calculations.&#xA;(it's also worth noting, of course, that for some purposes it's the perception that matters...)&lt;/p&gt;&#xA;" OwnerUserId="320" LastEditorUserId="320" LastEditDate="2015-02-04T11:50:21.007" LastActivityDate="2015-02-04T11:50:21.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="429" PostTypeId="2" ParentId="423" CreationDate="2015-02-03T15:04:19.403" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/424/88&quot;&gt;Fred's answer&lt;/a&gt; solves this problem using classical mechanics/kinematics. However, you can also solve this type of problem using an energy approach (which I have always found easier to do - Newton's laws just feel more intuitive). &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Newton's first law (conservation of energy)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The baseline potential energy (zero) is the truck's starting position, and the final energy state is the ending position. Assuming the truck comes to a stop at the top, the total energy - all of which is potential energy - of the truck at the ending position will be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$PE_f=M_{truck}\cdot g\cdot H$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H=D\cdot\tan\theta$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's first law (conservation of energy) says $KE_i = PE_f$ (assuming zero friction), so the formula to find the initial velocity needed for the truck to arrive at the top is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$KE_i=\frac{1}{2}m v_i^2=PE_f$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{i}=\sqrt{\frac{2PE_f}{m}}$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Newton's second law ($F=ma$)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The time of travel is found using Newton's second law ($F=ma$). The acceleration is the change in velocity over time. The final velocity is zero, so we know the change in velocity from the equation above ($\Delta v=v_i$). We also know the force $F$ acting on the truck is its mass times gravity. Therefore: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F=ma\rightarrow mg=m\frac{\Delta v}{\Delta t}\rightarrow \Delta t=\frac{v_i}{g}$&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-02-03T15:28:39.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="430" PostTypeId="5" CreationDate="2015-02-03T16:23:39.120" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-03T16:23:39.120" LastActivityDate="2015-02-03T16:23:39.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="431" PostTypeId="4" CreationDate="2015-02-03T16:23:39.120" Score="0" Body="For questions about how certain features of an aircraft affect its performance and function, such as engine type or wing configuration." OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-02-04T00:02:13.777" LastActivityDate="2015-02-04T00:02:13.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="432" PostTypeId="2" ParentId="414" CreationDate="2015-02-03T16:42:35.090" Score="4" Body="&lt;p&gt;You build the delay into your system. [This first requires your system to be repeatable. If your delay varies then this method won't work.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would collect lab data and map all of your actuators and their respective delays with the hydraulic line length, etc. Say you have 4 hydraulic actuators and the results show the following delays.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Act1: 30 ms&lt;/li&gt;&#xA;&lt;li&gt;Act2: 50 ms&lt;/li&gt;&#xA;&lt;li&gt;Act3: 200 ms&lt;/li&gt;&#xA;&lt;li&gt;Act4: 1 ms&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you wanted Act3 and Act4 to run at the 'same time' you would artificially add a 199ms delay to Act4. The timing of the events would look like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&amp;lt;0ms: Determine what you want to do. Move Act3 and Act4 by a certain amount.&lt;/li&gt;&#xA;&lt;li&gt;0 ms: Send the Act3 signal.&lt;/li&gt;&#xA;&lt;li&gt;199 ms: Send the Act4 signal&lt;/li&gt;&#xA;&lt;li&gt;200 ms: Act3 and Act4 move at the same time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is similar to how moder compression ignition fuel injectors are calibrated. Each actuator has a slightly different delay based on multiple environmental factors. By mapping out the different delays you can repeatably inject fuel when required.&lt;/p&gt;&#xA;" OwnerUserId="231" LastActivityDate="2015-02-03T16:42:35.090" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="433" PostTypeId="1" CreationDate="2015-02-03T16:58:00.103" Score="7" ViewCount="226" Body="&lt;p&gt;I'd like to build an exploratory robot that can fit within confined spaces&lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the robot moves along the inside of a 4&quot; pipe, it's likely to encounter some obstacles that need to be removed.  Typically, that will be tree roots up to 1/2&lt;sup&gt;&quot;&lt;/sup&gt; thick.  There could also be globs of tissue paper caught by the roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority of the pipe is made from clay, but a small portion of the pipe is cast iron.  The portion that has the obstacles is clay and was laid in the early 1960's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pipe to be traversed will likely have fluid (assume water) within it to varying degrees from partially full to completely full.  I'm initially planning on tethering the robot with coaxial cable and possibly an additional line like a steel cable.  The propulsion mechanism is &lt;a href=&quot;https://engineering.stackexchange.com/q/420/16&quot;&gt;still being defined&lt;/a&gt;, but I expect that the robot will have some ability to fix itself in place while clearing out an obstacle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical way of removing roots and other obstacles is through using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Drain_cleaner#Electric_drain_cleaners&quot; rel=&quot;nofollow noreferrer&quot;&gt;drain cleaning system&lt;/a&gt;.  The cutting head is usually a pronged piece of metal that can burrow through the obstacle.  The problem with this approach is that the pipe is damaged as part of the cleaning process and some amount of roots are left behind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another commercial approach that completely clears the line utilizes a spinning wire brush.  The downside of this approach is that pipe wall is severely damaged as part of the cleaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to devise a cutting mechanism that doesn't significantly damage the pipe but also removes the majority of the obstacles encountered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type of cutting, shearing, grinding, etc... mechanism should I consider in order to clear out the path in front of the robot as it explores the line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;It's a rather boring application, actually.  I want to see how badly the roots are creeping into the sewer line from my house.  Ideally, I'd like to clear out the roots too without damaging the clay pipe. &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-03-26T04:40:12.983" Title="Cutting mechanism for a robot in confined spaces" Tags="|mechanical-engineering|systems-design|cutting|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="434" PostTypeId="2" ParentId="420" CreationDate="2015-02-03T20:58:24.477" Score="4" Body="&lt;p&gt;You are looking to build a sewer robot, so you maybe want to look at other sewer robots. I have no diret experience with designing or handling sewer robots, but I can talk about the ones I saw at trade fairs.&lt;br&gt;&#xA;Some builds have sets of 3 or 4 wheels with a mechanism that pushes the weels against the pipe walls, or  - more typically - sets of 2 wheels. In the first case, the wheels are quite small. In the latter case, the wheels are sometimes shaped so that when looking at the robot from the front the cross section is more or less circular. &lt;a href=&quot;http://www.wasseraktiv.at/vorsorgen/pruefen/kanal-tv/650,kanal-roboter.html&quot; rel=&quot;nofollow&quot;&gt;Find a handful of pictures of the seconfd type here&lt;/a&gt;. The first type, with an X shape, &lt;a href=&quot;http://prokasro.de/en/products/uv-technology-en/view/productdetails/virtuemart_product_id/54/virtuemart_category_id/8&quot; rel=&quot;nofollow&quot;&gt;you can see here&lt;/a&gt;. I think it's far larger than 4&quot;. Not that this is not a classical sewer robot but for some pipe building application that I don't understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One small robot I have seen had a small pivotable arm at the front with a light source and a camera. At a branching in the pipes, the robot could push the arm into one of the branches. When the robot moved forward, it would be slide into this branch because of the arm.&lt;br&gt;&#xA;I've never seen tracked robots that would fit into a 4&quot; pipe - again, don't take my word for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt; Most hits for the term sewer robot where on german sites. Maybe the actual english term is different, and only germans insist on calling them sewer robots. &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2015-02-03T21:19:24.140" LastActivityDate="2015-02-03T21:19:24.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="435" PostTypeId="1" AcceptedAnswerId="436" CreationDate="2015-02-03T23:43:30.253" Score="21" ViewCount="8273" Body="&lt;p&gt;I recall reading in an old issue of &lt;em&gt;Model Railroader&lt;/em&gt; an article about railroad bridges.  In it, the author mentioned that you shouldn't have a model suspension bridge for a railroad track on your layout because no such arrangement would be prototypical.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his own words, &quot;suspension bridges and trains don't mix.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that the author probably wasn't an engineer (at least in the sense of this site), but his remark did get me thinking.  I am somewhat of a railroad buff, and I can't think of a suspension bridge for trains - the closest thing that comes to mind are the trolley tracks that were on the Brooklyn Bridge&lt;sup&gt;*&lt;/sup&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that there wouldn't be many suspension bridges for trains, since other bridge types would be better suited (cheaper, easier to build, etc.) but why are they so rare?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*&lt;/sup&gt;&lt;sub&gt;Bear in mind, though, I live in the US, and am pretty unfamiliar with foreign railroad infrastructure.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="32" LastEditorUserId="368" LastEditDate="2015-03-13T20:23:34.783" LastActivityDate="2017-01-14T18:17:35.127" Title="What makes suspension bridges unsuitable for railways?" Tags="|civil-engineering|bridges|rail|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="436" PostTypeId="2" ParentId="435" CreationDate="2015-02-04T01:20:03.500" Score="23" Body="&lt;p&gt;There are a few main reasons why suspension bridges aren't used for railroads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason is that suspension bridges are typically used where very long spans are needed. Trains are very heavy, especially when compared to lanes of highway traffic.  This means that long spans require very strong support structures, which in the case of suspension bridges are cables and towers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second reason goes along with the first; trains cause high dynamic loads as they move along the rail.  This can increase the vertical loads by 30%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third is that trains don't really have suspensions, especially freight trains. This means that any movement in the bridge itself has little opportunity to be dampened before it reaches the train. Suspension bridges are relatively flexible by design which makes transferred motion even more of a problem. You do not want the bridge to be rolling under a train!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of these are impossible engineering issues to overcome. But by the time that you have accommodated all of them, you might as well find a different location or build a truss bridge instead.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorDisplayName="user16" LastEditDate="2015-02-04T02:02:44.190" LastActivityDate="2015-02-04T02:02:44.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="437" PostTypeId="1" AcceptedAnswerId="439" CreationDate="2015-02-04T03:00:14.113" Score="8" ViewCount="1240" Body="&lt;p&gt;I'm a mechatronics engineering student currently on workplacement. I've put forward a research proposal to develop an improved mechanical gripper used for automated product handling. The company likes the proposal and wants to go ahead with it, but they want more detail in the proposal, specifically a budget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little lost in this sense, I'm not sure where to start. I'm not sure how much a prototype will cost, nor how many I will need. I have considered taking a small percentage of the overall losses caused by the current solutions but even those seem unjustifiably large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From other professional engineers experience, what would be the usual method to develop a budget.&lt;/p&gt;&#xA;" OwnerUserId="340" LastEditorUserId="368" LastEditDate="2015-02-13T23:46:30.307" LastActivityDate="2015-02-13T23:46:30.307" Title="How do I form a budget for a mechanical research project?" Tags="|mechanical-engineering|project-management|research|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="438" PostTypeId="1" CreationDate="2015-02-04T04:48:03.410" Score="11" ViewCount="1602" Body="&lt;p&gt;Given moisture content, Specific gravity of solids, initial volume, and weight. I'm asked to compute the moist unit weight, dry unit weight, and degree of saturation of this compacted soil. This is already done. This compacted soil sample was then submerged in water. After 2 weeks, it was found that the sample had swelled and its total volume had increased by 5%. Then I'm asked to compute the new unit weight and moisture content of the soil sample after 2 weeks of submersion in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moisture content and total volume are known to change, but what properties remain constant during the submersion? Can S(r) be taken as 1?&lt;/p&gt;&#xA;" OwnerUserId="343" LastEditorUserId="88" LastEditDate="2015-02-10T11:24:36.200" LastActivityDate="2015-02-10T11:24:36.200" Title="Soil swelling and mass/volume relationships" Tags="|civil-engineering|geotechnical-engineering|soil|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="439" PostTypeId="2" ParentId="437" CreationDate="2015-02-04T05:08:15.423" Score="13" Body="&lt;p&gt;It's important to remember that this budget is part of a proposal; it is inherently not set in stone, so it does not have to be perfect. You do not have to identify every expense that you will incur during the course of the project, but you need to show that you have considered, identified, and can estimate the major costs of the project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have one already, I'd highly recommend creating a timeline (possibly in the form of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Gantt_chart&quot;&gt;Gantt chart&lt;/a&gt;, for which you should be able to find multiple Excel templates or other programs to create one.) It sounds like you have a decently clear goal in mind (make sure you know what you mean by &quot;improved,&quot; though that's mostly unrelated to this question.) But you also need to have a good idea of how you'll get to that end result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some questions I would consider asking yourself&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What stages will this project require? (Research, design, building, testing, etc.)&lt;/li&gt;&#xA;&lt;li&gt;How long will it take me to complete each stage? OR How long can I allocate to each stage? (The second may be more relevant if you have a strict deadline and certain things that you know will take X amount of time [shipping times, special orders for unique parts, etc.])&lt;/li&gt;&#xA;&lt;li&gt;What could prevent me from staying on schedule, and how likely are these events? (Out of stock parts, design challenges that may cause problems)&lt;/li&gt;&#xA;&lt;li&gt;What will I need for each stage, and will this incur any cost? &lt;/li&gt;&#xA;&lt;li&gt;Are any of these expenditures particularly risky, and how does this affect the budget? (Components that are liable to break during assembly or testing, travel costs that may be volatile) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You don't have to have the answers to all of these questions, but you should consider them. Some of them are inherently unanswerable; some you will only be able to answer after years of experience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you create this general outline for the project, you should start to see where the budget naturally comes into existence. Obviously you'll need to plan for building prototypes. But are these off the shelf parts, or are they things that will need to be made new, and if so, how do you plan on making them? Is the test you're going to run potentially destructive, necessitating multiple prototypes? These are fairly obvious questions, but consider some slightly less obvious ones, especially if you're focused solely on the new product. What kinds of things is this going to grab, and will you need samples for the testing process? This is an existing tool that you're improving, will acquiring a current iteration help you to understand it better, and will the company provide that or do you have to buy it? Are there similar products in existence from other manufacturers that could help you out? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you have to look at the project from as many angles as possible. And my first point is important enough to re-iterate. This is a proposal, you are absolutely allowed to change it halfway through the project, assuming you have a compelling reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, note that this is how I approach a budget, and how I've done it on projects I've managed or participated in in the past. But this may not be exactly what this company wants. Don't be afraid to ask your supervisor or whoever you're presenting the budget to what they're looking for. You're a student, they know that, they don't expect you to know how to do everything. You're there to learn, so take advantage of that.&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-02-04T05:08:15.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="440" PostTypeId="2" ParentId="119" CreationDate="2015-02-04T09:09:23.960" Score="1" Body="&lt;p&gt;In the field of civil engineering, I don't think I've ever had comments on &quot;selling&quot; my ideas. There is certainly scope in my field for people of all levels to make technical suggestions, and those that are good get implemented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't say there is much &quot;selling&quot; to be done within internal civil engineering teams though. There is normally a problem to be solved, there may be meetings to discuss it, and the actions come out of the meeting. I get the impression that in your field you are more of a manufacturing/production engineering and hence only good ideas get prototyped? I don't think I can find an equivalent in civils.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-02-04T09:09:23.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="441" PostTypeId="2" ParentId="435" CreationDate="2015-02-04T10:03:53.617" Score="8" Body="&lt;p&gt;Light rail, and a floating rather than a suspension bridge, however I suspect many of the issues are the same:&#xA;&lt;a href=&quot;http://bulletin.pbworld.com/volumes/2014_04/taking_light_rail_over_floating_bridge.aspx&quot;&gt;http://bulletin.pbworld.com/volumes/2014_04/taking_light_rail_over_floating_bridge.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The bridge destined to host light rail is the Homer M. Hadley Memorial Bridge, one of the longest floating bridges in the world, which currently carries the Interstate 90 highway over Lake Washington. Two of the bridge’s existing lanes will be replaced with dual-track light rail in order to take the region’s light rail system over Lake Washington as part of Sound Transit’s East Link extension from Seattle to Redmond. While passenger rail systems have been developed that cross cable support structures and encounter multi-dimensional movements, there is no other bridge carrying passenger rail that accommodates the wide range and magnitude of movements that will be required of the Hadley Bridge.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it seems that development is ongoing in this area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Sorry, not enough reputation to add as a comment yet, but I thought it might be of interest)&lt;/p&gt;&#xA;" OwnerUserId="157" LastActivityDate="2015-02-04T10:03:53.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="443" PostTypeId="1" AcceptedAnswerId="445" CreationDate="2015-02-04T12:47:47.763" Score="33" ViewCount="25224" Body="&lt;p&gt;Some building are really tall, if you flush the toilet and the contents go into a pipe and straight down, there could be a lot of energy, potentially enough to cause harm to the sewer pipe at the end of the fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that in my home, the pipe goes straight down and then there is just a 90 degree bend. According to Wikipedia the calculations for &lt;a href=&quot;http://en.wikipedia.org/wiki/Terminal_velocity&quot;&gt;Terminal velocity&lt;/a&gt; have a lot of variables, but in essence things that are falling get to top speed quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When solids are falling straight down a drain pipe in a high rise building, how fast do they fall? What is the potential for the things that are falling to damage the pipes and how is this addressed when engineering the structure?&lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="368" LastEditDate="2015-05-01T20:01:55.893" LastActivityDate="2017-11-18T11:27:28.587" Title="How fast does solid waste fall in vertical drain pipes?" Tags="|structural-engineering|piping|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="444" PostTypeId="1" AcceptedAnswerId="692" CreationDate="2015-02-04T12:53:34.027" Score="5" ViewCount="1658" Body="&lt;p&gt;Bridges in the UK can be assessed (to standard BA34/90) using an HB vehicle (to standard &lt;a href=&quot;http://www.standardsforhighways.co.uk/dmrb/vol1/section3/bd3701.pdf&quot; rel=&quot;nofollow&quot;&gt;BD37/01&lt;/a&gt;). The HB vehicle is 16 wheel loads, each of 2.5kN times &quot;units&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For concrete bridges the number of units is normally defined as 45 units at Ultimate Limit State (ULS) and 0 units at Serviceability Limit State (SLS); although the number of units for SLS is 30 if the bridge spans transversely. Reference: BS5400-4:1990 Clause 4.2.2 as modified by BD24/92.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For steel bridges there doesn't seem to be an equivalent clause. So how many units should be used at ULS, and how many at SLS?&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2015-02-16T12:38:42.543" LastActivityDate="2015-02-16T16:17:42.013" Title="How many units of HB are used for steel bridges?" Tags="|civil-engineering|steel|bridges|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="445" PostTypeId="2" ParentId="443" CreationDate="2015-02-04T14:08:55.770" Score="34" Body="&lt;p&gt;&lt;em&gt;Please note&lt;/em&gt;: I'm not a building designer by trade, but I have had to investigate related questions for other reasons.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll let you in on a dirty secret about sanitation lines within buildings - The biggest concern is not about how fast things are flying, it's about maintaining air pressure and providing adequate ventilation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ecobuilding.org/code-innovations/case-studies/Heriot_Watt_University_reportJan2007fromStudorwebsite.pdf&quot; rel=&quot;noreferrer&quot;&gt;This guide&lt;/a&gt; presents a bit of a historical background to tall building design with respect to sanitation system design.  &lt;a href=&quot;http://www.cibse.org/getmedia/56352fa8-370c-4943-9638-77920939fe3b/The-control-of-air-pressure-within-tall-building-drainage-SoPHE-ppt-Compatibility-Mode.pdf.aspx&quot; rel=&quot;noreferrer&quot;&gt;This presentation&lt;/a&gt;&#xA;provides some more recent perspective and goes into some of the research that is refining current building standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traditional viewpoint is that waste forms an annular flow and has a terminal velocity in the range of 3 - 5 &lt;sup&gt;m&lt;/sup&gt;/&lt;sub&gt;s&lt;/sub&gt;.  Current research casts some doubt on those assumptions.  The contention is that drainage tends not to be steady state. And with multiple streams contributing to the outflow, the actual velocity may vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as mentioned, the real concern is making sure the air flow balances out and that water traps don't blow.  People aren't as concerned about what happens to the waste on its way out so long as they don't have to smell the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, for tall buildings, a secondary vent stack is used to provide adequate ventilating air to the fixtures in use.  Current building codes in the US and UK specify sufficiently wide drainage pipe such that sufficient balancing air flow will always be available.  The second presentation I linked goes into detail regarding different drain pipe sizes and considerations for very tall buildings such as skyscrapers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NXtIwm.jpg&quot; alt=&quot;secondary ventilation system&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The secondary provides sufficient air ventilation so that water traps at the lower levels don't get blown out by the pressure waves generated from drainage coming from the upper floors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach that's used is to separate the different building levels into different drainage zones.  This allows for smaller diameter pipe to be used while minimizing the risk that lower level water traps will get blown out from upper level drainage.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/MIXZ0m.jpg&quot; alt=&quot;zoned drainage&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yet another approach that's used is the use of &quot;Positive Air Pressure Attenuators&quot; (PAPA) such as the following.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/lO5Akm.jpg&quot; alt=&quot;pressure relief valves&quot;&gt;&lt;br&gt;&#xA;&lt;sub&gt;Note, this happens to be just one vendors product portfolio.  Other vendors for these devices exist.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One challenge with PAPAs is that their status with current building codes in tall buildings is unclear.  I personally have installed PAPAs in my residence and can attest to their abilities.  However, building codes are understandably slow to update and the long term reliability of PAPAs need to be demonstrated.  Again, from my own experience, using PAPAs requires modifying the system design from traditional approaches and requires additional considerations.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-02-04T14:08:55.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="446" PostTypeId="2" ParentId="443" CreationDate="2015-02-04T14:14:08.683" Score="19" Body="&lt;p&gt;This is not a factor in practice: Water (and entrained solids) directed into a vertical pipe segment adhere to the sides of the pipe.  I.e., there is no free-fall in a vertical waste pipe.  Rather, liquids swirl around the side of the pipe (producing a self-cleaning effect known as scouring).  Even if you managed to introduce a dense solid with insufficient water flow it would enter at an angle and bounce off the sides of the pipe, not approaching its terminal velocity.&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2015-02-04T14:14:08.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="447" PostTypeId="2" ParentId="437" CreationDate="2015-02-04T16:13:39.210" Score="4" Body="&lt;p&gt;I'd go about this in a different way than &lt;a href=&quot;https://engineering.stackexchange.com/a/439/61&quot;&gt;Trevor&lt;/a&gt;. I'd start with a bill of materials and other things you need. Material is obviously the physical stuff, other things means your time and time of others, lab time etc.&lt;br&gt;&#xA;List all the stuff you know you need and account for the small odds and ends that you can't know for certain, like nuts and bolts. The smaller a position is, the more it is ok to have only an estimate. If you need stuff you can't buy, make sure to do your research, for example if you need a custom machining job (that you can't do yourself). Group positions in a sensible way, like mech. components, electronic components, services etc.&lt;br&gt;&#xA;It's probably ok if you list your own working time and other work 'in house' only with hours, not with costs - your boss will know how to trasnlate hours into dollars. List positions you're unsure about, so you can discuss them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of this approach is that it's simpler than to create a detailed project schedule. The disadvantage is that you don't have a nice gant chart to impress your boss, also if your work involves lots of interfaces and deliverables with others in the company a detailed, realistic schedule is very important. If the project consists of you, a lab and a bunch of parts, my approach could be sufficient. &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-02-04T16:13:39.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="448" PostTypeId="1" AcceptedAnswerId="450" CreationDate="2015-02-04T16:38:13.967" Score="2" ViewCount="623" Body="&lt;p&gt;I am currently reading through a book on Software Engineering principles, with a lot of the content having to do with proper approaches to software development methodologies and project management approaches. After reading through the chapter on risk assessment and management, one of the review questions caught my eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asks me to to examine the GPS navigation system of a commercial aircraft as well as the elevator in a high-rise building, and to identify 3 hazards that could become linked to software failures. For the elevator one I foresee several hazards such as the Elevator Doors not opening, the system stopping the elevator between floors, the system failing to the point of &quot;dropping&quot; the elevator from the 217th floor, and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The commercial airline GPS system though is leaving me stumped. With a car you run the risk of out-dated maps attempting to take you through/over/across hazards or nonexistent routes. While trying to research, I found that apparently radar only works about 100-150 miles past the coast, from there pilots will radio in at checkpoints with location, altitude etc. Aside from the GPS outright failing (picture over the ocean, no landmarks at all or falling off course if the GPS is tied into the AutoPilot), I'm not seeing many other hazards that could arise from software failure in a commercial airline GPS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would anyone care to offer their thoughts on this? I feel as if I am either over looking something or way over thinking this.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how can issues such as in an aircraft GPS be prevented? With an elevator you have the Governor and &quot;drop-locks&quot;, but how do you avoid GPS failures/errors in an aircraft?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After thinking it over some more I put the following together:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Commercial Aircraft GPS Navigation&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hazard 1:&lt;/strong&gt; If the GPS outright fails, the plane can become lost due to unfamiliar terrain or lack of landmarks (picture over the ocean). Even with an old fashioned compass or use of the sun for direction, there are concerns of fuel quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preventative Measure 1:&lt;/strong&gt; While certainly not fool-proof, the best preventative measure I can think of is to ensure a backup GPS is available on all aircraft should the first one completely fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hazard 2:&lt;/strong&gt; Should the GPS navigation in the aircraft fail (but appear to be functioning correctly) there is the chance of 2 planes encountering a mid-air collision as they enter into the same flight path and altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preventative Measure 2:&lt;/strong&gt; Having either a second GPS system available to check against or by conducting check points after the flight path has been established, falling so far off course should be avoided. I am not familiar with aviation standards, but I like to assume that most aircraft have some kind of proximity alarm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hazard 3:&lt;/strong&gt; If the GPS is tied into the Auto-Pilot system (which I believe is the case with most commercial aircraft today), then a GPS failure (but seemingly functioning) event could cause the aircraft to rise and stall or to enter into a catastrophic path such as straight into a mountain side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preventative Measure 3:&lt;/strong&gt; Other than building in a method to force the auto-pilot function to disengage, I am unsure of what measure could be taken on this particular event (though it should not occur on a commercial aircraft with the pilots monitoring systems even if not actually “flying” the aircraft).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;High-Rise Elevator&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hazard 1:&lt;/strong&gt; There is the risk of the software failing and thus causing the doors to remain closed, trapping passengers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preventative Measure 1:&lt;/strong&gt; I believe most modern elevators have methods in place to prevent this, but my method would be to build a backup or exception catch which would allow passengers to open the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hazard 2:&lt;/strong&gt; Similar to the above, a software failure could lead an elevator to become stuck between floors – this is often seen in movies but does actually happen. Depending upon the amount of building material separating building floors, this can also lead to passengers being thoroughly trapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preventative Measure 2:&lt;/strong&gt; There are different issues which could cause this particular hazard situation – the cable could become snagged, cable locks deploy, etc. If this is due to governor, the best measure I think of is for a governor failure to go ahead and trigger the cable locks, keeping the elevator in place, but stopping it from plunging down (or up). From there, the standard call button can be used to contact emergency services.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hazard 3:&lt;/strong&gt; Then there is the most concerning hazard of a high-rise elevator: a software failure which causes the governor to malfunction and the elevator to drop straight down at the mercy of gravity and its own weight. Doing so from say the 82nd floor would almost assuredly result in expiration of the passengers on board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Preventative Measure 3:&lt;/strong&gt; In this particular scenario, the best measure I can think of is to build into the system where if the governor fails and the elevator begins to move [X] number of feet per second, engage the cable locks to stop the elevator before serious injury causing speed can be built up.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is there something further I'm overlooking? I feel as if I'm grasping at straws with my current ideas for preventative measures. Besides back-ups to the main system in most of the hazard situations, I'm not seeing many other measures that can be taken to prevent or handle their occurrence.&lt;/p&gt;&#xA;" OwnerUserId="351" LastEditorUserId="351" LastEditDate="2015-02-04T20:08:35.553" LastActivityDate="2016-09-12T18:39:14.497" Title="Possible Hazards of Software Controlled Systems (Ex. Airline GPS/Elevator)?" Tags="|mechanical-engineering|electrical-engineering|aerospace-engineering|aircraft-design|software|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="449" PostTypeId="1" CreationDate="2015-02-04T16:41:32.170" Score="18" ViewCount="1376" Body="&lt;p&gt;When using the &lt;em&gt;finite element method&lt;/em&gt;, I have always used either already meshed domains or very simple ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've heard, meshing complex geometries is often outsourced to specialized companies (as it is considered not to be an interesting part of the job). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering how it is done: is it partly automatic, should you have to define the points and connectivities by hand in some cases? What are the most commonly used criteria to ensure the mesh will fulfill the expectation of the customer? What are the trends: should we expect it to be fully automatic in the coming years? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;I recently found a partial answer to this question: Isogeometric Analysis (IGA). IGA can been seen as an extension of the finite element method in order to solve the problem of mesh generation by creating a mesh directly from the CAD. It uses the CAD spline description of the geometry to automatically build both the mesh and the finite element space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And one of the reason it has been developped is because the authors noticed that mesh generation is so painful that it takes most of the time to achieve in the industry, and mesh convergence is only rarely checked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method seems to be really interesting but not widely used since relatively new (10 years).&lt;/p&gt;&#xA;" OwnerUserId="352" LastEditorUserId="352" LastEditDate="2016-08-16T10:06:43.667" LastActivityDate="2017-03-19T09:14:57.767" Title="Meshing of complex geometrical domains" Tags="|modeling|finite-element-method|meshing|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="450" PostTypeId="2" ParentId="448" CreationDate="2015-02-04T20:16:40.720" Score="4" Body="&lt;p&gt;Commercial air traffic is in pretty constant contact with air traffic control, and GPS is not the primary method of tracking aircraft position. &lt;a href=&quot;http://www.bbc.com/news/world-asia-pacific-26544554&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt;, written in the wake of the MH370 disappearance, gives a pretty good rundown of how planes are tracked currently. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The essence of it is that most air traffic control systems still rely on radar, whether primary (bouncing radio signals off of objects) or secondary (using a transponder to get more accurate data and computing position/velocity using successive pings.) However, most modern aircraft has GPS installed on it, it's more for pilot reference than anything else. At this point, we have 3 systems to track position, one of which is entirely separate from the plane, so the entire tower would have to go down in order for it to fail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, considering the ubiquity of cell phones, if a plane's entire navigation system falls apart simultaneously, there will still be dozens or hundreds of GPS capable devices on board. In this sort of emergency, the GPS only needs to function well enough to get you below the cloud ceiling and into a position to fly VFR (using sight instead of instruments) and land safely at a close airfield. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And really, what this means is that the question is based on a flawed premise. If the GPS on a plane cuts out, the pilot probably goes &quot;Damnit, that sucks. Gotta get that fixed later.&quot; and continues about his day as if nothing happened, because he's relying on the transponder for position tracking. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-02-04T20:16:40.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="451" PostTypeId="2" ParentId="226" CreationDate="2015-02-04T20:28:43.323" Score="13" Body="&lt;p&gt;In Ye olden days DC generators were brushed commutated devices. They had a one or more stator windings and an armature winding. Field wound DC generators as well as motors were commonly connected in one of three methods: Series, Shunt and Compound. Without getting into details, each had its own set of strengths and weaknesses. But you only have to remember these two things: the voltage of a DC motor is dependent on its input shaft speed. Current is a function of torque. More voltage means more RPM's and more amps means more newton-meters (or foot-pounds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with all that, you need a constant speed source to get a constant voltage. And you need to ensure you have enough torque to satisfy the current demand of your load otherwise voltage drops off. Old automobiles had commutated generators. They couldn't regulate the voltage so they used a range of around 10-14 volts and used a relay that simply closed when the engines speed was within the voltage range. If the voltage went too low or too high, the relay opened. Primitive by today's standards. The Alternator in today's automobile uses a voltage regulation circuit that varies the armature current which changes the field strength based on the stators output voltage. Lower speed means more current to the armature and less current at higher speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how different were DC generators from motors? Not very different at all. If anything they mostly differed in mechanical design as they were to be coupled to a prime mover (steam, ICE, electric etc.). Though, in much larger dynamos they had adjustable commutator brushes to compensate for the shift in the commutation plane as a result of heavy load characteristics. A hand wheel would turn a worm gear which would advance or retard the commutation plane to bring the generator back into its normal operating parameters. You don't need to worry about this as I am sure you motor isn't megawatt capable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am guessing your motor is a permanent magnet type motor. Its nameplate RPM is what you need to spin the motor at to get the nameplate voltage. This means if you have a 12V motor that spins at 6000 RPM, you need 6000 RPM to get 12V. If you don't have a constant speed source you have no way to regulate the voltage. You would need a buck-boost switching regulator to get a constant voltage from your motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are using this for a renewable energy project like wind or hydro, a charge controller is usually designed for a wide input voltage swing via a buck/boost regulator. Solar panels are a close analogy to a permanent magnet DC generator, no internal voltage regulation and a varying amount if input energy. Sun might be shining bright one minute and a minute late, be blocked by a cloud. So the charge controller does its best to make a useful steady-voltage from its varying input. From there, use storage batteries to capture that power for later use and to act as a buffer for low input events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just for reference, an AC motor can also generate power if you spin it faster than its nameplate RPM, usually at synchronous speed. But again, no voltage regulation and a constant speed is needed. More trouble than its worth. Also of note: jet planes use a very elaborate mechanical speed regulator to produce constant shaft speeds which ensures a constant 60 or 400Hz AC frequency as the throttle is varied.&lt;/p&gt;&#xA;" OwnerUserId="359" LastEditorUserId="6767" LastEditDate="2020-04-13T13:37:05.047" LastActivityDate="2020-04-13T13:37:05.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="452" PostTypeId="1" AcceptedAnswerId="456" CreationDate="2015-02-04T21:38:02.410" Score="6" ViewCount="1304" Body="&lt;p&gt;I was wondering how to mount a small rotary encoder to a platform in order to measure how far a small treaded robot has traveled. We're using this tread/wheel set: &lt;img src=&quot;https://i.stack.imgur.com/SmpEM.jpg&quot; alt=&quot;track set&quot; /&gt;&#xA;We've tried epoxying the encoder shaft directly to the end of an extra gear wheel, but it broke off quite easily. I'm thinking about grinding down the encoder shaft until it can just squeeze into the axle hole of the drive wheel and then using the epoxy inside the wheel. It would then be driven off of the tread.&#xA;Like this:&lt;img src=&quot;https://i.stack.imgur.com/NiYor.jpg&quot; alt=&quot;My Idea&quot; /&gt;&#xA;However I'm not sure this would be the best idea, and other suggestions would be appreciated. Especially with keeping any assembly either coupled to the drive shaft or to a tread.&lt;/p&gt;&#xA;&lt;p&gt;Note: Its for an EE senior project, and none of us have much experience with construction skills. Its going to be the platform for an automated robot on Aruino.&lt;/p&gt;&#xA;" OwnerUserId="366" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-01-20T12:54:38.420" Title="Mounting and Coupling an Encoder" Tags="|mechanical-engineering|applied-mechanics|joining|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="453" PostTypeId="5" CreationDate="2015-02-04T22:23:41.327" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-04T22:23:41.327" LastActivityDate="2015-02-04T22:23:41.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="454" PostTypeId="4" CreationDate="2015-02-04T22:23:41.327" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-04T22:23:41.327" LastActivityDate="2015-02-04T22:23:41.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="455" PostTypeId="1" AcceptedAnswerId="467" CreationDate="2015-02-04T23:07:24.913" Score="7" ViewCount="1226" Body="&lt;p&gt;I am trying to emulate the resistance of a hand wheel which moves a table via a series of shafts and chains which turn an acme screw. I am switching to a CNC system that will allow the user to keep the old style crank for manual positioning. The crank will instead attach to an encoder which will command the CNC system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is encoders don't have shaft resistance or enough resistance. What I am looking for is a method to produce shaft resistance that feels sticky. You could quickly accelerate it but you have to put a bit of effort behind the motion and the crank would stop dead and hold its position when let go. I tried searching for such a device but perhaps I am searching for the wrong terms. In my head I picture one or more slotted discs sealed in a housing filled with a very viscous or sticky liquid. The only thing that comes close is a viscous shaft coupling. But they are couplers. I need a through bore or keyway whose housing can be mounted to a frame.&lt;/p&gt;&#xA;" OwnerUserId="359" LastActivityDate="2015-02-11T02:33:11.813" Title="Creating smooth shaft resistance" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="456" PostTypeId="2" ParentId="452" CreationDate="2015-02-04T23:10:12.973" Score="4" Body="&lt;p&gt;The flat on the encoder shaft would typically be used for &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_screw&quot; rel=&quot;nofollow&quot;&gt;set screw&lt;/a&gt;. If you are able to get a gear or wheel with the right size shaft then adding a screw should be relatively straightforward. This will not only hold the gear onto the encoder, but also give it some increased tortional strength which should be better than just epoxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are unable to get a gear to exactly fit the encoder shaft size then I would suggest being careful about how you reduce the size of the shaft or increase the size of the hole. You need to ensure that the axis of rotation remains central. If it can be properly turned on a lathe (or even drill chuck if necessary) I think you'll get a more even result than going at it with a file by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not expect your encoder to have a particularly robust bearing system within it, so it may not be suitable for use directly as a wheel axle, but the end decision really depends on your component specifications and the loading on your robot frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final point, in case you've not considered it, is the resolution you require from your encoder measurement. This will be affected by the gear ratio you choose from your motor. Higher ratios will give you a more precise measurement, but also a higher count in your software, so it really depends on your expected move distance and sotware performance.&lt;/p&gt;&#xA;" OwnerUserId="157" LastActivityDate="2015-02-04T23:10:12.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="457" PostTypeId="1" AcceptedAnswerId="458" CreationDate="2015-02-05T01:42:47.943" Score="6" ViewCount="124" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lab-on-a-chip&quot; rel=&quot;nofollow&quot;&gt;Labs-on-a-chip&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Integrated_circuit&quot; rel=&quot;nofollow&quot;&gt;silicon chips&lt;/a&gt; both have the potential - and in some cases have already reached the potential - to drastically scale down tasks, in these cases performing chemical reactions and performing computations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the appeal of silicon chips is - besides their size - is that they can be easily mass-produced. Labs-on-a-chip have tremendous potential, but they are in the early stages of development.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can labs-on-a-chip be mass-produced in the same way that silicon chips are mass-produced?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="110" LastEditDate="2015-02-08T19:06:06.233" LastActivityDate="2015-02-08T19:06:06.233" Title="Can the production rate of labs-on-a-chip ever reach the production rate of silicon chips?" Tags="|electrical-engineering|biomedical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="458" PostTypeId="2" ParentId="457" CreationDate="2015-02-05T02:20:24.417" Score="5" Body="&lt;p&gt;Production rates of lab-on-chip fluidic devices can &lt;em&gt;exceed&lt;/em&gt; the production rate of silicon ICs easily.  Some types of lab-on-chip devices can be fabricated via injection molding.  Of course, there are subsequent operations: assembly, QC.  But those can be automated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware of a device that's already being produced at a rate of 5 million units a year.  &lt;a href=&quot;http://www.abaxis.com/veterinary/products/chemistry-profiles.html&quot; rel=&quot;nofollow&quot;&gt;These ones.&lt;/a&gt;  They are not small enough to fit the &quot;chip&quot; category: the disk is about 100mm diameter.  Nevertheless, it follows the philosophy of lab-on-chip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time.  In terms of complexity, present cutting edge lab-on-chip devices are 8 to 10 orders of magnitude simpler than present cutting edge ICs.  Let me put it this way: if today's lab-on-chip were silicon ICs, they would be &lt;a href=&quot;http://en.wikipedia.org/wiki/Operational_amplifier#Internal_circuitry_of_741-type_op-amp&quot; rel=&quot;nofollow&quot;&gt;741 OpAmps&lt;/a&gt; from 1968.&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2015-02-05T02:29:50.667" LastActivityDate="2015-02-05T02:29:50.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="459" PostTypeId="1" AcceptedAnswerId="463" CreationDate="2015-02-05T02:22:02.780" Score="8" ViewCount="7689" Body="&lt;p&gt;I want to attach something small to the smooth metal shaft of a small motor (2-3 mm in diameter), like fan blades, but I want it to be temporary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I make it too loose, then the fan won't turn at all, but if I glue or solder it on, then it will be too permanent and I want to be able to take it off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to firmly attach something temporary to a smooth, small shaft that has low friction while still being able to remove it when it needs to be?&lt;/p&gt;&#xA;" OwnerUserId="116" LastEditorUserId="282" LastEditDate="2015-02-15T19:38:21.390" LastActivityDate="2015-02-15T19:38:21.390" Title="How do you attach something temporary to a smooth metal shaft?" Tags="|mechanical-engineering|materials|friction|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="460" PostTypeId="2" ParentId="459" CreationDate="2015-02-05T02:28:58.003" Score="8" Body="&lt;p&gt;Consider using something like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Set_screw&quot; rel=&quot;noreferrer&quot;&gt;set screw&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be seen in the left hand side of this image:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/d0pFss.jpg&quot; alt=&quot;set screw&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set screw is a small screw used to apply a high amount of friction and to keep a surrounding device in place.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another option would be a dissolvable adhesive like &lt;a href=&quot;http://gbp.com/coated/specialty-products/dissolvable&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt;.  The idea being that you can use the adhesive to keep the fan blades in place but then dip the component in water to get the blades to release later.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-02-05T02:35:41.640" LastActivityDate="2015-02-05T02:35:41.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="461" PostTypeId="2" ParentId="459" CreationDate="2015-02-05T02:30:08.597" Score="6" Body="&lt;p&gt;There are two methods that immediately come to mind.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Attach it with a set-screw.&lt;/li&gt;&#xA;&lt;li&gt;Use a weak adhesive. This would be strong enough to hold the items in place while in service, but weak enough that it could be twisted apart without damaging the individual parts.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-05T02:30:08.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="462" PostTypeId="2" ParentId="455" CreationDate="2015-02-05T04:51:48.983" Score="6" Body="&lt;p&gt;It sounds like you're looking for either a &lt;a href=&quot;http://en.wikipedia.org/wiki/Torque_limiter&quot; rel=&quot;nofollow&quot;&gt;torque limiter&lt;/a&gt; or a &lt;a href=&quot;http://en.wikipedia.org/wiki/Inertia_damper&quot; rel=&quot;nofollow&quot;&gt;rotary damper&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Torque limiter prevents rotational force from the driving shaft over a certain level affect the driven shaft. Kind of like audio cut off on a microphone. When the driving shaft exceeds this limit it will slip allowing it to move somewhat freely without affecting the driven shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Rotary dampner works by using paddles inside a closed container of liquid that can only pass liquid through paddles at a limited rate, thereby limiting the maximum velocity of the shaft.&lt;/p&gt;&#xA;" OwnerUserId="340" LastEditorUserId="340" LastEditDate="2015-02-11T02:33:11.813" LastActivityDate="2015-02-11T02:33:11.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="463" PostTypeId="2" ParentId="459" CreationDate="2015-02-05T05:44:04.880" Score="11" Body="&lt;p&gt;I'd recommend searching for shaft couplings. These are a great option for connecting two rotating shafts in a non-permanent fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set screw (as mentioned) can work in low load cases, but is often best with a flat spot cut into the shaft so that you've got more than just friction between the screw end and shaft to constrain things. The setscrew may also scratch or dent the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other designs of coupling will actually tighten and clamp to the shaft in question. They'll have a much better surface to keep static friction and may actually have more clamping force. Certainly better for a smooth shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further advantage of a coupling is that there are a number of more sophisticated designs that will transmit rotation even if the shafts are not well aligned. Options exist to allow small amounts of bending or translation.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-02-05T05:44:04.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="464" PostTypeId="1" AcceptedAnswerId="485" CreationDate="2015-02-05T07:30:11.323" Score="15" ViewCount="1586" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperboloid_structure&quot;&gt;Hyperboloid towers&lt;/a&gt; were very popular in the end of 19th and the first half of 20th centuries - water towers, powerline anchor towers, sometimes tall radio towers were built using this design. The claimed advantage is using less steel compared to other designs for the same strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are very rarely used nowadays (to such extent that old hyperboloid towers are treated as cultural heritage objects and protected by the state in some countries).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did they lose popularity? Is there any inherent defect it the design? Is steel not expensive anymore?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="153" LastEditDate="2015-02-05T09:49:08.433" LastActivityDate="2015-02-05T23:54:27.037" Title="Why are hyperboloid towers not popular anymore?" Tags="|steel|structures|engineering-history|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="465" PostTypeId="1" AcceptedAnswerId="1837" CreationDate="2015-02-05T09:10:13.490" Score="5" ViewCount="187" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zQtSjm.jpg&quot; alt=&quot;Spring on shaft with retaining clip&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to get another component like this but I can't seem to find any manufacturers / retailers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To describe it:&lt;br&gt;&#xA;It is spring mounted on a shaft and retained by bushing on the top and an enlarged shaft diameter bottom. E-clip on top is for securing shaft in place. Spring pushes the pin down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone knows what they are called, I would like to know.&lt;/p&gt;&#xA;" OwnerUserId="378" LastEditorDisplayName="user16" LastEditDate="2015-02-25T16:26:55.590" LastActivityDate="2015-02-25T16:26:55.590" Title="Name of spring actuated component" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="466" PostTypeId="2" ParentId="370" CreationDate="2015-02-05T09:19:34.000" Score="0" Body="&lt;p&gt;I would get a free 3-year &lt;a href=&quot;http://www.autodesk.com/education/free-software/inventor-professional&quot; rel=&quot;nofollow&quot;&gt;student version&lt;/a&gt; of Autodesk Inventor (because I am familiar with it, SolidWorks, CATIA, work as well). Then model the desk and perform &lt;a href=&quot;https://www.youtube.com/watch?v=hVBSrfXawmE&quot; rel=&quot;nofollow&quot;&gt;static analysis&lt;/a&gt;. The days of force diagrams on A0 sheets are long gone.&lt;/p&gt;&#xA;" OwnerUserId="379" LastEditorDisplayName="user16" LastEditDate="2015-02-15T13:29:04.983" LastActivityDate="2015-02-15T13:29:04.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="467" PostTypeId="2" ParentId="455" CreationDate="2015-02-05T10:06:59.847" Score="5" Body="&lt;p&gt;You're looking for a Rotary Viscous Damper like &lt;a href=&quot;http://ace-ace.com/wEnglisch/pages/Produkte/index.php?IdTreeGroup=268&quot; rel=&quot;noreferrer&quot;&gt;these&lt;/a&gt; - a device that provides viscous friction in rotary movement. Viscous friction is proportional to speed of movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See video of such a damper &lt;a href=&quot;https://www.youtube.com/watch?v=OE9Fm_KgRQY&quot; rel=&quot;noreferrer&quot;&gt;activated by a spring&lt;/a&gt; and a practical use - &lt;a href=&quot;https://www.youtube.com/watch?v=ynpAUtIOQZo&quot; rel=&quot;noreferrer&quot;&gt;slowly falling lid&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-02-05T10:06:59.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="468" PostTypeId="2" ParentId="457" CreationDate="2015-02-05T10:27:56.900" Score="0" Body="&lt;p&gt;It's doubtful - for the simple reason: a lab-on-chip requires a silicon chip to make sense of the readouts, so increase in production of labs-on-chip will automatically create increase of demand (and production) of silicon chips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an avenue which can overcome that trend - single-use, disposable labs-on-chip, say, preloaded with a dose of a marker substance, so you need to replace it after the measurement, attaching a new one to the same &quot;front-end&quot; silicone chip - say, medical samplers where the readout of given marker of given patient uses up one disposable lab-on-chip, which is inserted into a reader to give the readout. Still, for any reusable lab-on-chip, a silicon part will remain a necessity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still again, the technologies may blend - say, you could claim the production of thermal sensor devices outpaced the production of microcontrollers, simply because about every microcontroller on the market contains one or more thermal sensor built in, next to all the production of standalone sensors (which hasn't been outpaced). I can imagine in the future many generic chips including some lab-on-chip components, so along with regular production this may increase the number.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-02-05T10:27:56.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="469" PostTypeId="2" ParentId="452" CreationDate="2015-02-05T10:41:30.383" Score="6" Body="&lt;p&gt;That model of encoder is not intended for mechanical measurements. It's meant to be used as a fancy knob with high friction, for turning by hand, say, to pick a position from menu on a display, and trying to couple it to a mechanism for readout is a misguided attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need is an encoder based on slotted optocouplers or a reflective optocouplers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XDJi4.jpg&quot; alt=&quot;enter image description here&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/UdpVT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note you can print the code wheel on transparency, to get the desired number of pulses per turn. Put the wheel on the axle, attach the optocoupler to the main plate - or even extend the tracks a little and use them as &quot;code strip&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have trouble purchasing these, you can often find them in computer mice and cheapest inkjet printers.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-02-05T10:41:30.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="470" PostTypeId="1" AcceptedAnswerId="471" CreationDate="2015-02-05T10:51:03.150" Score="11" ViewCount="2186" Body="&lt;p&gt;Many people are familiar with the story of &lt;a href=&quot;https://en.wikipedia.org/wiki/Archimedes#Discoveries_and_inventions&quot; rel=&quot;nofollow&quot;&gt;Archimedes submerging a crown in water&lt;/a&gt; to identify if the volume of the crown was consistent with the volume of the same amount of gold.  The test was to see if a lighter cheaper metal (like silver) was included in crown.  If the crown maker substitutes silver for some of the gold, they make a sinister profit at the risk of their life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Platinum is heavier than gold, silver is lighter.  Using the values as of January of 2015, could &lt;strong&gt;any combination of metals&lt;/strong&gt; be used to create a crown that would have the same volume as a solid gold crown, with a cost effective benefit for the crown maker?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the values of metals in 287 BCE – 212 BCE, could it be cost effective for the crown maker?&lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="368" LastEditDate="2015-04-15T21:39:33.553" LastActivityDate="2015-04-15T21:39:33.553" Title="Could a gold/{second metal} crown be created that would pass the Archimedes test cost effectively?" Tags="|metallurgy|engineering-history|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="471" PostTypeId="2" ParentId="470" CreationDate="2015-02-05T11:21:54.990" Score="16" Body="&lt;p&gt;Yes. Especially considering gold and platinum prices as of today, Pt costs less than Au. - but let's earn much more with more modern solution and simultaneously slowly murder the king in a very nefarious plot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold is 40 $\$/g$ at 19.3 $g/cm^3$&lt;br&gt;&#xA;Platinum is 39 $\$/g$ at 21.45 $g/cm^3$&lt;br&gt;&#xA;Depleted Uranium is about 6 $\$/g$ at 19.1 $g/cm^3$ &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20110916103215AAzspaB&quot; rel=&quot;noreferrer&quot;&gt;(1)&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[sorry for the source, but at least it cites some references, so I believe the prices aren't that much off.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho_{Au} = 19.3 g/cm^3 \\ \rho_{Pt} = 21.45 g/cm^3 \\ \rho_{DU} = 19.1 g/cm^3 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume we want to produce $1cm^3$ of the new alloy of same density as gold. Our set of equations will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ V_{Pt} + V_{DU} = 1 \\ \rho_{Pt} V_{Pt} + \rho_{DU} V_{DU} = 1 \rho_{Au} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, changing sides and substituting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ V_{DU} = 1-V_{Pt} \\ \rho_{Pt} V_{Pt} + \rho_{DU} (1-V_{Pt}) = \rho_{Au} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho_{Pt} V_{Pt} - \rho_{DU} V_{Pt} = \rho_{Au} - \rho_{DU}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ (\rho_{Pt} - \rho_{DU} ) V_{Pt} = \rho_{Au} - \rho_{DU}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ V_{Pt} = { { \rho_{Au} - \rho_{DU} } \over { \rho_{Pt} - \rho_{DU}  } } $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ V_{Pt} = 0.2/2.35 = 0.085 cm^3 ; M_{Pt} = \rho_{Pt} V_{Pt} = 1,823 g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ V_{DU} = 1 - V_{Pt} = 0.915 cm^3 ; M_{DU} = \rho_{DU} V_{DU} = 17,476 g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing these by density of gold, we'll obtain masses that comprise 1 gram of the &quot;fake gold&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ m_{DU} = 0.905g \\ m_{Pt} = 0.095g $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now considering the prices,  0,095g Pt is 3.70 USD;  0.905g DU is 5.45 USD for a total of &lt;strong&gt;9.13USD for gram of the alloy, replacing 40USD for gram of gold&lt;/strong&gt;. That's less than one fourth the price.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course since the appearance of platinum-DU alloy won't fool anyone, you'd need to make some kind of &quot;skeleton&quot; to be plated with gold on the outside to hide the forgery and somewhat extend the king's life before he dies of brain cancer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Using the values of metals in 287 BCE – 212 BCE, you won't find depleted uranium on sale. But I remember a historian friend telling me the ancients didn't recognize platinum as a precious metal - miners would dig it up sometimes along with silver, and naming it &quot;young silver&quot;, believing it needs to rest in soil for a few generations more, would bury it to &quot;mature&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I believe, with platinum being considered nearly worthless, addition of some lighter metal, like copper to balance out the density gain, replacing majority of the crown with platinum would bring immense profits. If only the crown maker knew of the density of the &quot;young silver&quot;...&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2015-02-06T00:53:51.777" LastActivityDate="2015-02-06T00:53:51.777" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="472" PostTypeId="1" CreationDate="2015-02-05T11:22:04.933" Score="4" ViewCount="322" Body="&lt;p&gt;I have a liquid circuit which looks a bit like that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V4lct.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is under suction, and the flow rate through each unit should be about 10ml/min. However, we noticed that unit &lt;code&gt;n&lt;/code&gt; and the ones the furthest away from the pump and bottle were not getting enough flow. Any suggestions on how to make sure each unit gets about the same amount of flow? My initial thought was to add a second pump at the end of the line to help the liquid (water) go round the circuit. Another possibility is to separate the load of &lt;code&gt;n&lt;/code&gt; units into 2 circuits in parallel, &lt;code&gt;1&lt;/code&gt; to &lt;code&gt;n/2&lt;/code&gt; and &lt;code&gt;n/2 + 1&lt;/code&gt; to &lt;code&gt;n&lt;/code&gt; at the bottle/pump location so that both circuits will have a better chance of seeing the same amount of liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some precisions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there is a restrictor on the return path of each unit to restrict the flow and create a pressure differential to allow the liquid to go round&lt;/li&gt;&#xA;&lt;li&gt;the length from unit #1 to unit #n is about 3m, the pump and bottle are located between unit #4 and #5.&lt;/li&gt;&#xA;&lt;li&gt;the tubing diameter along that length is 4.8mm ID, and each teed off line in each unit is 1.6mm ID, with the restrictor 0.38mm ID (~60mm length)&lt;/li&gt;&#xA;&lt;li&gt;the pump is a diaphragm liquid pump&lt;/li&gt;&#xA;&lt;li&gt;there is a vent port with a filter on the bottle lid to allow for the pressure to vent out.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-02-05T15:26:12.610" Title="Ensuring even flow in liquid pumping in distribution circuit under suction" Tags="|pressure|fluid-dynamics|pumps|microfluidics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="473" PostTypeId="2" ParentId="472" CreationDate="2015-02-05T11:34:35.913" Score="5" Body="&lt;p&gt;Looks like a typical water based central heating system (apart from the scale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to balance the flow by restricting flow for the units that get too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done by partially closing a valve or adding a restrictor for those units. Better if there is some kind of feedback loop to auto restrict when too much water is going through.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-02-05T11:34:35.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="475" PostTypeId="1" AcceptedAnswerId="476" CreationDate="2015-02-05T13:21:08.610" Score="8" ViewCount="8221" Body="&lt;p&gt;When using weathering steel (COR-TEN) on a project, what sort of bolts should I be using?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have yet to see fixings made of weather steel; presumably because the way it uses a sacrificial layer to form a protective rust coating makes it unsuitable for precision applications (such as bolt threads; maintaining the structural area, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also aware that mixing different metals; even various steel alloys, can lead to accelerated corrosion due to anodic/cathodic behaviour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what type of bolts &lt;em&gt;should&lt;/em&gt; be specified with weathering steel?&lt;/p&gt;&#xA;" OwnerUserId="151" LastEditorDisplayName="user16" LastEditDate="2015-02-07T23:44:58.430" LastActivityDate="2017-08-14T15:06:11.313" Title="What bolts should I specify with Weathering Steel?" Tags="|steel|metallurgy|bolting|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="476" PostTypeId="2" ParentId="475" CreationDate="2015-02-05T13:31:18.467" Score="8" Body="&lt;p&gt;For structural applications (in the US), the most common bolt for weathering steel is ASTM A 325 Type 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 1 is a plain steel bolt that can be galvanized, but in this situation the zinc in the galvanizing will quickly be used trying to protect the rest of the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update for British bolts&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the only option for UK seems to be to get these same bolts in metric (M24) size.  References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cooperandturner.co.uk/news/19&quot; rel=&quot;nofollow&quot;&gt;Supplier&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://resource.npl.co.uk/docs/science_technology/materials/life_management_of_materials/publications/online_guides/pdf/weathering_steel_bridges.pdf&quot; rel=&quot;nofollow&quot;&gt;Weathering steel bridges&lt;/a&gt; (page 13 of pdf)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.steelconstruction.org/index.php?option=com_documents&amp;amp;task=downloadDocument&amp;amp;doc=68103&amp;amp;file=83763&quot; rel=&quot;nofollow&quot;&gt;Specific discussion&lt;/a&gt; about the A325 bolts meeting EN standards&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-02-19T14:11:12.453" LastActivityDate="2015-02-19T14:11:12.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="477" PostTypeId="1" AcceptedAnswerId="478" CreationDate="2015-02-05T13:33:45.353" Score="4" ViewCount="16072" Body="&lt;p&gt;What are the most popular and the most efficient ways to machine a fillet on an external edge of an object? If I were to use a CNC mill, what tools can I use for it, and how would the method change with the radius of the fillet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could imagine that internal fillets can be easily achieved by a ball-end to a mill, but I've never come across machining an external one.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="106" LastEditDate="2015-02-06T02:43:25.297" LastActivityDate="2015-02-06T02:43:25.297" Title="How do I manufacture a fillet?" Tags="|machining|manufacturing-engineering|cnc|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="478" PostTypeId="2" ParentId="477" CreationDate="2015-02-05T13:51:19.470" Score="3" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AQxSu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on whether that's one-off or something to repeat, a properly shaped bit with &quot;negative curve&quot; will be the most efficient (single pass) and neat (no grooves from multiple passes) solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a photo of the tool in question, but the one in the middle has some similar properties - it was used to create undercuts (-30 degrees edges) and its side edge curves in a way similar to what you'd need to obtain. So, simply grind a bit to the correct radius and pass it once (or in enough approaches as not to break the bit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you either don't grind your own bits or are unwilling to use up a new blank for something you'll use once, use a round tip bit, and have the CAM software calculate tracks for the right curve, then make it pass enough times so that the lines will be possible to remove with sandpaper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note you don't need many bits for different curves. Make one of fairly large curve and route it along the right circle (radius = bit radius - fillet radius) in the Z-axis , and it will be much more graceful to smooth out than created with a round tip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AMTPa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2015-02-05T16:00:57.833" LastActivityDate="2015-02-05T16:00:57.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="479" PostTypeId="2" ParentId="477" CreationDate="2015-02-05T14:17:23.110" Score="5" Body="&lt;p&gt;On a mass manufactured part, one solution is to cast the piece and shape the die (with an inside fillet) to form an outside fillet. Of course, if you want your casting to have an inside fillet, you'll need an outside fillet on your die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you're not casting, and working with a commonly machinable material, external corners would typically be rounded over using a tool called a 'radius cutter' or more formally, a 'corner rounding end mill.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lpfef.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particular cutter only cuts one radius, so if you have two fillet radii, you'd need a tool change. They are available in a range of sizes and made out of different materials with different coatings. High speed steel is probably the most common material.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-02-05T16:18:32.580" LastActivityDate="2015-02-05T16:18:32.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="480" PostTypeId="2" ParentId="472" CreationDate="2015-02-05T15:26:12.610" Score="1" Body="&lt;p&gt;Looks like you want a typical register, so that sum of the pipe length supply and return is the same for each unit. No guarantee that it works, though. I'll see if I make a drawing if the idea is unclear.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-02-05T15:26:12.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="481" PostTypeId="1" AcceptedAnswerId="482" CreationDate="2015-02-05T17:11:32.893" Score="9" ViewCount="1972" Body="&lt;p&gt;I am a complete beginner to hydraulic systems, and I've wanted to learn more about this area. I'm designing a hydraulic system that involves using hydraulics to push/pull objects using pistons. I have looked at what the basic requirements are for a hydraulic system, but there is one thing that escapes me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I come from an electronic background, and I noticed that the hydraulic pumps (for example, &lt;a href=&quot;http://uk.rs-online.com/web/p/hydraulic-gear-pumps/5115680/&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;) seem to lack a motor to drive the fluid. Am I wrong? If not, I've been looking everywhere for a motor that can/should be attached to said pump, but I cannot seem to find anywhere that sells them. Is it just a simple DC motor (with correct specs), or should there be a specific motor designed for hydraulic pumps?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking around, I came across &lt;a href=&quot;http://www.boschrexroth.com/dcc/Vornavigation/Vornavi.cfm?Language=EN&amp;amp;VHist=g97568,g96068,g101267&amp;amp;PageID=p148897&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, but looking through the specs, I don't see a power requirement, and being used to seeing power consumption in datasheets, I'm not even sure it is a motor!&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="77" LastEditDate="2016-03-21T00:13:41.407" LastActivityDate="2020-05-15T19:24:12.053" Title="Motor for a hydraulic pump in a hydraulic system" Tags="|electrical-engineering|pumps|power-electronics|hydraulics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="482" PostTypeId="2" ParentId="481" CreationDate="2015-02-05T20:06:05.763" Score="7" Body="&lt;p&gt;You are correct, getting oil pressure will require an electric motor as your source of rotation, potentially a gear box to slow that rotation down, and the pump itself. You'll also need a reservoir to store the non-pressurized oil and probably a filter to keep the oil clean (dirty oil will damage your components.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're new to the world of hydraulics, and want to keep things a little simpler for now, you might want to search for a 'Hydraulic Power Unit' which usually includes all of these components, already connected together. This way you just have a pressure line out, and a return line ('tank line') and don't have to be too concerned with RPM, torque, or any of the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to assemble the whole thing yourself, or no HPU's are available that work for your application, you'll want to find a motor with the same output as your pump's input (In your case a tapered shaft.) In your specific example, if you see their datasheet at &lt;a href=&quot;http://docs-europe.electrocomponents.com/webdocs/12c4/0900766b812c44d2.pdf&quot;&gt;http://docs-europe.electrocomponents.com/webdocs/12c4/0900766b812c44d2.pdf&lt;/a&gt; you can see that they offer these pumps in a range of different shaft types most of which conform to a standard interface (SAE J744, for example). You'll want to pick a motor with a compatible output shaft. You'll need to make sure that the RPM and Torque of the motor work with your pump and provide enough flow an pressure respectively for your hydraulic application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please remember that hydraulics can be very dangerous. There can be significant amounts or stored energy that you can't see. Think of the whole system as a giant line capacitor.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-02-05T20:06:05.763" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="483" PostTypeId="2" ParentId="470" CreationDate="2015-02-05T21:09:14.977" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Using the values of metals in 287 BCE – 212 BCE, could it be cost effective for the crown maker?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Of the metals known and used in antiquity (copper, gold, silver, lead, iron, tin, mercury, zinc), gold is by far the densest, at $19.30 \text{ g/cm}^3$; mercury is in second place at $13.53 \text{ g/cm}^3$.  Platinum may have been known, but it certainly wasn't workable: &lt;a href=&quot;http://www.sciencekids.co.nz/sciencefacts/metals/platinum.html&quot;&gt;it was described in 1557&lt;/a&gt; as a metal &quot;which no fire nor any Spanish artifice has yet been able to liquefy&quot;.  Today, you could make an imitation gold crown simply by gold-plating a tungsten ($19.25 \text{ g/cm}^3$) crown, but in the time of Archimedes, no known metal or alloy was dense enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The question specifies the Archimedes submersion test to compare densities; the difference between $19.3 \text{ g/cm}^3$ and $19.25 \text{ g/cm}^3$ is less than the uncertainty caused by the surface tension of water.)&lt;/p&gt;&#xA;" OwnerUserId="370" LastEditorUserId="106" LastEditDate="2015-02-05T23:30:53.793" LastActivityDate="2015-02-05T23:30:53.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="484" PostTypeId="2" ParentId="470" CreationDate="2015-02-05T21:43:37.443" Score="4" Body="&lt;p&gt;Using similar calculations to SF.'s, you can make a much cheaper fake gold bar with tungsten and iridium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Wolfram Alpha (prices may fluctuate):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gold costs &lt;a href=&quot;http://www.wolframalpha.com/input/?i=gold+price+in+dollars+per+gram&quot; rel=&quot;nofollow&quot;&gt;$\$40.71\text{/g}$&lt;/a&gt; and has a density of $19.3\text{ g/cm}^3$&lt;/li&gt;&#xA;&lt;li&gt;Tungsten costs &lt;a href=&quot;http://www.wolframalpha.com/input/?i=tungsten+price+in+dollars+per+gram&quot; rel=&quot;nofollow&quot;&gt;$\$0.05\text{/g}$&lt;/a&gt; and has a density of $19.25\text{ g/cm}^3$&lt;/li&gt;&#xA;&lt;li&gt;Iridium costs &lt;a href=&quot;http://www.wolframalpha.com/input/?i=570+dollars+per+troy+ounce+in+dollars+per+gram&quot; rel=&quot;nofollow&quot;&gt;$\$18.33\text{/g}$&lt;/a&gt; (price from &lt;a href=&quot;https://www.quandl.com/c/markets/rare-metals&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; since Wolfram Alpha doesn't have it) and has a density of $22.56\text{ g/cm}^3$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, first off notice that iridium is both cheaper and denser than gold, so we could just make a solid iridium bar with some empty space for a savings of about 55%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, using a mix of tungsten and iridium we can do &lt;em&gt;way&lt;/em&gt; better. Shamelessly stealing the equations from PT., we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_{Au} = 19.3\text{ g/cm}^3 \\&#xA;\rho_{W} = 19.25\text{ g/cm}^3 \\&#xA;\rho_{Ir} = 22.56\text{ g/cm}^3 \\&#xA;V_{Ir} = { { \rho_{Au} - \rho_{W} } \over { \rho_{Ir} - \rho_{W}  } } \\&#xA;V_{Ir} = { { 19.3\text{ g/cm}^3 - 19.25\text{ g/cm}^3 } \over { 22.56\text{ g/cm}^3 - 19.25\text{ g/cm}^3  } } = { { 0.05\text{ g/cm}^3 } \over { 3.31\text{ g/cm}^3 } } = 0.015\text{ cm}^3 \\&#xA;M_{Ir} = \rho_{Ir} V_{Ir} = 22.56\text{ g/cm}^3 \times 0.015\text{ cm}^3 = 0.34\text{ g} \\&#xA;V_{W} = 1 - V_{Pt} = 0.985\text{ cm}^3 \\&#xA;M_{W} = \rho_{W} V_{W} = { 19.25\text{ g/cm}^3 \times 0.985\text{ cm}^3 } = 18.95\text{ g} \\&#xA;m_{Ir} = 0.02\text{ g} \\&#xA;m_{W} = 0.98\text{ g} \\&#xA;\text{Price}_{Ir} = 0.02\text{ g} \times \$18.33/\text{g} = \$0.37 \\&#xA;\text{Price}_{W} = 0.98\text{ g} \times \$0.05/\text{g} = \$0.05$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving us a final price of $\$0.42/\text{g}$, or about 98% cheaper than gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, tungsten was first isolated &lt;a href=&quot;http://en.wikipedia.org/wiki/Tungsten#History&quot; rel=&quot;nofollow&quot;&gt;in 1781&lt;/a&gt; and iridium in &lt;a href=&quot;http://en.wikipedia.org/wiki/Iridium#History&quot; rel=&quot;nofollow&quot;&gt;1748&lt;/a&gt;, so unless the crown maker is an alchemist way ahead of their times, they probably wouldn't have access to either of them in a pure form.&lt;/p&gt;&#xA;" OwnerUserId="394" LastEditorUserId="394" LastEditDate="2015-02-06T17:09:15.857" LastActivityDate="2015-02-06T17:09:15.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="485" PostTypeId="2" ParentId="464" CreationDate="2015-02-05T23:54:27.037" Score="8" Body="&lt;p&gt;Meet &lt;a href=&quot;http://en.wikipedia.org/wiki/Vladimir_Shukhov&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vladimir Shukhov&lt;/a&gt;, a Russian architect who first developed hyperboloid structures. He was born in 1853, died in 1939, and created over 200 hyperboloid structures in the intervening years. He was the reason hyperboloids gained the popularity that they did. His first design, the first hyperbolic structure ever, was the &lt;a href=&quot;http://en.wikipedia.org/wiki/Shukhov_Tower_in_Polibino&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shukhov Tower in Polibino&lt;/a&gt;, pictured &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/a/a1/First_Shukhov_Tower_Nizhny_Novgorod_1896.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/a/a1/First_Shukhov_Tower_Nizhny_Novgorod_1896.jpg&quot; alt=&quot;Shukhov Tower in Polibino&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Shukhov_Tower&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another tower&lt;/a&gt; also bears the name of Shukhov, and it achieved great fame, too. Shukhov also built the &lt;a href=&quot;http://en.wikipedia.org/wiki/Adziogol_Lighthouse&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adziogol Lighthouse&lt;/a&gt;. In total, Shukhov designed and built 200 hyperboloid structures. He died in 1939, which could one reason for the decline in the popularity of hyperboloids. It isn't common for the death of one architect to essentially stop an entire movement (although &lt;a href=&quot;http://en.wikipedia.org/wiki/Antoni_Gaud%C3%AD&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaudí&lt;/a&gt; also worked with hyperboloids), but that may have been the case here. Shukhov was the reason hyperboloids were popular in the first place; without him, their popularity died down. Also, part of the reason that Shukhov was interested in hyperboloid structures was that &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyperbolic_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;hyperbolic geometry&lt;/a&gt; was being developed during the time when he first entered architecture. &lt;a href=&quot;https://books.google.com/ngrams/graph?content=hyperboloid&amp;amp;case_insensitive=on&amp;amp;year_start=1800&amp;amp;year_end=2000&amp;amp;corpus=15&amp;amp;smoothing=5&amp;amp;share=&amp;amp;direct_url=t4%3B%2Chyperboloid%3B%2Cc0%3B%2Cs0%3B%3Bhyperboloid%3B%2Cc0%3B%3BHyperboloid%3B%2Cc0%3B%3BHYPERBOLOID%3B%2Cc0&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Google n-gram&lt;/a&gt; for &quot;hyperboloid&quot; shows that the term's usage during the late 19th century was its peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final note, &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_hyperboloid_structures&quot; rel=&quot;nofollow noreferrer&quot;&gt;many hyperboloid structures&lt;/a&gt; were built after Shukhov's death, so their decline wasn't &lt;em&gt;that&lt;/em&gt; steep.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:49.940" LastActivityDate="2015-02-05T23:54:27.037" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="486" PostTypeId="1" AcceptedAnswerId="493" CreationDate="2015-02-06T03:40:51.830" Score="13" ViewCount="5838" Body="&lt;p&gt;I am involved right now in the fabrication of steel structures on the order of 5-50 feet long. Presently, we measure these structures with garden variety commercial tape measures. Most of the time, we work to a tolerance of about +/-1/16&quot; and don't have any problems. Recently, we're trying to make some items to a very high tolerance (at least +/- 1/32&quot;, approaching 1/64&quot; or .016&quot;) This tolerance is due to visual not mechanical criteria, but it is still very important to our management.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can we reliably measure these sorts of distances with that level of precision? I'm prepared to order some NIST traceable tape measures, but it's not clear to me if they'll really improve the situation. Are there other technologies or techniques that could practically be applied in a fabrication setting? Are any surveying tools accurate enough to solve the problem? Cost is obviously a factor, but we're more concerned with repeatability and robustness than price.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this tolerance will sound ridiculous to most, but I imagine that some other industries may have to perform similarly, perhaps large engines for ships or power plant components?&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="160" LastEditDate="2015-02-06T14:09:16.633" LastActivityDate="2015-12-15T14:19:56.573" Title="Measuring longer distances with high accuracy" Tags="|measurements|tolerance|manufacturing-engineering|surveying|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="487" PostTypeId="2" ParentId="459" CreationDate="2015-02-06T04:04:34.107" Score="10" Body="&lt;p&gt;A method which works extremely well and which does not require set screws and/or flats on the shaft and which can be added or removed easily and rapidly is a collet based connector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While any one of these will only work for a limited range of shaft sizes, units are available off the shelf for a wide range of shaft diameters and you can make or modify them yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one example - many styles are available.&lt;br&gt;&#xA;Assembled view at top.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the middle row the &quot;collet&quot; is the slit part in the middle.&lt;br&gt;&#xA;The nut at left is assembled onto the thread at right with the collet between them.&lt;br&gt;&#xA;A conical shape inside the nut compresses the collet jaws onto the shaft as the nut is tightened.&lt;br&gt;&#xA;The collet is fixed relative to shaft and nut and thread by friction.&lt;br&gt;&#xA;If desired the collect could be keyed or pinned to the nut but this is seldom necessary.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know how the threaded bolt at bottom is used - it is not essential for the above scheme but may optionally insert into the &quot;nut&quot; and provide a threaded shaft end at right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RRw6R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.prestwich.ndirect.co.uk/surfacedrive.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The above image is from this page&lt;/a&gt; where there are numerous other versions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While these versions are intended to attach items to the end of a shaft, the same principle allows a collar etc to be attached anywhere along a smooth rod - with a flange or bolted together discs or other means used to attach the &quot;payload&quot;. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From the Stack Exchange robotics page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://robotics.stackexchange.com/questions/4632/what-mechanical-parts-can-be-attached-to-dyj48-stepper-motor&quot;&gt;September 2014 question&lt;/a&gt; - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comes this version of the above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6nToC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above and similar are available from &lt;a href=&quot;http://www.hobbyexpress.com/collet_prop_adapter_for_6mm_shaft_m8_1201_prd1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for a range of shaft sizes and similar are available from many other sellers. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="-1" LastEditDate="2017-04-13T12:49:17.810" LastActivityDate="2015-02-06T04:04:34.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="488" PostTypeId="2" ParentId="486" CreationDate="2015-02-06T04:20:53.110" Score="5" Body="&lt;p&gt;Even the best metal tape measure is susceptible to significant thermal expansion over large distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try a laser measurement device ('electronic tape measure') instead:&#xA;&lt;a href=&quot;http://www.engineersupply.com/Laser-Measurers.aspx&quot; rel=&quot;noreferrer&quot;&gt;http://www.engineersupply.com/Laser-Measurers.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The laser distance measure, flat plates clamped to the object, and some shims of known thickness, should be all you need to precisely locate something over a large distance.&lt;/p&gt;&#xA;" OwnerUserId="400" LastActivityDate="2015-02-06T04:20:53.110" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="489" PostTypeId="1" CreationDate="2015-02-06T05:45:51.893" Score="6" ViewCount="2551" Body="&lt;p&gt;I know ceramics are generally considerably less dense than pure metals or alloys but sand is used in barbells for its probably cheapness, user-friendly and easy availability but its a decent density too. rammed sand-  1682 kg/m cubed using Engineering Toolbox compared to 2658 for steel chips.  So if you wanted the most dense for something to have great compressive strength- what would it be? &lt;/p&gt;&#xA;" OwnerUserId="401" LastActivityDate="2018-12-28T17:46:51.347" Title="What is the densest ceramic?" Tags="|materials|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="490" PostTypeId="2" ParentId="489" CreationDate="2015-02-06T08:00:50.450" Score="10" Body="&lt;p&gt;According to the following two sources, diamond is a ceramic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www-materials.eng.cam.ac.uk/mpsite/interactive_charts/strength-density/basic.html&quot; rel=&quot;noreferrer&quot;&gt;Ceramic Strength-Density Graphs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.virginia.edu/bohr/mse209/chapter13.htm&quot; rel=&quot;noreferrer&quot;&gt;Ceramics @ Virginia University&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph below, from &lt;a href=&quot;http://www-materials.eng.cam.ac.uk/mpsite/interactive_charts/strength-density/basic.html&quot; rel=&quot;noreferrer&quot;&gt;Ceramic Strength-Density Graphs&lt;/a&gt; shows diamond is the strongest ceramic, whereas zirconia is one of the most dense and strongest ceramics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/69BHU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The site &lt;a href=&quot;http://www.matweb.com&quot; rel=&quot;noreferrer&quot;&gt;matweb&lt;/a&gt; has a database of all sorts or materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By specifying ceramics with a compressive yields strength ranging between 100 MPa &amp;amp; 16 500 MPa &amp;amp; a density range between 2.5 &amp;amp; 22.6 $\text{g/cm}^3$, corundum, aluminium oxide, &amp;amp; alumina have the highest compressive yield strength &amp;amp; density, with a compressive yield strength of 3000 MPa &amp;amp; a density of 3.96 $\text{g/cm}^3$.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-02-06T08:00:50.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="491" PostTypeId="5" CreationDate="2015-02-06T09:30:54.790" Score="0" Body="&lt;p&gt;Questions about measurement, instrumentation, calibration and metrology. Covers both automatic and manual measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also questions about conversions between measurement units, and official standards for engineering and scientific measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also the tag of metrology.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="33" LastEditDate="2016-02-23T02:50:59.933" LastActivityDate="2016-02-23T02:50:59.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="492" PostTypeId="4" CreationDate="2015-02-06T09:30:54.790" Score="0" Body="Measurements, instrumentation, calibration and metrology." OwnerUserId="160" LastEditorUserId="160" LastEditDate="2015-02-09T17:12:39.473" LastActivityDate="2015-02-09T17:12:39.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="493" PostTypeId="2" ParentId="486" CreationDate="2015-02-06T10:07:10.417" Score="6" Body="&lt;p&gt;To get that sort of accuracy over that scale is not trivial and probably won't be cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For smaller size up to a few meters a portable CMM would be an option (&lt;a href=&quot;http://www.faro.com/products/metrology/faroarm-measuring-arm/overview&quot; rel=&quot;nofollow&quot;&gt;here's an example&lt;/a&gt;). These have accuracy on the order of 10 $\mu$m and are used for things like high end/F1 car manufacture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, CMM type instruments wouldn't be useful for anything larger than a few meters as it is limited by the arm length and must be fixed in place when measuring to get a sensible result. For larger pieces the best performing option would be so sort of laser tracker (&lt;a href=&quot;http://www.faro.com/products/metrology/faro-laser-tracker/overview&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;). These also have pretty good repeatability (~20$\mu$m).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The versions I've seen are phase shifting interferometers, which fire a laser onto a retro-reflector in a metal ball which is placed on the surface. They are used by people like Boeing/Airbus to check tolerances on plane manufacture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These aren't cheap the laser trackers are approximately $80,000. Conventional surveying laser rangefinders are much cheaper but I'm not sure they would have suitable accuracy.&lt;/p&gt;&#xA;" OwnerUserId="153" LastEditorUserId="153" LastEditDate="2015-02-06T14:40:14.653" LastActivityDate="2015-02-06T14:40:14.653" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="494" PostTypeId="1" AcceptedAnswerId="495" CreationDate="2015-02-06T12:29:24.593" Score="5" ViewCount="781" Body="&lt;p&gt;The job site requires welding both aluminum and steel (&lt;em&gt;each to themselves only&lt;/em&gt;) for different areas of the project.  The contractor has a dual purpose wire feed welder (aluminum and steel) but I am unsure that a dual purpose welder will be optimal for both types of metals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you get quality welds from a multiple metal wire feed welder, or do you need separate units for steel and aluminum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking to understand any potential differences between the three types (aluminum, steel &amp;amp; dual) of welding machines and the welds they produce.  Likewise, any testing results that indicate the strength of welds for both types of metal using both dedicated and dual purpose welding units would be useful. &lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="33" LastEditDate="2015-02-07T10:11:54.473" LastActivityDate="2016-09-08T07:38:26.507" Title="Can you get quality welds from a multiple metal wire feed welder, or do you need separate units for steel and aluminum?" Tags="|materials|metallurgy|welding|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="495" PostTypeId="2" ParentId="494" CreationDate="2015-02-06T14:44:10.977" Score="7" Body="&lt;p&gt;The short answer: it's not optimal, but may work. Hopefully if these welds are critical, they're being performed to some code (In the US, for most structural work they'd be AWS D1.1 and D1.2 respectively.) Aluminum is considered among the hardest metals to weld well, so quality control is especially important there. Overall, the wire feeder itself doesn't have too much impact on weld quality - a bunch of other factors including the power source, welder (person), joint preparation, consumable choice, and cleanliness of the welding environment will make a bigger difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel is an easier metal to feed through a welding lead, because it has a higher column strength, and is less likely to buckle. It's guided by a narrow tube ('liner') that runs through the cable, and is typically just pushed by the wire feeder (which may be built into the power supply or a standalone unit.) Because aluminum has a lower column strength, it is more likely to buckle, and thus not feed well. When this happens, the wire will stop feeding out of the gun, and just fill up the liner which then gets jammed and has to be cleaned out. This is a pain for the fabricator, but shouldn't affect weld quality other than maybe causing some extra starts and stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a short welding lead, and using large diameter aluminum wire, they may well be fine using the same wire feeder. If not, there are a number of other strategies. For small amounts of work, they may use a spool gun which is essentially a miniature wire feeder which the welder carries around. A small spool of aluminum wire is actually inside the gun, along with the drive rollers which only have to push the aluminum a few inches as a column before it gets to the weld. Another common strategy is a push-pull gun, where the wire feeder pushes the wire with to help it come off the spool, but additional drive rollers in the gun pull it to keep some tension on the wire as it goes through the liner. It is possible that they are using the same wire feeder, but changing from a conventional MIG gun to a push-pull gun when they switch from steel to aluminum. That would be a common practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, I wouldn't be worried about their choice of wire feeder. For the most part, it will work or it won't. I would be more concerned with their overall quality control system. Are welders working to a written procedure? Have welders been tested before they're set loose to weld? How and how often are finished welds inspected? If they are welding to a code, these answers will be dictated by it. If they are not welding to a code, you can get a rough idea of what level of QC they employ.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-02-06T14:44:10.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="496" PostTypeId="2" ParentId="494" CreationDate="2015-02-06T14:47:18.427" Score="6" Body="&lt;p&gt;This assumes US codes are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question of whether or not a quality weld can be produced needs to be proven through testing. Per AWS (American Welding Society) codes (D1.1, D1.5, etc) the welder (person) must be certified for the weld type (FCAW, SMAW, etc), the materials to be joined, and the position to be welded (flat, vertical, overhead, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to all of that, the actual weld to be used needs to have been tested to prove that a quality weld can be made in a perfect situation. This is through the PQR (procedure qualification record) and WPS (weld process specification) process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All welds are also tested. This can be visual all the way up to xray. At the very least all welds are supposed to be visually inspected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this testing and paper work is there so that you have the best possible chance of getting a good final product. It is a lot of work and expense, but it is the only way that you can have confidence in a weld without wondering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In answer to this specific question, if all of the testing and paperwork has been done correctly, the contractor's model of welding machine doesn't really matter.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-06T14:47:18.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="497" PostTypeId="2" ParentId="475" CreationDate="2015-02-06T14:57:21.617" Score="2" Body="&lt;p&gt;I assumed you'd use normal bolts, but use a nylon washer to electrically insulate the bolt from the weathering steel (in order to avoid bi-metallic corrosion).&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-02-06T14:57:21.617" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="498" PostTypeId="1" AcceptedAnswerId="500" CreationDate="2015-02-06T20:51:23.443" Score="10" ViewCount="1206" Body="&lt;p&gt;Rockets like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ariane_5&quot;&gt;Ariane 5&lt;/a&gt; weigh hundreds of tons, but with about 85% of that weight being fuel, the payload fraction is only about 3% (~10-20 tons).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Virgin Galactic is building &lt;a href=&quot;https://en.wikipedia.org/wiki/VSS_Voyager&quot;&gt;suborbital space planes&lt;/a&gt;, mostly for tourism purposes. They fly at about Mach 4, too slow to escape earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I wonder if an air-launched spacecraft in the next 20 years realistically could really fly us to the moon - that is, could they reach escape velocities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As side question in case they can: Would they be more or less fuel efficient than standard rockets such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Saturn_V&quot;&gt;Saturn V&lt;/a&gt;? How much payload would be realistically being transportable?&lt;/p&gt;&#xA;" OwnerUserId="261" LastEditorUserId="106" LastEditDate="2015-02-07T20:58:10.603" LastActivityDate="2019-02-20T03:02:40.087" Title="Could air-launched space planes reach escape velocities?" Tags="|aerospace-engineering|aircraft-design|propulsion|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="499" PostTypeId="2" ParentId="193" CreationDate="2015-02-06T23:27:14.300" Score="8" Body="&lt;p&gt;Another option to the radiant solution above would be under floor air distribution systems (UFAD) using displacement ventilation. These systems require a raised floor plenum which works well in theatre style auditoriums, where you often have this anyway. The plenum essentially becomes your supply duct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The systems are sized to condition the occupied zone only, say the first 2m over the outlet. So no matter how high you ceiling is, you'll end up heating only a 2m high volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike in radiant systems, which heat people directly and do not change the air temperature, the outside air component is treated. Whether this is an advantage or not depends on the required outside air volumes, where these volumes are introduced relative to the occupants (i.e. are you blowing cold air on someone?) as well as expected comfort levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Opportunities for energy savings in UFAD systems&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lower supply temperatures when heating, higher supply temperatures when cooling&lt;/li&gt;&#xA;&lt;li&gt;The conditioned volume is limited to the occupied zone no matter what the ceiling height is&lt;/li&gt;&#xA;&lt;li&gt;Supply velocity is lower, however supply volumes are increased; there might be some value in optimisation here and thinking about perimeters and internal zoning&lt;/li&gt;&#xA;&lt;li&gt;Lower pressure drop in supply&lt;/li&gt;&#xA;&lt;li&gt;Throw in some heat exchanger to capture the heat from the return air&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A few other things to note:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lower velocity and higher volumes taken together means that the number of outlets might increase and that these systems cannot handle the same peak loads as conventional systems; this will be important for perimeter zones. The building fabric must be optimised to limit peak loads (losses/gains through glazing)&lt;/li&gt;&#xA;&lt;li&gt;Lower velocities also means less problems with draft (in theory at least)&lt;/li&gt;&#xA;&lt;li&gt;If the acoustic zoning doesn't match the mechanical zoning, things will get a bit tricky; acoustic separation might require walls being taken down through to the plenum&lt;/li&gt;&#xA;&lt;li&gt;If you have a tiled flooring system, the outlet diffusers can be swapped around which has multiple benefits: for one, it provides a very flexible room layout that can be changed at any time; in addition, the systems introduce some flexibility for occupants to make themselves comfortable by relocating diffuser tiles&lt;/li&gt;&#xA;&lt;li&gt;Wikipedia has an article which provides a general overview of &lt;a href=&quot;https://en.wikipedia.org/wiki/Underfloor_air_distribution&quot; rel=&quot;noreferrer&quot;&gt;UFAD systems&lt;/a&gt; and one for &lt;a href=&quot;https://en.wikipedia.org/wiki/Displacement_ventilation&quot; rel=&quot;noreferrer&quot;&gt;displacement ventilation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is &quot;most efficient&quot; depends on many other things besides space heights. UFADs can be a good solution in certain situations but same is true for e.g. radiant systems. What you'd definitely like to avoid is ending up conditioning a 4m high volume of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few UFAD images from the wiki page to save you a click:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/gdkz7.png&quot; alt=&quot;UFAD concept section&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iOFq0.jpg&quot; alt=&quot;UFAD stratification&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="75" LastEditorUserId="75" LastEditDate="2015-02-07T03:42:11.730" LastActivityDate="2015-02-07T03:42:11.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="500" PostTypeId="2" ParentId="498" CreationDate="2015-02-07T00:15:11.510" Score="11" Body="&lt;p&gt;Believe it or not, we could have done this 50 years ago, if government funding hadn't been pulled from a project at the last minute. Frustratingly, after years of work by scientists, engineers and technicians, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Boeing_X-20_Dyna-Soar&quot; rel=&quot;noreferrer&quot;&gt;Boeing X-20 Dyna-Soar&lt;/a&gt; project was cancelled just after work had started on the actual spacecraft.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/7/71/NASA_Color_Dyna_Soar.jpg&quot; rel=&quot;noreferrer&quot;&gt;Here's&lt;/a&gt; an artist's impression of the X-20:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/7/71/NASA_Color_Dyna_Soar.jpg&quot; alt=&quot;X-20&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The X-20 was the result of a military program that aimed to develop an orbital spaceplane to be used for bombing and reconnaissance. It was designed to be launched into orbit and to stay there for a short while. Despite its small size - only 35 feet long - it would reach orbital speeds after launch, in theory. It managed to get to Mach 18 during practice glide tests.&lt;/p&gt;&#xA;&lt;p&gt;The X-20 was not designed to be air launched, but to be launched on top of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Titan_(rocket_family)#Titan_III&quot; rel=&quot;noreferrer&quot;&gt;Titan III&lt;/a&gt; missile. &lt;em&gt;However&lt;/em&gt;, a similar design - a precursor to the X-20, if you will - called &lt;a href=&quot;http://www.astronautix.com/craft/bomi.htm&quot; rel=&quot;noreferrer&quot;&gt;Bomi&lt;/a&gt; &lt;em&gt;was&lt;/em&gt; designed to be launched like this. &lt;a href=&quot;http://www.astronautix.com/graphics/b/boostgli.jpg&quot; rel=&quot;noreferrer&quot;&gt;Here&lt;/a&gt;&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt; is a comparison of Bomi (on the left), the X-20 (the two on the far right), and Robo, a related project:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CKHzd.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CKHzd.jpg&quot; alt=&quot;Bomi&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://www.astronautix.com/graphics/b/boostgli.jpg&quot; rel=&quot;noreferrer&quot;&gt;astronautix.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;There were two versions of Bomi: a suborbital one, with a maximum speed of Mach 4, and an orbital one, with a maximum speed of - well, orbital velocity. The latter is probably the one you're interested in. It would have been 23 feet long and would have had a payload of 34,000 kilograms - enough for two nuclear bombs.&lt;/p&gt;&#xA;&lt;p&gt;Both versions would have been launched on some sort of launcher - the larger vehicle which Bomi is shown attached to. This design might also be changed depending on whether the flight was to be orbital or sub-orbital.&lt;/p&gt;&#xA;&lt;p&gt;Bomi was eventually cancelled as funding was pulled for &lt;a href=&quot;http://www.astronautix.com/craft/dynasoar.htm&quot; rel=&quot;noreferrer&quot;&gt;Dyna-Soar&lt;/a&gt; (the X-20), which then suffered the same fate. But Dyna-Soar got past the glide-test stage (being dropped from a B-52), and almost actually made it to space. Had the resources been moved to Bomi, it could have succeeded.&lt;/p&gt;&#xA;&lt;p&gt;Could Bomi have escaped Earth orbit? With a bit of work, it could have. Think of how various rocket families have evolved. Different types can fulfill different missions. The Saturn V was the end result of smaller, suborbital and orbital rockets. If Bomi had been developed to the extent of the Apollo program, I think it's very likely that it could have made it out of Earth orbit.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$^1$&lt;/span&gt; This image appears to be in the public domain, as stated &lt;a href=&quot;http://www.astronautix.com/articles/conssite.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-02-16T16:30:17.970" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="501" PostTypeId="2" ParentId="119" CreationDate="2015-02-07T04:46:17.343" Score="2" Body="&lt;p&gt;I wouldn't be too stuck on the &quot;salesman&quot; term here. Engineering is (supposed to be) a (mostly) rational enterprise (sometimes). So the general approach would be to identify and understand technical and business related aspects/concerns and address them rationally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stabbing comfortably in the dark about your specific issue, I would say it is probably worth thinking about whether the most important aspects and impacts of your idea are clear in your peers' minds. Things like money, time/resourcing, potential to win projects could be considerations you'd want to address. The other answers provide already some good comments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to add the following that I think hasn't been mentioned yet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Timing in regards to impacted disciplines&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good idea / innovation may struggle if it means immediate, additional and potentially unpaid work for a partner discipline or colleague. E.g. (coming from a building design and construction background) if the building service engineers have essentially finished their design and your new idea would lead to substantial changes to their services, you have a problem because you are asking them either to work for free or to enter variation negotiations with the client. Same thing would happen no matter what stage they are if they have overclocked their hours for whatever reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible solutions could include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Push for your idea earlier in the project to give your colleagues time to digest and understand the impact of your idea&lt;/li&gt;&#xA;&lt;li&gt;Make your idea part of your team's bid (your idea might win the next project for the team)&lt;/li&gt;&#xA;&lt;li&gt;Is there an opportunity to influence the project brief?&lt;/li&gt;&#xA;&lt;li&gt;Consider that it might not happen in this project. Take your time to explain the idea to colleagues outside the context of the immediate project. Listen, ask questions, try to understand the reasons for the resistance (there might or might not be good reasons as already discussed by @TomAu and @jhabbot).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Resourcing impacts of partner disciplines&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your partner disciplines don't understand your idea, they might decide not to support it since they will have a hard time identifying how much resources and time they should allocate. Project time is often not the right time to fix this; people chase deadlines then (although sometimes project time is the only opportunity). Try to introduce ideas early on. Listen to your peers' concerns and offer help overcoming them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Understanding the design process within your context&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who pays for the project? Who benefits from the specific idea? Who is impacted during design, execution, operation, phase out? Are there implications to the project program or budget (for clients and/or any disciplines involved) if your idea is taken up? Is special resourcing required? Does it make someone obsolete?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Don't get stuck on your one idea and on this one colleague who keeps telling you to improve your salesmanship&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this particular idea doesn't fly today, let it go for now, wait for the next one, however keep the idea at the back of your head and mention it again when the time comes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't convince this one key person, find another one or wait for another one. In the meantime keep asking what they mean by &quot;salesmanship&quot;, if they have a good example for it, what they would have expected from you, if they could get you a mentor or if they could mentor you themselves. Their response to these questions might also help you understand their motivation and thinking.&lt;/p&gt;&#xA;" OwnerUserId="75" LastEditorUserId="75" LastEditDate="2015-02-09T10:17:37.293" LastActivityDate="2015-02-09T10:17:37.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="502" PostTypeId="1" AcceptedAnswerId="509" CreationDate="2015-02-07T06:28:30.450" Score="6" ViewCount="127" Body="&lt;p&gt;I have a proposed underfloor air distribution system (UFAD) for a one level office in a temperate climate. Typically the demand for heating and cooling in this climate is about even for offices. However the building is well insulated and has a fair amount of (shaded) low U-value glazing on the sun facing elevation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question boils down to whether the thermal mass of the slab will be beneficial overall or not. At the moment I tend to have insulation within the plenum, i.e. excluding the impact of the thermal mass to create a more controlled supply, instead of having the insulation under the slab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on some handwavy assumptions though, I could see some benefits of including the mass of the slab in order to reduce cooling loads (which I expect to be somewhat dominant due to the glazing and amounts of insulation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However what would happen once the slab is fully heated, i.e. runs out of capacity? Would I night purge it? Can someone share their thoughts/experiences? When I insulate the plenum, should I include or exclude the slab?&lt;/p&gt;&#xA;" OwnerUserId="75" LastEditorUserId="75" LastEditDate="2015-02-09T10:46:25.823" LastActivityDate="2015-02-09T10:46:25.823" Title="Where to insulate Underfloor air distribution system in a temperate climate" Tags="|mechanical-engineering|building-physics|hvac|energy-efficiency|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="504" PostTypeId="1" AcceptedAnswerId="510" CreationDate="2015-02-07T11:39:17.593" Score="0" ViewCount="143" Body="&lt;p&gt;The Coriolis effect impacts the way water tends to circle before going down the drain.  Many (most, all?) toilets are designed to work with the Coriolis effect.  I am considering a line of toilets that would flush opposite of this, this is strictly a marketing device.  But I am concerned that if the toilet design, forces the water to circle opposite of the Coriolis effect before going down the drain it could cause issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it problematic if we build toilets that flushes backwards? &lt;/p&gt;&#xA;" OwnerUserId="257" LastActivityDate="2015-02-07T15:20:41.263" Title="Is making a toilet flush backwards problematic?" Tags="|building-physics|waste-disposal|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ClosedDate="2015-02-09T17:15:12.593" ContentLicense="CC BY-SA 3.0" />
  <row Id="505" PostTypeId="1" CreationDate="2015-02-07T12:41:24.260" Score="4" ViewCount="1677" Body="&lt;p&gt;I purchased a Weller WSD161V soldering station (similar to &lt;a href=&quot;http://www.starelectronic.co.th/upload/images/Document/3.%20%E0%B8%AA%E0%B8%B4%E0%B8%99%E0%B8%84%E0%B9%89%E0%B8%B2%E0%B8%AD%E0%B8%B8%E0%B8%AA%E0%B8%B2%E0%B8%AB%E0%B8%81%E0%B8%A3%E0%B8%A3%E0%B8%A1/Hand%20Tools/Weller/WDD81V.pdf&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;, at least with regards to the pneumatic system I'm about to describe -- sorry, I was unable to find a manual for my specific model on the internet) with a &lt;a href=&quot;http://www.tme.eu/en/Document/5db584fd0c811610cf63bdad6da68bf4/OI_DSX_80_89421.pdf&quot; rel=&quot;nofollow&quot;&gt;DSX80&lt;/a&gt; desoldering iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This desoldering iron works like an automatic desoldering pump -- when pressing a button on the iron, it suctions the molten solder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To work, it needs (as described on the link above) a 35 l/min supply of dry, oil-free, 58-87 psi (4-6 bar) compressed air. I recall reading somewhere that there is a venturi inside the soldering station which generates the vacuum for the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to space and noise constraints, I decided to purchase a tankless air compressor -- specifically, the &lt;a href=&quot;http://www.einhell.com/com_en/shop/4862/4891/4908/4909/air-compressor-kit-bt-ac-180-kit.html&quot; rel=&quot;nofollow&quot;&gt;Einhell BT-AC-180&lt;/a&gt;. It is able to supply compressed air at a pressure of 8 bar, with a flow rate of 77 l/min at 4 bar and 63 l/min at 7 bar, so it should be plenty to supply my desoldering iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have a couple of questions -- keep in mind my expertise is in electronics, and I have very little knowledge of pneumatic equipment:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The compressor doesn't appear to have any kind of pressure regulator. Since my desoldering system only consumes 35 l/min of air, which is well above what the compressor can supply even at a pressure of 8 bar, I understand the compressor should well be capable of supplying the full 8 bar to the desoldering system. This would be out of spec for the system, and I'm afraid it might blow some kind of safety valve inside it. &lt;strong&gt;Does there exist some kind of pressure regulator I can couple to the output of the compressor, which will drop pressure to the safe 4-6 bar range for the desoldering system?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In looking for such a system, I was told by a salesman at a compressor shop that if a tankless compressor's flow is impeded, their internal safety valve will blow, so there needs to be a path for the compressed air to get out at all times. I've learned not to take salesmen at their word, and in this particular case I'm inclined to believe this is wrong, because the compressor in question comes with some kind of pistol which can cut out the flow of air -- I guess the manufacturer wouldn't include such an accessory if it would cause the compressor's safety valve to blow out. Still, since I recently purchased this compressor, there's still time to return it to the dealer if it is unfit for my purpose. &lt;strong&gt;Is this compressor compatible with the desoldering system, or should I return it and go for a compressor with a tank?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The desoldering system calls for dry air, and I was told by a salesman that condensation is an issue with compressors -- at first I found it weird since compressing air should increase (not decrease) its temperature, but now I believe I understand the effect, which should be similar to a capillary tube in refrigeration systems: when the flow of air is restricted, its pressure drops and so does the temperature, which may then form condensation. The salesman said the solution to this is adding a filter to the compressor output. &lt;strong&gt;Is my understanding of the phenomenon correct? Will a filter solve the problem? What kind of filter should I look for?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Taking into account my complete lack of experience with pneumatic systems, any other tips on how to connect the system, what extra equipment to purchase, etc. are welcome.&lt;/p&gt;&#xA;" OwnerUserId="410" LastEditorUserId="110" LastEditDate="2015-02-10T11:44:53.883" LastActivityDate="2020-04-26T01:28:37.513" Title="Tankless air compressor for electronics desoldering station" Tags="|mechanical-engineering|electrical-engineering|pressure|pumps|compressed-air|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="506" PostTypeId="2" ParentId="504" CreationDate="2015-02-07T12:46:42.490" Score="5" Body="&lt;p&gt;Per the ever informative &lt;a href=&quot;http://www.snopes.com/science/coriolis.asp&quot;&gt;Snopes&lt;/a&gt; &#xA;The hemisphere doesn't affect the spin of the water. This means that the water spin down the drain is controlled by the design of the drain. This shouldn't be hard to design for.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-07T12:46:42.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="507" PostTypeId="5" CreationDate="2015-02-07T13:24:00.807" Score="0" Body="&lt;p&gt;Welding is the process of physically combining two pieces together. This is usually accomplished by melting the components, but can also be done mechanically in the friction stir process. &lt;/p&gt;&#xA;" OwnerUserId="-1" LastEditorUserId="33" LastEditDate="2016-03-23T00:52:57.320" LastActivityDate="2016-03-23T00:52:57.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="508" PostTypeId="4" CreationDate="2015-02-07T13:24:00.807" Score="0" Body="Use this tag when the question deals with the process of making a weld. For questions about weld design or the use of welds, use the [welds]." OwnerUserId="-1" LastEditorUserId="33" LastEditDate="2016-03-23T00:52:57.320" LastActivityDate="2016-03-23T00:52:57.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="509" PostTypeId="2" ParentId="502" CreationDate="2015-02-07T15:05:00.200" Score="4" Body="&lt;p&gt;It depends &amp;mdash; can you run air through the plenum without simultaneously running the active heating/cooling system? If not, you'll never get the benefit of its thermal storage capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when heating, the plenum will always be hotter than the slab, putting heat into it. In order to extract heat from the slab, you need to be able to put building air that's cooler than the slab directly into the plenum. Yes, there will be some transfer of heat directly from the slab to the rooms through the floor, but that would only occur near the plenum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing applies, with the heat flow reversed, when cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are secondary issues, too. The transient response of the system (i.e., the ability to change the temperature of the building interior rapidly) will be slowed down by the thermal mass of the slab if the plenum is not insulated. This could be a problem in some situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when cooling, there could be condensation issues in the plenum. With active cooling, excess moisture in the air is removed at the expansion coils, and the system is designed to handle it. However, when doing passive cooling from the slab, the condensation will occur inside the plenum, which could cause maintenance and reliability issues.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-02-07T15:05:00.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="510" PostTypeId="2" ParentId="504" CreationDate="2015-02-07T15:20:41.263" Score="4" Body="&lt;p&gt;Per the &lt;a href=&quot;http://en.wikipedia.org/wiki/Coriolis_effect#Draining_in_bathtubs_and_toilets&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on the Coriolis effect&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Contrary to popular misconception, water rotation in home bathrooms under normal circumstances is not related to the Coriolis effect or to the rotation of the earth, and no consistent difference in rotation direction between toilet drainage in the Northern and Southern Hemispheres can be observed.&lt;/p&gt;&#xA;&lt;p&gt;Only if the water is so still that the effective rotation rate of the earth is faster than that of the water relative to its container, and if externally applied torques (such as might be caused by flow over an uneven bottom surface) are small enough, the Coriolis effect may determine the direction of the vortex.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can be forgiven for believing this to be true though.  Again from the Wikipedia article:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The idea that toilets and bathtubs drain differently in the Northern and Southern Hemispheres has been popularized by several television programs and films, including Escape Plan, Wedding Crashers, The Simpsons episode &amp;quot;Bart vs. Australia&amp;quot;, and The X-Files episode &amp;quot;Die Hand Die Verletzt&amp;quot;.[30] Several science broadcasts and publications, including at least one college-level physics textbook, have also stated this.[31][32]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="272" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-02-07T15:20:41.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="511" PostTypeId="1" CreationDate="2015-02-07T15:45:35.313" Score="1" ViewCount="394" Body="&lt;p&gt;What type of sensing element is used in Water Quality Monitoring meters such as those pictured below? I might not be using the correct search terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MYVR2.jpg&quot; alt=&quot;Water Quality Meter Example 1&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/gvFG3.jpg&quot; alt=&quot;Water Quality Meter Example 2&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/7U8uU.jpg&quot; alt=&quot;Water Quality Meter Example 3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these meters monitor Total Dissolved Solids (TDS) or pH levels in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect there is an electro-chemical sensor. I have not been able to locate any information on the sensing element or the mechanism. I intend to build a simple water quality meter using a simple microprocessor and respective sensing element.&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="368" LastEditDate="2015-03-05T22:32:41.833" LastActivityDate="2015-05-08T18:48:27.210" Title="What type of sensing element is used in Water Quality Monitoring Meters?" Tags="|electrical-engineering|civil-engineering|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="512" PostTypeId="2" ParentId="253" CreationDate="2015-02-07T16:09:01.173" Score="6" Body="&lt;p&gt;The choice of time step sets the bandwidth of the control loop.  The highest unity gain frequency (UGF) you can hope to achieve in the closed loop is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Nyquist_frequency&quot; rel=&quot;noreferrer&quot;&gt;Nyquist frequency&lt;/a&gt;&#xA;$$&#xA;f_N=\frac12 f_s=\frac{1}{2\,\Delta t}&#xA;$$&#xA;where $\Delta t$ is the sample time.  Practically, the UGF will be somewhat lower than this. This means that above this frequency your feedback will not be suppressing the disturbance fluctuations in your system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UGF also limits how much gain you can have at frequencies below but near the UGF.  For frequencies within an order of magnitude of the UGF, $\text{UGF}/10$, you won't be able to have a gain much higher than $\sim10$.  A gain of $10$ in the closed loop means that disturbance fluctuations at those frequencies are suppressed by a factor of 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,the choice of operating frequency is a practical one.  Faster systems are more expensive; slower systems may not provide enough disturbance suppression.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-02-07T16:09:01.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="513" PostTypeId="2" ParentId="438" CreationDate="2015-02-07T17:42:12.933" Score="2" Body="&lt;p&gt;Look at the typical diagram of soil showing soil/water/air:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2unKp.png&quot; alt=&quot;Soil&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Simple&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking simplistically about it the items that could change:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Soil mass can't change. No soil was added. It would be good to assume that no major chemical reactions occurred either.&lt;/li&gt;&#xA;&lt;li&gt;Water mass can change. It was sitting in water.&lt;/li&gt;&#xA;&lt;li&gt;Air can't increase if the sample was submerged. Once again ignore any major chemical reactions that might create gas.&lt;/li&gt;&#xA;&lt;li&gt;Mass and volume have a well defined ratio for each substance.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From these items, the only way that volume could increase would be if the volume of water increased. This would mean an increase in the volume of voids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the simple (maybe naive) way to think about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also where the &lt;a href=&quot;http://en.wikipedia.org/wiki/Atterberg_limits&quot; rel=&quot;nofollow noreferrer&quot;&gt;Atterberg Limits&lt;/a&gt; come into play. They define the water contents where the physical properties of the soil change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Complicated&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more complicated way to think about the system would be to consider chemical changes to the soil. Without getting too specific into items that I'm not qualified to explain, it is possible that chemical reactions could occur that cause the soil volume to increase by itself. Think about how rust is a chemical reaction that effectively causes steel to increase in volume. This would also change the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Including chemical reactions into the mix creates questions like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does it make any sense to compare the properties of this &lt;em&gt;new&lt;/em&gt; soil compound to the &lt;em&gt;old&lt;/em&gt; soil compound?&lt;/li&gt;&#xA;&lt;li&gt;Is the reaction reversible? e.g. Does drying the sample cause everything to go back to the original masses and volumes?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Without more constraints on what we are working with, it is hard to give a definitive answer.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-07T17:42:12.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="514" PostTypeId="1" AcceptedAnswerId="516" CreationDate="2015-02-07T17:46:07.500" Score="9" ViewCount="6596" Body="&lt;p&gt;I heard that it is unwise to choose cylindrical horizontal beam. My question is why?&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2015-02-07T23:53:25.307" Title="Why is it unwise to choose cylindrical horizontal beam" Tags="|civil-engineering|beam|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="515" PostTypeId="2" ParentId="511" CreationDate="2015-02-07T18:03:13.140" Score="4" Body="&lt;p&gt;Hand-held sensors measuring multiple properties may require more than one type of sensor element; it's important to know what exactly you need to measure. There is no one property (or fixed set of properties) that defines &quot;water quality&quot; though some (including the two you mention, pH and TDS) are more commonly used than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of pH and TDS, either can be determined by passing a current through the sample and measuring the voltage of the circuit. Rather than using a specialized sensor, the unit may simply use a battery with some fixed electrodes (so the distance traveled through the sample is known); your microprocessor board can handle the voltage measurement on its own. &lt;a href=&quot;http://www.who.int/water_sanitation_health/resourcesquality/wqmchap6.pdf&quot; rel=&quot;nofollow&quot;&gt;This water quality monitoring text&lt;/a&gt; gives the following expectation for commercial devices: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pocket-sized, battery-powered, portable meters that give readings with&#xA;  an accuracy of ± 0.05 pH units are suitable for field use.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This should be achievable with a &quot;DIY&quot; device, though calibration may be tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source above also points to an electrode-based method as being acceptable for dissolved oxygen (DO) measurements, though you haven't mentioned that as one of your target measurements. The electrodes required may work for pH as well, or you may require separate circuits; &lt;a href=&quot;http://www.hannainst.com/usa/prods2.cfm?ProdCode=HI%201285-5&amp;amp;id=005004&quot; rel=&quot;nofollow&quot;&gt;this probe&lt;/a&gt; measures pH and electrical conductivity (EC) with what looks like a single electrode. It's used in &lt;a href=&quot;http://www.hannainst.com/usa/prods2.cfm?ProdCode=HI%209813-5&amp;amp;id=005002&quot; rel=&quot;nofollow&quot;&gt;this device&lt;/a&gt;, which also measures TDS, and I'm guessing it does so by using the EC measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best option may be to find a pet store that specializes in aquarium supplies; they sell many different affordable sensor devices for home monitoring. You should be able to find manufacturer contact information if none of those devices can be modified, cannibalized or reverse-engineered to meet your needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should expect that TDS in the sample may affect pH readings, and vice versa, since you'll be measuring them in nearly the same way (conductivity and voltage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related Reading&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if you've already come across these in your research since you didn't elaborate in the question, but readers may get some useful information from the following sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://water.usgs.gov/owq/methods.html#field&quot; rel=&quot;nofollow&quot;&gt;USGS Water Quality Methods &amp;amp; Techniques - Field Methods &amp;amp; Equipment&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cramwetlands.org/about&quot; rel=&quot;nofollow&quot;&gt;California Rapid Assessment Method for wetland monitoring&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.epa.gov/fem/methcollectns.htm&quot; rel=&quot;nofollow&quot;&gt;US EPA Test Methods&lt;/a&gt; - These are sometimes very specific and you may find one that identifies particular sensor types. See also: &lt;a href=&quot;http://water.epa.gov/scitech/methods/cwa/&quot; rel=&quot;nofollow&quot;&gt;Clean Water Act Analytical Methods&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.iccontrols.com/files/TDS_Measurement.pdf&quot; rel=&quot;nofollow&quot;&gt;Total Dissolved Solids Measurement&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-05-08T18:48:27.210" LastActivityDate="2015-05-08T18:48:27.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="516" PostTypeId="2" ParentId="514" CreationDate="2015-02-07T18:12:45.387" Score="14" Body="&lt;p&gt;The strength of a horizontal beam is defined based on a property called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Section_modulus&quot;&gt;section modulus&lt;/a&gt;. Shapes that have more material distributed near the top and the bottom have a higher section modulus. This is why I-shaped beams or tall rectangles are common choices. Round shapes, on the other hand, have most of their material concentrated around the center of the shape, and very little at the top and the bottom, so they don't make very strong beams. That means a round beam would need a lot more material to achieve the same strength as a taller shaped beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could see this intuitively if you take a ruler or other piece of thin material you have handy and try to bend it. You'll notice that in the flat direction, it's very weak and easy to bend. In the wide direction though, it's quite strong. Even though the piece of material is the same size, having more of it at the top and bottom makes it stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this isn't too hard to understand, when a beam bends it can't stay perfectly straight. In order to form a curve it has to change shape slightly. The top of it gets compressed (pushed inwards) and the bottom part experiences a tensile stress (gets pulled apart.) As you move closer to the center, there is less stress of each type. In fact, when you get to the middle, the beam is neither getting pulled apart or pushed together, it is called the 'neutral axis.' So the top and bottom have to be strong to resist the compression and tension, but the center only has to be strong enough to keep everything attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you did really want to use a cylindrical shape for a horizontal beam (maybe for visual reasons or wind resistance) a round tube would be more efficient than a solid cylinder, because it has less material around the neutral axis. It still wouldn't be as efficient as a rectangular tube or traditional structural shape though.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-02-07T18:12:45.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="517" PostTypeId="2" ParentId="514" CreationDate="2015-02-07T19:03:14.683" Score="13" Body="&lt;p&gt;A cylindrical beam is probably not the best beam for given situation.  This doesn't mean that there are no specific situations where a cylindrical beam would be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pros&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can resist the same load in any direction or combination of directions. It is as strong in the horizontal direction as the vertical direction.&lt;/li&gt;&#xA;&lt;li&gt;No need to worry about or check the Lateral Torsional Buckling (LTB) limit state.&lt;/li&gt;&#xA;&lt;li&gt;Works well as beam-column (moments and axial loads together). &lt;/li&gt;&#xA;&lt;li&gt;The circular shape may have less surface area than other shapes when considering things like painted area or ice buildup.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cons&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Not as efficient (weight of material per strength) as an I-shape for loads in only one direction.&lt;/li&gt;&#xA;&lt;li&gt;Sections may be difficult to find or more costly than more common shapes.&lt;/li&gt;&#xA;&lt;li&gt;Connections at the ends and to other members coming in from the side are more complicated.&lt;/li&gt;&#xA;&lt;li&gt;Stiffeners under areas of concentrated loads are harder to design and fabricate.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unless one of the items in the &quot;Pros&quot; list greatly outweighs the &quot;Cons&quot;, a cylindrical beam is probably not the best option.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-07T19:03:14.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="518" PostTypeId="1" AcceptedAnswerId="534" CreationDate="2015-02-07T19:50:24.377" Score="5" ViewCount="251" Body="&lt;p&gt;I am familiar with the &lt;a href=&quot;https://engineering.stackexchange.com/questions/63/what-are-the-pros-and-cons-of-a-traffic-circle-versus-a-traffic-light-intersecti&quot;&gt;round-a-bout/traffic light pros and cons&lt;/a&gt;, however, I've been tasked with deciding on a solution for the given parameters below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, there is a two-way (four lane) road with a speed limit of 35 mph. Rough traffic counts for peak hours are 580 vehicles in the morning (primarily North bound), 660 vehicles mid-afternoon (North/South split), and 400 evening (primarily South bound).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Access to a new facility will create a need for an access road resulting in a 3-way intersection. Access road will be to the East and for the most part only North bound traffic will make right turns to enter the facility. Less than 10% of the North bound traffic is expected to use the facility at any given time. Once finished at this facility, it is expected that the traffic will leave via the same access road and make right hand turns to continue North bound. However, as much as 20% of this traffic will need to make a left turn to return South bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume a community with some, but very little experience with round-a-bouts, and a primary desire to keep North and South bound traffic flow steady.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Giving your opinions and reasons, would you consider a round-a-bout or traffic control device utilizing a camera system for the left turn lane on the access road to be better?&lt;/p&gt;&#xA;" OwnerUserId="278" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-02-10T13:26:14.197" Title="Best traffic control device/system for given situation" Tags="|civil-engineering|highway-engineering|traffic-intersections|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="519" PostTypeId="1" CreationDate="2015-02-07T20:05:11.193" Score="7" ViewCount="2189" Body="&lt;p&gt;The specifications for standard (ASTM A 325) bolts do not list minimum service temperatures. I am thinking of outdoor applications like Alaska or Canada (lows of -60F) and wondering if the cold temperatures would have an effect on the bolts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a type of bolt (&lt;a href=&quot;http://www.portlandbolt.com/technical/specifications/astm-a320/&quot; rel=&quot;nofollow&quot;&gt;ASTM A 320&lt;/a&gt;) that has a grade that specifies testing at low temperatures (-150F). Those bolts seem to be for cryogenic facilities. It seems that this is over-kill for most outdoor locations where the coldest ever recorded temperature on earth is about &lt;a href=&quot;http://en.wikipedia.org/wiki/Lowest_temperature_recorded_on_Earth&quot; rel=&quot;nofollow&quot;&gt;-128F&lt;/a&gt; in Antarctica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the cold a concern for standard structural bolts? Are there European standards that might apply?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="368" LastEditDate="2015-02-13T23:43:03.430" LastActivityDate="2015-02-13T23:43:03.430" Title="Are standard structural bolts acceptable for all cold climates?" Tags="|bolting|structural-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="521" PostTypeId="2" ParentId="119" CreationDate="2015-02-07T23:34:52.557" Score="1" Body="&lt;p&gt;As once engineering career evolves it is important to develop soft skills. Good communication skills will help persuade peers to accept once point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A presentation with extensive engineering analysis data that appeals to engineering design team member doesn’t always appeal to manufacturing or operations engineering team members. In such cases it is a good idea to tone down engineering analysis data and highlight the benefits to manufacturing or operation engineering departments.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As once engineering career evolves it is beneficial to understand engineering’s impact on other department such as marketing, accounting, manufacturing and operations. Building good relationships with members of other departments helps the process communicating complex technical information. These internal departments our internal clients and they need to be server the same or better then external clients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Established reputable companies in US recognize the importance of soft skills. These companies have programs to develop soft skill and good communication skills for their employees. Personality test such as Myers and Briggs help determine the personality characteristics of individuals, and training programs to develop any area deemed needing attention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0140157352&quot; rel=&quot;nofollow&quot;&gt;“Getting to Yes: Negotiating Agreement Without Giving In”&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/How_to_Win_Friends_and_Influence_People&quot; rel=&quot;nofollow&quot;&gt;“How to Win friends are Influence People”&lt;/a&gt; are too good books to read to help address concerns of the stated question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0140157352&quot; rel=&quot;nofollow&quot;&gt;Getting to Yes: Negotiating Agreement Without Giving In&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/How_to_Win_Friends_and_Influence_People&quot; rel=&quot;nofollow&quot;&gt;How to Win Friends and Influence People&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Myers%E2%80%93Briggs_Type_Indicator&quot; rel=&quot;nofollow&quot;&gt;Myers–Briggs Type Indicator&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2015-02-07T23:34:52.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="522" PostTypeId="2" ParentId="514" CreationDate="2015-02-07T23:53:25.307" Score="2" Body="&lt;p&gt;The use of a cylindrical horizontal beam is unwise in that it has added weight that offers no added strength.Therefore you are paying for unneeded material which is unwise,unless aesthetic concerns require the use of the cylinder shape.&#xA; Highly unusual lateral loads might make the use of a cylindrical beam an option but those forces are normally addressed with other design components.&lt;/p&gt;&#xA;" OwnerUserId="416" LastActivityDate="2015-02-07T23:53:25.307" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="523" PostTypeId="2" ParentId="14" CreationDate="2015-02-08T00:48:30.343" Score="5" Body="&lt;p&gt;Diurnal swings, i.e. temperature differences between day and night, are also an important consideration in hot climates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the temperature at night drops to something acceptable, you can use night purge to cool a thermal mass (e.g. concrete slabs or walls, phase change materials if you'd like to get fancy). This works well together with a well insulated envelop. In essence what @MarchHo talked about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand temperatures are high 24/7 (the tropics come to mind), you would avoid using thermal mass as it will stabilise the high temperatures and make it more difficult to cool down the building by other means. Your main source of passive cooling are breezes. Insulation to the roof and extensive shading will be important. If possible, the floor could be raised to allow breezes extracting heat from both sides. In addition to cooling loads, dehumidifications loads will be significant and are hard to deal with passively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is based on common rules of thumb; there might be some smart design solutions that achieve comfort in different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you can go underground as for example in &lt;a href=&quot;https://en.wikipedia.org/wiki/Coober_Pedy&quot; rel=&quot;nofollow&quot;&gt;Coober Pedy&lt;/a&gt;. Find some images &lt;a href=&quot;https://duckduckgo.com/?q=coober+pedy+underground+house&amp;amp;t=ffsb&amp;amp;iax=1&amp;amp;ia=images&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="75" LastEditorUserId="75" LastEditDate="2015-02-08T03:24:57.433" LastActivityDate="2015-02-08T03:24:57.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="524" PostTypeId="2" ParentId="494" CreationDate="2015-02-08T02:13:30.783" Score="1" Body="&lt;p&gt;Yes,you can get good quality welds of both steel and aluminum with the same welding machine. A wire welder or MIG gun {Metal  Inert  Gas} is normally loaded with a 30lb. reel of copper coated steel wire of various diameters{20-45 mm.} depending on the thickness of steel to be joined.To weld aluminum you will bypass the steel reel drive ,plug in an aluminum spool gun loaded with a 5lb. reel of aluminum wire,here again of varying diameters,and change the weld cover gas from the CO 2 mix to a straight argon cover gas.Years ago the standard wire welding machine would have also had to have an additional component attached, a high frequency rectifier,but todays wire welders commonly come factory ready to accept an aluminum spool gun.The operators experience and dexterity now comes into the picture,but a failed weld will not be the machines' fault.  &lt;/p&gt;&#xA;" OwnerUserId="416" LastActivityDate="2015-02-08T02:13:30.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="525" PostTypeId="1" CreationDate="2015-02-08T02:35:57.640" Score="1" ViewCount="726" Body="&lt;p&gt;Some TIG electrodes in use in the early 1970's were marked Thoriated Tungsten and Non-Thoriated Tungsten. What were the specific purposes or properties for these TIG welding electrodes?&lt;/p&gt;&#xA;" OwnerUserId="416" LastActivityDate="2015-05-08T07:20:52.333" Title="Thoriated tungsten TIG electrodes were in use 45 years ago.What were their specific use and problems?" Tags="|welding|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="526" PostTypeId="1" AcceptedAnswerId="528" CreationDate="2015-02-08T03:37:28.050" Score="3" ViewCount="6320" Body="&lt;p&gt;Weathering steel is not typically painted, because it produces a thin rust layer that protects the rest of the steel from further corrosion. Would painting weathering steel be an acceptable additional line of protection, or would this cause some unforeseen complication?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="368" LastEditDate="2015-02-13T23:41:59.130" LastActivityDate="2015-02-13T23:41:59.130" Title="Is it acceptable to paint weathering steel?" Tags="|materials|steel|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="527" PostTypeId="2" ParentId="525" CreationDate="2015-02-08T04:13:27.973" Score="5" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.twi-global.com/technical-knowledge/faqs/health-and-safety-faqs/faq-the-use-of-thoriated-tungsten-electrodes/&quot; rel=&quot;noreferrer&quot;&gt;TWI-Global&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;blockquote&gt;&#xA;    &lt;p&gt;TIG welding electrodes usually contain small quantities of other metallic oxides which can offer the following benefits:-&lt;/p&gt;&#xA;  &lt;/blockquote&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;facilitate arc starting&lt;/li&gt;&#xA;  &lt;li&gt;increase arc stability&lt;/li&gt;&#xA;  &lt;li&gt;improve current-carrying capacity of the rod.&lt;/li&gt;&#xA;  &lt;li&gt;reduce the risk of weld contamination&lt;/li&gt;&#xA;  &lt;li&gt;increase electrode life&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oxides used are primarily those of zirconium, thorium, lanthanum, yttrium or cerium. Additions are usually of order 1%-4%. All these oxides greatly improve arc initiation, especially when direct current (DC) welding is employed. &lt;strong&gt;Thorium oxide (thoria) has been used for many years having been found effective in terms of long life and thermal efficiency&lt;/strong&gt;. Zirconium oxide (zirconia) has been commonly used for alternating current (AC) TIG welding, normally for welding aluminium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because thorium is radioactive there are hazards associated with using it. The main cause of concern with thorium are the alpha particles it emits. Alpha particles cannot penetrate the skin, but if they enter the lungs or digestive track they can be carcinogenic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue with using thorium in electrodes is not so much about radiation exposure during welding, but the potential to inhale or ingest thorium laced when the tips of electrodes are ground to have a conical point. The conical point is required to maintain maximum arc stability during welding. Due to electrode erosion during welding, the tip needs to be frequently ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the grinding process thorium laced dust is produced. If not enough care is taken during the grinding of the electrodes (via dust masks &amp;amp; ventilation) there is a risk to the person grinding the electrodes in breathing in the thorium, and alpha particles, in the dust. This then increases the posibility of the person doing the grinding of developing cancer. The risk of cancer is low, but it appears to be enough of an issue for the Danish Welding Institute to recommend the phasing out of thoriated electrodes because non-radioactive alternatives are available. The linked article goes into more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.weldingtipsandtricks.com/thoriated-electrodes.html&quot; rel=&quot;noreferrer&quot;&gt;Welding Tips &amp;amp; Tricks&lt;/a&gt; pure tungsten electrodes are inferior to thoriated electrodes becaue they &quot;quiver and dance&quot; when the current is increased, leading to difficulty in producing a good weld. The site recommend lanthanated electrodes for those concerned about radiation exposure associated with thoriated electrodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar warning about the inhalation or ingestion of thorium laced dust can be found on these sites:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.diamondground.com/thoriated-tungsten-warning/&quot; rel=&quot;noreferrer&quot;&gt;Diamond Ground Products&lt;/a&gt; with recommendations from the American Welding Society &amp;amp; the Welding Institute&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.millerwelds.com/resources/articles/tungsten-electrode-guide&quot; rel=&quot;noreferrer&quot;&gt;Miller Welds&lt;/a&gt; gives a brief comparison between pure tungsten and thoriated electrodes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tungsten electrodes have the highest consumption rate of all electrodes. They provide excellent arc stability for AC balanced wave welding and good stability for sine wave welding. They are generally not used for DC welding because the arc are not as strong as those produced by tungsten alloy electrodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thoriated electrodes are apparently the most commonly used electrodes. They are preferred for longevity and ease of use. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thorium increases the electron emission qualities of the electrode, which improves arc starts and allows for a higher current carrying capacity. This electrode operates far below its melting temperature, which results in a considerably lower rate of consumption and eliminates arc wandering for greater stability. It also features a lower level of weld contamination than other electrodes.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Unlike pure tungsten, these electrodes are only for speciality type AC welding (thin gauge aluminium or materials less than .060-in.), but they are exceptional for DC electrode negative or straight polarity on carbon and stainless steel, nickel and titanium applications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-02-08T04:13:27.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="528" PostTypeId="2" ParentId="526" CreationDate="2015-02-08T04:48:41.897" Score="4" Body="&lt;p&gt;According to &lt;a href=&quot;http://ntl.bts.gov/lib/32000/32400/32427/FHWA-RD-92-055.pdf&quot; rel=&quot;nofollow&quot;&gt;Maintenance Coatings of Weathering Steel &lt;/a&gt;, published in 1995 by the US Department of Transportation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From page 11 of the report:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Painting&#xA;  of&#xA;  new&#xA;  uncontaminated&#xA;  weathering&#xA;  steel&#xA;  is&#xA;  generally&#xA;  not&#xA;  considered&#xA;  a&#xA;  problem.&#xA;  Test&#xA;  fence&#xA;  and&#xA;  laboratory&#xA;  data&#xA;  developed&#xA;  by&#xA;  the&#xA;  paint&#xA;  industry&#xA;  have&#xA;  indicated&#xA;  that&#xA;  conventional&#xA;  coating&#xA;  systems&#xA;  such&#xA;  as&#xA;  oil&#xA;  alkyds&#xA;  and&#xA;  epoxies&#xA;  will&#xA;  perform&#xA;  comparably&#xA;  on&#xA;  weathering&#xA;  steel&#xA;  and&#xA;  on&#xA;  carbon&#xA;  steel&#xA;  if&#xA;  the&#xA;  degree&#xA;  of&#xA;  surface&#xA;  preparation&#xA;  is&#xA;  equivalent. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The&#xA;  major&#xA;  problem&#xA;  faced&#xA;  by&#xA;  highway&#xA;  departments&#xA;  and&#xA;  other&#xA;  owners&#xA;  of&#xA;  weathering&#xA;  steel&#xA;  structures&#xA;  is&#xA;  protecting&#xA;  weathering&#xA;  steel&#xA;  that&#xA;  has&#xA;  corroded&#xA;  in&#xA;  the&#xA;  presence&#xA;  of&#xA;  chlorides&#xA;  and&#xA;  other&#xA;  contaminants.&#xA;  Conventional&#xA;  cleaning&#xA;  techniques&#xA;  such&#xA;  as&#xA;  dry&#xA;  abrasive&#xA;  blasting&#xA;  do&#xA;  not&#xA;  remove&#xA;  the&#xA;  chlorides,&#xA;  which&#xA;  apparently&#xA;  penetrate&#xA;  the&#xA;  bases&#xA;  of&#xA;  pits&#xA;  in&#xA;  the&#xA;  steel.&#xA;  The&#xA;  performance&#xA;  of&#xA;  standard&#xA;  highway&#xA;  coatings&#xA;  such&#xA;  as&#xA;  oil&#xA;  alkyd,&#xA;  epoxies,&#xA;  and&#xA;  zinc-rich&#xA;  systems&#xA;  over&#xA;  chloride&#xA;  contaminated&#xA;  steel&#xA;  has&#xA;  not&#xA;  been&#xA;  satisfactory. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally, the &lt;a href=&quot;http://www.dot.state.fl.us/officeofdesign/CPR/WeatheringSteelBridges.shtm&quot; rel=&quot;nofollow&quot;&gt;Florida Dept of Transport&lt;/a&gt; is recommending using: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;weathering steel for new steel bridges in suitable environments.  Use a single coat of inorganic zinc paint system for extremely aggressive environments.  Where higher aesthetics are required, use a 3 coat inorganic zinc paint system with clear coat finish.  The use of the 3 coat system with clear coat finish should be an exception&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-02-08T04:48:41.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="529" PostTypeId="2" ParentId="191" CreationDate="2015-02-08T06:31:51.603" Score="13" Body="&lt;p&gt;That is correct, there are a number of unwanted, or tramp, metals (Cu, Sn, Sb, As) that enter the recycling stream from, for example, car bodies that are ground into scrap without removing all the copper wiring, or tin-coated steel cans. Antimony and arsenic tend to creep in from low-quality and low-cost primary iron sources.&lt;/p&gt;&#xA;&lt;p&gt;The answer to the question is no. Recycled steel is mixed as evenly as possible from varied sources, its composition is measured, and then pure iron is added as needed to dilute the tramp metals to tolerable levels for resale or further processing, such as meeting a specific steel grade for a specific product or application. Stainless steels and other high-alloy grades which are known at recycling time are processed separately due to the value of Ni, Cr, etc.&lt;/p&gt;&#xA;&lt;p&gt;It is currently uneconomical to reprocess iron to remove tramp elements, and so it simply isn't done at all. Two books mention the process as a regular and economical one: (&lt;a href=&quot;https://books.google.com/books?id=LTw0JJWGgvEC&amp;amp;pg=PA284&amp;amp;lpg=PA284&amp;amp;dq=dilution+of+tramp+element&amp;amp;source=bl&amp;amp;ots=EGct9ZyZE4&amp;amp;sig=VmDLOfuAM4QNi1iHh7vkVUQ7kjg&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=4qLXVKPTGIWpgwSuvYGYBQ&amp;amp;ved=0CB4Q6AEwAA#v=onepage&amp;amp;q=dilution%20of%20tramp%20element&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;Minerals, Metals and Sustainability: Meeting Future Material Needs&lt;/a&gt;, p. 284, starting at &amp;quot;dilution&amp;quot;) and (&lt;a href=&quot;https://books.google.com/books?id=OXuMAQAAQBAJ&amp;amp;pg=PA27&amp;amp;lpg=PA27&amp;amp;dq=dilution+of+tramp+element&amp;amp;source=bl&amp;amp;ots=V-I42OrM45&amp;amp;sig=EfKyoPAL-v8pR5roejsk_Isnxsw&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=4qLXVKPTGIWpgwSuvYGYBQ&amp;amp;ved=0CEkQ6AEwCA#v=onepage&amp;amp;q=dilution%20of%20tramp%20element&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;Steel Production: Processes, Products, and Residuals&lt;/a&gt;, starting on p. 104, read until it isn't relevant anymore). The reason it is uneconomical is that the tramp elements react more weakly with oxygen than iron at constant temperature, so to remove them by oxidation would require oxidizing all of the iron first. The reason for this is thermodynamic, and predicated on the fact that among competing reactions, those with the largest decreases in free energy proceed virtually to completion prior to other reactions even starting, especially with large differences in free energy among the competing reactions. To determine which reactions have the largest decreases, an Ellingham diagram may be used.&lt;/p&gt;&#xA;&lt;p&gt;In the Ellingham diagram below, the horizontal axis is temperature, the vertical axis is change in Gibbs free energy. The lines running across the diagram at various angles correspond to free energy change caused by element oxidation reactions with oxygen, as a function of temperature. In our case, the diagram may be read by choosing a temperature of interest, and reading up from the bottom to find the first element to react with oxygen. For example, if we have steel with Fe, Mn, Sn, and Cu in it, we can see that at 1000K then Mn, Fe (to FeO), Sn and Cu are the order of largest to smallest drop in free energy.&lt;/p&gt;&#xA;&lt;p&gt;Granted, the temperature of interest is closer to 1900K (above the melting point of iron), but the general trends of each Gibbs free energy change function continue to the right on the diagram and iron remains below the tramp elements Cu, Sn, As and Sb at practical temperatures, and likely to their respective boiling points. As a result, removing tramps from Fe would require oxidizing effectively all of the iron first. And because Sn, Sb, As and Cu are slightly soluble in iron, they require separation via chemical reaction.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://image.slidesharecdn.com/ellinghamdiagram-120924085031-phpapp02/95/ellingham-diagram-1-728.jpg?cb=1348494684&quot; alt=&quot;Ellingham diagram.&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;One can see the solubility of tramps from their phase diagrams with iron, of which I have posted Sb-Fe below. The diagram has temperature against composition, with each contiguous 2D region composed of either one phase, or a mixture of the two phases to its left and right, which are in equilibrium at that combination of temperature and composition. At the bottom left we see that for small amounts of Sb and room temperature, there is a contiguous region which in this case denotes a single phase, or alpha-Fe (the kind we are familiar with). Because there is Sb present, and it is in a single phase, it must be dissolved in the iron. The same is true, with varying severity, of the other tramps.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0dV9I.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0dV9I.gif&quot; alt=&quot;Fe-Sb phase diagram.&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://www.himikatus.ru/art/phase-diagr1/Fe-Sb.gif&quot; rel=&quot;noreferrer&quot;&gt;himikatus.ru&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;As Chris H commented, there is a question also of when other alloying elements are controlled. Generally alloy addition is controlled as close to solidification as possible, to minimize alloy loss.&lt;/p&gt;&#xA;&lt;p&gt;Scrap is melted in bulk in an electric arc furnace. If the scrap stream is sufficiently mixed, then the tramp concentration may be estimated based on past usage and the primary iron is added prior to chemistry analysis to compensate for the estimate. The bulk is then melted, oxygen is removed via the addition of elements at the bottom of the Ellingham diagram, specifically Ca and Al, and the molten metal is transferred to one or more highly insulated ladles. The Ca and Al rapidly react with oxygen dissolved in the melt to create low-density oxide slag which floats and is removed mechanically. Chemistry is taken after this process, and if the tramps are sufficiently diluted, the metal is transferred to ladles. If not, sufficient primary iron is added to dilute the melt.&lt;/p&gt;&#xA;&lt;p&gt;Once in the ladle, additional alloying elements are added. They are not added earlier because of the Ellingham diagram: most alloying elements including Mn, Mo, Cr, V, C, etc. have greater free energy loss than Fe, and so react first. In other words, they fade. To avoid expensive alloy addition fading, they are added as close to the solidification process as possible. Additionally, by removing oxygen using Al and Ca first, there is less oxygen dissolved in the iron to react with the more expensive alloying elements. Once in the ladle, there is very little liquid-atmosphere interface turbulence, so the diffusion of new oxygen into the liquid iron is relatively slow. There is of course still a time limit, and holding a ladle for too long will cause alloy fading. After alloy addition, chemistry is checked, and then the ladle is poured.&lt;/p&gt;&#xA;&lt;p&gt;Edited to add sources.&#xA;Edited to add discussion of alloy control.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-02-18T12:09:38.383" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="530" PostTypeId="2" ParentId="191" CreationDate="2015-02-08T08:28:57.353" Score="6" Body="&lt;p&gt;To concur with the &lt;strong&gt;David Tweed&lt;/strong&gt; &amp;amp; &lt;strong&gt;starrise&lt;/strong&gt;, it is uneconomic to separate the individual metals in steel alloys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do so would first require the alloys to be crushed and ground to the size of the crystal grains within the alloys. Then some form of mineral/crystal selection process would need to be devised to segregate and separate the wanted from the unwanted, such as: froth flotation; maybe heavy media; possibly gravity separation methods such as shaking tables or spirals (but I doubt these would be successful as gravity separation methods rely of significant density &amp;amp; weight differences); though magnetic separation, as used in the mineral sands industry might be an option for some alloys. Even after this, there will always be a waste or tailings division where the really difficult alloy crystals will be collected in a dump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crushing, grinding and separation all cost money. These costs and a profit have to come from the steel alloys being recycled into there individual metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As of early February 2015, the value of a selection of metals is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gold USD 1233.30 per oz, USD 39.6515 per g or USD 39 651 510.84 per&#xA;tonne (yes, 39.651 million dollars per tonne)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Platinum USD 1220 per oz or USD 39 223 905.97 per tonne (39.2239 M$/t)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Silver USD 16.68 per oz or USD 536 274.38 per tonne (0.536 274 M$/t)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Cobalt USD 29 500 per tonne&lt;/li&gt;&#xA;&lt;li&gt;Nickel USD 14 965 per tonne&lt;/li&gt;&#xA;&lt;li&gt;Lead USD 1850 per tonne&lt;/li&gt;&#xA;&lt;li&gt;Steel Billet USD 500 per tonne&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the aptly named &lt;em&gt;precious&lt;/em&gt; metals, Au, Pt &amp;amp; Ag, the price source was &lt;a href=&quot;http://www.kitco.com/&quot; rel=&quot;noreferrer&quot;&gt;Kitco&lt;/a&gt;.&#xA;The price source for the base metals, Co, Ni, Pb &amp;amp; steel billet was &lt;a href=&quot;http://lme.com/&quot; rel=&quot;noreferrer&quot;&gt;LME&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iron ore is currently selling for approximately USD 65 per tonne as stated on &lt;a href=&quot;http://www.indexmundi.com/commodities/?commodity=iron-ore&amp;amp;months=60&quot; rel=&quot;noreferrer&quot;&gt;Index Mundi&lt;/a&gt; and &lt;a href=&quot;http://ycharts.com/indicators/iron_ore_spot_price_any_origin&quot; rel=&quot;noreferrer&quot;&gt;Y Charts&lt;/a&gt;. That's for an average grade of 60 per cent iron. The open cut iron mines in Australia and Brazil, operated by &lt;a href=&quot;http://www.riotinto.com/&quot; rel=&quot;noreferrer&quot;&gt;Rio Tinto&lt;/a&gt;, &lt;a href=&quot;http://www.bhpbilliton.com/home/Pages/default.aspx&quot; rel=&quot;noreferrer&quot;&gt;BHP-Billiton&lt;/a&gt; &amp;amp; &lt;a href=&quot;http://www.vale.com/EN/Pages/default.aspx&quot; rel=&quot;noreferrer&quot;&gt;Vale&lt;/a&gt; are quite happy to produce iron ore at that price. &lt;a href=&quot;http://lkabminerals.com/&quot; rel=&quot;noreferrer&quot;&gt;LKAB&lt;/a&gt; is likewise happy to produce magnetite iron ore from the Kiruna underground mine in Sweden for that price.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.macrobusiness.com.au/category/iron-ore-price/&quot; rel=&quot;noreferrer&quot;&gt;Macrobusiness&lt;/a&gt; has an article about the possibility of iron ore prices going down to USD 30 per tonne in 2015.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At prices such as 0.536 to 39.6 million dollars a tonne it is easy to see why the &lt;em&gt;precious&lt;/em&gt; metals are recycled. But at USD 500 a tonne for steel billet and USD 65 per tonne for iron ore there is no incentive to separate the alloying metals from steel alloys.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-02-08T08:28:57.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="531" PostTypeId="1" AcceptedAnswerId="536" CreationDate="2015-02-08T09:56:35.350" Score="22" ViewCount="8363" Body="&lt;p&gt;What is the reason that handlebars are installed on motorcycles and steering wheels are installed on cars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the way to use both handlebars and steering wheels are quite similar, but a steering wheel allows for much more rotation than a handle bar when you control a vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give me a reason why a massive vehicle should use a steering wheel and a light weight vehicle should use a handlebar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason maybe involve scientific reasons, safety reasons or designing reasons.&lt;/p&gt;&#xA;" OwnerUserId="424" LastEditorUserId="368" LastEditDate="2015-02-13T23:45:03.610" LastActivityDate="2022-04-20T21:41:31.997" Title="What is a reason that handlebars fit to a motorcycle and steering wheel fit to a car?" Tags="|mechanical-engineering|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="532" PostTypeId="1" CreationDate="2015-02-08T10:10:12.447" Score="9" ViewCount="3490" Body="&lt;p&gt;I have recently bought a flat in a 25-storey building. I wonder what is the practical limit for the number of storeys in a mass-produced residential buildings built from monolithic reinforced concrete?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we expect the number of storeys in typical buildings of this type to rise in the forthcoming years or this is the reasonable limit of the technology? In all references I saw so far it was claimed that this technology has &lt;em&gt;no limits&lt;/em&gt; on the height of the buildings. But I doubt this because all skyscrapers I know were built from steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If monolithic reinforced concrete has no limitations, why have very tall buildings not been built with this technology?&lt;/p&gt;&#xA;" OwnerUserId="426" LastEditorUserId="368" LastEditDate="2015-02-13T23:41:05.470" LastActivityDate="2017-03-23T06:50:06.280" Title="What is the practical limit on the number of storeys in residential buildings built with monolithic reinforced concrete?" Tags="|civil-engineering|structures|concrete|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="533" PostTypeId="2" ParentId="531" CreationDate="2015-02-08T10:42:06.067" Score="5" Body="&lt;p&gt;When steering a car, depending on the degree of turn, sometimes it is necessary to rotate the steering wheel through a large arc. It is easier to do this using a circular steering wheel, particularly when the wheel has to be turned hand over hand. This cannot be done easily with semicircular steering wheels, aircraft type steering wheels or if handlebars where used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The handlebars on a motorbike are rarely turned to the maximum arc angles that car wheels are turned. Also, on large radius curves motorbike riders use their body angle to lean into curves to assist steering. Car drivers however, only have the steering wheel to steer the car, moving their bodies doesn't do anything to the steering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motorbike riders are exposed to the elements while driving and they must balance on two narrow wheels and sit on a narrow seat. Slight turns of the handle bars can lead to graver consequences for motorbike riders than similar slight turns of a steering wheel in a car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When holding onto the handlebars the rider forms one apex of a triangle which increases the stability of the rider. If motorbikes had steering wheels riders would not have the same level of stability they do with handlebars. Additionally, trying to steer a steering wheel while balancing on two narrow wheels is asking for trouble. It would decrease the stability of the rider as different muscles in body are used when turning handle bars to turning a wheel in the same plane; particular muscles in the torso. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard configuration for mass produced cars is four wheels on the ground, this a stable and balanced configuration. Also, car drivers sit on wider seats, compared to motorbike riders. Cars drivers are in a stable driving situation when they drive. It is easier for car drives to control a car using a steering wheel than it would be if they had to use handlebars.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-02-08T10:42:06.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="534" PostTypeId="2" ParentId="518" CreationDate="2015-02-08T11:51:13.553" Score="4" Body="&lt;p&gt;A round-a-bout introduces a disruption in &lt;em&gt;any&lt;/em&gt; traffic, no matter which direction. In your case the only two &quot;collision&quot; directions - two left turns - are the least used ones, so  traffic lights working in a preference program with induction loops on the two possible left turns, plus uncontrolled right turn lanes would be optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lVDAa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crossing would stay in green light as long as there's no need for vehicles to take the turn. Only detectors would activate the phases for left turns (and only one would block both the north and the south road). You can either limit the rate at which the side exits can be activated by detectors on the main direction (better, more expensive) or by timing it, so that there must pass at least X time between the turn activations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If real estate is a premium, you can create about all of the infrastructure for this crossing by reducing the main road to one lane in each direction, dedicating the remaining lane to turns, inclusion, and the &quot;islets&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an actual map from one of our controllers with a similar situation. This one includes two zebra crossings as well. It uses more detectors to determine optimal timing of green light, and change signals even before given vehicle would have to stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/viW6S.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;(an interesting thing about this one, it uses detection loops on the main road but videodetection on the side road - the road was still due to be finished when the traffic lights were installed, and so we couldn't put induction loops in the gravel - currently induction loops would be better there (both more reliable and cheaper) except videodetection is already in place, so replacing it would only mean more cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if your investor is rich and wants something prestigious and memorable (and truly optimal for the traffic), get a round-a-bout with a viaduct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/anOTi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2015-02-08T11:57:39.880" LastActivityDate="2015-02-08T11:57:39.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="535" PostTypeId="1" CreationDate="2015-02-08T13:07:48.700" Score="3" ViewCount="2435" Body="&lt;p&gt;Tin Whiskers are a major problem in the design of electronic circuits and the manufacturing of electronic hardware. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0T5Zu.jpg&quot; alt=&quot;Tin Whiskers Example 1&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/V3QAm.gif&quot; alt=&quot;Tin Whiskers Example 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are Tin Whiskers? How can electronic products be design to mitigate the issue of Tin Whiskers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://video.stltoday.com/David-Gilbert-and-Tin-Whiskers-25779048?ndn.trackingGroup=91130&amp;amp;ndn.siteSection=stltoday&amp;amp;ndn.videoId=25779048&amp;amp;freewheel=91130&amp;amp;sitesection=stltoday&amp;amp;vid=25779048&quot; rel=&quot;nofollow noreferrer&quot;&gt;David Gilbert and Tin Whiskers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eetimes.com/document.asp?doc_id=1279227&quot; rel=&quot;nofollow noreferrer&quot;&gt;Understanding and mitigating tin whiskers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://electronicdesign.com/analog/don-t-let-tin-whiskers-destroy-your-design&quot; rel=&quot;nofollow noreferrer&quot;&gt;Don’t Let Tin Whiskers Destroy Your Design&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.calce.umd.edu/tin-whiskers/TINWHISKERMITIGATION.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mitigation Strategies for Tin Whiskers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.maximintegrated.com/en/app-notes/index.mvp/id/5250&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tin Whiskers Are Real and Complex&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-02-08T18:41:59.727" LastActivityDate="2015-02-21T21:57:17.590" Title="How to design electronics to mitigate tin whiskers?" Tags="|electrical-engineering|materials|manufacturing-engineering|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="536" PostTypeId="2" ParentId="531" CreationDate="2015-02-08T13:41:56.560" Score="27" Body="&lt;p&gt;Actually, handlebars and steering wheels are less similar than you might think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a two-wheeled vehicle is moving fast enough to balance, the front wheel is never turned more than a few degrees. The primary mechanism for steering is leaning the vehicle, not turning the front wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to turn right, you actually tug briefly on the &lt;em&gt;left&lt;/em&gt; side of the handlebar. This causes the wheels to track to the left of the center of mass, which in turn causes the bike to lean to the right. This lean is what causes the direction to change, while maintaining balance &amp;mdash; the total force on the bike's center of mass still passes through the contact patches between the tires and the road. During the turn, the front wheel is essentially straight with respect to the bike's frame, and it's the geometry of how the tires contact the road, aided by &lt;em&gt;slight&lt;/em&gt; tension on the right handlebar, that keeps the bike turning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To come out of the turn, you tug slightly &lt;em&gt;more&lt;/em&gt; on the right handlebar (not the left), which causes the tires to track to the right, bringing them into a more vertical alignment with respect to the center of mass. This causes the bike to stop leaning and stop turning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is so intuitive when using handlebars, most people don't even think about the details of what's actually going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only time you steer a bike by moving the handlebars by large amounts is when you are moving so slowly that you have one or both feet on the ground (holding the bike vertical) and are trying to maneuver in a tight space.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Additional points in response to comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An inexperienced bicyclist at low speeds will turn the handlebar wildly back an forth in an effort to maintain his balance, but I would hardly call that &quot;steering&quot;. The main reason a 2-wheeled vehicle stays upright is the gyroscopic effect of the rotating front wheel. If the bike should start to lean, the wheel will experience a force that steers it toward the direction of the lean, which corrects the lean. If the bike is moving slowly, this effect is greatly reduced, and the rider is required to use the handlebar to maintain balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, on a bicycle, the rider is typically 90% or so of the mass of the system (bike + rider), and an experienced rider can steer merely by shifting his weight without touching the handlebar at all. On a motorcycle, the rider might only be about 25% of the total weight or less, which makes using the handlebar pretty much mandatory.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2015-02-08T20:42:00.677" LastActivityDate="2015-02-08T20:42:00.677" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="537" PostTypeId="1" CreationDate="2015-02-08T15:57:31.027" Score="3" ViewCount="208" Body="&lt;p&gt;I am an electronics engineer trying to build one small underwater mechanical device (about 12 cubic cm) as part of a personal automation project. I have figured out the electronics for this, but I am stuck at one mechanical problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to make a 3D printable CAD model of the device. Please tell me what kind of mechanism I need, and explain its working principles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Description&lt;/strong&gt;: Each device will have one inlet and two outlets ($o_A$ and $o_B$). The inlet is at the top and the outlets $o_A$ and $o_B$ are at the bottom. In an underwater installation, these devices would be placed vertically one below other, interconnected with 3/4 inch PVC pipe. Outlet $o_A$ of the above device will be connected with the inlet of the below device with 3/4 inch PVC pipe.  Normally, the solids falling down from the inlet should flow down vertically through the bottom outlet, $o_A$. &lt;em&gt;But occasionally the solids needs to be diverted to the other outlet, $o_B$.&lt;/em&gt; How can this be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&#xA;I see from the comments that 'Diverter Valve' is what I was looking for. Please explain its working principles.&lt;/p&gt;&#xA;" OwnerUserId="430" LastEditorUserId="160" LastEditDate="2015-02-09T12:38:56.397" LastActivityDate="2015-02-19T04:14:15.227" Title="How to divert solids from a downward vertical pipe" Tags="|mechanical-engineering|marine-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="538" PostTypeId="1" AcceptedAnswerId="547" CreationDate="2015-02-08T18:01:10.593" Score="8" ViewCount="187" Body="&lt;p&gt;I want to look at the thermal heat plume of a human in still air in a room. So I have a tank of water with a heating coil/cylinder in it. How do I calculate what the heat output of my heating coil will need to be in order to mimic that of a human's ($100W$) in air? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/leqXv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on dimensional similarity between water:air, Reynolds' number and Reyleigh number. The Rayleigh number, which governs buoyancy is given by: $\dfrac{g\beta}{\nu \alpha \kappa}q x^4$, where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha=$ thermal diffusivity&#xA;$\beta=$ thermal expansion coefficient&#xA;$\kappa=$ thermal conductivity&#xA;$\nu=$ kinematic viscosity&#xA;$x$= distance from the heated surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't really understand if the Rayleigh number is the right value to be attempting to maintain similar between both scenarios. Any help would be much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So using the Q* formula to get similarity between the two media &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{Q_{air}}{\left(\rho C_pT_{\infty}g^{1/2}x\right)_{air}}=\dfrac{Q_{water}}{\left(\rho C_pT_{\infty}g^{1/2}x\right)_{water}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and rearranging to get Q_water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_{water}=Q_{air}\dfrac{\left(\rho C_pT_{\infty}g^{1/2}x\right)_{water}}{\left(\rho C_pT_{\infty}g^{1/2}x\right)_{air}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then substituting in: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AIR: $\rho=1.225kg/m^3$, $C_p=1.005kJ/kg\,K$, $T_{\infty}=21^{\circ}C$, $x=0.1m$, $Q_{air}=100W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WATER: $\rho=1000kg/m^3$, $C_p=4.19kJ/kg\,K$, $T_{\infty}=8^{\circ}C$, $x=0.1m$, $Q_{air}=?W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get $Q_{water}\simeq 1.2\times10^4W$... But this seems much too high. Can this be right?&lt;/p&gt;&#xA;" OwnerUserId="434" LastEditorUserId="434" LastEditDate="2015-02-27T09:35:26.513" LastActivityDate="2015-02-27T09:35:26.513" Title="Scale models of thermal plumes" Tags="|mechanical-engineering|fluid-dynamics|heat-exchange|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="539" PostTypeId="2" ParentId="532" CreationDate="2015-02-08T18:15:26.253" Score="12" Body="&lt;p&gt;Any limit is going to be hard to quantify.  There are a lot of factors that have to be weighed when choosing the basic material type. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is that the limit has already been chosen for each building.  This was done during the design by the architects and engineers that worked on the building. Some of these decisions might have depended on the technologies that were available at the time that the building was designed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the factors that would have been taken into account:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Cost of steel versus concrete&lt;/strong&gt; - The relative price of the materials has changed throughout history.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Strength of concrete available&lt;/strong&gt; - It used to be the case that concrete was limited to about 4,000 psi (27.6 MPa) compressive strength. Modern high strength concrete can be higher than 10,000 psi (69 MPa).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Strength of steel available&lt;/strong&gt; - Steel strengths have increased from 36 ksi (248 MPa) to 50 ksi (345 Mpa) and even 100 ksi (689 MPa).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Area of wall and column space required to support the upper floors&lt;/strong&gt; - Buildings are heavy.  As the building gets taller, there is more weight pressing down on the bottom floors. This increased force requires more area of material.  At some point, the usable space on the bottom floors is reduced more than is acceptable. Per unit of area, steel is stronger than concrete, so it will take less area to support the same load.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Stiffness of the building&lt;/strong&gt; - Very tall buildings sway as the wind blows on them. How much they move is controlled by the weight and stiffness of the building.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Future creep (shortening) of the building&lt;/strong&gt; - Both steel and concrete creep. That is they compress over time if a constant force is applied. The amount of creep is affected by age, strength or the material and forces acting on the material. In very tall buildings, this shortening needs to be accounted for in the design. A lighter building will need to accommodate less creep.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Seismic (earthquake) design&lt;/strong&gt; - Steel is a ductile material. Concrete is a brittle material.  In locations where high seismic forces are expected, steel may be required. It has the ability to undergo extreme deflections without complete failure.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Quality control&lt;/strong&gt; - Concrete will be poured onsite, and steel is typically fabricated offsite under controlled conditions. The anticipated quality of the end product or amount of oversight required to ensure a quality product are both a cost consideration.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are a lot of factors that go into the design of skyscrapers. Each item above has a cost associated with it. The end result is at least partially controlled by the estimated price. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern skyscraper designs sometimes include a concrete core that goes all or most of the way to the top.  This shows that there isn't much of a height limit to concrete construction as long as you are ok with a reduced usable volume.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-08T18:15:26.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="540" PostTypeId="2" ParentId="531" CreationDate="2015-02-08T18:34:22.027" Score="8" Body="&lt;p&gt;The car, being heavier, requires greater force to steer, which is provided by gearing the steering wheel down so that several turns are required from lock to lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were &lt;a href=&quot;http://en.wikipedia.org/wiki/Tiller#mediaviewer/File:1904_Cyklonette_von_oben_gesehen_-_Deutsches_Museum_Verkehrszentrum.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;cars with tiller steering&lt;/a&gt;, inherited from yachting practice no doubt (only backwards), where the length of the tiller gave enough leverage instead.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how their turning circle compared with their steering wheel contemporaries, or what sort of violence the driver did to a front seat passenger in the process, but the idea didn't exactly seem to catch on!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/1/1b/1904_Cyklonette_von_oben_gesehen_-_Deutsches_Museum_Verkehrszentrum.JPG/1024px-1904_Cyklonette_von_oben_gesehen_-_Deutsches_Museum_Verkehrszentrum.JPG&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:49.983" LastActivityDate="2015-02-08T20:09:06.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="541" PostTypeId="5" CreationDate="2015-02-08T18:40:28.557" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-08T18:40:28.557" LastActivityDate="2015-02-08T18:40:28.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="542" PostTypeId="4" CreationDate="2015-02-08T18:40:28.557" Score="0" Body="Questions on the design and materials used to connect pieces together using bolts." OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-02-09T17:12:18.877" LastActivityDate="2015-02-09T17:12:18.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="543" PostTypeId="2" ParentId="519" CreationDate="2015-02-08T18:44:58.210" Score="4" Body="&lt;p&gt;To answer your question bluntly, yes, the cold can be a concern for standard structural bolts, because the cold can be a concern for just about any metal or plastic. I can give some insight into why the cold is a factor, but I want to be clear that I can't make a recommendation on acceptable temperature ranges for the standard bolts, so if you can't find some data to ensure the proper operation of them, it may be best to use the low-temperature bolts to be safe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In metals, this phenomenon is known as the ductile-brittle transition, which occurs at a specific temperature based on the material and the strain rate. Note that this means it's dependent on how quickly you deform the material, not how much force you apply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two types of deformation, elastic (where the material returns to its original shape) and plastic (where the material permanently deforms.) On a molecular level, elastic deformation happens when the bonds between molecules in the material are stretched. Because no permanent change in the structure occurs, the material can retain its original shape. At a certain point though, the structure begins to change, often in the form of dislocation movement. Existing defects in the material begin to move, and this movement cannot be spontaneously reversed, causing the permanent change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ease and rate of dislocation movement is in part based on temperature. Temperature is a measure of energy, and if there's more energy in a material, some of that energy goes towards dislocation movement. This is important because dislocation motion can help prevent fracture. Cracks exist in all parts, they are impossible to avoid, and parts will fracture at those cracks because stress is concentrated there and the material is naturally weaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At these cracks, part of the energy goes into forming new surfaces (propagating the crack further into the part.) Most of the rest of the energy goes into moving the dislocations by plastically deforming the material. If the material is too resistant to deformation (too brittle), no energy will be dissipated by deformation, leaving more energy available for crack propagation. This is what causes metals to be more likely to fracture in lower temperatures. At a certain critical temperature, the material becomes more brittle than ductile, and its fracture toughness is significantly reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determining this temperature is not simple from a theoretical point, and it's usually done experimentally I believe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ocw.mit.edu/courses/materials-science-and-engineering/3-11-mechanics-of-materials-fall-1999/modules/frac.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; is what I used to refresh my memory from my molecular material behavior class, it's good some good info, but I wouldn't worry about getting bogged down in the math because it's incredibly complex for non-simplified cases. The explanations aren't as bad though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said at the top, the simple answer is yes, it will affect the bolt strength, and I think the standards are reluctant to give specific answers if the bolt isn't designed for a temperature range because bolt size, defect size, and type of loading can all affect this.  &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-02-08T18:44:58.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="544" PostTypeId="5" CreationDate="2015-02-08T18:54:25.023" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-08T18:54:25.023" LastActivityDate="2015-02-08T18:54:25.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="545" PostTypeId="4" CreationDate="2015-02-08T18:54:25.023" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-08T18:54:25.023" LastActivityDate="2015-02-08T18:54:25.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="547" PostTypeId="2" ParentId="538" CreationDate="2015-02-08T19:07:07.893" Score="8" Body="&lt;p&gt;Thermal plumes have been studied extensively for fire safety applications. Often you know the heat release rate $Q$ but little more. A dimensionless group called $Q^*$ (pronounced &quot;Q star&quot;) is used instead of more common parameters like the Reynolds number and Rayleigh number. This parameter can be thought of as the strength of the heat source at a particular distance. It correlates well for thermal plumes. You can derive this group by non-dimensionalizing the Navier-Stokes equations and setting dimensionless groups equal to 1 to define the characteristic length and velocity. For more information, check out &lt;a href=&quot;http://fire.nist.gov/bfrlpubs/fire97/PDF/f97109.pdf&quot; rel=&quot;nofollow&quot;&gt;Gunnar Heskestad's paper on this dimensionless group&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the fire modeling case, generally people ignore Prandtl number similarity and some other things, so they say the dimensionless temperature and velocity distributions are only functions of $Q^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most relevant parameters are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T^* \equiv \frac{T - T_\infty}{T_\infty}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q^* \equiv \frac{Q}{\rho c_p T_\infty (g x)^{1/2} x^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more explicit, if you know the temperature ($T$) as a function of height ($x$) above the hot object, you can find $T^*$ as a function of $Q^*$. $Q^*$ is like a dimensionless spatial coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking, your setup is not going to be exactly similar because your coil and a human are not geometrically similar (and the heat flux distribution on the coil is probably not similar either). In your photo, I assume the human would be lying down if any reasonable geometric similarity is desired. The far-field should be okay, and I'll assume this is what interests you [2].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also not exactly clear what quantity you are interested in. I assumed you want to get the temperature distribution in the plume, say, at a height $x_1$ above in reality which would be $x_2$ in your model. Correct me if this is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while I don't do experiments, I imagined your heating coil has an output of $W$, not heat flux. Let me know if I'm mistaken and I'll change my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the other parameters may or may not be valid in your case (it seems to be okay for fire safety [1]), so I'll do the analysis assuming it's not. You can skip the remainder if you want to assume the two mentioned parameters are all you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get the number of required groups from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Buckingham_%CF%80_theorem&quot; rel=&quot;nofollow&quot;&gt;Buckingham $\pi$ theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant parameters I've identified are $T$ (temperature at height $x$), $x$, $Q$, $g$, $\alpha$, $\beta$, $\nu$, $T_\infty$, $\rho$, and $c_p$. The Buckingham $\pi$ theorem suggests there will be 6 dimensionless groups here. (Assuming that I am not missing a parameter. I also need to check that the dimensional matrix is not rank deficient. For more details about dimensional analysis, I recommend reading &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0882756826&quot; rel=&quot;nofollow&quot;&gt;Dimensional Analysis and Theory of Models&lt;/a&gt; by Henry Langhaar.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the first 5 dimensionless groups are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T^* \equiv \frac{T - T_\infty}{T_\infty}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q^* \equiv \frac{Q}{\rho c_p T_\infty (g x)^{1/2} x^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Pr \equiv \frac{\nu}{\alpha}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Gr_x \equiv \frac{g \beta (T - T_\infty) x^3}{\nu^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho^* \equiv \beta (T - T_\infty)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fifth group is inspired by the Boussinesq approximation. In that approximation, the density difference is modeled as a temperature difference. Similarity in this parameter ensures that your density field is similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the remaining group, I needed to get a little creative. Similarity does not require this group takes any particular form, but it's best to stick with parameters with known physical meanings (or parameters which can be derived from governing equations, which usually have physical meanings). I can't think of anything good off hand, but the following works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Pi_6 \equiv \frac{g x}{c_p(T - T_\infty)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to match all of these for similarity. It should be clear that matching all of these will be a challenge. As I said, it appears to be common practice in fire safety to ignore everything but $T^*$ and $Q^*$. I don't know if this is because the other parameters don't matter, or if it's just for convenience. Sorry if this is not the answer you expected, but as with many things in engineering, the answer is not easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] I remembered later that the non-dimensionalization of the Navier-Stokes equations suggests that $Q^*$ is the only parameter in the solution. So perhaps $T^*$ and $Q^*$ are all you need, and the Buckingham $\pi$ approach just gives you superfluous parameters. I don't recall all the details of the non-dimensionalization, but if there is interest I'm sure I could reproduce it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] The theoretical argument that supports the use of $Q^*$ assumes the heat source is a point source. So it's really only correct far away, because the temperature goes to infinity at the point source in the model. This is because $Q^*$ goes to infinity at $x = 0$, as you can see from its definition. If you are developing a correlation, say $T^* = a (Q^*)^b$ where $a$ and $b$ are coefficients, you can get around this by defining a &quot;virtual origin&quot;, which will allow you to develop a correlation without singularities. Basically, instead of using $x$ you define use instead $x_\text{virtual} = x + x_\text{origin}$. That is, $Q^*$ is now written:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q^* \equiv \frac{Q}{\rho c_p T_\infty (g [x + x_\text{origin}])^{1/2} (x + x_\text{origin})^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You pick $x_\text{origin}$ such that your correlation fits better. It's another parameter in the correlation. If you know the surface temperature, you can pick $x_\text{origin}$ such that the surface temperature is what the correlation returns at $x = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as the argument supporting the use of $Q^*$ really makes the far-field assumption from the beginning, it's not clear that simply using a virtual origin is enough to make a correlation valid in the near-field (even if you have geometric similarity). I can't say whether or not the other factors I've identified factor in or not.&lt;/p&gt;&#xA;" OwnerUserId="49" LastEditorUserId="49" LastEditDate="2015-02-10T22:17:37.447" LastActivityDate="2015-02-10T22:17:37.447" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="548" PostTypeId="1" CreationDate="2015-02-09T01:08:12.050" Score="5" ViewCount="153" Body="&lt;p&gt;I'm a bit confused about all the different &quot;definitions&quot; of uncertainty?&lt;br&gt;&#xA;One can calculate &quot;uncertainty&quot; of a single measurement by simply checking how accurate the tool that is being used to do the measuring is.&lt;br&gt;&#xA;One can also calculate &quot;uncertainty&quot; of a group of measurements by finding the standard deviation which means that any more measurements made will have a 70% chance of being within one standard deviation of the &quot;true value&quot;.&lt;br&gt;&#xA;Finally, there is also standard error which is also a measure of &quot;uncertainty&quot;, but I'm not sure what it says.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which &quot;uncertainty&quot; do I have to find in the following problem and would I go about finding it?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particular 1.00-kg bag of rice consists of 2.00% &quot;supergrains&quot;, each of which contains 5ppm of arsenic and the rest of the grains contain no arsenic. What is the average concentration?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A grain of rice weighs 20 mg. What is the uncertainty in the concentration due to sampling if a 10.00 g sample is taken for analysis from this bag?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the entire contents of the bag were ground to spherical particles 100 micrometer in diameter, what would the uncertainty be if a 10.0 g sample were taken for analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part is pretty straight forward. I calculated that the average concentration of arsenic in this bag of rice is 100ppb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the next two questions I don't understand partly because as I said in the beginning, I'm not sure what &quot;definition&quot; of uncertainty they are asking for/about. How would I go about finding it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct any assumptions or definitions I've made if they are incorrect in any way. I'm really trying to learn and I know how important precision of language is.&lt;/p&gt;&#xA;" OwnerUserId="442" LastEditorUserId="95" LastEditDate="2015-02-09T11:57:10.190" LastActivityDate="2015-03-11T12:52:21.450" Title="Question about &quot;uncertainty&quot;" Tags="|measurements|statistics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="549" PostTypeId="1" CreationDate="2015-02-09T04:34:41.673" Score="3" ViewCount="290" Body="&lt;p&gt;I'm having some trouble using a motor driver board. I'm pretty sure its being driven properly since there is ~3V @ the output v.s. none when its off. However it should be supplying ~2A (rectifier bottleneck) of current, rather than topping out at 120mA. I've tried both outputs, both give the same results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using this motor driver: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00CAG6GX2&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.amazon.com/gp/product/B00CAG6GX2/ref=oh_aui_detailpage_o00_s00?ie=UTF8&amp;amp;psc=1&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/eDf0Q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have both output enable jumpers in place, and have tried with and without the 5V enable jumper in. I tie one logic input to 5V and the other to GND. The driver is powered through a 5V DC-DC converter (~2A verified). Am I doing something silly, or is it likely a dead H-bridge?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;UPDATE (moved from self-answer):&#xA;I'm sorry for not being more specific. It turns out that for the 3v motors I was attempting to drive the L298 is not the best solution since it has a few volts drop at high currents, leaving the motors with almost nothing across them if using a rectified 3V power source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The following is based upon using the L298 on its own, NOT with the original board. I've tried increasing the voltage supply to 6V, and this seems to be working. I intend to increase to a higher voltage, as well as using pwm across a smoothing capacitor in parallel with the motor in order to give a proper output voltage while using a large enough source to account for the drop across the internal transistors in the L298.&lt;/p&gt;&#xA;" OwnerUserId="366" LastEditorUserId="366" LastEditDate="2016-01-20T07:56:34.757" LastActivityDate="2016-01-20T07:56:34.757" Title="Motor Driver board: limited current" Tags="|electrical-engineering|power-electronics|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="550" PostTypeId="1" CreationDate="2015-02-09T05:32:06.610" Score="15" ViewCount="30945" Body="&lt;p&gt;I am from the mathematics StackExchange section, many of my students are engineering students at University. I was wondering what kind of calculus do you real engineers use? I have known two engineers. One from airplane design and another from metrology. The former used very very little calculus, some ODE's with constant coefficients by linearization. The latter used only basic math, no calculus, with some excel.  I want to be honest with any engineering student so they know what awaits them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a follow up question. Did you find it beneficial to have about four semesters of calculus? Maybe you do not use anything from it, but it does enhance your mathematical reasoning, which does have a positive externality on your engineering skills?  &lt;/p&gt;&#xA;" OwnerUserId="443" LastEditorUserId="191" LastEditDate="2015-02-11T22:17:40.750" LastActivityDate="2015-02-12T18:07:15.437" Title="What kind of math do engineers really use?" Tags="|education|" AnswerCount="14" CommentCount="7" ClosedDate="2015-02-27T20:14:39.637" ContentLicense="CC BY-SA 3.0" />
  <row Id="551" PostTypeId="2" ParentId="550" CreationDate="2015-02-09T07:14:41.417" Score="10" Body="&lt;p&gt;A little background (honest disclosure).  I started out getting my B.S./M.S. in Mech Eng. from a fairly practical/applied school before deciding to continue with a PhD at a more theoretical school. As a result, I don't claim to be a real engineer (my general experience is that academics working in engineering are usually mediocre engineers), but I've got a few thoughts that might be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my research, I find myself dealing with ODEs, PDEs, linear algebra (both applied and abstract) and that sort of thing. At times I've had to re-learn math concepts that I had forgotten or never learned in the first place. Whatever fraction of your students go into academia will be more likely to use calculus regularly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more applied activities, such as consulting projects or building race cars for a student completion. I find much less demand for those skills, although they are useful at times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases, calculus is more valuable for concepts than for actual computation. I'll want to know that one quantity is the integral of another in order to understand a problem, but that doesn't mean that I'm actually going to sit down and integrate an equation with a pencil and paper. Particularly, I think that understanding the basic ideas of differential equations can be extremely valuable across many disciplines (dynamic systems, heat transfer, electronics...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experiences that you describe are not unreasonable for quite a few reasons (not comprehensive list):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Many practical problems may be solved by analytically with higher mathematics. However the analytic solution, once known reduces the actual computation to simple arithmetic. In some cases it's not only easier to use the given solution, but actually required. In the case of various codes and standards, an engineer would expose themselves to liability if they deviated from a prescribed computation procedure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Numerical solutions to problems are increasingly easy to come by and are more broadly applicable than analytic solutions. It's often easier to throw a numerical method at an integral, ODE, PDE, series... rather than try to remember/derive the solution. Complex geometry, non-linear behavior etc. often mean that conventional methods are impractical or impossible. And, with a lot of modern software, the math is totally invisible to the user.  I've seen 1st year students with little experience quickly learn the tools to simulate stresses in complex load scenarios and compute transient heat conduction with non-linear boundary conditions (basically no math required).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There's a whole lot of empirical data that goes into engineering. Experiments and experience may be just as good or better than mathematics in some cases. I couldn't even start to calculate (from first principles) the coefficient of friction between two materials, but I can look it up in a book or measure it myself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-02-09T07:14:41.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="552" PostTypeId="2" ParentId="550" CreationDate="2015-02-09T09:06:53.987" Score="15" Body="&lt;p&gt;In my civil engineering degree we used ODEs for the relationship between force, moment and deflection. I don't remember using PDEs myself, but my brother-in-law (doing civils at a different university) used them for hydraulics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real life (as a bridge designer) I can't remember actually using calculus. University mainly concentrated on the theory and the mathematical models used, whereas in actual engineering design we have computer software that does all the calculation for us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is a lot of benefit to a theoretical and mathematical background at university - as a professional engineer you need to have a basic understanding to know whether the software is giving you a sensible answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(As an aside, as you mentioned Excel, I've used that a hell of a lot in real design.)&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-02-09T09:06:53.987" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="553" PostTypeId="2" ParentId="548" CreationDate="2015-02-09T11:03:34.860" Score="1" Body="&lt;p&gt;The problem is that the term &quot;uncertainty&quot; is colloquially used to refer to lots of different things where it probably shouldn't be. For a guide to definitions of various terms in metrology see the &lt;a href=&quot;http://www.bipm.org/en/publications/guides/&quot; rel=&quot;nofollow&quot;&gt;VIM&lt;/a&gt;. The &lt;a href=&quot;http://www.bipm.org/en/publications/guides/&quot; rel=&quot;nofollow&quot;&gt;GUM&lt;/a&gt; provides a somewhat practical guide to actually computing the uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if you compare a single measurement to some nominal &quot;true&quot; value, as in your first point. This is not the uncertainty. This would be the measure accuracy or error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty is a description of the spread of a set of measurements. It also requires that a distribution and confidence level of the measurements is given. Commonly a normal distribution at 95% (~2$\sigma$) is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard uncertainty, also sometimes called the standard error is just the standard deivation of the set of measurements, i.e. the uncertainty at one $\sigma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I your situation you are dealing with a binary situation (Supergrain or not), which is not what I'm very familiar with. In such a situation your distribution is bi-nominal, although for larger sample sizes you can use the central limit theorem to approximate to a normal distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case &lt;a href=&quot;http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; says the standard uncertainty is given by $\sqrt{\frac{1}{n} p(1-p)}$ where n is the number of grains and p the probability (2%). Scale this up to get whatever your desired confidence level is. For the third question I feel you are missing some information to work out how 100 $\mu$m balls effect the number of particles in your sample. Do you have some information on the density?&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-02-09T11:03:34.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="554" PostTypeId="2" ParentId="531" CreationDate="2015-02-09T11:17:40.507" Score="4" Body="&lt;p&gt;gearing!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of force required to repoint a car's front wheels at low speed is so high that there is a rack and pinion mechanism to allow it to be done by even a low-strength driver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't an issue with power steering - and there were some cars in the 1950s which had handlebars, HOWEVER it was discovered it was too easy for the driver to unintentionally make violent changes at speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with motorcycles, once a car is moving at speed, steering changes are minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason (safety), handlebar steering power-steering systems quickly stopped being used in cars in favour of a geared setup requiring large changes of wheel angle to effect changes of front wheel angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Safety is also why push-button automatic gearbox selectors were tried and then dropped. It's far easier to install lockouts on a lever to prevent cars being shifted from (D)rive to (R)everse whilst in motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With electronic systems, pushbutton selectors may make a comeback as complex mechanical interlocks are no longer needed and a solenoid will do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also possible that with speed-sensitive steering systems and electronic control, a handlebar setup may be safely deployed - however it is so non-traditional that it would probably be a major insurance risk due to drivers making catastrophic errors at speed.&lt;/p&gt;&#xA;" OwnerUserId="447" LastActivityDate="2015-02-09T11:17:40.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="555" PostTypeId="2" ParentId="549" CreationDate="2015-02-09T12:00:36.333" Score="2" Body="&lt;p&gt;Make sure that the 12V supply is capable of sourcing sufficient current. This the source that powers the output A &amp;amp; B.  The 5V input supply is most likely to power the logic circuitry. On PCB board 12V supply power supply should be routed to the H-bridge which will provide the voltage to both Output A and Output B as mentioned above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also when you are measuring current at output A or B make sure their a load. I suggest placing a 1K or 10K resistor. It is best that you select an appropriate load resistor representing the actual load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Product Description from Amazon&lt;/strong&gt;&#xA;Feature&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chip: L298N&lt;/li&gt;&#xA;&lt;li&gt;Logic voltage: 5V&lt;/li&gt;&#xA;&lt;li&gt;Logic current 0mA-36mA&lt;/li&gt;&#xA;&lt;li&gt;Storage Temperature: -20°c to +135°c&lt;/li&gt;&#xA;&lt;li&gt;Operating mode: H-bridge driver (dual)&lt;/li&gt;&#xA;&lt;li&gt;Drive voltage: 5V-35V&lt;/li&gt;&#xA;&lt;li&gt;Drive current: 2A (MAX single bridge)&lt;/li&gt;&#xA;&lt;li&gt;Maximum power: 25W&lt;/li&gt;&#xA;&lt;li&gt;Dimensions: 43x43x27mm&lt;/li&gt;&#xA;&lt;li&gt;Package includes: 1x L298N Module&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sparkfun.com/datasheets/Components/General/L298N.pdf&quot; rel=&quot;nofollow&quot;&gt;Full-Bridge Motor Driver Dual - L298N&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-02-15T09:00:44.113" LastActivityDate="2015-02-15T09:00:44.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="556" PostTypeId="1" AcceptedAnswerId="564" CreationDate="2015-02-09T12:34:54.493" Score="10" ViewCount="20654" Body="&lt;p&gt;I was reading about how AC to DC converters work with a step-down transformer and then a diode bridge to convert the lower, stepped down AC voltage into DC. What I don't understand is since the input AC appears to be connected to the primary coil of the transformer, how does the DC load affect the power used from the AC supply?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the DC load somehow feedback and lower the resistance of the primary coil so that more power can be drawn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there is no load on the DC side, does power still flow through the AC primary coil, and if so, why doesn't it just melt?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorDisplayName="user16" LastEditDate="2015-02-09T12:38:15.683" LastActivityDate="2023-08-11T02:54:23.517" Title="Does a transformer use power when output isn't under load?" Tags="|electrical-engineering|power|ac|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="557" PostTypeId="1" AcceptedAnswerId="560" CreationDate="2015-02-09T12:45:47.267" Score="13" ViewCount="8947" Body="&lt;p&gt;I have used a cheap multimeter to measure voltages in simple DC circuits, but I have seen pictures of them plugged straight into the mains and used to measure various home-built generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't the higher voltage fry the multimeter, and also in theory could a small cheapo multimeter be safely used to measure very high voltages? If you get the setting wrong on the dial, does this matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not planning to plug one in, nor would I recommend anyone who doesn't know what they're doing to do this either, I'm just wondering how it works.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="110" LastEditDate="2015-02-10T11:44:09.167" LastActivityDate="2017-01-08T01:03:29.077" Title="How does a multimeter protect itself from high voltages?" Tags="|electrical-engineering|measurements|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="558" PostTypeId="2" ParentId="557" CreationDate="2015-02-09T13:10:45.397" Score="7" Body="&lt;p&gt;Simple answer is good multimeters have good protection circuts. When the measuring source outside of the meters capabilities of the meter, the meter will give out of range message. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four main multimeters categories. They are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Category I&lt;/strong&gt;: used where equipment is not directly connected to the&#xA;mains (CAT I)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Category II&lt;/strong&gt;: used on single phase mains final sub-circuits (CAT II)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Category III&lt;/strong&gt;: used on permanently installed loads such as&#xA;distribution panels, motors, and 3-phase appliance outlets (CAT III)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Category IV&lt;/strong&gt;: used on locations where fault current levels can be very&#xA;high, such as supply service entrances, main panels, supply meters,&#xA;and primary over-voltage protection equipment (CAT IV)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is the user responsibility to us the appropriate multimeter for the appropriate task. Below is a Fluke 87 III multimeter. In the bottom right hand corner it states CAT III&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/67Av8.jpg&quot; alt=&quot;Fluke 87 III&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a mulitmeter which is in the el-cheapo classification and there is no visible notice of multimeter category classification. This is a situation where &lt;strong&gt;user be aware is very important&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EPJSa.jpg&quot; alt=&quot;Craftsman Meter&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both type of multimeter play in important role in the area of engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basic Protection&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most basic form of protection is a fuse. Depending on the price point and manufacture this protection can vary. A high quality multimeter might use a High Rupture Capacity (HRC) fuse where as a low quality meter would use a glass fuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High Rupture Capacity (HRC)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ez7JR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cylFz.jpg&quot; alt=&quot;HRC fuse&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glass Fuse&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/esE21.jpg&quot; alt=&quot;Glass fuse&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Advance Protection&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond the basic protection circuit there are many stages of other protection circuitry consisting of diode, Wire Wound Resistors, Zener Diodes, MOV (Metal Oxide Varistors) and thermistors (PTC – Positive Temperature Coefficient). The configuration of these components are depends on manufacture, multimeter category, price point other factors. &lt;a href=&quot;http://www.eevblog.com/2012/10/19/eevblog-373-multimeter-input-protection-tutorial/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; links explains the few advance circuits of the Fluke 27 multimeter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;El-cheapo multimeters don't have good protection thus is prone to damge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Multimeter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Multimeter&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://electroniclevel3.wikispaces.com/HRC+Fuses&quot; rel=&quot;nofollow noreferrer&quot;&gt;High Rupturing Capacity (HRC) Fuse&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.electricaltechnology.org/2014/12/hrc-fuse-high-rupturing-capacity-fuse-types.html#&quot; rel=&quot;nofollow noreferrer&quot;&gt;HRC Fuse (High Rupturing Capacity Fuse) and its Types&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Thermistor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thermistor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Resistor#Wire_wound&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wire wound Resistors&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Varistor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Varistor (MOV)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eevblog.com/2012/10/19/eevblog-373-multimeter-input-protection-tutorial/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Multimeter Input Protection Tutorial&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2017-01-08T01:03:29.077" LastActivityDate="2017-01-08T01:03:29.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="559" PostTypeId="2" ParentId="550" CreationDate="2015-02-09T13:34:11.613" Score="11" Body="&lt;p&gt;I originally wrote this as a comment attached to AndyT's answer but in response to dcorking's  comment I've decided to  expand here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I graduated nearly 30 years and my experience is similar to AndyT's.  After graduating I went straight into industry. Since graduating, I and everyone I have worked with or been associated with have never used and have never needed to use calculus in our day to day work as engineers. The types of engineers I have worked with include: civil, mechanical, ventilation, mining, electrical and environmental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During my career I have used some trigonometry, algebra and statistics, plus financial maths (NPV, IRR, etc.) for project evaluations, feasibility studies and sometimes when I had to write or review justifications for capital expenditure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I emerged into the real world, work desktop computers were starting to be used by engineers. My early career was a mixture of doing designs on paper and using computers. Eventually computers dominated and I ended up using computer design software and spreadsheets for my engineering and design work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between two-thirds and three-quarters of all the maths I learnt in university I have never used after I started working. I have since realised that much of the maths I was required to learn was an exercise in teaching me how to think and solve problems. The maths unit that I particularly found useless for my career, but had to study, was eigenvectors. I know that some engineers find eigenvectors indispensable. It was one unit I was happy to forget after I sat the exam!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineering courses need to be accredited by professional engineering societies, hence engineers are required to learn a lot of maths, just in case it's needed. When students start their courses they don't always know where they will end up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Research engineers and those involved with leading edge higher technologies use more of the maths and calculus they were taught.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can recall overhearing a conversation one my lectures was having with another student and he said that the only time he used calculus was in the 1950s when he was involved in the design of certain types of internal combustion motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing about engineer's in industry is they soon end up being managers – looking after people, money and ideas. A background knowledge of calculus is useful but now days computers do all the complex calculations for us. We plug in the number and interpret the results. We need to know the concepts of how the software works to make sure the software is not giving us rubbish. It's one of the reasons why engineering students need to study maths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can recall attending a students meet industry seminar when I was a student and an experienced engineer told everyone that while at university they needed to use scientific calculators, but as they progressed through their careers they will end up using calculators that only had addition, subtraction, multiplication and division keys.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-02-09T13:34:11.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="560" PostTypeId="2" ParentId="557" CreationDate="2015-02-09T13:42:36.223" Score="7" Body="&lt;p&gt;&lt;strong&gt;WARNING&lt;/strong&gt;: Note that &lt;strong&gt;some cheap meters are not suitable for use with 230 VAC AC mains.&lt;/strong&gt; Some meters may have AC voltage ranges able to conceptually measure to well above AC mains voltage BUT have internal componentry not certified, suitable or safe at eg 230 VAC. &lt;strong&gt;Use of such meters to measure such voltages is akin to a safer than usual game of &quot;Russian Roulette&quot; which still may end in death.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;_________________&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failures may occur to power dissipation in components or to voltage breakdown even when power dissipation is within bounds. &#xA;Voltage ranges are usually less stressed than most other ranges when overrated.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Auto-ranging meters start at the highest range and work down until the reading becomes a certain percentage of full scale on that range. Switching can be via eg MOSFETs used to either short resistors in dividers or to pick voltages from appropriate tapping points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cheap meters protection is limited.&lt;br&gt;&#xA;Current much over 200 mA on low current ranges will usually blow the internal fuse.&lt;br&gt;&#xA;Too high current on the high amps range (10A, 20A, will blow the shunt or a fuse if fitted.&lt;br&gt;&#xA;High voltage on lower voltage AC or DC volts or on current ranges will sometimes destroy the meter (Ask me how I know :-) ).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said: Over ranged inputs will not necessarily stress other than the input circuitry. Higher than minimum necessary input resistor wattages can protect short term.   Zener diodes or other clamps can stop high voltage getting into the circuitry proper.&lt;br&gt;&#xA;Very keen manufacturers can provide electronic switches. These can be a simple as a high voltage MOSFET in series with the input) which can be turned off when needed. This will add some error due to voltage drop but this can be controlled and designed for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So eg an overvoltage condition is applied to a current input, the input resistor starts to dissipate excessive power, the inside end of the resistor is clamped by a zener or TVS (transient voltage suppressor) and a fast MOSFET switch is triggered to disconnect the overload. The higher than necessary power dissipation rating of the input circuitry provided enough time for the protection to act. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Real world example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Intersil application note -&lt;br&gt;&#xA;&lt;a href=&quot;http://www.intersil.com/content/dam/Intersil/documents/an04/an046.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;AN046 Building a Battery Operated Auto Ranging DVM with the ICL7106&lt;/strong&gt;&lt;/a&gt; provides specific examples of the design of autoranging equipment and how the issues involved are addressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how the front end looks conceptually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CXqXD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here is how it ends up in practice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0WB5r.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.intersil.com/content/dam/Intersil/documents/an04/an046.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.intersil.com/content/dam/Intersil/documents/an04/an046.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-12-29T12:15:55.497" LastActivityDate="2015-12-29T12:15:55.497" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="561" PostTypeId="2" ParentId="556" CreationDate="2015-02-09T13:51:37.613" Score="4" Body="&lt;p&gt;A transformer offers resistance to AC current flow due to the magnetic field produced by the current flow. This &amp;quot;AC resistance&amp;quot; is termed &amp;quot;impedance&amp;quot; and is a function of number of turns, core material, air gap in core, core dimensions, and more.&lt;/p&gt;&#xA;&lt;p&gt;When there is no load, the applied AC voltage will cause &amp;quot;magnetising current&amp;quot; to flow. This will cause some losses due to eddy current losses in the core and copper losses due to resistance in the winding (&amp;quot;I squared R losses&amp;quot; as power = Current^2 x Resistance).&lt;/p&gt;&#xA;&lt;p&gt;These losses are relatively small compared to full load power, but not trivial at rest. A few percent of full load power would usually be good.&lt;/p&gt;&#xA;&lt;p&gt;When a DC load is applied, it loads the AC secondary circuit which is tightly couple by the core's magnetic fields to the primary winding. So the DC load resistance appears as if it is an AC impedance load on the primary side and input power increases to meet the load.&lt;/p&gt;&#xA;&lt;p&gt;If you apply DC (rather than AC) to a transformer winding there is no ongoing magnetic field change, there is no impedance due to the varying magnetic field and current is limited by the resistance which is low compared to the impedance that should be being generated. If the DC supply has enough muscle power the transformer &amp;quot;just melts&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="57" LastEditDate="2023-08-11T02:54:23.517" LastActivityDate="2023-08-11T02:54:23.517" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="562" PostTypeId="2" ParentId="550" CreationDate="2015-02-09T14:12:17.297" Score="8" Body="&lt;p&gt;&lt;em&gt;This is from the view of a civil engineer.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers typically don't use higher level math because the code specifications are written specifically to avoid the need.  You don't want a building or bridge to fail because an engineer didn't take an integral correctly. Wherever possible, the &lt;em&gt;hard&lt;/em&gt; math has been reduced to a simplified equation, a chart or a graph. This is done to limit the possible sources of errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complicated math is done and checked before it is placed in the codes. This way the engineer that uses the code later does not have to worry about it being correct. Usually, just referencing a code is enough to &quot;prove&quot; that an answer is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineering for the public is so controlled by codes and specifications that in some areas there is little math to actually be done. The answer is found in a table. The table was likely designed with lots of math input and university research, but a table was developed to eliminate the need to redo standard calculations on every project. This is even the case in seismic (earthquake) design. Unless a design is so special that a full computer model needs to be created, all of the complex interactions between the soil, the structure, and nearby faults is reduced to a simple horizontal load that is applied through the center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building codes and uncertainties in loads require factors of safety to be somewhat large compared to other professions. This means that a simplified method for solving a problem doesn't affect the end result much when compared to an &lt;em&gt;exact&lt;/em&gt; mathematical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of the day to day calculations that an engineer completes use the same sets of formulae with different inputs. This is why huge Excel spreadsheets can be created to do a lot of the work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't mean that higher level math and the theories that go behind it aren't useful though. All of those topics help to train an engineer's mind to visualize what is really going on. The topic on &lt;a href=&quot;https://engineering.stackexchange.com/questions/207/how-do-engineers-really-use-numerical-simulation&quot;&gt;numerical simulation&lt;/a&gt; speaks to this.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-02-10T01:45:03.940" LastActivityDate="2015-02-10T01:45:03.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="563" PostTypeId="2" ParentId="550" CreationDate="2015-02-09T14:19:59.323" Score="7" Body="&lt;p&gt;Depending on how you look at it, none and all of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cycle of doing something the hard way, learning a short cut and then moving on to advanced material repeats all the way through college.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example once I started taking Algebra, I stopped doing multiplication tables. College level math is the same way. After calculus most engineers take differential equations. At that point I really stopped doing calculus and started relying on tools to do it for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In controls work we use a lot of Laplace transforms to define a system. While I technically know the full theory behind the Laplace transform, I haven't done one by hand in almost a decade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while I haven't 'used' calculus since my 3rd-4th years of university everything I learned during them required a fundamentals of calculus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: An analogy of sorts. This is like asking someone on the 14th floor of a building how many times they use the 3rd floor. It may be never, but without the 3rd floor there wouldn't be a 14th floor either.&lt;/p&gt;&#xA;" OwnerUserId="231" LastEditorUserId="231" LastEditDate="2015-02-10T16:19:38.917" LastActivityDate="2015-02-10T16:19:38.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="564" PostTypeId="2" ParentId="556" CreationDate="2015-02-09T15:12:02.613" Score="10" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Does the DC load somehow feedback and lower the resistance of the primary coil so that more power can be drawn?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes.  It would be simpler to analyze an AC load though.  The diodes are not central to your question:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UBRVt.gif&quot; alt=&quot;AC voltage source connected to transformer connected to load resistance&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The impedance of RL is also transformed, so if you have a 10:1 transformer and RL is 2 Ω, the AC source will see the transformer as a 200 Ω resistor (&lt;span class=&quot;math-container&quot;&gt;$10^2⋅2$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;As the current in a coil changes, it creates a changing magnetic field.  In the case of a transformer with a load, however, the change in magnetic field creates a current in the secondary, which immediately creates its own changing magnetic field in the opposite direction, cancelling out the primary's field.  People tend to forget that &lt;a href=&quot;https://www.google.com/books/edition/Radio_Frequency_Electronics/9X5gy_mDamsC?hl=en&amp;amp;gbpv=1&amp;amp;dq=an%20ideal%20transformer%20has%20%22no%20magnetic%20field%22%20while%20operating&amp;amp;pg=PA166&amp;amp;printsec=frontcover&quot; rel=&quot;nofollow noreferrer&quot;&gt;an ideal transformer has &lt;strong&gt;no magnetic field&lt;/strong&gt; while operating&lt;/a&gt;.  Any change in either coil's field is immediately cancelled by a change in the other.&lt;/p&gt;&#xA;&lt;p&gt;The &amp;quot;feedback&amp;quot; is caused by the same effect.  The primary causes the secondary to change, and the secondary causes the primary to change in return.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;When there is no load on the DC side, does power still flow through the AC primary coil, and if so, why doesn't it just melt?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;With nothing connected to the secondary side, the secondary coil is open circuited and does nothing.  It's just some metal that happens to be nearby.  The circuit is now just an AC source driving the primary coil, which behaves as a lone inductor:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/coFHi.gif&quot; alt=&quot;AC voltage source connected to an inductor&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Ideal inductors do not consume any power; they just store energy temporarily in one half of the cycle and return it to the supply on the other half.  Real coils are not made of perfect conductors, though, and have some resistance, so the power consumed by the primary coil will be determined by the resistance of the wire.&lt;/p&gt;&#xA;&lt;p&gt;Also, it's not quite right to say &amp;quot;power still flow through the AC primary coil&amp;quot;.  &amp;quot;Current&amp;quot; is flowing through the primary, and the resistance of the primary to that current causes it to &amp;quot;dissipate energy&amp;quot; (or power) into the room.  &amp;quot;Power&amp;quot; is actually the &lt;em&gt;rate&lt;/em&gt; at which energy flows, and energy actually flows through the empty space between the wires, not in the wires themselves.  Once you understand this, a lot of things make much more sense.&lt;/p&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditDate="2022-05-09T21:51:24.527" LastActivityDate="2022-05-09T21:51:24.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="565" PostTypeId="2" ParentId="438" CreationDate="2015-02-09T17:26:23.977" Score="5" Body="&lt;p&gt;The given information describing a &lt;em&gt;compacted&lt;/em&gt; soil sample is as follows: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;initial moisture content, $\omega_{init}$&lt;/li&gt;&#xA;&lt;li&gt;specific gravity, $G_s$&lt;/li&gt;&#xA;&lt;li&gt;initial volume, $V_{init}$&lt;/li&gt;&#xA;&lt;li&gt;initial weight, $W_{init}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For completeness: the following information has already been determined: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;moist unit weight, $\gamma_{wet}$ using the relationship $\gamma_{wet}=\frac{W_{init}}{V_{init}}$&lt;/li&gt;&#xA;&lt;li&gt;dry unit weight, $\gamma_{d-init}$ using the relationship $\gamma_{d-init}=\frac{\gamma_{wet}}{1+\omega_{init}}$&lt;/li&gt;&#xA;&lt;li&gt;saturation, $S$ using the relationship $S=\frac{V_{water}}{V_{voids}}=\frac{V_{water}}{V_{init}-V_{solids}}=\frac{\frac{W_{init}\omega_{init}}{\gamma_w}}{V_{init}-\frac{\gamma_{d}V_{init}}{G_s\gamma_w}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(where $\gamma_w$ is the unit weight of water)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The problem is to determine the unit weight and the moisture content after the soil sample has been submerged and allowed to swell 5%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key detail for this problem is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This compacted soil sample was then submerged in water.... After two weeks... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A soil sample that has been submerged in water for two weeks can/should be assumed** to have become &lt;em&gt;saturated&lt;/em&gt; ($S=100\%$); i.e., all of the air in the void spaces has escaped, and the void space is now 100% filled with water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list of soil sample properties that can be assumed to remain constant after submersion is fairly short: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Specific gravity, $G_s$&lt;/li&gt;&#xA;&lt;li&gt;Weight of solids, $W_s$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of the other properties, such as saturation, unit weight, dry unit weight, moisture/water content, void ratio, etc. are dependent on the volume of voids and the amount of water in the soil. Both the amount of water (it was submerged) and the volume (it has swelled) have changed, so ALL of these properties will also change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once all of this has been recognized, the remaining portion of the problem is trivial: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;New wet unit weight: $\gamma_{new}=\gamma_{sat-new}=\frac{W_s+W_{w-new}}{V_{new}}=\frac{\gamma_{d-init}V_{init}+\gamma_w(V_{new}-V_{solids})}{V_{vew}}=\frac{\gamma_{d-init}V_{init}+\gamma_w(V_{new}-\frac{\gamma_{d}V_{init}}{G_s\gamma_w})}{V_{init}(1+5\%)}$&lt;/li&gt;&#xA;&lt;li&gt;New moisture content: $\omega_{new}=\frac{W_{w-new}}{W_{solids}}=\frac{\gamma_w(V_{new}-V_{solids})}{W_{solids}}=\frac{\gamma_w(V_{init}(1+5\%)-\frac{\gamma_{d}V_{init}}{G_s\gamma_w})}{\gamma_{d-init}V_{init}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Mechanism of Soil Swelling Behavior&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The simplified effective stress equation is as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma^{\prime}=\sigma-u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\sigma^{\prime}$ is the effective stress, $\sigma$ is the total stress, and $u$ is the pore water pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above equation assumes a static condition. However, when the simplified effective stress equation is imbalanced, a dynamic condition occurs and the soil must either consolidate (i.e, &quot;shrink&quot;), or swell. Swelling of soil occurs when the two sides of the simplified effective stress equation are not balanced, and: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is positive pore water pressure inside the void space of the soil, and &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;the &lt;em&gt;effective stress&lt;/em&gt; inside of the soil matrix is greater than the externally applied &lt;em&gt;total stress&lt;/em&gt; minus the &lt;em&gt;pore water pressure.&lt;/em&gt;&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Said another way: when a soil is compacted, some amount of &lt;em&gt;total stress&lt;/em&gt; is applied. Once equilibrium has been achieved, this &lt;em&gt;total stress&lt;/em&gt; is associated with some combination of &lt;em&gt;effective stress&lt;/em&gt; and &lt;em&gt;pore water pressure&lt;/em&gt;. If the &lt;em&gt;total stress&lt;/em&gt; changes, the previous combination of &lt;em&gt;effective stress&lt;/em&gt; and &lt;em&gt;pore water pressure&lt;/em&gt; within the soil matrix initially remains, but the imbalance this causes must dissipate over time. In order for the imbalance to dissipate, the voids must either increase in volume (swelling), or decrease in volume (consolidation), depending on the nature of the imbalance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the total stress has been removed/reduced. The &lt;em&gt;pore water pressure&lt;/em&gt; is &quot;pushing&quot; against the &quot;walls&quot; of the soil matrix pores (as always happens when $u&amp;gt;0$ - even when the simplified effective stress equation is balanced). Due to the reduction of &lt;em&gt;total stress&lt;/em&gt;, there is too much internal stress (i.e., &lt;em&gt;effective stress&lt;/em&gt;) being applied, and it must be relieved by a decrease in *pore water pressure * (i.e., an increase in volume). Or said another way, the applied total stress is not enough to stop the pores from expanding due to the pushing of the internal &lt;em&gt;pore water pressure&lt;/em&gt;. Therefore the soil will swell until this imbalanced condition is resolved. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;**The reasons for this assumption are somewhat complicated, and the assumption may not always be accurate. However, in general, the most conservative assumption for most soil mechanics/geotechnical problems is for the soil to be saturated. Therefore, if there is reason to believe the soil may be saturated, even if there is uncertainty, we almost always assume the soil is in fact saturated. &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2015-02-09T19:49:00.507" LastActivityDate="2015-02-09T19:49:00.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="566" PostTypeId="2" ParentId="55" CreationDate="2015-02-09T18:46:23.927" Score="4" Body="&lt;p&gt;Some materials are somewhat better than others in lower light levels BUT in a typical indoor situation, even an extremely optimised material simply lacks the availability of input energy to convert. eg Mono-crystalline silicon is somewhat better at very low light levels than Poly-crystalline but neither can overcome the fundamental lack of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full sunlight ~= 100,000 lux (lumens/metre^2).&lt;br&gt;&#xA;Room-light is say 250 lux (quite bright).&lt;br&gt;&#xA;A modern LCD monitor displaying all white and set to &quot;nice and bright&quot; will have a brightness &lt;strong&gt;at the screen surface&lt;/strong&gt; of about 250 lux. &#xA;The difference between full sunlight &amp;amp; this is 400:1.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good PV (photovoltaic) panel in bright sunlight will provide 150W per square metre delivered. (Over 200 W at the cell in the best cases). At 250 lux the BEST you can expect is around 500 milliWatts per square metre. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the 250 lux = 250 Watts/m^2 I mentioned above you'd need 250 Watts of &lt;strong&gt;actual light energy&lt;/strong&gt; shining on exactly one square metre. The very best LEDs available convert about 50% of their DC input to light. A good commercial product is very approximately 25% efficient. So you'd need to illuminate a one square metre area with &lt;strong&gt;1,000 Watts&lt;/strong&gt; DC input** of good commercial LED lighting to  get 500 milliWatts output. An about 2000:1 ratio of power in to power out. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;** eg a top &quot;150 Watt replacement&quot; bulb (Philips, Cree, ...) may have about 25 Watts of input. You'd need 1000 / 25 = 40 of these &quot;150 Watt replacement&quot; bulbs shining on a good 1 square metre PV panel to produce about 500 milliWatts output. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-02-09T18:58:45.410" LastActivityDate="2015-02-09T18:58:45.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="567" PostTypeId="2" ParentId="550" CreationDate="2015-02-09T19:36:31.773" Score="7" Body="&lt;p&gt;I agree, as discussed in a few of the other answers, that most of the time engineers do not directly use calculus (or other advanced math) very often in order to do their day to day job. And at the same time, having an understanding of it is vital for a good engineer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would add, though, that understanding advanced mathematics well enough to use it effectively can be extremely helpful in this present era in which advanced mathematical tools are readily available. For example, a program such as Mathcad allows the user to perform direct integration of a domain, and an engineer who understands how to use this properly can create extremely effective, accurate, and fast tools to solve routine problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a geotechnical engineer, one problem I may often find myself solving in which this ability turns out to be most useful is the primary settlement $S_p$ of a soil layer.  The settlement equation is simple: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_p=H_{\text{layer}}\varepsilon_v=H_{\text{layer}}\frac{\Delta e}{1+e_0}$$ where $\varepsilon_v$ is vertical strain and $e$ is the void ratio of the soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it turns out that $\Delta e$ is a stress-dependent quantity, and stress varies with depth (i.e., it is a function of depth, $z$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta e=C_c\log{\frac{\sigma^{\prime}_0+\Delta \sigma^{\prime}}{\sigma^{\prime}_0}}$$ where $C_c$ is the compression index (constant), and $\sigma^{\prime}$ is effective stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that in practice things are even worse since $e_0$ also varies with depth as well, but we often assume it to be constant when performing calculations in order to make things easier.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\sigma^{\prime}$ changes continuously with depth, the the usual way to do this problem is to just split the soil profile into 1 foot layers, and to use the effective stress at the center of each layer to find $S_p$ for that layer. Then you just add them up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a much better, and easier, way to do this is to simply &lt;em&gt;directly integrate&lt;/em&gt; using a tool like Mathcad! Instead of dividing a 15 foot soil column into 1 foot increments, and performing the same set of calculations at each of the 15 layers, all I have to do (a single time) is this: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Define pore water pressure as a function of depth, $z$ (simplest case): &#xA;$$u(z)=0$$ &lt;/li&gt;&#xA;&lt;li&gt;Define total stress as a function of depth, $z$: &#xA;$$\sigma_0(z)=\gamma_{\text{soil}}z$$&lt;/li&gt;&#xA;&lt;li&gt;Define effective stress as a function of depth, $z$: &#xA;$$\sigma^{\prime}_0(z)=\sigma_0(z)-u(z)$$&lt;/li&gt;&#xA;&lt;li&gt;Define effective stress increase as a function of depth, $z$ (simplest case is a constant increase): &#xA;$$\Delta\sigma^{\prime}(z)=1000\text{ psf}$$ &lt;/li&gt;&#xA;&lt;li&gt;Define change in void ratio as a function of depth, $z$: $$\Delta e(z)=C_c\log{\frac{\sigma^{\prime}_0(z)+\Delta \sigma^{\prime}(z)}{\sigma^{\prime}_0(z)}}$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And finally, find the total consolidation of the layer down to any depth $z=H_{\text{layer}}$ by &lt;em&gt;directly integrating&lt;/em&gt; the primary settlement equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_p=\int_{0}^{H_{\text{layer}}}{\frac{\Delta e(z)}{1+e_0}\text{d}z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach is quicker, more accurate, and easier than the method taught in your soil mechanics or foundations textbook. However, it requires an ability to understand and apply basic calculus in order to implement it properly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are loads of other examples (e.g., structural analysis of a beam in bending, groundwater flow, volumetric flow analysis of a watershed hydrograph, etc etc) in which direct integration would be a superior approach to that commonly used if the right tool is available. &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2015-02-11T20:35:58.703" LastActivityDate="2015-02-11T20:35:58.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="569" PostTypeId="1" AcceptedAnswerId="577" CreationDate="2015-02-09T22:48:52.813" Score="38" ViewCount="18973" Body="&lt;p&gt;The campus where I work has a long covered walkway (~.5 mile) which has several labeled pipes running under the roof (chilled water, fuel oil, air...). All of the pipes run dead straight &lt;em&gt;except&lt;/em&gt; for the natural gas lines, which have little loops spaced about every 250ft, as seen in the attached image (the lowermost, yellow line. There's another natural gas line hidden above all the others, which also does the same thing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line isn't branching at these points, and there doesn't seem to be any need to divert the pipe in order to support it.  I've looked at some building codes to see if I could find a reason (or even a requirement) to insert these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas as to what these are? It's driving me batty!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VhLi2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="463" LastEditorUserId="90" LastEditDate="2015-02-10T11:40:04.500" LastActivityDate="2021-01-27T05:54:16.403" Title="What is the purpose of these diversions in a natural gas line?" Tags="|mechanical-engineering|building-physics|pipelines|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="570" PostTypeId="1" CreationDate="2015-02-09T23:18:07.607" Score="9" ViewCount="634" Body="&lt;p&gt;I have a $2\text{ mm}$ sheet of &lt;a href=&quot;https://en.wikipedia.org/wiki/High-density_polyethylene&quot; rel=&quot;nofollow noreferrer&quot;&gt;HDPE&lt;/a&gt; at $280-320^{\circ}\text{ F}$. I want to pierce the skin of this sheet with a $2\text{ mm}$-high pin with a surface area of $3.33\text{ mm}^2$. The pin is also made of HDPE and has cooled completely. With the Young's modulus of HDPE being $0.8\text{ GPa}$, how much force do I need to apply to have the cooled HDPE pin reach the molten layer of the HDPE sheet at $280-320\text{ F}$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: I've been told that the pin is supposed completely weld to the layer of HDPE.  Based on the melt temp of the HDPE and the starting temp of the pin, I just don't think its possible in the 2 minutes before the material is quenched.  I've attempted to figure out the rate of heat transfer, but can't find the thermal contact conductance coefficient, or the thermal contact resistance.  Searching online turned up a tool &quot;CoCoE.exe&quot; but was having trouble running the program on my computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gcvgM.png&quot; alt=&quot;More Descriptive Image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="465" LastEditorUserId="465" LastEditDate="2015-02-11T22:15:11.247" LastActivityDate="2015-02-13T18:49:19.023" Title="How do I compute the force required to pierce HDPE?" Tags="|materials|temperature|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="571" PostTypeId="2" ParentId="390" CreationDate="2015-02-09T23:18:08.763" Score="3" Body="&lt;p&gt;The average and standard deviation from the above sample set are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Average&lt;/strong&gt; ($\mu$): 4.754V&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Standard Deviation&lt;/strong&gt; ($\sigma$): 0.142368&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mean&lt;/strong&gt;: 4.74&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;USL&lt;/strong&gt;: 5.5&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;LSL&lt;/strong&gt;: 4.5&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C$_{p}$&lt;/strong&gt; : 1.1706&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C$_{pl}$&lt;/strong&gt;: 0.5947&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C$_{pu}$&lt;/strong&gt;: 1.7466&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C$_{pk}$&lt;/strong&gt;: 0.5947 &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Lk4C.png&quot; alt=&quot;Histogram for Random Data&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/zfx7D.png&quot; alt=&quot;Histogram using Original Data&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Equations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Pr}(\mu - \sigma \leq x \leq \mu + \sigma) \approx 0.6827$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Pr}(\mu - 2\sigma \leq x \leq \mu + 2\sigma) \approx 0.9545$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Pr}(\mu - 3\sigma \leq x \leq \mu + 3\sigma) \approx 0.9973$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{p} = \frac{(USL - LSL)}{6\sigma}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{pu} = \frac{USL-\mu}{3*\sigma}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{pl} = \frac{\mu-LSL}{3*\sigma}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{pk} = min(C_{pu},C_{pl})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the above data, it appears that the process is between 1 sigma and 2 sigma capable. Approximately about 10% of the product could fall outside of the requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also a sample set of 30 is considered a generally acceptable sample size. The sample set in this problem is 15 which is on the low side.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Suggestions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Check with the vendor if they sorted the sample parts&lt;/li&gt;&#xA;&lt;li&gt;The power supply is not delivering the required 1A of current at full&#xA;load&lt;/li&gt;&#xA;&lt;li&gt;The current carrying conductor might not be capable to delivering the&#xA;current at full load&lt;/li&gt;&#xA;&lt;li&gt;The power supply might have some minor design issues too&lt;/li&gt;&#xA;&lt;li&gt;Also check the test system too&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The distribution appears to be a normal bell curve. A supplier Quality engineering might be able help address this issue. &lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="282" LastEditDate="2015-06-30T21:45:29.817" LastActivityDate="2015-06-30T21:45:29.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="572" PostTypeId="1" CreationDate="2015-02-09T23:52:46.737" Score="4" ViewCount="1571" Body="&lt;p&gt;Anyone used to doing housework has probably noticed that vacuum cleaners come in both &quot;bagged&quot; and &quot;bagless&quot; varieties. If you're of a certain age, you may have noticed when these &quot;bagless&quot; vacuums appeared back in the 1990s. Though I didn't recognize it at the time, these &quot;bagless&quot; vacuums are an example of a technology I would later study in my undergraduate career, called &lt;a href=&quot;http://en.wikipedia.org/wiki/Cyclonic_separation&quot; rel=&quot;nofollow noreferrer&quot;&gt;cyclonic separation&lt;/a&gt;. &lt;a href=&quot;http://www.theguardian.com/technology/2014/jan/17/dyson-cinetic-vacuum-cleaner-cyclone-flexible&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article from The Guardian&lt;/a&gt; talks a bit about the Dyson* company's introduction and application of cyclone technology in vacuum cleaners, and makes the claim that the newest models separate &quot;even the smallest of particles&quot; without filtration (i.e., entirely by cyclonic separation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given my experience with cyclones, this is quite an impressive claim. Though widely used to deal with coarse and medium-coarse particles, industrial cyclones are typically incapable of removing fine particulates, including many of the &quot;allergens&quot; that Dyson claims their vacuums now remove, without additional filtration of the exhaust.** In the area of air pollution control, cyclones are often regarded as &quot;pre-treatment&quot; devices for this reason, and because they are more durable and cost less to operate and maintain than e.g. baghouses, wet scrubbers or electrostatic precipitators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes this even more impressive to me is that these vacuums are portable; not only do they move around as you clean, the body of the vacuum tilts up and down. In handheld applications, from what I understand, they can even be inverted! I don't understand how a cyclone would continue to function in this situation. I would expect a sharp drop in control efficiency (or in plain terms, a lot of dust in the exhaust).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, my fourth edition copy of &lt;a href=&quot;http://books.google.com/books/about/Air_Pollution_Control.html?id=Xfi7bwAACAAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Air Pollution Control: A Design Approach&lt;/a&gt; (Cooper &amp;amp; Alley, 2011) describes cyclones exclusively in terms of a fixed vertical orientation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Typically, a particulate-laden gas enters tangentially near the top of&#xA;  the cyclone... The gas flow is forced into a downward spiral...&#xA;  Centrifugal force and intertia cause the particles to move outward,&#xA;  collide with the outer wall, and then slide downward to the bottom of&#xA;  the device. (p. 135)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this configuration, gravity causes the separated particles to fall into the hopper, where they are segregated from the air stream and can later be removed. &lt;a href=&quot;http://yosemite.epa.gov/oaqps/EOGtrain.nsf/12f4a7ba1594b00d85256b450052f92a/5d8c02256df2dec285256da3005cc39a/$FILE/SI%20445_7.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This EPA inspection course handbook&lt;/a&gt; shows three inlet configurations (two tangential and one axial); in each case, the contaminant falls downward into the hopper under the influence of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia cites &lt;a href=&quot;http://www.powderbulksolids.com/node/37772&quot; rel=&quot;nofollow noreferrer&quot;&gt;an article describing horizontally-oriented cyclones&lt;/a&gt; utilizing &quot;a powerful secondary air stream to direct collected material toward the collection hopper,&quot; something that is not included in the traditional vertical design. The article includes the following illustration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BX5pp.jpg&quot; alt=&quot;Horizontal-cyclone.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every industrial cyclone I've seen &quot;in the wild&quot; has been configured vertically—in fact, I had never heard of a horizontal cyclone before researching this question. In either case, the dimensions of the device must be carefully determined to provide optimal collection; the Cooper &amp;amp; Alley book gives the following theoretical equation for the diameter &lt;em&gt;d&lt;sub&gt;p&lt;/sub&gt;&lt;/em&gt; of the smallest particle that should be collected with 100% efficiency:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$d_p = \left[ {{9 \mu W} \over { \pi N_e V_i (\rho_p - \rho_g)}} \right] ^ {1/2}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$N_e$ is the number of effective turns in the outer vortex, in a tangential-top inlet configuration;&lt;/li&gt;&#xA;&lt;li&gt;$V_i$ is the gas inlet velocity; and&lt;/li&gt;&#xA;&lt;li&gt;$W$ is the width of the inlet duct.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The remaining parameters are density and viscosity of the gas and density of the particle (these do not depend on the design of the cyclone). The number of effective turns is determined by:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$N_e = {{1 \over H} \left[ L_b + {L_c \over 2} \right]}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$H$ is the height of the inlet duct;&lt;/li&gt;&#xA;&lt;li&gt;$L_b$ is the length of the cylindrical (upper) section of the cyclone body; and&lt;/li&gt;&#xA;&lt;li&gt;$L_c$ is the length of the conical (lower) section of the cyclone body.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Particles with diameter less than $d_p$ are collected with decreasing efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of all this background information is not to bore you, but to demonstrate how the dimensions of the cyclone—including height, length and width—determine the overall theoretical collection efficiency. These equations explicitly assume that the device is vertically oriented and &lt;strong&gt;implicitly assume that it is stationary&lt;/strong&gt;; the latter is never stated in the book, but every illustration and description is of fixed devices in industrial applications, so I think it's a very safe interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect the stationary aspect is important not just because it's standard in industry but because of the importance of maintaining cyclonic flow. One comment points out that accelerating the cyclone horizontally will help 50% and hurt 50% of the entrained particles. The issue I have with that is that under ideal operation, the claim here is near complete control; there is no further benefit of &quot;helping&quot; some of the particles. Any acceleration that counteracts the separation will add contaminants to the exhaust stream. Not only that, if the flow pattern changes, the number of effective turns &lt;em&gt;N&lt;sub&gt;e&lt;/sub&gt;&lt;/em&gt;  will change. A reduction there implies an increase in the cutoff diameter and a decrease in control efficiency, even if only temporarily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dyson's technology is proprietary; even if one of their engineers or researchers finds this question, I don't expect a detailed description of how their device works. Also, there are many other brands of cyclonic vacuum cleaner on the market today; I don't want to frame this question as specific to one brand's technology. What I'd like to know is &lt;strong&gt;how can a cyclone maintain efficient operation in conditions involving constant, unpredictable motion and variable orientation?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* Disclaimer: I have no relationship with this company or any of its vendors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** The implication here is that other models of cyclonic vacuum (including models sold by other manufacturers) do provide additional filtration after the cyclone. I can't tell how much of this is marketing copy designed to make Dyson's technology look better than it is.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-05-05T18:31:56.600" LastActivityDate="2015-05-05T18:31:56.600" Title="How do cyclonic vacuum cleaners maintain collection efficiency with constant motion and changes in orientation?" Tags="|fluid-dynamics|environmental-engineering|air-quality|cyclones|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="574" PostTypeId="2" ParentId="550" CreationDate="2015-02-10T00:33:21.653" Score="3" Body="&lt;p&gt;I'll give an example of calculus that I used today as a Software Engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were estimating the computational time of performing an operation on each of many groups of elements. The time taken for an individual group is proportional to the size of the group squared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're not sure of the distribution of the sizes of the groups, but depending on different algorithms we might use, we might be able to model them as normally distributed, power-law distributed, exponentially-distributed, etc., as well as influence the parameters of those respective distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the expected value of $X^2$ where $X$ is sampled from some distribution requires basic calculus knowledge :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, things like this pop up from time to time. I don't know that I've ever used it explicitly in terms of writing software that carries out calculations related to calculus, nor have I used it as an authoritative decision-making tool. Usually that's left up to &quot;try a few things and see what works best&quot; but it's definitely useful for basic whiteboard brainstorming or estimation. In this case, it let us theorize about what kind of distribution we hope will work best, and focus our efforts on trying that path out. I can certainly say that very basic fundamentals of calculus are useful to understand the dynamics of some software systems. Four semesters is probably overkill.&lt;/p&gt;&#xA;" OwnerUserId="467" LastEditorUserId="95" LastEditDate="2015-02-10T00:55:30.200" LastActivityDate="2015-02-10T00:55:30.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="576" PostTypeId="2" ParentId="550" CreationDate="2015-02-10T00:54:11.123" Score="0" Body="&lt;p&gt;When I was taking courses as a &quot;special student&quot; at Carnegie Mellon University in Pittsburgh (in the mid-1970s),&quot;engineering mathematics&quot; consisted of linear algebra, ordinary and partial differential equations, and &quot;special topics&quot; such as power series and fourier series solutions, as well as LaPlace transforms. This is a &quot;heavy&quot; engineering school, and many will have programs that are &quot;lighter.&quot;&lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2015-02-10T00:54:11.123" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="577" PostTypeId="2" ParentId="569" CreationDate="2015-02-10T01:06:44.057" Score="36" Body="&lt;p&gt;The loops are known as expansion loops. They need to be placed in pipelines to enable the pipelines to contend with thermal expansion and contraction and other forces that can affect the pipeline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are typically placed in gas pipelines, irrespective of when the gas is hot or cold - natural gas or steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following quote is from &lt;a href=&quot;http://petrowiki.org/Pipeline_design_consideration_and_standards&quot;&gt;Pipeline Design&lt;/a&gt;. It's near the end of the page under &lt;em&gt;Pipe Expansion and Supports&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Steel piping systems are subject to movement because of thermal expansion/contraction and mechanical forces. Piping systems subjected to temperature changes greater than 50°F or temperature changes greater than 75°F, where the distance between piping turns is greater than 12 times the pipe diameter, may require expansion loops.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Such loops are not only used in gas pipelines and are used in pipelines that convey other fluids such as oil. Pipes transporting liquid fluids can use other expansion control devices such as single slip expansion joints or bellow type expansion joints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.districtenergy.org/assets/pdfs/09CampConference/Proceedings/Tuesday/2PatnodePipeExpansion.pdf&quot;&gt;Accommodating Thermal Expansion in Pipes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://articles.compressionjobs.com/articles/oilfield-101/4336-pipelines-safety-joints-bends-shafer-valve?showall=1&quot;&gt;Pipeline Safety&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="49" LastEditDate="2015-02-11T22:18:11.970" LastActivityDate="2015-02-11T22:18:11.970" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="581" PostTypeId="2" ParentId="572" CreationDate="2015-02-10T05:34:27.887" Score="1" Body="&lt;p&gt;I'm betting that a big part of the discrepancy here lies in the duration of operation. Industrial separators are intended for long term continuous operation, while home cleaners can stand to shut down frequently to be emptied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The act of separating particles is accomplished centrifugally, neither gravity nor a secondary air stream are required. The issue at hand is how to deal with the excess of particles accumulating on the outer walls of the chamber. In the short term you don't need to do anything special to keep particles from getting into the exhaust but an excess of dust and lint will eventually clog the system. The industrial systems are specifically designed to deal with this problem in the long term (as you've indicated). The home systems can probably get away with intermittent loss of gravity before too much gunk collects but I don't think that you can run one of these home vacuums fully inverted 100% of the time and expect it to operate correctly for as long as it would if it were oriented normally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll add that (from a perusal of image searches), it appears that the designers made an effort to direct the air downward as it enters the first cyclone.  This may have an effect of pushing accumulated dirt in that direction (much as the secondary stream does in a horizontal design), so that gravity isn't the only force driving the dirt &quot;down&quot;.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-02-10T05:34:27.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="582" PostTypeId="1" CreationDate="2015-02-10T09:39:25.527" Score="2" ViewCount="1680" Body="&lt;p&gt;Normally, people use &lt;code&gt;Autoclaved Aerated Concrete (AAC)&lt;/code&gt; to build a concrete house because ACC is a good thermal insulation brick that easy to buy and not expensive&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And another way i've heard about insulation wall is &lt;code&gt;Cavity wall&lt;/code&gt;. IMO, it rarely find a house that use this kind of wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the main topic of discussion here might be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;code&gt;Autoclaved Aerated Concrete (AAC) wall&lt;/code&gt; vs &lt;code&gt;Cavity wall&lt;/code&gt; which one is the best thermal insulation wall?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that another method can be purpose if it better than &lt;code&gt;ACC wall&lt;/code&gt; and &lt;code&gt;Cavity wall&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="424" LastEditorUserId="368" LastEditDate="2015-02-13T23:08:53.300" LastActivityDate="2015-03-03T18:11:48.270" Title="What is the best insulation wall for a house? ACC wall, Cavity wall, etc" Tags="|structures|concrete|energy-efficiency|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="584" PostTypeId="2" ParentId="550" CreationDate="2015-02-10T12:12:29.333" Score="5" Body="&lt;p&gt;An electronics engineer here, who found the maths the most difficult part of his degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quite routinely have to use and manipulate complex numbers when doing RF engineering, circuit modelling and design.  They have also been useful when modelling ultrasonic propagation.  I have often wished that Excel handled complex numbers as a built-in type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An understanding of ODEs is vital when designing control and feedback systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Understanding the concepts of Fourier series, Laplace and Z-transforms and convolution has been necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing for me has been in knowing what maths is out there and being able to ask a mathematician for help when needed.  The mathematicians that I have consulted have invariably been delighted to help out with practical problems.&lt;/p&gt;&#xA;" OwnerUserId="477" LastActivityDate="2015-02-10T12:12:29.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="585" PostTypeId="1" AcceptedAnswerId="587" CreationDate="2015-02-10T13:06:50.863" Score="4" ViewCount="82" Body="&lt;p&gt;For Newtonian fluids, the sinking (or rising) speed of a particle is (reasonably) well known. What is a relationship for shear thinning fluids?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="106" LastEditDate="2015-02-16T00:25:44.103" LastActivityDate="2015-02-16T00:25:44.103" Title="Sinking speed of particle in shear thinning fluid?" Tags="|fluid-dynamics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="586" PostTypeId="2" ParentId="582" CreationDate="2015-02-10T14:33:47.093" Score="0" Body="&lt;p&gt;In the UK, cavity wall construction for houses is very common. The external skin is generally made from brick, with the internal skin made from &quot;breeze blocks&quot;. Breeze blocks are commonly made from aerated concrete!&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-02-10T14:33:47.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="587" PostTypeId="2" ParentId="585" CreationDate="2015-02-10T17:33:17.953" Score="4" Body="&lt;p&gt;Doing a brief review of the literature, for spherical particles at high Reynolds numbers, &lt;a href=&quot;http://books.google.com/books?id=ekIrBgAAQBAJ&amp;amp;lpg=PP1&amp;amp;ots=T2jDGPKMHP&amp;amp;lr&amp;amp;pg=PA87#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;it appears that you can use standard drag curves&lt;/a&gt; to get the drag coefficient  as long as you use the following Reynolds number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Re}_\text{PL} \equiv \frac{\rho V^{2-n} d^n}{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Reynolds number assumes the fluid follows a &lt;a href=&quot;http://en.wikipedia.org/wiki/Power-law_fluid&quot; rel=&quot;nofollow&quot;&gt;power-law&lt;/a&gt;. $n$ is the flow behavior index, which is less than 1 for a shear thinning fluid. $m$ in the above equation is $K$ on Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you just want the terminal velocity, you can plug this in the &lt;a href=&quot;http://en.wikipedia.org/wiki/Terminal_velocity#Physics&quot; rel=&quot;nofollow&quot;&gt;standard equations for terminal velocity&lt;/a&gt;, as all they assume is that you know the drag coefficient at the terminal velocity state. (You can derive them from a steady force balance or from the unsteady case. I recommend trying both just to get a feel for these sorts of problems.)&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2015-02-10T17:33:17.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="588" PostTypeId="2" ParentId="481" CreationDate="2015-02-10T19:09:52.547" Score="0" Body="&lt;p&gt;I would consider looking into BLDCs as motors, if they fit the hydraulic power required for your application, i.e. motor electric power = hydraulic power. I would pick a slightly bigger motor, with maybe 1.2 times the hydraulic power. Also, dependent on the hydraulic system you want to actuate (whether it's parallel or series, just like in electricity) and when it contains valves, you either need volumetric flow rate (increase/decrease motor speed), pressure (increase/decrease motor moment), or a certain combination of both. That is for the gear pump you mention.&lt;/p&gt;&#xA;" OwnerUserId="436" LastActivityDate="2015-02-10T19:09:52.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="589" PostTypeId="1" AcceptedAnswerId="14677" CreationDate="2015-02-10T19:12:02.523" Score="4" ViewCount="505" Body="&lt;p&gt;NIL (Nanoimprint lithography) and hot embossing seem to mean the same thing in many instances. Is hot embossing a subset technique of NIL? Or are they completely different methods?&lt;/p&gt;&#xA;" OwnerUserId="455" LastEditorUserId="33" LastEditDate="2016-02-23T03:58:32.007" LastActivityDate="2017-04-07T10:42:09.093" Title="What is the difference between NIL and hot embossing?" Tags="|lithography|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="590" PostTypeId="1" CreationDate="2015-02-10T20:21:35.900" Score="21" ViewCount="398" Body="&lt;p&gt;One of the counterpoints to the Thorium fuel cycle is that Protactinium, which is generated  in this cycle &lt;a href=&quot;http://en.wikipedia.org/wiki/Protactinium#In_nuclear_reactors&quot;&gt;degrades reactor efficiency and thus needs to be removed&lt;/a&gt;, at least from the liquid fluoride or molten salt reactors. However, as far as I can tell, Protactinium was not removed during the functioning of the first solid-fuel Thorium reactor, which was the third core used at Shippingport; or at least I can find no mention of Protactinium being removed (during operation) in the &lt;a href=&quot;http://www.inl.gov/technicalpublications/Documents/2664750.pdf&quot;&gt;official fuel report&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, question(s):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Quantitatively, how bad can reactor efficiency be degraded by Protactinium not being removed?&lt;/li&gt;&#xA;&lt;li&gt;To what degree does this degradation depend on the type and other parameters (geometry etc.) of the reactor?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="480" LastActivityDate="2017-01-26T00:59:07.287" Title="How much can Protactinium degrade the efficiency of the Thorium fuel cycle?" Tags="|nuclear-technology|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="591" PostTypeId="1" AcceptedAnswerId="592" CreationDate="2015-02-10T20:22:09.863" Score="6" ViewCount="1108" Body="&lt;p&gt;If I know the tensile strength of a pin, how do I compute the total strength of multiple pins?  That is, 2 objects are connected by multiple pins, evenly distributed, each with a height of $2mm$ and a cross sectional area of $3.33mm^2$.  For some reason the answer of just multiplying the tensile strength of each pin by the number of pins, doesn't make 100% sense to me.&lt;/p&gt;&#xA;" OwnerUserId="465" LastEditorUserId="33" LastEditDate="2015-02-10T21:49:51.800" LastActivityDate="2015-02-10T21:49:51.800" Title="How do I compute combined tensile strength?" Tags="|materials|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="592" PostTypeId="2" ParentId="591" CreationDate="2015-02-10T20:59:53.387" Score="3" Body="&lt;p&gt;The short answer is yes, it is that simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about it this way. The pins are connected to end plates of a given area. If all we're worried about is the tensile strength, we can place a one dimensional force on each end plate to put the system in pure tension. Now, split the system into a couple parts and do a force balance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we split it right down the middle, so that we're seeing the actual force on the top plate, and the internal force along each of the pins. These two forces have to be equal, otherwise the system would be in motion, meaning the pins have yielded or fractured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to calculate the stress in the pins, we simply divide the force applied across all the pins by the total area. Because in this case, the pins are all of even cross sectional area, the stress will be divided evenly between them. The implication of this is that none of the pins should fail before the others, and any difference in their tensile strength is due to manufacturing defects or variations, so we can just multiple the tensile strength of a single pin by the number of pins to get the overall tensile strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an important assumption here that the pins are distributed evenly on the plates. If the pins are lopsided, they're going to end up under some amount of bending, which complicates the stress calculations in them and adds to the stress that they see under the same applied load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we make the pins of variable cross-sectional area, then the load begins to concentrate in the larger pins, but it should still end up with the same stress in each pin because the load will distribute proportionally to the area, which is how we calculate stress anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where this gets more complex is if the pins are made of different materials or treated differently to change the tensile strength. This is essentially creating a composite material, and there we have to take into account volume fractions to calculate new yield and tensile strength numbers, but also consider which material fails first. For further reading I'd suggest chapter 3 in &lt;a href=&quot;http://web.mit.edu/course/3/3.225/book.pdf&quot; rel=&quot;nofollow&quot;&gt;this MIT book&lt;/a&gt; (that is a 128 page pdf, beware if you're on mobile). &lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="41" LastEditDate="2015-02-10T21:14:31.387" LastActivityDate="2015-02-10T21:14:31.387" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="593" PostTypeId="2" ParentId="570" CreationDate="2015-02-10T22:32:36.053" Score="1" Body="&lt;p&gt;After starting to redo the analysis, I've come to realize this is a fairly complicated process, and instead I'll outline what I would do to solve this if I had more time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, some assumptions I'd make:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;perfectly plastic behavior of the HDPE (i.e., the stress-strain curve is flat after yielding)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;thermal convection from the flat surface is irrelevant (though it is taken into account for the pin)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the pin has an adiabatic tip&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the metal is at a constant temperature&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the pin pierces the surface when the energy absorbed by a cylindrical portion of the material directly beneath it is exceeds the toughness of the material multiplied by the volume of cylindrical portion of the material (I can explain this differently if desired.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the contact resistance decreases as the HDPE melts (not sure how)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;some material properties can come from &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/pen.760350205/pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; and others from &lt;a href=&quot;http://www.matbase.com/material-categories/natural-and-synthetic-polymers/commodity-polymers/material-properties-of-high-density-polyethylene-hdpe.html&quot; rel=&quot;nofollow&quot;&gt;this data sheet&lt;/a&gt;, and &lt;a href=&quot;https://books.google.com/books?id=V9g2SPNV4AcC&amp;amp;lpg=PA52&amp;amp;ots=ZporCRLw-l&amp;amp;pg=PA53#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt; (this data may or may not be valid for the HDPE you have)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm not sure what value of the convective heat transfer coefficient for the pin is valid and would have to think more about this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.ineos.com/Global/Olefins%20and%20Polymers%20USA/Products/Technical%20information/INEOS%20Typical%20Engineering%20Properties%20of%20HDPE.pdf&quot; rel=&quot;nofollow&quot;&gt;heat of fusion comes from this data sheet&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given all of this information, you can write 1D heat transfer solver for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Stefan_problem&quot; rel=&quot;nofollow&quot;&gt;Stefan problem&lt;/a&gt; to compute the temperature and thickness of the solid skin. From there, you can use the temperature of the skin to find the yield stress. My simple model suggests that $F = \sigma_\text{y} A_\text{contact}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is likely zero force or nearly so, because the &lt;a href=&quot;http://www.dynalabcorp.com/technical_info_hd_polyethylene.asp&quot; rel=&quot;nofollow&quot;&gt;melting temperature of HDPE is about 266 F&lt;/a&gt; (&lt;a href=&quot;http://www.maropolymeronline.com/Properties/HDPE%20Prop.asp&quot; rel=&quot;nofollow&quot;&gt;it's reported a little higher here at 279 F&lt;/a&gt;), so the HDPE probably has already melted at the specified temperature. You'll meet the molten layer immediately upon touching the surface (at least if the temperature is uniform enough).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the temperature range is wrong, I think I can develop an estimate assuming that HDPE is a perfectly plastic material, that the impacter is flat, and some other things, and then change my answer. Let me know if you'd like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you want an estimate of the force required to &lt;em&gt;continue&lt;/em&gt; penetrating the medium, I can find an old terminal ballistics book I read that has some models for such things, but I don't know how accurate they are.&lt;/p&gt;&#xA;" OwnerUserId="49" LastEditorUserId="49" LastEditDate="2015-02-13T18:49:19.023" LastActivityDate="2015-02-13T18:49:19.023" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="594" PostTypeId="1" AcceptedAnswerId="599" CreationDate="2015-02-11T01:51:06.607" Score="3" ViewCount="114" Body="&lt;p&gt;I'm writing a presentation for people and I need an analogy to convey the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Person A wants to sell a liquid, so he fills a tank, measures volume&#xA;  and sends to Person B. Now Person B receives the liquid and wants to&#xA;  know if he received the volume he bought from Person A, so since this&#xA;  liquid is now stored in his tank, Person B measures the liquid volume.&#xA;  Person B measures won't match exactly Person A measures - due to&#xA;  losses and imprecisions in both measures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are oil tanks and I need to explain this to really varied audience that contains people unskilled in metrology. Also, Person A and B can be the same person in different places - which is actually the case that I'm studying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think an analogy for the tanks would be good, but I can't think anything! How can I explain this problem in a way that a lay audience will understand?&lt;/p&gt;&#xA;" OwnerUserId="488" LastEditorUserId="368" LastEditDate="2015-02-13T05:00:45.320" LastActivityDate="2015-02-13T05:00:45.320" Title="Communicating losses and imprecision in liquid measurement by analogy" Tags="|education|metrology|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="595" PostTypeId="1" AcceptedAnswerId="685" CreationDate="2015-02-11T02:28:29.930" Score="10" ViewCount="259" Body="&lt;p&gt;Movable span bridges (&lt;a href=&quot;http://en.wikipedia.org/wiki/Bascule_bridge&quot;&gt;bascule&lt;/a&gt; and &lt;a href=&quot;http://en.wikipedia.org/wiki/Vertical-lift_bridge&quot;&gt;lift&lt;/a&gt;) typically employ a counterweight to help reduce the amount of energy and size of motor required to move the span. Due to variations in the the construction of the bridge and counterweight, the theoretical balance of both the span and the counterweight may be off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a movable span is complete, how close to perfectly balanced do the span and counterweight need to be?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="368" LastEditDate="2015-02-13T23:08:04.637" LastActivityDate="2015-02-16T02:57:37.103" Title="How accurately are moveable span bridges balanced?" Tags="|civil-engineering|bridges|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="596" PostTypeId="2" ParentId="550" CreationDate="2015-02-11T05:08:55.617" Score="4" Body="&lt;p&gt;As a computational scientist, i work closely with engineers developing the software tools they use to solve different kinds of engineering problems.  My work relies heavily on partial differential equations and numerical analysis, for which integrals, derivatives, taylor series, limits, green's theorem, optimization, rates of change, etc... are all the basic tools i use every day of my life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, professional engineers are the tool users, while i see myself as a toolmaker.  An engineer can certainly use a tool without knowing much about the intricacies of how it was made... But to pick the right tool for the job at hand, you have to understand the wider variety of tools to choose from and their advantages / disadvantages.  The only way to understand the advantages of one numerical tool over another, you have to understand the building blocks of that tool.  For this, calculus is absolutely necessary.&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2015-02-11T05:08:55.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="597" PostTypeId="2" ParentId="590" CreationDate="2015-02-11T05:33:07.530" Score="1" Body="&lt;p&gt;The protactinium separation is a nice benefit of a liquid fluoride thorium reactors, made possible by the fact that the fuel (and the protactinium) are in liquid form.  That it easy to pump around and do chemistry-stuff with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Shippingport reactor was a solid fueled (thorium oxide) reactor with water as a coolant and moderator. So the protactinium would have been stuck in the fuel elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other fuel cycles (e.g. U-235) generate reactor poisons as well. These actually render solid fuel elements useless before all of the fuel has been consumed. It is possible to melt the fuel down and recover the useful fissionable material. This process has not enjoyed the level of adoption that it might otherwise due to politics, bureaucracy etc. Often, spent fuel is simply disposed of without reprocessing.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-02-11T05:33:07.530" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="598" PostTypeId="2" ParentId="595" CreationDate="2015-02-11T08:46:55.323" Score="4" Body="&lt;p&gt;However close the motor &amp;amp; drivetrain can compensate for. The more torque it can output the more you can get away with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example an elevator's counterweight is tuned to the mass of the elevator filled to half capacity and the motor will compensate for the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also during the operation of the bridge the mass and geometry can change; thermal expansion may move the center of mass; rain, dew, frost and grime will add mass to the span as will a new coat of paint.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-02-11T18:22:27.930" LastActivityDate="2015-02-11T18:22:27.930" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="599" PostTypeId="2" ParentId="594" CreationDate="2015-02-11T11:37:10.370" Score="6" Body="&lt;p&gt;I'm not sure of the specifics of your problem, it seems to be relatively straightforward to me. I have previoulsy considered addressing the subject of measurement errors with schoolchildren in the following manner - it may be of help to you and your audience or change in subject may confuse and distract them from your topic, that's up to you to decide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1 - give all the students and length of string (considerably longer than a metre) and ask them to put two knots in the string exactly a metre apart, based on their own estimation. The smart ones will use a length they already know as a guide - most common is their own height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2 - measure the lengths achieved against a ruler and see who got closest. A leaderboard of some kind can make this more fun. This can lead into a discussion of estimation in general. Plotting a scatter plot of the measurements might also be interesting and lead into a discussion about distributions (with enough students I think I'd expect a roughly normal distribution?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3 - now change the ruler for a tape measure and get everyone to remeasure someone elses string - they will inevitably get slightly different results. Write these down and look at the differences. Perhaps now also draw out a scatter plot of the differences. The change in results should bring up all sorts of questions to prompt your discussions on metrology. Examples could be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are the metre ruler and the tape measure the same length? Which of these is the most accurate measure of a metre? How can we test them?&lt;/li&gt;&#xA;&lt;li&gt;Why do two measures of the same thing come out different? Was there an element of judgement by eye on the part of the measurer? Did the string stretch? Did the tape measure stretch? Were the conditions in the room the same? Have the knots tightened or slipped?&lt;/li&gt;&#xA;&lt;li&gt;Was the level of precision used in recording the measurements the same? What is the right level of precision? Did we agree on what part of the knots we were measuring to?&lt;/li&gt;&#xA;&lt;li&gt;Is there a general trend in the difference results? Are they dominated by bias or random errors?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've done steps 1 and 2 with students, along with similar experiments on time and mass. (Generally they are roughly ok at estimating a metre, largely underestimate how long a second is, and are wildly inaccurate at estimating a kilogram). Step 3 is an extension that I've planned but not yet gotten round to trying out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you could also carry out similar experiments/demonstrations with your exact problem of liquids in containers as well.&lt;/p&gt;&#xA;" OwnerUserId="157" LastActivityDate="2015-02-11T11:37:10.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="600" PostTypeId="1" AcceptedAnswerId="606" CreationDate="2015-02-11T12:55:57.317" Score="4" ViewCount="6051" Body="&lt;p&gt;I've attempted to answer the following question and I need to know if I am on the right track:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;LOGIC GATES&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Question 3&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A manufacturing plant uses two tanks to store certain liquid chemicals&#xA;  that are required in a manufacturing process. Each tank has a sensor&#xA;  that detects when the chemical level drops to 25% of full. The sensors&#xA;  produce a HIGH level of 5V when the tanks are more than one-quarter&#xA;  full. When the volume of chemical in a tank drops to one-quarter full,&#xA;  or less, the sensor puts out a LOW level of 0V. It is required that a&#xA;  single green light-emitting diode (LED) on an indicator panel shows&#xA;  when &lt;strong&gt;both&lt;/strong&gt; tanks are &lt;strong&gt;less&lt;/strong&gt; than one quarter full.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3.1 Show the truth table&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3.2 Derive the Boolean expressions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3.3 Show how 3 NAND gates can be used to implement this&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(Consider the wording very carefully in this question)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's question 3.1 and 3.2. The last sentence of the main question is a bit tricky. It says &quot;Show when both tanks are less than 1/4 full.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my solution:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both inputs (namely Tank A and Tank B) need to be LOW in order to indicate a LOW output&lt;/li&gt;&#xA;&lt;li&gt;The only logic circuit to satisfy the above is an OR circuit.&lt;/li&gt;&#xA;&lt;li&gt;0 + 0 = 0; 0 + 1 = 1; 1 + 0 = 1; 1 + 1 = 1&lt;/li&gt;&#xA;&lt;li&gt;From the above I believe OR operations is correct, since A and B are both LOW and thus causing the output to be low.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I was considering AND Circuit&lt;/strong&gt; but then 0 x 1 = 0 and 1 x 0 = 0. These operations conflict with the question. &lt;strong&gt;The output must only show LOW when BOTH inputs are LOW&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore my solution for 3.1 and 3.2 would be using an OR gate. Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="499" LastEditorUserId="368" LastEditDate="2015-02-13T04:48:46.993" LastActivityDate="2015-02-13T04:48:46.993" Title="Logic gates for an indicator light to show the state of two sensors" Tags="|electrical-engineering|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="601" PostTypeId="2" ParentId="600" CreationDate="2015-02-11T13:20:32.617" Score="1" Body="&lt;p&gt;The question is actually incomplete, since it doesn't specify whether a logical &quot;low&quot; or a logical &quot;high&quot; lights the green LED. Usually in such cases, you must assume that you need a logical &quot;high&quot; to light the LED. Therefore, your logic needs to generate a logical &quot;high&quot; only when both inputs are logical &quot;low&quot;.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-02-11T13:20:32.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="602" PostTypeId="2" ParentId="600" CreationDate="2015-02-11T13:34:46.850" Score="1" Body="&lt;p&gt;First of all, be careful to only answer what you are asked. This will be relevant in your school work, but also during your professional career.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, your question &quot;Therefore my solution for 3.1 [...] would be using an OR gate.&quot; is irrelevant in the case of the 3.1 question where you are only asked the truth table. As its name implies, a truth table is a table, which you already have a part of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, usually in engineering, the answer to one question uses the output of the previous question. This scheme of thinking also applies to your professional life, where you divide your [big] problems into smaller ones more manageable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I do not directly answer the question, but being able to see if you're on the right track (and find your mistake if you're not) is also an ability of an engineer.&lt;/p&gt;&#xA;" OwnerUserId="154" LastActivityDate="2015-02-11T13:34:46.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="603" PostTypeId="2" ParentId="498" CreationDate="2015-02-11T13:48:25.000" Score="9" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Now, I wonder if an air-launched spacecraft in the next 20 years realistically could really fly us to the moon - that is, could they reach escape velocities?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Air launch to LEO: Done now&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Air launch to lunar orbit - yes, but at 20%-25% of LEO payload&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Air launch to Moon and back to LEO: Yes, but with about 5% of LEO payload&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is easy to overlook some practical realities when enthusing over paper-based systems.&lt;br /&gt;&#xA;Ratio of air-launched vehicle mass to winged return-to-base Mothership mass must not be overlooked. The Mothership size sets an upper limit to the space-vehicle mass. Increases above heavy lift aircraft payload masses may be possible with eg balloons, but this calls for some extremely specialised systems. Looking at the figures below it looks like manned lunar return to earth's surface is an unrealistically high expectation for air launched systems. Small unmanned craft to lunar orbit are practical.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The answer is &amp;quot;yes, obviously&amp;quot; as you can build a small-er lunar launcher than is usually used and you can conceivably build a means to air-launch it. eg Balloon launch can allow very substantial mass and has been proposed in various studies.&lt;/p&gt;&#xA;&lt;p&gt;The existence proof of the general concept comes in the form of the several &amp;quot;Orbital Sciences Corporation&amp;quot; air launched orbital vehicles. These are only used for LEO (low earth orbit) insertion but escape velocity would be achievable given a suitably small payload.&lt;/p&gt;&#xA;&lt;p&gt;The material below gives examples of what could be realistically achieved based on existing small air-launched LEO satellite launchers and the as it was then 2013 proposal from Orbital Sciences, Burt Rutan and Paul Allen.&lt;/p&gt;&#xA;&lt;p&gt;This demonstrates that a not insignificant air launch could deliver about 800 to 1000 pounds to lunar orbit - more with utter leading edge fuels and systems or even larger 'motherships'. This is uncomfortably smaller than what you realistically want to deliver one person to lunar orbit and back. While scaling is possible it's not looking attractive for multi-person lunar return flights.&lt;/p&gt;&#xA;&lt;p&gt;The advantages of aerial launch is not altitude gain as such but the significant gain in reduced air resistance, and the small gain in velocity. While the air launch velocity is a minor fraction of orbital velocity, a ground based launcher must add the initial velocity while supporting the maximum mass against gravity. This is minor compared to air resistance losses, but useful. Air resistance halves &lt;em&gt;about&lt;/em&gt; every 15,000 feet, and drag is inversely related to air density. And drag is proportional to velocity squared - so if you can start slower and higher it can help significantly. You will ultimately need very substantial &amp;quot;horizontal&amp;quot; velocity to orbit, but initially, getting up out of the thick lower atmosphere with minimal losses is extremely important.  The &amp;quot;mothership&amp;quot; has wings and air breathing engines and fuel is cheap compared to the cost of carrying it to high altitude and high velocities, so an air launched system provides gains in launch vehicle costs and capabilities in situations where it is reasonably possible to build a large enough &amp;quot;mothership&amp;quot;. For small LEO payloads it's eminently viable (and used), for very small one way lunar payloads its doable, but for lunar return, the mothership logistics start to become formidable.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Here is &lt;a href=&quot;https://www.youtube.com/watch?v=m7_yyvGxwrE&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;a video of the air launch of an XL Systems &amp;quot;Pegasus&amp;quot;&lt;/strong&gt;&lt;/a&gt;. This shows the action from just before launch until stage 1 burnout.&lt;/p&gt;&#xA;&lt;p&gt;The &amp;quot;next stage&amp;quot; of this capability as of May 2013 is shown here.&lt;br /&gt;&#xA;&lt;a href=&quot;http://www.nasaspaceflight.com/2013/05/stratolaunch-orbital-air-launch/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Stratolaunch and Orbital – The Height of Air Launch&lt;/strong&gt;&lt;/a&gt; . How this has been modified by more recent events I know not but this showed what was being planned in 2013 so is pertinent to your question.&lt;/p&gt;&#xA;&lt;p&gt;This launcher proposed a 13,500 pound payload to LEO.&lt;br /&gt;&#xA;That's not vast - but definitely provides useful payload&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7uPan.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The assignment of relative delta V's and fuel requirements to missions is too complex to allow simplistic answers that cover more than specific examples, but as a really really rough indication, the &amp;quot;delta-V&amp;quot; from LEO to lunar orbit is very approximately 40% of that required to reach LEO from the earth's surface. The table below provides velocity changes needed for various orbital and location transitions. This gives 3.9 km/s as the delta V needed from LEO to lunar orbit.&lt;/p&gt;&#xA;&lt;p&gt;The basic formula for calculating velocity change for a rocket is the (not surprisngly) &amp;quot;rocket equation: -&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;V = Isp x g x ln(M2/M1)&lt;/p&gt;&#xA;&lt;p&gt;Isp = specific impulse of fuel&lt;br /&gt;&#xA;M2 = start mass&lt;br /&gt;&#xA;M1 = end mass&#xA;g = gravitational constant (~~= 10 m/s/s)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Call M2/M1 = mass ratio = MR.&lt;/p&gt;&#xA;&lt;p&gt;Using a modest by modern standards Isp of 300, to produce a delta-V of say 4000 m/S requires a MR of about 3.7 or end mass ~= 1/3.7 = 27% of total.&lt;br /&gt;&#xA;So ABOUT 25% of the above 13,500 pounds could be delivered to lunar orbit&lt;br /&gt;&#xA;=~ 3375 pounds = 1.5 tons&lt;br /&gt;&#xA;~= 1.5 tonnes :-)&lt;/p&gt;&#xA;&lt;p&gt;This in turn could return about 840 pounds to LEO and a rather lesser amount back to earth. The table below is from &lt;a href=&quot;http://www.lr.tudelft.nl/index.php?id=29335&amp;amp;L=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Delft university page&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GkSKI.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Related:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.nz/search?q=pegasus+launcher&amp;amp;biw=1745&amp;amp;bih=977&amp;amp;tbm=isch&amp;amp;tbo=u&amp;amp;source=univ&amp;amp;sa=X&amp;amp;ei=2krbVK-xBYXgmAXxioHgBA&amp;amp;sqi=2&amp;amp;ved=0CCYQsAQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pegasus launcher pictures with links&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pegasus_(rocket)&quot; rel=&quot;nofollow noreferrer&quot;&gt;OSC Pegasus&lt;/a&gt; - 44 launches since 1990.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.spaceflight101.com/pegasus-xl-info.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pegasus XL&lt;/a&gt; - 443 kg to LEO so ABOUT 100 kg to lunar orbit.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.marketwatch.com/story/nasa-selects-orbitals-pegasus-rocket-to-launch-icon-ionospheric-science-satellite-2014-11-24&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Pegasus mission 2014&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.facebook.com/OrbitalSciencesCorp&quot; rel=&quot;nofollow noreferrer&quot;&gt;OSC Facebook page&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Inner system delta V chart&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://en.wikipedia.org/wiki/Delta-v_budget&quot; rel=&quot;nofollow noreferrer&quot;&gt;**Wikipedia - Delta-v budget&lt;/a&gt;&lt;br /&gt;&#xA;and also used in &lt;a href=&quot;https://space.stackexchange.com/questions/2046/delta-v-chart-mathematics&quot;&gt;this stack exchange post&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nkeWa.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-05-11T09:01:33.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="604" PostTypeId="2" ParentId="550" CreationDate="2015-02-11T14:42:51.497" Score="3" Body="&lt;p&gt;I have a bachelor's in computer engineering. I'm still early in my career (currently mostly software, but I'm trying to get more involved in the hardware aspect of things), but here's my experience:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I was wondering what kind of calculus do you real engineers use?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The single most-used topic for me both in school and elsewhere was the Fourier transform. It came up time and again in my electrical engineering classes, and I now work in telecommunications, where it's come up in various forms relatively often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, it's the concepts and background, and understanding the physical reality through the equations that has helped me most rather than the actual numbers and calculations (which I've seen very rarely outside school). Knowing how to blindly follow rules and do calculations can help to do well in school (depending on the professor), but in my experience it's more important to have a conceptual understanding and general idea of the behavior of circuits than to be able to calculate an exact numerical answer. At work we'd get the answer the quick way -- plug the numbers into a simulator. But if you have a conceptual understanding, you'll know what to expect, and notice when something's wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my experience I'd say that the most important thing is to understand well how the equations describe the physical system and be able to translate back and forth. That is, let the equations enhance your understanding of the physical system.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Maybe you do not use anything from it, but it does enhance your&#xA;  mathematical reasoning, which does have a positive externality on your&#xA;  engineering skills?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes! The ability to describe a physical system in mathematical terms, and then understand and predict its behavior is a skill I gained in school, and I believe is very important for any engineer.&lt;/p&gt;&#xA;" OwnerUserId="501" LastActivityDate="2015-02-11T14:42:51.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="605" PostTypeId="2" ParentId="600" CreationDate="2015-02-11T15:04:35.270" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Therefore my solution for 3.1 and 3.2 would be using an OR gate. Would I be correct on this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the LED turn &lt;strong&gt;ON&lt;/strong&gt; when the input for the LED is &lt;strong&gt;LOW&lt;/strong&gt;, then yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that it is the other way around (a HIGH state would turn the LED on). Then you would just invert the output (i.e. use a NOR gate)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an advice, &lt;a href=&quot;http://en.wikipedia.org/wiki/De_Morgan%27s_laws#Engineering&quot; rel=&quot;nofollow&quot;&gt;DeMorgan's law&lt;/a&gt; might become handy when you are solving this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overline{A+B} = \overline{A} \cdot \overline{B}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\overline{A \cdot B} = \overline{A} + \overline{B} $&lt;/p&gt;&#xA;" OwnerUserId="455" LastActivityDate="2015-02-11T15:04:35.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="606" PostTypeId="2" ParentId="600" CreationDate="2015-02-11T15:07:54.360" Score="7" Body="&lt;p&gt;So let's break this down, piece by piece.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The output must only show LOW when BOTH inputs are LOW. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We also know that the sensors read High (5V) when the tank volume is above 25%.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;| Tank 1 | Tank 2 | Monitor LED |  &#xA;|--------|--------|-------------|  &#xA;| High   | High   | Lit         |  &#xA;| High   | Low    | Lit         |  &#xA;| Low    | High   | Lit         |  &#xA;| Low    | Low    | Dark        |  &#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Please note&lt;/strong&gt;: For the sake of this explanation, we're assuming 0 means low and 1 means high.  So the logic diagram above is &quot;upside down&quot; from a traditional representation that starts low and iterates to high.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we look at a typical AND gate, we find that we're close but our logic is backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3kZKH.gif&quot; alt=&quot;AND gate logic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's introduce the NOT gate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/A7ir9.gif&quot; alt=&quot;NOT gate logic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we put the NOT gate after our AND gate, we end up with a NAND gate and that's &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; the logic diagram we want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3WySQ.gif&quot; alt=&quot;NAND gate image&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/bjHl8.gif&quot; alt=&quot;NAND gate logic&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dang! that brings us back to square one, right?  Nope, not quite.  What happens if we use two NOT gates and move them to the other side of our AND gate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXosN.png&quot; alt=&quot;Inverted NAND&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That seems to do the trick!  Our High signals become Low, and our Low signals become High.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll let you work out the algebraic expressions for that diagram; it should be trivial to translate now.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The fun part is trying to express this using just NAND gates.  The trick is to use the NAND gate as an inverter &lt;em&gt;and&lt;/em&gt; realize that our logic is backwards from the voltage that's present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we layout the three NAND gates like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BATVG.png&quot; alt=&quot;3 NAND gates&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if Tank 1's sensor is fed to both gates of &lt;code&gt;U1&lt;/code&gt;, and Tank 2's sensor is fed to both gates of &lt;code&gt;U2&lt;/code&gt;, we're using the first tier of NAND gates as inverters.  Looking at the logic diagram, we see that our green LED will be lit in the cases we want, and not lit in the case when both tanks are below 25%.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;| Tank 1 | Tank 2 | T1 V | T2 V | !T1 V | !T2 V | 2nd NAND | LED |&#xA;|--------|--------|------|------|-------|-------|----------|-----|&#xA;| High   | High   | 5    | 5    | 0     | 0     | 5        | Lit |&#xA;| High   | Low    | 5    | 0    | 0     | 5     | 5        | Lit |&#xA;| Low    | High   | 0    | 5    | 5     | 0     | 5        | Lit |&#xA;| Low    | Low    | 0    | 0    | 5     | 5     | 0        |     |&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You also asked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Therefore my solution for 3.1 and 3.2 would be using an OR gate. Would I be correct on this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at the truth table, with 0 for Low and 1 for High:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VinDV.gif&quot; alt=&quot;OR truth table&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that would work too.  But using an OR gate doesn't necessarily set you up to easily understand how to use the NAND gate configuration.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If nothing else, this exercise should help you understand the duality in being able to express positive and negative boolean logic.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-02-11T15:16:11.817" LastActivityDate="2015-02-11T15:16:11.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="607" PostTypeId="2" ParentId="550" CreationDate="2015-02-11T15:10:03.260" Score="1" Body="&lt;p&gt;If we have to answer this question very briefly, I would say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Engineers do use codes, and the applying code do not need calculus, but only calculation and software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Most engineers use codes written by others in their lifetime career.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Top ones write and modify codes and software, they use math. They make the complex problems simplified for others, put them in table, software, and arithmetic formulas. &lt;/p&gt;&#xA;" OwnerUserId="502" LastActivityDate="2015-02-11T15:10:03.260" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="608" PostTypeId="2" ParentId="532" CreationDate="2015-02-11T15:20:21.093" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In all references I saw so far it was claimed that this technology has no limits on the height of the buildings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This statement is more or less true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/539/88&quot;&gt;hazzey's answer&lt;/a&gt; has already done a good job of summarizing the &lt;em&gt;actual&lt;/em&gt; limitations of building height - i.e., the factors that, in any real application, control the decision of how many storeys to build a building. However, there is still the question of how high a structure &lt;strong&gt;could be&lt;/strong&gt;, assuming we were able to ignore all of these other factors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we make a simplifying (and very naive) assumption that the only limitation of the height of a structure is the compressive strength of the concrete itself, and also that the only load being carried by the concrete is the load resulting from the weight of the vertical monolithic concrete column above (there are no live loads, or load transfers; the building is essentially a massive block of reinforced concrete), the calculation is fairly straightforward. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Unit weight of concrete: $$\gamma_c=150\frac{\text{lbf}}{\text{ft}^3}$$&lt;/li&gt;&#xA;&lt;li&gt;Compressive strength of concrete (high performance concrete): $$f'_c=20,000\frac{\text{lbf}}{{\text{in}}^2}$$&lt;/li&gt;&#xA;&lt;li&gt;Stress carried by concrete at the bottom: $$f=H_{c}\gamma_c$$&lt;/li&gt;&#xA;&lt;li&gt;Set $f=f'_c$ and solve for maximum height: $$H_{max}=\frac{f`_c}{\gamma_c}=\frac{20,000\text{psi}}{150\text{pcf}}=19,200\text{ft}$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is so high (3.64 mi, or 5.85 km) that the acceleration due to gravity would be noticeably different at the top of the structure; the unit weight of concrete at the top would be roughly be 99.82% of what it is at the bottom - that is, about 149.73 pcf. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the incredible stress applied to the concrete would result in appreciable strains. One equation for the modulus of elasticity of high strength concrete (from ACI) is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_c=40,000\sqrt{f'_c}+1\times 10^6 \text{psi}=6,657\text{ksi}=45.9\text{GPa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Hooke's Law, the maximum strain at the bottom of the structure would be around 0.3%: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\varepsilon_{max}=\frac{f'_c}{E_c}=0.3\%$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the strain across the entire structure height, we simply integrate: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{0}^{H_c}\frac{f(z)}{E_c}\text{d}z=28.8\text{ft}$$ where $f(z)=\gamma_cz\cdot g(z)$ (gravity, $g$ is a function of height $z$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the reduced height of the structure after taking into account concrete strain would be around 19170 ft (3.63 mi, or 5.84 km). &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.constructionweekonline.com/article-9184-top-10-worlds-tallest-concrete-buildings/2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; from Contruction Week Online, at 92 storeys (423 m, or 1388 ft) Trump International Hotel and Tower is currently the world's tallest concrete building (by their definition), and it is the 9th tallest building in the world. This is around 7% of the height possible (as defined by the simplified analysis above). Although the simplified analysis ignores all sorts of practical considerations and includes no safety factors, it is at least somewhat instructive as to what might be possible using high performance reinforced concrete. &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-02-11T19:51:30.490" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="609" PostTypeId="1" AcceptedAnswerId="614" CreationDate="2015-02-11T15:33:08.663" Score="3" ViewCount="7115" Body="&lt;p&gt;In &lt;a href=&quot;https://engineering.stackexchange.com/questions/591/how-do-i-compute-combined-tensile-strength&quot;&gt;this question about combined tensile strength&lt;/a&gt;, all of the tensile bars are of the same material. But what if the case that they are made of different materials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we have $200$ bars of $A=10 (\rm mm^2)$. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$100$ bars of steel with $E=200 \, (\rm {GPa})$, at tensile limit,  $\sigma=200\, (\rm {MPa})$, $\eta=20 (\%)$. &lt;/li&gt;&#xA;&lt;li&gt;$100$ bars of aluminum with $E=60 \, (\rm {GPa})$, at tensile limit,  $\sigma=80\, (\rm {MPa})$, $\eta=10(\%)$. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is total strength (the maximum force capacity) of the bars? Assume that one end is fixed and the other is a rigid body moving in longitudinal direction, so that all bars have the same change in length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: We have to assume that the material is of plasticity, and will not hardening after yield. (The tensile limit is the same as the yield strength.) This may be beyond the text book of elasticity theory.&lt;/p&gt;&#xA;" OwnerUserId="502" LastEditorUserId="368" LastEditDate="2015-02-13T04:37:49.637" LastActivityDate="2017-04-10T18:13:45.137" Title="How to Compute Total Strength of Composite Tensile Bars" Tags="|mechanical-engineering|materials|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="612" PostTypeId="2" ParentId="449" CreationDate="2015-02-11T16:06:59.343" Score="1" Body="&lt;p&gt;(1) Is it partly automatic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is. And it could be totally automatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Should you have to define the points and connectivities by hand in some cases? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, except in a classroom homework. By the way, it is called node and element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) What are the most commonly used criteria to ensure the mesh will fulfill the expectation of the customer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be a book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) What are the trends: should we expect it to be fully automatic in the coming years? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is already automatic, but still need improvement.&lt;/p&gt;&#xA;" OwnerUserId="502" LastActivityDate="2015-02-11T16:06:59.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="613" PostTypeId="2" ParentId="609" CreationDate="2015-02-11T16:38:12.470" Score="2" Body="&lt;p&gt;Short answer: the applied stress will be distributed across the area of all of the bars in accordance with the material stiffness, and the tensile strength of the system (assuming no breakage) is simply the combined tensile strength of all of the materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force capacity of each material is as follows: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Steel: $$100\text{bars}\frac{10\text{mm}^2}{\text{bar}}200\text{MPa}=200\text{kN}$$ &lt;/li&gt;&#xA;&lt;li&gt;Alumninum: $$100\text{bars}\frac{10\text{mm}^2}{\text{bar}}80\text{MPa}=80\text{kN}$$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the total force capacity $P_t$ of the entire system is 280kN. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You didn't ask about Young's modulus but it was implied: The composite Young's modulus (and the corresponding stress-strain curve) of the combined system is found by simple area-averaging: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E_{comp}=\frac{E_sA_s+E_aA_a}{A_t}=130\text{GPa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_{comp}=\frac{P_t}{A_t}=140\text{MPa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\varepsilon_{comp}=\frac{\sigma_{comp}}{E_{comp}}=0.11\%$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of load carried by each material is determined by the combination relative stiffness and area, and will be consistent no matter the load (that is, &lt;em&gt;up to a certain point&lt;/em&gt;; see the additional comment below): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Steel: $$\frac{E_sA_s}{E_{comp}A_t}=76.9\%$$&lt;/li&gt;&#xA;&lt;li&gt;Aluminum: $$\frac{E_aA_a}{E_{comp}A_t}=23.1\%$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One principle to recognize here is that &lt;em&gt;stiffness attracts load&lt;/em&gt;. This is sometimes surprising to students, but it makes sense if you think about it. If you have some load attached to a steel cable, and then add a rubber band attached to the same load, the load in the steel cable will be slightly reduced. However, most of the load will remain in the steel cable (because of its higher stiffness). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, once the strength capacity of one of the materials is reached, 100% of any additional load will be taken up by the other material (regardless of its stiffness). In this case, the steel will reach its full capacity first (since it is more stiff, and the areas are equal), so the portion of load carried by the steel will go down (and the portion carried by aluminum will go up) beyond the point that the steel has reached full load carrying capacity (note that this ignores any additional strain required to mobilize the excess capacity of the aluminum, which would break the steel): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Steel (just before composite breakage): $$\frac{\sigma_sA_s}{\sigma_{comp}A_t}=69.2\%$$ &lt;/li&gt;&#xA;&lt;li&gt;Aluminum (just before composite breakage): $$\frac{\sigma_aA_a}{\sigma_{comp}A_t}=30.8\%$$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;EDIT: Things are more complicated than this, however, because as load is increased, the area decreases. And as Trevor Archibald points out in &lt;a href=&quot;https://engineering.stackexchange.com/a/614/88&quot;&gt;his answer&lt;/a&gt;, things get even more complicated if you allow for one of the materials to break. In the rubber band example, if the steel is about to break, but the rubber band has excess capacity, you &lt;em&gt;might&lt;/em&gt; be able to add a few more paper clips! &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-02-11T20:43:55.243" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="614" PostTypeId="2" ParentId="609" CreationDate="2015-02-11T17:19:41.593" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/592/41&quot;&gt;My answer&lt;/a&gt; of that question covers this a little bit, and the link fully explains it, but I can go into more detail so you don't have to wade through that giant PDF. I'll link it again here at the bottom, because I'm going to reference equations in it for this answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might imagine, the total strength of a composite such as the example above is a combination of the two materials. The proportions in which each material contribute to the overall strength are related to the volume fractions of the constituent materials. Because in this case, the lengths of the bars is always equal and the bars are of constant area along their length, we can simplify the volumes to areas. This means that we can compute the elastic modulus using a version of equation 3.1 in the book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_c=E_{St}A_{St}+E_{Al}A_{Al}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, St and Al are steel and alumin(i)um, respectively, the c subscript refers to the composite, and A is the area fraction, not the total area. So in the example problem, $A_{St}=0.5$ and $A_{Al}=0.5$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the stress in the bars, we simply use the definition of stress, given that we can measure the strain and we know the strain is the same in all bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_c=\epsilon E_c=\epsilon E_{St}A_{St}+\epsilon E_{Al}A_{St}=\sigma_{St} A_{St}+\sigma_{Al} A_{Al}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, though, it's important to note that we are seeing different stresses in the bars of different material, which is a natural result of them having the same strain but different elastic moduli. Unless the materials break at the same strain, one of them will fail first, at which point, the same load will be applied to the whole system, but only one type of bar will still be intact to carry that load. The total cross sectional area of the system is reduced as well though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to determine which material breaks first. Because we're applying a known strain rate here, we can find the ultimate strain of each material using the stiffness and ultimate strength. Once we know the strain at which the first material breaks, we plug that strain into the second equation above to see the total stress in the composite at this elongation. However, because the same force is being applied over a smaller area, the strength of the composite after the breakage of the first material (in this case, the steel) is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_{TS,c}=\sigma_{Al} A_{Al}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can compare this value of tensile strength to the value of tensile strength just before the steel breaks. If the value before steel fracture is greater, then the steel fracture results in the fracture of the entire system. If the value after steel fracture is greater, the system will continue to carry load until it reaches the tensile strength of just the aluminum bars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mit.edu/course/3/3.225/book.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mechanical Properties of Materials&lt;/a&gt;, David Roylance, 2008.&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-02-11T17:19:41.593" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="615" PostTypeId="2" ParentId="550" CreationDate="2015-02-11T19:20:37.647" Score="2" Body="&lt;p&gt;&lt;em&gt;This is written from the point of view of someone getting a PhD in mechanical engineering. My math background is somewhat comparable to (but definitely inferior to) that of PhD students in an applied math program.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As other have indicated, the answer to this question depends greatly on the particular engineer's work. In many cases, advanced math is genuinely useless. &lt;a href=&quot;https://engineering.stackexchange.com/a/562/49&quot;&gt;A civil engineer mentioned code based work as an example&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a PhD student working in computational fluid dynamics, I need a reasonably solid understanding of everything through PDEs. Math is a tool I use to solve problems, just like an experimentalist might consider a thermometer a tool. I develop mathematical models (usually solved by computers) for use by myself and other engineers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Topics covered in my undergraduate math education that I find useful in my work:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;integral, differential, and vector calculus (Basically all of it, though I admit I've only used Lagrange multipliers once or twice since undergrad)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;probability and statistics (the class I had was fairly dumbed down, however)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;differential equations (both ordinary and partial)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I also took an undergraduate complex analysis course which I found to be fascinating, though I must admit I've used nearly none of it since then. Some of the graduate math courses I've taken and found useful include asymptotic analysis, measure-theoretic probability (not so much for the measure theory, directly, but for thinking more carefully), and numerical PDEs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My undergrad differential equations background was fairly deficient, however. The basic ODE class must be hard to teach, because (roughly) 75% of the students in there don't need to know much about ODEs and the other 25% needs to know the subject well. (I could write a lot more on this subject, in particular, which areas I think were deficient.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to go on a bit of a tangent to address a related topic. There are a large number of engineers who believe advanced math is more useless to them than it actually is, and they often are quite vocal about it. Some engineers seem to go out of their way to avoid using any sort of math at all&lt;sup&gt;[1]&lt;/sup&gt;, even if it would be helpful. One company that has tried to recruit people from my research group &lt;em&gt;bragged&lt;/em&gt; that they don't do any math, as if that would entice us. To be honest, they became an inside joke. A lot of their work is code based, and while the codes tend to be conservative, they are not always correct or helpful in every case. When someone has to make an &quot;engineering judgment&quot;, I hope the judgment is based on an evidence-based mathematical model and not speculation. (I'm not sure why this opinion about the usefulness of advanced math exists, but I think it comes partly from the difficulty of the math and also ignorance.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers who don't use advanced math should at least be aware of the potential pitfalls of blindly using engineering software based on advanced math. Many engineers trust the software as if its result is infallible. I am funded by a government agency who produces a simulation software (and I help develop the software) and I recall one of their engineers being severely annoyed at users who claim to have discovered new physics: temperatures higher than the adiabatic flame temperature (the highest temperature possible in combustion due to the first law). What actually happened was that the simulation software did not use a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Flux_limiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;TVD&lt;/a&gt;&quot; scheme, and the developers assumed (perhaps implicitly) that people using the software would recognize when things go awry and add additional resolution. My impression is that they didn't want to make the software foolproof because it would slow things down dramatically, but apparently this problem cropped up so many times they added the foolproof algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not to say that advanced math is always necessary. While some engineers might consider it fun to overdo something with mathematical sophistication, if it's not necessary solve a problem, it's probably a waste of time.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;[1] Incidentally, the same is true for programming. For a class taught my by MS advisor, he specifically designed an assignment to be &quot;impossible&quot; to solve in Excel because it required solution of large linear systems of equations many times. By far the easiest way to do this would be writing a few dozen lines of code. He required people to turn in their code to receive credit. He still received spreadsheets! Apparently you can do this in Excel, but you needed to type in the matrix manually! Surely not easy or fun when you need a 500x500 matrix.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="49" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-02-12T18:07:15.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="616" PostTypeId="1" AcceptedAnswerId="2243" CreationDate="2015-02-11T23:21:32.310" Score="5" ViewCount="10725" Body="&lt;p&gt;I have a uniform load applied to the underside of a plastic sheet which is to be resisted by the combination of a bolt, washer, and nut as depicted below (the other end of the bolt is secured and will not budge). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B5unK.png&quot; alt=&quot;Diagram of nut, bolt, washer, and ASA plastic.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thickness of the Luran S 777k ASA (Acrylonitrile Styrene Acrylate) plastic is currently unknown, but to get the ball rolling let's just say it is 1/8&quot; thick (EDIT: confirmed). The plastic will bear directly on the square washer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For modeling purposes, I have decided to assume the plastic is totally fixed/rigid and that the load is being applied to the bolt. In this model, it is the washer that is bearing on the fixed plastic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect that the limiting factor of this system is going to be the pull-through capacity of the plastic as (I assume) the washer will tear it apart way before the washer or the bolt fail. How can I determine the pull-through capacity of this system? Is my assumption that the pull-through capacity of the plastic is the limiting factor valid? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My idea is to simply obtain the shear resistance of the plastic material online somewhere, and calculate the pull-through capacity like so:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Shear strength of the material is $\tau_{\text{ASA}}=??$&lt;/li&gt;&#xA;&lt;li&gt;Load-resisting cross-sectional area: $$A=4s_{\text{washer}}\cdot t_{\text{plastic}}$$&lt;/li&gt;&#xA;&lt;li&gt;Therefore the pull-through capacity of washer bearing on plastic is: $$P_t=\tau_{\text{ASA}}\cdot 4s_{\text{washer}}\cdot t_{\text{plastic}}$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, I'm concerned there may be other considerations at work here that I need to take into account. Is it really this simple? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks to hazzy for the simple but reassuring answer, and the points he made. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the limiting consideration for this particular problem seems to be fatigue/cyclic bending of the plastic around the washer, since in actuality the load is being applied at either end of the plastic section (instead of uniformly as I proposed in the model above). &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2015-03-30T17:07:27.623" LastActivityDate="2016-07-22T05:36:58.683" Title="Finding the pull-through capacity of a nutted bolt and washer in ASA plastic" Tags="|mechanical-engineering|materials|applied-mechanics|plastic|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="617" PostTypeId="1" AcceptedAnswerId="626" CreationDate="2015-02-11T23:58:32.340" Score="13" ViewCount="433" Body="&lt;p&gt;My understanding is that cavitation occurs in the flow of a liquid when the static pressure drops below the vapor pressure, even intermittently. So even if the &lt;em&gt;time-averaged&lt;/em&gt; static pressure (what you might measure) is above the vapor pressure, the pressure fluctuations from turbulence or other unsteadiness could be large enough to cause cavitation locally. So comparing the time-averaged static pressure against the vapor pressure isn't enough; you need to add some extra cushion to account for the pressure fluctuations. (This is my interpretation, not having read too deeply into this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in various books, websites, and journal articles I have seen two different types of dimensionless numbers for estimating whether the flow through a valve or nozzle cavitates. They are generally called the cavitation index or cavitation number. They take one of two forms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma = \frac{p_\text{in} - p_\text{vapor}}{p_\text{in} - p_\text{out}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma = \frac{p_\text{in} - p_\text{vapor}}{\tfrac{1}{2} \rho V^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p_\text{in}$ is the inlet pressure, $p_\text{out}$ is the outlet pressure, $p_\text{vapor}$ is the vapor pressure, $\rho$ is the liquid density, and $V$ is some characteristic velocity of the flow (say, in the nozzle case, the velocity at the outlet). Some forms of this number are inversions of the numbers above, but these aren't that different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What is the difference between these parameters?&lt;/em&gt; Based on energy conservation you can relate the pressure drop to the flow rate, but typically there is an empirical coefficient added in to account for non-idealities. Is there something else I am missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Is one form preferred over the other?&lt;/em&gt; Best I can tell whether to use one or the other depends on what sort of data you have (so, for flow over a turbine blade, the velocity form is preferred), but I've seen both even for nozzles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where can I get accurate data to predict cavitation based on these numbers?&lt;/em&gt; I've tried using some data on atomizer nozzles from various journal articles but generally they use different forms of the cavitation number. Some of the data suggests the flow through the nozzle will cavitate at the pressures I want, but other data for similar nozzles suggests it won't. I'm not sure what the source of the inconsistency is. My understanding could be faulty, the cavitation number model could be too simplistic, the data could be inaccurate, etc.&lt;/p&gt;&#xA;" OwnerUserId="49" LastEditorUserId="282" LastEditDate="2015-02-15T19:37:35.957" LastActivityDate="2015-02-15T19:37:35.957" Title="Estimating whether the flow through a valve or nozzle cavitates" Tags="|mechanical-engineering|fluid-dynamics|multiphase-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="619" PostTypeId="1" CreationDate="2015-02-12T02:33:14.837" Score="17" ViewCount="428" Body="&lt;h3&gt;Background&lt;/h3&gt;&#xA;&lt;p&gt;When I took my first class in controls (2006-ish as taught by an aerospace professor for a mix of ME and EE students) it was basically all done in Laplace-transform, transfer functions etc.&lt;/p&gt;&#xA;&lt;p&gt;More recently (2012) I took a graduate level controls class at a different school and it was almost all state-space. Really it was a bunch of abstract linear algebra proofs that happened to relate to observability and controllability. I chalked up the difference to the fact that it was for grad students who would be working on a more theoretical class of problems (almost no mention was made to relate this to any real system).&lt;/p&gt;&#xA;&lt;p&gt;Now, from talking to undergrads at the same school, I'm given to understand that state-space is the way that control theory is currently taught. Laplace methods are briefly covered but quickly dismissed as out-dated.&lt;/p&gt;&#xA;&lt;p&gt;I work in combustion and have no real idea what's going on.&lt;/p&gt;&#xA;&lt;h3&gt;Questions&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this an accurate indication of the way that controls is being taught these days?&lt;/li&gt;&#xA;&lt;li&gt;Either way, does that match with the state/foreseeable future of controls?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I'm also interested to know the merits of one method over the other.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-02-14T11:56:56.353" Title="Modern method for teaching controls" Tags="|education|control-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="620" PostTypeId="1" CreationDate="2015-02-12T03:17:10.380" Score="10" ViewCount="1870" Body="&lt;p&gt;What if any liability would an engineer have in the following situation? This is specifically in reference to engineering within the USA.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An engineer produces a design and plans that meet the standards and codes in effect at that time.  The design is sealed and submitted to the owner for bidding and construction. Because of funding constraints, the project is put on hold. Enough years pass that there have been code changes that would affect the design.  The owner then bids and completes the project as originally designed. This assumes that no engineer was later retained to look at the plans and update them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the engineer's liability for any issues that arise from using the original codes? This assumes that the later codes fix errors or make other needed improvements.&lt;/li&gt;&#xA;&lt;li&gt;Would it matter if the engineer was aware that the project was being constructed to the original plans after the delay? The engineer is no longer under contract.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-02-24T02:01:48.543" LastActivityDate="2015-02-24T02:54:18.503" Title="What is an engineer's liability for code changes that occur after submitting a design?" Tags="|liability|ethics|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="621" PostTypeId="2" ParentId="620" CreationDate="2015-02-12T04:09:32.993" Score="2" Body="&lt;p&gt;Where I live at least, the governing codes active at the time that a building is permitted apply to the project, regardless of when construction starts. As you probably know, even if a new code is published, it doesn't have the force of law until referenced by the appropriate authority having jurisdiction (AHJ.) In practice, this means there are often 1-3 years where a new code is available, but not yet binding. Of course, if an engineer is aware that the code has become more strict or closed a loophole, but chooses to follow the provisions of an older code still in effect, they may be acting unethically and open themself up to liability issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The written analysis that the engineer prepares will reference the edition of the code that the analysis is based on. As long as they aren't intentionally taking advantage of a loophole, I think clearly stating the code version used should be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how well this translates to industries where the code requirements are specified by customers rather than the government. I imagine there is a similar (if shorter) lag between the publishing of a new code and its common adoption. Another thing to consider is that many codes are published with fairly minor changes that wouldn't make a large difference in the acceptability of a design. When a code changes drastically, industry publications, vendors, and regulators tend to start active education campaigns to make people aware of the changes. For example, here in California, our Green Building Code just underwent a major revision. As a result lighting and control system vendors hosted a number of seminars to explain the impacts to specifiers and electricians (and advertise their products, of course.) This means that for a major code change, the industry becomes aware of the new requirements and would be more likely to catch any outdated designs.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-02-12T04:09:32.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="622" PostTypeId="2" ParentId="620" CreationDate="2015-02-12T04:09:56.827" Score="4" Body="&lt;p&gt;First, I think it's important to note that actual legal liability is going to vary case-by-case, it's never so simple as &quot;They were the engineer, it's their fault.&quot; In situations where something happens and lawsuits are drawn up, they're probably going to name every single company who is at all related, and all key people in those companies responsible for the design. However, I think I can give some general insight based on my experiences and what I've been told in legal liability training.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I lay the majority of the blame in this situation on the owner, who uses plans that are years old without having anyone qualified look at them and update them. I've seen (small) issues with this in my position. Drawings are completed and sent to another department for approval, and for whatever reason, they sit on an application engineer's desk for months, sometimes close to or upwards of a year. The drawings are then sent out as-is, but internal changes that have been made to drawing standards since the drawings were originally updated aren't complied with. These are minor issues, but it's easy to see this occurring on a larger time scale and with more serious consequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawings should be dated. That's fairly common practice, and this is one good reason why. We want to know when the drawings were made, to determine what things have changed since that design, whether that be standards, codes, laws, or simply interfacing parts. If the drawings are made to the codes and standards in place at that time, I don't see how the engineer can be held responsible for that design, unless it can be shown that the engineer had knowledge of the code needing a revision for safety issues. In this case, making the print fit the code isn't good enough, because we know the code won't prevent some issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It really comes down to the fact that, in the design process, every consideration towards safety needs to be taken, and where risk cannot be eliminated, it needs to be made clear what the risk is and how it can be mitigated by the end-user. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the second point, this sounds like a very gray area, and I really couldn't discuss any legal aspects of this. But I will say that I think, if an engineer knows that the design being built is out-of-date and could cause issues in some manner, they have an ethical responsibility to contact someone still affiliated with the project and apprise them of the situation. It may not be their legal responsibility, but I think it's clearly the right thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should almost go without saying that any engineer still affiliated with the project has the same obligation, only stronger, to say that the designs need to be updated and they cannot be used in their current form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give some context for this, the main product that my company manufactures can be extremely dangerous in its operating environment. Guarding is absolutely required unless the product is entirely shielded by other parts of the machine. The guarding itself may vary slightly from company to company, but its largely the same because it's governed by an ISO standard. However, this standard has changed over the years, and even the organization that creates the primary standard has changed. This means that we have old designs which feature old guarding that are still active in our system. We have one employee who is extremely knowledgeable about the industry and in particular the safety codes and even sits on some safety committees. He makes sure that new designs comply with the guarding standards and that old products are updated as necessary. Sometimes this means going so far as to tell the customer that we won't sell them this product if they won't let us update the guarding (newer guarding is in some cases more expensive). He isn't always the most popular person, but this is his job, and that's how we manage our code compliance in a world where codes and standards are not consistent. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-02-12T04:09:56.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="623" PostTypeId="2" ParentId="550" CreationDate="2015-02-12T05:04:32.250" Score="1" Body="&lt;p&gt;The answers all generally make valid points but I think they miss the real reason engineers take a pretty standard 2 year math curriculum: efficiency in learning the rest of their coursework.  The people that devised the original curricula were not interested in creating a &quot;liberal arts&quot; foundation where calculus would exercise your mind etc.  They wanted to train engineers, plain and simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to train engineers, you need to teach them subjects like mechanics, fluids, waves, etc.  To learn those different topics efficiently, you need calculus and linear algebra.  Sure you can replace a calculus argument by devising some very clever, elementary argument, but it's much better to give ONE argument via calculus that encompasses a variety of cases. Same thing goes for linear algebra.  For example, the concept of whether the nullspace of a linear system is trivial or not ties together quite nicely with the analogous concept in linear ODEs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could argue all day about whether learning this way makes a better engineer or not, but one thing is clear to anyone that's taught: this is a very efficient way of training engineers. And how well one understands the math being taught will have a direct effect on how well one understands the rest of the engineering curriculum.&lt;/p&gt;&#xA;" OwnerUserId="506" LastEditorUserId="506" LastEditDate="2015-02-12T05:15:08.117" LastActivityDate="2015-02-12T05:15:08.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="624" PostTypeId="2" ParentId="620" CreationDate="2015-02-12T06:15:38.150" Score="10" Body="&lt;p&gt;This is a question about the law, not engineering, and the law is &lt;strong&gt;entirely&lt;/strong&gt; dependent on the jurisdiction. Assuming that we are dealing with a common-law country like Australia, the UK or the US then liability can arise from three sources: the contract, statutory liability and the tort of negligence.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Contract&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Basically, you owe an obligation under a contract to do what it says in the contract; noting that this may contain implicit terms and terms read into the contract by statute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liability under the contract arises to one party when the other party breaches a term or condition of the contract. If you have complied with the contract then the delay cannot have caused you to stop complying.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Statute&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There may be laws in the relevant jurisdiction (which could be the jurisdiction of the owner, the engineer, the site or all of these) which impose a liability on the engineer. If this is an obligation to be read into a contract then see above. If this is a duty owed to the State, then it would generally be for the state to prosecute either criminally or civilly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most jurisdictions, compliance with a code or standard is at best a partial defence, so if you have not complied with the law initially then you would be liable. Most laws are not retrospective so it is difficult to see how the delay could change your liability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in circumstances where you know that the project is being constructed to obsolete codes and these may cause a breach of the new statute (particularly one going to WHS or environmental damage) there is probably an obligation to do what you can in the circumstances. Drawing this to the owners attention would probably suffice for this.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Negligence&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In order to establish negligence as a Cause of Action under the law of torts, a plaintiff must prove that the defendant:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;had a duty to the plaintiff, &lt;/li&gt;&#xA;&lt;li&gt;breached that duty by failing to conform to the required standard of conduct (generally the standard of a reasonable person), &lt;/li&gt;&#xA;&lt;li&gt;the negligent conduct was the cause of the harm to the plaintiff, and&lt;/li&gt;&#xA;&lt;li&gt;the plaintiff was, in fact, harmed or damaged.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A case bought where the engineer was in ignorance of the project proceeding would fail on the first limb. If the engineer was aware the owner would still find this a difficult hurdle - what duty does the engineer owe to the ex-client? If they could overcome this they would probably fall foul of 2 and 3 - the engineer could quite reasonably assume that the owner had engaged others to update the plans. Note that I have not even considered that there must be &lt;em&gt;actual&lt;/em&gt; harm or damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notwithstanding, if the engineer, on becoming aware, wrote to the owner to say &quot;Hey, I hope you are not using my plans because they are out of date&quot; then I would suggest that this would be all that a &quot;reasonable person&quot; could do.&lt;/p&gt;&#xA;" OwnerUserId="507" LastEditorUserId="90" LastEditDate="2015-02-12T10:15:50.850" LastActivityDate="2015-02-12T10:15:50.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="625" PostTypeId="1" CreationDate="2015-02-12T06:40:50.180" Score="2" ViewCount="1175" Body="&lt;p&gt;I am mainly interested in reducing drag in aircraft and I have an idea that I want to learn more about...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a superhydrophobic coating reduce drag in water-craft such as boats and submarines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is the case, could a superhydrophobic coating decrease the drag coefficient of aircraft making them more aerodynamic? Or would some other material that's phobic to air be needed for this, and does it exist?&lt;/p&gt;&#xA;" OwnerUserId="509" LastEditorUserId="95" LastEditDate="2015-02-12T23:14:23.270" LastActivityDate="2017-01-22T04:55:43.690" Title="Can superhydrophobic surfaces be used to reduce drag?" Tags="|fluid-dynamics|aerospace-engineering|aircraft-design|experimental-physics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="626" PostTypeId="2" ParentId="617" CreationDate="2015-02-12T08:57:30.557" Score="14" Body="&lt;p&gt;&lt;strong&gt;The difference between the two equations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cavitation number is the ratio of the  static pressure difference to the dynamic pressure difference.  So, if you want to use the first equation, you would need to take the pressure using a Pitot tube to measure the total pressure, whereas if you want to use the second equation you will need to measure the freestream velocity, but I would recommend measuring it upstream rather than downstream because of possible effects of acceleration and boundary layer growth.  Also, your $V$ should be $V_{in}$ such that it corresponds to the same location where $p_{in}$ is measured, because this equation is derived from Bernoulli's equation which says the energy is conserved along a streamline.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is one form preferred over the other?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all my experience working in cavitation research for many years, we have almost always used the latter equation you mentioned (although I have mainly been working in hydrofoils and propulsion systems). The reason is that we could get more accurate non-intrusive velocity measures using &lt;a href=&quot;https://en.wikipedia.org/wiki/Laser_Doppler_velocimetry&quot;&gt;Laser Doppler velocimetry (LDV)&lt;/a&gt; than by using an intrusive method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where can I get accurate data to predict cavitation based on these numbers?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is difficult to use experimental data to predict the cavitation number because of the differences in things like turbulence intensity and air nuclei content, which are difficult to match in reality with controlled laboratory methods.  Traditionally, in my circles, this is done by running some CFD analysis codes on your design.  There are two different approaches here: (1) compute the average mean flow using a RANS or LES technique, and (2) using a bubble dynamics code which will model the air nuclei, but requires a flowfield (either from experimental measures or from the from the CFD model). If you use a typical RANS CFD model to compute the flow-field, it should give you the the pressure coefficient which has a very similar definition to the cavitation number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_P = \frac{P-P_\infty}{\frac{1}{2}V^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are doing some CFD calculation on your nozzle, you should find the location of minimum pressure, and that is the place where cavitation should occur. You can infer the cavitation number from this pressure coefficient as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma = -C_P^{min}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_P^{min}$ is the minimum pressure in your nozzle.  I explain this in more detail in this &lt;a href=&quot;http://www.sname.org/communities1/resources/viewtechnicalpaper/?DocumentKey=7663cdf9-7481-4556-84ae-cbfabf204cb1&quot;&gt;paper&lt;/a&gt;. However, this will only give you an idea of the time averaged cavitation inception number.   Most people don't go to such detail in trying to get such an accurate prediction of cavitation inception, unless it is absolutely critical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get a more accurate number, you need to consider that cavitation inception requires three things to happen at the same time: (1) a local area of pressure which is below the vapor pressure of water, (2) an air nuclei which enters into that low pressure region, and (3) the air nuclei must be in the low pressure for a significant enough time that it basically rapidly grows, becomes unstable and hence collapses. The way people have been able to more accurately estimate this is by using a using a Lagrangian method that simulates sending air nuclei through an Eulerian CFD dataset.  Some of the real experts in this field are the people at Dynaflow-inc.com.  I might suggest taking a look at this paper: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chahine, G.L. &quot;Nuclei Effects on Cavitation Inception and Noise&quot;, 25th&#xA;  Symposium on Naval Hydrodynamics, St. John's, NL, Canada, Aug. 8-13,&#xA;  2004. &lt;a href=&quot;http://www.dynaflow-inc.com/publications/pdf_documents/2004/ONR-SNH25-Chahine-1.pdf&quot;&gt;PDF here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, if you don't want to go to all that trouble, I would recommend that you compute an estimate of the pressure fluctuations $p'$ based on the ambient turbulence intensity of your flow, and then subtract this value off of your mean pressure to get a better estimate of cavitation number.  You should be able to get this value out of the turbulence model if your using a RANS technique.  If you are looking at possible CFD techniques to use, unless you have a lot of money to spend, I might suggest looking into using OpenFOAM.&lt;/p&gt;&#xA;" OwnerUserId="511" LastEditorUserId="160" LastEditDate="2015-02-12T11:17:27.860" LastActivityDate="2015-02-12T11:17:27.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="627" PostTypeId="2" ParentId="625" CreationDate="2015-02-12T09:52:26.390" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Superhydrophobe&quot; rel=&quot;nofollow&quot;&gt;Superhydrophobic&lt;/a&gt; surfaces generally consist of a series of micro/nano scale pillars that increase the contact angle of water on the surface. Larger contact angle means water forms more spherical drops and can roll off more easily. Changing the chemical properties of the surface layer can also help change the contact angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These surfaces are optimized to effect the contact angle of water so will not have a significant effect on the drag of vehicles (except maybe boats) or radar detection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However surfaces with other patterns/shapes of structures can have these effects. See &lt;a href=&quot;http://www.bloomberg.com/bw/articles/2013-02-25/airplanes-with-sharks-skin-it-may-cut-fuel-burn&quot; rel=&quot;nofollow&quot;&gt;riblets&lt;/a&gt; or sharkskin suits for reducing drag in aircraft/swimmers and &lt;a href=&quot;http://en.wikipedia.org/wiki/Anti-reflective_coating#Moth_eye&quot; rel=&quot;nofollow&quot;&gt;moth-eye anti-reflection coatings&lt;/a&gt;, for light, not radar but that is probably just the size of features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common uses for super-hydrophobic coatings are in water repellant windscreens for cars and on waterproof jackets. &lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-02-12T09:52:26.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="629" PostTypeId="2" ParentId="298" CreationDate="2015-02-12T11:04:18.780" Score="0" Body="&lt;p&gt;The friction you are talking about when referencing the cloth would more correctly be refered to as rolling resistance. In a true sense, you need to increase (the coeficient of) friction to increase traction or increase the force acting normal to the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can increase the coeficient of friction by changing the wheel material, as you did with the cloth wrap but you found this increased rolling resistance once the vehicle was moving, thus reducing efficiency. An alternative would be to increase mass, thus increasing the force available to generate traction using the formula for traction GlenH7 provided $F_t = \mu_tmg$ where $m$ is mass and $g$ is gravitational acceleration. Increasing mass will negatively affect your acceleration so you could adjust your design to shift as much of the existing mass to the drive wheels as possible (see the dragster analogy, cars are very long, all the weight is at the back), just enough that the front wheel(s) only just maintains contact with the ground on acceleration (see dragster again).&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2015-02-12T11:04:18.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="630" PostTypeId="2" ParentId="620" CreationDate="2015-02-12T11:09:19.153" Score="1" Body="&lt;p&gt;The engineer's liability could either be to the owner or to the public (i.e. public safety).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main possibility for liability to the owner comes from the contract. The engineer should make sure that the contract states what standards apply - either by including dates of specific standards or referring to a set of standards &quot;as at the contract date&quot; (or similar wording). If the standards are updated during the design period, if it is a safety critical aspect, then the owner would need the design to meet the latest standard; but as the engineer isn't contracted to that latest standard, there would need to be an extension/alteration to the contract = additional fees for the engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of safety liability to the public, it all comes down to what the engineer knows. e.g. If the new standard was released during design, the engineer should be aware and should inform the client, and negotiate new fees (or refuse to continue the design if the owner won't agree). If, as in your situation, it is many years after the design was completed, then it would come down to whether the engineer knows that the building is now being built, and whether he could know that the original design has not been updated for the latest standards. It would be a hard task for someone to prove that the engineer knew exactly what was going on and deliberately took no action (i.e. informing the owner or a public safety body).&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-02-12T11:09:19.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="631" PostTypeId="2" ParentId="619" CreationDate="2015-02-12T12:32:01.573" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;Curriculum differs between universities&lt;/li&gt;&#xA;&lt;li&gt;Engineering curriculum differs between countries &lt;/li&gt;&#xA;&lt;li&gt;Curricula are updated regularly to meet new accreditation requirements&lt;/li&gt;&#xA;&lt;li&gt;Sometimes there is a distinct difference between graduate and&#xA;undergraduate programs from the same school&lt;/li&gt;&#xA;&lt;li&gt;Some professors are progressive and update courses more frequently&#xA;to keep up with advances&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Not too long ago I too followed a Linear Systems course in a graduate program. Based on your description it appears that I too might have had a similar experience. In this case I know for sure the Prof. had not updated the course for at least 10 years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Recommendation:&lt;/strong&gt; I suggest learning the modern methods to solve control systems problems. I also suggest that you become a member of respective professional society; in your case it might be ASE. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, some solve engineering problems using brute force methods. Qualified, skilled professional engineers use skill acquired through university studies as well as knowledge gained from being part of professional organizations to efficiently solve engineering problems. Part of the important value an engineer with undergraduate as well as graduate schooling brings to an organization is the ability to quickly recognize and solve problems efficiently.&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-02-14T11:56:56.353" LastActivityDate="2015-02-14T11:56:56.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="632" PostTypeId="2" ParentId="449" CreationDate="2015-02-12T15:32:17.613" Score="9" Body="&lt;p&gt;There are a number of techniques for meshing complex domains for Finite Element Analysis.  They generally fall into two categories: Structured vs. Unstructured. For structured meshes, basically the entire mesh can be mapped directly to a 3D array of XYZ coordinates, whereas unstructured grids cannot. There is a good description of the classifications with pictures here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Grid_classification&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Grid_classification&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within structured meshing, there are two specific types:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Structured meshes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Cartesian mesh - This is basically using hexahedral cubes to represent the elements.  A well known package that uses Cartesian meshing would be Cart3D. This is not really complicated, but the difficulty is defining where the cubes intersect the surface.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Body-fitted mesh - in body-fitted curvilinear meshes, they can be divided into: algebraic grids, or elliptic grids.  In either case the user has to define the points on the boundaries of the domain.  To generate points in the interior of the domain, algebraic grids usually use some variation of a technique called Hermite interpolation to generate the interior points.  Elliptic grids can produce curvilinear grids where basically all the gridlines are orthogonal, and are generally what is used when it comes to body-fitted meshes.  The interior points here are basically computed by solving an elliptic partial differential equation.  The defacto textbook for these types of body-fitted techniques is available online here: &lt;a href=&quot;http://www.erc.msstate.edu/publications/gridbook/&quot; rel=&quot;noreferrer&quot;&gt;http://www.erc.msstate.edu/publications/gridbook/&lt;/a&gt;.  The author of this book, is basically considered &quot;the father of grid generation&quot;, because he came up with the Elliptic mesh for mesh generation.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unstructured meshes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Since unstructured grids cannot be mapped to a 3D array, so they must also specify a connectivity mapping, which can relate which elements are related to other elements. The basic algorithm that is used is called &quot;Delauney triangulation&quot;, which is discussed in detail here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Delaunay_triangulation&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Delaunay_triangulation&lt;/a&gt;.  One of the popular books that covers this topic is called &quot;The Handbook of Grid Generation&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The basic algorithm here is, given an initial set of points on the boundary:&#xA;(1) Compute an initial triangulation, (2) Perform a quality check based on Ruppert's refinement algorithm       (&lt;a href=&quot;http://en.wikipedia.org/wiki/Ruppert%27s_algorithm&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Ruppert%27s_algorithm&lt;/a&gt;), (3) Insert or delete points based on Ruppert's algorithm such that the Tetrahedra that are generated have a minimum angle (e.g. 24 degrees).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To answer your question about criteria, what makes a good mesh has to do with a number of factors, but a couple of the most important factors are: (1) grid resolution (is there enough grid points to get the resolution required) and (2) the geometry of the elements (skew, minimum angle, aspect ratio, etc.).  This is discussed here: &lt;a href=&quot;http://en.wikipedia.org/wiki/Types_of_mesh&quot; rel=&quot;noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Types_of_mesh&lt;/a&gt; These both will effect the quality of a Finite Element solution.  There is another aspect of unstructured grid meshing called &quot;Advancing Front&quot; which is used to produce points near the boundary in the case of Fluid Dynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After saying all that, most techniques require some work up front and then are somewhat automatic as well.  In any type of mesh algorithm, the user is going to have to spend some time to define the geometry and some initial point distribution on the surface.  From my experience, body-fitted meshes take the most time.  Both the Delaunay triangulation and the Cartesian meshes are basically automatic in generating the points of the interior domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been doing much work in this field in the past few years, but the trend in the past was moving away from body-fitted grids to either unstructured Delaunay triangulations or Cartesian grids. Also, there have been some codes which can convert a cartesian mesh to a unstructured Delaunay mesh and vice-versa (e.g. Gambit).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think these meshing codes will ever be fully automatic, because some level of input is required to specify the geometry, which usually involves cleaning up a CAD model. More recently techniques have been developed to automate much of these tasks as well. Generating the interior points of the domain is pretty much completely automatic these days.  These modern grid generation systems are getting quite mature these days in terms of producing high quality grids.  One of the research areas in the past decade has been in the area of speeding up the grid generation by using parallel processing, and in the past few years parallel grid generation using multiple Graphical Processing Units (GPU).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an entire list of mesh generation software here: &lt;a href=&quot;http://www.robertschneiders.de/meshgeneration/software.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.robertschneiders.de/meshgeneration/software.html&lt;/a&gt;&#xA;These should fall under one of the three categories above.&lt;/p&gt;&#xA;" OwnerUserId="511" LastEditorUserId="511" LastEditDate="2015-02-14T07:31:42.613" LastActivityDate="2015-02-14T07:31:42.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="634" PostTypeId="2" ParentId="619" CreationDate="2015-02-12T16:54:36.923" Score="6" Body="&lt;p&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt;  I've studied control theory from a mathematics perspective, not an engineer's perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical Control Theory is based on linear systems and is also limited to them as well.  Linearization is helpful in many cases, but not entirely applicable in others.  Analysis tools (laplace transform, pole placement, root locus, routh hurwitz, etc...) are easiest to conduct on linear systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optimal Control Theory is the more modern approach because it can handle nonlinear systems directly without using the frequency domain.  The idea is that control can be achieved by minimizing a suitably chosen cost function over time.  Usually, this cost function is a function of the state space variables and is usually subject to constraints defined by the dynamics, boundary conditions, and/or feasible paths.  The word &quot;optimal&quot; doesn't mean that its necessarily better or superior to classical control theory.  Here, &quot;optimal&quot; means that control is obtained by finding a relative extrema (usually a minimum) of a function.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how the objective function is chosen, minimization can require either solving a linear or non-linear system of equations.  Fortunately, many numerical/analytical tools from optimization and/or calculus of variations can handle such problems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, minimizing a non-linear function can be a rather costly endeavor from a computational point of view.  If it is possible to analyze by some other means (i.e. classical control theory), then it may be more worthwhile to do so. &lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2015-02-12T20:23:38.390" LastActivityDate="2015-02-12T20:23:38.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="637" PostTypeId="1" AcceptedAnswerId="638" CreationDate="2015-02-12T20:42:33.570" Score="10" ViewCount="4433" Body="&lt;p&gt;A friend of mine is a structural engineer and he often complains to me about how little sleep he gets at night worrying over whether or not he accounted for all the little details involved in his design.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all the double and triple checking one might do, buildings are very complicated structures.  Sometimes its impossible to account for every possible situation.  Of course, one does to the best of their ability to try to account for everything.  But if something goes wrong, say a building either collapses or some other less severe problem causes injury or death, I would assume that the engineers can be held accountable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they covered by some sort of malpractice insurance in a similar fashion to doctors?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="368" LastEditDate="2015-04-09T23:27:07.653" LastActivityDate="2015-04-09T23:27:07.653" Title="Do structural engineers carry insurance for catastrophic errors?" Tags="|structural-engineering|liability|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="638" PostTypeId="2" ParentId="637" CreationDate="2015-02-12T21:38:46.383" Score="10" Body="&lt;p&gt;You might be interested in &lt;a href=&quot;http://www.nspe.org/resources/professional-liability&quot;&gt;the National Society of Professional Engineers' page of professional liability resources&lt;/a&gt;, which has a section for insurance. There are quite a few entities offering professional liability insurance for engineers; &lt;a href=&quot;http://www.nspe.org/sites/default/files/resources/pdfs/2014%20Liability%20Survey%20Results.pdf&quot;&gt;NSPE surveyed 18 of them last year&lt;/a&gt; with questions like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question #6: What percentage of your total book of&#xA;  [Architecture/Engineering] premium comes from firms with a revenue of:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Less than \$500,000&lt;/li&gt;&#xA;  &lt;li&gt;\$500,000 to \$5,000,000&lt;/li&gt;&#xA;  &lt;li&gt;More than \$5,000,000&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Professional liability insurance is a very important part of running a business, and engineering firms are no exception. Independent engineering contractors may also carry insurance. However, even in jurisdictions that require professional liability insurance in certain fields, there may be lesser-known alternatives or loopholes that allow the requirements to be circumvented. For example, &lt;a href=&quot;http://www.floridamalpractice.com/med20.htm&quot;&gt;the state of Florida does not require all practicing physicians to carry medical malpractice insurance&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As an alternative to having an actual malpractice insurance policy,&#xA;  Florida law also allows doctors to use other types of pre-arranged&#xA;  secured assets to cover claims in these amounts, like trust accounts,&#xA;  bank letters of credit, and similar arrangements. There is nothing&#xA;  inherently wrong with these other types of security, but they are&#xA;  rarely used.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...the law also has a loophole that allows doctors to carry no&#xA;  insurance at all.  If your doctor practices without insurance he&#xA;  should have a sign posted on the wall of his office advising his&#xA;  patients of that fact.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So when you ask if engineers are insured &quot;in a similar fashion to doctors,&quot; you need to understand that even for doctors, the situation varies from jurisdiction to jurisdiction. I'm not aware of any US state that requires individual engineers or engineering firms to carry professional liability insurance; that said, I studied in California and I've never worked for a firm practicing in another state, so such laws may very well exist. &lt;a href=&quot;http://www.nspe.org/resources/issues-and-advocacy/take-action/position-statements/disclosure-professional-liability&quot;&gt;To quote NSPE again&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Many firms carry and maintain professional liability insurance&#xA;  policies and/or use specific methods to transfer or mitigate&#xA;  professional liability risk exposure.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;NSPE encourages individual employed engineers and their companies to&#xA;  work in harmony to minimize professional liability risk exposure&#xA;  consistent with good professional practice and sound business&#xA;  judgment.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;... While insurance coverage is not the only way to protect the firm&#xA;  and employees, it is NSPE's position that companies that do not carry&#xA;  professional liability insurance should disclose that fact in writing&#xA;  to their employed engineers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a more in-depth treatment of the subject, I recommend auditing or enrolling in an online course, such as &lt;a href=&quot;http://ocw.mit.edu/courses/civil-and-environmental-engineering/1-133-masters-of-engineering-concepts-of-engineering-practice-fall-2007/&quot;&gt;Concepts of Engineering Practice from MIT OpenCourseWare&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-02-23T16:54:33.133" LastActivityDate="2015-02-23T16:54:33.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="639" PostTypeId="5" CreationDate="2015-02-12T22:22:16.967" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-02-12T22:22:16.967" LastActivityDate="2015-02-12T22:22:16.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="640" PostTypeId="4" CreationDate="2015-02-12T22:22:16.967" Score="0" Body="Questions about the control, measurement and regulation of air contaminants impacting public health and/or visibility. Do not use this tag for questions specific to greenhouse gases and climate impact." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-02-13T01:04:36.793" LastActivityDate="2015-02-13T01:04:36.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="642" PostTypeId="2" ParentId="637" CreationDate="2015-02-13T00:58:23.080" Score="5" Body="&lt;p&gt;Air's answer is pretty thorough already, and the liability issue is truly complex. In my liability training at work, we've been told of situations where engineers from every firm with any association to the machinery or structure in question are named in suits, not necessarily because they're all at fault, but (1) so that their fault can be determined by a judge and (2) because in a civil case, settling may be cheaper than paying for a lawyer to try and get the company/engineers dismissed from the suit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to do is point out a good example of a catastrophic failure, the level of research that can go into finding out what happened, and the consequences for the firms and engineers involved. One of my professors favorite case studies was the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyatt_Regency_walkway_collapse&quot;&gt;Kansas City Hyatt Regency walkway collapse&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, two walkways, at the fourth and second floor levels and directly above/below one another, collapsed during an event the hotel hosted, killing 114 people and injuring. It was the largest loss of life in a structural collapse in the US from the time that it happened in 1981 until the World Trade Center collapsed on Sept 11th. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a large investigation by the National Bureau of Standards, a link to which can be found &lt;a href=&quot;http://fire.nist.gov/bfrlpubs/build82/art002.html&quot;&gt;here&lt;/a&gt;. That PDF is 370 pages long, including 95 pages of appendices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of the investigation, the engineering firm and the engineers who signed off on the final drawings lost their licenses to practice, though it appears they avoided any criminal charges and that most of the civil damages were paid by the owners of the hotel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, all cases are different, so while they criminal charges were dismissed here, it's possible they wouldn't be in another case. I think the important thing to take away here is that a lot of the work engineers do can be dangerous if not done properly, and being involved on a project that ultimately injures someway means you are extremely likely to have to answer for your design choices at some point, even if they are sound and you have done nothing wrong. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-02-13T00:58:23.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="643" PostTypeId="1" CreationDate="2015-02-13T04:19:11.267" Score="7" ViewCount="3064" Body="&lt;p&gt;When analyzing a pile or drilled shaft as a beam-column, how do you determine the unbraced length of the portion that is in the ground? Typically, the braced points on a beam or column are well defined, e.g. there are connections at discrete locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pile in the ground will have some resistance to movement because of the soil. Is this resistance enough to consider the pile to be continuously braced (Lb = 0)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of considering the situation might be to look at the deflection diagram. Should the braced length be based off the first or second point where the deflection crosses zero? Or similarly where the moment diagram crosses zero?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="282" LastEditDate="2015-02-15T19:29:10.597" LastActivityDate="2017-07-12T16:50:11.617" Title="What is the unbraced length of a pile in soil?" Tags="|civil-engineering|structural-engineering|soil|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="644" PostTypeId="1" AcceptedAnswerId="647" CreationDate="2015-02-13T04:50:09.690" Score="7" ViewCount="400" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Maglev&quot;&gt;Magnetic levitation&lt;/a&gt; systems of transportation such as trains seem to be commonplace in early 2000's science fiction, and while some systems exist currently, they aren't exactly common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that this technology has grown unpopular (or at least has fallen into obscurity) in recent years and I'm curious as to why that may be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it primarily technical factors or perhaps technology issues that prevent wider adoption of mag-lev transportation systems?  Or is there something else that's slowed the rate of installation?&lt;/p&gt;&#xA;" OwnerUserId="340" LastEditorUserId="106" LastEditDate="2015-02-14T16:44:10.253" LastActivityDate="2015-02-14T18:11:50.190" Title="Could magnetic levitation transportation systems become a commonplace technology?" Tags="|mechanical-engineering|civil-engineering|transportation|" AnswerCount="2" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="645" PostTypeId="2" ParentId="644" CreationDate="2015-02-13T14:01:41.463" Score="3" Body="&lt;p&gt;Like many things, the driver is an economic one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maglev is an entirely practical proposition but must compete in cost-benefit with other systems - notably wheeled ones. It generally fails to do so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About 15+ years ago China investigated various means of providing high speed rail services. A &quot;demonstration&quot; Maglev system was constructed between Shanghai's Pudong airport and the edge of Shanghai proper. The 30 km line cost $1.2 billion to build. The train and track systems were a joint venture between Siemens and ThyssenKrupp and track and support facilities were built by Chinese companies. The train runs at around 350-400 kph typically (430 kph stated as daily peak velocity and in non-commercial tests a top speed of 501 kph was reached). The hope on the part of the suppliers was that this would serve as the first of a nationwide network of similar trains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is the sole example of its kind in China and is considered something of a white elephant - it stops on the city edge and users must either catch taxies or public road trnsport or walk to a nearby local underground rail station to complete their journey into the city. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of a national maglev network, the &quot;conventional&quot; wheeled D trains were introduced for long distance national services (200 + kph) and subsequently the even faster G trains - still wheeled and with top speeds in the 300-400 kph range. Even by D train a 4 hour  road trip takes about 2 hours by train. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite possible popular images to the contrary, China is not averse to implementing 'high tech' systems if it seems to make sense to do so. The fast comfortable convenient D and G trains are good examples of this, but it seems the TUV Maglev was 'a train too far'. Definitely marvellous but not sufficiently superior to what can be achieved by more &quot;usual&quot; means at greater cost benefit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.travelchinaguide.com/china-trains/types.htm&quot; rel=&quot;nofollow&quot;&gt;Chinese train types&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Shanghai_Maglev_Train&quot; rel=&quot;nofollow&quot;&gt;Wikipedia - The Shanghai Maglev - 上海磁浮示范运营线&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-02-13T14:01:41.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="646" PostTypeId="2" ParentId="459" CreationDate="2015-02-13T14:25:14.243" Score="2" Body="&lt;p&gt;If you have access to a 3D printer and appropriate software, you could 3D print a coupler.  You can adjust the radius of the hole to control how tight the coupler is to the shaft.  I do this for connecting motors to LEGO axles.  I have been surprised at how much friction I get from plastic against a smooth metal shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.thingiverse.com/thing:673563&quot; rel=&quot;nofollow&quot;&gt;http://www.thingiverse.com/thing:673563&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="532" LastActivityDate="2015-02-13T14:25:14.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="647" PostTypeId="2" ParentId="644" CreationDate="2015-02-13T15:34:13.200" Score="3" Body="&lt;p&gt;For a given method of transport to become the most popular, it needs to be the safest, cheapest, most efficient way to get from point A to point B, relative to comparable forms of transportation. In this case, the comparable transportation method is to use normal trains, which run on coal, electricity, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's compare the two:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Safety&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is sort of hard to compare because the two methods involve different ways of killing people. There are the old &quot;run-someone-over&quot; and &quot;hit-a-service-vehicle&quot; scenarios, and in fact &lt;a href=&quot;https://en.wikipedia.org/wiki/Lathen_train_collision&quot; rel=&quot;nofollow&quot;&gt;the only major (and fatal) accident on a maglev train&lt;/a&gt; occurred about eight and a half years ago. 23 people died when the train hit a service vehicle. However, the crash was attributed to human error, not the maglev technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting these types of accidents aside, maglev trains aren't any more dangerous that normal trains. People can die if they accidentally touch power lines, but normal electric trains pose the same risks. Fires aren't any more likely to be started, and fuel can't be spilled. So safety isn't an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cost&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, there &lt;em&gt;is&lt;/em&gt; a difference. In a Congressional report, &lt;em&gt;Report to Congress: Costs and Benefits of Magnetic Levitation&lt;/em&gt;, it was estimated that maglev tracks would cost 40-100 million dollars per mile, compared to 10 million dollars for high-speed rail (HSR). The cost can vary given the area the train is passing through. The report estimates that a maglev system would cost 1.92 times as much as an HSR system in rural areas, 1.22 times as much in suburban areas, 1.20 times as much in mountainous areas, and 1.13 times as much in urban areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's very interesting, because it shows that the technology might be accepted in some areas but not others. Maglev subway systems and &lt;a href=&quot;https://en.wikipedia.org/wiki/Elevated_railway&quot; rel=&quot;nofollow&quot;&gt;elevated railways&lt;/a&gt; could become more prevalent in cities but not in rural areas. It wouldn't take too much of a change in infrastructure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different routes lead to different costs per mile, with the cost of routes in the American Northeast Corridor being one and a half times as much as average. Fortunately, thus setup is an outlier, and is really due to the high population in the region and the replacement of the existing rail lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different methods range in price, too. The Congressional report also compares the programs of different countries. They vary quite a lot, though, and there doesn't seem to be a definite pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last thing: The report was written in 2005. Since then, the value of the dollar has changed and technology has improved. I think that the cost of maglev trains has gone down quite a lot since then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Efficiency&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://large.stanford.edu/courses/2010/ph240/ilonidis2/&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; (admittedly short) compares a German &lt;a href=&quot;https://en.wikipedia.org/wiki/Transrapid&quot; rel=&quot;nofollow&quot;&gt;Transrapid&lt;/a&gt; maglev train and the &lt;a href=&quot;https://en.wikipedia.org/wiki/ICE_3&quot; rel=&quot;nofollow&quot;&gt;ICE 3&lt;/a&gt; high-speed train, also developed in Germany. The energy usage is about the same, though at lower speeds Transrapid has the edge over ICE 3. There isn't data for higher speeds, though, which is odd, because the newest Transrapid models can go much faster than the data explains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.21stcenturysciencetech.com/articles/Summer03/maglev_3.pdf&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; shows that maglev trains are still many times more efficient than airplanes or cars. It's also a bit speculative (using maglev as a launch system for spacecraft?!), but it's fairly comprehensive. But it's a bit dated, like the Congressional report.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-02-13T15:34:13.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="649" PostTypeId="1" CreationDate="2015-02-13T23:53:30.853" Score="5" ViewCount="1496" Body="&lt;p&gt;During a recent search I discovered the existence of flexible solar panels. Besides the obvious differences of low weight and flexibility what are the high level technical differences between flexible and rigid solar panels? What is the typical Lux VS Current curves for such solar panels?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yAX8I.jpg&quot; alt=&quot;Flexible Solar Panel&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The available technical specification for the above solar panel are listed below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Technical Specifications:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Power Output: ~6V @ 1W (~160 mA)&lt;/li&gt;&#xA;&lt;li&gt;Output Type: DC Voltage&lt;/li&gt;&#xA;&lt;li&gt;Dimensions: 100mm / 3.94&quot; x 198.43mm / 7.81&quot; x 0.77mm / 0.03&quot;&lt;/li&gt;&#xA;&lt;li&gt;Operating temp range: +32 to +158°F (0 to +70°C)&lt;/li&gt;&#xA;&lt;li&gt;Weight: 27.93g&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://panasonic.net/energy/amorton/en/solar_battery/&quot; rel=&quot;noreferrer&quot;&gt;Amorphous Si Solar cells&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://solarmade.com/PowerFilm.htm&quot; rel=&quot;noreferrer&quot;&gt;Solarmade&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.powerfilmsolar.com/&quot; rel=&quot;noreferrer&quot;&gt;Powerfilm solar&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.adafruit.com/products/1485?gclid=COzFrb2K4MMCFUk2gQodyHoAVA&quot; rel=&quot;noreferrer&quot;&gt;Flexible 6V 1W Solar Panel&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.solbian.eu/index.php?option=com_k2&amp;amp;view=itemlist&amp;amp;layout=category&amp;amp;task=category&amp;amp;id=16&amp;amp;Itemid=239&amp;amp;lang=en&quot; rel=&quot;noreferrer&quot;&gt;Solbian&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt;: I am not endorsing any of these products.&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="368" LastEditDate="2015-08-10T16:10:26.947" LastActivityDate="2015-08-10T16:10:26.947" Title="What are the technical differences between flexible and rigid solar panels?" Tags="|electrical-engineering|solar-panel|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="650" PostTypeId="1" AcceptedAnswerId="656" CreationDate="2015-02-14T01:27:23.320" Score="1" ViewCount="1827" Body="&lt;p&gt;I've recently been learning about FET amplifier circuits and I'm looking for some information that can help explain why the calculated and simulated Gain Values can differ by up to 15% in some cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this image is an example of the first FET amp.&#xA; &lt;a href=&quot;http://www.electronics-tutorials.ws/amplifier/amp14.gif&quot; rel=&quot;nofollow&quot;&gt;http://www.electronics-tutorials.ws/amplifier/amp14.gif&lt;/a&gt;&#xA;which gives the following values&#xA;calculated Gain = -11.8&#xA;simulated Gain = -13.4&#xA;difference = 12.698%&#xA;any help explaining this difference will be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="534" LastEditorUserId="110" LastEditDate="2015-02-16T00:26:07.293" LastActivityDate="2015-02-16T00:26:07.293" Title="Percent (%) error between calculated gain and measured gain of FET amplifiers" Tags="|electrical-engineering|power-electronics|circuit-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="651" PostTypeId="1" AcceptedAnswerId="654" CreationDate="2015-02-14T03:58:12.383" Score="20" ViewCount="6637" Body="&lt;p&gt;I've always wondered why refrigerators don't have some of their parts located outdoors like an air conditioner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In warm weather, it seems like it would make sense to have the condenser outside like an AC unit to avoid heating the room. In cold weather, it seems like it would be much more efficient to have the condenser outside to cool down faster. Why is the condenser still indoors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, does this exist in some places? Have I personally just never seen it? Maybe for large commercial, refrigerator/freezers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, would the cost simply outweigh the benefit? Is it cheaper to have the AC cool the room down after the fridge has heated it slightly rather than install another condenser outside? In cold weather, is the fridge acting as an efficient space heater, so putting the condenser outside not really get you anywhere? I'd be interested to see if a cost/benefit analysis has been done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, is it just convenience? Is it simply much easier to just move the fridge in, plug it in, and be done? &lt;/p&gt;&#xA;" OwnerUserId="161" LastActivityDate="2022-10-26T12:20:08.170" Title="Why don't refrigerators have parts outdoors?" Tags="|mechanical-engineering|thermodynamics|refrigeration|" AnswerCount="6" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="652" PostTypeId="1" CreationDate="2015-02-14T05:37:02.783" Score="6" ViewCount="611" Body="&lt;p&gt;Why have thorium nuclear reactors not been ore fully developed?  Earth has more thorium than uranium.  Is it technical difficulties or the fact that uranium technology was developed first for military reasons?  Are there other reasons?&lt;/p&gt;&#xA;" OwnerUserId="398" LastEditorUserId="160" LastEditDate="2015-02-14T14:56:29.163" LastActivityDate="2021-03-30T12:37:52.917" Title="Why have thorium fueled nuclear reactors not been more fully developed?" Tags="|engineering-history|nuclear-technology|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="653" PostTypeId="1" CreationDate="2015-02-14T06:01:41.487" Score="10" ViewCount="924" Body="&lt;p&gt;There are two things involved: wave drag and boundary layer separation. The wave drag depends on the Mach number while the latter depends on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Reynolds_number&quot;&gt;Reynolds number&lt;/a&gt; of the flow. It's easy to maintain the incoming Mach number as it is independent of the geometry; however, the Reynolds number depends on the geometry of the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{Re} = \frac{\rho u d}{\mu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If air is used as a medium, then assuming that the flow is to be maintained at a constant Mach number, $\rho$ and $u$ will be fixed by the gas dynamic relations. $\mu$ is pretty much out of our hands, so the only non-fixed parameter is $d$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $d$ is much smaller for a model than for a real aircraft, the flow will have a lower $\text{Re}$ than a real aircraft would. This will give different flow separation characteristics for a model than for a real aircraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In subsonic testing, only thing that matters is $\text{Re}$, which can be fine tuned to match with the actual size by tweaking $u$ for given $d$. But in supersonic flow, we do not have that luxury, as $u$ is decided by the Mach number of incoming flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how are wind tunnel models used for design of aircraft, spacecraft and missiles? Are there correction techniques to predict flow separation better? Can the same techniques be used for dealing with CFD data?  &lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="368" LastEditDate="2015-04-18T21:48:32.383" LastActivityDate="2015-04-18T21:48:32.383" Title="How are wind tunnel models used when designing supersonic aircraft?" Tags="|fluid-dynamics|aerospace-engineering|aircraft-design|airflow|wind-tunnels|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="654" PostTypeId="2" ParentId="651" CreationDate="2015-02-14T06:27:11.133" Score="12" Body="&lt;p&gt;Electrical powered domestic refrigerators developed from &lt;a href=&quot;https://en.wikipedia.org/wiki/Icebox&quot;&gt;ice boxes&lt;/a&gt;, also known as cold closets, which were tin or zinc lined insulated cabinets that had a ice compartment. As the ice melted it kept the inside of the ice box cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For premises not connected to the electric grid kerosene refrigerators were developed and are still available, &lt;a href=&quot;https://www.thenaturalhome.com/servel400k.htm&quot;&gt;kerosene refrigerator&lt;/a&gt;. There are even refrigerators powered by &lt;a href=&quot;http://home.howstuffworks.com/refrigerator5.htm&quot;&gt;gas and propane&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.polarpowerinc.com/products/refrigerator/reliable-vaccine-ref.htm&quot;&gt;Alternatively powered refrigeration units&lt;/a&gt; are still produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With mains/grid powered refrigerators, as you allude to, the compressor and power unit of electrically powered refrigerators are part of the domestic refrigeration unit for convenience. It is possible to have the compressor and power unit outside the premises as occurs with &lt;a href=&quot;https://en.wikipedia.org/wiki/Air_conditioning&quot;&gt;split system air conditioners&lt;/a&gt; but it would be inconvenient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By having grid powered refrigerators as a compact unit they can be placed anywhere, inside or out, connected to a power outlet and they start operating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For refrigeration on an industrial scale, &lt;a href=&quot;http://www.coldroomsolutions.com/walk-ins.html&quot;&gt;cold rooms are used&lt;/a&gt;. Depending on their scale and portability, the compressor and power unit are on the outside.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-02-14T15:10:25.797" LastActivityDate="2015-02-14T15:10:25.797" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="655" PostTypeId="2" ParentId="652" CreationDate="2015-02-14T07:18:32.810" Score="5" Body="&lt;p&gt;As &lt;a href=&quot;https://physics.stackexchange.com/a/20037/4066&quot;&gt;I wrote&lt;/a&gt; over on a similar question on our sister Physics site, the German THTR-300 Thorium High-Temperature Reactor operated for about 16,000 hours and the &lt;a href=&quot;http://www.iaea.org/PRIS/CountryStatistics/ReactorDetails.aspx?current=92&quot; rel=&quot;nofollow noreferrer&quot;&gt;IAEA has data on it&lt;/a&gt;. So, a thorium-fuelled nuclear reactor has been developed and operated. So it's possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High-grade uranium ore hasn't been in short supply, and uranium ore is a tiny part of total costs of a nuclear reactor, so the fact that thorium is more common is irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commercial-scale electricity from nuclear fission may be many decades old, but there are only a few hundred generating reactors, across a large number of different designs. And that means, despite it being a multi-trillion euro/pound/dollar industry (at current prices), when it comes down to it the nuclear industry is still a series of prototypes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind that Thorium on its own is useless in a reactor - it just doesn't produce enough neutrons per fission. So you need uranium in there too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So not only do you need everything that a uranium reactor needs, you need all the processing for the thorium fuel and materials to handle its chemistry too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are enough complications as it is, without starting to play with a whole new fuel cycle. We've got a lot more experience of managing the uranium fuel cycle than the uranium/thorium fuel cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we'd be taking something that's already incredibly complicated, and complicating it further, for no obvious economic advantage. Bad plan.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2015-02-14T07:30:33.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="656" PostTypeId="2" ParentId="650" CreationDate="2015-02-14T07:21:55.817" Score="0" Body="&lt;p&gt;&quot;A simulation&quot; is a series of calculations which are hidden and which you supply input &#xA;parameters for. The simulation &quot;models&quot; the target device by embodying assumptions which may or may not be visible and/or explicitly stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A calculation&quot; (or a family of associated calculations) that can be used to arrive at a solution to the operation of a circuit is essentially a manual or visible simulator - you are &quot;modelling&quot; the circuit with numerical equivalents. Assumptions are usually explicit in the form of constants, transfer functions and formulae.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your statement is that the calculated and simulated values can differ by up to 15%. You did not mention the vitally important 3rd member of the family of explain how results from the other two compared in accuracy. The important 3rd member is &lt;strong&gt;reality&lt;/strong&gt;.&lt;br&gt;&#xA;ie you have calculated, simulated and &lt;strong&gt;actual&lt;/strong&gt; results.  The degree to which simulation and calculation differ from reality is what is important. Both differ from reality because the &quot;model&quot; of the device(s) that they use also differ from reality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At an informed guess I'd expect a simulation to be closer to reality because it embodied a &quot;model&quot; which more closely matched reality. It is easy for the simulation maker to use formulae, lookup tables, piece wise approximations and more.&lt;br&gt;&#xA;Calculations are liable to involve not much more than Use of much more than one or a number of formulae and so cannot match the theoretical and/or empirical detail of a simulation. &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-02-14T07:21:55.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="657" PostTypeId="2" ParentId="652" CreationDate="2015-02-14T07:52:49.360" Score="7" Body="&lt;p&gt;The Oak Ridge National Laboratory had an experimental &lt;a href=&quot;http://www.economist.com/news/science-and-technology/21600656-thorium-element-named-after-norse-god-thunder-may-soon-contribute&quot; rel=&quot;noreferrer&quot;&gt;thorium reactor&lt;/a&gt; in the 1960s, but the program was terminated in the 1970s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be partially be attributed to the needs of both the military and businesses, at the time. Unlike $^{235}{U}$, &lt;a href=&quot;https://en.wikipedia.org/wiki/Thorium&quot; rel=&quot;noreferrer&quot;&gt;thorium&lt;/a&gt; is not naturally fissile, it needs to be bombarded with slow neutrons to ultimately produce &lt;a href=&quot;http://en.wikipedia.org/wiki/Uranium-233&quot; rel=&quot;noreferrer&quot;&gt;$^{233}{U}$&lt;/a&gt;, which is fissile and can be used as a nuclear fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the issues for the military was that the products of the thorium fuel cycle cannot easily be used for nuclear weapons production, which was an aim of the nuclear powers at the time. $^{233}{U}$ can be used in nuclear weapons but it is contaminated by &lt;a href=&quot;http://en.wikipedia.org/wiki/Uranium-232&quot; rel=&quot;noreferrer&quot;&gt;$^{232}{U}$&lt;/a&gt;, which has a half life of 68.9 years, but is an intense gamma radiation emitter, which makes it hazardous to handle. This also makes it a problem in thorium based nuclear power systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason is business had invested heavily in uranium reactors and developed that technology and it wanted a return on its investment. Thorium reactors were perceived as a competing technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Research into thorium reactors is being done by &lt;a href=&quot;http://singularityhub.com/2012/12/11/norway-begins-four-year-test-of-thorium-nuclear-reactor/&quot; rel=&quot;noreferrer&quot;&gt;Norway&lt;/a&gt;, &lt;a href=&quot;http://blogs.telegraph.co.uk/finance/ambroseevans-pritchard/100026863/china-going-for-broke-on-thorium-nuclear-power-and-good-luck-to-them/&quot; rel=&quot;noreferrer&quot;&gt;China&lt;/a&gt;, &lt;a href=&quot;http://www.thorium.tv/en/thorium_reactor/thorium_reactor_1.php&quot; rel=&quot;noreferrer&quot;&gt;USA&lt;/a&gt;, &lt;a href=&quot;http://www.itheo.org/articles/world%E2%80%99s-first-thorium-reactor-designed&quot; rel=&quot;noreferrer&quot;&gt;India&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more information about thorium and thorium reactors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.thoriumenergyalliance.com/downloads/American_Scientist_Hargraves.pdf&quot; rel=&quot;noreferrer&quot;&gt;Liquid Fluoride Thorium Reactors&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.whatisnuclear.com/articles/thorium.html&quot; rel=&quot;noreferrer&quot;&gt;Pros &amp;amp; Cons of Thorium&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.world-nuclear.org/info/Current-and-Future-Generation/Thorium/&quot; rel=&quot;noreferrer&quot;&gt;Thorium - World Nuclear Association&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-02-14T15:26:10.590" LastActivityDate="2015-02-14T15:26:10.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="658" PostTypeId="2" ParentId="649" CreationDate="2015-02-14T07:57:07.250" Score="4" Body="&lt;p&gt;The main high-level technical difference is that flexible panels are pretty much useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're an absolutely tiny part of the market. A novelty. A toy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand why, have a think about what a solar panel does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It converts sunlight to electricity. To do that, in needs to capture sunlight. And to do that, it needs to be facing the sun. So, careful calculations are done to select the correct angle and alignment when designing PV plants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flexibility (beyond the minimum needed to avoid damage under high wind loading) is a useless attribute here. Once you've aligned your PV, the last thing you want it to do is to go flapping off in different directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And although PV is cheap (pretty much the cheapest power there is now, in the tropics and sub-tropics, once you've accounted for a sensible carbon price), it's not so cheap that you can afford to waste 80% of the potential generation because it's not facing the sun at the right times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make things worse, the longevity of PV depends on the encapsulation staying complete. If there are breaks in it, then water vapour and other chemicals get in, and the performance degrades quickly. Too much flexing will lead to breaks in the encapsulation.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2015-02-14T07:57:07.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="659" PostTypeId="1" CreationDate="2015-02-14T08:11:37.917" Score="12" ViewCount="23730" Body="&lt;p&gt;I'm looking for advice on how to attach a thermocouple to a metal surface in a fashion that's relatively temperature resistant (more-so than) glue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got some very fine gauge wire on hand to make the thermocouples and specialized spot welder that will join them to form a bead. My goal is to attach them to a 3-7 mm thick sheet some kind of metal. Could be steel, aluminum, copper etc. I don't have a strong preference on either the metal used or the exact thickness. I have access most commonplace metalworking equipment and can purchase basic tools and materials but can't justify the expense of a purpose-built welder for this kind of thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've tried two things (unsuccessfully):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Soldering: I used a gas torch to heat copper plate enough to get a nice puddle of solder on it and then tried to dip the (already welded) TC. The wire simply would not wet, it would just push the solder out of the way and couldn't make a connection.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;TIG welding: I tried to weld a both TC and bare TC wire to an aluminum plate by forming a small puddle on the plate and then attempting to dip the wire into the puddle while quickly extinguishing the arc. Mostly the wire just melted and &quot;ran away&quot; from the arc. Other times it just didn't fuse to the surface.  I also tried putting the wire in contact with the surface in advance and starting the arc on it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know that I've heard mention of doing this, but haven't been able to find any good information about how to actually do it. I could use anything from a definitive answer to comments on how to improve the techniques that I've attempted already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. I weld semi-regularly but have very little experience soldering (esp. with a torch), so don't assume a lot of skill or knowledge in that department.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2021-12-16T15:54:59.257" Title="Affixing a thermocouple to a surface" Tags="|temperature|joining|welding|" AnswerCount="11" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="661" PostTypeId="2" ParentId="659" CreationDate="2015-02-14T13:51:55.873" Score="1" Body="&lt;p&gt;If you aren't too worried about response speed, you might be able to embed the thermocouple in a thermal grease, then weld a piece of shim steel over the top to provide mechanical support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have only ever seen thermocouples welded directly to a substrate before with spot welding equipment.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2015-02-14T13:51:55.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="662" PostTypeId="1" AcceptedAnswerId="1823" CreationDate="2015-02-14T14:39:18.577" Score="11" ViewCount="2028" Body="&#xA;&#xA;&lt;p&gt;I was recently reading up on the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hyatt_Regency_walkway_collapse&quot; rel=&quot;noreferrer&quot;&gt;1981 Hyatt Regency walkway collapse&lt;/a&gt;.  It was a tragic loss of life and haunts many to this day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linked article does a better job of summarizing things, but here's a summarized version of what led up to the collapse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engineering firm (Gillum &amp;amp; Associates) designed the walkways to be supported on really long, threaded rods.  The company supplying the rods (Havens Steel) didn't like the specified rods because they were going to be expensive and likely to be damaged during construction.  Havens Steel sent an alternate design to Gillum &amp;amp; Associates which was used for the construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternate design didn't meet code, catastrophically failed, and led to the loss of many lives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Missouri Board of Architects, Professional Engineers, and Land Surveyors found Gillum &amp;amp; Associates guilty of negligence and revoked all of the engineering licenses assigned to Gillum &amp;amp; Associates and its employees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to understand better is &lt;em&gt;why&lt;/em&gt; Gillum &amp;amp; Associates was held liable for the disaster.  &lt;strong&gt;Wasn't it really Havens Steel's fault for proposing the alternate design to begin with?&lt;/strong&gt;  If Havens Steel had built according to the original design proposed by Gillum &amp;amp; Associates, then the disaster never would have occurred.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorUserId="18" LastEditDate="2015-04-14T20:26:59.493" LastActivityDate="2021-03-25T14:27:54.187" Title="Why were engineers negligent in the 1981 Hyatt Regency walkway collapse?" Tags="|licensure|ethics|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="663" PostTypeId="1" CreationDate="2015-02-14T15:56:56.020" Score="13" ViewCount="1433" Body="&lt;p&gt;Bolts and anchor rods that are very high strength ($F_u$ &gt; 150ksi) are not supposed to be galvanized.  This is because of concerns about &lt;a href=&quot;http://www.portlandbolt.com/technical/faqs/galvanizing-high-strength-bolts/&quot;&gt;hydrogen embrittlement&lt;/a&gt;. This includes ASTM A 490 bolts and &lt;a href=&quot;http://www.portlandbolt.com/technical/specifications/astm-a354/&quot;&gt;ASTM A 354 Gr. BD&lt;/a&gt; anchor rods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The project in question has very high lateral loads and is outdoors. Typical &lt;a href=&quot;http://www.portlandbolt.com/technical/specifications/astm-f1554/&quot;&gt;ASTM F 1554 Gr. 105&lt;/a&gt; anchor rods end up being very large in diameter. This is why ASTM A 354 Gr. BD rods (2.5 inch diameter) are specified. There is no space for more anchor rods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the project is outdoors, a corrosion protection coating is required. The only coatings that can be used are Zinc/Aluminum coatings such as &lt;a href=&quot;http://www.astm.org/Standards/F1136.htm&quot;&gt;ASTM F 1136 Grade 3&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a history of aluminum not being used in contact with concrete because of &lt;a href=&quot;http://www.theconcreteproducer.com/concrete/how-does-contact-with-aluminum-affect-concrete.aspx&quot;&gt;corrosion concerns&lt;/a&gt;. This seems to be specifically with solid aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to be concerned with the aluminum content of this coating reacting with the concrete that the anchor rods are embedded in?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="122" LastEditDate="2015-02-14T18:06:23.583" LastActivityDate="2016-04-12T15:49:35.813" Title="Can Zinc/Aluminum coated anchor rods be used in concrete?" Tags="|materials|concrete|corrosion|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="664" PostTypeId="1" AcceptedAnswerId="665" CreationDate="2015-02-14T16:25:28.327" Score="6" ViewCount="1600" Body="&lt;p&gt;I am working on an idea that involves a jet engine, but I don't quite know how the fuel system works. In a combustion engine like that in a car, you have nozzles that spray the fuel into the piston chambers (I think, correct me if I am mistaken), and then it's ignited when the piston compresses the chamber (sometimes using a spark plug, sometimes it doesn't). How is fuel mixed with air and ignited inside a jet engine? I know that compressed air comes from the intake system (the fan blades pull in and compress air and direct it into the part where it mixes with fuel), but how does fuel get mixed in?&lt;/p&gt;&#xA;" OwnerUserId="537" LastEditorUserId="106" LastEditDate="2015-02-14T18:06:52.243" LastActivityDate="2015-02-23T01:57:17.263" Title="How is fuel mixed with air in a jet engine?" Tags="|mechanical-engineering|fluid-dynamics|aircraft-design|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="665" PostTypeId="2" ParentId="664" CreationDate="2015-02-14T17:08:09.040" Score="7" Body="&lt;p&gt;The place where air and fuel are mixed is the &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Combustor&quot; rel=&quot;nofollow&quot;&gt;combustor&lt;/a&gt;&lt;/strong&gt;, also known as the &lt;strong&gt;flame holder&lt;/strong&gt;:&#xA;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/1/1c/Combustor_diagram_componentsPNG.png&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/1/1c/Combustor_diagram_componentsPNG.png&quot; alt=&quot;Combustor #1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;diffuser&lt;/em&gt; takes in the compressed air and slows it down (remember that aircraft with jet engines are traveling extremely fast). If the air goes too fast, it won't burn enough; if it goes too slow, it won't provide enough thrust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air goes in through gaps in the &lt;em&gt;liner&lt;/em&gt;. There are holes all along it, called &lt;em&gt;primary holes&lt;/em&gt;, &lt;em&gt;intermediate holes&lt;/em&gt; and dilution holes*. These further slow the air down, and help it enter the main part of the combustion chamber. However, most of the air goes through the &lt;em&gt;swirler&lt;/em&gt;, which mixes the air as it enters. The &lt;em&gt;dome&lt;/em&gt; is a related device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside this chamber is an &lt;em&gt;igniter&lt;/em&gt;, which creates the sparks needed for the fuel to burn. Next to it is the &lt;em&gt;fuel injector&lt;/em&gt;, a pipe which inserts fuel into the chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air follows a series of complex paths inside the combustor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/4/49/Combustor_diagram_airflow.png&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/4/49/Combustor_diagram_airflow.png&quot; alt=&quot;Combustor #2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture should give you a good idea of them. Now you probably understand the names of the holes into the main chamber! They cool the combusted air (entering through the swirler) as it goes out the other end and into the next (and rear) section of the engine.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-02-23T01:57:17.263" LastActivityDate="2015-02-23T01:57:17.263" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="666" PostTypeId="1" AcceptedAnswerId="2941" CreationDate="2015-02-14T18:31:57.687" Score="8" ViewCount="4520" Body="&lt;p&gt;I'm trying to construct a stainless steel high pressure containment vessel (pressure reactor) (2000 psi or so). Yes I realize that this is a don't try this at home type of thing, and I plan to put some lexan in front of it before I do any testing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking at 2&quot; stainless pipe with a wall thickness of .148&quot;(rated 4000 psi), and I'm trying to figure out the best way to seal the ends. I plan on welding one end cap on, but the other end needs to be removable. I'm not certain how to calculate how much pressure a threaded pipe end would be able tolerate, or if perhaps some sort of high pressure triclamp. I've seen some industrial pressure reactor vessels, but they are rates for something like 13,000 psi, way overkill, they are also super expensive. Looking for a DIY lower cost alternative. Any info would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="539" LastEditorUserId="282" LastEditDate="2015-02-15T19:38:15.563" LastActivityDate="2017-05-16T10:55:24.657" Title="How to seal a pressure vessel" Tags="|mechanical-engineering|pressure|piping|" AnswerCount="3" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="667" PostTypeId="1" AcceptedAnswerId="671" CreationDate="2015-02-14T19:23:55.787" Score="9" ViewCount="1177" Body="&lt;p&gt;This is inspired by the discussion and controversy surrounding the &lt;a href=&quot;https://en.wikipedia.org/wiki/Keystone_Pipeline&quot; rel=&quot;nofollow&quot;&gt;Keystone Pipeline&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main part of the Keystone Pipeline system is about 3,400 kilometers long, stretching across a large portion of the United States. The Keystone XL extension would add another long section to it. Altogether, the length of all the segments would be . . . oh, something very big, I imagine (though I acknowledge that none of the oil will travel through all sections).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the practical limit for how long an oil pipeline can be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the interest of narrowing this down, I'd like to focus on two sub-questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is the structural integrity of the pipeline at risk the longer it gets?&lt;/li&gt;&#xA;&lt;li&gt;Is there any possibility of adverse effects on the oil over longer spans?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="106" LastEditorUserId="6790" LastEditDate="2020-05-01T13:45:34.260" LastActivityDate="2020-05-01T15:09:48.460" Title="What is the maximum length of oil pipelines?" Tags="|mechanical-engineering|structural-engineering|pipelines|petroleum-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="671" PostTypeId="2" ParentId="667" CreationDate="2015-02-14T20:05:28.843" Score="13" Body="&lt;p&gt;The total length of the pipeline has little to do with the length that oil can be pumped in a pipe. This is because a pipeline is broken into many smaller segments between pumping stations. Stations are conveniently located either where required (see discussion below) or where another pipeline joins in.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pipelines are rarely one single pipe between point A and point B. They have lots of smaller pipelines connecting into them.  Each of these pipeline will have their own pump stations. These stations also help to divide the pipeline into segments. With segments, you can ensure that different oil is sent to specific locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think about the system from a logistics point of view. Any long pipeline is almost guaranteed to pass close to another location that either produces or consumes oil. It only makes sense to connect to these locations as the pipeline goes past.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Factors that affect the length that oil can be pumped in a single pipe&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a large topic. Everything that affects regular pipe design also effects pipelines.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Length of pipe&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Smoothness of pipe&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Temperature of the oil&lt;/strong&gt; - Oil in pipeline is usually heated because warmer oil has a lower viscosity and is easier to pump&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Elevation changes&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Number and degree of turns&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pressure capacity of pipe&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Diameter of pipe&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Allowable pressure drop between ends&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;A lot more...&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Structural Integrity of Pipe&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the pipe does not influence the strength of the pipe.  All pipe lengths in a pipeline will be welded and tested to the same requirements.  &lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-14T20:05:28.843" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="672" PostTypeId="1" CreationDate="2015-02-14T22:05:25.593" Score="3" ViewCount="198" Body="&lt;p&gt;I ran across an interesting combination of packaging for some medicine the other day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The medicine in question was a prescription migraine medication.  The exterior, cardboard package that it came in was fairly typical.  The interesting part was that the medicine's outer packaging consisted of a thick, foil envelope, and the inner packaging consisted of a thick, foil &lt;a href=&quot;http://en.wikipedia.org/wiki/Blister_pack&quot; rel=&quot;nofollow&quot;&gt;single blister pack&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which got me to thinking - both layers of packaging appeared to be equally effective at blocking light and moisture from seeping into the underlying pill.  Using such extensive packaging would appear to be more expensive that traditional packaging techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What properties of a chemical (pharmaceutical) intended for general usage (consumption) would merit the double layer of packaging and additional expense?  Or to be a bit more specific, what is it about a medication that causes it to need such extensive protections?  Or am I completely off-base, and it's more of a manufacturing concern in making sure the pill isn't damaged before being delivered to the consumer?&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-02-15T01:37:12.683" LastActivityDate="2015-02-16T23:10:36.823" Title="Understanding redundant pharmaceutical packaging" Tags="|manufacturing-engineering|chemical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="673" PostTypeId="2" ParentId="672" CreationDate="2015-02-15T00:40:26.157" Score="2" Body="&lt;p&gt;If it were a chemical protection issue it would most likely be sensitivity to oxidation or moisture damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other possibility is double protection against tampering, which has in the past lead to people dying as in the case of &lt;a href=&quot;https://en.wikipedia.org/wiki/Tylenol&quot; rel=&quot;nofollow&quot;&gt;Tylenol&lt;/a&gt; being laced with cyanide in Chicago in 1982. The other issue about tampering or hoax tampering is that it has been used in extortion attempts against pharmaceutical companies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of the Tylenol tampering case in 1982, the company involved introduced tripled sealed packaging. That crisis cost the company involved $100 million, at the time.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: After the question was altered to specify prescription medication&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree with GlenH7's comment below, after the question was altered, that being prescription medication would mollify the case for tampering, but not totally eliminate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other thing I can think of is to increase the shelf life of the medication if the outer foil envelope was hermetically sealed.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-02-15T02:02:32.983" LastActivityDate="2015-02-15T02:02:32.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="674" PostTypeId="1" CreationDate="2015-02-15T03:03:42.530" Score="3" ViewCount="477" Body="&lt;p&gt;There is a topic discussing the pros and cons of &lt;a href=&quot;https://engineering.stackexchange.com/questions/63/what-are-the-pros-and-cons-of-a-traffic-circle-versus-a-traffic-light-intersecti&quot;&gt;roundabouts&lt;/a&gt;. A less common intersection is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Single-point_urban_interchange&quot; rel=&quot;nofollow noreferrer&quot;&gt;Single Point Urban Interchange (SPUI)&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an intersection that seems to be mostly used on overpasses. What makes this intersection unique is that all of the different exits, on ramps, and crossing road lanes go through the same location, i.e. the &quot;single point&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1wHl4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When might a SPUI be beneficial, and when might other intersections be better?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-02-18T11:11:43.323" Title="When are Single Point Urban Interchanges used?" Tags="|civil-engineering|highway-engineering|traffic-intersections|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="675" PostTypeId="2" ParentId="370" CreationDate="2015-02-15T06:03:47.330" Score="1" Body="&lt;p&gt;What you are asking for is a statics analysis or really something that an engineer would learn in a &quot;Mechanics of Materials&quot; course. You need to know how much stress is placed on the desk members as a result of the 300 lb force and whether it can hold the load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have &lt;a href=&quot;http://www.cadwolf.com/Documents/The_Wolf/OverflowAnswers/Force_on_Desk&quot; rel=&quot;nofollow&quot;&gt;solved this problem for the cross beam support on the desk&lt;/a&gt;. However, the highest load will be seen on the support member when the load is directly above it and not when the load is at the end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analysis can be carried out for the remaining members but to do a thorough analysis you need to look at the connection points as those would be the likely choke points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The document linked above was done on a platform I am developing called CADWOLF. You can alter the load to see the resulting forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result of the load you described is a load of 74.49 lbf on the cross member that supports the desk and a reactionary force of 274.5 lbf at the point where the desk connects to the legs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The document describes the process of summing the forces and moments to get these results. This same process can be used with the loads on the cross member and the reactionary force to calculate the load on the cross member connecting the vertical legs to the lower horizontal legs. &lt;/p&gt;&#xA;" OwnerUserId="546" LastEditorDisplayName="user16" LastEditDate="2015-02-16T00:25:25.237" LastActivityDate="2015-02-16T00:25:25.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="676" PostTypeId="2" ParentId="653" CreationDate="2015-02-15T06:34:45.353" Score="6" Body="&lt;p&gt;In the Fluid Dynamics community about 40 years ago, the group was primarily divided into experimentalists and theorists.  However, at that time CFD was quite new, had to be run on expensive supercomputers, and untrusted.  It was quite common that a theorist or experimentalist would at best discount the results of the CFD, while others may totally disregard the CFD results as useless. In fact, my former PhD advisor Dr. David Whitfield, was one of the early pioneers of using CFD alongside aerodynamics experiments at the Arnold Engineering Development Complex (AEDC).  This &lt;a href=&quot;http://www.arnold.af.mil/library/factsheets/factsheet.asp?id=13140&quot;&gt;reference&lt;/a&gt; explains well the thinking about CFD in those days:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At AEDC, CFD was used to supplement wind tunnel testing, but according&#xA;  to Dr. Whitfield, not a lot of people believed in CFD in the early&#xA;  1970s. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;In fact,&quot; he said, &quot;my efforts to promote CFD within AEDC in the&#xA;  early 1970s probably got me kicked-out, or through, most mahogany&#xA;  doors. However, when CFD was used to explain the source of the flow&#xA;  angularity problem in the test section of 16T, and when AEDC Fellow&#xA;  Dr. John Adams' CFD group in VKF explained how a tunnel there was &#xA;  actually operating at Mach 12 and not Mach 16 as previously thought,&#xA;  CFD found new life.&quot; &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;I was told once that 'AEDC is a test-data place, and there is no&#xA;  place for CFD,'&quot; he explained. &quot;Our objective was to help those&#xA;  running the tunnels to be able to do their jobs better. I don't think&#xA;  AEDC should just be a 'test-data' place. Rather it should be a place&#xA;  for solutions and physical understanding of the problems, and this can&#xA;  be accomplished better by the mutual cooperation between those&#xA;  focusing on experiments and those focusing on numerics.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In those days, generally the designer would design a new prototype and send it to the wind tunnel to test, and maybe some CFD would be performed at the same time.  There would generally be many prototypes built and tested, which was very costly. One such experimental facility where I used to work charged $16,000 per day of testing.  On the other hand, with the development of robust open source CFD codes, such as OpenFoam, and cluster computers, CFD simulations are quite cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, over time CFD began to mature, and with the popularization of cluster computers became quite feasible to run cheaply. With more and more validations with experiments being published in such journals as the AIAA Journal, CFD models have begun to be trusted more and more.  Nowadays, the cost of running experiments is much more expensive than running CFD simulations.  Therefore, more CFD simulations are used in the initial design stages, with many iterations back and forth, and even these days CFD-based design optimization (CDO) is often used in the design process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, it is my understanding that wind tunnels are used these days primarily for the following reasons: (1) testing finalized prototypes, and (2) conducting fundamental research in supersonic flows, especially in order to develop more accurate numerical models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding achieving flow similarity, when you have two different non-dimensional numbers, such as the Reynolds Number and the Mach number, the experimentalist must choose which number is most important to match.  For subsonic flows, Reynolds number should be used, whereas for transonic and supersonic flows the Mach number should be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often times one is not able to match the Reynolds number of the actual prototype by using a model test in the wind tunnel.  Consider for example a 747 which has a Reynolds number of 2,000,000,000 (&lt;a href=&quot;http://physics.info/turbulence/&quot;&gt;reference&lt;/a&gt;).  It is almost impossible to produce a wind tunnel which can match these types of Reynolds numbers.  People have tried to increase the Reynolds number by decreasing the temperature and using low density gases at low temperatures.  For example, the European transonic wind tunnel (ETW) is one of the world's largest cryogenic wind tunnels, which uses nitrogen as cold as -196$^{\circ}$C, but only achieves a maximum Reynolds number of 50 million per meter.  With a max test section length of 9 meters, the maximum possible Reynolds number would be 450,000,000, still less than half that of a Boeing 747.  In these cases, people have developed scaling laws to deal with how to scale the results up to the larger Reynolds number.  The scaling primarily has to do with the thickness of the boundary layer, which also effects other things like skin friction, and ultimately lift and drag.  There was a special conference held at Princeton University in 2003 to discuss these issues.  The results of that conference was this book: &lt;a href=&quot;http://link.springer.com/book/10.1007/978-94-007-0997-3&quot;&gt;http://link.springer.com/book/10.1007/978-94-007-0997-3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="511" LastActivityDate="2015-02-15T06:34:45.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="679" PostTypeId="2" ParentId="650" CreationDate="2015-02-15T18:43:44.370" Score="0" Body="&lt;p&gt;If you measure the gain, there is usually some variability in the measurements. This means that you would have to measure enough times to get a statistic distribution for the gain. The computed value should be equal to the mean of the distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm also curious about: is the gain given in Decibels (dB)? &lt;/p&gt;&#xA;" OwnerUserId="436" LastActivityDate="2015-02-15T18:43:44.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="680" PostTypeId="1" CreationDate="2015-02-15T19:27:16.670" Score="3" ViewCount="1056" Body="&lt;p&gt;We have a 7075 Al pump housing that has currently undergone a T651 temper but requires a T751 treatment to alleviate stress corrosion cracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it necessary to re-stretch a 7075 aluminum part to relieve internal stresses after a T751 heat treatment (Solution heat treated and stretched to remove internal stresses then overaged/stabilized) when a T651 (Solution heat treated and stretched to remove internal stresses then artificially aged) heat treatment was performed just previously?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part will only be held at minimum over-age temperatures to alleviate SCC in order to maintain most of the tensile and yield strength of the material.&lt;/p&gt;&#xA;" OwnerUserId="553" LastEditorUserId="553" LastEditDate="2015-02-20T01:21:55.743" LastActivityDate="2015-02-20T01:21:55.743" Title="Is stress relieving by stretching necessary when performing a T751 heat treatment after a T651?" Tags="|metallurgy|aluminum|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="681" PostTypeId="1" CreationDate="2015-02-15T19:49:11.723" Score="12" ViewCount="654" Body="&lt;p&gt;I have had three houses with right angles in the layout, and in each one that configuration causes vortices to form that suck in leaves, snow, and any debris blowing by and keep it in the nook of the building.  Rough idea in the following sketch, with the prevailing wind from the left:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vdncx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What passive methods or designs can be used to disrupt and prevent such vortices?&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="282" LastEditDate="2015-02-15T19:57:22.340" LastActivityDate="2015-02-24T04:20:03.343" Title="Foiling vortex traps around buildings" Tags="|civil-engineering|architecture|building-design|airflow|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="685" PostTypeId="2" ParentId="595" CreationDate="2015-02-16T02:57:37.103" Score="6" Body="&lt;p&gt;This &lt;a href=&quot;https://heavymovablestructures.org/assets/technical_papers/64.pdf&quot; rel=&quot;noreferrer&quot;&gt;paper&lt;/a&gt; from the &lt;a href=&quot;https://heavymovablestructures.org&quot; rel=&quot;noreferrer&quot;&gt;Heavy Movable Structures&lt;/a&gt; group goes into detail about how roadway movable spans are balanced. It references AASHTO (&lt;a href=&quot;http://www.transportation.org/Pages/Default.aspx&quot; rel=&quot;noreferrer&quot;&gt;American Association of State Highway and Transportation Officials&lt;/a&gt;) manuals such as those below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://books.google.com/books?id=wSkIJaDO1ooC&amp;amp;lpg=SA2-PA2&amp;amp;ots=TxgSGZ89r8&amp;amp;dq=AASHTO%20movable%20spans&amp;amp;pg=SA1-PA14#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;AASHTO LRFD Movable Highway Bridge Design Specifications&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://books.google.com/books?id=IKwfVvt0FOAC&amp;amp;lpg=PT252&amp;amp;dq=how%20to%20balance%20a%20movable%20span%20bridge&amp;amp;pg=PT252#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;AASHTO Maintenance Manual for Roadways and Bridges&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The general guidelines are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The spans are balanced so that the dead load reaction is greater than 1,000 kips per girder when the span is down. This keeps the span down even if the locking mechanism isn't working.&lt;/li&gt;&#xA;&lt;li&gt;The balance is calculated multiple times by both the contractor and the engineer based on theoretical and fabricated weights.&lt;/li&gt;&#xA;&lt;li&gt;The counterweight is designed so that the weight can be adjusted from the theoretical. It must accommodate a 5% heavier or 3.5% lighter span.&lt;/li&gt;&#xA;&lt;li&gt;The counterweight can be adjusted by adding or removing individual 1 cu. ft. blocks. These blocks are typically concrete but can also be steel or lead.&lt;/li&gt;&#xA;&lt;li&gt;The final check of weight can be done with strain gauges.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-16T02:57:37.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="686" PostTypeId="1" AcceptedAnswerId="689" CreationDate="2015-02-16T05:50:12.400" Score="6" ViewCount="3030" Body="&lt;p&gt;I have a chicken coop I plan on modifying to automatically open and close the door at dawn and dusk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What sort of actuator or mechanism would be appropriate for operating the small, side-hinged door?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constant power is not available (solar) plus it needs enough holding force to keep foxes out.&lt;/p&gt;&#xA;" OwnerUserId="557" LastActivityDate="2020-06-18T07:25:37.507" Title="Electromechanically open and close a small, hinged door?" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="687" PostTypeId="1" CreationDate="2015-02-16T07:14:30.827" Score="2" ViewCount="1992" Body="&lt;p&gt;I have a brazed plate heat exchanger similar to &lt;a href=&quot;http://www.danfoss.com/Products/Categories/Group/HE/Plate-Heat-Exchangers/Brazed-Heat-Exchangers-with-1-pass/5a54aff6-ae09-428b-a7c3-5d05ddc00fad.html&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt;. I can regulate flow and temperature of water on the primary side of the exchanger to achieve the desired temperature of water on secondary side. The desired temperature and flow on the secondary side are not under my control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I optimize the flow and temperature on the primary side so that I get the lowest possible temperature for water exiting the heat exchanger, while still achieving the desired temperature on the secondary side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The transfer medium is chemically treated water, the maximum temperature is around 100 °C at 4 bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I increase the flow to some maximum value I will also increase the heat loss through the pipe walls. However, increasing the flow rate would allow me to lower the supply temperature, which decreases the heat loss. How would I calculate the best flow rate and supply temperature combination to minimize heat loss through the pipe walls?&lt;/p&gt;&#xA;" OwnerUserId="558" LastEditorUserId="368" LastEditDate="2015-02-18T21:29:03.320" LastActivityDate="2018-08-27T10:40:53.520" Title="Optimizing return temperature in a brazed plate heat exchanger" Tags="|mechanical-engineering|heat-exchanger|heat-transfer|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="688" PostTypeId="2" ParentId="674" CreationDate="2015-02-16T09:11:09.377" Score="2" Body="&lt;p&gt;Often it's a lack of real estate for any other interchange. As the interchange is very narrow compared to a partial cloverleaf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other option is then a diamond interchange. The diamond interchange is more economical because it doesn't need a particularly wide bridge (4-6 lanes depending on the traffic of the other road).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The single point interchange also has a large conflict area in the middle which is less safe than putting 2 intersections next to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one major plus is that left-turning traffic from both off ramps of the Highway can go together. The left turning traffic of the other road can go together if a contra-flowing setup is used.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-02-18T11:11:43.323" LastActivityDate="2015-02-18T11:11:43.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="689" PostTypeId="2" ParentId="686" CreationDate="2015-02-16T11:40:59.320" Score="9" Body="&lt;h2&gt;Rod &amp;amp; Nut drive&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A threaded rod and captive nut, with either rod or nut driven rotationally by a motor is liable to offer a good solution. Because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Power level is set by thread pitch and attachment point to the door. While &quot;the bigger the better&quot; always helps, almost any &quot;sensible&quot; size of motor should be able to be used. I say &quot;sensible&quot; to eliminate utterly tiny motors such as pager vibrator motors. But usually anything in the 100 milliWatt to 100 Watt range COULD be used. Lower wattage requires longer time.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The threaded rod provides positive locking at zero power with no prospect of &quot;overhauling&quot;. As long as the door will not flex under Foxy assault when pulled solidly into its frame then dinner is off.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Level of travel can be set to suit by length of rod and mounting. Any door size up to a full dometic house door could be handled in this manner - so a small coop door is well within capability.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A cordless drill motor + gearbox is liable to be an excellent drive unit. These can usually be operated from 12V (9V to 18V units) and usually have a two stage reduction gearbox. They usually have a reversing switch which is not useful in this context if remote operation is required. To use, dismantle the drill, bypass or remove the switch and feed voltage to the two motor wires directly. Reversing the polarity reverses direction. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;http://homepages.nildram.co.uk/~sbuniak/BarnDoor/Barn-door%20basic%20principles.htm&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Barn-door star tracker&lt;/strong&gt;&lt;/a&gt; which illustrates the principle well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6yclh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their calculations page is &lt;a href=&quot;http://homepages.nildram.co.uk/~sbuniak/BarnDoor/Barn-door%20basic%20calculations.htm&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt; - overkill in this case but potentially useful. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.nz/search?q=nut+threaded+rod+drive&amp;amp;espv=2&amp;amp;biw=1745&amp;amp;bih=977&amp;amp;tbm=isch&amp;amp;tbo=u&amp;amp;source=univ&amp;amp;sa=X&amp;amp;ei=zNjhVKUfltTyBYehgHA&amp;amp;ved=0CBsQsAQ&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;A zillion versions of how you might do this&lt;/strong&gt; - each image links to a webpage&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RF Link:&lt;/strong&gt; As a bonus, a radio link using 2 x Arduinos and 2 RF modules, with a range of 10's of metres to a few kilometres can be constructed for about $10 in components all up. Ask if interested. This also applies to the wiper based system below.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Wiper motor &amp;amp; mechanism&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A possible solution depending on power availability is a wind screen wiper motor and mechanism. These are made to sweep a wiper arm across a sometimes dirty windscreen with considerable drag force. Units made for truck use are substantially more powerful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical automotive unit is rated at 50-100 Watts at 12V but will operate at lower voltage with reduced power. I have some Indian made truck wiper units rated at about 300 Watts!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 12V motor can be operated from a very small lead acid battery - say 12 V x 1.2 Ah. These can be charged by solar power. The battery should be maintained at a constant 13.7 Volts. You can obtain dedicated regulators for this purpose - PB137 is similar to the standard LM317 but rated at 13.7 Volts out.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.digikey.com/product-search/en/integrated-circuits-ics/pmic-voltage-regulators-linear-ldo/2556290?k=pb137&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;PB137 in stock Digikey $1.27 in 1's&lt;/strong&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.st.com/web/en/resource/technical/document/datasheet/CD00001634.pdf&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;PB137 Data Sheet&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a wiper mechanism is liable to have substantial backlash and depending on which way the door swings, may allow entry by pressing on the door. If the door is external and swings closed into a frame then it is likely to be Fox proof (for many values of Fox).&lt;br&gt;&#xA;An electrically latching mechanism could be added. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wiper mechanisms often have auto-stop points using internal switches. These usually allow stopping only at one end of travel. A similar arrangement can be implemented with microswitches and diodes. A switch is arranged to break current when it is operated during motor operation but the switch has an is opposed diode connected across it. Initially the diode has no effect and the motor stops. When polarity is reversed the diode conducts and moves the mechanism out of the 'dead spot'. Be sure that switches used are rated for the desired current in DC motor operation - this is more demanding than AC operation at the same current. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have implemented systems using rod &amp;amp; nut &amp;amp; cordless drills, and also wiper motor &amp;amp; mechanisms. Wiper based is easiest if it meets the needs. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-02-16T12:11:35.077" LastActivityDate="2015-02-16T12:11:35.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="690" PostTypeId="2" ParentId="686" CreationDate="2015-02-16T11:42:23.780" Score="1" Body="&lt;p&gt;Following are few Suggestions base on the problem description:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use a bi-stable Solenoid. The magnates in the solenoid will do most of opening and closing of the door. The force required will determine the size to the bi-stable solenoid. The challenge will be powering the system. &lt;/li&gt;&#xA;&lt;li&gt;A hybrid power system might work best. A rechargeable Lithium Ion or Lithium Polymer battery coupled with some type of energy harvester is a good candidate. A super capacitor most likely will needed to provide a burst of power.&lt;/li&gt;&#xA;&lt;li&gt;Solar power energy harvester could trickle charge the recharge the power source during the daytime.&lt;/li&gt;&#xA;&lt;li&gt;Also some type of generator that can capture the energy generate by a chicken moving inside to coop is also a good option to power the system. This will need some creativity imagination and research.&lt;/li&gt;&#xA;&lt;li&gt;A Photocell can be used to determine the dawn and dusk&lt;/li&gt;&#xA;&lt;li&gt;Harvesting the vibration energy created by closing and open of the door event could be used to power a radio (6lowpan, Zibgee, BLE etc) to communicate door open and close events to a sensor hub or a locale server. &lt;/li&gt;&#xA;&lt;li&gt;The presence of a fox most likely will create movement in the chicken coop. Using a little bit of creativity this movement could be converted in to usable energy, to charge a super capacitor that will flash a LED, generate sound and communicate to base. Also consider using a radio as mentioned above to communicate the base which could active alarm to deter the fox.&lt;/li&gt;&#xA;&lt;li&gt;The vibration energy create by movement of chicken moving in and out of the chicken coop can be harvested. This can not only be a good energy source but also could with some &lt;a href=&quot;https://engineering.stackexchange.com/questions/390/how-can-the-vendor-be-approved-disapproved-based-on-validation-test-data/571#571&quot;&gt;statistical analysis&lt;/a&gt; be used to determine in all the chicken are in or out the coop.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.takano-sanki21.com/english/product/bistable.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two position bistable solenoid&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Current Market&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.flemingoutdoors.com/solar-kit-for-automatic-chicken-coop-door.html?ref=lexity&amp;amp;_vs=google&amp;amp;_vm=productsearch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Kit For Automatic Chicken Coop Door&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=xARNapqjWvg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Advance Automatic Chicken Coop Door Opener&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-02-18T23:11:17.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="691" PostTypeId="1" CreationDate="2015-02-16T11:51:18.483" Score="2" ViewCount="562" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Specific_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;Self-support length is the maximum length of a vertical column of the material (assuming a fixed cross-section) that could suspend its own weight when supported only at the top&lt;/a&gt; this is an important consideration in the construction of a &lt;a href=&quot;https://space.stackexchange.com/questions/2699/what-is-a-space-elevator&quot;&gt;space elevator.&lt;/a&gt;  While it is unlikely a strand one molecule wide to stretch from the surface of the Earth to space, it would be interesting to know what the longest self-support length a chain one molecule wide (single-chain polymers) could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A larger molecule gives a larger cross section, but it is also heavier.  Most space elevator concepts are &lt;a href=&quot;https://space.stackexchange.com/questions/6644/taper-ratio-for-partial-space-elevator&quot;&gt;tapered &lt;/a&gt; to provide additional strength where the load is greatest.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of single-chain polymers can provide the longest self-support length for a column one molecule wide, and what would that self-support length be?&lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="-1" LastEditDate="2017-04-13T12:58:55.687" LastActivityDate="2015-02-18T12:23:50.877" Title="What is the longest self support length a chain one molecule wide can be?" Tags="|chemical-engineering|structural-engineering|stresses|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="692" PostTypeId="2" ParentId="444" CreationDate="2015-02-16T16:17:42.013" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://www.standardsforhighways.co.uk/dmrb/vol1/section3/bd3701.pdf&quot; rel=&quot;nofollow&quot;&gt;BD37/01&lt;/a&gt; has the required loading in Chapter 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section 4.1 gives the HB loading required for various bridges uses. It doesn't segregate based on superstructure type; it only segregates based on roadway class: &lt;code&gt;Motorway and Trunk Road&lt;/code&gt;, &lt;code&gt;Principal roads&lt;/code&gt;, and &lt;code&gt;Other public roads&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that both steel and concrete bridges are designed to the same number of HB units.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-16T16:17:42.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="693" PostTypeId="1" AcceptedAnswerId="700" CreationDate="2015-02-16T20:34:29.427" Score="4" ViewCount="3600" Body="&lt;p&gt;Suppose I have a sealed enclosure, say &lt;a href=&quot;http://www.hoffmanonline.com/product_catalog/catalog_item_detail.aspx?cat_1=34&amp;amp;cat_2=2346&amp;amp;cat_3=78347&amp;amp;catIDs=78347,186&amp;amp;itemIDs=3239,4185&amp;amp;catalog_item=4185&quot; rel=&quot;nofollow&quot;&gt;this one from Hoffman&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The box is 16&quot;x12&quot;x6&quot;, made of galvanized steel with a wall thickness of .0635&quot; or .0785&quot; (gauge 16 or 14).  I'm planning to put some electronics in this box, which will generate an arbitrary amount of heat. How can I use this information to estimate how much the air inside the box will heat up above ambient outside the box?&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="110" LastEditDate="2015-02-17T04:02:51.327" LastActivityDate="2015-09-15T12:00:20.543" Title="How do I determine the thermal resistance of a metal box?" Tags="|mechanical-engineering|thermodynamics|temperature|heat-transfer|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="697" PostTypeId="2" ParentId="681" CreationDate="2015-02-16T23:04:23.973" Score="1" Body="&lt;p&gt;Build a fence or plant a hedge line to block off the right angle (connecting the two corners of the triangle that rubbish tends to collect in).  That way the wind, and the stuff it's carrying, should mostly slide around the structure.&lt;/p&gt;&#xA;" OwnerUserId="177" LastActivityDate="2015-02-16T23:04:23.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="698" PostTypeId="2" ParentId="672" CreationDate="2015-02-16T23:10:36.823" Score="1" Body="&lt;p&gt;When you're looking at packaging medicines, over the counter and prescription, you're not just looking at engineering requirements, but you're looking at legal requirements, both legislated (passed into law by a legislating body, like a parliament or congress) and court-imposed (the result of court or judge orders in connection with lawsuits).  You could also be looking at marketing decisions; even prescription medication packaging is often designed with marketing in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand it, therefore, you have to look at all of those.  They're likely to be different, and the combinations of all of them can often seem nonsensical or wasteful.  Companies will often keep their reasoning or rationale (or lack thereof) secret for good and bad reasons.&lt;/p&gt;&#xA;" OwnerUserId="177" LastActivityDate="2015-02-16T23:10:36.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="699" PostTypeId="2" ParentId="452" CreationDate="2015-02-17T00:00:59.710" Score="2" Body="&lt;p&gt;Another alternate solution would be to use a Hall Effect sensor. There are many ways to apply a hall effect sensor to measure distance. Most simple might be to attach the hall effect sensor to the motor and count the rotations. Then translated the number of rotation to distance. Below are few reference links that better explains how to use a hall effect sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some automotive application Hall Effect sensor is use to measure speed. Also hall effect sensor is used in clam-shell cell phone to detect open and close. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xRjGK.gif&quot; alt=&quot;Hall Effect Sensor Gear Tooth&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EkCG1.jpg&quot; alt=&quot;Hall Effect Sensor&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=S-haU2AqpKU#t=24&quot; rel=&quot;nofollow noreferrer&quot;&gt;RPM Counter&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://electromechhobby.blogspot.com/2012_12_01_archive.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;RPM Counting with Hall Effect Sensor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.sciepub.com/ajme/1/7/16/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Position Measurement with Hall Effect Sensors&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://sensing.honeywell.com/honeywell-sensing-sensors-magnetoresistive-hall-effect-applications-005715-2-en.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hall Effect Sensing and Application&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.hamlin.com/technical-detail-hall-sensor.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Test Geartooth Sensing using 55075 device&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.farnell.com/datasheets/1680391.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hall Effect Gear Tooth Sensors&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-01-20T12:54:38.420" LastActivityDate="2016-01-20T12:54:38.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="700" PostTypeId="2" ParentId="693" CreationDate="2015-02-17T03:33:24.463" Score="6" Body="&lt;p&gt;This will depend very much on the heat flow around the enclosure. There is a thing called the H factor which describes the heat transport properties between a surface and a fluid. The value of H varies with surface properties and flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to do your calculation you need a few different assumptions. Let's simplify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the air in the box is uniformly heated&#xA;2) air flows around five sides of the box the same way; no heat transfer takes place through the bottom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can consider the box a &quot;thermal resistance&quot; (the walls) in series with a heat sink (the air flowing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermal resistance (K/W) of the walls is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R=\frac{t}{kA}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your box, t = 1.6 mm, A = 3.4E5 mm$^2$, k = 17 K/W, so R = 2.8E-4 K/W : the walls themselves won't present much thermal resistance. So we need to look at the air around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat transfer coefficient is a strong function of air flow rate (this is why you have fans for cooling!). Let's make a simple assumption that there is &lt;em&gt;some&lt;/em&gt; natural air flow, but no forced convection. The heat transfer coefficient will then be at the lower end of the scale, or about 0.5 W/m^2/K. (See for example &lt;a href=&quot;http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html&quot; rel=&quot;nofollow&quot;&gt;this reference&lt;/a&gt;). With the &quot;used&quot; area of the box of 0.34 m$^2$ (12x16+(12+16+12+16)x6 in$^2$), you get about 1.5 K temperature rise inside the box for every Watt of power dissipated. A little fan around the outside could easily cut that by a factor 10...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So just to drive the point home explicitly: this calculation is conservative and &lt;strong&gt;VERY APPROXIMATE&lt;/strong&gt;; the number calculated can be easily an order of magnitude off from the number you get when you build your box, because there are SO many factors that will affect this. Controlling the air flow explicitly (versus relying on &quot;whatever air flow lies around&quot;) is really essential to get any kind of believable limits on your answer. This is reflected in the comments above...&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditDate="2015-02-17T15:58:06.843" LastActivityDate="2015-02-17T15:58:06.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="701" PostTypeId="2" ParentId="3" CreationDate="2015-02-17T03:47:58.973" Score="3" Body="&lt;p&gt;The presence of a swimming pool in a natatorium causes a few unique issues that most buildings do not experience:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;High humidity levels&lt;/strong&gt; - Water is everywhere and is frequently sprayed. Most buildings have ventilation systems that keep the humidity low.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;High chloride content&lt;/strong&gt; - Chlorine is used for a disinfectant. Chlorine will rapidly cause corrosion on bare steel.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Each of these items individually is bad for steel, but together they can cause serious issues. As this &lt;a href=&quot;http://www.steelconstruction.info/Standard_corrosion_protection_systems_for_buildings#Introduction&quot; rel=&quot;nofollow&quot;&gt;table&lt;/a&gt; shows, swimming pools are as harsh an environment as chemical plants and coastal areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To mitigate the corrosive nature of the environment, a &lt;a href=&quot;http://resource.npl.co.uk/docs/science_technology/materials/life_management_of_materials/publications/online_guides/pdf/indoor_swimming_pool_buildings.pdf&quot; rel=&quot;nofollow&quot;&gt;double protection system&lt;/a&gt; is employed. This system consists of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A layer of hot &lt;strong&gt;dip galvanizing&lt;/strong&gt; or &lt;strong&gt;zinc rich primer and 2 coats of epoxy&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;A field applied addition layer of &lt;strong&gt;epoxy&lt;/strong&gt; or &lt;strong&gt;acrylic urethane&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The top layer is essentially a first line of defense.  It will need to periodically be replaced. The galvanized or zinc layer is there to provide extra time when the top layers fail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These types of systems typically use contrasting colors for each coating layer. This way any reduction of protection due to a missing layer is readily visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Together this system can protect the steel structure for up to 50 years.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-17T03:47:58.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="702" PostTypeId="2" ParentId="680" CreationDate="2015-02-17T04:51:13.350" Score="2" Body="&lt;p&gt;&lt;em&gt;I am going to start out by saying that aluminum is a mess. There are too many designations and things that you can do to it after it solidifies.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To directly answer your question, what might necessitate an additional round of &lt;a href=&quot;http://www.matweb.com/reference/aluminumtemper.aspx&quot; rel=&quot;nofollow&quot;&gt;heat treating&lt;/a&gt; is directly affected by the desired final mechanical properties. Each of the designations that you listed is a complete heat treatment. Either one is chosen or the other is chosen, they are not done in a sequence other than what is already defined by the string of numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the 7075 chart on &lt;a href=&quot;http://www.tkmna.com/wcm/idc/groups/internet/documents/web_content/mdaw/mdiy/~edisp/d_004646.pdf&quot; rel=&quot;nofollow&quot;&gt;page 11 of this pdf&lt;/a&gt; the only difference between the -T6 and the -T7 is that the -T7 has a slightly better resistance to &lt;a href=&quot;https://en.wikipedia.org/wiki/Stress_corrosion_cracking&quot; rel=&quot;nofollow&quot;&gt;stress corrosion cracking&lt;/a&gt;. If this is what you are looking for, choose the -T7.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The reason that there are so many different alloys and heat treatments is that each combination produces a material with slightly different physical properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Properties of the two heat treatments that you listed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.matweb.com/search/DataSheet.aspx?MatGUID=4f19a42be94546b686bbf43f79c51b7d&amp;amp;ckck=1&quot; rel=&quot;nofollow&quot;&gt;7075-T651&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.matweb.com/search/DataSheet.aspx?MatGUID=6653b72914864cc0a0ff7adf5b720167&quot; rel=&quot;nofollow&quot;&gt;7075-T7351&lt;/a&gt; (slightly different from the one you listed)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Without &lt;strong&gt;very&lt;/strong&gt; detailed requirements (and lots of experimenting), it is almost impossible to say which alloy and heat treatment of aluminum will be the best for a given situation. You can always go with the most common types.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-17T04:51:13.350" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="703" PostTypeId="1" CreationDate="2015-02-17T05:22:41.173" Score="0" ViewCount="163" Body="&lt;p&gt;I am working on a hand-held prototype model in which I need to fix motor RPM below a certain displacement. For example, if overall displacement is within 5 mm then the motor should be running at a fixed RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After getting this done (I mean if displacement is &gt; 5 mm) I wish to work on sensing the linear acceleration of the hand-held prototype. I will use this data to control my motor's RPM again. Typical hand-held displacement range would be around 5 cm to 30 cm in 2 sec. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone suggested to use IMUs to sense acceleration and rotation but I am not sure how to design my prototype to utilize an IMO. How can I select an IMU and determine an appropriate sensitivity level? References to relevant literature would be very helpful.&lt;/p&gt;&#xA;" OwnerUserId="566" LastEditorUserId="368" LastEditDate="2015-02-17T17:29:15.907" LastActivityDate="2015-02-19T19:36:58.007" Title="How to sense displacement without using any camera?" Tags="|electrical-engineering|robotics|sensors|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="704" PostTypeId="2" ParentId="687" CreationDate="2015-02-17T10:26:07.140" Score="2" Body="&lt;p&gt;As stated the problem is probably not well enough constrained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the specification, the way you get minimum temperature out is to pump as much volume as you can through the primary and then regulate the input temperature to achieve desired output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. the more primary flow there is the lower delta temperature drop you need in the primary side. Pumping &quot;flat out&quot; may not have been what you had in mind. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Defining &#xA;$T_{p,in}$ = Temperature primary in&lt;br&gt;&#xA;$T_{s,out}$ = Temperature secondary out etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$kV_{in}(T_{p,in}-T_{p,out}) = V_{out}(T_{s,out}-T_{s,in})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k is an efficiency factor for your heat exchanger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. Energy removed from the primary stream is $V_{in}\Delta T_{in}$ and energy received by the secondary stream is $V_{out}\Delta T_{out}$, with a factor to account for losses and inefficiencies in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with a perfectly insulated lossless system there will be a temperature delta above ideal due to the thermal transfer resistance of the exchanger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that heating due to flow / pumping losses in the heat exchanger are ignored (or otherwise they add to the energy available to transfer). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{p,out}$ must always be greater than or equal to $T_{s,out}$ and may be greater than $T_{s,in}$ depending on design (but you hope not). For an ideal non counterflow exchanger $T_{p,out}$ is usually much greater than $T_{s,out}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If minimum $T_{p,out}$ is a genuine need then you want to use a counterflow heat exchanger.&#xA;A counterflow exchanger allows $T_{p,out}$ to approach $T_{s,in}$ and $T_{s,out}$ to approach $T_{p,in}$. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="41" LastEditDate="2015-02-17T14:08:33.723" LastActivityDate="2015-02-17T14:08:33.723" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="705" PostTypeId="2" ParentId="703" CreationDate="2015-02-17T12:31:14.807" Score="4" Body="&lt;p&gt;I can say using accelerometers to detect displacement won't work. These devices have so much of inherent noise that discerning very slow acceleration from remaining immobile is about impossible. You won't be able to detect acceleration of order of 0.1 mm/s&lt;sup&gt;2&lt;/sup&gt; and it takes just a several seconds at such acceleration to exceed your 5mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd have a hard time with a camera as well, especially if the device is hand-held. True the camera looking along X axis would work with Y,Z plane displacement in confined environments (relatively nearby objects in focus) but it would get easily confused by rotation. You'd need at least two cameras looking in different (and not just opposite) directions and environment that is conductive to image detection (you won't get much displacement measurement from the image of the sky). Also, rotation will be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will work with ~5mm distance ranges is frequency-based distance metering - preferably ultrasound, though infrared/visible light would be possible too (though obviously more difficult due to frequencies being much, much higher). Unfortunately plain &quot;reflective&quot; ultrasound distance sensors won't help much about rotation/tilt of the hand, distance from walls changing rapidly despite the device not moving. What you need are three ultrasound emitters (speakers), each of a different frequency, placed around the location where the measurement is to take place, and a sampler (microphone) in the &quot;handheld&quot;. By filtering given frequency you get signal from one emitter, and then by analyzing the change of phase of the wave in time intervals equal to the wave period, you have displacement of the device relative to the emitter. By triangulating the three distances you have total displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(also, a gyroscope will help you account for rotation.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One more word of note as to why accelerometers are so ungrateful to work with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your 3-axis accelerometer, the &quot;immobile&quot; [X,Y,Z] readouts are [0,0,-9.807] m/s&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That last factor changes with latitude, somewhat with longitudal location and altitude, not to mention internal properties of the accelerometer, like temperature or air pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cursory peek at &lt;a href=&quot;https://www.sparkfun.com/datasheets/Components/SMD/adxl335.pdf&quot; rel=&quot;nofollow&quot;&gt;some generic&lt;/a&gt; 3-axis accelerometer datasheet will give you a ballpark figure of maybe 1% precision. (0.3% non-linearity, 1% cross-talk, 300μg/√Hz noise RMS, thermal and other minor ones). Let's be overly generous and change it to 0.1%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.1% of 9.807m/s&lt;sup&gt;2&lt;/sup&gt; is 9.807mm/s&lt;sup&gt;2&lt;/sup&gt;. That's the error you can expect from your accelerometers - changing from immobile to nearly a centimeter per second over course of one second, sunk in the noise.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="106" LastEditDate="2015-02-19T19:36:58.007" LastActivityDate="2015-02-19T19:36:58.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1705" PostTypeId="1" AcceptedAnswerId="1727" CreationDate="2015-02-17T14:40:48.747" Score="10" ViewCount="744" Body="&lt;p&gt;My car has ultrasonic proximity sensors to help me park. I've noticed that when motorcycles whiz past me the proximity alarm goes off. I originally thought that the motorcycles were just too close, but now I have observed that that isn't the case; cars or other road users at a similar distance and speed do not set off the sensors. It doesn't seem to matter what speed I am traveling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not interested in troubleshooting some aspect of the alarm. I'm just trying to understand why this occurs. What special characteristic of a motorcycle or its movement causes it to be detected by the sensors when other road users are not detected?&lt;/p&gt;&#xA;" OwnerUserId="571" LastEditorUserId="282" LastEditDate="2015-03-06T19:28:50.770" LastActivityDate="2015-03-06T19:28:50.770" Title="How do ultrasonic proximity sensors detect motorcycles differently from cars and trucks?" Tags="|electrical-engineering|sensors|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1706" PostTypeId="1" AcceptedAnswerId="1723" CreationDate="2015-02-17T14:47:53.910" Score="20" ViewCount="5979" Body="&lt;p&gt;Bridges are designed for the loads that come from the vehicles that are expected to cross them.  This includes the weight the vehicle and any dynamic loads that may be introduced from movement of the vehicle. Dynamic loads may be from &quot;bouncing&quot; or from hitting a joint or pothole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially it would seem obvious that more load is applied to a bridge while the vehicles are in motion (weight of vehicle plus dynamic load). The dynamic loads are proportional to the travelling speed of the vehicles, but as vehicles go faster, they typically are spaced farther apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When vehicles are stopped, they typically are much more closely spaced than when they are moving.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can there be a situation where more load is on a bridge because of closely spaced &lt;strong&gt;parked&lt;/strong&gt; vehicles than from &lt;strong&gt;moving&lt;/strong&gt; vehicles that are farther apart?&lt;/li&gt;&#xA;&lt;li&gt;Are these two situations covered in bridge design?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastActivityDate="2017-01-17T16:20:01.133" Title="Does a roadway bridge experience more load when vehicles are parked or when they are moving?" Tags="|civil-engineering|bridges|structural-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1708" PostTypeId="2" ParentId="687" CreationDate="2015-02-17T15:23:48.030" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/704/41&quot;&gt;Russell&lt;/a&gt; covered this fairly well, but I want to expand on a couple topics he only briefly mentioned, and look at the analysis slightly differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you should have control over the direction of your flows as well. That is, you should be able to switch the inlet and outlet of the primary flow so that they're on the same side of the heat exchanger as the inlet and outlet of the secondary flow or vice versa. What will give you the lowest possible $T_{p,out}$ is a cross-flow arrangement (i.e., the primary inlet and secondary outlet on one side, and the primary outlet and secondary inlet on the other side.) There must be a temperature difference in the two streams at every contact point in the heat exchanger to have continuous heat transfer, but a cross flow arrangement allows the cold side output to be higher than the hot side output, which is not the case in a parallel flow arrangement, as shown in the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LBbvj.png&quot; alt=&quot;Counter flow (A) vs Parallel flow (B)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, finding the actual lowest possible $T_{p,out}$ is not as simple as doing a few calculations. You have to set up a system of equations that govern the behavior of the different sides of the heat exchanger as well as the boundary between the two flows and then use those to find the lowest primary outlet temperature using an optimization method. Optimization of thermal systems is a semester-long 400-level (undergrad senior, early grad student) college course. This is on the simpler side of those problems, but it's still not something trivial, because you either need a computer program to calculate a bunch of scenarios or use a more nuanced approach that takes less computing power but more thought and consideration, and is more in-depth than we can do here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we can make a few general statements. The heat that leaves the primary side is the same heat that is transferred to the secondary side. Since we can't change anything about the secondary side, we can only change how we transfer the heat from the primary side. In general, that equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_p=\dot{m_p}(h_{p,out}-h_{p,in})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Q$ is the heat transfer, $\dot{m_p}$ is the mass flow rate, and $h$ is the enthalpy. Enthalpy can be tricky to define if you don't have much experience in thermodynamics, but it's a general way to define the energy in a flow based on temperature, phase, pressure, and volume. For example, it can quantify the energy change when water changes to steam at 100°C. The point is, if we can't change $Q_p$, what we want to do is generally lower $h_{p,out}$. To do this, we can either lower the input temperature (thus lowering the enthalpy at the input, but with the same mass flow rate, the enthalpy difference must be the same) or we can decrease the mass flow rate. This will increase the enthalpy difference, and if the input enthalpy remains the same, the output will get lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note here that there's a limit to how low the output of the primary flow can be. It can't go lower than the input temperature of the secondary flow, otherwise the heat transfer would be reversed; this violates the 2nd Law of Thermodynamics, and &lt;a href=&quot;https://www.youtube.com/watch?v=6vxHkAQRQUQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this house, we obey the laws of thermodynamics!&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also some diminishing returns, because as you decrease the mass flow, you're going to see less effective heat transfer, so you just might not get your desired output on the secondary side. My advice would be to lower the input temperature to just about the output temperature you need on the secondary side and keep the flow rate at a good pace. &lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-02-17T15:23:48.030" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1709" PostTypeId="1" AcceptedAnswerId="1711" CreationDate="2015-02-17T20:05:05.127" Score="4" ViewCount="913" Body="&lt;p&gt;I'm giving SolidWorks a try and am following one of their tutorials on creating a part. When I create circles they are looking a bit &quot;boxy&quot; to me. My graphics card has 512MB of ram and I have checked the rendering settings that I found and everything seems to be OK - but I would have expected something more &quot;circular&quot; and less rough looking, but I'm new to working with CAD - is this normal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/koy1P.png&quot; alt=&quot;Boxy Edges&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ylBSY.png&quot; alt=&quot;Full screen View&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5umIB.png&quot; alt=&quot;Rendering Settings&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="43" LastEditorUserId="41" LastEditDate="2015-02-17T20:25:13.417" LastActivityDate="2015-02-17T21:22:15.590" Title="Are boxy edges normal in SolidWorks when working with a drawing?" Tags="|computer-aided-design|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1711" PostTypeId="2" ParentId="1709" CreationDate="2015-02-17T21:22:15.590" Score="10" Body="&lt;p&gt;Yes, this is normal. By default, SolidWorks renders curves on the screen using less than the highest possible level of detail that your monitor is capable of displaying. The point of this is to allow the screen to be redrawn quickly as you edit your design and change the view. The more detailed the curve, the longer it takes to redraw the screen; depending on your hardware, this could make the software feel unresponsive or &quot;laggy&quot; and make it harder to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can adjust the Image Options in the Document Properties tab to render curves with more detail; see &lt;a href=&quot;http://help.solidworks.com/2014/English/SolidWorks/sldworks/HIDD_OPTIONS_IMAGE_QUALITY.htm?id=b432045cef064e2db19d520f397724c7#Pg0&quot; rel=&quot;noreferrer&quot;&gt;documentation here&lt;/a&gt; and this screenshot from &lt;a href=&quot;http://help.solidworks.com/2014/English/api/swconst/DP_ImageQuality.htm&quot; rel=&quot;noreferrer&quot;&gt;the API help pages&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GsHJw.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slider at the top of the window controls how precisely curves are drawn:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;strong&gt;Low - High (slower) and Deviation:&lt;/strong&gt;  (For assemblies, available only if Apply to all reference part documents (below) is selected.)&#xA;  The slider controls the image quality resolution, and Deviation is the&#xA;  maximum chordal deviation in effect. Move the slider or type a value&#xA;  in Deviation. The slider setting and deviation value are coupled and&#xA;  are inversely proportional.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If all you're concerned about is confirming that your circle is actually a circle even when it is rendered with corners, you may not want to increase this setting, as it will have a negative impact on performance.&lt;/p&gt;&#xA;" OwnerUserId="368" LastActivityDate="2015-02-17T21:22:15.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1712" PostTypeId="1" CreationDate="2015-02-17T23:18:43.193" Score="2" ViewCount="1338" Body="&lt;p&gt;What is the purpose and benefit of the two capacitors on both the input and output pins of a LDO Regulator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kwxly.jpg&quot; alt=&quot;LDO Circuit&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="368" LastEditDate="2015-02-18T05:01:59.650" LastActivityDate="2015-02-28T03:27:02.030" Title="What is the purpose of the capacitors on LDO Regulator Circuit?" Tags="|electrical-engineering|circuits|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1713" PostTypeId="1" AcceptedAnswerId="1728" CreationDate="2015-02-17T23:44:38.207" Score="6" ViewCount="129" Body="&lt;p&gt;What with all the work done by the FAA recently on regulating drone usage, I thought it would be nice to turn my eyes to the sky for another, less-known type of aerial denizen: the &lt;a href=&quot;https://en.wikipedia.org/wiki/Airborne_wind_turbine&quot; rel=&quot;noreferrer&quot;&gt;aerial wind turbine&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has all the relevant information, so I won't rehash it here. The most important thing here is that aerial wind turbines - and I'm mostly interested in aerostat-based designs - are free-floating, just like certain types of tethered drones, which have given the FAA headaches because they're not like most aircraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did find &lt;a href=&quot;http://www.avweb.com/avwebflash/news/aerial_wind_turbine_204758-1.html&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; article, which says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Regulations and technological restrictions suggest it may not happen very soon, or at all, but some researchers believe aerial turbines will be tapping high-altitude winds for power generation sometime in our future and perhaps within the decade.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That said, companies currently working on the project seem very loosely formed. They are in the development phase and are aware of federal airspace restrictions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.faa.gov/uas/civil_operations/&quot; rel=&quot;noreferrer&quot;&gt;Here's&lt;/a&gt; the relevant FAA page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that FAA regulatory measures on drones cover aerial wind turbines, but I'm not positive, and I'd like to see some specific mention of them in regulations, if possible. I assume they're treated as (tethered) drones, but I haven't been able to find any direct mention.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Getting to the point: Aerial wind turbines are, in general, much smaller than typical wind turbines. There are obviously some factors that make it easier for wind turbines built on the ground to be larger (support, for instance). But larger turbine blades, while a source of trade-offs, can be much more helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What regulations (in the United States), if any, limit the size of aerial wind turbines?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-02-18T19:37:26.190" LastActivityDate="2015-02-18T22:07:29.253" Title="Do regulations in the United States limit the size of aerial wind turbines?" Tags="|aerospace-engineering|renewable-energy|regulations|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1714" PostTypeId="1" CreationDate="2015-02-18T01:21:20.347" Score="3" ViewCount="127" Body="&lt;p&gt;I have been experimenting with building various support tables around &lt;a href=&quot;http://spotweldinc.com/press_welder.html&quot; rel=&quot;nofollow&quot;&gt;press-type spot-welders&lt;/a&gt;.  These are relatively simple fixures; think of a tabletop with legs under it and devices like sliding gibs, pins, clamps, rollers, etc mounted to the table top to guide a specific set of parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is what material to use to build these tables.  Being in a metal-working shop, I usually reach for things like angle iron, sheet metal, and &lt;a href=&quot;http://www.unistrut.us/index.php?WP=S00_Channel&quot; rel=&quot;nofollow&quot;&gt;Unistrut&lt;/a&gt;.  This is fine for building most parts of the tables, but due to a phenomenon inherent in spot-welding, it is not smart to build steel pieces bridging over the copper electrodes and &lt;a href=&quot;http://spotweldinc.com/cables_shunts.html&quot; rel=&quot;nofollow&quot;&gt;shunts&lt;/a&gt;.  I cannot think of the name of the effect right now, but basically if you have a steel member passing through the magnetic field that surrounds the shunts and other parts that come off the secondaries of the transformer, it will reduce the effective current delivered at the ends of the electrodes.  I think the current drop is proportional to the mass of the steel (and presumably the distance), but I am having trouble finding reliable information on this effect.  If I knew its name, that would help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am exploring two solutions for building these tables.  One would be if I had a formula modeling the mass and distance of the steel relative to the current, I could (maybe) fit the steel structural members in a position where I will not get (much) interference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another solution would be to keep steel members out of it and to build those parts out of another material.  I considered aluminum angle, but I really do not know if this would behave like the steel in the induced magnetic field.  My gut tells me it would not, but I really do not know.  Another choice could be plastic bar and sheet although I am not eager to go there since that would cause more fasteners, countersinking, thicker material and/or more supports to get comparable strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do any of you know of a useful formula or rule of thumb that could guide me as to steel placement?  Do you know if aluminum would not have side-effects in this situation?  Do you have other ideas for materials that are easy to work with? Another idea: Is there a shielding that would help in this situation?&lt;/p&gt;&#xA;" OwnerUserId="580" LastEditorUserId="580" LastEditDate="2015-02-18T01:39:24.097" LastActivityDate="2015-03-03T19:05:50.370" Title="What materials or techniques can I use to reduce this electrical effect when building spotwelding tables?" Tags="|electrical-engineering|materials|welding|current|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1715" PostTypeId="1" AcceptedAnswerId="1724" CreationDate="2015-02-18T03:57:53.167" Score="6" ViewCount="2038" Body="&lt;p&gt;In the US, typical structural steel welding is covered by either AWS D1.1 or AWS D1.5 (bridges). Both of these codes cover a wide range of carbon steels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AWS has a separate code for welding stainless steels, AWS D1.6. There is some mention of welding to carbon steels, but this is mostly talking about choosing the filler metal and not the testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither D1.1 nor D1.5 cover stainless steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this specific situation, I am looking at a fillet weld between a carbon steel plate and a stainless steel plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the procedure for qualifying (PQR and WPS) a weld between the two materials? Does one code control over the other? Would the weld have to be qualified per both specifications?&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2023-06-02T17:47:15.913" Title="What is the process for qualifying a weld between structural steel and stainless steel?" Tags="|steel|structural-engineering|welding|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1716" PostTypeId="5" CreationDate="2015-02-18T05:07:28.630" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-02-18T05:07:28.630" LastActivityDate="2015-02-18T05:07:28.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1717" PostTypeId="4" CreationDate="2015-02-18T05:07:28.630" Score="0" Body="Use for questions primarily about circuit design, components, testing or other practical applications." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-02-18T05:07:28.630" LastActivityDate="2015-02-18T05:07:28.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1718" PostTypeId="1" AcceptedAnswerId="1719" CreationDate="2015-02-18T07:53:38.490" Score="11" ViewCount="32835" Body="&lt;p&gt;I've seen it in several data sheets - it is a measure of error of some kind, of course. The problem is I dont know the exact meaning. I've seen it in the context of repeatability, accuracy and linearity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is the following data sheet: &lt;a href=&quot;http://www.smcpneumatics.com/pdf/americansmc/ZSE40_ISE40/ZSE40_and_ISE40_Series_Digital_Switches.pdf&quot;&gt;smc data sheet&lt;/a&gt; (On page 3)&lt;/p&gt;&#xA;" OwnerUserId="46" LastActivityDate="2018-01-08T10:41:38.147" Title="What does &quot;$\pm$ 0.5% F.S.&quot; mean?" Tags="|nomenclature|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1719" PostTypeId="2" ParentId="1718" CreationDate="2015-02-18T08:17:31.650" Score="15" Body="&lt;p&gt;FS = FULL SCALE = maximum reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that the accuracy is such that the reading is &lt;strong&gt;probably&lt;/strong&gt; within + or - 0.5% of the &lt;strong&gt;FULL SCALE&lt;/strong&gt; reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very important and easily overlooked qualification of the result.&lt;br&gt;&#xA;If I have a reading of 1 Volt and the accuracy is +/- 0.5% it means that the actual result should lie in the range 1 - 0.5% x 1 to 1 + 0.5%  of 1&lt;br&gt;&#xA;= 0.995V to 1.005 V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However - if I measure the result on the 10V range then 0.5% of 10V = 0.5% of Full Scale&lt;br&gt;&#xA;= 0.05V.&#xA;So 1V +/- 0.5% of FS&lt;br&gt;&#xA;= 0.95V to 1.05V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the 100V range, 1V +/- 0.5% FS lies in the range&lt;br&gt;&#xA;0.5V to 1.50 V. !!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for specifying results in this manner is that the error experienced on a given range tends to largely be constant regardless of the actual reading. So, as the input gets smaller the error becomes increasingly large in proportion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So eg on a 100V range a reading of 0.5V  +/- 0.5% FS lies in the range&lt;br&gt;&#xA;0 to 1V !&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-02-18T08:26:28.980" LastActivityDate="2015-02-18T08:26:28.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1720" PostTypeId="2" ParentId="1712" CreationDate="2015-02-18T08:45:45.623" Score="4" Body="&lt;p&gt;The capacitors serve two distinct functions&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They serve as &quot;reservoir&quot; capacitors. If the regulator is somewhat&#xA;remote from the energy source then sudden current peaks will cause&#xA;$V_{in}$ to dip due to V = I x R drop in the wiring. $C_{in}$ serves&#xA;to provide current for such peaks. $C_{out}$ acts similarly. It is&#xA;charged to Vout_regulator in normal operation and if a heavy&#xA;transient load exceeds the regulator's ability $C_{out}$ acts as a&#xA;short term energy source. Both these roles are the same as for any&#xA;conventional 3 terminal regulator.&lt;/li&gt;&#xA;&lt;li&gt;They (usually just $C_{out}$) serve to provide &quot;loop stability&quot; or phase&#xA;margin in some cases. A 3 terminal regulator is a high gain&#xA;amplifier with negative feedback. Under certain circumstances the&#xA;phase shift in the amplifier can become such that the feedback is&#xA;converted to positive feedback and you have a quite capable&#xA;oscillator.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Adding $C_{out}$ modifies the &quot;transfer function&quot; of the overall amplifier (that we see as a regulator).&lt;br&gt;&#xA;The data sheet will specify capacitance values for $C_{out}$ which allow the overall system to be stable.&lt;br&gt;&#xA;As well as a range of capacitance values $C_{out}$ often requires a specific ESR (equivalent series resistance) range. Too much or too little ESR and again you have positive feedback and oscillation. The &quot;locus&quot; of ESR and capacitance values is always given in the datasheet when it matters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_{in}$ is not usually involved in stability calculations. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-02-28T03:27:02.030" LastActivityDate="2015-02-28T03:27:02.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1721" PostTypeId="2" ParentId="1712" CreationDate="2015-02-18T11:03:18.207" Score="1" Body="&lt;p&gt;The ESR of the Output capacitor is important for setting up the a pole in internal closed-loop system of LDO Regulator. This close loop is responsible for controlling the current allowed depending on the feedback voltage and reference voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ESR of capacitor helps maintain stability of the LDO control loop. The output capacitor also can help manage rapid changes in load current (transient responses). Using a larger value helps improve transient response of the LDO but this also increases start-up time of the LDO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also changes in output capacitor has impact in the input capacitor too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to place a good capacitor based on the data sheets recommended capacitance and ESR values. Ceramic capacitors using X5R or X7R dielectric is a good choice. This is because they have good temperature stability and good low voltage co-efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.analog.com/media/en/Other/Miscellaneous/PM_ldo_design_08451b.pdf&quot; rel=&quot;nofollow&quot;&gt;The Fundamentals of LDO Design and Application&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=cM7t1Mpu7s4&quot; rel=&quot;nofollow&quot;&gt;Linear and LDO regulators and Switch Mode Power Supply Tutorial&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-02-21T14:59:34.607" LastActivityDate="2015-02-21T14:59:34.607" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1722" PostTypeId="1" AcceptedAnswerId="2152" CreationDate="2015-02-18T12:02:45.663" Score="10" ViewCount="892" Body="&lt;p&gt;Most introductory books on control theory usually start their &lt;a href=&quot;http://en.wikipedia.org/wiki/State_observer&quot;&gt;state observation&lt;/a&gt; part by introducing the Luenberger observer, and after that they might continue by introducing the &lt;a href=&quot;http://en.wikipedia.org/wiki/Kalman_filter&quot;&gt;Kalman filter&lt;/a&gt;. When reading papers from academic journals I have also come across all kinds of fancy nonlinear estimation methods (sliding-mode observers, passive observers, etc...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, from my short experience, it seems like there is only one type of state-estimation method used outside academia: the Kalman filter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my observation correct or are other types also being used? If so, why is the Kalman filter so predominantly used over other state observers? Is it because it's fairly easy to implement?&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="6790" LastEditorDisplayName="user16" LastEditDate="2020-05-04T08:02:54.683" LastActivityDate="2020-05-04T08:02:54.683" Title="Types of state observers/estimators actively used in the industry?" Tags="|control-engineering|kalman-filters|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1723" PostTypeId="2" ParentId="1706" CreationDate="2015-02-18T13:49:05.383" Score="13" Body="&lt;p&gt;&lt;strong&gt;Parked vehicles vs moving vehicles&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Closely spaced parked (or slow moving) vehicles are definitely more onerous, as stated on page 89, Appendix 2.A, Clause 2.A.1 of &lt;a href=&quot;http://www.nra.co.za/content/TMH7_Parts_1_and_2.pdf&quot;&gt;the South African bridge design code TMH7&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;It is generally accepted and can readily be shown that except in the very small span range, the worst loading condition occurs under&#xA;  congested (bumper to bumper) conditions caused by a traffic blockage&#xA;  and that the dispersion of traffic at speed caused by increased&#xA;  vehicular inter-spacing, more than off-sets the effects of impact.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, this is only true by inspection for an unlimited number of vehicles of the same weight per unit length. If an unusual extremely heavy vehicle is crossing a bridge, then we can’t apply this rule, as we only have one of this vehicle and it can’t be bumper to bumper with itself. So we turn to bridge design codes to consider relevant configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bridge design codes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to bridge design every country seems to come up with their own bridge loading in their design code. I have read a &lt;em&gt;lot&lt;/em&gt; of these codes - a significant proportion of my job is programming these bridge loading codes into bridge design software. Most design codes have various loading configurations to be checked, although they rarely specify that they are for &quot;parked vehicles&quot; or &quot;moving vehicles&quot;. The closest I know of is the United Kingdom's &lt;a href=&quot;http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol3/section4/bd8611.pdf&quot;&gt;BD86&lt;/a&gt; standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BD86 is for assessment of existing UK bridges to carry exceptionally heavy vehicles (named SO or SOV loads). Clauses 3.20 to 3.25 gives two different configurations. Either (i) the SO/SOV moves at normal speed, with a dynamic factor, and normal loading is not allowed within 25m of the SO/SOV; or (ii) the SO/SOV moves at low speed, with no dynamic factor, but the normal loading is allowed closer (not within 5m).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about all the other design standards across the world? On the basis that I haven't heard of any bridges failing due to the standard design load being too low, I think it is a safe assumption that the design standards definitely consider the worst case. So, whatever it is that is designed for, it must represent the worst case out of either moving load or closely spaced static load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most countries' highway loading standards consider two configurations which must be designed for: a configuration of a single vehicle (or a single vehicle per lane), and a configuration of a uniformly distributed load (i.e. a load per area that is applied wherever it adds to the effect being designed for). This second configuration could be the closely spaced parked vehicles, or it could represent many lorries/trucks spaced apart but their weight averaged out over the total length. I suspect that it may in fact represent both of these situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important fact here is the relative weight of a lorry/truck compared to a car. In the UK, a normal lorry (i.e. one not requiring special permission for its journey) has a maximum weight of 44 metric tonne, and has a length of about 12m. This gives 3.7t/m. A typical car (I've picked the Vauxhall Astra) is about 2 tonne and 4.5m long giving 0.45t/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest that in a moving load situation you might get lorries/trucks fairly well spaced out, but in a traffic jam situation, when lorries and cars tend to mix, the lorry/truck spacing is about the same because the gaps are filled with cars. With the cars producing little load per length compared to a lorry/truck, the average load per unit length &lt;em&gt;may&lt;/em&gt; be about the same whether traffic is moving or stationery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it comes at it from a different angle, I believe AASHTO's Manual for Bridge Evaluation; commentary clause C6B.7.2.2 implies something like this. It says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;The probability of having a series of closely spaced heavy vehicles of the maximum allowable weight becomes greater as the maximum allowed weight for each unit becomes less. That is, it is more likely to have a train of light-weight vehicles than to have a train of heavy-weight vehicles.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;assume&lt;/em&gt; this can be applied to the parked/moving situation. That is, the more vehicles you have on the bridge at once, the lower the probable average weight of vehicle.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2015-03-03T13:10:52.827" LastActivityDate="2015-03-03T13:10:52.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1724" PostTypeId="2" ParentId="1715" CreationDate="2015-02-18T14:34:43.140" Score="4" Body="&lt;p&gt;According to &lt;a href=&quot;http://awo.aws.org/wp-content/uploads/2014/stainlessconference/Campbell-AnUpdateonAWSD16StructuralWeldingCode-Stainless%20Steel.pdf&quot; rel=&quot;nofollow&quot;&gt;this slideshow&lt;/a&gt; it looks like a weld between stainless and carbon or low-alloy steels would be within the scope of D1.6 entirely. It would not be prequalified so it would have to be qualified by testing. Table 4.2 (reproduced in the slide show) covers the test types if you want to qualify with a CJP, but you would need a full copy of D1.6 to see the acceptance criteria and specific geometry for each test. If the fillet weld is all you need to cover, you could use table 4.4 and qualify with a fillet weld only (again, geometry is going to have to come from the full code.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, this is a relatively young standard, and the third edition is expected out this year, so there's potential for some change.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-02-18T14:34:43.140" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1725" PostTypeId="2" ParentId="703" CreationDate="2015-02-18T16:03:03.480" Score="3" Body="&lt;p&gt;I think going optical is the only best solution.&#xA;How about handheld interferometer? The reflection from interfoerometers are the best, as they give nanometer precision. &#xA;If you could mount reflective surfaces on the object that moves, by simply capturing the reflected interferrometer patterns you can find the path difference and feedback to the displacement pretty instantaneously.  However it is restricted to only linear motion.&lt;/p&gt;&#xA;" OwnerUserId="587" LastActivityDate="2015-02-18T16:03:03.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1726" PostTypeId="2" ParentId="1722" CreationDate="2015-02-18T16:46:52.650" Score="6" Body="&lt;p&gt;I know that Luenberger and sliding-mode observers are used in field oriented control algorithms to estimate rotor position and speed of permanent magnet synchronous machines (PMSM).  &lt;a href=&quot;http://www.ti.com/lit/an/sprabq3/sprabq3.pdf&quot;&gt;Here is a white paper by TI&lt;/a&gt; where they mention the sliding-mode observer.  And &lt;a href=&quot;http://www.freescale.com/files/microcontrollers/doc/ref_manual/DRM148.pdf&quot;&gt;here is a white paper by Freescale&lt;/a&gt; that mentions the Luenberger observer.  &lt;a href=&quot;http://www.freescale.com/files/microcontrollers/doc/ref_manual/DRM099.pdf&quot;&gt;Here is another from Freescale&lt;/a&gt; talking about the sliding-mode observer.  &lt;/p&gt;&#xA;" OwnerUserId="198" LastActivityDate="2015-02-18T16:46:52.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1727" PostTypeId="2" ParentId="1705" CreationDate="2015-02-18T19:09:12.470" Score="4" Body="&lt;p&gt;It is likely that the noise of the motorcycle is causing the sensor to activate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://static.catalogs.rockwellautomation.com/pub-files/13676018925183f2e400f4c/pub/Ultrasonic-Sensing/Of/page.pdf&quot; rel=&quot;nofollow&quot;&gt;listing&lt;/a&gt; of the advantages and disadvantages of ultrasonic sensors lists one of the disadvantages as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While ultrasonics exhibit good immunity to background noise, these sensors are still likely to falsely respond to some loud noises, like the hissing sound produced by air hoses and relief&#xA;  valves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A loud motorcycle could fall into that same level of noise (sound pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other &lt;a href=&quot;https://dlnmh9ip6v2uc.cloudfront.net/datasheets/Kits/HRXL-MaxSonar-WR_Datasheet.pdf&quot; rel=&quot;nofollow&quot;&gt;sensor spec sheets&lt;/a&gt; that I have seen have phrases like:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Many acoustic noise sources will have little to no effect on the reported range...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This would likely be why the sensor is not going off all the time, but is triggered by loud noises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The noise from a motorcycle is about &lt;a href=&quot;http://www.gcaudio.com/resources/howtos/loudness.html&quot; rel=&quot;nofollow&quot;&gt;100dB&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-18T19:09:12.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1728" PostTypeId="2" ParentId="1713" CreationDate="2015-02-18T22:07:29.253" Score="3" Body="&lt;p&gt;I think you may be misinterpreting aerial wind turbines as a type of aircraft. The FAA section of the Code of Federal Regulations defines an aircraft as &lt;a href=&quot;http://www.ecfr.gov/cgi-bin/text-idx?SID=f5ac1e30d58cfc9185ccd1cb128c617b&amp;amp;node=pt14.1.1&amp;amp;rgn=div5#se14.1.1_11&quot; rel=&quot;nofollow&quot;&gt;a device that is used or intended to be used for flight in the air.&lt;/a&gt; I suppose the real debate comes down to what you define as &quot;flight,&quot; which the CFR does not deal with. While an aerial wind turbine certainly is supporting itself in the air through aerodynamics, it's still tethered to the ground and I wouldn't exactly call that &quot;flight&quot; in the same sense as it's being applied to the aircraft being regulated in these codes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is really more complex than anyone can answer right now, because I have a feeling that in the not-so-distant future (&amp;lt;10 years?) a company manufacturing aerial wind turbines will submit some sort of request for a ruling to the FAA on whether or not they're aircraft, and if they disagree, they'll file a suit and go to court over the definition of &quot;flight.&quot; But for now, let's assume they aren't aircraft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditional wind turbines are regulated by &lt;a href=&quot;http://www.ecfr.gov/cgi-bin/text-idx?SID=f5ac1e30d58cfc9185ccd1cb128c617b&amp;amp;node=pt14.2.77&amp;amp;rgn=div5&quot; rel=&quot;nofollow&quot;&gt;Title 14, Part 77&lt;/a&gt; of the CFR, and I think the aerial wind turbines would be regulated with the same code. They're merely an obstruction that is fairly high and has some capacity to move around with the wind, depending on how tightly they're tethered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ecfr.gov/cgi-bin/text-idx?SID=f5ac1e30d58cfc9185ccd1cb128c617b&amp;amp;node=pt14.2.77&amp;amp;rgn=div5#se14.2.77_117&quot; rel=&quot;nofollow&quot;&gt;Section 17&lt;/a&gt; describes an obstruction as anything higher than 499 feet above ground level, with lower height restrictions near airports, for obvious reasons. However, &lt;a href=&quot;http://www.ecfr.gov/cgi-bin/text-idx?SID=f5ac1e30d58cfc9185ccd1cb128c617b&amp;amp;node=pt14.2.77&amp;amp;rgn=div5#se14.2.77_19&quot; rel=&quot;nofollow&quot;&gt;Section 9&lt;/a&gt; states that the FAA must be notified of any construction going higher than 200 feet. The key here is that the regulations say absolutely nothing about whether or not you are legally allowed to build something over 200 feet. It simply says that you should notify the FAA, and they will make a determination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are below 500 feet, the code doesn't appear to be explicit on this fact, but it seems you'd find little argument with your request. Over 500 feet, they'll look into it a little more deeply, because that is officially an obstruction. But they're still free to say &quot;We don't think this will be an issue, go ahead.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So do the regulations explicitly limit the size? No. They make no limitations on what you can build, just that you have to get approval once you pass certain restrictions, dependent on your location and proximity to airports. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-02-18T22:07:29.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1730" PostTypeId="1" AcceptedAnswerId="1731" CreationDate="2015-02-18T02:05:22.357" Score="7" ViewCount="404" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Tsar_Bomba&quot; rel=&quot;nofollow&quot;&gt;Tsar Bomba&lt;/a&gt; packed a yield of 50 megatons into a package 2.1m in diameter by 8m long.  Assuming that the lead tamper modification was not used (which boosts the yield of the Tsar Bomba design to 100 megatons), how much would the yield scale as a multiple-stage thermonuclear warhead (assume that extra stages are added as-needed) was scaled up, say to 3.5m diameter and 20m length?&lt;/p&gt;&#xA;" OwnerUserId="964" OwnerDisplayName="Shalvenay" LastEditorUserId="131" LastEditDate="2015-02-19T03:30:58.213" LastActivityDate="2015-02-19T04:00:08.970" Title="How does thermonuclear warhead yield scale with size?" Tags="|nuclear-technology|nuclear-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1731" PostTypeId="2" ParentId="1730" CreationDate="2015-02-19T02:58:54.173" Score="5" Body="&lt;p&gt;Disclaimer: I am not a nuclear engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've read, nuclear weapon design is a very complex science. I don't think that you could reliably describe a scaling rule. What you &lt;strong&gt;could&lt;/strong&gt; do is look at some historical data and extrapolate from there. However, extrapolation is dangerous™ and you should not assume that something will continue to scale in the same way as it has for your existing data points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are (hopefully) unlikely to obtain empirical data from a 50+ megaton detonation, extrapolation is the best that we can do. &lt;a href=&quot;http://en.wikipedia.org/wiki/Nuclear_weapon_yield&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; has some information that we can use to perform these calculations. They do not list dimensions but they do list a number of masses. They also provide yield-per-mass numbers (6 MT/ton or 25 TJ/kg theoretical, 5.2 MT/ton practical). Based on the page you cited, the Tsar Bomba had a yield-per-mass of 8.89 TJ/kg. This is clearly significantly below the theoretical max. As with many engineering devices, it is likely that at the 50 MT scale you cannot achieve anything close to theoretical max efficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you &lt;strong&gt;assume&lt;/strong&gt; that you can maintain this level of efficiency for a device of a greater size (and therefore a greater mass) you can multiply 8.89 times the mass in kg of the hypothetical device and this would give you a reasonable &lt;a href=&quot;http://en.wikipedia.org/wiki/Scientific_Wild-Ass_Guess&quot; rel=&quot;noreferrer&quot;&gt;WAG&lt;/a&gt; concerning the yield. I would guess that your efficiency would continue to decline so you can account for that by rounding down your answer to make it a &lt;strong&gt;little&lt;/strong&gt; more accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="131" LastActivityDate="2015-02-19T02:58:54.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1732" PostTypeId="2" ParentId="433" CreationDate="2015-02-19T03:21:06.583" Score="3" Body="&lt;p&gt;You should be able to create a non-damaging cutting system if you integrate it with your drive system to control cutting element location. For example, if you use 3+ rubber wheels evenly spaced around the diameter of the device you should be able to reliably fix the device in a static location relative to the pipe walls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend 4+ high-speed grinding-type cutters place at say the 0, 90, 180, and 270 degree points. Grinding is slower than cutting but you can remove a larger area of debris more safely than with a cutting device. A grinding device also simplifies debris removal (should wash down the drain). As an alternative, you could have a single (optionally pointed) large grinding tip attached to the front of the device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case, allow at least a quarter-inch clearance between the cutters and the wall to prevent damage at turns. Make sure that the cutters do not extend too far in front of the wheels or you will need to increase your clearance to compensate based on anticipated pipe turns. You can adjust the clearance based on your tolerance for potential damage, but with clay pipes I would be &lt;strong&gt;very&lt;/strong&gt; conservative!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you want to avoid the additional engineering challenge of &lt;a href=&quot;http://en.wikipedia.org/wiki/Trap_%28plumbing%29&quot; rel=&quot;nofollow&quot;&gt;drain traps&lt;/a&gt;, you should plan to insert the device directly into the pipe below a toilet. Although, since you mention a 4&quot; pipe, perhaps you will be inserting directly into a cleanout on the target pipe.&lt;/p&gt;&#xA;" OwnerUserId="131" LastActivityDate="2015-02-19T03:21:06.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1733" PostTypeId="1" AcceptedAnswerId="1740" CreationDate="2015-02-19T03:24:30.607" Score="3" ViewCount="310" Body="&lt;p&gt;The situation is a steel baseplate that has to be anchored to concrete to resist lateral loads.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7yWzS.png&quot; alt=&quot;base plate with 8 holes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lateral load on the base plate is great enough that more than one or two anchor rods are required. The base plate has room for 8 anchor rods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The holes in the base plate will be somewhat over-sized per &lt;a href=&quot;https://www.aisc.org/DynamicTaxonomyFAQs.aspx?id=1840&quot; rel=&quot;nofollow noreferrer&quot;&gt;AISC&lt;/a&gt; to aid in setting the base plate over over the anchor rods. The anchor rods have already been cast into the concrete. The nuts on the anchor rods will not be tightened enough to consider the connections to be slip-critical. There will be some movement of the base plate required before it engages the anchor rods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variations in the placement of the anchor rods and the hole locations will mean that not all of the anchor rods will be engaged at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, how do you determine how many anchor rods can be considered to be effective in resisting the load? The number of anchor rods will directly effect the size of each anchor rod.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="368" LastEditDate="2015-03-05T22:47:15.907" LastActivityDate="2015-03-05T22:47:15.907" Title="How to determine the number of effective anchors in a base plate?" Tags="|structural-engineering|bolting|equipment-selection|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1734" PostTypeId="2" ParentId="389" CreationDate="2015-02-19T03:44:03.880" Score="7" Body="&lt;p&gt;As with any new technology, the cost is the big driver here. In addition, these devices produce electricity which is a form of energy that a typical internal combustion automobile can only utilize for ancillary equipment. This would effectively improve fuel efficiency but the gain would be relatively minor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers are generally reluctant to use expensive new technologies that are relatively untested when existing methods are sufficient for achieving the goal. In this case, most automobile manufacturers strive to produce a cost-competitive product. The people who are &lt;strong&gt;most&lt;/strong&gt; concerned with fuel efficiency will tend to consider a hybrid (or all-electric) vehicle instead of one powered by a standard internal combustion engine. Most other consumers would prefer to purchase a less-expensive vehicle over one that could slightly increase their fuel efficiency.&lt;/p&gt;&#xA;" OwnerUserId="131" LastActivityDate="2015-02-19T03:44:03.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1735" PostTypeId="2" ParentId="1730" CreationDate="2015-02-19T04:00:08.970" Score="3" Body="&lt;p&gt;I am also not a nuclear engineer, but here are my thoughts on it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My impression is that you might have trouble scaling a weapon in an arbitrary dimension but that they should scale basically linearly with mass assuming an acceptable configuration of the weapon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice you reach a maximum size for your h-bomb based on the input energy available and various mechanics that they try to keep secret.  However, to detonate an h-bomb you simply need an adequate energy flux, the source of that energy flux is irrelevant.  In practice it's an fission bomb because nothing less than this can provide the needed energy.  However, there's nothing about the fission part of it that's actually needed--a fusion detonator works even better.  (Besides, the fusion stage of an h-bomb actually contains a simplified fission device in it's heart--there is a rod of Pu-239 there, after the energy flux has crushed the sides of the fusion device in then the result gets crushed lengthwise, turning the rod into a critical mass and setting off the crushed lithium deutride.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you can simply stack up the fusion stages of h-bombs one after another, the energy flux from one sets off the next.  You can also arrange multiple fusion devices around the fission trigger.  Just watch your configuration that you don't destroy a stage before it's fired.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2015-02-19T04:00:08.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1736" PostTypeId="2" ParentId="537" CreationDate="2015-02-19T04:14:15.227" Score="3" Body="&lt;p&gt;As &lt;a href=&quot;https://engineering.stackexchange.com/users/86/dave-tweed&quot;&gt;Dave Tweed&lt;/a&gt; mentioned, this is a simple example of a diverter valve. It allows flow to be directed to one or the other output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.ductingsystems.com/parts/diverter_valve_manual_drawing.gif&quot; alt=&quot;Diagram from valve manufacturer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ductingsystems.com/nordfab-parts-manual-diverter-valve.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Valve manufacturer site&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="131" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-02-19T04:14:15.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1737" PostTypeId="1" CreationDate="2015-02-19T04:51:53.157" Score="10" ViewCount="1992" Body="&lt;p&gt;I hear that designing the thermal efficiency of such systems is very difficult. I'm not sure why, though, and I am interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, I bet the heat is somehow a function of the total power in the system. On the other hand, as individual bits are flipped, I imagine the heat migrates around the die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does heat move around the die, and how does this affect the cooling of the CPU? Are specific compensations made to accommodate the movement of heat?&lt;/p&gt;&#xA;" OwnerUserId="590" LastEditorUserId="368" LastEditDate="2015-07-02T22:37:23.190" LastActivityDate="2015-07-02T22:37:23.190" Title="How are the thermodynamics of CPUs and other chips handled?" Tags="|electrical-engineering|heat-transfer|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1738" PostTypeId="2" ParentId="389" CreationDate="2015-02-19T07:19:26.787" Score="14" Body="&lt;p&gt;There are good reasons why, although most of your links are from several years ago, very little has come of any of them. The economics and the engineering just aren't favourable. Part of the art of an engine is dumping the heat as quickly as possible; furthermore, heat is a low-quality form of energy, so once your energy is heat, you've already lost all your best opportunities to get more work done with it. Let's look at those two things in detail.&lt;/p&gt;&#xA;&lt;h3&gt;getting rid of the heat quickly is the name of the game&lt;/h3&gt;&#xA;&lt;p&gt;The exhaust is &lt;em&gt;designed&lt;/em&gt; to take heat and the products of combustion away from the engine. The radiator does a similar job of removing and dissipating the heat.&lt;/p&gt;&#xA;&lt;p&gt;Dumping heat quickly is crucial in a heat engine, as the efficiency depends on the temperature of the cold reservoir, and the delta to the higher temperature.&lt;/p&gt;&#xA;&lt;p&gt;So anything you put in the way, such as your proposed energy harvesting device, will slow the rate at which heat leaves the engine. This not only reduces efficiency (it seems bizarre that you've excluded efficiency as a consideration: why else would you look at energy harvesting anyway?)  It will also raise the equilibrium temperature of the car's working parts, shortening its life.&lt;/p&gt;&#xA;&lt;p&gt;If fuel is that valuable that energy harvesting looks attractive, then it's worth making the car more efficient in the first place, so that there was less waste heat: first, reduce the consumption of high-value energy, before trying to recycle low-value energy. And that brings me to ...&lt;/p&gt;&#xA;&lt;h3&gt;energy versus exergy&lt;/h3&gt;&#xA;&lt;p&gt;Heat is, in a large proportion of cases, a waste product. It's almost always the least useful form of energy. That's really what the Carnot efficiency limit is telling you: that to get any work out of low grade heat, you can only do so with very low efficiency; that is, almost all of the heat will stay as heat.&lt;/p&gt;&#xA;&lt;p&gt;When doing engineering with heat and other forms of energy, it's very useful to build up an intuition to distinguish between &lt;em&gt;energy&lt;/em&gt; (the thing measured in joules) and &lt;em&gt;exergy&lt;/em&gt; (the thing that gets work done). The form that energy is in, determines how much work it can do. Chemical energy, such as that in fuel, can do huge amounts of work efficiently - it has very high exergy. &lt;strong&gt;But the same amount of energy as heat can do much less work&lt;/strong&gt; - it has very low exergy. Internal combustion engines are so inefficient because they take a high-grade form of energy (chemical), and immediately convert into a low-grade one (heat).&lt;/p&gt;&#xA;&lt;p&gt;The humble alternator is very good at its job. Solid-state thermo-electric generators simply don't come close, not in cost, not in performance. Using the relative rotation of a conductor and a magnet is a very effective means of turning heat into electricity: that's why it's used in power stations, cars, bike dynamos, and many other uses.&lt;/p&gt;&#xA;&lt;p&gt;So although energy harvesting looks clever, it's bad engineering: its trying to fix one symptom, rather than address the underlying cause.&lt;/p&gt;&#xA;&lt;p&gt;If you want more work out of those joules, then get that work done &lt;em&gt;before&lt;/em&gt; those joules are in the form of heat.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-02-19T07:19:26.787" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1739" PostTypeId="1" CreationDate="2015-02-19T16:39:22.750" Score="1" ViewCount="921" Body="&lt;p&gt;I am designing a stainless metal base plate to mount an orientation critical sensor. The sensor has two 6H7 6mm deep holes to align it. One of the holes is a slot to allow for adjustment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensor manufacturer prescribes a 6g6 pin (loose fit) but that would result in a nonstandard transition fit on the hole in my plate if I'm correct. Could I simply use a spring dowel instead?&lt;/p&gt;&#xA;" OwnerUserId="270" LastEditorDisplayName="user16" LastEditDate="2015-02-26T18:32:12.630" LastActivityDate="2015-02-26T18:32:12.630" Title="Tolerance Fit with Spring Dowels" Tags="|mechanical-engineering|fasteners|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1740" PostTypeId="2" ParentId="1733" CreationDate="2015-02-19T21:41:54.880" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://www.aisc.org/uploadedFiles/Research/Research_Reports/Kanvinde%20-%20Shear%20Transfer%20in%20Exposed%20Column%20Base%20Plates.pdf&quot; rel=&quot;nofollow&quot;&gt;This report (large PDF warning)&lt;/a&gt; on base plate design says that the AISC &lt;em&gt;Steel Design Guide I&lt;/em&gt; instructs the use of plate washers fillet welded to the top of the base plate, with the inner diameter of the washer being much closer to the rod diameter, to ensure initial engagement of the rod. Specifically, I found that information in Sections 2.3.2 (pp.8-10), 3.3.1.2(p.21), and a visual example in Figure 3.9. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While that doesn't directly answer your question, it seems the answer is &quot;You can't determine that, so we put in other design elements to account for it.&quot;&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-02-19T21:41:54.880" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1741" PostTypeId="1" AcceptedAnswerId="2061" CreationDate="2015-02-19T23:37:42.743" Score="4" ViewCount="4697" Body="&lt;p&gt;Imagine I have an important axle in the gearbox of a motor. I want to support it with ordinary &lt;a href=&quot;https://en.wikipedia.org/wiki/Ball_bearing&quot; rel=&quot;nofollow&quot;&gt;ball bearings&lt;/a&gt; at several places along its length. I need both the outer and inner races to stay at fixed diameters at a fixed distance from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I &lt;em&gt;can&lt;/em&gt; change the number of balls in between the races. I can choose any number of different bearings (though each one has to have a certain diameter, so there's actually a limit). I could squeeze an arbitrary number in, so long as that number is below the maximum and can still support the inner and outer rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What determines the optimal number of balls used?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="368" LastEditDate="2015-02-20T00:11:09.203" LastActivityDate="2016-03-15T04:44:54.760" Title="What determines the optimal number of balls between races in a bearing?" Tags="|mechanical-engineering|bearings|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1742" PostTypeId="1" CreationDate="2015-02-20T01:07:56.827" Score="28" ViewCount="6229" Body="&lt;p&gt;&lt;em&gt;Subtitled:&lt;/em&gt; &lt;strong&gt;Ah, strong am I with the Force, but how do I measure it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all due apologies for paraphrasing &lt;a href=&quot;http://www.imdb.com/character/ch0000015/quotes&quot; rel=&quot;noreferrer&quot;&gt;Yoda&lt;/a&gt;, I am trying to measure a force.  Fortunately, it's an easier force to measure than &lt;a href=&quot;http://starwars.wikia.com/wiki/Midi-chlorian/Canon&quot; rel=&quot;noreferrer&quot;&gt;midi-chlorians&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I tutor others in karate, one of the things I emphasize is a focus on the technique behind the basic strikes.  Experience has shown that the proper technique can yield a lot more force behind any strike.  The problem I run into is that even though a student claims they can &quot;feel&quot; the difference in a strike being more powerful, they often revert back to their incorrect technique because they haven't convinced themselves that the difference is truly there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to design a system that measures the force behind a strike.  In particular, I need to identify some sensor at the point of impact that can be used.  National Geographic has done &lt;a href=&quot;http://en.wikipedia.org/wiki/Fight_Science&quot; rel=&quot;noreferrer&quot;&gt;something very similar&lt;/a&gt;, but the equipment they use, such as crash test dummies, is prohibitively expensive.  Likewise, I don't need the resolution or accuracy provided by that equipment.  Whether someone hit with 100 or 101 lb&lt;sub&gt;f&lt;/sub&gt; of force is immaterial.  I want to demonstrate when then go from 100 to 200.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What sensor or system could I use in order to measure the force at impact from various karate strikes?  Ideally, I could mount the device in a  &lt;a href=&quot;http://en.wikipedia.org/wiki/Focus_mitt&quot; rel=&quot;noreferrer&quot;&gt;focus mitt&lt;/a&gt;, or similar, and be able to attach the pad to an adjustable system so various strikes could be measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My principle design constraints are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;low(er) cost&lt;/li&gt;&#xA;&lt;li&gt;repeatable measurements&lt;/li&gt;&#xA;&lt;li&gt;range of 0–2,000 lb&lt;sub&gt;f&lt;/sub&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for a complete system design, which would include all the monitoring electronics.  I'm just trying to focus on a sensor that can provide a measurable output of some sort, such as voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that I plan on running wires to the device, so I already have an easy way to power the system.  I may go wireless at some point, but I would use a battery then and I want this question to remain focused on the sensor itself.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2016-04-06T00:28:20.000" LastActivityDate="2016-04-06T00:28:20.000" Title="How to quantify martial arts strikes?" Tags="|sensors|measurements|biomechanics|" AnswerCount="10" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1743" PostTypeId="1" CreationDate="2015-02-20T03:14:42.297" Score="4" ViewCount="330" Body="&lt;p&gt;&lt;strong&gt;Combustion Air&lt;/strong&gt; is the air that is required to provide complete combustion for a fuel source. Enough oxygen is required so that there is not an issue with &lt;a href=&quot;http://en.wikipedia.org/wiki/Carbon_monoxide_poisoning&quot; rel=&quot;nofollow&quot;&gt;carbon monoxide&lt;/a&gt;. This is an issue in houses and anywhere that has a furnace, water heater, boiler, etc. that isn't electric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the codes and references (&lt;a href=&quot;http://www.nationalboard.org/index.aspx?pageID=164&amp;amp;ID=191&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://www2.iccsafe.org/states/phoenix/Phoenix_IMC/PDFs/Chapter%207.pdf&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;http://www.comfort-calc.net/Combustion_Air_Calculation.html&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;) that I have seen use either the volume of the room that contains the equipment or an area of free-flowing vent from the outside in a ratio to the BTUs or Watts of the equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither of these measures seem to be directly related to the &lt;a href=&quot;http://en.wikipedia.org/wiki/Combustion#Chemical_equations&quot; rel=&quot;nofollow&quot;&gt;chemical equation&lt;/a&gt; of combustion.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Comparing the volume of the room doesn't seem to take into consideration a replenishment of the oxygen from anywhere.&lt;/li&gt;&#xA;&lt;li&gt;Specifying an area of vent seems to ignore the speed of air through the vent. It also doesn't seem to relate to the size of the chimney.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the logic behind either specifying a room volume or vent area for combustion air?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="41" LastEditDate="2015-02-20T03:21:50.113" LastActivityDate="2016-10-14T10:18:34.360" Title="What is the logic behind combustion air requirements?" Tags="|mechanical-engineering|combustion|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1744" PostTypeId="2" ParentId="1742" CreationDate="2015-02-20T04:05:12.420" Score="10" Body="&lt;p&gt;I think that an &lt;a href=&quot;http://www.pcb.com/Products.aspx?tx=52&quot;&gt;impact force sensor&lt;/a&gt; is what you're looking for. I'm not sure how expensive the linked sensors are so you could also use an &lt;a href=&quot;https://abieneman.wordpress.com/2010/04/04/punch-acceleration-sensor/&quot;&gt;acceleration sensor&lt;/a&gt; and do a little math. The second option is likely to be less accurate but I suspect that it will also be significantly less expensive!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know you said that you didn't want the complete system design but I figured I'd go for the extra credit in the second link. ;)&lt;/p&gt;&#xA;" OwnerUserId="131" LastActivityDate="2015-02-20T04:05:12.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1749" PostTypeId="2" ParentId="1742" CreationDate="2015-02-20T12:32:21.490" Score="4" Body="&lt;p&gt;These are few suggestions to develop the system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest using a polymeric bladder type approach used in Passage Occupancy Detection Systems in automotive industry. This &lt;a href=&quot;http://www.google.com/patents/US6578871&quot; rel=&quot;nofollow noreferrer&quot;&gt;patent&lt;/a&gt; describes the automotive application. The bladder system would be part of the focus mitt, which include a MEMS base differential pressure sensor. When impact force will be detected by the pressure sensor, the data will be processed by the simple microprocessor like a Atmel tiny, MSP430 or a PIC. In case analog pressure sensor is chosen then an Analog Front End (AFE) stage is required, but I2C or SPI options are available too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach will be to use an inertial measurement unit (IMU) which provides two to six degrees of freedom (DOF) couple with a microprocessor like the MSP430 or PIC. These IMU have Gyroscopes and accelerometers. This approach will give more data that can be used for later analysis. Also the communication is done via I2C or SPI to the microprocessor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S6E38.jpg&quot; alt=&quot;Pressure Sensors&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/h7FAd.jpg&quot; alt=&quot;IMU&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/CvEDF.jpg&quot; alt=&quot;PODS&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system will require simple power system like a simple rechargeable button battery. USB power charging system might be your best choice. The USB charging system will enable the data download option. Also some type of energy harvesting mechanism such as &lt;a href=&quot;http://www.mide.com/products/volture/piezoelectric-vibration-energy-harvesters.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;vibration&lt;/a&gt; could be used to generate energy required to power the system, similar to some of the new gym machines such as treadmills, elliptical bikes etc. This can be a great low power application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option would be to connect a wearable such as a fit bit that will not only provide the power source but also data storage, review and analysis. Fit bit will also give the ability connect to a Smart phone of real time analysis and feedback to the student. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going directly from focus mitt to Smart might be a great value added proposition. This can be achieve in many ways. Using Cortex ARM M0 micro with an on-chip BLE might be the best currently available option. Many vendors such as NXP, Freescale, and Cypress offer this option. It is even better is a Cortex ARM M0/M4 + Bluetooth Low Energy/Bluetooth Smart in a module. Below is an example of such device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KgWHq.jpg&quot; alt=&quot;BLE Module 1&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wlSG9.jpg&quot; alt=&quot;BLE Module 2&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/GYXSG.jpg&quot; alt=&quot;BLE Cortex MO block Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these vendors offer small form factor affordable development kit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally to address the VOC the students, providing real-time measurable feedback between correct and incorrect techniques will drive the students desire to attain perfection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.google.com/patents/US6578871&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vehicle occupant weight detection system with occupant position compensation&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www3.panasonic.biz/ac/e/control/sensor/pressure/ps/index.jsp&quot; rel=&quot;nofollow noreferrer&quot;&gt;PS Pressure Sensors&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sparkfun.com/products/10121?gclid=CN6GkMiE8sMCFdcYgQod4bIARg&quot; rel=&quot;nofollow noreferrer&quot;&gt;6 Degrees of Freedom IMU Digital Combo Board&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.seeedstudio.com/depot/Seeed-Micro-BLE-Module-w-CortexM0-Based-nRF51822-SoC-p-1975.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Seeed Micro BLE Module w/ Cortex-M0 Based nRF51822 SoC&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cache.freescale.com/files/rf_if/doc/data_sheet/MKW2xDxxx.pdf?pspll=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kinetis KW2x - 2.4 GHz RF, Analog Mixed Signal, IEEE® 802.15.4, Low Power Microcontrollers (MCUs)&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nxp.com/products/microcontrollers/key_feature/bluetooth/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Products and solutions for Bluetooth Smart connectivity&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cypress.com/?docID=50617&quot; rel=&quot;nofollow noreferrer&quot;&gt;Programmable System-on-Chip (PSoC®) + Bluetooth Low Energy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-02-21T12:21:08.963" LastActivityDate="2015-02-21T12:21:08.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1750" PostTypeId="2" ParentId="1742" CreationDate="2015-02-20T13:08:45.753" Score="10" Body="&lt;p&gt;I think what you want is a &lt;a href=&quot;https://www.google.com/#q=piezoelectric+pressure+sensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;piezoelectric pressure sensor&lt;/a&gt;.  You might be able to steal one from an old bathroom scale, but the sensors in a scale may only be good up to &amp;#126;100 lbs of force.  These piezoelectric transducers convert applied pressure to a small voltage across the two leads.  This pressure can be converted to force by measuring the surface area of the sensor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to read it out you will need to build an amplification stage, probably consisting of a low noise preamp stage and a more robust amplification stage.  The second part is a circuit which holds the maximum value since you are probably most interested in the maximum force applied. You can probably do both the second stage of amplification and any data processing with an &lt;a href=&quot;https://arduino.stackexchange.com/&quot;&gt;Arduino&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to be careful how you mount the piezoelectric transducer to the focus mitt.  You want to make it something that people can punch without transferring too much of the pressure past the sensor.  Once you have it built, you can calibrate it by placing known weights on top of it.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:09.290" LastActivityDate="2015-02-20T13:08:45.753" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1751" PostTypeId="2" ParentId="1742" CreationDate="2015-02-20T14:49:08.577" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://www.inventables.com/technologies/conductive-foam&quot;&gt;Conductive foam&lt;/a&gt; will change it resistance when compressed, so it makes a good, cheap force sensor, as long as you have a system to make many measurements in quick succession and remember the one with the least resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the force you want to measure, you can use several layers above each other, and because you can cut it into any shape, you're not restricted to the (small) surface of a specific sensor. &lt;/p&gt;&#xA;" OwnerUserId="600" LastActivityDate="2015-02-20T14:49:08.577" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1752" PostTypeId="2" ParentId="1742" CreationDate="2015-02-20T15:09:40.877" Score="14" Body="&lt;p&gt;You could just hit a pendulum and measure the angle to which it swings. I have found that three liter soda bottles filled with water and hanging from a rope are good for hitting. Knowing the weight of the bottle, a little trig will give the energy, from which you can get the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: If the transfer of energy from fist to target happens through a short enough time span, then the acceleration of the bottle forward will be approximately the centripetal acceleration through the bottom of the arc. The potential energy at the top of the swing is equal to the kinetic energy at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=ma \to \text{with substitutions} \to F=\frac{m\left(2\left(\frac{mgh}{m}\right)\right)}{r} = \frac{2mgh}{r} = 2mg(1-\cos(\theta))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=v^2/r $$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a$ = centripetal acceleration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v$ = speed along the arc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r$ = radius (rope length)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$PE=mgh$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$PE$=potential energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$ = mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g$ = acceleration due to gravity (about 9.80665 m/s&lt;sup&gt;2&lt;/sup&gt; or 32.1740 ft/s&lt;sup&gt;2&lt;/sup&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h$ = height&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$KE=0.5mv^2 \text {or } v^2=2KE/m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$KE$ = kinetic energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$ = mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v$ = speed along the arc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h=r-r(\cos(\theta))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h$ = height&#xA;$r$ = rope length&#xA;$\theta$ = the greatest angle of the rope from vertical &lt;/p&gt;&#xA;" OwnerUserId="601" LastEditorUserId="2246" LastEditDate="2016-03-22T19:45:18.640" LastActivityDate="2016-03-22T19:45:18.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1753" PostTypeId="1" AcceptedAnswerId="1754" CreationDate="2015-02-20T15:24:06.147" Score="7" ViewCount="4966" Body="&lt;p&gt;It may just be my perception, but it seems like water main breaks (at least in Pittsburgh PA) are more common in the winter during the cold weather. It may just that they are more news worthy in the winter (water+cold=ice &gt; news).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are water mains more likely to break in the winter? If so what can be done to limit or prevent the occurrence? &lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="368" LastEditDate="2015-02-26T16:42:22.803" LastActivityDate="2016-02-27T14:06:31.230" Title="Why do water mains break in the winter?" Tags="|pipelines|infrastructure|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1754" PostTypeId="2" ParentId="1753" CreationDate="2015-02-20T15:43:16.747" Score="15" Body="&lt;p&gt;Cold makes things shrink. Despite being buried &amp;mdash; presumably below the frost line &amp;mdash; both the ground outside the pipe and the water inside the pipe are much colder in winter than in summer. This causes the pipe segments to shrink, putting tension on the joints, and increasing the chances that a weak one will fail. By contrast, warm weather puts the joints under compression, helping to seal any leaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the ground does shift during freeze/thaw cycles (a.k.a. &quot;frost heaves&quot;), and this can cause additional stresses on pipes, even below the frost line &amp;mdash; especially if an old leak has washed out some of the soil supporting the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's really nothing that can be done that I know of to mitigate this for existing pipes. Any preventative measures must be put in place when the pipe is first installed. For example, putting a layer of crushed stone around a pipe helps to decouple it both mechanically and thermally from the surrounding soil.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2015-02-20T17:28:06.673" LastActivityDate="2015-02-20T17:28:06.673" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1757" PostTypeId="2" ParentId="1742" CreationDate="2015-02-20T18:10:52.130" Score="5" Body="&lt;p&gt;I agree 100% with Jeff, the advantage of this is that you could build it for cheap, very, very solid. The height it swings to will give you the energy with the formula $E=mgh$ where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E=energy in Joules,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m=mass in kg,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g=9.81m/s^2,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h=height difference in m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will be the total amount of energy transferred to the target. Unfortunately, this amount of energy is not necessarily representative of how destructive/fight-ending a strike is, as pushing it to that height slowly will require the same energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knock-out power is also represented by the acceleration of the head in the few split-seconds directly after impact, this is what causes a concussion. This could be measured with a circuit that measures the time taken for it to pass two closely spaced points, the first of which should be very close to the initial position of the pendulum. The strike should go through such that the fist/foot goes further than the last of these sensors. If these points are very closely spaced relative to the arc length of the pendulum, you can use $a=2s/t^2$ where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a=acceleration in m/s^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s=distance between sensors in m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t=time in s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have two aims, arguably in order of importance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) make it accelerate as fast as possible&#xA;2.) get it as high as possible&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you achieve aim 1 you may cause a concussion/broken rib, assuming you have enough of 2, meaning you would also need to push through the strike! If you achieve aim 2 with a poor aim 1, you will push away your opponent without causing any real damage. I guess I don't really need to expand on this as you are a karate instructor, apologies for going over things you probably already know, it's just best to be precise and complete. Caution, these values will not be &quot;real&quot; in the sense that due to the distribution of the mass through the moving pendulum you will have a more complicated  formula to calculate the real values. However, you will get numerical values that you can have higher or lower, and you could &quot;calibrate&quot; it by giving it a kick yourself, setting targets for your students. Also, you could try for example a snapping side kick and a thrusting front kick to show what values you need for the two numbers for different goals, ie (cracked rib/flash knockout) vs (create distance/cause them to fall down)&lt;/p&gt;&#xA;" OwnerUserId="606" LastEditorUserId="41" LastEditDate="2015-02-20T18:58:56.590" LastActivityDate="2015-02-20T18:58:56.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1758" PostTypeId="1" AcceptedAnswerId="1762" CreationDate="2015-02-20T19:34:43.573" Score="4" ViewCount="5253" Body="&lt;p&gt;I have a small &lt;a href=&quot;http://en.wikipedia.org/wiki/IP_Code&quot; rel=&quot;nofollow&quot;&gt;IP66&lt;/a&gt; &lt;a href=&quot;http://www.hammondmfg.com/1554FLP.htm&quot; rel=&quot;nofollow&quot;&gt;polycarbonate enclosure&lt;/a&gt; with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Oxygen_sensor#Zirconia_sensor&quot; rel=&quot;nofollow&quot;&gt;Zirconia-based oxygen sensor&lt;/a&gt; which produces significant amounts of heat (the internal sense element runs at 450C using a 1.5W heating element, sensor surface reaches 85C). The enclosure also contains sensitive analog electronics to read the sensor. The residual heat from the sensor is driving the ambient temperature in the enclosure up past 40C (measured on the outside of the enclosure front panel). The oxygen sensor and another gas sensor are only rated for ambient temperatures of 50C maximum, so I'm worried about keeping those in spec. I'm also concerned about needing temperature compensation for the analog circuits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I remove the excess ambient heat from the enclosure while still maintaining an IP66 rating?&lt;/p&gt;&#xA;" OwnerUserId="29" LastEditorUserId="29" LastEditDate="2015-02-24T18:19:28.290" LastActivityDate="2015-02-25T01:48:03.200" Title="How to remove heat from an IP66 enclosure?" Tags="|mechanical-engineering|electrical-engineering|heat-transfer|cooling|" AnswerCount="6" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1760" PostTypeId="1" AcceptedAnswerId="1808" CreationDate="2015-02-20T20:46:22.240" Score="7" ViewCount="786" Body="&lt;p&gt;I'm doing an investigation about building a bridge over a railway line and its &lt;a href=&quot;http://en.wikipedia.org/wiki/Overhead_line#Overhead_catenary&quot; rel=&quot;nofollow&quot;&gt;catenary system&lt;/a&gt;. I'm mainly interested in the electromagnetic environment on the bridge as it is affected by the presence of the catenary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this an issue that I need to be concerned with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no experience in either of these fields and that is exactly why I would like to get insight from people who do work in these particular fields. &lt;strong&gt;To answer some of the questions that have been raised in the comments section:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What order of distance, what voltage? AC/DC? | South Africa or ...? | I'd expect regulatory requirements for unrelated reasons would place you well outside EM interaction range.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ideally I would like the bridge to be of reasonable height such that the pedestrians wouldn't mind taking the stairs. I have freedom to reduce the height of the catenary system for exactly this reason. It is a 25 kV AC catenary system. What is the standard EM interaction range?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What electromagnetic enviroment? You haven't specified any. Will this bridge parallel a power line? Be near a large radio transmitter? What makes you think there are electromagnetic issues here at all?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Railway_electrification_system&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt;, AC high voltage lines &quot;give off&quot; electromagnetic radiation. The bridge is not close to anything really (no power line or large radio transmitter), it just goes over the train track and its catenary.&lt;/p&gt;&#xA;" OwnerUserId="610" LastEditorUserId="282" LastEditDate="2015-03-06T18:49:40.093" LastActivityDate="2015-03-06T18:49:40.093" Title="Building a steel footbridge over a railway line and its catenary" Tags="|electrical-engineering|civil-engineering|bridges|rail|" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1761" PostTypeId="2" ParentId="1753" CreationDate="2015-02-20T20:52:15.977" Score="1" Body="&lt;p&gt;Water mains don't freeze.  The water pressure is regulated along the way (water towers, etc), so temp doesn't affect pressure.  Frozen ground heaves, cold pipes contract and cold pipes are more brittle.&lt;/p&gt;&#xA;" OwnerUserId="611" LastActivityDate="2015-02-20T20:52:15.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1762" PostTypeId="2" ParentId="1758" CreationDate="2015-02-20T21:11:08.280" Score="7" Body="&lt;p&gt;Considering the three methods of moving heat:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Convection&lt;/strong&gt; - To keep the IP66 rating of the enclosure, you can't add any holes for exhaust fans.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Radiation&lt;/strong&gt; - At the temperatures that you are talking about, radiation will not be removing much heat.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Conduction&lt;/strong&gt; - This is a viable alternative since it could work through the walls of the enclosure.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can add fins on the outside of the enclosure and rely on heat being conducted through the walls of the enclosure to the external &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_sink&quot;&gt;heatsink&lt;/a&gt;. This will depend on the material that the enclosure is made from. If it is metal already, this could work as-is. If it is plastic, you may be able to retrofit a metal plate into a wall and still seal it to IP66.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the heat is outside of the enclosure, a fan could be added to help the heatsink function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option may be to have an internal heatsink and then move the heat through the walls of the enclosure via either &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_pipe&quot;&gt;heat pipes&lt;/a&gt; or a circulation closed-loop cooling system. The wall penetrations could be sealed to still meet the IP66 requirements.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-20T21:11:08.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1763" PostTypeId="1" AcceptedAnswerId="1764" CreationDate="2015-02-20T20:42:10.647" Score="3" ViewCount="1358" Body="&lt;p&gt;I am constructing an autonomous, unmanned boat for voyages of several months.  Traditional autopilots use a linear actuator to move the tiller but require several adjustments per minute. They will consume a lot of electricity and probably wear out mid-voyage.  We are looking at wind-vane designs to steer the boat.  Wind-vanes are able to steer for great distances without adjustment.  The boat will be approx. 5m in length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image below, the 'Course Setting' is done by manually rotating the wind-vane and dropping a pin to hold this position of the two discs.  We need a way to rotate the vane fairly precisely and then lock it into position.  We are using 12V DC.  After it is locked, it should draw no power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jhODM.gif&quot; alt=&quot;aux rudder trim tab&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the current design (not our drawing) would have any electromechanical mechanism hanging off of the end of the boat and attached to the rod of the trim tab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything we can use to turn and then lock the wind vane in place?  Preferably something that could be proofed against wind and water.&lt;/p&gt;&#xA;" OwnerUserId="613" OwnerDisplayName="Joseph Geddes" LastEditorUserId="368" LastEditDate="2015-03-05T22:41:45.537" LastActivityDate="2015-05-08T11:53:36.860" Title="Precise rotational movement and locking" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1764" PostTypeId="2" ParentId="1763" CreationDate="2015-02-20T20:55:08.540" Score="2" Body="&lt;p&gt;I am going to suggest that you use a worm-drive linear actuator.  This will allow you to have a linear movement that does your adjustment and then holds that adjustment when power is removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't use a linear actuator, you might have to build something yourself.  Regardless of how you do is, I'd still suggest that you use a worm gear system.  These can be made to be self-locking when the motor is turned off.&lt;/p&gt;&#xA;" OwnerUserId="686" OwnerDisplayName="Dwayne Reid" LastActivityDate="2015-02-20T20:55:08.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1765" PostTypeId="2" ParentId="1739" CreationDate="2015-02-21T04:33:35.677" Score="0" Body="&lt;p&gt;If there is a slot and a round hole for alignment, how would you slide it for adjustment if one of the pins is a slip fit into a round hole. All you will be able to do is rotate it around that one pin! I would take the sensor to a machine shop and mill a parallel slot to the existing slot so that you can slide the sensor on two pins fixed in your baseplate. I assume there are holes for the screws to bolt onto the plate, if not, have the machine shop put them in also! Cheers&lt;/p&gt;&#xA;" OwnerUserId="617" LastActivityDate="2015-02-21T04:33:35.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1766" PostTypeId="2" ParentId="1742" CreationDate="2015-02-21T08:49:33.503" Score="4" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Strain_gauge&quot; rel=&quot;nofollow noreferrer&quot;&gt;strain guage&lt;/a&gt; attached to a target would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike piezzo, they are very durable against impact type force (piezzo crystals tend to shatter when the pressure is applied rapidly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can be attached to pretty much arbitrary &quot;board&quot; which yields very little, so unlike pendulum types they will not falsify the results with force applied slower, or sink the energy into the pendulum's own flexibility. If you use a water bottle, as suggested in a different answer, &quot;stirring the water&quot; with the impact will sink a major part of the energy, only part of it going into propelling the bottle itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And by changing the material you attach them to, you can tune them in to pretty much any force - they measure the stress in the material and basing on the material's Young's Modulus, they can be adapted to sensing anything from sub-Newton forces, to loads on bridges and freight ship hulls.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/JLv4Y.jpg&quot; alt=&quot;Strain guages&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to approach using them practically? You pick some kind of plank, bar, or other surface that will serve as the hit target and will bend somewhat upon the impact, but strong enough that it won't snap, attached firmly with both ends. You glue the guage to the &quot;safe&quot; side, somewhere around the point of impact. Then you use some simple microcontroller (&lt;a href=&quot;http://uk.farnell.com/cypress-semiconductor/cy8ckit-049-42xx/prototype-board-cy8c4245axi-483/dp/2420489&quot; rel=&quot;nofollow noreferrer&quot;&gt;PSoC4&lt;/a&gt; comes to mind, easy to program and with all the remaining hardware needed on-board) and measure peak resistance changes. A simple program that picks peak resistance over certain threshold within given time frame and sends it over RS232 to the PC should be simple enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to calibrate it? Place the contraption horizontally, put a known static weight at the point of impact, measure the difference between the resistance unencumbered and with the weight. 1kg weight will exert 9.806 newtons force, and unless you went very fancy with the material, linear approximation of the Young's Modulus is pretty good; your readouts will be linearly proportional to the force, so a simple proportion will suffice to get the measurement in Newtons.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-02-21T08:49:33.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1767" PostTypeId="1" AcceptedAnswerId="1921" CreationDate="2015-02-21T09:55:56.060" Score="12" ViewCount="7238" Body="&lt;p&gt;According to BS5950, a beam section can be classified as plastic, semi-compact, compact or slender. For the same section area, a H-beam can take axial compression (without buckling) better than an I-beam, and as such, uses a different strut curve in the code:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JQFqH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I understand that a H-beam has a wider flange compared to an I-beam, but at what point, precisely, does this transition from I- to H- occurs? For example, is a 400x300 (depth x width) beam considered a H- or an I-beam?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extracted from BS5950 guide, the following table shows H-beams (also known as universal columns, some of which with depth greater than width. This is the reason why I don't believe the differentiation is so straight forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tRRXw.png&quot; alt=&quot;Section Property Table&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="620" LastEditorUserId="620" LastEditDate="2015-02-28T12:27:37.193" LastActivityDate="2015-03-04T00:42:40.843" Title="At what point does an I-beam becomes a H-beam?" Tags="|civil-engineering|steel|beam|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1770" PostTypeId="2" ParentId="1737" CreationDate="2015-02-21T13:40:52.100" Score="4" Body="&lt;p&gt;All the fundamental issues about the Thermodynamics of the heat sink design are well presented &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_sink&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (make sure not to miss the pretty CFD pics at the bottom of the page).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's not presented here is the larger flowfield structure inside the computer case.  In more recent years, with the push to get CPU speeds at 3+ GHz, there has been more work in designing (1) &lt;em&gt;ducted fans&lt;/em&gt; as well as (2) &lt;em&gt;flow ducts&lt;/em&gt; into the casing which pass air quickly in and out of the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Ducted fans&lt;/em&gt; produce more thrust (or move more air) than regular fans, because the duct causes less flow leakage around the tip which happens to be radially-speaking the highest velocity point of the fan. (This is a similar concept to wing tips on planes).  So, the blade tip is the place on the fan that can move air the fastest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding &lt;em&gt;flow ducts&lt;/em&gt; within the casing, the idea is to use the Bernoulli effect of a nozzle to accelerate the flow over the heat sink so that it can remove heat as fast as possible.  This has especially become popular for overclockers trying to reach speeds of 4+ GHz (e.g. see &lt;a href=&quot;http://www.overclockers.com/ducts-the-cheap-cooling-solution/&quot; rel=&quot;nofollow&quot;&gt;http://www.overclockers.com/ducts-the-cheap-cooling-solution/&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The desire to produce faster and faster CPU's really have pushed the need to design better cooling systems.  Topics such as liquid or nitrogen cooling are not discussed, but are also alternative methods to try to more efficiently cool the CPU, especially for overclocking at speeds above 5 GHz (e.g. see &lt;a href=&quot;http://www.tomshardware.com/reviews/5-ghz-core-i7-980x-overclocking,2665.html&quot; rel=&quot;nofollow&quot;&gt;http://www.tomshardware.com/reviews/5-ghz-core-i7-980x-overclocking,2665.html&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I leave you with something to think about... I once heard the heat produced by a CPU running at 10 GHz is equivalent to the heat of the sun.  There is a pretty good discussion on that topic here: &lt;a href=&quot;http://www.reddit.com/r/askscience/comments/ngv50/why_have_cpus_been_limited_in_frequency_to_around&quot; rel=&quot;nofollow&quot;&gt;http://www.reddit.com/r/askscience/comments/ngv50/why_have_cpus_been_limited_in_frequency_to_around&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="511" LastEditorUserId="511" LastEditDate="2015-02-22T07:45:27.397" LastActivityDate="2015-02-22T07:45:27.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1771" PostTypeId="2" ParentId="1737" CreationDate="2015-02-21T15:37:09.267" Score="4" Body="&lt;p&gt;The thermal system around a modern processor chip is indeed complicated and a major design focus.  For both electrical and economic reasons, it's good to make individual transistors in a processor small and close together.  However, the heat comes from these transistors.  Some is dissipated all the time just because they sit there with power applied.  Another component occurs only when they switch states.  These two can be traded off to some extent when the processor is designed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each transistor doesn't dissipate much power, but millions and millions (literally) crammed together in a small area do.  Modern processors would cook themselves in seconds to 10s of seconds if this heat wasn't actively and aggressively removed.  50-100 W is not out of line for a modern processor.  Now consider that most soldering irons run from less than that, and heat a chunk of metal with about the same surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution used to be to clamp a big heat sink onto the small die.  In fact, the heat sink was a integral part of the overall design of the processor.  The package has to be able to conduct the heat power from the die to the outside, where the clamped-on heat sink can conduct it further and eventually dissipate it to flowing air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is no longer good enough as the power density of these processors has gone higher.  High end processors now either contain some active cooling or a phase change system that moves heat from the die to radiating fins more efficiently than plain old conduction through aluminum or copper did with the old heat sinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases Peltier coolers are employed.  These actively pump heat from the die to someplace else where it is easier to couple to the air flow.  This comes with its own set of problems.  Peltiers are rather inefficient coolers, so the total power that needs to be gotten rid of is significantly larger than just what the die dissipates.  However, the active pumping action can help, even if the radiating fins eventually are much hotter.  This works because the aluminum or copper of the radiating fins can stand much higher temperatures than the semiconductor die can.  Silicon stops acting like a semiconductor at around 150&amp;deg;C, and real circuits need some operating margin below that.  However, heat sink fins can easily handle much higher temperatures.  A active heat pump makes use of this difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past there have been processors cooled with flowing liquid nitrogen.  This doesn't make economic sense for ordinary desktop PCs with today's technology, but heat management has been a important part of computer design pretty much since the beginnings of computers.  Even back in the 1950s, keeping all those vacuum tubes from melting each other was something that had to be carefully considered.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="282" LastEditDate="2015-07-02T18:53:39.420" LastActivityDate="2015-07-02T18:53:39.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1773" PostTypeId="2" ParentId="1739" CreationDate="2015-02-21T17:37:48.530" Score="0" Body="&lt;p&gt;You can also create a channel external to the sensor if you can't (don't want to) machine an additional slot in the sensor. Just take some 1/8 or 1/4 bar stock and cut pieces to mount on your base plate so that the outside edges of your sensor just barely slide inside the 2 bars. Then you can use the existing hole/slot on the sensor as purely mounting holes for screws to secure the sensor inside the 2 &quot;rails' formed by the bar stock pieces. This is all assuming that you only need to adjust the sensor location in one plane ( or two planes if need to adjust height off the stainless plate, just shim it up in that case).&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2015-02-21T17:37:48.530" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1774" PostTypeId="2" ParentId="1742" CreationDate="2015-02-21T17:52:56.583" Score="4" Body="&lt;p&gt;Here is a really cheap and cheerful way of showing the size of an impact. Take 2 boards of plywood, put a large compression spring between one and the other board so the spring is sandwiched but removable by at least one of the boards like an &quot;H&quot;. &#xA;Bolt this to a wall.&quot;H|&quot;&#xA;Get set size balls of playdo (kids putty) place on of these inside the spring (probably in a plastic bag or similar to save on mess), refit 2nd board and hit the board hard. The impact on the playdo will cause different deformation depending on how hard it was hit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make the board which is fixed to the wall have sides around it like a box, place 5 (shorter than the sides) compression springs (four corners and the middle holding the putty) you could probably have a piece of string to pull out the second board for ease of getting to the playdo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this too simplistic? it really depends how scientific you want the result and how good you are with a soldering iron as opposed to 6 bits of wood and 5 compression springs&lt;/p&gt;&#xA;" OwnerUserId="629" LastActivityDate="2015-02-21T17:52:56.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1775" PostTypeId="1" CreationDate="2015-02-21T19:27:58.557" Score="2" ViewCount="51" Body="&lt;p&gt;In 2008, &lt;a href=&quot;https://en.wikipedia.org/wiki/Large_Hadron_Collider#2008_quench_incident&quot; rel=&quot;nofollow&quot;&gt;an electrical fault led to a freaky magnetic quench&lt;/a&gt; at the Large Hadron Collider, which in turn released a lot of helium, damaging dozens of superconducting magnets in the process and setting the LHC back a full year. &lt;a href=&quot;http://www.symmetrymagazine.org/article/november-2008/explain-it-in-60-seconds-magnet-quench&quot; rel=&quot;nofollow&quot;&gt;Magnetic quenches happen often&lt;/a&gt; in particle accelerators, and helium is often released in the process, but the accident at the LHC was spectacular in scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cousin of the LHC is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Relativistic_Heavy_Ion_Collider&quot; rel=&quot;nofollow&quot;&gt;Relativistic Heavy Ion Collider&lt;/a&gt; at Brookhaven National Laboratory. Both are hadron accelerators and, of course, use superconducting magnets. The colliders are similar enough that I think they warrant a close comparison. I'm not familiar with the ins and outs of the magnets themselves, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the magnets at the RHIC as susceptible to damage as the magnets at the LHC?&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-02-21T19:27:58.557" Title="Could magnets at the RHIC be damaged like the ones at the LHC were in 2008?" Tags="|magnets|experimental-physics|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1776" PostTypeId="1" AcceptedAnswerId="1779" CreationDate="2015-02-21T21:28:27.043" Score="5" ViewCount="1626" Body="&lt;p&gt;I'm building a software application that uses GPS for a purpose related to roads.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know how many decimal places of GPS data should be stored to provide measurements that are accurate to within a few feet?&lt;/p&gt;&#xA;" OwnerUserId="630" LastEditorDisplayName="user16" LastEditDate="2015-02-22T16:32:10.123" LastActivityDate="2015-02-23T13:36:59.833" Title="How many decimal places of GPS should be stored to be accurate within a few feet?" Tags="|surveying|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1777" PostTypeId="2" ParentId="535" CreationDate="2015-02-21T21:57:17.590" Score="3" Body="&lt;p&gt;&lt;strong&gt;What are tin whiskers?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easy explanation is that &lt;em&gt;Tin Whiskers&lt;/em&gt;  are tiny threads of tin that grow out of lead-free tin soldier. They are a problem because they can cause electrical shorts at random times. They can begin to grow from weeks to years after a part is produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact causes of tin whiskers are not known, but their growth seems to be caused by internal stresses in the tin that work to push out the thin whiskers. Since the issues are caused by the internal structure of the tin, analyzing the basic material properties is hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tin is not the only material that has whiskers. Metallic whiskers have also been researched in zinc, cadmium, indium, silver, aluminum, and gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to mitigate tin whiskers?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this article from &lt;a href=&quot;http://www.eetimes.com/document.asp?doc_id=1279227&amp;amp;page_number=4&quot; rel=&quot;nofollow&quot;&gt;EE Times&lt;/a&gt; and this article from &lt;a href=&quot;http://electronicdesign.com/analog/don-t-let-tin-whiskers-destroy-your-design&quot; rel=&quot;nofollow&quot;&gt;Electronics Design&lt;/a&gt;, the following are strategies in order of mitigating the root cause to mitigating the effects:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Don't use pure tin.&lt;/li&gt;&#xA;&lt;li&gt;Use a matte tin finish instead of a bright tin finish.&lt;/li&gt;&#xA;&lt;li&gt;Anneal the tin after plating.&lt;/li&gt;&#xA;&lt;li&gt;Refinish the part with a hot-solder dip.&lt;/li&gt;&#xA;&lt;li&gt;Apply a conformal coating. These are special coatings that help to contain the whiskers as they form. They also can help to insulate board components from being shorted by whiskers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-21T21:57:17.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1778" PostTypeId="2" ParentId="1776" CreationDate="2015-02-21T22:03:33.607" Score="0" Body="&lt;p&gt;There are some complicated maths to calculate distances from GPS coordinates. The actual distance per degree (latitude and longitude are degrees) varies depending on where you are on the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you only need to know the maximum distance per degree anywhere on Earth. According to &lt;a href=&quot;http://en.wikipedia.org/wiki/Latitude&quot; rel=&quot;nofollow&quot;&gt;this wikipedia page&lt;/a&gt; this distance is just under $112km$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to store values accurate to $1m$, you need to store to a precision of $1/112000$, which is $0.00000893$. So storing six decimal places should give you the precision you need.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorDisplayName="user16" LastEditDate="2015-02-22T01:52:48.910" LastActivityDate="2015-02-22T01:52:48.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1779" PostTypeId="2" ParentId="1776" CreationDate="2015-02-21T22:03:45.947" Score="4" Body="&lt;p&gt;The circumference of the earth is about 25,000 miles &lt;em&gt;[40,075 km]&lt;/em&gt;, or 131.5 Mfeet.  That divided by 360 is 365 kfeet/degree &lt;em&gt;[111.3 km/degree]&lt;/em&gt;.  A degree value with 4 decimal places has a implied accuracy of 0.00005 deg, or 18 feet &lt;em&gt;[5.5 m]&lt;/em&gt;.  That's roughly the accuracy of a typical consumer grade GPS anyway, so there is little point going farther.  If you have a special (and expensive) surveying grade GPS, then you can use 5 decimal places and get about 1.8 foot &lt;em&gt;[0.5 m]&lt;/em&gt; numerical accuracy.  Each extra digits reduces the numerical error by a factor of 10.  Degrees with 6 fraction digits would be specifying location to about 2 inches &lt;em&gt;[about 5.5 cm]&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above was assuming the worst case where the 360 degrees are spread out over the entire circumference of the earth.  That is true for longitude at the equator, and latitude everywhere.  The numeric accuracy of longitude is what is computed above scaled by the cosine of the latitude.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="270" LastEditDate="2015-02-23T13:36:59.833" LastActivityDate="2015-02-23T13:36:59.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1782" PostTypeId="2" ParentId="1706" CreationDate="2015-02-22T15:26:48.137" Score="2" Body="&lt;p&gt;If you are talking about &lt;em&gt;vertical&lt;/em&gt; live load only, the case where you have traffic jam would be more governing to the design of the bridge &lt;em&gt;deck&lt;/em&gt;. The momentary increase &lt;em&gt;and&lt;/em&gt; &lt;strong&gt;decrease&lt;/strong&gt; in vertical force due to bumps from uneven roadway can be neglected since:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the magnitude would be small for a properly maintained roadway&lt;/li&gt;&#xA;&lt;li&gt;the magnitude would still be small for a badly maintained roadway as&#xA;drivers tend to slow down if they have difficulty driving straight&lt;/li&gt;&#xA;&lt;li&gt;such sporadic occurrences tend to cancel out globally as they don't occur in synchrony&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An established highway bridge design code such as &lt;strong&gt;BD37&lt;/strong&gt; &lt;em&gt;do not&lt;/em&gt; make any explicit consideration for the dynamic bumps that you mentioned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where you have vehicles moving on the bridge, traction and braking effects would generate a longitudinal &lt;em&gt;horizontal&lt;/em&gt; force that is governing to the design of the bridge &lt;em&gt;substructure&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through the proper use of load combinations (as shown in table below) and their corresponding load factors which have already been rigorously studied and tested, it is possible for all loading scenarios to be considered to achieve a safe design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GKm7F.png&quot; alt=&quot;Load Combination Table&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="620" LastEditorUserId="255" LastEditDate="2015-02-23T14:09:47.307" LastActivityDate="2015-02-23T14:09:47.307" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1783" PostTypeId="1" CreationDate="2015-02-22T17:53:20.543" Score="4" ViewCount="8330" Body="&lt;p&gt;I'm trying to tackle a mechanics problem to be solved using a finite-element analysis, but I'm stuck on one particular bit of theory first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done analysis using a linear-eleastic model, and elasto-plasticity with constant yield stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've got to do elasto-plasticity with increasing yield stress. I've been given a table of values to use with this: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;plastic strain ￼       0.00  0.01  0.02  0.03  0.04  0.05  0.06 &#xA;yield stress  (MPa) ￼  200   210   220   225   230   235   240&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I don't understand the concept of an 'increasing yield stress', I've only ever come across a material with a single yield point - where it yields...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can shed some enlightenment that would be great. Happy to provide more info if needed.&lt;/p&gt;&#xA;" OwnerUserId="642" LastEditorUserId="282" LastEditDate="2015-06-12T21:11:42.010" LastActivityDate="2015-06-12T21:11:42.010" Title="Increasing Yield Strength with Plastic Deformation" Tags="|mechanical-engineering|materials|applied-mechanics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1784" PostTypeId="2" ParentId="1783" CreationDate="2015-02-22T18:34:53.477" Score="5" Body="&lt;p&gt;The yield point of a material is determined by looking at its stress-strain graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/n2Qgp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from the image, some materials do not exhibit an obvious point where the material starts yielding. When this is the case, the &quot;yield point&quot; is determined by the 0.2% offset rule. This is shown by the dashed line in the graph.&#xA;As you can see from the graph, even after this yield point, the stress still increases as the strain increases.  This is the situation that you have described in your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This increase in stress will continue until the material breaks at its tensile strength. The graph below shows a material with an obvious yield point as can be seen from the first peak on the graph. This material also shows an increase in stress as the strain increases even past its yield point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rscNj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one of these materials experiences stress past its yield point, it will experience some elastic shortening, but it will not go back to its original length.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-22T18:34:53.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1785" PostTypeId="1" AcceptedAnswerId="1790" CreationDate="2015-02-22T19:21:19.357" Score="24" ViewCount="10004" Body="&lt;p&gt;Lightning strikes have been known to cause massive amounts of &lt;a href=&quot;http://www.lightningmaster.com/White-Papers/Lightning-Damage&quot; rel=&quot;noreferrer&quot;&gt;damage&lt;/a&gt;. The &lt;a href=&quot;http://www.lightningsafety.com/nlsi_lhm/lpts.html&quot; rel=&quot;noreferrer&quot;&gt;stats on a lightning bolt&lt;/a&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;current levels sometimes in excess of 400 kA, temperatures to 50,000 degrees F., and speeds approaching one third the speed of light&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These are massive numbers, but lightning protection systems are designed to draw the lightning away from the building or structure that they are protecting. Lightning protection systems can be simply thought of as lightning rods connected to the ground via cabling (downconductor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kndwb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NOAA &lt;a href=&quot;http://www.nws.noaa.gov/directives/sym/pd03041006curr.pdf&quot; rel=&quot;noreferrer&quot;&gt;specification for lightning protection&lt;/a&gt; requires that lightning rods be at least 0.5in (13mm) in diameter. The downconductor is a similar size copper cable (&lt;a href=&quot;https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes&quot; rel=&quot;noreferrer&quot;&gt;4/0 AWG or 12mm&lt;/a&gt;). The allowable amperage for this type of wire is only about &lt;a href=&quot;http://www.cerrowire.com/ampacity-charts&quot; rel=&quot;noreferrer&quot;&gt;250A&lt;/a&gt; for constant current.  I realize that this is more of a heat limit rather than a instantaneous current capacity limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this &lt;a href=&quot;http://www.lightningsafety.com/nlsi_lhm/conventionalLPT.pdf&quot; rel=&quot;noreferrer&quot;&gt;paper on lightning protection&lt;/a&gt; (page 28):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Positive feedback on the operation of a lightning protection system is seldom documented and most often not even noticed. Only in some rare cases can it be documented that a lightning protection system has been struck if it works properly and there is no damage. There is sometimes evidence at the strike termination point which can be noted during a careful inspection, but it is seldom cost effective for the owner of a lightning protection system to obtain the expertise necessary to conduct such a careful inspection.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can a seemingly small 0.5in (13mm) piece of metal handle a lightning strike with little or no visible damage much less without being completely destroyed?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="282" LastEditDate="2015-02-24T01:30:50.137" LastActivityDate="2017-03-08T22:37:18.537" Title="Why doesn't a lightning strike destroy the lightning rod?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1786" PostTypeId="2" ParentId="1783" CreationDate="2015-02-22T20:40:17.923" Score="2" Body="&lt;p&gt;From a materials science and engineering stand-point, materials which exhibit work-hardening change their structure at lattice and microstructural scales as plastic deformation proceeds in such a way as to resist further deformation. Virtually all metals exhibit work hardening, and thermoplastic polymers may also exhibit work-hardening (or work-softening) depending on temperature, strain-rate, and polymer chain structure and organization.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Metals&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;For metals, the general theory holds that plastic deformation occurs due to the motion of crystal lattice dislocations through the material. Dislocations and other lattice defects put the surrounding lattice into various states of stress. As a dislocation moves through the lattice, it may come near another defect, and the stress fields surrounding them will interact. If the stress fields increase as a result of moving the two defects closer together, it takes increasing effort to move the dislocation. In other words, the local lattice yield stress increases with increasing strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For constant environmental conditions, the non-dislocation defects have constant concentration, and this only explains the existence of a yield point. However, a number of proposed, and generally accepted, mechanisms exist for building up the concentration of dislocations within the metal as plastic deformation proceeds. The mechanisms include dislocation pileups on inclusions and grain boundaries, and dislocation sources pumping new dislocations into grains with increasing strain. These mechanisms increase the concentration of dislocations in metal grains, making further strain increasingly difficult. The result at large scales is work-hardening: as strain increases, stress needed to cause continued flow increases. Additionally, strain-to-failure decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metals which undergo work-hardening have increased yield strengths, but only at low homologous temperatures. Homologous temperature is current absolute temperature divided  by absolute melting temperature. An increase in work-hardening is associated with an increased level of energy storage in the material, which means an increase in free energy. Work-hardening, therefore, is unstable. Given enough time, or increased temperatures, dislocations will move in such a way as to reduce strain energy, which will reduce the yield strength toward the original, un-strained material strength. Exceedingly pure metals, and metals at sufficiently high homologous temperatures, can recover their yield strength as rapidly as they are deformed. As a result, there are a continuum of heat treatments available to control the properties of metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that work-hardening effects are partly anisotropic. I am unsure of the precise microstructural explanations for this, but I imagine it has to do with grains changing shape, different dislocation concentrations along different axes, and dislocation pileup occuring anisotropically. Uniaxial strain will harden a metal to a greater degree in the direction of strain, such as in wire drawing. Biaxial strain will usually harden a metal to a greater degree in the plane of forming than in the thickness direction, as with sheet-forming. It is worth noting that if the strain is tensile, the compressive yield strength of the material in that direction will decrease as the tensile yield strength increases, due to the Bauschinger effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above applies to face-centered cubic and body-centered cubic metals generally. Hexagonal close-packed metals do exhibit work-hardening, but tend to be more brittle, so be careful modeling HCP metals. You may find this &lt;a href=&quot;http://www.newagepublishers.com/samplechapter/001425.pdf&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; useful in understanding everything going on.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Polymers&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Thermoplastic polymers I know less about, but my understanding is that at a microstructural level the chains are randomly oriented and tangled in amorphous polymers, and in crystalline polymers the chains are laterally organized in crystalline spherulites interspersed by amorphous regions. If the polymer is strained slowly enough, or at high enough temperature, the chains have time to reorganize as they are strained. Generally, the chains will tend to organize in order to maximize the stress on each individual chain, so that the chains align length-wise along the direction of stress. As the fraction of chains aligned with the stress direction increases, the strength of the polymer increases. This process, referred to as polymer chain orientation, is exploited during blow molding of PET soft drink bottles to increase the local in-plane mechanical properties of the bottles. There is also an added benefit of decreasing carbon dioxide permeability in the thickness direction because there is less room between individual chains for gas to pass through. Less free volume in polymers is associated with less ductility and a tendency toward brittle fracture, among other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work-hardening in both thermoplastics and metals increases strength in the direction of stress and decreases further ductility, but there are a number of differences between them. First, the change in strength for thermoplastics is entirely anisotropic, and polymers weaken in directions other than the stress directions, whether uniaxial or biaxial. For metals, due to the different microstructural mechanisms, there is some increase in yield strength in non-stress directions. Second, dislocation density may be reduced via heat treatments without melting the metal, but to remove full polymer orientation virtually requires melting the polymer. Third, the exact chain structure -- i.e. length, side-groups, branching, tacticity, copolymerization -- of a given thermoplastic plays a role in how orientable the chains are, as well as virtually all of its other properties. With metals grain size, concentration of inclusions, and previous working history are the major factors in work hardenability. &lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-02-22T20:46:20.113" LastActivityDate="2015-02-22T20:46:20.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1787" PostTypeId="2" ParentId="1785" CreationDate="2015-02-22T21:21:54.120" Score="2" Body="&lt;p&gt;&lt;em&gt;[ This started as an answer.  But the calculation ended up 3 orders of magnitude short.&lt;br&gt;&#xA;So, this is more of a comment, I guess. ]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single lightning bolt has a finite amount of energy &lt;sup&gt;1&lt;/sup&gt;.  The ballpark number is $5 \cdot 10^9 \text{J}$, which is equivalent to 145 L of petrol&lt;sup&gt;2&lt;/sup&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that all of the energy of the lightning bolt is dissipated in the rod.  Let's say that out lightning rod in 12mm diameter and 10m long.  Density of copper is 8960 kg/m&lt;sup&gt;3&lt;/sup&gt;,&lt;br&gt;&#xA;so the mass of our rod is $10.13 \text{ kg}$.&lt;br&gt;&#xA;The specific heat of copper is $\text{0.385} \frac{\text{kJ}}{\text{kg °C}}$.  How much energy does it take to warm up the copper rod from 25°C to the melting point of copper, which is $1083 \text{°C}$?  It's $4.05 \cdot 10^5 \text{J}$.&lt;br&gt;&#xA;The heat of fusion for copper is $\text{213} \frac{\text{kJ}}{\text{kg}}$.  How much energy does it take to melt our rod?  It's $2.13 \cdot 10^6 \text{J}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made an assumption that all of the lightning's energy is dissipated in the rod.  But, I don't know of that's a sane assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;Even though it has very high peak power and current.&lt;/sub&gt;&lt;br&gt;&#xA;&lt;sup&gt;2&lt;/sup&gt; &lt;sub&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Harvesting_lightning_energy&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;&lt;/sub&gt;&lt;br&gt;&#xA;&lt;sup&gt;3&lt;/sup&gt; &lt;sub&gt;&lt;a href=&quot;http://pcdandf.com/pcdesign/index.php/magazine/10179-pcb-design-1507&quot; rel=&quot;nofollow&quot;&gt;Who were Preece and Onderdonk?&lt;/a&gt;  The article about early (1880, 1928) equations describing currents that cause wire to melt.&lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2015-08-09T06:44:30.020" LastActivityDate="2015-08-09T06:44:30.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1788" PostTypeId="1" AcceptedAnswerId="1791" CreationDate="2015-02-22T21:22:16.370" Score="7" ViewCount="1287" Body="&lt;p&gt;Measuring things like a martial artist's strike can be challenging because it is a complex motion and common vernacular will use imprecise or incorrect terms.  This question is an attempt to identify the applicable measurements (quantification) and the correct units behind those numbers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Background:&lt;/em&gt;&lt;br&gt;&#xA;In a &lt;a href=&quot;https://engineering.stackexchange.com/q/1742/16&quot;&gt;recent question of mine&lt;/a&gt;, I asked for sensors that could be used to measure a martial artist's strike, such as a kick or punch.  My primary intent was to demonstrate to the student how their technique improved their strike.  The answers there are great, and definitely addressed my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What led to this question is that I received a follow-up, clarifying comment asking if I wanted to measure force, energy, or power.  To which I responded that I simply needed something measurable to show the difference or improvement to the student.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a subsequent comment&lt;sup&gt;*&lt;/sup&gt; that got me thinking:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From what you write here, I would like to add that it is neither force nor pressure nor energy that you actually want to measure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*&lt;/sup&gt;&lt;sub&gt;The comment has since been deleted, but I started this question shortly after it was left.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What got me thinking about that second comment is while I've seen many references using units of force, there are other ways of quantifying a martial arts strike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.washingtoncitypaper.com/articles/39469/straight-dope-the-physics-of-punching-someone-in-the-face&quot; rel=&quot;nofollow noreferrer&quot;&gt;This reference&lt;/a&gt; prefers to use 'force' and condemns using 'pressure'&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.livestrong.com/article/1003413-lightweight-vs-heavyweight-punch-power/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This reference&lt;/a&gt; also uses force, but mixes up their terminology when describing acceleration and velocity, so their usage of terminology is suspect.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sciencebuddies.org/science-fair-projects/project_ideas/Sports_p020.shtml#background&quot; rel=&quot;nofollow noreferrer&quot;&gt;This reference&lt;/a&gt; focuses on velocity, momentum, and energy.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?feature=player_detailpage&amp;amp;v=sfnGkV6qmTw#t=34&quot; rel=&quot;nofollow noreferrer&quot;&gt;This reference&lt;/a&gt; uses $lb_F$ for their measurements.  Coincidentally, this is the series that started me down my original line of thought with the other question.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My question:&lt;/em&gt;&lt;br&gt;&#xA;What is the correct terminology for measuring (quantifying) the magnitude of a martial artist's strike versus another strike?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measurement should be applicable across a variety of strikes (punches, kicks, elbows, etc...) and valid for different practitioners.  Likewise, I'm not worried about defining good versus bad; I want to be able to quantify the strike.  This ties into basic process improvement - we measure the process we want to improve and compare the measured results.  &lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-06-22T18:26:55.010" Title="What was that?! Quantifying the impact of a martial arts strike" Tags="|measurements|biomedical-engineering|terminology|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1789" PostTypeId="1" CreationDate="2015-02-22T21:31:32.293" Score="2" ViewCount="173" Body="&lt;p&gt;Quoting from the &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_balloon&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A solar balloon is a balloon that gains buoyancy when the air inside is heated by the sun's radiation, usually with the help of black or dark balloon material.  The heated air inside the solar balloon expands and has lower density than the surrounding air. As such, a solar balloon is similar to a hot air balloon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What material would be the most efficient (minimize material mass, maximize payload, maximize heat retention) for weaving the skin of a solar hot air balloon? I want to consider thermal conductivity of the outer surface, radiance of the inner one, and structural strength. I want to achieve the 60 metric tons payload lifted to at least 200 meters over tropical sea level.  &lt;/p&gt;&#xA;" OwnerUserId="645" LastEditorUserId="645" LastEditDate="2016-01-04T18:22:53.247" LastActivityDate="2016-01-04T18:22:53.247" Title="Which material is most efficient for the skin of a solar balloon?" Tags="|materials|aerospace-engineering|thermodynamics|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1790" PostTypeId="2" ParentId="1785" CreationDate="2015-02-22T21:33:32.600" Score="32" Body="&lt;p&gt;The current limit specification for a wire is limited by the heat that current will produce, and how much heat the wire can dissipate before getting too hot.  &quot;Too hot&quot; depends on the circumstances.  You will see higher current ratings for the same type of wire in chassis wiring applications than the electrical code allows for home installations, for example.  This is mostly due to how hot too hot is.  The ultimate limit for extreme applications is that the conductor not melt.  Temperatures anywhere near that would be unsafe running along wooden supports inside a wall in a house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you say, the wire is rated for 250 A &lt;i&gt;continuous&lt;/i&gt;.  Lightning is decidedly not continuous.  1 ms is &quot;long&quot; for the time of the main lightning stroke.  There may be several strokes in one event, but the total time is still short, and the other non-main strokes will have significantly less current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the math.  You say the wire is 12 mm in diameter, so has a cross-sectional area of 113 mm&amp;sup2; = 113x10&lt;sup&gt;-6&lt;/sup&gt;m&amp;sup2;.  The resistivity of copper at 20&amp;deg;C is 1.68x10&lt;sup&gt;-8&lt;/sup&gt; &amp;Omega;m.  A 1 meter length of this wire therefore has a resistance of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1.68x10&lt;sup&gt;-8&lt;/sup&gt; &amp;Omega;m)(1 m)/(113x10&lt;sup&gt;-6&lt;/sup&gt; m&amp;sup2;) = 149 &amp;micro;&amp;Omega;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power with 400 kA thru this resistance is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(400 kA)&amp;sup2;(149 &amp;micro;&amp;Omega;) = 23.8 MW  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Times the 1 ms time the current is applied yields the energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(23.8 MW)(1 ms) = 23.8 kJ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of copper is 8.93 g/cm&amp;sup3;, and our 1 m length has a volume of 113x10&lt;sup&gt;-6&lt;/sup&gt; m&amp;sup3;, which is 113 cm&amp;sup3;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(113 cm&amp;sup3;)(8.93 g/cm&amp;sup3;) = 1010 g total mass of copper&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific heat of copper is 0.386 J/g&amp;deg;C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(23.8 kJ) / (0.386 J/g&amp;deg;C)(1010 g) = 61&amp;deg;C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means putting 400 kA through a 12 mm diameter copper wire for 1 ms will cause a temperature rise of 61&amp;deg;C.  That's a rather extreme value for a lightning strike.  The main stroke is usually substantially shorter than 1 ms, and the other strokes have substantially less current.  However, even with these numbers it shows that while the wire will certainly get toasty for a while, it is well within the wire's ability to handle without any structural failure.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2015-06-02T12:09:34.297" LastActivityDate="2015-06-02T12:09:34.297" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1791" PostTypeId="2" ParentId="1788" CreationDate="2015-02-22T22:18:43.890" Score="4" Body="&lt;p&gt;It all comes down to force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all know Newton's second law: $\vec{F}=m\vec{a}$&lt;sup&gt;1&lt;/sup&gt;. It says that the greater the force that is applied to an object, the greater its acceleration will be (and acceleration is really just the second derivative of position with respect to time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you measure force? One common method, used to measure the weight of an object, is to use a spring scale. You can either place an object on top of the spring or hang it from a spring. Be careful, though - when hanging it from a spring, be aware that the scale directly measures tension. Move in it some kinky direction and you'll have weird readings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for the force on a spring is $\vec{F}=k\vec{x}$, where $\vec{x}$ is the displacement&lt;sup&gt;2&lt;/sup&gt; and $k$ is the spring constant of the spring. So if you push against a spring, you can figure out the force applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned momentum. If you want to be snarky (but accurate), use &lt;em&gt;impulse&lt;/em&gt; - the change in momentum of an object (or system). Impulse can be written as $\vec{I}$ of $\vec{J}$. I've seen both conventions used. Regardless, though, impulse is defined as&#xA;$$\vec{I}=\vec{J}=\vec{F} \Delta t=\Delta \vec{p} =m\vec{v}_f-m\vec{v}_o$$&#xA;A greater force means a greater change in momentum. A force applied for a greater time interval means a greater change in momentum. By the way, divide all the terms by mass and you get that familiar kinematics equation:&#xA;$$\vec{a}t=\vec{v}_f-\vec{v}_o$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy is a weird choice. Unless you're punching your opponent straight into the air, you'll be transferring some chemical energy into kinetic energy. If you want to ignore your body's energy conversion, just say that the force does work on the opponent's body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back in high school (i.e. algebra-based) physics, work was defined as&lt;sup&gt;3&lt;/sup&gt;&#xA;$$\vec{W}=\vec{F}s$$&#xA;However, your fist is most likely not applying the same amount of force while it travels and is in contact with your opponent. So we have to use calculus:&#xA;$$\vec{W}=- \int_{x_o}^{x_f} \vec{F} dx$$&#xA;Mind the minus sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the energy transferred is equal to the work done by the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power is simply energy (here it's work) divided by time:&#xA;$$\vec{P}=\frac{\vec{W}}{t}$$&#xA;The relationships there are implied. But here, energy is equal to the work done. And that's really just a function of force, $F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So impulse, energy and power are all measures of force. Same goes for pressure. Take the proper derivatives and you'll see:&#xA;$$\frac{d\vec{p}}{dt}=\frac{d\vec{W}}{ds}=\frac{d^2\vec{P}}{dt \cdot ds}=A\vec{P}=\vec{F}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how well that addresses your question, though. All of these things are different entities, and what you measure depends on &lt;em&gt;how&lt;/em&gt; you measure it. If you measure how fast your opponent moves after the punch (and his mass), you can measure impulse or work&lt;sup&gt;4&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no &quot;right way&quot; or &quot;wrong way&quot; to measure the magnitude of the strike. In all these cases, the greater the quantity, the greater the force that has been applied. The only difference, for practical purposes, is that the force is divided amongst something (area), applied for a given amount of time (impulse), applied over a given distance (work), or applied over a given distance in a certain amount of time (power). That's it. Call the quantity $\vec{Q}$, and you get $\vec{Q} = \vec{F} \cdot d\text{whatever}$&lt;sup&gt;5&lt;/sup&gt;. Just pick your $d \text{something}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Also written as $F=m \frac{d^2x}{dt^2}$&#xA;&lt;br&gt;&#xA;&lt;sup&gt;2&lt;/sup&gt; Displacement is a vector and thus force is a vector. Same goes for acceleration.&#xA;&lt;br&gt;&#xA;&lt;sup&gt;3&lt;/sup&gt; I'm ignoring the $\cos \theta$ I really should throw in there, for simplicity.&#xA;&lt;br&gt;&#xA;&lt;sup&gt;4&lt;/sup&gt; $\vec{J}=m\vec{v}_f-m\vec{v}_o$ and $KE=\frac{1}{2}mv^2$. Take your pick.&#xA;&lt;br&gt;&#xA;&lt;sup&gt;5&lt;/sup&gt; Yeah, yeah. Force divided by $d \text{something}$ in the case of power, but multiplied by $d \text{something else}$, too.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-02-22T22:18:43.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1792" PostTypeId="2" ParentId="1788" CreationDate="2015-02-22T23:15:23.423" Score="9" Body="&lt;p&gt;The basic problem is that you haven't defined what a &quot;good&quot; kick or punch is.  Is the purpose make the opponent lose his balance?  To inflict the maximum pain?  To break bones?  To hit a particular part of the body to disable him?  Shatter the most wooden boards stacked on top of each other?  Something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these different outcomes will weigh various physical properties of the strike differently in coming up with a single number of effectiveness.  For example, getting someone to lose balance, even fall over, is mostly about impulse, which is force times time, but as long as the time is short enough.  Breaking bones is more about maximum force.  Inflicting a concussion (as is the goal of a knockout in boxing) is probably more about overall momentum, but also targeted in the right place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, your question can't be answered without a definition of how &quot;good&quot; is measured.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-02-22T23:15:23.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1793" PostTypeId="2" ParentId="1758" CreationDate="2015-02-23T08:36:39.820" Score="3" Body="&lt;p&gt;Just out of curiosity, if the enclosure is sealed what will an oxigen sensor measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would conduct the heat outwards via a metal structure. But cosidering the heat your sensor  generates, I would reconsider the polycarbonate choice. &lt;a href=&quot;http://en.wikipedia.org/wiki/Polycarbonate&quot; rel=&quot;nofollow&quot;&gt;Polycarbonate &lt;/a&gt; max operating temperature is about $115\,^\circ C$. A metal enclosure would conduct the heat much faster to the environment.&lt;/p&gt;&#xA;" OwnerUserId="270" LastActivityDate="2015-02-23T08:36:39.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1794" PostTypeId="1" AcceptedAnswerId="1802" CreationDate="2015-02-23T11:57:48.563" Score="7" ViewCount="14250" Body="&lt;p&gt;I am reading a novel set in the future. A planet has been found where there are extensive deposits of diamond. Presumably diamond is mined much like marble is here on earth (there is not a lot of detail on this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A character has a knife where the blade/shaft is a single piece of diamond. In 2015 here on Earth you can purchase &lt;a href=&quot;http://www.fieldandstream.com/blogs/gun-nuts/2013/03/diamond-blade-meridian&quot;&gt;diamond blade knives&lt;/a&gt;, but they seem to be all steel (no diamonds per comment by @mart). There are some &lt;a href=&quot;https://en.wikipedia.org/wiki/Diamond_blade&quot;&gt;saw blades that have pieces of diamond attached&lt;/a&gt; to a steel blade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming large quantities of reasonably priced materials, would it be possible to design and manufacture a functional knife made from a single section of diamond (all but the handle)?&lt;/p&gt;&#xA;" OwnerUserId="257" LastEditorUserId="330" LastEditDate="2015-02-23T18:22:45.417" LastActivityDate="2023-09-30T22:33:31.803" Title="Could a knife blade &amp; shaft be made of 100% diamond?" Tags="|materials|" AnswerCount="7" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1795" PostTypeId="2" ParentId="1794" CreationDate="2015-02-23T14:05:56.367" Score="5" Body="&lt;p&gt;The first criterion would be to obtain a diamond, or a fragment of a diamond, large enough from which a knife could be made. Just because such a diamond has not been found on Earth doesn't mean it can't exist elsewhere. Years ago, astronomers found &lt;a href=&quot;http://news.bbc.co.uk/2/hi/science/nature/3492919.stm&quot; rel=&quot;nofollow&quot;&gt;a diamond star&lt;/a&gt;, 4000 km across. &lt;a href=&quot;http://news.nationalgeographic.com/news/2011/08/110825-new-planet-diamond-pulsar-dwarf-star-space-science/&quot; rel=&quot;nofollow&quot;&gt;Another source&lt;/a&gt; claims a &quot;corpse of a star&quot; 55,000 km across (five times Earth's diameter) made of carbon and oxygen &quot;most likely crystalline in nature, like a diamond&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To shape a knife, with a combined handle, from a single diamond would require very careful grinding and polishing. This would require much effort and time, but it wouldn't be impossible. It's how gem diamonds are cut from raw diamonds. A lot a care would be required to prevent the diamond from shattering. It's a delicate procedure that requires skill, the right tools, the correct procedures and experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the knife has been shaped, the knife would need to be polished; something that has already been &lt;a href=&quot;https://en.wikipedia.org/wiki/Diamond_knife&quot; rel=&quot;nofollow&quot;&gt;done&lt;/a&gt; on a smaller scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One issue with a knife made entirely of diamond is how useful would it be and if it would shatter if it were dropped or if it struck another hard item with force, given that &lt;a href=&quot;https://en.wikipedia.org/wiki/Diamond&quot; rel=&quot;nofollow&quot;&gt;diamond&lt;/a&gt; is brittle.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="330" LastEditDate="2015-02-23T18:58:02.740" LastActivityDate="2015-02-23T18:58:02.740" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1796" PostTypeId="2" ParentId="1794" CreationDate="2015-02-23T14:28:26.847" Score="4" Body="&lt;p&gt;Working a large chunk of raw diamond into a knife blade is theoretically possible.  We could do it today on earth, although the process wouldn't be cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest problem I see is that while diamond will maintain a sharp edge, it is too brittle to make a useful knife blade except perhaps for specialty applications where the knife won't be subjected to other stresses on the blade, and absolute sharpness is really really important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other properties of diamond you might want to explore are that it is a very good conductor of heat, can withstand high temperatures, is transparent, and has a high index of refraction.  I would focus on exploiting these characteristics in a science fiction novel where the material itself is cheap and abundant.  There are a lot more interesting ways to use cheap diamond than making knife blades out of it.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-02-23T14:28:26.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1797" PostTypeId="5" CreationDate="2015-02-23T17:26:42.673" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-02-23T17:26:42.673" LastActivityDate="2015-02-23T17:26:42.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1798" PostTypeId="4" CreationDate="2015-02-23T17:26:42.673" Score="0" Body="Use this tag for questions about the legal and financial liability of engineers and engineering firms, including historical suits and judgments. Questions about ethical and licensing issues should consider using the &quot;ethics&quot; or &quot;licensure&quot; tags instead." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-02-23T17:26:42.673" LastActivityDate="2015-02-23T17:26:42.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1799" PostTypeId="2" ParentId="662" CreationDate="2015-02-23T17:34:50.633" Score="10" Body="&lt;p&gt;No, the responsibility was squarely with the engineers.  First, the original design couldn't be practically built.  That's not negligent, but not too smart.  The construction company &lt;i&gt;proposed&lt;/i&gt; a alternate design, which Gillum then didn't object to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard different versions, where Gillum failed to respond to Havens Steels proposal, which was then taken as implicit approval, or they outright approved it.  Either way, it was Gillum's responsibility to oversee the design and make sure it met the requirements and the code.  Apparently, even the original design was not up to code.  It's hard to see how Gillum and the engineers working for Gillum were not grossly negligent here.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-02-23T17:34:50.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1800" PostTypeId="5" CreationDate="2015-02-23T18:03:26.660" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-02-23T18:03:26.660" LastActivityDate="2015-02-23T18:03:26.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1801" PostTypeId="4" CreationDate="2015-02-23T18:03:26.660" Score="0" Body="Questions about engineering ethics as it relates to professional practice, especially: professional organizations and licensing bodies; ethics review boards; topics on the PE/FE exams; etc. Not to be confused with liability or personal/individual morality." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-02-23T18:03:26.660" LastActivityDate="2015-02-23T18:03:26.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1802" PostTypeId="2" ParentId="1794" CreationDate="2015-02-23T18:38:23.293" Score="10" Body="&lt;p&gt;&lt;strong&gt;You could work pure diamond into something shaped like a knife, but it wouldn't make a very good knife&lt;/strong&gt;. Diamonds are hard, but they're brittle. The blade would break long before its edge went dull, but this is arguably worse: you can resharpen a dull blade, but a broken blade (especially one not made of a material that can be melted and reforged) becomes essentially useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If this is for a story, then one possible subject for research might be the &lt;em&gt;macahuitl&lt;/em&gt;: a swordlike weapon used by the Aztecs&lt;/strong&gt;. The main body of this weapon was a simple shaft of wood with grooves running along its edge, but small blades of sharpened obsidian would be fitted into the grooves. A blade could chip or break, but the weapon itself would remain intact through most blows, while the scalpel-sharp obsidian could cut even better than metal (seriously: obsidian is dangerous stuff). Modern &quot;diamond knives&quot; work according to a similar principle: only the edge is made of diamond, while the main body is made of metal, or something else that's not so prone to breakage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the purposes of a story, I could imagine the diamond-knife concept being scaled up into a sort of &quot;neo-macahuitl&quot; that used diamond blades&lt;/strong&gt;. The main body of the weapon could be made of metal, hard rubber, or any other material of sufficient strength, and then small diamond blades (all diamond, or diamond-edged metal) could be fitted into it. It might even be possible to make the blades replaceable.&lt;/p&gt;&#xA;" OwnerUserId="659" LastActivityDate="2015-02-23T18:38:23.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1803" PostTypeId="2" ParentId="1758" CreationDate="2015-02-23T19:46:52.187" Score="1" Body="&lt;p&gt;40&amp;deg;C is really no big deal for most electronic components, especially when those components themselves don't dissipate much heat.  Electronics to amplify and buffer the output of a sensor, or convert it to digital and pass the data on, shouldn't require much power by themselves, certainly a lot less than the 1.5 W the sensor is already dissipating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to check the datasheets, of course, but most ordinary electronic components work over the 0-70&amp;deg; range at least.  For a little more money, you can get most semiconductor ICs in the &quot;commercial&quot; temperature range, which will usually extend well above 100&amp;deg;C, like 125&amp;deg;C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the short answer is that you haven't demonstrated there is a problem here to solve.  Just make sure to check all the datasheet, and calculate the worst case offset and gain errors using the full temperature range you expect.  But then again, that's standard practise anyway, and the datasheets will already spec things like offset voltage over the operating range of the part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd stay away from electrolytic capacitors, since their lifetime goes down significantly with temperature.  For a sensor amplifier and maybe digital converter, you shouldn't need them anyway.  Again, I don't see a problem here.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-02-23T19:46:52.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1804" PostTypeId="2" ParentId="1760" CreationDate="2015-02-23T20:10:47.243" Score="3" Body="&lt;p&gt;This sounds like a pretty broad question, but to get you started on the power line issue specifically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I live, in the US, OSHA requires 10 feet of separation between any uninsulated power line 600v - 50kV and any worker or equipment. This might be a good place to start. Don't forget to consider that the bridge will deflect downwards under its maximum loads, and the power cable will bounce around in windy conditions (and possibly when a train passes under.) These variables should be taken into account in your design. If the bottom of the bridge requires any kind of maintenance (repainting?) you may want to provide a separation distance that allows a crew to get access and work under it without getting within 10' of the power line.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To be clear, the separation distance as required here is about preventing an arc from jumping across the air gap, since air is an imperfect insulator. This distance is not intended to address interactions with sensitive electronics or other EMF-sensitive devices. If that's what your concern is about, please specify what kinds of devices you need to be able to work. It may also be possible to detail a robust insulating material between the power lines and the bridge, although in an outdoor environment, you'd need to protect against water forming a conductive path around the insulator.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-02-23T23:16:12.947" LastActivityDate="2015-02-23T23:16:12.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1805" PostTypeId="2" ParentId="1794" CreationDate="2015-02-23T21:08:58.390" Score="4" Body="&lt;p&gt;Methods already exist for creating sizeable artificial diamonds from more common forms of carbon. Some companies even offer a service where you can send the ashes of a beloved family member or pet to be compressed into a diamond for jewelry. It's not unreasonable to believe that this process could be scaled up and specialized to form other shapes like a knife. That won't mitigate any of the material limitations other answers have already detailed though. It'll just make it easy to get a knife-shaped diamond.&lt;/p&gt;&#xA;" OwnerUserId="664" LastEditorUserId="664" LastEditDate="2015-02-24T14:45:48.870" LastActivityDate="2015-02-24T14:45:48.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1806" PostTypeId="2" ParentId="1758" CreationDate="2015-02-23T22:31:21.500" Score="2" Body="&lt;p&gt;Could you mount Peltier cooling units to the outside to pull heat out of the enclosure?  I would think you'd want to have cooling fins &lt;em&gt;inside&lt;/em&gt; the unit to increase the rate of heat absorption.&lt;/p&gt;&#xA;" OwnerUserId="667" LastActivityDate="2015-02-23T22:31:21.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1807" PostTypeId="1" CreationDate="2015-02-23T22:41:20.700" Score="10" ViewCount="13499" Body="&lt;p&gt;The specifications for large structural projects typically call for the structure to have a specific design life. This can be 50 years, 100 years, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accommodating the design life for steel can be as simple as adding additional thickness to account for the &lt;a href=&quot;http://www.steelconstruction.info/Corrosion_of_structural_steel#Chlorides&quot;&gt;expected corrosion&lt;/a&gt; over this time period. This calculation would also take into consideration any variations based on coatings or type of steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;History has shown that unreinforced concrete structures can last &lt;a href=&quot;http://en.wikipedia.org/wiki/Concrete#History&quot;&gt;hundreds of years&lt;/a&gt;. The Romans have some examples of this such as the &lt;a href=&quot;http://en.wikipedia.org/wiki/Pantheon,_Rome&quot;&gt;Pantheon&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with reinforced concrete is that eventually the &lt;a href=&quot;http://en.wikipedia.org/wiki/Concrete_degradation&quot;&gt;reinforcing will corrode&lt;/a&gt;, expand in size, and cause the concrete to crack. There can also be issues with the aggregates that are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can a designer calculate and, by contract, guarantee the life of a reinforced concrete structure? &lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-16T18:24:28.303" Title="How is the design life of a reinforced concrete structure calculated?" Tags="|civil-engineering|structural-engineering|concrete|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1808" PostTypeId="2" ParentId="1760" CreationDate="2015-02-24T00:52:16.307" Score="7" Body="&lt;p&gt;&quot;standard EM interaction range&quot; depends a &lt;strong&gt;lot&lt;/strong&gt; on frequency, or its inverse which is wavelength. Your 25 kV catenary probably is 50 Hz or 60 Hz, same as your national grid. That means the wavelength is 6000 kilometers. That's far more than your dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, you can ignore the electro-magnetic aspects and treat this as an electro-static problem. Arcing is the main risk. Here there are two concerns that need tackling: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you prevent catenary movement? Some bridges have isolators beneath them to keep the catenary away from the bridge. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if the previous mechanism fails, will it fail safely? If the catenary would hit the structure, there will be a rather spectacular light show and the catenary might weld itself to the underside of your bridge. That will result in some high currents. Will the people on the bridge itself be safe? Metal may not be the best surface choice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also keep in mind that dry air is a much better isolator than wet air/rain. It's not just water forming a direct path, raindrops already are an issue. Your bridge probably should be designed to direct rainfall away from the catenary.&lt;/p&gt;&#xA;" OwnerUserId="669" LastEditorUserId="33" LastEditDate="2015-02-24T01:33:26.240" LastActivityDate="2015-02-24T01:33:26.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1811" PostTypeId="2" ParentId="433" CreationDate="2015-02-24T03:42:26.200" Score="2" Body="&lt;p&gt;I would suggest a Dremel-like cutting wheel positioned at the front which can be rotated around 360˚ such that the wheel stays parallel to the inner surface of the pipe. The small diameter of the cutting wheel will mean you can cut quite close without damaging the pipe. This is assuming you will have a camera and light on the front to see where to orient the cutter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach would be to have a hole-cutter type blade on the front, mounted just inside another slightly larger cylinder to prevent the blade from hitting the inside surface of the pipe as it moves when biting into roots. &lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-02-24T03:42:26.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1813" PostTypeId="2" ParentId="681" CreationDate="2015-02-24T04:20:03.343" Score="4" Body="&lt;p&gt;&lt;strong&gt;CFD&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ran this situation through a couple of simple 2D &lt;a href=&quot;https://en.wikipedia.org/wiki/Computational_fluid_dynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;CFD&lt;/a&gt; models. As you can see from the pictures below, one way to keep debris from gathering in this area is to increase the velocity of wind in the corner. To do this, you need to place an obstacle that will direct the wind &lt;em&gt;into&lt;/em&gt; the corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Original Situation:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WF2ls.png&quot; alt=&quot;Original Situation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Obstacle Added:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S8d6d.png&quot; alt=&quot;Obstacle Added&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with this, there will be a dead spot in the corner.  This should help to lessen it though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another benefit of this configuration is that it works just the same if the wind is coming form the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What kind of obstacle?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact design of the obstacle wouldn't need to be too involved. A planter or solid bench may be enough to make a difference. The main criteria are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Solid so that the wind is directed &lt;strong&gt;around&lt;/strong&gt; it.&lt;/li&gt;&#xA;&lt;li&gt;High enough that the wind moving &lt;strong&gt;over top&lt;/strong&gt; of it doesn't affect the wind moving around.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There will likely be some trial and error.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-24T04:20:03.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1814" PostTypeId="2" ParentId="1788" CreationDate="2015-02-24T05:21:01.217" Score="1" Body="&lt;p&gt;If I were tasked with measuring this I would also want to know what the purpose or desired outcome of each strike was. This is because you may want to measure different things for different types of strike.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A strike that is designed to cause damage must focus the force on a small area in order to maximize the pressure at that point - so for that type of strike you would want to measure the pressure.&lt;/li&gt;&#xA;&lt;li&gt;A strike that is designed to impart a force on the target (for example to wind them) - you would need measure the force applied at the area of impact.&lt;/li&gt;&#xA;&lt;li&gt;A strike that is designed to push the target away - you should measure how much of the force was transferred to the target and how much back into the striker (kinetic energy before and after the strike).&lt;/li&gt;&#xA;&lt;li&gt;A strike designed to startle and confuse the opponent may rely on speed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There may be other scenarios too, but the main ones will be interested in force and pressure. In the third bullet above, you would want to measure the kinetic energy of both parties before and after the strike, this could be done with a high-speed camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the confusingly contradicting references you cited may not be contradictory at all, they may just be focusing on measuring different things because of different types of strike (or different desired outcomes for the same strike types, in which case I imagine the technique may be different as well as what you want to measure).&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-02-24T05:21:01.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1815" PostTypeId="2" ParentId="1788" CreationDate="2015-02-24T07:09:05.047" Score="1" Body="&lt;p&gt;With a strike or kick, you move (more or less) you whole body, connect a hand or foot or whatnot to your target and if your technique is good, the target will deform far more than your body. I think this thought is crucial, as there are a lot of joints and squishy things between the part of your body hitting the target and your center of mass. This also means that a strike that may apply a lot of energy to a light focus mitt may be less effective when applied to a heavy bag, because in the latter case the squishy bits squish more. As we all know, the idea with striking someone is to have only little total and no plastic deformation in our squishy bits, while we want to plastically deform the target and move it someplace. Sometimes we will emphasize one over the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the target side, you have some deformation and some movement. The exact composition will depend on the speed of the strike and the weight of the target. Of course you could argue that the deformation is actually a form of movement or vice versa. Here, by movement I mean the target actually moves as a whole, and by deformation any 'internal' movements of the target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Deformation&lt;/strong&gt;&lt;br&gt;&#xA;There are probably elastic materials that yield a quantifiable result in how much you deform them. One rather cost effective way to show your students how much prgress they make with their strikes (or not) would be to stack up boards and see how many they can break. I'm not sure Karate is ready for such a novel idea.&lt;br&gt;&#xA;Another way would be to have a spring and measure its elastic deformation. If I were to design such a system, my first start would be a robust air baloon and a pressure sensor, rigged so that you get the pressure spike. Compressing gases is not ideally elastic (some energy gets lost as heat). But you could mount an air cushion on the 'student' side of your mitt. I think any classical mechanical spring will need an aditional cushion, and that would add uncertainties. I think in Karate there's a practice of sticking boards in the ground and punchiung them, the teacher checks how much the board 'gives' by holding a hand a few cm behind it. If the board doesn't reach the hand, still much the student has to practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Movement&lt;/strong&gt;&lt;br&gt;&#xA;You could build a pendulum and measure how far it swings. Or maybe accelerometers and som complex integration that lets you arrive at an energy. Or you could mount your spring onto a fixed object, for example by sticking a board in the ground. Then you eleminate the movement term and can focus on deformation.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-02-24T07:09:05.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1816" PostTypeId="1" CreationDate="2015-02-24T10:03:50.990" Score="4" ViewCount="105" Body="&lt;p&gt;I am wondering about the mathematical implications of concatenating very small windows from different trials.  All of the trials are the same size, and are for the same period of time after a stimuli. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, let's say my data has 3 channels, 6 trials, and 10 data points per trial ( size of data [3x10x6]).  What I want to build is a new data set with the same 3 channels, but with concatenating the windows of 2 datapoints for all of the trials, and then repeating this for the all of the possible number of windows.  In this example it would be 5 windows (size of new data [3x12x5]).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that any frequency analysis will be pointless due to dramatic changes in the signal, but what other implications to the data should I be concerned about? Would it still be reasonable to remove trending or denormalize this new data?&lt;/p&gt;&#xA;" OwnerUserId="672" LastEditorDisplayName="user16" LastEditDate="2015-02-24T17:49:35.287" LastActivityDate="2015-02-24T17:49:35.287" Title="Implications of concatenating EEG signals" Tags="|biomedical-engineering|signal|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1817" PostTypeId="1" CreationDate="2015-02-24T13:39:22.480" Score="7" ViewCount="704" Body="&lt;p&gt;A large turbine used in an electric plant can easily have 10+ meters in diameter and rotate at 70+ RPM. Should any asymmetry occur the rotating turbine will produce beat which will sooner or later destroy the plant. Crafting a turbine that large and perfectly symmetrical looks like a tough challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this problem solved? Why doesn't the beat from turbine asymmetry cause mechanical damage?&lt;/p&gt;&#xA;" OwnerUserId="374" LastActivityDate="2015-02-25T02:02:48.800" Title="How do large turbines deal with asymmetry and beat arising from those?" Tags="|dynamics|stresses|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1818" PostTypeId="2" ParentId="1817" CreationDate="2015-02-24T13:50:42.793" Score="4" Body="&lt;p&gt;Adjustable counterweights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wheels and tires of your car are not manufactured perfectly circular but after the tire gets put on (or you go for a rebalance) the mechanic will tack on weights to put the center of mass closer to the axle. The turbine can be balanced during normal maintenance in a similar way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember that 70 rpm is roughly 1.2 Hz. And the plant's structure will be engineered to have a natural frequency that is not a harmonic of the typical turbine speed.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-02-24T13:50:42.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1819" PostTypeId="2" ParentId="1817" CreationDate="2015-02-24T14:45:13.780" Score="14" Body="&lt;p&gt;It is done carefully.  Lots of other rotating machinery has the same problem, and whole systems exist just to deal with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, jet engines are usually smaller, but also usually spin much faster.  Balancing a jet engine is something that gets lots of attention at manufacturing, and again any time when it is put back together after having been sufficiently disassembled.  I have worked on metrology systems designed to measure a few &amp;micro;m non-roundness of circular parts a meter or more in diameter.  Large maintenance facilities have such systems, in addition to the manufacturers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method of balancing I saw used a stackup of multiple &quot;plates&quot;.  These plates were ring-like structures that would be part of the spinning system.  No plate could be manufactured exactly right.  The metrology system would accurately measure the asymmetries of each plate, then calculate what orientation the multiple plates have to be mounted together to make one larger object that was balanced to high precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also other techniques, like build it best you can, then balance afterwards.  Actually in practise multiple techniques are employed, with final assembly balancing always one of them.  This works a lot like dynamic balancing of tires.  The turbine is spun slowly, and the vibrations measured synchronous to the rotations.  From that a computer system calculates how much weight to add or remove where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common scheme I've seen is drilling holes or indentations in a certain area of metal left available for that purpose.  The computer will tell you where to drill and what diameter based on the analysis of the vibrations.  In another case we has a electronic product that spun as part of its normal operation.  In this case we left a few thru-hole solder pads at different angles from the shaft.  The vibration analyzer then suggested which pads to fill in with solder, which was done manually by a technician in this case.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This overall process of measure and adjust is repeated until the vibrations are below some specified level.  Despite the best measurements and algorithms, any one adjustment doesn't ever seem to null out the vibrations quite as much as it should in theory.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="41" LastEditDate="2015-02-25T02:02:48.800" LastActivityDate="2015-02-25T02:02:48.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1820" PostTypeId="2" ParentId="390" CreationDate="2015-02-24T15:19:03.620" Score="2" Body="&lt;p&gt;I agree with What Trevor said in a comment.  You specified 5V &amp;plusmn;10% (4.5-5.5 V) and that's what you got.  I don't see the problem here.  All the samples you got are within spec.  There may be other reasons to not use this vendor, but samples failing to meet specs is not one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, first article samples don't tell you that the vendor is OK, they only tell you whether the vendor is definitely not OK.  In the real world, you don't just qualify a vendor and then blindly buy from them after that.  It is a ongoing relationship.  The samples were all within spec, but then again, unless the vendor is really stupid, of course they were.  They can afford to manually check and hand-pick 15 items to send you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially since one out of 15 was quite close to the edge, you should have the vendor explain to you how their design or calibration procedure guarantees all the units will be in spec, how you know they will stay in spec over the expected lifetime, etc.  In short, it's still &lt;i&gt;your&lt;/i&gt; job to have some confidence that these sample are representative and that you know there is a system in place you believe in to deliver future product in spec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are really worried about this, then set up your own incoming inspection.  You could possibly do this with a random sampling from every lot received, or every unit, depending on how critical this is to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if you're really this worried, then your original specs are probably too loose.  Allowing the bare power supply to be &amp;plusmn;10% sounds unnecessarily high in the first place.  That's not pushing any limits, so tighter requirements like &amp;plusmn;5% shouldn't cost much more, if any.  Some electronic parts are specified for 5V &amp;plusmn;10%, so you can't have the bare supply eat up the entire error budget.  The voltage ulitimately delivered where it is used will have a wider tolerance than whatever it is at the supply.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-02-24T15:19:03.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1822" PostTypeId="2" ParentId="1763" CreationDate="2015-02-24T19:03:18.337" Score="1" Body="&lt;p&gt;One problem with the windvane pictured above is that it might require continous 360 degrees rotation around its mounting vertical axis, e.g. the case boat navigates in circles and/or wind direction rotates 360 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise the boat loses the advantage of having a wingsail being able to continously rotate in either direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That results in twisting of the bowden cable(s) when the actuator isnt mounted onto the windvane assembly, or twisting of the electric cables to the actuator when it does.  &lt;/p&gt;&#xA;" OwnerUserId="677" LastActivityDate="2015-02-24T19:03:18.337" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1823" PostTypeId="2" ParentId="662" CreationDate="2015-02-24T19:33:10.190" Score="11" Body="&lt;p&gt;I remember watching a long video about this in my senior seminar course. Scared the hell out of me.&lt;/p&gt;&#xA;&lt;p&gt;The Wex Law dictionary &lt;a href=&quot;https://www.law.cornell.edu/wex/negligence&quot; rel=&quot;nofollow noreferrer&quot;&gt;defines negligence&lt;/a&gt; as:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A failure to behave with the level of care that someone of ordinary prudence would have exercised under the same circumstances.  The behavior usually consists of actions, but can also consist of omissions when there is some duty to act (e.g., a duty to help victims of one's previous conduct).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Notice that negligence includes &lt;em&gt;&amp;quot;omissions when there is some duty to act&amp;quot;&lt;/em&gt;. The key thing to understand is, as the engineer on the project (i.e. the &amp;quot;engineer of record&amp;quot;), it was Gillum's duty to act in a way consistent with the standard of care required of them in their role as an engineer, and they failed to do so.&lt;/p&gt;&#xA;&lt;p&gt;The bottom line is &lt;strong&gt;NOT&lt;/strong&gt; whose drawing it was originally, or who came up with the design. Not only that (this might be controversial) but it may not have even legally mattered whether Gillum actually stamped the drawings with the design change (though the fact that they did stamp the drawings made the case against them a slam dunk). The stamp signifies something more fundamental: an attestation they were &lt;em&gt;aware&lt;/em&gt; and &lt;em&gt;cognizant&lt;/em&gt; of the design change.&lt;/p&gt;&#xA;&lt;p&gt;As the engineer of record on the project, it was the responsibility of Gillum &amp;amp; Associates - legally, professionally, ethically, and practically - to check those drawings, and to make it known that there was a problem on the project if one were found. But even if an engineer is not the engineer of record, if they make a statement in a professional capacity that a design is acceptable without having actually looked at the change, that is negligence.&lt;/p&gt;&#xA;&lt;p&gt;Some alternate scenarios:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;If&lt;/em&gt; Havens Steel and the contractor had made and implemented the change without Gillum knowing, there would have been no negligence on the part of Gillum.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;If&lt;/em&gt; Gillum had reviewed the drawing and sufficiently communicated to the manufacturer or the contractor that there was a problem, and yet the modified design was used anyway, it would have been a completely different case and although they would have certainly been part of the litigation, Gillum probably would have survived the storm.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;If&lt;/em&gt; Gillum had managed their liability more effectively by explicitly stating &lt;strong&gt;in the contract&lt;/strong&gt; that checking change orders was outside of their scope for the project, then it's possible that they might have escaped the episode since at that point the owner/contractor would have been deviating from the original design without approval. But even in that case, negligently stamping the design change would have doomed them.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditorDisplayName="user16" LastEditDate="2021-03-25T13:40:05.727" LastActivityDate="2021-03-25T13:40:05.727" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="1824" PostTypeId="1" AcceptedAnswerId="1835" CreationDate="2015-02-24T19:53:24.217" Score="5" ViewCount="496" Body="&lt;p&gt;I used LabView for a lot of my BME undergrad, but the labs focused mainly on things dealing electrical signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the real world I'm working mostly on ME projects, but we lack a good deal of experimentation equipment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the limitations of LabVIEW as it relates to ME experiments?&lt;/p&gt;&#xA;" OwnerUserId="465" LastEditorUserId="1586" LastEditDate="2015-06-19T20:23:51.720" LastActivityDate="2015-06-19T20:23:51.720" Title="LabView or another software for experiments" Tags="|mechanical-engineering|measurements|product-testing|labview|" AnswerCount="1" CommentCount="5" ClosedDate="2015-02-25T14:33:23.937" ContentLicense="CC BY-SA 3.0" />
  <row Id="1829" PostTypeId="1" AcceptedAnswerId="1844" CreationDate="2015-02-24T21:45:35.510" Score="5" ViewCount="9008" Body="&lt;p&gt;I have a metric thread pitch gauge that came in a tap and die set, it has pitches like $0.75$, $0.8$, $1.25$, etc. which is the distance between each thread. However there are two gauges that say $27$ and $28$ - I thought maybe it means $0.28\text{ mm}$ or possibly $0.28\text{ inches}$ if they threw in some non-metric ones, but it is neither of these (it's about $0.9\text{ mm}$) what are these?&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2016-06-16T18:36:24.120" Title="What does &quot;28&quot; and &quot;27&quot; mean on my metric thread pitch gauge?" Tags="|threads|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1831" PostTypeId="2" ParentId="251" CreationDate="2015-02-24T22:36:19.613" Score="5" Body="&lt;p&gt;Per ACI 318 13.2, two-way slabs are designed based on &quot;&lt;strong&gt;column strips&lt;/strong&gt;&quot; and &quot;&lt;strong&gt;middle strips&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To paraphrase the code:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;strong&gt;Column strip&lt;/strong&gt; is a design strip with a width on each side of a column centerline equal to 0.25L&lt;sub&gt;1&lt;/sub&gt; or 0.25L&lt;sub&gt;2&lt;/sub&gt;, whichever is less. Column strip includes beams if any.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;L&lt;sub&gt;1&lt;/sub&gt; and L&lt;sub&gt;1&lt;/sub&gt; are the span lengths in the two slab directions.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A &lt;strong&gt;Middle strip&lt;/strong&gt; is a design strip bounded by two column strips.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a simplified method in the code and has some criteria to meet. This method then allows you to accommodate holes in the slab by placing the missing reinforcement on the edges of the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how this compares to the Hilleborg method.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-24T22:36:19.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1832" PostTypeId="2" ParentId="659" CreationDate="2015-02-24T22:41:15.933" Score="1" Body="&lt;p&gt;I would use a thermal epoxy (&lt;a href=&quot;http://en.wikipedia.org/wiki/Potting_%28electronics%29&quot; rel=&quot;nofollow&quot;&gt;potting agent&lt;/a&gt;) for this. I have done this successfully with a thermistor before. Also I'm not sure, but you probably want to electrically isolate the thermocouple so that you don't lose voltage to an earthed component and thus get an incorrect reading. So a potting agent is great for this (high thermal conductivity, but not electrical conductivity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can buy split-bag packs from electronics suppliers, take off the clip so the liquids can mix and then pour/blob it on as needed. It then cures in about 24h. However if this is just a one-off application and you only need a small amount, this is probably a very expensive way to do this task, so find a local electronics hobbyist (try your local hackspace) and ask them for a small blob when they're doing a larger job with the stuff.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-02-24T22:41:15.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1833" PostTypeId="2" ParentId="1817" CreationDate="2015-02-24T23:35:07.473" Score="3" Body="&lt;p&gt;Additional counterweights may not be required. The individual blades on the rotor will have slightly different masses because of manufacturing tolerances. In aircraft engines, each blade is accurately weighed and the different masses are arranged around the shaft to improve the balancing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the rotor can be designed to run faster than its critical whirling speed in normal operation. In that situation, it tends to be self-centering, and the bearings can &quot;float&quot; on a film of oil so the rotor is spinning about its center of mass, not its geometrical center. The large unbalance forces only occur for a short time at startup and shutdown, when the rotor accelerates/decelerates through the whirling speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A related issue is that when the machine is shut down, the residual heat naturally rises and the temperature difference between top and bottom can bend the rotor out of balance. To avoid this the rotor may be continuously driven at a slow speed (e.g. 1 RPM) when the machine is nominally &quot;stopped&quot; until the rotor has cooled down, which may take several hours for a large turbine.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-02-24T23:35:07.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1834" PostTypeId="1" CreationDate="2015-02-24T23:59:22.283" Score="6" ViewCount="207" Body="&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/topics/aeronautics/features/amelia.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;AMELIA&lt;/a&gt; is the product of a partnership between NASA and the California Polytechnic State University. One of its innovations is circulation control. An engine is mounted over each wing, and high pressure air is redirected from the exhaust over the wing and through tiny slots, adding lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AMELIA is still in wind tunnel testing, but if it makes it to the prototype stage, it would have a carrying capacity of about 100 passengers. A good commercial comparison might be a low-seating version of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Boeing_737&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boeing 737&lt;/a&gt;, although it's &lt;em&gt;old&lt;/em&gt;. A newer (and perhaps better) comparison is with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Airbus_A318&quot; rel=&quot;nofollow noreferrer&quot;&gt;Airbus A318&lt;/a&gt;. Its cousin, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Airbus_A320_family#A319&quot; rel=&quot;nofollow noreferrer&quot;&gt;A319&lt;/a&gt;, is a similar aircraft that has been more popular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, assuming that a prototype/production version of AMELIA is manufactured, it might be pretty similar in size to these aircraft. They're all twinjets; AMELIA, though, has its engines on top of its wings. In theory, one could modify AMELIA to have more engines. This version would be different, just as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Boeing_747&quot; rel=&quot;nofollow noreferrer&quot;&gt;747&lt;/a&gt; is a completely different craft that the 737, or how the A318 can't compare to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Airbus_A380&quot; rel=&quot;nofollow noreferrer&quot;&gt;A380&lt;/a&gt;. Scaling it up, though, could be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the addition of two more engines&lt;sup&gt;1&lt;/sup&gt; interfere in any way with the airflow from the original engines and the subsequent lift produced? I doubt it, because engines on four-engine aircraft (e.g. the 747 and A380) don't seem to present any interference issues that impact performance. But the setup for AMELIA is pretty unique. Could adding an additional engine on either side interfere with airflow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/images/content/657865main_AMELIA_Sky1_Cal-Poly_5400x3600.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://www.nasa.gov/images/content/657865main_AMELIA_Sky1_Cal-Poly_5400x3600.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Image credit: NASA.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;For spacing considerations, imagine spacing like that on the A380. &lt;a href=&quot;http://www.airbus.com/aircraftfamilies/passengeraircraft/a380family/specifications/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; gives some basic specifications, while &lt;a href=&quot;http://www.airbus.com/fileadmin/media_gallery/files/tech_data/AC/Airbus-AC-A380-20131201.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; gives some drawings outright.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.023" LastActivityDate="2018-06-12T05:47:05.577" Title="Could adding extra engines to AMELIA disrupt airflow from the other engines?" Tags="|fluid-dynamics|aerospace-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1835" PostTypeId="2" ParentId="1824" CreationDate="2015-02-25T00:15:41.093" Score="3" Body="&lt;p&gt;LABVIEW can be easily be used for ME related experimentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One such example would be to use an actuating mechanism to exercise a user interface on a mechanical DUT, which dispense a specific material quantity into a holding container. The weight of the material is measured using a weighing scale which communicates to a computer or controller. Also the test system also includes a digital manometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PDmPN.jpg&quot; alt=&quot;Actuator&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/P6HLB.jpg&quot; alt=&quot;RS232 Communication&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LABVIEW commands the actuator via RS232, and the weight of the material is captured using a scale. The weight data is communicated back to LABVIEW via RS232 serial communication.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XA2iF.jpg&quot; alt=&quot;Scale&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/T9bSv.jpg&quot; alt=&quot;Manometer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example of using basic tools to develop fairly complex ME experiment,both electrical and mechanical tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hk549.jpg&quot; alt=&quot;Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Limitation of LABVIEW&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Labview is a very capable tool for bench top experimentation either in ME or EE environment. To the most part capability of the tools is limited by users understanding and experience of the LABVIEW software and other tools.&lt;/li&gt;&#xA;&lt;li&gt;Labview’s graphical programming environment doesn't blend with the traditional structured or OOP programming environment. Therefore maintenance and enhancement is a limitation in a traditional sense. National Instrument will argue against my opinion.&lt;/li&gt;&#xA;&lt;li&gt;Labview offers basic program structures such as for loops, if then else, and while loops. This is sufficient for typical basic bench top testing software. But Labview has limitation in implementing advance structure like a binary search tree or recursion. This can be done, but not too elegant.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With the growth in LABVIEW, it is almost necessary to have fairly modern computer to use LABVIEW software. But with currently a good enough computer can be purchased for a bargain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, limitation is mostly the skill of the LABVIEW user. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.miraiintertech.com/home/downloads/SCN-SCL-Manual2011.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mechatronics Cylinder&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cn.mt.com/cn/zh/home/supportive_content/product_documentation/installation_instructions/PS_Comm_Tst/jcr:content/download/file/file.res/Ser_Communication_Test_PS_scale_7-23-09.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Serial Communication Test (RS232) PS Scale&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://tmi.yokogawa.com/files/uploaded/BU7673_00E_015_1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Digital Manometers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-02-25T23:21:22.147" LastActivityDate="2015-02-25T23:21:22.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1836" PostTypeId="2" ParentId="63" CreationDate="2015-02-25T00:18:55.500" Score="1" Body="&lt;p&gt;Most of the pros regarding a traffic circle have been touched on here. It's also worth mentioning that traffic circles, despite their efficiency in decreasing accidents in heavily-congested areas, are truly harrowing -- often fatal -- for the blind &amp;amp; impaired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long, R. G., Guth, D. A., Ashmead, D. H., Emerson, R. W., &amp;amp; Ponchillia, P. E. (2005). Modern roundabouts: Access by pedestrians who are blind. &lt;em&gt;Journal of Visual Impairment &amp;amp; Blindness, 99&lt;/em&gt;(10), 611-621.&lt;/p&gt;&#xA;" OwnerUserId="684" LastActivityDate="2015-02-25T00:18:55.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1837" PostTypeId="2" ParentId="465" CreationDate="2015-02-25T00:53:02.867" Score="5" Body="&lt;p&gt;Perhaps a press-fit spring plunger?&#xA;&lt;img src=&quot;https://cdn.mscdirect.com/global/images/ProductImages/6515556-23.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some product codes to get you started:&#xA; - MCS: 05137617, 05137815, 65155566&#xA; - Grainger: 2YLK5, 2YLN2, 2YLL9&lt;/p&gt;&#xA;" OwnerUserId="684" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.043" LastActivityDate="2015-02-25T00:53:02.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1838" PostTypeId="2" ParentId="1758" CreationDate="2015-02-25T01:35:51.013" Score="3" Body="&lt;p&gt;Since your enclosure is IP66 you can pour a steady stream of water over it to absorb and carry the heat away.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-02-25T01:35:51.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1839" PostTypeId="2" ParentId="1758" CreationDate="2015-02-25T01:48:03.200" Score="2" Body="&lt;p&gt;You could pre-cool the gas entering the enclosure, the heat would then go into re-warming up this gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be done by passing the gas through a simple finned radiator-pipe in a temperature controlled environment (basically a pipe through a fridge/freezer). This might be a good way to do it if you want to control the temperature that the gas re-enters the chamber as you could control the temperature of the pre-cooler and monitor the temperature that goes back in.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-02-25T01:48:03.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1840" PostTypeId="1" AcceptedAnswerId="1845" CreationDate="2015-02-25T02:42:17.167" Score="5" ViewCount="427" Body="&lt;p&gt;I was thinking about fighter aircraft like the &lt;a href=&quot;http://en.wikipedia.org/wiki/Hawker_Siddeley_Harrier&quot;&gt;Harrier&lt;/a&gt; which have &lt;a href=&quot;http://en.wikipedia.org/wiki/Thrust_vectoring&quot;&gt;thrust vectoring&lt;/a&gt; and computer-controlled stabilisation nozzles. Also an &lt;a href=&quot;http://en.wikipedia.org/wiki/Dihedral_%28aeronautics%29#Anhedral&quot;&gt;anhedral angle&lt;/a&gt; on the (small) wings. The aircraft relies so much on the computer stabilisation, and flies like a brick if the engine fails, so why have those wings at all? Couldn't you get as much lift from wings of the same area that run parallel to the fuselage, that is much shorter (left-to-right) but deeper (front-to-back) wings? And wouldn't this create much less drag reducing the thrust requirements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having thought about this I realised that for fighters it's probably not a good idea as they would lose a lot of pitch-maneuverability, but for passenger jets, would a roll-stabilised approach with a short wing that runs the full length get the same lift while reducing drag? Is such a wing viable?&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-04-18T19:30:52.350" Title="Is a wing that runs the length of an aircraft viable?" Tags="|aircraft-design|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1841" PostTypeId="2" ParentId="1840" CreationDate="2015-02-25T03:44:50.093" Score="2" Body="&lt;p&gt;Thinking about my comment above, the fuselage length side fins could be replaced with smaller stabilising fins and the technology used for &lt;a href=&quot;https://en.wikipedia.org/wiki/Cruise_missile&quot; rel=&quot;nofollow&quot;&gt;cruise missiles&lt;/a&gt;, with the addition of thrust vectoring, adapted for passenger flying aircraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flight while airborne could be achieved, as proven by cruise missiles and other missiles. Take off could be ramp assisted if needed, it's partly why the British aircraft carriers had ramps at the bow end when the Harrier jump jets were in service. Landing could be an issue but that's where the thrust vectoring would be critical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other option is to replace the narrow side fins with &lt;a href=&quot;https://en.wikipedia.org/wiki/Variable-sweep_wing&quot; rel=&quot;nofollow&quot;&gt;variable sweep wings&lt;/a&gt; that could be retracted during flight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thrust vectoring functions would be critical to provide manoeuvrability when unexpected events occur, such a avoiding bad weather and encroachment into flying space by other aircraft - collision avoidance.   &lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-02-25T03:44:50.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1842" PostTypeId="1" CreationDate="2015-02-25T04:44:09.673" Score="4" ViewCount="277" Body="&lt;h1&gt;Motivation&lt;/h1&gt;&#xA;&lt;p&gt;Imagine that you are a recent graduate of an ABET-accredited engineering program in the United States. You have taken and passed the Fundamentals of Engineering exam and you are hunting for your first job as an engineer. You send out applications, sit for interviews and have the good fortune to receive two offers of employment. Both positions are interesting and offer reasonable compensation, benefits and responsibilities at established companies; however, only one of the two positions represents &lt;em&gt;qualifying experience&lt;/em&gt; for the purpose of becoming a licensed Professional Engineer (PE).&lt;/p&gt;&#xA;&lt;h2&gt;Problem:&lt;/h2&gt;&#xA;&lt;p&gt;&lt;strong&gt;What is the dollar value of that &lt;em&gt;qualifying experience&lt;/em&gt;?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;In other words, how can the applicant calculate the value of progress toward PE licensure as part of the &amp;quot;total compensation&amp;quot; the job offers, as is commonly done for benefits such as employer-subsidized health insurance, relocation assistance or gym memberships, so as to make a more informed decision about which offer to accept?&lt;/p&gt;&#xA;&lt;h1&gt;Discussion&lt;/h1&gt;&#xA;&lt;p&gt;For a position to represent &amp;quot;qualifying engineering experience,&amp;quot; &lt;a href=&quot;http://www.nspe.org/resources/licensure/resources/demonstrating-qualifying-engineering-experience-licensure&quot; rel=&quot;nofollow noreferrer&quot;&gt;The National Society of Professional Engineers specifies&lt;/a&gt; that it must involve:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A major branch of engineering&lt;/li&gt;&#xA;&lt;li&gt;Supervision by one or more qualified engineers&lt;/li&gt;&#xA;&lt;li&gt;The development of &amp;quot;technical skill and initiative in the application of engineering principles and sound judgment in reviewing such applications by others&amp;quot;&lt;/li&gt;&#xA;&lt;li&gt;A reasonably well-rounded experience in that branch of engineering&lt;/li&gt;&#xA;&lt;li&gt;Work of increasing complexity and responsibility&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I myself interviewed for two positions at the same organization, with the same title, shortly after finishing my undergraduate degree. One of the positions was supervised directly by a PE and the other was not, nor did it involve work that would be reviewed at some point by a PE—it was made very clear during the interviews whether or not each job represented qualifying experience toward a PE.&lt;/p&gt;&#xA;&lt;p&gt;In other cases, a recent graduate may entertain offers from both traditional engineering firms and companies that operate in related, but not strictly engineering, fields. For example, &lt;a href=&quot;http://en.wikipedia.org/wiki/Accenture&quot; rel=&quot;nofollow noreferrer&quot;&gt;Accenture&lt;/a&gt; recruits engineering graduates at my alma mater for non-engineering positions. Some engineering graduates have a background that qualifies them for software development positions. Others may have the option of pursuing positions in R&amp;amp;D or pure research.&lt;/p&gt;&#xA;&lt;p&gt;The best answers will thoroughly justify their analysis and cite references for any statistics used. I wouldn't discount a novel &amp;quot;back of the napkin&amp;quot; approach, though.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-02-26T18:41:27.597" Title="What is the dollar value of &quot;qualifying experience&quot; in an entry-level US engineering job?" Tags="|licensure|employment|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1843" PostTypeId="1" AcceptedAnswerId="1850" CreationDate="2015-02-25T09:47:35.883" Score="2" ViewCount="145" Body="&lt;p&gt;In submerged conditions galvanic corrosion effects exist between metals with dissimilar nobleness and the presence of seawater as electrolyte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does reducing the contact area by using an nonconducting paint &lt;em&gt;on the interface layer between the two metal parts&lt;/em&gt; help delay this process? Or is any the solid contact electrical route, e.g. via a bolt or so, enough to make the deterioration speed limited by the electrolytic process c.q. the surface area in contact with the electrolyte?&lt;/p&gt;&#xA;" OwnerUserId="270" LastActivityDate="2015-02-25T13:40:25.757" Title="Subsea galvanic corrosion protection by paint" Tags="|materials|steel|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1844" PostTypeId="2" ParentId="1829" CreationDate="2015-02-25T09:59:15.317" Score="6" Body="&lt;p&gt;28 is a standard Threads Per Inch dimension for 1/4 and 1/2 inch screws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{25.4 \,\text{[mm/Inch]}}{28\, \text{[Threads/Inch]}} = 0.9 \,\text{[mm metric pitch]}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 27 seems strange to me. But I usually use metric bolts.&lt;/p&gt;&#xA;" OwnerUserId="270" LastActivityDate="2015-02-25T09:59:15.317" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1845" PostTypeId="2" ParentId="1840" CreationDate="2015-02-25T10:07:03.470" Score="8" Body="&lt;p&gt;You mean very low &lt;a href=&quot;https://en.wikipedia.org/wiki/Aspect_ratio_%28aerodynamics%29&quot;&gt;aspect ratio (span to longitudinal length)&lt;/a&gt; wings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low aspect ratio has larger induced drag and glide very poorly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total drag coefficient of an aircraft can be expressed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_d = C_{D0}+ \frac{{C_L}^2}{\pi e AR}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_D$ is is the aircraft drag coefficient&lt;br&gt;&#xA;$C_{d0}$    is the aircraft zero-lift drag coefficient,&lt;br&gt;&#xA;$C_L$    is the aircraft lift coefficient,&lt;br&gt;&#xA;$\pi$    is pi,&lt;br&gt;&#xA;$e$    is the Oswald efficiency number&lt;br&gt;&#xA;$AR$    is the aspect ratio.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that drag is inversely proportional to the aspect ratio. So a low aspect ratio needs more thrust just to stay in the air. This means higher fuel costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the requirement for commercial aircraft is being able to recover after losing your engines. A good glide ratio (which needs high lift to drag) is essential to that.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-02-25T10:07:03.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1847" PostTypeId="1" AcceptedAnswerId="1851" CreationDate="2015-02-25T10:31:13.053" Score="5" ViewCount="170" Body="&lt;p&gt;&lt;strong&gt;The setup&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a cylinder inside another cylinder (placed with their symmetry axis horizontal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inner cylinder is about Ø100 mm in diameter and a centimeter or two in lenght/thickness. On the perimeter there is a thread. This cylinder is &lt;strong&gt;fixed&lt;/strong&gt; in rotational motion, and cannot rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer cylinder is much longer. It has a thread on the inside through half of its' lenght or so and this thread fits with the inner cylinder thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when rotating the outer cylinder, the inner cylinder will move back and forth (the inner cannot rotate but can move translationally).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My issue is that the outer cylinder might rotate many revolutions (it follows a bicycle wheel), and therefor at some point the inner cylinder will move &lt;em&gt;out of&lt;/em&gt; the thread of the outer cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a mechanical and smart method to let such a cylinder leave and &lt;em&gt;reenter&lt;/em&gt; the thread of the outer cylinder?&lt;/strong&gt; E.g. the outer cylinder would rotate in one direction untill the inner cylinder leaves the thread. Then when the outer cylinder is reversed in rotational direction, I want the inner cylinder to reenter the thread and start moving the opposite way.&lt;/p&gt;&#xA;" OwnerUserId="94" LastActivityDate="2015-02-25T17:57:41.363" Title="How to let a &quot;screw&quot; leave and reenter a thread" Tags="|mechanical-engineering|threads|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1850" PostTypeId="2" ParentId="1843" CreationDate="2015-02-25T13:40:25.757" Score="3" Body="&lt;p&gt;No solid contact is required at all, as the seawater provides the electrical route. In order to prevent galvanic corrosion, you would need to electrically isolate your two metals. That said, I can see that a solid electrical route would allow faster transfer of electrons, speeding up corrosion, and hence having no solid electrical route would delay corrosion.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-02-25T13:40:25.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1851" PostTypeId="2" ParentId="1847" CreationDate="2015-02-25T14:01:33.513" Score="7" Body="&lt;p&gt;&lt;strong&gt;Threads&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have the thread automatically reengage some thought will have to go to the exact type of thread used.  A coarse thread and maybe a slight taper will help the threads to properly engage while minimizing the chance of cross-threading or binding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mechanism&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You said that eventually the inner piece will disengage from the outer piece. This doesn't sound like the problem. Your problem is that you then want the pieces to reengage when the rotation is reversed. To do this, the inner part needs to have some force on it to keep it in contact with the outer part. This sounds like a perfect case for using a spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spring would start to compress as the inner piece neared the end of the outer piece. Once the inner piece is free, the spring would provide enough force to press the inner piece back against the outer piece. This way the threads could engage once the rotation is reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A downside to this is that there will be some wear between the threads of each piece. The spring is constantly pushing the two together even as the rotation is keeping the threads from engaging.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-02-25T14:01:33.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1852" PostTypeId="2" ParentId="1840" CreationDate="2015-02-25T14:07:26.347" Score="2" Body="&lt;p&gt;Helicopters are proof you don't need wings at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, short wings (left right), no matter how long (front back), are less efficient than wide wings.  You can see this from basic physics without having to understand anything about how wings actually work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a plane in straight, level, and steady flight.  The net force on the air is to push down on it by the weight of the plane.  That force is produced by imparting momentum downwards on the air immediately surrounding the plane as the plane flys by.  Momentum is mass x velocity.  In this sense stubby wings and wide wings are equivalent.  You can get the same momentum by pushing a little air a lot (stubby wing), or a lot of air a little (wide wing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, consider the power requirement.  Power is proportional to the &lt;i&gt;square&lt;/i&gt; of velocity times the mass.  Therefore, pushing a little air at high speed takes more power than pushing a lot of air at low speed.  Stubby wings transfer more power into the air for the same lift.  This extra power shows up as higher drag, which ultimately requires more pushing from the engines to overcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wide and thin wings are best for efficiency, but there are structural limits and other trade offs.  Note that wings of gliders (where efficiency is very important since the power comes from altitude loss) are very wide, but thin in the other two dimensions.  They also can't carry much payload, in part because the wings are too fragile to support it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything is a trade off.  Jet fighters have other important criteria, like maneuverability, high top speed, good cockpit visibility, small radar cross section, etc.  It is often useful to give up some efficiency in return for these other features.  It all depends on what the plane is supposed to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the F104 as a example of stubby and thin wings.  It was fast, but also very tricky to fly, with several pilots lost due to inability to control the plane.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="282" LastEditDate="2015-04-18T19:30:52.350" LastActivityDate="2015-04-18T19:30:52.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1853" PostTypeId="1" AcceptedAnswerId="1871" CreationDate="2015-02-25T15:21:54.267" Score="6" ViewCount="3782" Body="&lt;p&gt;As I was browsing &lt;a href=&quot;http://www.greenbookee.com/iso-965-1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISO 965-1&lt;/a&gt; it defines tolerances on a plenty of variables regarding metric threads, but it gives the pitch of the thread as a simple number from an allowed set ( 3 – 2 – 1,5 – 1– 0,75 – 0,5 – 0,35 – 0,25 – 0,2 ), and no tolerances on these are ever given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/59SoE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it so? Are they defined in some other standard or is there a specific reason why they aren't defined to a given tolerance? Specifically, when I'm using a threaded rod as a lead-screw actuator, I'd like to know what inaccuracy can I expect coming from imprecision of pitch - how accurately do 1000 turns of a M10x1 threaded rod convert to 1000mm travel distance of the nut (minus backslash etc)? &lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2015-02-26T21:34:59.907" LastActivityDate="2015-02-27T08:36:59.553" Title="Is the tolerance of pitch of a thread standarized?" Tags="|mechanical-engineering|threads|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1854" PostTypeId="2" ParentId="1847" CreationDate="2015-02-25T17:57:41.363" Score="0" Body="&lt;p&gt;The simplest solution would be to use a spring such that it exerts a small force to push the moving cylinder back in. As it 'unscrews' the spring will be tensioned (or compressed) and as the wheel starts spinning the other way, the small force will help the thread to re-enter.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-02-25T17:57:41.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1855" PostTypeId="2" ParentId="1842" CreationDate="2015-02-25T18:59:28.730" Score="0" Body="&lt;p&gt;Trying to put a dollar value on something like that is missing the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first question you should be asking yourself is whether you want to, and you think it's appropriate for your career objectives, to become a PE.  If you're a civil or structural engineer, then that is pretty much mandatory.  In other engineering fields it may be irrelevant.  For example, I've been a electrical engineer for 35 years and don't have a PE license.  It just not something the that matters for the kind of things I do and who I do them for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So forget about monetary value.  Do you want to be a PE or not?  If so, then you can consider activities that get you there as part of your education.  Did you get your engineering degree in the first place because you did the monetary analysis and decided you'd make more money that way as apposed to being, say, an auto mechanic?  No, of course not.  Or at least I hope not.  If you did, bail out now, you'll never be more than a mediocre engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Becoming a PE is part of your overall engineering education.  If you decide that's the direction you want to go in, because you have an aptitude, interest, and passion in the field, then take the job that fulfills that goal.  If you don't intend to become a PE, then take either job based on other criteria, like which one you have the best chance of learning from or being mentored by someone that you get along with and likes doing such things.  Money should NOT be much of a consideration this early in your career.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="110" LastEditDate="2015-02-26T03:00:30.720" LastActivityDate="2015-02-26T03:00:30.720" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1856" PostTypeId="2" ParentId="1739" CreationDate="2015-02-25T21:22:40.597" Score="1" Body="&lt;p&gt;You could use a standard Ø8 m6 dowel pin in a Ø8 H7 hole in your plate, but have the protruding end machined to Ø6 g6. That gives you your standard hole in the plate, an interference fit for the pin so it won't come out, your g6 fit for the sensor and a step so its impossible to assemble the pin the wrong way round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're worried about the pin passing through the plate, either put a counterbored hole in the plate with a step for the pin to sit on or make the pin from a drill blank or some other 'hard' steel with a Ø5 m6 section, the Ø6 end won't pass through the plate.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2015-02-25T21:22:40.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1857" PostTypeId="2" ParentId="1842" CreationDate="2015-02-25T22:28:26.740" Score="7" Body="&lt;p&gt;From what I was able to find (see &lt;a href=&quot;http://career-advice.monster.com/career-development/education-training/professional-engineer-license-pe/article.aspx&quot;&gt;reference 1&lt;/a&gt; and &lt;a href=&quot;http://news.thomasnet.com/imt/2013/07/15/is-a-professional-engineer-license-worth-it&quot;&gt;reference 2&lt;/a&gt;), it appears that a licensed PE is paid on average approximately $5,000 per year more than a non-licensed engineer.  Note that the amount holds up even with different organizations performing the salary survey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://news.thomasnet.com/imt/2013/07/15/is-a-professional-engineer-license-worth-it&quot;&gt;reference 2&lt;/a&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on 2010 survey findings from the National Society of Professional Engineers (NSPE), licensed engineers make an annual median salary of \$99,000, while those with no license or professional certification earn \$94,000 a year. More recently, according to a 2012 report from the American Society of Civil Engineers and the American Society of Mechanical Engineers, engineers with a PE license earn a median income of \$100,000, compared with the \$95,500 for unlicensed engineering professionals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Plugging those numbers into a &lt;a href=&quot;http://en.wikipedia.org/wiki/Present_value&quot;&gt;present value&lt;/a&gt; spreadsheet, I came up with a present value of approximately \$33k in favor of taking the role leading towards a PE. &lt;sub&gt;See assumptions below&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that your question pertains to a recent college graduate.  The present value would be less for someone who earns their PE license later on in their career.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Money obviously isn't the only reason to become a Professional Engineer (see &lt;a href=&quot;http://www.engineeringandleadership.com/4-reasons-why-im-applying-to-become-a-professional-engineer/&quot;&gt;other reasons 1&lt;/a&gt; and &lt;a href=&quot;http://www.nspe.org/resources/licensure/why-get-licensed&quot;&gt;other reasons 2&lt;/a&gt;).  But the economics of earning your PE are certainly attractive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those additional references cite career development and enhanced job security in the sense of it being easier to find another job should you become unemployed.  It's pretty difficult to assign a readily agreed upon value to those intangibles, but most would agree that they have some amount of value to them.  Even if you're having a hard time seeing how a PE license would lead to a greater salary, one of the references points out that having a PE may lead towards a role with greater responsibility.  More job responsibility often correlates with an increase in salary.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;My assumptions:&lt;br&gt;&#xA;* Work for 40 years (eg. N = 40)&lt;br&gt;&#xA;* No additional income for first 4 years (recognizing typical 4 years of work experience requirements)&lt;br&gt;&#xA;* Assumed an average of \$5000 additional per year of employment and that the salary differential remains consistent across all of the years of employment&lt;br&gt;&#xA;* Discount rate of 10% (conservative given historical market rate of 6 or 7%)&lt;br&gt;&#xA;* A 7% discount rate would yield ~\$49.5k and a 5% rate would yield \$68k&lt;br&gt;&#xA;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-02-25T22:47:39.423" LastActivityDate="2015-02-25T22:47:39.423" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1858" PostTypeId="1" CreationDate="2015-02-26T03:00:02.540" Score="11" ViewCount="1178" Body="&lt;p&gt;&lt;strong&gt;Lifting Body&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lifting_body&quot; rel=&quot;noreferrer&quot;&gt;lifting body&lt;/a&gt; design is to shape a vehicle's body in such a way as to produce lift without wings. Research has shown that this can be an efficient method of reducing drag while still providing lift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has typically done for aircraft or spacecraft:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5wItO.jpg&quot; alt=&quot;lifting bodies&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Could a similar approach help to make trains more efficient?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Passenger trains already look streamlined and aerodynamic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cmaiq.jpg&quot; alt=&quot;TGV&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Freight trains do not:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LzY9I.jpg&quot; alt=&quot;freight train&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air resistance is not the only form of resistance that trains must overcome. They must also overcome the resistance of their wheels on the track. This is where I would think that the energy savings from a lifting body design would come from. Any additional lift created by the lifting body design would reduce the friction between the wheels and rail, thus saving energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Train wheels have flanges on them, so they don't need traction to steer. The cars also don't need traction on the rails since the driving wheels are only on the engines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a lifting body create enough lift at typical train speeds to make a noticeable difference?&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2023-08-08T22:05:57.833" Title="Would a &quot;lifting body&quot; design be useful for a train car?" Tags="|aerospace-engineering|rail|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="1859" PostTypeId="2" ParentId="1858" CreationDate="2015-02-26T05:08:03.180" Score="13" Body="&lt;p&gt;I won't say it's not possible to make a noticeable difference.  But I would say that it's pretty darn unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lift and drag forces on any body generally depend on the velocity of the body ($v$), the density of the fluid ($\rho$), the area of the object ($A$) and a dimensionless coefficient ($C_L$ or $C_D$).  The applicable area can be a bit fuzzy in terms of whether you're talking about frontal area or planform area, but for a train this is going to basically be the area in the plane normal to the direction of travel (given that most of the train cars are simply in the wake of the locomotive).  The coefficients can vary a bit with velocity due to turbulence, but will generally be between 0 and 2. The functional form is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F_L = 1/2\rho A C_L v^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crudely estimating a freight train as being $4\, m$ tall and wide, moving at $16 \, m/s$ ($57 \, kph$).  This gives us that lift will be around $2500 N$ for the entire train (not enough to lift an ordinary automobile). This calculation is not meant to be extremely precise, but even it's off  by a factor of ten, lift would be minimal compared to the overall weight of the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, you might consider the merits of adding additional aerodynamic elements.  Here you get into a tradeoff between generating lift (to reduce downward force on the wheels) and generating drag.  While lift-to-drag ratios can get high (~50 in good cases) the rolling resistance of train wheels is actually very low ($C_{rr} \sim 0.00035$).  So although the lift will reduce the losses due to friction, it will increase the losses due to drag more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, reducing drag is good but providing lift is probably not worth it. In fact, reducing drag at the cost of adding some weight could be beneficial because wheels are so very good at bearing that load efficiently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B. I consulted &lt;a href=&quot;http://en.wikipedia.org/wiki/Rolling_resistance&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;http://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html&quot;&gt;engineering toolbox&lt;/a&gt; for $C_{rr}$ data. I could not verify sources for railroad wheels, but was able to confirm that values for road tires were correct via &quot;Fundamentals of Vehicle Dynamics&quot; by Gillespie. Any ideas on verification would be welcome.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-02-26T05:08:03.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1860" PostTypeId="2" ParentId="1858" CreationDate="2015-02-26T09:31:50.210" Score="10" Body="&lt;p&gt;You don't want to decrease the downward force on the wheels because that is your primary means of &lt;strong&gt;slowing down in an emergency&lt;/strong&gt;. It already takes a while to stop a train, don't make it any longer. Remember that kinetic friction (wheels locked) is proportional to the downwards force on the wheel and the brakes are spread along to each car.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Train wheels have flanges on them, so they don't need traction to steer. The cars also don't need traction on the rails since the driving wheels are only on the engines.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They &lt;strong&gt;do&lt;/strong&gt; need traction, wheels are slightly conical with the radius at the outside being larger than the inside and they are both coupled to the axle. This way when the train is off-center or in a turn the outside wheel has a larger effective radius thus will turn it back to the center of the track.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-02-26T10:21:24.057" LastActivityDate="2015-02-26T10:21:24.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1861" PostTypeId="1" CreationDate="2015-02-26T12:17:55.497" Score="24" ViewCount="8343" Body="&lt;p&gt;In a gas turbine engine there're multiple sets of blades - one set after another and combustion products pass all the sets and each set of blades gets some of the power. This increases utilization of power from burning gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile hydroelectric plants use turbines with a single set of blades and the typical usecase is where there's a channel for feeding water from an elevated reservoir and the turbine is at the bottom and water runs through the turbine and then just flows down the river. I assume there's still some mechanical power left unextracted when water flows out of the turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are water turbines not &quot;chained&quot; so that water exiting the first turbine drives the second turbine using the residual mechanical power?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="6790" LastEditDate="2020-07-31T21:39:26.550" LastActivityDate="2023-12-15T13:53:38.403" Title="Why don't hydroelectric plants use cascades of turbines instead of single turbines?" Tags="|mechanical-engineering|turbines|water-resources|renewable-energy|hydroelectric-dams|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1862" PostTypeId="2" ParentId="1861" CreationDate="2015-02-26T13:47:11.377" Score="3" Body="&lt;p&gt;Water is going to have to leave the turbine at a speed. That what you've referred to as its residual mechanical power. The thing is, the turbine has already slowed down the water as much as can reasonably be done, while still allowing the water to leave the plant and not flood it. So slowing it down further with an extra stage of turbine just isn't an option. If it could be slowed down further, then the first turbine would be designed to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are examples of turbines in series: there are rivers with more than one run-of-river hydro plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for most storage hydro, it's simplest just to extract as much of the kinetic energy as you can in one go. It's fewer things to maintain and manage. Chaining them in series would just reduce the energy available for the downstream turbines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, the energy you can retrieve is limited to height of the drop times weight of the water (times &lt;em&gt;g&lt;/em&gt;, gravity's acceleration), minus the kinetic energy of the water upon leaving the plant. (It can't leave with zero kinetic energy, as zero kinetic energy would mean that it didn't leave the plant at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding more turbines has no effect whatsoever on that equation. If the drop's the same, and the mass of water is the same, and the speed of the water leaving the plant is the same, then the amount of energy harvested is the same (assuming constant turbine efficiency).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, from your question, you're wondering why a hydro plant isn't more like a CCGT, with its multi-stage turbines. A hydro plant is much simpler, more efficient, and more effective than a CCGT. A CCGT has its complications because it's a thermal plant with highly-compressible fluids and a phase transition (water to steam). A hydro plant is just harvesting kinetic energy. A cascade of turbines doesn't offer anything other than complications to a hydro plant.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-02-27T11:37:54.607" LastActivityDate="2015-02-27T11:37:54.607" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="1863" PostTypeId="2" ParentId="1861" CreationDate="2015-02-26T13:53:00.807" Score="33" Body="&lt;p&gt;The exhaust gasses are compressible fluids, whereas liquid water is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an animation of how a gas turbine works: &lt;a href=&quot;https://www.youtube.com/watch?v=gqNtoy2x5bU&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=gqNtoy2x5bU&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the combustion stage, the gas and compressed air are mixed together, already at high pressure. The burning releases the energy stored in the gas, heating up the released gasses (exhaust). This would create an even higher pressure, so in order to prevent back-flow the combustion section is a larger volume to keep the pressure the same or lower. This large volume of high pressure gasses drives the turbine. As these high pressure compressed gasses pass through the first set of blades, &lt;strong&gt;the pressure reduces and the gases expand&lt;/strong&gt;. There is still some pressure left and more energy can be extracted with another set of blades, and another, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since liquid water is not compressible, it doesn't expand as the pressure reduces. This actually makes it much easier to extract the energy. You pass the water through a nozzle, reducing the high pressure inside the pipe to atmospheric pressure outside the nozzle, and accelerating the water up to a high velocity. This energy can then be extracted all in one go by the turbine, because the water is not expanding and energy escaping elsewhere. &lt;a href=&quot;http://en.wikipedia.org/wiki/Turgo_turbine&quot; rel=&quot;noreferrer&quot;&gt;Turgo turbines&lt;/a&gt; are actually very efficient at extracting this energy, up to 90%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why multiple stages are not needed in hydro plants. However you could still 'chain' them together in the literal sense. If you have a very large drop, you could put a series of small turbines at intervals down the drop, the released water from one going into another. However, the amount of energy available would not change from having a larger turbine at the very bottom and using higher-pressures.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="95" LastEditDate="2015-02-28T03:13:45.743" LastActivityDate="2015-02-28T03:13:45.743" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1864" PostTypeId="1" AcceptedAnswerId="1867" CreationDate="2015-02-26T15:06:37.470" Score="15" ViewCount="1197" Body="&lt;p&gt;In a program on &lt;a href=&quot;http://www.npr.org/2015/02/25/389008046/a-hard-look-at-the-risks-of-transporting-oil-on-rail-tanker-cars&quot; rel=&quot;noreferrer&quot;&gt;NPR&lt;/a&gt; that I was listening to, there was a bit about a bridge that from the description sounded to a layman as unsound and is still in use. The program described it as an old wooden railway (and I'm aware that has its own set of challenges) bridge with rotting timbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the United Sates, if a member of the public sees a bridge (railway, tramway, car/truck, foot, bike, etc...) that is of questionable soundness, what is the process for him or her to determine who us responsible for it, and if they (the responsible parties) should look at it and have it evaluated?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a particular agency that is responsible for regulating bridges and ascertaining their safety?&lt;/p&gt;&#xA;" OwnerDisplayName="user348" LastEditorUserId="33" LastEditDate="2015-03-20T01:29:50.627" LastActivityDate="2015-08-24T22:43:39.660" Title="How should the public raise questions about unsafe structures in the United States?" Tags="|civil-engineering|bridges|regulations|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1865" PostTypeId="2" ParentId="1864" CreationDate="2015-02-26T15:28:51.393" Score="6" Body="&lt;p&gt;Here in the US, I would contact the state department that manages roadways.  Those are referred to as the &quot;highway department&quot;, &quot;department of public works&quot;, and various other names.  Here in Massachusetts it's the Department of Transportation, often known as &quot;Mass DOT&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays state governments have web sites, so it should not be hard to at least find a general phone number for the department.  You have to be prepared to get bounced around to different people, but eventually, if you preservere, you should be able to talk to someone that has oversight responsibility for the structure in question but is low enough down to actually know what you mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, most likely what you think is a problem isn't.  Public infrastructure is reviewed and inspected on a schedule in most cases.  This schedule is designed so that major changes, like corrosion, won't happen fast enough between inspections to cause a problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sudden catastrophic things do happen, and there is always the unlikely remote chance that you are the one person that happened to notice, but probably not.  Most likely you will just be wasting their time, and may also be treated like that unless and until you can provide enough evidence of a real problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really think a piece of public infrastructure has a problem, take a picture of it.  That will help the person you are reporting the problem to evaluate quickly whether it's just something that looks bad to a layman or really is a problem they weren't already aware of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing you might do is show the picture to someone you know with the appropriate engineering background.  If they think it still could be a real problem, then go ahead and report it.  If not, then you haven't wasted the public servants' time, which ultimately you are paying for.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-02-26T15:28:51.393" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1866" PostTypeId="1" CreationDate="2015-02-26T15:33:32.127" Score="5" ViewCount="566" Body="&lt;p&gt;I have the opportunity to do a few metal 3D prints for just the material cost, and would like to experiment with the performance of heat sink designs (mostly for fun). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I have no experience designing heat sinks, I would say that the expected failure rate when just printing what I have in mind and testing it will be quite high, so I need a good way to evaluate the performance of a design before creating a prototype.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since this is just for my fun, I do not want to invest in pricey finite element simulation tools. I could not find anything free that is useful, since I think it has to take into account airflow and thermal radiation. How can I get an idea of how my design will perform before printing a prototype and without using finite element simulation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am planning on a 4 cm by 4 cm 120 W Peltier element (ceramic surface), the heat sink should be  stainless steel, though I am hoping the relative performance of two heatsinks doesn't change with a different material (one day I hope I can use copper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not so sure about temperature differential here but I am assuming 25 °C ambient temperature.&lt;/p&gt;&#xA;" OwnerUserId="706" LastEditorUserId="706" LastEditDate="2015-02-26T18:43:16.403" LastActivityDate="2015-02-26T21:25:48.177" Title="Evaluating performance of heat sink designs without finite element simulation" Tags="|heat-transfer|finite-element-method|3d-printing|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1867" PostTypeId="2" ParentId="1864" CreationDate="2015-02-26T15:51:03.940" Score="12" Body="&lt;p&gt;I can understand your concern. This is going to be difficult for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Finding out the agency that is responsible for the maintenance may be next to impossible for a layperson. Note that I said &quot;maintenance&quot;, because this might be a different entity than the owner due to agreements.&lt;/li&gt;&#xA;&lt;li&gt;Due to the number of complaints that the public agencies receive, it is very likely that your comment will get ignored. This often has nothing to do with whether or not your concern has merit.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Public vs. Private&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Public&lt;/em&gt; agencies will be the easiest to contact. In a city, calling the city emergency or non-emergency number will at least let you talk to someone who could forward your concern on to the appropriate agency. What is done after that is uncertain unless there is an active emergency situation, e.g. bridge span missing. In that case, the standard emergency procedures will be in effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple concern will likely get ignored. At the best a local worker will be dispatched in the next day or so to do a drive-by inspection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Private&lt;/em&gt; agencies might not have an easy method for interfacing with the public. Depending on the company, the headquarters might not be located in the same state as you. I am specifically thinking about Railroads in this instance. To find a contact number, it might involve sitting down and searching a website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, active emergencies will be handled by the local emergency (911) management. Simple concerns will likely be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inspections&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every agency (private or public) that I can think of has inspection requirements. These regulate the frequency and extent of inspections. For many typical bridges, an inspection is required every year. These are usually routine inspections which mean that at a minimum every area is visually inspected. Certain structure of concern might have more frequent requirements or more in depth inspections. These inspections might include actually measuring the thickness of steel plates or sounding for concrete spalls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it really an unsafe structure?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless the person reporting the concern has a background that would allow them to know what is and is not unsafe, it is likely that there isn't an issue. A lot of concerns are based on the visual condition of a structure. This can be very misleading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Easily visible conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;flaking paint&lt;/li&gt;&#xA;&lt;li&gt;rust&lt;/li&gt;&#xA;&lt;li&gt;chipped concrete&lt;/li&gt;&#xA;&lt;li&gt;rotted timber&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Whether or not these conditions affect the safety of the structure depend on the exact location within the structure and their extent. Most structures are designed with factors of safety that mean that almost any local issue will not cause the entire structure to be unsafe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if the condition has already been reported in an inspection, steps may have already been taken to ensure that the structure isn't loaded in a way to make it unsafe. For roadway bridges, this may mean posting vehicle load limits. For railway bridges, it may mean having trains go slower while crossing the bridge.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-02-26T15:56:58.457" LastActivityDate="2015-02-26T15:56:58.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1868" PostTypeId="5" CreationDate="2015-02-26T16:24:26.757" Score="0" Body="&lt;p&gt;The employment tag should not be used to ask about personal experiences or to ask polls. Questions should be specifically answerable with facts and references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that employment is very location specific, so include that information.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="33" LastEditDate="2016-02-23T03:27:15.640" LastActivityDate="2016-02-23T03:27:15.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1869" PostTypeId="4" CreationDate="2015-02-26T16:24:26.757" Score="0" Body="Questions about working in an engineering field, whether as an engineer, researcher, technician/paraprofessional, manager/executive, student or recent graduate. Be careful not to ask poll-type questions." OwnerUserId="368" LastEditorUserId="33" LastEditDate="2016-02-23T03:27:15.640" LastActivityDate="2016-02-23T03:27:15.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1870" PostTypeId="2" ParentId="1864" CreationDate="2015-02-26T18:09:00.143" Score="4" Body="&lt;p&gt;In addition to the other answers, I would advise you to not underestimate the utility of going to one of your elected representatives. Unless you live in Nebraska (which has a unicameral state legislature), you have two state representatives, a US House representative, and two US Senators that you are a constituent of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While part of their job may yelling at other elected officials, spewing rhetoric, and maintaining gridlock (bitterness? what bitterness?), helping constituents with problems is also a big part of their job, and it's one they'll probably be glad to do, because it's a good way for them to improve their public image and, less cynically, to actually help people. They obviously won't be able to go out and inspect the infrastructure themselves, but they should be able to help get you to the right part of the state or federal government for the issue that you've noticed, and to help ensure that you get a suitable answer, whether that be a new inspection, confirmation that an inspection was recently done, or learning that repairs may already be planned or in an approval process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The US House of Representatives has a website that allows you to &lt;a href=&quot;http://www.house.gov/representatives/find/&quot;&gt;determine who your representative is&lt;/a&gt;, and I wouldn't be surprised if your state legislatures have similar tools. I would guess that the state governments are the ones responsible for most of the infrastructure you'll see, so I would advise starting with your state representatives, as their smaller constituency will also result in your voice being more visible. And don't be afraid to reach out to more than one person, in more than one way. Call, email, send letters; this is the point of democracy and not enough people take advantage of it.&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-02-26T18:09:00.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1871" PostTypeId="2" ParentId="1853" CreationDate="2015-02-26T18:16:57.110" Score="3" Body="&lt;p&gt;I think the pitch is implied by the major and pitch diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tolerance of these $D_{maj}$  and $D_{pitch}$ are given in &lt;a href=&quot;http://en.wikipedia.org/wiki/ISO_965&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;ISO 965-2&lt;/code&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll attempt juggling with the math in the &lt;a href=&quot;http://en.wikipedia.org/wiki/ISO_metric_screw_thread&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISO metric screw wiki&lt;/a&gt;, (&lt;em&gt;please correct if wrong&lt;/em&gt;):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5esRY.png&quot; alt=&quot;http://en.wikipedia.org/wiki/ISO_metric_screw_thread&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(3/8)H=(D_{maj}-D_{pitch})\cdot(1/2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H=P\cos(30^\circ)=P\sqrt{3}/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P = \underbrace{(2/\sqrt{3})\cdot(8/3)\cdot(1/2)}_{1.5396}\cdot (D_{maj}-D_{pitch})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an M10x1.25 fine thread (could not find the M10x1.00) this would result to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;M10x1.25&lt;/code&gt;: $D_{maj}=[9.972...9.76]$ $D_{pitch}=[9.16...9.042]$  --&gt; $P=[0.92..1.43]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a $1000mm$ rod, this would be a nominal $800$ turns with a minimum of $698$ and a max of $1083$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite a range I think! I also found this picture of a mismatch failure &lt;a href=&quot;http://www.boltscience.com/pages/screw8.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;(src)&lt;/a&gt;:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/TvLEj.png&quot; alt=&quot;broken&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="270" LastEditorUserId="270" LastEditDate="2015-02-27T08:36:59.553" LastActivityDate="2015-02-27T08:36:59.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1872" PostTypeId="2" ParentId="1866" CreationDate="2015-02-26T19:30:03.853" Score="2" Body="&lt;p&gt;Some pointers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It's first about surface area! But don't put the fins too close together or there is not enough space for natural convection to start... I read somewhere that ~0.25&quot; (~6 mm) is a reasonable spacing.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Stainless steel is not a good choice. You want a metal with good thermal conductivity. Aluminum is just about perfect since it is also light. Copper would also work.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat sink design seems fairly mature to me, I'd copy other designs for a start.    &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="168" LastEditorUserId="368" LastEditDate="2015-02-26T20:21:11.343" LastActivityDate="2015-02-26T20:21:11.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1873" PostTypeId="2" ParentId="1866" CreationDate="2015-02-26T20:06:40.533" Score="5" Body="&lt;p&gt;First, to counter what some people are saying about printed metals, selective laser sintering can produce parts that are as thermally conductive as their base metals.  They are limited only by porosity, and state of the art machines can produce &lt;a href=&quot;http://dx.doi.org/10.4028/www.scientific.net/KEM.523-524.244&quot; rel=&quot;nofollow&quot;&gt;fully-dense (zero porosity) metal parts&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, you probably don't need to take radiation into account because you are at low temperature and airflow is going to make a much bigger contribution than radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, traditional heat sink design uses equations that calculate the temperature distribution along the fin from the root to the tip in order to come up with an &quot;effective area&quot; upon which an average convection coefficient can be imposed. This means that you aren't calculating the exact airflow, but rather using an average over the whole surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any good heat transfer textbook will have formulas for many different fin shapes, including fins with a non-constant fin area. I personally like &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471304581&quot; rel=&quot;nofollow&quot;&gt;Incropera and DeWitt&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, if you design sufficiently complicated fins, the standard equations will break down. You then have three choices: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Approximate your fins with similar fins.  This can lead to results that are close enough. &lt;/li&gt;&#xA;&lt;li&gt;Simplify your design to make it fit an established pattern &lt;/li&gt;&#xA;&lt;li&gt;Go to a full finite element simulation.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would advise using number 1 or number 2 because they might be more accurate. It is easy to make subtle mistakes in finite element models that completely invalidate your results&lt;sup&gt;1&lt;/sup&gt;, however if you do choose to go that route, you should know that there are free open-source FEM solvers (Elmer comes to mind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now some general tips:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As others have noted, stainless steel is a bad conductor of heat. Aluminum is probably a &lt;em&gt;much&lt;/em&gt; better choice, but unless you decide not to 3D print it, you probably can't use aluminum. This is the reason that I probably would print something else if I had access to a 3D printer&lt;/li&gt;&#xA;&lt;li&gt;The Coefficient of Performance of Peltier coolers is often less than one, so in order to get e.g. 20W of cooling power, you will need to put in more than 20W of electrical power. Your heat sink needs to dissipate the sum of the cooling power and the electrical power.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; You can model heat sink designs without using FEM and as long as your designs are not too crazy, you will probably get good results. Traditional &quot;by-hand&quot; calculations don't take into account radiation and assume an average airflow over the whole surface, which is usually good enough. I personally would avoid making heat sinks at all and use your time on the 3D printer for better things.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;Trust me, I'm a graduate student in Mechanical Engineering, and I've screwed up &lt;em&gt;so&lt;/em&gt; many models.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="709" LastEditorDisplayName="user16" LastEditDate="2015-02-26T21:25:48.177" LastActivityDate="2015-02-26T21:25:48.177" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1874" PostTypeId="2" ParentId="1834" CreationDate="2015-02-26T20:48:50.793" Score="1" Body="&lt;p&gt;No, but it would take a slight modification and I don't think they'll do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you have to understand why AMELIA is injecting high pressure air over the top of the wing. In aerodynamics, lift is produced as a result of &lt;a href=&quot;https://en.wikipedia.org/wiki/Circulation_%28fluid_dynamics%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;circulation&quot;&lt;/a&gt;, which means that the air above the wing travels further in the same amount of time than the air on the bottom of the wing.  The high pressure air from the engine, in this case, aids the air on the top of the wing because NASA has put a slot on the top of the wing (probably about 3/4 of the way back) to give more energy to the air flowing over the top of the wing. Since it has more energy, they can force it to take an even longer path (remember the air on top takes a longer path than the air below) without stalling. Stalling is when the air above the wing doesn't have enough energy to travel fast enough to reach the back of the wing at the same time as the air traveling beneath the wing, and so just gives up and separates from the wing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, the addition of a second engine doesn't preclude this in any way, however, it would mean that you have more exhaust air available. I predict (although I don't work for NASA) that the engineers would make the wings wider and have each engine provide high pressure air to its own section of the wing. This would give the same benefit, only scaled up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however, that I think it is quite unlikely that four engines will ever be used in this configuration. Modern two-engined craft are as resilient against engine failures as old four-engined craft and are now allowed to fly almost every commercial route (See &lt;a href=&quot;https://en.wikipedia.org/wiki/ETOPS&quot; rel=&quot;nofollow noreferrer&quot;&gt;ETOPS&lt;/a&gt;). Combined with the fact that two-engined planes can be made more efficient, I don't think that AMELIA will ever have four engines.&lt;/p&gt;&#xA;" OwnerUserId="709" LastEditorUserId="7083" LastEditDate="2018-06-12T05:47:05.577" LastActivityDate="2018-06-12T05:47:05.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1875" PostTypeId="2" ParentId="1861" CreationDate="2015-02-27T01:07:50.173" Score="6" Body="&lt;p&gt;What's missing so far is an explanation why you can't expand from high pressure to atmospheric in a single stage gas turbine. There are two types of gas turbines - impulse and reaction turbines. Both face the same problem but it's easier to understand in the impulse turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An impulse turbine accelerates the gas through a nozzle from high pressure P1 to a lower pressure P2, increasing its speed to V. The fast moving gas hits the turbine blades and gives up its momentum and kinetic energy, becoming slow moving gas at pressure P2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that for some value of pressure difference, the speed V reaches the speed of sound (in that gas at that temperature). At which point the turbine blades are highly inefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a very old book I can't find just now about steam turbines (same thing : steam is a gas!) efficiency started to fall off somewhere around Mach 0.5 which corresponded to a 40% pressure reduction in a stage. (The actual velocity can be found from Bernoulli equation) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives a way to find the number of stages you need to efficiently convert any given pressure ratio into shaft power. Given newer blade designs, Mach 0.5 may no longer be the upper limit but the same basic principle applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an aircraft jet engine, after several stages of subsonic acceleration, the hot gases escape through one last nozzle and may well exceed Mach 1 to provide thrust for the aircraft - but not very efficiently. (The SR71 Blackbird's engines transitioned to a different mode of operation - practically a ramjet - for Mach 3 operation)&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2015-02-27T01:07:50.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1877" PostTypeId="1" CreationDate="2015-02-27T05:36:45.817" Score="4" ViewCount="3068" Body="&lt;p&gt;I need to bend one circular iron pipe with 4 cm diameter and 2 mm thickness&#xA;as a guide rail for an electric stair lifter similar to this picture (the spiral bending part of stair):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZxVPe.jpg&quot; alt=&quot;stair-lift-wheelchair-user&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I bend the metal pipe as a continuous rail guide for the stair lifter device and a normal continuous stair? I tried ordinary bending devices like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PFDIp.jpg&quot; alt=&quot;http://www.galco.com/images/moreinfo/bender5.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made one easy bending multilayer pipe (which I used for house thermal piping &quot;The PEX-AL-PEX pipes&quot;) as a &lt;strong&gt;pattern for guiding the bending process&lt;/strong&gt;, but we could not make correct bending diameters and the metal pipe does not fit as a continuous stair. I also tried changing its form from a circular  to elliptical cross-section. What mathematical equations must I consider for bending it (what force vectors for 30 cm diameter with 40 cm height?) and how I could make it practical? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to bend my failed sample's metallic pipe with heat and hand pressure to make it fit to the pattern but finally its circular cross-section changed with a very hard process to make an ugly continues spiral sample stair as a guide rail!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way of welding the pipe so the circular cross-section doesn't change or some kind of professional pipe bending machine used for this application? I have no access to CNC machines or anything similar to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8gXIH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="715" LastEditorUserId="368" LastEditDate="2015-03-05T22:37:49.900" LastActivityDate="2015-03-05T22:37:49.900" Title="How to bend iron pipe into a spiral guide rail for stair lifters" Tags="|mechanical-engineering|piping|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1878" PostTypeId="1" CreationDate="2015-02-27T09:24:59.197" Score="9" ViewCount="1668" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Stainless_steel&quot;&gt;Stainless steel&lt;/a&gt; comes in several qualities.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;AISI 303 (1.4305) (Stainless Steel A1) approximately 304 but more machinable due to sulfur.&lt;/li&gt;&#xA;&lt;li&gt;AISI 304 (1.4301) (Stainless Steel A2) comprises of 18% chromium and 8% nickel.&lt;/li&gt;&#xA;&lt;li&gt;AISI 316 (EN 1.4401) (Stainless Steel A4) is 16% chromium, 10% nickel and 2% molybdenum.&lt;/li&gt;&#xA;&lt;li&gt;AISI 316L is the low carbon variant to enhance weldability.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For seawater &lt;a href=&quot;http://en.wikipedia.org/wiki/Marine_grade_stainless&quot;&gt;316 quality&lt;/a&gt; is often advised. I'd like to understand:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;By what process does molybdenum make the metal more resistant to corrosion in seawater?&lt;/li&gt;&#xA;&lt;li&gt;How do the material loss rates compare between these grades of stainless steel, as well as to ordinary structural steel (S235, S275 or S355)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="270" LastEditorUserId="368" LastEditDate="2015-02-27T17:22:09.387" LastActivityDate="2015-03-12T09:56:09.517" Title="Stainless steel corrosion resistance in seawater" Tags="|materials|structural-engineering|metallurgy|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1879" PostTypeId="2" ParentId="1877" CreationDate="2015-02-27T14:30:38.177" Score="3" Body="&lt;p&gt;Cold bending steel, especially with a relatively thin wall, is difficult to do accurately. Typically the standard for custom parts like this is that the shape is close enough that the person installing it can bend the pipe into the final curve with the use of clamps. Then they weld it in place so it keeps its shape. I'm going to assume that you're already done the first hard part - figuring out the radius and helical pitch that you need to create, although it's not entirely clear from your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using rolling benders ('three roll benders') like you've shown is one common way to solve this. One strategy for getting the right shape from those tools is to form your bend in multiple, lighter passes. That is curve it slightly once, more the next time, and more again until you get to your desired shape. Keep in mind that the same settings on the final pass will produce different curves if your start with a straight piece of pipe or one that has already been curved. Using multiple, lighter passes produces a more consistent curve throughout. You're probably aware already, but rolling benders always leave a flat segment at the end, so you need to start with a longer piece and plan to cut the two ends off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that still isn't accurate enough for your application, the next mechanized option would be to use a tool called a mandrel bender (see &lt;a href=&quot;http://metal.baileighindustrial.com/metalworking/mandrel-benders&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, for examples) which supports the pipe from the inside to prevent the section from crushing or deforming too much. As far as I know, these cannot form helixes, only single-plane bends, but sometimes those can be used at an angle to approximate the shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If neither of those are options for you, or you need a lower cost solution, manually heating and bending is possible, although be aware that too much heat can compromise the strength of the steel. In order to prevent the section deformation you're describing, people will sometimes fill the tube with dry sand and weld a cap on each end. This way, the tube is less likely to close in on itself as you heat it and bend it. If you do this, it is very important that the sand is dry, because wet sand will generate steam which can be very dangerous. For the same reason, the end caps should not form an airtight seal with the tube, leave part of the shape unwelded. Please be careful if you attempt this method.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-02-27T14:30:38.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1880" PostTypeId="1" AcceptedAnswerId="1881" CreationDate="2015-02-27T20:14:32.750" Score="13" ViewCount="514" Body="&lt;p&gt;I am now in charge of a product line my company has been shipping for a decade. One of the previous product engineers was... shall we say, less than conscientious about sustainability and proper documentation. We have shipped hundreds of units, of multiple design variants, under the &lt;em&gt;exact same part number&lt;/em&gt;. The manual presently reflects only one variant, meaning many users can't use it. And we've had multiple instances where a user has tried to reorder a unit by part number, only to find that what we ship them does not match what they already had.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, this is terrible. One does not change the specs or user interface of a product without also changing the part number. We will avoid such things in the future. But my question is about the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have, on paper, documentation indicating what design variant each serial number corresponds to. My thinking is to create a spreadsheet, and name each variant retroactively, so we can at least support users that call in or place reorders. We would then create proper manual(s) so that the user can, based on their serial number, understand the operation of the units they have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm just making up that solution. It occurs to me that there may be formal, industry-standard methods of dealing with such things. Is there a procedurally-correct way to handle my existing install base?&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="368" LastEditDate="2015-05-11T21:11:40.067" LastActivityDate="2015-05-11T21:11:40.067" Title="I have inherited a product with inconsistent designs sharing the same part number. What do I do?" Tags="|project-management|documentation|product-support|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1881" PostTypeId="2" ParentId="1880" CreationDate="2015-02-27T21:33:11.753" Score="11" Body="&lt;p&gt;I don't think there is any formal standard way out of this mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good news is that at least tracked the changes by serial number.  Inventing different models after fact seems pointless because the customers won't know what models they have.  From the customer view, these are all one model but vary by serial number.  Document it that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could create a different manual for different serial number ranges if the changes are large, or have occasional sections in the manual that only apply to particular serial number ranges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure that the customer can easily find the serial number of units they have.  This should be spelled out very clearly right in the beginning of the manual.  It should warn customers that there are significant variations between serial numbers, and how exactly to identify the serial number of their unit, and then consistently use that throughout the manual when differences have to be noted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be a good idea to put a chart in the front of the manual that briefly shows which features are present in which serial number ranges.  You could possibly then also name each variant, like A, B, etc, then refer to that in the rest of the manual.  You might even suggest to customers that they circle the variant they have, or leave a big obvious place to write their variant letter inside the front cover so they can flip back to it easily when confronted with sections that differ by variant.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="399" LastEditDate="2015-02-28T00:18:46.067" LastActivityDate="2015-02-28T00:18:46.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1882" PostTypeId="1" AcceptedAnswerId="1892" CreationDate="2015-02-27T23:57:58.973" Score="4" ViewCount="806" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Concorde&quot; rel=&quot;nofollow&quot;&gt;Concorde&lt;/a&gt; had a famously high &lt;a href=&quot;http://www.boeing.com/commercial/aeromagazine/aero_12/whatisaoa.pdf&quot; rel=&quot;nofollow&quot;&gt;angle of attack&lt;/a&gt; (and &lt;a href=&quot;https://www.aviation-history.com/theory/angle_of_attack.htm&quot; rel=&quot;nofollow&quot;&gt;pitch angle&lt;/a&gt;) when landing. This led to its famous &lt;a href=&quot;https://en.wikipedia.org/wiki/Droop-nose&quot; rel=&quot;nofollow&quot;&gt;droop nose&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/1/11/Concorde_landing_Farnborough_Fitzgerald.jpg&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/1/11/Concorde_landing_Farnborough_Fitzgerald.jpg&quot; alt=&quot;Concorde&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Concorde's Russia counterpart, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tupolev_Tu-144&quot; rel=&quot;nofollow&quot;&gt;Tupolev Tu-144&lt;/a&gt;, had the same issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the only two supersonic airliners to have been built, so there is, admittedly, a small sample size. The Tu-144 has also been derisively called &quot;the Concordski&quot;, as some claim that its designers stole plans from the Concorde. So there are clear similarities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are some supersonic airliners on the drawing board. Will they, too, have high angles of attack when landing? As a follow-up question: Will they, too, be fitted with droop noses?&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-04-16T15:12:35.020" Title="Do all supersonic airliners have a high angle of attack when landing?" Tags="|aerospace-engineering|" AnswerCount="4" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1883" PostTypeId="1" AcceptedAnswerId="1894" CreationDate="2015-02-28T00:30:05.980" Score="7" ViewCount="10692" Body="&lt;p&gt;I am trying to use a Piezo Electric Ceramic Disc Transducer for ultrasonic sensing measurements for fluid levels. A Piezo Ceramic Disc with a Resonant frequency of 215 KHz or 1 MHz is under consideration. These devices specify a Radial and Thickness mode vibration configuration.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3f84k.jpg&quot; alt=&quot;Piezo Electric Ceramic Disc Transducer&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Primary Question&lt;/strong&gt;: What is Radial and Thickness mode vibration configuration?&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Secondary Question&lt;/strong&gt;: Is Axial and Thickness mode vibration mean the same mode?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DlCir.jpg&quot; alt=&quot;Transducer1&quot; /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/IRdp3.jpg&quot; alt=&quot;Transducer2&quot; /&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Part Specifications&lt;/h2&gt;&#xA;&lt;p&gt;&lt;em&gt;Piezo Ceramic Disc 1 MHz,Thickness mode vibration&lt;/em&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dimensions: 15mm diameter x 2.1mm thickness&lt;/li&gt;&#xA;&lt;li&gt;Resonant frequency fr: 1 MHz±3 %&lt;/li&gt;&#xA;&lt;li&gt;Electromechanical coupling coefficient Kt:≥39%&lt;/li&gt;&#xA;&lt;li&gt;Resonant impedance Zm: ≤7.6 Ω&lt;/li&gt;&#xA;&lt;li&gt;Static capacitance Cs: 900pF±15%@1kHz&lt;/li&gt;&#xA;&lt;li&gt;Test Condition: 23±3 °C 40~70% R.H.&lt;/li&gt;&#xA;&lt;li&gt;fr, Zm, Kp =&amp;gt; Thickness mode vibration&lt;/li&gt;&#xA;&lt;li&gt;Cs =&amp;gt; LCR meter at 1KHz 1Vrms&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;em&gt;Piezo Ceramic Disc R 215 KHz, Radial mode vibration&lt;/em&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dimensions: 10mm diameter x 2mm thickness&lt;/li&gt;&#xA;&lt;li&gt;Resonant frequency fr: 215 KHz±5 KHz&lt;/li&gt;&#xA;&lt;li&gt;Electromechanical coupling coefficient Kp:≥60%&lt;/li&gt;&#xA;&lt;li&gt;Resonant impedance Zm: ≤6 Ω&lt;/li&gt;&#xA;&lt;li&gt;Static capacitance Cs: 450pF±15%@1kHz&lt;/li&gt;&#xA;&lt;li&gt;Test Condition: 23±3 °C 40~70% R.H.&lt;/li&gt;&#xA;&lt;li&gt;fr, Zm, Kp =&amp;gt; Radial mode vibration&lt;/li&gt;&#xA;&lt;li&gt;Cs =&amp;gt; LCR meter at 1KHz 1Vrms&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cyzLw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cyzLw.png&quot; alt=&quot;Piezoelectric Modes of Vibration&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.americanpiezo.com/knowledge-center/piezo-theory/vibration.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Modes of vibration for PIEZOELECTRIC elements&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2021-01-15T02:27:20.660" LastActivityDate="2021-01-15T02:27:20.660" Title="What is Radial and Thickness Mode Vibration in a Piezo Electric Ceramic Disc Transducer?" Tags="|mechanical-engineering|electrical-engineering|sensors|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="1884" PostTypeId="2" ParentId="1880" CreationDate="2015-02-28T02:10:02.880" Score="2" Body="&lt;p&gt;The formal way to &lt;strong&gt;avoid&lt;/strong&gt; such situation is called &lt;strong&gt;configuration management&lt;/strong&gt;. Even with such a formal process there are occasions that engineering run into such scenarios. One of the most recent is the &lt;a href=&quot;http://www.consumerreports.org/cro/news/2014/10/everything-you-need-to-know-about-the-takata-air-bag-recall/index.htm&quot; rel=&quot;nofollow&quot;&gt;Takata Air Bag Recall&lt;/a&gt;. Therefore there are plenty of examples to handle such scenarios, but one size doesn't fit all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following are few suggestions to address the current issue as well a avoid future situations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Your customer is king, so track down and document all the customers that are affected.&lt;/li&gt;&#xA;&lt;li&gt;Fix the design using sound engineering practices. Revise all engineering and other documentation to reflect the change.&lt;/li&gt;&#xA;&lt;li&gt;Notify all customers of the situation and intended corrective action. &lt;/li&gt;&#xA;&lt;li&gt;Develop and plan to recall defective material and replace the material with revised design. Automotive and Medical industry frequently have recalls. This might be a good place to start researching for ideas. &lt;/li&gt;&#xA;&lt;li&gt;Be proactive as oppose to be reactive. Depending on the industry and customer base the possibility of any liability and litigation need to be considered.  &lt;/li&gt;&#xA;&lt;li&gt;Implement a good configuration management system. This system should capture engineering change requests, implement changes and notify changes to customers (ECN - Engineering change notification)&lt;/li&gt;&#xA;&lt;li&gt;If not in place consider instituting a some type quality systems.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Configuration_management&quot; rel=&quot;nofollow&quot;&gt;Configuration management&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.iso.org/iso/iso_9000&quot; rel=&quot;nofollow&quot;&gt;ISO 9000 - Quality management&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-02-28T12:53:55.463" LastActivityDate="2015-02-28T12:53:55.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1885" PostTypeId="2" ParentId="1880" CreationDate="2015-02-28T03:33:11.850" Score="5" Body="&lt;p&gt;Tracking the various product versions by serial number is a good idea, but be careful to standardize that as well. Years ago while working for an electronics manufacturer in service and repair, I discovered two units; one was serial number 0024, the other was 00024. They had changed from a four-digit number to a five-digit one and in the process created a duplicate (except for the number of leading zeros).  To avoid confusion in the future we relabeled the four-digit one to be 0024A instead. Not the best answer perhaps, but it worked.&#xA;You may need to do something similar. The suggestion to document the various versions in one document with subsections and notes is a good one, if done properly.&lt;/p&gt;&#xA;" OwnerUserId="730" LastActivityDate="2015-02-28T03:33:11.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1886" PostTypeId="2" ParentId="1883" CreationDate="2015-02-28T03:34:39.630" Score="4" Body="&lt;p&gt;&lt;strong&gt;Radial&lt;/strong&gt;: Oscillatory waveform is between edge and centre.&lt;br&gt;&#xA;&lt;strong&gt;Thickness&lt;/strong&gt;: Oscillation is across disc faces so will be higher frequency all else being equal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial and thickness are probably equivalent terms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, while the relationship does not have to be this close&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Your thickness mode disc resonates at 1 MHz and is 2.1mm thick.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your radial mode disc is 10mm in diameter (not radius) and resonates at 215 kHz.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dia / thickness = 10:2.1 = 4.76:1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1 MHz / 215 kHz = 4.65:1&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="399" LastEditorUserId="110" LastEditDate="2015-02-28T14:30:36.503" LastActivityDate="2015-02-28T14:30:36.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1887" PostTypeId="2" ParentId="1861" CreationDate="2015-02-28T03:39:15.047" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/water_turbine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Water turbines&lt;/a&gt; are a major source of electric power.&#xA;A water turbine generally has only one rotor disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TR5LC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(from &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Water_turbine_(en_2).svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Old Moonraker at Wikipedia&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/gas_turbine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gas turbines&lt;/a&gt; are used in natural gas electric power generators, jet aircraft, and a few other vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gas turbine generally have lots of rotor disks, which can be divided into two groups: compressor rotor disks and turbine rotor disks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compressor section of a gas turbine needs lots of rotor disks, because reducing the number of rotor disks reduces efficiency by either&#xA;(a) increasing the pressure differential across each disk to keep the total compression ratio the same, reducing compression efficiency, or&#xA;(b) keeping the pressure differential across each disk the same, reducing the total compression ratio, which reduces the efficiency of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Brayton_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brayton cycle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water turbines don't need a compressor section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While in principle a gas turbine could have lots of rotor disks, in practice we find that aircraft turbines generally have only 1 or 2 rotor disks, and (bolted to the ground) natural gas turbines generally have only 1 or 2 or 3 rotor disks,&#xA;not that much different from water turbines which have only 1 rotor disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas turbines used in electric power generators are oil-powered or natural gas-powered electric generators and are designed to extract as much energy as possible as electric power; the thrust pushing against the bolts holding them on the ground is unnecessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/Uw90o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hitachi H-25&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(Hitachi H-25 from Russell Ray, Power Engineering)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://gasturbinespower.asmedigitalcollection.asme.org/data/Journals/JETPEZ/27198/071401_1_f1.jpeg&quot; rel=&quot;nofollow noreferrer&quot;&gt;100-kW micro gas turbine&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(100-kW micro gas turbine photo from &lt;a href=&quot;http://gasturbinespower.asmedigitalcollection.asme.org/article.aspx?articleid=1429992&quot; rel=&quot;nofollow noreferrer&quot;&gt;M. Cadorin et. al &quot;Analysis of a Micro Gas Turbine Fed by Natural Gas and Synthesis Gas: MGT Test Bench and Combustor CFD Analysis&quot;&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/9I7Hj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Siemens Gas Turbine 200&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;http://www.energy.siemens.com/nl/en/fossil-power-generation/gas-turbines/sgt-200.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Siemens Gas Turbine 200 (SGT-200) for industrial power generation&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/n8dg9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;GE gas turbine&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(from &lt;a href=&quot;http://spectrum.ieee.org/energywise/energy/fossil-fuels/ges-new-gas-turbines-play-nicely-with-renewables&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tekla Perry: &quot;GE’s New Gas Turbines Play Nicely With Renewables&quot;&lt;/a&gt;.)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mTSBr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(&lt;a href=&quot;http://www.integralpower.com/opra.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;OPRA's 2 MW class OP16 gas turbine&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GYq5G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(&lt;a href=&quot;http://www.rentechboilers.com/case_studies/amherst.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;natural gas or oil-powered Saturn 20 at Amherst College&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="721" LastEditorUserId="721" LastEditDate="2015-03-02T20:20:50.657" LastActivityDate="2015-03-02T20:20:50.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1888" PostTypeId="1" AcceptedAnswerId="1890" CreationDate="2015-02-28T04:25:51.817" Score="4" ViewCount="1713" Body="&lt;p&gt;I am involved in an upcoming product launch and we recently had a discussion about serial numbers. We will of course use variant-specific model identifiers and revision numbers, but couldn't see any good reason to include any leading zeros in serial numbers, or otherwise make them fixed-length with padding numbers, prefixes, characters, etc. But this seems to be a common practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything wrong with 1, 2, ... 10, 11, 12, ... 100, 101, 102, ... ? What possible advantages are lost by just starting at 1 and counting up?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="110" LastEditDate="2015-02-28T14:30:56.090" LastActivityDate="2015-02-28T14:36:24.850" Title="Is there a good reason for long and fixed-length serial numbers?" Tags="|documentation|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1889" PostTypeId="2" ParentId="1888" CreationDate="2015-02-28T05:10:43.757" Score="5" Body="&lt;p&gt;One advantage for fixed length serial numbers is that it's easier to do database searches, particularly if there's a pattern in the numbers. Try writing the code to interrogate a database for variable length numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With leading zeros in a list of numbers the number 2 always comes before 10,11,100 and any other multiple digit number beginning with a 1. It's easier to see numbers when they are in numerical and alphanumerical order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long numbers can make it difficult to create forgeries. There's also the psychological benefit for consumers who may question short serial numbers and whether the company has been around for a long time and whether its reputable. For this very reason some companies always start serial or batch numbers with a number greater than 1001 or 10,001.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long numbers become mandatory when large numbers of items are produced, such as operating system or software licences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the code and pattern used for serial numbers, long numbers can incorporate information such as model type, batch number, manufacturing plant (where multiple plants are used).&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="110" LastEditDate="2015-02-28T13:31:52.900" LastActivityDate="2015-02-28T13:31:52.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1890" PostTypeId="2" ParentId="1888" CreationDate="2015-02-28T13:46:59.337" Score="5" Body="&lt;p&gt;Long serial numbers allow manufacturers to put more information into the serial number beyond showing relative manufacturing order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a manufacturer has multiple plants or multiple critical suppliers, they may want to indicate those sources within the serial number in case they need to track down defects later on.  One common technique is to assign ranges of serial numbers to those sources.  Another technique is to put plant and supplier codes at fixed locations within the serial number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having low production runs tied to serial numbers can leak information to competitors.  During World War II, British scientists used the serial numbers from tanks in order to &lt;a href=&quot;http://www.theregister.co.uk/2014/05/20/german_tank_problem_and_leaky_data/&quot;&gt;fairly accurately predict the rate of production&lt;/a&gt;.  That information helped them make better determinations on how to deploy resources.  Likewise, competitors can use the knowledge of low production runs against a competitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the two, the ability to quickly track manufacturing source and parts suppliers is probably more pertinent.  In an era of global manufacturing, it's fairly routine to see recalls cascade across multiple industries because of an issue at a shared, upstream parts supplier.  Batteries and other electronic components tend to be most frequently recalled like this.  That's not to say the same information can't be linked within a database, but that requires a lookup beyond just examining the serial number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's invaluable for a manufacturer to be able to succinctly state &quot;this range of serial numbers&quot; or &quot;look for these codes within the serial number of the part&quot; as part of the recall process.  Customers don't want to have to type in serial numbers at a website to see if they are affected or not.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-02-28T13:46:59.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1891" PostTypeId="2" ParentId="1888" CreationDate="2015-02-28T14:36:24.850" Score="2" Body="&lt;p&gt;Fixed-width serial numbers are just easier to handle.  It also allows you to know where the serial number is within a overall product ID string without special rules to parse the string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the serial number is expressed in hexadecimal in the string, it can contain letters A-F in addition to the digits 0-9.  If the part ID string might also contain a suffix or prefix with these letters, then it becomes ambiguous with a variable-length serial number and without some delimiter.  Consider &quot;07933A&quot;.  Is the whole thing the HEX serial number, or is 07933 the serial number and &quot;A&quot; a suffix denoting a configuration option, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fixed-width serial numbers strings are also easier to see in a list, and sort on.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Serial number generation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You didn't ask about this, but you have to think carefully about what/how new serial numbers are generated.  This is probably more important than how exactly the serial number will be listed on the nameplate.  I have seen a number of screwups, usually as a result of giving manufacturing too much leeway, or too much information so that they &lt;i&gt;think&lt;/i&gt; they know how to create new serial numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I usually advise now is a dedicated piece of hardware that is pre-programmed with a range of serial numbers, then gives out those serial numbers sequentially.  When the range is used up, it refuses to give out another serial number.  The hardware must be returned to a central serial number management place to get updated with a new range, or swapped out entirely.  You don't give any documentation about the internals of this device to manufacturing, whether internal or contract, remote or local.  This device is a necessary part of the overall production process, usually part of the final test jig.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One advantage of this method is that it's easy to assign serial number ranges to different production lines or even different manufacturers.  Engineering will need to create internal prototypes and sometimes samples, so they have one or more of these units too.  Any engineer can grab the unit floating around engineering to assign a guaranteed unique serial number to a new prototype.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This dedicated hardware need not be big or expensive.  It can be as simple as a USB dongle or small device connected to the USB.  I've done this with a USB microcontroller and external EEPROM.  Again, production gets no documentation on this device other than how to use it.  They can be given multiple devices as spares, which is OK since each has its own unique serial number range.  We usually programmed these devices with a range of 65536 (2&lt;sup&gt;16&lt;/sup&gt;) serial numbers.  To update the serial number range required soldering a temporary jumper between two pads for that purpose, and running a special program on the host that only one person at the home office had.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Screwup #1&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Apollo Computer made workstations in the 1980s.  Each had to have a unique serial number.  Among other things, these serial numbers were used to allow each node to make its own globally unique ID numbers used in the file system.  The serial numbers were a hexadecimal string.  A bunch of years after production started, someone noticed that the serial number of their node was well beyond the number of nodes supposedly produced.  It turns out production was assigning serial numbers in decimal, even though each digit was really hexadecimal.  This was fixed by having them make new units by filling in the unused serial numbers with digits A-F in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case no real harm was done, except that it made tracking units by serial number ranges awkward.  Below a certain number, ranges were divided into decimal-only and hexadecimal.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Screwup #2&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Serial numbers were assigned by the production test system, which included a PC.  There were multiple production lines, and each PC was carefully assigned a particular range.  The contract manufacturer in China was given explicit instructions to notify us in case of any trouble with any of the production test systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One day someone noticed that some of the 10s of thousands of little gizmos received from the factory in China had duplicate serial numbers.  This was bad, because the larger system these gizmos were part of relied on the serial number to uniquely identify each gizmo, which was in large part the point of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that the factory in China had done something stupid with one of the production test PCs and messed up the data on it.  Instead of telling us like they were supposed to, they knew they could fix the problem on their own and we'd never know.  Their fix was to do a complete disk copy from the working system to the broken system.  From then on the two systems assigned duplicate serial numbers.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-02-28T14:36:24.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1892" PostTypeId="2" ParentId="1882" CreationDate="2015-02-28T14:53:09.547" Score="4" Body="&lt;p&gt;The basic problem is that a wing that works well in supersonic flight does not work well in low speed flight.  The principles of how higher pressure is created below the wing relative to above the wing is quite different between the two flight regimes.  Aircraft intended for supersonic cruise have to have wings that work well in supersonic flight.  However, these aircraft have to eventually land, which will be done at well below supersonic speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some compromises have to be made somewhere.  Note that this discrepancy between efficient cruise and not falling out of the sky at low landing speed is already a issue even with subsonic aircraft.  Allowing some reconfiguration of the wing by the use of flaps is a common solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gets trickier with a delta wing.  Part of the solution used in the Concorde was high angle of attack on landing, which required the famous droop nose so that the pilots could see the ground.  Other planes that are superosonic-capable have other mechanisms.  For example, the F-14 and F-111 have variable wings that reconfigure themselves between delta shape for supersonic flight, and more traditional wings for subsonic fight.  Particularly the F-111 was known for problems in this area, and this concept doesn't scale well to airliner size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these mechanisms are expensive, add weight, and complexity, so I'm sure creative engineers will continue to come up with new ways of dealing with this problem.  None of the solutions so far are optimum, so there is room for alternatives.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="33" LastEditDate="2015-02-28T17:19:27.447" LastActivityDate="2015-02-28T17:19:27.447" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1893" PostTypeId="2" ParentId="1882" CreationDate="2015-02-28T15:58:29.280" Score="4" Body="&lt;p&gt;The high angle of attack is required because the landing speed must be reduced to a value comparable to that of subsonic aircraft, mainly so that the SST can use the same runways. If the SST could have a longer runway for slowing down, it could land at a higher speed, with a more &quot;normal&quot; angle of attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer your question about the droop noses, I would say that there's no longer a need for this. Instead, the pilots will use a video system, similar to those already used on UAVs (unmanned aerial vehicles).&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-02-28T15:58:29.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1894" PostTypeId="2" ParentId="1883" CreationDate="2015-02-28T17:49:00.683" Score="5" Body="&lt;p&gt;There is an excellent &lt;a href=&quot;http://www.ultrasonic-resonators.org/misc/references/articles/Piezo_Technologies__%27An_Overview_of_the_Properties_of_Different_Piezoceramic_Materials%27.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;white paper&lt;/a&gt; from &lt;a href=&quot;http://www.piezotechnologies.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;piezotechnologies.com&lt;/a&gt; which is probably worth a read if you are new to working with piezoelectric devices.  Quoting from the section on resonance frequencies:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If an AC voltage is applied to a piezoceramic disc with a diameter several times its thickness, two series of resonances may be observed. The resonance at the lowest frequency corresponds to motion in the largest dimension of the ceramic, which in this case is in the radial direction.  Often, several harmonics will also be seen at higher frequencies.&lt;/p&gt;&#xA;&lt;p&gt;The first thickness resonance appears quite different from the radial harmonics that are found just below it. The impedance drops to a much lower value and the span in frequency is far wider. The thickness mode frequency often appears much “rougher” than the radial mode. This is because the harmonics of the radial mode are excited by the strong thickness mode.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, the radial mode is lower in frequency because it has a larger dimension. Since it is also the lowest frequency resonance it is much cleaner because it does not coincide with any other harmonics.  See the figure below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.piezotechnologies.com/Documents/White-Papers/Piezo-White-Paper-2-FINAL---Parameters.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O6Q7S.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The axial (thickness) mode on the other hand has a stronger resonance since the piezo-crystal is designed to be driven in that direction.  In addition, it overlaps with some of the higher harmonics of the radial mode and therefore has a 'dirtier' appearance.  See the figure below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ultrasonic-resonators.org/misc/references/articles/Piezo_Technologies__%27An_Overview_of_the_Properties_of_Different_Piezoceramic_Materials%27.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KAThx.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The full spectrum showing the narrower, weaker radial mode and its resonances together with the axial mode puts the relative broadness and strength of the axial mode in perspective.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ultrasonic-resonators.org/misc/references/articles/Piezo_Technologies__%27An_Overview_of_the_Properties_of_Different_Piezoceramic_Materials%27.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L72r7.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="110" LastEditDate="2021-01-15T02:24:42.320" LastActivityDate="2021-01-15T02:24:42.320" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="1895" PostTypeId="1" AcceptedAnswerId="1902" CreationDate="2015-02-28T17:51:46.813" Score="3" ViewCount="2068" Body="&lt;p&gt;I am specifying a product made in Australia, which is made of &quot;Grade 520 Steel.&quot; The manufacturer of the product lists a minimum yield stress of 520 N/mm² (75,400 psi) and a minimum ultimate stress of 650 N/mm² (94,300 psi) with a minimum elongation of 20% and Young's modulus of 205 kN/mm² (29,700 ksi.) They describe the steel as &quot;A fine grain micro alloyed carbon steel which is fully weldable.&quot; The raw steel form should be solid round bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that the plan checker wants a reference to a normative documents (ISO, EN, ASTM spec, etc.) for the material grade. I cannot find who (if anyone) defines  this grade of steel. What is the standard organization and standard number that would govern this material?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have, of course contacted the manufacturer, but what's above is all that they've been able to tell me so far. The only material specification '520' I've been able to find is a withdrawn ASTM tube spec. This component is significantly over-designed, so I'm not concerned with the actual properties, just finding the appropriate document. There are other products available made with ASTM materials, but for visual reasons we'd prefer the Australian product.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-03-01T03:58:00.310" Title="Normative Document for 'Grade 520' Steel" Tags="|materials|steel|standards|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1896" PostTypeId="2" ParentId="1753" CreationDate="2015-02-28T18:28:29.043" Score="2" Body="&lt;p&gt;As far as I know, there are two major factors, fatigue loading and low temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pipes buried below roads experience fatigue loading from traffic, surprisingly enough. Literally cars driving over the road surface imparts a load onto the pipes, even if it's a very small load. However, this happens many times every day and pipes stay in the ground for long periods of time. This means they experience millions of fatigue cycle loads over decades (50+ years isn't unusual for gas pipelines, at least).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add in thermal cycling thanks to the weather and metallic pipes (and plastic, to a lesser extent) will become brittle under low temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see where this is going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrosion caused by water is yet another factor etc etc etc and basically its all a surprisingly hostile environment for something which has to last so long.&lt;/p&gt;&#xA;" OwnerUserId="605" LastActivityDate="2015-02-28T18:28:29.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1897" PostTypeId="1" CreationDate="2015-02-28T18:29:16.730" Score="4" ViewCount="239" Body="&lt;p&gt;A certain product (link: &lt;a href=&quot;http://metro.co.uk/2015/02/27/wankband-is-first-gadget-that-harnesses-the-energy-of-masturbation-5081633/&quot; rel=&quot;nofollow&quot;&gt;Metro&lt;/a&gt;, somewhat safe for work) is currently trending on social media, that allows men to convert the kinetic energy of their more vigorous personal activities into electrical energy to recharge their phone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've had kinetic watches before, which re-charge by the day-to-day movement through wear; all of which have needed a few extra shakes periodically to keep them going- so it seems dubious to me that the band could really generate enough energy to charge my phone, which must require substantially more power...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would/does the W*nkBand (&lt;a href=&quot;http://www.pornhub.com/devices/wankband&quot; rel=&quot;nofollow&quot;&gt;product site&lt;/a&gt;: nsfw) work?&lt;/p&gt;&#xA;" OwnerUserId="151" LastEditorUserId="282" LastEditDate="2015-03-03T01:27:14.883" LastActivityDate="2015-03-03T01:27:14.883" Title="Does The W*nkBand Work?" Tags="|electrical-engineering|power-electronics|energy-efficiency|energy|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="1898" PostTypeId="2" ParentId="1897" CreationDate="2015-02-28T20:30:09.020" Score="5" Body="&lt;p&gt;Because this question has proven a little more derisive than I had anticipated (where's that engineering sense of fun/humour?)- I thought I'd attempt an answer as well, if only to prove it's an engineering question which can be answered using the engineering method; although hopefully someone using something more than Google will be able to chip in with a much more informed answer...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm definitely not an electronic engineer- but as I see it, sooner or later this question comes down to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E = CK$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$E$ is electric energy stored/used by the phone&lt;/li&gt;&#xA;&lt;li&gt;$C$ is a constant determined by the efficiencies of converting, storing and transfering the original kinetic energy to the electric energy.&lt;/li&gt;&#xA;&lt;li&gt;$K$ is the original kinetic energy generated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It turns out that kinetic energy harvesting is actually a major field of research and development (especially with the dawning advent of wearable tech) &lt;a href=&quot;http://www.extremetech.com/extreme/161079-kinetic-energy-harvesting-everyday-human-activity-could-power-the-internet-of-things&quot;&gt;This article&lt;/a&gt; states that intentionally shaking an object (which, we could call analogous with the band's usage) could generate something in the region of 3 milliwatts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://www.physicsforums.com/threads/best-efficiency-to-convert-kinetic-into-electric.288251/&quot;&gt;another forum&lt;/a&gt; (which I admit isn't the best reference); there are very few losses in the conversion of kinetic energy to electricity- so around about 95% of the 3 mW will be converted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lithium ion battery has around &lt;a href=&quot;http://en.wikipedia.org/wiki/Lithium-ion_battery&quot;&gt;80-90%&lt;/a&gt; charge/discharge efficiency; which means that 90% of 95% of the original 3 mW will make it to the phone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the energy given to the phone by the band is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E = (95 \% \times 90 \% ) \times 3\text{mW} = 2.565 \text{mW} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently an iPhone battery holds around &lt;a href=&quot;http://www.reddit.com/r/explainlikeimfive/comments/1ralgt/how_many_joules_do_i_need_to_power_a_cell_phone/&quot;&gt;5 watt hours&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the band would require:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{ 1 \text{W} \times 5 \text{hours} } { 2.565 \text{mW} } = 81 \text{days} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm guessing it was never designed to be the primary power source- and looking at the same reddit post quoted for the battery capacity- it seems that the minimum delivery to keep your phone going (i.e. standby power usage) is 21.8 mW; which means that there's a 7:1 ratio of generation to usage time (let alone the suggestion of the same article that 1W is required to register as 'charging').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From researching the answer to this question- I'm beginning to think it's just a hoax for a bit of marketing exposure- but I'll be the first to admit that electricity and its generation is far beyond my training as a structural engineer; so maybe someone will correct me?&lt;/p&gt;&#xA;" OwnerUserId="151" LastActivityDate="2015-02-28T20:30:09.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1899" PostTypeId="2" ParentId="1895" CreationDate="2015-02-28T20:57:55.403" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://www.astm.org/Standards/A615.htm&quot; rel=&quot;nofollow&quot;&gt;ASTM A615&lt;/a&gt; appears to standardize rebar using that grading designation: i.e. Grade 520 [MPa] (or Grade 75 [ksi]), but only provides testing to ensure that a particular material is at least some strength. The test does not specify composition, only that the material is plain carbon-steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://shop.bsigroup.com/ProductDetail/?pid=000000000030239498&quot; rel=&quot;nofollow&quot;&gt;BSEN10149&lt;/a&gt; is a British standard grading for steel alloys evidently, and some grades together with mechanical properties are noted on a &lt;a href=&quot;http://www.steelstrip.co.uk/structural_steels1.htm&quot; rel=&quot;nofollow&quot;&gt;commercial site&lt;/a&gt;, though none of the grades meet the properties you are looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.asminternational.org/&quot; rel=&quot;nofollow&quot;&gt;The American Society for Metals&lt;/a&gt; has a &lt;a href=&quot;http://www.asminternational.org/documents/10192/3466171/06117_Chapter%203B.pdf/a764507a-3499-4d23-b348-5536d31c0ba2&quot; rel=&quot;nofollow&quot;&gt;useful article&lt;/a&gt; on HSLA steel grades and associated ASTM standards, but none appear to be for bar material with the required yield strength, though they may be relevant if you intend to do your own testing and verification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sae.org/&quot; rel=&quot;nofollow&quot;&gt;The Society of Automotive Engineers&lt;/a&gt; has &lt;a href=&quot;http://papers.sae.org/780138/&quot; rel=&quot;nofollow&quot;&gt;an article&lt;/a&gt; on a grade 980X which according to [Wikipedia] has the required strength but not elongation at failure, and is not in bar form. There is also no associated testing standard that I can see. There may be one in the article but I do not have access.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible this alloy has no current standard, in which case it is highly recommended you verify the properties of the alloy for yourself. Vendor data is often fudged optimistically. Materials of a given category, such as microalloyed steels, always trade off strength for ductility, and the alloy you describe is above the tradeoff curve for all of the microalloyed steels I've seen in my brief search for this answer. That doesn't mean such an alloy doesn't exist, but it would make me cautious. Either the alloy is a new development that is specially made by this vendor, or they are erring optimistically in their marketing. In either case you should consider verifying the properties, prior to putting the materials in service, using the closest applicable testing standard available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: this answer is attempting to give the OP starting points. Unfortunately I don't have a single answer. Perhaps if the OP obtains their own answer from the vendor at some later time, they could post and accept it.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-03-01T00:24:27.020" LastActivityDate="2015-03-01T00:24:27.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1900" PostTypeId="1" AcceptedAnswerId="1901" CreationDate="2015-02-28T21:12:00.727" Score="2" ViewCount="523" Body="&lt;p&gt;&lt;strong&gt;Downdrag&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://www.dot.state.oh.us/Divisions/Engineering/Geotechnical/Geotechnical_Documents/2007%20Workshop%20Downdrag%20in%20Foundation%20Design.pdf&quot; rel=&quot;nofollow&quot;&gt;Downdrag&lt;/a&gt; in foundations occurs when soil layers consolidate and settle under additional load. This typically occurs when additional fill is placed on top of the existing soil. The actual downdrag force comes from the friction of the soil against piles as the soil moves down. It is assumed that soil settlement goes along with downdrag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Retaining Walls&lt;/strong&gt;&lt;br&gt;&#xA;The controlling load for cantilever retaining walls is usually the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lateral_earth_pressure&quot; rel=&quot;nofollow&quot;&gt;lateral earth pressure&lt;/a&gt;. Any vertical load on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Retaining_wall#Sheet_piling&quot; rel=&quot;nofollow&quot;&gt;sheet pile&lt;/a&gt; or &lt;a href=&quot;http://www.haywardbaker.com/WhatWeDo/Techniques/EarthRetention/SoldierPilesAndLagging/default.aspx&quot; rel=&quot;nofollow&quot;&gt;soldier pile&lt;/a&gt; is usually so small that it is ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This retaining wall is designed to retain additional fill over consolidating soil. The piles are embedded below the consolidating soil and will not settle. Would the downdrag or settling soil affect the calculation of the horizontal soil forces?&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-01T17:41:41.707" Title="Does downdrag affect cantilever pile retaining wall design?" Tags="|civil-engineering|geotechnical-engineering|retaining-wall|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1901" PostTypeId="2" ParentId="1900" CreationDate="2015-03-01T00:35:58.713" Score="1" Body="&lt;p&gt;It depends on the construction sequence and support conditions; however in my experience most walls like these are built by installing the piles and then excavating the open side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you've noted, consolidation settlement is a concern for piles because the 'downdrag' force effectively reduces their capacity as a percentage of the vertical resistance is now being used to counter the downdrag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Walls, however, provide capacity against lateral movement, and therefore loss of vertical capacity is insignificant. The consolidation, however, could effect the lateral pressures (depending on the construction method / support conditions; see note at end).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lateral force applied to the retaining structure is calculated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_h = K \sigma_v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sigma_h$ is is the resulting lateral earth pressure&lt;/li&gt;&#xA;&lt;li&gt;$K$ is the lateral earth pressure coefficient&lt;/li&gt;&#xA;&lt;li&gt;$\sigma_v$ is the applied vertical pressure at the calculation location&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the pile is installed in situ prior to excavation then the lateral earth pressure can be based on the 'at rest' condition (see note at end), as the soil is never allowed to displace enough to generate the full 'active' stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The at rest lateral earth pressure coefficient for solis is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_0 = (1 - \sin \phi ' ) \times OCR ^ {( \sin \phi ' )} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$K_0$ is the 'at rest' lateral earth pressure coefficient&lt;/li&gt;&#xA;&lt;li&gt;$\phi '$ is the effective stress friction angle of the soil being retained&lt;/li&gt;&#xA;&lt;li&gt;$OCR$ is the overconsolidation ratio ( &gt; 1 for overly consolidated soils)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As the soil consolidates, therefore, the at rest lateral earth pressure increases; and therefore the lateral loading that the wall must structurally resist and retain increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'at rest' condition can only be maintained so long as the retaining wall does not move. As the wall displaces, the lateral earth coefficient of the retained soil tends towards the smaller, 'active' coefficient (that is not proprotional to consolidation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is therefore a conservative assumption to say that the wall must sustain 'at rest' pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.co.uk/books?id=lYoS4VXJPJ4C&amp;amp;pg=PA206&amp;amp;lpg=PA206&amp;amp;dq=embedded%20retaining%20wall%20at%20rest&amp;amp;source=bl&amp;amp;ots=ZeFYLruEeD&amp;amp;sig=zq6c6HdLNYuEjKrVuKj3I52ql5U&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=o-DyVIuXIMfa7Abau4HgDg&amp;amp;ved=0CFUQ6AEwCQ#v=onepage&amp;amp;q=embedded%20retaining%20wall%20at%20rest&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Tomlinson&lt;/a&gt; suggests that $K_0$ should only be used when the movement of the structure is less than $5 \times 10^{-4} $ the retained height in normally consolidated soils; I would expect that limit to be lower for overly consolidated soils, as they require a larger strain to mobilse the 'active' condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted that, to the old British Standards at least, these kinds of walls are typically designed considering only the bending moment determined from the active / passive pressures, with an applied factor of safety (circa 1.4 - 1.5); potentially accounting for this.&lt;/p&gt;&#xA;" OwnerUserId="151" LastEditorUserId="151" LastEditDate="2015-03-01T17:41:41.707" LastActivityDate="2015-03-01T17:41:41.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1902" PostTypeId="2" ParentId="1895" CreationDate="2015-03-01T03:04:37.627" Score="4" Body="&lt;p&gt;I saw some material for tension rods that match the material properties that you gave, but these places do not list the steel used. It might be proprietary:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ronstanrigging.com/arch_w/systemrange.asp?RnID=704&quot; rel=&quot;nofollow&quot;&gt;Ronstan&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.daversteels.co.uk/bar_systems_load_capacity&quot; rel=&quot;nofollow&quot;&gt;DaverSteels&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A third manufacturer &lt;a href=&quot;http://www.macalloy.com/tension-rods&quot; rel=&quot;nofollow&quot;&gt;Macalloy&lt;/a&gt;, has a similar component that also doesn't list a material specification, but it does have an &lt;a href=&quot;http://www.macalloy.com/system/files/tension-rod-systems.pdf&quot; rel=&quot;nofollow&quot;&gt;independent certification&lt;/a&gt; of its material properties. This leads me to believe that this is a custom steel that is only guaranteed by its given properties and the agreements between the company and the foundry.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-01T03:58:00.310" LastActivityDate="2015-03-01T03:58:00.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1903" PostTypeId="1" AcceptedAnswerId="1940" CreationDate="2015-03-01T17:40:49.340" Score="14" ViewCount="4218" Body="&lt;p&gt;What happens to the working of petrol engine when it is emptied and filled with diesel or to  a diesel engine emptied and filled with petrol? Will the engine be able to operate and, if not, why not?&lt;/p&gt;&#xA;" OwnerUserId="739" LastEditorUserId="282" LastEditDate="2015-03-03T01:26:52.833" LastActivityDate="2019-02-27T16:16:25.043" Title="What happens when you put the wrong type of fuel in an internal combustion engine?" Tags="|mechanical-engineering|automotive-engineering|combustion|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1905" PostTypeId="1" AcceptedAnswerId="1909" CreationDate="2015-03-01T18:26:26.137" Score="7" ViewCount="6637" Body="&lt;p&gt;I'm thinking of developing a pocket 'power bank' that could provide both 6V and 12V - a charger to charge either a phone (with 6V) or a notebook computer (with 12V). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrical diagram is simple enough; switch attachment of the terminals like below.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CVHVv.png&quot; alt=&quot;schematic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;The schematics is not everything though. I'd like the charger to have two sockets, one providing 6V, the other - 12V, and to &quot;sense&quot; what to serve by having a plug in either of the sockets (let's say using both at a time is forbidden; I can place them in such a way that both plugs won't fit at a time, say, a slider that opens either of them but never both).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best if the &quot;sensing&quot; was done in the simplest, mechanical way (e.g. there are sockets that provide an extra &quot;sensing&quot; contact that shorts to mass if the plug is in), but if that would prove too difficult or mechanically complex, a more complex electronic solution would work too, providing it won't drain the batteries when nothing is plugged in. (and of course the circuitry must accept external charging current - accepting only one of the two, either 6 or 12V is fine). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course while just cutting one of batteries off for 6V would be the easiest solution, it would halve the capacity for 6V application and cause parasitic charging issue when switching to 12V after the 6V power has been depleted, so it's not really an acceptable solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how to build such a circuitry - that provides power from two batteries, in series or in parallel depending on which socket was used?&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2017-05-17T07:31:36.350" LastActivityDate="2017-05-17T14:02:17.840" Title="Switching batteries from series to parallel - technical solution" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="4" CommentCount="24" ContentLicense="CC BY-SA 3.0" />
  <row Id="1906" PostTypeId="1" AcceptedAnswerId="1936" CreationDate="2015-03-01T19:20:06.230" Score="10" ViewCount="1501" Body="&lt;p&gt;As a follow up to &lt;a href=&quot;https://engineering.stackexchange.com/questions/1864/how-should-the-public-raise-questions-about-unsafe-structures-in-the-united-stat&quot;&gt;this question&lt;/a&gt;, I wonder if structural integrity monitoring information can be provided real-time and to a sufficient degree that safety decisions can realistically be made from the information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put things in context, in the USA and other parts of the world federal, state, and local governments regularly receive structural integrity reports on infrastructure, including bridges. These reports are the results of assessments that are performed on a scheduled basis, and it can be many months or possibly years between assessments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bridges are an important part of federal, state, and local government economy since they provide critical infrastructure.  It is also important to protect historical land marks such as the Golden Gate Bridge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zrPSW.jpg&quot; alt=&quot;Golden Gate Bridge Picture 1&quot;&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Golden Gate Bridge, an iconic bridge in the United States&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;And the consequences of insufficient information can be quite severe.  For example, in August of 2007 the &lt;a href=&quot;http://en.wikipedia.org/wiki/I-35W_Mississippi_River_bridge&quot; rel=&quot;nofollow noreferrer&quot;&gt;I-35W Mississippi River bridge&lt;/a&gt; collapsed, which resulted in the loss of many lives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6h2Ab.jpg&quot; alt=&quot;bridge collapse 2007&quot;&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;I-35W after the collapse&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are there technologies that could provide real-time status such that they would help forewarn federal, state, and local governments of impending bridge failures? For example, is there something that could have predicted the I-35 collapse and notified the public in order to avoid the loss of life?  It seems like this &lt;a href=&quot;http://www.gizmag.com/wireless-bridge-sensor/19380/&quot; rel=&quot;nofollow noreferrer&quot;&gt;wireless sensor to monitor structural integrity of bridges&lt;/a&gt; would work, but I'm not certain as the domain is outside of my experience.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-03-04T00:18:16.120" Title="Can the structural integrity of bridges be measured in real-time for safety?" Tags="|civil-engineering|structural-engineering|bridges|sensors|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1907" PostTypeId="2" ParentId="1905" CreationDate="2015-03-01T21:12:52.623" Score="1" Body="&#xA;&#xA;&lt;p&gt;You have addressed many of the main points - it's really just &quot;a matter of engineering&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take 6V out from the top of the right hand battery and 12V from the 2nd contact down (12V output of DPDT switch) as shown on the diagram below, then you never get incorrect voltages at the output. The changeover to 12V mode can be triggered by inserting a plug in the 12V socket, as you suggest. If you wanted this to happen only when there was a need for 12V current draw you could connect a high resistance sense electronics powered by +6V that detected an actual load present. With some thought you could fully automate this so a 12V load connected to the 12V socket would auto-trigger 12V mode when a 12V load was detected and terminate once there was no load current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases a physical relay is still hard to better in terms of value for money and low voltage drop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 12V charging you can trigger the 6V/12V changeover based on charger voltage or a dedicated switch or with separate charging jacks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Battery balance should be reasonable for identical batteries which are equally charged initially. As you note, you must never draw current from only one battery or charge one battery - changing from 6V to 12V with imbalanced batteries is probably OK but changing from 12V to 6V will tend to destroy the switch with imbalance currents. A very small current limiting resistor in each battery lead would be wise - this can probably be just the wiring resistance with connections being made in such a way as to maximise interbattery resistance with the existing wiring. A resistance that limits real world maximum battery to battery imbalance current to say 5 to 10x relay rated current for &lt;strong&gt;very occasional&lt;/strong&gt; unexpected circumstances is probably OK. Resistance in the battery leads is of course undesirable in normal operation and should be minimised where possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f70oV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorDisplayName="user16" LastEditDate="2015-03-01T21:40:05.983" LastActivityDate="2015-03-01T21:40:05.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1908" PostTypeId="1" AcceptedAnswerId="1954" CreationDate="2015-03-01T21:58:32.903" Score="15" ViewCount="9477" Body="&lt;p&gt;On a recently launched Russian diesel-electric submarine, the rear propeller has two distinct features. You can see spheres at the base of every propeller blade:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vDXwO.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the trailing edge of the shaft has four adjoining fins in line with the axis of rotation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nb9C4.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most importantly, what is the purpose and function of the spheres, and what's beneath the veil in the first image? I believe that it may differ from the second image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's made of a different material, so possibly a sacrificial element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it to reduce noise? Is it to smooth flow or prevent cavitation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that in the first image the spheres are at the leading edge and in the second image the spheres are at the trailing edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a related &lt;a href=&quot;https://www.youtube.com/watch?v=hTrg2nhh0-E&quot; rel=&quot;noreferrer&quot;&gt;youtube video&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="742" LastEditorUserId="6790" LastEditDate="2020-05-02T19:35:38.827" LastActivityDate="2020-05-02T19:35:38.827" Title="Purpose of spheres and fins on submarine propeller" Tags="|mechanical-engineering|fluid-mechanics|propulsion|marine-engineering|naval-engineering|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1909" PostTypeId="2" ParentId="1905" CreationDate="2015-03-01T22:33:21.090" Score="0" Body="&lt;p&gt;I think I came up with an idea which should be trivial to implement and work without any active electronics. Since the design is to employ dedicated charging cables and I can take liberties with the sockets, I can use the plugs as &quot;keys&quot; that route the power as I like: I'd use 4-terminal socket (just a single one) and a &quot;6V cable&quot; and &quot;12V cable&quot; (obviously differing with plugs on the &quot;recipient&quot; end), which would connect to the terminals in a way I need. No &quot;sensing circuitry&quot; and the process of switching voltage is integrated into necessary activity of the user of the device.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rkBXr.png&quot; alt=&quot;schematic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;I'm still leaving the question open - maybe someone comes up with something even better?&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-03-01T22:33:21.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1910" PostTypeId="1" AcceptedAnswerId="25593" CreationDate="2015-03-01T23:00:33.383" Score="6" ViewCount="3388" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Turbo-compound_engine&quot; rel=&quot;noreferrer&quot;&gt;Turbo compound&lt;/a&gt; engines extract some exhaust energy by using it to power a turbine connected to the driveshaft.  Why are these engines not more widely used?&lt;/p&gt;&#xA;" OwnerUserId="398" LastActivityDate="2019-04-26T02:04:08.450" Title="Why are turbo compound engines not more widely used?" Tags="|mechanical-engineering|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1911" PostTypeId="2" ParentId="1905" CreationDate="2015-03-01T23:38:00.260" Score="3" Body="&lt;p&gt;You haven't told us the context this device will be used in, so it's hard to evaluate how appropriate different alternatives are.  Here is one possibility:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The battery is always 12 V.  For one thing, this simplifies charging.  The 12 V outlet simply wires to the battery leads, and all is well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 6 V outlet is actually driven by a buck converter from the 12 V battery.  Again, we don't know the context of this device, so it's hard to say whether the extra electronics is worth the other simplicities.  Buck converters from 12 V to 6 V can be small, cheap, and quite efficient.  Since according to your own answer you can use a special cable for each outlet with extra pins, you can use this mechanism to enable the buck converter so that it uses no quiescient power when 6 V isn't demanded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two main advantages of this approach is that no interlock is needed (mechanical intracacies are often more expensive to develop than a circuit) and charging doesn't have to deal with somehow sensing and adjusting to how the battery is configured.  The battery is always charged as a 12 V battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could take this concept further by using a higher voltage battery, like 18 V for example, and a buck converter for both 12 V and 6 V.  This has the side advantages that both outputs will be well regulated.  Another important issue you may have overlooked is not allowing the external load to drain the batteries to the point they become damaged.  with a circuit between the battery and your load, it can manage the battery and disconnect the load when the battery gets too low.  It can also do over-current sensing, or current limiting, or otherwise deal with a badly behaved load without something catching on fire.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question only asks for something simple, but there may be more issues to think about than are obvious at first glance.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-03-01T23:38:00.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1912" PostTypeId="1" CreationDate="2015-03-01T23:59:16.733" Score="8" ViewCount="505" Body="&lt;p&gt;In the US, clothes dryers are one of the leading causes of house fires: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 1998, CPSC estimates that there were approximately 15,600 clothes dryer fires resulting in 20 deaths, 370 injuries and $75.4 million in property damage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or &lt;a href=&quot;http://www.nfpa.org/safety-information/for-consumers/causes/dryers-and-washing-machines&quot;&gt;more recent numbers&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 2010, an estimated 16,800 reported U.S. non-confined or confined home structure fires involving clothes dryers or washing machines resulted in 51 civilian deaths, 380 civilian injuries and $236 million in direct property damage.&#xA;  Clothes dryers accounted for 92% of the fires; washing machines 4%, and washer and dryer combinations accounted for 4%.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A fairly &lt;a href=&quot;https://www.cpsc.gov//PageFiles/117075/dryer.pdf&quot;&gt;comprehensive study&lt;/a&gt; about &lt;a href=&quot;https://en.wikipedia.org/wiki/Lint_%28material%29&quot;&gt;lint&lt;/a&gt; ignition was completed by the Consumer Protection Safety Commission (&lt;a href=&quot;http://www.cpsc.gov/&quot;&gt;CPSC&lt;/a&gt;) in 2003. This study showed that the dryer's internal temperature greatly increased once the vent pipe's area was reduced by 50%. In homes, this reduction is due to lint gathering in the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.whirlpool.com/digitalassets/LTE5243DQ/Ventilation%20Specification_EN.pdf&quot;&gt;manual for a random dryer&lt;/a&gt; mentions that minimum air flow through the 4&quot; (10 cm) diameter vent pipe is 1,200 feet per minute (20 ft/s or 6.1 m/s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a concept of &lt;a href=&quot;http://www.engineeringtoolbox.com/sewage-piping-systems-d_568.html&quot;&gt;minimum velocity&lt;/a&gt; in drainage pipes and culverts to ensure that they are self cleaning, i.e. debris does not build up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously air and lint are different than water and sediment. I haven't been able to find any particle properties for lint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is the solution to lint clogging dryer vents just to increase the velocity of the air in the vent?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-02T00:05:50.743" LastActivityDate="2015-03-11T16:02:10.380" Title="Could house fires be prevented by increasing dryer vent speed?" Tags="|mechanical-engineering|airflow|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1913" PostTypeId="2" ParentId="1912" CreationDate="2015-03-02T00:17:21.150" Score="4" Body="&lt;p&gt;Taking the data you supplied at face value, a relatively simple solution is for the dryer to measure either the exhaust air flow or back pressure, and refuse to operate when out of spec.  Sensing these parameters can be done cheaply with today's electronics, especially considering such dryers already have a microcontroller in them to do the sequencing, run the user interface, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real question is whether consumers would put up with a nanny-dryer, and whether they'd be willing to pay the small extra cost.  The cost delta would be quite small, but in such a high volume highly competitive market where consumers buy largely on price, this might be a problem.  I expect manufacturers would strongly resist attempts to mandate this.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-03-02T00:17:21.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1915" PostTypeId="2" ParentId="1912" CreationDate="2015-03-02T02:19:57.540" Score="2" Body="&lt;p&gt;One of the issues of having an exhaust velocity that is too high is the increased possibility of injuring someone who might go past, particularly and eye or ear canal injury. A cover on the outlet would reduce air velocity and increase the chance of lint build up near the exhaust outlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue with increased air velocity is increased noise from the outlet and vibration of the duct. The owners and the neighbours could be upset. Additionally, increased air velocity may also increase power consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the ducts being made of metal, lint build up due to static charge will be less of a problem that if than if the ducts were made of plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Air intimates, the problem might lie with the installation and alignment of the duct. More bends and kinks can reduce cross-sectional area, increase shock losses, reduce air velocity and the carrying power of the air to carry the lint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Screws or rivets joining duct segments that protrude into the internal duct space would be issue as they would act as collection points for lint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suspicions are that a “set and forget” attitude and a total unawareness of the necessity for periodic maintenance and cleaning of the duct are a significant factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like the idea of a back pressure sensor that @Olin Lantrop discusses in his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other solution would be to alter the airflow system of driers to give a high speed short timed pulse of air before each drying cycle and at the end of each drying cycle to flush out accumulated lint or debris in the duct.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-03-02T02:19:57.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1919" PostTypeId="1" CreationDate="2015-03-02T07:52:19.157" Score="8" ViewCount="2279" Body="&lt;p&gt;First, a neat image &lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Communications_tower_reaching_for_the_clouds.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;from Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/pEK9Wl.jpg&quot; alt=&quot;lattice cell tower with triangular cross-section&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the cell tower has three vertical (or maybe slightly inclined to the inside) thicker pipes which are located at angles of a triangle and those three thicker pipes are interconnected with numerous thinner pipes and this yields a lattice tower with triangular horizontal cross-section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly you need at least three vertical pipes to make this happen. The design above with triangular cross-section is very popular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However sometimes cell towers are built with square cross-section - they use four vertical pipes located at angles of a square instead of three pipes. It looks like waste of steel and design complication - what can a tower with square cross-section do what one with triangular cross-section cannot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could imagine there're was a difference when a tower had notable load concentrated in one direction. Yet I'm asking about cell towers that carry rather lightweight equipment and mostly have to withstand wind loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are square cross-section cell towers used instead of triangular cross-section towers?&lt;/p&gt;&#xA;" OwnerUserId="374" LastActivityDate="2015-03-02T18:45:47.227" Title="Why build a cell lattice tower with square cross-section instead of triangular cross-section?" Tags="|civil-engineering|steel|structures|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1920" PostTypeId="1" AcceptedAnswerId="1923" CreationDate="2015-03-02T13:11:14.063" Score="2" ViewCount="254" Body="&lt;p&gt;I have a rf module transmitting from an antenna. As you see in Figure 1, I have a radio mast that sends waves in all directions. I don't want this. My plan for controlling wave direction is Figure 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I control RF wave direction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z5sj6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="752" LastEditorUserId="33" LastEditDate="2015-03-02T13:52:55.203" LastActivityDate="2015-03-02T14:36:40.447" Title="Control RF wave direction or path" Tags="|electrical-engineering|telecommunication|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1921" PostTypeId="2" ParentId="1767" CreationDate="2015-03-02T13:29:18.143" Score="10" Body="&lt;p&gt;BS5950-1:2000 Clause 1.3.23 defines an H-section as having &quot;&lt;em&gt;an overall depth not greater than 1.2 times its overall width&lt;/em&gt;&quot;, and Clause 1.3.25 defines an I section as having &quot;&lt;em&gt;an overall depth greater than 1.2 times its overall width&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that at exactly a ratio of 1.2, it would be an H section not an I section.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2015-03-02T13:46:54.927" LastActivityDate="2015-03-02T13:46:54.927" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1922" PostTypeId="2" ParentId="1920" CreationDate="2015-03-02T13:29:37.413" Score="1" Body="&lt;p&gt;There are a number of different designs for &lt;a href=&quot;http://en.wikipedia.org/wiki/Directional_antenna&quot; rel=&quot;nofollow noreferrer&quot;&gt;directional antennas&lt;/a&gt;, and which one is most useful to you depends on the particular RF frequency you will be operating at and how much directionality you want.  A simple &lt;a href=&quot;http://en.wikipedia.org/wiki/Yagi-Uda_antenna&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yagi-Uda&lt;/a&gt; with the generic design shown below is probably the easiest directional antenna to manufacture yourself.  This antenna uses a single driven dipole antenna with one (or possibly more) passive 'director' elements in front of it and one (or possibly more) slightly larger 'reflector' elements behind.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:A8-8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sqmrq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these directional antennas operate on the principle of &lt;a href=&quot;http://en.wikipedia.org/wiki/Phased_array&quot; rel=&quot;nofollow noreferrer&quot;&gt;phased arrays&lt;/a&gt; of smaller antennas.  In a phased array each antenna can be thought of as an individual dipole source.  The directionality happens because these individual dipoles &lt;a href=&quot;http://en.wikipedia.org/wiki/Interference_%28wave_propagation%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;interfere&lt;/a&gt; constructively in the direction of propagation and destructively in other directions.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-03-02T13:36:57.593" LastActivityDate="2015-03-02T13:36:57.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1923" PostTypeId="2" ParentId="1920" CreationDate="2015-03-02T14:36:40.447" Score="1" Body="&lt;p&gt;Directional antennas generally use one of two priciples: Reflection or phasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reflection works just like it does with optical mirrors, except that the wavelength is much much longer and therefore the construction of something that reflects may seem unintuitive.  Most of the world's really high gain* antennas are parabolic dishes, which focus the power in a relatively narrow beam.  For example, large radio telecscopes are like that.  Common rooftop UHF TV antennas also usually worked on this principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phasing means multiple elements with careful spacing and being fed with a specific phase shift of the carrier between them.  If arranged properly, the multiple elements all add to a wave going one direction, but cancel each other out going other directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider two vertical antennas that each radiate equally in all horizontal directions.  These antennas, I'll call them A and B, are spaced 1/4 wavelength apart.  B is fed with a signal that lags A by 1/4 wavelength.  The signal radiated by A in the direction of B will be in phase with B by the time it gets there.  The signals from the two antennas will then add going in the direction from A to B.  However, the signal emitted by B in the direction of A will find A 180&amp;deg; out of phase, so the two signals cancel.  This is a simple system that has a gain of 3 dB in the A to B direction compared to one of the antennas by itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phased arrays can contain more than just two phased emitters.  More emitters allows tighter control over the beam shape.  The Aegis radar is a good example of this technique taken much farther.  The large number of individual emitters all acting together form a narrow beam, which is narrow enough to focus on a single aircraft.  One advantage of this system is that the direction can be steered by electronics very quickly, much faster than a reflecting antenna can be mechanically re-aimed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Gain&quot; when applied to antennas doesn't mean the antenna can radiate more RF power than it receives electrically.  Ideally, antennas radiate all the electrical power they receive.  The difference is how concentrated it is in some directions.  Gain is how much a antenna can concentrate the power in its optimal direction relative to a standard emitter.  There are two references commonly used for 0 dB gain, a isotropic radiator and a dipole.  A dipole has a donut-shaped radiation pattern, so has gain along its plane relative to a isotropic radiator.  Both references are commonly used, so if it is not stated the statement is either ambiguous or it is implied by other context.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-03-02T14:36:40.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1924" PostTypeId="2" ParentId="1919" CreationDate="2015-03-02T14:44:01.383" Score="4" Body="&lt;p&gt;You are correct that a triangle is the most efficient basic shape. Once the overall design has some constraints, other factors may hold more sway over the final design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some of the design considerations that may make a square cross section preferable to a triangular one:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Limited ground area&lt;/strong&gt; - For a given ground area, more steel and thus more strength will be available if 4 legs are used instead of 3. The use of 3 legs may require the distance between each leg to exceed the available area on the ground.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ease of connecting accessories&lt;/strong&gt; - Notice how the lower transmitter on your picture is greatly skewed compared to the bar that it is attached to. Adding another side to the tower may make these connections more square. This will also affect the antennae at the top of the tower. The anticipated uses of the tower will help to determine this.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ease of making cross-brace connections&lt;/strong&gt; - Each of the cross braces need to be connected to the main legs of the tower. Typically there are multiple braces connecting at the same location. Making the tower cross section be square will help to make these connections at 90deg, which may be easier to construct or fabricate.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Shape of material used for legs&lt;/strong&gt; - Not all towers use round tubes for the vertical legs. With a round tube, a connection to the leg is equally difficult in any direction. If &quot;L&quot; or square shapes are used, connections are much easier if everything is square.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Proprietary designs&lt;/strong&gt; - Some manufactures may only build one shape or the other. They are used to the design that they typically make.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Steel weight&lt;/strong&gt; - Your argument for triangular cross sections being the most efficient is basically a steel weight consideration. The actual weight of material may not be that large of a factor in the final cost. Adding 25% to the weight of steel will not increase the final tower cost by 25%.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastEditorUserId="368" LastEditDate="2015-03-02T18:45:47.227" LastActivityDate="2015-03-02T18:45:47.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1925" PostTypeId="1" CreationDate="2015-03-02T15:27:04.853" Score="4" ViewCount="3006" Body="&lt;p&gt;Dynamic Mechanical Analysis, otherwise known as DMA, is a technique where&#xA;a small deformation is applied to a sample in a cyclic manner. This allows the&#xA;materials response to stress, temperature, frequency and other values to be&#xA;studied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;DMA works by applying a sinusoidal deformation to a sample of known geometry.&#xA;The sample can be subjected by a controlled stress or a controlled strain. For a&#xA;known stress, the sample will then deform a certain amount. In DMA this is done&#xA;sinusoidally. How much it deforms is related to its stiffness. A force motor is used&#xA;to generate the sinusoidal wave and this is transmitted to the sample via a drive&#xA;shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modulus values change with temperature and transitions in materials can be&#xA;seen as changes in the E’ (elastic modulus) or tan delta curves (tan delta = elastic modulus / loss modulus). This includes not only the glass&#xA;transition and the melt, but also other transitions that occur in the glassy or&#xA;rubbery plateau. The glass transition (Tg) is seen as a large drop (a decade or more) in the storage modulus when viewed on a log scale against a linear temperature scale.  A concurrent peak in the tan delta is also seen. The value reported&#xA;as the Tg varies with industry with the onset of the E’ drop, the peak of the tan&#xA;delta, and the peak of the E’ curve being the most commonly used. This is a typical output from a DMA scan in temperature with a ramp rate near 5 °C:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qwfuM.gif&quot; alt=&quot;DMA TG&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below shows the results from a &lt;a href=&quot;http://www.instron.com/en-us/our-company/library/glossary/d/dma-dynamic-mechanical-analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;DMA scan&lt;/a&gt;. &#xA;In black there is the elastic modulus, in gray the loss modulus and in brown the tangent delta or damping. What is the meaning of the two tangent delta peaks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/53xVm.png&quot; alt=&quot;DMA scan&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the material changing its property during the DMA scan? The x-axis is temperature, while modulus and tangent delta have different units on the y-axis. And what is the meaning of the increasing elastic modulus?&lt;/p&gt;&#xA;" OwnerUserId="755" LastEditorUserId="755" LastEditDate="2015-03-03T16:46:35.817" LastActivityDate="2020-11-01T23:46:11.177" Title="About glass transition temperature in a DMA scan of an unknown material" Tags="|materials|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1926" PostTypeId="1" AcceptedAnswerId="1927" CreationDate="2015-03-02T17:45:28.383" Score="58" ViewCount="36579" Body="&lt;p&gt;I thought my electric car charging unit uses 6.6&amp;nbsp;kW of power. However, I found the label and it actually says 6.6&amp;nbsp;kVA. When I saw this I thought something along the lines of...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Well, $ P=VI $, therefore kVA must be the same thing as kW... strange, I wonder why it's not labelled in kW.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So a quick Google search later, and I found &lt;a href=&quot;http://www.dieselserviceandsupply.com/Power_Calculator.aspx&quot;&gt;this page&lt;/a&gt;, which has a converter that tells me 6.6&amp;nbsp;kVA is actually just 5.28&amp;nbsp;kW. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Watt&quot;&gt;Wikipedia page for watts&lt;/a&gt; confirmed what I thought, that a watt is a volt times an ampere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what part of all this am I missing, that explains why kVA and kW are not the same?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="330" LastEditDate="2015-03-02T21:42:12.200" LastActivityDate="2021-07-16T19:05:23.513" Title="Why is kVA not the same as kW?" Tags="|electrical-engineering|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1927" PostTypeId="2" ParentId="1926" CreationDate="2015-03-02T17:54:45.867" Score="66" Body="&lt;p&gt;The problem is that the formula $P=I\ V$ is correct when dealing with DC circuits or with AC circuits where there is no lag between the current and the voltage.  When dealing with realistic AC circuits, the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/electric/powerac.html&quot;&gt;power is given by&lt;/a&gt; &#xA;$$&#xA;P=I\ V\ \cos(\phi),&#xA;$$&#xA;where $\phi$ is the phase difference between the current and the voltage.  The unit kVA is a unit of what is called 'apparent power' whereas W is a unit of 'real power'.  Apparent power is the maximum possible power attainable when the current and voltage are in phase and real power is the actual amount of work which can be done with a given circuit.   &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-03-02T17:54:45.867" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="1928" PostTypeId="2" ParentId="1926" CreationDate="2015-03-02T19:49:22.303" Score="11" Body="&lt;p&gt;When an AC line is driving an inductive or capacitive load, then the load will spend some of its time taking power from the source, but will also spend some of its time &lt;em&gt;feeding power back&lt;/em&gt; to the source.  In some contexts, a device which draws a total of 7.5 joules each second and returns a total of 2.5 joules may be regarded as though it was drawing 5 watts (especially if whenever the device is returning power some other load is ready to consume it immediately).  Something like a transformer, however, will suffer conversion losses not only during the part of the cycle when the load is drawing power, but will &lt;em&gt;also&lt;/em&gt; suffer losses during the part of the cycle when the load is feeding it back.  While a transformer would probably dissipate less heat driving the above load than one which drew 10 joules/second and returned zero, it would dissipate more than when driving a load which drew 7.5 joules/second and returned zero.&lt;/p&gt;&#xA;" OwnerUserId="397" LastActivityDate="2015-03-02T19:49:22.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1929" PostTypeId="1" CreationDate="2015-03-02T21:00:57.053" Score="7" ViewCount="7441" Body="&lt;p&gt;I was just wondering how a diesel engine starts, given that there are no spark plugs. How does it build up 'momentum' for that first explosion in the cylinder? Does it have something to do with the starter motor?&lt;/p&gt;&#xA;" OwnerUserId="762" LastEditorUserId="368" LastEditDate="2015-03-02T22:48:06.907" LastActivityDate="2016-06-17T14:50:50.273" Title="How does a diesel internal combustion engine start even though there are no spark plugs?" Tags="|mechanical-engineering|automotive-engineering|combustion|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1930" PostTypeId="2" ParentId="1929" CreationDate="2015-03-02T21:21:20.427" Score="7" Body="&lt;p&gt;Diesels are started by using a combination of &quot;glow plugs&quot; and a starter motor.  Once the diesel gets going, combustion is caused by compression of the mixture.  Initially, the cylinder walls can be so cold that they cool the mixture during compression enough to not have it ignite.  Each cylinder has a glow plug that provides a small spot of hot surface that is sufficient to ignite the compressed mixture, even when the cylinder walls have cooled it.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2015-03-02T22:48:34.777" LastActivityDate="2015-03-02T22:48:34.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1931" PostTypeId="2" ParentId="1926" CreationDate="2015-03-02T21:24:24.537" Score="30" Body="&lt;p&gt;Both watts and volt-amps come from the same equation, $P=IV$, but the difference is how they're measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get volt-amps, you multiply root mean square (RMS) voltage ($V$) with RMS current ($I$) with no regard for the timing/phasing between them. This is what the wiring and pretty much all electrical/electronic components have to deal with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get watts, you multiply instantaneous voltage ($V$) with instantaneous current ($I$) for every sample, then average those results. This is the energy that is actually transferred.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now to compare the two measurements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If voltage and current are both sinewaves, then $\text{watts} = \text{volt-amps} \times \cos(\phi)$, where $\phi$ is the phase angle between voltage and current. It's pretty easy to see from this that if they're both sine waves and if they're in phase ($\phi = 0$), then $\text{watts} = \text{volt-amps}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you're NOT dealing with sine waves, the $\cos(\phi)$ relationship &lt;em&gt;no longer applies&lt;/em&gt;! So you have to go the long way around and actually do the measurements as described here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might that happen? Easy. DC power supplies. They're everywhere, including battery chargers, and the vast majority of them only draw current at the peak of the AC voltage waveform because that's the only time that their filter capacitors are otherwise less than the input voltage. So they draw a big spike of current to recharge the caps, starting just before the voltage peak and ending right at the voltage peak, and then they draw nothing until the next peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course there's an exception to this rule also, and that is Power Factor Correction (PFC). DC power supplies with PFC are specialized switching power supplies that end up producing more DC voltage than the highest AC peak, and they do it in such a way that their input current follows the input voltage almost exactly. Of course, this is only an approximation, but the goal is to get a close enough match that the $\cos(\phi)$ shortcut becomes acceptably close to accurate, with $\phi \approx 0$. Then, given this high voltage DC, a secondary switching supply produces what is actually required by the circuit being powered.&lt;/p&gt;&#xA;" OwnerUserId="763" LastEditorUserId="763" LastEditDate="2015-03-05T14:36:21.013" LastActivityDate="2015-03-05T14:36:21.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1933" PostTypeId="2" ParentId="1906" CreationDate="2015-03-02T23:05:54.863" Score="3" Body="&lt;p&gt;Structural monitoring with sensors is indeed possible. The issues are the cost of monitoring, and which parts can actually be monitored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more critical a bridge is, the more likely it is to be worth paying the cost of monitoring (criticalness being mainly due to number of vehicles crossing per day, but also the availability of alternate routes and the length of it (i.e. number of vehicles which may be on it during collapse) play a part). A major bridge built these days is likely to have some sort of monitoring built into it, as noted in the article you linked, and an example of which is the &lt;a href=&quot;http://www.james-fisher.com/news-and-media/press-releases/strainstall-wins-contract-provide-structural-health-monitoring-new-forth-replacement-crossing/&quot; rel=&quot;nofollow&quot;&gt;Forth Replacement Crossing&lt;/a&gt; (now called the Queensferry Crossing). Some bridges have had systems retrofitted, such as the &lt;a href=&quot;http://www.forthroadbridge.org/acousticmonitoring&quot; rel=&quot;nofollow&quot;&gt;existing Forth Road Bridge&lt;/a&gt;, which has acoustic monitoring to listen for cables snapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Existing bridges are most likely to be fitted with monitoring if an assessment shows that there is cause for concern. If a bridge assessment finds that part of the bridge is deteriorating, it may recommend more frequent inspections, or the installation of monitoring. Because of the cost of monitoring, this is likely to only happen on major bridges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, could monitoring have prevented the I35-W collapse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly: I35-W, although not a small bridge, was not on the same scale as the Golden Gate Bridge or the Forth bridges. Monitoring was hence only likely to be installed if the bridge had been identified as deficient. And that hadn't happened because the routine bridge inspections and assessments had failed to notice that a small connecting plate was undersized. Your link to sensors seems to suggest that new cheaper sensors may now be available. I still can't see the cost as being affordable given the shear number of bridges which each authority has control over, but maybe over a long period (say 50 years as the budget might stretch to 2% of an authority's bridges per year - numbers plucked completely out of the air by me) it could happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly: I don't believe most monitoring is realtime. Although I'm sure monitoring systems have automatic alarms that go off if significant movement occurs, I struggle to believe that someone would have a particularly fast response to it. More than likely, a system alarm would result in someone going to have a look, maybe the same day maybe not. Mainly the systems are there to look at long term issues. Taking the Forth Road Bridge as an example - the acoustic monitoring is there to record the &quot;ping&quot; from a cable strand snapping. There are a few thousand individual strands making up the suspension cables. The acoustic monitoring is there so that someone can look at the results and say &quot;it looks like 63 have snapped in the past six months, maybe we should think about doing something to reduce the load&quot;. I don't think it's really there with the expectation that it will record 100 snapping in a day, resulting in the bridge being instantaneously closed. (The Forth Road Bridge closes due to high wind a few times a year - they have a proper system in place to close it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly: the &lt;a href=&quot;http://www.dot.state.mn.us/i35wbridge/ntsb/finalreport.pdf&quot; rel=&quot;nofollow&quot;&gt;official report into the collapse&lt;/a&gt; suggests that the failure was very sudden: there was maybe 10 seconds warning. Even with realtime monitoring, you need to a) receive results, b) determine that they're catastrophically bad (easy if you have the 100 pings in my Forth example, difficult if you have accelerometers as per your link, which might be telling you that the bridge moved 0.05 degrees) and c) get the police out there to close the bridge. Total time estimation: 10minutes if you're very lucky. Not 10 seconds. (And probably more like 3 months if you set up the monitoring for long term effects and weren't sure whether the 0.05 degrees was significant or not, because the system wasn't set up for that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Disclaimer - I've tried to put in facts and references where I can, but the actual conclusions I've come to are heavily based on my opinions and are not straight facts.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-03-02T23:05:54.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1934" PostTypeId="1" CreationDate="2015-03-02T23:24:33.797" Score="9" ViewCount="11789" Body="&lt;p&gt;At a high level both a linear and LDO regulators appear to be the same. Most block diagrams appear to be the same too. Most of the device packages are three terminal packages. What are the differences between linear and LDO regulators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XRsn4.jpg&quot; alt=&quot;Regulator&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="282" LastEditDate="2015-03-03T23:15:26.307" LastActivityDate="2022-12-16T17:18:30.187" Title="What are the differences between linear and low-dropout (LDO) regulators?" Tags="|electrical-engineering|power-electronics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1935" PostTypeId="2" ParentId="1934" CreationDate="2015-03-02T23:44:18.717" Score="6" Body="&lt;p&gt;At a very high level linear low-dropout regulators have a pass transistor, an operational amplifier, a resistor divider circuit feeding the operational amplifier and VREF. The operational amplifier is an error amplifier. In short the pass transistor acts as a variable resistor which controls the current flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S3VUI.png&quot; alt=&quot;Regulator Block Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Block diagram for linear &amp;amp; low dropout regulator&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dropout voltage = $V_{in}$ - $V_{out}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Linear Regulator:&lt;/strong&gt; The pass transistor is a Darlington transistor. The dropout voltage is around 1.5V–2.5V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Abnac.jpg&quot; alt=&quot;Linear Regulator&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Basic schematic for linear regulator&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is dropout voltage specification for LM7805 linear regulator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/79XmZ.jpg&quot; alt=&quot;LM7805 Droput Voltage&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Low Dropout Regulator:&lt;/strong&gt; The pass transistor is a PNP transistor. The dropout voltage is very low, approximately 0.7 V or even lower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mvaRE.jpg&quot; alt=&quot;LDO Regulator&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Basic schematic for low dropout regulator&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is dropout voltage specification for LT1965 low dropout regulator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uTJ5D.jpg&quot; alt=&quot;LT1965 low dropout voltage&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lit/an/snva020b/snva020b.pdf&quot; rel=&quot;noreferrer&quot;&gt;Linear Regulators: Theory of Operation and Compensation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.fairchildsemi.com/datasheets/lm/LM7805.pdf&quot; rel=&quot;noreferrer&quot;&gt;LM7805  Linear Voltage Regulator&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cds.linear.com/docs/en/datasheet/1965fb.pdf&quot; rel=&quot;noreferrer&quot;&gt;LT1965 Series Low Dropout Linear Regulator&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-09-07T02:03:44.023" LastActivityDate="2016-09-07T02:03:44.023" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="1936" PostTypeId="2" ParentId="1906" CreationDate="2015-03-03T02:37:23.427" Score="7" Body="&lt;p&gt;There are many different sensors that can be placed on a bridge to help &lt;a href=&quot;http://www.oregon.gov/ODOT/HWY/BRIDGE/docs/SHM_SCL_submission.pdf&quot; rel=&quot;noreferrer&quot;&gt;monitor its condition&lt;/a&gt;. After the I-35 bridge collapse, &lt;a href=&quot;http://mtri.org/bridgecondition/&quot; rel=&quot;noreferrer&quot;&gt;this area&lt;/a&gt; has seen a &lt;a href=&quot;http://www.govtech.com/transportation/Sensors-Wireless-Tech-Help-States-Monitor-Troubled-Bridges.html&quot; rel=&quot;noreferrer&quot;&gt;lot of publicity&lt;/a&gt;. Most of these &lt;a href=&quot;https://www.campbellsci.com/bridge-monitoring&quot; rel=&quot;noreferrer&quot;&gt;sensors&lt;/a&gt; were previously available, but the need wasn't as apparent to the public.&lt;/p&gt;&#xA;&lt;h3&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Structural_health_monitoring#Examples&quot; rel=&quot;noreferrer&quot;&gt;Sensors&lt;/a&gt;&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Accelerometers&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Strain Gauges&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Displacement Transducers&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Level Sensing Stations (Tilt Monitors)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Anemometers&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Temperature Sensors&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Dynamic Weight-in-Motion Sensors&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Corrosion Monitors&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Acoustic Crack Monitors&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Acoustic Cable Monitors&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;These sensors can be wired or wireless. Some are even &lt;a href=&quot;http://www.ecmag.com/section/miscellaneous/new-wireless-bridge-sensors-powered-passing-traffic&quot; rel=&quot;noreferrer&quot;&gt;powered by the vibrations&lt;/a&gt; from the passing vehicles.&lt;/p&gt;&#xA;&lt;h3&gt;Cost&lt;/h3&gt;&#xA;&lt;p&gt;Cost is a major factor in considering a bridge monitoring system. Even if the initial cost seems &lt;a href=&quot;http://esciencenews.com/articles/2011/07/29/averting.bridge.disasters.new.technology.could.save.hundreds.lives&quot; rel=&quot;noreferrer&quot;&gt;reasonable&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The sensors are small, wireless, rugged, and require practically no maintenance, he says. They are expected to last more than a decade, with each costing about \$20. An average-sized highway bridge would need about 500 sensors for a total cost of about \$10,000.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This doesn't account for the cost of continued monitoring of the sensors. Real-time monitoring is expensive and even less than real-time monitoring imposes a constant drain on Department of Transportation budgets that are already stretched too thin. If there was more money in the budget, the bridges would be in better condition.&lt;/p&gt;&#xA;&lt;p&gt;These bridges are already inspected regularly by trained inspectors that climb all over the structure. This was done on the I-35 bridge and it still collapsed.&lt;/p&gt;&#xA;&lt;p&gt;The cost and complexity of the systems mean that only a few very large structures would be fully monitored. The cost of these systems is likely more than the amount listed above. The value above is also just the cost of the individual sensors. A full study of the bridge would have to be done to insure that the sensors are placed in the locations where they would be most useful.&lt;/p&gt;&#xA;&lt;h3&gt;Location&lt;/h3&gt;&#xA;&lt;p&gt;Each sensor measures only a single data point. There would have to be a lot of thought and research to determine which sensors are the most beneficial.&lt;/p&gt;&#xA;&lt;p&gt;Even after the sensors are placed, a comprehensive criteria would have to be in place so that the people monitoring the sensor results would know what constitutes a &lt;em&gt;warning&lt;/em&gt;. There might be many false positives and maybe even false negatives.&lt;/p&gt;&#xA;&lt;p&gt;As an &lt;a href=&quot;http://www.futurity.org/5-years-after-i-35-collapse-big-leap-in-bridge-sensors/&quot; rel=&quot;noreferrer&quot;&gt;article&lt;/a&gt; points out:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Ultimately, adds Fu, real-time remote sensing will serve as a valuable supplement, but not fully replace human inspections. “You can’t put sensors everywhere,” he says.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;End Result&lt;/h3&gt;&#xA;&lt;p&gt;Even if all of these sensors were in place on the I-35 bridge and monitored in real-time, the actual &lt;a href=&quot;http://www.dot.state.mn.us/i35wbridge/ntsb/finalreport.pdf&quot; rel=&quot;noreferrer&quot;&gt;collapse happened so fast&lt;/a&gt; that traffic would not have been able to be stopped in time. The only difference would be that there would have been a lot more information for the investigators to go through.&lt;/p&gt;&#xA;&lt;p&gt;Think of a plane's &lt;a href=&quot;https://en.wikipedia.org/wiki/Flight_recorder&quot; rel=&quot;noreferrer&quot;&gt;Black Box&lt;/a&gt;, but for a bridge.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-03-03T02:37:23.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1937" PostTypeId="2" ParentId="1934" CreationDate="2015-03-03T14:26:07.367" Score="13" Body="&lt;p&gt;The difference is due to the strongest force in the universe: Marketing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both devices are linear regulators.  This means they take a high and varying voltage in, and produce a low and steady (&quot;regulated&quot;) voltage out.  These type of regulators are totally powered from the input voltage, and adjust how much current they allow to pass from the input to the output.  The concept is the same as a mechanical pressure regulator.  In that case there is a real valve inside that is opened more or closed more so that the output pressure is maintained.  Electronic regulators use transistors for much the same purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One parameter of a linear regulator is how low the input voltage can go while it can still control the output voltage within spec.  This parameter is often referred to as the &lt;i&gt;headroom&lt;/i&gt; or &lt;i&gt;dropout voltage&lt;/i&gt;.  Old regulators needed several volts.  The famous LM7805 needs about 2&amp;frac12; volts.  Since it's output is 5 V, that means its input needs to be at least 7.5 V else the output being 5 V is not guaranteed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back in neolithic times, cavemen would make a power supply by using a big-iron transformer to drop the line voltage, 4 diodes to take the absolute value, and a cap to smooth it out.  To get a regulated supply, they'd follow that with a linear regulator.  It didn't matter much that the linear regulator needed over 2 V headroom since there had to be significant slop built into the raw unregulated supply for various other reasons anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today there is much more pressure to make power supplies small, light, cheap, and sometimes efficient.  The strive for efficiency is actually not so much to not waste the power, but to not have to get rid of the heat from wasting the power.  Electronic components have advanced considerably since the caveman days, so today it's generally cheaper and smaller to make the power supply more efficient than to add the ability to dissipate a lot of heat.  Over 80% efficient is &quot;easy&quot;, 90% efficient reasonably attainable in a lot of mainstream circumstances, and close to 95% attainable when need be and efficiency is a prime priority.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a 7805 in this context.  Even at only 500 mA output (half its maximum capability), it would dissipate 1.3 W &lt;i&gt;at minimimum&lt;/i&gt;.  That can be a lot of heat to get rid of in a small and sealed plastic package.  You can't just plunk a surface mount part on a board if it's going to dissipate that much power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manufacturers of linear regulator ICs started paying attention to the dropout voltage.  A whole new class of linear regulators emerged that needed less than a volt, sometimes just a couple 100 mV, of headroom.  Of course these technological advances were accompanied by advances in marketing.  It was no longer good enough to simply publish a lower headroom spec.  This new class of regulator was to be brought forth as a new device altogether, called the [&lt;i&gt;insert trumpet blasts, picture of clouds parting, deep booming voice&lt;/i&gt;] &lt;b&gt;Low-Dropout Regulator&lt;/b&gt;, henceforth to be know as a &lt;b&gt;LDO&lt;/b&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, being a marketing term, nobody bothered to specify how low exactly &quot;low&quot; is.  Marketing has advanced even more since the dawn of the LDO, so nowadays pretty much every regulator everyone makes except anything called a 78xx or 79xx is called a &quot;LDO&quot;.  That's become the TLA for linear regulators to the point that them new kids (those that never had to walk to school barefoot in the snow uphill both ways, what's this world coming to?!) don't even know the difference anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to pick a number, I'd say anything that can work with less than 1 V headroom can call itself a LDO, preferably 500 mV, but as always, you have to &lt;b&gt;read the datasheet&lt;/b&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power supplies are largely not made with big-iron transformers anymore where you start with a raw unregulated voltage and linearly regulate it down to the voltage you actually want.  Today this would be done with a switcher all in one.  However, no matter how you make a regulated voltage, you still have a distribution problem if the device draws significant current.  There is also the ripple caused by the switching pulses which can cause trouble with some kinds of analog circuits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution is to have the switcher make a few 100 mV more than you want, then use LDOs at multiple points of use.  Let's say you're making local 3.3 V from a 3.8 V pre-regulated bus coming directly from a switcher.  At 100 mA, the LDO only dissipates 50 mW, which even a SOT-23 package can do easily.  This method not only guarantees that the voltage will be the regulated value despite some drop in the pre-regulated voltage by the time it gets to the point of use, but also produces a nice and clean supply since the local regulator will greatly attenuate the ripple from the pre-regulated supply.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="110" LastEditDate="2015-03-05T00:42:31.520" LastActivityDate="2015-03-05T00:42:31.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1938" PostTypeId="1" CreationDate="2015-03-03T16:05:21.133" Score="7" ViewCount="966" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1gTeC.jpg&quot; alt=&quot;Scrooge McDuck swimming pool&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I wanted to build a gold coins swimming pool [as in DuckTales], what &lt;em&gt;engineering&lt;/em&gt; (not security) concerns would I need to address to make such a pool feasible?  For example, since gold is much heavier than water, would I need to be concerned more about the overall weight that is supported by the base or the lateral pressures applied to the sides of the pools?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more details:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;He had metal walls. Would steel walls be sufficient, would I need something stronger like Titanium?&lt;/li&gt;&#xA;&lt;li&gt;Humans are, in nearly all cases, under $3$ meters tall. Even though Scrooge's depth gauge is at $90 ft$, for engineering reasons let's keep the pool to $3$ meters of depth. &lt;strong&gt;Note:&lt;/strong&gt; the question originally proposed a $30m$ x $30m$ x $30m$ pool, which I am now changing to a $30m$ x $30m$ x $3m$ pool.&lt;/li&gt;&#xA;&lt;li&gt;The density of the gold coins is near enough the density of pure gold, i.e. $19.3\frac{g}{cm^3}$&lt;/li&gt;&#xA;&lt;li&gt;The gold coins would be as tightly packed as they would be if you expect them to get jostled around often (e.g. swimming!). They would not be in neat stacks.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="781" LastEditorUserId="781" LastEditDate="2015-03-04T17:10:31.570" LastActivityDate="2015-03-04T17:10:31.570" Title="What concerns would I need to consider for building a swimming pool of gold coins?" Tags="|structures|structural-engineering|building-design|" AnswerCount="3" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1939" PostTypeId="2" ParentId="1906" CreationDate="2015-03-03T17:05:19.110" Score="6" Body="&lt;p&gt;Real-time monitoring can be done and has already been done. The technology is there and as pointed out by hazzey, is a matter of cost versus the risk involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a practising Engineer, I am/was involved in two deep excavation projects in two different countries where a cut-and-cover tunnel is/was constructed right under a major highway bridge. In both cases, traffic closure or diversion is not an option. The Building Authorities in both projects requested for real-time structural health monitoring as the consequences of a mishap can be catastrophic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both projects, an array of optical prisms together with robotic total stations are deployed to monitor bridge pier movements. The total stations are networked to a server where data is being uploaded continuously. The server itself is connected to the internet, where parties involved can download the data and process the information. In addition, some systems can generate SMS alerts in case where a predetermined level is breached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before any monitoring is done, an impact/damage assessment has to be carried out to determine:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the predicted soil movement (horizontal and vertical) at the foundation of the bridge based on the construction method&lt;/li&gt;&#xA;&lt;li&gt;the kind of forces that will be induced on the piles&lt;/li&gt;&#xA;&lt;li&gt;the structural capacity of the existing piles as determined from as-built drawings&lt;/li&gt;&#xA;&lt;li&gt;the existing load on the piles during service&lt;/li&gt;&#xA;&lt;li&gt;the reserve capacity of the piles&lt;/li&gt;&#xA;&lt;li&gt;the allowable lateral movement and settlement of the bridge foundation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is followed by a condition survey to assess the current state of health of the bridge. It usually includes visual inspection for cracks and measurements on bridge bearing movements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reinforced concrete and steel bridges, under normal circumstances, do not require real-time monitoring as they do not fail suddenly. Bridges that have been properly designed tend to fail in serviceability before they fail structurally, giving sufficient warning before they collapse. Also, regular inspection and maintenance of the bridges help to ensure that the bridges remain serviceable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With enough data, real-time monitoring can capture long-term movement trend and smooth out reading errors. In cases of fatal design, human error in interpreting survey records, or human error in assessment of the bridge structural capacity, real-time monitoring would not be of much help to avert a disaster.&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2015-03-03T17:05:19.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1940" PostTypeId="2" ParentId="1903" CreationDate="2015-03-03T17:28:52.467" Score="8" Body="&lt;p&gt;Putting diesel fuel in a gasoline engine is just about impossible. Diesel fuel nozzles are larger than gasoline nozzles, and modern gas caps are too small for diesel nozzles to fit into. However, if you managed to get it in there, the diesel fuel is too heavy and evaporates too slowly for the spark plugs to ignite it effectively. &lt;a href=&quot;http://auto.howstuffworks.com/fuel-efficiency/alternative-fuels/diesel-in-unleaded.htm&quot;&gt;One source&lt;/a&gt; says that it won't start at all, &lt;a href=&quot;http://alternativefuels.about.com/od/dieselbiodieselvehicles/a/gasolinindiesel.htm&quot;&gt;another source&lt;/a&gt; says that it probably won't start, but there's a chance it will, it'll just run horribly and end up as a smokey disaster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The converse is possible, since the gasoline nozzle is the smaller of the two. In this case, as noted in the second article linked above, gasoline would risk damage to the emissions system, and it will definitely lead to damage of the engine. Diesel fuel lubricates the engine as it is consumed. Gasoline, being thinner (part of the reason it evaporates easier), will not provide that lubrication and you'll start to see friction damage on engine components. Because it's also designed to combust differently, you'll see messed up timing, which can cause further damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that, in the increase in popularity of diesel cars and the lack of mindfulness of drivers, Volkswagen has &lt;a href=&quot;http://jalopnik.com/volkswagen-fixes-their-diesels-because-idiots-keep-putt-509309545&quot;&gt;altered its diesel tank gates&lt;/a&gt; to prevent gasoline from being put into the tank, because of the damage it can cause. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-03-03T17:28:52.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1941" PostTypeId="2" ParentId="582" CreationDate="2015-03-03T18:11:48.270" Score="1" Body="&lt;p&gt;An AAC wall has an &lt;a href=&quot;http://www.yourhome.gov.au/materials/autoclaved-aerated-concrete&quot; rel=&quot;nofollow&quot;&gt;insulation R rating of 1.43&lt;/a&gt;. A cavity brick wall has an R value of under 0.82.  So AAC isn't even double the insulation of cavity bricks, and at this point AAC may not be as inexpensive as cavity bricks + standard insulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many, many construction walls that have much greater R values, but as you've not given any information on your structure and environment it's impossible for us to weigh tradeoffs for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind, though, that a standard 2x4 framed wall with R-13 insulation will have an R value of 10 - 6 times greater than AAC. So unless you have special requirements that require concrete or brick walls, standard timber framing should be cheaper and better unless you're looking at basement or foundation walls.&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2015-03-03T18:11:48.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1942" PostTypeId="2" ParentId="572" CreationDate="2015-03-03T18:32:06.610" Score="4" Body="&lt;p&gt;You are essentially asking, &lt;em&gt;&quot;can a cyclone be designed that works without, or against gravity?&quot;&lt;/em&gt; and the answer is, of course, yes.  Most practical industrial designs can use gravity to increase efficiency or reduce cost, and the equations you've seen need this simplification, but a cyclone doesn't depend on gravity.  You just need to whip the air around fast enough that the larger particles can't make it into the smaller diameter exit.  Orientation is irrelevant when you've got high rotation rates, and the powerful air pumps used in the Dyson vacuums can produce very high rotation rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, the vacuums are multi stage cyclones.  Even my old Dyson has at least two stages, one big obvious cyclone, and then a bunch of tiny cyclones which are designed for the smaller particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notably, mine does have a two stage air filter after the cyclones, but I'm sure the newer models with their even smaller second or third stage cyclones and more powerful motors are even better with smaller particles.  I would be surprised if they could get a HEPA rating on a cyclone-only design, though, so I'd be wary of their marketing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the first stage, you'll find that inverting it has only one effect - the trash comes to the top rather than the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it still doesn't enter the smaller diameter exit tube for that stage.  The particles weigh too much, and the spinning wind prevents them from ever leaving the edges of the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while some industrial cyclones and those you've studied all rely on gravity for some portion of their rated design, that doesn't need to be the case at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, continuous changes in orientation simply don't matter that much. I'm sure they do affect it, but the speed of change from one orientation to another is very slow relative to the speed of the spinning in the cyclone. Energy is lost, perhaps a few more particles get through that wouldn't otherwise get through if it were held stationary, but they are caught by the next stage, and this would only happen to those particles that are closer to the size the next stage is designed for anyway.&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2015-03-03T18:32:06.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1943" PostTypeId="2" ParentId="1743" CreationDate="2015-03-03T18:52:22.190" Score="4" Body="&lt;p&gt;These are things that installers and builders can control and understand.  They aren't interested in running the equations themselves, they just have a customer who is asking for a product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the industry has settled on some basic standards, and as long as everyone builds according to code and the devices is rated for indoor use, then they only need either a big enough room, or a big enough vent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big enough room falls under safe working environment.  Assuming a specific length of work period, and sealed room starting with the average air mixture, can a person complete their work with the device running and at the end of the period still have a low enough CO2 and a high enough O2 that they are still well within the EPA limits.  This is long before headaches start and other physical or mental problems manifest from high levels of CO2 or low levels of O2 become apparent. Combined with an oxygen sensor (required in such devices) the gases in this size room will generally not exceed a limit unless a door is never opened and the room is truly sealed. And generally most rooms built to code have plenty of air migration, few are truly sealed.  Still, it's possible to cause problems with this setup, which is why the room size requirement is so very large, to the point where it'd be odd to have such a large room without multiple windows or doors. Even then, I'd still recommend venting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A big enough vent simply assures adequate CO2 and O2 exchange with an area of standard air mixture.  This exchange will occur even if the vent isn't powered, and without a pressure differential - similar to osmosis.  An area of high CO2 in air contact with an area of low CO2 will cause CO2 migration from the high concentration area to the low.  The rate of diffusion through the vent is directly related to the area of the vent.  Thus continuous operation with the proper vent area is possible without ever reaching dangerous levels. There is a tradeoff between vent size and heat loss, though, so it's worthwhile keeping the vent closer to the smallest acceptable size than the largest if the point is to heat the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these devices are (almost?) never rated for sleeping areas, and they advise users to leave the area if they develop headaches, one of the first signs of CO2 poisoning.&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2015-03-03T18:52:22.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1944" PostTypeId="2" ParentId="1714" CreationDate="2015-03-03T19:05:50.370" Score="4" Body="&lt;p&gt;The welder you are using forms a very large inductive loop when the contacts are welding. Any conductive materials you bring within that loop will form one half of a transformer, and they will &quot;steal&quot; some of the energy meant to go into the weld.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to break this effect you need to eliminate conductive paths within the magnetic field developed by the welding loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be simply choosing insulating materials (plastic, for instance). If you must use metal, then you can choose higher resistance metals.  Silver, copper, gold are bad choices.  Iron, steel, and stainless steel are a magnitude higher resistance than the above, but they are still pretty conductive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So split the steel table up, and use non conductive fastening methods to prevent the currents from moving from piece to piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single length of steel bar near the welding tip will have little effect.  But four lengths of steel electrically connected looping around the welding tip (ie, the tip goes through the center) will be highly coupled to the magnetic filed in the welder, and sap much more energy.  Insert a plastic shim on just one joint, though, and that metal loop becomes open and affects the welding process very little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you can use a plastic table (and there are many very durable plastic tables that would work for most industrial situations) then that might be your best bet.  If not, though, then simply placing electrical insulation in the table will prevent eddy currents that would alter the welder.  Consider asking another question if you want a particular design or table requirement critiqued for adjustments to prevent these currents.&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2015-03-03T19:05:50.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1945" PostTypeId="2" ParentId="38" CreationDate="2015-03-03T19:18:55.313" Score="3" Body="&lt;p&gt;SC-FDMA is encoded and transmitted from the handset where there are power tradeoffs.  This encoding is better than OFDMA for the handset for the following related reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Reduces peak to average power ratio&lt;/li&gt;&#xA;&lt;li&gt;Increases efficiency of the power amplifier&lt;/li&gt;&#xA;&lt;li&gt;Increases battery life&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So at the base station they can employ transmitters that can handle a higher peak transmit power, with lower efficiency as it has a significant amount of power available.  But on a mobile handset, this requires more expensive components, more space, and, probably most importantly, more power since peak transmissions are not as efficient as transmitting at a lower level but for a longer period of time (ie, higher peak to average ratio means lower energy efficiency, more heat, more power). The tradeoff is a little lower spectrum efficiency, but it's worth it for increased battery life, and particularly since most high speed data is going to the handset, not coming from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=dr4YQAfifKA&quot; rel=&quot;nofollow&quot;&gt;This short video&lt;/a&gt; explains further.&lt;/p&gt;&#xA;" OwnerUserId="784" LastEditorUserId="784" LastEditDate="2015-03-03T20:33:05.763" LastActivityDate="2015-03-03T20:33:05.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1946" PostTypeId="2" ParentId="1910" CreationDate="2015-03-03T19:36:28.957" Score="5" Body="&lt;p&gt;There are a lot of reasons, but the main reasons are increase cost and decreased reliability.  The power they produced also cost engine efficiency as they increased the back pressure in the exhaust line.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you design an engine you have a few choices.  For instance, you can either increase engine efficiency, or sap energy from the exhaust line.  If you employ variable valve timing, for instance, you can extract much more of that exhaust energy and put it into the crankshaft without a turbo.  If you instead add a turbo, you've added another set of gears, more moving parts, taken up more space, and added more weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the turbo might be slightly better at extracting energy than the variable valve timing, but does it offset the additional cost, weight, and reliability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So most engine manufacturers don't employ turbo unless there's a reason for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, some races have a limit to the number of cylinders and cylinder volume, and racers do care more about that extra 0.1% efficiency than they care about reliability since they have a team of mechanics working on the vehicle.  So it makes sense there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't make much sense on a standard consumer vehicle where the cost/weight tradeoffs relative to performance can be more effectively employed elsewhere in the engine.&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2015-03-03T19:36:28.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1947" PostTypeId="2" ParentId="1938" CreationDate="2015-03-03T20:02:15.860" Score="8" Body="&lt;p&gt;The weight of the gold would be about 52,110,000kg, over 900 meters square of floor area.  This is about 82.4psi of floor pressure, or, under standard gravity, 57.9kN/m^2 live load. This is only an order of magnitude greater than typical industrial (storage, assembly) buildings that accept &lt;a href=&quot;http://theconstructor.org/structural-engg/live-loads-for-different-building-floors/6993/&quot; rel=&quot;nofollow&quot;&gt;7.5kN/m^2 across the floor surface&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total weight of the gold is about 1/5 the weight of the empty empire state building, but it's about 8 times smaller in area (900 square meters for the gold vs 7,400 square meters for the empire state building).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, back of the napkin, it would be reasonable to assume that you'd be able to find a piece of land that could support the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical steel building techniques will be perfectly fine with sub 100 PSI support structures, though I'd suggest a concrete floor anyway.  No need for exotic materials, though I daresay you can afford them. Single wall pressure vessels can handle the pressure for the walls, if you can make your building circular rather than square, but if you insist it must be square then you'll probably need to have double walls with some diagonal bracing to keep the structure square and the walls straight in the center of the walls. While 80PSI isn't extraordinary, over 30 meters of wall it will exert exceptional force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is a quick back-of-the-napkin analysis, and I'm freely mixing pressure and weight to make determining whether it's feasible or impossible easier.  A more rigorous design and analysis would be required to actually build such a structure.&lt;/p&gt;&#xA;" OwnerUserId="784" LastEditorUserId="784" LastEditDate="2015-03-04T13:56:07.360" LastActivityDate="2015-03-04T13:56:07.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1948" PostTypeId="2" ParentId="1903" CreationDate="2015-03-03T21:30:40.360" Score="13" Body="&lt;p&gt;To understand how these fuels behave in internal combustion (IC) engines, you need to first understand the properties of the fuels and how they differ.&lt;/p&gt;&#xA;&lt;h1&gt;Fuel Properties&lt;/h1&gt;&#xA;&lt;p&gt;Both petrol (gasoline) and diesel are lighter-than-water organic liquids, most commonly produced from petroleum. Diesel requires much less effort to produce—you can &lt;a href=&quot;http://makezine.com/projects/make-03/backyard-biodiesel/&quot; rel=&quot;noreferrer&quot;&gt;make it yourself from cooking oil&lt;/a&gt; with only a few readily-available chemicals.&lt;/p&gt;&#xA;&lt;p&gt;Gasoline is lighter that diesel, with a density about three-quarters that of water, and highly volatile. That means it evaporates quickly at room temperature, which is why it has such a strong odor and is stored in closed containers &lt;a href=&quot;https://i.stack.imgur.com/l2c1g.jpg&quot; rel=&quot;noreferrer&quot;&gt;such as these&lt;/a&gt; when not fueling a vehicle directly. Apart from the odor, gasoline vapors are also very potent carcinogens and precursors for the formation of photochemical (&amp;quot;brown&amp;quot;) smog, which during one particularly bad incident &lt;a href=&quot;http://www.wired.com/2010/07/0726la-first-big-smog/&quot; rel=&quot;noreferrer&quot;&gt;mistaken for a chemical weapons attack&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Diesel, in comparison to gasoline, is not very volatile at all. This is partly because it's less refined; the closer the distillate is to crude oil, the more stable it tends to be. You can store it diesel an open container for short periods with no significant losses.&lt;/p&gt;&#xA;&lt;h2&gt;Flash Point&lt;/h2&gt;&#xA;&lt;p&gt;Closely related to the volatility of the liquid is its &lt;em&gt;flash point&lt;/em&gt;, the temperature at which the liquid's vapor pressure is high enough for its vapors to ignite in the presence of an ignition source. Below this temperature, the equilibrium state will be such that while you can burn the vapor, e.g. with a lit match, the combustion reaction would be localized. Above the flash point, the flame would cause a chain reaction that spreads throughout (&lt;em&gt;ignites&lt;/em&gt;) the volume of vaporized fuel.&lt;/p&gt;&#xA;&lt;p&gt;As you would expect, gasoline's high volatility results in a much lower flash point than diesel. &lt;a href=&quot;http://en.wikipedia.org/wiki/Flash_point#Examples&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia gives the values as −43 °C (gasoline) and 52 °C (diesel).&lt;/a&gt; For reference, that means that if you filled an indoor swimming pool with gasoline and let the room come to equilibrium, you'd have to cool that room down below −43 °C (&lt;a href=&quot;http://www.antarctica.ac.uk/about_antarctica/geography/weather/temperatures.php&quot; rel=&quot;noreferrer&quot;&gt;approximately Antarctic temperatures&lt;/a&gt;) before a lit match would &lt;em&gt;not&lt;/em&gt; cause the room to explode. And even at a temperature well below the flash point, that match would burn &lt;em&gt;really&lt;/em&gt; fast.&lt;/p&gt;&#xA;&lt;p&gt;On the other hand, fill the same swimming pool with diesel fuel, and you'd have to heat the room above 52 °C (&lt;a href=&quot;http://en.wikipedia.org/wiki/U.S._state_temperature_extremes&quot; rel=&quot;noreferrer&quot;&gt;approximately Death Valley temperatures&lt;/a&gt;) before a lit match would cause the room to explode.&lt;/p&gt;&#xA;&lt;p&gt;This is different from the autoignition temperature, also given in the Wikipedia article, which is the point at which you no longer need to light the match. Kaboom! Although gasoline has the lower flash point, it's diesel that has the lower autoignition temperature (256 °C, compared to 280 °C, from the WP article above).&lt;/p&gt;&#xA;&lt;p&gt;As we move on from talking fuels to talking engines, the #1 point to remember is that &lt;strong&gt;with an ignition source, gasoline ignites first; without an ignition source, diesel ignites first.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Engine Properties&lt;/h1&gt;&#xA;&lt;p&gt;&amp;quot;IC engine&amp;quot; is a broad category comprising many engine types and technologies. The main difference between conventional diesel and gas engines has to do with how the fuel is ignited (see #1 point above!): gasoline engines typically use &lt;em&gt;spark ignition&lt;/em&gt; and diesel engines use &lt;em&gt;compression ignition&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Injection&lt;/h2&gt;&#xA;&lt;p&gt;Before ignition, the fuel in the tank needs to be mixed with air. This can be done in a variety of ways and although the general concept is simple, the fuel properties have a great impact on how this mixture is achieved.&lt;/p&gt;&#xA;&lt;p&gt;Typically, the fuel is &lt;em&gt;atomized&lt;/em&gt; in the cylinder—sprayed through &lt;a href=&quot;https://i.stack.imgur.com/0sONy.jpg&quot; rel=&quot;noreferrer&quot;&gt;small nozzles called fuel injectors&lt;/a&gt;—producing many very small droplets of fuel suspended in air. (Older cars used &lt;a href=&quot;http://en.wikipedia.org/wiki/Carburetor&quot; rel=&quot;noreferrer&quot;&gt;a different technology&lt;/a&gt; to create the fuel-air mixture.) The more homogeneous the fuel-air mixture (or &amp;quot;charge&amp;quot;), the more efficiently and completely it will burn and the better the engine will operate overall.&lt;/p&gt;&#xA;&lt;p&gt;Our goal is to achieve a good fuel-air mixture quickly, ignite the fuel, extract its energy (in the form of useful work) and move on to the next stroke. This is how we get power (&lt;em&gt;work over time&lt;/em&gt;) out of the engine. Atomizing the liquid fuel both disperses it and increases its surface area, allowing it to evaporate more quickly. Gasoline's high volatility compared to diesel makes it much easier to completely vaporize, obtaining that homogeneous charge. This is not necessarily always a benefit, though—which leads us to consider the properties of the engines themselves.&lt;/p&gt;&#xA;&lt;h2&gt;Ignition&lt;/h2&gt;&#xA;&lt;p&gt;Spark ignition is perhaps the easier method to understand; it's comparable to the lit match analogy from earlier. The difference is that after vaporizing the fuel and mixing it with the air, the ignition source is electric spark rather than an open flame. Because the spark controls the exact time of ignition, we can make sure it happens at the best point during the piston's stroke to transfer power to the drive train. If ignition occurs at the wrong time, it could drive the piston in the wrong direction or burn the fuel incompletely. This is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Engine_knocking&quot; rel=&quot;noreferrer&quot;&gt;knocking&lt;/a&gt;, and it can cause permanent damage to the engine.&lt;/p&gt;&#xA;&lt;p&gt;Many additives have been used over the course of the past century to prevent gasoline from igniting without a spark, including lead and &lt;a href=&quot;http://www.epa.gov/mtbe/gas.htm&quot; rel=&quot;noreferrer&quot;&gt;MTBE&lt;/a&gt;. Other formulations are designed to be &lt;a href=&quot;http://www.consumer.ftc.gov/articles/0210-paying-premium-high-octane-gasoline&quot; rel=&quot;noreferrer&quot;&gt;used in high-performance gasoline engines&lt;/a&gt; (which have higher compression ratios) or to &lt;a href=&quot;http://www.arb.ca.gov/fuels/gasoline/gasoline.htm&quot; rel=&quot;noreferrer&quot;&gt;improve emission characteristics&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Compression Ignition&lt;/h3&gt;&#xA;&lt;p&gt;In a diesel engine, there is no spark; instead, the air in the cylinder is compressed rapidly, &lt;a href=&quot;https://physics.stackexchange.com/q/17948&quot;&gt;heating it up&lt;/a&gt; past the auto-ignition point of the fuel. Once there's enough thermal energy available to make the combustion reaction spontaneous, &lt;em&gt;kaboom!&lt;/em&gt; You get your ignition. But it does take &lt;strong&gt;a lot of compression&lt;/strong&gt; to reach that point—much more than is required for efficient operation of a spark-ignition engine. This is measured in terms of a ratio of the volume of the closed cylinder at the beginning and end of a stroke, also known as the &lt;em&gt;compression ratio&lt;/em&gt; of the engine.&lt;/p&gt;&#xA;&lt;p&gt;Without going into all the details of engine cycles, here's an animation that shows how the volume of the cylinder changes as the piston moves up and down:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://commons.wikimedia.org/wiki/File%3A4StrokeEngine_Ortho_3D_Small.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/d/dc/4StrokeEngine_Ortho_3D_Small.gif&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;&lt;a href=&quot;http://commons.wikimedia.org/wiki/File%3A4StrokeEngine_Ortho_3D_Small.gif&quot; rel=&quot;noreferrer&quot;&gt;4StrokeEngine Ortho 3D Small&lt;/a&gt; (&lt;a href=&quot;http://creativecommons.org/licenses/by-sa/3.0&quot; rel=&quot;noreferrer&quot;&gt;CC BY-SA 3.0&lt;/a&gt; or&#xA;&lt;a href=&quot;http://www.gnu.org/copyleft/fdl.html&quot; rel=&quot;noreferrer&quot;&gt;GFDL&lt;/a&gt;), by Zephyris (Own work), from Wikimedia Commons&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A typical gasoline-powered spark-ignition engine has a compression ratio of about 10:1, meaning the volume of the cylinder at the start of the compression stroke is 10 times the volume of the cylinder at the end of the stroke. Diesel-powered compression-ignition engines have higher compression ratios, typically around 17:1 but may be higher. &lt;strong&gt;Both gasoline and diesel compression ratios vary from engine to engine;&lt;/strong&gt; the thing to remember is that diesel engines have significantly &lt;em&gt;higher&lt;/em&gt; compression ratios than their equivalent gasoline counterparts.&lt;/p&gt;&#xA;&lt;h1&gt;Using the Wrong Fuel&lt;/h1&gt;&#xA;&lt;p&gt;So, what happens when you use petrol (gasoline) in a diesel engine, or vice versa? A lot of things—but generally, what people care about are:&lt;/p&gt;&#xA;&lt;h2&gt;1. Will the engine run?&lt;/h2&gt;&#xA;&lt;h3&gt;Diesel in a Gasoline Engine&lt;/h3&gt;&#xA;&lt;p&gt;As &lt;a href=&quot;https://engineering.stackexchange.com/a/1940/368&quot;&gt;Trevor notes&lt;/a&gt;, there are safety measures in place to prevent filling a gasoline vehicle with diesel fuel. Diesel is also thicker than gasoline, which can mean clogged fuel lines and injectors, but let's assume that the diesel actually makes it into the engine.&lt;/p&gt;&#xA;&lt;p&gt;Remember the #1 point from our discussion of fuels? When dealing with spark ignition, diesel is &lt;em&gt;harder&lt;/em&gt; to ignite than gasoline. It has a much higher flash point, so we need to raise its temperature quite a bit before the engine has any chance of running. The compression ratio of the spark-ignition engine is lower, which means the fuel doesn't get as hot. As a consequence, &lt;strong&gt;it's very likely that the engine won't run at all&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If the engine has a high enough compression ratio, and/or if it's a hot enough day—remember, the flash point of diesel is somewhere around the high end of Death Valley temperatures—then the engine will run, but not very well. It also may not run for very long; you may achieve ignition in some cylinders but not others, in some strokes but not others, and eventually either the fuel itself or its partially-burned leftovers are going to clog up some part of the system. Even when you do get ignition, you'll get less power because the RPMs are higher (diesel burns more slowly than gasoline) and the temperature is lower.&lt;/p&gt;&#xA;&lt;h3&gt;Gasoline in a Diesel Engine&lt;/h3&gt;&#xA;&lt;p&gt;This is in many ways a more interesting case—it's easier to fit a gasoline nozzle into a diesel vehicle at the pump and the thinner gasoline may not have as much trouble moving through a fuel system designed for thicker diesel, so we can be confident that the fuel will reach the engine.&lt;/p&gt;&#xA;&lt;p&gt;Now we're in the opposite situation from before. We have a fuel that not only is &lt;em&gt;much&lt;/em&gt; easier to ignite via spark than compression (remember, its autoignition temperature is &lt;em&gt;higher&lt;/em&gt; than diesel's), it's also formulated with additives to prevent it from igniting without a spark so that it won't damage the engines it's designed to work in. And so again, there is the possibility that &lt;strong&gt;the engine may not run at all&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;But that doesn't necessarily mean no ignition! I'm going to quote &lt;a href=&quot;http://www.straightdope.com/columns/read/2179/can-i-use-diesel-fuel-instead-of-regular-gas&quot; rel=&quot;noreferrer&quot;&gt;The Straight Dope&lt;/a&gt; for this part of the explanation:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Since gasoline is designed to be resistant to self-ignition, gasoline&#xA;in a diesel engine either won't ignite or will ignite at the wrong&#xA;time. Some diesel engines run leaner than gasoline engines (meaning&#xA;that the air-fuel mix has a higher proportion of air than a gasoline&#xA;engine). That increases the chances that the gasoline won't ignite and&#xA;that unburnt fuel will be sent into the hot exhaust system--where,&#xA;ironically, it could ignite, leading to possible exhaust damage.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This leads us to:&lt;/p&gt;&#xA;&lt;h2&gt;2. Will the engine be damaged?&lt;/h2&gt;&#xA;&lt;p&gt;Diesel engines are often perceived as &amp;quot;tougher&amp;quot;—they're built to withstand the much higher pressures that come with their high compression ratios and, at least in the US, they're generally marketed in larger vehicles and trucks. They deliver more torque, they move heavy loads, they're used in generators and industrial equipment; so it's ironic that putting gasoline in a diesel engine is &lt;em&gt;much more likely&lt;/em&gt; to cause major damage to the vehicle than the converse.&lt;/p&gt;&#xA;&lt;p&gt;In addition to the risk of gasoline vapors passing through the engine to explode in other systems,  if the engine &lt;em&gt;does&lt;/em&gt; manage to ignite the gasoline in the cylinder, &lt;a href=&quot;https://engineering.stackexchange.com/a/1940/368&quot;&gt;Trevor pointed out&lt;/a&gt; that the thinner gasoline will provide less lubrication to the engine; even at diesel RPMs, it won't take too long for the engine to tear itself apart without proper lubrication. The Straight Dope article also mentions loss of lubrication potentially damaging the fuel pump.&lt;/p&gt;&#xA;&lt;h1&gt;Bottom Line&lt;/h1&gt;&#xA;&lt;p&gt;Either way, it's possible to ignite the fuel and drive the pistons, meaning the engine could run—poorly. But it's much better for the vehicle if the engine does &lt;em&gt;not&lt;/em&gt; run—especially for the diesel vehicle, which is more likely to be damaged by the gasoline (and could theoretically explode!). We are not here to give out auto maintenance advice, but if you understand the problem from an engineering perspective, it should be clear that the best case scenario here is that you discover the problem before starting up the vehicle, so that the fuel tank can be drained and refilled with the correct fuel type.&lt;/p&gt;&#xA;&lt;p&gt;And remember, kids: &lt;a href=&quot;https://i.stack.imgur.com/fUAEk.gif&quot; rel=&quot;noreferrer&quot;&gt;Don't try this at home!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-03-03T21:59:06.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1949" PostTypeId="2" ParentId="1807" CreationDate="2015-03-04T00:15:56.110" Score="1" Body="&lt;p&gt;I can't answer this question in terms of a building structure; however, I can for a reinforced concrete pavement, which may still be some interest for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In line with the other answers, a trial pavement design is devised, which is then assessed against the design life. The load departed on the pavement is expressed in terms of a standard axle repetition over the design life. For example, a pavement might be designed to withstand 1x10&lt;sup&gt;5&lt;/sup&gt; standard axle repetitions over a 40-year design life. This is called the Design Repetitions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A trial pavement is selected, then a fatigue analysis is undertaken which determines the pavement's &lt;strong&gt;Allowable&lt;/strong&gt; Repetitions. The Design Repetitions are then divided by the Allowable Repetitions and if this value, defined as the Cumulative Damage Factor (CDF), is &amp;lt; 1.0, then your pavement will survive the design life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, CDF = n / N where n = Design Repetitions, N = Allowable repetitions. Note that you could do a back-calculation to determine the design life if you knew the other terms.&lt;/p&gt;&#xA;" OwnerUserId="767" LastEditorUserId="368" LastEditDate="2015-03-04T00:24:20.110" LastActivityDate="2015-03-04T00:24:20.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1950" PostTypeId="1" CreationDate="2015-03-04T03:30:10.537" Score="3" ViewCount="208" Body="&lt;p&gt;The &lt;a href=&quot;http://conceptengine.tripod.com/conceptengine/id3.html&quot; rel=&quot;nofollow&quot;&gt;Concept IC engine&lt;/a&gt; is a reciprocating engine that opens the exhaust valves soon after combustion begins.  The inventors claim that the exhaust acts similarly to a recoilless gun so that the piston is still pushed while the exhaust has more energy than a conventional engine, allowing more energy to be extracted by a turbo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a viable concept?&lt;/p&gt;&#xA;" OwnerUserId="398" LastEditorUserId="368" LastEditDate="2015-03-04T17:34:17.690" LastActivityDate="2015-04-14T15:37:25.013" Title="Is the concept IC Engine a viable concept?" Tags="|mechanical-engineering|combustion|turbines|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="1951" PostTypeId="2" ParentId="1938" CreationDate="2015-03-04T04:16:48.227" Score="3" Body="&lt;p&gt;&lt;strong&gt;Gold Properties&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.azom.com/article.aspx?ArticleID=1965&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\gamma=19.3\ g/cm^3 \\&#xA;yield\ strength=205MPa$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How Tall?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold is relatively soft. It is also relatively dense. At some point a stack of gold coins will stop being separate coins on the bottom. The stack will be so tall that the bottom coins are smashed together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that this height is about &lt;a href=&quot;https://physics.stackexchange.com/questions/39512/stacking-gold-bars&quot;&gt;1,000 meters&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good news! Your gold coins will remain coins; you won't have to swim through a solid blob of gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Horizontal Pressure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are going to assume that a pile of gold coins behaves somewhat like a pile of rocks. This angle of friction is 10 degrees per &lt;a href=&quot;https://what-if.xkcd.com/111/&quot; rel=&quot;nofollow&quot;&gt;XKCD&lt;/a&gt; and this &lt;a href=&quot;http://asoiaf.westeros.org/index.php/topic/78114-the-hobbit-a-long-expected-spoiler-movie-thread/page-20#entry3985923&quot; rel=&quot;nofollow&quot;&gt;random post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi=10^\circ$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;https://en.wikipedia.org/wiki/Lateral_earth_pressure#Rankine_theory&quot; rel=&quot;nofollow&quot;&gt;Rankine active earth pressure&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_a = tan^2(45-\frac{\phi}{2})=0.70$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that 70% of the vertical force of the gold is exerted laterally on the walls of the vault. This is a lot. A typical soil is less dense and exerts only about 30% of the lateral force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the height of the gold, the following forces would have to be resisted by the bottom of the wall:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3m of gold = 0.4 MPa (8,300 lb/sf)&lt;/li&gt;&#xA;&lt;li&gt;30m of gold = 4.0 MPa (83,000 lb/sf)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This would be a &lt;strong&gt;very&lt;/strong&gt; sturdy wall!&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-04T12:52:13.417" LastActivityDate="2015-03-04T12:52:13.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1952" PostTypeId="2" ParentId="1938" CreationDate="2015-03-04T06:50:43.790" Score="5" Body="&lt;p&gt;&quot;Swimming&quot; would be about impossible.&lt;br&gt;&#xA;&lt;strong&gt;IF&lt;/strong&gt; the gold conveyed buoyancy as water does, which it wouldn't, - at full density you'd &quot;float&quot; with about 5% immersed.&lt;br&gt;&#xA;If mean density due to packing was about 50% then you'd &quot;float at about 10% immersed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you dug/burrowed/dived (ouch) a hole and coins then piled in on top of you the mechanisms that cause buoyancy - by transferring force uniformly to all adjacent surfaces and balancing pressure across the body - would not apply, and it would be like trying to surface from a trench cave in, but worse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your body projected frontal area was about  0.5 m^2 then a 100mm / 4 inch layer of coins above you at full density would weigh about one ton/tonne.&lt;br&gt;&#xA;Even at half average gold density that's 500 kg. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Swimming in bricks may be less painful. &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-03-04T06:50:43.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1953" PostTypeId="2" ParentId="52" CreationDate="2015-03-04T13:09:07.073" Score="1" Body="&lt;p&gt;You need to describe your actual problem more, but it sounds like a solenoid is not the best answer here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you want more than two stable positions, a stepper motor may be more appropriate.  A stepper motor has many stable positions.  If you want linear motion, then some mechanical conversion from the rotary motion of the stepper is needed, like a rack and pinion for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory a stepper motor could be made as a linear motor over some limited travel distance, but I've never seen one.  This is probably because of low demand and cheaper ways to solve most such problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is a continuous motor with a mechanical linkage that prevents back-driving, like a worm gear.  If the motor is a brushless DC, then the firmware can keep track of position.  Otherwise, some sort of position sensor will be required.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-03-04T13:09:07.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1954" PostTypeId="2" ParentId="1908" CreationDate="2015-03-04T14:46:58.590" Score="10" Body="&lt;p&gt;In the video you've linked to, the spheres are seen on both the leading and trailing edges of the propeller:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fenwo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect they are intentional - there are a number of ways to attach a propeller without having to disturb that surface, or using flush caps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cavitation is caused by a drop in pressure.  This would be seen on the trailing edges of the propeller, and is worse on the outer edge of the propeller which is moving faster through water than the inner area.  I doubt these spheres have any effect on cavitation at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is either a sensor system, or noise reduction.  If sensors, it's possible that these could be used as a rearward doppler array, even on a moving propeller, though this seems unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More likely, though, is not just a passive noise reduction system, but an &lt;a href=&quot;http://www.armscontrol.ru/subs/snf/snf03221.htm&quot; rel=&quot;noreferrer&quot;&gt;active noise reduction system&lt;/a&gt;.  These may be transducers that emit sound intended to cancel out specific, otherwise unavoidable noise the submarine generates during normal operation.  Active noise suppression systems have been in use in silent submarines for decades now, but are, of course, kept secret.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is that these are caps that are covering something else, and are removed (manually or merely fall off by using the propeller in water) prior to active duty.  They may be hiding parts of the propeller design not intended to be public.  Given that they are made of a distinctly different material or coated differently in some photographs, this might be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, they might be there simply to throw off competitors and are red herrings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately &lt;a href=&quot;http://americanhistory.si.edu/subs/anglesdangles/taming.html&quot; rel=&quot;noreferrer&quot;&gt;the secretive nature of stealth submarine technology&lt;/a&gt; means we may not be able to definitively prove any of these hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2015-03-04T14:46:58.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1955" PostTypeId="1" CreationDate="2015-03-04T16:21:42.333" Score="7" ViewCount="477" Body="&lt;p&gt;What we learn in engineering school is often idealistic, with textbook examples showing typical prismatic steel beams with a plastic section classification or excavation in a perfect sand or clay geology. However, in the real world nothing is perfect and engineers will face complex problems not taught in school, where there is little referencing literature and where the design codes do not provide clear guidelines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the tasks I was assigned when I started out as an engineer was to design cofferdams for both abutments of a river bridge with a 30 degree skew across the river due to site constraints. As a result, the abutments at both side of the river are skewed. As a designer for the contractor, often we do not get to dictate the shape of the excavation. Material cost is always a source of great consternation, even more so in temporary works design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I built a 2D frame for each strut level to do a structural analysis for the odd-shaped cofferdam. However, as the excavation is on a 35 degree slope, the sheet pile wall on the land side has greater lateral earth pressure than the river side. Also, due to the parallelogram shape, I have difficulty balancing the forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YupHp.png&quot; alt=&quot;cofferdam strutting arrangement for an abutment&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How should you model the boundary conditions for such an irregularly-shaped cofferdam, where the lateral earth pressure acting at the strut level is not balanced?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="620" LastEditorUserId="368" LastEditDate="2015-03-04T17:02:10.267" LastActivityDate="2015-03-09T14:36:34.990" Title="Boundary condition for a parallelogram-shaped excavation on an embankment" Tags="|civil-engineering|structural-engineering|modeling|geotechnical-engineering|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1956" PostTypeId="5" CreationDate="2015-03-04T16:59:08.420" Score="0" Body="&lt;p&gt;Use this tag for questions about numerical or analytical modeling and simulation of engineered systems such as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;analytical solutions to partial differential equations (PDEs) &amp;amp; ordinary differential equations (ODEs)&lt;/li&gt;&#xA;&lt;li&gt;Discretization techniques (Finite Element, Finite Difference, Discrete Element, Finite Volume, etc.)&lt;/li&gt;&#xA;&lt;li&gt;General Purpose engineering modeling and computer aided engineering (CAE) questions&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="368" LastEditorUserId="33" LastEditDate="2016-02-23T02:55:27.270" LastActivityDate="2016-02-23T02:55:27.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1957" PostTypeId="4" CreationDate="2015-03-04T16:59:08.420" Score="0" Body="Use this tag for questions about numerical or analytical modeling and simulation of engineered systems." OwnerUserId="368" LastEditorUserId="2291" LastEditDate="2015-07-16T10:12:31.987" LastActivityDate="2015-07-16T10:12:31.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1958" PostTypeId="1" AcceptedAnswerId="1960" CreationDate="2015-03-04T19:50:21.837" Score="13" ViewCount="7203" Body="&lt;p&gt;...or in other words, why doesn't the material just stick to the screw, rotating in place, without progressing along its length?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simplest case, the answer is obvious: gravity. If it's a granular or liquid material and the screw is tilted to a side, it will just roll/slide/flow along the blade, to remain on the bottom side of the tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KE01J.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then, &lt;a href=&quot;http://en.wikipedia.org/wiki/Archimedes%27_screw#Variants&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A variant of the Archimedes' screw can also be found in some injection molding machines, die casting machines and extrusion of plastics, which employ a screw of decreasing pitch to compress and melt the material. Finally, it is also used in a specific type of positive displacement air compressor: the rotary-screw air compressor. On a much larger scale, Archimedes' screws of decreasing pitch are used for the compaction of waste material.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In such case the forces - pressure, shearing, viscosity, adhesion, all would outweigh gravity, and probably in some cases friction against the tube walls. For example, in injection molding, what is there to keep the half-molten plastic from forming a clumping mass, sticking to the screw and keeping rotating in place without any progress? When the pressure increases, which force prevents the material from backing up into area of lower pressure?&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2022-01-26T01:57:28.817" Title="What makes the material travel through Archimedes' Screw?" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1959" PostTypeId="2" ParentId="1958" CreationDate="2015-03-04T20:37:46.310" Score="6" Body="&lt;p&gt;The ones I've seen that actually cause significant pressure don't have a smooth cylinder wall.  The walls have grooves at right angles to the blade passing over them.  It's a tradeoff since the grooves also provide a leakage path around the blade.  But with a combination of a smooth and minimum-friction blade plus a grooved wall, the thing works.  Ideally the walls have maximum friction parallel to the blade and none perpendicular to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently we can achieve this well enough that such things actually work.  I remember wondering the same thing about my mother's old hand-cranked meat grinder, probably made in Germany in the early 1900s.  The blade and walls were made of the same metal, and the blade wasn't particularly polished or anything.  The walls did have grooves, which was enough to make the whole thing work.  It produced enough pressure to force the meat thru small openings at the end.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-03-04T20:37:46.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1960" PostTypeId="2" ParentId="1958" CreationDate="2015-03-04T21:08:49.617" Score="10" Body="&lt;p&gt;There's actually two kinds of screw conveying: Water lifting with screws happens just as in your gif, in the lower part of an inclined pipe we have water and the screw pushes 'packets' of water upwards. The water can't flow back because there's no path, the blades are not fully submerged. This mode is in principle also possible with solids.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/l1QoR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l1QoR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;em&gt;source: &lt;a href=&quot;http://www.wasser-wissen.de/lexikon/h/hebewerk.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wasser Wissen&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, screw conveyors for solids or the screws in extruders and meat grinders or certain dewatering presses work differently: Here it is the friction between the medium and the walls that prevents the material from corotating with the screw. This will be hard or impossible to achieve with low-viscosity liquids like water, as they will flow anywhere. Solids, on the other hand, can be moved in any direction - upward, downward etc.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-04-13T14:18:01.633" LastActivityDate="2016-04-13T14:18:01.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1961" PostTypeId="2" ParentId="1955" CreationDate="2015-03-05T02:21:19.510" Score="2" Body="&lt;p&gt;&lt;strong&gt;Unbalanced Forces&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted, the unbalanced soil heights and the skew create unequal forces in the struts. This is likely going to turn the short side into what is effectively a deadman. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short side will be pushing into the soil. This can mobilize the soil's passive resistance. Since passive resistance is greater than active soil forces, it may be more conservative to assume this lower active force. Also, the short wall may not deflect enough to mobilize the passive pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Modeling&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually modeling the entire excavation will require using cantilever piles. The excavation can't be just a rigid box since the forces are uneven. The piles will need to mobilize some of the passive reaction of the soil below the excavation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forces on the two opposite walls will need to be balanced. These forces can then be used to design the strut frame. The interaction of the opposite walls of the excavation and the soil is not linear, so this will not be easily converted to simple boundary conditions for the frame model. There is a lot of interaction between the deflections of the walls and the soil response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Construction Staging&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differing soil heights not only create modeling complexities in the final condition, they will also create issues at every depth of excavation. This is a point that is often overlooked in design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sheeting or lagging has to be able to support the soil and any additional loads as the excavation progresses. It is good to know where you want a waler to be located in the final condition, but it must be safe to be able to remove enough soil to be able to instal this waler. This means that the wall will be exposed down to a couple of feet below the waler location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even somewhat standard excavations can require 6 or more separate construction phases to be analyzed.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-09T14:36:34.990" LastActivityDate="2015-03-09T14:36:34.990" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="1962" PostTypeId="2" ParentId="1882" CreationDate="2015-03-05T09:54:58.257" Score="3" Body="&lt;p&gt;There are two main considerations here which causes the supersonic designs you mention to differ from more conventional, subsonic aircraft:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the supersonic aircraft employ delta wings to generate their lift. Delta wings are particularly thin wings with little shape to them, but rely on generating strong vorticies to produce lift. The thin profile makes them excellent at producing little drag, however strong vorticies are difficult to produce at slow speeds as viscous forces dominate.&lt;/li&gt;&#xA;&lt;li&gt;in order to reduce the drag contribution of shockwaves produced by the wings at high speeds, the designers have made the span narrow with a long chord, resulting in a very low aspect ratio. Low aspect ratios produce an awful lot of induced drag.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, given these considerations, the angle of attack must be large to produce enough lift at speeds slow enough to land at, and the design of the wings means that stall is less of a design consideration so these AoAs are viable; the drag penalty for flying at such high AoA makes up a smaller proportion of the total drag, so there is a smaller penalty for accepting the additional drag if it reduces landing speed and allows your aircraft to operate out of more airports around the globe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also of note: these aircraft fly at significantly higher speeds and somewhere between cruising and pulling up to a terminal gate you have to lose all that speed - using your wings as a giant air brake is a pretty good way of doing that without requiring any heavy braking system, indeed flaring is the primary mechanism of braking for most supersonic aircraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I didn't see your follow up question; almost certainly not, the droop nose is phenomanally heavy and complex to make it a desireable design feature, it was included out of necessity. These days I would almost guarantee a electronic HUD system which allows the pilots to &quot;see&quot; through the floor using cameras and a battery of other sensors to aid them. Such a system would require multiple failsafes in the event of a power outage, however even considering this it would be substantially simpler, cheaper and lighter - quite possibly safer also since the droop mechanism could fail, just as a camera could, and there is a limit to how many failsafes you could feasibly include on such a mechanism. By contrast, having an array of multiple cameras on independant circuits, each with an isolated backup power supply makes for a simple, easily expandable and very effective failsafe.&lt;/p&gt;&#xA;" OwnerDisplayName="user815" LastEditorDisplayName="user815" LastEditDate="2015-03-05T13:35:49.713" LastActivityDate="2015-03-05T13:35:49.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1963" PostTypeId="2" ParentId="1958" CreationDate="2015-03-05T12:21:22.623" Score="6" Body="&lt;p&gt;This is effectively an addition to other answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An original Archimedes screw for liquids does not work as most people imagine it does,&lt;/strong&gt;&lt;br&gt;&#xA;and the diagram does NOT show how an original Archimedes screw works for fluids. The diagram is valid - it's just of something else which works in a similar manner and has become synonymous with the original design in peoples' minds. And both versions do not work as most people imagine them to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an original Archimedes Screw the outer cylinder is integral with the &quot;screw&quot; and turns with the screw - there is no moving seal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a marble or solid object that cannot fit through the sealing gap is used the two screws work the same. When a fluid is used the difference is important. From the reference below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An analysis, using the lifting of marbles instead of water, is used in almost all nineteenth century texts. The lower end of the helical tube dips into a dish of marbles and scoops one up. The helix continues to revolve, and the marble is continually being lifted a short distance up an inclined plane. The frictional forces are small, and the marble keeps rolling down an infinite succession of inclined planes formed by the revolving helix. At the same time the marble resides at the local low spot on the helix, and is carried up the slope by forces perpendicular to its local motion. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The key point is that the &quot;payload&quot; sess a downwards ramp the whole way and simply &quot;runs downhill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good illustration of the principle is given on &lt;a href=&quot;http://physics.kenyon.edu/EarlyApparatus/Fluids/Archimedes_Screw/Archimedes_Screw.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;this page&lt;/strong&gt;&lt;/a&gt;&lt;br&gt;&#xA;They use tubing so there is no doubt about the &quot;seal&quot;.&lt;br&gt;&#xA;This image demonstrates what the fluid or other payload &quot;sees&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dMHyO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effectively the fluid sits in &quot;buckets at all times and there is no opportunity to escape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In typical original style screws the whole outer casing is sealed to the &quot;screw&quot; and the outer casing rotates with the screw. Consideratuiomn of a cross section shows that as with the tubing each batch of water sits in a container and NOT in a sloping tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in the stationary casing systems mentioned below the payload still usually sits in a &quot;bucket&quot; and runs forever downhill down a ramp as it ascends (!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crucial point of the original system is that the screw can be turned at any speed and even stopped and there is no leakage (apart from any that results from poor construction). Here is a &quot;modern&quot; real world device that works in this manner. This allows the device to be turned by hand at slow speed, or rotated intermittently, or with pauses, with no loss of fluid. Modern sealing techniques and relatively fast and consistent speeds of rotation allow systems where the casing is stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/03dUo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.math.nyu.edu/~crorres/Archimedes/Screw/applications/ScrewFarmer1.jpg&quot; rel=&quot;noreferrer&quot;&gt;Image from here&lt;/a&gt;&lt;br&gt;&#xA;From &lt;a href=&quot;http://www.math.nyu.edu/~crorres/Archimedes/Screw/Applications.html&quot; rel=&quot;noreferrer&quot;&gt;this very good page&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nature of the original design is clearly explained by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Archimedes&amp;#39;_screw&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia - Archimedes Screw&lt;/a&gt; page.&lt;br&gt;&#xA;While they suggest that whether the inner and outer were sealed in original designs, even a brief consideration of the available facts shows that it was indeed the case. viz&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Depictions of Greek and Roman water screws show them being powered by a human treading on the outer casing to turn the entire apparatus as one piece, which would require that the casing be rigidly attached to the screw.&quot; - Wikipedia&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="399" LastEditorUserId="33" LastEditDate="2015-03-05T13:57:20.387" LastActivityDate="2015-03-05T13:57:20.387" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="1964" PostTypeId="1" CreationDate="2015-03-05T18:31:05.143" Score="5" ViewCount="644" Body="&lt;p&gt;Assuming that there are no constraints on cost, land to develop, or supply of materials, is there a theoretical limit on how tall a structure can be built?  What are the main physical constraints that limit the tallness of structures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I emphasize &lt;em&gt;structure&lt;/em&gt; because i don't want to limit the discussion to skyscrapers or other structures that would also have biological (human) occupancy constraints.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Update&lt;/em&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the comments, I want to further impose the constraint that the structure must support its own weight.  I realize that this question is a bit theoretical in scope and that real engineers always have to work with cost/material/land constraints in practice.  I'm just wondering, if we eliminated those particular constraints, is the tallness of a self-supporting structure infinitely unbounded, theoretically?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="106" LastEditDate="2015-03-06T03:35:11.243" LastActivityDate="2023-12-08T15:09:56.177" Title="Is there a theoretical bound to how tall a structure can be built?" Tags="|civil-engineering|structures|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="1966" PostTypeId="1" CreationDate="2015-03-05T21:16:45.320" Score="5" ViewCount="201" Body="&lt;p&gt;I need to select a precise flow meter for the following application condition:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Flow range: 0-150 liter per hour&lt;/li&gt;&#xA;&lt;li&gt;Medium: Diesel&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are so many meters on the market that I am a bit lost. How do I determine whether a flow meter is acceptable for this application? If there are many acceptable flow meters, why should I choose one over the other?&lt;/p&gt;&#xA;" OwnerUserId="819" LastEditorUserId="368" LastEditDate="2015-03-05T22:19:16.730" LastActivityDate="2018-02-26T10:16:18.453" Title="How to select a low flow rate diesel flow meter" Tags="|mechanical-engineering|petroleum-engineering|equipment-selection|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1967" PostTypeId="5" CreationDate="2015-03-05T22:30:09.457" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-03-05T22:30:09.457" LastActivityDate="2015-03-05T22:30:09.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1968" PostTypeId="4" CreationDate="2015-03-05T22:30:09.457" Score="0" Body="Questions about how to choose the correct size, grade, options, etc. for a known piece of equipment, given a well-defined use case." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-03-05T22:45:38.320" LastActivityDate="2015-03-05T22:45:38.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1969" PostTypeId="1" CreationDate="2015-03-05T23:03:22.067" Score="11" ViewCount="1505" Body="&lt;p&gt;When using a &lt;a href=&quot;http://en.wikipedia.org/wiki/Galvanic_anode&quot;&gt;sacrificial zinc anode&lt;/a&gt;, is there a limit to the distance that its protection will extend? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The protection works based on electrical conductivity, so distance would seem to affect the performance of the anode based on resistance. On the other hand, there doesn't seem to be a large current induced, so the any voltage drop would be very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article only mentions that the total number of anodes is usually equally spaced. Does this mean that one large anode in one place protects an entire structure the same as ten smaller anodes equally spaced?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-11-11T02:00:55.160" LastActivityDate="2015-11-11T02:00:55.160" Title="Will a sacrificial anode protect any length of metal that it is connected to?" Tags="|materials|corrosion|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1970" PostTypeId="1" AcceptedAnswerId="1976" CreationDate="2015-03-05T23:16:05.160" Score="2" ViewCount="383" Body="&lt;p&gt;Does the dead weight tester technology provide capability for an accurate, low pressure calibration standard in the range of zero to 150 cm H&lt;sub&gt;2&lt;/sub&gt;O, and if so is there a manufacturer that provides such an instrument?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what is the best primary standard for that range of pressure in terms of accuracy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect for dead weight testers there might be a lower limitation due to the balance of gravitational force and frictional force. I am looking to measure gauge pressure.&lt;/p&gt;&#xA;" OwnerUserId="820" LastEditorUserId="33" LastEditDate="2015-11-11T01:59:55.027" LastActivityDate="2015-11-11T01:59:55.027" Title="Dead weight pressure tester for very low positive pressures" Tags="|measurements|pressure|metrology|standards|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1973" PostTypeId="1" CreationDate="2015-03-06T01:40:28.707" Score="4" ViewCount="190" Body="&lt;p&gt;The question relating to &lt;a href=&quot;https://engineering.stackexchange.com/questions/1906/can-the-structural-integrity-of-bridges-be-measured-in-real-time-for-safety&quot;&gt;measuring structural integrity of bridges&lt;/a&gt; brought awareness to six categories of &lt;a href=&quot;http://www.oregon.gov/ODOT/HWY/BRIDGE/docs/SHM_SCL_submission.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;structural health monitoring of bridges&lt;/a&gt;. They are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bridge foundations&lt;/li&gt;&#xA;&lt;li&gt;Concrete superstructures&lt;/li&gt;&#xA;&lt;li&gt;Movable bridges&lt;/li&gt;&#xA;&lt;li&gt;Steel fatigue&lt;/li&gt;&#xA;&lt;li&gt;Structural dynamics&lt;/li&gt;&#xA;&lt;li&gt;Corrosion protection&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V9fE4.jpg&quot; alt=&quot;gusset plates&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a good sensing element to measure &lt;strong&gt;bridge corrosion&lt;/strong&gt;? What characteristics in this sensing element make it a good candidate for corrosion monitoring? Is an &lt;a href=&quot;https://engineering.stackexchange.com/questions/1883/what-is-radial-and-thickness-mode-vibration-in-a-piezo-electric-ceramic-disc-tra&quot;&gt;ultrasonic transducer&lt;/a&gt; a good candidate? &lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2023-06-29T07:29:57.087" Title="What is a good sensing element to monitor corrosion in a bridge structural health monitoring system?" Tags="|electrical-engineering|civil-engineering|bridges|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1974" PostTypeId="2" ParentId="1973" CreationDate="2015-03-06T03:02:28.897" Score="4" Body="&lt;p&gt;&lt;strong&gt;Ultrasonic Thickness Gauge&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most common methods of measuring steel plate thickness is an &lt;a href=&quot;http://www.steelconstruction.info/Weathering_steel#Monitoring_of_steel_thickness&quot; rel=&quot;nofollow&quot;&gt;ultrasonic thickness gauge&lt;/a&gt;. These gauges can be used while performing a bridge inspection to determine changes in steel plate thickness between inspections.  These are also used to confirm the actual thickness of a nominally specified plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only shows corrosion changes between inspections. It doesn't continuously monitor corrosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Continuous Measurements&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous monitoring of steel corrosion is an interesting topic of research for strands of suspension bridges. There are a &lt;a href=&quot;http://www.fhwa.dot.gov/publications/research/infrastructure/structures/bridge/14024/index.cfm&quot; rel=&quot;nofollow&quot;&gt;lot of methods&lt;/a&gt; that have been researched, these mostly rely on detecting minor changes in electrical, magnetic, or acoustic changes in the steel strands. The types studied are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Linear polarization resistance (LPR) sensors.&lt;/li&gt;&#xA;&lt;li&gt;Coupled multiple array sensors (CMASs).&lt;/li&gt;&#xA;&lt;li&gt;Bimetallic (BM) or galvanic sensors.&lt;/li&gt;&#xA;&lt;li&gt;Temperature-relative humidity sensors.&lt;/li&gt;&#xA;&lt;li&gt;Acoustic emission (AE).&lt;/li&gt;&#xA;&lt;li&gt;Magnetostrictive (MS).&lt;/li&gt;&#xA;&lt;li&gt;Main magnetic flux method (MMFM).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Concrete Reinforcing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also useful to measure and detect any corrosion occurring in the reinforcing steel embedded in concrete. These types of &lt;a href=&quot;http://www.sciencedaily.com/releases/2010/04/100421111355.htm&quot; rel=&quot;nofollow&quot;&gt;monitors&lt;/a&gt; are useful to determine corrosion so that remediation measures can be planned before more visible damage has occurred. In concrete, this damage is usually in the form of spalled concrete.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-06T03:02:28.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1975" PostTypeId="2" ParentId="1964" CreationDate="2015-03-06T04:05:38.063" Score="15" Body="&lt;p&gt;&lt;strong&gt;Pedantic Answer First&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You specifically eliminated most constraints that would make this anything other than an theoretical exercise. You did specify that it had to be &quot;built&quot; though. By building something, you have to physically move the top piece into place. This means that the entire structure will have been slightly taller than the final condition, i.e. the last &quot;block&quot; would have been &quot;placed&quot;. This also implies that &quot;something&quot; was used to place the last &quot;block&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other Things We Are Throwing Out the Window&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gravity decreases as the distance away from the Earth's center increases.&lt;/li&gt;&#xA;&lt;li&gt;We don't need to worry about pesky things like wind, so lateral force is out.&lt;/li&gt;&#xA;&lt;li&gt;The foundation for this structure is adequate. This means that the Earth is made out of something stronger than our structure.&lt;/li&gt;&#xA;&lt;li&gt;We aren't allowed to do anything that lightens our building. No thrusters, balloons, etc.&lt;/li&gt;&#xA;&lt;li&gt;Buckling is not a problem. This structure can be any shape that we want.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To the Question at Hand&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://physics.stackexchange.com/questions/51829/materials-with-a-high-compressive-strength-to-weight-ratio&quot;&gt;question from physics.SE&lt;/a&gt; gets right to the issue. We need the material that has the best strength to density ratio. We need a strong strong material that is light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that question, diamond is our material. Also from the ceramics chart &lt;a href=&quot;http://www-materials.eng.cam.ac.uk/mpsite/interactive_charts/strength-density/basic.html&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A Diamond Tower Then&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now just need to see how tall a tower of diamond can be before the bottom layer crushes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Material_properties_of_diamond&quot;&gt;Properties&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\gamma = 3.515 g/cm^3\\&#xA;\text{Strength} = 225 GPa$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;a href=&quot;http://www.wolframalpha.com/input/?i=225%20GPa%20%2F%20%289.8%20m%2Fs%5E2%20*%203.515%20g%2Fcm%5E3%29%20as%20kilometers&quot;&gt;Wolfram is doing my math&lt;/a&gt; correctly for me, the answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$6,530 Km$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or about the radius of the Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But Wait! There is More!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far this has been about a solid tower that is the same shape the whole way through. The limiting factor was the stress on the bottom layer. What if we make the bottom wider than the top? Why not a pyramid of diamond? We can make this taller!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So a Pyramid of Diamond&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pyramid gets us taller. What limits the size of the base? Nothing, because we have no limits. So as the base gets larger, it eventually gets so large that it wraps around and touches itself from the other side of the Earth (sorry everyone.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we don't stop there. Spheres are self supporting, so let's turn this structure into a sphere that covers all of the Earth.  We now can go even higher! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What have we created?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A Star Made of Diamond&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Astronomers found a &lt;a href=&quot;http://www.space.com/26335-coldest-white-dwarf-star-diamond.html&quot;&gt;star that is made entirely of diamond&lt;/a&gt;. But why stop there? The only thing that limits a star's size is the mass that turns it into a Black Hole. We can argue about stopping here or whether larger black holes still count, but you see where we are going...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Largest Structure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have now seen that the largest/tallest structure is star made of diamond. (With a squishy Earth center.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What Went Wrong?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how did we get from &quot;Tallest Structure?&quot; to &quot;Star Made of Diamond&quot;? We didn't have any limits on what was possible. When you throw the constraints out the window, &lt;strong&gt;anything&lt;/strong&gt; is possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are no longer in the realm of Engineering. We are now in Theoretical Physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineering is using science to work within constraints.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-06T13:41:04.157" LastActivityDate="2015-03-06T13:41:04.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1976" PostTypeId="2" ParentId="1970" CreationDate="2015-03-06T07:03:23.460" Score="2" Body="&lt;p&gt;In this situation, you might do just as well to use a U-tube manometer.  Is some respects, this is a sort of dead-weight tester where the fluid itself is the dead weight. In this device pressure is determined from the difference in height between two connected volumes of fluid. Usually this is just a transparent tube bent into a 'U' shape, placed upright in front of a convenient means of measuring the relative heights of the free surfaces of the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/u1Szx.gif&quot; alt=&quot;U-tube Manometer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One end of the tube is connected to the pressurized volume, while the other is left open.  The pressure difference may be calculated from the height difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P = \rho g h $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\rho$ is the density of the liquid ($\sim 1000 \, kg/m^3$ for water) and $g$ is gravitational acceleration ($9.81 \, m/s$ in most places). Or, if you need your measurements in cm of water, simply use water as the working fluid and the height difference will be exactly the quantity you want. This is basically why &quot;length units of liquid&quot; can be used as a unit of pressure*.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a standard manometer is not sensitive enough, you can tilt it off of vertical by some known angle.  As a result, the liquid will travel a larger, more easily measured, distance along the tubes to achieve the same difference in height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite the seemingly primitive technology involved, measurements of this kind can be quite accurate when performed correctly. There's a nice &lt;a href=&quot;http://www.nist.gov/calibrations/upload/pmc-2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;document from NIST&lt;/a&gt; that goes into detail about how to maximize the accuracy of these devices. Also a &lt;a href=&quot;https://books.google.com/books?hl=en&amp;amp;lr=&amp;amp;id=FzgXBQAAQBAJ&amp;amp;oi=fnd&amp;amp;pg=PA229&amp;amp;dq=ruthberg+%22pressure+measurements+for+the+range%22&amp;amp;ots=dprESfnvpb&amp;amp;sig=GyEALrIq4xvbPJow9U80JoYFD7Y#v=onepage&amp;amp;q=ruthberg%20%22pressure%20measurements%20for%20the%20range%22&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;review by Ruthberg&lt;/a&gt; specifically for low pressure measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I don't mean to endorse these units of pressure. The pascal ($Pa$ or $N/m^2$) is the correct SI unit of pressure derived from the newton and the meter.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2015-03-06T16:43:07.107" LastActivityDate="2015-03-06T16:43:07.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1977" PostTypeId="1" AcceptedAnswerId="1978" CreationDate="2015-03-06T09:02:15.897" Score="31" ViewCount="11083" Body="&lt;p&gt;I work in the middle of London, in an area full of large office blocks. Across the road from my office they have started construction of a large building (10 stories plus). Over the last few weeks, diggers have dug a large (and vertical walled) hole. Lorries have taken the resulting dirt and old concrete away, leaving a very neat hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the last day or so, the lorries have returned with new dirt (or the old dirt crushed) and diggers have been putting it back in the hole (and compacting it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why put the dirt back? Surely leaving the hole deeper would allow for deeper basement (or digging it shallower would be cheaper)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a structural engineer, so this is all lost on me, but I'm fascinated.&lt;/p&gt;&#xA;" OwnerUserId="824" LastEditorUserId="368" LastEditDate="2015-03-19T20:59:12.353" LastActivityDate="2019-05-13T11:07:27.057" Title="Why dig out and then fill in before building a large structure?" Tags="|structural-engineering|geotechnical-engineering|building-design|foundations|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1978" PostTypeId="2" ParentId="1977" CreationDate="2015-03-06T09:19:11.080" Score="29" Body="&lt;p&gt;This is to make sure they know what the foundation is made of. For all they knew there may have been an old tunnel underneath that would have collapsed when the new building is put on top. London is built on top of an old marsh, this type of soil is very prone to sinking and uneven settling, digging down and reinforcing the foundation alleviates that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also ensures the foundation is uniform under the building to avoid a new tower of Pisa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the age of the city it may have been to scour the land for potential archaeological finds (fervently hoping they would come up empty).&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-03-06T10:43:56.817" LastActivityDate="2015-03-06T10:43:56.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1979" PostTypeId="1" CreationDate="2015-03-06T11:01:29.017" Score="9" ViewCount="1447" Body="&lt;p&gt;I am trying to calculate the force required to close the valve on an annular blow-out preventer. A rubber ring (called a packing unit) closes around the pipe when a force is applied from underneath, as shown &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/5c/Annular_BOP.svg/2000px-Annular_BOP.svg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/5c/Annular_BOP.svg/2000px-Annular_BOP.svg.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual valve has the hydraulic fluid coming into a chamber that goes the around the outside of the main body, which then lifts a piston which applies the force to the packing unit, as shown &lt;a href=&quot;https://i.stack.imgur.com/iLNH1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iLNH1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iLNH1.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://geologie.vsb.cz/DRILLING/drilling/theory/theory_html_a2129a3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;geologie.vsb.cz&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the documentation for the bop it says that the hydraulic pressure required to close or open the valve is 3000 PSI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, how do I calculate the force to close the valve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you just multiply 3000 by the area the pressure is acting on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$3000*6894.7*(\pi R_o^2 - \pi R_i^2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$R_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_o$&lt;/span&gt; are the inner and outer radius of the ring, shown in the picture in the second link. Units of these are meters. &lt;span class=&quot;math-container&quot;&gt;$6894.7$&lt;/span&gt; is to convert from PSI to N/m&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this gives a HUGE force. Am I doing something wrong in the maths? Or am I interpreting the documentation wrong. Does a 3000 PSI closing pressure not mean that you have to apply 3000 PSI of pressure to the hydraulic fluid acting on the  outer ring of the piston? Maybe it means that 3000 PSI is needed to squish the packing unit so it closes? In this case what information do I need and how do I calculate the force required to close the valve?&lt;/p&gt;&#xA;" OwnerUserId="825" LastEditorUserId="6767" LastEditDate="2019-07-14T09:33:07.647" LastActivityDate="2019-07-14T09:33:07.647" Title="Calculate force to lift piston from a given closing pressure" Tags="|mechanical-engineering|hydraulics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="1980" PostTypeId="1" AcceptedAnswerId="1982" CreationDate="2015-03-06T13:11:42.143" Score="3" ViewCount="1238" Body="&lt;p&gt;Whenever I've come across notation for specifying how much reinforcing steel is in a member, it is typically by giving the area of steel as a percent of the cross-sectional area of the member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eO22J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in this picture, you could say that the reinforcing steel is &lt;code&gt;0.02&lt;/code&gt; or &lt;code&gt;2%&lt;/code&gt;. You could also specify the reinforcing steel given a schedule, e.g. &lt;code&gt;#8 @ 12&quot; O.C&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently came across a piece of software that was asking for the reinforcing steel of a floor slab in units of &lt;code&gt;mm^2/mm&lt;/code&gt;. I cannot figure out how they came across those units. How could I calculate the reinforcement in these units given I know how much steel is in the slab in the other conventions I listed earlier?&lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="33" LastEditDate="2015-03-07T19:50:03.700" LastActivityDate="2015-03-07T19:50:03.700" Title="How do units of mm^2/mm represent the reinforcing steel area in a slab?" Tags="|civil-engineering|structural-engineering|reinforced-concrete|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1982" PostTypeId="2" ParentId="1980" CreationDate="2015-03-06T13:55:53.063" Score="7" Body="&lt;p&gt;Your second bar description is what you need to compare it to:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;#8 @ 12&quot; O.C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A #8 bar is 0.79 in&lt;sup&gt;2&lt;/sup&gt;. This gives us $\frac{0.79in^2}{12in}$ or in&lt;sup&gt;2&lt;/sup&gt; over in. This is the similar to your mm&lt;sup&gt;2&lt;/sup&gt;/mm. (Obviously the units are not interchangeable!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method of describing reinforcing is useful if you don't care about the actual size of rebar used. It could be #8 at 12&quot; or #6 at 6.625&quot;.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-07T13:00:24.753" LastActivityDate="2015-03-07T13:00:24.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1983" PostTypeId="1" AcceptedAnswerId="1984" CreationDate="2015-03-06T14:41:51.470" Score="1" ViewCount="2417" Body="&lt;p&gt;We are working on some equipment that is going to be attached to existing, 3,000 psi reinforced concrete floor slabs. In some locations, the best way to do this is to through-bolt and 'sandwich' the slab with a piece of threaded rod (5/8&quot; rod in our case.) On the top side of the slab, our equipment has a large steel bearing surface. On the bottom side, we expect we'll need steel plate washers. We were thinking of using square washers cut from 1/2&quot; A36 steel. What I don't know is how big of a square would be 'standard.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a chart or standard practice for what size plate washer to use on what size fastener? I couldn't find one in AISC, but may have missed it, or maybe it's in an ACI standard? If there's no prescriptive chart, what checks would typically be done to size it by calculation?&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="33" LastEditDate="2015-11-11T01:54:48.250" LastActivityDate="2015-11-11T01:54:48.250" Title="Sizing plate washers for through-bolting to concrete" Tags="|structural-engineering|standards|fasteners|reinforced-concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1984" PostTypeId="2" ParentId="1983" CreationDate="2015-03-06T15:13:23.807" Score="1" Body="&lt;p&gt;There are some standard square washer sizes. These may not specifically be for your configuration, but they may help to decide on a size to start with.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;ASTM F 436 Square Washers are found in the AISC Steel Manual on the table that shows other bolt and washer diameters. These are specifically for beveled washers and the sizes given are only minimums. You can see the sizes &lt;a href=&quot;http://www.portlandbolt.com/products/washers/hardened-beveled/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. For a 5/8&quot; rod, the washers are 1.75&quot; square.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Square washers are also used in timber construction. You can find some typical sizes &lt;a href=&quot;http://www.portlandbolt.com/products/washers/square-plate/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. These washers are 2.5&quot; square for 5/8&quot; rod.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Custom washers might be the way to go. Since none of the washer types listed above are specifically for bolting to concrete, a custom size might be required.  The references above give a starting point, but the final size might be controlled by the allowable bearing strength on concrete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Making a square washer is a simple machining task, so going the custom route may not add a lot of cost.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-06T15:13:23.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1985" PostTypeId="2" ParentId="1966" CreationDate="2015-03-06T16:07:12.113" Score="2" Body="&lt;p&gt;Typically it sounds like you need a &lt;strong&gt;&lt;a href=&quot;http://www.omega.com/subsection/positive-displacement-flowmeters.html&quot; rel=&quot;nofollow&quot;&gt;positive displacement flow meter&lt;/a&gt;&lt;/strong&gt; as you have a viscous fluid in the process and they are not very suitable to turbine style flow meters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Positive displacement flow meters come in a wide range of fluid viscosity, but there is a  series of questions that needs to be answered before you choose the right one. The best thing would be for you to talk to an application engineer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The web page I pointed you before is from Omega Engineering - they carry these products and you should be able to get an engineer on chat to answer your questions. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="828" LastEditorUserId="828" LastEditDate="2015-03-06T18:09:01.617" LastActivityDate="2015-03-06T18:09:01.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="1986" PostTypeId="2" ParentId="1929" CreationDate="2015-03-06T16:56:25.490" Score="0" Body="&lt;p&gt;1) Glow plugs. These have already been covered in another answer. It is a small, resistive plug that current flows through and it heats up. Older ones are just dumb ceramic heaters that control current with V=IR. The newer ones are PWM controlled to use less energy and heat the cylinder air to the right temp. &lt;a href=&quot;https://www.fcpeuro.com/products/vw-diesel-glow-plug-beetle-golf-jetta-passat-touareg-0250202023?gclid=COWQ1P2YlMQCFQYFaQodKE4AKg&quot; rel=&quot;nofollow&quot;&gt;Example&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Intake heaters. Some larger engines don't have space in the head for glow plugs or they wouldn't work reliably in the environment. These engines have intake air heaters. They heat the air prior to it going into the cylinder. &lt;a href=&quot;http://sinisterdiesel.com/i-16445095-sb-grid-heater-delete-for-dodge-cummins-1998-2007-5-9l.html?gclid=CJ-6veqYlMQCFQyUaQodVHAADg&quot; rel=&quot;nofollow&quot;&gt;Example&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Ether. Ether has a much lower autoignition temperature. On very large diesel engines (such as those used in mining trucks) the factory or 3rd party ECM will dose an appropriate amount of ether to assist in getting the engine started. &lt;a href=&quot;http://www.quickstart-ether.com/products&quot; rel=&quot;nofollow&quot;&gt;Example&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="231" LastActivityDate="2015-03-06T16:56:25.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1987" PostTypeId="1" AcceptedAnswerId="10606" CreationDate="2015-03-06T17:37:53.700" Score="2" ViewCount="243" Body="&lt;p&gt;I am interested in a microcontroller for a low power application. I have been advise to look at MSP430 or Microchip PIC microcontrollers. I also wonder if &lt;a href=&quot;http://www.arm.com/products/processors/cortex-m/cortex-m0.php&quot; rel=&quot;nofollow&quot;&gt;ARM-Cortex-M0&lt;/a&gt; is a good choice too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a high level the system will have two analog sensors, few GPIO to control LED, and actuation mechanisms. The systems is intended to be powered with standard batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the critical parameters that warrant attention in researching for a suitable low power micro controllers?&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="1586" LastEditDate="2016-07-13T19:34:53.440" LastActivityDate="2016-12-04T02:25:25.623" Title="How to select a micro controller for a low power application?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1988" PostTypeId="1" AcceptedAnswerId="1993" CreationDate="2015-03-06T19:19:02.307" Score="5" ViewCount="1014" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sinkhole&quot; rel=&quot;noreferrer&quot;&gt;Sinkholes&lt;/a&gt; have been known to occur in the &lt;a href=&quot;http://www.cnn.com/2014/08/05/world/asia/seoul-skyscraper-sinkholes/&quot; rel=&quot;noreferrer&quot;&gt;middle of cities&lt;/a&gt; or &lt;a href=&quot;http://corvettemuseum.org/enews/backups/media/sinkhole.htm&quot; rel=&quot;noreferrer&quot;&gt;other locations&lt;/a&gt; where they affect buildings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TxpQYm.jpg&quot; alt=&quot;sinkhole&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dep.state.fl.us/geology/geologictopics/sinkhole.htm&quot; rel=&quot;noreferrer&quot;&gt;Some areas&lt;/a&gt; are more prone to sinkholes than other areas because of the presence of old mines or limestone bedrock. Even in these areas where sinkholes are more common, humans continue to build buildings and even &lt;a href=&quot;http://indianapublicmedia.org/news/sinkholes-discovered-monroe-airport-17956/&quot; rel=&quot;noreferrer&quot;&gt;airports&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do engineers prevent or mitigate sinkholes? Is it as low-tech as dumping in rock until no more fits?&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-26T05:28:35.020" Title="How do engineers deal with sinkholes?" Tags="|geotechnical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1989" PostTypeId="2" ParentId="659" CreationDate="2015-03-06T20:32:29.280" Score="0" Body="&lt;p&gt;How about Ceramic Cement? When properly mixed and used OB-400 is good to 1425C (2600F).     Soldering a TC to metal surfaces takes skill and patience. Even more of the above when TIG welding. Have you considered using a &lt;a href=&quot;http://www.omega.com/subsection/washer-assemblies-self-adhesive.html&quot; rel=&quot;nofollow&quot;&gt;self-adhesive thermocouple&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="828" LastActivityDate="2015-03-06T20:32:29.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1990" PostTypeId="2" ParentId="1977" CreationDate="2015-03-06T23:05:37.770" Score="21" Body="&lt;p&gt;There are a few reasons why the footprint of large buildings are excavated and replaced.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;All of the old building must be removed -&lt;/strong&gt; Typically, older buildings will have been built with weaker materials or less stringent building standards than are currently used. Depending on the age of the existing structure, complete plans may not even exist. This means that enough material needs to be removed to ensure that the soil can be assured to have the required strength for the new structure. Also this ensures that new piles will not be driven into old concrete, steel, or masonry.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pollutant or Hazardous Waste removal -&lt;/strong&gt; Especially if the old building was a factory (recently or at one time), there might be hazardous materials in the soil. Environmental regulations have not always been as strict as they are now. A lot of bad things were just buried and forgotten.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Historical artifacts -&lt;/strong&gt; If the area is thought to have had any sort of history, digging below existing buildings may uncover some of this history. Historical societies love this, but owners and contractors hate the delay and added cost.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Groundwater removal -&lt;/strong&gt; If groundwater is close to the surface or close to the basement of the new building, additional excavation may be required so that a sealing layer of concrete or clay can be placed. This can seal the site so that it isn't a muddy mess while the rest of the construction finishes. Dewatering wells may also be installed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-06T23:05:37.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1991" PostTypeId="2" ParentId="659" CreationDate="2015-03-06T23:51:03.563" Score="2" Body="&lt;p&gt;I ended up finding a solution that works (haven't tried the cements suggested by others yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick was to use a spot welder rather than a tig welder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the correct power settings, that stuck the TC to the surface without damaging either component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not exceptionally strong (I used very fine TC wires), but it should result in conditions that are as close as possible to what they would be if the TC were not there.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-03-06T23:51:03.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="1993" PostTypeId="2" ParentId="1988" CreationDate="2015-03-07T00:31:31.727" Score="5" Body="&lt;p&gt;Whether it's natural subsidence, like sinkholes or human induced subsidence like the collapse of underground engineered chambers or mining subsidence the two ways of dealing with it are backfill or leaving it alone and the enforcement of an exclusion zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where backfill is used for the remediation of subsidence it's generally loose rock fill, because it's the cheapest form of backfill and excavations aren't going to be established against the backfill, particular in civil situations. In mining, the subsidence backfill can be loose rock, sand or tailings, or cemented rock, sand or tailings; depending on circumstance, what materials are available and how much the company is prepared to pay.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: 26 March 2015&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across this picture of a sinkhole being backfilled with concrete on the &lt;a href=&quot;http://www.msn.com/en-au/news/world/giant-sinkholes/ss-AA9Xam4?ocid=mailsignout#image=5&quot; rel=&quot;nofollow noreferrer&quot;&gt;MSN news website&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cQYBD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Additional Information 26 March 2015&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One group of &lt;a href=&quot;http://www.rembco.com/news/sinkholes-101/&quot; rel=&quot;nofollow noreferrer&quot;&gt;geotechnical contractors&lt;/a&gt; in the US advocates excavation and backfilling of sinkhole where the bedrock in no deeper than 4.5 m (15 ft). For deeper holes it recommends grouting and for very deep holes it recommends cap grouting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://efotg.sc.egov.usda.gov/references/public/WI/527.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Karst Sinkhole Treatment&lt;/a&gt; document by the Natural Resources Conservation Service recommends establishing of a buffer zone around the hole and backfilling. Depending on circumstances the backfill will include loose rock, concrete and if necessary geotextile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When floods hit &lt;a href=&quot;http://www.sott.net/article/263232-Flood-leaves-70-sinkholes-in-Calgary&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calgary&lt;/a&gt;, in Canada, in June 2013, creating numerous sinkholes, the holes were backfilled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.sinkholeguide.com/sinkhole-faqs.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sinkhole Guide&lt;/a&gt; recommends backfilling sinkholes with,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;“native earth materials or concrete. Broken limestone rip-rap or a&#xA;  concrete plug in the bottom of the sinkhole often helps create a&#xA;  stable foundation for the fill. Above that, add clayey sand to form a&#xA;  barrier that will help to prevent water from seeping downward through&#xA;  the hole and enlarging it further. Lastly, add sand and top soil, and&#xA;  landscape to surrounding conditions. Additional fill may be necessary&#xA;  over time, but most holes eventually stabilize.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the &lt;a href=&quot;http://www.fhwa.dot.gov/engineering/geotech/hazards/mine/workshops/kdot/kansas04.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;US Department of Transportation&lt;/a&gt;, fly ash in grout has been used to backfill sinkhole/subsidence holes in abandoned mines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In North Dakota, the &lt;a href=&quot;http://www.psc.nd.gov/docs/amlarticles/structural-monitoring-at-a-pressurized-grout-project.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Public Service Commission&lt;/a&gt; backfills subsidence hole in abandoned mines in that State. &lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-03-26T05:28:35.020" LastActivityDate="2015-03-26T05:28:35.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1994" PostTypeId="1" CreationDate="2015-03-07T02:38:45.757" Score="3" ViewCount="5023" Body="&lt;p&gt;When loaded vertically, a beam experiences compression in its top flange and tension in its bottom flange. The top (compression) flange would tend to buckle vertically much like a column, however, the web prevents such buckling from taking place. But if there is enough compression with no lateral support, the flange and web would start to buckle horizontally together. The bottom (tension) flange restrains such movement, causing the beam to twist. This is phenomenon known as &lt;a href=&quot;https://www.youtube.com/watch?v=wUkvxHqPbTY&quot; rel=&quot;nofollow noreferrer&quot;&gt;lateral torsional buckling&lt;/a&gt; (&lt;strong&gt;LTB&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DkcGR.jpg&quot; alt=&quot;lateral torsional buckling&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A laced strut is a built-up strut made of two (or sometimes three equally spaced) beams joined together by a system of diagonal lacings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mQikx.jpg&quot; alt=&quot;laced strut&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Globally, a laced strut is not susceptible to LTB as its section modulus in the lateral direction is greater than that in the vertical direction. However, the same could not be said of the individual beam component that made up the strut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a 20m strut has lacings spaced at 1m intervals, what would be the effective length used to calculate LTB for the &lt;em&gt;individual&lt;/em&gt; beam component and &lt;strong&gt;why&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2015-03-09T14:30:14.790" Title="Effective length for lateral torsional buckling in a laced strut" Tags="|steel|structural-engineering|beam|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1995" PostTypeId="1" AcceptedAnswerId="1996" CreationDate="2015-03-07T11:25:26.880" Score="9" ViewCount="3076" Body="&lt;p&gt;Is the smaller gear (pinion) always mounted to the input shaft when meshed with a bigger gear that is mounted on the output shaft? Are there places where the bigger gear drives the smaller gear?&lt;/p&gt;&#xA;" OwnerUserId="739" LastEditorUserId="33" LastEditDate="2015-03-07T13:19:25.147" LastActivityDate="2015-04-14T16:57:24.730" Title="Mechanism of gear rotation" Tags="|mechanical-engineering|gears|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="1996" PostTypeId="2" ParentId="1995" CreationDate="2015-03-07T12:23:28.437" Score="8" Body="&lt;p&gt;Two meshed gears are used to transfer rotational drive between two shafts.&lt;br&gt;&#xA;The relative speeds of rotation are inversely proportional to the number of teeth on each gear.  That is -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Input}_{\text{RPM}} / \text{Output}_{\text{Rpm}} = \text{Output gear}_{\text{teeth}} / \text{Input gear}_{\text{teeth}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if it is desired that the output shaft rotate more slowly than the input shaft then the output gear is larger. But, if it is desired that the output shaft rotate faster than the input shaft then the output gear is smaller.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the relationship above becomes obvious &quot;by inspection&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the arrangement shown below, for every complete (360 degree) rotation of the small gear the large gear rotates only part of a turn. The large gear has a lower RPM rate than the small gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;IF the small gear was the DRIVING or INPUT gear then the large DRIVEN or OUTPUT gear would be turning more slowly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IF the large gear was the DRIVING or INPUT gear then the small DRIVEN or OUTPUT gear would be turning more rapidly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which arrangement is used depends on whether an increase or decrease in RPMs is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque or &quot;twisting force&quot; is inversely proportional to speed.&lt;br&gt;&#xA;That is the slower turning shaft will have proportionately more torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b0eYI.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagram from &lt;a href=&quot;http://en.wikipedia.org/wiki/Gear_ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia - Gear ratio&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Look at the examples below and you will see how the gear size relates to relative shaft speed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.technologystudent.com/gears1/gears5.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Several animated examples&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.technologystudent.com/gears1/gears7.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;3 meshed gears animated example&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://graphicdesign.stackexchange.com/questions/31584/how-can-i-create-working-gears-in-illustrator&quot;&gt;Animated 2 to 1 speed example&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;1:1 and 1:2 example from &lt;a href=&quot;https://biology.stackexchange.com/questions/3186/given-atp-synthases-structure-how-can-3-33-protons-ultimately-synthesize-one-a&quot;&gt;Essentially unrelated stack exchange biology question&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G29hJ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="-1" LastEditDate="2017-04-13T12:48:15.570" LastActivityDate="2015-03-07T18:19:39.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1997" PostTypeId="2" ParentId="1995" CreationDate="2015-03-07T12:55:08.457" Score="9" Body="&lt;p&gt;There are many examples of low input speed to high output speed gearing:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;kitchen &quot;salad spinner&quot;&lt;/li&gt;&#xA;&lt;li&gt;old-fashioned hand-cranked siren&lt;/li&gt;&#xA;&lt;li&gt;electrical generator attached to a tractor PTO (power takeoff) &amp;mdash; also wind-powered generators&lt;/li&gt;&#xA;&lt;li&gt;the &quot;overdrive&quot; gear in any vehicle transmission&lt;/li&gt;&#xA;&lt;li&gt;the speed regulator on a steam engine, or in many kinds of clockwork mechanisms&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Indeed, any sort of spring- or weight-driven clock works this way. The spring or weight is used to apply torque to the slowest-moving gear in the mechanism, and the escapement (e.g., balance wheel or pendulum) at the other end of the system of gears regulates the speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, it's more efficient to use a belt drive for this kind of speed change. For example, an old-fashioned &quot;spinning wheel&quot; for making thread.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2015-03-07T18:06:09.950" LastActivityDate="2015-03-07T18:06:09.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1998" PostTypeId="1" AcceptedAnswerId="2000" CreationDate="2015-03-07T19:12:07.967" Score="13" ViewCount="15570" Body="&lt;p&gt;Pressure treated lumber is specified for many exterior applications because of its resistance to insect damage and fungal rot. But how does it compare to untreated wood, mechanically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider a rim joist supporting the ground floor of a residential structure with a pier and post foundation. If the joist has been damaged by rot in a location that is impractical to fully protect from exposure, I might be tempted to replace the joist with a pressure treated member of the same nominal dimension (in &lt;em&gt;addition&lt;/em&gt; to proper flashing) for added protection in that location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is an existing structure, by far the easiest approach is to use a member with the same dimension to replace the rotted joist. However, this relies on the new member meeting the same load-bearing requirements as the old member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building codes should provide enough wiggle room that in this particular example, there's not much of a safety concern for the homeowner. After all, the rotted joist had not failed, and it would definitely have less strength than the member was originally rated for. In practice, treated and untreated members may be manufactured from different wood species with different mechanical properties to begin with; for the purposes of this question, assume the species is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does pressure treatment result in a member with more or less strength in tension, compression or torsion? Does it affect the durability of the wood* &lt;em&gt;in ways not related to rot or insect damage?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;* Not the fasteners; that's a different issue that's pretty well-covered online. See &lt;a href=&quot;http://www.strongtie.com/productuse/ptwoodfaqs.html&quot;&gt;this page from Simpson&lt;/a&gt;, for example.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-03-09T16:29:12.847" LastActivityDate="2016-03-16T16:02:43.217" Title="How does pressure treatment affect the mechanical properties of lumber?" Tags="|structural-engineering|wood|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="1999" PostTypeId="2" ParentId="1994" CreationDate="2015-03-07T21:28:32.017" Score="5" Body="&lt;p&gt;&lt;strong&gt;Beams Only&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, Lateral Torsional Buckling (LTB) is only applicable to beams. Beams are members that are loaded so that they have moment. This is what the first image shows. The members in the second image are not beams.  They are struts that are loaded primarily axially. They are holding the walls of the excavation apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that LTB is not a controlling consideration for the struts as pictures. The rest of this answer does have some portions that would be applicable to both columns (struts) and beams though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lateral Torsional Buckling&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the question, LTB occurs when the compression flange of a beam starts to rotate out of plane (buckle). Buckling is controlled for LTB of a beam the same way that it is controlled in a column. It is controlled by bracing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lacing of two members together effectively makes the built-up member act as a whole. The controlling properties are now the combined section properties instead of the individual properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Individual codes might use different equations or methods to check for LTB, but the basic mechanics are the same. The distance between braces on the built-up member is what matters for a given section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lacing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In laced members, the lacing is the smaller bars that go between the main members. This is typically in a zigzag pattern. The lacing is there to tie the members together to make them act as one. For an individual member, the lacing effectively braces it at each location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bracing must be designed so that it can transfer shear from one member to the other. In the US, &lt;a href=&quot;http://msc.aisc.org/globalassets/modern-steel/steel-interchange/2002/2002v09_si_web.pdf&quot; rel=&quot;nofollow&quot;&gt;codes&lt;/a&gt; require the lacing to be designed for about 2% of the axial load in the build-up member. &lt;a href=&quot;http://www.fhwa.dot.gov/bridge/steel/pubs/if12052/volume04.pdf&quot; rel=&quot;nofollow&quot;&gt;AISC and AASHTO&lt;/a&gt; have limits on the spacing of the lacing connections based on the buckling of the compression flange of the member between lacing points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the lacing is designed as required, the built-up member will act like one section and the individual components do not need to be analyzed individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bracing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lacing is not considered external bracing. The lacing bars are an efficient alternative to providing solid cover plates to create a built-up member. Bracing is external to the member as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the US, AISC differentiates between &lt;em&gt;nodal&lt;/em&gt; bracing and &lt;em&gt;relative&lt;/em&gt; bracing. These differ in that nodal bracing provides rigid brace points that do not move, while relative bracing only ensures that two brace points do not move relative to each other. There are different formulae that account for the bracing of a column using each type of brace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your question, the individual members will be braced at the 1 meter intervals and, this would be relative bracing.  This distance is too far apart to be considered &lt;em&gt;lacing&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-09T14:30:14.790" LastActivityDate="2015-03-09T14:30:14.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2000" PostTypeId="2" ParentId="1998" CreationDate="2015-03-07T21:32:14.420" Score="8" Body="&lt;p&gt;Pressure treatment does have a small, but documented effect on the strength of the member, particularly if it is 'incised' (has slots or holes cut into it as part of the pressure treating process.) If you're working to American codes, according to the &lt;a href=&quot;http://www.awc.org/helpoutreach/faq/faqFiles/Pressure_treat_valu.php&quot;&gt;American Wood Council&lt;/a&gt;, pressure treated wood is limited to a maximum duration factor of 1.6. This wouldn't matter for your example of fixing a house, because the duration factor would already be much smaller. Effectively, this just restricts the use of pressure treated lumber for resisting impact loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, if it is incised which is often the case, the NDS has a derating factor of 20% for the strength of the member except for compression perpendicular to the grain, which is not derated at all. Note that there is also a 'wet service factor' for any member that will be exposed to moisture, regardless of any treatment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure treated lumber intended for structural applications should have a research report that covers usage, labeling, and special instructions. These often have useful information. For example, here are the research reports for &lt;a href=&quot;http://www.icc-es.org/Reports/pdf_files/ESR-1081.pdf&quot;&gt;two&lt;/a&gt; &lt;a href=&quot;http://www.icc-es.org/Reports/pdf_files/ESR-3038.pdf&quot;&gt;random&lt;/a&gt; brands. (Warning: PDF links.)&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-03-07T21:32:14.420" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2001" PostTypeId="1" AcceptedAnswerId="2003" CreationDate="2015-03-07T22:40:33.667" Score="9" ViewCount="5108" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Epicyclic_gearing&quot; rel=&quot;noreferrer&quot;&gt;planetary gear&lt;/a&gt; is a gear that rotates not around an axle but around another gear known as a sun gear, or star gear. I have designed prototype systems in the past that used a single planetary gear rotating around a rotating gear. However since then almost every system I've seen using planetary gears have used multiple gears, most commonly in sets of &lt;a href=&quot;http://forums.autodesk.com/autodesk/attachments/autodesk/78/316168/1/sun%20and%20planets.gif?nobounce&quot; rel=&quot;noreferrer&quot;&gt;3&lt;/a&gt;, &lt;a href=&quot;https://d2t1xqejof9utc.cloudfront.net/screenshots/pics/2260e59ecf6198bdcdcb266aeae6cf64/medium.jpg&quot; rel=&quot;noreferrer&quot;&gt;4&lt;/a&gt;, &lt;a href=&quot;http://www.bncgears.com/file/0_2011_06_24_194353.gif&quot; rel=&quot;noreferrer&quot;&gt;5&lt;/a&gt; and &lt;a href=&quot;http://www.intechpower.com/Portals/62516/images/planetary_gear.gif&quot; rel=&quot;noreferrer&quot;&gt;6&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mechanical advantages and disadvantages of using multiple planetary gears and what is the optimum number of gears?&lt;/p&gt;&#xA;" OwnerUserId="340" LastEditorUserId="340" LastEditDate="2015-03-07T23:34:58.803" LastActivityDate="2015-03-08T00:08:32.283" Title="What is the benefit of additional planetary gears?" Tags="|mechanical-engineering|dynamics|statics|gears|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2002" PostTypeId="1" AcceptedAnswerId="2010" CreationDate="2015-03-07T22:40:44.457" Score="2" ViewCount="3095" Body="&lt;p&gt;Chainsaws use oil to lubricate the chain. The usage is quite minor, probably less than 1cc per minute of work. The oil is normally supplied from a refillable compartment to the chain only when the chainsaw is running.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the delivery managed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pump? The capillary effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's a pump, I'd be interested in details - I mean, I have a hard time imagining a pump this slow (&amp;lt;1cc/min) supplied from the high-RPM engine... or does it use an extensive gearbox to slow down? A separate motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[note: I mean the oil used for chain lubrication, not mixed with fuel for the two-stroke engine of motorsaw]&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-03-20T21:33:27.963" Title="How is oil supplied in chainsaws?" Tags="|mechanical-engineering|pumps|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2003" PostTypeId="2" ParentId="2001" CreationDate="2015-03-08T00:08:32.283" Score="4" Body="&lt;p&gt;&lt;strong&gt;Advantages&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll start off by quoting from the &quot;Benefits&quot; section of &lt;a href=&quot;https://en.wikipedia.org/wiki/Epicyclic_gearing#Benefits&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The load in a planetary gear train is shared among multiple planets, therefore torque capability is greatly increased. The more planets in the system, the greater the load ability and the higher the torque density.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The more the merrier. The same idea is covered &lt;a href=&quot;http://www.gearsolutions.com/article/detail/5837/epicyclic-gearing-a-handbook&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Torque Splits&lt;/strong&gt; &lt;br&gt;&#xA;   When considering torque splits one assumes the torque to be divided among the planets equally, but this may not be a valid assumption. Member support and the number of planets determine the torque split represented by an “effective” number of planets. This number in epicyclic sets constructed with two or three planets is in most cases equal to the actual number of planets. When more than three planets are used, however, the effective number of planets is always less than the actual number of planets.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Let’s look at torque splits in terms of fixed support and floating support of the members. With fixed support, all members are supported in bearings. The centers of the sun, ring, and carrier will not be coincident due to manufacturing tolerances. Because of this fewer planets are simultaneously in mesh, resulting in a lower effective number of planets sharing the load. With floating support, one or two members are allowed a small amount of radial freedom or float, which allows the sun, ring, and carrier to seek a position where their centers are coincident. This float could be as little as .001-.002 inches. With floating support three planets will always be in mesh, resulting in a higher effective number of planets sharing the load.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The single-planet arrangement doesn't change the torque capability, but the multiple-planet arrangement does. You can increase the number of planets without limit and still continue to increase it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage of using more planets is less energy loss. Wikipedia claims that there is an energy loss of only about 3% per stage, but it doesn't back that up with an inline citation, nor does it specify the number or arrangement of planets. &lt;a href=&quot;http://www.cs.berkeley.edu/~sequin/CS285/2011_REPORTS/CS285%20final%20paper_Eric&amp;amp;Jessie.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; gives the same feature as a &quot;typical energy loss.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disadvantages&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious one here is that the more planets you add, the more complex the system becomes. If there's a major failure, it could be amplified by having more planets (alternatively, it could be argued that some arrangements provide redundancy).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is there an optimal number of gears? It really depends on your application. Complexity can be a huge issue, and with complexity comes cost. Repairs, in turn, can be expensive if something goes drastically wrong. If the application involves a great load and torque, then more gears can be helpful.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-03-08T00:08:32.283" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2004" PostTypeId="1" AcceptedAnswerId="2005" CreationDate="2015-03-08T00:17:04.670" Score="3" ViewCount="539" Body="&lt;p&gt;In common US usage, the heights of tall things are sometimes converted to a &quot;number of &lt;a href=&quot;https://en.wikipedia.org/wiki/Storey&quot; rel=&quot;nofollow&quot;&gt;stories&lt;/a&gt;&quot;. The thinking is that people can better compare heights to similar tall buildings that they might have seen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The US conversion is usually:&#xA;$$\text{Number of stories} = \text{Round}\Big(\frac{\text{height in feet}}{10}\Big)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that something similar is used in metric countries like:&#xA;$$\text{Number of storeys} = \text{Round}\Big(\frac{\text{height in meters}}{3}\Big)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming that like most things that are common usage, this conversion is not correct in practice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a more accurate height of a building story?&lt;br/&gt;&#xA;How does this cause confusion when comparing the height of a building in &quot;stories&quot; to the actually number of stories?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-08T00:22:46.473" LastActivityDate="2015-03-08T00:48:12.563" Title="Is a building story actually 10ft (3m)?" Tags="|structures|metrology|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2005" PostTypeId="2" ParentId="2004" CreationDate="2015-03-08T00:48:12.563" Score="2" Body="&lt;p&gt;There's a handy-dandy table &lt;a href=&quot;http://www.ctbuh.org/TallBuildings/HeightStatistics/HeightCalculator/tabid/1007/language/en-GB/Default.aspx&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                                           Office      | Residential/hotel | Function Unknown &#xA;                                                                             or Mixed-Use &#xA;floor-to-floor height (f)                  3.9m        | 3.1m              | 3.5m &#xA;Entrance lobby level floor-to-floor height 2.0f = 7.8m | 1.5f = 4.65m      | 1.75f = 6.125m &#xA;Number of mechanical floors above ground   s/20        | s/30              | 2/25&#xA;(excluding those on the roof)&#xA;&#xA;Height of mechanical floors                2.0f = 7.8m | 1.5f = 4.65m      | 1.75f = 6.125m &#xA;Height of roof-level mechanical            2.0f = 7.8m | 2.0f = 6.2m       | 2.0f = 7.0m &#xA;areas / parapets / screen walls&#xA;&#xA;H = Building height&#xA;f = Typical occupied floor-to-floor height&#xA;s = Total number of stories&#xA;&#xA;m = meters&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These numbers are just averages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this, it is possible to derive the height of an office building:&#xA;$$H_{\text{office}}=3.9s+11.7+3.9(s/20)$$&#xA;The same can be done for a residential building:&#xA;$$H_{\text{residential}}=3.1s+7.75+1.55(s/30)$$&#xA;The page then gives some comparisons of real buildings and the variation from the formula. However, while the &quot;aggregate&quot; variation for residential buildings (for example) comes out to 0.36%, a better indicator of accuracy would be to take the absolute value of the variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary: The average height depends on the function of the building (i.e. office vs. residential). But there are, of course, deviations from these figures.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-03-08T00:48:12.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2006" PostTypeId="1" AcceptedAnswerId="2007" CreationDate="2015-03-08T08:43:33.363" Score="11" ViewCount="3577" Body="&lt;p&gt;I have got a spinning top from a Canadian manufacturer, and I am deeply amazed by how neatly it spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interest in physics has led me to try to find out at what top speed I can spin the top and how that compares to other tops made from other materials and other shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have glued a little sticker to it and recorded it with my phone's camera but the frame rate is far too slow to be able to count the number of rotations over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have access to specialized equipment but I can measure or approximate the volume, mass, density, temperature and probably also the moment of inertia. How do I measure the angular speed of the top? &lt;/p&gt;&#xA;" OwnerUserId="844" LastEditorUserId="847" LastEditDate="2015-03-09T04:56:27.317" LastActivityDate="2017-01-07T20:28:51.750" Title="Practical setup to measure angular velocity" Tags="|measurements|applied-mechanics|" AnswerCount="5" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2007" PostTypeId="2" ParentId="2006" CreationDate="2015-03-08T13:20:47.247" Score="7" Body="&lt;p&gt;There are several ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most direct is to paint half of the top white and the other black, then point something with a light sensor at it.  AC amplify the output a little, then determine the frequency.  This could be with a deliberate frequency counter, a scope that displays frequency directly, or a microcontroller you program to find the period and ultimately show you the frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A off the shelf method is called a &lt;i&gt;Strobe Tach&lt;/i&gt;.  This isn't a direct frequency measurment, but can work pretty well nonetheless.  It is a strobe that flashes at a calibrated rate that you can adjust.  You adjust the stobe flashing to appear to &quot;freeze&quot; the spinning top, then look at what frequency the strobe is set to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another way it to put the top on something coupled to a microphone.  There will inevitably be some imbalance, which will cause vibration at the rotation frequency.  This can be amplified and the frequency determined like the light signal in the first suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can magnetize the spinning thing, then you can use a magnetic pickup to get a similar signal than the light sensor or microphone would produce.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-03-08T13:20:47.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2009" PostTypeId="2" ParentId="2006" CreationDate="2015-03-08T16:36:27.323" Score="12" Body="&lt;p&gt;Probably the easiest practical method to measure the frequency of a spinning top with everyday equipment is to analyze the sound created by the motion and look for the characteristic frequency. This can be done with a spectrum analyzer app that should be available for free for most smartphones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Place the spinning top close to the microphone of the phone and record the sound with the spectrum analyzer app. You should see several peaks in the frequency spectrum that slowly move towards lower frequencies while the top is spinning down. The lowest one should be the rotation frequency, and the other ones are harmonics at integer multiples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get a clear signal you need the top to make a measurable sound. Placing it on top of a hollow resonator might help. Also, if the top spins on a surface that has some roughness this will significantly enhance the sound, but it might disturb what you want to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AI4Ot.png&quot; alt=&quot;spinning top&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture shows a quick test I just did. I used an iPhone app called &quot;Spectrum Analyzer&quot; which has a nice spectrogram mode. As you can see (though the scale is admittedly hard to read) my top created a sound whose lowest frequency component started out at a bit over 100 Hz. So the angular velocity was roughly 600 rad/s.&lt;/p&gt;&#xA;" OwnerUserId="846" LastEditorUserId="33" LastEditDate="2015-03-08T17:31:04.877" LastActivityDate="2015-03-08T17:31:04.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2010" PostTypeId="2" ParentId="2002" CreationDate="2015-03-08T21:59:22.003" Score="6" Body="&lt;p&gt;Delivery is through a small nozzle that fills a slot behind the bar that leads into a hole in the bar which then leads to the chain. This slot is so that when adjusting the tightness and move the bar it will still deliver oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pump is driven by a worm gear which is coupled to the sprocket that drives the chain or directly to the drive shaft (depends on the model). For those that are coupled to the chain sprocket you can turn the chain to pump oil manually, for the other type you can run the engine. This lets you ensure the chain is oiled before you fire it up after changing the chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=PNyOmxwO4gs&quot; rel=&quot;nofollow noreferrer&quot;&gt;This video&lt;/a&gt; show a disassembly down to where the worm gear is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following picture comes from &lt;a href=&quot;http://www.google.com/patents/US5236314&quot; rel=&quot;nofollow noreferrer&quot;&gt;patent US5236314&lt;/a&gt; and shows the pump assembly and mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lobrb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Component 9 is a shaft that is rotationally coupled to the shaft driven by the worm gear but will oscillate as it rotates because of groove 12 and the cam 14. It is pulled back by washer 15 down to a point defined by settable screw 18. So when slot 21 is near the intake the chamber will expand and when it reaches the discharge side the chamber will contract expelling the oil.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2016-03-22T21:24:47.350" LastActivityDate="2016-03-22T21:24:47.350" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2011" PostTypeId="1" CreationDate="2015-03-08T22:13:45.600" Score="5" ViewCount="265" Body="&lt;p&gt;I'm trying to see if I can find any literature on this specific question, but a few days of scholarly digging has been fruitless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a 5&quot; round pipe with a bluff body inserted, is there an upper bound for the Reynolds number at which the von Karman Vortex Street is no longer discernible? It seems to me that the vortices would be shed so close to the bluff body that they would be covered up by eddies in a turbulent flow (Say, R&lt;sub&gt;e&lt;/sub&gt; &gt;=100,000).&lt;/p&gt;&#xA;" OwnerUserId="847" LastEditorUserId="33" LastEditDate="2015-03-09T01:03:28.923" LastActivityDate="2015-03-21T03:53:51.627" Title="Upper bound for von Karman Vortex Street?" Tags="|fluid-dynamics|fluid-mechanics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2012" PostTypeId="2" ParentId="2006" CreationDate="2015-03-08T22:43:55.547" Score="2" Body="&lt;p&gt;You can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Wagon-wheel_effect&quot; rel=&quot;nofollow&quot;&gt;wagon wheel&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Stroboscopic_effect&quot; rel=&quot;nofollow&quot;&gt;stroboscopic&lt;/a&gt; effect and count how many times the perceived rotation reversed direction until it is stopped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lets you get the RPM without a super slow motion camera.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-03-08T22:43:55.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2013" PostTypeId="2" ParentId="2011" CreationDate="2015-03-09T00:41:33.450" Score="1" Body="&lt;p&gt;I have a limited understanding of the subject, having investigated it for a specific project  some while ago.  That said - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the general 'opinion' is that well below Re = 100,000 turbulence effects will dominate and that Vortex Street formation will not be detectable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This excellent 2013 article &lt;a href=&quot;http://forums.x-plane.org/index.php?showtopic=69180&quot; rel=&quot;nofollow&quot;&gt;The Kármán Vortex Street - Chuck Bodeen&lt;/a&gt; wanders around the subject in general and has some specific comments relevant to your question plus some relevant links.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along the way he notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This means that the vortex street begins to form at the “critical” Reynolds Number (about 40), and that, for Re above 400, turbulence in the flow shatters the regular pattern of the “street”. Another source puts the critical number at 90 and some say 100. The Reynolds number beyond which turbulence destroys the vortex street may be even higher than 400 in certain cases. A video from the Earth System Sciences laboratory at the University of California, Irvine, shows that vortex streets can form behind cylinders at Reynolds numbers around 2,000, but that turbulence dominates at Re 15,000.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, to both confuse and inform, he cites this study &lt;a href=&quot;http://www.personal.psu.edu/faculty/c/x/cxc11/papers/New_ANTENNA.pdf&quot; rel=&quot;nofollow&quot;&gt;Vortex Shedding from a Ground&#xA;Tracking Radar Antenna and 3D Tip&#xA;Flow Characteristics&lt;/a&gt; &#xA; which investigates vortex shedding at well over Re=100,000. &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-03-09T00:41:33.450" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2014" PostTypeId="2" ParentId="2011" CreationDate="2015-03-09T00:50:08.197" Score="1" Body="&lt;p&gt;This is from examples given in the &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/An_Album_of_Fluid_Motion&quot; rel=&quot;nofollow&quot;&gt;An Album of Fluid Motion&lt;/a&gt;&lt;/em&gt; 5th, 1997.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of a web address to get these images, so I will just repeat what is said about a couple of the images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Figure on Table of contents -&lt;/strong&gt; &lt;em&gt;Karman vortex street behind a circular cylinder&lt;/em&gt; &quot;The Reynolds number is about 300, which is near the upper limit for stability.&quot; &lt;br&gt;&#xA;&lt;strong&gt;Figure 47 -&lt;/strong&gt; &lt;em&gt;Circular cylinder at R=2,000&lt;/em&gt; &quot;At this Reynolds number one may properly speak of a boundary layer ... turbulent wake.&quot; &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this, the Vortex Street disappears into the turbulence somewhere after &lt;strong&gt;R&lt;sub&gt;e&lt;/sub&gt;= 2,000&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-09T00:50:08.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2015" PostTypeId="1" AcceptedAnswerId="2027" CreationDate="2015-03-09T02:03:45.357" Score="12" ViewCount="12559" Body="&lt;p&gt;When looking at thread descriptions, one of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Screw_thread#Lead.2C_pitch.2C_and_starts&quot; rel=&quot;noreferrer&quot;&gt;basic properties&lt;/a&gt; is always the number of thread starts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HmBHi.png&quot; alt=&quot;wikipedia thread starts&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I could tell, all of the major standard bolt threads are single-start. This includes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Unified Standard (UNC, etc.)&lt;/li&gt;&#xA;&lt;li&gt;National Pipe Thread (NPT, NPS)&lt;/li&gt;&#xA;&lt;li&gt;British Standard&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I only found one standard thread that can also come in a multiple-starts: &lt;a href=&quot;http://www.amesweb.info/Screws/AcmeScrewNutThreadDimensions.aspx&quot; rel=&quot;noreferrer&quot;&gt;ACME&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the reasons why single-start threads are so common and multiple-start threads are rare? I am specifically interested in bolts and other fasteners.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2020-11-21T17:45:24.043" Title="Why are most standard bolt threads single start?" Tags="|threads|fasteners|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2016" PostTypeId="2" ParentId="2015" CreationDate="2015-03-09T02:15:43.087" Score="9" Body="&lt;p&gt;For any given size of fastener and given thread pitch, a single-start thread gives you the greatest mechanical advantage in terms of the torque required to achieve a given tension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from acme threads that are often used on leadscrews for mechanical motion (e.g., CNC machinery), the only other place I have seen multiple-start screws is on self-tapping screws used to fasten plastic cases together. In such applications, the speed of assembly is more important than the tension, which is limited by the plastic anyway.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-03-09T02:15:43.087" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2017" PostTypeId="1" AcceptedAnswerId="2020" CreationDate="2015-03-09T02:39:12.847" Score="13" ViewCount="17547" Body="&lt;p&gt;I've heard the term &lt;em&gt;spiral curve&lt;/em&gt; used to describe a section of highway that is more aesthetically pleasing to the driver's eye. However, I believe I've driven on enough road to say that I can't definitively tell the difference between any given curve other than how &quot;sharp&quot; it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain how one can determine if a section of curve on a highway is classified as a &quot;spiral&quot; curve, and are there other advantages besides &quot;making it look good&quot;?&lt;/p&gt;&#xA;" OwnerUserId="848" LastEditorUserId="33" LastEditDate="2015-03-09T04:10:53.733" LastActivityDate="2016-03-23T16:52:07.227" Title="What is a Spiral Curve, and How is it Different from a Normal Curve?" Tags="|civil-engineering|highway-engineering|surveying|rail|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2018" PostTypeId="2" ParentId="2017" CreationDate="2015-03-09T04:03:20.767" Score="9" Body="&lt;p&gt;It isn't a question of being &quot;aesthetically pleasing to the eye&quot;, but rather of comfort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a curve that was simply a circular arc, upon entering that curve, you would immediately feel a sideways centripedtal force that is proportional to your speed and inversely proportional to the radius of the curve. In other words, you would have to suddenly jerk the steering wheel to a position that would allow you to follow the curve, and just as suddenly jerk it back to straighten out again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of this, highway curves have an &quot;easement&quot;, which means that they follow a shape in which you gradually steer into the curve and then gradually steer out of it, giving a gradual increase and decrease in the sideways force that is felt. Mathematically, the shape of such a curve is a logarithmic spiral.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-03-09T04:03:20.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2019" PostTypeId="2" ParentId="2002" CreationDate="2015-03-09T04:11:36.317" Score="2" Body="&lt;p&gt;I sent this question along to a forester of 20+ years and this was his response:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So...... back in the day......... the first saws had a manual oiler.&#xA;  This was a younger lumberjack in training who operated the oil can. He&#xA;  held this job until one of the more senior lumberjacks cut off&#xA;  something vital and he was promoted.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The next generation of oilers was still manual. There was a thumb&#xA;  plunger on the case. It was located near the center of the case and&#xA;  was operated by the same thumb/hand which operated the throttle. The&#xA;  theory being that you could give one good squeeze and cut/oil at the&#xA;  same time. the operator was also able to titrate the oil use to the&#xA;  dryness or wetness of the wood.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One of the posters was correct in that there have been worm gears used&#xA;  also. These are very effective in that they are throttle based. they&#xA;  work as fast as the saw requires and in theory you don't burn up bars&#xA;  and chains. There is an adjustment screw which has numbers from 1-5&#xA;  with the preset usually being a 2. I believe that this is the current&#xA;  style in use. The output is based on a motor speed of 9000 rpm. There&#xA;  are some other pump styles in use but these are the most reliable.&#xA;  They are mechanical. They leak on the floor when not in use. Gravity!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="368" LastActivityDate="2015-03-09T04:11:36.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2020" PostTypeId="2" ParentId="2017" CreationDate="2015-03-09T04:27:53.280" Score="12" Body="&lt;p&gt;&lt;strong&gt;Spiral&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spiral curve is a geometric feature that can be added on to a regular circular curve. The spiral provides a gradual transition from moving in a straight line to moving in a curve around a point (or vise-verse).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of a spiral is about making the road or track follow the same form that the vehicle naturally takes. In a car, you don't go directly from going straight to fully turning. There is a transition area where you slowly turn the steering wheel. Lateral acceleration is slowly increased as the spiral is entered, or it is slowly decreased as the spiral is exited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image below, the spiral is the red portion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VwEjh.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;br&gt;&lt;sub&gt;Wikipedia Image&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hghways&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On highways, the lanes are wide enough that you can drive a spiral just by moving from one side of the lane to the other. Sometimes this is referred to as, &quot;Straightening out the curve.&quot; For this reason, spirals are not used on all curves in highways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Railroads&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One industry where spirals are used extensively is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Track_transition_curve&quot; rel=&quot;noreferrer&quot;&gt;railroad industry&lt;/a&gt;. Cars on a road have the freedom to move from one side of the lane to the other, a train does not have that luxury. A train is guided rigidly by the rails and has no room to deviate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On railroad tracks, all but very low speed curves have spirals on both sides of the curve. This provides a gentle transition in lateral acceleration. This greatly improves passenger comfort. Spirals are used on freight lines because they also reduce the forces on the track components themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One disadvantage of the use of spirals is that it increases the amount of space required for each curve. Spirals effectively lengthen the curve. &lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-09T04:55:40.223" LastActivityDate="2015-03-09T04:55:40.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2021" PostTypeId="2" ParentId="1742" CreationDate="2015-03-09T08:18:25.953" Score="7" Body="&lt;p&gt;I'm going to offer a slightly less helpful answer by poking holes in the answers of others to explain why they aren't ideal, then offering an incomplete, poorly thought through answer of my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The majority of answers here suffer from one of three stumbling blocks:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;they require unnecessarily large jigs to be feasible with complexities which make them undesirable (pendulum, I'm looking at you here)&lt;/li&gt;&#xA;&lt;li&gt;they ignore Newton's Laws of reactionary forces in their implication.&lt;/li&gt;&#xA;&lt;li&gt;the solution is absurdly complex for the task at hand.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Ok, so starting with the pendulum, first of all, very novel solution, in theory a very good one, but there are two issues here: how do you mount the pendulum? Does the instructor hold the end of the string high above their head and hope they don't get hit? Or do you use a swing-like A-frame that you might find at a playground? If so, you may as well just go to a playground, stand on a swing and have them hit you/a pad, it's a free solution. If you want something more technical and you do figure out how to suspend the pendulum, how do you guarantee that they hit it squarely, in the plane in which you measure height, without wasting a bucket of energy spinning the milk carton? If you ever used a tyre swing as a kid you'll know that swinging it in a plane of your choosing, and without spinning, is tricky. The lighter the mass of the carton the more you are going to see instantaneous force only, which could be what you want, I don't fight so I don't know, but I do know that the heaviest blow will be backed up a heap of energy, not a very brief impulse of intense momentum. Standardising between blows will also be a challenge as the amount of contact time with the carton will greatly affect the height it rises to (energy = force x distance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything which you hold in your hands is going to require you to not move; unless you are Superman you are going to move a bit, especially as your students improve, so the force which is registered is going to tail off as you provide less resistance to the blows, which means your students are going to see deminishing returns on improving their technique. Hardly ideal. The only fix to this is to mount this to something rigid and have them hit that. I would suggest a wall or similar if you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conductive foam? How complex is this getting? It's a nice theoretical solution but I can't imagine that being the simplest installation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the solution - any solution must meet three criteria: it's got to be cheap and cheerful, it's got to have repeatable results, and it has to be calibrated somehow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building on a variant of @ToyB's solution, a leaf spring is the simplest fix here, mounted on a board to something like a wall. At each end the spring should have a slot cut with a bolt through it to attach the spring to the board whilst allowing the spring to deflect. Attach a strain gauge to the back and connect it up to a simple voltmeter. You could buy all of the necessary components for \$10. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calibrate by placing a range of weights on the spring (just place the spring, on its board, on a table to do this) and measure the voltages. To add a step to the complexity but improve the solution, swap the simple voltmeter for a simple datalogger and record the voltage curve that you get as you hit the spring, then use Excel or similar and a lookup table with your voltage-force relationship to give your students a complete force curve of their hits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could even have a few spring options for people/blows of different hits to improve your accuracy. I'd imagine you could achieve all of this for less than \$50.&lt;/p&gt;&#xA;" OwnerDisplayName="user815" LastEditorDisplayName="user16" LastEditDate="2015-03-09T12:40:54.073" LastActivityDate="2015-03-09T12:40:54.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2022" PostTypeId="1" CreationDate="2015-03-09T09:24:47.627" Score="3" ViewCount="629" Body="&lt;p&gt;I want to ask about the torque experienced by a axial flow turbine, in relation to the pitch angle or angle of attack. &#xA;If you can suggest an equation, will greatly help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm new to technical terms, please bear with me) &lt;/p&gt;&#xA;" OwnerUserId="850" LastActivityDate="2015-03-16T18:38:57.070" Title="Torque in axial flow turbine" Tags="|turbines|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2025" PostTypeId="1" CreationDate="2015-03-09T18:20:26.630" Score="13" ViewCount="6427" Body="&lt;p&gt;It is well known that dust gets deposited on a rotating ceiling fan because formation of a boundary layer over fan blades. The air layer next to blade surface is stagnant due to no-slip condition and the presence of a strong viscous force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I modify the blade design to prevent the blades from accumulating dust? Is it possible to make the blade's surface so smooth as to prevent deposition?&lt;/p&gt;&#xA;" OwnerUserId="739" LastEditorUserId="368" LastEditDate="2015-03-09T20:13:34.697" LastActivityDate="2015-07-01T19:24:29.913" Title="How to prevent deposition of dust on ceiling fan blades" Tags="|mechanical-engineering|fluid-dynamics|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2027" PostTypeId="2" ParentId="2015" CreationDate="2015-03-09T19:08:16.557" Score="11" Body="&lt;p&gt;As Dave Tweed points out, the ratio of torque to tension is lower the lower the lead angle is. Since the important measure of bolt tightness is generally the tension in the bolt, we want to achieve that minimum pretension with the least effort possible. Assuming we have to maintain a certain shear area of the thread (so the the threads are stronger than the bolt when fully engaged) having two starts means we double the lead angle and greatly increase the amount of force required in the wrench to tighten the fastener appropriately. On its own though, this isn't the end of the world in practical applications because a big enough torque arm, (or a shear wrench) makes this just a matter of using a bigger motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger problem is that we want bolt threads to be self-locking. That is, we wouldn't want the pretension in the bolt to cause it to loosen. Imagine if the bolt had 10 starts, and therefore a very steep helix for the threads - no matter how hard we tighten the bolt, the pretension will immediately loosen the bolt when we let go of the wrench. This is because the lead angle lets so much force transfer into the rotation of the bolt (or nut) that it can overcome the friction between the internal and external threads. This would make the bolt not very effective without an external locking device. By contrast, standard single start fasteners are often pretensioned (or simply made snug tight) and trusted to self-lock based on their shallow lead angle. In situations with high vibration or thermal cycles, additional locking elements may be used, but they aren't normally required if the bolt is properly tightened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why single start threads are typically used for fasteners (things that aren't supposed to move) but multiple-start threads aren't uncommon for leadscrews (which are supposed to move freely, or have an external brake.)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Screws which form their own mating thread in another material are often two-start, as they have an additional resistance to unscrewing from the compression and roughness of the substrate they displaced around the threads. This is the case with some sheet metal screws, and also most wood screws.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-03-09T19:16:57.637" LastActivityDate="2015-03-09T19:16:57.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2028" PostTypeId="2" ParentId="1912" CreationDate="2015-03-09T23:33:06.837" Score="1" Body="&lt;p&gt;Just for shiggles I'm imagining a perfect dryer duct with no kinks that fits recommended installation parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that influencing the vent speed would:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Force slightly more particles through the lint trap&lt;/li&gt;&#xA;&lt;li&gt;Simply cause the lint clog to collect further from the dryer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Preventative maintenance aside (annually cleaning your ducts), I think a better solution would be to design a lint trap with a larger surface area and a smaller mesh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this doesn't necessarily require an increase of fan speed, it would require an increase in pressure to overcome the resistance of the finer mesh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A secondary option would be an additional lint trap that was ionized to catch the fine particles that sneak past the first trap. It seems like half the work is already done here, as dryers tend to cause static-y clothes.&lt;/p&gt;&#xA;" OwnerUserId="847" LastEditorUserId="847" LastEditDate="2015-03-10T01:25:00.490" LastActivityDate="2015-03-10T01:25:00.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2029" PostTypeId="1" AcceptedAnswerId="2033" CreationDate="2015-03-10T05:29:20.740" Score="5" ViewCount="14848" Body="&lt;p&gt;I have an 80 g·cm motor with a rotational frequency of 15,000 rpm. I want to lift a weight of 2 kg at a speed of 0.5 m/s. How do I calculate the gear ratio required?&lt;/p&gt;&#xA;" OwnerUserId="858" LastEditorUserId="368" LastEditDate="2015-03-10T16:28:38.930" LastActivityDate="2015-03-10T17:32:42.410" Title="How do I calculate the gear ratio to lift a weight at a constant speed?" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2030" PostTypeId="2" ParentId="2029" CreationDate="2015-03-10T08:05:06.687" Score="6" Body="&lt;p&gt;If I understand correctly the problem is like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PjnkN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity of the load is $R\omega=0.5=R\underbrace{\frac{\pi}{30}\frac{15000}{n}}_\omega$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $n$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n=1000 \pi  R$$&lt;/p&gt;&#xA;" OwnerUserId="859" LastActivityDate="2015-03-10T08:05:06.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2031" PostTypeId="1" CreationDate="2015-03-10T11:39:24.483" Score="5" ViewCount="161" Body="&lt;p&gt;I want to be able to measure the strain response in a material composed of 2 lamina, bottom one (0) top one half (0) half (90), and see the differences in the unidirectional versus the bidirectional area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying with the Lamina function, and using a negative pressure at one edge where I want the load, and bounding the other edge with boundary conditions. The error it gives me is &lt;code&gt;(Lamina can only be used with plane stress elements)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I get past this error? Do I need to use a different function?&lt;/p&gt;&#xA;" OwnerUserId="860" LastEditorUserId="7461" LastEditDate="2017-05-14T17:13:50.253" LastActivityDate="2017-10-29T04:17:26.013" Title="How do I model a longitudinal tensile strain in a laminate composite using Abaqus software?" Tags="|mechanical-engineering|materials|simulation|abaqus|" AnswerCount="1" CommentCount="1" ClosedDate="2017-10-30T01:29:56.137" ContentLicense="CC BY-SA 3.0" />
  <row Id="2032" PostTypeId="1" CreationDate="2015-03-10T13:20:56.560" Score="11" ViewCount="877" Body="&lt;h1&gt;Background&lt;/h1&gt;&#xA;&lt;p&gt;I am working with a small team on a university project to build a greenhouse heat sink. This will pass warm air at the top of the greenhouse down through a chamber underground filled with a material to absorb and store the warm air. We have two prototype greenhouses; one will act as a control for baseline measurements and the other will have the heat sink.&lt;/p&gt;&#xA;&lt;h2&gt;Set-up&lt;/h2&gt;&#xA;&lt;p&gt;I have built several temperature sensors and loggers for the final prototype, but some preliminary tests are being made on various materials:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Granite chips between 15-25mm, irregular shape&lt;/li&gt;&#xA;&lt;li&gt;Tempered glass fractured into small pieces 7-15mm approx., atleast 2 sides are flat&lt;/li&gt;&#xA;&lt;li&gt;Concrete fragments 30-80mm, irregular shape - &lt;strong&gt;test not completed&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;These were placed in a 5 L box. The box has a small fan and piping at the bottom to blow air into the chamber and release the air through a number of 6 mm holes in the pipe at the base of the box.  The top of the box is sealed except for a vent which has the same diameter as the tube with the fan.  A PT1000 temperature sensor is also inserted into the centre of each material to capture measurements every second.  Here is an image of the test box:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5TOTnm.jpg&quot; alt=&quot;test box set up&quot; /&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Procedure&lt;/h2&gt;&#xA;&lt;p&gt;The free air space was calculated on a smaller sample of both materials to give a rough figure of 42% for the granite and 43% for the glass. Two tests were then performed on the granite and then the glass:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Both cooled outside for a several hours to about 5.5 °C, then brought into the room and left for 1 hour with the fan on. The temperature was recorded as the material warmed up to room temperature.&lt;/li&gt;&#xA;&lt;li&gt;After the first test the materials were then placed in a freezer and cooled down to -20 °C, the temperature was recorded again.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h2&gt;Results&lt;/h2&gt;&#xA;&lt;p&gt;As can be seen below, the glass exhibits a lag in both data sets, warming up and cooling down, after which the temperature change becomes more linear. Whereas the granite shows a more linear change in temperature throughout.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Glass Warming (x-axis seconds, y-axis temperature)&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/GzS8Vm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Glass Cooling (x-axis seconds, y-axis temperature)&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/f3SOnm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Granite Warming (x-axis seconds, y-axis temperature)&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/aIiSQm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Granite Cooling (x-axis seconds, y-axis temperature)&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yxOSJm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Questions&lt;/h1&gt;&#xA;&lt;p&gt;We are discussing the results at the moment and I am interested in expert opinions of the data we collected. The data is interesting and we are interpreting it correctly. Specifically:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;The shape of the glass fragments allows for a more interlocking shape, which could restrict airflow more, but wouldn't this still have a more linear temperature change?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Could the glass data be due to minor thermal expansion changes in the material?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The glass has a lower thermal conductivity rating than the granite, is this the reason for the lag?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="583" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-07-26T20:20:50.897" Title="Comparing the performance of glass and granite in a heat storage application" Tags="|mechanical-engineering|materials|heat-transfer|" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2033" PostTypeId="2" ParentId="2029" CreationDate="2015-03-10T14:32:44.740" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have a 80gcm motor with a rpm of 15000.&lt;br&gt;&#xA;  I want to lift a weight of 2 kg at a speed of 0.5 m/s.&lt;br&gt;&#xA;  How do I go about calculating the gear ratio required for this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Firstly - is it possible?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, is there enough input power available for the desired output power?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To within about 2% a very handy formula applies - it can be derived conventionally and seen that several factors happen to cancel nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watts = kg x metres x RPM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;80 gram∙cm = 0.080 kg x 0.01 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for input W = 0.080 kg x 0.01 m x 15000 = 12 Watts.&lt;br&gt;&#xA;This is the maximum Wattage you can deliver if properly geared at 100% efficiency&lt;br&gt;&#xA;(we should be so lucky).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Desired power = Force x distance per unit time&lt;br&gt;&#xA;Watts = Joules/sec = mg∙d/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;= 2 kg x $g$ x 0.5 m/s = 2 x 9.8 x 0.5 = 9.8 Watts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to work at all overall efficiency needs to be at least 9.8/12&#xA;or greater than about 82%.&lt;br&gt;&#xA;That's potentially doable but also potentially difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now to the actual problem.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following assumes that the output weight or force is taken from the end of a radius of the driven &quot;gear&quot;. If output is instead taken from eg a windlass drum at lower diameter to the driven gear the ratios will be scaled based on the relative diameters. Ignore that for now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque_in x RPM_in = Torque_out x RPM_out at 100% efficiency&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or RPM_out = Torque_in x RPM_in  / Torque_out at 100% efficiency &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RPM out = 0.080 kg x 0.01 m x 15000 RPM  / (2 kg x 0.5 m) = 12 RPM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So gear ratio = 15000/12 = 1250:1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specification of not just output Torque but actual force (2 kg x $g$) constrains the actual output pulley size if output is taken off at the pulley radius.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-03-10T17:32:42.410" LastActivityDate="2015-03-10T17:32:42.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2034" PostTypeId="2" ParentId="2025" CreationDate="2015-03-10T14:40:34.707" Score="3" Body="&lt;p&gt;If the blades are conductive, or accommodate a conductive coating, a small static charge should repel dust by imparting a charge to the particles, which would then expel/repel them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charge could be periodically alternated since some dust is already ionized and would thus be attracted to one charge, but repelled by the opposite.&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2015-03-10T14:40:34.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2036" PostTypeId="2" ParentId="2006" CreationDate="2015-03-08T09:43:51.880" Score="1" Body="&lt;p&gt;I would imagine a nice crude formula might be of help $T_p = 4\pi^2 I / mgrT_s$. So if the spin speed is too fast for you, the precession speed will be inversely proportional and very slow. Here you can see the effect of the material through its mass, but you also have to approximate the moment of inertia, as well as the perpendicular distance between the spin axis and the precession axis&lt;/p&gt;&#xA;" OwnerDisplayName="Ali Moh" LastActivityDate="2015-03-08T09:43:51.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2037" PostTypeId="2" ParentId="2006" CreationDate="2015-03-08T09:51:48.527" Score="1" Body="&lt;p&gt;The answer could be to get hold of a high-speed LED stroboscopic light. How easy/expensive this will be depends on the angular speed you are dealing with. There are even some phone apps which will go to moderate speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You tune the speed of the stroboscope to make the mark on your spinning top appear stationary. But this will be difficult because the speed will not be constant.&lt;/p&gt;&#xA;" OwnerDisplayName="Rob Jeffries" LastActivityDate="2015-03-08T09:51:48.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2038" PostTypeId="1" CreationDate="2015-03-10T22:17:26.477" Score="2" ViewCount="94" Body="&lt;p&gt;I've heard that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Clock_of_the_Long_Now&quot; rel=&quot;nofollow&quot;&gt;10,000 year clock&lt;/a&gt; will use multiple power sources (heat differential, mechanical weight, etc.), but how are they all integrated?&lt;/p&gt;&#xA;" OwnerUserId="863" LastEditorUserId="106" LastEditDate="2015-03-10T22:48:01.073" LastActivityDate="2015-03-10T22:48:01.073" Title="Power system of the 10,000 year clock" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2039" PostTypeId="1" CreationDate="2015-03-11T03:17:07.470" Score="4" ViewCount="1169" Body="&lt;p&gt;I have not found any information relating to using electicity to ionize air with temperature, but I am sure that there should be some information available. &#xA;I am thinking of the nozzle of modern jet engine (about 2000℃)considering I don't have any sophisticated equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to measure or even calculate the energy(Joule) needed to maintain ionization of air at high temperature(2000℃)?&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="865" LastEditDate="2015-03-17T05:31:13.313" LastActivityDate="2018-10-10T03:00:13.833" Title="How to calculate the energy needed to ionize air at a specific temperature?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2040" PostTypeId="2" ParentId="2025" CreationDate="2015-03-11T08:02:25.157" Score="6" Body="&lt;p&gt;&lt;em&gt;Assuming&lt;/em&gt; your explanation of dust accumulation on a blade is correct my suggestion would be to add slots to the blade to create a cascade of aerofoil sections. Doing this would allow you to consistently reenergise your BL and limit its growth, and jumping from this design to a serated TE for turbulent flow wouldn't be a much greater step of complexity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frankly I am a tad sceptical of the explanation, since whilst a no-slip condition would be enforced at the blade surface with the air, no such condition exists between the dust and the blade surface, so the shear force acting across the layer of dust is the only force acting on it (yes there is friction with the blade but if the particles are remotely spherical this ought to be small). This shear force becomes significant when you consider the thickness of a layer of dust which a human might notice compared to boundary layer thickness at that Re number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate your much bigger issue is that when stationary, dust will always settle on your blades. I doubt even a static charge would help you here as you can't easily ionise the dust in the air and you're just as likely to attract dust as repel it if you can't ionise it (which is why old TVs tended to accumulate a layer of dust on the glass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternative solution: design your fan with an absurdly overpowered motor so you can spin it at 1000s RPM and use the centrifugal force to spin the dust off of it. &lt;em&gt;Edit: a couple of other points here, thinking about it, there is also a pressure force encouraging airflow towards the blade tips, and it would be trivial to use the high RPM to produce a high frequency vibration through the blades, indeed this is likely to be a natural side effect, which will help to create turbulence in the BL and also potentially &quot;bounce&quot; the dust particles off of the blades and into faster moving flows. The higher RPM also helps increase Re number, further decreasing BL thickness and encouraging turbulence.&lt;/em&gt; Just remember to make sure the blades never ever fail; could be messy. &lt;em&gt;This would actually be a half decent solution to the problem if it weren't so absurd/impractical/expensive/dangerous.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternative alternative solution: include a feather duster in the box.&lt;/p&gt;&#xA;" OwnerDisplayName="user815" LastEditorDisplayName="user815" LastEditDate="2015-03-11T10:39:20.070" LastActivityDate="2015-03-11T10:39:20.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2041" PostTypeId="1" AcceptedAnswerId="3765" CreationDate="2015-03-11T13:05:31.940" Score="6" ViewCount="265" Body="&lt;p&gt;I have an welded steel mesh, like in  the following image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qT2lx.jpg&quot; alt=&quot;steel mesh&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Material: steel&lt;/li&gt;&#xA;&lt;li&gt;Diameter of the wires: 4mm&lt;/li&gt;&#xA;&lt;li&gt;Mesh: 15x15cm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I found the following information about the welds:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;the minimal relative strength of welds in shear: F&lt;sub&gt;ms&lt;/sub&gt; = 4.4kN&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I want cut out an rectangle from this mesh (dimensions are on the following image) and want hang it on the wall, &lt;strong&gt;suspended by &lt;em&gt;two&lt;/em&gt; screws&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want hang to the welding points some weights. (on the image the drop-like things) :), in two scenarios:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;uniformly attach to &lt;em&gt;each welded point&lt;/em&gt; some weight (like the blue &quot;drops&quot;)&lt;/li&gt;&#xA;&lt;li&gt;attach one big weight to the center (the red drop)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8VFs9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the weight I can hang on the mesh (in both scenarios) without significant structural deformations (2% elongation)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't want have broken welds. Will the welds fracture before the bars yield or vice versa? At what load will the first of these failure modes occur?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also, i'm looking for some &quot;guide&quot; how it should be calculated - if it is possible to do in easy (read: &quot;idealized&quot;) form, without specialized software. I would be happy to calculate it myself, but need some &quot;how to&quot;. Understand than using finite-elemets or such will give most precise result, but is here some &quot;easier/simpler&quot; way?&lt;/p&gt;&#xA;" OwnerUserId="866" LastEditorUserId="368" LastEditDate="2015-03-11T20:53:01.470" LastActivityDate="2016-03-04T17:09:29.727" Title="Hanging a steel mesh without significant deformation" Tags="|mechanical-engineering|steel|structural-engineering|" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="2042" PostTypeId="1" CreationDate="2015-03-11T13:14:02.960" Score="2" ViewCount="244" Body="&lt;p&gt;I've been trying to compare popular materials for bicycle frames in terms of their mechanical properties. One thing I noticed is that it's hardly possible to find a data sheet specifying the impact toughness of a material. I've only found it for some carbon-fibre composites and nothing else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is impact toughness not a commonly listed material specification? Is it considered unimportant? Are the measurement and testing standards unclear? Are there any other reasons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm specifically referring for aluminium 6061 and 7005, steel 4140 and 4130, Titanium 3Al-2.5V and some carbon fibre composites. I could find the data for carbon fibre, but none for any of the metal alloys.&lt;/p&gt;&#xA;" OwnerUserId="369" LastEditorUserId="368" LastEditDate="2015-05-22T19:03:58.960" LastActivityDate="2016-01-25T18:44:10.270" Title="What is the impact toughness for some common bicycle frame materials?" Tags="|mechanical-engineering|materials|bicycles|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2043" PostTypeId="2" ParentId="1912" CreationDate="2015-03-11T15:40:31.863" Score="2" Body="&lt;p&gt;There is a product on the market today that is specifically aimed to address this issue - it is the &lt;a href=&quot;http://www.fantech.net/Support/FAQs/Dryer-Exhaust/&quot; rel=&quot;nofollow&quot;&gt;dryer booster fan&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I installed one of these on my dryer (I was seeing excessive lint buildup in my ducts which were borderline for size and length) and it made a big difference: not only do the ducts stay cleaner - but the clothes dry significantly more quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this experience, it does seem to me that the built in fan in most dryers is underpowered - a variable speed fan that measures the flow rate (and maintains it constant) would clearly solve this problem without some of the drawbacks mentioned in other answers. If you added an alarm when back pressure exceeds a certain value it would be even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is almost certainly &quot;yes&quot;. But the appliance business is cut-throat: it would be hard to justify the additional cost (unless insurance companies cotton on and provide a premium discount for this safety feature). I have no objective measurement of the cost savings due to faster drying times - just anecdotal evidence.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;disclosure&lt;/strong&gt; the only relationship I have with the linked site is that I bought one of their fans to solve the problem described in my house&lt;/p&gt;&#xA;" OwnerUserId="565" LastEditorUserId="565" LastEditDate="2015-03-11T16:02:10.380" LastActivityDate="2015-03-11T16:02:10.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2044" PostTypeId="1" AcceptedAnswerId="2045" CreationDate="2015-03-11T16:42:38.097" Score="8" ViewCount="2342" Body="&lt;p&gt;I need to replace the fan on my graphics card. The original fan is exactly like &lt;a href=&quot;http://www.dhgate.com/product/original-for-sapphire-apistek-graphics-card/190676310.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; but there is another fan that seems to be a &lt;a href=&quot;http://www.dhgate.com/product/original-sauter-ga51s2l-12v-0-13a-blade-diameter/230471693.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;new revision&lt;/a&gt; of the same model number from the same manufacturer (NNTG vs. NNTM). The main difference seems to be the shape of the blades:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uSkiv.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither I nor the DHGate seller can find any data about the fans (CFM, noise) so that's why I ask here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make up my mind I need an answer to both questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Which blade design is likely to cool the GPU the most?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Which one is likely to be the quietest?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The safe choice would be to get the exact same one (NNTG) but if the new design can be more efficient with the same noise, or as efficient but quieter, I'd prefer the new one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I don't want to buy an aftermarket fan (even if the prices are similar), I just want a replacement for the old one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2015-04-26 : Finally I bought both and, unlike what I said in the comments, I found a way to quickly switch the fan without taking out the whole heatsink assembly, allowing me to test both fans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite the general consensus that NNTM was supposed to cool the GPU better, temperature with NNTG is actually 5°C cooler than with NNTM (after 10 minutes of running glxgears in full-screen). Maybe the motor doesn't match the power needed for the redesigned blades of NNTM, I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, if you need to replace a GA512SL fan for a RadeonHD like I did, buy the NNTG model :-)&lt;/p&gt;&#xA;" OwnerUserId="870" LastEditorUserId="870" LastEditDate="2015-04-26T19:21:23.380" LastActivityDate="2015-04-26T19:21:23.380" Title="How does the shape of CPU/GPU fan blades affect their performance and loudness?" Tags="|mechanical-engineering|cooling|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2045" PostTypeId="2" ParentId="2044" CreationDate="2015-03-11T16:50:05.910" Score="5" Body="&lt;p&gt;Hopefully some helpful observations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Both fans have the same power rating&lt;/li&gt;&#xA;&lt;li&gt;Blade size differs I would have thought given the power rating this would equate to the smaller blades having a faster rpm, the larger blades can have a lower rpm for the same air flow.&lt;/li&gt;&#xA;&lt;li&gt;Rotational speed can be a factor in noise as faster generally equates to noisier, I say generally as not always the case.  Larger fans generally have a lower rotational speed and produce less noise.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So to summarise I think both probably have similar air flow specifications, but the smaller bladed fan has a faster rpm, so is probably more noisy than the larger bladed version.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="368" LastEditDate="2015-03-11T18:08:37.930" LastActivityDate="2015-03-11T18:08:37.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2046" PostTypeId="1" CreationDate="2015-03-11T19:39:51.347" Score="7" ViewCount="15555" Body="&lt;p&gt;How would you go about solving the following problem. I believe Bernoulli's equation needs to be employed, but I'm not sure how.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the magnitude of the force that needs to be applied to a piston of a 20ml syringe with 1cm diameter tube to drain it in 20 seconds through a 40mm length needle of 0.2mm inner diameter. The fluid inside the syringe is water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Force = ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume of syringe = 20 ml = 0.00002  m^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Syringe diameter = 0.01 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Length of needle = 0.04 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needle diameter = 0.0002 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time to drain syringe = 20 s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluid density of water at 20 degrees celcius = 998.21 kg/m^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dynamic viscosity of water at 20 degrees celcius = 0.001002 Pa.s&lt;/p&gt;&#xA;" OwnerUserId="871" LastEditorUserId="871" LastEditDate="2015-03-11T22:01:43.343" LastActivityDate="2019-06-01T00:36:56.297" Title="Force required to empty a syringe" Tags="|mechanical-engineering|fluid-dynamics|fluid-mechanics|pipelines|piping|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2047" PostTypeId="1" CreationDate="2015-03-11T20:32:03.330" Score="5" ViewCount="3397" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QDCKA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am designing a two dimensional cam profiles.  I want to use the &quot;modified sine&quot; method for drawing the position and angle changes.  (see attached sketch).   The modified sine curve is actually a combination of cycloidal curve at the first and last 1/8 of the curve and a sine curve in the middle 7/8 of the curve.  It is easily employed when the terminal velocities are zero.  However,  often times it is necessary for a cam profile to simply go from one velocity (perhaps zero) to a constant terminal velocity. The terminal velocity is simply an angle on the displacement diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The profile is defined by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y=&#xA;\begin{cases}&#xA;\frac h{4+\pi}\left(\pi\frac\theta\beta-\frac14 \sin \left(4 \pi \frac\theta\beta \right) \right), &amp;amp; 0\lt\theta\lt\frac18\beta \\[2ex]&#xA;\frac h{4+\pi}\left(2+\pi\frac\theta\beta-\frac94 \sin \left(4\pi\frac\theta{3\beta}+\frac\pi3 \right) \right), &amp;amp; \frac18\beta\lt\theta\lt\frac78\beta \\[2ex]&#xA;\frac h{4+\pi} \left(4+\pi\frac\theta\beta-\frac14 \sin \left(4\pi\frac\theta\beta \right) \right), &amp;amp; \frac78\beta\lt\theta\lt\beta&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum velocity which can be achieved is at $45\deg \left( \frac\pi4 \right)$ therefore, only the first half of the curve is usable for my need. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;as an example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what method would you employ to design a curve that would go from point $(0,0)$ at angle zero, to the point $(3,2)$ slope $30$ degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What coefficients $h$ and $\beta$ in the above equations will create a curve such that the slope at point $(3,2)$ is equal to $\frac{30}{180}\pi$?&lt;/p&gt;&#xA;" OwnerUserId="874" LastEditorUserId="874" LastEditDate="2015-03-13T15:00:32.980" LastActivityDate="2015-03-16T03:16:29.193" Title="How to calculate a modified sine curve?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2049" PostTypeId="1" CreationDate="2015-03-11T21:38:24.180" Score="10" ViewCount="3516" Body="&lt;p&gt;While considering &lt;a href=&quot;https://engineering.stackexchange.com/q/2041/368&quot;&gt;this recent question&lt;/a&gt;, I started wondering how to determine the strength of the welds in the steel mesh. As AndyT points out in &lt;a href=&quot;https://engineering.stackexchange.com/questions/2041/hanging-a-steel-mesh-without-significant-deformation#comment3116_2041&quot;&gt;his comment&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The welds at the connections are only intended to keep the bars at&#xA;  right angles during handling - they are not intended to take any load.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The intent of the manufacturer, of course, doesn't stop someone from using the product to support a load anyway. &lt;strong&gt;Just what sort of load &lt;em&gt;can&lt;/em&gt; this type of weld support? What is/are the likely failure mode(s)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This weld has different geometry and loading than fillet welds (which are the only type I'm at all familiar with). I think it's properly called a &lt;em&gt;flare-v groove&lt;/em&gt; weld, though most of the examples I can find show that as joining a round bar to a flat bar or plate. The geometry is different enough that I suspect those are not equivalent, from a design standpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aws.org/technical/d3/D1.1_2000_Section2_Design.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This AWS technical manual&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;2.3.3.2 Effective Weld Size (Flare Groove).&lt;/strong&gt; The effective weld size for flare groove welds when filled flush to the surface of a&#xA;  round bar, a 90° bend in a formed section, or a rectangular tube shall&#xA;  be as shown in Table 2.1, except as permitted by 4.10.5.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Table 2.1 indicates the effective weld size is either 1/2 or 3/8 the radius of the outside surface of the weld. I'm still not sure this applies to the crossed bars geometry, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was learning about fillet welds, it was always assumed the weld would fail first in shear along its throat. Given that a square mesh has no cross-bracing and there is a single point of contact between the members (rather than a line or plane of contact), can I afford to make that assumption?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially in the case where there are only two rods rather than a mesh grid, this weld could be loaded in quite the variety of configurations. Let's consider only the cases of pure torsion, pure tension and pure shear. Assume the weld fails before the members.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-03-12T13:44:15.070" Title="How do I calculate the strength of a weld joining two crossed rods?" Tags="|mechanical-engineering|welds|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2050" PostTypeId="1" AcceptedAnswerId="2056" CreationDate="2015-03-11T21:51:51.500" Score="8" ViewCount="3790" Body="&lt;p&gt;On old bridge plans I have occasionally seen a note to &quot;burr threads after installation&quot;. This was done to keep the nuts of bolted connection from loosening all the way and falling off of the bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this practice can be seen in this &lt;a href=&quot;https://books.google.com/books?id=bxAkAAAAMAAJ&amp;amp;pg=PA1524&amp;amp;lpg=PA1524&amp;amp;dq=bridge%20burr%20threads&amp;amp;source=bl&amp;amp;ots=B3fbOBL1C5&amp;amp;sig=e_QglH2tkXRRS7bhha4sFVPXtKo&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=HrAAVfPsJIKXgwSh44CACQ&amp;amp;ved=0CEoQ6AEwBg#v=onepage&amp;amp;q=bridge%20burr%20threads&amp;amp;f=false&quot;&gt;Google Books&lt;/a&gt; excerpt from &lt;em&gt;Bridge Engineering, Volume 2&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nuts on the floorbeam hangers have a tendency to work loose; this we generally remedy by putting on check nuts where the thread is long enough to permit; if not, we burr the thread after adjusting.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That book is from 1916, but I have seen this on much more recent bridge plans. Sometimes the note is along the lines of &quot;intentionally damage threads ...&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The practice of damaging the threads is still an option per this &lt;a href=&quot;http://www.aisc.org/DynamicQuestion.aspx?Grpid=6&amp;amp;QueId=1760&amp;amp;ste=_Preview&amp;amp;id=2100&amp;amp;type=0&quot;&gt;AISC response&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't such a big issue for normal bolted connections where the bolt could be cut (or otherwise destroyed) and replaced with a new one. This is a problem for anchor bolt connections where one end of the anchor bolt is embedded in concrete. These bolts can not be easily replaced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intention of damaging the bolt thread is to keep the nut from working loose. The expectation seems to be that with enough torque the nut would effectively re-tap the bolt as it is removed.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it feasible to back the nut off of the bolt without destroying the bolt? &lt;/li&gt;&#xA;&lt;li&gt;Would there be any confidence that the anchor bolt could be reusable after this procedure is complete? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-12T12:38:12.717" LastActivityDate="2015-03-12T12:38:12.717" Title="Can nuts safely be removed from bolts where the threads have been intentionally damaged?" Tags="|structural-engineering|bridges|bolting|threads|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2051" PostTypeId="2" ParentId="2046" CreationDate="2015-03-11T22:44:06.017" Score="5" Body="&lt;p&gt;You can get a minimum bound from energy balance alone.  This is as if the fluid has no viscosity, so the force you have to apply over the distance is only due to the kinetic energy required to expell the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter of the tube is 1 cm, so the area is .785 cm&amp;sup2;.  That means the plunger travel distance is 25.5 cm = 0.255 m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluid is squeezed down to 200 &amp;micro;m diameter, which is a crossectional area of 31.42x10&lt;sup&gt;-9&lt;/sup&gt; m&amp;sup2;.  The volume of fluid is 20 ml = 20x10&lt;sup&gt;-6&lt;/sup&gt; m&amp;sup3;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(20x10&lt;sup&gt;-6&lt;/sup&gt; m&amp;sup3;)/(31.42x10&lt;sup&gt;-9&lt;/sup&gt; m&amp;sup2;) = 637 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is how far the 200 &amp;micro;m stream has to travel in 20 seconds, for a speed of 31.8 m/s.  20 ml of water has a mass of 20 g, or 0.020 kg.  The total kinetic energy that is therefore imparted on the fluid is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;frac12;(0.020 kg)(31.8 m/s)&amp;sup2; = 10.1 J&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can solve for the force required over the plunger travel distance to impart this energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(10.1 J)/(0.255 m) = 39.8 N = 8.95 pounds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's actually a lot more than I expected before working it out.  It would be interesting to see how much higher the force is when viscosity of the fluid is taken into account.  It might be possible that kinetic energy is actually the dominant effect for something with relatively low viscocity like water.  Obviously the force would go way up for something thick and gloppy, probably to the point where a typical syringe couldn't handle the pressure to acheive the expulsion time of 20 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hmm, that's a interesting point.  Let's see what the pressure is.  The area of 0.785 cm&amp;sup2; is 0.123 in&amp;sup2;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(8.95 pounds)/(0.123 in&amp;sup2;) = 73 PSI&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the pressure inside the syringe required to expell the fluid just due to the kinetic energy requirement alone.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Added&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There is yet another effect at work that makes the minimum required force higher, still without invoking viscosity.  The speed won't be the same for every part of the flow thru the narrow tube of the needle.  The flow will be laminar, so the outer edges will be slower with the highest speed in the center.  The average still needs to be what was calculated above, but the power will be higher because it scales with the square of the speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is the same as the ratio between the RMS flow rate and the average flow rate.  For example, for a linear profile from edge to center, RMS is 22.5% higher than the average.  Of course that's a rather unreasonable profile, but it illustrates the concept.  I picked the shape of a half-sine as a close-enough profile.  This means the flow speed is 0 at the edges and smootly peaks in the middle.  Perhaps someone more familiar with fluid dynamics can tell us what the real profile is, but I expect this will come close enough for the purpose of increases the energy requirement due to the spread of flow speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was too lazy to do the 2D integrals, so I had the computer do the integrals numerically for me.  The RMS of the sine peak profile is 17.9% higher than the average.  That means the 10.1 J calculated before needs to be increased by this amount.  That comes out to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force = 46.9 N = 10.5 pounds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure = 86 PSI&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, this is without the additional force needed to overcome viscosity of the liquid.  The only properties of the liquid this relies on are its density, and that flow thru a 200 &amp;micro;m diameter pipe will be laminar.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2015-03-13T21:58:26.400" LastActivityDate="2015-03-13T21:58:26.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2056" PostTypeId="2" ParentId="2050" CreationDate="2015-03-12T07:57:07.597" Score="9" Body="&lt;p&gt;I asked a friend who is a highly competent EE by day - but who restores super sized steam engines by night and collects olde heavy metal engines etc and has much experience with old large corroded items. His comments:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It really depends on the specific situation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mild steel bolts in good condition which have not been too excessively deformed, and where the nuts are in reasonable condition, it is often possible to &quot;re-form&quot; the damaged threads by applying sufficient torque to back the nut off.  In some case you can encourage this to happen by carefully grinding the deformed end of the bolt to get it back to something close the original nominal thread diameter.  You can also increase your chances if you apply a penetrating lubricant (and any lubricant is better than none).  Applying heat may also help if that is possible  - and if you can apply heat and then apply lubricant at a temperature where it is not broken down the lubricant will tend to be drawn into the thread as further cooling occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the nut is in poor condition or the shaft of the bolt has corroded and become &quot;waisted&quot; your probability of success will be pretty low.  If just the nut is corroded you may be able to split the nut with chisel, but you then have the problem of how to reform the threads.  If sufficient bolt end exists you may be able to use a die to cut new threads, but you will then have to use a bush under the new nut to take up the length of the remaining old thread as the new thread is unlikely to run into the old properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases you may be able to reduce the diameter of the bolt and cut a new thread at the reduced diameter.  If you are willing to make you own nut, you have full freedom as to the diameter you choose  - at the expense of a non-standard thread for your successors to deal with when its their turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For high tensile bolts your job is much harder and I think the chance of success pretty low  - and you will probably need more specialised equipment.  A carbide shell type cutter which fits over the bolt thread may allow the nut to be cut away and perhaps a die grinder could be used to clean up the thread so it can be re-cut (but even HSS dies will struggle with high tensile steel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are special repair kits available that work a bit like concrete anchors.  They have a nut with a segmented sleeve which has sawtooth ridges on the inside and is slightly tapered.  You slip it over the (suitably prepared) bolt end and then tighten the nut.  As the nut tightens it compresses the sleeve which in turn bites into the bolt.   Don't ask me what they are called or who makes them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also some process used in civil engineering that allows a failed steel tie in a blind hole to be repaired in-situ.  It must be some kind of welding process but I have no idea how it might work.  I believe that it was used to do some repair work on the bridge over the Grafton gully [in Auckland, New Zealand].  I think it is also be used to extend existing concrete bridge structures when the roadway underneath the bridge must be widened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Ken Mardle.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-03-12T07:57:07.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2057" PostTypeId="2" ParentId="1878" CreationDate="2015-03-12T09:56:09.517" Score="4" Body="&lt;p&gt;First, I'd like to point out that 316/316L may not necessarily be recommended for permanent immersion (intermittent immersion is okay) in seawater [1].  The rule of thumb is that you need at least 12% chromium in order to form a passive layer that protects the base metal from corrosion [2, chapter 1 page 4].  Molybdenum is added to improve pitting resistance.  For permanent immersion in seawater, the recommendation is at least 6% Mo [1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for material loss rates, it will depend on various factors such as the chemical environment (e.g. pH, chlorides, sulfides), temperature, pressure, etc.  Additives can substantially change properties related to corrosion resistance.  Ref [2, chapter 3 figure 1] provides an explanation for the role of various additives / alloying elements.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.worldstainless.org/Files/issf/non-image-files/PDF/Marineapplications.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.worldstainless.org/Files/issf/non-image-files/PDF/Marineapplications.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;J. R. Davis, &lt;em&gt;Corrosion of Weldments (#04182G)&lt;/em&gt;. ASM International, Jan 2006. &lt;a href=&quot;http://www.iranpejohesh.com/wp-content/uploads/2012/05/Corrosion-of-Weldments.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.iranpejohesh.com/wp-content/uploads/2012/05/Corrosion-of-Weldments.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="876" LastActivityDate="2015-03-12T09:56:09.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2058" PostTypeId="2" ParentId="2049" CreationDate="2015-03-12T13:44:15.070" Score="7" Body="&lt;p&gt;The welds on welded wire mesh are for more than just holding the wires together while handling. There are specific weld strength requirements for welded wire. The design of welded wire reinforcing uses the number of cross wires to determine development length. This is the reason why a weld strength is required. You can't have the welds breaking while force is applied. It is because of this mechanical anchorage that welded wire mesh can be considered continuous with only one square overlap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Weld Specifications&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two different specifications that might apply depending on the material to be welded. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rebar mats would follow AWS D1.4.&lt;/li&gt;&#xA;&lt;li&gt;Welded wire should follow ASTM A185 (plain wire) or ASTM A497 (deformed wire)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Welded Rebar - AWS D1.4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AWS D1.4 is the welding specification that covers welding of reinforcing bars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what the weld is called between two crossing bars. It isn't called out in AWS D1.4, but a detail for the weld is shown in ACI 318 for the welding of cross bars in corbels. The detail from the commentary is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eEBGrm.png&quot; alt=&quot;cross bar weld.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also per ACI 318 Section 11.9.6(a), the weld is required to develop the full yield strength of the bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Welded Wire Reinforcing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Welded wire gets a complete fusion as shown in this image from the &lt;em&gt;&lt;a href=&quot;http://rebar.ecn.purdue.edu/wwr/Library%5CWRI%20Manual%20of%20Standard%20Practice.pdf&quot; rel=&quot;noreferrer&quot;&gt;Manual of Standard Practice: Structural Welded Wire Reinforcement&lt;/a&gt;&lt;/em&gt; by the Wire Reinforcement Institute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QtTBDm.jpg&quot; alt=&quot;welded wire reinforcing weld&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weld shear strength is specified by either ASTM A185 (plain wire) or ASTM A497 (deformed wire) depending on the type of wire. For all but the thinnest wire, the required weld strength is 35,000 psi (240 MPa). This puts the weld strength at about half of the yield strength of the wire (65,000 psi or 70,000 psi).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Strength in Other Directions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weld strength requirements as shown above are only for shear. When used in concrete, this is the only direction that force should be applied. It would be difficult to determine how these welds would perform in peeling, torsion, or rotation about the crossing location.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-12T13:44:15.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2059" PostTypeId="1" AcceptedAnswerId="2078" CreationDate="2015-03-12T16:08:06.983" Score="3" ViewCount="4206" Body="&lt;p&gt;I was doing a question involving a T-Beam question wrong since I was selecting the wrong area to use for $Q_{max}$ when trying to maximise the shear stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\large{\tau}= \frac{VQ}{It}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend argued that the area both above and below the Neutral Axis should be equal but despite trying multiple times, I have been unable to get the same result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this statement correct?&lt;/strong&gt; (and if not, why)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem by argument of equal values both above and below the beam that perhaps it would be correct, even if the beam is unsymmetrical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HugpN.png&quot; alt=&quot;T-Beam&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="782" LastEditorUserId="90" LastEditDate="2015-03-13T02:16:58.377" LastActivityDate="2015-03-16T13:43:10.863" Title="Is the Qmax of an T-Beam equal if calculated both above and below the neutral plane?" Tags="|structural-engineering|beam|statics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2060" PostTypeId="2" ParentId="2031" CreationDate="2015-03-12T19:01:54.160" Score="3" Body="&lt;p&gt;Without knowing more about your problem the first thing that you need to do is make sure that you are using the right geometry and elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use that representation you need to use:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2D geometry&lt;/li&gt;&#xA;&lt;li&gt;Mesh it as shells&lt;/li&gt;&#xA;&lt;li&gt;Assign a plain stress assumption&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="885" LastEditorUserId="33" LastEditDate="2015-03-12T19:17:16.827" LastActivityDate="2015-03-12T19:17:16.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2061" PostTypeId="2" ParentId="1741" CreationDate="2015-03-13T02:31:47.910" Score="6" Body="&lt;p&gt;The number of balls in a bearing affects a couple of different properties of the bearing:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bearing Speed&lt;/li&gt;&#xA;&lt;li&gt;Bearing Strength&lt;/li&gt;&#xA;&lt;li&gt;Bearing Life&lt;/li&gt;&#xA;&lt;li&gt;Bearing Cost&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;Full Complement&lt;/strong&gt; bearing means that the bearing has as many ball as can physically fit. Having the maximum number of balls means that there isn't enough space left for a cage. The cage keeps the balls from touching each other. When the balls are allowed to touch and rub, the maximum speed of the bearing is reduced. A full complement of balls increases the load capacity by 30% over a bearing with fewer balls in a cage. (&lt;a href=&quot;http://web.mit.edu/2.75/fundamentals/FUNdaMENTALs%20Book%20pdf/FUNdaMENTALs%20Topic%2010.PDF&quot; rel=&quot;noreferrer&quot;&gt;Reference&lt;/a&gt; page 16)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Number of Balls&lt;/strong&gt; affects the load on each individual ball. The equation for the force per length unit of bearing diameter is:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{F}{ZD^2}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: $Z=\text{Number of balls}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of balls directly affects the load on each bearing and thence the life of the bearing. (&lt;a href=&quot;http://nhbb.com/files/catalog_pages/HiTech_Catalog.pdf&quot; rel=&quot;noreferrer&quot;&gt;reference&lt;/a&gt; page 46)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bearing Life&lt;/strong&gt; is controlled by a number of factors. These factors are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Type&lt;/li&gt;&#xA;&lt;li&gt;Race material&lt;/li&gt;&#xA;&lt;li&gt;Ball material&lt;/li&gt;&#xA;&lt;li&gt;Load&lt;/li&gt;&#xA;&lt;li&gt;Speed&lt;/li&gt;&#xA;&lt;li&gt;Lubrication&lt;/li&gt;&#xA;&lt;li&gt;Cleanliness&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of these are factors that affect the number on cause of bearing failure: &lt;strong&gt;Fatigue&lt;/strong&gt;. (&lt;a href=&quot;http://nhbb.com/files/catalog_pages/HiTech_Catalog.pdf&quot; rel=&quot;noreferrer&quot;&gt;Reference&lt;/a&gt; page 45)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final consideration is &lt;strong&gt;Cost&lt;/strong&gt;. Using standard, off-the-shelf bearings are the cheapest. If one of these bearings can be used in your project, it is quite possible that this the best solution. This is even if the project has to be redesigned slightly to accommodate the bearing. Anything that makes your bearing &lt;em&gt;special&lt;/em&gt; is likely to &lt;a href=&quot;http://www.nmbtc.com/bearings/white-papers/bearing-designing-to-lower-cost/&quot; rel=&quot;noreferrer&quot;&gt;cost a lot more&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the end, &lt;strong&gt;there may not be a definable optimum number of balls&lt;/strong&gt;. The best bearing will likely be the one that meets your &lt;strong&gt;speed&lt;/strong&gt;, &lt;strong&gt;load&lt;/strong&gt; and &lt;strong&gt;life&lt;/strong&gt; requirements while being available for a &lt;strong&gt;reasonable price&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="368" LastEditDate="2015-03-13T03:22:53.640" LastActivityDate="2015-03-13T03:22:53.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2062" PostTypeId="1" AcceptedAnswerId="2067" CreationDate="2015-03-13T04:00:33.250" Score="8" ViewCount="584" Body="&lt;p&gt;I am building a device that needs to be inside of a microwave oven but somehow needs to communicate with another device outside of the microwave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I reflect microwaves so they don't come into contact with the electronics in the device, while still allowing the device to transmit and receive data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency of the microwaves in a microwave oven is pretty much the same frequency that is used in &lt;a href=&quot;http://en.wikipedia.org/wiki/Bluetooth&quot; rel=&quot;nofollow&quot;&gt;Bluetooth&lt;/a&gt;. I was thinking the only possible way of doing this is to have an antenna that is small enough that it can slip outside of a microwave oven through the door (but small enough so the door is still able to close properly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I use some material to protect the electronics of the device from microwaves? Is there a way to have this device communicate with another device outside of the microwave oven without the microwaves interfering with it?&lt;/p&gt;&#xA;" OwnerUserId="889" LastEditorUserId="368" LastEditDate="2015-05-06T15:50:53.173" LastActivityDate="2015-05-06T15:50:53.173" Title="Protecting device from microwaves in microwave oven.. but allowing Bluetooth?" Tags="|mechanical-engineering|electrical-engineering|materials|telecommunication|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2063" PostTypeId="1" CreationDate="2015-03-13T07:59:48.463" Score="9" ViewCount="795" Body="&lt;p&gt;Just wondering what the difference is between 2D flood modelling programs. I am primarily considering whether HECRAS 5.0 (2D - currently in Beta) would be an acceptable alternative for simple 2D flood depth and velocity studies when compared to TUFLOW and MIKE Flood?&lt;/p&gt;&#xA;" OwnerUserId="891" LastEditorUserId="368" LastEditDate="2015-03-13T18:28:16.497" LastActivityDate="2019-08-23T23:02:30.327" Title="Differences between 2D hydrological modelling programs eg. HECRAS 5.0 (2D), TUFLOW, MIKE Flood" Tags="|civil-engineering|hydrology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2064" PostTypeId="2" ParentId="2062" CreationDate="2015-03-13T09:04:32.550" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a way to have this device communicate with another device outside of the microwave oven without the microwaves interfering with it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not with most local wireless communication protocols because they use the same frequency band as a microwave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However on lower power settings the microwave will turn off periodically. That will be your chance to communicate.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-03-13T09:04:32.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2066" PostTypeId="2" ParentId="2062" CreationDate="2015-03-13T12:17:58.087" Score="7" Body="&lt;p&gt;With regard to protecting active electronics inside the oven, you have the choice of either &lt;em&gt;absorbing&lt;/em&gt; the microwaves (e.g., water jacket) or &lt;em&gt;reflecting&lt;/em&gt; the microwaves (any high-conductivity metal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any sort of reflector for microwaves inside an oven must be designed very carefully, because there's a good chance of creating unwanted &quot;hot spots&quot; and/or areas of high field strength (causing arcing).&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-03-13T12:17:58.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2067" PostTypeId="2" ParentId="2062" CreationDate="2015-03-13T13:01:21.803" Score="11" Body="&lt;p&gt;&lt;strong&gt;Your best option for wireless communication in this situation is some kind of optical link.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Microwave_oven#Principles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Consumer microwaves&lt;/a&gt; typically operate at a frequency of 2.45 GHz and &lt;a href=&quot;http://en.wikipedia.org/wiki/Bluetooth&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bluetooth&lt;/a&gt; operates from 2.4 GHz to 2.485 GHz, so any shield you design for one will be equally effective for the other.  You could design a Faraday cage which gives you enough suppression at 2.45 GHz but still allows enough power at higher frequencies (&lt;a href=&quot;http://en.wikipedia.org/wiki/Extremely_high_frequency&quot; rel=&quot;nofollow noreferrer&quot;&gt;EHF&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Terahertz_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;THF&lt;/a&gt;) to communicate.  However, there aren't many devices which operate at those frequencies, and designing one yourself is not a simple feat.  An antenna will likely suffer from the same problem.  If it transmits at 2.4 GHz it will also receive at that frequency, and the load caused by this will likely also destroy your device.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, you should build a Faraday cage around your device &lt;a href=&quot;https://physics.stackexchange.com/a/89090/35024&quot;&gt;which suppresses 2.45 GHz but still allows light to pass through&lt;/a&gt;.  A cage with 0.5 cm holes should give you more than enough suppression since the wavelength of 2.45 GHz is ~12 cm.  Building an optical link may sound complicated, but the hardware is fairly cheap and you might be able to steal some of it from a television remote.  The &lt;a href=&quot;http://en.wikipedia.org/wiki/PSK31&quot; rel=&quot;nofollow noreferrer&quot;&gt;amateur radio&lt;/a&gt; community has some experience building DIY optical communications systems which you can draw on.  You can obtain hardware for the encoding and decoding using either the &lt;a href=&quot;http://en.wikipedia.org/wiki/PSK31&quot; rel=&quot;nofollow noreferrer&quot;&gt;PSK31&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Radioteletype&quot; rel=&quot;nofollow noreferrer&quot;&gt;RTTY&lt;/a&gt; protocols from the same set of devices they use for radio communications.&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2015-03-17T16:18:55.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2068" PostTypeId="1" AcceptedAnswerId="3930" CreationDate="2015-03-13T18:26:07.460" Score="7" ViewCount="3002" Body="&lt;p&gt;Suppose I want to ship my product via UPS, or whatever other professional carrier service. What vibration and shock forces should I design my product to withstand? And how can I effectively test my product against those forces?&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2015-08-13T15:56:09.113" Title="Are there standard vibration and shock specs for shipment of a product?" Tags="|vibration|product-testing|shock|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2069" PostTypeId="1" AcceptedAnswerId="2088" CreationDate="2015-03-13T19:11:53.423" Score="5" ViewCount="893" Body="&lt;p&gt;A common method of repairing old &lt;a href=&quot;http://en.wikipedia.org/wiki/Rivet#High-strength_structural_steel_rivets&quot; rel=&quot;noreferrer&quot;&gt;riveted structures&lt;/a&gt; is to &lt;a href=&quot;http://www.state.nj.us/turnpike/documents/DCA2011SS-13-PAGES.pdf&quot; rel=&quot;noreferrer&quot;&gt;replace any damaged rivets with new high strength bolts&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When this is done, the new bolts are typically installed like they would be on a new structure which results in a bolt pretension. My concern is that the one bolt location might be effectively slip-critical while the rest of the existing rivets are something less than slip-critical. This may concentrate stresses on the one bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When new high strength bolts are installed within an array of existing rivets, is there a concern with stress concentrations resulting from the increased clamping force from the bolt?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-14T18:34:23.593" Title="Are stress concentrations a concern when replacing rivets with high strength bolts?" Tags="|structural-engineering|bolting|rivets|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2070" PostTypeId="2" ParentId="2068" CreationDate="2015-03-13T20:38:43.863" Score="4" Body="&lt;p&gt;When I have received fragile content in the past it has always come with a &lt;a href=&quot;http://www.shockwatch.com/products/impact-indicators/shockwatch-label/&quot; rel=&quot;nofollow&quot;&gt;Shockwatch label&lt;/a&gt; attached to the side.  I'm not 100% sure, but I believe there is some sort of guarantee by the shipper to keep the forces below the level registered by the label (a guarantee which most certainly comes at added cost).  Looking up the specs in the link above, these labels activate if the shock parameters are above:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Force: 25-100 G&lt;/li&gt;&#xA;&lt;li&gt;Duration: 0.5-50 ms&lt;/li&gt;&#xA;&lt;li&gt;Tolerance: $\pm$15%&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These ranges indicate the range of labels available and the tolerance indicates the tolerance once you choose a set of parameters.  If you can design the packaging to handle 100 G for 50 ms, then you should be in the clear.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-03-13T20:38:43.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2071" PostTypeId="1" AcceptedAnswerId="2092" CreationDate="2015-03-13T21:26:35.323" Score="12" ViewCount="857" Body="&lt;p&gt;I was reading up on &lt;a href=&quot;https://en.wikipedia.org/wiki/Fly-by-wire&quot; rel=&quot;noreferrer&quot;&gt;fly-by-wire&lt;/a&gt; development, and I saw a short section about &lt;a href=&quot;http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20070013704.pdf&quot; rel=&quot;noreferrer&quot;&gt;fly-by-wireless&lt;/a&gt; technology. It seems like a great idea, with the potential to lower costs, weight and complexity. I can see a possible scenario where it could be an issue, though:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Two aircraft are very close together (e.g. on a runway or flying in formation).&lt;/li&gt;&#xA;&lt;li&gt;One pilot transmits commands through the aircraft's fly-by-wire system to other parts of the aircraft. &lt;/li&gt;&#xA;&lt;li&gt;The other aircraft accidentally receives the signal because it is so close.&lt;/li&gt;&#xA;&lt;li&gt;Things get very bad very quickly.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The thing is, I haven't been able to find any technical specifications regarding fly-by-wireless systems, and I have no idea if the transmission would be powerful enough to reach the other craft, nor if it would then be interpreted as actual data sent from &lt;em&gt;that&lt;/em&gt; aircraft's pilot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this cross-interference between fly-by-wireless systems possible? If so, how can it be mitigated?&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="368" LastEditDate="2015-05-06T15:50:40.037" LastActivityDate="2015-12-22T03:36:44.723" Title="Is interference between aircraft an issue for fly-by-wireless technology?" Tags="|electrical-engineering|aerospace-engineering|emc|telecommunication|product-testing|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2072" PostTypeId="2" ParentId="1807" CreationDate="2015-03-13T22:44:49.957" Score="10" Body="&lt;p&gt;Design life can be one of two different things, and they aren't interchangeable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reference to '100 years design life' might mean that it's designed for a '1-in-100-years' loading case (wind load, or tidal surge, or whatever).  This is solely about a means of quantifying the magnitude of loading. It is actually nothing whatever to do with the durability of the structure, it's about the strength of the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asks about a different matter - the durability, and specifically the durability of reinforced concrete.  It is quantified normally by reference to past experience in the particular environment telling you what the critical deterioration mechanism is likely to be, and then either reference to a standard solution, or a calculation of the lifetime for that mechanism.  The calculations are normally to some degree empirical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 'standard' structure, with a 'normal' exposure condition, 'normal' concrete characteristics, and 'normal' design life requirements, there will be standard solutions in the relevant design code, that probably simply defines the quantity of cover that will satisfy the design life.  What constitutes 'normal' will depend upon the  jurisdiction of the design code - different blends of cement are available in different parts of the world, and what is 'normal' for a national design code in a wholly-temperate country won't be 'normal' in the tropics or polar regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in a structure in the splash zone in the Arabian peninsula, frost attack is not going to be a problem, but physical salt attack (or salt weathering) will be.  Frost attack is where water freezing in pores and cracks expands and breaks the concrete.  Salt weathering is where salty water is wicked up and evaporates at such a rate that salt crystals grow within the pores and break the concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a designer strays outside what their local design rules consider 'normal', or if the environment is especially aggressive, or the durability requirements are unusually onerous, then a specific calculation will be required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common failure of reinforced concrete is that metallic reinforcement starts to corrode.  Steel in concrete does not corrode because the concrete is very high pH, and steel in a high-pH environment is 'passivated' and does not corrode.  However, slowly over time carbon dioxide from the atmosphere diffuses into the concrete and neutralises it.  If you know the characteristics of your concrete, you can predict how fast that happens (by reference to empirical experience).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What normally actually triggers the corrosion (at least in marine or other salty environments - eg road salt), however, is chloride attack, where chloride ions diffuse in from the surface.  Once the concentration of the chloride ions at the surface of the bar reaches a critical value, corrosion will soon take hold.  You can calculate this, if you assume a concentration of chloride at the surface (from empirical data), and know the characteristics of the concrete (either empirical data, or by testing how fast chloride ions diffuse through it, but beware that as the concrete ages, its characteristics change, and you need to allow for that), and know the critical threshold (from empirical data).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a handy free program that does this calculation for you called &lt;a href=&quot;http://www.life-365.org&quot; rel=&quot;nofollow&quot;&gt;Life-365&lt;/a&gt;, and it comes out of an American Concrete Institute committee. It does the chloride diffusion calculation for you, draws graphs and stuff, and if you're in the USA it even has the empirical data you need built-in so you don't need to look up what the local conditions are. (I use the program, but am not associated with it otherwise). The manual to the program has more detailed discussion of the science behind it, but the best thing is you can just play with it and see what effect changing something has on the life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do the calc and you don't get enough life, then either you can put the reinforcement deeper (so it takes longer for the chloride to diffuse to it), or you make the concrete more resistive to chloride diffusing through it, or you use bar that needs a higher threshold value of chloride (stainless, say), or you surface treat the bar, or the concrete, or you put in galvanic or electro-chemical systems, or corrosion inhibitors, or something else.  Lots of this stuff comes back to empirical data - they've tested it, and have test data that shows it will prevent corrosion for n years if you put in x amount of whatever.&lt;/p&gt;&#xA;" OwnerUserId="902" LastEditorUserId="902" LastEditDate="2015-03-16T18:24:28.303" LastActivityDate="2015-03-16T18:24:28.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2073" PostTypeId="2" ParentId="1706" CreationDate="2015-03-13T23:10:25.913" Score="5" Body="&lt;p&gt;Which is more significant depends upon the bridge in question, certainly its length and the characteristics of the loading applied.  For this discussion I am assuming highway traffic loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question refers to dynamic effects, and it is worth noting that this is more than just impact from striking a pothole.  If an elastic single span simple beam is instantaneously loaded with a force, the peak deflection that results is twice the deflection under the same force in the static case, and this effect is unrelated to the effect of instantaneous spikes in applied loading due to vehicles hitting surface irregularities (potholes etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most highway bridges, I believe the dynamic effect is more significant.  Fast-but-spaced-out traffic is more onerous than slow-and-bunched traffic.  However, this conclusion is based upon the observation that there are more short bridges (spans of up to tens of metres) than long bridges - there is not a general principle that gives one universal answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's relatively simple to determine that for a very short bridge whether the traffic is queuing or not is irrelevant - if the bridge is shorter than one vehicle, only one vehicle (or one axle) will be on the bridge, so whether there is a queue or not does not influence the number of vehicles loading the structure or the loading.  Conversely it's easy to imagine that a very long bridge (several hundreds of metres long), one vehicle hitting a pothole will have a negligible effect, since even if the load from that one vehicle doubles instantaneously, if there are hundreds of vehicles on the deck it won't have a proportionately great effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In UK practice, trunk road bridges are designed and assessed to Highways Agency documents (so-called 'BD's and 'BA's).  The vehicular highway loading is in two 'flavours' - HA is 'normal' traffic, and HB is an arbitrary loading used to examine the characteristics of the bridge under abnormal loading.&#xA;HA loading for design is defined in &lt;a href=&quot;http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol1/section3/bd3701.pdf&quot; rel=&quot;nofollow&quot;&gt;BD37&lt;/a&gt; and the derivation includes allowance for impact - read Appendix A: &quot;the impact effect of an axle on highway bridges can be as high as 80% of the static axle weight and an allowance of this magnitude was made in deriving the HA loading&quot;, though the influence of impact reduces as the span becomes greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traffic queues not only give rise to nose-to-tail bunching, they potentially give rise to vehicles squeezing closer side-to-side.  In the BDs this is referred to as 'lateral bunching', which is where more vehicles crowd onto the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol1/section3/bd3701.pdf&quot; rel=&quot;nofollow&quot;&gt;BD37&lt;/a&gt; allows for both impact and lateral bunching simultaneously - ie, it assumes that you have a tightly-packed traffic jam that's also travelling at high speed.  This obviously doesn't happen, but is what the code encapsulates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When assessing existing structures, however, the UK standards don't apply both effects together.  &lt;a href=&quot;http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol3/section4/bd2101.pdf&quot; rel=&quot;nofollow&quot;&gt;BD21&lt;/a&gt; is the code for assessing structures.  Clause 5.23 specifically addresses this question (UDL and KEL are two component parts of HA loading):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The HA UDL and KEL have been derived using a lateral bunching factor to take into account the possibility that, in slow moving situations, more lanes of traffic than the marked or notional lanes could use the bridge. Probabilistic analysis shows that maximum impact effects, which occur at high speeds, should not be considered together with maximum lateral bunching. Comparison of the effects of alternative traffic speed and bunching situations have led to the conclusion that high speed high impact effect with no lateral bunching is the most onerous criterion for bridge loading. The HA UDL and KEL are therefore to be adjusted in order to eliminate the lateral bunching factor by dividing by&#xA;the following Adjustment Factor (AF)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The adjustment factor is a relatively large number for loaded lengths up to 20m, but tails off to 1.0 (ie divide by 1.0, so don't change the value) at 40m loaded length.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not possible to make definitive statements from this (ie, you can't say &quot;below 20m it's dynamic, above 40m it's traffic queues&quot;), because the loading is to some degree empirical and derived after a probabilistic analysis, and includes guesswork (described as &quot;by estimation&quot; in the standard) with allowance for the findings of sensitivity analysis.  Again, BD37&lt;a href=&quot;http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol1/section3/bd3701.pdf&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; Appendix A has discussion of this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;For long loaded lengths, the main factors affecting the loading are the traffic flow rates, percentage of heavy vehicles in the flows, frequency of occurrence and duration of traffic jams and the spacing of vehicles in a jam.  These parameters were determined by studying the traffic patterns at several sites on trunk roads, by load surveys at other sites and, where the required data was unobtainable, by estimation. A statistical approach was adopted to derive characteristic loadings from which nominal loads where obtained. Sensitivity analyses were carried out to test the significance on the loading of some of the assumptions made.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that the above is a bit non-rigorous with respect to spans.  The length that matters with respect to load derivation is the 'loaded length', which is not always the same thing as the span.  For a single simply-supported span, if you're examining the bending effect, the two are synonymous, but very many bridges are more complex than that (eg, multiple continuous spans, or integral abutments etc).  Loaded length is the length over which the element of structure is loaded, and when designing you should choose the length that gives rise to the most onerous effect for the element you are designing for doing the calculations with respect to that particular element.  This is often the whole length of a span, but there are cases where (eg) loading a shorter length has a greater effect, particularly in continuous structures, partly because the shorter the loaded length used, the greater the intensity of load to be applied.&lt;/p&gt;&#xA;" OwnerUserId="902" LastEditorUserId="902" LastEditDate="2015-03-17T13:46:00.383" LastActivityDate="2015-03-17T13:46:00.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2074" PostTypeId="1" AcceptedAnswerId="2075" CreationDate="2015-03-14T01:30:55.137" Score="8" ViewCount="258" Body="&lt;p&gt;I am a programmer facing an electronic question, so I thought here is the place to ask!&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I have a magnetic field sensor which provides me with magnetic field values (XYZ axis) 250 times a second. &lt;/li&gt;&#xA;&lt;li&gt;An electronic circuit with a programmable microprocessor controls a coil, which can change the magnetic field close enough to this sensor. Yet, the two devices are not strictly positioned, so I cannot relay on positioning measurement.&lt;/li&gt;&#xA;&lt;li&gt;I wish to send 2 different types of signals from the circuit to the sensor, which can withstand strong noises on the magnetic field, and be shorter than half a second (Yes, seconds - we are in 250 Hz !)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Current solution is to transmit a 17Hz square wave, then 12Hz, then 17Hz for signal A, reversing the order for signal B. But in order to detect these signals it is needed to send long enough waves, which brings the signals to length around 1.5 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question here is: Is there a way, let's say a pattern, like musical pattern that can be used to signal faster, and still be reliable?&lt;/p&gt;&#xA;" OwnerUserId="903" LastEditorUserId="33" LastEditDate="2015-03-14T01:40:28.497" LastActivityDate="2015-03-14T02:17:37.947" Title="Best way to send a signal in a very low frequency (250Hz)?" Tags="|electrical-engineering|signal|magnets|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2075" PostTypeId="2" ParentId="2074" CreationDate="2015-03-14T02:17:37.947" Score="7" Body="&lt;p&gt;With a sample rate of 250 Hz, you could easily send 100 bits/second or more. In 0.5 second, that would be on the order of 50-60 bits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, what bit patterns should you select that are readily distinguishable from each other and also from outside noise and interference? A common solution is to use pseudorandom bit sequences, also known as &lt;a href=&quot;http://en.wikipedia.org/wiki/Barker_code&quot;&gt;Barker Codes&lt;/a&gt;, that have the desired low cross-correlation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A DSP technique known as &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Matched_filter&quot;&gt;matched filter&lt;/a&gt;&quot; can be used to detect the codes.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-03-14T02:17:37.947" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2076" PostTypeId="1" CreationDate="2015-03-14T04:13:05.630" Score="4" ViewCount="1500" Body="&lt;p&gt;I came accross this word tesla turbine while reading about the applications of boundary layer flow but couldn't find much details of such turbine anywhere.&lt;/p&gt;&#xA;" OwnerUserId="904" LastActivityDate="2021-10-04T10:52:00.007" Title="What is a tesla turbine and how is it different from other turbines(like pelton or francis turbine)?" Tags="|mechanical-engineering|fluid-mechanics|turbines|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2078" PostTypeId="2" ParentId="2059" CreationDate="2015-03-14T09:14:48.157" Score="5" Body="&lt;p&gt;The answer to the headline question is yes, it is equal.  It physically has to be equal, since you can't have two different values of a particular directional stress at one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quoted equation is for shear stress, where&#xA;V = total shear force at the location in question;&#xA;Q = first moment of area 'beyond' the point considered;&#xA;t = thickness in the material perpendicular to the shear;&#xA;I = Moment of Inertia of the entire cross sectional area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem seems to be the assumption that the cross-sectional area above and below the neutral axis is equal (ie, the embedded question &quot;the area both above and below the Neutral Axis should be equal ... is this statement correct&quot; to which the answer is no)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neutral axis is the line about which the first moment of area each side is equal, not the line about which the areas each side are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying that equation above and below the neutral axis, V an I are properties of the whole section so identical wherever you do the calculation on this cross-section, t is a function of the level at which the calculation is done so identical in both applications, and Q is equal by definition (that's the definition of the neutral axis), so the answer has to be identical since all the numbers substituting into the equation are identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: for the diagram shown, if T=10, W=60, H=70, then the tee shown has a cross-section of 1200, and both flange and stem each have area 600.  However, the neutral axis is NOT at the bottom surface of the flange (ie, is not where there is equal area above and below).  In this example, the neutral axis is at 22.5 from the top surface, or 12.5 below the level at which the areas are equal either side.  So there's 725 area above the neutral axis, and 475 below the neutral axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating for above the neutral axis, there are two rectangular regions,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Q = 10 x 60 x 17.5 + 12.5 x 10 x 6.25 = 11281.25&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Calculating for below the neutral axis,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Q = 47.5 x 10 x 23.75 = 11281.25&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I = 552500 and t = 10 at the neutral axis, so whatever value of V is used, $\tau$ is the same.&lt;/p&gt;&#xA;" OwnerUserId="902" LastEditorUserId="902" LastEditDate="2015-03-16T13:43:10.863" LastActivityDate="2015-03-16T13:43:10.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2079" PostTypeId="2" ParentId="2076" CreationDate="2015-03-14T10:35:01.750" Score="7" Body="&lt;p&gt;While &amp;quot;conventional turbines operate by imparting impulse or reaction energy from an incoming fluid using &amp;quot;buckets&amp;quot; or spiral surfaces on the turbine &amp;quot;runner&amp;quot;, a &amp;quot;Tesla Turbine&amp;quot; is a bladeless or &amp;quot;drag turbine&amp;quot;. A Tesla turbine's disks are bladeless and essentially smooth and the &amp;quot;drag&amp;quot; from boundary layer interaction of the fluid passing over the essentially smooth disk surfaces imparts energy from the fluid flowing across the disk surface.&lt;/p&gt;&#xA;&lt;p&gt;A simplistic view of boundary layer effect is that in a stream of fluid flowing across a surface, immediately adjacent to the surface the fluid is stationary, at some distance away from the surface the fluid is at or close to its mean velocity and the velocity varies from zero to mean across the intervening &amp;quot;boundary layer&amp;quot;.  Moving from a simplistic to &amp;quot;somewhat accurate&amp;quot; model of boundary later effects can take as much of your life as you are willing to give it - but good approximations are available.&lt;br /&gt;&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Boundary_layer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;br /&gt;&#xA;&lt;a href=&quot;https://www.google.co.nz/search?q=boundary+layer&amp;amp;biw=1536&amp;amp;bih=860&amp;amp;tbm=isch&amp;amp;tbo=u&amp;amp;source=univ&amp;amp;sa=X&amp;amp;ei=kRcEVZyRCdTp8AWqnoDYAg&amp;amp;ved=0CEQQsAQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Many useful glimpses ...&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;In practice a Tesla Turbine consists of two or more rotating disks mounted on one shaft with fluid injected between the disks.   Fluid is injected (usually) tangentially at the disk outer edge and spirals towards the centre as it imparts energy by &amp;quot;drag&amp;quot; to the disks either side of it that it is flowing between.&#xA;The inverse of the turbine is the Tesla pump, which imparts energy to the fluid by boundary layer drag as the disk rotates as fluid is supplied to the disk surface.&lt;/p&gt;&#xA;&lt;p&gt;As with most of Tesla's legacy, there are some good practical useful ideas involved &lt;strong&gt;and/but&lt;/strong&gt; there is a lot of hype rubbish and misinformation surrounding them.&lt;/p&gt;&#xA;&lt;p&gt;A Tesla Turbine or pump is able to achieve reasonably good efficiency compared to an impulse or reaction transfer equivalent, &lt;strong&gt;providing&lt;/strong&gt; flow is essentially laminar across the disk surfaces. When flow is turbulent efficiency falls. Laminar flow is maintained only at relatively low interdisk spacings which also limits the absolute flow rate or power available for a given size. Many devices which use the basic principle claim laminar flow while not achieving it. This includes a relatively recent Kickstarter project which claimed laminar flow and quite a lot of other &amp;quot;magic stuff&amp;quot; but just reading their own provided material revealed the contradictions.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Phoenix [Tesla] Turbine builder club - &lt;a href=&quot;http://phoenixnavigation.com/ptbc/data.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;USEFUL CHARTS &amp;amp; DATA&lt;/strong&gt;&lt;/a&gt; - Much from Tesla directly.  USE WITH CARE. His comment cited from that site confirms my above comments about throughput and efficiency, without mentioning laminar flow per se.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&amp;quot;Owing to a number of causes affecting the performance, it is difficult to frame a precise rule which would be generally applicable, but it may be stated that within certain limits, and other conditions being the same, the torque is directly proportionate to the square of the velocity of the fluid relatively to the runner and to the effective area of the disks and, inversely, to the distance separating them. The machine will, generally, perform its maximum work when the effective speed of the runner is one-half of that of the fluid; but to attain the highest economy, the relative speed or slip, for any given performance should be as small as possible. This condition may be to any desired degree approximated by increasing the active area of and reducing the space between the disks.&amp;quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://phoenixnavigation.com/ptbc/links.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Their links page&lt;/a&gt; - MANY Tesla Turbine links, and more - useful.&lt;/p&gt;&#xA;&lt;p&gt;instr&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Tesla_turbine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia - TTs&lt;/a&gt; - reasonable.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I've only skimmed it quickly but it looks like quite a good practical feel for the TT could be obtained from this 1950 PhD thesis - typewritten and rather hard to read, but looks useful.&lt;br /&gt;&#xA;&lt;a href=&quot;https://ia800409.us.archive.org/12/items/leaman/leaman.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;THE DESIGN, CONSTRUCTION AND INVESTIGATION OF A TESLA TURBINE&lt;/strong&gt;&lt;/a&gt;&lt;br /&gt;&#xA;[Link updated October 2021] 1950 Thesis. 89 pages&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://peswiki.com/index.php/Directory:Tesla_Turbines&quot; rel=&quot;nofollow noreferrer&quot;&gt;PESWIKI&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Build a 15,000 Rpm Tesla Turbine Using Hard Drive Platters&lt;br /&gt;&#xA;&lt;a href=&quot;http://www.instructables.com/id/Build-a-15,000-rpm-Tesla-Turbine-using-hard-drive-/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Instructables&lt;/a&gt; DIY using hard disk platters.&lt;br /&gt;&#xA;Far better than most instructables pages.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://poseidon01.ssrn.com/delivery.php?ID=047093101091071100065126094101080067010037073076067052094080122127029026113084013126099054038127042010098069085113105002067094098048031000085085002097005066096080109004086041085030081091066072005091026124099002086015027103071116088002086120022111003078&amp;amp;EXT=pdf&amp;amp;INDEX=TRUE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Design, Construction and Simulation of Tesla Turbine&#xA;&lt;/a&gt; 9 pages - 2021&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://digital.library.unt.edu/ark%3A/67531/metadc67979/m2/1/high_res_d/thesis.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;TESLA TURBINE TORQUE MODELING&#xA;FOR CONSTRUCTION OF A DYNAMOMETER AND TURBINE&lt;/a&gt; - MSc thesis, 2011. 92 pages&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ijeat.org/wp-content/uploads/papers/v8i4/D6031048419.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Design, Manufacturing and Testing of a Hybrid&#xA;Tesla Turbine&lt;/a&gt; 4 pages 2019&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=THE+DESIGN%2C+CONSTRUCTION+AND+INVESTIGATION+OF+A+TESLA+TURBINE&amp;amp;rlz=1C1CHBF_enNZ834NZ839&amp;amp;oq=THE+DESIGN%2C+CONSTRUCTION+AND+INVESTIGATION+OF+A+TESLA+TURBINE&amp;amp;aqs=chrome..69i57j69i64l2.696j0j4&amp;amp;sourceid=chrome&amp;amp;ie=UTF-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Many links&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2021-10-04T10:52:00.007" LastActivityDate="2021-10-04T10:52:00.007" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2080" PostTypeId="2" ParentId="2071" CreationDate="2015-03-14T11:25:37.200" Score="8" Body="&lt;p&gt;Modern communications systems are able to achieve essentially any desired data rate and degree of message integrity in any well-enough-defined environment. It's &quot;just a matter&quot; of the degree of effort, complexity and $ that are needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Designing a system to meet any desired data rate, number of users, and integrity is thus &quot;just a matter of engineering&quot;. Proximity, signal strength, interfering signals, ... are 'just part of the system specification'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things go bad when the specification and reality agree to differ and/or when properly failsafe design is not used. &quot;Defining the problem&quot; is arguably the hardest and most important part of such a system.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-03-14T11:25:37.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2081" PostTypeId="1" AcceptedAnswerId="2105" CreationDate="2015-03-14T12:39:34.807" Score="8" ViewCount="3311" Body="&lt;p&gt;Trucks and smaller vehicles with diesel engines run on diesel fuel which looks more or less like kerosene - less flammable than gasoline, low viscosity liquid. However larger diesel engines &lt;a href=&quot;https://en.wikipedia.org/wiki/W%C3%A4rtsil%C3%A4-Sulzer_RTA96-C&quot; rel=&quot;nofollow&quot;&gt;like this 120 thousand horsepower marine engine&lt;/a&gt; have more or less the same design but use fuel oil which looks very different from diesel fuel - much higher viscosity and I'd guess igniting fuel oil at room temperature is a challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does it happen that engines of the same design use two different fuels? If one of them is superior to the other then why won't they all stick to that superior fuel?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="6790" LastEditDate="2020-05-02T19:34:30.470" LastActivityDate="2020-05-02T19:34:30.470" Title="Why do some diesel engines run on diesel fuel and others on fuel oil?" Tags="|mechanical-engineering|combustion|engines|diesel|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2082" PostTypeId="1" AcceptedAnswerId="2090" CreationDate="2015-03-14T12:58:17.830" Score="7" ViewCount="2176" Body="&lt;p&gt;Suppose I have two parts with holes aligned and I have to install a solid rivet into the &quot;combined&quot; hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the rivet diameter must be slightly smaller than the hole diameter, otherwise it simply won't fit. So I heat this slightly smaller rivet, insert it into the hole and then deform its tail so that the rivet now has two heads and sit properly in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume I follow all the procedures and best practices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the deformation only affect the tail or does it also make the middle of the rivet expand and fill the gap between the hole walls and the rivet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if I wait till the rivet cools down and then grind the rivet head off - will it leave its place easily or will I have to force it out?&lt;/p&gt;&#xA;" OwnerUserId="374" LastActivityDate="2015-03-16T15:58:56.750" Title="Do solid rivets fully fill the holes when installed properly?" Tags="|structural-engineering|steel|rivets|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2083" PostTypeId="1" AcceptedAnswerId="2089" CreationDate="2015-03-14T13:08:37.043" Score="5" ViewCount="844" Body="&lt;p&gt;AFAIK when you heat up a piece of steel to high temperature and let it cool down slowly (as opposed to shock cooling by say submerging it into water) steel gets softer - that's called tempering. Steel getting softer means it can bear less load without getting deformed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we connect parts of steel skyscraper with steel rivets. We heat them up in fire and then put them into holes and deform their tails. While the rivet is inserted inside a rather cool surrounding construction (and being deformed) it cools down which more or less resembles tempering process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it looks like installing rivets leaves them tempered and so rather soft and prone to deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this process of rivets getting softer and &quot;weaker&quot; accounted for?&lt;/p&gt;&#xA;" OwnerUserId="374" LastActivityDate="2016-02-05T22:32:11.520" Title="How is steel tempering addressed when installing pre-heat rivets?" Tags="|structural-engineering|steel|rivets|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2084" PostTypeId="2" ParentId="2071" CreationDate="2015-03-14T14:19:25.577" Score="9" Body="&lt;p&gt;To add to what Russell said, it would take a very stupidly designed protocol to allow the comands from one aircraft to control the other, even assuming that cross-reception is good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a everyday example, think of a bunch of people talking on cell phones standing near each other.  One person accidentally receiving another's conversation just doesn't happen.  This has to be deliberately designed into the protocols, but that's no different from designing the landing gear robust enough to hold the weight of the aircraft.  It's just something you'd do as a normal part of the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My fear about this kind of system is not cross-communication, but overpowering interference.  You have to make sure that the other nearby aircraft's transmissions didn't overwhelm your receivers to the point where they can't hear your signals anymore.  That shouldn't be too hard to do either assuming everyone plays nice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, guarding against deliberate outside interference would not be so easy.  The other aircraft, if designed to do so, could deliberately radiate orders of magnitude more power to your receivers than your own transmitters do.  The protocols would make it essentially impossible for them to take over your aircraft, but your controls would no longer function.  This is basically like a DoS (denial of service) attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt we will see these kinds of systems widely deployed, at least in commercial and military aircraft, until they are considered safe from this kind of attack.  Otherwise, a terrorist with a steerable parabolic dish antenna could cause some serious trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other alternatives to using lots of heavy copper cable for communication, like fiber optics and various kinds of multiplexing.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2015-03-17T16:02:54.527" LastActivityDate="2015-03-17T16:02:54.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2085" PostTypeId="1" AcceptedAnswerId="2093" CreationDate="2015-03-14T14:57:35.473" Score="4" ViewCount="2214" Body="&lt;p&gt;I was once told that the speed limits on banked highway curves (specifically off-ramps) were determined by assuming zero friction between the car and the road, such that as long as you stayed on the correct path of travel, the banking of the curve would prevent a vehicle from sliding outwards. That is, the angle of the banking would mean that the portion of your gravity vector normal to the road surface provides enough centripetal force without counting on your wheels to provide any 'sideways' friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a civil engineer and have no familiarity with the relevant codes, but I was wondering if this is true, or a myth. If it's not true, is there a simple formula used to set these speed limits, or is it a much more complicated procedure/judgment?&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-03-16T00:55:00.293" Title="How are maximum speed limits determined for banked curves?" Tags="|civil-engineering|highway-engineering|transportation|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2086" PostTypeId="2" ParentId="2081" CreationDate="2015-03-14T15:00:20.577" Score="4" Body="&lt;p&gt;Simple economics. Marine engines consume enormous amounts of fuel, so in order to reduce operating costs, they use the cheapest, least desirable sludge that the oil refineries can produce.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-03-14T15:00:20.577" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2088" PostTypeId="2" ParentId="2069" CreationDate="2015-03-14T18:24:22.627" Score="2" Body="&lt;p&gt;According to AISC J1.9 rivets can be considered as sharing the load with high strength bolts, specifically in joints designed as slip-critical as per J3. The apply this section to both new work and alterations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The commentary on this section points out that this is because the ductility of the rivets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you do not need to be concerned with the distribution of load across the various fasteners as long as your holes are appropriately sized for slip critical connections. What's not clear to me is if the faying surface requirements of RCSC for slip critical connections would apply.These would obviously be hard or impossible to verify when replacing just some fasteners in an existing connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second page of &lt;a href=&quot;http://msc.aisc.org/globalassets/modern-steel/steel-interchange/2004/2004v09_si_web.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; Modern Steel Construction article suggests that while the riveted connection is not slip critical, simply tightening the new bolts as if it were is enough to consider the bolts and rivets to share the load, but won't make the joint slip-critical as a whole.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.boltcouncil.org/files/2ndEditionGuide.pdf&quot; rel=&quot;nofollow&quot;&gt;RCSC Design Guide&lt;/a&gt; (PDF link) for bolted and riveted connections has a more detailed explanation in section 14.2.2, which suggests that in connections that are under significant load, the AISC rule may be a simplification.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-03-14T18:34:23.593" LastActivityDate="2015-03-14T18:34:23.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2089" PostTypeId="2" ParentId="2083" CreationDate="2015-03-14T19:48:48.230" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://law.resource.org/pub/us/cfr/ibr/003/astm.a502.1976.pdf&quot;&gt;ASTM A502&lt;/a&gt; covers rivets for structural situations. The linked version of the standard is from 1976, but it is probably safe to assume that riveting technology hasn't changed much since then. The latest &lt;a href=&quot;http://www.astm.org/Standards/A502.htm&quot;&gt;ASTM A502&lt;/a&gt; standard is from 2009 though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rivets are made from what is typical structural steel to begin with. This means that weathering, &quot;Grade 3&quot; rivets are made from &lt;a href=&quot;http://www.astm.org/Standards/A588.htm&quot;&gt;ASTM A588&lt;/a&gt; steel. A588 is a common designation for structural steel with atmospheric corrosion resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual rivet isn't heated to a high enough temperature to change the steel structure as this &lt;a href=&quot;http://mysite.du.edu/~jcalvert/tech/rivets.htm&quot;&gt;article on rivets states&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Rivets were heated to 950°F - 1050°F, handbooks say. This is not hot enough for an austenitic transition, so it must only have been for causing thermal expansion...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Putting all of this together means that rivets start out with the material properties of standard structural steel and aren't heated enough to change those properties.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-14T19:48:48.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2090" PostTypeId="2" ParentId="2082" CreationDate="2015-03-14T20:32:37.370" Score="7" Body="&lt;p&gt;The short answer is: &lt;strong&gt;Yes, as the rivet head is formed, the shank of the rivet is also deformed to fill the hole.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Just like structural bolts, rivets are installed in holes that are 1/16 inch larger in diameter than the rivet. This clearance allows for manufacturing imperfections and clearance for easily placing the rivet in the hole.&lt;/p&gt;&#xA;&lt;p&gt;A US military specification for riveting (&lt;a href=&quot;http://www.engineersedge.com/MIL-R-47196A.pdf&quot; rel=&quot;noreferrer&quot;&gt;MIL-R-47196A&lt;/a&gt;) states in 3.3.3.3:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The driven rivet shall completely fill the hole...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The figures in that MIL Spec show that rivets that do not completely fill the hole are one of the ways that a rivet can be unacceptable.&lt;/p&gt;&#xA;&lt;p&gt;An image of this from a &lt;a href=&quot;https://mechanicalinfo.wordpress.com/tag/cold-riveting/&quot; rel=&quot;noreferrer&quot;&gt;web page on riveting&lt;/a&gt; shows this as well:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Vu11.jpg&quot; alt=&quot;Rivet diagram&quot; /&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Video of Hydraulic Riveting&lt;/h3&gt;&#xA;&lt;p&gt;You can also see &lt;a href=&quot;https://www.youtube.com/watch?v=I9Q5OHYDbvo&quot; rel=&quot;noreferrer&quot;&gt;video of a hydraulic rivet on YouTube&lt;/a&gt; that shows just how much deformation is involved with riveting.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-03-16T15:58:56.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2091" PostTypeId="1" AcceptedAnswerId="2140" CreationDate="2015-03-14T21:32:54.417" Score="7" ViewCount="1726" Body="&lt;p&gt;Some roadway intersections that are near railroad tracks have signs that light up when a train is approaching. These signs warn that certain turns are not allowed because of the train. One of these signs is shown below from the &lt;a href=&quot;http://www.edmonton.ca/transportation/ets/safety_security/lrt-traffic-signs-and-signals.aspx&quot; rel=&quot;noreferrer&quot;&gt;City of Edmonton&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WDkw5.jpg&quot; alt=&quot;No right turn on train sign&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that normal train crossing signals are the responsibility of the railroad and that traffic signals are the responsibility of the highway department. This wouldn't normally seem like a big problem, but these are two completely different systems and organizations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously there is some way that the two owners coordinate and communicate the train warning information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the train warning signal get passed from the railroad signal to the traffic signal? Is this as simple as a wire run from the train signal that is &quot;high&quot; while lights are on? Is there standard way that this connection is done?&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2018-11-07T23:45:18.610" Title="How are train crossing signals and highway traffic signals coordinated?" Tags="|rail|traffic-intersections|traffic-light|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2092" PostTypeId="2" ParentId="2071" CreationDate="2015-03-14T21:44:00.663" Score="10" Body="&lt;p&gt;Cross-interference between aircraft is a high unlikely event because all commercial aircraft designs have to pass &lt;a href=&quot;https://en.wikipedia.org/wiki/DO-160&quot; rel=&quot;nofollow noreferrer&quot;&gt;DO-160&lt;/a&gt; environmental testing requirements. Among the &lt;a href=&quot;https://en.wikipedia.org/wiki/DO-160&quot; rel=&quot;nofollow noreferrer&quot;&gt;DO-160&lt;/a&gt; testing specification is EMI/EMC Testing. These tests include Radiated Emission, Interference and Immunity Testing.  Part of these tests are to answer &quot;&lt;em&gt;Two aircraft are very close together&lt;/em&gt;&quot; and &quot;&lt;em&gt;The other aircraft accidentally receives the signal because it is so close.&lt;/em&gt;&quot; Below is a picture of an &lt;a href=&quot;https://en.wikipedia.org/wiki/Anechoic_chamber&quot; rel=&quot;nofollow noreferrer&quot;&gt;anechoic chamber&lt;/a&gt; used to test aircraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/e1kA4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fly-by-wire allows the aircraft control system to access and command monitoring and control systems. The monitoring and control systems have a specific address similar to an internet protocol (IP) address. This enables the aircraft control system to address specific devices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aircraft testing protocols are similar to automobile testing protocols except more stringent. But all engineering designs are not foolproof. To mitigate risk most engineers use a process called &lt;a href=&quot;https://en.wikipedia.org/wiki/Failure_mode_and_effects_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;DFMEA (Design Failure Mode Effect Analysis)&lt;/a&gt;. Unfortunately there are still a few unfortunate incidents. The most recent examples are &lt;a href=&quot;https://en.wikipedia.org/wiki/Malaysia_Airlines_Flight_370&quot; rel=&quot;nofollow noreferrer&quot;&gt;the loss of MH370&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/2009%E2%80%9311_Toyota_vehicle_recalls#Accelerator_pedal_recall&quot; rel=&quot;nofollow noreferrer&quot;&gt;Toyota vehicle recalls&lt;/a&gt; for &lt;a href=&quot;http://en.wikipedia.org/wiki/Sudden_unintended_acceleration&quot; rel=&quot;nofollow noreferrer&quot;&gt;sudden unintended acceleration&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.rtca.org/store_product.asp?prodid=759&quot; rel=&quot;nofollow noreferrer&quot;&gt;DO-160F Environmental Conditions and Test Procedures for Airborne Equipment&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.emcintegrity.com/military-and-aerospace/mil-std-461&quot; rel=&quot;nofollow noreferrer&quot;&gt;MIL-STD-461  EMC Integrity: EMC emissions and immunity&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-12-22T03:36:44.723" LastActivityDate="2015-12-22T03:36:44.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2093" PostTypeId="2" ParentId="2085" CreationDate="2015-03-15T01:45:00.377" Score="7" Body="&lt;p&gt;This information comes from a &lt;a href=&quot;http://www.iowadot.gov/design/dmanual/02a-02.pdf&quot;&gt;design document&lt;/a&gt; by the Iowa DOT (US). It might not apply everywhere in the world, but the considerations are probably universal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarity, the amount of banking of a turn is typically called &lt;em&gt;super elevation&lt;/em&gt;. At least in the US, this is given as a percentage for roadways.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;strong&gt;maximum&lt;/strong&gt; super elevation is &lt;strong&gt;8%&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The typical &lt;strong&gt;high-speed&lt;/strong&gt; super elevation is &lt;strong&gt;6%&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The typical &lt;strong&gt;urban&lt;/strong&gt; super elevation is &lt;strong&gt;4%&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;6%&lt;/strong&gt; typical super elevation was chosen for:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This reduces the risk of slow moving vehicles sliding down a superelevated roadway during winter conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Speed&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Design speed of a highway is not always the same as the posted speed. Drivers have this habit of not following the speed limit. Designers know this and make the roadway safe at these increased speeds as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are only two ways that speed will affect the curve radius. Either:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The design speed has been set and the geometry limits are based off of that speed.&lt;/li&gt;&#xA;&lt;li&gt;The geometry has been set and the design speed is reduced to still provide adequate levels of safety.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Curve Radius&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the super elevation and design speed are taken into consideration to determine the minimum radius of the curve. These also consider the maximum side friction from both the vehicle's tires and the comfort of the driver. These values are listed in tables for easy look up or in a spreadsheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friction of the vehicle's tires on the road is incredibly variable. It depends on the weather, condition of the road, condition of the tires, etc. Because of this, the values used for design are very conservative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other factor controlling curve design is driver comfort. It is surprising how ofter roadway design is controlled by human factors. In this case, the typical driver will be concerned about the amount of horizontal acceleration they are feeling and slow down before the vehicle is in danger of sliding sideways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Result&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the end, there are limits on the super elevation of curves that are based on low friction. But this is the controlling factor at low speed and not high speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These super elevations are no where near the amount that would be required for no side friction. As you can imagine you never know when there will be a traffic jam on a road because of a wreck. You don't want trucks to tip over toward the center of the curve if they are forced to stop!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum curve radius is then determined based on looking up the information in tables. Often there are other concerns that also might after the curve such as sight distance or vertical curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How it works in design&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting the roadway cross section details (of which super elevation is one) is usually done last in the sequence of design. The sequence usually goes like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Set design details, e.g. design speed, road type, number of lanes, etc.&lt;/li&gt;&#xA;&lt;li&gt;Set the horizontal alignment. This would be where the curve radius is set. This is also where geometric constraints are found that might require reducing the design speed.&lt;/li&gt;&#xA;&lt;li&gt;Set vertical alignment. This stage can also reveal more geometric constraints.&lt;/li&gt;&#xA;&lt;li&gt;Set details of cross section, e.g. super elevation, cross slopes, embankment slopes, ditches, etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-16T00:55:00.293" LastActivityDate="2015-03-16T00:55:00.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2094" PostTypeId="1" AcceptedAnswerId="2095" CreationDate="2015-03-15T03:24:33.990" Score="9" ViewCount="939" Body="&lt;p&gt;I'm an artist building a project which one could think of as a &quot;sliding door&quot; – a very narrow panel running on grooved wheels riding in tracks at top and bottom. It will be motorized and driven by a stepper motor, timing belt and pulleys. I will probably end up driving it from top and bottom too so it doesn't rack. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the gist of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/drVuw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt; does it matter if the load is on the top set of wheels or the bottom set? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this is a dead simple &quot;more efficient to push or to pull&quot; type question but the answer is not obvious to me.&lt;/p&gt;&#xA;" OwnerUserId="896" LastActivityDate="2015-03-16T16:28:18.150" Title="&quot;Sliding door&quot; - is top weight bearing more efficient?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2095" PostTypeId="2" ParentId="2094" CreationDate="2015-03-15T05:57:58.623" Score="9" Body="&lt;p&gt;Driving from the top is better. If you think about it on an intuitive level, when you pull from the top, the weight of the unit 'wants' to stay under the wheels, so your door tends to stay straight up and down. If you pull from the bottom, the weight of the door is tempted to lag behind and tip over, causing the whole mechanism to bind. This is because the center of mass of the door lags slightly behind the driven end as the door starts and stops moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If everything were perfectly sized and aligned, this wouldn't be a problem, but in practice, driving from the bottom will be more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to keep in mind is the risk of over-constraint. If the tracks at the top and the bottom are not very well lined up with each other, the wheels are likely to get stuck. In entertainment, we build mechanisms like this all the time (we call them 'elevator doors' or 'sliders.') We would typically detail this so the top is driven, and supported by two or more wheels on a track that is strong enough to support the whole door. At the bottom, we'll run a tab (probably made of steel) through a slotted block (usually made of plastic - UHMW for example.) The tab at the bottom is just to keep the door from getting pushed in a way that would twist the track at the top, it doesn't actually support any of the weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sketch of what that solution looks like in cross section. It can be scaled to any size, but the dimensions shown are easy to do with minimal tooling. &lt;img src=&quot;https://i.stack.imgur.com/yFta6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do something similar, you can make the slot bigger than the tab, reducing the risk of binding. One other strategy, if you have enough space above and below the door, is to use a &lt;a href=&quot;https://www.mayline.com/sft277/651.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;mayline&lt;/a&gt; &lt;a href=&quot;https://books.google.com/books?id=A30rBgAAQBAJ&amp;amp;pg=PA402&amp;amp;lpg=PA402&amp;amp;dq=mayline%20cable%20rigging&amp;amp;source=bl&amp;amp;ots=wyptc4dVsW&amp;amp;sig=iYt5n2TLvxDNo5w2eA1TyJvSCkE&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=LR4FVf3cMYOuogSl7oA4&amp;amp;ved=0CEIQ6AEwAw#v=onepage&amp;amp;q=mayline%20cable%20rigging&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;mechanism&lt;/a&gt; which uses two opposing cables to keep the door square. These are more common for things that move vertically, but can be used horizontally as well.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-03-16T16:28:18.150" LastActivityDate="2015-03-16T16:28:18.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2096" PostTypeId="1" AcceptedAnswerId="7000" CreationDate="2015-03-15T06:35:49.177" Score="8" ViewCount="1052" Body="&lt;p&gt;In designing mechanical parts in an assembly, sometimes the mating surfaces are raised faces. I've been told it's to provide a better seal. However, for piping (flanges), it seems to be the consensus that a raised face flange does not provide any extra seal functionality, but has more to do with the manufacturing method (forged vs cast).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question doesn't necessarily apply just to piping flanges, but to designing parts in general. Is it desirable to use a raised face on mating parts, and when (always, sometimes, etc)?&lt;/p&gt;&#xA;" OwnerUserId="140" LastActivityDate="2016-01-19T21:41:03.990" Title="Why raised face for mating parts?" Tags="|mechanical-engineering|design|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2098" PostTypeId="2" ParentId="2096" CreationDate="2015-03-15T12:54:31.730" Score="5" Body="&lt;p&gt;Surfaces on a two part mated assembly that is required to provide sealing must be designed so that the seal acts as a seal. While that sounds obvious, it seems to have escaped too many designers and designs which attempt to rely on properties of the jointing materials are too common. While such designs may work when new, &quot;properties of the jointing material&quot; is often time variant, and Murphy ensures that in most cases the variation is in an undesirable direction. (One exception, in many cases, are the &quot;silicone rubbers&quot;. If these seal initially and use materials properly matched to the jointed materials, lifetimes of 20+ years are not uncommon.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of sealing against a pressure difference a simplistic perspective is that the force per area at the seal surfaces from the material in the pipe must be lower than the force per area exerted by all sources of sealing pressure. That sentence may sound strange but is written that way because there may be more than one source of sealing pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious source of sealing pressure is the bolts retaining the join.&lt;br&gt;&#xA;An obvious way of ensuring that bolting &quot;psi&quot; (lbf/in^2, Pa, ...) is maximised is by minimising the area so there is &quot;more pressure to go around&quot; for a given bolting force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Raised flanges allow formal sealing areas to be designed and area to be minimised.&lt;br&gt;&#xA;Without such raised areas sealing psi is reduced. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Related:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another less obvious and exceedingly useful means of exerting sealing pressure is to use the internal pressure to create its own sealing forces. This is the principle used by &quot;O rings&quot; which use pressure differential to deform a flexible ring in such a way that the sealing psi exceeds the leakage psi from the same pressure source. While O rings can be easily designed by following standard rules and tables, they in fact involve very black magic indeed, and numerous low cost designs that appear to employ O rings in fact just use squashed rings of flexible material. The difference between a true O ring seal and a bit of squashed rubber is proper design and manufacture and a relatively small extra $ outlay. The difference in performance is incomparable. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-03-17T03:42:08.980" LastActivityDate="2015-03-17T03:42:08.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2100" PostTypeId="2" ParentId="2096" CreationDate="2015-03-15T21:57:25.180" Score="7" Body="&lt;p&gt;In many cases, it's done for the sake of precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, an engine block is often sand cast. All mating surfaces in the casting master are made oversize, with the intention being that they will later be milled down to their final size. Since milling is orders of magnitude more precise than casting, this yields a more consistent result &amp;mdash; &lt;em&gt;where required&lt;/em&gt; &amp;mdash; while maintaining the lower overall cost of the casting process.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-03-15T21:57:25.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2101" PostTypeId="1" AcceptedAnswerId="2106" CreationDate="2015-03-15T23:34:27.277" Score="10" ViewCount="2101" Body="&lt;p&gt;I found the following comments on direct and indirect control in &lt;a href=&quot;http://www.elsevierscitech.com/pdfs/JPC_SurveyPrize.pdf&quot;&gt;&quot;Survey on iterative learning control, repetitive control and run-to-run control.&quot; by Wang, Gao, and Doyle&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are two application modes to use the learning-type control. First, the learning-type control method is used to determine the control signal directly, and this kind of learning-type control is called direct learning-type control. Second, there is a local feedback controller in each cycle and the learning-type control is used to update the parameter settings of the local controller, so this kind is called indirect learning-type control. The methods that can be used for designing direct learning-type control and indirect learning-type control will be discussed in Section 4 and in Section 5, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the difference between direct and indirect learning control? My understanding is that in indirect control you can alter the control parameters and the input signal as opposed to just the input signal in direct control; is this correct? I also do not understand the significance of a &quot;local&quot; controller.&lt;/p&gt;&#xA;" OwnerUserId="920" LastEditorUserId="368" LastEditDate="2015-03-16T16:09:51.190" LastActivityDate="2015-03-16T19:25:11.497" Title="What is the difference between direct and indirect learning control?" Tags="|mechanical-engineering|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2102" PostTypeId="2" ParentId="2022" CreationDate="2015-03-16T01:14:20.910" Score="2" Body="&lt;p&gt;This topic is much larger than just a simple equation involving the angle of attack of turbine blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been many university papers writen about topics involving turbine design and properties. Two of these are: &lt;a href=&quot;http://dave.ucsc.edu/physics195/thesis_2011/thesis_viscuso.pdf&quot; rel=&quot;nofollow&quot;&gt;angle of attack&lt;/a&gt; and &lt;a href=&quot;http://www.diva-portal.org/smash/get/diva2:441336/FULLTEXT01.pdf&quot; rel=&quot;nofollow&quot;&gt;tidal turbine&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blades are designed as airfoils, so the details of the blades are important. The speed of the fluid as well as the speed of the turbine rotation combine to determine if the blades stall. In addition to this, the Reynolds number is another important factor as with all fluid modeling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading through these two papers shows that even with equations and computer models, physical model testing is still required. Looking at these papers might be a place to start looking for background information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically this is a &lt;strong&gt;very&lt;/strong&gt; big topic that can not be reduced to a single equation.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-16T13:23:41.913" LastActivityDate="2015-03-16T13:23:41.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2103" PostTypeId="2" ParentId="2047" CreationDate="2015-03-16T03:16:29.193" Score="2" Body="&lt;p&gt;I would use a Hermite interpolation. It uses the following four functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ h_1 =  2s^3 - 3s^2 + 1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ h_2 = -2s^3 + 3s^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ h_3 = s^3 - 2s^2 + s $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ h_4 = s^3 -  s^2 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And combines them like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ output = &#xA;(h_1 * startPoint) +&#xA;(h_2 * endPoint) +&#xA;(h_3 * gradientIn) +&#xA;(h_4 * gradientOut)&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value $ s $ in the four functions is your interpolating parameter, as it goes from $ 0 $ to $ 1 $ the $ output $ goes from your $ startPoint $ $ (0, 0) $ to your $ endPoint $ $(3, 2) $. Your $ gradientIn $ wasn't specified, but looks to be $ 0 $ and $gradientOut $ is as you specified: $tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} $ so removing the terms that will multiply by zero (start point and gradient in):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ x_s = (h_2 * x_{end}) + (h_4 * tan(\frac{\pi}{6})) = (h_2 * 3) + (h_4 * \frac{1}{\sqrt{3}}) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ y_s = (h_2 * y_{end}) + (h_4 * tan(\frac{\pi}{6})) = (h_2 * 2) + (h_4 * \frac{1}{\sqrt{3}}) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more information about this type of interpolation curve, here is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Cubic_Hermite_spline&quot; rel=&quot;nofollow&quot;&gt;mathematical description&lt;/a&gt; and a more &lt;a href=&quot;http://cubic.org/docs/hermite.htm&quot; rel=&quot;nofollow&quot;&gt;functional description&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-03-16T03:16:29.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2104" PostTypeId="1" AcceptedAnswerId="2946" CreationDate="2015-03-16T04:13:59.720" Score="4" ViewCount="578" Body="&lt;p&gt;In case of Uranium, things are lot easier, because $UF_6$ is gaseous from around 60 $\,^{\circ}\mathrm{C}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, that in most cases, plutonium isotope separation is not needed, because there is no need to separate its fissile isotopes. But there is an exception: &lt;a href=&quot;https://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator&quot; rel=&quot;nofollow noreferrer&quot;&gt;radioisotope thermoelectric generators&lt;/a&gt; (used to power remote devices such as spacecraft and unmanned polar equipment) need ${^{238}}Pu$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, Plutonium can be separated from the depleted fuel cells with &lt;a href=&quot;https://engineering.stackexchange.com/questions/26/how-does-nuclear-fuel-reprocessing-work&quot;&gt;chemical methods&lt;/a&gt;, but how to separate ${^{238}}Pu$ from the result?&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2019-01-13T23:49:55.843" Title="How to separate isotopes of Plutonium?" Tags="|nuclear-technology|chemical-engineering|nuclear-reprocessing|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2105" PostTypeId="2" ParentId="2081" CreationDate="2015-03-16T10:45:13.710" Score="13" Body="&lt;p&gt;Have you looked at the size of one of those maritime diesel engines? They are larger than your car and need to deliver a lot of power to move and power the ship. That takes a lot of fuel so it's cheaper to burn more cheaper fuel even if it is of inferior quality. The bigger size also lets it use wider fuel lines so the viscosity is less of an issue. You also need to heat fuel oil to be able to pump it around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cars use diesel because emission regulations and fuel economy. The narrow regulations of diesel fuel lets the manufacturers design the engine to burn &lt;em&gt;just&lt;/em&gt; that fuel as efficiently and as cleanly as possible. There is also a size constraint on car engines. So more viscous fuel would be an issue and take up precious space for wider fuel channels and heating elements to keep the oil liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cost per ton per mile just works out better to use diesel for trucks and cars while fuel oil works out better for the large ships.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2017-01-19T12:23:51.117" LastActivityDate="2017-01-19T12:23:51.117" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2106" PostTypeId="2" ParentId="2101" CreationDate="2015-03-16T11:07:22.667" Score="7" Body="&lt;p&gt;I have only skimmed through the paper. The terminology is put forward by the paper to organize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 6 in the paper is the key (bottom of page 1595). That is a diagram of indirect learning control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the learning controller takes two inputs: one is the system output, and the other is the output of the 'local controller'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the local controller? The paper says: &quot;In principle, any real-time feedback control law can be chosen as the local control.&quot; It seems that the local controller is a real-time feedback controller, such as a PID controller (&lt;a href=&quot;https://en.wikipedia.org/wiki/PID_controller&quot;&gt;proportional-integral-derivative controller&lt;/a&gt;), but it could be more complex, or even a simple open loop controller. The feedback is shown in the diagram, where the output of the system is fed into the local controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is the other input to the local controller? That is the output of the learning controller. For the purpose of this paper, it is whatever input the designer considers to be useful. The examples in the paper range from setting the duration of a batch process, through setting the references of a PID controller, to updating parameters of advanced controllers such as model-based or neural network controllers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper describes direct learning control as an arrangement without the intervening, 'local', real-time controller. In other words, the learning controller is connected directly to the process or system under control. &lt;strong&gt;Update:&lt;/strong&gt; This direct connection will modify the setpoint of a real-time feedback controller.&lt;/p&gt;&#xA;" OwnerUserId="160" LastEditorUserId="160" LastEditDate="2015-03-16T19:25:11.497" LastActivityDate="2015-03-16T19:25:11.497" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2107" PostTypeId="1" AcceptedAnswerId="2193" CreationDate="2015-03-16T16:49:13.207" Score="7" ViewCount="1587" Body="&lt;p&gt;I have a vibration motor contained in an object. The motor looks similar to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ww4Eq.jpg&quot; alt=&quot;Typical vibration motor&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the motor is running, I want to simulate the movement of the container object that results from the vibration of the motor. I'm able to model the parts/assemblies in SolidWorks. How can I model or simulate the movement of the motor and the container object?&lt;/p&gt;&#xA;" OwnerUserId="931" LastEditorUserId="368" LastEditDate="2015-03-16T17:50:08.713" LastActivityDate="2015-03-23T16:06:12.747" Title="How can I model or simulate motion caused by an internal vibrating motor?" Tags="|mechanical-engineering|vibration|modeling|simulation|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2108" PostTypeId="2" ParentId="2022" CreationDate="2015-03-16T18:38:57.070" Score="2" Body="&lt;p&gt;Answering this question simply is not feasible, you are asking a question which is enormously difficult to answer without a large number of parameters. The biggest issue which you face is that there are many, many correct solutions to this question without constraints, but only one physically correct solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is similar to the question &quot;how big does my radiator need to be to cool my engine?&quot;: the question is simple enough, you might even know the power which you need to extract to prevent overheating, but for a given duct there is a whole range of mathematically correct inlet/outlet velocities and pressures, but only one physically correct solution. Our problem has not been sufficiently constrained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To nudge you in the correct direction, there are two big factors to consider here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;what happens to the flow passing through your turbine?&lt;/li&gt;&#xA;&lt;li&gt;how does the design of your turbine affect the torque at the shaft?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first question can be greatly simplified by imposing the maximum efficiency we can, or perhaps some factor of that, and assume that this is feasible. This is governed by &lt;a href=&quot;http://en.wikipedia.org/wiki/Betz&amp;#39;s_law&quot; rel=&quot;nofollow&quot; title=&quot;Betz&amp;#39;s Law&quot;&gt;Betz's Law&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you apply this constraint, it had ought to be trivial to obtain the power produced at your turbine using basic information about the flow. You know that for this to be achieved you must have zero loses in the flow as a result of your turbine (beyond the energy extracted of course), so for a given number of blades there is only one solution. Now create a model in which the physics impose zero loses at this solution; the closer to a real world model this is, the more accurate your answer will be. Now vary AoA to see how your loses change within your model. Your model should account for separation but there are plenty of 2D open source codes you could try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above method will not get you the answer, as it assumes that the solution you start with is optimum and works from there, but it will give you an idea of how AoA changes affect your torque in an arbirary system, and it is simple enough that most people with basic knowledge of fluid dynamics and some time could do it.&lt;/p&gt;&#xA;" OwnerDisplayName="user815" LastActivityDate="2015-03-16T18:38:57.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2109" PostTypeId="1" CreationDate="2015-03-16T18:43:23.253" Score="6" ViewCount="2551" Body="&lt;p&gt;&lt;strong&gt;Pavement Design&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In roadway pavement design, the (US) controlling design load is usually the &lt;a href=&quot;http://www.pavementinteractive.org/article/equivalent-single-axle-load/&quot; rel=&quot;nofollow&quot;&gt;Equivalent Single Axle Load&lt;/a&gt; (ESAL). This number attempts to quantify how much bending stress a given vehicle causes in the pavement. Trucks have a much higher ESAL load than cars. In this case &quot;truck&quot; means &quot;semi, big rig, freight carrier&quot; and not a family pickup truck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical values are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Car Axle = 0.0002 or 0.0003 ESAL&lt;/li&gt;&#xA;&lt;li&gt;Legal Truck Axle = ~2 ESAL (wide range based on weight)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By this measure, a typical car or pickup truck doesn't place any appreciable stress on the road pavement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pavement Life&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two sections of road that are designed and built exactly the same and a certain period of time.  This period of time could be 5 year, 10 years, 20 years, etc.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Road driven on only by typical family or commuter vehicles. Basically only passenger vehicles (not busses) and not hauling goods or construction materials.&lt;/li&gt;&#xA;&lt;li&gt;Road with no vehicle traffic, but experiences the same weather conditions as (1). This would also include salt application in areas where that it common.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you be able to tell which section of pavement was driven on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would light weight vehicles cause so little additional damage to the pavement that it would be indistinguishable from natural weathering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corollary is that trucks cause all of the pavement damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be clear, I am not concerned with visual things that you would be able to use to differentiate, e.g. oil stains, rubber marks, etc.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="6790" LastEditDate="2018-05-07T11:42:37.667" LastActivityDate="2018-05-07T11:42:37.667" Title="Does passenger car traffic actually damage roads?" Tags="|civil-engineering|highway-engineering|roadway|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2111" PostTypeId="2" ParentId="2109" CreationDate="2015-03-16T21:11:27.147" Score="7" Body="&lt;p&gt;Yes, there would be a noticeable difference.  Normal passenger cars don't really stress the road surface much.  That's not so much the issue.  As a analogy, you can have 100 people walk across a small plank, and it holds fine, but even 10 people on the plank at the same time will stress it to breaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cars in normal operation don't much hurt a normal road surface, but the road surface will degrade on its own.  Eventually a point is reached where normal cars cause significant additional damage.  For example, consider a small rock that is part of the asphalt mix coming loose over time.  Cars driving over it can dislodge the rock and randomly move it around until it ends up on the side of the road.  Without cars, the loose rock would have stayed in place, at least a lot longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This small rock principle multiplies over time.  A dislodged or two small rock creates a weak spot, so ordinary car traffic can accelerate more material getting dislodged at the boundary.  This causes the boundary of a defect to grow more rapidly with car traffic than without.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there is a flip side too.  Car traffic kills plants growing in cracks.  Any abandoned road, even after just one growing season, will have weeds growing in cracks.  Cracks are inevitable, and so are wind-blown or bird-dropped seeds.  Roots from these plants grow, dislodging road material, and trapping windblown dirt, making more soil.  This causes more plants to grow, more soil to be made etc.  In places with enough rain to be generally green if left alone, a road disappears under soil remarkably quickly if left alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the difference will be quite noticeable because completely different forces are at work tearing down a active road versus a abandoned one.  The main state road near my house in Massachusetts was last repaved 20 years ago.  It now has lots of cracks and has developed lots of potholes in the last few weeks due to the spring conditions.  However, I've seen roads abandoned for only 10 years hardly look like a road anymore unless you are on top of it and see the patches of asphalt in between the significant clumps of weeds.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2015-07-06T23:01:03.657" LastActivityDate="2015-07-06T23:01:03.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2112" PostTypeId="1" AcceptedAnswerId="2291" CreationDate="2015-03-17T02:19:35.227" Score="5" ViewCount="6957" Body="&lt;p&gt;&lt;strong&gt;Construction Joints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construction joints in concrete footings or walls are places where the concrete pour can be stopped so that everything doesn't have to be poured at once. Later, once the concrete has hardened, the next part of the concrete can be placed against the previous pour. These locations are sometimes planned for in the design, and special details are added to the plans to show how the construction joint is to be formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical detail of one of these joints from &lt;a href=&quot;http://epg.modot.org/index.php?title=751.34_concrete_pile_cap_non-integral_end_bents&quot; rel=&quot;nofollow noreferrer&quot;&gt;MODOT&lt;/a&gt; is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QqZXw.jpg&quot; alt=&quot;Typical construction joint details&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These joints have a &quot;key&quot; formed in them so that when the next batch of concrete is placed, there isn't a flat shear plane across the entire wall thickness. The &quot;key&quot; is supposed to help to transfer shear across the joint. In the US, the blockout is typically formed by placing a 2x4 (2 in x 4 in nominally) board in the fresh concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes a rubber or plastic waterstop is cast into the concrete at these locations to help keep water from flowing through the joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Shear key design&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I have seen this detail used many times, I have never seen a calculation done to design it. This exact same detail is used no matter what the thickness of the wall is or the shear force that may be acting on the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you design (or check) a shear key like this?  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do you only consider shear-friction across the joint, ignoring the key?&lt;/li&gt;&#xA;&lt;li&gt;Do you only consider shearing of the concrete in the key?&lt;/li&gt;&#xA;&lt;li&gt;Is it some combination of the two?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastEditorUserId="122" LastEditDate="2015-03-17T05:35:55.630" LastActivityDate="2015-04-02T15:40:29.887" Title="Design behind typical keyed construction joints in concrete walls" Tags="|civil-engineering|structural-engineering|reinforced-concrete|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2114" PostTypeId="1" AcceptedAnswerId="2118" CreationDate="2015-03-17T16:01:04.487" Score="13" ViewCount="18827" Body="&lt;p&gt;We have a simple class 1 lever:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin {array}{c}&#xA;\text {5,000 kg} \\&#xA;\downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \, \downarrow \\&#xA;==================== \\&#xA;\hphantom {==} \triangle \hphantom {==============} \\&#xA;\vdash  \text{1 m} \dashv \vdash \hphantom {======} \text{4 m} \hphantom {======} \dashv \\&#xA;\end {array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lever ($===$) is 5 m long. The fulcrum ($\triangle$) is 1 m from one end of the lever. The lever has a object sitting uniformly upon it weighing 5,000 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I compute the upward force that needs to be exerted at the end of the 1 m side of the lever to keep the lever stationary? $F = (W \times X)/L$ is simple when the weight is applied at the very end of the lever. But what happens if the weight is distributed along the lever?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our final goal is to tether the free end (on the 1m side) to keep the lever level and we need to know how strong the tether should be.&lt;/p&gt;&#xA;" OwnerUserId="941" LastEditorUserId="41" LastEditDate="2015-03-18T14:03:23.883" LastActivityDate="2017-11-12T17:16:46.850" Title="How to calculate lever force when lever has uniformed distributed load?" Tags="|mechanical-engineering|statics|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2115" PostTypeId="2" ParentId="2114" CreationDate="2015-03-17T16:40:40.153" Score="4" Body="&lt;p&gt;The effect of any bit of force along a lever is proportional to its distance from the fulcrum.  This nice linear relationship works out so that for a rigid mass, you can simply model it as a point mass at its center of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For weight effects (force due to mass and gravity), it's purely the horizontal distance from the fulcrum to the center of mass that matters.  If you define X to the right and Y up in your diagram, then the Y coordinate of the mass is irrelevant.  Note, however, that when the lever moves, the X coordinate of the mass also moves, especially when it is not right on the lever arm.  For small movements of the lever, you my be able to ignore this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put more mathematically, the torque on the fulcrum is the vector from the fulcrum to the center of mass, cross the gravitational force on that mass.  Since the latter is always down (-Y) in this example, only the X component of the vector to the mass matters in getting the toque magnitude.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2015-03-19T23:16:11.830" LastActivityDate="2015-03-19T23:16:11.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2116" PostTypeId="2" ParentId="2114" CreationDate="2015-03-17T16:43:01.323" Score="8" Body="&lt;p&gt;In any continuous situation, you simply use integration.  The linear mass density of your block is $\lambda=\frac{m}{\ell}=$1000 kg/m.  Now you can express the torque due to an infinitesimal slice of the rod of width $dx$ at position $x$ as&#xA;$$&#xA;d\tau=(\lambda dx) * x * g&#xA;$$&#xA;where $x$ is measured from the fulcrum.  Finally, you just sum up all of the little torques from each infinitesimal slice with integration.&#xA;$$&#xA;\tau = \lambda g\int_{-1}^{4}x\ dx=7.5\ g\lambda=73.5\ \text{kN*m}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-03-17T18:39:29.327" LastActivityDate="2015-03-17T18:39:29.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2118" PostTypeId="2" ParentId="2114" CreationDate="2015-03-17T17:25:16.063" Score="9" Body="&lt;p&gt;Since the mass is 5k kg and the lever is 5m, this makes it quite easy to simplify because it is exactly 1k kg per m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leftmost 2k kg (2m) of the mass has its center of mass exactly above the fulcrum so can be ignored as it provides no contribution to the moment. This leaves 3k kg (3m) spread from 1m to 4m on the right side. The center of mass will therefore be at 2.5m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it's super-simple, assuming you want the moment when the lever is level (i.e. when gravity is pulling straight down, perpendicular to the lever):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{torque} = rF = rmg $$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ r $ is the radius (distance) in m (2.5).&lt;/li&gt;&#xA;&lt;li&gt;$ m $ is the mass in kg (3000).&lt;/li&gt;&#xA;&lt;li&gt;$ g $ is the acceleration due to gravity in $ \text{ms}^{-2} $ (9.80665).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$ \text{torque} = 2.5 * 3000 * 9.80665 = 73549.875 \text{ Nm} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your edit/update indicates that you are looking for the upwards force at the 1m end, this will be the torque (from above) divided by the distance (1m). Which is therefore 73549.875 N.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="95" LastEditDate="2015-03-17T18:18:42.933" LastActivityDate="2015-03-17T18:18:42.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2119" PostTypeId="2" ParentId="2114" CreationDate="2015-03-17T18:18:51.613" Score="5" Body="&lt;p&gt;To answer the new question, which it really rather different to the original question, you will require a 7500 g N downward force at the left-hand tip to balance forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking moments about your support (which is now, indeed, a pivot):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{\text{LHS free end}} * 1 = 5000*g * 1.5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{\text{LHS free end}} = 7500*g \text{ N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, yes, you can treat your distributed load as a point load acting at the centre of the beam. You can prove this my solving this by integration of the distributed load.&lt;/p&gt;&#xA;" OwnerDisplayName="user815" LastEditorUserId="106" LastEditDate="2015-03-18T00:15:46.153" LastActivityDate="2015-03-18T00:15:46.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2120" PostTypeId="1" AcceptedAnswerId="2127" CreationDate="2015-03-17T20:04:10.807" Score="11" ViewCount="394" Body="&lt;p&gt;I have seen several block diagrams with diagonal inputs, drawn with an arrow through the block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tHTpI.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;a href=&quot;https://www.google.com/search?q=indirect+adaptive+control&amp;amp;biw=1550&amp;amp;bih=720&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ei=CYYIVbfkFImegwT41oHwCA&amp;amp;sqi=2&amp;amp;ved=0CAcQ_AUoAg#imgdii=_&amp;amp;imgrc=gULVia1RSqWyPM%253A%3BgD-PX1e3g17aiM%3Bhttp%253A%252F%252Fieeexplore.ieee.org%252Fiel5%252F5%252F20157%252F931488%252F1104583.gif%3Bhttp%253A%252F%252Fieeexplore.ieee.org%252Fiel5%252F5%252F20157%252F931488%252F931488.plain.html%3B444%3B246&quot; rel=&quot;noreferrer&quot;&gt;and others like that (via Google image search)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is difference between this and a typical signal input? What are the dotted lines signifying?&lt;/p&gt;&#xA;" OwnerUserId="920" LastEditorUserId="63" LastEditDate="2015-03-17T22:55:15.547" LastActivityDate="2015-03-18T12:15:06.077" Title="What is the significance of a diagonal input in a control diagram?" Tags="|mechanical-engineering|electrical-engineering|control-engineering|systems-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2121" PostTypeId="1" CreationDate="2015-03-18T02:52:54.793" Score="6" ViewCount="180" Body="&lt;p&gt;Soil freezes naturally all of the time. It is at this point that dirt work usually stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Occasionally soil is &lt;a href=&quot;https://en.wikipedia.org/wiki/Ground_freezing&quot; rel=&quot;nofollow&quot;&gt;intentionally frozen&lt;/a&gt; to improve its structural properties. This process &lt;a href=&quot;http://www.pbs.org/wgbh/nova/next/tech/artificial-ground-freezing/&quot; rel=&quot;nofollow&quot;&gt;has been used notably in Boston's Big Dig and the Fukushima Daiichi nuclear plant&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When soil is intentionally frozen so that tunnels or vertical shafts can be excavated, the structural properties of the frozen soil are important. It would seem logical that the frozen water would help to stiffen the soil, but how are the actual structural properties of the soil determined?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Do all frozen soils end up with the same properties? i.e. the presence of ice controls the properties.&lt;/li&gt;&#xA;&lt;li&gt;Are the frozen soil properties related to the unfrozen soil properties in any way?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-08-18T01:27:46.513" LastActivityDate="2015-09-15T16:23:32.923" Title="How are the structural properties of intentionally frozen soil determined?" Tags="|civil-engineering|geotechnical-engineering|soil|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2122" PostTypeId="2" ParentId="2114" CreationDate="2015-03-18T09:04:08.833" Score="4" Body="&lt;p&gt;A uniformly distributed load can be considered to act in its centre. Working in kg and m:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clockwise moment about the left hand end = 5000 * 2.5 = 12500&#xA;Anticlockwise moment about the left hand end = F * 1 (where F is the reaction at the fulcrum)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These must be equal for it to be balanced, giving F=12500kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resolving vertically (the total downward force must equal the total upward force), taking T as the reaction on the tether: T + 5000 = 12500, therefore T = 7500kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or converting into N (as you say you want a force, and kg is mass not force) then T = 7500 * 9.81 =  73575N = 73.6kN&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2015-03-18T12:43:20.050" LastActivityDate="2015-03-18T12:43:20.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2123" PostTypeId="2" ParentId="2121" CreationDate="2015-03-18T10:21:41.257" Score="-3" Body="&lt;ol&gt;&#xA;&lt;li&gt;take core sample&lt;/li&gt;&#xA;&lt;li&gt;stick in freezer&lt;/li&gt;&#xA;&lt;li&gt;run tests&lt;/li&gt;&#xA;&lt;li&gt;?&lt;/li&gt;&#xA;&lt;li&gt;profit!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One important factor is the water content of the soil, freezing dry sand won't do much compared to freezing a waterlogged riverbank.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-03-18T10:21:41.257" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2124" PostTypeId="1" AcceptedAnswerId="2128" CreationDate="2015-03-18T10:44:46.117" Score="7" ViewCount="1188" Body="&lt;p&gt;I think, it could lead to, for example, very environment friendly &lt;em&gt;and&lt;/em&gt; very efficient urban traffic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I also think, it required very precise part construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do such things exist? At least in plan?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Extension:&lt;/em&gt; Clarification added, waiting for acceptable answer again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Extension #2:&lt;/em&gt; Despite the &quot;common sense&quot;, diesel engines aren't so simple. Although there is no spark, there is a part which injects the fuel in exactly the perfect moment into the piston. This happens with around 2000atm pressure. Maybe this would be hard in case of gases.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditDate="2016-01-21T22:35:49.157" LastActivityDate="2016-01-22T18:22:51.087" Title="Does a diesel-like (ignition by compression) engine running on gaseous fuel exist?" Tags="|mechanical-engineering|combustion|diesel|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2125" PostTypeId="1" AcceptedAnswerId="2129" CreationDate="2015-03-18T10:48:17.857" Score="13" ViewCount="1895" Body="&lt;p&gt;I am new to mechanical engineering, although I have a scientific background (postgrad in Mathematics), and I (mostly) code for a living.&lt;/p&gt;&#xA;&lt;p&gt;I have an idea about creating a mechanical device; I envisage it will entail gears, linkages and actuators.&lt;/p&gt;&#xA;&lt;p&gt;I have a rough idea where things will fit but I would like to be able to test and tweak the design in software before building the actual device. As a point of clarification, when I say &amp;quot;test,&amp;quot; I mean view via animation, for example, whether two items will collide when in motion, or if there is sufficient leeway between them as they move past each other.&lt;/p&gt;&#xA;&lt;p&gt;This allows me to do the design and simulation testing of the parts, before finalising the design and then building the physical system from the design.&lt;/p&gt;&#xA;&lt;p&gt;The stages are:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Build the 3D design in software&lt;/li&gt;&#xA;&lt;li&gt;Run simulation to see if it &amp;quot;works,&amp;quot; if not fix design and iterate&lt;/li&gt;&#xA;&lt;li&gt;Build physical system from design that &amp;quot;works&amp;quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I have figured out that the system consists of three subsystems working together.&lt;/p&gt;&#xA;&lt;p&gt;So, I would like to design and test each sub-component before integrating them into the complete system.&lt;/p&gt;&#xA;&lt;h3&gt;My question then is this:&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this how design is done in the real world?&lt;/li&gt;&#xA;&lt;li&gt;What are the pros and cons of the scheme I have planned?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I am intending to use FreeCAD to do the design and testing.&lt;/p&gt;&#xA;" OwnerUserId="932" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-07-13T16:20:03.450" Title="How is a mechanical system designed and tested without building a physical model?" Tags="|mechanical-engineering|systems-design|computer-aided-design|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2126" PostTypeId="2" ParentId="2124" CreationDate="2015-03-18T11:40:24.843" Score="4" Body="&lt;p&gt;The only ones I know of are pilot-oil gas motors. The ones I know burn biogas or methane plus a small amount of oil, typically rape oil. The oil is neccesary for combustion but the biggest part of the power comes from the gasous fuel. Typically, the lower limit for oil consumption is about 5% (by energy content), the engines can run on oil solely. Efficencies are advertised as 45-50% for a 500kW stationary engine used as a CHP plant (with utilization of exhaust pressure). This is better than gas-only engines. One producer is Schnell Motoren, maybe you can find further information &lt;a href=&quot;http://www.schnellmotor.de/en/products/dual-fuel-chp-unit.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;on their site&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-03-18T11:40:24.843" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2127" PostTypeId="2" ParentId="2120" CreationDate="2015-03-18T12:15:06.077" Score="6" Body="&lt;p&gt;The diagonal input generally means that it changes some parameter of the block, such as its gain, without otherwise becoming part of the signal that the block is processing.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-03-18T12:15:06.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2128" PostTypeId="2" ParentId="2124" CreationDate="2015-03-18T12:24:40.930" Score="9" Body="&lt;p&gt;It depends on what you mean by &quot;diesel-like&quot;.  If you mean an engine with compression ignition running on gaseous fuel, then the answer is no.  The high autoignition temperature of fuels such as natural gas prohibits sparkless ignition of pure natural gas&lt;sup&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0196890404000330&quot;&gt;1&lt;/a&gt;&lt;/sup&gt;. @mart is right that compression ignition engines running on gaseous fuels need a pilot fuel which ignites the mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, diesel engines have been modified to run on gaseous fuels by adding spark ignition&lt;sup&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0360319904000710&quot;&gt;2&lt;/a&gt;&lt;/sup&gt;. These engines are &quot;diesel-like&quot; in that they have high compression ratios (and were made from actual diesel engines for purposes of research convenience) so again, it depends on what you mean by &quot;diesel-like&quot;.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-03-18T12:24:40.930" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2129" PostTypeId="2" ParentId="2125" CreationDate="2015-03-18T12:31:25.080" Score="12" Body="&lt;p&gt;&lt;strong&gt;Answer: Yes&lt;/strong&gt;, this is exactly how it is done in the real world. What you have described is what I do in my job to check systems in CAD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you indicated you would like me to step through my design process, I have detailed it below.  Note that most of this does not involve CAD.  CAD is invaluable, but only if you are prepared to get out the pencil and paper first.  Also note that this is just &lt;em&gt;my&lt;/em&gt; design process, it is by no means the only way to go about things.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Preparation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;When I begin any design, I start by figuring out general parameters like what space will it need to fit into, what it needs to interface with, what the input and output will be.  Lets say (for sake of a concrete example) that I'm making a machine that takes in a long pice of pipe and cuts it into predetermined sections.  My first steps would be to determine the size and material of pipe that I want to input, what lengths the machine should cut the pipe into, what method I'm going to use to cut the pipe, and how much space I have on the shop floor for said machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would then proceed to make a really rough sketch of what I'm thinking.  In the pipe cutter example, it might be as simple as a box labeled &quot;cutter&quot; and a line labeled with the pipe weight diameter and length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to do some math to figure out what forces, speeds, etc will be required.  Since you said you come from a mathematics background, I won't dwell on this too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I usually make another sketch here, except now I have some feel for how big my components will need to be because I know the forces involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After I do the math, I look for commercial parts that fulfill my needs and (if possible) download CAD files or mechanical drawings for those parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I have some commercial parts, I'll make another sketch, this time showing the relative positions of my commercial parts with their interfaces labeled so I know what kind of support structure I have to build.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;3D Modeling&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;At this point, I finally break out the CAD package.  I start by making 3D models of whatever commercial parts I have that don't have 3D models available online and then proceed to my support structure parts and finlly assemble all the parts into an assembly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the part you are probably very curious about if you have never used CAD before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 3D CAD, you should draw each part (usually either something that comes as a commercial part or is made from a single piece of material in its own part file. Once you have part files, you can make what is called an assembly. An assembly allows you to select several part files and define the linkages between them.  CAD programs have no sense of &quot;collisions&quot; so you have to tell the program which faces are aligned with other faces, what distance apart they are, etc.  Each part starts out with six degrees of freedom and each constraint reduces the degrees of freedom by some amount.  Specifying two faces are parallel removes one degree of freedom, a mate (two surfaces are coincident) removes two or three, etc.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;FreeCad&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I haven't used FreeCad, so I can't comment specifically on what the pros and cons will be, but my guess is this: the pros will be that it is free and you will see what dimensions work and what doesn't work, the cons will be that commercial software would have been easier to use.&lt;/p&gt;&#xA;" OwnerUserId="709" LastEditorUserId="709" LastEditDate="2015-03-18T21:53:42.050" LastActivityDate="2015-03-18T21:53:42.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2130" PostTypeId="2" ParentId="2124" CreationDate="2015-03-18T13:00:26.603" Score="4" Body="&lt;p&gt;On &lt;a href=&quot;http://www.marinelink.com/news/article/first-w-rtsil-marine-df-engines-ordered/307642.aspx&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt;, &lt;a href=&quot;http://www.wartsila.com/products/marine-oil-gas/engines-generating-sets/dual-fuel-engines/wartsila-31df&quot; rel=&quot;nofollow&quot;&gt;Wartsila&lt;/a&gt; states it has made diesel-gas engines and that the first ones have been operating for 70 000 hours.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="33" LastEditDate="2016-01-22T01:59:35.557" LastActivityDate="2016-01-22T01:59:35.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2131" PostTypeId="1" AcceptedAnswerId="2135" CreationDate="2015-03-18T14:52:43.163" Score="6" ViewCount="13259" Body="&lt;ol&gt;&#xA;&lt;li&gt;Honey dippers are often made of wood.  Is there a property of honey that drives this?&lt;/li&gt;&#xA;&lt;li&gt;Honey dippers are shaped like ovoids with lateral grooves. Is there something about the viscosity of honey that caused this decision?  Is this design somehow superior to using a spoon?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Wooden_honey_dipper_(5536641494).jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JOelXm.jpg&quot; alt=&quot;honey dipper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="948" LastEditorUserId="110" LastEditDate="2015-03-18T23:02:04.067" LastActivityDate="2015-03-18T23:02:37.477" Title="What drives the honey dipper design?" Tags="|fluid-dynamics|design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2132" PostTypeId="2" ParentId="2081" CreationDate="2015-03-18T15:42:35.583" Score="5" Body="&lt;p&gt;A diesel engine for a car needs a fuel which is liquid even in winter. This fuel should contain a very small amount of sulfur to limit air pollution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The marine bunker oil is not liquid at room temperature, it has to be heated to about 50 °C before pumping out of the tank and to about 130 to 140 °C before injecting it into the cylinders. It contains a lot of sulfur, a lot of harbours do not allow it to be used inside the harbour because of the air pollution. The ship needs an extra fuel tank with diesel oil for the harbour and for the cold start of the engine. The bunker oil has to be pre heated after a cold start before it could be used.&lt;/p&gt;&#xA;" OwnerUserId="950" LastActivityDate="2015-03-18T15:42:35.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2133" PostTypeId="1" AcceptedAnswerId="2138" CreationDate="2015-03-18T19:18:08.730" Score="14" ViewCount="4122" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Scenario:&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Suppose the emergency brake button in the cab of a train is not working. Would the emergency brake cords / buttons in the passenger cars still work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is inspired by &lt;a href=&quot;https://scifi.stackexchange.com/q/84144/23802&quot;&gt;this question on the SciFi StackExchange&lt;/a&gt;, asking why the passengers in the &lt;a href=&quot;https://www.youtube.com/watch?v=GYOYewO_Veg&quot; rel=&quot;nofollow noreferrer&quot;&gt;train fight scene from the movie SpiderMan 2&lt;/a&gt; did not simply pull the emergency brakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that particular scene, a R46 City Subway Car was rendered &quot;unstoppable&quot; by ripping out the speed control lever in the car, which also happened to disable the emergency brake button (see 16 seconds into the video).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With movies, we suspend our disbelief in order to enjoy the story that is presented.  But the above SciFi question got me to thinking about how emergency brake systems are designed for trains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given trains significant mass and momentum when moving, it seems like there would be multiple, redundant safety systems to provide braking capability for the train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;My Question:&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Is there a common safety design used for the braking system of trains?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that design account for portions of the system failing and allowing other portions to compensate for the failed components?  (ie. would the passenger car emergency brakes still work?)&lt;/p&gt;&#xA;" OwnerUserId="959" LastEditorUserId="-1" LastEditDate="2017-04-13T12:43:39.343" LastActivityDate="2015-12-14T13:32:01.560" Title="If the emergency brake in a train is broken, do the passenger car brakes still work?" Tags="|mechanical-engineering|rail|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2135" PostTypeId="2" ParentId="2131" CreationDate="2015-03-18T22:30:09.613" Score="7" Body="&lt;p&gt;Honey dippers work by using the viscosity of the liquid to raise a small amount of the liquid from a reservoir.  Spoons work by trapping the small amount of liquid in a gravity well.  Spoons therefore will work even when the liquid has low viscosity.  The honey dipper method only works with appreciable amounts of liquid when that liquid has high viscosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given enough time, the liquid will flow off the honey dipper.  The viscosity lengthens that time, allowing you to move and turn the instrument to compensate for the slowly creeping flow before it separates and drips - until you want it to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material only needs to allow the thick liquid to adhere, so something hydrophobic or that resists adhesion, like teflon, would be a bad choice.  In the specific case of the liquid being honey, the material of course has to be food-safe.  Wood was the obvious choice long ago because it was available, cheap, and the technology for working it well established.  Other materials available 100s of years ago would have worked too, like walrus tusk or ivory, but those were much more expensive.  Metals are not a good choice due to corrosion and possible chemical reactions.  Still, the same metals used for cutlery should have worked as well, but wood was much cheaper, and unlike the cutlery, the higher strength was not needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape is to provide a large surface area for the volume.  This traps more of the liquid in the slow-flowing boundary layer.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="110" LastEditDate="2015-03-18T23:02:37.477" LastActivityDate="2015-03-18T23:02:37.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2136" PostTypeId="2" ParentId="2133" CreationDate="2015-03-18T23:16:08.330" Score="7" Body="&lt;p&gt;Short answer: I suspect that ripping out the speed control lever would have immediately applied the emergency brake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In every safety system I have ever designed, emergency stop buttons are &lt;a href=&quot;https://en.wikipedia.org/wiki/Switch#Contact_terminology&quot;&gt;normally closed switches&lt;/a&gt; so that nothing will operate unless the switch conducts current.  Therefore if the wires to the emergency stop were severed, the brakes would come on.  The brakes would also come on if the brake vacuum pump was damaged, because the air brakes in trucks and trains will be on unless the engine is actively turning them off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I have never worked on train brakes, I'm basing my answer off of my personal knowledge of truck brakes and aircraft flight controls.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-03-18T23:16:08.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2137" PostTypeId="2" ParentId="2133" CreationDate="2015-03-19T00:33:21.940" Score="1" Body="&lt;p&gt;The braking system in a train is designed so that the loss of pressure at any point along the trainline will trigger the brakes -- an end-of-train-device, distributed-power locomotive, cab car/cabbage, or a passenger car emergency brake handle can still operate the brakes (service &lt;em&gt;and&lt;/em&gt; emergency, or emergency only, depending on what's operating it) even if the head-end's brake valve is inoperable for some reason (say, a mechanical jam).  Furthermore, the brake valve lever on a road-going locomotive in the US (whether it uses a control console or a traditional control stand) is mechanically separate from the throttle (speed control) and reverser levers -- you can still apply the brakes if the reverser is pulled or if the throttle is jammed.&lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2015-03-19T00:33:21.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2138" PostTypeId="2" ParentId="2133" CreationDate="2015-03-19T00:44:40.873" Score="20" Body="&lt;p&gt;&lt;strong&gt;Train Brakes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common brakes on trains are &lt;a href=&quot;http://en.wikipedia.org/wiki/Railway_air_brake#Modern_systems&quot;&gt;air brakes&lt;/a&gt;. As the name implies, these work off of air pressure. The braking power isn't controlled in the way that you would immediately think of though. They do not work like car brakes where the harder you press on the brake pedal, the harder the pressure goes through the lines to the brake cylinders. They work the opposite. The &lt;strong&gt;less&lt;/strong&gt; pressure in the line, the &lt;strong&gt;more&lt;/strong&gt; braking force is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fail-safe&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rail brakes are designed to be &lt;strong&gt;fail-safe&lt;/strong&gt;. That is, when a failure occurs, the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fail-safe&quot;&gt;safe operation&lt;/a&gt; happens. In a train, all of the cars have air lines that are attached together. This forms one long, continuous line from the locomotives through all of the other cars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This air line is used to both fill the air reservoirs on each can and control the amount of braking. This is accomplished through an ingenious valve that works solely on air pressure. As mentioned above, once all of the reservoirs on each of the cars has been filled, a &lt;strong&gt;decrease&lt;/strong&gt; in the air pressure in the air line causes the brakes to be engaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beauty of this system is that is something bad happens and the air pressure drops, the brakes are engaged automatically.  This means that if there is a rupture in the air line, or cars become uncoupled, the brakes are fully engaged without any other action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, if there is very low or no pressure in the line, a specially amount of braking force is applied called &lt;strong&gt;emergency braking&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Manual Brakes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The emergency braking force is enough to stop the train or a single run away car, but due to leakages, the braking force will eventually reduce. This is why train cars also have a mechanical brake. It is engaged by turning the big brake wheel are the end of the car. These are also the brakes that must be engaged to keep cars from rolling away if they are left alone in a siding.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-20T00:41:50.947" LastActivityDate="2015-03-20T00:41:50.947" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2139" PostTypeId="1" CreationDate="2015-03-19T01:18:51.880" Score="4" ViewCount="559" Body="&lt;p&gt;&lt;strong&gt;Emissions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Emissions from vehicles have an associated social cost for everyone. We hear a lot about CO&lt;sub&gt;2&lt;/sub&gt;, but CO&lt;sub&gt;2&lt;/sub&gt; is not the only thing that comes out of a tailpipe. From &lt;a href=&quot;http://en.wikipedia.org/wiki/Exhaust_gas#Internal-combustion_engines&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, emissions components of petrol and diesel engines are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nitrogen&lt;/li&gt;&#xA;&lt;li&gt;Carbon dioxide&lt;/li&gt;&#xA;&lt;li&gt;Water vapor&lt;/li&gt;&#xA;&lt;li&gt;Oxygen&lt;/li&gt;&#xA;&lt;li&gt;Nitrogen oxides&lt;/li&gt;&#xA;&lt;li&gt;Carbon monoxide&lt;/li&gt;&#xA;&lt;li&gt;Particulate matter&lt;/li&gt;&#xA;&lt;li&gt;Hydrocarbons&lt;/li&gt;&#xA;&lt;li&gt;Sulfur dioxide&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My impression is that CO&lt;sub&gt;2&lt;/sub&gt; gets the most attention because climate change is a hot topic. This doesn't mean that it is the emission component that has the highest cost to society though.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Do any of the components above have higher costs to society than CO&lt;sub&gt;2&lt;/sub&gt; per same unit of measure (e.g. gram, kilogram, or ton)?  &lt;br&gt;&lt;br&gt; &#xA;  If they have higher costs per unit of measure, do they have a higher cost than CO&lt;sub&gt;2&lt;/sub&gt; based on their component percentage in typical emissions? The component may be very costly per ton but it makes up a very small amount of typical emissions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that &lt;em&gt;cost&lt;/em&gt; is kind of hard to define in these cases. It does seem possible though, because in reports on greenhouse gas a monetary figure is always given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Costs&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The US EPA (Environmental Protection Agency) has published information on &lt;a href=&quot;http://www.epa.gov/climatechange/EPAactivities/economics/scc.html&quot; rel=&quot;nofollow&quot;&gt;The Social Cost of Carbon&lt;/a&gt;. From the &lt;a href=&quot;http://www.epa.gov/climatechange/Downloads/EPAactivities/scc-fact-sheet.pdf&quot; rel=&quot;nofollow&quot;&gt;SCC Fact Sheet&lt;/a&gt;, the cost social cost of a ton of CO&lt;sub&gt;2&lt;/sub&gt; in 2015 is between \$12 and \$116 depending on the discount rate used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Florida Department of Transportation has also produced computer software that estimates the benefits of transit usage. This is the Trip Reduction Impacts for Mobility Management Strategies (&lt;a href=&quot;http://www.nctr.usf.edu/clearinghouse/software.htm&quot; rel=&quot;nofollow&quot;&gt;TRIMMS&lt;/a&gt;) software. As a part of this, a co-benefit of reduced CO&lt;sub&gt;2&lt;/sub&gt; emissions is estimated at $50/metric ton. (&lt;a href=&quot;http://www.epa.gov/otaq/stateresources/policy/420r11014.pdf&quot; rel=&quot;nofollow&quot;&gt;TRIMMS Supplemental Report page 16&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The costs above are for CO&lt;sub&gt;2&lt;/sub&gt;, but I assume that similar studies have been done for other components of emissions.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="368" LastEditDate="2015-03-20T20:37:28.830" LastActivityDate="2015-03-20T21:17:54.563" Title="Which component of vehicle tailpipe emissions has the greatest social cost?" Tags="|chemical-engineering|combustion|transportation|environmental-engineering|air-quality|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2140" PostTypeId="2" ParentId="2091" CreationDate="2015-03-19T01:35:57.190" Score="7" Body="&lt;p&gt;The answer is slightly more complex than what the OP proposes -- this is due to two factors:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The interconnect must be &lt;em&gt;fail safe&lt;/em&gt; -- i.e. if the interconnect circuit fails, the system must be able to detect this and report a problem to both highway and railroad maintenance crews.  To this end, special supervised circuits are used, using two relays in opposite states to detect a failure of any single relay to operate correctly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The interconnect must trigger the traffic light preemption cycle sufficiently before the railroad crossing signals trigger to allow queues to clear off the track before the train arrives.  This may require both extended advance warning times (in excess of the 20 second regulatory minimum) and the use of advance warning outputs from the railroad grade crossing predictor -- just about all constant warning time predictors support advance preemption, but older DC or AC/DC three-track-circuit systems and Audio Frequency Overlay detectors may not.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Furthermore, there are special programming concerns on the highway side as well -- the grade crossing preemption cycle must be set up correctly to provide sufficient track clearance green before it blocks all movements over the grade crossing, either using red arrows or illuminated blank-out signs to prohibit turning movements onto the tracks, while also allowing pedestrians to clear from the railroad crossing (this can be critical in geometries where the railroad crossing bisects or is in close proximity to the intersection).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preemption reservice (where the traffic signal controller handles a second preemption request that closely follows the first) and extra logic on both sides (highway and rail) may be necessary when multiple tracks or switching movements are involved, and coordinated signalization and preemption are needed when multiple traffic signals are in close proximity to the grade crossing.  Diagonal crossings (either cutting multiple approaches or bisecting the intersection) pose even more challenges, sometimes requiring multiple grade crossing predictors to be interconnected to each other as well as to the highway traffic signal controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more gory detail on the intricacies of this than you ever need, the FHWA has an &lt;a href=&quot;http://safety.fhwa.dot.gov/hsip/xings/com_roaduser/07010/sec04b.cfm#i&quot; rel=&quot;nofollow noreferrer&quot;&gt;excellent handbook chapter on this topic&lt;/a&gt;.  In particular, Exhibit 3 in the pre-emption sidebar (reproduced below) provides a sample diagram of a fail-safe interconnect relay circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LfOSR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LfOSR.jpg&quot; alt=&quot;FHWA Highway-Rail Grade Crossing Handbook, Exhibit 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="964" LastEditorUserId="964" LastEditDate="2018-11-07T23:45:18.610" LastActivityDate="2018-11-07T23:45:18.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2142" PostTypeId="1" CreationDate="2015-03-19T09:27:53.680" Score="1" ViewCount="3372" Body="&lt;p&gt;I am trying to calculate the effect of insulation and weather conditions on a cottage. The cottage is made from a timber frame while the panels are of 4mm plywood. The weather conditions vary throughout the year between 0 and 30 Celsius. Cottage is around 25 m2 with 2.2 m height, consider as single room.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; What is the indoor temperature range if the insulation material used has an R-2.0?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; If we are to install a timber furnace, how much timber do we need to maintain the room at optimal conditions during the winter ? How long will it take to reach the optimal temperature ? (assume optimal is 25 Celsius)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="971" LastEditorUserId="368" LastEditDate="2015-03-19T17:56:43.603" LastActivityDate="2015-07-06T23:01:12.190" Title="How to calculate indoor temperature?" Tags="|heat-exchange|hvac|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2143" PostTypeId="1" CreationDate="2015-03-19T12:34:24.747" Score="4" ViewCount="426" Body="&lt;p&gt;Is it possible to make a homemade Q-Thruster assuming you cover all the safety requirements? I know the one made by (&lt;a href=&quot;http://cannae.com/&quot; rel=&quot;nofollow&quot;&gt;http://cannae.com/&lt;/a&gt;) uses a magnetron and resonate cavity to bounce the microwaves, and honestly seems easy to build. All I would need is to get the magnetron from a microwave and build a resonate cavity chamber to bounce the microwaves.&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="282" LastEditDate="2015-03-19T22:50:08.737" LastActivityDate="2015-03-19T22:50:08.737" Title="Is it possible to make your own Q-Thruster?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2144" PostTypeId="2" ParentId="2143" CreationDate="2015-03-19T13:00:14.807" Score="3" Body="&lt;p&gt;Yes, you could.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if it failed to work in the same manner that theirs reportedly did, you'd be in trouble, as NASA would not test it for you :-). If yours DID work then NASA may be interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ie there was considerable debate as to whether it really works - despite it having &quot;been shown&quot; to do so by NASA testing. Updates referenced below indicate that NASA is still working on it, albeit with a limited budget, and that results are still positive, as of February this year (2015).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yours failed to work you'd be on your own.&lt;br&gt;&#xA;However, it's conceivable that NASA and others MAY be willing to provide guidance.&lt;br&gt;&#xA;(This is an alteration of my initial more negative conclusion).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Good Wikipedia overview supplied by &lt;strong&gt;Greg Mueller&lt;/strong&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Quantum_vacuum_plasma_thruster&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Quantum vaccuum plasma thruster&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excellent summary of NASA results with some pictures and charts - February 6th 2015.&lt;br&gt;&#xA;&lt;a href=&quot;http://nextbigfuture.com/2015/02/update-on-emdrive-work-at-nasa.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Update on EMDrive work at NASA Eagleworks&lt;/strong&gt;&lt;/a&gt; - some good links&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.libertariannews.org/wp-content/uploads/2014/07/AnomalousThrustProductionFromanRFTestDevice-BradyEtAl.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Online copy of NASA report claiming US government copyright waiver&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20140006052.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA test summary - 2013&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA paper abstract &lt;a href=&quot;http://arc.aiaa.org/doi/abs/10.2514/6.2014-4029&quot; rel=&quot;nofollow noreferrer&quot;&gt;Anomalous Thrust Production from an RF Test Device Measured on a Low-Thrust Torsion Pendulum&lt;/a&gt; - for $ paper. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reported results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/faoYc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hgree.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-03-19T13:16:43.067" LastActivityDate="2015-03-19T13:16:43.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2145" PostTypeId="2" ParentId="2142" CreationDate="2015-03-19T13:24:45.323" Score="3" Body="&lt;p&gt;The indoor temperature in the unheated case will be just the low pass filtered outdoor temperature.  In steady state, therefore, the indoor temperature will be the same as the outdoor - at first glance.  However, sunlight heating one or two sides or coming in through a window can make a big difference.  Basically, there are too many variables to give a reasonable prediction in the real-life case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we can do the math on the parts you have told us.  I think those values are highly suspect, since the overall effective insulation will be quite different from just what the walls are rated as, but I'll use your numbers anyway.  This is less of a useful answer to your particular problem than to show how to analyze something like this if you did actually have all the pertinent information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common &quot;R value&quot; rating is how many &amp;deg;F rise needs to be across one square foot to transfer 1 BTU per hour.  You say to approximate the room as a square box that has 25 m&amp;sup2; floor and ceiling area, and 2.2 m wall height.  The box therefore is 5 x 5 x 2.2 m in dimension.  It's not clear whether the floor should count towards the heat loss.  Let's ignore it for this analysis to simplify things.  Each wall is (5 m)(2.2 m) = 11 m&amp;sup2;, and you have 4 of them, so 44 m&amp;sup2; for the walls.  That plus the 25 m&amp;sup2; roof is a total area of 69 m&amp;sup2; = 743 ft&amp;sup2;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to find the temperature difference that is to be maintained across this surface.  You say you want to maintain 25&amp;deg;C = 77&amp;deg;F (wow, that's wastefully warm!) inside, but that's only half the spec.  We also need to know how cold it is on the outside of that wall, which you say is 0&amp;deg;C = 32&amp;deg;F worst case.  That means the temperature difference is 45&amp;deg;F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we've defined the problem as how much heat power does it take to maintain 45&amp;deg;F across 743 ft&amp;sup2; of R-2 insulation.  From the description of R-value above, you can see that each square foot will transfer 22.5 BTU/h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(22.5 BTU / h ft&amp;sup2;)(743 ft&amp;sup2;) = 16.7 kBTU/hour&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you want a heater that can put out 15,000-20,000 BTU/hour.  This is a spec that should be directly given for any commercially made heater.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="282" LastEditDate="2015-07-06T23:01:12.190" LastActivityDate="2015-07-06T23:01:12.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2146" PostTypeId="2" ParentId="2133" CreationDate="2015-03-19T13:59:50.767" Score="4" Body="&lt;p&gt;I was trained to operate diesel electric locomotives for the Canadian Pacific Railway about 20 years ago. I'm not sure if the technology has been updated, likely not that much. The other posts about how the brakes work are correct. Air is pumped from the locomotives to fill reservoirs on each car to be able to release the brakes. Any large loss of pressure along the train applies the brakes. From the locomotive there are 3 ways to initiate emergency breaking. Press the engineer emergency brake button or let the timer expire on the engineer emergency brake button. (setup so if something happens to the engineer it will automatically stop the train). There is a also an emergency brake lever for the conductor to pull on his side of the cabin. The last way is change the direction of the throttle form forward to reverse or reverse to forward. All of these will release the air pressure and cause all the brakes to be applied both the train car brakes and locomotive brakes. &lt;/p&gt;&#xA;" OwnerUserId="977" LastActivityDate="2015-03-19T13:59:50.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2147" PostTypeId="1" AcceptedAnswerId="2233" CreationDate="2015-03-19T18:21:05.420" Score="6" ViewCount="1593" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Jewel_bearing&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeweled bearings&lt;/a&gt; use a metal spindle with a jewel lined pivot hole (pictured below).  They have been important in the manufacture of mechanical timepieces since the early 1700's.  Their importance lies in the fact that they can be machined to very high accuracy, they have low and predictable friction, and their hardness gives them very low wear.  They are known for operating in sealed environments for decades with no servicing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the advent of synthetic crystal manufacturing in the early 1900's jeweled bearings became the standard for every important bearing in mechanical timepieces.  Every quality mechanical watch manufactured these days, for instance, has at least 17 jeweled bearings for the 'going train'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{150px}$&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Watch_jewel_bearing_and_capstone.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wDmS2m.png&quot; alt=&quot;Jeweled bearing from wikimedia commons&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are there any other common applications for jeweled bearings besides mechanical timepieces?&lt;/li&gt;&#xA;&lt;li&gt;If so, what are the largest jeweled bearings used for?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-03-19T19:59:51.797" LastActivityDate="2016-03-23T18:16:44.033" Title="Are jeweled bearings commonly used for applications other than mechanical timepieces?" Tags="|mechanical-engineering|friction|bearings|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2148" PostTypeId="2" ParentId="2071" CreationDate="2015-03-20T00:55:56.690" Score="5" Body="&lt;p&gt;No, crossed communications would be a total non-issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it like a home wireless network. All your devices at home can talk to each other on your WiFi network, and the same is true for your neighbour. But your devices can't talk to your neighbour's devices because they are on a different wireless network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such wireless networks in planes would have to be far more robust than your typical home network, as there are other potential issues:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Accidental interference from another craft that doesn't cross-communicate between craft, but potentially blocks communication on one or both networks.&lt;/li&gt;&#xA;&lt;li&gt;Deliberate interference (jamming) of the communications by for example a terrorist on the plane or nearby.&lt;/li&gt;&#xA;&lt;li&gt;Deliberate interception or spoofing of communications to hi-jack the plane. This would require hacking into the secure network and would therefore be much harder than jamming.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some ideas to mitigate these problems might be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Directional antennae on the wireless equipment. This could help to make jamming or interception much harder because other signals would be blocked so any would-be hacker would need to have their equipment physically located directly between the antennae.&lt;/li&gt;&#xA;&lt;li&gt;Multi-part encryption keys and certificates in the devices and stored on a card held by the pilot. This would allow the equipment to verify commands were coming from a verified source (the pilot/cockpit) and also ensure that nobody else could listen in or intercept/spoof the commands.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-03-20T00:55:56.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2149" PostTypeId="1" AcceptedAnswerId="2170" CreationDate="2015-03-20T02:28:52.070" Score="6" ViewCount="595" Body="&lt;p&gt;I'm creating a light up dancing pole. This requires a translucent material that is strong enough to support the weight and force of a human being, even when it's hollowed out to so LEDs can go inside. The material must also be reasonably affordable, so something like diamond is out of the question.  I figured that &lt;strong&gt;polycarbonate&lt;/strong&gt; was my best bet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some searching and found a polycarbonate tube with the right dimensions (8ft length, ~45mm outer diameter).  Here are the exact specs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Length: 8 ft &lt;/li&gt;&#xA;&lt;li&gt;Outer Diameter: 1.750 in.&lt;/li&gt;&#xA;&lt;li&gt;Inner Diameter: 1.500 in.&lt;/li&gt;&#xA;&lt;li&gt;Wall Thickness: 0.125 in&lt;/li&gt;&#xA;&lt;li&gt;Weight: 0.332 lb/ft&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is whether this pole can support the torque of an average-strength 150lb human dancer.  Not someone who is using all their bodily strength to destroy the thing, but apply force at different angles and forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be secured to the ground by bolting it to a weighted platform.  My concern is whether the pole could potentially break.&lt;/p&gt;&#xA;" OwnerUserId="985" LastEditorUserId="160" LastEditDate="2015-03-20T14:13:24.350" LastActivityDate="2015-09-03T14:31:26.717" Title="Can polycarbonate dance pole support the torque of a dancer?" Tags="|mechanical-engineering|materials|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2150" PostTypeId="1" CreationDate="2015-03-20T04:23:00.027" Score="2" ViewCount="2096" Body="&lt;p&gt;I have made a VDG generator with a rubber band as the belt and a glass roller. It doesn't seem to work. I think the rubber band may be conductive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of using other materials for the band and the roller, such as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;+------------+--------------------------+&#xA;| Roller     | Band                     |&#xA;+------------+--------------------------+&#xA;| Glass      | Rubber from a balloon    |&#xA;| Glass      | Strip from a plastic bag |&#xA;| Human Hair | Rubber from a balloon    |&#xA;| Glass      | Silk                     |&#xA;| Glass      | Felt                     |&#xA;+------------+--------------------------+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which of these pairs would work, and why?&lt;/p&gt;&#xA;" OwnerUserId="986" LastEditorUserId="368" LastEditDate="2015-03-20T21:37:50.517" LastActivityDate="2020-08-20T03:44:05.773" Title="What materials can I use for the roller and band of a DIY Van de Graaff generator?" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2151" PostTypeId="1" CreationDate="2015-03-20T04:25:41.153" Score="9" ViewCount="3692" Body="&lt;p&gt;Many materials have a specific heat which varies with temperature, especially as the temperature change grows large.  How does one calculate the heat energy that an object receives in this case?  Can we simply use the specific heat capacity at the beginning temperature or the end temperature?&lt;/p&gt;&#xA;" OwnerUserId="865" LastEditorUserId="282" LastEditDate="2015-03-23T23:47:14.627" LastActivityDate="2015-04-29T06:56:12.023" Title="How can one calculate the change in heat energy when the specific heat varies with temperature?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|energy|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2152" PostTypeId="2" ParentId="1722" CreationDate="2015-03-20T04:29:21.747" Score="8" Body="&lt;p&gt;I can only speak for the industry I have worked in (heavy machinery). I have only seen Kalman filters used in practice as observers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the data sources in heavy machinery tend to be quite noisy (pressure or accelerometer sensors). Kalman filters (as compared to simpler Luenberger observers) provide better resilience when faced with high noise levels. I have seen them overall behave more robustly than Luenberger observers. I have seen extended Kalman filters used for sensor fusion in nonlinear systems as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The various fancier methods may often have increased computing requirements which make implementation harder in embedded microprocessors. Additionally, the general popularity of Kalman filters means that there is a higher chance managers and engineers outside of controls areas have at least heard of them before. This sort of &quot;brand recognition&quot; can help when selling a solution internally in a large company that isn't focused on controls. Along with this, the support in various libraries or in packages such as Simulink/Matlab is quite old and has been stress tested considerably already.&lt;/p&gt;&#xA;" OwnerUserId="325" LastEditorUserId="325" LastEditDate="2015-03-20T04:35:09.760" LastActivityDate="2015-03-20T04:35:09.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2153" PostTypeId="2" ParentId="2151" CreationDate="2015-03-20T05:21:49.177" Score="6" Body="&lt;p&gt;Neither. In this sort of situation, there's no &quot;simple&quot; linear solution; you need to use integral calculus to add up the incremental heat absorbed at each temperature along the way. The only time that this calculation becomes a simple multiplication is when the quantity being integrated (the specific heat) is a constant over the range of the integration.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-03-20T05:21:49.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2154" PostTypeId="2" ParentId="2150" CreationDate="2015-03-20T07:04:52.957" Score="2" Body="&lt;p&gt;As well as mechanical properties and lack of conductivity you need a means of creating the charge to be transferred. The following table gives the relative amount and the polarity of charge generated by various combinations of material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are may such lists available but this looks better than many, and includes some comments on 'incorrect&quot; information available elsewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Santoprene rubber belt with a polyurethane roller looks like an interesting combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alpha Lab Inc's &lt;a href=&quot;http://www.trifield.com/content/tribo-electric-series/&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;The TriboElectric Series&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polyurethane foam on Teflon looks worth trying!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When researching materials, the final comment on the above referenced page is a warning about inaccurate data and may be worth noting:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A triboelectric series table has been circulating on the internet, and it contains various inaccuracies. Though attribution is rarely given, it appears to be mostly from a 1987 book. It lists air as the most positive of all materials, polyurethane as highly negative, and various metals being positive or negative, apparently based on their known chemical electron affinities, rather than on electrostatic experiments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The page provides additional details to back their warning.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Related formulae: &lt;a href=&quot;http://www.trifield.com/content/electrostatic-formulas/&quot; rel=&quot;nofollow&quot;&gt;Electrostatic Formulas for Force, Voltage, Discharge Time etc. on Charged Samples or Surfaces&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorDisplayName="user16" LastEditDate="2015-03-22T00:14:25.733" LastActivityDate="2015-03-22T00:14:25.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2155" PostTypeId="2" ParentId="2151" CreationDate="2015-03-20T09:56:13.567" Score="5" Body="&lt;h2&gt;Neither.&lt;/h2&gt;&#xA;&lt;p&gt;As has already been pointed out, this is not trivial to do, but here is a suggested method:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;accurately measure out a certain quantity of fuel, then burn that fuel and use a material with a very constant or otherwise well known specific heat capacity to determine how much energy your test piece is receiving through time by recording it's temperature.&lt;/li&gt;&#xA;&lt;li&gt;use the same quantity of fuel, in the same apparatus, with a test piece of identical geometric properties but a different material and repeat the experiment. This time you assume the energy which your test piece receives based on step 1 and use the recorded temperature to determine the specifc heat capacity of the material.&lt;/li&gt;&#xA;&lt;li&gt;now that you have the specific heat capacity curve for this material, use it like any other material but integrate your curve over the temperature span you measure to determine the quantity of heat energy absorbed.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;This method isn't perfect, it relies on linear superposition which isn't perfectly valid for temperature as some factors of heat exchange have a non-linear dependancy, but it's not a bad method for &amp;quot;calibrating&amp;quot; your material at a basic level.&lt;/p&gt;&#xA;" OwnerDisplayName="user815" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-03-20T09:56:13.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2156" PostTypeId="2" ParentId="2151" CreationDate="2015-03-20T11:35:31.937" Score="10" Body="&lt;p&gt;In a similar vein to my answer about &lt;a href=&quot;https://engineering.stackexchange.com/a/2116/272&quot;&gt;calculating the lever force in a continuous situation&lt;/a&gt;; you need to use integration.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start by taking the standard heat law that you are familiar with&#xA;$$&#xA;\Delta Q=c\ m\ \Delta T&#xA;$$&#xA;and replacing the $\Delta$s with differentials:&#xA;$$&#xA;dQ=c(T)\ m\ dT.&#xA;$$&#xA;This new equation reads: For an infinitesimal (very tiny) change in temperature, I get an infinitesimal (very tiny) change in the heat. In the limit of infinitesimals everything is linear so this simple linear equation still holds.  Now you simply sum up all of the infinitesimal changes in the heat flux using integration&#xA;$$&#xA;\Delta Q=m\int_{T_i}^{T_f}c(T)\ \ dT.&#xA;$$&#xA;If you don't actually want to do the integration, that's ok.  Matlab will have no problem doing this for you, and the Matlab approach works even if you don't have an analytic function to describe $c(T)$ (i.e. you only have data).  If you don't have access to Matlab, then use &lt;a href=&quot;https://www.python.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Python&lt;/a&gt;.  It's free, open source, and incredibly powerful.&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-03-20T20:27:08.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2157" PostTypeId="1" AcceptedAnswerId="2160" CreationDate="2015-03-20T14:51:32.553" Score="5" ViewCount="780" Body="&lt;p&gt;One of the places I frequent recently had the curb and sidewalk rebuilt by the city.  Previously, there was a roughly continuous transition from the street to the rest of the private drive next to the building.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the rebuild, the sidewalk was leveled out which created two sloped areas coming from the street.  The problem with this new design is that even at reasonable speeds, vehicles are somewhat bottomed out from the up-down-up motion created by the two slopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it matters, this is a transition point from a public street and sidewalk onto private property for a non-residential building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This is what the entryway looks like now&lt;/em&gt;:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9MGik.png&quot; alt=&quot;double slope picture&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What it used to look like&lt;/em&gt;:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/RGjvV.png&quot; alt=&quot;previous version of entryway&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note, I have exaggerated the slopes in the illustrations in order to clarify what I'm asking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a common term used in roadway design that describes the double slope?&lt;/strong&gt;  I would like to be able to research the term to see if there was any code violation.  At a minimum, I'd like to use the correct terminology when I ask the city about rectifying the problem. &lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-03-21T03:38:07.550" Title="Term for double slope from street" Tags="|civil-engineering|terminology|roadway|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2158" PostTypeId="1" CreationDate="2015-03-20T16:08:53.700" Score="8" ViewCount="656" Body="&lt;p&gt;I am planning to use &lt;a href=&quot;http://en.wikipedia.org/wiki/Polydimethylsiloxane&quot; rel=&quot;noreferrer&quot;&gt;polydimethylsiloxane&lt;/a&gt; (PDMS) to create a phantom for mimicking tissue for imaging. For this, I plan on using &lt;a href=&quot;http://www.sigmaaldrich.com/catalog/product/aldrich/761036?lang=en&amp;amp;region=US&quot; rel=&quot;noreferrer&quot;&gt;SYLGARD® 184&lt;/a&gt;, which is a viscous liquid. When adding other components to it, air bubbles will be introduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have learned that one way (available to me) to remove bubbles / de-gas the solution is by using a vacuum pump to create negative pressure. What negative pressure is sufficient to remove the air from the gel-like solution?&lt;/p&gt;&#xA;" OwnerUserId="991" LastEditorUserId="368" LastEditDate="2015-03-26T18:33:06.423" LastActivityDate="2018-08-19T21:17:50.997" Title="Degassing silicone solution with negative pressure" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2159" PostTypeId="2" ParentId="2157" CreationDate="2015-03-20T17:18:29.033" Score="0" Body="&lt;p&gt;I'd be surprised if there's a common term for the double slope. Based on some &lt;a href=&quot;http://www.4ishgd.valencia.upv.es/index_archivos/55.pdf&quot; rel=&quot;nofollow&quot;&gt;internet searching for terms used in the US&lt;/a&gt;, the slope is called a &lt;strong&gt;grade&lt;/strong&gt;, but I've not found anything particular to your situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd suggest possible wording for your discussion with the city would be &lt;em&gt;&quot;the multiple changes in grade tend to result in the underside of vehicles scraping on the driveway&quot;&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-03-20T17:18:29.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2160" PostTypeId="2" ParentId="2157" CreationDate="2015-03-20T19:52:32.077" Score="4" Body="&lt;p&gt;&lt;strong&gt;Name&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure that it has a specific name. It is just, &quot;the way driveways are when there is a sidewalk.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Geometry&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like most things related to roadway design in the US:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It has been studied. There is a well named report available called, &lt;a href=&quot;http://onlinepubs.trb.org/onlinepubs/nchrp/nchrp_w151.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geometric Design of Driveways&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;It can vary based on local codes. See report above.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The report linked above studied driveway geometry requirements from around the US and tried to do some modeling as well. The graphical results are shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eqBKC.png&quot; alt=&quot;Driveway Recommendations&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion that I got from skimming through the report was that lots of driveways have bad geometry; drive slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless there is a local code that the drive is in violation of, it might not be something that the city is willing to fix.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-20T19:52:32.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2161" PostTypeId="2" ParentId="2139" CreationDate="2015-03-20T20:21:07.257" Score="5" Body="&lt;h1&gt;The easy question&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Do any of the components above have higher costs to society than CO&lt;sub&gt;2&lt;/sub&gt;&#xA;per same unit of measure (e.g. gram, kilogram, or ton)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Yes, absolutely.&lt;/strong&gt; This should come as no surprise given the composition of the exhaust—it's mostly inert gases. Water vapor, CO&lt;sub&gt;2&lt;/sub&gt; and nitrogen gas are all &lt;em&gt;products&lt;/em&gt; of the various reactions taking place in the vehicle's engine and exhaust system. An ideal combustion reaction between air and a simple hydrocarbon has water vapor and CO&lt;sub&gt;2&lt;/sub&gt; as its only products; these are much less reactive than either the hydrocarbon fuel or the oxygen and less reactive &lt;em&gt;usually&lt;/em&gt; implies less societal cost.&lt;/p&gt;&#xA;&lt;p&gt;CO&lt;sub&gt;2&lt;/sub&gt; gets the attention that it does because of its role in climate change, which is driven by its ubiquity and abundance as a component of anthropogenic emissions. CO&lt;sub&gt;2&lt;/sub&gt; does not contribute significantly to acid rain; it has no acute* or chronic health impacts; being odorless and colorless, it has no nuisance or aesthetic impact. It's simply a &amp;quot;repeat offender&amp;quot; in terms of society's industrial development because of our reliance on combustion as an energy source.&lt;/p&gt;&#xA;&lt;p&gt;So, recognizing that the societal costs associated with CO&lt;sub&gt;2&lt;/sub&gt; emissions are entirely due to its role in climate change, and recognizing that several of the other exhaust constituents contribute to public health, aesthetic and direct economic costs, it is enough to show that any of these has &lt;em&gt;at least as much&lt;/em&gt; climate impact as CO&lt;sub&gt;2&lt;/sub&gt;. And that is common knowledge in the field—if you look at &lt;a href=&quot;https://www.ipcc.ch/publications_and_data/ar4/wg1/en/ch2s2-10-2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this table of direct global warming potentials (GWP) from  the IPCC&lt;/a&gt;, you'll see that both methane and nitrous oxide (both &lt;a href=&quot;http://rael.berkeley.edu/sites/default/files/very-old-site/Climatic_Change.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;trace components of vehicle exhaust&lt;/a&gt;) have significantly higher GWP** than CO&lt;sub&gt;2&lt;/sub&gt;.&lt;/p&gt;&#xA;&lt;p&gt;That answers your question but it doesn't tell us anything particularly &lt;em&gt;useful&lt;/em&gt; because pollutants in the real world are emitted at different concentrations.&lt;/p&gt;&#xA;&lt;h1&gt;The hard question&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If they have higher costs per unit of measure, do they have a higher&#xA;cost than CO&lt;sub&gt;2&lt;/sub&gt; based on their component percentage in typical&#xA;emissions? The component may be very costly per ton but it makes up a&#xA;very small amount of typical emissions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Your first question was somewhat trivial to answer; your second is not. There is no typical emissions in a global sense because vehicle engine, fuels and emissions technologies are so varied. There is no typical impact because geography and weather are driving factors. My tentative answer, noting that &lt;em&gt;I do not consider myself an expert in this area,&lt;/em&gt; is that &lt;strong&gt;we probably don't know&lt;/strong&gt;. And the reason for this is not that the impacts of these other pollutants are unknown but that the &lt;em&gt;cost&lt;/em&gt; of those impacts varies in so many ways, e.g.:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The value of lost worker productivity, the cost of health care and access to health care vary hugely throughout the world.&lt;/li&gt;&#xA;&lt;li&gt;Comparing short-term and long-term costs must be done on an equal basis, but what form that equal basis takes is always to some degree arbitrary or subjective.&lt;/li&gt;&#xA;&lt;li&gt;Chronic health impacts are a mess to unravel; there are so many different influences at work over the course of a person's lifetime, and &lt;em&gt;not&lt;/em&gt; independently of one another.&lt;/li&gt;&#xA;&lt;li&gt;Aesthetic and nuisance impacts are extremely difficult to quantify; usually reduced to implications for the tourism industry, which ignores the impact to permanent residents.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Translating impact to social cost&lt;/h2&gt;&#xA;&lt;p&gt;Take the example of (ground level) ozone. It's not a component of vehicle exhaust but its precursors are and its concentration is regulated under the &lt;a href=&quot;http://www.epa.gov/air/criteria.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;National Ambient Air Quality Standards&lt;/a&gt; because during the 20th century the U.S. experienced photochemical smog events so severe that &lt;a href=&quot;http://www.wired.com/2010/07/0726la-first-big-smog/&quot; rel=&quot;nofollow noreferrer&quot;&gt;residents mistook them for chemical warfare attacks&lt;/a&gt;. The main precursors of ground level ozone are NOx and reactive organic gases, also known as volatile organic compounds (VOC) and &lt;a href=&quot;http://www.epa.gov/otaq/models/nonrdmdl/nonrdmdl2005/420r05015.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;divided up in various ways under other acronyms&lt;/a&gt; that you don't necessarily need to understand, but should be aware of. In any case, these are both components of vehicle exhaust with their own social costs to consider (in the case of NOx, mainly acid rain and acute health impacts; in the case of VOC, mostly chronic health impacts). In combination, they react to form ozone, and &lt;em&gt;this&lt;/em&gt; impact is &lt;em&gt;very&lt;/em&gt; important—but cannot be attributed to either pollutant alone. Here's a figure showing ozone concentration as a function of NOx and VOC concentrations:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/N8fPF.jpg&quot; alt=&quot;nox-voc-ozone&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Notice the &amp;quot;VOC Limited&amp;quot; and &amp;quot;NOx Limited&amp;quot; regions of the graph. These indicate conditions where there is already plenty of one precursor, and adding more of that precursor will not appreciably change the resulting ozone concentration.&lt;/p&gt;&#xA;&lt;p&gt;Now, consider the implications of this for your question. How would you quantify the impact of one ton of NOx? Are you going to assume some ambient VOC concentration? If so, you'll have to work hard to justify it, and it will be highly dependent on location. NOx is mainly anthropogenic*** but VOC comprises many different compounds, &lt;a href=&quot;http://www.borenv.net/BER/pdfs/ber5/ber5-095.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;some of which are emitted from nice, clean forests&lt;/a&gt;. (Thanks a lot, &lt;a href=&quot;http://en.wikipedia.org/wiki/FernGully:_The_Last_Rainforest&quot; rel=&quot;nofollow noreferrer&quot;&gt;FernGully&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;p&gt;Is it impossible? No, I could give you a number right now, made of equal parts sunshine dust and horse manure. More importantly, someone—almost certainly not an engineer, note—could spend a whole lot of time and money to do an analysis with some amount of merit and utility that &lt;em&gt;might&lt;/em&gt; be compatible enough with the EPA's dollar-value analysis of CO&lt;sub&gt;2&lt;/sub&gt; emissions to reasonably compare pollutants. To be honest, I don't think there's any need for such an analysis of the other constituents of vehicle exhaust, and if one exists I'm certainly not aware of it.&lt;/p&gt;&#xA;&lt;h2&gt;What's special about CO&lt;sub&gt;2&lt;/sub&gt;?&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I know that &lt;em&gt;cost&lt;/em&gt; is kind of hard to define in these cases. It does&#xA;seem possible though, because in reports on greenhouse gas a monetary&#xA;figure is always given.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Despite the challenges, someone must have at least &lt;em&gt;tried&lt;/em&gt; to quantify all these costs—academics need to do research, after all, and where we can identify impacts, we should be able to translate them into costs &lt;em&gt;somehow&lt;/em&gt;, right?&lt;/p&gt;&#xA;&lt;p&gt;In theory, sure. In practice, the scope of the US EPA's economic analysis of the &amp;quot;Social Cost of Carbon&amp;quot; (SCC) is (to my knowledge, again disclaiming the status of &amp;quot;expert&amp;quot;) unique. Partly this is because of the huge role of CO&lt;sub&gt;2&lt;/sub&gt; in &lt;em&gt;both&lt;/em&gt; human industry &lt;em&gt;and&lt;/em&gt; anthropogenic climate change. Partly it's by necessity, as a result of political, economic and social obstacles—this analysis is a way of justifying legislation, motivating industry self-regulation and influencing consumer choices. Market-based solutions, i.e., putting a dollar cost on everything, are currently in vogue, which drives funding for this type of research.&lt;/p&gt;&#xA;&lt;p&gt;On the other hand, there are no emission trading programs for uncontroversial air toxics like &lt;a href=&quot;http://www.epa.gov/airtoxics/hlthef/butadien.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;1,3-Butadiene&lt;/a&gt;, so the sort of research performed on its impacts is not geared toward producing a raw dollar value that could be directly compared to the SCC numbers. Never mind doing the same thing for every VOC species that comes out of your tailpipe.&lt;/p&gt;&#xA;&lt;p&gt;I'll give you another reason that CO&lt;sub&gt;2&lt;/sub&gt; is a special case: It's really hard to control. In fact, controlling the &lt;em&gt;other&lt;/em&gt; components of emissions usually involves producing &lt;em&gt;more&lt;/em&gt; CO&lt;sub&gt;2&lt;/sub&gt;, via more complete combustion in the engine and &lt;a href=&quot;http://www.arb.ca.gov/research/seminars/mooney/mooney.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;catalytic conversion in the exhaust system&lt;/a&gt;. If we could produce a car that would emit nothing &lt;em&gt;but&lt;/em&gt; CO&lt;sub&gt;2&lt;/sub&gt; and water, that would be a wonderful thing. Meanwhile, not much happens to the CO&lt;sub&gt;2&lt;/sub&gt; after it's emitted; some of it dissolves into the oceans, but for the most part it persists in the carbon cycle.****&lt;/p&gt;&#xA;&lt;p&gt;Other air pollutants have been simpler to control and/or less persistent in the atmosphere. Lead has been eliminated from gasoline in most parts of the world (&lt;a href=&quot;http://www.lead.org.au/fs/Leaded_Petrol_Possibly_Still_Sold_for_Road_Use_20110617.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;but not all&lt;/a&gt;). The social costs therefrom are related to chronic health impacts; these would be calculated (by economists, note—not engineers) by taking into account things like the cost of medical care, what type would be received, the probability of receiving it, the value of lost productivity to the economy—all of this depends on where you are in the world. Similarly, the social costs due to particulate emissions are localized (with &lt;a href=&quot;http://www.epa.gov/blackcarbon/2012report/Chapter2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;one notable exception&lt;/a&gt; that I don't believe is factored into the SCC analysis).&lt;/p&gt;&#xA;&lt;h2&gt;Partial answers?&lt;/h2&gt;&#xA;&lt;p&gt;If you limit the scope of your inquiry, there &lt;em&gt;is&lt;/em&gt; research out there that you might find helpful. For example, you could take a look at &lt;a href=&quot;http://www.arb.ca.gov/research/apr/past/01-335.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this 2006 analysis of air pollution control regulations in California&lt;/a&gt;. It's organized by regulation, and the applicable mobile source regulation is on page 48. The ex ante analysis actually gives some numbers:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The incremental cost of the Rule was estimated at \$65 per vehicle. The&#xA;cost effectiveness of the proposed regulations was estimated at \$1.20&#xA;per pound of NMHC and \$0.10 per pound of CO reduced. ... The cost&#xA;effectiveness of benzene reduction, attributing the entire cost to the&#xA;benzene benefit, was from \$4.0 to \$31 million per cancer case reduced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Keep in mind, this is &lt;em&gt;not&lt;/em&gt; the social cost you're seeking, but the &lt;em&gt;predicted net cost of compliance&lt;/em&gt; with the particular regulation. And if you keep reading to the ex post facto analysis, you read that the rule &amp;quot;did not lend itself well to ex post cost estimation&amp;quot; because automakers don't want to disclose process and accounting information they consider proprietary and it's impossible to isolate the effects of one regulation in an environment that sees new regulations coming into effect constantly. That's not what makes your question difficult to answer, but I think the challenges are analogous.&lt;/p&gt;&#xA;&lt;p&gt;And that's after narrowing the scope of the inquiry in several dimensions to a single state, a single regulation, one component of social cost and two categories of pollutant.&lt;/p&gt;&#xA;&lt;h2&gt;An alternative answer&lt;/h2&gt;&#xA;&lt;p&gt;It's important to recognize that passenger vehicle emissions control is one of the &lt;a href=&quot;http://www.arb.ca.gov/research/health/healthup/feb08.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;major success stories of air quality management (see especially pp. 8-10)&lt;/a&gt;. As tailpipe emissions get cleaner and cleaner, CO&lt;sub&gt;2&lt;/sub&gt; would seem to be more and more the pollutant with the highest cost. But it's not as simple as that, because &lt;a href=&quot;http://onlinepubs.trb.org/onlinepubs/trnews/trnews259billioncars.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the number of tailpipes is also increasing&lt;/a&gt;; note also the predicted changes in fleet composition in Figure 1.&lt;/p&gt;&#xA;&lt;p&gt;With all this in mind, and thinking to the future, the &amp;quot;easy answer&amp;quot; to your question might be that &lt;strong&gt;no sexy dollar-value analysis is required to show that CO&lt;sub&gt;2&lt;/sub&gt; emissions are the single most important component of motor vehicle emissions by far&lt;/strong&gt; and will remain so long into the future. They are the hardest for us to deal with, except by &lt;a href=&quot;http://www.arb.ca.gov/msprog/zevprog/zevprog.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;radically changing the way we power our vehicles&lt;/a&gt;. And that's only a first step; it shifts the carbon emissions to another sector that's &lt;a href=&quot;http://www.bloomberg.com/news/articles/2015-03-18/texas-city-pulls-plug-on-fossil-fuels-with-shift-to-solar-power&quot; rel=&quot;nofollow noreferrer&quot;&gt;just as much in need of transformation&lt;/a&gt;, which in turn shifts emissions to manufacturing, etc. The idea is that, rather than simply playing a shell game with the emissions, we realize incremental overall improvements with every move.&lt;/p&gt;&#xA;&lt;p&gt;And that's the biggest difference between CO&lt;sub&gt;2&lt;/sub&gt; and these other pollutants. We literally &lt;em&gt;destroy&lt;/em&gt; hydrocarbons, NOx and CO in vehicle exhaust systems, which are already very close to 100% efficient, and could always be made more efficient (at greater cost)—but we end up producing CO&lt;sub&gt;2&lt;/sub&gt; as a result. We can shift away from burning high-sulfur fuels and &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydrodesulfurization#Process_description&quot; rel=&quot;nofollow noreferrer&quot;&gt;remove what sulfur we can't avoid&lt;/a&gt;—but this process requires energy, and as long as that energy comes from combustion, we're &lt;em&gt;again&lt;/em&gt; increasing CO&lt;sub&gt;2&lt;/sub&gt; emissions.&lt;/p&gt;&#xA;&lt;p&gt;In a sense, all roads lead to CO&lt;sub&gt;2&lt;/sub&gt;, despite the greater impact of some other exhaust components, pound-for-pound. It's effectively the terminal atmospheric cost of any pollutant we're capable of controlling, and will remain so even as we make improvements in sequestration technology and power generation. That might be ultimately the biggest reason we put the effort into translating its impact into a market cost, when we don't go to the same lengths for other pollutants.&lt;/p&gt;&#xA;&lt;h1&gt;Further Reading&lt;/h1&gt;&#xA;&lt;p&gt;This is already a very long answer, so here's a limited selection of links to resources that I anticipate various types of reader might find useful.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For the research-minded individual, I suggest browsing &lt;a href=&quot;http://www.arb.ca.gov/research/research-results.php?category=Economic%20Analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;the California Air Resources Board's list of Economic Analysis research projects&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Those more interested in policy might prefer &lt;a href=&quot;http://www.iea.org/publications/freepublications/publication/policy-pathways---energy-efficiency-in-urban-transport-systems.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Tale of Renewed Cities: A policy guide on how to transform cities by improving energy efficiency in urban transport systems&lt;/a&gt; (PDF).&lt;/li&gt;&#xA;&lt;li&gt;The layperson may be interested in reading &lt;a href=&quot;https://global.oup.com/academic/product/two-billion-cars-9780199737239?cc=us&amp;amp;lang=en&amp;amp;&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two Billion Cars: Driving Toward Sustainability&lt;/a&gt; or, if you prefer, in &lt;a href=&quot;http://www.npr.org/templates/story/story.php?storyId=99375553&quot; rel=&quot;nofollow noreferrer&quot;&gt;listening to an interview with one of its authors&lt;/a&gt; (17:39).&lt;/li&gt;&#xA;&lt;li&gt;Finally, anyone in the US considering a vehicle purchase in the near future could do worse than to seek guidance from the &lt;a href=&quot;http://www.driveclean.ca.gov/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Buyers Guide to Cleaner Cars&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;* Within at least an order of magnitude of current atmospheric levels.&lt;/p&gt;&#xA;&lt;p&gt;** It's worth mentioning here that warming potential is not the only facet of anthropogenic climate change, which also includes, e.g., ocean acidification—mainly driven by CO&lt;sub&gt;2&lt;/sub&gt;. That said, of the climate impacts we are able to predict, warming is both the most well-studied and associated with the greatest societal costs, by far.&lt;/p&gt;&#xA;&lt;p&gt;*** Non-combustion sources include fertilizer manufacturing and agricultural activity. With some amount of rhetorical contortion, it's possible to consider some of the latter to be non-anthropogenic emissions.&lt;/p&gt;&#xA;&lt;p&gt;**** The carbon cycle is much more sensitive to anthropogenic emissions than the water cycle; carbon doesn't simply precipitate out as a liquid! A small amount will dissolve into the oceans, but the processes that remove atmospheric carbon are orders of magnitude slower and less effective, as compared to water vapor.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-03-20T21:17:54.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2162" PostTypeId="1" AcceptedAnswerId="2168" CreationDate="2015-03-21T01:23:41.477" Score="9" ViewCount="1937" Body="&lt;p&gt;Magnetic bearings work by levitating a rotating shaft so that it is not in contact with its supports. This greatly reduces the friction of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all of the literature that I have seen on magnetic bearings, the bearings are described as &quot;low friction&quot; and not &quot;no friction&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_bearing#Design&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... they do not suffer from wear, have low friction ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.synchrony.com/knowledge/how-magnetic-bearings-work.php&quot;&gt;Synchrony&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... they do not suffer from wear, they have low friction ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.steorn.com/bearings/performance/&quot;&gt;Steorn&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... A low-friction bearing ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.calnetix.com/resource/magnetic-bearings/advantages-magnetic-bearings&quot;&gt;Calnetix&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... extremely low friction and wear ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would seem that there wouldn't be any friction since the magnets are keeping the rotating shaft from touching anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does the friction come from in magnet bearings?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-21T12:51:12.867" LastActivityDate="2015-10-18T19:15:57.547" Title="What causes friction in magnetic bearings?" Tags="|mechanical-engineering|friction|magnets|tribology|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2166" PostTypeId="1" CreationDate="2015-03-21T07:36:12.390" Score="1" ViewCount="935" Body="&lt;p&gt;I am looking to buy a centrifugal monoblock pump for my laboratory. The suction and discharge sizes are mentioned as 12mm x 12mm. Does anybody know what does it mean? Does it mean 12mm OD, or does it mean 1/2&quot; BSP?&lt;/p&gt;&#xA;" OwnerUserId="994" LastEditorUserId="2015" LastEditDate="2015-12-11T18:56:26.843" LastActivityDate="2015-12-11T18:56:26.843" Title="Pump suction and discharge sizes" Tags="|mechanical-engineering|fluid-mechanics|pumps|hydraulics|piping|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2167" PostTypeId="2" ParentId="100" CreationDate="2015-03-21T09:37:23.923" Score="4" Body="&lt;p&gt;What about this.  Create the drop as usual, but use the hottest water you can in order to slow the creation of the stresses which of course will still occur.  Here's the critical step...... decrease the depth of the water with experimentation and finally, release the drop right at the water surface which should, to some extent, reduce the length of the tail or practically eliminate it.  The drop will fall at a much reduced rate considering the semi-weightless condition in the water.  Another thing to consider would be to snip the drop just before it drops.  By snipping the drop just before it drops, the tail, which cools much quicker than the head, is practically eliminated and so the head with its internal stresses are not threatened by the brittle tail.&lt;/p&gt;&#xA;" OwnerUserId="995" LastEditorUserId="995" LastEditDate="2015-03-22T07:17:07.227" LastActivityDate="2015-03-22T07:17:07.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2168" PostTypeId="2" ParentId="2162" CreationDate="2015-03-21T14:22:07.487" Score="4" Body="&lt;p&gt;The bearings themselves may have ideally 0 friction, but there will still be &lt;em&gt;some&lt;/em&gt; losses of rotational energy in the system as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the vacuum was perfect, anything that conducts electricity even a little becomes a transformer secondary in the presence of changing magnetic fields.  Conducting material in the rotating thing will dissipate power due to eddy currents caused by any external magnetic field, including the earth's.  Clearly there are strong magnetic fields in the magnetic bearing.  Even a small asymmetry in that field will cause a apparent alternating magnetic field to the spinning part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works in reverse too.  If the spinning part is magnetized, then anything on the outside that is conductive will dissipate some power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small imbalances and asymmetries in the magnetic fields of both the fixes stuff and the rotary stuff are very hard to avoid.  Conducting materials are also hard to avoid.  In the end, the best engineering tradeoff is usually to accept some small losses.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="33" LastEditDate="2015-10-18T19:15:57.547" LastActivityDate="2015-10-18T19:15:57.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2169" PostTypeId="2" ParentId="2162" CreationDate="2015-03-21T14:43:40.690" Score="5" Body="&lt;p&gt;There are two types of losses in magnetic bearings, windage and electromagnetic losses. Windage or aerodynamic loss is the dissipation of rotational energy due to the viscosity of air or other gases trapped between the rotating and stationary portions of the system. These effects are more significant in high speed applications and clearly non-existent in a vacuum sealed chamber. The electromagnetic losses can come from hysteresis and/or eddy currents in the magnets. These effects have to do with a variation in the magnetic flux due to the rotating ferromagnetic materials. Below are some references I have found useful. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/3642004962&quot; rel=&quot;nofollow&quot;&gt;Magnetic Bearings: Theory, Design and Application to Rotating&#xA;Machinery&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.femm.info/dmeeker/pdf/01333140.pdf&quot; rel=&quot;nofollow&quot;&gt;Effect of Magnetic Hysteresis on Rotational Losses in Heteropolar&#xA;Magnetic Bearings&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="920" LastEditorUserId="110" LastEditDate="2015-03-22T13:00:48.627" LastActivityDate="2015-03-22T13:00:48.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2170" PostTypeId="2" ParentId="2149" CreationDate="2015-03-21T14:55:35.927" Score="6" Body="&lt;p&gt;CAVEAT: I do not have any design experience in plastic, but by using first principle, this is how I would tackle the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing we need to know is what are the typical forces exerted on a pole during pole dance. Assuming that the dancer is spinning at not more than &lt;code&gt;45&lt;/code&gt; degrees from the axis of the pole, then the maximum centripetal force acting on the pole would be equal to the weight of the dancer (&lt;code&gt;150lb&lt;/code&gt; or &lt;code&gt;68kg&lt;/code&gt;). Given that the pole is &lt;code&gt;8ft&lt;/code&gt; or &lt;code&gt;2.44m&lt;/code&gt; in length, the maximum bending moment induced at the base of the free cantilevering pole would be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m = 68kg, a = 9.81m/s², L = 2.44m&#xA;M = F x L&#xA;  = m x a x L&#xA;  = 3.3 kNm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Polycarbonates, in general, have ultimate tensile strength in the range of &lt;a href=&quot;http://en.wikipedia.org/wiki/Polycarbonate&quot;&gt;55-75MPa&lt;/a&gt;. This is one-fourth of the tensile strength of structural steel. From the technical specifications in &lt;a href=&quot;http://www.eplastics.com/Plastics_Material_Selection_Guide.htm&quot;&gt;eplastic.com&lt;/a&gt;, the polycarbonate material specified has an ultimate tensile strength of &lt;code&gt;9500psi&lt;/code&gt; or &lt;code&gt;65.5MPa&lt;/code&gt;. Assuming this is also the yield strength for the plastic material, then the elastic section modulus can be calculated as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Do = 44.5mm, Di = 38.1mm, py = 65.5MPa&#xA;Z = π(Do⁴ - Di⁴)/32Do&#xA;  = 3.97cm³&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And the moment capacity of the pole can be conservatively calculated as such:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mc = pyZ&#xA;   = 0.26kNm&#xA;   &amp;lt; M&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Based on the above, the polycarbonate dance pole would &lt;strong&gt;not&lt;/strong&gt; be strong enough to support the force generated by a dancer.&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2015-03-21T14:55:35.927" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2171" PostTypeId="1" CreationDate="2015-03-21T18:02:54.360" Score="5" ViewCount="1116" Body="&lt;p&gt;An important factor in specifying the efficiency of photo-voltaic (PV) cells is called the fill factor (FF). &lt;a href=&quot;http://www.pveducation.org/pvcdrom/solar-cell-operation/fill-factor&quot; rel=&quot;nofollow&quot;&gt;It is defined&lt;/a&gt; as the ratio of the maximum power of the cell to the product of the open-circuit voltage $V_{oc}$ to the short-circuit current $I_{sc}$.&#xA;$$&#xA;P_{max} = V_{oc} * I_{sc} * FF&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is $FF$ constant in this equation?  I've found online that &#xA;$$&#xA;FF = \frac{V_{mp}*I_{mp}}{V_{oc}*I_{sc}}.&#xA;$$&#xA;But it wasn't specified if those values are under STC or not.  If they aren't, then how would I get $V_{mp}$ and $I_{mp}$?&lt;/p&gt;&#xA;" OwnerUserId="1000" LastEditorUserId="368" LastEditDate="2015-07-06T04:25:53.613" LastActivityDate="2015-07-06T04:25:53.613" Title="Is the fill factor of a photo-voltaic cell constant?" Tags="|electrical-engineering|energy-efficiency|renewable-energy|photovoltaics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2172" PostTypeId="1" CreationDate="2015-03-21T18:43:29.927" Score="5" ViewCount="1147" Body="&lt;p&gt;Normally, so called shelf papers are used or special coatings called kiln washes are used to prevent molten glass from adhering to the kiln shelf while it is being fired.  This is expensive and time consuming.  Can a different material be used that will not allow the molten glass to stick to it, such as a type of graphite?  This would also improve the appearance of the backside of the finished glass project.&lt;/p&gt;&#xA;" OwnerUserId="995" LastActivityDate="2021-10-31T16:33:41.257" Title="Will molten glass stick to a graphite shelf inside a kiln?" Tags="|materials|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2173" PostTypeId="1" CreationDate="2015-03-21T22:18:26.457" Score="2" ViewCount="1217" Body="&lt;p&gt;Mostly interested in understanding how engineers determine if an engineering process is stable and capable, particular in design for manufacturing (DFM) frame work for components. Some example components to consider are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Flow rate for Solenoid Valve&lt;/li&gt;&#xA;&lt;li&gt;Torque for a Brushless DC Motor&lt;/li&gt;&#xA;&lt;li&gt;Capacitance for a Capacitor&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note: There are infinite amount of examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does an engineer confirm if a vendor has a stable and capable process (incoming material)? To be cost competitive it is important to have the least amount (optimal) manufacturing test and inspection processors yet ensure the end product meets specification.  How does an engineer ensure the final product meet specification (outgoing material)?&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="282" LastEditDate="2015-03-23T22:45:21.693" LastActivityDate="2015-04-06T21:53:08.373" Title="How to determine if an engineering process is stable and capable?" Tags="|statistics|product-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2174" PostTypeId="2" ParentId="2171" CreationDate="2015-03-22T07:46:34.543" Score="2" Body="&lt;p&gt;Power output of a PV panel is driven by several exogenous factors: the amount &lt;em&gt;and spectrum&lt;/em&gt; of light on it; the presence &lt;em&gt;and distribution&lt;/em&gt; of any shadows; the voltage applied to the panel; and the temperature of the panels - which is driven partly by the ambient temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've got the power rating of the panel then you can calculate the fill factor: it's the power rating divided by the open-circuit voltage and then divided by the short-circuit current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've got the power rating, and you know you're applying maximum-power-point-tracking to the voltage, and you know the ambient temperature, then the PV datasheet will usually give you a coefficient for how quickly max power changes with changes in temperature. The datasheet may also have several curves for various combinations of insolation and temperature. Together, these will give you a decent first estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole thing is a non-linear system, and in the end, you'll only know the power out under specific conditions for a specific system by running it through those conditions.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2015-03-22T07:46:34.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2175" PostTypeId="2" ParentId="2171" CreationDate="2015-03-22T14:15:41.210" Score="2" Body="&lt;p&gt;Fill Factor is important for PV (photovoltaic) cell and panel designers but of less direct importance to users. What is usually more important are the practical parameters which result from a given fill factor. Reduced fill-factor cells will also be found to have lower efficiency of conversion, requiring larger panel areas per Watt out. Actual insolation levels are usually variable enough that variations based on fill factor are in many cases liable to be swamped by other variables. Insolation levels plus panel cleanliness, and variations across the day and season and local air conditions all add up to a produce very variable and uncertain output level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reputable manufacturers provide Vmp. Imp, Voc and Isc.&lt;br&gt;&#xA;They also specify coefficients of change per degree C of Vmp, Imp, Wmp at 25 C.&lt;br&gt;&#xA;Given all these parameters one could theoretically calculate panel output in given insolation conditions. Even after allowing for the factors mentioned above typical non MPPT installations often achieve substantial mismatch between panel Wmp operating point and load which further reduce available output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parameters are almost always at 25C panel temperature and AM1.5 insolation conditions. Fill Factor is specified by some but often isn't, and can be easily calculated from the supplied parameters.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-03-22T14:15:41.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2176" PostTypeId="1" CreationDate="2015-03-22T16:25:14.997" Score="3" ViewCount="1497" Body="&lt;p&gt;The reality is that sci-fi &lt;a href=&quot;http://en.wikipedia.org/wiki/KITT&quot; rel=&quot;nofollow noreferrer&quot;&gt;KITT&lt;/a&gt; is almost here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OtUuH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently I read articles &lt;a href=&quot;http://www.cnet.com/news/delphi-car-driving-itself-across-the-country/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Delphi car driving itself across the country&lt;/a&gt; as well as &lt;a href=&quot;http://www.nytimes.com/2015/03/20/business/elon-musk-says-self-driving-tesla-cars-will-be-in-the-us-by-summer.html?_r=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tesla cars will be self-driving this summer&lt;/a&gt;. As of 2013 four states (Nevada, Florida, California, and Michigan) have enacted some type of legislation to permit autonomous vehicles. This indicates legislators recognizing the emerging technology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bU5Qc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also these autonomous vehicle have many sensors to gather data real-time which are processed by complex software algorithm. Radar technology is one such used in front and &lt;a href=&quot;http://www.conti-online.com/www/automotive_de_en/themes/passenger_cars/chassis_safety/adas/rcta_en.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;rear cross traffic alert&lt;/a&gt; systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bgHcx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among the many potential advantages, few traffic collision is one of them. Likewise there are many obstacles too. Among them are disclaimer such as the one by Toyota &quot;&lt;a href=&quot;https://www.youtube.com/watch?v=HpcMxbrE1LQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Do not rely exclusively on rear cross traffic alert or blind spot monitoring systems. These are driver assist devices and are not a substitute for safe and attentive driving&lt;/a&gt; &quot;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the reliability of systems such as &lt;strong&gt;rear cross traffic alert system&lt;/strong&gt;, in a self-driving car the primary obstacle impeding the success of self-driving car with human as fall back option?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.kpmg.com/US/en/IssuesAndInsights/ArticlesPublications/Documents/self-driving-cars-are-we-ready.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Self-Driving Cars: Are we ready? - KPMG&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cnet.com/videos/tesla-cars-will-be-self-driving-this-summer/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tesla cars will be self-driving this summer&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.edn.com/design/automotive/4368069/Automobile-sensors-may-usher-in-self-driving-cars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Automobile sensors may usher in self-driving cars&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://abcnews.go.com/Technology/driverless-car-begin-cross-country-trip-sunday/story?id=29807224&quot; rel=&quot;nofollow noreferrer&quot;&gt;Driverless Car to Begin Cross-Country Trip Sunday&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Sudden_unintended_acceleration&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sudden unintended acceleration&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-03-29T16:19:00.227" LastActivityDate="2016-02-18T08:04:37.243" Title="Is the reliability of automotive sensor technology impeding the success of self driving car?" Tags="|sensors|automotive-engineering|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2178" PostTypeId="2" ParentId="2176" CreationDate="2015-03-22T23:32:23.890" Score="3" Body="&lt;p&gt;It's hard to know what all of the remaining technical obstacles are. Most of the obvious technical challenges have been solved already &lt;em&gt;for a single self-driving car in amongst human-driven cars&lt;/em&gt;. If self-driving cars were to become normal, this raises some possible unknown issues:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What would be the effect of every car using LIDAR, RADAR, SONAR to understand their surroundings?&lt;/li&gt;&#xA;&lt;li&gt;Would those 'active' direction and range sensors cause problems/interference with other cars or other equipment when densely packed together on busy streets?&lt;/li&gt;&#xA;&lt;li&gt;Would co-ordination between vehicles become required/desired to prevent undue hesitation or incorrect action when vehicles see other vehicles from different vendors (with different rules and assumptions about traffic behaviour)?&lt;/li&gt;&#xA;&lt;li&gt;Once pervasive (and therefore a worthwhile target), could the 'active' sensors be confused, interfered with, or possibly spoofed with malicious intent (insurance scams, causing collisions, murder, terrorism, etc.)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My personal opinion is that as self-driving cars become more widespread, we will need to use 'passive' sensors (i.e. cameras) a lot more and infer the environment using image processing (computer vision) of visible light and other (non-visible) electromagnetic radiation where appropriate. While it may be possible to filter out interference, prevent interference with other (non-cars) equipment, protect against deliberate signal manipulation, etc. I think that computer vision will ultimately be the simplest and most reliable way for the car's computer to 'see' and understand its environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technical advances needed in this area are mostly related to parralelisation of processing multiple inputs in a timely fashion. Standardisation of output from processed images (such as stereoscopic depth analysis, material/texture analysis, object recognition and tracking, etc.) will allow for more modular and parallelised systems as well as interoperability with off-the-shelf components of a larger system.&#xA;Using more than two cameras for radial depth perception rather than the standard two-camera stereoscopic analysis. These are some of the advances needed, if computer vision will end up being the primary input for self-driving cars.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-03-22T23:32:23.890" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2179" PostTypeId="1" CreationDate="2015-03-23T00:30:05.637" Score="2" ViewCount="641" Body="&lt;p&gt;I was reading up on &lt;a href=&quot;https://en.wikipedia.org/wiki/Cable_car_(railway)&quot; rel=&quot;nofollow&quot;&gt;cable cars&lt;/a&gt;, and it struck me as odd that they only use one cable. While the system isn't likely to fail, and there are contingency measures (e.g. brakes) in case something happens while a car is going downhill, it seems odd that the designers of, say, &lt;a href=&quot;https://en.wikipedia.org/wiki/San_Francisco_cable_car_system&quot; rel=&quot;nofollow&quot;&gt;the San Francisco cable car system&lt;/a&gt; didn't add in at least a second cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is because of turning (explained &lt;a href=&quot;http://cablecarmuseum.org/archive/Anat/Anat.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) - if there were two cables, one would have to go faster for a given turn (unless the &quot;let go&quot; method was used).&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-03-23T21:07:46.780" Title="Why don't cable cars use multiple cables?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2180" PostTypeId="1" AcceptedAnswerId="2181" CreationDate="2015-03-23T03:59:28.190" Score="3" ViewCount="1605" Body="&lt;p&gt;In Ernest Cline's novel &lt;a href=&quot;http://en.wikipedia.org/wiki/Ready_Player_One&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Ready Player One&lt;/em&gt;&lt;/a&gt;, the main character lives in the &quot;stacks&quot; - a dystopian vision of what a trailer park may look like in the future.  The &quot;stacks&quot; are primarily composed of trailerhouses stacked&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... at least fifteen mobile homes high (with the occasional RV, shipping container, Airstream trailer, or VW microbus mixed in for variety).  In recent years, many of the stacks had grown to a height of twenty units or more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The trailers on the bottom level rested on the ground, or on their original concrete foundations, but the units stacked above them were suspended on a reinforced modular scaffold, a haphazard metal latticework that had been constructed piecemeal over the years.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There's also a great picture from the cover which helps visualize:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TJXsY.jpg&quot; alt=&quot;Ready Player One cover art&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm curious how feasible this actually is.&lt;/strong&gt;  Can a typical mobile home support that much weight?  Can standard steel scaffolding provide enough support?  Can the &lt;em&gt;ground&lt;/em&gt; support that much weight without a better foundation than typical for a motor home?&lt;/p&gt;&#xA;" OwnerUserId="318" LastActivityDate="2015-03-23T15:22:09.003" Title="How high can you stack mobile homes?" Tags="|civil-engineering|structures|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2181" PostTypeId="2" ParentId="2180" CreationDate="2015-03-23T09:36:18.410" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can a typical mobile home support that much weight?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, most mobile homes are just flimsy wooden constructions where emphasis is on weight saving. If you want to stack cheap housing blocks I suggest stacking steel old shipping containers. They also come preequipped with anchor points in the corners.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can standard steel scaffolding provide enough support?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Standard haphazard steel scaffold, probably not. Properly engineered lattice work yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the ground support that much weight without a better foundation than typical for a motor home?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, the higher you go the higher the tipping momentum is, this will put more force on one side. This could be stabilized with tension wires between the individual towers. This will widen the effective base. (and provide a place for people to hang their laundry.)&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="86" LastEditDate="2015-03-23T12:28:51.300" LastActivityDate="2015-03-23T12:28:51.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2182" PostTypeId="1" CreationDate="2015-03-23T09:39:22.903" Score="6" ViewCount="1343" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&lt;p&gt;I participate every year in a huge water balloon battle. The clash happens between two universities so you can imagine that the commitment, at least for some people, is quite high. To give you an idea last year some 40000 (thats 40e3) water baloons were thrown, plus there were some home made war machines, namely a catapult, a trebuchet, some &lt;a href=&quot;http://en.wikipedia.org/wiki/Arquebus&quot; rel=&quot;nofollow noreferrer&quot;&gt;arquebuses&lt;/a&gt; running on compressed air and such. Safety is of course the primary concern and nothing more than water/water balloons has ever been thrown.&lt;/p&gt;&#xA;&lt;h2&gt;Actual Question&lt;/h2&gt;&#xA;&lt;p&gt;I want to build a water cannon and I would love some advice/ideas. What do I mean by water cannon? I mean a machine capable of throwing a lot of water, from 1 to 3 litres, at some 20 metres &lt;em&gt;at least&lt;/em&gt;, in a shape like a cannonball rather than a stream.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Is it possible for a &amp;quot;blob&amp;quot; of water to maintain its shape after being shot in the air at some speed, at least for a limited range?&lt;/strong&gt; I see two challenges: accelerating the water and &amp;quot;keeping it together&amp;quot; while in the air. The latter can be achieved with a plastic container (such as a balloon), while the first part may be trickier. Balloons tend to rip and explode if accelerated too much, while leaving the balloon out may make the second challenge impossible.&lt;/p&gt;&#xA;&lt;p&gt;The design I am considering could have a huge tank connected to a smaller one that serves as the &amp;quot;firing chamber.&amp;quot; You fill the firing chamber, then close the line between the tank and the firing chamber, then you open another line that launches the water out through a barrel. The idea is to use compressed air but my main concern is that the water will not maintain a &amp;quot;ball&amp;quot; form while in the air.&lt;/p&gt;&#xA;&lt;h2&gt;Constraints&lt;/h2&gt;&#xA;&lt;p&gt;The best solution for me would be one that works within the constraints for the water balloon battle.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;About 100 € material cost.&lt;/li&gt;&#xA;&lt;li&gt;No fire is allowed in the propulsion system, compressed air is fine but no electric compressor is allowed so efficiency is a must. Our arquebus could fire a lemon sized balloon some 50 m away on about 2 atm.&lt;/li&gt;&#xA;&lt;li&gt;Safety is absolutely the primary concern. For no reason at all there can be risks for the operators or the opponents. We experimented with a firing shell with our arquebus because friction with the barrel would destroy the balloon, and the shell was secured to the barrel with many redundant wires to be sure not to fire it too.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Workforce and time are pretty unlimited if the project is cool. We have some very high manual skills and quite a lot of common and not so common tools.&lt;/p&gt;&#xA;" OwnerUserId="48" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2022-02-08T05:17:01.327" Title="Is it possible to fire a ball of water that holds its shape?" Tags="|mechanical-engineering|fluid-dynamics|pressure|projectiles|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2183" PostTypeId="2" ParentId="2125" CreationDate="2015-03-23T13:36:13.213" Score="1" Body="&lt;p&gt;CAD is essential for determining Geometrical interference and special design.  If Finite Elements are used in the CAD package, the data can be used for stress and strain calculations (determining if the system will withstand the loads).  This can get complicated since if a system is going to be dynamic, there are static loads and dynamic loads.  Dynamic loads need to be calculated when the model is subjected to varying loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to mechanical strength, there are other parameters that need to be determined by using a simulation.  Suppose you are designing a suspension system.  The spring constant of the spring and the damping constant of the damper need to be selected using a simulation.  A mathematical simulation can be developed using either Simulink or coding it up in a computer language.  Using expected force profiles and test conditions the parameters can be determined using the simulation.&lt;/p&gt;&#xA;" OwnerUserId="1010" LastActivityDate="2015-03-23T13:36:13.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2184" PostTypeId="1" AcceptedAnswerId="2188" CreationDate="2015-03-23T13:36:47.700" Score="7" ViewCount="7504" Body="&#xA;&#xA;&lt;p&gt;It's a movie trope that snipers are spotted because someone catches a glimpse of a reflection of the scope. It seems to me that this is a serious design flaw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that a good number of scopes have &quot;lids&quot; to protect against dirt, etc., when not in use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought is this: Why is the front-most (nearest the target) lens/glass so close to the end of the pipe (as in ╞═════╪╡)? Why isn't that piece of glass hidden &lt;em&gt;way&lt;/em&gt; inside the pipe (as in ╞═════╪════╡)? That &lt;em&gt;extra&lt;/em&gt; bit of pipe could then block out any light/reflections apart from what is in the view of the shooter. It would also make the scope longer, but that can't be &lt;em&gt;that&lt;/em&gt; much of a downside, can it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there &lt;em&gt;are&lt;/em&gt; such scopes, how common are they? Or, put the other way around, given that reflections are undesired, how is this actually dealt with &lt;em&gt;outside of the movies&lt;/em&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1011" LastEditorUserId="1011" LastEditDate="2016-02-11T10:20:14.967" LastActivityDate="2020-12-14T08:00:43.267" Title="Must rifle scopes have revealing reflections?" Tags="|optics|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2185" PostTypeId="2" ParentId="245" CreationDate="2015-03-23T13:48:04.393" Score="-3" Body="&lt;p&gt;If you have access to Matlab, the beam can be modeled as a collection of Finite Elements.  Then the system can be represented with a Mass Matrix, Spring Constant Matrix, and a Damping Constant Matrix.  The set of Forces also can be represented as a Force Vector. By solving the equations you have stress(sigma), strain(epsilon), and deflection (delta).&lt;/p&gt;&#xA;" OwnerUserId="1010" LastActivityDate="2015-03-23T13:48:04.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2186" PostTypeId="2" ParentId="2107" CreationDate="2015-03-23T13:55:11.607" Score="0" Body="&lt;p&gt;The vibrator is a rotating object with an eccentric center of mass.  It can be lumped at a given offset.  The force is calculated using F = mrw^2.&lt;/p&gt;&#xA;" OwnerUserId="1010" LastActivityDate="2015-03-23T13:55:11.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2187" PostTypeId="2" ParentId="2182" CreationDate="2015-03-23T13:56:41.573" Score="4" Body="&lt;p&gt;You don't need anything that elaborate to launch water balloons quite nicely.  I was actually kicked out of a dorm in college due to a &quot;water balloon incident&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that surgical tubing makes a great rubber band.  It can stretch to about 4x its rest length, and is pretty efficient at releasing the stored energy back.  We used about 2.5 feet of doubled-up tubing on each side of a basket.  That means when stretched all the way back, the tubing was 10 feet long on each side.  That's quite a few pounds of pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the basket we used the largest plastic funnel available in the local grocery store.  Cut holes on opposite sides of the wide part for each of the surgical tubing strings.  The funnel forms a nice shape for holding the water balloon, and the hole in the small end lets air in so that the balloon can exit the basket without getting stuck by air suction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simple and cheap device worked very well.  We were shooting about 1 pound water balloons, which is about as massive as you should get with something going that fast.  We could easily shoot up and over a wing of the dorm that was 4 stories above where we were shooting from.  We had the two people holding the loose ends of the surgical tubing sitting on a concrete wall, with the third person below on the ground.  The launch angle was fairly close to optimal, and I'd say we could shoot about 75 meters, maybe even a bit more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another time we were shooting from the balcony of a dorm on a hill, probably about 20 m from the street.  Even though the launch angle was only moderately up, we could shoot into the courtyard behind a building across and down the street a bit, probably about 100 m total horizontal distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With two people holding the ends on level ground, you won't achieve launch angle for maximum distance, but you probably don't actually want to optimize for that anyway.  You could always make some rig to hold the ends of the tubing and allow rotation of the whole mechanism for aiming and the like.  It depends on how fancy you want to get.  Three people is very cheap, mobile, sets up easily, and can do fire-and-run easily.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="110" LastEditDate="2015-04-07T12:40:53.563" LastActivityDate="2015-04-07T12:40:53.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2188" PostTypeId="2" ParentId="2184" CreationDate="2015-03-23T14:13:19.547" Score="3" Body="&lt;p&gt;This is not just something that happens in the movies.  Look up some history on the famous sniper Carlos Hathcock and his encounter with the enemy sniper &quot;Cobra&quot;.  Each knew the other was in the area.  Hathcock saw a flash of light and fired at it.  This turned out to be a reflection from Cobra's scope.  The bullet went right through the scope into Cobra's eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, I don't know for real, but certainly there is a tradeoff between length and clunkiness and the possibility of greater reflection from the front element of the lens.  Modern lens coating should be able to mitigate reflections a good amount without requiring a extra long barrel.  Keep in mind that the barrel would have to be flared to match the outer angle of the field of view through the scope.  Something that large has its own drawbacks in a sniping situation.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2015-03-23T15:16:08.097" LastActivityDate="2015-03-23T15:16:08.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2189" PostTypeId="2" ParentId="2184" CreationDate="2015-03-23T14:35:00.690" Score="1" Body="&lt;p&gt;Lenses always reflect part of the light, so lense optics do too. The reflection is strongest for light shining into the telescope along the line of sight. As a rule of the thumb, if a lightsource can be seen through the telescope, someone standing at the lightsource has a chance to see the reflection of this lightsource. This effect is common to all lense optices and is exploited in products to find hidden cameras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd still need to re-research the reasons lenses always in this specific way, it's been a decade since the relevant physics class.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2015-03-23T20:38:51.433" LastActivityDate="2015-03-23T20:38:51.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2190" PostTypeId="1" CreationDate="2015-03-23T14:39:25.807" Score="4" ViewCount="2480" Body="&lt;p&gt;I'm trying to make a lifting mechanism that will support variable weights, similar to this product: &lt;a href=&quot;http://www.humanscale.com/products/product.cfm?group=quickstand&quot; rel=&quot;nofollow&quot;&gt;http://www.humanscale.com/products/product.cfm?group=quickstand&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the quickstand there is a screw that adjusts the spring force between 4 and 11 pounds, which appears to be constant throughout its 17 inch stroke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of spring are they using? I've searched for a long time but there doesn't seem to be any off-the-shelf solution to accomplish this effect. There are adjustable gas springs, but that's done with gas bleeding which is obviously a one time adjustment.&lt;/p&gt;&#xA;" OwnerUserId="1012" LastEditorUserId="33" LastEditDate="2016-01-27T02:07:11.237" LastActivityDate="2023-06-08T23:00:16.037" Title="What kind of spring exerts a constant force but is also adjustable?" Tags="|mechanical-engineering|applied-mechanics|springs|" AnswerCount="4" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2192" PostTypeId="2" ParentId="2184" CreationDate="2015-03-23T15:40:52.253" Score="3" Body="&lt;p&gt;It is possible to buy anti reflection devices for your rifle scope, such as the one shown below (found &lt;a href=&quot;http://www.midwayusa.com/product/634257/sunguard-killflash-ard-anti-reflection-device-nightforce-56mm-objective-not-benchrest?cm_vc=ProductFinding&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.midwayusa.com/product/634257/sunguard-killflash-ard-anti-reflection-device-nightforce-56mm-objective-not-benchrest?cm_vc=ProductFinding&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/unh2dm.jpg&quot; alt=&quot;Anti-reflective device&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why they don't come with this to begin with I can only guess, but you will get less light in the scope when using devices like this, so that may be a reason.&lt;/p&gt;&#xA;" OwnerUserId="1016" LastEditorUserId="272" LastEditDate="2015-03-25T17:11:54.720" LastActivityDate="2015-03-25T17:11:54.720" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2193" PostTypeId="2" ParentId="2107" CreationDate="2015-03-23T16:06:12.747" Score="2" Body="&lt;p&gt;@Narada is correct, but I their answer is a bit short on detail, so I will expand on it.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Calculating Force&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;First, you can calculate the force that the motor exerts on the containing object from the acceleration of the mass at the end of the motor shaft.  As the mass rotates, it accelerates toward the motor shaft at a rate $\omega^2 y$, where $\omega$ is the angular velocity and $y$ is the distance from the motor shaft to the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_centroids&quot; rel=&quot;nofollow&quot;&gt;centroid&lt;/a&gt; of the mass. If we assume that the mass is a perfect semi-circular prism, then $y = \frac{4R}{3\pi}$, where $R$ is the radius of the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the acceleration is known, the force can be computed by Newton's Second Law, $F = ma = m\omega^2 y = \frac{4mR\omega^2}{3\pi}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direction of said force is always directed toward the motor axis, so as it spins it alternates between purely horizontal force and purely vertical force (or two different horizontal directions if the motor is mounted standing up).  We can decompose the force into vertical and horizontal forces by using sines and cosines, and realizing that the angle of the mass is equal to $\omega t$, where $t$ is the elapsed time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{x} = \frac{4mR\omega^2}{3\pi} \cos{(\omega t)}$$&#xA;$$F_{y} = \frac{4mR\omega^2}{3\pi} \sin{(\omega t)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $F_x$ is the horizontal force and $F_y$ is the vertical force.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Simulating Vibration&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;If you have the Solidworks simulation software, you can apply the $F_x$ and $F_y$ loads at the point of the vibration motor and Solidworks should give you the vibration behavior of the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately, if you just want to know how far the vibration motor will move your object and the vibration motor is mounted close to the center of the object, you can just use Newton's laws of motion and treat your whole object as some mass with an applied force.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that the angular velocity must be the same for the vibration motor and the object (assuming a rigid body with no other vibration motors) one can see that $m_{motor}y_{motor} = m_{obj}y_{obj}$. Since all but $y_{obj}$ are known, you can easily get an estimate of how far the object will move. The object will alternately move to $\pm y_{obj}$ in both directions.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-03-23T16:06:12.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2194" PostTypeId="2" ParentId="2182" CreationDate="2015-03-23T17:33:06.560" Score="0" Body="&lt;h1&gt;Options&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It is probably possible to do something similar to what you want. It really depends on the range that you are looking for and how concerned you are about it being a &quot;ball&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fire Hose&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If a fire hose can shoot a stream of water, then all you need is a really short stream of water. It might not be a &quot;ball&quot; like you are wanting, but it would get water across the field. To get the range, you might need to have it be more of a stream than a short burst.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Laminar Fountain&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To have a compact &quot;package&quot; of water stay together, you will need a special nozzle and water system. Something similar to this is used in fountains. It looks like the phrase is &lt;strong&gt;jumping jets&lt;/strong&gt; or &lt;strong&gt;laminar fountain&lt;/strong&gt;. These are fountains that keep their shape and have a very abrupt start and end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two controlling factors for these types of fountains:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sharp start and stop mechanism.&lt;/li&gt;&#xA;&lt;li&gt;Laminar flowing water. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Start and stop&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rotating disk with a hole in it could create a repetitive on/off pattern. This could be powered by a cordless drill or something similar. The rate could be adjusted spacing the holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Laminar water&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keeping the flowing water laminar is going to be the biggest problem. Laminar flow is defined by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Reynolds_number&quot; rel=&quot;nofollow&quot;&gt;Reynolds Number&lt;/a&gt; being below 2100. For the situation of water flowing in a tube:&#xA;$$R_e = (2.475×10^6)*v*D &amp;lt; 2100 \\&#xA;where: \\&#xA;v = velocity (m/s)\\&#xA;D = diameter (m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The requirement for the flow to be laminar is going to restrict the speed of the water leaving the tube. This is going to directly affect the range that water will shoot. It may be in the end that there just isn't as much range as you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out some DIY laminar fountain tutorials for various ways that these streams have been created. A couple that I found via a web search are, &lt;a href=&quot;http://scuttlebots.com/2014/06/17/laminar-fountains-what-are-they/&quot; rel=&quot;nofollow&quot;&gt;Laminar Fountains&lt;/a&gt; for the math and &lt;a href=&quot;http://laminar.forumotion.com/&quot; rel=&quot;nofollow&quot;&gt;The Laminar Project Forum&lt;/a&gt; for other ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Water Pressure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to have limits on how you can create the water pressure. You mention that compressed air is an option, so moving forward with a pressure tank for the water seems like the most reasonable way to go.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-03-24T23:45:07.443" LastActivityDate="2015-03-24T23:45:07.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2195" PostTypeId="2" ParentId="2182" CreationDate="2015-03-23T19:30:25.673" Score="3" Body="&lt;p&gt;You totally changed the question by editing it, so now I'm giving a different answer from before.  I'll let the original answer stay because what it says is still valid, although no longer as relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, you can't fire &quot;balls&quot; of water.  Liquid moving through air doesn't work that way.  Any ball more than drop size would break up, and drops have too little mass for their air resistance to be shootable the distance you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ball of water is held together by its surface tension.  In a weightless environment, you can have balls of water that stay together.  However, this doesn't work anymore as the ball moves through the air.  As the size of a ball increases, the air pressure forces on it also increase, but the surface tension force stays the same.  Then there is also the speed of the ball.  As the ball size gets bigger, the terminal velocity also gets bigger, and the air forces on the ball go with the square of the velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this means that balls of liquid moving through air are inherently limited in size.  That's what a &quot;drop&quot; is.  There is a reason that raindrops only get to a certain size.  A maximum size raindrop in a cloud will grow as it hits other micro drops of cloud on the way down.  If it grows too big, the air forces on it get larger due to its size alone.  At the same time the speed goes up and the air force goes up with the square of that speed.  All these factor together limit water at 1 atm of air to about 50 &amp;micro;l in size.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="110" LastEditDate="2015-04-07T12:41:13.593" LastActivityDate="2015-04-07T12:41:13.593" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2196" PostTypeId="2" ParentId="2179" CreationDate="2015-03-23T21:07:46.780" Score="2" Body="&lt;p&gt;I wouldn't be surprised if the brakes on a cable car had multiple contours.  But, I don't see why brakes alone would be insufficient.  The brakes are sufficient for stopping at traffic lights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the gripping mechanism fails, then the cable will break and stop.  A team of horses will tow the cable car to the depot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, &lt;a href=&quot;http://www.huffingtonpost.com/2013/04/13/san-francisco-cable-car-accidents_n_3077129.html&quot; rel=&quot;nofollow&quot;&gt;SF cable cars are not the safest public transport&lt;/a&gt;.  The system was designed in 1890s when safety concerns were more relaxed.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2015-03-23T21:07:46.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2197" PostTypeId="2" ParentId="2173" CreationDate="2015-03-23T22:06:12.697" Score="1" Body="&lt;p&gt;One possible method to determine stability and capability is using control charts. Below is an example control chart representing an unstable process. The data points are represented in blue, which include few data points above the upper control limit (UCL - Purple line) and below the low control limit (LCL - Red line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vdCsN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is data for the above chart. The data is Apple stock price for 15 days. Data is for &lt;strong&gt;illustration&lt;/strong&gt; purpose only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{| l | l | l | l | l | l | l | l | l | l |}&#xA;\hline&#xA;&amp;amp;	&amp;amp;	&amp;amp;	&amp;amp;   &amp;amp;	&amp;amp;	&amp;amp;	\bar{X} Chart &amp;amp; \bar{X} Chart &amp;amp; \bar{X} Chart\\	&#xA;Date &amp;amp;		Open &amp;amp;	High &amp;amp;	Low &amp;amp;	Close &amp;amp;	\bar{X} &amp;amp;	R &amp;amp;	LCL &amp;amp;	CL &amp;amp;	UCL \\ \hline&#xA;3/19/2015 &amp;amp;	126.7&amp;amp;	127.2&amp;amp;	125.4&amp;amp;	125.5&amp;amp;	126.2&amp;amp;	1.8&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\ &#xA;3/18/2015 &amp;amp;	126.0&amp;amp;	128.6&amp;amp;	125.3&amp;amp;	127.4&amp;amp;	126.7&amp;amp;	2.7&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/17/2015 &amp;amp;	125.9&amp;amp;	127.2&amp;amp;	124.6&amp;amp;	127.0&amp;amp;	126.2&amp;amp;	2.6&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/16/2015 &amp;amp;	123.8&amp;amp;	124.9&amp;amp;	123.3&amp;amp;	125.4&amp;amp;	124.4&amp;amp;	2.0&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/13/2015 &amp;amp;	124.9&amp;amp;	125.9&amp;amp;	122.5&amp;amp;	123.5&amp;amp;	124.2&amp;amp;	3.3&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/12/2015 &amp;amp;	123.3&amp;amp;	125.9&amp;amp;	122.6&amp;amp;	125.4&amp;amp;	124.3&amp;amp;	3.2&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/11/2015 &amp;amp;	126.7&amp;amp;	126.2&amp;amp;	124.1&amp;amp;	123.7&amp;amp;	125.2&amp;amp;	3.0&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/10/2015 &amp;amp;	126.4&amp;amp;	127.2&amp;amp;	123.8&amp;amp;	124.5&amp;amp;	125.4&amp;amp;	3.4&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/9/2015 &amp;amp;	127.9&amp;amp;	129.5&amp;amp;	125.0&amp;amp;	127.1&amp;amp;	127.4&amp;amp;	4.5&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/6/2015 &amp;amp;	128.4&amp;amp;	129.3&amp;amp;	126.2&amp;amp;	126.6&amp;amp;	127.6&amp;amp;	3.1&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/5/2015 &amp;amp;	128.5&amp;amp;	128.7&amp;amp;	125.7&amp;amp; 126.4&amp;amp;	127.3&amp;amp;	2.9&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/4/2015 &amp;amp;	128.1&amp;amp;	129.0&amp;amp;	126.8&amp;amp;	128.5&amp;amp;	128.1&amp;amp;	2.2&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/3/2015 &amp;amp;	126.9&amp;amp;	128.0&amp;amp;	125.5&amp;amp;	127.3&amp;amp;	126.9&amp;amp;	2.4&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;3/2/2015 &amp;amp;	126.2&amp;amp;	127.2&amp;amp;	125.3&amp;amp;	126.0&amp;amp;	126.2&amp;amp;	1.9&amp;amp;	124.1&amp;amp;	126.1&amp;amp;	128.2 \\&#xA;	 &amp;amp;	&amp;amp;	&amp;amp;	&amp;amp; &amp;amp;	126.1&amp;amp;	2.8  \\			&#xA;	 &amp;amp;	&amp;amp;	&amp;amp;	&amp;amp; &amp;amp;	\bar{\bar{X}} &amp;amp;	\bar{R} \\ \hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Steps for the Chart&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$\bar{X}$, Average of observation&lt;/li&gt;&#xA;&lt;li&gt;$R$, Range , Max - Min&lt;/li&gt;&#xA;&lt;li&gt;$\bar{\bar{X}}$, $\bar{R}$, Average of $\bar{X}$ and Average of Range&lt;/li&gt;&#xA;&lt;li&gt;The center line of $\bar{X} Chart$ is $\bar{X}$&lt;/li&gt;&#xA;&lt;li&gt;LCL and UCL lines for $\bar{X} chart$ use $\bar{X}$ equation&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Below is a control chart representing a &lt;strong&gt;stable&lt;/strong&gt; process. All data points are between LCL and UCL. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yaYKO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; Note: The data has been adjusted to illustrate a stable process &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the specification limits (Upper specification limit: USL and low specification limit: LSL) are outside of both the LCL and UCL then the process is considered a &lt;strong&gt;capable&lt;/strong&gt; process.&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="282" LastEditDate="2015-04-06T21:53:08.373" LastActivityDate="2015-04-06T21:53:08.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2198" PostTypeId="5" CreationDate="2015-03-24T00:02:41.193" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-03-24T00:02:41.193" LastActivityDate="2015-03-24T00:02:41.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2199" PostTypeId="4" CreationDate="2015-03-24T00:02:41.193" Score="0" Body="Questions related to sensors or sensor-based systems and their applications. " OwnerUserId="36601" LastEditorUserId="36601" LastEditDate="2022-12-28T15:50:07.697" LastActivityDate="2022-12-28T15:50:07.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2200" PostTypeId="2" ParentId="2151" CreationDate="2015-03-24T00:21:28.750" Score="4" Body="&lt;p&gt;I'd try and fit the material to a model.&lt;br /&gt;&#xA;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Debye_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;Debye model&lt;/a&gt; is the &amp;quot;standard&amp;quot;. (sorry the wiki article is a bit over the top.)  In the Debye model the material can be fit with one &amp;quot;Debye temperature&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Edit upon request.  (though, I would trust the wiki article over my answer.)&#xA;At high temperatures, (but not too high) materials have a heat capacity that is equal to 3kT * N, where N is the number of atoms.  (It's only atoms and not electrons that count for heat capacity, which is interesting...)  As the temperature drops atoms stop shaking so much and some of the vibrational modes &amp;quot;freeze out&amp;quot;.  The modes are at such a high energy that there is not enough thermal energy to excite them. The Debye temperature is a rough measure of where modes freeze out, and the heat capacity starts to decrease.&lt;/p&gt;&#xA;" OwnerUserId="168" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-03-25T00:14:39.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2201" PostTypeId="2" ParentId="2184" CreationDate="2015-03-24T01:00:18.707" Score="11" Body="&lt;h3&gt;The lenses which make up the scope will always reflect some light (&amp;#126;0.5%).&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;All optical materials will reflect some amount of light.  This is due to the fact that light travels slower inside of the medium than it does in the surrounding air.  The slowing down of a wave always results in some reflection; in optics it is governed by the &lt;a href=&quot;http://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fresnel equations&lt;/a&gt;, but there are analogous reflections in &lt;a href=&quot;http://en.wikipedia.org/wiki/Signal_reflection&quot; rel=&quot;nofollow noreferrer&quot;&gt;electronic circuits&lt;/a&gt; and &lt;a href=&quot;http://www.acs.psu.edu/drussell/Demos/reflect/reflect.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;waves traveling along a string&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Partial_transmittance.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vZ3FG.gif&quot; alt=&quot;Partial Transmission&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;http://en.wikipedia.org/wiki/Optical_coating&quot; rel=&quot;nofollow noreferrer&quot;&gt;techniques which use destructive interference between multiple reflections&lt;/a&gt; at successive boundaries to give optical devices very, very low reflectivities at a specific wavelength.  However, for a rifle scope it is important that the reflectivity be low for the entire visible range.  Browsing the &lt;a href=&quot;http://www.edmundoptics.com/technical-resources-center/optics/anti-reflection-coatings&quot; rel=&quot;nofollow noreferrer&quot;&gt;coatings offered by Edmunds&lt;/a&gt; shows that a good coating can reduce the reflectivity down to about 0.5% across the entire visible range. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The lens tube cannot simply be extended without degrading the image quality.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;There is a lot of information on hoods on &lt;a href=&quot;https://photo.stackexchange.com/search?q=lens+hood&quot;&gt;photography.SE&lt;/a&gt; (&lt;a href=&quot;https://photo.stackexchange.com/questions/399/why-are-some-lens-hoods-petal-shaped-and-others-not&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://photo.stackexchange.com/questions/26752/could-i-get-a-lens-hood-that-suits-all-lenses/26753#26753&quot;&gt;here&lt;/a&gt; for instance), but I'll briefly summarize.  The length of the hood is limited by the field of view.  Essentially, light from every point in the scene needs to be able to reach &lt;strong&gt;every part of the front lens.&lt;/strong&gt;  I.E. it is not enough for the light to reach only the center portion of the lens.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, simply extending the tube beyond the front of the objective lens is not an option because light from the edge of the scene would not reach the edge of the lens.  The hood must have a radius which is larger than the objective lens itself before it can be extended; otherwise you are throwing away light from part of the scene.  Pay attention next time you see a professional photographer (or enthused amateur) with a lens hood and you will immediately notice this.  The most effective hood is actually a cone with an angle set by the field of view which will start to make the scope very large and bulky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.flickr.com/photos/colinsd40/4179924452/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vis3vm.jpg&quot; alt=&quot;Lens Hood from Flikr&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The reflection is less noticeable than you might imagine.&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Since the objective lens of the scope is a focusing lens, the outer surface is curved like a diverging mirror.  This means that the light redirected toward the enemy sniper is significantly less intense than it would be for a flat mirror.  For a flat mirror the entire surface of the mirror redirects light from the sun to the enemy's eye, while a curved mirror has a much smaller area which gives the correct angle between the enemy's eye and the sun.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="-1" LastEditDate="2017-04-13T12:44:06.987" LastActivityDate="2015-03-25T17:11:05.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2202" PostTypeId="1" CreationDate="2015-03-24T08:01:43.147" Score="2" ViewCount="172" Body="&lt;p&gt;What material is used for the bottom window of 3D printer that the cured resin layer does not stick to and that it is transparent to the uv (365 nm) light?&#xA;And what material is used for the top plate where the model should stick.&lt;/p&gt;&#xA;" OwnerUserId="1028" LastActivityDate="2016-09-06T20:13:16.203" Title="What material for DLP/SLA 3D printer bottom window?" Tags="|materials|3d-printing|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2203" PostTypeId="1" CreationDate="2015-03-24T09:19:37.440" Score="1" ViewCount="86" Body="&lt;p&gt;An &lt;a href=&quot;http://en.wikipedia.org/wiki/Organic_Rankine_cycle&quot; rel=&quot;nofollow&quot;&gt;organic rankine cycle&lt;/a&gt; engine is a steam engine that uses, instead of water, a fluid with a higher vapor pressure. This allows a rankine cycle type process at lower temepratures, like the waste heat of a motor or geothermal energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect the mechanical part of the ORC engine to be significantly easier to build, and thus cheaper, because the temperatures are lower and so thermal expansion is less of an issue. However, I don't know enough about constructing pistons or turbines to be sure. Can someone shed light on this?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-03-24T09:19:37.440" Title="Is it significantly less challenging to build an ORC piston or turbine than the same for a water steam engine?" Tags="|mechanical-engineering|turbines|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2204" PostTypeId="1" AcceptedAnswerId="2277" CreationDate="2015-03-24T14:14:22.033" Score="2" ViewCount="4106" Body="&lt;p&gt;I was wondering, what is the strongest : a rectangle or an arc (same width/height) ? &lt;img src=&quot;https://i.stack.imgur.com/4JPEA.png&quot; alt=&quot;screenshot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you have to imagine the screenshot in 3D : you have 4 walls and a paved layer (the rectangle) or a dome&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason why the &lt;em&gt;rectangle&lt;/em&gt; would be stronger : same size as the dome but more raw materials&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason why the &lt;em&gt;dome&lt;/em&gt; would be stronger : the vertical pressure is directed to the walls -&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Keystone_(architecture)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Keystone&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the 4 &quot;walls&quot; are infinitely strong&lt;/li&gt;&#xA;&lt;li&gt;the dome / rectangle has a weight negligible relatively to the force applied&lt;/li&gt;&#xA;&lt;li&gt;&lt;b&gt;The dome / rectangle is not infinitely strong, it can bend / break given the vertical pressure&lt;/b&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;EDIT&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am making a plastic box and people will walk on it (box size is size of a foot). for some reasons walls must be thin and the box must be pretty thin too and we wondered what shape is stronger / is less likely to bend in its center&lt;/p&gt;&#xA;" OwnerUserId="1029" LastEditorUserId="1029" LastEditDate="2015-03-26T16:05:04.970" LastActivityDate="2023-06-21T12:54:20.100" Title="Mechanically, is an arc stronger than a rectangle?" Tags="|structural-engineering|" AnswerCount="3" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="2205" PostTypeId="2" ParentId="2204" CreationDate="2015-03-24T16:14:42.863" Score="0" Body="&lt;p&gt;A inverted &lt;a href=&quot;https://en.wikipedia.org/wiki/Catenary&quot; rel=&quot;nofollow&quot;&gt;catenary arch&lt;/a&gt; will not need any tensile strength to support the force provided that it is uniform across the entire top. However all that force will be transfered to to walls and out to the side. It needs a firm anchor to avoid the outward bulge that would destroy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the rectangle has a similar anchor then you can think away all the material that is not part of a catenary arch. and see above. The normal sag would then be counteracted by the force at the anchor points&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-03-24T16:14:42.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2207" PostTypeId="1" AcceptedAnswerId="2218" CreationDate="2015-03-24T23:51:19.043" Score="11" ViewCount="8224" Body="&lt;p&gt;I am mounting solar panels on top of a recreational vehicle (RV). To reduce drag, I was planning on building a small ramp in front of the panels to deflect wind before it hits the flat panel and mounts. Here is a rough sketch (red is brackets, black is panel, purple is ramp):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IKLCI.png&quot; alt=&quot;Solar Panel&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone told me that there would be MORE drag if I did it this way than if I didn't put the deflector/ramp.  I will do the work even if there will be VERY LITTLE difference between drag avoided by putting in a ramp and not having a ramp at all, but I definitely don't want to do it if it will INCREASE drag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the space under the panel will be minimal.  I am going to try to get them as close as possible to the roof.  The panels are about 2&quot; thick and the space between will be roughly 1/4-1/2 inch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would adding a ramp in front of the panel affect drag on the vehicle? Would it increase, decrease or stay the same?&lt;/p&gt;&#xA;" OwnerUserId="1033" LastEditorUserId="368" LastEditDate="2015-03-26T15:48:18.167" LastActivityDate="2022-07-07T12:38:58.447" Title="How would adding a wind deflector in front of roof-mounted solar panels affect drag on an RV?" Tags="|automotive-engineering|airflow|solar-panel|drag|" AnswerCount="4" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2208" PostTypeId="2" ParentId="2207" CreationDate="2015-03-25T00:29:00.023" Score="2" Body="&lt;p&gt;I can remember this being discussed during a fluids lecture in my student days a long time ago, when the subject of car roof racks and truck fuel scoops was raised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The panel will increase the drag of the RV and reduce its designed streamlined characteristics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ramp as you are planning will reduce the drag of the panel but will not restore the original streamlining characteristics of the RV. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The affect of the ramp will be similar to fuel scoops attached to trucks, see the picture below. They reduce fuel consumption reducing drag. Such devices have been installed on trucks in since the 1980s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YJYPh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/cPeqF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/CSFd5.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/4jD0J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/lm9no.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/b8EnK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/2dXaP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Ha0aS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yuoID.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-03-27T00:45:11.150" LastActivityDate="2015-03-27T00:45:11.150" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2209" PostTypeId="2" ParentId="2204" CreationDate="2015-03-25T02:52:11.537" Score="3" Body="&lt;p&gt;Your question is very broad, but at its heart is question about how various shapes transfer load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rectangular Beam&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a solid beam is loaded, it will experience moments and shears. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moments are primarily resisted by the material furthest away from the centroid of the section. This leaves the material at the center of the beam under-utilized. This is the reason that steel beams are commonly shaped like an &quot;H&quot; on its side. Solid beams are not efficient with regard to moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear is transferred through the entire section of a solid beam, so the entire section is utilized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Arch&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arches primarily transfer load through axial forces. These forces utilize the entire area of the section. &lt;strong&gt;It is for this reason that an arch is the more efficient of the two shapes.&lt;/strong&gt; The entire area of the section is employed to transfer the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Only in comparison&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comparison above only works if the supports are not controlling. Arches require their supports to resist large horizontal forces. They basically convert a portion of the vertical forces to horizontal forces. There is no conversion of this type in beams, so the beam supports are simpler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An Example&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One notable example of an efficient arch is the dome of the basketball stadium at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Construction_of_Assembly_Hall_%28Champaign%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;University of Illinois&lt;/a&gt;. &lt;a href=&quot;http://www.concretecontractor.com/concrete-construction-projects/assembly-hall/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This structure&lt;/a&gt; is a concrete dome that is 3.5 inch thick at its thinnest.  The structure has steel strands wrapped around its base that help to resist the horizontal forces. The dome is self-supporting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/15kUb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an &lt;a href=&quot;http://zaielacademic.net/java_applets_apps/algorithms.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;interesting article&lt;/a&gt; on the efficiency of arches that mentions the dome above.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-25T02:52:11.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2210" PostTypeId="1" CreationDate="2015-03-25T04:26:30.860" Score="8" ViewCount="285" Body="&lt;p&gt;A &lt;a href=&quot;http://www.purdue.edu/newsroom/releases/2015/Q1/theoretical-study-suggests-huge-lava-tubes-could-exist-on-moon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;study&lt;/a&gt; from Purdue University suggests the Moon contains a number of wide underground lava tubes, up to 5000 m wide and about 1500 m high, in which subterranean cities could be constructed. Such tubes would protect potential residents from cosmic radiation and from the extremes of temperature on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XlucI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to constructing anything in such tubes the inside of the lava tubes would need to be inspected for integrity of the rock and where localized weaknesses were found these would need to be fixed. The risk of a rock or a wedge of rock dislodging from the backs/roof/ceiling or walls of the tube and falling on people, plant, equipment or buildings would need to be to mitigated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What would be the best way to stabilize any local weak spots in the internal surface of the lava tubes?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, here are some common tools and techniques for adding stability to natural and engineered structures:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tensioned cable bolts&lt;/li&gt;&#xA;&lt;li&gt;Rock bolts&lt;/li&gt;&#xA;&lt;li&gt;Shotcrete&lt;/li&gt;&#xA;&lt;li&gt;Shotcrete, steel mesh and bolts&lt;/li&gt;&#xA;&lt;li&gt;Concrete lining via form work, as opposed to shotcrete&lt;/li&gt;&#xA;&lt;li&gt;Polymer coatings&lt;/li&gt;&#xA;&lt;li&gt;Ceramic coatings&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="90" LastEditorUserId="368" LastEditDate="2015-03-25T17:38:55.187" LastActivityDate="2015-06-20T13:30:49.407" Title="Localized stabilisation of Lunar lava tubes" Tags="|civil-engineering|geotechnical-engineering|reinforcement|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2211" PostTypeId="2" ParentId="2207" CreationDate="2015-03-25T06:19:33.297" Score="4" Body="&lt;p&gt;Stress on panel mounts needs considering.&lt;br /&gt;&#xA;This can be greatly reduced (but not 100% eliminated) by a &amp;quot;ramp&amp;quot; that protects the panel from air impact in the forward direction.&lt;/p&gt;&#xA;&lt;p&gt;Flat plate drag for a panel will be &amp;lt;= classic drag equation result.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_{lost}= D \cdot V $$&lt;/span&gt;&lt;br /&gt;&#xA;= power required to accelerate all opposing air out of the path of the object from rest to velocity &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; as the vehicle passes.&lt;br /&gt;&#xA;Lower &lt;span class=&quot;math-container&quot;&gt;$C_d$&lt;/span&gt; improves on this.  Maximum Cd will probably approach but  be less than 1. Reduced can be small but non zero. (A Prius has Cd ~= 0.3. Lower can be achieved in this case).&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$D = \frac{1}{2}  C_D \rho  A  V^2 $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; = force in Newton on panel (or ramp)&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; = projected frontal area m^2&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; = velocity m/S&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; = Air density &lt;span class=&quot;math-container&quot;&gt;$\approx$&lt;/span&gt; 1.2 &lt;span class=&quot;math-container&quot;&gt;$\frac{kg}{m^3}$&lt;/span&gt; at STP&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$C_D$&lt;/span&gt; = Drag coefficient relative to flat plate drag &lt;span class=&quot;math-container&quot;&gt;$0 &amp;lt; C_D \leq 1$&lt;/span&gt;&lt;br /&gt;&#xA;Filling in the values gives:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D = 0.6  A  V^2$$&lt;/span&gt;&#xA;e.g. a 1000 mm wide x 100 mm tall panel with &lt;span class=&quot;math-container&quot;&gt;$C_D = 1$&lt;/span&gt; at 100 km/h &lt;span class=&quot;math-container&quot;&gt;$\approx$&lt;/span&gt; 28 m/s  may cause up to:&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$0.6 \cdot 1 \cdot 1 \cdot 0.1 \cdot 28^2 = 47 N$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_{lost} = D \cdot V = 47 \cdot 28 = 1.3 kW $$&lt;/span&gt;&#xA;Not vast but noticeable.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2022-07-07T12:38:58.447" LastActivityDate="2022-07-07T12:38:58.447" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="2212" PostTypeId="1" CreationDate="2015-03-25T09:13:27.730" Score="3" ViewCount="566" Body="&lt;p&gt;A while back, I asked &lt;a href=&quot;https://softwarerecs.stackexchange.com/questions/17467/free-pid-software-piping-and-instrumentation&quot;&gt;this question&lt;/a&gt; for a free &lt;a href=&quot;https://en.wikipedia.org/wiki/Piping_and_instrumentation_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Piping and Instrumentation Diagram (P&amp;amp;ID)&lt;/a&gt; software on softwarerecs.SE. I described the core features of a P&amp;amp;ID software thusly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where a proper P&amp;amp;ID software differs from a drawing tool like Visio or Diag is that the P&amp;amp;ID tool can output a parts list listing all components and connectors. Components and connecting lines can have additional specifications (like the power and type of a motor, or the size and material and pressure level of a pipe). This specifications can then be written to the material/component list.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Confession time: I've never worked with a P&amp;amp;ID software, this is just what I think (from poring over plans and counting valves) would be really helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What core features do working engineers need from a modern P&amp;amp;ID tool?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to clairfy:&lt;/strong&gt; I'm not looking for a specific product, I'm looking for the core features that make a tool great.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="-1" LastEditDate="2017-04-13T12:50:05.980" LastActivityDate="2016-04-19T13:22:11.077" Title="What makes a good piping and instrumentation diagram tool?" Tags="|process-engineering|computer-aided-design|pi-diagram|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2213" PostTypeId="1" CreationDate="2015-03-25T09:47:54.960" Score="3" ViewCount="629" Body="&lt;p&gt;I received a tricky bonus question today in my engineering class: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WT6fk.png&quot; alt=&quot;Bonus Question&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've begun by equating all moments and forces to zero, about an arbitrary point, which I labeled at $C$. Finally, I attempted to solve for $C_x$ and $C_y$ without examining $M_c$. However, my answers for $C_x$ and $C_y$ were incorrect, and I'm unsure as to the process for solving $M_c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I solve for the reactions at the base of a T-frame that is subject to various point loads, including a fixed tether at one end of the top bar?&lt;/p&gt;&#xA;" OwnerUserId="1036" LastEditorUserId="368" LastEditDate="2015-03-25T19:47:26.150" LastActivityDate="2015-03-26T15:14:59.207" Title="How do I determine the reactions at the base of a tethered T-frame?" Tags="|mechanical-engineering|statics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2215" PostTypeId="2" ParentId="2213" CreationDate="2015-03-25T11:51:34.420" Score="4" Body="&lt;p&gt;The reactions at $C$ can be determined by resolving vertically, horizontally, and for moments. For equilibrium the total vertical force, horizontal force and moment must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (slight) complication in this question is the inclined force $T_{BD}$. This needs to be resolved into a vertical and a horizontal force. You can use trigonometry to work out the angle of the force to the vertical (I will call it $\theta$). The vertical component is then $T_{BD} \times \cos (\theta)$ and the horizontal component is $T_{BD} \times \sin (\theta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once $T_{BD}$ has been resolved into components, summing vertical forces, horizontal forces should be relatively easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For moments: the horizontal component of $T_{BD}$ will be acting clockwise about $C$ at a lever arm of 600 mm; the vertical component will be acting clockwise about $C$ with a lever arm of 150 mm.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternative method for moments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find this way more complicated, but it is valid. Instead of resolving Tbd into horizontal and vertical components, you can treat it was one force. The lever arm is the minimum distance from $C$ to a passing through $B$ and running in the direction of $T_{BD}$, i.e. it is measured as the perpendicular distance from $T_{BD}$ to $C$. Note that this is not necessarily the distance from $C$ to $B$!&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="106" LastEditDate="2015-03-25T21:25:01.040" LastActivityDate="2015-03-25T21:25:01.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2216" PostTypeId="2" ParentId="2207" CreationDate="2015-03-25T12:50:47.440" Score="0" Body="&lt;p&gt;If I've interpreted the sketch correctly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the panel sits above the roof, with clearance underneath it&lt;/li&gt;&#xA;&lt;li&gt;the ramp height is equal to the clearance height&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am guessing that the detractor's thought process went like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the air which previously would have hit the front of the panel will&#xA;still hit the front of the panel&lt;/li&gt;&#xA;&lt;li&gt;the air which would have gone under the panel will now hit the ramp&lt;/li&gt;&#xA;&lt;li&gt;therefore all the ramp does is add drag, as it doesn't affect the air&#xA;which hits the panel&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm not an expert on fluid dynamics, but the rising air from the ramp would clearly affect the air which would have hit the panel. I don't know whether this effect is greater or lesser than the additional drag of air no longer being able to pass under the panel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of a better solution - you should try and remove vertical faces, therefore extend the ramp so that the top of the ramp coincides with the top of the panel. I can state that this will have less drag than your design (assuming I interpreted your sketch correctly), but I'm afraid I don't know whether it's better than nothing at all.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-03-25T12:50:47.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2217" PostTypeId="1" AcceptedAnswerId="2222" CreationDate="2015-03-25T16:48:06.667" Score="7" ViewCount="4497" Body="&lt;p&gt;I have a PCB of approximately 4&quot; x 4&quot;. It has a single long spiraling trace on it. The trace is 50 mil 4 oz copper, so nominally it's good for something like twenty amps before it overheats. But this board overheats much faster. My supposition is that the windings being so close to each other has a compounding effect on the heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, presumably I could increase the surface area of the PCB, and thereby dissipate the same power with less temperature rise. My question is, how does one calculate such? What is the relationship between the power dissipated by my copper plane, its temperature rise, and its surface area? Assume still air.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2015-03-26T13:24:44.880" Title="How can I calculate the temperature rise of a circuit board for a given power dissipation and area?" Tags="|electrical-engineering|heat-transfer|temperature|power|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2218" PostTypeId="2" ParentId="2207" CreationDate="2015-03-25T16:49:46.083" Score="10" Body="&lt;p&gt;The concept of what you're doing is sound, and as Russell McMahon notes the efficiency gains could be significant enough to justify the change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd strongly suggest that you consider adding a ramp to the back edge as well. Drag force is very sensitive to the downstream (rear) end of a body as well You get some positive pressure at the front of the vehicle, but you also get strong negative pressure at the rear.  This is made much worse by the development of turbulence and flow separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conceptually, what you want to do is reduce the cross-section at the back end to reduce the area over which that negative pressure is applied.  You can do that by tapering the back end (a reverse ramp in your case).  The only trick is that you have to do so pretty gradually to avoid a phenomenon called flow separation. this is basically the air stream tumbling over itself and generating vortexes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So go ahead with the ramp in the front, but add a very shallow (~10 degrees) ramp on the back as well.  That'll give you the best improvement for minimal investment.  Also, try to get the ramps to fit up right to the edges of the solar panels. You want the path that the air follows to be as smooth as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some experimental evidence and great pictures, check &lt;a href=&quot;http://www.ara.bme.hu/oktatas/letolt/Vehicleaerodyn/Vehicleaerodyn.pdf&quot;&gt;this presentation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.5 shows small improvements in rounding the front of a van but they quickly level off (no benefit for making the front end smoother)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.11,17,18 shows flow separation at the rear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.18 (upper right) shows an optimal point for reducing the read x-section. Tapering too sharply undoes the benefit of reducing area.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-03-25T16:49:46.083" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2220" PostTypeId="1" CreationDate="2015-03-25T20:45:14.847" Score="10" ViewCount="507" Body="&lt;p&gt;I have an ASTM A311-04 1.25&quot; diameter round 1050 steel bar that exhibits a tensile strength reading of 155,227 PSI. Normally, we receive material in the 13-133,000 PSI range. The bar is machined to a 5/8-18 UNF thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have experienced instances where the threads of these machined parts get damaged when installing a prevailing torque nut. The nut is hand tightened and then driven tight with a pneumatic gun. Threads are being cracked off from the shaft during the installation of the prevailing torque nuts. This is a new phenomenon and seems to be centered around threads machined in steel that has this &quot;high&quot; tensile strength (155,227 PSI). The nuts and assembly process have been ruled out as root cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the material tensile strength create a situation where the threads become susceptible to damage? In tests, it appears as though the threads are being pulled off the shaft and getting embedded in the threads of the nut. The hardness is 33 HRC if that is of interest.&lt;/p&gt;&#xA;" OwnerUserId="1040" LastEditorUserId="33" LastEditDate="2015-03-27T19:50:36.290" LastActivityDate="2016-01-20T15:09:01.387" Title="Are threads machined from high tensile steel more susceptible to damage?" Tags="|mechanical-engineering|materials|steel|machining|threads|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2221" PostTypeId="2" ParentId="2217" CreationDate="2015-03-25T22:52:53.953" Score="4" Body="&lt;p&gt;Base on a quick calculation using online max current calculator for trace width properties, I believe your traces are undersized for 20A of current. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strip line current calculations (Internal Traces) for 50 mil 4 oz copper is &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Max Current: 3.76 amps&lt;/li&gt;&#xA;&lt;li&gt;Trace Temperature: 35.0 °C&lt;/li&gt;&#xA;&lt;li&gt;Resistance: 0.0000640 ohms&lt;/li&gt;&#xA;&lt;li&gt;Voltage Drop: 0.000241 volts&lt;/li&gt;&#xA;&lt;li&gt;Power Dissipation: 0.000904 watts&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Microstrip max current calculations (External Traces) for 50 mil 4 oz copper is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Max Current: 7.51 amps&lt;/li&gt;&#xA;&lt;li&gt;Trace Temperature: 35.0 °C&lt;/li&gt;&#xA;&lt;li&gt;Resistance: 0.0000640 ohms&lt;/li&gt;&#xA;&lt;li&gt;Voltage Drop: 0.000480 volts&lt;/li&gt;&#xA;&lt;li&gt;Power Dissipation: 0.00361 watts&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eeweb.com/toolbox/external-pcb-trace-max-current&quot; rel=&quot;nofollow&quot;&gt;External PCB Trace Max Current&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eeweb.com/toolbox/internal-pcb-trace-max-current&quot; rel=&quot;nofollow&quot;&gt;Internal PCB Trace Max Current&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://circuitcalculator.com/wordpress/2006/01/31/pcb-trace-width-calculator/&quot; rel=&quot;nofollow&quot;&gt;PCB Trace Width Calculator&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mcalc.sourceforge.net/&quot; rel=&quot;nofollow&quot;&gt;Microstrip Analysis/Synthesis Calculator&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.rogerscorp.com/documents/1526/acs/Temperature-Rise-Estimations-in-Rogers-High-Frequency-Circuit-Boards-Carrying-Direct-or-RF-Current.pdf&quot; rel=&quot;nofollow&quot;&gt;Temperature Rise Estimations in Rogers High Frequency Circuit Boards Carrying Direct or RF Current&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.smpspowersupply.com/ipc2221pcbclearance.html&quot; rel=&quot;nofollow&quot;&gt;Printed Circuit Board (PCB) Spacing / Clearance vs. Voltage&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-03-25T23:04:18.567" LastActivityDate="2015-03-25T23:04:18.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2222" PostTypeId="2" ParentId="2217" CreationDate="2015-03-26T13:24:44.880" Score="6" Body="&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The copper side with the traces is modeled as a sheet of copper rather than traces.&lt;/li&gt;&#xA;&lt;li&gt;The body is thin enough that thermal conductivity within the body is unimportant, and the entire device is considered to be at a uniform temperature.&lt;/li&gt;&#xA;&lt;li&gt;Only the two broad surfaces contribute to the heat loss, the sides are neglected.&lt;/li&gt;&#xA;&lt;li&gt;The surroundings, including the air and radiative syncs, are at a uniform temperature $T_s$&lt;/li&gt;&#xA;&lt;li&gt;Thermal coefficients: &lt;a href=&quot;http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html&quot; rel=&quot;noreferrer&quot;&gt;$\epsilon_{cu}$=0.78&lt;/a&gt;, &lt;a href=&quot;http://www.ti.com/lit/an/snva183b/snva183b.pdf&quot; rel=&quot;noreferrer&quot;&gt;$\epsilon_{pcb}$=0.50&lt;/a&gt;, &lt;a href=&quot;http://www.thermal-wizard.com/tmwiz/convect/natural/hup-isot/hup-isot.htm&quot; rel=&quot;noreferrer&quot;&gt;$h_{up}=7.25\ \frac{\text{W}}{\text{m}^2\text{K}}$&lt;/a&gt;, &lt;a href=&quot;http://www.thermal-wizard.com/tmwiz/convect/natural/hdp-isot/hdp-isot.htm&quot; rel=&quot;noreferrer&quot;&gt;$h_{down}=3.63\ \frac{\text{W}}{\text{m}^2\text{K}}$&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Under these assumptions we can estimate the temperature of the board by simply equating heat flows.  The heat coming in per unit time is from &lt;a href=&quot;http://en.wikipedia.org/wiki/Joule_heating&quot; rel=&quot;noreferrer&quot;&gt;Joule heating&lt;/a&gt; from the current running through the copper and is given by &#xA;$$&#xA;q_{in}=I^2R.&#xA;$$&#xA;The heat flowing out has two escape mechanisms; &lt;a href=&quot;http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html&quot; rel=&quot;noreferrer&quot;&gt;radiative heat transfer&lt;/a&gt; to the surroundings which is given by &#xA;$$&#xA;q_{rad}=\epsilon\sigma A(T^4-T_s^4)&#xA;$$ &#xA;and &lt;a href=&quot;http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html&quot; rel=&quot;noreferrer&quot;&gt;convective heat transfer&lt;/a&gt; to the air which is given by &#xA;$$&#xA;q_{conv}=hA(T-T_s).&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we just equate the heat flows &#xA;$$&#xA;\begin{align}&#xA;q_{in}&amp;amp;=q_{out}\\&#xA;I^2R&amp;amp;=A\left[\sigma(T^4-T_s^4)(\epsilon_{cu}+\epsilon_{pcb})+(T-T_s)(h_{up}+h_{down})\right]&#xA;\end{align}&#xA;$$&#xA;we can rearrange this to look like a quartic equation&#xA;$$&#xA;\sigma(\epsilon_{cu}+\epsilon_{pcb})T^4+(h_u+h_d)T -&#xA; \left[\frac{I^2R}{A}+\sigma(\epsilon_{cu}+\epsilon_{pcb})T_s^4+(h_u+h_d)T_s\right]=0.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't easy to solve analytically, but Mathematica would have no problem.  I put it into Python and numerically found the minimum of the absolute value.  I assumed a total resistance of $R=1\Omega$ so your actual results may vary.  The results are shown below.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QmWPd.png&quot; alt=&quot;Board Temperature&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the Python code also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;import scipy.optimize as opt&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;# Define the function with some extra variables&#xA;def tempOpt( t, ts, i):&#xA;    sigma = 5.67e-8&#xA;    ec = 0.78&#xA;    ep = 0.50&#xA;    hu = 7.25&#xA;    hd = 3.63&#xA;    r  = 1&#xA;    a  = 0.0103&#xA;    e = ec + ep&#xA;    h = hu + hd&#xA;    out = sigma*e*t**4 + h*t - i**2*r/a - sigma*e*ts**4 - h*ts&#xA;    return out&#xA;&#xA;# Decide the ranges for the current and temperature&#xA;currents = np.linspace( 0.1, 20, 50)&#xA;temps = [0, 20, 40, 60, 80, 100]&#xA;&#xA;# Calculate the value&#xA;resDict = dict()&#xA;for temp in temps:&#xA;    resVec = np.zeros( np.shape( currents))&#xA;    cnt = 0&#xA;    for current in currents:&#xA;        # Define a new function of one variable and a minimum at zero&#xA;        # Don't forget to convert between Kelvin and Celcius&#xA;        def tempNow( t):&#xA;            return abs( tempOpt( t+273.15, temp+273.15, current))&#xA;        # Find the minimum with a reasonable guess&#xA;        guess = temp + 30&#xA;        resVec[cnt] = opt.fmin( tempNow, 30)&#xA;        cnt += 1&#xA;    # Store results in dictionary&#xA;    resDict[temp] = resVec&#xA;&#xA;# Plot&#xA;plt.figure(1)&#xA;plt.clf()&#xA;for temp in temps:&#xA;    plt.plot( currents, resDict[temp], lw=2)&#xA;plt.xlabel('Current (A)')&#xA;plt.ylabel('Board Temp ($^\circ$C)')&#xA;leg = plt.legend( temps, loc=2)&#xA;leg.set_title( '$T_s$ ($^\circ$C)')&#xA;plt.title('Board Temperature (R=1$\Omega$)')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-03-26T13:24:44.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2224" PostTypeId="1" CreationDate="2015-03-26T15:13:06.250" Score="4" ViewCount="487" Body="&lt;p&gt;Does a metal-halide lamp fixture use power if the bulb is burnt out?  If so, how many Watts would a  277 V, 1,000 W fixture draw with the lamp burnt out?  Transformer is in-line between the 480 V 3-phase supply and the light switch.&lt;/p&gt;&#xA;" OwnerUserId="1048" LastEditorUserId="368" LastEditDate="2015-03-26T15:32:47.890" LastActivityDate="2015-04-17T20:12:17.873" Title="Power consumed by a transforming fixture with the light bulb burnt out" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2226" PostTypeId="2" ParentId="2220" CreationDate="2015-03-26T18:05:23.003" Score="8" Body="&lt;p&gt;First, I want to say that I'm suspect of that material actually being ASTM A311 (though I don't have a copy of that standard available to verify.) Looking on Matweb, &lt;a href=&quot;http://www.matweb.com/search/DataSheet.aspx?MatGUID=6f57ae20090e4a4c875a050124ab06a4&quot;&gt;this material&lt;/a&gt; is the closest I found in the 1050 steels. &lt;a href=&quot;http://www.matweb.com/search/QuickText.aspx?SearchText=a311%201050&quot;&gt;This search&lt;/a&gt; turns up quite a few results, but from what I can tell of A311, it's for cold-drawn and stress relieved bars, but doesn't include quenched and tempered bars. &lt;a href=&quot;http://www.astm.org/DIGITAL_LIBRARY/TOPICS/PAGES/section01_stds.htm&quot;&gt;ASTM's page&lt;/a&gt; seems to suggest A434 is for Q&amp;amp;T bars. So my first thought is that your supplier is doing something weird.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Independent of whether or not you've got the right material, as was noted in the comments, increasing strength generally decreases ductility. The material may yield at a higher strength, but after it yields, there's less elongation before the material fractures. This can be more prevalent in situations where there are high stress concentraters, such as screw threads. Fracture mechanics is a big messy area, but the key here is that a more brittle (less ductile) material is less resistant to crack growth because it can't dissipate enough energy during its deformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the extreme, think of a piece of glass and a rubber band. Glass won't deform as easily, but once you start to move it, it breaks fairly easily. A rubber band stretches with no problem, but you have to stretch it a lot farther before it breaks. I think that's what you're seeing with your threads. There may technically be more strength there, but any defects in the material will weaken it more than a less strong, more ductile material. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-03-26T18:05:23.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2227" PostTypeId="1" CreationDate="2015-03-27T03:34:10.273" Score="10" ViewCount="1363" Body="&lt;p&gt;I am trying to define the material properties of a shape memory alloy material in ANSYS APDL V15. There are material properties for Superelastic materials (SUPE option) and material properties for Shape memory effect materials (MEFF option) depending on the option taken. Even after taking the right option for shape memory effect, the data shown is not the right model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My code for defining the material with the trigger is as follows:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;/PREP7  &#xA;MP,EX,2,47.2E9&#xA;MP,PRXY,2,0.33&#xA;TB,SMA,2,1,7,MEFF,,  &#xA;TBDATA,1,0,315.15,100E6,4.4E6,0.05,2.0E10  &#xA;TBDATA,7,0  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The output windows confirms that I took the memory effect option by saying:  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;DEFINE DATA TABLE  SMA   FOR MATERIAL     2  &#xA;Shape Memory Alloys OPTION  MEMORY EFFECT  &#xA;WITH A MAXIMUM OF  1 TEMPERATURES AND    7 DATA POINTS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, the material model still displays the material constants as being &quot;SIG-SAS, SIG-FAS, SIG-SSA, SIG-FSA, EPSILON, ALPHA&quot; for a total of 6 constants. These correspond to the SUPE option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the name and number of constants correspond to the Superelastic model and not the Shape memory effect, which I should have triggered using the MEFF option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this happening? How do I define this material using the constants for the Shape memory effect, rather than the constants for the Superelastic model?&lt;/p&gt;&#xA;" OwnerUserId="254" LastEditorUserId="368" LastEditDate="2015-08-24T15:44:11.760" LastActivityDate="2019-07-25T21:04:33.523" Title="ANSYS APDL Shape memory effect option of SMA not working" Tags="|mechanical-engineering|finite-element-method|ansys|alloys|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2228" PostTypeId="1" AcceptedAnswerId="2230" CreationDate="2015-03-27T04:21:19.403" Score="6" ViewCount="882" Body="&lt;p&gt;Why is it that a second order system with an additional pole or zero can be approximated into the general second order system for analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was asked during our lecture and I am still looking for the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my reading I've only learned that when an additional pole or zero is added to the system, the step response of this system starts to resemble the step response of the original system. I am not sure if this is the answer to the question.&lt;/p&gt;&#xA;" OwnerUserId="1054" LastEditorUserId="272" LastEditDate="2015-03-27T12:13:11.257" LastActivityDate="2015-03-27T12:29:40.347" Title="Second order system with additional pole" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2229" PostTypeId="1" AcceptedAnswerId="2231" CreationDate="2015-03-27T06:25:38.760" Score="9" ViewCount="1049" Body="&lt;p&gt;If you look at the design documentation in our projects (Biogas and waste-to-energy plants), you find out what we plan to build but not what other options were considered and why they were discarded. This information rests only in the engineers head. Occasionally, a plant will be in a conceptual stage for years, sometimes switching between engineers. So I find myself revisiting an option that a coworker considered and ultimataly discarded last year.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a good way to track design decisions, the 'whys' and 'why-nots'? I'd strongly prefer an approach that's tried, tested and used elsewhere in industry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional information: My company will be pursuing ISO 9001 certification, an approach that fits with this QA regime would be prefered.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="110" LastEditDate="2015-03-29T21:04:19.020" LastActivityDate="2015-03-30T19:00:31.207" Title="What's a professional way of tracking design history in plant engineering?" Tags="|design|project-management|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2230" PostTypeId="2" ParentId="2228" CreationDate="2015-03-27T12:29:40.347" Score="5" Body="&lt;p&gt;I think there must be some other restrictions placed on your professor's comment.  Adding a pole to a general second order system makes it a third order system and the dynamics can change dramatically.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example I can think of in which adding a pole or zero does not change the system dynamics dramatically is when the pole or zero is at much higher frequency than the pair of poles of the second order system.  In this case, the dynamics only change at very short timescales.  The step response, for example, would only change at the very beginning and the damping or oscillatory behavior later in the response would remain essentially the same.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, I wouldn't fret too much about making approximations of such simple systems.  Once you are actually working with them in the field you will likely use numerical software to understand them, and such software has no need to make any approximations for a third order system.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-03-27T12:29:40.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2231" PostTypeId="2" ParentId="2229" CreationDate="2015-03-27T17:01:23.463" Score="3" Body="&lt;p&gt;I think this depends on how much information you want, and how formally you want to track it. It sounds like currently, none of this information is written down anywhere, which is obviously the first problem. But putting down information for a solution that you don't intend to pursue is, at best, a less-than-optimal use of time. You'll have to define which ideas are fleshed out enough to be recorded. If you have a meeting and a dozen ideas are put on the table, but half are almost immediately discarded, do you want to make notes of what those ideas were, and why they were discarded? That's a lot of effort for what was probably a very low-level analysis, but they're still design decisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems more intuitive to me to only do this for ideas which reach the stage of having some actual work done on them. This way, you've already got something tangible, and this just has to be filed away in a project file, which I'd hope is something you already have. Because this system really will only be generating reference materials for future projects, I don't think it would be a good idea to create a whole new system for archiving these materials, or overhauling an existing system. Find a way to integrate it into your current design management system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one key I would make sure you have is a way of classifying your different designs by their key parameters. I don't know much about your field, but I'd assume there are certain design parameters that each project has to meet (physical size, capacity, type of plant, etc.). Make sure these are easily visible somewhere, so that when starting a new project, you can identify old projects that are similar in various regards. If you have these discarded ideas stored in those folders, you'll be able to analyze their utility to your current project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I also want to warn in general about going too deep into this. Again, I work in a different industry, where projects are much shorter, and as such, there are many more of them, but some products have been in existence in some form for decades, and there are 15-20 revisions of them. Maintaining a revision history for actual design changes that were implemented is very important. Knowing what the customer was given in the past, when changes were made, and why they were made is key to not repeating past mistakes and correctly servicing old designs. But when you catalog designs that never were fully realized, you're adding non-essential data on top of essential data, and there's only so much you can sort through before things start to get lost. It sounds like you're looking for a replacement for solid communication and good experience. When these projects change hands, the engineers involved have to communicate all the relevant information. I understand wanting to make sure you know what has been considered, but I'd advise you to temper that desire so as not to flood your records with unnecessary data. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-03-27T17:01:23.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2232" PostTypeId="2" ParentId="2229" CreationDate="2015-03-27T21:29:29.837" Score="1" Body="&lt;p&gt;&lt;em&gt;This &lt;a href=&quot;https://engineering.stackexchange.com/a/2231/33&quot;&gt;answer from Trevor&lt;/a&gt; gives a pretty good over all philosophy, but I want to get more at the details.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First, figure out why the alternate records are not being kept.&lt;/strong&gt; This likely started out as either a foresight issue or a storage (paper or computer) issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At one time in the past, someone thought that the alternates would not be useful, so they were thrown out or deleted. This is likely a company culture problem. Now that this has been identified as an issue, it can be raised company-wide. Keeping records that have already been created is easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the original engineers thought that the alternatives might be useful, they might have not kept the documents because of the cost of storage (physical or computer). If this is still the case, this problem needs to be resolved now. If this is no longer the case, the word needs to be spread throughout the company that keeping records is not a costly proposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second, look at the alternative comparison method.&lt;/strong&gt; There are two categories of alternative consideration: thought and calculation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thought comparisons are ideas that discarded quick enough that they never make it to paper in anything other than concept form. If these ideas were discarded so quickly the first time, then they probably don't need to be recorded. The effort to discredit them is so trivial that they will just be discredited again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternatives that actually made it to the calculation stage should be kept in some form. If calculations or plans were done the first time, then there is already something to save. Even if the idea proves to be a dead-end, the work has already been done, so put it in the file! As long as it is dated, it can be referenced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Write a memo&lt;/strong&gt; (to the file if not to someone in particular) when a decision is made. This saves time in the future, but it adds a step to the process. This is something to work on along with improved documenting in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Last, date everything!&lt;/strong&gt; Try to work with the systems that are already in place to keep the final records, just add a date to everything. Even if other methods of organization fail, the order of alternatives can be rebuilt using the dates on the items.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-03-27T23:17:09.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2233" PostTypeId="2" ParentId="2147" CreationDate="2015-03-28T03:10:57.817" Score="7" Body="&lt;p&gt;Typically jewel bearings (as you've described) are not used in large scale industrial equipment. The main reason jewel bearings are used is their inherent hardness. The bearing material has very little wear over the life of the bearing. Most of the wear occurs to the shaft riding in the bearing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In industry, you typically want to avoid wearing the shafting under any circumstance. An example is power generation turbine shafting. The turbine shaft will typically ride in a journal bearing in which the journal is hydrodynamically lubricated via forced lubrication. The bearing material is normally made of a softer material than the shaft. Under normal operations the shaft rides on a thin film of oil, separating it from the bearing material. In the event of a loss of lubrication or a breakdown of the oil film, the shaft will contact the bearing material. Since it is typically much easier and less costly to replace damaged bearing material versus replacing an entire turbine shaft and rebalancing said shaft, the bearings are normally designed such that the bearing material is sacraficial. When the hard shaft material contacts the soft bearing material, the bearing material will heat up and be wiped away instead of scoring the shaft. A jewel bearing would do the opposite, scoring and wearing the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other applications of jewel bearings, aside from watch movements, include gyroscopes, turbine flow meters, and other assemblies which are typically subjected to small loads and do not generate large amounts of heat which require might forced lubrication to remove the heat.&lt;/p&gt;&#xA;" OwnerUserId="1052" LastEditorUserId="33" LastEditDate="2015-03-28T11:59:13.607" LastActivityDate="2015-03-28T11:59:13.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2234" PostTypeId="1" CreationDate="2015-03-28T04:02:35.417" Score="3" ViewCount="9964" Body="&lt;p&gt;I am trying to build a worm gear based on a single axis solar tracker for a worm gear box. I need some guidance choosing a worm gear &amp;amp; drive motor for the application. These are the steps I have explored so far - if I am wrong could someone please give links to material which gives a good explanation so I can learn and understand it better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.anaheimautomation.com/products/stepper/stepper-motor-item.php?sID=24&amp;amp;pt=i&amp;amp;tID=75&amp;amp;cID=19&quot; rel=&quot;nofollow&quot;&gt;Motor planning to choose&lt;/a&gt;; &lt;a href=&quot;http://www.bonfiglioli.com/media/filer_public/6c/62/6c62d02b-0529-43a8-b52a-1467ba0d8bdb/br_cat_vf-w_nem_uk_r00_0.pdf&quot; rel=&quot;nofollow&quot;&gt;Worm gear selection &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General equations for worm gear &amp;amp; motor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) $Efficiency = \frac{\text{output Power of worm gear}}{\text{input Power of worm gear}} = \frac{P_o}{P_\text{Input}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) $\text{Output power} = P_o = 2*\pi*\text{Output speed} * \text{Output torque}/60$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Output power in kW&lt;/li&gt;&#xA;&lt;li&gt;Output speed in rev/min&lt;/li&gt;&#xA;&lt;li&gt;Output torque in Nm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;3) $\text{Input power} = P_\text{Input} = 2*\pi*\text{Input speed} * \text{Input torque}/60$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) $\text{Net Torque}= T_\text{output} + T_\text{Input} + T_\text{holding} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to drive a 250 kW plant I have assumed 15 kW per row, this gives me 16 rows with 1 column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worm gear has to drive 16 rows with coupling with single column item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each panel weighs 20 kg and there are 60 panel per row which gives 1200 kg.&#xA;Each row has a structural weight around 700 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a net weight of 1900 kg per row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the load on the worm gear would be 1900*16 = 30400 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Based on my calculations above, how can I calculate the exact motor &amp;amp; drive motor required. I need some specific relation here.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1063" LastEditorUserId="1063" LastEditDate="2015-03-31T04:44:07.003" LastActivityDate="2017-07-28T03:24:07.360" Title="Worm gear &amp; motor selection calculation" Tags="|mechanical-engineering|electrical-engineering|gears|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2235" PostTypeId="1" CreationDate="2015-03-28T15:44:22.360" Score="5" ViewCount="1140" Body="&lt;p&gt;My washing machine broke down recently, and they replaced the motor. The old motor was not broken,&#xA;as far as I could tell; the sensor that told the CPU that the drum was balanced was what was out of&#xA;order. At any rate now I have a perfectly decent rotor and stator, and I started thinking about how to&#xA;recycle them. Apropos of not much of anything, I started thinking about a one‐man hovercraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have so far;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The washer service manual specifies a max RPM. I assume this is completely useless because it refers to&#xA;the motor and drum assembly and I’m not going to dismantle my washing machine to weigh the drum. I can call the repair guy and find out how much the drum weighs, but I'll only do so if I find that the calculation takes too long (I don't want to bother him on the weekend).&lt;/li&gt;&#xA;&lt;li&gt;The required lift force is equal to a &gt; 9.8*(mass of hovercraft + mass of rider), where a is some&#xA;acceleration over a time T. Where T is simply the amount of time necessary to move vertically from rest to a&#xA;position above the ground equal to the height of the deployed skirt. In other words solve&#xA;$Y=Y_{0}+V_{0}T+0.5aT^{2}$ for a known Y, zero initial velocity, and $a=\frac{F}{m}$.&lt;/li&gt;&#xA;&lt;li&gt;The friction and air resistance are negligible, however since increasing horizontal component of motion&#xA;reduces vertical component, thrust should be higher than what is required to simply achieve lift by some&#xA;factor.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Questions;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can I evaluate the power of the motor? I have not been able to find specs online so far. Does it even make sense to talk about the power of the motor without knowing the power&#xA;supply? Is there a maximum and minimum power output? Alternately, is power even a relevant&#xA;parameter, or should I be concerned solely with RPM?&lt;/li&gt;&#xA;&lt;li&gt;The friction and air resistance are negligible, however since increasing horizontal component of motion&#xA;reduces vertical component, thrust should be higher than what is required to simply achieve lift by some&#xA;factor.&lt;/li&gt;&#xA;&lt;li&gt;What factor must I exceed vertical thrust by in order to move horizontally?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1067" LastEditorUserId="421" LastEditDate="2016-07-17T06:06:00.397" LastActivityDate="2016-07-17T06:06:00.397" Title="Hovercraft design parameters" Tags="|mechanical-engineering|electrical-engineering|fluid-mechanics|dynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2236" PostTypeId="2" ParentId="2235" CreationDate="2015-03-28T17:43:37.827" Score="10" Body="&lt;p&gt;Designing the lift system for a hovercraft isn't actually about the mass and the acceleration, it's about the &lt;em&gt;pressure&lt;/em&gt; required to lift the hull and the &lt;em&gt;rate&lt;/em&gt; at which the air leaks out from under the hull and needs to be replaced. Together, these determine the lift &lt;em&gt;power&lt;/em&gt; required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Horizontal motion is a completely separate issue, which requires knowledge about the friction of the skirt, aerodynamic losses, the slopes you want to be able to handle, the mass of the craft and the acceleration you want to be able to achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it may be possible to use a single motor for both functions, in conjunction with ducts and baffles, it's often simpler overall to use two separate motors.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-03-28T17:43:37.827" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2237" PostTypeId="1" CreationDate="2015-03-28T21:24:31.907" Score="8" ViewCount="812" Body="&lt;p&gt;&lt;strong&gt;Computer Models&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computer modeling is used in various engineering fields. I am specifically considering structural analysis or finite element analysis (FEA). Sometimes models are used to speed up repetitive calculations that could be done by hand. Sometimes models are used to perform calculations that are not easy or even possible to do by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Checking&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few standard methods for checking the results of computer models. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Run verification models and confirm that the results match a previously calculated answer.&lt;/li&gt;&#xA;&lt;li&gt;Run simple models that can be checked by hand calculations.&lt;/li&gt;&#xA;&lt;li&gt;Test physical models.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The problem with the first two checking methods above is that they either only check specific situations or they only check the simple parts of the program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physical model method can be expensive for full size models and scale models may not always give the same results as the full size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves a gap in what results can be checked. For any complicated model, there is no easy way to check that the results from the program are correct. The engineer must trust that the software produced correct results from the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comparison Check&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model could be input into two different programs (made by different companies). The assumption is that if the results of the two models are similar enough, then the results should be correct for the model used. This would not catch any errors in creating the original model, but it would catch errors in the software implementation.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Could two separate programs be used to check the &quot;correctness&quot; of the results from the model?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Would using this method of comparing a model in two separate programs provide the same level of assurance in the results as any of the other checking methods?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;What could be the disadvantages of using this procedure?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastEditorUserId="1051" LastEditDate="2015-04-10T23:56:35.793" LastActivityDate="2016-03-02T21:44:51.777" Title="Can two different modeling programs be used to confirm the results of each other?" Tags="|modeling|software|finite-element-method|" AnswerCount="6" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2238" PostTypeId="2" ParentId="2237" CreationDate="2015-03-29T13:55:08.993" Score="7" Body="&lt;p&gt;Yes, getting a second opinion can be useful.  This is done routinely in weather forecasting where exact solutions are unknown, and there is some judgement about how to apply various factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be less wiggle room in something like a finite element mesh stress analysis because the iterative equations for solving it will be basically the same no matter who wrote the software.  The real issue is not solving the mesh as much as creating a good enough mesh in the first place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way therefore to get multiple opinions is to vary the mesh parameters.  Hopefully you still get pretty much the same answer.  If you make the mesh 2x finer and get a significantly different answer, then that's a strong clue the original mesh wasn't detailed enough.  You also then don't know for sure the next mesh is detailed enough without making a even more detailed one and getting about the same answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course nowadays mesh generation itself is somewhat automated and adaptive.  This is no longer about just the physics of solving the mesh, but include heuristics about when and how to subdivide.  Different software can vary in this regard, so running two different programs with the same initial data can be useful.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="272" LastEditDate="2015-04-07T00:14:50.187" LastActivityDate="2015-04-07T00:14:50.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2239" PostTypeId="2" ParentId="2229" CreationDate="2015-03-29T14:36:46.453" Score="0" Body="&lt;p&gt;Following are three general methods how white goods, consumer goods and OEM manufactures I have associated with document ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Front End (Conceptualization) idea documentation&lt;/strong&gt;&lt;br&gt;&#xA;Create project numbers for ideas considered big, impact full, and ground breaking but not currently considered for implementation. Document these ideas in most cases on a one or two page document and before shelving the project. With the growth in search engine technology and software these ideas can be stored in a searchable database. With a transitional nature of the current work force I found this a good way the document the idea. With the growth in technology and implementation method the detail documentation of technology is less relevant because in $t+1$, it is high likely that better and improve technology will be available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Design reviews (NPD - New Product Development)&lt;/strong&gt;&lt;br&gt;&#xA;Design reviews another good process that can help document engineering design decisions. This tends to be more technical documentation, ideas that are rejected tend not to be well documented. In most organizations &lt;strong&gt;design reviews&lt;/strong&gt; are part of &lt;strong&gt;phase gate model&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back end (In or close production) idea documentation&lt;/strong&gt;&lt;br&gt;&#xA;I found organizations using an existing &lt;strong&gt;change management&lt;/strong&gt; system to document new ideas. In this system the engineer propose the idea in a simple one or page document. The idea is reviewed and accepted or shelved for future. This particular organization used the SAP for change management, thus system as less apt to technology base searching but did sever well for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, most ideas are shelved for the following reason&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lack of financial resources to execute the idea/suggestion &lt;/li&gt;&#xA;&lt;li&gt;Lack of human resources to execute the idea/suggestion&lt;/li&gt;&#xA;&lt;li&gt;Doesn't fit into the current project plan &lt;/li&gt;&#xA;&lt;li&gt;The market is not yet ready to embrace new technology&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore it good the track ideas, because when above barriers are resolved the ideas become realistic solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Change_management&quot; rel=&quot;nofollow&quot;&gt;Change management&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Design_review&quot; rel=&quot;nofollow&quot;&gt;Design review&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Phase%E2%80%93gate_model&quot; rel=&quot;nofollow&quot;&gt;Phase–gate model&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-03-29T22:32:36.513" LastActivityDate="2015-03-29T22:32:36.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2240" PostTypeId="1" AcceptedAnswerId="2497" CreationDate="2015-03-29T18:19:15.250" Score="2" ViewCount="1236" Body="&lt;p&gt;In order to reduce operational cost and improve the safety and reliability of aircraft's the aerospace industry and government is increasingly evaluating new ideas such structural heath monitoring systems for aircraft's. Such a system will enable continuous monitoring, inspection and detection of damage to the aircraft with minimal human involvement. This will also help the airline industry to improve aircraft maintenance scheduling process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m5IwE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Figure: Early Diagnosis for Microcracks in Aircraft &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently &lt;a href=&quot;http://www.delta.com/content/www/en_US/about-delta.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Delta airlines&lt;/a&gt; and an aircraft manufacture has teamed up with &lt;a href=&quot;http://www.sandia.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sandia Laboratory&lt;/a&gt; to install sensors in aircraft to help understand flaw detection. Research also indicates that ultrasonic sensors are most widely used sensing element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aR4GL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;br&gt;&#xA;Which part of an aircraft can most benefit from structural health monitoring (SHM) system? Is it the fuselage, wings, landing gear, flaps, rudder, elevator, vertical stabilizer or another component? Why is this component the prime candidate? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://phys.org/news/2014-09-in-flight-sensor-health-safer-flights.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;In-flight sensor tests a step toward Structural Health Monitoring for safer flights&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.compositesworld.com/articles/structural-health-monitoring-composites-get-smart&quot; rel=&quot;nofollow noreferrer&quot;&gt;Structural health monitoring: Composites get smart&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ndt.net/search/docs.php3?id=4595&amp;amp;content=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Early Diagnosis for Microcracks in Aircraft&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cnn.com/2015/12/01/asia/air-asia-crash-report/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Air Asia QZ8501: Pilot response to equipment malfunction caused crash&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-12-01T12:11:26.897" LastActivityDate="2015-12-01T12:11:26.897" Title="Which part of an aircraft can most benefit from a structural health monitoring (SHM) system?" Tags="|electrical-engineering|aerospace-engineering|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2242" PostTypeId="1" CreationDate="2015-03-30T01:44:40.237" Score="5" ViewCount="5563" Body="&lt;p&gt;&lt;strong&gt;Spring Design&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://springipedia.com/compression-general-design.asp&quot; rel=&quot;noreferrer&quot;&gt;spring design&lt;/a&gt;, there a a variety of parameters to consider. Some of the parameters are solely &lt;a href=&quot;http://www.industrial-springs.com/online-shop/compression-springs/tryk_c_galv&quot; rel=&quot;noreferrer&quot;&gt;physical&lt;/a&gt; (wire diameter,  coil diameter, length, etc.) and some are determined from the physical dimensions in combination with the &lt;a href=&quot;http://www.isckc.com/spring-materials.html&quot; rel=&quot;noreferrer&quot;&gt;material properties&lt;/a&gt; (spring constant, allowable stress, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Spring Set&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One physical characteristic of a spring is whether or not it will experience &lt;em&gt;set&lt;/em&gt;. This is a permanent deformation of the spring after it has been compressed. Whether or not &lt;em&gt;set&lt;/em&gt; is an issue depends on the level of stress that the spring undergoes versus the &lt;a href=&quot;http://www.springsandthings.com/pdf/Compression-new.pdf&quot; rel=&quot;noreferrer&quot;&gt;material's tensile strength&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common  rule of thumb is that set is not a concern if the spring only experiences 40% or less of its tensile stress. If the spring experiences between 40% to 60% of its tensile stress, then the set can be compensated for during manufacturing. Above 60% of the tensile stress, set cannot be controlled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that spring set can be compensated for is that it occurs after only a few cycles and then the full amount of set has been reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Creep&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Creep is a separate phenomenon from set. Set happens immediately, but creep happens with constant load over time. Since the metal is under stress, it should exhibit creep. Creep is also affected by temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know how a spring is affected by long-term constant load. I have not been able to find any information about creep in springs or how to calculate it during spring design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the long-term creep of a spring determined?&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2021-05-07T23:04:07.160" Title="How is the long-term creep of a spring determined?" Tags="|mechanical-engineering|springs|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2243" PostTypeId="2" ParentId="616" CreationDate="2015-03-30T02:17:29.107" Score="2" Body="&lt;p&gt;&lt;strong&gt;Short Answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it should be as simple as taking the perimeter of your washer multiplied by the thickness of the plastic. The shear stress over this area should be your maximum tension on the bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Longer Justification&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The washer should be considered to be rigid in relationship to the plastic. This will set the minimum area of shear as the perimeter of the washer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the plastic likely exhibits an amount of plastic deformation before failure, the actual load at failure will likely be larger than the value calculated by the simple method. Much like designing a concrete slab or block shear in a bolted connection, there are many different possible failure planes, but there is a minimum conservative failure plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design should take into consideration this minimum possible failure plane. If this is assumed, then the design will be conservative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.styrolution.com/Product/Luran-S_Luran-S-777K_SKU400700190193.html&quot; rel=&quot;nofollow&quot;&gt;data sheet&lt;/a&gt; of the material shows that there isn't a large amount of strain at yield (3.3%). This is less than most structural metals, so you shouldn't have to worry about the effects of large deformations on the calculations.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-03-30T02:17:29.107" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2244" PostTypeId="1" AcceptedAnswerId="2250" CreationDate="2015-03-30T06:08:09.233" Score="2" ViewCount="2437" Body="&lt;p&gt;This one should be easy to answer by looking at the thermodynamics, but I find it not to be. Say we have a natural gas stream, ~700 N∙m&lt;sup&gt;3&lt;/sup&gt;/h, that we want to compress from 16 bar to 250 bar. What will the power consumption of the compressor be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that stymies me is that the power uptake is not path independent, but mostly depends on how many cooling steps you have in between. Or so it seems to me. You don't need to do the math for me, I just provided the numbers as an example. I would appreciate a detailed explanation of how to arrive at an approximate real world value.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="368" LastEditDate="2015-03-30T15:52:12.627" LastActivityDate="2015-03-31T08:41:10.890" Title="Typical power consumption of large compressor" Tags="|mechanical-engineering|gas|compressors|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2245" PostTypeId="1" CreationDate="2015-03-30T11:40:24.797" Score="15" ViewCount="5961" Body="&lt;p&gt;A lot of discussion about the modernisation of electricity systems is about &quot;inertia&quot;. This is usually a qualitative discussion about how turbines (in hydro, coal &amp;amp; gas plants) with lots of kinetic energy in the form of angular momentum and fast responsiveness provide voltage and frequency stabilisation at the scale of quarter-cycle (5ms in 50Hz grids) to a small number of seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, discussions often stall because it's quite rare to see this &quot;inertial response&quot; quantified, and its source identified. As I understand it, the system itself has very low electrical capacitance, so I guess most of the inertial response comes from the rotation of turbines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is inertial response quantified for national electricity systems and what are some typical values of system inertia?&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="368" LastEditDate="2015-03-30T15:36:47.023" LastActivityDate="2021-01-08T17:58:58.323" Title="Quantifying inertia on the electricity grid" Tags="|electrical-engineering|power-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2246" PostTypeId="2" ParentId="2237" CreationDate="2015-03-30T12:25:10.910" Score="2" Body="&lt;p&gt;&lt;strong&gt;Answer from the point of view of a Design Engineer&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Checking the results of one program against another will give you some level of certainty that the results are correct. It is unlikely to give you 100% certainty, but then that level of certainty is hard to achieve.&lt;/p&gt;&#xA;&lt;p&gt;A big issue I see is being able to transfer the model from one piece of software to another. Although model import/export is being improved by most software companies (because of BIM), I wouldn't expect all features of a model to be exportable. Geometry is relatively easy to import/export as the exchange file only needs to contain coordinates. But e.g. member end releases are likely to be stored very differently by different software, so unless/until a universal file exchange format is agreed, I suspect a lot of effort would be required to completely rebuild a model in the second software program.&lt;/p&gt;&#xA;&lt;p&gt;Based on my own experience, errors in results are far more likely to come from incorrect input of data or incorrect assumptions than they are from poorly written software. The time and effort in using independent software to verify an answer is therefore probably not a good use of your time.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Answer from the point of view of a Software Engineer&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Verifying software against other software isn't taken as justification that your software is correct. It is far better to find published data/results that can be used to verify that software gives correct answers. Imagine a sales meeting where a software company is trying to sell their software to an engineering company:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Engineer: How do we know that your software is correct?&lt;/p&gt;&#xA;&lt;p&gt;Software salesman: Well, we checked it against our competitor's software and got the same answer.&lt;/p&gt;&#xA;&lt;p&gt;Engineer: So you're saying that your competitor is sufficiently better than you that his software is the yardstick against which you measure your software? Sounds like we should buy his software instead!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="255" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-03-30T12:25:10.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2247" PostTypeId="2" ParentId="2237" CreationDate="2015-03-30T12:46:41.417" Score="4" Body="&lt;p&gt;I think this is a good practice overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using two different softwares, you may be able two avoid two kind of errors:&#xA;1) errors that come from an inaccurate software (which should not be overlooked),&#xA;2) errors that come from the lack of habit of the user with the software (hidden options, default settings...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the softwares are different enough, the odds of getting two times the same wrong results are low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the errors that come from a bad modeling choice cannot be avoided this way. So I'd say the main disadvantage is to over-trust the results because they have been confirmed by two softwares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do think that it is better to master one software, running all kind of test cases (comparison with academic results for instance), than using several softwares and having only an average knowledge. Moreover, I think it is best to know the basics of FEM analysis, and using only two &quot;one-click-to-run&quot; softwares is a bad practice because users are likely to reproduce modeling errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I am writing as an electromagnetism/thermal FEM analysis user (no other domains).&lt;/p&gt;&#xA;" OwnerUserId="861" LastEditorUserId="861" LastEditDate="2015-03-30T20:43:36.610" LastActivityDate="2015-03-30T20:43:36.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2248" PostTypeId="2" ParentId="2237" CreationDate="2015-03-30T13:42:54.570" Score="2" Body="&lt;p&gt;I concur with the other answers here, that this in general can be a good idea and will help ensure the accuracy of simulation results. In terms of how good it is in relation to the other verification methods, I'd say previously known results and physical tests are both better options if feasible, but hand calculations may require over-simplification if the model is sufficiently complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I really want to point out is something that hasn't been addressed on the final point: potential weaknesses of this practice. Using two different FEA packages may catch a peculiarity of one package that causes an error, provided you can identify which analysis is correct and which one is off. However, there are some general limitations to FEA overall, regardless of method or implementation. Sharp corners and other stress concentrators that cause singularities in the model aren't something that will change much from package to package, these will always be weak spots. This is where engineering knowledge and intuition is required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done simulations on parts I know stand up easily to certain stresses, and the model shows that the internal stress is 10x the yield strength; this is obviously incorrect, because it's on an involute spline pattern and the FEA software doesn't like that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, it should be obvious that changing software doesn't eliminate user error. If you make an error in the model or parameters, that error will screw you up no matter what you use to analyze it. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-03-30T13:42:54.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2249" PostTypeId="2" ParentId="2245" CreationDate="2015-03-30T14:45:40.903" Score="10" Body="&lt;p&gt;&lt;a href=&quot;http://enbalatalks.enbala.com/2013/05/07/grid-inertia-an-unseen-hazard-with-renewable-generation/&quot;&gt;This blog post&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt; identifies the two main sources of inertia within the power grid:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Classic&quot; generation, typically steam turbines&lt;/li&gt;&#xA;&lt;li&gt;Large industrial motors&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your understanding is correct in that overall system capacity is comparably low and provides a negligible effect to the system inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a reliability perspective, system inertia is a good thing.  The large rotational mass providing system inertia slows the decline in frequency should there be a sudden change in the generation or load of the system.  System inertia helps prevent protective load shedding mechanisms from kicking in by providing time for compensating control systems to adjust generation to the changing environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inertia has become a greater subject of interest as newer renewable generation technologies have increased their footprint on electric grids.  Newer renewable technologies connect their generation source to the electric grid through power inverters which do not provide any inertia to the rest of the system.  Likewise, renewable technologies are enabling the retirement of older generation technologies which results in less system inertia being available.  This decline in inertia is compounded by a decrease in large industrial motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;&lt;sub&gt;&lt;em&gt;Please note, this source is a bit biased as they sell a product related to grid inertia&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ccc.centers.ufl.edu/sites/default/files/files/background.pdf&quot;&gt;This presentation&lt;/a&gt; goes into some of the details regarding how system inertia is calculated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mechanical dynamics are modeled by the second-order differential equation:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$J\frac{d^2\theta}{dt^2}=T_m - T_e$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta$: angle (rad) of the rotor with respect to a stationary reference.&lt;br&gt;&#xA;$J$: moment of inertia.&lt;br&gt;&#xA;$T_m$: mechanical torque from the turbine.&lt;br&gt;&#xA;$T_e$: electrical torque on the rotor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, you would need to sum the inertia provided by all of the major contributing sources.  This is obviously a non-trivial exercise as generation schedules vary as do production schedules for large industries.  You also have to take into account the preferred ramp rate of the generators which will vary based upon fuel source.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To provide a negative answer to your question - I think it's these aspects that makes it so hard to discuss system inertia in a quantified manner.  There are too many variables and the environment is dynamic.  You could perhaps identify the inertia for a small region, but certainly not for the region of a typical balancing authority or at a national scale.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some concluding thoughts:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pessimist might argue that system reliability is doomed due to the decrease in overall system inertia and that we'll see more brownouts and blackouts as part of upgrading the overall power grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That outlook is likely a bit too bleak though.  Balancing authorities can require that more &lt;a href=&quot;http://en.wikipedia.org/wiki/Operating_reserve&quot;&gt;spinning reserves&lt;/a&gt; be available, which can provide fast(er) response generation for localized imbalances within the grid.  Likewise, national level energy committees can provide compensation in the arbitrage market for fast voltage and frequency providers such as grid scale bulk electric storage systems (BES).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, those changes won't come for free - it takes fuel to provide spinning reserves, and grid scale BES aren't cheap.  But the challenges are surmountable even if decisions have to be made based upon empirical evidence.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-03-30T16:05:08.087" LastActivityDate="2015-03-30T16:05:08.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2250" PostTypeId="2" ParentId="2244" CreationDate="2015-03-30T16:59:26.693" Score="5" Body="&lt;p&gt;The work is surely path dependent; you can estimate a fast passable value in two way.&#xA;Isothermal, as in the case when several stages with intercoolers are used&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(SI Units)&#xA;KWiso = 2.78*10^-4 * (VolFlowRate) * P1 * LN( P2/P1 )&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And Adiabatic with costant ratio of specific heats (k)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(SI Units)&#xA;KWad = 2.78*10^-4 * k /( k-1 ) * (VolFlowRate) * P1 * [(P2 / P1)^(k/(k-1)) -1]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Dividing this ideal one for an Efficiency value that is 75-87 % for centrifugal and 80-91 % for Axial. You can also improve the value using polytropic (n) stage efficiency&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;PolyEff = [(k-1)/k] / [(n-1)/n]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you want a more accurate interstage cooling calculation consider every stage as a different compressor and get a big balance.&lt;/p&gt;&#xA;&lt;p&gt;Reference: Perry 's Chemicals Engineer Handbook Section 10-45&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oavoq.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="1080" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-03-31T08:41:10.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2251" PostTypeId="2" ParentId="2244" CreationDate="2015-03-30T17:15:05.883" Score="1" Body="&lt;p&gt;Maybe I didn't get the problem it sounds a little bit simple, but as I see you would have to calculate using a PVT correlation (Peng-Robinson for exemple) the power your transformation requires in a ideal system (ΔS=0, the power required for this transformation is path independent) and then you can divide the value you got by the efficiency of your compressor, this will change with your equipment, but 0,8 is a good value, having more cooling stages will give you more efficiency, with less colling stages you will have a lower value for efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the following equations with a PVT correlation to calculate the power required by the transformation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dU=CvdT+[T*(∂P/∂T)−P]dV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dS=(Cv/T)dT+(∂P/∂T)dV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dH=dU+d(PV)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cv=Cp-R&lt;/p&gt;&#xA;" OwnerUserId="1075" LastEditorUserId="1075" LastEditDate="2015-03-30T17:26:26.323" LastActivityDate="2015-03-30T17:26:26.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2252" PostTypeId="2" ParentId="2237" CreationDate="2015-03-30T17:35:37.780" Score="6" Body="&lt;p&gt;&lt;em&gt;I write this from the perspective of an engineer who develops simulation software.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the practice described is bad, and I recommend you do not use two different softwares to &quot;confirm&quot; the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, two different modeling softwares can not be used to confirm much anything other than their similarity. Two softwares could easily both get two similar but wrong answers, especially if they use similar models. I can think of at least one instance where this is definitely the case, and Trevor Archibald mentions another. I would be more impressed by two softwares which use different modeling techniques getting similar results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This subject is called the verification and validation of computer models, and it has a fairly vast literature. I'll offer a sketch of the basics. Verification is comparing a model against an &quot;exact&quot; solution (which could be a hand calculation, or something more complex), that is, checking the math of the software. The assumptions behind the exact solution could be wrong, but at the very least you'd want to make sure the software gets the math part right. Validation is comparing a model against an experiment. This allows you to check whether the model you are using is accurate, which is something verification can't do for you.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The problem with the first two checking methods above is that they either only check specific situations or they only check the simple parts of the program.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a real problem facing software developers and users. There are established ways to handle it that are much better than comparing two different softwares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is that you can never test every possible case. Your software might pass case A, but case A doesn't involve physics X, Y, or Z, and that makes you totally off case B. So, what you'd want are a large number of checks that cover at least all of the basic features you want to check. Many softwares have &quot;V&amp;amp;V suites&quot; which are basically exactly that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of verification, there are numerous options. You could generate new exact solutions for different cases. Sometimes this alone is adequate. However, as you've noticed, oftentimes what you can do by hand is limited to very simple cases. For the more general cases, you could use a technique called the method of manufactured solutions (Google it). This requires programming and can get messy, but allows you to test basically anything you could think of. (The messiness part can be handled via a library like &lt;a href=&quot;https://github.com/manufactured-solutions/MASA&quot; rel=&quot;nofollow&quot;&gt;MASA&lt;/a&gt;, by the way.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'd like to point out that contrary to what Olin Lathrop suggests, with the method of manufactured solutions, you can generate what are, for the purpose of testing, exact solutions. They are not &quot;exact&quot; in the strict sense, because they do not exactly satisfy the equations the software is solving without modification. But they satisfy equations which are very close, and the difference is accounted for to make the test rigorous. This technique is not very popular at the moment, but it has been used to test things which previously were thought to be difficult to test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of validation, you could look for more experimental data or run your own experiments.&lt;/p&gt;&#xA;" OwnerUserId="49" LastEditorUserId="49" LastEditDate="2016-03-02T21:44:51.777" LastActivityDate="2016-03-02T21:44:51.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2253" PostTypeId="1" CreationDate="2015-03-30T18:06:16.157" Score="5" ViewCount="705" Body="&lt;p&gt;The first photo is an example of a light curtain. I am certain that every individual beam is from an individual laser and receiver pair on both cases. Would it be possible to have one laser beam perhaps be directed from the top or side of the case to create a contiguous light &quot;plane&quot;. Hopefully the picture below could help you visualize what I have in mind. The receiver end would have to be able to detect fluctuations of the laser beam at a reasonable resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eIbKX.png&quot; alt=&quot;Light Curtain&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iljvC.png&quot; alt=&quot;Laser beam -&amp;gt; planar beam&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1084" LastEditorUserId="1084" LastEditDate="2015-03-30T20:52:40.733" LastActivityDate="2017-01-01T21:41:02.883" Title="Do contiguous laser curtains exists? and if not are there any lenses that could spread a concentrated beam into a &quot;planar&quot; beam" Tags="|optics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2254" PostTypeId="2" ParentId="2229" CreationDate="2015-03-30T19:00:31.207" Score="0" Body="&lt;p&gt;Others have made a lot of good comments about documentation in general, but I want to suggest a particular class of software that will help tremendously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find &lt;a href=&quot;https://en.wikipedia.org/wiki/Revision_control&quot; rel=&quot;nofollow&quot;&gt;version control software&lt;/a&gt; to be excellent for these sorts of issues. It's not too common outside of software development, unfortunately, but it's only a matter of time before it catches on in other fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example: My research group has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Apache_Subversion&quot; rel=&quot;nofollow&quot;&gt;Subversion&lt;/a&gt; repository that most people use (A popular alternative software is &lt;a href=&quot;https://en.wikipedia.org/wiki/Git_%28software%29&quot; rel=&quot;nofollow&quot;&gt;git&lt;/a&gt;). Using version control helps ensure continuity of the research projects after someone leaves, because most of the files relevant to a project are there, along with a detailed log of what work they did at what time, and (if they do it right), their rationale. I'll elaborate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a folder on your computer that contains all of the files&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything is automatically dated when it's committed (the version control term for what others might call &quot;uploading&quot;). The commit message acts as a short memo (as Mahendra Gunawardena suggested, you might want longer and more official memos for bigger decisions). You describe the change to the project files, say &quot;We decided that approach X is not feasible. We are going to try Y instead. Deleted the files associated with approach X and created a new CAD model for approach Y.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, later, if you decide that approach X is actually the right thing and you want to dig up the deleted files, it's not hard. You can roll back any part to older version with ease and continue from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some additional benefits to this approach. First, it makes sharing files with others pretty easy. They just check out the repository (a version control term which could either mean the current version, or all versions) and then regularly update it (there are commands to do that). Basically, everyone is synchronized. Second, you can get a backup (older versions) for basically free. With that being said, this doesn't mean you don't need to keep normal backups, just that you have another option for restoring accidentally lost files. I highly recommend backing up the entire repository as well.&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2015-03-30T19:00:31.207" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2255" PostTypeId="2" ParentId="329" CreationDate="2015-03-30T22:02:30.220" Score="1" Body="&lt;p&gt;Any heat exchanger has different values for the viscosity along its length, since it depends on the fluid's temperature. You should use an average value for the viscosity. Surely if it varies too much it is going to be harder to estimate this value, but it can be done. You can do it by knowing the shear rate in the different regions of your stream, and then prorate how much of your fluid goes though each region (in percentage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to keep in mind that with the best Nusselt number correlation you will have at least a 20% error, so it is always a little uncertain&lt;/p&gt;&#xA;" OwnerUserId="1075" LastEditorUserId="368" LastEditDate="2015-03-30T23:33:15.970" LastActivityDate="2015-03-30T23:33:15.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2256" PostTypeId="2" ParentId="2182" CreationDate="2015-03-30T22:02:42.893" Score="4" Body="&lt;p&gt;This precise question has been researched in detail. As Olin Lathrop suggested, there is a maximum drop size. So, &quot;balls&quot; of water will tend to break up into smaller droplets. How big will the largest stable droplet be? The answer is that it depends on the surface tension of the water, how fast the droplet is launched, the drag coefficient, and atmospheric air density. Most of these you have little control of, though you can get a &lt;a href=&quot;http://www.engineeringtoolbox.com/water-surface-tension-d_597.html&quot; rel=&quot;nofollow&quot;&gt;modest increase (~6%) in surface tension&lt;/a&gt; by using nearly freezing water (which may be advantageous in this situation for other reasons too). Just don't use actually frozen water, as while that would solve the breakup problem, it would be extremely dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an equation that gives you an idea of how things vary (from the book &lt;a href=&quot;http://books.google.com/books?id=R5MH86P-hLcC&amp;amp;lpg=PP1&amp;amp;pg=PA31#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Atomization and Sprays by A. Lefebvre&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;D_\text{max} = \frac{8 \sigma}{C_\text{D} \rho_\text{a} U^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D_\text{max}$ is the maximum diameter, $\sigma$ is the surface tension of the water, $C_\text{D}$ is the drag coefficient of the droplet, $\rho_\text{a}$ is the atmospheric air density, and $U$ is how fast the droplet is launched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this equation, you can see that higher surface tension is better, and faster launches hurt the drop size. Plugging in some plausible numbers, I get a maximum droplet size on the order of a centimeter assuming the launch velocity is 10 m/s. These droplets are not very big but may be okay for your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a related point, as you can see in the book I linked to, no, the &quot;ball&quot; won't maintain its shape. The book gives a series of illustrations to show possible distortions the drop will go through prior to breaking up (&lt;a href=&quot;http://books.google.com/books?id=R5MH86P-hLcC&amp;amp;lpg=PP1&amp;amp;pg=PA30#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;e.g.&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water balloon acts similarly to a fluid with extremely (impossibly) high surface tension. The tension in the balloon itself is pretty similar to surface tension. So, I think it is in your best interest to keep using balloons. You can avoid popping for a compressed air cannon with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sabot&quot; rel=&quot;nofollow&quot;&gt;sabot&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="49" LastEditorUserId="49" LastEditDate="2015-03-31T19:48:11.610" LastActivityDate="2015-03-31T19:48:11.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2258" PostTypeId="2" ParentId="329" CreationDate="2015-03-31T03:44:46.343" Score="2" Body="&lt;p&gt;First, as Arthur notes, even the best Nusselt number correlations are often as much as 20% off, so don't expect any analytical method to give results that are much better than approximate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, there are ways to compute the Reynolds number for shear thinning fluids.  &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0377025704000473&quot; rel=&quot;nofollow&quot;&gt;Rudman, Blackburn, et al&lt;/a&gt; suggest using the effective viscosity for the mean wall shear stress.  If you have a fluid that can be modeled as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Herschel%E2%80%93Bulkley_fluid&quot; rel=&quot;nofollow&quot;&gt;Herschel-Bulkley Fluid&lt;/a&gt;, then the Reynolds number equation takes the following form.&#xA;$$ Re = \frac{\rho \bar{U} D}{\eta_w} $$&#xA;$$ \eta_w = K^{\frac{1}{n}} \frac{\tau_w}{(\tau_w - \tau_y)^{1/n}} $$&#xA;$$ \tau_w \approx \frac{4}{D} \frac{\Delta p}{L} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors note that other formulations of the Reynolds number exist for shear thinning flows, and there is no &quot;perfect&quot; Reynolds number for shear thinning flows, but state that this formulation takes the dynamics near the walls of the pipe into account well.&lt;/p&gt;&#xA;" OwnerUserId="709" LastEditorUserId="709" LastEditDate="2015-03-31T18:50:35.783" LastActivityDate="2015-03-31T18:50:35.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2260" PostTypeId="1" AcceptedAnswerId="2261" CreationDate="2015-03-31T11:24:20.767" Score="0" ViewCount="218" Body="&lt;p&gt;I had a &lt;a href=&quot;http://www.omegaeng.cz/ppt/pptsc_eng.asp?ref=PX26&amp;amp;Nav=preb02&quot; rel=&quot;nofollow&quot;&gt;PC Mountable Wet/Wet Differential Pressure Sensor&lt;/a&gt;&#xA;I used it to measure a pressure difference , I don't know how to convert its output to pressure ! &lt;/p&gt;&#xA;" OwnerUserId="1092" LastEditorUserId="368" LastEditDate="2015-04-01T03:50:37.437" LastActivityDate="2015-08-31T19:51:41.957" Title="Px26-015 pressure sensors output" Tags="|electrical-engineering|pressure|sensors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2261" PostTypeId="2" ParentId="2260" CreationDate="2015-03-31T11:39:55.037" Score="1" Body="&lt;p&gt;Well, it looks like your sensor is +/-5psid (based on the link you provided), so looking at the datasheet, we have (assuming a 10VDC excitation) 16.7mV at 1psi and 50mv at 5psi. Assuming the sensor is linear (the datasheet states that linearity is +/-0.25% of full scale), you have the following relationship:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P = V * 0.1201 - 1.006&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where &lt;code&gt;P&lt;/code&gt; is expressed in psi and &lt;code&gt;V&lt;/code&gt; in mV. This does not take into account any possible hysteresis (0.2% of full scale according to the datasheet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an accurate reading, you should probably do your own calibration by applying a series of known pressures (measured with a calibrated instrument, e.g. manometer) and reading the sensor output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did I arrive at the relationship? It's fairly easy, you're fitting a straight line to two points, so you're looking for a slope and an offset:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1psi = 16.7mV * slope + offset&#xA;5psi = 50mV * slope + offset&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That's 2 linear equations with 2 unknowns (&lt;code&gt;slope&lt;/code&gt; and &lt;code&gt;offset&lt;/code&gt;), which is easily solved to give the values mentioned above. That's really basic maths though, I would expect any engineer or even technician to know how to do that, which is why I hadn't included the details on how to calculate the values in the first place.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2015-03-31T14:49:08.913" LastActivityDate="2015-03-31T14:49:08.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2262" PostTypeId="2" ParentId="2253" CreationDate="2015-03-31T12:23:21.507" Score="5" Body="&lt;p&gt;This could be done with a laser (and I'll give you one idea of how), but honestly you will probably be better off with an array of infrared LEDs instead of a laser.  The infrared LEDs can be treated in much the same way over distances of a few meters.  You would collimate each LED with a small lens and then put another lens in front of the receivers (i.e. &lt;a href=&quot;https://engineering.stackexchange.com/a/331/272&quot;&gt;photodiodes&lt;/a&gt;) on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to do it with a laser, then you will likely want to use one or more &lt;a href=&quot;http://en.wikipedia.org/wiki/Laser_diode&quot; rel=&quot;nofollow noreferrer&quot;&gt;laser diodes&lt;/a&gt;.  They are cheap and reliable and well suited to a simple measurement like power detection.  One feature of laser diodes is that the beam produced by them is elliptical.  For many applications this is a disadvantage, but for yours it is useful.  The beam coming out from the laser diode will be highly divergent and you will need to collimate it with two &lt;a href=&quot;http://en.wikipedia.org/wiki/Cylindrical_lens&quot; rel=&quot;nofollow noreferrer&quot;&gt;cylindrical lenses&lt;/a&gt;.  This will allow you to keep the beam narrow in one direction while allowing it to expand significantly in the other.  Theoretically there isn't an upper limit to the ratio of the two beam sizes, but I suspect that a practical upper limit is ~100:1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:CollimatingLensSVG.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PQz4P.png&quot; alt=&quot;Collimating Light&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a ratio of 100:1 isn't good enough for you, then you can use a series of beamsplitters such as I've tried to sketch below.  In this case you would collimate the laser and send it in from the top of your transmitter bar and a series of evenly spaced beamsplitters would send part of the beam off toward the receiver bar.  They could be made to essentially overlap, but you need to be careful not to have too much overlap lest you have significant interference effects (this is one reason that incoherent infrared light might be better).  The beamsplitters would need to have properly chosen reflectivities to give the beams an equal amount of power.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5kMs9m.png&quot; alt=&quot;Series of Beamsplitters&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there is the issue of detection.  For this you would need an array of &lt;a href=&quot;https://engineering.stackexchange.com/a/331/272&quot;&gt;photodiodes&lt;/a&gt; on the receiver side.  Photodiodes typically aren't that large, so you would need to focus the light down onto each one.  This would require a pair of cylindrical lenses which are essentially the same as the ones you used to collimate the laser on the other side.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-03-31T12:35:17.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2263" PostTypeId="2" ParentId="448" CreationDate="2015-03-31T17:36:17.663" Score="1" Body="&lt;p&gt;The question relates to the functions of GPS related software, and not the loss of GPS signals or functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the algorithmic software errors related to GPS related computations or communications must be reviewed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This handling occurs in two different levels:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;GPS position calculation (which is conducted by a GPS-receiver card)&lt;/li&gt;&#xA;&lt;li&gt;GPS handling on the Aircraft system side, which reads position data from the GPS position card (or device).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;1) For the position calculation card, refer to patents and the calculation methodologies of GPS position. Functionally, software error (due to hardware or due to software program) can occur at this level, and may cause erroneous position calculation. Handling of GPS Satellite almanacs, or World Coordinate system conversion could be potential failure scenarios. I do not mean that these failures could happen in a particular device. Rather, these can be possible software errors to consider and verify against.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) GPS data is used by some autopilots and also for navigation systems. In order to use the current position of the aircraft, the flight control system talks to the GPS card or position Equipment (can be GPS/INS integrated, or not). While talking to the GPS Equipment, software errors could hypothetically occur at &#xA;2.a) interface handling (message failure)&#xA;2.b) mis-interpretation (message is misunderstood, e.g. the position is given as 34.5323423 but it is parsed as 34deg53'23&quot;, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this answers the chain of thoughts that the author of the book was askig for: &quot;to identify 3 hazards that could become linked to software failures&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, these scenarios would be caught by the use of complex system development process guidebooks (i.e. SAE-ARP-4754A (Guidelines For Development Of Civil Aircraft and Systems), or DO-178C Software Considerations in Airborne Systems and Equipment Certification, etc.) or by reusing existing and proven software/equipment.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastEditorUserId="33" LastEditDate="2015-03-31T17:56:57.060" LastActivityDate="2015-03-31T17:56:57.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2264" PostTypeId="1" CreationDate="2015-03-31T18:58:34.357" Score="4" ViewCount="200" Body="&lt;p&gt;Right now, changes to BOMs are executed through verbal or email instructions to an employee who serves as BOM gatekeeper. This means paper trails are almost non-existent. Who changed what and why is almost impossible to track down. I'm planning to establish an engineering change order system to correct this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an issue. There are BOM changes that require engineering approval: swapping parts around, changing PCB revisions, etc. But there are BOM changes that don't need engineering approval: wire lengths, lug sizes, pre-approved alternate parts, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously our BOM gatekeeper could just sort the changes into those piles intelligently. But is there a standard automated way to accomplish the same end? To define certain changes that can be made, and others that can't without engineering approval?&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="110" LastEditDate="2015-04-01T00:07:37.433" LastActivityDate="2016-04-24T13:59:26.413" Title="How can I arrange my ECO system to enforce the principle of least privilege?" Tags="|project-management|documentation|procedure|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2265" PostTypeId="1" AcceptedAnswerId="6290" CreationDate="2015-03-31T20:08:01.130" Score="18" ViewCount="8358" Body="&lt;p&gt;...or what type of motor is used there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this type of motor - usually powered with low-voltage AC (~12V), but at times with 230V, in several appliances that require very slow rotation and sometimes a fair momentum - a color-shifting lamp, the microwave plate, an ice cream mixer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The funny property of it is it picks the start direction at random and keeps spinning in that direction until switched off - but I never faced a situation when it would get stuck in the &quot;unstable balance&quot; position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is this type of motor and why does it behave that way?&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-05-06T20:15:57.433" Title="Why does the microwave plate start in a random direction?" Tags="|electrical-engineering|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2266" PostTypeId="1" AcceptedAnswerId="2276" CreationDate="2015-03-31T20:39:13.563" Score="-5" ViewCount="136" Body="&lt;p&gt;What general specifications are there in building a storm shelter in the American Midwest?  I believe the major considerations are tornados and severe thunderstorms, as well as flooding along river valleys.  &lt;/p&gt;&#xA;" OwnerUserId="177" LastActivityDate="2015-04-01T18:09:36.513" Title="Building a storm shelter" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2015-11-08T13:33:35.803" ContentLicense="CC BY-SA 3.0" />
  <row Id="2267" PostTypeId="2" ParentId="2264" CreationDate="2015-03-31T22:27:02.850" Score="3" Body="&lt;p&gt;The main purpose of a ECO process is to communicate changes to team members, other departments, customer and suppliers. ECO is industry accepted process that has now been automated using various tools. All affected parties need to provide input for the system to work effectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many global organization require multiple levels of approvals. In one such organization the changes are classified as priority 1 and priority 2. The change request would be reviewed at most at three levels. The first level change was chaired by the Chief Engineer, and the project manager, all affected engineers, and engineering managers would discuss change. An approved change would be assigned a priority number. A priority 1 change would be routed to next level where the affected manufacturing plants would discuss and provide input. Note: In this organization most manufacturing plants were global. Next an approved change would precede to the next level for approval mostly for funding and customer communications. The change owner champions the change and is responsible for maintaining the documentations. Some content from &lt;a href=&quot;https://engineering.stackexchange.com/questions/1880/i-have-inherited-a-product-with-inconsistent-designs-sharing-the-same-part-numbe/1884#1884&quot;&gt;this&lt;/a&gt; post can be helpful too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DUgkr.jpg&quot; alt=&quot;ECO-Global Company&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; This organization was global with over 200,000 employees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technology has made the process efficient. Some of the available software platforms are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://go.sap.com/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;SAP&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.oracle.com/partners/en/products/applications/agile/get-started/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Agile Product Lifecycle Management&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Product_lifecycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Product lifecycle management&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.acmpglobal.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Change management professionals&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-04-24T13:59:26.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2268" PostTypeId="2" ParentId="2265" CreationDate="2015-04-01T03:45:40.677" Score="6" Body="&lt;p&gt;Three types of motor may be used, both of which could do this. One of these (the Synchronous motor) is wat is used here and is a subset of the Brushless DC motor. (A misnomer as there is no pure DC used in the motor proper in a BLDCM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual motor type is a synchronous motor, identified correctly by jpa. &#xA;The synchronous motor is a special-case of the BLDCM (Brushless DC motor) that I describe below. In the general case a BLDCM generates an AC field from a DC source - either a fixed frTequency field that the rotor follows at fixed speed, OR from a variable frequency a source whose frequency is based on the current rotor speed and applied in such a way that the rotor &quot;chases&quot; the field which is derived from its own motion. (Phase lead/lad allows speed change - another subject). In the synchronous motor seen here there is a coil with winding axis vertical when motor sits flat on a surface. The coil connects to (in this case low voltage AC via a transfomer from) the AC mains so alternatively produces N-S or S-N magnetisation along its axis. Poles are created by adding plates with multiple radial tabs - each tab is a pole. As the coil changes NS, SN, NS the alternative tabs are all N or all S and as the field changes the NSNSNS... patterm moves in steps around the circumference.  The rotor has N and S permanent magnet poles. These initially aligh in opposite phase wrt the stator poles and when these reverse polarity the rotor is attracted AND repulse to a position one tab away. However, if fully symmetrical, a N pole on the rotor could be attracted to the S to its &quot;left&quot; or the S to it's right. Ince rotating it will have a preference for the pole in its direction of motion but, as startup, could go either way. And does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stator pole polaritiesreverses succesively &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NSNSNS ...&lt;br&gt;&#xA;SNSNSN ...&lt;br&gt;&#xA;NSNSNS ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotor follows stator changes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) From here &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   NS     &amp;lt;- rotor in position 3-4&#xA; SNSNSNSN &amp;lt;- Stator&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(2a) To here is valid&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  NS      &amp;lt;- rotor moves left to position 2-3&#xA; NSNSNSN  &amp;lt;- Stator changes polarity from (1) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(2b) But, so is: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    NS     -&amp;gt; rotor moves right to position 4-5  &#xA; NSNSNSNSN &amp;lt;- Stator changes polarity from (1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In this case there is no DC - the field is suppluied from AC mains and the rotor &quot;chases&quot; the rotating AC field. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Motor types:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Most usual in the past  - Traditionally a &lt;strong&gt;&quot;shaded pole&quot; motor&lt;/strong&gt; may be used where a &quot;bodge&quot; is used to distort the magnetic field from a field winding in such a way that a rotating magnetic &quot;vector&quot; is produced that the rotor follows. A magnetic shunt is produced with a turn of conductor at the airgap in the steel core that the field coil is wound on. When power is first applied the rotor position relative to the airgap will cause it to be jerked in one or other direction and once motion has started the rotating field that results reinforces that motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shaded pole motors are simple, cheap, and have been around for almost ever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excellent &lt;a href=&quot;https://www.youtube.com/watch?v=wHhhj8IHq5c&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;laymans introduction to shaded pole motors&lt;/strong&gt;&lt;/a&gt; - You tube video. 8 minutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Shaded-pole_motor&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Shaded Pole Motors - Wikipedia&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jMopE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) A &lt;strong&gt;brushless DC motor&lt;/strong&gt; (BLDCM) may be used.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The synchronous motor described above is a special case simple subset of a BLDCM. In both cases a permanent magnet rotor follows a rotating AC field. In a 'true' BLDCM the fe\ield is usually generated electronically by switching DC. In these simple synchronous motors the rotating field is supplied from the AC mains via a transformer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motors that need a clean fast start use magnetic sensors that give absolute feedback on direction and speed. Motors that must rotate the right way (eg disc drive motor) may use sensorless systems that derive back EMF voltages from the motor windings BUT circuitry is included to check rotation and adjust the powering if direction starts wrongly. Systems that do not care about direction and that want lowest cost just use a sensorless system and accepts what comes. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2016-10-31T11:01:18.327" LastActivityDate="2016-10-31T11:01:18.327" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2270" PostTypeId="1" CreationDate="2015-04-01T08:33:50.883" Score="12" ViewCount="1546" Body="&lt;p&gt;A neat &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/3/35/Sideling_Hill_cut_MD1.jpg/1024px-Sideling_Hill_cut_MD1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;image from Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mcNyLl.jpg&quot; alt=&quot;cut with steps&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There're clearly visible steps on the cut wall. This makes the cut wider at the top and so requires more work so those steps have some serious reason behind them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the purpose of these steps?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="90" LastEditDate="2015-04-02T01:47:07.840" LastActivityDate="2015-07-28T17:01:38.770" Title="Why do earth cuts sometimes have steps on their walls" Tags="|civil-engineering|geotechnical-engineering|roadway|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2271" PostTypeId="2" ParentId="2212" CreationDate="2015-04-01T09:01:21.193" Score="1" Body="&lt;p&gt;Axsys may be what you are looking for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bentley.com/en-us/products/bentley+axsys+process/features-list.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.bentley.com/en-us/products/bentley+axsys+process/features-list.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can link your drawings to your simulation files (HYSYS, PRO/II etc). I.e. Axsys can read relevant data (heat exchanger sizing, pipes, valves etc) and then allows you to generate datasheets, equipment lists etc. So if you need to make a design change you can just update your simulation file and it will propogate to your drawings (PFD and P&amp;amp;ID) and equipment datasheets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty convenient if you'll be going through several design iterations. That said, I haven't used it professionally so I can't say how well it works. And, its not free.&lt;/p&gt;&#xA;" OwnerUserId="1103" LastActivityDate="2015-04-01T09:01:21.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2272" PostTypeId="2" ParentId="2270" CreationDate="2015-04-01T10:43:01.807" Score="2" Body="&lt;p&gt;They provide a terrace on each level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which result in a more stable pyramid shape of the remaining hill that is less likely to collapse onto the road. And it removes/alleviates the need for a strong retaining wall.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-04-01T10:43:01.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2273" PostTypeId="2" ParentId="2125" CreationDate="2015-04-01T11:01:52.997" Score="1" Body="&lt;p&gt;I work for a fortune 500 mechanical engineering company.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of product development in an mechanical system is broadly: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that you have a complete specification of things like forces, temperatures, pressures etc. and can begin designing straight away. The first thing is to start generating design concepts that might or might not meet the specification. This is a combination of selecting the geometry that will accomplish the task, the material that will enable ease of manufacture and performance at all operating conditions, with the right structural integrity properties to last for the specified design life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This design is completed in a CAD package, where the solid components are modeled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, assemblies and production drawings can be made, but first we must validate our design with some analysis. This can be basic math and physics to show the dimensions will work within the design envelope, basic kinematics and kinetics, or mass balances for a design that processes a fluid for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More accurate analysis requires the use of sophisticated methods implemented in software known as CAE (computer aided engineering) or ALD (analysis led design)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For studying stresses and strains, temperatures, and to some extent fluids, the finite element method is used to calculate local properties based on a meshed model and some boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For studying fluid interaction with the geometry and looking at pressures, temperatures, and other fluid and thermal properties, computational fluid dynamics is used. This is most commonly an implementation of the finite volume method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You alluded to simulate and iterate. That is essentially how the process is done. However, statistically verified methods are used using design of experiments theory to produce optimized design configurations.&lt;/p&gt;&#xA;" OwnerUserId="1104" LastEditorUserId="33" LastEditDate="2015-04-01T12:42:13.493" LastActivityDate="2015-04-01T12:42:13.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2274" PostTypeId="2" ParentId="2270" CreationDate="2015-04-01T12:00:10.283" Score="20" Body="&lt;p&gt;The main reason for the stepped arrangement of the slope is &lt;a href=&quot;http://www.cedengineering.com/upload/Rock%20Engineering%20-%20Rock%20Slopes.pdf&quot; rel=&quot;noreferrer&quot;&gt;slope stability&lt;/a&gt; and safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such walls/cuttings are divided into two components:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;sub vertical walls between two flat horizons&lt;/li&gt;&#xA;&lt;li&gt;flat sections between sub vertical walls&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By having a stepped arrangement, material and hence weight, is removed from above the toe of slope (where the bottom part of the slope meets the ground). This reduces the pressure at the toe of the slope and assists with the &lt;a href=&quot;http://www.dot.ca.gov/hq/esc/geotech/geo_manual/Rock%20Cut%20Slope%20Design_June%202013.pdf&quot; rel=&quot;noreferrer&quot;&gt;stability&lt;/a&gt; of the slope. This is critical because should the toe fail, the whole wall/slope will fail. To further assist with toe pressure reduction and slope stability, the sub vertical parts of the wall generally have a slope themselves. The maximum angle of each wall component is determined by a geotechnical analysis, which considers things such as: rock mass rating, joint orientation, joint persistence, conditions of joints (joint infill, joint friction, presence of slippery minerals such as talc), &lt;a href=&quot;http://web.mst.edu/~rogersda/umrcourses/ge341/The%20Rock%20Quality%20Designation%20%28RQD%29%20in%20Practice.pdf&quot; rel=&quot;noreferrer&quot;&gt;RQD&lt;/a&gt; (Rock Quality Designation), rock strength. In stronger and more component rock the angle can be steep. In weaker and less component rock, the wall angle will be shallower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on local terminology and whether the slope is for civil or mining purposes, the flat sections of the stepped slope are called benches, catch benches or berms. In addition to stepping the slope back into the in-situ rock, the benches catch smaller rocks and rubble that can fall off the wall from time to time. The wider the benches the better, firstly due to the increased offset of the wall and pressure reduction at the toe, but also the increased capacity to catch rocks falling from the slope above.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="2246" LastEditDate="2015-07-28T17:01:38.770" LastActivityDate="2015-07-28T17:01:38.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2275" PostTypeId="2" ParentId="2242" CreationDate="2015-04-01T17:20:39.800" Score="3" Body="&lt;h2&gt;Your question is insufficiently constrained.&lt;/h2&gt;&#xA;&lt;p&gt;The short answer is that you haven't provided enough information to give a complete answer: creep is complicated enough for a single material, however there are a variety of mechanisms depending on the material, temperature, normalised stress, time etc etc. In a &amp;quot;generic metal&amp;quot; the below diagram is a fair place to start for determining which mechanism you are working with, but those breakpoints will still change with the material. Obviously something like aluminium stops behaving like a solid metal &lt;strong&gt;well&lt;/strong&gt; below 3000 deg C.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9k1fn.gif&quot; alt=&quot;Temperature-stress dependence in metals&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Some nickel superalloys experience &lt;em&gt;less&lt;/em&gt; creep at higher temps, although these are exotic and &lt;em&gt;extremely&lt;/em&gt; expensive.&lt;/p&gt;&#xA;&lt;p&gt;If your question truely is &lt;em&gt;&amp;quot;How is this determined for a specific spring of a specific material?&amp;quot;&lt;/em&gt; then the answer is disappointingly simple: put your spring in a constant temp oven for a range of very long times and a range of loads and measure what happens. It doesn't matter by which mechanism it is creeping, all you want to know is how it will respond in time to that load at that temp.&lt;/p&gt;&#xA;&lt;p&gt;Edit: following your comment I feel I understand that what you are asking slightly better so should revise the answer accordingly. Whilst it is always best to test your parts creep is particularly nasty in that you can't get your answers quickly and it tends to be expensive. A material would help more than anything else here, but here is a suggested design method (I am not familiar with the design of springs to alleviate creep, this is just a common sense set of suggests from a fellow engineer):&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Decide yourself the operating conditions you expect it to take: load, duration of load and temperature. You ought to know this all.&lt;/li&gt;&#xA;&lt;li&gt;(optional) Decide what kind of spring response you want and design the spring accordingly. Different springs respond to load fundamentally differently.&lt;/li&gt;&#xA;&lt;li&gt;Find some equation for estimating the stress distribution in a spring; you haven't specified what type of spring you are designing but based on the mention of compressive load I would assume a steel coil spring but it could be one of many. If that is the case then I don't have an equation for the stress distribution, which is likely to be complex as coil springs probably have a combination of axial and shear loads when they deform. Either way, find some stress distribution, I'd imagine because of continuity a mean stress would do which at first glance would just be a function of angle of coil relative to direction of compression.&lt;/li&gt;&#xA;&lt;li&gt;Decide whether creep is seriously a consideration for you. Most springs operate no where near the UTS of the material they are made from and if you are orders smaller then I'd question whether creep is something you need to worry about at all. Your car sits on its springs for decades and doesn't lose appreciable ride height. Even if you do get some change, does it influence on your design at all if your spring grows fractionally shorter over the space of a decade?&lt;/li&gt;&#xA;&lt;li&gt;Decide on how important weight saving is to you and choose your material to give you the desired spring constant whilst reducing cost as best you can.&lt;/li&gt;&#xA;&lt;li&gt;Use a table similar to the one below to find the region of creep which you are operating in, given the material, and calculate your rate of creep from an equation which applies to that region and constants appropriate to your material. You can find this all online.&lt;/li&gt;&#xA;&lt;li&gt;Figure out whether this rate is acceptable.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Please consider that the first step that one would take to alleviate creep (decreasing the normalised stress) will change the characteristics of your spring which likely restricts you to a material change or a different spring design.&lt;/p&gt;&#xA;" OwnerDisplayName="user815" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-04-01T19:01:39.887" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2276" PostTypeId="2" ParentId="2266" CreationDate="2015-04-01T18:09:36.513" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://www.fema.gov/&quot; rel=&quot;nofollow&quot;&gt;FEMA (Federal Emergency Management Agency)&lt;/a&gt; publishes some specifications for the design of safe rooms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per &lt;a href=&quot;https://www.fema.gov/safe-rooms&quot; rel=&quot;nofollow&quot;&gt;their guidance&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.fema.gov/media-library/assets/documents/2009&quot; rel=&quot;nofollow&quot;&gt;FEMA P-320&lt;/a&gt; is for residential safe rooms&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.fema.gov/media-library/assets/documents/3140&quot; rel=&quot;nofollow&quot;&gt;FEMA P-361&lt;/a&gt; is for community and small business safe rooms&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And according to &lt;a href=&quot;https://www.fema.gov/safe-rooms/frequently-asked-questions-tornado/hurricane-safe-rooms#Q18&quot; rel=&quot;nofollow&quot;&gt;their FAQ&lt;/a&gt;, there are restrictions regarding potential flood risks.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Per FEMA P-361, flood hazards should be considered when designing a residential safe room. Flood loads acting on a structure containing a safe room are strongly influenced by the structure’s location relative to the flood source. Tornado or hurricane residential safe rooms should be located outside of the following high-risk flood hazard areas:  &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Flood hazard areas subject to high-velocity wave action (Zone V areas) and Coastal A Zones  &lt;/li&gt;&#xA;  &lt;li&gt;Floodways  &lt;/li&gt;&#xA;  &lt;li&gt;Any areas subject to storm surge inundation associated with any modeled hurricane category, including coastal wave effects&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;More information on these siting restrictions can be found in FEMA P-361. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should also look at &lt;a href=&quot;https://www.fema.gov/safe-rooms/frequently-asked-questions-tornado/hurricane-safe-rooms#Q20&quot; rel=&quot;nofollow&quot;&gt;their answer&lt;/a&gt; regarding other applicable standards:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;FEMA P-361 provides the design criteria used with common building codes and standards to design a building. This means that the underlying building code (such as the International Building Code or International Residential Code) applies and building construction must comply with the many items that are regularly governed by code requirements. Standards such as ASCE 7 also apply but are used in conjunction with the safe room design criteria described in FEMA P-361 to produce a structure capable of resisting loads much higher than those for normal buildings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So you are correct that the major considerations are tornadoes and severe thunderstorms.  Their considerations for flooding are fairly common sense as they can be summarized with &quot;don't build where it will flood&quot;, but they certainly provide more guidance than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that the guidelines are applicable outside of the North American Midwest.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-04-01T18:09:36.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2277" PostTypeId="2" ParentId="2204" CreationDate="2015-04-01T18:12:03.327" Score="4" Body="&lt;p&gt;The answer to your question is very simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the arch and the rectangle are made of the same material, and the rectangle height and width are equivalent to the arch rise and span, and they are loaded with the same load in the same place, and the weight of the material is neglected, the rectangle will always have at least as as much load capacity as the arch, and could have more depending on the material properties. Always. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a result of the principle of superimposition. If you overlay your arch on top of your rectangle, you'll see that the rectangle, in a manner of speaking, &lt;em&gt;contains&lt;/em&gt; the arch: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b3yiC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extra material above and below the arch, given your assumptions (the most important of which, aside from using the same material and dimensions, is ignoring self weight of the material), cannot negate the load carrying capacity of the arch &quot;inside&quot; of it. Therefore, the rectangle will have at least as much capacity as the arch. &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2015-04-02T11:52:53.213" LastActivityDate="2015-04-02T11:52:53.213" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2278" PostTypeId="1" CreationDate="2015-04-01T20:46:05.217" Score="8" ViewCount="4766" Body="&lt;p&gt;I have been thinking about hydroelectric plants and I have been wondering why they are built the way they are.  As in understand it, a stream of water is sprayed at a turbine, located at the bottom of a dam (Pelton Turbine.)  This does not seem to be a very effective way to convert the energy.  What are the factors that make this method effective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method would be to place the water into buckets, then connect these buckets to a belt system and use the belt to power the generator.  This method is similar to how  water wheel works.  What are the draw backs to this method that make it less effective than the turbine.&lt;/p&gt;&#xA;" OwnerUserId="1109" LastEditorUserId="6790" LastEditDate="2020-07-31T21:04:15.340" LastActivityDate="2020-07-31T21:04:15.340" Title="Why do hydroelectric plants use turbines instead of an efficient 'water wheel' configuration?" Tags="|electrical-engineering|hydraulics|power|turbines|hydroelectric-dams|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="2279" PostTypeId="1" CreationDate="2015-04-01T21:29:29.987" Score="4" ViewCount="5193" Body="&lt;p&gt;How can I plot such a graph with Excel or Matlab by a uniform load q, and how can I see where the stress becomes 0.1q?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loaded area and the equation is on the picture shown below, equation is highlighted in yellow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0Zode.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Zodem.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/znYKu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/znYKum.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PwzXg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PwzXgm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1110" LastEditorDisplayName="user16" LastEditDate="2015-04-03T16:48:24.793" LastActivityDate="2016-03-25T05:21:37.150" Title="Stress Bulbs for strip &quot;continuous&quot; footings" Tags="|civil-engineering|geotechnical-engineering|foundations|" AnswerCount="1" CommentCount="4" ClosedDate="2016-03-27T19:23:02.160" ContentLicense="CC BY-SA 3.0" />
  <row Id="2280" PostTypeId="1" AcceptedAnswerId="2284" CreationDate="2015-04-01T20:22:55.213" Score="3" ViewCount="299" Body="&lt;p&gt;I am trying to figure out what kind of device is being used to stop and start the flow of liquid inside this device. Below is a link to the device. It looks too small to be a solenoid or a pump. How else would they do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard somewhere it may be a Venturi system but how do you electronically control a Venturi system that is that small?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wunderbar.com/dispensing/beverage-dispensing/liquor-dispenser/skyflo&quot; rel=&quot;nofollow&quot;&gt;http://www.wunderbar.com/dispensing/beverage-dispensing/liquor-dispenser/skyflo&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1134" OwnerDisplayName="JoeTaylor" LastEditorUserId="110" LastEditDate="2015-04-02T00:04:25.603" LastActivityDate="2015-04-02T01:33:26.140" Title="What type of liquid flow control could this device use?" Tags="|electrical-engineering|valves|liquid|actuator|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2281" PostTypeId="2" ParentId="2278" CreationDate="2015-04-01T21:52:03.943" Score="19" Body="&lt;p&gt;Actually the turbine method is very efficient. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Turgo_turbine&quot;&gt;Turgo turbine&lt;/a&gt; is up to 90% efficient at extracting the energy from the water. Large-scale hydro plants using &lt;a href=&quot;http://en.wikipedia.org/wiki/Francis_turbine&quot;&gt;Francis Turbines&lt;/a&gt; can be as high as 95% efficient (&lt;a href=&quot;http://www.mpoweruk.com/hydro_power.htm&quot;&gt;see here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrical generators typically use high RPMs, so you need speed as well as torque to drive it round. Low RPM generators are possible too, but less efficient. Energy available in hydro power comes from the 'head' (height drop) and the 'flow' (amount of water per second). So in low-head situations, you need much more flow to get the same energy. In these cases, you won't get high pressures (and therefore high speeds) without gearing (efficiency losses there too), so a water-wheel approach is feasible. In fact there are &lt;a href=&quot;http://blogs.worldwatch.org/revolt/an-advance-in-low-head-hydro-technology-opens-new-sites-for-potential-electricity-generation/&quot;&gt;low-head generators similar to how you describe&lt;/a&gt; - the bucket-and-belt approach - as well as &lt;a href=&quot;http://www.small-hydro.com/Programs/Innovative-Technologies/121-1-Very-Low-Head-Turbine-Generator.aspx&quot;&gt;more traditional low-head turbines&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low-head situations are harder to engineer (economically), because high flows are needed so the whole installation (channels, bearings, generators, etc.) must be much stronger and bigger. It's cheaper (per kW of generating capability) to use high-head situations with smaller/faster turbine wheels at higher pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these are the main reasons you see mostly turbines instead of other types of hydro plant. As technology improves and low-head locations become cheaper to exploit, we will see more of them.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-04-01T21:52:03.943" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2282" PostTypeId="2" ParentId="2280" CreationDate="2015-04-01T22:20:36.757" Score="0" Body="&lt;p&gt;While I don't know exactly how that system works, there is no technical need to stop/start the flow by electronic means at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could be multiple reservoirs, for example 2x 25ml reservoirs and 1x 50ml one. When the bottle is tipped upside-down, one or both (depending on user selection) of the 25ml reservoirs empties into the drink, while the 50ml one refills from the bottle. When the bottle is tipped the right way up again the 50ml reservoir refills both 25ml ones (it doesn't matter if it is left half empty, it will always refill completely when the bottle is poured again).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronics in that system could then simply detect whenever the bottle is poured and which measure was selected, and transmit that information. The flow control could be purely mechanical.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-04-01T22:20:36.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2283" PostTypeId="2" ParentId="2278" CreationDate="2015-04-02T00:55:31.303" Score="14" Body="&lt;p&gt;There are two different ways for the water flow to transmit energy. One is the pressure difference between the water entering and leaving the turbine. The other the kinetic energy change of fast flowing water entering the turbine and slow flowing water leaving it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your bucket idea seem to be like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Pelton_wheel&quot;&gt;http://en.wikipedia.org/wiki/Pelton_wheel&lt;/a&gt; which works by extracting the kinetic energy from a fast water stream and is efficient for low flow rates. If the buckets on the wheel are moving at half the speed of the water, the theoretical efficiency is 100%, but if the speeds are in a different ratio the efficiency can be much less. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the water supply has a high pressure (e.g. at the bottom of a deep dam) it is more efficient to use the pressure difference directly in a design like a Francis Turbine, rather than converting the high pressure into a high flow velocity (which would waste a lot of energy in turbulence in the water and friction against the walls of the pipes) and then using the energy that is left to drive a Pelton wheel or some other &quot;bucket&quot; design.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-04-02T00:55:31.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2284" PostTypeId="2" ParentId="2280" CreationDate="2015-04-02T01:33:26.140" Score="5" Body="&lt;p&gt;I'm going to cheat by referencing the patent. ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Automatic Bar Controls owns the Wunderbar site and has a patent on the product. The patent is &lt;a href=&quot;https://www.google.com/patents/US8925769?dq=8925769&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ei=GZUcVZjgFYGqNteEgdAE&amp;amp;ved=0CB4Q6AEwAA&quot;&gt;US008925769&lt;/a&gt;. The actual detail drawings don't seem to be available there though, so you have to get it straight from the US Patent Office &lt;a href=&quot;http://pdfpiw.uspto.gov/.piw?PageNum=0&amp;amp;docid=08925769&amp;amp;IDKey=47085CEF0D39&amp;amp;HomeUrl=http%3A%2F%2Fpatft.uspto.gov%2Fnetacgi%2Fnph-Parser%3FSect2%3DPTO1%2526Sect2%3DHITOFF%2526p%3D1%2526u%3D%2Fnetahtml%2FPTO%2Fsearch-bool.html%2526r%3D1%2526f%3DG%2526l%3D50%2526d%3DPALL%2526S1%3D8925769.PN.%2526OS%3DPN%2F8925769%2526RS%3DPN%2F8925769&quot;&gt;listing&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name of the patent is: &lt;strong&gt;Wireless spout and system for dispensing&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Embodiments of the present invention provide a pouring device for a container for the dosing of liquid. The pouring device has various features intended to ease use, including an improved removal and attachment system and a system to identify and visually illustrate selected pour sizes. In a specific embodiment, the pouring device has a colored light or LED indicator system that allows the user to quickly and easily confirm the selected pour size. Embodiments may also include one or more features that ease attachment and removal of the pouring device to a liquid container, such as a replaceable cork system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The specific part is Claim 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(c) an electrically operated valve disposed within the spout housing for selectively clamping the conduit so that a registerable amount of liquid is dosed;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That seems to answer the question of how it works. It physically clamps off a tube to stop the flow of fluid. More specifically from the description in the patent:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The silicone tube can be easily squeezed to stop the liquid flow. This can be done by a stepping motor, a motor with gearbox, or any other motor or appropriate mechanism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-04-02T01:33:26.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2285" PostTypeId="1" CreationDate="2015-04-02T11:25:53.593" Score="6" ViewCount="1748" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/AN-94&quot; rel=&quot;nofollow&quot;&gt;AN-94&lt;/a&gt; is an assault rifle of Russian design chambered in 5.45 caliber. It is claimed to be state-of-art in Russian small arms development.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important design differences (relative to common designs like AKM) include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Internal receiver, chamber, bolt, bolt carrier, barrel and gas tube/piston are all integrated into one independent assembly - called firing unit.&lt;/li&gt;&#xA;&lt;li&gt;Stock, external receiver form what is called &quot;Effect Envelope&quot; and firing unit can move along the barrel axis relative to this envelope - which is supposed to increase accuracy by delaying recoil until the &lt;strong&gt;second&lt;/strong&gt; round is fired.&lt;/li&gt;&#xA;&lt;li&gt;The firing unit includes a rammer connected to the bolt carrier by a cable and pulley system. The rammer is moving forward as the bolt moves backwards, to accomodate for pre-feeding sequence, by stripping the next round from the magazine and moving it just in behind the chamber, where it's being subsequently fed by returning bolt.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The gun specification claims that when firing full-auto the first two rounds are being fired at 1800 rpm speed, but all subsequent ones at 600 rpm. Why is the time between the first two shots shorter? I can't find anything that would differentiate between the first two and all the subsequent rounds? I would understand if the gun was either keeping the same cyclic rate or rather shoot in a continuous stream of two-round bursts with mean rate of 600rpm but neither is the case. Can someone explain this to me please?&lt;/p&gt;&#xA;" OwnerUserId="1114" LastEditorUserId="1560" LastEditDate="2015-10-21T12:23:41.597" LastActivityDate="2015-10-21T12:23:41.597" Title="Why does AN-94 have higher burst fire than its cyclic rate?" Tags="|mechanical-engineering|firearms|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2286" PostTypeId="2" ParentId="2224" CreationDate="2015-04-02T11:58:42.260" Score="0" Body="&lt;p&gt;I found the below link to be very helpful, they have provided clearly the power required for metal halide lamp fixture. &#xA;&lt;a href=&quot;http://www.hoveyelectric.com/hovey-electric-power-blog/bid/85127/How-To-Calculate-Lighting-Costs-For-Metal-Halide-Fixtures&quot; rel=&quot;nofollow&quot;&gt;http://www.hoveyelectric.com/hovey-electric-power-blog/bid/85127/How-To-Calculate-Lighting-Costs-For-Metal-Halide-Fixtures&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link provides this list of typical metal halide power consumption:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;175 Watt Metal Halide  Fixture = 205 Actual Watts  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;250 Watt Metal Halide Fixture = 290 Actual Watts  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;400 Watt Metal Halide Fixture = 458 Actual Watts  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;1,000 Watt Metal Halide Fixture = 1,081 Actual Watts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1116" LastEditorUserId="368" LastEditDate="2015-04-17T20:12:17.873" LastActivityDate="2015-04-17T20:12:17.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2287" PostTypeId="1" CreationDate="2015-04-02T12:53:56.237" Score="20" ViewCount="2623" Body="&lt;p&gt;Here's how a typical earthfill carrying a roadway joins a typical bridge&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/09xSh.jpg&quot; alt=&quot;earthfill joins a bridge&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parts of the earthfill slopes are greenish - that's grass - and parts are whitish - that's concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earthfill slope is fully covered in concrete where it joins the bridge but the rest of the earthfill is only partially (lower part only) covered in concrete, the rest of the slope is grass-covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This photo is of a 20+ meters high earthfill and lower earthfills' slopes are usually not covered in concrete at all except where they join bridges - the join is usually fully covered regardless of how high the earthfill is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's so special in that part where the earthfill joins a bridge that this part is fully covered in concrete while the rest of the earthfill is either partially covered or not covered at all?&lt;/p&gt;&#xA;" OwnerUserId="374" LastActivityDate="2015-04-02T14:09:50.613" Title="Why are earthfills fully covered in concrete where they join bridges?" Tags="|civil-engineering|bridges|geotechnical-engineering|roadway|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2288" PostTypeId="2" ParentId="2287" CreationDate="2015-04-02T14:06:13.837" Score="25" Body="&lt;p&gt;This is called &lt;strong&gt;Slope Paving&lt;/strong&gt;. (In the US at least.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is done for two reasons at bridge abutments. These reasons are related:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Steeper slopes are possible.&lt;/li&gt;&#xA;&lt;li&gt;Protection from undermining abutments and piles.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Steep Slopes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The soil that is used to create the earthen fills will only stand up naturally to a certain steepness. By placing a layer (~4in) of concrete over the soil, the slope can be steeper than would naturally work. The concrete reinforces the soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steeper slopes are beneficial because they reduce the footprint of the embankment fill. This reduces the length of the spans by allowing the bridge abutments to be closer to the roadway underneath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Abutment Protection&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concrete slope paving also helps to protect the soil from eroding under the bridge abutments. The abutments (and piles) rely on having soil around them. This soil needs to be kept in place to ensure that the bridge performs as designed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some locations also put drains for the bridge deck near the abutments. It is simplest to let these drain pipes empty on the slope under the bridge. If the slope was not paved, the water would quickly erode the soil slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area under the bridge is partially shaded, so it can be difficult for vegetation to grow there. Natural vegetation would help with erosion, but if no plants grow, the bare soil would erode quickly. This concern is eliminated by covering the slope with concrete.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-04-02T14:06:13.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2289" PostTypeId="2" ParentId="2287" CreationDate="2015-04-02T14:09:50.613" Score="2" Body="&lt;p&gt;I remember a plenty of older bridges that have these places with soil as well. It's always barren earth with no vegetation whatsoever due to shortage of light and total lack of rainfall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's often harboring homeless, rowdy youth or kids, who seriously erode the slope with their feet. There's often trash, broken flasks, discarded refuse of homeless sifting through their &quot;loot&quot;. Generally, unpleasant, dirty, barren spots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Covering them with concrete, especially &quot;grid&quot; slabs with gravel filling the holes, making walking or sitting quite unpleasant (if much safer) - solves some of these problems. Considering there won't be any foliage, there are hardly any disadvantages.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-04-02T14:09:50.613" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2290" PostTypeId="2" ParentId="2285" CreationDate="2015-04-02T15:27:16.253" Score="4" Body="&lt;p&gt;The hammer mechanism is quite complex; it can &quot;choose&quot; to fire another round either as soon as its chambered or only in the forward position of the internal (moving) receiver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In single fire mode, it hits the round only once, when it's in the forward position, and doesn't do it again until both the barrel with the next round is in the forward position and the trigger was pressed again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2-bullet burst, the &quot;hit as soon as the round is chambered&quot; is enabled - after the first shot, the next one is fired when only the next bullet is chambered, but then the mechanism returns to inert forward position, awaiting next press of the trigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In full auto mode pressing the trigger sets the release of hammer ASAP, but it is reset after the second shot (when the barrel moves backwards) and doesn't &quot;return&quot; until next press of the trigger - after the first two rounds, the gun operates similarly as in single fire mode, except it fires as soon as the inner receiver reaches the forward position while the trigger is depressed, no need to release it first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, this behavior is achieved by a very complex construction of the trigger mechanism. New round is not chambered while the previous one (or just the shell) is still inside, so the mechanical &quot;logic&quot; that enables the hammer to fire in different position determines whether new rounds are chambered at &quot;double rate&quot; or only after a full cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://world.guns.ru/assault/rus/an-94-abakan-e.html&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to think of it is that AN-94 is two guns: a fairly standard Kalashnikov derivative, except its bolt carrier, chamber and barrel are replaced with a second, recoil-operated gun. While the Kalashnikov normally chambers the round on its forward movement and fires it when locked in the forward position, this second gun chambers the round on backwards movement (thus need for the rammer, otherwise it would be unable to strip it from the cartridge), and fires it as soon as it's chambered, still during its backward movement. &quot;The Kalashnikov part&quot; operates normally for single and burst mode, but the fire selector may allow the &quot;inner gun&quot; to shoot once, while moving after the first &quot;kalashnikov&quot; shot. (and also, the chambering of consecutive rounds after the first occurs following the &quot;inner gun procedure&quot; (backwards), not the &quot;kalashnikov procedure&quot; (forward motion), but that's merely a quirk.)&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2015-04-02T16:43:16.867" LastActivityDate="2015-04-02T16:43:16.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2291" PostTypeId="2" ParentId="2112" CreationDate="2015-04-02T15:40:29.887" Score="3" Body="&lt;p&gt;Typically one does not consider the key when checking a slab - footing interface. This is based on ACI318 Section 15.8.1.4, which discusses the transfer of lateral forces to footings and directs the reader to Section 11.6. 11.6 which says this,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;11.6.3 A crack shall be assumed to occur along the shear plane considered.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, you do not consider the key to be acting in any way structurally because you are to assume that a crack will form at the plane defining the intersection of the wall and footing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section 11.6.4 goes on to discuss how the nominal shear-friction capacity of the cross section is calculated, what coefficient of friction to use, and how much the concrete should be intentionally roughened at the cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience, these keys are typically added (by the request of the architect) to combat water intrusion at the concrete joint.&lt;/p&gt;&#xA;" OwnerUserId="129" LastActivityDate="2015-04-02T15:40:29.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2294" PostTypeId="2" ParentId="33" CreationDate="2015-04-03T03:01:14.277" Score="8" Body="&lt;p&gt;Multi-way bridges are rare for river crossings for the reasons you describe; however, they are seen as a component of highway interchanges, usually when left entrances/exits are present or access ramps have been &quot;braided&quot; into an existing system interchange, or where a SPUI has been built over an existing divided highway.  Viaducts over railroad tracks or highways are also built in this configuration on occasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this is the &lt;a href=&quot;https://goo.gl/maps/uC5kx&quot;&gt;intersection of Russel Blvd. and Gravois Ave. in St. Louis, MO&lt;/a&gt;, where a highway intersection is partially undercut by Interstate 55; another good example (in SPUI form), is at &lt;a href=&quot;https://goo.gl/maps/d2a3C&quot;&gt;Hampton Ave and I-64, also in St. Louis&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even more extreme forms of this can be found in Atlanta, GA, where the Downtown Connector freeway was tunneled under two intersections: &lt;a href=&quot;https://goo.gl/maps/0vJNa&quot;&gt;Baker St NE and Piedmont Ave NE&lt;/a&gt; and &lt;a href=&quot;https://goo.gl/maps/rH46k&quot;&gt;Memorial Dr SE and Capitol Ave SE&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, New York calls itself home to a true four-way bridge, the little-known &lt;a href=&quot;https://goo.gl/maps/K3wlb&quot;&gt;Macomb's Dam Bridge&lt;/a&gt; in the Bronx (the four-way part comes on the east approach, where it intersects ramps from the Major Deegan Expressway that are elevated in order to pass over trackwork to the south).&lt;/p&gt;&#xA;" OwnerUserId="964" LastEditorUserId="964" LastEditDate="2015-04-03T03:09:40.567" LastActivityDate="2015-04-03T03:09:40.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2295" PostTypeId="1" AcceptedAnswerId="2299" CreationDate="2015-04-03T16:09:37.653" Score="19" ViewCount="4780" Body="&lt;p&gt;I know that the rubber on car and truck tires wear, and the road concrete wears out. I wondered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While steel is hard and elastic, it still causes friction (interaction&#xA;between molecules) and therefore abrasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we have on average 20-30 wagons with four axles with a full load of 50-60 tons on each wagon and a traction vehicle. How many train passes can a railway rail endure before it must be replaced? What is the equivalent time frame in which this number of passes is achieved on average?&lt;/p&gt;&#xA;" OwnerUserId="1137" LastEditorUserId="272" LastEditDate="2015-04-07T12:05:20.807" LastActivityDate="2017-04-21T14:45:27.743" Title="How many train passes can railway tracks endure?" Tags="|civil-engineering|materials|rail|transportation|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2296" PostTypeId="2" ParentId="2224" CreationDate="2015-04-03T16:36:10.140" Score="1" Body="&lt;p&gt;yes the MH lamp fixture could consume power even when lamp is burnt. This power will be at 277V but will also depend on the resistive circuit pathway through the fixture.&lt;/p&gt;&#xA;" OwnerUserId="1138" LastActivityDate="2015-04-03T16:36:10.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2297" PostTypeId="1" AcceptedAnswerId="2330" CreationDate="2015-04-03T17:39:29.163" Score="9" ViewCount="318" Body="&lt;p&gt;A lot of research has been devoted to creating electrical devices that emulate biological sensors, including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Visual:  Cameras, color/light intensity sensors&lt;/li&gt;&#xA;&lt;li&gt;Auditory: Microphones, ultrasonic sensors&lt;/li&gt;&#xA;&lt;li&gt;Tactile:  Pressure sensors, temperature sensors&lt;/li&gt;&#xA;&lt;li&gt;Balance:  Gyroscopes, accelerometers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, I have yet to find a comprehensive sensor/processing algorithm to detect and interpret odors.  Certainly, there are &quot;olfactory&quot; sensors which are dedicated to a specific purpose, such as carbon monoxide detectors, and other hazardous gas detectors.  But I have yet to find a general purpose sensor/processing algorithm that can readily detect and interpret odors within the range and resolution of a human nose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do such sensors/algorithms exist?  If so, what are they and how do they work?  If not, what are the primary obstacles to developing them?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="282" LastEditDate="2015-04-06T22:39:16.247" LastActivityDate="2015-10-22T00:48:03.953" Title="Sensors / processing algorithms to emulate a human's sense of smell" Tags="|electrical-engineering|sensors|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2299" PostTypeId="2" ParentId="2295" CreationDate="2015-04-03T21:31:18.883" Score="14" Body="&lt;p&gt;&lt;em&gt;This is an admittedly North American response.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MGT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the US, how much traffic goes over a given track in a year is measured in Million Gross Tons (MGT) e.g. 1 MGT = 2 000 000 000 lbs [spaces instead of commas to be world-friendly]. This is a measure of the total weight of cargo and vehicles but not necessarily the number of individual trains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rail Life&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The life of a typical railroad rail is between &lt;a href=&quot;http://www.aar.com/pdfs/Part1_WRI_Management.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1300 MGT and 380 MGT&lt;/a&gt; depending on if the rail is on a straight (tangent) track or in a curve. There is more friction in a curve. Also, the rail on the low side of a super-elevated curve wears faster than the rail on the high side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Friction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this great difference in life of the rail, areas that have lots of curves employ machines that &lt;a href=&quot;http://www.aar.com/pdfs/Part4_FrictionControl_SFK.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;add lubrication to the rail&lt;/a&gt;. This lubrication is applied after the locomotives pass so that their traction capability is not reduced. Some photos of these &lt;a href=&quot;http://www.lbfoster-railtechnologies.com/Friction_Modifiers.asp#mstto=&quot; rel=&quot;nofollow noreferrer&quot;&gt;machines&lt;/a&gt; are below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/F8gSa.png&quot; alt=&quot;LB Foster machine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UZui2.png&quot; alt=&quot;LB Foster Rail Lubrication&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Number of Trains&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the US, the largest/heaviest trains carry only one commodity. A typical example of this is a so called &quot;coal unit train&quot;. This is a 100-car train loaded with coal where each car weighs 286,000 lbs. Assuming an average of 500 MGT of life:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{Number of Trains} = \frac{500*1e6}{(100*\frac{286,000}{2000})} \approx 35 000 \text{ trains}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A high capacity line might have 25 trains a day on one track, so the life span in years would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{35000}{25*365} \approx 4 \text{ years}$$&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="10979" LastEditDate="2017-04-21T14:45:27.743" LastActivityDate="2017-04-21T14:45:27.743" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2300" PostTypeId="1" AcceptedAnswerId="18440" CreationDate="2015-04-03T21:31:26.793" Score="20" ViewCount="440530" Body="&lt;p&gt;&lt;strong&gt;Given:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thermodynamics text reads as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In SI units, the force unit is the newton ($N$), and it is defined as the force required to accelerate a mass of $1\cdot kg$ at a rate of $1\cdot\frac{m}{s^2}$. In the English system, the force unit is the pound-force ($lbf$) and is defined as the force required to accelerate a mass of $32.174\cdot lbm$ (1 slug) at a rate of $1\cdot\frac{ft}{s^2}$. That is...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1\cdot N =  1\cdot kg\times1\cdot\frac{m}{s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1\cdot lbf =  32.174\cdot lbm\cdot\times1\cdot\frac{ft}{s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all practical purposes, such as at STP conditions or close to it like when we have a rounded off sea-level acceleration due to gravity of $32.2\frac{ft}{s^2}$ $(101\cdot kPa)$, can I just think of the $lbf$ in the following way...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W=1\cdot lbf=1\cdot lbm \times 32.174\cdot\frac{ft}{s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that for the weight of an object having a mass of $1\cdot kg$ (also at sea-level) in SI units as...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W=9.81\cdot N=1\cdot kg\times9.81\cdot\frac{m}{s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes or no and why?&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="282" LastEditDate="2015-04-11T19:12:28.517" LastActivityDate="2022-02-02T17:18:14.110" Title="Pound-force (lbf) vs Pound-mass (lbm)" Tags="|mechanical-engineering|thermodynamics|international|" AnswerCount="10" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2301" PostTypeId="2" ParentId="2300" CreationDate="2015-04-03T22:09:37.670" Score="4" Body="&lt;p&gt;1 pound mass is that mass that weighs one pound in 1 g gravity.  For most practical cases, a pound mass and a pound weight define the same amount of stuff on the surface of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To define a pound mass we rearrange Newton's law of F = mA to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m = F/A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then plug in the particulars to get pound mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 pound mass = (1 pound force)/(32.174 ft/s&amp;sup2;)&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-04-03T22:09:37.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2302" PostTypeId="1" AcceptedAnswerId="2310" CreationDate="2015-04-03T23:18:11.360" Score="2" ViewCount="5767" Body="&lt;p&gt;&lt;strong&gt;Given:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem in my thermodynamics text is stated as follows...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the mass and the weight of the air contained in a room whose dimensions are $V=$ $15ft$ x $20ft$ x $20ft$. Assume the density of the air is $\rho=0.0724\cdot\frac{lbm}{ft^3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First find the mass...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m=\rho\times V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m=0.0724\cdot\frac{lbm}{ft^3}\times 6000\cdot ft^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=434.3\cdot lbm$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now find the force acting on the air due to gravity. This is the weight of the air assumed at sea-level...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W=m\times g$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W=434.3\cdot lbm\times32.174\cdot\frac{ft}{s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=13976\cdot lbf$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it hard to believe that in an average size room the air weighs a whopping $14,000\cdot lbf$. Did I do something wrong in my calculations or is this correct? If this is correct perhaps we earthlings living on the surface of the earth are the real &lt;em&gt;extremophiles&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="282" LastEditDate="2015-04-06T23:15:10.143" LastActivityDate="2015-04-06T23:15:10.143" Title="Mass and weight of air in a room" Tags="|mechanical-engineering|thermodynamics|pressure|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2303" PostTypeId="1" AcceptedAnswerId="2309" CreationDate="2015-04-03T23:56:48.727" Score="2" ViewCount="3181" Body="&lt;p&gt;&lt;strong&gt;Given:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acceleration of high-speed aircraft is sometimes expressed in g’s (in multiples of the standard acceleration of gravity). Determine the upward force, in $N$, that a $70\cdot kg$ man would experience in an aircraft whose acceleration is $6\cdot g’s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I account for the normal acceleration due to gravity by subtracting $1\cdot g$ from the upward acceleration of the aircraft. This is what we would see if we had drawn a free body diagram (not shown)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=ma=m\times(6g-1g)=70\cdot kg\times5\times9.81\frac{m}{s^2}=3433.5\cdot N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution in Text:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the solution in the text does not do this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=ma=m\times6g=70\cdot kg\times6\times9.81\frac{m}{s^2}=4120.2\cdot N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the net upward acceleration not $5\cdot g's$?.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="41" LastEditDate="2015-04-09T18:17:56.433" LastActivityDate="2018-01-25T11:20:59.497" Title="Aircraft force in g's experienced by pilot" Tags="|aerospace-engineering|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2304" PostTypeId="2" ParentId="1969" CreationDate="2015-04-04T01:43:37.090" Score="4" Body="&lt;p&gt;There isn't necessarily a distance limit. Basically you are preventing galvanic corrosion via use of zinc which has a higher electric potential than the metals you are trying to protect. The corrosion occurs because you have two dissimilar metals (different potentials) in the precense of an electrolye (similar to a galvanic cell or battery). The electrolytic solution will serve as a means of transporting ions from the more reactive metal to the less reactive metal. A voltage difference is set up between the metals and the potential across the solution will carry the ions as discussed. What sets the &quot;distance&quot; which you've discussed is the mobility of the ions in the electrolyte solution. If the electrolyte solution has a high resistivity the ions will not readily be transported. In this case, you will see little to no benefit from the zinc. For high resistivity electrolyte solutions, typically a voltage (a DC current is normally used) is impressed on the system to overcome the high resistivity of the electrolytic solution. When used in conjection with zinc anodes, this ensures the zinc will corrode vice the metals you are trying to protect. You typically only see this on very large structures, such as ocean freightliners or pipelines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, one zinc anode will protect a large area, so long as the resistivity of the electrolyte solution isn't terribly high. If the resitivity is high, you will need to use more anodes, more tightly spaced. The zincs are evenly spaced to ensure the resistance from each zinc to the surrounding area is roughly equivalent, otherwise, you will see one zinc corrode faster than the others. If the structure is very large or your solution is of low conductivity, you may need to use impressed current for protection. It is difficult to say how many anodes you need without knowing the conductivity of the solution, the surface area you are concerned with, and the metals you are attempting to protect.&lt;/p&gt;&#xA;" OwnerUserId="1052" LastActivityDate="2015-04-04T01:43:37.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2305" PostTypeId="1" CreationDate="2015-04-04T01:50:03.970" Score="7" ViewCount="1080" Body="&lt;p&gt;I am working on an indirect ammonia chiller that is used to cool the ice in an arena. The system consists of 3 reciprocating compressors, an evaporative condenser with a vfd on the fan, a float throttling valve and a plate heat exchanger acting as the evaporator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With one compressor operating, the condensing pressure of the ammonia ranges from 140 psi to 180 psi depending on the speed of the fan. When the condensing pressure is higher, the compressor is doing more work. When the fan speeds up, the head pressure lowers which corresponds to  a drop in compressor work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that lowering the condensing pressure reduces compressor work and am familiar with the pressure-enthalpy chart that is used to describe the refrigeration cycle. My question is, how does increasing the fan speed cause a drop in compressor work? The fan kicks on and rejects more heat from the gas, but what is the process or mechanism that occurs as you follow the refrigeration cycle that results in the compressor lowering the condensing pressure?&lt;/p&gt;&#xA;" OwnerUserId="1143" LastEditorUserId="368" LastEditDate="2015-04-08T04:44:17.613" LastActivityDate="2016-04-08T20:35:11.250" Title="How does a compressor know to lower condensing pressure in chiller?" Tags="|mechanical-engineering|hvac|compressors|" AnswerCount="2" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2306" PostTypeId="1" AcceptedAnswerId="2308" CreationDate="2015-04-04T04:30:51.033" Score="1" ViewCount="164" Body="&lt;p&gt;&lt;strong&gt;Given:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem in my thermodynamics text reads as follows...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vacuum gauge connected to a tank reads $5.4\cdot psi$ at a location where the barometric reading $P_{Hg}= 28.5\cdot in$. Determine the absolute pressure in the tank. Take $\rho_{Hg} = 848.4\frac{lbm}{ft^3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for a vacuum gauge is as follows...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{vac}=P_{atm}-P_{abs}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewrite as...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{abs}=P_{atm}-P_{vac}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine atmospheric pressure but first convert $ft$ units to $in$ since that is how we traditionally define pressures in English units...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{atm}=\rho g h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho=848.4\cdot\frac{lbm}{ft^3}\times\frac{ft^3}{(12in)^3}=.4910\cdot\frac{lbm}{in^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g=32.174\cdot\frac{ft}{s^2}\times\frac{12in}{ft}=386.0\cdot\frac{in}{s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{atm}=.4910\cdot\frac{lbm}{in^3}\times386.0\cdot\frac{in}{s^2}\times28.5\cdot in=5402\cdot\frac{lbm}{in\cdot s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atmospheric pressure obtained cannot be right because the scalar value $5402$ is no where near $14.7$ and the units obtained ($\frac{lbm}{in\cdot s^2}$) look completely wrong. Shouldn't they be $\frac{lbf}{in^2}$? Where did I go wrong? After this is solved I know how to obtain the absolute pressure in the tank so I shall stop here.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="282" LastEditDate="2015-04-06T21:33:04.130" LastActivityDate="2015-04-06T21:33:04.130" Title="Determine abs pressure in a tank where the atm and vac pressures are given" Tags="|mechanical-engineering|thermodynamics|pressure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2307" PostTypeId="1" CreationDate="2015-04-04T05:11:56.343" Score="6" ViewCount="2958" Body="&lt;p&gt;As it is well known, a vacuum balloon using the materials we have at our disposal is not possible, because of the sheer force they have to resist from the air outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked this question on Physics stack exchange and haven't got an answer yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an idea and I want to know if there is anything that makes it impossible.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Build a balloon out of graphene which is very light and it is conductive too.&lt;/li&gt;&#xA;&lt;li&gt;Since graphene is not airtight, cover graphene (that is not yet inflated) with a good electrical insulator which is also airtight. &lt;/li&gt;&#xA;&lt;li&gt;Now make the graphene store electric charge. &lt;/li&gt;&#xA;&lt;li&gt;This would exert force from inside as the electric charges on the graphene repel each other. &lt;/li&gt;&#xA;&lt;li&gt;This causes the balloon to inflate while having a vacuum inside.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now my question is there any theoretical problems (not engineering problems) with this model?&lt;/p&gt;&#xA;" OwnerUserId="1140" LastEditorUserId="282" LastEditDate="2015-04-06T22:28:18.653" LastActivityDate="2022-12-03T02:34:37.307" Title="Can vacuum balloon become possible with this idea?" Tags="|materials|aerospace-engineering|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2308" PostTypeId="2" ParentId="2306" CreationDate="2015-04-04T05:20:16.957" Score="2" Body="&lt;p&gt;$ \rho = 26.3 \ slugs/ft^3 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho  g = \gamma= 847\ lb_f/ft^3$   &quot;specific weight&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ ft^3 = (12 in)^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ P = \rho g  h = \frac{847 lb_f}{(12 in)^3}\cdot 25.8 in$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is the difference between a lb mass and lb force.&lt;/p&gt;&#xA;" OwnerUserId="1145" LastEditorUserId="1145" LastEditDate="2015-04-04T06:06:35.787" LastActivityDate="2015-04-04T06:06:35.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2309" PostTypeId="2" ParentId="2303" CreationDate="2015-04-04T06:27:11.917" Score="6" Body="&lt;p&gt;The acceleration is given. However, the upward force is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6mg = F_{up} - mg$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{up} = 7\cdot mg$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of no acceleration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 = F_{up,normally} - mg ==&amp;gt; F_{up,normally} = mg$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we neglect the gravitational force experienced every day by this man, then we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6mg = F_{up}$&lt;/p&gt;&#xA;" OwnerUserId="1145" LastEditorUserId="1145" LastEditDate="2015-04-04T07:55:57.800" LastActivityDate="2015-04-04T07:55:57.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2310" PostTypeId="2" ParentId="2302" CreationDate="2015-04-04T06:50:50.870" Score="1" Body="&lt;p&gt;A pound force is defined as the force required to accelerate a slug at 1 ft/s^2.&#xA;The density of air is $\rho = 0.0724 \ lb_m/ft^3 = 0.0724/32.2 \ slugs/ft^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight of the air is &#xA;$\rho V g = 0.0724/32.2 \ slugs/ft^3 \cdot32.2 ft/s^2\cdot 6000 ft^3 = 0.0724\cdot 6000 \ slugs\ ft/s^2 = 434.4 lb_f$  &lt;/p&gt;&#xA;" OwnerUserId="1145" LastActivityDate="2015-04-04T06:50:50.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2311" PostTypeId="1" AcceptedAnswerId="2318" CreationDate="2015-04-04T07:24:34.600" Score="5" ViewCount="425" Body="&lt;p&gt;&lt;strong&gt;Given:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A problem in my thermodynamics text reads as follows...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The barometer of a mountain hiker reads $13.8$ $psia$ at the beginning of a hiking trip and $12.6$ $psia$ at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of $.074\cdot\frac{lbm}{ft^3}$ and take $g = 31.8\cdot\frac{ft}{s^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our solution begins by solving for the change in air pressure and using this to determine elevation climbed...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta P=P_1-P_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$=(13.8-12.6)\frac{lbf}{in^2}=1.2\frac{lbf}{in^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta P=\rho g h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rewrite as...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h=\frac{\Delta P}{\rho g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But first we needed to convert $ft$ units to $in$ since that is how we traditionally define pressures in the English system...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho=.074\frac{lbm}{ft^3}\times\frac{ft^3}{(12in^3)}\times.03108\frac{slugs}{lbm}=1.33\times10^{-6}\frac{slugs}{in^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g=31.8\frac{ft}{s^2}\times\frac{12in}{ft}=381.6\frac{in}{s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now solve for $h$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h=\frac{\Delta P}{\rho g}=\frac{1.2\cdot lbf}{1.33\times10^{-6}\frac{slugs}{in^3}\cdot381.6\frac{in}{s^2}}=2364\frac{lbf\cdot in^2\cdot s^2}{slugs}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer in text:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h=2363\cdot ft$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scalar values are almost identical but the units just don't fit. Where did I go wrong?&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="282" LastEditDate="2015-04-06T23:13:24.810" LastActivityDate="2015-04-06T23:13:24.810" Title="Determine elevation from the difference in air pressure" Tags="|mechanical-engineering|thermodynamics|pressure|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2312" PostTypeId="2" ParentId="2264" CreationDate="2015-04-04T12:24:14.320" Score="1" Body="&lt;p&gt;Smaller organizations naturally have simpler ECO processors. In one such organization with about 500 employees, the change owner submitted the change to the change management board. The board consisted of VP manufacturing Engineering, VP of Research and Development, VP of Quality, representative of other affected departments. These included engineer, documentation, procurement and supply chain management. The change leader coordinated the meeting. Approved changes are submitted to documentation management system for revision updates or to create new initial documentation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the PCB were manufactured by contract manufactures (CM) and assembly of the final product was done in house. The organization was operating in a heavily regulated industry. Therefore all changes to product had to be approved. Though the documentation was complicated the process needless to say was simple and effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Software tools currently available for smaller organizations are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.omnifysoft.com/&quot; rel=&quot;nofollow&quot;&gt;Omnifysoft&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ptc.com/product-lifecycle-management/windchill&quot; rel=&quot;nofollow&quot;&gt;Windchill&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-04-04T12:29:49.633" LastActivityDate="2015-04-04T12:29:49.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2315" PostTypeId="1" CreationDate="2015-04-04T15:20:25.937" Score="3" ViewCount="3007" Body="&lt;p&gt;I am doing my final year engineering project on Magnetic Field Assisted Combustion and was curious to see what people thought about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Companies sell rare earth magnetic arrangements to be attached to fuel lines of gas burners and they are said to improve combustion efficiency but why exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have performed a number of experiments using a standard butane/propane gas burner with some magnets manufactured by one of said companies and have had some contradicting results. With lower strength magnets, heat transfer unexpectedly slowed down but with a much stronger arrangement, heat transfer rate was increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the burn out time of the same amount of gas took 8 minutes less with the magnets in place around the fuel line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read a number of journals on similar subjects but even within these, the actual reason for the increase in heat output is still not known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on the subject would be massively appreciated and possibly give me some other areas to investigate that I have not already thought of.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastActivityDate="2015-04-04T15:47:19.023" Title="Do magnetic fields applied to a fuel line have an effect on heat output?" Tags="|thermodynamics|heat-transfer|combustion|energy-efficiency|magnets|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2316" PostTypeId="2" ParentId="2315" CreationDate="2015-04-04T15:41:19.040" Score="11" Body="&lt;p&gt;If you can replicate what you report above under controlled conditions then you can retire for life next year. Really. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overwhelming opinion by all reputable test houses or suitably competent careful and honest individuals is that for combustion improvement by treating the **fuel* with magnets you'd be just as well off rubbing on snake oil - or drinking it. Really. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Magnets are the universal cure / improver / energy source / you name it.&lt;br&gt;&#xA;I am &lt;strong&gt;not&lt;/strong&gt; saying that you are dishonest OR that you did not think you saw the result that you report.&lt;br&gt;&#xA;But, it is overwhelmingly likely that you have missed something crucial that affects the result.&lt;br&gt;&#xA;Having this happen is a part of everyday engineering testing. Spotting it is the trick.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many months CERN (who should be as real as they come) were reporting faster than light transit speeds for some particles. They checked everything they could think of scrupulously before they reported their results. In due course a non FTL explanation was found. If CERN can manage this, so can the rest of us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have had factory engineers give me written reports containing data that showed processes were running at &gt; 100% efficient when there was no suggestion or expectation that this was the case. I knew with certainty that something was wrong with their data. They didn't know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is not the answer you wanted and I also know that some people will assure you that you can treat fuel &quot;magnetically&quot; and improve combustion efficiently. And, it &lt;strong&gt;MAY&lt;/strong&gt; be true. But &quot;everyday Science&quot; says that it's &quot;extremely unlikely&quot; and the necessary path for you is to&lt;br&gt;&#xA;Replicate your results, then &#xA;Have somebody else replicate your results and then, if it still seems to be working,&lt;br&gt;&#xA;Arrange &quot;double blind&quot; testing so that there is no way to fool yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that use of magnets to affect a combustion process in some way (rather than pretreating the fuel) is potentially more likely to be valid - still unlikely enough that very great care is needed, but ionised gases can be affected by magnetic fields so there is some basis for possible changes. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-04-04T15:47:19.023" LastActivityDate="2015-04-04T15:47:19.023" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2317" PostTypeId="2" ParentId="2307" CreationDate="2015-04-04T23:57:41.593" Score="3" Body="&lt;p&gt;Let's completely neglect material strength and just consider air pressure and the force from electric charges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Atmospheric_pressure&quot; rel=&quot;nofollow&quot;&gt;Air pressure&lt;/a&gt; is 101325 Pascals, or 101325 Newtons per square meter. That means that the net force on one square meter of material must be equal to that force in order to prevent the balloon from breaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm completely abusing notation here, but let's give the material a charge density of $\mathbb{Q}$ Coulombs per square meter - in other words, on one square meter of material lies $Q$ Coulombs. The balloon has a radius of $r$, meaning that (assuming it's a sphere) the surface area is $4 \pi r^2$, and the total charge is $4 \pi r^2\mathbb{Q}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force on this one square meter from the electric charge is, therefore,&#xA;$$F=k\frac{4 \pi r^2 \mathbb{Q}Q}{r^2}=4 \pi k \mathbb{Q}Q$$&#xA;For the balloon to work, this force must be equal to 101325 Newtons. This means that $\mathbb{Q} \approx 9.47 \times 10^{-4}$ Coulombs per square meter. Not too bad. Seems plausible.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-04-04T23:57:41.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2318" PostTypeId="2" ParentId="2311" CreationDate="2015-04-05T01:52:20.797" Score="5" Body="&lt;p&gt;I'm glad we got rid of that obsolete system of obscure units decades ago. The metric system is so much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the definition of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Slug_%28mass%29&quot;&gt;Slug&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${1 Slug = 1 {lb}_f.{s}^2/ft}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting that into the units you got, the ${slugs}$, ${lb}_f$ &amp;amp; ${s}^2$ go &amp;amp; ${ft}$ comes in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only mistake you made was to exclude ${in}^2$ in your final calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure in US units is pounds per &lt;em&gt;square inch&lt;/em&gt;, not pounds as you had it. Correct for this and you end up with units of feet in the final calculation.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-04-05T01:52:20.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2319" PostTypeId="1" AcceptedAnswerId="2320" CreationDate="2015-04-05T06:48:07.117" Score="2" ViewCount="443" Body="&lt;p&gt;I was recently on the UCLA campus (University of California Los Angeles) and I saw the Bunche Hall. I love the design and the architecture of the building. However I saw a little (what I would call) a closet with no door on the side but it was just three walls with a roof, extremely small, underneath the stairs, outdoors, a bit below the floor level. There is a small alley and then you take two stairs down and the closets face a fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't even know what to call it. My question is, from architectural/structural engineering point of view, what is this building device? Why is it there? What is its purpose? What is it even called? The two closets reminded of telephone booths but there was nothing but dirt, cobwebs, and leaves in there. Were they once used for something? I am posting the pictures below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OQ4Qx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/P3sd2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/SiPDQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PRdEV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ZlF3T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/AjQwR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1154" LastEditorUserId="33" LastEditDate="2015-04-12T00:47:04.070" LastActivityDate="2015-04-12T00:47:04.070" Title="What's the purpose of these small closet-like-rooms-without-a-door?" Tags="|civil-engineering|structural-engineering|building-physics|building-design|architecture|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2320" PostTypeId="2" ParentId="2319" CreationDate="2015-04-05T08:01:34.693" Score="12" Body="&lt;p&gt;I do not know, I've never been there and I have no other information other than what is in your question. However, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd say that best guess, until somebody who knows more presents, is your suggestion of now-unused telephone booths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the wall are two plates. These are at the same height in each case and consistent with connecting a telephone. Other equipment may be equally possible (eg: ATM machines, but seem less likely). More details about these plates and what is under them is probably the best clue apart from finding somebody who knows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/leCzv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other wall marks suggesting mountings and possibly a shelf on the right hand wall in the right cubicle, but high up marks in the left hand one are not so obviously relevant,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lights were almost certainly identical (diffuser missing at right) indicating downwards illumination in both cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7LQE8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reality? I don't know.&#xA;But, if I &lt;em&gt;had&lt;/em&gt; to bet your life on it, I'd guestimate telephones. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-04-05T14:21:44.357" LastActivityDate="2015-04-05T14:21:44.357" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2321" PostTypeId="1" AcceptedAnswerId="2548" CreationDate="2015-04-05T09:58:48.483" Score="5" ViewCount="3496" Body="&lt;p&gt;I know that in case of wedge we have a 2D behavior which creates a discontinuity across the shock, but I am unable to understand why this discontinuity is not present in the case of a conical surface. I have a vague idea that the continuity equation ensures this as, due to increase in the flow area, the velocity must increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But still it would be great help if I could get a proper explanation, I couldn't find one in books and net.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to know which 2D constraints are actually relieved which causes the 3D relaxation effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PPd6l.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:-&lt;/strong&gt; I think i need to reiterate the question, I wanted to ask why does the supersonic flow bend toward the surface in case of being obstructed by a cone unlike in case of wedge, and as per its reasoning i have also read that the flow is more &lt;strong&gt;free&lt;/strong&gt; in case of cone, due to presence of 3 dimensions, commonly known as 3D relaxation, so it would be nice if someone could comment upon it too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VJHP9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1155" LastEditorUserId="272" LastEditDate="2015-04-06T16:57:38.697" LastActivityDate="2015-04-26T14:59:09.160" Title="Explanation of supersonic flow differences between a 2D wedge and a 3D cone" Tags="|mechanical-engineering|fluid-dynamics|shock|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2322" PostTypeId="1" AcceptedAnswerId="2325" CreationDate="2015-04-05T11:59:37.073" Score="3" ViewCount="2763" Body="&lt;p&gt;What are the construction procedures that allow the mitigation of potential problems due to water in a retaining wall?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific: I'm talking about reinforced concrete walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cases of water are: if I'm studying a reinforced concrete wall such that the soil is not saturated, and it rains so that the soil is saturated or if the water table level rises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are possible solutions? I'm looking for general solutions.&lt;/p&gt;&#xA;" OwnerUserId="1156" LastEditorUserId="33" LastEditDate="2015-04-05T16:29:37.127" LastActivityDate="2023-12-18T17:35:45.957" Title="What are the solutions to water problems in a retaining wall" Tags="|civil-engineering|retaining-wall|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2323" PostTypeId="2" ParentId="2303" CreationDate="2015-04-05T12:48:56.543" Score="4" Body="&lt;p&gt;The question can't be answered as stated.  Acceleration is a vector quantity, but only its magnitude is given.  In other words, &lt;em&gt;in which direction&lt;/em&gt; is the aircraft accelerating by 6 g?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the question only asked about the &lt;i&gt;upward force&lt;/i&gt; on the pilot, only the vertical component of the acceleration is relevant.  It therefore makes a huge difference whether the acceleration is vertical or horizontal.  In the pure horizontal case, for example, the upward force on the pilot is only due to gravity, (70 kg)(9.8 m/s&amp;sup2;) = 686 N.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="282" LastEditDate="2015-04-07T00:15:21.903" LastActivityDate="2015-04-07T00:15:21.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2325" PostTypeId="2" ParentId="2322" CreationDate="2015-04-05T17:39:19.607" Score="2" Body="&lt;p&gt;&lt;strong&gt;Design&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In design, the presence of water behind the wall can have two effects. It can cause additional force to be applied to the wall from hydrostatic pressure, or its presence can cause the soil characteristics to change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing soil properties has the most effect on clayey soils. Relatively dry clay will have cohesion and reduce the force on the wall. Wetter clay will lose this cohesion and apply more load to the wall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way that the wall is designed will effect how the water needs to be handled. If the wall was designed to for the effects of water all the way up the wall, then nothing may need to be done. If it was designed for only a certain height of water, then this must be assured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Removing water&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the volume of water that is expected and the source, a few different methods of water removal can be employed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weep hole&lt;/li&gt;&#xA;&lt;li&gt;Perforated drain pipe&lt;/li&gt;&#xA;&lt;li&gt;Surface water diversion&lt;/li&gt;&#xA;&lt;li&gt;Dewatering wells&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Weep holes&lt;/strong&gt; are small holes through the face of the retaining wall. These allow any water that is trapped behind the wall to slowly drain out. They are best for small volumes of water. They are often installed in all walls as a minimal level of assurance that water will not be trapped behind the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Perforated drain pipes&lt;/strong&gt; are usually installed at the base of a wall in combination with free draining fill or geotechnical fabric. They can drain larger volume of water, but the connection details and where the outflow is located need to be investigated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Surface water diversions&lt;/strong&gt; may include ditches or paved areas that keep surface water from running down a slope and being contained behind the wall. This method helps where ground water is less of a concern when compared to runoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dewatering wells&lt;/strong&gt; are only an option that is considered where the wall must be built below the groundwater level and where the force from the water can not be designed for. The wells have pumps that lower the ground water elevation in the area. These can pump large quantities of water, but they require constant monitoring and power.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-04-05T17:39:19.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2326" PostTypeId="1" CreationDate="2015-04-05T19:33:26.270" Score="5" ViewCount="192" Body="&lt;p&gt;I would like to construct a hollow beam using either 2&quot;x4&quot; or 2&quot;x6&quot; structural lumber.&lt;sup&gt;1&lt;/sup&gt;  My thought is to take 4 boards and butt them against each other in order to form a square beam&lt;sup&gt;2&lt;/sup&gt; with a hollow core.  The boards would be glued and mechanically fastened with screws in order to keep the beam from separating.  Likewise, the beam would be clamped under significant pressure while the glue dries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about the resulting moment of inertia for the beam.  My understanding is that moment equations for beams assume a homogenous material.  My concern is that's an assumption that I'm likely approximating but not actually meeting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I okay to use the standard moment of inertia equations as is?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there some sort of de-rating that I should apply to the beam since it is a composite material instead of a homogenous material?  If I should de-rate, how would I apply that within the standard calculations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;Please note, these are nominal dimensions.  Actual dimensions are approximately $\frac{1}{2}&quot;$ smaller.&lt;/sub&gt;&lt;br&gt;&#xA;&lt;sup&gt;2&lt;/sup&gt;&lt;sub&gt;Nominally, that would be a 6&quot; or 8&quot; square, but would be 5&quot; or 7&quot; once actual dimensions are taken into account.&lt;/sub&gt;&lt;br&gt;&#xA;&lt;sup&gt;3&lt;/sup&gt;&lt;sub&gt;Please also note that these beams will not be used in a load bearing manner in which anyone could be injured.  I'm building a pergola and I want to run electrical wire through the center of the beams.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorUserId="90" LastEditDate="2015-04-06T02:00:45.613" LastActivityDate="2020-10-09T22:08:47.433" Title="Effects of constructed tubular/annular beam upon moment of inertia" Tags="|mechanical-engineering|structural-engineering|beam|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2327" PostTypeId="1" CreationDate="2015-04-05T19:58:17.343" Score="3" ViewCount="1096" Body="&lt;p&gt;I found this &lt;a href=&quot;http://i.kinja-img.com/gawker-media/image/upload/s--dsYdp705--/18c7saub4nw9ggif.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;gif&lt;/a&gt;, and I was wondering how much force is applied on the pin due to the spring from the top? Also, is there an industry standard for the amount of force exerted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2D4nN.gif&quot; alt=&quot;lock tumbler animation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1157" LastEditorDisplayName="user16" LastEditDate="2015-04-05T20:41:45.773" LastActivityDate="2015-05-16T07:23:12.193" Title="How much force does a lock's pin have due to the spring?" Tags="|mechanical-engineering|springs|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2329" PostTypeId="1" AcceptedAnswerId="2337" CreationDate="2015-04-06T03:33:21.427" Score="3" ViewCount="395" Body="&lt;p&gt;I am trying to come up with a way to restrict beer flow using a valve that does not cause foam. I have found that a solenoid valve causes a change in the direction the beer must flow and this sharp change through a different size opening creates foam quickly. I have thought about a ball valve or pinch valve but I need it to be electric, work very quickly and have a small form factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know if anything like what I created in this video exist? &lt;a href=&quot;https://youtu.be/iE6qe7045Ck&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/iE6qe7045Ck&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea would be electromagnets that cause the metal pieces to slide in one direction or the other opening and closing the flow of beer. The idea is keeping the opening in the valve the same diameter as the inner diameter of the tubing I would be using (3/8&quot;). It would need to handle at most 14 psi and at least 12 psi. (Average beer lines are 9-10psi)&lt;/p&gt;&#xA;" OwnerUserId="1160" LastEditorUserId="843" LastEditDate="2015-04-06T14:35:35.930" LastActivityDate="2015-04-07T11:56:28.227" Title="Does this exist? -Sliding Fluid Restrictor -Solenoid Valve -Ball Valve" Tags="|mechanical-engineering|electrical-engineering|fluid-mechanics|valves|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2330" PostTypeId="2" ParentId="2297" CreationDate="2015-04-06T07:29:14.277" Score="2" Body="&lt;p&gt;Odor assessment is usually performed by human sensory analysis using &lt;a href=&quot;http://en.wikipedia.org/wiki/Chemoreceptor&quot; rel=&quot;nofollow&quot;&gt;chemosensors&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A chemoreceptor, also known as chemosensor, is a sensory receptor that&#xA;  transduces a chemical signal into an action potential.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Recently I have also heard of &lt;a href=&quot;http://www.phonearena.com/news/Breakthrough-sensor-to-give-our-smartphones-a-sense-of-smell_id48089&quot; rel=&quot;nofollow&quot;&gt;a sensor from Honeywell that could potentially be used in smart phones&lt;/a&gt;. These sensors are also called &lt;a href=&quot;http://en.wikipedia.org/wiki/Electronic_nose#Working_principle&quot; rel=&quot;nofollow&quot;&gt;electronic noses&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bio-electronic noses use olfactory receptors - proteins cloned from&#xA;  biological organisms, e.g. humans, that bind to specific odor&#xA;  molecules. One group has developed a bio-electronic nose that mimics&#xA;  the signaling systems used by the human nose to perceive odors at a&#xA;  very high sensitivity: femtomolar concentrations.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The more commonly used sensors for electronic noses include&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;metal–oxide–semiconductor (MOSFET) devices - a transistor used for amplifying or switching electronic signals. This works on the&#xA;  principle that molecules entering the sensor area will be charged&#xA;  either positively or negatively, which should have a direct effect on&#xA;  the electric field inside the MOSFET. Thus, introducing each&#xA;  additional charged particle will directly affect the transistor in a&#xA;  unique way, producing a change in the MOSFET signal that can then be&#xA;  interpreted by pattern recognition computer systems. So essentially&#xA;  each detectable molecule will have its own unique signal for a&#xA;  computer system to interpret.&lt;/li&gt;&#xA;  &lt;li&gt;conducting polymers - organic polymers that conduct electricity.&lt;/li&gt;&#xA;  &lt;li&gt;polymer composites - similar in use to conducting polymers but formulated of non-conducting polymers with the addition of conducting&#xA;  material such as carbon black.&lt;/li&gt;&#xA;  &lt;li&gt;quartz crystal microbalance - a way of measuring mass per unit area by measuring the change in frequency of a quartz crystal resonator.&#xA;  This can be stored in a database and used for future reference.&lt;/li&gt;&#xA;  &lt;li&gt;surface acoustic wave (SAW) - a class of microelectromechanical systems (MEMS) which rely on the modulation of surface acoustic waves&#xA;  to sense a physical phenomenon.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1116" LastEditorUserId="368" LastEditDate="2015-04-24T22:08:28.680" LastActivityDate="2015-04-24T22:08:28.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2331" PostTypeId="2" ParentId="2326" CreationDate="2015-04-06T07:46:44.977" Score="7" Body="&lt;p&gt;Moment of inertia is not a material property, it is a geometric property. Regardless of the material you are dealing with, a member that is 2&quot; wide by 10&quot; tall in cross section has the same moment of inertia. Therefore, you should use the standard equations for calculating your beam section's moment of inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address the portion of your question regarding wood's non-isotropic (anisotropic) material properties. The governing wood design codes/standards (National Design Specification, NDS, for Wood Construction in the US) suggest ignoring wood's inherent anisotrpoic nature and to, instead, use a conservatively lower value for these properties. The most applicable of these properties for structural wood design is the modulus of elasticity, values for which are compiled in the supplement to NDS list on a per species basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And with respect to &quot;de-rating&quot; the material properties since you are creating a built-up member, no, you should not modify your material properties in any manner. You have not changed your material properties have not changed. Instead, you need to calculate the shear force being transferred between all your members and design your dowel fasteners (nails) or glue appropriately. This is done by calculating the shear flow $(q = \frac{VQ}{I})$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, you can complete your analysis/design by properly calculating the moment of inertia of your built-up member, using the appropriate material properties, and properly sizing your fasteners.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="282" LastEditDate="2015-04-06T23:14:17.657" LastActivityDate="2015-04-06T23:14:17.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2332" PostTypeId="2" ParentId="2224" CreationDate="2015-04-06T09:25:10.947" Score="2" Body="&lt;p&gt;The simple answer is &quot;Yes, it may use power&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How &lt;em&gt;much&lt;/em&gt; power will depend entirely on the design of the ballast or control gear, and could range from a negligible power draw for smart electronics that recognise that the lamp is failed and just sit there, to something more substantial for gear that keeps trying to restart the lamp. If this answer is important, I suggest asking for information from the manufacturer of the fixture or the control gear.&lt;/p&gt;&#xA;" OwnerUserId="320" LastActivityDate="2015-04-06T09:25:10.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2333" PostTypeId="1" AcceptedAnswerId="6695" CreationDate="2015-04-06T09:38:22.343" Score="6" ViewCount="2564" Body="&lt;p&gt;I'm currently looking around for a new bike and I've noticed lots of load-bearing carbon parts (frames, cranksets, forks etc.) with an interesting finish like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cj8gy.jpg&quot; alt=&quot;Carbon crankset&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I was wondering &lt;strong&gt;what this type of carbon cloth is called&lt;/strong&gt; and &lt;strong&gt;why it's used&lt;/strong&gt;. I've seen it used on parts (e.g. forks) where the forces are predictable (laterally during cornering or longitudinally during braking) and thus unidirectional or woven cloth would appear to be the best option from a specific stiffness perspective. It does look pretty cool and thus presumably has marketing value but other than that, I'd appreciate any ideas!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tBtZc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1161" LastEditorUserId="1161" LastEditDate="2015-12-29T11:44:11.597" LastActivityDate="2020-09-21T15:46:52.033" Title="Forged carbon fibre finish" Tags="|mechanical-engineering|materials|" AnswerCount="5" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2334" PostTypeId="2" ParentId="2329" CreationDate="2015-04-06T11:22:08.460" Score="0" Body="&lt;p&gt;I am not a Mechanical Engineer. But may I suggest a faucet cartridge. Attached are two images. These with a combination of stepper motor might work for you. The cartridges are small and is design to be used with water. Just a thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vAL6a.jpg&quot; alt=&quot;faucet cartridge external &quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LXJol.png&quot; alt=&quot;faucet cartridge internal &quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-04-06T22:33:36.490" LastActivityDate="2015-04-06T22:33:36.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2335" PostTypeId="1" CreationDate="2015-04-06T13:34:48.480" Score="3" ViewCount="199" Body="&lt;p&gt;If I run Google search for &quot;roof&quot; the pictures are typically of medium or dark gray, medium or dark green, medium or dark brown, brick red and other not very light colors - very rare are clay or sand coloured or similar colours roofs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neat image &lt;a href=&quot;http://upload.wikimedia.org/wikipedia/commons/8/8d/Song_bird_perched_on_asphalt_shingle_roof.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;from Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E8qnf.jpg&quot; alt=&quot;not the lightest roof in the world&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roofing materials nowadays are typically engineered (plastic coated steel, etc), not natural materials such as straw, and can have whatever color you please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like light coloured roofs absorb much less sunlight and so heat up notably less and this should prolong their lifetime. In my experience the key reason of dark ruberoid degradation is evaporation of resin because of sun radiation - it loses elasticity, basically turns into cardboard and then inevitably tears into pieces. I assume similar degradation processes happen in all materials that are either coated in plastic or rely on elastic resin for their properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are lighter coloured roofs not more popular? Is there any engineering reasoning behind this, or is it simply a design aesthetics choice? Does a light color roof provide any significant benefit from durability standpoint?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="374" LastEditDate="2015-04-08T13:37:08.937" LastActivityDate="2015-04-08T13:37:08.937" Title="Are light coloured roofs more durable? If so, why aren't they more common?" Tags="|civil-engineering|materials|thermodynamics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2336" PostTypeId="2" ParentId="2335" CreationDate="2015-04-06T14:09:56.600" Score="5" Body="&lt;p&gt;You're quite right that it's generally beneficial to paint roofs white. It can significantly reduce cooling costs and can reduce heat island effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason that I can see to avoid it would be in cold climates, where a black roof provides a bit of very cheap solar heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is simply down to ignorance and aesthetics. Many individual home owners simply don't know how beneficial something so simple can be. Some don't believe it when you tell them. Some refuse to do it because they don't like the way it looks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more insight, look for news stories related to attempts by U.S.Secretary of Energy Steven Chu to promote this idea. He presented a strong, science-based case for painting roofs and roads. Reports form the time will give a good sense of the attitudes and politics involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arguably, this isn't really an engineering problem.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-04-06T14:09:56.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2337" PostTypeId="2" ParentId="2329" CreationDate="2015-04-06T16:07:10.377" Score="3" Body="&lt;p&gt;What you are describing is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gate_valve&quot; rel=&quot;nofollow&quot;&gt;gate valve&lt;/a&gt; and you are correct in assuming that this would not cause foam. Unfortunately, I do not think that gate valves are manufactured in the small size that you need.  If your tubing is 3/8&quot; ID, that is approximately equal to schedule 40 1/4&quot; nominal pipe size.  I have never seen a gate valve smaller than 3/4&quot; nominal pipe size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, if I were trying to accomplish the task you describe, I would find a beer tap and attach a solenoid to the handle.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-04-06T16:07:10.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2339" PostTypeId="1" AcceptedAnswerId="2358" CreationDate="2015-04-06T21:45:19.850" Score="6" ViewCount="2808" Body="&lt;p&gt;The ISO 9223 standard indicates that there should be no corrosion at temperatures below 0 degrees celsius. Independent researchers have however proposed to lower the minimun temperature stated in the standard to lower values in order to account for the actual corrosion observed in Nordic climates. What is the explanation for the observed corrosion at temperatures below the freezing point?&lt;/p&gt;&#xA;" OwnerDisplayName="user1164" LastEditorUserId="843" LastEditDate="2015-04-08T14:31:14.293" LastActivityDate="2015-04-08T14:31:14.293" Title="How does corrosion occur at a temperature below 0 degrees Celsius?" Tags="|materials|corrosion|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2340" PostTypeId="1" AcceptedAnswerId="2352" CreationDate="2015-04-06T23:20:08.780" Score="6" ViewCount="37882" Body="&lt;p&gt;What are the recommended or relevant standards for US city planning with regards to depth ranges for various utilities such as water, natural gas, and power, as well as for subway or rail, etc...?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, water and gas lines should be buried well below the frost line to prevent damage from ground movement.  What standard(s) dictate those depths?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am primarily interested in standards applicable to major cities such as Atlanta, Boston, Chicago, Dallas, Denver, New York, San Diego, Seattle, Washington, etc...  My assumption is that there are enough similarities between the cities that a standard would have been developed.&lt;/p&gt;&#xA;" OwnerUserId="1165" LastEditorUserId="33" LastEditDate="2016-01-25T20:28:48.370" LastActivityDate="2023-09-05T22:55:45.670" Title="How deep are utilities typically buried?" Tags="|civil-engineering|design|standards|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2341" PostTypeId="1" CreationDate="2015-04-06T23:49:17.030" Score="5" ViewCount="313" Body="&lt;p&gt;If you have taken basic statics classes you'll remember the free body diagrams used to calculate the forces inside a mechanical system; such as a beam structure consisting of pivots, rollers, fulcrums and fixed points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One area of study that is not covered in great detail however is the addition of lateral moving parts in these systems, such as pistons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the system that I am calculating a piston is conected to a secondary beam at a pivot, the secondary beam is connected at the opposite end to a primary beam by a pivot. and the primary beam is connected to a fixed position by a pivot again. The purpose of this system is work as a mechanical gripper pulling in and pushing out on the primary beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how to calculate the forces acting in the system giving the changing shape of the system given a known force value of the piston.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I apologise for not uploading an image to provide clarity, but I am unforetunately unable to do so at work. Additionally I apologise for being vague, I am just looking for a starting point to determine the rest, researching the topic so far has been a futile effort.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="340" LastActivityDate="2015-04-06T23:49:17.030" Title="How to account for pistons inside a free body diagram?" Tags="|mechanical-engineering|structures|dynamics|" AnswerCount="0" CommentCount="3" ClosedDate="2016-03-05T00:47:32.177" ContentLicense="CC BY-SA 3.0" />
  <row Id="2342" PostTypeId="2" ParentId="2329" CreationDate="2015-04-07T08:09:51.833" Score="3" Body="&lt;p&gt;&lt;a href=&quot;http://www.globalspec.com/learnmore/flow_control_flow_transfer/valves/pinch_valves_multiturn&quot; rel=&quot;nofollow&quot;&gt;Look at pinch valves&lt;/a&gt;: A small piece of tubing that can be pinched shut pneumatically or mechanically (solenoid). When open, you have the full diameter of your tubing. The only part that touches the medium is the tube, and you don't have any sliding seals. The latter part is important, because you don't want to mess up your beer with sealant grease and you don't want $CO_{2}$ to bubble along the seals. Also it seems you are looking for a DIY solution, so a simple build will be your friend.&lt;br&gt;&#xA;I don't know about foaming. You could test by having beer flow through tubing and pinch it with your fingers.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="110" LastEditDate="2015-04-07T11:56:28.227" LastActivityDate="2015-04-07T11:56:28.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2343" PostTypeId="1" AcceptedAnswerId="2348" CreationDate="2015-04-07T11:30:24.463" Score="3" ViewCount="3190" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EYwcJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the previous examples I've seen, there is a connection rod, for example, between a and c and then another rod from c to one of the pins. However, in this case, the rod ac is directly connected to another pin? I assume there is no relative velocity in this case. Is this assumption correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: By pins, I was referring to O2 and O8. I've just realised that C is a slider. I thought it was a rigid 'pin' like O2 and O8. Sorry for the confusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think I know how to approach this problem now. Va is perpendicular to OA. Vc is a horizantal line from O2. Vc/a is perpendicular to the rod AC starting from Va.Then I need to intersect Vc and Vc/a to find the magnitude of Vc. Is this approach correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: I'm running out of time. I quickly sketched this to check if I'm on the right track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uQbYW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1167" LastEditorUserId="110" LastEditDate="2015-04-08T01:25:35.273" LastActivityDate="2020-12-27T23:10:51.163" Title="How to draw the velocity and acceleration diagrams if there is no connecting rod between the pins?" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2345" PostTypeId="2" ParentId="2300" CreationDate="2015-04-07T12:17:23.590" Score="1" Body="&lt;p&gt;Absolutely, yes you can. In fact, &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/slug.html&quot; rel=&quot;nofollow&quot;&gt;the mass of a slug is derived from the acceleration due to gravity&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2015-04-07T13:15:30.607" LastActivityDate="2015-04-07T13:15:30.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2347" PostTypeId="1" CreationDate="2015-04-07T16:52:52.187" Score="7" ViewCount="1136" Body="&lt;p&gt;Most reasonably sized commercial aircraft rely on &lt;a href=&quot;http://en.wikipedia.org/wiki/Jet_engine&quot; rel=&quot;nofollow noreferrer&quot;&gt;jet engines&lt;/a&gt; for propulsion.  Upon landing these aircraft use a number of means to come to a stop with the two most important being (I believe) mechanical brakes and jet engine &lt;a href=&quot;http://en.wikipedia.org/wiki/Thrust_reversal&quot; rel=&quot;nofollow noreferrer&quot;&gt;thrust reversal&lt;/a&gt;.  This reverse thrust is generated by opening some small doors on the outside of the engine (shown below).  How efficient is the thrust reversal process?  For specificity, let's define the efficiency to be reverse thrust divided by forward thrust for a given engine speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:F-GTAR_Air_France_(3698209485).jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i9I8jm.jpg&quot; alt=&quot;Thrust reversers on A320&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-05-13T16:01:38.297" Title="How efficient are jet engine thrust reversers?" Tags="|mechanical-engineering|engines|thrust|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2348" PostTypeId="2" ParentId="2343" CreationDate="2015-04-07T20:36:37.283" Score="1" Body="&lt;p&gt;Draw a vertical line at &lt;em&gt;C&lt;/em&gt;. The instant center of rotation of the connecting rod &lt;em&gt;AC&lt;/em&gt; is located somewhere along this line. To find the exact location, extend &lt;em&gt;OA&lt;/em&gt; to intersect this line (below the slider). Now imagine a CW rotation about this point &lt;em&gt;K&lt;/em&gt; will move the slider to the right and the crank perpendicular to &lt;em&gt;OA&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3wPVE.png&quot; alt=&quot;pic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the relative rotation of &lt;em&gt;DB&lt;/em&gt; to &lt;em&gt;AC&lt;/em&gt; is at &lt;em&gt;B&lt;/em&gt;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Instant_centre_of_rotation#Relative_centre_of_rotation_for_two_contacting_planar_bodies&quot; rel=&quot;nofollow noreferrer&quot;&gt;law of instant centers&lt;/a&gt; means that body &lt;em&gt;DB&lt;/em&gt; rotates about a point along the line &lt;em&gt;KB&lt;/em&gt;. Since it is connected to &lt;em&gt;O8&lt;/em&gt; at &lt;em&gt;D&lt;/em&gt; the rotation center of &lt;em&gt;DB&lt;/em&gt; is at &lt;em&gt;M&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, velocity at &lt;em&gt;C&lt;/em&gt; is perpendicular to &lt;em&gt;KC&lt;/em&gt;, velocity at &lt;em&gt;A&lt;/em&gt; is perpendicular to &lt;em&gt;O2A&lt;/em&gt;, velocity at &lt;em&gt;D&lt;/em&gt; is perpendicular to &lt;em&gt;O8D&lt;/em&gt; &lt;strong&gt;and&lt;/strong&gt; &lt;em&gt;MD&lt;/em&gt; and finally velocity at &lt;em&gt;B&lt;/em&gt; is perpendicular to &lt;em&gt;KB&lt;/em&gt; &lt;strong&gt;and&lt;/strong&gt; &lt;em&gt;MB&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastEditorUserId="1051" LastEditDate="2015-04-08T06:14:51.713" LastActivityDate="2015-04-08T06:14:51.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2349" PostTypeId="1" CreationDate="2015-04-07T21:46:05.980" Score="9" ViewCount="7153" Body="&lt;p&gt;Let's say a car at 60 mph requires roughly 20 hp to maintain its speed (i.e., to overcome rolling resistance and drag).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this car makes 240 hp and gets about 34 mpg (not too unrealistic), how is it getting 34 mpg if the engine is only at 0.05 efficiency? I know modern automobiles are supposed to have an efficiency limit near 25-30% &lt;a href=&quot;http://en.wikipedia.org/wiki/Carnot_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;thanks to Mr. Carnot&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rl5zP.gif&quot; alt=&quot;Contour plot of efficiency vs. angular velocity at various load factors. Efficiency increases with load factor with a global maximum at 100% load near 3300 rpm of approximately 0.3.&quot;&gt;&lt;br&gt;&#xA;Chart from &lt;a href=&quot;http://ratherlightvehiclecompany.com/fueleconomy.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like gearing must be part of the answer but I'm having a hard time understanding just how the gearing allows the engine to generate 20 hp while clearly be at a higher efficiency than indicated on the chart for low engine loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I'm not getting exactly what the engine loading is referring to? &lt;/p&gt;&#xA;" OwnerUserId="1175" LastEditorUserId="33" LastEditDate="2016-02-11T02:01:33.943" LastActivityDate="2022-02-03T23:45:40.263" Title="How can &quot;mpg&quot; be high under low engine load?" Tags="|mechanical-engineering|automotive-engineering|fuel-economy|" AnswerCount="10" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2350" PostTypeId="2" ParentId="2349" CreationDate="2015-04-08T00:02:02.400" Score="18" Body="&lt;h1&gt;Thermodynamic efficiency vs. fuel economy&lt;/h1&gt;&#xA;&lt;p&gt;When you cite an efficiency of 25-30% for an internal combustion energy, you're talking about the thermodynamic efficiency of the engine. This is, at the theoretical level, based on a temperature differential. It has nothing directly to do with the fuel.&lt;/p&gt;&#xA;&lt;p&gt;When you cite a fuel economy of 34 &lt;em&gt;miles per gallon&lt;/em&gt;, you're now talking about something that depends a great deal on other factors—for example, the &lt;em&gt;energy density of the fuel&lt;/em&gt;. How much extractable energy is there in a gallon of gasoline? How about a gallon of antimatter? A gallon of chocolate milk?&lt;/p&gt;&#xA;&lt;p&gt;Many engines can accept different fuels, or fuel blends, with different energy densities, without a significant change in their thermodynamic efficiency. For example, ethanol is blended with gasoline, but has an energy density about 30% lower than gasoline; a gallon of one is not equal to a gallon of the other.&lt;/p&gt;&#xA;&lt;p&gt;To operate an engine at a certain thermodynamic efficiency means having a temperature differential; to maintain that temperature differential, you need to add energy at some rate. Getting the same thermodynamic efficiency when your fuel has a lower energy density simply means increasing the fuel delivery rate (&lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\dot{m}$&lt;/span&gt;) so that the energy delivery rate (&lt;span class=&quot;math-container&quot;&gt;$q_{in}$&lt;/span&gt;) stays the same. This ignores the various nuances of how engines burn different fuels but, generally speaking, there is no direct relationship between the thermodynamic efficiency and the fuel economy.&lt;/p&gt;&#xA;&lt;h1&gt;Significance of engine loading&lt;/h1&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I feel like gearing must be part of the answer but I'm having a hard&#xA;time understanding just how the gearing allows the engine to generate&#xA;20 hp while clearly be at a higher efficiency than indicated on the&#xA;chart for low engine loading.&lt;/p&gt;&#xA;&lt;p&gt;Maybe I'm not getting exactly what the engine loading is referring to?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;When the car is not accelerating, engine loading comes from whatever forces are acting against the motion of the engine. Internal friction (pistons, crankshaft, transmission, etc.), external friction (tires on road surface), drag, gravity (when going uphill). &amp;quot;Load&amp;quot; means how much power is required of the engine for the car to have some speed and acceleration.&lt;/p&gt;&#xA;&lt;p&gt;As you point out, when a vehicle is cruising on the highway, it only needs a small percentage of its total available power output to maintain speed. Unless we're talking &lt;em&gt;really&lt;/em&gt; high speeds and/or an &lt;em&gt;exceptionally&lt;/em&gt; gutless wonder of a car, cruising on the highway just isn't a high-loading situation. Your confusion seems to come from the fact that vehicles get &lt;em&gt;better&lt;/em&gt; fuel economy when cruising at highway speeds than they do while accelerating.&lt;/p&gt;&#xA;&lt;p&gt;The key thing to realize is that &lt;strong&gt;getting better fuel economy doesn't imply that the engine is operating at a higher thermodynamic efficiency&lt;/strong&gt; because there are many other factors that go into fuel economy. The thermodynamic efficiency of the Carnot cycle is only one of those factors. Another factor is the efficiency of the combustion reaction (which is technically not part of the Carnot cycle). Another is how much power is being used to accelerate the vehicle (useful work) versus how much is being lost to friction, drag and conduction (waste heat, &lt;span class=&quot;math-container&quot;&gt;$q_{out}$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;h1&gt;Calculating fuel economy&lt;/h1&gt;&#xA;&lt;p&gt;Consider the following relationship—where does engine loading come in?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text {fuel economy (mpg)} = \dfrac {\text {speed (mph)}} {\text {flow rate (gal/h)}} = \dfrac {v} {Q}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In an idealized situation, with no drag on the vehicle, minimal internal and external friction, driving on a horizontal plane, the power required to maintain any speed is effectively zero. This means the load on the engine (when not accelerating) is also effectively zero. The Carnot efficiency is irrelevant at this point, but it would be very low as well. However, the fuel economy would be &lt;em&gt;enormous&lt;/em&gt; because you have some &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; in the numerator with nearly zero &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; in the denominator.&lt;/p&gt;&#xA;&lt;p&gt;The opposite situation is even easier to demonstrate; you can do it at home in your own car. Just floor the accelerator with the transmission in neutral. (Don't actually do this.) Instant high load scenario as you accelerate the hell out of that crankshaft but &lt;span class=&quot;math-container&quot;&gt;$v = 0$&lt;/span&gt;, so your fuel economy is zero.&lt;/p&gt;&#xA;&lt;p&gt;Realistic scenarios are more complicated but the long story made short is that &lt;strong&gt;the combustion reaction that occurs in the cylinder of the engine is much less efficient during periods of very high acceleration (i.e., near max load).&lt;/strong&gt; More fuel passes through the engine unburned, or only partly burned, meaning you didn't extract as much energy from the same gallon of fuel. You're still going somewhere and your engine is operating at a higher thermodynamic efficiency due to the load placed on it, but in terms of fuel economy, that benefit is reduced by the cost of lower combustion efficiency. It is even possible, when combustion efficiency is very poor, for that cost to outweigh the benefit of high loading entirely. (This might be expected if the vehicle has been very poorly maintained. In reality, so much depends on the age of the car, the quality of its control units, the gear in which you are accelerating, that it's hard to make an exact prediction for a general scenario.)&lt;/p&gt;&#xA;&lt;p&gt;The other thing I want to mention is that you have to consider where your power is going. &amp;quot;High engine loading&amp;quot; just means that a lot of the power the engine is capable of producing is being demanded; it doesn't tell you where the power is going. If it's going to fight drag, which increases as the square of velocity, then that's wasted power and wasted fuel. You can deliver it very efficiently but if it's not adding to the speed* of the vehicle, it's not contributing to fuel economy. It only looks efficient when you draw your system boundary around the engine and ignore the purpose of the car.&lt;/p&gt;&#xA;&lt;p&gt;* Technically, we should consider elevation as well, but fuel economy is typically calculated in terms of horizontal path distance. Gains and losses due to change in elevation are either assumed to cancel out overall or accounted for with some coarse correction factor.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-02-10T17:15:22.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2351" PostTypeId="1" CreationDate="2015-04-08T01:43:50.467" Score="8" ViewCount="676" Body="&lt;p&gt;I have a question about air conditioning and the probable modes of failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently installed a &lt;a href=&quot;https://nest.com/thermostat/life-with-nest-thermostat/&quot; rel=&quot;nofollow&quot;&gt;Nest thermostat&lt;/a&gt; in my home. I've noticed that it seems to prefer to have the air conditioner's compressor on for multiple short durations instead of one long period. This got me thinking...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) This is going to kill the starter capacitor of the air conditioner more quickly than normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I wonder why they made the control system like this... It MAY extend the life of the compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my thought process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The AC compressor is effectively a non-linear constant-Q source (i.e. you pump energy into it (by compressing the freon) at a rate (which is technically a function of the temperature differences involved) and it heats up until the energy coming in (from the compression) equals the energy going out (by the blower)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to run this continuously, the compressor itself would get really, really hot, always running at its maximum temperature (i.e. the crossover point between the efficiency-drop of the compressor at the high temperature vs the convection-cooling of the compressor). If you work it in smaller increments, you can keep the temperature down of the compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The starter capacitor is an annoying $100 capacitor... but the compressor dying can effectively total the AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: how much of the wear on the compressor is due to over-heating? Is this actually going to make an appreciable difference, or is Nest just unnecessarily eating my starter capacitor?&lt;/p&gt;&#xA;" OwnerUserId="1177" LastEditorUserId="272" LastEditDate="2015-04-08T16:54:55.240" LastActivityDate="2015-09-12T14:55:31.350" Title="Air Conditioning Modes of Failure" Tags="|mechanical-engineering|thermodynamics|hvac|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2352" PostTypeId="2" ParentId="2340" CreationDate="2015-04-08T02:09:14.750" Score="2" Body="&lt;p&gt;First, if you care about a certain city, look up the information. This should be information that is readily available. Big cities are usually go about publishing this information online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tunnels (subway or vehicle) are different enough that they don't have standard depths. They are designed to be far enough below ground to miss utilities and basements, unless they can't be for some other reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, below is the information for a random location that I found first (&lt;a href=&quot;http://www.co.washington.tx.us/default.aspx?Washington_County/Utility%20Rules&quot; rel=&quot;nofollow&quot;&gt;Washington County, Texas&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Depth of Underground Lines - The depth of underground lines shall be as specified herein for each type of utility. Where placements at such depths are impractical or where unusual conditions exist, the department shall specify other protection as may be appropriate in lieu of the depth of bury required for the particular utility line. Any and all buried utility lines will be placed at a minimum depth of 36&quot;. A ny deviation from the specified depth must be requested in writing and approved before Commissioners Court.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;High pressure gas and liquid petroleum lines will be constructed no less than forty-eight inches (48&quot;) lower than the lowest part of the drainage or bar ditch, and the drainage is to be considered at least two feet (2') below the center of the roadway. &lt;/li&gt;&#xA;  &lt;li&gt;Fiber Optic lines will be constructed no less than forty-eight inches (48&quot;) lower than the lowest part of the drainage or bar ditch, and the drainage is to be considered at least two feet (2') below the center of the roadway.&lt;/li&gt;&#xA;  &lt;li&gt;Communications cable will be constructed no less than thirty-six inches (36&quot;) lower than the lowest part of the drainage or bar ditch, and the drainage is to be considered at least two feet (2') below the center of the roadway.&lt;/li&gt;&#xA;  &lt;li&gt;Water Lines will be constructed no less than thirty-six inches (36&quot;) lower than the lowest part of the drainage or bar ditch, and the drainage is to be considered at least two feet (2') below the center of the roadway; crossings to be encased.&lt;/li&gt;&#xA;  &lt;li&gt;Underground Power line crossings and longitudinal shall be encased (placed in conduit) and buried a minimum of thirty-six inches (36&quot;) under roadway ditches, and sixty inches (60&quot;) below the pavement surface.&lt;/li&gt;&#xA;  &lt;li&gt;Cable television and copper cable communication lines shall have a minimum depth of cover twenty-four inches 24&quot;) under ditches or 18 inches beneath the bottom of the pavement structure, whichever is greater.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Warning! - Do not rely on these for digging!&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.call811.com/default.aspx&quot; rel=&quot;nofollow&quot;&gt;Call 811 to have utilities marked before you dig!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-04-08T02:09:14.750" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2353" PostTypeId="1" CreationDate="2015-04-08T02:33:08.577" Score="1" ViewCount="1935" Body="&lt;p&gt;I am trying to design a simple plastic electronic enclosure for a PCB and I wanted to learn how to get the enclosure as ready as possible for manufacturing. I basically wanted to understand the process that a mechanical engineer would go through to make sure the design of the enclosure is good enough for manufacturing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have read about stress/strain and wanted to find out what other things I should know when designing plastic enclosures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a lot of enclosures such as the one below, seem to have added extrusions on the internal sides, I would like to understand the functionality behind these - are these to do with stress and strain making the product more resistant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas of the processes involved by a mechanical engineer to make a manufacture ready plastic enclosure would be great&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SIamV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1178" LastActivityDate="2016-07-15T19:09:59.240" Title="Designing electronic enclosures for manufacturing" Tags="|mechanical-engineering|manufacturing-engineering|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2355" PostTypeId="1" AcceptedAnswerId="2370" CreationDate="2015-04-08T02:45:00.753" Score="3" ViewCount="3321" Body="&lt;p&gt;I want to build a small ice cannon that uses pressurized air (either through an air compressor or a CO&lt;sub&gt;2&lt;/sub&gt; cartridge) to shoot small ice pellets at my coworkers. I plan to use a thermoelectric cooler to freeze the water but I don't understand how to spec it.&lt;/p&gt;&#xA;&lt;p&gt;For example, what are the key parameters of &lt;a href=&quot;https://www.sparkfun.com/products/10080&quot; rel=&quot;nofollow noreferrer&quot;&gt;this product&lt;/a&gt; that will let me know if the cooler will achieve what I'm hoping it to?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h2&gt;Thermoelectric Cooler - 40x40mm&lt;/h2&gt;&#xA;&lt;p&gt;COM-10080 In Fritzing Library&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Description:&lt;/strong&gt; Thermoelectric coolers (TEC or Peltier) create a temperature differential on each side. One side gets hot and the other&#xA;side gets cool. Therefore, they can be used to either warm something&#xA;up or cool something down, depending on which side you use. You can&#xA;also take advantage of a temperature differential to generate&#xA;electricity. The thermal tape listed below works very well to attach&#xA;heat sinks to the hot side.&lt;/p&gt;&#xA;&lt;p&gt;This Peltier works very well as long as you remove the heat from the&#xA;hot side. After turning on the device, the hot side will heat quickly,&#xA;the cold side will cool quickly. If you do not remove the heat from&#xA;the hot side (with a heat sink or other device), the Peltier will&#xA;quickly reach stasis and do nothing. We recommend using an old&#xA;computer CPU heatsink or other block of metal to pull heat from the&#xA;hot side. We were able to use a computer power supply and CPU heatsink&#xA;to make the cold side so uncomfortable we could not hold our finger to&#xA;it.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Features:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;40 x 40 x 3.6mm&lt;/li&gt;&#xA;&lt;li&gt;lmax - 7A&lt;/li&gt;&#xA;&lt;li&gt;Umax - 15.4V&lt;/li&gt;&#xA;&lt;li&gt;Qcmax - 62.2W&lt;/li&gt;&#xA;&lt;li&gt;Tmax - 69C&lt;/li&gt;&#xA;&lt;li&gt;1.7 Ohm resistance&lt;/li&gt;&#xA;&lt;li&gt;127 thermocouples&lt;/li&gt;&#xA;&lt;li&gt;Max Operating Temp: 180°C&lt;/li&gt;&#xA;&lt;li&gt;Min Operating Temp: -50°C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If I want to freeze 10 mL of water from room temperature within a certain time period, how would I spec out the cooler? What are the base equations that govern how long it takes to freeze the water?&lt;/p&gt;&#xA;" OwnerUserId="1179" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-05-06T05:07:20.327" Title="How do I spec out a thermoelectric cooler?" Tags="|mechanical-engineering|thermodynamics|cooling|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2356" PostTypeId="2" ParentId="2353" CreationDate="2015-04-08T03:02:49.540" Score="2" Body="&lt;p&gt;To design such a box, it would need to be approached from different perspectives. But the final criteria would be is it legal, and does it satisfy other requirements. For example, does it satisfy CE mark compliance requirements for safety and EMI, and will it do the job the user wants it for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CE requirements and so on are often assessed by specialist bodies who put it through rigorous checks. This involves many specialists and specialist equipment. This is a large topic in itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then after that you have considerations such as choice of materials (including composites), strength, durability, cost, methods of manufacture and so on. These will most likely be the easier problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm afraid you are embarking on an extremely difficult problem which requires a multi-disciplinary approach and years and years of experience. The phrase &quot;A box is a box&quot; is not true. A box, in this case, is Pandora's box.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2015-04-08T03:02:49.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2358" PostTypeId="2" ParentId="2339" CreationDate="2015-04-08T12:16:48.490" Score="5" Body="&lt;p&gt;As Dan mentioned in the comments, the presence of salt (and other pollutants) can alter the freezing point of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.iso.org/iso/catalogue_detail.htm?csnumber=53499&quot; rel=&quot;nofollow&quot;&gt;ISO 9223 abstract&lt;/a&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;key factors in the atmospheric corrosion of metals and alloys. These are the temperature-humidity complex, pollution by sulfur dioxide and airborne salinity.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All of which are factored into the &quot;time of wetness&quot;, which can increase in the presence of salinity (and other pollutants, acknowledged in the ISO standard itself) as found in observations made in Antarctica published in &lt;em&gt;&lt;a href=&quot;http://au.wiley.com/WileyCDA/WileyTitle/productCd-0470080329.html&quot; rel=&quot;nofollow&quot;&gt;Uhlig's Corrosion Handbook&lt;/a&gt;&lt;/em&gt;, (p. 330):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;..that in the presence of marine salt, liquid water monolayers could form under ice layers resulting in high corrosion rates in temperatures well below 0C&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sulfur dioxide is also identified in the ISO abstract, this a common pollutant from vehicles, industry etc. which according to can chemically react with water vapour, forming sulfuric acid, the basis of &lt;a href=&quot;http://www.epa.gov/acidrain/what/&quot; rel=&quot;nofollow&quot;&gt;acid rain and acid snow&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerDisplayName="user1184" LastEditorDisplayName="user1184" LastEditDate="2015-04-08T12:38:16.293" LastActivityDate="2015-04-08T12:38:16.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2359" PostTypeId="1" CreationDate="2015-04-08T13:23:59.587" Score="5" ViewCount="1682" Body="&lt;p&gt;I am trying to prove the equation&#xA;$$&#xA;\frac{\bar{P}}{V}=\frac{1}{2}E_0^2\sigma_{AC}&#xA;$$&#xA;which can be rewritten as&#xA;$$&#xA;\begin{align}&#xA;  \frac{\bar{P}}{V} &amp;amp;= \frac{1}{2}E_0^2\sigma_{AC}\\&#xA;  &amp;amp;=\frac{1}{2}E_0^2\ \omega\ \epsilon_0\ \epsilon^{''}_r\\&#xA;  &amp;amp;=\frac{1}{2}E_0^2\ \omega\ \epsilon_0\ \epsilon_r^{'}\ \tan(\delta)&#xA;\end{align}&#xA;$$&#xA;Here $\bar{P}$ stands for the time-averaged power loss which satisfies the equation&#xA;$$&#xA;\bar{P}=\frac{1}{T}\int_0^T U\ I\ dt,&#xA;$$&#xA;where $T=\frac{2\pi}{\omega}$ is the time period, $U=U_0 e^{j\omega t}$ is the complex sinusoidal voltage, and $I=j\omega\epsilon^{'}_rC_0U + \omega\epsilon^{''}_rC_0U$. The instructions say to use&#xA;$$&#xA;\begin{align}&#xA;U_0 &amp;amp;= E_0h\\&#xA;C &amp;amp;= \epsilon_r\epsilon_0\frac{A}{h}\\&#xA;V &amp;amp;= A\ h \\&#xA;\sigma_{AC}&amp;amp;=\omega\epsilon_0\epsilon^{''}_r=\omega\epsilon_0\epsilon^{'}_r\tan(\delta)\\&#xA;\tan(\delta) &amp;amp;= \frac{\epsilon^{''}_r}{\epsilon^{'}_r}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The problem I face is after solving the main integral part, which is like:&#xA;$$&#xA;\epsilon^{''}_r*(F(T) - F(0)) + j*\epsilon^{'}_r((F(T) - F(0))&#xA;$$&#xA;where $F(t) = e^{2j\omega t}$ and I neglected all the constants for simplicity.&#xA;$F(T)$ is equal to $\exp(j*4*\pi)$ which is 1, making $(F(T)-F(0))$ zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of root mean squaring both $U$ and $I$ to begin with, but this gives out $\sqrt{\epsilon^{''2}_r+\epsilon^{'2}_r/2}$ term which doesn't seem to lead to the proof result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a hunch that I maybe missing some elementary calculation basics. I would really appreciate any help that you can offer.&lt;/p&gt;&#xA;" OwnerUserId="1185" LastEditorUserId="272" LastEditDate="2015-04-09T11:05:14.363" LastActivityDate="2016-03-28T04:42:17.700" Title="Mathematical expression for power loss in a dielectric" Tags="|electrical-engineering|power|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2360" PostTypeId="1" CreationDate="2015-04-08T14:31:27.597" Score="2" ViewCount="236" Body="&lt;p&gt;I have this example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fiv8I.png&quot; alt=&quot;Image of shaft dimensions and forces (momentum)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$M_k$ = 80 Nm&lt;/li&gt;&#xA;&lt;li&gt;$d_1$ = 30 mm&lt;/li&gt;&#xA;&lt;li&gt;$d_2$ = 20 mm&lt;/li&gt;&#xA;&lt;li&gt;$a$ = 600 mm&lt;/li&gt;&#xA;&lt;li&gt;$b$ = 500 mm &lt;/li&gt;&#xA;&lt;li&gt;$G$ = 8,5e4 MPa&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I want to know the twist at free end of this shaft. I am using the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\phi = \dfrac{M_k \cdot a}{G \cdot \dfrac{\pi d_1^4}{32}} +  \dfrac{M_k \cdot b}{G \cdot \dfrac{\pi d_2^4}{32}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gives me 0,037 rad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the result in the book is 0,013 rad. Where is the error?&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="368" LastEditDate="2015-04-08T22:25:13.987" LastActivityDate="2015-04-08T23:16:29.970" Title="Twist of two diameter shaft" Tags="|mechanical-engineering|stresses|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2361" PostTypeId="2" ParentId="2355" CreationDate="2015-04-08T14:35:03.633" Score="1" Body="&lt;p&gt;You ultimately need to know the totaly electrical engergy required to go into the cooler per ice pellet.  However, even without calculating this, it seems fairly obvious that too much energy will be required for reasonable battery operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the energy, follow these steps:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Assume a starting temperature for the water, like 20&amp;deg;C.  Using the specific heat of liquid water, calculate the energy required to cool it down to 0&amp;deg;C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Now look a the heat released by phase change from liquid to solid for your mass of water, and add that to the total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Look up or guess the efficiency of your Peltier cooler at, say, 25&amp;deg;C in and 0&amp;deg;C out.  Divide the total energy required by this value to get the electrical input energy.  Note that Peltier coolers are very inefficient devices, most likely well under 10% in this application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives you the theoretical minimum required electrical energy input.  There will be various losses in the system and other inefficiencies.  For example, the cooler won't cool just the ice and nothing around it.  And, no insulation is perfect so there will be some heat coming from ambient that needs to be removed, in addition to removing the heat from the cooling and solidifying water.  All around as a rough SWAG I'd probably figure at least 4x more energy required overall than just the minimum from the basic physics of cooling the water thru a Peltier device.  Ultimately only real measurements will get you good numbers because there are too many variables that are too hard to quantify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at how big the Peltier needs to be, what you need to do to remove heat from its hot side, the size of the battery you need to support this, and realize why this is a totally unrealistic idea.  It is conceivably possible with a device that sits on a desk and is plugged into the wall.  But, shooting hard pellets, like ice, around a office sounds like a fundamentally dumb and irresponsible idea in the first place.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-04-08T14:35:03.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2362" PostTypeId="2" ParentId="1950" CreationDate="2015-04-08T17:40:00.730" Score="3" Body="&lt;p&gt;Reading through the site I would say no. Using the exhaust gas as input for the turbine will have horrible efficiencies for powering everything he listed. Also, on the &quot;How turbochargers work&quot; page he stated that he uses the compressed air from the turbocharger to directly power a turbine. A turbine and a compressor can be thought of as opposite cycles of the same mechanism...one puts work in and changes pressure, the other changes pressure and gets work out. You would at best be getting the same air and energy as not having a turbo, but since there are mech losses that wouldn't be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum, I don't think the heated exhaust would provide enough power for the turbo, water pump, etc as he suggests, and his compressor/turbine combo wouldn't work as he describes (unless I misinterpreted what he stated).&lt;/p&gt;&#xA;" OwnerUserId="1187" LastActivityDate="2015-04-08T17:40:00.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2363" PostTypeId="1" AcceptedAnswerId="2366" CreationDate="2015-04-08T21:53:54.640" Score="24" ViewCount="10700" Body="&lt;p&gt;In some cars, I have noticed the rear wheels are located at the extreme rear of the vehicle. However, I have noticed that the rear wheels of buses are always located about 1/4th of the way forward from the rear. What is the reason for this?&lt;/p&gt;&#xA;" OwnerUserId="1189" LastEditorUserId="368" LastEditDate="2015-04-08T22:31:52.303" LastActivityDate="2016-03-26T05:00:10.017" Title="Why are rear wheels not placed at the extreme rear of a bus?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="8" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2364" PostTypeId="2" ParentId="2363" CreationDate="2015-04-08T22:24:24.533" Score="16" Body="&lt;p&gt;I think one of the reasons is structural. The bottom on the bus is like a bridge with long steel beam running the length, and if the supports are at the ends the stresses and deflections would be too high. By moving the wheels more towards the center of gravity in reduces the flexure of the center of the bus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition the driveshaft can be shorter which means less flexure and less wear.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2015-04-08T22:24:24.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2366" PostTypeId="2" ParentId="2363" CreationDate="2015-04-08T22:40:00.600" Score="32" Body="&lt;p&gt;Some ideas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Wheel Load Distribution&lt;/strong&gt;: The load is greater on the rear wheels providing the power; more force on the front ones bring no benefit and would provide less traction.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Better manoeuvrability&lt;/strong&gt; from having a shorter wheelbase.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Better Ground Clearance&lt;/strong&gt; in some conditions, especially for bumps and or up a increasing slope for instance.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Better Driving&lt;/strong&gt;: The front wheels now turn around a point closer to the C.G. than with the rear wheels. Not good with vehicle dynamics but this appears better than the rear end 'trailing' behind.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Structural&lt;/strong&gt;: As some people have pointed out, it's better to balance the weights on either side of the axle (think beams). You have a (basically) point load for the engine and a distributed load for the rest of the bus. Rather than placing it all on one side, you can distribute it to reduce the ultimate loading.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="782" LastEditorUserId="782" LastEditDate="2015-04-10T15:23:08.810" LastActivityDate="2015-04-10T15:23:08.810" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2367" PostTypeId="2" ParentId="2360" CreationDate="2015-04-08T23:16:29.970" Score="2" Body="&lt;p&gt;The problem is not your formula. If your diagram is correct and you have copied the values correctly, then the answer indeed is 0,037 rad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I suggest that you do the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Check that you copied the diagram correctly&lt;/li&gt;&#xA;&lt;li&gt;Check that you copied the correct values&lt;/li&gt;&#xA;&lt;li&gt;Check that you didn't misread the answer key&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you still can't determine what is wrong, talk to your professor or TA. Textbooks have errors, and if you are in a class of students who are all working on this problem, they might already know about the issue.  If they don't already know, they'll probably be grateful that you alerted them to the error.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-04-08T23:16:29.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2368" PostTypeId="2" ParentId="2363" CreationDate="2015-04-09T02:47:06.773" Score="8" Body="&lt;p&gt;The answer is a combination of factors. Weight has a great deal to do with the positioning of the rear axle. If you assume that the bus has a full load of passengers, you'll want the power to the rear wheels closest where there is weight. Larger buses have dual-axles/wheels to accommodate the extra weight, but the positioning of the axles often remains the same. Some heavy-duty dual-wheeled trucks have the rear axle closer to the rear of the vehicle to compensate for the loads they will carry. (trailers, recreational vehicles, etc).  Conventional passenger buses need a shorter wheelbase, and a heavy-duty single rear axle. It's a compromise between weight, ride and fuel economy.&lt;/p&gt;&#xA;" OwnerUserId="1190" LastActivityDate="2015-04-09T02:47:06.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2369" PostTypeId="2" ParentId="306" CreationDate="2015-04-09T03:44:30.163" Score="6" Body="&lt;p&gt;Part of the answer is that the dynamics of pneumatic tires are well understood. In basic suspension geometry design, your caster and kingpin angles contribute to the creation of pneumatic trail, which provides a return to center force when cornering.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same deformation of the tire contact patch when presented with aggressive steering input, or acceleration or braking, allows the tire to grip more than if it was solid, and increase the response time of the whole vehicle to a particular input. There is a sweet spot there, as a flat tire gives no response, but a rigid tire will break traction too easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not to say that these things couldn't be overcome by an active suspension, but the suspension would need to actively deform in directions that are not currently considered. State of the art computer controlled suspensions very quickly change the response profile of the shock absorber, but they don't actively adjust the steering or camber angles of the wheel.&lt;/p&gt;&#xA;" OwnerUserId="1192" LastActivityDate="2015-04-09T03:44:30.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2370" PostTypeId="2" ParentId="2355" CreationDate="2015-04-09T04:18:57.850" Score="2" Body="&lt;p&gt;There seems to be two questions here, how to evaluate thermoelectrics and how much cooling is required to freeze water. I'll lead off with some instruction on heat transfer and thermoelectrics and then discuss your design question.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Heat Transfer, Really Quickly&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;When heat conducts through a solid, the heat flow is proportional to the temperature difference such that $q = K \Delta T$, where $q$ is the heat rate (usually in Watts, energy moved per unit time); $K$ is the conductance, a property of the solid; and $\Delta T$ is the temperature difference&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Heat sink manufacturers will specify an effective conductance of their heat sinks, usually in &amp;deg;C/W&lt;/li&gt;&#xA;&lt;li&gt;The &quot;R-value&quot; of insulation is the inverse of conductance in really weird units&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Thermoelectrics&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Thermoelectric datasheets contain values that are easy to measure (for the manufacturers) but not easy to put into system models (for you).  The first step is to get the Seebeck coefficient, $\alpha$, and the thermal conductance, $K$, from the data provided. These properties are important because they feed into the standard model of thermoelectrics&lt;sup&gt;1&lt;/sup&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ q_{c} = \alpha I T_c - \frac{1}{2} I^2 R - K (T_h - T_c) $$&#xA;$$ q_{h} = \alpha I T_h + \frac{1}{2} I^2 R - K (T_h - T_c) $$&#xA;$$ V = \alpha (T_h - T_c) + IR $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $q$ is heat rate (more on that in the design section), $I$ is current, $R$ is electrical resistance, $T_h$ and $T_c$ are the cold and hot side temperatures, respectively and $V$ is the applied voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K$ can be found easily by dividing Qcmax by Tmax (which properly should be called $\Delta T_{max}$).  Tmax is the maximum temperature difference the module can sustain when the hot side is held at a reference temperature.  At this point, all of the electrical power input is being used to pump the heat which was conducting through the device, i.e. $q_{c,max} = K \Delta T_{max}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seebeck coefficient can be found by using Qcmax.  At max heat pumping, the temperature difference has to be zero (from the first equation), so $\alpha I_{max} T_c = q_{c,max} + \frac{1}{2} I_{max}^2 R$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Designing a System&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ywOfw.png&quot; alt=&quot;Thermoelectric System&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your system, I imagine, will look a bit like this.  Your thermoelectric will need to be able to pump out both the heat that comes in from the outside (how much will depend on how well you insulate your compartment) and whatever heat is released by the water during the freezing process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted, it takes 88 J/g of water to freeze water. If you multiply 88J/g of water by 10g (at 1g/mL) and divide by the number of seconds you want it to take, you will have your heat rate requirement $q$ to freeze the water.  As Olin said, add in some amount of heat that will conduct through the walls and you have an estimate of $q_c$, the amount of heat the thermoelectric needs to move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the heat sink that you buy will have some conductance rating $C$ in &amp;deg;C/W, we know that the hot side of the thermoelectric $T_h$ will be equal to $C q_h + T_{ambient}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have enough information to solve the problem.  Effectively, what we need to ask is whether there is a current $I$ that will satisfy the three equations below for the known $q_c$, $T_c$, $\alpha$, $R$, and $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ q_{c} = \alpha I T_c - \frac{1}{2} I^2 R - K (T_h - T_c) $$&#xA;$$ q_{h} = \alpha I T_h + \frac{1}{2} I^2 R - K (T_h - T_c) $$&#xA;$$ T_h = Cq_h + T_{ambient} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although you might be able to solve those equations by hand, I would just plug them into Excel and use the solver functionality to see if it is possible.  If it isn't possible, try adjusting your $q_c$ requirement, e.g. by freezing the ice slower or adding a second thermoelectric module (which would cut $q_c$ per module in half.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-04-09T04:18:57.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2372" PostTypeId="2" ParentId="306" CreationDate="2015-04-09T11:42:20.870" Score="3" Body="&lt;p&gt;jhabbott lists some terrific reasons for continuing to use pneumatic tires, here are some additional reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pneumatic tires are cheap to manufacture&lt;/li&gt;&#xA;&lt;li&gt;Pneumatic tires get the job done with little maintenance.&lt;/li&gt;&#xA;&lt;li&gt;Pneumatic tires' failures are easily repaired in the field.&lt;/li&gt;&#xA;&lt;li&gt;Similar tires can be used across a wide variety of vehicles and even vehicle types.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In contrast, when active suspensions fail they are expensive, not field repairable, and require specialized parts.&lt;/p&gt;&#xA;" OwnerUserId="449" LastActivityDate="2015-04-09T11:42:20.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2376" PostTypeId="2" ParentId="2347" CreationDate="2015-04-09T15:04:52.090" Score="3" Body="&lt;p&gt;The answer depends on factors like the bypass ratio of the engine and the design of the thrust reverser, e.g. bucket doors at the back of the engine or vanes to deflect the bypass airflow from the fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the OP's proposed measurement (reverse thrust / forward thrust), the efficiency is also strongly dependent on the engine speed. At maximum thrust the reverser may be 50% or 60% efficient, but at lower power levels it may be less than 10%.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-04-09T15:04:52.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2377" PostTypeId="2" ParentId="2363" CreationDate="2015-04-09T15:54:35.340" Score="6" Body="&lt;p&gt;There are lots of factors but I'd say the two major factors are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Drive Shaft Length&lt;/strong&gt; - A very long drive shaft causes various engineering problems so there is some pressure to minimize the length.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Maneuverability&lt;/strong&gt; - Longer wheelbases cause problems mostly in making turns in tight quarters. The longer the wheelbase the bigger the difference between the track of the front wheels and the rear wheels. So the fronts go around the corner fine but the rear wheels go up over the curb and take out the light pole. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1206" LastActivityDate="2015-04-09T15:54:35.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2378" PostTypeId="2" ParentId="2363" CreationDate="2015-04-09T16:07:37.147" Score="5" Body="&lt;p&gt;Think about how you would support a long, flexible beam (or a 30cm ruler, if you prefer). If you put the supports at the ends, the beam will sag in the middle; but if you put the supports at 1/4 and 3/4 of the length, the weight of the beam between the supports will be counterbalanced by the weight of the beam outside the supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are a lot of other factors involved in designing a vehicle, but the tiny cars with wheels right at the end are the oddities.&lt;/p&gt;&#xA;" OwnerUserId="1207" LastActivityDate="2015-04-09T16:07:37.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2379" PostTypeId="2" ParentId="2347" CreationDate="2015-04-09T16:47:10.803" Score="1" Body="&lt;p&gt;I agree that reverse thrust is not nearly as effective (&quot;efficient&quot; might not be the correct term) as forward thrust, and I'm surprised that it could be as high as 50%. The previous responder (alephzero) did say that it is probably only that efficient at full power, and in my experience full reverse thrust is rarely used - very noisy. However, on big advantage of using reverse thrust is that it eliminates the residual forward thrust that is otherwise present. Just that can make a big difference on a slippery runway.&lt;/p&gt;&#xA;" OwnerUserId="1208" LastEditorUserId="272" LastEditDate="2015-05-13T16:01:38.297" LastActivityDate="2015-05-13T16:01:38.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2380" PostTypeId="2" ParentId="2303" CreationDate="2015-04-09T17:17:36.790" Score="2" Body="&lt;p&gt;Olin answered it correctly. The pilot, with no acceleration, experiences 1 g of force. Add 6 g's upward acceleration and he will experience 7 g's of force. The DIRECTION of the acceleration must be specified in order to give an answer to the question. I would bet that the question assumes an upward acceleration, but it doesn't say.&lt;/p&gt;&#xA;" OwnerUserId="1208" LastActivityDate="2015-04-09T17:17:36.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2381" PostTypeId="2" ParentId="2300" CreationDate="2015-04-09T17:25:43.097" Score="4" Body="&lt;p&gt;There seems to be some confusion here. In the English (or American) system the &quot;official&quot; measure of mass is the slug. Turns out that 32.2 lbm = 1 slug. So to plug into the equation F=MA you can use M in slugs, A in ft/sec and F in lbf. And, as someone said, at &quot;standard&quot; gravity 1 lbm exerts 1 lbf on its support (its weight). If you are going to do any significant calculations, it's best, in my opinion, to get rid of all lbm designations and convert everything to slugs.&lt;/p&gt;&#xA;" OwnerUserId="1208" LastActivityDate="2015-04-09T17:25:43.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2382" PostTypeId="1" AcceptedAnswerId="2401" CreationDate="2015-04-09T17:45:20.373" Score="11" ViewCount="1709" Body="&lt;p&gt;Most people have had the experience of moving the tab of an &lt;a href=&quot;https://en.wikipedia.org/wiki/Beverage_can#Stay-on-tab&quot; rel=&quot;nofollow noreferrer&quot;&gt;aluminum can&lt;/a&gt; back and forth until it breaks off. It usually only takes a few complete back and forth motions before the tab breaks off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6PljV.jpg&quot; alt=&quot;Aluminum can with stay tab&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the root cause of the tab breaking off?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible causes seem to be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A fatigue fracture.&lt;/li&gt;&#xA;&lt;li&gt;An overstressing of the metal.&lt;/li&gt;&#xA;&lt;li&gt;A result of plastic deformation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But which one is it?&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2016-01-27T02:08:36.187" LastActivityDate="2016-01-27T02:08:36.187" Title="Why do beverage can tabs break when bent?" Tags="|mechanical-engineering|materials|mechanical-failure|fatigue|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2383" PostTypeId="1" AcceptedAnswerId="2397" CreationDate="2015-04-09T17:46:10.807" Score="6" ViewCount="106" Body="&lt;p&gt;I was thinking about this &lt;a href=&quot;https://www.kickstarter.com/projects/voltera/voltera-your-circuit-board-prototyping-machine/description&quot; rel=&quot;nofollow&quot;&gt;kickstarter project&lt;/a&gt; where they print circuit boards, and later they show it printing solder paste.  I've seen other liquid or paste dispensing systems like this too, maybe for medical or chemical uses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how do you make a system that dispenses a known amount of paste like that?  I get that I can run a motor and put pressure on top of a syringe and goo will come out of it. Maybe a stepper with some gearing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a feedback mechanism?  It seems too small to have some kind of flow meter there.  Are these kinds of things just open loop and you calculate the volumes ahead of time and just crank down the motor accordingly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must have to get it to some baseline first though, like a home position where paste is at the bottom of the nozzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just curious how these things work.&lt;/p&gt;&#xA;" OwnerUserId="1210" LastEditorUserId="282" LastEditDate="2015-04-11T17:20:08.193" LastActivityDate="2015-04-11T17:20:08.193" Title="controlled depositing of goo with a syringe" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2385" PostTypeId="2" ParentId="2333" CreationDate="2015-04-09T18:39:08.443" Score="3" Body="&lt;p&gt;Reading the campagnolo website it does seem like they are carbon fibre. I would guess that they used a prepreg moulding method to create the hub. Moulding prepreg allows fibres in all directions so you get maximum bending and torsion resistance, as well as great strength-weight characteristics. If you sand it down you won't get the fibre look that cfrp is famous for, and you can give it a finish for the shine.&lt;/p&gt;&#xA;" OwnerUserId="1187" LastActivityDate="2015-04-09T18:39:08.443" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2386" PostTypeId="2" ParentId="2382" CreationDate="2015-04-09T18:48:34.820" Score="1" Body="&lt;p&gt;Fatigue is the stress on a material due to cyclic loading. The best comparison is a rubber band. When you pull it over and over it stretches out because the fatigue is causing the elasticity to wear out and it becomes more and more plastic until the band snaps. While technically the the tab is breaking because of plastic deformation, fatigue fracture would be more correct be the plastic deformation doesn't cause a fracture until the tab has been loaded significantly more times than the life-cycle intends.&lt;/p&gt;&#xA;" OwnerUserId="1187" LastActivityDate="2015-04-09T18:48:34.820" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2387" PostTypeId="2" ParentId="651" CreationDate="2015-04-09T19:04:30.183" Score="8" Body="&lt;p&gt;The cost of installing and maintaining a split refrigeration system for cabinet style refrigerators would swamp the cost of the refrigerator itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large refrigerated rooms, or whole house cooling systems, the cost of the entire system is greater than the cost of the labor and maintenance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given mass manufacturing, though, it doesn't make sense to sell a $400 refrigerator that requires a lot more money to install, including holes, routing pipes, placement of an internal and external unit, etc.  Further, it would decrease reliability and require more maintenance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, there is actually little efficiency to be gained.  It's such a small space to cool that the little heat increase in the summer isn't large, and is offset by reduced heating in the winter (when a refrigerator with an external unit would actually have to work harder).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does make sense (and is more efficient) for much larger cooling requirements, but not for small refrigerator sized consumer appliances.&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2015-04-09T19:04:30.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2388" PostTypeId="1" CreationDate="2015-04-09T22:10:13.433" Score="3" ViewCount="432" Body="&lt;p&gt;Consider the Robin boundary condition for the diffusion/heat equation $\mathrm{u_t=a(t)u_{xx}+f(x,t)}$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{-k(t)u_x(0,t)=h(t)u(0,t)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{u_x(0,t)+\frac{h(t)}{k(t)}u(0,t)=0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathrm{k(t)}$ thermal conductivity and $\mathrm{h(t)}$ heat tranfer coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Question: Is it possible that the ratio $\mathrm{h(t)/k(t)}$ to be constant? Could anyone please help me? I have really no idea. &lt;/p&gt;&#xA;" OwnerUserId="1216" LastEditorUserId="110" LastEditDate="2015-04-11T14:58:36.083" LastActivityDate="2015-04-11T14:58:36.083" Title="Ratio between heat transfer coefficient and thermal conductivity" Tags="|fluid-mechanics|chemical-engineering|cooling|superconduction|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2389" PostTypeId="2" ParentId="2388" CreationDate="2015-04-09T22:37:33.740" Score="2" Body="&lt;p&gt;It certainly is possible for the ratio to be constant or very approximately constant. Indeed, it's very much possible for $k$ and $h$ to be constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As written, they're only specifically functions of time and could take on just about any form you choose (including a constant value).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would not commonly encounter $k(t)$, where conductivity is some prescribed function of time (can't think of a way to physically do it off the top of my head). It's more likely to find $k(u(t))$, which is almost always true, but we neglect that if the change is small enough, as the problem becomes non-linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constant $h$ is pretty easy to find, as long as the driving force behind the external flow is constant and any buoyant effects are negligible. Buoyancy will, again make the problem non-linear.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-04-09T22:37:33.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2390" PostTypeId="1" CreationDate="2015-04-09T23:22:23.457" Score="3" ViewCount="11135" Body="&lt;p&gt;My background is more Chemistry and Biology, not Physics, or specifically engineering with fluid dynamics, so bear with me! Also, this is a little long, so I will do my best to make it easy to follow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Problem:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to derive a formula to calculate air flow rates (in CFM) based on the static pressure (inches W.C.) of the HEPA filter I am blowing through for a laminar flow hood that I am building. &lt;a href=&quot;http://en.wikipedia.org/wiki/Laminar_flow_cabinet&quot; rel=&quot;nofollow&quot;&gt;Link&lt;/a&gt; I am trying to figure this out because many HVAC application fans have ratings of their output at certain static pressures, many, many more do not, and it would be really helpful to beable to derive this information on my own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Information I have available to use:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Standard fan information, from the manufacturer (Brake Horsepower, Fan wheel dimensions, RPM &amp;amp;/or CFM).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Volume of the Plenum Chamber, &lt;a href=&quot;http://en.wikipedia.org/wiki/Plenum_chamber&quot; rel=&quot;nofollow&quot;&gt;Link&lt;/a&gt;, in this case &gt;= 4 ft$^3$. (12&quot; x 24&quot; x 24&quot; as it is generally accepted that the size of an adequate plenum must be &gt;= the volume of the filter).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Static Pressure of the HEPA filter, in my case 1&quot; w.g., and although I don’t know how it relates to anything or what it means, I also have the resistance range of, 0.69.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The desired volume/rate of air on the intake of the filter, calculated out to 400 CFM @ 1.00&quot; w.g. SP. ( Based upon this formula (Desired FPM Output) x (Volume of HEPA Filter) = Required CFM Input. &lt;a href=&quot;http://www.fungifun.org/English/Flowhood#match&quot; rel=&quot;nofollow&quot;&gt;Link&lt;/a&gt;. --The desired output of the entire unit, 100-200 FPM, ideally.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions developed from research &amp;amp; attempts to solve:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Some fans are rated at 0.00” SP all the way up through +1.5” SP. What are the standard conditions for these specs? I.E. in chemistry you have SC = 1.0 ATM @ 0$^{\circ}$C. If a fan is rated at 500 CFM without any information given about Static Pressure, what would be the static pressure? Is it 0.00” w.c.? above, or below?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have stared at the Affinity, or Fan Laws, &lt;a href=&quot;http://www.greenheck.com/library/articles/10&quot; rel=&quot;nofollow&quot;&gt;Link,&lt;/a&gt; for countless hours now. Specifically Law #2. SP$_2$ = SP$_1$( CFM$_2$ / CFM$_1$)$^2$. I was hoping that this would solve my problems, but I have been trying to solve for CFM$_1$, since I know I want SP$_2$ to equal 1.0” and I know I want CFM$_2$ to equal 400 @ SP$_2$. I have been operating under the assumption that a fan, at it’s highest RPM/CFM is operating at 0.0” SP. But then you can’t calculate for the obvious reasons, so this has to be wrong. So back to standard conditions for fans.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1217" LastEditorUserId="33" LastEditDate="2015-04-10T10:55:20.337" LastActivityDate="2017-05-02T22:18:25.910" Title="CFM in relation to Static Pressure" Tags="|mechanical-engineering|pressure|airflow|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2391" PostTypeId="2" ParentId="2363" CreationDate="2015-04-10T01:35:35.927" Score="10" Body="&lt;p&gt;Front axle load is an important consideration.  It's hard to carry a lot of weight on an axle that steers, as you are effectively limited to single wheels, and you want to minimise the power assistance needed to actually turn the steering wheel. &lt;/p&gt;&#xA;" OwnerUserId="1218" LastActivityDate="2015-04-10T01:35:35.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2392" PostTypeId="2" ParentId="2390" CreationDate="2015-04-10T01:40:16.740" Score="1" Body="&lt;p&gt;I don't know the whole answer to your question, but I can tell you why you are confused about the fan laws. Those laws are used if you want to change the motor attached to the fan, they don't tell you the shape of the fan curve itself, which is the locus of possible operating points for a given fan and motor combination.  You are correct that at maximum flow rate, static pressure is zero.  Similarly, at maximum static pressure, flow rate is zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you actually want is to know the relation between the static pressure (usually called head, $H$) and the flow rate $Q$ (what you have called CFM).&#xA;I don't know the typical shape of fan curves, but if I were trying to do what you are doing, I'd probably just assume the fan curve was approximately $H = H_{max} - AQ^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make that assumption, then given that you know the maximum head and maximum flow rate from the data sheet, you can estimate what the flow rate will be at your operating pressure. Unfortunately, the only way to know for sure is to experiment, unless someone else here knows the real fan curve equation.  Even then, keep in mind that engineering calculations, especially those dealing with compressible fluids, can easily by 20% off. Give yourself a generous safety factor.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-04-10T01:40:16.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2393" PostTypeId="2" ParentId="2349" CreationDate="2015-04-10T02:12:56.147" Score="6" Body="&lt;p&gt;The plot in the question does not apply to the situation being asked.  The power capacity of an automotive engine is dynamically changed while driving by both altering the RPM and by throttling the airflow in the intake.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak efficiency of a car engine is somewhere midrange, with inefficiencies under both high and low loadings. At 10% loading, the efficiency is still better than half what it is at peak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But say then we're getting 10% efficiency at 34mpg.. There's no inconsistency there.  It just means if you could somehow achieve 100% efficiency, you'd get 340mpg under that load scenario. &lt;/p&gt;&#xA;" OwnerUserId="1218" LastActivityDate="2015-04-10T02:12:56.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2394" PostTypeId="1" CreationDate="2015-04-10T02:48:19.587" Score="3" ViewCount="209" Body="&lt;p&gt;Most of the breast pumps I've examined utilize a very similar mechanical design; that is, they use lead-screw motors to produce linear actuation of a piston-like device inside a cavity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some breast pumps also offer electrical controls for both speed and suction strength.  From the lead-screw design, I can easily understand how to control the speed of suction.  But I'm not quite sure how suction strength is controlled.  How is this typically done in lead-screw actuated piston-like air pumps?  &lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="368" LastEditDate="2015-04-10T03:31:51.450" LastActivityDate="2015-04-10T04:25:08.397" Title="How to control suction strength on a breast pump?" Tags="|mechanical-engineering|pumps|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2395" PostTypeId="2" ParentId="2394" CreationDate="2015-04-10T04:25:08.397" Score="3" Body="&lt;p&gt;I don't know how this &lt;strong&gt;is&lt;/strong&gt; done but the following sounds likely and easily achievable.&lt;br&gt;&#xA;It's how I would approach implementing a solution initially if I had no other information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt; IF you can reverse the lead screw then piston position is correlated to suction and you can simply control stroke length. So motor speed affects suction profile and piston position affects pressure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Pump stroke will affect suction.&lt;br&gt;&#xA;If there is no significant volume input into the controlled space then piston position provides the same suction level.  This would allow open-loop control, but results would be uncertain due to variation in enclosed space with different users and pressure sensors are cheap and easy to use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pressure sensor allows actual suction to be measured and rate and magnitude to be controlled. Simple feedback should allow full pressure/time profile to be controlled easily. Ideally the desired pressure time / pressure profile is generated in software and pump speed and direction adjusted to suit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably a simple enough requirement that very basic hardware could be used. A &quot;ramp&quot; signal is generated - maybe a triangle wave or perhaps modified to a shape that suits. Magnitude of waveform = desired pressure.&lt;br&gt;&#xA;Adjusting the analog magnitude could use a user controlled potentiometer or digitally controlled equivalent. Pressure is adjusted to track this with desired:actual pressure controlling lead screw positioning. Oscillator speed adjusts rate.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single cheap quad opamp (eg LM324) would suffice.&lt;br&gt;&#xA;Or for the very keen even a hex package of schmitt inverters (eg 74xx14).&lt;br&gt;&#xA;But a microcontroller solution is liable to be as low cost all up and more flexible in results.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-04-10T04:25:08.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2396" PostTypeId="2" ParentId="2390" CreationDate="2015-04-10T05:15:30.967" Score="3" Body="&lt;p&gt;As I understand it, you would like to develop an equation for the flow rate on the exit side of the HEPA filter based on the static pressure the fan produces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for the pressure versus flow rate fan curve will be unique to every fan. Influencing factors will be whether the fan is axial or centrifugal and for centrifugal fans, whether the fan blades are forward curving, backward curving or radial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other factors that will influence the exit flow rate from the HEPA filter will be associated with the design of your system:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;shock losses – basically caused by the abruptness of changes in air&#xA;direction &amp;amp; the lack of streamlining&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;friction from materials in the air pathway due to material roughness&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;back pressure, if it occurs&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;the resistance to airflow caused by the filter &amp;amp; anything else&#xA;blocking the path of the air flow&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To get a better answer you will need to include a diagram of your system, with dimensions.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-04-10T12:35:03.850" LastActivityDate="2015-04-10T12:35:03.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2397" PostTypeId="2" ParentId="2383" CreationDate="2015-04-10T08:26:31.710" Score="7" Body="&lt;p&gt;I do this type of things quite a lot at work, and yes they are typically syringe-actuated device, where you either apply a pneumatic pressure inside the syringe barrel for a certain amount of time (such as this &lt;a href=&quot;http://www.loctite.co.uk/loctite-4087.htm?nodeid=8802622210049&quot;&gt;glue dispensing machine&lt;/a&gt;) or move the piston inside the syringe by a certain amount at a certain speed, using some kind of electrical drive (stepper motor + gearing) and a lead screw type mechanism (like this &lt;a href=&quot;http://www.syringepump.com/NE-1000.php&quot;&gt;syringe pump&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, there is no feedback mechanism, they operate open-loop, but are calibrated beforehand, by dispensing onto a balance and using gravimetry to work out the volume dispensed per unit of movement of the plunger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you say, you have to make sure, the system is primed before use so you would typically dispense a little bit to waste before using it for its designed purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge with viscous liquids (and therefore pastes and assorted goos), is you have to control the speed of the plunger quite carefully, otherwise you may over-dispense, and you may cavitate the pump if you refill too quickly. You would typically dispense and refill quite slowly in these cases.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-04-10T08:26:31.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2398" PostTypeId="1" AcceptedAnswerId="2400" CreationDate="2015-04-10T13:47:48.143" Score="8" ViewCount="1933" Body="&lt;p&gt;I've been studying the design of turbopumps and their impellers, all the design equations and what-not; this question concerns the contours of the impeller vanes. I'm going to attempt to enter aerospace engineering in the future, so I figured I might as well start getting exposure to the concepts now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently attempting to design an impeller in CAD to test my knowledge of the equations as well as get better at CAD, but one of the facets of the design rather baffles me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see here (&lt;a href=&quot;http://my.fit.edu/~dkirk/4262/Liquid%20Rocket%20Engine%20Design.pdf&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;, page 208):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NLDxC.png&quot; alt=&quot;Impeller radial contour&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation of the radial contour of the impeller (that is, concerning $\beta_1$ and $\beta_2$ should be relatively easy given the other content in the paper, but radial impellers in turbochargers and other such equipment almost always appear to have a &quot;lip&quot; or forward facing top edge near the eye of the volute or casing inlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of what I'm referring to:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iCYFO.jpg&quot; alt=&quot;CAD-modeled impeller&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can tell that the vanes are backwards-curved, but near the top of the impeller vane closer to the eye, the curvature reverses and flares forward to approach near-flat slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the purpose of this axial contour (forgive me if that isn't the proper term)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there specific design parameters associated with that contour? The Rocketdyne text above didn't seem to make any mention of the axial profiles of the impeller, other than the diagram to the left describing the dimensioning of the shrouds.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Where can I find an explanation and source of design information for how to design the contour? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1229" LastEditorUserId="272" LastEditDate="2015-04-10T14:06:28.680" LastActivityDate="2018-10-21T23:01:19.877" Title="Contours of radial-flow impeller vane" Tags="|mechanical-engineering|fluid-dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2400" PostTypeId="2" ParentId="2398" CreationDate="2015-04-10T15:18:12.787" Score="6" Body="&lt;p&gt;I should have a little more overview than I have of pumps but I don't. What you refer to seems to be called an &lt;strong&gt;entrance vane&lt;/strong&gt; or an &lt;strong&gt;impeller guide vane&lt;/strong&gt;. The purpose is to help the water gain the required rotational velocity:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... in connection with a centrifugal impeller it is desirable to provide an &lt;strong&gt;axially extending entrance-vane section having vanes curved such that their radius of curvature gradually decreases in the direction of the fluid flow there through&lt;/strong&gt;. &lt;a href=&quot;http://www.google.com/patents/US2384265&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; [Patent, 1945]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The water approaches the impeller with a flow direction along the axis. Without curving it inwards, the water would have a very abrupt change from going straight to moving with the blade, with a considerable waste of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, by curving the blades inwards, you 'scoop' the water up. The water will be pushed along the the slope of the vain and will help turn the axial motion that it had to rotational, increasing the pump efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This design is not easy to machine and hence only used where necessary, like rocket engines where you need a good deal of pressure. Most applications are just not worth the trouble.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;With this construction, the entrance vane section comprises a hub portion having a plurality of vanes extending radially therefrom, with each of the vanes having a rather complicated twisted profile. Obviously, such an entrance vane section is difficult to cast and/or machine. Accordingly, it is an object of this invention to provide a novel construction of such an impeller to facilitate its fabrication.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="782" LastEditorUserId="77" LastEditDate="2018-10-21T23:01:19.877" LastActivityDate="2018-10-21T23:01:19.877" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="2401" PostTypeId="2" ParentId="2382" CreationDate="2015-04-10T16:15:54.760" Score="14" Body="&lt;p&gt;Every time you bend aluminium below the recrystallisation temperature, the macroscopic grains become smaller: this is known as cold work. The sides of the tab are either stretched or compressed. The effect is the same as the rolling example below. This is indeed a plastic deformation: the tab stays in place and doesn't bend back and stays in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FMECN.png&quot; alt=&quot;cold work&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.boredofstudies.org/wiki/Personal_and_Public_Transport&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you've just done is to make it harder for those grains to slip over each other, making the material harder and increasing it strength. You have however also made it a lot less ductile (and more brittle). If you bend a spoon, it's hard to bend it back to the correct shape. This is because that area has been cold worked and is stiffer than the surrounding metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mq4D0.jpg&quot; alt=&quot;graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://practicalmaintenance.net/?p=1135&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same happens with your aluminium tab, except that there comes a point where it snaps. You can think of it like the beam below: You're bending it with a fixed deflection that corresponds to a certain angle θ. At some point as your ductility drops, you're going to go past the strain limit and it will snap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NoJS7.jpg&quot; alt=&quot;bending&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.doitpoms.ac.uk/tlplib/beam_bending/bend_moments.php&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it might appear to be fatigue, from an engineering perspective is really not the correct term to use. Fatigue is used to describe problems typically appearing after several thousand load cycles.&lt;/p&gt;&#xA;" OwnerUserId="782" LastEditorUserId="782" LastEditDate="2015-04-10T21:05:18.863" LastActivityDate="2015-04-10T21:05:18.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2403" PostTypeId="1" CreationDate="2015-04-10T20:30:53.960" Score="5" ViewCount="112" Body="&lt;p&gt;I am thinking about building small plane with a pulsejet but the engine would need to be roughly 2 cm in diameter. Is it even possible to fit something to store gas on that sized plane? At least for like 10 seconds of running.&lt;/p&gt;&#xA;" OwnerUserId="1236" LastEditorUserId="41" LastEditDate="2015-04-10T21:22:17.680" LastActivityDate="2018-08-06T10:00:56.540" Title="How to store gas on little pulsejet plane?" Tags="|aircraft-design|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2404" PostTypeId="2" ParentId="2237" CreationDate="2015-04-10T21:17:39.643" Score="1" Body="&lt;p&gt;The boundary conditions and modeling technique will greatly influence the results. What I suggest is running a simplified/idealized version (like planar, or axisymmetric) and a full solid one and comparing the two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with two different FEA software is that under the hood the solvers are going to be largely the same. The differences observed would be from different convergence criteria maybe, or from different assumptions on how boundary conditions are applied. You are not checking the model, but the ability of each solver to know it has reached a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think FEA results should be validated by first common sense and hand calculations, then by similar but different models (solids instead of beams for example) and finally by physical testing to see if parts fail &lt;em&gt;where&lt;/em&gt; and &lt;em&gt;how&lt;/em&gt; the FEA predicts. The &lt;em&gt;when&lt;/em&gt; a part will fail is more difficult because it is infuenced by manufacturing processes, material variations and resudial stresses.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2015-04-10T21:17:39.643" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2405" PostTypeId="1" CreationDate="2015-04-11T07:36:57.233" Score="13" ViewCount="14127" Body="&lt;ol&gt;&#xA;&lt;li&gt;Here is a typical definition of air pressure:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Air pressure is caused by the weight of the air molecules above. Even tiny air molecules have some weight, and the huge numbers of air molecules that make up the layers of our atmosphere collectively have a great deal of weight, which presses down on whatever is below.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;And yet, all sources I've seen state that air pressure is equal in all directions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;1 &amp;amp; 2 seem contradictory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does air pressure from above not crush us? The answer I see consistently given is that an equal air pressure from below balances it out. But if a car were resting on me from above and crushing me, then another car pressing against me from below would not relieve that pressure -- it would only increase the pressure I would feel! If I were in an enclosed closet, and one of the walls were to press in against me, and then the opposite wall would also press in against me, the second wall would not &quot;balance&quot; things out, but rather only increase the pressure I would feel!&lt;/p&gt;&#xA;" OwnerUserId="1238" LastEditorUserId="282" LastEditDate="2015-04-11T17:55:13.843" LastActivityDate="2021-07-19T18:36:41.130" Title="Why is air pressure in all directions?" Tags="|mechanical-engineering|pressure|" AnswerCount="8" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2406" PostTypeId="2" ParentId="2405" CreationDate="2015-04-11T08:14:44.770" Score="3" Body="&lt;p&gt;&lt;em&gt;I've tried to split up the questions a bit, drop a comment if I've missed something.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Air pressure is caused by the weight of the air molecules above. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is indeed correct.  The air pressure is proportional to how much air is above it: You have less on a high mountain than at sea level. The diagram shows this in practise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Boss3.png&quot; alt=&quot;atmosphere pressure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Air pressure is equal in all directions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is also true: It will push equally in all directions. If it would be unequal, it would try to reach equilibrium. The air molecules will be subject to both the force of gravity pulling it towards earth (compressing it) and the force of other molecules, pushing it away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rT75v.gif&quot; alt=&quot;Sour![][2]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://electron6.phys.utk.edu/101/CH5/buoyancy.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And yet, all sources I've seen state that air pressure is equal in all directions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For some little point in the atmosphere, this would be true. There would be equal force acting on it in all directions.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1 &amp;amp; 2 seem contradictory.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a very small difference for, say, a small cubed container since the bottom will have a tiny bit higher pressure from the air above it than the top side and the pressure will be marginally higher. However that decrease in pressure with altitude will take place both inside and outside the box. In general, the pressure difference can be ignored for almost all applications.&lt;/p&gt;&#xA;" OwnerUserId="782" LastEditorUserId="782" LastEditDate="2015-04-11T16:30:53.493" LastActivityDate="2015-04-11T16:30:53.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2407" PostTypeId="2" ParentId="2405" CreationDate="2015-04-11T10:25:10.443" Score="2" Body="&lt;p&gt;Pressure is given by the formula,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${P = {\rho}gh}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${\rho}$&lt;/span&gt; = density&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${g}$&lt;/span&gt; = gravitational constant&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${h}$&lt;/span&gt; = height/depth&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Pressure at any &lt;strong&gt;point&lt;/strong&gt; below the upper boundary of fluids, such as air and water, is uniform in all directions due to the fluid molecules being in constant motion and continually bumping into one another. Pressure increases with the depth of the fluid due to the amount of fluid above it, but any point on a horizontal plane will have the same pressure.&lt;/p&gt;&#xA;&lt;p&gt;Compare this to rock in the Earth's crust and mantle. Ignoring tectonic stresses, the pressure in the vertical direction is still given by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${P = {\rho}gh}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, because of the solid nature of rock, molecules are not rapidly moving and they do not continually bump into one another. Consequently, pressure in the lateral direction is not equal to the pressure in the vertical direction and pressure/stress in rock is not uniform in all directions.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.rocscience.com/assets/resources/learning/hoek/Practical-Rock-Engineering-Chapter-10-In-Situ-and-Induced-Stresses.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This source&lt;/a&gt; gives the lateral pressure/stress as being related to the vertical pressure/stress.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${{\sigma}_h = k{\sigma}_v  = k{\rho}gh}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="-1" LastEditDate="2021-07-19T16:25:01.460" LastActivityDate="2021-07-19T16:25:01.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2408" PostTypeId="2" ParentId="2405" CreationDate="2015-04-11T13:50:39.293" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is air pressure equal in all directions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Imagine what it would mean for a thin flat piece of metal if the pressure were not equal from above and below.  There would be more pressure pushing down from the top than pushing up from below which would equate to a net force.  This force would start to accelerate the piece of metal downwards; there would be no equilibrium.  Now forget about the piece of metal.  Without it there would be air molecules rushing down from the pressure gradient.  They would actually rush down until they equalized the pressure gradient and stopped moving.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why does air pressure from above not crush us? The answer I see consistently given is that an equal air pressure from below balances it out.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This isn't quite correct.  The pressure is not simply equal from above and below with your body being a zone of different pressure.  Rather, your entire body is at the same pressure as the surroundings.  To understand the difference, think about a tank from which some of the air can be evacuated (a vacuum tank).  When the tank is full of air at equal pressure to the surroundings the lid can be removed easily.  If you seal the container, pump some of the air out, and then try to remove the lid you will find that it is very tightly stuck.  This is because there is a strong force on the lid caused by the pressure gradient between inside and outside.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that your body is at atmospheric pressure is actually very important to the way it functions.  If you were to be thrown out of a spaceship where the pressure is near zero, all of the gases (oxygen being an important one) would evaporate out of the fluids in your body.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-04-11T13:50:39.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2409" PostTypeId="1" AcceptedAnswerId="2410" CreationDate="2015-04-11T16:55:05.160" Score="5" ViewCount="1643" Body="&lt;p&gt;I want to pull some weight horizontally and for that i have settled for a linear actuator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sao5w.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actuator should pull some weight as it moves back and forth as the gif shows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are linear actuators i could find but they are too bulky with heavy motors and a lot of lubrication and the noise they make is not music to my ears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found electrical actuators to be a perfect fit but i do not know how they work for instance this actuator: &lt;a href=&quot;http://www.kollmorgen.com/en-us/products/linear-actuators/rodless-actuators/r4-series/&quot; rel=&quot;nofollow noreferrer&quot;&gt;R4 Series&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1243" LastEditorUserId="421" LastEditDate="2016-01-16T19:10:26.370" LastActivityDate="2017-11-27T12:35:16.377" Title="How does a rodless electrical linear actuator work?" Tags="|mechanical-engineering|linear-motors|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2410" PostTypeId="2" ParentId="2409" CreationDate="2015-04-11T17:04:35.857" Score="7" Body="&lt;p&gt;I found these images. Looks like they are driven by a motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Belt Driven Model&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KWSao.png&quot; alt=&quot;Belt Driven Model&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Screw Driven Model&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/99O5N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest reviewing &lt;a href=&quot;http://www.kollmorgen.com/en-us/products/linear-actuators/rodless-actuators/r4-series/_literature/r3-r4_drive-shaft-installation_en-us_reva/&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; document&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="282" LastEditDate="2015-04-11T17:11:06.453" LastActivityDate="2015-04-11T17:11:06.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2411" PostTypeId="2" ParentId="2409" CreationDate="2015-04-11T17:59:00.090" Score="2" Body="&lt;p&gt;For a minimum of moving parts, you can get linear stepper motors. With the right kind of controller (microstepping), the motion can be very smooth and quiet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have in the past worked with linear actuators from &lt;a href=&quot;http://www.nipponpulse.com/&quot; rel=&quot;nofollow&quot;&gt;Nippon Pulse&lt;/a&gt;, driven by a motor controller from &lt;a href=&quot;http://www.technosoftmotion.com/en/&quot; rel=&quot;nofollow&quot;&gt;Technosoft&lt;/a&gt;. Very precise and powerful, but also very expensive!&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-04-11T17:59:00.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2412" PostTypeId="2" ParentId="2349" CreationDate="2015-04-11T20:41:03.290" Score="-1" Body="&lt;p&gt;To summarize, engine efficiency is determined, in descending order, by:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Thermodynamics:&lt;/strong&gt; &lt;em&gt;Second Law of Thermodynamics (Heat Engines)&lt;/em&gt;. It is     impossible to extract an amount of heat QH from a hot reservoir and use it all to do work W . Some amount of heat QC must be exhausted to a cold reservoir. This precludes a perfect heat engine.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Energy content of fuel:&lt;/strong&gt;&#xA;&lt;em&gt;The specific energy content of a fuel&lt;/em&gt; is the heat energy obtained when a  certain quantity is burned (such as a gallon, litre, kilogram). It is sometimes called the heat of combustion. Its theoretical value is determined from the &lt;em&gt;Gibbs free energy&lt;/em&gt; which is a thermodynamic potential that measures the &quot;usefulness&quot; or process-initiating work obtainable from a thermodynamic system at a constant temperature and pressure (isothermal, isobaric)(in function of the specific engine).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Specific internal combustion engine efficiency:&lt;/strong&gt; efficiency of a modern internal combustion engines is determined mainly by &lt;em&gt;air-fuel ratio&lt;/em&gt;, &lt;em&gt;compression ratio&lt;/em&gt;, &lt;em&gt;valve control systems&lt;/em&gt;, &lt;em&gt;engine temperature management&lt;/em&gt;,   mostly taken care of by closed loop control systems consisted of sensors (air volume,temperature, lambda sensors) connected to microcomputer. These together with the , &lt;em&gt;gear box&lt;/em&gt; , &lt;strong&gt;determine optimal engine performance&lt;/strong&gt; at specific &lt;em&gt;throttle position&lt;/em&gt;, &lt;em&gt;car load&lt;/em&gt; and &lt;em&gt;inclination&lt;/em&gt;.(&lt;em&gt;drag coefficient&lt;/em&gt; and &lt;em&gt;tires size&lt;/em&gt; should be considered as well.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="983" LastActivityDate="2015-04-11T20:41:03.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2413" PostTypeId="1" CreationDate="2015-04-11T21:16:04.727" Score="4" ViewCount="603" Body="&lt;p&gt;I'm building a small hobby CNC machine out of aluminium with the end goal of milling plastic, aluminium and possibly some harder stuff like steel (although if you saw the machine you'd laugh me out of the room).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine isn't as rigid as I would've liked in it's current incarnation. My solution to this would be to use a carbide endmill (~6-8mm or so) at high RPM (~8k) so the tool pressure is reduced and the machine doesn't flex. I've learned that HSS cutters will dull if the chip load isn't high enough, hence the use of carbide as it's a lot harder. I've tried researching this but all I can find are calculators and charts for chip loads for different brands of end mills. My spindle is 3-phase and will happily spin up to 24k RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I use a carbide cutter at high speed to reduce chip load? Do I risk breaking the cutter by doing this?&lt;/p&gt;&#xA;" OwnerUserId="1246" LastEditorUserId="41" LastEditDate="2015-04-16T17:41:06.980" LastActivityDate="2021-03-02T22:16:36.037" Title="High spindle speed, low tool pressure" Tags="|mechanical-engineering|machining|cnc|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2414" PostTypeId="2" ParentId="2044" CreationDate="2015-04-11T21:32:50.850" Score="3" Body="&lt;p&gt;Factors determining fan airflow (heat reduction): &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Blade pitch.&lt;/li&gt;&#xA;&lt;li&gt;Blade shape and size.&lt;/li&gt;&#xA;&lt;li&gt;Motor power and RPM.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In your case 3 is the same, 1 and 2 in the &quot;newer version are bigger&quot;, so it will cool better.&#xA;Main source of noise in fans is coming from the bearing  and the electromotor (same in both cases) and from the blades which are slightly larger. Considering that our hearing works on logarithmic scale, I highly doubt that the increase of blades's size in the second model will double the noise, i.e. make it possible to discern from the first. Based on that, draw your conclusion :).&lt;/p&gt;&#xA;" OwnerUserId="983" LastActivityDate="2015-04-11T21:32:50.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2415" PostTypeId="2" ParentId="2405" CreationDate="2015-04-11T21:56:49.523" Score="4" Body="&lt;p&gt;Air pressure is exerted on the surface of a body by air molecules hitting the surface and being reflected. Each of these reflections (gazillions of which happen per second) transfers a little impulse on the surface, which macroscopically means a permanent force (per unit of area). Why do the air molecules bounce and hit all the time? Either because the air is moving at large (aka. &quot;wind&quot;), or because they bounce around irregularly (aka. &quot;temperature&quot;). The latter kind of movement knows no preferred direction and therefore the pressure is the same no matter what orientation the test surface has. The very fact that there is no net movement (wind) is expressed by the fact that the same force acts on the back side of a thin surface as on the front side (so there is no net force).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then how come the air pressure is related to the weight of air above us? In &lt;em&gt;equilibrum&lt;/em&gt; the force caused by air pressure from below on an imaginary horizontal surface is just enough to keep the air column above it &quot;in place&quot;, whoich means that it equals the weight. We need not always have equilibrum, but if we don't then the stringer of the forces causes accelaration and movement - until equilibrum is reached.&lt;/p&gt;&#xA;" OwnerUserId="1249" LastActivityDate="2015-04-11T21:56:49.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2419" PostTypeId="1" AcceptedAnswerId="2421" CreationDate="2015-04-12T05:46:26.693" Score="6" ViewCount="4825" Body="&lt;p&gt;I understand how a circle (or sphere if you want 3D) is the best shape for holding a vacuum inside a container, but what if you wanted to have a large positive pressure &lt;em&gt;inside&lt;/em&gt; of the container instead of outside? Would a circle/sphere still be the best shape as far as material required to hold X amount of pressure? &lt;/p&gt;&#xA;" OwnerUserId="1253" LastEditorUserId="368" LastEditDate="2015-04-12T17:06:34.363" LastActivityDate="2016-04-16T13:50:08.100" Title="Is a circle the strongest 2D shape for containing internal pressure?" Tags="|mechanical-engineering|pressure|geometry|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2420" PostTypeId="1" AcceptedAnswerId="2450" CreationDate="2015-04-12T05:59:16.773" Score="5" ViewCount="2978" Body="&lt;p&gt;&lt;strong&gt;When we apply a vertical shear force Sy and the structure is symmetric about x axis, then why is it logical to have the position of shear center at the location of intersection of line of action of shear force and x axis?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we go by basic definition then the moment about the shear center due to shear forces should be &lt;strong&gt;0&lt;/strong&gt;, so the line of action of external shear force is logical, but when we consider moment about x- axis, rather than getting cancelled, &lt;strong&gt;I &lt;em&gt;think&lt;/em&gt; they get added&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;eg. This is the situation.&lt;/strong&gt;&#xA; &lt;img src=&quot;https://i.stack.imgur.com/s4xHw.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now look at the final solution depicting the shear flows&lt;/strong&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/1P7mO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;So as I suggested, instead of moments being equal and opposite, I think they exactly equal in magnitude and direction. So where am I getting a wrong interpretation?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:-&lt;/strong&gt; I tried to keep the question general and explained accordingly, to be specific the question is as follows:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The thin-walled single cell beam shown in Fig. 20.11 has been idealized into a combination&#xA;of direct stress carrying booms and shear stress only carrying walls. If the&#xA;section supports a vertical shear load of 10 kN acting in a vertical plane through booms&#xA;3 and 6, calculate the distribution of shear flow around the section.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Boom areas: B1 =B8 =200mm2, B2 =B7 =250mm2, B3 =B6 =400mm2, B4 =&#xA;B5 =100mm2.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x is horizontal and y is vertical&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It is a closed section beam. An idealized version of an airplane wing&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1155" LastEditorUserId="33" LastEditDate="2016-01-26T03:33:59.000" LastActivityDate="2019-05-22T19:51:04.043" Title="Logic behind location of shear centre" Tags="|mechanical-engineering|structural-engineering|aerospace-engineering|structures|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2421" PostTypeId="2" ParentId="2419" CreationDate="2015-04-12T07:42:38.230" Score="15" Body="&lt;p&gt;Despite being difficult to manufacture, a sphere is the best shape for a  &lt;a href=&quot;https://en.wikipedia.org/wiki/Pressure_vessel&quot; rel=&quot;nofollow noreferrer&quot;&gt;pressure vessel&lt;/a&gt;, but due to the manufacturing difficulties are more costly to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;https://www.asme.org/engineering-topics/articles/pressure-vessels/pressure-vessels-ensure-safety&quot; rel=&quot;nofollow noreferrer&quot;&gt;American Society of Mechanical Engineers website&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Theoretically, a spherical pressure vessel has approximately twice the strength of a cylindrical pressure vessel with the same wall thickness&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8KjRw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Other &lt;a href=&quot;http://www.wermac.org/equipment/pressurevessel.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;sources&lt;/a&gt; agree with this, adding:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Spherical pressure vessels is preferred for storage of high pressure fluids. A sphere is a very strong structure. The even distribution of stresses on the sphere's surfaces, both internally and externally, generally means that there are no weak points. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-04-12T07:51:18.197" LastActivityDate="2015-04-12T07:51:18.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2422" PostTypeId="2" ParentId="2419" CreationDate="2015-04-12T16:06:56.700" Score="2" Body="&lt;p&gt;Due to the fact that the pressure differential is the same everywhere around the surface of a sphere, it must be the strongest, therefore hold the greatest pressure, and withhold the greatest pressure, too. A circle won't hold pressure, as it is only 2 dimensional, and everything would escape!&lt;/p&gt;&#xA;" OwnerUserId="1256" LastEditorUserId="110" LastEditDate="2015-04-12T16:19:45.100" LastActivityDate="2015-04-12T16:19:45.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2423" PostTypeId="1" AcceptedAnswerId="2487" CreationDate="2015-04-12T18:09:56.273" Score="18" ViewCount="805" Body="&lt;p&gt;I have been searching for this answer for awhile. I've read numerous texts and even watched some lectures online, but often times this is never explained and just given. The viscous stress term in the Navier-Stokes equations looks like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\nabla \cdot \tau = \nabla \cdot \mu \left(\nabla\vec{u} + (\nabla\vec{u})^T\right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the term $\nabla \cdot \mu \nabla\vec{u}$ is easy enough to understand as it is just velocity diffusion, but I have a hard time coming up with a physical interpretation of the term $\nabla \cdot \mu (\nabla\vec{u})^T$. After I expanded this term I ended up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\nabla \cdot \mu (\nabla\vec{u})^T =&#xA;\begin{pmatrix}&#xA;\frac{\partial}{\partial x} \nabla \cdot \vec{u} \\&#xA;\frac{\partial}{\partial y} \nabla \cdot \vec{u} \\&#xA;\frac{\partial}{\partial z} \nabla \cdot \vec{u} &#xA;\end{pmatrix}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which seems to imply that this effect is not present in a divergence-free velocity field, but I still can't come up with or find any physical intuition about what this term actually means. Does anyone understand what this term physically represents?&lt;/p&gt;&#xA;" OwnerUserId="1258" LastEditorUserId="282" LastEditDate="2015-04-16T17:39:15.623" LastActivityDate="2015-04-20T00:04:09.090" Title="What is the physical interpretation of the second term in the viscous stress tensor in the Navier-Stokes equations?" Tags="|mechanical-engineering|fluid-dynamics|fluid-mechanics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2424" PostTypeId="1" CreationDate="2015-04-13T00:56:55.240" Score="2" ViewCount="1774" Body="&lt;p&gt;Those sitting in front and mid part doesn't experience much shock, while those those sitting in back does. Why?&lt;/p&gt;&#xA;" OwnerUserId="1189" LastActivityDate="2015-04-13T02:03:44.127" Title="Why people experience more jerk or shock sitting in the back seats of a bus?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2425" PostTypeId="2" ParentId="2424" CreationDate="2015-04-13T02:03:44.127" Score="6" Body="&lt;p&gt;Two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The suspension in the rear of the bus is firmer because it has to carry more load&lt;/li&gt;&#xA;&lt;li&gt;The area behind the wheels of a bus acts as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lever#Classes_of_levers&quot;&gt;third-class lever&lt;/a&gt;, amplifying the movement due to roadway bumps.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-04-13T02:03:44.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2426" PostTypeId="2" ParentId="2305" CreationDate="2015-04-13T02:26:25.563" Score="0" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: I still think that I'm on the right track, but I have realized some errors that I made (specifically I have drawn my diagram with the wrong area shaded for work, I labeled the diagram h instead of T, and I skimmed over the specific volume ratio of the compressor) so until I get this straightened out, consider the answer as tentative at best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand, it works as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;By increasing the fan speed, the heat rejected by the evaporator is increased&lt;/li&gt;&#xA;&lt;li&gt;The increased heat rejected can only be accomplished by &quot;widening&quot; the points on the evaporation line&lt;/li&gt;&#xA;&lt;li&gt;The increased quality at the output of the evaporator causes the compressor to compress to a lower pressure.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That may not be right, but it seems to make sense on the diagram I drew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nTJEB.jpg&quot; alt=&quot;Vapor-compression refrigeration cycle&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="709" LastEditorUserId="709" LastEditDate="2015-04-14T05:03:57.883" LastActivityDate="2015-04-14T05:03:57.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2427" PostTypeId="1" AcceptedAnswerId="2456" CreationDate="2015-04-13T02:50:23.227" Score="3" ViewCount="345" Body="&lt;p&gt;Consider a balloon, it is a flexible, expanding pressure vessel made of synthetic rubber. I am trying to design a system that will use hemispheric pockets of neoprene to grip an object, much like the one found in this journal:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;H.  Choi and M.  Koç, 'Design and feasibility tests of a flexible gripper based on inflatable rubber pockets', International Journal of Machine Tools and Manufacture, vol. 46, no. 12-13, pp. 1350-1361, 2006. (Behind a paywall, abstract available &lt;a href=&quot;http://libra.msra.cn/Publication/6847499/design-and-feasibility-tests-of-a-flexible-gripper-based-on-inflatable-rubber-pockets&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Although I am not entirely certain where to start, I am dealing with a high pressure (8 bar) and need to work on a small scale (max radius of 40mm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Values I have recieved from equations found online suggests thickness values in meters and radius expansion reaching the hundrends of meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What equations do I need to calculate a solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are neoprene, latex and chloroprene viable materials to use or should I find alternatives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a hemisphere that best pocket shape I can use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, how can I calculate the amount of expansion?&lt;/p&gt;&#xA;" OwnerUserId="340" LastEditorDisplayName="user16" LastEditDate="2015-04-13T12:32:00.360" LastActivityDate="2015-04-14T10:54:01.260" Title="How to design a hemispheric flexible pressure vessel?" Tags="|pressure|thermodynamics|compressed-air|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2428" PostTypeId="1" CreationDate="2015-04-13T05:47:22.893" Score="4" ViewCount="540" Body="&lt;p&gt;Zinc-coated roofing steel loses its zinc coat after about ten years because of being exposed to elements and it has to be painted. Zinc-coated powerline posts lose zinc too and experience corrosion, however they are built of thick parts which are &quot;eaten up&quot; slowly and once enough cross-section is lost they are replaced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I've recently seen a culvert assembled from pieces of corrugated steel with coating looking like zinc. Parts were something like 5 millimeters thick and connected with numerous bolts and the culvert diameter was about 1,5 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water passing through a culvert bring sand and other particles which surely grind the surface. I assume this should remove the coating from the submerged part of the culvert within several years and then corrosion will start and eat the not so thick steel rather quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is abrasion by solid particles addressed in culverts made of galvanized steel?&lt;/p&gt;&#xA;" OwnerUserId="374" LastActivityDate="2015-04-13T17:29:49.317" Title="How does galvanised steel culvert withstand abrasion from solid particles in the stream?" Tags="|civil-engineering|steel|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2429" PostTypeId="1" AcceptedAnswerId="2430" CreationDate="2015-04-13T06:33:37.157" Score="7" ViewCount="487" Body="&lt;p&gt;I am trying to find some information regarding the design of plastic hinges that can move freely but also hold a position in any degree, like in this little camera: &lt;a href=&quot;https://www.kickstarter.com/projects/podolabs/podo-the-first-stick-and-shoot-camera&quot;&gt;https://www.kickstarter.com/projects/podolabs/podo-the-first-stick-and-shoot-camera&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the considerations of design and if there's any kind of reference.&lt;/p&gt;&#xA;" OwnerUserId="1262" LastActivityDate="2015-04-15T23:57:25.623" Title="Information regarding plastic hinges that can hold the position" Tags="|mechanical-engineering|plastic|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2430" PostTypeId="2" ParentId="2429" CreationDate="2015-04-13T10:28:52.063" Score="4" Body="&lt;p&gt;A hinge bending with the material rather than mechanically is known as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Living_hinge&quot; rel=&quot;nofollow&quot;&gt;living hinge&lt;/a&gt;. Consideration are placed upon the number times it will be bent during the lifetime and what sort of forces it will experience. However, cannot see how this could this could be kept in place without some additional reinforcement; I am pretty sure there is a sheet piece of metal in there to keep it in shape.&lt;/p&gt;&#xA;" OwnerUserId="782" LastActivityDate="2015-04-13T10:28:52.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2431" PostTypeId="2" ParentId="2428" CreationDate="2015-04-13T10:36:41.343" Score="1" Body="&lt;p&gt;Are you sure it wasn't aluminium? The differences when searching for images is not that big. Another option is that it has been coated with some polymer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aluminum pipes are also corrugated and can be constructed as either&#xA;  structural plates or from a single piece of aluminum. Due to their&#xA;  high resistance to corrosion aluminum pipes do not develop rustlines&#xA;  and are often used in high corrosion environments such as saltwater&#xA;  applications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="782" LastActivityDate="2015-04-13T10:36:41.343" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2433" PostTypeId="1" AcceptedAnswerId="2457" CreationDate="2015-04-13T13:19:19.540" Score="4" ViewCount="319" Body="&lt;p&gt;I have to draw two custom pipes, 1 1/4&quot; &amp;amp; 2&quot;, with on one end male tapered pipe thread and on the other end parallel pipe thread. The connections are Parallel female thread with a tapered male thread, with Teflon tape as sealant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm familiar with the metric tread, but the British/ Whitworth thread system is new for me. After a search on the internet and the standards: DIN 3858 / DIN ISO 228, i have the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;According to DIN 3858, tapered thread largest available size is 1 1/2&quot;. Why are there couplings on the market that are larger? In my case 2&quot;.&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is the designation of &quot;Pipe thread DIN 3858 - R 2&quot; correct?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;What is the minimal thickness of the pipes when it concerns tapered thread? Are there guidelines for this?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For the 2&quot; connection i selected a OD = 60.3 mm, t = 2 mm. (probably too thin)&lt;/li&gt;&#xA;&lt;li&gt;For the 1 1/4&quot; connection i selected a OD  42.4 mm, t = 2.6 mm.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1077" LastEditorUserId="1077" LastEditDate="2015-04-14T06:29:15.327" LastActivityDate="2015-04-14T09:31:14.503" Title="British tapered pipe thread; Wall thickness? &amp; Thread designation tpd?" Tags="|mechanical-engineering|threads|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2435" PostTypeId="1" AcceptedAnswerId="2436" CreationDate="2015-04-13T14:49:51.510" Score="3" ViewCount="147" Body="&lt;p&gt;Are there any industry standards (&lt;a href=&quot;http://en.wikipedia.org/wiki/Specification_(technical_standard)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineering Specifications&lt;/a&gt;) for the minimum / maximum space required for the electrical box that holds a regular AC wall plug?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dr0USs.jpg&quot; alt=&quot;Type B grounded duplex outlet&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The problem I am trying to solve is quite simple - are / is there any local (USA, Canada, or EU) industry standards that require, or set, dimensions in the Height, Width, and Depth (in inches or centimeters) of the space required to be used by the electrical box for an AC wall plug as shown below:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0SVhHm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1157" LastEditorDisplayName="user16" LastEditDate="2015-04-13T15:52:43.657" LastActivityDate="2015-04-13T16:49:50.020" Title="Industry standard for AC wall electrical box space" Tags="|electrical-engineering|standards|specifications|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2436" PostTypeId="2" ParentId="2435" CreationDate="2015-04-13T16:49:50.020" Score="3" Body="&lt;p&gt;The normative reference for the design of wall boxes for electrical work in the US is NEMA OS-1 for metal boxes and NEMA OS-2 for nonmetallic boxes. This, in turn limits the size that components to go into these boxes (including outlets) may be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want to buy the standard, you can find a good bit of information from it posted from various manufacturers, &lt;a href=&quot;http://www.emersonindustrial.com/en-US/documentcenter/EGSElectricalGroup/products_documents/commercial_products/commercial_box/switch_outlet_box_covers/switch_outlet_box_covers/AEC_MC007_Switch_and_Outlet_Boxes_and_Covers.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; is one example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page A3 of &lt;a href=&quot;http://www.hubbell-rtb.com/literature_pdf/Hubbell-RTB_Catalog.pdf&quot; rel=&quot;nofollow&quot;&gt;this document&lt;/a&gt; explains the limitations on how full the boxes can be, as dictated by NEC 314.16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, there are many different sizes and configurations of these boxes, but they have fairly standard front-facing features so that various outlets, switches, or other devices can attach uniformly. When you buy an industrial outlet, the cut sheet will usually tell you how deep of a box it requires, since that's the most variable dimension between boxes.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-04-13T16:49:50.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2437" PostTypeId="1" AcceptedAnswerId="2463" CreationDate="2015-04-13T16:57:14.100" Score="4" ViewCount="10592" Body="&lt;p&gt;Many states require that a PE applicant have some number of years of &quot;progressive engineering experience&quot;. Are there details as to what qualifies? For example, if an engineer is working in sales and not design, I would tend to think that would not count. But that's just my impression. Are there defined rules? General guidelines or standards of practice?&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2015-04-30T01:30:58.610" Title="What constitutes progressive engineering experience?" Tags="|licensure|" AnswerCount="4" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2438" PostTypeId="2" ParentId="2428" CreationDate="2015-04-13T17:29:49.317" Score="5" Body="&lt;p&gt;The design and establishment of a culvert, or any structure, is influenced by the materials available, the environmental conditions that the culvert will withstand, water pH, soil type  and the amount of money available for the project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing lasts for ever and all engineering structures have an anticipated/designed service life, after which, if possible the service life of the structure may be extended by some form of refurbishment or the structure is replaced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By way of example, this &lt;a href=&quot;http://www.corrugatedsteelpipe.org/index.html&quot;&gt;manufacturer&lt;/a&gt; of steel pipes, makes pipes that can be used for culverts and the service life of each the product depends on the different type of coatings applied to the pipes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Zinc – 25 years&lt;/li&gt;&#xA;&lt;li&gt;Aluminium – 75 years&lt;/li&gt;&#xA;&lt;li&gt;Asphalt &amp;amp; polymer laminate – 100 years&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As the website state:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;choice of coatings depends on the soil conditions and other site or application factors&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If galvanised steel is being used in a culvert there is already an expectation that it's service life will be less than if it were constructed from something else such as aluminium or concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cpisupply.com/CPI%20Products/2%20Corrugated%20Metal%20Pipe%20Design%20Guide.pdf&quot;&gt;This site&lt;/a&gt; has a design guideline for culvert design. Point 4 states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Determine which abrasion level most accurately describes the typical storm event (2 year storm). The expected stream velocity and associated abrasion conditions should be based on a typical flow and not a 10 or 50-year design flood&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and Point 5 states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;determine whether the structural gage for the selected material is sufficient for the design service &#xA;  life&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If galvanised steel has been used to construct a culvert it's design life has been determined and accepted and deemed suitable for the application based on factors such as these.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-04-13T17:29:49.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2439" PostTypeId="1" CreationDate="2015-04-13T17:50:54.493" Score="7" ViewCount="5859" Body="&lt;p&gt;I don't have experience in mechanical engineering so I came here to ask for help. Here is sketch of one simple construction I want to realize:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3rhl4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have 5 cm thick wooden plate, 50 cm long and 35 cm wide. I will drill a hole through the plate for 40–50 cm long shaft, which will carry a fixed gear driven by a DC motor to rotate the shaft. The shaft must be able to carry a load of at least 30 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can I &quot;fix&quot; this gear shaft to carry a load without falling down? &lt;a href=&quot;http://cdn3.volusion.com/vyfsn.knvgw/v/vspfiles/photos/am-2392-2.jpg?1420269442&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is one shaft I found&lt;/a&gt;; I don't see way to use it in this situation. How would I mount a gear on this shaft?&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="368" LastEditDate="2015-04-14T22:32:10.990" LastActivityDate="2015-04-14T22:32:10.990" Title="How to mount a load-bearing gear shaft through a wooden plate" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2440" PostTypeId="1" AcceptedAnswerId="2441" CreationDate="2015-04-13T16:27:56.593" Score="3" ViewCount="894" Body="&lt;p&gt;Given the information on swords &lt;a href=&quot;http://quantumchymist.blogspot.co.uk/2014/02/why-arent-more-blades-made-of-titanium.html?m=1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When you need to place a great deal of strength into a thin blade, steel remains the best all around choice. Titanium blades will be simultaneously weaker and softer than similar steel blades.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the material (LDHEA) described &lt;a href=&quot;http://www.tandfonline.com/doi/full/10.1080/21663831.2014.985855#tabModule&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarized here from the Applications section of the Wikipedia article on Scandium, as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An alloy called Al20Li20Mg10Sc20Ti30 has been shown to be as strong as titanium, light as aluminium, and hard as ceramic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In combination with other materials, how could LDHEA be used to make a good shortsword?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps steel for the flat and LDHEA for the edge and point?&lt;/p&gt;&#xA;" OwnerUserId="1215" OwnerDisplayName="Malandy" LastEditorUserId="368" LastEditDate="2015-04-14T19:53:22.883" LastActivityDate="2015-04-14T19:53:22.883" Title="How do I use this &quot; Low-Density, High-Hardness, High-entropy Alloy&quot; to make a good shortsword?" Tags="|materials|metals|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2441" PostTypeId="2" ParentId="2440" CreationDate="2015-04-13T17:24:29.440" Score="5" Body="&lt;p&gt;As a HEMA practitioner and a physicist, I can tell you that different swords need different properties and it is unlikely one material will be the end-all for a sword material. Depending on the sword, you may actually want different parts of the blade to have different properties. &lt;a href=&quot;https://www.google.com/search?q=rapier&amp;amp;oq=rapier&amp;amp;aqs=chrome..69i57j0l5.795j0j7&amp;amp;sourceid=chrome&amp;amp;es_sm=91&amp;amp;ie=UTF-8&amp;amp;safe=high&quot;&gt;Rapiers&lt;/a&gt;, for instance, are much less rigid than hand-and-a-half swords (or &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Longsword&quot;&gt;longsword&lt;/a&gt;&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said earlier, different properties are desired for different parts of the blade. For instance, a high hardness on your edge is great, because you can deliver more cuts before it dulls. However, high hardness in your flat does not really help you, because it usually means brittleness, and your blade &lt;em&gt;will&lt;/em&gt; come into contact with opponents' blades and potentially break or shatter, leaving you more-or-less defenseless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lighter blades are not always better. Sure, a lighter blade can &lt;em&gt;ideally&lt;/em&gt; be moved faster, but the human body can only move so fast and benefits only so much from it. Cutting swords, however, need a certain degree of heft to them to enable their cuts. Trying to cut someone with a rapier, for instance, is usually laughable because it simply lacks the same momentum as a tulwar, katana, longsword, or any other sword. (Rapiers have a tendency to simply bounce off in such attempts to cut.) Also, some weight to your sword makes it harder to deflect, which is another problem rapiers classically encounter when attempting to cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the ideal blade, without regard to form, is made of a layering of both hard and soft metals, as seen in &lt;a href=&quot;http://en.wikipedia.org/wiki/Wootz_steel&quot;&gt;wootz&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Damascus_steel&quot;&gt;damascus&lt;/a&gt; steel. Many blacksmiths, when manipulating material properties for a sword, attempt to get a hard edge and soft flat. This material seems promising, but we need to know more about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How brittle is it? Dr. Carl Koch, the senior author for that paper, &lt;a href=&quot;http://nextbigfuture.com/2014/12/new-alloy-is-as-light-as-aluminum-as.html&quot;&gt;says&lt;/a&gt; &quot;we think it’s tougher – less brittle – than ceramics.&quot; That bodes well for it as a sword material. At the very least, it would make a good material for the edge and points of swords. As noted earlier, the flexibility of this material may eliminate it for certain swords or recommend it for others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lighter material lets your sword be larger, which means you could (in theory) have swords that were formerly limited to &lt;a href=&quot;http://images2.fanpop.com/image/photos/8700000/The-Same-Sword-Brothers-inuyasha-8711240-448-243.jpg&quot;&gt;anime and video games&lt;/a&gt;. You also run into the problem of simply larger size not being what you want. For instance, a &lt;a href=&quot;http://en.wikipedia.org/wiki/Zweih%C3%A4nder&quot;&gt;two-handed&lt;/a&gt; sword is not used in the same way as a &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Messer_(weapon)&quot;&gt;Messer&lt;/a&gt;&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In review&lt;/strong&gt;, it seems like this material is promising, and &lt;em&gt;could be good&lt;/em&gt;, but hardness and strength-to-weight are not the only properties which make a good sword material. Toughness (brittleness) is an important factor which we do not know.&lt;/p&gt;&#xA;" OwnerUserId="11631" OwnerDisplayName="PipperChip" LastActivityDate="2015-04-13T17:24:29.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2442" PostTypeId="2" ParentId="2439" CreationDate="2015-04-13T18:59:33.730" Score="5" Body="&lt;p&gt;The general name for what you're trying to make is a 'thrust bearing.' That means something that allows a shaft to move relative to another part, with the load being axial (down the shaft) instead of radial (perpendicular to the shaft.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many types of thrust bearings, but I'm going to guess that you're looking for a simpler, lower cost solution. I'm just basing that on the fact that your plate is made out of wood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One simple method is to use a device called a shaft collar. &lt;img src=&quot;https://i.stack.imgur.com/a4qWF.jpg&quot; alt=&quot;Shaft Collar&quot;&gt; which is just a round clamp  you can attach to your shaft to keep it from falling through the hole. Shaft collars are made in many different styles, but all operate on the same principle. It should be easy to buy one for your size of shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that if your machine runs very fast, or for more than a few minutes, the shaft collar will slowly drill through your wooden plate. To solve this, you would want to put something between the shaft collar and the wooden plate. The cheapest solution is to use a piece of plastic - UHMW is a fairly good material for this, and is pretty affordable. They also make bronze thrust washers specifically for this purpose, which are surprisingly affordable. For example, look at McMaster Carr &lt;a href=&quot;http://www.mcmaster.com/#5906k513/=wqjylt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Part #5906K513&lt;/a&gt;. Stacking one or even two of these between the collar and the wood would make the bearing last for a pretty long time as long as your machine isn't moving very fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your machine does move very fast, you'll want to look into rolling element thrust bearings (for example, &lt;a href=&quot;http://www.mcmaster.com/#catalog/121/1187/=wqjz5d&quot; rel=&quot;nofollow noreferrer&quot;&gt;McMaster #5909K44&lt;/a&gt;,) which work much better and last longer, but are a bit more expensive. Rolling element bearings are also much less tolerant to abuse (dirt, dust, impact) and some require lubrication. You may also want to look at getting a shoulder machined into your shaft in that case, rather than using a shaft collar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget to check that the wood is strong enough, 30kg of suspended load is nothing to sneeze at.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-04-13T19:29:48.460" LastActivityDate="2015-04-13T19:29:48.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2443" PostTypeId="2" ParentId="2439" CreationDate="2015-04-13T19:04:36.423" Score="2" Body="&lt;p&gt;There are a couple issues here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the shaft you found is a keyed shaft, meant to be used with something with the inverse shape on the mating hole. This isn't necessarily a bad thing, as some gears will have keys or keyways on them, to ensure that the gear doesn't rotate relative to the shaft. The alternative is sizing the bore on the gear and the shaft as a &lt;a href=&quot;http://en.wikipedia.org/wiki/Interference_fit&quot; rel=&quot;nofollow noreferrer&quot;&gt;press fit&lt;/a&gt;, but that will be harder to assemble, and if you don't have the right tool, you risk damaging the shaft, the gear, or both. However, it allows you to use a simple round shaft, which will make the next part easier to deal with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as getting the shaft to not move up and down relative to the plate, you're probably looking at some sort of flange and some bearings. If you want the shaft to rotate relative to the plate, you'll definitely want a &lt;a href=&quot;http://en.wikipedia.org/wiki/Rolling-element_bearing#Roller_bearings&quot; rel=&quot;nofollow noreferrer&quot;&gt;roller bearing&lt;/a&gt; in there. To keep the shaft from falling through, the easiest way will be to attach a washer or something similar to the shaft. Make sure the joining method is strong enough (welding would be great, but you probably don't have access to this. Epoxy might work, but it might not be strong enough.) Then, to make sure the flange will rotate on top of the plate, use a &lt;a href=&quot;http://en.wikipedia.org/wiki/Thrust_bearing&quot; rel=&quot;nofollow noreferrer&quot;&gt;thrust bearing&lt;/a&gt; between the flange and the plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://engineering.stackexchange.com/a/2442/41&quot;&gt;Ethan mentions&lt;/a&gt;, 30kg is a quite a bit of weight, so be sure to check that all the components in your assembly are strong enough to do their part. You'll need a good size motor to be able to turn that much weight, and you'll also want to make this stuff as carefully as possible, as poor cuts and assembling will just add to the losses in the system.&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-04-13T19:04:36.423" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2444" PostTypeId="2" ParentId="2437" CreationDate="2015-04-13T19:34:33.420" Score="2" Body="&lt;p&gt;This answer is based on requirements for the state of Illinois; I'm going to make a reasonable assumption that other states in America have similar guidelines, as most states have a license by comity procedure that may involve obtaining references, but accepts the certification from other states as proof of experience and testing requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dsps.wi.gov/Licenses-Permits/Engineer/PElicense&quot; rel=&quot;nofollow&quot;&gt;Illinois Administrative Code on Approved Experience&lt;/a&gt; calls out the different ways in which you can reach the 4 year experience requirement to become a licensed Professional Engineer. Various equivalents are prescribed for obtaining graduate degrees or participating in coop programs during a bachelor's degree. Particularly relevant to this question are Sections (5) and (6), which notes that the length and validity of experience must be verified by either a licensed PE or someone who is practicing engineering (e.g. my supervisor was able to be verified by the Engineering Manager for our company, even though he is not a PE), and that the work must be &quot;in the practice of professional engineering&quot; as defined in Section 4(o) &lt;a href=&quot;http://www.ilga.gov/legislation/ilcs/ilcs3.asp?ActID=1344&amp;amp;ChapterID=24&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This obviously leaves it open to a few people just deciding it's okay to lie, but I'm sure if the certification board is at all suspect, they can request further proof in the form of a resume, employment records, more references, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll also qualify this by saying that I only feel comfortable extending this to the US, and possibly Canada, as I believe there are similar reciprocity agreements in place as exist between states. For the rest of the world, I'm not as familiar. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-04-13T19:34:33.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2445" PostTypeId="2" ParentId="2427" CreationDate="2015-04-13T20:15:40.553" Score="-2" Body="&lt;p&gt;Bicycle tires have pressures way below 8 bars. Pressure is a minor issue, look for standard pipes, pipe closures or joints in a hardware shop, HomeDepot etc.&lt;/p&gt;&#xA;" OwnerUserId="1270" LastActivityDate="2015-04-13T20:15:40.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2447" PostTypeId="1" AcceptedAnswerId="2448" CreationDate="2015-04-13T21:21:38.260" Score="11" ViewCount="1780" Body="&lt;p&gt;Here is &lt;a href=&quot;http://s.hswstatic.com/gif/gear-worm.jpg&quot; rel=&quot;noreferrer&quot;&gt;an image of a worm gear&lt;/a&gt;. What is english word for mechanical element (two of them will be necessary for a worm gear) in which the worm shaft will &quot;lie&quot; and rotate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one photo of the element where it is part of a housing; I need it as separate mountable element (I will mount it on plate):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Htdevm.jpg&quot; alt=&quot;unknown element in worm gear housing&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="368" LastEditDate="2015-04-13T22:28:52.863" LastActivityDate="2015-04-14T01:29:14.917" Title="What is the mechanical element that holds a worm gear shaft in place called?" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2448" PostTypeId="2" ParentId="2447" CreationDate="2015-04-13T21:29:54.263" Score="15" Body="&lt;p&gt;Anything that holds a shaft and allows it to rotate is called a &quot;bearing&quot;. It could be a simple metal bushing or a complete assembly with inner and outer races holding balls or rollers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bearing that is designed to be mounted to a surface is sometimes called a &quot;&lt;a href=&quot;http://en.wikipedia.org/wiki/Pillow_block_bearing&quot;&gt;pillow block&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-04-13T21:29:54.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2449" PostTypeId="2" ParentId="2437" CreationDate="2015-04-13T22:59:38.183" Score="1" Body="&lt;p&gt;Every engineer who interviewed for a job could be classified as a sales engineer. Most engineering do perform internal or external sales functions within organization. In my opinion Sales Engineer is just title that for human resources and other management organizational structures. A good discussion about salesmanship can be found &lt;a href=&quot;https://engineering.stackexchange.com/questions/119/in-engineering-fields-how-often-do-you-find-yourself-being-asked-to-be-a-sales/124#124&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any engineer who has pass FE exam and performance any of the following functions listed below is eligible to sit for a PE exam. The functions are planning, scheduling, budgeting, contracting, supervision, project control, risk assessment, design, construction, evaluation, consultation, engineering, surveying, research and development, operations and management. Most sales engineers get involved in the above functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an engineer’s work experience is obtained under the guidance and supervision of the PE is an absolute plus.  But I think each state has their own procedure. It is best that you review the requirement for your state. Below are two excerpts from State of New Jersey and State of Tennessee. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New Jersey PE Work Experience Requirements&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: New Jersey is emphasizing working under the supervision of a PE &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jxzN0.jpg&quot; alt=&quot;New Jersey PE Work Experience Requirements]&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tennessee’s  PE Work Experience Requirements&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YyXZ1.jpg&quot; alt=&quot;Tennessee’s  PE Work Experience Requirements&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ncees.org/licensing-boards/&quot; rel=&quot;nofollow noreferrer&quot;&gt;List of licensing boards&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ncees.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;National Council of Examiners for Engineering and Surveying&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.state.nj.us/lps/ca/pels/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;State Board of Professional Engineers and Land Surveyors&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.tn.gov/regboards/ae/index.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;State of Tennessee, Board of Architectural and Engineering Examiners&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-04-14T00:34:37.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2450" PostTypeId="2" ParentId="2420" CreationDate="2015-04-14T00:01:55.087" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;why is it logical to have the position of shear center at the location&#xA;  of intersection of line of action of shear force and x axis?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That statement isn't logical. I think you have misunderstood how the shear centre is defined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear centre is the point such that an applied force passing through the S.C. does not cause any &lt;strong&gt;rotation&lt;/strong&gt; of the section. In other words, if you apply a shear force through the shear centre, it does not cause any &lt;strong&gt;torsion&lt;/strong&gt; in the beam but only &lt;strong&gt;bending&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the S.C. depends only on the &lt;strong&gt;geometry&lt;/strong&gt; of the beam section, not on the applied loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can apply a shear force at any point on the beam. If the force does not pass through the S.C, you need to replace it by an equal shear force through the S.C, plus a moment about the S.C. You can then find the deflections and stresses due to bending (caused by the force through the S.C.) and torsion (caused by the moment about the S.C.) separately, and add them together to get the total deflections and stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your aircraft wing example, the shear forces caused by the aerodynamic lift and drag will act through the &lt;strong&gt;centre of pressure&lt;/strong&gt; of the wing, and the centre of pressure is usually &lt;strong&gt;not&lt;/strong&gt; the same point as the shear centre. The aerodynamic forces will therefore cause a combination of bending and twisting in the wing. &lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2015-04-14T00:06:58.840" LastActivityDate="2015-04-14T00:06:58.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2451" PostTypeId="2" ParentId="2447" CreationDate="2015-04-14T01:29:14.917" Score="6" Body="&lt;p&gt;Not only does the shaft have a &quot;bearing&quot; surface but the ends of the shaft also need something to &quot;bear&quot; against due to the &quot;thrust&quot; of the shaft in either direction.  So I would add the term &quot;thrust bearing&quot; to the mix some where in there ... this can be as simple as adding a single steel ball for a bearing surface to each end of the shaft to deal with the worm gear's lateral movement unless you have dealt with this in some other way. &lt;/p&gt;&#xA;" OwnerUserId="1275" LastActivityDate="2015-04-14T01:29:14.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2452" PostTypeId="1" CreationDate="2015-04-14T01:31:56.607" Score="4" ViewCount="1190" Body="&lt;p&gt;I have two wooden plates with the same dimensions (5 cm thick, 50 cm long and 35 cm wide). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CWSk2.jpg&quot; alt=&quot;sketch of plate orientation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between plates must be at least 10 cm. Plate P&lt;sub&gt;1&lt;/sub&gt; is fixed but P&lt;sub&gt;2&lt;/sub&gt; rotates about the y-axis. Angle alpha can go from 0 degrees (when the plates are parallel) to at least 90 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iJOv5m.jpg&quot; alt=&quot;sketch of distance between plates&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I realize this design in practice? Which mechanical elements should I use? It must be strong, there will be load of max. 30 kg hanging on plate P&lt;sub&gt;2&lt;/sub&gt;. There must be some mechanism that will connect these two plates and enable P&lt;sub&gt;2&lt;/sub&gt; to rotate but when one sets a desired angle between plates, the mechanism must hold the angle constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/koSMym.jpg&quot; alt=&quot;sketch of angle between plates&quot;&gt; &lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="368" LastEditDate="2015-04-14T19:20:03.243" LastActivityDate="2015-04-14T19:25:32.847" Title="How can I hold an adjustable angle between two load-bearing plates?" Tags="|mechanical-engineering|steel|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2453" PostTypeId="2" ParentId="2429" CreationDate="2015-04-14T05:03:08.457" Score="3" Body="&lt;p&gt;The type of hinge used in products to hold displays in place at various angles and for similar tasks are called generically &lt;strong&gt;torque hinges&lt;/strong&gt;. There are many different constructions. Design would be a typical mechanical engineering problem- coefficients of friction, tolerances and so on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they have discrete positions where they prefer to stay they may be called &quot;detent hinges&quot;. &lt;/p&gt;&#xA;" OwnerUserId="1276" LastActivityDate="2015-04-14T05:03:08.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2454" PostTypeId="2" ParentId="2452" CreationDate="2015-04-14T06:14:57.263" Score="3" Body="&lt;p&gt;My First thought is to use a standard hinge, with the hinge itself on the intersecting lines. The plates can mounted with an offset to create your desired gap.  Secondly, you need a positioning mechanism. You can use a rotary detent mechanism, for example two disks with radial grooves that interlock in each other when bolted together and can rotate when the bolt is loosened up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not choose a friction mechanism at the rotation axis. The load is large, which creates a strong moment.&lt;/p&gt;&#xA;" OwnerUserId="1077" LastActivityDate="2015-04-14T06:14:57.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2455" PostTypeId="1" CreationDate="2015-04-14T06:45:10.530" Score="3" ViewCount="1833" Body="&lt;p&gt;I am using a BeagleBone Black, Revision B with a &lt;a href=&quot;http://www.4dsystems.com.au/product/Beagle_Bone_LCD_Capes/4DCAPE-43/&quot; rel=&quot;nofollow noreferrer&quot;&gt;4D Systems LCD touch screen&lt;/a&gt;. The screen worked initially with a stock Angstrom or Debian distribution but I want to use Android instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to use all of the pre-built Android images from Texas Instruments for the 4D screen, but I can not boot anymore. There is nothing on the screen. I tried to boot on the 4D screen via HDMI. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am writing the image to the SD card with &lt;code&gt;./mkmmc-android.sh /dev/sdb&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional details:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;4D System LCD is of model 4DCAPE-43T&lt;/li&gt;&#xA;&lt;li&gt;Android distribution is 4.2.2 (Jelly Bean)&lt;/li&gt;&#xA;&lt;li&gt;Ubuntu is the host system&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://e2e.ti.com/support/embedded/android/f/509/p/414724/1473912#1473912&quot; rel=&quot;nofollow noreferrer&quot;&gt;Android Pre-built Image can not boot&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6MM7O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9eD82.png&quot; alt=&quot;BBB_Ubuntu_Screen_OP&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This is a screen capture from above e2e.ti.com forum link&lt;/p&gt;&#xA;" OwnerUserId="1278" LastEditorUserId="110" LastEditDate="2015-04-27T02:43:33.113" LastActivityDate="2015-04-27T02:43:33.113" Title="Can not boot BeagleBone Black with 4D LCD touch screen using pre-built Android images" Tags="|electrical-engineering|software|embedded-systems|" AnswerCount="1" CommentCount="1" ClosedDate="2015-04-14T11:55:08.100" ContentLicense="CC BY-SA 3.0" />
  <row Id="2456" PostTypeId="2" ParentId="2427" CreationDate="2015-04-14T09:14:00.650" Score="3" Body="&lt;p&gt;This is fully doable. This is the in the higher range as bicycle tires (although with the outer tire as well and not the inner tube alone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright, here goes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What equations do I need to calculate a solution?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The pressure in a tube can can be modelled by the following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Hoop Force:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_H = \frac{pr}{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Axial Force&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_A = \frac{pr}{2t}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is a hemisphere that best pocket shape I can use?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is a design question, although when inflated it should ideally be cylindrical to prevent extra stress. This is a bike tyre picture from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I9Nsj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are neoprene, latex and chloroprene viable materials to use or should I find alternatives?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Neoprene I imagine will leak air, although slowely. Latex and chloroprene should work. I would suggest that you get bike parts to do this if you are trying to build a prototype.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Additionally, how can I calculate the amount of expansion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your system is a little dynamic here so it will get bigger as you fill it with pressure.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Rewrite the equations above with $\epsilon E $ to replace $\sigma$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;With a constant pressure and E-modulus you should be able to get the final strain.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="782" LastEditorUserId="782" LastEditDate="2015-04-14T10:54:01.260" LastActivityDate="2015-04-14T10:54:01.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2457" PostTypeId="2" ParentId="2433" CreationDate="2015-04-14T09:31:14.503" Score="1" Body="&lt;p&gt;After some discussion with an experienced college and a continued search, I'am able to answer my own question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DIN 3858 is valid for  type C stud ends as in DIN 3852-2 and type Z tapped holes with parallel internal thread as in DIN 3852-2. Example Type C stud &lt;a href=&quot;http://mdmetric.com/4100/Chapter%20D%20UK.pdf&quot; rel=&quot;nofollow&quot;&gt;-click-, page 19&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case i'm looking for pipe threads and not thread on a stud. For this case EN 10226-1:2004 is valid. Also a references was available in DIN 3858. This guidance show the full pipe thread range until 6&quot;.  This explains directly the bigger available sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wall thickness questions can be answered with  some common sense. The wall thickness must be larger than the complete thread heigth incl the height created by the taper. Say roughly 2 time thread thickness.&#xA;From the steel pipe catalog, i even found some process pipes specifically for 2&quot; &amp;amp; 1 1/4&quot;. 60.3 x 3.65 &amp;amp; 48.3 x 3.25.  These fit with properly with the thread and my selection was wrong in conclusion. These were thin wall pipe suitable for compression couplings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It took me some time, but some good lessons are learned for future work.&lt;/p&gt;&#xA;" OwnerUserId="1077" LastActivityDate="2015-04-14T09:31:14.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2458" PostTypeId="2" ParentId="2455" CreationDate="2015-04-14T11:12:33.790" Score="3" Body="&lt;p&gt;Base on the information provide try shorting the jumpers on the 4D Systems LCD Display. As best as I know Android LCD cape only recognizes the LCD at a specific address.  I believe there is EEPROM information in the LCD cape that is required to complete the android boot up process (Verification required for this sentence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Below is picture to help locate the jumpers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bASPs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bASPst.jpg&quot; alt=&quot;4D Systems Jumpers&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Below are excerpts from the 4D Systems LCD 4DCAPE-43T data sheet for&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AiS7H.jpg&quot; alt=&quot;EEPROM Shunt&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/lrZ5T.jpg&quot; alt=&quot;EEPROM Jumpers&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumptions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Beaglebone Black + 4D System LCD worked with stock Angstrom or Debian distribution&lt;/li&gt;&#xA;&lt;li&gt;4D System LCD is of model 4DCAPE-43T&lt;/li&gt;&#xA;&lt;li&gt;Android distribution is 4.2.2 (Jelly Bean)&lt;/li&gt;&#xA;&lt;li&gt;Ubuntu is the host system&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An image of the Beaglebone black + 4D Systems with Angstrom distribution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qqD9t.jpg&quot; alt=&quot;Beaglebone black + 4D System with Angstrom&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Debugging Tips&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Make sure the you have good micro SD Card, that is 4GB. I have had issue with poor quality SD Cards&lt;/li&gt;&#xA;&lt;li&gt;Configure and use minicom to log in as root&lt;/li&gt;&#xA;&lt;li&gt;If you are using external power supply make sure the supply is rated at sufficiently (800 mA will work, it is best to be able to source at least 1A or 2A)&lt;/li&gt;&#xA;&lt;li&gt;Connect both the external and USB power supply for boot-up&lt;/li&gt;&#xA;&lt;li&gt;Monitor the boot of process using a FTDI serial debugging cable&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Serial Debugging Cable&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HakBf.jpg&quot; alt=&quot;FTDI Debugging Cable&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;External Power Supply&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oUX2d.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GZCdZ.jpg&quot; alt=&quot;External Power Supply&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can purchase a serial debugging cable from adafruit, sparkfun, or other DIY online stores. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://forum.4dsystems.com.au/forum/forum-aa/4d-systems-hardware/beaglebone-display-modules-and-capes/4498-boot-android-ti-jb-on-beaglebone-black-from-sd-card&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boot android (TI-JB) on beaglebone black from SD card&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.4dsystems.com.au/productpages/4DCAPE-43/downloads/4DCAPE-43(T)_datasheet_R_1_9.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;4D 4.3” LCD CAPE Beagle Bone Black 4.3” LCD CAPE&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://elinux.org/Beagleboard:BeagleBone_Black_Accessories&quot; rel=&quot;nofollow noreferrer&quot;&gt;BeagleBone Black Accessories&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://derekmolloy.ie/beaglebone/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Embedded Systems - Derek Molloy&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://eewiki.net/display/AOA/BeagleBone+Black&quot; rel=&quot;nofollow noreferrer&quot;&gt;BeagleBone Black&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-04-16T19:27:32.800" LastActivityDate="2015-04-16T19:27:32.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2459" PostTypeId="2" ParentId="2413" CreationDate="2015-04-14T15:25:59.937" Score="2" Body="&lt;p&gt;Before we get into calculating the best feeds and speeds for your goals, there are a few more concepts we need to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chipload: Chip Thickness per Tooth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While feedrates are specified in length units per minute, the more important measurement is something called &quot;Chipload&quot;. Thing of a chip as looking something like a comma in cross section, or perhaps an apostrophe. One starts big and gets smaller at the end. The other starts small and gets bigger at the end. We'll ignore that difference for a moment, though it is important as we shall see later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chipload is a measurement that is independent of spindle rpm, feedrate, or number of flutes that tells how hard the tool is working. That's a very useful thing, as you could imagine. Hence, manufacturers and machinists typically like to talk about chipload for a particular tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that a tool with more flutes (cutting edges) has to be fed faster to maintain a particular chipload. Since each tooth is going to take a cut every rotation, a tooth has only a fraction of a rotation in which to cut a chip that reaches the chipload thickness. During the time it takes to rotate the next tooth to start cutting, the tool has to have moved far enough to shave off a chip that is thick enough. Hence, tools with more flutes can be fed faster. A 4 flute endmill can be feed twice as fast as a 2 flute, all other things being equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do Tools Break from Too Much Chipload?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can imagine that forces simply become too great if a tool tries to take too much &quot;bite&quot; by having too much workload. This can chip or break the cutter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a second issue that comes from too much chipload--the chips get bigger and eventually can't get out of the cutter's way. Beginning machinists probably break more tools because they don't get the chips out of the way fast enough than because the force of the feed is breaking the tool. If the cutter is down in a deep slot, the chips have a particularly hard time getting out of the way. We use air blasts, mists, and flood coolant to try to clear the chips out of the way, but if they're way down a hole or slot, it makes it that much harder, and we have to reduce speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More exotic tools are available with 5, 6, 10 or more flutes. Experienced hands will tell you that if you're profiling (where there's lots of chip clearance) steel and aren't using 5 or 6 flutes, you're leaving money on the table. Let's run the numbers in G-Wizard. Suppose we're profiling some mild steel--1020 or some such. We're going to profile the outside of a part, so there's plenty of clearance. Cut depth will be 1/2&quot;, cut width 0.100&quot;, and we'll use a 1/2&quot; TiAlN Endmill. Here are the numbers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;4 Flute: 3158 rpm, 29.8 IPM. MRR is 1.492 cubic inches/minute. A little over 1 HP.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;5 Flute: Same rpms, now 37.3 IPM. MRR = 1.865. 1.3 HP. That's 30% faster cutting.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;6 Flute: Now 44.8 IPM. MRR = 2.238. 1.6 HP. 60% faster than the 4 flute case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1282" LastActivityDate="2015-04-14T15:25:59.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2461" PostTypeId="2" ParentId="2452" CreationDate="2015-04-14T16:32:35.997" Score="-1" Body="&lt;p&gt;I think that easiest way to solve the problem of rotating of plate P&lt;sub&gt;2&lt;/sub&gt; will be using a mechanism like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0uTUr.jpg&quot; alt=&quot;hinge mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use it, join plate P&lt;sub&gt;1&lt;/sub&gt; to an additional plate with length of 10 cm (height and width depends on the dimensions of mechanism) like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HO1dD.png&quot; alt=&quot;sketch of additional plate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After join of &quot;unknown element&quot; with P1 and after placing P2 on &quot;unknown element&quot;, construction will look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/znTFG.png&quot; alt=&quot;sketch of linkage between plates&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still need to find way to prevent plate P&lt;sub&gt;2&lt;/sub&gt; from falling, I don't think that screws will be enough to attach this mechanism to the plates.&lt;/p&gt;&#xA;" OwnerUserId="1268" LastEditorUserId="368" LastEditDate="2015-04-14T19:25:32.847" LastActivityDate="2015-04-14T19:25:32.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2462" PostTypeId="2" ParentId="1995" CreationDate="2015-04-14T16:57:24.730" Score="2" Body="&lt;p&gt;Any mechanical clock or watch relies on the motive power applied to the large gear (the &quot;wheel&quot;) which drives the smaller one (the &quot;pinion&quot;).  Thus the weight in a longcase clock is suspended by a cord, rope or chain from the &quot;great wheel&quot; (usually making a rotation every 12 hours) and the rate of rotation is geared up to the escape wheel (which often has the seconds hand mounted on it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the tooth form is usually different when gearing up : friction is critically important in a clock, transmitting high forces is usually less so (and is usually catered for by making the great wheel thicker than the others). So the teeth are usually cycloidal in form, where the deep part of the slot in a tooth is approximately rectangular, which means the base of a pinion tooth is undercut. This is a fundamentally weaker tooth form, especially since pinions may have as few as 6 teeth, but runs freely with little friction and zero pressure angle (see below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example&#xA;&lt;img src=&quot;https://i.stack.imgur.com/4yqOL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from &lt;a href=&quot;http://www.ryerson.ca/~v7chan/cncproject/gear/Cycloid%20Gear%20Calculator.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An extreme case is the lantern pinion&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6KFCB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;(from &lt;a href=&quot;http://www.donaldsonworkshop.com/chapdetail.php?project=clock&amp;amp;chapter=08ClickWheelsPinions.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;) where the pinion tooth is completely undercut!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You never lubricate the teeth of a clock or watch wheel : that only adds viscosity (i.e. friction), wasting power, and does nothing to eliminate wear. This is because the contact surfaces of the teeth roll over each other, there is no sliding motion involved. (The pivots, unless running in ballraces, do need lubrication. John Harrison employed ballraces for a prototype marine chronometer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, while gearing speed down also involves contact surfaces rolling over each other, the purpose is usually to amplify force, and to do that with the least material, a stronger tooth form is required. This is normally an &lt;a href=&quot;http://en.wikipedia.org/wiki/Involute_gear&quot; rel=&quot;nofollow noreferrer&quot;&gt;involute tooth form&lt;/a&gt;, where each tooth is wider at the base, like a wedge. &#xA;&lt;img src=&quot;https://i.stack.imgur.com/puHpt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the teeth press each other outwards as well as turning each other, at an angle known as the pressure angle (usually 20 degrees in modern gears, formerly 14.5 degrees). Thus the axles are pushed apart, increasing friction on the pivots and requiring a stronger gearbox. (The animation on the Wikipedia page exaggerates the pressure angle). Traditionally, involute pinions are only cut down to 12 teeth, with the 20 degree PA making for more friction but stronger teeth  with wider roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So : yes, gearing can be used to increase rotational speed, but it usually requires a different tooth form, otherwise it loses a lot of power to friction.&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2015-04-14T16:57:24.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2463" PostTypeId="2" ParentId="2437" CreationDate="2015-04-14T17:21:48.753" Score="3" Body="&lt;p&gt;Along with the &lt;a href=&quot;http://www.nspe.org/resources/licensure/what-pe&quot; rel=&quot;nofollow&quot;&gt;National Society of Professional Engineers&lt;/a&gt;, most professional engineering licensing boards within the United States use either of the following terms:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Qualifying engineering experience&lt;/li&gt;&#xA;&lt;li&gt;Progressive engineering experience&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And from what I have seen of European licensing requirements, their terminology is similar.  For example, &lt;a href=&quot;http://www.feani.org/site/index.php?id=261&quot; rel=&quot;nofollow&quot;&gt;FEANI&lt;/a&gt; uses the term &quot;valid professional experience.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could just as easily rephrase the requirement to state &quot;engineering experience that is progressive&quot; if that helped the phrase make more sense to you.  Grammatically, it's easier to say &quot;progressive engineering experience.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does &lt;code&gt;{qualifying | progressive | valid}&lt;/code&gt; engineering experience actually mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;obvious&quot; part is that the work experience has to be Engineering related.  &lt;a href=&quot;http://mn.gov/aelslag/progexpeng.pdf&quot; rel=&quot;nofollow&quot;&gt;This pdf from the Minnesota government&lt;/a&gt; provides a really good overview of potential activities ranging from application of theory to management, communication skills, and understanding social implications of decisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineers.texas.gov/downloads/serinfo_ws.pdf&quot; rel=&quot;nofollow&quot;&gt;Guidance from the state of Texas&lt;/a&gt; is also pretty explicit in focusing on what work the candidate actually performed instead of more general descriptions.  &lt;a href=&quot;http://ef.engr.utk.edu/ef402-2011-01/gen/pe.php&quot; rel=&quot;nofollow&quot;&gt;Guidance from the state of Tennessee&lt;/a&gt; is pretty good at summarizing expectations regarding work experience:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Work Experience&lt;/strong&gt; After passing the FE exam, you will continue your journey toward professional licensure by gaining engineering experience. Many jurisdictions have specific requirements about the type of experience you need to gain. Most require that you gain experience under the supervision of someone who is already licensed, and that your experience involve increasing levels of responsibility. Once you begin work, contact your licensing board to find out what experience is needed and talk with professional engineers in your company to find out how you can gain this experience.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key in those is to demonstrate how the work experience of the candidate became progressively more difficult as they gained experience.  For example, an Engineer with multiple years of experience is not still doing the exact same level or scope of work that they performed when they first started on the job.  In other words, there is a &lt;em&gt;progression&lt;/em&gt; in the level of work expected from the Engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we look at those terms again, we can draw some conclusions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Qualifying&lt;/strong&gt; refers to the work being supervised by another PE and implies &lt;em&gt;progressive&lt;/em&gt; experience.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Progressive&lt;/strong&gt; refers to work that requires more and more skill in order to perform the task while also implying &lt;em&gt;qualifying&lt;/em&gt; experience.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Valid professional experience&lt;/strong&gt; manages to capture the essence of both &lt;em&gt;qualifying&lt;/em&gt; and &lt;em&gt;progressive&lt;/em&gt; but has &lt;a href=&quot;http://www.feani.org/site/fileadmin/PDF_Documents/EUR_ING/Final_Working_Exper_CRITERIA_-_WEB_Form_Dec_2013.doc&quot; rel=&quot;nofollow&quot;&gt;supplementary documentation to explain that&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Covering some of your related questions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... if an engineer is working in sales and not design, I would tend to think that would not count. But that's just my impression. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Sales&quot; is a tricky aspect.  FEANI implied in some cases that sales experience could qualify.  From a US perspective, my belief is that sales experience is unlikely to qualify unless it was more of design-sell type role.  The key would be &quot;a demonstrated ability to apply knowledge, skills and attitudes for achieving observable results.&quot;&lt;br&gt;&#xA;&lt;sup&gt;*&lt;/sup&gt;&lt;sub&gt;That's courtesy of the &lt;a href=&quot;http://www.feani.org/site/fileadmin/PDF_Documents/EUR_ING/Final_Working_Exper_CRITERIA_-_WEB_Form_Dec_2013.doc&quot; rel=&quot;nofollow&quot;&gt;FEANI supplementary guidance&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there defined rules? General guidelines or standards of practice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based upon the above references, I would say there are guidelines and standards of practice. But the field of Engineering continues to evolve and strictly defined rules would be more limiting than useful.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-04-14T17:21:48.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2464" PostTypeId="1" CreationDate="2015-04-14T18:56:43.950" Score="12" ViewCount="322" Body="&lt;p&gt;A typical optimal control problem with state &lt;span class=&quot;math-container&quot;&gt;$x(t)$&lt;/span&gt; and control &lt;span class=&quot;math-container&quot;&gt;$y(t)$&lt;/span&gt; can be expressed as &lt;span class=&quot;math-container&quot;&gt;$$\max_{x(t), y(t)} \int_0^{t_1} f(t,x(t), y(t)) dt$$&lt;/span&gt; subject to &lt;span class=&quot;math-container&quot;&gt;$x'(t)= g(t, x(t), y(t))$&lt;/span&gt; and boundary conditions for &lt;span class=&quot;math-container&quot;&gt;$x(t)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I want to solve a problem that looks very similar, but the dynamics are &lt;span class=&quot;math-container&quot;&gt;$$x'(t)= g(t, x(t), y(t), z(x(t)) ),$$&lt;/span&gt; where, the sequence &lt;span class=&quot;math-container&quot;&gt;$z(\cdot)$&lt;/span&gt; is to be chosen although it's argument is the state.&lt;/p&gt;&#xA;&lt;p&gt;I'm not sure where one can look for solutions to this non-standard problem. How can I approach this problem?&lt;/p&gt;&#xA;" OwnerUserId="1285" LastEditorUserId="1832" LastEditDate="2021-03-18T17:23:13.757" LastActivityDate="2021-03-18T17:23:13.757" Title="How do I solve an optimal control problem where the dynamics depend on some function of the state?" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="2465" PostTypeId="1" CreationDate="2015-04-14T19:14:36.193" Score="1" ViewCount="1263" Body="&lt;p&gt;&quot;LPG&quot; is the term used here for a mixture of Propane &amp;amp; Butane sold for public use. The exact ratio varies depending on where you buy it and I think also with the season and is not known to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on a personal project, I have a gas tank (Liquefied petroleum gas) with a pressure of 4-5 bar (about 50 to 80 psi). Gas in it is a combination of liquid and gas (you can hear the liquid inside if you shake it).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need some kind of a device/filter that would convert it to gas at the output (instead of liquid) with a small pressure so I can burn it cleanly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using an industrial vaporizer that is designed to produce gas from LPG, but it gave me something stronger than I need, and when it burns it seems that it is &quot;not clean&quot;. (Maybe the industrial gas that is used with the vaporiser is a different mix of gases than domestic LPG?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, what I need is to build something that works like a large version of  a cigarette lighter, like this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tGsfW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tGsfWt.jpg&quot; alt=&quot;Cigarette lighter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, in a lighter like in the photo there is liquid and gas, and a wick that turns produce &quot;clean&quot; fire, without all the side effects that the vaporizer creates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am missing a lot of knowledge here, and I would like to know if someone can guide me to how should I do this, what are the names and definitions for the stuff I need etc,&#xA;Any knowledge regarding to what I should do, or how, or where can I find it will be mostly appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I used the phrase &quot;home gas&quot; which is common in my state, I meant Liquefied petroleum gas (LPG).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the vaporizer I used now (it belongs to someone who knows what he is doing, I do not plan on killing myself :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ptUg4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ptUg4m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'll try to re-describe my problem,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get &quot;clean slow flame&quot; such as is produced by a cigarette lighter.&#xA;Instead, what I am getting, regardless of the vaporizer pressure setting, burns  like it's not clean - it acts as if some of the gas is not really burning or properly vaporising when it comes out.   &lt;/p&gt;&#xA;" OwnerUserId="1286" LastEditorUserId="6790" LastEditorDisplayName="user16" LastEditDate="2020-05-01T17:38:21.020" LastActivityDate="2021-08-06T15:56:32.010" Title="How can I turn home gas (LPG) from liquid to gas?" Tags="|gas|hvac|combustion|petroleum-engineering|vaporization|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2466" PostTypeId="5" CreationDate="2015-04-14T22:10:52.447" Score="0" Body="&lt;p&gt;Hydraulics can be about using pressurized fluid to do work or about the large scale flow of fluid.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="33" LastEditDate="2016-02-23T03:04:31.313" LastActivityDate="2016-02-23T03:04:31.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2467" PostTypeId="4" CreationDate="2015-04-14T22:10:52.447" Score="0" Body="Questions about the flow and use of fluids." OwnerUserId="368" LastEditorUserId="33" LastEditDate="2016-02-23T03:04:31.313" LastActivityDate="2016-02-23T03:04:31.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2469" PostTypeId="2" ParentId="505" CreationDate="2015-04-15T05:49:23.503" Score="3" Body="&lt;p&gt;For 1. and 2., you can solve the problem by having an adjustable valve from the output line to free air. This way you let some of the compressors output flow escape, leaving enough of flow and pressure for the soldering station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To operate:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fully open the &quot;release valve&quot; before starting the compressor.&lt;/li&gt;&#xA;&lt;li&gt;Monitor air pressure with a gauge, ideally next to the soldering station but if the compressor has one on output it'll probably work also.&lt;/li&gt;&#xA;&lt;li&gt;Slowly close the &quot;release valve&quot; until the pressure is at the level you want.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As for 3., there are various kinds of filters for removing condensation water from compressed air, but I'm not able to recommend any specific kind.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2015-04-15T05:49:23.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2470" PostTypeId="1" CreationDate="2015-04-15T07:37:19.433" Score="4" ViewCount="6513" Body="&lt;p&gt;I am looking for a material that can be used in a non commercial environment that will allow all (or nearly all) light to penetrate but will be heat resistant (By heat resistant I mean it will not allow the heat to penetrate to the other side, not that the materiel itself won't get hot - although they may be dependent on each other).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To better understand my requirements, I am trying to illuminate a room with 40,000 lux of light using a few metal halide (MH) bulbs, for a scientific experiment. The bulbs will consume 1800 - 2400 watt which means it will produce a lot of heat. Since real people will partake in the experiment (sitting about 6-12 inches from the light source in order to increase the amount of lux reaching the subject) I am concerned the heat produced by the bulbs will be too difficult to bear. Therefore, I want to place some material (such as a glass or plastic) in between the subject and the light source so maximum light penetrates with minimum heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to do this?&lt;/p&gt;&#xA;" OwnerUserId="1291" LastEditorUserId="33" LastEditDate="2016-02-04T14:04:58.613" LastActivityDate="2016-11-16T14:01:51.390" Title="What is the Best Heat Resistant Translucent Material?" Tags="|mechanical-engineering|materials|thermodynamics|heat-transfer|" AnswerCount="5" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2471" PostTypeId="2" ParentId="2465" CreationDate="2015-04-15T07:38:53.693" Score="7" Body="&lt;p&gt;Lighters usually use Butane.&lt;br&gt;&#xA;Gas bottles use some or all Propane.&lt;br&gt;&#xA;Propane pressures are MUCH higher.&lt;br&gt;&#xA;A plastic 'tank' filled with Propane would probably explode.&lt;br&gt;&#xA;As size increases the container strength must be increased disproportionately.&lt;br&gt;&#xA;As temperatures increase pressures rise FAR faster than changes in absolute temperature.   &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;We understand from comments you have added that your source is a Propane or LPG tank - such as these.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.thepropanepeople.com/uimages/site_images/Tanks.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://streatfuel.files.wordpress.com/2011/08/lp-cylinders-me.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eQ4cT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A plastic cigarette lighter MAY use all Butane or even &quot;longer chain&quot; hydrocarbons and will be designed to work at low pressures. Also, because they are only used briefly they do not need to take account of the large heat energy required to vaporise a liquified gas continuously. Without some means of providing thermal input the temperature of the vaporiser will drop until the liquid will no longer vaporise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some intermediate temperature the liquid may only partially vaporise and the output may contain a mixture of gas plus liquid droplets. As the liquid droplets enter the air they will absorb ambient heat energy and expand and may change the air:gas mixture ratio and make the flame unstable. If this happens you probably need an improved way of adding heat energy to the vapouriser. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was doing this I would TRY heating the gas feed as near the outlet as possible, &lt;strong&gt;bearing in mind the safety concern mentioned below.&lt;/strong&gt; One method for testing would be to use a towel soaked in hot water, and/or a water bath containing hot water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the vaporiser you have tried works with the liquified gas that it is intended for and not with your gas then your gas may be different in some manner. The most likely difference is probably the Propane / Butane ratio. You &lt;strong&gt;may&lt;/strong&gt; have 100% Propane, or 100% Butane, of some mix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If cylinders used with eg barbecues are user with a Propane / Butane mix and are regularly refilled before being empty, there is a tendency for Propane to be released preferentially and for Butane to form an increasing fraction of the bas output. The 'cure' for such Butane concentration is to fully drain tanks occasionally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;WARNING:&lt;/strong&gt; Note that 'playing with' a gas generator that you do not understand well is potentially dangerous and possibly even life threatening. People have died from unexpected release and subsequent ignition of quite small quantities of butane. (A friend of a friend of mine died from gas release from a small disposable butane gas cylinder that he thought was empty.)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Propane has the potential to be even more dangerous due to the higher potential pressures. Propane has a MUCH higher vapor pressure than Butane.&lt;br&gt;&#xA;At room temperatures Propane pressure is about 500% of that for Butane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At temperatures you may experience on a really hot day Propane pressures &lt;strong&gt;WILL&lt;/strong&gt; be more than 10 x Butane pressure at room temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say your gas is at 4 or 5 Bar pressure. If you put it in another container it &lt;strong&gt;MUST&lt;/strong&gt; still be at the same pressure. If you put it somewhere hotter pressure could double. If you make a big &quot;lighter&quot; it will be even weaker than a small lighter BUT pressures will be much higher. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt; :-(&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Propane / Butane &amp;amp; mixtures - pressure versus temperature.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vdJJ8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-04-17T18:59:56.620" LastActivityDate="2015-04-17T18:59:56.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2472" PostTypeId="2" ParentId="2470" CreationDate="2015-04-15T07:44:33.640" Score="3" Body="&lt;p&gt;Infra-red will be the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's going to be a trade-off between weight, light transmission, cost, thermal conductivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best bet will be looking for so called filter &quot;gels&quot; as termed in the lighting industry, particularly stage lighting. They tend to be developed for precisely your requirements; to absorb infra-red and dissipate it as heat, while leaving the colour spectrum intact. They are basically plastic sheets with tuned optic/thermal properties. IR heat shields I believe are what you might want. LEE filters are the renowned manufacturer. Talk to them&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2015-04-15T07:44:33.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2474" PostTypeId="2" ParentId="2470" CreationDate="2015-04-15T08:20:16.830" Score="0" Body="&lt;p&gt;I would suggest either &lt;a href=&quot;http://en.wikipedia.org/wiki/Polycarbonate&quot; rel=&quot;nofollow&quot;&gt;polycarbonate&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Poly%28methyl_methacrylate%29&quot; rel=&quot;nofollow&quot;&gt;perspex/plexiglas/acryclic&lt;/a&gt;. They both have good thermal properties, with a thermal conductivity of the order of 0.2 W/m/K, and are good substitutes for glass in terms of optical properties. The thicker the sheet, the better the thermal insulation, but depending on the size, you may need to provide some reinforcement so that the plastic sheet doesn't bend under its own weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on where you are, there are various people who sell this type of engineering materials. I tend to use &lt;a href=&quot;http://www.edplastics.co.uk/&quot; rel=&quot;nofollow&quot;&gt;Engineering Design Plastics&lt;/a&gt;, because they're local (for me), but there are others.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-04-15T08:20:16.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2475" PostTypeId="1" CreationDate="2015-04-14T16:46:19.883" Score="5" ViewCount="2752" Body="&lt;p&gt;I have to make a dual blackboard like the one here, &lt;a href=&quot;http://www.ucd.ie/physics/lhcb/contact/blackboard.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ucd.ie/physics/lhcb/contact/blackboard.jpg&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2H2Of.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2H2Ofm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My biggest concern is what sort of mechanisms I should use to allow for smooth and easy movement of the blackboards.  The boards need to be brought up and down, and be able to stay in place once they've been moved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do have an idea on how to do it with weighted cords and pulleys, but I believe there are people with more experience in this area.  I'm not as certain about the mechanisms to keep the boards in place though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help identify the parts I need to make one of the board to move up and down? Also, which parts do I need in order to make the boards stay in place after being moved? &lt;/p&gt;&#xA;" OwnerDisplayName="Noor" LastEditorDisplayName="user16" LastEditDate="2015-04-15T13:29:19.840" LastActivityDate="2015-04-15T18:49:34.720" Title="Mechanisms for moving dual blackboards" Tags="|mechanical-engineering|design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2476" PostTypeId="2" ParentId="2351" CreationDate="2015-04-15T12:44:16.013" Score="1" Body="&lt;p&gt;I don't know for sure without more details of your setup, but I have two possible explanations (both of which may be partially or wholly responsible for the behavior you see):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Your thermostat doesn't directly control the compressor in your cooler. It tells your heating/cooling system when to come on and go off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case what could be happening is the thermostat is telling your cooler to come on, and the cooler control system is deciding how much to run the compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physical heat exchange interfaces (on both the hot and cold sides) can only transfer so much heat in a given amount of time, but the compressor is capable of higher performance than the minimum required to maintain this level of heat exchange. This would be by design, so that it can quickly come up to maximum performance when switched on. If you notice that it comes on for a bit longer initially, then tops up the pressure with short bursts after a while, this is probably what is happening.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The smart-thermostat has learned about a time lag in your property between enabling heating/cooling and registering that change. It is therefore cautiously applying heating/cooling to avoid large swings one way or the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the location of the thermostat and your heating/cooling system(s), and the layout of your building, it may take quite some time after switching on the heating/cooling before any effect is noticed by the thermostat. It may have learned this when first installed and applies heating/cooling for a short time, then waits to feel the results of this before deciding if it needs some more.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-04-15T12:44:16.013" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2477" PostTypeId="2" ParentId="2475" CreationDate="2015-04-15T13:28:01.313" Score="5" Body="&lt;p&gt;The ones I've seen worked the same way old windows did.  The boards slide loosley in grooves.  On each side, there is a counterweight in the wall, and a rope over a pulley at top.  The rope attaches to the board on one end, over the pulley, and to the top of the counterweight on the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The counterweights can be quite large.  Each one is roughly half the weight of the sliding board.  They were usually round torpedo-shaped pieces of iron with a hole at top sortof like the eye of a needle.  That is for attaching the rope.  The whole weight could be a couple feet long and 2-3 inches in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, it's not clear whether the boards are supposed to move independently with a fixed board behind in the lower position, or it's just the two boards you show.  If just the two boards, then they can be counterweights for each other.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-04-15T13:28:01.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2478" PostTypeId="1" AcceptedAnswerId="2481" CreationDate="2015-04-15T13:33:17.973" Score="14" ViewCount="30069" Body="&lt;p&gt;Is there a method for plating steel with stainless steel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, is it chemical, electrical, or electrochemical? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a quick search on the internet but was unable to find a service. I'm interested in applying a food safe finish to something that would otherwise be cost prohibitive to make out of solid stainless steel.&lt;/p&gt;&#xA;" OwnerUserId="539" LastEditorDisplayName="user16" LastEditDate="2015-04-15T13:38:26.783" LastActivityDate="2021-11-21T05:46:58.763" Title="Is stainless steel plating possible?" Tags="|materials|steel|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2480" PostTypeId="2" ParentId="2470" CreationDate="2015-04-15T10:43:32.680" Score="8" Body="&lt;p&gt;Illuminate the room indirectly via dichroic reflectors at 45 degrees to each transparent wall. These will reflect light at 90 degrees (i.e. into the room) but allow the heat to pass straight through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(NB you may want watercooled heatsinks behind the reflectors, or some other adequate arrangement to dispose of the heat.)&lt;/p&gt;&#xA;" OwnerUserId="435" OwnerDisplayName="Brian Drummond" LastActivityDate="2015-04-15T10:43:32.680" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2481" PostTypeId="2" ParentId="2478" CreationDate="2015-04-15T14:25:56.857" Score="16" Body="&lt;p&gt;Keeping things simple, &lt;a href=&quot;https://en.wikipedia.org/wiki/Steel&quot; rel=&quot;noreferrer&quot;&gt;steel&lt;/a&gt; is an alloy of iron and carbon, whereas &lt;a href=&quot;https://en.wikipedia.org/wiki/Stainless_steel&quot; rel=&quot;noreferrer&quot;&gt;stainless steel&lt;/a&gt; is essentially an alloy of iron, carbon and chromium or iron, carbon chromium and nickel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All forms of steel, whether they be ordinary iron and carbon alloy or stainless steel are &lt;a href=&quot;http://metals.about.com/od/properties/a/Steel-Production.htm&quot; rel=&quot;noreferrer&quot;&gt;made&lt;/a&gt; from a melt in furnaces. Because of this stainless steel cannot be plated to ordinary steel by chemical means either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stainless steel can be welded to ordinary steel but a TIG welder is required, but this wouldn't suit your purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot dipping is unlikely to be an option due to the melting temperatures of ordinary steel and stainless steel. Depending on the type of stainless steel, the temperatures will be similar or higher for stainless steel. This would damage the main item made from ordinary steel.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my initial answer I stated that electroplating stainless steel onto steel was not possible. Thanks to references to scientific papers supplied by @starrise and @Jaroslav Kotowski, it appears a form of stainless steel can be electro-deposited onto copper and stainless steel items. There was no mention of depositing onto ordinary steel. A deposit 23 um (0.023 mm, 0.9055 thousandths of an inch) thick was deposited in one set of experiments. I haven't found any references that claim the process have been commercialized.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-04-16T13:37:38.500" LastActivityDate="2015-04-16T13:37:38.500" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2482" PostTypeId="2" ParentId="2470" CreationDate="2015-04-15T14:34:35.843" Score="2" Body="&lt;p&gt;To start, ordinary window glass will probably do just fine in this situation, but there is a small chance that local heating will cause it to crack. I'd probably test that first before moving on to anything more exotic, which will be harder to find/replace and cost more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my lab we primarily use Pyrex (Borosilicate glass) glass as window material to let in large amounts of light while blocking air flow. It handles heat quite well and is transparent from the visible spectrum and into the short IR. I've also had success recently with Gorilla glass, which is more expensive, but comes in very thin sheets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fuzed quartz will also handle the heat easily, but is will cost more than the Pyrex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd caution against plastics, many of them absorb strongly in the IR (which your lamps are producing in addition to visible light). In our work we use strong lamps to radiantly heat plastics to the point of melting and ignition. One or two 500 W lamps a few inches away will get the plastic melting and bubbling within minutes. The Pyrex in between, which is nearly touching the lamps, barely notices the heat even once the plastic catches fire.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-04-15T14:34:35.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2483" PostTypeId="2" ParentId="2475" CreationDate="2015-04-15T18:49:34.720" Score="0" Body="&lt;p&gt;Converted from a prior comment:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boards in such arrangements usually slide on pairs of tracks, one behind the other. Cords from one board go up over a top mounted pulley and down to the other board.&lt;br&gt;&#xA;If tracks do not provide good alignment at least two set of cords are needed - one set at each end.&lt;br&gt;&#xA;If boards can be slid up without jamming then a central cord may be adequate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boards need to be close to equal weight. Weight can be added if needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many ways to lock boards in place are available, but rare earth magnets at bottom and a steel plate on board would work well.&lt;br&gt;&#xA;This can provide as positive a lock as is needed. Pulling down on the top board would pull the bottom board free from the magnets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A modern version of a board system like this could use electric motors - either operating cables or using &quot;lead screws&quot; made from eg cheap industrial threaded rod. The amount of use that such systems would get would make basic nut-on-rod movement systems viable.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-04-15T18:49:34.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2484" PostTypeId="2" ParentId="2478" CreationDate="2015-04-15T19:39:23.207" Score="6" Body="&lt;ol&gt;&#xA;&lt;li&gt;Stainless steel is an alloy (contains different phases of iron along with different alloying elements)&lt;/li&gt;&#xA;&lt;li&gt;Electroplating will only allow you you to deposit 'Fe' atoms without any of the Stainless steel properties.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you can do a teflon coating, or if mild steel is not your only choice of base material, you may have other options.&lt;/p&gt;&#xA;" OwnerUserId="1297" LastEditorUserId="368" LastEditDate="2015-04-15T20:36:22.490" LastActivityDate="2015-04-15T20:36:22.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2485" PostTypeId="1" CreationDate="2015-04-15T22:35:43.427" Score="9" ViewCount="247" Body="&lt;p&gt;&lt;strong&gt;Project background:&lt;/strong&gt;&#xA;I am working on a project where i need to weld plastic using ultrasonic transducers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&#xA;Since I'm still learning engineering, I have a hard time figuring out what all the factors are in order to choose the right transducer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I've done so far:&lt;/strong&gt;&#xA;I have searched the internet to find information regarding this subject, but due to (what seems to me) the wide area of knowledge needed, I haven't found anything that I properly can comprehend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basically:&lt;/strong&gt; What do I need to take into account when choosing a transducer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Not-so-basically:&lt;/strong&gt;&#xA;I suspect that the properties of the material certainly affects the welding process, so here comes a sidequestion: what properties of the material do I need to know in order to calculate/decide which transducer to use?&lt;/p&gt;&#xA;" OwnerUserId="1299" LastEditorUserId="282" LastEditDate="2015-04-16T17:42:47.207" LastActivityDate="2015-04-16T17:42:47.207" Title="What type of ultrasonic transducer would be suitable for ultrasonic welding?" Tags="|mechanical-engineering|friction|welding|vibration|plastic|" AnswerCount="0" CommentCount="10" FavoriteCount="0" ClosedDate="2016-04-19T00:51:56.230" ContentLicense="CC BY-SA 3.0" />
  <row Id="2486" PostTypeId="2" ParentId="2429" CreationDate="2015-04-15T23:57:25.623" Score="0" Body="&lt;p&gt;Here is an idea for you if you can't find an off-the-shelf product that fits your need. There are plastic coolant hoses that can be bent and stay in any position desired to direct the coolant flow for machine tool applications. The hoses are built from articulated plastic ball and socket joints. The hoses have the detent property due to the tight fit(friction) of the ball inside the socket. So just make your own hinge using two of these hose sections as pivot points and attach a plastic plate between the two pivots. It will be a little &quot;bulky&quot; but if you have no size constraints it won't matter. You will have a nice hinge that will stay where you position it, just like the coolant hoses themselves.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2015-04-15T23:57:25.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2487" PostTypeId="2" ParentId="2423" CreationDate="2015-04-16T00:23:12.620" Score="12" Body="&lt;p&gt;You should not separate those two terms in search for physical interpretation. The term $\nabla\vec{u}+\left(\nabla\vec{u}\right)^T$ is the strain rate tensor $\dot{\gamma}$. The momentum flux (or stress) due to the fact we have a flowing fluid is accounted for by the whole term $\mu\left(\nabla\vec{u}+\left(\nabla\vec{u}\right)^T\right)$. In the NS equation both terms can be thought of as force densities (force per unit volume). You are right, that the second term is zero for incompressible flows (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: The complete derivation of the strain rate tensor is complex and it might be out of scope here. If you are interested I have found that a good resource is Introduction to Fluid Mechanics by Whitaker. Briefly, lets accept that the tensor $\nabla \vec{u}$ represents the strain rate and solid like rotational motion. Any tensor can be decomposed in the following way:&#xA;$$\nabla \vec{u} = \frac{1}{2}\left(\nabla\vec{u}+\left(\nabla\vec{u}\right)^T\right)+\frac{1}{2}\left(\nabla\vec{u}-\left(\nabla\vec{u}\right)^T\right)$$&#xA;The first term is typically called the strain rate tensor, is symmetric, and it can be shown that it includes no rigid rotational motion. The second term is typically called the vorticity tensor, its is skew symmetric, and it can be shown that it does not contribute to the rate of strain and that it represents rigid like rotational motion.&lt;/p&gt;&#xA;" OwnerUserId="1031" LastEditorUserId="1031" LastEditDate="2015-04-20T00:04:09.090" LastActivityDate="2015-04-20T00:04:09.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2488" PostTypeId="1" CreationDate="2015-04-16T01:08:23.367" Score="4" ViewCount="77" Body="&lt;p&gt;How do CAD designers account for the fact that a blank may have to be flipped to a different side in order to allow the piece to be completely milled through a CNC system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example, let's use this car made from a rectangular block of pine:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aW0RT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aW0RTt.jpg&quot; alt=&quot;pinewood derby &amp;quot;leap frog&amp;quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that three planes need to be surfaced in order to mill the object.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The bottom plane of the block has a slight concave shape&lt;/li&gt;&#xA;&lt;li&gt;The side plane has two hollowed out shapes that would not be accessible from either the top or bottom planes&lt;/li&gt;&#xA;&lt;li&gt;The top plane has a slightly concave inflection applied to the slope &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And there are several cuts that go all the way through the blank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems easy enough to mill the bottom shape first, including any through cuts from the bottom.  This work on the bottom wouldn't significantly affect the overall shape of the blank, and there shouldn't be any complications moving to the next plane.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after you complete the bottom, milling either the side or the top would appear to create problems for flipping the blank to the final plane.  Specifically, the outline of the original blank would have changed significantly enough that prior mill registrations would be thrown off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that this is a somewhat common problem as complex shapes are fairly commonplace.  And I'm more interested in the general approach to solving this problem than I am about how the example car would be designed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how does a CAD designer account for the fact that the edges of the blank will be changing during milling and that those changes may (will) complicate the milling process? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Disclaimer&lt;/em&gt;: My knowledge of CAD design and CNC manufacturing is fairly limited.  My knowledge comes primarily from what I picked up at University and has been supplemented by advanced DIY projects.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-04-16T02:44:28.760" Title="Accounting for blank registration within CAD design" Tags="|design|computer-aided-design|cnc|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2490" PostTypeId="1" CreationDate="2015-04-16T02:22:18.193" Score="2" ViewCount="69" Body="&lt;p&gt;I'm having problems trying to use chamfer on an edge. Typically I see yellow previews of how cuts/extrusions will appear, but this is not the case for my first time using chamfer. I have tried looking at tutorials in chamfer, I keep clicking on an edge, but the purple arrows that say flip sides only appears. On most YouTube tutorials the yellow indicators appear after; why don't I see the same thing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vOIQq.png&quot; alt=&quot;solidworks screen shot&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1302" LastEditorUserId="41" LastEditDate="2015-04-16T17:39:55.247" LastActivityDate="2015-04-16T17:39:55.247" Title="Why don't I see a yellow preview indicator when using the chamfer tool?" Tags="|computer-aided-design|solidworks|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2491" PostTypeId="2" ParentId="2488" CreationDate="2015-04-16T02:44:28.760" Score="3" Body="&lt;p&gt;First, we should recognize that parts aren't just blindly thrown onto a table and a generic mill used to cut whatever shape we want. Each part requires the proper tooling and fixturing. On a production level, the tooling is probably going to be a standard set of tools that you select from for most parts. Depending on what kind of parts you are making, fixturing can shared between parts, but custom parts may require custom fixturing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, at my company, we have different series for our products, and the parts in one series have similar geometry on their key features that allows us to use the same chucks and blocks to hold them during machining processes. However, I recently designed a new part with a new forging that still technically fits into one of our larger series, but will require brand new custom fixturing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, most parts of any complexity won't be able to be fully machined in one operation, and even if it's possible, it doesn't mean it's the best way to get the part to function as intended. This means switching tools while cutting the part from the same direction, and changing the way the part is held to be able to access different sections of the part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To this end, in the course of designing a part, we try to ensure that the machining operations of the same type and on the same side or in the same section of the part can be done at once, i.e. without having to handle the part. This saves machining time, but it also ensures that we use our fixturing as efficiently as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last thing to note, more from a manufacturing engineering point of view, is that the parts aren't necessarily machined entirely on one machine. What this really means is that if we are making 100 parts, we can do the first machining operation of the part 100 times. Then, we reset the machine and do the second operation 100 times. We reset again and do the third operation 100 times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully, at this point you can see how these things come together. We have a rough intent for the order of operations during design, often executed by creating the features in that order in CAD, so that we have an idea of what the part will look like at each step. These operations are grouped into sets that can be performed with the part chucked in the same manner, and then fixturing is created for each set. The first set of operations may only require simple fixturing because it may use a standard blank like a simple round bar. Subsequent operations may be more complex to implement, but we know what the part will look like, and we can plan the fixturing based on that. Unless we're doing one-piece orders, we can do all of one group for a whole order on one machine, then do the next group on a separate machine, or reconfigure the original machine to do the next operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a workflow perspective, I should note that this process is not entirely handled by the design engineers. As a design engineer, I have a good handle on what is and isn't possible, but for special cases, I'll go to the manufacturing engineers or the machinists and ask for their input, see what will make their job easier, and whether what I want to do can actually be done. They will be the ones to actually plan out the order of operations and design/order fixturing and tooling, but communication is key here, as is often the case.&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-04-16T02:44:28.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2493" PostTypeId="2" ParentId="2478" CreationDate="2015-04-16T06:21:24.687" Score="4" Body="&lt;p&gt;There is a method for plating stainless steel with an Fe-25Ni-16Cr alloy which uses electroplating. The results from &lt;a href=&quot;https://www.researchgate.net/publication/231241559_Electroplating_of_Stainless_Steel&quot; rel=&quot;nofollow&quot;&gt;the article&lt;/a&gt; give a maximal electroplated layer thickness of 23 um. The properties of the electroplated alloy seem to be of similar quality to their reference sample of SAE 316 stainless steel. Electroplating stainless steel onto substrates is challenging specifically due to the required element chromium, which gives stainless steel its resistance to oxidative degradation. It is evidently difficult to create a stable, three element solution of Fe, Ni, and Cr due to the lack of a well-developed chemical complex for delivering chromium to the cathode. A quick Google Scholar search turns up 19 citations for that article, including one patent, which may be of further use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There does not appear to be a commercial entity selling this particular service as of yet, that I could find. The typical turn-around time for materials processes from initial research to practical commercial application is on the order of 20 years, so you may have to wait some time yet (~2030) for this particular process.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2015-04-16T06:21:24.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2494" PostTypeId="1" AcceptedAnswerId="2505" CreationDate="2015-04-16T06:48:03.313" Score="3" ViewCount="2992" Body="&lt;p&gt;I have an aluminum square tube (diameter 1/2&quot;) that I would like to attach to a flat aluminum surface which has a circular hole of smaller diameter in it for water to flow through at low pressure. Would welding a flange to the square tube (with screw holes for going into the surface) be the best option, or are there better ways for making this kind of attachment? This is not a high pressure system so I don't think there will be any complications there.&lt;/p&gt;&#xA;" OwnerUserId="1303" LastActivityDate="2015-04-16T19:09:02.520" Title="Best way to attach a square tube to a flat surface?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2495" PostTypeId="2" ParentId="2494" CreationDate="2015-04-16T06:56:24.283" Score="1" Body="&lt;p&gt;Yes, attaching the two parts with a flange is the first thing that comes to mind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few other options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weld, solder, or epoxy the tube directly to the flat surface.&lt;/li&gt;&#xA;&lt;li&gt;Solder the square tube to a fitting whose outer diameter is the same size or slightly smaller than the hole.&lt;/li&gt;&#xA;&lt;li&gt;Reshape the end of the square tube to be round, and attach using one of the previously-mentioned methods.&lt;/li&gt;&#xA;&lt;li&gt;Seal the end of the square tube drill a hole the same size as the other hole, and join the two with appropriately-sized fittings (e.g., plumbing fittings).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Edit: You mentioned in a comment that the connection needs to be rigid, but otherwise you could clamp a hose around the square tube and attach the other end of the hose to a fitting installed through the hole in the flat aluminum surface.&lt;/p&gt;&#xA;" OwnerUserId="1277" LastEditorUserId="1277" LastEditDate="2015-04-16T07:14:27.223" LastActivityDate="2015-04-16T07:14:27.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2496" PostTypeId="1" AcceptedAnswerId="2499" CreationDate="2015-04-16T07:57:00.897" Score="6" ViewCount="4511" Body="&lt;p&gt;Recently I observed installing new storm sewage along a city street. Pipes have plain surface (not corrugated) and are made of some plastic looking like HDPE (polyethylene) and their diameter is around 0,5 meters and the walls are around 40 millimeters thick. That's a lot of material. I assume the same structural strength could be achieved if the pipes were made corrugated and with much thinner walls. I saw corrugated plastic pipes for storm sewage once and so they surely exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would plain surface pipes be preferred over corrugated pipes for this scenario?&lt;/p&gt;&#xA;" OwnerUserId="374" LastActivityDate="2015-04-16T09:08:45.537" Title="Why prefer plain pipes over corrugated pipes for storm sewage and culverts?" Tags="|civil-engineering|piping|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2497" PostTypeId="2" ParentId="2240" CreationDate="2015-04-16T08:25:29.763" Score="4" Body="&lt;ul&gt;&#xA;&lt;li&gt;First you need a part that &lt;strong&gt;exceeds its fatigue limit under expected stresses&lt;/strong&gt; because you need slow coalescence of cracks occurring uniformly in the measured area, not a local defect which might occur outside the sensor placement and resistant to modelling based on measurements from other pieces of the part.  Steel/titanium etc parts stressed below their fatigue limit won't tell you anything.&lt;/li&gt;&#xA;&lt;li&gt;Parts that are easy to access can't lower maintenance costs as much as &lt;strong&gt;difficult to reach&lt;/strong&gt; structures.&lt;/li&gt;&#xA;&lt;li&gt;Since you're trying to lower the cost without compromising safety of the procedures, the logical step is to put a monitor on the parts expected to show signs of fatigue first and for which fatigue is well understood.  You want agreement with the model, proving that you can design the sensor, not that the sensor happened to find or not find crack growth.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Based on these considerations, problems I would expect are of interest are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The fuselage.  Compression cycles are a different fatigue mechanism than time spent in the air, requiring modelling and checks.&lt;/li&gt;&#xA;&lt;li&gt;Landing gear attachment point.  Landing hardness and its effects on the fatigue of connecting structures can potentially not match airframe life &lt;em&gt;or&lt;/em&gt; duty cycles, requiring checks.&lt;/li&gt;&#xA;&lt;li&gt;Wings.  Wings encounter a large number of cycles and require access panels to measure internal structures.  Fatigue in the middle of a large beam is probably relatively predictable.&lt;/li&gt;&#xA;&lt;li&gt;Engine pylon attachment points.  Compressor stalls, vibration in the engine, and awkward landings/maneuvers can all result in uncommon forces that would be difficult to treat based on empirical data, requiring checks&lt;/li&gt;&#xA;&lt;li&gt;The tail surfaces that are far off the ground are a good place to perhaps save time on lifts.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Wings and fuselage are going to be best to demonstrate the sensor engineering principles.  Pylons and landing gears (excepting any steel parts) would present the most insight perhaps.&lt;/p&gt;&#xA;" OwnerUserId="1305" LastEditorUserId="1305" LastEditDate="2015-04-27T06:20:40.780" LastActivityDate="2015-04-27T06:20:40.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2498" PostTypeId="2" ParentId="2478" CreationDate="2015-04-16T08:28:40.887" Score="1" Body="&lt;p&gt;You also can use mechanical plating (I'm not sure it's the right name) by basically rolling two sheets of metal together (&lt;a href=&quot;http://www.google.de/imgres?imgurl=http%3A%2F%2Fwww.wickeder.de%2Ftypo3temp%2Fpics%2Fcaee557c32.jpg&amp;amp;imgrefurl=http%3A%2F%2Fwww.wickeder.de%2F&amp;amp;h=373&amp;amp;w=970&amp;amp;tbnid=j2SfFL32ASitrM%3A&amp;amp;zoom=1&amp;amp;docid=8DX_XpgHfcTrqM&amp;amp;ei=u3EvVaujO434asH_gGA&amp;amp;tbm=isch&amp;amp;iact=rc&amp;amp;uact=3&amp;amp;dur=1063&amp;amp;page=1&amp;amp;start=0&amp;amp;ndsp=45&amp;amp;ved=0CGMQrQMwFA&quot; rel=&quot;nofollow&quot;&gt;Image&lt;/a&gt;) or &lt;a href=&quot;http://en.wikipedia.org/wiki/Explosion_welding&quot; rel=&quot;nofollow&quot;&gt;Explosion welding&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="405" LastEditorUserId="86" LastEditDate="2015-04-16T11:28:33.393" LastActivityDate="2015-04-16T11:28:33.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2499" PostTypeId="2" ParentId="2496" CreationDate="2015-04-16T09:08:45.537" Score="7" Body="&lt;p&gt;You are correct in your assumption that corrugations give pipes strength and hence such pipes can have thinner walls that smooth walled pipes made of the same material when used for the same application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pipelines are made to transport fluids or slurries. When it comes to the flow of fluids or slurries in a pipe, one of the import considerations is the smoothness, or roughness, of the internal wall of the pipes. Smooth walled pipes, or as you called them “plain pipes” have a small surface roughness on their internal walls. The corrugations of corrugated pipes are a very large roughness. This will increase the resistance to flow within the pipes and increase the turbulence of flow in the pipe. Increased turbulent flow, particularly of slurries, can lead to increased wear of pipes. The smoother the flow in pipes the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes, the flow of sewage in pipes is not a continuous steady stream. As the spasmodic flow subsides the sewage fluid and solids will collect in the valleys of the corrugations. If the corrugated pipes are made of steel, this  will lead to a higher probability of the pipes corroding more quickly in these locations. The other result will be the presence of decomposing sewage and fluid trapped in the corrugations which will lead to the creation of a habitat for harmful bacteria and the production of noxious gases. The result of which would be an increased risk to public health. To prevent this, such pipes will need periodic flushing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With smooth walled pipes there will be less accumulation of sewage based items in the pipe as it will tend to form a narrower line along the center of the base of the pipe. Corrugations run transverse to the pipe, allowing for more material to collect in bottom of the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wall thickness of any pipe will be determined by the strength properties of the material from which the pipe is made and the pressures that the pipe must withstand; both internal and external pressure. The deeper a pipe is buried the greater the soil/ground pressure that it must withstand. HDPE is a weak material, a wall thickness of 40 mm for a 500 mm diameter pipe means it will have to withstand high pressures but it also helps the pipe keep its shape.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-04-16T09:08:45.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2500" PostTypeId="1" AcceptedAnswerId="2519" CreationDate="2015-04-16T12:26:33.823" Score="5" ViewCount="8447" Body="&lt;p&gt;I have a 8-pole synchronous machine (as a generator), which is rated to produce 1000 kVA. Its synchronous reactance is 0.4 Ohms.&#xA;It is connected to a diesel engine with the grid frequency 60 Hz.&#xA;The load is resistive type, 500 kW, 480 V. &#xA;I need to calculate the speed, power factor and power angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Speed can be calculated by: $n = 120 \cdot \frac{f}P$  , where $f$ is frequency (60 Hz) and $P$ are poles (8)  ... and we get the result 900 rpm.&lt;/li&gt;&#xA;&lt;li&gt;Power factor can be calculated by: $\mathrm{PF} = \cos(\mathrm{PA})$, where $\mathrm{PF}$ is power factor, $\mathrm{PA}$ is power factor angle (is that the same as power angle?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, how can I calculate power factor and power angle?&lt;/p&gt;&#xA;" OwnerUserId="1311" LastEditorUserId="5136" LastEditDate="2016-04-29T10:22:59.367" LastActivityDate="2016-04-29T10:22:59.367" Title="Synchronous machine power factor and angle" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2501" PostTypeId="2" ParentId="1882" CreationDate="2015-04-16T15:12:35.020" Score="1" Body="&lt;p&gt;A long trapezoidal wing grows significantly in lift coefficient and area through flaps/slats and Fowler flaps, respectively.  Delta wings, especially those using elevons, essentially have no capability to employ high-lift devices.  A confluent design driver for delta wings is that they stall only at very high angles of attack; a vortex sheet rolls up in front of both edges and makes the wing have a high effective camber after the rest of the airflow goes over the sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawback of using vortex lift is the longer landing gear and very bad lift-to-drag ratio, approximately 3.  This additional drag is what made losing two engines on the Concorde unrecoverable at the takeoff speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as nose-droop on future designs, I would expect to see the use of cameras and monitors instead.  Triple redundant, battery-backed electronics are likely lighter weight than a mechanical hinge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fun fact: it's impossible to generate lift at supersonic speeds without emitting external shockwaves, some towards the ground.  Because we know that any aircraft will generate lift and that the creation of shockwaves is a source of aerodynamic (non-isentropic compression/expansion) drag, a low lift co-efficient given by a large wing implies some aerodynamic savings at cruise.&lt;/p&gt;&#xA;" OwnerUserId="1305" LastActivityDate="2015-04-16T15:12:35.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2502" PostTypeId="2" ParentId="498" CreationDate="2015-04-16T15:53:15.680" Score="4" Body="&lt;p&gt;Start your mental model by assuming a rocket flight path.  A velocity/altitude vs time chart for the Space Shuttle:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EAAnX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EAAnX.jpg&quot; alt=&quot;space shuttle flight profile&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://www.aerospaceweb.org/question/aerodynamics/maxq/maxq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;aerospaceweb.org&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;Jet engines have better &lt;span class=&quot;math-container&quot;&gt;$Isp$&lt;/span&gt; than rockets.  Let's put jet engines on our rocket.  The Falcon 9 puts out about 1.1m lbs of thrust, so we can use a GE-90 to add 120,000 lb, doubling the acceleration at sea level.  Elon Musk said that the Falcon 9 costs about \$54m per entire rocket.  The GE-90 costs about $24m.  Oops.  We added 50% cost to the system (not including integration or developing a recovery system) and the thrust drops off rapidly with altitude.&lt;/p&gt;&#xA;&lt;p&gt;Let's use an F-414 instead.  It costs about \$4m and can be useful up to about Mach 2 with a properly designed inlet, and the speed really helps us develop ram pressure, which feeds the afterburner ramjet style.  We get 26,000lbs of thrust for only \$4m and longer burn, better but not stellar.  The rocket we're lifting still has to be gigantic, so we're not &lt;em&gt;that&lt;/em&gt; well off yet.&lt;/p&gt;&#xA;&lt;p&gt;Pure ramjets present the dilemma of dead weight at liftoff, adding more rocket at the slowest phase of acceleration, so maybe we can't win there either.  Ramjets only overtake rockets in &lt;span class=&quot;math-container&quot;&gt;$Isp$&lt;/span&gt; at about Mach 0.5 and can't generate full thrust for a while because they will blow air out the front if they add too much fuel until ram pressure is high enough.&lt;/p&gt;&#xA;&lt;p&gt;So..  air-breathing engines don't generate a ton of thrust per dollar and have a low speed range.  Wings lift at a rate of about 16:1, so we can use our engines to accelerate slowly and fly to 40,000ft and Mach 1.  This won't save a ton of rocket weight because it's at about 1/25th of it's final velocity and one minute of drag.  Let's say we cut weight 20% and only need to carry 900,000lb.&lt;/p&gt;&#xA;&lt;p&gt;A 747-8 carries 308,000lb of cargo and costs about \$350,000,000.  Let's say that costs and cargo scale linearly, we're at least looking at a \$700,000,000 launcher, a far cry from \$54m, amortized over the number of launches, but so is the development cost, which for the 747-8 was \$3.7bn.  Again, scaling linearly, we need about \$8bn to spread over a lot of launches.  SpaceX recently raised \$1bn from Google and Fido, not quite enough.&lt;/p&gt;&#xA;&lt;p&gt;There lies the dilemma of launching payloads with air-breathing aircraft.  Either you need a vastly cheaper, higher thrust-to-weight jet engine that develops thrust at zero velocity, or you go back to rockets and recovery techniques like ULA and SpaceX are developing.&lt;/p&gt;&#xA;&lt;p&gt;Many have tried to assume longer air-breathing flight paths at ever increasing speeds, but you start using scramjets, pre-coolers, heat management, and it never seems to get smaller, perform over a large enough envelope, or reach a high enough speed to matter for the eventual rocket anyway.&lt;/p&gt;&#xA;" OwnerUserId="1305" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-02-20T03:02:40.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2503" PostTypeId="1" CreationDate="2015-04-16T08:33:34.123" Score="5" ViewCount="1002" Body="&lt;p&gt;I have six known elastic constants of &lt;a href=&quot;http://en.wikipedia.org/wiki/Tetragonal_polycrystalline_zirconia&quot; rel=&quot;noreferrer&quot;&gt;TZP&lt;/a&gt;. What is the relationship between the known and the unknown constants? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Known constants (GPa):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_{11}=327\\ c_{12}=100\\ c_{13}=62\\  c_{33}=264\\ c_{44}=59\\ c_{66}=64$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find the others? Is $c_{22}=c_{11}$? Does TZP exibhit transverse isotropy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the relationships between the tensor components and the coordinate axes given by: $c_{11}$ for x-direction, $c_{22}$ for y and $c_{33}$ for z; $c_{23}$ (4) for yz, $c_{31}$ (5) for zx, and $c_{12}$ (6) for xy? &lt;/p&gt;&#xA;" OwnerUserId="1318" OwnerDisplayName="Jimmy" LastEditorUserId="368" LastEditDate="2015-04-16T17:03:54.183" LastActivityDate="2018-06-26T17:19:39.843" Title="What are the elastic constants for tetragonal zirconia polycrystal (TZP)?" Tags="|materials|stiffness|elastic-modulus|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2504" PostTypeId="1" CreationDate="2015-04-16T17:55:42.033" Score="2" ViewCount="943" Body="&lt;p&gt;A bimetallic strip is a band of two metals attached in parallel that expand at different rates in response to temperature changes, thus bending the band, often used for measuring temperature. A bike project I'm attempting would be greatly improved if I had a band that bent in response to moisture changes. Specifically, brake/rim dust being fed back into the braking system is considered the top factor for rim wear in wet weather riding, and I've attached a prototype sponge to my brake pads to clean the rim when I brake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works well but I don't have the sponge on all the time since it would just be rubbing a dry rim, and it's a pain to engage, so if I want to make it useful to more than just tinkerers it needs to auto-engage. There are a lot of other issues to consider even if I had a material that bent when wet, but this is the thing I know the least about how to design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to tinker with an arm that closes a 2-5mm gap when wet from rain. A separate sponge or squeegee will deal with brake dust primarily, but some possible contamination from brake dust and road salts should be expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest thing I've been able to find is &lt;a href=&quot;http://arstechnica.com/science/2013/01/polymer-film-generates-mechanical-energy-just-by-getting-wet/&quot; rel=&quot;nofollow&quot;&gt;a polymer sheet that dances around on a wet surface&lt;/a&gt;, which tells me it could be done, but there's got to be a simpler way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even using something like a strip of &lt;a href=&quot;http://www.greenstreak.com/div3/Waterstops/swellstop.asp&quot; rel=&quot;nofollow&quot;&gt;swellstop&lt;/a&gt; (a rubber+sand compound for sealing concrete) glued to a non-expanding rubber might work, but I suspect swellstop doesn't contract when dry, and probably doesn't dry quickly either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's relatively easy to find materials that expand when wet, but seems hard to find ones that return to their original shape when dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I really have to resort to &quot;smart&quot; polymers? Is anyone familiar with a material that is less exotic that could fill my need?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if I can't find one, what's my next step? I'm a dog flying an airplane when it comes to designing something like this from scratch, do I need a team of PhDs? Where would I even start?&lt;/p&gt;&#xA;" OwnerUserId="1314" LastEditorUserId="33" LastEditDate="2016-01-26T03:29:00.333" LastActivityDate="2016-01-26T03:29:00.333" Title="What materials could be used to make a strip that bends when wet?" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2505" PostTypeId="2" ParentId="2494" CreationDate="2015-04-16T19:09:02.520" Score="2" Body="&lt;p&gt;If it needs to be removable, then yes, attaching a flange will be the best choice. If so, you'll need to put some sort of seal between the flange and the existing aluminum surface. The faying surfaces of aluminum will not, on their own, form a good seal. There are many types of seals, but an elastomeric gasket or O-ring will probably be your best bet. If you use an O-ring, it is best to put a groove in one or both of the mating parts so it is properly supported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for making the watertight aluminum-to-aluminum connection itself, brazing will probably be your best bet. Soldering is possible, but won't be as strong or reliable. Welding is possible, although a 1/2&quot; tube is likely to have a thin wall, so you'd need a skilled TIG welder to make the joint both strong and watertight. Brazing is the best compromise because it will be reasonably strong, and doesn't take too much skill to make watertight. There will also be less of a chance of melting through the wall or collapsing the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that can make this kind of joint easier is running the tube through the flange, so it sticks out maybe 1/4&quot; from the other side of the flange. This reduces the risk of deforming the tube as you heat it, and it's easy to grind or machine the extra tube off when you're done welding/brazing. If you can keep the gap reasonably small between the tube and the thickness of the flange, you'll also get some help from capillary action to make your weld stronger and more watertight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do keep in mind that some alloys of aluminum are significantly effected by heat, and heat as low as 400 degrees Fahrenheit can soften the aluminum permanently. If strength or hardness is important to you, you'll want to pay attention to what alloys you're using.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-04-16T19:09:02.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2506" PostTypeId="2" ParentId="2504" CreationDate="2015-04-16T20:12:33.317" Score="3" Body="&lt;p&gt;I do not know of any material which will meet your needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I can offer a potential solution.  You stated that your sponge is &quot;a pain to engage&quot; and I think that is your real problem.  If you can make your sponge easier to engage and disengage, then you don't need the exotic material that you are looking for. I don't know what type of brakes or bicycle you have, so I won't try to offer specific solutions on how to engage or disengage your sponge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to consider measuring how much the sponge interferes with your normal riding. Measuring the power required to turn the wheel against the sponge is quite simple. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Get a small &lt;a href=&quot;https://en.wikipedia.org/wiki/Spring_scale&quot; rel=&quot;nofollow&quot;&gt;spring scale&lt;/a&gt; and hook it on the end of one of your spokes with the sponge engaged and the wheel stationary&lt;/li&gt;&#xA;&lt;li&gt;Measure the length from the axle to the scale. Call it $l$.&lt;/li&gt;&#xA;&lt;li&gt;Slowly pull on the scale until the wheel starts to turn and note the force it took to engage. Call that force $F$. Make sure your are holding the scale perpendicular to the spoke.&lt;/li&gt;&#xA;&lt;li&gt;Measure the diameter of the wheel. Call it $d$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We can now find the torque exerted on the wheel by the sponge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T = Fl $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the torque, the power required to pedal against your sponge, as a function of bicycle speed is given by the following equation.&#xA;If your force ($F$) is in Newtons, your lengths ($l$ and $d$) are in meters, and your speed ($S$) is in meters per second, the answer will be in Watts. If your units are different, just use Wolfram Alpha and it will convert your units for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P = 2TS/d $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that $P$ will be small compared to the overall effort of riding a bike (50W to 100W) and if it is, you don't need to worry at all!&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-04-16T20:12:33.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2507" PostTypeId="1" AcceptedAnswerId="2522" CreationDate="2015-04-16T23:36:50.343" Score="7" ViewCount="1340" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chemical_oxygen_generator&quot;&gt;Chemical oxygen generation (COG)&lt;/a&gt; is used in commercial airliners to supply emergency oxygen to passengers and crew if the plane is depressurized. Weight for weight it produces far more oxygen than compressed oxygen tanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read the original paper on COG, several sites from oxygen candle manufacturers and the US Federal Aviation Administration document on oxygen equipment (riveting I assure you). However, most of the rest of my knowledge is from my own background as a physicist; I want to know how these methods compare from an engineering perspective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides reduced weight, are there any other advantages of COG in aircraft? Does it have added safety benefits when compared to high pressure gaseous oxygen storage? The oxygen candle burns extremely hot when it goes off but I'd imagine high pressure gas canisters could be far more dangerous and prone to exploding if damaged.&lt;/p&gt;&#xA;" OwnerUserId="1317" LastEditorUserId="368" LastEditDate="2015-04-17T16:36:18.643" LastActivityDate="2016-02-10T17:09:47.430" Title="What are the advantages of chemical oxygen generation over compressed tank storage?" Tags="|aerospace-engineering|aircraft-design|chemical-engineering|compressed-gases|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2508" PostTypeId="2" ParentId="2503" CreationDate="2015-04-17T01:01:42.903" Score="1" Body="&lt;p&gt;&lt;strong&gt;Generalized Linear Elasticity Tensor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, stress and strain can each be modeled by a 2nd-rank tensor with $3\times 3=9$ elements. A full linear relationship between the stress and strain tensors requires each element to be paired with every other element, giving a 4th-rank tensor with $9\times 9=81$ elements. By symmetry of the stress tensor, i.e. $\sigma_{ij}=\sigma_{ji}$, every element $C_{ijkl}=C_{jikl}$ in the full elasticity tensor. The number of unique elements is now 54, since one of every three elements has a symmetric partner. Likewise by symmetry of the strain tensor, i.e. $\varepsilon_{ij}=\varepsilon_{ji}$, every element $C_{ijkl}=C_{ijlk}$ in the full elasticity tensor. The number of unique elements is now 36, again a reduction of one-third. There is an additional strain-energy symmetry because the strain energy contribution of one elastic constant depends on $C_{ijkl}\varepsilon_{ij}\varepsilon_{kl}$, and reversing the order to $C_{klij}\varepsilon_{kl}\varepsilon{ij}$ should have no effect on the total strain energy, giving $C_{ijkl}=C_{klij}$. The number of unique elements is now 21.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Voigt Reduced Elasticity Tensor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Voigt notation, the 21 elements may be expressed in a symmetric $6\times 6$ 2nd-rank tensor or matrix, which relates a $6$-element 1st-rank stress tensor or vector to a similar strain vector. The elements of the stress and strain vectors are $\sigma_{ii}\rightarrow\sigma_{i}$ and $\varepsilon_{ii}\rightarrow\varepsilon_{i}$ for axial stresses and strains, and $\sigma_{23}\rightarrow\sigma_{4}$, $\sigma_{31}\rightarrow\sigma_{5}$ and $\sigma_{12}\rightarrow\sigma_{6}$ for shear stresses, with corresponding doubled shear strains $2\varepsilon_{23}\rightarrow\varepsilon_{4}$, $2\varepsilon_{31}\rightarrow\varepsilon_{5}$ and $2\varepsilon_{12}\rightarrow\varepsilon_{6}$. I will refer to all shear stresses as $\tau_{i}$ and strains as $\gamma_{i}$. The coefficients then go from 4 subindices to 2 subindices, each ranging from 1 to 6, and I will write with lower case as $c_{ij}$, as the OP has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hookean Elasticity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming Hookean linear elasticity, 9 of the 21 elements of the Voigt tensor immediately vanish, specifically those relating shear stresses ($\tau_{ij}$) to axial strains ($\varepsilon_{kl}$) and axial stresses ($\sigma_{ij}$) to shear strains ($\gamma_{kl}$), leaving 12 elements. The elements thus eliminated are 14, 15, 16, 24, 25, 26, 34, 35, and 36.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three of the remaining 12 relate shear stress in one sense to shear strain in another sense (e.g. $\tau_{12}$ to $\gamma_{23}$), and thus also vanish under Hookean linear elasticity. The elements thus eliminated are 45, 56, and 46.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tetragonal Symmetry in Elasticity&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only 9 elements remain, in three categories:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial: $c_{11},c_{22},c_{33}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial due to Poisson effect: $c_{12},c_{13},c_{23}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear: $c_{44},c_{55},c_{66}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By symmetry of the tetragonal lattice, lattice parameters $a_{1}=a_{2}\neq a_{3}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_{11} = c_{22}$ because the elastic response must be the same in direction $1$ and $2$ due to the lattice symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_{13} = c_{23}$ because the Poisson effect is similarly symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_{44} = c_{55}$ because the elastic shear response must be the same in the sense of direction pairs $1,3$ ($c_{44}$) and $2,3$ ($c_{55}$), since directions $1$ and $2$ are symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c_{33}$ and $c_{66}$ do not share any symmetries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves 6 unique elements, which are the ones you have, i.e. $c_{11},c_{12},c_{13},c_{33},c_{44},c_{66}$.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="-1" LastEditDate="2018-06-26T17:19:39.843" LastActivityDate="2018-06-26T17:19:39.843" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2509" PostTypeId="2" ParentId="2507" CreationDate="2015-04-17T02:49:51.983" Score="4" Body="&lt;p&gt;From personal experience of refuge systems, oxygen generating systems are preferred over medical oxygen gas tanks because of space requirements, portability and the ease of relocation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When emergency refuge systems are designed and installed, either underground, in mines or tunnelling projects or at industrial plants like oil refineries and chemical plants, the critical aspects are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the number of people that will require refuge&lt;/li&gt;&#xA;&lt;li&gt;the duration of survivability, prior to being rescued&lt;/li&gt;&#xA;&lt;li&gt;robustness of the refuge system&lt;/li&gt;&#xA;&lt;li&gt;prevention of damage to the oxygen system from events such as:&#xA;corrosive elements, explosions, cave-ins and entombment.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To supply the emergency oxygen requirements for a group of people a large number of oxygen tanks are required and the plumbing required to connect all the tanks with piping, valves and pressure regulators can be considerable. All this requires a lot of space. It also requires a lot time to connect and disconnect and reconnect when the refuge system is relocated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen generating systems are more compact, they require less space, require less time to maintain and they are portable. Refuge system using oxygen generating devices can be repositioned quickly and with minimal effort compared. The &lt;a href=&quot;http://www.minearc.com.au/&quot; rel=&quot;nofollow&quot;&gt;MineARC&lt;/a&gt; website shows the refuge portable chambers they make, by way example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you state in your question weight for weight, oxygen generators produce more oxygen than gas canisters. It is for this reason that oxygen generators are also used for emergency purposes in submarines and large aircraft: they are compact, portable and they offer the occupants of such craft a longer duration of survivability which in these craft is critical.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an after thought, for the pilots it might be worth considering an oxygen generator that is not a candle and does not create the same amount of heat a candle would. Drager and MSA make portable oxygen supply devices for use in emergencies that allow people to pass through a hazardous atmosphere to reach a safe location. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The devices use potassium superoxide as the oxygen source. It reacts with the water vapour in a person's breathe to generate oxygen via the chemical reaction,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${4KO}_2 + {2H}_2{O} = {4KOH} + {3O}2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.draeger.com/sites/en_uk/Pages/Chemical-Industry/Draeger-Oxyboks-K.aspx&quot; rel=&quot;nofollow&quot;&gt;Drager&lt;/a&gt; &lt;a href=&quot;http://www.draeger.com/sites/assets/PublishingImages/Products/cin_oxyboks_k/Master-Documents/oxyboks-k25-pi-9046597-en.pdf&quot; rel=&quot;nofollow&quot;&gt;Oxyboks&lt;/a&gt; and &lt;a href=&quot;http://www.msha.gov/S&amp;amp;HINFO/USERNOTE/PORTPAK.pdf&quot; rel=&quot;nofollow&quot;&gt;MSA&lt;/a&gt; equivalent offer an oxygen supply that will last between 8 minutes and 2 hours, depending on breathing rate. In an aircraft, this time could be expanded by having a larger supply of potassium superoxide.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-04-17T14:53:18.193" LastActivityDate="2015-04-17T14:53:18.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2510" PostTypeId="1" AcceptedAnswerId="11628" CreationDate="2015-04-17T04:34:05.567" Score="5" ViewCount="15153" Body="&lt;p&gt;I am designing a system where a beam is acted on by a horizontal force. I intend to hold the beam in place with a set of screws, although I am uncertain how much force they will be able to withstand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What method can I use to determine the holding force of screws?&lt;/p&gt;&#xA;" OwnerUserId="340" LastActivityDate="2016-09-28T01:25:41.290" Title="How to calculate screw pullout strength?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2511" PostTypeId="1" AcceptedAnswerId="2512" CreationDate="2015-04-17T09:11:43.427" Score="6" ViewCount="1690" Body="&lt;p&gt;Electronic cigarette elements that heat the fluid are usually termed &quot;atomizers.&quot; They are usually just some nichrome wire or similar wrapped around a wick that becomes saturated with the nicotine fluid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the process that &quot;vaporizes&quot; the fluid? Does the element just heat the fluid until it evaporates? Or something else, e.g., is the majority of fluid sort of spattered off the wick due to a smaller proportion of it violently evaporating? Or some other process/phenomenon? Is the term &quot;atomizer&quot; a scientific term in this context, or just something adopted for electronic cigarettes?&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="272" LastEditDate="2015-04-17T18:30:39.253" LastActivityDate="2015-04-17T18:30:39.253" Title="Are electronic cigarette &quot;atomizers&quot; just evaporating the fluid?" Tags="|thermodynamics|vaporization|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2512" PostTypeId="2" ParentId="2511" CreationDate="2015-04-17T11:21:17.933" Score="8" Body="&lt;p&gt;Yes, it evaporates the fluid as you describe, using a wicking substance (usually silica fibers or wire wool) and a heating coil. Note that it doesn't boil the fluid to achieve evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of &lt;a href=&quot;http://en.wikipedia.org/wiki/Evaporation#Factors_influencing_the_rate_of_evaporation&quot;&gt;factors that influence the rate of evaporation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Surface area&lt;/strong&gt;. Evaporation happens at the boundary between the liquid and air, so the greater surface area you achieve, the greater the possible rate of evaporation. The wick draws the liquid by &lt;a href=&quot;http://en.wikipedia.org/wiki/Capillary_action&quot;&gt;capillary action&lt;/a&gt; and allows it to spread out along its fibers, creating a large surface area in a small space.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Temperature&lt;/strong&gt;. The state-change from liquid to gas requires energy, this comes from heat in the surrounding environment. That's why when you blow on your damp hands they get cold. The heating coil ensures there's a plentiful supply of heat energy to help with this state-change.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Low Pressure&lt;/strong&gt;. Air pressure pushes against the surface and makes it harder for evaporation to happen. Due to &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli%27s_principle&quot;&gt;Bernoulli's principle&lt;/a&gt;, the moving air-flow from the drag lowers the air pressure and helps with evaporation.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Low Concentration&lt;/strong&gt;. If the air is already saturated with vapor (from the liquid or other substances already in the air) then it will be harder for evaporation (new vapor going into the air). The flow of fresh air from outside going across the wick ensures maximum evaporation. You may notice that when it is humid, you get much less of the cigarette vapor per drag.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Regarding the term &lt;em&gt;atomizer&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It usually means converting a liquid into very small particles (for example by means of vibration or a nozzle), small enough to be suspended in air. However, the goal of this is sometimes to cause evaporation (it creates a large surface area on all the tiny droplets, making evaporation easier), for example in an ultrasonic humidifier. A counter example (where the goal is not evaporation) would be a spray-paint nozzle. So that is the typical use of the term, I speculate that the reason e-cigarette manufacturers have adopted this term is the shared goal of vaporization along with the fact that &lt;em&gt;atomizer&lt;/em&gt; sounds pretty cool.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-04-17T11:21:17.933" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2513" PostTypeId="2" ParentId="2511" CreationDate="2015-04-17T12:28:56.540" Score="6" Body="&lt;p&gt;Vapor is something that forms over any liquid and does not require the liquid to be above its boiling temperature.  The atmosphere is a great example; it contains about &lt;a href=&quot;https://www.google.com/search?q=how+muych+water+vapor+in+the+atmoshpere&amp;amp;oq=how+muych+water+vapor+in+the+atmoshpere&amp;amp;aqs=chrome..69i57j0.7306j0j7&amp;amp;sourceid=chrome&amp;amp;es_sm=93&amp;amp;ie=UTF-8#q=how+much+water+vapor+in+the+atmosphere&amp;amp;spell=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;37 billion gallons&lt;/a&gt; of water in the form of vapor, but the temperature of (most of) the atmosphere is well below the boiling point of water.  The goal of a vaporizer is simply to increase the rate of this process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand how it does this you need to know a little bit about &lt;a href=&quot;http://en.wikipedia.org/wiki/Vapor_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;vapor pressure&lt;/a&gt;.  A liquid at a given temperature will have some of its molecules which are moving fast enough to escape the surface of the liquid and end up in the vapor phase above the liquid.  If you put the liquid in a sealed container, this process will continue until the &lt;em&gt;partial pressure&lt;/em&gt; of the vapor reaches the vapor pressure of the liquid.  What is important about this process is that it is only the pressure of the substance itself (i.e. partial pressure) which matters, even if the rest of the air above the substance is at a higher pressure than the vapor pressure, the liquid will still vaporize until its partial pressure reaches the vapor pressure.  The vapor pressure increases with increasing temperature until it reaches the atmospheric pressure at the boiling point of the liquid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Vapor_pressure.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1QeuJ.png&quot; alt=&quot;Vapor pressure illustration and plots&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that background, we can now discuss how the vaporizer in an e-cigarette works.  The amount of ambient vapor (in the surrounding air) of the nicotine solution in e-cigarettes is very low so turning it into vapor is simply a matter of letting it sit for long enough.  The amount of time it would take is too long for a smoker to wait though so they use two tricks to speed up the process.  1) The fluid is heated so that the vapor pressure increases, and 2) a wick is used to drastically increase the surface area between which the air and fluid interact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heating of the fluid has the desirable side effect of making the cloud of vapor visible too.  The best way to understand this is to think about a boiling pot of water.  The water is converting to a gas at the surface and rising away from the pot, but you can't see water in its gaseous stage.  You can, however, see steam because the water gas is mixing with the cool air above the pot and small droplets of water are condensing, much like a cloud.  The same thing happens in an e-cigarette.  Local heating in the vaporization area heats the air along with it and produces a vapor which you can't see.  When this vapor mixes with the cool air as you inhale, it condenses to tiny droplets which give it the desirable appearance of smoke.  So, when you blow out after puffing you are actually blowing out a little nicotine filled cloud.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the term &lt;a href=&quot;http://en.wikipedia.org/wiki/Atomization&quot; rel=&quot;nofollow noreferrer&quot;&gt;atomizer&lt;/a&gt;, it does have a semi-scientific definition.  An atomizer is something which creates a fine suspension of liquid droplets in a gas.  This is exactly what I described to you above; the vaporizer turns the liquid to vapor which then becomes a suspension as the vapor saturated air cools.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-04-17T14:29:31.293" LastActivityDate="2015-04-17T14:29:31.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2514" PostTypeId="1" CreationDate="2015-04-17T14:54:04.973" Score="5" ViewCount="313" Body="&lt;p&gt;I recently saw a wind turbine that I'm trying to identify the type or category of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mD1wK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mD1wKt.png&quot; alt=&quot;turbine view 1&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pTTBj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pTTBjt.png&quot; alt=&quot;turbine view 2&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JQGTe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JQGTet.png&quot; alt=&quot;turbine view 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the turbine looks like two squirrel cage fans mounted side by side on a fairly tall ~40' pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blades of the turbines rotate along the horizontal axis of turbine.  Essentially, they work like &lt;a href=&quot;http://en.wikipedia.org/wiki/Paddle_wheel&quot; rel=&quot;nofollow noreferrer&quot;&gt;paddle-wheels&lt;/a&gt; of a steamboat.  There are shrouds in front of the blades to direct the airflow, which obscures the view of the blades themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/930lS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/930lSt.png&quot; alt=&quot;turbine rotation called out&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Note, rotational direction may be in the opposite direction of what's drawn.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know name of the type or category of this type of wind turbine as I would like to research it some more.  Most of the images and types I was able to find so far have been the &quot;windmill&quot; style turbine with a central housing and 3 or 4 blades coming off of the center.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-04-17T16:41:52.273" LastActivityDate="2015-04-24T21:54:41.413" Title="Identifying category of wind turbine" Tags="|turbines|wind-power|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2516" PostTypeId="1" AcceptedAnswerId="5726" CreationDate="2015-04-17T15:08:22.973" Score="11" ViewCount="1240" Body="&lt;p&gt;The design of a retaining wall commonly involves determining the lateral earth pressure using either Rankine theory or Coulomb theory. Both theories involve mobilising the shear resistance of a triangular wedge of soil extending for a considerable distance away from the base of the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a double-walled cofferdam, such as the one in the picture below, the short distance between the two walls would prevent such failure wedge from extending all the way down to the bottom. In which case, how does one go about determining the earth pressure from the sand fill material in between the two walls?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Ddfe.jpg&quot; alt=&quot;double-walled cofferdam&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2015-10-16T15:05:30.177" Title="How to determine the lateral earth pressure in a double-walled cofferdam?" Tags="|civil-engineering|geotechnical-engineering|soil|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2517" PostTypeId="2" ParentId="2514" CreationDate="2015-04-17T15:53:47.147" Score="3" Body="&lt;p&gt;The design looks like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Savonius_wind_turbine&quot; rel=&quot;nofollow&quot;&gt;Savonius turbine&lt;/a&gt;, but horizontally mounted, instead of the typical vertical mounting that you see with this design. I'm really not sure why this would be mounted this way either, as one of the key benefits of the Savonius design is that it's never not facing the wind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom of the Wikipedia article mentions that the horizontal mounting could be used to keep the turbine aloft if it's not rigidly supported, because the rotation will generate lift. However, this does have a rigid support, so I'm not sure what the thinking is behind the orientation.&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-04-17T15:53:47.147" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2518" PostTypeId="2" ParentId="2514" CreationDate="2015-04-17T16:19:16.720" Score="1" Body="&lt;p&gt;It is a Savonius Turbine. &lt;a href=&quot;http://www.reuk.co.uk/VAWT-Vertical-Axis-Wind-Turbine.htm&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; provides additional information about Savonius turbines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A Savonius is a type of vertical axis wind turbine (VAWT)&#xA;  generator invented in 1922 by Sigurd Johannes Savonius from&#xA;  Finland....&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The Savonius is a drag-type VAWT which operates in the same way as&#xA;  a cup anemometer....&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Savonius wind turbines typically only have an efficiency of around&#xA;  15% - i.e. just 15% of the wind energy hitting the rotor is turned&#xA;  into rotational mechanical energy....&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Savonius turbines do not scale well to kW sizes, but for a small&#xA;  project they are typically the easiest and cheapest wind turbine&#xA;  generators to build yourself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1116" LastEditorUserId="368" LastEditDate="2015-04-24T21:54:41.413" LastActivityDate="2015-04-24T21:54:41.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2519" PostTypeId="2" ParentId="2500" CreationDate="2015-04-17T17:50:21.653" Score="5" Body="&lt;p&gt;There is a difference between the power factor and power angle calculations. Under ideal conditions, for a purely resistive load, there is a phase difference between the voltage and current. But we always have reactive components present, so there is an angle between the current and voltage. The cosine of the angle between the current and voltage is called the power factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power angle is the angle between a generator's internal voltage and its terminal voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rapidtables.com/electric/Power_Factor.htm&quot; rel=&quot;nofollow&quot;&gt;This link at Rapid tables&lt;/a&gt; provides more detail on how to calculate the power factor:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Power Factor is equal to the real or true power P in watts (W)&#xA;  divided by the apparent power |S| in volt-ampere (VA)....&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For sinusoidal current, the power factor PF is equal to the absolute&#xA;  value of the cosine of the apparent power phase angle φ (which is also&#xA;  is impedance phase angle):&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$PF = |\cos φ|$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;PF&lt;/em&gt; is the power factor.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$φ$ is the apparent power phase angle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nptel.ac.in/courses/IIT-MADRAS/Electrical_Machines_II/pdf/2_6.pdf&quot; rel=&quot;nofollow&quot;&gt;This document&lt;/a&gt; discusses the power angle:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The phase angle $\delta$ between the terminal voltage $V_T$ and the&#xA;  excitation voltage $E_f$ in Fig. 53 is usually termed the torque&#xA;  angle. The torque angle is also called the load angle or power angle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1116" LastEditorUserId="368" LastEditDate="2015-04-24T22:18:00.087" LastActivityDate="2015-04-24T22:18:00.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2520" PostTypeId="1" AcceptedAnswerId="2576" CreationDate="2015-04-17T18:14:56.800" Score="5" ViewCount="1888" Body="&lt;p&gt;In a typical finite elements analysis package, a modal analysis gives the N first natural modes, and it is possible to get the equivalent stress and total deformations for each of these frequencies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what amplitude did the solver use to get these results? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't we obtain a gain instead, for displacements and stresses and for each frequency?&lt;/p&gt;&#xA;" OwnerUserId="1326" LastEditorUserId="272" LastEditDate="2015-04-17T20:19:57.973" LastActivityDate="2015-04-22T04:48:01.243" Title="What amplitude is used for a finite element modal analysis excitation?" Tags="|mechanical-engineering|finite-element-method|simulation|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2521" PostTypeId="1" AcceptedAnswerId="2541" CreationDate="2015-04-17T19:32:28.337" Score="3" ViewCount="268" Body="&lt;p&gt;In my mechanical vibrations class we studied the method to orthonormalize a set of differential equations by the mass matrix (principle coordinates). This is where you take the matrix of eigenvectors from the un-damped system and normalize it by the mass matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying the mass matrix by the modal matrix gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X'MX = I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And multiplying the stiffness matrix by the modal matrix gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X'KX = W^2(ii)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where W^2(ii) is the matrix of squared natural frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is it appropriate to try and orthonormalize the damping matrix in the same fashion? &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X'CX = 2*zeta*omega(ii)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Should this transform the system's damping into principle coordinates?&lt;/p&gt;&#xA;" OwnerUserId="1174" LastActivityDate="2015-04-18T16:53:31.730" Title="Orthonormalization of Damping Matrix -- modal analysis" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2522" PostTypeId="2" ParentId="2507" CreationDate="2015-04-17T19:36:21.790" Score="4" Body="&lt;p&gt;One advantage these chemical systems may have over compressed tanks is how well they are suited for intermittent use and long-term storage (as opposed to regular, continuous operation). The &lt;a href=&quot;https://www.faa.gov/regulations_policies/handbooks_manuals/aircraft/amt_airframe_handbook/&quot; rel=&quot;nofollow&quot;&gt;US FAA Aviation Maintenance Technician Handbook&lt;/a&gt; points out this advantage &lt;a href=&quot;https://www.faa.gov/regulations_policies/handbooks_manuals/aircraft/amt_airframe_handbook/media/ama_Ch16.pdf&quot; rel=&quot;nofollow&quot;&gt;on page 16-5&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sodium chlorate chemical oxygen generators also have a long shelf&#xA;  life, making them perfect as a standby form of oxygen. They are inert&#xA;  below 400 °F and can remain stored with little maintenance or&#xA;  inspection until needed, or until their expiration date is reached.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Compressed gas cylinders, on the other hand, require regular maintenance and testing whether in use or in storage. Pressure vessels are prone to metal fatigue and corrosion over extended periods. Exposure to extreme temperatures can accelerate fatigue or cause an already-weakened cylinder to fail; COG systems, in comparison, are somewhat less vulnerable to temperature extremes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As part of an emergency system, an oxygen cylinder would experience infrequent use and mechanical fatigue would be less of a concern than corrosion.* Corrosion is driven mainly by the reactivity and partial pressure of the stored gas and by the presence of moisture in the cylinder. Proper selection of fittings such as valves and regulators is critical to minimize the potential for a galvanic reaction between the fitting and the cylinder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that pressure vessels support a &lt;em&gt;difference&lt;/em&gt; of pressure between the inside and the outside of the vessel. During a decompression event, the outside pressure drops rapidly while the inside pressure stays constant, leading to a rapid increase in stress on the vessel. This has two implications: first, the capacity of the tank must be reduced to account for the lower atmospheric pressure at operating altitude; second, a worn tank is most likely to explode just when you need it the most—not at all a desirable quality in a safety device. (COG systems do not, to my knowledge, explode, though &lt;a href=&quot;http://lessonslearned.faa.gov/ll_main.cfm?TabID=4&amp;amp;LLID=10&amp;amp;LLTypeID=2&quot; rel=&quot;nofollow&quot;&gt;they can start fires when handled improperly&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to strict inspection and maintenance requirements for compressed gas cylinders. They have to be rotated in and out of service every so often to go through requalification testing and potentially reconditioning or condemnation. For some idea of how complicated these requirements are, take a look at &lt;a href=&quot;http://www.aviationpros.com/article/10387217/high-pressure-gas-cylinders-maintenance-requirements&quot; rel=&quot;nofollow&quot;&gt;this 2003 Aircraft Maintenance Technology article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this, of course, represents a cost to the operator—service contracts, training, etc. In &lt;a href=&quot;http://www.slate.com/articles/business/moneybox/2014/12/cheap_airlines_why_americans_will_suffer_worse_service_on_flights_in_order.html&quot; rel=&quot;nofollow&quot;&gt;an industry that competes mainly on ticket prices&lt;/a&gt;, cost is king, and reduced maintenance costs would be an attractive selling point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also nice to be able to relate the safety system cost directly to the occasions when you need to deploy the system. Going back to the initial point about suitability for intermittent, emergency applications: You're paying maintenance costs for that cylinder whether there's a depressurization or not. The cost of the oxygen itself is probably minor in comparison. So reducing the frequency of incidents by an order of magnitude may not change the cost very much. On the other hand, only paying for the COG system when you use it means the airline can reduce its costs by reducing depressurization incidents. This isn't so much an engineering concern, and I am speculating a bit here; the point is, businesses often don't make decisions in the same way that engineers do, and that's something to keep in mind when comparing design alternatives.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* You might think that pure oxygen is not a corrosive gas (perhaps because we breathe oxygen) but the definition of &quot;corrosive&quot; can vary a bit with context. For example, &lt;a href=&quot;http://www.alspecialtygases.com/files/Design_and_Safety_Handbook_3001.5.pdf&quot; rel=&quot;nofollow&quot;&gt;the Air Liquide Design and Safety Handbook (p. 2)&lt;/a&gt; uses this definition:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;Corrosive Gases&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;These are gases that corrode material or tissue on contact, or in the&#xA;  presence of water. ... Due to the probability of irritation and damage&#xA;  to the lungs, mucous membranes and eye tissues from contact, the&#xA;  threshold limit values of the gas should be rigidly observed. Proper&#xA;  protective clothing and equipment must be used to minimize exposure to&#xA;  corrosive materials.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The emphasis here is on personal safety and health impacts and in that context, oxygen isn't a corrosive gas. But in the context of storing a gas in a pressurized cylinder, the emphasis is on the potential for the stored gas to react chemically with the walls or components in a way that weakens them, and in that context, pure oxygen definitely counts as &quot;corrosive.&quot;&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-02-10T17:09:47.430" LastActivityDate="2016-02-10T17:09:47.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2523" PostTypeId="1" CreationDate="2015-04-17T20:19:14.393" Score="1" ViewCount="361" Body="&lt;p&gt;I'm wondering if somebody could explain the 4 position method a bit more understandably. I am trying to follow the explanation in &lt;a href=&quot;http://momath.org/mathmonday/math-monday-linkages-part-4-four-bars-four-positions/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. What I fail to understand is how does the actual fourbar linkage in this case look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QlZM2.png&quot; alt=&quot;M bar four positions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: Situation described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the image and description I gather that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;F is one of the ground joints (pink) &lt;/li&gt;&#xA;&lt;li&gt;The pink Circle is length of that link (link 3)&lt;/li&gt;&#xA;&lt;li&gt;P&lt;sub&gt;23&lt;/sub&gt; and P'&lt;sub&gt;23&lt;/sub&gt; are location of next link joint, and thus the length of that link is F to P'&lt;sub&gt;23&lt;/sub&gt; minus the pink circles radius (link3)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However what I don't understand from the description is where the other ground joint is. Does it happen to be &lt;strong&gt;M'&lt;/strong&gt;? But that seems awfully short when looking at the demonstration piece? What about &lt;strong&gt;P&lt;sub&gt;13&lt;/sub&gt;&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="368" LastEditDate="2015-04-17T20:50:46.703" LastActivityDate="2015-04-17T22:04:05.890" Title="4 bar mechanism how does this 4 point method work" Tags="|mechanical-engineering|linkage|kinematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2524" PostTypeId="2" ParentId="2523" CreationDate="2015-04-17T22:04:05.890" Score="2" Body="&lt;h1&gt;Answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The other ground link is not shown and neither is the floating link. This diagram only shows&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 4 desired output link positions&lt;/li&gt;&#xA;&lt;li&gt;The 6 poles generated from those positions&lt;/li&gt;&#xA;&lt;li&gt;The process involved to obtain 1 link&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your first two points are correct, however, P&lt;sub&gt;23&lt;/sub&gt; and P'&lt;sub&gt;23&lt;/sub&gt; do not form the next link joint. P&lt;sub&gt;23&lt;/sub&gt; is the &lt;em&gt;pole&lt;/em&gt; of the rotation from DC to CE and P'&lt;sub&gt;23&lt;/sub&gt; is the reflection of P&lt;sub&gt;23&lt;/sub&gt; through an invisible line which passes through P&lt;sub&gt;13&lt;/sub&gt; and P&lt;sub&gt;12&lt;/sub&gt;.  P'&lt;sub&gt;23&lt;/sub&gt; and P'&lt;sub&gt;24&lt;/sub&gt; are used solely to find M'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not very familiar with Burmester's Theory, but I imagine that the process to find the other links involves picking another set of poles, generating another set of mirrored points, and then using another set of circles like M and M' to find the next ground link.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Supplementary Material&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Some other reference reading that may clear things up.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://momath.org/home/math-monday-linkages-part-3-four-bars-two-or-three-positions/&quot; rel=&quot;nofollow&quot;&gt;Another earlier article by the same author on four-bar linkages&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Burmester%27s_theory&quot; rel=&quot;nofollow&quot;&gt;The Wikipedia page for Burmester's Theory&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You may also want to download the GeoGebra file attached to the article and &lt;a href=&quot;http://www.geogebra.org/&quot; rel=&quot;nofollow&quot;&gt;GeoGebra&lt;/a&gt; itself. You can play around with it and see how the diagram was generated.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-04-17T22:04:05.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2525" PostTypeId="2" ParentId="2500" CreationDate="2015-04-17T22:22:08.590" Score="0" Body="&lt;p&gt;If the load is purely resistive, the power factor is 1 and the power angle is zero by definition.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-04-17T22:22:08.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2528" PostTypeId="1" CreationDate="2015-04-18T00:26:32.603" Score="14" ViewCount="1268" Body="&lt;p&gt;In &quot;Structures: or Why Things Don't Fall Down&quot; James Gordon talks about trees being prestressed with the outside in tension and the inside in compression. Why does it make a tree stronger to have it in tension on the surface and compression at the core and how do I read the graph in the first picture below? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in the third pic it talks about doing the reverse with a concrete beam. How do you create tension using rebar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fDadV.jpg&quot; alt=&quot;photo from book&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FKbiM.jpg&quot; alt=&quot;photo from book&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/xNXSN.jpg&quot; alt=&quot;photo from book&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1336" LastEditorUserId="272" LastEditDate="2015-04-18T02:47:44.067" LastActivityDate="2015-04-21T13:04:23.777" Title="Why does prestressing in trees make them stronger?" Tags="|materials|structures|stresses|wood|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2529" PostTypeId="2" ParentId="2528" CreationDate="2015-04-18T01:21:21.880" Score="12" Body="&lt;h1&gt;How To Read the Graphs&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The graphs are plots of &lt;a href=&quot;http://www.engineeringtoolbox.com/stress-strain-d_950.html&quot; rel=&quot;nofollow&quot;&gt;stress&lt;/a&gt; versus position.  Stress is the force per unit area exerted on a material.  Positive values are tension and negative values are compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first (leftmost) plot shows the normal stress vs. position that would be expected from bending if there was no pre-stress.  The middle plot shows the pre-stress on the tree. The rightmost plot shows the &lt;a href=&quot;https://en.wikipedia.org/wiki/Superposition_principle&quot; rel=&quot;nofollow&quot;&gt;superposition&lt;/a&gt;, or sum, of the bending stress and the pre-stress.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Why Trees are Prestressed&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In mechanical engineering, strength is defined as the maximum stress that a material can tolerate before failing. Every material has unique strength limits, called their &lt;em&gt;ultimate strength&lt;/em&gt;. In some materials this is the same in tension and compression but in others the ultimate tensile and compressive strengths are different.&#xA;Most metals have approximately the same tensile and compressive strength, but materials like wood and concrete fall into the latter category, where the tensile and compressive strength are significantly different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article you provided states that the ultimate tensile strength of wood is greater than its ultimate compressive strength. If the tree were not prestressed, the side away from the wind would always fail first, because as shown by the first plot, the maximum tensile and compressive strengths are the same in bending without pre-stress. By having tensile pre-stress on the outside of the tree, the tree can reduce the maximum compressive stress and so stand up to stronger winds.  Note that as a consequence of Newton's Third Law, there must be both tensile and compressive pre-stress to balance the forces.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;In Concrete&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In concrete, the situation is different because concrete has essentially zero tensile strength.  Prestressed concrete avoids failure by having the steel rebar in a state of tensile pre-stress and the concrete in a state of compressive pre-stress.  This can be accomplished either by stretching the rebar while the concrete is being poured, or by pouring concrete around cables coated to prevent the concrete from sticking to the cable. In the latter case, the cables are tightened to prestress the concrete after it has cured.&lt;/p&gt;&#xA;" OwnerUserId="709" LastEditorUserId="255" LastEditDate="2015-04-21T13:04:23.777" LastActivityDate="2015-04-21T13:04:23.777" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2531" PostTypeId="2" ParentId="280" CreationDate="2015-04-18T09:24:44.573" Score="3" Body="&lt;p&gt;That is a really nice thought experiment! In general I would argue that you only need to know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Total-Pressure ( $p_t$ of your supply)&lt;/li&gt;&#xA;&lt;li&gt;Static/Ambient Pressure ( $p_\infty$into which you are expanding)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Taking into account the uncertainties (or tolerances) you will know whether your flow will have reached $M=1$ in the throat (or how likely it is that it has not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{p_\infty}{p_t}\leq 0.528 \quad, \text{assuming two-atomic-gas with } \gamma=1.4 \text{ in } \frac{p^*}{p_t}=\left(\frac{2}{\gamma+1}\right)^{\gamma/(\gamma-1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the equations for 1D steady compressible flow there is only one solution so the only way for the flow not to reach sonic speed would be a large total pressure loss so that the critical ratio is never reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as thrust is concerned the answer to your question is a bit more complicated since different set-ups (over/under expanded) or geometries (e.g. dual-bell).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as measurement is concerned you might want to have a look on acoustic air speed measurement systems.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2015-04-18T09:24:44.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2532" PostTypeId="2" ParentId="336" CreationDate="2015-04-18T09:41:38.220" Score="0" Body="&lt;p&gt;In case you only have the resources to perform one simulation only I would agree with @Subodh and use $ k-\omega\:SST$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you can afford multiple simulations I would use different models and compare. This way you can identify the influence of the turbulence model in your particular application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you clarify if you are looking for an optimal velocity distribution or if you are more interested in separations?&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2015-04-18T09:41:38.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2533" PostTypeId="2" ParentId="2423" CreationDate="2015-04-18T13:14:06.153" Score="3" Body="&lt;p&gt;I agree with @sturgman one should not look at individual parts but try to understand it in ints context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the very basic version of the Navier-Stokes-Equation (using &lt;a href=&quot;http://en.wikipedia.org/wiki/Einstein_notation&quot; rel=&quot;nofollow&quot;&gt;Einstein-Notation&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho\frac{\mathrm{D}u_i}{\mathrm{D}t} = \rho k_i + \frac{\partial}{\partial x_i} \left( -p + \lambda^*\frac{\partial u_k}{\partial x_k}\right) + \underbrace{\frac{\partial}{\partial x_j} \left( \eta \left[ \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right] \right)}_{\nabla \,\cdot\, (\eta \, \left[ (\nabla\vec{u})+(\nabla\vec{u})^\mathrm{T}\right])} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underbraced part in its original can be rewritten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial}{\partial x_j} \left( \eta \left[ \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right] \right) = \eta \left( \frac{\partial^2u_i}{\partial x_j \partial x_j} + \frac{\partial}{\partial x_i}\left[\frac{\partial u_k}{\partial{x_k}} \right]\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which leads to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho\frac{\mathrm{D}u_i}{\mathrm{D}t} = \underbrace{\rho k_i}_{\text{I}} - \underbrace{\frac{\partial p}{\partial x_i}}_{\text{II}} +\underbrace{(\lambda^* + \eta)\frac{\partial}{\partial x_i}\left[\frac{\partial u_k}{\partial x_k}\right]}_{\text{III}} + \underbrace{\eta \left[ \frac{\partial^2u_i}{\partial x_j \partial x_j}\right]}_{\text{IV}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In symbolic notation this should look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho\frac{\mathrm{D}\vec{u}}{\mathrm{D}t} = \rho\vec{k} - \nabla p + (\lambda^* + \eta)\nabla(\nabla\cdot\vec{u}) + \eta\nabla\cdot\nabla\vec{u}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part $\text{III}$ is not always shown like this depending on the way the Newtonian stress tensor was introduced. Since $\lambda^*$ is a fluid property which is very hard to measure but varies only little, &lt;em&gt;Stokes Hypothesis&lt;/em&gt; sets it to $-2/3 \eta$ (Which is technically only true for monoatomic gases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part $\text{III}$ describes a feature of a fluid where the atomic structure of the fluid-molecule can absorb energy, it is sometimes referred to as pressure-viscosity. Whereas Part $\text{IV}$ describes the resistance of the flow when sheared, part $\text{III}$ describes the resistance of a fluid-volume when it is &quot;isobarically&quot; expanded or compressed.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-04-18T16:23:16.420" LastActivityDate="2015-04-18T16:23:16.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2534" PostTypeId="1" AcceptedAnswerId="2535" CreationDate="2015-04-18T15:17:38.743" Score="47" ViewCount="6963" Body="&lt;p&gt;We sell products that attach to a motor drive's DC bus. We also formerly sold diode kits that let you hook one product up to multiple drives. We stopped selling those diode kits because they were unreliable with modern hardware, and we had better solutions. My customer tells me he wants to keep using the old diode kits, because he can have his repair techs disconnect power to one drive and have it replaced, while the others are still powered on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I maintain that this is an unsafe practice, because diodes are not safety-rated devices. (See question &lt;a href=&quot;https://electronics.stackexchange.com/questions/165389/is-it-safe-to-treat-a-system-downstream-of-a-reverse-biased-diode-as-de-energize/165390#165390&quot;&gt;here&lt;/a&gt;.) But my customer is rather insistent. What is my ethical obligation in this case?&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2016-05-31T14:34:38.357" Title="My customer wants to use my products to do something unsafe. What is my ethical obligation?" Tags="|electrical-engineering|ethics|sales|safety|" AnswerCount="9" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2535" PostTypeId="2" ParentId="2534" CreationDate="2015-04-18T15:38:08.193" Score="33" Body="&lt;p&gt;I'm going to set aside the potential legal liability aspects of your question.  In part because you didn't ask about them, but also because liability will vary based upon jurisdiction.  Obviously, consult an attorney familiar with the relevant law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethically, I think you've done everything that you're obligated to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have contacted the customer and notified them of an unsafe situation or configuration of the product.  And you have (strongly) advised them that they need to discontinue using the older product in the unsafe configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the customer remains insistent about using the product in an unsafe manner, you could present a brief piece showing to them why it's unsafe.  As noted in the answers to your EE question, if the current or voltage exceeds the reverse limits of the diode, bad things will occur.  Sometimes a customer needs to see the numbers behind a recommendation in order to understand the magnitude of what could go wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's say you have already done that and the customer &lt;em&gt;still&lt;/em&gt; insists on using the product in a questionable manner.  There's honestly not a whole lot more you can do.  You can't break into their facility and confiscate the products.  Likewise, you can't take them to a court of law and force an injunction against them.  Potentially, you could contact your local safety administration (OSHA in the US) and notify them of your concerns.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all of those steps are going way above and beyond what you're obligated to do in this particular case.  This is especially so when there is a safe usage for the product along with an unsafe approach.  And any of those actions are likely to irreparably damage your relationship with that client.  Damaging the relationship will impair your credibility with them and make it less likely that they'll listen to your concerns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your obligation is to lay it out to them in unambiguous terms that you believe they need to stop using the product in their &quot;preferred&quot; manner and that your firm will no longer provide any support whatsoever regarding future use of that product in that configuration.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-04-18T17:38:43.220" LastActivityDate="2015-04-18T17:38:43.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2536" PostTypeId="2" ParentId="2534" CreationDate="2015-04-18T15:40:33.763" Score="3" Body="&lt;p&gt;An Engineering Change Notification (ECN) is an industry accepted practice to inform vendor of parts change over, which avoid the ethical obligation issue addressed in the question. An Engineering Change Order (ECO) process includes ECN steps which is briefly discuss &lt;a href=&quot;https://engineering.stackexchange.com/questions/2264/how-can-i-arrange-my-eco-system-to-enforce-the-principle-of-least-privilege&quot;&gt;here&lt;/a&gt; under product life cycle management. An ECN will allow customers to make the necessary business changes in timely manner including initiating of  life time parts purchases. Also process allow vendors to make appropriate design changes without disrupting business.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following are common scenario to initiate an ECN in the context of this question.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lack of demand for parts in current economic/business condition&lt;/li&gt;&#xA;&lt;li&gt;Regulations such as &lt;a href=&quot;http://en.wikipedia.org/wiki/Restriction_of_Hazardous_Substances_Directive&quot; rel=&quot;nofollow noreferrer&quot;&gt;ROHS&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Waste_Electrical_and_Electronic_Equipment_Directive&quot; rel=&quot;nofollow noreferrer&quot;&gt;WEEE&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Changes in raw material or vendors&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A side note: Beside life time part buy situations, some organizations make radical business decision. For example US government has directed vendors to continue manufacturing parts, mostly to maintain strategic military equipment. Also &lt;a href=&quot;https://www.rocelec.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rochester Electronics&lt;/a&gt; an organization that manufacture part obsoleted by large semiconductor companies such as Intel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making business or engineering decision based on ethics is a gray area and situation dictates the decision. Therefore more information is the form of suppliers business, vendors business, industry and other information is beneficial to advice on how to address the ethical obligation component of the question. Below are few examples of widely known recent and past situations where ethical obligations have been addressed.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Dr. Kent Brantly and Nancy Writebol, two well know Ebola patients were given experiments medication at Emory University Hospital in Atlanta Georgia USA. It is considered unethical to provide experimental medication to patients, this a good example violation of generally accepted ethical practices to save the lives of two patients.&lt;/li&gt;&#xA;&lt;li&gt;There were some US based manufactures that moved production to other parts of the world because the certain chemical were labeled toxic in by Environmental Protection Agency (EPA), thus the use in USA was banned. The destination manufacturing location didn't have such regulations. If a chemical is toxic in USA it is toxic in any other part of the world. Based on the available information, in my opinion this practice is unethical.&lt;/li&gt;&#xA;&lt;li&gt;Asian honey that is banned in Europe and USA due to harmful antibiotics and heavy metal are been sold in USA through illegal mean. Based on the available information, in my opinion is unethical. &quot;&lt;a href=&quot;http://www.foodsafetynews.com/2011/08/honey-laundering/#.VTVt2NL4_Gi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Asian Honey, Banned in Europe, Is Flooding U.S. Grocery Shelves&lt;/a&gt;&quot; article highlights situation. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;More related to engineering &quot;&lt;a href=&quot;https://engineering.stackexchange.com/questions/662/why-were-engineers-negligent-in-the-1981-hyatt-regency-walkway-collapse&quot;&gt;1981 Hyatt Regency walkway collapse&lt;/a&gt;&quot; and &quot; &lt;a href=&quot;http://en.wikipedia.org/wiki/Space_Shuttle_Challenger_disaster&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space Shuttle Challenger disaster - O-ring&lt;/a&gt;&quot; are two well know case studies that address engineering ethics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary making decision on ethical obligation is a gray area. Whatever the decision is, it should be a decision that you can be comfortable for the rest of your life. Based on the information you have provided I suggest issuing a ECN, let your customer make the best decision for the customer business.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-04-20T21:33:37.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2537" PostTypeId="2" ParentId="2534" CreationDate="2015-04-18T15:46:28.533" Score="5" Body="&lt;p&gt;Tell him to add a proper (double pole) shutoff switch and educate the techs to use it when replacing motors. (Or make a mechanical interlock so the exposed contacts are never live)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will make it actually safe because when it's disconnected the accessible contacts will not be connected to the power source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general try to give a safe alternative. Often when such careless practices occur it's because they are convenient. Your goal then is to provide safety with similar convenience in operation.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2016-05-31T14:34:38.357" LastActivityDate="2016-05-31T14:34:38.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2538" PostTypeId="2" ParentId="2534" CreationDate="2015-04-18T16:08:20.680" Score="4" Body="&lt;p&gt;I think this somewhat depends on how your company does business. If this is a situation where you simply have a catalogue of products and customers buy products, determining wholly on their own what they need for their project, there's not really much you can do. You sell what you want to sell, make sure that the product is sound from your end, and put safety notices on parts stating what type of usage they can safely handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it sounds more like you work with the customer, basing what you sell on the application they're using it for, and determining together what will best suit their needs. In this case, for what you've already sold them, telling them you don't condone the manner in which they're using it is probably the limit of your responsibility. It's unreasonable for you to go out to their site and physically take the parts away from them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For future sales though, if you know they are going to be using the product in a manner for which it was not designed, I think you do have an ethical (and potentially legal, depending on your location) obligation to tell them you won't sell them the product for that application. On the legal side, that's more a &quot;cover your ass&quot; move, but better safe than sorry. People looking at the bottom line may not like it, but the lost sale will be less than the cost of a court settlement when something goes wrong and your parts are involved (because it doesn't matter who is at fault, everyone &lt;em&gt;involved&lt;/em&gt; will be named in the suit.)&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-04-18T16:08:20.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2539" PostTypeId="2" ParentId="2534" CreationDate="2015-04-18T16:08:33.163" Score="12" Body="&lt;p&gt;Almost every manufactured product can be abused in an unsafe way by a creative enough user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the customer asked your opinion on doing something which is evidently unsafe, I would suggest (on the grounds of common sense, not on the laws of any particular country) that you put your objection in writing, so there can be no future argument in court about &quot;who said what to whom&quot;. Beyond that, you can't stop people being foolish, so don't lose any sleep over it.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-04-18T16:08:33.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2540" PostTypeId="2" ParentId="2520" CreationDate="2015-04-18T16:28:24.283" Score="5" Body="&lt;p&gt;The amplitudes of the modes from a vibration analysis are arbitrary. Often they are &quot;mass-normalized&quot; which is mathematically convenient for using them in a subsequent step in the analysis. There may be an option to scale the in other ways, for example &quot;engineer's normalization&quot; where the largest deflection is set to 1.0, or the option to scale a specified degree of freedom or stress component to any amplitude you choose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may know the amplitude of vibration by testing the structure, or by monitoring it in use with strain gauges or accelerometers. If so you can scale the analysis results to match what you know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to calculate the actual amplitudes and stresses for a given set of loads, you need to do a steady state response analysis, and/or a transient response analysis, whichever is appropriate to the loading and operating conditions.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-04-18T16:28:24.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2541" PostTypeId="2" ParentId="2521" CreationDate="2015-04-18T16:53:31.730" Score="2" Body="&lt;p&gt;The one-word answer is &quot;maybe&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real structures, damping is often nonlinear, and hard to model from first principles. Also, a physically realistic but &quot;arbitrary&quot; damping matrix does not have nice mathematical properties - for example the mode shapes will be complex (i.e. different parts of the structure will vibrate out of phase with each other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, in class (and in many real world situations) you will learn about models of damping that are mathematically convenient but which don't have much basis in physics. Your &quot;modal damping&quot; model based on $2\zeta\omega$ is a good choice for lightly damped structures. Instead of trying to model or measure the $N^2$ terms of the complete damping matrix, you only need to estimate $m$ damping coefficients for the $m$ modes in the frequency range of interest (and often $m \ll N$). You will probably also learn about the Rayleigh damping model, which arbitrarily assumes the damping matrix is of the form $a M^\alpha + b K^\beta$ where $a$, $b$, $\alpha$, $\beta$ are scalar parameters, and most often $\alpha = \beta = 1$.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-04-18T16:53:31.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2544" PostTypeId="2" ParentId="653" CreationDate="2015-04-18T18:13:16.407" Score="1" Body="&lt;p&gt;From my experience experiments are only used to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;validate numerical methods&lt;/li&gt;&#xA;&lt;li&gt;resolve flow-features which are not captured correctly by CFD (e.g. unsteady flow, differences in length and time scales, fluid-structure interaction)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As @Wes said the quality and accuracy of modern CFD is so high combined with the computing power of modern clusters that the conduction of simple experiments is normally not worth it anymore.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2015-04-18T18:13:16.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2545" PostTypeId="2" ParentId="2534" CreationDate="2015-04-18T19:41:38.480" Score="1" Body="&lt;p&gt;The question you have to answer for YOURSELF (not your company) is will you be able to sleep well if one of these guys gets fried? Since the customer has disclosed his unsafe intention, I think that you should not  feel obligated to provide the part. Yes, he may just go elsewhere to get the part, but that is on him. &lt;/p&gt;&#xA;" OwnerUserId="1345" LastEditorUserId="1345" LastEditDate="2015-04-18T21:57:45.117" LastActivityDate="2015-04-18T21:57:45.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2546" PostTypeId="2" ParentId="166" CreationDate="2015-04-18T22:41:53.707" Score="5" Body="&lt;p&gt;HDE 226868 already mentioned the transistor in detail, so I will add many, many technological advances and theoretical backgrounds for the evolution of the computer.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Metallurgy&lt;/strong&gt;&lt;br&gt;&#xA;For the mass-market you need cheap and reliable methods to produce the desired electronic elements. For mass production you need integrated circuits and that means, especially if you go down to small dimensions, you need homogenous high-quality base material.&#xA;Only with high-grade silicon wafers this was possible and while the &lt;a href=&quot;https://en.wikipedia.org/wiki/Czochralski_process&quot; rel=&quot;noreferrer&quot;&gt;Czochralski process&lt;/a&gt; from 1916 is cheap it did not give such good results for silicon like &lt;a href=&quot;https://en.wikipedia.org/wiki/Zone_melting&quot; rel=&quot;noreferrer&quot;&gt;zone melting&lt;/a&gt; which was invented 1950/51.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Boolsche Algebra und Cybernetics&lt;/strong&gt;.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Boolean_algebra&quot; rel=&quot;noreferrer&quot;&gt;The logical algebra developed by Boole&lt;/a&gt; allowed to build computers on binary states. If you ever had the pleasure to try using decimal systems on an electronic machine (BCD), you see how unbelievably easier it is to implement logical and numerical functions. With progressing complexity, it was also necessary to invent the necessary background for developing and controlling complex electronic system, so cybernetics as a new study began to develop.&#xA;It really cannot be underestimated what Babbage and later von Neumann did: Instead of building specialized models for tasks they invented the idea of a programmable machine: Provide simple building blocks of commands and express your solution with this simple commands instead of rewiring the machine (Perhaps there are still some people here who literally &quot;reprogrammed&quot; computers by rewiring parts !). Allow conditional execution. Treat data and program as unit. The general-purpose computer was born.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Photography and Chemistry&lt;/strong&gt;&#xA;Only photography was able to provide people to allow scaling down developed solutions cheaply, robustly and efficiently. Without photo lithography and the development of chemicals allowing the etching of microscopic structures, computers could be small, but they would be extremely expensive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1137" LastActivityDate="2015-04-18T22:41:53.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2547" PostTypeId="1" AcceptedAnswerId="2553" CreationDate="2015-04-19T05:49:23.140" Score="2" ViewCount="129" Body="&lt;p&gt;This is the top part of a basic sit/stand base I'm making.  It is a monitor platform and sliding keyboard tray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NPFC4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NPFC4t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of the large base with 2 rods on each end &lt;a href=&quot;http://www.codeovereasy.com/2013/09/diy-adjustable-desk-for-under-25/&quot; rel=&quot;nofollow noreferrer&quot;&gt;(like this design see video at bottom of page)&lt;/a&gt; I'm looking for another way to raise and lower the desk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one issue with the base design, is that in the lowered position, the platform is already 4 inches up off the desk and will then have to sit on top of the base another 3.5 inches or so. The addition of a base anymore than an inch or so would be too high...(unless the platform does &lt;em&gt;not&lt;/em&gt; have to sit on the base)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for some sort of electronic or manual jack that could fit under the over-hanging portion of the platform. Perhaps a small one on each of the 4 corners.... or maybe 1 longer one for each side? (The platform is 33&quot;X18&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to raise the desk about 12 inches from where the top sits now. Also will be in an office setting so can't be too loud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any type of devices that could be engineered... or purchased... or somewhere in between, that can work for this application or have been use for something similar in the past?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other random ideas I have had: some sort of accordion or scissor lift that attaches across to the bottom of each vertical leg of the platform....or maybe air bags?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a close up of the leg and edge hanging over.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LxgO4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LxgO4t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update1&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or for a more manual design, I came up with scissor sliding rods with rails. On one side, it would have a sliding rail attached to the top of the over hang, and then a rail attached to some sort of base - that would fit beneath the over hang:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2WoGV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has something like this ever been proven to work decent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem with the &lt;a href=&quot;https://www.youtube.com/watch?v=XDU43UTL2Yg&quot; rel=&quot;nofollow noreferrer&quot;&gt;scissors as a whole unit&lt;/a&gt; is that the top would need to be attached to the legs of the platform...and not sure how many inches it would add..in addition, the whole desk would move forward and back, along with up and down (and I have no idea what the kid is saying...probably some thing like I'm 8 and smarter than you).&lt;/p&gt;&#xA;" OwnerUserId="1351" LastEditorUserId="1351" LastEditDate="2015-04-19T18:52:22.697" LastActivityDate="2015-04-19T22:55:20.000" Title="Miniature electric or manual jacks?" Tags="|mechanical-engineering|power-electronics|hydraulics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2548" PostTypeId="2" ParentId="2321" CreationDate="2015-04-19T11:24:28.883" Score="5" Body="&lt;p&gt;The answer to this question is complicated since the flow is fundamentally different.&lt;/p&gt;&#xA;&lt;p&gt;When supersonic flow hits a &lt;strong&gt;wedge&lt;/strong&gt; it is abruptly turned by one &lt;em&gt;oblique shock&lt;/em&gt;. After this all streamlines are parallel since in this 2D-flow-scenario the geometry/flow-area does not change anymore.&lt;/p&gt;&#xA;&lt;p&gt;When supersonic flow (think of a stream tube/cylinder) hits a &lt;strong&gt;cone&lt;/strong&gt; a differential volume element of the stream tube will constantly increase in size while it follows the cone surface. This changes the pressure and is called &lt;em&gt;relaxation&lt;/em&gt;. Due to this volume increase of the stream tube the &lt;strong&gt;cone flow&lt;/strong&gt; has two features/regimes which the &lt;strong&gt;wedge flow&lt;/strong&gt; cannot have:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The stream lines after the shock are curved (bent) so that they align with the cone geometry.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Certain flow-settings will result in a &lt;em&gt;sonic-line&lt;/em&gt; within the flow downstream of the shock. This means that the supersonic flow after the oblique shock is decelerated until it is subsonic see figure below taken from &lt;a href=&quot;http://naca.central.cranfield.ac.uk/reports/1955/naca-report-1242.pdf&quot; rel=&quot;noreferrer&quot;&gt;Naca Report 1242&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nbgXE.png&quot; alt=&quot;Naca Report 1242&quot; /&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-04-19T11:24:28.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2549" PostTypeId="2" ParentId="2534" CreationDate="2015-04-19T11:48:43.777" Score="5" Body="&lt;p&gt;I have tried, but have not found a way to respond to this question that is not opinion based. By their nature, a person's ethics are just that; a person's ethics. They are usually deeply held, personal beliefs. Other words for ethics might be morals, values, conscience, etc., again, these are not necessarily universal in their importance. They are personal. In addition, how a person behaves in the workplace, is still usually going to reflect their own personal beliefs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main point that jumped out at me from the question, was that it was asked at all. The fact that it was asked, means &quot;ethical obligation&quot; is something important to the person who asked the question. Not everyone would give it the same amount of thought, or priority. The fact that it was asked in a public forum, as though there has been some real conflict about what action to take, means it may be weighing rather heavily on Mr. Collings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer would be to do something we don't do nearly enough, and that is to think things all the way through, to their various conclusions. For example, what would a &quot;worst case scenario&quot; look like? Or a best case scenario? Then, as Mr. Christianson suggested, you would have to ask yourself if those conclusions were ok with you, and matched closely to your own, personally arrived at value's. If yes, then you're good to go. If not, you may need to look further for your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also keep in mind that what meets the legal, or professional definition of ethical, or unethical behavior, is not always going to meet our own definition of those same terms, nor should they. In fact, it has been said that facing these kinds of moral dilemmas, is what builds character. To be honest, I just made that up. It's actually just my opinion. &lt;/p&gt;&#xA;" OwnerUserId="1355" LastActivityDate="2015-04-19T11:48:43.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2550" PostTypeId="1" CreationDate="2015-04-19T15:58:18.063" Score="3" ViewCount="1610" Body="&lt;p&gt;I have this aggregate gradation curve in hand, how can I say whether its gap, well or uniformly graded?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EWGPJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1110" LastActivityDate="2015-04-20T16:20:04.610" Title="Classification of aggregates" Tags="|civil-engineering|materials|geotechnical-engineering|concrete|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2551" PostTypeId="1" AcceptedAnswerId="5438" CreationDate="2015-04-19T16:52:08.270" Score="5" ViewCount="1489" Body="&lt;p&gt;How small/compact can a pump or other mechanism be, that would allow compressing gas (or liquid) to pressures of excess of 10GPa, continuously?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughput may be minuscule, of order of milligrams per hour; even lower is acceptable. Also, for &quot;size of the device&quot; let us count all the necessary infrastructure: if it requires energy of a nuclear power plant to operate, it's not compact. Let's say we have some 30 watt of energy &quot;for free&quot;; anything above that goes out of our &quot;compactness budget&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Input gas is pressurized to &quot;reasonable&quot; levels - order of 30KPa; if more pressure is required, the container goes into &quot;compactness budget&quot; with at least 20kg of the gas inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's abundant heat sinking capacity, but any extra heating goes out of our &quot;budget&quot;. Operating (ambient) temperature - choose anything between 40K and 400K as you like. It should have operational lifetime of at least a year, preferably ~10 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(clarifying: &quot;continuous&quot; - we can assume a small &quot;buffer&quot; container on the output side, emitting the pressurized gas at rate that is the average of the intake speed, so short &quot;discontinuity&quot; like piston motion is fully acceptable - simply keep the &quot;buffer&quot; above 10GPa with losses of order of assumed average input (1mg/hour, or less if required, but fairly constant).)&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2015-04-19T16:58:47.913" LastActivityDate="2015-11-14T23:18:52.790" Title="Gigapascal pressures in a small package" Tags="|pressure|pumps|compressors|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2552" PostTypeId="2" ParentId="2550" CreationDate="2015-04-19T22:24:02.650" Score="5" Body="&lt;p&gt;The flat &quot;table&quot; area in the middle of the curve indicates that particles of roughly 1–4 mm diameter are not well-represented in your aggregate. This gap in your particle size distribution is why it's called gap-graded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you want to look at the slope of the curve in different regions to tell whether particles are accumulating in the sieves of approximately that size. Since there's a positive slope from about 0.2–1 mm, you know your aggregate contains a decent mix of sands of different grades. Since there's no significant slope from 1–4 mm, you know your aggregate doesn't contain much fine gravel. Then, since there's a positive slope after about 10 mm, you know your aggregate contains a decent mix of medium and coarse gravels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A well-graded aggregate has a smoother gradation curve, indicating that there are some particles in the mix of every different size. For instance, take a look at these figures from the &lt;a href=&quot;http://onlinemanuals.txdot.gov/txdotmanuals/pdm/hot_mix_asphalt_concrete_pavement_mixtures.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pavement Design Guide of the Texas Dept. of Transportation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PkkJO.gif&quot; alt=&quot;dense graded mix figures&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure on the right is a &lt;a href=&quot;http://www.commandalkonconnect.com/2014/01/13/concrete-mix-design-art-and-science-the-0-45-power-chart/&quot; rel=&quot;nofollow noreferrer&quot;&gt;0.45 power curve&lt;/a&gt;; the scale of the x-axis is the 0.45th power of the sieve opening, where your gradation curve uses a logarithmic scale of the actual particle diameter. It's a bit different way to see the data but the basic shape you're looking for is the same—a smooth curve that levels off at the tails but not in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also specific criteria for determining whether your mix is well-graded; see &lt;a href=&quot;http://en.wikipedia.org/wiki/Soil_gradation#Calculating_the_Coefficients_of_Uniformity_and_Curvature&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia on soil gradation&lt;/a&gt; for an overview. Judging by your graph, $C_u \approx 19$ and $C_c \approx 8$, which also indicates your aggregate is not well-graded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uniformly-graded is pretty easy to identify on a size distribution graph because it's primarily composed of one size of particle; what you see on the graph is a flat tail on the left followed by a sudden jump to near 100% at or around the particle size that dominates the mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at Figure 2 in &lt;a href=&quot;http://www.pavementinteractive.org/article/gradation-and-size/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article from Pavement Interactive&lt;/a&gt; for a visual comparison of the general shape of the curve for different classifications of aggregate. (Note that these are again 0.45 power curves.)&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-04-20T16:20:04.610" LastActivityDate="2015-04-20T16:20:04.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2553" PostTypeId="2" ParentId="2547" CreationDate="2015-04-19T22:55:20.000" Score="1" Body="&lt;p&gt;Put the fixed point of the rail on the back on both the upper &lt;em&gt;and&lt;/em&gt; lower rails in the video you linked they are on opposite sides leading to a side to lateral movement. &lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-04-19T22:55:20.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2554" PostTypeId="1" CreationDate="2015-04-20T05:45:44.890" Score="2" ViewCount="578" Body="&lt;p&gt;I would like to design a waveguide circulator.  I tried to design one but, my attempt was not successful.  My design used a cylindrical cavity with ferrite as region. I then connected 3 waveguides in 0,120 and -120 directions. Upon applying the excitation, I got a pathetic result.  I have referred to some more resources but can not resolve this problem.  Has anybody tried this before?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My design and its parameters are shown below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/38GnC.jpg&quot; alt=&quot;Waveguide circulator design&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Frequency of operation: 915 MHz&lt;/li&gt;&#xA;&lt;li&gt;Dielectric permittivity of material(ferrite): 14.2&lt;/li&gt;&#xA;&lt;li&gt;The height of cylinder: 5.5 mm&lt;/li&gt;&#xA;&lt;li&gt;Radius of cylinder 30 mm&lt;/li&gt;&#xA;&lt;li&gt;The waveguide cross section a = 20 mm&lt;/li&gt;&#xA;&lt;li&gt;The waveguide cross section b = 2.5 mm&lt;/li&gt;&#xA;&lt;li&gt;The length of waveguide l = 50 mm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1362" LastEditorUserId="272" LastEditDate="2015-04-20T11:56:37.760" LastActivityDate="2015-04-20T11:56:37.760" Title="How to design a 915 MHz waveguide circulator?" Tags="|electrical-engineering|rf-electronics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2556" PostTypeId="2" ParentId="2534" CreationDate="2015-04-20T13:46:16.580" Score="3" Body="&lt;p&gt;&lt;strong&gt;An (almost) safe solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the information available from this question &lt;strong&gt;PLUS&lt;/strong&gt; the highly relevant prior question &lt;a href=&quot;https://electronics.stackexchange.com/questions/165389/is-it-safe-to-treat-a-system-downstream-of-a-reverse-biased-diode-as-de-energize/165390#165390&quot;&gt;Is it safe to treat a system downstream of a reverse-biased diode as de-energized?&lt;/a&gt; there is a method that is still not 100%-regulation-passing safe, but one which is liable to have an extremely good chance of ensuring that nobody gets hurt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the diode isolators are intended to ENSURE that the de-energised circuit is not powered and remains at earth potential while being worked on, &quot;&lt;strong&gt;make it so&lt;/strong&gt;&quot;. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Connect the two output terminals at &quot;tongue here&quot; together robustly with a suitably solid conductor and terminations and &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Connect the combined circuit equally robustly to ground. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the isolation method works then nothing will be damaged.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the isolation diodes leak in a minor and expected way the current will be shunted to ground and all shall still be sweetness and light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the isolation diodes fail to any significant extent - do you sell ringside tickets? - there will be some pyrotechnics and some things may die but, if you/they have implemented the clamp correctly, it will not be people. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would &lt;strong&gt;consider&lt;/strong&gt; working on such a circuit, provided that I had personally checked how it was done first, but I'd treat the circuit as potentially live to the maximum extent possible. I have worked on known live circuits with a greater risk factor than this one poses - but fully knowingly and voluntarily. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Liability and warning:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IANAL BUT:&lt;/strong&gt;&lt;br&gt;&#xA;Here &quot;you&quot; is whoever is held responsible by authorities. It may or may not include &quot;you&quot; the manufacturer.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is utterly essential ethically and probably legally that people working on this equipment are completely aware of all aspects of what is being done and why.&lt;br&gt;&#xA;If somebody dies and the worker was aware what was being done and why and the various implications in complete detail, you can expect regulatory prosecution - but MAY escape a manslaughter or equivalent prosecution.&lt;br&gt;&#xA;If somebody dies and the worker was &lt;strong&gt;NOT&lt;/strong&gt; aware what was being done and why and the various implications in complete detail you do not deserve to escape a manslaughter or equivalent prosecution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linemen are used to working on circuits which should be disconnected and which should stay connected and which probably are and will. Earthing sticks and similar are commonly used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BgMOe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Checklist:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: The related tasks must have a checklist associated and the last 5 items on it should be.&lt;br&gt;&#xA;-Remove all the shorting and earthing connections.&lt;br&gt;&#xA;-Check that all the shorting and earthing connections have been removed&lt;br&gt;&#xA;-Check that ALL the shorting and earthing connections have been removed&lt;br&gt;&#xA;-Check that &lt;strong&gt;ALL&lt;/strong&gt; the shorting and earthing connections have been removed    &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;- HAVE YOU CHECKED THAT ALL THE SHORTING AND ...&lt;/strong&gt;&lt;/h2&gt;&#xA;" OwnerUserId="399" LastEditorDisplayName="user16" LastEditDate="2015-04-23T22:08:43.747" LastActivityDate="2015-04-23T22:08:43.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2557" PostTypeId="1" AcceptedAnswerId="2559" CreationDate="2015-04-19T19:12:42.510" Score="6" ViewCount="1076" Body="&lt;p&gt;Is it possible to convert the energy you would normally lose when braking your bicycle to rotational energy by attaching an extra wheel which would start to rotate once you use your brakes? E.g., when facing a red light after riding down a hill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could restore that energy, you could regain part of your velocity without as much effort after stopping at the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am quite sure I'm not the first one who had this idea (as I heard about some projects trying to achieve the same thing but then with cars) so I wonder what the technical difficulties are and why I have not seen bicycles like this on the road?&lt;/p&gt;&#xA;" OwnerUserId="1369" OwnerDisplayName="Gert" LastEditorUserId="368" LastEditDate="2015-04-20T17:04:35.210" LastActivityDate="2018-03-11T11:14:21.160" Title="What are the technical obstacles for using flywheel-based regenerative braking on bicycles?" Tags="|mechanical-engineering|bicycles|regenerative-braking|flywheels|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2558" PostTypeId="2" ParentId="2557" CreationDate="2015-04-19T19:28:58.260" Score="6" Body="&lt;p&gt;I'm not a bicycle mechanic or expert, but I would say the main drawbacks of a flywheel system are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Complication of maintenance - the more moving parts, the more often your bike will break down, and the harder it is to repair.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Price - more moving parts means more expense.  You need a whole extra gear system to run a flywheel!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Weight/danger - flywheels are designed to store rotational kinetic energy ($KE_{rot} = \frac 1 2 I \omega^2$, where $I \propto m$), and to store a lot of rotational kinetic energy, you either need high rotational speeds, high mass, or both.  High rotational speeds mean a lot of wear and tear, or more danger to the user (legs, straps, shoes, etc getting caught in a high RPM system).  High mass is annoying for bicyclists trying to lug an extra 15 pounds up a hill.  Either way, it's an issue.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Counterproductive - A lot of bicyclists use their bikes to exercise, so it's a bit counterproductive to have an extra system designed to give them less exercise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, that's not to say bicycle flywheels aren't neat - they sure are, and in many ways they're a great idea.  But you asked about the drawbacks.&lt;/p&gt;&#xA;" OwnerDisplayName="Brionius" LastActivityDate="2015-04-19T19:28:58.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2559" PostTypeId="2" ParentId="2557" CreationDate="2015-04-19T21:36:14.590" Score="9" Body="&lt;p&gt;In Europe hybrid electric bikes are becoming quite popular - you pedal and a small battery + motor gives you an extra push. Such bikes include regenerative braking: when you brake gently the motor operates &quot;in reverse&quot; and charges the battery back up. Given that all the hardware is already there it is not a big additional engineering challenge. It is especially useful in hilly cities - you regain gravitational energy. Your flywheel is a lot of complexity : and the speed matching is particularly challenging since you need to be able to accelerate the flywheel regardless of its speed and the speed of the bike. This requires a continuously variable gearing mechanism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrical storage is just much simpler.&lt;/p&gt;&#xA;" OwnerUserId="565" OwnerDisplayName="Floris" LastActivityDate="2015-04-19T21:36:14.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2560" PostTypeId="2" ParentId="2557" CreationDate="2015-04-20T23:15:33.460" Score="4" Body="&lt;p&gt;Another difficulty would be the gyroscopic effect - given the speed of the flywheel, the forces would be enormous, coming down that Alpine pass, you'd probably flip over at the first hairpin bend (switchback in the USA).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could be solved with dual contra-rotating flywheels, but that just adds more complexity on an already too complex solution. The hybrid electric bike is probably a simpler solution, as well as allowing other ways of charging.&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2015-04-20T23:15:33.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2561" PostTypeId="1" CreationDate="2015-04-21T04:19:00.787" Score="2" ViewCount="160526" Body="&lt;p&gt;Is 1 kg/s of air flow rate equivalent to 1 m&lt;sup&gt;3&lt;/sup&gt;/s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am calculating supply air flow rate into a zone for an air conditioning system. The simulation software gives the result in m&lt;sup&gt;3&lt;/sup&gt;/s but the mathematical formula that I used takes in kg/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to convert the flow rate, and if so, how do I convert between kg/s and m&lt;sup&gt;3&lt;/sup&gt;/s?&lt;/p&gt;&#xA;" OwnerUserId="1375" LastEditorUserId="368" LastEditDate="2015-04-21T14:58:21.267" LastActivityDate="2019-07-02T09:36:09.943" Title="Converting air flow rate between kg/s and m^3/s" Tags="|mechanical-engineering|hvac|airflow|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2562" PostTypeId="2" ParentId="2561" CreationDate="2015-04-21T04:59:59.723" Score="8" Body="&lt;p&gt;You need to consider the density of the air, which varies with temperature and air pressure. At 15 degrees Celsius, at sea level, the density of air is 1.225 ${kg}/{m^3}$. The table &lt;a href=&quot;https://en.wikipedia.org/wiki/Density_of_air&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; gives air densities at 5 degree intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now density is mass divided by volume,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\rho} = m/v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, to get the volume flow rate (in ${m}^3/s$), for a know mass flow rate, divide the mass flow rate (in $kg/s$) by the density (in $kg/{m}^3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, a flow rate of $1 kg/s$ is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 / 1.225 = 0.8163  {m}^3/s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the mass flow rate for a know volume flow rate multiple the volume flow rate by the air density.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="41" LastEditDate="2015-04-21T13:48:42.110" LastActivityDate="2015-04-21T13:48:42.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2563" PostTypeId="2" ParentId="2561" CreationDate="2015-04-21T05:09:49.377" Score="5" Body="&lt;p&gt;HVAC-Systems usually give a &lt;em&gt;Volume-Flow-Rate&lt;/em&gt;, hence $m^3/s$. If you need &lt;em&gt;Mass-Flow-Rate&lt;/em&gt; ($kg/s$) you simply need to multiply with the density ($\rho$) of the fluid. The density can be calculated using the &lt;em&gt;ideal gas law&lt;/em&gt; (see &lt;a href=&quot;http://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \rho_s = \frac{p_s}{R T_s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please observe that you need &lt;em&gt;static&lt;/em&gt; values for the pressure ($p_s$, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Pressure&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;) and temperature ($T_s$, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Temperature&quot; rel=&quot;noreferrer&quot;&gt;3&lt;/a&gt;). The &lt;em&gt;specific gas constant&lt;/em&gt; ($R$, see &lt;a href=&quot;http://en.wikipedia.org/wiki/Gas_constant#Specific_gas_constant&quot; rel=&quot;noreferrer&quot;&gt;4&lt;/a&gt;) for air is depending on the humidity of the air but $287.058 J/kg/K$ is a good starting point. For low velocities static and absolute values are so close to each other that you should be able to use the absolute/ambient pressure and Temperature. However, the density is changing with pressure and temperature, neglecting either will decrease the accuracy of the measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my experience the HVAC system has a measurement-error ($5\%$) which is larger than the error which is made by assuming static=total. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;All values in SI units!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-04-21T05:30:29.817" LastActivityDate="2015-04-21T05:30:29.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2564" PostTypeId="1" CreationDate="2015-04-21T06:48:41.130" Score="11" ViewCount="1969" Body="&lt;p&gt;I have built (several) discrete Extended Kalman Filters (EKF). The system model I am building has 9 states, and 10 observations. I see that most of the states converge except one. All except 1-2 of the EKF state estimate appears to drift. Since the EKF is dependent on all the states being convergent, the rest of the states are very erroneous after the divergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I check the observability of the EKF? Do I simply check the rank of the measurement Jacobian and see if it's less than the maximum rank of the measurement Jacobian?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;After adding more measurements in my simulation, I was able to get things to converge. However, my question about observability still remains!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Issue:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ground truth and EKF estimate graphs can be &lt;a href=&quot;https://i.stack.imgur.com/pKhyr.png&quot; rel=&quot;noreferrer&quot;&gt;found here&lt;/a&gt; or see below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The model is quite non-linear between time steps 400-600 hence some divergence of some of the states&lt;/li&gt;&#xA;&lt;li&gt;Figure/State 6 is the one that seems to be diverging&lt;/li&gt;&#xA;&lt;li&gt;Please ignore the &quot;sensor readings&quot; plots for Figures 8/9&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Things I have tried:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I know for linear state space systems you can use the &lt;a href=&quot;http://en.wikipedia.org/wiki/Observability#Definition&quot; rel=&quot;noreferrer&quot;&gt;Cayley Hamilton Theorem&lt;/a&gt; to check observability.&lt;/li&gt;&#xA;&lt;li&gt;I have tried to check the Innovation/measurement residual &lt;code&gt;e&lt;/code&gt; and all innovations converge to 0&lt;/li&gt;&#xA;&lt;li&gt;I've also tested different inputs and they don't seem to affect the convergence of the diverging state(s)&lt;/li&gt;&#xA;&lt;li&gt;I've tuned the EKF without any sign of convergence for the diverging state(s)&lt;/li&gt;&#xA;&lt;li&gt;Graphs for another &lt;a href=&quot;https://i.stack.imgur.com/mFMc3.png&quot; rel=&quot;noreferrer&quot;&gt;input signal:&lt;/a&gt; or see below&lt;/li&gt;&#xA;&lt;li&gt;After talking to a colleague, he suggested that I investigate another issue that could be that there is an observation that is linearly dependent on 2 states, e.g. &lt;code&gt;y = x1 + x2&lt;/code&gt;. There is an infinite number of values that could satisfy the same &lt;code&gt;y&lt;/code&gt;, but shouldn't observability capture this issue as well?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please let me know if there's anything else I can provide.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Ground truth &amp;amp; EKF estimate graphs:&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for larger view&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HxziY.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aayWK.jpg&quot; alt=&quot;image a&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vQGyw.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MuxFW.jpg&quot; alt=&quot;image b&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/u8ZpM.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jpu5Y.jpg&quot; alt=&quot;image c&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iSTNt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zFzMf.jpg&quot; alt=&quot;image d&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Vl3gm.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I4mZh.jpg&quot; alt=&quot;image e&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HKPwH.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FBzYi.jpg&quot; alt=&quot;image f&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/83xix.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mTX1r.jpg&quot; alt=&quot;image g&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gLP6f.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y5V7t.jpg&quot; alt=&quot;image h&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lDENR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xfjU6.jpg&quot; alt=&quot;image i&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Additional input signal:&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for larger view&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/amPke.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tkaPs.jpg&quot; alt=&quot;image m&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TjO8k.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZopG4.jpg&quot; alt=&quot;image n&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UZ4Do.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5hDlv.jpg&quot; alt=&quot;image o&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JoUtf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OYIUc.jpg&quot; alt=&quot;image p&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CxOeE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D0BDu.jpg&quot; alt=&quot;image q&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fCDMD.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wP6Z2.jpg&quot; alt=&quot;image r&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/zyp6p.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kYLxj.jpg&quot; alt=&quot;image s&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/22kqK.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2QOTd.jpg&quot; alt=&quot;image t&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SyYS4.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xWaCi.jpg&quot; alt=&quot;image u&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1377" LastEditorUserId="6790" LastEditorDisplayName="user16" LastEditDate="2020-05-04T08:03:01.937" LastActivityDate="2020-05-04T08:03:01.937" Title="Observability using the Discrete Extended Kalman Filter (EKF)" Tags="|electrical-engineering|control-engineering|control-theory|optimal-control|kalman-filters|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2567" PostTypeId="1" CreationDate="2015-04-21T13:56:07.623" Score="4" ViewCount="534" Body="&lt;p&gt;I am performing a number of simple heat transfer experiments with the application of a magnetic field around the fuel line to evaluate the effects.  To attempt to quantify some of the heat losses in the system I have worked out how many moles of water will be produced by combustion of a butane-propane gas mixture in air. I am hoping to find a relationship between that mass of water and the energy lost from the system by producing that water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the relationship between the HHV and LHV but that seems to only deal with water vapour that has been condensed into a liquid so I'm unsure if that is any use.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastEditorUserId="368" LastEditDate="2015-04-21T15:06:30.083" LastActivityDate="2015-04-21T15:11:55.063" Title="What relationship is there between mass of water vapour produced in combustion and the energy lost through producing this water?" Tags="|mechanical-engineering|thermodynamics|chemical-engineering|heat-transfer|combustion|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2568" PostTypeId="2" ParentId="2567" CreationDate="2015-04-21T14:30:11.247" Score="5" Body="&lt;p&gt;If you get complete combustion of any hydrocarbon fuel, your products will be primarily $CO_2$ and $H_2O$. Other products will be present in quantities too small to be relevant to the heat losses that you're looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two gases will change the heat capacity, conductivity and density of the flame and your exhaust gases. So heat heat transfer computations that you perform regarding convection and conduction should change accordingly. Note that these properties vary with temperature over the range that you're interested in (300–2,000 K, roughly), so it's best not to just pick one value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total energy from the fuel should be computed from the heat of combustion ($\Delta H_C$) of the gas being burnt. From balancing the chemical equations you can determine the amount of water produced for every mole of fuel.  The heat stored in the water depends on it's temperature (look this up in a table or compute the integral of $c_p$ plus the heat of vaporization). The ratio $H_{H_2O}/\Delta H_c$ tells you how much energy went into the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These gases do thermally radiate as well and you may want to consider their emissivity, although the flame sheet itself (where combustion is still in process) is likely to be a more significant radiator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HHV-LHV difference is meaningful if you are trying to extract as much heat as possible from your exhaust gases. For example, after a turbine engine has extracted work from the combustion products, they'll be too cold to produce mechanical work, but could be used to heat a building. It just has to be cold enough for the water to condense so that the heat of vaporization can be released ( &amp;lt;100 °C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to find most of the relevant thermal properties online (in the &lt;a href=&quot;http://webbook.nist.gov/chemistry/&quot; rel=&quot;noreferrer&quot;&gt;NIST Chem Web Book&lt;/a&gt; in the entries for 'fluid properties.' Any standard text of combustion (Law, Turns, Glass etc.) will usually have tabulated info (based on your project you should already have consulted at least one of those heavily).&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="368" LastEditDate="2015-04-21T15:11:55.063" LastActivityDate="2015-04-21T15:11:55.063" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2569" PostTypeId="2" ParentId="2567" CreationDate="2015-04-21T14:41:52.660" Score="2" Body="&lt;p&gt;You need to consider the latent heat of vaporisation of the water produced. As from this excerpt from this &lt;a href=&quot;http://www.clarke-energy.com/2013/heating-value/&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Due to high combustion temperatures, this water takes the form of steam which stores a small fraction of the energy released during combustion as the latent heat of vaporisation; in simple terms, as heat energy stored in the vapourised ‘state’ of water.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The total amount of heat liberated during the combustion of a unit of fuel, the HHV or HCV, includes the latent heat stored in the vapourised water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Latent_heat&quot; rel=&quot;nofollow&quot;&gt;latent heat of water&lt;/a&gt; vaporisation is 2260 kJ/kg.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="368" LastEditDate="2015-04-21T15:06:49.103" LastActivityDate="2015-04-21T15:06:49.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2570" PostTypeId="2" ParentId="2561" CreationDate="2015-04-21T16:42:28.770" Score="1" Body="&lt;h2&gt;Constant Mass Flow:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{m} = \rho A V $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hence mass flow is equal to density times area times velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assuming you have an area of $1$ m^2 with a velocity of $1$ m/s, air with a density of $1.225$ kg/m^3 will equate to a mass flow of $1.225$ kg/s. For incompressible flows, the mass flow rate is constant.&lt;/p&gt;&#xA;" OwnerUserId="782" LastActivityDate="2015-04-21T16:42:28.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2571" PostTypeId="1" AcceptedAnswerId="2735" CreationDate="2015-04-21T18:29:41.173" Score="10" ViewCount="1403" Body="&lt;p&gt;Is there a specific chemical that is used frequently in the production of liquid crystal displays? During my internet research so far, it seems as if the specific chemical composition of the liquid crystal does not matter very much. &lt;a href=&quot;http://www.sigmaaldrich.com/materials-science/material-science-products.html?TablePage=16378837&quot;&gt;This&lt;/a&gt; website gives a list of apparently commonly used LCs. Are there one or two chemicals that are used in practically all common LCDs?&lt;/p&gt;&#xA;" OwnerUserId="844" LastEditorUserId="844" LastEditDate="2015-04-22T15:02:57.930" LastActivityDate="2015-05-07T08:59:59.300" Title="What liquid crystals are used most frequently in displays?" Tags="|chemical-engineering|optics|liquid|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2572" PostTypeId="1" AcceptedAnswerId="3205" CreationDate="2015-04-21T19:11:45.827" Score="4" ViewCount="3248" Body="&lt;p&gt;Oblique cutting generates transverse cutting force which reduces tool life. Oblique cutting involves principal and auxiliary cutting edges in the cutting action which increases friction and advances wear in the cutting tool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Orthogonal cutting doesn't have those complexities, but oblique cutting is found to be used in more often in an industrial setting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are oblique cuts made instead of orthogonal cuts when it is more efficient to make an orthogonal cut?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VTLB2.png&quot; alt=&quot;Orthogonal and oblique cutting&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="739" LastEditorDisplayName="user16" LastEditDate="2015-04-22T13:48:38.333" LastActivityDate="2015-06-16T11:19:52.110" Title="Oblique cutting implemented in Industrial applications" Tags="|mechanical-engineering|materials|mechanical-failure|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2573" PostTypeId="1" CreationDate="2015-04-21T20:53:56.630" Score="5" ViewCount="2520" Body="&lt;p&gt;There are a number of sources that confirm blood pressure cuff size can have a significant impact upon the measured reading.  (See &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/6123760&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pubmed/6732409&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;http://www.mayoclinic.org/diseases-conditions/high-blood-pressure/expert-answers/blood-pressure-cuff/faq-20058337&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;, &lt;a href=&quot;http://thebloodpressureblog.blogspot.com/2009/06/why-your-doctors-office-blood-pressure.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;4&lt;/a&gt;)  The most &lt;a href=&quot;http://www.bloodpressure-drs-practical-guide.com/bloodpressurecuff.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;succinct description&lt;/a&gt; I have seen is this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The error that is introduced by too small a blood pressure cuff is a reading that is too high and the error can be 50 mm and this is not rare. ...&lt;br&gt;&#xA;  Too large a cuff introduces the opposite error, an artificially low reading but not nearly the magnitude of the too small cuff error. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Disclaimer, I haven't reviewed the material too closely on the site I quoted, so it's likely they're trying to sell something.  But their summary is in alignment with other non-commercial sites I've found.&lt;/em&gt; &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to understand is &lt;em&gt;why&lt;/em&gt; the blood pressure cuff size affects the measured reading.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my simplistic understanding, a blood pressure cuff squeezes against the brachial artery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xkrbz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xkrbzs.png&quot; alt=&quot;brachial artery&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compression of the brachial artery temporarily stops blood flow and allows the practitioner to measure when flow commences again.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the cuff is large enough to stop blood flow, I would like to understand the physiological principles that are affecting the read measurement.  And I think this is a valid engineering question as I'm interested in the interaction between the measuring device (blood pressure cuff) and the system being measured (blood pressure via the brachial artery).  In other (much bigger) words, what biomedical engineering principle is at work to cause the measured blood pressure to differ?&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-04-23T23:24:20.470" Title="Why does blood pressure cuff size matter?" Tags="|design|biomedical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2574" PostTypeId="2" ParentId="2573" CreationDate="2015-04-21T22:11:00.593" Score="3" Body="&lt;p&gt;I would guess that there is more than one reason to that (However, I am speculating, since I could not find any technical research on it). But in general the question is what effects can lead to a change in “pressure-reading”:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to compress the artery in a way, that the blood flow is stopped the arm needs to be compressed as well. For simplicity one could think of it as reducing the diameter/cross-area of the arm. When inflating the cuff the inner side of the cuff moves inward (and thereby reduces the area, and thereby compressing the artery). Applying the same pressure to two cuffs (one small one large). The larger one will extend more and thereby reduce the area to a larger extend (Since assuming the cuffs have the same thickness but differ in length, there is more fabric for the larger one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since one needs some way of “calibrating” the &lt;em&gt;cuff-pressure&lt;/em&gt;-reading with the &lt;em&gt;blood-pressure&lt;/em&gt; there will be a standard-cuff which allows for a calibration/correlation of both pressures. However, I think one will also need a standard-arm to provide the calibration. Therefore, I would argue, that the cuff size might play a role in the pressure reading, it is however not the only influence. For example I would guess that also the size/area-ratio between cuff and arm-cross-area will play a role in the correlation, so that measuring blood-pressure on children might require a smaller cuff.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2015-04-21T22:11:00.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2575" PostTypeId="2" ParentId="2573" CreationDate="2015-04-22T01:57:15.570" Score="4" Body="&lt;p&gt;The first clue is in your reference to the succinct description: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You may need a large cuff if your arm size is 11 or 12 inches. By 13 inches you do need a large cuff. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The brachial artery is protected by the muscles and fat in the upper arm. The larger the girth of the upper arm the thicker the muscles and fat protecting the artery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pressure applied to the upper arm to squeeze the brachial artery must be applied through the thickness of muscle and fat (soft tissues) between the artery of the surface of the arm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the aim of the blood pressure cuff is to squeeze the brachial artery to temporarily prevent blood flow, if a small cuff is used on a large-girth arm, with thick muscles and/or lots of fat, a higher pressure will need to be applied than when a larger cuff is used. The opposite occurs when a large cuff is used on a small-girth arm, with small muscles and very little fat: less pressure is needed to squeeze the brachial artery shut because of the lower resistance offered by the arm muscles and fat. Hence to apply the correct pressure to squeeze the brachial artery, the correct size of cuff needs to be used.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="368" LastEditDate="2015-04-23T23:24:20.470" LastActivityDate="2015-04-23T23:24:20.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2576" PostTypeId="2" ParentId="2520" CreationDate="2015-04-22T04:48:01.243" Score="7" Body="&lt;p&gt;The short answer is, there is no amplitude used. Even more important though, is the fact that &lt;strong&gt;the displacements and stresses shown in the results of a modal analysis cannot be used to say anything about the physical behavior of the part in absolute terms.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic equation of motion is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[M][\ddot{U}]+[B][\dot{U}]+[K][U]=F(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M$, $B$,and $K$ are matrices for mass, dampening, and stiffness, respectively, and $U$ is the displacement. The material properties are known, and we are solving for the displacement. These are specified for the individual nodes created when meshing an FEA model. For a modal analysis, we ignore the damping effects, and assume there are no loads present. This essentially poses the question &quot;If we constrain the part in a certain manner but apply no load to it, what are the possible ways in which it will vibrate?&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the equation of motion when we neglect the damping and apply no load (F=0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[M][\ddot{U}]+[K][U]=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are trying to solve this equation for non-zero values of $U$; that is, points at which the inertial forces and spring forces of the material are equal. In this idealized case of no damping, the part could theoretically vibrate through these points forever after an initial displacement/force is applied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In solving the idealized equation above, we also let &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$[\ddot{U}]=\lambda[U]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda$ is an eigenvalue. This is a consequence of harmonic motion, which makes sense when you think about it a bit. If a part is oscillating between two positions and the points of the part take linear paths between their two positions, the acceleration vector will always be a linear multiple of the displacement vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the final form of the equation, we see that we only have 4 distinct terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda[M][U]+[K][U]=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $M$ and $K$ are known, we're just looking for a displacement matrix for which a constant $\lambda$ exists. But given the nature of this equation, it's easy to see that for a matrix $[V]=A[U]$ where $A$ is an arbitrary constant, $\lambda_U=\lambda_V$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, it's useful to note that the units of $\lambda$ are $s^{-2}$, meaning that $\lambda=\phi^2$ with $\phi$ being the resonant frequency of the vibration mode in question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as I said at the top, there is no amplitude used to the actual main calculations in a modal analysis. To display the results and give the &quot;displacement&quot; values in the results, there is some standard amplitude or normalization done, but those numbers are not what you're looking for in a modal analysis, and they shouldn't be used in absolute form; how you &lt;em&gt;can&lt;/em&gt; use them is in determining the proportionality of displacements between nodes. If you see point A has a displacement of 2mm when point B has a displacement of 1mm, you know that 2:1 ratio will always exist for this vibration mode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine how much a part will actually vibrate, you have to do a full vibrational analysis, defining the loads and their frequencies yourself, using the knowledge you've gained from the modal analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two good references:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Modal_analysis_using_FEM&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article on &lt;em&gt;Modal Analysis using FEM&lt;/em&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://machinedesign.com/archive/vibration-analysis-designers&quot; rel=&quot;noreferrer&quot;&gt;A take on vibration and modal analysis as it applies to design&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-04-22T04:48:01.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2577" PostTypeId="1" AcceptedAnswerId="2581" CreationDate="2015-04-22T06:36:20.740" Score="4" ViewCount="1047" Body="&lt;p&gt;I'm building a press for cider. I'm going to be using a 24&quot; x 24&quot; press board. I'm looking to deliver 40 psi to the cider. How can I determine the amount of weight required, in tons? What sort of equation would I use?&lt;/p&gt;&#xA;" OwnerUserId="1387" LastEditorUserId="368" LastEditDate="2015-04-24T15:04:45.350" LastActivityDate="2015-04-24T15:04:45.350" Title="How do I calculate the weight needed on a press board to deliver a certain amount of pressure?" Tags="|mechanical-engineering|pressure|hydraulics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2578" PostTypeId="1" CreationDate="2015-04-22T06:59:03.303" Score="5" ViewCount="185" Body="&lt;p&gt;I noticed that when a group of people enter in an hydraulic lift/elevator often when the last person steps in the lift/elevator has a small but noticeable recoil: the lift goes down a few mm then it compensates. &#xA;This creates a recoil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that regardless of the number of persons on the lift/elevator this effect triggers when several people enter the cabin in succession and always happens when the last one steps in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hypothesis is that a sensor perceives the increased pressure but does nothing while the pressure is building,  instead when the pressure is stabilised because no more person enters it gives a bit of extra oil to the piston and we experience the recoil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my idea correct? Or is there a better explanation? &lt;/p&gt;&#xA;" OwnerUserId="1388" LastEditorUserId="1388" LastEditDate="2015-04-22T09:37:45.657" LastActivityDate="2015-04-22T14:19:55.920" Title="Recoil of hydraulic lift/elevator" Tags="|pressure|building-physics|hydraulics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2579" PostTypeId="2" ParentId="2577" CreationDate="2015-04-22T08:15:36.070" Score="2" Body="&lt;p&gt;The calculation you made in your comment is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$11.52 \:\mathrm{US} \:\mathrm{tons} = 23,040 \:\mathrm{lb_f}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{23,040 \:\mathrm{lb_f}}{24″ \times 24″} = 40 \:\mathrm{psi}$&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="368" LastEditDate="2015-04-22T20:20:01.907" LastActivityDate="2015-04-22T20:20:01.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2580" PostTypeId="2" ParentId="2578" CreationDate="2015-04-22T09:52:43.500" Score="4" Body="&lt;p&gt;When an elevator stops it will set a brake to immobilize the elevator cabin. This way a sudden leak doesn't drop the cabin while the door is open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the controller prepares to move the elevator cabin, it will release the brake which allows the cabin to move again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an ideal system, there will be no recoil.  However, air bubbles and elasticity of materials will allow the piston to compress a bit under the weight of passengers entering the cabin. This is then detected by a sensor and compensated for.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-04-22T14:05:09.970" LastActivityDate="2015-04-22T14:05:09.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2581" PostTypeId="2" ParentId="2577" CreationDate="2015-04-22T11:30:56.587" Score="10" Body="&lt;p&gt;The obvious place to start is &quot;what does psi mean?&quot; It's an abbreviation for &quot;pounds per square inch&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a 24&quot; x 24&quot; board, then we have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$24~in\times24~in=576~in²$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get 40psi, i.e. 40 pounds for each square inch, we would need &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$40~\frac{lbs}{in²}*576~in² = 23040 ~lbs$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've asked for an answer in tons, so we need to convert 23040 pounds to tons. Assuming you're using the US short ton, there are 2000 pounds in a ton. So we need &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$23040 ~lbs/2000 ~\frac{lbs}{ton} = 11.52~US~tons$$&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="41" LastEditDate="2015-04-22T16:35:26.380" LastActivityDate="2015-04-22T16:35:26.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2582" PostTypeId="2" ParentId="2578" CreationDate="2015-04-22T13:46:37.790" Score="1" Body="&lt;p&gt;The sinking of the cabin in a hydraulic lift is a result of the oil compressing in the cylinder (or piston), due to the extra weight (there should be no air in the system). This will be more profound when more oil is in the cylinder, i.e. at high buildings at the top floors. Also keep in mind that the piston travel that you experience is dependent on the &lt;a href=&quot;http://www.mitsubishielectric.com/elevator/overview/elevators/systems.html&quot; rel=&quot;nofollow&quot;&gt;suspension&lt;/a&gt;.&#xA;This is &quot;dynamically&quot; compensated by a system that detects if the cabin has moved from the floor level and is activated if the cabin moved beyond a limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, the cabin is held in place by a mechanism if the load difference expected is quite significant (ex. loading by a forklift). If the customer pays, you may find it in smaller lifts, but it is extremely rare.&lt;/p&gt;&#xA;" OwnerUserId="1125" LastEditorUserId="33" LastEditDate="2015-04-22T14:19:55.920" LastActivityDate="2015-04-22T14:19:55.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2583" PostTypeId="1" AcceptedAnswerId="2585" CreationDate="2015-04-22T15:04:50.520" Score="6" ViewCount="85" Body="&lt;p&gt;I'm dealing with some hydrocarbons (diesel, gas oil, kerosene, ethanol) where I'm putting them in a chamber and shining light through them to hit a sensor on the other side in order to measure their colour. Unfortunately what I'm finding is that the lenses and reflective surfaces in the chamber that I'm using for these optical measurements are being stained by the products and subsequently affecting the measurements. When the chamber is emptied, small amounts of product remain clinging to the internal surfaces. As this dries a residue is left which remains on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried a couple of different materials: stainless steel reflectors, transparent Grilamid plastic lenses, a stain-resistant transparent potting and regular glass. All of them end up with a residue that clings to the surface that needs to be wiped clean. Unfortunately, for my application I'm not going to be able to get access to the internal surface to clean it so I need to find a way to prevent this residue from clinging to the internal surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main restriction for me is that the system needs to basically be 100% maintenance free, and unfortunately I can't control the fluid coming in and out and therefore can't rinse the system out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there are any products (both transparent and reflective) that are basically completely &quot;hydrophobic&quot; but for hydrocarbons and will prevent the product clinging at all to the surface and staining the lenses and reflectors?&lt;/p&gt;&#xA;" OwnerUserId="1391" LastEditorUserId="1391" LastEditDate="2015-04-22T22:19:18.193" LastActivityDate="2015-04-22T22:19:18.193" Title="How to avoid dyed liquids staining reflective/transparent surfaces" Tags="|materials|optics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2584" PostTypeId="2" ParentId="2583" CreationDate="2015-04-22T15:46:14.780" Score="2" Body="&lt;p&gt;You will want to rinse the chamber between uses with something that dries clear and is a non-polar solvent. A typical solvent like pentane or diethyl-ether should do the trick though keep in mind the fire hazard.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-04-22T15:46:14.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2585" PostTypeId="2" ParentId="2583" CreationDate="2015-04-22T15:54:02.210" Score="2" Body="&lt;p&gt;A super-hydrophobic coating would work well for you because super-hydrophobic materials are self-cleaning.  &lt;a href=&quot;http://www.sandia.gov/&quot; rel=&quot;nofollow&quot;&gt;Sandia National Labs&lt;/a&gt; has a nice report on super-hydrophobic coatings which can be found &lt;a href=&quot;http://www.sandia.gov/research/research_development_100_awards/_assets/documents/2008_winners/Superhydrophobic_SAND2008-2215W.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (associated patent &lt;a href=&quot;http://www.google.com/patents/US7485343&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).  In it they report the invention of a coating who's hydrophobicity can be varied by application of UV light.   The &lt;a href=&quot;http://en.wikipedia.org/wiki/Contact_angle&quot; rel=&quot;nofollow&quot;&gt;contact angle&lt;/a&gt; of their coating is greater than 150$^\circ$, and they reference two other patents which reach contact angles of 169$^\circ$ and 172$^\circ$ which are both very high, the maximum is 180$^\circ$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, all of this research is fairly recent and doesn't seem to have found its way to industry yet.  I can't find many coating houses who sell super-hydrophobic optical coatings.  &lt;a href=&quot;http://www.aculon.com/super-hydrophobic-coatings.php?_kk=superhydrophobic%20coatings&amp;amp;_kt=18bf94de-5d58-4823-9fe3-53bac7ca1fbf&amp;amp;gclid=COu7_NqgisUCFXMF7AodmVAAbw&quot; rel=&quot;nofollow&quot;&gt;Aculon's website&lt;/a&gt; says that they sell one, but they don't publish the contact angle, and the picture they show makes it appear to only be ~90$^\circ$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aculon also claims to sell an &lt;a href=&quot;http://www.aculon.com/oleophobic-coatings.php?gclid=CKOdjMGhisUCFdcYgQodkhYAKg&quot; rel=&quot;nofollow&quot;&gt;oleophobic&lt;/a&gt; (oil repelling) coating.  This may work for you, but you would need to test a sample with the materials you plan to use.  The video on their website shows someone trying to write on it with a Sharpie and it just won't take to the surface.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with all of these coatings is longevity.  Rain-X is an excellent example of a hydrophobic coating, but it must be reapplied once a month or so.  My eyeglasses came with a hydrophobic coating which was very impressive when they were new.  Now that they are 10 months old, everything sticks to them again.  So, whatever you decide to build, you should be able to service it regularly.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-04-22T15:54:02.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2589" PostTypeId="1" AcceptedAnswerId="2590" CreationDate="2015-04-23T09:56:08.000" Score="15" ViewCount="7732" Body="&lt;p&gt;I am designing a metal plate that will be laser-cut (or machine-cut) and then folded. I want to know how to size the pre-folded plate in order to get the right dimensions after folding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y5Pq3.png&quot; alt=&quot;Diagram of a folded 2mm aluminium plate with holes and window&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My actual part is not exactly like this (I simplified it for ease of drawing) but it shows what I want to achieve. In this case it's a 2mm aluminium plate, the red arrows show the inner dimensions after folding that I want to specify and achieve. The holes must also line up and the window should be correctly placed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively I would expect some compression along the inner part of the folds and stretching on the outer parts - ideally along the centre of the plate - but I don't know if this is what will happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the red arrows are 100mm each, should the plate be 300mm? I'm guessing not, so how do I calculate the radius of curvature that will be achieved and if I need to add (or remove) material at the folds in order to achieve my required dimensions?&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2017-05-20T16:15:21.360" Title="How do I size metal plates to get the correct dimensions after folding?" Tags="|machining|metal-folding|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2590" PostTypeId="2" ParentId="2589" CreationDate="2015-04-23T11:21:22.917" Score="17" Body="&lt;p&gt;Your assumption is right! A 300mm long plate with two foldings won't do! This is because you need to take into account the bend allowance and the bend compensation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why is so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram of what's going on:&#xA;&lt;img src=&quot;https://www.oemmfginc.com/wp-content/uploads/2010/08/image_thumb.png&quot; alt=&quot;Bend allowance, bend deduction diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When you bend a material, part of it will extend (the external part of the bend), while another part will retract (the internal part).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line (in the thickness of the plate) where the dimension doesn't change is called the Neutral line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The neutral line is usually located between a third and a half of the material thickness (from the interior to the exterior of the bend). Meaning this line will hold its dimension, while the top surface (inner bend surface) will contract a little bit, and the bottom surface (outer bend surface) will expand a little bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia has a nice &lt;a href=&quot;http://en.wikipedia.org/wiki/Bending_%28metalworking%29#Bend_allowance&quot;&gt;little part&lt;/a&gt; on the calculation required, taking into account the angle of the bend, and the thickness of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the formula for Bend Allowance:&#xA;$BA = A \left( \frac{\pi}{180} \right) \left( R + K \times T \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, &lt;code&gt;BA&lt;/code&gt; is the Bend Allowance, &lt;code&gt;A&lt;/code&gt; is the bend angle in degrees, &lt;code&gt;R&lt;/code&gt; is the inside bend radius, &lt;code&gt;K&lt;/code&gt; is the ratio between the distance from the inside face to the Neutral line to the thickness of the material, usually $\frac{1}{3}$, and &lt;code&gt;T&lt;/code&gt; is the thickness of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is the formula for the Bend Deduction: $BD = 2 \left(R + T \right) \tan{ \frac{A}{2}} - BA$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, &lt;code&gt;BD&lt;/code&gt; is the Bend Deduction, &lt;code&gt;R&lt;/code&gt; is the inside bend radius, &lt;code&gt;A&lt;/code&gt; is the bend angle in degrees, &lt;code&gt;R&lt;/code&gt; is the inside bend radius, &lt;code&gt;T&lt;/code&gt; is the thickness of the material and &lt;code&gt;BA&lt;/code&gt; is the Bend Allowance.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In your case, you want to calculate the distance from the internal faces, not just for the straight part of the metal plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, you just have to add the thickness of the material to your dimensions, i.e, your A length (from the picture above) is $(New Length) = (Old Length) + t * 2 $.&lt;/p&gt;&#xA;" OwnerUserId="154" LastEditorUserId="154" LastEditDate="2015-04-23T15:40:24.323" LastActivityDate="2015-04-23T15:40:24.323" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2591" PostTypeId="1" CreationDate="2015-04-23T14:02:03.637" Score="4" ViewCount="640" Body="&lt;p&gt;I am currently building a mini boat-like robot to clean the water from floating trash. I used two servo motors to move the loader that I have installed in front of the boat, and two 3 V DC motors installed in the back to allow the boat to move across the water. I also used Arduino to program my servos and motors. When I tried to test my robot in water, the two DC motors did not push the boat forward. Why didn't the air push my boat?&#xA;Here are the materials that I used to build my robot:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Two 1 L water bottles&lt;/li&gt;&#xA;&lt;li&gt;A small plastic box to hold the electric circuits.&lt;/li&gt;&#xA;&lt;li&gt;Another small plastic box to serve as a trash collector.&lt;/li&gt;&#xA;&lt;li&gt;Two plastic propellers (taken from a mini fan).&lt;/li&gt;&#xA;&lt;li&gt;A toy loader.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The total weight of the boat is around 700 g. The picture below is the robot I am working on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nxbWQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1400" LastEditorUserId="272" LastEditDate="2015-04-23T17:02:09.703" LastActivityDate="2015-04-24T10:08:38.330" Title="Why My Mini Boat Does Not Move Forward?" Tags="|mechanical-engineering|fluid-dynamics|propulsion|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2592" PostTypeId="2" ParentId="2591" CreationDate="2015-04-23T14:09:28.733" Score="6" Body="&lt;p&gt;Your fans are too small. If you hold 1 and make it go max speed you will barely feel the force it applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead consider a drive-train going into the water and turning a screw. Because water is much more viscous and dense you get much better acceleration.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-04-23T15:29:42.580" LastActivityDate="2015-04-23T15:29:42.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2594" PostTypeId="1" AcceptedAnswerId="2615" CreationDate="2015-04-23T21:00:11.883" Score="5" ViewCount="561" Body="&lt;p&gt;I need to release pressure of around 50psi, after 3-4 seconds of initiating, from a vessel using a mechanical switch. I had considered a pneumatic solenoid, but I want to forgo the inconvenience of charging the device with electricity in addition to the air, as well as save the space that would be needed for the bulky solenoid, circuitry, and battery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I therefore need what would essentially be a &quot;mechanical-fuse&quot; that releases pressure after a delay when triggered. I considered a screw-type piston that gets forced along a track by the air in the vessel itself, which at a set psi could guarantee a specific time of release. I also need the system to be no longer than 5 inches along the axis of released pressure. Does a &quot;screw-fuse&quot; sound like a viable idea? Does anyone have any other sources for a mechanical fuse of this type, or an idea of their own?&lt;/p&gt;&#xA;" OwnerUserId="1405" LastEditorUserId="1405" LastEditDate="2015-04-24T03:45:24.520" LastActivityDate="2015-04-26T04:31:48.950" Title="Are there any precedent designs for a mechanically-delayed pressure release valve?" Tags="|mechanical-engineering|pressure|compressed-air|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2595" PostTypeId="1" AcceptedAnswerId="2603" CreationDate="2015-04-24T06:16:10.763" Score="4" ViewCount="444" Body="&lt;p&gt;A quote from my textbook :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Turbomachinery flows are steady only in a time-averaged sense; that is, the flow is periodic,&#xA;  with a period equal to the time taken for a blade to move a distance equal to the spacing&#xA;  between adjacent blades. Despite the unsteadiness, in elementary analysis all variables are&#xA;  assumed to have steady values.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you please explain this 'unsteadiness'? What effect does a blade moving a distance equal to the spacing between adjacent blades has on making the flow periodic?&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2015-04-24T07:13:08.767" LastActivityDate="2015-04-24T15:11:50.987" Title="What makes a flow periodic in Turbomachines?" Tags="|mechanical-engineering|fluid-dynamics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2596" PostTypeId="2" ParentId="2595" CreationDate="2015-04-24T06:42:08.193" Score="4" Body="&lt;p&gt;Hopefully this picture will help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OYFI3m.jpg&quot; alt=&quot;http://www.cd-adapco.com/&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can observe that the stream (blue) is divided by blades. This separation of the initial flow makes the flow unsteady and can be compared to pulses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impulse vs reaction turbines comparison:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rgEAQm.png&quot; alt=&quot;Turbines impulse vs reaction - wiki&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1193" LastEditorUserId="272" LastEditDate="2015-04-24T11:52:03.840" LastActivityDate="2015-04-24T11:52:03.840" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2597" PostTypeId="2" ParentId="2595" CreationDate="2015-04-24T08:28:12.107" Score="2" Body="&lt;p&gt;I think what is meant in your textbook could be also induced velocities by a stator (as meaning of &quot;adjacent blades&quot;). &#xA;Think of a multi-stage turbine with multiple rotor-stator stages: the flow will always be affected by the rate of turn of the turbine, which makes it unsteady (time dependent), because the rate of turn is a function of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with no stator, the inflow to your turbomachine will almost never be fully homogeneous, resulting in a variation in velocity over the duration of one turn. This means a blade of your turbomachine will experience different forces in one rotation due to altered velocities and pressures.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2015-04-24T08:28:12.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2598" PostTypeId="2" ParentId="2591" CreationDate="2015-04-24T10:08:38.330" Score="3" Body="&lt;p&gt;Your air propellers did push your boat, but the delivered thrust was not enough! &#xA;As ratchet freak an Dan said, the density of the medium in which the propeller is working is of great importance, as the delivered thrust $T$ is roughly calculated by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T = c_{th} \frac{\rho}{2}v^2A_{prop}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;nevermind the factor $c_{th}$ but you easily see that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T \sim \rho&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and as the density $\rho$ of water is approximately 1000 times higher than the density of air, a propeller in water will generate significantly more Thrust given the same circumstances as for the same propeller in air.&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2015-04-24T10:08:38.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2599" PostTypeId="1" CreationDate="2015-04-24T12:02:06.910" Score="6" ViewCount="108" Body="&lt;p&gt;I've noticed that quite a few commercial nuclear reactors have an extended outage (they close down for several months) some time late in their third decade, and the maintenance works during the closure involve significant investment. This cost has been high enough in some cases that the reactor has permanently shut down rather than undergo the works (e.g. the Vermont Yankee, Dec 2014) because the business case just didn't stack up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the works involved, in terms of scope and cost; how much do these vary depending on the type of reactor (PWR, BWR, ...)?&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-04-24T17:10:25.013" LastActivityDate="2017-07-23T18:15:25.483" Title="Summary of major outage management works needed in the 4th decade of a nuclear reactor?" Tags="|maintenance|cost-engineering|nuclear-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2601" PostTypeId="1" CreationDate="2015-04-24T13:24:26.293" Score="5" ViewCount="170" Body="&lt;p&gt;I am trying to make a model of non-stationary heating on a plate, where I must use an explicit numerical method to solve the model. I decided the plate will be made of steel (0,5% C) and I found that the thermal diffusivity coefficients $k/\rho c_p$ are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha_{steel} = 14.74 \times 10^{-6} \:\mathrm{m^2/s}$$&#xA;$$\alpha_{air} = 1.9 \times 10^{-5} \:\mathrm{m^2/s}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have to implement a heat source but I don't know what coefficient I should use for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it's a 2D problem, my solving code looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def heat_source(x, y, matrica):  &#xA;if (x &amp;gt;= 0.8 and x &amp;lt;= 1.2) and (y &amp;gt;= 0.8 and y &amp;lt;= 1.2):&#xA;    if matrica[10,10]&amp;lt;300.:     &#xA;        return 280 + randn(1)*20&#xA;    else:&#xA;        return 0.&#xA;else:&#xA;    return 0.0 &#xA;&#xA;dmx = (M0[iY,iX+1] + M0[iY,iX-1] - 2.0 * M0[iY,iX])/dx**2.0 # conduction in x-direction&#xA;dmy = (M0[iY+1,iX] + M0[iY-1,iX] - 2.0 * M0[iY,iX])/dy**2.0 # conduction in y-direction&#xA;M_new[iY,iX] = M0[iY, iX] + k_diffusion*dt*(dmx+dmy) + dt*k_heating*heat_source(X[iX],     Y[iY], M0)  - dt*k_cooling*(M0[iY,iX]-T_ok)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Am I using the right coefficients? If yes, what coefficient should I use for the heat source?&lt;/li&gt;&#xA;&lt;li&gt;What's the difference between non-stationary and stationary source?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1411" LastEditorUserId="1411" LastEditDate="2015-04-25T19:18:23.340" LastActivityDate="2018-08-25T22:01:32.577" Title="What coefficients should I use when modeling non-stationary heating of a steel plate?" Tags="|mechanical-engineering|heat-transfer|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2602" PostTypeId="1" CreationDate="2015-04-24T14:15:05.723" Score="4" ViewCount="529" Body="&lt;p&gt;I have these 2 problems to solve (on the image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 9 stores that want different quantities of a product. The first store asked for 62 units, the second one 10 units, etc etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supplier sent a total of 138 units (case 1). I have to distribute those 138 units to all the stores in a fair way. Each box sent from the supplier has 2 units and it's better to use multiples of 2 units for each store, so that it's not necessary to dismantle the box. If it's not possible, we can dismantle the box but just in the last case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case 2, there are some stores who will get less than what they asked. That's ok. I just need to distribute them in a fair way so one store doesn't get everything and the others nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a way to do this but it's really complex and full of errors so I'm trying to find an easier way like a formula or something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XD0Ii.png&quot; alt=&quot;Table&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What I'm doing is (I'm going to give an example for the first line and Case2):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Giving a box to every store so there is no risk to have one store receiving nothing.&lt;/li&gt;&#xA;&lt;li&gt;Calculate: NecessityByPriority = Necessity-1box = 62-2 = 60&lt;/li&gt;&#xA;&lt;li&gt;Calculate: PriorityEachStore: 60/sum(NecessityByPriority) =  60/72 = 83,3333%&lt;/li&gt;&#xA;&lt;li&gt;Calculate: PriorityUnits = PriorityEachStore*(Stock-9box) = 83,3333%*(58-18) = 33,3333 [the 9box corresponde to the number of units given in the first step: 9stores*2units each]&lt;/li&gt;&#xA;&lt;li&gt;Calculate: NumberBoxes = round(PriorityUnits/(Unit/box)) = round(33,3333/2) = 17&lt;/li&gt;&#xA;&lt;li&gt;Calculate: NumberUnits = NumberBoxes*(Unit/box) = 17*2 = 34&lt;/li&gt;&#xA;&lt;li&gt;Calculate the final quantity: NumberUnits+1box[given in step 1] = 34+2 = 36&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the case given there is not a great error but I am going to show you one with error above in a simulator I have done in Excel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KpoiR.png&quot; alt=&quot;Table2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here there were 3 units that remained. This is due to rounding. I can add a line to verify if there is something remaining and if so to redistribute. But I was looking for an easier way without all this formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have another process that calculates with the same formulas, everything for the first store. Then makes a 2nd iteration and recalculates everything except the first store. And on on until there is no more store. The problem is that here there is only 9 stores but sometimes there are hundreds.&lt;/p&gt;&#xA;" OwnerUserId="1413" LastEditorUserId="33" LastEditDate="2016-01-31T03:25:03.067" LastActivityDate="2016-01-31T14:25:59.427" Title="How to fairly distribute items when there is more demand than supply?" Tags="|project-management|engineering-managment|industrial-engineering|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2603" PostTypeId="2" ParentId="2595" CreationDate="2015-04-24T14:39:16.070" Score="3" Body="&lt;p&gt;The words &lt;strong&gt;unsteady&lt;/strong&gt; and &lt;strong&gt;periodic&lt;/strong&gt; have different meaning depending on the context they are used in. My answer is mainly with regards to axial or radial turbomachines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say our machine is running at a specific operating point. The rotation of the blades (the shaft) does not change. The massflow-rate through the system is constant.&#xA;Even though this operating-point is &lt;em&gt;steady&lt;/em&gt; the flow inside of the machine does &lt;em&gt;not&lt;/em&gt; have to be steady.&#xA;For practical purposes one usually designs every blade row in its specific (relative) &lt;em&gt;frame of reference&lt;/em&gt;. This means the designer 'sits' on the air-foil.&#xA;Now we assume further: steady flow- and boundary conditions. Doing this makes the aero and mechanical design feasible. When designing the blades and vanes for a turbomachine one will not design 'every' blade but assume that the flow around the blades will be the same for all blades of one blade-row. (Even though this assumption is usually not valid at the boundaries of the operating range it is a very practical approach to get the design-process started.) This means the flow is believed to be &lt;strong&gt;steady&lt;/strong&gt; and &lt;strong&gt;periodic&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However! There is one thing called the 'Unsteadiness Paradox' &lt;a href=&quot;https://i.stack.imgur.com/KOQnc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;. It basically boils down to the fact that a turbomachine &lt;strong&gt;has&lt;/strong&gt; to be unsteady in order to work. But the &lt;em&gt;unsteadiness&lt;/em&gt; in this case refers to the rotation of the blade-row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KOQnc.png&quot; alt=&quot;Turbomachine&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure shows a sketch from Greitzer. If one assumes the turbo-machine to be 'black box' which just adds or extracts work than it is not possible to describe the turbo-machines analytically. It only works when adding a stationary and rotating black-box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without any simplifications of the flow and the geometry it would not be possible to design a turbomachine and predict its operating characteristics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things have to be simplified:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Aerodynamics&lt;/li&gt;&#xA;&lt;li&gt;Geometry&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In case of the aerodynamics the unsteadiness (eddies and turbulence) in the flow is usually averaged out. In case of the geometry one assumes &lt;strong&gt;periodicity&lt;/strong&gt; (see the figure below from Oxford University):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/E45q4.png&quot; alt=&quot;multi-passage-simulation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the version (a) only one passage was simulated and simply copied (all passages look the same). In version (b) all passages were simulated showing slight differences under certain operating conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second simplification to the geometry is the introduction of &lt;strong&gt;mixing planes&lt;/strong&gt; so the rotating and stationary part of the turbomachine can be calculated separately (see figure below from Cambridge University):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/64vi5.png&quot; alt=&quot;multi-row-simulation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unsteadiness which is introduced by the relative motion of blades and vanes (right) is averaged-(smeared)-out (left).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These simplifications are necessary in order to have fast optimisation and development of the turbomachine and are later corrected and checked by experiments or high fidelity simulations if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KOQnc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&#xA;Dean, R.C., &quot;On the Necessity of Unsteady Flow in Fluid Machines&quot;, ASME J. Basic Eng. March 1959, pp 24-28&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-04-24T15:11:50.987" LastActivityDate="2015-04-24T15:11:50.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2604" PostTypeId="2" ParentId="2602" CreationDate="2015-04-24T16:56:28.877" Score="5" Body="&lt;p&gt;&lt;strong&gt;What is &lt;em&gt;Fair&lt;/em&gt;?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your definition of &lt;em&gt;fairness&lt;/em&gt; is the quantity that you are really trying to optimize (maximize). There can be multiple methods to optimize for this, but first you have to define how this will be calculated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to describe some possible methods below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Maximize Number of Full Orders&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method assumes that the happiness of each consumer is weighted the same. Because of this, you want to maximize the number of full orders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Order the consumers from least requested to most requested and start filling orders from the top. This will give you the most completely filled orders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Maximize the Biggest Orders&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This methods assumes that the consumers that ordered the most are the most worthwhile consumers to keep supplied. You want these important consumers to be happy because you don't care about the smaller ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Order the consumers from most requested to least requested and start filling orders from the top. This will make the biggest consumers the happiest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fulfill the Same Percentage of All of the Orders&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks to be the method that you tried to use.  This one is more complicated, because you will end up with partial units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum all of the orders and give a weight to each order based on the individual order divided by the sum of all orders. This is the percentage of the total available that each consumer will get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be considered the most &lt;em&gt;fair&lt;/em&gt;, but it isn't necessarily the best from a business standpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Choose what to Optimize&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to decide on what is to be optimized in order to be able to optimize the distribution. In this answer I have made some assumptions about the situation. I mostly came at the question from a business point of view, but the answer could be completely different if you were looking at supplying machines or if there were costs to not supplying some consumers. All of these issues will change the final solution.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-04-24T16:56:28.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2605" PostTypeId="1" AcceptedAnswerId="2606" CreationDate="2015-04-24T19:27:27.740" Score="5" ViewCount="130" Body="&lt;p&gt;I want to embed a wire inside a plastic and heat it to specific temperatures (80$^\circ$C 100$^\circ$C 130$^\circ$C 180$^\circ$C). I want to find a material that is resistive enough for resistive heating, but also I want the wire's resistivity to vary fairly strongly with temperature - this way I can know the actual temperature that I'm getting by looking at the resistance during the heating process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that I cannot seem to find a material with the properties I want. I tried nichrome wire and carbon fibers, and while they are resistive enough to give me heat, their resistance does not change enough (for my desired temperature ranges) to be measurable with my setup.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="1415" LastEditDate="2015-04-25T08:32:04.000" LastActivityDate="2015-04-25T08:32:04.000" Title="Material which can be used for resistive heating and simultaneous temperature measurement" Tags="|electrical-engineering|materials|heat-transfer|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2606" PostTypeId="2" ParentId="2605" CreationDate="2015-04-24T20:23:37.500" Score="4" Body="&lt;p&gt;Just about anything you can find will exhibit a temperature dependence on its resistivity.  The first hit looking up nichrome says it's temperature dependence is 0.004/&amp;deg;C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better model for resistivity is the Steinhardt-Hart(sp?) equation, but you probably don't need something that elaborate.  Since you only want to control the temperature to 4 discrete values, you can figure out ahead of time what the total resistance of a particular diameter and length of wire will be at each of those 4 temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To measure the resistance, you need to measure the voltage across the wire and the current thru it, then divide.  With the right analog electronics, these values can be presented to A/D inputs of a microcontroller, which can compute the resistance, determine the temperature, then vary the current via a controller accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd probably implement the resistance to temperature function as a piecewise-linear table lookup.  That's quick to compute and can model any non-linearities without lots of cycles.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-04-24T20:23:37.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2609" PostTypeId="1" AcceptedAnswerId="2611" CreationDate="2015-04-25T08:40:28.170" Score="11" ViewCount="7097" Body="&lt;p&gt;Why are tremors of an earthquake felt most on the upper floors of a building in comparison to lower floors? Does this have something to do with a third class lever?&lt;/p&gt;&#xA;" OwnerUserId="739" LastEditorUserId="33" LastEditDate="2016-02-23T02:19:49.440" LastActivityDate="2016-02-23T02:19:49.440" Title="Why are tremors of an earthquake felt more on upper floors?" Tags="|mechanical-engineering|civil-engineering|applied-mechanics|seismic|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2610" PostTypeId="1" CreationDate="2015-04-25T11:39:48.287" Score="3" ViewCount="1904" Body="&lt;p&gt;Is it possible to build interferometer, for linear displacement measurements, using old CD/DVD drive parts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am talking about using parts such as beam-splitter, detector, laser and frame where it all mounted if possible…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the XY stage on a microscope with piezo-motors. Interferometer should be the best linear encoder to be used as a feedback loop for proper driving of piezos. There are commercial linear encoders based on Michelson interferometer, but they are too expensive. Old CD/DVD drives appeared to be extremely cheap source of: lasers, laser drivers, 50/50 mirrors, detectors, detector drivers, maybe anything else.&lt;/p&gt;&#xA;" OwnerUserId="1420" LastEditorDisplayName="user16" LastEditDate="2015-05-07T22:17:04.427" LastActivityDate="2015-05-07T22:17:04.427" Title="Building laser interferometer from CD/DVD drives parts" Tags="|mechanical-engineering|measurements|optics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2611" PostTypeId="2" ParentId="2609" CreationDate="2015-04-25T12:16:44.653" Score="10" Body="&lt;p&gt;First, your statement is not universally true.  Motion of the bottom of a building can be greater or lesser in upper floors.  It depends on the type of motion, how free the building is to move in other axes, where you are relative to the building center of mass, and whether the motion hits any resonant frequencies of the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the motion excites a resonant frequency, then the motion will probably be larger as you go higher.  Picture one of those cat toys that has a suction cup on one end, a flexible stick coming out of it, and a ball of soft material at the other.  If you were to move the suction cup end a little sideways, you can get much larger movement of the ball if your motion is at the right frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show a reduced motion above the ground level, imagine a tall thin box sitting on a piece of paper on a table.  If you move the paper back and forth quickly so as to move the bottom of the box while letting it pivot as it wants to, the middle of the box will move least.  This is because the box is acting like a lever about its center of mass.  On the &quot;floor&quot; that matches the center of mass, the motion is smaller than at the bottom.  The motion then gets larger again going up from there.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-04-25T12:16:44.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2612" PostTypeId="1" AcceptedAnswerId="2698" CreationDate="2015-04-25T12:17:45.800" Score="13" ViewCount="16027" Body="&lt;p&gt;I am making a device for measurements. I would like to measure the distance within accuracy of 1 mm. Range could be 2 cm to 15 cm . I looked at &lt;a href=&quot;https://www.sparkfun.com/datasheets/Sensors/Infrared/GP2D120XJ00F_SS.pdf&quot; rel=&quot;noreferrer&quot;&gt;Proximity Sensors&lt;/a&gt; but the readings displayed by these sensors are not steady.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to measure the thickness of the plate (carbon steel). The two sensors will be mounted on a structure. The sensors will give me the distance of the surface from the sensor. Then I will calculate the thickness of the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are possible types of &lt;strong&gt;low cost sensors&lt;/strong&gt; I can use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eYc3e.png&quot; alt=&quot;Sensor Arrangement&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1338" LastEditorUserId="41" LastEditDate="2015-05-01T20:17:30.323" LastActivityDate="2022-09-19T10:55:36.110" Title="Distance sensors with accuracy of 1 mm?" Tags="|mechanical-engineering|measurements|sensors|tolerance|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2613" PostTypeId="2" ParentId="2602" CreationDate="2015-04-25T12:34:59.703" Score="0" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question has been answered using lean manufacturing engineering concepts. I have used Toyota Manufacturing as basis for the solution.  The idea can be applied to any other manufacturing engineering process such as pharmaceutical, consumer electronics, textile, or any other similar industry.  Since this is an engineering forum, I have attempted to answer in an engineering frame work. The question could be answered in a pure business frame work too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Response&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supplying raw material to manufacturing plant such Toyota Assembly plant has very similar correlation to supplying merchandise to a store. It is critical that all required parts be in stock in an assembly plant to a manufacture product, similar to having merchandise on shelf in a store to allow the consumer to make purchase. It is said that GM would lose over half millions US dollar if the Covert plant in Bowling Green Kentucky is not in operation for an hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S8OJt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A product/raw material required for the assembly of Toyota automobile is coolant temperature sensor. I have assumed that all the assembly plants listed below require a coolant temperature sensor. The fictitious volumes per month are also listed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{| l | l | c |}&#xA;\hline&#xA; Toyota \ Manufacturing &amp;amp;  Model &amp;amp;  Average\ Sensors\\&#xA;&amp;amp; &amp;amp; required\ per\ week\\	&#xA;\hline&#xA;Mississippi &amp;amp; Corolla &amp;amp; 7500\\ &#xA;Kentucky &amp;amp; Camry &amp;amp; 750 \\&#xA;Texas &amp;amp; Tacoma &amp;amp; 75 \\&#xA;Indiana &amp;amp; Sienna &amp;amp; 8 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using data engineering or information engineering statistical data tools and chart like individual moving range charts (I-MR)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/s3DkG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statistical tools such as the above will help determine the average pull rate (demand), standard deviations, and other information to determine minimum amount to product/raw material required to successfully operate the engineering manufacturing plant. Similar concept can be adopted to understand the demand in the stores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the supply side, demand averages, upper and lower control limit (LCL) is used to distribute the raw material to engineering manufacturing locations. The LCL will help determine with some level of confidence the minimum level of required raw material as well as strategy to distribute additional raw material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;br&gt;&#xA;If information such as standard deviation, UCL, LCL, median and range are added to the above Toyota Manufacturing Plant Temperature sensor weekly requirement table a great deal of insight to plant operation can be made available how best to distribute supplies. A combination of Data Engineering, Information Engineering, Process Engineering and statistics can greatly benefit toward optimization of the distribution process.  &lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-04-29T00:12:35.003" LastActivityDate="2015-04-29T00:12:35.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2614" PostTypeId="1" AcceptedAnswerId="2616" CreationDate="2015-04-26T04:07:32.957" Score="3" ViewCount="207" Body="&lt;p&gt;I have a simulation software which shows energy consumption at 30 minutes timestep. If a system consumes 0.06 kWh at 6am, and 0.09 kWh at 630am, how to calculate the energy consumption between 6-7am? Can I assume it is the average of both timestep? Does finer timestep, eg. 10 minutes interval returns a different result?&lt;/p&gt;&#xA;" OwnerUserId="1375" LastActivityDate="2015-04-26T07:38:21.530" Title="Calculating kWh for different timestep" Tags="|mechanical-engineering|electrical-engineering|hvac|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2615" PostTypeId="2" ParentId="2594" CreationDate="2015-04-26T04:31:48.950" Score="3" Body="&lt;p&gt;There are three possibilities here that come up offhand, between my own mind and ratchet freak's comment:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The choke that ratchet freak mentioned -- this is the simplest approach if you are OK with a slow decay of pressure&lt;/li&gt;&#xA;&lt;li&gt;The dampened piston that ratchet freak also mentioned, where the mechanical switch pulls out a lock from the piston to allow pressure to retract it against the calibrated force from the damper&lt;/li&gt;&#xA;&lt;li&gt;A clockwork mechanism that is triggered by the switch and provides the three-to-four second delay by letting a spring unwind&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The main issue with the second and third mechanisms is that they would require some sort of manual or automatic reset functionality (a reset could be triggered by pressure building up to a certain point).&lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2015-04-26T04:31:48.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2616" PostTypeId="2" ParentId="2614" CreationDate="2015-04-26T04:32:43.497" Score="4" Body="&lt;p&gt;I think you are getting power and energy confused, so I'll quickly define them before giving an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Energy&lt;/strong&gt; energy is a measure of how much potential for work exists.  If you had a battery, it contains a certain amount of energy. Once that energy is used, the battery is &quot;dead&quot; and must be recharged (or thrown away). This is usually measured in Joules (J) or kilowatt-hours (kWh).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Power&lt;/strong&gt; is the rate at which energy is being used. For example, for a fan to blow air it must be continuously consuming energy.  One joule per second is called a Watt (W).  1000W is a kilowatt (kW).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an appliance consumed a constant 1000W for one hour, it would have consumed 1kWh.  1kWh could also be consumed by an appliance consuming a constant 100W for 10hours: in general the energy is the integral of the power over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your software returns energy consumption values (kWh) for 30 minute timesteps, that means that it is (presuming it is designed sanely) integrating the power over every 30 minute timestep.  For your specific example, that means that between 6 and 7am the system consumed 0.15kWh of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 10 minute timestep would have resulted in 6 values that added up to 0.15kWh, assuming that you have set up your model to achieve step-size independence. If you set your timestep to 10 minutes and you get a different value, then you should probably view your 30 minute results with suspicion, but that is an entirely different topic.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-04-26T04:32:43.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2617" PostTypeId="2" ParentId="2614" CreationDate="2015-04-26T07:38:21.530" Score="1" Body="&lt;p&gt;In SI units the measure of &quot;amount&quot; of electrical energy is the Joule.&lt;br&gt;&#xA;Energy is a measure of &quot;work done&quot;.&lt;br&gt;&#xA;Energy can be (and often is) expressed in terms of 'rate of doing work'.&lt;br&gt;&#xA;The rate at which work is done or at which energy is supplied or utilised is also named &quot;Power&quot;.&lt;br&gt;&#xA;Work done = 'rate of doing work' x time period involved&lt;br&gt;&#xA;= Power x time (involved)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 1 Joule of energy = 1 Watt of power for 1 Second = 1 W.s &#xA;1 Joule is a small amount of energy compared to typical usage so a larger multiple is used of 1 kWh&lt;br&gt;&#xA;= 1000 Watts work rate x 3600 seconds&lt;br&gt;&#xA;= 3,600,000 Joule = 1 MJ = 1 kWh &lt;/p&gt;&#xA;&#xA;&lt;p&gt;kWh (KiloWatt hours) is a typical unit of &quot;amount&quot; for electrical energy use for eg heaters, motors, lighting etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO if you operate at a power level of 120 Watts from 6am to 6.30 am you will consume 120 Watts  x 1/2 hour = 60 Watt hours = 0.060 kWh. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;180 Watts from 6:30am to 7am = 180 x 1/2 Hour = 90 Wh = 0.090 kWh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As kWh = amount of energy, successive figures can be added.&lt;br&gt;&#xA;So 0.060 kWh + 0.090 kWh = 0.150 kWh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Key summary:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy = &lt;strong&gt;Joule&lt;/strong&gt; or Watt-hour (W.h) or KiloWatt hour (&lt;strong&gt;KWh&lt;/strong&gt;)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy = Watts of power x period used &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy = W1.T1 + W2.T2 + .... WnTn&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-04-26T07:38:21.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2619" PostTypeId="1" CreationDate="2015-04-26T18:40:16.087" Score="1" ViewCount="323" Body="&lt;p&gt;I am working in telecom company which is willing to install EMC telephone line filters and EMC power line filters on single facility that has all the equipment. I have been reading about filters issue from last two to three weeks. I have seen that all the manufacturers that are building these type of filters are providing leakage current data for &lt;strong&gt;EMC power line filters&lt;/strong&gt; for single phase and three phase power line filters but no such specs for telephone line and data line filters. &#xA;My query is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can power line filter, telephone line filter and data line filter be install on same single facility even though power line filters have leakage current?&lt;/li&gt;&#xA;&lt;li&gt;How I can separate the power line filters from other filters on single facility so that telephone line does not get noise from leakage current of power line filter due to the ground loop?&lt;/li&gt;&#xA;&lt;li&gt;Will separate ground/earth for each filer will eliminate the noise problem or not?&lt;/li&gt;&#xA;&lt;li&gt;Is there any other solution that I can use to keep telephone lines secure from noise of leakage current of power line filters.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="368" LastEditDate="2015-11-20T01:08:59.403" LastActivityDate="2015-11-20T01:08:59.403" Title="EMC filter selection and installation while there is power line filter leakage current" Tags="|electrical-engineering|emc|electronic-filters|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2620" PostTypeId="1" CreationDate="2015-04-26T21:48:17.690" Score="2" ViewCount="1473" Body="&lt;p&gt;I have an 4 stroke spark ignition engine simulator which takes user inputs and outputs performance data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need rpm, torque and brake specific fuel consumption (bsfc) to plot the bsfc map. I am able to plot torque and bsfc against rpm respectively without any problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a contour plot requires x(rpm), y(torque) and z(bsfc) where z is a function of x and y, how can I express bsfc in terms of rpm and torque or can I take advantage of the simulator to produce raw data required to plot the map saving me trouble in deriving the equation. If its the latter, how can I obtain bsfc data? What parameters should I change or keep constant?&lt;/p&gt;&#xA;" OwnerUserId="823" LastActivityDate="2015-04-26T21:48:17.690" Title="What data do I need to plot a BSFC map" Tags="|mechanical-engineering|thermodynamics|combustion|modeling|simulation|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2621" PostTypeId="1" CreationDate="2015-04-26T22:49:10.877" Score="7" ViewCount="805" Body="&lt;p&gt;I am a mason contractor whom specializes in Rumford Fireplaces. I am helping design a fireplace that had a problem fitting in a single 18&quot; x 18&quot; flue pipe. So I installed two 13&quot; x 13&quot; pipes adjacent to each other.&lt;br&gt;&#xA;They start at the smoke chamber tight together with a smooth as possible connecting partition.  I'm cutting out the connecting flue walls some 12&quot; down from the chimney top so as to form a plenum in order to avoid pressure variations from wind on one side more than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two flue pipes will be equal in height and length and have the same amount of masonry surrounding them to keep temperatures even.  They will be centered in the smoke chamber and their total area is 256 $in^2$, which is nearly equal to 1/9th of the 48&quot; x 48&quot; fireplace opening as required by all codes.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My masonry teacher built a design like this long ago as have I, and I have had no problems. However, I am being told on certain blogs that this design will cause turbulence and double the drag resistance.  I'm also told that I should have used a single 13&quot; x 20&quot; flue with a inside area of 192 $in^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that having my pipes connected at the top and bottom prevents turbulence from pressure and flow differences.  And that having more cross sectional area from the combined flues should easily compensate for the added drag of the extra 12&quot; of smooth tile partition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also believe that in a smoke chamber, a intermittent vortex exists below the damper blade and never rises into the flue area because the flow changes to laminar flow before entering the pipes and the parting divider.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, I don't use double flues.  However, I really can't see how two pipes connected at both ends and being larger in combined area will hurt flow via turbulence to a point of where you would see enough flow loss to generate eddy's in the flow.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;If anyone can help verify if two flue pipes is sufficient or not in this design, I'd appreciate it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize my thoughts: I don't believe the air flow will be negatively affected from using two flue pipes instead of a single flue pipe because of the connection between the two pipes and having a larger overall combined area within the pipes.&lt;/p&gt;&#xA;" OwnerUserId="1431" LastEditorUserId="33" LastEditDate="2016-02-17T21:23:35.753" LastActivityDate="2019-02-11T01:42:55.607" Title="One large fireplace flue versus two smaller ones" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|masonry|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2622" PostTypeId="1" AcceptedAnswerId="2632" CreationDate="2015-04-26T23:39:46.737" Score="5" ViewCount="265" Body="&lt;p&gt;This may fall a little under chemistry processes but I felt it has enough pertaining to aerospace to put it here. Basically, I'm in the process of attempting to develop a way to derive $I_{sp}$ as pertaining to rocket engines rather than rely on charted information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since specific impulse is essentially the exhaust gas velocity working against gravitational force,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;I_{sp} = \frac{v_e}{g_0}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...it stands to reason that the ideal exhaust gas velocity equation can be substituted in here, giving something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;I_{sp} = \frac{\sqrt{\frac{TR}{M}\cdot\frac{2\gamma}{\gamma-1}\cdot(1-\frac{\rho_e}{\rho}^{\frac{\gamma - 1}{\gamma}})}}{g_0}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious problem here is that this is the &lt;em&gt;ideal&lt;/em&gt; exhaust gas velocity, so this is a sort of &quot;perfect-universe&quot; $I_{sp}$. Because most rocket engines use either hydrocarbon or hydrogen/oxygen propellants, water vapor is a major fraction of this exhaust. And as is usually taught early on in chemistry courses, water vapor is a textbook failure of ideal gas behaviors because of its intermolecular forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is - is there a &quot;real&quot; specific impulse formula; something like the van der Waals equation for this application? &lt;/p&gt;&#xA;" OwnerUserId="1229" LastEditorUserId="110" LastEditDate="2015-04-27T01:34:51.410" LastActivityDate="2015-09-25T15:24:39.467" Title="Calculating real or more accurate specific impulse" Tags="|aerospace-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2623" PostTypeId="1" AcceptedAnswerId="2625" CreationDate="2015-04-27T04:40:18.970" Score="6" ViewCount="936" Body="&lt;p&gt;I have a structure like such:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CZvAm.png&quot; alt=&quot;Structure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to my lack of knowledge of ANSYS, I have made the singular distributed loading (GI) represented as two distributed loadings (GH and HI).&#xA;Would this be an accurate model?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iFnkL.png&quot; alt=&quot;Model&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like there should be some bending in the middle (HE to EB).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I model it is a singular 240kN force on the center points I get:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WL5eo.png&quot; alt=&quot;Model with Singular force&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which of these would be more accurate?&lt;/p&gt;&#xA;" OwnerUserId="1434" LastEditorUserId="33" LastEditDate="2015-04-28T14:57:57.980" LastActivityDate="2015-04-28T14:57:57.980" Title="Is a distributed load in two parts equal to a full distributed load?" Tags="|structural-engineering|structures|modeling|stresses|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2625" PostTypeId="2" ParentId="2623" CreationDate="2015-04-27T06:27:16.510" Score="4" Body="&lt;p&gt;I would expect the modeling as a single load to be accurate.  Force per linear area is the same expressed either way.  You could look at a linear load on a single beam and just add more points of integration analytically and try it in ANSYS to see it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HE and BE segments will undergo buckling as its deformation mechanism after modest compression.  The single load would logically be larger in aggregate since it is also applied to the small area supported directly by HE, but an eyeball examination says that this will be negligible and not affect the prediction that buckling is what you watch for in HE and BE.  Are G, I, D, and F constrained in the model or free to move?  Could affect buckling strength.&lt;/p&gt;&#xA;" OwnerUserId="1305" LastEditorUserId="1305" LastEditDate="2015-04-27T11:55:52.997" LastActivityDate="2015-04-27T11:55:52.997" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2626" PostTypeId="1" AcceptedAnswerId="2638" CreationDate="2015-04-27T07:12:50.343" Score="3" ViewCount="182" Body="&lt;p&gt;Some sources say that higher pressure in hydrocyclones implies better performance it leads to a finer filtration. But why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because water is almost incompressible and the particles can be pressured and their density increased? &lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="368" LastEditDate="2015-05-05T18:31:35.447" LastActivityDate="2015-05-05T18:31:35.447" Title="Hydrocyclone effectiveness vs. absolute pressure" Tags="|mechanical-engineering|fluid-dynamics|pressure|cyclones|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2627" PostTypeId="2" ParentId="2602" CreationDate="2015-04-27T11:40:35.187" Score="2" Body="&lt;p&gt;First, do away with &quot;one for everyone&quot; as the primary ones will be left with empty stock and the marginal ones with unsold ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distribute according to your &quot;Same Percentage of All of the Orders&quot; method, rounding it down; assigning maybe 95% of the number you'd be able to assign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then dispatch the remaining units to stores that issue individual requests, per-request basis, first-come, first-served, depending on their individual demand - actual orders from the customers, not just &quot;want to have it in stock&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;necessity&quot; distribution is subject to random fluctuations; you optimize basing on predicted demand that is subject to randomness when it comes to the actual sales. By delaying distribution from the &quot;reserve cache&quot; until there's a confirmed demand you fill these gaps in demand, making sure that all stores actually run out of stock at the same moment and in the final phase they all run on the reserve.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-04-27T11:40:35.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2628" PostTypeId="2" ParentId="2623" CreationDate="2015-04-27T12:49:09.683" Score="6" Body="&lt;p&gt;Splitting a uniform load into separate pieces that are still continuous will have no effect.  This is frequently done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as your question about bending in HE and EB, there shouldn't be any bending because all of the forces are balanced. A sum of the moments at H or E will show that the moments from the beams on either side are opposite and equal. That means that the resulting moment at the joint is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your added question, a single point load is not a replacement for the distributed load. The distributed load causes bending in the horizontal members. The distributed load also distributes load to the vertical members on each side. The single concentrated load focuses all of the force in the central column.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-04-28T01:10:39.257" LastActivityDate="2015-04-28T01:10:39.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2629" PostTypeId="2" ParentId="2609" CreationDate="2015-04-27T14:04:57.487" Score="6" Body="&lt;p&gt;Harmonic coupling with the building's natural vibration frequencies, dissipation of energy in a massive building, and the variety of vibration modes that can experience coupling could result in more or less displacement and more or less velocity at lower floors or upper floors.  There's no universal relationship between either velocity or displacement between upper and lower floors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, earthquakes have two prominent types of waves, S waves and P waves.  S waves move at a 45 degree angle to the propagation and will slowly fall behind P waves to the point that the earthquake actually separates into two phases.  Having experienced this personally, it's actually easy to tell if you were at the epicenter or far away because the building motion changes and the quake has two distinct peaks.  You don't need a seismometer for a large enough quake far enough away.  The importance of the two wave types is that they can excite unique vibration modes, twisting vs bending, so there isn't even a universal answer for one building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tall building is the easiest way to visualize how the velocity and frequency at the ground and upper floors are not necessarily related.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foundation is usually relatively fixed, though some designs attempt to de-couple ground movement from structure movement through the use of mechanical bearings of various sorts where the structure meets the foundation.  At the fixed foundation, a tall building moves according to the earthquake.  Heavy though the foundation may be, the earth is still heavier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the upper floors, if the building is in harmony with the motion, it will be as if pushing a light pole at its natural frequency.  You will get it to break most likely and the motion at the top will be severe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In tall buildings, typically this frequency is much larger than the earthquake frequency, so while some energy does accumulate, very little of it is conserved because of all the destructive interference and dissipation in the large number of structural connections.  The dangerous buildings occur right around ten floors.  Shorter than this, the building wants to vibrate faster than the quake.  Taller than this, the building wants to vibrate more slowly than the quake.  At extreme heights, energy dissipation becomes more pronounced as there is just too much mechanical loss through hysteresis for 2n or higher vibration modes to have much effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's really cool video of buildings that built up energy swaying after an earthquake.  Note that the lateral movement at the top is large, but you can imagine that the velocities at the top related to the buildup are not extreme relative to the fixed foundations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=g0cz-oDfUg0&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=g0cz-oDfUg0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1305" LastActivityDate="2015-04-27T14:04:57.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2630" PostTypeId="2" ParentId="2623" CreationDate="2015-04-27T14:23:41.040" Score="3" Body="&lt;p&gt;Adding just a slight notion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your experience would always expect the beam HE to buckle, because it is a Eulerian buckling beam (don't know if it's also called like this in English).&#xA;The numeric only sees an evenly distributed load, so there will be no buckling... Try to alter your forces to the left and right of H a bit and you will get a moment which will cause HE to bend (If its support is rigid).&lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2015-04-27T14:23:41.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2632" PostTypeId="2" ParentId="2622" CreationDate="2015-04-27T15:08:38.013" Score="4" Body="&lt;p&gt;There is no general formula for isp that would provide accurate values all the propellant combinations and nozzle expansion ratios one might use. The &lt;a href=&quot;http://www.grc.nasa.gov/WWW/CEAWeb/ceaHome.htm&quot; rel=&quot;nofollow&quot;&gt;NASA Chemical Equilibrium (CEA) program&lt;/a&gt; can provide the information you are looking for and would be an answer to your question.  It is useful for any of the common propellant systems and many exotic propellant systems, e.g., fluorine,  metals, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extracted from nasa CEA webpage: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;CEA is a program which calculates chemical equilibrium product concentrations from any set of reactants and determines thermodynamic and transport properties for the product mixture. Built-in applications include calculation of theoretical rocket performance (isp at any nozzle expansion ratio), Chapman-Jouguet detonation parameters, shock tube parameters, and combustion properties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The code in fortran is freely available (check web site).  This code is the &quot;bible&quot; for the rocket industry.  I used this 50 years ago while working on the Apollo program and working on other rocket systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program is very fast. You could set CEA up to be a call from your application or you could run a bunch of cases with the propellants you are most interested in and call a curvefit to the data for the specific situations you are looking at.&lt;/p&gt;&#xA;" OwnerUserId="1435" LastEditorUserId="3120" LastEditDate="2015-09-25T15:24:39.467" LastActivityDate="2015-09-25T15:24:39.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2633" PostTypeId="1" AcceptedAnswerId="2635" CreationDate="2015-04-27T18:16:46.357" Score="10" ViewCount="1461" Body="&lt;p&gt;If a permanent room or platform is suspended from a larger structure (say, on a cable that is attached to a hook on a track in a ceiling) and designed to be in constant but not necessarily repetitive motion, does the load that the room/platform places on the larger structure count as a dead load or a live load?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, it is permanent and won't ever be removed (like a dead load,) but on the other hand, it moves (like a live load.) I'm not talking about the loads that would be incurred by its motion, only by its weight.&lt;/p&gt;&#xA;" OwnerUserId="1436" LastEditorUserId="368" LastEditDate="2015-05-05T18:24:05.450" LastActivityDate="2015-05-05T19:20:30.293" Title="Is a permanent platform that is hanging and moving a dead load or a live load?" Tags="|structural-engineering|structures|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2634" PostTypeId="1" CreationDate="2015-04-27T18:47:04.313" Score="3" ViewCount="5897" Body="&lt;p&gt;I have a formula to compute the deflection of a beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I have computed the deflection I get a vector with non-negative elements representing the deflection in each position on the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I am plotting the deflection vector I get a downwards parabola since the elements are positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I found this plot somewhere else&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uKsfA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the plot illustrates how the beam is bending at positions where loads are placed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct to get negative deflections? I wonder if I need to multiply my elements in my deflection vector by -1 in order to obtain the same plot as shown here.&lt;/p&gt;&#xA;" OwnerUserId="1437" LastActivityDate="2015-04-27T19:25:17.823" Title="Plotting deflection of a beam" Tags="|mechanical-engineering|materials|structural-engineering|beam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2635" PostTypeId="2" ParentId="2633" CreationDate="2015-04-27T19:04:02.317" Score="9" Body="&lt;p&gt;&lt;strong&gt;What is the difference between a Live or Dead Load?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Live Load is typically a moving load. This can be vehicles on a bridge or occupants and vehicles in a building. Dead loads are (relatively permanent loads).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are simple ways to differentiate the type of load, but why do we care?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic answer is, &quot;Because the &lt;em&gt;Code&lt;/em&gt; cares.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why does the Code differentiate between a Live Load and a Dead Load?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two reasons why loads are sorted into groups:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Different Load Combinations are described&lt;/li&gt;&#xA;&lt;li&gt;Different Load Factors are applied&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The load combinations are standard groupings of loads that can occur simultaneously. The load factors take into account how accurately a load can be calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most Codes that use them, Live Loads have a higher load factor than Dead Loads. This is because it is easier to calculate the Dead Loads than it is to calculate the Live Loads. Especially with vehicles, you have little control over the weight of the actual vehicle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Live Loads have more variability.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is the only reason why there is a difference between what gets placed in the two groups. The live loads have a higher factor of safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What other factors apply?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is that a moving load has the ability to create an additional Impact Load.  If this is the case in this situation, then a separate Impact Load should be included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the end result?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, it may not matter which category gets assigned to the load. The main criteria will be how confident you are in the loading that will be imposed on the structure from the moving platform. If the load isn't likely to change, then you could consider it as a Dead (Permanent) Load. If it is likely to change, then be more conservative in calculating the load or apply a higher factor to it. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All codes will allow you to be more conservative if you choose to be.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2015-05-05T19:20:30.293" LastActivityDate="2015-05-05T19:20:30.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2636" PostTypeId="2" ParentId="2634" CreationDate="2015-04-27T19:25:17.823" Score="6" Body="&lt;p&gt;&lt;strong&gt;Deflection&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A deflection is a movement in a direction. It is up to you to determine your frame of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like the calculations that you have done yourself (all positive) are shown for a force also in the positive direction. This makes sense logically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deflections shown in the graph that you attached (all negative) may be in a different coordinate system where the force is acting in the negative direction. It is up to you to determine whether you are calculating in the frame reference of the force or from the outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember that it is reasonable to get both positive or negative deflections in the same component. A beam with two supports and a cantilever could show negative deflection at the free end and positive deflection between the supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 1 - Gravity on a Beam&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SHNmH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deflection is negative from the world-view but positive from the view of the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example 2 - Cantilever&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZZdyk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deflection is opposite for the free end (right side) when compared to the portion under the arrow.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-04-27T19:25:17.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2637" PostTypeId="1" CreationDate="2015-04-27T19:31:41.083" Score="3" ViewCount="997" Body="&lt;p&gt;I am trying to figure out how to calculate either the distance or angle of 2 Idler Gears in a Timing Gear Configuration like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MH8YH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been able to eyeball things, so they are approximate. I would like to come up with an equation to get an exact number for either dimension A or B. I came up with an angle for A of approx 44.7 deg where all the teeth mesh.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to use CAD to figure out how this needs to be done. Now I just need to figure out how to calculate all of this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ltjyv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically what I ended up doing is all the angles are multiples of the tooth angle for each gear. Next I took a line at the tangent points of each gear perpendicular to a connecting line between each angle. See the blue lines in the above picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this problem the tooth angle can not be used to calculate the spacing directly because of the gear mesh as seen on the bottom of the 22 tooth gear as seen in the top picture. The meshing actually takes place off of the tooth angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to clarify I am using 64 pitch gears, the diametrical pitch is calculated using standard gear equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dp = N/P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = Number of teeth&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P = Pitch&lt;/p&gt;&#xA;" OwnerUserId="1438" LastEditorUserId="1438" LastEditDate="2015-04-29T21:13:56.023" LastActivityDate="2015-04-29T21:13:56.023" Title="How to find angle of timing gears" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2638" PostTypeId="2" ParentId="2626" CreationDate="2015-04-27T20:00:22.537" Score="1" Body="&lt;p&gt;The use of the word &lt;strong&gt;performance&lt;/strong&gt; might be misleading since a connection to &lt;em&gt;pressure loss&lt;/em&gt; is quite reasonable. However, in this context I guess &lt;strong&gt;performance&lt;/strong&gt; is connected to the filtration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using higher pressure implies in deed a higher pressure loss but it also results in a higher circumferential velocity. Since the main working principle of a hydrocyclone is to divide a solid-liquid suspensions by density a higher circumferential velocity will increase the radial acceleration. Since small particles will encounter higher drag du to their smaller Reynoldsnumber it will take longer for them to travel to the outside wall (based on &lt;a href=&quot;https://books.google.de/books?id=8dj5IJbEwu4C&amp;amp;lpg=PP2&amp;amp;ots=3jH8p-NrNE&amp;amp;lr&amp;amp;pg=PR4#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Svarovsky (1981)&lt;/a&gt;,  residence time theory ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/686U0.png&quot; alt=&quot;hydrocyc&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to have a &lt;em&gt;better&lt;/em&gt; filter a higher radial acceleration is needed which (all other things kept equal) asks for higher pressure to produce higher circumferential velocities.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2015-04-27T20:00:22.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2639" PostTypeId="2" ParentId="2619" CreationDate="2015-04-27T23:59:26.510" Score="1" Body="&lt;p&gt;The leakage current specification for power line filters is provided for applications which are sensitive to earth leakage, such as medical equipment.  It can be safely ignored in a system where all inter-equipment and out-bound interconnects are balanced and run over twisted pair (noise incoming on a system earth reference is common mode noise to a properly balanced input, and is at a low enough frequency that the DC CMRR of the input should still mostly hold).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your questions in order:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Yes -- there is no problem with this.  The leakage current of a powerline filter must be &amp;lt;5mA anyway, otherwise you'll see GFCI/RCD nuisance tripping all over the place, and &lt;em&gt;nobody&lt;/em&gt; wants to have the shame of building an ersatz GFCI tester now, do they?&lt;/li&gt;&#xA;&lt;li&gt;TL;DR -- stop worrying about the &quot;ground loop&quot;, it won't break anything in a reasonably well-designed system.  For the gory details, I highly recommend &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0470189304&quot; rel=&quot;nofollow&quot;&gt;Henry Ott's fantastic work, &lt;em&gt;Electromagnetic Compatibility Engineering&lt;/em&gt;&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Earth itself is a poor conductor.  It's the equipotential &lt;em&gt;between the filters&lt;/em&gt; that counts; this is typically accomplished with a reference plane or reference grid in the equipment area/room.  See the linked reference for more details!&lt;/li&gt;&#xA;&lt;li&gt;Unless you're in some sort of alternate universe where the Bell System never existed, your telephone lines are balanced twisted pairs -- which, when applied correctly, have excellent rejection properties at 60Hz.  So, I would not be concerned in the least about &quot;back coupling&quot; unless you hear hum; if you are hearing hum, I'd check for imbalances in the lines before I started diddling with random ground-related things.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="964" LastActivityDate="2015-04-27T23:59:26.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2640" PostTypeId="1" CreationDate="2015-04-28T03:18:48.990" Score="7" ViewCount="2065" Body="&lt;p&gt;I am making a waterproof case for my Galaxy S5 and/or my Raspberry Pi. I plan to make the main body from (or at least decorate it with) wood, though I can use other, stronger materials where needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking of using some controller (as opposed to the touch screen) while underwater if that would make it easier to avoid common points of failure. &lt;strong&gt;My question is, where do normal waterproof cases usually fail (leak)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My design criteria are, from most important to least important:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Durability / Maximum Depth&lt;/li&gt;&#xA;&lt;li&gt;Usability Underwater (mostly using the camera)&lt;/li&gt;&#xA;&lt;li&gt;Cost&lt;/li&gt;&#xA;&lt;li&gt;Aesthetics&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If I took my phone scuba diving and dove to the maximum depth of about 33 meters, there would be about 43 psi acting on the case. Would the failure point then become the case itself cracking under pressure?&lt;/p&gt;&#xA;" OwnerUserId="1440" LastEditorUserId="368" LastEditDate="2015-04-30T22:35:09.673" LastActivityDate="2016-02-10T10:28:58.587" Title="What are the common failure points of waterproof cases?" Tags="|mechanical-engineering|wood|waterproofing|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2641" PostTypeId="1" CreationDate="2015-04-28T04:19:26.107" Score="5" ViewCount="376" Body="&lt;p&gt;I've resolved a relative displacement tensor into a strain tensor and a rotation tensor, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the strain tensor is:&#xA;$$&#xA; \varepsilon_{i,j} =\begin{pmatrix} 0.2 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0.8 &amp;amp;0.4\\0&amp;amp;0.4&amp;amp;0.4\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the rotation tensor is:&#xA;$$&#xA;\omega_{i,j} = \begin{pmatrix} 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp;-0.2\\0&amp;amp;0.2&amp;amp;0\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would these conditions physically change a small cube (with respect to the Cartesian cooridates (x,y,z) ?&lt;/p&gt;&#xA;" OwnerUserId="1407" LastActivityDate="2015-11-18T12:37:36.500" Title="Physical interpretation of the relative displacement tensor?" Tags="|materials|fluid-mechanics|stresses|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2642" PostTypeId="2" ParentId="2640" CreationDate="2015-04-28T07:47:15.027" Score="2" Body="&lt;p&gt;Wood isn't a particularly waterproof material (it absorbs water &amp;amp; humidity), I would go for plastic or metal. To ensure the &quot;waterproofness&quot; of the case, I would suggest using an o-ring or similar foam gasket under compression, this is the best way to ensure it is sealed. Normally with o-rings, you would aim for a 20% compression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is the question of whether you want it to be a case as in a closed box, or still want to be able to use the device when in the case. If the latter, then that puts a lot more constraints one the design, meaning you probably need to have touch-sensitive glass, openings &amp;amp; covers for connectors, etc... which makes it a lot more complex.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-04-28T07:47:15.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2643" PostTypeId="2" ParentId="2633" CreationDate="2015-04-28T12:41:49.730" Score="4" Body="&lt;p&gt;Since you say a permanent room or platform, I'm assuming you're expecting it to be occupied and used for whatever function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hazzey does an excellent job of describing the differences between dead and live loads, so I won't go into that here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it is going to be occupied and used, I wouldn't lump the entire load into just dead or just live:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily calculate the weight of the structure, you know the materials being used and you can find out their weights. Place this under dead load, and factor accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you have an idea of what will be going on in this room/platform, but you can't give as exact a number as above. Find out the appropriate loads for your use case. Place these under live loads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add the dead and live loads and boom, you have the weight you need. Since the room will have people on it, and people are unpredictable, placing the entire object under dead load seems irresponsible to me. However, separating the loads will turn out to be more economical than factoring the entire weight as a live load.  &lt;/p&gt;&#xA;" OwnerUserId="3" LastActivityDate="2015-04-28T12:41:49.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2644" PostTypeId="2" ParentId="2637" CreationDate="2015-04-28T13:01:54.220" Score="1" Body="&lt;p&gt;You can solve this by modeling the gears as abutting circles with their cicumferences proportional to the number of teeth.  It will be easier to then work with the radii instead of circumferences directly, but the rest is just basic geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have to decide one more constraint than pictured.  There are several ways to express this degree of freedom, but one is how far apart the two middle gears are.  Or, you could specify how far apart the top and bottom gears are.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-04-28T13:01:54.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2645" PostTypeId="2" ParentId="2640" CreationDate="2015-04-28T14:22:27.593" Score="2" Body="&lt;p&gt;For a waterproof container to be useful, you have to be able to open it and put something inside.  The most likely location for failure is obviously at the points where the case opens up.  This is usually dealt with by using a rubber gasket which compresses slightly when the case is closed, creating a continuous seal between the two sides.  The most likely points of failure of the rubber gasket are at any places where it makes a sharp bend, such as the four corners for a rectangular case.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-04-28T14:22:27.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2646" PostTypeId="2" ParentId="2640" CreationDate="2015-04-28T15:44:35.690" Score="10" Body="&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;  The design of genuinely waterproof containers is an arcane art but is reasonably doable if certain rules are adhered to which codify the underlying &quot;magic&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failures occur when the basic sealing rule discussed below is violated. I.e. pressure on all sealing surfaces exceeds differential pressure at all times.  In many products the failure is inbuilt due to lack of formal design, and is just waiting for conditions where the requisite 'rule' is not met. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest and usually best method is the use of a properly designed O ring. &#xA;The typical cause of any failures is that people use squashed rubber rings that look like O-rings but which are not designed and so fail to meet the basic rule in some situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;_________________________&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Requirement for sealing:&lt;/strong&gt; To be and to remain waterproof a minimum requirement is that ALL locations along the sealing surfaces must either be bonded with an impervious material or exert a pressure (force per unit area) that is in excess of the pressure difference between inner and outer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A coating material (e.g. &quot;grease&quot;) may be used to help meet this &quot;rule&quot; but the use of a coating material in an undesigned manner is no guarantee of sealing success. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;O-rings - the usual easiest and best solution:&lt;/strong&gt; The most easily designable means of providing a seal is to use an O-ring.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An O-ring typically looks like, &lt;strong&gt;and&lt;/strong&gt; is, a circular rubber ring of round ring cross section. However many circular rubber rings of round ring cross section &lt;strong&gt;LOOK&lt;/strong&gt; like O-rings, &lt;strong&gt;but aren't&lt;/strong&gt;. I.e. to work as an O-ring certain well defined constraints need to be met and it is very likely that if a round rubber ring seal is not formally designed then it will not work in the manner that an O-ring does under all circumstances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An O-ring uses the pressure differential in such a way that the ring moves and/or distorts in a manner that produces more pressure on the sealing surfaces than the pressure differential does. I.e. the resisting pressure is caused by the target sealing pressure and is designed to always be greater. Designing a seal that works like this from first principles is a formidable task as it depends on a number of dimensions and material properties. Designing one by using standard tables is relatively straight forward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An analogy:&lt;/strong&gt; A simple analogy for O-ring operation that is incomplete but useful is to imagine a panicking crowd of people trying to get through a door that must be opened towards the crowd in order to allow people through. Increasing crowd pressure on the door simply makes it harder to open. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;O-ring design is most easily achieved by using parameters chosen from standard tables. These include groove width and depth, ring material diameter, ring loop diameter, groove inner diameter, ring wall angle, surface finish, ring and groove material a and a bit more. Simply follow standard 'cookbook' procedures and an O-ring will usually work extremely well indeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Maintaining design integrity:&lt;/strong&gt;   The Challenger 'Space Shuttle' was destroyed due to an O-ring failure in an SRB (solid rocket booster) joint. However, the ring was well designed and able to handle the task assigned to it. The failure occurred because the task changed in an uncontrolled manner. Pressure on the overall joint system allowed the mechanical separations between the joint sides to increase and to adopt an angle rather than being parallel. Hot gases were then able to escape past the O-ring, causing it to be progressively destroyed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly in a DIY case - not only do the O-ring &amp;amp; housing need to be designed to work together but the designed mechanical relationships of the joint parts need to be maintained within specification. If a case can flex under impact or force on the case such that the joint clearances or angles exceed design limits then leakage may occur. I.e. a vital part of the design is ensuring that worst case joint &quot;excursions&quot; are known and not exceeded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two main types of O-ring - radial seal and face or axial seal.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A radial seal O-ring is more common and easier to design well as sealing forces are generated by the process of sliding the two parts of the join together - any fasteners or screws etc play no part in creating the sealing force - they simply &#xA;serve to hold the pieces in correct relative proximity.  &lt;/li&gt;&#xA;&lt;li&gt;A face or axial seal O-ring system effectively &quot;squashes&quot; a ring between two plates. It has advantages in the sort of surfaces that can easily be sealed BUT relies on fasteners, screws, etc. to pull the two sealing surfaces together in order to provide adequate sealing force. While this is not hard to design it does depend on correct assembly and no failure of the compression system.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some products employ or seem to employ both types of O-ring in the one joint - two tubular parts are slid together with an O-ring between them and the outer part is screwed down onto a face seal. In designs that do this in cheap consumer equipment it seems likely that the radial O-ring does most of the work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;_____________&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Real world products:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low cost ABS housing Asian manufactured consumer products are not usually expected to have an IP66+ design sealing rating. But, it can be achieved. &#xA;I was involved with the design of the two main types of portable solar powered lights &lt;a href=&quot;http://public.fotki.com/RussellMc/atw/bogo/sl2africa01/?view=roll#17&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;seen here&lt;/strong&gt;&lt;/a&gt; - roll up from start point for large light examples, down for small light.  The larger lights were built first with input from a Chinese designer who was good at making things that looked good but resistant to certain suggestions re: sealing. The sealing worked 'well enough' when assembled correctly  but could have been far better. The small light uses a true O-ring front seal and an &quot;almost O-ring&quot; battery box seal (less critical) (and the factory manager managed a design change along the way because he 'knew better'). The all important front seal works superbly. The lights are 100% tested to 5 metre equivalent immersion depth during manufacture and can be relied on to stay dry in any use. (As they float, getting one to 5+ metres deep is usually unattractive to users :-)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The switches use a magnet operated reed switch and are 100% waterproof and extremely reliable in operation (by design). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fgmgZ.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I can and will add table references, design links and photos to this answer if it looks like people find it useful. If it receives little interest it's not worth the time (and I must do other things at this stage).&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="368" LastEditDate="2015-04-30T22:44:13.397" LastActivityDate="2015-04-30T22:44:13.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2647" PostTypeId="1" CreationDate="2015-04-28T16:43:38.197" Score="2" ViewCount="114" Body="&lt;p&gt;I have some questions regarding Electromagnetic compatibility (EMC) filtering  and power line filter leakage.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What will be the effect of leakage current of power line filter (2A)  when both power line and telephone line filters are installed on same enclosure having common ground and no RCD or GFCI device installed in the facility.  Will it create a hum in the phone lines?   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What should be the line to ground capacitance of a power line filter for a 50Hz telephone line filter?    &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What should be the line to ground capacitance of a telephone line filter?   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1430" LastEditorUserId="368" LastEditDate="2015-11-20T01:08:44.477" LastActivityDate="2015-11-20T01:08:44.477" Title="Electromagnetic compatibility (EMC) filter and power line filter leakage" Tags="|electrical-engineering|emc|electronic-filters|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2648" PostTypeId="2" ParentId="2633" CreationDate="2015-04-28T17:08:17.183" Score="2" Body="&lt;p&gt;A moving load should always be classified as a live load even when its weight is known and remains the same. A dynamic body of mass has an extra degree of freedom. This positional uncertainty calls for a higher factor of safety. Your &quot;permanent&quot; room is not much different from a &quot;permanent&quot; mobile trailer in that both are capable of movement and that the gravitational load do not vary much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A dead load, on the other hand, is permanent in both magnitude and position. I.e. it is a static load. A static load, being predictable, requires a lower factor of safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would further argue that the classification of load depends on what you are designing. For the hook itself, you may even go for a higher safety factor by classifying it under crane load by reason that failure can result in catastrophic consequences to anyone below or inside the suspended room.&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2015-04-28T17:08:17.183" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2649" PostTypeId="2" ParentId="2640" CreationDate="2015-04-28T21:44:47.367" Score="5" Body="&lt;p&gt;The major factors surrounding failure points of waterproofing are, as has been said, the ingress of water around the main case opening and any I/O areas. For the main case opening, this ingress is usually mitigated by an O-ring. Due to the deformation of the O-ring under pressure, hard, 90$^\circ$ bends usually are the primary point of failure. This is the reason why most case seals are given a fairly wide-radius (with respect to the channel itself) rounding on the corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give perspective for better seals, high-vacuum connections usually use copper for gaskets; CF flanges have a &quot;knife-edge&quot; that bites into the soft copper gasket, creating a leak-proof seal at high-vacuum, large-differential pressures. If you intend for &lt;em&gt;deep&lt;/em&gt; immersion and have access to the manufacturing needed, this may be the route to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the issue of I/O interfacing, most deep immersion systems use fully-isolated bulkhead connectors, where the cable itself does not penetrate through a hole in the case. The actual transmission, for example, would be through a female-to-female connector electrically connected and then &quot;potted&quot; using thermoplastic, epoxy, etc. into the wall of the bulkhead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the issue of &lt;em&gt;phones&lt;/em&gt; specifically, the problem here is that while any sort of touch-screen layer can be made arbitrarily ingress-resistant, the layer of touch-transmittable material is often quite thin and thus not suitable to much more than casual water immersion (i.e. low water pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, the issue of wood itself as a waterproofed case material; certain, denser hardwoods are notably less porous than others, but this can be increased by applying a layer of sealer such as painted-on epoxy or polyurethane inside the case. Still, FWIW I wouldn't be putting my phone in a wood case and throwing it in the pool.&lt;/p&gt;&#xA;" OwnerUserId="1229" LastEditorUserId="368" LastEditDate="2015-04-30T22:49:13.807" LastActivityDate="2015-04-30T22:49:13.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2650" PostTypeId="2" ParentId="2151" CreationDate="2015-04-29T06:56:12.023" Score="3" Body="&lt;p&gt;If you have an equation  $Cp=f(T)$, the problem is simple (as long as the integration does not make any problem) since $$\Delta Q=m\int_{T_i}^{T_f}Cp(T)\ \ dT$$ as Chris Mueller answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us admit that you only know $Cp(T_i)$ and $Cp(T_f)$. So, interpolate linearly to get $$Cp(T)=Cp(T_i)+\frac{Cp(T_f)-Cp(T_i)}{T_f-T_i} (T-T_i)$$ and, integrating, you will then get $$\Delta Q=m\, \frac{Cp(T_f)+Cp(T_i)}2 \,(T_f-T_i)$$ which shows that you just need to use the average value of the known $Cp$'s.&lt;/p&gt;&#xA;" OwnerUserId="1135" LastActivityDate="2015-04-29T06:56:12.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2651" PostTypeId="1" AcceptedAnswerId="2653" CreationDate="2015-04-29T08:05:42.903" Score="9" ViewCount="13172" Body="&lt;p&gt;An idea that keeps popping up in my head, that I don't know enough structural/civil engineering to know if it's good from a technical point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a piece of &lt;a href=&quot;http://en.wikipedia.org/wiki/Polyethylene&quot; rel=&quot;nofollow&quot;&gt;polyethelene&lt;/a&gt; (PE) piping, fill it with concrete, let the concrete bind. My understanding is that under compression, a concrete (or other) pillar will 'want' to shear apart in a plane 45° to the direction of the compression (assuming we don't bend our pillar). This is an outward movement. This shear force could be contained by the surrounding pipe, since PE is quite good in tension. Othe materials for the pipe might work the same, I say PE because it's quite corrosion resistant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know everyone is building concrete pillars with rebar inside, I don't claim my idea is superior. I'm not interested in understanding why everyone is building pillars the way everyone is building them, I want to understand the flaws and limitations in my idea. Some thoughts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's useful to think as pressure in the pipe, as we have that data easily available - if we use PN10, it can take 10 bar, etc., and we don't need to think about the thickness of the pipe and it's yield strength. But how would the compressive force translate into pressure on the pipe? The concrete is no liquid, so pressue will be less than compressive load / pipe area. How much? I think understanding this will tell us how much, or little, our pillar will carry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible issue is that PE is quite smooth. The pipe can't take forces via friction and in my mind that translates to point loads at the places where the concrete happens to deform most.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="368" LastEditDate="2015-05-11T21:14:52.297" LastActivityDate="2017-03-30T18:36:57.903" Title="Would a PE-pipe filled with concrete make a good pillar?" Tags="|structural-engineering|columns|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2652" PostTypeId="2" ParentId="2651" CreationDate="2015-04-29T09:32:38.627" Score="5" Body="&lt;p&gt;Compared to steel, polyethylene (PE) is a weak material. Steel is an order of magnitude stronger than PE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.maropolymeronline.com/Properties/HDPE%20Prop.asp#Tensile%20Strength%20at%20Break&quot;&gt;This site&lt;/a&gt; from the US gives a tensile yield strength for PE of 3800-4800 psi, which is 26-33 MPa and this &lt;a href=&quot;http://www.marleypipesystems.co.za/images/pdfdownloads/productbrochures/hdpe-physical-properties.pdf&quot;&gt;other site&lt;/a&gt; from South Africa gives a tensile yield strength of 26 MPa ($N/{mm}^2$ is the same as $MPa$). Whereas this site lists the tensile strengths of various steels used in making pipe as ranging from 48-90 ksi, or 311-620 MPa which significantly stronger than PE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to using PE as a mould for a pillar it needs to be determined if the PE pipe is strong enough to support the weight &amp;amp; the resultant pressure from the wet concrete just after the pipe has been filled without the pipe deforming, usually by bulging out at the bottom. Once the concrete has set, the pipe is superfluous. In some situations it could be detrimental as the pipe loosens and allows moisture to collect in the gap between the set concrete a the pipe. PE is very susceptible to deformation from heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue with PE pipes is that most are not as straight as they need to be to act as moulds for significant concrete pillars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give tensile strength to a concrete pillar, a specially designed steel cage is inserted in to the centre of the mould, for the full length of the pillar prior to the mould being filled with concrete.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-04-29T09:32:38.627" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2653" PostTypeId="2" ParentId="2651" CreationDate="2015-04-29T13:01:51.283" Score="8" Body="&lt;p&gt;Your idea could be ok, but depending on the situation it has a major flaw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PE pipe would basically be there to contain the wet concrete, while we wait for the concrete to cure (i.e. to set and strengthen). This is the situation where we need to worry about pressure from the concrete on the pipe. Once the concrete is cured, however, the concrete on its own is completely capable of supporting a lot of vertical force. We design concrete columns based on the compressive strength of concrete; we don't need to add anything else to stop the shearing failure you describe. And the horizontal force on the pipe caused by vertical force on the concrete is unlikely to be an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why do we put reinforcement inside concrete columns? The answer is that a column doesn't just take vertical load, it normally takes horizontal forces and bending moments too. And these forces create tension. Concrete is not very good in tension. So we add steel as steel &lt;em&gt;is&lt;/em&gt; good in tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we use the PE pipe to take the tension? I haven't looked into this in detail. In theory, yes. But you'd a) need thick enough PE and b) need some way to transfer stress between the PE and the concrete. As you say, PE is smooth, so without the friction to transfer the stress we're probably not going to meet point b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in conclusion, your PE pipe would work well as something to construct a concrete column which will take vertical compressive loads only. But as currently proposed it's not actually going to add to the load carrying capacity of the column.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-04-29T13:01:51.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2654" PostTypeId="1" CreationDate="2015-04-29T09:42:59.217" Score="4" ViewCount="621" Body="&lt;p&gt;Some modern cars, including models by BMW, Ford and Audi, have a system on the panel that tells the driver the next gear to use while driving, for optimal fuel economy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the vehicle decide which gear provides the best fuel economy? What data does it require to make that indication in real time?&lt;/p&gt;&#xA;" OwnerUserId="1467" OwnerDisplayName="NG_21" LastEditorUserId="368" LastEditDate="2015-04-30T21:00:52.463" LastActivityDate="2015-04-30T21:00:52.463" Title="How does the optimal-gear indicator work?" Tags="|automotive-engineering|gears|fuel-economy|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2656" PostTypeId="2" ParentId="2190" CreationDate="2015-04-29T21:26:47.297" Score="2" Body="&lt;p&gt;Took something like that apart today.  It uses cables and pulleys to support the monitor and a tension spring to oppose the monitors weight.  A screw at the other end of the spring can increase tension as required to over come the monitors weightand friction in the mechanism holds the monitor in place at the desired height.&lt;/p&gt;&#xA;" OwnerUserId="1461" LastActivityDate="2015-04-29T21:26:47.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2657" PostTypeId="2" ParentId="2654" CreationDate="2015-04-30T00:15:37.983" Score="5" Body="&lt;p&gt;In a modern car with electronic engine control, fuel injection, etc, the amount of engine power is always dependent on what the ECU is doing in response to the driver's control inputs like throttle position, brake pressure, etc. If the ECU also knows what gear you are in, it is straightforward to calculate whether you could produce the same amount of power more efficiently (i.e. with lower fuel burn) in a different gear. Automatic transmission cars have been doing that &quot;calculation&quot; for decades (originally with a simple mechanical &quot;computer&quot; instead of electronics), to decide when to shift gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big thing that is missing from the systems I'm seen is the lack of situational awareness of what is going on outside the car, which is a severe limitation on the usefulness of the systems. You probably already know that following a line of traffic at a steady 30 mph on a level road is more fuel-efficient in 5th gear than in 2nd or 3rd, but what you can see through the windows might be telling you that 2nd or 3rd is a much better choice in the real world situation.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-04-30T00:15:37.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2658" PostTypeId="1" AcceptedAnswerId="2660" CreationDate="2015-04-30T00:15:41.107" Score="6" ViewCount="4027" Body="&lt;p&gt;Here is the link to the current CA PE Exam application: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bpelsg.ca.gov/pubs/forms/ceapp.pdf&quot;&gt;http://www.bpelsg.ca.gov/pubs/forms/ceapp.pdf&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page 2, which is the list of engagements/references, states: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For each engagement claimed as qualifying experience, list the name of the person who will serve as your reference. Individuals serving as references must be licensed as Professional Engineers in California in the discipline of licensure for which you are applying or legally exempt from licensure. Individuals serving as references must have been appropriately licensed, or exempt from licensure, during the period of the engagement. YOU MUST LIST A MINIMUM OF FOUR REFERENCES. At least one of the applicants licensed references must be from someone who is or was in a supervisory capacity over the applicant, for each engagement for which the applicant desires credit. References must be from individual legally authorized to practice civil engineering in the state or country where the projects are located.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This statement is very confusing. First it says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Individuals serving as references must be licensed as Professional Engineers in California in the discipline of licensure for which you are applying or legally exempt from licensure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then later on it says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;References must be from individual legally authorized to practice civil engineering in the state or country where the projects are located.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, which is it: do the references have to be CA PE's, or not? Can someone tell me for sure? &lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2015-06-24T19:29:19.920" Title="Do references for the California PE application have to be California PE's?" Tags="|licensure|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2659" PostTypeId="2" ParentId="2437" CreationDate="2015-04-30T01:30:58.610" Score="0" Body="&lt;p&gt;The state of California has some specific guidance as to what is considered experience. (&lt;a href=&quot;http://www.bpelsg.ca.gov/applicants/faq_eng.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;California FAQ&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Qualifying  engineering  work  experience  is  that  experience  in  the  appropriate branch of engineering which has been gained while performing professional level engineering tasks under the direction of a person authorized to practice in the branch of engineering in which the applicant is seeking licensure.  There is no limit to the amount of such qualifying experience which will be accepted by the Board,  provided  that  the  experience  meets  the  other  requirements  indicated herein.  Applied engineering research is considered to be an engineering task, which may constitute qualifying experience.&lt;/p&gt;&#xA;&lt;p&gt;Work in management, proposal writing, contract administration, estimating, sales, and other peripheral areas, however, is presumed to contain little or no element of  qualifying  experience,  and  therefore  an  applicant  must  provide  a  detailed explanation of what portions of such work are actually qualifying and why the Board's presumption is not correct, if the applicant expects to obtain any credit for this type of work.  Such peripheral experience will then be evaluated on a partial credit basis as applicable to each applicant's particular situation.  Thus, the actual credit allowed may range from near zero to a substantial amount.&lt;/p&gt;&#xA;&lt;p&gt;Subprofessional  work  such  as  work  normally  performed  by  a  drafter  or  a technician is not qualifying.  Nor is construction inspection qualifying.  However, work as a field engineer may be qualifying.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-04-30T01:30:58.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2660" PostTypeId="2" ParentId="2658" CreationDate="2015-04-30T01:42:29.173" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;http://www.bpelsg.ca.gov/applicants/faq_eng.pdf&quot;&gt;California FAQ&lt;/a&gt; (see page 41 of the PDF) does a better job of explaining this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In general, civil engineer applicants are required to provide references &#xA;  from  licensed  civil  engineers  or  from  persons  otherwise  authorized  to practice  civil  engineering.    ...  [A]ll  civil  engineer applicants verifying California work experience must provide references from  persons  who  are  licensed  as  civil  engineers  in  California.  ... If a civil engineer applicant has had  qualifying  work  experience  outside  of  the  United  States,  the references  for  that  experience  must  be  from  persons  authorized  to practice civil engineering in accordance with the laws of the country in which  the  experience  took  place.   ... For  civil  engineering  work  experience  which  was  gained  in California or any other place where such work is required to be under the jurisdiction  of  a  licensed  civil  engineer,  at  least  one  of  the  applicants licensed  references  must  be  from  someone  who  is  or  was  in  a supervisory capacity over the applicant, for each engagement for which the applicant desires credit. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also on the instruction form for the &lt;a href=&quot;http://www.bpelsg.ca.gov/applicants/civilpe_errf_instructions_2010.pdf&quot;&gt;Reference Form&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For work in California, references must be California Civil &#xA;  Engineers.  For work out of state/country, references must be licensed in that state/country. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-04-30T01:42:29.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2661" PostTypeId="1" CreationDate="2015-04-30T04:26:02.057" Score="2" ViewCount="157" Body="&lt;p&gt;For one of our engineering reports we had to make a model car out of recyclable materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One section of the report is we have to discuss some aspects of Physics that will be taken into consideration when making the model car. The model car cannot be powered by electricity or battery and it has to travel at least one metre. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only think of inertia(diameter of wheels) and torque that can affect the speed and distance of the car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other aspects of Physics that can affect the model car?&lt;/p&gt;&#xA;" OwnerUserId="1466" LastEditorUserId="1466" LastEditDate="2015-04-30T04:34:39.067" LastActivityDate="2015-04-30T20:59:50.107" Title="Help understanding performance of a Model Car" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2662" PostTypeId="2" ParentId="2661" CreationDate="2015-04-30T06:10:59.770" Score="2" Body="&lt;p&gt;The mass of the car will affect its acceleration, momentum (kinetic energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friction of the wheels against the ground and friction of the wheel bearings and/or axle bearing will impact both speed and distance, depending on whether the wheels are fixed to the axle and the axle rotates or the axle is fixed and the wheels rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power of your motive force will be very important.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-04-30T06:10:59.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2663" PostTypeId="1" AcceptedAnswerId="2665" CreationDate="2015-04-30T06:57:52.027" Score="7" ViewCount="2109" Body="&lt;p&gt;Why is a Non-causal PD Controller:&#xA;$C(s) = K_ds + K_p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a causal PD Controller is:&#xA;$C(s) = (K_ds + K_p)/(\tau s + 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference being the denominator. $\tau$ is the time constant and is &quot;&lt;em&gt;sufficiently small and does not perturb the other poles substantially&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1423" LastEditorUserId="33" LastEditDate="2015-04-30T12:41:44.937" LastActivityDate="2015-08-23T16:35:36.143" Title="Causal and Non-Causal PD Controller" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2664" PostTypeId="1" AcceptedAnswerId="2667" CreationDate="2015-04-30T10:44:18.687" Score="11" ViewCount="1269" Body="&lt;p&gt;I am currently doing a project which has led me to my first encounter with IR cameras, and I am therefore quite curious as to how they operate. Specifically, I would like to know the following&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How is heat converted to an electrical signal (current or voltage)?&lt;/li&gt;&#xA;&lt;li&gt;How is the &lt;em&gt;spectral bandwidth&lt;/em&gt; of an IR camera so well defined?&lt;/li&gt;&#xA;&lt;li&gt;Why are IR cameras so much more expensive than color video cameras? (Color cameras have IR suppressors, right?)&lt;/li&gt;&#xA;&lt;li&gt;How are 'regular' IR cameras different from radiometric cameras?&lt;/li&gt;&#xA;&lt;li&gt;What is the difference between IR cameras which can detect temperatures up to, say 1000 °C, as compared to IR cameras which can detect temperatures up to 400 °C?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="859" LastEditorUserId="368" LastEditDate="2015-05-05T18:09:06.313" LastActivityDate="2015-05-05T18:09:57.027" Title="How does a microbolometer (IR camera) work?" Tags="|electrical-engineering|thermodynamics|heat-transfer|optics|temperature|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2665" PostTypeId="2" ParentId="2663" CreationDate="2015-04-30T11:15:26.793" Score="7" Body="&lt;p&gt;Because with the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C(s) = K_ds + K_p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you have a non-proper transfer function. In process engineering, you always need to have &lt;a href=&quot;http://en.wikipedia.org/wiki/Proper_transfer_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;proper transfer functions&lt;/a&gt;, where the degree of the numerator is less than or equal to the degree of the denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A work around is to introduce an &quot;approximate derivative&quot;, which is the second form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C(s) = (K_ds + K_p)/(\tau s + 1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $\tau$ becomes too large, it changes the dynamics of the transfer function significantly, and therefore the overall closed-loop response. This also helps reduce the amount of noise generated by the derivative term (acts as a low-pass filter, see &lt;a href=&quot;http://en.wikipedia.org/wiki/PID_controller#Noise_in_derivative&quot; rel=&quot;nofollow noreferrer&quot;&gt;PID controllers&lt;/a&gt; on Wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, however, derivative gains are rarely used, PI controllers being the preferred implementation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The causality aspect is because in the non-causal PD controller, you have more zeroes than poles, see &lt;a href=&quot;https://dsp.stackexchange.com/questions/10204/why-more-poles-than-zeroes&quot;&gt;this discussion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="-1" LastEditDate="2017-04-13T12:47:29.797" LastActivityDate="2015-08-23T16:35:36.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2666" PostTypeId="2" ParentId="2610" CreationDate="2015-04-30T11:23:05.087" Score="2" Body="&lt;p&gt;Theoretically, it should be perfectly possible. However, there are some practical considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extraction of components: I've never taken apart a DVD drive but there are lots of  videos and guides on the internet saying it is possible. They can provide better information on how to do that bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laser Safety: DVD-drives have relatively powerful lasers, particularly for DVD-writer/RW. A quick internet search says they are &lt;strong&gt;Class 3 lasers&lt;/strong&gt; (They are normally safe as you cannot access them). &lt;strong&gt;This could easily blind you. I strongly recommend you look up and follow appropriate laser safety procedures&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General set-up: The simplest approach would be to make a simple Michelson interferometer. Using one mirror as the reference arm and attaching another to the surface you are measuring. You will probably want a photo-diode or similar to measure the output signal and possibly a beam expander, at least for visualization purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance measurement: The simplest way to determine distance from an interferometer is via fringe counting. This requires that the distance changes in a continuous fashion. If you try and measure a step or other discontinuity you will loose the count. The response time of your photo-diode may also place a limit on the speed at which you can change the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coherence: The biggest problem I can see is the coherence length of the laser. Laser diodes have relatively short coherence length. Therefore your signal quality will degrade as the distance between reference and object arm changes. Therefore, try and make sure you reference arm is reasonably well matched with the object arm length initially. In any case, this will likely restrict the possible measurement range to something on the order of a few cm.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-04-30T11:23:05.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2667" PostTypeId="2" ParentId="2664" CreationDate="2015-04-30T11:59:24.080" Score="12" Body="&lt;p&gt;There are a lot of questions in your question, and it should probably be broken up into multiple different questions.  I don't want to wait until that happens though so I'll address the ones that I know the answers to.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;How is heat converted to an electrical signal (current or voltage)?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Microbolometer&quot; rel=&quot;noreferrer&quot;&gt;microbolometer&lt;/a&gt; is just a special case of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Bolometer&quot; rel=&quot;noreferrer&quot;&gt;bolometer&lt;/a&gt; which contains a material whose resistance is very sensitive to its temperature.  The change in resistance caused by heating from incident electromagnetic (EM) radiation is read out by a circuit similar to what you will find in a voltmeter.  These devices can be designed to be sensitive to incredibly low amounts of power, and generally also have a high dynamic range.  The ones I've used in the laser industry are sensitive all the way from 10 mW to 100 W, a dynamic range of 10&lt;sup&gt;4&lt;/sup&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;How is the spectral bandwidth of a camera so well defined? &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Bolometers are known for having incredibly broad spectral bandwidths.  Since the device actually measures the heat deposited by the EM radiation, the bandwidth of the detection material itself (usually either &lt;a href=&quot;http://en.wikipedia.org/wiki/Amorphous_silicon&quot; rel=&quot;noreferrer&quot;&gt;amorphous silicon&lt;/a&gt; or &lt;a href=&quot;http://en.wikipedia.org/wiki/Vanadium_oxide&quot; rel=&quot;noreferrer&quot;&gt;vanadium oxide&lt;/a&gt;) is defined by the wavelengths at which it absorbs.  The bandwidth of the microbolometer detectors must therefore be defined with external optics which reject or absorb the other wavelengths.  My guess would be that they use an absorbing &lt;a href=&quot;http://www.edmundoptics.com/optics/optical-filters/bandpass-filters/infrared-ir-bandpass-filters/3387/&quot; rel=&quot;noreferrer&quot;&gt;IR bandpass filter&lt;/a&gt; in front of the detector surface.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;Why are IR cameras so much more expensive than color video cameras? (Color cameras have IR suppressors, right?)&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know exactly, but the ability to manufacture these things &lt;em&gt;en masse&lt;/em&gt; only became possible in the last few years while &lt;a href=&quot;http://en.wikipedia.org/wiki/Charge-coupled_device&quot; rel=&quot;noreferrer&quot;&gt;charge-coupled device (CCD)&lt;/a&gt; detectors have been in mass production since the 1980's.  You are correct that CCD detectors incorporate an IR filter, but the underlying materials are only sensitive down to ~1-2 &amp;mu;m so they don't work in the deep IR like the microbolometers do.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;5&quot;&gt;&#xA;  &lt;li&gt;What is the difference between IR cameras which can detect temperatures up to, say 1000 °C, as compared to IR cameras which can detect temperatures up to 400 °C?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;All warm bodies emit radiation across the full range of the EM spectrum.  The spectral content of the radiation emitted at a given wavelength is given very nearly by &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/bbrc.html&quot; rel=&quot;noreferrer&quot;&gt;Plank's blackbody curve&lt;/a&gt; (shown below).  One of the key features of this curve is that the peak of the emitted radiation shifts towards longer wavelengths with lower temperatures.  The peak of the emitted radiation is given by &lt;a href=&quot;http://en.wikipedia.org/wiki/Wien%27s_displacement_law&quot; rel=&quot;noreferrer&quot;&gt;Wien's law&lt;/a&gt; which is&#xA;$$&#xA;\lambda_{max}=\frac{b}{T}&#xA;$$&#xA;where $b$ is Wien's constant ($b=2.8977721\cdot10^{3}\:\mathrm{m*K}$) and the temperature $T$ is in units of Kelvin.  From this you can calculate that the peak wavelengths of the temperatures you ask about are: $\lambda_{1000}=2.3\ \mu \text{m}$ and $\lambda_{400}=4.3\ \mu\text{m}$.  So, detectors which are designed to be sensitive to those different temperatures are simply tuned (probably by tuning the bandpass filter in front) to be more sensitive at the different peak wavelengths.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:Wiens_law.svg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VKNeq.png&quot; alt=&quot;Blackbody curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="368" LastEditDate="2015-05-05T18:09:57.027" LastActivityDate="2015-05-05T18:09:57.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2669" PostTypeId="2" ParentId="2621" CreationDate="2015-04-30T14:26:06.017" Score="3" Body="&lt;p&gt;I am attacking this problem from the point of view of the flow of the gas itself through the chimney. So what I am looking at is whether using two vents vs one increases speeds of the flow to the point where it is inherently turbulent as determined by the critical Reynolds number. To do this I make some assumptions and base calculations. The sources I used for most of this are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.woodheat.org/how-chimneys-work.html&quot; rel=&quot;nofollow&quot;&gt;How Chimneys Work &lt;/a&gt;: General info for my education!&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.diva-portal.org/smash/get/diva2:523302/FULLTEXT01.pdf&quot; rel=&quot;nofollow&quot;&gt;Characterization of residential chimney conditions for flue gas flow measurements&lt;/a&gt;: Typical data.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ie00082a034&quot; rel=&quot;nofollow&quot;&gt;Critical Reynolds number for Newtonian flow in rectangular ducts&lt;/a&gt;: Calculation of Reynolds number.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nfpa.org/research/fire-protection-research-foundation/reports-and-proceedings/other-research-topics/impact-of-chimney-top-appurtenances-on-flue-gas-flow&quot; rel=&quot;nofollow&quot;&gt;Impact of Chimney-top Appurtenances on Flue Gas Flow&lt;/a&gt;: Correlation for flow rate of flue gas.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.rumford.com/index.htm&quot; rel=&quot;nofollow&quot;&gt;Rumford Fireplaces&lt;/a&gt;: Information about Rumford fireplaces.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The first thing I did is calculate the critical Reynolds number for the flow on the two different ducts. I assume the ducts where $11.3\times 11.3$ inches for the &lt;em&gt;square&lt;/em&gt; one and $17.2\times 11.2$ inches for the &lt;em&gt;rectangular&lt;/em&gt; one. The Reynolds number for a rectangular duct is defined as:&#xA;$$&#xA;Re=\frac{D_{h}\rho&amp;lt;v&amp;gt;}{\mu}&#xA;$$&#xA;where $D_h$ is the hydraulic diameter, $\rho$ is density, $\mu$ is viscosity, and $&amp;lt;v&amp;gt;$ is the average velocity of the flue gas. The hydraulic diameters are 11.2 and 13.6 in for the square and rectangular vents, respectively. Using Ref. 3 above the critical Reynolds numbers for laminar/turbulent transition are 1688 and 1548 for square and rectangular ducts, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assuming that I have the flow rate that yields the critical Reynolds number for the &lt;em&gt;rectangular&lt;/em&gt; duct, that the flow rate remains the same when moving to the two &lt;em&gt;square&lt;/em&gt; ducts, and that density and viscosity of the flue gas do not vary between the two choices, the effect of moving to the two &lt;em&gt;square&lt;/em&gt; vents results in obtaining a Reynolds number of 966, which is lower than the critical value for that piping system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This calculation suggests that all things remaining equal, it is unlikely that the flow will become turbulent because of switching to two smaller vents. Instead, switching to the two &lt;em&gt;square&lt;/em&gt; ducts should decrease the velocity of flow and therefore the Reynolds number. However, this does not take into account any other effects like the induced disturbances due to the wall in between the two ducts and at the end of the chimney. Although you seem to have taken precautions to deal with those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I did not perform calculations to determine whether the flow rate will actually stay the same. I noted that in some of the references above it is explained that the flow rate will depend on the area of the vents among other factors.&lt;/p&gt;&#xA;" OwnerUserId="1031" LastEditorUserId="1031" LastEditDate="2015-04-30T14:36:27.540" LastActivityDate="2015-04-30T14:36:27.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2670" PostTypeId="1" CreationDate="2015-04-30T14:49:13.997" Score="8" ViewCount="608" Body="&lt;p&gt;I am looking to use FEA (specifically, Creo Simulate) to re-create the results of a static strength test on a part my company will be testing in our lab. Because the lab results are the ones we will be using to determine the utility and capability of the part, I'm doing this mostly as an exercise, to strengthen my FEA skills and to see how close to reality I can get the results of the analysis to be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a couple different reasons (write access to model files, investigation of the bolt functionality of Creo, more realistic application of load) I am attempting to model part of the test stand in addition to the part of interest. The part itself has a flange with a bolt pattern as part of its design, and we are using that bolt pattern to attach it to an adapter. The adapter will attach to the output of the motor, with the torque being applied through a key that sits between the adapter and the motor output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below I have a picture of the keyway in the adapter for clarity. The part of interest is attached to the back side of this adapter, the key fits along the entire length of the keyway, and the adapter connects to the test stand output using the four larger bolt holes (these don't transmit any significant torque, they simply keep the assembly together.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BBzzAm.png&quot; alt=&quot;Flange with keyway&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My issue is determining how best to model the applied load on the walls of the keyway. My intuition tells me that the load will vary along the length of the keyway, as what we're really applying is a torque that gets distributed throughout the key. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Given that $T=r \times F$, I'd expect there to be an inverse relationship between the force I should apply and the position along the keyway at which I apply it. Does the fact that this line of action is not radial complicate things? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In addition, I have two coordinate systems defined, the standard Cartesian, and a cylindrical one with the z-axis going through the center hole of the adapter as you might expect. Will applying the load in one coordinate system or the other significantly alter the result, and is one preferable? Will the distributed force equation be easier to define in one coordinate system?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-04-30T21:23:29.047" Title="How do I best model a torque applied over a non-radial line?" Tags="|mechanical-engineering|finite-element-method|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2671" PostTypeId="2" ParentId="2661" CreationDate="2015-04-30T20:59:50.107" Score="1" Body="&lt;p&gt;The second paragraph seems to be asking about some physics-based appoach for powering the car, and driving it for 1 meter (at least); and asking for some other aspects of the model-car design that is based on physics-knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part can be the conceptual design of the car. And the second part can be more related to the detail design of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conceptually, the car can be driven by the following methods, where energy is not coming from batteries or electricity.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mechanical power: gears and springs that can be found in toy-cars.&lt;/li&gt;&#xA;&lt;li&gt;inertia release: based on Newton's laws, something (fluid, or balls) can be released towards the rear of the car, to move the car forward. You could design a simple mechanism (e.g. 1 Liter of pet bottle with a hole at the rear-bottom-corner, throwing the liquid towards the back of the car). or some other mechanism, that releases a few balls by the help of gravity towards the back. Gravity can be used to pull the balls downwards, and some sloped-road can be built to guide them towards the back.&lt;/li&gt;&#xA;&lt;li&gt;chemical reaction: soda + Mentos, kind of rocket science, but the force is good. although not much computable beforehand and more of a quick-explosion rather than a slow and adjustable thrust.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;After the conceptual design, other physical knowledge comes into play. But that depends on the conceptual design, and there are possibly tens or hundreds of physical relationships and dependencies between the selected design features, design characteristics, materials and the overall performance of the vehicle. Most importantly, weight of the car and the main friction (tire-ground, or inter-mechanism friction) will play a huge role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you edit your question to give more details, and maybe with some sketches of your possible design, we could elaborate more on this interesting and thought provoking project.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2015-04-30T20:59:50.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2672" PostTypeId="2" ParentId="2670" CreationDate="2015-04-30T21:23:29.047" Score="4" Body="&lt;p&gt;I would ignore the fact that the keyway makes the actual forces slightly off-center at least when calculating the forces themselves. The forces can be placed at their true locations once they are calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also assume that the load is continuously distributed from the center to the outer edge. This assumes that the key is machined with a tight enough tolerance (or deflects/deforms enough) to equalize the distribution. This also assumes that a given amount of rotation produces more force at the extremes of the key than at the center (where it is zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, it is a matter of finding the maximum force per unit length at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gUq0Qm.png&quot; alt=&quot;Simplified Forces&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P = \frac{1}{2}\frac{T}{\frac{2}{3}R} \\ &#xA;and \\&#xA;P = \frac{1}{2}R *w_\text{max}\\&#xA;gives:\\&#xA;w_\text{max} = \frac{T}{\frac{2}{3}R^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I have gone through this, I have the nagging feeling that I am missing the point of your question, so let me know if I missed it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that I am used to working with models where the finer details of the exact load placement does not affect the final results, so I may be making assumptions that are not correct for your model.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-04-30T21:23:29.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2673" PostTypeId="1" AcceptedAnswerId="6391" CreationDate="2015-04-30T22:50:14.330" Score="10" ViewCount="734" Body="&lt;p&gt;I'm trying to design a bioreactor to produce citric acid with whey and Aspergillus Niger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step in the process would be to put whey into the reactor with some dextrose (approx. 10%). Then, this solution should be sterilized, and I was thinking of using batch steam sterilization by direct injection of the steam in the reactor.  The sterilization should be done at 121°C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to calculate the sterilization time and the amount of steam required for a given reactor volume (approx. 600 m&lt;sup&gt;3&lt;/sup&gt;).  I was thinking of calculating it as if I had a 600 m&lt;sup&gt;3&lt;/sup&gt; autoclave, but I doubt the approximation holds. In that case, the death kinetics would be given by the &lt;a href=&quot;http://www.rpi.edu/dept/chem-eng/Biotech-Environ/Projects00/sterilize/batch.html&quot;&gt;Arrhenius equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA; k = Ae^{\frac{-E}{RT}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd need to determine the activation energy, $E$, the pre-exponential factor $A$, or some approximation for $k$. Then, I'd just have to integrate the equations for the heat and hold energy required:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\nabla heat &amp;amp;= ln(\frac{N_0 V_0}{N_1 V_1}) = \int_{t_0}^{t_1} k(t) dt \\&#xA;\nabla hold &amp;amp;= ln(\frac{N_1 V_1}{N_2 V_2}) = k(t2_-t_1) dt&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out $t_1$ and $t_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I approaching this problem correctly?&lt;/p&gt;&#xA;" OwnerUserId="1475" LastEditorUserId="41" LastEditDate="2015-05-01T16:52:29.990" LastActivityDate="2015-12-02T16:23:28.730" Title="Steam sterilization of whey in a bioreactor" Tags="|chemical-engineering|process-engineering|steam|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2674" PostTypeId="1" CreationDate="2015-05-01T17:12:00.280" Score="3" ViewCount="1286" Body="&lt;p&gt;I am designing a 4'x12' trailer frame.  8' of it will support weight, the balance will be the tongue.  My area of concern is bending at the ends of the long supports.  There will be a wheel assembly located 4' from the frames end that will support all but the tongue weight of 100 -120 lbs.  I plan to use commercial steel tubing so the moment of inertia ($I$) and Section Modulus ($Z$) are known quantities.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need is a formula that is suited to this design - ie. something that will tell me how much the frame might bend under load.  I have some knowledge of simple design as well as an AISC and Machinery's handbooks  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Its going to be a 4' x 8' with a 4' tongue, the tongue weight should be 50 to 80 lbs. I'm designing it to keep the weight below 900 lbs.  I want it to have 15&quot; wheels.   I'm using components for a design weight of 2000 lbs. and plan on using axle-less wheel hubs.  There will be a close to a uniform load spaced over the wheels.  The wheel assembly will be off center to improve tracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a suitable &quot;Beam Diagram&quot; to ensure I stay within a reasonable deflection.   I have seen them designed with 2&quot; x 2&quot;, and 2&quot; x 3&quot; tubing.&lt;/p&gt;&#xA;" OwnerUserId="1486" OwnerDisplayName="Ernest Filomarino" LastEditorDisplayName="user16" LastEditDate="2015-05-03T12:26:05.850" LastActivityDate="2017-04-09T07:41:16.170" Title="Calculating deflection for a home built trailer" Tags="|mechanical-engineering|steel|design|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2676" PostTypeId="1" AcceptedAnswerId="3763" CreationDate="2015-05-02T09:20:04.410" Score="4" ViewCount="195" Body="&lt;p&gt;I'm trying to develop a thermoelectric (Peltier tile-based) beverage cooling system. Ideally, I'd also like a device to solve the warm beer problem if the thermodynamics give me any confidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I'll detail my thought flow in the theory sense and then I'll get to my specific setup that I had in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if one immerses a 100 W, 10% efficient Peltier tile in a glass of water at 298 K (let's say, containing 10 mol exactly), if the container system is assumed to be adiabatic in nature, to drop the temperature to 274 K (roughly 35 °F) requires:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Q = m \, c_p \, \Delta T = 180.2 \times 4.1813 \times 24 = 18083.3 \:\mathrm{J}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, since the efficiency gives the nominal &quot;heat transfer&quot; power to be 10 W:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;t = \frac{18083.3\:\mathrm{J}}{10\:\mathrm{J/s}} = 1808.33 \:\mathrm{s}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the cooling would take about 30 minutes to chill the glass down, much more when you consider that a glass of water doesn't adhere closely to adiabatic behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual setup (poorly-drawn hand sketch follows):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FNid4.jpg&quot; alt=&quot;The experimental setup&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, obviously there are four sets of values to consider here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The specific heat and thermal conductivity of the beverage itself&lt;/li&gt;&#xA;&lt;li&gt;The specific heat and thermal conductivity of the glass container holding the beverage, easily &lt;em&gt;approximated&lt;/em&gt; by making the assumption that the glass can be characterized by fused silica&lt;/li&gt;&#xA;&lt;li&gt;The specific heat and thermal conductivity of the water transfer fluid, well-established&lt;/li&gt;&#xA;&lt;li&gt;The specific heat and thermal conductivity of the 6061 aluminum working chamber, which is again well-established&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My problem is how to model the system and get model calculations for the transfers between each step, so, for example, solving the equation so I know if I use a given cumulative wattage of Peltier tile, it'll reduce the temperature of the system by a given amount in a certain amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As before, I'm willing to concede, for the purposes of discussion, that this system can be assumed to be adiabatic with respect to the external environment; that is, no heat is absorbed from the environment during the cooling process.&lt;/p&gt;&#xA;" OwnerUserId="1229" LastEditorUserId="1293" LastEditDate="2015-07-31T19:52:53.027" LastActivityDate="2015-07-31T19:52:53.027" Title="Determining behavior of heat exchange system" Tags="|mechanical-engineering|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2677" PostTypeId="1" AcceptedAnswerId="2678" CreationDate="2015-05-02T10:03:48.727" Score="5" ViewCount="471" Body="&lt;p&gt;I have 3-axis accelerometer and 3-axis gyroscope sensor data coming from my bike trip. Until now I have extracted only basic information like overspeeding, hard acceleration and sudden turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What additional motion characteristics can I infer from this data? I also have x, y coordinates.&lt;/p&gt;&#xA;" OwnerUserId="1467" LastEditorUserId="368" LastEditDate="2015-05-05T18:02:51.197" LastActivityDate="2015-05-05T18:05:11.463" Title="What kind of motion information can be extracted from 3-axis accelerometer and 3-axis gyroscope sensor data" Tags="|mechanical-engineering|electrical-engineering|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2678" PostTypeId="2" ParentId="2677" CreationDate="2015-05-02T13:17:32.553" Score="7" Body="&lt;p&gt;This information you have, in theory, allows complete reconstruction of location and orientation of the bike along the entire path.  This is basically what is referred to as &lt;i&gt;inertial navigation&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the problem in real life is that there will be some error on all the signals.  Since determining location and orientation require integrating the signals you have, these errors accumulate over time.  Note that position is the second integral of acceleration.  After some time, the resulting location and orientation data will be so far off as to be useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small and cheap sensors like you can put on a bicycles, the integrals may only be useful for a few seconds, tens of seconds at most.  I once worked on a device that tracked head motion during a golf swing, and it likewise had 6-axis acceleration and rotation rate data.  The useful integration time window was about 2 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to really answer the question, you can only use this kind of data for characterizing short term conditions or events.  If used in conjunction with something that is long-term stable, like GPS, then you can use it to fill in the high frequency information between GPS readings.  However, for an ordinary bike ride, the extra position accuracy is of little use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this data on its own for its high frequency information, like smoothness/roughness of the road, number of larger pothole incidents, how wobbly/steady the bike is, how much it sways side to side when you are pedaling, when and how hard you braked, etc.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2015-05-05T18:05:11.463" LastActivityDate="2015-05-05T18:05:11.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2680" PostTypeId="1" AcceptedAnswerId="2711" CreationDate="2015-05-02T13:48:56.413" Score="6" ViewCount="575" Body="&lt;p&gt;I need to connect two flat bars of metal with some kind of bearing that doesn't increase the joint thickness by more than 2-3mm (the less the better), as shown here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9nH2w.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9nH2wm.png&quot; alt=&quot;two bars of 50x5 mm cross-section fastened near their ends on a 30- to 180-degree swivel with axis of rotation parallel to the 5 mm dimension&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are almost no axial loads but the peak lateral (shearing) stress on the bearing would be of order of 2000 N with 800 N sustained being the norm and the directions of the stress varies (although axial respective to the lower beam would be most common and strongest). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a serious restriction on thickness of the joint - the present 10 mm is already pushing the envelope; another 2-4 mm would be acceptable though unwelcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bars will be steel or duraluminum. There are extra braces, etc., which prevent buckling and limit stress in the perpendicular direction to negligible levels, but the two bars need to carry axial stress and lateral stress &quot;in the thick direction&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moderate friction is acceptable - 5 N·m friction torque is still a reasonable value. Rotation is moderately fast, at most ~1 rev/s, usually much less. Also the range of rotation is limited; only 30-180 degrees angle between the two bars is needed (and even a little less than that if necessary). Lifetime doesn't need to exceed 1000 hours; less for the prototype.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will be prototyped using a common DYI workshop without specialized machinery like CNC devices, also preferably no very expensive solutions - of course purchase of a pre-made bearing fitting the criteria would be perfectly acceptable if the price is not excessive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of bearing or joint would fulfill these prerequisites?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Let me add a picture explaining the general layout of forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BoZKh.png&quot; alt=&quot;drawing explaining layout of the forces&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general situation any common bolt, bushing (with flanged axis) etc. would be sufficient. Here the problem is getting it all to be narrow enough not to extend way out of the narrow plane occupied by the two bars, and not fall apart.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2015-05-05T08:09:34.153" LastActivityDate="2015-05-06T20:53:50.687" Title="How to design a narrow bearing that supports shear direction load" Tags="|mechanical-engineering|bearings|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2683" PostTypeId="1" CreationDate="2015-05-03T12:01:17.840" Score="4" ViewCount="3037" Body="&lt;p&gt;How does Force Effect &lt;a href=&quot;https://forceeffect.autodesk.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://forceeffect.autodesk.com&lt;/a&gt; do it?  Given a 10m long bar angled as the hypotenuse of a 3, 4, 5 triangle with an xy pin constraint a (0,0), another xy constraint at (4,3), and a vertical force of 100N applied down at (10,6) Force Effect calculates the x and y reaction forces below at C and A. I am not as interested in this particular answer in how to get to it by a codeable algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BNcS8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the equations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum Moments about C = 0 =&gt;  Y$_{ca}$(R$_{ax}$)  + X$_{ca}$(R$_{ay}$) +  X$_{cb}$(F$_{1}$) = 0 =&gt;  3 R$_{ax}$ - 4 R$_{ay}$ = -4(-100)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum Moments about A = 0 =&gt; Y$_{ac}$(R$_{cx}$) + X$_{ac}$(R$_{cy}$) + X$_{ab}$(F$_{1}$) = 0 =&gt;  - 3 R$_{cx}$ + 4 R$_{cy}$ = -8(-100)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum Forces Y = 0 =&gt; R$_{ay}$ + R$_{cy}$ + F$_{1}$ = 0 =&gt; R$_{ay}$ + R$_{cy}$ = -(-100)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum Forces X = 0 =&gt; R$_{ax}$ + R$_{cx}$ = 0 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which could translate to a 4x4 matrix:  Ax = b&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;|  3.   -4.    0.    0. |   | Rax |  =  | 400 |&#xA;|  0.    0.   -3.    4. |   | Ray |  =  | 800 |&#xA;|  0.    1.    0.    1. |   | Rcx |  =  | 100 |&#xA;|  1.    0.    1.    0. |   | Rcy |  =  |   0 |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And attempting to solve for x = inv(A)* b gives the following error in SciLab (or MatLab)&#xA;--&gt;inv(A)&#xA;       !--error 19 &#xA;Problem is singular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the matrix is not invertible- whatever that means!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’d be open to some sort of substitution method to solve it if that is any easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not homework.  This is a simplified version of a real-world problem of a large free body being held in place by two fixed pins.  It is not dynamic – nothing is moving.&lt;/p&gt;&#xA;" OwnerUserId="1487" LastEditorUserId="1487" LastEditDate="2015-05-03T21:27:40.680" LastActivityDate="2015-07-15T15:10:28.767" Title="How to solve an overconstrained static free body diagram?" Tags="|mechanical-engineering|software|statics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2684" PostTypeId="1" CreationDate="2015-05-03T15:41:28.040" Score="6" ViewCount="373" Body="&lt;p&gt;A friend of mine that doesn't use the internet much runs a commercial laundry. They use an old 'Manlove Tullis' laundry calendar (c. 1980) that looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;strong&gt;(click to enlarge photos)&lt;/strong&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zwy2R.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zwy2Rm.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/DJINR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DJINRm.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/kH2NT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kH2NTm.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, it is a machine with 3 rollers (2 visible in the photo) on a hot plate. Garments (e.g. bedsheets) go in wet, and come out ironed flat, and dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hot bed runs on steam fed by 6 pipes (2 for each roller) underneath the machine, coming from a big boiler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of days ago 1 of the pipes broke off/burst under the machine with wear &amp;amp; tear over the years; where the pipe is joined by a coupling to the hot bed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that over the years with repeated super-heating and cooling some metal fatigue has crept in, or it became so brittle that it finally failed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After allowing the machine and the pipes to cool down for a day, he went under the machine to try to fix it; but the pipe is seized to the [possibly brass] coupling. The coupling affixes to the hot bed and it appears it cannot be removed because it is both seized and in an inaccessible part of the machine. It is a 22mm coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pipe is steel, and could be cut off and a new one joined by welding it or something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Underneath the machine, the working area is only 50cm in height at most, and it is 10ft x 15ft / 3m x 5m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From how it's been explained to me, when you're under the machine lying on the floor; the actual pipe/coupling that needs removing is within a small cavity above you. Something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kbM5z.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The space is very limited and the angle very awkward; there are also some health &amp;amp; safety issues with operating a welding torch under there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional coupling could be added, but this introduces another mode of failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has just installed a copper stop end, temporarily, so that he can use the machine with 2 out of 3 rollers operating at 100% and 1 operating at 50%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boiler operates at 8 Bar / 115 PSI &amp;mdash; therefore the laundry calendar also operates at 115 PSI. From reading online I understand that the pipework is more than adequate. It is also regularly inspected and serviced.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;My question is:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;How can we get this pipe and/or coupling off and a new one on?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The person I'm posting this question for is very knowledgeable on the subject, but for the first time I've seen, he's stuck here. Three engineers have been out on-site and don't know what to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has said that &quot;it either needs to be welded or a proper coupling added; but it cannot be cut off because it is inaccessible&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1489" LastEditorUserId="368" LastEditDate="2015-06-05T18:49:27.500" LastActivityDate="2016-12-19T06:11:29.810" Title="Pipe seized to inaccessible coupling - how to attach a new pipe?" Tags="|piping|steam|industrial-engineering|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2685" PostTypeId="2" ParentId="2683" CreationDate="2015-05-04T01:17:25.903" Score="3" Body="&lt;p&gt;your system of equations is underdetermined, which means that you don't have enough information to solve it. You can proof it by calculating the determinant. For a unique solution it has to be unequal zero.&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/System_of_linear_equations&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/System_of_linear_equations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For static problems the determination can be assessed by counting the degrees of freedoms and the degrees of constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real world problem is solvable. You need more equations, which you can use to get enough information to describe a determined problem. You need the elastic modulus of the material which your beam consists of. Everything afterwards is rather a mathematical methodology&#xA;&lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory#Statically_indeterminate_beams&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory#Statically_indeterminate_beams&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical authors in the field of mechanics like Timoshenko cover the theme in more depths in their books.&lt;/p&gt;&#xA;" OwnerUserId="1490" LastActivityDate="2015-05-04T01:17:25.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2686" PostTypeId="1" AcceptedAnswerId="2687" CreationDate="2015-05-04T04:18:36.493" Score="5" ViewCount="240" Body="&lt;p&gt;I am rebuilding a mini soil compactor roller. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I replaced the old 8 hp Kubota diesel engine unit with a new Dongfeng unit. I want to weld the coupling shaft, which is built from 2&quot; pipe with rubber damper at one end, to the hydraulic motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I ensure the shaft is balanced after we weld it to the flywheel? We would like to do this cheaply if possible.&lt;/p&gt;&#xA;" OwnerUserId="1491" LastEditorUserId="368" LastEditDate="2015-05-05T17:34:53.150" LastActivityDate="2015-05-05T17:34:53.150" Title="Balancing flywheel and coupling shaft at home" Tags="|mechanical-engineering|flywheels|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2687" PostTypeId="2" ParentId="2686" CreationDate="2015-05-04T08:23:50.450" Score="3" Body="&lt;p&gt;With some low friction bearings you can spin it on the shaft and wait for it to come to a rest and mark the low point, do this a few times and you may see them gathering at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then drill out a portion of the wheel where it is heaviest and repeat.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-05-04T08:23:50.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2688" PostTypeId="1" CreationDate="2015-05-04T11:38:04.393" Score="1" ViewCount="5598" Body="&lt;p&gt;I have following example (the question is at the bottom):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sl1U6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load distribution $q$ is constant.  I add condition of zero bend in center of the beam, $w_B = 0$, and the assumption $F_{RA} = F_{RC}$.  So I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{RA} + F_{RB} + F_{RC} = q\cdot 2 L$$&#xA;$$F_{RB} = 2 q L-2 F_{RA}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide beam to two sections (from left)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I. $x_1\in\{ 0;L\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M(x_1) = F_{RA} \cdot x_1 - q \frac{x_1^2}{2}$$&#xA;$$\frac{\partial M(x_1)}{\partial F_{RB}}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;II. $x_2 \in\{0;L\}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M(x_2) = F_{RA} (L + x_2) + F_{RB}\cdot x_2 - q \frac{(L + x_2)^2}{2}$$&#xA;$$\frac{\partial M(x_2)}{\partial F_{RB}}=x_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use condition $w_B = 0$&#xA;$$w_b = 0 = \frac{1}{E\cdot I_z} \left ( \int_0^L M(x1) \cdot \frac{\partial M(x_1)}{\partial F_{RB}} dx_1  + \int_0^L M(x2) \cdot \frac{\partial M(x_2)}{\partial F_{RB}} dx_2 \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = \int_0^L M(x2) \cdot \frac{\partial M(x_2)}{\partial F_{RB}} dx_2 = \\ \int_0^L \left ( F_{RA} (l + x_2) + F_{RB}\cdot x_2 - q \frac{(l + x_2)^2}{2} \right ) \cdot x_2 dx_2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute for $F_{RB}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_0^L \left ( F_{RA} (l + x_2) + (2 q L-2 F_{RA})\cdot x_2 - q \frac{(l + x_2)^2}{2} \right ) \cdot x_2 dx_2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for $F_{RA}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{RA} = \frac{1}{4} q l$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It's this result correct? It seems suspicious to me. I supposed $F_{RB}\geq F_{RA}.$&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1186" LastEditorUserId="1186" LastEditDate="2015-05-06T05:18:52.550" LastActivityDate="2019-12-28T18:28:58.183" Title="Beam with constant load distribution on three supports - force distribution to supports" Tags="|mechanical-engineering|structures|statics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2690" PostTypeId="1" AcceptedAnswerId="2691" CreationDate="2015-05-05T05:09:47.133" Score="12" ViewCount="4238" Body="&lt;p&gt;I have been wondering about this question for quite some time. Assuming an ideal case, the energy from photons hitting solar cells is converted into electric energy as described by the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$RI^2t=W\equiv E=\hbar\nu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\nu$ is the frequency of photons. Using a lens won't increase the frequency of photons, thus no extra electricity is generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in thinking that no extra electricity will be generated by solar cells when a lens is used to focus light onto them?&lt;/p&gt;&#xA;" OwnerUserId="1301" LastEditorUserId="368" LastEditDate="2015-08-10T16:12:50.903" LastActivityDate="2017-05-08T17:39:07.457" Title="Will more electricity be generated by using a lens to focus sunlight onto solar cells?" Tags="|electrical-engineering|optics|photovoltaics|solar-energy|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2691" PostTypeId="2" ParentId="2690" CreationDate="2015-05-05T05:52:08.410" Score="15" Body="&lt;p&gt;Your equation is partly correct. You've calculated the energy per photon ($\hbar \nu$), but you've neglected the &lt;strong&gt;number&lt;/strong&gt; of photons. That's why the units don't match (power is energy per unit time, while you've only got energy for each photon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal power (energy per unit time) depends on the area of the solar panel, $A_p$, the number of photons striking it per unit time ($\Phi$) and the energy of each photon, $E$, such that $W_{Ideal} =A_p \cdot \Phi \cdot E$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lens or mirror can focus light (a flux of photons) onto a small area. Under really ideal conditions, the area of the lens ($A_L$), would replace $A_p$ in the formula above. So, if the lens is larger than the solar panel, it can capture a larger flux of photons and direct them to the panel, increasing the power. &lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="368" LastEditDate="2015-05-05T17:12:54.400" LastActivityDate="2015-05-05T17:12:54.400" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2692" PostTypeId="2" ParentId="2690" CreationDate="2015-05-05T12:20:51.987" Score="10" Body="&lt;p&gt;Yes, increasing the illumination on a solar cell by using lenses or mirrors increases the electric power output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are limiting factors.  The efficiency of a solar cell goes down with temperature.  The current stays roughly proportional to the photon flux, but the open circuit voltage goes down as the semiconductor junction is heated.  Still, more flux yields more power, although not quite linearly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep going, and the solar cell gets so hot that the semiconductor it's made from no longer acts like a semiconductor.  That's around 150&amp;deg;C for silicon.  If you can keep the cell cool, you can hit it with higher photon flux.  However, other non-linear effects start to get in the way and you start getting diminishing returns at high flux levels.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-05-05T12:20:51.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2693" PostTypeId="1" AcceptedAnswerId="2694" CreationDate="2015-05-05T18:31:05.723" Score="5" ViewCount="4128" Body="&lt;p&gt;I have made a Python program in which I have computed the deflection of a beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am plotting the deflected beam and am now asked to show the length of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the beam is not deflected, it is easy to show the length of the beam but when the beam is deflected, I guess the length of the beam is longer than the undeflected beam. How can I compute the length of the deflected beam? Is it correct to distinguish between the length of the deflected and undeflected beam?&lt;/p&gt;&#xA;" OwnerUserId="1437" LastEditorUserId="272" LastEditDate="2015-05-05T19:03:51.330" LastActivityDate="2016-03-31T16:41:06.290" Title="Does the length of a beam change upon deflection?" Tags="|mechanical-engineering|structural-engineering|structures|beam|statics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2694" PostTypeId="2" ParentId="2693" CreationDate="2015-05-05T19:58:29.563" Score="8" Body="&lt;p&gt;The dimensions of the beam and magnitude of the deflection are important here. In most structural applications, it's reasonable to assume the length of a beam is unchanged by a small deformation. One of the basic assumptions of beam theory is that there is some internal surface of the beam called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Neutral_axis&quot; rel=&quot;nofollow&quot;&gt;neutral axis&lt;/a&gt; that holds no tension or compression stress, meaning no change in length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with any approximation, we know the truth is more complex. The beam will lengthen slightly on one side of the neutral axis/surface and shorten slightly on the other. The question becomes, is your application sensitive enough to a change in length that you need to take into account a very slight amount of lengthening and shortening along the top and bottom surfaces of the beam? If you only want to draw a beam with small deformations so that the user can see an illustration, you don't need to consider this complexity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example of when you would need to look at the change in length of any part of the beam is if you have a very tight tolerance between one end of your beam and some other element that must not come into contact with the beam. In that case you might have good reason to make sure that a very slight elongation along the top fibers of the beam doesn't cause it to scrape or bind up against the neighboring element when the beam is deforming under load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that beam theory assumes there is no loading in the longitudinal direction; otherwise the member would be considered a beam-column.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-06-09T20:19:36.513" LastActivityDate="2015-06-09T20:19:36.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2695" PostTypeId="1" AcceptedAnswerId="2697" CreationDate="2015-05-05T20:20:10.200" Score="14" ViewCount="5815" Body="&lt;p&gt;If I use an ultrasonic sensor will it detect the water level? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about a product to read water level on water boxes (common in Brazil). I researched about instrumentation for this measure, and I think that an ultrasonic sensor is the best option. Will the water correctly reflect  ultrasound and not change the normal measurements against a solid obstacle? &lt;/p&gt;&#xA;" OwnerUserId="1503" LastEditorUserId="282" LastEditDate="2015-05-07T22:17:38.787" LastActivityDate="2018-01-01T18:08:00.703" Title="Can I use a ultrasound sensor to measure water level?" Tags="|electrical-engineering|measurements|sensors|control-engineering|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2696" PostTypeId="1" CreationDate="2015-05-05T22:58:17.850" Score="7" ViewCount="149" Body="&lt;p&gt;A contractor working for me on a steel construction project is using a steel product they refer to as 'laser plate' which is marketed as specifically as having good performance on laser cutters. This doesn't appear to be a trade name, but a colloquial term. Looking at the material, it has little to no mill scale, but doesn't look quite as clean or shiny as other plate I've seen sold as 'hot rolled, pickled and oiled.' The contractor assured me that the plate conforms to ASTM A36, which is all my project requires, and I'm requesting a mill certificate to confirm that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided it is certed for A36, are there any special concerns I should have about this special steel? In general do steels marketed for lasercutting have a different chemical composition, or do they just have the mill scale removed? The closest source I've been able to find is &lt;a href=&quot;http://qa.arcelormittalna.com/documents/en/Inlandflats/TechnicalPaperStudy/LASER_BEAM_CUTTING_OF_STEELS.pdf&quot;&gt;here, (PDF)&lt;/a&gt; which suggests that lower silicon may have some influence but mostly just in how it affects mill scale adhesion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, I'm only asking about the type of material. I'm familiar with concerns relating to laser cutting as a process and am not concerned about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plates they are using range from 3/16&quot; thick to 1/2&quot; thick and the plates will be welded according to AWS D1.1. &lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2019-02-08T16:16:08.803" Title="Special Considerations for Steel Plate Marketed for Laser Cutting" Tags="|materials|structural-engineering|steel|metallurgy|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2697" PostTypeId="2" ParentId="2695" CreationDate="2015-05-05T23:58:27.503" Score="13" Body="&lt;p&gt;Technology used in depth finder, in marine application is mostly like is the best to measure the water from the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To measure water level from the bottom of tank one could use a piezo electric ceramic transducer combined with an Analog Front End (AFE) and a micro-controller to measure water level. The diagram below best explains the configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hUq5N.jpg&quot; alt=&quot;Fluid Level Sensing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use a piezo electric ceramic transducer from Steminc, TDC1000 AFE from Texas Instrument and a MSP430 micro-controller also from Texas Instrument. There might be other configurations, but currently I am only aware of this configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Piezo Electric Transducer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lc1yE.jpg&quot; alt=&quot;Transducer 1&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/V1ze7.jpg&quot; alt=&quot;Transducer 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Analog Front End (AFE)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kldQ7.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SEnXP.jpg&quot; alt=&quot;Analog Frond End&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Limitation&lt;/strong&gt;: The thickness and material of the tank might be an issue. For the most part above configuration work with plastic. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.steminc.com/pzt/en/piezo-disc&quot; rel=&quot;noreferrer&quot;&gt;Steminc - piezo electric transducer&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/11308/110&quot;&gt;Piezoelectric matching/backing layer materials&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sensortech.ca/site/index.cfm?DSP=Page&amp;amp;ID=57&quot; rel=&quot;noreferrer&quot;&gt;Piezoelectric Modes of Vibration&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/1883/what-is-radial-and-thickness-mode-vibration-in-a-piezo-electric-ceramic-disc-tra&quot;&gt;What is Radial and Thickness Mode Vibration in a Piezo Electric Ceramic Disc Transducer?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/general/docs/lit/getliterature.tsp?genericPartNumber=tdc1000&amp;amp;fileType=pdf&quot; rel=&quot;noreferrer&quot;&gt;TDC1000 Ultrasonic Sensing Analog Front End (AFE) for Level Sensing&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/186312/how-to-measure-output-of-the-ultrasonic-transducer&quot;&gt;How to measure output of the ultrasonic transducer?&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-01-01T14:54:30.193" LastActivityDate="2018-01-01T14:54:30.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2698" PostTypeId="2" ParentId="2612" CreationDate="2015-05-06T00:28:49.757" Score="7" Body="&lt;p&gt;Answering the question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What are possible types of low cost sensors I can use?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are several types of sensors that can provide millimeter level accuracy. &quot;Low cost&quot; is a very relative term, so you'll need to do some shopping around based on your specific budget.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Optical sensors-&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Included here are those of the type you listed, though it's a very cheap sensor meant more to provide a &quot;yeah something is in front of me, about yay-far-away&quot;. There are also &lt;a href=&quot;http://www.dimetix.com/Products/EN/_FRM_FLS_C_EN.html&quot; rel=&quot;noreferrer&quot;&gt;laser sensor&lt;/a&gt; systems which can provide millimeter level accuracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ultrasonic Sensors-&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most ultrasonic sensors have relatively low accuracy; in the 25.4mm range. But some &lt;a href=&quot;http://www.walkerindustrial.com/Baumer-Ultrasonic-Distance-Measuring-Sensor-with-b-p/unck-09u6914-d1.htm&quot; rel=&quot;noreferrer&quot;&gt;higher end sensors&lt;/a&gt; have sub-millimeter accuracy. Some are &lt;a href=&quot;http://www.bonanza.com/listings/GOWE-Ultrasonic-Thickness-Gauge-1-2-220mm-Steel-Resolution-0-1mm-Misc-/232916608?gpid=68416460341&amp;amp;gpkwd=&amp;amp;goog_pla=1&amp;amp;gclid=CjwKEAjw1KGqBRC55bru-sa7zCcSJAAxsBf5sjieKkg127jleXWaxs2BCYBF6z3lNRQbgHl4cD28uhoCWbvw_wcB&quot; rel=&quot;noreferrer&quot;&gt;stand alone units&lt;/a&gt; with high accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Eddy current sensors-&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These measure induced eddy currents and &lt;a href=&quot;http://www.micro-epsilon.com/displacement-position-sensors/eddy-current-sensor/eddyNCDT_3300/index.html&quot; rel=&quot;noreferrer&quot;&gt;provide incredible accuracy&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mechanical measurement-&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be implied by your diagram, but it's not specified. &lt;a href=&quot;http://www.amazon.com/b?node=256411011&quot; rel=&quot;noreferrer&quot;&gt;Calipers&lt;/a&gt; are an obvious choice for measuring thickness. For a semi-automated approach you can also use linear displacement sensors. &lt;/p&gt;&#xA;" OwnerUserId="1505" LastActivityDate="2015-05-06T00:28:49.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2699" PostTypeId="1" AcceptedAnswerId="3710" CreationDate="2015-05-06T02:29:30.643" Score="6" ViewCount="8135" Body="&lt;h3&gt;Weight of Structural Steel&lt;/h3&gt;&#xA;&lt;p&gt;When calculating the weight of a steel structure, I usually calculate the weight of all of the various members and plates first. I then add an additional percentage to account for the weight of nuts and bolts. This percentage is usually between 2% and 5% of the total weight.&lt;/p&gt;&#xA;&lt;p&gt;It makes sense that bolts will add additional weight beyond just the weight of the members. It is also complicated and time consuming to do it the &amp;quot;proper&amp;quot; way, e.g. deduct the weight of the hole and add the weight of the correct length and size of bolt. This is done with a spreadsheet or by hand and without the aid of specialized steel detailing software. Another complication is that furnished bolt lengths are partially determined by the fabricator.&lt;/p&gt;&#xA;&lt;h3&gt;The Problem&lt;/h3&gt;&#xA;&lt;p&gt;My company has always done this. The one time that a third party reviewer asked why the steel weight was ~5% higher, we were unable to cite a code reference where this was called for.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How is the weight of fasteners typically included in steel weight calculations?&lt;/strong&gt; This can either be for a bid weight or a design dead load weight.&lt;/p&gt;&#xA;&lt;p&gt;As additional information, in all welded construction, the weight of bolts might be so minor that it is negligible. Some of my projects use completely bolted connections including built-up members.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-03-11T22:00:52.387" Title="Is there a standard percentage to increase structural steel weight to account for bolts?" Tags="|structural-engineering|steel|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2701" PostTypeId="2" ParentId="2355" CreationDate="2015-05-06T05:07:20.327" Score="0" Body="&lt;p&gt;Calculate latent heat Q= ML , using amount of water you want to freeze . Divide by time you want to want the cannon to be made. You Get cooling capacity . Add losses to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make trials of TEC from calculator below. &#xA;&lt;a href=&quot;https://viveksilwal.wordpress.com/2015/04/15/28/&quot; rel=&quot;nofollow&quot;&gt;https://viveksilwal.wordpress.com/2015/04/15/28/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once this is done, Install a software Ansys Icepak, you can learn it in a week, there you can simulate TEC easily. Analyse your system there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although thermoelectric is very less efficient , but I think a good heat sink and tec combination can make you freezed balls.&lt;/p&gt;&#xA;" OwnerUserId="1508" LastActivityDate="2015-05-06T05:07:20.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2703" PostTypeId="2" ParentId="2612" CreationDate="2015-05-06T09:21:41.057" Score="2" Body="&lt;p&gt;Various solutions are possible depending on your exact requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Contacting system&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably the most obvious system, consisting of 1 or 2 lever arms which are deflected by the plate. By measuring the deflection the size could be determined. The main drawback of this is that it is a contacting system which may be a problem depending on your setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Video system&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another alternative which would be simple to implement is a video system. Mount a camera above your sample and image the plate as it passes through. By either calibrating the camera before hand or mounting a line scale beneath the plate the size can be determined. Resolution will depend on resolution of the camera and scale, although I think 1 mm shouldn't be hard to achieve. This is probably what I would do if you didn't want a contact system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Laser rangefinder&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These may be possible depending on what you consider low cost. Most systems are based on time of flight, which limits the resolution to ~1 mm. Additionally most cheap systems I have seen have a minimum range of at least 5 cm, which may be too large for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Interferometer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also may be possible. However, simple fringe counting methods have difficulties when measuring steps, so will probably not be suitable for measuring a plate (unless its infinite). Frequency scanning methods are possible, but probably don't fall under low cost, measurement speed may also be an issue depending on sample length and throughput required.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-05-06T09:21:41.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2704" PostTypeId="1" CreationDate="2015-05-06T10:06:21.757" Score="4" ViewCount="4551" Body="&lt;p&gt;I have designed a circular, 800 mm (31.5&quot;) diameter tripod table. The top is made of stone and weighs 40 kg (88 lbs). The height of the table is 750 mm (29.5&quot;).  There is a central column with tripod feet lower down which touch the floor 108 mm (4.25&quot;) in from the outer edge of the table top if looking in plan. The fulcrum created by two of the feet is a line 240 mm (9.5&quot;) to the table edge at its widest point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: How much weight can be put on the edge of the table top before the table tips over, and how much can this weight increase if the feet touchdown 90 mm (3.5&quot;) from the outer edge? (In this case the fulcrum line would now be approx 230 mm (9&quot;))&lt;/p&gt;&#xA;" OwnerUserId="1510" LastEditorUserId="368" LastEditDate="2015-05-11T21:15:50.347" LastActivityDate="2019-01-07T20:29:31.900" Title="How to calculate a tripod table's stability?" Tags="|statics|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2705" PostTypeId="1" AcceptedAnswerId="2710" CreationDate="2015-05-06T12:49:24.117" Score="5" ViewCount="2277" Body="&lt;p&gt;I'm looking for a power source that is able to deliver 12V at 15-30A (yes, amps). I was looking at high capacity batteries, but looking at the specs, they either say they have a lower discharge voltage than 12V, or they don't say anything at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do such power sources exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping to use this in a hydraulic system. The pumps I've seen require 12V/24V and consume around 10-15A minimum. I've been looking at Hydraulic Power Units (HPUs), and their specs say that power consumption is really high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading around, it seems most, if not all, hydraulic systems are powered by small diesel engines. I would really like to avoid that, if it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such high capacity power sources do not exist, what would be my alternative?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Apologies for the above sounding like a &quot;find things for me&quot; question. This was not the intention. Maybe I haven't actually understood fully how hydraulic systems actually are designed. Also, I'm not very good with words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to hydraulic systems, and I am in the process of designing a portable hydraulic system, similar to ones used in vehicles, like construction vehicles. However, my hydraulic system will not be part of a vehicle. I realise vehicles use their engines and some form of petrol/diesel to electricity generator to obtain the power required, but I cannot see a way of obtaining that same power without using a vehicle's engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about getting that same power without the use of a vehicle? Ideally, the system can be moved around without it being attached to AC wall electricity, but that can be left until a later date.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Details/Background of my project:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to build a hydraulic exoskeleton that uses as little electrical components as possible. I have gone for hydraulic over pneumatic as hydraulic packs a lot more punch than pneumatic. Most of the sites online sell the valves, pumps, pistons etc, but I have previously found it difficult to find a way of powering these pumps, up until I asked on here &lt;a href=&quot;https://engineering.stackexchange.com/questions/481/motor-for-a-hydraulic-pump-in-a-hydraulic-system&quot;&gt;a while back&lt;/a&gt;, where I was introduced to HPUs. From what I have understood (do correct me if I'm wrong), I need a way of powering these HPUs to generate the necessary RPM and flow rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found the pistons I would like to use, the hydraulic fluid, how much to use, the reservoir tanks, the filtering system, etc, but I am stuck on the power source. I just have no idea how to harness the power I need for this. This is where I started looking at ultra-high capacity batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest, I don't even know which would be best, a DC- or AC-input motor. My thought was that the constant forward drive of the motor required a DC source, hence the batteries. The link I provided in the comments below points to AC motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll keep adding details if and when they are required. As I said, I'm not good with words and I'm new to hydraulics.&lt;/p&gt;&#xA;" OwnerUserId="391" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-05-08T00:06:53.367" Title="Hydraulic power source" Tags="|electrical-engineering|power-electronics|hydraulics|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2706" PostTypeId="2" ParentId="2704" CreationDate="2015-05-06T12:50:29.707" Score="3" Body="&lt;p&gt;I think (with some simplification) it should be 23 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is why:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;legs are weightless&lt;/li&gt;&#xA;&lt;li&gt;table's mass is a pointmass in the center of the table&lt;/li&gt;&#xA;&lt;li&gt;table will topple over the connecting line between two legs&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;First we do a balance of momentum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\Sigma M &amp;amp;= 0 \\&#xA;\Sigma M &amp;amp;= F \cdot (400 mm -d) - 40 kg \cdot d \\&#xA;\rightarrow F &amp;amp;=  \frac{40kg\cdot d}{400mm-d}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is $d$? It is the (at the moment) unknown distance to the fulcrum line (considering, the table will not topple over one, but over two legs).&#xA;Let's calculate that $d$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the table's legs have an angle of 120° between each other. We want the height of the triangle composed of two legs and the connection line between them. &#xA;Our rectangular triangle is built by the height of the former triangle, half of the connection line and a leg's distance to the center (which is 400mm - 108mm = 292mm and now the hypotenuse).&#xA;The angle is split in half, so its 60°.&#xA;We can now calculate the height (now the adjacent leg to the angle) by&#xA;$$&#xA;d = cos (60°) \cdot 292mm =146mm&#xA;$$&#xA;We can now solve the momentum balance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F =  \frac{40kg\cdot 146mm}{400mm - 146mm}=22.992kg&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correct me if I'm wrong. (some pictures would help...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the new configuration it could be increased by 2.314kg (you can calculate that yourself by now... :P ) &lt;/p&gt;&#xA;" OwnerUserId="1409" LastActivityDate="2015-05-06T12:50:29.707" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2707" PostTypeId="2" ParentId="475" CreationDate="2015-05-06T13:00:19.830" Score="3" Body="&lt;p&gt;WR grade bolts are available in the UK in 1/4 inch imperial size HSFG (when imported from America) and also in M24 TCB's, both of which are used regularly in bridge steelwork in the UK.  &lt;/p&gt;&#xA;" OwnerUserId="1515" LastActivityDate="2015-05-06T13:00:19.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2708" PostTypeId="2" ParentId="2704" CreationDate="2015-05-06T13:23:59.100" Score="0" Body="&lt;p&gt;I'll assume you apply the load at a single point at the worst position (on the edge equidistant between two legs). The real loading may be higher than this depending on distribution and position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case you have a simple lever problem, with the load and part of the table balanced by the rest of the table (assuming the legs are relatively light).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fraction of the table on the loaded side of the table will be a &lt;a href=&quot;http://en.wikipedia.org/wiki/Circular_segment&quot; rel=&quot;nofollow&quot;&gt;circular segment&lt;/a&gt;. Therefore, the fraction is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f=\frac{\theta-\sin{\theta}}{2\pi} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\theta$ is in radians and is given by&#xA;$$ \theta =2\arccos{\frac{d}{R}} $$&#xA;$d$ is the distance from the centre to the fulcrum and $R$ is the radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming uniform mass density of the table each segment will act through its center of mass. Therefore there are three components to balance. The table will tip if.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(R-d)L + \frac{R-d}{2}fM &amp;gt; \frac{d+R}{2}(1-f)M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rearranging&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L&amp;gt;(\frac{R+d}{R-d}\frac{1-f}{2}-\frac{f}{2})M $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for your table $R=400$ mm, $d=160$ mm, $M=40$ kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$theta =1.59 rads$$&#xA;$$f=0.094$$&#xA;$$L_{max}=40.4kg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've probably made some calculation or algebra error here so I would check it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this model the height of the table is irrelevant. I reality lateral loads on the table are non-zero, e.g. someone pushing sideways. I which case height will have a significant effect.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-05-06T13:23:59.100" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2710" PostTypeId="2" ParentId="2705" CreationDate="2015-05-06T14:11:20.403" Score="5" Body="&lt;p&gt;Pretty much all electric motors demand a lot more power for starting from a dead stop than for generating motion in a steady state. There are a whole lot of schemes for smoothing out this 'inrush current' demand, either from the electrical side or from the mechanical side (by decoupling the load and gradually re-connecting it once the motor spins up.) The problem with HPUs designed for use on vehicles is that they are designed with the assumption that the power source has a battery already there to provide the starting current for the engine starter. These starting batteries can typically provide hundreds of amps for a few seconds - much more than a modest HPU will require.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because they usually have such a large power source available, and because all of the strategies for reducing inrush current have some cost and complexity associated, vehicle HPUs are generally designed with no intention to minimize the inrush current requirements. This is made further more complicated as solenoids and relays are often used for the control systems on mobile hydraulics, which have inrush currents of their own, often applied at the same time as the pump starting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While automotive batteries can provide extremely high currents (for a short time) most AC to DC power supplies can only deliver maybe 2 times as much power momentarily as they can in the steady state. This means that if you connect a vehicle hydraulic system to a AC to DC power supply rated for the same current that the HPU needs in the steady state, you likely won't have enough current to start the system. As a result, the pump will be starved for power and you may hear a whine or just notice it getting very hot without turning. Additionally, any relays of solenoids will likely 'chatter.' This is because on their own, there is enough power form the power supply to actuate the solenoid, but as soon as the motor is connected, the voltage drops too low and the solenoid opens again, restarting the cycle. Both of these conditions will damage your HPU and should not be sustained for more than a second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest strategy is to provide an automotive battery (typically lead-acid,) like the HPUs are designed to run off of. Depending on the size of your HPU and the length of time you need it to run on a charge will dictate the size of the battery you need. Automotive batteries are produced in a range of sizes designer for everything from starting very small cars to powering entire RVs for days without running the engine. They can also easily be combined in parallel. You mention that voltage form these batteries can sometimes drop below their 12 volt nominal value. That is true as they loose charge, but most HPUs can handle a significant voltage drop below nominal (10-20%) before performance is seriously effected. If the use will be steady state, a battery can be provided in addition to an Ac to DC power supply, which will maintain the voltage and keep the battery charged, just like a battery in a car is fed by an alternator once the engine is running.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you must power your HPU without a battery at all, you need to plan for your AC to DC power supplies to have 2 or 3 times as much available current as the nameplate current demand of the HPU, or implement your own inrush limiting power supply (typically some combination of chokes and capacitors, but I don't know much about how they work.) Power supplies in the range of 100-200A at 12VDC are commercially available. If you need more than this, you can choose power supplies designed to be put into a parallel master-slave configuration and connect two or three power supplies together. One manufacturer of such supplies is MeanWell.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-05-06T14:17:35.137" LastActivityDate="2015-05-06T14:17:35.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2711" PostTypeId="2" ParentId="2680" CreationDate="2015-05-06T14:29:26.193" Score="2" Body="&lt;p&gt;This is one solution for a flush mounted bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use a 5 mm wide ball bearing on one leg of it. A ball bearing is preferred here since it more suited for minor thrust loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prevent the bearing from working its way out, it can be cryogenically chilled in liquid nitrogen, and then pressed into place. When the bearing warms up to room temperature it would expand tightly against the metal flat bar causing it to be locked in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same can also be done with the shaft that rides in the bearing, affix the shaft solidly to one flat bar by cryogenically chilling the shaft and pressing into the flat bar. Then press the entire assembly into the inner race of the ball bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heating the steel flat bar instead of chilling the shaft or bearing can also work for press fitting, as in this case you are expanding the hole beforehand instead of shrinking the part. &#xA;But to secure the the shaft to the inner race of the bearing by heating, you need to heat the bearing itself, but overheating the bearing will cause it to lose its temper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to secure the assembly even more you can peen around the outside of the bearing and on the ends of the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... if weaker attachment methods are acceptable, you can try assembling the parts with red loctite or cyanoacrylate super glue instead of press fitting.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastEditorUserId="1459" LastEditDate="2015-05-06T15:15:45.187" LastActivityDate="2015-05-06T15:15:45.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2712" PostTypeId="5" CreationDate="2015-05-06T15:12:23.220" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-05-06T15:12:23.220" LastActivityDate="2015-05-06T15:12:23.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2713" PostTypeId="4" CreationDate="2015-05-06T15:12:23.220" Score="0" Body="Use this tag for non-technical questions about interpersonal communication in the engineering profession." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-05-06T15:12:23.220" LastActivityDate="2015-05-06T15:12:23.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2714" PostTypeId="1" AcceptedAnswerId="2715" CreationDate="2015-05-06T15:19:00.057" Score="8" ViewCount="636" Body="&lt;p&gt;In &lt;a href=&quot;http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20000036534.pdf&quot;&gt;this NASA document&lt;/a&gt;, it is mentioned on page 221 (239 of the pdf) that a &quot;23 tol gear box&quot; was used. What does that mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A google search for &quot;23 tol gear box&quot; (with quotes) comes up completely empty, and &quot;tol gear box&quot; shows only irrelevant results. A search for just tol doesn't come up with any convincing results either. I can't rule out that it is a typo of some sort. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the context I presume it to mean that the gearbox output runs 23 times as fast as the powered side, but I have to be certain. &lt;/p&gt;&#xA;" OwnerUserId="1517" LastActivityDate="2015-05-06T15:47:39.733" Title="What does &quot;tol&quot; mean in a gear box?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2715" PostTypeId="2" ParentId="2714" CreationDate="2015-05-06T15:24:33.647" Score="8" Body="&lt;p&gt;Oh, I get it. The typo is a missing space. It's supposed to mean &quot;23 to 1&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1517" LastEditorUserId="1517" LastEditDate="2015-05-06T15:47:39.733" LastActivityDate="2015-05-06T15:47:39.733" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2717" PostTypeId="5" CreationDate="2015-05-06T15:48:33.763" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-05-06T15:48:33.763" LastActivityDate="2015-05-06T15:48:33.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2718" PostTypeId="4" CreationDate="2015-05-06T15:48:33.763" Score="0" Body="Use this tag for questions about communications technology and its applications." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-05-06T15:48:33.763" LastActivityDate="2015-05-06T15:48:33.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2719" PostTypeId="1" CreationDate="2015-05-06T16:13:06.967" Score="3" ViewCount="338" Body="&lt;p&gt;I am computing the deflection of a beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the length of the beam is 10 m and I have load at the position 4 m from left with force 5 N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UaeZC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I add another load at the exact same position with force 4 N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I just sum up the forces and write 9 N instead of 5 N?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a school exercise in which I'm told to plot the deflected beam and show all the loads. The only clever way to show the loads is to plot arrows at the positions and annotate the arrows with the load force. But if I have many loads at the same position, it will be quite messed up.&lt;/p&gt;&#xA;" OwnerUserId="1437" LastEditorUserId="368" LastEditDate="2015-05-11T21:16:13.047" LastActivityDate="2017-05-03T23:41:29.350" Title="Sum up the loads at same position on a deflected beam" Tags="|structural-engineering|beam|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2720" PostTypeId="2" ParentId="2719" CreationDate="2015-05-06T16:25:25.777" Score="2" Body="&lt;p&gt;Yes, you can simply add the loads together. Imagine taking an actual beam, and stacking some weights on that point. Then add more weights on top of that. There's no way to tell the difference between 9 N, 5+4 N, 3+6 N, etc. Make sure you're reading the problem correctly though, because that sounds like an almost trivial case. Beam deflection problems will often have distributed loads, which are different. For some purposes (end reactions only) you can treat those as equivalent point loads, but for others (internal force/moment reactions) you have to consider the distribution itself. &lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="41" LastEditDate="2015-05-07T14:52:10.477" LastActivityDate="2015-05-07T14:52:10.477" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2721" PostTypeId="1" AcceptedAnswerId="2724" CreationDate="2015-05-06T19:44:03.523" Score="7" ViewCount="2178" Body="&lt;p&gt;Note: This may be UK-specific, I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I overheard part of a conversation about 3-phase electricity, one guy was saying that mains generators use 3 coils (or something) and as it spins it produces 3 different waves of alternating current that are 120 degrees apart. That's why the power lines often have three (or six) sets of cables, one for each phase. He said that if different phases are electrically connected then bad things happen, and sometimes if only one phase goes down, every third house on a street could lose power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This raised a few questions for me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean all power stations have to be in sync with each other? If so how is this achieved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it true about every third house, I've never seen this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean we effectively have 3 grids overlaid on top of each other?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="95" LastEditDate="2015-05-06T20:26:41.653" LastActivityDate="2015-05-07T00:40:59.137" Title="Does a 3-phase electricity grid need to be synchronised and are the phases split between houses?" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2722" PostTypeId="5" CreationDate="2015-05-06T19:49:25.350" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-06T19:49:25.350" LastActivityDate="2015-05-06T19:49:25.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2723" PostTypeId="4" CreationDate="2015-05-06T19:49:25.350" Score="0" Body="For questions about solar generation and solar panel design and manufacturing." OwnerDisplayName="user1184" LastEditorDisplayName="user1184" LastEditDate="2015-05-06T21:04:06.153" LastActivityDate="2015-05-06T21:04:06.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2724" PostTypeId="2" ParentId="2721" CreationDate="2015-05-06T20:24:03.957" Score="11" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does that mean all power stations have to be in sync with each other? If so how is this achieved?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, they do.  Each generating station receives a timing signal from their Energy Management System that essentially tells the generator if it is synchronized to the grid frequency or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grid frequency is primarily driven from a standard and consistent time source, such as from an atomic clock.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it true about every third house, I've never seen this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kinda true if you really stretch things, but not really.  See below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this mean we effectively have 3 grids overlaid on top of each other?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Definitely not.  It's one grid using three phases. &lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The basics of what you were told are essentially correct.  For a host of reasons, it is more efficient and practical to transfer AC power in three phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, if the phases cross and connect with each other, &lt;code&gt;Bad Things&lt;/code&gt; really do happen.  At a minimum, it will trip the protective relays and shut the area down for a while.  The momentary arc that's generated when they cross can be quite spectacular though and can easily destroy large portions of trees that were silly enough to have their branches provide the phase crossing mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extent of &lt;code&gt;Bad Things&lt;/code&gt; also depends upon the voltage level of the line and amount of current being carried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't see every third house going dark because of the way distribution laterals are setup.  There are two main approaches to feeding distribution loads:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All three phases transferred along the distribution lateral (aka balanced)&lt;/li&gt;&#xA;&lt;li&gt;A single phase transferred along the distribution lateral.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The single phase per lateral approach is easier to put in place but is less efficient because the loads tend not to balance well across the various laterals coming out of the distribution substation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easier to balance out distribution laterals by transferring all three phases along the length of the line.  However, power is delivered to blocks of houses and each block is fed off a single phase from the distribution line. You can verify this if you have overhead lines by looking for the step down transformer (it's a big metal can hanging off of the power pole) and verifying that only one phase feeds into the can.  &lt;sup&gt;2, 3&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that residential service in North America is almost always single phase.  The primary driver behind this is that most residential grade motors are single phase, and the demand simply isn't there to justify the additional equipment expense from delivering two phases to a residential drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;I'm explicitly ignoring things like regional interconnections such as what is seen in North America.  Then you could argue for there being multiple grids, but that's not really germane to your question.&lt;/sub&gt;&lt;br&gt;&#xA;&lt;sup&gt;2&lt;/sup&gt; &lt;sub&gt;You can identify if you have a 3 phase distribution feed in your neighborhood by looking at the wires at the very top of the distribution power pole.  Three wires spread across arms at the top of the pole generally mean a 3 phase service in the area.&lt;/sub&gt;&lt;br&gt;&#xA;&lt;sup&gt;3&lt;/sup&gt;&lt;sub&gt;If you see a cluster of metal cans and all three phases feeding into those cans then you've likely spotted a capacitor bank which is used to condition the power on the line.  It's not used for stepping down the voltage for residential service.&lt;/sub&gt;   &lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-05-07T00:40:59.137" LastActivityDate="2015-05-07T00:40:59.137" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2725" PostTypeId="1" AcceptedAnswerId="2745" CreationDate="2015-05-06T20:51:48.677" Score="12" ViewCount="11110" Body="&lt;p&gt;In order to make this a manageable question, let's add a few simplifications.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The dust particles can be well described as uniform spheres of radius $R$ and density $\rho$.  &lt;/li&gt;&#xA;&lt;li&gt;The space is enclosed and there is no bulk flow, i.e the air is still in a macroscopic sense.&lt;/li&gt;&#xA;&lt;li&gt;The air is at the &lt;a href=&quot;http://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure&quot;&gt;standard temperature and pressure (STP)&lt;/a&gt;; $T=20\ ^\circ\mathrm{C}$ and $P=1\ \mathrm{atm}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Under these conditions, what is the settling time for dust particles?  At what size/density does Brownian motion of the air become important? &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-05-07T23:36:15.113" Title="How long does it take for dust to settle out of the air?" Tags="|fluid-mechanics|air-quality|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2726" PostTypeId="2" ParentId="2680" CreationDate="2015-05-06T20:53:50.687" Score="2" Body="&lt;p&gt;I recommend a sliding contact bearing. It's easy to manufacture and seems to do the job for a prototype, especially if it has to be cheap. Steel seems to be the better choice in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather use two thinner bars for one of the links, so you get a double lap. This will reduce the moment on the bearing and yields a more even contact stress distribution. So if you can really spend 3 mm more, you can split into two 4 mm bars instead of one 5 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a drill press and a reamer to achieve small clearence. All pieces of the bearing point can be drilled and reamed together. But for the middle bar, the inner one, you may use a press fitting. Thus the bolt will only move on the outer bars where a transition fitting shall be a good choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you want to invest in bronze sleeves for the bearings. It will lower the friction and result in longer lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P. S.: If you have to chill something to a low temperature, it's often enough to use dry ice (CO2) and ethanol or aceton. This yields about -75 °C. It's not LN2 but often you don't need -195 °C.&lt;/p&gt;&#xA;" OwnerUserId="1490" LastActivityDate="2015-05-06T20:53:50.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2727" PostTypeId="1" CreationDate="2015-05-06T22:31:09.343" Score="1" ViewCount="2253" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SZJoD.jpg&quot; alt=&quot;whole problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 2D plate with heat source at one part and I need to calculate the heat generation. The plate is made of aluminum (k = 212 W/m°C). I have watched &lt;a href=&quot;https://www.youtube.com/watch?v=u9W8CGQy_ZU&quot; rel=&quot;nofollow noreferrer&quot;&gt;this example video&lt;/a&gt;, but I'm not sure if I can apply it to my case. Some data: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Length of the plate: 1 m &lt;/li&gt;&#xA;&lt;li&gt;Width: 1 m &lt;/li&gt;&#xA;&lt;li&gt;Surface being heated: 0.16 m^2 &lt;/li&gt;&#xA;&lt;li&gt;T&lt;sub&gt;max&lt;/sub&gt; = 500 °C&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The heat source is placed in the center of the plate and the boundaries of the plate are insulated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the heat generated as in &lt;a href=&quot;http://geodynamics.usc.edu/~becker/teaching/557/problem_sets/problem_set_fd_2dheat.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this example, equation (4)&lt;/a&gt;. You can see Q is being divided by ρc&lt;sub&gt;p&lt;/sub&gt; and thus the units of Q must be W/m&lt;sup&gt;3&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1411" LastEditorUserId="33" LastEditDate="2016-01-30T21:49:21.980" LastActivityDate="2016-01-30T21:49:21.980" Title="How to calculate heat generation on 2D plate?" Tags="|mechanical-engineering|thermodynamics|fluid-mechanics|heat-transfer|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2728" PostTypeId="2" ParentId="2721" CreationDate="2015-05-06T22:56:12.053" Score="11" Body="&lt;p&gt;I can't speak for the UK, but the &quot;every third house&quot; is certainly not true in Denmark. Here the standard arrangement is that &lt;em&gt;every house&lt;/em&gt; gets &lt;em&gt;all three phases&lt;/em&gt; at 230 V. Inside the house some of the circuits are fed from each phase, so in some error conditions you can experience that the floor sockets are dead but the ceiling lights work, or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some high-powered appliances such as stoves are often constructed to be connected to all three phases at once -- the difference between two different live phases produces AC with an effective voltage of 400 V instead of the 230 V between a live phase and neutral.&lt;/p&gt;&#xA;" OwnerUserId="1520" LastActivityDate="2015-05-06T22:56:12.053" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2729" PostTypeId="2" ParentId="2721" CreationDate="2015-05-06T23:42:32.740" Score="3" Body="&lt;p&gt;In North America, you generally don't see 3 phase power being delivered to houses.  The transformers (whether they're up in a pole, or in a green box on the ground) take apart the 3-phase 1100V power delivered to them and step it down to 110V.  For 220V, two feeds are connected appropriately, and 220V is delivered to each breaker box.&#xA;In commercial and industrial settings, you can get three phase power delivered.  We have big air compressors and hoists that run on 3 phase 220V.  A motor has to be designed to run that kind of power.  We have had outages on one and on two of the three phases.  Bad things happen.  Things draw WAY more amperage than they should from the remaining power and smoke and unhappiness results.  In our old building, this happened it was because it was old and a circuit-breaker for that purpose wasn't installed.  In our new building, the electrician was incompetent and didn't install it.  Large installations of fluorescent lights will use 3 phase 367V power.  The outage in the new building wrecked the lighting so that afterward, 2 of the 5 bulbs in each fixture wouldn't light.  All the ballasts needed to be replaced.&#xA;We also have transformers in the building that step up the 3 phase 240V delivered to 3 phase 400V and 600V for some specialized equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in residential neighborhoods, what gets delivered to the house is one phase power 110V on two buses that, when wired correctly can give you 220V power.  And if all three phases aren't being delivered correctly to the transformer, a circuit-breaker in that trips and nobody gets any power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burning down people's home because of power snafus isn't good karma.  The power company ensures that it won't be their fault!  &lt;/p&gt;&#xA;" OwnerUserId="1522" LastEditorUserId="110" LastEditDate="2015-05-07T00:01:35.263" LastActivityDate="2015-05-07T00:01:35.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2730" PostTypeId="2" ParentId="2725" CreationDate="2015-05-07T01:11:30.137" Score="3" Body="&lt;p&gt;For an individual particle, you could use &lt;a href=&quot;http://en.wikipedia.org/wiki/Stokes%27_law&quot; rel=&quot;nofollow&quot;&gt;Stokes' law&lt;/a&gt;:&#xA;$$v_{\text{terminal}}=\frac{2gR^2(\rho_{\text{particle}}-\rho_{\text{air}})}{9\mu}$$&#xA;where $\mu$ is the dynamic viscosity of the air, which is &lt;a href=&quot;http://en.wikipedia.org/wiki/Viscosity#Air&quot; rel=&quot;nofollow&quot;&gt;easy to calculate&lt;/a&gt;. Using Stokes' law assumes you know the initial height of a dust particle, and may not be convenient for large quantities because each particle may start at a completely different position - in other words, it is not a model of a large system of particles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found some &lt;a href=&quot;http://www.cdc.gov/niosh/topics/aerosols/pdfs/Aerosol_101.pdf&quot; rel=&quot;nofollow&quot;&gt;more accurate data&lt;/a&gt; for particles of different radii, given in half-lives; slightly more data is &lt;a href=&quot;http://www.tbh.eu/en/useful-information/settling-times.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A graph of settling time for coal, iron and cement is given &lt;a href=&quot;http://conveyor-dynamics.com/wp-content/uploads/2014/01/ROCKY-Dust-Settling-References.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, further illustrating the non-linear, inverse-exponential relationship between dust radii and settling time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of settling is applied &lt;a href=&quot;http://www.mpia.de/~mordasini/slidesws1123/L9dustdynamics.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; to solar nebulae. I'm not sure exactly how many of the formula can be applied here, but some may be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t=\frac{\rho_{\text{dust}}}{\rho_{\text{air}}}\frac{R}{v_{\text{thermal}}}$$&#xA;where&#xA;$$v_{\text{thermal}}=\sqrt{\frac{8k_BT}{\pi \mu m_{\text{particle}}}}$$&#xA;Set the conditions to STP and you could have a better answer, though the realms of application are vastly different!&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-05-07T23:36:15.113" LastActivityDate="2015-05-07T23:36:15.113" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2731" PostTypeId="1" CreationDate="2015-05-07T02:03:18.207" Score="5" ViewCount="103" Body="&lt;p&gt;Is there data available to quantify the effect on pavement by light tracked vehicles with rubber inserts.  One example of such a vehicle is the &lt;a href=&quot;http://en.wikipedia.org/wiki/M113_armored_personnel_carrier&quot; rel=&quot;nofollow noreferrer&quot;&gt;M113 armored personnel carriers&lt;/a&gt; shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/File:US_M113_in_Samarra_Iraq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Dysfm.jpg&quot; alt=&quot;M113 armored personnel carrier&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1526" LastEditorDisplayName="user16" LastEditDate="2015-05-07T17:24:24.440" LastActivityDate="2015-05-07T17:24:24.440" Title="Effect on pavement by light tracked vehicles with rubber inserts" Tags="|mechanical-engineering|civil-engineering|highway-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2732" PostTypeId="2" ParentId="2265" CreationDate="2015-05-07T06:27:08.243" Score="5" Body="&lt;p&gt;It is a &lt;a href=&quot;http://en.wikipedia.org/wiki/Synchronous_motor&quot; rel=&quot;noreferrer&quot;&gt;synchronous AC motor&lt;/a&gt;. It will spin at a precise rate relative to the AC frequency (50 Hz or 60 Hz). This is useful to keep the spin rate constant under varying loads, such as in a microwave oven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From above Wikipedia link:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A single-phase (or two-phase derived from single phase) stator winding is possible, but in this case the direction of rotation is not defined and the machine may start in either direction unless prevented from doing so by the starting arrangements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2015-05-07T06:27:08.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2733" PostTypeId="1" CreationDate="2015-05-07T08:07:12.480" Score="2" ViewCount="3153" Body="&lt;p&gt;In a &lt;a href=&quot;https://engineering.stackexchange.com/questions/2265/why-does-the-microwave-plate-start-in-a-random-direction&quot;&gt;recent question of mine about microwave plates&lt;/a&gt;, jpa &lt;a href=&quot;https://engineering.stackexchange.com/a/2732/203&quot;&gt;correctly identified&lt;/a&gt; the type of motor used for microwave plates as a specific variant of AC synchronous motor, equipped with gears that reduce the speed to some 5RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iRYwo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iRYwom.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A single-phase (or two-phase derived from single phase) stator winding is possible, but in this case the direction of rotation is not defined and the machine may start in either direction unless prevented from doing so by the starting arrangements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I am unclear on - what is the principle of operation of such motors?&lt;br&gt;&#xA;What construction property of the motor is responsible for the start direction being unpredictable but the motor never being stuck and unable to start?&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2022-08-02T16:25:21.803" Title="How does single-phase AC synchronous motor work?" Tags="|electrical-engineering|ac|motors|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2734" PostTypeId="1" AcceptedAnswerId="2736" CreationDate="2015-05-07T08:23:58.640" Score="3" ViewCount="9158" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&#xA;I am working on a project that involves heating air inside a small box, then measuring the temperature over time.  The purpose of the box is to be a basic test chamber for PID experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Initial Test&lt;/strong&gt;&#xA;I made a 10 cm * 10 cm * 10 cm box from 3 mm plywood. I then placed a small ~1 W heater inside with a temperature probe and observed the temperature rise... very slowly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposed New Prototype&lt;/strong&gt;&#xA;I am going to construct a new prototype once I decide how to calculate the approximate time to warm the air inside the box to a set-point.  Some specifications and assumptions are laid out below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ~1 W heater consists of 4 parallel 100 ohm resistors and a 5 V power supply (there is also a 330 ohm resistor and an LED for visual indication).  I don't want to change this heater design, as the intention is that this is easy to replicate and uses readily accessible 5 V power sources and electronic components.&lt;/li&gt;&#xA;&lt;li&gt;Currently no fan has been used but I have a small 20 mm, 5 V fan on order, so will integrate this into the design at some point.&lt;/li&gt;&#xA;&lt;li&gt;The box will be used at room temperature between 21–23 °C&lt;/li&gt;&#xA;&lt;li&gt;Ideally a temperature rise up to 30 °C (or more) would be possible over a 5-10 minute time interval&lt;/li&gt;&#xA;&lt;li&gt;The new proposed box size is 5 cm * 5 cm * 5 cm, 8 times smaller than the previous prototype.  This is open to change depending on the previous 2 requirements of maximum temperature rise and time taken to achieve that change.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am not after exact timings - I am aware that losses through the box will have an effect. But if this effect is minimal, then a simplified approximate solution is preferable as some design parameters may change slightly.  I am looking for any guidance and calculations that will save me time instead of having to make different boxes and learn by trial and error.&lt;/p&gt;&#xA;" OwnerUserId="583" LastEditorUserId="368" LastEditDate="2015-05-08T17:55:27.567" LastActivityDate="2015-09-03T17:57:05.763" Title="Heating the air in a small box" Tags="|thermodynamics|heat-transfer|heat-exchange|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2735" PostTypeId="2" ParentId="2571" CreationDate="2015-05-07T08:59:59.300" Score="4" Body="&lt;p&gt;The article &lt;a href=&quot;http://www.ijestr.org/IJESTR_Vol.%201,%20No.%207,%20July%202013/Liquid%20Crystal%20Display.pdf&quot; rel=&quot;nofollow&quot;&gt;Liquid Crystal Display: Environment &amp;amp; Technology&lt;/a&gt; (2013) provides a detailed list of chemicals often used in LCDs. There are a myriad of chemicals that are used, each serving a specific purpose for the functionality of LCDs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for liquid crystals (LCs) themselves:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The composition of&#xA;  LC includes a bicyclohexyl compound (35-50% by weight), a cyclohexyl phenyl compound (15-25% by weight), a&#xA;  bicyclohexyl phenyl compound (20-25% by weight), and a cyclohexyl biphenyl compound (15-20% by weight).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article states that there are typically 10-25 compounds used to manufacture LCs, with many mixtures used, depending on the application the LCD is used for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variations in mixtures result in differing  chemical&#xA;properties of alkyl or alkoxy side chains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The back light unit has an interesting component:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The backlight unit (BLU) major constituent of LCD contains hazardous mercury to operate. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerDisplayName="user1184" LastActivityDate="2015-05-07T08:59:59.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2736" PostTypeId="2" ParentId="2734" CreationDate="2015-05-07T09:12:14.483" Score="7" Body="&lt;p&gt;A simple approach would be to just consider the heat added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific heat equation gives the temperature change due to heating as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=mc\Delta T$$ where $Q$ is heat added, $m$ is mass. $c$ is the specific heat of the material in question (~1.005kJ/kgK for air at room temperature) and $\Delta T$ is the temperature change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 5x5x5 cm box will have an air mass of ~0.15 g (air density ~1.2 kg/m$^3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 1W (1 J/s) heater this gives a temperature change of &#xA;$$\Delta T = \frac{Q}{mc}= \frac{1}{0.00015*1005}= 6.6 K/s$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will suggest this is much higher than you observed with the larger box (even accounting the increased volume) which suggests there is a factor not being considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two factors leap to mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Heat losses from the box. Unless your box is well insulated it will lose heat via radiation and conduction. However for temperatures near room temp. I would expect these loses to be low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Heating the box. Inevitably if you heat the air in the box this will transfer to the box itself. The box has much larger weight so will require more energy to heat (probably 100x or more). Intuitively I would think the transfer of heat from the air to the box would be slow compared to the heating of the air. But I am not an expert in this so may be wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More likely I think the problem may be that most of the heat is going into the box initially if the heater is in contact with it, due to much much better thermal conductivity of wood than air. This is what I would look at to improve the rate of heating.&lt;/p&gt;&#xA;" OwnerUserId="153" LastEditorUserId="153" LastEditDate="2015-05-07T09:51:11.177" LastActivityDate="2015-05-07T09:51:11.177" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2737" PostTypeId="2" ParentId="2733" CreationDate="2015-05-07T12:16:38.133" Score="2" Body="&lt;p&gt;The wikipedia link in the answer you dismissed to ask this question gives you the answer: &quot;A single-phase (or two-phase derived from single phase) stator winding is possible, but in this case the direction of rotation is not defined and the machine may start in either direction unless prevented from doing so by the starting arrangements &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When AC is applied, the motor is unstable until it starts rotating, so it starts rotating (maybe a butterfly in China flaps its wings).  It could (in theory) stay stationary, but only in the manner that if you stand a broomstick on end, it could stay balanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe a better example is a piece of paper - stand a piece of paper on edge, and it falls one way or the other.  Once it starts falling one way, it keeps falling that way unless you do something like blow or push it back the other way.  The motor is unstable, it starts rotating one way or another, and once it starts rotating, it keeps going.  You can push it back the other way - if you take one of these and wedge something in the mechanism so it can't continue, it will reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want it to only rotate one way you simply build a ratchet or other clever mechanical in the mechanism so if it starts backwards it immediately reverses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this sort of fishtank circulation pump:&#xA;&lt;a href=&quot;http://www.tunze.com/149.html?&amp;amp;user_tunzeprod_pi1%5Bpredid%5D=-infoxunter052&quot; rel=&quot;nofollow&quot;&gt;http://www.tunze.com/149.html?&amp;amp;user_tunzeprod_pi1%5Bpredid%5D=-infoxunter052&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, a propeller that runs backwards doesn't do what you want, but if the motor starts backwards, what happens is that rather than pushing water out the front, the rotor pulls itself forwards out of its housing.  When it does that it then hits a carefully positioned part of the housing in front of the propeller, and bounces off.  It's now turning in the opposite direction, and carries on doing so.  When it's turning in that (correct) direction, it's pushed back in the housing, clear of the thing that it bounced off, so continues to turn in that direction.&lt;/p&gt;&#xA;" OwnerUserId="902" LastActivityDate="2015-05-07T12:16:38.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2738" PostTypeId="2" ParentId="2704" CreationDate="2015-05-07T13:11:25.490" Score="2" Body="&lt;p&gt;The question is not quite internally consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table is 400mm radius, and the feet are apparently on a 292mm radius pitch circle (since they are 108mm in from the edge).  By simple trig, the fulcrum line is therefore 292cos(60) from the circle centre, which is 146 from the centre, which is 254mm from the edge (the question states this is 240mm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the legs and top are substantially rigid (ie, small deflections - they aren't bending so much that their shape in plan changes) the mass of tabletop acts at the centre,  so the moment of mass of the top is 40kg x 146mm = 5840 kgmm about the fulcrum line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that much moment will balance the table (put it on the point of toppling over the fulcrum).  So the mass we can put there is 5840/254 = 22.99kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 90mm from the edge, feet pitch circle is now 310mm radius, teh fulcrum is 155mm from the centre and 245mm from the edge, so we can carry 40x155/245 = 25.31kg, ie 2.32kg more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That was the easy way, and is already advanced in one answer (Knigge46's), but you can do the stabilising moment calc the hard way if you like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an arbitrary strip of tabletop, parallel to the fulcrum line, width delta-x.  By pythagoras the length of this strip is 2(r^2-x^2)^.5 where r is tabletop radius and x is perpendicular distance from the strip to the centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let f be the offset between table centre and fulcrum line. Let d be the areal density of the top.  Thus the moment of this strip about the fulcrum line is d . 2(r^2-x^2)^.5 . (x-f) delta-x, which you can summate across the full table top, or change from a summation to an integration simply by changing the delta-x to a dx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do that integration for x from -r to r, you get -5.840kgm, or 5840 kgmm, the same answer as treating the table top as a point mass at the centre&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to do the stabilising and destabilising contributions from the top you can do that by integrating over different ranges:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;stabilising, (-r &amp;lt;= x &amp;lt; f) gives -6.986kgm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;destabilising (beyond the fulcrum) (f &amp;lt; x &amp;lt;= r) gives 1.146kgm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So net effect of the top is stabilising by (-6.986 + 1.146 ) = -5.840kgm, again.&lt;/p&gt;&#xA;" OwnerUserId="902" LastActivityDate="2015-05-07T13:11:25.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2739" PostTypeId="2" ParentId="2733" CreationDate="2015-05-07T15:21:45.927" Score="2" Body="&lt;p&gt;Most low-power, single-phase synchronous motors are &lt;a href=&quot;http://en.wikipedia.org/wiki/Shaded-pole_motor&quot; rel=&quot;nofollow&quot;&gt;shaded-pole motors&lt;/a&gt;, in which a turn of copper wire is placed around some fraction of each of the poles for the main coil. This creates a phase shift in the magnetic field that causes the motor to turn in a particular direction at start-up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the rotor can be a simple squirrel cage rotor that is powered by induction, or it can be a permanent magnet. The former type is used in applications in which some &quot;slip&quot; is acceptable. The latter type is used in electric clocks for &lt;em&gt;strictly&lt;/em&gt; synchronous operation.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-05-07T15:21:45.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2740" PostTypeId="1" CreationDate="2015-05-07T15:58:10.183" Score="5" ViewCount="12203" Body="&lt;p&gt;I'm trying to make a simple Venturi tee which will have air flow through it while sucking some water from down below (like paint sprayers do).  I want the device to pull the water to a height of 1 meter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means that I need 0.1 atmospheres (10&lt;sup&gt;4&lt;/sup&gt; Pascals). The dynamic pressure equation states that $P_{a}=\frac12\rho v^2$, meaning that $v=2P_a/\rho$.  The density of air is $1.22 kg/m^3$,  $P_a$ is 10&lt;sup&gt;4&lt;/sup&gt; so  $v=128\ \mathrm{m/s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I experimented a bit with my air pump and found out that with a nozzle of ~5 mm in diameter I can fill a 10 litre bag with air in ~7 sec. That should mean that the velocity is around 60 m/s.  With those results I took a hose of 16 mm in diameter, stuck my pump nozzle inside (from one side) , made a small hole juxatoposed to the nozzle from above (maybe 3-5 mm diameter) and stuck a straw in it.  When I turn the apparatus on, hovewer, the straw (placed in a bottle of water) lifts the liquid for no more than several millimeters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does my Venturi pump not work? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AA3oL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1536" LastEditorUserId="368" LastEditDate="2015-05-08T18:10:46.827" LastActivityDate="2016-02-13T17:13:24.300" Title="Why is my Venturi pump much weaker than expected?" Tags="|mechanical-engineering|pressure|fluid-mechanics|pumps|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2741" PostTypeId="2" ParentId="2727" CreationDate="2015-05-07T16:02:41.877" Score="2" Body="&lt;p&gt;This is not an answer to your question. This is an answer to what information you need to determine in order to solve a heat problem such as this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, a heat conduction problem in 2 dimensions needs 4 constraints, 2 in each direction.  I will use the example in section 2 of the PDF you linked (&lt;a href=&quot;http://geodynamics.usc.edu/~becker/teaching/557/problem_sets/problem_set_fd_2dheat.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://geodynamics.usc.edu/~becker/teaching/557/problem_sets/problem_set_fd_2dheat.pdf&lt;/a&gt;) to illustrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drew a diagram of the 2D heat conduction that is described in the problem.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/mKQ9d.jpg&quot; alt=&quot;Problem Schematic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have mentioned before that you wish to solve the problem using an explicit finite-difference method.&#xA;The first step would be to discretize the problem area into a matrix of temperatures.&#xA;For the example linked, elements 1.5km wide by 2km high were chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary conditions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Perfect insulation (zero heat flux) on the left and right sides.  In an explicit solver, all Neumann (fixed first derivative) conditions are represented by creating an extra row of &quot;fictitious&quot; temperatures which are outside the problem area.  Since the heat flux is zero, the fictitious temperatures are held equal to the temperatures at the edge of the problem zone.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fixed temperature (Dirichlet BC) at the upper surface for all time.  The surface of the problem area will always be 0&amp;deg;C.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Initial temperature at lower surface of 1300&amp;deg;C, with the central point begin held at 1500&amp;deg;C for all times &amp;gteq;0. Note I have denoted the Heaviside function as $u(t)$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once you get the boundary conditions set up correctly, explicit solvers become simple plug and chug to solve the temperature distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to find out how much heat is supplied by the constant temperature heat source, you will need to use Fourier's law with the approximate derivatives given by the explicit finite-difference method.&#xA;Assuming the heat is applied at a point $T_{i,j}$, the drawing below shows how to compute the heat flux per unit depth.&#xA;Obviously, you would need to know how far in an out of the page this problem extends to compute the heat flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FJKaG.jpg&quot; alt=&quot;Compute heat per unit depth&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-05-07T16:02:41.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2742" PostTypeId="1" AcceptedAnswerId="2743" CreationDate="2015-05-07T17:39:16.427" Score="7" ViewCount="667" Body="&lt;p&gt;I understand the difference between a #3 finish on a piece of stainless steel compared to a #4 finish being that the surface roughness is different.  Specifically, a #4 finish is classified as having an Ra value of less than 25 micro-inches where as a #3 finish has an Ra value of less than 40 micro-inches.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basis of my question is if I have two cabinets made from 304 SST, one has a #3 finish, one with a #4 finish in the same outdoor environment, which will rust faster?  Are there any other concerns between the two cabinets as far as performance of the cabinet (welds, hinges, bends, etc.)? Have there been any studies conducted? &lt;/p&gt;&#xA;" OwnerUserId="1539" LastEditorUserId="33" LastEditDate="2015-05-07T18:42:35.520" LastActivityDate="2015-05-08T18:27:14.133" Title="Will stainless steel with a #3 finish corrode faster than a #4 finish?" Tags="|materials|steel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2743" PostTypeId="2" ParentId="2742" CreationDate="2015-05-07T18:05:02.240" Score="5" Body="&lt;h3&gt;Surface Finish&lt;/h3&gt;&#xA;&lt;p&gt;I have read about a couple of corrosion concerns with stainless steel surface finish. I don't have any direct experience with differing rates of corrosion, but below are some considerations:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The smoother the finishes are less likely to collect dirt and water on the surface. Each of these will help to increase the local rate of corrosion. Some smooth (mirror-like) finishes are even considered &amp;quot;self cleaning&amp;quot; from an architectural stand point.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is related to #1 above. When finishes have a defined brush pattern, the brush lines should be oriented vertically.  This helps to allow water and dirt to flow downward instead of collecting.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h3&gt;Fasteners and Hinges&lt;/h3&gt;&#xA;&lt;p&gt;Fasteners and hinges will have some of the same issues. Any location where water and dirt will collect will likely show increased corrosion.&lt;/p&gt;&#xA;&lt;p&gt;Another consideration is galvanic corrosion. This is increased corrosion that occurs when two different types of metals are in contact. This can be common if hinges and fasters aren't also made of stainless steel.&lt;/p&gt;&#xA;&lt;h3&gt;References&lt;/h3&gt;&#xA;&lt;p&gt;A manufacturer's website that has more information about the different finishes can be found &lt;a href=&quot;http://www.aalco.co.uk/datasheets/Stainless-Steel-Surface-Finishes_135.ashx&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In aggressive environments, such as in the nuclear or offshore industries, a correctly polished stainless steel surface has a better resistance to corrosion than a surface that is roughly or badly polished.  A smooth surface is less susceptible to an accumulation of deposits and stainless, which often become focal points for localised corrosion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-05-07T19:40:40.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2744" PostTypeId="2" ParentId="2742" CreationDate="2015-05-07T19:32:03.773" Score="4" Body="&lt;p&gt;In addition to hazzey's answer I'd just like to add that &lt;strong&gt;passivation&lt;/strong&gt; of stainless is very important to reduce the risk of corrosion, especially in welded products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically the steel is cleaned then dipped into various chemical baths following welding. This allows the steel to restore its chromium-oxide rich &quot;passive&quot; layer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the application it may be possible to remove surface contaminants from welds with a stainless wire brush rather than using the chemicals but this will affect the surface finish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the pickling chemicals used in the passivation process can dull a very shiny finish. The #3 and #4 brushed finishes may be affected so it would be worth consulting the passivation company before going ahead with it. For high shine finishes it is worth considering electro-polishing.&lt;/p&gt;&#xA;" OwnerUserId="1541" LastEditorUserId="368" LastEditDate="2015-05-08T18:27:14.133" LastActivityDate="2015-05-08T18:27:14.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2745" PostTypeId="2" ParentId="2725" CreationDate="2015-05-07T23:25:57.723" Score="5" Body="&lt;p&gt;Solid particle settling time in air depends mainly on the size of the particle. Different forces become significant depending on what size range you're talking about, so it's hard to give an answer that's both concise and accurate.&lt;/p&gt;&#xA;&lt;p&gt;I'll do my best to synthesize the important points rather than parrot a reference; that said, where practical applications in the field of air quality are concerned, the text I recommend is &lt;a href=&quot;http://www.waveland.com/browse.php?t=380&amp;amp;pgtitle=Air%20Pollution%20Control%20by%20Cooper-Alley&quot; rel=&quot;noreferrer&quot;&gt;Air Pollution Control by Cooper &amp;amp; Alley&lt;/a&gt;. In particular, I'm going to pull many of the details for this answer from Section 3.3: Particulate Behavior in Fluids.&lt;/p&gt;&#xA;&lt;h2&gt;Gravitational Settling Overview&lt;/h2&gt;&#xA;&lt;p&gt;Dust doesn't behave like &lt;a href=&quot;http://en.wikipedia.org/wiki/Galileo%27s_Leaning_Tower_of_Pisa_experiment&quot; rel=&quot;noreferrer&quot;&gt;Galileo's bocce balls&lt;/a&gt;; small particles of different sizes fall at different rates. &lt;strong&gt;For solid particles, variation in settling velocity is due mainly to the influence of drag forces.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;You might expect that Brownian motion would &amp;quot;juggle&amp;quot; very small particles around, keeping them from settling. Sufficiently small dust particles &lt;em&gt;can&lt;/em&gt; remain entrained indefinitely but, practically speaking, that has more to do with the air never being perfectly still than it does with Brownian motion. In the context of air quality, we care about Brownian motion mainly when considering impaction (e.g., on water droplets in a &lt;a href=&quot;http://www.epa.gov/ttncatc1/dir1/cs6ch2.pdf&quot; rel=&quot;noreferrer&quot;&gt;PM wet scrubber&lt;/a&gt;) or deposition (e.g., &lt;a href=&quot;http://www.flow3d.com/wp-content/uploads/2014/08/Particulate-Matter-Modeling-in-Near-Road-Vegetation-Environments.pdf&quot; rel=&quot;noreferrer&quot;&gt;on foliage near roadways&lt;/a&gt;). Neither of these mechanisms are relevant to the case of pure gravitational settling.&lt;/p&gt;&#xA;&lt;p&gt;In fact, when a solid particle gets small enough to start considering the motion of discrete air molecules, we find that it actually settles a bit &lt;em&gt;more&lt;/em&gt; quickly than &lt;a href=&quot;http://en.wikipedia.org/wiki/Stokes%27_law&quot; rel=&quot;noreferrer&quot;&gt;Stokes' law&lt;/a&gt; implies. This is when we apply the experimentally-determined &lt;a href=&quot;http://en.wikipedia.org/wiki/Cunningham_correction_factor&quot; rel=&quot;noreferrer&quot;&gt;Cunningham slip correction factor&lt;/a&gt; to reduce the Stokes drag coefficient. The correction factor in air is related to the particle diameter &lt;span class=&quot;math-container&quot;&gt;$d_p$&lt;/span&gt; and the &lt;a href=&quot;http://en.wikipedia.org/wiki/Mean_free_path#Mean_free_path_in_kinetic_theory&quot; rel=&quot;noreferrer&quot;&gt;mean free path&lt;/a&gt; &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; by:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$C = 1 + 2.0 \frac{\lambda}{d_p} \left [ 1.257 + 0.40 \exp(-0.55 \frac{d_p}{\lambda}) \right ]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;As for what &amp;quot;small enough&amp;quot; actually means, the Cooper &amp;amp; Alley text says:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For particles smaller than 1 micron, the slip correction factor is&#xA;always significant, but rapidly approaches 1.0 as particle size&#xA;increases above 5 microns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That could be justification enough to spare yourself the time or processing cycles required to calculate the correction factor when all you're concerned with are relatively large particles.&lt;/p&gt;&#xA;&lt;h2&gt;Equation of Motion&lt;/h2&gt;&#xA;&lt;p&gt;We can derive an equation of motion in one dimension as follows.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Apply Newton's second law to the particle in terms of its relative velocity in the fluid.*&#xA;&lt;span class=&quot;math-container&quot;&gt;$$m_p v_r' = F_g - F_B - F_D$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Stokes' law gives the drag force in terms of the viscosity of the fluid and the velocity and diameter of the particle; the buoyant force is equal to the weight of the displaced fluid.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$m_p v_r' = m_p g - m_{air} g - 3 \pi \mu d v_r$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Divide by the mass of the particle.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v_r' = g - \frac{m_{air}}{m_p} g - \frac{3 \pi \mu d}{m_p} v_r$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Express mass as the product of volume and density, where the volume of the particle and the volume of displaced air are the same.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v_r' = g - \frac{\rho_{air}}{\rho_p} g - \frac{3 \pi \mu d}{\rho_p V} v_r$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Using &lt;span class=&quot;math-container&quot;&gt;$V_{sphere} = \frac{1}{6} \pi d^3$&lt;/span&gt;, simplify the drag force term and move it to the left side.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v_r' + \frac{18 \mu}{\rho_p d^2} v_r = (1 - \frac{\rho_{air}}{\rho_p}) g$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;This is a linear ODE with a known coefficient (at STP) representing the following &lt;em&gt;characteristic time&lt;/em&gt; for settling particles:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\tau = \frac{\rho_p d^2}{18 \mu}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The characteristic time is a useful parameter for comparing the behavior of different systems of particles dispersed in fluids, similar to how the Reynolds number can be used to identify when different systems will have similar flow regimes. Applying the Cunningham slip correction factor gives the slip-corrected time &lt;span class=&quot;math-container&quot;&gt;$\tau' = C \tau$&lt;/span&gt; and the equation of motion that I'll use in the next section:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v_r' + \frac{v_r}{\tau'} = (1 - \frac{\rho_{air}}{\rho_p}) g$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;* The coordinate system for this example is defined such that the falling velocity is positive.&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Terminal Velocity&lt;/h2&gt;&#xA;&lt;p&gt;For a solid particle falling in air, &lt;span class=&quot;math-container&quot;&gt;$\dfrac{\rho_{air}}{\rho_p}$&lt;/span&gt; is close to zero. Under that assumption, setting &lt;span class=&quot;math-container&quot;&gt;$v_r' = 0$&lt;/span&gt; in the equation of motion gives the terminal settling velocity of the particle:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v_t = \tau' g$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Using that terminal velocity, the solution of the equation of motion can be expressed as:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{v_r}{v_t} = 1 - e^{-t \over \tau'}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;By the time &lt;span class=&quot;math-container&quot;&gt;$t = 4 \tau'$&lt;/span&gt;, the particle has already reached 98% of its terminal velocity. If you calculate the characteristic time for dust particles, you'll see that this takes only fractions of a second; dust particles spend most of their settling time falling at terminal velocity. The velocity itself varies significantly with particle diameter, but &lt;strong&gt;it can take anywhere from hours to days for &lt;a href=&quot;http://www.epa.gov/pmdesignations/faq.htm#0&quot; rel=&quot;noreferrer&quot;&gt;fine particulates&lt;/a&gt; to settle just a few meters&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Larger Dust&lt;/h2&gt;&#xA;&lt;p&gt;This is all well and good for smaller dust, but what about the bigger stuff that gets in your eyes and makes you cough? Well, bad news from Cooper &amp;amp; Alley:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;For a particle larger than 10–20 microns settling at its terminal&#xA;velocity, the Reynolds number is too high for the Stokes regime&#xA;analysis to be valid. For these larger particles, empirical means are&#xA;required to obtain the settling velocity...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&amp;quot;Empirical means&amp;quot; is a nice way of saying &lt;em&gt;figure it out yourself&lt;/em&gt; or else get used to reading charts that plot fitted curves with ugly decimal exponents to the results of previous experimentation.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-05-07T23:25:57.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2746" PostTypeId="2" ParentId="2695" CreationDate="2015-05-08T01:17:21.647" Score="6" Body="&lt;h3&gt;Yes&lt;/h3&gt;&#xA;&lt;p&gt;To answer your specific question, yes, you can use an ultrasonic sensor to measure fluid level. &lt;a href=&quot;https://engineering.stackexchange.com/a/2697/33&quot;&gt;Mahendra's answer&lt;/a&gt; describes that. You mentioned that you feel that an ultrasonic sensor is the best solution, but I wanted to add make sure that you were aware of some of the other methods that are used to measure fluid levels.&lt;/p&gt;&#xA;&lt;h3&gt;Measuring Techniques&lt;/h3&gt;&#xA;&lt;p&gt;This list of fluid level measuring techniques comes from an appropriately named website, &amp;quot;&lt;a href=&quot;http://www.sensorsmag.com/sensors/leak-level/a-dozen-ways-measure-fluid-level-and-how-they-work-1067&quot; rel=&quot;noreferrer&quot;&gt;A Dozen Ways to Measure Fluid Level and How They Work&lt;/a&gt;&amp;quot;.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sight Glass&lt;/li&gt;&#xA;&lt;li&gt;Floats&lt;/li&gt;&#xA;&lt;li&gt;Hydrostatic Devices&lt;/li&gt;&#xA;&lt;li&gt;Bubblers&lt;/li&gt;&#xA;&lt;li&gt;Load Cells&lt;/li&gt;&#xA;&lt;li&gt;Strain Gauges&lt;/li&gt;&#xA;&lt;li&gt;Magnetic Level Gauges&lt;/li&gt;&#xA;&lt;li&gt;Capacitance Transmitters&lt;/li&gt;&#xA;&lt;li&gt;Magnetostrictive Level Transmitters&lt;/li&gt;&#xA;&lt;li&gt;Ultrasonic Level Transmitters&lt;/li&gt;&#xA;&lt;li&gt;Laser Level Transmitters&lt;/li&gt;&#xA;&lt;li&gt;Radar Level Transmitters&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;There is no need to repeat the details of how each of these methods work, but this is a comprehensive list of technologies that could be considered.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-05-08T01:17:21.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2747" PostTypeId="1" CreationDate="2015-05-08T03:34:03.870" Score="4" ViewCount="196" Body="&lt;p&gt;I want to know if there are some fairly common thermoplastics that allow water to pass through (by diffusion) at a relatively high rate (on the order of days). For example, PET water bottles leak water by diffusion but it's extremely slow - it will take 100 years to let the water through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of asking my question is &quot;what are some thermoplastics with very high diffusion coefficient for water?&quot;. I thought PMMA (Acrylic) was one but in my experiment it was too slow.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="272" LastEditDate="2015-05-08T12:01:33.243" LastActivityDate="2015-05-11T15:36:22.917" Title="Thermoplastics that are permeable to water - high diffusion coefficient" Tags="|materials|experimental-physics|plastic|waterproofing|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2751" PostTypeId="2" ParentId="1763" CreationDate="2015-05-08T11:53:36.860" Score="1" Body="&lt;p&gt;I believe the simplest modification to the drawing that meets the requirements is to replace the top disk in the coupler with a standard spur gear. Replace the bottom disk with a worm gear engaged with the upper gear. This would allow the top to be rotated to the desired position, while the bottom only would rotate small amounts so you wouldn't have to worry about twisted cables.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-05-08T11:53:36.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2752" PostTypeId="1" AcceptedAnswerId="2760" CreationDate="2015-05-08T13:06:06.750" Score="0" ViewCount="2113" Body="&lt;p&gt;I am a high school student wondering if the pulley system described below will work, and if so, how to go about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I have a small system where as you pull back an object, it causes two lengths, labeled $d$ and $p$, to move forward. $d$ is dependent on many factors other than just the pulley system, but $p$ is just growing linearly as the object is pulled back. I've determined that the relationship between $d$ and $p$ is: $$p=\sqrt{A+Bd^2}-C$$&#xA;Where $A$, $B$, and $C$ are positive constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram to visualize how $p$ is connected to the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JhFO0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length $p$ is the distance from the smaller block to the edge of the base. As you pull the object back, the string is pulled which pulls the spring and increases the length $p$. My question is, is there any sort of pulley system or just general construction I can insert between the object you pull and the pulley which is connected to the smaller block such that as you pull the object back, $p$ does not increase at a linear rate, but instead increases at the exact same rate of $d$.&lt;/p&gt;&#xA;" OwnerUserId="1548" LastEditorUserId="272" LastEditDate="2015-05-08T13:18:35.603" LastActivityDate="2015-05-09T13:52:53.957" Title="How to design and build a pulley system for a given purpose?" Tags="|mechanical-engineering|pulleys|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2754" PostTypeId="2" ParentId="2234" CreationDate="2015-05-08T13:26:11.210" Score="1" Body="&lt;p&gt;As mentioned in the comments, the weight of the solar panels is not the issue as the weight can be balanced such that it contributes negligible torque to the system. Also mentioned in the comments is that the driving mechanism should be for each panel as this simplifies the mechanics and will increase the reliability and ease of maintenance of the system. So the real question is &quot;What driving and holding torque do I need for an individual panel?&quot; To answer this you need to know the wind load. How much torque would high winds for the area produce on panels of your size?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this is not a particularly easy calculation. I suggest opening a new question to ask about this (preferably after looking up the wind-speed so you can give the relevant info, including panel size) as it differs significantly from the current question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your concerns about power, most motor and worm gear applications will be used in much much higher speed systems, and thus will easily be able to handle the power requirements if they can handle the torque. As such, once you find the wind torque, you can size your worm gear(s) for that torque and then the motor for the input torque.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-05-08T13:26:11.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2755" PostTypeId="1" CreationDate="2015-05-08T19:34:36.577" Score="4" ViewCount="1972" Body="&lt;p&gt;I have a split-type inverter air conditioning unit in one room placed near the ceiling. Assume that the cooling capacity is met, and I'm using the correct CFM of fans. I would like to share this air conditioner with the adjacent rooms. My plan is to install two fans for each room to be cooled. There will be four fans in all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;: Are the following implementations correct?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I install one exhaust fan to blow cold air from the colder room near&#xA;the floor to the living room (hotter). This is because cold air&#xA;tends to settle to the floor and this makes it easier for cold air&#xA;to be moved from one room to another.&lt;/li&gt;&#xA;&lt;li&gt;I put one exhaust fan at the higher point (directly in front of&#xA;split type air con) which is pulling hot air from the other room&#xA;back to the colder room. My thinking is that in the other room which&#xA;is hotter originally, hot air rises. So putting the exhaust fan at&#xA;the higher point makes it easier to move hot air from the hot room&#xA;to the colder room.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thus, #1 and #2 complete the heat exchange cycle and cools the other room. I repeat the same implementation for the other room. What do you think, does this make sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some people suggest the reverse. This way putting the exhaust fan to pull cold air to the hotter on the top (near the ceiling) instead of near the floor. Since its easier for cold air to falls to the floor thus more efficient. Is this a better scenario?&lt;/p&gt;&#xA;" OwnerUserId="1552" LastEditorUserId="282" LastEditDate="2015-05-12T22:11:44.797" LastActivityDate="2015-05-12T22:11:44.797" Title="Where to place fans for maximum effectiveness in sharing air conditioning between rooms?" Tags="|heat-transfer|hvac|airflow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2756" PostTypeId="1" CreationDate="2015-05-08T21:06:35.387" Score="12" ViewCount="510" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jaalr.png&quot; alt=&quot;standard thermal furnace&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the standard design for a thermal furnace used in the Clauss process, which converts H&lt;sub&gt;2&lt;/sub&gt;S to SO&lt;sub&gt;2&lt;/sub&gt;.  The main problem with the furnace is that gas mixing is rather poor and results in only a 60% conversion rate.  This in turn increases downstream equipment costs to handle the impurities.  A design improving the mixing of the gases is highly sought after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H&lt;sub&gt;2&lt;/sub&gt;S and O&lt;sub&gt;2&lt;/sub&gt; is fed separately into the reactor.  The combustion reaction starts and increases temperature to about 1400 °C.  The choke point in the centre of the reactor is there to force the gases to better mix on either side of it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;What I have done thus far&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have a design modification on the injectors that allowed much greater mixing, with inspiration taken from fuel injectors in motor vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OQvKx.png&quot; alt=&quot;modified injectors&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not include the choke point in this drawing.  It was merely done to test the validity of the concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The twice-angled injectors provide horizontal as well as radial velocity to the inlet gases.  This causes a swirling effect on the fluid, improving mixing by about 60%.  Mixing is here defined as the homogeneity of the outlet product distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantages are two fold: The gas particles need to travel further due to the swirling, increasing the time they stay in the reactor.  Thus, a larger conversion is also achieved, or viewed from a different perspective, a smaller reactor is needed to achieve the same conversion as the standard unit, dramatically reducing costs.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The question&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wish to exploit certain fluid dynamics phenomena to improve upon the mixing.  Eddy formation, for example, is used in the choking section. What else can be done to improve mixing?  What features can be added/removed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Explain your proposed design in words, no need for actual modelling.&lt;br&gt;&#xA;Of course, it would help me to see the idea, but it is not necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have access to Fluent in which I simulate these designs and compare them to the standard unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am still eager to see what you can come up with.&lt;/p&gt;&#xA;" OwnerUserId="1553" LastEditorUserId="1553" LastEditDate="2015-05-09T18:32:55.537" LastActivityDate="2015-05-11T08:00:23.587" Title="Exploiting fluid dynamics for the optimal design for gas mixing in a furnace" Tags="|fluid-mechanics|chemical-engineering|finite-element-method|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2757" PostTypeId="2" ParentId="2747" CreationDate="2015-05-09T07:36:52.747" Score="1" Body="&lt;p&gt;&lt;strong&gt;Polyamide / Nylon&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our subsea work we generally avoid this in applications in contact with water/seawater as it is hygroscopic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot give you exact values as there are many polyamide grades and some are less hygroscopic than others, so you'll need to research further. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would definitely be worth considering as a potential candidate for your experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;I have made an assumption in this answer so it's best if I state it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hygroscopic materials absorb water. I have assumed that they will also let water pass through them if a pressure difference exists across the material. &lt;/p&gt;&#xA;" OwnerUserId="1541" LastEditorUserId="1541" LastEditDate="2015-05-09T07:55:02.860" LastActivityDate="2015-05-09T07:55:02.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2759" PostTypeId="2" ParentId="2684" CreationDate="2015-05-09T09:32:28.610" Score="1" Body="&lt;p&gt;This doesn't answer the &quot;how can we get the coupling off&quot; question, but explores the welding option further. It seems that welding looks like one of the main options if the coupling is seized in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the space restrictions it might be worth looking into hiring a portable orbital welding unit. This assumes that there is a pipe stub coming out of the seized coupling that could be prepared and welded to. The orbital welders are able to be used in tight spaces without much clearance behind; however, they do require careful setting of parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im not saying this is a definite solution, but it's worth looking into and possibly contacting welding unit manufacturers about parameters needed and whether it would be suitable for your application.&lt;/p&gt;&#xA;" OwnerUserId="1541" LastEditorUserId="1541" LastEditDate="2016-12-19T06:11:29.810" LastActivityDate="2016-12-19T06:11:29.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2760" PostTypeId="2" ParentId="2752" CreationDate="2015-05-09T13:52:53.957" Score="2" Body="&lt;p&gt;If I understand your question correctly, then you want to know if there is some mechanism which can convert a movement of distance $d$ to a movement of distance $p$, given that you want $p=\sqrt{A+Bd^2}-C$.  If $d$ is caused by pulling on a rope, then the easiest way I can think to achieve your objective is a spring-loaded &lt;a href=&quot;https://en.wikipedia.org/wiki/Cam&quot; rel=&quot;nofollow&quot;&gt;cam&lt;/a&gt;.  You would need to relate $d$ to the angle of turn of the cam and then make the radius of the cam the appropriate function of angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this depends on you having a low requirement for range of motion. You could of course, use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Block_and_tackle&quot; rel=&quot;nofollow&quot;&gt;block and tackle&lt;/a&gt; to amplify the movement, but you will run up against size and frictional limitations.&#xA;Additionally, it may be difficult to fabricate your cam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the details of your project, but it might be worthwhile to look at the range of $d$ and $p$ that you expect and see if you can come up with a linear relation that approximates your desired relation to the desired degree of accuracy.  It is much easier to design linear mechanisms and you may find that your equation is needlessly complex.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-05-09T13:52:53.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2761" PostTypeId="2" ParentId="2755" CreationDate="2015-05-09T15:02:06.317" Score="1" Body="&lt;p&gt;I suspect that the reverse method is more efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Efficiency is energy required to reach a certain thermostat temperature. A lower amount of energy required to reach the desired thermostat temperature means a more efficient design. &lt;/li&gt;&#xA;&lt;li&gt;Like most households, the thermostat is at head height.&lt;/li&gt;&#xA;&lt;li&gt;The direction of the fans does not impact energy used at a given airflow rate.&lt;/li&gt;&#xA;&lt;li&gt;The air-conditioned room is room A and the unconditioned room is room B&lt;/li&gt;&#xA;&lt;li&gt;Your first scenario, with the lower going from A-&gt;B and the upper going from B-&gt;A is scenario (1). The reverse is scenario (2).&lt;/li&gt;&#xA;&lt;li&gt;Room A is always uniformly at the thermostat temperature because the cool air from the air conditioning unit is at the top of the room and as the cooled air sinks it mixes evenly.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In scenario (1), Room A is at a uniform temperature and Room B has a temperature gradient. The magnitude of the temperature gradient will be related to the fan airflow. A higher airflow will create a lower gradient.&#xA;In scenario (2), Room A and Room B are both (the same) uniform temperature throughout because the falling cool air mixes throughout the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to achieve a desired thermostat temperature in Room B, scenario (1) will need a lower thermostat temperature in Room A or a higher fan airflow rate than in scenario (2) because the temperature gradient will mean the thermostat in Room B will always be hotter than in Room A for scenario (1).&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2015-05-09T15:02:06.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2762" PostTypeId="2" ParentId="2756" CreationDate="2015-05-09T18:32:15.717" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Preliminary results&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I added a conical structure before the choking point to separate the flow.  Basically cutting the fluid.  This cone is attached with 4 supports.  This configuration increased mixing by a ridiculous amount.  I achieved a near linear product distribution.  However, I have not run temperature or structural analysis on this cone to validate whether or not it can with stand the temperatures or load applied to it.  This cone was added to the standard structure.  Further analysis should be done with the cone and twice angled injectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sinusoidal wall was added to furnace to aid in Eddy formation on the boundaries.  This increased the linearity of the product distribution, but lower the conversion rate, which I do not understand at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To aid in the modelling process, a simpler reaction is used.  Benzene and oxygen is fed into the reactor at 600 Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The legend off all of the following images range from 0%(clear) to 100%(red).&#xA;All scenarios that were run, used exactly the same operating conditions and the overall length of the reactors stayed constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting conversion looks as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wkZtU.png&quot; alt=&quot;Standard&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting average conversion at the outlet was found to be 40.09%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the addition of the conical structure, the conversion increased to 43.43% and looks as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zVsOB.png&quot; alt=&quot;Cone&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major improvement in conversion was seen when two choke points were added.  The conversion that was found: 78.46%.  Which is almost double compared to the standard reactor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iFkUC.png&quot; alt=&quot;Double Choke&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next iteration, involved adding rounded features to the reactor.  This resulted in a final conversion of 78.57%, which is not a large increase by any measure.  But, it can be cheaply done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yz9au.png&quot; alt=&quot;Double Choke Fillet&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two cones where added into the double choke design so that the geometry of the cone may aid in eddy formation in the compartments.  The result was as expected, and a conversion of 85.35% was found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GbApL.png&quot; alt=&quot;Double Choke Double Cone&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The previous design was modified with rounding similar to that of a previous design.  This resulted in a conversion of 86.71%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jttj4.png&quot; alt=&quot;Double Choke Double Cone Fillet&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These experiments of mine show that there are improvements to be made to this archaic design (literately, it came from the early 90's)  by exploiting certain phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently in the process of combing the double choke, double cone, rounded design with the twice angled injectors.&lt;/p&gt;&#xA;" OwnerUserId="1553" LastEditorUserId="368" LastEditDate="2015-05-09T19:06:09.417" LastActivityDate="2015-05-09T19:06:09.417" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2763" PostTypeId="2" ParentId="2756" CreationDate="2015-05-09T23:08:22.457" Score="5" Body="&lt;p&gt;As far as I understand your question you are looking for a way to diffuse/mix two gases into each other. The process is very hard to simulate &quot;correctly&quot; because of the characteristics of the equations. However, it is quite unlikely that you will have a worse mixing than predicted because the models usually underestimate the turbulent mixing processes. Your biggest problem could be the pressure loss connected to the turbulent mixing depending on the operating pressure of your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good thing is mixing is needed in a lot of applications maybe you can get some ideas: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA investigated a lot of impingement mixing processes for their liquid-liquid combustion-chambers:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/i2RJH.jpg&quot; alt=&quot;Nasamixing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GE, Pratt, and Rolls-Royce have investigated the most efficient way to mix bypass and core-flow of their jet-engines:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WKdfT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally flying and stationary gas turbine manufactures have experimented a lot to come up with a very efficient (fast) way to mix air and fuel:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WnGFq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the idea behind all the examples is to increase the surface of the two fluids. On a molecular level the mixing/diffusion itself can only be increased by increasing the temperature. So at a given temperature mixing can only be increased by increasing the mixing-surface and let the molecular happen on a bigger scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, especially shear-layers and swirling flows which are used in the last example are very hard to simulate because the turbulence models often do not capture the physics good enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A short summary or comment to your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I wish to exploit certain fluid dynamics phenomena to improve upon the mixing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is only one basic phenomenon of mixing and that is diffusion. In order to increase the mixing while having a constant diffusion is by increasing mixing-surface-area which is done most efficiently by swirling shear layers.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-05-10T06:27:23.003" LastActivityDate="2015-05-10T06:27:23.003" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2764" PostTypeId="1" AcceptedAnswerId="2828" CreationDate="2015-05-10T07:34:37.737" Score="8" ViewCount="798" Body="&lt;p&gt;I have a Fisher and Paykel Washing Machine (model number WL80T65C, well past warranty) which got rained on (a lot) and one of the capacitors blew up. I've done an amateur replacement of the capacitor which might work but when taking the circuit apart I forgot to take a photo until half way through. I've managed to place everything except for two cables, one of which is Green/Yellow and coming from the mains. This is Australia so I presume it's a ground cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only hint I can find as to which is the ground and which is the other is the sad face that's printed on the circuit board, as shown in the photo attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DQhWF.jpg&quot; alt=&quot;Rude frowny face mocking me&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cables connect to where the blue marks are, one on the left and one on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is a frown face in a PCB a standard thing and does it tell me which is ground or are Fisher and Paykel mocking me? Is there a place where I can look for the schematics of this thing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here's a better photo of the sad face and its friend, taken in the day time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: It's become apparent that the component was ground and it didn't matter which way around I fixed the cables but I'm still curious as to what the sad face means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AjUHi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1559" LastEditorUserId="1559" LastEditDate="2015-05-13T03:09:17.633" LastActivityDate="2020-02-26T15:58:06.030" Title="What does a Sad Face on a circuit board mean?" Tags="|electrical-engineering|circuits|embedded-systems|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2766" PostTypeId="1" AcceptedAnswerId="2767" CreationDate="2015-05-10T20:28:06.510" Score="1" ViewCount="280" Body="&lt;p&gt;I was always taught that electricity flows through the path of least resistance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem that I'm having is that I've seen/done this:&#xA;Take 3 wires have one wire break off into the 2 other wire (like a fork in the road) and put electricity through the first single wire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I take a multi-meter and test the end of each of the wires that forked from the original wire. Both wires show current/energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be possible, this happens even if I use 2 wires with different conductivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ending output might differ between the wires, but both wires are getting energy, even though one of them is more resistant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain this to me?&lt;/p&gt;&#xA;" OwnerUserId="1564" LastEditorUserId="110" LastEditDate="2015-05-10T23:02:41.790" LastActivityDate="2015-05-10T23:02:41.790" Title="Path of least resistance misunderstanding" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2767" PostTypeId="2" ParentId="2766" CreationDate="2015-05-10T20:39:25.710" Score="8" Body="&lt;p&gt;Ohm's law states that current through a resistor is voltage over resistance. Resistors in parallel will see an equal voltage. This means that each resistor will get some current through them. The one with lower resistance will get more. But they will both conduct current and the total current through the circuit will be the sum of the current through each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that you can work out that formula for resistance of 2 different resistors in parallel is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{R_{tot}}  = \frac1{R_1}+\frac{1}{R_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise to get a 5 $\Omega$ resistance from 2 10 $\Omega$ resistors would be much more finicky as under your assumption even the slightest change in resistance would cut off the slightly stronger resistor.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-05-10T20:39:25.710" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2768" PostTypeId="1" AcceptedAnswerId="7764" CreationDate="2015-05-10T21:57:28.567" Score="1" ViewCount="325" Body="&lt;p&gt;This may be a trivial problem, but I'm having a tough time with conceptualization.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using a code for modeling flow out of a tank.  I'm not experienced in fluid dynamics; I'm more experienced with chemical aspects of the problem.  I want to use pressure boundary conditions (a constant pressure at the bottom of the tank and atmospheric pressure at the tank outlet).  The code is for incompressible flow and I must input the pressure values in terms of pressure/density (units of m2 s-2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is I don't understand which densities to use for each boundary condition.  At the bottom of the tank, I'm guessing that I should use the tank fluid density (or some weighted average if there is more than one kind of fluid layer in the tank).  But what should I use at the top (where the boundary condition is the pressure of the atmosphere)?  When I divide atmospheric pressure by the density of air, I get a value that is higher than the bottom boundary of the tank and flow goes the wrong way (down, not up to the outlet).  When I divide atmospheric pressure by the same density that I use for the bottom boundary (the density of the fluid in the tank), my output shows pressure values that are lower than atmospheric.  It seems like I should get this, but I don't.  Any help would be appreciated, but unfortunately, I'm not allowed to post the code.&lt;/p&gt;&#xA;" OwnerUserId="1416" LastActivityDate="2016-03-04T14:45:56.840" Title="Conceptual problem: how to define pressure boundary conditions for stratified fluid in units of pressure/density?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2769" PostTypeId="2" ParentId="2740" CreationDate="2015-05-11T00:31:04.133" Score="7" Body="&lt;p&gt;The problem is that the airspeed drops as soon as the airflow leaves the nozzle. It is not the same as a water-jet expelling into air (in that case the density of the water is much greater than that of air so the water holds its velocity much better). The low-pressure air will suck in air from around it in the wider pipe and thus become turbulent around the outside edges of the cylindrical stream within a very short distance as it leaves the nozzle, and slow right down. The airflow will still be fast in the center of the cylindrical stream, but not at the edges where your intake is. You need to position the intake of your straw &lt;em&gt;inside&lt;/em&gt; the nozzle, at its narrowest point.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/kkGdL.jpg&quot; alt=&quot;Modified version of the diagram from the original post&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have drawn a zoomed-in part of your diagram and shown an extension to the nozzle in blue, also an extension to your intake in orange so that it opens inside the fastest part of the airflow. If you can't modify it in this way, at the very least place the intake right up as close as possible to the exit-point of the nozzle, but ideally you want the intake to come in just before the air exits the nozzle. Additionally, make the 'straw' and intake about half the diameter of the nozzle exit, as it will lift the water much more easily.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-05-11T00:31:04.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2771" PostTypeId="1" AcceptedAnswerId="2773" CreationDate="2015-05-11T07:32:39.047" Score="4" ViewCount="18596" Body="&lt;p&gt;I've recently discovered a very awesome software tool for the synthesis and analysis of mechanisms.  The tool is called GIM, from Faculty of Engineering in Bilbao, Department of Mechanical Engineering, University of the Basque Country (UPV/EHU)  &quot;The software is intended for educational purposes, in particular destined to the field of kinematic analysis, motion simulation and synthesis of planar mechanisms&quot; as well as for the &quot;static analysis of mechanical structures.&quot;  I'm doing some personal dynamic art work with six bar linkages (not for profit) so I think I'm okay here.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ehu.eus/compmech/software/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link to download GIM software&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=0xMHisJR384&quot; rel=&quot;nofollow noreferrer&quot;&gt;Four Bar Tutorial Video&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=2DkrrYgSLMc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Synthesis Sample Video&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I will say the interface wasn't totally intuitive.. after a few minutes in the users guide I was able to figure out the double click thing.  (Duh!  Read the messages on the bottom bar.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do have one issue that is driving me crazy.  I'm actually trying to easily compare different design alternatives. I'd just like to display coupler point output path, but I can't figure out how to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/09GIG.jpg&quot; alt=&quot;four bar mechanism with motion paths displayed&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anybody else used this tool?  Anybody know the secret to just showing a single output path?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and yes I'm aware of the &lt;a href=&quot;http://www.saltire.com/HTML5/Mechanisms/4%20Bar%20Linkage.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.saltire.com software&lt;/a&gt; and that would work for displaying output in a nice format, but I won't have the synthesis and full analysis tools available. &lt;/p&gt;&#xA;" OwnerUserId="1567" LastEditorUserId="6767" LastEditDate="2019-05-26T20:52:17.100" LastActivityDate="2019-05-26T20:52:17.100" Title="GIM Mechanism Software (Kinematics / Analysis / Synthesis)" Tags="|mechanical-engineering|software|linkage|kinematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2772" PostTypeId="2" ParentId="2756" CreationDate="2015-05-11T07:51:48.267" Score="1" Body="&lt;p&gt;rul30 put it best:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Basically, the idea behind all the examples is to increase the surface of the two fluids. On a molecular level the mixing/diffusion itself can only be increased by increasing the temperature. So at a given temperature mixing can only be increased by increasing the mixing-surface and let the molecular happen on a bigger scale.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One way to do this are &lt;a href=&quot;http://en.wikipedia.org/wiki/Static_mixer&quot; rel=&quot;nofollow noreferrer&quot;&gt;static mixers&lt;/a&gt;. A static mixer is a series of blades, typically helical, that are inserted in a pipe. The blades will 'cut' and turn the fluid so that different volume elements are in contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will, however, not be able to model one in 2D. There are different types - Helical:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/mZGK5.jpg&quot; alt=&quot;Helical statix mixer&quot;&gt;&lt;br&gt;&#xA;X Element:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/1Xa5K.jpg&quot; alt=&quot;X Element Statix mixer&quot;&gt;&lt;br&gt;&#xA;and others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picking the right mixer is probably a science in itself, at first glance I only found articles about their use in adhesives production - these mixers are often used in liquid-liquid and liquid-gas application. &lt;a href=&quot;http://www.stamixco-usa.com/mixer-selection&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page advises yet another type for gas-gas application&lt;/a&gt;, a &lt;a href=&quot;http://www.stamixco-usa.com/corrugated-plate&quot; rel=&quot;nofollow noreferrer&quot;&gt;corrugated plate&lt;/a&gt; type. Static mixers for gas-gas mixing are also used in exhaust gas treatment, this could be one avenue of further research. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pictures: &lt;a href=&quot;http://schumacher-verfahrenstechnik.de/Statische-Mischer.34.0.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schumacher Verfahrenstechnik&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2015-05-11T08:00:23.587" LastActivityDate="2015-05-11T08:00:23.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2773" PostTypeId="2" ParentId="2771" CreationDate="2015-05-11T08:42:39.810" Score="3" Body="&lt;p&gt;Okay, it took me way too long to figure this out.  I'm thinking this is just not a very intuitive interface.  Here's the fix for anybody else who makes it here. Click #1, Click #2 (carefully!), then check the box at #3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sAwHL.jpg&quot; alt=&quot;4 bar display output path&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1567" LastActivityDate="2015-05-11T08:42:39.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2774" PostTypeId="1" CreationDate="2015-05-11T11:38:11.963" Score="8" ViewCount="15447" Body="&lt;p&gt;I have cropping/cutting wedges with a thickness of 4 mm cutting either side of a round mild steel bar (17.5 mm). Imagine two wedges being driven together with a round mild steel bar in the middle (similar to the design of bolt cutters)&#xA;These wedges are operated using 2 hydraulic cylinders (6&quot; diameter, 2 1/2 travel) (one for each cutting wedge) to force them together (against the mild steel). The cropping wedges are made from BS 970 304S12 steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine the force required to crop this round bar?  What equations/formulas are relevant?&lt;/p&gt;&#xA;" OwnerUserId="1571" LastEditorUserId="272" LastEditDate="2015-05-11T11:59:37.773" LastActivityDate="2018-03-06T07:05:57.883" Title="Force required to crop/cut a mild steel bar" Tags="|mechanical-engineering|steel|hydraulics|cutting|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2775" PostTypeId="2" ParentId="2684" CreationDate="2015-05-11T14:32:07.300" Score="4" Body="&lt;p&gt;The way of removing couplings in the worst case scenario, where all other alternatives has failed, is to physically destroy the coupling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut/grind 2 slots the entire length of the coupling on opposite sides, usually one on top and one on the bottom. The slots need to be deep enough that it is at least flush with the pipe its screwed onto, pretty much as close to the threads as possible. The closer it is to the threads the easier this will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill the top slot with anti-seize or some good EP lubricant. Get a sharp steel wedge (also covered with anti-seize) and hammer it into the slot, work slowly from one end of the slot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the coupling is cast iron it will likely crack right off. If its mild steel or some other soft material it will slowly open up, and you can probably unscrew it after it loosens up only a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the steel wedge, grinding down a cold chisel can work too, just taper the chisel a bit more to make it easier to hammer into the cut groove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The anti seize prevents the steel wedge from galling against the coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cut opposite the split cut is to give the coupling a place to bend.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2015-05-11T14:32:07.300" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2776" PostTypeId="2" ParentId="2747" CreationDate="2015-05-11T15:36:22.917" Score="1" Body="&lt;p&gt;I recall (I think) there is a company called Poron that makes porous plastics with various pore sizes. You can &quot;dial in&quot; the diffusion rate you want by selecting the right thickness of material and pore size.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2015-05-11T15:36:22.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2777" PostTypeId="1" CreationDate="2015-05-11T16:37:46.200" Score="19" ViewCount="54939" Body="&lt;p&gt;To head off the &quot;why not use a bigger motor&quot; question, I'm a high school mentor for an FTC (First Tech Challenge) robotics team, with a bit of a soft mechanical / hardware background.  They're somewhat limited in what they're allowed to use, motors in particular.  Up to 8x 1.5 ft-lb 12v DC motors, max.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This year they came across the problem of needing more lifting power, which gearing could help with, but to get the lift they needed cost a lot of speed, which in a very short (~2 minutes) competition, was a costly trade off when there was a need to go up and down frequently.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibilities I've considered (and haven't had time to mess with yet, but I'm trying to research and get input from engineers better than myself..)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Having two motors directly driving the same gear on the shaft seems like the first obvious answer, but error prone (slipping out of sync and grinding potentially.)&lt;/li&gt;&#xA;&lt;li&gt;Two motors driving two separate pulleys on the driven shaft, which should theoretically gain power without the gear jamming issues.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm planning on having the kids experiment and test, but as my ME experience is extremely weak, I'm soliciting input. &lt;/p&gt;&#xA;" OwnerUserId="1573" LastEditorUserId="272" LastEditDate="2015-05-11T16:51:07.040" LastActivityDate="2024-02-29T16:20:26.197" Title="Effective ways to drive a single shaft with multiple motors?" Tags="|mechanical-engineering|gears|robotics|motors|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2778" PostTypeId="2" ParentId="2777" CreationDate="2015-05-11T16:49:54.793" Score="22" Body="&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Differential_(mechanical_device)&quot; rel=&quot;noreferrer&quot;&gt;differential&lt;/a&gt; is a mechanical device designed to do exactly what you propose.  It will allow the two motors to spin at slightly different rates while still combining the power.  The most common use of a differential is in the  drivetrain of an automobile in which it is used to power both wheels from one engine while still allowing the wheels to spin at different rates, i.e. acting in the reverse sense from the application you are looking for.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Differential_(Manual_of_Driving_and_Maintenance).jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xcYJJm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2016-01-15T13:48:13.997" LastActivityDate="2016-01-15T13:48:13.997" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2779" PostTypeId="2" ParentId="2777" CreationDate="2015-05-11T18:56:14.500" Score="23" Body="&lt;p&gt;While a mechanical differential does what you ask, you don't really need it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can connect two identical electric motors together on the same shaft.  There is no &quot;slipping out of sync&quot; because there isn't a issue of sync in the first place.  Drive the two motors the same and both will develop close to the same torque.  One will have slightly more torque than the other, but the two torques still add.  There is no harm in a little mismatch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the worst case, you drive one motor at full and the other not at all.  The undriven motor will just add a little frictional load to the shaft as long as its electrical connections are left open.  As long as you drive each motor hard enough so that it would rotate at that shaft speed with no load, it's not going to slow anything down.  You'd have to drive the two motors significantly differently for one to add torque to the shaft while the other is adding drag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes that you don't have a closed loop controller that is trying to regulate motor speed.  As long as your controller is only changing the effective equivalent voltage the motor sees, directly connecting the two motors is fine.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-05-11T18:56:14.500" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2780" PostTypeId="1" AcceptedAnswerId="2785" CreationDate="2015-05-11T20:59:23.277" Score="4" ViewCount="23377" Body="&lt;p&gt;I want to affix a cantilever to wall. I will support the other end of the cantilever with a strut made of wood, that attaches to some point on the wall below the cantilever, as shown in this sketch (click for full resolution):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z3Cj2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z3Cj2m.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At what angle will the strut provide the greatest vertical strength/support for the free end of the cantilever?&lt;/p&gt;&#xA;" OwnerUserId="1574" LastEditorUserId="1574" LastEditDate="2015-05-12T20:12:15.857" LastActivityDate="2023-10-18T07:24:19.603" Title="What angle - for a strut - provides the greatest vertical strength/support for a cantilever?" Tags="|structures|statics|stresses|wood|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2781" PostTypeId="5" CreationDate="2015-05-11T21:13:36.033" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-05-11T21:13:36.033" LastActivityDate="2015-05-11T21:13:36.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2782" PostTypeId="4" CreationDate="2015-05-11T21:13:36.033" Score="0" Body="Use this tag for questions about providing technical support for engineered goods and services." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-05-11T21:13:36.033" LastActivityDate="2015-05-11T21:13:36.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2783" PostTypeId="1" CreationDate="2015-05-11T22:46:13.303" Score="2" ViewCount="116" Body="&lt;p&gt;What order of magnitude would common strains take on within concrete bridges? I am looking for types of strains that develop over longer periods of time as well as strains due to dynamic loading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia states the following on their &lt;a href=&quot;http://en.wikipedia.org/wiki/Creep_and_shrinkage_of_concrete&quot; rel=&quot;nofollow&quot;&gt;page on concrete creep and shrinkage&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Changes of pore water content due to drying or wetting processes cause significant volume changes of concrete in load-free specimens. They are called the shrinkage (typically causing strains between 0.0002 and 0.0005, and in low strength concretes even 0.0012) or swelling (&amp;lt; 0.00005 in normal concretes, &amp;lt; 0.00020 in high strength concretes).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suppose this is a type of strain that develops over longer periods of time. How about strains due to earthquakes?&lt;/p&gt;&#xA;" OwnerUserId="1576" LastEditorUserId="272" LastEditDate="2015-05-12T01:14:00.260" LastActivityDate="2016-09-17T20:43:52.743" Title="Typical strains in concrete bridges" Tags="|civil-engineering|structural-engineering|reinforced-concrete|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2784" PostTypeId="1" CreationDate="2015-05-11T23:30:06.367" Score="3" ViewCount="282" Body="&lt;p&gt;I am working on a project where I need to determine the density of a balsa wood strip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the length and height of the strip. The two measures left to take are the weight and the thickness of the strip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To achieve this, I am using an Arduino Uno. The weight measure must be accurate to 1 g; I achieved this with an &lt;a href=&quot;http://www.dfrobot.com/image/data/SEN0160/hx711_english.pdf&quot; rel=&quot;nofollow&quot;&gt;HX711&lt;/a&gt;. In the thickness measurement I have problems, because it has to be accurate to 1 mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strips are between 17 cm thick and 1 cm thick. I have tried using an IR sensor, ultrasonic sensor and laser sensor but all of them give me bad readings. This is because the environment where the machine has to work is in a balsa factory, where there is noise and dust. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I get this measurement as accurate as I need?&lt;/p&gt;&#xA;" OwnerUserId="1578" LastEditorUserId="368" LastEditDate="2015-05-12T16:28:35.167" LastActivityDate="2015-05-23T16:45:05.460" Title="Measure the distance to a balsa wood strip with 1 mm accuracy" Tags="|mechanical-engineering|measurements|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2785" PostTypeId="2" ParentId="2780" CreationDate="2015-05-12T01:02:16.150" Score="8" Body="&lt;h2&gt;Assumptions&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The angle between the wall and the strut is $\theta$&lt;/li&gt;&#xA;&lt;li&gt;$a$ is the depth of the table top&lt;/li&gt;&#xA;&lt;li&gt;$P$ is the weight on the table top, applied at the edge furthest from the wall&lt;/li&gt;&#xA;&lt;li&gt;The strut will fail when it buckles, which implies $F_{\text{max}}=\frac{\pi^2EI}{L^2}$ where $L$, $E$ and $I$ are the length, the elastic modulus, and the moment of area, respectively, of the strut&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Analysis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The axial force on the strut will be $F=\frac{P}{\cos\theta}$.&#xA;The length of the strut will be $L=\frac{a}{\sin\theta}$.&#xA;Combining both equations with the equation for buckling we have: $(EI)_{\text{required}}=\frac{Pa^2}{\pi^2\sin^2\theta \cos\theta}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$EI$ is the stiffness of the strut.&#xA;The most efficient strut will be one for which $(EI)_{\text{required}}$ is minimized.&#xA;The lowest $(EI)_{\text{required}}$ occurs when $\sin^2\theta \cos\theta$ is maximized and that is when $\theta=\sin^{-1}\sqrt{\frac{2}{3}}$ so the most efficient angle is $\theta\approx54.7^{\circ}$&#xA;&lt;img src=&quot;https://i.stack.imgur.com/K74XZ.jpg&quot; alt=&quot;strut&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1579" LastEditorUserId="5136" LastEditDate="2016-04-29T13:01:06.597" LastActivityDate="2016-04-29T13:01:06.597" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2786" PostTypeId="2" ParentId="2784" CreationDate="2015-05-12T06:03:49.740" Score="0" Body="&lt;p&gt;If the balsa can be measured by contact based methods you can use an arm and roller or roller and plunger or various other displacement based means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If contact based means are not acceptable you should have said so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many things missing from your description which MAY be relevant. As you do not know a good solution you also cannot know how relevant various aspects are and you should maximise what you tell people if you want good answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full dimension range? mass range? rate of feed range? how handled? other ?&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-05-23T16:45:05.460" LastActivityDate="2015-05-23T16:45:05.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2787" PostTypeId="1" AcceptedAnswerId="2791" CreationDate="2015-05-12T06:54:03.450" Score="1" ViewCount="4245" Body="&lt;p&gt;If you look at the man-made objects around you, you will see that almost all of them are either rectangular, or parallel and perpendicular. There are a few circles here and there, but the overwhelming trend in basically everything humans make now is the rectangle. This is true from the computer user interface that you're looking at now and the typography in it all the way up to the shape of building blocks in your city, and if your city is completely planned, to the shape of large sections of it as well, and the fields around it. Even some administrative borders are also rectangular, which seems like the only geometric trend with these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product and graphic designers apparently only follow this trend and make everything rectangular because that's what it looked like when they started to work with a particular medium. They didn't probably set the trend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could argue that rectangles are easier to pack together, but triangles and hexagons are just as easy to pack, and there are numerous other shapes, polygonic or curved, that pack perfectly well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of rectangles even applies to the third dimension and there most things are box like, our fridges, houses etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a list were to be prepared of the primary motivations for such rectangular practise, then what would top that list ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the question also applies to the relevance of such motivations in the current age of 3D printing, and modular and adaptive architecture. &lt;/p&gt;&#xA;" OwnerUserId="1581" LastEditorUserId="1663" LastEditDate="2015-05-19T12:32:19.280" LastActivityDate="2015-05-19T12:32:19.280" Title="Why are so many things around us rectilinear?" Tags="|mechanical-engineering|structural-engineering|design|engineering-history|geometry|" AnswerCount="2" CommentCount="4" ClosedDate="2015-05-12T13:55:40.950" ContentLicense="CC BY-SA 3.0" />
  <row Id="2789" PostTypeId="2" ParentId="2784" CreationDate="2015-05-12T08:42:02.080" Score="5" Body="&lt;p&gt;The thickness is pretty easy.  I'd add a rotary encoder on a cam shape, push the cut balsa under the cam till you see max reading.  You'll have to create a lookup table for angle vs observed height against a standard. Cam should return to zero via gravity (its a law!) but you can add a torsion spring if necessary.  You will have to choose materials, finishes and thicknesses (weights) so as not to dent the balsa wood.  Travel is from left --&gt; right. Obviously you can NEVER back up the travel of the balsa wood in the sketch below.  That will immediately cause a dent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9DmP9.png&quot; alt=&quot;thickness gauge&quot;&gt;&#xA;Dirty factory environment, the weight is much more difficult.  What about a postage type of scale?  something with an accurate strain gauge measuring system inside?  I'm thinking a scale where the &quot;table&quot; is either a metal mesh or metal mesh with bearings so the balsa can ride over it. Mesh because its less likely to be affected by dirt and accumulated crud.  Bearings if you need it.  Obviously you will have to pick length that works, so automation doesn't adversly affect scale performance.  Pick up item via 'fork' that doesn't interfere with scale table (motion for fork lifting device via planar 4 bar mechanism?) &lt;/p&gt;&#xA;" OwnerUserId="1567" LastEditorUserId="1567" LastEditDate="2015-05-12T09:35:29.713" LastActivityDate="2015-05-12T09:35:29.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2790" PostTypeId="2" ParentId="2787" CreationDate="2015-05-12T08:43:24.823" Score="6" Body="&lt;p&gt;In ancient times &lt;a href=&quot;https://en.wikipedia.org/wiki/Roundhouse_%28dwelling%29&quot;&gt;roundhouses&lt;/a&gt; were constructed and are still constructed in predominantly lesser developed countries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One issue with roundhouses is, once constructed they can't be extended laterally and retain the same shape. Rectangular buildings can be easily extended and most of the original structure can remain - knock down one wall and extend while the other three walls remain unaffected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of usable floor space the corners of triangles usually become wasted or unusable space, whereas the corners of rectangular floored rooms can be used to their maximum potential easily. Also for the same amount of floor space provided by a rectangular room/building would require a larger triangular room/building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constructing rectangular rooms or buildings is easy because the tools required to ensure the corners are right angles are easily made and readily available - such a builders squares or set squares.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The construction of a structures with more sides than a rectangle or square requires beams and other materials to be cut at specific angles for which tools need to be made; this adds to the cost of construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, by using more than four walls, more wall materials are required; which adds even more cost. Also, more walls means more time required to construct which means more cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Circular rooms or buildings are more difficult to make because systems have to be devised to ensure true circles are constructed and such structures are more time consuming to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to structural analysis prior to construction, circles, ellipses and any shape with a curve are difficult to analyze compared to triangles and rectangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rectangular structures are easier and cheaper to construct that structures using different floor shapes.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-05-12T08:43:24.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2791" PostTypeId="2" ParentId="2787" CreationDate="2015-05-12T11:11:40.613" Score="4" Body="&lt;h3&gt;Buildings&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The primary reason is gravity. When humans first started building things structures that were perpendicular to the ground stayed up more. Stacking rocks for example, the straighter you could get it the more stable it would be, therefore the more 'square' the rocks you used, the easier it would be to get it straight. This has not changed, while it's possible to build walls at an angle, these need to be stronger and have more support. Actually if you build a long enough 'rectangular' structure, the top wouldn't be flat, it would bend with the curvature of the Earth. However, at scales we deal with on a day-to-day basis, the easiest shape to build with is rectangular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triangular structures can also be strong, but you get less usable space inside the structure for the same amount of material used.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Human devices&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Gravity comes into it again, though indirectly. Also geometry plays a big role in what shape our human devices are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to gravity, there are lots of things to see on the surface of Earth, such as trees, rocks, wild animals, food, etc. but not all that much up in the air and downwards there is just the ground. As we evolved, it was more important to see things to the left and right than up and down, such as predators or potential food. So we evolved two eyes, arranged horizontally, giving us a better field of vision in the horizontal plane than the vertical one. Our arms and hands also evolved to allow us to manipulate things horizontally more than vertically. This means if you're looking at something or interacting with it (such as your monitor and keyboard) it's much easier to do so in a horizontal direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An arc shape (similar to the shape a car windscreen wiper makes) might actually be more ideal than a rectangle for interacting with, but it is more difficult to manufacture, so ease of manufacturing also comes into it. We do have curved interfaces, but generally when we need to pack lots of things into a small yet accessible space, such as a car dashboard, or an air-traffic-controller's desk, where function is more important than simplicity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rectangles are much less universal in smaller human devices. Circles can also be simple to make, for example mugs/vases/bowls on a potters wheel, table-legs on a lathe. This is where geometry comes in. If you want to enclose something, for example hold tea in a mug, using the same amount of material you can hold more with a circular shape than a square one. However if the purpose is to spread things out, to make them accessible (for example words on a page or items on a table) moving them away from each other at right angles will get them further apart for a shorter move-distance. So the function plays a much bigger role than physical viability for smaller items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Design for aesthetics does of course also affect things, but they need to remain functional otherwise they would become useless, so this limits what can be done in this regard and guides the shape of things around us.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-05-12T11:11:40.613" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2792" PostTypeId="1" AcceptedAnswerId="2793" CreationDate="2015-05-12T11:45:01.943" Score="4" ViewCount="1660" Body="&lt;p&gt;I'm trying to get my bearings (a somewhat heavy-handed expression) on pump design and I'm finding in the various books I've procured on the subject that there is very little consideration given to the output pressure developed by a pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there isn't much detail to this question, I apologize; I'm still in the far-below-basic level of understanding. Obviously, head is a key parameter of pump design, but my concerns are with output pressure. The reason being, I'm considering for example rocket engine design where the key parameters of pump requirements are chamber pressure and flow rate. I understand that flow rate varies with head by the characteristic Q-H curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how does pump head vary with or affect the output pressure developed by a pump? Are the two concepts related, or are they independent of one another? If they are related, how would I go about figuring out the required head parameters from the pressure and flow requirements?&lt;/p&gt;&#xA;" OwnerUserId="1229" LastEditorUserId="41" LastEditDate="2015-05-12T12:34:48.447" LastActivityDate="2015-05-12T12:34:48.447" Title="Output pressure of centrifugal pump" Tags="|mechanical-engineering|fluid-mechanics|pumps|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2793" PostTypeId="2" ParentId="2792" CreationDate="2015-05-12T12:31:31.650" Score="7" Body="&lt;p&gt;Head and pressure are effectively the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's impossible to look at a pump and say &quot;this pump will provide 10 bar of pressure.&quot; One of my favorite aspects of fluids has always been how interconnected an entire fluid system is, and how it stabilizes itself on its own. So the pressure that's actually supplied depends on the rest of the system: the losses in the pipe, the inlet and outlet conditions of the fluid to the system, the fluid itself, and how much power you're supplying to the pump. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully you've come across the Bernoulli equation somewhere in your reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P+{\rho}gz+{\frac12}{\rho}v^2=constant$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Remember that this can only be used for incompressible flows, but since we're talking about pumps and probably only common liquids at relatively low speeds, the incompressible assumption is fine. It only becomes an issue when $Mach&amp;gt;0.3$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the form above, we see that the terms together will output a constant in units of pressure. However, we can rewrite to be in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P}{\rho g}+z+\frac{v^2}{2g}=constant$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the constant has units of length (more specifically, units of length of a particular fluid, e.g. mmHg), and it's referred to as the head. But this term is fundamentally no different from the constant we get in the first equation, it's just expressed differently. $H=P_{total}/{\rho g}$, but because this is an imcompressible flow, the density is constant, and we know that $g$ won't change significantly unless we're moving the fluid over a massive height differential. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-05-12T12:31:31.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2794" PostTypeId="1" CreationDate="2015-05-12T12:37:55.880" Score="4" ViewCount="417" Body="&lt;p&gt;I want to experiment with geothermal cooling, and have an r-10 insulated workshop which I'd like to cool during the worst of the summer heat this year. The workshop is about 400 square feet (37 m&lt;sup&gt;2&lt;/sup&gt;), and 11' (3.3 m) ceilings, with a concrete floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep costs down, I'm not interested in investing in a heat pump.  I'd like to try laying long horizontal loops of pipe in the nearby ground 4–6' down (1.2–1.8 m), run water through them, and then through a radiator in the shop I can blow air through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ground around here is the typical 55 °F (13 °C) average temperature, and in this area is mostly heavy dense clay.  The water table will be many feet below the loops in the area I plan to lay them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I could do this experimentally, lay an arbitrary amount of pipe, and take measurements (inflow/outflow temp difference and flow rate), I'm wondering what equations and engineering knowledge I should be looking at to give me rough back-of-the-napkin numbers on how much pipe I'm looking at to produce a certain temperature differential in the shop.  &lt;/p&gt;&#xA;" OwnerUserId="784" LastEditorUserId="368" LastEditDate="2015-05-12T16:18:17.500" LastActivityDate="2015-06-28T08:57:31.077" Title="How much pipe would I need in the ground to cool an outbuilding?" Tags="|cooling|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2796" PostTypeId="1" AcceptedAnswerId="2831" CreationDate="2015-05-12T13:18:28.760" Score="4" ViewCount="108" Body="&lt;p&gt;I am working with a very simple 3-DOF damped LTI spring-mass system. As an exercise I am altering two parameters simultaneously (stiffness and damping coefficients connecting mass 1 to mass 2) to see which combination minimizes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Settling time&lt;/li&gt;&#xA;&lt;li&gt;Overshoot&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The objective is to determine which parameter has the greater effect on the performance, either the stiffness or the damping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: How can I determine if changing stiffness or damping is having more of an effect on the aforementioned criteria ? Is there a way to quantify how they contribute to the performance ?&lt;/p&gt;&#xA;" OwnerUserId="1174" LastEditorUserId="68" LastEditDate="2015-05-16T02:40:44.440" LastActivityDate="2016-07-17T03:25:13.870" Title="Dominant parameters in the dynamical response of a linear oscillator" Tags="|dynamics|springs|vibration|stiffness|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="2797" PostTypeId="2" ParentId="2796" CreationDate="2015-05-12T15:14:02.747" Score="1" Body="&lt;p&gt;The main effect of changing the stiffness will be on the natural frequencies of the system, the effect on the damping will be secondary in my opinion. There are plenty of damping models available out there (e.g. hysteretic damping), and modelling damping is more of an art than a science, but the most basic damping model is viscous damping, where the damping coefficient is proportional to the relative velocity of the masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an in-depth study of damping models see &lt;a href=&quot;http://www-g.eng.cam.ac.uk/dv_library/Theses/sondiponthesis.pdf&quot; rel=&quot;nofollow&quot;&gt;Damping Models for Structural Vibration&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-05-12T15:14:02.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2798" PostTypeId="1" AcceptedAnswerId="2799" CreationDate="2015-05-12T16:00:54.030" Score="4" ViewCount="3408" Body="&lt;p&gt;I want to fix a wire rope horizontally between two trees with a T-connector in the middle of the wire rope to hang a tire swing from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;tree-------wire rope------???-----------------------tree&#xA;  I                        I                         I&#xA;  I                        I                         I&#xA;  I                        I                         I&#xA;  I                        I                         I&#xA;  I                        I                         I&#xA;  I                    tire swing                    I&#xA;  I                                                  I&#xA;  I                                                  I&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I looked for a T-shaped connector but have not found a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of connection can I use to hang the tire swing from the middle of the wire rope? I need something that will not deteriorate during the use of the swing.&lt;/p&gt;&#xA;" OwnerUserId="1585" LastEditorUserId="368" LastEditDate="2015-05-12T16:06:50.283" LastActivityDate="2021-08-10T01:33:22.480" Title="How can I make a tee connection on a wire rope?" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2799" PostTypeId="2" ParentId="2798" CreationDate="2015-05-12T16:10:28.233" Score="13" Body="&lt;p&gt;What you're describing is known as a 'low angle bridle.' The problem with this configuration, where a flexible cable runs nearly horizontally and a load is applied in the middle, is that the tension in the top cable gets very big. In fact, to have the top cable be perfectly horizontal would require infinite force at each end, since its own weight is always going to make it sag a little. Just getting pretty close to horizontal requires a really large force. Because it's generally a bad idea, I don't think you'll find a piece of hardware that looks exactly like what you're imagining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if your trees are tall enough, you can make a more typical bridle (Think a Y shape) where the top wire rope goes up as well as out. In conventional rigging, we try to keep the angle between the top two legs below 90 degrees, although there are some exceptions. If you go that route, you'll want the top piece to actually be two separate pieces of wire rope. Connecting the two bridle legs to the vertical drop you'll want to use either an &lt;a href=&quot;http://en.wikipedia.org/wiki/Shackle#Pin_shackle&quot;&gt;anchor shackle&lt;/a&gt; if it's temporary, or a pear shaped link if you can swage the cable in place. (Obviously since your load moves, make sure to &lt;a href=&quot;http://en.wikipedia.org/wiki/Safety_wire#Mousing&quot;&gt;mouse&lt;/a&gt; the anchor shackle if you go that route!)&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="368" LastEditDate="2015-05-12T18:48:35.983" LastActivityDate="2015-05-12T18:48:35.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2801" PostTypeId="2" ParentId="2798" CreationDate="2015-05-12T18:58:33.800" Score="3" Body="&lt;p&gt;Why does it have to be dead horizontal?  With a bit of slack, the cross-rope can be part of the swing as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wire ropes and trees are not a good combination anyway - you'll ring-bark the trunk with a wire rope and kill the top of the tree.  Use a regular rope instead (polyprop if you don't want it rotting).  With a regular rope, I'd tie an Alpine butterfly and hang the swing off that.&lt;/p&gt;&#xA;" OwnerUserId="1590" LastActivityDate="2015-05-12T18:58:33.800" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2802" PostTypeId="1" AcceptedAnswerId="2806" CreationDate="2015-05-12T19:30:03.300" Score="9" ViewCount="6686" Body="&lt;p&gt;I don't understand how to use the information in a stepper motor datasheet to understand how much torque it can generate.  I was going to do a simple board and try to lift a small weight as an example but I realize I need to know how much torque the motor can generate. I'll probably have a little spool out there to hold the string for the weight, which probably also changes how much torque it can deliver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I see in this &lt;a href=&quot;http://www.mercurymotion.com/products/zlbjmd/st-24byj.pdf&quot; rel=&quot;noreferrer&quot;&gt;datasheet&lt;/a&gt; is &quot;in traction torque&quot; and I'm not sure what that means.   I don't see any graphs or other information that relate how much input voltage/power will give me how much torque.  What happens if I run it at 5 V instead of 12 V, for instance, or if I current-limit it to 1 mA?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just trying to learn what to look for and how to read the datasheet properly.&lt;/p&gt;&#xA;" OwnerUserId="1210" LastEditorUserId="368" LastEditDate="2015-05-12T23:42:30.987" LastActivityDate="2015-05-13T02:40:47.123" Title="Reading stepper motor datasheets to get torque and speed" Tags="|mechanical-engineering|electrical-engineering|linear-motors|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2803" PostTypeId="2" ParentId="2802" CreationDate="2015-05-12T19:45:28.733" Score="1" Body="&lt;p&gt;I would say that the information on the datasheet it incomplete, you need at least a torque vs. speed graph for the rated voltage, and whether it is being driven at half steps, quarter steps, etc... You can then scale the torque based on the voltage applied. See for example &lt;a href=&quot;http://www.wantmotor.com/ProductsView.asp?id=156&amp;amp;pid=75&amp;amp;sid=80&quot; rel=&quot;nofollow&quot;&gt;this datasheet&lt;/a&gt; which provides that sort of data.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="368" LastEditDate="2015-05-12T23:44:09.350" LastActivityDate="2015-05-12T23:44:09.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2805" PostTypeId="2" ParentId="2802" CreationDate="2015-05-12T22:02:57.440" Score="4" Body="&lt;p&gt;You are considering it from an electrical engineering perspective. Speed vs torque.  It's more a digital motor with a rotor position that can be easily controlled (read head on floppy drives).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a brushless motor and its strongest torque (traction or holding) occurs when the coils are energized.  They have fractional hp ratings (&gt;34.3 mN·m).  To do something in the real world, you typically need some gearing.  This one does: 64 5.625° steps.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I extracted the following speed torque curve from &lt;a href=&quot;http://www.allaboutcircuits.com/vol_2/chpt_13/5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stepper Motors&lt;/a&gt;, which is a fairly good resource to understand the theory.  with the rearing, I'd expect the holding torque up to 120 Hz (120 revolutions/sec).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gtguj.png&quot; alt=&quot;Stepper speed torque curve&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can do something, but not much.  I'd check out Brushless DC (BLDC) motors if you want to do something.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="368" LastEditDate="2015-05-12T23:46:42.697" LastActivityDate="2015-05-12T23:46:42.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2806" PostTypeId="2" ParentId="2802" CreationDate="2015-05-12T22:42:51.293" Score="3" Body="&lt;p&gt;@StainlessSteelRat linked to a very good resource for &lt;a href=&quot;http://www.allaboutcircuits.com/vol_2/chpt_13/5.html&quot; rel=&quot;nofollow&quot;&gt;stepper motors at All About Circuits&lt;/a&gt;, but I fear he didn't address your questions.  I'll go through your question line by line.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'll probably have a little spool out there to hold the string for the weight so that probably changes how much torque it can deliver too.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, the sentence above is wrong. The radius of the spool will not change the torque. It will change the weight you can lift, but only because $Fr=T$, where $F$ is the force available for lifting, $T$ is the torque output, and $r$ is the radius of the spool. BTW, this method works fairly well, I have done it myself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Anyway all I see in this datasheet is &quot;in traction torque&quot;, and I'm not sure what that means&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on the torque curve from All About Circuits, it is probably almost equal to the holding torque.  As you see from that curve, torque is fairly constant at the low end of motor speed.  All About Circuits also mentions that in stepper applications, the speed of the stepper motor should be gradually accelerated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What happens if I run it at 5V instead of 12V for instance. Or if I current limited it to 1mA.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you run stepper motors at a higher voltage (within rated limits), the current, and therefore holding torque, will go up.  Another way to look at it is that you can run the motor faster for the same torque.  I don't have any equations here, but increased current leads to increased force in electromagnets. If you current limit the solenoid, the holding torque will go down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last note: @am304 makes an important note about half/quarter stepping.  The torque will reduce for half/quarter stepping because the magnets in the stepper motor are acting in opposing directions, reducing the net torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry I don't have any equations, but experiments are fun right?&lt;/p&gt;&#xA;" OwnerUserId="709" LastEditorUserId="709" LastEditDate="2015-05-13T02:40:47.123" LastActivityDate="2015-05-13T02:40:47.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2807" PostTypeId="1" AcceptedAnswerId="2808" CreationDate="2015-05-12T23:40:40.827" Score="9" ViewCount="4997" Body="&lt;p&gt;The term Systems Engineer has always interested me as it generally involves many different fields of engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Systems Engineer - Interdisciplinary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition I'm familiar with is usually defined as an interdisciplinary profession of engineering, where the engineer has experience in a number of fields and uses this during the design of a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Systems Engineer - IT related&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After searching for jobs here in Australia, I've found many systems engineer positions. However almost all of them are relating to IT jobs, referring to the programming of computer systems at the system level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why is there a conflict of these two definitions?&lt;/li&gt;&#xA;&lt;li&gt;Which one was first?&lt;/li&gt;&#xA;&lt;li&gt;And is Systems Engineer no longer valid in the interdisciplinary sense?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please forgive any mistakes and assumptions I have made, I'm new and ready to learn.&lt;/p&gt;&#xA;" OwnerUserId="1594" LastEditorUserId="368" LastEditDate="2015-05-13T20:05:43.330" LastActivityDate="2015-10-02T18:39:02.950" Title="Is systems engineer an IT profession or an interdisciplinary field of engineering?" Tags="|engineering-history|terminology|systems-engineering|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2808" PostTypeId="2" ParentId="2807" CreationDate="2015-05-13T00:45:34.583" Score="10" Body="&lt;p&gt;Yes, you are correct that there are different definitions of Systems Engineer that vary by company. In fact, different business units of the same company may even use the term differently.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;http://careers.stackoverflow.com/jobs/78569/senior-systems-engineer-systems-architecture-bookingcom?searchTerm=systems%20engineer&quot;&gt;job posting on Stack Overflow Careers from Booking.com has a Senior Systems Engineer - Systems Architecture role&lt;/a&gt;. This role has responsibilities such as taking &quot;ownership of IT objectives set from business priorities and strategic IT development areas&quot;, providing &quot;technical leadership for end-to-end development and implementation&quot;, &quot;research, test, and implement prototype systems&quot;, &quot;conduct hand-over to operational technical teams for further development and systems integration&quot;, and &quot;design and architect E-commerce and business support systems&quot;.&lt;/li&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;http://careers.stackoverflow.com/jobs/77809/senior-systems-engineer-operations-sauce-labs-inc?searchTerm=systems%20engineer&quot;&gt;job posting on Stack Overflow Careers from Sauce Labs Inc has a Senior Systems Engineer, Operations role&lt;/a&gt; with responsibilities that include maintaining &quot;a close working relationship with the other engineering teams&quot;, &quot;systems administration, engineering, and design&quot;, &quot;working with real hardware&quot; (and an enumeration of networking and data storage hardware), &quot;proactive R&amp;amp;D&quot;, creating &quot;NOC runbooks, procedures, documentation, and diagrams of the environments you manage&quot;, and even helping &quot;build out new data centers around the globe&quot;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://lmig.taleo.net/careersection/lmigcorp/jobdetail.ftl?job=61414&amp;amp;lang=en&quot;&gt;Liberty Mutual, an insurance company, has a job posting for a Principle Software/Systems Engineer role&lt;/a&gt; that is &quot;responsible for dealing with the overall performance&quot; of one of their systems, functions &quot;as a technical lead in program design, coding, testing, debugging, and documentation related to automation capabilities&quot;, is &quot;directly responsible for projects or sub-projects&quot; and &quot;may lead projects&quot;, and is &quot;responsible for quality assurance review and the evaluation of new and existing software products&quot;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.utc.com/Careers/Job/Systems-Engineer-Wilson-North-Carolina/569BR&quot;&gt;UTC Aerospace Systems has a job posting for a Systems Engineer&lt;/a&gt; with responsibilities including the development of equipment installation layouts, interface control documents, aircraft electrical system schematics, and serving &quot;as a technical customer liaison engineer for the design engineering staff&quot;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.utc.com/Careers/Job/Senior-Systems-Engineer-Beverly-Massachusetts/1449BR&quot;&gt;UTC Building &amp;amp; Industrial Systems has a posting for a Senior Systems Engineer&lt;/a&gt; that is more IT-oriented, providing administration, maintaining IT policies, ensuring disaster recovery processes are &quot;efficient and operational&quot;, and participating in on-call rotation for support.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://jobs-boeing.com/washington/systems-engineering/jobid7568463-systems-engineer-2_3-jobs&quot;&gt;Boeing Systems Engineers&lt;/a&gt; apply &quot;an interdisciplinary, collaborative approach to plan, design, develop and verify a lifecycle balanced system of systems and system solutions&quot;, perform requirements development, ensure interfaces, perform analysis &quot;to optimize system architecture&quot; and for &quot;ffordability, safety, reliability, maintainability, testability, human systems integration, survivability, vulnerability, susceptibility, system security, regulatory, certification, product assurance and other specialties quality factors &quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As for why there is conflict, you'd probably have to ask an HR representative or hiring manager at these companies. I do notice a trend, though. At companies that create a product that is intensive in both hardware and software, the role of a systems engineer is an interdisciplinary engineer and has involvement across the engineering product development process. However, at companies that are software-intensive or provide a service, the term tends to be used more for network engineering or system administration roles, but it does appear that there is a multidisciplinary approach between some combination of software engineering, network engineering, system administration, database administration, or project management. In both cases, knowledge stretching across two or more domains or fields is common, but those fields may not be typically thought of as engineering fields. I can't say that this is a guarantee, but in about 10-15 minutes of searching, that's what I've noticed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first definition tends to be the one I associate with the title of Systems Engineer. In fact, there is a &lt;a href=&quot;http://sebokwiki.org/wiki/Guide_to_the_Systems_Engineering_Body_of_Knowledge_(SEBoK)&quot;&gt;Guide to the Systems Engineering Body of Knowledge&lt;/a&gt; that says that systems engineering &quot; is an interdisciplinary approach and means to enable the full life cycle of successful systems, including problem formulation, solution development and operational sustainment and use&quot;. They have a page that provides &lt;a href=&quot;http://sebokwiki.org/wiki/Systems_Engineering_(glossary)&quot;&gt;a deeper definition and short discussion, as well&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as which one comes first, I think it's also your interdisciplinary definition. At least, &lt;a href=&quot;http://en.wikipedia.org/wiki/Systems_engineering#History&quot;&gt;according to Wikipedia&lt;/a&gt;, the term &quot;systems engineering&quot; dates to the 1940s. In fact, the whole page on Wikipedia appears to generally align with the SEBOK definition and your definition.&lt;/p&gt;&#xA;" OwnerUserId="2" LastEditorUserId="2" LastEditDate="2015-05-13T01:43:37.353" LastActivityDate="2015-05-13T01:43:37.353" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2809" PostTypeId="2" ParentId="2807" CreationDate="2015-05-13T00:46:04.583" Score="7" Body="&lt;p&gt;I don't believe there is any conflict but variation in how &lt;strong&gt;Human Resource&lt;/strong&gt; choose to define system engineer positions within specific organizations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my opinion that System Engineer has a broad definition mostly related to the specific industry.  In my experience system engineer is an interdisciplinary professional of engineering as describe in your question. Depending on the industry the interdisciplinary knowledge will widely vary. But in almost all case apart from engineering skills a system engineering is expected to have some knowledge in project management and other business related functions. In my experience systems engineers in all cases work closely with customers, vendor and is the glue between other engineering disciplines. A solid understanding of &lt;a href=&quot;https://engineering.stackexchange.com/questions/2264/how-can-i-arrange-my-eco-system-to-enforce-the-principle-of-least-privilege/2267#2267&quot;&gt;product life cycle management&lt;/a&gt; is essential for a System Engineer. Following are few examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;System Engineer – Healthcare industry - Medical Devices&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;System Engineer was expected have some knowledge of biomedical, software, electrical, and mechanical engineering. More importantly the systems engineer was expected to have deep understanding of regulatory requirements, project management (Scope, Cost and Schedule Management), quality systems and medical device manufacturing system. The system engineer was the primary contact with the customer and vendors on technical issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;System Engineer – Automotive industry&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;System engineer was expected have some knowledge in electrical, mechanical and embedded software engineering. In this industry the system engineer was expected to have deep understanding of project management (Scope, Cost and Schedule Management), federal regulatory requirements, quality systems, test system and manufacturing system. System engineer oversaw all &lt;a href=&quot;https://engineering.stackexchange.com/questions/2534/my-customer-wants-to-use-my-products-to-do-something-unsafe-what-is-my-ethical/2536#2536&quot;&gt;engineering change request (ECN)&lt;/a&gt;, project design reviews and was the primary contact for both customer and vendors on technical issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;System Engineer – IT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on my limited knowledge in this field I believe the system engineer - Information Technology might be required to have some knowledge in computer hardware (networking, peripherals, etc ), computer software (operating systems, programming languages, etc), and some level project management (Scope, Cost and Schedule Management). It is most likely that the system engineer is the primary contact to the customer and vendors on technical issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally system engineers in other industries might have similar interdisciplinary responsibilities some of which &lt;a href=&quot;https://engineering.stackexchange.com/questions/2807/is-systems-engineer-an-it-profession-or-an-interdisciplinary-field-of-engineerin/2808#2808&quot;&gt;@Thomas Owens&lt;/a&gt; has captured in his response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Systems_engineering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Systems engineering&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.incose.org/AboutSE/WhatIsSE&quot; rel=&quot;nofollow noreferrer&quot;&gt;What is Systems Engineering?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.mitre.org/publications/systems-engineering-guide/systems-engineering-guide/the-evolution-of-systems&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Evolution of Systems Engineering&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pmi.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Project Management Institute - Professional Body&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pdma.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Product Development and Management Association&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-05-13T10:32:40.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2810" PostTypeId="2" ParentId="2807" CreationDate="2015-05-13T00:51:57.130" Score="1" Body="&lt;p&gt;The unfortunate thing about the words engineering and engineer is that there are no universal uniform abiding standards for when they should be used and to whom they should apply. Many professional engineering bodies would like the term engineer to be reserved for people who have university qualifications in engineering. It is partly why in some parts of the world professional (university qualified) engineers need to sit and pass registration examinations for their field of engineering before they can work as engineers.&lt;/p&gt;&#xA;&lt;p&gt;The word &lt;a href=&quot;http://www.oxforddictionaries.com/definition/english/engineer&quot; rel=&quot;nofollow noreferrer&quot;&gt;engineer&lt;/a&gt; derives from the Medieval Latin &lt;em&gt;ingeniare&lt;/em&gt; 'contrive, devise'. The Oxford English Dictionary has the following definitions for Engineer:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A person who designs, builds, or maintains engines, machines, or structures.&lt;/p&gt;&#xA;&lt;p&gt;A person qualified in a branch of engineering, especially as a professional: an aeronautical engineer&lt;/p&gt;&#xA;&lt;p&gt;A person who controls an engine, especially on an aircraft or ship.&lt;/p&gt;&#xA;&lt;p&gt;North American - A train driver.&lt;/p&gt;&#xA;&lt;p&gt;A skilful contriver or originator of something: the prime engineer of the approach&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So in addition to train drivers who call themselves engineers there are also automotive mechanics (people who repair and service cars and trucks) who do likewise and sound recording specialists who are called sound engineers.&lt;/p&gt;&#xA;&lt;p&gt;So the fact that there are two different job roles that use the title  Systems Engineer does occur, you just have to be more careful of what is required when you see such jobs advertised.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-05-13T00:51:57.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2811" PostTypeId="2" ParentId="2807" CreationDate="2015-05-13T01:23:34.933" Score="6" Body="&lt;p&gt;Systems Engineering pre-dates IT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classical Systems Engineering has roots in aerospace industry (for better, or for worse).  Projects were were getting multidisciplinary, and complex, and required multiple contractors to complete.  So appeared a need for a kind of engineer to keep track of various aspects (such as weight, for example) on a relatively high level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.incose.org/&quot;&gt;International Council on Systems Engineering&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;NASA Systems Engineering Handbook (&lt;a href=&quot;http://web.stanford.edu/class/cee243/NASASE.pdf&quot;&gt;1995 edition&lt;/a&gt;, &lt;a href=&quot;http://www.acq.osd.mil/se/docs/NASA-SP-2007-6105-Rev-1-Final-31Dec2007.pdf&quot;&gt;2007 edition&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="63" LastActivityDate="2015-05-13T01:23:34.933" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2812" PostTypeId="1" AcceptedAnswerId="2813" CreationDate="2015-05-13T04:59:24.827" Score="2" ViewCount="2585" Body="&lt;p&gt;I wish to calculate air leakage in a piston cylinder arrangement. One side is pressurized and the other is at atmospheric pressure. All steel construction. No piston rings or lubrication. The piston oscillates at about 50 to 80 times per sec, amplitude 5 mm, and is used as an indentation marking tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the piston diameter, cylinder diameter, piston length, air pressure and roughness. I need to calculate the leakage in liters per minute to determine: &lt;strong&gt;If the clearance is increased, how much increase in piston length is required for the same leakage?&lt;/strong&gt; Less clearance gives more air efficiency, but is difficult to manufacture.&lt;/p&gt;&#xA;" OwnerUserId="1598" LastEditorUserId="368" LastEditDate="2015-05-14T16:41:04.147" LastActivityDate="2015-05-16T13:14:43.403" Title="Calculating leakage in an all steel piston cylinder arrangement" Tags="|mechanical-engineering|fluid-mechanics|pistons|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2813" PostTypeId="2" ParentId="2812" CreationDate="2015-05-13T08:32:10.457" Score="3" Body="&lt;p&gt;If the piston is in (slight) clearance inside the cylinder, it's a bit like calculating the flow rate through a capillary, with the difference in diameter being the capillary diameter and the contact length between the piston and the cylinder the capillary length. There are various online calculators to help you with the actual calculation, such as &lt;a href=&quot;http://www.tlv.com/global/TI/calculator/air-flow-rate-through-orifice.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;. If you're after the theory, it's based on &lt;a href=&quot;http://en.wikipedia.org/wiki/Bernoulli%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bernoulli's principle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length doesn't actually come into account in Bernoulli's equation, only the pressure and flow rate, if you neglect the difference in altitude. If you want to take account of the length, then you also need to look at the friction and things start to get messy. The most common approach is generally the &lt;a href=&quot;http://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Darcy-Weisbach equation&lt;/a&gt;. It's expressed as pressure loss, but you can rearrange the equation to work out the fluid velocity and therefore the flow rate. Careful which friction factor you use, the Darcy factor is 4 times the so-called Fanning friction factor. For laminar flow, it is generally approximated as $\frac{64}{R_e}$ where $R_e$ is the Reynolds number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air properties are also dependent on pressure and temperature, but assuming dry air at ambient temperature at standard atmospheric pressure (not exactly true, but it won't be far off) and a pressure drop of 1 bar between the piston chamber and the atmosphere, I get a flow rate of ~20.4 mL/min with a 1mm length and ~1.7mL/min with a 12mm length. I checked &lt;em&gt;a posteriori&lt;/em&gt; that the assumption of laminar flow was valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Darcy-Weisbach equation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure loss form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta P = f_D * \frac{L}{D} * \frac{\rho u^2}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re-arrange to give flow rate, given that $Q = A * u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = A * \sqrt{\frac{2}{\rho} * \frac{\Delta P * D}{f_D * L}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Friction factor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume flow is laminar (need to check &lt;em&gt;a posteriori&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_D = \frac{64}{R_e} = 64 * \frac{\nu A}{QD}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives (using $\nu = \mu / \rho$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = \frac{A * \Delta P * D^2}{L} * \frac{1}{32 \mu}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Numerical application&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D = 0.04e-3;    % [m], diameter&#xA;L = 1e-3;       % [m], length&#xA;dP = 1e5;       % [Pa], pressure difference&#xA;mu = 1.846e-5;  % [kg/(m*s)], dynamic viscosity of dry air at 1atm and 300K&#xA;rho = 1.2922;   % [kg/m^3], air density at standard conditions for P and T&#xA;A = pi*D^2/4;   % [m^2], cross area&#xA;Q = A*dP*D^2/(L*32*mu);     % [m^3/s], flow rate&#xA;Re = Q*D/(A*(mu/rho));    % [-], Reynolds number&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PAEi3.png&quot; alt=&quot;Flow rate&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a flow rate of ~20.4 ml/min and a Reynolds number of ~758, which is less than 2000, so the assumption of laminar flow is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change the length to 12mm, and you get a flow rate of ~1.7 ml/min and a Reynolds number of ~63.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="1324" LastEditDate="2015-05-16T13:14:43.403" LastActivityDate="2015-05-16T13:14:43.403" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="2814" PostTypeId="1" AcceptedAnswerId="2824" CreationDate="2015-05-13T08:59:46.440" Score="6" ViewCount="2158" Body="&lt;p&gt;I am trying to understand the difference in the definition of &amp;quot;fault tolerance&amp;quot; and &amp;quot;robustness&amp;quot; when it comes to system design.&lt;/p&gt;&#xA;&lt;p&gt;For instance, what would be the main design considerations or what would be the definition of the terms when looking at engineering problems?  In particular, how do those terms apply to mechanical or electrical power engineering?&lt;/p&gt;&#xA;&lt;p&gt;I found some definitions but they were centered on computational engineering. The document didn't give much insight in how to define those terms in other engineering disciplines.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Robustness: The capability to cope with unknown errors during execution and provide the system services all the same.&lt;/p&gt;&#xA;&lt;p&gt;Fault-tolerance: The capability to tolerate a certain set of errors defined during the development of a system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://ti.tuwien.ac.at/ecs/teaching/courses/hwswcode_vu_WS_2011/protocols/9-fault-tolerance&quot; rel=&quot;noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What would be the difference between a fault tolerant and robust design?&lt;br /&gt;&#xA;Can a hard distinction be made?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Which techniques or design principles are applied to ensure a fault tolerant design, and which are applied to ensure a robust design?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1600" LastEditorUserId="-1" LastEditorDisplayName="user16" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-05-26T19:39:44.387" Title="What is the difference between fault tolerance and robustness of a system?" Tags="|mechanical-engineering|electrical-engineering|mechanical-failure|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2819" PostTypeId="5" CreationDate="2015-05-13T14:16:54.360" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-05-13T14:16:54.360" LastActivityDate="2015-05-13T14:16:54.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2820" PostTypeId="4" CreationDate="2015-05-13T14:16:54.360" Score="0" Body="Systems Engineering is a multidisciplinary field that combines lifecycle planning and risk analysis with core features of other engineering disciplines. It includes understanding physical systems, organizational planning, computer modeling, and information management. It is also commonly used to describe a position in information technology which involves managing multiple technology stacks and overseeing computer networks." OwnerUserId="1443" LastEditorUserId="1443" LastEditDate="2015-05-13T15:23:49.420" LastActivityDate="2015-05-13T15:23:49.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2821" PostTypeId="1" AcceptedAnswerId="2909" CreationDate="2015-05-13T14:33:36.993" Score="8" ViewCount="2013" Body="&lt;p&gt;Most modern aircraft engines, such as the one depicted below taken from &lt;a href=&quot;http://en.wikipedia.org/wiki/Turbofan&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, are composed of several compressor stages which are driven by a turbine (or several), and a combustion chamber in-between, in order to increase the temperature of the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Turbofan_operation.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MRrIB.png&quot; alt=&quot;Modern Turbofan Sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, manufacturers and designers focus on increasing compression ratios as well as combustion temperatures for efficiency enhancement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, under simplifying assumptions such as perfect gas, no energy losses or friction, and constant inlet temperature and velocity: How is the efficiency of this thermodynamic cycle evaluated? How can one quantify the efficiency gain from a pressure or temperature increase?&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="106" LastEditDate="2015-05-13T23:09:34.177" LastActivityDate="2015-06-07T05:58:17.593" Title="Efficiency in a gas turbine or aircraft engine" Tags="|mechanical-engineering|thermodynamics|aerospace-engineering|aircraft-design|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2824" PostTypeId="2" ParentId="2814" CreationDate="2015-05-13T18:58:22.847" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;What would be the difference between a fault tolerant and robust design?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The most important difference is that robustness takes into account external factors. A more robust system will function in spite of some conditions that would impair the normal function of a less robust system. More robust systems might be referred to as sturdy, heavy-duty, perhaps even overbuilt; less robust systems might be referred to as delicate or fine-tuned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I own an &lt;a href=&quot;http://www.collectorsweekly.com/pocket-watches/hamilton&quot;&gt;antique watch&lt;/a&gt;. It works very well under normal operating conditions but its parts are somewhat delicate. If I drop it on the ground, the glass and/or some internal components of the watch are very likely to break (based on experience, I'm afraid). If I put it through the washing machine, the soapy water is very likely to rust and corrode some internal components. I could dent or scratch the metal case without a great deal of effort; although this probably wouldn't affect the mechanical function of the watch, it would certainly impair its aesthetic function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In comparison, most modern watches wouldn't stop functioning if you dropped them from your wrist or pocket. Their windows may be acrylic, mineral crystal or sapphire, any of which is more durable than the glass in my pocket watch. Their cases may use a harder metal or polymer that resists dents and scratches better than my pocket watch. They can be designed to be waterproof. They can use &lt;a href=&quot;http://en.wikipedia.org/wiki/Quartz_clock&quot;&gt;quartz movements&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all qualities that make a watch &lt;strong&gt;more robust, but not more fault-tolerant&lt;/strong&gt;. None of the failures I described above are really the fault of the watch or its components; they are caused entirely by user error. Unlike robustness, &lt;strong&gt;fault-tolerance is concerned with faults that are expected to occur, especially within components of the system.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://en.wikipedia.org/wiki/Residual-current_device&quot;&gt;ground fault circuit interrupter&lt;/a&gt; (GFCI) is a great example of fault-tolerant design. A ground fault occurs when the current in a circuit is allowed to flow directly to the ground. The classic example involves using a hair dryer in the bathtub, but there are many ways ground faults can occur that don't involve user error.* A residential electrical system is usually in place for decades; we do expect that at some point, ground faults are expected to occur, and except in cases of user error,** they occur within the components of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the circuit runs directly from a main line, a ground fault can seriously threaten life and property. With the addition of a fuse on the circuit, if a ground fault causes the current in the circuit to exceed some upper limit (overload), the fuse burns out and current stops before anything important melts or catches fire. Circuit breakers do the same thing but unlike fuses, they don't have to be replaced when they're triggered; they can be reset at the electrical panel, which can mean less downtime and expense. GFCI circuit breakers additionally protect against leakage, which involves smaller currents that can still be dangerous even if they wouldn't melt a conductor or start a fire. These technologies make the system increasingly &lt;strong&gt;more fault-tolerant&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Which techniques or design principles are applied to ensure a fault tolerant design, and which are applied to ensure a robust design?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Both of these qualities exist on a spectrum.&lt;/strong&gt; If someone points at a design and says, &quot;That's a robust design,&quot; what they mean is that it's relatively robust in comparison to some reference point they have in mind. Different people have different reference points, so I think it's better to be explicit about what comparison you're making. Note that in the first part of this answer, I didn't say my antique watch is not robust—just that it is &lt;em&gt;less robust than&lt;/em&gt; modern watches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a pretty broad topic, but I'll take a stab at a summary treatment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Robustness&lt;/strong&gt; can come from many places, but is usually closely related to cost. Take fasteners as an example: stainless vs. galvanized; screws vs. bolts; the smallest size rated for your load, or the next size up. In each case it's a trade-off between robustness and cost. As long as either option meets the basic requirements of the design, this is a business decision more than an engineering decision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone is telling you that you have to produce a robust design, ask them, in comparison to what, and by what metric? You need a specific definition of &quot;robust,&quot; whether it's comparative (&quot;make it more water-resistant than our competitor's product&quot;) or quantitative (&quot;make it water-resistant to 15 m&quot;). Once you have that definition, you can forget about the word &quot;robust.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fault-tolerance&lt;/strong&gt; is something that can require a bit more imagination and due diligence on your part. Managers who lack experience with the specific product or technology you're working with, especially those without technical backgrounds, may not be able to tell you what faults to expect. They'll say something like, &quot;The user should be able to replace broken components without interrupting operation of the system,&quot; and you'll have to do the work to figure out:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In what ways can each component fail?&lt;/li&gt;&#xA;&lt;li&gt;How likely is any component to fail, in each of the ways it can fail?&lt;/li&gt;&#xA;&lt;li&gt;What other components will be affected when a component fails in a particular way and what will be the overall impact on the system?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you gather this information, be sure to consider the assumptions your design relies on. All assumptions break down at some point, and knowing how your system will behave in those circumstances will help you to identify faults that could occur. Once you've identified what faults to expect, it's usually more straightforward to figure out how much it will cost to mitigate particular risks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some general principles:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If a process needs to maintain some minimum capacity, and a critical component is likely to fail at some point, make the component redundant (e.g., more engines than needed for basic flight) or provide a fallback (e.g., the ability to perform an &lt;a href=&quot;http://www.airsafe.com/events/noengine.htm&quot;&gt;unpowered landing&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Always use tested products and components, even if it means you have to do the testing yourself.&lt;/li&gt;&#xA;&lt;li&gt;Provide a mechanism for your customers/clients/users to give feedback about failures as they occur. You'll hardly ever anticipate every possibility and being able to retrofit or adapt an existing design is important.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* Stripping too much insulation during installation; pests chewing on wires; running wires through studs without a nail plate; cutting flex conduit improperly/not deburring; etc.&lt;br&gt;&#xA;** You could also argue for expanding the system to include users as internal components—especially children and pets—but that's another discussion altogether.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-05-14T17:38:25.047" LastActivityDate="2015-05-14T17:38:25.047" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2825" PostTypeId="2" ParentId="2821" CreationDate="2015-05-13T22:06:17.077" Score="2" Body="&lt;p&gt;One thing increased compression ratios improve upon is that it will increase the burn speed and burn completeness, reducing the percent of unburnt fuel/particles going out the exhaust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increased compression also allows higher expansion ratios for the exhaust turbine, which will increase power. This is similar to how compression ratios affect standard piston based engines.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2015-05-13T22:06:17.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2826" PostTypeId="1" AcceptedAnswerId="2829" CreationDate="2015-05-14T02:56:45.873" Score="25" ViewCount="87019" Body="&lt;p&gt;This is a continuation of me trying to understand torque and stepper motors in my other &lt;a href=&quot;https://engineering.stackexchange.com/questions/2802/reading-stepper-motor-datasheets-to-get-torque-and-speed&quot;&gt;question&lt;/a&gt;.  I'm trying to understand the torque a motor would be required to generate to lift a small weight, and the formulas involved.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The first part of my question is to verify if I am calculating this correctly:&lt;/p&gt;&#xA;&lt;p&gt;Let's say I have a 450 g mass (roughly one pound) then the force of gravity pulling it down is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\begin{align}&#xA;F &amp;amp;= ma \\&#xA;&amp;amp;= 0.450 \:\mathrm{kg} * 9.8 \:\mathrm{m}/\mathrm{s}^2 \\&#xA;&amp;amp;= 4.41 \:\mathrm{N} \\&#xA;\end{align}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If I have a stepper motor with a spindle for my string that pulls up my motor with a radius of 5 cm.  I think my torque needed would be:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\begin{align}&#xA;T &amp;amp;= Fr \\&#xA;&amp;amp;= F * 0.05 \\&#xA;&amp;amp;= 0.22 \:\mathrm{Nm} \\&#xA;\end{align}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So now if I want to move that mass I need to find a stepper motor that can output more than 0.22 Nm of torque, right?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The follow-on to my question is that if I want to see how fast I can move it then I need to look at a Torque speed curve, right?&lt;/p&gt;&#xA;&lt;p&gt;My confusion is this: do I have to ensure that I'm moving slow enough to get the torque I need, or does that curve say if you need this torque you won't be able to go above this speed because the motor won't let you?&lt;/p&gt;&#xA;" OwnerUserId="1210" LastEditorUserId="1210" LastEditDate="2021-08-27T19:15:26.697" LastActivityDate="2021-08-27T19:15:26.697" Title="Understanding required torque for a motor lifting a weight" Tags="|mechanical-engineering|motors|torque|stepper-motor|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="2828" PostTypeId="2" ParentId="2764" CreationDate="2015-05-14T12:13:22.570" Score="7" Body="&lt;p&gt;I would venture that the sad faces are a type of &lt;a href=&quot;http://en.wikipedia.org/wiki/Pictogram&quot; rel=&quot;noreferrer&quot;&gt;pictogram&lt;/a&gt; to warn &quot;untrained technicians&quot; about the electrical shock hazard present at those connections.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Designers often have to worry about conveying meaningful information in a small amount of space to diverse groups that may not speak a common language.  This is particularly true for appliances that are sold internationally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PCB looks like it connects to and draws either 230V at 4A or 120V at 6.3A, so you've got almost a kiloWatt of surprise waiting for you if you're handling the board in an unsafe manner while plugged in.  That's not necessarily lethal, but certainly an unpleasant shock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.iso.org/iso/graphical-symbols_booklet.pdf&quot; rel=&quot;noreferrer&quot;&gt;ISO universal symbol&lt;/a&gt; to warn for an electricity hazard looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/r9pz7.png&quot; alt=&quot;ISO Electricity Warning&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I suspect that sort of a symbol may be easily overlooked by someone who isn't trained and simply wants to get their appliance working again.  It could also be misconstrued as meaning only  &quot;mains connect here.&quot;  Or it's possible that the PCB designer wasn't aware of the ISO symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, a sad face does a pretty decent job of a) catching your attention while looking at the board and b) attempting to warn you to be careful around those points.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorUserId="709" LastEditDate="2015-05-14T13:24:39.350" LastActivityDate="2015-05-14T13:24:39.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2829" PostTypeId="2" ParentId="2826" CreationDate="2015-05-14T12:28:11.077" Score="27" Body="&lt;p&gt;You have the right concept, but slipped a decimal point.  5 cm = 0.05 m.  The gravitational force on your 450 g mass is 4.4 N as you say, so the torque just to keep up with gravity is (4.4 N)(0.05 m) = 0.22 Nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that is the absolute minimum torque just to keep the system in steady state.  It leaves nothing for actually accellerating the mass and for overcoming the inevitable friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the real torque required, you have to specify how fast you want to be able to accellerate this mass upwards.  For example, let's say you need at least 3 m/s&amp;sup2;.  Solving Newton's law of F = ma:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(0.450 kg)(3 m/s&amp;sup2;) = 1.35 N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That, in addition to the 4.4 N just to balance gravity means you need 5.8 N upwards force.  At 0.05 m radius, that comes out to a torque of 0.28 Nm.  There will be some friction and you want some margin, so in this example a 0.5 Nm motor would do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that torque isn't the only criterion for a motor.  Power is another important one.  For that you have to decide what's the fastest speed you want to be able to pull the mass upwards at.  Let's say 2 m/s for sake of example.  From above, we know the highest upwards force is 5.8 N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5.8 N)(2 m/s) = 11.6 Nm/s = 11.6 W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After accounting for some losses due to friction and leaving a little margin, the motor should be rated for about 15 W minimum.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-05-14T12:28:11.077" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="2831" PostTypeId="2" ParentId="2796" CreationDate="2015-05-14T16:46:11.730" Score="2" Body="&lt;h1&gt;Equations of Motion&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;From what I understand from the question and your comments, you can write the equations of motion of the 3-dof oscillator in the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M \frac{d^{2}x}{dt^{2}} + C \frac{dx}{dt} + K x = f$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the vector $x$ represents the displacements, $M$, $C$ and $K$ are the mass, damping and stiffness matrix respectively, and $f$ is some forcing vector (e.g. impact, step function, etc.). Therefore, this type of equation may be classified as a Linear Time-Invariant (LTI) system.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Damping models&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There are several ways to solve this type of LTI systems, for instance numerical time integration, and as mentioned in the response by @am304, different models to represent the damping matrix $C$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rayleigh damping is indeed one form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = aM + bK$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where parameters $a$ and $b$ are chosen arbitrarily. You could also use modal damping, which is similar and may be written in the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V' C V= 2 x_iW$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $W$ is a diagonal matrix containing the linear eigenvalues of your system, $V$ the associated eigenvectors and $x_i$ is the damping coefficient (which may vary from mode to mode but can be considered as a single constant as a first approximation). The main advantage of this method is that the damping carries with it the inherent dynamics of your linear system and is diagonal in the modal domain.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Performance&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In terms of performance, with respect to the overshoot and the settling time, they will be both dependent on the damping term as well as the stiffness, since the two quantities are related in the two modeling strategies here proposed. However, for a given stiffness matrix, the solution will reach a steady-state with a specific amplitude, and the overshoot and settling time will solely depend on $x_i$ as shown in the picture below (Last mass displacements for: $x_i=0.2$ (blue), $x_i=0.5$ (red) and $x_i=0.8$ (black)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nTSen.jpg&quot; alt=&quot;Last mass displacements for: xi=0.2 (blue), xi=0.5 (red) and xi=0.8 (black)&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="209" LastEditorUserId="1832" LastEditDate="2016-07-17T03:25:13.870" LastActivityDate="2016-07-17T03:25:13.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2832" PostTypeId="1" CreationDate="2015-05-14T16:46:46.883" Score="29" ViewCount="8893" Body="&lt;p&gt;Is it physically possible to build an air compressor with no moving parts? I'm envisioning a thermodynamic cycle capable of compressing air with no moving parts and operated in a stationary way. There is no limitation on the compression factor, provided it is significantly greater than one (1.1, 2, 100...) but the design must be realizable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zero moving parts is a big constraint. You can interpret it as the absence of pistons, crankshafts and other complex mechanisms that will wear out over time. If some moving parts are needed, what are the minimum required moving parts, with the lowest maintenance requirements?&lt;/p&gt;&#xA;" OwnerUserId="1616" LastEditorUserId="368" LastEditDate="2015-05-20T16:21:03.107" LastActivityDate="2023-07-18T00:20:50.597" Title="Is it possible to build an air compressor with no moving parts?" Tags="|mechanical-engineering|thermodynamics|fluid-mechanics|" AnswerCount="11" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2833" PostTypeId="2" ParentId="2832" CreationDate="2015-05-14T21:43:47.163" Score="21" Body="&lt;p&gt;It is possible to make a air pump/compressor with absolutely no moving parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make a small, electrically non-conductive chamber and stick 2 electrodes into it. Pulse an arc through the electrodes so that the chamber pressure rises and falls rapidly. Use tesla air flow check valves (because they have zero moving parts) for intake and exhaust. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the arc forms across the chamber, the enclosed air will become superheated and expand mostly out of the exhaust port due to the tesla valves, then the chamber cools and pulls in fresh air through the other tesla valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can also be done with any pulse-able heat source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some type of sealing check valve is used to replace the tesla valve, you can maintain high levels of compression too.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastEditorUserId="1459" LastEditDate="2016-01-14T14:50:31.003" LastActivityDate="2016-01-14T14:50:31.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2834" PostTypeId="2" ParentId="2832" CreationDate="2015-05-14T22:08:03.533" Score="3" Body="&lt;p&gt;Interpreting &quot;moving parts&quot; as meaning that every solid part of the device has to be rigid, and the requirement as the capability to deliver a constant stream of air at a non-trivially greater-than-ambient constant pressure, I suspect the answer is a qualified no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also assuming that fuels and working fluids don't count as &quot;parts.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems reasonable to assume that any remotely practical air compressor will operate under conditions where the ideal gas law applies, so $p \propto \frac{NT}{V}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach would be to imitate a typical air compressor and try eliminating as many moving parts as possible. For example, something like a &lt;a href=&quot;http://en.wikipedia.org/wiki/Hydraulic_ram#Construction_and_principle_of_operation&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydraulic ram&lt;/a&gt; could eliminate pistons, impellers, screws, etc. and allow us to extract energy from a moving body of water to compress air, but it still requires valves. A valveless pump as seen in &lt;a href=&quot;https://www.youtube.com/watch?v=EsaawMQ5JbY&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; requires a special rotating piston. A basic &lt;a href=&quot;http://en.wikipedia.org/wiki/Siphon&quot; rel=&quot;nofollow noreferrer&quot;&gt;siphon&lt;/a&gt; has no moving parts at all and it can create pressure if you enclose the lower reservoir, but is totally impractical as part of an air compressor—and even if it weren't, you'd still need some sort of valve to deliver the pressurized air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach is to manipulate temperature, which sounds like what you've got in mind. It's easy enough to generate heat without moving parts; a burner or an electric coil will do it. But how do you get around the valve issue? In order for the pressure to build, you need an enclosed space, and once you have pressure, the air needs to exit that enclosed space. If you wanted to get creative, you could try something like a diaphragm with an aperture that only opens when the diaphragm has expanded; the pressure will then make its own exit. But an expanding and contracting solid diaphragm or bladder also seems like a moving part, to me. It might be more durable than other types of moving part, I suppose, but then again it might not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To produce a constant stream of pressure then you need a holding tank and the magnitude of your delivery pressure will be reduced significantly based on the upper limit of pressure you can develop in the holding tank and how quickly you can develop it. The Tesla valves suggested in &lt;a href=&quot;https://engineering.stackexchange.com/a/2833/368&quot;&gt;netduke's answer&lt;/a&gt; are very clever but they're really differential flow-limiting devices; I don't see them being able to develop and hold pressure in a tank that you could release on demand for pneumatic power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the reason it's a &quot;qualified&quot; no is this. In theory, if you accept that your air compressor may be totally impractical for most purposes, you don't count elastic deformation as movement and you &quot;cheat&quot; a few times with valves and regulators, then yeah. You can create a device that compresses air in a tank, and then do with it what you will. In practice, it sounds like a poor idea that doesn't scale well, but it's an interesting exercise to play with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another qualification is that you might get a completely different answer in a microfluidics context.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-05-14T22:08:03.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2836" PostTypeId="1" AcceptedAnswerId="2837" CreationDate="2015-05-15T02:44:28.287" Score="5" ViewCount="428" Body="&lt;p&gt;This is a follow up question to my question &lt;a href=&quot;https://engineering.stackexchange.com/questions/2826/understanding-required-torque-for-a-motor-lifting-a-weight/2829#2829&quot;&gt;here&lt;/a&gt; about torque and stepper motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There Olin explains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note also that torque isn't the only criterion for a motor. Power is&#xA;  another important one. For that you have to decide what's the fastest&#xA;  speed you want to be able to pull the mass upwards at. Let's say 2 m/s&#xA;  for sake of example. From above, we know the highest upwards force is&#xA;  5.8 N.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(5.8 N)(2 m/s) = 11.6 Nm/s = 11.6 W&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After accounting for some losses due to friction and leaving a little&#xA;  margin, the motor should be rated for about 15 W minimum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm unclear about what kind of Watts we're talking about here, is this just Watts in the mechanical sense or in the electrical sense.  In other words should I be thinking about needing to deliver 15W as in 15V @ 1A?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also wondering about controlling the amount of power delivered to the motor.  In steady state I guess it would be Input Voltage divided by motor coil resistance?  But to change the position where the current and voltage are changing the current must be the input voltage / (reactance of the coil plus the coil resistance).  But the reactance changes with frequency... so do I have to look at the rise time of the input voltage to understand the frequency involved and then try to calculate from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a little lost on that point, but trying to read more and understand.&lt;/p&gt;&#xA;" OwnerUserId="1210" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-05-15T11:53:18.687" Title="Power and motor current" Tags="|mechanical-engineering|motors|stepper-motor|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2837" PostTypeId="2" ParentId="2836" CreationDate="2015-05-15T11:53:18.687" Score="4" Body="&lt;p&gt;In the text you quoted, I was talking about mechanical power out of the motor.  This is what the power rating of a motor, given no other qualifications, means.  A &quot;15 W motor&quot; therefore can put out 15 W of mechanical power under the right conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No motor is 100% efficient, so more electrical power than that will need to be supplied to the motor.  For example, if the motor is 85% efficient at a particular operating point where it produces 15 W of mechanical power, it will require (15 W)/85% = 17.7 W of electrical power.  The remaining 2.7 W that doesn't come out in the form of mechanical power will be dissipated as heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your electrical model of the a motor isn't quite right.  Think of it as a resistor in series with a voltage source.  Yes, coils are in fact inductive, but the time constant of the inductance with the resistance in the system is usually far less than the mechanical time constant and so isn't relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resistance is fixed and is the actual DC resistance of the coils.  The voltage source is a function of the motor's speed, and always &lt;i&gt;opposes&lt;/i&gt; the applied voltage when the motor is spinning in the direction that it is being pushed by the applied voltage.  This voltage source represents the motor acting as a generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque produced by the motor is proportional to the current thru it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you first apply a voltage to a stantionary motor, the voltage source is 0 and all the applied voltage is across the resistance.  This causes a large current to flow, called the &lt;i&gt;stall current&lt;/i&gt;.  As the motor speeds up, the voltage source opposes some of the applied voltage.  That leaves less votlage across the resitor, which reduces the current, which reduces the torque.  Eventually a balance is reached where the current drops to the point where it produces just enough torque to keep the motor going at that speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a unloaded motor, little torque is required to keep it going, and the internal voltage source opposes most of the applied voltage.  There is little voltage left across the resistor, so the current drawn by the motor is much lower than the stall current.  If the motor is mechanically loaded, then a higher torque is required to keep it going, so the steady state speed will be lower, the internal voltage source lower, more voltage across the resistance, and the current higher.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-05-15T11:53:18.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2838" PostTypeId="1" CreationDate="2015-05-15T15:54:24.930" Score="5" ViewCount="255" Body="&lt;p&gt;I am designing an experiment where the hydraulic hose is going to be subjected to relatively high internal pressure for prolonged period of time and it is absolutely imperative that the rate of permeation is minimised. I have found one great book that deals with gases but nothing on liquids or liquids under elevated pressure. I am unable to find any data regarding the permeation rates of hydraulic fluids and/or water through the hydraulic hoses or more generally through various polymers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of time is in years and the pressure in tens of MPa so I am assuming that the permeation will not be negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine the permeation rate of a fluid through the hose before running my experiment and minimise the permeation rate if it is not negligible?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastEditorUserId="368" LastEditDate="2015-05-18T15:42:14.073" LastActivityDate="2015-05-28T15:40:40.033" Title="Minimising permeation rate of hydraulic fluid through hydraulic hose" Tags="|fluid-mechanics|pressure|applied-mechanics|hydraulics|plastic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2840" PostTypeId="1" CreationDate="2015-05-15T20:54:17.800" Score="7" ViewCount="340" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On May 12, 2015, an Amtrak train derailed in Philadelphia, PA, killing 8 and injuring 200 people. I saw a story on the incident where multiple NTSB spokespeople have said at this and other train derailments that &lt;a href=&quot;https://www.fra.dot.gov/Page/P0621&quot; rel=&quot;nofollow&quot;&gt;Positive Train Control&lt;/a&gt; (PTC), had it been installed,&#xA;&lt;a href=&quot;http://www.npr.org/sections/thetwo-way/2015/05/14/406652406/positive-train-control-the-tech-that-couldve-prevented-amtrak-derailment&quot; rel=&quot;nofollow&quot;&gt;could have prevented&lt;/a&gt; the derailments from happening, particularly in incidents where human error is the key factor, rather than technical failures or track obstructions. PTC uses GPS and transponder technology to avoid collisions and derailments by ensuring that trains operate within speed limits and do not cross improperly-switched tracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the first link points out, the Rail Safety Improvement Act of 2008 (passed in the wake of the &lt;a href=&quot;http://en.wikipedia.org/wiki/2008_Chatsworth_train_collision&quot; rel=&quot;nofollow&quot;&gt;Chatsworth train collision&lt;/a&gt;) mandates that much of the nation's rail system have PTC installed by Dec. 31, 2015. However, the rail industry has been lobbying to &lt;a href=&quot;http://www.railwayage.com/index.php/ptc/ptc-relief-on-the-way.html?channel=40&quot; rel=&quot;nofollow&quot;&gt;extend the deadline&lt;/a&gt;, claiming that they will be unable to have PTC installed and operational on all required lines, and be forced to either cease operations, or operate illegally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is required to implement Positive Train Control on existing rail lines in the United States? Does implementation require track improvements, engine/car improvements, or both? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are the systems that are currently used or in development inter-compatible? For instance, a commuter rail system near me uses existing BNSF and Union Pacific lines (among others) in addition to lines that it owns itself, and Amtrak operates on some of these lines as well. Do all trains operating on these lines need identical PTC systems, or can the commuter rail develop a different but compatible system than the freight rail?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How long have complete rollouts of PTC taken, and what have they cost? (This is a fact-based way of answering the more debatable question 'Is 7 years a reasonable period of time to allow for PTC development and implementation?' and to examine what kind of budgetary strain the unfunded mandate places on smaller rail systems.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="41" LastEditorUserId="41" LastEditDate="2015-05-18T15:49:51.177" LastActivityDate="2015-05-19T23:12:02.023" Title="What is required to implement Positive Train Control on railroads in the US?" Tags="|civil-engineering|rail|transportation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2841" PostTypeId="1" AcceptedAnswerId="2842" CreationDate="2015-05-16T00:42:28.253" Score="4" ViewCount="5287" Body="&lt;p&gt;I'm designing a solenoid valve for an automated irrigation project that I'm working on. After having looked at some designs, I have a question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.ytimg.com/vi/SwqM8zpmAD8/0.jpg&quot; alt=&quot;example design&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most of the designs I've seen, the orifice is smaller than the pipe diameter. Wouldn't this reduce the flow rate? If this is the case and if I'm using such a design, do I make that opening bigger?&lt;/p&gt;&#xA;" OwnerUserId="1636" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.107" LastActivityDate="2018-08-21T18:01:06.453" Title="Solenoid valve pressure drop" Tags="|fluid-mechanics|pressure|valves|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2842" PostTypeId="2" ParentId="2841" CreationDate="2015-05-16T02:02:12.510" Score="4" Body="&lt;p&gt;A valve that is smaller than the pipe diameter will reduce the flow rate, but so will any valve. When calculating what sort of pressure/head loss (pressure and head are equivalent) a system will have, there are two contributing factors: major losses and minor losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_{total}=h_{major}+h_{minor}$$&#xA;$$h_{major}=\lambda \left( \frac{l}{d_h} \right) \left( \frac{v^2}{2g} \right)$$&#xA;$$h_{minor}=\xi \left( \frac{v^2}{2g} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because these losses are expressed as head, they are in units of length. $\lambda$ is the friction factor, which is a complex number that varies based on the type of flow (laminar, turbulent, transitional) and then the Reynolds number and tube roughness. That's a topic worth exploring elsewhere, but isn't directly related to your question. $l$ is the length of straight pipe, $d_h$ is the hydraulic diameter, which for a circular pipe is the actual diameter. As pipe shape changes, there are different expressions for hydraulic diameter. $v$ and $g$ are flow velocity and gravitational acceleration, as you would expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equation is the one we're really looking at though. This is how you calculate the head loss across the valve. $\xi$ is the loss coefficient, and $v$ and $g$ are the same as above. The only way you can get loss coefficients is by &lt;a href=&quot;http://www.engineeringtoolbox.com/minor-loss-coefficients-pipes-d_626.html&quot; rel=&quot;nofollow&quot;&gt;looking them up&lt;/a&gt; or by calculating them empirically, if values don't exist for your components. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have to account for the change in overall diameter of the pipe though. This can be done a couple ways, depending on how the pipe diameter changes. If it's sudden, it's probably best to treat it as another pipe feature which will cause a pressure drop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a more gradual change, you will have some head loss, but the bigger effect will be the narrowing diameter increasing the velocity of the flow. This is a fairly trivial result of conservation of mass. The mass flow rate in and out must be the same, and if we assume constant density (reasonable for liquid water at speeds under 450 m/s {1000 mph}), then we'll see that velocity varies inversely with cross-sectional area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{m}_{1}=\dot{m}_{2}$$&#xA;$$\dot{m}=\rho \dot{V}$$&#xA;$$\dot{V}=Av$$&#xA;$$\therefore A_{1} v_{1}=A_{2} v_{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This $v_{2}$ is the inlet velocity of the valve, and if the velocity increases, so will the minor loss across the valve. So yes, increasing the diameter of the valve will make the pressure drop go down, but you can't eliminate the pressure drop entirely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, standard designs are standard for a reason. The valves constrict a bit, but they don't bulge out more than necessary from the rest of the pipe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Most equations referenced from &lt;a href=&quot;http://www.engineeringtoolbox.com/total-pressure-loss-ducts-pipes-d_625.html&quot; rel=&quot;nofollow&quot;&gt;Engineering Toolbox&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-05-16T02:02:12.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2843" PostTypeId="2" ParentId="2814" CreationDate="2015-05-16T05:46:17.230" Score="0" Body="&lt;p&gt;Here is my attempt at a short and sweet distinction between the two concepts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fault tolerant:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many mistakes of other people can I handle before I fail myself?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Robust:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many new scenarios can I be useful in (or new inputs can I accept that I have never seen before) that I wasn't originally designed for?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1638" LastActivityDate="2015-05-16T05:46:17.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2844" PostTypeId="2" ParentId="2327" CreationDate="2015-05-16T07:23:12.193" Score="2" Body="&lt;p&gt;&lt;strong&gt;Hooke's Law&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force of applied by the spring to the pin, will depend on how much the spring is compressed. This can be calculated by using Hooke's Law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qcEZi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finding k (Using assumptions)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find find &lt;strong&gt;k&lt;/strong&gt;, for demonstration we will just assume a few properties of the spring based on what we can see in the gif.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.acxesspring.com/spring-constant-calculator.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spring Constant Calculator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/UJdB0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k = 0.136 N/mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Finding X maximum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For demonstration we will use 3 mm for the maximum compression of the spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculating Maximum Force From Spring&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the assumed properties and plugging them into Hooke's Law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = k X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = (0.136)*(3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = 0.408 N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Industry Standards&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This type of lock is a &lt;strong&gt;Tumbler Lock&lt;/strong&gt; and the spring force is most likely not regulated or standardized in industry, but other contributors may have more information.&lt;/p&gt;&#xA;" OwnerUserId="1594" LastActivityDate="2015-05-16T07:23:12.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2845" PostTypeId="1" CreationDate="2015-05-16T10:50:44.230" Score="5" ViewCount="597" Body="&lt;p&gt;A steam injector effectively raises fresh water pressure above boiler pressure, otherwise it would not work. Replacing the steam by compressed air and condensation heat by a combustion, can an ejector be able to compress more air than consumed to a higher pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main objective is emulate an steam boiler feedwater injector. Take a bit of compressed air from a pressurized bottle and use it in some way to inject more air in the same bottle at a same (a little higher) pressure. In other words, an air compressor running from a jet pump.&lt;/p&gt;&#xA;" OwnerUserId="1616" LastEditorUserId="368" LastEditDate="2015-05-18T15:50:56.440" LastActivityDate="2015-05-18T15:52:42.580" Title="Can a boiler feedwater injector be used to compress air?" Tags="|fluid-mechanics|thermodynamics|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2846" PostTypeId="1" CreationDate="2015-05-16T11:39:24.560" Score="7" ViewCount="1118" Body="&lt;p&gt;Why are copper pipes used in counter-flow heat exchangers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I enhance heat transfer by using a pipe made of any other material?&lt;/p&gt;&#xA;" OwnerUserId="1641" LastEditorUserId="368" LastEditDate="2015-05-18T16:01:16.650" LastActivityDate="2015-05-18T21:56:18.893" Title="Can I use a pipe of any material other than copper in a counter-flow heat exchanger?" Tags="|materials|thermodynamics|heat-exchanger|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2847" PostTypeId="2" ParentId="2846" CreationDate="2015-05-16T11:49:28.263" Score="2" Body="&lt;p&gt;No, not any material.  Only materials with a higher thermal conductivity than copper will increase the rate of heat transfer.  In some cases, you may get better heat transfer by using a stronger material that could be formed into a thinner pipe, but you're generally better off going for thermal conductivity.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-05-16T11:49:28.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2848" PostTypeId="2" ParentId="2845" CreationDate="2015-05-16T12:17:25.313" Score="3" Body="&lt;p&gt;I'm not sure if an injector will work in this situation.  In the case of a steam injector, the specific volume of the mixture is &lt;em&gt;lower&lt;/em&gt; than that of the inlet steam, which is partly how the outlet pressure is able to increase above the inlet pressure.  If you have a combustion reaction between the air and some kind of fuel, the specific volume of the reaction products will be much &lt;em&gt;higher&lt;/em&gt; than the inlet air, so you may not be able to build up enough pressure in the outlet stream to make it work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to instead look into turbo pumps like the kind used on liquid fueled rocket motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both the motive fluid and the suction fluid are air, then this will work, though the efficiency may not be as great as a steam-water injector.  The inlet air will induce flow in the suction air, and the resulting outlet flow will have pressure and velocity somewhere in-between that of the inlet and suction streams.  So, if the suction air is at atmospheric pressure, the outlet flow will be higher than atmospheric.  Whether or not the pressure will be high enough to meet your needs depends on the pressure and flow rate of the inlet air.  A &lt;em&gt;pneumatic vacuum&lt;/em&gt; uses this same principle, you may want to look into those as well.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="368" LastEditDate="2015-05-18T15:52:42.580" LastActivityDate="2015-05-18T15:52:42.580" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2849" PostTypeId="2" ParentId="2846" CreationDate="2015-05-16T12:57:53.810" Score="8" Body="&lt;p&gt;What we usually care about when selecting a material for a heat exchanger is its thermal conductivity. The higher the thermal conductivity, the easier it'll be to transfer heat between the two fluids. This means that we'll either (a) be able to create a larger temperature difference using a material with a higher thermal conductivity in a heat exchanger of a given size or (b) use a smaller heat exchanger for a given temperature difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html&quot;&gt;This table&lt;/a&gt; shows a list of thermal conductivities of common materials in $\frac{W}{mK}$. Unfortunately, it can't be sorted in order of highest to lowest, but if you scan through it, you'll see that $k_{Cu}=400 \frac{W}{mK}$, which is the highest they list, apart from diamond ($1000 \frac{W}{mK}$). While a heat exchanger made of diamond would be really cool, it would also be really expensive. The next highest thermal conductivities after copper are for silver and gold, which are also not cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;, there is one important case where copper is unacceptable to use, despite its high thermal conductivity and relatively low cost. As &lt;a href=&quot;http://www.ukcopperboard.co.uk/literature/pdfs/Installation-Tips/Guide-to-suitability-with-chemicals.pdf&quot;&gt;this document&lt;/a&gt; (2 page PDF) shows, copper reacts poorly with ammonia, which still sees common usage as a refrigerant (known as R717), particularly in industrial applications, where a large quantity of refrigerant is needed (it's cheap) and the toxicity (biological, not environmental) is not an issue. You won't find it in your refrigerator, but it's probably what cools the ice rink closest to you (or it will be soon.) So in this case, the fact that we'd be selecting a material with a lower thermal conductivity is preferable to using copper and having the ammonia corrode through the system. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-05-16T12:57:53.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2850" PostTypeId="2" ParentId="2832" CreationDate="2015-05-16T15:18:10.237" Score="0" Body="&lt;p&gt;Assuming this question is a theoretical one, the answer can be heating the air. It is similar to afterburners in military jets. &lt;a href=&quot;http://en.wikipedia.org/wiki/Afterburner&quot; rel=&quot;nofollow&quot;&gt;See: Wikipedia\afterburner&lt;/a&gt; You do not have to add the combusting fluid into your flow, if you use channels, as in a domestic water heater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle of the Afterburner, as related to increasing the pressure of an airflow is quoted hereby: &lt;em&gt;&quot;The afterburner then injects fuel downstream of the turbine and reheats the gas. In conjunction with the added heat, the pressure rises in the tailpipe and the gas is ejected through the nozzle at a higher velocity. The mass flow is also slightly increased by the addition of the fuel.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1097" LastEditorUserId="1097" LastEditDate="2015-05-22T19:09:21.510" LastActivityDate="2015-05-22T19:09:21.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2851" PostTypeId="2" ParentId="2832" CreationDate="2015-05-16T20:06:42.167" Score="13" Body="&lt;p&gt;You won't attain very high flow nor very high pressure (plus you must control the gas outflow), and it's a bit of cheating on the &quot;no moving parts&quot; but you can make a compressor with no moving &lt;em&gt;solid&lt;/em&gt; parts. Plus it's very wasteful on energy - you'd be far better off converting the energy to electricity and using a classic compressor, but if for some reason you want to get pressurized air with no mechanical parts, that would be it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cOjLY.png&quot; alt=&quot;Water flow based air pump&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a strong flow of water and a large altitude difference. Take a rapid, strong flow of water. Using either the &lt;a href=&quot;http://en.wikipedia.org/wiki/Venturi_effect&quot; rel=&quot;noreferrer&quot;&gt;Venturi effect&lt;/a&gt; or some other way aerate the water - mix it with air/gas bubbles. As the water travels down the pipe rapidly, the bubbles travel upwards slower than the water flow - they are carried downwards. Despite rapid flow, the outlet of the pipe is somewhat constricted and as result, the water pressure grows with the height of the water column, and with the water column pressing down, the pressure in the bubbles grows as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the pipe turns sideways. The bubbles are no longer dragged downwards, and so they move towards the upper side of the pipe, eventually escaping the water flow and amassing in a reservoir above the upper edge of the pipe; they are at the same pressure as the water - which must yet to pass a constriction, so its pressure is fairly high. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it won't work with narrow pipes and low flow as water viscosity would reduce the downward speed and distribution of pressure. And the amount of air taken out must be controlled because - unlike in normal compressors - the pressure is maintained at all times, but the volume of the gas drops, and if you exhaust it, you will start drawing water. And obviously there are better uses for a high-volume, high-pressure water flow than compressing some air. It's wasteful, because most of the water's energy is lost. Still - the concept is sound; scaling the height of the downwards pipe you can attain quite reasonable pressures, at about 1 bar per 10m. And the only physical moving part is the outlet valve which does not need to move during operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Trompe&quot; rel=&quot;noreferrer&quot;&gt;Trompes&lt;/a&gt; are very similar and were once used to provide compressed air to furnaces.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorDisplayName="user16" LastEditDate="2015-09-18T19:50:44.300" LastActivityDate="2015-09-18T19:50:44.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2852" PostTypeId="1" AcceptedAnswerId="2906" CreationDate="2015-05-17T00:38:37.407" Score="12" ViewCount="2643" Body="&lt;p&gt;&lt;strong&gt;tl;dr:&lt;/strong&gt; After an extended conversation with an old-timer, I realized a few things:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The single most valuable measurement for the majority of people will be water-depth-in-well.&lt;/li&gt;&#xA;&lt;li&gt;The second most valuable will be water-flow-from-well.&lt;/li&gt;&#xA;&lt;li&gt;The &amp;quot;bubbler&amp;quot; solution discussed below has another major weakness (in addition to the frailty of air pumps): the introduction of oxygen into the well water will cause oxide formation, leading to mineral encrustation of not only the opening of the tubing, but extending all the way up inside to wherever its normal level would be. He knows because he's had to deal with something almost exactly analogous and it was a major hurdle. Larger size tubing will slow  down the process, but eventually the tubing will be blocked.&lt;/li&gt;&#xA;&lt;li&gt;We are reexamining the solution that uses a bladder-in-tank with differential pressure sensor. He had specific ideas about how to do this that sound doable (but there are still some details to be dealt with).&lt;/li&gt;&#xA;&lt;li&gt;Oh, and he solved the tank problem in about 10 seconds. Put a pressure sensor on the pipe from the tank to the pressure pump. Ignore the spikes that happen when the pump kicks in, and we have exactly the pressure reading we want with cheap, well-understood sensors. Sheesh! It was so obvious once he said it I almost kicked myself.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I thank all of you for your ideas and your analysis. If anyone is interesting in seeing how the project unfolds, keep an eye on &lt;a href=&quot;http://waterunderground.net/&quot; rel=&quot;noreferrer&quot;&gt;waterunderground.net&lt;/a&gt;. It's pretty empty at the moment, but should have more content in a month or so.&lt;/p&gt;&#xA;&lt;h3&gt;Backstory&lt;/h3&gt;&#xA;&lt;p&gt;I am designing an open-sourced well &amp;amp; water-usage monitoring system for people in Northern California. The goal is to be able to measure water flow from well-to-tank, tank-to-house, and tank-to-irrigation, plus monitor the water depth in the tank and the well. Our current target parts cost is under \$200 for a system including CPU, 3 flow sensors, and 2 pressure sensors, although we think we may be able to get it closer to \$100 after a few design iterations.&lt;/p&gt;&#xA;&lt;p&gt;We appear to have the flow sensor portion solved now that we finally have a supplier of Female G1 =&amp;gt; U.S. 1&amp;quot; slip adapters to integrate cheap Hall effect sensors into a standard U.S. piping environment. The depth measurement solution is not so straightforward.&lt;/p&gt;&#xA;&lt;p&gt;I'm asking for a sanity check on my reasoning here before I go off and start buying stuff that is wrong, either in size, type, or altogether.&lt;/p&gt;&#xA;&lt;h3&gt;Problem statement&lt;/h3&gt;&#xA;&lt;p&gt;I need a &lt;em&gt;low-cost&lt;/em&gt; way to measure the depth of 2 columns of water with moderately decent accuracy, say +/- 5%. Although our own property is site Alpha 1, we would like a solution that scales up, or down, for other properties with similar needs.&lt;/p&gt;&#xA;&lt;p&gt;We have:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A 3,000 gal storage tank that is approx. 8.5' of water when full. Other tanks are of similar height +/- 5'.&lt;/li&gt;&#xA;&lt;li&gt;A water well. Our own well is 75' deep w/ 37' of water. Other wells in the area are as shallow as 30' w/ 15' of water, or as deep as 300' w/ 70+' of water.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;We have the following criteria:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;No more than \$30 for the tank and (hopefully) no more than \$50 for the well. Lower costs would be great.&lt;/li&gt;&#xA;&lt;li&gt;Solution must integrate &lt;em&gt;in some manner&lt;/em&gt; (handwave) with an Arduino, BeagleBone Black, or similar low-cost controller.&lt;/li&gt;&#xA;&lt;li&gt;A continuous readout is desirable, but something that triggers every 15, 30, or &amp;lt;whatever&amp;gt; minutes would be acceptable.&lt;/li&gt;&#xA;&lt;li&gt;No electronics/electrical systems &lt;em&gt;in&lt;/em&gt; the well or tank.&lt;/li&gt;&#xA;&lt;li&gt;No metal in the well or tank, with the possible exception of material used to weigh down the tubing that goes into the water.&lt;/li&gt;&#xA;&lt;li&gt;The solution should work reasonably well (no pun intended) for wells from 35' deep w/ 15' of water, up to wells 300' deep w/60+' of water.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Amongst several solutions considered so far, our current front-runner is a &amp;quot;bubbler&amp;quot;, as described in &lt;a href=&quot;http://www.sensorsmag.com/sensors/leak-level/a-dozen-ways-measure-fluid-level-and-how-they-work-1067&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A bubbler-type level sensor is shown in Figure 3. A dip tube having&#xA;its open end near the vessel bottom carries a purge gas (typically&#xA;air, although an inert gas such as dry nitrogen may be used when there&#xA;is danger of contamination of or an oxidative reaction with the&#xA;process fluid) into the tank. As gas flows down to the dip tube's&#xA;outlet, the pressure in the tube rises until it overcomes the&#xA;hydrostatic pressure produced by the liquid level at the outlet. That&#xA;pressure equals the process fluid's density multiplied by its depth&#xA;from the end of the dip tube to the surface and is monitored by a&#xA;pressure transducer connected to the tube.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We are planning on using:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A 1/4&amp;quot; to 3/8&amp;quot; open-ended tube weighted down (or better yet, zip-tied to the well's up-pipe) to hang a short distance above the bottom (we can get closer in the tank, but wells tend to silt up so that will be within a couple of feet). The small down-tube is a strong point in favor of this approach because almost nothing is going into the well itself.&lt;/li&gt;&#xA;&lt;li&gt;Some (cheap) source of air pressure sufficient (300+ kPa) to blow all of the water out of the tube in the well. Once the value from the sensor plateaus it means we're blowing bubbles and we can convert pressure to feet of water.&lt;/li&gt;&#xA;&lt;li&gt;At the top we tee the tube into a differential pressure sensor, such as the &lt;a href=&quot;http://cache.freescale.com/files/sensors/doc/data_sheet/MPX5500.pdf&quot; rel=&quot;noreferrer&quot;&gt;Freescale MPX5500DP&lt;/a&gt;, which can handle up to 500 kPa, which translates to approx. 160' of water. They have a slightly more accurate one (the 5100 series) for shorter columns, such as in the tank. We selected the differential sensor to allow for varying atmospheric pressure.&lt;/li&gt;&#xA;&lt;li&gt;The specifics of the Arduino turning the air pump on/off have not been decided, but I believe it will be straightforward once we know what kind/size of pump we are trying to control.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Note: although we can easily calibrate the reading from the tank sensor, the well may be more problematic. In our own case we have a way to use a drop-line to directly measure the well depth and water column height, in other cases this may be difficult.&lt;/p&gt;&#xA;&lt;h3&gt;Questions&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there anything about this approach that is fundamentally flawed?&lt;/li&gt;&#xA;&lt;li&gt;Will temperature changes (primarily in the tank, not so much in the well) make any real difference here?&lt;/li&gt;&#xA;&lt;li&gt;Other than the volume of air needed for different diameters of tubing, will a pump have to work harder to achieve a given pressure if we use a larger or smaller down-tube?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h3&gt;Update to answer questions:&lt;/h3&gt;&#xA;&lt;p&gt;User null asked if there was unnecessary redundancy in the system; wouldn't just the depth in the tank be sufficient? Not really. Each of the measurements gives us some information the others do not. Although there is some overlap in what is being measured, I see that as an opportunity for a sanity check on the system.&lt;/p&gt;&#xA;&lt;p&gt;For example, if the measured flow from the well does not have a fairly close correlation (shifted in time because of the tank) with the combined flows to the house and irrigation system, then &lt;em&gt;something&lt;/em&gt; is out of whack.&lt;/p&gt;&#xA;&lt;p&gt;Combining the flow-from-well chart with the well-water-depth chart can give critical information about the well's &lt;a href=&quot;http://en.wikipedia.org/wiki/Groundwater_recharge&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;recharge rate&lt;/em&gt;&lt;/a&gt;. If recharge is dropping off, then we have some &lt;em&gt;serious&lt;/em&gt; trouble coming toward us.&lt;/p&gt;&#xA;&lt;p&gt;Finally, if our well-water-depth is dropping &lt;em&gt;and we aren't using that much water&lt;/em&gt; then it could mean that one of our neighbors, say the the 300 acre vineyard about 1/2 mile up the hill, is over-pumping. Unfortunately, California is the only state without any regulation of below-ground water, so we can't stop them, only get ready to order a 3,500 gallon load of water for $175 a pop.&lt;/p&gt;&#xA;" OwnerUserId="1609" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2021-08-13T03:13:02.463" Title="Low cost, moderately accurate water depth measurement" Tags="|civil-engineering|fluid-mechanics|sensors|pumps|water-resources|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="2854" PostTypeId="2" ParentId="2846" CreationDate="2015-05-17T08:57:57.283" Score="7" Body="&lt;p&gt;If your question means, &amp;quot;for a given surface area and wall thickness, is there anything that performs better than Copper&amp;quot;, then the answer is -&lt;br /&gt;&#xA;&amp;quot;Unless you are prepared to make your heat exchanger from diamond or silver, then, no.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;However, if you change your question very slightly to &amp;quot;Is some other material than Copper sometimes more effective&amp;quot; then the answer is sometimes &amp;quot;yes&amp;quot;. At least Aluminum &amp;amp; Steel are potential starters in some cases and other materials may be.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Effectiveness&amp;quot; may be measured in a range of ways depending on requirement.&lt;/p&gt;&#xA;&lt;p&gt;A CFHE (Counter Flow Heat Exchanger) of a given thermal performance may be more &lt;strong&gt;cost effective&lt;/strong&gt; if made from some form of iron or steel with a TC of 15% to 20% of copper. This would require a wall_area/wall_thickness figure  5 times greater than for Copper. If size and weight do not matter or matter little then just making it larger with a substantially lower material bill may be acceptable. (eg In many HVAC situations where space is not utterly critical and additional weight is acceptable, a galvanised iron duct heat exchanger would be acceptable if cost was significantly lower overall).&lt;/p&gt;&#xA;&lt;p&gt;Table 1, below, is a sorted subset of the table in the reference cited by Trevor. I discontinued it when thermal conductivities dropped to almost certainly impractical levels.&lt;br /&gt;&#xA;It can be seen that steels have TCs 5+ x worse than Copper.&lt;br /&gt;&#xA;Aluminum is about half as good.&lt;/p&gt;&#xA;&lt;p&gt;Table 2 below gives specific gravities for some metals.&lt;br /&gt;&#xA;Copper and Aluminum have SGs of 8.96 and 2.7 respectively.&#xA;ie for a piece of metal of given dimensions Cu weighs 8.96/2.7 = 3.3+ times as much.&lt;/p&gt;&#xA;&lt;h3&gt;Aluminum&lt;/h3&gt;&#xA;&lt;p&gt;&lt;strong&gt;Simplistically,&lt;/strong&gt; an equal performance Al heat exchanger with the same metal thickness would need to be twice as big in area as for Cu - but IT WOULD weigh 2 / 3.3 =~ 60% of the weight of a Cu heat exchanger.  So if weight was more important than size and corrosion control was adequate then Al may be &amp;quot;better than&amp;quot; Cu in that case. Even better - in the unlikely event that you could thin the metal down to half copper thickness so that thermal resistance was the same  as for Cu, then the weight would be roughly 1/2 x 1/3.3  =~ 15% of the weight of Cu. In practice the simplistic case is complicated by various second order effects (e.g. fabrication probably needs different methods), but it may well be possible to build an Al CFHE at lower mass than a functionally equivalent Copper one.&lt;/p&gt;&#xA;&lt;h3&gt;Steel or iron&lt;/h3&gt;&#xA;&lt;p&gt;Steel typically has very slightly lower density than Copper.&lt;br /&gt;&#xA;Thermal conductivity of iron in Table 1 is shown as 20% of that of Copper and various steels typically lie in the 10% - 15% range. It seems likely that a  steel based material with a TC of around 20% of that of Cu could be produced.&lt;br /&gt;&#xA;Such a material has no prospect of competing with Cu in a CFHE on a weight or volume basis - but has a good chance of winning on a cost basis, depending on various weightings. Steel is very readily available in a wide range of thicknesses and sheet sizes, and is more easily joined and sealed industrially than copper. Copper solders well. It can be welded or brazed. Steel is generally not as solderable but the lower thermal conductivity helps in soldering and it can be spotwelded, welded, brazed and more.&lt;/p&gt;&#xA;&lt;p&gt;In HVAC applications where size and weight are not major considerations steel is likely to be able to be cost effectively used for CFHE use.&lt;/p&gt;&#xA;&lt;p&gt;A few other materials in Table 1 have superior thermal conductivities to steel but are liable to be unattractive due to price or other reasons.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8KStW.jpg&quot; alt=&quot;Table 1&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/u62JI.jpg&quot; alt=&quot;Table 2&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-05-18T21:56:18.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2859" PostTypeId="1" AcceptedAnswerId="2861" CreationDate="2015-05-18T03:38:48.707" Score="16" ViewCount="11034" Body="&lt;p&gt;I once meditated on the fact that a car, in downhill, would consume more fuel when in neutral to keep the engine on, rather than in gear. To keep the engine on when in gear, in fact, you mostly need &lt;em&gt;the gravity&lt;/em&gt; and enough thrust to keep it rotating, and therefore no fuel (just motor oil).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I'd expect cars to do but I'm probably missing something. How's actually the story? And if it's not like this, would/could my idea work somehow? (note: patent pending :P)&lt;/p&gt;&#xA;" OwnerUserId="1654" LastEditorUserId="33" LastEditDate="2015-05-19T17:59:03.593" LastActivityDate="2015-05-19T17:59:03.593" Title="While going downhill, does my car consume less fuel when in neutral or in gear?" Tags="|automotive-engineering|fuel-economy|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2861" PostTypeId="2" ParentId="2859" CreationDate="2015-05-18T12:28:53.200" Score="7" Body="&lt;p&gt;Note: This is specific to gasoline based engines.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming your car was produced after the ~1990's it is equipped with a &lt;a href=&quot;http://en.wikipedia.org/wiki/Engine_braking#Legal_implications&quot; rel=&quot;nofollow&quot;&gt;Deceleration Fuel Cut-Off&lt;/a&gt; (DFCO) system which cuts the fuel flow to the engine when the gas pedal is not depressed and the car is in gear.  In this case it is more fuel efficient to stay in gear because you are letting gravity run the engine instead of fuel.  In addition, it saves wear on your brakes because the partial-vacuum created in the engine leads to &lt;a href=&quot;http://en.wikipedia.org/wiki/Engine_braking&quot; rel=&quot;nofollow&quot;&gt;engine-braking&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many states in the US it is actually required that you stay in gear when going downhill so that you have better control over the vehicle.  To quote &lt;a href=&quot;http://en.wikipedia.org/wiki/Energy-efficient_driving#Burn_and_coast&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article on energy-efficient driving&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Coasting with a vehicle not in gear is prohibited by law in most US states. An example is Maine Revised Statues Title 29-A, Chapter 19, §2064 &quot;An operator, when traveling on a downgrade, may not coast with the gears of the vehicle in neutral.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="272" LastEditorUserId="368" LastEditDate="2015-05-18T17:03:15.013" LastActivityDate="2015-05-18T17:03:15.013" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2862" PostTypeId="2" ParentId="2859" CreationDate="2015-05-18T14:27:57.880" Score="5" Body="&lt;p&gt;The answer depends a great deal on the control system around the engine, and how fast you are going down the hill relative to the gear you are in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a basic engine system, like 1960s or earlier, at idle throttle setting the fuel usage would go up a bit with engine speed.  The higher engine speed would make a higher vacuum, which would pull in more fuel-air mixture.  If the downhill speed and gearing come out to the same engine speed as idle, then the fuel usage would be the same as idle.  If you used a higher gear so that the engine would go slower, then it might use a little less fuel.  However, there isn't a lot of engine speed room below idle.  You run the risk of the engine bucking, which is not a good idea to subject the power train to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To meet late 1960s pollution limits, some cars were equipped with a &quot;decel valve&quot;.  This sort of effectively stepped on the gas for you when the engine was undergoing rapid deceleration, or was externally driven (rapid deceleration is being &quot;externally&quot; driven from the momentum of the engine).  The reason was that this condition with low fuel input would cause more pollutants to be emitted.  The short term expedient was to give it more fuel.  In such a case, running the engine faster than idle due to coasting down a hill in a sufficiently low gear would definitely use more fuel than idling in neutral down the hill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With more modern control systems, it's hard to know.  Some cars detect this condition and effectively shut down the engine.  In that case, you are better off with the engine engaged than in neutral.  For example, my Honda Civic hybrid routinely shuts down the engine and re-starts it as the conditions dictate.  Going down a hill uses no fuel at all, and actually charges the battery a bit depending on speed and gear.  Many modern cars are also capable of shutting off some of the cylinders under light load.  In that case it again would take less fuel to keep the engine engaged than to let it idle.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2015-05-18T17:06:16.690" LastActivityDate="2015-05-18T17:06:16.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2863" PostTypeId="1" CreationDate="2015-05-18T18:56:18.260" Score="6" ViewCount="261" Body="&lt;p&gt;I am trying to estimate a transfer function for a pan - tilt unit which is not as easy as I hoped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its input parameter is a velocity, but it is constrained to move from 0°  - 359° and the other way around. Saying in another way, it is not capable of continuously pannning 360°. When it hits 359° it will not move in same direction, but has to move in the other direction. Pan can be moved with max velocity $\approx \frac{\pi}{2}$ rad/s and for the tilt is the max velocity $\approx$ 0.3 rad/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same goes for the tilt, it is just limited to move within a more narrow field.  Both the pan and tilt part is driven by stepper motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of test can be performed to identify the system? I know its position and the velocity it moves with at all time.  My idea was to change input direction when the position reaches a limit, but it would just mess up my sine sweep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system ID has to be used to apply a proper controller by using pole placement, and make the system act as i want it.  The system has to be able to track  a face. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input to the system is a Pixel displacement which by using some simple math can be converted into angular displacements =&gt; the input is the angular displacement..  The PTU has to center itself such that the camera mounted at the top keeps the face at the center at all time. &lt;/p&gt;&#xA;" OwnerUserId="1659" LastEditorUserId="1659" LastEditDate="2015-05-19T07:28:23.020" LastActivityDate="2016-04-13T13:12:06.213" Title="System identification on a system with constraints" Tags="|mechanical-engineering|control-engineering|stepper-motor|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2864" PostTypeId="2" ParentId="2863" CreationDate="2015-05-18T22:56:51.220" Score="0" Body="&lt;p&gt;This partially an exploration of the question and partially a partial answer, but is not suited to the comment format.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is &quot;obvious enough&quot; but provides two simple rules to allow you to make (drastic) position adjustment decisions in the rotational direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For desired position = P and&lt;br&gt;&#xA;range of legitimate positions are  0 &amp;lt;= P &amp;lt;= 359 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate desired P. Then - &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If P_modulo_360 &amp;lt; P adjust P by -360 degrees. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If P &amp;lt; 0 adjust P by + 360 degrees.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Otherwise set camera to P by following whatever rules are established. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the tilt direction, the rule is -&lt;br&gt;&#xA;&quot;If tilt is outside range then you cannot get there from here&quot; :-(. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It seems that you CANNOT do what you want to do in all cases - eg if the face cannot be centred in the vertical axis by tilting because end of travel is reached then there is no way of centring it and you need to 'declare an exception' - probably by sounding an alarm to bring the situation to the operator's attention. However, when the &quot;end of travel&quot; is reached rotationally you can turn 359 degrees (or less) in the opposite direction to continue on the other side of the end-stop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what you are asking is to identify  situations where 'other action' is required. ie you say&lt;br&gt;&#xA;&quot;...  The PTU has to center itself ... keeps the face at the centre at all times.&quot; Two different actions are required for rotate and for tilt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tilt is &quot;easiest&quot; because there is nothing you can do to point the camera correctly if the end of travel is reached. You can only bring the situation to an operator's attention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a rotate unit that can assume absolute positions from 0 to 359 degrees and&lt;br&gt;&#xA;camera position in degrees is &quot;P&quot;.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This portion is obvious but 'sets the scene':&lt;br&gt;&#xA;If while eg rotating clockwise you logically reach a hard boundary at absolute position = 359 degrees, while intending to reach absolute position P = &quot;360 + X&quot; degrees, you can instead rotate anticlockwise to P - 360 degrees.&lt;br&gt;&#xA;Position will now be P = 360 + X - 360 = X as desired.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To achieve this transition &quot;automatically&quot; you simply compare P with P_modulo_360.&lt;br&gt;&#xA;If P_modulo_360 = P all is well.&lt;br&gt;&#xA;Any time (P_modulo_ 360) &amp;lt; P you need to adjust P by -360 degrees .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempts to pass the end stop while travelling anti clockwise  attempts to set P &amp;lt; 0.&lt;br&gt;&#xA;Any time P &amp;lt; 0 you need to adjust P by + 360 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transiting between positions either side of the rotational stop obviously plays havoc with smooth tracking and is to be avoided if at all possible.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-05-18T22:56:51.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2865" PostTypeId="1" AcceptedAnswerId="3217" CreationDate="2015-05-19T01:51:04.870" Score="7" ViewCount="154" Body="&lt;p&gt;If I take a large thin plate and I uniformly drill a very large number of holes (diameter &quot;d&quot;) very close together (minimum spacing &quot;s&quot;), what will be the stiffness of this new plate? When I say &quot;stiffness&quot; I mean the derivative of average macroscopic displacement with respect to stress. With no holes it would be Young's modulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Circle_packing#Uniform_packings&quot;&gt;Here&lt;/a&gt; is a way to visualize how the drills are to positioned, i.e. close packed circles with space in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the question I want to answer. Of course, the stiffness will depend on which direction the stress is applied - for example, if I am pulling along the &quot;grain&quot; it will be stiffer than if I turn the plate 45 degrees and pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about doing some sort of analytic model (found a few papers but they are hard to understand). Or a finite element model. However, I am interested in how the parameters &quot;d&quot; and &quot;s&quot; (see above) change the stiffness for each type of circle packing pattern. I am stuck now, what would you recommend as the best way to approach this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if this problem has already been solved (wouldn't be surprised) and someone could point me to a reference that would work too.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastEditorUserId="272" LastEditDate="2015-05-19T02:19:30.107" LastActivityDate="2015-06-16T23:35:54.550" Title="Elasticity of a drilled plate" Tags="|mechanical-engineering|stiffness|elastic-modulus|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2866" PostTypeId="2" ParentId="2832" CreationDate="2015-05-19T01:59:12.247" Score="0" Body="&lt;p&gt;This is easily done using a supersonic flow. Either heat addition or shock waves.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2015-05-19T01:59:12.247" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2867" PostTypeId="2" ParentId="2863" CreationDate="2015-05-19T05:05:17.930" Score="2" Body="&lt;p&gt;&quot;Design of experiments&quot; could help you achieve the minimum number of tests to come up with a mathematical model (transfer functions) of your setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try two methods: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First is the BLACK BOX approach&lt;/strong&gt; (you have no clue what the camera is doing).&#xA;You could divide the whole movable region into sections, elevationwise and azimuthwise. By doing different SysId tests in intersections of those areas, you would find how &lt;em&gt;location&lt;/em&gt; changes the dynamic behavior of your system. The difference might be a result of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;CG of the camera, which is not perfectly at the center and changes the inertial load on the servo motors.&lt;/li&gt;&#xA;&lt;li&gt;differing servo specs at different angular positions (due to mechanism design)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;After doing the sys-ID tests in those different positions, you would come up with a &quot;non-linear model&quot; in the sense that for different positions your model would be somewhat different. (I suspect that elevationwise difference would be more than the azimuthwise difference).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your test inputs would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;square signals (with different amplitudes),&lt;/li&gt;&#xA;&lt;li&gt;ramp commands (with different speeds),&lt;/li&gt;&#xA;&lt;li&gt;sine sweeps (with differing frequencies).&#xA;Remember that, your position &lt;em&gt;should not&lt;/em&gt; hit the &lt;em&gt;walls&lt;/em&gt; (limits), in order to not ruin the data. To do that, you need to define your limits beforehand. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example: Suppose you are doing sine sweep at Azimuth, so elevation limits are no concern (for this case). And suppose you are doing it around 300degrees. The magnitude of your sine-sweep can be 5 degrees (and with different frequencies), 10 degrees, but not 60 degrees, as probably the camera would hit the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a good practice to design your tests in a table format (Excel or otherwise), and use them as inputs to your &quot;test conducting code&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your test table might have the following data fields:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Azimuth_0 (starting point of the test-case) &lt;/li&gt;&#xA;&lt;li&gt;Elevation_0 (starting point of the test-case)&lt;/li&gt;&#xA;&lt;li&gt;Type_of_test (0,1,2, 0: square (sudden command), 1: ramp (different speeds), sine (sinusoidal) )&lt;/li&gt;&#xA;&lt;li&gt;amplitude (the range to go)&lt;/li&gt;&#xA;&lt;li&gt;Frequency (for sine inputs, 0 if it's not a sine sweep)&lt;/li&gt;&#xA;&lt;li&gt;Speed (for ramp inputs, 0 if it's not a ramp input)&lt;/li&gt;&#xA;&lt;li&gt;direction (0 for azimuth, 1 for elevation)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So at the end of allowable range, your model might not be great towards hitting the limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After getting your different systems, you could design separate control gains for separate regions, and may blend or just switch the gains as you switch from one region to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SECOND Approach can be a more functional one: the GRAY BOX approach&lt;/strong&gt;, where you know what the camera is doing when commanded.&#xA;&lt;a href=&quot;http://scholarworks.rit.edu/theses/8052&quot; rel=&quot;nofollow&quot;&gt;Here's a useful article&lt;/a&gt; describing the System Identification of a Pan-Tilt camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps to define your experiments. Good luck.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastEditorUserId="1097" LastEditDate="2015-05-19T07:24:29.970" LastActivityDate="2015-05-19T07:24:29.970" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2868" PostTypeId="2" ParentId="126" CreationDate="2015-05-19T05:25:32.440" Score="3" Body="&lt;p&gt;In addition to the above answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&quot;Standardization&quot;, using standard parts throughout the company for similar functions.&lt;/li&gt;&#xA;&lt;li&gt;Increasing commonality between products (like common power boards, common interface boards, common mechanical parts, etc).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These would also cut down on design and test efforts, as well as decrease the diversity of inputs and hence optimizing the purchase prices.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Cost-cut targets can be given for both developers and the purchasing departments, so that the teams would try to optimize the design and the buy-price.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1097" LastEditorUserId="1097" LastEditDate="2015-05-19T05:31:34.043" LastActivityDate="2015-05-19T05:31:34.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2870" PostTypeId="1" CreationDate="2015-05-19T07:24:33.127" Score="2" ViewCount="1402" Body="&lt;p&gt;I have a stepper motor with 200 steps per revolution and a pulleys configuration as following: 3 pulleys with 100 mm diameter each on 1184 mm PCD and are equally spaced (120 degrees) from each other. A large ring with a diameter of 1074 mm is placed on the center on the pulleys configuration (center of ring to center of each pulley is 592 mm). A belt of 3626.5 mm length and 30 mm width is positioned around the pulleys and contacts the ring. The motor is installed vertically on the bottom of one of the pulleys (concentric on the same axis). I want to know how many steps it takes for the motor in order to have a full revolution on the large ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a simple drawing of the configuration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z6Eoy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RMVxM.png&quot; alt=&quot;Added for Clarification&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1664" LastEditorUserId="1664" LastEditDate="2015-05-21T08:15:36.870" LastActivityDate="2015-05-21T08:32:30.263" Title="How to calculate the steps needed for a motor to rotate a circular object in a pulley configuration?" Tags="|mechanical-engineering|motors|stepper-motor|pulleys|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2871" PostTypeId="2" ParentId="2870" CreationDate="2015-05-19T08:02:17.970" Score="2" Body="&lt;p&gt;Assuming a perfect transmission, the amount the pulley needs to turn for a full revolution of the large ring is governed by the &quot;gear ratio&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta_{ring} = 2 \pi * 1074 / 100 = 67.48$ rad or 10.74 rev.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you are full stepping, this is 2,148 steps. If you are half-stepping, it's 4,296 half-steps or if quarter-stepping, 8,592 quarter-steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, it's possible that teeth are jumped, depending on how well the mesh is between the pulleys and the belt, and the belt and the ring, in which case you'll get some slipping and won't get quite a full revolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like there is a 5mm radial gap between the ring and the pulleys, so I would adjust the calculations as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{ring} = 1074/2 = 537$ mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{pulley} = 50$ mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta_{ring} = 2 \pi * (537+2.5) / (50+2.5) = 67.57$ rad or 10.276 rev.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you are full stepping, this is 2,055 steps. If you are half-stepping, it's 4,111 half-steps or if quarter-stepping, 8,221 quarter-steps.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2015-05-21T08:32:30.263" LastActivityDate="2015-05-21T08:32:30.263" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2872" PostTypeId="2" ParentId="126" CreationDate="2015-05-19T11:54:01.190" Score="1" Body="&lt;p&gt;Look at the &lt;i&gt;whole&lt;/i&gt; product cost, not just the electronic parts.  It's quite possible that those represent a relatively small part of the overall cost.  For example:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;What is the cost of assembly?  Your picture indicates reasonably modern parts that are amenable to modern volume production techniques.  Are these really being used, and used effectively?  Are you producing these devices in sufficiently large batches so that the manufacturer can use up whole reels of parts, for example.  Have a talk with your manufacturer and ask him what is causing him pain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;What yield is the raw assembly process getting.  If too low, find out what the one or two most common problems are and see if a small re-design can fix it.  This might be as simple as making some pads larger, or smaller, or better thermally balanced.  Again talk to the manufacturer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;What is the cost of test/calibration?  If this is labor intensive, perhaps you can create a automated test jig to reduce the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;What is the field cost?  Having to send a technician to fix a problem for even just 1 in 10000 units will dwarf savings of using a few cheaper capacitors, for example.  Find out what's failing and re-design to address it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My boss at my first job out of school had a nice way of visualizing failure costs.  If it takes $500 to do a field repair because of a cap that fails in one out of 500 units, think of it as sending out the unit with a dollar bill wrapped around the cap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Once you've gone thru all of the above, &lt;i&gt;then&lt;/i&gt; you can take a quick look at the BOM cost.  This is often the smallest part of the problem, but too often the first place inexperience engineers go to save money.  It can also work in reverse.  $.25 extra for a over-specified electrolytic cap that reduces expensive field failures may well be a cost saving measure.  Then also consider the hidden and less quantifiable costs of field failures, like customer dissatisfaction and loss of reputation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-05-19T11:54:01.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2873" PostTypeId="1" AcceptedAnswerId="2874" CreationDate="2015-05-19T12:02:10.323" Score="22" ViewCount="15419" Body="&lt;p&gt;If you have a look at the car's wheels, you'll notice that they have holes which can be of different forms (mostly circular or rectangular).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do they have such holes?  Doesn't that reduce the stiffness of the wheels?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XjBCgm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="303" LastEditorUserId="272" LastEditDate="2015-05-19T12:57:19.330" LastActivityDate="2016-04-26T03:07:23.847" Title="Why do car wheels have holes?" Tags="|mechanical-engineering|automotive-engineering|wheels|" AnswerCount="6" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2874" PostTypeId="2" ParentId="2873" CreationDate="2015-05-19T12:54:27.333" Score="27" Body="&lt;p&gt;Car wheels have holes mostly due to weight and cost considerations. Each hole is a chunk of material that you aren't wasting and weighing down the wheel with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another bonus, the holes help with cooling the brakes by allowing airflow between the inside and outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape and size of the holes are calculated to have a minimal impact on the structural integrity of the wheel.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="5683" LastEditDate="2016-03-21T08:25:48.677" LastActivityDate="2016-03-21T08:25:48.677" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2875" PostTypeId="2" ParentId="2873" CreationDate="2015-05-19T15:58:47.653" Score="17" Body="&lt;p&gt;Mainly to reduce weight. A car's handling characteristics are improved by keeping the 'unsprung weight' (the weight of the car not isolated from the ground by springs i.e. the wheels, axles, hubs, brake disks, calipers, etc.) as low as possible.  Holes in the wheels reduce this weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lower weight helps the unsprung portions of the car to follow the bumps and dips of the road more closely. &lt;/p&gt;&#xA;" OwnerUserId="1670" LastEditorUserId="1670" LastEditDate="2015-05-20T10:00:13.313" LastActivityDate="2015-05-20T10:00:13.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2876" PostTypeId="1" CreationDate="2015-05-19T17:33:50.583" Score="2" ViewCount="313" Body="&lt;p&gt;I am planning on having some column caps made (for Lally columns) and it seems to me the best approach would be to have them cast from ductile iron. The picture below shows the basic design:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8DY33.png&quot; alt=&quot;Lally column cap&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course steel is an alternative, but I don't trust welds and I doubt the same strength can be achieved by welding. Note that those are 1/2&quot; ribs, so it would be expensive to have steel cut and welded for the fabrication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any counter-argument to using ductile iron?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="33" LastEditDate="2015-05-19T18:24:37.210" LastActivityDate="2016-02-07T01:06:02.150" Title="Is ductile iron a good material for a support column cap?" Tags="|structural-engineering|steel|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2877" PostTypeId="2" ParentId="126" CreationDate="2015-05-19T17:48:41.350" Score="0" Body="&lt;p&gt;In addition to the above,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Consider arrayed components instead of discretes. i.e., a 4-unit 1k SMT 1206 for an IC's pull-ups instead of 4pcs 1k 0805. Saves board space and assembly time also.&lt;/li&gt;&#xA;&lt;li&gt;Weigh the expense of going double-sided with any possible reduction in PCB footprint.&lt;/li&gt;&#xA;&lt;li&gt;If out-sourcing PCB fab, component assembly, final assembly... call around. Discuss larger orders.&lt;/li&gt;&#xA;&lt;li&gt;If in-sourcing the previous, look at tools, jigs, fixtures, supplies, binning, lighting, proper seating and work area... anything that can help improve speed and efficiency.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1673" LastActivityDate="2015-05-19T17:48:41.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2879" PostTypeId="2" ParentId="2876" CreationDate="2015-05-19T18:24:01.243" Score="3" Body="&lt;h3&gt;Forces&lt;/h3&gt;&#xA;&lt;p&gt;Without providing the forces that the cap is supposed to resist, comparing two materials is rather difficult. That being said, I am going to try to provide some information that might cause you to rethink your position.&lt;/p&gt;&#xA;&lt;h3&gt;General Comparisons&lt;/h3&gt;&#xA;&lt;p&gt;You make some comments that don't seem to make sense.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Cost of fabrication&lt;/strong&gt; will likely be a main consideration. Casing a ductile iron piece requires making a sand mold for each piece. This will be more expensive than the completely common welding and cutting required to manufacture the cap from steel plate and pipe. Any machine shop can weld this piece, only specialty foundries will cast pieces. If you are looking at very large number of parts, the unit price may eventually favor casting.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Strength of metal&lt;/strong&gt; is also a consideration. You don't mention what material strength you require, but any typical strength of ductile iron will be matched by commonly available grades of plate steel.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Welding&lt;/strong&gt; is not an issue. I don't know why you &amp;quot;don't trust welds and ... doubt the same strength can be achieved by welding.&amp;quot; This is a very vague statement. Common welds use 70 ksi filler metal. This will certainly be adequate.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h3&gt;Conclusion&lt;/h3&gt;&#xA;&lt;p&gt;To answer your direct question, there probably isn't a reason why ductile iron couldn't work in your situation, but you haven't made very convincing arguments as to why it might be better. In a cost comparison, a ductile iron casting will probably be more expensive unless a very large number will be made. At that point you will need to get cost comparisons to know for sure.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-05-20T18:01:59.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2880" PostTypeId="2" ParentId="2873" CreationDate="2015-05-19T20:03:37.413" Score="5" Body="&lt;p&gt;The holes in wheels serve a few purposes. They reduce the weight of the wheel itself, although not by much. The holes in those particular wheels actually appear to be adding rigidity and strength to the wheel. The extra folds in the steel make it stronger than if it was just flat. The holes may also help prevent the build up of brake dust. I believe Ratchet Freak is correct about airflow.&lt;/p&gt;&#xA;" OwnerUserId="1681" LastActivityDate="2015-05-19T20:03:37.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2881" PostTypeId="2" ParentId="2873" CreationDate="2015-05-19T22:34:27.247" Score="4" Body="&lt;p&gt;It's for airflow to allow for extra cooling. In most situations, that extra airflow isn't going to help. But when you're doing heavy braking coming off of a mountain, it can make the difference between your being able to brake and your brakes failing from over-heating.&lt;/p&gt;&#xA;" OwnerUserId="1689" LastEditorDisplayName="user16" LastEditDate="2015-05-20T15:00:46.383" LastActivityDate="2015-05-20T15:00:46.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2882" PostTypeId="2" ParentId="2840" CreationDate="2015-05-19T23:12:02.023" Score="4" Body="&lt;p&gt;Speaking personally (I work for a US Class I as a software developer, working with a system that will interface closely with the PTC implementation at my company and knowing developers who have worked on parts of that PTC implementation):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The main physical hardware improvements required to implement AAR-type PTC on a rail system are fourfold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. The existing wayside infrastructure must be brought up to a point where it can be reasonably interfaced to the PTC system -- relay interlocking, for instance, must be replaced with solid-state, microprocessor interlocking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. The PTC-specific infrastructure must be fitted to the wayside.  This includes switch position detection and signal system interfacing, as well as the telecommunications (radio towers and backhaul) infrastructure required for PTC data radios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. PTC systems must be fitted to all locomotives operating over the network -- this includes the onboard computers, data radios, and displays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d. Finally, a back-office system must be put in place to pass authority, switch and signaling, train, and track bulletin information back and forth between the trains on the system, the wayside units, and other components (such as the dispatching system) as well as foreign railroads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atop this, there is also the need to gather extremely precise and accurate data about the permanent way (where track is located) and equipment mounted to it (such as switches, signals), as well as designate locations along the track (stations, mileposts, etal) precisely.  This is not easy -- while track geometry recording cars capture much of this data, they are only run periodically, and are vulnerable to operator error as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The design of the current PTC systems is &lt;em&gt;intended&lt;/em&gt; to be intercompatible; the Association of American Railroads has set communication standards for use in the PTC system to try to ensure this.  However, we won't know what bugs will shake out until we test this in the field, of course.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you are asking about an individual PTC implementation, the first adopter of this technology was, ironically, Amtrak, who has the &lt;a href=&quot;http://en.wikipedia.org/wiki/Advanced_Civil_Speed_Enforcement_System&quot; rel=&quot;nofollow&quot;&gt;Advanced Civil Speed Enforcement System&lt;/a&gt; or ACSES in place on most of the Northeast Corridor in order to allow high-speed Acela operations there; this was deployed in 2000.  ACSES uses much of the same technology as will be deployed in other PTC installations, and can be bridged to foreign railroad PTC systems; however, it is not quite identical in protocol to future PTC deployments, and is designed to overlay over existing coded cab signals instead of replacing them.  However, the NEC lacks the volume of traffic, the amount of hardware, and the quantity of inter-railroad interchange that the freight lines of the US have, which makes the problem significantly simpler for Amtrak to handle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="964" LastActivityDate="2015-05-19T23:12:02.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2883" PostTypeId="2" ParentId="2873" CreationDate="2015-05-19T23:16:53.837" Score="3" Body="&lt;p&gt;Yes I agree, the &quot;moment of inertia&quot; is a factor in making &quot;spoked&quot; wheels, the holes in pressed wheels will reduce weight, and allow circulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth is, for this kind of wheel, it is &lt;strong&gt;largely cosmetic.&lt;/strong&gt; &#xA;It also makes them easier to manually handle (finger holes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this type of wheel it would not make a lot of difference if they were not there. But even if they were discs milled from billet magnesium alloy they can be a lot thinner between the hub and the rims, in the same way an I-beam profile is thinner than a square beam. &lt;/p&gt;&#xA;" OwnerUserId="1506" LastActivityDate="2015-05-19T23:16:53.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2884" PostTypeId="1" AcceptedAnswerId="2886" CreationDate="2015-05-20T04:58:27.930" Score="1" ViewCount="140" Body="&lt;p&gt;In my engineering class we must make a machine that can sort 15 marbles with 3 different types in less than 2 minutes. I plan on having one marble go over/under a light sensor with a flashlight on the opposite side so it can read its reflection, then have a gear with three cups spin according to what the light sensor value is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think that the light sensor will be able to read the value and spin in less than 1 second? Or should I put something to stop the marble while it is being read?&lt;/p&gt;&#xA;" OwnerUserId="1691" LastEditorUserId="272" LastEditDate="2015-05-20T11:58:31.533" LastActivityDate="2015-05-20T16:58:06.357" Title="How fast can a VEX robotics light sensor operate?" Tags="|electrical-engineering|sensors|optics|robotics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2886" PostTypeId="2" ParentId="2884" CreationDate="2015-05-20T09:22:54.307" Score="1" Body="&lt;p&gt;The light sensor should not be a problem. Depending on the system you use this should work in milliseconds or less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanical spinning of the cup will probably be the limiting factor. 1s may be possible but balls probably drop faster than this if you don't have a gate. I would probably add a gate or use a different method to control the balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other obvious issue to me is how are the balls flowing in. If they are near continuous do you need to be able to separate two balls of the same type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you should try and design a flexible system so that you can add and remove things like a gate if they are needed or not as there will likely be some issues you haven't thought of yet.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-05-20T09:22:54.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2887" PostTypeId="2" ParentId="2870" CreationDate="2015-05-20T10:00:55.703" Score="2" Body="&lt;p&gt;You provided a lot of redundant information which is moot to the answer. Whether the pulleys drive the ring through contact, the belt, how the belt goes etc is all moot to the answer. The only thing that matters is the gear ratio between a pulley and the ring (since the motor drives the pulley directly, the gear ratio between the motor and the pulley is 1.) The only thing that could change is direction of rotation - if the belt makes &quot;8&quot; style double loop or the transmission is through contact and not belt, it would rotate the ring in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one full turn of the pulley, the belt travels the distance of its circumference. The same distance of travel is covered by any point on the circumference of the ring. Since circumferences are directly proportional to radii and diameters, we can compare these for the gear ratio instead; the ratio will be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1074mm/100mm = 10.74&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means for 10.74 rotations of the pulley the ring makes one turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now simply multiply this by number of steps of the motor per turn of the pulley:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;10.74*200 steps = 2148 steps&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result will be slightly different if you take into account some grooves, teeth, slippage, backslash, stretching of the belt and other factors, but the idea remains the same - the distance of travel of the belt is the distance of travel of the point where the belt makes efficient contact with a wheel. The ratio of circumferences of the circles drawn by these points is your gear ratio; times $n_{steps}$ you get the number of steps.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-05-20T10:00:55.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2889" PostTypeId="2" ParentId="2852" CreationDate="2015-05-20T10:44:54.273" Score="1" Body="&lt;p&gt;Depending on accuracy and &lt;em&gt;sustained&lt;/em&gt; purity of water (doesn't need to be particularly pure or dirty, just keep the same level of purity), a very cheap system would be two wires exposed to water (e.g a double wire with isolation stripped on one side), immersed in the well/container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need is to measure the resistance between the two wires; apply fixed voltage through a resistor, measure voltage drop between the wires. &lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2jbD3.png&quot; alt=&quot;schematic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;The water, allowing current flow between the wires at varied distance makes them create varied resistance depending on how far they are immersed. Callibrate the system making measurements for specific depths. Both Arduino and BeagleBone have ADC on board, and the components (other than the boards) will be below $3. This &lt;em&gt;will&lt;/em&gt; fail though if the water purity changes as the change in water resistance will utterly thwart the fine readings of wire resistances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be bypassed with a circuit similar to this one, but keeping the wires in isolation (including the immersed tips; some hot glue perhaps?) and in a higher distance from each other (e.g. a &lt;a href=&quot;http://en.wikipedia.org/wiki/File:Ladder_line.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ladder Line&lt;/a&gt; twin-lead wire) - but in this case you need a somewhat more complex circuit - an LC frequency generator with the two wires acting as a capacitor. The water level will act as dielectric changing the capacitance of the line, and you need to measure the frequency changes in software. Still, the circuit board shouldn't be more than $15 or so.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-05-20T10:44:54.273" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2890" PostTypeId="1" CreationDate="2015-05-20T13:42:23.193" Score="6" ViewCount="602" Body="&lt;p&gt;&lt;strong&gt;Problem statement&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an engineering project in school, we are checking the stability in the longitudinal direction of an apartment building, taking into account the dynamic effect of earthquakes. For this, we are making a two-dimensional finite element model in Matlab. We have to assume that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is a plane stress condition in the columns &lt;/li&gt;&#xA;&lt;li&gt;There is plane strain condition in the floors and walls.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The stability of the building is guaranteed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In the transverse direction: membrane action in the columns and the walls.&lt;/li&gt;&#xA;&lt;li&gt;In the longitudinal direction bending stiffness of the connections between the columns, the floors and walls.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The picture attached shows an apartment building with properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nine bays (each 5 m width and 0.25 m thick)&lt;/li&gt;&#xA;&lt;li&gt;Six floors (each 3.5 m height and 0.25 m thick)&lt;/li&gt;&#xA;&lt;li&gt;Depth of the building is 12 m.&lt;/li&gt;&#xA;&lt;li&gt;The building is supported by twenty columns (each 1.6 m width and 5 m height)&lt;/li&gt;&#xA;&lt;li&gt;The thickness of the columns has still to be determined but should lie between 0.2 m and 0.8 m.&lt;/li&gt;&#xA;&lt;li&gt;E = 35 GPa&lt;/li&gt;&#xA;&lt;li&gt;Coefficient of Poisson = 0.25&lt;/li&gt;&#xA;&lt;li&gt;Mass density = 2500 kg/m³&lt;/li&gt;&#xA;&lt;li&gt;Finishing layer of 350 kg/m²&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Plane strain and plane stress can be modeled by means of beam elements, provided that in the elements with plane strain condition adjusted values E' and ν' for the Young’s modulus and the coefficient of Poisson are provided:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E'=E/(1-ν^2) and ν'=ν/(1-ν)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are struggling with the implementation of the plane stress and plane strain conditions.  We have tried to model this as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Plane strain: We modelled the building with a thickness of 1 m for the superstructure. Therefore, we adjusted the mass density from 2500 kg/m³ to 12 m*2500 kg/m³.  In this way the mass of each element is obtained by multiplying by the thickness and length of the element.&lt;/li&gt;&#xA;&lt;li&gt;Plane stress: We doubled the stiffness of the column to include both columns at  the base. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This implementation gives not the results we are hoping for. Are there any suggestions for improvement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QFCzO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1700" LastActivityDate="2015-09-12T15:55:37.270" Title="Dynamics of structures: combined plane stress and plane strain conditions" Tags="|civil-engineering|structural-engineering|structures|dynamics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2891" PostTypeId="2" ParentId="2884" CreationDate="2015-05-20T16:58:06.357" Score="1" Body="&lt;p&gt;A sensor's switching frequency will usually be in the order of milliseconds, so it definitely won't be a problem. However the highest delay you will have in the system is turning to the right cup to dispense the last marble that was sensed. Attach a two bit absolute encoder on the spinning gear to always find the shortest path to the right cup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the marble feeder you can use a vertical tube stack of the marbles to be sensed. Two gates can be used along the tube feeder, one that will dispense the marbles one by one from first to last to a chamber for them to be sensed, and a second one to dispense it in the corresponding cup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This task should be a fast one to solve using this setup. Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="1704" LastActivityDate="2015-05-20T16:58:06.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2893" PostTypeId="1" CreationDate="2015-05-21T07:18:44.533" Score="3" ViewCount="2313" Body="&lt;p&gt;I have a carbon fiber composite component that is often and repeatedly subjected to stress. When it was brand new, everything looked normal. After a week of use, there are what seem to be bubbles developing under the clear coat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Why would there be bubbles developing under the clear coat? Wouldn't the manufacturing process of heating and vacuuming bagging have eliminated most of the air within the composite? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Does the presence of these bubbles indicate a manufacturing defect? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suspicions are that these bubbles are merely a cosmetic defect as I can see them, which indicates that they are not in between the layers of unidirection fiber which actually gives the structure its strength. On the other hand, the presence of the bubble does make me wonder - are there possibly large voids within the structure? From what I understand voids weaken carbon fiber composites, correct? &lt;/p&gt;&#xA;" OwnerUserId="1715" LastActivityDate="2015-05-21T13:14:03.037" Title="Bubbles in carbon fiber composite" Tags="|materials|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2895" PostTypeId="1" AcceptedAnswerId="2901" CreationDate="2015-05-21T09:03:46.213" Score="8" ViewCount="1034" Body="&lt;p&gt;A neat image &lt;a href=&quot;http://en.wikipedia.org/wiki/British_Rail_Mark_3#/media/File:BR_Mk.IIIa_TSO_No.12604_%288074749189%29.jpg&quot; rel=&quot;noreferrer&quot;&gt;from Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/c8JHcl.jpg&quot; alt=&quot;railway carriage side looks neat&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the railway carriage side looks like it is crafted from a single sheet of metal (if we ignore the doors and the windows and some minor parts like that small thing near the carriage number) - no seams, no rivets, no anything just a 20+ meters wide and about 3 meters tall single sheet of metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it indeed possible and reasonable to craft a carriage side of such large single sheet or should I expect that there're neatly hidden seams somewhere?&lt;/p&gt;&#xA;" OwnerUserId="374" LastActivityDate="2015-05-21T21:25:18.223" Title="Is this railway carriage side likely made of single metal sheet?" Tags="|materials|rail|metals|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2896" PostTypeId="2" ParentId="2895" CreationDate="2015-05-21T09:12:12.860" Score="8" Body="&lt;p&gt;It's possible, aluminum sheeting comes in &lt;a href=&quot;https://www.google.be/search?q=aluminum%20roll&amp;amp;num=100&amp;amp;safe=off&amp;amp;tbm=isch&quot;&gt;large rolls&lt;/a&gt;; You take one that is wide enough and you can cut out a single sheet long enough for the length of the carriage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's probably attached to the structure using welds on the inside&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However seams can be very well hidden under the paint after welding them together and sanding it down.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-05-21T09:12:12.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2897" PostTypeId="2" ParentId="2893" CreationDate="2015-05-21T13:14:03.037" Score="2" Body="&lt;h3&gt;Bubbles&lt;/h3&gt;&#xA;&lt;p&gt;Since you mentioned that the &amp;quot;bubbles&amp;quot; appear to be under the clear coat, it is likely caused by a failure of the bond between the clear coat and the material underneath. This is especially likely since you said that the part is subjected to repeated stressing. The clear coat, carbon fiber and epoxy are different materials and have different physical properties.  The stress causes each material to act differently.&lt;/p&gt;&#xA;&lt;p&gt;This doesn't mean that there were bubbles to begin with. The clear coat is very good at transmitting light, so discontinuities will show up readily. You would have been able to see them initially if they were there.&lt;/p&gt;&#xA;&lt;h3&gt;Manufacturing defect&lt;/h3&gt;&#xA;&lt;p&gt;It is really hard to tell from your description if this is a manufacturing defect or not. It could be, but it also could be caused by using the part in a way that it wasn't designed for.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-05-21T13:14:03.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2898" PostTypeId="2" ParentId="2873" CreationDate="2015-05-21T13:29:49.910" Score="19" Body="&lt;p&gt;Many of the answers so far have mentioned that part of the purpose of the holes is weight reduction, but most of them don't express why weight reduction in the wheels is important.  There are two major reasons; the first (also mentioned by Steve Ives) is that the suspension systems in vehicles operate better if the 'unsprung' mass is kept as low as possible, and the second (not mentioned so far) is that &lt;strong&gt;shaving weight from the wheels contributes more significantly to performance than shaving weight from the rest of the vehicle&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why this is true, consider the energy that the engine must put into the vehicle to get it moving at a speed $v$:&#xA;$$&#xA;E=\frac12 m_tv^2+\frac12 I\omega^2&#xA;$$&#xA;For the wheels we can express the angular velocity in terms of the linear velocity as $\omega=\frac{v}{r}$ where $r$ is the radius of the wheel.  The moment of inertia of the wheel can be expressed as $I=\eta\ m_dr^2$ where $m_d$ is the mass of the wheel.  Here $\eta$ is a number between $\frac12$ and 1, but is closer to 1 since a wheel is closer to a &lt;a href=&quot;http://en.wikipedia.org/wiki/List_of_moments_of_inertia&quot;&gt;thin hoop than a solid disk&lt;/a&gt;. Sticking all of this back in gives&#xA;$$&#xA;\begin{align}&#xA;  E&amp;amp;=\frac12 m_tv^2+\frac12\eta\ m_dr^2\frac{v^2}{r^2}\\&#xA;  &amp;amp;=\frac12 (m_s+m_d)v^2+\frac12\eta\ m_dv^2\\&#xA;  &amp;amp;=\frac12[m_s+(1+\eta)m_d]v^2,&#xA;\end{align}&#xA;$$&#xA;where I have used $m_s$ as the non-rotating mass of the vehicle.  So, you can see that &lt;strong&gt;shaving mass from the wheels is equivalent to shaving a factor of $1+\eta\simeq 2$ as much mass from the non-rotating parts of the car&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is an additional, relatively minor, effect due to angular momentum for which it is advantageous to reduce the weight of the wheels.  Due to conservation of angular momentum, the body of the car will tend to roll toward the outside of a turn when the wheels are rotated to initiate the turn.  Reducing the moment of inertia of the wheels reduces their angular momentum and thereby reduces the amount of body roll upon steering.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-05-21T13:29:49.910" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2899" PostTypeId="2" ParentId="410" CreationDate="2015-05-21T15:06:45.503" Score="2" Body="&lt;p&gt;The focal distance of the laser lens and the distance from the target (i.e. table) that an optical mouse will work with are two different things. You ask for the former, but it seems that you want the later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer &lt;strong&gt;the question that you asked&lt;/strong&gt;: Every mouse is different, but you can expect on the range of millimeters. If you can remove the lens from the mouse, then hold it with pliers and move it closer and further from a white sheet of paper under the sun. The height at which the sunlight makes a point on the paper is the focal distance. On a cheap mouse laser lens that I tried, I got about half a centimeter judging by eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer &lt;strong&gt;the question that you meant to ask&lt;/strong&gt;: Every mouse is different, but you can expect on the range of millimeters to just a few centimeters. My nice modern Microsoft Curve wireless mouse fails to work at more than 2-3 millimeters above the mousepad. My Logitech G500 works a good centimeter above the mousepad, and I remember when optical mice first came out how I could hold the mouse a good 5 cm above the desk and still have it detect the motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I recommend trying to get the oldest optical mouse that you can, the sort with the bright red laser.&lt;/strong&gt; By my own testing, these seem to work furthest from the targeting surface.&lt;/p&gt;&#xA;" OwnerUserId="449" LastActivityDate="2015-05-21T15:06:45.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2900" PostTypeId="2" ParentId="2852" CreationDate="2015-05-21T15:26:08.723" Score="2" Body="&lt;p&gt;I think your actual goal is to measure the volume of water in the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least for the tank, you could apply strain gauges to the base of the tank. More water in the tank means more weight, which in turn means a different amount of strain. The exact relationship depends on the base and how you apply the gauge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage is that you do not have to put anything inside the tank. The disadvantages are that this will not work for the well.&lt;/p&gt;&#xA;" OwnerUserId="1385" LastActivityDate="2015-05-21T15:26:08.723" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2901" PostTypeId="2" ParentId="2895" CreationDate="2015-05-21T18:16:09.673" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it indeed possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and reasonable to craft a carriage side of such large single sheet&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It can be reasonable, but typically isn't.  To form an entire side of the carriage from one sheet of steel would require a forming press that is gargantuan. However, one could instead have three more reasonably sized presses and then weld the sections together. Further, this could be more flexible - changing out the press plates is easier and faster, storing smaller press plates is cheaper and easier, if the machine breaks down you can still use the other presses, whereas if one piece of the gargantuan press breaks down, the entire operation might be waiting for the repair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a whole host of issues that follow the same reasoning - if a panel is damaged, it's expensive and time consuming to rework, and affects a much larger part of the carriage than a single damaged panel would.  The pieces are much harder to work with, move, and fasten, etc, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's very unlikely that it's a single seamless sheet of steel or aluminum.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;or should I expect that there're neatly hidden seams somewhere?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You should expect seams, and in fact, as &lt;a href=&quot;https://engineering.stackexchange.com/questions/2895/is-this-railway-carriage-side-likely-made-of-single-metal-sheet#comment4955_2895&quot;&gt;ratchet freak pointed out&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The bodyshell is [...]of full monocoque construction with an all-welded mild steel stressed skin,&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Monocoque is a technique where the skin or surface of the construction serves an integral support purpose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seams, therefore, may also provide support as ribs, and so you may find that having seams provides an added benefit, or perhaps reduces the need for internal supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seams, of course, are not visible for both aerodynamic and visual appeal purposes, and it's relatively easy to hide weld lines with grinding, sanding, and polishing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So without actual proof of manufacturing process, I believe you can safely assume smaller sheets with invisible seams are more likely than single seamless formed sheets.&lt;/p&gt;&#xA;" OwnerUserId="784" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-05-21T21:25:18.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2903" PostTypeId="2" ParentId="2832" CreationDate="2015-05-22T05:22:37.153" Score="5" Body="&lt;p&gt;The high-voltage construction in an &lt;a href=&quot;http://en.wikipedia.org/wiki/Ionocraft&quot; rel=&quot;noreferrer&quot;&gt;ionocraft&lt;/a&gt; provides air flow without any moving parts. The pressure difference is very small, but combining multiple stages will increase it.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2015-05-22T05:22:37.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2904" PostTypeId="1" AcceptedAnswerId="2907" CreationDate="2015-05-22T11:47:17.743" Score="3" ViewCount="112" Body="&lt;p&gt;I have noticed when examining performance of various heating elements in different atmospheres that in a vacuum maximum temperatures are lower than the in air temperatures. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Typical heating element alloys are Nickel-Chromium and Iron-Chromium-Aluminum)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As Olin Lathrop says below the evaporation of the element becomes significantly increased in a vacuum. I found this interesting quote in an old book:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This idea of working in a vacuum brought with it new difficulties as&#xA;  regards the construction of a furnace. The ordinary platinum&#xA;  resistance furnace proved of little value, as, owing to diminution of&#xA;  pressure, the volatility of the platinum foil, at high temperatures,&#xA;  was increased to such an extent that it was destroyed in a short time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2015-05-27T18:40:31.090" LastActivityDate="2015-05-27T18:40:31.090" Title="Why do heating elements have lower rated temperatures in vacuums?" Tags="|thermodynamics|temperature|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2905" PostTypeId="1" AcceptedAnswerId="4199" CreationDate="2015-05-22T12:54:26.960" Score="10" ViewCount="5168" Body="&lt;p&gt;I have just begun studying mechanics of materials and I am struggling to understand intuitively &lt;strong&gt;how to select the area&lt;/strong&gt; in first moment of area calculations. I was hoping someone has a relatively easy explanation.&lt;/p&gt;&#xA;&lt;p&gt;The problem arises when calculating the shear stress &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; at a specific point in a beam due to a specified shear force. For &lt;span class=&quot;math-container&quot;&gt;$\tau_{xy}$&lt;/span&gt; the calculations seem to be the same:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5eI5E.jpg&quot; alt=&quot;Example Problem&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\tau_{xy}$&lt;/span&gt; due to &lt;span class=&quot;math-container&quot;&gt;$V(x)$&lt;/span&gt; at point A requires the calculation of the first moment of area, &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt;, shaded here:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/FxZFBm.jpg&quot; alt=&quot;Area 1&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, then the problem requires me to find &lt;span class=&quot;math-container&quot;&gt;$\tau_{xz}$&lt;/span&gt; due to &lt;span class=&quot;math-container&quot;&gt;$V(x)$&lt;/span&gt; at point B. The shaded area under is the area we have to use, and my question is why?&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tAXdZm.jpg&quot; alt=&quot;Area 2&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I know this is probably a very banal question for you guys, but I just really want to understand this, and browsing the web has led me nowhere.&lt;/p&gt;&#xA;" OwnerUserId="1721" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-01-27T02:14:27.057" Title="How do you select the correct area to consiser for first moment of area calculations?" Tags="|mechanical-engineering|materials|beam|stresses|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2906" PostTypeId="2" ParentId="2852" CreationDate="2015-05-22T13:32:17.797" Score="2" Body="&lt;p&gt;An alternative could be a barometric chip enclosed in a waterproof container with a membrane, weighed down to rest at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.adafruit.com/products/1603&quot; rel=&quot;nofollow&quot;&gt;barometer circuit for Arduino&lt;/a&gt; is available from Adafruit for under $10. If you go for standalone chip, you can reduce the price even more. It communicates over I2C, so you can attach it to BeagleBone as well. Your worst headache now is an enclosure that is entirely waterproof but doesn't isolate the inside from pressure changes - some kind of flexible membrane would be necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accuracy would be somewhat affected by weather (air pressure) with about +/- 0.5m inaccuracy though it may be nullified by a second barometer on the surface, measuring the air pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As usually, the device would need to be callibrated in software, individually, by submerging it to two known depths and recording the readouts as fixed points, letting it extrapolate from there.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-05-22T13:32:17.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2907" PostTypeId="2" ParentId="2904" CreationDate="2015-05-22T13:41:16.807" Score="3" Body="&lt;p&gt;One reason is because in vacuum there is no convection to help remove the heat.  All heat loss from the element is either by radiation, or via conduction thru the supports and wires.  Technically this only means lower maximum heater power, but that may be dumbed down to temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason is due to evaporation of the heating element.  At high enough temperatures, this can matter.  With 1 atm ambient pressure, evaporation is greatly reduced at low partial pressures, like heater element metals have at glowing temperatures.  In vacuum, even a little pressure is greater than ambient (because ambient is theoretically 0), so there can be significantly faster evaporation, reducing the life of the heating element.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-05-22T13:41:16.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2908" PostTypeId="1" CreationDate="2015-05-22T17:21:13.943" Score="14" ViewCount="237" Body="&lt;p&gt;I am working with a set of alloys that must first be ground up in a ball mill at cryogenic temperatures. It is important that no contamination makes it into the material and careful steps are taken to ensure this end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately when we ran a set of samples through our RBS lab and then again through PIXE we saw that we had Fe and Cr contamination. At first it was thought that it came from the cutting tools used to prepare the samples so we ran another sample that was cut using EDM. The results were the same. The only other contact the material has with a stainless steel is from the ball milling step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have been using a 440C for the cup and balls but it would seem a 316L may be a better fit for this application. I know 440C does not typically do well at cryo-temps but that is what other labs have been using so we didn't expect for there to be problems of this type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some things to consider for the new cup and ball material are machinability, cost, availability, cryogenic properties, vibration and fatigue resistance, and ability to be sealed (cups are filled in an inert Ar atmosphere). Another possibility is a heat treatment of the current 440C materials, however I am unsure what the best approach would be in that regard.&lt;/p&gt;&#xA;" OwnerUserId="553" LastEditorUserId="553" LastEditDate="2015-05-22T17:30:49.970" LastActivityDate="2016-04-27T20:26:31.290" Title="Looking for best material for a cryogenic ball mill cup and balls" Tags="|materials|cryogenics|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2909" PostTypeId="2" ParentId="2821" CreationDate="2015-05-22T18:39:51.573" Score="5" Body="&lt;p&gt;Gas turbines are modeled using Brayton cycle which in the simplest case consits of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Isentropic compression (in a compressor) &lt;/li&gt;&#xA;&lt;li&gt;Constant-pressure heat addition (combustion chamber) &lt;/li&gt;&#xA;&lt;li&gt;Isentropic expansion (in a turbine)&lt;/li&gt;&#xA;&lt;li&gt;Constant pressure heat rejection&#xA;&lt;img src=&quot;https://i.stack.imgur.com/KT5G5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And since efficiency is defined as Net output work / Heat input, Efficiency can be easily related to temperature of cycle states as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fAqiQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Processes 1-2 and 3-4 are isentropic, and P2 = P3 and P4 = P1. Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sRRqJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally efficiency can then be related to compression ratio as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MEQN6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However most gas turbines do not operate on this simple theoretical ideal conditions as isentropic compression&amp;amp; expansion, constant pressure heat addition, single stage compression and single stage expansion. And in such cases the modeling and efficiency analysis are far more complex than the ideal cycle.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Abbreviations:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$w_{net}$: mechanical net work, Turbine power minus compressor power&lt;/li&gt;&#xA;&lt;li&gt;$q$: heat, addition or rejection. $q_{in}$ = heat addition and $q_{out}$ = heat rejection &lt;/li&gt;&#xA;&lt;li&gt;$k$: specific heat ratio $Cp/Cv$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2015-06-07T05:58:17.593" LastActivityDate="2015-06-07T05:58:17.593" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2910" PostTypeId="1" CreationDate="2015-05-22T20:27:19.793" Score="0" ViewCount="256" Body="&lt;p&gt;What does the word &quot;pararail&quot; mean? I would like to know the meaning. Search over web does not help much.&lt;/p&gt;&#xA;" OwnerUserId="426" LastActivityDate="2015-05-23T09:28:11.180" Title="What does the word &quot;pararail&quot; mean?" Tags="|terminology|" AnswerCount="1" CommentCount="6" ClosedDate="2015-05-23T14:34:44.707" ContentLicense="CC BY-SA 3.0" />
  <row Id="2911" PostTypeId="1" CreationDate="2015-05-23T01:56:58.217" Score="3" ViewCount="1798" Body="&lt;p&gt;Why is the word &quot;mileage&quot; used for vehicles, instead of efficiency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any difference between mileage and efficiency? &lt;/p&gt;&#xA;" OwnerUserId="1729" LastEditorUserId="368" LastEditDate="2015-07-02T22:40:43.957" LastActivityDate="2016-01-12T21:29:25.437" Title="Why the word &quot;mileage&quot; is used for vehicles, instead of efficiency?" Tags="|mechanical-engineering|automotive-engineering|terminology|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="2912" PostTypeId="1" CreationDate="2015-05-23T04:10:28.997" Score="5" ViewCount="147" Body="&lt;p&gt;When the ultrasound transducer emits a signal and the signal comes back to the receiver how does the software or hardware determine where it came from? I understand it finds distance or depth by delay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example when you see an ultrasound of a baby and it has the 2D spread, how does it determine if the reflected wave is from the right or left side of the spread.&lt;/p&gt;&#xA;" OwnerUserId="1731" LastActivityDate="2015-06-04T11:20:24.020" Title="How does an ultrasound device determine where the reflection is coming from?" Tags="|electrical-engineering|measurements|optics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2914" PostTypeId="2" ParentId="2911" CreationDate="2015-05-23T08:52:42.347" Score="4" Body="&lt;p&gt;&lt;strong&gt;Short:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fuel efficiency was long ago typically measured in&lt;br&gt;&#xA;&lt;strong&gt;miles&lt;/strong&gt; traveled per gallon of fuel&lt;br&gt;&#xA;= &lt;strong&gt;miles&lt;/strong&gt; per gallon&lt;br&gt;&#xA;= &lt;strong&gt;m&lt;/strong&gt;pg&lt;br&gt;&#xA;= &lt;strong&gt;mileage&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays we tend to use litres per 100 kilometre (BUT we don't say litreage). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both measures of efficiency have merit - the old way (mpg) stated what you could achieve with what you had.   The new way (l/100 km) states what you need to achieve what you want to do. Either method is good enough once people are used to it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Longer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vehicle's efficiency of fuel use is usually stated in one of two ways&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Distance traveled on a given amount of fuel&lt;/li&gt;&#xA;&lt;li&gt;Amount of fuel used to travel a given distance&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both are legitimate means of conveying differences in efficiency but one is proportional to the inverse of the other so not only does one increase as the other decreases but the way that inverse proportionality works, one &quot;squashes the results together&quot; at one end of the measurement range and the other does it at the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have noted, distance in the UK was traditionally measured in miles and fuel volume was measured in gallons. Efficiency statements were usually of the form in (1) above ie distance per standard amount of fuel.&lt;br&gt;&#xA;The resultant logical units are miles per gallon or mpg.&lt;br&gt;&#xA;&quot;Mileage&quot; is just a &quot;short hand&quot; way of expressing mpg efficiency.&lt;br&gt;&#xA;A modern small car may achieve mileages of 30 to 50 mpg. A truck may achieve a mileage or fuel economy of 5 to 10 mpg, the heavier the truck the lowert the mpg and a 60 tonne/ton + main battle tank in enthusiastic may approach the gallons per mile range.  So the mpgs between a very carefully driven small car and a medium car driven at motorway speeds may be 50 and 25 mpg - with a difference of 50-25 = 25 mpgh difference. However a tank transporter may achieve say 4 mpg but the main battle tank it carries may achieve 2 mpg. The ratio is again 2:1 but the difference is 4-2 = 2 mpg.&lt;br&gt;&#xA;So a 2:1 difference may be 25 mpg or 2 mpg. This is OK as long as it is understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my country at least (New Zealand) we used to use method 1. above but it is now more common to use form 2. As distance units are kilometres and fuel volumes are measured in litres you MIGHT expect that we would measure economy in litres per kilometre. We don't. A very economical small car may achieve efficincies of 0.05 litres per kilometre or 50 ml per km. A heavy truck may achieve 300 ml per km. These are perfectly sensible units to use BUT mls ( = millilitres, = 1/1000 th of a litre) are not what people are used to thinking oin when buying fuel. and, while a typical fuel tank may contain 50,000 ml of fuel, most people are not overly aware of this. So it was desired to have a unit that used litres rather than ml. And so, the measure used is litres per 100 km. So the 50 ml/km car now achieves 5l/100km and the 300 ml/km lorry now achieves 30l/100km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do NOT generally talk about &quot;literage&quot; instead of &quot;mileage&quot; even though it would make as much sense to do so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we still get &quot;compression&quot; of results at one end of the range  - it's just at the other end. A medium car around town may achieve 10 l/100km. A small car on the open road may achieve 5 l/100 km. The ratio is 2:1. The difference is 10l-5l = 5l/100 km difference. But a tank transporter may achieve say 100l per 100 kilometre and a Main Battle Tank 200 l/00 km. The ratio in duel consumption is again 2:1 but now the difference is 200 -100  = 100 l/100 km difference.&lt;br&gt;&#xA;So 2 cars with 2:1 difference in fuel economies may achieve 5l/100km difference in fuel consumption.  2 heavy vehicles with 2:1 ratio of fuel efficiency may have a 100 l/100km difference in consumptions. Again these results are not a problem as long as people understand them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interesting change occurs with the change away from mileage (= mpg) to l/100 km. It was common to talk about &quot;fuel economy&quot; -the result was in miles per unit of fuel used, and went up as economy improved. We now till talk about fuel economy with the measure being litres per special-unit of distance and the result goes DOWN as economy improves. If we are talking about &lt;strong&gt;fuel&lt;/strong&gt; economy it arguably makes sense to have a measure which is expressed in terms of fuel used.  Arguably :-). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with .   &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="282" LastEditDate="2015-06-01T22:39:29.130" LastActivityDate="2015-06-01T22:39:29.130" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2915" PostTypeId="2" ParentId="2910" CreationDate="2015-05-23T09:28:11.180" Score="2" Body="&lt;p&gt;If this was a quiz show and you had one guess, the safest offering may be 'The suspension system in a vintage Scorpion snowmobile&quot;, &lt;strong&gt;BUT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Para rail&quot; and &quot;pararail&quot; are 'made up' words usually used as product names and/or trademarks for a range of things by a range of manufacturers over a range of years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google's ngram viewer swears it does not know the word - usually a goodish sign that a term is a specialist one.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here are some of the available &quot;pararails&quot;:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tactical-life.com/gear/suarez-international-advanced-aug-forward-para-rail-one-source-tactical-new-product/#suarez-international-advanced-aug-forward-para-rail-_-new-product-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Steyr AUG Rifle foward grip slide &lt;strong&gt;pararail&lt;/strong&gt;&lt;/a&gt; - the slide rail mid picture.&lt;br&gt;&#xA;Guess: Paratroop rail &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hTjQv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.vintagesnowmobile.com/cgi-bin/yabb2/YaBB.pl?num=1210039557&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Pararail&lt;/strong&gt; suspension on a vintage scorpion snowmobile&lt;/a&gt;&lt;br&gt;&#xA;Guess: Wazzat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also &lt;a href=&quot;http://www.vintagesledders.com/forums/index.php?topic=966.20&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;br&gt;&#xA;and &lt;a href=&quot;http://classicsnowmobiles.blogspot.co.nz/2015_02_01_archive.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W2nUp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://wssafety.com/products/rooftop-guardrails-safety-rail-2000/architectural-series/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Same name&lt;/a&gt; same type of product different company. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.all-liftsystems.com/products/safetyfall-protectionguardrail-systemsmodular-railing-systems/railing-kits/pararail-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pararail rooftop fall protection system&lt;/a&gt;&lt;br&gt;&#xA;Guess: Parapet rail&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sny9H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.nz/search?q=pararail&amp;amp;biw=1776&amp;amp;bih=1063&amp;amp;tbm=isch&amp;amp;tbo=u&amp;amp;source=univ&amp;amp;sa=X&amp;amp;ei=U0NgVYj9L4v48QWc8ICgBQ&amp;amp;ved=0CC4QsAQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Many maybes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-05-23T09:28:11.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2917" PostTypeId="1" AcceptedAnswerId="2919" CreationDate="2015-05-23T14:01:24.447" Score="8" ViewCount="2781" Body="&lt;p&gt;System with compressed air seems more expensive and harder to build and integrate neither using simple electromagnetic drives to open public transport doors.&lt;/p&gt;&#xA;" OwnerUserId="303" LastActivityDate="2015-05-23T21:52:41.193" Title="Why does public transport uses compressed air to open doors instead of electromotors?" Tags="|transportation|motors|compressed-air|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2918" PostTypeId="2" ParentId="2917" CreationDate="2015-05-23T14:38:25.577" Score="7" Body="&lt;p&gt;They are smoother and if something gets caught in between them then the maximum force it can exert is capped depending on the pressure and the piston area. This means that a trapped limb is less likely to get crushed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an emergency (a typical concern for public transport) you only need to open a valve (pulling the emergency open will do that) and push to open them. This is simpler and more reliable than disconnecting a electronic actuator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a last point most public transport vehicles already have an existing pneumatic system for the brakes and suspension (as demonstrated by all the hissing a typical bus does) to tap into.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-05-23T21:52:41.193" LastActivityDate="2015-05-23T21:52:41.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2919" PostTypeId="2" ParentId="2917" CreationDate="2015-05-23T14:49:20.427" Score="13" Body="&lt;p&gt;Do you mean, for example, a bus entry door? It may be as simple as the &quot;service factor&quot; for motors (&quot;electromagnetic devices&quot;) as compared to pneumatics. Inexpensive motors with sintered bronze bushings wear out, and better ones with actual bearings must be routinely lubricated. Air-operated cylinders are cheap and last a very long time, with the air typically containing traces of added oil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two are really not all that different, and this has been a hot-button topic for a long time. Electrical systems require a generator/alternator, regulator, wiring, control device, and actuator. Pneumatic systems require a compressor, tank/oiler, tubing, valve, and piston. Electrical systems (wiring) can short or open; pneumatic systems can plug or leak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.automationworld.com/motion-control-systems/how-decide-between-pneumatic-and-electric-actuators&quot;&gt;This article&lt;/a&gt; gives some pros and cons between the two, with interesting cost-analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electro-Mechanical&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Provide precise control and positioning.&lt;/li&gt;&#xA;&lt;li&gt;Help adapt machines to flexible processes.&lt;/li&gt;&#xA;&lt;li&gt;Low operating cost.&lt;/li&gt;&#xA;&lt;li&gt;Most economical when deployed in a moderate scale in processes where their performance advantages can be a benefit and when the electronics are separate from the actuator to segment and minimize replacement costs.&lt;/li&gt;&#xA;&lt;li&gt;Stepper motors are an economical choice for accurate positioning at lower speeds. However, steppers may lose synchronization with the controller when employed open loop without an encoder or if they are undersized for an application. &lt;/li&gt;&#xA;&lt;li&gt;Servos, by definition, are closed loop and provide superior performance at high speeds, albeit at a higher cost.&lt;/li&gt;&#xA;&lt;li&gt;Components of an electric actuator include the mechanical actuator that translates motor rotation to linear speed and thrust, the motor, an electronic driver or amplifier to power the motor, and a controller to control motion. The total cost for these components ranges from 800 USD to 3,000 USD and up.&lt;/li&gt;&#xA;&lt;li&gt;While component costs of electric actuators are high, operating costs are low. High component costs often deter the use of electric actuators because savings in operating costs compared to pneumatics are often not adequately considered or are outright ignored.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pneumatic Cylinders&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Provide more force and speed per unit size than any other actuator except hydraulic. &lt;/li&gt;&#xA;&lt;li&gt;Force and speed are easily adjustable and are independent of each other.&lt;/li&gt;&#xA;&lt;li&gt;Are most economical when the scale of deployment matches the capacity of the compressor.&lt;/li&gt;&#xA;&lt;li&gt;Small compressors are efficient and economical when used for a small number of devices.&lt;/li&gt;&#xA;&lt;li&gt;Low component costs.&lt;/li&gt;&#xA;&lt;li&gt;High operating costs (install, replacement cylinders, electricity for compressor (76%))&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Added: In addition to the points @ratchet freak has made, mechanical break-away (limit) systems are possible for motors; however they are seldom used. Typically the motor is sized to generate torque below a threshold value as to not cause harm. However, two issues arise: 1. This is a mechanical linkage, and would not be as repeatable, reliable, or self-correcting as a (pressure regulated) pneumatic system. And 2. If the motor failed, the actuator is usually locked due to mechanical gearing (thus relying on break-away linkage.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to implement a &lt;a href=&quot;https://www.parker.com/portal/site/PARKER/menuitem.7100150cebe5bbc2d6806710237ad1ca/?vgnextoid=f5c9b5bbec622110VgnVCM10000032a71dacRCRD&amp;amp;vgnextfmt=EN&amp;amp;vgnextdiv=&amp;amp;vgnextcatid=4491073&amp;amp;vgnextcat=POWERROD+LINEAR+MOTOR+CYLINDER&amp;amp;Wtky=#&quot;&gt;cylinder as a linear motor&lt;/a&gt;; essentially a solenoid with two or more coils. But it would be heavy, use lots of copper wire, and use an elaborate drive system (especially if to limit maximum thrust.)&lt;/p&gt;&#xA;" OwnerUserId="1673" LastEditorUserId="1673" LastEditDate="2015-05-23T16:08:34.683" LastActivityDate="2015-05-23T16:08:34.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2920" PostTypeId="1" AcceptedAnswerId="7759" CreationDate="2015-05-23T15:44:27.637" Score="3" ViewCount="2470" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a washing process after a crystallization we want to know how much filter area we would need and if said area will work for our fixed volume flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{V_F}$ has a given value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Velocity $u$ should be the     superficial velocity and is at a fixed value we found in our mechanical engineering script.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\frac{\dot{V_F}}{u} = A$ we can calculate our needed  filter area. Question is, will that area work for our volume flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\dot{V_F} = \frac{A \cdot \Delta p}{\eta_F(\alpha_w \cdot L )+ f_M}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can then calculate $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta p = \Delta p_{filter medium} + \Delta p_{filter cake}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta p_{filter cake} = \frac{(1-\epsilon)^2}{\epsilon^3 \cdot x^2_{32}} \cdot 150 \cdot L \cdot \eta_F \cdot u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we insert $L$ in the equation for $\Delta p_{filter cake} $ and we get the exact $\Delta p_{filter cake}$ that we originally assumed when calculating $L$. Actually we expect to get a different value in order to find all variables via iteration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhere we made a mistake in our assumptions I think, but I can't see where right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\dot{V_F}$ is a function of $L$ however we can't change $\dot{V_F}$. And we insert it anyway in our equation for $L$ so I think that assumption is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;English is not my mother tongue, so please excuse any spelling / grammar mistakes. Furthermore this is not a given tasks, but a problem we stumbled upon during our group project (designing a plant).&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="368" LastEditDate="2015-11-20T01:15:23.100" LastActivityDate="2016-03-04T12:17:11.417" Title="Filter Area for given flow rate" Tags="|fluid-mechanics|chemical-engineering|process-engineering|fluid-filtration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2923" PostTypeId="1" CreationDate="2015-05-23T18:34:32.193" Score="2" ViewCount="114" Body="&lt;p&gt;Take a horizontal bar 10' pinned on the right end, so that there is a hinge point on the right end. Space requirements dictate that the bar must be cut in down to 5' (cut in middle and keep the left side, discard the right side) and yet it still must pivot as it did before. To accomplish this a mechanism must be constructed underneath/around the shorter bar, but must not extend out to where the original hinge point was, as that would defeat the purpose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mechanism is relatively straight forward to construct out of curved track that the bar rides along, where the curved track has a radius equal to the distance from the bar to the original hinge point. &lt;strong&gt;However I want to know if it's possible to construct a mechanism only out of linkages and straight track.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have created such a device in solidworks using only linkages and straight track that gets very close to simulating the external hinge point, but is not perfect. There is only a very small range of motion where the simulated hinge point is relatively stationary; for the rest of the motion the hinge point is not stationary at all. This leads me to believe this is not possible to construct; for the mechanism to perfectly simulate the external hinge point for a small range of motion, it must also perfectly simulate it for all range of motion. This is similar to how no two different functions, when graphed, truly overlap. If the functions perfectly overlap at all, then you know that they must be the same function. (This is not true for negative power function, which partially overlap with their positive power function counterparts. However, these functions are discontinuous, whereas the motion of linkages and track is continuous...) &lt;/p&gt;&#xA;" OwnerUserId="1735" LastActivityDate="2015-05-23T18:34:32.193" Title="Mechanical device to simulate hinge axis external/off of part" Tags="|mechanical-engineering|kinematics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2924" PostTypeId="1" AcceptedAnswerId="2930" CreationDate="2015-05-23T19:54:09.733" Score="3" ViewCount="17073" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iUTmzm.jpg&quot; alt=&quot;image from the measurement result&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the current should be higher than 0.67 mA, because I know phones use current in the order of 1A. Also the charger is able to support up to 1A, it doesn't make sense that the charger would have such a superior current than the phone will draw. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I connected the multi-meter in series with the phone charger. That was done by stripping the red wire then cutting it, and then I held the 2 rods of the multi meter at both ends. The phone indicated that it was charging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the phone indicated that it was charging and the battery was not full, I would take out the possibility that the phone disconnected the charger automatically. Also, I'm not sure about the idle state of the phone. The phone was turned on, and it has about a 3 inch screen. The battery is 1800 mAh. It still doesn't make sense that even in &quot;idle&quot; state it will draw such a small current while charging. Please let me know if I need to provide any more information about the measuring method, I didn't elaborate much on it because I thought it was simple, but I totally understand the possibility of a small mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another picture of the way of measurement, the reading is 0.45mA (from a different phone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rxM3b.jpg&quot; alt=&quot;2nd measurement picture&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="33" LastEditDate="2015-05-26T12:56:49.103" LastActivityDate="2022-03-12T07:24:42.487" Title="How much current is this phone drawing?" Tags="|electrical-engineering|measurements|rf-electronics|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2925" PostTypeId="1" CreationDate="2015-05-23T22:20:04.220" Score="2" ViewCount="194" Body="&lt;p&gt;I have a project that includes some standard AISC clevises on threaded rods in pure tension. I am trying to find guidance on how best to address the possibility that the rod could rotate with respect to the clevises over time, causing the clevis to eventually disconnect from the rod. My belief, and that of the engineer on the project, is that this is not a likely problem unless the rod was expected to ever see near-zero loads, and we have specified a jamb nut in any case, just to be conservative. All the same, I am wondering if there is any code or documented standard practice that would back us up for situations like this where pretension is not controlled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, a code would answer this question in the general case, but some specifics of our application are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material is ASTM A36 for both the rods and the clevises. The rods theoretically experience around 3 kips of tension under dead load only, and about 10 kips under the code maximum live load (controlling). They only experience a load reversal (go into compression) during a code-level seismic event. The threads on the rod are right hand at both ends (not turnbuckle style.) The clevis is AISC #3 size with a 1&quot; DIA rod. Both clevises are themselves restrained against rotation. The axis of the rod is plumb. At installation, we will be able to verify that each rod is substantially loaded, but not necessarily that they are sharing the load equally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at some other clevis assemblies like this used for diagonal bracing of frames, and they don't have even a jamb nut, but I doubt they experience the same sustained tension, and they aren't in the same orientation.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-05-26T04:45:09.570" LastActivityDate="2015-05-26T04:45:09.570" Title="Thread-locking on AISC Clevises" Tags="|structural-engineering|steel|threads|fasteners|aisc|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2926" PostTypeId="1" CreationDate="2015-05-24T05:02:58.607" Score="0" ViewCount="137" Body="&lt;p&gt;I have seen institutions giving two different degrees to almost same courses. Some institutions give Bachelor/Master in engineering and some give Bachelor/master in Technology. So is their any underlying difference between Engineering and  Technology?  I am a student pursuing B-tech course. And in my semester subjects,  there is a subject humanities where we have to learn what's engineering and what's technology both!  But I didn't understand the difference as there is no such clear definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have checked the Wikipedia articles and read the ABET definition of Engineering which says: The profession in which a knowledge of the mathematical and natural sciences gained by study, experiences and practices are applied with judgement to develop ways to utilize economically the materials and forces for the benefit of mankind. (Accreditation Board For Engineering And Technology, 2002)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a Technologist also do the same thing. In what way are they different?  I have also checked the post for which it was informed as a duplicate post, and there the aspect on which they talked about is different from my question. I am not interested, being a technologist/engineer where will I get job. My question is specific to this point,  on what basis an &lt;strong&gt;engineer and technologist are defined?  what are the extra things that an engineer do or technologist don't and the vice versa.&lt;/strong&gt;&#xA;I also checked the answers of that supposed original post,  the highest voted up answer has some basic faults. One thing I will like to mention technicians and technologist are &lt;strong&gt;not the same&lt;/strong&gt; It was first defined in french and for completely different meaning. So I am looking for the particular differences an engineer and engineering has with technologist and technology. &lt;/p&gt;&#xA;" OwnerUserId="1738" LastEditorUserId="1738" LastEditDate="2015-05-29T11:23:29.843" LastActivityDate="2015-05-29T11:23:29.843" Title="What is the course content difference between conventional engineering and engineering technology programs?" Tags="|engineering-history|" AnswerCount="0" CommentCount="3" FavoriteCount="0" ClosedDate="2015-05-24T13:51:25.797" ContentLicense="CC BY-SA 3.0" />
  <row Id="2927" PostTypeId="1" CreationDate="2015-05-24T11:00:03.850" Score="1" ViewCount="272" Body="&lt;p&gt;I need equations to calculate these cryogenic problems: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Generally how much energy (joules) is required to convert room temperature (21 °C) nitrogen gas into 1 liter LN2 using Joule–Thomson effect ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a more energy-efficient way to do that ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How much will cost to convert 1 liter vaporized nitrogen into liquid again in it's very low temperature right after vaporization above −195.79 °C ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1741" LastActivityDate="2015-05-24T11:15:32.120" Title="Liquifying nitrogen using Joule–Thomson effect" Tags="|cryogenics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2928" PostTypeId="2" ParentId="2927" CreationDate="2015-05-24T11:15:32.120" Score="1" Body="&lt;p&gt;If I am not mistaken the answer to your first question would be none. $\mu_{JT} = \frac{\partial T}{\partial p}|_H$ shows that that the conversion takes place under circumstances with no change in enthalpy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not mean however that you do not need energy for the process. However I can't answer that because you did not specify what pressures you will be working with. I assume you want to have liquified nitrogen at 1 atm. What that will cost you depends on the pressure you start with and how you achieve that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple instructions on the web in order to liquify nitrogen at low costs. You might want to check them out. &lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2015-05-24T11:15:32.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2929" PostTypeId="2" ParentId="2924" CreationDate="2015-05-24T12:30:40.650" Score="1" Body="&lt;p&gt;If the meter is correctly connected between the charger and the phone then it will show the current that the phone as a whole is drawing. However, that does not mean that it necessarily shows the current that the phone's internal  electronics are drawing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may not be obvious, but, current INTO the phone and current presently being used by the phone can easily be different. See below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To attain reasonable battery life the power (actually voltage) applied to a LiIon / LiPo (Lithium Ion / Lithium Polymer) battery &lt;strong&gt;MUST&lt;/strong&gt; be removed once the battery is fully charged. Leaving the charge voltagfe applied to the battery will substantially reduce its lifetime. So, once charged, a phone will disconnect charger from battery. What happens then is up to the designer.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Phone may run from battery.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Phone may run from charger.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Phone may run from battery &amp;amp; charger combined.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the phone runs wholly from battery (maybe at a few mA) the battery voltage will fall. when it reaches a predefined level the charger will &quot;kick back in&quot; and recharge the battery. While the battery is discharging the phone will input whatever current the designed has cased the charger front end to draw in those circumstances. I'd probably aim at much lower than 0.67 mA but that's a believable figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phone direct from charger makes great sense but does not seem to be what is happening in your case. Phone from some mix  - 0 mA or more could happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See below for likely minimum real world current draws.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Cellphone current draw covers an immensely wise range, with competent designers making significant efforts to minimise it where possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best case I have seen phones with standby times quoted in weeks, maybe 2 or 3 in some cases, although  this would be unusual.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A week is 168 hours long.&lt;br&gt;&#xA;0.67 mA x 168 hours = 113 mAh&lt;br&gt;&#xA;3 weeks at that rate = 339 mAh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A battery capacity of&lt;br&gt;&#xA;500 mAh = a very small cellphone battery&lt;br&gt;&#xA;1000 mAh = typicaly non-&quot;smart&quot; phone.&lt;br&gt;&#xA;500 mAh = typical smart-phone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is unlikely that the phone always draws 0.67 mA&lt;br&gt;&#xA;= over a month with even a 500 mAh battery)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, even when in standby a phone will perform various functions periodically. These increase current drain. How much idle current a phone draws depends on peripherals activated, background tasks running and more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is entirely possible that some of the time some phones draw that little current.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You need to tell us exactly how the charger is connected. &#xA;A circuit diagram would be even better. &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-05-24T12:30:40.650" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="2930" PostTypeId="2" ParentId="2924" CreationDate="2015-05-24T15:47:21.943" Score="7" Body="&lt;p&gt;This answer is based on extra information provided by the OP in the form of a photo. The claimed currents were far too low due to a misunderstanding in how to read the meter's 10A range display.&lt;/p&gt;&#xA;&lt;p&gt;The prior answer would have been correct if the information provided by the OP re currents being at sub-mA levels had been correct. As such it is useful to readers seeking to understand the implications if the supplied information had been correct.&lt;/p&gt;&#xA;&lt;p&gt;A later suggested anonymous edit (August 2017) wrongly described what can be determined by measuring phone input current. Note that the OP includes battery charging current in his query re phone current draw.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The phone is drawing 670 mA&lt;/p&gt;&#xA;&lt;p&gt;The meter is on the 10 Amp range.&lt;br /&gt;&#xA;This is because the 3 jacks at bottom are (probably) labelled some thing like&lt;br /&gt;&#xA;Common / 10 Amps / V Ohm mA&lt;br /&gt;&#xA;from left to right.&lt;br /&gt;&#xA;The Red lead is in the common socket and the black lead in the 10A socket.&lt;br /&gt;&#xA;This causes the meter to be on the 10A range even though the selector dial is set to 20 mA. Moving the dial may change the decimal point position but the input sockets are hard wired to the internal circuitry.&lt;/p&gt;&#xA;&lt;p&gt;Note that the display reads &amp;quot;-&amp;quot;.&lt;br /&gt;&#xA;This is either due to leads being swapped as above or possibly due to charger +ve entering meter on -ve lead.&lt;/p&gt;&#xA;&lt;p&gt;The coloured ellipses on the photo highlight the points commented on above.&lt;/p&gt;&#xA;&lt;p&gt;Investigate. Report back.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vkCi6.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Subsequently provided photo confirming the meter connections.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fCUrN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fCUrN.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2022-03-12T07:24:42.487" LastActivityDate="2022-03-12T07:24:42.487" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2932" PostTypeId="1" CreationDate="2015-05-25T13:00:03.187" Score="3" ViewCount="50" Body="&lt;p&gt;I've been reading &lt;a href=&quot;http://info.tek.com/www-wi-fi-overview-of-the-physical-layer-and-transmitter-measurements-primer.html&quot; rel=&quot;nofollow&quot;&gt;this summary&lt;/a&gt;(info.tek.com) of Wifi standards from Tektronix, which seems to do a pretty decent job of explaining things, but has left me with a couple of questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For access points that support multiple standards, how are the beacons transmitted? I would guess they use BPSK, but it's not clear whether this would be DSSS or OFDM (e.g. 802.11n devices don't support DSSS as far as I can see) &lt;/li&gt;&#xA;&lt;li&gt;When both the access point and end device support multiple standards (b/g/n seems pretty common) how do they decide which standard to use? I guess they let other devices know what they support through broadcasts/probe requests, so do they just default to the one offering highest data rates?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1748" LastEditorUserId="282" LastEditDate="2015-06-01T23:47:44.453" LastActivityDate="2015-06-01T23:47:44.453" Title="Wifi modulation and standard interoperability" Tags="|electrical-engineering|telecommunication|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2933" PostTypeId="1" CreationDate="2015-05-25T14:04:56.753" Score="6" ViewCount="4092" Body="&lt;p&gt;I'm designing 2 objects that should be attached to each other and removed when it's necessary. So I though about snap-fit as a solution. There is a lot of types of snap-fit but I'm planing to use a circular one do you know what are the parameters that should be respected to make a good snap fit? I am using solidworks.&lt;/p&gt;&#xA;" OwnerUserId="1749" LastEditorUserId="9" LastEditDate="2015-05-25T17:21:01.853" LastActivityDate="2023-07-29T08:02:00.963" Title="Snap-fit design" Tags="|mechanical-engineering|design|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2934" PostTypeId="1" AcceptedAnswerId="2947" CreationDate="2015-05-26T00:51:54.240" Score="8" ViewCount="7628" Body="&lt;p&gt;I'm working on a project that uses a strong stepper motor to power a wheel that needs to be able to &quot;coast&quot; freely at times. The motor is too stiff when not engaged, and along with my gearing ratio, makes it impossible to get the free-spinning coast** that I want from the wheel when the motor is not powered.  I only need the wheel to turn in a single direction, so I've been playing around with a design (see image) where the  motor slides just a bit to engage via a small set of gear tracks to pull itself into the other gear.   I was thinking that I would simply just reverse direction when I wanted to pull the motor away to enable free-spinning. Unfortunately, it doesn't work so well. The gear does engage, but the little bit of play it has when it slides causes it to make contact with the very last tooth of the tracks at the same time. This causes it to make a grinding noise, and certainly isn't efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how I could get this to do what I want? I've been searching online, but haven't found anything and I am not sure of how to technically describe what I'm trying to do in a few search terms.  So, I'd be grateful for any links to similar designs as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** Sorry, I forgot to add an important detail that will explain why I need to find a way to disengage the motor... The motor will turn the wheel forward, but when disengaged, the wheel needs to be able to turn freely in either direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/35K2j.png&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it's two days later, and this is what I'm going to try next:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RniiF.png&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It consists of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A drive gear (in red) that turns counterclockwise, with a one-way clutch bearing mounted to it. The clutch bearing turns freely clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an arm with an intermediate gear that will make contact with the large gear when engaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a spring attached to the arm. The spring pulls the arm down, to engage the gear. The important thing is that the spring is NOT strong enough to overcome the resistance of the idle stepper motor. So the  arm will not move unless assisted by the stepper motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's how I hope this will work, assuming we start with it not engaged:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stepper motor turns on, counterclockwise. The movement, in addition to the spring, engages the intermediate gear. The clutch bearing prevents the arm from dis-engaging the intermediate gear, yet allows the stepper motor to turn the gears freely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To disengage, we simply reverse direction of the stepper motor just a few degrees. The clutch bearing will lock, pulling the arm away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have yet to prototype this, as my 3D printer is down for a few days. But I would be interested in feedback and whether anyone thinks this will work.&lt;/p&gt;&#xA;" OwnerUserId="1754" LastEditorUserId="1754" LastEditDate="2015-05-28T18:19:54.603" LastActivityDate="2018-06-21T20:49:06.687" Title="How can I get a stepper motor to engage with a freely spinning wheel?" Tags="|gears|stepper-motor|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2935" PostTypeId="1" AcceptedAnswerId="21808" CreationDate="2015-05-26T06:40:26.810" Score="10" ViewCount="317" Body="&lt;p&gt;I am working on modeling concrete fibers (metallic fibers) as a mathematical model. My work is for my thesis. I am a PhD student of numerical analysis, but I am working on a real tunnel project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have had issues with the distribution of fibers in the concrete. I am trying to find a way to develop a stochastic differential equation for the fibers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following questions:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there any mathematical model for concrete fiber? (not statistical model)    &lt;/li&gt;&#xA;&lt;li&gt;Is there any technical information available for how concrete fibers behave?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1755" LastEditorUserId="282" LastEditDate="2015-05-27T23:06:36.800" LastActivityDate="2018-05-21T17:46:11.073" Title="Developing a stochastic differential equation model for concrete fibers" Tags="|civil-engineering|materials|modeling|concrete|" AnswerCount="1" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2936" PostTypeId="2" ParentId="2934" CreationDate="2015-05-26T08:45:54.720" Score="1" Body="&lt;p&gt;This solution seems a bit complicated. Since you say it only has to go in one direction, have you considered a ratchet mechanism as an alternative, such as those used on the pedals of a bike? This would only kick in if the motor speed exceeded that of the gear and would otherwise just coast along.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not 100% clear on your idea, but perhaps it would work better if it was friction rather than contact based (e.g. made contact with a rubber slider) and was designed to work only as long as there would be no contact, having no impact when it would be engaged.&lt;/p&gt;&#xA;" OwnerUserId="782" LastActivityDate="2015-05-26T08:45:54.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2937" PostTypeId="2" ParentId="2933" CreationDate="2015-05-26T08:59:30.453" Score="3" Body="&lt;p&gt;The main questions here are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What material are you using?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How much load is there on the joint?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How many times does it need to be removable?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Snap fit may be a good idea, but it's difficult to elaborate further when I don't know the usage case: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There has to be a compromise between being able to connect the pieces and the strength of a the joint. A stiff joint is strong but could be impossible to push into place or break it if excess force is applied.&lt;/li&gt;&#xA;&lt;li&gt;I would try to design not to carry any ordinary structural load on the snap fit and rather unload the connected members independently. Since the joints have a bit of weakness they are more difficult to predict the strength of and will be easier to break. The snap-fit in my eyes should ideally only hold the object from sliding apart in that sense.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="782" LastActivityDate="2015-05-26T08:59:30.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2938" PostTypeId="2" ParentId="2934" CreationDate="2015-05-26T12:51:43.350" Score="1" Body="&lt;p&gt;Stepper motors have significant &lt;i&gt;cogging&lt;/i&gt;.  If that is getting in the way, then a stepper is probably not appropriate here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at using a DC motor with a shaft encoder, or some way of knowing position.  Then it's a matter of a servo control loop to drive the DC motor to get the desired position.  If you specify a low-cogging motor, the mechanical resistance when left open circuit can be quite small.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-05-26T12:51:43.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2940" PostTypeId="1" AcceptedAnswerId="2943" CreationDate="2015-05-26T15:40:08.587" Score="5" ViewCount="3327" Body="&lt;p&gt;I was trying to find the moment of area about point A on the following figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mN8FS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prefer to use integration, since it can be extended to different shapes, and did this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\displaystyle Q_z=\int_{-25}^{25}\int_{-23}^{35}y \, dy \, dz=-17400 \:\mathrm{mm}^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the solution I have been given states that $Q_z=17400 \:\mathrm{mm}^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This made me wonder about two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the meaning of the sign (if there is any)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What have I done wrong in my calculation (aka &quot;What am I not understanding here?&quot;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1721" LastEditorUserId="368" LastEditDate="2015-05-26T21:31:21.640" LastActivityDate="2015-05-31T16:23:18.263" Title="First Moment of Area - Meaning of Sign/Sense" Tags="|mechanical-engineering|geometry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2941" PostTypeId="2" ParentId="666" CreationDate="2015-05-26T16:04:26.183" Score="9" Body="&lt;p&gt;I have done this before so would like to share some of what I know. Super critical CO2 has some unique properties, for one it is extremely difficult to keep in a closed container since it has near zero surface tension. As such, if a standard threaded fitting is used and the fluid comes into contact with it you can be fairly certain it will leak (and usually badly). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Welding stainless of this thickness properly requires a powerful TIG welder, this in turn calls for a certain level of skill in the operator. If you have both of these then please disregard this comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option all together is carbon steel. It is inexpensive (compared to your stainless piece), comes in a variety of sizes (DOM is a good place to start), welds nicely with a basic MIG, and is easier to machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless sizing and material type are critical, although engineering this piece with an adequate factor of safety will keep things together. Once again this requires a certain level of skill, if you don't have it then this is not a safe project to tackle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next you will need correctly rated and calibrated pressure gauges. Installation is straightforward although their importance can not be under estimated. Two is the minimum in case one is not reading properly (stuck needle for example). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calibrated pressure relief valves are also essential, and once again two is the minimum in case of failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you manage to build a well engineered container then the next step is getting the liquid in it. I suggest first packing it with dry ice (can be found at nearly any supermarket) and then sealing the unit. A bottle of CO2 can be used to maintain pressure although their regulators are intended for low pressure output. If you are unsure of how to perform modifications safely on one of these setups then this project should not be carried out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally what I am getting at is this, a project of this type brings up safety concerns even for a veteran engineer. Playing with extreme pressures and super critical fluids in a home built container is no easy task and could easily result in catastrophic failure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I certainly can understand the scientific curiosity/challenge aspect(s) of a project like this although there are still other concerns that need addressed beyond what I wrote here. If your skill sets are not in place to at minimum perform the operations listed thus far then I suggest leaving this project alone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not saying you do not have the ability to acquire said skills, just that this is not the type of thing to learn and try them out on.&lt;/p&gt;&#xA;" OwnerUserId="553" LastActivityDate="2015-05-26T16:04:26.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2942" PostTypeId="2" ParentId="2814" CreationDate="2015-05-26T19:39:44.387" Score="0" Body="&lt;p&gt;&lt;strong&gt;Robustness&lt;/strong&gt; (of a system):&#xA;The system keeps its expected performance (not failing) even when the real operation conditions (external to the system) are not the same as the ones assumed in the design. The broader is the range around the expected conditions more robust is the system. As a simple example you can consider a car that was designed to run over relatively smooth asphalt and you happen to use the car in a very poorly maintained road. The more robust is the car design, less will be the impact of the poorly maintained road on the performance (for example comfort of passengers) or expectation of failure (kilometers run before breaking) of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fault tolerant&lt;/strong&gt; (system):&#xA;The system keeps the expected performance (not failing) even when one of its components (which are internal to the system) fail. For this analysis is important to know the criticality of each component and also the likelihood of failure of that component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make a system fail-tolerant you make each component less critical (one plane with more than one engine and that don't need all engines working in order to fly is fault-tolerant on engine failures). A plane with just one engine is not fault-tolerant on the regard of engine failures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is not practical to make the system fail-tolerant on regard of a specific component, you make that component more robust, so the likelihood of that component to fail is reduced (which means that component can cope a wide range of operating conditions when you look at it as a system).&lt;/p&gt;&#xA;" OwnerUserId="1579" LastActivityDate="2015-05-26T19:39:44.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2943" PostTypeId="2" ParentId="2940" CreationDate="2015-05-26T23:48:07.580" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the meaning of the sign (if there is any)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no physical meaning associated with the negative sign in your result; it just means you got the math wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral definition of the first moment of area is $\iint_A {r(y,z) \, dA}$, or the sum over the entire cross-sectional area of the product of a differential area element and its perpendicular distance from the axis of interest.* Both area and distance in this sense are conceptually non-negative scalars, regardless of which side of the axis you're on, so your answer must always be non-negative as well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What have I done wrong in my calculation (aka &quot;What am I not understanding here?&quot;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The sign of your answer is not the only problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're going to go through the integration process, you must be careful to use the proper distance function $r(y,z)$, which in this case is the &lt;a href=&quot;http://en.wikipedia.org/wiki/Euclidean_distance&quot; rel=&quot;nofollow&quot;&gt;Euclidean distance&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case where your axis of interest is parallel to one coordinate axis, your distance function is invariant with respect to that axis. If you define your coordinate axes as shown in the figure, with the origin at the centroid of area, then the coordinates of point $A$ are $(-23,20)$ and your distance function is simply the one-dimensional Euclidean distance between the differential area element with height $y$ and the height $y_A$ of the line parallel to the z-axis that contains the point $A$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r(y,z) = |y-y_A| = |y+23|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your integral is now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_z = \iint\limits_A |y+23| \,dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that's left, recognizing that $dA=dy\,dz=dz\,dy$, is to determine the limits of integration. With the origin at the centroid, as in the figure, the limits of integration are the bounds of the area:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_z = \int\limits_{-25}^{25} \int\limits_{-35}^{35} |y+23| \,dy\,dz$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to say &lt;em&gt;exactly&lt;/em&gt; what your mistakes were without seeing your step-by-step calculations. Perhaps when you split up the absolute value into its piecewise definition, you forgot the second term? An alternative approach would be to redefine the origin at $A$, in which case your distance function would be nearly correct, but you'd have had to to recalculate the y-limits of integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer will be the same no matter where you put the origin.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* The equation I have written here is the general two-dimensional form, where the axis of interest is not necessarily parallel to either coordinate axis.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-05-31T16:23:18.263" LastActivityDate="2015-05-31T16:23:18.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2944" PostTypeId="2" ParentId="84" CreationDate="2015-05-27T04:32:37.353" Score="3" Body="&lt;p&gt;I've worked quite a bit with vacuum pumps in the past and they certainly could get your slurry out of the tank however a complete rig can run in the $10,000 range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another less expensive option is an airlift pump. This is nothing more than a tube that has compressed air blowing through it which in turn 'pushes' the slurry to the top of the pipe. It does so because the volume in the pipe is substantially less dense than the slurry surrounding the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will not carry the fluid vertically very high so would require multiple steps to make it 4m. Certainly a disadvantage, however these things are cheap and don't break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option 3 would be to use a basic trash pump. They are inexpensive and can handle a 4m lift, however getting them primed can be difficult. Options on this include a check vale system (but they 'leak' and lose prime over time), a valve system (also 'leaky'), or a backflow from the sump tank so the system can be filled with liquid before firing the trash pump. The backflow is probably the easiest to perform, and would allow for easy priming of the pump with only the most basic of training.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is a diaphragm pump, they work extremely well and can lift your slurry without issue if you purchase one with the correct rating. The only drawback is they consume vast quantities of compressed air and are not terribly efficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally you could consider a diaphragm pump as a primer for a standard centrifugal pump. This would allow for intermittent use of the diaphragm pump for priming purposes and the use of an efficient trash pump for primary fluid transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together, a trash/centrifugal pump would seem ideal with some type of priming system. Backflow from the slurry container would be the easiest but a diaphragm pump would do nicely as well. These units are inexpensive, easy to work on, pump a great deal of liquid quickly, and can safely handle 'dirty' slurries if need be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="553" LastEditorUserId="553" LastEditDate="2015-05-27T18:29:33.137" LastActivityDate="2015-05-27T18:29:33.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2945" PostTypeId="1" CreationDate="2015-05-27T04:45:17.283" Score="1" ViewCount="139" Body="&lt;p&gt;I have two &lt;a href=&quot;http://en.wikipedia.org/wiki/Business_telephone_system&quot; rel=&quot;nofollow&quot;&gt;PBX telephones&lt;/a&gt;. Each one has a line cables to the houses, but there is no connection between these two pbx telephones, neither wire nor wireless. How can I join between them wirelessly? What equipment should I buy?&lt;/p&gt;&#xA;" OwnerUserId="1767" LastEditorUserId="282" LastEditDate="2015-05-28T01:08:03.277" LastActivityDate="2015-09-25T02:33:18.837" Title="How to join between two PBX telephones wirelessly?" Tags="|telecommunication|wireless-communication|consumer-electronics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2946" PostTypeId="2" ParentId="2104" CreationDate="2015-05-27T05:11:21.267" Score="4" Body="&lt;p&gt;Pu238 is not usually isotopically separated from spent nuclear fuel for exactly the reason you pointed out, it would be very difficult. Instead most of what we have comes from one of two different processes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is bombardment of Np237 (also made in a nuclear reactor) with neutrons. It will become Np238 and then undergo a beta emission (fancy way of saying it spits out an electron from one of the neutrons in the nucleus) forming your Pu238.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other option is to bombard Am241 with neutrons to produce Am242, (excluding the metastable nuclei) this in turn will do another beta decay (like our Np238) and turn into Curium-242 ( Cm242 ) . Cm242 is unstable and will decay by alpha emission (fancy way of saying it coughs out a helium nucleus) thus forming our Pu238.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that is about it, there are other paths to Pu238 but these are the easiest to perform so far. Eventually we will run a thorium fuel cycle, when that happens there will be an abundance of this radioisotope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="553" LastEditorUserId="-1" LastEditDate="2019-01-13T23:49:55.843" LastActivityDate="2019-01-13T23:49:55.843" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="2947" PostTypeId="2" ParentId="2934" CreationDate="2015-05-27T11:46:30.263" Score="1" Body="&lt;p&gt;You can drive a lightly resisted gear on a swivel that rotates on the same axis as the stepper axle. Similar to a &lt;a href=&quot;http://en.wikipedia.org/wiki/Sun_and_planet_gear&quot; rel=&quot;nofollow&quot;&gt;sun&amp;amp;planet configuration&lt;/a&gt; but the planet can rotate around its own axle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the stepper is turning it will turn the planet into the wheel where the gear on the end meshes with it. To disengage you turn the stepper in the other direction for a bit to clear the planet off the wheel.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-05-27T11:57:26.733" LastActivityDate="2015-05-27T11:57:26.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2948" PostTypeId="1" CreationDate="2015-05-27T13:13:45.523" Score="12" ViewCount="3902" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KHwER.png&quot; alt=&quot;nano model structural&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to analyze this model that is stressed and deformed. I fixed the lowest of points in the x, y, and z directions, and I pulled top of points upward (y direction). When I run the model in Ansys I got the following error: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Check for insufficiently constrained model.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TK3J4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to cut the model to make it smaller, and analyzed the model like this. I fixed all points in the x and z directions. The lowest of elements are all fixed in x, y, and z. This didn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tHLKb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then cut the model to one element. It worked.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Ioni0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;How can I analyze this model?&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;What is the error and how do I need to set the boundary conditions?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1771" LastEditorUserId="33" LastEditDate="2015-11-12T02:16:34.193" LastActivityDate="2016-03-11T04:07:16.680" Title="Insufficiently constrained structural model in Ansys" Tags="|mechanical-engineering|modeling|ansys|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2949" PostTypeId="1" AcceptedAnswerId="2959" CreationDate="2015-05-27T14:29:06.947" Score="1" ViewCount="1549" Body="&lt;p&gt;What is the minimum number of TET elements you need, in order to fully fill a cube(HEX element)?&lt;/p&gt;&#xA;" OwnerUserId="1584" LastEditorUserId="1097" LastEditDate="2015-05-30T18:41:04.733" LastActivityDate="2015-06-03T19:39:25.247" Title="Minimum number of tetra elements required, to represent a cube?" Tags="|finite-element-method|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2950" PostTypeId="2" ParentId="2945" CreationDate="2015-05-27T15:31:01.563" Score="1" Body="&lt;p&gt;There are many possible answers. In your place I did the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I would buy 2 &lt;a href=&quot;http://en.wikipedia.org/wiki/Analog_telephone_adapter&quot; rel=&quot;nofollow&quot;&gt;analog telephone voip adapters&lt;/a&gt; (at most some tens of dollars). Putting them to the phones, from now you have IP connection.&lt;/li&gt;&#xA;&lt;li&gt;Buy two &lt;a href=&quot;http://en.wikipedia.org/wiki/Wireless_router&quot; rel=&quot;nofollow&quot;&gt;wireless routers&lt;/a&gt; (some tens of dollars). Connect them with static IP addresses and routes.&lt;/li&gt;&#xA;&lt;li&gt;Optionally, if you also don't have power, then you can buy &lt;a href=&quot;http://en.wikipedia.org/wiki/Solar_panel&quot; rel=&quot;nofollow&quot;&gt;solar panels&lt;/a&gt; (thousands of dollars) and a &lt;a href=&quot;http://en.wikipedia.org/wiki/Uninterruptible_power_supply&quot; rel=&quot;nofollow&quot;&gt;UPS&lt;/a&gt; (hundreds of dollars). But the first is costly, maybe a mechanical battery charger would be significantly cheaper (some hundred dollars).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The whole thing would cost around some hundreds dollars if you don't have power, but yo can solve this even below \$100 if you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main disadvantage of this system is that the range of the wireless router is short.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2015-05-27T15:31:01.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2951" PostTypeId="1" CreationDate="2015-05-27T15:56:38.620" Score="4" ViewCount="162" Body="&lt;p&gt;I have an HDPE pin, 2 mm high, 2 mm diameter, that I impress into a flat plaque of hot HDPE.  It is not hot enough to melt the pin, but still hot enough that the material will shrink 4% in 24 hours.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I calculate the force required to remove the pin from this plaque?  The coefficient of friction being used is 0.24, and an expected value for the force might be 3600 N.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that while the pin is pulled from the plaque, due to elongation the diameter toward the top of the pin becomes smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My end goal is to figure out a formula that I can use to approximate the expected pull force of this pin given varying heights, diameters, and if the pin is hollow or solid on the inside.  We cannot vary the shrinkage value.&lt;/p&gt;&#xA;" OwnerUserId="465" LastEditorUserId="368" LastEditDate="2015-06-02T17:47:20.443" LastActivityDate="2015-06-02T17:47:20.443" Title="Calculating a pull force between two pieces of plastic" Tags="|materials|statics|friction|plastic|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2953" PostTypeId="1" AcceptedAnswerId="2955" CreationDate="2015-05-27T16:11:54.837" Score="14" ViewCount="915" Body="&lt;p&gt;Each power plant has a load gradient characteristic, describing to what extent it can change the output of its generators in a given timeframe. Reading the literature, I know that, for example, a gas turbine power plant has a vastly higher load gradient than a one that uses coal as its primary energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I cannot find a description which parts of the plant contribute the most to the load gradient. I can imagine that a gas-powered plant can change its output faster because there's no mill involved as in a coal power plant, but where does its gradient exactly come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically: What constitutes the load gradient of&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a gas turbuine power plant,&lt;/li&gt;&#xA;&lt;li&gt;a Lignite (brown coal) power plant&lt;/li&gt;&#xA;&lt;li&gt;a bituminous coal power plant&lt;/li&gt;&#xA;&lt;li&gt;a nuclear power plant?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And why is there a difference between lignite and bituminous coal power plants?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd very happy to read up on the details if one could name a book. I've read several books that describe how these plants work, but fail to mention what creates or limits their load gradient.&lt;/p&gt;&#xA;" OwnerUserId="1775" LastEditorUserId="1775" LastEditDate="2015-05-28T13:42:48.810" LastActivityDate="2015-05-28T13:42:48.810" Title="What limits a power plant's load gradient?" Tags="|electrical-engineering|thermodynamics|power-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2955" PostTypeId="2" ParentId="2953" CreationDate="2015-05-27T19:34:51.567" Score="4" Body="&lt;p&gt;A gas turbine plant basically uses jet engines optimized for shaft power instead of exhaust push to drive the generators.  These can be controlled quickly because there is little stored energy.  You reduce the fuel flow rate, and the turbine pretty much has to slow down quickly, within a few seconds, taking tens of seconds to reach the new steady state.  The combustion chamber is small, and the little heat stored in its walls and the turbine blades is small compared to the overall power flowing thru the device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A large coal or oil fired plant has a large boiler with water/steam pipes running thru it.  These things take some time to heat up and cool down.  Then there is also a lag from less heat into the boiler before the less steam makes its way to the turbine.  Pressures and flow rates also have to be adjusted so that the liquid to gas phase change continues to happen in the right region of the pipes in the boiler.  Simply shutting off the fuel to the boiler abruptly can possibly cause damage.  All this takes more time than for a jet engine to slow down or speed up due to a change in its fuel flow.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-05-27T19:34:51.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2956" PostTypeId="2" ParentId="2911" CreationDate="2015-05-27T21:46:44.767" Score="2" Body="&lt;p&gt;Curious about the question, I did a little bit of research on mileage and efficiency. Efficiency of a vehicle is measured in various different forms and context matters. Efficiency in motor sports is quite different from efficiency as relates to general consumers. Fuel efficiency or mileage is what most consumers interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mileage as it is related to fuel efficiency most likely has root to 1973 oil embargo. The PEW article on &lt;a href=&quot;http://www.pewtrusts.org/en/research-and-analysis/fact-sheets/2011/04/20/driving-to-545-mpg-the-history-of-fuel-economy.&quot; rel=&quot;nofollow&quot;&gt;“Driving to 54.5 MPG: The History of Fuel Economy”&lt;/a&gt; tracks “Average estimated fuel economy by model year&quot; starting 1975,  which most likely has some relationship to the &lt;a href=&quot;http://1973%20oil%20crisis&quot; rel=&quot;nofollow&quot;&gt;1973 oil embargo&lt;/a&gt;. The &lt;a href=&quot;http://en.wikipedia.org/wiki/Corporate_Average_Fuel_Economy&quot; rel=&quot;nofollow&quot;&gt;Corporate Average Fuel Economy (CAFE)&lt;/a&gt; standard passed by USA congress in 1975 in response to the early 1970’s oil price shock also supports the above claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needless to say these days, automotive manufactures use mileage as a marketing tool to promote new model automotive, which is an efficient method to promote new vehicle models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding to the discussion, new hybrid vehicle platforms lead by Toyota Prius I believe has redefined fuel efficiency or mileage to higher level, which brings me to a question. With the introduction of &lt;a href=&quot;http://www.plugincars.com/cars&quot; rel=&quot;nofollow&quot;&gt;electric cars&lt;/a&gt; lead by &lt;a href=&quot;http://www.teslamotors.com/&quot; rel=&quot;nofollow&quot;&gt;Tesla Motors&lt;/a&gt; is how long will mileage (mpg) be relevant? Is miles per watt (mpw) the new efficiency barometer for automotive?  &lt;/p&gt;&#xA;" OwnerUserId="282" LastActivityDate="2015-05-27T21:46:44.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2957" PostTypeId="2" ParentId="2838" CreationDate="2015-05-27T22:00:33.033" Score="1" Body="&lt;p&gt;I have worked on several pieces of equipment in the past so would like to pass along some information I know about their hydraulic systems as it applies to your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equipment is very expensive so to make up for that companies run them as much as possible. Machinery ran at some sites (e.g. mines) go for 24 hours a day and are only 'down' when performing maintenance.  The equivalent for hydraulic hoses service time in regards to permeability failure is easily years (which can be seen by the hours meter and maintenance logs of the equipment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Permeation for your case is likely negligible, here is why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most high pressure hydraulic systems (e.g. like that found on a loader) operate at several thousand psi (10's of MPa like your example). They also have to deal with several instances of 'surging' (sudden increase in pressures) throughout their service life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluids also heat up as the machine is used so the hoses have to deal with an elevated temperature environment. This lowers the viscosity increasing permeability through the hose material and of course any chemical reactions that may occur will now happen more quickly. It is not unusual to have temperatures above 80C (not recommended but this is common). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Working machinery also has to contend with constant hose movement (if it is polymer based instead of steel it is because there is movement). This in turn means the hoses also have to fight abrasion where they are tied to and rub against the chassis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These hoses are also located outdoors where they see all kinds of weather conditions and temperatures fluctuations from frigid cold to over 120F while contending with spilled diesel, motor oil, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Permeation does occur in some instances but it is usually only seen on hoses that are put into service for the wrong application (e.g. wrong pressure rating). Lifetime of the hose (from a permeability standpoint) is easily several years of service life and it is not unusual to see them last substantially longer in this regard (even in harsh environments).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your experiment I assume this is a static hose rated for these pressures and fluid types. I am also assuming this is in an ambient indoor temperature environment with clean hydraulic fluid and no hose movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case it should be expected that any issues with permeability will be at minimum when compared to the standard service seen by these hoses. Good quality hoses are engineered to last for a long duration in much more severe environments than your lab. If it is installed properly there really should not be a concern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest making a trip to your local Catapillar, John Deere, Case, etc., etc. dealer and speak with one of their hose technicians. Just be sure to convert your requirements to standard (psi, inches, feet, etc.) and they will be happy to answer any further questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="553" LastEditorUserId="553" LastEditDate="2015-05-28T15:40:40.033" LastActivityDate="2015-05-28T15:40:40.033" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2958" PostTypeId="2" ParentId="2905" CreationDate="2015-05-27T22:08:34.463" Score="0" Body="&lt;p&gt;V causes no shear stress at point B.&#xA;The 1st moment of area refers to the area above the point you're studying, in the negative direction of the shear stress applied. You can see that at point B, that area is inexistent, and so the shear stress due to V is equal to zero.&lt;/p&gt;&#xA;" OwnerUserId="1777" LastActivityDate="2015-05-27T22:08:34.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2959" PostTypeId="2" ParentId="2949" CreationDate="2015-05-28T01:00:40.947" Score="3" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BmlXi.png&quot; alt=&quot;5 tet elements&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: 5 tetrahedral elements can form a cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=XAhgw2Z4T2M&quot; rel=&quot;nofollow noreferrer&quot;&gt;This video&lt;/a&gt; shows how to decompose a cube into five tetrahedra. Not saying it can't be done with less, but I can't figure out how :-)&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1324" LastEditDate="2015-06-03T19:39:25.247" LastActivityDate="2015-06-03T19:39:25.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2962" PostTypeId="1" CreationDate="2015-05-28T15:19:18.253" Score="2" ViewCount="198" Body="&lt;p&gt;I have developed a second order transfer function model for a permanent magnet dc motor in terms of speed and voltage. I had performed open loop no load experiments on the dc motor and derived the parameters. When I experimentally apply a 12 V to the dc motor the speed that I get does not match with the simulation results obtained from the developed model. How do one validate the open loop transfer function model? Should I use the loaded condition values of voltage and speed for validation&lt;/p&gt;&#xA;" OwnerUserId="1786" LastActivityDate="2015-05-28T15:19:18.253" Title="Validation of transfer function" Tags="|motors|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2963" PostTypeId="1" CreationDate="2015-05-28T19:26:46.260" Score="3" ViewCount="967" Body="&lt;p&gt;I've been working with TSOP IR receivers and emitters on robots since the past year. With multiple robots communication we have signal collision algorithms and such. We still faced a lot of interference from the fluorescent bulbs. We have tried covering the receiver from the top. I need a deeper understanding of the datasheet of Vishay TSOP 34156 receiver. This is the datasheet: &lt;a href=&quot;https://www.pololu.com/file/0J19/TSOP34156.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.pololu.com/file/0J19/TSOP34156.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are my questions :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What does transmission distance mean in terms of the receiver? As far as I know it doesn't transmit anything.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is directivity? Angle of half transmission distance? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Figure 9 : max envelope duty cycle vs burst length?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;&lt;strong&gt;Burst length should be 6 cycles/burst or longer.&#xA;After each burst which is between 6 cycles and 70&#xA;cycles a gap time of at least 10 cycles is necessary.&#xA;For each burst which is longer than 1.8 ms a corresponding&#xA;gap time is necessary at some time in the&#xA;data stream. This gap time should have at least same&#xA;length as the burst.&#xA;Up to 2200 short bursts per second can be received&#xA;continuously.&lt;/strong&gt;&quot;&#xA;What is burst length exactly?What exactly do you mean by a cycle? I have               tried to find the information on cycles unsuccessfully. &#xA;This is how my communication protocol looks like.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JfwYH.jpg&quot; alt=&quot;IR communication protocol(not to scale)&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Legend for the image : signature bit(high) : 1200 uS&#xA;                       interval(low) : 300uS&#xA;                       zero (high)   : 300uS&#xA;                       one (high)    : 600uS&#xA;I transmit total of 32 bits consisting of 1s and 0s and intervals in between them excluding the signature wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of cycles what does this mean? Where do I stand? within the limits or I have to send shorter bursts of data? Is the TSOP34156 good for a swarm robotic purpose or are there better IR receivers out there that I do not know of?&lt;/p&gt;&#xA;" OwnerUserId="1790" LastEditorUserId="282" LastEditDate="2015-06-04T00:11:05.400" LastActivityDate="2015-06-04T00:11:05.400" Title="TSOP IR receiver" Tags="|electrical-engineering|wireless-communication|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2964" PostTypeId="1" CreationDate="2015-05-28T20:49:21.880" Score="6" ViewCount="29166" Body="&lt;p&gt;I'm having a hard time finding an answer from the internet. What does starting torque mean? &lt;/p&gt;&#xA;" OwnerUserId="1791" LastEditorUserId="272" LastEditDate="2015-05-29T01:54:12.237" LastActivityDate="2017-08-24T02:27:12.890" Title="What is Starting Torque?" Tags="|mechanical-engineering|motors|torque|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2965" PostTypeId="2" ParentId="2964" CreationDate="2015-05-28T21:34:47.063" Score="6" Body="&lt;p&gt;Is the torque required to make something that is not turning start turning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can think of starting torque as an instantaneous value (the torque just before the motor starts moving). You can also thing as the torque function from that moment until the motor reaches the desired speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque provided by a motor is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{motor}(0) = T_{static friction}\\T_{motor}(t&amp;gt;0)=T_{dynamic\ friction}+J\dot{\omega} +K\omega^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega$ is the angular speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{\omega}$ is the angular acceleration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J$ is the polar moment of inertia of the motor rotor plus load (as seen from the motor)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K$ is constant (in a simple model) that represents the viscous forces&lt;/p&gt;&#xA;" OwnerUserId="1579" LastEditorUserId="1579" LastEditDate="2015-05-28T22:08:18.287" LastActivityDate="2015-05-28T22:08:18.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="2966" PostTypeId="1" AcceptedAnswerId="3530" CreationDate="2015-05-28T21:40:02.537" Score="9" ViewCount="2327" Body="&lt;p&gt;I am working on a compressed air energy storage system. The size and weight of the system are heavily constrained; nothing should (ideally) exceed a few pounds. For this reason, I would like to store the energy (compress the gas) and extract the energy (make the gas do work) with the same mechanism &lt;strong&gt;in a rotary fashion&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially what I need is a &lt;strong&gt;rotary&lt;/strong&gt; air compressor that, when air is forced through in the opposite direction, doubles as a pneumatic motor. I'm working with fairly high pressures (I'm estimating a few hundred psi) but low volume. In my search, I have found a plethora of compact rotary air compressors and rotary pneumatic motors, but there is hardly comment on what systems would work as both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, it seems very intuitive that an air compressor could have these properties, but I don't want to jump to any conclusions. I have looked at several compressors, and the most applicable to my situation seem to be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Centrifugal compressor&lt;/li&gt;&#xA;&lt;li&gt;Axial flow compressor&lt;/li&gt;&#xA;&lt;li&gt;Rotary screw compressor&lt;/li&gt;&#xA;&lt;li&gt;Rotary vane compressor&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The centrifugal compressor is ideal, but it seems the least likely in my eyes to be reversible, at least with any efficiency. I also looked at pneumatic motors, of which there were fewer available. Most applicable seemed to be the:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rotary vane motor&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other systems, such as the pietro motor, were obviously not applicable in my light weight, compact application. The correlation between the rotary vane compressor and the rotary vane motor is promising, but I would like to know about any options I have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What rotary gas compression systems can double as motors powered by the gas they compress?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; &#xA;The answer most likely lies in the similarity between a radial (centripetal) turbine and a centrifugal compressor. &lt;/p&gt;&#xA;" OwnerUserId="1792" LastEditorUserId="1792" LastEditDate="2015-08-02T23:08:51.540" LastActivityDate="2015-10-06T13:09:59.980" Title="Air compressor that doubles as a pneumatic motor?" Tags="|mechanical-engineering|motors|compressed-air|energy-storage|compressed-gases|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="2967" PostTypeId="1" AcceptedAnswerId="3254" CreationDate="2015-05-28T22:46:41.087" Score="2" ViewCount="833" Body="&lt;p&gt;What would the adhesive strength of chewing gum be? Would it be enough to support a car bumper? I'm more interested in how much tensile strength chewing gum has. I'm working on a project regarding the used cars movie, and in the movie, the salesman repaired a bumper that was unconnected with chewing gum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bumper I'm thinking of is the one featured in the beginning of the movie, a '57 Chevy sedan. I want to see if it is even plausible for chewing gum to support a car bumper, so assume a fixed load of just the bumper. Since it was the '50s I would guess the bumper would be 25 pounds at least.&lt;/p&gt;&#xA;" OwnerUserId="1793" LastEditorUserId="282" LastEditDate="2015-06-04T00:31:07.813" LastActivityDate="2015-06-19T13:58:03.567" Title="Adhesive strength of chewing gum" Tags="|materials|automotive-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2968" PostTypeId="1" AcceptedAnswerId="2971" CreationDate="2015-05-29T01:14:45.657" Score="6" ViewCount="436" Body="&lt;p&gt;I'm working on designing a Cold Spray system that uses a convergent-divergent nozzle to achieve a 800 m/s exhaust velocity.  I'm assuming 400 psi nitrogen at room temperature is provided at the inlet of the nozzle.  The throat area is 0.0859 square inches (tiny, I know).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;  \dot{m} &amp;amp;= \rho\ v\ A \\&#xA;  \rho &amp;amp;= 31.920\ \frac{\text{kg}}{\text{m}^3} \text{(N}_2\text{ at 400 psi)},\\&#xA;  v &amp;amp;= 353.575\ \frac{\text{m}}{\text{s}} \text{(speed of sound at 400 psi)},\\&#xA;  A &amp;amp;= 5.54*10^{-5}\ \text{m}^2,\\&#xA;  \Rightarrow\qquad\dot{m}&amp;amp;=0.625\ \frac{\text{kg}}{\text{s}}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;assuming $M = 1$ in the throat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This figure seems absurdly large.  Can someone shed some light on the situation?&lt;/p&gt;&#xA;" OwnerUserId="1794" LastEditorUserId="272" LastEditDate="2015-05-29T01:46:30.873" LastActivityDate="2015-05-29T03:24:56.430" Title="Convergent-Divergent Nozzle Design from Exhaust Velocity" Tags="|mechanical-engineering|aerospace-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2969" PostTypeId="2" ParentId="2964" CreationDate="2015-05-29T02:18:04.623" Score="8" Body="&lt;p&gt;There are two different aspects to start-up torque.  One is associated with the motor and the other with the system which you are trying to drive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The start-up torque of a mechanical system is the minimum torque required to get it moving from a standstill.  the start-up torque of a motor is the maximum torque which it is capable of producing from a standstill.  If the start-up torque of your motor is not greater than the start-up torque of your mechanical system, then the motor will not be able to get the system moving.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below shows the torque curve for a typical AC motor.  The start-up torque, although not marked, is the torque value at the far left of the plot (at zero speed).  If your mechanical system has a higher start-up torque than this, then this motor will not be able to get it running even if the torque required to keep it running is lower than the full load torque.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Torque-Speed_Curve_for_a_typical_AC_motor.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RS4Pp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-05-29T02:18:04.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2971" PostTypeId="2" ParentId="2968" CreationDate="2015-05-29T03:24:56.430" Score="9" Body="&lt;p&gt;You're not that far off. In your analysis, you've just neglected the fact that the gas will accelerate as it passes through the converging nozzle.  This is basically converting potential energy, in the form of pressure, into kinetic energy, velocity. As a result, the gas will no longer be at 400 psi when it reaches the throat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compact form to compute mass flow through a choked nozzle is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{m} = C_d A \sqrt{\gamma \rho P \left( \frac{2}{1+\gamma} \right)^{\frac{\gamma+1}{\gamma-1}}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$ is the ratio of specific heats (1.4 for nitrogen), and $C_d$ is an dimensionless coefficient that depends on the efficiency of the nozzle design (let's guess 0.5 for now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the above, I got $0.18 \, kg\cdot s^{-1}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a sanity check I compared this result to manufacturer reported values for flow control nozzles that I use in my lab. When I scaled up their values for a $.0252 \, in.$ nozzle at $100 \, psig$ up by area and pressure, I got $0.15 \, kg\cdot s^{-1}$. Seems reasonable.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-05-29T03:24:56.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2972" PostTypeId="2" ParentId="2963" CreationDate="2015-05-29T12:16:08.787" Score="2" Body="&lt;p&gt;You asked too many questions in one, so I'll just briefly make a few key points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the problem these devices have.  The dynamic range of light levels they have to work with is enormous, and the actual signal can be small relative to the ambient light level.  One way they deal with this is to high pass filter the received light level.  That essentially subtracts off the ambient level, hopefully leaving the small variations at the carrier frequency that are the signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when the light level changes abruptly, is that signal or a change in ambient level?  How long is long enough to declare the current state the new normal?  The answer is these devices have some sophisticated filtering to try to detect the weak carrier signal superimposed on what could possibly be a much stronger background.  This is one reason each device variant is specified for a particular carrier frequency, usually around 40 kHz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The device therefore is looking for changes from the recent average condition.  This means any data you send needs to be in bursts that can't last too long, and must have gaps between the bursts that can't last too long either.  The bursts also need to be long enough so that short spikes can be rejected as not being a valid carrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The datasheet is telling you that you must send bursts that are at least 6 carrier cycles long, but not longer than 70 cycles, and that you need at least 10 cycles gap between any two bursts.  I usually use 10/10 burst/gap minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That still leaves you lots of ways of encoding data.  You usually leave the gaps fixed at 10 cycles, and encode data in the 10-70 cycles you get for a burst.  Keep in mind that the output won't resolve burst length down to a cycle or two, but it should be quite possible to detect the difference between 10 and 20 cycle bursts, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many schemes for encoding data using this low level mechanism, but that's beyond the scope of this answer.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Added in response to comments&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One cycle refers to one &lt;i&gt;carrier cycle&lt;/i&gt;.  Since you are using the version that works with 56 kHz carrier, the easiest way to make the carrier is to drive the transmitting IR LED with a 56 kHz square wave.  1 / 56 kHz = 17.9 &amp;micro;s, which is the period of one carrier cycle.  For a square wave, that means the LED will be on half that time and off half that time, or 8.93 &amp;micro;s for each phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A burst is a sequence of carrier cycles in a row.  A 10-cycle burst is therefore ten 8.93 &amp;micro;s IR pulses in a row, separated by 8.93 &amp;micro;s of no light emitted between each pulse.  For the gap between pulses, you need to leave the equivalent of 10 carrier cycles with no light emitted, which means a 179 &amp;micro;s gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the reciever is built, it is more sensitive to signals received straight into the front.  Sensitivity falls off as the IR comes in more from the side.  The datasheet tells you the angle at which the effective transmission distance goes is down by a factor of 2.  Note that since light level falls off with the square of the distance, this is the angle at which the received signal strength is 1/4 of what it would be if the signal was coming straight face-on.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2015-05-29T18:38:46.573" LastActivityDate="2015-05-29T18:38:46.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2973" PostTypeId="2" ParentId="2964" CreationDate="2015-05-29T13:09:50.333" Score="6" Body="&lt;p&gt;&quot;Starting torque&quot; and &quot;locked-rotor torque&quot; are the same thing.  So if you run across the later term, you can treat it as a synonym for &quot;starting torque.&quot;  The NEMA MG-1 definition of locked-rotor torque is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;the minimum torque which [a motor] will develop at rest for all&#xA;  angular positions of the rotor, with rated voltage applied at rated&#xA;  frequency.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="198" LastActivityDate="2015-05-29T13:09:50.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2974" PostTypeId="1" AcceptedAnswerId="2975" CreationDate="2015-05-28T23:54:37.397" Score="4" ViewCount="882" Body="&lt;p&gt;I'm planning to build a laser shooting target. Something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RXQlM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RXQlMm.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the long term I want to build something advanced like a professional grade target, in which the user is able to tell exactly where was the shot, but for now I'd be happy with a basic one like in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically whenever a laser shot hits the black area it counts as a shot inside.&#xA;I'm just unsure about how can I do it since I don't want it to easily confuse a shot with regular light, and if possible I would like to use optics side-effects to avoid cross-shooting (whenever one shoots someone else target, while misaligned with it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin with, one has to be 10 meters away from the laser target, with a variation of a meter at most.  The targets are held at a meter and a half high.  The laser of a gun, in conformance with the regulations, has a ~3 mW peak during 15-20 ms of activity. The black area is about 7 cm in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the official specifications for a laser target in terms of light, but I found a brand with the specifications of &lt;a href=&quot;http://www.apeom.cz/products/pentathlon/&quot; rel=&quot;nofollow noreferrer&quot;&gt;one target&lt;/a&gt; (end of the page). It states &quot;Sun resistance: 70k Lux.&quot; Does that mean that it is able to distinguish a laser shot from the ambient light at up to 70k lux ambient light? What is a simple way to understand the amount of light that represents?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea so far its having a photo sensitive transistor in a PCB bolted to the back of the target.  An aluminium sheet in the front with a circle in the middle of it, and a smoked acrylic sheet in its back (to fill the circle). And either a fresnel or a biconvex lens right next to the acrylic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both the biconvex lens and the acrylic converge the frontal light to a point (the point where the photo transistor stays), so that spot becomes mostly light coming exactly from the front of the target, and becomes sort of immune to most ambient light (since it will refract to the points around the transistor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems I see so far are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Which one (fresnel/biconvex) has the shortest light converging point?&lt;br&gt;&#xA;(I would prefer the target as slim as possible)&lt;/li&gt;&#xA;&lt;li&gt;If one shoots the edge of the valid area, it's not a straight shot. Assuming one is perfectly aligned with the target, and the weapon is at its height, it would already become a ~90.2° incision. Being a little misaligned (half meter) and being half a meter lower than the target while shooting the center would represent a ~90.4° incision. Being misaligned and shooting the edge opposed to the misalignment direction would be an even more tilted shot. Is that little tilt enough to make the lens focus out of the transistor? It seems like a really small deviation.&lt;/li&gt;&#xA;&lt;li&gt;Is it worth it to get a sheet of red cellophane between the acrylic and the transistor? It would filter the non-red light, wouldn't it?&lt;/li&gt;&#xA;&lt;li&gt;To avoid counting lights turning on or a cloud passing by as valid shots, can I can count 30 ms from the moment in which the light peak happens, and see if it has disappeared?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So basically the main question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would either of these two lenses solve my problem, by converging the light coming from one direction into one point? Would it be practical? Are there better alternatives (like leaving the transistor alone without focusing; mirrors; etc.)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;Floris guess was probably right. The signal seems to be modulated. I managed to have a shot at the spec sheet of one of the guns:&lt;img src=&quot;https://i.stack.imgur.com/s8i2C.jpg&quot; alt=&quot;&quot;&gt;&#xA;Those graphs mean that with UIPM 2014 signal it pulses 10 times in 15.6 milisecs?&#xA;Should I get my target also recognising those pulses?&#xA;I bought a BH1750FVI light sensor module, can those kind of sensors recognise differences of light in fractions of miliseconds?&#xA;&lt;strong&gt;EDIT2:&lt;/strong&gt;&#xA;Found the official regulations:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yyXN1.png&quot; alt=&quot;Regulations&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/JAtLp.png&quot; alt=&quot;Regulations&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/nKmRO.png&quot; alt=&quot;Regulations&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/4aixX.png&quot; alt=&quot;Regulations&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Together with what was already answered, it should be enough to help anyone with the same question.&lt;/p&gt;&#xA;" OwnerUserId="1799" OwnerDisplayName="Claudiop" LastEditorUserId="1799" LastEditDate="2015-05-30T03:05:19.263" LastActivityDate="2015-05-30T03:05:19.263" Title="Focusing light from a laser shot into one point" Tags="|electrical-engineering|optics|photonics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2975" PostTypeId="2" ParentId="2974" CreationDate="2015-05-29T01:46:09.547" Score="5" Body="&lt;p&gt;Briefly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to distinguish your signal from any others (and incidentally get a massive gain in signal to noise ratio) is to use modulation. The simplest modulation scheme just turns the laser on and off very quickly (a few MHz) and your detection circuit (synchronized with the switching of the laser) ADDs detected signal when the laser is on, and SUBTRACTS when the laser is off. This is the principle of lock-in amplifiers. You can build a simple one with cheap CMOS switches, and they are incredibly effective. Any signal not at the &quot;detected&quot; frequency will be sometimes added and sometimes subtracted - so averaging over a few 1000 cycles it will be reduced to almost nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Filtering light from wavelengths you are not interested in helps as well - this is why your IR remote has a dark red window, for example (on the receiver as well...)&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Further thoughts&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Having read your question again, and in light of the comments below, here is a bit more detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general for a situation like you describe (where you want to detect a &quot;signal&quot; (the laser hit) in the presence of &quot;noise&quot; (the other guns, ambient light, the sun...) you have to look for the ways in which one signal is different from the others. This comes down to three things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Spatial difference&lt;/li&gt;&#xA;&lt;li&gt;Wavelength difference&lt;/li&gt;&#xA;&lt;li&gt;Temporal (modulation) difference&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let's take each of these in turn. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Spatial difference&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You know roughly where the gun is positioned that is pointing at the target. We can use that to eliminate &quot;cross fire&quot;. Let's look at this figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QcnUm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see - if you have an on-axis aperture near the focal plane of your lens, then light that comes in off-axis will be rejected. How big that aperture needs to be follows directly from the geometry of the target - shooter - neighbor configuration. You already said your shooter would be approximately 10 m ± 1 m from the target - and if you are trying to keep the target &quot;thin&quot; (maybe a focal length of 20 cm) the focal plane would be very close to the focusing distance of the lens: $\frac{10\cdot0.2}{10-0.2}=20.4 cm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle subtended by the aperture at the center of the lens will be the acceptance angle - any lasers shooting outside that range will be blocked by the aperture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3slJU.png&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the choice of lens. A (plastic) Fresnel lens will be thinner, lighter and cheaper than another kind of lens - and sufficient for this application. They don't suffer much from spherical aberration, so it is possible to have a relatively large lens with a short-ish focal length. These may be hard to find, but &lt;a href=&quot;http://www.teachersource.com/product/fresnel-lenses/microscopes-magnification&quot; rel=&quot;nofollow noreferrer&quot;&gt;I found a very cheap 7&quot;x10&quot; lens&lt;/a&gt;. You could cut it to size and stack two or three on top of each other to give you the focal length you need (the shorter the focal length, the thinner the target; but otherwise there is not much reason to spend the money, certainly for a proof of principle). If your target area is only 38 mm, there is a cheap 50 mm biconvex lens available on the same site: &lt;a href=&quot;http://www.teachersource.com/product/biconvex--biconcave-lens-sets/light-color&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.teachersource.com/product/biconvex--biconcave-lens-sets/light-color&lt;/a&gt; . Disclaimer: I am not associated with this site or the products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General note about lenses: fresnel is thinner (and often cheaper) than conventional; little to be gained in this application from plano-convex vs biconvex (although plano would allow the front of the target to be flat).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Wavelength difference&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Lasers are &quot;monochromatic&quot;, meaning their energy covers a very narrow range in the spectrum. This means that if you have the right filter, you can block a lot of &quot;non-laser&quot; light. The narrower the filter, the better the relative blocking. For example, you can get a &lt;a href=&quot;http://www.omegafilters.com/products/filters/bandpass/635bp5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;635 nm filter with 5 nm FWHM&lt;/a&gt; - that's centered on a &quot;better&quot; red laser pointer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x0PEs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That will cut out an enormous amount of external light - if you imagine (for estimating) that sunlight is uniform between 400 and 800 nm, then cutting out everything that is not within 5 nm of the laser cuts the sunlight contribution by 80 x ($\frac{5}{400}$). It would be expensive - but shows it can be done. How well your &quot;red acetate sheet&quot; will work is something you would have to experiment with...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important reason to consider a filter is &lt;em&gt;saturation&lt;/em&gt;: in the next step we will look at modulation, but that can only work if the detector is still linear over the range of intensities that are incident. If the detector (or electronics) are saturated, there will be no change in signal when the input intensity changes - so you would not be able to detect the small signal on top of the large one. By filtering &quot;a good chunk&quot; of the sunlight out, you make the system less sensitive - reduce the need for a massive dynamic range.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Modulation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The final point - and the one I started this post with - concerns modulation. In its simplest form, you can look for a sudden increase in intensity. For example, if you put a high-pass filter (series capacitor, shunt resistor) on the output of your photocell you would only see a pulse if the intensity suddenly increases. This would happen with a laser pulse, but not with sunlight. You could feed that signal into a comparator, and it would detect the pulse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More complex systems could use pulse modulation - you can learn about this, and get kits for very little money, from &lt;a href=&quot;https://learn.sparkfun.com/tutorials/ir-communication&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://learn.sparkfun.com/tutorials/ir-communication&lt;/a&gt; . The problem with the transmitters on these systems is that throw a wide beam (so you don't have to &quot;aim&quot; at the TV). Replacing them with a narrow beam transmitter (like a laser diode) you should be able to re-use most of the other components. This would allow you to use several guns (each programmed with their own &quot;code&quot;) and tell them apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as I mentioned earlier, the really cool system uses a lock-in amplifier scheme; using a cheap quartz oscillator you should be able to detect the signal regardless of phase (derive the quadrature signal from the oscillator and have two detection channels; then add the square of these signals together to have intensity independent of phase). As long as your detector does not saturate, this method is incredibly sensitive: it used to be one of my favorite optics demonstrations to show how you can &quot;see&quot; a small LED light up from across the room when the oscilloscope shows &quot;nothing but noise&quot;. There is a nice description in &lt;a href=&quot;http://www.phys.utk.edu/labs/modphys/lock-in%20amplifier%20experiment.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article - especially part D&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="565" OwnerDisplayName="Floris" LastEditorUserId="565" LastEditDate="2015-05-29T19:29:13.577" LastActivityDate="2015-05-29T19:29:13.577" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="2977" PostTypeId="1" CreationDate="2015-05-29T20:33:12.517" Score="4" ViewCount="173" Body="&lt;p&gt;I am trying to build an alarm that works in an opposite way to a security alarm in a shop. In shops RF tags on items set off an alarm as you walk through a detector. I wanted to build alarm that sounded as you move out of range of a detector, rather than into range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tags on high value items seem to be simple coils of wire encased in a thin plastic sleeve, they don't appear to have a power source.  I presume they emit a small signal that is enough to trigger the walk through loop.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically my gran keeps leaving her keys on the table and walking out the front door and getting locked out. I figured if she had a tag on her key fob that would trigger an alarm on her wrist or bag when she was more than 2 or 3 metres from her keys might solve the problem and save her a fortune in locksmiths!&lt;/p&gt;&#xA;" OwnerDisplayName="David Weaver" LastEditorUserId="272" LastEditDate="2015-05-29T21:30:19.497" LastActivityDate="2015-05-30T01:31:21.770" Title="How can I build an anti proximity alarm?" Tags="|electrical-engineering|wireless-communication|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2978" PostTypeId="2" ParentId="2977" CreationDate="2015-05-30T01:31:21.770" Score="6" Body="&lt;p&gt;Reverse the logic for the RF tag detector and alarm. As you state, RF detectors and alarms activate when the RF tag is within range of the detector. Change the logic of the detector-alarm system to sound the alarm, as the default situation, when the tag is out of range. When the tag is in range the alarm turns off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a similar logic to that of dead man switches for train locomotives.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-05-30T01:31:21.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2980" PostTypeId="1" CreationDate="2015-05-30T13:23:16.150" Score="6" ViewCount="612" Body="&lt;p&gt;If you look at a tire from the rear, you'll notice that the tire isn't quite a cylinder. It has edges that are slightly beveled. Sometimes this effect is clearly apparent, but other times, as in many car tires, it's hard to see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand the use on motorcycles, since they are typically angled on turns, but cars are, in general, not designed to take corners on two wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do cars have beveled tires?&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2016-04-29T23:40:13.920" Title="Why do cars have beveled tires?" Tags="|automotive-engineering|wheels|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2981" PostTypeId="2" ParentId="2980" CreationDate="2015-05-30T13:34:13.550" Score="6" Body="&lt;p&gt;When a car turns the sideways force will pull the outer side under the rim. A angle sticking out will get worn off quite quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if a tire isn't properly inflated then the walls will bulge over and put more force on the outer side. Again a corner will get worn off rapidly under these circumstances. &lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-05-30T13:34:13.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2982" PostTypeId="1" CreationDate="2015-05-31T06:59:44.197" Score="7" ViewCount="3859" Body="&lt;p&gt;I noticed this pressure gauge on the Piccadilly Line of London Underground yesterday. It was located on the side of a row of seats. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/iMGqgm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Close-up of the gauge at hand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Grtfx.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Grtfxt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;click for larger version of image&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure gauge seemed to randomly fluctuate between 0.5 and 1.5 bar when the train was in motion, with no significant correlation with the acceleration or the speed of the train. However, when the train was in stations and the door opened, the gauge reads out as ~3 bar. The close-up was taken when the train was stopped at a station. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this pressure gauge measuring, and why is it helpful to have it displayed in such a prominent location in direct sight?&lt;/p&gt;&#xA;" OwnerUserId="164" LastEditorDisplayName="user16" LastEditDate="2015-05-31T12:29:36.540" LastActivityDate="2015-05-31T12:29:36.540" Title="What is this pressure gauge on the London Underground measuring?" Tags="|mechanical-engineering|pressure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2983" PostTypeId="2" ParentId="2982" CreationDate="2015-05-31T07:06:30.487" Score="6" Body="&lt;p&gt;This appears to be the pressure gauge measuring the brake pressure of the brake cylinders. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://www.whatdotheyknow.com/request/purpose_of_barometer_on_tube_tra&quot;&gt;Transport For London&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The gauges that the requestor is describing are the Brake Cylinder&#xA;  Pressure Gauges that we have on every car.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;These indicate the air pressure in the brake cylinders in real-time&#xA;  and are used by operational and maintenance staff for monitoring&#xA;  status and fault-finding.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="164" LastActivityDate="2015-05-31T07:06:30.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2986" PostTypeId="1" AcceptedAnswerId="2988" CreationDate="2015-05-31T12:08:08.527" Score="3" ViewCount="300" Body="&lt;p&gt;My question is &quot;what is the need of using auxiliary carry?&quot;  I mean 'AC' flag is set when D3 generates a carry. Then why not different flags for D2 or D4? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why suddenly we are treating D3 carry in a special way? &lt;/p&gt;&#xA;" OwnerUserId="1738" LastEditorUserId="63" LastEditDate="2015-08-20T14:20:41.393" LastActivityDate="2023-05-17T21:53:23.520" Title="What is the need of having a separate flag as Auxiliary carry in microprocessor 8085?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2987" PostTypeId="1" CreationDate="2015-05-31T13:05:06.087" Score="3" ViewCount="12571" Body="&lt;p&gt;I understand that the model for a first order system is &#xA;$$G(s)=\frac{k_m*e^{-\tau s}}{T_ms+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case $\tau=3$ and the system gain $k_m=10$, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to calculate the transfer function and after that I can calculate the poles by solving for $s$ in the transfer function's denominator. Am I on the right track? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the transfer function given this graph? I don't have access to a book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/M5ahM.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1813" LastEditorUserId="368" LastEditDate="2015-06-02T17:26:58.650" LastActivityDate="2015-06-02T17:26:58.650" Title="How to compute the transfer function from a response graph" Tags="|control-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2988" PostTypeId="2" ParentId="2986" CreationDate="2015-05-31T13:55:17.293" Score="4" Body="&lt;p&gt;It is used for BCD arithmetic. Each decimal requires 4 bits. &lt;/p&gt;&#xA;" OwnerUserId="1579" LastEditorUserId="1579" LastEditDate="2015-05-31T13:57:35.793" LastActivityDate="2015-05-31T13:57:35.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2991" PostTypeId="1" AcceptedAnswerId="2997" CreationDate="2015-05-31T17:59:30.210" Score="14" ViewCount="702" Body="&lt;p&gt;Some laser applications consist simply of concentrating light into a small spot.  Two example applications are laser welding and cutting. In these cases a CO&lt;sub&gt;2&lt;/sub&gt; laser is often used which needs a regulated power supply, a water cooling system, and a supply of CO&lt;sub&gt;2&lt;/sub&gt; gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do these applications use a laser instead of a simpler (i.e. incoherent) light source such as an AC powered arc lamp?&lt;/p&gt;&#xA;" OwnerUserId="1616" LastEditorUserId="272" LastEditDate="2015-05-31T23:43:51.560" LastActivityDate="2015-11-08T23:09:40.047" Title="Why are lasers used for concentrated light applications instead of incoherent light sources?" Tags="|optics|lasers|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="2992" PostTypeId="1" AcceptedAnswerId="3012" CreationDate="2015-05-31T19:05:56.117" Score="6" ViewCount="1218" Body="&lt;p&gt;Lasers have a wide range of applications, but can be very costly to produce at high power/intensity.  Since light is nothing more than a narrow band of the electromagnetic spectrum, it makes me wonder if the same high power/intensity applications can be achieved at a lower cost by concentrating  other wavelengths (e.g. Infrared, radio, etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cursory google search revealed nothing of this sort, which makes me suspect that i this is not possible.  But i just want a clear answer as to whether other electromagnetic waves (besides light) can be concentrated and propagated as a beam invthe same way as a laser.  If possible, i'd also like to understand if/why larger wavelengths hinder our ability to concentrate these waves into high power/intensity applications.&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="2088" LastEditDate="2016-01-11T12:34:44.423" LastActivityDate="2022-03-22T03:36:07.580" Title="Can all electromagnetic spectra be focused like a laser?" Tags="|electrical-engineering|lasers|photonics|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="2993" PostTypeId="1" CreationDate="2015-05-31T19:20:02.267" Score="7" ViewCount="1307" Body="&lt;p&gt;It is like a cam that has an off-center hole, and it rotates in its fixture with a knob to lock it in place. The purpose is that you can loosen the knob and rotate the piece to slightly raise and lower the shaft that is connected on the other end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this have a name? Is it a commonly-available part or is it something that has to be custom machined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a printing press, used to adjust the vertical position of the impression cylinder a couple thousandths of an inch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/er0fp.jpg&quot; alt=&quot;Mechanism&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1816" LastEditorUserId="446" LastEditDate="2015-07-08T20:33:34.247" LastActivityDate="2016-01-05T00:19:21.167" Title="What is the name of this mechanism for raising and lowering the impression cylinder of a printing press?" Tags="|mechanical-engineering|machining|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="2995" PostTypeId="2" ParentId="2992" CreationDate="2015-05-31T20:09:35.137" Score="2" Body="&lt;p&gt;From the comment of @Dan and the linked Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Maser&quot; rel=&quot;nofollow&quot;&gt;masers&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the laser was developed, Townes and Schawlow and their colleagues at Bell Labs pushed the use of the term optical maser, but this was largely abandoned in favor of laser, coined by their rival Gordon Gould. In modern usage, devices that emit in the X-ray through infrared portions of the spectrum are typically called lasers, and devices that emit in the microwave region and below are commonly called masers, regardless of whether they emit microwaves or other frequencies.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Gould originally proposed distinct names for devices that emit in each portion of the spectrum, including grasers (gamma ray lasers), xasers (x-ray lasers), uvasers (ultraviolet lasers), lasers (visible lasers), irasers (infrared lasers), masers (microwave masers), and rasers (RF masers). Most of these terms never caught on, however, and all have now become (apart from in science fiction) obsolete except for maser and laser.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From this, everything is either called a &lt;em&gt;maser&lt;/em&gt; or a &lt;em&gt;laser&lt;/em&gt;. Also, let's ignore the debate about whether all of the letters need to be capitalized or not.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-05-31T20:09:35.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2996" PostTypeId="1" CreationDate="2015-05-31T22:46:56.397" Score="6" ViewCount="377" Body="&lt;p&gt;If you inject cold and hot air in outlets of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Vortex_tube&quot;&gt;vortex tube&lt;/a&gt;, do you obtain a greater pressure at inlet?&lt;/p&gt;&#xA;" OwnerUserId="1616" LastEditorUserId="90" LastEditDate="2015-06-01T07:33:09.693" LastActivityDate="2019-03-06T05:25:19.400" Title="Is vortex tube reversible?" Tags="|thermodynamics|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2997" PostTypeId="2" ParentId="2991" CreationDate="2015-06-01T00:15:49.037" Score="11" Body="&lt;p&gt;There are many reasons why highly monochromatic light, such as that emitted by a laser, is useful for delivering a large amount of power to a small spot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, incoherent light sources such as a lamp are &lt;strong&gt;extended sources&lt;/strong&gt; which means that they are emitting light from a piece of material which takes up a finite amount of space.  When focusing this light to a point, the focal spot is limited by the size of your source multiplied by the magnification of your imaging system.  This may sound like a small effect, but if you want to focus the light to a spot size which is on the order of the wavelength ($~1\ \mu\text{m}$) it becomes important.  Lasers, on the other hand, act like true point sources and can be imaged to spot sizes smaller than the wavelength of the light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second issue with incoherent light sources is that they generally &lt;strong&gt;emit light in all directions.&lt;/strong&gt; So, even if you can generate an equivalent amount of optical power, it is much more difficult to gather it all up into a collimated source which is needed prior to focusing to a spot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final advantage of lasers is their &lt;strong&gt;monochromaticity&lt;/strong&gt;.  This is useful because the wavelength can be precisely chosen to match a particular application.  CO&lt;sub&gt;2&lt;/sub&gt; lasers, for instance, emit at $10.6\ \mu\text{m}$; which is absorbed well in &lt;a href=&quot;https://www.epiloglaser.com/how-it-works/laser-material-compatibility.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;a wide range of materials&lt;/a&gt; but not all that well in metals. Nd:YAG lasers and its harmonics (the three wavelengths labeled as 'solid-state laser' in the image below) are, however, absorbed well in metals and can be used for metal processing.  With incoherent sources the light is emitted over a broad range of wavelengths which may or may not exhibit desirable properties in the material you are trying to process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.laserfocusworld.com/articles/2011/04/laser-marking-how-to-choose-the-best-laser-for-your-marking-application.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VDA2L.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-11-08T23:09:40.047" LastActivityDate="2015-11-08T23:09:40.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="2998" PostTypeId="2" ParentId="2987" CreationDate="2015-06-01T00:42:18.103" Score="1" Body="&lt;p&gt;The response of a linear time invariant system can be split up into a steady state and transient response. For a stable system (all poles of the system have negative real parts) the transient response will go to zero when the time goes to infinity and thus the response would only contain the steady state response. If you would apply some periodic input signal to the system the steady state response would also contain the frequencies present in the input, however for a step response, of step size 1, the steady state would just be a constant defined by the gain of the transfer function of the system at $s=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steady state response in the given graph seems to tends to 10, so $\left|G(0)\right|=10$ indeed yields $k_m=10$. Your value for the delay $\tau$ is correct as well, assuming the step function jumps to one at $t=0$, be remember to add the units of seconds to this parameter (same units as the time axis). Similar if the y-axis of the figure would have been given an unit, then $k_m$ would require to have this unit as well, but in this case it is dimensionless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transient response can be found with the help of the &lt;a href=&quot;http://en.wikipedia.org/wiki/Characteristic_equation_(calculus)&quot; rel=&quot;nofollow&quot;&gt;poles of the system&lt;/a&gt;, namely for distinct poles the transient response, $y_t(t)$, can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y_t(t)=\sum C_i e^{p_it},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where each $p_i$ is a pole of the system and all $C_i$ are constants, which can be solved for with boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will leave finding the pole of the system up to you, but if you have any questions feel free to ask.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2015-06-01T00:42:18.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3000" PostTypeId="1" AcceptedAnswerId="3054" CreationDate="2015-06-01T13:07:21.207" Score="1" ViewCount="618" Body="&lt;p&gt;I'm interested in the usage of a self-Locking hex nut with a nylon insert (ISO 7040 or DIN 985). &#xA;How is it used correctly? Is there a prescribed usage, or a norm that describes how it should be used? &lt;/p&gt;&#xA;" OwnerUserId="80" LastActivityDate="2015-06-04T14:13:33.207" Title="How is a self-locking hex nut (ISO 7040) used?" Tags="|civil-engineering|bolting|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3001" PostTypeId="2" ParentId="2992" CreationDate="2015-06-01T14:09:06.043" Score="6" Body="&lt;p&gt;You might be interested to learn that not only are there microwave (and IR, and X-ray ... ) lasers, but that some of these &quot;devices&quot; can even occur naturally in gas nebulae in space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the wikipedia article on &lt;a href=&quot;http://en.wikipedia.org/wiki/Megamaser&quot;&gt;Megamasers&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A megamaser is a type of astrophysical maser, which is a naturally&#xA;  occurring source of stimulated spectral line emission.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is currently a lot of research on x-ray lasers, which have existed for decades but are now being brought down to &lt;a href=&quot;http://spie.org/x94083.xml&quot;&gt;bench / desktop sizes&lt;/a&gt;. These devices generate pulses of coherent x-rays at 10 – 20 nm wavelengths.  &lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2015-06-01T14:09:06.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3002" PostTypeId="2" ParentId="2987" CreationDate="2015-06-01T14:35:34.997" Score="1" Body="&lt;p&gt;As already figured out, $ \tau = 3$ and $ k_m = 10 $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time constant $ T_m $ is the time (ignoring the delay) the first-order system takes to reach $63.2 \%$ of its final value. From the plot it seems to be around $2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the transfer function appears to be $ \frac{10 e^{-3 s}}{2 s+1} $&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2015-06-01T14:35:34.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3003" PostTypeId="1" CreationDate="2015-06-01T17:44:31.863" Score="5" ViewCount="5251" Body="&lt;p&gt;I want to buy some magnets, but the ones that I have found on online shops are mostly grade N38. They are pretty cheap, but they need to be as strong as the ones found in a hard drives. I am afraid hard drive magnets are N52 (or at least a higher grade than N38).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a minimum grade that must be used in hard drives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;maybe for the possible heat? that there should be used N42SH or something?&lt;/p&gt;&#xA;" OwnerUserId="1823" LastEditorUserId="1823" LastEditDate="2015-06-05T10:38:49.090" LastActivityDate="2015-06-05T10:38:49.090" Title="What grade of Neodymium magnets are commonly used in hard drives?" Tags="|magnets|computer-hardware|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3004" PostTypeId="2" ParentId="1834" CreationDate="2015-06-01T18:20:32.660" Score="0" Body="&lt;p&gt;Disrupt: no. Effect somehow: yes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;New engines would over-power the aircraft, so you would have to choose smaller main (or inner) engines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With more engines, engine maintenance tasks would take longer, and reliability figures (mean time to between failures, etc) would decrease to some degree. It would also aeroelastically effect the wing (which must be considered).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a proof-of-concept demonstration (on paper), there's no need to increase the complexity of the aircraft by adding more engines. If someone needs to flight-prove this concept, then adding more engines can be postponed till there's a major benefit from having the extra engines.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2015-06-01T18:20:32.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3005" PostTypeId="1" AcceptedAnswerId="3007" CreationDate="2015-06-01T19:52:11.580" Score="-1" ViewCount="245" Body="&lt;p&gt;I'm using a 10/100/1000 BASE-T fiber transceiver. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this mean?&lt;/p&gt;&#xA;" OwnerUserId="1825" LastEditorUserId="272" LastEditDate="2015-06-01T20:14:34.483" LastActivityDate="2015-06-02T17:47:42.787" Title="What is the &quot;10/100/1000 BASE-T&quot; laser/transceiver rate unit signifying?" Tags="|lasers|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3006" PostTypeId="2" ParentId="2996" CreationDate="2015-06-01T19:57:23.997" Score="1" Body="&lt;p&gt;No; air will not flow from low pressure to high pressure, at least not in a vortex tube.  The momentum equation for an inviscid fluid flowing at steady-state in one dimension is &#xA;$$&#xA;\frac{\partial \left( \rho u^2 \right)}{\partial x} = -\nabla p&#xA;$$&#xA;where $\rho$ is the fluid density, $u$ is velocity, and $\nabla p$ is the pressure gradient.  The negative sign in front of the pressure gradient signifies that, in the absence of any other forces, the flow direction will always be opposite that of the pressure gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your hypothetical reverse vortex tube, you want the outlet pressure to be higher than the cold/hot inlet pressures, but you also want the flow direction to be from the inlet to the outlet.  These two conditions contradict each other via the momentum equation.  In other words, you want the flow direction to be in the &lt;em&gt;same&lt;/em&gt; direction of the pressure gradient, which is not allowed (without some internal machinery like a compressor, etc.)&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-06-01T20:48:39.307" LastActivityDate="2015-06-01T20:48:39.307" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3007" PostTypeId="2" ParentId="3005" CreationDate="2015-06-02T02:43:31.707" Score="5" Body="&lt;p&gt;The 10/100/1000 refers to the transmission rate in megabits per second&lt;br&gt;&#xA;= millions of bits per second.&lt;br&gt;&#xA;Also may be shown as Mbps or Mb/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1000 Mb/s is a data rate typically achieved by modern LAN interfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10 and 100 mb/s are the maximum rates achieved by older standards. &#xA;A 10/100/1000 label is indicating that the device operates at the 1000 Mb/s rate where it is interfacing with equally capable equipment, but is also able to work at the 10 or 100 Mb/s rate and standards when the equipment it is interworking with is limited to those rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As hazzey notes, Base-T does not apply to fibre based equipment. It is a designation indicating that twisted pair is used as the transmission medium.&lt;br&gt;&#xA;The original &lt;a href=&quot;http://en.wikipedia.org/wiki/Ethernet&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;&quot;Ethernet&quot; LAN interface&lt;/strong&gt;&lt;/a&gt;  (now part of IEEE 802.3-xxx) used coaxial cable and operated at 10 Mb/s.  When twisted pair interfaced  alternatives were introduced, the name BASE-T was introduced for twisted pair rather than coaxial cable interconnected equipment. 10 Base-T was used to designate 10 Mb/s twisted pair equipment.&lt;br&gt;&#xA;&quot;Just to confuse&quot; designations 10Base5 and 10Base2 were designations relating to thin-coax 10 Mb/s Ethernet in the 1980s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Gigabit_Ethernet&quot; rel=&quot;nofollow&quot;&gt;Wikipedia - Gigabit Ethernet&lt;/a&gt; - &#xA;nominally re 1000 Mb/s systems but covers the historical development.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ethernet &lt;a href=&quot;https://www.perle.com/products/10-100-1000-media-converter.shtml&quot; rel=&quot;nofollow&quot;&gt;Fibre and twisted pair interfaces&lt;/a&gt;  exist but that is not what you are dealing with.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Related&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Ethernet_over_twisted_pair&quot; rel=&quot;nofollow&quot;&gt;Ethernet over twisted pairs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/10BASE2&quot; rel=&quot;nofollow&quot;&gt;Wikipedia - 10BASE2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/10BASE5&quot; rel=&quot;nofollow&quot;&gt;Wikipedia - 10BASE2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.nz/search?q=ethernet&amp;amp;biw=1280&amp;amp;bih=789&amp;amp;tbm=isch&amp;amp;tbo=u&amp;amp;source=univ&amp;amp;sa=X&amp;amp;ei=URRtVYrrEqfGmQW7-4KQAg&amp;amp;ved=0CFAQsAQ#tbm=isch&amp;amp;q=ethernet+base-10&quot; rel=&quot;nofollow&quot;&gt;10BASE2 &amp;amp; 10BASE5 eternet&lt;/a&gt; and&lt;a href=&quot;https://www.google.co.nz/search?q=ethernet&amp;amp;biw=1280&amp;amp;bih=789&amp;amp;tbm=isch&amp;amp;tbo=u&amp;amp;source=univ&amp;amp;sa=X&amp;amp;ei=URRtVYrrEqfGmQW7-4KQAg&amp;amp;ved=0CFAQsAQ#tbm=isch&amp;amp;q=ethernet+10base5&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://www.google.co.nz/search?q=ethernet&amp;amp;biw=1280&amp;amp;bih=789&amp;amp;tbm=isch&amp;amp;tbo=u&amp;amp;source=univ&amp;amp;sa=X&amp;amp;ei=URRtVYrrEqfGmQW7-4KQAg&amp;amp;ved=0CFAQsAQ#tbm=isch&amp;amp;q=ethernet+10base2&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.nz/search?q=ethernet&amp;amp;biw=1280&amp;amp;bih=789&amp;amp;tbm=isch&amp;amp;tbo=u&amp;amp;source=univ&amp;amp;sa=X&amp;amp;ei=URRtVYrrEqfGmQW7-4KQAg&amp;amp;ved=0CFAQsAQ&quot; rel=&quot;nofollow&quot;&gt;Ethernet&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorDisplayName="user16" LastEditDate="2015-06-02T17:47:42.787" LastActivityDate="2015-06-02T17:47:42.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3009" PostTypeId="2" ParentId="2996" CreationDate="2015-06-02T11:05:15.713" Score="0" Body="&lt;p&gt;No. What you will do is simply mix your gasses, arriving at the mean temperature. You loose all the potential useful work in the temperature difference without producing any work. In a way it's hard to explain, in another way terribly easy: there's simply no mechanism, no process there that might transform this heat difference into a higher pressure at the outlet.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-06-02T11:05:15.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3010" PostTypeId="1" CreationDate="2015-06-02T11:28:52.947" Score="6" ViewCount="5226" Body="&lt;p&gt;Plastic pipes look superior - no corrosion, lightweight, easy to install, a bit flexible. Yet there're lots of cases when culverts and drainage pipes are made of steel (corrosive) and reinforced concrete (very heavy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can other pipe materials do that plastic cannot in drainage and culverts?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="272" LastEditDate="2015-06-02T11:58:21.650" LastActivityDate="2016-06-30T15:48:32.470" Title="Why prefer any other pipes over plastic pipes for drainage and culverts?" Tags="|civil-engineering|materials|piping|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3011" PostTypeId="2" ParentId="3010" CreationDate="2015-06-02T12:34:50.343" Score="5" Body="&lt;p&gt;Although it doesn't corrode in the same say that steel does, plastic also degrades over time.  The common corrugated steel pipes are galvanized, so take some time to corrode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concrete is usually used to support larger soil pressure from the outside.  Corrugated steel pipe can sometimes get squished by the pressure of the soil, especially when it's a bit unstable and the pipe is buried deeply.  For large diameters, concrete is cheaper, especially the deeper it is buried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue to consider especially for small shallow culverts is that light is actually a liability.  When the soil acts like a very slow-moving fluid, light pipes rise over time due to bouancy.  You can sometimes see this where a corrugated steel pipe was buried only a foot below the surface of a dirt road.  After some years, the top of the pipe is at the road surface, where it gets damaged by the traffic.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-06-02T12:34:50.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3012" PostTypeId="2" ParentId="2992" CreationDate="2015-06-02T13:16:58.007" Score="6" Body="&lt;p&gt;Yes, other electromagnetic radiation can be focused, and routinely is.  Basically, everything up to x-rays can be focused relatively easily.  Once you get to energies so high that they go right thru most material, using these materials to focus such rays gets tricky or impossible.  For example, I don't think we know of a way to focus gamma rays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a cosmic scale, gravity can be used to bend the path of radiation, which works for all wavelengths.  It is currently beyond our capabilities to construct devices that do this in any meaningful way, but we can and have made use of the ones that occur naturally.  Distant galaxies are sometimes used by astronomers as &lt;i&gt;gravitational lenses&lt;/i&gt; to see more clearly things that are even more distant behind these galaxies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most electromagnetic radiation, there are two ways we can change its propagation path, refraction and reflection.  For visible light, this is done with lenses and mirrors, respectively.  Finding a structurally suitable material that is transparent at other wavelengths (like glass is at visible wavelengths) can be difficult, and most focusing of wavelengths far from the visible spectrum is usually done with reflection rather than refraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of reflection used to focus non-visible radiation abound.  Old UHF TV roof antennas were commonly &lt;i&gt;corner reflectors&lt;/i&gt;.  These has a series of rods that formed a V-shaped pocket pointed in the direction of the transmitter.  The actually antenna was a split rod or sometimes a series of rods inside the V.  Other antenna systems use unconnected conducting rods and other conducting objects to reflect the radiation to increase its flux at the actual receiving (or transmitting, antennas work equally both ways) element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large radar dishes or radio-telescopes are examples of extreme focusing of radiation, usually in the GHz range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At long enough wavelengths, yet another method can be used to make a focused beam of radiation.  This is done by controlling the phase of multiple emitters.  As a simplistic example, consider two vertical radio antennas spaced 1/4 wavelength apart.  Antenna A is fed with a signal that is 90&amp;deg; ahead of that to antenna B.  Imagine a wavefront emitted from A.  After 1/4 cycle it reaches B, which is now emitting the same signal.  The two wavefronts emitted by A and B add, and the signal appears strong to someone receiving in the B direction.  Now imaging a wavefront leaving B.  When it gets to A, A is emitting 1/2 wavelength ahead, which is 180&amp;deg; out of phase.  The two signals therefore cancel each other in the A direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This basic principle, called a &lt;i&gt;phased array&lt;/i&gt;, can be applied to a large number of individual transmitters (or receivers, works both ways).  With lots of individual elements in the phased array, a very narrow beam can be emitted (or again, the overall antenna can listen in a narrow direction).  A great example of this is the Aegis radar system originally deployed on the Arleigh Burke class of destroyers.  The large hexagonal patches on the front and sides are actually arrays of many small radar antennas, with the phase shift of each being controllable dynamically.  This means the beam can be re-directed very quickly electronically without requiring a mechanical sweep of a physical antenna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This principle can also be used to steer a reflected or transmitted beam.  At optical wavelengths, this is usually called a &lt;i&gt;diffraction grating&lt;/i&gt;.  A series of parallel scratches separated by a fraction of a wavelength is etched onto a mirror.  At the right combination of wavelength and reflection angle, the reflections from each of the peaks are either in phase and therefore add, or are out of phase and cancel each other.  Since the angle they are in phase at is a function of wavelength, such diffraction gratings can be used to separate light by its wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These parallel scratches can also be applied to the surface of transparent objects.  When you look thru one at a thin vertical slit of white light aligned with the scratches, you see it spread out horizontally into its component colors.  Some spectrometers work on this principle.  I have heard of one effort to use this effect to focus x-rays.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2015-06-02T17:14:58.903" LastActivityDate="2015-06-02T17:14:58.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3013" PostTypeId="1" CreationDate="2015-06-02T13:18:56.723" Score="6" ViewCount="512" Body="&lt;p&gt;Inspired by &lt;a href=&quot;http://edition.cnn.com/2015/05/30/travel/paris-love-locks-bridges-feat/index.html&quot; rel=&quot;noreferrer&quot;&gt;this story&lt;/a&gt; about Paris authorities ordering removal of thousands of &quot;love lock&quot; padlocks from bridge railings. This is not a DIY SE question because it's not DIY scale in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose we have several dozen meters of valuable metalworks (perhaps some one hundred years old forged or cast railings) and so we cannot just melt the whole thing and produce new railings. We have to remove the padlocks and place the railings back. And we'd be happy to do all of that over a week or so, not waste a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that we're not required to work on the bridge and can remove the railings for a while (and install cheap looking temporary railings for that period).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be an adequate technological process for that? Cutting with boltcutters? Cutting with angle grinder? Cutting with gas welder? Anything else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iPOof.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iPOofm.jpg&quot; alt=&quot;locks on railing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorDisplayName="user16" LastEditDate="2015-06-02T13:31:12.607" LastActivityDate="2022-06-24T19:09:27.307" Title="What would be a practical way for removing thousands of padlocks from valuable bridge railings?" Tags="|metals|cutting|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3015" PostTypeId="2" ParentId="3013" CreationDate="2015-06-02T14:16:53.320" Score="0" Body="&lt;p&gt;cutting with oxygen torches or angle grinder will likely cause damage to the railings. Heat damage from the torch and abrasion damage from the grinder spray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any of the non boron carbide/standard metal shackles, some type of powered bolt cutters could be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The locks with the super hard shackles like boron carbide, chill it with liquid nitrogen and shatter the shackle with pneumatic air hammers/chisels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything else can be plasma cut.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2015-06-02T14:16:53.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3016" PostTypeId="2" ParentId="3013" CreationDate="2015-06-02T14:41:22.813" Score="8" Body="&lt;h3&gt;Just do it&lt;/h3&gt;&#xA;&lt;p&gt;Since the primary requirement is to not harm the existing structure, anything that is done will have to be manual.  Humans are good about dealing with varying conditions. Each lock type, shape, and location is different.&lt;/p&gt;&#xA;&lt;p&gt;We don't trust robots to excavate archeological sites for the same reason.&lt;/p&gt;&#xA;&lt;h3&gt;Do no harm&lt;/h3&gt;&#xA;&lt;p&gt;Even with humans manually cutting the locks, the method used needs to be easy enough to control and contain so that the structure isn't damaged.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Grinders&lt;/strong&gt; have blades that are relatively large and will need to be used in a confined space. There is a danger of accidentally cutting the structure.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Plasma cutters&lt;/strong&gt; have a real risk of over cutting through the lock and damaging the structure.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Gas cutting torches&lt;/strong&gt; have the same drawbacks as plasma cutters.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mechanical bolt cutters&lt;/strong&gt; can be positioned accurately, and it is easy to see exactly what will be cut. They can be hydraulically powered to lessen the physical exertion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;It would be relatively quick and safe to cut the locks manually. If any individual lock is difficult, it can be skipped.  These few locks can be dealt with later once there is more space. This will keep the production level up.&lt;/p&gt;&#xA;&lt;h3&gt;A different solution?&lt;/h3&gt;&#xA;&lt;p&gt;The question (and my answer to this point) has focused on &lt;em&gt;cutting&lt;/em&gt; the lock off. These are likely cheap and common locks that would be easy for a locksmith to remove.&lt;/p&gt;&#xA;&lt;p&gt;An experienced locksmith with the right tools (bump key, vibrating gun thing, etc.) may be able to walk down the line actually &lt;em&gt;unlocking&lt;/em&gt; the locks. This would be by far the easiest route and it might even be quicker. Even if it only worked for a subset of the locks, it would be something easy to try.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-06-02T16:59:21.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3017" PostTypeId="2" ParentId="3003" CreationDate="2015-06-02T15:13:05.673" Score="3" Body="&lt;p&gt;The chances of all hard drives using the same grade of magnet is very low.  Different designs call for different grades of magnets.  It is possible that some hard drives use N52 magnets but in general there is no way to tell.  If it is really important for you to know the grade, you can send magnets out to be tested.  They would could use something like a DC Hysteresis Graph test to determine the magnet's properties.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could get a rough idea by picking something up with the magnet and then comparing that to a known magnet.  For example, if you have a magnet that you know is N52, you could try picking up increasingly heavy objects with the N52 magnet and your unknown magnet and see if they can pick up the same mass of object.  If they both pick up a 3 kg object but can't pick up a 3.25 kg object, for example, you might be reasonably sure that your unknown magnet is a N52 magnet.  If the unknown magnet can only pick up 1 kg, however, then you could be reasonably sure it isn't N52.  This test would require both magnets to be the same size and shape.&lt;/p&gt;&#xA;" OwnerUserId="198" LastEditorUserId="198" LastEditDate="2015-06-02T15:18:50.623" LastActivityDate="2015-06-02T15:18:50.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3018" PostTypeId="1" AcceptedAnswerId="3020" CreationDate="2015-06-02T17:38:46.190" Score="11" ViewCount="6553" Body="&lt;p&gt;Lift produced by an airplane wing is related to airspeed - this much is clear; a plane moving too slowly will stall. But what is that relation? Linear? Quadratic? Exponential? I don't need the exact equation, which is surely quite complex, just the character of the relation.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="1097" LastEditDate="2015-06-03T07:09:14.933" LastActivityDate="2015-06-03T21:56:40.247" Title="How is lift related to airspeed?" Tags="|fluid-mechanics|aerospace-engineering|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3019" PostTypeId="2" ParentId="2612" CreationDate="2015-06-02T17:53:23.370" Score="1" Body="&lt;p&gt;You haven't stated whether you want to get sensor readings into a computer or not. Also, it's not clear how frequent you need to do this task, and how much setup time is allowed per measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to do some &lt;strong&gt;Quality inspection&lt;/strong&gt; (of whether the metallic plate is within tolerance bounds), you might want to consider simple or advanced image processing. for simple image processing, take a picture from front view and measure the distance between one end of the metal plate to the other end, or put a measurement paper behind the metal plate and take or observe the picture for measurement. If you have a human operator, this would probably be a simple way, just the accuracy requirement might be missed, depending on the setup of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Proximity sensors&lt;/em&gt; and other types of sensors might be effected by the movement of the metal plate, are you sure that it is not moving?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are sure about the alignment quality (that two surfaces are parallel), then you could push the plate to one side and use a regular micrometer or other measurement device. &lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2015-06-02T17:53:23.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3020" PostTypeId="2" ParentId="3018" CreationDate="2015-06-02T18:17:59.687" Score="10" Body="&lt;h1&gt;Quadratically.&lt;/h1&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.grc.nasa.gov/WWW/K-12/FoilSim/Manual/fsim0005.htm&quot; rel=&quot;noreferrer&quot;&gt;NASA page #1&lt;/a&gt; says that, as an approximation,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\text{Lift}\propto v^2$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; is the airspeed.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/problems/Dalette/LiftvsAirspeed_ans.htm&quot; rel=&quot;noreferrer&quot;&gt;NASA page #2&lt;/a&gt; says that, for a better calculation,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\text{Lift}=av^2+bv+c$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; are constants. This is odd because it implies that lift is not merely proportional to &lt;span class=&quot;math-container&quot;&gt;$v^2$&lt;/span&gt;, because there is the &lt;span class=&quot;math-container&quot;&gt;$bv$&lt;/span&gt; term in there, as well as the &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; term.&lt;/p&gt;&#xA;&lt;p&gt;But there is a better equation, given &lt;a href=&quot;http://www.av8n.com/how/htm/4forces.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; (and &lt;a href=&quot;http://www.langleyflyingschool.com/Pages/CPGS+4+Aerodynamics+and+Theory+of+Flight+Part+1.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;), among other places:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\text{Lift}=\frac{1}{2}\rho v^2 \times \text{lift coefficient} \times \text{area}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is air density.&lt;/p&gt;&#xA;&lt;p&gt;This is similar to the drag formula.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-06-03T00:09:18.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3021" PostTypeId="2" ParentId="3018" CreationDate="2015-06-02T18:54:16.533" Score="8" Body="&lt;p&gt;As has already been mentioned, the primary relationship is that lift goes with the square of the airspeed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you some intuition as to why this is, consider what a wing does.  As it moves along, it deflects air downwards.  Lift is the upwards force from imparting downward momentum on the air the wing passes thru.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Momentum is mass times velocity, and force is momentum per time.  As the airspeed increases, both more air per unit time is pushed downward, and it is pushed downward faster.  Put another way, the force is (mass / time)(velocity), with both mass/time and velocity roughly proportional to the airspeed, so the force is proportional to the square of the airspeed.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-06-02T18:54:16.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3022" PostTypeId="2" ParentId="2688" CreationDate="2015-06-02T20:02:34.880" Score="2" Body="&lt;p&gt;This problem can be simplified by observing the symmetry. This allows us to solve only one side, considering it as a beam of span &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; which is fixed and pinned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m6yRg.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, reactions are really just concentrated loads, so lets replace that pinned support with a concentrated vertical load &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;. To obtain &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;, we first take a look at how the beam deforms without the pinned support (i.e. just the cantilever). A simple cantilever has a total vertical displacement at the end equal to &lt;span class=&quot;math-container&quot;&gt;$d_q = \dfrac{q\cdot L^4}{8EI}$&lt;/span&gt;. Now, I lied. A reaction isn't just a concentrated load. It's also knowledge of the true displacement. We know that the end of the cantilever actually has a displacement equal to 0. Therefore, the concentrated load &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; has to create a displacement that cancels out the value we calculated above.&#xA;Now, a cantilever with a concentrated load at its extremity has a vertical displacement equal to &lt;span class=&quot;math-container&quot;&gt;$d_P = \dfrac{P\cdot L^3}{3EI}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to cancel out, we need &lt;span class=&quot;math-container&quot;&gt;$d_q = d_P$&lt;/span&gt;. Therefore, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\dfrac{q\cdot L^4}{8EI} = \dfrac{P\cdot L^3}{3EI} \therefore P = \dfrac{3qL}{8}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JjJhm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;span class=&quot;math-container&quot;&gt;$F_{RA} = F_{RC} =\dfrac{3qL}{8}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_{RB} = 2\cdot\dfrac{5qL}{8}$&lt;/span&gt; (it's times two, once for each side of the real beam).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It anyone's interested, the figures were created with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D analysis tool.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:28:58.183" LastActivityDate="2019-12-28T18:28:58.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3023" PostTypeId="2" ParentId="326" CreationDate="2015-06-02T21:55:58.773" Score="2" Body="&lt;p&gt;I don't think there are any engineering issues with them, I just think there is no economic or operational niche for them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have the operational overhead of aircraft so why not just use a plane to begin with? They can only be used over relatively flat terrain or still water but is there anyplace in the world where the necessary terrain and economic need for ground hugging fast transport exists? I doubt it. E.g. I can't see them working as reliable ferries in populated areas for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also see WIGs as one of those 90% great, 10% deal breaking suck technologies. The deal breaking suck is the limitation on terrain which even over water, will be unpredictable. Just a bit of wave toss up and suddenly you might as well be flying. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem, rarely discussed is that these things are LOUD because the big ones are fully revved jets operating right on the ground. Read up on the first hand accounts of Soviet test observers on the Caspian sea. No way your getting those operating near populated areas these days. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also suspect that over land they have a similar problem to hovercraft in that the powerful downdrafts that keep them lofted turn almost anything they fly over that isn't nailed down into a projectile. Imagine a huge WIG flying up over a gravel beach. It'd be like a a machine gun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It only take a one big negative to render a design or technology useless and I think WIGs have several big negatives. &lt;/p&gt;&#xA;" OwnerUserId="1837" LastActivityDate="2015-06-02T21:55:58.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3024" PostTypeId="1" AcceptedAnswerId="3060" CreationDate="2015-06-02T23:21:04.027" Score="2" ViewCount="1096" Body="&lt;h3&gt;Summary&lt;/h3&gt;&#xA;&lt;p&gt;A short way to phrase this question is:&lt;br&gt;&#xA;&lt;em&gt;Shear&lt;/em&gt; describes the translation of one part of a body relative to another given an applied force.&lt;br&gt;  What is the equivalent rotational term?  That is,&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What&lt;/em&gt; describes the rotation of one part of a body relative to another given an applied moment of force?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h3&gt;In more detail:&lt;/h3&gt;&#xA;&lt;p&gt;I'm trying to find the right word for a particular type of distortion.  I have an assembly of items connected to be approximately solid.  I mount one flat to a wall, and put some force or moment onto the opposite flat side so that the assembly distorts.  In particular, I want to correctly describe the relative rotation of the two flats (in any direction), and I want this term to describe only the rotation, and exclude any translation (so that if it only shears without twisting I'd like the quantity to be zero).&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hL8dm.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;I thought of &lt;strong&gt;torsion&lt;/strong&gt;, but definitions of torsion always seem to refer to a bar or an elongated axis of some type.  Is a bar required for &amp;quot;torsion&amp;quot; to be meaningful?  I thought of &lt;strong&gt;twist&lt;/strong&gt;, but that doesn't necessarily imply a dynamic quantity, so, for example, one could freeze in a twist, and something can be twisted without forces applied.  I don't think &lt;strong&gt;bend&lt;/strong&gt; is correct because that could just refer to translation.  Is there a better word that I'm not thinking of, or even a concise phrase?&lt;/p&gt;&#xA;&lt;p&gt;If there is a word that matches my needs, a reference that I could review would be very helpful.&lt;/p&gt;&#xA;&lt;hr&gt;&#xA;&lt;h3&gt;Why &amp;quot;bending&amp;quot; is not the word I need&lt;/h3&gt;&#xA;&lt;p&gt;Below is a sketch of two objects that are bending along the same curve (both by intuition and the equations), yet the rotation of the ends is different in each.  I'm interested in a word or phrase for describing the rotation of the ends (or a tendency for the ends to rotate or not, etc).&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LS3iD.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also, according to wikipedia, &lt;a href=&quot;http://en.wikipedia.org/wiki/Bending&quot; rel=&quot;nofollow noreferrer&quot;&gt;bending&lt;/a&gt; &amp;quot;characterizes the behavior of a slender structural element&amp;quot; under load, but I'm not interested in slender elements (and only drew the above to illustrate bending).  The first is the common way a beam would bend, but not everything bends like a beam, and I am hoping to find a word that characterized the particular difference illustrated here.&lt;/p&gt;&#xA;&lt;h3&gt;Things that don't bend like normal beams&lt;/h3&gt;&#xA;&lt;p&gt;A reasonable model is a highly anisotropic material, rigid in one axis but less so in the others.  For example, a bundle of parallel fiber optic cables.&lt;/p&gt;&#xA;&lt;p&gt;I think another would be a tube filled with water at high pressure, like a fire hose.  (I don't know how this bends, but I doubt it bends like a beam.)&lt;/p&gt;&#xA;&lt;p&gt;Here's a sketch of a structure that doesn't bend like a beam, in that the side parallel to the wall stays parallel, even though the rest of the structure bends in a normal way.  The black dots are meant to represent rotating joints.  Even if this item were long and slender I think the ends would stay parallel.  The point of the third picture is that you don't necessarily get to see what's on the inside (but it could still be important to describe that the two walls don't rotate with respect to each other in response to an applied force).&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JnHsI.png&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="1838" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-06-04T23:26:04.990" Title="How to describe the rotation of opposite ends of a body due to an applied moment" Tags="|mechanical-engineering|statics|terminology|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3025" PostTypeId="2" ParentId="3024" CreationDate="2015-06-03T00:24:34.930" Score="1" Body="&lt;p&gt;In &lt;a href=&quot;http://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow&quot;&gt;Euler-Bernoulli beam theory&lt;/a&gt;, this is known as &lt;em&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Bending&quot; rel=&quot;nofollow&quot;&gt;bending&lt;/a&gt;&lt;/em&gt;, or, if the &quot;beam&quot; continues to oscillate, &lt;em&gt;dynamic bending&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this specific case - assuming that &quot;up&quot; is &quot;up&quot; in real life - this is called the &lt;em&gt;first vertical bending&lt;/em&gt;. It is dynamically shown here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/4/43/Beam_mode_3.gif&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/4/43/Beam_mode_3.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternate terms include &lt;em&gt;deflection&lt;/em&gt; (given &lt;a href=&quot;http://paws.wcu.edu/radams/Intro_to_Beam_Theory.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;) and &lt;em&gt;bending action&lt;/em&gt; (given &lt;a href=&quot;http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect10.d/IAST.Lect10.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;), though I'm less certain of the applicability of the latter.&lt;/p&gt;&#xA;" OwnerUserId="106" LastActivityDate="2015-06-03T00:24:34.930" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="3026" PostTypeId="2" ParentId="3018" CreationDate="2015-06-03T01:39:35.643" Score="3" Body="&lt;p&gt;In addition to the basic equation ($L = CL * 0.5 * \rho * V^2 * \text{Ref Area}$), the coefficient of lift varies with speed if the speed range of concern is large enough, e.g. the &lt;strong&gt;Reynolds number&lt;/strong&gt; changes considerably (maybe 2-3 times). The Reynolds number affects $CL$, especially for higher $CL$ values.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastEditorUserId="106" LastEditDate="2015-06-03T21:56:40.247" LastActivityDate="2015-06-03T21:56:40.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3027" PostTypeId="1" CreationDate="2015-06-03T03:26:56.343" Score="6" ViewCount="1229" Body="&lt;p&gt;I have an FEM model that I am doing a modal analysis of. I wanted to check that how I am computing the physical displacement is the correct way, as I've read a lot of about normalizing modes, participation factors, effective masses, etc. and I'm not 100% sure on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got the various mode shapes and frequencies but now I want to compute how much actual displacement I get if I apply a harmonic force to it. However, I don't want to compute this using the FEM tool as I want to compute a displacement in another tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I have made my FEM model using units m/kg/N/s and computed a set a modal mass normalized modes, ${\phi_i}$, whose respective amplitudes are $q_i$. The &lt;em&gt;physical&lt;/em&gt; displacement of my model would then be:&#xA;$$&#xA;    U(x,y,z) = \sum_i  q_i\  \phi_i(x,y,z)&#xA;$$&#xA;Does this have units of meters as my modes are mass normalized or do I need another scaling factor here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual mode amplitudes I compute with:&#xA;$$&#xA;    q_i = \frac{F_i}{-\omega^2 + j\ C\ \omega - \omega_i^2}&#xA;$$&#xA;Here $\omega$ is the frequency I'm looking at, $C$ some damping constant, $\omega_i$ the resonance frequency of the mode. The actual force I apply in my other tool is some factor of the generalized modal force, $F_i$. Which is just a projection of the forces I choose to apply in units of Newtons into the various modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the above sound correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, do FEM models typically output using modal mass normalization just because the calculations of displacements are straight forward like this?&lt;/p&gt;&#xA;" OwnerUserId="1840" LastEditorUserId="90" LastEditDate="2015-07-14T05:14:11.357" LastActivityDate="2015-07-14T05:14:11.357" Title="FEM modal analysis computing displacements" Tags="|modeling|finite-element-method|modal-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3029" PostTypeId="2" ParentId="3018" CreationDate="2015-06-03T09:15:40.310" Score="-1" Body="&lt;p&gt;Lift is proportional to the square of velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vMQc4.gif&quot; alt=&quot;Diagram for relationship of Lift and Velocity&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1846" LastActivityDate="2015-06-03T09:15:40.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3030" PostTypeId="1" AcceptedAnswerId="3223" CreationDate="2015-06-03T09:26:54.690" Score="5" ViewCount="429" Body="&lt;p&gt;I have the utilization rates of several machines for each week over a year. These differ per week because of the occurrence of machine failures and changes of orders. Meaning that one week a machine could be a bottleneck and the other week not. I want to rank these machines based on the utilization and so determine their criticality. However if I calculate the mean utilization, information gets lost. The same is when I count the events the utilization is bigger than a threshold. For example; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;machine A has a utilization of 50% and 14 weeks the utilization &gt; 90%&lt;/li&gt;&#xA;&lt;li&gt;machine B has a utilization of 70% and 7 weeks the utilization &gt; 90% &lt;/li&gt;&#xA;&lt;li&gt;machine C has a utilization of 89% and 6 weeks the utilization &gt; 90%&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In my opinion machine A is more critical than machine B. So ranking based on a threshold value seems reasonable. However, when comparing machine B and C this is not that clear. Does anyone know a statistical/mathematical method to get a good comparison?&lt;/p&gt;&#xA;" OwnerUserId="1847" LastEditorUserId="282" LastEditDate="2015-06-06T15:21:52.227" LastActivityDate="2015-06-17T13:45:15.440" Title="How to compare utilization rates?" Tags="|statistics|industrial-engineering|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3031" PostTypeId="1" CreationDate="2015-06-03T11:15:08.297" Score="3" ViewCount="270" Body="&lt;p&gt;Let's suppose I have a water jet as source, and this jet goes through a tube and is equally divided into 4 tubes. Is there a way to control the flow/pressure at each of those tubes separately by adding a different nozzle at the end of each tube or controllable valves at the beginning of each tube? Or is it only possible to get relative pressures at each tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how can something like this be implemented?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An application would be having only one horizontal tube (source of water) connected to 4 vertical tubes used as fountaines and where I want to control the height of the water output of each fountain. &lt;/p&gt;&#xA;" OwnerUserId="1849" LastEditorUserId="282" LastEditDate="2015-06-03T23:28:49.297" LastActivityDate="2015-06-29T19:44:42.143" Title="Water pressure divider" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3032" PostTypeId="2" ParentId="643" CreationDate="2015-06-03T12:46:45.557" Score="2" Body="&lt;p&gt;$\require{cancel}$&#xA;A common method was devised by Davisson and Robinson (1965). By this method, what you need to determine is the &quot;length of fixity&quot;, which is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_f = 1.8\left(\dfrac{EI}{n_h}\right)^{0.20}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $EI$ is the pile's stiffness and $n_h$ is the soil's coefficient of horizontal subgrade modulus. This is the relevant length for a buckling analysis, even if your pile is longer than $L_f$. Obviously, if your pile has an uncovered length ($L_u$) , that needs to be added to $L_f$, in which case your buckling length is $L_f + L_u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that this equation is not set to a specific set of units, as can be seen by dimensional analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L_f = [L] = 1.8\left(\dfrac{EI}{n_h}\right)^{0.20} = \left(\dfrac{\left[\dfrac{\cancel F}{\cancel{L^2}}\right][L^{\cancel{4}2}]}{\left[\dfrac{\cancel F}{L^3}\right]}\right)^{\frac{1}{5}} = [L]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://docsdrive.com/pdfs/medwelljournals/jeasci/2007/1394-1398.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; explains it in greater detail. &lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-07-12T16:50:11.617" LastActivityDate="2017-07-12T16:50:11.617" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3033" PostTypeId="2" ParentId="3013" CreationDate="2015-06-03T13:04:02.403" Score="1" Body="&lt;p&gt;The &lt;strong&gt;portable angle grinder&lt;/strong&gt; that I saw them using on TV looked very effective.&lt;br&gt;&#xA;In the hands of a competent workman the risk of damage is minimal. &#xA;With a cutting blade this is more a &quot;geared metal saw&quot; than a grinder per se.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blade size could be selected to be only as large as needed for most locks.&lt;br&gt;&#xA;Use of a diamond cutting blade that was as durable as possible would minimise the need for blade changes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I suspect that the tool they were using was liable to be as good as most, another one which is very uncommon in this sort of application suggests itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A carborundum grit or sapphire or other media air powered cutter&lt;/strong&gt; could be as capable as desired. These are used in industrial cutting applications for metal up to many times the thickness of the largest padlock.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this would require a significant air supply, a portable compressor of any desired size could be provided. This may be more expensive to setup than most other options but would provide close to &quot;hot knife through butter&quot; cutting and precision cutting location once established.   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A guide could be provided to fit through a padlock hasp so no cutting would happen for any grit that got past the 'target', &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A recovery tube with suction could be arranged to recover most of the cutting agent and &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A hood that swung over the cutting area to minimise back spray.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All this would take some design and setup - but not excessive compared to the task and, once running it would be very fast.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note - bolt cutters:&lt;/strong&gt; Many locks, even quite cheap ones, are hardened. A suitably huge bolt cutter will tackle many hardened locks but the weight and jaw size gets very very large and it would be very impractical for the number of locks involved. - I heard an estimate of 45 tons on the initial bridge alone.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-12-04T11:35:23.577" LastActivityDate="2015-12-04T11:35:23.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3034" PostTypeId="2" ParentId="3030" CreationDate="2015-06-03T13:31:10.707" Score="1" Body="&lt;p&gt;Measuring utilization by itself is masking some of the information you're attempting to glean from this environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your description, there are at least three cases where a machine may not be put to use:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Machine is broken&lt;/li&gt;&#xA;&lt;li&gt;Change order has been placed, waiting on re-tooling or equivalent&lt;/li&gt;&#xA;&lt;li&gt;No work for machine to produce&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In addition to tracking overall utilization, I would recommend tracking outages based upon their categories.  It's not enough to say &quot;Machine A had a 50% utilization rate&quot; because you don't know &lt;em&gt;why&lt;/em&gt; the machine was sitting idle for half of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, tracking utilization is merely tracking the symptoms of the problems within your overall process.  Tracking outages will allow you to identify what the problems actually are and address those.  For example, by knowing outage rates then you may see that change orders are causing more down time than machine breakages.  Or you may see that the machines break too often and are impacting overall productivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you won't be able to glean those insights until you track the individual outage causes.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-06-03T13:31:10.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3035" PostTypeId="2" ParentId="3031" CreationDate="2015-06-03T14:12:14.573" Score="2" Body="&lt;p&gt;Yes, what you want is a device called a regulator. This will limit the pressure at a given spot in the pipe to a fixed amount. Regulators are commonly available for water at output pressures from 0-2,000 PSI, and probably higher if you look harder. Keep in mind that a regulator can only reduce pressure, not increase it, so if you need the four output pressures to be equal, you need to set the output pressure lower than the input pressure so that the regulators constrict the output, distributing the pressure through all four branch lines. The regulator can go at the beginning or end of each branch line.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-06-03T14:12:14.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3036" PostTypeId="2" ParentId="3018" CreationDate="2015-06-03T14:17:13.863" Score="1" Body="&lt;p&gt;It's all based in air pressure. Lifting requires velocity because it needs to create pressure gradients between the wing profile, so the air that is lower in pressure in the bottom part of the wing tend to equalize with the pressure in the upper part. When you stall, that means that you are no longer creating pressure between the upper and lower part of the wing profile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why you have different wing profile geometries, so that you can make different pressure ranges in different velocity scales.&lt;/p&gt;&#xA;" OwnerUserId="1853" LastEditorUserId="33" LastEditDate="2015-06-03T17:21:12.437" LastActivityDate="2015-06-03T17:21:12.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3037" PostTypeId="2" ParentId="3030" CreationDate="2015-06-03T14:54:47.933" Score="2" Body="&lt;p&gt;I don't know of any mathematical or statistical techniques that would help you analyse machine utlization data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the additional information you've given in the comments, my understanding is that your aim is the minimize the cost of keeping spare parts for the machines based on machine usage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical factor is going to be the duration of usage for each machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using your example data, you are correct in surmising that Machine A is the critical machine because it was heavily used for 14 weeks of the year, whereas the other two machines were only heavily used for 7 and 6 weeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your ranking should be based on the expected periods of heavy utilization, with machines being used for the longest period being the critical ones for which spare parts are kept. The overall utilization for the year is irrelevant for your study.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-06-03T14:54:47.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3038" PostTypeId="1" AcceptedAnswerId="3145" CreationDate="2015-06-03T15:09:56.840" Score="4" ViewCount="205" Body="&lt;p&gt;If you have a prestressed concrete beam or slab and are jacking a cable from both ends, does the order of operations alter the result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a symmetric curve, the tension diagram along the cable's span after friction losses is always shown as symmetric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zmWaO.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the case regardless of whether the ends were jacked simultaneously or one after the other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ends are jacked simultaneously then the symmetric friction loss makes perfect sense. However, if the cable is jacked first from the left and then from the right, I'd expect the tension profile to be asymmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When jacking from the left, the cable is pulled and on its left end the cable force is equal to $P_0$, the jacking force. On the right side, the cable force is $P_0 - \Delta P$, where $\Delta P = P_0\left(1-e^{-(\mu\alpha+kL)}\right)$ is the friction losses. The left side is then anchored (disregard the anchorage slip losses for a moment). The right side is then jacked. Effectively, this jacking adds a force of $\Delta P$ to the right end, so that it too is now at $P_0$. However, wouldn't the left end also be affected with an increment equal to $\Delta Pe^{-(\mu\alpha+kL)}$, such that the diagram is asymmetric?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about if one considers the fact that when the left side is anchored, anchorage slip losses occur? So the diagram should go: friction losses from jacking the left side, anchorage slip losses from anchoring the left side, friction losses from jacking the right side and then anchorage slip losses from anchoring the right side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;After a few honest-to-God minutes freaking out thinking I was the biggest idiot on Earth I realized that what @sanchises commented doesn't rule out my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cable in my question does remain in static equilibrium, as can be seen here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KzP5V.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When jacking from the left, the cable suffers at its left extremity a force of $P$. Friction losses along the way ($\Delta P$) cause the right extremity to end up with $P - \Delta P$. The right extremity is then jacked, which effectively applies a force of $\Delta P$. Friction losses also apply here, now with a value of $\Delta\Delta P$, such that the left anchor is additionally jacked with $\Delta P - \Delta\Delta P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in the left anchor with a force of $P + \Delta P - \Delta\Delta P$ and the right anchor with only $P$. These are not, however, the only forces acting on the cable: the friction losses are what keeps it in static equilibrium. Globally, we have to the left $$P + \Delta P - \Delta\Delta P +  \Delta\Delta P = P + \Delta P$$ and to the right $$\Delta P + P - \Delta P + \Delta P = P + \Delta P$$ both of which equal the total jacking force (first $P$ on the left and then $\Delta P$ on the right).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2015-06-11T19:01:07.060" LastActivityDate="2015-06-11T20:01:56.420" Title="Is the order of operations in jacking a cable relevant?" Tags="|civil-engineering|structural-engineering|beam|concrete|prestressed-concrete|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3039" PostTypeId="2" ParentId="3013" CreationDate="2015-06-03T15:41:56.267" Score="0" Body="&lt;p&gt;A handheld battery, pneumatic, or hydraulic would be the best bet.  There is very little risk of damage to the railing as compared to some of the other suggested methods and while you lose some time when compared to a grinder or torch, the accuracy outweighs it, given the value of the railings.  This would also eliminate the need to remove the railing from the site and eliminates the need for human strength (cutting a few thousand bolts would get tiring).&lt;/p&gt;&#xA;&lt;p&gt;One suggested option is a battery powered handheld rebar cutter:&#xA;&lt;a href=&quot;https://www.genscoequip.com/admin/products/attachments/bc16_bolt_cutter_copy1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.genscoequip.com/admin/products/attachments/bc16_bolt_cutter_copy1.pdf&lt;/a&gt;&#xA;Multiple batteries would be required though and you would likely need 3-4 people, each with their own.&lt;/p&gt;&#xA;" OwnerUserId="1831" LastEditorUserId="38346" LastEditDate="2022-06-24T19:09:27.307" LastActivityDate="2022-06-24T19:09:27.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3040" PostTypeId="2" ParentId="3024" CreationDate="2015-06-03T16:20:48.983" Score="1" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;First picture&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand the picture and explanation correctly, you are looking for a word to describe the change from &quot;A&quot; to &quot;A+dA&quot;. In which case the word you need, in a generic sense, is &lt;strong&gt;rotation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want something more specific to your particular situation, you are looking for the &lt;strong&gt;rotation of one plane relative to another&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Laughably, after writing this and re-reading the original question, &lt;strong&gt;relative rotation&lt;/strong&gt; is actually used in the description! Is there any reason why you didn't think was the correct term?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Second picture&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top and bottom beams in the middle part are &lt;strong&gt;bending&lt;/strong&gt;. For the overall structure &lt;em&gt;plane sections have not remained plane&lt;/em&gt;. This would take the generalised description of &lt;strong&gt;distortion&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Summary&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have asked: &lt;strong&gt;What describes the rotation of one part of a body relative to another given an applied torque?&lt;/strong&gt; The simple, generic, answer is &lt;strong&gt;distortion&lt;/strong&gt;. But you are complicating things by looking at a &lt;em&gt;body&lt;/em&gt; rather than a &lt;em&gt;member&lt;/em&gt;. For a &lt;em&gt;member&lt;/em&gt;, the rotation caused by a moment is, quite simply, &lt;strong&gt;bending&lt;/strong&gt;. But you have rejected this because you are looking at a larger structure which has complicated relative stiffnesses etc. This means you lose the nice precise words and have to rely on generic ones instead. Hence why &lt;strong&gt;distortion&lt;/strong&gt; is probably the most accurate, but isn't specific to rotation.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2015-06-04T09:34:34.597" LastActivityDate="2015-06-04T09:34:34.597" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3041" PostTypeId="1" AcceptedAnswerId="3042" CreationDate="2015-06-03T19:08:43.163" Score="8" ViewCount="9167" Body="&lt;p&gt;I'm building a pizza oven and, to do so, I bought a whole load of engineering bricks (not &lt;a href=&quot;http://www.travisperkins.co.uk/Ibstock-Brick-Class-B-Perforated-Engineering-Red/p/123759&quot;&gt;these&lt;/a&gt; but similar). They've got five holes vertically through them, each hole is 2x6x6.5 cm (80 ml) and the walls, both end and dividing, are 2 cm thick by 6.5 cm high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the oven is fired up it will reach ~500 °C. If we use some good solid concrete as masonry for the bricks, which will seal the ends of the holes, I'm worried that as the air heats up, the pressure inside will turn them into a sort of ceramic bomb that will make cooking pizza a rather more exciting experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that likely to happen? How can I calculate the pressure that will be exerted on the bricks by 80 ml of air at 500 °C?&lt;/p&gt;&#xA;" OwnerUserId="1856" LastEditorUserId="33" LastEditDate="2016-02-17T21:23:46.447" LastActivityDate="2021-01-03T21:20:15.233" Title="Would engineering bricks explode if used to build an oven?" Tags="|mechanical-engineering|thermodynamics|pressure|mechanical-failure|masonry|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3042" PostTypeId="2" ParentId="3041" CreationDate="2015-06-03T19:29:12.160" Score="9" Body="&lt;p&gt;The pressure rise is easy enough to calculate from the ideal gas law&#xA;$$&#xA;PV=nRT.&#xA;$$&#xA;In your case $V$, $n$, and $R$ are fixed so you can convert this to a simple ratio&#xA;$$&#xA;P_2=P_1\frac{T_2}{T_1}.&#xA;$$&#xA;Be careful here to use an absolute temperature scale (usually measured in Kelvin).  From the numbers you give, the pressure will rise to about 2.6 atm assuming that the trapped air was initially at 1 atm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition we can calculate the energy released by a gas if it were to expand by blowing out the walls without changing temperature.  The work done in an &lt;a href=&quot;http://en.wikipedia.org/wiki/Isothermal_process#Calculation_of_work&quot;&gt;isothermal process&lt;/a&gt; is given by&#xA;$$&#xA;W_{A\rightarrow B}=-\int_{V_a}^{V_b}p\ dV = -nRT\ln\left(\frac{V_b}{V_a}\right),&#xA;$$&#xA;where the last line assumes that it follows the ideal gas law in the process. If the initial volume is $V_a=80\ \text{mL}$ at 500 °C, then the final volume is given by $V_b=2.6*V_a=208\ \text{mL}$.  Putting in the rest of the constants gives an energy release of $\sim 22\ \text{J}$ which is &lt;strong&gt;less energy than is released by a typical firecracker&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I make no guarantees of your safety, but I certainly don't think that this will cause your structure to explode.  On top of the fact that the pressure isn't particularly high, heating to 500 °C is a relatively slow process and I suspect that the air will find its way out through cracks or holes before it hits the full pressure.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="368" LastEditDate="2015-06-05T19:03:42.570" LastActivityDate="2015-06-05T19:03:42.570" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3043" PostTypeId="1" CreationDate="2015-06-03T20:14:10.013" Score="2" ViewCount="2581" Body="&lt;p&gt;If an I-shaped section is loaded in two directions (in both weak and strong axes), how can I find the maximum load-carrying capacity of them beam? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an interaction equation in AISC 360-10, but it's for beams which are also loaded axially. Can that interaction equation be used for only flexure in two directions, taking the axial force to be 0?&lt;/p&gt;&#xA;" OwnerUserId="1110" LastEditorUserId="368" LastEditDate="2015-06-05T19:26:57.530" LastActivityDate="2015-08-26T18:08:11.323" Title="Biaxial flexure of steel I-beams" Tags="|structural-engineering|steel|aisc|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3044" PostTypeId="2" ParentId="3043" CreationDate="2015-06-03T20:49:43.147" Score="8" Body="&lt;p&gt;The simple answer is: &quot;yes, you can use the given equations with P&lt;sub&gt;r&lt;/sub&gt;=0.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three reasons for my answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The code section (AISC 360, Chapter H) doesn't specifically state that P&lt;sub&gt;r&lt;/sub&gt; cannot be equal to zero.&lt;/li&gt;&#xA;&lt;li&gt;The equations in code section H1 are referenced by other sections that apply to both compression and tension (e.g. H2), so there must be some continuity to the equation when going from tension to compression.&lt;/li&gt;&#xA;&lt;li&gt;The graphs in the commentary on Chapter H show graphs that go from zero axial force with moment to axial force with zero moment.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-06-03T20:49:43.147" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3045" PostTypeId="2" ParentId="3030" CreationDate="2015-06-03T22:19:45.203" Score="1" Body="&lt;p&gt;Simplest way to study machine utilization is the chart the trends. I used excel to generate three sets of data to approximately match data as specified by the question&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;machine A has a utilization of 50% and 14 weeks the utilization &gt; 90% &lt;/li&gt;&#xA;&lt;li&gt;machine B has a utilization of 70% and  7 weeks the utilization &gt; 90%&lt;/li&gt;&#xA;&lt;li&gt;machine C has a utilization of 89% and  6 weeks the utilization &gt; 90%&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Below is trend graph for the set of random data&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IeS3u.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following are statistical information for the above set of random data.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The averages for random A,B and C data set are: 55,67, and 84&lt;/li&gt;&#xA;&lt;li&gt;The median for random A, B, and C data set are: 60, 72, and 88 &lt;/li&gt;&#xA;&lt;li&gt;The standard deviation for random A, B, and C data set are: 28, 32,&#xA;and 17&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Looking at the trend Machine A a drop in utilization as highlighted by Root Cause 1. This type of information can help as drive an investigation. Example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An obvious question would was preventive maintenance (PM) performed&#xA;around week 13&lt;/li&gt;&#xA;&lt;li&gt;Was there a change in operator at week 13 on machine A&lt;/li&gt;&#xA;&lt;li&gt;Does the machine utilization track power outage&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Beside standard deviation, averages and trend analysis, &lt;a href=&quot;https://engineering.stackexchange.com/questions/390/how-can-the-vendor-be-approved-disapproved-based-on-validation-test-data/571#571&quot;&gt;histograms&lt;/a&gt;, &lt;a href=&quot;https://engineering.stackexchange.com/questions/2173/how-to-determine-if-an-engineering-process-is-stable-and-capable/2197#2197&quot;&gt;control charts&lt;/a&gt; and bar graph are two other simple but powerful statistical tools that you might want to consider. &lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-06-03T22:19:45.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3047" PostTypeId="1" AcceptedAnswerId="3082" CreationDate="2015-06-04T02:10:11.593" Score="5" ViewCount="8348" Body="&lt;p&gt;All these processes sound like structural change caused by heating.&lt;/p&gt;&#xA;" OwnerUserId="602" LastEditorUserId="368" LastEditDate="2015-06-05T19:15:55.567" LastActivityDate="2015-06-07T07:10:20.433" Title="What's the difference between annealing, curing and sintering?" Tags="|process-engineering|metallurgy|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3048" PostTypeId="1" CreationDate="2015-06-04T04:46:27.043" Score="2" ViewCount="132" Body="&lt;p&gt;There are several companies producing equipment that produce drinking water from the air. They use dessicants, hygroscopic brine, or deliquescent salts, to adsorb water from atmospheric humidity.  Then they use a heat source to evaporate the adsorbed water to be condensed and funneled to a collection point. Some systems use solar photovoltaics (PV) to run fans and controls, and solar heat for de-adsorption. One system claims to use the process's downward flow to create a lower pressure for de-adsorption, which makes it more efficient at lower temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for information on electronic controls, the core of which is probably a delta-humidity-inside/outside sensors; the best adsorbant material (quantity and configuration), and the material to use for the condensation surface.&lt;/p&gt;&#xA;" OwnerUserId="1860" LastEditorUserId="33" LastEditDate="2015-06-04T12:39:39.453" LastActivityDate="2015-06-04T21:54:49.763" Title="What are the best materials to use for producing drinking water from the Air?" Tags="|electrical-engineering|chemical-engineering|process-engineering|water-resources|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3050" PostTypeId="2" ParentId="2912" CreationDate="2015-06-04T11:20:24.020" Score="5" Body="&lt;p&gt;There are several types of ultrasound images. The most familiar for medical imaging is probably the B-mode type. This type of device uses a linear array of transducers to generate sound waves and to detect the reflected signal which is then processed to form a 2D cross-section or planar image of the tissues being scanned. When a sound wave passes into an area of different density or acoustic impedance part of it is reflected. The greater the difference in impedance, the stronger the return. The time delay and magnitude of the return signal (echo) are processed to determine the brightness of each image pixel. The processing algorithms are too complicated for this venue but in short the magnitude of the return gives an indication of the edges or interfaces between different tissues and structures while the delay time gives the distance or depth within the body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern devices use phased array transducers to aim and focus sound pulses. Pulsed Doppler is used to determine frequency shifts which can be used along with the time delay, and the assumption that sound travels at a constant velocity of 1540 m/s, to calculate relative motion which can be displayed as a moving image using directional Doppler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, each pulse is either aimed in a specific direction or produced by one of an array of transducers and the reflection is from that same direction.&lt;/p&gt;&#xA;" OwnerUserId="1862" LastActivityDate="2015-06-04T11:20:24.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3051" PostTypeId="2" ParentId="3041" CreationDate="2015-06-04T13:12:38.073" Score="10" Body="&lt;p&gt;I'd be less worried about the shards and more about the structural integrity of the oven. When the bricks crack they lose most of their structural integrity. This may lead to the oven collapsing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's much safer to line the inside with firebricks. They are solid and more expensive but you can be sure they won't crack under the heat. They will also protect the wall behind them.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-06-04T13:12:38.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3053" PostTypeId="2" ParentId="3000" CreationDate="2015-06-04T13:29:32.410" Score="3" Body="&lt;p&gt;Nylon insert locking nuts, also called elastic stop nuts or nyloc nuts, are nuts with a plastic insert of an ID smaller than the screw to which it is applied.  Nyloc nuts are useful in applications where vibration or insufficient clamping loads may cause the nut to loosen and back off.  These nuts are extremely common and fairly reliable in tension, though they excel in shear loaded scenarios. They are not useful in higher temperature applications (&gt;250F), as the elastic material can degrade, and these nuts are typically avoided for repeated use due to wear and hardening of the elastic material.  &lt;/p&gt;&#xA;" OwnerUserId="1831" LastActivityDate="2015-06-04T13:29:32.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3054" PostTypeId="2" ParentId="3000" CreationDate="2015-06-04T14:13:33.207" Score="2" Body="&lt;p&gt;The biggest consideration is that the thread must fully engage the nylon ring. If your bolts are too short, the nylon won't do it's job by deforming to the shape of the threads. Depending on how critical your application is, in many industries they are considered single-use as the nylon gets less effective the more times the nut is installed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For very long studs, power driving the nut can generate a lot of heat, melting the nylon insert. Because of this some people insist on only installing nylocs using hand wrenches, although this is only a practical concern in very rare circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the application, nylon insert nuts don't perform well in high temperature environments, or in the presence of chemicals that attack the specific plastic used. It's also worth noting that they are not the most robust of thread-locking options, so in very high vibration applications, or applications where the bolt forms a pivot, they may not be advisable. All-steel locknuts, or pinned nuts may be a a better option in such cases.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-06-04T14:13:33.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3055" PostTypeId="1" CreationDate="2015-06-04T14:17:21.237" Score="2" ViewCount="187" Body="&lt;p&gt;Consider a piece of material of infinite extent (i.e. no boundaries) with a point source of heat at its center which oscillates at frequency $\omega$ about the mean temperature $\langle T\rangle$.  If I place a temperature sensor somewhere else in the material at a distance $r$ from the heat source, what will the transfer function between the source and the sensor be? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From experience I expect it to be a single pole low-pass filter of the form&#xA;$$&#xA;  \frac{T_{sns}}{T_{src}}=\frac{1}{1+\tau s},&#xA;$$&#xA;where $s=i\omega$.  I also expect that the time constant $\tau$ will be a function of the thermal conductivity $\kappa$, the distance between the source and the sensor $r$, and maybe the heat capacity $Q$.  Naive dimensional analysis of these quantities shows that &#xA;$$&#xA;\tau\propto \frac{Q}{\kappa\ r}&#xA;$$&#xA;has the correct units of seconds.  Can anyone give a more rigorous answer?&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-06-04T14:17:21.237" Title="Thermal transfer function of an isotropic material" Tags="|thermodynamics|heat-transfer|transfer-function|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3057" PostTypeId="1" CreationDate="2015-06-04T19:21:02.937" Score="4" ViewCount="104" Body="&lt;p&gt;As you may know Steel plants and Tile Factories use a lot of water. I want to know how do these factories recycle water vapor which is mixed with flue gases to reuse it?&#xA;Do they take advantage of Heat Exchangers or other type of condensers?&#xA;If they do what kind of them?&lt;/p&gt;&#xA;" OwnerUserId="906" LastEditorUserId="33" LastEditDate="2015-11-12T01:59:03.310" LastActivityDate="2015-11-12T01:59:03.310" Title="What procedure do steel plants use to recycle water vapor?" Tags="|process-engineering|recycling|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3058" PostTypeId="1" AcceptedAnswerId="3065" CreationDate="2015-06-04T21:27:52.493" Score="1" ViewCount="2113" Body="&lt;p&gt;There are multiple algorithms out there for generating synthetic EKG signals (for test and validation of heart rate estimation algorithms). All you plug in is the desired heart rate, and the algorithm outputs an EKG signal with the desired heart rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to create a similar algorithm - but for pulse oximetry. I would like to implement a signal generator which generates a synthetic infrared and red (photodiode) signal based on a user-specified/pre-defined SPO2 rate. I then want to take the synthetic red and infrared signal and let them be processed by an SPO2 estimation algorithm. If the algorithm outputs an SPO2 estimate which is close to the pre-defined (i.e. the true SPO2) SPO2 value, then I know that the algorithm works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about creating such a signal generator? &lt;/p&gt;&#xA;" OwnerUserId="1868" LastEditorUserId="368" LastEditDate="2015-06-05T19:53:06.553" LastActivityDate="2015-06-05T19:53:06.553" Title="How to generate a synthetic infrared and red LED signal for SPO2 calculation" Tags="|biomedical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3060" PostTypeId="2" ParentId="3024" CreationDate="2015-06-04T22:15:07.897" Score="3" Body="&lt;p&gt;What you're describing is a subset of the general concept of rigidity (or stiffness). I think the measurement you're looking for is the inverse of rotational stiffness (&lt;a href=&quot;http://en.wikipedia.org/wiki/Stiffness#Rotational_stiffness&quot; rel=&quot;nofollow noreferrer&quot;&gt;as defined by Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A body may also have a rotational stiffness, &lt;em&gt;k&lt;/em&gt;, given by&#xA;  $$k=\frac {M} {\theta}$$&#xA;  where&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;em&gt;M&lt;/em&gt; is the applied &lt;a href=&quot;http://en.wikipedia.org/wiki/Moment_(physics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;moment&lt;/a&gt;&lt;br&gt;&#xA;  &lt;em&gt;θ&lt;/em&gt; is the rotation&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could say something like, &lt;strong&gt;&quot;The rotational stiffness of the assembly is very high, such that the rotation of the free end with an applied bending moment is nearly zero.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to come up with a term for the inverse of rotational stiffness, you might get some mileage out of &quot;skew,&quot; as in, &quot;The end of the assembly opposite and parallel to the wall will not tend to skew with respect to the wall when any force is applied.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Informally, this use of &quot;skew&quot; may just make sense to the reader; formally, it's a bit tricky to justify, and an audience of engineers may second-guess it. You could argue that there are more parallel lines between parallel planes than there are between intersecting planes, because all parallel lines between intersecting planes must be parallel with the line of intersection, therefore intersecting planes are more &quot;skewed&quot; because there are fewer ways to pair lines between them that are &lt;em&gt;not&lt;/em&gt; skew. &lt;a href=&quot;http://en.wikipedia.org/wiki/Skew_lines&quot; rel=&quot;nofollow noreferrer&quot;&gt;You'd be on shaky ground with the mathematicians, though.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's my direct answer to your vocabulary/documentation question. The rest of this will be a bit more free-form, starting with a response to &lt;a href=&quot;https://engineering.stackexchange.com/questions/3024/twisting-or-torsion-of-a-non-bar-vocabulary#comment5274_3040&quot;&gt;your comment on AndyT's answer&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;my assembly doesn't really have clear surfaces (think of two spheres&#xA;  stuck together). So instead of, say, &quot;torsional rigidity&quot;, I'd need&#xA;  something like &quot;given any two virtual surfaces, one close to the&#xA;  binding surface and one close to the free end, when a force is applied&#xA;  to the free end, the two virtual surfaces will experience only a small&#xA;  relative rotation&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You'll need to at least define these virtual surfaces initially. One of them should be easy, since it's a plane defined by a wall. If the wall is curved, you might use the plane having a normal vector that averages the axes of the fasteners used to mount the assembly, which passes through some &quot;centroid of mounting&quot; (call this the &quot;virtual wall&quot; if you like). A more complex geometry will demand a more complex explanation; efforts to reduce the complexity of that explanation are unreasonable if and when they impair the effectiveness of the explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beams and shafts aside, the difference between rotational stiffness and &lt;a href=&quot;http://en.wikipedia.org/wiki/Torsion_(mechanics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;torsional rigidity&lt;/a&gt; has to do with the nature of the forces applied. Rotational rigidity corresponds to  bending moments, while torsional rigidity corresponds to twisting moments; in a torsional context, your virtual plane(s) would be &lt;a href=&quot;http://en.wikipedia.org/wiki/Plane_of_rotation#Plane_of_rotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;planes of rotation&lt;/a&gt;. The same would not be true in a bending context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an analogy, consider the motion of a &lt;a href=&quot;http://en.wikipedia.org/wiki/Rubik%27s_Cube&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rubik's Cube&lt;/a&gt;. When you twist one of the layers of the cube, the angle through which the layer moves is within the plane of the cube. Now, imagine twisting a softball-sized lump of rubber in the same way that you would twist a Rubik's Cube. Because the rubber is solid, the half in one hand is constrained relative to the half in the other hand, and the moment you apply deforms it. How much it deforms depends on its &lt;em&gt;torsional stiffness&lt;/em&gt;; the internal stresses are shear stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, consider the motion of an accordion. The two ends of the accordion are initially parallel but, when you hold the accordion in your two hands, you can manipulate it so that the ends are not parallel (because of its flexible diaphragm). Now recall that lump of rubber. Holding it in the same grip as before, you can apply a bending moment to deform the rubber in a different way, corresponding to bending rather than twisting. How much it deforms now depends on its &lt;em&gt;rotational stiffness&lt;/em&gt;; the internal stresses are normal stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a slender beam or shaft, torsion is only practically relevant with respect to a single axis, and bending is only practically relevant with respect to the other two (i.e., the major and minor axes of a beam). If we have a lump of rubber, bending and torsion could occur &lt;em&gt;and be significant&lt;/em&gt; in any direction. We don't have the advantage of simple equations developed for classical beams and shafts (or even &lt;a href=&quot;http://en.wikipedia.org/wiki/Timoshenko_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;less-simple methods&lt;/a&gt; used for, e.g., short beams), but &quot;torsion&quot; and &quot;bending&quot; are still meaningful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define virtual planes, the angle between the two planes quantifies the relative rotation you're after. &lt;a href=&quot;https://engineering.stackexchange.com/questions/3024/twisting-or-torsion-of-a-non-bar-vocabulary#comment5322_3024&quot;&gt;Ethan48's suggestion of using the term &quot;dihedral angle&quot;&lt;/a&gt; is technically correct, but probably more abstract/general than you need—identify the planes and say the angle between them is $\theta$. If you know that angle to be much smaller than $M$, don't even go through the trouble of naming the inverse of rotational stiffness; just claim that $\theta \approx 0°$, justify it with &quot;high rotational stiffness&quot; and be done with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, if you go through the trouble to formally define virtual planes, that points fixed to the plane when the assembly is unloaded must remain collinear so that the virtual plane doesn't warp into a curved surface under loading. (Do you want to measure the angle between curved surfaces? I don't!)&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-06-04T22:15:07.897" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3061" PostTypeId="1" AcceptedAnswerId="3062" CreationDate="2015-06-05T03:34:50.300" Score="9" ViewCount="5875" Body="&lt;p&gt;The obvious answer is a compass. I have an old (1952) transit that has a compass which seems to be accurate to 15'. However, that doesn't seem too incredibly accurate at all. Furthermore, most modern digital theodolites I see don't appear to have a compass (correct me if I'm wrong). I'm curious how a surveyor actually finds north with high accuracy. Are there digital theodolites or other instruments with compasses? Are north-south lines set out with GPS? I'd appreciate any insight.&lt;/p&gt;&#xA;" OwnerUserId="1870" LastEditorUserId="282" LastEditDate="2015-06-12T21:12:16.073" LastActivityDate="2022-01-08T03:08:05.827" Title="What instrument does a surveyor use to accurately find North?" Tags="|measurements|surveying|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3062" PostTypeId="2" ParentId="3061" CreationDate="2015-06-05T04:22:49.553" Score="9" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gyrotheodolite&quot;&gt;Gyrotheodolites&lt;/a&gt;  are an older, but accurate way to find &lt;a href=&quot;https://en.wikipedia.org/wiki/True_north&quot;&gt;true north&lt;/a&gt;. They can be used on the Earth's surface, but these days they tend to be used for surveying underground, in tunneling and mining applications. GPS surveying methods are replacing gyrotheodolite methods for this application on the Earth's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gyrotheodolites are &quot;surveying instruments composed of a gyroscope attached to a theodolite&quot;. There is more information about gyrotheodolites &lt;a href=&quot;http://www.digplanet.com/wiki/Gyrotheodolite&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When used underground, they are the best way to establish the initial surveying reference points underground. Depending of the extend of the underground infrastructure, occasionally, gyrotheodolites are used for check surveys of the underground reference points. After the initial reference points have been established conventional theodolites or total stations are used underground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For underground infrastructure that can only be accessed by vertical shafts they are the best way to establish surveying reference points underground. The other way is to drop two plumb bob lines down the shaft and get the lateral co-ordinates of the plumb bob lines on the surface from known surface survey stations. Leaving the plumb bob lines untouched surveyors go underground by another shaft and survey the two plumb bob lines from underground and then survey what will be the reference point for that level underground. After making corrections for the sway of the plumb bob lines the lateral co-ordinates for the underground reference point can be calculated with good accuracy. Once that is done knowing the direction of north, underground, becomes possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tunnels that connect with the Earth's surface via an &lt;a href=&quot;https://en.wikipedia.org/wiki/Adit&quot;&gt;adit&lt;/a&gt;, or similar portal, and are horizontal or sub-horizontal can be surveyed using conventional theodolites or total stations, with the first reference points taken directly from the surface via line of sight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One limitation of gyrotheodolites is they can't be used at either the North or South Pole &quot;where the Earth's axis is precisely perpendicular to the horizontal axis of the spinner and the meridian is undefined&quot;. Because of this gyrotheodolites tend not to be used within 15 degrees of either pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For surveying on the Earth's surface GPS surveying equipment can be used such the &lt;a href=&quot;http://www.trimble.com/Survey/GNSS-Surveying-Systems.aspx&quot;&gt;GNSS system&lt;/a&gt; provided by Trimble (not advertising, just using it as an example).&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-06-05T08:45:12.983" LastActivityDate="2015-06-05T08:45:12.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3065" PostTypeId="2" ParentId="3058" CreationDate="2015-06-05T12:17:17.180" Score="5" Body="&lt;p&gt;Short:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1.) Mindless and without understanding method that would work well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is as quick and easy as is likely to be possible, should work 'well enough' and would be 'easy enough' to implement without any in depth understanding past ensuring that simulated sensor output characteristics (impedance etc) match those of the target sensor.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Obtain known good SPO2 meter and sensor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Monitor data signals from LEDs and record&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Connect sensor to &quot;patients&quot; or test samples of choice and adjust SPO2 level.   (Patients can easily 'adjust' SPO2 level over a wide range by controlled breathing. Ask me how I know :-) ).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At a range of SPO2 levels log waveforms and store in simulator&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;'Play back' signals as required.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(2.) The path to understanding - much harder.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;strong&gt;&quot;Real World sensor Outputs&quot;&lt;/strong&gt;  below for a set of raw and processed sensor output signals for IR and visible red LEDs in an SPO2 measurement environment. Using the related evaluation  kit (or an equivalent circuitry of your choice) would allow examination of SPO2 sensor waveforms at various SPO2 levels. From these it would be &quot;simple enough&quot; to provide analog levels corresponding to equivalent sensor outputs either at selected SPO2 levels or, potentially, continuously variable ones. Available on ebay. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted below, the transferability of sensor signals from one system to others from other manufacturers is far from certain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Below I have provided detailed information from a single supplier. I have no involvement with the company or their products. They seem to &quot;know their stuff&quot; and examples show that TI have used their equipment in their development work. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplistically, SPO2 is measured by comparing the attenuation of two wavelengths of red light by a portion of the body which has blood flowing through it. The wavelengths are chosen such that one is attenuated in a manner which is a function of Oxygen content when it passes oxygenated haemoglobin and the attenuation of the other is independent of haemoglobin oxygenation percentage. Oxygenated haemoglobin attenuation is a function of blood vessel optical path length and haemoglobins saturation versus Oxygen partial pressure &quot;transfer function&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Haemoglobin saturation versus oxygen partial pressure.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uFrQS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxygen-unnaffected signal provides a reference attenuation against which the affected signal can be compared to deduce oxygen saturation level, after taking the factors mentioned above into account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a much more complete and competent explanation see &lt;a href=&quot;http://www.apmkr.com/bio-device/transmittance_oximeter.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not too surprisingly, it's not that easy. SPO2 assessment is a black art - far less straight forward than one might expect when considering the method used. I base this claim on an interesting and detailed Hewlett Packard report which I read many years ago, which said as much, and then set out to demonstrate how empirical the art is in practice. That this is still the case to some extent is amply demonstrated by the myriad internet  pages and ads comparing brand x with the seller's brand Y and showing how unreliable the other brand is under various conditions. For example: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Brand A &amp;amp; Band B versus two of &quot;&lt;em&gt;ours&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/P1NK7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And again ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FG1gA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - if I was required to build a SPO2 sensor simulator, in the absence of known good sensor signal data, I would take as many reputable SPO2 meters as I could find and observe the signals produced by their sensors under various conditions and displayed SPO2 values. An issue is that sensors and SPO2 meters may be to some extent matched. A sensor pair which works with brand A SPO2 meter may implement assumptions which make it unsuited for use with a brand B meter.  Determining to what extent this is true is one of the first steps in evaluating algorithm performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, at least one company seems keen to help you do this ...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Real world sensor outputs:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start - super cheap complete pulse-oximeters on ebay for about $US20 all up. Unbelievable - and performance totally unknown&#xA;&lt;a href=&quot;http://www.ebay.com/itm/CMS50DL-SPO2-Monitor-OXYMETRE-SATUROMETRE-POULS-METRE-ECG-OXYMETER-OM1-/120971428378?pt=LH_DefaultDomain_71&amp;amp;hash=item1c2a75821a&quot; rel=&quot;nofollow noreferrer&quot;&gt;CMS50DL SPO2 Monitor OXYMETRE SATUROMETRE POULS METRE ECG OXYMETER OM1&lt;/a&gt; - the &quot;OM1&quot; in the title is probably a hat-tip / pretend similarity to an APM model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buy one. Test. IF it works (IF) buy several.&lt;br&gt;&#xA;Pull apart, bring out LED signals, use, play ... . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and - much more &quot;real&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;APMKOREA &lt;a href=&quot;http://www.apmkr.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Home Page&lt;/a&gt; - easier to look at than describe - sensors evaluation boards, application notes, waveform photos, ... . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;APM provide both PPG (essentially pulse waveform) and SPO2 measurement equipment. By the time I realised they were overlapping I had somewhat mixed PPG and SPO2 references as they overlap. Sorting out which is which I leave as &quot;an exercise for the student&quot; as anyone interested enough to follow this up will probably find it all of interest and relevance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Likely of most relevance -&lt;/strong&gt; APM's ICOM1 (RS232) and ICOM2 (USB) &lt;a href=&quot;http://www.apmkr.com/bio-device/SpO2_module_detail.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pulse oximetry modules&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.apmkr.com/bio-device/reflectance_oximeter4.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;SPO2 and PlethysmoGraphy discussion&lt;/a&gt; looks useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.apmkr.com/bio-device/SpO2_module_detail.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;SPO2 modules&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;APMKOREA &lt;a href=&quot;http://www.apmkr.com/spo2_oximeter_blood_sensor.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;&quot; Leading PhotoPlethysmoGraphy Sensor Technology&quot;&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Offer a wide range of emitters, detectors, reflective and transmissive (through tissue) sensors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus an &lt;a href=&quot;http://www.apmkr.com/bio-device/PPG%20with%20evaluation%20kit.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;sensor evaluation and development board&lt;/a&gt; which will work with onboard reflective sensors or external transmissive or reflective ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qPvwa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And finally&lt;/strong&gt;, the heart of what looks helpful is these example signals - presumably available from their evaluation systems for SPO2 levels of your choice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z1GuJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z1GuJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Larger version here - MUCH easier to read&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.apmkr.com/bio-device/pd_oximeter.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Appears to be very useful application guide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Nonin &lt;a href=&quot;http://www.nonin.com/Accurate-Pulse-Oximeter&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Acurate Pulse Oximeter&quot;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ecostore &lt;a href=&quot;http://www.echostore.com/wireless-oximeter-cms50e.html&quot; rel=&quot;nofollow noreferrer&quot;&gt; Wireless pulse oximeter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.anesthesiaweb.org/lung-function.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lung function and anaesthesia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-06-05T12:17:17.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3066" PostTypeId="1" AcceptedAnswerId="3079" CreationDate="2015-06-05T15:35:15.687" Score="2" ViewCount="712" Body="&lt;p&gt;I'm building a model of a long vehicle with 6 axles and I want to be able to steer it. How do I calculate the amount each axle should steer to minimise dragging of some wheels round sideways as it turns?&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-06-08T16:37:48.887" Title="How to calculate the steering ratios on multiple axle vehicles?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3067" PostTypeId="2" ParentId="3066" CreationDate="2015-06-05T15:40:42.180" Score="1" Body="&lt;p&gt;All axles of the wheels should end up crossing in the same point. That will be the center of your turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have that point you simply draw the line from the center of the wheel to that point and get the angle compared to the Right vector &lt;code&gt;acos(dot(right, desiredAxle))&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This provides the angles for each wheel separately. If there is a complex steering mechanism then you can use those values to approximate the best steering angle to set.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-06-05T15:40:42.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3068" PostTypeId="2" ParentId="3047" CreationDate="2015-06-05T21:23:20.133" Score="3" Body="&lt;p&gt;Knowing the context here might be useful. These terms have nothing to do with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Annealing&lt;/em&gt; is warming a metal, usually steel, to give it a soft, easy shaped type of crystalline structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Curing&lt;/em&gt; is the process of letting a metal sit to work out internal stresses that may have formed during a heat treatment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Sintering&lt;/em&gt; is reducing a metal to a powder and then pressing it together under heat and pressure to form a solid mass. Certain metals, like tungsten, have such a high melting point that it is easier to sinter them into shape rather than cast them. Note that sintering actually technically refers to the process of creating the metal particles, not the pressing.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2015-06-05T23:32:40.853" LastActivityDate="2015-06-05T23:32:40.853" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3069" PostTypeId="2" ParentId="2993" CreationDate="2015-06-05T21:34:06.343" Score="2" Body="&lt;p&gt;It is called a &lt;em&gt;trunnion adjustment&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2015-06-05T21:34:06.343" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3071" PostTypeId="1" CreationDate="2015-06-06T02:36:04.900" Score="1" ViewCount="713" Body="&lt;p&gt;I am recalling from a turbo machinery course that a series of rotors and stators in a turbo machine operate like this. The air enters the machine and the rotors speed up the air so it slams into the stators and increases pressure (because it loses velocity so there is a corresponding increase in pressure according to Bernoulli equation)  and repeats this several times before leaving the turbo machine (please correct me if the general idea is incorrect).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With fans in HVAC systems, there is a corresponding pressure drop across a damper that modulates flow. If there is a constant speed can that has a modulating damper on the discharge side to control flow, what do the pressure changes look like on the fan side of the damper and the discharge side of the damper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe the two scenarios are not similar but I think I am confused as to why air loses pressure across a damper.&lt;/p&gt;&#xA;" OwnerUserId="1143" LastEditorUserId="282" LastEditDate="2015-06-19T20:32:30.167" LastActivityDate="2015-09-18T04:54:10.090" Title="Pressure drop across a damper vs rotor stator pressure increase" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3073" PostTypeId="1" AcceptedAnswerId="3166" CreationDate="2015-06-06T06:37:45.013" Score="4" ViewCount="112" Body="&lt;p&gt;I've heard that the part of a tire which is towards the inner side of a vehicle has material which is more wear resistant than the part of a tire on the outer side of the vehicle.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, I have difficulty with identifying the inner side of a tire. Is there a difference in the material properties of tires from the inner side to the outside?&lt;/p&gt;&#xA;" OwnerUserId="908" LastEditorUserId="90" LastEditDate="2015-06-06T06:50:43.557" LastActivityDate="2015-06-13T01:26:08.567" Title="Are tires symmetrical &amp; are their material properties uniform?" Tags="|mechanical-engineering|materials|automotive-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3075" PostTypeId="1" CreationDate="2015-06-06T14:27:58.580" Score="10" ViewCount="4679" Body="&lt;p&gt;If roads were made of removable parts, we'd probably have fiber cable internet everywhere by now I'd say.&lt;br&gt;&#xA;It'd be so much easier to install e.g. new electricity and water infrastructure and to maintain them if roads where made like that. We'd be able to redefine the whole architecture of a neighborhood if necessary, without the need to take existing roads into account. We could even make an urban area a landscape again with ease if e.g. a city wanted more green areas somewhere. We could make machines that piled road parts on top of each other making maintenance as easy as pie.&#xA;So why do we still pave roads? There's probably some aspect of civil engineering I'm not taking into account..&lt;/p&gt;&#xA;" OwnerUserId="1885" LastEditorUserId="272" LastEditDate="2015-06-08T23:52:49.970" LastActivityDate="2017-07-07T16:42:52.050" Title="Why do we pave roads instead of making them from removable parts?" Tags="|civil-engineering|engineering-history|highway-engineering|roadway|infrastructure|" AnswerCount="7" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="3076" PostTypeId="2" ParentId="3075" CreationDate="2015-06-06T19:12:04.163" Score="8" Body="&lt;p&gt;You mean going back to &lt;a href=&quot;http://en.wikipedia.org/wiki/Cobblestone&quot;&gt;cobblestone roads&lt;/a&gt;? Many sidewalks are often paved with &lt;a href=&quot;http://en.wikipedia.org/wiki/Road_surface#Other_surfaces&quot;&gt;pavers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the challenges is maintaining a level surface. Asphalt holds itself together like a sheet and a dip in the surface below will pull on the sides delaying the onset of potholes. while separate parts will start moving under the vibration and frost heaving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the surface is uneven cars going over them will be loud. In &lt;a href=&quot;https://www.google.be/maps/@50.856833,4.691301,3a,75y,137.12h,62.24t/data=!3m4!1e1!3m2!1ses80kFmDor8V5iZwN7ucAQ!2e0&quot;&gt;this street&lt;/a&gt; there is a busline going through and each time the bus runs through those living there know it by the noise and vibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laying pavers is a very labour-intensive work, each stone must be layed and put into place individually. Before you can even start doing that you need to flatten and compact the surface below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However lets go in between small pavers and a single sheet of asphalt: concrete slabs. &lt;a href=&quot;https://www.google.be/maps/@50.893911,4.593633,3a,75y,254.93h,52.44t/data=!3m4!1e1!3m2!1sVTNSCb80XP2Kq7zwq09ctg!2e0&quot;&gt;This road&lt;/a&gt; is a good example. However replacing those isn't as easy as you think it is. You need to close down the entire lane break up the old slab using a jack hammer clean up the debris and crane in the new slab. Then they need to fill the joint.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-06-06T19:12:04.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3077" PostTypeId="1" AcceptedAnswerId="3204" CreationDate="2015-06-06T23:35:28.910" Score="2" ViewCount="761" Body="&lt;p&gt;My apologies for a likely very simple question. But what is this nut which I have on a micrometer head called, and what is the tool for removing it called?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PjMOA.jpg&quot; alt=&quot;micrometerhead&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1888" LastEditorUserId="272" LastEditDate="2015-06-08T12:05:02.263" LastActivityDate="2015-06-16T10:40:50.723" Title="What is this nut on a micrometer head and what is the tool for removing it called?" Tags="|mechanical-engineering|nomenclature|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3078" PostTypeId="1" AcceptedAnswerId="3084" CreationDate="2015-06-07T01:35:19.467" Score="5" ViewCount="2975" Body="&lt;p&gt;I have some university work to do based on a real world application with five or &lt;a href=&quot;http://en.wikipedia.org/wiki/Six-bar_linkage&quot; rel=&quot;nofollow&quot;&gt;six bar mechanism&lt;/a&gt;, but I am unable to find such a device. Are there real applications of such mechanisms? I've considered some, such as the trigger of a gun and a locking plier, but they are all &lt;a href=&quot;http://en.wikipedia.org/wiki/Four-bar_linkage&quot; rel=&quot;nofollow&quot;&gt;four bar mechanisms&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1853" LastEditorUserId="272" LastEditDate="2015-06-08T12:02:54.363" LastActivityDate="2016-02-08T18:47:35.097" Title="Applications of Five-Bar Mechanisms" Tags="|mechanical-engineering|kinematics|linkage|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3079" PostTypeId="2" ParentId="3066" CreationDate="2015-06-07T02:56:21.957" Score="7" Body="&lt;p&gt;What you are talking about is called &lt;a href=&quot;http://en.wikipedia.org/wiki/Ackermann_steering_geometry&quot;&gt;Ackerman steering&lt;/a&gt; and is pretty easy to get approximately right, but requires non-linear proportional steering control to get fully right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put an imaginary focus/center point of steering somewhere to the side of the vehicle, in the plane of the axis of rotation for each wheel, then the axis through the center of each wheel should all meet at that point. The math is simple trigonometry, but building a mechanical steering system that gets this right for many wheels for all degrees of steering is hard. (For a rover robot I built, I ended up driving each wheel using a servo, because math for servo controllers is easy :-)&lt;/p&gt;&#xA;" OwnerUserId="1889" LastEditorUserId="95" LastEditDate="2015-06-08T16:37:48.887" LastActivityDate="2015-06-08T16:37:48.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3080" PostTypeId="1" AcceptedAnswerId="3093" CreationDate="2015-06-07T03:00:57.307" Score="10" ViewCount="8556" Body="&lt;p&gt;I am affixing a servo motor with a machined enclosure to a base plate with drilled holes. The current design uses eight M-2.5 bolts in a figure U pattern (3-2-3) around the enclosure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tightening that many bolts/screws takes time, and the small gauge makes them finickier than larger-gauge bolts (missed nut thread starts, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to change this design to use fewer, bigger, bolts, what would be the pros/cons of this? Is the amount of position slop significantly different between, say, three M-6 bolts, and eight M-2.5 bolts? Is there a formula for the affixing force that N bolts of size S will provide?&lt;/p&gt;&#xA;" OwnerUserId="1889" LastActivityDate="2018-11-09T14:01:00.403" Title="Many small bolts, or few large bolts?" Tags="|bolting|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3081" PostTypeId="1" CreationDate="2015-06-07T05:17:58.267" Score="1" ViewCount="43" Body="&lt;p&gt;If a bangle is made out of gold and silver, connected with each-other would there be a corrosion happening? If so can it be explained using the galvanic series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also do those metals undergo oxidation under normal conditions?&lt;/p&gt;&#xA;" OwnerUserId="1890" LastEditorUserId="33" LastEditDate="2015-06-07T11:50:16.193" LastActivityDate="2015-06-07T11:50:16.193" Title="Corrosion of a galvanic couple made with silver and gold" Tags="|materials|chemical-engineering|corrosion|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3082" PostTypeId="2" ParentId="3047" CreationDate="2015-06-07T07:10:20.433" Score="4" Body="&lt;p&gt;&lt;strong&gt;Annealing&lt;/strong&gt; is a heat-treatment process for stress-relief of a material. Annealing facilitates reduction of internal stresses and total elastic energy stored in inter-atomic bonds within a treated material. The term is used for appropriate heat-treatments of metals, ceramic glasses, and high-performance polymers. An example would be a cold-rolled steel billet annealed so that it can be worked further into final products. The process of annealing is used whenever internal stress are unacceptable, such as prior to a machining step. Machining changes the distribution of internal stress by removal of material, which is rebalanced by the machined part deforming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Curing&lt;/strong&gt; is a heat-treatment process for accelerating a chemical reaction, and is generally used in the context of polymeric and polymeric composite materials. Thermoset polymers which do not set in a reasonable time at room temperature are cured at higher temperatures in a curing oven. An example would be a pre-impregnated or prepreg carbon-fiber mat draped over a mold and then cured so that it retains the shape of the mold. The monomers react much more rapidly at the curing temperature, and form a thermoset polymer, hardening the material prepreg mat. Personally, I have never heard curing applied to any other class of material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sintering&lt;/strong&gt; is a heat-treatment process for causing a powdered material to become a monolithic bulk material by diffusion between individual powder particles. Sintering may occur in either or both of the solid and liquid states. Consider a slurry mixture of alumina powder and wax binder pressed into shape. If the binder is melted off, the alumina powder that remains is only mechanically bonded. That green form may then be sintered to bind the alumina particles together into a monolithic bulk in the same shape as the green form. The same process may conceivably be applied to any material that can undergo diffusion at reasonable rates and temperatures, though it is most commonly used in the context of ceramic and metallic materials, especially high-melting-point refractory metals. Personally, I have never heard it applied to polymers, as their melting points (if such exists) are so low they are simply processed in the liquid state. The process of sintering is used where (1) the material's melting temperature is unreasonably high or (2) parts of acceptable quality may be produced more rapidly by sintering than by casting or machining, such as in low-stress, complex-shaped steel gears.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2015-06-07T07:10:20.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3083" PostTypeId="2" ParentId="3080" CreationDate="2015-06-07T09:31:36.827" Score="-3" Body="&lt;p&gt;Its better always to use more small bolts than few bigger ones because if some of them fail, its better to have more of them.&lt;/p&gt;&#xA;" OwnerUserId="1892" LastActivityDate="2015-06-07T09:31:36.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3084" PostTypeId="2" ParentId="3078" CreationDate="2015-06-07T10:21:59.553" Score="6" Body="&lt;p&gt;Six bar linkages are pretty easy to find. Six bar is the go to guy when a four bar does not offer enough movement flexibility. Bigger European style hidden hinges are nearly all six bar mechanisms, essentially they are two four bar mechanisms that share part of the structure. These are common in many uses but the sixbar hinge you can probably find at home with no special effort usually in kitchen/other cabinets. Sixbar linkages are nearly as common as fourbar linkages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Five bar linkages are a bit problematic. They have a freedom of 2. And they can not be connected in multi loop configurations as the only config possible is one loop. Since two degrees of freedom is A bit bad design they tend to be avoided. Sometimes you see these with a cam/Gear connection serving as link 5. I have seen powershears that have this property, they are easily mistaken as fourbars as the fifth bar is of limited movement. Sometimes also you see these in robotics setups where corner connected fourbar would not fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm not at home so I don't have my catalogue with me. So no pictures and details.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="2918" LastEditDate="2016-02-08T18:47:35.097" LastActivityDate="2016-02-08T18:47:35.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3085" PostTypeId="1" AcceptedAnswerId="3086" CreationDate="2015-06-07T11:08:09.377" Score="5" ViewCount="617" Body="&lt;p&gt;I want to create a solar powered kettle but first, I need to know how much power will be required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means, I think, I'll need to know how much water I want to bring up to a certain temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand how to work this out, and I'm frankly stumped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to work this out, if I knew I wanted to bring 0.5litres to 60 degress C, how much power would I need to achieve this within 30 minutes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My end goal here is to find a suitable solar panel powerful enough.&lt;/p&gt;&#xA;" OwnerUserId="1894" LastEditorUserId="41" LastEditDate="2015-06-08T20:27:07.393" LastActivityDate="2015-06-08T20:27:07.393" Title="Solar powered kettle - how much power for 1/2 litre" Tags="|mechanical-engineering|electrical-engineering|thermodynamics|solar-panel|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3086" PostTypeId="2" ParentId="3085" CreationDate="2015-06-07T12:15:36.910" Score="8" Body="&lt;p&gt;The amount of water and how much you want to heat it only tells you the &lt;i&gt;energy&lt;/i&gt; required, not the power.  Power is energy per time, so it matters how fast you want to get there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you want to heat 500 ml of water by 60&amp;deg;C in 30 minutes.  Do the math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific heat of water is 4.18 J/g&amp;deg;C.  The total energy required is therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(500 g)(60&amp;deg;C)(4.18 J/g&amp;deg;C) = 125 kJ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This energy spread out over 30 minutes (1800 seconds) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(125 kJ)/(1800 s) = 70 W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;However&lt;/b&gt;, that is only the power that has to go into the water.  Your 500 ml of water sitting in some container at near boiling temerature is most likely loosing more than that to the environment.  Put another way, your heater has to not only provide enough power to raise the temperature of the water, but also to overcome losses of this heat to the enviroment.  At only 70 W into the water, the heat loss will be significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a lot of care spent on insulation, maybe 150 W is enough.  It's really hard to guess since we have no idea what your mechanical system looks like.  A few 100 W would be better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, consider that 30 minutes is a long time to wait for a cup of boiling water.  If the system can just do that in full sun, then it won't work in anything less than ideal circumstances.  500 ml of water is also not a lot of stuff to heat.  I think you are grossly underspecifying your heater.  You may get 500 ml to boil in 30 minutes, but I expect you won't be happy with the performance of the heater overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure you get about 1 kW per square meter of full sunlight, so a few 100 W doesn't really take a large collector.  However, you have to pay attention to the aligment, and focus the rays on a small spot roughly the size of the bottom of the kettle.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2015-06-08T16:49:38.327" LastActivityDate="2015-06-08T16:49:38.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3087" PostTypeId="1" AcceptedAnswerId="3088" CreationDate="2015-06-07T14:56:01.723" Score="8" ViewCount="1667" Body="&lt;p&gt;From what I understand, a control systems engineering job is almost always a senior level position. What kind of entry level jobs do most engineers have before they are qualified to actually design control systems? Is a PE license necessary for this kind of work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently working on a masters degree in electrical and computer engineering. I am mostly interested in control theory and that is the subject of my thesis.&lt;/p&gt;&#xA;" OwnerUserId="1895" LastEditorUserId="368" LastEditDate="2015-06-07T16:00:39.733" LastActivityDate="2020-02-26T09:25:13.537" Title="How does one become a control systems engineer?" Tags="|control-engineering|control-theory|employment|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3088" PostTypeId="2" ParentId="3087" CreationDate="2015-06-07T16:03:05.307" Score="10" Body="&lt;p&gt;To a large extend the answer depends on the industry. For example if you work for a large aerospace company it might take a while before you can get chance to get involved in designing a control systems. In this case you will be most likely a team member design a control system. On the contrary if you happen to work for smaller company you will mostly get to design smaller control system early in your career.  Your choice is, do you want be a large fish in a small pond or small fish in a big pond. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For either choice you might want to get practical experience in area such as &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Measurement systems including sensor characteristics and technologies&#xA;including pressure, motion, flow, ultrasonic sensors etc&lt;/li&gt;&#xA;&lt;li&gt;Signals, Transmission and networking including EMI/EMC, communication&#xA;protocols, cross talk, routers&lt;/li&gt;&#xA;&lt;li&gt;Control elements such as motor controls, pressure relieving devices,&#xA;valves etc &lt;/li&gt;&#xA;&lt;li&gt;Code Standards and Regulation such as ANSI, NEMA, and OSHA&lt;/li&gt;&#xA;&lt;li&gt;Implementation methodologies including management of change (Scope,&#xA;ECN/ECO, Cost, Time)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You also might want to look at &lt;strong&gt;&lt;a href=&quot;https://www.isa.org/templates/three-column.aspx?pageid=53025&quot; rel=&quot;nofollow noreferrer&quot;&gt;Control Systems Engineer (CSE) Licensure&lt;/a&gt;&lt;/strong&gt; exam as well as some of the requirements for current control engineering entry level jobs. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Below are some engineering SE post that discuss some aspects of the above&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.isa.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Society of Automation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/2071/is-interference-between-aircraft-an-issue-for-fly-by-wireless-technology&quot;&gt;Is interference between aircraft an issue for fly-by-wireless technology?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/2264/how-can-i-arrange-my-eco-system-to-enforce-the-principle-of-least-privilege&quot;&gt;How can I arrange my ECO system to enforce the principle of least privilege?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/2297/sensors-processing-algorithms-to-emulate-a-humans-sense-of-smell&quot;&gt;Sensors / processing algorithms to emulate a human's sense of smell&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-06-08T11:38:26.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3089" PostTypeId="1" CreationDate="2015-06-07T19:25:28.533" Score="4" ViewCount="169" Body="&lt;p&gt;In particular, say I have a dynamic model of the inverted pendulum / cart as described &lt;a href=&quot;http://ctms.engin.umich.edu/CTMS/index.php?example=InvertedPendulum&amp;amp;section=ControlStateSpace&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. Instead of using the error difference in force to input into my controller, I want to take into consideration a motor driver (voltage input) to a DC motor for the force.  Would the best method then to analytically describe the force relationship to the voltage input?  If so, then would I need to include a voltage state variable so that I can build a controller based on that signal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My motivation for asking is that I feel I have a fundamental disconnect with controls engineering.  In my studies I've gone over classical and modern theory without the benefit of actually applying theory to application.  Most of what I participate in deal with electromechanical systems, but most books, articles, etc I come across don't really delve into the physical interpretation of the plant input.  I would appreciate any feedback on the matter.&lt;/p&gt;&#xA;" OwnerUserId="1898" LastEditorUserId="282" LastEditDate="2015-06-26T20:50:15.670" LastActivityDate="2016-05-03T13:58:40.667" Title="What is the best method to model the input of an electro-mechanical system for control design and simulations?" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3092" PostTypeId="2" ParentId="3089" CreationDate="2015-06-08T13:27:12.640" Score="4" Body="&lt;p&gt;There are various levels of plant models you can develop, depending on what you want to do. The model you mention is probably suitable for most of the cases in terms of designing a controller. However, it seems that in your case the level of abstraction is too high, so my suggestion would to break the plant model further into sensors + plant + actuators, which would allow you to model the details of the actual motor actuation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various tools to do this, starting from first principles (pen &amp;amp; paper) and implementing in something like &lt;a href=&quot;http://uk.mathworks.com/products/matlab/&quot; rel=&quot;nofollow&quot;&gt;MATLAB&lt;/a&gt;/&lt;a href=&quot;http://uk.mathworks.com/products/simulink/&quot; rel=&quot;nofollow&quot;&gt;Simulink&lt;/a&gt;. If you want something slightly higher-level, where you can just drag and drop standard models, you can use something like &lt;a href=&quot;http://uk.mathworks.com/products/simscape/&quot; rel=&quot;nofollow&quot;&gt;Simscape&lt;/a&gt;, &lt;a href=&quot;http://uk.mathworks.com/products/simelectronics/&quot; rel=&quot;nofollow&quot;&gt;SimElectronics&lt;/a&gt; and &lt;a href=&quot;http://uk.mathworks.com/products/simmechanics/&quot; rel=&quot;nofollow&quot;&gt;SimMechanics&lt;/a&gt;, which are all add-on to Simulink.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-06-08T13:27:12.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3093" PostTypeId="2" ParentId="3080" CreationDate="2015-06-08T15:30:33.833" Score="10" Body="&lt;p&gt;It's really not as simple as a rule of thumb - there are a lot of factors in each application. I'm going to assume that your bolt application is a fairly traditional situation where you are bolting one piece of material to another (one shear plane) not a more complex sandwich (Isolation pads, transition plates, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most bolted connections, the bolts are intended to provide a clamping force normal to the faying surfaces to allow a large friction force to develop between the two materials being bolted. As such, while we pretty much always check that the bolts can hold the load in shear, for design of the connection for performance, the clamping action is a greater consideration. If your faying surfaces are very flat and clean, and your two materials are very stiff, you can imagine that a single, large bolt would suffice for any problem as the clamping force would apply equal friction across the entire faying surface. One problem with using a single bolt is that if the joint does slip, it could slip in a direction that loosens the nut against the bolt, leading to a catastrophic failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, usually our two surfaces are somewhat flexible, dirty, and not flat. Because of this, a bolt only successfully applies a clamping force for a small area around itself, so joints that resist a moment (like most motor mounts) will not be very effective with a single bolt. Instead, adding more bolts, farther apart from each other, creates 'moment couples' where because of the distance between each bolt, the actual slip resistance required at each bolt is less. In general, for connections resisting a moment, you want to maximize the overall size of the bolt pattern within reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are, of course, a bunch of other factors. As you suggest, since the absolute tolerance is larger on larger bolts, they do generally require more sloppy holes, meaning they won't inherently provide alignment that is as good as smaller bolts. However, if you align your components independently (by measuring or with a jig,) and tighten the bolts, you can still keep the component in the right place just as well. Conversely, because holes for smaller bolts generally are less oversized, aligning a pattern of many small bolts requires much more precise machining of your parts than aligning a couple of larger bolts. This is because of the smaller oversizing factor primarily, but is compounded by the fact that the more holes you have, the more likely your worst case is to happen (where two holes are just within tolerance in the opposite directions of each other.) Of course, you could drill unusually oversized holes for smaller bolts, but you would find that the head (and/or washer) would not have very much contact with the base material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as cost, for modestly sized parts the costs of machining the parts almost certainly costs more than the cost of the fasteners themselves, so a few larger bolts would be a better option - slightly more expensive bolts, but fewer holes to drill. The size of a hole to drill has much less impact on cost than the time to locate a new hole, especially if it deep enough to require multiple steps (like a spotting drill or center drill) and therefore a tool change. In addition, depending on your scale, materials, and thickness, sometimes smaller holes are actually more expensive as they have to be drilled less aggressively to avoid tool breakage. Two big exceptions to this statement would be if your pieces are being mass produced by casting, injection molding, or a similar volumetric process, or if they are being cut by a profiling process like waterjet or laser cutting, where linear inches are the main driver of cost. AS you point out, the time to assemble the device is mostly governed by the number of bolts rather than their size - for a given length of thread - a large bolt is actually faster to tighten. So this also favors fewer, larger bolts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a formula governing the clamping force, it's nothing too special. Once you establish the pretension on each bolt as installed, you simply multiply that by the static coefficient of friction for your faying surface combination. The hard part is establishing the pretension that you will accomplish in each bolt - there are formulas that will give you tension as a function of torque, lead angle, and materials, but they are known not to be very accurate. The best way to find this value would be by direct measurement after tightening the bolts using the same method you will use in production (torque, feel, turn-of-the-nut, etc.)&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-06-08T17:36:48.247" LastActivityDate="2015-06-08T17:36:48.247" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3094" PostTypeId="2" ParentId="3075" CreationDate="2015-06-08T19:13:56.830" Score="1" Body="&lt;p&gt;Pavers or some similar pavement is very labor intensive. While a person places each brick a machine will place the vast majority of asphalt (HMA) or concrete (PCC). While there are machines that can place large quantities of brick at once they are not widely in use (That I've seen anyway).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paver roads are also very maintenance intensive. If not well maintained then something normal like snowplowing over a road with pieces can be difficult or dangerous as an individual brick can catch the plow and be &quot;shot&quot; out away from the plow. The driver has to be careful and the road has to be kept in good shape. &#xA;Overall this makes PCC or HMA roads more cost effective. Roads really are pieced together anyway. They're just blended/layered in a way that makes it harder to notice. In most concrete roads (once the bad part is broken out) dowel rods are inserted into adjacent sections of concrete thereby preventing too much vertical shift but still allowing for the pieces to remain separate. &#xA;As to movement of roads. Generally roads are built on Right of Way (ROW). ROW is property that is for the common good. This is where all the utilities run (or in an easement) and the roads. The reason you can't just move a road is because it's in the ROW which is property set apart from a private parcel nearby. Meaning you don't generally own to the middle of the street. You own some feet back from the street and the City &quot;owns&quot; the rest. That makes it slightly more difficult to rearrange roads etc. Not to mention the cost of relocating the utilities. A previous project I did cost almost $500k to take power lines and put them underground for a 1/2 mile. Those power lines did not change their location except for elevation. &lt;/p&gt;&#xA;" OwnerUserId="1909" LastEditorUserId="1909" LastEditDate="2015-06-12T15:09:36.353" LastActivityDate="2015-06-12T15:09:36.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3095" PostTypeId="2" ParentId="3075" CreationDate="2015-06-08T19:51:55.490" Score="0" Body="&lt;p&gt;Something that was touched on but not fully addressed is the cost of building roads as they are today. I believe the cost of building a new road is about $1 million per mile, which seems like a great deal of money until you really think about all the planning and work put into a road. One million USD per mile is &lt;em&gt;incredibly&lt;/em&gt; efficient, so it would have to take something cheaper than that to contend with the current system.&lt;/p&gt;&#xA;" OwnerUserId="3" LastEditorUserId="368" LastEditDate="2015-06-09T16:36:05.580" LastActivityDate="2015-06-09T16:36:05.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3099" PostTypeId="1" AcceptedAnswerId="3104" CreationDate="2015-06-08T23:54:06.047" Score="28" ViewCount="16763" Body="&lt;p&gt;So I got myself questioning what could be worse for the driver... a collision of two identical cars at equal speed (frontal crash) or the same car with the same speed crashing through a wall? The first case I see it would double the impact, but also it will absorb the energy into the other car structure, otherwise, in a solid and rigid wall, all the energy would come back to the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which situation is worse for the passengers?&lt;/p&gt;&#xA;" OwnerUserId="1853" LastEditorUserId="272" LastEditDate="2015-06-08T23:57:43.260" LastActivityDate="2019-06-15T15:44:29.957" Title="Which is Worse: Car vs. Car or Car vs. Wall?" Tags="|automotive-engineering|safety|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3100" PostTypeId="2" ParentId="3099" CreationDate="2015-06-09T00:12:39.857" Score="11" Body="&lt;p&gt;In the limit of the cars being identical and the wall being immutable, I would argue that the two situations are the same based on symmetry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the collision of the two cars with no wall.  Conservation of momentum implies that the end result is both cars at a stand-still.  If they hit each other perfectly head on, the vehicles will buckle and absorb energy identically.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine placing a completely inflexible wall between the two cars as they collide.  Nothing about the situation changes; the cars still end up at a stand-still and absorb energy in the same way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you instead consider a wall which collapses when hit by the car, then hitting the wall is safer.  Neglecting the added danger of flying bricks and a building on top of your car, that is.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-06-09T20:14:57.230" LastActivityDate="2015-06-09T20:14:57.230" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3101" PostTypeId="1" AcceptedAnswerId="3103" CreationDate="2015-06-09T01:44:08.630" Score="1" ViewCount="7603" Body="&lt;p&gt;The dimensions of the container for the water is 6.06m by 2.44m the depth of water is at 1m equating to 14.786m^3 or 14,786 ℓ of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sytems is a closed system with a pump circulating the water 150 ℓ/min @ 1100 bar. How long would it take for the water to heat up from 20*C to 30*C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would the time change if the depth of water increase to 2m (ie. 29.572 ℓ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q:&lt;/strong&gt; What formula(s) do I need to use to find the time required for the water heat up from 20*C to 30*C. Also what other variable is required for me to calculate this time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers will be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditDate="2015-06-10T00:44:00.163" LastActivityDate="2015-06-10T07:14:44.937" Title="Calculating time to heat up water at flow rate of 150 ℓ/min. @ 1100 bar" Tags="|thermodynamics|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3103" PostTypeId="2" ParentId="3101" CreationDate="2015-06-09T03:40:31.783" Score="3" Body="&lt;p&gt;These two very simple to use formulas will probably allow easy order of magnitude scoping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Time to heat with given heating power&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;T = V x 1000 cc/l x 4.17 x K / W seconds&#xA;or T = 4170 V.K/W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Power required to heat in time T&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power = W = = V x 1000 cc/l x 4.17 x K / T Watts &#xA; or W = 4170 V.K/T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;T = seconds&lt;br&gt;&#xA;V = litres&lt;br&gt;&#xA;K = degrees K (or C) rise&lt;br&gt;&#xA;W = Watts heating power&lt;br&gt;&#xA;4.17 = Specific heat of water over range 20 C to 30 C adjusted for mean water density so units are J/cc/K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is increased by any losses&lt;br&gt;&#xA;eg electric kettle uninsulated is 85% - 95% efficient &#xA;Pumping energy adds partially to heating. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The key factor is the amount of heat stored in water per cc per degree K (or degree C) rise in temperature. Once this is known all else can be calculated.&lt;br&gt;&#xA;A more useful than many table of &lt;a href=&quot;http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;water thermal properties&lt;/strong&gt; may be found &lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt; on the well-worth-bookmarking &quot;Engineering Toolbox&quot; site. Rather than giving just a few figures at a few selected temperatures this gives properties at a wide range of temperatures, so it can be established if temperature has a significant effect on the relevant properties in the context of the current problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific heat of water is an average of about 4.18 Joule per gram* per degree K across the 20 C to 30C temperature range, or 4.17 Joule per cc per degree K. The 0.01 J/unit/K is not going to make much difference to your result. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The per cc and per gram figures differ slightly because the density of water is very slightly less than 1 g/cc across this range. The specific heat will be &lt;strong&gt;about&lt;/strong&gt; 4.18 Joule per gram and 4.17 J/cc per degree K. The usual value given is 4.182 J/g/K, often without reference to temperature range.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other useful (and necessary) facts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;One Joule is provided by 1 Watt of heating per second or&lt;br&gt;&#xA;1 J =  1W/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 litre = 1000 cc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 metre^3 = 1000 litre &#xA;And 1 litre = 10 x 10 x 10 cm or 100 x 100 x 100 mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total_energy = SH x cc x delta-C  = SH x litres x 1000 x delta-C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power = energy /second = Total_energy / seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure matters little here except as it may affect water density.&lt;br&gt;&#xA;Look up compressibility of water to see how much this matters.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Doubling water head affects volume (of course) but has an insignificant effect on pressure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Power input&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy in Volume of V litres =&#xA; V x 1000 cc/l x SG g/cc x SH  x delta_T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V = 14,786 (user figure)&lt;br&gt;&#xA;delta_T = 10 K (user figure)&lt;br&gt;&#xA;SG = 1 (actual figiure rolled into SH - see text)&lt;br&gt;&#xA;SH = 4.17 J/cc/K (SG is included in this - see text)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy = 14,786 x 1000 x 1 x 4.17 x 10 = 616,576,000 J  or W.s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can heat it in 1 second using a 617,000! kW heater.&lt;br&gt;&#xA;This may be hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or in one hour with 616,576/3600 seconds = 171,000 Watts of heat (+ losses)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or using a 3kW element as used in domestic hot water heaters.&lt;br&gt;&#xA;(Max size available is usually 3 kW - larger for special uses).&#xA;616576000/3000  =~ 57 hours! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To heat it in and '8 hour day' = 616.6 MJ/(8 x 3600) = 21.4 kW.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Heat capacity of tank may be significant.&#xA;Insulation of tank will be significant.&lt;br&gt;&#xA;Other losses may occur.&lt;br&gt;&#xA;But that gives you an idea.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-06-10T07:14:44.937" LastActivityDate="2015-06-10T07:14:44.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3104" PostTypeId="2" ParentId="3099" CreationDate="2015-06-09T07:24:45.027" Score="30" Body="&lt;p&gt;From the point of view of the driver of a car, impacting another car is &lt;strong&gt;about as bad&lt;/strong&gt; as crashing against an ideal wall (a wall with zero deformation whatsoever).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were a &lt;a href=&quot;https://en.wikipedia.org/wiki/Reflection_%28mathematics%29#Reflection_through_a_hyperplane_in_n_dimensions&quot; rel=&quot;noreferrer&quot;&gt;plane reflection&lt;/a&gt; between the two cars, then &lt;em&gt;vs. Car&lt;/em&gt; would be &lt;strong&gt;exactly&lt;/strong&gt; equal to &lt;em&gt;vs. Wall&lt;/em&gt; (the contact points between both cars would all be on the same plane, due to reflection, so each car could be considered a wall for the other). But this plane reflection does not exist:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XEt3R.png&quot; alt=&quot;Cars crash from above&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we have instead is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotational_symmetry#n-fold_rotational_symmetry&quot; rel=&quot;noreferrer&quot;&gt;2-fold rotational reflection&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the left part of the car is heavier than the right part. The left and right parts will get crushed differently, with the left part of each car going further than if there had been an immovable wall. Heavy parts of each car will &lt;strong&gt;slide&lt;/strong&gt; beside each other, with a lot of the energy absorbed by steel deformation, and a longer distance between point of impact and final point, thus lesser deceleration. In this scenario, if you happen to sit on the heavy side you are lucky, but if you happen to sit on the light side it might be worse than a wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, rather than all forces having the same direction, some of the energy will be converted into &lt;strong&gt;rotation&lt;/strong&gt;, which can be either a good or bad thing depending on where you sit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, cars have a few hard &lt;strong&gt;structural beams&lt;/strong&gt; (or parts that can be considered as beams) and most of the rest is softer. If hitting a wall, deceleration is immense as soon as a beam touches the wall. If hitting another car, the beams will probably enter the other car's soft parts. Here again, distance between point of impact and final position will be longer, thus a less violent deceleration. This is especially true at very high speed, with beams of each car piercing through most of the opposite car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, crashing into an ideal wall is probably &lt;em&gt;a bit worse&lt;/em&gt; than crashing into another car, but better drive safely and avoid crashes :-)&lt;/p&gt;&#xA;" OwnerUserId="425" LastEditorUserId="425" LastEditDate="2015-06-15T02:36:40.500" LastActivityDate="2015-06-15T02:36:40.500" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3105" PostTypeId="2" ParentId="3099" CreationDate="2015-06-09T10:06:04.150" Score="5" Body="&lt;p&gt;The answer will depend on the wall, and on the other car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the comparison to an ideal, immovable wall. In the case of a hypothetical collision between a &lt;a href=&quot;https://en.wikipedia.org/wiki/Humvee&quot;&gt;Humvee&lt;/a&gt; (2500 KG) and a &lt;a href=&quot;https://en.wikipedia.org/wiki/Volkswagen_Beetle&quot;&gt;VW Beetle&lt;/a&gt; (850 KG), for the Humvee &lt;strong&gt;&quot;vs. car&quot; is better&lt;/strong&gt;, whereas for the Beetle &lt;strong&gt;&quot;vs. wall&quot; is better&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the comparison to a paper thin, soft wall. For both the Humvee and Beetle &lt;strong&gt;&quot;vs. wall&quot; is better&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="449" LastActivityDate="2015-06-09T10:06:04.150" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3106" PostTypeId="1" AcceptedAnswerId="3108" CreationDate="2015-06-09T11:47:12.390" Score="1" ViewCount="276" Body="&lt;p&gt;I would like to get a simple estimation of the parameters for a humidity chamber system. The humidity chamber has one constant inflow ($k_1$) and and outflow which depends on the humidity of a neighbor humidity chamber (diffusion, modeled as $k_2(RH_{ext} - RH)$. $RH$ is the humidity of the chamber, $RH_{ext}$ is the humidity of the external chamber. I have the following system description (first order):&#xA;$$&#xA;\frac{dRH}{dt} = k_1 + k_2(RH_{ext} - RH)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have measurement data for $RH$ (output) and $RH_{ext}$ (input). $k_1$ could be seen as a second (constant) input. I would like to estimate the parameters $k_1$ and $k_2$ in a simple manner (maybe least-square based) using my (discrete) measurement data in Matlab, but I don't know how I could proceed. Use the system identification toolbox? Other ways?&lt;/p&gt;&#xA;" OwnerUserId="1925" LastEditorUserId="272" LastEditDate="2015-06-09T12:29:28.917" LastActivityDate="2015-06-09T13:35:02.313" Title="Parameter Estimation for First Order System" Tags="|measurements|modeling|control-theory|systems-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3108" PostTypeId="2" ParentId="3106" CreationDate="2015-06-09T13:35:02.313" Score="1" Body="&lt;p&gt;It sounds like you have &lt;a href=&quot;http://uk.mathworks.com/products/matlab/&quot; rel=&quot;nofollow&quot;&gt;MATLAB&lt;/a&gt;, so it's a simple linear fit:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;From your vector $RH(t)$, compute $\frac{d RH}{dt}$ by numerical differentiation&lt;/li&gt;&#xA;&lt;li&gt;Create a matrix $A$ with ones on the first column and the values of $RH_{ext}(t) - RH(t)$ on the second column&lt;/li&gt;&#xA;&lt;li&gt;Create a matrix $B$ with the values of $\frac{d RH}{dt}$ (use 0 as the initial value)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your result vector is given $A$ \ $B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;dRH = [0; diff(RH)./diff(t)];&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;A = [ones(size(RH)) RH_ext-RH];&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;k = A\dRH; % k(1) is k1, k(2) is k2&lt;/code&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-06-09T13:35:02.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3109" PostTypeId="2" ParentId="2693" CreationDate="2015-06-09T14:10:53.593" Score="6" Body="&lt;p&gt;Adding to @Air's answer, there's also the issue of boundary conditions. A simple span where neither support allows for axial displacements will have a slight gain in length, including along the &quot;neutral axis&quot;. This is because, in this case, the &quot;neutral axis&quot; will hold a tensile stress. Since the beam deforms, the neutral axis changes from a horizontal straight line to a polynomial curve, which obviously has a greater length going from A to B. This increase in length of the neutral axis implies in a uniform tensile stress along the entire beam (in addition to the transversely linear stress profile due to bending). This tensile stress is usually not taken into account since the increase in length (and therefore the requisite tensile stress) is very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if one of the supports allows for axial displacements, then there will be no increase in length of the neutral axis, since the support will simply move slightly inwards to compensate. Obviously, in the case of the cantilever there is also no change in neutral axis length since the free end simply moves closer to the fixed end. This does, however, imply that a horizontal beam of length $L$ will, once loaded, have a horizontal length of $L - \Delta$ (and a vertical displacement such that the total length equals $L$).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-06-09T14:10:53.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3110" PostTypeId="1" AcceptedAnswerId="3162" CreationDate="2015-06-09T14:31:28.457" Score="5" ViewCount="2334" Body="&lt;p&gt;When studying the strength of materials at the introductory level, we often make five simplifying assumptions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Material is homogeneous and isotropic.  &lt;/li&gt;&#xA;&lt;li&gt;Material obeys Hooke's law  &lt;/li&gt;&#xA;&lt;li&gt;Body is assumed to be prismatic  &lt;/li&gt;&#xA;&lt;li&gt;Effect of self-weight is neglected  &lt;/li&gt;&#xA;&lt;li&gt;Load is assumed as static load.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Under what circumstances is assumption 4 valid?  When can the effect of self-weight be neglected and when must it be included?&lt;/p&gt;&#xA;" OwnerUserId="21" LastEditorUserId="272" LastEditDate="2015-06-09T17:07:41.070" LastActivityDate="2015-06-12T19:14:59.563" Title="When can the effect of self-weight be neglected?" Tags="|mechanical-engineering|materials|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3112" PostTypeId="2" ParentId="3075" CreationDate="2015-06-09T14:45:11.830" Score="7" Body="&lt;p&gt;I think that there are a few things that you aren't considering.&lt;/p&gt;&#xA;&lt;h3&gt;Subgrade&lt;/h3&gt;&#xA;&lt;p&gt;Most roads have at least 12 in (300 mm) of compacted aggregate (rock) under the actual pavement. If the existing soil is very weak, then this depth of aggregate subbase can be increased. Sometimes the soil needs to be strengthened by mixing in cement.&lt;/p&gt;&#xA;&lt;p&gt;This means that just removing road panels doesn't completely remove the road structure.&lt;/p&gt;&#xA;&lt;h3&gt;Pavement Joints&lt;/h3&gt;&#xA;&lt;p&gt;If you have ever driven over a concrete road that has joints, you will notice that the ride is rougher and the road noise is louder. Each time a vehicle travels over a joint in the road, there is a bump. To help minimize this, a layer of asphalt is sometimes placed over the concrete. This creates a smoother ride by creating a flatter surface. This is another strike against pavement panels.&lt;/p&gt;&#xA;&lt;h3&gt;Frequency of change&lt;/h3&gt;&#xA;&lt;p&gt;Part of the proposal seems to be based on the thought that roads will be replaced or moved frequently. No matter how easy it is to change the land use from road to green space, it will be wasteful to spend time doing so. I don't know of any area where this type of change would happen often enough to justify the work required. Land use is planned in advance for a reason. There are also already ways to make temporary roads for special circumstances.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-06-09T14:45:11.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3113" PostTypeId="2" ParentId="3075" CreationDate="2015-06-09T15:20:41.507" Score="18" Body="&lt;p&gt;This is very common misunderstanding. When we look a a road, we see few centimeters thick layer of bituminous surface, and we think about &quot;the road&quot; as &quot;the asphalt/tarmac.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look how a road is build (or rebuilt), you can see that &quot;the road&quot; is actually over half a meter (or more) deep, consists of layers of sand and gravel and has complex rainwater draining system. And this is the most expensive part of the road: the soft, movable and adaptable interface which connects hard surface to soft soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even &quot;removable&quot; roads, made of cobblestone or concrete panels require exactly same understructure. Even &quot;trivial&quot; sidewalks require identical understructure (on a smaller scale). If you ever see amateur-made sidewalks, created by simply laying flat stones or concrete panels on the ground, pay close attention to how individual panels fit the ground. When new, they're fine. But after few years, with usage and water flushing some ground away, they become notoriously unstable. Temporary walkways set up on concerts or other mass events often exhibit same problems within a day (as they're used quite heavily).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Optimizing costs of reassembling surface is optimizing the least expensive part of a road.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also other reasons, e.g. paving is mechanized process and very efficient, joints suck and make very bad driving and walking experience, road is a structure tightly tailored to local needs (width, curve radii, bank, elevation) and it's not possible to make a prefabricated road that would cover every possible application. Perhaps the last problem is what completely obliterates your proposed benefits: instead of saying &quot;let's have a road here&quot; we'll be more like: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it possible to lay our road around here? &lt;/li&gt;&#xA;&lt;li&gt;Nope, we don't have this radius. &lt;/li&gt;&#xA;&lt;li&gt;Maybe between those buildings there? &lt;/li&gt;&#xA;&lt;li&gt;Sorry, it's 5 cm too wide, it won't fit without violating building code. &lt;/li&gt;&#xA;&lt;li&gt;Up that hill? &lt;/li&gt;&#xA;&lt;li&gt;We'll have to either flatten this hill to 0.001% or make it more steep to 0.002%, because roads come only in those elevations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This will lead to &lt;strong&gt;more restrictions&lt;/strong&gt; on road placement, not less.&lt;/p&gt;&#xA;" OwnerUserId="1931" LastEditorUserId="368" LastEditDate="2015-06-09T16:39:13.123" LastActivityDate="2015-06-09T16:39:13.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3114" PostTypeId="2" ParentId="3075" CreationDate="2015-06-09T15:28:13.320" Score="4" Body="&lt;p&gt;As some of the other posters have mentioned a road has several built up layers (hopefully). The drainage system is usually very important. Bad quality roads are usually because the drainage is poor or non-existent. Roman roads were sometimes as much as 10 feet thick, which is why many are still in use today, 2000 years later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to create a temporary road using steel. Using steel mesh and bar you can create a road that will just lay on dirt. It is EXTREMELY expensive, but in the future if metal become cheaper than rock, metal roads could become a reality. Currently, they are only used in specialized military situations.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2015-06-09T15:28:13.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3115" PostTypeId="1" AcceptedAnswerId="7121" CreationDate="2015-06-09T15:43:24.007" Score="3" ViewCount="1044" Body="&lt;p&gt;I have been &lt;a href=&quot;http://www.lab-quip.co.uk/dynamic-cone-penetrometer-dcp&quot; rel=&quot;nofollow&quot;&gt;reading about DCP&lt;/a&gt; and watched &lt;a href=&quot;https://www.youtube.com/watch?v=UadgMNG-LWA&quot; rel=&quot;nofollow&quot;&gt;a video&lt;/a&gt; of how it is done (the video is not in English). But there are some things that are still unclear and I would greatly appreciate some help and clarification. I have these questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a standard number of times that the hammer must be dropped?&lt;/li&gt;&#xA;&lt;li&gt;What is recorded after an $x$ number of hammer drops?&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lat/long coordinates of where test occurred?&lt;/li&gt;&#xA;&lt;li&gt;How far the metal head penetrated the ground (in cm)?&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;What happens to that data? What does it correlate with to give us the strength of the ground? What is that strength measured in?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1932" LastEditorUserId="272" LastEditDate="2015-06-09T17:09:30.580" LastActivityDate="2016-01-28T02:37:28.880" Title="Dynamic Cone Penetrometer (DCP) results and correlation" Tags="|geotechnical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3116" PostTypeId="2" ParentId="3099" CreationDate="2015-06-09T16:52:48.650" Score="6" Body="&lt;p&gt;I'm guessing you haven't seen the &lt;a href=&quot;https://www.youtube.com/watch?v=r8E5dUnLmh4&quot; rel=&quot;nofollow&quot;&gt;Mythbusters episode&lt;/a&gt; on this.&#xA;Then, see &lt;a href=&quot;http://www.wired.com/2010/05/mythbusters-energy-explanation/&quot; rel=&quot;nofollow&quot;&gt;this writeup&lt;/a&gt; that explains why they are correct, since Mythbuster's explanation leaves something to be desired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, it's exactly the same. In the 2 car example, each car has the same amount of energy (as each other, and as the car in the 1 car example) since they are all going the same speed (and have the same mass). In the two car crash, the total energy in the crash is doubled, however, the energy is distributed equally between the two cars. Therefore, the energy for each car is identical to a single car hitting a ridged wall. The tests in Mythbusters illustrates this.&lt;/p&gt;&#xA;" OwnerUserId="1933" LastEditorUserId="191" LastEditDate="2015-11-09T09:24:26.773" LastActivityDate="2015-11-09T09:24:26.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3117" PostTypeId="2" ParentId="1908" CreationDate="2015-06-09T18:42:32.483" Score="2" Body="&lt;p&gt;When fluid flow increases in velocity (because of smaller area of blade root) the pressure decreases and can be low enough to produce cavitation. The spheres produce turbulence changing the flow characteristics, keeping the pressure up.  Hub tip pressure drops as flow fills in after the hub and causes cavitation.   The flat hub blades disrupt the flow and keep pressure up.  Several designs are used in both commercial and military applications.  Search Google for propeller cavitation images for good diagrams.&lt;/p&gt;&#xA;" OwnerUserId="1938" LastEditorUserId="1951" LastEditDate="2015-06-12T01:14:55.993" LastActivityDate="2015-06-12T01:14:55.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3119" PostTypeId="2" ParentId="659" CreationDate="2015-06-10T09:28:47.540" Score="2" Body="&lt;p&gt;For an easily removable option you can use Kapton tape. It is heat resistant and works well for holding a thermocouple temporarily in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wrap the wires in Kapton for electrical isolation.&lt;/p&gt;&#xA;" OwnerUserId="1947" LastActivityDate="2015-06-10T09:28:47.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3122" PostTypeId="1" AcceptedAnswerId="3124" CreationDate="2015-06-10T12:11:25.607" Score="2" ViewCount="695" Body="&lt;p&gt;Yesterday, I tried to find constants $k_1$ and $k_2$ for the following system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\mathrm d RH}{\mathrm dt}=k_1+k_2(RH_{ext}-RH).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have some first computations for $k_1$ and $k_2$ and would like to test how good a simulation with these parameters works. I think I could also write this as state space model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x}=-k_2x+\left[\begin{matrix} k_1 &amp;amp; k_2\end{matrix}\right]\left[ \begin{matrix} u_1 \\ u_2\end{matrix} \right],$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y = x,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x = RH$, $u_1=1$ ($const.$) and $u_2=RH_{ext}$. I have measurement data for $RH_{ext}$ in Matlab (vector), a time vector $t$ with the same length and would like to do a simulation for $RH$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions how I could proceed? Simulink is possible, but just a script-based solution would be even better.&lt;/p&gt;&#xA;" OwnerUserId="1925" LastEditorUserId="282" LastEditDate="2015-06-30T00:51:50.133" LastActivityDate="2015-06-30T00:51:50.133" Title="Simulation of First Order System (Matlab &amp; Simulink)" Tags="|measurements|modeling|simulation|systems-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3124" PostTypeId="2" ParentId="3122" CreationDate="2015-06-10T15:11:56.307" Score="1" Body="&lt;p&gt;Yes, if you have values for &lt;code&gt;k1&lt;/code&gt; and &lt;code&gt;k2&lt;/code&gt;, data for &lt;code&gt;RH_ext&lt;/code&gt; and &lt;code&gt;t&lt;/code&gt;, your suggestion would work although I think you have your matrices wrong. You probably need to define &lt;code&gt;x&lt;/code&gt; as &lt;code&gt;RH - k1/k2&lt;/code&gt; to get rid of the constant term:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = -k2;&#xA;B = k2; % x_dot = -k2*x + k2*u&#xA;C = 1;&#xA;D = 0; % y = x&#xA;sys = ss(A,B,C,D);&#xA;y = lsim(sys,RH_ext,t);&#xA;RH_sim = y + k1/k2;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2015-06-11T06:45:58.657" LastActivityDate="2015-06-11T06:45:58.657" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3126" PostTypeId="5" CreationDate="2015-06-10T15:40:39.797" Score="0" Body="&lt;p&gt;This tag should be used to categorize questions that are mainly about solving problems related to research in engineering, from the perspective of researchers and academics.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Appropriate uses&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This tag is best applied to questions that are asked from the perspective of a researcher and/or best answered by a researcher. Here are a few examples of topics that might be good fits for the tag:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Obtaining or using funding&lt;/li&gt;&#xA;&lt;li&gt;The process of peer review&lt;/li&gt;&#xA;&lt;li&gt;The role of research in education/academia/industry&lt;/li&gt;&#xA;&lt;li&gt;Collaboration between researchers/institutions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Inappropriate uses&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This tag isn't needed for questions that are only indirectly related to the research field; for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Questions that are about a specific paper or finding should be tagged with the discipline and any applicable concepts/technologies.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.meta.stackexchange.com/q/261/368&quot;&gt;Are academic guidance questions on-topic?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="368" LastEditorUserId="6767" LastEditDate="2018-10-29T09:56:54.813" LastActivityDate="2018-10-29T09:56:54.813" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3127" PostTypeId="4" CreationDate="2015-06-10T15:40:39.797" Score="0" Body="Questions about performing research in the field of engineering." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-06-10T15:40:39.797" LastActivityDate="2015-06-10T15:40:39.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3128" PostTypeId="1" CreationDate="2015-06-10T16:52:45.547" Score="2" ViewCount="319" Body="&lt;p&gt;I need to design some backlash free rotational joints for slow and intermittent movements (device for human-computer interaction) for my uni project. Possibly loaded in any direction; the loads won't exceed ~100 N (fyi just to get the sense of scale) and I'd like to use plain bearings, as they should take less space and possibly have less stick-slip and smoother motion than rolling bearings. There's, for instance, that company based in Cologne, Germany that's got a really big variety of bearings - but then again, their off-the shelf preloaded bearings are really hard to move on the shaft (checked it personally), which is definitely not what I'm wanting. I was told by a salesman, that basically, as the bearings are press-fit, you can slightly alter the clearance by changing the hole tolerance. (As an example -  with recommended tolerances, a joint with 5 mm shaft would have between 0,02 mm and 0,10 mm of radial play, so choosing a more strict tolerance one can get eg. 0,01 to 0,05 mm play. Side note, perhaps can be useful for others - I was told that changing the hole tolerance generally relates 1-to-1 to the change in fitted bearing inside diameter tolerance, but I haven't checked it for myself and it's word-of-mouth, not in the catalogues).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Below is a half-cross-section of a proposed play-free rotational joint: part 3 (with a fixed axis) is connected with axial bearing 5. Part 1, together with two press-fitted flange bearings 4, revolves around horizontal axis. It is all pressed against each other to eliminate backlash by nut 2. I think I can estimate friction caused by axial bearings, simply by defining the torque for screwing the nut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1O5kD.jpg&quot; alt=&quot;Half cross-section of play-free joint&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are there any examples of design or scientific articles regarding play-free (preloaded) plain bearing (rotational) joints? Do you perhaps know from experience, how would such bearings (esp. plastic ones, esp. - I know I'm wanting too much - on carbon shaft) behave when given no or very little radial clearance? (by 'behave' I mean friction and 'feel' in the first place). I assume it should increase, but to what extent? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Are such designs used at all, with this intent in mind? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess it is rather a matter of experience than pure calculation - but I'm lacking information on both - I'd really appreciate any hint.&lt;/p&gt;&#xA;" OwnerUserId="1877" LastEditorUserId="1586" LastEditDate="2016-03-01T23:28:09.493" LastActivityDate="2016-03-01T23:28:09.493" Title="Plain bearings (plastic) in backlash free joints" Tags="|mechanical-engineering|bearings|" AnswerCount="0" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3129" PostTypeId="2" ParentId="3110" CreationDate="2015-06-10T17:14:35.570" Score="0" Body="&lt;p&gt;you can always include gravity as an additional load (for instance, for a constant cross-section beam, a uniform one), so - as @joojaa already mentioned - it is good to omit it and keep things tidy and clean for better understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sidenote: for many structures the loads from structure self weight is (and should be) much smaller than the intended load. Maybe a plane won't be a best example, but still (oversimplification warning) - B747 weighs around 180 tonnes empty, take-off weight can be up to ~400 tonnes (after wiki). The wings weigh about 80 tonnes (&lt;a href=&quot;http://www.answerbag.com/q_view/93809&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;), so the load apart from gravity is four times bigger than weigh. Of course, weight is not negligible, but still considerably smaller than load.&lt;/p&gt;&#xA;" OwnerUserId="1877" LastEditorUserId="1877" LastEditDate="2015-06-11T11:21:25.307" LastActivityDate="2015-06-11T11:21:25.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3130" PostTypeId="1" AcceptedAnswerId="3131" CreationDate="2015-06-10T18:10:07.387" Score="4" ViewCount="80" Body="&lt;p&gt;I have this differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{\eta (t)}=\frac{-2\eta(t)}{C(\widehat{R}+\Delta R)}+\frac{K^2 L}{C E^2 (\widehat{R}+\Delta R)}+K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, my book says, correspond to a LTI system, asymptotically stable with a constant input:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{K^2 L}{C E^2 (\widehat{R}+\Delta R)}+K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says also that the steady state value of the variable $\eta (t)$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tilde{\eta (t)} = \frac{K C(\widehat{R}+\Delta R)}{2} + \frac{K^2 L}{2E^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book doesn't say how to get this, but I reached the same result solving the differential equation (after a change a variable it can be solved separating variables) and calculating the limit for $t\rightarrow\infty$.&#xA;My question is: is it really necessary to solve the differential equation or is there any property I can use to avoid it? I'm asking this because looking at the expression of the steady state value you can see that is equivalent to the ratio of the input and the coefficient of $\eta(t)$.  Is it random or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="1957" LastEditorUserId="272" LastEditDate="2015-06-11T12:30:56.617" LastActivityDate="2015-06-11T12:30:56.617" Title="Steady state value, a possible shortcut?" Tags="|control-theory|mathematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3131" PostTypeId="2" ParentId="3130" CreationDate="2015-06-10T18:53:47.257" Score="4" Body="&lt;p&gt;Think for a moment about what it means for $\eta(t)$ to have reached its &lt;strong&gt;steady&lt;/strong&gt; state.  It means precisely that $\dot{\eta}(t)=0$.  If you plug that into your first equation and solve for $\eta(t)$ you get your solution as you've already observed.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-06-10T18:53:47.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3132" PostTypeId="1" AcceptedAnswerId="3134" CreationDate="2015-06-10T19:32:53.503" Score="1" ViewCount="65" Body="&lt;p&gt;One antonym word I can think of is &quot;derating&quot;.  If I say &quot;derate motor for pump X&quot;, it can mean &quot;select a motor with lower power output, since the pump doesn't need a more powerful motor&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about selecting a more powerful motor?  How can I say &quot;I want to ____ the motor for this pump&quot; and mean &quot;up&quot; the motor, &quot;bump up&quot; the motor, &quot;choose-a-stronger&quot; motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a phrase or single word, that can be used to speak of this concept clearly.  For example when explaining the concept to the sales team or to the customer (user of the engineering application).&lt;/p&gt;&#xA;" OwnerUserId="1958" LastActivityDate="2015-06-10T23:33:31.927" Title="Is there a universally accepted way to refer to choosing a more powerful motor for an application?" Tags="|motors|sales|communication|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3133" PostTypeId="2" ParentId="231" CreationDate="2015-06-10T19:35:09.870" Score="1" Body="&lt;p&gt;You are correct that the Froude number (Fr) is very important for wave resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://engineering.stackexchange.com/a/293&quot;&gt;answer given by nivag&lt;/a&gt; regarding the speed-length ratio (also known as &quot;hull speed&quot;) is not correct. That limit is often quoted, but it is a myth that it is somehow impossible for displacement hulls to surpass it. Ships can travel faster than that ratio implies, but for conventional ships the energy requirement is usually prohibitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thin hulls (like rowing shells) are able to operate easily beyond the ratio of&#xA;$1.34 L$. At Olympic level, rowing shells for example, operate at Fr&#xA;between about 0.45 and 0.7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from skin-friction, you will also need to consider &quot;form drag&quot;. This component can be important for stubby hulls (i.e. those with a small length to beam ratio, L/B) at low Fr. A tug boat will have greater form drag than a rowing shell at the same Fr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hull has a transom (i.e. cut-off) stern there will also be a large component of resistance when the transom is not running fully dry. In that case there is a lot of eddying, and possible wave-breaking behind the stern at low Froude numbers. At higher Fr the transom is dry, and wave resistance and form drag are much lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you tell us a little more about the principal proportions of the boat (e.g.&#xA;displacement weight, length, beam and draft) we might be able to offer more advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hull is fairly slender, say L/B &gt; 5, you can try some free software to&#xA;estimate the total resistance (viscous + wave). See, for example, &lt;a href=&quot;http://www.cyberiad.net/michlet.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Michlet&lt;/a&gt; and &lt;a href=&quot;http://www.cyberiad.net/flotilla/flotilla.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flotilla&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water depth can also affect wave resistance. In this case the depth-based&#xA;Froude number plays an important role, very much like the Mach number in aerodynamics. Michlet and Flotilla will both allow you to vary water depth and to see the effect on wave resistance and wave patterns.&lt;/p&gt;&#xA;" OwnerUserId="1908" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-06-10T20:15:36.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3134" PostTypeId="2" ParentId="3132" CreationDate="2015-06-10T23:33:31.927" Score="3" Body="&lt;p&gt;Quite sure the word used is &quot;oversized&quot;. When you say to an engineer that a motor is oversized for that application (for that pump) the engineer will not think the motor dimensions are too big (and will not fit) but he will understand as the motor rated values would exceed the minimum requirements by an unexpected high amount. The context of the conversation will clarify if you mean &quot;oversized&quot; as a good thing (heavy-duty) or as a bad thing (waste of money, energy, space or even security compromise in the case the motor can destroy the pump for example).&lt;/p&gt;&#xA;" OwnerUserId="1579" LastActivityDate="2015-06-10T23:33:31.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3135" PostTypeId="2" ParentId="231" CreationDate="2015-06-11T01:36:10.557" Score="2" Body="&lt;p&gt;&quot;Hull speed&quot; is actually the ratio of speed to the square root of length.&#xA;To make things even more confusing, length is in feet, and speed is in&#xA;knots. That's how the constant 1.34 arises. (ProTip: Let's never speak of &#xA;it again!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wave resistance ($R_w$) begins its rapid rise at a Froude number (Fr) of &#xA;about 0.35. Below that Fr, $R_w$ is usually small compared to the skin-friction &#xA;and other hydrodynamic drag components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for sake of example, let the wave resistance coefficient be defined as &#xA;$C_w = R_w/(0.5 \rho U^2 S)$, where $\rho$ is water density, $U$ is ship speed,&#xA;and $S$ is the (static) wetted surface area of the hull. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In deep water, $C_w$ increases roughly like Fr to the 6th power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The depth-based Froude number is $F_h = U/\sqrt{g h}$, where $g$ is &#xA;gravitational acceleration, and $h$ is water depth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For finite depth water, $C_w$ can increase almost like Fh to the 10th power&#xA;as $F_h \rightarrow 1$. Once through (the critical value) $F_h =1$, wave resistance&#xA;begins to decrease, and it can be lower than in deep water for the same &#xA;length-based Froude number (Fr). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_h &amp;lt; 1$ is usually referred to as sub-critical; $F_h &amp;gt; 1$ is super-critical, &#xA;and (roughly) $0.9 &amp;lt; F_h &amp;lt; 1.1$ is trans-critical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the trans-critical regime, the hull also experiences forces and moments&#xA;that significantly change its attitude with respect to the undisturbed free-surface of water.&#xA;The trim and heave of a hull is known as &quot;squat&quot;. This phenomenon is difficult &#xA;to predict accurately. It can have some effects on resistance but, more importantly, in &#xA;shallow water there is also a danger of the ship grounding against the sea-bed. &#xA;This can cause large losses of income, and there have also been fatalities attributed&#xA;to the phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wave patterns for finite depth are quite interesting...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cyberiad.net/gallery/shipwakes/hw8fd/fh080.gif&quot; rel=&quot;nofollow&quot;&gt;fh080.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cyberiad.net/gallery/shipwakes/hw8fd/fh090.gif&quot; rel=&quot;nofollow&quot;&gt;fh090.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cyberiad.net/gallery/shipwakes/hw8fd/fh099.gif&quot; rel=&quot;nofollow&quot;&gt;fh099.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cyberiad.net/gallery/shipwakes/hw8fd/fh101.gif&quot; rel=&quot;nofollow&quot;&gt;fh101.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cyberiad.net/gallery/shipwakes/hw8fd/fh110.gif&quot; rel=&quot;nofollow&quot;&gt;fh110.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cyberiad.net/gallery/shipwakes/hw8fd/fh120.gif&quot; rel=&quot;nofollow&quot;&gt;fh120.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $F_h$ enters the trans-critical regime, wave patterns change dramatically.&#xA;The angle of the V-shape opens out and becomes 90 degrees at $F_h = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sub-critical speeds, transverse waves (those perpendicular to the ship's&#xA;track) are apparent. &#xA;In super-critical flow, transverse waves disappear. (In short, they cannot keep&#xA;up with the ship).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DISCLOSURE: These patterns were made using my (free) program Flotilla.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More patterns can be found at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cyberiad.net/wakeimages.htm&quot; rel=&quot;nofollow&quot;&gt;www.cyberiad.net/wakeimages.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1908" LastEditorUserId="33" LastEditDate="2015-09-30T00:59:36.667" LastActivityDate="2015-09-30T00:59:36.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3139" PostTypeId="1" AcceptedAnswerId="3140" CreationDate="2015-06-11T14:09:29.923" Score="-1" ViewCount="586" Body="&lt;p&gt;The four sides of the &lt;em&gt;Great Pyramid of Giza&lt;/em&gt; are oriented to the four cardinal directions of the compass (north, west, east, south).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Did I.M. Pei do the same with the &lt;em&gt;Louvre Pyramid&lt;/em&gt;?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1963" LastEditorUserId="368" LastEditDate="2015-06-11T15:35:41.220" LastActivityDate="2015-06-11T15:35:41.220" Title="Is the Louvre Pyramid oriented to the four cardinal points of the compass?" Tags="|engineering-history|architecture|" AnswerCount="1" CommentCount="0" ClosedDate="2015-06-11T22:52:20.687" ContentLicense="CC BY-SA 3.0" />
  <row Id="3140" PostTypeId="2" ParentId="3139" CreationDate="2015-06-11T14:51:56.610" Score="4" Body="&lt;p&gt;It doesn't appear so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zGrdW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;(screen shot &lt;a href=&quot;https://www.google.be/maps/@48.860611,2.337644,745m/data=!3m1!1e3&quot; rel=&quot;nofollow noreferrer&quot;&gt;from google maps&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-06-11T14:59:29.107" LastActivityDate="2015-06-11T14:59:29.107" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3141" PostTypeId="1" AcceptedAnswerId="3144" CreationDate="2015-06-11T16:21:06.690" Score="1" ViewCount="333" Body="&lt;p&gt;$\frac{0.25}{s^2+0.5s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I use formulas for 2nd order systems in &lt;a href=&quot;http://www.ece.unm.edu/course/ece345/lectures/LectureNotes4.pdf&quot; rel=&quot;nofollow&quot;&gt;this pdf&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, how can I understand if this sytem is underdamped, damped, etc?&lt;/p&gt;&#xA;" OwnerUserId="1964" LastEditorUserId="9" LastEditDate="2015-06-12T09:22:47.343" LastActivityDate="2015-06-12T17:31:00.007" Title="How to calculate of Second-order system Frequency and Time domain values?" Tags="|control-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3142" PostTypeId="1" CreationDate="2015-06-11T16:52:27.947" Score="1" ViewCount="248" Body="&lt;p&gt;I'm working in a project that consists of something similar to a motorized skate. I know how to do it, but I don't know what motor to use. I need that the skate to be able to move 90 Kg to 20 Kph (12 miles per hour) or more if it's possible. I also need the motor as cheap as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read things about motors but we can say that I don't know anything about them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of motor and what facts do I need? What transmission ratio should I use between the motor and the wheels? What size and type of wheels? Any special type of transmission?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the basic mechanics, but I don't know what motor should I use, my requirements are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Move 90 Kg to 20 Kph (if the motor for that is very expensive only 15 kph)&lt;/li&gt;&#xA;&lt;li&gt;I don't mind the acceleration &lt;/li&gt;&#xA;&lt;li&gt;I think that it should work with battery (I'm not sure)&lt;/li&gt;&#xA;&lt;li&gt;The skate is thought for city, but may have to go up small slopes&lt;/li&gt;&#xA;&lt;li&gt;As cheaper as possible&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't have the budget to buy numerous motors and try them all, so I need to be sure that it will work before I purchase one.&lt;/p&gt;&#xA;" OwnerUserId="1965" LastEditorUserId="272" LastEditDate="2015-06-12T15:11:10.727" LastActivityDate="2015-06-12T15:11:10.727" Title="How can I choose a motor for my project?" Tags="|mechanical-engineering|motors|" AnswerCount="0" CommentCount="5" ClosedDate="2015-06-11T18:16:36.847" ContentLicense="CC BY-SA 3.0" />
  <row Id="3143" PostTypeId="1" AcceptedAnswerId="3159" CreationDate="2015-06-11T17:20:10.830" Score="6" ViewCount="1691" Body="&lt;p&gt;I was wondering how the values of thermodynamic properties as enthalpy, entropy, internal energy, specific heat, specific volumes ... etc, of gases and liquids are calculated using programming libraries such as &lt;code&gt;cantera&lt;/code&gt; and &lt;code&gt;thermopy&lt;/code&gt; as an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are steam tables made for instance? How do we get all these values of properties if not experimentally?&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2015-06-13T10:28:08.043" Title="How are thermodynamics properties of gases and liquids calculated numerically?" Tags="|mechanical-engineering|thermodynamics|numerical-methods|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3144" PostTypeId="2" ParentId="3141" CreationDate="2015-06-11T19:15:06.337" Score="6" Body="&lt;p&gt;No, you cannot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you already guessed, it's very difficult to fit&#xA;$$\dfrac{0.25}{s^2+0.5s}$$&#xA;into the described format&#xA;$$\dfrac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$$&#xA;There is simply no value for $\omega_n$ and $\zeta$ where this equation holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably know that transfer functions in the Laplace (s) domain have a direct relation to differential equations. I come from a mechanical engineering background, so I'll take a mechanics example. The 'second order system' you describe is a model for a basic forced mass-spring-damper: i.e., $m\ddot{x}+d\dot{x}+kx=F$. Here, a dot indicates a derivative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m5GgE.png&quot; alt=&quot;Image courtesy Wikipedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image courtesy &lt;a href=&quot;http://commons.wikimedia.org/wiki/File:Mass-Spring-Damper.png&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;. I use $d=B$, since $d$ stands for &lt;strong&gt;d&lt;/strong&gt;amper (not sure why anyone would use $B$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can take the Laplace transform easily, knowing that a derivative corresponds to a multiplication with $s$.&#xA;$$mX(s)s^2+dX(s)s+kX(s)=F(s)$$ or, rearranging to get the transfer function&#xA;$$\dfrac{X(s)}{F(s)}=\dfrac{1}{ms^2+ds+k}$$&#xA;Hey, look at that! That's our second order system, but now a physical interpretation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how is your transfer function different? It has $k=0$! In other words, it does not have a spring, just a damper and a force! You can now easily see that this thing will never oscillate when given a step response - so a natural frequency $w_n$ is silly to describe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can translate this mass/spring/damper idea to an inductor (mass), resistor (damper) capacitor (spring) -problem (assuming a circuit in series). Just imagine this circuit without a capacitor - it will not oscillate!&lt;/p&gt;&#xA;" OwnerUserId="946" LastEditorUserId="946" LastEditDate="2015-06-11T19:31:30.923" LastActivityDate="2015-06-11T19:31:30.923" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3145" PostTypeId="2" ParentId="3038" CreationDate="2015-06-11T20:01:56.420" Score="1" Body="&lt;p&gt;After the discussion with @sanchises in the comments, the answer is that, yes, the order of operations in jacking does affect the final tension diagram of the cable. A symmetric cable simultaneously jacked will have a symmetric tension diagram, while the same cable jacked from one side and then the other will not. The diagram actually ends up as a line almost parallel to the tension diagram after the jacking of the left side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cmAUP.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cable in this example is jacked from the left side (grey line) and loses around 15% of its tension at the right anchor ($\Delta P$). It is then jacked from the right side by that same amount, adding to the tension profile according to the yellow line, which results in the blue line, with the left anchor being incremented by $0.15\cdot0.85=13\%$.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-06-11T20:01:56.420" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3146" PostTypeId="2" ParentId="1908" CreationDate="2015-06-11T22:46:03.243" Score="-1" Body="&lt;p&gt;This diagram illustrates cavitation locations.&#xA;&lt;a href=&quot;http://www.boatdesign.net/forums/attachments/props/37159d1257771030-prop-doing-bad-cav_varie.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.boatdesign.net/forums/attachments/props/37159d1257771030-prop-doing-bad-cav_varie.jpg&lt;/a&gt;&#xA;The covered prop may look like this:&#xA;&lt;a href=&quot;https://d2t1xqejof9utc.cloudfront.net/screenshots/pics/0a97825d2248893edcf274662567b552/medium.JPG&quot; rel=&quot;nofollow&quot;&gt;https://d2t1xqejof9utc.cloudfront.net/screenshots/pics/0a97825d2248893edcf274662567b552/medium.JPG&lt;/a&gt;&#xA;The fins are integral to the hub/spinner and turn with the prop. The pitch is less than the blade root.&lt;/p&gt;&#xA;" OwnerUserId="1951" LastActivityDate="2015-06-11T22:46:03.243" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3147" PostTypeId="2" ParentId="1908" CreationDate="2015-06-11T22:57:45.740" Score="0" Body="&lt;p&gt;The props are on different subs.  A transition piece covers the gap between the after hull and propeller.  In the first pic it is almost cylindrical, the second decidedly conical.  The russians backfit the second pic hub designs to several older class designs, even the tandem 4-blade props on the Victor class.  Also seen in some Chinese subs.&lt;br&gt;&#xA;Longer hub taper produces more gradual flow so in itself reduces cavitation onset. The blades appear welded to hub.  Relatively cheap fix compared to some designs.&lt;/p&gt;&#xA;" OwnerUserId="1951" LastEditorUserId="1951" LastEditDate="2015-06-12T00:11:30.383" LastActivityDate="2015-06-12T00:11:30.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3148" PostTypeId="1" CreationDate="2015-06-12T01:13:16.990" Score="3" ViewCount="565" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fhgkG.jpg&quot; alt=&quot;problem sketch&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm designing a system to determine pressure drops across thin samples while simultaneously measuring velocity (and through multiplying by area, volumetric flowrate). I need a way of determining how accurate my setup is. The big catch is that what drives the flow is a pressure regulator, so flowrate is entirely dependent on what pressure loss I experience. If I had anything more than my phone this second I would generate and attach more professional pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving down the line is the following. First, a pressure regulator downsizes the pressure from the source to whatever value of pressure I want. Then it goes through my clamped sample (in my case I'm dealing with feathers most of the time but it's fairly versatile). Following the test section the cross sectional area of the pipe diverges to double the diameter and then extends for a few inches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three static ports. The first is before the test section. The second is after the test section. The third is after the pipe diverges. My manometer will test between 1 and 2 as well as 2 and 3. 1 and 2 will indicate the pressure loss across the feather/sample. 2 and 3 indicate the static pressure decrease due to the divergence which will allow me to approximate velocity through the Bernoulli equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in essence: how do I validate my manometer readings between both cases? I want to be able to say with confidence that my setup works. &lt;/p&gt;&#xA;" OwnerUserId="1866" LastEditorUserId="1293" LastEditDate="2015-06-12T06:05:55.187" LastActivityDate="2015-06-18T12:26:16.520" Title="How to verify results from a manometer/venturimeter?" Tags="|fluid-mechanics|pressure|measurements|airflow|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3149" PostTypeId="1" AcceptedAnswerId="3160" CreationDate="2015-06-12T03:28:32.003" Score="2" ViewCount="139" Body="&lt;p&gt;I bought a germicidal UVC florescent lamp online and i would like to put it in a an 20 inch box for sanitation uses. After going to the hardware store I couldn't find any light that is compatible with the T8 light bulb other then this 15 Watt fixture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.homedepot.com/p/Lithonia-Lighting-1-Light-Flush-Mount-White-Ceiling-Fluorescent-Closet-Light-CUC8-15-120-LP-S1-M4/202193137&quot; rel=&quot;nofollow&quot;&gt;http://www.homedepot.com/p/Lithonia-Lighting-1-Light-Flush-Mount-White-Ceiling-Fluorescent-Closet-Light-CUC8-15-120-LP-S1-M4/202193137&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum current it can support is 0.27A. The UVC bulb function at 0.6 A. &#xA;Here is the link for the bulb:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.1000bulbs.com/product/7342/GERM-3000008.html&quot; rel=&quot;nofollow&quot;&gt;https://www.1000bulbs.com/product/7342/GERM-3000008.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the the questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Would their be too much current drawn that the heat my damage the lamp if I install it this way? (even though I would be using it for 15 min. maximum each time)&lt;/li&gt;&#xA;&lt;li&gt;If get a new lamp that has compatible power with the closet light above, would the lamp give the specific light wavelength required to get ultra-violet radiation (250nm)? Because I know that different ballasts would give different light temperatures.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1736" LastEditorUserId="9" LastEditDate="2015-06-12T14:02:41.460" LastActivityDate="2015-06-12T17:18:05.903" Title="What will happen if I install a 46 watt fluorescent bulb on a 15 watt light fixture?" Tags="|electrical-engineering|power-electronics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3150" PostTypeId="1" AcceptedAnswerId="3173" CreationDate="2015-06-12T03:52:35.640" Score="4" ViewCount="6149" Body="&lt;p&gt;I'd like to know which stainless steels are good for heating on an induction stove. Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Induction_cooking#Cookware&quot; rel=&quot;nofollow&quot;&gt;notes&lt;/a&gt; that the surface &lt;s&gt;resistance&lt;/s&gt; resistivity is a good figure of merit. However, that article only lists the surface &lt;s&gt;resistance&lt;/s&gt; resistivity for two types of stainless steel -- the better being 432. However, I cannot find 432 from my favorite supplier, although they do have 430.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will 430 have a similar surface resistance? Is the a resource where I can find the surface resistance of other stainless steels? I have searched around but only found results for the two stainless steels that wikipedia lists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Looking at the &lt;a href=&quot;http://dx.doi.org/10.1109/TIA.1973.349892&quot; rel=&quot;nofollow&quot;&gt;IEEE article&lt;/a&gt; that underlies the wikipedia article, I think that the wikipedia article should use &quot;surface resistivity&quot; instead of &quot;surface resistance&quot;. I was using the terms in the wikipedia article. Sorry for any confusion. I will edit the wikipedia article if appropriate.&lt;/p&gt;&#xA;" OwnerUserId="1966" LastEditorUserId="282" LastEditDate="2015-06-18T01:07:19.010" LastActivityDate="2015-06-18T01:07:19.010" Title="What stainless steels are good for induction heating?" Tags="|materials|steel|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3152" PostTypeId="1" AcceptedAnswerId="3155" CreationDate="2015-06-12T04:54:03.063" Score="2" ViewCount="161" Body="&lt;p&gt;What is safe and what is acceptable in the context of Engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically speaking we are in the process of designing a test rig which will pump water. What is a maximum safe temperature for the water to achieve without infringing employee safety. ie water at ${90}$ $^oC$ will burn so clearly that's not safe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q:&lt;/strong&gt; How can I calculate/find in order to justify the value of the maximum safe water temperature to operate the test rig.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1586" LastEditDate="2015-06-17T17:14:29.003" LastActivityDate="2015-06-17T17:14:29.003" Title="How does one define 'safe' in engineering?" Tags="|safety|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3153" PostTypeId="5" CreationDate="2015-06-12T05:46:27.173" Score="0" Body="" OwnerUserId="1293" LastEditorUserId="368" LastEditDate="2016-01-05T22:43:00.293" LastActivityDate="2016-01-05T22:43:00.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3154" PostTypeId="4" CreationDate="2015-06-12T05:46:27.173" Score="0" Body="Questions concerning problems which cannot be exactly solved, or where the exact solution may be much more difficult to acquire than a solution by numerical methods." OwnerUserId="1293" LastEditorUserId="368" LastEditDate="2016-01-05T22:43:00.293" LastActivityDate="2016-01-05T22:43:00.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3155" PostTypeId="2" ParentId="3152" CreationDate="2015-06-12T07:05:18.630" Score="5" Body="&lt;p&gt;For something to be safe there must be no risk of injury, harm, disability or death to anyone when something is being assembled, used or in the future as a result of the something being used. Where anyone is defined as any person - whether they be employee, government or military official or member of the public.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Local laws may qualify this by setting limitations. In the English style legal system, and variations of it, everyone has an obligation of due care to not cause harm, injury, disability or death to anyone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some legal jurisdictions will have codes, guidelines and best practice procedures that must be adhered to. All such codes and regulations are based on studies and investigations that have been conducted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When engineering a system the users of the system must be considered. A safe hot water temperature for small children and people metal disabilities will be lower than a that used exclusively by adults and people without mental disabilities because small children and people with mental disabilities may not have the presence of mind to recognize dangers like adults without mental disabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the US there are &lt;a href=&quot;http://inspectapedia.com/plumbing/Hot_Water_Temperature_Laws.php&quot;&gt;hot water temperature laws&lt;/a&gt;.&#xA;In Canada, the &lt;a href=&quot;https://canadasafetycouncil.org/home-safety/heated-debate-about-hot-water&quot;&gt;Canada Safety Council&lt;/a&gt; has information on its website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other sources of information as well:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://australiansolarandair.com.au/scalding.html&quot;&gt;Companies informing customers in Australia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://r.duckduckgo.com/l/?kh=-1&amp;amp;uddg=http%3A%2F%2Fwww.beama.org.uk%2Fdownload.cfm%2Fdocid%2F84B7267F-2635-410A-8AFBF3D46F889E5C&quot;&gt;Companies informing customers in the UK&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.plumbingengineer.com/content/what-are-safe-hot-water-temperatures&quot;&gt;Professional associations&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Standard associations for each country&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a quick guide, the table &lt;a href=&quot;http://www.australiansolarandair.com.au/scalding.html&quot;&gt;on this website&lt;/a&gt; lists the time required for hot water at various temperatures to cause a scald injury.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-06-12T07:05:18.630" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3156" PostTypeId="1" CreationDate="2015-06-12T11:11:57.997" Score="6" ViewCount="195" Body="&lt;p&gt;I wish to remove water and nitrogen from my reactor effluent in order to decrease downstream equipment sizes and subsequently capital costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stream composition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${H_2S}$.................5.6%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${SO_2}$.................3.3%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${S_2}$.................9.2% &lt;/p&gt;&#xA;&#xA;&lt;p&gt;${COS}$.................0.9%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${CS_2}$.................0.1%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${CO_2}$.................1.3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${H_2}$.................1.4%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;${H_2O}$.................34.5%&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;${N_2}$.................43.7%&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing these components would result in a approximately 90% less flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Restrictions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The sulphur &lt;strong&gt;MUST&lt;/strong&gt; stay above melting point, $\pm$130 $^\circ$C at 1 atm.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;ZERO&lt;/strong&gt; sulphur compounds must leave the water or nitrogen streams. (meaning that if these compounds are in the water or nitrogen streams, they have to be removed at a later stage as well.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I have done thus far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the Clausius-Clapeyron equation to solve for a pressure where the dew point of water is higher than the melting point of sulphur.  I found that at 3.5 atm of pressure and approximately 130 $^\circ$C, liquid water forms but sulphur stays liquid, and density separation can be performed with the added bonus that sulphur is insoluble in water.&#xA;However, I cannot think of way to remove the nitrogen or a cheaper,more efficient way to remove the water.&lt;/p&gt;&#xA;" OwnerUserId="1553" LastEditorUserId="272" LastEditDate="2015-06-12T11:56:15.110" LastActivityDate="2015-06-13T10:51:58.483" Title="Removing water and nitrogen from reactor effluent" Tags="|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3158" PostTypeId="2" ParentId="3152" CreationDate="2015-06-12T14:35:35.410" Score="5" Body="&lt;p&gt;A common refrain in these kind of discussions is &quot;there is no such thing as safe, only safer.&quot; Nothing we make is perfectly safe in an absolute manner - we don't design buildings to withstand asteroid impacts and we don't design cars to survive falling off a bridge. Instead, when we say safe, we usually mean &quot;safe enough&quot; - the level of risk is acceptable based on an understanding of all the ways we anticipate it could fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard part of this is defining the acceptable level of risk for a given situation. In many fields, a governing body has already defined the acceptable level of risk and written a code/standard/regulation that we get to follow. This saves everyday engineers and designers from having to spend a lot of time deciding what is and isn't acceptable. For people working in a field that doesn't have such standards, or people developing the standards, a risk assessment (however formal) is required to decide what risks can be tolerated. These assessments involve many factors, including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;What are the consequences of failure?&lt;/strong&gt; If your water is just warm enough to cause discomfort, you may be able to tolerate that risk. If it is so hot that it could cause permanent injury, your risk tolerance should be much lower.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;How likely is a failure?&lt;/strong&gt; If the water is in an open container that frequently splashes on people, you should keep it at a cooler temperature than if it's in an insulated tank and could only escape in a catastrophic equipment failure.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;What are the options for mitigating this risk?&lt;/strong&gt; If using room-temperature water just means that your experiment will take a little more time, it may be prudent to just not heat the water at all. If your experiment can only be done at high temperatures, then you might try to mitigate the risk by not allowing employees near the hot water, or requiring them to wear special insulating clothing.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So getting to a specific number depends on a variety of other aspects of your design. Certainly one conservative approach would be to follow whatever regulations are in place for tap water in your jurisdiction. Keeping the water in the neighborhood of 38 degrees celsius would be one reasonable option - about what is comfortable for people to touch. Of course if your process requires employees to keep their hands in the water for prolonged periods of time, even this may be too hot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if the water is piped and insulated and employees will not be exposed to it directly, you could consider following standards established for steam heat or perhaps industrial process piping if they more closely match your application. In broad strokes, these alternate guidelines will allow you to keep your water at higher temperatures in exchange for building more robust piping systems with additional safety features to reduce the frequency of ruptures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without more specifics of your application, system, and jurisdiction, we can't conclusively offer a specific number, but hopefully this gives you some pointers for finding the right value.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-06-12T14:35:35.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3159" PostTypeId="2" ParentId="3143" CreationDate="2015-06-12T15:59:15.267" Score="4" Body="&lt;p&gt;The answer is using a thermodynamic equation of state (EOS). An EOS is one that relates the free energy of a substance to its physical properties (temperature and density for example). Once such an equation is developed all other thermodynamic properties can be derived from them (via thermodynamic relationships). To see an example for water, checkout the The International Association for the Properties of Water and Steam website (&lt;a href=&quot;http://www.iapws.org/&quot; rel=&quot;nofollow&quot;&gt;http://www.iapws.org/&lt;/a&gt;). Here they have their latest release on water properties: &lt;a href=&quot;http://www.iapws.org/relguide/IAPWS-95.html&quot; rel=&quot;nofollow&quot;&gt;http://www.iapws.org/relguide/IAPWS-95.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now water is the most common and most used substance so the model is very detailed with a lot of parameters so as to match experimental conditions almost perfectly. Other substances are typically correlated with more general EOSs. A thermodynamics textbook would be a good place to learn more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, you are probably familiar with the most famous of these equations. The ideal gas equation of state is useful for the prediction of the thermodynamic properties for gases at low pressures.&lt;/p&gt;&#xA;" OwnerUserId="1031" LastActivityDate="2015-06-12T15:59:15.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3160" PostTypeId="2" ParentId="3149" CreationDate="2015-06-12T17:18:05.903" Score="1" Body="&lt;p&gt;Well that's a lot of light (and heat) for fixture rated for a 15W lamp. And of course, UVC light is not good for skin and eyes. Here is some &lt;a href=&quot;http://www.americanairandwater.com/lamps.htm&quot; rel=&quot;nofollow&quot;&gt;more info&lt;/a&gt; about UVC lamps.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Actually, no, the opposite. The lamp will attempt to draw more current from this ballast than it can provide, causing the ballast to get warmer than normal. Meanwhile, the lamp will not produce the full quantity of light it is supposed to, since it is not getting the full power it is rated for.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Furthermore, UVC lamps appear to have some &lt;a href=&quot;http://www.light-sources.com/?q=germicidal-uvc/products/electronic-ballasts-uvc-germicidal-lamps&quot; rel=&quot;nofollow&quot;&gt;unique qualities&lt;/a&gt; which may make them incompatible with traditional fluorescent fixtures. Namely, &lt;em&gt;&quot;[UVC Lamps] generate harmonic and transient voltages and currents. They exhibit a seemingly strange characteristic – the voltage decreases with increasing current. This process is known as negative AC resistance.  Therefore, gas discharge devices, such as germicidal lamps, are inherently unstable; once lit, the current increases without limit unless other circuit elements - ballasts - are used to restore and maintain operation.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;As long as a new lamp is a &quot;UVC&quot; type, rated for 15W, it may emit light just fine (but of course, be a &quot;smaller&quot; lamp, and produce less light.) However, if UVC lamp operation is incompatible with the ballast in the fixture (designed for traditional fluorescent lamps) then it might not work at all, or destroy itself, or the ballast.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that since UVC is a rather unusual light frequency, some materials may be incompatible with it, such as the plastic cover for a traditional fluorescent lamp. I'm not sure, but UVC light could damage this cover, or not pass through it at all. Note that most &lt;a href=&quot;http://www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Daps&amp;amp;field-keywords=t8+uvc+light+fixture&amp;amp;rh=i%3Aaps%2Ck%3At8+uvc+light+fixture&quot; rel=&quot;nofollow&quot;&gt;UVC fixtures&lt;/a&gt; seem to have no covers over the lamps at all. At 46W per lamp, they are going to get a lot warmer than 15W lamps also, so the added airflow may be necessary simply to cool them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advice is to get a lamp and fixture designed for the purpose you have in mind. It may be more expensive, but it will be more robust and operate correctly, without fear it may overheat or fail catastrophically.&lt;/p&gt;&#xA;" OwnerUserId="1673" LastActivityDate="2015-06-12T17:18:05.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3161" PostTypeId="2" ParentId="3141" CreationDate="2015-06-12T17:31:00.007" Score="0" Body="&lt;p&gt;You can initially consider the system to be $$\frac{\frac{0.25}{\epsilon}\epsilon}{s^2+0.5s+\epsilon}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives $\omega =\sqrt{\epsilon }$ and $\zeta =\frac{0.25}{\sqrt{\epsilon }}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the limit $\epsilon \to 0$, $\omega \to 0$ and $\zeta \to \infty$. From the latter we conclude that it is damped to the hilt.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2015-06-12T17:31:00.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3162" PostTypeId="2" ParentId="3110" CreationDate="2015-06-12T19:14:59.563" Score="3" Body="&lt;p&gt;(Warning: this answer is skewed towards static applications - dynamic applications like the self-weight of a moving cable or a vehicle are much more complicated.) In actual applications, self weight is usually significant and it's fairly rare that you can neglect it for a real-world system. However, there are a number of ways to address self-weight with varying degrees of accuracy depending on the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are only a few applications where self-weight can be written off completely as insignificant based on engineering judgement. These would be situations where the self weight is a very small portion of the total weight. One example that comes to mind would be an item hung from wire rope. The weight of the wire rope may well be 5 lbs for a load of 5,000 lbs. I would be comfortable ignoring self weight in that scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason that in academic exercises, it's tempting to ignore self-weight, is that it adds an iterative step to design. You size a beam for example to meet a loading requirement (and if you're clever, an estimate for the self-weight) but then once you pick the beam and know the true self-weight, you have to re-evaluate the structure. This is a cumbersome process, and especially when working towards a specific right answer as in a textbook problem, it can take many iterations to find the perfect beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When checking a design, it is necessary to validate it against your acceptance criteria (probably dictated by relevant codes) including the self-weight, but that doesn't mean that you have to follow this iterative process throughout the entire design. One option is to add an estimate of your self-weight to the other downward loads you are designing for, and process the entire design based on that increased loading criteria. Then at the end, you would add in the appropriate gravity loads, and return your other loads to their proper values for final checks. Because self-weight acts on specific members, rather than spread out evenly, it's important that your initial estimate is conservative. This method saves a lot of time when manually calculating things, but only if your self-weight estimate was conservative enough. If you can't afford to be modestly conservative in your design, then it's likely that your design won't pass the final checks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the era of computer-based structural modeling, of course, each iteration is nearly free, so there is little advantage to adding this extra step. Further complicating the discussion is the fact that when seismic loading governs, it is also influenced by the dead weight, so there is a second-order effect of any approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the short answer to your question is: it is almost never OK to completely ignore self-weight, but there are some tricks to reduce it's impact on the design process.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-06-12T19:14:59.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3163" PostTypeId="1" AcceptedAnswerId="3164" CreationDate="2015-06-12T19:51:41.537" Score="1" ViewCount="95" Body="&lt;p&gt;I need to install an actuators to LABVIEW base test system because the old actuators are defective. We cannot purchase the old actuators because they are obsolete. The old actuators communicated using the COM ports 1 and 8. The COM port are hard coded in the LABVIEW program. The new actuators communicate via a USB interface. But COM ports 1 &amp;amp; 8 not available. The systems states that they are &quot;in use&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LABVIEW based test systems is running on a Window 7 machine. Is there a way to point the new actuator USB interfaces to COM port 1 and 8? &lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="368" LastEditDate="2015-06-15T22:38:01.720" LastActivityDate="2017-02-02T20:09:28.057" Title="Unable to install Actuators to a LABVIEW based Engineering Test System" Tags="|software|labview|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3164" PostTypeId="2" ParentId="3163" CreationDate="2015-06-12T20:36:57.600" Score="1" Body="&lt;p&gt;What you might need to do is to remove the &quot;in use&quot; port and then assign the respective COM port. Follow is a suggested set of steps.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Open command prompt (Start &gt; Run) and type: &lt;strong&gt;cmd&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;li&gt;At the prompt type: &lt;strong&gt;set devmgr_show_nonpresent_devices=1&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Followed by C:&gt; devmgmt.msc   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t4r14.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5vToz.jpg&quot; alt=&quot;Administrator Window&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Next in the device manager view &gt; Show hidden files  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TU0Jd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BxWIG.jpg&quot; alt=&quot;Device Manager-1&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Expand COM by clicking on the &quot;+&quot; and uninstall the COM ports&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kwCG1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A8S48.jpg&quot; alt=&quot;Device Manager-2&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;From device manager select the COM port that the device is currently attach  to and right click   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Okm4J.jpg&quot; alt=&quot;COM Port Configuration&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Select Properties -&gt; Port Setting -&gt; Advance -&gt; Assign the desired  COM port from the drop down  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bg4gM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FDT4I.jpg&quot; alt=&quot;Advance Setting for COM port&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know how it works out for you&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="282" LastEditDate="2015-06-30T21:44:30.877" LastActivityDate="2015-06-30T21:44:30.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3165" PostTypeId="1" AcceptedAnswerId="3167" CreationDate="2015-06-12T20:48:46.163" Score="1" ViewCount="701" Body="&lt;p&gt;I have been learning the program embedded systems using Keil µVision 4.7 free version from Mentor graphic. So far all the learning has be using projects that have been already pre-defined. This has worked well so far. I would like to take the next step. Below are some of the several tutorial that I have tried with out much success. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.keil.com/support/man/docs/uv4/uv4_creating_apps.htm&quot; rel=&quot;nofollow&quot;&gt;Creating Applications using Keil µVision IDE from Mentor Graphic&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=Z_BwDGf7FTs&quot; rel=&quot;nofollow&quot;&gt;Create a new project using Keil uVision4 for STM32 (tutorial)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.circuitstoday.com/getting-started-with-keil-uvision&quot; rel=&quot;nofollow&quot;&gt;Getting Started with Keil uVision&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What the minimum step to create a new an Embedded System project using Kiel µVision (Including defining the micro controller, and even setting up the debugger) so issues describe below are avoided?&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="282" LastEditDate="2015-07-06T22:17:52.350" LastActivityDate="2015-07-06T22:17:52.350" Title="How to create Embedded System Program from Scratch using Keil µVision?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="1" ClosedDate="2015-06-15T17:02:18.557" ContentLicense="CC BY-SA 3.0" />
  <row Id="3166" PostTypeId="2" ParentId="3073" CreationDate="2015-06-12T23:51:16.440" Score="4" Body="&lt;p&gt;There are &lt;a href=&quot;https://www.kaltire.com/whats-difference-directional-asymmetrical-symmetrical-tread-patterns/&quot; rel=&quot;nofollow&quot;&gt;Tread pattern differences&lt;/a&gt;, which allows part of the tire to make good contact with the ground for performance (grip). Part of the tire with wide channels to evacuate water (typically called ribs), and sipes so the tire wears properly, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An Asymmetrical tire is often referred to in the tire world as an inside outside tire, where a directional tires typically have a V pattern with the v point toward the hood of the car, and the 2 ends facing the rear. To answer the question of rubber on the surface or thread of the tire being different, this is incorrect it remains the same across the thread, while the rubber used for the sidewall in some cases is different. &lt;a href=&quot;http://swampertires.blogspot.com/&quot; rel=&quot;nofollow&quot;&gt;This describes the different terms typically used.&lt;/a&gt; while if you search for Goodyear you can find a tire with Kevlar in it which is stronger side wall. The inner side of the tire leads me to believe you might have a worn suspension part in the front causing premature tire wear, or that the alignment of the vehicle is off, I would post more links however I do not have the rep it appears.&lt;/p&gt;&#xA;" OwnerUserId="1973" LastEditorUserId="33" LastEditDate="2015-06-13T01:26:08.567" LastActivityDate="2015-06-13T01:26:08.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3167" PostTypeId="2" ParentId="3165" CreationDate="2015-06-13T00:10:17.570" Score="1" Body="&lt;p&gt;Here are some steps for a ARM Cortex-M4&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Close your current project in keil µVision, menu: projects-&gt;close project&lt;/li&gt;&#xA;&lt;li&gt;Select &quot;New uVision Project...&quot; from &quot;Project&quot; menu&#xA;&lt;img src=&quot;https://i.stack.imgur.com/q7ilH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;open an empty new one, menu: projects-&gt;create a new µVision project&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ChA09.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;Select a Target processor&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qufIE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;You get a popup that asks if you want to copy the startup assembler file into your project startup startup_xxxx.s.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yNVps.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;now you see the empty project with just the startup file included.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/YQ0yB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/li&gt;&#xA;&lt;li&gt;Project is ready for embedded software development. It is good idea to rename &quot;Target 1&quot; and &quot;Source Group 1&quot; that describes the project&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Right click on Target 1 &gt; Select Manage Project Items -&gt; Update Project Targets, Groups and Files&#xA;&lt;img src=&quot;https://i.stack.imgur.com/n2xdW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;8&quot;&gt;&#xA;&lt;li&gt;In order to startup our micro-controller, we need to add 'system_.c' file to our project. Keil already has this file defined. To add manually, Right-click on our &quot;Startup&quot; group and select &quot;Add Existing Files to Group 'Startup'...&quot;:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zhp4H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;a.  Keil already have common startup files for some microcontrollers. They are stored at /Keil/ARM/Startup folder&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/z8RX9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;b.  From there, for my LaunchPad I will move to TI folder, and then TM4C123 folder:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/NduA0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;c.  And there i can see my system_TM4C123.c file:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/DXjmK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;now you need to add the copied file into the project. project-&gt;manage-&gt;components,.. or simply click the icon with the three coloured boxes&#xA;add there the startup_TM4C123.c file. you should then see it listed on the left window bar.&lt;/li&gt;&#xA;&lt;li&gt;now create your xxx.c file containing the main procedure.&#xA; you need to add it to the project again, like you did in the previous step.&lt;/li&gt;&#xA;&lt;li&gt;and now you need to set some registers in the controller. for now you can simply do that at the beginning of the main procedure.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note: Most the screen shots were created by screen name vasily.sib on piazza forum. I copied these improved them for my purpose.&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-06-15T23:31:33.510" LastActivityDate="2015-06-15T23:31:33.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3170" PostTypeId="2" ParentId="3143" CreationDate="2015-06-13T10:28:08.043" Score="0" Body="&lt;p&gt;If you assume the gas you are dealing with is a semi-perfect gas (e.g. the heat capacity is not a function of pressure, and the gas follows ideal gas law), then you can calculate all properties (in the order of enthalpy, internal energy, entropy and gibbs free energy) if you can measure the specific heat capacity at constant pressure (cp, sometimes, a fitted polynomial) base on the basic relations. E.g.&#xA;$dh=cpdT$, $u=h-RT$, $ds=(dh-vdp)/T$, etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steam table I believe is calculate from EOS as suggested by @sturgman. The vapor liquid saturation line can be deduced (though not very accurately) using equal area rule if the EOS is cubic.&lt;/p&gt;&#xA;" OwnerUserId="1975" LastActivityDate="2015-06-13T10:28:08.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3171" PostTypeId="2" ParentId="3156" CreationDate="2015-06-13T10:51:58.483" Score="2" Body="&lt;p&gt;Due to my lack of reputation I can only answer instead of commenting which I would have prefered. Thus my explanation will rather cover some basic ideas and suggestions rather than an elaborate separation process. Also you should not I have not graduated yet and I'm eager to learn myself. So instead of downvoting I would like to receive constructive comments on what I got wrong or what I could improve. Mainly my point is to get a discussion started and see if we could use some ideas to help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What do we start with?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I see you already have a two phase system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vapor Phase&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$H_2S$&lt;/li&gt;&#xA;&lt;li&gt;$SO_2$&lt;/li&gt;&#xA;&lt;li&gt;$COS$&lt;/li&gt;&#xA;&lt;li&gt;$CS_2$&lt;/li&gt;&#xA;&lt;li&gt;$H_2$&lt;/li&gt;&#xA;&lt;li&gt;$N_2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Liquid Phase&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S_2$&lt;/li&gt;&#xA;&lt;li&gt;$H_2O$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At your given Temperature and Pressure. Please note that I did not have access to all VLE data of all your components, you will have to check that for varying pressures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1st Separation Step&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would start with separating those two phases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2nd Separation Step&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_2$ has the lowest boiling point of all components in your vapor phase. Except for $H_2$. You could lower the temperature  to -60,2 °C and then $H_2$ and $N_2$ would stay in vapor state, all other components would be liquified. This would allow for an easy phase separation. Sidenote: I'm no chemist however  $H_2$ and $N_2$  react to ammonium, which has a higher boiling point. However you did not consider reactions of the components before so I just assume this might not be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3rd Separation Step&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you pointed out $S_2$ and $H_2O$ have a significant difference in density. So you could use a mixer/settler setup e.g. to separate them. You would then have to combine the $S_2$ stream with your other $N_2$ free stream. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't really see yet, is how you get your initial reactor effluent. From what I know there is no way $S_2$ would be in gasous state unless you operate under very high temperaturs or under a vacuum. In both cases that would have been a crucial information here to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answer helps you a little bit. &lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2015-06-13T10:51:58.483" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3173" PostTypeId="2" ParentId="3150" CreationDate="2015-06-13T16:14:24.347" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;... that article only lists the surface resistance for 430 &amp;amp; 432 stainless steel -- the better being 432. However, I cannot find 432 from my favorite supplier, although they do have 430. Will 430 have a similar surface resistance?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;br&gt;&#xA;430 and 432 are listed as having almost identical resistivities - see below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface resistivity by itself is not an adequate figure of merit.&lt;br&gt;&#xA;Inductive heating is a function of AC skin depth and resistivity. At lower frequencies skin depth increases and higher resistivity materials are required to produce the desired load. As frequency rises skin depth decreases and total material resistance increases due to less depth x area volume and lower resistivity materials are better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recommendations to use stainless steels is related to the typical frequencies of current induction cookers. This is mainly limited by the economics of currently available high power electronic drivers. As driver costs fall with time frequencies will be able to be increased, reducing skin depth and allowing lower resistivity metals to be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;430 and 432 are listed as having almost identical resistivities - see below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Care should be taken to distinguish between resistIVITY and resistANCE.       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ResistIVITY is a property of the material and expresses the resistance between opposite faces of a unit cube of material but does not tell you actual resistance that results when dimensions are other than a unit cube.&lt;br&gt;&#xA;Resistivities are usually expressed either in microOhm.inch or microOhm.centimetre.   I'll call these mOin and mOcm respectively.    &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The reason for the units of Ohms x length is that, starting with a unit cube resistance increases with increasing thickness and decreases with increasing face area. So R for some other shape = Resistivity x t/A Ohms.&lt;br&gt;&#xA;To have units of Ohms, resistivity must cancel the t/A, &#xA;so units of resistivity = Ohms x A/t = Ohms x length.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;ResistANCE is the electrical resistance to current flow in Ohms of the actual heating material used in the shape/dimensions used. The pot base offers Ohmic resistance to induced current as a result of material resistance for a slab the shape of the base and a depth controlled by the skin depth at the frequency used.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Resistivities are usually expressed either in microOhm.inch or microOhm.centimetre.   I'll call these mOin and mOcm respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 mOin = 2.54 mOcm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Links to many Stainless steel data sheets &lt;a href=&quot;http://www.aksteel.com/markets_products/stainless.aspx&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;430 - 24 micro-Ohm.Inch (mOi)&#xA;From &lt;a href=&quot;http://www.aksteel.com/pdf/markets_products/stainless/ferritic/430_Stainless_Steel_DS_201503.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;430 ultraform (added Titanium) -  also 24 mOin (so suspect)&lt;br&gt;&#xA;From &lt;a href=&quot;http://www.aksteel.com/pdf/markets_products/stainless/ferritic/AK%20430%20ULTRA%20FORM%C2%AE%20Stainless%20Steel%2020130617.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;432 - 24.5 mOin&lt;br&gt;&#xA;304 - 29 mOin&lt;br&gt;&#xA;Both from your wikipedia link &lt;a href=&quot;http://www.aksteel.com/markets_products/stainless.aspx&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;304 - 69 &amp;amp; 72 mOcm  (about 28 mOin)&lt;br&gt;&#xA;347 - 72 mOcm  (~= 28 mOin)&lt;br&gt;&#xA;316 - 75 mOcm  (~= 30 mOin)&lt;br&gt;&#xA;All from  &lt;a href=&quot;http://hypertextbook.com/facts/2006/UmranUgur.shtml&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;br&gt;&#xA;(suspiciously) Identical figures &lt;a href=&quot;http://eddy-current.com/conductivity-of-metals-sorted-by-resistivity/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Skin effect &amp;amp; skin-depth:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Skin_effect&quot; rel=&quot;nofollow&quot;&gt;Skin effect - Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rfcafe.com/references/electrical/skin-depth.htm&quot; rel=&quot;nofollow&quot;&gt;Skin deth formula &amp;amp; notes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://chemandy.com/calculators/skin-effect-calculator.htm&quot; rel=&quot;nofollow&quot;&gt;Skin effect depth calculator&lt;/a&gt; - does not have an entry for stainless steel - but see simple associated formula. Skin depth scales with 1/(sqrt(relative permeability))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depth ~= Sqrt(Resistivity/(Pi x Freq x absolute permitivity)) see above reference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ecee.colorado.edu/~ecen3400/Chapter%2020%20-%20The%20Skin%20Effect.pdf&quot; rel=&quot;nofollow&quot;&gt;Textbook chapter&lt;/a&gt; 7 page pdf.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-06-14T08:40:32.277" LastActivityDate="2015-06-14T08:40:32.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3174" PostTypeId="1" AcceptedAnswerId="3179" CreationDate="2015-06-13T22:07:23.317" Score="9" ViewCount="151887" Body="&lt;p&gt;So i had this experience when driving in a cold night, where my windshield just got all blurry. I knew that on the drivers manual they say you should turn the hot air and soon it will go off, but my friend that was on the passenger seat said the cold air will work too. We turned the AC off and re-experienced the fact. I got the chronometer and and got the time of both situations, defog with hot and cold air and there were really small diference btween them. How can i explain this in terms of thermodynamics? &lt;/p&gt;&#xA;" OwnerUserId="1853" LastEditorDisplayName="user16" LastEditDate="2015-06-13T23:15:28.140" LastActivityDate="2018-04-14T14:41:36.033" Title="Which is better: Defog a car windshield with cold or hot air?" Tags="|thermodynamics|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3175" PostTypeId="2" ParentId="3174" CreationDate="2015-06-14T02:53:44.107" Score="4" Body="&lt;p&gt;My experience over many years has been similar, except I find, the cold air from the air conditioner directed to the windscreen demisted the windscreen faster than hot air similarly directed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently it is winter were I am. On cold humid nights when the car heater is turned on, the &lt;strong&gt;inside&lt;/strong&gt; of the windows mist/fog up. By turning on the air conditioner while the heater is on allows for the car interior to be warm and for the windows to be mist/fog free, particularly if air directed to the windscreen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While living in a tropical climate I had a similar experience, but in reverse; warm humid air, turn on the air conditioner and the &lt;strong&gt;outside&lt;/strong&gt; of the windscreen misted/fogged up. That was easily fixed by clearing it with the wiper blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all due to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Dew_point&quot; rel=&quot;nofollow&quot;&gt;dew point&lt;/a&gt; of the air and water vapour mixture in contact with the car windows. “The dew point is the temperature at which water vapour condenses into liquid at the same rate at which it evaporates”. Sufficiently change the temperature of the air in contact with the windows by either heating it or cooling it and the dew point changes. Water vapour, from the air no long condenses on the windows and the windows do not mist/fog up and what mist/fog condensed out onto the windows evaporates back into the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens on car windows because they are the only part of the car which are in contact with the interior air of the car and the exterior atmospheric air. They are the barriers between the two. Car roofs have lining which acts as insulation and another barrier. Car doors and pillars have air gaps in them which acts as another form of weak insulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to which is the better way to demist car windows,it depends on your circumstances at the time and whether you want to be cold or warm at the same time.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-06-14T02:53:44.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3176" PostTypeId="1" CreationDate="2015-06-14T06:04:45.820" Score="4" ViewCount="207" Body="&lt;p&gt;What type of mechanism can I use to connect 2 coaxial pipes of unequal diameters, where the outer pipe can slide over the inner pipe? I was thinking about splines, is there any better way...?&#xA;Also, if I want the connection to be leakproof, what are the options that I have ?&lt;/p&gt;&#xA;" OwnerUserId="1980" LastEditorUserId="1980" LastEditDate="2015-06-14T06:58:12.143" LastActivityDate="2021-06-11T05:02:30.207" Title="Connection for 2 sliding coaxial pipes" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3179" PostTypeId="2" ParentId="3174" CreationDate="2015-06-14T12:30:44.437" Score="12" Body="&lt;p&gt;The reason the cooled air appears to work about as well as the heated air is that the cooled air is also de-humidified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the fastest de-misting, you want warm and dry air.  Car owner manuals often tell you to turn on the air conditioner and the defroster at the same time in these conditions.  The air conditioner cools the air, which forces it to dump much of whatever moisture it contained.  The heater then warms the air again, but without adding any moisture back.  The result has high capacity for obsorbing more moisture, so quickly removes the condensation from the inside of the windshield.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-06-14T12:30:44.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3180" PostTypeId="1" CreationDate="2015-06-14T19:59:16.040" Score="9" ViewCount="1929" Body="&lt;p&gt;If I understand correctly, both air conditioners and dehumidifiers conventionally exhaust waste heat through their exhaust air, though both also produce exhaust water. I'd like to know,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are air conditioners or dehumidifiers currently being produced that exhaust heat through their exhaust &lt;em&gt;water&lt;/em&gt;? &lt;/li&gt;&#xA;&lt;li&gt;If not, are there engineering or physical reasons why this is not feasible?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Use case: I currently live where the weather is regularly both hot and humid, and my current dwelling is configured such that my only feasible options for reducing internal heat and humidity are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;an energy-inefficient external air conditioner (located outside of, but connected to, my dwelling) that also seems inefficient at humidity reduction;&lt;/li&gt;&#xA;&lt;li&gt;an energy-efficient internal dehumidifier that exhausts&#xA;&lt;ul&gt;&#xA;&lt;li&gt;waste heat into the dwelling&lt;/li&gt;&#xA;&lt;li&gt;waste water into an open reservoir (which I filter for drinking).&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What I'd like would be an internal unit like my dehumidifier, but which would instead:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;use its wastewater for cooling;&lt;/li&gt;&#xA;&lt;li&gt;exhaust that warm water into a closed, insulated reservoir;&lt;/li&gt;&#xA;&lt;li&gt;{sound alarm, shutoff} when the reservoir filled (as it does now) or overheated.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would then remove and empty the reservoir outside.&lt;/p&gt;&#xA;" OwnerUserId="1984" LastEditorUserId="368" LastEditDate="2015-06-15T17:23:28.570" LastActivityDate="2015-06-30T20:46:35.833" Title="Can air conditioners/dehumidifiers exhaust their waste heat in their exhaust water?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|hvac|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3181" PostTypeId="1" CreationDate="2015-06-14T20:56:56.063" Score="0" ViewCount="6148" Body="&lt;p&gt;I recently signed a lease for a high-rise an apartment unit that directly overlooks the rooftop of lower floors.  That is, my apartment is on a smaller floor plate than the units below it, and I could theoretically open the window and walk out onto the roof of the lower floors.  Unfortunately, I don't have any pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roof has several small vents that look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/5lqyG.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And some larger ones sort of like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QQuyFm.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't remember what kinds of pipes I saw, if any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I have to worry about the air coming from the roof vents?  Would there be any bathroom/sewage ventilation on a rooftop like this?&lt;/p&gt;&#xA;" OwnerUserId="1985" LastEditorUserId="368" LastEditDate="2015-06-15T17:40:00.257" LastActivityDate="2017-03-09T23:04:02.120" Title="What type of exhaust is vented from the rooftop of this apartment?" Tags="|hvac|building-design|drain-waste-vent|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3182" PostTypeId="2" ParentId="3181" CreationDate="2015-06-14T21:50:10.753" Score="0" Body="&lt;p&gt;Found an answer in the plumbing code:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;906.2&lt;/strong&gt; Each vent shall terminate not less than ten (10) feet (3,048 mm) from,&#xA;  or not less than three (3) feet (914 mm) above, an openable window,&#xA;  door, opening, air intake, or vent shaft, or not less than three (3)&#xA;  feet (914 mm) in every direction from any lot line, alley and street&#xA;  excepted.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1985" LastActivityDate="2015-06-14T21:50:10.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3183" PostTypeId="1" AcceptedAnswerId="3216" CreationDate="2015-06-14T23:45:36.407" Score="8" ViewCount="10235" Body="&lt;p&gt;I need to find the emission power of a WiFi mobile phone antenna. I have checked the IEEE 802.11 specifications but it says that the power transmission is variable and depends on the mobile phone fabricant and power saving options. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also searched for information about some specific mobile phones but the datasheets available on the Internet seem quite weak and do not say anything about WiFi power transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a complete datasheet of mobile phones? How can I determine the transmission power in the WiFi antenna of a specific mobile phone? &lt;/p&gt;&#xA;" OwnerUserId="1316" LastEditorUserId="282" LastEditDate="2015-06-15T22:17:56.997" LastActivityDate="2019-04-14T19:03:59.860" Title="How can I find the emission power of a WiFi mobile phone antenna?" Tags="|electrical-engineering|power-electronics|wireless-communication|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3184" PostTypeId="1" AcceptedAnswerId="3188" CreationDate="2015-06-15T09:15:38.480" Score="3" ViewCount="1103" Body="&lt;p&gt;I have to put together a small solar system for a college project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have solar cells with an output of 3.65A and 0.6V. Now if I paralel two groups of two cells connected in series I will double the amps and volts. And the output power will be like 8W. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, do a 1.2V panel really work? Can I charge a 12V battery? I mean the panel will be 8W and high amperage low voltage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to have a minimum 12V panel to charge battery? This means I will have to put in series 20 cells to obtain 12V. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there high voltage low amperage solar cells so I don't need to put together so many cells? &lt;/p&gt;&#xA;" OwnerUserId="1989" LastEditorUserId="282" LastEditDate="2015-06-16T01:01:53.400" LastActivityDate="2015-06-16T01:01:53.400" Title="8W Solar panel high amperage low voltage" Tags="|electrical-engineering|solar-panel|photovoltaics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3185" PostTypeId="2" ParentId="3180" CreationDate="2015-06-15T09:43:36.927" Score="1" Body="&lt;p&gt;I can provide some information about a solution that you would have to build / modify yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what a quick research yielded I understand that you cannot use the wastewater for drinking directly. Feel free to look up greywater. As you pointed out, you want to filter it first. I strongly recommend to use a biomembrane-based filter and make sure that the filter will generate potable water. You might want to read &lt;a href=&quot;http://grist.org/green-living-tips/ask-umbra-can-i-reuse-the-water-from-my-dehumidifier/&quot; rel=&quot;nofollow&quot;&gt;this Q&amp;amp;A from grist.org about reusing water from a dehumidifier&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second point I want to address is the use of your the heat generated and the cold water. Air to water heat transfer is not very feasible since the heat transfer is rather poor. So firstly I would see it's not very efficient to try and heat the condensed water with hot air. Which would make up my next point, you wouldn't see a great effect in using the condensed water for cooling anything. You shouldn't be able to significantly chill your rooms with it. What I could think of very quickly is that you could cool water with it, although the efficiency here wouldn't be astonishing either. Given a certain $\Delta T$ of 10–20 K you would need for heat transfer to take place and given the fact that you would operate with condensed water of a temperature around 5 °C you would be able to cool something down to 15 °C best case scenario. Maybe 10 °C. This is not significantly lower the the cold tap water I get around here so I guess it's not worth the trouble.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="368" LastEditDate="2015-06-15T17:20:04.620" LastActivityDate="2015-06-15T17:20:04.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3187" PostTypeId="1" AcceptedAnswerId="3191" CreationDate="2015-06-15T10:35:18.380" Score="5" ViewCount="2651" Body="&lt;p&gt;I have &lt;a href=&quot;http://www.allpumpsdirect.co.uk/files/1806869/uploaded/FLJ%20Duplex%202%20Tech%20Data.pdf&quot; rel=&quot;nofollow&quot;&gt;specifications for a positive displacement pump&lt;/a&gt; where, in the data table on the last page, the given flow rate is referred to in the column header as 'Open Flow.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the maximum flow generated by the pump?&lt;/p&gt;&#xA;" OwnerUserId="1992" LastEditorUserId="368" LastEditDate="2015-06-15T15:53:58.283" LastActivityDate="2015-06-15T15:59:49.850" Title="What does 'Open Flow' mean for a pump flow rate?" Tags="|mechanical-engineering|fluid-mechanics|pumps|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3188" PostTypeId="2" ParentId="3184" CreationDate="2015-06-15T11:18:53.213" Score="4" Body="&lt;p&gt;If you want to charge a 12 V battery, you need a bit more than 12 V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to achieve that is to put enough individual cells in series.  Another way is to have the panel produce a lower voltage and use a electronic circuit called a &lt;i&gt;boost converter&lt;/i&gt;.  These convert low voltage at high current to high voltage at low current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're clever, you can build in optimal use of the solar array and battery charge management into the boost converter.  However, if the panel can only put out 8 W, which is 670 mA at 12 V, then you can just size the battery so that it can take 670 mA indefinitely without harm.  Or, if this is a lead-acid battery, just have the boost converter produce 13.6 V when it can, but never more.  A &quot;12V&quot; car batter, for example, can take that indefinitely.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-06-15T11:18:53.213" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3190" PostTypeId="2" ParentId="3187" CreationDate="2015-06-15T11:51:18.137" Score="3" Body="&lt;p&gt;Pump flow rates are usually useless without the curves given in the diagrams. The flow rate and the pressure are linked. You will have to find the operating point for your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you know the pressure you want to achieve on the pressure side (e.g., hydrostatic pressure) and you know what pressure drop occurs in your pipe. With this information you can the select a pump that achieves that pressure at your desired flow rate. Open flow is the flow rate the pump achieves with a pressure of 0 on the pressure side or, in other words, with a pressure head of 0 m. I think the latter is more comprehensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take Curve 5 from &lt;a href=&quot;http://www.allpumpsdirect.co.uk/files/1806869/uploaded/FLJ%20Duplex%202%20Tech%20Data.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;your data sheet&lt;/a&gt;, for example; say you want to pump a fluid (water) from reservoir 1 to reservoir 2 and this results in a geodetical height difference of 16 m. In the case of water, 16 m would approximately result in 1.6 bar pressure. This is your static pressure drop. You can then add the dynamic pressure drop resulting from the fluid travelling through your pipes. This is a function of the flow rate. With those information the purple curve can be calculated and then you can see your operating point (blue). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uv8Vw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uv8Vwm.jpg&quot; alt=&quot;curve 5&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some ways to move the operationg point as you wish, e.g., using a bypass.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="368" LastEditDate="2015-06-15T15:59:49.850" LastActivityDate="2015-06-15T15:59:49.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3191" PostTypeId="2" ParentId="3187" CreationDate="2015-06-15T12:03:03.050" Score="4" Body="&lt;p&gt;The open flow is the flow rate through the pump where no pressure is developed. That is, if you have fluid coming in and out horizontally, and the input and output hoses are open to the outside air (not a closed system developing pressure. This is the most flow the pump will ever create.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually though, we use pumps to either lift fluids vertically, or to pressurize a closed system. In order to do either of these, you need a higher pressure on the outlet side of the motor than on the inlet side. In these cases, the pump will run slower as it has to exert more pressure. The curves on the second page of your link show you for each bar of pressure you need to develop, how much the flow rate will be reduced. In the case of a pressure piping system, you'll know the pressure rise you need directly. If you're using the pump to push fluid uphill, the pressure will depend on the vertical rise (usually referred to as 'head') and the density of the fluid. There will also be frictional losses which may be worth accounting for depending on the length of your piping system. For water, each meter of head is roughly .1 bar, closer to .09 for oil.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-06-15T12:03:03.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3192" PostTypeId="1" CreationDate="2015-06-15T12:14:47.110" Score="2" ViewCount="174" Body="&lt;p&gt;I'm buffing the years of built up dirt off a aluminum grain tipper trailer, using a woolen buffing pad mounted on a angle grinder. I'm going in a upwards &amp;amp; downwards motion with great success, however, that's the pattern it's leaving. 1st I thought I'll finish by doing a 2nd, lighter buff or polish in a sideways left/right direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I've remembered advice of cleaning a motorbike frame, to only work in one direction (probably with the frame in the case of a motorbike).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I buff and polish aluminum to get the best finish, without leaving a pattern? (Directional buffing/polishing in particular.)&lt;/p&gt;&#xA;" OwnerUserId="1995" LastEditorUserId="368" LastEditDate="2015-06-15T16:11:18.150" LastActivityDate="2015-09-15T08:06:08.597" Title="Buffing/Polishing bulk Aluminum" Tags="|mechanical-engineering|aluminum|surface-preparation|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3193" PostTypeId="2" ParentId="3180" CreationDate="2015-06-15T13:24:55.963" Score="0" Body="&lt;p&gt;What you could do is place a heat exchanger in the (warm) exhaust air flow, then run your (cool) waste water through it before discharging the water outside.  I'm envisioning a recycled radiator from a car, with a small pump to get the water up from the dehumidifier into the upper opening of the radiator.  You definitely need to dump the water outside though, since it will slowly leak heat back into your house if you store it inside.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-06-15T13:24:55.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3194" PostTypeId="1" CreationDate="2015-06-15T15:51:39.063" Score="6" ViewCount="883" Body="&lt;p&gt;In my mechanical engineering training , I've noticed that many constitutive relations (e.g. fourier law, poiseuille law, hooke's law) involve a flux which increases as the material parameter increases.  In heat transfer and fluid flow, this is called a &quot;conductivity&quot;.  In solid mechanics, the stress increases as some elastic moduli increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this trend, I found it perplexing that in electrical engineering the preference is to use resistance rather than conductance, with current increasing as resistance decreases in ohm's law.  Is there an advantage to using resistance instead of conductivity in electrical engineering in general?  Or is it just traditional notation handed down the great minds who came (and published) before us?&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2015-06-26T13:08:05.463" Title="Why do we prefer resistance over conductivity in describing Ohm's law?" Tags="|electrical-engineering|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3195" PostTypeId="1" CreationDate="2015-06-15T15:56:35.787" Score="4" ViewCount="6341" Body="&lt;p&gt;I need to know the difference in seawater density at 28 and 98 degrees Fahrenheit for control system synthesis. How can I calculate the density for a given temperature?&lt;/p&gt;&#xA;" OwnerUserId="920" LastEditorUserId="368" LastEditDate="2015-06-15T16:16:17.203" LastActivityDate="2015-06-15T20:47:39.777" Title="How do I calculate the density of seawater at different temperatures?" Tags="|control-engineering|water-resources|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3196" PostTypeId="1" CreationDate="2015-06-15T16:01:28.183" Score="2" ViewCount="1750" Body="&lt;p&gt;What forces are relevant when it comes to a davit lifting arm with a support? I'm not sure how to take the weight of the davit into account or how to calculate the max stress/strain on the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The davit is set on a beam and it lifts a 979 lb lid off of a tank. There is a support coming from the vertical piece about three-fourths of the way up that reaches over to the horizontal piece about half way. &lt;/p&gt;&#xA;" OwnerUserId="1997" LastEditorUserId="110" LastEditDate="2015-06-15T23:45:32.560" LastActivityDate="2019-12-28T18:28:41.173" Title="How to rate a davit beam?" Tags="|structural-engineering|steel|beam|applied-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3197" PostTypeId="2" ParentId="3195" CreationDate="2015-06-15T16:06:05.933" Score="6" Body="&lt;p&gt;The density depends not only on temperature, but also on salinity, see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Seawater&quot; rel=&quot;noreferrer&quot;&gt;wikipedia page on seawater&lt;/a&gt;, which has this neat diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rIoFu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you have seawater of constant salinity, you can use the above graph to get your answer by drawing a vertical line at the corresponding salinity value. The intersections of that line with the lines of equal density will give you your data points of density vs. temperature. Read the value of the density from the line you are intersecting and find the corresponding y-value to get the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have your data points of density vs. temperature, you can then graph it and interpolate it for the desired temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, assuming a constant salinity of 35 PSU, I get the following data points:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;density_35psu = [1.028 1.0275 1.027 1.0265 1.026 1.0255 1.025];&#xA;temp_35psu = [2 6.5 9.5 12.4 15.4 17.3 19.55];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which gives the following graph (I have fitted a second-order polynomial and extrapolated to the temperatures of interest):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yYaoL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gives me a seawater density of 1.0284 g/cm^3 at 28°F and 1.0197 g/cm^3 at 98°F.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2015-06-15T20:47:39.777" LastActivityDate="2015-06-15T20:47:39.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3198" PostTypeId="2" ParentId="3195" CreationDate="2015-06-15T16:27:32.203" Score="3" Body="&lt;p&gt;You could use the Equation of State for seawater. Sources and detailed information on this topic are to be found &lt;a href=&quot;http://www.ocean-sci.net/5/91/2009/os-5-91-2009.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://www-pord.ucsd.edu/~ltalley/sio210/readings/gill_appendix3_ppsw.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://tos.org/oceanography/archive/23-3_millero.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that the equation apparently does not cover temperatures below freezing point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Equation of State for Seawater&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(\rho - \rho^0) = AS_A + BS^{1.5}_A + CS^2_A$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho^0$ is the densitiy of pure water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = a_0 + a_1T + a_2T^2 + a_3T^3 + a_4T^4 + a_5T^5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B = b_0 + b_1T+ b_2T^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C=c_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Units&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[ \rho ] = \frac{kg}{m^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$[ T ] = °C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimentally-determined parameters for calculating $A$ and $B$ can be taken from Table 2 in &lt;a href=&quot;http://www.ocean-sci.net/5/91/2009/os-5-91-2009.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Ocean Science paper&lt;/a&gt; (&lt;em&gt;F. J. Millero and F. Huang&#xA;, 2009)&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qQPa3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table shown here is found in the &lt;a href=&quot;http://www.ocean-sci.net/6/379/2010/os-6-379-2010.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;corrigendum&lt;/a&gt; to the linked paper, not the paper itself. &lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2015-06-15T17:32:10.923" LastActivityDate="2015-06-15T17:32:10.923" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3199" PostTypeId="5" CreationDate="2015-06-15T17:43:35.237" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-06-15T17:43:35.237" LastActivityDate="2015-06-15T17:43:35.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3200" PostTypeId="4" CreationDate="2015-06-15T17:43:35.237" Score="0" Body="Questions about the drainage and ventilation portions of plumbing systems, also known as the DWV system of a building." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-06-15T17:43:35.237" LastActivityDate="2015-06-15T17:43:35.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3201" PostTypeId="2" ParentId="3181" CreationDate="2015-06-15T18:07:24.047" Score="1" Body="&lt;p&gt;Modern waste vents are usually bare plastic pipes extending a short distance up through the roof, with no cap. Compare the vents in this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GjQdV.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Judging by the apparent age of the neighborhood, the waste vent on the left is probably nothing more than a stub of &lt;a href=&quot;https://en.wikipedia.org/wiki/Acrylonitrile_butadiene_styrene&quot; rel=&quot;nofollow noreferrer&quot;&gt;ABS&lt;/a&gt; pipe. PVC (white) is also sometimes used. Older homes can have metal (copper, cast iron) or even clay vent pipes, but they're no more complex, just a different material. These pipes drop straight down into your drain system, so there's no harm done by rainwater or small debris falling into the vent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, vents for appliances are designed specifically to protect the system from any ingress of water, dirt, debris, small animals, and so on. They're also usually attached to larger-diameter ducts, as opposed to smaller-diameter vent pipes, because they need to allow for more airflow. Sewer ventilation is very low-flow, it's just preventing the gradual buildup of gases produced through natural processes, so it doesn't need a large pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vent in the middle is larger than the waste vent because a furnace or boiler puts out combustion byproducts at a much higher rate than decomposition produces sewer gas. You can see that the attic fan on the right has the largest duct by far, and that makes sense because it's attached to a system that's designed specifically to move large volumes of air through the home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also reasons to be particular about how you vent appliances, since this can impact how well they operate, home energy efficiency, etc. There's no reason to be particular about how your sewer pipes are vented, as long as the sewer gas can't flow back into the home, which is achieved easily with &lt;a href=&quot;https://en.wikipedia.org/wiki/Trap_(plumbing)&quot; rel=&quot;nofollow noreferrer&quot;&gt;traps&lt;/a&gt; in the system, far from the vents. All that's needed at the vent end is a simple, inexpensive, unobtrusive pipe stub and a good seal with the roof.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-06-15T18:35:40.010" LastActivityDate="2015-06-15T18:35:40.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3202" PostTypeId="2" ParentId="3196" CreationDate="2015-06-15T18:20:32.330" Score="2" Body="&lt;p&gt;If I understood your description correctly, the structure is something like this (ignore the dimensions and loads themselves, they're simply representative of the relative values. Also, I didn't apply the self-weight, which can be applied as a simple uniform load along all the members):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vldXp.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This being the case, your structure is going to present all of the classic forces (axial force on the column and diagonal, shear and bending on all elements other than the diagonal). In this purely fictitious example, here are the normal, shear and bending diagrams, as well as the deformed configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SQa4Z.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the forces, one can easily obtain the stresses and strains (via linear elastic theory) and compare them to the beam strength and code requirements (also, make sure you have the correct code. Many times specific structures have specific codes, so a standard &quot;steel&quot; code may not be sufficient for a davit). Depending on the flexibility of the structure, a second-order analysis may be necessary. This may very well be the case for a davit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All figures were created with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis program.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:28:41.173" LastActivityDate="2019-12-28T18:28:41.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3204" PostTypeId="2" ParentId="3077" CreationDate="2015-06-16T10:40:50.723" Score="7" Body="&lt;h1&gt;Short Answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The nut on it is called a &lt;strong&gt;slotted lock nut&lt;/strong&gt; or &lt;strong&gt;locking bearing nut&lt;/strong&gt;. You can remove it with a &lt;strong&gt;hook spanner&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Long Answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I guess your micrometer is Japanese and thus probably doesn't conform DIN standards but for slotted lock nuts you can check &lt;strong&gt;DIN 981&lt;/strong&gt; or &lt;strong&gt;DIN 1804&lt;/strong&gt; and for hook spanner you can check &lt;strong&gt;DIN 1810&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why use Lock Nut&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The main reason is screw loosening. With time all screws get loose (no pun intended) and on a device such as micrometer that loose fit will cause errors. That nut, locks bearing into place. Below is a similar Lock Nut for bearings from Misumi USA, you can check &lt;a href=&quot;http://us.misumi-ec.com/pdf/fa/2012/p1_0917.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;datasheet here&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/uee2u.gif&quot; alt=&quot;Hard Locking Bearing Nut Misumi USA&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Information about lock nuts and hook spanners from SKF but I don't have enough rep to post more links.&lt;/p&gt;&#xA;" OwnerUserId="879" LastActivityDate="2015-06-16T10:40:50.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3205" PostTypeId="2" ParentId="2572" CreationDate="2015-06-16T11:19:52.110" Score="2" Body="&lt;h1&gt;Short answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Orthogonal cutting increases pressure and decreases tool life. Thus isn't used if aim isn't cutting. Lathe operation is a shaping operation, not a cutting operation most of the time. &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Long answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Orthogonal cutting isn't suitable because&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Cutting forces act on a smaller area of cutting surface on tool if orthogonal cutting is employed so it increases wear on tool tip. However oblique cutting distribute cutting forces to greater area thus increases tool life. &lt;/li&gt;&#xA;&lt;li&gt;You usually don't want corners on your part. Corners reduce strenght and you want chamfers.  &lt;/li&gt;&#xA;&lt;li&gt;Chip flow is another problem in orthogonal cutting. &lt;/li&gt;&#xA;&lt;li&gt;It's not possible to obtain perfect tool tip for orthogonal cutting, so addition of ploughing force increases total cutting force.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, if you're trying to cut off a part you'll use orthogonal cutting. Below are some examples of cutting tool shapes&#xA;&lt;img src=&quot;https://i.stack.imgur.com/okdvD.jpg&quot; alt=&quot;Cutting tool shapes&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check &lt;a href=&quot;https://books.google.com.tr/books?id=r8paflRca90C&amp;amp;lpg=PP1&amp;amp;hl=tr&amp;amp;pg=PA76#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fundamentals of Metal Cutting and Machine Tools&lt;/a&gt; pages 117-139 which are accessible in Google Books.&lt;/p&gt;&#xA;" OwnerUserId="879" LastActivityDate="2015-06-16T11:19:52.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3206" PostTypeId="1" AcceptedAnswerId="3209" CreationDate="2015-06-16T11:29:21.090" Score="1" ViewCount="1016" Body="&lt;p&gt;I have been assigned to build a ramp for a small car. I have only one A2 sheet of paper, scissors and a stapler. What would be the best slope profile, in order to move the car as far as possible after leaving the ramp, using only gravity? Any other ideas on construction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The car is 1:43 size. The weight is 100 g. The car must stay on the track and there is no initial push.&lt;/p&gt;&#xA;" OwnerUserId="1584" LastEditorUserId="1586" LastEditDate="2015-06-17T17:48:11.797" LastActivityDate="2015-06-17T17:48:11.797" Title="How to build a paper ramp for a small car?" Tags="|applied-mechanics|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3207" PostTypeId="1" AcceptedAnswerId="3210" CreationDate="2015-06-16T11:58:58.760" Score="1" ViewCount="104" Body="&lt;p&gt;I have the task to connect 6 input sources of fluid with 9 output (sinks) sources. Each input should be connected to each output. Each input should be able to distribute the fluid to only 1 output at a time. I have straight pipes, Y connections and valves, that I can use. I should use the MINIMUM amount of valves and Y connections as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a standard approach to reducing the number of components required for a design like this?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is the best I could get to so far:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ZUt7z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1584" LastEditorDisplayName="user16" LastEditDate="2015-06-16T15:08:24.437" LastActivityDate="2015-06-16T16:21:56.387" Title="Distribute 6 inputs to 9 outputs with pipes" Tags="|fluid-mechanics|pipelines|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3209" PostTypeId="2" ParentId="3206" CreationDate="2015-06-16T15:05:32.663" Score="5" Body="&lt;p&gt;So there's a number of concerns that you want to think about.  Some of these you'll have to determine experimentally, others you'll be able to calculate and then iteratively improve based upon your observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be understood / obvious that the higher up you can get the car, then the faster it will be able to go, and ultimately travel the furthest distance.  The challenge though is transferring the vertical potential energy into horizontal motion.  So the taller and longer of a ramp you can make, then the furthest your car will end up traveling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a fixed, rigid plate to form the initial ramp.  Run some experiments to determine the steepest angle the plate can form but still allow the car to roll smoothly off of the ramp.  Having the car crash at the end of the ramp doesn't do you any good and wastes any energy that was converted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know your angle of attack for the ramp, you'll want to start experimenting with various ramp designs.  This is where the principles of beam deflection start coming into play.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd look into C and I (or H) beams as they ought to provide optimal stiffness at the edges of the beams while minimizing the amount of material required.  An H beam design may allow you to join two beams together and fasten them in areas where the car won't be impacted by the staples.  Another advantage of those beams is that they'll provide rails to keep the car on the track.  That said, you'll need to fold under the edges of the H beam where the ramp meets the horizontal surface the car will travel along.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, look into triangular beams to create the struts that will support and lift your ramp up into the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to do some of the beam calculations ahead of time, it's pretty easy to search for Young's modulus of elasticity for paper.  That parameter can be affected by paper weight and coating, so that's research you'll want to do on your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the angle of attack and your preferred beam design will allow you to lay things out on the sheet of A2 paper that you have.  This represents your final constraint inasmuch as you have a fixed amount of material you can use.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-06-16T15:27:59.373" LastActivityDate="2015-06-16T15:27:59.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3210" PostTypeId="2" ParentId="3207" CreationDate="2015-06-16T15:28:11.840" Score="1" Body="&lt;p&gt;As mentioned in the comments, I think it is possible to do it with 15 valves (1 per inlet and 1 per outlet) and 13 Y connections as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0lP8U.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-06-16T15:28:11.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3211" PostTypeId="2" ParentId="2210" CreationDate="2015-06-16T15:49:47.283" Score="2" Body="&lt;p&gt;The natural way to stabilize it is to pressurize it. The normal atmospheric pressure is ~ 100 kPa. To do that a gigantic air-tight &lt;em&gt;liner&lt;/em&gt; should be inserted within the lava tube. It would be a little larger than cavern on order to transfer pressure loads to rock walls and sealing. A tricky equilibrium must be designed between gravitational induced rock pressure and internal gas pressure. It is a challenging engineering task but looks more realistic than large scale engineering works the under lunar surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a benefit, this solution will provide a large habitable space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A precedent exists in terrestrial engineering practice during the construction of &lt;a href=&quot;http://www.quora.com/How-and-when-were-the-NYC-subway-tunnels-built-between-boroughs-especially-underwater&quot; rel=&quot;nofollow noreferrer&quot;&gt;subway tunnels&lt;/a&gt; under the river bed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture provided by question shows very 'artificial' cross-section. The beautiful FEA analysis may be repeated with more realistic cross-section and internal pressure included. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ma115.jpg&quot; alt=&quot;terrestrial lava tube &quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terrestrial lava tube crossection. The flat floor of some lava tubes is usually a result of later sediments.&lt;/p&gt;&#xA;" OwnerUserId="2008" LastEditorUserId="2008" LastEditDate="2015-06-20T13:30:49.407" LastActivityDate="2015-06-20T13:30:49.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3212" PostTypeId="2" ParentId="3207" CreationDate="2015-06-16T16:09:35.613" Score="3" Body="&lt;p&gt;I recommend approaching problems like this by starting with a base case and working up from there until you see a pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For piping, the base case is a straight case with one input flowing to one output. Depending on the exact specs of the problem, this requires either 0 or 1 valves; it does not require any branching.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I------O   (or, I--▷◁--O)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Add an input to the base case:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  I--▷◁--Y---O&#xA;         /&#xA;I--▷◁--+&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Add an output to the base case:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I---Y--▷◁--O&#xA;     \&#xA;      +--▷◁--O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Add both an input and an output to the base case:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  I--▷◁--Y---Y--▷◁--O&#xA;         /     \&#xA;I--▷◁--+       +--▷◁--O&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;https://i.stack.imgur.com/qVj3W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;nicer-looking diagram&lt;/a&gt;, if you prefer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can see from looking at these simple cases is that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A single input or output requires no valves.&lt;/li&gt;&#xA;&lt;li&gt;If there is more than one input or output, each requires one valve.&lt;/li&gt;&#xA;&lt;li&gt;Each input or output beyond the first requires one branch.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since your problem has multiple inputs and multiple outputs, the minimum number of valves is equal to the total number of inputs and outputs and the minimum number of branches is equal to two fewer than the total number of inputs and outputs.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-06-16T16:21:56.387" LastActivityDate="2015-06-16T16:21:56.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3214" PostTypeId="1" CreationDate="2015-06-16T19:20:13.657" Score="3" ViewCount="175" Body="&lt;p&gt;I am trying to create a set-up where I can radially vibrate a 1x1 cm sample using a piezoelectric disc transducer. I know that at the $f_{min}$ of a PZT transducer, for the disc geometry, there is a radial vibration. How can I tune this vibration if I want to vibrate my sample radially at different frequencies?&lt;/p&gt;&#xA;" OwnerUserId="2012" LastEditorUserId="421" LastEditDate="2016-06-08T23:11:49.180" LastActivityDate="2016-06-08T23:11:49.180" Title="How to tune the radial mode vibration of a PZT disc transducer?" Tags="|mechanical-engineering|electrical-engineering|sensors|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3215" PostTypeId="1" AcceptedAnswerId="3233" CreationDate="2015-06-16T21:30:06.937" Score="3" ViewCount="828" Body="&lt;p&gt;I work in consultancy and I am designing a product for a customer. The customer has requested a specific PSU. I have contacted this PSU manufacturer and they request 6 L/min of coolant flow and minimum 3 Bar. I now have to specify a suitable pump to deliver these fluid characteristics to the PSU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have confirmed this is not a typo for &quot;maximum pressure,&quot; but no reasoning for specifying a minimum pressure is forthcoming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My confusion is why the flow rate AND minimum pressure are specified. The manufacturer does not know what pressure differential I require to generate the specified flow around my bespoke circuit - so why specify a minimum pressure AND a flow rate?&lt;/p&gt;&#xA;" OwnerUserId="1992" LastEditorUserId="368" LastEditDate="2015-06-17T13:47:49.040" LastActivityDate="2015-06-17T19:59:25.507" Title="Why is minimum pressure specified on a PSU water cooling circuit?" Tags="|fluid-mechanics|pressure|power-electronics|cooling|computer-hardware|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3216" PostTypeId="2" ParentId="3183" CreationDate="2015-06-16T23:13:55.760" Score="8" Body="&lt;p&gt;Not too sure if you will be able to find datasheet for specific cell phone. These are proprietary information and mobile phone manufactures usually don’t share this type of information in public for various reasons. Depending on the business needs you should contact the vendors directly and seek datasheet, which might be obtained under a non-disclosure agreement (NDA).  But there are a couple of other ways that you can obtain similar data. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Access Public Data from Federal Communications Commission (FCC)&lt;/strong&gt;&lt;br&gt;&#xA;If the mobile phone is sold in USA, then the mobile phone manufacture has to file a report with the FCC. This is public information, thus accessible through FCC web site. These reports will give you transmission power levels per the FCC test criteria, which might not be exactly what you are seeking, but you will get close. You will need the FCCID for the particular mobile phone which you should be able to find.  Below is output power specification for a &lt;strong&gt;&lt;a href=&quot;http://www.samsung.com/global/microsite/galaxys4/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Samsung Galaxy S4&lt;/a&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jC22D.jpg&quot; alt=&quot;Samsung Galaxy S4&quot;&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;IEEE 802.11 output power levels for Samsung Galaxy S4 (FCC)&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://transition.fcc.gov/oet/ea/fccid/&quot; rel=&quot;nofollow noreferrer&quot;&gt;FCCID Search form&lt;/a&gt; is a good place to start. You might be able obtain similar infomation for European Union (EU) too.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Test using &lt;a href=&quot;https://engineering.stackexchange.com/questions/2071/is-interference-between-aircraft-an-issue-for-fly-by-wireless-technology/2092#2092&quot;&gt;Semi-anechoic chambers&lt;/a&gt;&lt;/strong&gt;&lt;br&gt;&#xA;This might be tedious and expensive, but can be done. There are a plethora of EMC test facilities that will glad to help you get this data. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Test using a GTEM chamber&lt;/strong&gt;&lt;br&gt;&#xA;This is less expensive, and can be done much more easily as long as you have access to the chamber and spectrum analyzer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KMANn.png&quot; alt=&quot;GTEM Chamber with Spectrum Analyzer&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Specification from Wifi Module&lt;/strong&gt;&lt;br&gt;&#xA;But you might want to review widely available wifi modules. This might be the most practical method because you have access to datasheets with tolerances etc. Here are a few examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sparkfun.com/products/13252&quot; rel=&quot;nofollow noreferrer&quot;&gt;WiFi Module - ESP8266&lt;/a&gt; has a +19.5dBm output power in 802.11b mode&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ww1.microchip.com/downloads/en/DeviceDoc/70005171A.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;RN171 from microchip&lt;/a&gt; has 12.0dBm output power &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ww1.microchip.com/downloads/en/DeviceDoc/70686B.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;MRF24WG0MA from microchip&lt;/a&gt; has +18dBm output power&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.st.com/web/en/resource/technical/document/datasheet/DM00102124.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;SPWF01SA from ST microelectronics&lt;/a&gt; has 18.3dBm output power&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OBCWT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chances are a wifi module with similar transmission power capabilities are in mobile phones.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.fcc.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Federal Communications Commission USA&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://transition.fcc.gov/oet/ea/fccid/help.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;FCC ID Help&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://apps.fcc.gov/oetcf/eas/reports/ViewExhibitReport.cfm?mode=Exhibits&amp;amp;RequestTimeout=500&amp;amp;calledFromFrame=N&amp;amp;application_id=D%2Fk2nPoaPtEKFGvMm0DjGQ%3D%3D&amp;amp;fcc_id=A3LSPHL720T&quot; rel=&quot;nofollow noreferrer&quot;&gt;Samsung Galaxy S4 FCC Report&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cellphones.procon.org/view.resource.php?resourceID=003054#FCCdatabase&quot; rel=&quot;nofollow noreferrer&quot;&gt;20 Highest and Lowest Radiation Cell Phones&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.fcc.gov/encyclopedia/faqs-wireless-phones&quot; rel=&quot;nofollow noreferrer&quot;&gt;FCC Wireless Phones FAQS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pctestlab.com/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;PCTEST Engineering Laboratory Inc&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/2071/is-interference-between-aircraft-an-issue-for-fly-by-wireless-technology/2092#2092&quot;&gt;Is interference between aircraft an issue for fly-by-wireless technology?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://transition.fcc.gov/pshs/techtopics/techtopics10.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;License-Exempt Wireless Applications for Public Safety&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/186494/how-to-find-out-which-certificates-fcc-ce-etc-are-necessary-for-a-product&quot;&gt;How to find out which certificates (FCC, CE etc.) are necessary for a product?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-08-21T18:57:52.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3217" PostTypeId="2" ParentId="2865" CreationDate="2015-06-16T23:35:54.550" Score="4" Body="&lt;p&gt;A simple first order approach would be to treat the plate like a composite material, with the holes acting as a medium with no modulus.  The &lt;a href=&quot;https://en.wikipedia.org/wiki/Rule_of_mixtures&quot; rel=&quot;nofollow&quot;&gt;rule of mixtures&lt;/a&gt; , treating the &quot;holes&quot; as fibers with 0 modulus, would yield a modulus of 0.  So, the &lt;a href=&quot;http://www.doitpoms.ac.uk/tlplib/fibre_composites/stiffness.php&quot; rel=&quot;nofollow&quot;&gt;Semi-Emperical Halpin Tsai&lt;/a&gt; would be better:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \eta = \frac{\frac{E_f}{E_m} -1}{\frac{E_f}{E_m}+1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_c = \frac{E_m(1+\eta f)}{(1-\eta f)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $ E_f = 0 $, we have $ \eta = -1 $, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_c = E_m(\frac{1-f}{1+f}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the square packing arrangement, the circles occupy 78.54% of the area.  So, the &quot;composite&quot; would be ~12% of the original modulus.  Again, this would be a first order approximation to save you from running 50 finite element models.  Then, run your FEA to watch out for the stress concentrators for final design.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-06-16T23:35:54.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3219" PostTypeId="2" ParentId="3194" CreationDate="2015-06-17T01:48:24.240" Score="4" Body="&lt;p&gt;I think your electrical/mechanical analogy is not quite right. If you think about DC electricity and mechanical statics, it seems reasonable that voltage (&quot;electromotive force&quot;) is the analogy of mechanical force, and it is then tempting to say Ohm's law $RI = V$ is the analogy of $Ku = F$. But if you try to combine mechanical springs in series and parallel, and similarly electrical resistances, that &quot;analogy&quot; soon falls apart. Really, I think it is just a coincidence that two simple equations look similar to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider an AC electrical circuit and steady-state mechanical dynamics, and write the electrical equation in terms of charge not current, the real analogy becomes clear: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanical: $(-M\omega^2 + Ci\omega + K)ue^{i \omega t} = Fe^{i \omega t}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrical: $(-L\omega^2 + Rj\omega + 1/C)qe^{j \omega t} = Ve^{j \omega t}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the analogy of electrical resistance is really mechanical damping, since they are the only things each equation that dissipate energy. Electrical capacitance is the analogy of mechanical compliance (the inverse of stiffness), and electrical inductance is the analogy of mechanical inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct analogy of mechanical statics $Ku = F$ is electrostatics: $(1/C)q = V$.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-06-17T01:48:24.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3220" PostTypeId="1" AcceptedAnswerId="3256" CreationDate="2015-06-17T02:28:36.597" Score="3" ViewCount="622" Body="&lt;p&gt;I have here a heating element that broke when it was used to ramp a furnace to 750 C, breaking at 740. The element had previously been used for higher temperatures and appears that it may have possibly shorted. The shape of the element is custom, as well as the furnace, and so we had a custom element machined with a specific wind, but this new element is straight. I have few tools and am limited in buying tools; So, whats the best way I can give the coil this shape? Some things to consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The previous coil broke at what appears to be a weak point. It's possible that it shorted, but its also possible that it just gave out. Therefore, with shaping it, there needs to be as few of strained points as possible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It needs to remain planar, which is our largest problem right now. I attempted to curl it around a cylinder, and it now spirals upward (we have more coil if we need to start over.) If it doesn't remain planar, shorting will be likely.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When we attempted to spiral around a cylinder, it gives a good general shape, but will generally expand after releasing tension.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Below are two photos of what we are dealing with. The poorly coiled lustrous one is the new, and the burnt black piece in the ceramic is the old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CoRPI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/3YMvj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2016" LastActivityDate="2015-06-19T19:35:39.850" Title="How do I manually shape this heating element?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3221" PostTypeId="1" AcceptedAnswerId="3225" CreationDate="2015-06-17T08:36:55.010" Score="11" ViewCount="631" Body="&lt;p&gt;I recently built the basic construction for a crane out of lightweight material. Basically the head of the crane contains an electric servo motor which spins the head of the crane by pushing itself away from the tower (the rotor is fixed to the axis which is fixed to the tower. The axis isn't fixed to the head in any way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All electronics required to operate the motor and more are inside the head. The only thing I should extend to the rest is a 240 V cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the head is allowed to make multiple rotations (preferably limitless), &lt;strong&gt;how do I avoid snapping my cable by rotating it too often?&lt;/strong&gt; I'm aware such solutions exist, but I have no experience in this area so I have no idea what to ask for specifically.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;/---------------------\&#xA;|                     |&#xA;|        motor        |&#xA;\----------|----------/&#xA;           |&#xA;___________|___________&#xA;\          |          /&#xA; \         |         /&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1070" LastEditorUserId="1070" LastEditDate="2015-06-19T22:15:31.437" LastActivityDate="2015-08-12T11:22:32.147" Title="Avoid snapping an electric cable from too much rotation" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3222" PostTypeId="2" ParentId="3215" CreationDate="2015-06-17T08:44:12.593" Score="5" Body="&lt;p&gt;You did not specify what materials you are working with and neither the dimensions. I made some assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the coolant circuit is likely driven by a pump. Therefore you need to know the flow rate $\dot{Q}$ the pressure needed on the pressure side. You can get more information &lt;a href=&quot;https://engineering.stackexchange.com/questions/3187/what-does-open-flow-mean-for-a-pump-flow-rate&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific setup to cool the PSU you have pressure drops along your tubings, curvatures and through the height differences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep things simple I just assumed a height difference of 20 cm and no differences in pressure in different reservoirs where the coolant might be pumped. Then the only significant pressure drop is the result of friction pressure loss. I also did not account for any curvatures. For $\rho$ I assumed water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta p_r = \lambda \cdot \dfrac{L}{d} \cdot \dfrac{\rho}{2} \cdot u^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed rubber tubings with the darcy friction factor formulae $\lambda$ = 0.02&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot below shows 3 different assumptions in lenghts of the tubing and inner diameter.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$L$ = 2 m and  $d$ = 0.01 m&lt;/li&gt;&#xA;&lt;li&gt;$L$ = 2 m and  $d$ = 0.005 m&lt;/li&gt;&#xA;&lt;li&gt;$L$ = 10 m and  $d$ = 0.005 m&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HNdYR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HNdYRm.png&quot; alt=&quot;Pressure versus flowrate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pumps have characteristic curves as well but the pressure decreases with increasing $\dot{Q}$. The intersect of the two curves is the operating point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NHW1N.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NHW1Nm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.ctgclean.com/tech-blog/2012/03/pumps-pump-performance-curves/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ctgclean&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now either the pump is fixed and you need to adjust your pressure drop along the tubing accordingly or vice versa. Therefore the specification of a minimum pressure instead of maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that shed some light on your question.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-06-17T08:52:33.437" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3223" PostTypeId="2" ParentId="3030" CreationDate="2015-06-17T08:56:12.930" Score="1" Body="&lt;p&gt;The problem is a multi-criteria decision analysis (MCDA), as I want to rank the machines based on several (available) criteria. With initially the following criteria: downtime costs and utilization. However with utilization there was the following problem: if the events are counted that surpasses a threshold or the average utilization is used, both situations would lead to loss of information. As it is an MCDA, both criteria can be added, thus the criteria are now:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;downtime costs per time unit&lt;/li&gt;&#xA;&lt;li&gt;average utilization per time span&lt;/li&gt;&#xA;&lt;li&gt;number of events the utilization that surpasses a threshold (e.g. 95%)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As these factors are not directly commensurable and differ in importance, a weighted method is needed. To select the appropriate method 7 guidelines are used (Guitouni and Martel, 1998);&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Guideline G1&lt;/strong&gt;: Determine the stakeholders of the decision process. If there are many decision makers (judges), one should think about group decision making methods or group decision support systems (GDSS).&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Guideline G2&lt;/strong&gt;: Consider the Decision Maker (DM) `cognition' (DM way of thinking) when choosing a particular preference elucidation mode. If he is more comfortable with pairwise comparisons, why using tradeoffs and vice versa?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Guideline G3&lt;/strong&gt;: Determine the decision problematic pursued by the DM. If the DM wants to get an alternatives ranking, then a ranking method is appropriate, and so on.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Guideline G4&lt;/strong&gt;: Choose the Multi Criteria Analysis Problem (MCAP) that can handle properly the input information available and for which the DM can easily provide the required information; the quality and the quantities of the information are major factors in the choice of the method.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Guideline G5&lt;/strong&gt;: The compensation degree of the MCAP method is an important aspect to consider and to explain to the DM. If he refuses any compensation, then many MCAP will not be considered.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Guideline G6&lt;/strong&gt;: The fundamental hypothesis of the method are to be met (verified), otherwise one should choose another method.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Guideline G7&lt;/strong&gt;: The decision support system coming with the method is an important aspect to be considered when the time comes to choose a MCDA method.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Based on all guidelines except guideline G6, this results in the following suitable methods: Analytic Hierarchy Process (AHP) or Promethee II. However, AHP has the assumption that inner and outer criteria are independent. A correlation test showed that there exists significant correlation between criteria 2 and 3, thus the criteria are not independent. Promethee II is therefore the appropriate method for ranking the machines given my situation. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Guitouni, A., Martel, J.-M., 1998. Tentative guidelines to help choosing an appropriate MCDA method. Eur. J. Oper. Res. 109, 501–521.&lt;/p&gt;&#xA;" OwnerUserId="1847" LastEditorUserId="368" LastEditDate="2015-06-17T13:45:15.440" LastActivityDate="2015-06-17T13:45:15.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3224" PostTypeId="1" CreationDate="2015-06-17T09:38:23.630" Score="4" ViewCount="165" Body="&lt;p&gt;I'm planning to build up a big &lt;a href=&quot;https://en.wikipedia.org/wiki/Tipi&quot; rel=&quot;nofollow&quot;&gt;tipi&lt;/a&gt; with some friends for relaxing on nice summer evenings.  The tipi is going to be placed on the shore of a medium sized river, specifically the Wertach in Germany (&lt;a href=&quot;https://de.wikipedia.org/wiki/Wertach_%28Fluss%29&quot; rel=&quot;nofollow&quot;&gt;Wikipedia to Wertach (German)&lt;/a&gt;).&#xA;We also want some electricity there, to run a small Stereo and some light which together use about 100 W.  Therefore we thought about using a waterwheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to build a waterwheel generator for about 400 EUR? And if it is, how could we do that? I already read that we could use a scooter engine for that.&lt;/p&gt;&#xA;" OwnerUserId="2017" LastEditorUserId="272" LastEditDate="2015-06-17T11:55:36.630" LastActivityDate="2015-06-17T11:55:36.630" Title="DIY Waterwheel which generates 100 W?" Tags="|electrical-engineering|fluid-mechanics|power-electronics|" AnswerCount="0" CommentCount="4" ClosedDate="2015-06-17T11:31:12.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="3225" PostTypeId="2" ParentId="3221" CreationDate="2015-06-17T12:07:23.563" Score="13" Body="&lt;p&gt;The device you are looking for is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Slip_ring&quot; rel=&quot;noreferrer&quot;&gt;slip ring&lt;/a&gt; such as the one shown below.  These devices use internal components which maintain contact as they spin (old ones used brushes) allowing them to turn by an unlimited amount in either direction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mercotac.com/?gclid=CP-b7NnWlsYCFYUXHwod-y0Afw&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YJyZ6.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-06-17T13:41:21.440" LastActivityDate="2015-06-17T13:41:21.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3227" PostTypeId="1" AcceptedAnswerId="3229" CreationDate="2015-06-17T14:34:06.153" Score="0" ViewCount="2896" Body="&lt;p&gt;How do we start a nuclear chain reaction? For example, if we wanted to do it this weekend, what would we need, and what are the basic steps? I'm not looking for a text book response or highly detailed information. I'm only curious about the most basic components. For example, we need to take A and B and submerge it into C? That's all I'm asking. If you had a daughter, and she asked, &quot;What happens in a nuclear power plant?&quot;, what would you tell her?&lt;/p&gt;&#xA;" OwnerUserId="2020" LastEditorUserId="2020" LastEditDate="2016-04-21T01:19:35.323" LastActivityDate="2016-04-21T01:19:35.323" Title="How does a nuclear reactor initiate nuclear fission?" Tags="|nuclear-technology|nuclear-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2015-06-17T23:03:30.807" ContentLicense="CC BY-SA 3.0" />
  <row Id="3228" PostTypeId="2" ParentId="3220" CreationDate="2015-06-17T15:14:25.383" Score="0" Body="&lt;p&gt;Edit: Classic engineer behavior, not answering the question that was asked. These are instructions for making the coil, not getting it into the shape you want. Sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had to do something like this for my senior design project. After determining the characteristics of the coil (length, diameter, and pitch) I used a lathe to fashion an appropriately sized threaded rod. I drilled a hole through the rod near the top of the threads large enough to accommodate the resistive wire that I was winding the heating element out of. I installed this jig in a lathe with the aforementioned hole close to the chuck and I turned the rotational speed of the lathe way down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fed the end of resistive wire, which was on a spool, through the aforementioned hole to secure the wire allowing it to wind. As I turned on the lathe I maintained constant tension on the wire as it wound around the threaded rod. After winding a sufficient length I (turned off the lathe and) clipped the end near the hole and removed the bit that was securing the wire to the . After releasing the tension from the wire it relaxed a little and slipped off the threaded shaft quite easily! If your wire doesn't relax as much then you might have to 'unscrew' it from the shaft, but it should come off. After that I did some finishing operations on the freshly-minted coil. I was able to produce several near-perfect coils using this technique!&lt;/p&gt;&#xA;" OwnerUserId="2014" LastActivityDate="2015-06-17T15:14:25.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3229" PostTypeId="2" ParentId="3227" CreationDate="2015-06-17T16:02:45.503" Score="10" Body="&lt;p&gt;Nuclear fission is spontaneous. In fact, if you have a critical mass of fuel assembled, you need to take explicit steps to keep it from going into chain reaction, typically by adding neutron-absorbing material, AKA &quot;control rods&quot; to the assembly. When you remove the control rods, the reactor starts up by itself.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-06-17T16:02:45.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3231" PostTypeId="1" CreationDate="2015-06-17T19:25:12.460" Score="3" ViewCount="266" Body="&lt;p&gt;I live in a part of Mexico that can get really hot in the summer. I've got an un-airconditioned bedroom on the other side of a wall from a room cooled by a powerful Mirage mini-split AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the only communication between the two rooms currently is the bedroom doorway which opens from a shared hallway.  This is insufficient because of the dead air volume above the doorway (the ceiling is inclined similarly to that over the AC unit). Please see pictures for reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea is to bore two holes 8-10 inches in diameter through the wall as shown in the detail shot. The purpose would be to establish a convection loop, where the hot air from top of the bedroom would flow out into the area of the AC unit (and be entrained in the cool air flow from it). To compensate for the reduction in pressure, cold air would replace the ventilated hot air by flowing in through the hallway and into the bedroom doorway. I understand that this scheme, if it works, will probably put an additional load on the AC. That's acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't done any math on this idea (indeed, I'm not sure what math to do), but am just guessing at the size of the holes. Intuitively, it seems feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on whether this would be effective before I get out my chisel and hammer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Overview shot to demonstrate problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/RHoG7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Detail shot, with proposed holes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0dZSs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, many thanks to all who responded, and particularly for the exhaustive analysis provided by Olin. The hole is being cut now, and the design has changed due to some unforeseen realities in the masonry. It will now be a rectangular opening about 12&quot; x 20&quot;, with the top corner about 6&quot; from the ceiling and wall. I'll report back on the outcome, and whether a fan is necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks again!&lt;/p&gt;&#xA;" OwnerUserId="2024" LastEditorUserId="2024" LastEditDate="2015-06-23T00:44:18.830" LastActivityDate="2015-06-23T11:11:15.240" Title="Convective fix for air conditioning dilemma?" Tags="|heat-transfer|hvac|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3233" PostTypeId="2" ParentId="3215" CreationDate="2015-06-17T19:47:11.277" Score="4" Body="&lt;p&gt;The pressure requirement could also be to raise the temperature at which the water boils.  The boiling point of water depends on the pressure it's at; higher pressure raises the boiling point, as the following image shows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZPggh.png&quot; alt=&quot;Image taken from www.engineeringtoolbox.com&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, water at atmospheric pressure boils at 100 C, but water at 3 bars boils at about 130 C.  When water starts to boil, it becomes less effective at removing heat.  So, the PSU internals might operate hotter than 100 C, which requires that the water be under pressure to prevent boiling.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Critical_heat_flux&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt; on critical heat flux for a more detailed explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, coolant reaching its critical heat flux was a major factor that doomed reactor #4 at the Chernobyl nuclear power plant.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-06-17T19:59:25.507" LastActivityDate="2015-06-17T19:59:25.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3234" PostTypeId="1" AcceptedAnswerId="3235" CreationDate="2015-06-17T20:57:46.077" Score="6" ViewCount="129" Body="&lt;p&gt;My North American facility is looking at a major shut down this fall.  In the area, three other large facilities are also doing either shutdowns or expansions in the fall.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A last minute project that I have been assigned requires a couple of weeks of machining custom components.  This is an issue as none of the local machine shops are likely to be able to fit the work within their schedule in the required time frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I think my only option is to start looking beyond the local shops.  What should one look for when trying to source non-local machining?  Are there any red flags to look for when talking to a shop?  &lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorDisplayName="user16" LastEditDate="2015-06-17T21:39:43.563" LastActivityDate="2015-06-19T20:02:24.147" Title="Going distant in sourcing machining" Tags="|machining|project-management|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3235" PostTypeId="2" ParentId="3234" CreationDate="2015-06-18T00:11:02.533" Score="6" Body="&lt;p&gt;I've dealt with this in many ways.  Ultimately, here is what I look for when I have to go beyond local sourcing of custom parts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Responsiveness.  You can have the best website in the world, but unless I know you are there for me when I need to change anything, I can't rely on you.&lt;/li&gt;&#xA;&lt;li&gt;Questioning.  The best vendors I have for custom pieces ask questions and explain to me every step of the way, so I know exactly what's going on, and what they might want to change.&lt;/li&gt;&#xA;&lt;li&gt;Compliance with my specifications.  Ultimately, if it is really complicated, I draw up a specification.  If they question my spec, that's a good thing.  If they quote and except themselves from my spec, that's a bad thing.&lt;/li&gt;&#xA;&lt;li&gt;Openness to visit.  Your local machine shop is usually willing to let you come in for a visit.  Your non-local machine shop should do the same.&lt;/li&gt;&#xA;&lt;li&gt;Quote comes in with an honest lead time.  Ultimately, if it isn't there on time, it isn't there.  Make sure they state lead time and are willing to back it up.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="282" LastEditDate="2015-06-19T20:02:24.147" LastActivityDate="2015-06-19T20:02:24.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3236" PostTypeId="2" ParentId="3234" CreationDate="2015-06-18T00:12:01.780" Score="3" Body="&lt;p&gt;At very highly level the three areas that need to manage are scope, cost and time also known as the triple constraints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scope:&lt;/strong&gt;&lt;br&gt;&#xA;Make sure the sponsor clearly defined the project. It is highly unlikely that the sponsor will document the scope. Therefore you need to document the scope and get some type of agreement from the sponsor. An e-mail might be a good start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure all drawings and other documents are clear with sufficient details. Recently I encountered there was a situation where specification state, use ABS (Acrylonitrile Butadiene Styrene), without any other details. These types of situations create confusion, delay, and unnecessary costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cost:&lt;/strong&gt;&lt;br&gt;&#xA;Make sure your sponsor, you and the vendor clearly understand the project related cost. It is good to request for quotes from multiple sources, and understand all the deliverable. Considering the current situation expect pay more than the market price. It is good idea to have about a 10% buffer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Timing:&lt;/strong&gt;&lt;br&gt;&#xA;Make sure the machine shop clearly understand the project timing. Don’t push the machine shops, they may agree to timing that they are unable to fulfill. Make an attempt to have regular communication with the vendor. This is very important. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be a good idea to seek recommendations from your current local machine shops. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pmi.org/default.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Project Management&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://pm.stackexchange.com/&quot;&gt;Project Management Stack Exchange&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:46:30.233" LastActivityDate="2015-06-18T22:51:14.117" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3237" PostTypeId="2" ParentId="3148" CreationDate="2015-06-18T00:52:10.507" Score="1" Body="&lt;p&gt;The easiest setup is a bucket on the outlet of your pipe, and a stopwatch.  Start the stopwatch when liquid starts flowing in, and stop it when the bucket is full.  Then your flow rate is 1 bucket / time, which can be converted into useful units.  Use this to check against your device and you should be good to go.  Once you have it calibrated against 5-10 data points you can be confident that you are calibrated.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-06-18T00:52:10.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3238" PostTypeId="1" AcceptedAnswerId="3241" CreationDate="2015-06-18T07:10:50.970" Score="4" ViewCount="872" Body="&lt;p&gt;Trivially you can prevent rocking with many vehicle by eliminating the suspension. But is it possible to allow a floaty suspension that glides over potholes, without also having to rock back and forth, with some kind of stabilization? I know that it's possible in buildings, but having a fixed platform really helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because the rocking motion is one of the primary sources of carsickness, and a bus trip this evening left me feeling rather queasy. This seems like something that could be improved, especially in such expensive vehicles.&lt;/p&gt;&#xA;" OwnerUserId="2028" LastActivityDate="2015-06-18T18:57:04.190" Title="Is it possible to design a bus that doesn't rock?" Tags="|mechanical-engineering|applied-mechanics|automotive-engineering|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3239" PostTypeId="2" ParentId="3238" CreationDate="2015-06-18T08:43:22.793" Score="8" Body="&lt;p&gt;You could have active suspensions, which could keep the vehicle level or even roll it into the corner (see for example &lt;a href=&quot;http://www.arnaudmiege.co.uk/vehicle/phd.html&quot;&gt;this PhD thesis&lt;/a&gt;), but they are typically prohibitively expensive and not feasible in practice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Semi-active suspensions, which can lock the anti-roll bar solid just ahead of a turn, are sometimes a good compromise between performance and feasibility.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-06-18T08:43:22.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3240" PostTypeId="2" ParentId="3148" CreationDate="2015-06-18T12:26:16.520" Score="1" Body="&lt;p&gt;You could buy/borrow an anemometer like in the picture below and hold it against the outlet of the test setup.  That would give you the approximate velocity of the flow, which you could easily convert to mass flow rate using $\dot m = v \cdot A \cdot \rho $ where $v$ is velocity, $A$ is area, and $\rho$ is density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GjO6l.jpg&quot; alt=&quot;anemometer&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-06-18T12:26:16.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3241" PostTypeId="2" ParentId="3238" CreationDate="2015-06-18T12:59:28.527" Score="4" Body="&lt;p&gt;There are in fact cars available with this type of feature. For example, the Mercedes S class is scanning the road ahead to counter imperfections in the road with its active suspension: &lt;a href=&quot;https://en.wikipedia.org/wiki/Active_Body_Control#Magic_Body_Control&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Active_Body_Control#Magic_Body_Control&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2031" LastActivityDate="2015-06-18T12:59:28.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3243" PostTypeId="2" ParentId="3238" CreationDate="2015-06-18T15:38:16.863" Score="1" Body="&lt;p&gt;Active suspensions have been around for 15-20 years in motorsport, so F1 and other regulated events have strict rules on what is and isn't allowed.  It's fairly well understood.  That said, it's understood well enough that the cost/benefit hasn't really worked out for production vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem isn't just the initial cost, it's also maintenance and robustness.  Whatever actuators you use, they need to be in the wheel well, and that's a hellish tough environment.  You also need sensors to check distance from the road (typically on the four corners of the vehicle), and again the underside of the vehicle is a tough place to live.  It's OK on the track, but when you add all the mud, stones, water and small furry animals that appear on the road, vulnerable things show that they're vulnerable.  If your kit isn't 99% sure to survive past the manufacturer's warranty, the manufacturer isn't going to go for it.&lt;/p&gt;&#xA;" OwnerUserId="2035" LastActivityDate="2015-06-18T15:38:16.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3244" PostTypeId="1" CreationDate="2015-06-18T18:05:24.883" Score="3" ViewCount="141" Body="&lt;p&gt;Let's say I have $3$ DC lines input as $10 V (x A), 24 V (y A), 6 V (z A)$. Now I want to convert the output to say $16$ VDC at $k A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of converter can I use?&lt;/p&gt;&#xA;" OwnerUserId="2036" LastEditorUserId="1293" LastEditDate="2015-06-19T12:38:29.760" LastActivityDate="2015-06-19T14:07:55.077" Title="If I have multiple different DC voltage inputs, how do I get a stabilized output at an intermediate voltage?" Tags="|electrical-engineering|circuit-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3245" PostTypeId="2" ParentId="3238" CreationDate="2015-06-18T18:38:04.730" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;http://en.wikipedia.org/wiki/Citro%C3%ABn_Xantia&quot; rel=&quot;nofollow noreferrer&quot;&gt;Citroën Xantia Activa&lt;/a&gt; was one of the earliest mass production active suspension vehicles I know of.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Activa active anti-roll bars&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;In 1994, the Activa technology was introduced, which is an extension to the Hydractive II suspension, where two additional spheres and two hydraulic cylinders are used together with computer control to eliminate body roll completely. This technology is more broadly known as active suspension, and the Xantia Activa has exceptional road holding comparable to true sports cars. It employs active anti-roll bars.&lt;/p&gt;&#xA;&lt;p&gt;In the Swedish magazine Teknikens Värld's moose test the 1999 model of Xantia V6 Activa still holds the record speed through the manoeuvre - faster than the Porsche 996 GT2.&lt;/p&gt;&#xA;&lt;p&gt;UK Models of the Activa came fitted with a XU10 2 litre turbocharged engine also fitted to the Citroën XM 2.0CT and Peugeot 605 SRi. It produced 150 bhp and 171 lb ft of torque and was a 'low-blow' type for smooth power delivery rather than outright bhp.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So it's very possible to create a flat platform.&lt;/p&gt;&#xA;&lt;p&gt;Anecdotally, friends as kids used to get car sick more often in Citroën cars with hydraulic suspension than normally sprung cars, my guess is that the ride makes you feel sea sick.&lt;/p&gt;&#xA;" OwnerUserId="2037" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-06-18T18:57:04.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3246" PostTypeId="2" ParentId="3244" CreationDate="2015-06-18T19:57:56.623" Score="2" Body="&lt;p&gt;If you are trying to get from 6VDC or 10VDC up to 16VDC, you can use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Boost_converter&quot; rel=&quot;nofollow&quot;&gt;boost converter&lt;/a&gt;, and if you're trying to get from 24VDC to 16VDC you can use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Buck_converter&quot; rel=&quot;nofollow&quot;&gt;buck converter&lt;/a&gt; or a &quot;linear regulator&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia pages I've linked you to will explain everything in more detail, but briefly, a linear regulator works by &quot;burning off&quot; the extra voltage in the form of heat through a resistor voltage divider or Zener diode and a buck converter works by quickly switching the load's access to the power supply. If you could imagine blinking a light on and off quickly, the average output is some dimmer value (lower voltage) than the input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no passive boost converter for DC; this relies solely on switching activity to excite an electrical storage component (inductor or capacitor) with the goal of raising (boosting) the voltage to a higher value than the input voltage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These systems exist on-a-chip, so you would have to set specific output current requirements and pick a voltage source to find the exact solution suitable for you; you can search &lt;a href=&quot;http://www.mouser.com/&quot; rel=&quot;nofollow&quot;&gt;Mouser&lt;/a&gt; or &lt;a href=&quot;http://www.digikey.com/&quot; rel=&quot;nofollow&quot;&gt;Digikey&lt;/a&gt; to find the exact parts once you set your requirements.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2015-06-19T14:07:55.077" LastActivityDate="2015-06-19T14:07:55.077" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3247" PostTypeId="1" AcceptedAnswerId="3253" CreationDate="2015-06-19T00:54:54.150" Score="7" ViewCount="252" Body="&lt;p&gt;My project team and I delivered the design of a 14ha urban infrastructure development in Sydney for a Client. The project was to hand over the development at a Tender level of design, which would be tendered upon by a number contractors. The succesful bidder would then take on the rest of the project as a Design and Construct contract, and complete the works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a part of our deliverables, we issued all engineering plans in pdf and AutoCAD dwg, and all design files including the 12d data terrain models and design strings. We have now had a request from the winning contractor to provide the entire 12d '.project' file, which includes all functions, modifiers, appliers and range files. Providing this would save the contractor a significant amount of time, even considering all of the design strings and models which have already been issued.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our involvement in the project has been completely handed over, and we consider all of our responsibilities departed. There are a number of people in our team who consider it not appropriate to provide the .project file, whilst there are a number of people who don't see it as an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone run into this issue before, or has anyone got any guidance on whether it should be provided? Are there any risks in sending the information? Does anyone have any thought's on where our Client's ownership of the data ends and our Intellectual Property begins?&lt;/p&gt;&#xA;" OwnerUserId="767" LastEditorUserId="272" LastEditDate="2015-06-19T12:06:09.423" LastActivityDate="2015-06-19T18:14:35.673" Title="Sending out a 12d Civil Design Model" Tags="|civil-engineering|modeling|design|computer-aided-design|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3248" PostTypeId="2" ParentId="3231" CreationDate="2015-06-19T03:20:29.707" Score="1" Body="&lt;p&gt;Some math, if you care:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally you want 12 air changes per hour for best effect.  So, say your room is a standard 20' x 20' x 10' (or if you like metric, 6m x 6m x 3m).  Ideally, at 12 changes per hour, you need to move 800 cfm through the holes.  For (2) 8 inch holes, that works out to about 1,100 ft/min - which is in the normal range for ducting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However much heat wants to move the air, it will.  But this size hole is adequate for the application.  If you get a fan (as suggested), make sure it can move at least 400 cfm through each hole, with a delta P of 0.25 inches of water minimum.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-06-19T03:20:29.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3250" PostTypeId="2" ParentId="3231" CreationDate="2015-06-19T12:31:03.867" Score="6" Body="&lt;p&gt;Those two holes by themselves will have little effect.  The pressure difference between the two ends of the holes will be so small that only a little air will flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to go thru all this trouble, then put a hole at the bottom too, with a fan that actively blows cold air from the bottom of the living room into the bedroom.  Hot air will then find its way out of the bedroom, with one of the paths being the holes at top.  However, the obvious answer is to put another air conditioner in the bedroom.  That will work well, be much less trouble, and probably cost less in the long run considering what the holes in the wall will do to the value of the house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see how much pressure difference there will be across the holes at top just due to passive convection.  Let's say the height is 8 feet.  A column of that height 1 square inch in crossection has a volume of 96 in&amp;sup3; = 1.57 liters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air is mostly N&lt;sub&gt;2&lt;/sub&gt; and O&lt;sub&gt;2&lt;/sub&gt;, with molecular masses of 28 and 32, respectively.  There is more N&lt;sub&gt;2&lt;/sub&gt;, so let's say the molecular mass of air is 29, which means the mass of one mole of air is 29 grams.  At 0&amp;deg;C = 273&amp;deg;K and 1 atm, a mole occupies 22.4 l.  That means the density is (29 g)/(22.4 l) = 1.295 g/l.  This means the mass of our 1 square inch column is 2.033 g at 273&amp;deg;K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the cold air is 68&amp;deg;F = 20&amp;deg;C = 293&amp;deg;K and the hot air is at 90&amp;deg;F = 32&amp;deg;C = 305&amp;deg;K.  The cold air mass per square inch of column is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(273&amp;deg;K / 293&amp;deg;K) 2.033 g = 1.894 g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the same for the hot column:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(273&amp;deg;K / 305&amp;deg;K) 2.033 g = 1.819 g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is therefore 0.075 g.  On earth, that has a weight of 0.000168 pounds, so the air pressure difference due to 68&amp;deg;F and 90&amp;deg;F over 8 feet is only 0.00017 PSI.  The flow rate that will cause thru your 8 inch holes simply won't be enough to make a meaningful difference.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-06-19T12:31:03.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3251" PostTypeId="2" ParentId="3231" CreationDate="2015-06-19T12:46:43.367" Score="2" Body="&lt;p&gt;Agree with @Olin Lathrop that the holes should be top and bottom rather than side-by-side.  However, unless your bedroom door must remain shut and has a tight seal, you could probably get away with just one vent with an integral fan, like the one shown in the picture below.  Put it anywhere on the wall where it's convenient, and the air leaking around the door will complete the circuit.  You can always add a dedicated return vent later if you need to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wl3Wr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-06-19T12:46:43.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3252" PostTypeId="1" AcceptedAnswerId="3314" CreationDate="2015-06-19T13:35:22.110" Score="5" ViewCount="458" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fieak.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fieakm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I have to solve the differential equation for radial disks, and therefore I need the shear force distribution. The disk is rotationally symmetric.&lt;/p&gt;&#xA;&lt;p&gt;I want to use the superpositions principle and set 3 areas.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$0 \le r \le D_i / 2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$D_i / 2 &amp;lt; r \le k / 2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$k/2 &amp;lt; r \le D / 2$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fGjnv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fGjnvm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The shear forces are for area 1&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sum F_z = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_{rz1} \pi r - \frac{\pi r^2}{2}p = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_{rz1} = \frac{p r}{2}p$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Area 2&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$0 =  -F_{rz2} \pi r + F_s \pi k $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_{rz2} = F_s \frac{k}{r} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Area 3&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$0 =  F_{rz3} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The solution confirms &lt;span class=&quot;math-container&quot;&gt;$F_{rz1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_{rz3}$&lt;/span&gt; for me. However it states that only after &lt;span class=&quot;math-container&quot;&gt;$F_{rz3}$&lt;/span&gt; is known one can solve for &lt;span class=&quot;math-container&quot;&gt;$F_{rz2}$&lt;/span&gt; for the area &lt;span class=&quot;math-container&quot;&gt;$D_1/2 &amp;lt; r \le D / 2$&lt;/span&gt;&#xA;The solution should be &lt;span class=&quot;math-container&quot;&gt;$F_{rz2} = F_s \frac{k}{2r}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Did I cut correctly?&lt;br /&gt;&#xA;What am I missing here?&lt;/p&gt;&#xA;&lt;p&gt;I know realized I had to insert &lt;span class=&quot;math-container&quot;&gt;$k/2$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; and then I get the correct solution for&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$0 =  -F_{rz2} \pi r + F_s \pi k/2 $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_{rz2} = F_s \frac{k}{2r} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However the question that remains is, why the hint to extend the area? Why is it not possible to calculate &lt;span class=&quot;math-container&quot;&gt;$F_{rz2}$&lt;/span&gt; before &lt;span class=&quot;math-container&quot;&gt;$F_{rz3}$&lt;/span&gt; when it apparantly is possible?&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-06-24T16:28:46.160" Title="Shear Force Distribution on disk" Tags="|mechanical-engineering|mathematics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3253" PostTypeId="2" ParentId="3247" CreationDate="2015-06-19T13:50:42.730" Score="8" Body="&lt;p&gt;Before I start, let me disclose that I don't know a lot about the details of infrastructure projects or your file format, but I think the general considerations are fairly universal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of considerations when providing your digital model to a contractor, and in my work I have decided to offer them my model in some situations, but not in others. Here are the main factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Linear Time.&lt;/strong&gt; By providing a model, you can probably save the project significant linear time by reducing the amount of work the contractor needs to in order to  prepare their shop drawings, allowing the work to start and finish sooner. This is probably the most compelling reason to provide the model. It's worth noting that in some industries where file formats and workflows aren't well standardized, sometimes this efficiency can't actually be realized as things have to be re-drawn or re-modeled anyway for the target platform.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Risk.&lt;/strong&gt; If there is a discrepancy between the model and your drawing, there should be language that instructs the contractor to follow the drawing. However, you have no control over how the contractor uses your model, and they may uncover an error in the model that you never noticed or didn't care about because it is never shown in the drawings. Depending on the contract language, these errors could expose you and your employer to additional liability.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Proprietary information.&lt;/strong&gt; By giving the contractor your design as a complete model, rather than just output drawings, you make it much easier for them to copy designs you may have developed over multiple projects and re-appropriate them on other jobs. While a well-written contract will make that clearly illegal, it's very hard to translate that into recovering financial losses. In addition, any custom objects (blocks, toolsets, plug-ins, etc.) you have created for the software may be transmitted. Again, this kind of intellectual property is hard to control once it is out in the world.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Costs.&lt;/strong&gt; Unless someone made a major mistake, the contractor should have bid the job assuming they had to work off of your drawings (the contract documents) and do or redo any ancillary work required to develop their shop drawings and fabricate/build the job. Likewise, your contract was just to produce the drawings, and the model is simply a tool you used. You may have created a model that is good enough for your drawings, but making it good enough to build off of would cost you money. For the added quality expectation, and the added risk, it would not be unreasonable to charge a fee for providing the model. Even if your model is perfect, the contractor may have questions about some part of it, or ask you to convert it to another format, or otherwise have modest requests that mean you need to devote time to the project that you didn't bid on.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By way of example for the discrepancy issues, when making a drawing for a machined part, I might not draw every feature to scale in my model out of laziness. While this is a bad practice, if I override the dimension and the output drawing clearly conveys the correct information, I have met my contract requirement in the design/engineering role. However, if I then provide the model, and the machine shop doesn't check the written dimension, the piece would be fabricated wrong. Each party will feel confident that this is the other party's fault. If I had never provided two redundant sources of information, blame would have been much easier to assign. This is the same reason that traditionally, we don't dimension segments as well as the whole that they add up to. We try to avoid opportunities for information we provide to be internally inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's obviously not directly applicable to your project, but here in the US, the &lt;a href=&quot;http://www.aisc.org/WorkArea/showcontent.aspx?id=25256&quot; rel=&quot;nofollow&quot;&gt;AISC Code of Standard Practice&lt;/a&gt; (AISC 303-10) answers contract questions like these for steel projects when they aren't clearly addressed by a specific contract. See section 4.3 for the details, but the general points are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The contractor can only use the model with permission of the engineer, and only for the one job (not to extrapolate to future jobs)&lt;/li&gt;&#xA;&lt;li&gt;The contractor has to acknowledge that the model is not a contract document (with an exception for projects where the original engineering deliverable was a digital model in lieu of drawings) and is superseded by information on the drawings.&lt;/li&gt;&#xA;&lt;li&gt;The contractor is still responsible for verifying all information&lt;/li&gt;&#xA;&lt;li&gt;The contractor has to remove information not relevant to their scope of work.&lt;/li&gt;&#xA;&lt;li&gt;(In the commentary:) The engineer may choose to charge the contractor a fee for use of the model, without selling the intellectual property rights to the contractor.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the end, the decision is very job specific. On a fast job with a team that has good existing relationships, it may be worth it to provide the model for free as a good will gesture. On very large projects with lots of locked up risk, it might be wise to only provide what you are contracted to. There may exist a middle ground where providing your model is an option, but only for a price.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-06-19T18:14:35.673" LastActivityDate="2015-06-19T18:14:35.673" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3254" PostTypeId="2" ParentId="2967" CreationDate="2015-06-19T13:58:03.567" Score="3" Body="&lt;h1&gt;Short Answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Yes it can hold. Because there are 10 brackets on a '57 Chevy front bumper and some more on rear bumper but not all of them are always under load conditions. A broken bracket or lost screw can be compensated with chewing gum.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Long Answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I haven't seen an actual '57 Chevy Sedan however with some chewing gum can hold bumper for a time if its used for lost screws on side brackets. Because of the design philosophy it seems there are a lot redundancy in bumper brackets. Modern bumpers are expected to break easily but in 50's they were sturdier. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Front Bumper&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Below is the assembly image I've found from &lt;a href=&quot;http://www.trifive.com/forums/showthread.php?t=28049&quot; rel=&quot;nofollow noreferrer&quot;&gt;TriFive.com&lt;/a&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/8pTd3.png&quot; alt=&quot;&amp;#39;57 Chevy Sedan Front Bumper Assembly&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see the side bumper screws are under shear only under acceleration. So a chewing gum will hold it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If brackets in the middle are loose some chewing gum can hold it for time again there are 10 brackets. But I'm not sure you can reach them without loosening sides. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rear Bumper&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Rear bumper assembly is from TriFive.com again.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/UlSvp.png&quot; alt=&quot;Rear bumper &amp;#39;57 Chevy&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If bumper sagged because of a few screws got loose or bracket legs are broken chewing gum can hold for some time. However again it won't hold for long. &lt;/p&gt;&#xA;" OwnerUserId="879" LastActivityDate="2015-06-19T13:58:03.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3255" PostTypeId="2" ParentId="1979" CreationDate="2015-06-19T19:21:57.027" Score="1" Body="&lt;p&gt;Roark's Table 15.3, Case 1e has a formula on the rubber ring, just to check your math if you want to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the inside radius to compress to 0, we'd need a compression force to equal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\text{Total Closing Force}}{\text{Volume of Rubber Ring}} = \frac{3E}{R_o(\frac{R_i(1-\nu)}{R_o} + \frac{2R_o(2+\nu)}{R_o+R_i})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know your properties of your rubber, but this would be the force required to close the rubber down to a 0 inside radius.  As $R_o \to 2R_i$ (which is what the pictures look like), and for most rubbers, $\nu = 0.5$, so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{Total Closing Force} = (\text{Volume of Rubber Ring}) . \frac{18E}{23 R_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, if the rubber has a somewhat normal modulus (&lt;a href=&quot;http://www.engineeringtoolbox.com/young-modulus-d_417.html&quot; rel=&quot;nofollow&quot;&gt;Engineering Toolbox&lt;/a&gt; cites values on the order of 10 - 100MPa, I'll use 100 MPa below) - the closing force gets crazy high.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, the action of the wedge reduces this force.  Most of this force is taken by the cylindrical wall due to the angle.  So, the pressurizing force would be $\cot(Big.Angle)$ times this force, for whatever that angle is.  If we run with 60 degrees, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Total Pressurizing Force} = 40.52\  \text{MPa}\ (\frac{\text{Volume}}{R_i})$$&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="1293" LastEditDate="2015-09-25T11:06:38.473" LastActivityDate="2015-09-25T11:06:38.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3256" PostTypeId="2" ParentId="3220" CreationDate="2015-06-19T19:25:25.147" Score="3" Body="&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; Just read the comments, #3 might be the best option, as #1 won't work, and #2 wont shape it, just hold it in place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First thing I am assuming is that this stays in the ceramic as it is heated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple things I would try:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fit the coil in the ceramic like the bottom picture is, but with the new coil, then put something heavy on top and let it sit there for a long while.  Hopefully that would work, but no promises.  &lt;/li&gt;&#xA;&lt;li&gt;Find a thin, non-conducting material that can withstand at least twice the temperature that it will be subjected to (Possibly even more ceramic), and put a plus shape over the heating element to hold it in place.  &lt;/li&gt;&#xA;&lt;li&gt;Bend the coil bit by bit to get it to stay in the right shape.  start by bending it to where it should be, then releasing it.  If it does not stay, bend it past the point you want it at.  Keep bending it farther until it stays in the place you want it.  You may need to do very small sections at a time, so this will take a very long time.  Unfortunately, this may cause a little weakness in the metal.&lt;/li&gt;&#xA;&lt;li&gt;Carve a wood shape that matches the ceramic and use that to form the spring using method 1 &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Good Luck, I hope you can get something to work!&lt;/p&gt;&#xA;" OwnerUserId="2047" LastEditorUserId="2047" LastEditDate="2015-06-19T19:35:39.850" LastActivityDate="2015-06-19T19:35:39.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3257" PostTypeId="1" AcceptedAnswerId="3559" CreationDate="2015-06-19T20:43:22.043" Score="3" ViewCount="496" Body="&lt;p&gt;When testing a pressurized pipe with off-standard flanges, often times the materials available to fabricate a blind flange of the correct size are limited to what is around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made it a habit of referring to Roark's formula's for stresses and strains, and going to the table for large deflection circular plates for these.  I use the area inside the bolt circle.  They have three boundary conditions available:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Simply supported (neither fixed nor held)&lt;/li&gt;&#xA;&lt;li&gt;Fixed but not held (no edge tension)&lt;/li&gt;&#xA;&lt;li&gt;Fixed and held.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Bolted connections are unusual - especially since I have a full face flanges.  I've designed to both simply supported and fixed but not held.  The simply supported works, the fixed but not held does not.  However, due to the flatness out to the OD of the face, it seems like the rotation of the edge should be 0, so it should be fixed but not held.  I'd like to make these thinner, so it would be nice if someone can demonstrate an accurate boundary condition (or perhaps that 80% of the fixed + 20% of the simply supported would be conservative via FEA).  I'd like to see better research into the mechanics of design for blind flanges (which is rarely covered in the context of the Boiler and Pressure Vessel Code, or elsewhere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main question is: what boundary condition for the edge should I use?&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="368" LastEditDate="2015-07-09T19:50:22.783" LastActivityDate="2015-09-12T00:52:00.967" Title="What boundary condition should I use for the edge of a blind flange?" Tags="|mechanical-engineering|piping|bolting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3258" PostTypeId="2" ParentId="3071" CreationDate="2015-06-20T04:06:41.473" Score="2" Body="&lt;p&gt;You're seeing things correctly - but you're not looking at the big picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our shop, we've got a fan that is rated for 0.036 psi of static pressure and 8000 cfm for dust control. It's 36&quot; in diameter.  The fan also has a damper a few feet after it for speed control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I close the damper completely,  the duct in front of the damper will experience a pressure rise, almost to 0.034 psi. The duct behind the damper remains at 0 psig. Since this is the fan's top pressure it can push, and the pressurized air in the duct is pushing back on the fan, almost no air flows through the duct. Instead, the fan is mostly trying to keep the space between the fan and the damper pressurized.  Think of this as potential energy (or voltage), and the damper is acting as a resistor removing that potential energy, with no current. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I open the damper,  the duct behind the damper still remains at 0 psi, but since I don't have a damper blocking the way,  the air can come rushing out the fan at all 8000 cfm. The duct before the damper is nearly at 0 psi as well. So,  the whole system is operating at a lower pressure, less than the 0.03 psi. There is no potential energy drop across the damper. But,  instead of working like an electrical &quot;battery&quot;, the fan operates on a &lt;a href=&quot;http://www.greenheck.com/library/articles/10&quot; rel=&quot;nofollow&quot;&gt;fan curve&lt;/a&gt; and the pressure in the entire ducting system drops, rather than keeping the same &quot;voltage&quot; like you would be used to in an electrical system. But the energy is there - just converted from static (pressure) energy to dynamic energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stators operate the same as the damper.  Imagine if another fan grabbed that pressurized air between the fan and the damper and pushed that against another damper. Repeat as often as you want and you have high pressure air. The same concept, just two different applications. &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-06-20T04:06:41.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3259" PostTypeId="1" AcceptedAnswerId="3260" CreationDate="2015-06-20T08:23:30.653" Score="2" ViewCount="243" Body="&lt;p&gt;I have machined parts that are made of aluminum and anodized black.  The reason that I chose aluminum is because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It can easily be machined and is low cost and is common, so easy to buy&lt;/li&gt;&#xA;&lt;li&gt;It is light weight compared to steel&lt;/li&gt;&#xA;&lt;li&gt;It can take heat&lt;/li&gt;&#xA;&lt;li&gt;It is easy to finish with oxidation&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8CmIt.jpg&quot; alt=&quot;1![2![3&quot;&gt;&lt;br&gt;&#xA;The parts are used on a CNC machine so being lightweight is important so that the machine can move faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I need my part to be significantly lighter weight, so would like to switch materials, but not pay a lot to get a prototype made.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there another metal that is a good substitute?  &lt;/li&gt;&#xA;&lt;li&gt;Can a plastic laminate be used? For example, plastic parts that I can get machined then add a high temperature tape to reflect heat?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2053" LastEditorUserId="368" LastEditDate="2015-07-09T19:46:20.877" LastActivityDate="2015-07-09T19:46:20.877" Title="Lightweight substitutes for machined aluminum parts" Tags="|mechanical-engineering|aluminum|cnc|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3260" PostTypeId="2" ParentId="3259" CreationDate="2015-06-20T11:37:57.373" Score="1" Body="&lt;p&gt;I'm not sure how your part looks, but I can tell you about temperature and plastics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've blasted &lt;a href=&quot;https://en.wikipedia.org/wiki/Phenol_formaldehyde_resin&quot; rel=&quot;nofollow&quot;&gt;phenolic&lt;/a&gt; composites with &lt;strong&gt;600 °F (310 °C)&lt;/strong&gt; and haven't had a problem. There are high temperature epoxies and vinyl esters that get to lower temperatures - and those are a lot more commonly available.  I wouldn't touch them with anything over &lt;strong&gt;300 °F (150 °C)&lt;/strong&gt; though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All composites range from &lt;em&gt;0.05 - 0.07 pci (1.38 g/cc - 1.97 g/cc)&lt;/em&gt;, depending on if you use more reinforcement, or more resin.  The phenolic can be as strong as an epoxy laminate, in testing.  For glass reinforcement, you can guarantee breaking strengths of &lt;strong&gt;25 ksi (200 MPa)&lt;/strong&gt;.  If you designed your parts with a professional, orienting the fibers, you can get above &lt;strong&gt;500 MPa&lt;/strong&gt;.  Wikipedia cites breaking strengths going into the GPa range, but there is no guarantee your part will have that - those are lab sample strengths, not actual usable part strengths.  Bear in mind fiberglass is fatigue resistant (better than your aluminum!), but only if the part is designed with a factor of safety of 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps you understand the world of high temperature composite substitutes.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="368" LastEditDate="2015-07-09T19:42:08.400" LastActivityDate="2015-07-09T19:42:08.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3263" PostTypeId="1" AcceptedAnswerId="3272" CreationDate="2015-06-21T19:05:01.283" Score="5" ViewCount="94" Body="&lt;p&gt;It is possible with a microphones array to isolate a sound coming from a direction. Is it possible with an array of photo diodes to isolate a light signal coming from one direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example having 4 photo-diodes oriented at the same direction, spaced by 2 cm, and facing lamps at the ceiling, is it possible to isolate the amplitude of each light source reaching the photo-diodes from different directions (different lamps). What would be the algorithm or the method used for doing such thing?&lt;/p&gt;&#xA;" OwnerUserId="1849" LastEditorUserId="1849" LastEditDate="2015-06-22T07:58:18.293" LastActivityDate="2015-06-22T15:07:07.953" Title="Light signal isolation" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3264" PostTypeId="1" CreationDate="2015-06-22T06:28:06.297" Score="3" ViewCount="1244" Body="&lt;p&gt;I need to design a simple gearbox for homework. I figured out to locate 2 sliding spur gears on the input shaft and 2 fixed spur gears on the output shaft. I prefer to use key joints, as they are the simplest possible.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How can the sliding gears be constrained? Will a long key do?&lt;/li&gt;&#xA;&lt;li&gt;How can the fixed gears be fixed into place? Will another design of key joint work?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="379" LastEditorUserId="282" LastEditDate="2015-06-22T22:41:00.803" LastActivityDate="2015-10-03T13:32:23.013" Title="How to limit axial movement for a key joint?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3265" PostTypeId="1" CreationDate="2015-06-22T07:50:34.630" Score="7" ViewCount="676" Body="&lt;p&gt;Almost every railroad and vehicle road landfill has its slopes overgrown with grass which is quite good at preventing erosion. However it takes some time (a year or more) for the grass to grow and emit enough roots to form a reliable mesh that would hold the earthfill surface and rains can cause enough damage to the earthfills before grass grows and becomes efficient against erosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are earthfills not eroded by rainfall before they are overgrown with grass?&lt;/p&gt;&#xA;" OwnerUserId="374" LastActivityDate="2017-10-27T12:25:57.097" Title="How are earthfills not eroded before they are overgrown with grass?" Tags="|civil-engineering|geotechnical-engineering|highway-engineering|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3266" PostTypeId="2" ParentId="3265" CreationDate="2015-06-22T08:22:35.723" Score="5" Body="&lt;p&gt;The slope angle of such slopes is constructed to be less than the &lt;a href=&quot;https://en.wikipedia.org/wiki/Angle_of_repose&quot; rel=&quot;nofollow noreferrer&quot;&gt;angle of repose&lt;/a&gt; for the material used in the slope. The angle of repose is the great angle to which granular materials can be dumped without slumping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z9yAt.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The higher the cohesiveness of the material, the higher the angle of repose. Stability of such a slope will be improved by compacting the material during construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to know the angle of repose for different materials is via testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shallower the slope angle, the slower the speed of water run-off, resulting in reduced erosion capabilities of water on slopes. Another method that assists in slope stabilization is to use &lt;a href=&quot;https://en.wikipedia.org/?title=Geotextile&quot; rel=&quot;nofollow noreferrer&quot;&gt;geotextiles&lt;/a&gt; to protect the surface of the slope. Additionally, slopes can be sprayed with a mixture of grass seeds and bitumen. The grass seeds develop roots and stabilize the surface while the bitumen hold the soil in place. Over time the bitumen breaks down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In arid regions, an initial stabilization is achieved by ensuring a quick growing grass develops a network of roots to lock in the soil. The grass is then deprived of water and allowed to die, but the network of roots remains. Once this happens, the long term stabilizing grass seed is allowed to grow on the slope, establishing it own network of roots.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="368" LastEditDate="2015-06-22T16:36:13.547" LastActivityDate="2015-06-22T16:36:13.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3267" PostTypeId="2" ParentId="3265" CreationDate="2015-06-22T08:36:10.247" Score="4" Body="&lt;p&gt;They lay turf pads down to kickstart the grass coverage. Even the loose pads will have better erosion protection than bare soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other options include laying down a sheet with holes for plants/shrubery to grow through, the sheet will let enough rain through to let the plants grow but prevent the hillside from eroding until the planted shrubs are robust enough to hold the soil.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-06-22T08:36:10.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3268" PostTypeId="1" AcceptedAnswerId="3271" CreationDate="2015-06-22T12:26:39.303" Score="9" ViewCount="12061" Body="&lt;p&gt;Every scientific calculator I've come across has a DRG button that controls whether trigonometric functions use units of degrees, radians or grads yet I've never seen or heard of any system that actually uses grads. When is the grads setting useful?&lt;/p&gt;&#xA;" OwnerUserId="2060" LastEditorUserId="368" LastEditDate="2015-06-22T15:06:30.170" LastActivityDate="2020-01-17T15:17:17.737" Title="What are grads used for on a scientific calculator?" Tags="|measurements|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3269" PostTypeId="2" ParentId="2774" CreationDate="2015-06-22T12:35:40.007" Score="5" Body="&lt;p&gt;This is a simple shear force problem. However there are a few considerations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You don't need 2 hydraulic cylinders. One should be mobile, the other should be stationary. Below is a simple and quick figure to show how they should align.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/B0cGB.png&quot; alt=&quot;Shear mechanism&quot;&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You should check &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_(sheet_metal)&quot; rel=&quot;noreferrer&quot;&gt;Sheet Metal Shears&lt;/a&gt; they will give you an idea.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Below is the simple equation for shear. &lt;strong&gt;T&lt;/strong&gt; is shear stress, it is around 400MPa for 0.3%C steels. If you don't have shear stress values you can use ultimate tensile stress values and multiply by 0.6-0.7, for an approximation. Use cross section area of your bar for &lt;strong&gt;A&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/8vPYj.gif&quot; alt=&quot;Shear force equation&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;304 steels aren't a great idea for this kind of work. It can't be hardened by heat treatment, just by cold working. So you won't have long life from your tool. Consider a martensitic stainless steel such as 410, you can harden it and have much higher life. If you don't need corrosion resistance then consider a heat treatable steel. However if you must use 304, try hardening it by hammering, or if you have rolling. That will increase tool life.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="879" LastActivityDate="2015-06-22T12:35:40.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3270" PostTypeId="2" ParentId="3268" CreationDate="2015-06-22T12:55:25.413" Score="7" Body="&lt;p&gt;Gradians are commonly used in land surveying. I only know this because I made the mistake of asking the same question of a Civil Engineer.   &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-06-22T12:55:25.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3271" PostTypeId="2" ParentId="3268" CreationDate="2015-06-22T13:02:03.833" Score="13" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gradian&quot;&gt;Grads&lt;/a&gt; were developed by the French as part of the metric system and have been referred to as the metric degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some countries in Europe surveyors use grads instead of degrees. The other use is by French artillery units who have used grads for decades.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="272" LastEditDate="2015-06-22T13:50:50.400" LastActivityDate="2015-06-22T13:50:50.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3272" PostTypeId="2" ParentId="3263" CreationDate="2015-06-22T15:07:07.953" Score="6" Body="&lt;p&gt;This concept in general as it applies to microphones is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Beamforming&quot;&gt;beamforming&lt;/a&gt;, and is most commonly applied to sound and radio waves. It is mostly aimed at wave forms that carry signals (ie change significantly over time,) but similar concepts could be used in your example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general terms, beamforming on the receiving side comes in two flavors. Either you know the location of the signal you want to receive, and you phase-align your sensors to be more sensitive to that direction, or you know what your signal 'looks' like, and you can locate it based on phase differences in your array of sensors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are proposing the second application, so the difficult thing for you is knowing how to distinguish one light source from all of the others. For example, if you had only one light in a dark room, it would be easy to identify your target signal. In your situation with multiple lights, if they all produce the same wavelengths, it will be nearly impossible to distinguish them if they are all on at the same time. If they produce different colors, you could possibly distinguish them, but you would still have a different problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for beamforming based on phase shift to work, you have to be able to identify each received waveform, and index them against each other. There are two problems here. The first, inherent problem is that conventional light bulbs produce essentially the same wave form forever, dimming very slowly, until they burn out. The problem with this is that since the waveform has no distinct changes in it, you can't index it to find time delays. when you see the same signal from two sensors, you might notice that they are 180 degrees out of phase, but you wouldn't be able to tell if they are just half a wavelength different, or 1000.5 wavelengths different. In order to tell how much more the signal has been delayed traveling to one sensor, you need a distinguishable change in the waveform (for example the light getting brighter or changing color.) The second, more pragmatic problem is that I'm not aware of any photodiode that is so responsive that you could actually observe the waveform of light. I'm not an electronics expert though, so they may exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need a strategy to make the waveform of the light source less uniform, and ideally to add a much lower frequency signal which is easier to detect. Assuming you have control over the lights, this should be easy to do. At its simplest, you could turn on one light at a time, and watch for the first light to hit your sensors. Since you know how far apart your sensors are, and you can easily identify the relative delays between the first and last sensor the light hits, you can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Multilateration&quot;&gt;Multilateration&lt;/a&gt; to locate the source. If it's not an option to only have one source on at a time, there are other ways to make the source identifiable such as perhaps turning it from 100% brightness to 80% brightness or changing it from white to blue at intervals. How significant these changes will be, and how robustly your system will detect them will depend on the sensitivity of your instruments and the degree of background noise.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-06-22T15:07:07.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3273" PostTypeId="1" AcceptedAnswerId="3276" CreationDate="2015-06-22T17:37:19.177" Score="0" ViewCount="1071" Body="&lt;p&gt;&lt;a href=&quot;https://youtu.be/WOoUVeyaY_8?t=284&quot; rel=&quot;nofollow noreferrer&quot;&gt;In this video&lt;/a&gt;, the video emphasises the very low recoil level of the AA-12 fully automatic shotgun. In particular, at &lt;a href=&quot;https://youtu.be/WOoUVeyaY_8?t=300&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;this point in the video&lt;/strong&gt;&lt;/a&gt; the presenter uses one hand to relatively lightly grip the gun while firing 300 rounds per minute. Note that the presenter does not brace the shotgun against his shoulder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below is a frame from the video at 5m:03 - it shows the gun in&quot;mid burst&quot; - two cartridge cases can be seen in the air, one having just been ejected and the other is closer to the camera. The users 'stance' appears to be intended to show the effortlessness of holding the gun while firing. If he is in fact using typical 12 gauge shells (which he implies that he is), holding a &quot;normal&quot; shotgun like this when firing would result in it being torn from his hands. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AB2cV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the video the presenter claims that the shotgun exhibits, for all intents and purposes, zero recoil. He remarks that there is a spring system in the buttstock that counters the momentum of the recoil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of two ways that the spring system could work:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The first recoil (or something else) compresses the spring, and then the spring extends upon the generation of the second recoil. &lt;/li&gt;&#xA;&lt;li&gt;Each recoil compresses a stiff spring that extends more slowly than the recoils compress it. Assuming a sufficiently quick firing rate over a sufficiently protracted period, complete compression of the springs would obtain; consequently, the user would need to abstain from firing the shotgun until the springs extended, lest s/he absorb the recoil herself. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, it seems to me that those systems would work only if the user braced the shotgun against something (i.e. his/her shoulder). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the simplest spring system that could produce recoil absorption comparable to that exhibited in the video?&lt;/p&gt;&#xA;" OwnerUserId="2064" LastEditorUserId="399" LastEditDate="2015-06-23T15:50:45.170" LastActivityDate="2015-06-23T15:50:45.170" Title="What is the simplest spring system that could produce recoil absorption comparable to that exhibited by the AA-12 assault shotgun?" Tags="|mechanical-engineering|springs|firearms|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3274" PostTypeId="2" ParentId="1788" CreationDate="2015-06-22T18:26:55.010" Score="1" Body="&lt;p&gt;A martial artist's live strikes can be quantified between practitioners using &lt;strong&gt;force applied as pounds per square inch&lt;/strong&gt;.  The basic premise is (of course) force equals mass times acceleration.  The importance is pairing with the &lt;strong&gt;area of contact&lt;/strong&gt; used with which it is delivered - &lt;strong&gt;inverse square law&lt;/strong&gt;.  These two are the main pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course wild complications in live striking - like transverse compression waves and [insert just about anything here].  In basis, transferring force can be qualitatively assessed through trained observation and the particular study of &lt;strong&gt;change of momentum, relative position and angle of attack relative to centre-mass&lt;/strong&gt;.  (Radial and longitudinal effects should be gestured rather than demonstrated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that there are some factors (not including deformation and shear stress) which cannot be accurately repeated in simulation and do require quantification/demystification - but is much harder to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, practitoners seek the obvious quantification of a strike:  &lt;strong&gt;timing&lt;/strong&gt; and &lt;strong&gt;accuracy&lt;/strong&gt;.  It should take into account the effectiveness of a technique arising as a result of competing motions.  A trained student's &lt;strong&gt;degree of automaticity&lt;/strong&gt; should also be assessed. &lt;/p&gt;&#xA;" OwnerUserId="2065" LastActivityDate="2015-06-22T18:26:55.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3275" PostTypeId="2" ParentId="3268" CreationDate="2015-06-22T18:50:13.343" Score="0" Body="&lt;p&gt;Grade is often used for inclination, such as looking at how steep a road is.  The units are not as &quot;crazy&quot; as they seem, when you think of grade as a percentage.  In other words, a hill can go down at anything from 0% (flat) to 100% (45 deg angle).  (rise/run)&#xA;Based on some comments added below, it is not clear if the G on this calculator is actually grade or not.&lt;/p&gt;&#xA;" OwnerUserId="2066" LastEditorUserId="2066" LastEditDate="2015-06-23T02:27:54.557" LastActivityDate="2015-06-23T02:27:54.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3276" PostTypeId="2" ParentId="3273" CreationDate="2015-06-22T21:16:21.620" Score="2" Body="&lt;p&gt;The simplest, and possibly only, spring only system that would work like that would be made of nonexisteum as the energy would still have to be dealt with, and the power level exceeds what is seen in the video.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This video &lt;a href=&quot;http://www.thefirearmblog.com/blog/2015/04/14/aa-12-fully-automatic-shotgun/&quot; rel=&quot;nofollow&quot;&gt;TFB&lt;/a&gt; provides another firing demo which gives a more realistic impression of the recoil level experienced in practice. The shooters take care to place the butt to their shoulders in a conventional manner, even after having fired a number of shots, and the recoil reaction on this bodies can clearly be seen. While the recoil may be substantially lower, it is nowhere nearly as low as &quot;zero recoil&quot; - and also is much lower than you'd expect from the initially cited video, suggesting that he used very low energy loads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a spring plus 'other things' to spread the energy pulses and absorb as much energy as possible would work - and demonstrably does, as the (US, not Russian) AA12 is an entirely real gun.&lt;a href=&quot;https://en.wikipedia.org/wiki/Atchisson_Assault_Shotgun&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;The original AA-12 was a 1972 US design&lt;/strong&gt;&lt;/a&gt; by Maxwell Atkinson (&lt;a href=&quot;https://www.google.co.nz/search?q=aa-12+shotgun&amp;amp;biw=1422&amp;amp;bih=876&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ei=6XCIVdrBBcS0mAXoqrWoDw&amp;amp;ved=0CAYQ_AUoAQ&amp;amp;dpr=0.9&quot; rel=&quot;nofollow&quot;&gt;images&lt;/a&gt;). It has since been developed by 'Military Police Systems' into the current version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Atchisson_Assault_Shotgun#History&quot; rel=&quot;nofollow&quot;&gt;Wikipedia advises&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the bolt flies back after firing to cycle another round, around&#xA;   80% of what would normally be felt as recoil is absorbed by a&#xA;   proprietary gas system. A recoil spring grabs another 10%, leaving the&#xA;   final recoil a remarkable 10% of the normal recoil for a 12-gauge&#xA;   round—so you can point the AA-12 at a target and unload the full&#xA;   magazine without significant loss of accuracy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on the notes below it seems likely that the &quot;proprietary gas system&quot; not only absorbs and stores the energy initially, but must also dissipate energy - presumably by heat transfer via compression. Power level to be dealt with during during firing at full rate of 300 shells/minute is about 150 Watts, or about 600 Joule for a maximum of under 4 seconds for an 18 round drum magazine. 600 Watt.seconds (600 J) is enough to make the gun 'somewhat warm' but low enough to be dealt with realistically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the level to be dealt with by the user is 10% of total, as claimed that's a continuous rate of about 15 Watts. That's a level of power input that people would exert for short periods on many hand operated units that take noticeable user power (many not so well known today) - sirens, power generators, window winders ... .  Well below the level needed to crank a car by hand (remember those ? :-) ). 12 gauge load energy can vary significantly. Looking at the effort taken in the video I'd guess his loads were not up to maximum power level for 12 gauge. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall the user + gun must absorb energy from the explosive charge as they apply an equal and opposite reactive force to the accelerating projectile. The velocity of the overall mass (user + gun) will be low (fortunately) compared to that of the projectile(s). The timing and shape of the reaction can be altered by springs (gas or mechanical) and moving mass but the &quot;reactive energy&quot; must be absorbed either as heat + user absorbtion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watching the video portion in question it appears that the gun barrel dips during the period during and after firing rather than rising as I would have expected. This could be achieved by accelerating an internal mass using pre-stored energy (gas, springs, ...) to counterbalance the impulse peak. The energy still needs to be absorbed but with suitable design could be spread across the available time period linearly or in a manner that best suits the users muscle response. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy involved varies with load used and can differ by more than 5:1 - so what you see in the video and what you experience with a maximum load can vary considerably. This reference &lt;a href=&quot;http://www.chuckhawks.com/12gauge.htm&quot; rel=&quot;nofollow&quot;&gt;The popular 12 gauge&lt;/a&gt; indicates that the upper limit that most &#xA;recreational shooters can tolerate from a 12 gauge shotgun is 21 foot-pounds of recoil energy*. That translates to about 3 kg.m or slightly under 30 Watt seconds.&lt;br&gt;&#xA;That's not as arcane a unit as it sounds because - the 300 rounds/minute firing rate = 5 rounds/second or a continuous  power level at 30 W.s per shell of 30 x 5 = 150 Watts. Providing a sustained 150 Watts is a substantial workload. Most users on an exercise bike would be unhappy to maintain that level for more than a few minutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I took the recoil energy, projectile mass and muzzle velocity figures they quoted on that page and compared calculated kinetic energy of the projectile with recoil energy. I had expected them to correlate approximately linearly but the claimed recoil energy increases much more rapidly than I'd expect from the calculated projectile energy. I tried various values of x in 0.5 x m x V^x without getting a reasonable fit. Using impulse (delta mV) is worse. Interesting. Maybe a better understanding of acceleration profile and burn dynamics is needed. Or, no doubt easier, just measure it. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Related:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.gizmag.com/aa-12-combat-shotgun-frag-12-automatic/11393/&quot; rel=&quot;nofollow&quot;&gt;Gizmag&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://world.guns.ru/shotgun/usa/atchisson-aa-12-e.html&quot; rel=&quot;nofollow&quot;&gt;World guns.ru&lt;/a&gt; More history and tehnical details.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-06-23T14:42:21.273" LastActivityDate="2015-06-23T14:42:21.273" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3277" PostTypeId="2" ParentId="3265" CreationDate="2015-06-22T21:31:49.647" Score="5" Body="&lt;p&gt;There are a few methods that are used to control erosion before permanent vegetation is established. The exact method used can depend on location (also country), season, preference, governing laws, etc.&lt;/p&gt;&#xA;&lt;h3&gt;Mulching&lt;/h3&gt;&#xA;&lt;p&gt;This is almost always done. &lt;a href=&quot;https://en.wikipedia.org/wiki/Mulch&quot; rel=&quot;noreferrer&quot;&gt;Mulch&lt;/a&gt; is frequently straw but can also be ground up newspaper. This helps to initially keep the soil from eroding while plants are growing. It also somewhat helps retain moisture in the soil so that plant will grow easier.&lt;/p&gt;&#xA;&lt;p&gt;The mulch is sometimes pressed into the soil with tracked equipment, or a type of netting is placed with the mulch. Both of these help to keep the mulch from blowing away.&lt;/p&gt;&#xA;&lt;p&gt;Hydro mulching is also a method that is frequently combined with &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydroseeding&quot; rel=&quot;noreferrer&quot;&gt;hydroseeding&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Seeding&lt;/h3&gt;&#xA;&lt;p&gt;The choice of seed is also a factor. Many times a portion of the seed used is a variety specifically chosen because it is quick growing. This plant may not be a species that will survive the winter, but it serves as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Cover_crop&quot; rel=&quot;noreferrer&quot;&gt;cover crop&lt;/a&gt; while the permanent vegetation grows.&lt;/p&gt;&#xA;&lt;h3&gt;Mechanical Means&lt;/h3&gt;&#xA;&lt;p&gt;In addition to the effects of mulch, other measures can be taken to slow down the speed of water flowing down a slope. Slow moving water is more likely to soak into the soil. Water that doesn't soak in will be moving slow enough that it won't cause erosion.&lt;/p&gt;&#xA;&lt;p&gt;A few of these methods are:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Grade the top of slope so that rain water isn't funneled to one point. Counterintuitively, this means making sure that there isn't a small berm at the top of slope. A small berm will eventually fail and cause water to be directed to one spot, causing erosion.  Without a berm, any water trying to flow down the slope is more evenly distributed.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Install terraces or &lt;a href=&quot;https://en.wikipedia.org/wiki/Fiber_roll&quot; rel=&quot;noreferrer&quot;&gt;fiber rolls&lt;/a&gt; staked across the slope. These help to keep the water from flowing in a straight line down the slope and causing erosion.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mats or wattles of natural material can be placed. These function similarly to mulch, but may withstand higher water velocities in areas likely to flood.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rock of various sizes can be placed. Plants will eventually grow through the rock.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is usually not recommended, but in the past, an &lt;a href=&quot;https://en.wikipedia.org/wiki/Asphalt#Asphalt_emulsion&quot; rel=&quot;noreferrer&quot;&gt;asphalt emulsion&lt;/a&gt; was sprayed on slopes to keep them from eroding. The thought was that plants would take root as the asphalt formed cracks.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-06-22T21:31:49.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3278" PostTypeId="1" CreationDate="2015-06-22T22:02:50.120" Score="5" ViewCount="161" Body="&lt;p&gt;I get the impression that a lot of old buildings fail to meet modern earthquake safety norms, and need to be rebuilt or retrofitted.  But what has changed in the last few decades that would make a building constructed today better than an older building?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you asked why modern automobiles are safer, I would say that the primary driving force (pardon the pun) is technology — crumple zones, seat belts, airbags, antilock brakes, traction control, distance sensors, alertness alarms — coupled with decreasing acceptance of drunk driving and public outrage over &lt;a href=&quot;https://en.wikipedia.org/wiki/Ford_Pinto#Design_flaws_and_ensuing_lawsuits&quot; rel=&quot;nofollow&quot;&gt;Ford Pintos&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you asked why aviation is safer, I would also credit technology (e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Traffic_collision_avoidance_system&quot; rel=&quot;nofollow&quot;&gt;TCAS&lt;/a&gt;), but mostly an accumulation of experience from all past mishaps.  For example, after &lt;a href=&quot;https://en.wikipedia.org/wiki/Tenerife_airport_disaster&quot; rel=&quot;nofollow&quot;&gt;Tenerife&lt;/a&gt;, the industry adopted standard radio phraseology.  After &lt;a href=&quot;https://en.wikipedia.org/wiki/Germanwings_Flight_9525&quot; rel=&quot;nofollow&quot;&gt;Germanwings Flight 9525&lt;/a&gt;, most airlines will not allow pilots alone in the cockpit again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what drives improvements in seismic safety?  Exotic base-isolation schemes aside, have technology and construction techniques changed much in the last few decades?  &lt;strong&gt;Was there anything preventing earlier builders from making structures as strong as they could be, or today's builders from making them stronger than they currently are?&lt;/strong&gt;  Is it a matter of decreasing willingness to risk human lives as life expectancy increases?  Increased economic prosperity allowing money to be spent on more concrete and rebar?   A tendency for government regulation to grow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is the premise of the question flawed to begin with?&lt;/p&gt;&#xA;" OwnerUserId="446" LastEditorUserId="33" LastEditDate="2016-02-23T02:20:08.693" LastActivityDate="2016-02-23T17:26:29.993" Title="What drives improvements in seismic safety?" Tags="|civil-engineering|geotechnical-engineering|safety|seismic|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3279" PostTypeId="2" ParentId="3231" CreationDate="2015-06-22T23:12:11.410" Score="2" Body="&lt;p&gt;I did something similar in an apartment, where there were air conditioners in the bedrooms at either end of the apartment and a living room and kitchen between them.  I mounted two side by side window fans in the tops of the doorways between the one bedroom and the living room and the other bedroom and the kitchen.  Similar to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00008XET9&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;this&quot;&gt;this&lt;/a&gt;.  I'm the only one in the family tall enough to have to duck going through the doorway, and I did not live there, so it was not a problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fans were usually run to take the higher level hot air from the central rooms into the bedrooms, forcing the cool air in the bedrooms back out through the lower level.  Going the other way was also possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the house in question, this would include the shared hallway in the conditioned space, which may be acceptable.  It would require the doors to be open, which may not be acceptable.  It would force air into the hallway from the bedroom at the top of the doorway, say, and allow return at a lower level.  It's not completely efficient because of short circuiting, but it should work to some level.  The advantage is that it's cheap and easy.  I just used a wooden bracket below the top of the doorway as a support for the fan.  I did have the advantage of rectangular doorways, however.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tXAgJ.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="998" LastEditorUserId="998" LastEditDate="2015-06-23T11:11:15.240" LastActivityDate="2015-06-23T11:11:15.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3280" PostTypeId="2" ParentId="3278" CreationDate="2015-06-22T23:55:22.613" Score="8" Body="&lt;p&gt;This is a pretty huge topic, but I can summarize a couple major factors.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;More population, more damage, more awareness&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The effect of &lt;a href=&quot;https://en.wikipedia.org/wiki/1906_San_Francisco_earthquake&quot; rel=&quot;nofollow&quot;&gt;a disaster that kills thousands of people&lt;/a&gt; and causes property damage in the billions of dollars (current equivalent) is felt much more intensely than that of a &lt;a href=&quot;https://en.wikipedia.org/wiki/1812_Wrightwood_earthquake&quot; rel=&quot;nofollow&quot;&gt;disaster of similar magnitude&lt;/a&gt; that affects a much smaller population. Note that the magnitudes indicated on these articles are not the same measurements; &lt;a href=&quot;http://booksite.elsevier.com/9780123735768/casestudies/01~Appendix_1.pdf&quot; rel=&quot;nofollow&quot;&gt;6.9 M&lt;sub&gt;L&lt;/sub&gt; is actually bit more than 6.9 M&lt;sub&gt;W&lt;/sub&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The twentieth century saw huge growth in world population and, more importantly, in population density in and around urban areas. Earthquakes are localized natural disasters that will affect as many people and structures as exist wherever they happen to occur. The 1906 quake in San Francisco, and &lt;a href=&quot;http://earthquake.usgs.gov/earthquakes/world/events/1960_05_22.php&quot; rel=&quot;nofollow&quot;&gt;other&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/?title=1985_Mexico_City_earthquake&quot; rel=&quot;nofollow&quot;&gt;large&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/1989_Loma_Prieta_earthquake&quot; rel=&quot;nofollow&quot;&gt;quakes&lt;/a&gt; throughout the 20th century, had huge impacts on the world's awareness of earthquake danger, both because they affected more and more people and because modern communication technology allowed their effects to be recorded and broadcast to an unprecedented degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To go way back, consider that when the &lt;em&gt;entire city&lt;/em&gt; of Pompeii was buried by the eruption of Mount Vesuvius, &quot;only&quot; about ten thousand people were killed. &lt;a href=&quot;https://en.wikipedia.org/wiki/Historical_urban_community_sizes&quot; rel=&quot;nofollow&quot;&gt;Pompeii was not a very large city for that time&lt;/a&gt; but a very large city still would have had less than a million people living in it, whereas &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_metropolitan_areas_by_population&quot; rel=&quot;nofollow&quot;&gt;modern urban areas have populations in the tens of millions&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Seismology is a young field&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The basic field of science that underlies seismic design and engineering is only about 150 years old; see &lt;a href=&quot;http://earthquake.usgs.gov/hazards/about/workshops/thailand/downloads/CSMpp1_History.pdf&quot; rel=&quot;nofollow&quot;&gt;A Brief History of Seismology (USGS)&lt;/a&gt; for an overview.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's possible to make a building strong without understanding the underlying science. You don't have to know what makes a building fall down to recognize that when a quake hits, more buildings of one type fall down than of another type. What you can't really do is agree upon and implement the type of evidence-based standards that really impact seismic safety on a broad scale. Architects want to do what looks good, governments want to do what's cheap, project managers want to do what's easy, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we had to understand a bit of what was going on; then, when we figured out what we needed to measure, we had to collect data. Earthquakes are not exactly easy to simulate when you're still trying to work out the basic theory involved. So it took some time, just as with aviation technology, before we knew what to do to have the best impact.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Has technology changed much in the past few decades?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Maybe not the technology itself so much as our understanding of the technology. &lt;a href=&quot;https://en.wikipedia.org/wiki/1994_Northridge_earthquake#Building_code_changes&quot; rel=&quot;nofollow&quot;&gt;The 1994 Northridge quake&lt;/a&gt; is an example of what I mean; we've used a lot of different methods and technologies for a long time without having the empirical evidence required to understand which of them work better than others, or why. There's nothing particularly difficult or high-tech about bracing cripple walls, bolting structures to their foundations or adding reinforcing elements to concrete, once you figure out just how much of a difference it makes (and &lt;a href=&quot;http://earthquake.usgs.gov/earthquakes/states/us_damage_eq.php&quot; rel=&quot;nofollow&quot;&gt;where it's most needed&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-02-23T17:26:29.993" LastActivityDate="2016-02-23T17:26:29.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3281" PostTypeId="5" CreationDate="2015-06-23T10:08:15.443" Score="0" Body="&lt;p&gt;Machining is a process of shaping that is done by removing material from raw material. It is also called subtractive manufacturing to distinguish from additive manufacturing.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Types of machining processes&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Basic machining processes are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Turning&lt;/strong&gt;, work piece is rotated and cutting tool is used to remove material. Lathe is the name of the machine used in this process.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Milling&lt;/strong&gt;, work piece is stationary a rotating cutting tool is used to remove material. Milling machine is the name of the machine used in this process.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Drilling&lt;/strong&gt;, work piece is stationary and a rotating cutting tool is used to cut holes or shape holes in the part. Drilling presses are used for drill operation but milling machines and lathes can also be used for drilling.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are also other machining operations such as boring and grinding or operations that use pressured fluid that contains abrasive particles to remove material such as waterjet cutter or electron discharge machining (EDM) that uses electron discharges to shape material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preparation of surface finishes and tolerances can also be included in this tag.&lt;/p&gt;&#xA;" OwnerUserId="879" LastEditorUserId="33" LastEditDate="2016-02-23T02:58:32.330" LastActivityDate="2016-02-23T02:58:32.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3282" PostTypeId="4" CreationDate="2015-06-23T10:08:15.443" Score="0" Body="Questions about shaping material with removing material from raw material, i.e. subtractive manufacturing. Posts may have [mechanical-engineering] and relevant material tag such as [steel] and if needed process tags such as [milling]." OwnerUserId="879" LastEditorUserId="33" LastEditDate="2016-02-23T02:58:32.330" LastActivityDate="2016-02-23T02:58:32.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3283" PostTypeId="1" CreationDate="2015-06-23T11:56:27.783" Score="6" ViewCount="1015" Body="&lt;p&gt;What are typical forging loads for various metals and configurations? I know that the typical test for a metal is to upset a 0.750&quot; high x 0.500&quot; diameter cylinder to a height of 0.5&quot;, but could not find any actual tables of what kind of load it takes to do this for various metals. In theory, I guess it would actually be plot of load x temperature for each metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm mainly concerned with upsetting loads and  backward extrusion loads.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="368" LastEditDate="2015-06-23T16:06:51.117" LastActivityDate="2015-06-23T16:06:51.117" Title="What are typical forging loads?" Tags="|metals|forging|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3284" PostTypeId="2" ParentId="3283" CreationDate="2015-06-23T13:23:49.577" Score="7" Body="&lt;h1&gt;Short answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Forging loads are dependent on total deformation and die type in addition to temperature. So that kind of table is not possible. But you can check flow stress curves such as given below from &lt;a href=&quot;http://nsmwww.eng.ohio-state.edu/FlowStressModule/Curves/curves.html&quot; rel=&quot;noreferrer&quot;&gt;Ohio State&lt;/a&gt; for 403 steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uuCQM.jpg&quot; alt=&quot;Flow stress vs temperature and strain&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Long answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It's not very easy to give forging loads because they are not linear. Forging loads change with strain. Below are two formulas to calculate forging loads. They are both for open forges, i.e. material flow isn't restricted, if you have a closed die things get fairly complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forging load is given below. Where $\mu$ is coefficient of friction, $\sigma_{f}$ is Flow Stress, h and d are height and diameter of the part at that deformation.&#xA;$$F= \sigma_{f}\frac{\pi}{4}d^{2}\left ( 1+\frac{\mu d}{3h} \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can calculate forging force for non homogeneous deformation is given below. Where $d_{0}$ and $h_{0}$ are initial diameter and height and $A_{f}$ is final area and $\sigma_{u}$ is ultimate tensile stress. &#xA;$$F=\sigma_{u}\left ( 1+\frac{\mu d_{0}}{3h_{0}} \right)A_{f}$$&lt;/p&gt;&#xA;" OwnerUserId="879" LastActivityDate="2015-06-23T13:23:49.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3285" PostTypeId="5" CreationDate="2015-06-23T15:10:42.020" Score="0" Body="&lt;p&gt;Structures can be both large and small. Structures are not necessarily buildings or bridges.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical Structural Engineering concerns include, for a given loading on a structure or structural element, ensuring that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The &lt;strong&gt;stresses do not exceed a specified value&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;strong&gt;structure does not deflect beyond a specified value&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;strong&gt;structure does not buckle&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;The &lt;strong&gt;natural frequencies of the structure are not excited&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Sometimes, the given loading is unusual - such as wind or seismic loading. In these cases is it often appropriate to use additional tags to categorize the question.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="33" LastEditDate="2016-02-23T02:45:49.160" LastActivityDate="2016-02-23T02:45:49.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3286" PostTypeId="4" CreationDate="2015-06-23T15:10:42.020" Score="0" Body="Questions related to the integrity, design requirements, or analysis of structures.  " OwnerUserId="2015" LastEditorUserId="33" LastEditDate="2016-02-23T02:45:49.160" LastActivityDate="2016-02-23T02:45:49.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3287" PostTypeId="5" CreationDate="2015-06-23T15:12:52.507" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-06-23T15:12:52.507" LastActivityDate="2015-06-23T15:12:52.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3288" PostTypeId="4" CreationDate="2015-06-23T15:12:52.507" Score="0" Body="Questions relating to or the engineering application of a device that must withstand a loading via deflection, stresses, resistance to buckling, or natural frequencies." OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-06-23T15:56:14.667" LastActivityDate="2015-06-23T15:56:14.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3289" PostTypeId="2" ParentId="3273" CreationDate="2015-06-23T15:16:02.043" Score="5" Body="&lt;p&gt;There is a pervasive misconception that recoil can be &quot;absorbed&quot; by such things as springs and proprietary gas systems.  This is not true; eventually all the recoil produced by launching the bullet must be absorbed by the shooter.  The only thing that a spring/gas system can do is spread the recoil over a longer period of time, so that the force acting on the shooter's body is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics terminology, recoil is analogous to momentum, not energy.  Momentum is the integral of force with respect to time, $P=\int{F \cdot dt}$, and momentum is a conserved quantity; i.e. it can't be destroyed.  The bullet and escaping propellant gas gain momentum during the launch because they're subjected to a huge force for a very short period of time.  As a consequence of Newton's third law of motion, the gun and shooter gain the same amount of momentum as the bullet, but in the opposite direction.  To dissipate this momentum, you can either absorb a lot of force in a short period of time, or absorb a lower amount of force over a longer period of time (preferable).  For automatic weapons, an ideal recoil system would dissipate the momentum to the shooter in such a way that they would just feel a constant force pushing on them; i.e. no impulses from the shots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question, the spring system to do this would be a progressive-rate spring combined with a reactive damper.  This type of setup is used on some high-end car suspensions, but I imagine it could be adapted for firearms as well.  The trick would be to adjust the reaction of the spring and damper so that they reduce the force exerted on the shooter to a minimum.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-06-23T15:16:02.043" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3290" PostTypeId="5" CreationDate="2015-06-23T15:18:24.357" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-06-23T15:18:24.357" LastActivityDate="2015-06-23T15:18:24.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3291" PostTypeId="4" CreationDate="2015-06-23T15:18:24.357" Score="0" Body="Questions about the theory and applications of structural elements designed to resist bending." OwnerUserId="2015" LastEditorUserId="368" LastEditDate="2015-06-23T15:33:06.810" LastActivityDate="2015-06-23T15:33:06.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3292" PostTypeId="5" CreationDate="2015-06-23T15:20:36.920" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-06-23T15:20:36.920" LastActivityDate="2015-06-23T15:20:36.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3293" PostTypeId="4" CreationDate="2015-06-23T15:20:36.920" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-06-23T15:20:36.920" LastActivityDate="2015-06-23T15:20:36.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3294" PostTypeId="2" ParentId="2658" CreationDate="2015-06-23T16:11:27.840" Score="1" Body="&lt;p&gt;As a side note (and to add to the confusion) if you are NOT a civil engineer, you can use any state for your reference, so long as you have at least one California reference.  I think that has to do with the seismic activity - since civil engineers deal with it more, they want to make sure your work has been reviewed by someone who has to deal with the seismic code.  I happen to have a Mechanical California license - and while some of my references were from California, several were not.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-06-23T16:11:27.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3296" PostTypeId="1" CreationDate="2015-06-23T16:21:59.470" Score="1" ViewCount="116" Body="&lt;p&gt;In a steam turbine, is it possible to run the starting ejector of the surface condenser along with the main ejector?&#xA;Will it improve the efficiency of vacuum generation?&lt;/p&gt;&#xA;" OwnerUserId="2074" LastEditorUserId="272" LastEditDate="2015-06-23T16:57:31.620" LastActivityDate="2015-06-23T16:57:31.620" Title="Steam Turbine - is it possible to run starting ejector and main ejector simultaneously for better efficiency?" Tags="|mechanical-engineering|turbines|steam|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3297" PostTypeId="1" CreationDate="2015-06-23T16:28:34.757" Score="1" ViewCount="13195" Body="&lt;p&gt;I have a customer that originally bought 0.0351&quot; thick sheet steel, CSA grade, G90 zinc coating.  They moved to 0.0236&quot; Grade80, G90 to cut costs.  The thinner, harder steel satisfied their customer's strength requirements.  Their customer would now like a heavier zinc coating (G110) for corrosion protection but our customer wants to keep the total metal thickness the same meaning the base metal thickness will need to decrease.  Strength is determined by the base metal thickness, not the zinc coating!  What is the formula to determine how thin I can go and stay within the strength requirements of my customer's customer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now-  Here is where you are going to beat me up.  I do not know the strength requirements yet of my customer's customer.  If I assume that requirement to be 'x', I could still calculate strength.  &lt;/p&gt;&#xA;" OwnerUserId="2075" LastEditorUserId="272" LastEditDate="2015-06-23T17:00:25.537" LastActivityDate="2015-06-23T17:41:34.567" Title="What thickness of steel sheet do I need to achieve a given strength?" Tags="|materials|steel|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3298" PostTypeId="2" ParentId="3297" CreationDate="2015-06-23T17:41:34.567" Score="2" Body="&lt;p&gt;The requisite thickness is a function of the applied loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your sheet suffers only tension, then the calculation is trivial: the total applied tensile force divided by the sheet's cross-sectional area has to be less than the allowable (&lt;strong&gt;not characteristic&lt;/strong&gt;) stress in the steel. This assumes the force is evenly applied along the sheet. If, however, the load is applied only at a given point, a more detailed analysis is required (finite elements are your friends). Since the stress is a function of area, the strength of your sheet is proportional to the thickness (halve the thickness, halve the strength).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sheet suffers only compression, then you need to check how the sheet will collapse: by stress (unlikely, only if a &lt;strong&gt;very&lt;/strong&gt; short sheet) or by buckling. The allowable compression force by stress is obtained in the same way as in tension: allowable stress times cross-sectional area (assuming distributed load). By buckling the absolute maximum value is given by Euler:&#xA;$$ F = \dfrac{\pi^2EI}{(kL)^2}$$&#xA;where $EI$ is the sheet's stiffness and $kL$ is the effective length (a function of the boundary conditions). This is the absolute maximum, but no code will ever allow you to get close to this value. Since this is a function of the sheet's moment of inertia, the strength is a function of the orientation of the sheet. The lower value is obviously given for the sheet's &quot;weak axis&quot;, so here the strength is also linearly proportional to thickness (double the thickness, double the strength).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sheet suffers only bending, then the orientation does matter: assuming the sheet is parallel to the floor, then the stress is a function of the cube of the thickness (double the thickness, eight times the strength... half the thickness, an eighth of the strength).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order to know whats the requisite thickness for your customer's customer, you need to know precisely what your customer's customer needs the piece for.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-06-23T17:41:34.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3300" PostTypeId="1" AcceptedAnswerId="3307" CreationDate="2015-06-23T22:02:27.137" Score="6" ViewCount="1934" Body="&lt;p&gt;We've had unusually dry weather and our well has run so low that we had to call in a water hauling company to refill it a few days ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water man told me that even though he dumped in ~ 1000 gallons, a great part of it will seep out of the well to the surrounding (drier) aquifer, the same way groundwater makes its way inside the well in normal conditions. Now a few days later, the water is low again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking some kind of drinking-water-safe plastic poly sheet sown or welded together at the seam would create a long and flexible tubing that could be progressively immersed inside of the well (with weight at the end) to effectively waterproof it on the inside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The well could then be refilled to the top and hold all the water that it's containing, instead of slowly seeping it out. Under normal conditions, the sleeve can be removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The well is 2 feet wide, 15 feet deep. 2 ft diameter, 3 ft long concrete pipe sections stacked on top of each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also another well on the site (very low water as well) that's 4 feet wide, 15 feet deep - a much better candidate for the concept as it could hold twice as much water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this sound doable? What are the problems I might face?&lt;/p&gt;&#xA;" OwnerUserId="2080" LastEditorUserId="282" LastEditDate="2015-06-26T20:49:08.387" LastActivityDate="2015-06-26T20:49:08.387" Title="Can I seal the inside of a water well to use it as water reservoir when the well is dry?" Tags="|pressure|plastic|water-resources|waterproofing|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3301" PostTypeId="1" CreationDate="2015-06-23T22:11:07.110" Score="5" ViewCount="878" Body="&lt;p&gt;I have designed a custom part that I need, but it is so simple I'm sure there must be off-the-shelf versions available. However I don't know what it's called to search for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nBQnm.png&quot; width=&quot;280&quot; alt=&quot;Threaded tube with flange at one end&quot;/&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rj7bY.jpg&quot; width=&quot;280&quot; alt=&quot;Flanged nut thing&quot;/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image on the left is the custom part I designed, it's basically a short threaded tube with a flange at one end. The tube is 10mm long and 8mm outer diameter, with an M7x0.75 thread. Note that although it's quite dark inside the tube in the picture and difficult to see, the tube is open at both ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image on the right is a kind of flanged nut type thing I found in a leather strap binding, it's similar to what I need only smaller and also too short, so I know I'm not the only person to want a part like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have searched for &quot;flanged nut&quot;, and &quot;threaded insert&quot;, among other terms, but I can't seem to find anything like this. Is it a thing? What's it called?&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="33" LastEditDate="2015-10-20T00:53:36.750" LastActivityDate="2015-10-24T19:51:02.603" Title="Is there a standard name for a flanged threaded tube?" Tags="|mechanical-engineering|fasteners|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3302" PostTypeId="2" ParentId="3301" CreationDate="2015-06-23T23:05:43.033" Score="1" Body="&lt;p&gt;There are lots of different parts that are similar to this. If you found it binding a leather book together, it was probably paired with a male threaded fastener with a matching head. If so, the pair would be called either a 'binding post' or a 'sex bolt.' They come in many different sizes and materials.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-06-23T23:05:43.033" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3303" PostTypeId="1" AcceptedAnswerId="3305" CreationDate="2015-06-23T23:38:25.450" Score="6" ViewCount="437" Body="&lt;p&gt;I'm trying to do some processing of optical emission spectra from sputtering plasmas, and am confused by what NIST means by &lt;a href=&quot;http://www.nist.gov/pml/data/handbook/index2.cfm&quot;&gt;'easily reversed' in their notation&lt;/a&gt;. Anyone have a clue?&lt;/p&gt;&#xA;" OwnerUserId="2084" LastEditorUserId="1586" LastEditDate="2015-06-24T17:53:35.503" LastActivityDate="2015-06-24T17:53:35.503" Title="What does 'easily reversed' mean in the NIST spectral database?" Tags="|optics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3305" PostTypeId="2" ParentId="3303" CreationDate="2015-06-24T01:37:01.043" Score="3" Body="&lt;p&gt;This means that the line in question can be apparent or not depending on whether the spectrum has been absorbed for other reasons. Or as &lt;a href=&quot;https://en.wikipedia.org/wiki/Spectral_line#Opacity_broadening&quot; rel=&quot;nofollow&quot;&gt;Wikipedia states&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...the reabsorption near the line center may be so great as to cause a self reversal in which the intensity at the center of the line is less than in the wings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, there is a paper that takes this topic more in depth: &lt;a href=&quot;http://epsppd.epfl.ch/Praha/98icpp/f089pr.pdf&quot; rel=&quot;nofollow&quot;&gt;Effect of Self-Reversed Spectral Lines and the Temperature of the Switching Arc Plasma&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-06-24T01:37:01.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3306" PostTypeId="1" AcceptedAnswerId="3310" CreationDate="2015-06-24T02:06:03.477" Score="2" ViewCount="7063" Body="&lt;p&gt;Is there a formula to calculate the grip generated by a tyre on a vehicle for a given surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 13,000 kg machine pulling nearly 200 N of weight. I need to figure out if we need to put any more weight onto the machine so that it has enough traction to start the initial pull. It's going to use pneumatic tyres and the surface is concrete.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditDate="2015-07-27T05:47:23.270" LastActivityDate="2015-07-27T05:47:23.270" Title="Calculating the grip of a tyre for a given surface" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3307" PostTypeId="2" ParentId="3300" CreationDate="2015-06-24T02:56:54.120" Score="5" Body="&lt;p&gt;I see two ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I've had great experience in my pressure vessel design with a 2.3 mm thick liner of polypropylene or polyethylene. In a pinch,  you can weld PVC, but I don't recommend it. Thinner stuff works as well, but the 2.3 mm allows you to place a weld bead on the edge using standard weld rod and plastic welding tools. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If not,  for a pretty reasonable DIY project,  you can always coat the inside with resin, with a bit of fiberglass. (3 oz / sq ft of glass reinforcement should line it perfectly.)  In terms of difficulty, think of &lt;em&gt;putting 3 coats of wall paper on the inside of this well&lt;/em&gt; In the USA, drinking water approved containment services should be NSF listed. A great resin for coating the inside of the tank would be Reichold Dion 9102-00 Impact. It's NSF approved for coating any storage container or piping system. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By the way,  all vinyl esters usually have BPA, if you're worried about that. Fortunately, they make FDA approved polyesters, which do not have BPA, but those aren't NSF approved. &lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-06-24T16:21:24.733" LastActivityDate="2015-06-24T16:21:24.733" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3308" PostTypeId="2" ParentId="3300" CreationDate="2015-06-24T03:07:59.050" Score="7" Body="&lt;p&gt;Well liners are generally made to order. One of your issues will be to find a company that can make what you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to ensure the liner is the correct size and that it does not tear when inserted into the well, while being filled with water, or when it is being removed from the well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention your wells are cased with concrete pipes. They will mostly be smooth on the inside of the well. However, there may be rough spots or chunks of concrete missing, or the joints may be rough and have sharp edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every insertion and removal of the liner risks tearing or weakening the liner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option, if the have the space, is to place a water tank on the surface and re-plumb the downpipes from all the roofs to discharge rain water (when it falls) into the tank. During droughts the tank can be fill by water brought in by a tanker.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-06-24T03:07:59.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3310" PostTypeId="2" ParentId="3306" CreationDate="2015-06-24T11:37:07.247" Score="7" Body="&lt;p&gt;According to the &lt;a href=&quot;http://www.engineeringtoolbox.com/friction-coefficients-d_778.html&quot;&gt;engineering toolbox&lt;/a&gt; the coefficient of static friction between a tire and pavement ranges between $\mu_{wet}=0.2$ for a wet surface and crappy tire and $\mu_{dry}=1$ for a dry surface and high quality tire.  You can use this to calculate the total amount of force that the tires will be able to apply with &#xA;$$&#xA;F_{tire}=\mu F_N=\mu Mg\cos\theta&#xA;$$&#xA;where $\theta$ is the slope of the road, with $\theta=0$ being a flat road. The total amount of force required of the tires to keep your load from rolling backward (with slightly more needed to get it moving forward) is given by &#xA;$$&#xA;F_{load}=(W+Mg)\sin\theta&#xA;$$&#xA;where $W$ is the weight of the load.  So, the maximum slope that your vehicle will be able to operate on is given by the point at which the backwards force of the load becomes equal to that which the tires can apply;&#xA;$$&#xA;\theta=\tan^{-1}\left(\frac{\mu Mg}{W+Mg}\right)=11.3^\circ,&#xA;$$&#xA;assuming that the pavement is wet and your tires aren't particularly great.  This doesn't account for any small undulations in the pavement or bumps in the road, but your tractor to load weight ratio is so high that you shouldn't have any problems.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-06-24T18:04:44.630" LastActivityDate="2015-06-24T18:04:44.630" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3311" PostTypeId="1" AcceptedAnswerId="3313" CreationDate="2015-06-24T13:16:54.330" Score="5" ViewCount="654" Body="&lt;p&gt;I've been working with SC magnet quench simulations. The conductor is made from NbTi in a Cu channel with a 5:1 ratio of Cu:NbTi. Opera3d has a quench program with an example that I began working from. In a quench scenario using a solenoid made from the conductor I described above heat propagates at different speeds throughout the bulk of the solenoid. For this case, heat conduction along the conductor (azimuthal) is fastest. Axial and radial propagation are much slower, with radial being the slowest.&lt;/p&gt;&#xA;&lt;p&gt;Opera's documentation documentation has the following description regarding the anisotropic thermal conductivity:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Nonlinear anisotropic thermal conductivity properties are defined by three expressions based on the functions Cu_Kappa(T), Bulk_Kappa_r(T), and Bulk_Kappa_z(T).&lt;/p&gt;&#xA;&lt;p&gt;For the azimuthal thermal conductivity of the bulk material,the conductivity of copper, Cu_Kappa(T), will be scaled by the copper factor. It is assumed that this is significantly higher than the conductivity of NbTi and dominates the thermal conduction in this direction. The bulk properties radially and axially are to be taken directly from the &lt;strong&gt;tables of measured values.&lt;/strong&gt; &lt;em&gt;(Note that the data used in this example are fictitious, but with similar characteristics to real materials)&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, the values used in the example are claimed to be taken from measured values.&lt;/p&gt;&#xA;&lt;p&gt;My question is this: Can I create the radial and axial thermal conductivities from the thermal conductivities of NbTi and Cu along with their proportions in the conductor?&lt;/p&gt;&#xA;&lt;p&gt;Here is what I've done so far in trying to find the answer:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Google searches. There are many articles and web pages which describe anisotropic &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt;. But all the information I've found assumes you already know the &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt; values.&lt;/li&gt;&#xA;&lt;li&gt;Hand calculations. I made the following model and tried to think of the problem in terms of thermal resistivity, which I then assumed I could treat like electrical resistances, i.e. use rules for parallel and series resistance to calculate the effective resistance.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tR8xv.png&quot; alt=&quot;Conductor Unit Cell&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;There is one more piece to this, however, and that is the insulation. Around the unit cell there is polyester insulation. It is 0.27 mm and 0.25 mm thick the edges that run in the axial and radial directions, respectively.&lt;/p&gt;&#xA;&lt;p&gt;I am going to try accounting for this by summing the materials in their respective direction, each scaled by a packing factor.&lt;/p&gt;&#xA;&lt;p&gt;Does this sound like a good approach? I've spent a bit of time on this already and have no idea if the answer to that approach will resemble the physical model.&lt;/p&gt;&#xA;" OwnerUserId="2076" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-06-24T15:28:03.967" Title="Calculating Anisotropic Thermal Conductivity for two Materials" Tags="|thermodynamics|magnets|superconduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3312" PostTypeId="2" ParentId="3301" CreationDate="2015-06-24T13:56:13.363" Score="4" Body="&lt;p&gt;I've always referred to this style of female fastener as a &quot;T-Nut&quot;&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-06-24T13:56:13.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3313" PostTypeId="2" ParentId="3311" CreationDate="2015-06-24T15:20:08.640" Score="1" Body="&lt;p&gt;You are not too far off with parallel and series resistors method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Rule_of_mixtures&quot; rel=&quot;nofollow&quot;&gt;Rule of mixtures&lt;/a&gt; for composite design works in the same fashion - by treating each of the &lt;em&gt;loads&lt;/em&gt; as either springs in parallel (when loaded in the direction of the fibers), or as springs in series (when loaded opposite to the direction of the fibers).  (See this nice &lt;a href=&quot;http://www.doitpoms.ac.uk/tlplib/bones/derivation_mixture_rules.php&quot; rel=&quot;nofollow&quot;&gt;explanation&lt;/a&gt; if the Wikipedia explanation seems a bit difficult).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result works for determining a wide variety of composite properties.  From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\frac{f}{k_f} + \frac{1-f}{k_m})^{-1} &amp;lt; k_{composite} &amp;lt; fk_f + (1-f)k_m $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be used to determine bounds on your mixture.  I'd really run the analysis with both values, and take the most conservative case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that doesn't work, I use &lt;a href=&quot;http://www.lib.ucdavis.edu/dept/pse/resources/fulltext/HDBK17-3F.pdf&quot; rel=&quot;nofollow&quot;&gt;MIL HDBK 17-3F&lt;/a&gt; for analysis.  Page 223 shows the equation, which has a slightly larger lower bound by placing additional emphasis on fiber conduction.  The insulation can then be added to the radials, again using the rule of mixtures (it's a pretty useful rule).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I can't help you on Opera3D parameters.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-06-24T15:28:03.967" LastActivityDate="2015-06-24T15:28:03.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3314" PostTypeId="2" ParentId="3252" CreationDate="2015-06-24T16:28:46.160" Score="1" Body="&lt;p&gt;When working with superposition method, any way of approaching the disc works - so long as you account for the reaction forces properly.  In this context, it is important to ensure that the sum of the forces is always 0, so unknown reactions (like your simple support in the middle) can be factored in appropriately.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-06-24T16:28:46.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3315" PostTypeId="1" AcceptedAnswerId="3343" CreationDate="2015-06-24T17:03:32.877" Score="7" ViewCount="2059" Body="&lt;p&gt;Our component sourcing organization purchases large amount of electronic components from suppliers. I believe this is through a bid process. Engineering provides the sourcing organization specification, and they purchase these parts for the plant to use in manufacturing. The issue is that recently the plant has been receiving a increasing amount of capacitors which might be defective. This is causing our products to fail. I have checked the few new incoming capacitors using fluke multimeter, and the measurements are with in specification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not practical to continuously inspect these capacitors, and I have been tasked to recommend the solution. How does one &lt;strong&gt;technically&lt;/strong&gt; address similar issues? &lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="282" LastEditDate="2015-06-26T21:37:58.113" LastActivityDate="2015-06-30T21:46:09.487" Title="How to identify low quality electronic components early?" Tags="|electrical-engineering|statistics|quality-engineering|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3317" PostTypeId="1" AcceptedAnswerId="3318" CreationDate="2015-06-24T18:22:12.483" Score="4" ViewCount="6697" Body="&lt;p&gt;I'm doing a static analysis on a uniformly loaded cantilever beam in ANSYS APDL. I'm using BEAM188 with all the default options, so it uses linear form. for the element behaviour, which I think means the bending moments and tensions are linear within the element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to plot the bending moments along the beam with an area contour, for that I'm using:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ETABLE,MI,SMISC,3 &lt;/li&gt;&#xA;&lt;li&gt;ETABLE,MJ,SMISC,16   &lt;/li&gt;&#xA;&lt;li&gt;PLLS,MI,MJ,1,0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This are the axial stresses due to bending, it looks very similar to the bending moment, it shows discontinuities between elements and for each element it shows the same value on both nodes.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/53s55.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/53s55m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does MI=MJ? Shouldn't there be a linear variation of the moment in the elements? Does it make sense for it to be discontinuous between elements? Or is this just a bad plot and I should plot the values on the nodes I get with PRESOL,M as opposed the plot each I and J node values for each element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results overall seem good, but I'm not sure if this is a correct representation of the behaviour of the elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry for such a basic question. I hope I wasn't too confusing.&lt;/p&gt;&#xA;" OwnerUserId="2093" LastEditorUserId="282" LastEditDate="2015-06-30T00:51:42.453" LastActivityDate="2016-09-15T11:17:28.090" Title="Results for Bending Moment using BEAM188 on ANSYS APDL" Tags="|mechanical-engineering|finite-element-method|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3318" PostTypeId="2" ParentId="3317" CreationDate="2015-06-24T18:47:27.990" Score="3" Body="&lt;p&gt;&lt;strong&gt;Short Answer:&lt;/strong&gt; The finite element method means uniform loads are not treated as uniform - and varying moments are not treated as varying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an interesting highlight of the differences between the finite element method and the analytical method.  A uniformly distributed beam can't be treated uniformly on a finite element.  Instead, it breaks up the load into two separate loads on each end, as it only analyzes each node.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1qgEi.jpg&quot; alt=&quot;Breaking up the beam&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between each of the nodes, it treats the load as constant.  Only at each node, does it perform a new &quot;sum of the forces and moments = 0&quot;, &lt;strong&gt;and spits out the brand new moments - hence the disjointed graph&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, it will treat shear loads as constant, moment loads as linear, rotations as quadratic, and displacements as cubic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZNDEk.png&quot; alt=&quot;Beam comparison between single element and theory&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, analytical theory will treat shear loads as linear, moments as quadratic, rotations as cubic, and displacements as quartic.  This is why you need so many elements.  A disconnected cubic will, with enough elements, resemble a quartic - just like a flight of stairs, with enough stairs, resembles an inclined plane.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-06-24T20:47:09.983" LastActivityDate="2015-06-24T20:47:09.983" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3319" PostTypeId="2" ParentId="2658" CreationDate="2015-06-24T19:29:19.920" Score="3" Body="&lt;p&gt;As it turns out, I have been approved to sit for the Civil PE exam in California this fall even though none of my references were CA PEs. So I suppose that answers my question. &lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2015-06-24T19:29:19.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3320" PostTypeId="1" CreationDate="2015-06-24T20:02:19.843" Score="-1" ViewCount="559" Body="&lt;p&gt;Me and a team are designing a plane and one of the goals is to have a variable pitch prop, however, I'm a bit lost as to how to do that. A lot of the factors depend on power output of the engine and prop length, but at this point I don't know where to start. I understand that as speed and altitude increases, pitch increases accordingly. The degree to which they change as altitude and velocity increases is what I need help with. (for a variable pitch propeller)&lt;/p&gt;&#xA;" OwnerUserId="2095" LastEditorUserId="2095" LastEditDate="2015-06-25T01:50:36.480" LastActivityDate="2015-06-25T03:37:41.737" Title="Calculating amount of propeller pitch for plane" Tags="|mechanical-engineering|fluid-mechanics|aerospace-engineering|" AnswerCount="1" CommentCount="0" ClosedDate="2015-08-14T21:41:47.803" ContentLicense="CC BY-SA 3.0" />
  <row Id="3321" PostTypeId="1" AcceptedAnswerId="3333" CreationDate="2015-06-25T00:47:24.340" Score="4" ViewCount="10030" Body="&lt;p&gt;If I have an electric vehicle on an incline (m=17,300 kg) I want to figure out how much energy it would generate by regenerative braking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used $a= g \cdot \sin({\theta}) $ and $f=m\cdot a$.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;at 05 deg incline, acceleration is 0.85 m/s/s; this creates a force of 11114.972 N&#xA;at 10 deg incline, acceleration is 1.70 m/s/s; this creates a force of 22145.352 N&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So bearing these values in mind if I don't want to accelerate and maintain a set speed I will need to create a braking force of 11114.972 N at 5 deg incline - this in turn will keep the machine at the same speed (ignoring friction, bumpy/uneven surface, etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I figure out the Watts generated for the regenerative braking force of 11114.972 N?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The DC motor (I think it's 3 phase) is a 105 kW motor rated at 350 V and 300 A, max torque is 190 Nm @ 285 V drawing 285 A. (That's all the information I have for the motor.) Efficiency of the motor would be 0.85 (Assumed).&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditDate="2015-08-07T00:31:11.793" LastActivityDate="2018-07-24T19:11:41.850" Title="How can I determine the power generated by a regenerative braking system?" Tags="|electrical-engineering|automotive-engineering|regenerative-braking|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3322" PostTypeId="2" ParentId="3321" CreationDate="2015-06-25T01:07:37.787" Score="0" Body="&lt;p&gt;The simple answer is based upon your speed.  $P = F*v$, so if you're going at 2 m/s, &lt;a href=&quot;http://www.wolframalpha.com/input/?i=11.2kN+*+2+m%2Fs+&quot; rel=&quot;nofollow&quot;&gt;Wolfram&lt;/a&gt; gives you 22.4 kW.  Quite a bit of energy to start with!  Of course, your final design will eat up some of this power due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Energy_conversion_efficiency&quot; rel=&quot;nofollow&quot;&gt;energy conversion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-06-25T01:07:37.787" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3323" PostTypeId="2" ParentId="3320" CreationDate="2015-06-25T03:37:41.737" Score="2" Body="&lt;p&gt;There are a bunch of questions to clarify what you are looking for.  It is not clear what you are trying to control? Are you using a piston engine?&lt;br&gt;&#xA;Are you trying to maximize airspeed?  Minimize fuel burn?&lt;br&gt;&#xA;For a typical piston engine, the pilot can set the manifold pressure with a throttle control, and can set the RPM.  A mechanical control system then changes the prop pitch, to keep the desired RPM (on some very old aircraft the pilot actually sets the pitch).  The pilot uses a table to choose the settings, that is provided in the Pilot Operating Manual for the aircraft.  For example, here is part of the table from a Cessna 206, at 2000 ft  &lt;a href=&quot;https://i.stack.imgur.com/Ybek0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wK5dO.png&quot; title=&quot;source: imgur.com&quot; /&gt;&lt;/a&gt;&#xA;To use this table, first the pilot has to know the outside temperature (as an example assume 11C).  Then, for a given RPM (such as 2400) and a given Manifold Pressure (MP) such as 24&quot; you can see the the engine generates 69% BPH (Brake Horse Power) and the airplane flies at 132 KTAS (Knots True Airspeed) and burns 15.6 GPH (gallons per hour).   If the pilot wants to save fuel, then he or she can reduce MP.  If the pilot wants to reduce noise, and engine wear, then it might be right to reduce RPM.  There are additional tables for different altitudes.  This table was probably generate based on flight test data.&lt;/p&gt;&#xA;" OwnerUserId="2066" LastActivityDate="2015-06-25T03:37:41.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3324" PostTypeId="1" AcceptedAnswerId="3337" CreationDate="2015-06-25T09:02:53.780" Score="17" ViewCount="2924" Body="&lt;p&gt;I was reading about &lt;a href=&quot;https://en.wikipedia.org/?title=CD-ROM&quot;&gt;how CD-ROMs work&lt;/a&gt; and I came to this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although it might seem simplest to use a pit to record a 0 and a land&#xA;  to record a 1, it is more reliable to use a pit/land or land/pit&#xA;  transition for a 1 and its absence as a 0,so this scheme is used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, I'd like to know why?&lt;/p&gt;&#xA;" OwnerUserId="2100" LastEditorUserId="272" LastEditDate="2015-06-25T12:06:26.733" LastActivityDate="2016-01-11T14:31:56.757" Title="Why is it more reliable to use the land/pit transition in a CD-ROM?" Tags="|optics|computer|computer-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3325" PostTypeId="2" ParentId="3315" CreationDate="2015-06-25T09:04:08.230" Score="1" Body="&lt;p&gt;Check some component randomly, low quality components produce heat immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can imagine, the liquid inside the capacitor will only dry out if the capacitor isn’t perfectly sealed and/or if the capacitor is exposed to high temperatures the definition of “high temperature” for us is any temperature above the standard room temperature of 25° C or 77° F.&lt;/p&gt;&#xA;" OwnerUserId="2101" LastEditorDisplayName="user16" LastEditDate="2015-06-25T16:06:27.167" LastActivityDate="2015-06-25T16:06:27.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3326" PostTypeId="1" CreationDate="2015-06-25T09:29:03.513" Score="7" ViewCount="21224" Body="&lt;p&gt;We have a rolling machine with 3 rollers; the bottom has two rollers with diameters of 168 mm and one top roller with a diameter of 219 mm.  I want to calculate the required force to bend a 10 mm thick plate which is 2,500 mm wide.  The rolling machine is connected to a 10 HP motor with a gear box and the rollers have an rpm of 8.&lt;/p&gt;&#xA;" OwnerUserId="2102" LastEditorUserId="272" LastEditDate="2015-06-25T12:04:47.973" LastActivityDate="2017-12-30T02:26:44.653" Title="How can I calculate the force required to bend a 10 mm thick plate in a rolling machine?" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3327" PostTypeId="2" ParentId="3324" CreationDate="2015-06-25T09:29:32.283" Score="11" Body="&lt;p&gt;Because of optical interference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the laser is fully in a pit or fully on land then it gets reflected cleanly and will be picked up by the sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, on the transition, half the laser is in the pit and half is in the land and the reflections will interfere with each other and the sensor won't pick anything up. The pit depth and laser color (wavelength) is picked so that this happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the sensor cannot detect whether it is seeing a land or pit. It can only detect when it changes.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="368" LastEditDate="2015-06-26T14:51:57.237" LastActivityDate="2015-06-26T14:51:57.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3328" PostTypeId="1" CreationDate="2015-06-25T09:37:57.297" Score="7" ViewCount="813" Body="&lt;p&gt;Recently I've begun thinking about experimenting with the construction of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tesla_turbine&quot;&gt;Tesla turbine&lt;/a&gt; for steam use. The key issue I can find is that most of the examples out there of such turbines use either optical discs or HDD disks to make the actual rotors. In the case of HDD disks, these work well because they're fairly stiff, but the use limits the size of the turbine to well, the size of the largest common HDD platter (3.5&quot;) - that is, unless someone can find me a link to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hard_disk_drive#Form_factors&quot;&gt;obsolete 8&quot; platter size&lt;/a&gt;. This obviously installs an upper practical boundary on the power level of such a turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the key issues against Tesla turbine development historically was the inability to develop both thin and stiffened rotors; while building it thick solves the stiffness issue, it drops overall efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm taking the time to actually learn some more about engineering - how is &quot;stiffness&quot; quantified in engineering, and what affects it (other than thickness)? What materials would be good candidates for something such as a Tesla turbine rotor?&lt;/p&gt;&#xA;" OwnerUserId="1229" LastActivityDate="2021-02-01T04:15:11.977" Title="Stiffening of Tesla turbine rotors" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3330" PostTypeId="2" ParentId="3315" CreationDate="2015-06-25T10:27:39.570" Score="1" Body="&lt;p&gt;One has to recall, on this question, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Capacitor_plague&quot; rel=&quot;nofollow&quot;&gt;capacitor plague&lt;/a&gt; problem that the computer industry experienced for a number of years. The issue with this is I'm not sure if there was ever a way to discover these effects prior to failure, and the main mitigation method was a supply-chain change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it does give perspective. As Aravinthkumar (answerer here) suggests, random component testing does weed out a fairly good number of bad components, however this will only detect components that are &lt;em&gt;immediately&lt;/em&gt; defective. It will fail to detect components that like the capacitor plague, simply have a premature lifespan, but with no other obvious issues. Also, like you yourself said, the components are usually within specification, so no red flags are raised there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I suggest you do is run your own product, continuously. If it's a microprocessor-based product, develop some sort of protocol much like a benchmark that is designed to stress-test the system at maximum allowable power. Cycle the power randomly (a lot of capacitors fail on initial startup, under &lt;em&gt;inrush current&lt;/em&gt; situations). If you're in a market segment where you don't expect your devices to be run continuously (i.e. 24h/day or close to), by doing so yourself you can detect if the components meet the specification for your product lifecycle, as well as make your determination a good deal faster than it would take the end-users to come to the same conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pushing the components a little harder might make issues come to light quicker. As it is good engineering practice to build a margin of safety into circuits (since we're talking capacitors, presumably power supplies), push the circuit harder (i.e. current draws) within the margin of safety, not so much as to cause failure but more so than you would expect from the end user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any rate, once you've determined that failure in your current component supply is probable (that is, higher than the manufacturer's failure rate, or your company's own arbitrary failure rate), select a different manufacturer or device line, etc.&lt;/p&gt;&#xA;" OwnerUserId="1229" LastActivityDate="2015-06-25T10:27:39.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3331" PostTypeId="1" AcceptedAnswerId="3802" CreationDate="2015-06-25T10:42:30.957" Score="2" ViewCount="593" Body="&lt;p&gt;In lab we have Struers Tegrapol-21 grinding and polishing machine. I use it for sample preparation. Usually I put 6 samples at once and prepare them with sample holder. However today I need to prepare 3 samples. I checked &lt;a href=&quot;http://www.struers.com/resources/elements/12/101288/TegraSystem%20brochure%20English.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;brochure&lt;/a&gt; and it's not clear. &lt;a href=&quot;http://www.struers.com/resources/elements/12/282889/TegraPol-21%20-25%20-31%20-35%20TegraForce-3%20-5%20TegraDoser-1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Manual&lt;/a&gt; wasn't helping either.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand why do I use two different type of holders. Ease of loading is an issue however I'm looking to learn is there a difference in grinding/polishing performance. Why and when I should choose one over the other? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is photo of single holder (left), sample holder(right) and Tegrapol-21.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/J26QU.jpg&quot; alt=&quot;Tegrapol-21&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can I use sample holder for 3 samples?&lt;/li&gt;&#xA;&lt;li&gt;Can I use single holder for 3 samples? &lt;/li&gt;&#xA;&lt;li&gt;If both of them is possible which one is preferred?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="879" LastEditorUserId="879" LastEditDate="2015-06-29T10:06:53.703" LastActivityDate="2015-08-04T18:12:11.940" Title="How to use sample holder on Metallography specimen preparation machine Struers Tegrapol-21" Tags="|materials|metals|grinding|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3332" PostTypeId="2" ParentId="3194" CreationDate="2015-06-25T11:09:22.880" Score="2" Body="&lt;p&gt;It seems most people here are tackling the subject from the standpoint of analogies, so as an electrical engineer who's trying to learn mechanical (the other way around) I'll take a different perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's more that &lt;em&gt;resistance&lt;/em&gt; is much more readily usable in electrical engineering than &lt;em&gt;conductance&lt;/em&gt;, from a practical standpoint. There is no doubt, the use of resistance gives way to many interesting derivations of Ohm's law (i.e. $P = I^2 R$) and this use of resistance rather than conductance allows for calculation of many practical, but mundane things, such as power dissipation of circuit elements. Here, there are two reasons to decreasing such dissipation - first, the increase in efficiency, and second, the decrease in heat generation, which is by far the driving factor, especially in the semiconductor era. There is no reason why you can't perform the equivalent operation $P = \frac{I^2}{G}$, but in practicality you'll almost never see that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By virtue of that fact, the calculations are made somewhat simpler or more readily-understandable using resistance (to the extent that people don't like reciprocals). But, the formulas generally favor resistance and its AC/complex counterpart &lt;em&gt;impedance&lt;/em&gt;, rather than &lt;em&gt;conductance&lt;/em&gt; and &lt;em&gt;admittance&lt;/em&gt;, although both varieties do exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems I finally hit the nail on the head in the comment above, but resistance has the foothold in EE because of not only historical perspective (people understood the concept of resistance first, conductance came rather second), but because the end goal of design is to release an optimal system. Because of this, one works to cut out or reduce &lt;em&gt;losses&lt;/em&gt; rather than improve &lt;em&gt;gains&lt;/em&gt;. It should also fit that if one is building an LED circuit, for example, it is easier to understand to fit the LED using a current-limiting 50 ${\Omega}$ resistor rather than the equivalent &quot;20 mS conductor&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to finally answer your question, I suspect that a large part of it is tradition, granted that both concepts are simply right- and left-handed versions of the same principle. But, that tradition through Ohm's law and DC principle, at least has a grounding point.&lt;/p&gt;&#xA;" OwnerUserId="1229" LastEditorUserId="1229" LastEditDate="2015-06-26T11:34:26.337" LastActivityDate="2015-06-26T11:34:26.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3333" PostTypeId="2" ParentId="3321" CreationDate="2015-06-25T11:30:48.993" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;If I have an electric vehicle on an incline (m=17,3oo kg) I want to figure out how much energy it would generate by regenerative braking ... and maintain a set speed.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Change in potential energy per time is as good an approach as using kinetic energy.&lt;br /&gt;&#xA;The answers &lt;strong&gt;should&lt;/strong&gt; be the same :-).&lt;/p&gt;&#xA;&lt;p&gt;E= mgh&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;  h = change in vertical axis  &#xA;  m = mass&#xA;  g = gravitational acceleration.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;In one second power = energy numerically.&lt;br /&gt;&#xA;Say g = 10 (more than good enough for 'playing')&lt;/p&gt;&#xA;&lt;p&gt;1 kph = 1000 metres in 3600 seconds.&lt;br /&gt;&#xA;1 kph = 0.278 m/s (or 1 m/s = 3.6 kph)&lt;/p&gt;&#xA;&lt;p&gt;V_vertical / V_vehicle = sin(slope angle)&#xA;V_vertical = V_vehicle  x sin(slope_angle)&lt;/p&gt;&#xA;&lt;p&gt;Mass  = 17300 kg&lt;br /&gt;&#xA;First calculate for 1 metre/second vertical rate of change to get a general feel.&lt;br /&gt;&#xA;Power = energy change in one second at constant V&lt;br /&gt;&#xA;P =  mgh = 17300 x 10 x 1 = 173,000 Joule/1s&lt;br /&gt;&#xA;= 173 kW at 1 m/s !!!!&lt;/p&gt;&#xA;&lt;p&gt;At your example 45 degrees and 15 kph&lt;/p&gt;&#xA;&lt;p&gt;V_vertical = 15 x sin(45) = 10.6 kph&lt;br /&gt;&#xA;10.6 kph = 10.6/3.6 m/s = 2.95 m/s vertically.&lt;br /&gt;&#xA;So power = 2.95 x 173 kW = 510 kW.&#xA;Change g to 9.8 instead of 10 and I get 500 kW&lt;br /&gt;&#xA;= your result.&lt;/p&gt;&#xA;&lt;p&gt;This tells you that under the stated conditions your motor would be &lt;strong&gt;trying&lt;/strong&gt; to regenerate at about 5 x it's rated power. It is almost certain that it would not achieve this, and quite likely that it would be damaged if you ran it in that manner. 500 kW (or even 100 kW) of magic smoke would be very impressive indeed!.&lt;/p&gt;&#xA;&lt;p&gt;It also tells you that the motor could not drive the vehicle up a 45 degrees slope at that rate.  Maximum speed would be about 105 kW/500 kW x 15 kph = 3 kph&lt;/p&gt;&#xA;&lt;p&gt;This is not a vast surprise when you consider what you are asking for.&lt;br /&gt;&#xA;Braking at constant V at 15 kph is equivalent to lifting a 17 ton vehicle ~= 10+ cars vertically at 3 m/S&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;________________________&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Useful rule of thumb&lt;/h3&gt;&#xA;&lt;pre&gt;&lt;code&gt;Energy  ~= kg x metres lifted vertically x 10 x efficiency &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;This is good for energy required to lift mass OR power delivered by falling masses.&lt;/p&gt;&#xA;&lt;p&gt;Power is energy per unit time so power can be used instead of energy is rates are in meters per second rather than total metres moves.&lt;/p&gt;&#xA;&lt;p&gt;So, for water wheel&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Power = litres/second x head in metres x 10 x efficiency &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;For pump&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Power = litres/second x head in metres x 10 / efficiency &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Note that efficiency Z = power out / power in in both cases, but&lt;/p&gt;&#xA;&lt;p&gt;Efficiency for water wheel = Power delivered  / Power from water flow&lt;br /&gt;&#xA;Efficiency for pump = Power in outlet water flow / Power used to drive pump&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-06-25T11:50:50.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3335" PostTypeId="2" ParentId="3315" CreationDate="2015-06-25T12:13:05.057" Score="3" Body="&lt;p&gt;You do not say what sort of capacitors are involved.&lt;br&gt;&#xA;Tests will vary with capacitor type.&lt;br&gt;&#xA;Electrolytics can be tested for eg leakage, ESR, behaviour as they approach rated voltage, ... .&lt;br&gt;&#xA;Ceramic capacitors that are &quot;of unknown origin&quot; may well exhibit temperature dependant variations not matched to they specifications.&lt;br&gt;&#xA;[Tantalum capacitors may explode with flame smoke sound and smell, but, hey , they do that anyway, so that's no guide :-) ] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is extremely easy to address by pre-emptive action - except for the part that involves getting the accountant to agree.  Once they realise the real cost of using bad parts agreement should follow.  Traditionally used ratios for the cost of locating a faulty part in&lt;br&gt;&#xA;incoming inspection and in-production testing / final test / customer fault is&lt;br&gt;&#xA; 1:10:100&lt;br&gt;&#xA;Find it early / don't let them happen at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution - &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;ONLY EVER buy known quality brand components from a known reputable sales channel.&lt;/strong&gt;   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Almost all the time you will get what you expect and if you don't you will have right of redress with the supplier. This may not extend to cost of repair but should at least cover cost of replacement. That is not enough as the cost of field replacements is often immense compared to component cost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a starting guide I say 'if you can buy Panasonic, do so'  - it may involve price and availability - and really this is liable to apply to a range of reputable manufacturers. Note - that is NOT an &quot;ad&quot; for Panasonic - I have no relationship with them  except as a happy customer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, you do not have to buy from them but, if a brand is sold by a major reputable reseller then it is very probably of acceptable quality when used within specification. I use Digikey as a first check, but a range of otrhers are equally good.  Taking &lt;a href=&quot;http://www.digikey.com/product-search/en/capacitors/aluminum-capacitors/131081?k=capacitor&quot; rel=&quot;nofollow&quot;&gt;aluminium electrolytic capacitors from Digikey&lt;/a&gt; as an example. Digikey list 12 manufacturers on that page. Most of the brands listed probably manufacture some or most of their capacitors in China. China is perfectly capable of making absolutely superb capacitors (and a;lmost any other product as well). But if you buy Chinese product via sales channels that you do not know you can trust, or if you buy brands you have not heard of, or products labelled with the logos of brands that you have heard of, but from sales channels whose integrity is unknown to you you will sooner or later find yourself &quot; &lt;em&gt;... receiving a increasing amount of capacitors which might be defective.&lt;/em&gt; &quot;&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-06-25T12:13:05.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3336" PostTypeId="2" ParentId="3326" CreationDate="2015-06-25T12:16:45.213" Score="4" Body="&lt;p&gt;The deflection of a beam, $\delta$ mm in terms of the force is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = \frac{WL^3}{48EI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $I = \frac{1}{12}b*h^3$ – in your case h is 10 mm, b = 2500 mm.  E is the &lt;a href=&quot;http://www.engineeringtoolbox.com/young-modulus-d_417.html&quot; rel=&quot;nofollow&quot;&gt;modulus&lt;/a&gt;, and based on your material – for steel it is 200 GPa.  Using this, and the knowledge of the unspecified L (length between the bottom rollers), you can calculate how much you will bend the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to keep the plate permanently bent, you will need to exceed the &lt;a href=&quot;https://en.wikipedia.org/wiki/Yield_(engineering)&quot; rel=&quot;nofollow&quot;&gt;elastic limit&lt;/a&gt; of the material – so you need to bend it so the stresses exceed the yield strength at the temperature you're working the material.  For your case, that's when:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma = \frac{WLh}{8I}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exceeds your yield strength – which for steel runs at 250 MPa.  The fun about permanent bending – you had to exceed some sort of $\delta_{min}$ to meet this yield strength, then your sheet will continue bending as you apply more force, to a full $\delta$.  However, when you remove the piece from the machine, it will &quot;spring back&quot; – the worst it will spring back will be $\delta – \delta_{min}$ – so you usually have to overshoot by a bit, check to see if it's bent enough, and bend it a second time.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="368" LastEditDate="2016-09-23T16:00:47.753" LastActivityDate="2016-09-23T16:00:47.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3337" PostTypeId="2" ParentId="3324" CreationDate="2015-06-25T14:21:36.097" Score="21" Body="&lt;p&gt;To elaborate on ratchet freak's answer; CD-ROMs work by sensing the intensity of the light reflected from the CD as it is spinning.  More light being reflected stands for a 1 and less light being reflected stands for a 0 (or vice versa).  One way to encode the information would be to have highly reflective surfaces for the 1's and dark patches for the 0's.  Printing dark patches at the size of the laser spot size is actually used in &lt;a href=&quot;https://en.wikipedia.org/wiki/Compact_disc#Writable_compact_discs&quot; rel=&quot;noreferrer&quot;&gt;writable CDs&lt;/a&gt;, but the technique suffers from degradation over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead &lt;a href=&quot;https://en.wikipedia.org/wiki/Compact_disc#Manufacture&quot; rel=&quot;noreferrer&quot;&gt;commercial CDs&lt;/a&gt; rely on the property of interference to create dark patches.  When the laser beam is in the transition region from land to pit, part of the laser beam reflects from the land and part from the pit.  The depth of the pit is &lt;a href=&quot;https://en.wikipedia.org/?title=CD-ROM&quot; rel=&quot;noreferrer&quot;&gt;roughly 1/4 of a wavelength&lt;/a&gt; such that the portion of the beam which reflects from the pit has picked up an extra $\lambda/2$ of phase.  This phase shift causes the beam falling on the photodetector to appear dark due to destructive interference.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The system used to detect this induced intensity modulation is a very simple one.&lt;/strong&gt;  It essentially consists of shining a laser at the disk and putting a photodiode on the reflected beam.  Actually detecting the depth of the pits would require an interferometer which is significantly more complicated.  This is because the interferometer would need a reference mirror whose position was stable to significantly better than the depth of the pits, ~250 nm.  This is a difficult task to achieve inside of a disc drive which has a bunch of moving components (such as a disc spinning at 500 RPM) and may even be mounted in a vehicle which is moving and shaking around.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hk-phy.org/articles/cdrom/cdrom_e.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wOhON.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-06-26T15:28:09.550" LastActivityDate="2015-06-26T15:28:09.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3338" PostTypeId="2" ParentId="3194" CreationDate="2015-06-25T14:40:56.723" Score="2" Body="&lt;p&gt;I would suggest, without any analogy, that resistance is a more obvious and clear concept to explain and understand.  How hard is it to push electronics through?  While conductance seems less intuitive, at least to me.  Additionally, it is not unusual to have 0 (or at least close enough to assume 0 for most purposes) of resistance.  For example for typical analog circuits the wires, or other conductors, can be assumed to have 0 resistance.  While working with conductance, you end up with a bunch of infinite values floating around.  That can make the math a bit more sloppy.&lt;/p&gt;&#xA;" OwnerUserId="2066" LastActivityDate="2015-06-25T14:40:56.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3340" PostTypeId="2" ParentId="3194" CreationDate="2015-06-25T20:47:29.820" Score="4" Body="&lt;p&gt;The laws that you mention (e.g. Fourier law, Poiseuille law, Hooke's law), which involve a flux which increases as &lt;strong&gt;&lt;em&gt;a&lt;/em&gt;&lt;/strong&gt; material parameter increases, are interesting examples.  Considering the original version of &lt;em&gt;Ohm's Law&lt;/em&gt; was also phrased in terms of conductance, not resistance.  This implies that in &lt;strong&gt;experimental physics&lt;/strong&gt;, it's easier to watch something that's easy to change, and plot a linear relationship to something that changes, then it is to watch something that's difficult to change, and plot it against something that's easy to change.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes sense.  You pull on a bar, and it grows.  It's easy to change the &quot;grow&quot; by a set amount and stop, but difficult to get an exact pull force.  You heat one end of the bar, and the heat spreads. It's easy to measure and you can stop when you reach a certain temperature. But you don't know how much heat you had to put in to make that happen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also easy to apply different exact voltages to a wire, and watch the current change. Note the slope of that experiment would be the &lt;em&gt;conductivity&lt;/em&gt;, not the &lt;em&gt;resistance&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are you doing these experiments?  So some day, you can make a bar that doesn't break when the load makes it grow.  But, for now, you plot the thing that you changed v. the thing that also changed.  And you call the slope &quot;E&quot;, or &quot;k&quot;, or &quot;$\sigma$&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in &lt;strong&gt;engineering&lt;/strong&gt;, it's different.  While I use Hooke's law to figure out that bending my beam will lead to a displacement, inevitably I'm dividing by EI.  Because I don't want my beam to deflect that much, I need a very large EI - but a small inverse would be better.  I really wish I didn't have to divide - when I use a calculator it's easier to type $x_1 * x_2 * x_3$ v. $ x_1 / (x_2 * x_3) * x_4 $ - I'd rather we rate materials by 1/E, then by E, and the inverse of section modulus as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When calculating out fluid flow in a pipe, I'm always sizing my pumps not by dividing by the &quot;flowability&quot; of pipes, but adding up the friction factor resistance. I'm sizing my moving system to overcome this resistance, not to find out how my smooth pipes are helping the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In heat transfer, I'm always dividing by the thermal conductivity.  They even invented &lt;a href=&quot;https://en.wikipedia.org/wiki/R-value_(insulation)&quot; rel=&quot;nofollow&quot;&gt;R-value&lt;/a&gt; on insulation so you could figure out the change in temperature between surfaces easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In electrical engineering, they didn't start with &lt;em&gt;resistors&lt;/em&gt; - those are a manufactured device.  So, they listed the value that mattered more to the engineers, and less to the physicist - the &lt;strong&gt;resistance&lt;/strong&gt;.  Unfortunately for Mechanical Engineers, we started with pipes and bars, so we keep with the same notation.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-06-26T13:08:05.463" LastActivityDate="2015-06-26T13:08:05.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3341" PostTypeId="2" ParentId="2933" CreationDate="2015-06-26T09:20:50.153" Score="7" Body="&lt;p&gt;Below is simple table for designing snap-fit joints from &lt;a href=&quot;http://fab.cba.mit.edu/classes/S62.12/people/vernelle.noel/Plastic_Snap_fit_design.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bayer Snap-Fit Joints for Plastics a Design Guide&lt;/a&gt;, please consult the pdf for more info. As I understand your joint is C cross section with 1st type. You should also check for spherical joint clearance should be y/2.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/1IkzI.png&quot; alt=&quot;Snap-Fit Design&quot; /&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;y = (permissible) deflection (=undercut),&lt;/li&gt;&#xA;&lt;li&gt;E = (permissible) strain in the outer fiber at the root; in formulae: E as absolute value = percentage/100,&lt;/li&gt;&#xA;&lt;li&gt;l = length of arm,&lt;/li&gt;&#xA;&lt;li&gt;h = thickness at root,&lt;/li&gt;&#xA;&lt;li&gt;b = width at root,&lt;/li&gt;&#xA;&lt;li&gt;c = distance between outer fiber and neutral fiber (center of gravity),&lt;/li&gt;&#xA;&lt;li&gt;Z = section modulus&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Z = I c$$&lt;/span&gt;,&#xA;where&lt;/li&gt;&#xA;&lt;li&gt;I = axial moment of inertia,&lt;/li&gt;&#xA;&lt;li&gt;Es = secant modulus,&lt;/li&gt;&#xA;&lt;li&gt;P = (permissible) deflection force,&lt;/li&gt;&#xA;&lt;li&gt;K = geometric factor.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;In order to use this formulas you need to consult material specific allowable strain limits (E) and secant modulus for appropriate temperature. If your material has distinct yield point, I guess it will have if it's not reinforced with fiber &lt;span class=&quot;math-container&quot;&gt;$0.7\epsilon_{y}$&lt;/span&gt; can be taken. Even though please check design guidelines. Below are some more resources for your design.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mit.edu/2.75/resources/random/Snap-Fit%20Design%20Manual.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;BASF Snap-Fit Design Manual&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sabic-ip.com/webtools/toolsHomePage.faces&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sabic online tool (Needs Registration)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="879" LastEditorUserId="19906" LastEditDate="2021-08-08T11:23:42.183" LastActivityDate="2021-08-08T11:23:42.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3342" PostTypeId="1" CreationDate="2015-06-26T10:53:07.300" Score="5" ViewCount="866" Body="&lt;p&gt;Consider a transmission for a tandem rotor helicopter with two engines and two main rotors such as the one below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a7OFj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a7OFj.png&quot; alt=&quot;CH-47 Powertrain&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both engines are connected with sprag (one-way) clutches such that an engine's speed can underrun the rest of the system (e.g. engine failure), but normally they move in sync. The two rotors are always synchronized so they don't collide with each other as they spin (in opposite directions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that we know the moments of inertia for each engine and each rotor. There are accessories (not shown in the picture) which are driven by the same synchronizing shaft and there are of course mechanical losses, but we can neglect all those for simplicity. There is, however, a speed change between the engines and the rotors that needs to be taken into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In modeling this system, since each engine can underrun the rotors, I need to integrate the local accelerations to get the speed of the rotors and each engine separately. Generally, my question is: &lt;em&gt;what torques and inertias does each &quot;see&quot; for the purposes for the purpose of finding an acceleration from unbalanced torques&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case that both engines are operating and driving (i.e. engaged clutches):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would each component (engine 1, engine 2, and the rotors collectively) &quot;see&quot; all four torques and all four inertias (the four being two rotors and two engines)?&lt;/li&gt;&#xA;&lt;li&gt;Or would each engine be unaffected by the torque and inertia of the opposite engine (i.e. due to the one-way clutch) and only &quot;see&quot; itself and the rotors (or maybe some portion thereof)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the case when an engine is being shut down and is no longer driving:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would the shut down engine only see its own inertia and small and diminishing torque (and then you'd have to add some losses to actually get it to decelerate)? &lt;/li&gt;&#xA;&lt;li&gt;Would the rotors &quot;see&quot; the torques and inertias of the rotors and the one remaining driving engine?&lt;/li&gt;&#xA;&lt;li&gt;What about the engine that's still driving - the rotors and itself?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2114" LastEditorUserId="2114" LastEditDate="2015-09-15T13:02:32.123" LastActivityDate="2015-09-15T13:02:32.123" Title="Coupling in a twin engine, tandem rotor power train" Tags="|mechanical-engineering|aerospace-engineering|modeling|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3343" PostTypeId="2" ParentId="3315" CreationDate="2015-06-26T17:01:25.147" Score="2" Body="&lt;p&gt;In the mechanical world, if you have a million bars, and pull on them until they yield, you'll get a wide variety of failures.  These failures, if you counted them with how many failed at what strength, you usually end up with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow&quot;&gt;normal distribution&lt;/a&gt;, though to make the math easy we usually fit it into a &lt;a href=&quot;https://en.wikipedia.org/wiki/Log-normal_distribution&quot; rel=&quot;nofollow&quot;&gt;log-normal distribution&lt;/a&gt;.  When dealing with a part who has a shorter life, and not a property failure (which may be the case here), you should use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Weibull_distribution&quot; rel=&quot;nofollow&quot;&gt;weibull distribution&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to the example, you'll find that a higher percentage yield at a point - the mean, and the rest scatter around the mean.  Some - a small amount, will be out of tolerance because you are testing a million.  3 of those million probably would yield at a load so low that even if I used that one piece in my design to a reasonable factor of safety, my design would have failed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in real design, I don't actually load it with a set load either - these loads vary as well.  As a result, I can identify the loading as a probability as well.  This can then lead to figuring out the probability of failure - the probability that the load is greater than the strength of the bar.  This type of design is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Probabilistic_design&quot; rel=&quot;nofollow&quot;&gt;Probabilistic Design&lt;/a&gt; - which even has a nice diagram showing this example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I run my probabilities, I need to make sure my bars don't slip up.  So, I run testing on every batch to see that it falls within my original distribution.  I make sure using comparative probability techniques that everything I get exceeds the bare minimum the engineers need to make sure the part doesn't fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This principle is the foundation for the very broad field of &lt;a href=&quot;https://en.wikipedia.org/wiki/Reliability_engineering&quot; rel=&quot;nofollow&quot;&gt;Reliability Engineering&lt;/a&gt;.  Using this thought that failure is inevitable - but the probability of failure can be minimized, is the basis for many Quality Control Programs, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Six_Sigma&quot; rel=&quot;nofollow&quot;&gt;Six Sigma&lt;/a&gt;.  For you it comes down to a few things with capacitors - lifetime and meeting the specified capacitance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I would approach your capacitor problem:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Enforcing your manufacturer to have an &lt;a href=&quot;https://en.wikipedia.org/wiki/ISO_9000&quot; rel=&quot;nofollow&quot;&gt;ISO quality&lt;/a&gt; system.  This will mean that if you invest in some testing of your own to find an optimal manufacturer, they will be making the part with the same quality as when you tested it.&lt;/li&gt;&#xA;&lt;li&gt;Run some &lt;a href=&quot;https://en.wikipedia.org/wiki/Highly_accelerated_stress_test&quot; rel=&quot;nofollow&quot;&gt;accelerated stress testing&lt;/a&gt; if lifetime is the problem.  Perform a test using your complete assemblies and a test using just your capacitors.  Find out what kind of lifetime you have - you'll need quite a few samples, but you can fit it into a Weibull or Normal Distribution.  Your manufacturer may have run these tests for you - in that case, include your reliance on the manufacturer's testing in the specification.&lt;/li&gt;&#xA;&lt;li&gt;Run some capacitance, testing with the multimeter, as well some destructive (over-excess voltage), leakage, overheating, etc testing.  Again, you'll need a few samples, but then you can have a probability of failure for what you are testing.  Also like before, if they provide this to you, include it in the specification so purchasing won't accidentally undermine you.&lt;/li&gt;&#xA;&lt;li&gt;Run these values through the engineers to determine the probability of failure.  You may find weak points you need to design around.  Redesign (if needed), and certainly re-qualify your specifications.  You want your specifications to ensure your final design meets the testing requirements&lt;/li&gt;&#xA;&lt;li&gt;Use &lt;a href=&quot;https://en.wikipedia.org/wiki/Control_chart&quot; rel=&quot;nofollow&quot;&gt;control charts&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Run_chart&quot; rel=&quot;nofollow&quot;&gt;run charts&lt;/a&gt;, and any of the other &lt;a href=&quot;https://en.wikipedia.org/wiki/Seven_Basic_Tools_of_Quality&quot; rel=&quot;nofollow&quot;&gt;7 Basic tools&lt;/a&gt; for all new parts.  These lets you keep track of new parts as they come in without checking every single part, and lets you know when parts are failing prematurely - to alert or discontinue working with your vendor.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, &lt;a href=&quot;https://en.wikipedia.org/wiki/Quality_Engineering&quot; rel=&quot;nofollow&quot;&gt;Quality Engineering&lt;/a&gt; is a very broad topic - I've really only scratched the surface even with this very large post.  But I think this will help you with your capacitor problem. &lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-06-26T18:24:35.003" LastActivityDate="2015-06-26T18:24:35.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3344" PostTypeId="2" ParentId="100" CreationDate="2015-06-26T18:54:31.330" Score="7" Body="&lt;p&gt;Perhaps you could form a spheroid of molten glass in free fall, then quench it with a cold gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest a cold gas instead of a liquid because you can't &quot;drop&quot; it into a liquid in free fall, and splashing it with a liquid rapidly enough to quick-freeze the exterior would probably involve asymmetric forces that would distort the sphere, whereas a gas would exert equal pressure on all sides. It would have to be some very cold gas! I don't know whether a heavy gas like argon increase thermal conduction, or something like hydrogen or helium might work better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tail doesn't seem like a necessary feature. Seems to me it's formed before the quench by the viscosity of the dripping glass, not the passage through the water. The tail isn't rapidly extruded from the blob of rapidly cooling glass; it's already present, formed by gravitation / stretching before the quench, and just cools in that tail shape.&lt;/p&gt;&#xA;" OwnerUserId="2118" LastEditorUserId="368" LastEditDate="2015-08-19T04:06:38.683" LastActivityDate="2015-08-19T04:06:38.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3345" PostTypeId="2" ParentId="3315" CreationDate="2015-06-26T19:36:57.147" Score="2" Body="&lt;p&gt;Looking for good and reliable suppliers might be one of best solutions. Vendors such as Digi-key, mouser, Newark always supply high quality electronic components. But understandably they come at a higher cost. If directly purchased from a manufacture it might be good idea to request for test data from suppliers to validate that product is meets specification. Most Suppliers have the test data readily available and if the product meets specification they will gladly share the data almost immediately. If the capacitors are purchased through an intermediary then you could request for sample data. A sample size of 30 pieces is good sample size to perform statistical analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively a small sample set of components can be tested and the measured data can be analyzed using basic statistics as part of incoming inspection. If the capacitors are SMT this might become tricky because randomly sampling across the SMT reel will cause issues for the SMT pick and place machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless after obtain this data consider performing some basic statistical analysis such as averages, standard deviation, and process capability analysis. This type of simple first order analysis will give a good indication if the incoming material is with in specification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is 30 randomly generated data point to illustrate the example basic analysis. The data points are based on 10µF capacitor with +/- 10% tolerance. Some data points are marginally below or above the lower or high tolerance limits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{| c | c | c | c | c | c | c | c | c | c |}&#xA;\hline&#xA;9.40 &amp;amp; 10.00 &amp;amp; 10.10 &amp;amp; 11.30 &amp;amp;10.10&amp;amp; 8.80 &amp;amp;9.60 &amp;amp;9.20 &amp;amp;11.50 &amp;amp;8.90 \\&#xA;\hline&#xA;8.60&amp;amp; 11.30&amp;amp;  10.00   &amp;amp;11.10&amp;amp;    10.60&amp;amp;11.10    &amp;amp;10.40&amp;amp; 9.10&amp;amp;   11.50&amp;amp; 8.90\\&#xA;\hline&#xA;8.70&amp;amp; 10.70&amp;amp; 11.20&amp;amp; 10.30&amp;amp; 10.30&amp;amp; 11.50&amp;amp; 10.80&amp;amp; 9.70&amp;amp; 9.50&amp;amp; 10.80 \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Average&lt;/strong&gt; ($\mu$): 10.17   &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Standard Deviation&lt;/strong&gt; ($\sigma$): 0.93  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C$_{p}$&lt;/strong&gt; = 0.35   &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C$_{pl}$&lt;/strong&gt; = 0.41  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C$_{pu}$&lt;/strong&gt; = 0.29  &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;C$_{pk}$&lt;/strong&gt; = 0.29  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fxJv2.jpg&quot; alt=&quot;Data1&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/a3QQT.jpg&quot; alt=&quot;Data 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base on the example data it appears that the capacitor value distribution is very wide. Looking are such data analysis certain characteristics such as the capability of the process and incoming material can be ascertained. This type analysis can help make simple decisions about the capacitors such as whether to accept or reject the material. &lt;a href=&quot;https://engineering.stackexchange.com/questions/390/how-can-the-vendor-be-approved-disapproved-based-on-validation-test-data/571#571&quot;&gt;How can the vendor be approved/disapproved based on validation test data?&lt;/a&gt; post closely related to the topic and some the equations discussed in greater detail.  Hope this response helps address the issue.&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-06-30T21:46:09.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3348" PostTypeId="1" CreationDate="2015-06-26T22:43:48.497" Score="26" ViewCount="116610" Body="&lt;p&gt;I have an Arduino board with a 9 degree of freedom sensor, from which I must determine the pitch, yaw, and roll of the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of one set of data from the 9-DOF sensor:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Accelerometer (m/s)&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\text{Acc}_{X}$ = -5,85&lt;/li&gt;&#xA;&lt;li&gt;$\text{Acc}_{Y}$ = 1,46&lt;/li&gt;&#xA;&lt;li&gt;$\text{Acc}_{Z}$ = 17,98&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Gyroscope (RPM)&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\text{Gyr}_{X}$ = 35,14&lt;/li&gt;&#xA;&lt;li&gt;$\text{Gyr}_{Y}$ = -40,22&lt;/li&gt;&#xA;&lt;li&gt;$\text{Gyr}_{Z}$ = -9,86&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Magnetometer (Gauss)&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\text{Mag}_{X}$ = 0,18&lt;/li&gt;&#xA;&lt;li&gt;$\text{Mag}_{Y}$ = -0,04&lt;/li&gt;&#xA;&lt;li&gt;$\text{Mag}_{Z}$ = -0,15&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How can I calculate pitch, yaw, and roll from these data?&lt;/p&gt;&#xA;" OwnerUserId="2122" LastEditorUserId="7083" LastEditDate="2018-07-31T22:55:53.617" LastActivityDate="2022-01-17T14:20:27.780" Title="Calculating pitch, yaw, and roll from mag, acc, and gyro data" Tags="|mechanical-engineering|electrical-engineering|sensors|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="3349" PostTypeId="2" ParentId="100" CreationDate="2015-06-26T23:21:43.433" Score="20" Body="&lt;p&gt;Prince Rupert's drops are an example of a tempered silica glass component: its surface has been cooled more rapidly than its interior. Tempering of glasses is important because it lends toughness to the glass, i.e. an ability to resist fracture under load, which explains why a drop can be hit with a hammer and survive. Silica glass, as is common with other ceramic materials, exhibits unstable crack propagation when its fracture strength is exceeded by its stress state. Unlike with most alloys, ceramics exhibit very little, or no, plastic deformation. When they reach their elastic limit they fracture. So if you stress a silica glass component too hard, it fractures rapidly and all at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A glass component may be tempered by cooling its exterior more rapidly than its interior so that there is a non-uniform residual stress distribution in the component. Specifically, because the exterior solidifies first, its density increases and volume decreases first, drawing material outward from the interior. Then, as the interior solidifies with less remaining material, it pulls inward on the exterior. The resulting stress state is tension in the interior and compression in the exterior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cracks only propagate when there is a tensile stress across the crack. If there is a residual compressive stress across the crack, it will remain closed unless stressed in tension. Because the compressive stress must be overcome before the crack opens, it takes a greater tensile stress to propagate a crack through a tempered glass component than an un-tempered component. If such a crack propagated past the neutral-stress surface between the exterior and interior of the component, the crack tip would be in tension due to the residual stress state of the interior. Such a crack would begin propagating in an unstable fashion as all of the residual stresses are released, resulting in an explosion of glass shards, as they all undergo elastic recovery from the non-uniform stress distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From all of this, it should be apparent that a &quot;perfectly&quot; spherical, tempered glass component is theoretically possible, as it is only required that the exterior of the glass cools more rapidly than the interior to obtain the required non-uniform stress distribution, while maintaining the desired shape. A combination of gravity and viscosity are the cause of the tail in a traditional Prince Rupert's drop. Therefore, removing each of those components, such as with a drop formed in free-fall by free-surface surface-tension relaxation of a &quot;floating&quot; blob of glass, can result in a sphere of viscous glass. Relaxation may take a long time and the glass must be kept viscous the entire time. The next step is cooling the sphere rapidly without disturbing its shape, which is admittedly difficult. Spraying it with fluids would cause ripples in the surface, and submersion would require moving it infinitesimally slowly, which would cause the wrong kind of non-uniform stress distribution. Exposing it to the vacuum of space might be sufficient, but I haven't done any calculations of the radiated heat loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The desired setup would likely be a radiation oven in the vacuum of space, with a blob of glass floating in it, with no relative velocity. The oven melts the glass, which relaxes into a sphere. The oven is turned off, the door is opened and the oven moves rapidly away from the sphere. The sphere emits radiation, cooling the surface more rapidly than the interior (or so we hope), and the glass is tempered, resulting in a Prince Rupert's Space Drop.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2015-06-26T23:21:43.433" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3350" PostTypeId="2" ParentId="3348" CreationDate="2015-06-26T23:33:05.430" Score="23" Body="&lt;p&gt;Pitch, roll and yaw are defined as the rotation around X, Y and Z axis. Below as a picture to illustrate the definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8IuOw.png&quot; alt=&quot;Roll Pitch and Yaw&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a previous project I used a ADXL345 Accelerometer from Analog Devices to calculate roll and pitch. Below are the equations used to calculated roll and pitch. I have made some of source code available for public use. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;accelerationX = (signed int)(((signed int)rawData_X) * 3.9);&#xA;accelerationY = (signed int)(((signed int)rawData_Y) * 3.9);&#xA;accelerationZ = (signed int)(((signed int)rawData_Z) * 3.9);&#xA;pitch = 180 * atan (accelerationX/sqrt(accelerationY*accelerationY + accelerationZ*accelerationZ))/M_PI;&#xA;roll = 180 * atan (accelerationY/sqrt(accelerationX*accelerationX + accelerationZ*accelerationZ))/M_PI;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Complete source code can be found &lt;a href=&quot;https://github.com/mahengunawardena/BeagleboneBlack_I2C_ADXL345/blob/master/ADXL345Accelerometer.cpp&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base on the above definitions&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;yaw = 180 * atan (accelerationZ/sqrt(accelerationX*accelerationX + accelerationZ*accelerationZ))/M_PI;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;a href=&quot;http://www.gnu.org/software/libc/manual/html_node/Mathematical-Constants.html&quot; rel=&quot;noreferrer&quot;&gt;M_PI&lt;/a&gt; = 3.14159265358979323846 it is constant defined in math.h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some references including Arduino base source code that might help you.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Flight_dynamics_(fixed-wing_aircraft)&quot; rel=&quot;noreferrer&quot;&gt;Flight dynamics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://github.com/mahengunawardena/BeagleboneBlack_I2C_ADXL345/blob/master/ADXL345Accelerometer.cpp&quot; rel=&quot;noreferrer&quot;&gt;Beaglebone black ADXL345 source code&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=bd7ULY1frA4&quot; rel=&quot;noreferrer&quot;&gt;Adxl345 Beaglebone black Embedded QT Demo&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://theccontinuum.com/2012/09/24/arduino-imu-pitch-roll-from-accelerometer/&quot; rel=&quot;noreferrer&quot;&gt;Arduino IMU: Pitch &amp;amp; Roll from an Accelerometer&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-04-23T01:00:22.197" LastActivityDate="2016-04-23T01:00:22.197" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3351" PostTypeId="2" ParentId="3089" CreationDate="2015-06-26T23:35:07.553" Score="3" Body="&lt;p&gt;I understand your motivation, and I've also seen the disconnect that can exist between the complexities of modern control theory and the relative simplicity of the control schemes implemented in many engineering applications. The answer by am304 gives some great suggestions. I would add to that the following points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Be very clear about which parts of your model are a representation of the plant (and therefore only useful for simulation) and which are a representation of the controller that you wish to practically implement. It may sound obvious but it can be easy to confuse the two when you're sitting behind a Simulink screen.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Think about the units you're working in at each point in the system. If you know what sensor you will be using to measure output then this will specify a conversion factor, presumably between angle and voltage. You may need to include appropriate scaling factors on your desired position to get a sensible error signal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a DC motor torque typically scales with current, and speed with voltage. For actuator type positioning applications torque control (and therefore current control) is usually desirable, so we can think of the motor as a current to torque converter. Unfortunately voltage modulation is generally the easiest motor input to control; however, current feedback is easy to achieve by placing sense resistors in the current path. Therefore a common technique is to implement a fast acting inner current control loop (based on an error signal between a desired current and the resistor sensed current), surrounded by a slower acting torque control loop (based on position error). This is referred to as cascade control and, provided the inner loop is sufficiently faster than the outer loop, allows the two controllers to be tuned relatively independently.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you don't fancy cascade control then you have two options for more complicated state-space methods. Either include a model of the torque output that the motor will achieve for a given voltage input within your control structure (although be clear that this is different to your actual plant model in simulation), or take a current measurement into your controller as well as a position measurement. This second approach is effectively combining the two cascaded controllers discussed above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hope that provides some useful information to get you started.&lt;/p&gt;&#xA;" OwnerUserId="157" LastEditorUserId="157" LastEditDate="2016-05-03T13:58:40.667" LastActivityDate="2016-05-03T13:58:40.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3352" PostTypeId="2" ParentId="2905" CreationDate="2015-06-27T01:25:07.553" Score="-1" Body="&lt;p&gt;Due to your shear force, the entire beam will attempt to shear.  (i.e. change from a nice long rectangular beam (regardless of cross section):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; --------------------------------- -&amp;gt; Pull on the top flange&#xA; |&amp;lt;-----L-----------------------&amp;gt;| - Due to shear across the cross section&#xA; --------------------------------- &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;to a parallelogram:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; ---------------------------------&#xA;/                                /&#xA;---------------------------------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(see &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_stress&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; diagram on Wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pull on the top flange is what causes a shear stress in the horizontal portion of the beam.  Since the beam has a giant hole, the shearing force can't grow straight down.  Instead, it has to travel out, through the top part, and down the beam.  This path of the shearing force is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_flow&quot; rel=&quot;nofollow&quot;&gt;shear flow&lt;/a&gt;.  As you can see in the definition of shear flow, it only happens in sections made of thin plates (like yours!).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I beams will have shear flow that starts on the edges and makes it's way to the middle.  This &lt;a href=&quot;http://www.assakkaf.com/courses/enes220/lectures/lecture15.pdf&quot; rel=&quot;nofollow&quot;&gt;lecture&lt;/a&gt; from the University of Maryland shows the shear flow really well, especially on page 10.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-06-29T14:58:46.237" LastActivityDate="2015-06-29T14:58:46.237" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3353" PostTypeId="1" AcceptedAnswerId="3357" CreationDate="2015-06-27T13:31:11.017" Score="3" ViewCount="2024" Body="&lt;p&gt;Normally, in investment casting, the mould is preheated before pouring the liquid metal in to reduce thermal shock from rapid cooling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a vacuum, the heat transfer due to gas convection is reduced to zero, which will also reduce the thermal shock. Does that mean that, if performing investment casting in a vacuum, I do not have to preheat the mould?&lt;/p&gt;&#xA;" OwnerUserId="2126" LastEditorUserId="368" LastEditDate="2015-06-29T17:11:44.987" LastActivityDate="2015-06-29T17:11:44.987" Title="Is it necessary to preheat an investment casting mould when working in a vacuum?" Tags="|metallurgy|metals|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3355" PostTypeId="1" CreationDate="2015-06-27T18:54:14.277" Score="7" ViewCount="937" Body="&lt;p&gt;I am looking to design a diaphragm pump for a device.  I know that I would need to achieve a (negative) vacuum pressure of ~ 350-400 mmHg.  I know from standard rotary pump design the pump rate should be dictated by stroke volume and speed, and assume this is the same for a diaphragm pump, with stroke length perhaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the ultimate achievable vacuum pressure? I am looking for an explanation of the idealized equations for calculating this property of a diaphragm pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be for pumping a water based fluid, although slightly more viscous. &lt;/p&gt;&#xA;" OwnerUserId="2130" LastEditorUserId="368" LastEditDate="2015-06-29T17:07:32.993" LastActivityDate="2015-08-28T17:32:31.763" Title="Calculating the ultimate achievable vacuum pressure of a diaphragm pump" Tags="|mechanical-engineering|pumps|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3356" PostTypeId="1" AcceptedAnswerId="5815" CreationDate="2015-06-27T23:22:44.917" Score="4" ViewCount="379" Body="&lt;p&gt;I'm trying to understand the importance of protective coatings on steel structures like the Golden Gate Bridge. For example, if we stopped maintenance on the Golden Gate Bridge today and left it packed with empty cars, what would be its expected lifetime (until a car falls into the water) and the variance of that lifetime? If we additionally removed all of the paint/coatings on any steel parts, what would be the expected lifetime and variance? I'm hoping for a rough answer from an expert in the field.&lt;/p&gt;&#xA;" OwnerUserId="2132" LastEditorUserId="2132" LastEditDate="2015-07-07T07:52:10.787" LastActivityDate="2015-10-20T08:19:24.970" Title="How do protective coatings affect the lifetime of bridges?" Tags="|civil-engineering|bridges|corrosion|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3357" PostTypeId="2" ParentId="3353" CreationDate="2015-06-28T02:20:03.767" Score="5" Body="&lt;h3&gt;Convection, Conduction, Radiation&lt;/h3&gt;&#xA;&lt;p&gt;Of the three modes of heat transfer, only one is affected by the vacuum. As you noted in your question, gaseous convection should be eliminated. That being said, the thermal shock is present as soon as the hot material hits the mold. As soon as the two materials come into contact, the heat transfer will be by conduction. This is the same whether in a vacuum or not.&lt;/p&gt;&#xA;&lt;p&gt;The mold should still be heated.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-06-28T02:20:03.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3358" PostTypeId="1" AcceptedAnswerId="3362" CreationDate="2015-06-28T02:57:28.337" Score="10" ViewCount="12873" Body="&lt;p&gt;I recently discovered that one of my roommates had attempted to boost the WiFi signal by cutting up a soda can and placing it near the router, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xisb7.png&quot; alt=&quot;Airport router with carved up soda can&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My roommate was probably following an internet guide like &lt;a href=&quot;http://www.wikihow.com/Make-a-Wi-Fi-Booster-Using-Only-a-Can&quot; rel=&quot;noreferrer&quot;&gt;this one&lt;/a&gt;, which advises shaping the modified can &quot;like a radar dish&quot; and placing it around the external antenna of the router. However, our router has no external antenna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will either configuration actually boost the router's reception? I don't know too much about antennae, but I suspect that the second configuration works more like a shield.&lt;/p&gt;&#xA;" OwnerUserId="2133" LastEditorUserId="368" LastEditDate="2015-06-29T16:58:32.940" LastActivityDate="2017-05-11T01:29:51.997" Title="Is it possible to &quot;boost&quot; a WiFi signal using an aluminum soda can?" Tags="|electrical-engineering|signal|wifi|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3359" PostTypeId="2" ParentId="2794" CreationDate="2015-06-28T08:57:31.077" Score="2" Body="&lt;p&gt;The numerical answer depends a lot on the other numerical data that is missing.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What sort of a summer is this? I mean, length of days (geolocation of the house), solar radiation vs time (is it 1000KW per square meter?)&lt;/li&gt;&#xA;&lt;li&gt;Are there already heat isolators in place? Is the building covered by a roof and is there air or space in between the ceiling and the roof-tiles? or is the roof cover just touching the internal air?&lt;/li&gt;&#xA;&lt;li&gt;how much power could you invest into the pump? If it is infinite, than with enough mass flow rate, you could use a simple 2 meter pipe, and as long as you can heat the water in the pipes, you would be giving away the heat to the ground. Like trying to heat the Earth.&lt;/li&gt;&#xA;&lt;li&gt;how much heat conduction is there, between the pipe and the soil? This could be best tested.&lt;/li&gt;&#xA;&lt;li&gt;how much heat can you subtract from the radiator? You cannot heat the ground more than you can cool the air. So a generous radiator is a must (which is probably simpler than piping).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As the numbers would follow the above clarifications, experimentation is the only valid method for determining the required piping and the radiator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point is, would you be trying to cool the workshop as a whole, or is it only the working point (one person standing) that you would like to cool. In the second case, using a spray-ventilator could work as well (if it is safe and no side-effects for the type of work you are doing). For the first case, cooling as a whole, isolating the sunrays from the building is the number 1 approach to take. A dark coated surface can rise to 80degrees Celcius under the sun, which essentially becomes a wonderful heater, and if you can put shadow-panels over your building, you would get rid of the sun's radiation and only deal with the hot air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One last point: why not employ some solar panels, and use their power to power your cooling solution?&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2015-06-28T08:57:31.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3360" PostTypeId="2" ParentId="3358" CreationDate="2015-06-28T10:06:34.763" Score="2" Body="&lt;p&gt;All of these cans are made up of Aluminum. Aluminum is a good conductive material and a dense material. More dense material means more capability of conductance. They can be used as antennas to transmit the signals but I am not sure whether this thing improves the WiFi signal strength as I had not tested it by myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one thing that makes sense to me is as it is made up of aluminum. It can be used to catch the signal and send it though and it is inexpensive as you see. It is just an extra layer that just increases the range a little. But you cannot say its a good replacement to those boosters available in the market. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had not done much research about it but I can say this with my common sense.&lt;/p&gt;&#xA;" OwnerUserId="2134" LastEditorUserId="282" LastEditDate="2015-06-30T00:31:47.317" LastActivityDate="2015-06-30T00:31:47.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3361" PostTypeId="2" ParentId="3358" CreationDate="2015-06-28T10:17:02.700" Score="3" Body="&lt;p&gt;This &lt;strong&gt;might&lt;/strong&gt; work to boost the signal in one direction, but it will reduce it in other directions. Antennas are often built with elements (like your drink can) that couple to the primary radiator. These act as &quot;Directors&quot; or &quot;Reflectors&quot;. You see directors on a classic TV antenna - All the cross beams along the main beam of the antenna. You see reflectors on Cell Phone towers, where a tall corner sits behind the radiator to direct the signal in small angle.&#xA;These both act to increase the signal strength in a particular direction. This is at the expense of signal strength in other directions. A correctly placed can like in your picture &lt;strong&gt;could&lt;/strong&gt; act as a reflector, increasing signal strength in the direction it faces, at the expense of signal strength in other directions. Whether this works depends critically on the size of the reflector, and the position of the reflector relative to the primary antenna. &#xA;Bear in mind this is much more likely to just reduce the signal by interacting with the antenna impedance, and reducing the quality of the match between the radio and the antenna. Like the comment above says - Experimenting will tell you which one you have. &lt;/p&gt;&#xA;" OwnerUserId="2135" LastActivityDate="2015-06-28T10:17:02.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3362" PostTypeId="2" ParentId="3358" CreationDate="2015-06-28T11:38:37.437" Score="13" Body="&lt;p&gt;What you show is a crude reflector.  Yes, you can use reflectors around a antenna to get more power in a particular direction at the expense of other directions.  This can be useful when the other antenna is in a known direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are several problems with this:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;It may violate the law.  You don't give your location (remember that filling in your profile isn't really for you, it's a courtesy for us), but here in the US the FCC will take a dim view of you exceeding allowed radiated signal strength in any direction, whether the strength in other directions is lower or not.  In reality, you are unlikely to get caught, especially if the signal is weak by the time it leaves your property.  This is not something you should do in an apartment building, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;The reflector may have the opposite effect.  Imagine a reflector &amp;frac14; wavelength from the antenna.  The signal from the antenna, bouncing off the reflector, and back to the antenna will be &amp;frac12; wavelength out of phase with the signal leaving directly from the antenna in the opposite direction.  These two waves will cancel, &lt;i&gt;reducing&lt;/i&gt; signal strength in the desired direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;At small reflector sizes, you won't get much reflection and more refraction instead.  The wavelength of 2.4 GHz is about 125 mm or 5 inches.  You need a reflector several times that to get good reflective effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;The shape you show makes little sense for what appears to be your intent, which is more signal strength to the right.  Disregarding the small size of the reflector, phase issues, etc, the curvature is way too much help focus a beam to the right.  To do that, you want a parabaloid shape with the antenna at the focus.  Your reflector may be somewhat parabolic, but with that curvature the focus point will be close to the left corner of the white box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't believe every hairbrain idea you see on the internet.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2015-06-29T16:59:46.417" LastActivityDate="2015-06-29T16:59:46.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3363" PostTypeId="1" CreationDate="2015-06-28T11:43:12.467" Score="4" ViewCount="343" Body="&lt;p&gt;I am reading a scientific paper in which an algorithm for evaluating an electromyography signal is described and I just don't get the meaning of the word &quot;window&quot; in the following part of a sentence: (&lt;a href=&quot;http://eudl.eu/pdf/10.4108/icst.mobihealth.2014.257817&quot; rel=&quot;nofollow&quot;&gt;First full paragraph of page 4&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Fig. 5 shows ... signals ... over two windows of length N = 500 &quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is window supposed to mean here?&lt;/p&gt;&#xA;" OwnerUserId="2136" LastEditorUserId="368" LastEditDate="2015-06-29T16:52:23.363" LastActivityDate="2022-10-18T10:34:16.703" Title="What is a &quot;window of length N&quot; in a signal processing algorithm?" Tags="|electrical-engineering|signal|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3364" PostTypeId="2" ParentId="3363" CreationDate="2015-06-28T13:28:42.637" Score="4" Body="&lt;p&gt;&quot;Window&quot; is often used to mean a successive set of samples processed as a batch.  In this case it appears that the algorithm works on batches of 500 samples at a time.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-06-28T13:28:42.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3365" PostTypeId="2" ParentId="3363" CreationDate="2015-06-28T17:50:45.350" Score="3" Body="&lt;p&gt;The term &lt;em&gt;time-window&lt;/em&gt; or often just &lt;em&gt;window&lt;/em&gt; is used to mean a short period of time during which something is possible, or during which you observe something, or during which you look for something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the paper introduces this concept earlier on when it says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For that, the recorded EMG-data were rectified and smoothed with a root mean square (RMS) function, working on a 250 ms time-window.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The full quote from the paper you mentioned says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fig. 5 shows the CCF the healthy (left) and the spastic hemiparetic lower limb EMG signals of s&lt;sub&gt;TA&lt;/sub&gt; and s&lt;sub&gt;GL&lt;/sub&gt; over two windows of length N = 500.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this case you can see from the graphs in Fig. 5 that the window-length is 500ms (0.5 seconds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.etymonline.com/index.php?term=window&quot; rel=&quot;nofollow&quot;&gt;one online etymology dictionary&lt;/a&gt;, the term was first used with this meaning during the space-program to describe the short period of time during which a launch was possible, the &lt;em&gt;launch window&lt;/em&gt;. And the English idiom &lt;em&gt;window of opportunity&lt;/em&gt; came from this use. Note that in this context the &lt;em&gt;window&lt;/em&gt; is defined by what's possible, or by some physical constraints that change over time, but in the paper you mentioned the &lt;em&gt;window&lt;/em&gt; seems to be deliberately constrained by the experimenters, probably to get better accuracy after smoothing/filtering the signal, or to recognise patterns that happen within their defined window while excluding matching patterns that happen over longer periods.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-06-28T17:50:45.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3367" PostTypeId="1" AcceptedAnswerId="3382" CreationDate="2015-06-28T21:49:59.043" Score="7" ViewCount="1156" Body="&lt;p&gt;I'm interested in metalworking as a hobby, to make small parts for my numerous other hobbies.  I'm looking at those cheap, typically Chinese import mini-lathes and mini-mills.  I've read about their limitations as compared to larger machines; this question is not about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, &lt;strong&gt;what are the pros and cons of a combo mini lathe/mill as opposed to separate machines?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2140" LastEditorUserId="368" LastEditDate="2015-06-29T16:32:44.747" LastActivityDate="2017-05-20T01:55:29.350" Title="What is the difference between using separate lathe and mill or a combination machine?" Tags="|mechanical-engineering|machining|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3368" PostTypeId="1" CreationDate="2015-06-29T03:16:08.723" Score="0" ViewCount="867" Body="&lt;p&gt;I am wondering if, at all, the wave equation, or the heat equation, is ever used in engineering? I think this equation is of theoretical, rather than, practical interest. For example, by solving the wave equation over a circular membrane we get all the possible standing waves on it. This is a theoretical result. But when engineers construct things do they ever rely on this equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to see an explicit example in engineering when the heat equation (or wave equation) PDE is set up and explicitly solved by the eigenvalue expansion.&lt;/p&gt;&#xA;" OwnerUserId="443" LastEditorUserId="368" LastEditDate="2015-06-30T15:19:23.933" LastActivityDate="2015-07-01T17:19:41.443" Title="PDE's in Engineering" Tags="|mathematics|" AnswerCount="1" CommentCount="11" ClosedDate="2015-06-29T16:22:15.097" ContentLicense="CC BY-SA 3.0" />
  <row Id="3369" PostTypeId="1" CreationDate="2015-06-29T04:52:39.703" Score="5" ViewCount="12932" Body="&lt;p&gt;Since I am unable to change the pitch of the blades on my ceiling fan, I am considering buying a larger one. The current one is certainly efficient enough as that is why I bought it in the first place (CFM/watt); however, I need more airflow.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Altering the pitch would have been ideal but that is not an option.  This ceiling fan has hidden blade connections.  &lt;strong&gt;How can I modify the fan blades to increase the airflow produced by the fan?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, does the idea of winglets have merit? What would be the effect of winglets on a ceiling fan vs an airfoil?  Or, would notching the trailing edge as suggested in &lt;a href=&quot;https://engineering.stackexchange.com/a/2040/368&quot;&gt;this answer&lt;/a&gt; help increase airflow?&lt;/p&gt;&#xA;" OwnerUserId="2143" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2019-04-28T12:58:22.730" Title="How can I increase the CFM of my ceiling fan?" Tags="|mechanical-engineering|airflow|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3370" PostTypeId="2" ParentId="3369" CreationDate="2015-06-29T07:39:05.023" Score="1" Body="&lt;p&gt;Ceiling fan blades are usually rather crude airfoils - you won't find one in a NACA catalog :-). In many cases a certain amount of violence may be wrought on them without too much fear of disaster. Whether this is so with your one depends on the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I write this I'm visualising some blades which are probably findable in my dungeon. If any of this sounded useful but unclear a photo or few may be able to eventuate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some the blade to hub transition consists of a flat sheet steel section and a degree of &quot;pitch&quot; (or even ALL the pitch) may have been imparted at this point by the simple expedient of &quot;twisting&quot; the transition piece. If so, you may be able to introduce some more twist by the same means. A long jawed vice grip, or a builders clamp or some similar tool with a say 3&quot; / 75 mm or longer jaw can be slipped over the blade close to the hub and twisted (add handle extension to taste to allow less force and smoother action). If doing this (or anything else in life), decide the worst case result and whether you are prepared to accept it if it happens, and don't start if not. If it does look possible then breakage odds should be low. Snapped blade is possible and damaged hub shouldn't be. Bending off-hub with blade mount in a vice removes hub damage risk but, hey, where's the fun? :-). Balance will be impaired and tuning will be needed. (I could have said &quot; ... may be ...&quot; but Murphy seldom deals in &quot;may&quot; if worse is possible.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the blade does not seem amenable to pitch shifting but the mount angle at the blade-mount join affects pitch, you may be able to shim the mount closest to the forward edge so that it becomes more forward. Longer bolt with washers may work. Nuts (correct size or larger) can act as fat washers if you need significant thickness and are not embarrassed or scornful of bush engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd think there was a moderate chance that adding leading edge extensions - initially at least with say cardboard and &quot;sellotape&quot;, would have a chance of working usefully well. You'd probably find that these only needed to be on the outer half of each blade length but longer is OK if desired. This has the advantage of potentially looking really terrible even if it works well - although this can be altered with enough work. Thin aluminum sheet is liable to last longer.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="368" LastEditDate="2015-06-29T16:19:44.723" LastActivityDate="2015-06-29T16:19:44.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3372" PostTypeId="2" ParentId="3368" CreationDate="2015-06-29T12:47:29.390" Score="3" Body="&lt;p&gt;The basic partial differential equations are relied on in many ways.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If I have something already in my book of equations (Roark's Formula's), then I use that book for solving of the plate.  These results were derived from the original equation and put in a more useful format&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I don't have it in my book (for example, recently I had to find the equation for a Navier-style plate with two free edges and two simply supported edges with a concentrated rectangular load in the center), then I turn back to the original partial differential equation - the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kirchhoff%E2%80%93Love_plate_theory&quot; rel=&quot;nofollow&quot;&gt;plate equation&lt;/a&gt;.  In this case, I found a solution online using scholar article searches. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No matter the solution that arrives, certain aspects are maintained as best as possible - conservation of energy, momentum, mass, etc. Each equation usually has a specific aspect that needs to be observed and utilized in even a numerical or FEA solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-07-01T17:19:41.443" LastActivityDate="2015-07-01T17:19:41.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3373" PostTypeId="2" ParentId="3355" CreationDate="2015-06-29T13:19:35.897" Score="3" Body="&lt;p&gt;Since you are looking for a vacuum that's less than full vacuum, What you are looking for is the &lt;a href=&quot;http://www.engineeringtoolbox.com/npsh-net-positive-suction-head-d_634.html&quot; rel=&quot;nofollow&quot;&gt;NPSH&lt;/a&gt; - Net Positive Suction Head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, when pumping matter in the liquid state to a vacuum pressure, the liquid state can boil at the reduced pressure to form gas.  These &quot;bubbles&quot; are known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Cavitation&quot; rel=&quot;nofollow&quot;&gt;cavitation&lt;/a&gt;, and they are known for destroying mechanical equipment.  At the vacuum pressure you are looking for, the water will boil at 70-80 Celsius, so as long as your system is under that temperature you should be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking over the pump diagram, you should see two diagrams - the normal pressure v. flow diagram you are used to, and the NPSH diagram.  This will tell you if your pump can withstand the vacuum requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, due to cavitation, you can't achieve full vacuum using water.  You would need to start with a gas.  And yes, vacuum pumps that go to as close as possible to full vacuum exist.  However, you can't use them to pump liquid.  Instead, you can use them like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Vacuum_sewer&quot; rel=&quot;nofollow&quot;&gt;vacuum sewer system&lt;/a&gt;, where the vacuum pump pumps the air out of a large pressure vessel that is connected to the water you want to move.   &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-06-29T13:19:35.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3374" PostTypeId="1" AcceptedAnswerId="3376" CreationDate="2015-06-29T14:03:34.793" Score="3" ViewCount="15972" Body="&lt;p&gt;There is a room on the south side of the building, which doesn't have any ventilation vents, but does have a window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Opening a window doesn't cause good air circulation (by which I mean that the air in the room is still hot, and the air on the street is still cold).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one create good air circulation in a room --with a single window- without using a fan or air conditioner?&lt;/p&gt;&#xA;" OwnerUserId="303" LastEditorUserId="368" LastEditDate="2015-06-29T20:14:22.107" LastActivityDate="2019-05-22T16:34:13.387" Title="How does one create a draft in a room (with a single window)?" Tags="|fluid-mechanics|hvac|airflow|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3375" PostTypeId="1" AcceptedAnswerId="3389" CreationDate="2015-06-29T14:06:55.073" Score="3" ViewCount="204" Body="&lt;p&gt;What type of a mechanical device can I use and program to apply a controllable force within the range of what a human can to an area about the size of a fingertip? It is desirable for the applied force to be precise and known.&lt;/p&gt;&#xA;" OwnerUserId="2146" LastActivityDate="2015-06-29T20:08:00.743" Title="Applying an adjustable force to a small area" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3376" PostTypeId="2" ParentId="3374" CreationDate="2015-06-29T14:16:28.173" Score="3" Body="&lt;p&gt;For simple ventilation, opening a single window may not mean a lot of circulation.  Air has to come in and out of the same window, so it just leads to eddies and turbulence next to the window, with &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_conduction&quot; rel=&quot;nofollow&quot;&gt;conduction&lt;/a&gt; doing most of the work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Opening two windows allows a pathway for the air to flow in and out, and allows the air to mix in, which creates an &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_exchanger&quot; rel=&quot;nofollow&quot;&gt;open heat exchanger&lt;/a&gt;, fueled by the more effective &lt;a href=&quot;https://en.wikipedia.org/wiki/Convection&quot; rel=&quot;nofollow&quot;&gt;convection&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-06-29T14:16:28.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3377" PostTypeId="2" ParentId="3375" CreationDate="2015-06-29T15:06:18.307" Score="3" Body="&lt;p&gt;There are many possible options, but possibly this simplest would be a cylinder driven by compressed air. Since compressed air pressure is easy to measure, and your cylinder area will be constant, it is easy to know how much pressure is being applied. Compressed air is also nice because the amount of pressure can be easily and quickly controlled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for something that already exists, there are a number of manipulators for robotic automation that are readily available, if relatively expensive.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-06-29T15:06:18.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3378" PostTypeId="1" CreationDate="2015-06-29T15:21:35.250" Score="4" ViewCount="1162" Body="&lt;p&gt;There are high-rate anaerobic digestion systems where you remove substrate from the digester, let it settle in a tank, and pump back the settled sludge (possibly not all of it) to ensure a higher sludge retention time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem with settling of sludge is that the forming of gas bubbles prevents settling. So before the settling tank, there's a degassing system. This is a basic PFD of such a process:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/j7T3J.png&quot; alt=&quot;PFD of anaerovic contact process&quot;&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://mebig.marmara.edu.tr/Enve424/Chapter7.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do these degasifiers work? What are their operating principles?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2015-06-30T07:14:05.840" LastActivityDate="2017-10-25T09:35:43.763" Title="How do wastewater sludge degassing systems work?" Tags="|civil-engineering|environmental-engineering|anaerobic-digestion|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3379" PostTypeId="5" CreationDate="2015-06-29T15:22:30.177" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-06-29T15:22:30.177" LastActivityDate="2015-06-29T15:22:30.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3380" PostTypeId="4" CreationDate="2015-06-29T15:22:30.177" Score="0" Body="Questions about the application of anaerobic digestion processes for biogas recovery, waste treatment and other engineered systems." OwnerUserId="61" LastEditorUserId="368" LastEditDate="2015-06-29T15:45:54.247" LastActivityDate="2015-06-29T15:45:54.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3381" PostTypeId="1" AcceptedAnswerId="3427" CreationDate="2015-06-29T15:28:41.657" Score="5" ViewCount="3195" Body="&lt;p&gt;I am interested in finding a way to calculate the force necessary to do a backwards extrusion. Typically these would be for forgings 0.5 to 3&quot; in diameter (or square) of different kinds of steel. The diagram below shows a typical type of forging I might want to do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FCVagm.png&quot; alt=&quot;hex die&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be die formed in A2 steel by backwards extrusion using a hex-shaped punch. The forging would be done with the work piece at its forging temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want a way to calculate the force (forging load) necessary do such extrusions.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2015-06-29T16:12:40.573" LastActivityDate="2015-07-02T10:58:45.827" Title="How can I calculate backward extrusion force" Tags="|mechanical-engineering|forging|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3382" PostTypeId="2" ParentId="3367" CreationDate="2015-06-29T15:37:50.973" Score="5" Body="&lt;p&gt;The problem with a combo is that they are flimsy machines. Also, the quality is usually below that of a dedicated machine, sometimes way below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend getting a mini-lathe and a drill press and mastering them. You can do 95% of everything you would do on a mill with those two items. In general a lathe is a more versatile, more precise machine. To make a mill that is as solid as a lathe requires double the weight (and cost).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical &quot;mini-mills&quot; for sale are really just cheap drill presses with a junky x-y table bolted to the bottom. You are much better off spending the money on good quality drill press. You can always add your own X-Y table.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2015-06-29T15:37:50.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3383" PostTypeId="2" ParentId="3378" CreationDate="2015-06-29T15:45:31.540" Score="1" Body="&lt;p&gt;The most common degassing system I've seen use the same basic principles as an activated charcoal system (activated charcoal, set between two gas-housing plenums, in a non-reactive (typically fiberglass) structure).  The pressure of the gas pushes the bubbles out through the filter, then to either a methane compressor or to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gas_flare&quot; rel=&quot;nofollow&quot;&gt;gas flare&lt;/a&gt; - in either case the hydrogen sulfide must be removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, activated charcoal can sometimes absorb the methane - which is not desired.  While you could rely on saturation to bring out the methane in an activated charcoal system or impregnate the carbon to not react with the methane, most system I've seen rely on &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_sulfide#Removal_from_fuel_gases&quot; rel=&quot;nofollow&quot;&gt;iron oxide&lt;/a&gt; with a double displacement reaction between the oxygen and the sulfur to form fool's gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The iron oxide can be regenerated, as shown in the article - so when you design the pressure vessel, make sure it can be fully flooded to handle the regeneration load.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="368" LastEditDate="2015-06-29T16:05:37.470" LastActivityDate="2015-06-29T16:05:37.470" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3384" PostTypeId="5" CreationDate="2015-06-29T19:08:29.360" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-06-29T19:08:29.360" LastActivityDate="2015-06-29T19:08:29.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3385" PostTypeId="4" CreationDate="2015-06-29T19:08:29.360" Score="0" Body="Questions relating to the various laws of thermodynamics, transformation or movement of a form of energy to another form, primarily but not limited to heating or cooling of a fluid." OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-06-29T20:16:00.203" LastActivityDate="2015-06-29T20:16:00.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3386" PostTypeId="5" CreationDate="2015-06-29T19:18:05.260" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-06-29T19:18:05.260" LastActivityDate="2015-06-29T19:18:05.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3387" PostTypeId="4" CreationDate="2015-06-29T19:18:05.260" Score="0" Body="Questions relating to the transfer of heat via the mechanism of convection, conduction, or radiation, in forms of systems (or your device) attempting to obtain thermodynamic equilibrium with another system (the environment or your other device)" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-06-30T00:32:43.923" LastActivityDate="2015-06-30T00:32:43.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3388" PostTypeId="2" ParentId="3031" CreationDate="2015-06-29T19:44:42.143" Score="0" Body="&lt;p&gt;I think the Bernoulli Equation is what you are looking for. Basically, the total energy in one point of the pipe is equal another point. Think about a giant water tank with water leaving a small hole in the bottom. The velocity the water is traveling at the top is very slow while the velocity leaving the hole is high. However, their volumetric flow rate is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I think you could add up the individual arms and set them equal to the main line. This assumes that the mainline is equally spaced between all the arms.&lt;/p&gt;&#xA;" OwnerUserId="2148" LastActivityDate="2015-06-29T19:44:42.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3389" PostTypeId="2" ParentId="3375" CreationDate="2015-06-29T20:08:00.743" Score="2" Body="&lt;p&gt;I agree the compressed air is the way to go. There are many vendors that sell small cylinders and even have bumpers available for the end of the actuator rod that could approximate a fingertip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For control, you don't mention any requirements, but if they're simply manually controlled on/off application of force, a regulator with some simple manual valving should do the trick, just make sure you have a way to vent the air in the cylinder when you're trying to release the force. Keep in mind that unless you control the flow into the cylinder, the force will be applied quite suddenly unless you use a flow control valve or orifice as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've done for applications like this is to use an I/P transducer, controlled by computer which would lets me precisely control the pressure applied to the cylinder by varying the current flowing in a loop, typically 4-20mA. (There are also voltage controlled versions, but I have no experience with them) This way, you could ramp the pressure at any rate you wish and the process is much more repeatable.     &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-06-29T20:08:00.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3390" PostTypeId="2" ParentId="3180" CreationDate="2015-06-30T01:40:36.310" Score="1" Body="&lt;p&gt;As a professional engineer, the first step is to say that you should not modify your dehumidifier at home.  The refrigerant in use has a high global warming potential, and emissions could have serious disasters on the environment.  Also, the high pressures in the cycle can be very dangerous.  Not to mention if refrigerant does leak into your water tank you may not even be able to filter out the contamination.  Finally if the machine is improperly designed, you could tear up your compressor or all sorts of things.  So, if you are interested, get a professional to design this system for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, if you were to say obtain a &lt;em&gt;custom built, custom designed&lt;/em&gt; dehumidifier, it is possible.  Looking at the typical refrigeration cycle (image linked from Wikipedia):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MKG5j.png&quot; alt=&quot;Look up refrigeration cycle on WIkipedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, your dehumidifier has the fan in the opposite direction.  It pulls air across the evaporator first, then across the warm condenser.  It has a pretty high efficiency because that cool air can remove a lot of the excess heat from the condenser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;custom built&lt;/strong&gt; dehumidifier would not blow the cold air across the condenser, but immerse the condenser directly into your water storage tank.  With that, the only limitation on the heat you could hold would be that the condenser would still need to turn the refrigerant from vapor to liquid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out how warm you can get it, select a refrigerant and find the boiling point at the compressor's operating pressure.  &lt;a href=&quot;https://en.wikipedia.org/wiki/1,1,1,2-Tetrafluoroethane&quot; rel=&quot;nofollow noreferrer&quot;&gt;R-134a&lt;/a&gt;, a very common refrigerant, typically runs at around 18 bar.  This &lt;a href=&quot;http://www.ohio.edu/mechanical/thermo/property_tables/R134a/R134a_PresSat.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;saturation table&lt;/a&gt; states that at 1800 kPa, the refrigerant boils at &lt;strong&gt;62.9 Celsius&lt;/strong&gt;.  So, that would be the maximum temperature your tank could reach before the refrigerator stopped working.  Due to thermodynamic inefficiency, heat exchanger inefficency, etc. you probably would get 10 - 20 C below that, but it would certainly be warmer than your air.  Ideally, with a custom designed process, you could reach really warm temperatures - again talk to a design professional.  Do not modify your home system to immerse the condenser.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-06-30T01:40:36.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3391" PostTypeId="1" CreationDate="2015-06-30T02:13:13.043" Score="2" ViewCount="6023" Body="&lt;p&gt;I have a 43-ton tractor. The tractive effort force [N] (TEF) required for a given rolling resistance (RR) coefficient of 0.06 is determined by the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${TEF}=(c + Grad\%)mg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where Grad% is the gradient percentile of incline, c is the RR coefficient, m is the mass of the tractor and g is the acceleration due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 5% incline the TEF is 46,401.3 N. So the tractor will require 46,401 N of force to move up a 5% incline. As the gradient increases the TEF increases as well - which makes sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But say for example I have 275 kW engine delivering max torque at 2300 RPM = 240.8 rad/s (P= τω; τ= P/ω = 275,000/2300 RPM = 275,000/240.8 = 1142.02 Nm &amp;lt;- is this calculation even right?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I find the maximum towing capacity of the tractor at a given gradient?&lt;/strong&gt; I.e, how much additional mass can I pull?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight I can pull will go towards &quot;0&quot; as the angle/gradient of incline increases. This is because the engine will be using more power going up hill then on a flat incline.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="368" LastEditDate="2015-07-09T18:59:26.257" LastActivityDate="2015-10-03T17:32:42.117" Title="How much load can a tractor tow uphill?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3392" PostTypeId="1" AcceptedAnswerId="3393" CreationDate="2015-06-30T06:09:49.630" Score="6" ViewCount="6924" Body="&lt;p&gt;If I have a vehicle (V1) with one 100 kW engine and an identical vehicle (V2) with 2 of these 100 kW engines, will this mean that V2 will have double the torque of V1 and double the top speed of V1?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Suspected Answer&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;It's not like in a series circuit in which, when you have two identical batteries in series, the the voltage doubles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing that there will be some increase (but not double) in torque, but the top speed would remain the same between V1 and V2 whilst pulling an identical load.  I suspect that the difference between V1 and V2 will be that V2 may be quicker to achieve the top speed, but the top speed won't increase by much.  I.E. V1's max speed pulling 30 kg will be 20 kph but might take 10 seconds, but V2s max speed pulling 30 kg will be 20 kph but might take 8 seconds.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1586" LastEditDate="2015-06-30T18:13:16.923" LastActivityDate="2015-07-06T10:10:54.797" Title="Does having 2 engines together increase anything?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3393" PostTypeId="2" ParentId="3392" CreationDate="2015-06-30T11:19:48.150" Score="9" Body="&lt;p&gt;The simple answer is yes. In Australia they have freight trucks that they call road trains. They have a second engine in the middle primary to start up which they switch off when they reach a certain speed limit. They use this type of topology when transporting freight across the outback. I thing most of them use a 400 hp Cummins diesel engine in the middle. Application is already in use but not exactly as per your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this type of topology  is very common in the locomotive space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some images to enjoy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0IqIG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/B0CF8.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/ip900.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/XwRtk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A 600 hp (450 kW) 19 L (1,200 cu in) Cummins engine powers the prime&#xA;  mover, while a 400 hp (300 kW) Cummins engine is installed in the rear&#xA;  trailer of the B-double, driving through an automatic transmission,&#xA;  giving a total of 1,000 hp (750 kW). Weights of 460 t (453 long tons;&#xA;  507 short tons) are achieved with ore loading in side-tipper bodies on&#xA;  a 100 km (62 mi) round trip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As these trucks operate on private property, they are not subject to&#xA;  governed weight and length rulings, but instead are used in the most&#xA;  efficient way possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also to below is an image from a patent to indicate active interest in you idea by the automotive - Power-train interest group. I have attached the patent for your reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kUp85.png&quot; alt=&quot;power trailer&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Road_train&quot; rel=&quot;noreferrer&quot;&gt;Road train&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cummins.com/&quot; rel=&quot;noreferrer&quot;&gt;Cummins Engine&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.google.com/patents/US20050109548&quot; rel=&quot;noreferrer&quot;&gt;Multi-combination vehicle incorporating a power trailer&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=8YAPxOFfBd4&quot; rel=&quot;noreferrer&quot;&gt;Longest load ever pulled by a Scania truck?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.truck.net.au/system/files/industry-resources/Future%20strategy%20for%20road%20supply%20and%20charging%20FINAL%20FOR%20WEB.pdf&quot; rel=&quot;noreferrer&quot;&gt;A future strategy for road supply and charging in Australia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-06-30T22:30:23.743" LastActivityDate="2015-06-30T22:30:23.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3394" PostTypeId="1" AcceptedAnswerId="3424" CreationDate="2015-06-30T13:02:46.953" Score="2" ViewCount="107" Body="&lt;p&gt;In a Vacuum Metallurgy (Casting), a mold needs to be heated (500 to 700 degrees) to reduce thermal shock on the metal surface. Normally, a gas fired oven is employed for such job, but in a vacuum environment how do you achieve the same?&lt;/p&gt;&#xA;" OwnerUserId="2126" LastEditorUserId="33" LastEditDate="2015-07-02T12:40:47.227" LastActivityDate="2015-07-02T12:40:47.227" Title="Heating Sand/Ceramic/Refractory Material under vacuum" Tags="|metallurgy|metals|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3395" PostTypeId="2" ParentId="3392" CreationDate="2015-06-30T13:28:26.623" Score="4" Body="&lt;p&gt;There is no simple answer. What is the top speed limiting factor? For a truck, it's mostly handling so no amount of power can increase that. For a sports car, it's mostly aerodynamic resistance. While the aerodynamic resistance (more precisely it's called &lt;a href=&quot;https://en.wikipedia.org/wiki/Drag_%28physics%29&quot; rel=&quot;nofollow&quot;&gt;drag&lt;/a&gt;) increases proportionally to velocity squared, the power needed to overcome it increases proportionally to the velocity cubed. So 8x the power would yield 2x top speed and 2x more power means a 1.26x faster vehicle (when considering aerodynamics alone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque will be definitely double, if the transmission and wheels can handle the doubled amount. At wost, you can add secondary set of entire propulsion set, just as they did with &lt;a href=&quot;https://en.wikipedia.org/wiki/Citro%C3%ABn_2CV#.22Sahara.22_four-wheel_drive&quot; rel=&quot;nofollow&quot;&gt;Citroen 2CV Sahara&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also research trainsets, as there are some real-life examples of almost identical vehicles differentiated only by amount of power plants.&lt;/p&gt;&#xA;" OwnerUserId="1931" LastEditorUserId="1931" LastEditDate="2015-07-06T10:10:54.797" LastActivityDate="2015-07-06T10:10:54.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3396" PostTypeId="1" AcceptedAnswerId="3408" CreationDate="2015-06-30T15:21:46.420" Score="4" ViewCount="271" Body="&lt;p&gt;I am looking to complete a materials science/engineering-based Master's degree in the UK. Those which I have been looking at, particularly Leeds and Manchester, do not seem to be accredited by the &lt;a href=&quot;http://www.iom3.org/&quot; rel=&quot;nofollow&quot;&gt;IOM3&lt;/a&gt; or other institutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in becoming chartered/incorporated in the future. If I were to complete an unaccredited course, what impact would the lack of accreditation have on my career prospects in the field of materials science/engineering?&lt;/p&gt;&#xA;" OwnerUserId="307" LastEditorUserId="282" LastEditDate="2015-07-02T18:29:26.740" LastActivityDate="2015-07-02T18:29:26.740" Title="What are the career implications of choosing a non-accredited postgraduate degree?" Tags="|materials|education|licensure|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3397" PostTypeId="1" AcceptedAnswerId="3407" CreationDate="2015-06-30T18:20:20.170" Score="4" ViewCount="2581" Body="&lt;p&gt;The rubber casing for my iPhone cable has frayed and split near the phone port, and I would like to seal it somehow to stop the split from spreading further. Would Krazy Glue be an effective sealant? I've heard that superglue can eat away at some materials.&lt;/p&gt;&#xA;" OwnerUserId="2091" LastEditorUserId="33" LastEditDate="2015-11-13T02:25:33.643" LastActivityDate="2019-06-23T05:11:17.023" Title="Can Krazy Glue be used as a sealant?" Tags="|materials|adhesive|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3398" PostTypeId="5" CreationDate="2015-06-30T18:51:19.343" Score="0" Body="&lt;p&gt;A mechanical linkage is a system of connected bodies. The connections are called joints, that come in many flavors. Linkages allow you to design various path movements, speed profiles and force changes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tag could be used for questions asking about the design or use of various mechanisms as described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest possible linkage is a four bar mechanism, because a triangle is fully defined. Another simple mechanism is the crank mechanism that can be thought of as a special case of the four bar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sXKqB.png&quot; alt=&quot;A four bar mechanism&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: A simplified schematic of a four bar mechanism. Links A-B, B-C, C-D and the ground connection form four bars. The circles represent rotational joints.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="6767" LastEditDate="2018-11-13T13:50:17.743" LastActivityDate="2018-11-13T13:50:17.743" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3399" PostTypeId="4" CreationDate="2015-06-30T18:51:19.343" Score="0" Body="A mechanical linkage is an system of connected bodies. The connections are called joints, that come in many flavors. Linkages allow you to design various path movements, speed profiles and force changes. The simplest possible linkage is a four bar mechanism and its special case the crank mechanism." OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2015-06-30T19:40:57.843" LastActivityDate="2015-06-30T19:40:57.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3400" PostTypeId="2" ParentId="3397" CreationDate="2015-06-30T19:06:20.413" Score="2" Body="&lt;p&gt;A quick search of &lt;a href=&quot;https://www.spilltech.com/wcsstore/SpillTechUSCatalogAssetStore/Attachment/documents/ccg/CBOOM.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PVC&lt;/a&gt;, the main coating for most cables, corrosion information turns up some answers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cyanoacrylate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cyanoacrylate&lt;/a&gt;, the main component of Krazy glue is not listed in most corrosion guides.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Acrylate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Acrylates&lt;/a&gt;, the main type of material that consist of most &quot;super glues&quot;, are not well supported for PVC (see the corrosion guide above).&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://web.archive.org/web/20170215054923/http://www.oatey.com/msds/msds62-purple-primer-pvc-cpvc2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PVC primer&lt;/a&gt; and bonder is typically sold at hardware stores - this would be ideal as it is designed to work and bond PVC together.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="20918" LastEditDate="2019-06-23T05:11:17.023" LastActivityDate="2019-06-23T05:11:17.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3401" PostTypeId="2" ParentId="3397" CreationDate="2015-06-30T20:16:58.760" Score="1" Body="&lt;p&gt;You can use cyanoacrylate but I would use a &quot;toughened&quot; super glue that has a rubber modifier in its chemical formula to make it less brittle. Also , super glue vapors have a tendency to blush surfaces with a white film so I would carefully mask off every surface except the repair area with masking tape before you apply any glue. You can use super glue accelerator to instantly cure the glue, very convenient for building up super glue to any thickness you like. &lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2015-06-30T20:16:58.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3402" PostTypeId="1" AcceptedAnswerId="3412" CreationDate="2015-06-30T20:29:08.570" Score="5" ViewCount="1460" Body="&lt;p&gt;How can I &quot;tie&quot; pressure change due to a mass leakage from the flow?&#xA;In a &quot;closed&quot; system I'd use $pV=nRT$ but this is an &quot;open&quot; system...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My open system is an horizontal (no gravity issue) pipe (constant sections) in which flows a gas. We may think pipe walls as permeable walls. Hence we have one incoming flow to the pipe inlet, and two outgoing flows: one from the pipe outlet and the other from the pipe wall&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the simplest case: Mach&amp;lt;0.30 and density is almost constant. To be complete, leakage is a known variable and I need to find outlet pressure&lt;/p&gt;&#xA;" OwnerUserId="2157" LastEditorDisplayName="user16" LastEditDate="2015-07-01T13:44:34.877" LastActivityDate="2015-07-09T00:14:55.737" Title="Pressure drop due to mass loss" Tags="|fluid-mechanics|gas|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3403" PostTypeId="2" ParentId="3180" CreationDate="2015-06-30T20:46:35.833" Score="1" Body="&lt;blockquote&gt;Are air conditioners or dehumidifiers currently being produced that exhaust heat through their exhaust water?&#xA;If not, are there engineering or physical reasons why this is not feasible?&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is unlikely a mainstream consumer unit would do this.  These units have to be designed to work in a variety of conditions, including hot and dry.  In a case like that, there would be very little water available, for cooling, so the system would has to be designed to work with just air cooling anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extra structures to augment the cooling system with the condensed water, when present, would make the unit more complex, larger, and more expensive.  It's not worth it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in humid conditions, there just isn't a lot of water in the air relative to the amount of air moving around.  Think of the many cubic feet of air that go thru a dehumidifier for just one drop of water to condense out.  Do the math:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets compare the heat in one cubic foot of air to the amount one drop of water could absorb.  I've watched a dehumidifier in a damp basement, and it certainly wasn't dripping one drop per cubic foot or air moving thru.  However, 1 cubic foot is 28.3 liters.  At 30&amp;deg;C and 1 atm, this cubic foot of air has a mass of about 33 grams.  In chemistry class we usually figured there were 20 drops of water per ml, so one drop has a mass of 1/20 g, or 50 mg.  The cubic foot of air is therefore 660 times more massive than the drop of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to stop here since it seems pointless to continue.  I haven't looked up the relative heat capacity of liquid water versus gasseous nitrogen and oxygen.  But even is that of water is several times more a few degrees increase of the water drop simply isn't going to amount to much of a decrease in the air.  And of course you can't transfer the heat from the air to the water 100% efficiently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even with a high condensation rate, this little bit of extra cooling just isn't worth the extra complexity, size, and materials to make use of it.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-06-30T20:46:35.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3404" PostTypeId="1" CreationDate="2015-06-30T23:46:05.370" Score="1" ViewCount="326" Body="&lt;p&gt;I am in search of a flat head machine screws with an 82 degree countersink that have 1/4-20 UNC thread, but have a countersunk head that is characteristic of a 5/16-18 standard screw. That is, I want a standard machine screw with an oversized head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been trying to search for these, but no search term I've tried works. If such a fastener existed, what would it be called? Can anyone confirm that they do or do not exist commercially?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I know of course that I could fabricate them, but for a number of reasons, that's not desirable.)&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-07-01T01:53:10.047" Title="Flat Head Machine Screws with Undersized Threads" Tags="|threads|fasteners|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3405" PostTypeId="2" ParentId="3404" CreationDate="2015-07-01T01:53:10.047" Score="2" Body="&lt;p&gt;I am not aware of anyone that makes such a beast nor have I seen one in the wild. What I have seen is head collars. collars are a tapered washer normally made of nylon but also available in metal that are designed to accommodate mismatched head taper. As they have no threads they can be used with multiple screws. Somewhere I have a box of them for #8 screws. Not what you were looking for but it might work. Also there are tapered lock washers.&lt;/p&gt;&#xA;" OwnerUserId="803" LastActivityDate="2015-07-01T01:53:10.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3406" PostTypeId="2" ParentId="3402" CreationDate="2015-07-01T02:00:05.817" Score="2" Body="&lt;p&gt;Divide both sides of the equation by time. This allows you to use flow rates, but does not account for friction, so you will have to account for that with your standard hydraulic formulas.&lt;/p&gt;&#xA;" OwnerUserId="803" LastActivityDate="2015-07-01T02:00:05.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3407" PostTypeId="2" ParentId="3397" CreationDate="2015-07-01T02:21:02.450" Score="3" Body="&lt;p&gt;Where a cable enters a connector is a point strain, point strains are remarkably prone to failure. I would not use glue at all particularly a rigid glue like super glue. I would use dental floss or string, and caulk or a curable plastic putty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin take a length of string about two foot long and fold it in half place the edge of your connector (the par directly opposite from where the cable enters the connector) in the fold of the string. Tye a couple knots around the connector ending with each end of the string  next to the cord, but on opposites sides. A drop or two of glue to hold the string to the connector body may be needed. Push the cable gently into the connector to relieve strain. alternating the two string tails, tie a series of half hitches up the cable. Mask off the connector edge. Cover the connector body and the cord end that is tied with caulk. Let cure. Remove the mask, trim the excess caulk cut the string from the connector edge and trim the string ends. Will outlast your phone.&lt;/p&gt;&#xA;" OwnerUserId="803" LastActivityDate="2015-07-01T02:21:02.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3408" PostTypeId="2" ParentId="3396" CreationDate="2015-07-01T04:19:48.530" Score="3" Body="&lt;p&gt;Try asking the &lt;a href=&quot;http://www.iom3.org/&quot; rel=&quot;nofollow&quot;&gt;IOM3&lt;/a&gt; &amp;amp; other accrediting institutions why the courses you are interested in are not accredited and if there is a possibility that they could be accredited in the near future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The courses may not cover material that such institutions consider vital for professionals engaged in that field and which would be required for membership of the institutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to study one of those courses you may need to study the missing subjects elsewhere prior to being eligible to for membership of the institutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding employment prospects, it's safer to study an accredited course. In times of economic downturns or gluts of professionals in particular fields employers can be very discriminating. Anyone without an accredited qualification can more easily have their application disregarded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not just job prospects in the UK that you need to consider; materials science/engineering is an international profession. Larger companies track the performance of universities and the courses they offer and they know which universities offer courses that are of most interest to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Companies also prefer to employ people with accredited qualifications so as to minimize any legal issues should there be a problem arising from their products or services and the personnel responsible for those products or services did not have an accredited qualification.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="282" LastEditDate="2015-07-02T18:29:20.640" LastActivityDate="2015-07-02T18:29:20.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3409" PostTypeId="1" AcceptedAnswerId="4085" CreationDate="2015-07-01T05:09:35.830" Score="3" ViewCount="707" Body="&lt;p&gt;I've had trouble with leaky pipe flanges, since my flanges are made of plastic. The problem comes down to the fact that sealing a flange is based on torque. Torque relies on bolt friction, which can change even between different bolt diameters. Therefore, it's a terrible indicator of the clamping force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other way to measure flange sealing force than by using a torque wrench? I'm interested in any method that works, whether it's a known system or something crazy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideal solutions would take around the same time to measure as the torque wrench (maybe double), can be used to check flanges that are currently in operation as well as new installations, and can work with a wide variety of flange, bolt and gasket materials.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="368" LastEditDate="2015-07-09T00:18:38.303" LastActivityDate="2015-08-20T00:26:15.783" Title="Measuring pipe bolt torque without a torque wrench" Tags="|mechanical-engineering|piping|bolting|threads|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3410" PostTypeId="1" CreationDate="2015-07-01T09:23:10.200" Score="11" ViewCount="485" Body="&lt;p&gt;There's a TV programme called &quot;How It's Made&quot; which often shows automation of mass-produced machined parts. Very often, it seems that metal parts are machined (especially drilled) by rotating the &lt;em&gt;part&lt;/em&gt; against a stationary tool rather than holding the part and drilling into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the advantage of doing it this way? I've never been able to work it out.&lt;/p&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="368" LastEditDate="2015-07-09T00:19:43.803" LastActivityDate="2017-05-18T15:04:58.570" Title="Why are mass-produced metal parts machined by rotating the part against a stationary tool?" Tags="|machining|manufacturing-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3411" PostTypeId="2" ParentId="3410" CreationDate="2015-07-01T09:49:20.023" Score="8" Body="&lt;p&gt;You can be sure the resulting surface is round. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you used a router and rotated the bit around the piece then you would get inaccuracies due to the movement being harder to control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a lathe-like setup the cutting tool is just a chisel that is cheaper to replace and refurbish when it's worn down than a router bit.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-07-01T09:49:20.023" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3412" PostTypeId="2" ParentId="3402" CreationDate="2015-07-01T11:50:55.023" Score="3" Body="&lt;p&gt;You can't use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow&quot;&gt;ideal gas law&lt;/a&gt; in an open system, but you can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Bernoulli&amp;#39;s_principle&quot; rel=&quot;nofollow&quot;&gt;Bernoulli's equation&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the conservation of energy: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_{in} = E_{out} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have one source in and two sources out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_{in} = E_{out} + E_{leak} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy is the sum of the enthalpy and kinetic energy (since the pipe is horizontal):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (h_1 + v_1^2/2)*m_1 = (h_2 + v_2^2/2)*m_2 + m_{leak} * (h_3+ v_3^2/2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition, the enthalpy is the internal energy plus $P/\rho$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for some important assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The velocity of the leak far from the pipe would be 0 (i.e., it's reached the atmosphere and not causing tornadoes).&lt;/li&gt;&#xA;&lt;li&gt;The leak's final pressure would be atmospheric pressure (via the same assumptions)&lt;/li&gt;&#xA;&lt;li&gt;The entire system is under roughly constant temperature, so the internal energy, u of the gas at all three points is the same&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$ (u + P_1/\rho + v_1^2/2)*m_1 = (u + P_2/\rho + v_2^2/2)*m_2 + m_{leak}*(u + P_{atm}) $$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Now we assume constant flow.  The pressure, internal energy and velocity at the inlet should be the same with respect to time.  Same with the outlet and leak.  So, differentiating with respect to time, the only thing that changes at each point is the mass:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$ P_{in} = P_{out} + P_{leak} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (u + P_1/\rho + v_1^2/2)*\dot{m}_1 = (u + P_2/\rho + v_2^2/2)*\dot{m}_2 + \dot{m}_{leak}*(u + P_{atm}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know by conservation of mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m_{in} = m_{out} + m_{leak} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which works when taking the derivative with respect to time.  So, we can remove the u terms.  Finally, it's easier to express mass flow as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{m} = \rho A V $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, substituting in the mass flow, and multiplying the entire equation by the constant density to convert from head to pressure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(P_1 +\rho\frac{V_1^2}{2})\rho AV_1 = (P_2 + \rho \frac{V_2^2}{2})\rho AV_2 + \dot{m}_{leak}(P_{atmosphere})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the units are Watts*density - to make your mass loss easier to solve.  While you could solve for the leak in this equation, using your knowledge of the pressure outside and the pressure at two places, you would still need to know at least one of the velocities to solve for the leak.  The other velocity, you could solve for because of conservation of mass again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{m}_{in} = \dot{m}_{out} + \dot{m}_{leak} $$&#xA;$$ \rho V_1 A = \rho V_2 A + \dot{m}_{leak} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, I did make these assumptions: &lt;em&gt;constant density&lt;/em&gt;, and &lt;em&gt;no loss due to pipe friction&lt;/em&gt; under the assumption the leaks are so light that the temperature didn't change, and so density didn't change - just decrease the velocity.  Ultimately, to figure out the velocities, you need to know the flow rate - so the only way to tell is with flowmeters, not pressure gauges.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="368" LastEditDate="2015-07-09T00:14:55.737" LastActivityDate="2015-07-09T00:14:55.737" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3413" PostTypeId="1" CreationDate="2015-07-01T14:03:01.583" Score="4" ViewCount="3161" Body="&lt;p&gt;I am building a machine for a project that should slowly tip a screen forward about 95 degrees, hold for a couple of seconds, and then slowly return to vertical.  The design I am working on uses two stepper motors connected to the screen with a belt drive.  I can change the gear ratio of this drive to anything up to 2.8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to select the smallest motor that will provide enough power for this application.  I have the torque - speed charts for the motors, as well as the weight and moment of inertia of the screen.  Is there a formula that I can use to calculate if the motor will be strong enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The screen will begin vertical, rotate forward about 95 degrees, and then return to vertical. The weight of the screen assembly is ~5.7 lbs. It will rotate about an axis ~2.75 inches below the center of mass. The screen assembly consists of a display surrounded on all sides by 1&quot; t-slotted aluminum.  The axis of rotation runs through the center of the bottom aluminum rail.  There are two small stepper motors mounted just below this rail as well.The aluminum frame is 11.75&quot; wide by 10.0625&quot; tall.  The motors and mounting plates extend an additional 1.66&quot; down.  Specs for the family of motors I plan to use can be found at &lt;a href=&quot;http://automationdirect.com/static/specs/surestepmotors.pdf&quot; rel=&quot;nofollow&quot;&gt;automationdirect.com&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2163" LastEditorUserId="2163" LastEditDate="2015-07-01T17:24:27.463" LastActivityDate="2015-08-05T23:33:09.487" Title="How to calculate if a motor is strong enough" Tags="|mechanical-engineering|gears|motors|stepper-motor|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3415" PostTypeId="1" AcceptedAnswerId="3437" CreationDate="2015-06-23T18:04:15.747" Score="6" ViewCount="807" Body="&lt;p&gt;I have a small issue that is causing a lot of problems. I'm on a &lt;a href=&quot;https://en.wikipedia.org/wiki/Formula_SAE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Formula SAE&lt;/a&gt; team and our battery is consistently shorting between our positive terminal and our chassis. Our chassis is the grounding point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our battery is underneath our seat, our seat is made of carbon and therefore when the driver moves shorts occur. I've tried using tape, rubber covers and plastic boxes. The box cracks under the weight of the driver and the tape and covers fall off during a long race.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I prevent shorts with a solution that wont break, nor fall off?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case a short does occur, we use breakers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a sketch of the position of the battery under the seat. At point one, when we are static, the seat does not touch the battery, but under high acceleration, the driver will rub on to the battery terminals. This knocks off tape and covers that are on it. In the course of a long race, eventually shorts start to occur after the tape/covers have been rubbed off. We noticed that the issues happened on tight turns and moments of high acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GgLnr.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2165" OwnerDisplayName="Adam" LastEditorUserId="368" LastEditDate="2015-07-09T19:30:00.870" LastActivityDate="2015-07-13T09:04:30.800" Title="How to prevent shorts between exposed battery terminals and racing vehicle chassis" Tags="|electrical-engineering|automotive-engineering|battery|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3416" PostTypeId="2" ParentId="2025" CreationDate="2015-07-01T19:24:29.913" Score="2" Body="&lt;p&gt;The fundamental problem central to the question is the management of dust which is inevitable in any inhabited space as most of the dust consist of shed human skin cells. Although it might seem desirable, rejecting the dust by any means merely decreases the air quality in the space; using charge repulsion results in the deposition of dust elsewhere with potentially equally undesirable results and until it &quot;sticks&quot; to something, it still diminishes the air quality.  Thus, managing rather than fighting the problem becomes the only option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest is to use a cotton/spandex cover such as these (Outer Space Ceiling Fan Blade Covers by Fan Blade Designs -these are mostly polyester so they will hold a static charge unfortunately) and a pillow case.  Roll the pillow case (the longer king size work best) like one would a sock and carefully put it over the cover.  Remove both but DO NOT SEPARATE and proceed in turn to each blade.  Place both items as one into a washing machine and select a long cycle.  Stop the cycle when the water has reached the top and separate the two underwater so as to maintain the air quality of the space.  A plastic bag can be used temporarily when access to the washer is not direct; however, still place both in the bag, do not separate them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further improve the quality of air in the room, a velcro patch may be sown into the portion of the fan blade cover that faces upward and a block shaped ceiling fan air filter such as those made by BestAir can be attached to it.  This will capture a great deal of dust and can easily be removed with the previous method without modification.  It is my understanding that neither will compromise airflow in the summer and that while the block may disrupt airflow during the winter, it does not do so to such an extent as to alter the basic function of the fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine this solution is far from what is desired in as much as it is both low tech and only elegant in appearance rather than substance.  Nevertheless, it is effective, inexpensive, and home decor friendly.  This last may seem insubstancial to a single person; however, anyone either married or in a long term committed relationship will tell you, it is not an idle consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other options; however, these do not involve the fan directly while still only managing the dust in the air albeit in an entirely different manner without eliminating the cleaning.  Whilst the cleaning is made more convenient and cleaner in terms of the air in the room, the original condition has not changed; nevertheless, having tried a number of options, I am convinced that no method involving repelling the dust will work without making dust stick unintentionally to other surfaces and making these surfaces harder to clean resulting in no net benefit.&lt;/p&gt;&#xA;" OwnerUserId="2143" LastActivityDate="2015-07-01T19:24:29.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3417" PostTypeId="1" CreationDate="2015-07-01T21:39:18.547" Score="2" ViewCount="914" Body="&lt;p&gt;I am writing some code for the analysis of 3D truss structures. The program is a typical implementation of the direct stiffness method for truss elements. While testing it and comparing results to other FEM programs, I encountered the following problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I have a joint with zero-force truss elements (for example a joint where 3 non-coplanar truss elements meet), the forces in these elements do not equal zero, which contradicts the joint equilibrium equation. This only happens for space trusses. When I consider plane trusses and add zero-force joints (for example a joint where two non-colinear truss elements meet), everything works out okay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot see what causes this problem. Perhaps some additional form of static condensation is needed for the 3D-case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought process is that if the force in an element equals zero, that means that $u_{i} - u_{j} = 0$, where $u_{i}$ and $u_{j}$ are the axial displacements of that element, expressed in its local coordinate system. Simply stated, no elongation means no force. Of course, the whole element can be displaced in a rigid-motion-manner only, if at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem seems to be the following. Assume you have a 3D truss defined, with no zero-force members. Attach 3 elements to it (i.e. a tetrahedron, which is stable by itself) so that the entire truss is stable again. Assume the joint where these elements meet is unloaded, hence all the forces in the added elements must equal zero. Of course, the other displacements and forces in the truss are not affected by this - it just acts the same as without the added three elements. Hence, the truss would somehow need to &quot;know&quot; that it must be displaced in such a manner that none of the three members is elongated (or contracted), i.e. they can only be displaced as a whole, rigidly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems pretty weird  and it hints that perhaps such situations should indeed be treated before the analysis. Then again, the direct stiffness method is, as far as I know, completely general and should be able to calculate any truss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further on, for 2D trusses this situation does not seem to arise at all; the zero force members are always recognized correctly by the direct stiffness method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an example of a truss where I encounter the problem. The image represents the element and node labeling of the truss. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From equilibrium of node 10013 (the upmost node), it follows that the forces in elements 112, 114 and 116 must equal zero. But, the program only calculated zero force in element 114. The forces in elements 112 and 116 are of order of magnitude similar to the other forces in the truss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I remove elements 112, 114 and 116, all the forces in the truss are calculated correctly when compared to another FEM solver (up to 3% difference, as mentioned above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ozQ0Q.jpg&quot; alt=&quot;3D truss example&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2168" LastEditorUserId="368" LastEditDate="2015-07-09T00:06:49.337" LastActivityDate="2015-07-15T22:53:40.113" Title="FEM 3D truss analyzer produces forces that do not sum to zero at joints" Tags="|civil-engineering|structures|finite-element-method|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="3418" PostTypeId="2" ParentId="2911" CreationDate="2015-07-01T23:55:22.653" Score="1" Body="&lt;p&gt;Miles per gallon or km per litre is probably more easy for consumer to comprehend. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Efficiency requires a unit of &lt;em&gt;kg x meter / joule&lt;/em&gt; where the mass is the payload of the vehicle and meter the distance traveled on 1 joule worth of fuel. &#xA;E.g. Passenger km per kWh and Cargo load km per kWh are measurements of this but is seldom used other then by professionals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some example data estimates in kWh/ Passanger km for city traffic:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Light MC 0,40  &#xA;Car 0,65 &#xA;Bus 0,30 &#xA;Tram 0,12&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www5.goteborg.se/prod/Lokalforvaltningen/dalis2.nsf/vyFilArkiv/Transport_2011.pdf/$file/Transport_2011.pdf&quot; rel=&quot;nofollow&quot; title=&quot;Source&quot;&gt;Commuter report&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1697" LastActivityDate="2015-07-01T23:55:22.653" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3420" PostTypeId="1" AcceptedAnswerId="3433" CreationDate="2015-07-02T02:12:59.257" Score="3" ViewCount="4268" Body="&lt;p&gt;&lt;em&gt;The question&lt;/em&gt;: Consider $1\ kg$ of air at $32\ C$ that is expanded by a reversible polytropic process with $n=1.25$ until the pressure is halved. Determine the heat transfer.&lt;br&gt;&#xA;Specific heat constant volume for air is $0.1786\  kJ/kg.K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is: $17.02\  kJ$ heat added&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My work so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{T_{1}}{T_{2}} = \frac{P_{1}}{P_{2}}^{\frac{n-1}{n}}$$&#xA;    $$n = 1.25$$&#xA;    $$\frac{P_{1}}{P_{2}} = 0.5$$&#xA;    $$T_{1} = 32+273$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd get $T_2 = 350.352$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Q = m*Cv*(T_2-T_1)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get $8.00\ kJ$ something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What am I doing wrong?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorDisplayName="user16" LastEditDate="2015-07-02T14:14:43.737" LastActivityDate="2015-07-02T14:14:43.737" Title="The Heat Added in a Reversible Polytropic Process" Tags="|mechanical-engineering|heat-transfer|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3421" PostTypeId="2" ParentId="3369" CreationDate="2015-07-02T04:25:00.597" Score="1" Body="&lt;p&gt;The motor is likely an &lt;a href=&quot;http://www.mqitechnology.com/downloads/motor_design/MotorDesign-10-CeilingFanMotor.pdf&quot; rel=&quot;nofollow&quot;&gt;induction type&lt;/a&gt;, which by design, &quot;slips.&quot; So the more load placed on it, the slower it will spin. Which means, if you increase the blade pitch angle, this will produce more load for the motor, resulting in lower RPM, and probably a very similar amount of airflow overall. Of course, if you could install a more powerful motor, more air would flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One interesting thing about page 10 of that link. If you were to essentially put the motor on a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dynamometer&quot; rel=&quot;nofollow&quot;&gt;dynamometer&lt;/a&gt; and produce a curve of it's torque vs. RPM, then you could calculate at what RPM the maximum power is produced, and adjust the blade shape, size, and pitch to suit. But if the engineer that designed the fan already did this, any improvement would be minor. And loading a small motor too heavily will cause it to reach a higher temperature and likely fail sooner. There is only so much power available from this motor to move air, and no &quot;weird trick&quot; will make it produce remarkably more airflow. Yes, you could &quot;airfoil&quot; the leading edges, plane-down the trailing edges, notch it, add &quot;tips&quot; or all sorts of other things, but these will not produce any remarkable difference as the motor is the limiting factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a practical solution, could a second fan be added? Or this fan upgraded to a more powerful model? A squirrel-cage fan moves lots more air (and use a lot more power), but that topology doesn't lend itself well to ceiling use.&lt;/p&gt;&#xA;" OwnerUserId="1673" LastActivityDate="2015-07-02T04:25:00.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3422" PostTypeId="2" ParentId="3381" CreationDate="2015-07-02T06:00:04.837" Score="1" Body="&lt;p&gt;We could model your problem this way: the equipment creates shear stresses along the shapes of the extrusion, hence the extrusion will be created for a force which exceeds your materials' sheer resistance. Denote this resistance by T. Hence it follows that F = T*A = T*6*d*l, where d is the depth of the extrusion, and l is the length of the hexagon side. So, you only need to look up the steel shear resistance T and you can solve this.&lt;/p&gt;&#xA;" OwnerUserId="2168" LastActivityDate="2015-07-02T06:00:04.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3423" PostTypeId="1" AcceptedAnswerId="3428" CreationDate="2015-07-02T06:01:49.487" Score="4" ViewCount="614" Body="&lt;p&gt;I major in microelectronics. I just want to apply a fixed and adjustable force to my flexible sample.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;'Fixed' means that the force could be hold for given time without change.&lt;/li&gt;&#xA;&lt;li&gt;'Adjustable' means that the force would be changed after my setting or its value could be changed with time and rule.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It would be better if the force was horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sample can endure the force with value about 10 N. And I hope the force could be adjustable between 0 and 10 N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a stepper motor with load cell may be the solution but I don't know how to realize it exactly. The only thing I can think of is to hang different balance weight to one side of my sample. I know some kind of universal testing machines could also get this done but they are too expensive.&lt;/p&gt;&#xA;" OwnerUserId="2171" LastEditorUserId="368" LastEditDate="2015-07-09T00:02:02.887" LastActivityDate="2015-07-09T00:02:02.887" Title="How can I apply a fixed and adjustable force to my flexible sample?" Tags="|mechanical-engineering|mechanical-failure|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3424" PostTypeId="2" ParentId="3394" CreationDate="2015-07-02T07:44:05.567" Score="2" Body="&lt;p&gt;Ok, After Some Wandering on Google, I came to know the process,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One Idea is to heat mold by radiation, A ceramic heater could be suitable for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the way it is actually done is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First preheat the mold in a gas fired oven, and in parallel start vacuum induction furnace to melt the material in different chamber, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;once mold and material is ready to be pour, put mold in a vacuum chamber for casting, close the chamber and starts molding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=WXFRRg8YMT0&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=WXFRRg8YMT0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this video gives nice explanation of the process.&lt;/p&gt;&#xA;" OwnerUserId="2126" LastActivityDate="2015-07-02T07:44:05.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3426" PostTypeId="1" CreationDate="2015-07-02T09:36:20.653" Score="0" ViewCount="39" Body="&lt;p&gt;Sugar beets are an interesting substrate for biogas production due to their high content of easily digested sugars and starches. However, they lack macronutrients to sustain the bacteria population. Reports of mono fermentation (no co-substrates) on a lab scale exist, the &lt;a href=&quot;http://biogas-infoboard.de/pdf/Linke%20aus%205.%20Bio-Solar%20Biogas-Facht-3.pdf&quot; rel=&quot;nofollow&quot;&gt;longest running experiment I'm aware of&lt;/a&gt; ran for 240 days using an SBR for solids/bacteria retention (which somewhat alleviates the need for macronutrients).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any plant digesting only sugar beets on a large scale (&gt;10 t/d), where it is confrmed that no co-substrates are used? I want to stress confirmed, because the fact that a reference sheet does not mention a co substrate does not mean it's not there.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="33" LastEditDate="2015-07-02T12:45:04.440" LastActivityDate="2015-07-02T12:45:04.440" Title="Has large scale mono fermentation of sugar beets been achieved?" Tags="|environmental-engineering|biogas|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3427" PostTypeId="2" ParentId="3381" CreationDate="2015-07-02T10:58:45.827" Score="3" Body="&lt;p&gt;You can find formulas for Backward extrusion in &lt;a href=&quot;https://books.google.com.tr/books?id=_ZVqPgAACAAJ&amp;amp;printsec=frontcover&amp;amp;dq=editions:ISBN0470505923&amp;amp;hl=tr&quot; rel=&quot;nofollow&quot;&gt;Groover's Principles of Modern Manufacturing 4th edition&lt;/a&gt; section 17.5.2. I summarized required formula but if you need more info ask or check the book. Also you can &lt;a href=&quot;http://www.egr.msu.edu/~pkwon/me477/bulkforming2.pdf&quot; rel=&quot;nofollow&quot;&gt;check these lecture notes&lt;/a&gt; I googled.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reduction ratio&lt;/strong&gt;: $r_x=\frac{A_0}{A_f}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_0$ and $A_f$ are initial and final area in mm respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In bacwards extrusion, effect of friction can be ignored so &lt;strong&gt;true strain&lt;/strong&gt; is $\varepsilon = ln r_x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Average flow stress&lt;/strong&gt; is $\overline{Y_f}=\frac{K\varepsilon^n}{1+n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;K&lt;/strong&gt; is strength coefficient and &lt;strong&gt;n&lt;/strong&gt; is strain hardening exponent. There are a few examples on &lt;a href=&quot;https://en.wikipedia.org/wiki/Strain_hardening_exponent&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt; but you'll probably get A2 steel's info from datasheet or materials books. (K is in MPa so $\overline{Y_f}$ is in MPa)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pressure required&lt;/strong&gt; for indirect extrusion is $p=\overline{Y_f}\varepsilon_x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Force required&lt;/strong&gt; is $F=pA_0$&lt;/p&gt;&#xA;" OwnerUserId="879" LastActivityDate="2015-07-02T10:58:45.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3428" PostTypeId="2" ParentId="3423" CreationDate="2015-07-02T11:49:35.067" Score="2" Body="&lt;p&gt;With a spring and a (accurate) actuator you can create one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One end of the spring is the force applicator the other end is moved by the actuator so the spring compresses a certain amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by &lt;a href=&quot;https://en.wikipedia.org/wiki/Hooke%27s_law&quot; rel=&quot;nofollow&quot;&gt;Hooke's law&lt;/a&gt; you can control the force that the spring applies to your test piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works best if the test piece doesn't deform.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-07-02T11:49:35.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3430" PostTypeId="2" ParentId="3394" CreationDate="2015-07-02T12:08:50.483" Score="2" Body="&lt;p&gt;A second idea for added control, if needed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your molders can accomplish it, you could also bury a &lt;a href=&quot;https://en.wikipedia.org/wiki/Heating_element&quot; rel=&quot;nofollow&quot;&gt;heating coil&lt;/a&gt; in the mold before you harden it.  The heating cable would need to be able to withstand the liquid metal temperatures, so something like &lt;a href=&quot;https://en.wikipedia.org/wiki/Tungsten&quot; rel=&quot;nofollow&quot;&gt;tungsten&lt;/a&gt; would be required.  This would also allow some control of the process after pouring - if you needed to get to a different &lt;a href=&quot;https://en.wikipedia.org/wiki/Steel#Material_properties&quot; rel=&quot;nofollow&quot;&gt;metallurgical state&lt;/a&gt;, for example.  The heater near the edges would keep the edges hot while the inside cools.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-07-02T12:08:50.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3431" PostTypeId="1" AcceptedAnswerId="3434" CreationDate="2015-07-02T12:19:03.220" Score="5" ViewCount="2402" Body="&lt;p&gt;The Bond number is a dimensionless number typically used to analyze cases where two fluids of different densities are in contact and subject to gravity only.  I'm analyzing a system where gas and liquid are entrained in a pipe (think garden hose with trapped air pockets), and both are subject to high accelerations, much higher than gravity.  I'd like to modify the equation for the Bond number to use my known value of acceleration in place of gravitational acceleration, but I can't find any published literature where someone has done this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to do is this: $$Bo=\frac{\Delta \rho a L^2}{\sigma}$$ where $a$ has replaced $g$ in the traditional equation: $$Bo=\frac{\Delta \rho g L^2}{\sigma}$$  Anyone have experience or at least comments on my approach?&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-11-23T16:52:25.683" Title="Bond number for accelerating flow" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3432" PostTypeId="1" AcceptedAnswerId="3435" CreationDate="2015-07-02T12:43:50.503" Score="5" ViewCount="1183" Body="&lt;p&gt;So one of the proposed suspension systems that will be used on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hyperloop&quot; rel=&quot;nofollow&quot;&gt;hyperloop&lt;/a&gt; include the externally pressurized air cushions. These cushions lift (or at least help lift) the capsule and reduce drag when the capsule is in motion. Also, the tube itself is supposed to have a very low pressure. But won't 100s of capsules moving in the tube releasing significant amounts of compressed air pressurize the tube and thereby significantly increase drag? I suppose this could be solved by perpetually depressurizing the tube, but I did not see any information on it. Does anyone know how this problem was addressed or am I just missing something here?&lt;/p&gt;&#xA;" OwnerUserId="2177" LastEditorUserId="33" LastEditDate="2015-07-02T20:17:56.280" LastActivityDate="2015-07-02T20:17:56.280" Title="How does the pressurized suspension of the Hyperloop not affect the tube pressure?" Tags="|fluid-mechanics|transportation|drag|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3433" PostTypeId="2" ParentId="3420" CreationDate="2015-07-02T12:55:26.590" Score="2" Body="&lt;p&gt;You can't use the specific heat constant of constant volume - a polytropic process isn't constant volume or constant pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, what is derived is the specific heat of polytopic expansion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_n = c_v\frac{n-k}{n-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where k is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_capacity_ratio&quot; rel=&quot;nofollow&quot;&gt;heat capacity ratio&lt;/a&gt;.  Air is considered a diatomic ideal gas, so k = 1.4.  When I lookup air's specific heat for volume via &lt;a href=&quot;http://www.engineeringtoolbox.com/specific-heat-capacity-gases-d_159.html&quot; rel=&quot;nofollow&quot;&gt;engineering toolbox&lt;/a&gt; is $c_v = 0.718 kJ/kg*K$, so I'm using that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, here's how I would approach it (note you did have p2/p1 backwards as well):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_2 = T_1(\frac{p_2}{p_1})^{1-1/n} = 305K*(0.5)^{0.2} = 265.52K $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ c_n = 0.718\frac{-0.15}{.25} = -0.4308 \frac{kJ}{kg*K} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Q = m c_n (T_2 - T_1) = 1kg * (-0.4308)*(265.5K - 305K) = 17.02 kJ $$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source for equations - the ever useful &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1591264146&quot; rel=&quot;nofollow&quot;&gt;MERM&lt;/a&gt;.  Don't take the PE exam without it.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-07-02T12:55:26.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3434" PostTypeId="2" ParentId="3431" CreationDate="2015-07-02T13:23:20.003" Score="5" Body="&lt;p&gt;So a search on youtube leads to this &lt;a href=&quot;https://www.youtube.com/watch?v=cXsvy2tBJlU&quot;&gt;incredible video&lt;/a&gt; documenting some experiments aboard the international space station of bubbles in a state of 0 gravity.  From the appearance it looks like if the acceleration is reduced to 0, then the bubbles act like a low bond number - the surface tension forces are dominating in the video.  (In this case it would correspond to a Bond number of 0)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the same philosophy of &lt;a href=&quot;https://en.wikipedia.org/wiki/Dimensional_analysis&quot;&gt;dimensional analysis&lt;/a&gt; that all dimensionless numbers are the ratio of forces, it would seem like you have two points that infer that acceleration is something you could scale using the Bond number.  So, I would say this would be an acceptable practice of dimensional analysis.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-07-02T13:23:20.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3435" PostTypeId="2" ParentId="3432" CreationDate="2015-07-02T13:24:52.930" Score="6" Body="&lt;p&gt;The air used for the air cushions will come from the air still in the tube pressurized by the pod itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the system remains closed (all air that is released is sucked out of the environment).&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-07-02T19:59:10.487" LastActivityDate="2015-07-02T19:59:10.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3436" PostTypeId="2" ParentId="3415" CreationDate="2015-07-02T13:56:47.660" Score="2" Body="&lt;p&gt;Tilt the battery at a slight angle away from the seat so that the seat doesn't touch the terminals.  You should be able to get away with small tilt angles without having the battery leak (assuming lead-acid), at least during use.  You may want to remove the battery in between uses though.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-07-02T13:56:47.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3437" PostTypeId="2" ParentId="3415" CreationDate="2015-07-02T14:02:09.200" Score="2" Body="&lt;p&gt;1st suggestion would be to find a sturdier cover for the battery. However, it sounds like it might just be badly positioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you move it, rotate it, flip it (if it's an AGM type that will work at any angle) to put the terminals away from harm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failing that, can you put some solid plastic or other insulator over the top of it, snugly, so it can't be touched. Nylon chopping boards are inexpensive at Ikea, drill a couple of holes for the terminals etc., layer a couple of boards to create a light, strong, solid block of insulator on top of the battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3-D printing something might work but it'd have to be very solid around the terminals.&lt;/p&gt;&#xA;" OwnerUserId="2180" LastEditorUserId="2180" LastEditDate="2015-07-13T09:04:30.800" LastActivityDate="2015-07-13T09:04:30.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3438" PostTypeId="1" CreationDate="2015-07-02T14:37:07.777" Score="5" ViewCount="117" Body="&lt;p&gt;I am designing a 3-stage platform like that of a 3D printer. Three stages correspond to X, Y, and Z axis. There is a nozzle with fixed height about the z-stage platform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stages are pretty accurate with a resolution of 1 micron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I need to adjust the height of the z stage so that the distance between the nozzle and the stage surface is about 10±1 microns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already checked &lt;a href=&quot;http://www.ia.omron.com/products/family/3120/&quot; rel=&quot;nofollow&quot;&gt;some&lt;/a&gt; of the high-resolution distance sensors, but they are way too expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if there are any alternate methods to achieve that.&lt;/p&gt;&#xA;" OwnerUserId="1996" LastEditorUserId="282" LastEditDate="2015-07-02T18:29:01.737" LastActivityDate="2015-07-02T18:46:12.380" Title="High-resolution Z stage distance measurement" Tags="|measurements|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3439" PostTypeId="2" ParentId="3438" CreationDate="2015-07-02T14:58:05.820" Score="2" Body="&lt;p&gt;Not going to happen. Your limiting factor here is not the sensor, so you can forget about that. Any kind of normal bearing or gear mechanism can get nowhere near 1 micron. Even CNC mills and lathes costing $500,000+ cannot get that kind of precision and they have bearings that cost thousands of dollars each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A micron is very small distance. Just to give you some idea, a precision grinder costing big bucks has a vertical adjustment graduated in 0.0001&quot; inches which means the mechanism has to be precise at least to 0.00005&quot; to deliver that granuality (ten thousandths of an inch). A micron is even smaller than that.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2015-07-02T18:46:12.380" LastActivityDate="2015-07-02T18:46:12.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3440" PostTypeId="2" ParentId="3423" CreationDate="2015-07-02T15:15:10.673" Score="0" Body="&lt;p&gt;You may want to look at a related question: &lt;a href=&quot;https://engineering.stackexchange.com/questions/3375/applying-an-adjustable-force-to-a-small-area&quot;&gt;Applying an adjustable force to a small area&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the answers to this question, a small piston powered by air pressure may work. It has the ability to adjust the pressure as needed and will also be useful for keeping the same pressure even if your sample deflects under the load.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-07-02T15:15:10.673" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3441" PostTypeId="1" AcceptedAnswerId="3444" CreationDate="2015-07-02T15:26:59.073" Score="3" ViewCount="145" Body="&lt;p&gt;Due to the fact that pipe flanges rely on bolts, and bolts rely on friction, it seems obvious to me, as a manufacturer of piping, to recommend that bolt torques need to be checked every so often, to ensure piping doesn't leak.  I've got a few questions about this, but here is the first one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How often do you normally check torques on bolts at a plant?  In other words, what is a good industry standard preventative maintenance schedule when it comes to threaded fasteners that are loaded below yield?  I would prefer that answers provide a reputable source for reference.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="368" LastEditDate="2015-07-02T15:41:03.650" LastActivityDate="2015-07-02T18:50:51.103" Title="How often do you need to check torques on bolted flanges?" Tags="|mechanical-engineering|piping|bolting|maintenance|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3442" PostTypeId="2" ParentId="218" CreationDate="2015-07-02T15:51:22.163" Score="5" Body="&lt;p&gt;In the world of plastic piping, the formula is different, because the material doesn't &lt;a href=&quot;https://en.wikipedia.org/wiki/Yield_(engineering)&quot; rel=&quot;nofollow&quot;&gt;yield&lt;/a&gt;.  For &lt;a href=&quot;https://en.wikipedia.org/wiki/Isotropy#Materials_science&quot; rel=&quot;nofollow&quot;&gt;isotropic&lt;/a&gt; plastics, B31.3 shows piping as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p = \frac{2St}{D-t} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where D, t and S remain the same as above.  However, the allowable strength (S) is given by an applicable ASTM specification, which functions the same as the yield stress - but is not always based on the materials ultimate strength.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Composite piping, being composed of &lt;a href=&quot;https://en.wikipedia.org/wiki/Transverse_isotropy&quot; rel=&quot;nofollow&quot;&gt;Orthotropic&lt;/a&gt; laminates, doesn't have a well defined strength - the material is designed with the pipe.  In these cases, the original assumption that pipes are &lt;strong&gt;&lt;em&gt;tested until failure&lt;/em&gt;&lt;/strong&gt; is absolutely correct.  B31.3 states again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ p = \frac{2SFt}{D-t} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where a new factor, F is introduced.  S is obtained from the &lt;a href=&quot;http://www.astm.org/Standards/D2992.htm&quot; rel=&quot;nofollow&quot;&gt;Hydrostatic Design Basis&lt;/a&gt; - and it is essentially an &lt;a href=&quot;https://en.wikipedia.org/wiki/Fatigue_(material)#High-cycle_fatigue&quot; rel=&quot;nofollow&quot;&gt;S-N Curve&lt;/a&gt; for that particular sequence of lamination.  F allows conversion between the two tests - 0.5 for the static test, 1.0 for the dynamic test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ASME is currently reviewing this method - and this is an exciting new area of development for them as they are generating a &lt;a href=&quot;http://gasapps.gastechnology.org/webroot/app/gtiishome/files/ASMEStandardsCommittee.pdf&quot; rel=&quot;nofollow&quot;&gt;new piping standard&lt;/a&gt; to relieve the industry of the expensive and extensive HDB testing requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quality testing mandated by ASTM D2996 / ASTM D2992 ensures the piping is made the same way - any change in the formula requires a re-test.  Using this method, composite piping is typically designed for a 50 year life-cycle.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-07-02T16:18:01.657" LastActivityDate="2015-07-02T16:18:01.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3443" PostTypeId="1" AcceptedAnswerId="3456" CreationDate="2015-07-02T18:31:29.740" Score="2" ViewCount="4410" Body="&lt;p&gt;I have been reading about embedded systems, and particularly the reset modes. As I understand a microcontroller can have several reset modes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is brownout reset (BOR) and power on reset (POR)? What is the difference between BOR and POR?&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="368" LastEditDate="2015-07-08T23:57:58.750" LastActivityDate="2015-08-30T05:10:54.467" Title="What is the difference between BOR and POR micro-controller reset modes?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3444" PostTypeId="2" ParentId="3441" CreationDate="2015-07-02T18:50:51.103" Score="1" Body="&lt;p&gt;There is no answer because there are too many variables, amount of vibration and thread geometry being just two of them. A joint could last for 10,000 years or work loose tomorrow. There is no way to know.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2015-07-02T18:50:51.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3445" PostTypeId="2" ParentId="3409" CreationDate="2015-07-02T18:55:01.117" Score="-1" Body="&lt;p&gt;Without knowing the application details it is impossible to say what might work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, anything involving a pressurized fluid requires a taper fitting to be secure. Using cheap half measures like gaskets will involve leaks and more maintenance.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2015-07-02T18:55:01.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3446" PostTypeId="1" CreationDate="2015-07-02T19:06:31.933" Score="3" ViewCount="93" Body="&lt;p&gt;I'm trying to make a project with keil uVision 5, for DA14580 (Dialog Semiconductor). &#xA;&lt;a href=&quot;http://www.keil.com/dd/chip/6853.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.keil.com/dd/chip/6853.htm&lt;/a&gt;&#xA;But in Select Device dialog there's no device from Dialog Semiconductor.&#xA;As you see in the photo, i dodn't found Dialog Semiconductor MCUs.&#xA;&lt;strong&gt;The Question is: How can i start a Project with some selected mcu?&lt;/strong&gt;&#xA;All examples that i can found is for TI microcontrollers or Atmel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OHUT5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Device missing from Pack Installer too:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/rdmlS.png&quot; alt=&quot;enter image description here&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only information about the device that I'm looking at is in device database in File Menu:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/YpkQx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to &quot;Add&quot; the device but, when I'm creating new project for a specific Device, (Photo No1) I cann't find the device in the list.&lt;/p&gt;&#xA;" OwnerUserId="2150" LastEditorUserId="7083" LastEditDate="2016-10-25T12:52:54.770" LastActivityDate="2016-10-25T12:52:54.770" Title="Keil Dialog Semiconductor" Tags="|electrical-engineering|embedded-systems|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3447" PostTypeId="1" CreationDate="2015-07-02T19:16:57.990" Score="3" ViewCount="381" Body="&lt;p&gt;I am in need of a device that extends freely without much resistance but compresses slowly.&#xA;I can provide a door closer analogy.The door should open fast but close slowly....&#xA;Do you think this can be achieved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically my idea is to come up with a design for Wickets setup in Cricket.For people who have no idea what the game is...basically I need a setup for a 28 inch long wooden cylindrical stump of 1.5 inch radius to selfstand and withstand forces of a ball hitting it.Once the ball,say baseball, hits it, it should spring backwards and then comeback stand straight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this I am replacing the bottom 2 inch section with heavy metal so that it stands on its own and connecting this 2 inch section to the top one through a spring to handle the impact of ball.&lt;img src=&quot;https://i.stack.imgur.com/hQ8N1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2183" LastEditorUserId="2183" LastEditDate="2015-07-06T19:25:47.023" LastActivityDate="2015-07-06T19:25:47.023" Title="Spring - Piston with different extension and compression rate" Tags="|applied-mechanics|springs|shock|pistons|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3448" PostTypeId="2" ParentId="3447" CreationDate="2015-07-02T19:59:52.157" Score="3" Body="&lt;p&gt;Door closer is a great mechanism - at it's heart is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dashpot&quot; rel=&quot;nofollow noreferrer&quot;&gt;dashpot&lt;/a&gt; - and a check valve.  If you have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Check_valve&quot; rel=&quot;nofollow noreferrer&quot;&gt;check valve&lt;/a&gt; in parallel to the dashpot, that accomplishes the desired function.  Watch below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BJ24c.jpg&quot; alt=&quot;Dashpot and check valve&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is your mechanism at rest.  You could add a spring to the top of the dashpot to force it to go back to the fully closed state when finished - but that would clutter up this illustration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you pull on the piece:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z2KHY.jpg&quot; alt=&quot;Pulling on dashpot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluid flows through the check valve, (as the flow pushes on the ball it forces the spring down, allowing plenty of flow).  Note check valves work in a variety of ways, but the ball and spring is shown here.  Since the fluid flows easily, the device can extend as far as the dashpot allows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jlu7w.jpg&quot; alt=&quot;Pushing on dashpot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When pushing the mechanism back, the check valve doesn't allow the fluid to flow.  As a result, the load must push back slowly as the highly viscous fluid moves around the plunger to let it sink back in to place.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-07-02T20:09:27.150" LastActivityDate="2015-07-02T20:09:27.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3449" PostTypeId="2" ParentId="3447" CreationDate="2015-07-02T20:03:41.250" Score="4" Body="&lt;p&gt;The first design that comes to mind is a pneumatic cylinder with a large one-way valve and a slow two-way release on the back end. I've attached quick sketch below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vcyEu.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea here is that the big valve on top can let air in freely - letting the device extend with little resistance. Pushing the piston back in, however, the large valve closes, forcing air out the much smaller release valve on bottom. The compression rate is determined by the diameter of this valve and the force pushing on the piston.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have to do a little research to find out if there's something like this on the market, but if not it wouldn't be too hard to modify a single-acting air cylinder to achieve the same purpose.&lt;/p&gt;&#xA;" OwnerUserId="2048" LastActivityDate="2015-07-02T20:03:41.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3450" PostTypeId="1" AcceptedAnswerId="3462" CreationDate="2015-07-02T20:42:34.323" Score="23" ViewCount="7203" Body="&lt;p&gt;I have seen a number of structures within rivers, which resemble steps, and which allow the water to cascade down them instead of flowing naturally down the course of the river. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1: River Avon in Bath&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rPrQj.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rPrQjm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2: River Seine in Paris (from &lt;em&gt;Les Miserables&lt;/em&gt; (2012))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cOL4a.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cOL4am.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the purpose of these structures?&lt;/p&gt;&#xA;" OwnerUserId="164" LastEditorUserId="272" LastEditDate="2015-07-02T21:29:04.757" LastActivityDate="2016-09-10T06:35:46.603" Title="What is the purpose of these &quot;steps&quot; in rivers?" Tags="|civil-engineering|hydrology|flow-control|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3451" PostTypeId="2" ParentId="3450" CreationDate="2015-07-02T21:02:02.110" Score="22" Body="&lt;p&gt;The 'steps' are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Weir&quot;&gt;weirs&lt;/a&gt;, and they are used for a number of flow control reasons in rivers.  In the case of the images you posted, they are probably being used to stabilize the grade (slope) of the river.  When used for this application they are sometimes called &lt;a href=&quot;https://en.wikipedia.org/wiki/Drop_structure&quot;&gt;drop structures&lt;/a&gt; which &quot;pass water to a lower elevation while controlling the energy and velocity of the water as it passes over.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In rather simplified terms, they slow down the flow of a river by dissipating the energy picked up from travelling downhill as heat and sound instead of allowing it to speed up the flow of the water.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-07-03T15:23:58.747" LastActivityDate="2015-07-03T15:23:58.747" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3452" PostTypeId="2" ParentId="3423" CreationDate="2015-07-02T21:23:47.500" Score="0" Body="&lt;p&gt;The poor man's solution would be to simply stack weights onto your sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While that seems low tech, you need surprisingly little different weights, which you can usually buy in sets, to get a massive range of forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The applied force is guaranteed to be independent of deflection, and this solution is also rather cheap compared to building specialized rigs.&lt;/p&gt;&#xA;" OwnerUserId="946" LastActivityDate="2015-07-02T21:23:47.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3453" PostTypeId="1" CreationDate="2015-07-02T22:41:11.517" Score="5" ViewCount="1709" Body="&lt;p&gt;I have two solar panels, a &lt;a href=&quot;http://pt.enat.pt/media/8119/datasheet_SANYO_HIT_EN.pdf&quot; rel=&quot;nofollow&quot;&gt;Sanyo HIT 240HDE4&lt;/a&gt; (35.5 V and 6.77 A at max power) and a &lt;a href=&quot;http://eu-solar.panasonic.net/fileadmin/user_upload/downloads/technical_documents/NEW_VBHN240SJ25_PEWEU_EN.pdf&quot; rel=&quot;nofollow&quot;&gt;Panasonic (SANYO) HIT VBHN245SJ25&lt;/a&gt; (44.3 V and 5.54 A at max power).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These panels are mismatched in both voltage and current, so aren't ideally suited to being connected either in series or parallel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I add a voltage regulator/converter to one of these panels to bring the voltage in line with the other, therefore allowing me to connect them in parallel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I bring the 240 W panel up to 44.3 V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I can connect both to my MPPT controller and not waste energy. &lt;/p&gt;&#xA;" OwnerUserId="2186" LastEditorUserId="368" LastEditDate="2015-08-10T16:11:47.883" LastActivityDate="2015-08-10T16:11:47.883" Title="How can I connect mismatched solar panels in parallel?" Tags="|electrical-engineering|solar-panel|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3454" PostTypeId="1" CreationDate="2015-07-03T01:47:44.520" Score="0" ViewCount="2010" Body="&lt;p&gt;For a gasoline internal-combustion engine, used in accordance with the manufacturer's instructions (specified quality oil, fuel, etc. used), what changes the efficiency of the motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example at higher RPM, more heat generated due to friction would mean it isn't as efficient as at lower RPM, since some energy is lost as heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engine's alignment would also have an impact on the efficiency of the engine &amp;ndash; for example if a car's motor is mounted upside down (i.e., 180&amp;deg; rotated from the camshaft's axis of rotation) it won't be as efficient as it would be the right side up.  So there is some relationship between engine tilt/misalignment and efficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another factor could be the temperature: hot engine vs cold engine vs overheating engine.  Also does ambient temperature affect the efficiency &amp;ndash; e.g., if the same engine is running in Antarctica vs the Sahara desert?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other factors might have an influence on the engine's efficiency (disregarding fuel types E10, U95, U98, etc.), and are they also valid for diesel motors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been looking for references with Google's aid, but I haven't found anything that is conclusive.&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="77" LastEditDate="2015-07-04T00:55:22.557" LastActivityDate="2015-07-04T00:55:22.557" Title="What affects an internal-combustion engine's efficiency?" Tags="|thermodynamics|automotive-engineering|motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3455" PostTypeId="1" AcceptedAnswerId="3459" CreationDate="2015-07-03T02:07:55.673" Score="0" ViewCount="7169" Body="&lt;p&gt;I am answering this question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Steam enters the superheaters of a boiler at a pressure of 25 bar and&#xA;  dryness of 0.98% and leaves at the same pressure at a temperature of&#xA;  370 °C. Calculate the heat energy supplied by the superheaters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer is supposed to be 405.51 J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Properties of Steam are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 25 bar and 370 °C: $h=3171.8\ \mathrm{kJ/kg}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h_{f}=962.11\ \mathrm{kJ/kg}$, $h_{fg}=1841.01\ \mathrm{kJ/kg}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h=h_f+xh_{fg}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h=962.11+0.98*1841.01=4070.474\ \mathrm{kJ/kg}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q=m(h_2-h_1)=4070.474-3171.8=898.674$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="368" LastEditDate="2015-07-08T23:45:16.837" LastActivityDate="2015-07-08T23:46:12.040" Title="Calculating the heat energy supplied by a superheater" Tags="|thermodynamics|steam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3456" PostTypeId="2" ParentId="3443" CreationDate="2015-07-03T04:43:14.220" Score="2" Body="&lt;p&gt;When the V&lt;sub&gt;dd&lt;/sub&gt; drops below a brown out threshold voltage, BOR will hold the microcontroller in reset state. Not all devices have BOR detection, but most do, and some have multiple voltage thresholds to select from.&lt;br&gt;&#xA;Between a BOR and Power On Reset the whole range of startup voltages can be covered to protect for proper operation after a power drop at the V&lt;sub&gt;dd&lt;/sub&gt; line.&lt;/p&gt;&#xA;" OwnerUserId="2189" LastEditorUserId="2189" LastEditDate="2015-08-30T05:10:54.467" LastActivityDate="2015-08-30T05:10:54.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3457" PostTypeId="1" CreationDate="2015-07-03T04:48:17.230" Score="3" ViewCount="181" Body="&lt;p&gt;We are about to commence a large project in Astana, Kazakhstan worth several hundred million dollars. The design of Stormwater quantity, quality, and hydrology will be a large component of our works, and I am having an issue locating rainfall Intensity-Frequency-Duration (IFD) tables for the region. Similar to those which one might use from the Bureau of Meteorology on any site in Australia and are crucial to the design of stormwater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, has anyone completed drainage design for this region, or can anyone provide pointers on where I might find IFD tables for this region? More generally, does anyone have pointers on designing stormwater systems in freeze-thaw conditions.&lt;/p&gt;&#xA;" OwnerUserId="767" LastActivityDate="2015-07-03T04:48:17.230" Title="Engineering in Kazakhstan and Rainfall Intensity" Tags="|civil-engineering|hydrology|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3459" PostTypeId="2" ParentId="3455" CreationDate="2015-07-03T09:52:55.650" Score="1" Body="&lt;p&gt;You just got the math wrong&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_1 = 962.11 + (0.98*1841.01) = 2766.2998 \ \mathrm{kJ/kg}$$&#xA;$$q  = (h_2-h_1) = 3171.8 - 2766.2998 = 405.5 \ \mathrm{kJ/kg}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I omitted your $m$ in final equation since we are calculating per kg.)&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="368" LastEditDate="2015-07-08T23:46:12.040" LastActivityDate="2015-07-08T23:46:12.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3460" PostTypeId="2" ParentId="435" CreationDate="2015-07-03T13:02:22.407" Score="6" Body="&lt;p&gt;If you look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/%C3%98resund_Bridge&quot; rel=&quot;noreferrer&quot;&gt;Øresund Bridge&lt;/a&gt; they state that the bridge was designed with heavy rail traffic in mind:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A girder and cable-stayed design was chosen to provide the specific rigidity necessary to carry heavy rail traffic, and also to resist large accumulations of ice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The trains run in a deck below the road trafic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the article in wikipedia lacks a reference for this statement, I can't give the actual reason. But it seems that a suspension bridge is too wobbly. Personally I have heard that trains on a suspension bridge would push a wave of bridge in front of them, tearing the tracks and the bridge apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly suspension bridges are often rather steep out to the middle, something trains do not like, trains need quite a bit less incline than what a car or a lorry can handle.&lt;/p&gt;&#xA;" OwnerUserId="2194" LastActivityDate="2015-07-03T13:02:22.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3461" PostTypeId="1" CreationDate="2015-07-03T13:02:38.310" Score="2" ViewCount="67" Body="&lt;p&gt;Due to the fact that pipe flanges rely on bolts, and bolts rely on friction, it seems obvious to me that in deflecting to meet the load, the nut will immeasurably turn.  So my question is - how much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, given a time based varying load on a given threaded fastener, what would be the rotation of the nut?  Has there been any experiments done to show an equation between a simply varying load (like a sine wave in the axial direction) and the rotation of a threaded nut?&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-07-03T13:02:38.310" Title="What is the load response of a threaded fastener?" Tags="|mechanical-engineering|stresses|threads|deflection|deformation|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3462" PostTypeId="2" ParentId="3450" CreationDate="2015-07-03T14:37:18.980" Score="20" Body="&lt;p&gt;The two photos in the post show the same structure: &lt;strong&gt;Pulteney Weir&lt;/strong&gt;, downstream of &lt;a href=&quot;https://en.wikipedia.org/wiki/Pulteney_Bridge&quot; rel=&quot;noreferrer&quot;&gt;Pulteney Bridge&lt;/a&gt; on the River Avon in Bath. The shot of the &quot;Seine&quot; in &lt;em&gt;Les Miserables&lt;/em&gt; was &lt;a href=&quot;http://www.imdb.com/title/tt1707386/locations&quot; rel=&quot;noreferrer&quot;&gt;filmed on location&lt;/a&gt; in Bath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pulteney Weir was designed by architect &lt;a href=&quot;http://www.theguardian.com/news/2003/jul/14/guardianobituaries.artsobituaries&quot; rel=&quot;noreferrer&quot;&gt;Neville Conder&lt;/a&gt;, and built between 1968 and 1972. It's one component of the Bath Flood Prevention Scheme, which was carried out after the disastrous &lt;a href=&quot;http://www.bathintime.co.uk/image/1116261/floods-on-the-lower-bristol-road-bath-december-1960&quot; rel=&quot;noreferrer&quot;&gt;flood of December 1960&lt;/a&gt;. The Bath in Time website has photos of the &lt;a href=&quot;http://www.bathintime.co.uk/image/1099732/fishing-on-the-pulteney-weir-c-1960s&quot; rel=&quot;noreferrer&quot;&gt;old weir&lt;/a&gt; (which descended in a single step) and of the new weir under construction (&lt;a href=&quot;http://www.bathintime.co.uk/image/317625/flood-prevention-barriers-being-constructed-at-pulteney-weir-c-1968&quot; rel=&quot;noreferrer&quot;&gt;1968&lt;/a&gt;, &lt;a href=&quot;http://www.bathintime.co.uk/image/818458/panorama-of-the-pulteney-weir-scheme-under-construction-21-july-1971&quot; rel=&quot;noreferrer&quot;&gt;1972&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An upstream view of the weir shows that there's more to the structure than the horseshoe-shaped steps: these meet a mid-river artificial island, and a channel to the left of the island terminates in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sluice&quot; rel=&quot;noreferrer&quot;&gt;sluice gate&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uewv2.jpg&quot; alt=&quot;upstream view of the weir&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://wikimapia.org/3675688/Pulteney-Weir-Bath#/photo/965114&quot; rel=&quot;noreferrer&quot;&gt;Photo by GeodesyMike from Wikimapia&lt;/a&gt;, licensed under CC-BY-SA.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with any &lt;a href=&quot;https://en.wikipedia.org/wiki/Weir&quot; rel=&quot;noreferrer&quot;&gt;weir&lt;/a&gt;, it has multiple purposes: to control the level of the river above the weir for navigation, fishing, and irrigation; to preserve water in long periods of drought; to hold flood waters back and release them gradually downstream; and to control the speed and force of the water, preventing the river from scouring the banks and damaging the foundations of the bridge and nearby buildings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=8k0o_nwjm5M&quot; rel=&quot;noreferrer&quot;&gt;This video&lt;/a&gt; shows the weir in operation during flood. You can see that the shape of the weir directs the force of the water to the centre of the river, preventing it from damaging the banks.&lt;/p&gt;&#xA;" OwnerUserId="2195" LastActivityDate="2015-07-03T14:37:18.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3463" PostTypeId="2" ParentId="435" CreationDate="2015-07-03T16:06:15.213" Score="13" Body="&lt;p&gt;Possibly the reputation of railway suspension bridges was fatally damaged by the poor performance of the very first one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1830, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stockton_and_Darlington_Railway&quot; rel=&quot;noreferrer&quot;&gt;Stockton and Darlington Railway&lt;/a&gt; built a suspension bridge over the River Tees, designed by &lt;a href=&quot;https://en.wikipedia.org/wiki/Samuel_Brown_(Royal_Navy_officer)&quot; rel=&quot;noreferrer&quot;&gt;Samuel Brown&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XDABi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Illustration by William Miller, from &lt;em&gt;The North Eastern Railway&lt;/em&gt; by William Tomlinson, 1915.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intended to carry trains of 150 tons, testing showed that the bridge towers sustained damage when the load exceeded 66 tons. &lt;a href=&quot;http://www.thenorthernecho.co.uk/history/11778931.The_world_s_first__and_worst__railway_suspension_bridge/?ref=ar&quot; rel=&quot;noreferrer&quot;&gt;Chris Lloyd describes the occasion&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the first engine, pulling 16 coal trucks, edged onto the bridge, the deck wobbled and shook, and the pillar on the Yorkshire side swayed and cracked. As the train neared the centre, the deck rose up in the middle, creating a mini-mountain, with eight trucks going up the Durham slope while, simultaneously, the other eight rolled down the Yorkshire side. The coupling in the middle snapped, and while the front eight wagons and the engine proceeded across to Yorkshire, the eight rear wagons ran away, speeding down the hump into Durham.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Trains were thereafter limited to four wagons at a time connected &quot;&lt;a href=&quot;https://archive.org/stream/northeasternrail00tomlrich#page/n221/mode/2up&quot; rel=&quot;noreferrer&quot;&gt;by means of chains with couplings which kept them 9 yards apart&lt;/a&gt;.&quot; In 1844 traffic was transferred to a cast iron girder bridge designed by &lt;a href=&quot;https://en.wikipedia.org/wiki/Robert_Stephenson&quot; rel=&quot;noreferrer&quot;&gt;Robert Stephenson&lt;/a&gt;, and the suspension bridge was demolished in 1880.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, two modern suspension bridges carry railways over very long spans:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The double-decker &lt;a href=&quot;http://www.jb-honshi.co.jp/english/technology/shimotsui.html&quot; rel=&quot;noreferrer&quot;&gt;Shimotsui-Seto Bridge&lt;/a&gt; (opened 1988) carries a road and a railway over a span of 940 m between Honshū and Hitsuishijima in Japan.&lt;/li&gt;&#xA;&lt;li&gt;The double-decker &lt;a href=&quot;https://en.wikipedia.org/wiki/Tsing_Ma_Bridge&quot; rel=&quot;noreferrer&quot;&gt;Tsing Ma Bridge&lt;/a&gt; (opened 1997) carries a road and a railway over a span of 1,377 m between Tsing Yi and Ma Wan islands, Hong Kong. The book &lt;a href=&quot;https://books.google.co.uk/books?id=sS7MBQAAQBAJ&amp;amp;pg=118&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;Structural Health Monitoring of Long-Span Suspension Bridges&lt;/em&gt;&lt;/a&gt; by You Lin Xu and Yong Xia describes in detail how the bridge engineers ensure this operates safely.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FGn1U.jpg&quot; alt=&quot;Tsing Ma Bridge&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Tsing_Ma_Bridge.jpg&quot; rel=&quot;noreferrer&quot;&gt;Public domain image of Tsing Ma Bridge by Snaveed, 2006.&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="2195" LastActivityDate="2015-07-03T16:06:15.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3464" PostTypeId="2" ParentId="400" CreationDate="2015-07-03T21:09:43.593" Score="2" Body="&lt;p&gt;For a low tech and somewhat primitive method, given a metal tank with access to the bottom you could simply tap on the outside of the tank with a metal or other hard object (not hard enough to dent or damage it, just enough to generate a sound.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start at the bottom and begin tapping repeatedly moving upwards until you hear the echoing sound you are making change pitch. I'm thinking that it will be higher pitched while tapping at the sediment layer and drop to a lower pitch when you hit the liquid layer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is similar to how a carpenter will tap along a wall until the pitch is higher. That's how you can tell when you have found a stud for hanging items on instead of just the dry wall.&lt;/p&gt;&#xA;" OwnerUserId="2190" LastEditorDisplayName="user16" LastEditDate="2015-07-04T00:54:29.043" LastActivityDate="2015-07-04T00:54:29.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3465" PostTypeId="2" ParentId="3453" CreationDate="2015-07-03T21:44:39.450" Score="3" Body="&lt;p&gt;While it is possible to transform the output of either panel to match the other, the cost of doing so and the gains made are such that it is unlikely to be worthwhile. The characteristics of your MPPT converter and your wiring length and diameter (and so resistance) may have a greater bearing. Placing them in series about halves the effective current compared to parallel connection so reduces wiring copper loss to 25% for the same wiring. Some converters are limited in voltage range they will accept and series connection may not be an option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not yet have the MPPT converter then using a per panel microinverter to produce mains AC at the panel may be the best solution. This allows combining outputs directly and the high voltage reduces wiring losses. Some microinverters will tolerate the voltage of either of your panels and others will work only with one or the other. Note that the MPPT Vin range and maximum Vin range may not match and it is possible to have an inverter operate but not offer MPPT.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you already have the inverter the a better solution may be to change the relative mounting angles to gain a larger &quot;spread&quot; of power during the day or year and thereby obtain more energy overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voltage mismatch is 35.5/44.3 = 80.1%&lt;br&gt;&#xA;The current mismatch is 5.54/6.77 = 81.8%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These values are so close for practical purposes that series or parallel connection is about as good power wise. Voltage and power have a negative change with increasing temperature and current usually has a small positive change but this affects both panels is not going to be enough to make a useful difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use a converter to up or down scale the voltage from either panel but unless this was done with an MPPT converter you would probably lose most of the 20% difference in the converter. If this was done with a 2nd MPPT converter you would be better off using the two converters to drive the common load. Depending on what the load was the two converters may &quot;fight&quot; so this needs to be done with due care.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly the best practical solution is to  arrange the panels so that one is pointed &quot;suboptimally&quot; so that it receives more sun at a period in the day or season than it would when optimally pointed? If the panels are fixed in position then the angle change could either be seasonally related so the off-angle panel was optimised for e.g. Autumn rather than Summer, or daily related so it got maximum insolation either earlier or later in the say than would usually be optimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If skewing panels I'd expect at 1st glance that placing them in series and deoptimising the 35.5 V/6.77 A panel would work best. (Working through the options MAY produce a different result). If the 6.77 A panels was off angled so it made about 5.5 A at solar noon in optimum conditions then it would make more output at some other time. The disadvantage of this is that if you e.g. optimise for morning then you get even less in the afternoon and so on. SO some half optimised result may work best overall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could spend many fascinating hours trying to get this right - and its probably not very valuable. Both panels are rated at about 240 Watts. If you lose 20% of one it's about 50 Watts loss peak or 10% of total max power, and less overall. You could add an extra 50 Watt panel of suitable rating or just keep them cleaner than most people usually do, or arrange so they can be adjustably tilted slightly seasonally and change the angle a few times a year.&lt;br&gt;&#xA;(Daily angle changes are much more effort intensive.)&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="368" LastEditDate="2015-07-08T23:54:26.940" LastActivityDate="2015-07-08T23:54:26.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3466" PostTypeId="2" ParentId="3454" CreationDate="2015-07-03T22:05:19.010" Score="2" Body="&lt;p&gt;Presumably we are talking about efficiency in terms of shaft/mechanical output energy as a fraction of chemical energy in the fuel being burned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important thing to note is that piston engines are most efficient in practice when run at full design load.  At lower loads they are less efficient because (to first order) pumping and friction losses are constant with each stroke, but power output is reduced.  (I.e., you could say that they aren't making &quot;full use&quot; of the combustion chamber volume, but they are incurring the full costs of cycling it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the remaining factors I'm mostly making educated guesses:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Again noting that pumping and friction losses are constant, in general one would assume that &lt;strong&gt;higher air density&lt;/strong&gt; will produce higher efficiency since it produces more power output per cycle.  So to first order the engine would be most efficient running in cold air at low altitudes.&lt;/li&gt;&#xA;&lt;li&gt;Assuming the engine is responsible for its own cooling presumably it will lose less energy to cooling when run at low RPMs and in a cold environment where, in the limit, it requires &lt;strong&gt;no active cooling&lt;/strong&gt; at all.&lt;/li&gt;&#xA;&lt;li&gt;Cold air also allows for &lt;strong&gt;higher compression&lt;/strong&gt; ratios, which increase efficiency.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Overheating is bad&lt;/strong&gt;: Once an engine gets critically hot it has to retard timing to avoid detonation, which reduces both power output per stroke and compression.  And in the limit, even assuming the lubricant doesn't experience temperature degradation, an overheating engine will experienced increased friction as moving parts expand and the metal begins to become sticky as it approaches its melting point.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Too cold can be bad&lt;/strong&gt;: When the engine is cold it may run at reduced efficiency only because the friction is higher until lubricants reach their minimum design operating temperature.&lt;/li&gt;&#xA;&lt;li&gt;Airflow turbulence and friction become more significant at high engine RPMs.  So holding all else equal I suspect more efficiency can be extracted from &lt;strong&gt;lower-RPM&lt;/strong&gt; piston engines.&lt;/li&gt;&#xA;&lt;li&gt;I would question whether the engine's orientation is a factor for fuel-injected gas engines &amp;ndash; especially for direct-injection engines.  If the lubricant system isn't designed to work upside down of course you have a problem, but otherwise I think the combustion cycle happens at high enough pressure that gravity isn't a factor.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that a lot of the factors that sap efficiency disappear if an engine can be tuned for and run at a fixed load in a stable atmosphere.  Requirements like responsive, variable power output in wide temperature ranges require a lot of efficiency tradeoffs.&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2015-07-03T22:05:19.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3467" PostTypeId="1" CreationDate="2015-07-03T23:12:26.023" Score="3" ViewCount="719" Body="&lt;p&gt;The discovery of &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon_nanotube&quot; rel=&quot;nofollow&quot;&gt;nanotubes&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Graphene&quot; rel=&quot;nofollow&quot;&gt;graphene&lt;/a&gt; has pushed the limit&#xA;of material strength high enough that building a space elevator&#xA;could be managable. But most of the new materials are created&#xA;from carbon or other non-metallic substances (silicon, boron).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to know: What is the strongest known &lt;em&gt;metallic&lt;/em&gt;&#xA;component and what advancements had been made in this regard ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the former question I was too imprecise because I erronously&#xA;assumed that it may be deduced what I mean with strongest metallic&#xA;material. So you get know my complete list of conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;After I perused the literature, I decided now that I want the&#xA;material with the highest &quot;tensile toughness&quot; meeting the other&#xA;conditions (in J m^-3). This solves elegantly the problem of&#xA;desired ductility because strong brittle and weak ductile&#xA;materials are ruled out.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;While I suppose the first condition will leave only metals,&#xA;in case it does not: Allowed are alloys and half-metals (B,Si&#xA;etc.) as long as the end product shows the characteristic metallic&#xA;luster if polished.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The material does exist and it can be provably synthesized on Earth.&#xA;Theoretical materials are not allowed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The material exists at standard NIST conditions: 293.15 K and&#xA;a pressure of 101 325 Pa. It exists in Standard Dry Air.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is also stable during a human lifetime, it does not dissolve, break&#xA;apart or lose more than 5% of its strength when stored in Standard&#xA;Dry Air.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given unlimited resources (money and processing) I expect that I&#xA;would get a 10 g ingot of the material within a week.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are the material parameters (Yield strength, compressive strength,&#xA;shear strength, tensile strength) in contrast to e.g. V2A steel ?&lt;/p&gt;&#xA;" OwnerUserId="1137" LastEditorUserId="1137" LastEditDate="2015-07-05T18:39:11.743" LastActivityDate="2019-04-09T18:26:04.330" Title="What is the strongest known *metallic* material which can be used on Earth?" Tags="|materials|metallurgy|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3468" PostTypeId="1" AcceptedAnswerId="3471" CreationDate="2015-07-04T05:40:35.293" Score="0" ViewCount="9279" Body="&lt;p&gt;I am trying to solve the problem:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What is the lifting force of in kN for a 10 m diameter spherical balloon with helium inside at 101 kPa and 320 K surrounded by air at 101 kPa and 298.15 K?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The given answer is 5.28 kN.&lt;/p&gt;&#xA;&lt;p&gt;I start from the Buoyancy Formula:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_{b} = (p_{air} - p_{gas})*g*V$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;And Ideal Gas Law:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$PV = mRT$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P/RT = m/V$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Density of air: &lt;span class=&quot;math-container&quot;&gt;$101/0.287*298.15=1.18$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Density of He: &lt;span class=&quot;math-container&quot;&gt;$101/0.287*320=1.10$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Volume of helium balloon: &lt;span class=&quot;math-container&quot;&gt;$(4\pi/3)*(10/2)^3=(500/3\pi)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Substitute:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_b = (1.18 - 1.10)*9.81*(500/3\pi)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I get 0.4101 N; am I missing something or is the given answer wrong?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="3505" LastEditDate="2023-05-17T19:18:08.297" LastActivityDate="2023-05-17T19:18:08.297" Title="What is the lifting force of a helium balloon in air?" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="3470" PostTypeId="1" CreationDate="2015-07-04T09:28:43.363" Score="0" ViewCount="234" Body="&lt;p&gt;I am trying to calculate the velocity in a pipe with length $L$ and diameter $D$, which is connected to bottom of a pressurized vessel. The vessel dimensions are known and the level of the liquid inside the vessel is known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to figure out the mass flow rate as a function of pressure inside the vessel.  We can ignore the friction losses in the pipe, and the vessel is filled with water.  Also, how can I find new pressure inside the vessel after the valve has been open for 1 second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G39g4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5qYwe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5qYwem.png&quot; alt=&quot;Pressurised water tank&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2202" LastEditorUserId="272" LastEditDate="2015-07-10T12:19:30.367" LastActivityDate="2015-07-10T12:19:30.367" Title="Homogeneous equilibrium model" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3471" PostTypeId="2" ParentId="3468" CreationDate="2015-07-04T10:49:32.077" Score="4" Body="&lt;p&gt;&lt;strong&gt;Firstly&lt;/strong&gt;, Gas constant of $He$ equals $2077 \ \mathrm{J/kg\cdot K}$ (engineering toolbox), so density of $He$ becomes $101/2.077*320 = 0.152\ \mathrm{kg/m^3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Secondly&lt;/strong&gt;, $Volume = 523.6 \ \mathrm{m^3}$ not as you calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thirdly&lt;/strong&gt;, you need to subtract balloon weight from the buoyancy force&#xA;$$F_b = \rho_{air}*g*V$$&#xA;$$F_b = 1.18 *9.81*523.6 = 6061 \ \mathrm{N}$$&#xA;$$F_{weight} = \rho_{He} Vg =  0.152*523.6*9.81 =  780.75 \ \mathrm{N}$$&#xA;$$F_{net} = F_b - F_{weight} = 6061 - 780.75 = 5280\ \mathrm{N} = 5.28\ \mathrm{kN}$$&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="368" LastEditDate="2015-07-08T23:35:54.917" LastActivityDate="2015-07-08T23:35:54.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3472" PostTypeId="1" CreationDate="2015-07-05T00:06:22.767" Score="-1" ViewCount="120" Body="&lt;p&gt;I work for a non-profit pre-engineering education company that provides an after school program to teach students the basic principles of engineering. We currently have a challenge involving shoe design that I have been asked to improve so that it better conveys engineering concepts and can be scorable at the competition day. Currently, the challenge provides essentially no engineering concepts.&#xA;Currently students are given a box of materials (like foam and string) and asked to make shoes. Proposals should be scorable with a rubric, preferably.&lt;/p&gt;&#xA;" OwnerUserId="2206" LastActivityDate="2016-10-04T01:42:08.730" Title="How can a Shoe Design Challenge better implement engineering concepts?" Tags="|education|" AnswerCount="2" CommentCount="2" ClosedDate="2015-07-07T00:02:30.903" ContentLicense="CC BY-SA 3.0" />
  <row Id="3473" PostTypeId="1" AcceptedAnswerId="3474" CreationDate="2015-07-05T10:09:05.673" Score="3" ViewCount="8607" Body="&lt;p&gt;I am solving the problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A gas bubble rising from the ocean floor is 1 inch in diameter at a depth of 50 feet. Given that specific gravity of seawater is 1.03, the buoyant force in lbs being exerted on the bubble at this instant is nearest to:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The given answer is 0.020 lbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start from the equation for buoyancy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_b=(P_{fluid}-P_{bubble})Vg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g=32.2\:\mathrm{ft/s^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V=(4/3)\pi(1\:\mathrm{in}/2)^3 * (1 \:\mathrm{ft}/12 \:\mathrm{in})^3=3.03*10^{-4}\:\mathrm{ft^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{fluid}=sp.gr(saltwater)*densityH_2O=1.03*62.4=64.272\:\mathrm{lb_m/ft^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P_gas:&#xA;Pressure, $P=62.4*50\:\mathrm{ft}=3120\:\mathrm{lb/ft^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$PV=mRT$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{bubble}=m/V=P/RT=3120/53.34*492R (STP)=0.11889\:\mathrm{lb_m/ft^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I substitute all of the values I only get 0.607 lbs. What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="368" LastEditDate="2015-07-08T23:15:35.387" LastActivityDate="2021-07-02T03:32:57.320" Title="Calculating the buoyant force of a rising bubble" Tags="|fluid-mechanics|thermodynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3474" PostTypeId="2" ParentId="3473" CreationDate="2015-07-05T10:52:00.903" Score="4" Body="&lt;p&gt;The buoyant force is equivalent to the weight of the fluid that the bubble displaces, not the difference of the fluid's and body's weight.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_b = \rho V g$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V = \frac{4}{3} \pi (1/2)^3 = \frac{1}{6} \pi\:\mathrm{in^3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V = 3,03 \cdot 10^{-4}\:\mathrm{ft^3}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\rho = 1.03 \cdot \rho_{water} = 64.272\:\mathrm{\frac{lbs}{ft^3}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;And now you simply multiply &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; and get &lt;span class=&quot;math-container&quot;&gt;$ \sim 0.02$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;s&gt;As a matter of fact the information about the depth is irrelevant as the force is independent of the depth. &lt;/s&gt;&#xA;As per comments: Generally speaking the depth is not relevant for the force as long as the object immersed is incompressible. In the case of a bubble, which is compressible, the volume will decrease with increasing depth. Hence the buoyant force will decrease. Also the object needs to be fully immersed for the first statement to be true.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2021-07-02T03:32:57.320" LastActivityDate="2021-07-02T03:32:57.320" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="3475" PostTypeId="1" AcceptedAnswerId="3490" CreationDate="2015-07-05T11:12:49.380" Score="2" ViewCount="6701" Body="&lt;p&gt;I am solving the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Air having an initial pressure of 6516 kPa and an initial volume of 0.113 m&lt;sup&gt;3&lt;/sup&gt; is compressed adiabatically to a final volume of 0.057 m&lt;sup&gt;3&lt;/sup&gt;. Calculate the work done by the gas as it compresses to a final pressure of 17237 kPa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The given answer is -175.9 kJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started from: $$PV^k = C$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work is therefore: $$W = \int\limits_1^2 P\,dV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(P_2V_2 - P_1V_1)} {(1-k)} =\frac{(17237*0.057)-(6516*0.113)} {(1-1.4)} = 615.025 \:\mathrm{kJ}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my relation correct? What am I missing and why is the answer negative?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="368" LastEditDate="2015-07-08T23:02:50.623" LastActivityDate="2016-12-31T19:38:02.030" Title="Why do I get negative work done for adiabatic compression of air?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3476" PostTypeId="2" ParentId="3472" CreationDate="2015-07-05T11:35:10.093" Score="3" Body="&lt;p&gt;Designing a shoe is as much of a engineering task as doing anything else. Just because there are all sorts of other designers nowadays does not mean that shoes are not engineering worthy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optimization makes nearly any task engineering worthy. Simple one dimensional goal such as weight, cost etc is easy to score. Engineering is not just about easy optimisations so some more complex is better. Some theories put engineering into the crossfire of contradictory parameters. So if you ask simultaneously cost, weight, waterproofness etc then it's even more engineeringly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another engineering thing is documentation, manufacturability and repeatability. After all there's not much point in engineering a solution if the solution will just be forgotten and cannot be improved upon, or worse cannot be built. So perhaps the students may build a prototype but get graded for a &quot;production&quot; version built by a third party based on the specification. (I have used this strategy and it works well)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally engineering is also about the process. So how they do things is as important as what you achieve. For this you could specify goals that have to be solved before you can go forward...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on..&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorDisplayName="user748" LastEditDate="2016-10-04T01:42:08.730" LastActivityDate="2016-10-04T01:42:08.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3477" PostTypeId="2" ParentId="3472" CreationDate="2015-07-05T13:35:41.000" Score="3" Body="&lt;p&gt;First, designing a shoe absolutely &lt;i&gt;is&lt;/i&gt; engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, there is no simple way to measure the &quot;goodness&quot; of a shoe.  If there were, there'd be only one model of shoe out there.  You therefore have to define a set of criteria to which the shoes will be measured.  This lets the kids know what to aim at, and you a way of scoring their results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could score based on a &quot;cost&quot; system, for example.  You already give them each the same set of materials to work with.  Put a cost on each item.  For example, each inch of string costs 1 unit, each square inch of foam 10 units, etc.  Then you add other costs that are really measures of how well different criteria are met.  For example, each gram of final mass might cost one unit, failure of a puncture test of the sole might cost 50 units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, overall I don't like the shoe concept since there are way too many criteria for goodness in ordinary life, and many of them difficult to measure, highly subjective, or emotional.  I would pick something different in the first place.  For example, give them a box of materials as you do now, but tell them they need to build a bridge that will span two tables.  You could score this based on cost of material, but also measure performance by what maximum load the bridge can sustain over a fixed defined span, how wide a span for a fixed defined load, etc.  The materials could be as simple as a few boxes of wooden toothpicks, a bottle of Elmer's glue, and 3 sheets of copier paper, for example.  Or simply specify they can use as many of whatever common brand of wooden toothpicks are available in your area, but the final weight will be part of the scoring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many many options.  You don't need to pick one that will make it difficult for the kids to know what to do and for you to measure the results.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-07-05T13:35:41.000" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3478" PostTypeId="1" AcceptedAnswerId="3479" CreationDate="2015-07-05T13:50:10.037" Score="12" ViewCount="14806" Body="&lt;p&gt;In fluid kinematics I can't understand the meaning of these terms : vorticity and circulation.&lt;br&gt;&#xA;Can somebody give me a description of these terms so that a lay person can understand them easily?&lt;/p&gt;&#xA;" OwnerUserId="21" LastEditorUserId="1293" LastEditDate="2016-07-23T07:46:11.400" LastActivityDate="2020-08-24T10:42:54.803" Title="What is the meaning of vorticity and circulation?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3479" PostTypeId="2" ParentId="3478" CreationDate="2015-07-05T16:23:15.590" Score="11" Body="&lt;p&gt;There are fundamental types of motion (or deformation) for a fluid element: translation, rotation, linear strain and shear strain. Usually all these types of motion occur concurrently which makes the analysis of fluid dynamics somehow difficult.&lt;/p&gt;&#xA;&lt;p&gt;One can express the rate of translation vector mathematically by the velocity vector &lt;span class=&quot;math-container&quot;&gt;$\vec{V}$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\vec{V} = u\vec{i} + v\vec{j} + w\vec{k}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;When it comes to expressing the rate of rotation of a fluid element it becomes quite challenging, Why? because a fluid element translates and deforms as it rotates, imagine an initially rectangular fluid element that starts to rotate while each line of the rectangule having a different angular velocity than the other. You can check White's book for the complete derivation but we can express the rotation vector &lt;span class=&quot;math-container&quot;&gt;$\vec{\omega}$&lt;/span&gt; for now as follows:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\vec{\omega} = \frac{1}{2} [(\frac{\partial w}{\partial y} - \frac{\partial v}{\partial z})\vec{i}&#xA;+ \frac{1}{2} (\frac{\partial u}{\partial z} - \frac{\partial w}{\partial x})\vec{j}&#xA;+ \frac{1}{2} (\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y})\vec{k}]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Putting it simply as half the curl of the velocity vector (just a mathematical manipulation):&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\vec{\omega} = \frac{1}{2}\vec{\nabla}\times  \vec{V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, let us define a vector which is called &lt;strong&gt;Vorticity vector&lt;/strong&gt; which is twice the angular velocity (we just got rid of the silly &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}$&lt;/span&gt;) and we might call it &lt;span class=&quot;math-container&quot;&gt;$\xi $&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\vec{\xi} = \vec{\nabla}\times \vec{V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Okay, enough with the math. What does it mean?&lt;/p&gt;&#xA;&lt;p&gt;For an arbitary point in a flow field:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any fluid element (particle) that occupy that point having a &lt;strong&gt;non-zero vorticity&lt;/strong&gt;, that point is called &lt;strong&gt;rotational&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Vice versa, Any fluid element (particle) that occupy that point having a &lt;strong&gt;zero vorticity&lt;/strong&gt;, that point is called &lt;strong&gt;irrotational&lt;/strong&gt; which means particle is not rotating.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/J0awM.png&quot; alt=&quot;enter image description here&quot; /&gt;&#xA;Flow from A to B is rotational (has voriticity) while flow from A to C is irrotational (has no vorticity).&lt;/p&gt;&#xA;&lt;p&gt;You can find many examples for rotational flows such as in wake regions behind blunt bodies and flow through turbomachines.&lt;/p&gt;&#xA;&lt;p&gt;According to this &lt;a href=&quot;http://www.ess.uci.edu/%7Eyu/class/ess228/lecture.4.vorticity.all.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;lecture&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;• Circulation and vorticity are the two primary measures of rotation&#xA;in a fluid.&lt;/p&gt;&#xA;&lt;p&gt;• Circulation, which is a scalar integral quantity, is a macroscopic&#xA;measure of rotation for a finite area of the fluid.&lt;/p&gt;&#xA;&lt;p&gt;• Vorticity, however, is a vector field that gives a microscopic&#xA;measure of the rotation at any point in the fluid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2020-08-24T10:42:54.803" LastActivityDate="2020-08-24T10:42:54.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3480" PostTypeId="2" ParentId="3467" CreationDate="2015-07-05T19:52:20.643" Score="4" Body="&lt;p&gt;I'm going to assume by &quot;tensile toughness&quot; that the questioner means &quot;integral of the space under a tensile-test curve.&quot; However, lacking straightforward access to that data for most materials, I will instead substitute the product of engineering strain at failure and engineering ultimate tensile strength. Two candidate materials, both steels:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Class     Grade      UTS (MPa)  STF (%)    Toughness (J m^-3)&#xA;---------- ---------- ---------- ---------- ------------------&#xA;Eglin      EG-5       ~1850       ~16.5     305&#xA;TWIP       *see below ~1000       ~62       620&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the Eglin steel, the properties are reported for low strain-rate. The alloys were invented for high-ground-penetration bomb noses. They tend to have slightly higher toughness at higher strain rates, but for comparison purposes we chose the lower strain rate here. See &lt;a href=&quot;https://data.epo.org/publication-server/pdf-document/EP04704052NWB1.pdf?PN=EP1594997%20EP%201594997&amp;amp;iDocId=7218467&amp;amp;iepatch=.pdf&quot; rel=&quot;nofollow&quot;&gt;this pdf&lt;/a&gt; for more information.  Eglin steels are designed to retain strength at higher temperatures, as demonstrated in the table in the pdf linked above. Their composition is 85-90% Fe, 5% Ni, 1.5-3.5% Cr, 1.25% Si, and small quantities (&amp;lt;1%) of C, Mn, Mo, W, V, and Cu. For these steels it is critical that there is minimal P, S, and N, and there must be as close to zero oxygen as possible. Even very small quantities of these embrittling elements can reduce properties considerably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the TWIP (&lt;strong&gt;TW&lt;/strong&gt;inning-&lt;strong&gt;I&lt;/strong&gt;nduced &lt;strong&gt;P&lt;/strong&gt;lasticity) steel, I had difficulty finding a uniform table of grades and properties, so I am using data from page 121 of &lt;a href=&quot;http://cdn.intechopen.com/pdfs-wm/13349.pdf&quot; rel=&quot;nofollow&quot;&gt;this pdf&lt;/a&gt;. The values given here are for lower strain rates, unlikely to be seen in their intended automotive energy-absorption use, but useful for comparison. The grade is not known to me, but the composition is in the article. TWIP steels have quite impressive mechanical properties, and have low material cost, requiring about 75-80% Fe, 15-25% Mn, ~ 0.5% C, and varying Al and Si. They are not particularly corrosion resistant and have fairly low heat tolerance. If the temperature gets close to or above the austenitization temperature (~730 °C), the retained-austenite microstructure will eventually be lost, removing the special properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary answer&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TWIP steels can have a UTS*STF of over 600 J m^-3, and&#xA;Eglin steels can have UTS over 1800 MPa with STF over 15%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from theoretical results of very recent ICME work worldwide (well over 2000 MPa, but no production process yet), these two metals are apex examples of high damage tolerance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that I haven't considered shape-forming processes. Some materials may not be formable into arbitrary shapes. There are castable Eglin steels, but TWIP requires a precise working process to form the microstructure, so it is not castable or even generally forge-able. It is not clear if Aermet, as noted in the comments, is castable, but it is age-hardened. Age-hardened materials have limited service temperatures, and in the case of Aermet that would seems to be somewhere below 500 °C.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="368" LastEditDate="2015-07-08T23:38:49.387" LastActivityDate="2015-07-08T23:38:49.387" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3482" PostTypeId="1" CreationDate="2015-07-05T23:37:30.473" Score="6" ViewCount="234" Body="&lt;p&gt;I am a photo technician working in photographic labs to repair printing machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to better understand how these machines operate.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Blue laser gun&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Laser gun type = 473 $nm$ blue dpss. &lt;/li&gt;&#xA;&lt;li&gt;On label = maximum output 50 $mW$, 473 $nm$, type 3b laser. &lt;/li&gt;&#xA;&lt;li&gt;Actual output power according to the manufacturer specification = 102 $\mu W$. &lt;/li&gt;&#xA;&lt;li&gt;Now when I read with the power meter = 40.6 $\mu W$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Green laser gun&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;laser gun type = 532 $nm$ blue dpss.&lt;/li&gt;&#xA;&lt;li&gt;on label = maximum output 100 $mW$, 532 $nm$, type 3b laser.&lt;/li&gt;&#xA;&lt;li&gt;actual output power according to the manufacturer specification = 706 $\mu W$.&lt;/li&gt;&#xA;&lt;li&gt;now when I read with the power meter = 279 $\mu W$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a 473 $nm$ dpss module outputs power at the exit window at 102 $\mu W$, what is the power of diode?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a 532 $nm$ dpss module outputs power at the exit window at 705 $\mu W$, what is the power of diode? You can say approximately, according to your experience, just a starting point to start replacing experiment of laser diodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this dpss laser modules the pump laser diode portion and the frequency doubling crystal portion are separated by a chamber. There is c-mount and to five 9mm laser diodes are used. I know how to replace / align a diode inside the gun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to replace the original diodes with the new one for the both modules due to week output. Suppliers discontinue this product and they are not supporting it any more. My question is what is the output power of the diode in side the laser guns? Approximately what is the diode power to start replacing diode?&lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="90" LastEditDate="2015-07-06T03:16:06.270" LastActivityDate="2015-07-07T15:49:46.107" Title="Laser diode input/output ratio" Tags="|lasers|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3483" PostTypeId="1" CreationDate="2015-07-05T23:51:22.213" Score="5" ViewCount="998" Body="&lt;p&gt;I have some laser diodes. I want to know how to measure the wavelength of laser diodes ranging from visible to near infrared? (300 nm to 1100 nm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type of instrument/meter do I need to use? spectrometer, spectrophotometer, wavelength meter, reflectometer, oscilloscope or what and how?    &lt;/p&gt;&#xA;" OwnerUserId="2215" LastEditorUserId="272" LastEditDate="2015-07-06T12:37:54.417" LastActivityDate="2015-07-06T12:37:54.417" Title="How to measure a laser diode's Wavelength" Tags="|lasers|metrology|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3484" PostTypeId="1" AcceptedAnswerId="3485" CreationDate="2015-07-06T00:26:33.297" Score="4" ViewCount="9407" Body="&lt;p&gt;I am trying to find the enthalphy of Helium if its internal energy is 200 kJ/kg. The answer is given as 333.42 kJ/kg, but I keep getting the wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin {align}&#xA;H &amp;amp;= U + PV \\&#xA;H &amp;amp;= U + mRT \\&#xA;\end {align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Universal Gas Constant, $R_u = 8.314 \:\mathrm{\dfrac{J}{K \cdot mol}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Individual Gas Constant, $R_i = 8.314 \:\mathrm{\dfrac{J}{K \cdot mol}} \div 4 \:\mathrm{\dfrac{g}{mol}} = 2.0785 \:\mathrm{\dfrac{kJ}{kg \cdot K}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin {align}&#xA;H &amp;amp;= 200 \:\mathrm{kJ/kg} + 1\:\mathrm{kg}*2.0785 \:\mathrm{kJ/kgK}*273.15 \:\mathrm{K} \\&#xA;H &amp;amp;= 767.74 \;...\\&#xA;\end {align}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is my answer different?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="368" LastEditDate="2015-07-06T23:53:44.240" LastActivityDate="2015-07-06T23:53:44.240" Title="Finding the enthalphy of Helium given internal energy" Tags="|thermodynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3485" PostTypeId="2" ParentId="3484" CreationDate="2015-07-06T01:09:00.343" Score="6" Body="&lt;p&gt;First, notice that the problem statement does not specify a temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The internal energy and enthalpy are given by:&#xA;$$ U = m c_vT $$&#xA;$$ H = m c_pT $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H = \frac{c_p}{c_v}U $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a monatomic gas, such as helium, the ratio $c_p/c_v$ is very nearly $5/3$.  It follows that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H = \frac{5}{3} \times 200 = 333.33$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very close to your expected answer.  If you were to use the precise values of $c_p$ and $c_v$ given by your textbook, I expect that you will get your textbook's answer of 333.42.&lt;/p&gt;&#xA;" OwnerUserId="2158" LastEditorUserId="2158" LastEditDate="2015-07-06T06:34:46.063" LastActivityDate="2015-07-06T06:34:46.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3486" PostTypeId="2" ParentId="3483" CreationDate="2015-07-06T01:28:31.247" Score="4" Body="&lt;p&gt;You want a wavelength meter if you have one or otherwise a spectrometer.  In more detail:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;wavelength meter&lt;/strong&gt; uses an interferometer to measure wavelength.  It can measure wavelengths very accurately.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;spectrometer&lt;/strong&gt; measures a spectrum, that is intensity &lt;em&gt;vs&lt;/em&gt; wavelength.  The wavelength at the peak intensity is what you are looking for. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;spectrophotometer&lt;/strong&gt; is used to measure the transmission spectrum of some material.  This is not what you want.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;reflectometer&lt;/strong&gt; measures reflectivity.  Again, not what you want.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Oscilloscopes&lt;/strong&gt; measure electrical voltage &lt;em&gt;vs&lt;/em&gt; time.    Again, not what you want.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2158" LastActivityDate="2015-07-06T01:28:31.247" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3487" PostTypeId="2" ParentId="3484" CreationDate="2015-07-06T06:28:55.500" Score="3" Body="&lt;p&gt;Following your method: You can't assume that temperature of Helium at that particular state is $273.15\ K$, but we can calculate it as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u = C_{v}\Delta T$$&#xA;$$\Delta T = \frac{u}{C_v} = \frac{200}{3} = 66.66667\ K$$&#xA;$$h = C_{p}\Delta T = 5*66.66667 = 333.3 \ kJ/kg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As John said, Values from your text book of $C_p$ and $C_v$ should yield a value closer to your expected answer.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2015-07-06T06:34:01.137" LastActivityDate="2015-07-06T06:34:01.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3488" PostTypeId="1" CreationDate="2015-07-06T08:29:38.027" Score="3" ViewCount="1431" Body="&lt;p&gt;I'm trying to design a single wheel trailer for my underpowered VW, much like &lt;a href=&quot;https://s-media-cache-ak0.pinimg.com/736x/bf/4d/b2/bf4db20abff1f2b07023362e9cd084f3.jpg&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike a conventional trailer, it is attached to the car at two points. To prevent the wheel from dragging, it has a &lt;a href=&quot;http://images.thesamba.com/vw/gallery/pix/765928.jpg&quot; rel=&quot;nofollow&quot;&gt;swivelling mechanism&lt;/a&gt;. Those are hard to come by, so I'm going to build one myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a mechanical engineer by trade, so the questions I have are probably quite basic:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;With caster wheel wobble in mind, should the wheel swivel freely, or is some friction desirable?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What kind of bearings should be used for the swivel point? There is definitely an axial load, but is the radial load considerable?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What would be the most effective geometry? How much trail/rake? Is this something I just have to experiment with?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2218" LastEditorUserId="282" LastEditDate="2015-12-07T10:51:20.297" LastActivityDate="2016-05-05T14:41:55.727" Title="How to design a swivel connection for a single wheel trailer?" Tags="|mechanical-engineering|design|geometry|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3490" PostTypeId="2" ParentId="3475" CreationDate="2015-07-06T11:15:07.213" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is my relation correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, it's correct. And here is the derivation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mechanical work is defined as:&#xA;$$W = \int_{V_i}^{V_f}P\,dV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiplying by $\frac{V^{k}}{V^{k}}$:&#xA;$$W = \int_{V_i}^{V_f}\frac{PV^{k}}{V^{k}}\,dV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $PV^{k}$ is constant, we can safely put it out the integration, yielding:&#xA;$$W = PV^k\int V^{-k}\,dV = PV^k\left[\frac{V^{-k+1}}{1-k}\right]^{V_f}_{V_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, putting $V_f = V_2$ and $V_i = V_1$:&#xA;$$W = \frac{(P_2V_2 - P_1V_1)} {(1-k)}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;why is the answer negative?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's just a sign convention, if work is done by a system work is positive, if work is done on a system work is negative (as in your case).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;what am I missing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I tried to solve the problem I got the same result as yours. Are you sure there is nothing missing in the problem description?&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="368" LastEditDate="2015-07-08T23:05:15.730" LastActivityDate="2015-07-08T23:05:15.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3492" PostTypeId="1" AcceptedAnswerId="3746" CreationDate="2015-07-06T15:40:46.590" Score="4" ViewCount="460" Body="&lt;p&gt;While deciding what tool (library) to use for a simulation of a hybrid gas turbine plant, I found that different libraries/tools of thermodynamic properties yield different values than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a simple test to calculate $c_p$ of air at $300\ K$(same goes for enthalpy by the way) only for simplification and comparison and here are the results.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Source              |    Cp at 300 K (J/kg.K)&#xA;--------------------------------------------&#xA;Thermopy  (python)  |   1004.89&#xA;Cantera   (python)  |   1010.06&#xA;CoolProps (python)  |   1006.627&#xA;EES Software        |   1005&#xA;Engineering Toolbox |   1005&#xA;Incropera's book    |   1007&#xA;Cengel's book       |   1007&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What do you think is the reason for the difference in values?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently the one closest to the properties sheets from Incropera and Cengel is &lt;code&gt;CoolProps&lt;/code&gt; but &lt;strong&gt;is this level of accuracy sufficient?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case might not be convincing for $c_p$, but it did occur to me once while calculating adiabatic flame temperature of some fuels and the error were in hundred Celsius range! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link : &lt;a href=&quot;https://gist.github.com/anonymous/3ec84b03c80928a5b5c0&quot; rel=&quot;nofollow&quot;&gt;used Python code&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2015-07-07T01:24:45.890" LastActivityDate="2015-07-30T18:53:44.893" Title="Different calculations for values of thermodynamic properties" Tags="|mechanical-engineering|thermodynamics|simulation|numerical-methods|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3494" PostTypeId="2" ParentId="3413" CreationDate="2015-07-06T22:51:25.503" Score="1" Body="&lt;p&gt;I'm going to assume from your use of the word &quot;slowly&quot; in the question that speed is not going to be an issue for you; therefore this is simply a question of torque. It sounds like the screen will be passing through the point which will generate the greatest amount of leverage, so lets calculate the torque at that point:&#xA;2.6 kg (~5.7 lbs) on the end of a 0.07 m (~2.75 inches) lever requires 1.79 Nm of torque, and through your maximum gear ratio of 2.8 this reduces to 0.64 Nm at the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we've not allowed for any friction or inefficiency in the system then we should ensure that the motor is capable of comfortably exceeding this figure. The third motor up in the range you specified (STP-MTR-17060(D)) may be sufficient at 0.81 Nm, but you may wish to go a little higher in the range to give some greater margin for friction. If you are using two motors then somewhere near the smallest from that range may be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have a candidate motor then you can use the torque speed curves to return to our speed assumption and calculate how quickly your screen will actually move (remembering to take into account the gear ratio) at the motors top speed. If this turns out to be unacceptably slow then you should immediately look at selecting a larger motor and perhaps adjusting the gear ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the top speed is close to what you would consider an acceptable average for the whole movement then you also need to factor in the effects of inertia. Extra torque above the holding torque is required to accelerate the inertias involved (your screen, gearing and motor) up to the top speed, so again you may wish to look at stepping up to a larger motor in the range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally you will want to consider the voltage and current requirements of the motors, as achieving the torque we've calculated is reliant upon the necessary current being available.&lt;/p&gt;&#xA;" OwnerUserId="157" LastActivityDate="2015-07-06T22:51:25.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3495" PostTypeId="1" CreationDate="2015-07-07T01:54:55.153" Score="3" ViewCount="1691" Body="&lt;p&gt;These days the city is repaving the main road close to my office. Looks like the rainwater drainage system is being upgraded and curbs and drains are added to manage the flow of rain water. Few casual google searches helped understand some of basic cross section for a new asphalt road. An image of the search is attached below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/H74EO.jpg&quot; alt=&quot;Cross Section&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a picture from the current repaving of the road, where an excavator is scoping out the dirt and trucked out. Then the dirt is replaced by different types of crushed stone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KUMaA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KUMaAm.jpg&quot; alt=&quot;Road Picture 1&quot;&gt;&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CAzeY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CAzeYm.jpg&quot; alt=&quot;Road Picture 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt; What is the engineering explanation for all this additional rework? To begin with was the road sinking? To the casual user the road appeared to be perfectly in good shape, thus a waste of tax payer money. How can the additional cost be justified?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://shropshirehistory.com/roads/romanroads.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roman Roads&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.dot.state.mn.us/bridge/pdf/hydraulics/drainagemanual/chapter%208.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Storm drainage systems&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.coventrytownship.com/includes/roadside_ditch.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roadside ditches? - Coventry Township&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="272" LastEditDate="2015-07-10T12:22:07.780" LastActivityDate="2015-08-12T03:19:44.727" Title="Are crushed cement stones and additional digging necessary when repaving a road?" Tags="|civil-engineering|roadway|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3496" PostTypeId="2" ParentId="3488" CreationDate="2015-07-07T02:28:41.197" Score="0" Body="&lt;p&gt;I have a trailer just like this one for my wife who has trouble backing an ordinary trailer.&#xA;To prevent wheel wobble at speed , I use a motor bike steering damper.&#xA;The trailer is attached to the conventional tow bar with a spreader attachment to provide 2 fixing points and can pivot up and down.&lt;/p&gt;&#xA;" OwnerUserId="2222" LastActivityDate="2015-07-07T02:28:41.197" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3497" PostTypeId="2" ParentId="3495" CreationDate="2015-07-07T03:13:08.737" Score="4" Body="&lt;p&gt;During my student days, ages ago, during one of my civil engineering lectures the lecturer told everyone that roads were designed and built for a 20 year life. After which the road could be dug up and new foundation materials laid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a prolonged period, and particularly with roads that are heavily utilized, foundation materials can move and need to be replaced and compacted anew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The replacement of the foundation materials in your example may be an opportunistic upgrade brought about by upgrading the drainage system. Instead just upgrading the drainage now and ripping up the road &quot;a few years later&quot; do everything now. It also means the road and the new drainage system will blend in with each other and ensure the drainage system functions as designed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if work for the upgraded drainage system involves deep digging it may damage the edge of the foundation of the road which would require the road foundation to be relaid.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-07-07T03:13:08.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3498" PostTypeId="2" ParentId="3378" CreationDate="2015-07-07T06:35:26.803" Score="0" Body="&lt;p&gt;I've found one build myself (sadly lost the link): Reactor column with bed material, sludge flows through the column top down, the reactor is kept at a suitable underpressure. I don't know if and how the bed handles clogging by sludge.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="368" LastEditDate="2015-07-07T16:47:03.880" LastActivityDate="2015-07-07T16:47:03.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3499" PostTypeId="1" CreationDate="2015-07-07T10:09:11.907" Score="3" ViewCount="64" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sequencing_batch_reactor&quot; rel=&quot;nofollow&quot;&gt;Sequencing batch reactors&lt;/a&gt; let sludge settle before drawing off the supernatant. For this, decanter arms are often used. An &lt;a href=&quot;http://www.fluidynecorp.com/webres/File/FloatingDecanter.pdf&quot; rel=&quot;nofollow&quot;&gt;example is here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge is to draw off the supernatant without creating turbulence, which would disturb the sedimented sludge and lead to sludge in the supernatant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm willing to compromise on the amount of supernatant I draw off, say only 20% of fill level. Can I use a simple standing pipe for this, or will I get excessive sludge?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="368" LastEditDate="2015-07-07T16:35:43.850" LastActivityDate="2015-07-07T16:35:43.850" Title="SBR-like reactors without decanters?" Tags="|fluid-mechanics|process-engineering|environmental-engineering|waste-water-treatment|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3500" PostTypeId="1" CreationDate="2015-07-07T10:13:01.703" Score="4" ViewCount="138" Body="&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;A steel cone is heated by a nichrome ribbon to 45–50 °C. I placed a Type K thermocouple over the surface of the metal to measure the heat transferred over the metal surface. The thermocouple reads about 45 to 50 °C, which is expected. The issue is when I put a plastic cap (enclosure) over the metal cone, the temperature drastically decreases.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Procedure&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There is a metal cone made of mild steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nichrome ribbon wire of 0.09 mm thickness is glued as a vertical strip from the tip to bottom on the metal surface of cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the opposite side of the metal surface a beaded thermocouple wire of 40 awg (0.08 mm) is glued from the tip to bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of them are placed on the metal using a superglue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A plastic cap that has a similar cone shape is put to the metal. The air gap between the metal and the plastic structure is 0.1 mm. In this gap the Nichrome wire and a Type K thermocouple are placed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;plastic cap -&gt;Thermocouple(or Nichrome wire)-&gt;glue(Superglue or Loctite)-&gt;Metal surface.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Observations&lt;/h1&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Without the plastic cap, the thermocouple measures about 45 °C after 4 to 5 minutes of heating.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When the plastic cap is added, the temperature increases up to 40 °C and struggles to reach 45 °C. The approximate time it takes to reach 45 °C is 10 to 12 minutes. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h1&gt;Questions&lt;/h1&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How is it possible to ensure that the thermocouple is measuring the temperature of the metal and not of the plastic?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We do not want the temperature measurement to change when the plastic cap is inserted. Is there any quick way of heating the plastic side and also getting the right temperature measurement?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am open to any suggestions to improve this topology.&lt;/p&gt;&#xA;" OwnerUserId="2230" LastEditorUserId="368" LastEditDate="2015-07-08T22:24:21.603" LastActivityDate="2015-07-08T22:54:35.740" Title="Should the temperature of a heated metal cone decrease in a plastic enclosure?" Tags="|thermodynamics|heat-transfer|temperature|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3502" PostTypeId="1" CreationDate="2015-07-07T14:27:23.230" Score="6" ViewCount="1850" Body="&lt;p&gt;Let's assume we have a pump and it can draw as much power as it needs. As the pump speeds up, it creates vacuum in the intake. At some point it draws the gas/liquid so fast, the intake pressure is so close to 0. At that point, no matter how much power we put into the pump, the flow rate doesn't increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured out it strongly depends on the intake size and shape, the type and design on of the pump etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On what other factors does it depend? Is there a way to calculate or at least estimate the maximum? Is it even a valid question?&lt;/p&gt;&#xA;" OwnerUserId="2231" LastEditorUserId="368" LastEditDate="2015-07-08T22:40:54.440" LastActivityDate="2016-01-12T23:25:37.253" Title="How do I calculate the max flow rate for a pump with no power restrictions?" Tags="|mechanical-engineering|fluid-mechanics|pressure|pumps|airflow|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3503" PostTypeId="2" ParentId="3502" CreationDate="2015-07-07T15:21:24.927" Score="3" Body="&lt;h3&gt;What you are describing is known as &amp;quot;choked flow&amp;quot;.&lt;/h3&gt;&#xA;&lt;p&gt;I am only familiar with choked flow in a gas, however a brief wikipedia snoop shows that liquids are limited by a different phenomenon, which makes physical sense if you understand the mechanism of choked flow in a gas.&lt;/p&gt;&#xA;&lt;p&gt;In its simplest form, it depends only on minimum cross sectional area and fluid properties.&lt;/p&gt;&#xA;" OwnerDisplayName="user815" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-07-07T15:21:24.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3504" PostTypeId="2" ParentId="3502" CreationDate="2015-07-07T15:36:18.417" Score="7" Body="&lt;p&gt;This phenomenon is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Cavitation&quot;&gt;cavitation&lt;/a&gt; for pumping liquids - i.e. for anything that can change phase.  If you are pumping gas, then the scenario really does come from choked flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you've got a compressor and you're running it as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Vacuum_pump&quot;&gt;vacuum pump&lt;/a&gt; to evacuate a tank - at some point the tank's pressure on the inside is so low, you can't pump any more air out.  This shows your question is a valid question.  It's a slightly different scenario, but it's easier to analyze using a real compressor at it's maximum suction pressure then your mythical high power compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Using a piston-style compressor analogy&lt;/strong&gt;, the maximum that the pump could remove would be based upon the max lift of the valve.  If you can't compress the gas to the point that the outlet valve will open to prevent air from coming back into the tank, then that's it - that's your maximum suction.  No matter how much power you put on it, you can't have more suction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other compressor designs work on the different principles, but the analogy holds - rotary turbines ultimately start with a gas at a pressure, then compress it with stators, rotate it again, compress it, etc. in cycles.  However, at the back end of the compressor is air (or in the case of a jet engine, combusted fuel) pushing back.  So long as the compression from that last stage of the turbine can push back on the air coming in (and all the way down the line), the compressor will work.  Beyond that point it will not.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-07-07T16:38:51.287" LastActivityDate="2015-07-07T16:38:51.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3505" PostTypeId="2" ParentId="3482" CreationDate="2015-07-07T15:49:46.107" Score="3" Body="&lt;p&gt;Conversion from the diode wavelength to the fundamental wavelength is generally a highly efficient process.  Conversion efficiencies in the high 90 percent range isn't unheard of for 1064 nm generation, which is the fundamental wavelength of the green laser.  Conversion to 946 nm, the fundamental wavelength of the blue laser, can also be highly efficient but is generally less so than for 1064 nm.  To be safe, I would assume an efficiency of 70% for 946 nm and 80% for 1064 nm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second harmonic generation process can also be highly efficient when used with pulsed lasers, but is significantly less so with CW lasers.  This is especially true for the low powers with which you are working.  For your lasers, I would guess the efficiency is something like 5%.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Guestimate Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Blue Laser: $70\% * 5\% = 3.5\%$&lt;/li&gt;&#xA;&lt;li&gt;Green Laser: $80\% * 5\% = 4.0\%$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-07-07T15:49:46.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3506" PostTypeId="1" AcceptedAnswerId="3507" CreationDate="2015-07-07T15:56:36.713" Score="2" ViewCount="141" Body="&lt;p&gt;My Lego model has a winch on the front. It's a simple spindle with a reel on it with a rubberised cog for winding at one end and two cogs at the other (as shown below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a technical name for the cog/gear on the blue spindle that stops the cog on the grey spindle spinning freely? Is this a common mechanism in real machinery?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, assuming this relies on friction and that the effectiveness will be reduced with wear and tear, how would that be mitigated in real equipment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VE37S.jpg&quot; alt=&quot;The gears/cogs in question&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2233" LastActivityDate="2015-07-07T17:35:18.870" Title="Gear that prevents another gear spinning freely" Tags="|gears|pulleys|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3507" PostTypeId="2" ParentId="3506" CreationDate="2015-07-07T16:49:53.763" Score="2" Body="&lt;p&gt;I would say that you have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Ratchet_(device)&quot; rel=&quot;nofollow&quot;&gt;ratchet&lt;/a&gt;.  Ratchets do fail.  The &lt;a href=&quot;https://en.wikipedia.org/wiki/Brownian_ratchet&quot; rel=&quot;nofollow&quot;&gt;Brownian Ratchet&lt;/a&gt; is a simple thought experiment done by thermodynamic physicists that shows that ratchets need to fail - otherwise you could extract energy from absolutely nothing - a contradiction of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Laws_of_thermodynamics&quot; rel=&quot;nofollow&quot;&gt;Laws of Thermodynamics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wear and tear on ratchets are mitigated via lubrication of parts of course, but bigger systems rely on things like &lt;a href=&quot;https://en.wikipedia.org/wiki/Governor_(device)&quot; rel=&quot;nofollow&quot;&gt;governers&lt;/a&gt; for real mechanical control instead of a ratchet.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-07-07T17:35:18.870" LastActivityDate="2015-07-07T17:35:18.870" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3508" PostTypeId="2" ParentId="3500" CreationDate="2015-07-07T21:07:35.593" Score="3" Body="&lt;p&gt;For plastic &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermoforming&quot; rel=&quot;nofollow&quot;&gt;thermoforming&lt;/a&gt;, or heating behavior, this kind of behavior would be expected.  Quite simply, when you form the plastic you typically heat up the plastic rather than the metal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you were working with &lt;a href=&quot;https://en.wikipedia.org/wiki/Low-density_polyethylene&quot; rel=&quot;nofollow&quot;&gt;LDPE&lt;/a&gt;, which is the only plastic I can think of that even changes shape at those temperatures.  Each material has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_capacity&quot; rel=&quot;nofollow&quot;&gt;heat capacity&lt;/a&gt; - 452 J/kg K for steel, and the LDPE has 1800 J/kg K - 3400 J/kg K.  For this answer, I'm using 2600 J/kg K.  The heat capacity is how much energy it would take to increase 1 kg of the material by 1 K (or 1 degree Celsius), assuming the system is completely insulated.  As you can see, the LDPE is larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Using a fluid mechanics analogy,&lt;/strong&gt; your wire is a hose, supplying energy to whatever you need it to warm up.  It will supply your system with so many Joules of heat per second - that's the Watts of electricity used.  Each of your materials is like a bucket that needs to be filled up.  But for some materials, it takes a lot more energy for them to register a single degree change of temperature - think of it as a large diameter bucket v. a narrow diameter bucket - to fill the larger diameter bucket to the same height requires a lot more water.  Same concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the steel (say a 1 kg cone) you need 9.04 kJ of thermal energy to raise it from 25 °C to 45 °C.  With a 1 kW wire, that's 9.04 seconds - no time at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding the mass of say 2 kg plastic onto the 1 kg cone, it now takes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2\:\mathrm{kg} * 2600\:\mathrm{\frac{J}{kg\cdot K}} + 1\:\mathrm{kg}* 452\:\mathrm{\frac{J}{kg\cdot K}})* (45\:\mathrm{°C} - 25\:\mathrm{°C}) = 113.04 \:\mathrm{kJ}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is 12 times the energy to heat it to the same temperature.  As a result, it takes 12 times as long - plus, while your original cone heated so fast you didn't have to worry about losses, the plastic now will take longer when heat transfer effects are taken into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, always heat the plastic, and form it around the metal - not the other way around.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="368" LastEditDate="2015-07-08T22:54:35.740" LastActivityDate="2015-07-08T22:54:35.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3509" PostTypeId="2" ParentId="3356" CreationDate="2015-07-07T23:01:58.457" Score="6" Body="&lt;p&gt;The goal of using protective coating/paint is to reduce corrosion of the metal framework. The degree of corrosion varies from the top to sea level and depends on several parameters such as salinity of rain/ocean, humidity, avg. temperature, the acidity of rain, so on so forth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle behind protective coatings may differ in terms of layers, but often, a primer containing a less noble metal is applied directly onto the metal whereto a top and even a middle layer is added, which contributes with resistance to UV light, moisture, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit 1 - additional info &lt;/em&gt;&lt;p&gt;&lt;/p&gt;&#xA;If this was insufficient, here is a detailed explanation of the principles of coatings. Here is what I known from courses and through my work at Hempel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The composition of paint is build from four components:&#xA; 1. a binder, &#xA; 2. a solvent, &#xA; 3. pigments and &#xA; 4. additives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The binder&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The binder functions as an adhesive and is non-volatile. The transition (of the paint) from liquid to a solid is determined by the binder as well as other parameters such as chemical, weather and corrosive resistivity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The solvent&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The solvent contributes with low visocity so that the coating me be applied more easily. Solvents are volatile compounds, that during the curing, evaporates leaving the 'film' behind. The film consist of intertwined molecules forming a polymer. While common, solvents are not a necessity, as the binder may contribute with a proper viscosity. Notice that water can be used as a solvent (waterborne paints), however a compound must be present in order to mingle the dispersed particles. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Pigments&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pigments are added to achieve a desired color. An other function of filler material is to reduce production cost. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Additives&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Additives contributes with product-enhancing attributes such as&#xA;extending shelf life, reduce curing time, improve moisturization of&#xA;pigments, etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Coatings are classified into two groups on basis of curing: physical drying or chemical curing. In brief, paints containing evaporating solvents are classified as physical drying, in contrary to paints that requires oxidative processes, as these are classified as chemically curing. This happens as gas species such as $\text{O}_2$, $\text{H}_2\text{O}$ and $\text{CO}_2$ causes oxidation which starts the polymerization process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;h3&gt;Corrosive protecting coatings&lt;/h3&gt;To briefly describe why corrosion happens, one must look at the chemistry of oxidation. It is energetically more favorable for a metal to be oxidized. I. e. in the presence of water or air, the following reaction will eventually occur.&#xA;\begin{equation}&#xA;\text{Fe} + \left[\text{O}\right] \rightarrow \text{Fe}^{++}+2\text{e}^-&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are generally 3 ways used to prevent the above reaction.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Formation of physical barriers that retards the diffusion of water and $\text{O}_2$. &lt;strong&gt;Barrier effect.&lt;/strong&gt;&#xA;Different compounds are used to achieve this effect: epoxy/bitumen/tar produces a barrier with low permeability for the water. Large pigments can produce an effect similar to, where the pigments are aligned parallel to the surface, resulting in a longer path the water must travel before reaching the metal surface.&lt;/li&gt;&#xA;&lt;li&gt;Ensure contact between a less noble metal and the metal-surface thus achieving a galvanic effect. &lt;strong&gt;Galvanic effect.&lt;/strong&gt;&#xA;To protect iron, metallic zinc is often used in abundance. This however requires the zinc throughout is in contact with it self and the surface. Zincprimers are not used in submerged surfaces, as the zinc salts are easily soluble. &lt;/li&gt;&#xA;&lt;li&gt;or ensure the water reacts with compounds inside the paint forming a inhibiting surface-layer. &lt;strong&gt;Inhibitor effect&lt;/strong&gt;. Pigments in used to achieve the inhibitor effect are often basic or water soluble, such may be various zinc derivates.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2238" LastEditorUserId="2238" LastEditDate="2015-07-12T17:59:28.430" LastActivityDate="2015-07-12T17:59:28.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3511" PostTypeId="2" ParentId="2032" CreationDate="2015-07-08T01:20:30.580" Score="1" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;My hypothesis&lt;/em&gt;&lt;/strong&gt; is that the glass had a plateau instead of the granite because glass is reflective to infrared lighting instead of granite - therefore shielding the &lt;em&gt;mostly&lt;/em&gt; radiative heat transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumptions:&lt;/strong&gt; I found a 5L box online with 340mm x 200mm x 125mm dimensions - which, with the insulated bottom, leads to a 0.203 square meter surface area for the box.  Based on some calculations, and using the &lt;a href=&quot;http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html&quot; rel=&quot;nofollow&quot;&gt;emisstivity&lt;/a&gt;s given here, is that during the &quot;heating cycle&quot;, over the course of the 1600 seconds of plateau, the glass would have losing heat due to radiation at a rate of 22W - &lt;a href=&quot;http://www.wolframalpha.com/input/?i=22+W+*+1600+s+%2F%28%28+heat+capacity+of+glass%29*7.182kg%29&quot; rel=&quot;nofollow&quot;&gt;Wolfram&lt;/a&gt; tells me that should have been about a 6.53K change, but the box didn't undergo that change.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the experiment was watching for a 15K change total, this is a &lt;strong&gt;significant&lt;/strong&gt; portion of the heat transfer.  Therefore the fan really is only doing a small fraction of the thermal work and radiation is picking up a significant portion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;a href=&quot;https://en.wikipedia.org/wiki/Infrared&quot; rel=&quot;nofollow&quot;&gt;infrared spectrum&lt;/a&gt;, where most of this heat would be lost, glass and granite seem to behave very differently.  &lt;a href=&quot;https://www.google.com/search?q=granite+infrared&amp;amp;espv=2&amp;amp;biw=1366&amp;amp;bih=643&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ei=LHecVdP-CMyu-AGnh4HgBA&amp;amp;ved=0CAcQ_AUoAg#tbm=isch&amp;amp;q=granite+infrared+camera+heating&amp;amp;imgrc=TmcHrGs4nTW-xM%3A&quot; rel=&quot;nofollow&quot;&gt;Granite&lt;/a&gt; seems somewhat transparent in the linked image. This is based on the fact that the edges in the image are blurry - if it were opaque the edges of the piping would be crisp at the hot spots (such as in the linked glass video) - but I'm &lt;em&gt;not&lt;/em&gt; an expert in radiation properties of materials.  &lt;a href=&quot;https://www.youtube.com/watch?v=wsjjdwLFNIM&quot; rel=&quot;nofollow&quot;&gt;Glass&lt;/a&gt; not only blocks the infrared radiation in the video, but according to the video seems to reflect the radiation.  It makes sense, that's how greenhouses work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would imply that since the sensor is directly in the middle of the box of material the layers of glass continuously reflected back any heat transfer (envision a steak with layers of well done and rare) - stalling out the process.  The granite did not have this effect and therefore proceeded to radiate in approximately uniform fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without further experiments it is hard to come to a definitive conclusion.  Further experiments removing radiation effects would prove out the hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-07-08T14:43:59.747" LastActivityDate="2015-07-08T14:43:59.747" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3512" PostTypeId="1" CreationDate="2015-07-08T03:08:04.087" Score="5" ViewCount="135" Body="&lt;p&gt;I am interested in setting up a controller which can modulate the output of 2 hotpads based on input from an infrared thermometer. The hotpads have their own dials, but I am considering controlling them by voltage instead. So my question is, what is the simplest method for modulating voltage (or rotating dials) based on temperature? One problem in particular that I am unsure about is how to account for the gradual propagation of heat through my system, since I am heating a large (2' diameter, 1&quot; thick) rotating aluminum disk. I have a laptop at my disposal, and minimal funding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would prefer to avoid wiring my own sensors, so any recommendations you have on how to achieve this would be great. I could handle a breadboard at best with my experience, and I have only an introductory level understanding of electromagnetism... But I can code in C, Java, and Matlab, and I have experience assembling PCs, so I am hoping this is within my abilities.&lt;/p&gt;&#xA;" OwnerUserId="2240" LastEditorUserId="368" LastEditDate="2015-07-08T22:31:14.587" LastActivityDate="2015-07-09T17:02:16.160" Title="Help with closed loop heat controller (for a non engineer)" Tags="|mechanical-engineering|control-engineering|sensors|circuits|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3513" PostTypeId="2" ParentId="3512" CreationDate="2015-07-08T10:54:07.817" Score="6" Body="&lt;p&gt;Since you seem to not be too familiar with control systems, electronics, and wiring things, I recommend a simple threshold detector that controls a relay.  The relay then switches the heating element on or off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless the IR thermometer has a output signal for this purpose, it won't be useful as the sensing element.  However, there are various devices that produce a electronic signal as a function of temperature.  For your temperature range, thermocouples or platinum RTDs is what you should look at.  Thermocouples need more complicated circuitry to produce a useable signal, whereas RTDs are much simpler to use.  Thermocouples go to higher temperatures though.  I'd go with a RTD unless the temperature is too high.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-07-08T10:54:07.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3514" PostTypeId="2" ParentId="3512" CreationDate="2015-07-08T11:59:18.313" Score="1" Body="&lt;p&gt;The feedback topology for thermal systems is fairly easy when the heat source is continuously adjustable, which I believe yours is.  This is because the transfer function of simple thermal systems has a single pole, just like a standard &lt;a href=&quot;https://en.wikipedia.org/wiki/RC_circuit#Transfer_functions&quot; rel=&quot;nofollow&quot;&gt;RC circuit&lt;/a&gt;.  In this case, a stable closed loop controller can be made by simply connecting the sensor directly to the actuator (assuming both are linear) with a variable gain stage in between.  After doing so, the gain can be increased until just below where the loop starts to oscillate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For long-term tracking you should also consider adding a switchable &lt;a href=&quot;http://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/differentiator-integrator-circuits/&quot; rel=&quot;nofollow&quot;&gt;integrator&lt;/a&gt; before the gain stage in your feedback electronics.  It is important that it be before the gain stage so that the DC portion of the feedback signal doesn't get hit with infinite gain when the integrator is switched on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this can be implemented in Matlab if you have the right packages and are capable of getting the signals into and out-of the computer.  It is probably easier to build the rather simple circuitry yourself though.  Alternatively, you could spend the money on a commercial &lt;a href=&quot;https://en.wikipedia.org/wiki/PID_controller&quot; rel=&quot;nofollow&quot;&gt;PID controller&lt;/a&gt; and set the D gain to zero.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="368" LastEditDate="2015-07-08T22:35:50.823" LastActivityDate="2015-07-08T22:35:50.823" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3517" PostTypeId="1" AcceptedAnswerId="3523" CreationDate="2015-07-08T13:31:04.777" Score="5" ViewCount="145" Body="&lt;p&gt;Does anyone have a feel for the flatness of Portland Cement when poured in a standard/typical way? Say, for example, that I wanted to pour a cement &quot;cube&quot; (20&quot;x 18&quot;x 19.25&quot;) in a mold. How flat would the top surface be without any special procedures? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background:&#xA;I am going to mount a forceplate on top of this cement block and I am trying to determine if I need to specify the flatness tolerance in the drawing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there somewhere I can look for this information?&lt;/p&gt;&#xA;" OwnerUserId="1174" LastEditorUserId="272" LastEditDate="2015-07-08T14:54:41.623" LastActivityDate="2015-07-10T02:28:45.430" Title="Typical flatness of Portland cement" Tags="|concrete|tolerance|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3520" PostTypeId="2" ParentId="3517" CreationDate="2015-07-08T14:47:16.943" Score="2" Body="&lt;p&gt;Understanding &quot;without any special procedures&quot; as meaning simply dropping the cement in a pile, it would probably end up with an inclination approximately equal to its &lt;a href=&quot;https://en.wikipedia.org/wiki/Angle_of_repose&quot; rel=&quot;nofollow&quot;&gt;angle of repose&lt;/a&gt;. Unfortunately, this value is highly variable, and a quick search has resulted in many different values:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;15 or 20 degrees if fine or coarse &lt;a href=&quot;http://www.eng-tips.com/viewthread.cfm?qid=151371&quot; rel=&quot;nofollow&quot;&gt;(1)&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;40-44 degrees &lt;a href=&quot;http://www.eng-tips.com/viewthread.cfm?qid=151371&quot; rel=&quot;nofollow&quot;&gt;(1)&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;39 degrees &lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20061110035600AADp8M7&quot; rel=&quot;nofollow&quot;&gt;(2)&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Can't trivially be defined &lt;a href=&quot;http://forum.bulk-online.com/showthread.php?1451-Inclination-Angle-for-Cement-Conveying&amp;amp;p=5320#post5320&quot; rel=&quot;nofollow&quot;&gt;(3)&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-07-08T14:47:16.943" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3521" PostTypeId="2" ParentId="3512" CreationDate="2015-07-08T16:59:20.757" Score="3" Body="&lt;p&gt;In my experience getting data from a rotating component is tricky without resorting to some kind of radio transmitter or a passing the data through a  slip ring, which would probably require amplification of the data signal due to the noise generated by the slip ring... it can be done though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be cautious about using an IR thermometer for anything other than a reference measurement, i.e., about how hot is that thing. There are too many factors affecting the measurement, mainly the emissivity of the surface. Maybe others have better experiences, but I've never had much success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen reasonably good results that might fit your needs and budget using a magnet and a Hall effect sensor. Place the magnet on the wheel and since the strength of the magnet will decline as the temperature increases, so will the voltage from the HE sensor. You'll only get the measurement once/rev though so you'll have to look at the peak values and calibrate from there. Another benefit is that you'll also have a tachometer so you can measure the rotational speed if you need to. This shouldn't cost more than a few $$, but might be more complicated than what you're looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to understanding how heat the disk will heat up, you could do some modeling using the software of your choice...I like Ansys, but if you don't have access to something like that, it's probably not in your budget. There are other, less costly solutions. If you have enough symmetry, you could probably even do it in Excel or a similar program. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also take a number of temperature measurements over the surface of the disk to see how it heats up and cools down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, you want to have a good idea of how hot the disk is an a given location when you only know how hot it is in one location.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2015-07-09T17:02:16.160" LastActivityDate="2015-07-09T17:02:16.160" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3522" PostTypeId="1" CreationDate="2015-07-08T18:51:25.153" Score="8" ViewCount="212" Body="&lt;p&gt;I was trying to understand the convolution better and its mathematical properties and interpretations in engineering (specially in the context of computer vision). Recall the convolution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ s(t) = (x * w)(t) = \int x(a) w(t-a) da$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the first argument (to the convolution) $x$ is usually called the input but the second argument $w$ (to the convolution) is usually called the &quot;&lt;strong&gt;kernel&lt;/strong&gt;&quot;. However, in computer vision and convolutional neural networks, the second argument is usually called a &quot;&lt;strong&gt;template&lt;/strong&gt;&quot; (maybe the image of an edge or a wheel, or some part of an object). However, in other areas, I think is signals and systems, its usually called a &quot;&lt;strong&gt;filter&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a computer software engineer, I believe that naming is extremely important because it gives us the power to think about specific concepts. Having bad names can lead to sloppy thinking. Therefore, I was assuming that these technical names probably were chosen with these ideas in mind. Does someone know or understand why these names have been used to the second argument of the convolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific names that I am aware of are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Kernel (from pure maths?)&lt;/li&gt;&#xA;&lt;li&gt;Filter (signals and systems?)&lt;/li&gt;&#xA;&lt;li&gt;Template (Computer vision/ Machine Learning)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Not sure if I am missing any, but I'd like to understand these naming better and possibly (hopefully) understand intuitively better what the convolution operator does and its interpretation in engineering and mathematics.&lt;/p&gt;&#xA;" OwnerUserId="2244" LastEditorUserId="2244" LastEditDate="2015-07-09T02:55:02.223" LastActivityDate="2015-07-09T12:19:58.227" Title="What are the meanings of the second argument to the convolution?" Tags="|electrical-engineering|signal|mathematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3523" PostTypeId="2" ParentId="3517" CreationDate="2015-07-08T19:35:04.083" Score="3" Body="&lt;p&gt;If you live in a country that uses ACI standards, you might consider specifying flatness using an F&lt;sub&gt;L&lt;/sub&gt; or F&lt;sub&gt;F&lt;/sub&gt; number.  ACI 117, Commentary Table R4.8.4 has this information, with flatness ranging from &quot;Conventional&quot; to &quot;Super Flat.&quot;  For the dimensions you have listed, specifying anything beyond &quot;Flat&quot; is probably overkill since your forceplate will likely cause the underlaying slightly-unflat cement to consolidate to the face of the plate anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what I said above may be overkill for what you're doing.  I'd just tell them to overfill and then strake it off using the top of the form as a guide, just like what you'd do scooping flour for a cookie recipe.  That should get you what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to clarify - if the cement is a dry mix, my comment about the consolidation of the cement is valid.  However, if this is a cement mix with water that is allowed to cure, the consolidation comment is not strictly applicable.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorDisplayName="user16" LastEditDate="2015-07-10T02:28:45.430" LastActivityDate="2015-07-10T02:28:45.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3524" PostTypeId="1" AcceptedAnswerId="3528" CreationDate="2015-07-08T20:47:36.880" Score="-3" ViewCount="113" Body="&lt;p&gt;I've been shopping around for gears, pulleys and other transmission components for building a milling machine over the past few weeks. Many of these parts are so rudimentary and ubiquitous, yet the prices are very high. A ribbed pulley, for instance, about 3 inches in diameter often sells for over $30 online. Many of these parts are either stamped or cast, making them cheap to manufacture. What factors into the cost of these components? &lt;/p&gt;&#xA;" OwnerUserId="2248" LastEditorUserId="2248" LastEditDate="2015-08-08T11:44:33.357" LastActivityDate="2015-08-08T11:44:33.357" Title="What factors into the high cost of transmission components?" Tags="|mechanical-engineering|linkage|" AnswerCount="1" CommentCount="4" ClosedDate="2015-07-09T01:16:02.900" ContentLicense="CC BY-SA 3.0" />
  <row Id="3525" PostTypeId="2" ParentId="2696" CreationDate="2015-07-08T21:03:07.827" Score="4" Body="&lt;p&gt;CO&lt;sub&gt;2&lt;/sub&gt; lasers are not able to cut highly-reflective materials, for reasons which should be obvious to anyone who has shone a laser pointer into a mirror.  It may be that the &quot;laser plate&quot; is just duller than regular plate, or has some other surface finish conducive to laser cutting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the steel is certified as A36, I can't see how there will be chemical concerns that would interfere with weldability in regards to D1.1.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-07-08T21:03:07.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3528" PostTypeId="2" ParentId="3524" CreationDate="2015-07-09T00:50:57.540" Score="3" Body="&lt;p&gt;In such cases you don't pay for the rubber, but for the administrative and storing costs. Manufacture a such rubber and then try to sell it on the ebay. With taxes, all of the law regulations, with post costs, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you only want a rare screw to repair a device, you can easily pay even tens of dollars for that. Why? Producing a kilo of such screws would only cost fewer as a single dollar. But storing a thousands of similar boxes full with rare screws, and then sent them on post to countries thousands of kilometers away, it costs much more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, there is another effect, which an already deleted answer said. They won't calculate the price on the costs plus a profit, but they will see how is it priced on the ebay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a highly unfortunate feature of the global society.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make significantly cheaper your costs if you pay a lot of them, and know the market (so you can buy it from cheaper manufacturers), and pay from them directly. But you can only do that if you aren't a hobbyist, but also a large manufacturer.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2015-07-09T00:50:57.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3529" PostTypeId="2" ParentId="3522" CreationDate="2015-07-09T12:19:58.227" Score="6" Body="&lt;p&gt;I think the diverse range of names for the second argument arises from the fact that the convolution operation is so useful in so many different fields.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is helpful to recall what the convolution operation does before addressing the specific terms.  Quoting from &lt;a href=&quot;http://mathworld.wolfram.com/Convolution.html&quot;&gt;Wolfram Mathworld&lt;/a&gt;, &quot;a convolution is an integral that expresses the amount of overlap of one function $g$ as it is shifted over another function $f$.&quot;  Expressed another way, convolution is a mathematical way of checking to see how much of one function exists in another function, as the two are slid past each other.  Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Convolution#Derivations&quot;&gt;visual examples of convolution&lt;/a&gt; has nice illustrations of how it works.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Kernel:&lt;/strong&gt; This is the most general term, and it arises from mathematics.  In mathematics, an &lt;a href=&quot;https://en.wikipedia.org/wiki/Integral_transform&quot;&gt;integral transform&lt;/a&gt; is a general transformation &lt;a href=&quot;http://mathworld.wolfram.com/IntegralTransform.html&quot;&gt;defined by&lt;/a&gt; $$g(\alpha)=\int_a^b f(t)K(\alpha,t)dt.$$ The function $K(\alpha,t)$ in this integral transform is called the kernel.  The convolution operation is just a sub-class of this more general transformation and so the second function is rightfully called the kernel.  Unfortunately, I don't know the origin of the term kernel in the general integral transform.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Filter:&lt;/strong&gt; In digital signal processing a &lt;a href=&quot;https://en.wikipedia.org/wiki/Matched_filter&quot;&gt;matched filter&lt;/a&gt; &quot;is obtained by correlating a known signal, or template, with an unknown signal to detect the presence of the template in the unknown signal.&quot;  In this sense the second function acts as a filter for the first function, telling you which parts of the first have the properties of the second. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Template:&lt;/strong&gt; This one is least familiar to me, but I think you can see how it arises from the same place as the term 'filter'.  The template is an a-priori known signal for which you are searching in the unknown signal.  The convolution of the two tells you what parts of the unknown signal have the same characteristics as the template.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-07-09T12:19:58.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3530" PostTypeId="2" ParentId="2966" CreationDate="2015-07-09T16:20:45.203" Score="1" Body="&lt;p&gt;I would recommend a &lt;strong&gt;forward inclined centrifugal system&lt;/strong&gt;, such as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Centrifugal_fan#Forward-curved&quot; rel=&quot;nofollow&quot;&gt;forward curved fan&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power input/output of any device, where fluid comes in/leaves with fluid rate $Q$ at $V$ and enters/exits at an angle $\theta$ at velocity $U$, would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal P = (V-U)(1-\cos(\theta))\rho QU$$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have this device compress the gas, the power input runs in reverse.  In both cases the angle helps.  See the &lt;a href=&quot;https://en.wikipedia.org/wiki/Centrifugal_fan#/media/File:Velocity_Triangle_for_Forward_Facing_Blade.png&quot; rel=&quot;nofollow&quot;&gt;velocity triangle&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real heart of this will comes down to putting some good valves on the openings.  $U$ is a double edged sword - while it ups your power, if $V$ isn't very high compared to $U$ nothing is really happening.  Don't forget $Q$ depends on $V$ or $U$, depending on how you look at it.  The key to modifying this is to throttle your inlet opening down (whichever way you run) to a very small opening to have the highest $V$ possible, while keeping the outlet carefully controlled to not constrict $Q$ or $U$ beyond what is necessary to keep $V/U$ decent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps using this as a first stage in a two stage rotary compressor could also help - the second stage is a true rotary compressor to really boost the pressure, but this assists the second stage to increase pressure beyond atmospheric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately no device on the market will be built to this strange service - but by having a fairly symmetric rotary system with carefully controlled inputs should yield some decent results.  I would definitely consult with a custom fan manufacturer.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-07-09T16:20:45.203" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3532" PostTypeId="1" CreationDate="2015-07-09T19:10:58.677" Score="3" ViewCount="522" Body="&lt;p&gt;I am planning to replace the material that is used for Cricket Stumps.  Cricket is a game similar to baseball.  One difference is the presence of 3 successive stumps between the hitter and catcher.  The ball will hit the stumps if the hitter misses it.  The combined weight of the stumps is 7 lbs.  Each stump is a 28 inch long wooden cylindrical stump of 1.5 inch radius weight 2.3 lbs. &#xA;&lt;em&gt;The stump can be hollow inside.&lt;/em&gt;  As of now the regular stumps are made of solid wood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am in search of a lightweight alternative for the wooden stump.  The material should be able to withstand the force of the ball which typically travels at 80 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any cheap and lightweight plastic material that can drastically reduce the weight of the stump and still withstand the forces mentioned above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4j5bp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4j5bpm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CNZFY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0YeVF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2183" LastEditorUserId="272" LastEditDate="2015-07-09T19:33:06.473" LastActivityDate="2015-07-11T16:56:16.083" Title="Best plastic composite alternative to wood" Tags="|materials|plastic|wood|" AnswerCount="2" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3533" PostTypeId="2" ParentId="3532" CreationDate="2015-07-09T21:09:07.843" Score="6" Body="&lt;p&gt;For impact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma_{max} = \frac{v}{V}E(1+\sqrt{\mu + \frac{2}{3}}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V = \sqrt\frac{gE}{\delta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g$ is Gravitational constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E$ is Young's modulus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta$ is the weight per unit volume&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu $ is the ratio of ball mass to bar mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v$ is velocity of impacting ball&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per &lt;em&gt;Roark's Formula's for stresses and strains 7th edition&lt;/em&gt;, this can be shown that as $\mu$ approaches $\infty$, then $\sigma_{max}$ will approach 2 times the case if the mass was placed on the bar and allowed to deflect under gravity.  In that case, we'll run with 2 times the ball mass (163 gm) and treat it as a fixed beam. for simpler equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With round pegs we therefore have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = \frac{2mgL^3}{3EI} $$&#xA;$$\sigma = \frac{2mgLr}{I}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need the deflection to stay the same, so the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bail_(cricket)&quot;&gt;bails&lt;/a&gt; will move the same.  We need to figure out how much that is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 7/3 pounds per peg, and a source saying they are willow wood, it is 15% air.  That works to the inside being 0.581 inches radius hollow. So I'd calculate $\delta = 0.002 in$ for the wood.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying for a really thin material, $I = \pi r^3t$, so to match deflection we have $Et = 0.120 Msi*in$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a good fiberglass composite (the cheapest of all of them), the modulus would be 1.5 Msi, so the thickness should be a mere 0.08in - a  drastic weight savings (1 pound per wicket).  The stress would be also acceptable levels.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-07-09T21:25:38.473" LastActivityDate="2015-07-09T21:25:38.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3534" PostTypeId="1" AcceptedAnswerId="3537" CreationDate="2015-07-09T21:26:55.567" Score="3" ViewCount="171" Body="&lt;p&gt;Over in &lt;a href=&quot;http://chat.stackexchange.com/transcript/message/22676589#22676589&quot;&gt;Programmers chat&lt;/a&gt; we were discussing whether a gold skeleton would be feasible. That got me thinking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to build a Terminator (T-800) using pure gold as the endoskeleton, would the skeleton be able to support the &lt;a href=&quot;https://scifi.stackexchange.com/q/32239/31563&quot;&gt;weight of the terminator&lt;/a&gt;? If so, what type of force would cause it to deform under stress? Would it be able to handle normal human tasks such as running, walking, or jumping? What about the superhuman feats of strength the terminator performs in the movies?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of this calculation, assume a T-800 weighs an extra 30% more than a human its size. Based on what Google says Arnold's size is and a rounding slightly, that assumption is 2m tall and 150kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question &lt;a href=&quot;https://physics.stackexchange.com/q/39512/57109&quot;&gt;Stacking gold Bars&lt;/a&gt; indicates that gold has quite a bit of strength given its reputation as a soft metal. However, that question talks about the compressive strength: also, I am not an engineer and unable to calculate the types of stresses a skeleton would undergo and whether those would be greater than the tensile strength of gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, assume the same or similar motors and supports: there is very little information out there, but clearly this machine is capable of supporting great loads compared to humans, throwing grown men across rooms like dolls, etc. Only the skeleton is gold: motors, electronics, etc. could remain as other metals (steel, titanium, copper).&lt;/p&gt;&#xA;" OwnerDisplayName="user608" LastEditorUserId="-1" LastEditDate="2017-04-13T12:43:39.343" LastActivityDate="2015-07-10T04:58:47.320" Title="Would a T-800 with a gold skeleton deform under stress?" Tags="|materials|stresses|metals|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="3536" PostTypeId="2" ParentId="3415" CreationDate="2015-07-10T00:23:58.217" Score="0" Body="&lt;p&gt;A piece of thick plastic or timber will make a lid that will solve this or use a gel battery and lay it on its side&lt;/p&gt;&#xA;" OwnerUserId="2222" LastActivityDate="2015-07-10T00:23:58.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3537" PostTypeId="2" ParentId="3534" CreationDate="2015-07-10T04:58:47.320" Score="4" Body="&lt;p&gt;The following table gives some of the material properties of four metals: gold, aluminium, titanium and iron. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Metal    Tensile      Shear    Bulk    Young's Brinell   Density&#xA;         Strength     Modulus  Modulus Modulus Hardness&#xA;           MPa         GPa      GPa     GPa      GPa      g/cm^3&#xA;Gold       120          27      180      79    188-216    19.300&#xA;Aluminium  125-300      26       76      70    160-550     2.700&#xA;Titanium   210          44      110     116    716-2770    4.506&#xA;Iron       150-430      82      170     211    200-1180    7.874 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From the table, gold has tensile and shear strengths similar to that of aluminium but its bulk density is more than double that of aluminium and slightly stronger than that of iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Young's modulus for gold is similar to that of aluminium, so both metals have similar level of stiffness and from the Brinell hardness numbers gold is a soft metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An endoskeleton made of gold could support itself but having shear and tensile properties similar to aluminium it would not be able to withstand large external forces that would try to destroy it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue with using gold as the main metal for the endoskeleton is moving the endoskeleton and other body parts would require a lot of energy and strong motors due to its weight as gold is more than twice as dense as iron.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-07-10T04:58:47.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3538" PostTypeId="1" AcceptedAnswerId="3540" CreationDate="2015-07-10T07:04:24.203" Score="3" ViewCount="616" Body="&lt;p&gt;Would the body of a normal round pot shaped pressure cooker (built to cook foods at 15 psi) be able to withstand being evacuated down to high-vacuum pressures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently it has been done before: &lt;a href=&quot;http://www.instructables.com/id/Pressure-Cooker-Vacuum-Chamber/&quot; rel=&quot;nofollow&quot;&gt;i.e. this instructable&lt;/a&gt; but was this person just lucky it didn't implode?&lt;/p&gt;&#xA;" OwnerUserId="2201" LastEditorUserId="368" LastEditDate="2015-07-10T15:59:32.660" LastActivityDate="2016-09-13T11:41:34.460" Title="Would a structurally sound pressure cooker be able to withstand an internal vacuum?" Tags="|mechanical-engineering|structural-engineering|pressure|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3539" PostTypeId="2" ParentId="3538" CreationDate="2015-07-10T14:19:48.393" Score="2" Body="&lt;p&gt;It MAY work, but there are no guarantees as a shape that is designed to place the body in tension throughout inder pressure will exhibit stresses under vacuum which there is no needs to design for in normal operation.&lt;br&gt;&#xA;ie it MAY work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the lid will need to be adapted in many cases as it is usually made to seal under positive pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thin steel tin can that is pressurised to well above it's design rating will round out at the ends but if placed under partial vacuum will flatten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking an extreme example - a PET 1.5 litre bottle can be pressurised to about 150 psi (ask me how I know :-) *) - at which point it will swell and may explode. Few will reach 200 psi. BUT the same will collapse partially under a few psi of vacuum and increasingly so with reducing pressure. You can flatten one with &quot;breath suction&quot;. [*How I know: Water rockets!]. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Testing:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a larger pressure vessel rated to several atmospheres + you could place your target pressure cooker therein and subject it to multiple atmospheres of external pressure. That will show you what sort of safety margin you have , subject to the pressure capability of the external chamber. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-07-10T18:08:45.033" LastActivityDate="2015-07-10T18:08:45.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3540" PostTypeId="2" ParentId="3538" CreationDate="2015-07-10T15:08:46.707" Score="3" Body="&lt;p&gt;Going with &lt;a href=&quot;http://www.walmart.com/ip/Presto-16-Quart-Aluminum-Pressure-Cooker/10771204&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; pressure cooker, made of aluminum, 15&quot; diameter, 12&quot; tall.  I'm giving this a similar thickness throughout.  At 8.4# weight, that leads to 0.094&quot; thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying Roark's 7th edition, Table 15.2, Case 20, using the approximation listed, formula, the 0.094&quot; thick shell can withstand 82 psi of external pressure without buckling.  Even with comfortable safety factors in place, this seems plausible.  Only when the shell gets really thin (I calculate 0.05&quot;, or anything lighter than &lt;a href=&quot;https://en.wikipedia.org/wiki/Sheet_metal&quot; rel=&quot;nofollow&quot;&gt;16 gauge&lt;/a&gt; for this particular pressure cooker) would the shell buckle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that since the bottom is flat (i.e. shell with infinite radius), then if it can handle 15 psi from the inside, it can handle 15 psi from the outside.  Also, the top is flat on this one as well, so it works the same.  Only for some models with curved tops would I need to look at the top as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, it's definitely possible for full vacuum, but I can't legally recommend it.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-07-10T15:08:46.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3541" PostTypeId="1" AcceptedAnswerId="3567" CreationDate="2015-07-10T16:30:23.910" Score="8" ViewCount="2315" Body="&lt;p&gt;I have a strip of stainless steel encastree'd at one end to which is applied a constant pressure on one side, and I need to know what the deflection equation y = f(x) is at equilibrium. &#xA;If the deflection was small I would be able to use one of the very well known deflection formulas which assume vertical loads, but it isn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to work out the bending moment M=f(x) using integrals to use:&#xA;$$y(x)=\int\int_{beam} \frac{M}{EI}dx^2$$&#xA;But I'm getting more and more confused. What is the right approach to solving this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZdihL.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZdihLm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1326" LastEditorUserId="368" LastEditDate="2015-07-13T17:21:41.447" LastActivityDate="2015-07-14T14:44:57.240" Title="Large deflection of a cantilever beam with distributed normal load" Tags="|mechanical-engineering|beam|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3542" PostTypeId="1" AcceptedAnswerId="3543" CreationDate="2015-07-10T16:32:20.300" Score="3" ViewCount="556" Body="&lt;p&gt;I am answering the following question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Gas initially at a pressure of 101.325 kPa and temperature of 60 °C&#xA;  undergoes the following cycle:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;1-2: Adiabatic Compression  through compression ratio of 4.5:1&lt;/li&gt;&#xA;  &lt;li&gt;2-3: Heating at constant volume through a pressure ratio of 1.35:1&lt;/li&gt;&#xA;  &lt;li&gt;3-4: Constant entropy expansion to initial pressure&lt;/li&gt;&#xA;  &lt;li&gt;4-1: Constant pressure cooling to initial volume&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;If $Cp = 1\ \mathrm{\frac{kJ}{kg\ K}}$ and $Cv = 0.678\&#xA;&amp;gt; \mathrm{\frac{kJ}{kg\ K}}$ for the gas, determine the thermal&#xA;  efficiency for the cycle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The given answer is 41.41%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it sounds pretty much like the Otto cycle, I try using the formula with compression ratio $r_k = 4.5$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = Cp/Cv = 1.47$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$e = 1-(1/(r_k^{k-1})) = 50.68\% $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? Is it not an Otto cycle?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="368" LastEditDate="2015-07-13T17:35:15.427" LastActivityDate="2015-07-13T18:21:47.733" Title="Thermal efficiency of a given complete cycle" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3543" PostTypeId="2" ParentId="3542" CreationDate="2015-07-10T18:59:20.783" Score="5" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What am I doing wrong? Is it not an Otto cycle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No, it's not an Otto cycle. It would be if it wasn't for the last process, heat rejection (cooling) in Otto cycle is performed at constant volume not constant pressure.&lt;/p&gt;&#xA;&lt;p&gt;This should be the &lt;span class=&quot;math-container&quot;&gt;$P-v$&lt;/span&gt; diagram of your engine (sorry for the lousy image):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/He5SS.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, naturally you can't apply Otto's efficiency to this cycle.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Side note:&lt;/strong&gt; compression ratio in Otto cycle &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is not &lt;span class=&quot;math-container&quot;&gt;$\frac{P_2}{P_1}$&lt;/span&gt; as in your calculations, it's defined as the ratio between maximum and minimum volume in the cycle &lt;span class=&quot;math-container&quot;&gt;$\frac{V_{max}}{V_{min}} = \frac{V_4}{V_3}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-07-13T18:21:47.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3544" PostTypeId="1" AcceptedAnswerId="3546" CreationDate="2015-07-11T02:12:58.180" Score="-1" ViewCount="2855" Body="&lt;p&gt;I am solving the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The tank of an air compressor has a volume of 3 cubic feet and is&#xA;  filled with air at a temperature of 40 °F. If a gauge on the tank&#xA;  reads 150 psig, what is the mass of the air in the tank?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The given answer is 1.78 lbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just $PV = mRT$, so I try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$    T = 40+460 = 500\ \mathrm{°R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$    P = (150\ \mathrm{psi_g} + 14.7\ \mathrm{psi_{atm}})*144 = 23,716.8\ \mathrm{\frac{lbf}{ft^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$    R = 53.34\ \mathrm{\frac{ft*lbf}{lbm*°R}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{PV}{RT} = m =  \frac{23,716.8\ \mathrm{\frac{lbf}{ft^2}}*3\ \mathrm{ft^3}}{53.34\ \mathrm{\frac{ft*lbf}{lbm*°R}}*500\ \mathrm{°R}} = 2.67\ \mathrm{lbm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the problem set wrong or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="368" LastEditDate="2015-07-13T17:32:15.783" LastActivityDate="2021-07-17T07:09:48.760" Title="Finding the mass in the tank of an air compressor from gage pressure" Tags="|mechanical-engineering|compressors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3545" PostTypeId="2" ParentId="3495" CreationDate="2015-07-11T02:22:51.937" Score="3" Body="&lt;p&gt;When a road needs to be repaired there are a few typical procedures that are used depending on what is wrong with the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the base (aggregate) of the road is replaced, it means that there was something structurally deficient with the road. The base is there to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Provide drainage so that water doesn't collect under the road surface. &lt;/li&gt;&#xA;&lt;li&gt;Replace weak soil.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the aggregate layer isn't thick enough, too much water may be collecting under the road surface and causing weak spots. Also, if the layer isn't thick enough the load from vehicle wheels is not able to distribute over a large enough area. This may be over stressing the soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direct answer to your question is that, yes, there was something wrong with the road. They aren't digging it up for the fun of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signs of a weak subgrade may be easily over looked if you aren't specifically aware of them. The signs start out as cracks and progress to pot holes as things get worse.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-07-11T02:22:51.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3546" PostTypeId="2" ParentId="3544" CreationDate="2015-07-11T07:45:03.343" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible the problem set is wrong? or am I missing something?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably, your calculations are correct I calculated $m$ to be $2.6697$ $lbm$ myself.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2015-07-11T07:45:03.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3547" PostTypeId="1" AcceptedAnswerId="3747" CreationDate="2015-07-11T16:18:46.103" Score="13" ViewCount="25542" Body="&lt;p&gt;I am contemplating buying a new car. However, the approach to the underground garage in my apartment has a 90 degree frustrating turn. Given the dimensions of the approach and car, what is the max turn circle for the car to fit the garage and turn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/A7A8w.jpg&quot; alt=&quot;garage and car dimensions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/qoVZt.jpg&quot; alt=&quot;more legible dimensions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;given Ackerman steering and the overhanging front part of the car I believe you can use Pythagoreas' theorem to get R min and R max. delta R should be less than the shortest path in the pathway, ie 2.5m. unfortunately the result does not seem plausible. feedback would be greatly appreciated.&lt;img src=&quot;https://i.stack.imgur.com/Gj2A4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="2267" LastEditorUserId="2267" LastEditDate="2015-07-20T15:24:22.767" LastActivityDate="2021-09-04T16:30:52.410" Title="How much clearance does a car need when turning a corner?" Tags="|automotive-engineering|" AnswerCount="4" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="3548" PostTypeId="2" ParentId="3532" CreationDate="2015-07-11T16:56:16.083" Score="1" Body="&lt;p&gt;You could use polycarbonate tubing, it is used for safety glasses and other unbreakable items and is lightweight. It is not stress crack resistant though so if you need to drill holes in in it there will be micro-cracks where you drill the hole. It is then advisable to anneal the plastic if you want the finished product to last in an outdoor environment.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2015-07-11T16:56:16.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3550" PostTypeId="1" AcceptedAnswerId="3552" CreationDate="2015-07-12T03:57:26.417" Score="2" ViewCount="1014" Body="&lt;p&gt;Air in the cylinder of a Diesel Engine is at $30^o\ $$C$ and $138\text{ kPa}$ in compression. If it is further compressed to 1/18th of its original volume. Which of the following most nearly equals the work done in compression of the displacement volume of the cylinder is $14.2$ Liters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer is $11\ \text{kJ}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_1 = 30^o\ C + 273.15 = 303.15\ K$$&#xA;$$P_1 = 138\ kPa$$&#xA;$$V_d = 14.2\ L*(1\ m^3/1000L) = 0.0142\ m^3$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Attempt 1: Using isentropic relation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$r_k = 18 = V_1/V_2$$&#xA;$$V_1 = \frac{mRT}{P_2} = \frac{1*0.287*303.15}{138} = 0.63046\ m^3$$&#xA;$$V2 = 0.03502578502\ m^3$$&#xA;$$P2 = \frac{mRT}{V2}=\frac{287*303.15}{0.0350257} = 2484\ Pa$$&#xA;for air:&#xA;$$k = 1.4$$ &#xA;$$W = \frac{(P_2V_2 - P_1V_1)}{k - 1} = \text{(A very very small value)}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Attempt 2: Using $r_k$ and displacement volume:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$$r_k = \frac{(V_d + V_c)}{V_c}$$&#xA;$$V_d = 14.2L = 0.0142\ m^3$$&#xA;$$V_c = 8.3529411*10^{-4} = V_2$$&#xA;$$V_1 = V_c + V_d = 0.01583529\ m^3$$&#xA;$$P_1V_1^k = P_2V_2^k $$&#xA;$$\frac{138*0.01583529^{1.4}}{(8.3529411*10^{-4})^{1.4}} = P2 = 8487.512\ kPa$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plug in everything into the work function&#xA;$$W = \frac{P_2V_2 - P_1V_1}{k-1} = 12.1 \ kJ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, I'm getting close but I don't know what I'm doing wrong... Any hints? And why doesnt' method 1 work?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="90" LastEditDate="2015-07-12T22:52:53.083" LastActivityDate="2015-09-12T01:52:12.130" Title="Methods for determining Work of Diesel Engine's Isentropic Process" Tags="|mechanical-engineering|thermodynamics|engines|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3551" PostTypeId="2" ParentId="3550" CreationDate="2015-07-12T12:47:16.513" Score="1" Body="&lt;p&gt;Considering that the following is already sufficient to solve the &quot;intensive&quot; problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Air in the cylinder of a Diesel Engine is at $30\,{}^o\mathrm{C}$ and $138\,\mathrm{kPa}$. If it is further compressed to $1/18\,$th of its original volume, calculate the work done on the system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;the compression ratio is $\varrho=v_1/v_2=18$, and the barometric ratio $\beta=P_2/P_1$ is rapidly derived from the isoentropic relation (given that $\gamma=1.4$ is temperature independent):&#xA;$$&#xA;P_1v_1^\gamma=P_2v_2^\gamma\to \frac{P_2}{P_1}=\left(\frac{v_1}{v_2}\right)^\gamma\to \beta=\varrho^\gamma&#xA;$$&#xA;this can be directly used inside the work equation. The latter is derived from the integral of $P\,\mathrm{d}v$ on Clapeyron's Diagram (isoentropic shaft work on a closed system), using the intensive relation, as demonstrated below:&#xA;\begin{align}&#xA; w=-\int_{v_1}^{v_2}P\,\mathrm{d}v&amp;amp;=-\int_{v_1}^{v_2}\frac{P_1v_1^{\gamma}}{v^\gamma}\,\mathrm{d}v\\&#xA;&amp;amp;=P_1v_1^{\gamma}\int_{v_1}^{v_2}-v^{-\gamma}\,\mathrm{d}v\\&#xA;&amp;amp;=P_1v_1^{\gamma}\frac{1}{\gamma-1}\left(v_2^{1-\gamma}-v_1^{1-\gamma}\right)\\&#xA;&amp;amp;=RT_1\frac{1}{\gamma-1}\left(v_2^{1-\gamma}/v_1^{1-\gamma}-1\right)\\&#xA;&amp;amp;=RT_1\frac{1}{\gamma-1}\left(\varrho^{\gamma-1}-1\right)&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the result will describe an intensive quantity, the extensive result will depend on the chemical substance contained (and not consumed) inside the system. That is derived from the displaced volume and the compression ratio, namely rearranging the terms to highlight $\varrho$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the compression work done, given the displacement volume of the cylinder as $\Delta V=14.2\,\mathrm{L}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta V=V_1-V_2\to\frac{\Delta V}{V_2}=\frac{V_1-V_2}{V_2}=\varrho-1\to V_2=\frac{\Delta V}{\varrho-1}&#xA;$$&#xA;from here the initial volume is known (substituting back the $V_2$ term):&#xA;$$&#xA;V_1=V_2+\Delta V=\frac{\Delta V}{\varrho-1}+\Delta V=\frac{14.2\,\mathrm{L}}{17}+14.2\,\mathrm{L}=15.03\,\mathrm{L}=15.03\times10^{-3}\,\mathrm{m^3}&#xA;$$&#xA;substituting all the first intensive parameters inside the Ideal Gas EoS, the constant molar quantity of air is known:&#xA;$$&#xA;n=\frac{P_1V_1}{RT_1}=\frac{15.03\times10^{-3}\,\mathrm{m^3}\cdot138\times10^{3}\,\mathrm{Pa}}{8.314\,(\mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}})\cdot303.15\,\mathrm{K}}=0.823\,\mathrm{mol}&#xA;$$&#xA;Knowing that the extensive parameter $W$ is expressed as $nw$, and from the previous equation $w=13.72\,\mathrm{kJ\,mol^{-1}}$, the requested value is $W=0.823\,\mathrm{mol}\cdot 13.72\,\mathrm{kJ\,mol^{-1}}=11.29\,\mathrm{kJ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, even if the result is different from the first given, there can be one explanation. Since in your first method you considered an initial mass of air equal to $1\,\mathrm{kg}$, this is contradictory with respect to the Equation of State. Because the initial (and constant) molar quantity of air depends to the initial extensive state $(T_1,P_1,V_1)$, then the only variable left $V_1$ is derived if the displacement volume and compression ratio are known.&lt;/p&gt;&#xA;" OwnerUserId="2274" LastEditorUserId="2274" LastEditDate="2015-09-08T19:56:50.877" LastActivityDate="2015-09-08T19:56:50.877" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3552" PostTypeId="2" ParentId="3550" CreationDate="2015-07-12T15:14:30.003" Score="2" Body="&lt;p&gt;Your second attempt is correct, you just had two mistakes: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you re-calculated $V_1$ you'll find that it equals $0.01503$ not&#xA;$0.0158$ $m^3$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;same issue with $P_2$.&#xA;$$P_2 = \frac{138*10^{3}*0.01503^{1.4}}{(8.3529411*10^{-4})^{1.4}}= 7.889*10^6\ Pa$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Substituting in isentropic work equation:&#xA;$$W = \frac{P_2V_2 - P_1V_1}{k-1} = \frac{(7.867*10^6 * 8.3529411*10^{-4}) - (138*10^3*0.01503)}{1.4-1}=11.2 \ kJ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is very close to your problem set answer, I think it was a multiple choice question as he is asking about which answer &lt;em&gt;nearly equals&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="2274" LastEditDate="2015-09-12T01:52:12.130" LastActivityDate="2015-09-12T01:52:12.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3554" PostTypeId="1" AcceptedAnswerId="3556" CreationDate="2015-07-13T16:43:00.907" Score="4" ViewCount="872" Body="&lt;p&gt;Almost all of the new Snapdragon CPU series (610, 810) have an overheating problem. This overheating problem may occur in the first few hours of using mobile phones that are equipped with the CPU, as we can see from the various reviews from around the web on Xiaomi mi4i, LG G4, Sony Xperia Z3+, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume Qualcomm, a billion dollar company, has a quality control department working on (or detecting) this issue before they release the CPU. How does such a significant issue get past QC without being detected way before the launch date?&lt;/p&gt;&#xA;" OwnerUserId="2280" LastEditorUserId="368" LastEditDate="2015-07-13T17:10:23.910" LastActivityDate="2015-07-14T12:15:50.647" Title="How do products like the overheating Snapdragon CPU make it through quality control?" Tags="|quality-engineering|computer-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3556" PostTypeId="2" ParentId="3554" CreationDate="2015-07-13T18:23:27.627" Score="1" Body="&lt;p&gt;There is no good reason to think that Qualcomm's QC people did not detect the problem. It's what you do about it that is the problem.&lt;br&gt;&#xA;That there is or ever was a problem is denied by Qualcomm, some users and some independent testers. But at least one reasonably well documented test set indicates that when running benchmark programs the processors tend to become more heat affected sooner than previous Qualcomm processors and than competitive alternatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High performance ICs generate substantial thermal output and all things being equal , output increases approximately linearly with throughput (as a major factor in energy loss is related to capacitive node charge / discharge, with energy dissipation being proportional to switching cycles per second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In new device aimed at being near the front in a competitive market, the manufacturer may strive for raw output first and efficiencies second. Temperature relates to both thermal output and the ability to handle the heat. A device that is pushed towards the upper limits of a technology may be less efficient and require the device user to take more extensive efforts to cool it. In the finite space available 'corners may be cut'. Some manufacturers (eg Samsung) have gone to newer lower lithographic width devices to achieve the needed efficiencies. Qualcomm chose not to and as the devices are made by a third party, the drive to be faster sooner apparently swamped the awareness of the need to be better cooled. Some prospective users have addressed the problem by &quot;baling out&quot;. Others by denying the problem exists as long as possible and then &quot;writing software&quot;  to fix it.   This is no doubt a painful learning experience. The project manager is unlikely to have the same problem again as (one can guess) he probably now works somewhere else, and Qualcomm will no doubt take greater notice of their QC department.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The &quot;answer&quot; very much deep-ends on who you believe.&lt;br&gt;&#xA;I'm not an expert at all on this. I asked Google - and found that a lot of people are not experts but there does seem to be at least one good looking answer available as to what is really happening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems they pushed too far too fast, used a fab house who was good but not as good as their cooperative opposition (Samsung) and allowed themselves to be less 'hands on' over the results until too late. (Almost*) ALL multicore processors in top end phones run into thermal speed limiting with time. Some limit more than others. The 810 is worst than most. (The Snapdragon 801 in eg the HTC One M8 at 2.3 Ghz seems to be an exception to the general rule). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following does not so much advise &quot;how&quot; but rather what did or didn't happen and how transparent people were about it and what is really happening and when it was known. The following are almost in 'order I found them'. I could have reordered by date but its interesting to see that the date and what is said largely do not correlate well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.forbes.com/sites/jaymcgregor/2015/05/06/qualcomm-finally-speaks-out-about-samsung-and-snapdragon/&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; on May 6th Qualcomm deny that there is a problem in commercial products, as opposed to preproduction prototypes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.techradar.com/news/phone-and-communications/mobile-phones/snapdragon-10-overheating-lg-g-flex-2-htc-one-m9-1290204&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; on April 4th various user and manufacturere's spkesmen are being less than traansparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.theinquirer.net/inquirer/news/2413064/sony-admits-snapdragon-810-is-causing-xperia-z3-overheating-issues&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; on June 15th !!! Sony are claimed to have admitted to problems with the 810 in their Z3+ after users demonstrated the ability to crash it by recording video. Sony advice a software fix will be provided. A software fix is a rather suspect way of fixing thermal issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.theinquirer.net/inquirer/news/2391263/samsung-to-ditch-qualcomm-snapdragon-chip-in-galaxy-s6-smartphone&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; on January 29th Qualcomm adumbrate that Samsung had decided not to use the 810 and to use an internal Samsung processor instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BUT&lt;/strong&gt; on  April 3oth &lt;a href=&quot;http://www.theinquirer.net/inquirer/news/2406556/qualcomm-theres-nothing-wrong-with-our-snapdragon-810-chip&quot; rel=&quot;nofollow&quot;&gt;the same source&lt;/a&gt; Qualcomm indicate that samsung may manufacture the snapdragon for the in future, and they say a decisin by LG to use a qualcomm hex core processor instead of the octal core snapdragon was made a year ago and is not due to any overheating issues and probably says that they will limit power use by reducing processing speeds for selected applicaations (where selected might mean &quot;all&quot;. . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And/but on Feb 5th it is advised &lt;a href=&quot;http://www.technobuffalo.com/2015/02/05/qualcomm-reportedly-fixes-its-overheating-snapdragon-810-processor/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; that there never was a problem and BESIDES it has already been fixed, maybe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which leads to the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Best picture you'll get, probably:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fairly definitively, on Feb 24th &lt;a href=&quot;http://arstechnica.com/gadgets/2015/04/in-depth-with-the-snapdragon-810s-heat-problems/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; some actual thermal tests. Nice time versus processor speed tests for  range of processors. What happened? Qualcomm pushed things a bit far. the processor was made for them by TSMC in Taiwan. The device uses a number of ARM Cortex cores. How these are implemented varies with manufacturer. Samsung are using a smaller transistor process than TSMC. The smaller lower power process is really needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They say:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In short, chips throttle, but the 810 throttles more than most, and it's severe enough that the 810 is actually slower than the 801 or 805 in some CPU-bound tasks over the long haul. The Exynos 7 Octa, which has similar specs on paper, is much better in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, Qualcomm has implied to us several times that its use of ARM Cortex CPU cores was a stopgap measure—Apple got the 64-bit A7 chip to market around a year before anyone expected it to. Chips are designed over a period of two or three years, so using the ready-made Cortex cores were the quickest way to get a 64-bit response to market.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results, unfortunately, don’t look great. It might be because the 810 is using a 20nm TSMC manufacturing process instead of the 14nm Samsung process used for the Exynos, or it might be that Samsung has more experience working Cortex CPU cores into its designs. Whatever the reason, our testing of real phones with these SoCs in them shows that 810-based phones are slower and have worse battery life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qualcomm's next major flagship is the Snapdragon 820, the first to use its custom-designed 64-bit “Kryo” architecture. Rumor has it that the chip will be made on the same 14nm Samsung process as the Exynos 7 Octa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A return to its own CPU cores plus a newer manufacturing process should hopefully mean a return to the kind of performance and battery life we’ve gotten from Qualcomm-based phones in years past. All signs point to the 810 being a one-time slip-up and not the start of a trend—let’s hope that those signs are accurate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;AND&lt;/strong&gt; just to add balance, on March 2nd, &#xA;&lt;a href=&quot;http://semiaccurate.com/2015/03/02/behind-fake-qualcomm-snapdragon-810-overheating-rumors/&quot; rel=&quot;nofollow&quot;&gt;these people&lt;/a&gt; - &quot;SemiAccurate&quot; -  say there is no problem never was a problem, it's all made up and here are the lab tests to prove it. They claim to have investigated in depth and it was essentially a FUD campaign by a Korean phone manufacturer whose name begins with Samsung. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I liked the time/temperature graphs. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-07-14T12:15:50.647" LastActivityDate="2015-07-14T12:15:50.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3559" PostTypeId="2" ParentId="3257" CreationDate="2015-07-13T23:32:20.003" Score="1" Body="&lt;p&gt;Given the clamping force of the bolts, I would think that Fixed and Held would be appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to test this is to calculate the edge tension (I don't recall if  Roark's has formulations for this) and compare this to the clamping force of the bolts (i.e., the static frictional force between the flanges due to the clamping will need to be greater that the edge tension).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not certain that this case really meets the criterion of &quot;large deflections&quot; for the blind flange.  The ones I've seen used are pretty beefy in thickness compared to the span and have almost no deflection.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-07-13T23:32:20.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3561" PostTypeId="1" CreationDate="2015-07-14T03:07:27.677" Score="2" ViewCount="565" Body="&lt;p&gt;How can I calculate the engine pressure in units of Bars by using torque, the cylinder volume, motor efficiency, and the gear reduction ratio?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My values are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Torque = 1063 Nm&#xA;Hydraulic Motor Cylinder Volume = 2.56 ℓ [litres]&#xA;Gear Reduction = 1 &#xA;Overall Efficiency = 0.76 [76%]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q:&lt;/strong&gt; Do I need any more values to calculate the engine pressure caused by the torque generated? If not how can I calculate the pressure in units of Bars?&lt;/p&gt;&#xA;" OwnerUserId="1228" LastEditorUserId="1228" LastEditDate="2015-07-16T22:54:42.293" LastActivityDate="2015-07-16T22:54:42.293" Title="Calculating engine pressure from measureable quantities" Tags="|pressure|automotive-engineering|engines|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3562" PostTypeId="2" ParentId="3027" CreationDate="2015-07-14T04:17:02.583" Score="3" Body="&lt;p&gt;As modal analysis is performed in the frequency domain, your modes and amplitudes are also in the frequency domain.The modal amplitudes ($q_i$) are scaled by the same amount as the applied load. If you want to know the displacement at each mode, then just multiply $q_i$ by whatever the magnitude of the load (for the mode) is scaled by. The maximum amplitude at that frequency is $q_i$ (e.g. the largest displacement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to know the displacements with respect to time, then you would need to perform an inverse Fourier transform on the amplitude and phase at each frequency, giving you the displacement with respect to time at all spatial locations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question about the units and normalized mass, perform basic dimensional analysis to see if the right hand side of your equation gives you units of mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Left hand side: $U$: units of $m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right hand side: $q$: $N/(kg/s^2)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as ($C$ generally has units of $kg/s$ and $\omega$ has units of $1/s$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi$: dimensionless (as normalized modal mass means no dimensions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $1$ $N$ $=$ $1$ $kg*m/s^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q$ has units of: $m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore to answer your question, the units work out, and $U$ is in $m$. &lt;/p&gt;&#xA;" OwnerUserId="2291" LastEditorUserId="90" LastEditDate="2015-07-14T04:51:40.560" LastActivityDate="2015-07-14T04:51:40.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3565" PostTypeId="1" CreationDate="2015-07-14T11:36:52.440" Score="0" ViewCount="198" Body="&lt;p&gt;I've seen videos of &lt;a href=&quot;https://en.wikipedia.org/wiki/Theo_Jansen&quot; rel=&quot;nofollow noreferrer&quot;&gt;Theo Jansen's&lt;/a&gt; linkage driven &lt;a href=&quot;http://www.strandbeest.com/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;walkers&lt;/a&gt;, and they all walk really slowly. Is it because they want to showcase leg movement or these things actually can't move fast?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Jansen-Strandbeest_crop.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8F4hEm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2294" LastEditorUserId="272" LastEditDate="2015-07-14T11:46:29.190" LastActivityDate="2015-07-14T21:10:30.690" Title="How fast can Jansen's linkage driven walker go?" Tags="|mechanical-engineering|linkage|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3566" PostTypeId="1" AcceptedAnswerId="3626" CreationDate="2015-07-14T12:29:28.053" Score="4" ViewCount="227" Body="&lt;p&gt;I am designing an NFC device but am still a little unsure about the principles behind NFC/RFID. From what I understand:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A primary coil constantly emits a 'carrier frequency' magnetic&#xA;field.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This field induces an e.m.f. on a passive secondary coil which is in&#xA;a secondary circuit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The secondary circuit comprises a system that converts data into a&#xA;modulation signal, which is realised by a modulating impedance in&#xA;the second circuit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This impedance causes something in the primary circuit to change,&#xA;thus receiving data.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is step 4. that I am most confused by. Sources I've read simply say the load impedance is &quot;felt&quot; by the primary coil, whatever that means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what is the function of having a separate transmit and receive antenna on NFC transceivers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a physicist I have very little prior exposure to electrical engineering or system design!&lt;/p&gt;&#xA;" OwnerUserId="2296" LastEditorUserId="282" LastEditDate="2015-07-19T21:40:53.950" LastActivityDate="2019-01-31T13:27:38.820" Title="How do NFC/RFID devices transmit data?" Tags="|electrical-engineering|wireless-communication|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3567" PostTypeId="2" ParentId="3541" CreationDate="2015-07-14T14:16:41.083" Score="4" Body="&lt;p&gt;For large deformation, there are changes to the strain-displacement relation and vertical equilibrium, but horizontal equilibrium remains unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let N be the axial force and M the bending moment, for finite rotations, these are defined as:&#xA;$$&#xA;N=EA\epsilon^o &#xA;\\&#xA;M=EI\kappa&#xA;$$ where&#xA;$$\epsilon^o = \frac{du}{dx} + \frac{1}{2}\left(\frac{dw}{dx}\right)^{2}&#xA;$$&#xA;and&#xA;$$&#xA;\kappa = -\frac{\frac{d^{2}w}{dx^{2}}}{\left[1 + \left(\frac{dw}{dx}\right)^2\right]^{3/2}}&#xA;$$&#xA;so vertical equilibrium is given by:&#xA;$$&#xA;\frac{d^2M}{dx^2} + N\frac{d^{2}w}{dx^2} + P = 0&#xA;$$&#xA;where $w$ is the vertical deflection&#xA;The solution is then the solution to this differential equation dependent upon your set boundary conditions&lt;/p&gt;&#xA;" OwnerUserId="2291" LastEditorUserId="2291" LastEditDate="2015-07-14T14:44:57.240" LastActivityDate="2015-07-14T14:44:57.240" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3568" PostTypeId="2" ParentId="2890" CreationDate="2015-07-14T15:17:27.250" Score="1" Body="&lt;p&gt;In order to incorporate both plane strain and plane stress in your model the stiffness and material properties should not be changed. The only difference is how you define element stiffness matrix. You still assemble the global stiffness matrix for the entire problem, but this global stiffness matrix is assembled from the individual plane strain or plane stress elements (as indicated in the assignment). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the type of elements (solid,shell,beam etc.) you can derive the element stiffness matrix from the shape functions given in many finite element texts or other &lt;a href=&quot;https://scicomp.stackexchange.com/questions/10678/basic-explanation-of-shape-function&quot;&gt;online forums&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows you to build a Finite Element model that incorporates both plane strain and plane stress conditions for individual elements without changing the material properties or performing the stiffness scaling that was discussed.&lt;/p&gt;&#xA;" OwnerUserId="2291" LastEditorUserId="-1" LastEditDate="2017-04-13T12:53:56.163" LastActivityDate="2015-07-14T15:17:27.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3569" PostTypeId="2" ParentId="245" CreationDate="2015-07-14T16:00:11.477" Score="2" Body="&lt;p&gt;A similar question has been asked on the site with an answer posted &lt;a href=&quot;https://engineering.stackexchange.com/questions/3541/large-deflection-of-a-cantilever-beam-with-distributed-normal-load/3567#3567&quot;&gt;here&lt;/a&gt; which shows the defining differential equations for large deformation of a beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question was posed for uniform loading on a cantilever beam, but the solution can be extended to generalized loading and boundary conditions. &lt;/p&gt;&#xA;" OwnerUserId="2291" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-07-16T14:15:33.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3570" PostTypeId="1" CreationDate="2015-07-14T16:00:48.677" Score="7" ViewCount="782" Body="&lt;p&gt;Post-tensioned prestressed concrete beams suffer from a multitude of losses. When jacked, they suffer friction losses and their tension diagram is, for example (to simplify the drawing, I'm using the linearized version of the friction loss equation here. It also goes without saying this is all horribly out of scale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VowRC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram obviously represents a beam prestressed from both ends. Also, the cable's path is clearly non-symmetric, with the right side probably being excessively curved, while the left is probably far smoother. The dotted lines simply demostrate what the digram would be if the cable were prestressed from only one end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the cable is then anchored to the concrete, the anchorage slips into the concrete, causing losses. These losses, however, are limited by friction to an area around the anchorage since the cable has by this time already been grouted. We can define the point where anchorage slip losses end as $X$. An anchorage slip of $\delta$ is by definition equal to&#xA;$$&#xA;\delta = \int\limits_0^X\Delta\epsilon_p(x) dx = \dfrac{1}{E_p}\int\limits_0^X\Delta\sigma_p(x) dx\\&#xA;\therefore A_pE_p\delta = \int\limits_0^X\Delta F(x) dx&#xA;$$&#xA;where $x=X$ is defined as the point where the anchorage slippage losses end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We therefore have a relation between the constant $A_pE_p\delta$ and the area between the force diagram before and after these losses. All we need to do now therefore is find $X$ such that the equality is satisfied. For simple cable paths there are some analytical methods, but they aren't relevant to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the slippage losses are small enough, we just need to calculate the losses for each side, such that $A = B = A_pE_p\delta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dexe4.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That case, however, is quite simple. From now on, I am less sure of whether what I'm saying is correct, so please correct me if I'm mistaken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, the losses are such that one side interferes with the other, it is easier to find an area $A+B+C = 2A_pE_p\delta$, where $A=B$, point $D$ is defined as the point where the losses first encountered each other and $C$ is simply a uniform loss along the entire cable that occurs once point $D$ is reached:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wwy0M.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is: &lt;strong&gt;is this a correct calculation of the anchorage slip losses in case the anchorages are released simultaneously? Is there a simpler (but generic) method?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The other question is &lt;strong&gt;what if the anchorage slippage losses are not simultaneous?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the left side we may find $X$ and reach the following result, where the area $A = A_pE_p\delta$:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PkqDY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the right side I'd consider calculating the losses on the friction diagram considering that the cable was only pulled from the right side, so we get the area $B = A_pE_p\delta$:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/knN0d.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then throw that on top of the results for the left side and get the following, where $C$ is the area overlapping with $A$ and $D$ is the part of $B$ after the end of $A$ ($C$ is contained within $D$ which is a part of $B$):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/59yfX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;strong&gt;what now?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should the area $D$ be transposed to accompany the force profile after the left anchor settled?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/T9U6p.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? Is there an easier (but generic) method to solve this case as well? Throughout this, I've made use of the assumption that the friction (and therefore anchorage slip) losses are linear, which they are not. Other than the mathematical headache of dealing with exponential equations, is there anything here that simply won't work using the losses' true exponential form?&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2015-07-14T18:06:39.883" LastActivityDate="2017-01-02T10:24:58.460" Title="(Non)simultaneous prestressing anchorage slip losses" Tags="|civil-engineering|structural-engineering|concrete|prestressed-concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3571" PostTypeId="1" AcceptedAnswerId="3577" CreationDate="2015-07-14T17:41:09.513" Score="9" ViewCount="1602" Body="&lt;p&gt;The common description for a continous PID-controller is written like this:&#xA;$$y(t)=K_p⋅e(t)+K_i\int_0^t e(τ)dτ+K_d\dfrac{de(t)}{dt}$$&#xA;The best value of the constants $K_p$, $K_i$ and $K_d$ for a given controlled system will depend on its time constant(s), be it a $\text{PT}_1$ system, or $\text{PT}_2$ system, etc. ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you do if the time constant of such a system is variable. Lets say, it varies between $T_a$ and $T_b$ ($T_a &amp;lt; T_b$). How do you design the PID-constants?&lt;/p&gt;&#xA;" OwnerUserId="46" LastActivityDate="2015-12-02T12:42:44.480" Title="How do you set up a PID-Control if the time constants of the controlled system are variable?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3572" PostTypeId="1" CreationDate="2015-07-14T19:52:38.747" Score="3" ViewCount="632" Body="&lt;p&gt;I have a mini-lathe with a 500W 3/4 HP variable speed motor, that is run by a controller at two different gearings. However, even at the LOW gearing I find the torque to be a little inadequate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to upgrade this motor to brushless and get more torque/HP?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uzAiX.jpg&quot; alt=&quot;MM7x14 motor&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="282" LastEditDate="2015-07-19T21:40:40.897" LastActivityDate="2015-07-19T21:40:40.897" Title="Upgrading to a brushless motor" Tags="|mechanical-engineering|motors|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3573" PostTypeId="2" ParentId="3565" CreationDate="2015-07-14T20:57:31.250" Score="-1" Body="&lt;p&gt;Short:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aren't they great ? :-) !!!!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From below - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot; ... it is logical to expect them to typically act in the ethereal beautiful whimsical way that they do. ...&quot; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Observers &quot;answer&quot; - no claims for certain correctness made, but 'seems logical':&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=OhE4n5x9fuE&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;The video cited&lt;/strong&gt;&lt;/a&gt; shows much faster action than many of his earlier creations. Without going back and looking at earlier designs the impression given is that he has increased active wind collection area to increase available power and energy levels and that operation speed is approaching &quot;reasonably fast&quot; for the construction methods and materials used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, I imagine that 'slow' better suits his wishes, but, the operation is essentially series asynchronous ripple-through switching, with lots of potential for slop and hysteresis, leading to non trivial minimum forces being needed before a given joint or lever will perform its action or reverse a prior action.&lt;br&gt;&#xA;With force levels varying above and below and never vastly more than the magnitude that is necessary to 'make something happen' it is logical to expect them to typically act in the ethereal beautiful whimsical way that they do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if speed of action was an aim then it would be possible to develop wind powered devices that specifically target the points above that lead to slow action. While it would be immensely hard to achieve consistent repetition rates of say sub-second order, it should be possible to apply available energy in a manner which significantly increases speed of operation.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These devices are essentially &quot;wind mill driven&quot; withe the aerodynamic input surfaces distributed across the structure.   By increasing drive surfaces to obtain more power and energy input (related but not identical) and directing power so that force &amp;amp; energy:losses on a given motion member were high, much 'crisper', if not exactly lightning fast, action would be achievable.&lt;br&gt;&#xA;But, then, where's the fun? :-)&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-07-14T21:10:30.690" LastActivityDate="2015-07-14T21:10:30.690" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3574" PostTypeId="1" AcceptedAnswerId="3575" CreationDate="2015-07-15T01:58:10.700" Score="3" ViewCount="9213" Body="&lt;p&gt;I am solving the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A Pipe 200 mm outside diameter and 20 m length is covered with a layer&#xA;  of 70 mm thick insulation having thermal conductivity of 0.05 W/m·K&#xA;  and a thermal conductance of 10 W/m&lt;sup&gt;2&lt;/sup&gt;·K at the outer surface. If the&#xA;  temperature of the pipe is 350 °C and the ambient temperature is 15&#xA;  °C, calculate the external surface temperature of the lagging.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have calculated the thermal resistances as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ R_{pipe} = \frac{1}{A \times h_o} = \frac{1}{ (\pi \times 200^2 \times (1/10^3)^2 \times ()} = 3.183098862  }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ R_{insulation} = \frac{1}{2\pi KL} = \frac{1}{ (2\pi \times 0.05 \times 20)}   = 0.04776312933 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat transfer is supposed to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ Q = \frac{\triangle T}{ R_{total}} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean by &quot;temperature of the lagging&quot;? The change in temperature from the outside and inside already given.&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="368" LastEditDate="2015-07-15T16:54:54.280" LastActivityDate="2022-10-11T11:11:15.830" Title="Calculating the temperature of the lagging around an insulated pipe" Tags="|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3575" PostTypeId="2" ParentId="3574" CreationDate="2015-07-15T06:55:48.260" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What does it mean by &quot;temperature of the lagging&quot; ?&#xA;   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Air's comment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The lagging is not really the insulation layer itself but the cladding around the insulation layer. The insulation has no real structural properties to speak of; lagging not only holds the insulation in place and protects it from the environment (including insects and rodents), it can also provide a lower-emissivity surface for the insulation layer, further reducing heat transfer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;however, since there is no given $k$ (thermal conductivity) value for the lagging I'll assume that the required temperature is the outer surface temperature of the insulation layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt; &#xA;Your $R_{pipe}$ term should be the resistance due to convection heat transfer from insulation surface to the ambient air:&#xA; &#xA;$$ R_{pipe} = \frac{1}{A_o  h_o} =  \frac{1}{\pi D_{ins} L h_o} = \frac{1}{\pi *  0.17 * 2 * 20 *  10} = 0.00468\ \text{°C/Watt}$$&#xA;and since the thermal insulation is a cylinder with inner and outer radius of $r_1$ and $r_2$ respectively, its thermal resistance should be calculated as follows:&#xA; &#xA;$${ R_{ins} = \frac{ln(\frac{r_2}{r_1})}{2\pi kL} = \frac{ln(\frac{170\ mm}{100\ mm})}{ (2\pi \times 0.05 \times 20)}   = 0.0844\ \text{°C/Watt} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q = \frac{\triangle T}{ R_{total}} = \frac{350 - 15}{ 0.00468+0.0844} = 3760.664 \text{ Watt}$$&#xA; &#xA;finally temperature of outer surface of insulation:&#xA; &#xA;$$ \triangle T = Q\times  R_{pipe} = 3760.664* 0.00468=17.6\ \mathrm{°C} = T_{outer} - T_{ambient}$$&#xA;  &#xA;$$T_{outer} = T_{ambient} + \triangle T = 15 + 17.6 =  32.6\ \mathrm{°C}$$&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2015-07-15T17:46:35.443" LastActivityDate="2015-07-15T17:46:35.443" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3577" PostTypeId="2" ParentId="3571" CreationDate="2015-07-15T12:54:08.697" Score="8" Body="&lt;p&gt;One way would be to implement some form of adaptive control. If your range of time constants is small and known, you could use something called &lt;a href=&quot;http://www.controleng.com/search/search-single-display/back-to-basics-how-gain-scheduling-works/52cd5aac6a456dfb6b1352494e4dd169.html&quot; rel=&quot;nofollow&quot;&gt;&quot;gain scheduling&quot;&lt;/a&gt;&#xA;where you determine before hand all the time constants you'll be dealing with (hopefully it is finite) and use if/then logic to define P I and D. It can be challenging to make sure you have covered enough variability to ensure stability and performance through the range. A good success story for gain scheduling is the Chinook helicopter. It can be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have a feasible prediction for what the time constants will be, you could look into using Model Reference Adaptive Control(MRAC). In this control scheme you have a reference model (your ideal system) with your chosen PID controller. The MRAC minimizes the error between what the plant is actually doing and what your reference model is doing. In this way you force your changing plant to act like your LTI model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could try using Model Identification Adaptive Controller (MIAC). Here the control scheme does system identification in real time and uses an update law for your controller. This one requires the most advanced skill of the three ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your system is changing time constants over time, it is no longer LTI. This means you need to either do gain scheduling (pretty easy if you know the range of time constants) or system identification with update law for your PID.&lt;/p&gt;&#xA;" OwnerUserId="1174" LastEditorUserId="1174" LastEditDate="2015-12-02T12:42:44.480" LastActivityDate="2015-12-02T12:42:44.480" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3578" PostTypeId="1" CreationDate="2015-07-15T15:05:18.850" Score="2" ViewCount="1932" Body="&lt;p&gt;I know that a gear reduction would decrease the RPM if the motor ran at constant speed, but in reality its speed is a function of the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first tried turning a wheel with a direct drive system, but the motor started getting warm (burning off the lubricant inside) and was spinning much slower than without a load because it didn't have the torque required. I know that a gear reduction would cause less stress (torque) on the motor, which in turn, would allow it to [possibly] go faster. I am essentially looking for the highest speed I can get, so how would I determine the proper gear reduction ratio to use? &lt;/p&gt;&#xA;" OwnerUserId="2164" LastEditorUserId="272" LastEditDate="2015-07-15T16:53:02.553" LastActivityDate="2021-09-03T18:15:11.857" Title="How do you calculate the best gear reduction for max speed?" Tags="|gears|motors|torque|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3579" PostTypeId="2" ParentId="2683" CreationDate="2015-07-15T15:10:28.767" Score="2" Body="&lt;p&gt;The problem is relatively simple if the supports have infinite stiffness. The way to then solve it conceptually is to apply the principle of superposition to: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) system with a support at $C$, applied force at $B$, and no constraint at $A$ $\rightarrow$ get deflection $\delta_{A1}$ at $A$; &#xA;(b) system with force $F_A$ at $A$, reaction force at $C$, and no constraint at $B$ $\rightarrow$ get deflection $\delta_{A2} = k{F_A}$ of $A$ as a function of $F_A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you must have: $\delta_{A1} + \delta_{A2} = 0$ for the superposition, so you can solve for $F_A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For real world problems with finite elasticity I concur with &lt;strong&gt;mark_r&lt;/strong&gt; - consult Wikipedia and Timoshenko!&lt;/p&gt;&#xA;" OwnerUserId="2220" LastActivityDate="2015-07-15T15:10:28.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3580" PostTypeId="2" ParentId="3578" CreationDate="2015-07-15T16:17:03.493" Score="1" Body="&lt;p&gt;What you really need to do is measure the power required to turn the wheel at various RPMs.  Then, match the power output of your motor to this data.  So, for example, if your motor generates 1 HP, see what RPM value corresponds to 1 HP of required power (maybe leave yourself some margin and use 0.9 HP instead), then set your gear ratio accordingly to get these RPMs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard part is, of course, measuring the power at different RPMs.  You may have to do this by trial and error.  You already know that direct drive (1:1 ratio) is too fast, so maybe try something like 3:2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring the rotational speed of the wheel is easy with a non-contact optical tachometer: &lt;img src=&quot;https://i.stack.imgur.com/GkQiG.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;I bought one a few years ago for less than $50.  The RPMs that the motor is rated to is usually stamped on the motor somewhere.  Once you measure the maximum RPMs that you can turn the wheel at (based on the power of your motor), the drive ratio is just the rated motor RPM divided by the measured wheel RPM.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-07-15T19:41:22.643" LastActivityDate="2015-07-15T19:41:22.643" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3581" PostTypeId="2" ParentId="370" CreationDate="2015-07-15T16:53:40.583" Score="3" Body="&lt;p&gt;Since you want to know what happens with a load applied to the corner of the desk, I'm going to simplify this question into two dimensions, assuming that the leg on that corner resists the load alone. Considering the fact that the rigidity of the steel members is orders of magnitude larger than that of the wooden desktop, this probably isn't too far from the truth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also going to assume that the desk is made of magical materials that don't have self-weight and that the desk is otherwise empty of other loads, just to keep things simple. Also, as others have mentioned, this is effectively impossible to do without some knowledge of statics. I can't give an entire lesson here, but I'm going to explain things as best I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure you effectively have is the following (removing the tail end of the desk after the foot, which is irrelevant, and the diagonal at the base of the foot, which just complicates matters and doesn't actually change the relevant internal stresses):&#xA;&lt;img src=&quot;https://i.stack.imgur.com/G3b8v.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular case can actually be resolved by hand, so here goes:&#xA;The load at the very edge of the table is $300\text{lb}$ and is $12\text{in} = 1\text{ft}$ from the diagonal. That means that the beam has to withstand a bending moment of $M=300\cdot1=300\text{ft-lb}$ and a shear force equal to the applied load of $Q=-300\text{lb}$ (negative because it's pointing down).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we are at the point where the diagonal starts to help the horizontal beam, so we need to figure out how much force goes to each of them. For this, we have to look a bit ahead and notice that the horizontal beam meets the column at another pinned joint (those &quot;balls&quot; in the figure). These joints allow parts to rotate relative to each other, which (and this is something you learn in statics) means that the bending moment at that point is zero. Since there are no other external loads applied along those $20\text{in}$ (between the horizontal bar's connection with the diagonal and with the column), the shear force must be constant along that stretch. And since the shear force is the derivative of the bending moment, the moment must vary linearly. And since the diagonal is pinned (&quot;ball&quot; connection) to the horizontal, it didn't steal any of the moment. That means that the horizontal beam goes from a bending moment of 300 at the beginning of the diagonal to zero at the column. The constant shear force along that stretch is therefore equal to the tangent of that linear variation, which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q = \dfrac{300\text{ft-lb}}{20\text{in} = \frac{5}{3}ft} = 180\text{lb}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, going back to the connection between the horizontal and the diagonal, we now know that the horizontal beam went from a shear force of $-300\text{lb}$ to $+180\text{lb}$. That means that the diagonal must have applied a vertical force equal to $+480\text{lb}$ onto the horizontal. However, since the diagonal is pinned on both ends and has no external loads applied on it, it can only contain axial loads. That means that those $480\text{lb}$ are actually just a component of the force actually applied by the diagonal. The horizontal component can be easily found by the tangent and is equal to $480\cdot\frac{20}{5} = 1920\text{lb}$. The total axial force on the diagonal can be found by Pythogoras: $\sqrt{480^2+1920^2} = 1979\text{lb}$, and is of &lt;strong&gt;compression&lt;/strong&gt;. Meanwhile, the horizontal component of this force has to be restrained by the horizontal beam, which therefore suffers a &lt;strong&gt;tension&lt;/strong&gt; of $1920\text{lb}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that's left now is the column. Since the horizontal beam is suffering a tension of $1920\text{lb}$, that needs to be absorbed by the column, which transforms that tension into a shear of $1920\text{lb}$. That shear, however, is cancelled out by the connection with the diagonal, which applies the same force (but onto a different side, therefore with a different sign... &lt;strong&gt;&lt;em&gt;statics&lt;/em&gt;&lt;/strong&gt;). Between those points, however, the shear is alive and well. And where there's shear, there's bending moment. A constant shear of $1920\text{lb}$ over $5\text{in}$ generates a bending moment of $1920\cdot\frac{5}{12}=800\text{ft-lb}$. Between the base of the column and the connection of the diagonal, there no longer is any shear, so the moment is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the horizontal beam had a shear of $+180\text{lb}$ which gets transmitted to the column as an axial &lt;strong&gt;tension&lt;/strong&gt; of equal value (that part of the column is being stretched, not squished!). However, after the connection with the diagonal, which also dumps its horizontal component of $-480\text{lb}$ (it was positive at the top because it pointed up. Here it points down, so it is negative). Therefore, between the base and the diagonal, the column suffers a compression of $300\text{lb}$, which makes sense since that part of the column would have to withstand the entire external load which was applied at the edge of the table. If its compression weren't equal to that applied load, something would be wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, you end up with a structure undergoing the following (click to expand):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wRkvY.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wRkvY.gif&quot; alt=&quot;internal forces&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, knowing the internal forces is not sufficient to know whether your desk will support it. That, however, is highly dependent on where you live and which codes apply (and I'm sure desks don't have to follow structural codes, but I'm sure there's some relevant code) and can't adequately be answered here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, for tension and shear there's usually little mystery to it. For tension, divide the tensile force by the cross-section area and compare that stress to the steel's strength (the weakest A500 is 45ksi), with some safety factor (allowable stress design often uses 60% of the steel's strength). For shear, divide the shear force by the &quot;shear area&quot;, which in your case is equal to the area of the &quot;vertical&quot; sides of the cross-sections. This gives you the shear stress, which should be compared to the steel's strength (allowable stress design uses 40% of the tensile strength).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bending and compression, however, are more complicated due to the risk of buckling and need to be done by the relevant codes. If one ignores buckling (one &lt;strong&gt;really&lt;/strong&gt; shouldn't), then it's just a matter of getting the relevant stress and comparing it to the strength again. For compression, that's the same as for tension. For bending, divide the bending moment by the elastic modulus to get the maximum tension/compression stress (see below) and compare against the allowable stress as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma = \dfrac{6Mh_1}{b_1h_1^3-b_2h_2^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, for what it's worth, the diagonal at the base of the foot might be relevant for buckling analysis, though if I had to guess I'd say the upper diagonal aiding the horizontal beam would be the controlling member (for buckling).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2015-07-15T18:03:31.007" LastActivityDate="2015-07-15T18:03:31.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3582" PostTypeId="1" AcceptedAnswerId="3756" CreationDate="2015-07-15T17:42:40.840" Score="4" ViewCount="755" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mOgz0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mOgz0m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho, p_o, \alpha, F_g, h, b$ are given. $F_T$ is wanted. There is a plate that can rotate around $A$. The plate measures b perpendicularly to the screen surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pressure gradient&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(z) = \rho g (2h - z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I neglect $p_0$ since it it's the same pressure on both sides of the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pressure Force&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_p = \int_{l_0}^{l_1} \int_0^b \! p \, \mathrm{d}y\mathrm{d}l$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sin(\alpha)=\dfrac{dz}{dl}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$dl=\dfrac{dz}{sin(\alpha)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_p = \int_{l_0}^{l_1} \int_0^b \! p \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I can integrate from $0$ to $h$ since I substituted $l$ with $z$ and I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_p = b \rho g \dfrac{3h^2}{2sin(\alpha)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the pressure force that works on the plate. To calculate its torque it has to be multiplied with its lever $l_a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find $l_a$ I have to regard the actual torque which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{p} \cdot l_a =  \int l \cdot p \, \mathrm{d}A$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_p \cdot l_a = \int_{l_0}^{l_1} \int_0^b \! l \cdot p \, \mathrm{d}y\mathrm{d}l$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again substitute $dl$ with $dz$ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_p \cdot l_a = \int_{l_0}^{l_1} \int_0^b \! l \cdot p \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume I can now insert insert $0$ for $l_0$ and $h$ for $l_1$ again. And at this point I'm unsure but I think since I integrate over $dz$ again I can insert $\frac{z}{sin(\alpha)}$ for $l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_p \cdot l_a = \int_{0}^{h} \int_0^b \! z \cdot p \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inserting and integrating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b \rho g \dfrac{3h^2}{2sin(\alpha)} \cdot l_a = \int_{0}^{h} \int_0^b \! \frac{z}{sin(\alpha)} \cdot \rho g (2h-z) \, \mathrm{d}y\mathrm{d}z \frac{1}{sin(\alpha)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b \rho g \dfrac{3h^2}{2sin(\alpha)} \cdot l_a = b \rho g \frac{1}{sin^2(\alpha)} \left[h^3-\frac{1}{3}h^3 \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{3h^2}{2} \cdot l_a = \frac{1}{sin(\alpha)} \left[h^3-\frac{1}{3}h^3 \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l_a = \frac{1}{sin(\alpha)} \left[\frac{2}{3}h-\frac{2}{9}h \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l_a = \frac{4h}{9sin(\alpha)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very unsure with how I substituted and inserted. I would appreciate someone to review if I did it correctly and maybe elaborate on what I got wrong if possible.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="368" LastEditDate="2015-07-30T16:04:05.273" LastActivityDate="2015-07-31T14:55:32.710" Title="Calculating pressure force on inclined area" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3583" PostTypeId="2" ParentId="3417" CreationDate="2015-07-15T22:53:40.113" Score="4" Body="&lt;p&gt;Just to let you know, I solved the problem. The element stiffness matrices had an error due to a typo, actually they were not even symmetric! Moral of the story: it's good to test for symmetry when you write a function which produces an element stiffness matrix. Anyway, thanks to all for your efforts and replies. &lt;/p&gt;&#xA;" OwnerUserId="2168" LastActivityDate="2015-07-15T22:53:40.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3585" PostTypeId="1" AcceptedAnswerId="3589" CreationDate="2015-07-16T02:41:15.393" Score="1" ViewCount="6244" Body="&lt;p&gt;Given the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the energy transfer rate across a 6 in wall of firebrick with a temperature difference across the wall of 50 &amp;deg;C. The thermal conductivity of the firebrick is $0.65\ \frac{\text{Btu}}{\text{hr ft }^\circ\text{F}}$ at the temperature of interest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is 369 W/m&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I used the following approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ x = 6in =&amp;gt; 0.5ft }$$&#xA;$${ \triangle T = 50^oC =&amp;gt; 122^oF }$$&#xA;$${ k = 0.65Btu/hr-ft-F }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ \frac{Q}{A} = \frac {k\triangle T}{x} = 158.6Btu/hr*ft^2 =&amp;gt; 500W/m^2 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I did not calculate the correct answer.  Is there a step that I am missing?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorDisplayName="user16" LastEditDate="2015-07-16T16:43:19.337" LastActivityDate="2020-12-14T06:23:35.943" Title="Problem with Energy Transfer Rate Conversion" Tags="|thermodynamics|heat-transfer|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3586" PostTypeId="5" CreationDate="2015-07-16T06:02:39.603" Score="0" Body="&lt;p&gt;Geotechnical Engineering is a branch of Civil Engineering focused on the engineering properties and behavior of natural and engineered geomaterials (soils and rocks), as well as the design and construction of foundations, earth dams and levees, retaining walls, embankments, underground structures and tunnels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all construction projects involve geotechnical engineering as most structures are built on, in or of geomaterials. Additionally geotechnical engineering involves understanding soil(or rock)-structure interaction for mitigation of natural hazards such as earthquakes and landslides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more in depth description and definition can be found at the Colorado Association of Geotechnical Engineers &lt;a href=&quot;http://www.cagecolorado.org&quot; rel=&quot;nofollow&quot;&gt;website&lt;/a&gt; or at &lt;a href=&quot;http://www.whatisgeotech.org&quot; rel=&quot;nofollow&quot;&gt;what is geotech?&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2291" LastEditorUserId="33" LastEditDate="2016-02-23T03:02:11.770" LastActivityDate="2016-02-23T03:02:11.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3587" PostTypeId="4" CreationDate="2015-07-16T06:02:39.603" Score="0" Body="Geotechnical Engineering is a branch of Civil Engineering focused on the engineering properties and behavior of natural and engineered geomaterials (soils and rocks), as well as the design and construction of foundations, earth dams and levees, retaining walls, embankments, underground structures and tunnels." OwnerUserId="2291" LastEditorUserId="2291" LastEditDate="2015-07-16T11:52:09.527" LastActivityDate="2015-07-16T11:52:09.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3588" PostTypeId="1" AcceptedAnswerId="3590" CreationDate="2015-07-16T06:25:14.473" Score="1" ViewCount="2344" Body="&lt;p&gt;For an assignment, I was given:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the heat loss per linear ft from $2$ $in$ nominal pipe. ($2.375$ $in$ outside diameter covered with $1$ $in$ of an insulating material having an average thermal conductivity of $0.0375$ $Btu/hrft^oF$. Assume that the inner and outer surface temperatures of the insulation are $380^oF$ and $80^oF$ respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is $116$ $Btu/lb^oF$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I used the formula from conduction through pipes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ Q = \frac{\triangle t}{R_{total}} = \frac{\triangle t}{ \frac{ln \frac{r_2}{r_1} }{2 \times \pi \times k \times L} }  = \frac{\triangle t}{ \frac{ln \frac{d_2}{d_1} }{2 \times \pi \times k \times L} } }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ \triangle t = 380-80= 300^oF }$$&#xA;$${ d_2 = 3.375 }$$&#xA;$${ d_1 = 2 }$$&#xA;$${ k = 0.0375 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I get close, but not correct.  I calculated it to be: $135.090464$ $Btu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorDisplayName="user16" LastEditDate="2015-07-16T16:40:34.217" LastActivityDate="2024-03-17T09:49:58.767" Title="Heat Loss per Linear Ft" Tags="|heat-transfer|pipelines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3589" PostTypeId="2" ParentId="3585" CreationDate="2015-07-16T08:02:00.973" Score="1" Body="&lt;p&gt;Your C to F conversion is correct for a different question.&lt;br&gt;&#xA;The Celsius and Fahrenheit scales have their zero points offset by a factor equal to 32 F degrees and this has to be allowed for when converting from one measurement system to the other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case you wish not to convert between two &quot;scales&quot; but to express the delta difference in degrees F rather than degrees C. So the ratio 9:5 is relevant and the offset of 32 degrees F between the two scales is irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short version of answer is: Doh! :-)&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-07-16T08:02:00.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3590" PostTypeId="2" ParentId="3588" CreationDate="2015-07-16T08:18:04.870" Score="1" Body="&lt;p&gt;I get 115.706...&lt;br /&gt;&#xA;= 116&#xA;ie their answer.&lt;/p&gt;&#xA;&lt;p&gt;What you are doing wrong (in this problem and your other recent one) is going too fast and not visualising how the data given translates into the real world situation.  You are applying formulae correctly and seem to have a good understanding of what is involved to turn given parameters into correctly formulated expressions. Now all you have to do is slow down a bit and think about what you are doing. Do that and you'll do well ! :-)&lt;/p&gt;&#xA;&lt;p&gt;The pipe is described as&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2 inch nominal pipe.&#xA;(2.375 inch outside diameter covered with 1 inch of an insulating material).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;One interpretation of that would be a pipe of 2.375 inch finished OD and 1 inch thick insulation under the outer surface. That gives a 0.375 inch internal pipe ID. Doesn't sound likely.&lt;/p&gt;&#xA;&lt;p&gt;A second interpretation is a 2.375 inch ID internal pipe with 1 inch of external insulation over it for an external OD of ???&lt;/p&gt;&#xA;&lt;p&gt;A third interpretation is the one you used.&lt;/p&gt;&#xA;&lt;p&gt;I used the 2nd one and get the correct answer.&lt;br /&gt;&#xA;Look at the 2nd version above.&lt;br /&gt;&#xA;What is the OD?&lt;br /&gt;&#xA;plug in the results and see what you get.&lt;/p&gt;&#xA;&lt;p&gt;Short answer: Doh!!! :-)&lt;br /&gt;&#xA;Try to aim at no doh!s - we all manage them but they can be minimised with due care.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2024-03-17T09:49:58.767" LastActivityDate="2024-03-17T09:49:58.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3591" PostTypeId="1" AcceptedAnswerId="3592" CreationDate="2015-07-16T09:24:31.383" Score="2" ViewCount="705" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/370/how-do-i-calculate-the-forces-on-a-desk-and-its-legs&quot;&gt;This question about calculating loads on parts of a desk&lt;/a&gt; made me finally ask my first question on SE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What generally applicable standards are there for office furniture (desks, let's say), that define the loads that they must be able to survive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In aerospace, the regulations ask for all structural parts to comply with all foreseeable loads in the complete operational scenarios during the entire life of the aircraft (static loads, dynamic loads, fatigue loads, etc); and 1.25 or 1.5 times that load is the ultimate load that each structure must be able to bear (even if for a few seconds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know open source materials, and real life experiences if available.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-07-17T15:14:30.287" Title="What loads are office desks required to withstand?" Tags="|mechanical-engineering|standards|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3592" PostTypeId="2" ParentId="3591" CreationDate="2015-07-16T13:33:11.400" Score="6" Body="&lt;p&gt;In the United States at least, the design of mass-produced furniture if covered under ANSI/BIFMA (Business and Institutional Furniture Manufacturer's Association) &lt;a href=&quot;https://www.bifma.org/store/ViewProduct.aspx?id=1375047&quot;&gt;X5.5&lt;/a&gt;. I don't have a copy of the standard to give you specific values, but in broad strokes the requirements for consumer goods like this (where the consequences of failure are much less than in an airplane or a building) codes focus less on design and calculation and more on testing of a physical sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important aspect for furniture (not unlike airplanes) is that in addition to handling a maximum load, it needs to be able to handle many cycles of loading and unloading. Because of this, the design may well be governed more by the number of cycles the desk needs to be tested to than the maximum weight it needs to support one time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another major distinction is that in the US, manufacturers are not required by law to conform to the standard. Unlike airplanes, where the FAA has the authority to require conformance with applicable standards, furniture is not centrally regulated. As a way of controlling quality though, many large purchasers like schools or corporations will require conformance on all products that they buy, causing some manufacturers to develop products that conform. The difference of course is that it is perfectly legal (and common) to sell a desk that has not been tested by the standard.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-07-16T13:33:11.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3597" PostTypeId="1" CreationDate="2015-07-17T14:02:51.010" Score="2" ViewCount="344" Body="&lt;p&gt;I am working on a project and I am defining the torque specs. I used the initial torque spec as a baseline and then went out on the floor and tested it using multiple torque wrenches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The components I am testing are &lt;a href=&quot;http://www.semikron.com/download/assets/gecont/325.pdf&quot; rel=&quot;nofollow&quot;&gt;thyristor/diode modules&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It calls for 45 in.lb to heatsink and 26 in.lb to terminals. Even backing off with the the 15% allowable tolerance, I am below the low end of 38 in.lb to the sink and it feels like its about to shear. I am encountering this with all the components I am testing. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Are the torque values I found through empirical testing sufficient? Should I be confident in applying them to a standard? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I am assuming the quality of screw, type of material the sink is made out of, all effect how much torque should be applied?&lt;/p&gt;&#xA;" OwnerUserId="2327" LastEditorUserId="368" LastEditDate="2015-07-17T15:21:22.327" LastActivityDate="2015-07-17T15:21:22.327" Title="Accuracy of torque specs?" Tags="|mechanical-engineering|electrical-engineering|power-electronics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3598" PostTypeId="1" CreationDate="2015-07-17T15:29:09.673" Score="1" ViewCount="6448" Body="&lt;p&gt;I have a set of kitchen knives which I've owned for 18 years. These knives were sold as &quot;never need sharpening&quot;. I assumed that to be complete nonsense at the time BUT it has proved to be true! These knives have never once been sharpened in 18 years and are used every day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible - and given that it is possible, why are most knives still made to be sharpened regularly? They were not at all expensive so that can't be the reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/y6yqJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I've just found the claim from the original packaging: &quot;These Fusion Infinity knives are fused with tungsten carbide giving them an extremely hard edge. There is no need to sharpen these knives.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question becomes resolves down to &quot;Why doesn't everyone do this?&quot; None of the answers have tried to address this angle yet.&lt;/p&gt;&#xA;" OwnerUserId="2162" LastEditorUserId="2162" LastEditDate="2015-07-17T21:22:44.240" LastActivityDate="2015-07-19T20:29:45.743" Title="Knives that never need to be sharpened" Tags="|steel|metallurgy|" AnswerCount="2" CommentCount="10" ClosedDate="2015-07-19T01:34:40.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="3599" PostTypeId="2" ParentId="3598" CreationDate="2015-07-17T17:28:43.567" Score="2" Body="&lt;p&gt;One possible explanation to your observation is that these knives come in a special holder, which contains a sharpener at each position you can insert a knife.  Each knife is sharpened a little when it is removed or replaced in the holder.  The advertising claim only meant that you won't have to deliberately sharpen them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a set of knives like this, and this method works very well.  You can feel the grinding action when you remove or insert a knife into the holder.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-07-17T17:28:43.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3600" PostTypeId="5" CreationDate="2015-07-17T19:54:40.333" Score="0" Body="&lt;p&gt;Optics is a branch of physics describing the behaviors and properties of light (electromagnetic waves). Questions regarding the properties of light, it's interactions with matter and the instruments used to detect and deflect (or concentrate) light should be posted here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some Relevant Categories:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;geometric optics (ray theory)&lt;/li&gt;&#xA;&lt;li&gt;physical optics (diffraction &amp;amp; interference)&lt;/li&gt;&#xA;&lt;li&gt;Quantum optics (photonic behavior)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Lenses and mirrors would apply.&lt;/p&gt;&#xA;" OwnerUserId="2291" LastEditorUserId="33" LastEditDate="2016-02-23T03:00:01.587" LastActivityDate="2016-02-23T03:00:01.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3601" PostTypeId="4" CreationDate="2015-07-17T19:54:40.333" Score="0" Body="Optics is a branch of physics describing the behaviors and properties of light (electromagnetic waves). Questions regarding the properties of light, it's interactions with matter and the instruments used to detect and deflect (or concentrate) light should be posted here. " OwnerUserId="2291" LastEditorUserId="2291" LastEditDate="2015-07-17T20:57:26.620" LastActivityDate="2015-07-17T20:57:26.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3602" PostTypeId="5" CreationDate="2015-07-17T20:04:37.510" Score="0" Body="&lt;p&gt;When a system is in static equilibrium it experiences no acceleration (the net force on the system is zero). Questions regarding statics problems should be asked here.&lt;/p&gt;&#xA;" OwnerUserId="2291" LastEditorUserId="33" LastEditDate="2016-02-23T02:52:23.833" LastActivityDate="2016-02-23T02:52:23.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3603" PostTypeId="4" CreationDate="2015-07-17T20:04:37.510" Score="0" Body="Statics is branch of engineering concerned with the analysis of forces and moments on physical systems in static equilibrium." OwnerUserId="2291" LastEditorUserId="2291" LastEditDate="2015-07-17T20:56:47.763" LastActivityDate="2015-07-17T20:56:47.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3604" PostTypeId="2" ParentId="3598" CreationDate="2015-07-17T20:50:07.873" Score="4" Body="&lt;p&gt;I've found a couple things about these knives (and &quot;stay-sharp&quot; knives in general).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the knives you have appear to be treated with tungsten carbide on the edges per &lt;a href=&quot;http://www.dotmaison.com/fusionknives_p_55821.html&quot; rel=&quot;nofollow&quot;&gt;this website&lt;/a&gt; and &lt;a href=&quot;http://www.britishblades.com/forums/showthread.php?138996-Richardson-Sheffield-Fusion-Knives-Sharpening&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;.  Since TC is incredibly hard, it dulls very slowly, hence why it is used for heavy-duty metal cutting operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the edge of the blade probably has micro-serrations that act like a tiny saw, effectively sawing through the cut material when you drag the knife through it (see &lt;a href=&quot;http://www.bladeforums.com/forums/showthread.php/196014-quot-Never-Sharpen-quot-kitchen-knives&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).  Realistically, this is what happens with a &quot;properly&quot; sharpened knife anyway at the microscopic level, though the serrations you'll find on a &quot;stay-sharp&quot; knife are relatively much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should also note that is it possible to sharpen those knives you have, but you'll probably need something like a diamond plate to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In regards to your edited question: the tungsten carbide edge, though durable, will eventually dull and it is a pain to properly sharpen again. Chefs will send their knives out for professional sharpening every so often, and I would imagine a service would charge extra for doing TC (this is a guess).  Otherwise, between sharpenings, a chef will want to just swipe his knives on a honing steel a few times between uses to touch the edge up (this is not actually sharpening, but is merely straightening out the burr on the edge).  The honing steel a chef uses is harder than the knife blade, so it is able to straighten the burr without being damaged.  With TC, since it is harder than the honing steel, it would scratch and gouge a honing steel and have no effect on the knife's edge.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-07-19T20:29:45.743" LastActivityDate="2015-07-19T20:29:45.743" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3605" PostTypeId="1" AcceptedAnswerId="3630" CreationDate="2015-07-17T21:56:42.583" Score="1" ViewCount="525" Body="&lt;p&gt;If I have a flat span of steel material,  3.0m x 150mm wide and 5mm thick, and apply an increasing downward force to the middle, at some point it will fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the profile of that piece of steel was T shaped instead of a flat span, it is much stronger.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the Safe Working Load (SWL) for different profiles calculatable by a specific formula?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming so, otherwise how could you design things like cranes, does the same formula apply to different materials e.g. aluminium v steel and different thicknesses of the same material?&lt;/p&gt;&#xA;" OwnerUserId="2331" LastEditorUserId="2246" LastEditDate="2015-07-20T14:34:14.310" LastActivityDate="2015-07-21T09:14:55.833" Title="How much strength does adding a T piece add?" Tags="|mechanical-engineering|materials|structures|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3606" PostTypeId="1" AcceptedAnswerId="3608" CreationDate="2015-07-18T00:11:36.530" Score="1" ViewCount="9836" Body="&lt;p&gt;I am solving the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A liquid-to-liquid counterflow heat exchanger is used to heat a cold fluid from 120 °F to 310 °F. Assuming that the hot fluid enters at 500 °F and leaves at 400 °F, calculate the log mean temperature difference for the heat exchanger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The given answer is 232 °F&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ LMTD = \frac{\triangle max - \triangle min }{ ln( \frac {\triangle max}{\triangle min} ) } }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ \triangle max = 500-120 = 380 }$$&#xA;$${ \triangle min = 400-310= 90}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plug them all together and I got 201.3383331.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? &lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="368" LastEditDate="2015-07-18T22:47:57.513" LastActivityDate="2021-01-19T14:37:24.627" Title="Log Mean Temperature Difference of a Counterflow Heat Exchanger" Tags="|heat-transfer|heat-exchanger|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3607" PostTypeId="2" ParentId="3605" CreationDate="2015-07-18T04:30:08.110" Score="1" Body="&lt;p&gt;One can determine the section properties of a shape by using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallel_axis_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;parallel axis theorem&lt;/a&gt;, as is taught in an elementary mechanics of materials course.  I have found the &lt;a href=&quot;http://www.amesweb.info/SectionalPropertiesTabs/SectionalPropertiesTbeam.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;following website&lt;/a&gt; that will calculate the section properties of a T-shape when one inserts the proper input.  Note that this website also lists the formulations behind the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yours could possibly have flange buckling issues given the slenderness of the flanges, but it's hard to tell without running any calculations.  The following snippet from the American Institute of Steel Construction (AISC) 14th Edition &lt;em&gt;Steel Construction Manual&lt;/em&gt; specifies the limiting b/t ratio that separated nonslender and slender compression elements.  This may not be applicable depending in your jurisdiction since you are in Australia.  However, the &lt;a href=&quot;http://steel.org.au/key-issues/compliance/standards-and-design/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Australian Steel Institute&lt;/a&gt; should have a similar provision codified in its code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XQoS0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulations for section properties of the shape are material-independent, so long as the material can be considered isotropic, homogenous, and elastic (steel and aluminum are, concrete is not).&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-07-19T20:58:46.493" LastActivityDate="2015-07-19T20:58:46.493" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3608" PostTypeId="2" ParentId="3606" CreationDate="2015-07-18T04:55:07.180" Score="4" Body="&lt;p&gt;$\triangle max$ and $\triangle min$ are not defined as maximum and minimum temperature differences in a heat exchanger. Quoting from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The LMTD is a logarithmic average of the &lt;strong&gt;&lt;em&gt;temperature difference between the hot and cold feeds at each end&lt;/em&gt;&lt;/strong&gt; of the double pipe exchanger.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/35kev.png&quot; alt=&quot;Temperature distribution of HEX inlets and outlets&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ \triangle T_{max} = 400-120 = 280 \text{ F}}$$&#xA;$${ \triangle T_{min} = 500-310= 190\text{ F}}$$&#xA;$${ LMTD = \frac{\triangle T_{max} - \triangle T_{min} }{ ln( \frac {\triangle T_{max}}{\triangle T_{min}} ) } } = \frac{280-190}{ln( \frac {280}{190})}= 232\text{ F}$$&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2015-07-18T08:10:19.737" LastActivityDate="2015-07-18T08:10:19.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3609" PostTypeId="1" AcceptedAnswerId="3610" CreationDate="2015-07-18T14:44:59.970" Score="7" ViewCount="294" Body="&lt;p&gt;Pressure drop across a porous medium can be described by the Forchheimer extension to Darcy’s law:&#xA;$$-\frac{dp}{dx} = \frac{\mu}{K_1}\cdot v + \frac{\rho}{K_2}\cdot v^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K_1$ and $K_2$ are permeability coefficients that depends on medium geometry (Pitz-Paal&#xA;et al., 1996).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just came across this paragraph in a paper studying the flow stability in porous media in solar volumetric receivers, which I couldn't digest at all:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the flow through a porous sample, the mass flow density is determined by the pressure difference between the two sides of the sample. &lt;strong&gt;&lt;em&gt;The pressure drop is produced by a blower. Instability occurs when a pressure drop causes different mass flow densities&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What blower? and what makes pressure drop cause different mass flow? (Keeping in mind previous equations)&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2015-09-30T17:28:56.053" LastActivityDate="2015-09-30T17:28:56.053" Title="What is &quot;flow instability&quot; in porous media?" Tags="|mechanical-engineering|fluid-mechanics|porous-medium|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3610" PostTypeId="2" ParentId="3609" CreationDate="2015-07-18T17:30:25.827" Score="4" Body="&lt;p&gt;So, I've only been able to find the &lt;a href=&quot;https://www.researchgate.net/publication/224988031_Theoretical_and_Numerical_Investigation_of_Flow_Stability_in_Porous_Materials_Applied_as_Volumetric_Solar_Receivers&quot; rel=&quot;nofollow&quot;&gt;abstract of the paper&lt;/a&gt; but it seems to outline the basic concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The device in question is forcing air through a porous ceramic medium in order to dissipate the heat that the medium absorbs from solar radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the blower. Some sort of device is necessary to cause the flow to occur (it's not going to happen spontaneously). In this case the authors have used a blower (substantially similar to a fan), in order to either force air towards one side of the medium or draw air away from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Instability&quot; rel=&quot;nofollow&quot;&gt;Instability&lt;/a&gt; can have have somewhat different meaning depending on context, but generally refers to a system with a tendency to deviate from a steady (not varying in time) solution. A classic example is an inverted pendulum. When perfectly vertical, the pendulum has 0 force and should remain in position (steady). However, any small deviation results in forces that drive the pendulum away from that steady state (instability). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on a &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0038092X06000314#&quot; rel=&quot;nofollow&quot;&gt;later paper&lt;/a&gt; by that group, they found that their porous flow system, which &quot;should&quot; exhibit a single, steady uniform velocity across the surface, became unstable when a small part of the medium was heated.  Specifically, they found that if the system at steady state was heated in one spot, the flow velocity in that region would drop, reducing the cooling effect in that region and allowing it to be heated further by the solar flux. The hot spot did not dissipate, allowing the system to return to the uniform, steady state.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2015-07-18T17:30:25.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3611" PostTypeId="2" ParentId="3609" CreationDate="2015-07-18T18:09:55.267" Score="1" Body="&lt;p&gt;In the unstable sense, the reactor will become a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fluidized_bed_reactor&quot; rel=&quot;nofollow&quot;&gt;fludizied bed reactor&lt;/a&gt;.  i.e. the different flow densities will cause some spots to flow more than others.  This means that &quot;wide&quot; openings, with more fluid through them, will get &quot;wider&quot;, and narrow opening, with less fluid through them won't open at all, causing them to get utilized even less until they are ignored altogether.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all of these cases, the original equations do not apply.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-07-18T18:09:55.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3612" PostTypeId="1" AcceptedAnswerId="3613" CreationDate="2015-07-18T18:23:49.867" Score="2" ViewCount="87" Body="&lt;p&gt;I work for a small engineering and manufacturing company. To date our production process for each product has been defined in each critical aspect by the design engineer, then handed to the production department and rarely looked at again by engineering. Many noncritical aspects are undefined and left to the variable judgement of the assembly workers. I'm considering assigning an engineer to the task of analyzing and documenting the entire production chain of a product for consistency and quality purposes, and to look for cost saving improvements. What discipline of engineering would this be considered, if any?&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2015-07-19T12:41:19.233" Title="What engineering discipline deals with ensuring that a manufacturing and testing process is clearly defined and followed?" Tags="|documentation|procedure|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3613" PostTypeId="2" ParentId="3612" CreationDate="2015-07-18T19:17:36.170" Score="2" Body="&lt;p&gt;Most engineering discipline are capable of defining manufacturing and test process. It is the management function to ensure the processes are followed.  Generally depending on the size and structure of the organization most defining  manufacturing and testing process are mostly handled by manufacturing test and process engineering teams. If a &lt;a href=&quot;http://asq.org/index.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quality&lt;/a&gt; organization exists this team gets heavily involved in ensuring the implementation and monitoring of processes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the event of the quality related issue the quality organization gets involved in root cause and analysis of the quality related issue. In event of issues with the manufacturing process the changes are made the to process through the use of a change management process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also periodic quality audit are established to ensure if the manufacturing and and test process are followed per the definition. This is an proactive measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question and response is loosely tied to one of your previous questions &lt;a href=&quot;https://engineering.stackexchange.com/questions/2264/how-can-i-arrange-my-eco-system-to-enforce-the-principle-of-least-privilege/2267#2267&quot;&gt;How can I arrange my ECO system to enforce the principle of least privilege?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are statically process control, six sigma - continuous improvement that can address some of the above mentioned issues.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Quality Engineer with direction from management as well as significant support from manufacturing and design engineering can address the this issue.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://asq.org/manufacturing/why-quality/overview.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quality In Manufacturing&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/3315/how-to-identify-low-quality-electronic-components-early&quot;&gt;How to identify low quality electronic components early?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-07-19T12:41:19.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3614" PostTypeId="1" CreationDate="2015-07-19T07:07:25.803" Score="0" ViewCount="4097" Body="&lt;p&gt;Is there an electric circuit that can multiply the input current? if yes, what is that called, and how to make it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My input is DC from a small DC motor which I use as a generator.&lt;br&gt;&#xA;My input current is 300 milliamps and I need 1200 milliamps as output. &#xA;Is there some way to increase the output to 1200 milliamps ?&lt;/p&gt;&#xA;" OwnerDisplayName="user2338" LastEditorUserId="399" LastEditDate="2015-07-23T11:12:40.077" LastActivityDate="2016-12-24T00:37:12.097" Title="Is there current multiplier circuit?" Tags="|electrical-engineering|circuit-design|circuits|" AnswerCount="4" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3616" PostTypeId="1" CreationDate="2015-07-19T11:39:14.037" Score="6" ViewCount="35439" Body="&lt;p&gt;I have a submersible pump and it's clearly stated in the user manual that it should not be operated without water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what will happen if it's not adhered to? The obvious answer is that it will malfunction. But I want to know what causes the malfunction to occur. I am trying to use it for some other purpose that does not require it to be submerged.&lt;/p&gt;&#xA;" OwnerUserId="2340" LastEditorUserId="110" LastEditDate="2015-07-19T12:07:35.860" LastActivityDate="2019-12-27T09:48:55.257" Title="What will happen if I run a submersible pump without water?" Tags="|mechanical-engineering|fluid-mechanics|pumps|turbomachinery|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3617" PostTypeId="1" AcceptedAnswerId="3622" CreationDate="2015-07-19T11:51:35.880" Score="3" ViewCount="3006" Body="&lt;p&gt;I have some tomato plants on my balcony (4th floor in a 7 story building). The balcony is a recessed one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qamnt.jpg&quot; alt=&quot;Photo of recessed balkonies&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The building has lightning protection on top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To support them, I put a 2 m long aluminum rod into each pot. At the tip of the rods I put some isolating tape, so that it covers the tip of the rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will these rods attract lightning?&lt;/p&gt;&#xA;" OwnerUserId="2341" LastEditorUserId="110" LastEditDate="2015-07-19T12:37:16.463" LastActivityDate="2016-11-17T00:18:31.047" Title="Will lightning strike an aluminum rod in a pot on my balcony?" Tags="|electrical-engineering|building-design|safety|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3618" PostTypeId="2" ParentId="3616" CreationDate="2015-07-19T12:13:31.067" Score="3" Body="&lt;p&gt;The pump can overheat and the electric motor can burn out.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-07-19T12:13:31.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3619" PostTypeId="2" ParentId="3617" CreationDate="2015-07-19T13:22:22.187" Score="3" Body="&lt;p&gt;The paths of lightning can not be predicted with certainty, so it becomes a question of probability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd say that your rods are quite unlikely to be hit by lightning, considering that you are 3 floors down from the top, and there appear to be other metal structure between you and the sky.  I'm assuming the railings in your picture are metal.  If lightning were to hit the side of the building at these balconies, then I'd expect the railings to be hit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is absurd to think that a few mils of insulation at the tips of the rods will make any difference.  If you're really this worried about it, why not use wooden rods instead of aluminum?&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-07-19T13:22:22.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3620" PostTypeId="2" ParentId="3616" CreationDate="2015-07-19T13:31:06.620" Score="9" Body="&lt;p&gt;The motor will spin much faster than it does normally when loaded by the pump moving water.  This higher speed may wear out something much quicker than during normal operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system should be able to handle the higher speed for a few seconds, but most likely it was not designed to operate at that speed long term.  The extra robustness probably wasn't built into the friction points to be able to handle it, especially when there is no water flowing and thereby removing heat.  Obvious failure points include the packing that seals around the shaft to keep the water out of the motor, the bearings, and possibly the motor brushes if it has any.  The system may also not be balanced for the high speed, putting additional stress onto the bearings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that is NOT a likely failure cause is the motor overheating.  Since it is spinning faster, it will draw less current, so the heating in the windings will be less.  Unless the water flow is designed to cool the motor, this is not the issue.  The ones I've seen simply have a motor sitting atop a pump, with no special water flow around the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with any device, when you violate the specs, you can't be surprised when it stops working, perhaps in spectacular ways.  If the specs say the unit is intended to work only when water is present, then that's what it's intended to do.  You can speculate on why that might be, but ultimately you don't really known.  Running it without water is a Bad Idea.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-07-19T13:31:06.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3621" PostTypeId="2" ParentId="3614" CreationDate="2015-07-19T13:39:44.093" Score="1" Body="&lt;p&gt;Your specifications are very loose, so yes, there are circuits and even passive devices that can produce a higher current out than in.  Such a circuit is called a &lt;i&gt;amplifier&lt;/i&gt;, and a &lt;i&gt;transformer&lt;/i&gt; is a example of such a passive device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how to make either of these, that is beyond the scope of this question.  There should be much information on these and other devices out there.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-07-19T13:39:44.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3622" PostTypeId="2" ParentId="3617" CreationDate="2015-07-19T14:01:11.770" Score="8" Body="&lt;p&gt;&lt;a href=&quot;https://what-if.xkcd.com/16/&quot; rel=&quot;nofollow noreferrer&quot;&gt;xkcd&lt;/a&gt; has a fantastic description of how lightning works. &lt;a href=&quot;https://www.youtube.com/watch?v=6MUYsIjTKvk&quot; rel=&quot;nofollow noreferrer&quot;&gt;This video&lt;/a&gt; is also quite good (especially as of 0:56, but the rest is worth watching as well) since it shows lightning in super slow motion, so you can see it &quot;searching&quot; for a path down to earth. The important thing here is to learn that lightning searches for a path in steps. Xkcd mentions 60-meter steps and the video mentions 50 yards (at 1:00), which for our purposes are close enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that for your rods to be struck by lightning, they need to be the path of least resistance within 60&amp;nbsp;m (or whatever) of the step leader (the searching tip of the lightning). For this to happen, it's pretty safe to assume that the step leader would basically have to be on (or very close to) the same horizontal plane as your rods and probably around 60&amp;nbsp;m away. This is because if it were coming from above, then the building's lightning rod would take care of it. If it were coming at something of an angle, then it would probably strike one of the floors above you (any exterior electrical sockets would do the trick or maybe even the railings as @Olin mentions in &lt;a href=&quot;https://engineering.stackexchange.com/a/3619/1832&quot;&gt;his answer&lt;/a&gt;). And if it were on the same horizontal plane but very close, then once again, there's probably a better route on one of the floors (maybe yours) through electrical sockets or railings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically, out of an infinity of possible points where the step leader may be, only a very, very small subset of those points would really risk your rods getting struck by lightning. So you're probably safe. Or, as Olin points out, just use wooden rods.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-11-17T00:18:31.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3623" PostTypeId="1" AcceptedAnswerId="5953" CreationDate="2015-07-19T07:33:37.257" Score="2" ViewCount="178" Body="&lt;p&gt;Is it possible to use an IR sensor to differentiate accurately between carbon fiber and a human hand kept on the carbon fiber surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to move the sensor from one end of the carbon fiber object (a rectangular surface) to the other and be able to detect a significant difference in reflected intensity when the sensor passes over the hand. Will this work? If not, what are alternative sensors that I could use? The sensor will be approximately 10 cm above the carbon fiber surface.&lt;/p&gt;&#xA;" OwnerUserId="2344" OwnerDisplayName="Rustin" LastEditorUserId="9" LastEditDate="2015-07-20T13:50:02.960" LastActivityDate="2015-10-29T01:26:07.233" Title="IR sensor with carbon fibre" Tags="|materials|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3624" PostTypeId="1" AcceptedAnswerId="3625" CreationDate="2015-07-19T20:37:25.677" Score="2" ViewCount="597" Body="&lt;p&gt;Referring to von Guiricke's early experiment to hold vacuum using &lt;a href=&quot;https://en.wikipedia.org/wiki/Magdeburg_hemispheres&quot; rel=&quot;nofollow&quot;&gt;Magdeburg hemispheres&lt;/a&gt;, what is the best known shape to hold say 10 atmosphere vacuum (external pressure) using &lt;em&gt;any material or material combination&lt;/em&gt; without buckling its walls? I.e., using the least amount of material so that the vacuumed vessel is as lightweight as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Material choice relaxed in question as better materials are now available compared to Magdeburg test time.&lt;/p&gt;&#xA;" OwnerUserId="1270" LastEditorUserId="1270" LastEditDate="2015-07-20T18:11:58.937" LastActivityDate="2015-07-20T18:11:58.937" Title="What is the most lightweight geometry for a vacuum holding vessel?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3625" PostTypeId="2" ParentId="3624" CreationDate="2015-07-19T21:46:44.583" Score="4" Body="&lt;p&gt;When I design pressure vessels to hold vacuum, here are the preferences I usually give to customers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Spherical is always the best in terms of minimizing the thickness, but not the most economical in therms of amount it can hold or for being friendly to the manufacturer&lt;/li&gt;&#xA;&lt;li&gt;Ellipsoidal (2:1) heads on a shell whose length is equal to the diameter is second best.  It comes out a bit oblate compared to the sphere (1.5:1), but per given packing area holds plenty of volume.  The ellipsoidal heads work out to equal the thickness requirements of the shell, but is not structurally as efficient as the sphere.&lt;/li&gt;&#xA;&lt;li&gt;Avoid Flat areas - especially flat bottoms.  If you need it to be flat on the bottom, weld a skirt to the bottom.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-07-19T21:46:44.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3626" PostTypeId="2" ParentId="3566" CreationDate="2015-07-19T22:14:35.113" Score="0" Body="&lt;p&gt;The transmitter to receiver coupling is similar to how a transformer works.  for a ideal transformer, whatever impedance is connected to the secondary appears as the impedance of the primary divided by the square of the turns ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, imagine a 7 V AC source and a 1:2 stepup transformer.  If you connect a 3 &amp;Omega; resistor directly to the 7 V, it will draw 2.3 A.  If you instead put the transformer between the source and the resistor, the resistor now sees 14 V applied to it.  It will draw 4.7 a from the secondary, which means 9.3 A must be going into the primary, and the resistance seen by the 7 V source is 3/4 &amp;Omega;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to get back to your question, surely you can see that a circuit could detect this 750 m&amp;Omega; on it's 7 V supply if designed to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coupling between transmitter and receiver is not like a ideal transformer, but changes in impedance presented to the secondary (the coil in the receiver) still cause apparent changes in the impedance of the primary (the coil in the transmitter).  It's a small matter of engineering to properly detect the pulses of different impedance the receiver is putting on its receiving coil, turn them into 1s and 0s, and decode this digital stream to get the data payload encoded onto it by the receiver.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-07-19T22:14:35.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3627" PostTypeId="1" AcceptedAnswerId="3631" CreationDate="2015-07-20T04:43:09.243" Score="27" ViewCount="3876" Body="&lt;p&gt;Making extremely accurate machines seems like a chicken and egg problem.  How do you make one if you don't already have one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, how was the first &lt;a href=&quot;https://en.wikipedia.org/wiki/Indexing_head&quot;&gt;indexing head&lt;/a&gt; made without an indexing head to make the indexing plate?  How was the first lathe made without a lathe to make the spindle?  Is it an iterative process where you start with something hand-made, and each machine can make one slightly better than the last?  I suppose that &lt;a href=&quot;https://www.mathsisfun.com/geometry/constructions.html&quot;&gt;geometric constructions&lt;/a&gt; are the source of accuracy of some primitive machines.  But how do you get from there to something like the precision screw required for an accurate lathe or micrometer?&lt;/p&gt;&#xA;" OwnerUserId="2140" LastEditorDisplayName="user16" LastEditDate="2015-07-23T12:29:51.663" LastActivityDate="2018-11-07T15:47:31.260" Title="How are increasingly accurate machines made?" Tags="|mechanical-engineering|engineering-history|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3628" PostTypeId="2" ParentId="44" CreationDate="2015-07-20T06:30:08.690" Score="-1" Body="&lt;p&gt;For steam installation on steam mains, thermodynamic type steam traps are the best suited. If sized correctly, they will solve the issue of water hammer. At the same time they will waste a minimum amount of steam during operation. &#xA;You can get more information here- &lt;a href=&quot;https://www.forbesmarshall.com/fm_micro/news_room.aspx?Id=seg&amp;amp;nid=176&quot; rel=&quot;nofollow&quot;&gt;Steam Trap Guide&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2346" LastActivityDate="2015-07-20T06:30:08.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3630" PostTypeId="2" ParentId="3605" CreationDate="2015-07-20T10:42:02.193" Score="4" Body="&lt;p&gt;In answer to the question &lt;strong&gt;Is the &quot;SWL&quot; for different profiles calculatable by a specific formula?&lt;/strong&gt; - Yes, although it’s not one formula, there are lots of different things to consider, all with sets of formulae depending on the exact situation. Calculating Safe Working Loads (SWL) (or, indeed, starting from a desired working load and calculating the required structural member) is a large part of what structural engineers do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A beam can “fail” in many different ways. Most likely for your given case would be exceeding the plastic capacity of the beam in bending, such that a hinge forms in the middle of the beam, causing significant deformation which then pulls the beam off one of its supports.&#xA;The plastic capacity of a metal beam is relatively easy to calculate, you can use the formula&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;s = M/Z&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where s is the yield stress of the material, M is the applied bending moment, and Z is the plastic section modulus of your chosen profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;s is material dependant, so changing the material will change this. You can get various grades of steel, with yields from 230 to 260 MPa. Aluminium can be supplied with yields from 15 to 20 MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M depends on the applied loading. In your case it is due to a force F applied in the middle of a 3.0m span, which gives a maximum moment of 0.75F. (Calculating bending moments is a whole separate topic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z depends on your profile. &lt;a href=&quot;https://en.wikipedia.org/wiki/Section_modulus#Plastic_section_modulus&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; gives some formulae for some basic sections. (Calculating section modulae is a big topic too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula will give your theoretical maximum load, and not your SWL. The SWL also takes into account factors of safety, which are used to ensure that the applied load doesn't exceed the capacity. There are design standards which specify these factors (which standards apply depend on the situation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your headline question &lt;strong&gt;“How much strength does adding a T piece add?”&lt;/strong&gt; - it all depends on the dimensions of the T piece. And, as alluded to by grfrazee, as increase the depth of the T you run the risk of the failure method changing to lateral torsional buckling.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-07-20T10:42:02.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3631" PostTypeId="2" ParentId="3627" CreationDate="2015-07-20T15:02:54.207" Score="8" Body="&lt;p&gt;Before the extremely accurate machines were made, they first had extremely accurate measuring devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll use the &quot;first lathe&quot; as an example. The bed of the lathe needs to be very flat, as the rest of the machine can be calibrated with it. One way it can be made that way is by putting the not-straight bed up against a much flatter surface plate, and scraping off the high spots on the lathe bed until they meet perfectly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flat surface plate must have come from someplace, and from what I recall they make it by taking at least 3 somewhat flat plates and lapping them with each other. Eventually all the plates will end up very flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a flat enough surface you can make make extremely straight measuring instruments, pair it up with a simple surface gauge and you can start producing items with extremely parallel surfaces. Rolling a round bar on the plate can check how straight it is, and the surface gage can detect taper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now with the straight round bar you can use that to align the axis of rotation to the bed of the lathe, thus creating the first rudimentary lathe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accurate lathe bearings were probably done the babbitt bearings way where you hold the rotating shaft/part in place and cast the bearing from molten lead around the bar, then you forcefully twist the bar creating a virtually zero clearance bearing. Fairly accurate if done properly and it can deal with slight misalignment of the hole drilled in the lathe head and the axis of rotation.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastEditorUserId="1459" LastEditDate="2015-07-21T12:58:21.850" LastActivityDate="2015-07-21T12:58:21.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3632" PostTypeId="1" CreationDate="2015-07-20T15:57:22.793" Score="4" ViewCount="28834" Body="&lt;p&gt;I am planning to make a portable charger for my smartphone. My input current is 300 milliamps and I am not sure if that will be enough to charge my phone's battery. What is the minimum current required to charge a typical smartphone battery?&lt;/p&gt;&#xA;" OwnerDisplayName="user2338" LastEditorUserId="368" LastEditDate="2015-07-20T16:32:32.903" LastActivityDate="2020-09-28T11:33:19.440" Title="What is the minimum current required to charge a smartphone battery?" Tags="|electrical-engineering|battery|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3633" PostTypeId="1" CreationDate="2015-07-20T18:27:45.877" Score="3" ViewCount="2742" Body="&lt;p&gt;With a normal spring, you compress it using a linear force to store energy and then it decompresses and releases the energy, again in a form of linear force. Is there a mechanical mechanism that stores energy by rotating force and releases energy by rotating force?  It doesn't have to be spring operated, but I think it's the only way to work, with springs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want it to work for many many rotations (to store energy) and to produce many many rotations! what's the best mechanism for this?&lt;/p&gt;&#xA;" OwnerUserId="2353" LastEditorUserId="272" LastEditDate="2015-07-21T12:10:05.560" LastActivityDate="2018-04-16T14:14:29.370" Title="Rotary energy storage mechanism" Tags="|mechanical-engineering|springs|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3634" PostTypeId="2" ParentId="245" CreationDate="2015-07-20T18:46:59.277" Score="0" Body="&lt;p&gt;You have not mentioned what load you intend to apply for so much of bending / twisting deformation. Fiberglass (S-glass) has around 2 percent elongation and is used in Elastica/ pole vaulting/ Bow Archery/ automobile propeller shaft and similar large deformation applications, is a material choice that can be considered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since twisting is large,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1) Unsymmetrical section for eccentric placement of shear center for a channel beam&lt;/li&gt;&#xA;&lt;li&gt;2) Flexible elastomer interleaving between &lt;em&gt;composite&lt;/em&gt; plies or suitable resin system choice&#xA;could be indicated.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The analysis requires consideration of in-plane/ bending forces occurring together. Interaction formulae involving EI/GJ are also used for structural dimensioning. With ANSYS you should use large deformation analysis.&lt;/p&gt;&#xA;" OwnerUserId="1270" LastEditorUserId="1270" LastEditDate="2015-07-20T21:06:31.177" LastActivityDate="2015-07-20T21:06:31.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3635" PostTypeId="2" ParentId="3633" CreationDate="2015-07-20T19:21:06.663" Score="5" Body="&lt;p&gt;Both a torsion bar and a &quot;coil&quot; spring do what you want.  The torsion bar is intended for rotation of only a fraction of a circle, and generally has a high spring constant.  The coil spring can be designed for a number of rotations, generally with a lower spring constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at any old windup watch or clock and most likely the energy storage mechanism is a coil spring.  Some old clocks are powered by dropping weights, but these are usually not &quot;wound&quot; to add the energy.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-07-20T19:21:06.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3639" PostTypeId="1" AcceptedAnswerId="3641" CreationDate="2015-07-21T01:47:14.253" Score="0" ViewCount="200" Body="&lt;p&gt;I am doing B.E. in Mechatronics Engineering. I am due to take a subject called 'Sensors and instrumentation'. So far I only have the syllabus for the course. I decided to do some reading in advance of the course starting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The syllabus refers to &quot;low pass sensors&quot; and &quot;band pass sensors&quot;.&lt;br&gt;&#xA;When I web search for these terms I find many mentions of low pass &lt;em&gt;filters&lt;/em&gt; and band pass &lt;em&gt;filters&lt;/em&gt;, but nowhere on the Net are &quot;low pass &lt;em&gt;sensors&lt;/em&gt;&quot; and &quot;band pass &lt;em&gt;sensors&lt;/em&gt;&quot; mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the terms &quot;low pass sensors&quot; and &quot;band pass sensors&quot; have their own meanings and, if so, what are they?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have been googling for the past five days for these two words but none of the results satisfactorily explained the meaning of 'low pass sensor' or 'band pass sensor'.&lt;/p&gt;&#xA;" OwnerDisplayName="user2338" LastEditorUserId="68" LastEditDate="2015-07-23T19:59:37.460" LastActivityDate="2015-07-23T22:39:18.493" Title="What is meant by the terms &quot;low pass sensor&quot; and &quot;band pass sensor&quot; in a course syllabus?" Tags="|electrical-engineering|sensors|power-electronics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3640" PostTypeId="2" ParentId="3632" CreationDate="2015-07-21T02:58:55.127" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;My input current is 300 milliamps and I am not sure if that will be enough to charge my phone's battery.*&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It will probably charge acceptably.&lt;br /&gt;&#xA;This varies with product but in most cases modern cellphones and other products which use 1 or 2 cell LiIon (Lithium Ion) batteries will charge from sources that supply less than maximum rated charge current.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Rather than considering the literal minimum current which would &amp;quot;just&amp;quot; cause charging to occur, I decided to take your question to mean more like &amp;quot;Can I charge smartphone batteries at some small but not trivial rate without significant problems&amp;quot;. ie I did not think you wanted to know if you could charge them at 10 or 100 or even 1000 microamps. The batteries themselves would probably charge at currents somewhat above the self discharge rate, but as that is probably under 100 uA&lt;/em&gt;*, it does not have a lot of relevance in most charging scenarios.&lt;br /&gt;&#xA;**eg 1400 mAh x 10-% say pa = 140 mAh.&lt;br /&gt;&#xA;140 mAh/8765 hours/year ~= 16 uA self discharge rate. Even at 50% self discharge per year a 1400 mAh battery self discharge rate is under 100 uA.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.microchip.com/stellent/groups/designcenter_sg/documents/market_communication/en027883.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The reference cited by Air&lt;/a&gt; is an excellent one but may be hard to follow if you re not technically experienced. The basic explanation below should tell you the minimum that is needed to meet your requirement.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;When there is 'more than enough' charging energy available most devices with LiIon (Lithium Ion) batteries charge at a rate which is set by the charger, not by the battery. The rate used is usually the maximum allowed by the battery manufacturer.  As this part of the charge cycle is carried out under constant current conditions it is usually referred to with the abbreviation CC mode (for &amp;quot;constant current). The manufacturer's maximum allowed rate is usually based on battery capacity.   Typically the maximum allowed charge rate is the &amp;quot;C/1&amp;quot; rate.&lt;br /&gt;&#xA;C = charge rate where mA of charge = battery mAh numerically.&lt;br /&gt;&#xA;So for eg a 1200 mAh battery, C = 1200 mA.&lt;/p&gt;&#xA;&lt;p&gt;Almost all modern cellphones and many other LiIon powered devices use the &amp;quot;USB&amp;quot; 5 Volt input charging standard. When the available charge current is less than the maximum some devices will not charge, but the majority of &amp;quot;5 volt input&amp;quot; devices will accept whatever current is available.&lt;/p&gt;&#xA;&lt;p&gt;A typical smartphone battery has a capacity of ABOUT 1500 mAh. This would have C = 1500 mA = max charge current. The phone will charge the battery either at C if ample energy is available or at the lower available rate until a predefined battery voltage is reached (usually 4.2V). It will then usually change to a constant voltage mode and the current will decrease with time under battery chemistry control.&lt;/p&gt;&#xA;&lt;p&gt;When the battery changes from CC (constant current) mode to CV (constant voltage mode)  the battery is usually about 70% to 80% &amp;quot;full&amp;quot; (technically 70% to 80% SOC = State of charge). The exact amount varies with situation - and if charging has been at much less than C (say 300 mA instead of 1200 mA in his case) the SOC (state of charge) will be higher than when charged at C. So, when charging at say 25% of max rate it will take less than 4 times as long to achieve a full charge as the slow final CV rate will have less capacity to input.&lt;/p&gt;&#xA;&lt;p&gt;It is &lt;strong&gt;absolutely vital&lt;/strong&gt;  when making a charger that you do not exceed Vbattery = Vmax - usually 4.2 Volts. If you are providing a 5V feed to the existing charger circuitry (usually inside the phone) then it will handle this. If you are connecting directly to the battery you should either use an 'off the shelf' LiIon charger IC or learn quite a lot about what is required to keep your battery alive when charging.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Added:&lt;/p&gt;&#xA;&lt;p&gt;A low capacity charger may charge a battery to a HIGHER final charge state and somewhat shorten battery life. This is because a charger will usually terminate when Ibatteruy is say at I = C/2 or C/4 or maybe C/10 for road warrior level charging. If the charger expects say a 1000 mAh battery then C/4 = 250 mA.&lt;br /&gt;&#xA;But if charging a 2000 mAh battery the 250 mA represents C/8 which is perhaps 2-5% higher charge level.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;###References:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Battery University&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://batteryuniversity.com/learn/article/charging_lithium_ion_batteries&quot; rel=&quot;nofollow noreferrer&quot;&gt;Charging Lithium Ion batteries&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://batteryuniversity.com/learn/article/how_to_prolong_lithium_based_batteries&quot; rel=&quot;nofollow noreferrer&quot;&gt;Prolonging LiIon battery lifetimes&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://batteryuniversity.com/learn/article/lithium_based_batteries&quot; rel=&quot;nofollow noreferrer&quot;&gt;LiIon basics &amp;amp; background&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://batteryuniversity.com/learn/article/the_li_polymer_battery_substance_or_hype&quot; rel=&quot;nofollow noreferrer&quot;&gt;LiIon / LiPo - comparisons &lt;/a&gt; useful as much for user discussion over some years.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Wikipedia&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lithium-ion_battery&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lithium Ion batteries&lt;/a&gt; - useful for general information and large number of linked references.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2020-09-28T11:33:19.440" LastActivityDate="2020-09-28T11:33:19.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="3641" PostTypeId="2" ParentId="3639" CreationDate="2015-07-21T07:56:52.750" Score="6" Body="&lt;p&gt;You are combining two concepts.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;A sensor&lt;/strong&gt; is any device that responds to a desired signal in some manner to produce an output that is used to detect the presence of the desired signal or to allow some desired attribute of the signal to be measured - such as its amplitude, frequency, rate of change, relative phase, etc. Even colour, temperature, dimensions, current/voltage/power ... .  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The terms &lt;strong&gt;low-pass&lt;/strong&gt; and &lt;strong&gt;band-pass&lt;/strong&gt; relate to the processing of a sensed signal so that only a portion of the signal is used to provide the sensor output. It may seem counter intuitive to ignore part of a signal, but this is useful when the signal may be accompanied noise or by other interfering or unwanted signals. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the above context, a low-pass sensor is a sensor, intended to detect or measure a specific signal or condition, that accepts the lower frequency components of the signal but rejects input above a selected frequency. A band-pass sensor is one which processes only signals with frequencies within a given range of frequencies, and rejects signal higher or lower in frequency than the selected band of frequencies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As examples:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A low pass sound sensor may be a microphone plus amplifier plus a filter set to exclude frequencies above say 4000 Hz. Such a sensor would respond to speech signals and the fundamental sounds of many musical instruments but would not be affected by high frequency noise signals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bandpass doppler radar may have the range of frequencies that it acts on limited to a range of Doppler frequencies corresponding to velocities from the speed of a person walking slowly up to about 100 mph. It would therefore tend to ignore signals from eg a door being very slowly closed (V &amp;lt;&amp;lt; walking speed) or to signals from reflections from aircraft (V &gt; 100 mph). &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-07-21T08:07:16.380" LastActivityDate="2015-07-21T08:07:16.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3643" PostTypeId="2" ParentId="3605" CreationDate="2015-07-21T09:14:55.833" Score="-1" Body="&lt;p&gt;There are better ways of &lt;em&gt;uniformly&lt;/em&gt; stiffening a plate e.g., by the &lt;em&gt;Sandwich&lt;/em&gt; principle. A somewhat compromised version is the &lt;em&gt;Isogrid&lt;/em&gt;, these are the distributed I-beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both methods, effective bending rigidity is increased by making plate central parts hollow and so lighter in weight. Stresses are reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please google under these names. In place of 5 mm is a thick plate, you can use 1/8 inch plate and tack weld on isogrid cells or at least orthogonal ribs/grills of same 1/8 inch thickness, say 3 inch cell sizes. The webs are like bottle crates. Stiffness could go up by 4/5 times. The extra fabrication is worth it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If alternate material choice is ok, even a 1-inch plywood sheet can be bonded with 1/16 thick steel sheets usind a good adhesive on either side for reduced deformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When improving designs, detail is important.&lt;/p&gt;&#xA;" OwnerUserId="1270" LastActivityDate="2015-07-21T09:14:55.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3644" PostTypeId="1" AcceptedAnswerId="3645" CreationDate="2015-07-21T09:48:56.713" Score="15" ViewCount="3138" Body="&lt;p&gt;What is this tiny rotating piece of hardware that I found inside my Samsung GE1202? It has E9C written in blue on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H4kSJm.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H4kSJm.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2360" LastEditorUserId="1586" LastEditDate="2015-07-28T00:24:16.493" LastActivityDate="2015-12-13T14:02:18.333" Title="What is this small rotating part from a cell phone?" Tags="|electrical-engineering|motors|consumer-electronics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3645" PostTypeId="2" ParentId="3644" CreationDate="2015-07-21T10:08:05.090" Score="20" Body="&lt;p&gt;These types of motor configurations are used in the cellphone's vibrate/buzzer function. Here is a Youtube video demonstrating the vibrate function and two other references:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=WWgN20Xx-5A&quot; rel=&quot;noreferrer&quot;&gt;How it works:Cell phone vibration motor?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.strivemobile.com/iphone-5-5s-vibrator-motor-replacement?gclid=CK2Xi7n868YCFVYTHwodULwDFA&quot; rel=&quot;noreferrer&quot;&gt;iPhone 5 / 5S Vibrator Motor Replacement&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.precisionmicrodrives.com/vibrating-vibrator-vibration-motors&quot; rel=&quot;noreferrer&quot;&gt;Vibration Motors by Precision Microdrives&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T6xx1.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wg3ur.gif&quot; alt=&quot;diagram of motor with eccentric rotating mass&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are called Eccentric Rotating Mass (ERM) vibration motors. This type of motor technology is also used in haptic feedback technology.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Rotating_unbalance&quot; rel=&quot;noreferrer&quot;&gt;Rotating unbalance&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.precisionmicrodrives.com/haptics-haptic-feedback-vibration-alerting&quot; rel=&quot;noreferrer&quot;&gt;Haptic Feedback&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=8&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CEwQFjAHahUKEwi3hqvPgOzGAhUCGh4KHTUuDcU&amp;amp;url=http%3A%2F%2Fwww.ti.com%2Flit%2Fpdf%2Fsszb151&amp;amp;ei=hB2uVbe0LYK0eLXctKgM&amp;amp;usg=AFQjCNEPZ6ZhM3YADI4Gz6fW79eeCbNKvA&amp;amp;sig2=OYt3Pcxu6ndfrV04FFbVWQ&quot; rel=&quot;noreferrer&quot;&gt;Haptics Solutions for ERM and LRA Actuators&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eetimes.com/document.asp?doc_id=1278948&quot; rel=&quot;noreferrer&quot;&gt;Haptics technology: picking up good vibrations&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-12-13T14:02:18.333" LastActivityDate="2015-12-13T14:02:18.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3646" PostTypeId="2" ParentId="3633" CreationDate="2015-07-21T12:05:29.143" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Mainspring&quot; rel=&quot;nofollow noreferrer&quot;&gt;mainsprings&lt;/a&gt; of mechanical timepieces provide an excellent example of how you can accomplish what you want.  They are a type of &lt;a href=&quot;https://en.wikipedia.org/wiki/Torsion_spring&quot; rel=&quot;nofollow noreferrer&quot;&gt;torsion spring&lt;/a&gt; made from a metal ribbon which can typically be wound 6 or 7 full turns.  The images below show what some of these look like when unwound and wound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Watch_mainsprings.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IXnuV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Alarm_clock_mainspring.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R6WdR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-07-21T12:42:16.543" LastActivityDate="2015-07-21T12:42:16.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3648" PostTypeId="1" AcceptedAnswerId="3653" CreationDate="2015-07-21T14:43:24.177" Score="0" ViewCount="2195" Body="&lt;p&gt;Would it be possible to power a small LED light using the headset jack of a regular smartphone, and if so, how would you connect the LED light with an audio jack. How would such an implement look like. I'm not very versed in electrical terminology so simplified explanations are appreciated. &lt;/p&gt;&#xA;" OwnerUserId="1613" LastActivityDate="2018-12-16T19:24:17.297" Title="Powering LED light using smartphone headset jack" Tags="|electrical-engineering|audio-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3649" PostTypeId="1" AcceptedAnswerId="4190" CreationDate="2015-07-21T15:41:57.677" Score="9" ViewCount="360" Body="&lt;p&gt;I need to design an automatic shifting system, for a &lt;a href=&quot;http://students.sae.org/cds/formulaseries/about.htm&quot; rel=&quot;nofollow&quot;&gt;FSAE car&lt;/a&gt;. Clutch and shifts are electronically engaged and disengaged using a &lt;a href=&quot;http://pi-innovo.com/&quot; rel=&quot;nofollow&quot;&gt;Pi Innovo&lt;/a&gt; engine control unit (ECU). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue I am worried about is downshifts, I would like to rev-match, for quick, clean downshifts and clutch life. &#xA;There is no possibility of my team using an electronic throttle, therefore I can not &quot;blip&quot; the throttle electronically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea is when a downshift is needed, I send a signal to FULLY disengage the clutch electronically, then when the engine RPM reaches what is consider matched, I send a signal to shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would allowing my engine to rev while the clutch is fully disengaged damage the clutch? Also, how is important is rev-matching? We already replace our clutch after a week of racing, would not rev-matching be that much of an issue?&lt;/p&gt;&#xA;" OwnerUserId="2165" LastEditorUserId="2246" LastEditDate="2015-07-21T20:49:50.397" LastActivityDate="2015-08-27T07:48:10.857" Title="Automatic Shifting without Electronic throttle" Tags="|electrical-engineering|automotive-engineering|control-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3650" PostTypeId="2" ParentId="3627" CreationDate="2015-07-21T17:31:26.640" Score="6" Body="&lt;p&gt;This was an actual thought process that was spoken out loud by Feynman in his famous paper &lt;a href=&quot;https://en.wikipedia.org/wiki/There%27s_Plenty_of_Room_at_the_Bottom&quot; rel=&quot;noreferrer&quot;&gt;There's plenty of room at the bottom&lt;/a&gt;, in 1959.&lt;/p&gt;&#xA;&lt;p&gt;Here's the related part:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A hundred tiny hands&lt;/p&gt;&#xA;&lt;p&gt;When I make my first set of slave &amp;quot;hands&amp;quot; at one-fourth scale, I am going to make ten sets. I make ten sets of &amp;quot;hands,&amp;quot; and I wire them to my original levers so they each do exactly the same thing at the same time in parallel. Now, when I am making my new devices one-quarter again as small, I let each one manufacture ten copies, so that I would have a hundred &amp;quot;hands&amp;quot; at the 1/16th size.&lt;/p&gt;&#xA;&lt;p&gt;Where am I going to put the million lathes that I am going to have? Why, there is nothing to it; the volume is much less than that of even one full-scale lathe. For instance, if I made a billion little lathes, each 1/4000 of the scale of a regular lathe, there are plenty of materials and space available because in the billion little ones there is less than 2 percent of the materials in one big lathe.&lt;/p&gt;&#xA;&lt;p&gt;It doesn't cost anything for materials, you see. So I want to build a billion tiny factories, models of each other, which are manufacturing simultaneously, drilling holes, stamping parts, and so on.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The whole paper is worth digesting completely, as it writes about several potential manufacturing processes, i.e:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Miniaturization by evaporation&lt;/li&gt;&#xA;&lt;li&gt;Rearranging the atoms&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1097" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-07-21T17:31:26.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3651" PostTypeId="2" ParentId="3644" CreationDate="2015-07-21T20:07:53.907" Score="6" Body="&lt;p&gt;That is a micro vibrating motor to produce the &quot;buzz&quot; effect from your phone. &#xA;I used much &lt;a href=&quot;http://www.chicagovibrator.com/Store/tabid/84/c/truck-vibrators/Default.aspx&quot;&gt;much larger vibrating motors&lt;/a&gt;, air driven, on the bottom of my Class 7 dump truck. Mounted under the bed, they would vibrate the bed, loosening the material so it would flow out cleanly. &lt;/p&gt;&#xA;" OwnerUserId="2368" LastEditorUserId="272" LastEditDate="2015-07-22T12:13:11.630" LastActivityDate="2015-07-22T12:13:11.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3652" PostTypeId="2" ParentId="3633" CreationDate="2015-07-21T22:47:15.070" Score="1" Body="&lt;p&gt;For &quot;many many rotations&quot;, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pneumatic_motor&quot; rel=&quot;nofollow&quot;&gt;pneumatic motor&lt;/a&gt; can act as both a compressor and motor. Spinning the motor causes air to be forced through a tube, one-way valve, and storage tank. Opening the valve allows the compressed air in the tank (potential energy) to flow back through the tube and motor, spinning it in reverse. As the number of rotations increases, the pressure increases, acting to stop the axle for a given torque. This presents a rotational limit, based on the volume of compressed gas, size of motor, size of tank, etc. For fast rotary motion this could work, but for slow motion, the pneumatic motor may &quot;leak&quot; and store little or no energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &quot;many many many rotations&quot;, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Brushed_DC_electric_motor&quot; rel=&quot;nofollow&quot;&gt;permanent magnet motor&lt;/a&gt;/generator -&gt; DC rectifier -&gt; battery (or supercapacitor) may work to store considerably more energy. Similar to the pneumatic concept, slow motion will not produce much charge. But the density of stored potential energy is much higher in batteries than it is air tanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are significant losses in both of these, so there won't be enough energy stored to return the axle to it's original starting position.&lt;/p&gt;&#xA;" OwnerUserId="1673" LastActivityDate="2015-07-21T22:47:15.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3653" PostTypeId="2" ParentId="3648" CreationDate="2015-07-21T23:03:12.403" Score="4" Body="&lt;p&gt;This is a very interesting question. I would say, &lt;strong&gt;yes and no&lt;/strong&gt;. The headphone jack is not designed for this at all, and you &lt;strong&gt;won't get much power out of it&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;a href=&quot;https://www.ifixit.com/Answers/View/97418/what%27s+the+iPhone+3.5+mm+output+jack+power+supply+output&quot; rel=&quot;nofollow&quot;&gt;preliminary research&lt;/a&gt; indicates that an iPhone 4 headphone jack may support up to &lt;strong&gt;60 mW&lt;/strong&gt; of power output. This, however would be &lt;strong&gt;AC power&lt;/strong&gt;, like sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I think you could put a &lt;a href=&quot;http://www.electronics-tutorials.ws/diode/diode_6.html&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;full bridge rectifier&lt;/strong&gt;&lt;/a&gt; in to make the signal only in the positive direction, then &lt;strong&gt;smooth the peaks&lt;/strong&gt; of the sine wave to a dc signal (approximately) &lt;strong&gt;with a capacitor&lt;/strong&gt;. The you &lt;strong&gt;have a DC source&lt;/strong&gt;. Then it's as simple as running an LED off of a battery. The thing is, you're not going to have much power. Otherwise, I would still have to say it is &lt;strong&gt;technically possible&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, to turn it on you would basically just be playing a sine wave beep. To turn it off, just stop playing the sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I ask &lt;strong&gt;what you were hoping to do&lt;/strong&gt; with this LED? I don't think you could make a very good flashlight with it, if that's what you mean.&lt;/p&gt;&#xA;" OwnerUserId="2371" LastActivityDate="2015-07-21T23:03:12.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3655" PostTypeId="2" ParentId="54" CreationDate="2015-07-22T09:56:27.233" Score="5" Body="&lt;p&gt;Carnegie, through their CETO device, and others have already looked at using &lt;strong&gt;wave power to directly pressurise water for reverse osmosis&lt;/strong&gt;: a wholly mechanical process rather than converting to electricity and back again (giving potential efficiency savings). Two challenges: first, there aren't many places in the world with a really huge wave resource (UK, Portugal being two that come to mind); and second, it's proven very hard to get wave machines working reliably. That's surmountable, but challenging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other significant potential development will seem counter-intuitive, and the key to unlocking it is to consider the wider system, rather than just the desalination process. That development is to &lt;strong&gt;move to lower-efficiency desalination processes&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's because lower-efficiency processes can have much lower capital costs. The advantage of that is that they can then be operated for a smaller proportion of the time, without taking a big hit on the cost per cubic metre of desalinated water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why would you want to run the desalination for a smaller proportion of the time? Because places that depend on desalinated water have a lot of sunlight. Which makes PV power cheap. But PV has a generation profile that only partly matches demand. There'll be times of insufficient power, and times of excess power. That excess power is really really cheap. And that's a great time to run desalination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a combined energy and water system that has a lot of PV and a lot of low-capex, lower-efficiency desalination can work very well. In effect, the desalinated water acts as a form of virtual storage. All electricity systems need storage somewhere in the system. For some countries, that's in the form of storage hydro. For others, it's in the form of gas-holders, coal bunkers or biomass bunkers. Those stores are pre-generation stores. In other systems, there's storage post-generation, in the form of low-grade thermal storage: when energy is going to be used as low-grade heat, it makes sense to store it in that form, as such storage is very cheap and very scalable. Similarly, storage of desalinated water is very cheap and very scalable. It acts as a time-buffer, a flexible delay mechanism, between the supply of PV electricity, and the demand for the energy service itself - in this case, desalinated water.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-07-22T17:07:49.477" LastActivityDate="2015-07-22T17:07:49.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3657" PostTypeId="1" AcceptedAnswerId="3667" CreationDate="2015-07-22T18:10:00.393" Score="0" ViewCount="1121" Body="&lt;p&gt;Searched, but didn't find this discussed here previously. I'm upgrading a hobby-level CNC router (a Shapeoko 3) to have wider belts (it came w/ 6 mm GT2 (2 mm)), I'm going up to 9 mm (otherwise same specs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ordered new pulleys and belting from SDP/SI, but when I went to install them was concerned that the end of the motor shaft (6.35 mm diameter NEMA23) is inset from the end of the pulley by almost 5 mm (4.88 mm). With the previous pulleys, they were flush with the motor end (but probably should have been slightly off to allow the belt to run true).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that too great a distance? What would be the maximum distance that the pulley could be off the shaft? What guidelines are there for this? How is it determined? (I've tried searching but haven't been able to find anything – am I not using the correct terminology?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is more important, that the belt run true or the pulley be on the shaft? There's only a couple of inches between the idlers (pairs of flanged bearings on M6 bolts) and a great deal of engagement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pulley in question is SDP/SI part # A 6A51-020DF0908.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="368" LastEditDate="2015-07-22T23:43:31.913" LastActivityDate="2015-07-23T17:04:43.640" Title="Guidelines for motor shaft length and pulley engagement" Tags="|mechanical-engineering|pulleys|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3658" PostTypeId="1" CreationDate="2015-07-23T06:47:07.893" Score="3" ViewCount="1290" Body="&lt;p&gt;In a single point cutting tool (right cut) what is the inclination angle ( lambda) in an Orthogonal rake system? I'm having difficulty in visualizing it. Please explain it in terms of ASA system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IH81J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="21" LastEditorUserId="272" LastEditDate="2015-07-23T11:52:22.970" LastActivityDate="2015-07-23T11:52:22.970" Title="What is the inclination angle in an Orthogonal rake system?" Tags="|mechanical-engineering|machining|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3659" PostTypeId="2" ParentId="3614" CreationDate="2015-07-23T11:26:31.510" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My input is DC from a small DC motor which I use as a generator.&lt;br&gt;&#xA;  My input current is 300 milliamps and I need 1200 milliamps as output. &#xA;  Is there some way to increase the output to 1200 milliamps ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With out using external energy input you can transform voltage or current but you can not increase the power level.  As power = voltage x current, that means, if voltage is increased then available current is increased, then available voltage but decrease.  That is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Power in &gt;= Power out so&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Vin x Iin &gt;= Vout x Iout&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Usually power out will be less than power in due to inefficincies in the conversion process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, imagine that your generator makes 300 ma at 6 Volts.&lt;br&gt;&#xA;Power in = 300mA x 6V = 1800 mW.&lt;br&gt;&#xA;Power out must be &amp;lt;= 1800 mW.&lt;br&gt;&#xA;For Iout = 1200 mA, Vout must be&lt;br&gt;&#xA;Vout &amp;lt;=  1800 mW / 1200 mA = 1.5V &#xA;So at 1200 mA out the voltage out is &amp;lt;= 1.5V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conversion could be carried out by a 'buck converter' which reduces output voltage and increases output current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good buck converters can be in the range 90% - 95% efficient so in this example, at 90% efficient:&lt;br&gt;&#xA;Input = 6V, 300 mA&lt;br&gt;&#xA;Iout = 1200 ma.&lt;br&gt;&#xA;VOut 6V  / 4 x 90% = 1.35V&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you want more current AND more power you need to &lt;a href=&quot;http://bit.ly/Ycannachangethelawsofphysics&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;&quot;change the laws of physics&quot;&lt;/strong&gt;&lt;/a&gt; &#xA;(Recommended :-). (Reference to 'change the laws of physics at 2m:25s but some (but not all) will find the whole 3m33s worth watching (once, anyway).)))&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-07-23T11:34:32.050" LastActivityDate="2015-07-23T11:34:32.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3661" PostTypeId="1" AcceptedAnswerId="3894" CreationDate="2015-07-23T13:47:44.550" Score="6" ViewCount="3882" Body="&lt;p&gt;Recently, tube steel manufacturers in the US have begun using the new ASTM A1085 specification for forming tube steel (a.k.a. hollow structural section or HSS) shapes, as opposed to the existing A500 specification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've read (&lt;a href=&quot;https://www.aisc.org/WorkArea/showcontent.aspx?id=33730&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.atlastube.com/astm-a1085&quot;&gt;here&lt;/a&gt;), the A1085 material spec has the following benefits over the existing A500 spec:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;tighter tolerances (no need to take reduced wall thickness for design)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;identical yield strengths for all types of tube members (as opposed to A500, in which the yield strength differs for round and rectangular shapes and depending on the grade)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;set maximum yield stress of 70 ksi (useful for seismic applications, though I am not well versed in this area)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;standard Charpy V-notch requirements corresponding AASHTO Zone 2 (I assume useful for the transportation industry for fatigue reasons)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These benefits come with a small premium - I've read 7% to 10% increased cost depending on the mill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I primarily work as a structural engineer in the nuclear industry.  Most of the work I do is in services, where we perform mostly small modifications to plants (i.e., no design of large buildings for the most part, but smaller supports, platforms, and the like).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have no need to restrict the maximum yield strength of the material and I have no fatigue concerns, is there any benefit to specifying the newer A1085 tube steel material over the existing A500 specification?  For practicing engineers in non- or light-seismic zones, have you realized any benefit using the new A1085 spec?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-07-23T13:58:03.197" LastActivityDate="2015-08-12T01:22:56.593" Title="Tube Steel - ASTM A1085 vs. A500" Tags="|structural-engineering|steel|specifications|columns|hss|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3662" PostTypeId="5" CreationDate="2015-07-23T13:54:58.247" Score="0" Body="&lt;p&gt;Hollow Structural Section (HSS) or tube steel members are structural steel members created by rolling sheet steel into rectangular, square, or circular profile shapes.  These are primarily used in the United States or other countries that follow US conventions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HSS are used primarily as structural columns, though they also find use as structural beams.  Since they are closed sections, they are inherently strong in torsional resistance and also exhibit less tendency to lateral torsional buckling as compared to wide-flange beams (&lt;a href=&quot;https://en.wikipedia.org/wiki/I-beam&quot; rel=&quot;nofollow&quot;&gt;W-shapes&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details of connections to HSS members is typically complicated due to the inside of the member being inaccessible.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="33" LastEditDate="2016-02-23T03:29:51.463" LastActivityDate="2016-02-23T03:29:51.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3663" PostTypeId="4" CreationDate="2015-07-23T13:54:58.247" Score="0" Body="Questions dealing with Hollow Structural Sections (HSS), also known as tube steel shapes. These can be round, square, or rectangular." OwnerUserId="2246" LastEditorUserId="33" LastEditDate="2016-02-23T03:29:51.463" LastActivityDate="2016-02-23T03:29:51.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3664" PostTypeId="1" AcceptedAnswerId="3666" CreationDate="2015-07-23T14:21:05.247" Score="7" ViewCount="340" Body="&lt;p&gt;I'm running an FEA on a vessel that has a &quot;splashing water&quot; effect with vibrations.  The problem I'm really coming up against is &lt;em&gt;what kind of Young's Modulus and Poisson's ratio to use for the fluids?&lt;/em&gt; - air and water.  I have been successful in using the bulk Modulus, and treat it as uncompressible - i.e. Poisson's ratio of 0.5, but I never used it for vibrational analysis, just static analysis.  &lt;strong&gt;Are these still good assumptions in a dynamic analysis?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-08-04T09:39:49.057" Title="Modelling splashing water in FEA" Tags="|mechanical-engineering|modeling|vibration|finite-element-method|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3665" PostTypeId="2" ParentId="3661" CreationDate="2015-07-23T15:09:47.203" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Set maximum yield stress of 70 ksi (useful for seismic applications, though I am not well versed in this area)&lt;/p&gt;&#xA;&lt;p&gt;If I have no need to restrict the maximum yield strength of the material and I have no fatigue concerns, is there any benefit to specifying the newer A1085 tube steel material over the existing A500 specification? For practicing engineers in non- or light-seismic zones, have you realized any benefit using the new A1085 spec?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;While I deal primarily in the realm of structural plastics - I have had experience in high seismic zones and will say a set yield strength is of a huge benefit.  The &lt;em&gt;new&lt;/em&gt; seismic code says that if the concrete floor fails first before the steel fails, that the seismic loads need to be 2.5x higher.  This will prevent the ground from ripping up under the part, possible damaging embedded piping, electrical systems, etc.  Instead, if the hold down points are considered &lt;em&gt;the weak point&lt;/em&gt; - so the structure falls over without tearing up the ground, then seismic loads are reduced drastically (A 2.5x multiplier can be HUGE).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Without this effect, I see no reason to embed a deliberate weak point into the structure.&lt;/strong&gt;  So, in light seismic zones while designing heavy storage tanks I have simply added the seismic multiplier to my anchoring design and ran with it.  In heavy seismic zones, the need for a deliberate weak point becomes obvious and that weak point needs to be carefully controlled - this material sounds like it would have a distinct advantage - 10% additional cost v. 250% extra loading.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-07-23T15:18:07.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3666" PostTypeId="2" ParentId="3664" CreationDate="2015-07-23T16:34:36.533" Score="5" Body="&lt;p&gt;This may be only a partial answer since I don't have any idea what to do for the Young's modulus of the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the paper &lt;em&gt;&lt;a href=&quot;http://authors.library.caltech.edu/48022/1/15.full.pdf&quot; rel=&quot;noreferrer&quot;&gt;Dynamic Pressures on Accelerated Fluid Containers&lt;/a&gt;&lt;/em&gt; by G. W. Housner.  This is referenced by ASCE 4-98 &lt;em&gt;Seismic Analysis of Safety-Related Nuclear Structures&lt;/em&gt;, Commentary Section C3.1, for analysis of hydrodynamic loads on tanks.  In the paper, Housner states that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3gsUS.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3gsUS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read the rest of the paper at the link above, if interested.  It seems that your assumption of incompressibility of the fluid (and correlating Poisson's ratio of 0.5) is reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might try doing some searching for computational fluid dynamics (CFD) modeling in regards to the Young's modulus question.  Other than that, I have no good suggestion where to find that information.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-07-23T16:34:36.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3667" PostTypeId="2" ParentId="3657" CreationDate="2015-07-23T17:04:43.640" Score="1" Body="&lt;p&gt;If I understand you correctly, you're worried about the radial load on the motor shaft due to belt tension when the pulley is located axially 5mm further from the motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not the loading scenario you've described is OK depends on many factors; the shaft dia, material properties of the shaft, the motor bearings, the speed of the shaft, tolerable shaft run-out, the belt tension as well as the axial location of the pulley.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The better motor suppliers will specify an allowable radial load at a specific axial location on the shaft, typically the center halfway between the motor bearing and the end of the shaft. That doesn't mean you have to load the shaft radially through that point, but the further away from the motor the radial load is applied, the less radial load is acceptable. I'm guessing that you don't have this information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most significant consequences of applying a radial load further away from the motor will be increased radial load on the motor bearings and increased shaft deflection. The increase in bearing load will reduce the life of the bearings and the deflection will cause greater shaft vibration. How significant these are really depends on your application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that you're using a typical toothed drive belt, which shouldn't require a lot of tension to function properly, so the effect &lt;em&gt;should&lt;/em&gt; be relatively small. If you're concerned, you could measure the effect with a dial indicator mounted to the motor, measuring the radial deflection of the shaft in line with the direction of the belt pull with no tension on the belt and compare that to the same measurement made with the belt properly tensioned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without more information, I would say to try it because the alternative of intentionally misaligning the belt in an attempt to keep the radial load closer to the motor is going to result in increased belt wear and unwanted friction.  &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-07-23T17:04:43.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3668" PostTypeId="1" AcceptedAnswerId="3669" CreationDate="2015-07-23T19:02:23.740" Score="4" ViewCount="3285" Body="&lt;p&gt;Is there a calculation / rule of thumb that can determine a containers rough maximum negative pressure (vacuum) from it's rated internal pressure (assuming a single layer construction so there is no additional bracing preventing expansion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example for a 50 litre stainless steel keg (cylindrical with convex end caps) with an integral 580 +/- 140 psi bursting disc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be valid to assume that the weakest point (excluding the bursting disc) is calculated to withstand 400 psi of internal pressure (minus approx 14 psi from external atmosphere), directed outward so placing the material under tension. By reversing pressure differential (My logic says the weakest point still remains the same) the weak spot would come under a compressive force and the maximum pressure under vacuum would be a function of the ratio of that particular stainless steal grades tensile:compressive strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&#xA;I just realized the bursting disc pressure is not relevant since that only indicates when a rupture would occur. In the reverse situation deformation into a flat can would occur well before then. However if instead of the bursting pressure you were to apply the 'Maximum working pressure' of 43.5 psi which presumably implies no deformation does the above apply?&lt;/p&gt;&#xA;" OwnerUserId="2201" LastEditorUserId="2246" LastEditDate="2015-07-23T22:50:51.503" LastActivityDate="2016-12-14T14:38:17.313" Title="Calculating maximum internal pressure from maximum rated external pressure?" Tags="|structural-engineering|pressure|stresses|mechanical-failure|pressure-vessel|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3669" PostTypeId="2" ParentId="3668" CreationDate="2015-07-23T19:14:01.653" Score="5" Body="&lt;p&gt;Vacuum pressure on a vessel is quite different than positive pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under positive pressure, the shell of the vessel is essentially under a uniform membrane tension due to the shell wanting to expand outward from the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under vacuum (negative) pressure, the opposite is true, where the shell is in compression.  While a cylinder is stronger in this type of compression than, say, a flat plate, one cannot directly correlate the vacuum pressure capacity with the positive pressure capacity.  This is due to the buckling failure mode of the thin shell under compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://engineering.stackexchange.com/questions/3538/would-a-structurally-sound-pressure-cooker-be-able-to-withstand-an-internal-vacu/3540#3540&quot;&gt;this recent relevant SE Question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; As an additional illustration for the buckling failure mode, I present the following image from &lt;em&gt;Buried Pipe Design&lt;/em&gt;, 3rd Ed., by Moser &amp;amp; Folkman.  The design equations presented in that text for piping are not strictly applicable to your case, but the theory is similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4QoXU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4QoXU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-07-24T01:35:16.107" LastActivityDate="2015-07-24T01:35:16.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3670" PostTypeId="2" ParentId="2966" CreationDate="2015-07-23T20:21:25.313" Score="0" Body="&lt;p&gt;I'm not an expert on air compressors or motors, but from my limited knowledge I think as you say that the centrifugal compressor would be the best for compression and a Tesla Turbine would be ideal as the motor. I think it should be possible to mount them on the same shaft but in separate airtight chambers, with some valves such that when the turbine is in operation the air is pumped out of the compressor chamber, and vice versa, so as not to cause undue resistance from the other impeller. Alternatively a clutch/grip mechanism that selects which one turns with the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a device could be considered a both-ways compressor/motor. To try to do it both ways with an impeller optimised for one of those scenarios seems like you will always be inefficient in the other.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-07-23T20:21:25.313" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3671" PostTypeId="2" ParentId="3627" CreationDate="2015-07-24T03:21:18.997" Score="0" Body="&lt;p&gt;Since many modern tools today are produced using (an earlier copy of) the &quot;same&quot; tools, &quot;How was the first one made?&quot; can be seen as a kind of &lt;a href=&quot;http://c2.com/cgi/wiki?ChickenAndEggProblem&quot; rel=&quot;nofollow&quot;&gt;chicken and egg problem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to bootstrap accurate machines from relatively coarse and crude tools is described in the book &quot;Build Your Own Metal Working Shop From Scrap&quot; by &lt;a href=&quot;http://en.wikipedia.org/wiki/David_J._Gingery&quot; rel=&quot;nofollow&quot;&gt;David J. Gingery&lt;/a&gt;, part of the &lt;a href=&quot;http://kk.org/cooltools/archives/12534&quot; rel=&quot;nofollow&quot;&gt;Build Your Own Metal Working Shop Series&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;http://web.archive.org/web/20070223174024/http://riters.com/Machine/index.cgi/Machine/Gingery&quot; rel=&quot;nofollow&quot;&gt;Gingery machines&lt;/a&gt; and &lt;a href=&quot;http://morgandemers.com/category/lathe/page/3/&quot; rel=&quot;nofollow&quot;&gt;Morgan Demers lathe project&lt;/a&gt; and &lt;a href=&quot;http://www.instructables.com/id/Build-Your-Own-metalworking-Lathe---Part-I/&quot; rel=&quot;nofollow&quot;&gt;Build Your Own (metalworking) Lathe&lt;/a&gt;. More distantly related: &lt;a href=&quot;http://c2.com/cgi/wiki?SelfAssembly&quot; rel=&quot;nofollow&quot;&gt;Self Assembly&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="721" LastActivityDate="2015-07-24T03:21:18.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3672" PostTypeId="2" ParentId="3668" CreationDate="2015-07-24T11:37:54.433" Score="1" Body="&lt;blockquote&gt;Is there a calculation / rule of thumb that can determine a containers rough maximum negative pressure (vacuum) from it's rated internal pressure (assuming a single layer construction so there is no additional bracing preventing expansion).&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clearly not, as consideration of a few different cases makes obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a mylar balloon.  It can handle some positive internal pressure, but it's ability to withstand negative pressure without collapsing is essentially 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, consider a spherical steel tank.  It can certainly take some negative pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the relationship between maximum positive and negative pressure in these two cases is quite different, making the answer to your question &quot;No&quot;.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-07-24T11:37:54.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3673" PostTypeId="2" ParentId="3668" CreationDate="2015-07-24T13:34:09.610" Score="0" Body="&lt;p&gt;I ran through once for atmospheric fiberglass tanks designed to hold specific gravity 1.2 (which corresponds to Hydrochloric acids, but can hold a wide variety of things).  These tanks were sized for New York City Seismic Loading:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A 4' diameter x 4' tall atmospheric tank can withstand 27&amp;quot; w.c. of vacuum,&lt;/p&gt;&#xA;&lt;p&gt;An 8' diameter x 8' atmospheric tank can withstand 4&amp;quot; w.c. of vacuum.&lt;/p&gt;&#xA;&lt;p&gt;A 12' diameter x 12' atmospheric tank can withstand 2&amp;quot; w.c. of vacuum.&lt;/p&gt;&#xA;&lt;p&gt;A 4' diameter x 8' tall atmospheric tank can withstand 14&amp;quot; w.c. of vacuum,&lt;/p&gt;&#xA;&lt;p&gt;An 8' diameter x 16' atmospheric tank can withstand 2.4&amp;quot; w.c. of vacuum.&lt;/p&gt;&#xA;&lt;p&gt;A 12' diameter x 24' atmospheric tank can withstand 1&amp;quot; w.c. of vacuum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Bottom Line:  While materials, thickness, etc. make the relationship between internal pressure design and external pressure design a blur, once you know what one vessel of a particular design and style can hold for both, &lt;strong&gt;it's pretty easy to scale both internal/external design parameters&lt;/strong&gt; (and even Height-to-Diameter Ratio!) and find out what another similar tank will do.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-07-24T13:34:09.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3674" PostTypeId="1" AcceptedAnswerId="3675" CreationDate="2015-07-24T14:55:40.963" Score="5" ViewCount="1578" Body="&lt;p&gt;I know that when a motor runs it generates torque and that torque can be used to do useful work. On the other hand, the motor needs strong support that absorbs the reaction torque. In our case let us assume that that support is provided by a workshop floor on which the motor is firmly attached (the workshop floor is essentially the Earth).  The Earth receives the reaction torque and being the massive object it is, it doesn't move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us imagine that we took our motor into space where there is no gravitational field.  What would happen if we tried to run the motor?  Assuming the motor is powered by a battery pack. The battery pack and its control electronics are neatly packed around the stator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the motor rotate at all?  Would the rotor and the stator rotate in opposite directions? Would there be a transfer of energy from the batteries to rotational mechanical energy?&lt;/p&gt;&#xA;" OwnerUserId="2398" LastEditorUserId="272" LastEditDate="2015-07-24T15:43:20.997" LastActivityDate="2015-07-25T08:46:48.487" Title="What would happen if we tried to run a motor in space when it is not attached to anything to provide support to it?" Tags="|electrical-engineering|motors|torque|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3675" PostTypeId="2" ParentId="3674" CreationDate="2015-07-24T15:32:16.780" Score="5" Body="&lt;p&gt;Initially &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume an ideal motor with no mechanical losses and &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Operation in a perfect vacuum.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Call the two portions rotor and stator - with all attached parts such as controller and power pack forming part of one or the other.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Start at rest relative to a &quot;fixed&quot; frame of reference. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Rotor and stator will rotate in opposite directions with the 'momentum of each portion being equal and opposite. If you manage to produce identical moment of inertia in rotor and stator they will rotate with equal and opposite rotational velocities. In the moments of inertia are in the ratio N:1 they will accelerate and, and will rotate with velocities of 1:N relative to the fixed frame of reference. The energy content in rotor and stator relative to the zero energy level initially will be equal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are non idealities such as bearing drag and eddy current or other losses they may affect rate of acceleration or final velocity depending on how the motor is controlled. A &quot;brushless dc motor&quot; whose speed is controlled by sensors or field sensing so that a fixed RPM rate is achieved will rotate so that differential rotor-stator speed is what the controller sets, and relative rotation rates relative to the fixed reference frame will relate to the relative moments of inertia as above. A motor such as a series wound &quot;universal&quot; motor whose speed is set only by losses (such as is typically used in vacuum cleaners) will increase in speed until losses from bearings, eddy current, copper losses etc equal power in. In such an environment an unloaded motor would usually easily accelerate to an immensely high speed and &quot;self-dismantle&quot;. (Even everyday vacuum cleaner motors which depend on the aerodynamic loading of their fans will usually rev to destruction if operated without the fan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rotational drive it is possible to introduce rotational realignment around the axis of rotation but not translation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A motor with offset mass relative to the axis of rotation will do interesting things but still imparts equal and opposite inertial variations to the two parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you now &quot;add air&quot; all bets are off  - you can achieve arbitrary translation and velocity in desired (or undesired :-) ) directions because you have introduced an independent medium which reactive force can be developed against. You can also arrange energy losses differentially between the two portions so that eg a rotor could be spun up and the stator braked to rest while &quot;floating&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-07-25T08:46:48.487" LastActivityDate="2015-07-25T08:46:48.487" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3676" PostTypeId="2" ParentId="3674" CreationDate="2015-07-24T15:33:11.613" Score="2" Body="&lt;p&gt;Newton's third law states that for every action, there is an equal and opposite re-action. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case you presented on earth, the stator and the rotor exert equal electromagnetic force on each other, but while the rotor is free to rotate, the stator is held in equilibrium by the bolts exerting an equal force in the opposite direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In space, all the same laws apply, but now there is no force holding the stator in place. When the equal force is applied to each, they will move in opposite directions with accelerations proportional to their relative masses, per Newton's second law: F=ma.&lt;/p&gt;&#xA;" OwnerUserId="2048" LastActivityDate="2015-07-24T15:33:11.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3677" PostTypeId="1" AcceptedAnswerId="3680" CreationDate="2015-07-24T15:35:05.160" Score="4" ViewCount="4615" Body="&lt;p&gt;This is a schematic for a plastic piece that has an oval hole on it. I understand that the Ø12 part of the measurement is for a diameter, but I do not understand what the &lt;strong&gt;&quot;X25L&quot;&lt;/strong&gt; at the end is indicating nor do I understand the &lt;strong&gt;&quot;2-&quot;&lt;/strong&gt; at the beginning. What is the meaning of those numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3MEBC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3MEBCm.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2399" LastEditorUserId="255" LastEditDate="2015-07-31T19:51:04.320" LastActivityDate="2015-07-31T19:51:04.320" Title="How does a dimension like &quot;2-Ø12X25L&quot; describe an oval hole?" Tags="|mechanical-engineering|drafting|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3678" PostTypeId="2" ParentId="3674" CreationDate="2015-07-24T15:41:38.677" Score="4" Body="&lt;p&gt;The stator and the rotor would rotate relative to each other in the same way as if one end were attached to the Earth.  If your frame of reference was attached to either the stator or the rotor, you would see the things attached to the other part rotating in the same way as if you were standing on Earth.  The only difference is that, in space, you would need to consider the conservation of angular momentum to determine how much the two portions rotate with respect to some external reference frame such as the stars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a motor to spin a flywheel is one of the ways that spacecraft &lt;a href=&quot;https://en.wikipedia.org/wiki/Attitude_control&quot; rel=&quot;nofollow&quot;&gt;rotate while in space&lt;/a&gt;.  Interestingly, the people who designed the Voyager spacecrafts neglected to account for this effect in the spinning tape recorders that were used to record data.  They ended up having to engage external thrusters every time the recorders were in use in order to maintain the correct pointing of the spacecraft.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your questions more directly:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Would the motor rotate at all? Would the rotor and the stator rotate in opposite directions?&lt;/strong&gt; Yes and yes.  The motor produces the same amount of torque between the rotor and stator (for a given load and speed) no matter what each is attached to.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Would there be a transfer of energy from the batteries to rotational mechanical energy?&lt;/strong&gt; Yes, rotational energy would be put into the object attached to the stator as well as the object attached to the rotor.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-07-24T17:16:20.400" LastActivityDate="2015-07-24T17:16:20.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3679" PostTypeId="2" ParentId="3677" CreationDate="2015-07-24T15:42:35.810" Score="1" Body="&lt;p&gt;The &lt;strong&gt;25L&lt;/strong&gt; refers to the length of the slot. It looks like you have two slotted holes, so the &lt;strong&gt;2-&lt;/strong&gt; means two separate holes. Having the entire drawing visible would help ascertain for sure if this is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, on structural (building) drawings, holes like this would be called out as &quot;12mm DIA. X 25mm SLOT, TYP,&quot; but as this is a mechanical drawing, I am not particularly up to snuff on the correct drafting standards.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-07-30T01:31:33.720" LastActivityDate="2015-07-30T01:31:33.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3680" PostTypeId="2" ParentId="3677" CreationDate="2015-07-24T15:45:19.940" Score="9" Body="&lt;p&gt;The &quot;2-&quot; refers to the fact there are two of these holes somewhere on the print. Rather than making that same note every time the hole appears, you can say &quot;2-&quot; or &quot;6-&quot; and leave it to be assumed that all similar holes are identical to the dimensioned one. (At my company, we use &quot;2x&quot; or &quot;6x&quot;, but this person may not have wanted to do this considering there's another 'x' in the note, but some proper spacing should easily clarify that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Ø12X25L is the full dimension of the hole size. I actually don't like the diameter symbol in front of the twelve, because it's not really a circle, but it's still relatively clear. In any case, the twelve is the diameter of the semicircles at each end, and thus also the width of the slot. The 25L is the length of the slot. Because it's not a circle, you do need two dimensions to determine its size. (Don't confuse this with bolt sizes, which I often see as M12x80L, where M12 is the bolt size and 80L is the actual length. Oval bolts are a bad idea generally.)&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-07-24T15:45:19.940" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="3681" PostTypeId="2" ParentId="277" CreationDate="2015-07-24T18:01:37.957" Score="1" Body="&lt;p&gt;A second option would be simply to have a small container that you fill up and cap (like a bucket with an airtight lid).  Then distribute your piping with perforations at will.  Arrange your system so the bucket is elevated enough that the soil can swap the water for gas (like a do-it-yourself water tank system):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mBGub.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mBGub.jpg&quot; alt=&quot;Aqua Globe Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So long as you make the cap on the tank airtight, water can only leave when the soil emits gas to displace the water in the tank.  Then you just fill the bucket up, cap the lid, and allow the water to dispense when the plants need it. &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-07-24T18:01:37.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3683" PostTypeId="1" CreationDate="2015-07-25T17:31:43.977" Score="4" ViewCount="431" Body="&lt;p&gt;I'm modelling a heat exchanger inside a channel. Talking about convective one:&lt;br&gt;&#xA;Geometry is simple, square of 1.5 cm but only one side exchanges heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TFrUn.png&quot; alt=&quot;[1]&quot;&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;problem&quot; is that only one side exchanges heat with environment, the others are adiabatic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I think best correlation to use is a simple internal correlation (e.g. Gnielinski), rather than an external correlation because this doesn't take into account boundary layer which can't grow up since there is the &quot;roof&quot; of the channel (characteristic length is no more a constant) making Nusselt number constant!&#xA;&lt;img src=&quot;https://i.stack.imgur.com/hep6J.png&quot; alt=&quot;[2]&quot;&gt; &#xA;but I'm not sure about this... I've never handled a problem like this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;an example of my &quot;fear&quot;:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$Re_D = 61000 $, certainly turbulent flow&lt;br&gt;&#xA;  using Gnielinski (internal flow correlation) I get $Nu_D = 125$&lt;br&gt;&#xA;  and since $L/D=3000&amp;gt;&amp;gt;60$ the most of the pipe is in &lt;em&gt;Fully Developed region&lt;/em&gt; &gt;that ensure me to &lt;strong&gt;average Nu&lt;/strong&gt; == &lt;strong&gt;local Nu&lt;/strong&gt;:&lt;br&gt;&#xA;  $Nu_{average}=Nu_D=125$  &lt;/li&gt;&#xA;  &lt;li&gt;while if I use an external correlation, at the end of the pipe, I get:&lt;br&gt;&#xA;  $Re_L = 2e+8 $, certainly turbulent flow&lt;br&gt;&#xA;  $Nu_L=0.0296*Re_L^{4/5}*Pr^{1/3} = 1.1e+5$&lt;br&gt;&#xA;  $Nu_{average}=(0.037*Re_L^{4/5}-871)*Pr^{1/3} = 134384$  &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;this is the &lt;em&gt;numerical proof&lt;/em&gt; about my doubt&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Some additional data:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;flowing fluid is air at $ T=700 [°C]$ , $p=10 [bar]$ , $m=36.56 [g/s]$ ;&lt;br&gt;&#xA;  these result in $\rho=3366.18 [g/m^3]$ , $u_{mean}=48 [m/s]$ , $\mu=4e-5 [Ns/m^2]$ , $Pr=0.734$ , $k=6.8e-2 [W/mK]$&lt;/li&gt;&#xA;  &lt;li&gt;walls are ceramic-made: adiabatic ones have infinite thickness while the only one which exchanges heat is 1 [mm] thick&lt;/li&gt;&#xA;  &lt;li&gt;channel is 50 [m] long and hydraulic diameter is 1.5 [cm]  &lt;/li&gt;&#xA;  &lt;li&gt;ambient is at constant temperature of 675 [°C], we can think to a fluid in phase-transition, so with a negligible thermal resistance, so don't worry about ambient...&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;EDIT: I want to share an online PDF which try to talk about it &lt;a href=&quot;http://kntu.ac.ir/DorsaPax/userfiles/file/Mechanical/OstadFile/dr_avami/8.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;(here)&lt;/a&gt; inter alia at pagg. 46 and 264&lt;/p&gt;&#xA;" OwnerUserId="2157" LastEditorUserId="2157" LastEditDate="2015-09-16T13:19:36.197" LastActivityDate="2015-09-16T13:19:36.197" Title="Which correlation for heat exchange in a partially adiabatic channel?" Tags="|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3684" PostTypeId="1" CreationDate="2015-07-25T17:38:15.807" Score="4" ViewCount="20" Body="&lt;p&gt;I am looking to lengthen the life of 321 stainless Steel electrodes (arcing through quartz glass).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I seal off the electrodes from oxygen (thermally conductive potting compound) will this increase the high-temp / high-voltage corrosion rate?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought keeping out humidity would help long term, (current electrodes are rusted out even though they are stainless), but I read about how the SS metal may need oxygen to heal the surface?  &lt;/p&gt;&#xA;" OwnerUserId="2405" LastEditorUserId="33" LastEditDate="2015-07-26T00:59:39.953" LastActivityDate="2015-07-26T00:59:39.953" Title="321 stainless electrode corrosion factors? (arc erosion resistantance)" Tags="|corrosion|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3685" PostTypeId="1" AcceptedAnswerId="24313" CreationDate="2015-07-25T23:03:44.690" Score="5" ViewCount="2180" Body="&lt;p&gt;I need to select a solid body in ANSYS Design Modeler, with script, in a design with multiples bodies. With the following command I select all the solid bodies, but this is not what I need.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ag.bodyPick;&#xA;ag.gui.SelectAll();&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm creating a script to model two spur gears engaged, first creating one tooth then making circular pattern to create the entry gear, with the following script I create the pinion but when I'm going to make the circular pattern for the gear, the pattern command give me an error, because I select all the solid bodies including the pinion, and I need to pick only the solid of the gear tooth to make the circular pattern them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my script:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ag.m.ClearAllErrors();&#xA;ag.gui.setUnits(ag.c.UnitMillimeter, ag.c.UnitDegree, ag.c.No);&#xA;&#xA;var p = new Object();&#xA;&#xA;//Plane&#xA;p.Plane  = agb.GetActivePlane();&#xA;p.Origin = p.Plane.GetOrigin();&#xA;p.XAxis  = p.Plane.GetXAxis();&#xA;p.YAxis  = p.Plane.GetYAxis();&#xA;&#xA;p.Sk1 = p.Plane.NewSketch();&#xA;p.Sk1.Name = &quot;Pinion&quot;;&#xA;p.Sk2 = p.Plane.NewSketch();&#xA;p.Sk2.Name = &quot;Wheel&quot;;&#xA;&#xA;//PINION&#xA;with (p.Sk1)&#xA;{&#xA;p.Sp1 = SplineBegin();&#xA;with(p.Sp1)&#xA;{&#xA;    SplineFlexibility = agc.Yes;&#xA;    SplineXY(-1.76564478418, 24.6869398366);&#xA;    SplineXY(-1.51020858813, 24.7038816792);&#xA;    //......... spline coordinates of the pinion profile&#xA;    //......... removed because character limit&#xA;    SplineXY(5.26099091644, 24.1843849328);&#xA;&#xA;    SplineFitPtEnd();&#xA;    }&#xA;    p.Ln2 = Line(-1.76564478418, 24.6869398366, -1.24843570599,17.4554120057);&#xA;    p.Ln3 = Line(3.71989256718, 17.1000701545, 5.26099091644, 24.1843849328);&#xA;    p.Cr4 = ArcCtrEdge(&#xA;              0, 0,&#xA;              3.71989256718, 17.1000701545,&#xA;              -1.24843570599, 17.4554120057);&#xA;    p.Ext1 = agb.Extrude(agc.Add, p.Sk1, agc.DirNormal, agc.ExtentFixe, 34.0, agc.ExtentFixed, 0.0, agc.No, 0.0, 0.0);&#xA;}&#xA;agb.Regen();&#xA;&#xA;&#xA;&#xA;var PF1 = ag.gui.CreateSpot();&#xA;ag.listview.ActivateItem(&quot;Type&quot;);&#xA;ag.listview.ItemValue = &quot;Construction Point&quot;;&#xA;ag.listview.ActivateItem(&quot;Definition&quot;);&#xA;ag.listview.ItemValue = &quot;Manual Input&quot;;&#xA;ag.listview.ActivateItem(&quot;FD8, X Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD9, Y Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD10, Z Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;agb.Regen();&#xA;&#xA;var PF2 = ag.gui.CreateSpot();&#xA;ag.listview.ActivateItem(&quot;Type&quot;);&#xA;ag.listview.ItemValue = &quot;Construction Point&quot;;&#xA;ag.listview.ActivateItem(&quot;Definition&quot;);&#xA;ag.listview.ItemValue = &quot;Manual Input&quot;;&#xA;ag.listview.ActivateItem(&quot;FD8, X Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD9, Y Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD10, Z Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;34.0&quot;;&#xA;agb.Regen();&#xA;&#xA;var LF1 = agb.LinePt();&#xA;LF1.AddSegment(PF1.GetPoint(1, 0),PF2.GetPoint(1, 0), 0);&#xA;agb.Regen();&#xA;&#xA;&#xA;var Pat = ag.gui.CreatePattern();&#xA;ag.listview.ActivateItem(&quot;Pattern Type&quot;);&#xA;ag.listview.ItemValue = &quot;Circular&quot;;&#xA;&#xA;ag.listview.ActivateItem(&quot;Geometry&quot;);&#xA;ag.bodyPick;&#xA;ag.gui.PickFilter(5, true);&#xA;ag.gui.PickFilter(11, false);&#xA;ag.gui.PickFilter(11, false);&#xA;ag.gui.SelectAll();&#xA;ag.listview.ItemValue = &quot;Apply&quot;;&#xA;&#xA;ag.listview.ActivateItem(&quot;Axis&quot;);&#xA;ag.edgePick;&#xA;ag.gui.SelectAll();&#xA;ag.listview.ItemValue = &quot;Apply&quot;;&#xA;ag.listview.ActivateItem(&quot;FD2, Angle&quot;);&#xA;ag.listview.ItemValue = &quot;16.3636363636&quot;;&#xA;ag.listview.ActivateItem(&quot;FD3, Copies (&amp;gt;0)&quot;);&#xA;ag.listview.ItemValue = &quot;21.0&quot;;&#xA;agb.Regen();&#xA;&#xA;//GEAR    &#xA;with (p.Sk2)&#xA;{&#xA;  p.Sp1 = SplineBegin();&#xA;  with(p.Sp1)&#xA;  {&#xA;    SplineFlexibility = agc.Yes;&#xA;    SplineXY(1.9304744152, 30.2623156532);&#xA;    //......... spline coordinates of the gear profile&#xA;    //......... removed because character limit&#xA;    SplineXY(-5.79031304774, 30.3548985987);&#xA;&#xA;    SplineFitPtEnd();&#xA;  }&#xA;  p.Ln2 = Line(1.9304744152, 30.2623156532, 0.209834175566,173.751999593);&#xA;  p.Ln3 = Line(-0.629381853016, 173.762062957, -5.79031304774, 30.3548985987);&#xA;  p.Cr4 = ArcCtrEdge(&#xA;              0, 191.250741537,&#xA;              -0.629381853016, 173.762062957,&#xA;              0.209834175566, 173.751999593);&#xA;  p.Ext2 = agb.Extrude(agc.Add, p.Sk2, agc.DirNormal, agc.ExtentFixe, 34.0, agc.ExtentFixed, 34.0, agc.No, 0.0, 0.0);&#xA;}&#xA;&#xA;var PF3 = ag.gui.CreateSpot();&#xA;ag.listview.ActivateItem(&quot;Type&quot;);&#xA;ag.listview.ItemValue = &quot;Construction Point&quot;;&#xA;ag.listview.ActivateItem(&quot;Definition&quot;);&#xA;ag.listview.ItemValue = &quot;Manual Input&quot;;&#xA;ag.listview.ActivateItem(&quot;FD8, X Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD9, Y Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;77.5&quot;;&#xA;ag.listview.ActivateItem(&quot;FD10, Z Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;agb.Regen();&#xA;&#xA;var PF4 = ag.gui.CreateSpot();&#xA;ag.listview.ActivateItem(&quot;Type&quot;);&#xA;ag.listview.ItemValue = &quot;Construction Point&quot;;&#xA;ag.listview.ActivateItem(&quot;Definition&quot;);&#xA;ag.listview.ItemValue = &quot;Manual Input&quot;;&#xA;ag.listview.ActivateItem(&quot;FD8, X Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD9, Y Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;77.5&quot;;&#xA;ag.listview.ActivateItem(&quot;FD10, Z Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;34.0&quot;;&#xA;agb.Regen();&#xA;&#xA;var LF2 = agb.LinePt();&#xA;LF2.AddSegment(PF3.GetPoint(1, 0),PF4.GetPoint(1, 0), 0);&#xA;agb.Regen();&#xA;&#xA;var Pat = ag.gui.CreatePattern();&#xA;ag.listview.ActivateItem(&quot;Pattern Type&quot;);&#xA;ag.listview.ItemValue = &quot;Circular&quot;;&#xA;&#xA;ag.listview.ActivateItem(&quot;Geometry&quot;);&#xA;ag.gui.PickFilter(5, true);&#xA;ag.gui.PickFilter(11, false);&#xA;ag.gui.PickFilter(11, false);&#xA;ag.bodyPick;          //&amp;lt;- Here is where I need to select only the gear body&#xA;ag.gui.SelectAll();&#xA;ag.listview.ItemValue = &quot;Apply&quot;;&#xA;&#xA;ag.listview.ActivateItem(&quot;Axis&quot;);&#xA;ag.edgePick;         //&amp;lt;- Here is where I need to select only the gear rotation axis for the circular pattern&#xA;ag.gui.SelectAll();&#xA;ag.listview.ItemValue = &quot;Apply&quot;;&#xA;ag.listview.ActivateItem(&quot;FD2, Angle&quot;);&#xA;ag.listview.ItemValue = &quot;2.74809160305&quot;;&#xA;ag.listview.ActivateItem(&quot;FD3, Copies (&amp;gt;0)&quot;);&#xA;ag.listview.ItemValue = &quot;130.0&quot;;&#xA;agb.Regen();&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2211" LastEditorDisplayName="user16" LastEditDate="2015-08-05T12:20:08.063" LastActivityDate="2018-10-23T18:27:03.457" Title="What is the script function to select Solid Body in ANSYS Design Modeler?" Tags="|mechanical-engineering|design|computer-aided-design|ansys|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3686" PostTypeId="1" CreationDate="2015-07-26T02:23:43.867" Score="2" ViewCount="1662" Body="&lt;p&gt;I was asked the question: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A band brake is installed on a drum rotating at 250rpm and a diameter&#xA;  of 900mm. The angle of contact is 1.5pi radians and one end of the&#xA;  brake band is fasted to a fixed pin while the other end to the brake&#xA;  artm is 150m from the fixed pin. The coefficient of friction is 0.25&#xA;  and the straight brake arm is 1000mm long and is placed perpendicular&#xA;  to the diameter bisecting the angle of contact. What should be the&#xA;  width in milimeters of a steel band with maximum tensile stress of&#xA;  55MPa and 3.0mm thick will be used?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer is 37mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I honestly see no way to solve this without any indication of power or force of some sort... here's my attempt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ S_{max} = \frac{F_1}{w\times t} }$$&#xA;$${ 55 Mpa = \frac{F_1}{w\times 3.0 mm} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ \frac{F_1}{F_2} = e^{(\theta \times friction)} }$$&#xA;$${ \frac{F_1}{F_2} = e^{(1.5rad \times 0.25)} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to find F1 seems to be the flaw. There are just no formulas I know of which doesn't require either power, force, torque?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ T = F \times d }$$&#xA;$${ P = 2\pi T N }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the problem set flawed? What am I missing? Is there another way?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastActivityDate="2015-07-26T02:23:43.867" Title="Band Brake placed Perpendicular to the Diameter bisecting the Angle of Contact" Tags="|mechanical-engineering|design|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3688" PostTypeId="1" CreationDate="2015-07-26T09:06:23.533" Score="5" ViewCount="2520" Body="&lt;p&gt;I would like to learn mechanical engineering as a side project. But, I would like to learn in a way similar to programming. Learn by doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course I understand I can't build things automatically just yet as I have no 3D-printer and they aren't advanced enough for most things (right?)... but Autodesk has simulation software which may do what I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I learn mechanical engineering by using AutoCAD for real projects?&lt;/p&gt;&#xA;" OwnerUserId="2410" LastEditorUserId="368" LastEditDate="2015-08-05T21:01:54.147" LastActivityDate="2019-09-23T20:48:54.673" Title="Can I use AutoCAD to learn mechanical engineering by DOING (like programming)?" Tags="|mechanical-engineering|computer-aided-design|3d-printing|" AnswerCount="7" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3689" PostTypeId="2" ParentId="3688" CreationDate="2015-07-26T11:51:36.230" Score="18" Body="&lt;p&gt;AutoCAD is a computerised drafting package that lets people draw pictures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanical engineering is more than just knowing how to draw pictures in two or three dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Amongst other things&lt;/strong&gt;, mechanical engineers are required to know about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Strength of materials&lt;/li&gt;&#xA;&lt;li&gt;Energy&lt;/li&gt;&#xA;&lt;li&gt;Power&lt;/li&gt;&#xA;&lt;li&gt;Torque&lt;/li&gt;&#xA;&lt;li&gt;Stresses&lt;/li&gt;&#xA;&lt;li&gt;Bending moments&lt;/li&gt;&#xA;&lt;li&gt;Couples&lt;/li&gt;&#xA;&lt;li&gt;Torsion&lt;/li&gt;&#xA;&lt;li&gt;Efficiency&lt;/li&gt;&#xA;&lt;li&gt;Thermodynamics&lt;/li&gt;&#xA;&lt;li&gt;Heat &amp;amp; heat transfer&lt;/li&gt;&#xA;&lt;li&gt;Psychrometry&lt;/li&gt;&#xA;&lt;li&gt;Fluids &amp;amp; fluid mechanics&lt;/li&gt;&#xA;&lt;li&gt;Hydraulics&lt;/li&gt;&#xA;&lt;li&gt;Pneumatics&lt;/li&gt;&#xA;&lt;li&gt;Tribology&lt;/li&gt;&#xA;&lt;li&gt;Kinematics&lt;/li&gt;&#xA;&lt;li&gt;Statics &amp;amp; dynamics&lt;/li&gt;&#xA;&lt;li&gt;Structural analysis&lt;/li&gt;&#xA;&lt;li&gt;Mechanical cycles like the Carnot cycle&lt;/li&gt;&#xA;&lt;li&gt;Gears &amp;amp; gear teeth design&lt;/li&gt;&#xA;&lt;li&gt;Crank shafts&lt;/li&gt;&#xA;&lt;li&gt;Vibration&lt;/li&gt;&#xA;&lt;li&gt;Manufacturing processes&lt;/li&gt;&#xA;&lt;li&gt;A good knowledge of chemistry, maths and physics is definitely&#xA;required&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You won't learn any of this from AutoCAD or any other drafting package. Such software may help you develop drafting skills.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All forms of engineering need to be studied at universities or colleges to ensure those who study engineering are qualified and competent. It's not something you pick up by reading a book or learning how to use one or more software packages.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="272" LastEditDate="2015-07-26T12:09:36.340" LastActivityDate="2015-07-26T12:09:36.340" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3690" PostTypeId="1" CreationDate="2015-07-26T12:00:36.483" Score="1" ViewCount="85" Body="&lt;p&gt;I've been struggling with this for days trying to make sense of a rpm vs torque chart I've seen recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chart states the engine load was measured 100% 70% 40% 0% and at 1000 2000 3000 4000 rpm. How is this done if using a chasis dyno?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attached the chart for reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's my question, but below I'll elaborate on my thinking and hopefully there will be a clear error in my understanding...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think part of my problem is that I'm unsure what the throttle is doing when these charts are produced using the dyno.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right in thinking that to get the 100% load points over each rpm, the test is started with the engine idling then pushing the throttle to 100%? The RPM will increase rapidly up to redline and the engine torque at each RPM can be estimated working back from wheel RPM and transmission ratio etc. They measure at 4 discrete RPM though, suggesting they might have control over the dyno resistive force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h8FOo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h8FOo.png&quot; alt=&quot;Engine Load vs RPM&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2411" LastEditorUserId="3633" LastEditDate="2016-06-03T19:16:08.853" LastActivityDate="2016-06-03T19:16:08.853" Title="How can different torque levels be measured at a single rpm using a chassis dyno?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3691" PostTypeId="2" ParentId="3688" CreationDate="2015-07-26T12:23:48.250" Score="6" Body="&lt;p&gt;As Fred alluded to in the above post, mechanical engineering is a vast topic. It takes time and commitment to become a mechanical engineer. Also &lt;strong&gt;AutoCAD or 3D-printing alone is not mechanical engineering&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AutoCAD and 3D-printers are tools used by mechanical engineers. By the way, there are few good self-taught mechanical engineers. Following a four-year engineering program is the conventional approach to becoming a mechanical engineer. But alternatively there are plenty of free online material to help become a mechanical engineer or learn a specific area of mechanical engineering. Below are few from &lt;a href=&quot;http://www.edx.org&quot; rel=&quot;nofollow&quot;&gt;edx&lt;/a&gt; and &lt;a href=&quot;http://www.coursera.org&quot; rel=&quot;nofollow&quot;&gt;coursera&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edx&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.edx.org/course/thermodynamics-iitbombayx-me209-1x-0&quot; rel=&quot;nofollow&quot;&gt;Thermodynamics&lt;/a&gt; from Indian Institute of Technology at Bombay&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.edx.org/course/introduction-aerodynamics-mitx-16-101x-0&quot; rel=&quot;nofollow&quot;&gt;Introduction to Aerodynamics&lt;/a&gt; from Massachusetts Institute of Technology&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.edx.org/course/introduction-aeronautical-engineering-delftx-ae1110x-0&quot; rel=&quot;nofollow&quot;&gt;Introduction to Aeronautical Engineering&lt;/a&gt; from Delft University of Technology&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.edx.org/course/dynamics-mitx-2-03x&quot; rel=&quot;nofollow&quot;&gt;Dynamics&lt;/a&gt; from Massachusetts Institute of Technology&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coursera&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.coursera.org/learn/engineering-mechanics-statics/home/info&quot; rel=&quot;nofollow&quot;&gt;Introduction to Engineering Mechanics&lt;/a&gt; from Georgia Institute of Technology&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.coursera.org/course/fluidpower&quot; rel=&quot;nofollow&quot;&gt;Fundamentals of Fluid Power&lt;/a&gt; from University of Minnesota&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ttps://www.coursera.org/course/introthermodynamics&quot; rel=&quot;nofollow&quot;&gt;Introduction to Thermodynamics&lt;/a&gt; from University of Michigan&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.coursera.org/learn/engineering-mechanics-statics-2&quot; rel=&quot;nofollow&quot;&gt;Applications in Engineering Mechanics&lt;/a&gt; from Georgia Institute of Technology&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="368" LastEditDate="2015-07-28T17:51:23.927" LastActivityDate="2015-07-28T17:51:23.927" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3692" PostTypeId="2" ParentId="3688" CreationDate="2015-07-26T12:23:55.990" Score="8" Body="&lt;p&gt;&lt;b&gt;No&lt;/b&gt;.  AutoCAD isn't going to teach you mechanical engineering any more than a compiler will teach you software engineering.  You have to actually &lt;i&gt;learn&lt;/i&gt; some things, then you will pick up experience by doing.  Also AutoCAD is a tool for describing mechanical parts in machine-readable form.  It's not going to &lt;i&gt;design&lt;/i&gt; a part for you, or teach you how.  Without a sufficient background in math and physics and other fields, you will have no way of knowing how to synthesize a design that meets a set of specs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowledge and experience are two different things.  Without up front knowledge, learning by experience and making mistakes will be very slow, and will never show you the things that you could have learned that you didn't bump into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a reason it takes a four year degree minimum to become a mechanical (or any other type) of engineer.  There is much much more to being a engineer than knowing how to run a few tools.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-07-26T12:23:55.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3693" PostTypeId="1" CreationDate="2015-07-26T15:46:29.557" Score="4" ViewCount="436" Body="&lt;p&gt;There are plenty of springs out there which have either a little PVC tube or a foam inside, when I first saw the tube I thought it was probably to prevent the metal spring from bending, but that wouldn't explain the foam that comes with others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an analysis for this materials within the springs?, in which situations shall I use this materials within the springs?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;PVC Tube:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ul8f2t.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ul8f2t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Foam:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M5yWy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M5yWy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WoV4e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WoV4e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="831" LastEditorUserId="831" LastEditDate="2015-07-27T15:27:54.440" LastActivityDate="2015-07-27T16:47:29.140" Title="What is the [PVC] tube or foam in metal springs for?" Tags="|mechanical-engineering|springs|mechanical-failure|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3695" PostTypeId="2" ParentId="3693" CreationDate="2015-07-26T18:14:59.353" Score="4" Body="&lt;p&gt;I know of at least one case where such material was added inside the spring to dampen it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spring will be resonant at some frequency.  If there is enough component of that frequency in whatever is driving the spring, then it could oscillate or &quot;ring&quot;.  This can be undesirable because it will subject whatever it is connected to to forces at this frequency, the center can move substantially sideways, and it can cause audible noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something squishy (like the foam) in the middle of the spring will dampen such oscillations, making any resonant ringing die down quickly if the spring happens to be excited at that frequency.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something rigid (like the tube) is probably just to keep the spring from buckling or oscillating sideways.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorDisplayName="user16" LastEditDate="2015-07-27T11:27:34.920" LastActivityDate="2015-07-27T11:27:34.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3696" PostTypeId="2" ParentId="3688" CreationDate="2015-07-27T00:51:45.613" Score="2" Body="&lt;p&gt;I think that there's a problem with your analogy.  Autocad is to engineering as is programming - to actually running your code with meaningful results. Otherwise it's just drawing.  3D printing is as the name implies, drawing.  In three dimensions, rather than two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that you're looking at hobby level engineering if you're categorizing it as a &quot;side project&quot;.  That can be rewarding and complex too.  If you're interested in CAD and 3D printing, design and print some widget.  &lt;strong&gt;Then&lt;/strong&gt;, get it to work.  Repeat process until it does work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples off the top of my drink addled head:-&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Design some lego, then get it to fit and stick together like the real&#xA;stuff.  Your issues will revolve around matching the scale of the&#xA;pieces to the repeatability /precision of the printer.  Get into that&#xA;and see how real Lego manage production quality. &lt;/li&gt;&#xA;&lt;li&gt;Design and build a    model arch bridge.  I mean out of individual&#xA;printed pieces,    including a keystone.  Man's been building them&#xA;for a 100,000 years.     You can get into what makes an arch stand&#xA;up, and whether circular or    parabolic shapes are best.&lt;/li&gt;&#xA;&lt;li&gt;Print something like a car powered by a    rubber band.  Try to&#xA;incorporate real metal bearings for the axles    (buy them dirt cheap&#xA;off eBay).  Or don't and see it it's feasible to    print axles that&#xA;will work without discrete bearings.  How will the    rubber band&#xA;turn the wheels, and how can it overcome axle friction?     You'll&#xA;have to investigate how two printed surfaces can slide past    each&#xA;other - printed stuff looks rough to me.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know that the above examples are trivial, but if you get into it, real engineering techniques can be discovered.  After all, it's pretty useless printing a million angels on a pin head, but dwell on the engineering know how that lets you do it...&lt;/p&gt;&#xA;" OwnerUserId="2417" LastActivityDate="2015-07-27T00:51:45.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3697" PostTypeId="1" CreationDate="2015-07-27T02:26:45.943" Score="3" ViewCount="55" Body="&lt;p&gt;I am creating a network to share files between a mobile device and multiple PCs directly through a LAN. The purpose of which is to use smart phones to capture files from remote devices which are outside the range of the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current plan is to use a portable router that could be brought on site and act as a bridge between the mobile client and the PCs to allow for the passing of the desired files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to LAN and router network and have had issues when trying to use the router as a bridge as so I was wondering if it would be better to create the network with the device in router, repeater and bridge modes? What would be the reasons for this? Or is there a better way to create a network?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: for this application the mobile device needs to be able to talk to multiple PCs at one time so WiFi direct is not a valid option. And there is nowhere on site to leave a permanent router which is why a portable router is a suggested option.&lt;/p&gt;&#xA;" OwnerUserId="2419" LastEditorUserId="90" LastEditDate="2015-07-27T05:41:37.357" LastActivityDate="2015-07-27T05:41:37.357" Title="Viability of creating a wireless LAN bridge between mobiles and PCs using a portable router?" Tags="|electrical-engineering|software|telecommunication|" AnswerCount="0" CommentCount="7" ClosedDate="2018-12-23T16:09:37.600" ContentLicense="CC BY-SA 3.0" />
  <row Id="3699" PostTypeId="1" CreationDate="2015-07-27T12:05:19.637" Score="3" ViewCount="250" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Phase_rule&quot; rel=&quot;nofollow&quot;&gt;Gibbs Phase Rule&lt;/a&gt; indicates that for a two phase, single component thermodynamic system we will have one independent intensive parameter. &#xA;Given that the Degree of Freedom is $1$ means that fixing one intensive parameter would fix the entire state of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any property $x$ is just a function of one other property say $y$, i.e. $x=f(y)$. This gives for two phase system $P=f(T)$, but why isn't $v=f(T)$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know $v$ can have a value anywhere between $v_f$ and $v_g$. Are there any assumptions for the Phase Rule, other than equilibrium, which would explain this?&lt;/p&gt;&#xA;" OwnerUserId="2426" LastEditorUserId="2274" LastEditDate="2015-09-28T14:10:08.603" LastActivityDate="2015-10-28T14:38:19.550" Title="Applicability of Gibbs' Phase Rule" Tags="|thermodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3700" PostTypeId="2" ParentId="3699" CreationDate="2015-07-27T12:37:47.853" Score="2" Body="&lt;p&gt;Specific Volume is a intensive property too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Gibbs you get 1 degree of freedom for your single component, 2 phase mixture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = N - P + 2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = 1 - 2 + 2 = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not yet fixed what you have to use as your degree of freedom. For your example, at a fixed specific volume v (which equals $\frac{1}{\rho}$) you can either change T or p. But keep in mind that T(p,v) and p(T,v). So a change in pressure results in a change in Temperature, too and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that clear up your confusion about the rule?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iO2wm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iO2wmm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6nQlN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6nQlNm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://engineering-references.sbainvent.com/thermodynamics/tv-pv-diagrams.php#.VbnllfntlBc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the diagramms. If you set $v$ you have 'locked' you state. You see now that you cannot change $p,T$ independently. And regarding your comment for example you can see that $T(v)|_p$. &lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2015-07-30T08:55:20.660" LastActivityDate="2015-07-30T08:55:20.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3701" PostTypeId="2" ParentId="3693" CreationDate="2015-07-27T14:03:44.500" Score="1" Body="&lt;p&gt;To elaborate on Olin's answer, the foam (and also the tube if it is pneumatic) is likely used to add some damping to the mechanical resonance of the spring.  The phenomenon of resonance in a spring-mass system is illustrated in the figure below.  In that figure $\omega_0$ is the resonance frequency of the system which is &lt;a href=&quot;https://en.wikipedia.org/wiki/Mechanical_resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;determined by the spring constant and the sprung mass&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For frequencies well below the resonant frequency the amount of motion of is equal to the amount of the push.  For frequencies well above the resonant frequency the amount of motion is much less than the amount of the push (this property is exploited in e.g. vehicle suspensions).  But, for frequencies near the resonance the amount of motion can be &lt;strong&gt;significantly greater&lt;/strong&gt; than the amount of the push.  We have all experienced this when playing with a weight on a string or a glass of liquid; there is a certain frequency at which a very small input can create a very large motion of the mass/liquid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Resonance.PNG&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HBG6Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Too much amplification at the resonance can literally tear the device apart (see e.g. the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tacoma_Narrows_Bridge&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tacoma Narrows bridge&lt;/a&gt;). This amount of amplification at the resonance is determined by the damping in the system.  More damping means less amplification and vice versa.  Damping is created in the system by inserting something which removes energy at the resonance.  Foam is used in some of your examples because it is a very lossy material which efficiently converts mechanical motion into heat.  Pneumatic tubes can be used (and often are in vehicle suspensions) to do work on a gas by forcing it through a small opening and thus, converting mechanical motion into heat.  The tubes can also be used to restrict the motion of the spring to be only in the direction of the spring.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-07-27T15:06:31.527" LastActivityDate="2015-07-27T15:06:31.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3703" PostTypeId="1" AcceptedAnswerId="3707" CreationDate="2015-07-27T14:34:30.440" Score="1" ViewCount="280" Body="&lt;p&gt;How can I calculate the propulsive force produced by ejection of a gas? I suppose it depends on the gas, the velocity with which it ejects, the amount of gas, and some other parameters, like the geometry of the nozzle for example. I would like a calculator to play with the gas density, the gas velocity etc, and see the various propulsive forces generated.&lt;/p&gt;&#xA;" OwnerUserId="2353" LastEditorUserId="368" LastEditDate="2015-07-28T18:13:16.187" LastActivityDate="2015-07-28T18:13:16.187" Title="How to calculate propulsive force of gas ejection" Tags="|aerospace-engineering|gas|propulsion|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3705" PostTypeId="2" ParentId="3693" CreationDate="2015-07-27T16:47:29.140" Score="1" Body="&lt;p&gt;Depending on the mechanics of the design, the material is primarily there for damping. The design that compresses urethane (typically) foam uses the material's damping properties to convert the motion energy into heat. Other designs rely instead on friction to provide damping and similarly convert kinetic energy into heat. When I worked on designing suspension elements for washing machines, we used both methods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while a spring provides increased resistance with increased displacement, a damper provides increased resistance with increased velocity. Most elastomeric material provide both damping and stiffness.    &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-07-27T16:47:29.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3706" PostTypeId="2" ParentId="3688" CreationDate="2015-07-27T17:01:42.087" Score="3" Body="&lt;p&gt;As a working professional ME with 18 years in the field, I'll tell you that there's a great deal of difference between what will work in AutoCAD (or any other such software package) and what will work in the real world. That's why we still review designs and build prototypes before going into production. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't tell you the number of times I've had to get the designers who draw up stuff and the people who have to make the stuff in the same room to hash out how to make something work. The designers will pull up their model and say something like, &quot;But it works in the model...&quot; while the people from production show them why it doesn't work in the real world. Then, after everyone sees the same problem, the work starts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want to goof around and make stuff for your 3D printer, that's great. I'd suggest a solid modeling package like Autodesk Inventor that will let you export .stl files directly and do assemblies, etc. It's expensive for commercial use, but they're supportive of people learning on their own, so I think you can get a license for little or no $$. Still, even becoming an expert user of that, or any CAD package won't make you into an engineer. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="368" LastEditDate="2015-07-28T17:53:30.183" LastActivityDate="2015-07-28T17:53:30.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3707" PostTypeId="2" ParentId="3703" CreationDate="2015-07-27T17:07:10.920" Score="1" Body="&lt;p&gt;It's pretty basic conservation of momentum kind of problem. A converging diverging nozzle is used to accelerate the gas such that it's supersonic leaving the nozzle. These problems are pretty well-defined. There are lots of calculators out there that let you monkey around with varying different parameters without getting into a full on mathematical understanding of compressible flow. Here's one: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.purdue.edu/~wassgren/applet/java/cdnozzle/&quot; rel=&quot;nofollow&quot;&gt;https://engineering.purdue.edu/~wassgren/applet/java/cdnozzle/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-07-27T17:07:10.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3708" PostTypeId="2" ParentId="3688" CreationDate="2015-07-27T17:58:38.683" Score="3" Body="&lt;p&gt;One thing that other people don't seem to have touched on is the utility of the simulation packaged in AutoCAD. It's a type of &lt;a href=&quot;https://en.wikipedia.org/wiki/Finite_element_method#Basic_concepts&quot; rel=&quot;nofollow&quot;&gt;Finite Element Analysis&lt;/a&gt;, which is an incredibly useful tool, but one that must be wielded very carefully. You can model a part, load it into an FEA package, apply constraints and forces as the part would see in the real world, and yet there is no guarantee that the results you get from that simulation have any correlation to real world outcomes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Understanding how to set up and run the simulation takes training of its own, and even with a well-designed simulation, understanding the results takes independent engineering knowledge. The software I primarily work in is Creo, and Creo's FEA package has a warning message that actually tells you that it thinks something is up with the results of the simulation, but it doesn't know exactly what, and that you need to use your engineering knowledge to analyze the results for accuracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, don't let any of this (or anything anyone else has said) stop you from playing around in AutoCAD. It doesn't matter how good of a concept an engineer can come up with, if they can't put it on paper or communicate it to someone else, it's worthless. However, if you want to learn by doing, realize that the end product of mechanical engineering is a machine. What you need to be doing is working with machines. Take apart existing things, find things that you can build yourself, and examine all of this stuff to try and understand how things were made and why they were designed they way they were. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-07-27T17:58:38.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3709" PostTypeId="1" CreationDate="2015-07-28T01:35:06.403" Score="5" ViewCount="516" Body="&lt;p&gt;&lt;em&gt;Background to the question&lt;/em&gt;: We have been importing lug bolts under a certain tariff heading which is under dispute :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crux of the question is whether the lug bolt ( which we have ascertained is in fact a screw ) is deemed to be fully threaded or not. The definition of this would determine the tariff heading that would be applied. Namely a separate tariff heading for a screw that is fully threaded and a different one for a screw that is partially threaded.  The question here is because of the design of this screw, namely with a seat is it considered to be fully threaded or not? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attached below a picture / link of the type of lug bolt I am talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://shop.mcgard.com/products/automotive/lug-nuts-lug-bolts/lug-bolts&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7zXlo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2433" LastEditorDisplayName="user16" LastEditDate="2015-07-28T01:43:14.447" LastActivityDate="2016-02-11T20:15:34.213" Title="Lug bolts - fully threaded or partially threaded?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3710" PostTypeId="2" ParentId="2699" CreationDate="2015-07-28T07:07:04.533" Score="6" Body="&lt;p&gt;&lt;a href=&quot;http://www.dot.ca.gov/hq/esc/techpubs/manual/bridgemanuals/bridge-design-aids/page/bda_11.pdf&quot; rel=&quot;nofollow&quot;&gt;Caltrans Bridge Design Aids, Section 11&lt;/a&gt; adds 3% for welds and bolts:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;Furnish Structural Steel (Bridge) and Erect Structural Steel (Bridge)&lt;/h3&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Segregate by type of steel and estimate in kilograms.&lt;/li&gt;&#xA;  &lt;li&gt;Include bearings (except PTFE Spherical Bearings), anchor bolts, shear connectors, and expansion dams (except where expansion dams are&#xA;  galvanized or embedded in concrete).&lt;/li&gt;&#xA;  &lt;li&gt;Ignore small, non-repetitive cuts, copes, and bolt holes.&lt;/li&gt;&#xA;  &lt;li&gt;Add 3% for welds and bolts.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is applicable to new bridges -- retrofits may have a much higher percentage of bolt weight.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="368" LastEditDate="2015-07-28T18:06:16.147" LastActivityDate="2015-07-28T18:06:16.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3713" PostTypeId="1" AcceptedAnswerId="3716" CreationDate="2015-07-28T11:41:34.847" Score="0" ViewCount="254" Body="&lt;p&gt;I wish to investigate an active vehicle braking method for use in extreme situations. Very high accelerations are acceptable in situations where this would prevent the much worse accelerations present in a collision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of a system that will use ejected gas/liquid to reposition the position of a car and/or assist in braking. We know that ejected mass (gas or other) create a propulsive force. I was thinking to utilize this mechanism to create a force that will decelerate the car. Or maybe change its position/trajectory (eg. when it moves out of the road in a steep turn).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would such system be viable?&lt;br&gt;&#xA;What parameters would such system have?&lt;br&gt;&#xA;eg to decelerate a car travelling at 30m/s within 3 meters &#xA;(about 15g deceleration!).&lt;br&gt;&#xA;How much mass should we eject and at which speed?&lt;br&gt;&#xA;Have such systems being proposed/researched/built?&lt;/p&gt;&#xA;" OwnerUserId="2353" LastEditorUserId="399" LastEditDate="2015-07-28T15:06:49.633" LastActivityDate="2015-07-28T17:06:04.797" Title="Vehicle emergency high-deceleration braking and steering system" Tags="|automotive-engineering|projectiles|" AnswerCount="4" CommentCount="7" FavoriteCount="0" ClosedDate="2015-07-28T17:13:09.673" ContentLicense="CC BY-SA 3.0" />
  <row Id="3714" PostTypeId="2" ParentId="227" CreationDate="2015-07-28T12:11:03.503" Score="2" Body="&lt;p&gt;While designing a relief the valve flow rate is the deciding variable. To calculate its value in two phase flow different models can be used like HEM, drift flux, separated flow, etc.  For this particular application (where two phase flow is because pressure drop across valve) HEM is best suited. I would also suggest to check for choking in the two-phase flow while designing relief valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I have no practical experience in area.&lt;/p&gt;&#xA;" OwnerUserId="2426" LastEditorUserId="272" LastEditDate="2015-07-28T12:48:28.063" LastActivityDate="2015-07-28T12:48:28.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3715" PostTypeId="2" ParentId="3713" CreationDate="2015-07-28T12:44:34.990" Score="2" Body="&lt;p&gt;First, what you are asking about is basically what a rocket engine does.  The physics is a momentum balance.  The total momentum the car looses is equal and opposite of the momentum of whatever you eject out the front.  You can answer the last question yourself using just high school physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, decelerating a car from 30 m/s to 0 in 3 meters is absurd.  At best that would happen within 200 ms, which means 150 m/s&amp;sup2;, which is over 15 g.  Neither you nor the car are going to survive that.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-07-28T12:44:34.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3716" PostTypeId="2" ParentId="3713" CreationDate="2015-07-28T13:33:26.117" Score="2" Body="&lt;p&gt;I gave this idea some thought some years ago -= catalysed by the events depicted in the video linked at the end.&lt;br&gt;&#xA;What you are describing is essentially a rocket.&#xA;The concept is obviously possible as it's been done long ago.&#xA;Whether it is practical in any real world situations is the main point to be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is hereby proclaimed that for the balance of this answer.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;g = 10 m/s/s&lt;br&gt;&#xA;(Multiply this by 0.98 if you wish)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 kgf is the &quot;weight&quot; of 1 kgm of material in a 1g gravitational field.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So  1 kgm &quot;weighs&quot; 1 kgf = 10 Newton&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Units of kgf.s/kgm = s (seconds) !!! :-)&#xA;(Only valid for &quot;rocket scientists) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a rocket to be useful and practical in your scenarios requires it to be &quot;safe&quot;, reliably able to be operated instantaneously at any time, suitably powerful and suitably compact and low mass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many potential solutions but one which seems most likely to be a practical contender is a HP (Hydrogen Peroxide) catalytic decomposition rocket. In such a system pressurised HP is expelled via a catalytic decomposition promoter (often a silver wire screen) and hence into a classic convergent divergent rocket motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A catalytic HP system has the advantages of being RELATIVELY low temperature, nothing but water and Oyxygen are expelled, it can be stored ready for use without need for cyrogenic cooling and is relatively non corrosive =(although cleanliness is utterly essential). The rocket is able to be &quot;throttled&quot; over most of its range and the thrust can be easily controlled in level and started and stopped at will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monopropellant Hydrogen Peroxide is not an especially high energy per volume or mass propellant. The standard &quot;figure of merit&quot; is Isp = specific impile which can be thought of as the amount of time in seconds that a unit mass of propellant can exert a force equal to its own weight in seconds. This can be expressed as eg lbf.s/lbm (pounds force x seconds per pound mass) and a bit of sloppy laziness allows people to cancel the lbf/lbm and call the resulting units &quot;seconds&quot;. In fact it's really (lbm. x m/s/s x s / (lbm)  = m/s = velocity!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anon - an Isp of 100 is not too hard and 200 is getting rather hard. Lets goo with 120 to start.  1 kg of HP will provide 120 kg force = say 1200N for 1 second. A little plugging in to formulae shows that you are going to need a significant mass of HP compared with vehicle mass to get very large accelerations. eg your suggested 30 m/s in 3m would (as Olin says) be bone crushing - but in fact is entirely achievable using ABOUT the same amount of propellant as doing it at a more sensible rate - see below.  BUT lets look at what it takes to apply 1g of acceleration (10 m/s/s) for 1s to a say 1000 kg car.&lt;br&gt;&#xA;1g acceleration to a 1 tone car takes 1000 kg x g = 10,000 N.&#xA;Working in kgf, with an Isp of 120 to get 1000 kgf thrust for one second requires 1000/120 = 8.5  kg of HP.&lt;br&gt;&#xA;HP density is usefully greater than 1 but call this 8 litres.&lt;br&gt;&#xA;Taking the original 30 m/s vehicle - to stop it at 1g requires 30 m/s / 10 m/s/s = 3 seconds of thrust so about 25 kg and 25 litres of HP.&lt;br&gt;&#xA;That's about 25/1000 = 2.5% of the vehicle mass - which is not insignificant BUT also 'useful' for something which can either accelerate the vehicle over about 0 - 100 kph in about 3 seconds OR bring it to a standstill in the same distance - even on mud, snow, ice or while in the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above I said that stopping the vehicle in 3 metres was impractical.  That's essentially true, but it's not impossible.  And a little consideration shows that you notionally need no more propellant to do do As long as you can get the same Isp you just need to consume the same amount of propellant at a higher rate. That requires a higher maximum capacity rocket motor, greater flow rates so bigger &quot;pluming&quot; bigger forces etc- but it's doable.&lt;br&gt;&#xA;To stop from 30 m/s (= 108 kph) in 3 metres requires&lt;br&gt;&#xA;a = V^2/(2 d) = 900/6 = 150 m/s/s = 15g (as Olin said)&lt;br&gt;&#xA;Time to stop is t = V/A = 30/150 = 0.2 seconds.&lt;br&gt;&#xA;Doing a 0-100kph ~= 0.60 mph in 0.2 s makes by long long log ago neighbours best toys look like toys indeed - and would make any occupant very sore indeed. BUT not necessarily dead - accelerations of that order over such short periods are survivable given proper containment - and extending it out to say 7.5 metres and 0.5 seconds and 6g has you well inside fighter aircraft and below peak moon-rocket levels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More useful than applying gross front/back velocity changes might be applying forces to allow steering authority in skids - or for a race car which is sliding across a ploughed surface towards a tyre wall - or a concrete wall. Quite modest amounts of propellant could achieve quite useful results. Under computer control such a system has the ability to extricate a vehicle for otherwise irrecoverable situations. Also, without due care, to insert them into them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Properly done such a system could even allow a vehicle to &quot;fly&quot; for short periods - &lt;a href=&quot;http://media.armadilloaerospace.com/2002_09_28/firstMannedFlight.mpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;like this&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Armadillo Aerospace hydrogen peroxide mono-prop 1st manned flight)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zmJPk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zmJPk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-07-28T13:53:57.340" LastActivityDate="2015-07-28T13:53:57.340" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3719" PostTypeId="5" CreationDate="2015-07-28T16:29:47.943" Score="0" Body="&lt;p&gt;A pressure vessel is a container designed to resist a uniform pressure load exerted by a fluid medium. This fluid can be atmospheric air for a vacuum chamber, or high-pressure steam in a heat exchanger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure vessels are typically designed to a safety code, such as the American Society of Mechanical Engineers Boiler and Pressure Vessel Code.&lt;/p&gt;&#xA;" OwnerUserId="9212" LastEditorUserId="1832" LastEditDate="2016-12-14T18:06:14.977" LastActivityDate="2016-12-14T18:06:14.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3720" PostTypeId="4" CreationDate="2015-07-28T16:29:47.943" Score="0" Body="Containers that are designed specifically to retain internal pressure or a vacuum" OwnerUserId="9212" LastEditorUserId="9212" LastEditDate="2016-12-14T18:06:23.777" LastActivityDate="2016-12-14T18:06:23.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3721" PostTypeId="2" ParentId="3713" CreationDate="2015-07-28T16:46:07.477" Score="2" Body="&lt;p&gt;Yes, this has been done for other vehicles.  The Soyuz spacecraft uses parachutes to slow, but then fires solid rocket motors just before landing to slow down.  According to &lt;a href=&quot;https://www.nasa.gov/mission_pages/station/structure/elements/soyuz/landing.html&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; the engines fire 1 sec before touchdown at 24 ft/s.  And &lt;a href=&quot;http://www.spaceflight101.com/soyuz-spacecraft-information.html&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt; says that landing is at 3 m/s (9.8 ft/s).  If these are both correct, the Soyuz decelerates with a delta-v of 14 ft/s in 1 sec just before touchdown.&lt;/p&gt;&#xA;" OwnerUserId="2066" LastActivityDate="2015-07-28T16:46:07.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3722" PostTypeId="2" ParentId="3709" CreationDate="2015-07-28T16:59:53.863" Score="7" Body="&lt;p&gt;From a purely functional perspective, I'd argue that the seat is irrelevant to whether a fastener is fully threaded. The seat is &lt;em&gt;functionally&lt;/em&gt; part of the head, not part of the shank, and the shank is all that's relevant here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, consider that the largest diameter of the seat significantly exceeds the O.D. of the fastener. This indicates that the seat is forged from the wire rod (as is the head) rather than cut or rolled (as are the threads).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the function of the seat is to mate with an external surface (like the head) rather than with internal threads. This is true even if that external surface is not perfectly flat, or if the hole is counter-bored; the force on the head or seat acts to push the fastener out of the hole, while the force on the threads acts to pull the fastener into the hole (when used as a screw).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may also be unthreaded chamfers at either end of the shank of a &quot;fully-threaded&quot; bolt, for ease of starting nuts and/or to make sure failure doesn't occur right at the joint of the head and shank. The presence of these small chamfers doesn't prevent a fastener from being &quot;fully-threaded.&quot; So I think the engineering perspective is clearly that the &quot;lug bolts&quot; (which you correctly point out are used as screws, despite the name) you've shown above are &lt;strong&gt;fully-threaded fasteners&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-02-11T20:15:34.213" LastActivityDate="2016-02-11T20:15:34.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3723" PostTypeId="2" ParentId="3713" CreationDate="2015-07-28T17:06:04.797" Score="1" Body="&lt;p&gt;15 g might seem like a lot, and it is, if the 15g is sustained as it would be in something like an aircraft's aerobatic maneuver. 15g isn't unreasonable as a limit to what humans can take in a short duration event like a crash, Indy car drivers have sustained horizontal impact forces many times that without significant injury. However, consider how those drivers are restrained to avoid injury, 5 pt harness, helmet, arm restraints, head restraints and so on. The level of G-force that humans can tolerate depends on the individual, their training and also the direction in which the G-force is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an interesting read on human tolerance of impact forces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ftp.rta.nato.int/publiC/PubFullText/RTO/EN/RTO-EN-HFM-113/EN-HFM-113-06.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ftp.rta.nato.int/publiC/PubFullText/RTO/EN/RTO-EN-HFM-113/EN-HFM-113-06.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-07-28T17:06:04.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3725" PostTypeId="2" ParentId="3688" CreationDate="2015-07-29T07:40:57.413" Score="1" Body="&lt;p&gt;As a current Mechanical Engineering student I have been working at a company for 3 years mainly doing 3D design. This ranges from 3D modelling of very complex assemblies all the way down to the simple pipes and sheetmetal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most you will learn is the way assemblies should be assembled in order, welding standards, sheetmetal standards and possibly the strength of structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to be a good mechanical Engineering you need to have knowledge in fluid mechanics, thermodynamics etc. &lt;/p&gt;&#xA;" OwnerUserId="2446" LastActivityDate="2015-07-29T07:40:57.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3726" PostTypeId="1" CreationDate="2015-07-29T08:28:26.730" Score="6" ViewCount="1032" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Sabatier_reaction&quot;&gt;Sabatier reaction&lt;/a&gt; converts CO&lt;sub&gt;2&lt;/sub&gt; to methane. Can a vehicle that burns methane recover it (at least some part) from the exhaust gases via the Sabatier reaction? I am interested to know how this would be implemented, and whether such a vehicle has ever been built or prototyped.&lt;/p&gt;&#xA;" OwnerUserId="2353" LastEditorUserId="163" LastEditDate="2015-11-08T08:45:20.273" LastActivityDate="2015-11-08T12:17:45.510" Title="Can the Sabatier reaction power a vehicle?" Tags="|automotive-engineering|chemical-engineering|energy|energy-storage|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3727" PostTypeId="2" ParentId="3726" CreationDate="2015-07-29T10:15:35.623" Score="14" Body="&lt;p&gt;According to Wikipedia: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It [Sabatier reaction ] involves the reaction of hydrogen with carbon dioxide at elevated temperatures (optimally 300–400 °C) and pressures in the presence of a nickel catalyst to produce methane and water. $$ CO_2 + 4H_2 \rightarrow CH_4 + H_2O + \text{Energy} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as you can see from the previous chemical reaction, to obtain 1 KMole of $CH_4$ you should have 4 Kmole of Hydrogen as reactants and a reaction medium of 300–400 °C (with god knows how much of catalyst you need).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing to a regular vehicle internal combustion engine running on Methane:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hydrogen is not available in ambient air to the degree that could&#xA;allow a sustainable Sabatier reaction (it's not Oxygen), meaning you need a&#xA;separate Hydrogen tank along side with your engine (not to mention&#xA;the explosive nature of Hydrogen).&lt;/li&gt;&#xA;&lt;li&gt;Preheating the reactants to 300–400 °C would require an external&#xA;heating system if done using hydrocarbon fuel, it would require an&#xA;additional combustion chamber and a heat exchanger.&lt;/li&gt;&#xA;&lt;li&gt;Higher heating value of Methane = 889 kJ/Mole and HHV of Hydrogen =&#xA;286 kJ/Mole (Wikipedia), so theoretically speaking, the Sabatier reaction&#xA;would provide you with 1 Mole of CH4 having heating value of 889 kJ by&#xA;supplying the reactor 4 Moles of H2 that had a heating value of 4 *&#xA;286 = 1144 kJ !!&lt;/li&gt;&#xA;&lt;li&gt;The resulting Methane and water have a ratio of 1:1, introducing this&#xA;mixture with this amount of water to the&#xA;combustion chamber will give you a VERY low efficiency combustion&#xA;process, So naturally you need a separation mechanism to lower water content as possible before entering combustion chamber.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So has anyone researched/built a vehicle that burns methane and it recovers it (at least some part) from the exhaust gases via Sabatier reaction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I really have no idea but from the previous points I don't see it as feasible at all as a conventional internal combustion engine running on Methane.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2015-07-29T10:46:08.807" LastActivityDate="2015-07-29T10:46:08.807" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3729" PostTypeId="1" AcceptedAnswerId="3742" CreationDate="2015-07-29T14:47:09.493" Score="5" ViewCount="835" Body="&lt;p&gt;I'm designing a compact mechanism for the deployment of a mast. The idea is that the linear actuator for the deployment is on the other side of the pivot, and a slider increases the lever arm to enable the actuator to deploy the mast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ran through the maths (shown in details below), and wrote the code for the simulation. I want to know how the actuator force and the lever length relate, and how the force varies with the angle. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/KLVsB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KLVsB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I run the code I get huge spikes in the force (see below, I capped the force to 10^6 to better see the pattern), but for example I don't see why the mechanism would jam between 40 and 80°. For some reason the cross product of HP with OP (what I call the signed lever arm of F) goes through zero but I see no reason why it would given the current dimensions and angle range set. How can I fix it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wBrvl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wBrvl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the force capped at 200N:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UVzIY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UVzIY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the denominator of F:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3a7Rh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3a7Rh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Annex&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;close all&#xA;clear all&#xA;&#xA;%See diagram for description of variables, units are metres&#xA;a = (2+10)*10^-3;&#xA;b = a;&#xA;c = 165*10^-3;&#xA;d = 70*10^-3;&#xA;h = 180*10^-3;&#xA;&#xA;m = 8; %Mass of the mast in kg&#xA;g = 9.81; %Acceleration of gravity in m/s²&#xA;&#xA;%Range of the input variables&#xA;thetaMin = 0*pi/180;&#xA;thetaMax = 100*pi/180;&#xA;tMin = 10*10^-3;&#xA;tMax = 100*10^-3;&#xA;&#xA;%Vectors&#xA;theta = thetaMin:(thetaMax-thetaMin)/100:thetaMax;&#xA;t = tMin:(tMax-tMin)/100:tMax;&#xA;&#xA;%Solve for F&#xA;F = zeros(length(t), length(theta)); %Actuator force&#xA;L = F; %Length of linear actuator&#xA;denominator = F; %Denominator of F, or &quot;signed lever arm&quot;&#xA;for i = 1:length(t)&#xA;    for j = 1:length(theta)&#xA;        M = [cos(theta(j)) sin(theta(j)) 0;&#xA;             -sin(theta(j)) cos(theta(j)) 0;&#xA;             0 0 1];&#xA;&#xA;        HP = M*[-t(i);&#xA;                a;&#xA;                0];&#xA;&#xA;       OP = [-b;&#xA;              h;&#xA;              0]+HP;&#xA;&#xA;       L(j,i) = norm(OP);&#xA;&#xA;        HG = M*[c;&#xA;                -d;&#xA;                0];&#xA;&#xA;        W = [0;&#xA;             -m*g;&#xA;             0];&#xA;&#xA;        weightMoment = HG(1)*W(2)-HG(2)*W(1); %cross(HG,W)&#xA;        forceLever = 1/norm(OP)*(HP(1)*OP(2)-HP(2)*OP(1)); %1/norm(OP)*cross(HP,OP)&#xA;&#xA;        denominator(j,i) = forceLever;&#xA;&#xA;        F(j,i) = -weightMoment/forceLever;&#xA;    end&#xA;end&#xA;&#xA;%Plot results&#xA;[T,THETA] = meshgrid(t,theta);&#xA;&#xA;figure(1)&#xA;surface(T*1000,THETA*180/pi,F)&#xA;ylabel('Angle (deg)')&#xA;xlabel('Slide length (mm)')&#xA;zlabel('Force (N)')&#xA;title('Actuator force')&#xA;&#xA;figure(2)&#xA;surface(T*1000,THETA*180/pi,L*1000)&#xA;ylabel('Angle (deg)')&#xA;xlabel('Slide length (mm)')&#xA;zlabel('Actuator length (mm)')&#xA;&#xA;figure(3)&#xA;surface(T*1000,THETA*180/pi,denominator*1000)&#xA;ylabel('Angle (deg)')&#xA;xlabel('Slide length (mm)')&#xA;zlabel('Signed lever arm')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Maths: solve&#xA;$$\vec{moment}(H)=\vec{HP}\times \vec{F}+\vec{HG} \times \vec {W}=\vec{0}$$&#xA;With&#xA;$$\vec{F}=\frac{\vec{OP}}{OP}F$$&#xA;$$\vec{HP}=&#xA;\left[ {\begin{array}{cc}&#xA;cos(\theta) &amp;amp; sin(\theta) \\&#xA;-sin(\theta) &amp;amp; cos(\theta)&#xA;\end{array} } \right]&#xA; \left( {\begin{array}{cc}&#xA;-t \\&#xA;a &#xA;\end{array} } \right)&#xA;$$&#xA;$$\vec{HG}=&#xA;\left[ {\begin{array}{cc}&#xA;cos(\theta) &amp;amp; sin(\theta) \\&#xA;-sin(\theta) &amp;amp; cos(\theta)&#xA;\end{array} } \right]&#xA; \left( {\begin{array}{cc}&#xA;c \\&#xA;-d &#xA;\end{array} } \right)&#xA;$$&#xA;$$\vec{OP}=\vec{OH}+\vec{HP}$$&#xA;$$\vec{W}=\left( {\begin{array}{cc}&#xA;0 \\&#xA;-mg &#xA;\end{array} } \right)$$&#xA;$$\vec{OH}=\left( {\begin{array}{cc}&#xA;-b \\&#xA;h &#xA;\end{array} } \right)$$&lt;/p&gt;&#xA;" OwnerUserId="1326" LastEditorUserId="1326" LastEditDate="2015-07-30T09:17:33.870" LastActivityDate="2015-07-30T09:47:08.030" Title="Discontinuity in mechanism simulation: is the mechanism jammed or is my analysis wrong?" Tags="|mechanical-engineering|modeling|mathematics|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3730" PostTypeId="1" CreationDate="2015-07-29T16:11:28.740" Score="-2" ViewCount="83" Body="&lt;p&gt;Assuming we have 2 identical cars. What is the total energy consumed by the two cars to start and accelerate up to a specific speed versus the total energy for one of them to accelerate up to a specific speed having the other tied to it? Or the fuel consumed for two vehicles to run on a road at 50mph and the fuel consumed for one vehicle tied  to the other. I basically don't want an exact Joules number or Fuel litres. I need to know the difference between each of the cases. Which is most economical and by how much?&lt;/p&gt;&#xA;" OwnerUserId="2353" LastEditorUserId="2353" LastEditDate="2015-07-29T17:03:26.110" LastActivityDate="2015-07-29T17:46:00.743" Title="seperate engines working vs single engine" Tags="|automotive-engineering|energy-efficiency|fuel-economy|" AnswerCount="1" CommentCount="8" ClosedDate="2015-07-30T11:25:55.597" ContentLicense="CC BY-SA 3.0" />
  <row Id="3731" PostTypeId="1" CreationDate="2015-07-29T16:35:25.387" Score="0" ViewCount="81" Body="&lt;p&gt;Currently, I have to design a final test for a product. I want to summarize the results in a test report. The report is automatically generated and gives a summary (quantitatively) of which specifications are met and which are not met. If all the specifications are met, the test contains an &quot;OK.&quot; Contained are product ID, diagrams, key figures, some explanatory paragraphs, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such a report have a specific technical term in quality engineering? With this term, I may be able to learn about best practices, etc., regarding such reports.&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="368" LastEditDate="2015-07-29T19:08:52.207" LastActivityDate="2015-07-29T19:08:52.207" Title="What is the technical term for a document that contains the results of a final test for a specific product?" Tags="|terminology|quality-engineering|industrial-engineering|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3732" PostTypeId="2" ParentId="3731" CreationDate="2015-07-29T17:39:51.707" Score="2" Body="&lt;p&gt;In most organizations, how you refer to a test in the report depends on the purpose of the test and in turn that will be reflected in the established document naming system. For example if the purpose of your test is validation, the report template would be named something like Validation Report - (Project XYZ) or something like that. Typically all of this should be set up in the quality management system. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-07-29T17:39:51.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3733" PostTypeId="2" ParentId="3730" CreationDate="2015-07-29T17:46:00.743" Score="2" Body="&lt;p&gt;In your scenarios, there are two places that the energy ends up: as kinetic energy in the vehicles themselves (0.5 &amp;times; mass &amp;times; velocity squared) and as losses (rolling resistance, air resistance, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two possibilities regarding your two scenarios.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The two scenarios occur in the same amount of time. This means that the total energy is the same in the two scenarios, and that in the two-engine scenario, the engines are operating at less than half their capacity. In this case, the energy consumed would be based only on the relative efficiency of the engines at the two different power levels.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The single-engine scenario takes more time to reach the final speed than the two-engine scenario. In this case, while the final kinetic energy is the same, you have more losses while getting there, and therefore more energy input will be required, regardless of the engine efficiency numbers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-07-29T17:46:00.743" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3734" PostTypeId="1" AcceptedAnswerId="3735" CreationDate="2015-07-29T19:27:19.243" Score="9" ViewCount="2671" Body="&lt;p&gt;I am an engineering student and I took my second dedicated controls class last semester. It was cool math and whatnot, but after all was said and done: I still have no idea how to make an actual working control system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, there was no tie-in to reality at all; it was all Laplace-domain mathematics that somehow affected our result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know is: how do I go from real-space, to controls-space, and back to real-space? To help with understanding, I have prepared an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have platform that is supported by a vertical rod. This rod can freely move up and down, or rather, it would be able to if it was not held in position by a motor (either a servo or a gear). The job of this motor is to keep the platform at the same height. E.g., if I place a mass on the platform, the motor must be able to adjust to it and respond accordingly such that after the response, the height of the platform is the same as before. (Of course, with a specified rise time, maximum overshoot, yada yada...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might sound trivial to you controls engineers out there, but I really don't know how to do this. I get that it is a closed loop system, but what I don't know is what all these laplace-space block diagrams correlate to in the real world. How do I get the block diagram for this system, and how do I build a controller inside of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of this example is to extend it to other problems, like airplane control, but I figure if I don't understand this I probably won't be able to do those other (cooler) things!&lt;/p&gt;&#xA;" OwnerUserId="2456" LastEditorUserId="368" LastEditDate="2015-07-29T20:03:01.717" LastActivityDate="2021-10-30T19:12:10.203" Title="How can I approach the application of control theory to real control systems?" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3735" PostTypeId="2" ParentId="3734" CreationDate="2015-07-29T20:03:22.863" Score="9" Body="&lt;p&gt;I hope I can help you in some way, your question bounces around a bit so I'm guessing a little on what you are wanting to know. This is how you come up with a controller that is implementable from scratch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Make the model (&quot;plant&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The #1 preliminary skill a controls engineer needs is to be able to make a mathematical model of a system, whether it be chemical, mechanical, electrical etc. For example, what are the equations of motion for your example above with the motor and the vertical platform? When making the model, you may also want to decide if you'll be implementing &lt;a href=&quot;https://en.wikipedia.org/wiki/State_space&quot;&gt;state space&lt;/a&gt; or classical control methods later on. I'm guessing your controls class covered at least the classical control methods (PID, phase lag, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, a working model of a plant (system) that a control engineer can work with is typically the transfer function (laplace relation between input and output) or the state space model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Decide what kind of control you want to implement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deciding how to control your modeled system can depend on several factors; like what measurements you can make available to the controller, what performance criteria you need to achieve etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Design the controller&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where you do the math, you draw the block diagram and evaluate how the controlled plant will perform. This is what you've been learning in class. This is why you are an engineer, because of the mathematics you know how your system will perform before you build it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Implement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you just need to find the hardware/software to implement your designed controller. For example, if you're going to do full state feedback, you need sensors to measure the states, amplifiers to add/subtract signals, and the required input ability to make your plant behave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall you need to model the system, know what kind of control to use, and then use the techniques from class to analyze how it will pan out. Implementing is just about knowing what kind of hardware to use. &lt;/p&gt;&#xA;" OwnerUserId="1174" LastEditorUserId="1174" LastEditDate="2015-07-29T20:13:30.890" LastActivityDate="2015-07-29T20:13:30.890" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3736" PostTypeId="1" CreationDate="2015-07-29T22:21:50.500" Score="2" ViewCount="1705" Body="&lt;p&gt;I am trying to find out information about the heat treatment processes of ferrous metals as part of a project which involves making a screwdriver. This includes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hardening, Annealing, Tempering, Carbonising and Quenching&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to the atomic structure of the metal at each stage of the heat treatment process?&lt;/p&gt;&#xA;" OwnerUserId="2460" LastEditorUserId="2460" LastEditDate="2015-08-01T01:50:55.597" LastActivityDate="2016-01-13T02:09:52.630" Title="Heat treatment processes for ferrous metals" Tags="|metallurgy|metals|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3737" PostTypeId="1" AcceptedAnswerId="3738" CreationDate="2015-07-29T22:55:47.283" Score="5" ViewCount="783" Body="&lt;p&gt;Is there a term used in systems engineering that describes a situation in which the design of one subsystem (call it &quot;A&quot;) affects the the requirements for another subsystem (call it &quot;B&quot;) &lt;em&gt;while &lt;strong&gt;at the same time&lt;/strong&gt; the design of B affects the requirements for A&lt;/em&gt;? &lt;/p&gt;&#xA;" OwnerUserId="2461" LastEditorUserId="272" LastEditDate="2015-07-29T23:14:35.627" LastActivityDate="2015-07-30T00:39:49.887" Title="Is there a term for a circular dependency in system design and engineering" Tags="|terminology|systems-engineering|systems-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3738" PostTypeId="2" ParentId="3737" CreationDate="2015-07-29T23:27:57.843" Score="9" Body="&lt;p&gt;If for some reason &quot;circular dependency&quot; doesn't work—it seems clear enough to me—you could also say that subsystems A and B are &quot;interdependent&quot; in their design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are essentially synonyms and in either case you may have to go on to explain exactly what you mean and why. But at least it's a bit more concise and less awkward a construction than &quot;circularly dependent&quot; if you need to keep referring to that aspect.&lt;/p&gt;&#xA;" OwnerUserId="368" LastActivityDate="2015-07-29T23:27:57.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3739" PostTypeId="2" ParentId="3737" CreationDate="2015-07-30T00:39:49.887" Score="5" Body="&lt;p&gt;Another term that would be applicable is &lt;strong&gt;&lt;em&gt;mutually reliant&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-07-30T00:39:49.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3740" PostTypeId="1" CreationDate="2015-07-30T06:56:47.297" Score="4" ViewCount="5113" Body="&lt;p&gt;I am taking a civil engineering course that we are constantly asked to tackle questions that are not really covered in the course. I have been looking for suitable reference materials but with no success. Here is one of the questions that I am stuck at:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Find the shear and bending moment diagram of the following rigid&#xA;  frame:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/arUL7.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have tried working out the forces but they seem to be wrong after checking with a free program called &quot;2D frame analysis.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I approach this problem?&lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="368" LastEditDate="2015-07-30T15:38:25.790" LastActivityDate="2019-12-28T18:14:53.383" Title="Drawing the shear and bending moment diagram of a 2D rigid frame" Tags="|mechanical-engineering|civil-engineering|statics|" AnswerCount="1" CommentCount="4" ClosedDate="2015-07-30T15:34:27.740" ContentLicense="CC BY-SA 3.0" />
  <row Id="3741" PostTypeId="1" CreationDate="2015-07-30T08:56:52.313" Score="2" ViewCount="713" Body="&lt;p&gt;The AASHTO &lt;em&gt;LRFD Bridge Specification&lt;/em&gt;, Section 10.5.3.3 says &quot;shaft lateral resistance&quot; should be checked at the strength limit state (i.e. factored loads). How is shaft lateral resistance determined using an LPile model? Is it the lateral load required to form a plastic hinge in the shaft?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm dealing with a very short shaft (light pole foundation) am I likely to be governed by service limit state deflections &amp;amp; overall stability instead?&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2246" LastEditDate="2015-07-30T16:17:28.950" LastActivityDate="2015-07-30T16:17:28.950" Title="How is drilled shaft lateral resistance calculated using LPile?" Tags="|structural-engineering|geotechnical-engineering|foundations|aashto|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3742" PostTypeId="2" ParentId="3729" CreationDate="2015-07-30T09:47:08.030" Score="3" Body="&lt;p&gt;If you plot the x values of OP and HP, they cross zero, which seems unphysical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking more closely, you are doing your rotations backwards, i.e. you are doing our rotations in +theta whereas you want to go in -theta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is something I often do too. To fix, just replace theta with -theta in M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a figure of F when you do this which looks much more reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wUPCu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wUPCu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-07-30T09:47:08.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3743" PostTypeId="2" ParentId="3741" CreationDate="2015-07-30T14:04:44.187" Score="1" Body="&lt;p&gt;Looking at the &lt;a href=&quot;http://www.ensoftinc.com/main/products/lpile6/m_lpile6.html&quot; rel=&quot;nofollow&quot;&gt;LPile website&lt;/a&gt;, it looks like they plot the strength of the pile in terms of a &lt;em&gt;p-y&lt;/em&gt; curve.  From the website,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nonlinear lateral load-transfer from the foundation to the soil is modeled using &lt;em&gt;p-y&lt;/em&gt; curves generated internally using published recommendations for various types of soils. Special procedures are programmed for computing &lt;em&gt;p-y&lt;/em&gt; curves for layered soils and for rocks. Alternatively, the user can enter manually any other externally generated &lt;em&gt;ip-y&lt;/em&gt; curves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, your lateral strength is based on the loading on the pile and the lateral deflection of the pile under load.  I think this is analogous to a &lt;em&gt;P-M&lt;/em&gt; curve one would generate for a concrete column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength limit state then depends on your lateral soil support and the strength of the pile, so it's not really possible to say whether or not a plastic hinge forms in the pile without further information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a short pile length, I would agree that you're likely limited by deflection and stability since the pile doesn't have a lot of bite into the soil depth.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-07-30T14:04:44.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3744" PostTypeId="2" ParentId="3740" CreationDate="2015-07-30T14:48:15.390" Score="4" Body="&lt;p&gt;This is an isostatic structure, which means it can be &quot;trivially&quot; solved (once one is more experienced) by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to do is to calculate the reactions. In this case you can see that all forces are horizontal and only &lt;span class=&quot;math-container&quot;&gt;$R_A$&lt;/span&gt; can withstand horizontal forces, so we already know that &lt;span class=&quot;math-container&quot;&gt;$X_A$&lt;/span&gt; will be equal to the negative sum of all external forces (defining positive as to the right):&#xA;&lt;span class=&quot;math-container&quot;&gt;$$X_A = -10 - 3\cdot4 = -22\text{kN}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the vertical reactions we need to calculate the moment around a given point, which must be equal to zero. I choose point &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$M_A = -3\cdot4\cdot\frac{4}{2} - 10\cdot5 + Y_B\cdot4 = 0\\&#xA;\therefore Y_B = 18.5\text{kN}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the sum of vertical loads must be null, this means that &lt;span class=&quot;math-container&quot;&gt;$Y_A = -18.5\text{kN}$&lt;/span&gt;. Knowing all the reactions we can now calculate the internal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start at point C. Here we have a concentrated transverse force, which means we'll have a constant shear force of &lt;span class=&quot;math-container&quot;&gt;$10\text{kN}$&lt;/span&gt; from C to D. At point C the bending moment will be null, while at point D it will be equal to the product of the force and its lever arm, so &lt;span class=&quot;math-container&quot;&gt;$M_{D,C} = 10\cdot1 = -10\text{kNm}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$M_{D,C}$&lt;/span&gt; is the moment at D on the side of C. It is negative because I arbitrarily chose counter-clockwise rotations as positive.). Axial forces here will clearly be null.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's go to point B. The reaction gives us a concentrated shear force of &lt;span class=&quot;math-container&quot;&gt;$18.5\text{kN}$&lt;/span&gt; along B to D. Just as above, the bending moment at D will therefore be &lt;span class=&quot;math-container&quot;&gt;$M_{D,B} = 18.5\cdot4 = +74\text{kNm}$&lt;/span&gt; (the force is trying to rotate point D counter-clockwise, therefore the moment is positive). Once at D, the shear force will become an axial force along A to D, which means this segment will experience tension of &lt;span class=&quot;math-container&quot;&gt;$18.5\text{kN}$&lt;/span&gt;. The same does not apply to the shear force generated at D because the support at B can't restrain that horizontal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's go to point A. The negative vertical reaction gives us an axial tensile force which we saw before, so that tells us so far, so good. The negative horizontal force gives us a concentrated shear force which is diluted by the distributed loads along A to D. The shear force at D will be equal to the sum of the (negative) reaction with the distributed load, so &lt;span class=&quot;math-container&quot;&gt;$-22 + 3\cdot4 = 10\text{kN}$&lt;/span&gt;, which is equal to the concentrated force at C, which means there is no shear discontinuity at D between A and C. This is also a good sign since there is no reason for there to be a discontinuity since there are no horizontal loads or reactions coming from B. The moment at D will be equal to the partial moments due to the reaction and the distributed load, so &lt;span class=&quot;math-container&quot;&gt;$M_{D,A} = -22\cdot4 + 3\cdot4\cdot\frac{4}{2} = -64\text{kNm}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how the total moment around D is null: &lt;span class=&quot;math-container&quot;&gt;$-10+74-64 = 0\text{kNm}$&lt;/span&gt;. This must always be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this we have all the internal forces of the structure, as can be seen here (figures created with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool. It is set to omit the bending moment sign):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v7Hzo.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v7Hzo.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:14:53.383" LastActivityDate="2019-12-28T18:14:53.383" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="3745" PostTypeId="1" AcceptedAnswerId="3889" CreationDate="2015-07-30T16:14:07.703" Score="5" ViewCount="18662" Body="&lt;p&gt;Occasionally I have to design T-connections between rectangular tube steel (HSS) members where I have a flare-bevel weld between the corner of the tube and the connecting flat member.  If the weld stress is high enough, a reinforcing fillet may be required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per AWS D1.1-2010, Section 2.4.2.7, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LBR1B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LBR1B.png&quot; alt=&quot;Section 2.4.2.7&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective throat is illustrated by Figure 3.3 and Annex A (pasted below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F5Tf1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F5Tf1.png&quot; alt=&quot;Figure 3.3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vfo49.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vfo49.png&quot; alt=&quot;Annex A&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus far, I have been unable to easily calculate the minimum effective throat of a flare-bevel with reinforcing fillet due to the complex geometry of the weld throat.  The best method I have found so far is to draw it in AutoCAD to scale and figure it out from there, but this is relatively time-consuming.  It is not immediately evident to me which plane of the weld is the minimum throat since the contour of the weld changes along its width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone seen a formulaic (i.e., programmable) methodology for determining the effective throat of a flare-bevel with reinforcing fillet?  I mostly work my calculations in &lt;a href=&quot;http://www.ptc.com/engineering-math-software/mathcad&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mathcad&lt;/a&gt;, so having something I can program into steps would be highly beneficial to my workflow.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2017-02-02T20:59:26.943" Title="Effective Throat of Flare-Bevel Weld with Reinforcing Fillet" Tags="|structural-engineering|welds|hss|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3746" PostTypeId="2" ParentId="3492" CreationDate="2015-07-30T18:53:44.893" Score="1" Body="&lt;p&gt;&lt;strong&gt;What do you think is the reason for the difference in values?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's likely that they have calculated the coefficients of whatever model they are using from different data sets (or sub-sets). It's also possible that they are using different models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;is this level of accuracy sufficient?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That depends on what you're using it for. In general you probably have an acceptable level of error in your output. To get an idea of whether your input accuracy is acceptable you can do a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sensitivity_analysis&quot; rel=&quot;nofollow&quot;&gt;sensitivity analysis&lt;/a&gt; on your simulation to see what kind of accuracy you need for your inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess however, is that a variation of less than 1% will probably not be a significant source of error in your output compared to the simplifications of your model, and thus the sensitivity analysis is unnecessary for the case of errors in $C_p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not know the accuracy of the rest of the package, so I'd recommend running a sensitivity analysis on your model, and an error estimation of each of your inputs to give you an idea of how much accuracy you need and if the package you use can deliver that accuracy.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-07-30T18:53:44.893" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3747" PostTypeId="2" ParentId="3547" CreationDate="2015-07-30T20:13:20.323" Score="11" Body="&lt;p&gt;To slightly generalize I'll reform the question slightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ridged 2-D body (car) has a line $l$ that moves with it.&#xA;The car can be linearly transformed as long as the instantaneous center of rotation lies along $l$ at least distance $R$ away from a point $c$ that also moves with the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case point $c$ lies in the center of the rear axle and $l$ lies on the rear axle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine the car's domain is limited to a quarter plane with edges $A$ and $B$. It initially is placed against $A$, far from $B$ with $l$ perpendicular to $A$, and the goal is to translate the car so that it is  against $B$ far from $A$ while minimizing the maximum distance from the closest edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($A$ and $B$ can be placed a inch away from the actual walls to prevent scratches and allow for non-idealized vehicle movement.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Reversals allowed&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The solution is to advance the car along $A$ until it is an infinitesimal distance from $B$ (using an infinite turning radius to travel in a straight line) Then rotate about the tightest turning radius until in contact with $B$ Then rotate about the tightest turning radius on the opposite side until back in contact with $A$. This results in linear movement in the opposite direction but rotation in the same direction. These two steps can be repeated (infinitely) until $l$ is perpendicular to $B$ at which point it can advance away from $A$ in a straight line. From a macro perspective this looks like the car sliding along $A$ until it reaches $B$, then rotating while maintaining contact with both walls and finally advancing along $B$. This solution is independent of turning radius but involved infinite reversals.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;No reversals&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now lets further constrain our translations so that center of rotation must be further from $A$ and $B$ than $c$. (This removes the usefulness of backing up)&#xA;Now the middle of the optimum strategy is obvious: turn at the maximum turning radius, but how do you minimize the distance to the wall approaching and exiting this strategy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You remain in contact with the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you approach the wall and see you're just about to clear it, rather than continuing to turn you can gradually increase the turning radius to remain in contact with the wall. To remain in contact with the wall means the line between the contact point and the center of rotation is perpendicular to the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we can get the position of the center of rotation while in the minimum turning radius portion of the turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oANXk.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oANXk.png&quot; alt=&quot;COR possition&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_{rear}=\sqrt{{O_{rear}}^2+(R_{min}+W)^2}$$&#xA;$$D_{front}=\sqrt{(O_{front}+WB)^2+(R_{min}+W)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This point fully defines the most interesting portion of the turn allowing one to see if any obstacle on the other side would be struck. To clear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bUZaB.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bUZaB.png&quot; alt=&quot;Tight corner diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{(D_{rear}-b)^2+(D_{front}-a)^2} \leq R_{min}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it makes a difference if you're going forward or backward. To see if you'd clear both directions you'd have to test with a and b reversed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed in the diagram above I've set $a=5.9m$ and $b=3.3m$. In this case it was because while the thick arc defined by drawing and equations above might be the most interesting portion of the curve, it is not going to be the limiting factor when $a$ and $b$ are not flipped. So we need to extend that curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end points can be found using similar triangles, from there, the curve will just be an tangent exponential decay to a distance $W$ from the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XoWTR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XoWTR.png&quot; alt=&quot;Diagram of swept closest point with added exponential sections&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these curves we can define a function $C$ to tell us whether the vehicle would clear an object placed at $(a,b)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C(a,b) =&#xA;  \begin{cases} &#xA;      \hfill \sqrt{(D_{rear}-b)^2+(D_{front}-a)^2} \leq R_{min}   \hfill &amp;amp; \text{ if } a \leq a_{check} \text{ and } b \leq b_{check} \\&#xA;      \hfill W+W_{rear}e^{\frac{(a_{check}-a)O_{rear}}{(R_{min}+W)W_{rear}}} \leq b   \hfill &amp;amp; \text{ if } a &amp;gt; a_{check} \text{ and } b \leq b_{check} \\&#xA;      \hfill W+W_{front}e^{\frac{(b_{check}-b)(O_{front}+WB)}{(R_{min}+W)W_{front}}} \leq a   \hfill &amp;amp; \text{ if } a \leq a_{check} \text{ and } b &amp;gt; b_{check} \\&#xA;      \hfill true   \hfill &amp;amp; \text{ if } a &amp;gt; a_{check} \text{ and } b &amp;gt; b_{check} \\&#xA;  \end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{check}=D_{front}-O_{rear}\frac{R_{min}}{D_{rear}}$$&#xA;$$b_{check}=D_{rear}-(O_{front}+WB)\frac{R_{min}}{D_{front}}$$&#xA;$$W_{front}=D_{front}-(R_{min}+W)\frac{R_{min}}{D_{rear}}-W$$&#xA;$$W_{rear}=D_{rear}-(R_{min}+W)\frac{R_{min}}{D_{front}}-W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to solve this system backwards to get the maximum $R_{min}$ that would allow passage requires making a few observations and assumptions. First we'll assume that you want to be able to drive around the corner in ether direction that means we'll swap $a$ and $b$ for whichever scenario is worse. If The front corner is further from the fixed axle than the rear corner (as is the case for all front steering vehicles I know of) then a &amp;lt; b is the tighter scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one could use a numerical method to find the $R_{min}$ that gave equality for the second inequality. If $a\geq a_{check}$ then you're done. If not, then find the $R_{min}$ that gives equality for the first inequality.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Glossary&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$W$ - Width of car&lt;/li&gt;&#xA;&lt;li&gt;$WB$ - Wheel Base&lt;/li&gt;&#xA;&lt;li&gt;$O_{front/rear}$ - Overhang of the front/rear&lt;/li&gt;&#xA;&lt;li&gt;$R_{min}$ - minimum distance between the center of rotation and the car&lt;/li&gt;&#xA;&lt;li&gt;$a$ - distance from outside wall to inside corner&lt;/li&gt;&#xA;&lt;li&gt;$b$ - distance from outside wall to inside corner&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Plugging in&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;With the numbers given It turns out that the maximum $R_{min}$ is slightly under $6.6m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4oqH3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4oqH3.png&quot; alt=&quot;Diagram of car just clearing the corner.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you might have to fold the right mirror in.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="1122" LastEditDate="2015-08-04T13:10:05.213" LastActivityDate="2015-08-04T13:10:05.213" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="3749" PostTypeId="1" CreationDate="2015-07-30T23:35:39.353" Score="5" ViewCount="2853" Body="&lt;p&gt;I came across this while reading on third angle orthographic projections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object has the exact same front and top view - both views are trapeziums of exact shape and size (As shown in the image). Also there is a hidden circle center line right thorough the middle of both of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to complete the projection as I cannot visualize this object - so I can't work out the right and left views. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The single image represents both the front and top views - I can't work out the left and right views.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qq7bb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qq7bb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2468" LastEditorUserId="33" LastEditDate="2015-07-31T02:26:53.337" LastActivityDate="2015-08-01T11:32:18.843" Title="Orthographic (engineering) drawing with same top and front view?" Tags="|mechanical-engineering|design|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3750" PostTypeId="2" ParentId="3749" CreationDate="2015-07-31T00:54:55.670" Score="5" Body="&lt;p&gt;I'm not convinced that the line is for a hidden feature. I would imagine the line indicates that entire part is revolved about the indicated axis. Naturally if the part has cylindrical symmetry, the top and front views can be made identical by proper orientation of the body relative to the drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming my understanding of the drawing is correct, I imagine the part to be a right conical frustum, which would look something like &lt;a href=&quot;http://mathworld.wolfram.com/ConicalFrustum.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this MathWorld entry&lt;/a&gt;, or as seen below. Simply rotate the 2D image 90 degrees counter-clockwise to achieve a proper orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e9v8U.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e9v8U.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if there is a hidden feature, then the part could look like a right square pyramidal frustum with a hidden hole, as shown here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w0IRA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w0IRA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-07-31T01:08:04.940" LastActivityDate="2015-07-31T01:08:04.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3751" PostTypeId="2" ParentId="3749" CreationDate="2015-07-31T05:23:13.920" Score="3" Body="&lt;p&gt;A simple way to approach a problem like this is to imagine that you are starting out with a block of material larger than your desired object. This is the &quot;blank&quot; out of which you'll fabricate, in your mind's eye, the 3D object with the maximum volume that could possibly satisfy your 2D projections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine you have somehow projected the front view onto the front face of your blank—or, whichever view onto the corresponding face. After projecting the front view image onto the front face of the blank, you must remove all the material that falls outside of the image. When you remove material from the face, you remove everything behind it, as though you're shooting straight through from front to back, parallel to the path of the projector beam. (Alternatively, if you like shop tools, think of tracing the projected shape and then setting your blank face-up on a band saw or drill press to remove material.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to see that projecting a perfect circle onto the front face of your blank would lead you to cut out a cylinder. In this first step, your result is simply an extrusion of the projection perpendicular to its surface, whatever its shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second step (and beyond, if applicable) is where it gets interesting. In your mind, reorient yourself so that you're looking from the perspective of your next view, and project the corresponding image onto that face. Repeat the process of drilling out/cutting away anything outside of the projected image, being careful to go straight through, perpendicular to line of sight in your new orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you go through this process with identical trapezoids as the top and front view, you get the right square pyramidal frustrum (or, in slightly less technical language, &lt;a href=&quot;http://mathworld.wolfram.com/TruncatedSquarePyramid.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;truncated square pyramid&lt;/a&gt;) from &lt;a href=&quot;https://engineering.stackexchange.com/a/3750/368&quot;&gt;starrise's answer&lt;/a&gt;, without the hidden hole. This is the maximum possible volume that satisfies the constraints of your top and front view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not, however, the &lt;em&gt;only&lt;/em&gt; possible volume. You can poke all sorts of skew holes through the object; you can round the corners, and if you round them enough, you'll get a truncated cone. One actual component of less-than-maximum volume that your trapezoidal views bring to mind is a pipe flange, something like this but with more prominent ribs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7RGXq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7RGXq.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach is more than just a mental exercise. If you wanted, using an actual block of wood, an overhead projector and some transparency sheets, you could actually manufacture that max-volume square pyramid using your trapezoid shape. Or, you could replicate the cover illustration of &lt;a href=&quot;https://en.wikipedia.org/wiki/G%C3%B6del,_Escher,_Bach&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gödel, Escher, Bach&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7m5IQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7m5IQ.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above image, the top/bottom and side views are projected through the finished component, revealing the orthogonal shapes that you would have to cut from a blank to produce a (maximum volume) reproduction of that component.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-07-31T15:41:27.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3752" PostTypeId="1" AcceptedAnswerId="3753" CreationDate="2015-07-31T08:31:21.547" Score="-1" ViewCount="549" Body="&lt;p&gt;I'm looking for a material with a Young's modulus of less than 100 Pa. I've looked at elastomers, but I haven't had any luck.&lt;/p&gt;&#xA;" OwnerUserId="2474" LastEditorUserId="272" LastEditDate="2015-07-31T11:53:40.697" LastActivityDate="2015-07-31T12:54:07.673" Title="Material with very small Young's modulus" Tags="|stresses|biomechanics|elastic-modulus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3753" PostTypeId="2" ParentId="3752" CreationDate="2015-07-31T12:20:35.247" Score="5" Body="&lt;p&gt;You are asking for an unrealistic material.  If you peruse the &lt;a href=&quot;http://www.engineeringtoolbox.com/young-modulus-d_417.html&quot; rel=&quot;nofollow&quot;&gt;engineering toolbox&lt;/a&gt; you'll find that the Young's modulus of typical materials is measured in 10&lt;sup&gt;9&lt;/sup&gt; Pa so the value you are asking for is 7 orders of magnitude smaller than typical materials.  Even the Young's modulus of the rubber in rubber bands is ~10&lt;sup&gt;7&lt;/sup&gt; Pa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another simple thought experiment will convince you of how unreasonable such a material is.  The change in altitude required to change the pressure by 100 Pa is 10 meters which is about 3 building stories.  Since Young's modulus is defined by &#xA;$$&#xA;\frac{\Delta \ell}{\ell}=\frac{P}{E},&#xA;$$&#xA;where $E$ is Young's modulus, the strain on your material would be 1 if you carried it from the top of a 3 story building to the ground floor.  This means that the material would compress to nothing from such a minor change in elevation.  Real materials would, of course, be well into the non-linear regime at that point.&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-07-31T12:54:07.673" LastActivityDate="2015-07-31T12:54:07.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3754" PostTypeId="2" ParentId="2464" CreationDate="2015-07-31T14:08:00.913" Score="3" Body="&lt;p&gt;Why would $z$ need to be external to $g$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g'(t,x(t),y(t))=g(t,x(t),y(t),z(x(t)))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now use $g'$ as $g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g$ can be any arbitrary function, so any function $z$ could just be incorporated into $g$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your restriction $h$ mentioned in the comment section. Any restrictions on the control input could be enforced via the cost function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{new}(t,x(t),y(t))=f_{old}(t,x(t),y(t))-C\,h(x(t),y(t))^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $C$ is sufficiently large to guarantee values of $h$ close enough to zero but not so large that numerical errors in $h$ would dominate the original $f$.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="3633" LastEditDate="2016-05-02T15:34:31.233" LastActivityDate="2016-05-02T15:34:31.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3755" PostTypeId="1" AcceptedAnswerId="3760" CreationDate="2015-07-31T14:14:40.423" Score="1" ViewCount="105" Body="&lt;p&gt;I have a wheel I CNC milled &lt;a href=&quot;http://www.lowes.com/pd_304090-210-304090.0___?productId=3122447&amp;amp;pl=1&amp;amp;Ntt=foam&quot; rel=&quot;nofollow&quot;&gt;out of foam&lt;/a&gt; and I need to mount a motor to it. I designed and printed a part that would attach the motor to the foam, but I anticipated that the wheels would be wood back then. It connects to the wheel using screws, but it does not seem like screws would hold well in foam because the foam can't hold threads. Is there any other way to do this? I was thinking of using longer screws with nuts on them, but I am open to other ideas. &lt;/p&gt;&#xA;" OwnerUserId="2164" LastEditorUserId="1174" LastEditDate="2015-07-31T18:45:03.223" LastActivityDate="2015-07-31T18:45:03.223" Title="How do you mount a polystyrene foam wheel to a motor?" Tags="|mechanical-engineering|materials|motors|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3756" PostTypeId="2" ParentId="3582" CreationDate="2015-07-31T14:55:32.710" Score="2" Body="&lt;p&gt;It looks like, from your math, you attempt to find the total normal force due to the pressure on the incline $F_p$, and then try to find a distance $l_a$ from point A such that if $F_p$ was applied there the torque would be equivalent to the torque caused by the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what you would use $l_a$ for, but if you're trying to equlibrate torques, it's easier to just calculate the torque due to the pressure (which you do in order to find $l_a$ anyway)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm going to start from your calculations and note the errors in bold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Throughout your calculation you've used the center dot to indicate scalar multiplication, while also using implied multiplication. Generally if the center dot is not everywhere, then it implies the dot product of vectors, so I've removed all instances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual torque is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau =  \int \vec{l} \times \vec{n} \, p \, \mathrm{d}A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\vec{n}$ is the vector normal to the surface, and $\vec{l}$ is the vector to the integrating point from axis A. As these vectors will always be perpendicular we can simplify using $l$ as the distance to point A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau =  \int l \, p \, \mathrm{d}A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = \int_{l_0}^{l_1} \int_0^b \! l \, p \, \mathrm{d}y\mathrm{d}l$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I'd integrate over y to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = b\int_{l_0}^{l_1}  \! l \, p \, \mathrm{d}l$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we can delay that as you did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we'll change our integration variable from $l$ to $z$ This step requires substituting the differential &lt;strong&gt;and changing the bounds&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$z=l\,\sin(\alpha)$$&#xA;$$\mathrm{d}z=\mathrm{d}l\,\sin(\alpha)$$&#xA;$$\mathrm{d}z \frac1{\sin(\alpha)}=\mathrm{d}l$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = \int_{z_0}^{z_1} \int_0^b \! l \, p \, \frac{1}{\sin(\alpha)} \mathrm{d}y\mathrm{d}z $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now inserting $0$ for $z_0$ and $h$ for $z_1$ makes sense &lt;strong&gt;but we can't substitute $z$ for $l$&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = \int_{0}^{h} \int_0^b \! l \, p \, \frac{1}{\sin(\alpha)} \mathrm{d}y\mathrm{d}z $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to substitute $l$ for $z\frac1{\sin(\alpha)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = \int_{0}^{h} \int_0^b \! \frac{z}{\sin(\alpha)} \, \rho g (2h-z) \frac{1}{\sin(\alpha)}\, \mathrm{d}y\mathrm{d}z $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = b \rho g \frac{1}{\sin^2(\alpha)} \left[h^3-\frac{1}{3}h^3 \right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while your substitutions weren't valid in some intermediate states the final torque was correct.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-07-31T14:55:32.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3757" PostTypeId="1" AcceptedAnswerId="3758" CreationDate="2015-07-31T15:14:27.970" Score="1" ViewCount="483" Body="&lt;p&gt;Temporary structures are usually constructed in a less precise manner than permanent structures. This is more so in a braced deep excavation where the walls are in constant motion as the excavation progresses downwards. During the assembly of walers and struts, there could be a small gap between the strut-waler connection due to fabrication imperfections. How does one decide whether the fillet weld connecting the strut to the waler need to be designed against compression due to possibility of such a small gap?&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2015-08-02T15:38:31.223" Title="Should a welded connection between strut and waler be designed to take compression load?" Tags="|structural-engineering|welds|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3758" PostTypeId="2" ParentId="3757" CreationDate="2015-07-31T15:23:09.200" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How does one decide whether the fillet weld connecting the strut to the waler need to be designed against compression &lt;strong&gt;due to possibility of such a small gap?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In our office, we generally design welds to take compression load due to the exact reason you're specifying.  Unless you can ascertain that you have full bearing between the two pieces (i.e., both pieces are milled to mate perfectly), I would say you should check compression on the weld.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a code reference, the closest I can find is from the AISC 360-10 specification, section J1.1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;em&gt;required strength&lt;/em&gt; of the connection shall be determined by &lt;em&gt;structural analysis&lt;/em&gt; for the specified &lt;em&gt;design loads&lt;/em&gt;, consistent with the type of construction specified, or shall be a proportion of the required strength of the connected members when so specified herein.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-08-02T15:38:31.223" LastActivityDate="2015-08-02T15:38:31.223" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3759" PostTypeId="1" CreationDate="2015-07-31T15:34:14.940" Score="3" ViewCount="212" Body="&lt;p&gt;I recently encountered the following practical problem. We have a system of several layers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a neodymium magnet; a disc, dimensions and grade to be determined&lt;/li&gt;&#xA;&lt;li&gt;1 mm of 925 sterling silver; we can suppose that this is a disc, diameter 15 mm&lt;/li&gt;&#xA;&lt;li&gt;0.5 mm of paper or tissue (cotton)&lt;/li&gt;&#xA;&lt;li&gt;1 mm of the same silver disc&lt;/li&gt;&#xA;&lt;li&gt;again, a neodymium magnet&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How can I estimate the attraction force between these two magnets in this system based on their dimensions and grade? I found several estimators online (&lt;a href=&quot;https://www.kjmagnetics.com/calculator.asp&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;, for example) but they all assume that there is nothing between the magnets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even being able to estimate &quot;X mm in diameter and Y mm in thickness would yield at least Z grams of pull&quot; would be great.&lt;/p&gt;&#xA;" OwnerUserId="2476" LastEditorUserId="33" LastEditDate="2016-01-27T02:57:36.693" LastActivityDate="2021-02-19T20:06:20.683" Title="How is the attractive force between two magnets changed when they are separated by non-metalic solids?" Tags="|magnets|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3760" PostTypeId="2" ParentId="3755" CreationDate="2015-07-31T16:06:33.157" Score="0" Body="&lt;p&gt;Using a longer screw with fender washers may work. You could try soldering a pin or something to the washers to avoid slippage. Then just clamp with a longer bolts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be concerned that adhesives may eat away at the foam. You'd have to try a couple different kinds first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have a motor with a threaded shaft?&lt;/p&gt;&#xA;" OwnerUserId="2477" LastActivityDate="2015-07-31T16:06:33.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3761" PostTypeId="2" ParentId="3755" CreationDate="2015-07-31T16:50:14.607" Score="1" Body="&lt;p&gt;I would design a hub with a flange to fit against the side of the wheel and use an adhesive that's chemically compatible with the foam and has good bonding strength to metal. 3M has some solutions that we've used for bonding foam sheets. You'll want to take advantage of the fact that adhesives are strongest in shear rather than tension/peel. You could also have an end plate with adhesive that would bolt through the foam wheel creating a &quot;sandwich&quot;. The torque should be transmitted by the adhesive surfaces since the bolts will tear up the foam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a mount that is removable, use an adhesive that's soluble.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-07-31T16:50:14.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3762" PostTypeId="1" CreationDate="2015-07-31T16:59:27.480" Score="5" ViewCount="625" Body="&lt;p&gt;How can I track a device's location (x,y,z coordinates) as well as the direction the device is facing (north, south, up, down, etc.)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard accelerometers are not precise enough for very small scale (personal hand-held devices) and tend to drift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, satellite GPS is not available in the location where this device would be used.&lt;/p&gt;&#xA;" OwnerUserId="2478" LastEditorUserId="368" LastEditDate="2015-07-31T19:57:08.600" LastActivityDate="2015-08-01T00:11:48.643" Title="Tracking location and direction without GPS?" Tags="|electrical-engineering|measurements|sensors|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3763" PostTypeId="2" ParentId="2676" CreationDate="2015-07-31T17:55:00.683" Score="1" Body="&lt;p&gt;There are two wattage ratings for a peltier plates. One is the power consumed, the other is the heat transported across the chip &lt;strong&gt;when there is a temperature difference of zero&lt;/strong&gt;. An important note is that the amount of heat transported across the plate is dependent on the temperature difference across the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fIOiJ.gif&quot; alt=&quot;Peltier graph&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image courtesy of:&#xA;&lt;a href=&quot;http://www.heatsink-guide.com/peltier.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.heatsink-guide.com/peltier.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the graph above as an example, if the hot side of the peltier plate is not attached to a heat sink then it will quickly heat up to 70 degrees hotter than the beverage unit and heat will stop flowing outward. At that point if this plate was 10% efficient it would still be producing out about 400W of heat. That heat would be produced on the hot side of the chip* but that means that without a heat sink of some sort the hot side would quickly become even hotter than 70 degrees above the beverage unit and heat would start to flow back into the beverage unit, heating it rather than cooling it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the heat sinks on your beverage &quot;cooling&quot; unit will be very, very important to how well your system cools (or heats) your beverage, and without any information on them my guess is that your system will heat your beverage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*The heat is actually produced throughout the chip but it gets transferred out along with the other heat and the graph is compensated to account for this extra heat transfer, and only show the heat transferred away from the cool side.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-07-31T17:55:00.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3765" PostTypeId="2" ParentId="2041" CreationDate="2015-07-31T19:30:59.470" Score="3" Body="&lt;p&gt;The weld points will definitely be the limiting factor (think about a cargo net, where both the strands and joints are flexible, but usually the strands are straight and the joints are at whatever angle they need to be at to let the strands be straight.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the &quot;shear&quot; strength is listed in linear kN I'm going to assume that that is the effective strength of one joint resisting a force that would slide one bar along the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll further assume that the steel used has a tensile yield strength of 370 MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a good assumption is that the weld joints can be approximated as having a circular cross section. We can estimate the diameter of the cross section:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\frac{\sigma}2\,\pi\,r^2$$&#xA;$$d=\sqrt{\frac{8F}{\sigma\,\pi}}=\sqrt{8\frac{4.4\,\mathrm{kN}}{370\,\mathrm{MPa}\,\pi}}=5.5\,\mathrm{mm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, that's unreasonable...so either those welds are made of some super alloy, or the maximum shear load of those welds is less than 4.4 kN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:(&lt;/strong&gt; Actually according to &lt;a href=&quot;https://engineering.stackexchange.com/a/2058/1122&quot;&gt;this answer&lt;/a&gt; the welds can be significantly larger than the wires so this might not actually be unreasonable. &lt;strong&gt;)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the maximum &lt;em&gt;tensile&lt;/em&gt; load a 4 mm diameter cylinder of 1018 steel could hold before yielding is about 4.4 kN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's proceed as if the diameter of the weld is 1 mm. Each joint could then handle some torque $\tau$ according to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau=\frac{\pi\,\sigma_y\,d^3}{24} \approx 50 \,\mathrm{N\cdot mm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are approximately 250 joints so the maximum total torque is approximately 62 N·m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque produced by the weight be calculated via Virtual Work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elongation of a ridged mesh $\epsilon$ is approximately half the radians of deformation $\gamma$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravitational potential energy gained from a uniformly distributed mass by stretching is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=\frac12 m\,g\,h\,\epsilon= \frac14\,m\,g\,h\,\gamma$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau=\frac{dE}{d\gamma}=\frac14\,m\,g\,h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for mass gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m=250\frac{4\,\tau}{g\,h}\approx 2\,\mathrm{kg}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This indicates that if the strength of the welds is close to 1018 Steel, and the weld diameter is close to 1 mm then it wouldn't even come close to holding up its own weight (8 kg) when hung at a 45 degree angle. However, if the weld points are 2 mm in diameter then it could hold 15 kg allowing it to support its own weight with a safety factor of 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were my project I would do destructive testing on one weld joint to see how much torque it takes to yield it. That would allow you to plug in for $\tau$ and get a reasonable estimate for the maximum load.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-03-04T17:09:29.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3766" PostTypeId="2" ParentId="3762" CreationDate="2015-07-31T21:31:35.710" Score="3" Body="&lt;p&gt;There are several way to address your question, on such way is to use a&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Magnetometer to track North, West, East, South etc&lt;/li&gt;&#xA;&lt;li&gt;Barometric Pressure Sensor to measure up &amp;amp; down (elevation)&lt;/li&gt;&#xA;&lt;li&gt;Accelerometers to track motion/speed&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Depending on your application our sensing elements can be added to improve accuracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also take a look a the concept dead reckoning. Algorithm developed using &lt;a href=&quot;https://en.wikipedia.org/wiki/Dead_reckoning&quot; rel=&quot;nofollow noreferrer&quot;&gt;dead reckoning&lt;/a&gt; can address the lack of GPS satellite signal. Dead reckoning is commonly used in automotive navigation application as well as some enterprise handheld applications. I also responded to a similar post on electrical engineering stackexchange, and link is listed below.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/180648/how-can-i-measure-distance-travelled-in-cm-without-gps/180659#180659&quot;&gt;How can I measure distance traveled in cm without gps?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Dead_reckoning&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dead reckoning&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://learn.sparkfun.com/tutorials/bmp180-barometric-pressure-sensor-hookup-/measuring-weather-and-altitude&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barometric Pressure Sensor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.freescale.com/webapp/sps/site/prod_summary.jsp?code=MAG3110&quot; rel=&quot;nofollow noreferrer&quot;&gt;High Accuracy, 3D Magnetometer&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2015-08-01T00:11:48.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3767" PostTypeId="1" CreationDate="2015-08-01T07:13:33.050" Score="3" ViewCount="567" Body="&lt;p&gt;Membership of which professional organizations for electrical and electronic engineers are recognized globally?&lt;/p&gt;&#xA;" OwnerUserId="2483" LastEditorUserId="282" LastEditDate="2015-08-02T14:59:51.273" LastActivityDate="2015-08-02T14:59:51.273" Title="Membership of professional societies/associations for electrical and electronic engineers" Tags="|electrical-engineering|licensure|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3768" PostTypeId="1" CreationDate="2015-08-01T08:48:20.477" Score="2" ViewCount="146" Body="&lt;p&gt;I'm a student of Integrated Water Management, and looking at a geothermal bore that is producing potable water for a town's water supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bore water reaches the surface at $50$ $^oC$, and flows at 1.6 million liters per day.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's suggested a vapour expander engine could convert this heat to useful work, but such an engine may not commercially exist.  Nevertheless, I'd like to calculate the energy potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help with the Excel formula I'd need to calculate this?&lt;/p&gt;&#xA;" OwnerUserId="2484" LastEditorUserId="90" LastEditDate="2015-08-01T11:01:29.570" LastActivityDate="2015-08-01T14:58:50.157" Title="Calculate energy (kW, or kJ) from Vapour Expander Generator" Tags="|renewable-energy|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3769" PostTypeId="2" ParentId="3768" CreationDate="2015-08-01T09:53:59.890" Score="2" Body="&lt;p&gt;Any engine that works at $50$ $^oC$ on the hot side is going to have a low efficiency.    This may not matter but needs to be noted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theoretical max possible $\eta$ (efficiency) is that defined by the Carnot cycle.  Depending on the actual implementation the cycle used may be different, but this sets an upper possible theoretical limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume $T_{cold}$ = $25$ $^oC$. Colder may be possible but cold side thermal sinking can be challenging - as will be seen, an enormous amount of &quot;waste&quot; heat energy will be present. Not all of this will flow via the cold sink - but a significant amount will. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta_{carnot} = \frac{\triangle T}{T_{hot}}$$       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $T_{hot}$ at $50$ $^oC$ = $273$ $+$ $50$ $=$ $323$ $K$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\triangle T$ $=$ $(50-25)$ $=$ $25$ $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\eta_{carnot}$ $=$ $25/323$ $=$ $7.7\%$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At high $T_{hot}$/$T_{cold}$ ratios it is possible in practice to get &gt; $50\%$ $\eta_{carnot}$ but as $\triangle T$ drops and $T_{hot}/T_{cold}$ is small, only a small fraction of the Carnot efficiency is possible. Here with $\eta_{carnot}$ $=$ $7.7\%$ I'd guess that much more than $2\%$ efficiency would be hard to achieve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1.6$ x $10^6$ $l$ x $sg$ $=$ $1$ x $4.2$ $kJ/kg$ x $\triangle T$ x $2\%$   =~~~ $1.3$ x $10^8$ $J$ $=$ $3,360$ $MJ/day$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Power$ $=$ $E/day /24 hours /360$ $=$ $~$ $39$ $kW$&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/E%26OE&quot; rel=&quot;nofollow&quot;&gt;E&amp;amp;OE&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's certainly useful - but it comes from a lot of water -  - 18.5 litre/second!&lt;br&gt;&#xA;It's possible that better alternatives may exist.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Home heating, agriculture, &quot;fish raising&quot; - eg &lt;a href=&quot;http://www.hukaprawnpark.co.nz/&quot; rel=&quot;nofollow&quot;&gt;prawns&lt;/a&gt;, &lt;a href=&quot;http://www.aquacultureoftexas.com/aquaponics.htm&quot; rel=&quot;nofollow&quot;&gt;mixed aquaponics&lt;/a&gt; and more. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-08-01T14:58:50.157" LastActivityDate="2015-08-01T14:58:50.157" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3770" PostTypeId="2" ParentId="3767" CreationDate="2015-08-01T11:07:33.800" Score="0" Body="&lt;p&gt;The purpose, structure and the value proposition of each of these organizations differ between been each other yet recognize each other credentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if once credentials are US base and has a need to acquire similar credentials for Engineers Australia, then one would have to submit the necessary paper work to &lt;a href=&quot;https://www.engineersaustralia.org.au/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineers Australia&lt;/a&gt; for evaluation and acceptance. Under the auspicious of the &lt;a href=&quot;http://www.ieagreements.org/Washington-Accord/Washington-Accord-Overview.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Washington Accord&lt;/strong&gt;&lt;/a&gt; most credentials are cross listed. I can confirm that my US based Electrical Engineering credentials were accepted by Engineering Australia in accordance with the Washington Accord. Therefore I believe that my credentials will be accepted by any professional engineering organization that is part of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Washington_Accord&quot; rel=&quot;nofollow noreferrer&quot;&gt;Washington Accord&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore if your electrical engineering credentials are accepted by one body who is a part of the Washington Accord then your electrical engineering credentials are almost certain to be accepted by any other body that is part of the Washington Accord. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The following are the signatory accreditation bodies of the Washington Accord.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Australia - (&lt;a href=&quot;https://www.engineersaustralia.org.au/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineers Australia&lt;/a&gt;, 1989)&lt;/li&gt;&#xA;&lt;li&gt;Canada - (&lt;a href=&quot;https://www.engineerscanada.ca/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineers Canada&lt;/a&gt;, 1989)&lt;/li&gt;&#xA;&lt;li&gt;Taiwan - (&lt;a href=&quot;http://www.ieet.org.tw/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Institute of Engineering Education Taiwan&lt;/a&gt;, 2007)&lt;/li&gt;&#xA;&lt;li&gt;Hong Kong - (&lt;a href=&quot;http://www.hkie.org.hk/eng/homepage.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Hong Kong Institution of Engineers&lt;/a&gt;, 1995)&lt;/li&gt;&#xA;&lt;li&gt;India - (&lt;a href=&quot;http://www.nbaind.org/views/Home.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;National Board of Accreditation&lt;/a&gt;, 2014)&lt;/li&gt;&#xA;&lt;li&gt;Ireland - (&lt;a href=&quot;https://www.engineersireland.ie/home.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineers Ireland&lt;/a&gt;, 1989)&lt;/li&gt;&#xA;&lt;li&gt;Japan - (Japan Accreditation Board for Engineering Education, 2005)&lt;/li&gt;&#xA;&lt;li&gt;Korea - (&lt;a href=&quot;http://www.abeek.or.kr/htmls_kr/en/index.jsp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Accreditation Board for Engineering Education of Korea&lt;/a&gt;, 2007)&lt;/li&gt;&#xA;&lt;li&gt;Malaysia - (Board of Engineers Malaysia, 2009)&lt;/li&gt;&#xA;&lt;li&gt;New Zealand - (&lt;a href=&quot;http://www.ipenz.org.nz/ipenz/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Institution of Professional Engineers New Zealand, 1989&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Russia - (Russian Association for Engineering Education, 2012)&lt;/li&gt;&#xA;&lt;li&gt;Singapore - (Institution of Engineers Singapore, 2006)&lt;/li&gt;&#xA;&lt;li&gt;South Africa - (&lt;a href=&quot;https://www.ecsa.co.za/default.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineering Council of South Africa,&lt;/a&gt; 1999)&lt;/li&gt;&#xA;&lt;li&gt;Sri Lanka - (&lt;a href=&quot;http://www.iesl.lk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Institution of Engineers, Sri Lanka, 2014&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Turkey - (MÜDEK, 2011)&lt;/li&gt;&#xA;&lt;li&gt;United Kingdom - (&lt;a href=&quot;http://www.engc.org.uk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineering Council UK&lt;/a&gt;, 1989)&lt;/li&gt;&#xA;&lt;li&gt;United States - (&lt;a href=&quot;http://www.abet.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ABET&lt;/a&gt;, 1989)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are few post here in engineering stackexchange related to similar discussion.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/6/what-is-the-process-of-transferring-an-engineering-license-from-the-united-state&quot;&gt;What is the process of transferring an engineering license from the United States to Germany?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Washington_Accord&quot; rel=&quot;nofollow noreferrer&quot;&gt;Washington Accord&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sydney_Accord&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sydney Accord&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ieagreements.org/Agreement-Constitutions-Version-1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Professional Engineers Agreement, IPEA&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-08-02T11:54:55.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3772" PostTypeId="2" ParentId="3749" CreationDate="2015-08-01T11:32:18.843" Score="0" Body="&lt;p&gt;Starrise is right.   The 'dot-dash' line is standard notation for a centre line and the 'trapezium' is part of the standard notation to indicate first angle or third angle projection.   The end view is two concentric circles - positioned to the right of the 'trapezium' indicates first angle projection, positioned to the left indicates third angle projection.   See the Wikipedia entry 'multiview orthographic projection'.&lt;/p&gt;&#xA;" OwnerUserId="2485" LastActivityDate="2015-08-01T11:32:18.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3774" PostTypeId="1" CreationDate="2015-08-01T16:38:39.963" Score="5" ViewCount="318" Body="&lt;p&gt;I am working on a flat-foldable conical structure and I would like to carry out some dynamic tests on it (deployment to get the displacements and accelerations...). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have designed the folded geometry of this conical structure on Soliworks. This CAD file (geometry) is composed of a significant number of surperimposed surfaces bodies (triangular facets). Obtaining this file is quite complicated and involves a equations and relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to use Ansys Workbench and the module Explicit Dynamic, but I have a two issues to figure out and I don't know whether it is possible.&#xA;First, when I import the CAD geometry in Ansys Workbench, the MechanicalDesign module recognizes connections between the superimposed surfaces but not between the edges of the facets. I do want any connection between the surfaces so I simply deleted them, but I do know how to create relation between the edges. I do not know if there is a way to fix this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody has an idea or better solution to carry out this test ? As a student in Engineering school in another department than Structure and Material, I have only a basic background in the field of Finite Element and simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, my CAD file approximate (10^-15) the angles and lengths of the surfaces bodies (facets) and I know it can have an impact on the deployment of structure at the initial position... &lt;/p&gt;&#xA;" OwnerUserId="2486" LastEditorUserId="110" LastEditDate="2015-08-02T11:50:02.587" LastActivityDate="2015-08-02T11:50:02.587" Title="Explicit Dynamic ANSYS - Origami structure" Tags="|mechanical-engineering|structural-engineering|dynamics|simulation|ansys|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3775" PostTypeId="2" ParentId="3767" CreationDate="2015-08-01T20:46:24.547" Score="2" Body="&lt;p&gt;From an Australian perspective:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I have a Bachelor's degree in electrical engineering.&lt;/li&gt;&#xA;&lt;li&gt;I am a graduate member of Engineers Australia.&lt;/li&gt;&#xA;&lt;li&gt;I am a member of Institute of Electrical and Electronic Engineers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;My degree&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;My bachelor's degree in engineering was received from a university whose engineering course is accredited by Engineers Australia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the course is accredited by Engineers Australia, and because Engineers Australia is a signatory of the &lt;a href=&quot;http://www.ieagreements.org/Washington-Accord/&quot; rel=&quot;nofollow&quot;&gt;Washington Accord&lt;/a&gt;, my engineering degree is also recognised by &lt;a href=&quot;http://www.ieagreements.org/Washington-Accord/signatories.cfm&quot; rel=&quot;nofollow&quot;&gt;all other signatories to the Washington Accord&lt;/a&gt;. &lt;strong&gt;This means my engineering degree is recognised globally.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Engineers Australia&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Since I hold a bachelor's degree in engineering, and I work in Australia, I am eligible to be a member of Engineers Australia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Membership of this organisation is specific to my country, and is tied to some legislation which is also specific to my country.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Australian law requires one to be a Engineers Australia &lt;em&gt;Chartered Professional Engineer&lt;/em&gt; (CPEng) to perform certain duties. This doesn't immediately transfer to other countries, where you need to attain the local equivalent of CPEng to meet that country's requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly the other local Australian registers, such as the &lt;em&gt;Registered Professional Engineer, Queensland&lt;/em&gt; (RPEQ) and &lt;em&gt;National Professional Engineers Register&lt;/em&gt; (NPER) are tied to Australian laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where equivalents to CPEng, RPEQ, NPER exist overseas, there &lt;strong&gt;may&lt;/strong&gt; be a &lt;em&gt;mutual recognition agreement&lt;/em&gt; which automatically grants the foreign version if you have the Australian version, and vice versa. See the Engineers Australia document &lt;a href=&quot;https://www.engineersaustralia.org.au/sites/default/files/shado/Member%20Services/International/ea_trade.pdf&quot; rel=&quot;nofollow&quot;&gt;Supporting the Professional Mobility of Engineers in Australia and the United States.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;The IEEE&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Since I hold a bachelor's degree in electrical engineering, I am eligible to be a member of the IEEE. &lt;strong&gt;This is an internationally recognised membership&lt;/strong&gt; - my IEEE membership stays the same no matter what country I am living in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my membership of the IEEE doesn't mean much in Australia. It does not grant any special privileges or duties in Australia, and IEEE is not relevant to any Australian laws. Membership of Engineers Australia is considered much more important than IEEE.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2015-08-01T20:46:24.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3776" PostTypeId="1" CreationDate="2015-08-02T04:57:26.787" Score="3" ViewCount="566" Body="&lt;p&gt;There's a rooftop we recently discovered with a great view, and it's easily accessible for us. Obviously we'd like to go up there a bit more, particularly in summer with a couple of fold up chairs or something. I was thinking of getting someone in to look at it properly to assess the structural integrity of the roof, but I figured it might help posting on here first to get an early opinion from anyone willing to offer some advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roof feels solid. As with any metal roof, there's obviously a few minor creeks here and there, but as a whole it does feel strong with 3 people up there. It's also reassuring to note that workers go up there quite often, and hence the walkways that have been setup on it so far. We want to set something up similar to the current walk ways - a slab of plywood for instance that will distribute the weight evenly across the vertical seams when standing or sitting on the roof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roof area is 6.25x8.50m and here are some photos we took:&#xA;&lt;img src=&quot;https://imgkk.com/i/1r7h.jpg&quot; alt=&quot;a&quot;&gt;&#xA;&lt;img src=&quot;https://imgkk.com/i/7qt5.jpg&quot; alt=&quot;b&quot;&gt;&#xA;&lt;img src=&quot;https://imgkk.com/i/k4m1.jpg&quot; alt=&quot;c&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advice anyone gives on here will not be taken with absolute authority - I understand there's only so much you can do with the limited data I've provided. All advice will be considered general and not relied upon, but whatever thoughts you have to share would be greatly appreciated. I'm particularly interested in comments of absolute rejection to us going up there, along with a reason of substance explaining why it's &lt;em&gt;profusely&lt;/em&gt; dangerous for us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone also talked about running the structure in the opposite direction to the support beams. Would this be better, from the point of view of safety?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="2491" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.133" LastActivityDate="2015-08-26T15:13:19.113" Title="First glance: structural integrity of this roof" Tags="|structural-engineering|structures|safety|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3777" PostTypeId="1" AcceptedAnswerId="3779" CreationDate="2015-08-02T07:23:43.680" Score="1" ViewCount="3743" Body="&lt;p&gt;For a Board Exam Review: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A triple thread worm has a diameter of 3 inches. The wheel has 25&#xA;  teeth and a pitch diameter of 5 inches. Material for both the worm and&#xA;  the wheel is of phosphor Bronze. Compute the Helix Angle&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer is 0.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helix Angle can be obtained by trigonometry:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SOxup.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SOxup.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ \tan(H) = \frac{\pi \times D_{worm}}{L} }$$&#xA;$${ L = N_t * P_a }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where L is the lead, N is the number of threads and P is the Axial Pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ L = 3threads * 3in }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well thats where I stopped. I don't know how to find the diameter of the worm gear without it being explicitly given to me. I tried other formulas but they were more of the diameter of the wheel instead of the gear. Is there a formula for solving the diameter of the gear?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="2170" LastEditDate="2015-08-02T09:15:28.567" LastActivityDate="2015-08-02T14:02:43.160" Title="Finding the diameter of a Triple Thread Worm Gear" Tags="|design|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3778" PostTypeId="2" ParentId="3776" CreationDate="2015-08-02T13:59:59.150" Score="1" Body="&lt;p&gt;As you mention, this is a question that cannot possibly be answered in any definite way without an on-site inspection and, even then, it would be a (much more) educated guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, in broad terms we can state that it's probably structurally fine to do so, so long as you don't overcrowd the roof. What exactly &quot;overcrowd&quot; means in practical terms I can't possibly say, but for an easy low-bar number of what is obviously acceptable, you can use the maximum number of workers you've seen up there at a time. This, however, is without a doubt highly conservative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better guess can be made if one knows the country and date (roughly) the building was constructed, from which one can check the relevant codes of the time for the design load of roofs. The current Brazilian codes, for instance, give three different design loads for roofs (prior to safety factors):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;With access to the public: $3\text{kN/m}^2$ (around $4\text{pax/m}^2$)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Without access to the public: $2\text{kN/m}^2$ (around $2.5\text{pax/m}^2$)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inaccessible: $0.5\text{kN/m}^2$ (less than $1\text{pax/m}^2$)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For an idea of what these population densities mean in practical terms, see &lt;a href=&quot;http://www.gkstill.com/Support/crowd-density/CrowdDensity-1.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. $2.5\text{pax/m}^2$ is already quite a party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Brazil, such a roof would have to be designed for at least $2\text{kN/m}^2$, arguably (since you apparently had no issue getting up there) even $3\text{kN/m}^2$ (obviously, without express knowledge, you'd have to assume $2\text{kN/m}^2$). These values change from country to country and from time to time (revisions to codes may alter the values), so knowledge of the country and approximate time of construction is absolutely necessary to make any sort of educated guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding aspects of &quot;living safety&quot; (no clue what the actual term would be), where there are fold up chairs there is alcohol, and given that this roof has no railings, that is arguably not the best of combinations...&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-08-02T13:59:59.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3779" PostTypeId="2" ParentId="3777" CreationDate="2015-08-02T14:02:43.160" Score="0" Body="&lt;p&gt;If I get your descriptions right you have everything you need given. You can use the pitch diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$tan(H) = \frac{\pi \cdot d_p}{N_t \cdot P_a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$tan(H) = \frac{\pi \cdot 5\:\mathrm{in}}{25 \cdot 3\:\mathrm{in}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$tan(H) = \frac{5\pi}{75}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$tan(H) = \frac{\pi}{15}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H = arctan \left( \frac{\pi}{15}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H = ~ 0.20 \:\mathrm{rad}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There you go. Please remember to include the units if you post an answer. An angle of 0.2 could have been degree, too. At least I wasn't sure until after I calculated it.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2015-08-02T14:02:43.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3780" PostTypeId="1" AcceptedAnswerId="3784" CreationDate="2015-08-03T05:27:15.603" Score="2" ViewCount="4073" Body="&lt;p&gt;Reviewing for a National Board Exam: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many 5/16 inch holes can be punched in one motion in a steel plate&#xA;  made of SAE 1010 Steel 7/16 inch thick using a force of 55 tons. The&#xA;  ultimate strength for shear is 50 ksi and use 2 Factor of Safety&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer is 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a common strength of materials question given among problem sets in engineering. Here is my attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ S_{plate} = \frac{F}{N\times d \times t} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ 50ksi = \frac{ 55tons \times\frac{2204.622lb_f}{1ton} }{N \times \frac{5}{16}in \times \frac{7}{16}in} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get N = 17...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think what I'm getting wrong is the conversion of tons; I've tried UK Tons, US Tons but nothing. Is there a correct conversion factor?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastActivityDate="2023-12-20T04:49:06.403" Title="Number of Holes punched through a Plate of SAE 1010 Steel" Tags="|steel|stresses|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3781" PostTypeId="1" AcceptedAnswerId="3783" CreationDate="2015-08-03T05:49:38.307" Score="3" ViewCount="737" Body="&lt;p&gt;We know that demand for electricity is not constant. Power plants are not operated under full load conditions. What if power plants were operated under full load condition and excess electricity was stored in capacitors? I have read that capacitors of 10 kF capacitance are available. Could the use of such capacitors make it reasonable to operate power plants under full load at all times?&lt;/p&gt;&#xA;" OwnerUserId="2494" LastEditorUserId="368" LastEditDate="2015-08-04T23:18:46.530" LastActivityDate="2020-03-15T22:01:39.913" Title="Can large capacitors allow power plants to run at full load and store excess electricity?" Tags="|energy-storage|power-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3782" PostTypeId="2" ParentId="3781" CreationDate="2015-08-03T06:11:00.100" Score="2" Body="&lt;p&gt;Capacitors typically store power over short periods of time, seconds or minutes.  This is too short to be useful for the electrical power grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically peak power demand occurs during the day, often mid-afternoon.  Excess power is available at night.  To be useful, then, the energy must be stored for many hours.  There are many possible technologies for storing energy but few are economical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most practical energy storage approach at the moment is &lt;a href=&quot;https://en.wikipedia.org/wiki/Pumped-storage_hydroelectricity&quot; rel=&quot;nofollow&quot;&gt;pumped water storage&lt;/a&gt;.   Excess electricity is used to pump water up to a reservoir.  When power demand is high, the gravitational energy released when the water flows back downhill is used to generate electricity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For capacitors to compete for practical power grid energy storage, they will need to have lower cost and higher leakage resistance.&lt;/p&gt;&#xA;" OwnerUserId="2158" LastEditorUserId="2158" LastEditDate="2015-08-03T17:28:01.243" LastActivityDate="2015-08-03T17:28:01.243" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3783" PostTypeId="2" ParentId="3781" CreationDate="2015-08-03T08:57:10.797" Score="6" Body="&lt;p&gt;Capacitors are currently &quot;rather too costly&quot; for this purpose.&lt;br&gt;&#xA;I've revised my estimated cost after some more research but you appear to be in the 'well over one hundred thousand dollars&quot; range!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dear way: If you were to assemble a 10kF 150 volt capacitor from available smaller capacitors now it would cost around 1 million dollars and store about 30 kWh of electricity - worth maybe 5 to 15 dollars retail depending where you are and a lot less wholesale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using 160 VDC rated parts, 1700 of &lt;a href=&quot;http://www.digikey.com/product-search/en?pv14=8&amp;amp;FV=fff40002%2Cfff8000c&amp;amp;k=supercapacitor&amp;amp;mnonly=0&amp;amp;newproducts=0&amp;amp;ColumnSort=0&amp;amp;page=1&amp;amp;quantity=0&amp;amp;ptm=0&amp;amp;fid=0&amp;amp;pageSize=25&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;these&lt;/strong&gt;&lt;/a&gt; would cost about $2,000,000 (really), weigh about 9 tones and occupy about 17 cubic metres (!).  You'd get a discount for quantity but overall they are not viable&#xA; &lt;a href=&quot;http://www.maxwell.com/images/documents/160vmodule_ds_3000246-5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Datasheet here&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;One of 1700:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n5ATZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5ATZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Doing it yourself&quot; is hardly cheaper..&lt;br&gt;&#xA;In 1000 quantity &lt;a href=&quot;http://www.digikey.com/product-detail/en/BCAP3400%20P285%20K05/1182-1038-ND/4896903&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt; 3400 &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt;F, 2.8fV cost a mere   53 dollars each. Each series string = 150/2.85 = 53 capacitors.&lt;br&gt;&#xA;Capacitance in series divides by number of caps so C per spring = 3400/53 = 64F.&lt;br&gt;&#xA;Number of strings = 10,000/64 = 156.&lt;br&gt;&#xA;Total caps = caps/string x strings = 53 x 156 = 8268&lt;br&gt;&#xA;Cost = 8268 x 53  = 438,204 dollars&#xA;That's better than one million plus above - but that's a lot of mounting to do AND balancing circuitry will be essential.&lt;br&gt;&#xA;1 to 2 million starts to look almost good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using smaller voltage high capacity units &quot;off the shelf &#xA;Alternatives include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Flow batteries&quot; using eg Vanadium oxide liquid in various oxidation levels. &quot;Tanks&quot; of liquid are pumped through a cell to &quot;charge&quot; and stored in another tank for subsequent discharge. Large trial systems exist. Not yet mainstream. May or may not &quot;make it&quot; commercially. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium Ion - as being used in Tesla cars and their Powerbank home storage technology. JUST becoming economic with careful timing of charge and discharge to buy cheap power and use it at peak cost periods. Large commercial installations exist in eg Germany - costs are about break even overall so far with advantages of continuity being a factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium Titanium (a LiIon variant) - coming - used in eg Suzuki LEAF along with traditional LiIon cells. Dearerthan than standard LiIon but immensely fast charge rates and can have 5,000 - 10,000 usage cycles with due care.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Molten salt - used to store energy thermally and make power off peak - eg station at Guila-Bend near Phoenix. This is solar thermal heated but electric heating could be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pumped storage as John mentioned. Some use but not very common. Efficiency overall about 60%. UK has a large system used for grid leveling applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flywheels - investigated for many decades - good in theory but to get acceptable energy densities you need large masses )(tons) rotating at 10s of thousands of RPM. Mechanical failure is not pretty. So far many have tried but there are no significant working systems.&lt;br&gt;&#xA;Flywheel added comment - March 2020: There are some flywheel storage systems available used for peaking load control with about 10 kWh capacity per flywheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others exist, but that gives a feel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2020-03-15T22:01:39.913" LastActivityDate="2020-03-15T22:01:39.913" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="3784" PostTypeId="2" ParentId="3780" CreationDate="2015-08-03T12:31:46.353" Score="4" Body="&lt;p&gt;The shear force required should depend upon the shear area: &#xA;$$N \pi \biggl(\frac{5}{16}''\biggr) \biggl(\frac{7}{16}''\biggr)$$&#xA; or number times circumference times height, since the side-walls of $N$ cylinders are resisting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would drop your answer by almost exactly the right amount. Short tons could take it the rest of the way. Doing the math quickly with short tons gets me $N = 5.122$, which of course we can't have part of a same-sized hole, so we need the floor, or $5$, the desired solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With long tons we get $N=5.634$, which has the same floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out by idkfa, I did not consider the safety factor. However, as the question asks about a production process and not an application for the end product, it isn't clear that the safety factor is relevant, in the same way that the SAE designation of the material is (mostly) irrelevant.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-08-04T00:55:54.783" LastActivityDate="2015-08-04T00:55:54.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3785" PostTypeId="1" AcceptedAnswerId="3797" CreationDate="2015-08-03T14:54:44.720" Score="2" ViewCount="150" Body="&lt;p&gt;I have seen several products on the internet marketed as 'ball screws' that have both the screw itself and the special ball bearing nut. They are intended to have a motor mounted on one end of the screw to rotates it, causing the nut to be moved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm looking for the particular kind of ball screw where the nut is part of a gear, and mounted via axial bearings, so that as it is rotated it remains stationary and the screw is extended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these things sold in complete units without any machining necessary? What are they called/where can I find one?&lt;/p&gt;&#xA;" OwnerUserId="2498" LastActivityDate="2015-08-21T17:35:49.520" Title="Name of the ball screw nut used in pistons?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3786" PostTypeId="2" ParentId="100" CreationDate="2015-08-03T15:37:31.107" Score="1" Body="&lt;p&gt;No tail in zero gravity. As long as the material is maintained in a heated environment you would have a &quot;near perfect&quot; sphere as long as pressure and temperature and absence of gravity is constant. &#xA;Cooling would result in similar uniform stresses to the Rupert&quot;s Drop although the effect of the tail would be missing.  Any distortion would result in a &quot;flaw&quot; and impact the uniform stress and the Rupert's Drop effect would not exist. In a perfect idea, you'd end up with a &quot;yourname&quot; sphere.&lt;/p&gt;&#xA;" OwnerUserId="2499" LastActivityDate="2015-08-03T15:37:31.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3788" PostTypeId="1" CreationDate="2015-08-03T18:31:48.830" Score="1" ViewCount="68" Body="&lt;p&gt;I have a ceramic nozzle type structure sort of like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://3.imimg.com/data3/LK/LH/MY-1731595/collector-nozzles-500x500.jpg&quot; alt=&quot;a&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer casing is joined with clay material with the inner ceramic nozzle. What we do is, we attach the clay in the casing, put the nozzle in base and press the casing with the help of a air press. But the thing is, we are not getting uniformity in this. The outer diameter formed with the casing is irregular. Can someone help here?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SUGGESTED ONLY - PLEASE CHECK:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Tundish#Casting&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Tundish nozzle - CASTING - here&lt;/strong&gt;&lt;/a&gt;  = Ceramic nozzle inside outer shell - the ceramic IS attached to the base of a &quot;Tundish&quot; flow smoothing reservoir for molten steel which accepts molten steel from &quot;ladles&quot; and delivers it via these nozzles to molds.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that the ceramic does not form smoothly against the outer shell when pressed, &lt;/p&gt;&#xA;" OwnerUserId="2503" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.160" LastActivityDate="2015-08-07T18:20:32.920" Title="How to attach a shell covering to a flat piece with clay as adhesive?" Tags="|mechanical-engineering|steel|pressure|" AnswerCount="0" CommentCount="3" ClosedDate="2015-08-04T16:21:23.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="3789" PostTypeId="1" CreationDate="2015-08-04T09:39:07.893" Score="1" ViewCount="48" Body="&lt;p&gt;For a National Board Exam Review:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An air conditioning plant with a capacity of $400$ $kW$ of refrigeration&#xA;  has an evaporating and condensing temperature of $3$ $^oC$ and $37$ $^oC$&#xA;  respectively. If it uses Refrigerant 12, what volumetric rate of flow&#xA;  under suction condition?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer is $0.172$ $m^3/s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ V = mv}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ COP = \frac{T_{min}}{T_{max} - T_{min}} = \frac{Q_A}{W} = \frac{h_1 - h_4}{h_2 - h_1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ COP = \frac{ (3 + 273) }{37 - 3} = \frac{400}{W} = \frac{400}{h_2 - h_1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ W = h_2 - h_1 = 49.24859678 kJ/kg }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've viewed every formula in the book and many other books. I cannot find a way to find m, for mass. even so... it's also hard to find the specific volume of freon without a given temperature. How am I supposed to go about this?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="90" LastEditDate="2015-08-04T11:01:31.807" LastActivityDate="2015-08-04T11:01:31.807" Title="The Volumetric Flowrate Under Suction Condition" Tags="|refrigeration|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3790" PostTypeId="2" ParentId="3664" CreationDate="2015-08-04T09:39:49.057" Score="3" Body="&lt;p&gt;Section 2.1.14 of the Abaqus Example Problems Manual is 'Water sloshing in a baffled tank', which might be useful (even if you don't use Abaqus at your own institution, the manuals are hosted by many institutions that allow public access).   The Abaqus water model uses an equation of state (EOS).   If you download the input file from the manual, you can cut and paste the material section or convert it to suit the code that you're using.&lt;/p&gt;&#xA;" OwnerUserId="2507" LastActivityDate="2015-08-04T09:39:49.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3791" PostTypeId="1" CreationDate="2015-08-04T09:56:40.687" Score="10" ViewCount="36383" Body="&lt;p&gt;I have been attempting to solve a question where there is a point force acting on the hinge of a beam. Here is the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I7pM3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I7pM3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure how to deal with the 2 kN point force at $C$ ($C$ and $E$ are the hinges). If I split the beam into three parts, $\overline{AC}$, $\overline{CE}$, and $\overline{EG}$, I don't know where that 2 kN force should go. If I include it in both of the equilibrium equations of $\overline{AC}$ and $\overline{CE}$, then the sum of $F_y$ will be imbalanced. I believe this problem is statically determinate but I am just stuck at this point. I don't want to attach my workings here yet as I really would like to tackle it myself with a little bit of clarification and help. &lt;/p&gt;&#xA;" OwnerUserId="2465" LastEditorUserId="368" LastEditDate="2015-08-04T15:49:14.597" LastActivityDate="2019-12-28T18:08:30.833" Title="How to deal with a point force acting right on the hinge of a beam?" Tags="|statics|beam|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3793" PostTypeId="2" ParentId="3791" CreationDate="2015-08-04T12:42:33.343" Score="4" Body="&lt;p&gt;I'll assume you know how to find the reactions but you're just unsure of the two hinges at C and E as that seems like your main concern. If you're not sure how to calculate the reactions I can add this later. I have used &lt;a href=&quot;http://skyciv.com/beam-calculator-features&quot; rel=&quot;nofollow noreferrer&quot;&gt;SkyCiv Beam&lt;/a&gt; to find the reactions:&lt;a href=&quot;https://i.stack.imgur.com/CE6Xv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CE6Xv.png&quot; alt=&quot;Reactions of the beam as found by SkyCiv Beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see these reactions balance just fine:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$$\sum F_y = 11 + 10 + 5 - (6+2+6+2\times6) = 26 - 26 = 0\text{ kN}&#xA;\\&#xA;\sum M_A = -32 +6(2)+2(4)+6(5)+12(11)- 10(8) -5(14)= 0\text{ kN.m}$$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now it doesn't really matter whether you choose to include the 2 kN point load at hinge C on member AC or CE. Just include it in the free body diagram (FBD) for one member or the other (NOT both!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's make the 2 kN point load at C act on the right end of member AC, not the left end of member CE. Remembering that a moment can NOT be supported at the hinge C:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PsWIl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PsWIl.png&quot; alt=&quot;Member AC&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$$\sum F_y = 0\\&#xA;11 - 6 - 2 + H_C  = 0\\&#xA;\therefore H_C  = 3\text{ kN}$$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now consider member CE (again no moment at C or E). The force Hc needs to be in the opposite direction as that found in the FBD for member AC:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9wZpB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9wZpB.png&quot; alt=&quot;Member CE&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$$\sum F_y = 0\\&#xA;H_C + H_E -6 = 0\\&#xA;3 + H_E - 6  = 0\\&#xA;\therefore H_E = 3\text{ kN}$$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Lastly consider member EG to confirm that it all balances fine (again the force at E needs to be opposite that in the FBD for member CE):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9E2Pp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9E2Pp.png&quot; alt=&quot;Member EG&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;$$\sum F_y = -H_E + 10 + 5 - 12 = -3 + 10 + 5 - 12 = 0 \text{ }\checkmark$$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's look at the shear force diagram (SFD) below and understand why it doesn't really matter which member the 2 kN point load acts on. We solved earlier that at point C the shear force was Hc = 3 kN. As you can see in the SFD there are TWO values at point C (x=4m): 5 kN and 3 kN. Obviously the difference between these values is the 2 kN point load. If we had added the point load in our diagram for member CE instead of member AC then we would have solved the shear force at point C to be Hc = 5 kN. So you can include it in either member and it will be correct - just don't include it in both members.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0BqFX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0BqFX.png&quot; alt=&quot;Shear Force Diagram (SFD)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://skyciv.com/beam-calculator-features&quot; rel=&quot;nofollow noreferrer&quot;&gt;SkyCiv Beam&lt;/a&gt; is pretty handy for analyses like this and it's a good way to check your logic, answers and working out. It will also solve the bending moment diagram (BMD) if you need it plus deflection, stress among others.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-08-04T23:24:45.290" LastActivityDate="2015-08-04T23:24:45.290" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3794" PostTypeId="2" ParentId="3264" CreationDate="2015-08-04T12:58:09.437" Score="1" Body="&lt;p&gt;If you've got two fixed gears on the output shaft and two sliding gears on the input shaft then your gears will not be in constant mesh. Are you aware of this? This is okay if you plan to change gears while your gearbox is not transmitting torque. However during operation this is rarely the case. I think the following is a better design and is usually how automotive transmissions such as motorcycle gearboxes work anyway: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gears on the output shaft are fully fixed axially and torsionally. The gears on the input shaft are fixed axially but are not fixed torsionally and are in constant mesh with the output shaft gears. The input shaft gears don't need a key then. In between the input shaft gears add a sliding selector gear on a key which transmits torque via dogs on the sides of the input shaft gears.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastActivityDate="2015-08-04T12:58:09.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3795" PostTypeId="1" CreationDate="2015-08-04T14:15:12.237" Score="6" ViewCount="1313" Body="&lt;p&gt;I am developing a model for the respiratory system of a neonate (newborn child) for a BME project by using pressure/volume data of the lungs. To model these, I plan to use a rubber balloon. However. the calculations are more difficult than I thought (especially because rubber behaves non-linearly and with a hysteresis effect, but I will ignore these effects).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will assume the balloon is perfectly spherical, that it behaves linearly and that the strain is dependent on the radius relative to the original radius ($R_0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the equations for the basis of my model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;V = \frac{4}{3} \pi R^3_1 \\&#xA;R_1 = \left( \frac{3V}{4\pi} \right)^\frac{1}{3} \\&#xA;\sigma = \frac{pR_1}{2t} \\&#xA;\epsilon = \frac{R_1 - R_0}{R_0} \\&#xA;E = \frac{\sigma}{\epsilon}&#xA;  = \frac{pR_0R_1}{2t(R_1-R_0)}&#xA;  = \frac{pR_0}{2t\left(1-\dfrac{R_0}{\left(\frac{3V}{4\pi}\right)^\frac{1}{3}}\right)} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My conclusion is: $$E \propto \frac{p}{V^\frac{1}{3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="2509" LastEditorUserId="5365" LastEditDate="2016-03-08T09:40:05.363" LastActivityDate="2016-03-09T01:00:07.097" Title="Calculate elasticity of rubber balloon" Tags="|pressure|biomedical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3796" PostTypeId="2" ParentId="2991" CreationDate="2015-08-04T14:16:54.767" Score="3" Body="&lt;p&gt;This is just an addition to the answer of Chris Mueller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you think about lasers, you always think about an aperture with lot of mirrors, lenses and optics in general. Lets say that you manage to create a focused (wide spectrum) beam at one point, now you want to bring it to the point of application. With a wide spectrum it won't work well, as the beam will lose focus for the different wavelengths. while passing through the optics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you a picture of the optical phenomena look at this famous &lt;a href=&quot;http://en.wikipedia.org/wiki/The_Dark_Side_of_the_Moon&quot; rel=&quot;nofollow&quot;&gt;album cover&lt;/a&gt;. You will notice that the exiting light is spread out(along the spectrum) and lost focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could refocus it, but that is impractical. For other considerations refer to the answer of Chris Mueller.&lt;/p&gt;&#xA;" OwnerUserId="1600" LastActivityDate="2015-08-04T14:16:54.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3797" PostTypeId="2" ParentId="3785" CreationDate="2015-08-04T14:48:03.177" Score="1" Body="&lt;p&gt;The thing is called a traveling screw (TS) linear actuator. &#xA;You should be able to get them as complete units from the companies making them.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2015-08-04T14:48:03.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3798" PostTypeId="1" CreationDate="2015-08-04T15:01:50.713" Score="5" ViewCount="569" Body="&lt;p&gt;I've been given a problem which I've attempted to solve, and hope some fellow engineers could possibly see if my approach is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are asked to calculate a column's axial deformation due to an applied load of F = 25 kN and the force due to the column's own weight. We are also given that ρ = 7800 kg/m&lt;sup&gt;3&lt;/sup&gt;, E = 200 GPa and g = 9.81 m/s. It's vertical length is L = 1 m and its radius varies vertically as r = (0.03)/(1+y).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5pR4y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5pR4y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the approach that I took where I first found the force due to the column's weight as a function of y:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(y)=\int_0^y \rho g A(y) \, dy$$&#xA;$$F(y)=\int_0^y \rho g \pi \left (\frac{0.03}{1+y}  \right )^2\, dy$$&#xA;$$F(y)=(0.03)^2 \rho g \pi \int_0^y \frac{dy}{(1+y)^2}$$&#xA;$$F(y)=(0.03)^2 \rho g \pi \left (\frac{y}{1+y}  \right )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then put this into the axial deformation equation and integrated to get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta_{w}=\int_0^L \frac{F(y)dy}{A(y)E}$$&#xA;$$\delta_{w}=\int_0^1 \frac{(0.03)^2 \pi \rho g \left (\frac{y}{1+y}  \right )dy}{\pi \left (\frac{0.03}{1+y}  \right )^2 E}$$&#xA;$$\delta_{w}=\int_0^1 \frac{(0.03)^2 \pi \rho g y(1+y)^2dy}{(0.03)^2 \pi E (1+y)}$$&#xA;$$\delta_{w}=\frac{\rho g}{E}\int_0^1 y(1+y)dy$$&#xA;$$\delta_{w}=\frac{5 \rho g}{6 E}$$&#xA;$$\delta_{w}=\frac{5 (7800)(9.81)}{6 (200)(10^9)}$$&#xA;$$\delta_{w}=3.18825(10^{-7})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the applied load I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta_{F}=\int_0^L \frac{Fdy}{A(y)E}$$&#xA;$$\delta_{F}=\frac{F}{E}\int_0^L \frac{dy}{\pi \left (\frac{0.03}{1+y}  \right )^2}$$&#xA;$$\delta_{F}=\frac{F}{(0.03)^2 \pi E}\int_0^L (1+y)^2 dy$$&#xA;$$\delta_{F}=\frac{7 F}{3 \pi (0.03)^2 E}$$&#xA;$$\delta_{F}=\frac{7 (25)(10^3)}{3 \pi (0.03)^2 (200)(10^9)}$$&#xA;$$\delta_{F}=1.03156(10^{-4})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, adding the two deformations I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta_{total}=\delta_{w}+\delta_{F}$$&#xA;$$\delta_{total}=-1.0347(10^{-4})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the negative sign is due to the fact that the forces are compressive. Is this result for the deformation of the beam accurate? My main point of concern is whether I calculated $F(y)$ correctly.&lt;/p&gt;&#xA;" OwnerUserId="2510" LastEditorUserId="2510" LastEditDate="2015-08-05T04:37:49.837" LastActivityDate="2016-03-03T13:32:11.273" Title="Axial deformation of variable cross-section member" Tags="|structural-engineering|applied-mechanics|columns|deformation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3799" PostTypeId="2" ParentId="3391" CreationDate="2015-08-04T16:40:02.097" Score="2" Body="&lt;p&gt;The power that the engine can deliver goes through a transmission before reaching the wheels. I believe most tractors are geared such that the maximum torque that can be applied to the wheels exceeds the maximum torque a road surface can apply to the wheel. This means it is likely that your tractor is traction limited rather than power limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum tractive force can be approximated by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_t=\mu W \sqrt{1-\frac1{Grad\%}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $W$ is the weight supported by the driving wheels and $\mu$ is the friction coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two wheel drive tractors typically $55\%$ to $70\%$ of the weight will be on the drive tires when on level ground without pulling anything. However, the weight will shift when pulling and with changes in incline. The pulling alone can sometimes shift up to $90\%$ of the weight to the rear tires. With a steep incline this may actually cause the tractor to nose up. In that case, the limiting factor is nosing up, rather than traction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friction coefficient for tires on asphalt is around $1$. For dirt, especially loose dirt, the dirt actually shears internally before the tire/dirt interface shears. This results in a &quot;friction coefficient&quot; that is dependent on contact area, and is why tractors typically have large tires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force required to pull something can be calculated from the formula you gave in your question, where $m$ is the mass of both the tractor and the thing you'd like to pull. (this requires the thing your pulling to have a similar rolling resistance)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note however, that that formula is valid only for shallow hills, and become less and less accurate as the hills get steeper. A $5\%$ grade is still well within that accurate range though.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-08-04T16:40:02.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3801" PostTypeId="2" ParentId="3791" CreationDate="2015-08-04T17:09:24.400" Score="4" Body="&lt;p&gt;While this beam presents five constraints (&lt;span class=&quot;math-container&quot;&gt;$X_A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$Y_A$&lt;/span&gt;,  &lt;span class=&quot;math-container&quot;&gt;$M_A$&lt;/span&gt;,  &lt;span class=&quot;math-container&quot;&gt;$Y_F$&lt;/span&gt;,  &lt;span class=&quot;math-container&quot;&gt;$Y_G$&lt;/span&gt;), it is in fact statically determinate. A statically indeterminate structure is one where there are more unknowns (constraints, in this case) than there are static equilibrium equations. Usually one has three equations: &lt;span class=&quot;math-container&quot;&gt;$\sum F_X = 0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\sum F_Y = 0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\sum M_? = 0$&lt;/span&gt; (where &lt;span class=&quot;math-container&quot;&gt;$?$&lt;/span&gt; is any arbitrary point). Hinges, however, give us an additional equation each: &lt;span class=&quot;math-container&quot;&gt;$\sum M_{h\pm} = 0$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$h\hspace{-2pt}\pm$&lt;/span&gt; is one side of the hinge (left or right), such as in this question. This is different from the global null bending moment equation which considers all forces to either side of the hinge. Adding the two additional equations given by the hinges at &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; to the three global equilibrium equations, we therefore have as many equations as we have contraints (5), and can therefore solve this problem by the traditional means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, there is a &lt;strong&gt;much easier&lt;/strong&gt; way of doing this which is entirely hands-on, &lt;strong&gt;without computational aides&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this hands-on approach, one needs to observe the double hinge in the span &lt;span class=&quot;math-container&quot;&gt;$\overline{CE}$&lt;/span&gt;. This means that the bending moment at &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; must be null, much like with a simply supported beam (a more in-depth explanation of why this comparison is valid can be seen at the end).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's replace that beam with the following pieces (notice that the loads at &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; are left blank for now):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3l4T2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3l4T2.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the beam representing &lt;span class=&quot;math-container&quot;&gt;$\overline{CE}$&lt;/span&gt; is trivial. For now all we need are the reactions, which are equal to &lt;span class=&quot;math-container&quot;&gt;$3\text{kN}$&lt;/span&gt; at each support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now get those reactions and toss them down to the other pieces, remembering that at &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; there is also the concentrated &lt;span class=&quot;math-container&quot;&gt;$2\text{kN}$&lt;/span&gt; force, which must be added. We therefore have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K18TI.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K18TI.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other pieces are also isostatic and can be trivially solved (assuming one knows how to obtain internal forces of isostatic structures). The resulting internal forces are (I changed the support at &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; just to make that piece stable for horizontal forces, which changes nothing in this case):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H4Xfz.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H4Xfz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Composing these diagrams, they are identical to those obtained by the original beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UmyGR.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UmyGR.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple reason why the comparison can be made between those double-hinges and a simply supported beam is because this is the basic principle behind Gerber beams (which is basically what &lt;span class=&quot;math-container&quot;&gt;$\overline{CE}$&lt;/span&gt; represents). They are beams which rest on other beams (see example &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/8/8b/3_spans_Gerber%27s_beam_%28support-hinge-support%29_%28real%29.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, where the beams to the right and left are Gerber beams) and which can therefore be &quot;lifted&quot; from the rest of the structure, solved, and then have their reactions distributed to the rest of the structure. One doesn't have to worry about the influence of external forces or the neighboring beams transmitting shear forces due to the fact that the bending moment must be null at each extremity of the Gerber beam. This means that the integral of the shear along the Gerber beam must be null, which can only occur if only the loads within the beam and the reactions at its extremities are considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program I used for these diagrams was &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2-D frame analysis tool.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:08:30.833" LastActivityDate="2019-12-28T18:08:30.833" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="3802" PostTypeId="2" ParentId="3331" CreationDate="2015-08-04T18:12:11.940" Score="1" Body="&lt;p&gt;I personally have the tegrapol 11 and I use it with the tegraforce 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our units sample mover plate holds three specimens simultaneously (concept is the same though). Google tegrapol 11 and tegra force 1 for images if this is unclear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many times I run my sample mover plate both with only one and only two specimens at a time merely leaving the last hole(s) empty.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If I were working with this unit and I were concerned with using the sample mover plate sans all 6 holes being filled: I would distribute the three specimens (skip every other hole).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I would NOT use the single sample mover plate for three specimens.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As with 2. I personally would use the six-holed sample mover plate for three specimens.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2513" LastActivityDate="2015-08-04T18:12:11.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3803" PostTypeId="1" AcceptedAnswerId="3819" CreationDate="2015-08-05T08:58:25.907" Score="6" ViewCount="8090" Body="&lt;p&gt;Steel seems to be a very complex material, not least because of the different phases and microstructures. At the moment, I'm especially interested in the bainite formation that happens when austenitic steel is cooled rapidly (but not rapidly enough to drive the martensite formation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: What is the mass density (or range of values, or functions describing the densities in relation to specific conditions) of these two microstructures?&lt;/p&gt;&#xA;" OwnerUserId="2518" LastEditorUserId="272" LastEditDate="2015-08-05T12:04:49.403" LastActivityDate="2015-08-05T22:52:27.510" Title="Densities of different phases of steel (austenite, bainite)" Tags="|materials|steel|metallurgy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3804" PostTypeId="1" CreationDate="2015-08-05T10:02:50.207" Score="4" ViewCount="581" Body="&lt;p&gt;In what way does the electrical conductivity $\sigma$ (or conversely, resistivity $\rho = 1/\sigma$) depend on the salinity $S$ of the saline- (or sea-) water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know they must be in some way proportional since the more free charge carriers (ions in this case) available, the more electric current is possible.&lt;/p&gt;&#xA;" OwnerUserId="2296" LastActivityDate="2015-08-05T12:00:58.247" Title="How does conductivity $\sigma$ of salt-water depend on salinity?" Tags="|electrical-engineering|chemical-engineering|current|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3805" PostTypeId="2" ParentId="3795" CreationDate="2015-08-05T10:08:30.197" Score="1" Body="&lt;p&gt;You need to be careful with the subscripts, e.g in the equation for $\sigma$, which $R$ do you want?   However, what you seem to be saying is: $$E \propto \frac{p}{R}$$ and, if you take $p$ and $R$ as increments from initial values, that seems reasonable.   Getting the constant of proportionality might be problematic.&lt;/p&gt;&#xA;" OwnerUserId="2519" LastActivityDate="2015-08-05T10:08:30.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3807" PostTypeId="1" AcceptedAnswerId="3829" CreationDate="2015-08-05T10:41:00.317" Score="3" ViewCount="287" Body="&lt;p&gt;I want to algorithmically process &lt;strong&gt;energy meter data&lt;/strong&gt;. The energy meter measures a heat or power producer or a heat or power consumer (but not both, so the measured energy will always have a positive sign). No additional information is known about the energy system (like maximum load) neither about the type of energy meter - only data stored in a database can be accessed. Processing will be done by an algorithm looking at data for a given time interval (no live processing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, data are &lt;strong&gt;weakly monotonic&lt;/strong&gt; of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:00 20.78 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:05 30.80 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:10 73.99 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:20 82.30 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:25 82.30 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:30 83.44 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy produced or consumed for a given period is simply the difference of the energy meter counts. So far, so good. However, the algorithm has to deal with the following three problems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;1. Outliers &quot;above&quot; have to be detected as invalid data.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:00 20.78 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:05 30.80 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2015-04-01 00:10 500 kWh&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:20 82.30 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:25 82.30 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:30 83.44 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;2. Outliers &quot;below&quot; have to be detected as invalid data.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:00 20.78 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:05 30.80 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2015-04-01 00:10 20 kWh&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:20 82.30 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:25 82.30 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:30 83.44 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In unlikely cases, there might be several consecutive outliers above or below or a combination of both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;3. A reset of the energy meter has to be detected automatically.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:00 20.78 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2015-04-01 00:05 30.80 kWh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2015-04-01 00:10 3.99 kWh&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2015-04-01 00:20 12.30 kWh&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2015-04-01 00:25 12.30 kWh&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2015-04-01 00:30 13.44 kWh&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a reset, counting starts a again from another level (a reset is simply a level shift). The level the counting starts from after the reset is often zero, but can also be any other positive number. A reset can occur at an arbitrary point in time (usually not too often).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my eyes, problems 1. - 3.  seem ubiquitous in measurement engineering and must have been already addressed. Nevertheless, I couldn't find any literature on this topic. Does anybody know about existing solutions to this problem? All help will be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2521" LastEditorUserId="163" LastEditDate="2015-08-07T00:29:46.890" LastActivityDate="2015-08-07T00:29:46.890" Title="How to automatically process energy meter data considering outliers and meter reset?" Tags="|measurements|energy|metrology|" AnswerCount="2" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="3808" PostTypeId="2" ParentId="3807" CreationDate="2015-08-05T11:47:21.933" Score="0" Body="&lt;p&gt;Discarding outliers is easy, you simply discard (or ignore) them.  The hard part is deciding what is a outlier and what is valid data.  This comes down to defining what is possible and likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One obvious check in your case is against the maximum power the system can consume or produce.  If the delta between two readings exceeds that, then something is wrong somewhere.  For example, if the system being measured can't consume more than 200 kW, then the meter can't increase by more than 17 kWh in a 5 minute period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the maximum power production can be different from the maximum power consumption.  Maybe your system can consume up to 200 kW, but can't ever produce more than 50 kW.  That means in 5 minutes the possible valid range is +16.7 kWh to -4.2 kWh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may know other things about your system, which you can use to detect invalid readings.  For example, even though its power consumption is bounded by -50 to +200 kW, maybe it can't change faster than 10 kW per 5 minutes.  Or, if the power is being produce by solar cells, then getting power production at night must be wrong.  Surely there are things you know your system can't do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, it all comes down to defining what valid is.  For example, you say the third reading in case 2 is wrong?  How do you know that?  If it's possible for the system to generate 10.8 kWh during a 5 minute interval, then consume 62.3 kWh during the next 5 minutes, then case 2 may be all correct readings after all.  You haven't said anything about your system to tell this is not the case.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-08-05T11:47:21.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3809" PostTypeId="2" ParentId="3804" CreationDate="2015-08-05T12:00:58.247" Score="5" Body="&lt;p&gt;Measuring the conductivity of water is actually one of the standard ways of measuring its salinity.  The figure below (taken from &lt;a href=&quot;http://www.eos.ubc.ca/ubcgif/iag/foundations/properties/resistivity.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this University of British Columbia page&lt;/a&gt;) shows that the relationship over many orders of magnitude is &#xA;$$&#xA;\sigma\propto S,&#xA;$$&#xA;where $\sigma$ is the conductivity and $S$ is the salinity.  They further note that the conductivity also varies with temperature and the relationship is given approximately by &#xA;$$&#xA;\frac{1}{\sigma(T)}=\frac{\frac{1}{\sigma_{18 ^\circ C}}}{1+0.025(T-18\ ^\circ C)}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Fioh.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Fioh.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fondriest has a &lt;a href=&quot;http://www.fondriest.com/environmental-measurements/parameters/water-quality/conductivity-salinity-tds/&quot; rel=&quot;nofollow noreferrer&quot;&gt;long tutorial on environmental conductivity measurements&lt;/a&gt; which is very informative.  It does an excellent job of explaining the environmental factors which affect water salinity and therefore measurements of water conductivity.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-08-05T12:00:58.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3810" PostTypeId="1" CreationDate="2015-08-05T13:57:39.373" Score="2" ViewCount="136" Body="&lt;p&gt;I am thinking about using a permanent magnet as an input device for my smartphone. I have been reading some papers about it, but I really want to understand how things work. Using the integrated compass I can read the field strength along three axes. Given these readings, what property of the magnetic field could I use for gesture recognition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do anything I have to save the magnetic field strength along the three axes when no magnets are close to the smartphone. Then compare it with other readings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I place a magnet close to the smartphone in a particular position the readings will change. I can tell if the magnet is in that position when I read the same values along the three axes. This is because the magnet changes the &quot;overall&quot; magnetic field read by the magnetometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can simulate a click when a magnet in a particular position is twisted/curled because the angle of the polar coordinates changes of about 180 degrees. This is because magnets have a north and south pole, so twisting it changes the pole and as a consequence the angle of the polar coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I would like to simulate a movement along an axis (like a slider).&lt;/strong&gt; I can tell if I am at the beginning or at the end of it by comparing the magnetic field strength to the ones previously detected in those positions. How can I tell if I am in the middle or in any other point? The obvious solution would be to save the readings in all those points but I hope there is something about the magnetic field I am missing. Is there a law or a correlation between the points I can use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are some readings I recorded of seven equidistant points along the x axis:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     x,      y,      z&#xA;-39.55, -51.19, -32.67&#xA;-26.01, -41.83, -32.95&#xA;-19.94, -34.08, -33.28&#xA;-16.43, -26.62, -32.72&#xA;-15.98, -22.6,  -33.07&#xA;-15.93, -20.72, -32.96&#xA;-16.85, -18.87, -33.11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The best thing I have thought so far is to approximate the movement with a linear function but things doesn't work that well (for example the 4th and 7th position have almost the same value along x but they are in two different positions so I can't tell which one is the end and which the middle). This is one example, depending on where I am I can find myself with very different problems. How should I cope with it? Magnetic field is less strong as the magnet gets away from the magnetometer; what kind of function can describe this situation?&lt;/p&gt;&#xA;" OwnerUserId="2523" LastEditorUserId="368" LastEditDate="2015-08-05T21:26:16.840" LastActivityDate="2015-08-05T21:26:16.840" Title="Using magnetic field readings for linear gesture recognition" Tags="|magnets|computer-engineering|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3815" PostTypeId="1" AcceptedAnswerId="3820" CreationDate="2015-08-05T17:02:16.623" Score="3" ViewCount="174" Body="&lt;p&gt;My drafting program outputs a .dwf format file. In it, arcs are encoded as &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(Circle 602331619,844923003 537072960 32715,65588)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where my so-far affirmed interpretation is &lt;code&gt;(Circle, X-center,Y-center Angle Unknown1,Unknown2)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image for the arc is shown below (highlighted in Gold):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7HBwn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7HBwn.png&quot; alt=&quot;The arc in concern is the highlighted on.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, each arc has its center and radius given. Are the pair of points at the end the starting and ending angles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take &lt;code&gt;(32715,65588)&lt;/code&gt; as &lt;code&gt;(3.2715 rad,6.5588 rad)&lt;/code&gt;, the output starts making sense. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="2525" LastEditorUserId="2525" LastEditDate="2015-08-06T20:14:46.097" LastActivityDate="2015-08-06T20:14:46.097" Title="AutoCAD .DWF Circle Output Meaning" Tags="|civil-engineering|drafting|autocad|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3816" PostTypeId="1" CreationDate="2015-08-05T18:09:55.583" Score="3" ViewCount="10081" Body="&lt;p&gt;Elevators were invented &lt;a href=&quot;https://en.wikipedia.org/wiki/Elevator#Pre-industrial_era&quot; rel=&quot;nofollow&quot;&gt;several centuries ago&lt;/a&gt;. How is it possible that in the 21st century, there are still &lt;a href=&quot;http://www.theshorthorn.com/news/campus/about-reports-made-of-people-trapped-in-elevators/article_194a978e-128f-11e2-913f-001a4bcf6878.html&quot; rel=&quot;nofollow&quot;&gt;a number of cases of people getting stuck in elevators&lt;/a&gt;? I have gotten stuck in an elevator myself before, and have anxiety about entering an elevator ever since. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would like to know, what causes elevators to get stuck to the point that it requires outside intervention? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't (or shouldn't) it automatically recalibrate itself? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it more of a mechanical issues or electrical issue? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does an elevator getting stuck indicate a serious issue with the elevator that require maintenance, or is it something that can happen to even newly installed elevators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will we ever be able to fully prevent elevators from getting stuck?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Additional Questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Occasionally, I may hear a loud bang when riding on an elevator. Although luckily the elevator does not get stuck. Why does that happen?&lt;/li&gt;&#xA;&lt;li&gt;In some elevators, a person stepping into the elevator will occasionally cause it to wobble up and down, as if it is trying to realign itself with the floor. This tends to happen more often if the person is heavy in their footsteps. Is there a reason to worry in this situation?&lt;/li&gt;&#xA;&lt;li&gt;Some times, an elevator will start beeping even though it did not exceed capacity. After the door closes, the beeping stops and it proceeds as normal. Why does this happen?&lt;/li&gt;&#xA;&lt;li&gt;In what case (if any) would getting stuck in an elevator be life threatening? (Assuming the person remains perfectly calm and will only attempt to leave the elevator under the guidance of trained personnel)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2528" LastEditorDisplayName="user16" LastEditDate="2015-08-05T18:40:35.797" LastActivityDate="2015-08-05T18:40:35.797" Title="Why do elevators get stuck?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="0" CommentCount="4" FavoriteCount="0" ClosedDate="2015-08-05T18:36:42.540" ContentLicense="CC BY-SA 3.0" />
  <row Id="3817" PostTypeId="5" CreationDate="2015-08-05T21:12:38.580" Score="0" Body="&lt;p&gt;This tag should be used to categorize questions that pertain specifically to Autodesk's &lt;a href=&quot;https://en.wikipedia.org/wiki/AutoCAD&quot; rel=&quot;nofollow&quot;&gt;AutoCAD&lt;/a&gt; and (preferably) no other CAD tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions about using CAD tools in general, or where the asker's use of AutoCAD is entirely coincidental to the problem at hand, should use &lt;a href=&quot;http://engineering.stackexchange.com/tags/computer-aided-design&quot;&gt;the more general computer-aided-design tag&lt;/a&gt;. In some cases, both of these tags can be applied to the same question.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="33" LastEditDate="2016-02-23T03:22:09.507" LastActivityDate="2016-02-23T03:22:09.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3818" PostTypeId="4" CreationDate="2015-08-05T21:12:38.580" Score="0" Body="Questions specific to AutoCAD, as opposed to general computer-aided drafting and design issues." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-08-05T21:12:38.580" LastActivityDate="2015-08-05T21:12:38.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3819" PostTypeId="2" ParentId="3803" CreationDate="2015-08-05T22:52:27.510" Score="5" Body="&lt;p&gt;There is a lot more going on in this question than appears at first glance. Density of austenite is fairly straightforward: it is approximately the atom-weighted sum of the face-centered cubic densities of the substitutional constituents as the microstructure consists of a single phase. In other words, Fe, Mo, V, etc. The interstitial constituents, i.e. C, N, add mass as they do not replace Fe lattice sites but instead fit between them, increasing mass without increasing volume, therefore increasing density. However in austenite there is at most 2% by weight of carbon in solution at $1130\:\textrm{C}$ as shown in the phase diagram at the bottom. The density of pure iron is $7870\:\textrm{kg}/\textrm{m}^3$ (from Google), and the density increase is probably close to about 2% of the density of graphite, $2250\:\textrm{kg}/\textrm{m}^3$ (from Google), or about $45\:\textrm{kg}/\textrm{m}^3$, for a total of about $7915\:\textrm{kg}/\textrm{m}^3$ for the density of fully saturated austenite at room temperature. It isn't clear how you would produce such an unstable material, but we can at least estimate in theory. There are slight volume changes in the FCC lattice caused by the substitutional and interstitial atoms, but without atomistic models these changes are challenging to predict. They may be determined experimentally, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For bainite, there are two phases: ferrite and cementite. The ferrite phase consists of virtually all of the substitutional microconstituents in a body-centered cubic structure, so its density may be calculated by atom-weighted average. There is also a trivial quantity of carbon in solution at equilibrium, increasing density as before. The density of the microstructure is then the volume-weighted average of the densities of the two phases. Plain-carbon steel with no other alloying additions should have ferrite density very close to that of pure iron, or about $7870\:\textrm{kg}/\textrm{m}^3$ (from Google), and cementite has a theoretical density of approximately $7640\:\textrm{kg}/\textrm{m}^3$ (&lt;a href=&quot;http://iicarbide.com/archives/IIC_iron_carbide_manufacturing_process_rev_01.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;). For steels the maximum amount of cementite is approximately 32% by volume, from a tie-line construction on the phase diagram. The maximum carbon concentration in a steel, by definition, is close to 2.14 weight percent, whereas in cementite is 6.67 weight percent. So the volume-weighted density would be a minimum of approximately $7795\:\textrm{kg}/\textrm{m}^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These numbers may be adjusted for temperature using coefficients of linear expansion cubed for the volume change (assuming linearity). They may also be adjusted by considering the densities of substitutional alloy additions. Finally I would caution that the numbers are theoretical for austenite, as plain-carbon steel, pure, retained austenite should be practically impossible to produce in bulk. I'd also caution that there are more complex phenomena to consider such as phase changes, diffusion, as well as other thermodynamic and kinetic phenomena that govern how steels form microstructures and phases, and which will definitely have an effect on the densities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9MD97.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9MD97.jpg&quot; alt=&quot;FeC Phase Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2015-08-05T22:52:27.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3820" PostTypeId="2" ParentId="3815" CreationDate="2015-08-06T01:06:43.210" Score="3" Body="&lt;p&gt;The file format for *.dwf files is defined. A reference for the various drawing commands can be found &lt;a href=&quot;http://paulbourke.net/dataformats/whip/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Specifically, the definition of the &lt;code&gt;Draw Circle/Circular Arc/Circular Wedge function&lt;/code&gt; is &lt;a href=&quot;http://paulbourke.net/dataformats/whip/DrwCircArcWdg.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;From that link an arc is described by:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;X,Y,R Start,End&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;X,Y - Center point (in logical coordinates) of the circle to be drawn&lt;/p&gt;&#xA;&lt;p&gt;R - Radius (in logical coordinates) of the circle to be drawn&lt;/p&gt;&#xA;&lt;p&gt;Start, End - The angles (in 360/65,536ths of a degree) that describe a &amp;quot;pie-slice&amp;quot; of the full circle to be rendered; legal values range from 0 to 65,536&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-08-06T01:06:43.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3824" PostTypeId="1" AcceptedAnswerId="3825" CreationDate="2015-08-06T06:07:46.417" Score="5" ViewCount="189" Body="&lt;p&gt;In an air-conditioned house equipped with an air extractor that removes for instance wet, smelly, etc. air from the bathrooms and kitchen, what are sensible cost-effective systems to prevent:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Heat loss during the winter by bringing in cold air and extracting hot air from inside the house&lt;/li&gt;&#xA;&lt;li&gt;Heat gain during the summer by bringing in hot air and removing the chilled air inside&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That is, I would like to minimize the energy cost of extracting chilled/heated air by somehow recycling that energy before the air is dumped outside.&lt;/p&gt;&#xA;" OwnerUserId="2535" LastEditorUserId="1586" LastEditDate="2015-08-07T00:29:53.493" LastActivityDate="2015-08-07T00:29:53.493" Title="How to minimize heat loss in an air extractor?" Tags="|heat-transfer|hvac|airflow|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3825" PostTypeId="2" ParentId="3824" CreationDate="2015-08-06T11:13:31.027" Score="5" Body="&lt;p&gt;What you want is called a &lt;i&gt;heat exchanger&lt;/i&gt;.  Imagine two long air tubes with a thin wall between them.  The air exiting the house travels in one tube, and entering air in the other, but in opposite directions.  Over the length of the tubes, heat transfers thru the thin wall.  Ideally, by the time the house air gets to the far end where the outside air comes in, it's at the outside air temperature, and vice versa for the outside air at the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, there are various ways to construct heat exchangers, and these things are either far from ideal, or very bulky and expensive.  The cheapest commercial ones are usually cross flow instead of opposite flow.  This is much easier to construct, but the efficiency is lower.  Cross flow units in series increases efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the time, the payback for a heat exchanger for something like a bathroom vent is too long to make sense.  Price out some units and do the math before jumping into this.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-08-06T11:13:31.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3826" PostTypeId="2" ParentId="3798" CreationDate="2015-08-06T11:29:46.117" Score="1" Body="&lt;p&gt;Your integration limits should be y to 1 to get force at height y. For example at y=1m force due to self weight is zero, and deformation is again zero; Again when y=0, self weight is maximum. I will leave the further work outs to you.&lt;/p&gt;&#xA;" OwnerUserId="1676" LastActivityDate="2015-08-06T11:29:46.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3828" PostTypeId="2" ParentId="3342" CreationDate="2015-08-06T14:36:41.073" Score="1" Body="&lt;p&gt;In general, for all ridged bodies &lt;strong&gt;constrained to rotate at the same speed&lt;/strong&gt;, their inertias and the torques acting on them can be added together:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i \tau_i=\dot\omega \sum_i I_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\tau_i$ represents the torques acting eternally on the system (like the torque produced by an engine, or the resistance of a propeller by the wind)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also model the system as system of equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau_i + \sum_j \tau_{ij} = \dot\omega_i I_i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\tau_{ij}$ is an anti symmetric matrix representing the torque at the interface between components $i$ and $j$. The sprag clutches ensure that the torque between the engine and the rest of the system cannot be negative (ideally. However in reality there will be a small drag torque).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate accelerations while both engines are engaged you can use the first equation, then use the second system of equations to determine the torque on the clutches. If the torque is negative for one engine, then that engine would no longer be constrained to rotate at the same speed. That means it can't be included in the first equation so you'd have to calculated it again, but this time exclude that engine. Then use the first equation on just the engine to determine engine acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the engine rotational velocity ever exceeds that of the rest of the system then the clutch would reengage so now the velocities are constrained together so they must be treated like a system again (and to bring them back to exactly the same rotational velocity you could take the inertia weighted average of rotational velocity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: if some ridged bodies are constrained to rotate a speed proportional to each other (eg. gears), the same equation can be used, but the inertia, torques, and speeds must be multiplied/ divided by the ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If constraints:&#xA;$$\omega_i=r_i\omega$$&#xA;exist, then:&#xA;$$\sum_i \tau_i r_i=\dot\omega \sum_i I_i r_i$$&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-08-06T14:36:41.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3829" PostTypeId="2" ParentId="3807" CreationDate="2015-08-06T18:36:35.413" Score="2" Body="&lt;p&gt;There are two ways to do it.&lt;/p&gt;&#xA;&lt;h3&gt;The old way&lt;/h3&gt;&#xA;&lt;p&gt;The traditional way is to develop a set of somewhat arbitrary rules based on the errors you manually classify. You filter out non-monotonicity (easy), identify resets (easy), and try to spot other bad values (trickier). That gives you a set of values to mark as missing, and then you analyse the rest of the data. This method is not well grounded in theory, but you will have the (somewhat unsatisfactory) defence that: &amp;quot;it's how lots of other people do it&amp;quot;.&lt;/p&gt;&#xA;&lt;h3&gt;Best practice&lt;/h3&gt;&#xA;&lt;p&gt;The best-practice way to do it is to write down the probability of everything relevant, and then apply Bayes' Theorem to work out what the most likely real time series was, given your recorded observations.&lt;/p&gt;&#xA;&lt;p&gt;You start with a prior distribution for the rate of use of energy, based on preceding work.&lt;/p&gt;&#xA;&lt;p&gt;And then create probability distributions for the ways that errors can happen: a meter reset, a dropped decimal point in recording; a dropped digit; a completely junk reading. Add in a distribution for the measurement error of the meter itself: they've usually got either a datasheet or an accredited standard which has an error range defined.&lt;/p&gt;&#xA;&lt;p&gt;The statistics should account for things like a real usage spike and a reset coinciding. You might need to specify a joint distribution if they are linked: for example, a power cut could conceivably result in a meter reset &lt;em&gt;and&lt;/em&gt; a power surge, as things like heaters, fridges and freezers would all come back on at full power when power is restored.&lt;/p&gt;&#xA;&lt;p&gt;And then you calculate a posterior distribution for actual energy use, which is the thing you're interested in.&lt;/p&gt;&#xA;&lt;h3&gt;Pros and cons&lt;/h3&gt;&#xA;&lt;p&gt;The second method has the advantage of being rooted in rock-solid theory. It is, however, quite a lot of work to set up the distributions; and in pretty much every real-world case, there isn't an analytic solution, so you have to find for a numerical solution (e.g. using markov-chain monte-carlo). Software packages such as &lt;a href=&quot;http://mc-stan.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Andrew Gelman's STAN&lt;/a&gt; will do that part of the work for you.&lt;/p&gt;&#xA;&lt;h3&gt;Before you start, chart&lt;/h3&gt;&#xA;&lt;p&gt;Either way, start by charting your raw data. The eye will pick up informative patterns.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-08-06T19:59:34.150" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3830" PostTypeId="1" CreationDate="2015-08-06T19:59:10.987" Score="3" ViewCount="439" Body="&lt;p&gt;I have the following system (see picture). When calculating gain and phase margins, should I consider $L=GK$ to be the open loop transfer function? If so, what difference does the transfer function $H$ applied on the reference $r$ make to the gain and phase margins? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0hNUo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0hNUo.jpg&quot; alt=&quot;Closed loop system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2539" LastEditorUserId="272" LastEditDate="2015-08-07T01:14:09.757" LastActivityDate="2015-08-12T15:33:40.327" Title="Gain and phase margins of a system" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3833" PostTypeId="2" ParentId="3830" CreationDate="2015-08-06T23:43:48.643" Score="3" Body="&lt;p&gt;No, you do not need to consider $H$ when thinking about the loop stability.  Doing the loop algebra shows why:&#xA;$$&#xA;y= KG(\eta H-y)\qquad\rightarrow\qquad y=\left(\frac{KG}{1+KG}\right)\ \eta H&#xA;$$ &#xA;The part in parentheses is just the standard transfer function of the loop.  The concept of gain and phase margin comes from considering that the overall gain becomes infinite (or very large) if $KG$ equals (or gets close to) -1. Writing&#xA;$$&#xA;KG=Ae^{i\phi}&#xA;$$&#xA;shows that this can happen if the gain $A$ is too close to 1 when the phase $\phi$ is too close to 180&amp;deg;.  $H$ and $\eta$ play no role in this consideration.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-08-06T23:43:48.643" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3834" PostTypeId="1" AcceptedAnswerId="3876" CreationDate="2015-08-07T13:05:36.557" Score="1" ViewCount="306" Body="&lt;p&gt;I am a beginner in Electronics. I am trying to make a small robot controlled by an Arduino. The problem is that I do not know how to calculate the power required to rotate a structure with a servo motor placed on its axis (see picture below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My servo motor is a Dynamixel MX-28 T with an average stall torque of 2.5 Nm at 12 V. The product page for the servo can be found &lt;a href=&quot;http://www.trossenrobotics.com/dynamixel-mx-28-robot-actuator.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and the specification sheet for it is &lt;a href=&quot;http://support.robotis.com/en/product/dynamixel/mx_series/mx-28.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think it is possible to rotate it using this servo motor?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Informations:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Structure weight: 2,2 kg (2200 g) (include battery-arduino-servo)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structure Dimension: 900mm/120mm/120mm (see picture below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structure type: Aluminum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Servo weight: 72 g&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;total weight 2.2 kg will be distributed fairly. 1.1 kilo on the top and on the bottom 1.1 kilo&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jhJwzm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jhJwzm.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/RQwWT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RQwWTm.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(click for full resolution)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Eld1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Eld1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2548" LastEditorUserId="2548" LastEditDate="2015-08-10T11:55:45.647" LastActivityDate="2015-08-10T21:12:13.393" Title="The couple of my servo motor is it powerful enough?" Tags="|mechanical-engineering|electrical-engineering|stepper-motor|robotics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3836" PostTypeId="1" CreationDate="2015-08-07T17:35:33.537" Score="4" ViewCount="512" Body="&lt;p&gt;A recent work project involved a number of &lt;a href=&quot;https://en.wikipedia.org/wiki/Conversion_of_units&quot; rel=&quot;nofollow&quot;&gt;unit conversions&lt;/a&gt; between various &lt;a href=&quot;https://en.wikipedia.org/wiki/System_of_measurement&quot; rel=&quot;nofollow&quot;&gt;systems of measurements&lt;/a&gt;.  Due to the scale and critical nature of the project, we wanted to ascertain that our conversion factors were as accurate as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While only tangentially related, what I couldn't determine was who held final authority in declaring what the &quot;one true&quot; conversion factors should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;http://www.nist.gov/pml/wmd/metric/upload/SP1038.pdf&quot; rel=&quot;nofollow&quot;&gt;NIST provides a set of conversions&lt;/a&gt; and does not make explicit reference to other standards organizations.  This article on &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Bureau_of_Weights_and_Measures&quot; rel=&quot;nofollow&quot;&gt;International Bureau of Weights and Measures (BIPM)&lt;/a&gt; implies that the BIPM is responsible for declaring those conversion factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a single organization responsible for declaring and verifying the accuracy and precision of conversions?&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-08-07T19:50:17.190" LastActivityDate="2015-08-11T13:21:19.743" Title="What organization specifies unit of measure conversions?" Tags="|measurements|standards|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3837" PostTypeId="2" ParentId="80" CreationDate="2015-08-07T17:55:49.510" Score="2" Body="&lt;p&gt;When there is a change of temperature of the reactor contents, which sensor detects the change first?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quicker the effects of a change can be detected, the more responsive the control system and (generally speaking) the smoother the control. Following this principle using whichever sensor is closest to the jacket inlet is preferred, as it will reduce the lag caused by heat transfer through the fluid bulk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you should consider employing cascade control. This uses two controllers, one for the reactor temperature (primary controller) and one for the jacket temperature (secondary controller). The output of the reactor temperature is used as the setpoint for the jacket controller, and the  output from jacket controller is used to control the position of the jacket valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This eliminates much of the heat transfer lag, and would provide more responsive control.&lt;/p&gt;&#xA;" OwnerUserId="2551" LastEditorUserId="2551" LastEditDate="2015-08-07T22:06:46.487" LastActivityDate="2015-08-07T22:06:46.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3838" PostTypeId="2" ParentId="3836" CreationDate="2015-08-07T18:54:32.123" Score="3" Body="&lt;p&gt;In the US the National Institute of Standards and Technology (NIST) was established in 1901 by &lt;a href=&quot;https://www.law.cornell.edu/uscode/text/15/272&quot; rel=&quot;nofollow noreferrer&quot;&gt;U.S. Code, Title 15, Ch. 7, sec. 272&lt;/a&gt;.  Section (b) of that code states that, among other things, the functions of the Secretary and Institute is&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(2) to develop, maintain, and retain custody of the national standards of measurement, and provide the means and methods for making measurements consistent with those standards;&lt;/p&gt;&#xA;&lt;p&gt;(3) to compare standards used in scientific investigations, engineering, manufacturing, commerce, industry, and educational institutions with the standards adopted or recognized by the Federal Government and to coordinate the use by Federal agencies of private sector standards, emphasizing where possible the use of standards developed by private, consensus organizations;&lt;/p&gt;&#xA;&lt;p&gt;(9) to assure the compatibility of United States national measurement standards with those of other nations;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;so the US government has assigned NIST to maintain those standards.&lt;/p&gt;&#xA;&lt;p&gt;Interestingly, the length of the inch/foot/yard was redefined (slightly) &lt;a href=&quot;https://en.wikipedia.org/wiki/United_States_customary_units#Units_of_length&quot; rel=&quot;nofollow noreferrer&quot;&gt;in 1959&lt;/a&gt; to make it more consistent with the metric system.  &amp;quot;The U.S., the United Kingdom and other Commonwealth countries agreed on this definition, and so it is often termed international measure.&amp;quot;  So, the current conversion between the imperial and metric units of length was agreed upon by the various governing bodies in the countries which use the imperial system.&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-08-07T19:16:43.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3839" PostTypeId="2" ParentId="3683" CreationDate="2015-08-07T19:56:23.940" Score="2" Body="&lt;p&gt;The type of your boundary conditions - specifically the adiabatic walls - is the only thing that would prevent you to use the conventional Nusselt number correlations for turbulent flow inside pipes (Gnielinski, Dittus-Boelter and Sieder-Tate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since heat is only transferred from only one side of the duct and as I suggested before in comments, you can treat that side as &quot;forced convection over a flat plate&quot; problem. I see that you are worried about the effect of boundary layer on the solution, so I made a little experiment to investigate that approach I suggested.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Simplified Problem Setup&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I didn't solve your exact problem but I proposed a simple setup only to investigate the flat plat approximation and here are my assumptions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - I took a tiny part of the duct (10 cm), not the whole 50 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 - I replaced ceramic material with copper, because simply I don't know which ceramic material you have in mind (and I don't have the thermal properties of ceramics), and just for simplification I used copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 - To simplify the problem even further, I assumed that the lower part of the solid copper material is at fixed temperature of 25 C (Since we are only investigating the heat transfer coefficient inside the duct, there is no need to make the problem more complex by introducing the heat transfer coefficient of the ambient air).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Methodology&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Total heat transfer from hot air in duct to the lower plate $Q$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Q = m^oC_p (T_{f_o}  - T_{f_i}) = UA_s\triangle T$$&#xA;where $T_f$ is air flow temperature, subscripts o and i represents outlet and inlet respectively, $U$ is the overall heat transfer coefficient , $A_s$ is the surface area of the lower side of the duct and $\triangle T$ is the difference between average temperature of flow inside duct and the temperature of the lower side of the copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U$ is calculated as follows:&#xA;$$ \frac{1}{U} =  \frac{1}{h_i} + \frac{L}{K} $$&#xA;where $h_i$ is the heat transfer coefficient of air flow inside the duct and $K$ is the thermal conductivity of the copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to calculate $h_i$:&#xA;$$ Nu = \frac{h_iL}{K} = 0.037 Re^{0.8} Pr^{\frac{1}{3}} $$&#xA;since at first iteration $T_{f_o}$ is unknown, all properties are calculated at $T_{f_i}$ only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Python 2.7 code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from __future__ import division&#xA;import CoolProp.CoolProp as coolprop&#xA;&#xA;nRe = lambda rho, v, L, mu: (rho*v*L) / mu&#xA;nPr = lambda cp, mu, k: (cp*mu) / k&#xA;nNu = lambda Re, Pr: 0.037 * (Re**0.8) * (Pr**(1/3))&#xA;Rho = lambda T: coolprop.PropsSI('D','P',60*100000,'T',T+273,'Air.mix')&#xA;Mu  = lambda T: coolprop.PropsSI('V','P',60*100000,'T',T+273,'Air.mix')&#xA;Cp  = lambda T: coolprop.PropsSI('C','P',60*100000,'T',T+273,'Air.mix')&#xA;K   = lambda T: coolprop.PropsSI('L','P',60*100000,'T',T+273,'Air.mix')&#xA;&#xA;duct_length = 0.1 # 10 cm&#xA;plate_thickness = 0.001 # 1 mm&#xA;duct_height = 0.015 # 1.5 cm&#xA;mass_flow = 0.03 # 30 g/s&#xA;surface_temp = 25 # C&#xA;&#xA;# first iteration&#xA;Tavg = 700&#xA;velocity = mass_flow/(Rho(Tavg) * (duct_height**2))&#xA;Re = nRe(Rho(Tavg), velocity, duct_length, Mu(Tavg))&#xA;Pr = nPr(Cp(Tavg), Mu(Tavg), K(Tavg))&#xA;Nu = nNu(Re, Pr)&#xA;h = (Nu * K(Tavg))/duct_length&#xA;U = ( (1/h) + (0.001/398) )**-1&#xA;Tfo = Tavg - ( (U * (duct_height  *duct_length) * (Tavg - surface_temp)) / (mass_flow*Cp(Tavg)) )&#xA;&#xA;# iteration loop&#xA;i = 0&#xA;while i &amp;lt; 30:&#xA;    Tavg = (700 + Tfo) / 2&#xA;    velocity = mass_flow/(Rho(Tavg) * (duct_height**2))&#xA;    Re = nRe(Rho(Tavg), velocity, duct_length, Mu(Tavg))&#xA;    Pr = nPr(Cp(Tavg), Mu(Tavg), K(Tavg))&#xA;    Nu = nNu(Re, Pr)&#xA;    h = (Nu * K(Tavg))/duct_length&#xA;    U = ( (1/h) + (0.001/398) )**-1&#xA;    Tfo = Tavg - ( (U * (duct_height  *duct_length) * (Tavg - surface_temp)) / (mass_flow*Cp(Tavg)) )&#xA;    i += 1&#xA;&#xA;print 'Outlet temperature = {0} C'.format(Tfo)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Result: Outlet temperature $T_{f_o}$ = 668.42 C&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;CFD Simulation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The reason for the 10 cm part was just to simplify the CFD geometry and lower computation time. Air was treated as an ideal gas and I used a RANS equations turbulence model (k-epsilon) and here are the results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature distribution at the center of the duct (left side is the inlet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gzzij.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gzzij.png&quot; alt=&quot;contour_1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature profile of duct outlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gz3J5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gz3J5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, the simulation resulted in average outlet temperature $T_{f_o}$ = &lt;strong&gt;684.57 C&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Please keep in mind that I didn't solve your exact problem, I just made a simplified setup to investigate the accuracy of the flat plate assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a quite big difference between our assumption and the CFD of the actual case (16.15 C), so if this level of accuracy is not sufficient for you as preliminary calculations (I don't think it is), then you should set up an experimental apparatus to get a Nusselt correlation for the actual case, you won't find much help with the available correlations with those boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&#xA;(I'll double check my calculations and get back if there were any errors.)&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2015-08-08T07:36:30.903" LastActivityDate="2015-08-08T07:36:30.903" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="3841" PostTypeId="2" ParentId="3830" CreationDate="2015-08-07T20:48:09.677" Score="0" Body="&lt;p&gt;To compute the gain and phase margins, we first need to determine the loop gain from $r\to y$. The block diagram is equivalent to the following one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gcftb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gcftb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is now straightforward to compute the loop gain from $r\to y$ and it turns out as $L=\frac{H K G }{H}$, and with pole-zero cancellations simplifies to $L=K G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $H$ does not affect the stability margins, because it cancels itself. However, $H$ does affect stability. If you want to determine whether it is minimum phase or not, use  $L=\frac{H K G}{H}$ before cancellation. For example, if $H$ is non-minimum phase and $G$ and $K$ are minimum phase and the gain and phase margins turn out to be positive, just don't overlook that you are dealing with a non-minimum phase system.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2015-08-12T15:33:40.327" LastActivityDate="2015-08-12T15:33:40.327" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3842" PostTypeId="1" AcceptedAnswerId="3847" CreationDate="2015-08-07T22:25:28.387" Score="2" ViewCount="115" Body="&lt;p&gt;In a sensor project I'm involved with, we have a requirement to bypass the airflow from one sensor while it does some further measurements. To do this, I implemented a flap driven by a &lt;a href=&quot;https://www.pololu.com/product/1053&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard servo&lt;/a&gt; which rotates the flap through approximately 90 degrees, opening one port and closing another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to try and improve this with a better, less leaky, mechanism. The first thing I've thought of is a cylinder rotating within another:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WIIaq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WIIaq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green outer cylinder contains three ports to which the inlet pipe and two outlet pipes (~20mm dia) are connected. The inner cylinder (~30mm dia) rotates to select one or other of the outlet pipes - thus it has either a really elongated inlet port or four ports. I figure I can place the servo in the centre - this dictates the inner cylinder diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an existing off-the-shelf product that would do this for me or will I have to try and 3-D print it (or find two waste pipes that will fit appropriately)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a reasonably way to do airflow selection or will it turn out to be even more leaky than the airflap (which wasn't particularly well done as it happens - it can be improved)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: As an addition in the OP rather than as an answer/comment: This is another way of doing this. Once you know &quot;pneumatic&quot; is the term you can add terms like &quot;servo valve&quot; to google image search and get all sorts of neat ideas (including little lego valves!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pU2l0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pU2l0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2555" LastEditorUserId="2555" LastEditDate="2015-08-08T21:30:30.627" LastActivityDate="2015-08-08T21:30:30.627" Title="Airflow selection mechanism" Tags="|mechanical-engineering|design|airflow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3843" PostTypeId="1" CreationDate="2015-08-08T00:29:33.513" Score="4" ViewCount="33" Body="&lt;p&gt;I currently need to paint a steel structure with galvanizing primer.  The primer is an epoxy with &gt;90% Zinc content.  The TDS for the primer states that intimate contact with the substrate is needed to achieve full effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense to me...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the paint spec and certification for the paint mandates an iron phosphate coating as pretreatment for the primer.  Does this not prevent intimate contact?  Or am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="2557" LastEditorUserId="90" LastEditDate="2015-08-08T01:31:15.893" LastActivityDate="2015-08-08T01:31:15.893" Title="Pretreatment for galvanizing primer" Tags="|corrosion|manufacturing-engineering|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="3845" PostTypeId="2" ParentId="3836" CreationDate="2015-08-08T02:09:23.097" Score="3" Body="&lt;p&gt;Internationally, the &lt;a href=&quot;https://en.wikipedia.org/wiki/General_Conference_on_Weights_and_Measures&quot; rel=&quot;nofollow&quot;&gt;General Conference on Weights and Measures&lt;/a&gt; (&lt;em&gt;French: Conférence générale des poids et mesures - CGPM&lt;/em&gt;) is the senior of the three Inter-governmental organizations established in 1875 to define standards for weights and measures. The other two organizations are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the International Committee for Weights and Measures (&lt;em&gt;French: Comité&#xA;international des poids et mesures- CIPM&lt;/em&gt;) and&lt;/li&gt;&#xA;&lt;li&gt;the International Bureau of Weights and Measures (&lt;em&gt;French: Bureau&#xA;international des poids et mesures - BIPM&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The conference meets in Sèvres (south-west of Paris) every four to six years. In 1960 the 11th CGPM approved the system of &lt;a href=&quot;https://en.wikipedia.org/wiki/International_System_of_Units&quot; rel=&quot;nofollow&quot;&gt;SI Units&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each country has its own organization for establishing national standards, of which weights and measures is one, and their standards are legally binding for their country. Use the conversion factors set by the standards organization for the country in where the project will be realised/established.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-08-08T02:09:23.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3846" PostTypeId="1" AcceptedAnswerId="3848" CreationDate="2015-08-08T02:39:39.027" Score="3" ViewCount="3727" Body="&lt;p&gt;Power cords for laptop computers have a transformer built into them.  The power supply has a transformer that steps down mains level power to a lower voltage and rectifier that converts the current from AC at the mains to DC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power surges con occur for a variety of reasons, such as a nearby lightning strike.  If a damaging power surge were to occur, would the power supply prevent the power surge from damaging the laptop computer?&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorDisplayName="user16" LastEditDate="2015-08-08T11:54:57.560" LastActivityDate="2015-08-08T12:13:45.447" Title="Evaluating the probability of damage to a laptop computer from power surges" Tags="|power-electronics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3847" PostTypeId="2" ParentId="3842" CreationDate="2015-08-08T03:41:32.587" Score="7" Body="&lt;p&gt;Are you bound to the shape/configuration of part you constructed in your question? If not, a simple 3-way &lt;strong&gt;pneumatic solenoid valve&lt;/strong&gt; would make this very simple and leak proof(relatively speaking). They are inexpensive too.(eBay) You would have an input and the valve will allow flow through one outlet and then energize the solenoid and you will get flow through the second outlet. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2015-08-08T03:41:32.587" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3848" PostTypeId="2" ParentId="3846" CreationDate="2015-08-08T12:07:14.693" Score="1" Body="&lt;p&gt;An occupational hazard of mine is working with a lot of people who own laptop computers.  I also happen to live in the central plains of the United States, and we get our fair share of thunderstorms.  They're not as pretty as the lightning storms of Phoenix, AZ, but they're generally a lot more powerful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anecdotally, I've heard of multiple times where the power supply (aka &quot;the brick&quot;) for the laptop computer becomes a sacrificial component and will be destroyed by a power surge whereas the laptop itself is just fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard of similar stories with desktop computers, but on a much more rare basis.  Generally, damaging surges will take out the power supply and the mainboard with a desktop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laptop computers have an additional block of circuitry that directs power to the battery for charging purposes and / or directing power to the rest of the laptop circuitry.  I would hazard a guess that this additional charging circuitry provides an additional measure of shunting and helps protect laptop computers from damaging power surges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between laptops and desktops, I have probably heard of an equal number of instances of laptop power supplies and desktops being destroyed.  I have heard of the actual laptop being destroyed less frequently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your most likely risk is losing the laptop power supply due to the damaging power surge.  I'm not certain how to evaluate that overall risk as almost everyone I know connects their electronics through protective power strips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you're not completely risk-free of the laptop being damaged.  Damaging power surges are notoriously fickle and can have incredible variance in intensity.  Lightning strikes vary in intensity; the distance between the strike and the equipment will vary; and the number of connected circuits between the strike and your equipment will also vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a protective standpoint, one of the best things you can do is put a surge protector block within the circuit panel supplying mains power to your dwelling.  From there, make sure that any sensitive electronics are powered through surge protecting power strips.  Those two measures will provide you a significant reduction of risk at a modest price.  And if you're really concerned, power everything down and disconnect the power strip when a storm is rolling through.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-08-08T12:13:45.447" LastActivityDate="2015-08-08T12:13:45.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3850" PostTypeId="1" AcceptedAnswerId="3851" CreationDate="2015-08-08T13:50:08.957" Score="4" ViewCount="95" Body="&lt;p&gt;Given a successful deuterium-tritium fusion reactor such as that conceived by ITER, could the world's energy needs theoretically be met without the input of any petroleum or natural gas products?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deuterium is currently typically produced by the girdler sulfide process from seawater which requires a natural gas input, although methods involving electrolysis or distillation are known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium is currently mined, although there are methods known of extracting it from seawater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the extractions of these elements be powered exclusively from the energy produced by fusion reactors in the future? This would leave seawater and the reserve of lithium from the earth's crust as the only fuel sources (or just seawater, if we are extracting the lithium from it). &lt;/p&gt;&#xA;" OwnerUserId="2563" LastEditorUserId="2563" LastEditDate="2015-08-08T14:30:30.290" LastActivityDate="2015-08-09T05:37:57.180" Title="Could future fusion energy exist without the input of any petroleum or natural gas products?" Tags="|renewable-energy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3851" PostTypeId="2" ParentId="3850" CreationDate="2015-08-08T15:14:15.553" Score="5" Body="&lt;p&gt;The energy cost of obtaining Deuterium and Lithium is minimal compared to the energy released by the fusion process. There are many references that provide relevant information but a sensible one to refer to is the official ITER site &amp;quot;Fusion for energy&amp;quot;. &lt;a href=&quot;http://fusionforenergy.europa.eu/understandingfusion/merits.aspx&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Fusion for Energy&lt;/strong&gt;&lt;/a&gt; (F4E) is the European Union’s Joint Undertaking for ITER and the Development of Fusion Energy.&lt;/p&gt;&#xA;&lt;p&gt;On their &lt;a href=&quot;http://fusionforenergy.europa.eu/understandingfusion/merits.aspx&quot; rel=&quot;noreferrer&quot;&gt;understanding fusion page&lt;/a&gt; they say:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A 1,000-megawatt electric fusion power plant would consume around 100 kg of deuterium and three tonnes of natural lithium in a year whilst generating 7 billion kilowatt-hour. To generate the same amount of electricity, a coal-fired power plant would need around 1.5 million tonnes of coal.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If used to fuel a fusion power station, the lithium in one laptop battery, complemented with half a bath of water, would produce the same amount of electricity as burning 40 tonnes of coal.&lt;br /&gt;&#xA;(eg ~= 8000 dollars at 5c/kWh wholesale or 40,000 dollars at NZ 25c/kWh retail).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However what is required is Tritium - Lithium is simply an easily obtained precursor. Obtaining Tritium from Lithium is the &amp;quot;hard part&amp;quot; and while it is proposed that this step be incorporated in the ITER system, and is liable to succeed if ITER succeeds, the process is at this stage &amp;quot;theoretical and experimental&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Tritium can be, and currently is, bred from Lithium using fast neutrons from existing reactors, the amount obtained in this manner is enough to initiate a national or international fusion energy system but nowhere near enough to sustain it. What is essential is the &amp;quot;breeding&amp;quot; of the essential Tritium from Lithium using a suitable source of fast neutrons from the ITER process itself&lt;/p&gt;&#xA;&lt;p&gt;ITER's own site, in the section &lt;a href=&quot;https://www.iter.org/mach/tritiumbreeding&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;Tritium Breeding&lt;/strong&gt;&lt;/a&gt; says&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;... While deuterium can be extracted from seawater in virtually boundless quantities, the supply of tritium is limited, estimated currently at twenty kilos. .... tritium can be produced within the tokamak when neutrons escaping the plasma interact with a specific element—lithium—contained in the blanket. This concept ... is an important one for the future needs of a large-scale fusion power plant.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;ITER will procure the tritium &amp;quot;fuel&amp;quot; necessary for its expected 20-year lifetime from the global inventory&lt;/strong&gt;. But for DEMO, the next step on the way to commercial fusion power, about 300g of tritium will be required per day to produce 800 MW of electrical power. No sufficient external source of tritium exists for fusion energy development beyond ITER, making the successful development of tritium breeding essential for the future of fusion energy.&lt;/p&gt;&#xA;&lt;p&gt;ITER will provide a unique opportunity to test mockups of breeding blankets, called Test Blanket Modules (TBM), in a real fusion environment. Within these test blankets, viable techniques for ensuring tritium breeding self-sufficiency will be explored.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;______________________&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How Tritium is obtained at present:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Construction &amp;amp; Operation of a Tritium Extraction Facility at the Savannah River Site&lt;br /&gt;&#xA;Final Environmental Impact Statement&#xA;&lt;a href=&quot;http://energy.gov/sites/prod/files/EIS-0271-FEIS-01-1999.pdf&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;DOE/EIS-0271  1999&lt;/strong&gt;&lt;/a&gt; - 140 page pdf&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;__________________________&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;With respect to generating hydrocarbons from &amp;quot;other sources of cheap energy&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;Yes. The process could be totally &amp;quot;geologically stored hydrocarbon&amp;quot; free, which I think is the 'spirit' of what you are asking.&lt;/p&gt;&#xA;&lt;p&gt;Once you achieve &amp;quot;energy too cheap to meter&amp;quot; [tm] (2nd time lucky perhaps) you can produce hydrocarbons from current biological waste - or from newly grown feedstock if you so desired. Various people have demonstrated 'just about anything organic' to petroleum products converters. Videos of some such are available on you-tube - caveat emptor as ever, and I knew a man in Louisiana who I have lost touch with who was doing just this several years ago, and having a lot of trouble interesting anyone in the process.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-08-09T05:37:57.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3852" PostTypeId="1" CreationDate="2015-08-08T21:22:42.700" Score="0" ViewCount="611" Body="&lt;p&gt;Lets take the substance to be water. To my knowledge , the only condition imposed on a velocity profile to distinguish between laminar and turbulent force is the &lt;strong&gt;&lt;em&gt;Reynold's number&lt;/em&gt;&lt;/strong&gt;. For lets say I have been given this velocity vector : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;V = (xy) i + (yz) j + (xz) k&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how will I determine whether water, with this velocity profile , can have a laminar flow or not ?&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2015-08-09T19:45:05.870" Title="How to determine whether a particular velocity vector will result in a laminar flow?" Tags="|fluid-mechanics|flow-control|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3853" PostTypeId="2" ParentId="3852" CreationDate="2015-08-09T01:12:47.953" Score="3" Body="&lt;p&gt;You are correct, the Reynolds number is what determines whether a flow is laminar or turbulent. The formula for the Reynolds number is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Re = \frac{\rho vL}{\mu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho$ is the density, $v$ is the velocity of the flow, $\mu$ is the dynamic viscosity of the fluid, and $L$ is the characteristic length associated with the flow. The first three figures can be determined from the info you have in that velocity vector and by knowing the fluid. But the characteristic length depends on where your flow is. You have to know something about the pipe you're in, the plate you're flowing against, the airfoil you're going over, etc. in order to determine the Reynolds number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the Reynolds number is proportional to that length, knowing the other three values really doesn't do much, unless they create a number so disproportionately large that the length would have to be impossibly small to make the flow laminar. But you also need to know something about the geometry (essentially the same thing that would give you the characteristic length) to know where the points of laminar flow and turbulent flow start with respect to the Reynolds number. The transitional values are different with the flow in a pipe and the flow over an airfoil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the real answer is, you can't know whether it's laminar or turbulent with the information you've provided.&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-08-09T01:12:47.953" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3854" PostTypeId="1" AcceptedAnswerId="3857" CreationDate="2015-08-09T04:21:47.880" Score="-2" ViewCount="214" Body="&lt;p&gt;It does make sense centrifugal compressors exist, because gases can compress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all know that liquids don't compress like gas. Then why do centrifugal or reciprocating pumps exists? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See even a the Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Centrifugal_pump&quot; rel=&quot;nofollow&quot;&gt;Centrifugal Pumps&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2571" LastEditorUserId="9" LastEditDate="2015-09-26T19:16:21.843" LastActivityDate="2015-09-26T19:16:21.843" Title="Why do pumps really exists?" Tags="|fluid-mechanics|pumps|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3856" PostTypeId="2" ParentId="3852" CreationDate="2015-08-09T09:16:42.097" Score="4" Body="&lt;p&gt;I am afraid the question you are asking might not even be a valid question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because of the &lt;strong&gt;correlation&lt;/strong&gt; between the &lt;strong&gt;Reynolds-Number&lt;/strong&gt; ($Re$) and the occurrence of &lt;strong&gt;turbulence&lt;/strong&gt;. $Re$ is not a fluid mechanic switch (turning turbulence on and off) it is more a parameter which allows for describing and clustering flow phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Re$ describes the ratio between &lt;strong&gt;inertial&lt;/strong&gt; and &lt;strong&gt;viscous&lt;/strong&gt; forces. This ratio has a direct influence on the stability of the flow. Flows with high viscosity (hence low $Re$) will be able to stay laminar while being subjected to small perturbations. The same perturbations will make a flow with low viscosity turbulent. High $Re$ will be able to &lt;em&gt;damp out&lt;/em&gt; perturbations (in velocity or pressure) because they are spread out fast enough. Maybe you can think of viscosity as the ability to conduct &lt;em&gt;velocity perturbations&lt;/em&gt; into the surrounding fluid the same way as high thermal conductivity will conduct &lt;em&gt;thermal perturbations&lt;/em&gt; into the surrounding fluid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is a re-occurring theme here at &lt;em&gt;engineering SE&lt;/em&gt; the answer is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it depends&lt;/li&gt;&#xA;&lt;li&gt;it's a bit more complicated &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please describe the setting in more detail:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;free flow vs. internal flow&lt;/li&gt;&#xA;&lt;li&gt;steady vs. unsteady&lt;/li&gt;&#xA;&lt;li&gt;symmetries&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and maybe provide a rough idea on the velocity profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the most tricky part of the $Re$ is the length-scale $L$ which has to be chosen carefully to enable the similarity. The length-scale is easy to find in simple 2D-flows like a pipe or a plate. It already gets tricky when you combine both.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-08-09T19:45:05.870" LastActivityDate="2015-08-09T19:45:05.870" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3857" PostTypeId="2" ParentId="3854" CreationDate="2015-08-09T09:33:04.360" Score="4" Body="&lt;p&gt;hm...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume the term &lt;em&gt;does not compress&lt;/em&gt; is what kicked of this question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Right?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task of a pump is to store energy in the fluid (potential energy in the form of pressure). This stored energy can be used for different purposes (e.g. @Fred comment). As you already mentioned there is a quite interesting feature of fluids. Some are compressible some aren't. This means whereas some will change their volume under pressure some won't.* &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might have guessed this influences the design of pumps. Depending on the fluid and its characteristics (one of which is compressibility, you might have a look for the Mach-Number, $Ma$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*) As pointed out by @Reza-Afzalan this statement is a simplification or rather exaggeration since all fluids are compressible but the volume change of fluids is usually negligible small. &lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-08-11T05:34:37.083" LastActivityDate="2015-08-11T05:34:37.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3858" PostTypeId="1" CreationDate="2015-08-09T13:43:05.577" Score="0" ViewCount="427" Body="&lt;p&gt;I have a simple question: in the research literature on wireless communications I often see the term &quot;Ergodic Capacity&quot; of a channel. What does it mean?&lt;/p&gt;&#xA;" OwnerUserId="2572" LastEditorUserId="2572" LastEditDate="2015-08-11T20:46:02.510" LastActivityDate="2017-07-24T03:27:12.933" Title="What is the meaning of 'ergodic capacity' of a wireless communication channel?" Tags="|electrical-engineering|wireless-communication|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3859" PostTypeId="2" ParentId="3854" CreationDate="2015-08-09T14:47:04.963" Score="3" Body="&lt;p&gt;All real  materials are compressible. If you look at thermodynamics tables, even small change in pressure causes density change in fluid. But as an engineer, when working on pumps you can omit small and effectless changes in density of fluids. So pumps, compressors, and fans all do the same thing: they compress material and create flow.&lt;/p&gt;&#xA;" OwnerUserId="2404" LastEditorUserId="2246" LastEditDate="2015-08-22T17:27:34.373" LastActivityDate="2015-08-22T17:27:34.373" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3863" PostTypeId="1" AcceptedAnswerId="3864" CreationDate="2015-08-10T01:58:17.750" Score="4" ViewCount="8083" Body="&lt;p&gt;Current gear manufacturing methods for gears usually depend on power tools, or computerised machines, or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, watchmakers and clockmakers at one point in history needed to make gears by hand. So what tools did they use? How did they make gears (especially really tiny ones)?&lt;/p&gt;&#xA;" OwnerUserId="2577" LastEditorUserId="368" LastEditDate="2015-08-10T16:33:49.310" LastActivityDate="2018-03-29T23:46:37.100" Title="How were gears manufactured before power tools, CNC, etc" Tags="|gears|engineering-history|manufacturing-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3864" PostTypeId="2" ParentId="3863" CreationDate="2015-08-10T02:45:00.747" Score="2" Body="&lt;p&gt;Watchmakers would have (and still do) make the tiny gears for watches by hand using any number of jeweler's saws, needle files, and tiny drills.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A jeweler's saw is pictured below (source: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0037CQ8ZO&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amazon&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BrDoO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BrDoO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needle files (source: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B000OVHZJ4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amazon&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HOkA8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HOkA8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drill bits (source: &lt;a href=&quot;http://watchmaking.weebly.com/drill-bits.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adventures in Watchmaking&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bwUXQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bwUXQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The website &lt;a href=&quot;http://watchmaking.weebly.com/introduction.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adventures in Watchmaking&lt;/a&gt; has a good walkthrough of the author's experience in making a watch without having done so before, if you're interested.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-08-10T02:45:00.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3865" PostTypeId="1" AcceptedAnswerId="3867" CreationDate="2015-08-10T12:25:25.637" Score="5" ViewCount="1937" Body="&lt;h1&gt;The Tibetan Bridge Type&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I have a question regarding the bridge type called &quot;Tibetan Bridge&quot;.&#xA;Recently close by my home town a new tourist attraction was built, the &lt;a href=&quot;http://www.highline179.com/en/guinness-world-record-for-the-highline179/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Highline 179&lt;/a&gt; bridge. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/BVm9t.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BVm9t.jpg&quot; alt=&quot;Highline 179 from Wikipedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was planned as the longest suspension bridge for pedestrians in the world, but before completion it was already outclassed by &#xA;the Skypark Bridge in Sochi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the Bridge is denoted as the longest Tibet-style pedestrian suspension bridge and I could not find information what a Tibet-style or Tibetan bridge is.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Other Tibetan Bridges&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I could find other Tibetian bridges (e.g. Tibetan Bridge Claviere), but I could not find any information what Tibet style could be from an engineering perspective. The Wikipedia article on bridge types does not list this particular type.&lt;/p&gt;&#xA;" OwnerUserId="2583" LastEditorUserId="33" LastEditDate="2015-11-14T05:27:48.630" LastActivityDate="2015-11-14T05:27:48.630" Title="What is a Tibetan Style Bridge?" Tags="|civil-engineering|bridges|architecture|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3866" PostTypeId="2" ParentId="3834" CreationDate="2015-08-10T12:50:07.933" Score="1" Body="&lt;p&gt;If the structure is ballanced around the pivot point, as you seem to indicate, then in theory it only takes enough torque to overcome whatever friction or other external resistances there are.  There is no inherent torque required just to move a mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does take a minimum torque to &lt;i&gt;accellerate&lt;/i&gt; a object.  To know this minimum torque, you have to decide what the object's moment of inertia is about the pivot point, how fast you want it to rotate, and how long you are willing to wait to get it to that speed.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-08-10T12:50:07.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3867" PostTypeId="2" ParentId="3865" CreationDate="2015-08-10T13:43:01.023" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Suspension_bridge#Precursor&quot; rel=&quot;nofollow&quot;&gt;The section on the precursor to modern suspension bridges, in the wikipedia article that you mentioned&lt;/a&gt;, talks about bridges of the &quot;Tibetan saint and bridge-builder Thangtong Gyalpo&quot; and points out that they &quot;did not include a suspended deck bridge which is the standard on all modern suspension bridges today&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I confess I am only guessing, and haven't found anything definitive to back this up, but I infer that a &quot;Tibet-style&quot; suspension bridge is hence one which doesn't have a &lt;em&gt;suspended bridge deck&lt;/em&gt;. What does this mean? In simple terms it means that the vertical cables (hangars) in a Tibet-style bridge are all the same length, such that the bridge deck has a significant sag in it. Modern bridges, with their suspended decks, have main cables with a significant sag, but varying lengths hangars such that the bridge deck is made (relatively) flat.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-08-10T13:43:01.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3868" PostTypeId="1" CreationDate="2015-08-10T13:47:52.457" Score="6" ViewCount="880" Body="&lt;p&gt;How can I measure changes in liquid level of the order of 50 μm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tank is cylindrical with a radius of about 12.5 mm and a total volume of about 45 ml. The fluid is water-based. The technique needs to be optical or similar because no external body can touch the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked at ultrasonic sensors but they don't seem to reach such high precision. Now I'm checking laser sensors as well, but it seems that they don't satisfy the required precision either.&lt;/p&gt;&#xA;" OwnerUserId="2437" LastEditorUserId="368" LastEditDate="2015-08-10T16:30:12.923" LastActivityDate="2017-03-06T20:37:02.747" Title="Precision measurement of liquid levels" Tags="|fluid-mechanics|sensors|optics|microfluidics|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="3870" PostTypeId="2" ParentId="3409" CreationDate="2015-08-10T15:51:01.737" Score="2" Body="&lt;p&gt;Although i know is's not the right answer, i would like to mention that using a micrometer to measure bolt length out of nut can be helpful, because that is proportional to torque, so for same installations one can use it to check if all bolts have the same load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;another option when a torque meter is not helpful is to use smart bolts. they have internal strain sensor. check e.x &lt;a href=&quot;http://www.smartbolts.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.smartbolts.com/&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="2404" LastActivityDate="2015-08-10T15:51:01.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3872" PostTypeId="1" CreationDate="2015-08-07T11:27:40.573" Score="3" ViewCount="210" Body="&lt;p&gt;I have data from a vibration experiment on a structure. This data was obtained by attaching an accelerometer to a FFT analyser, which records the data at a high sampling rate (typically in kHz range) and can also perform real-time FFT of the recorded data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/37YER.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/37YER.jpg&quot; alt=&quot;sample signal and its FFT&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image above shows the recorded signal below and the FFT above. The FFT shows all the dominant frequencies in the original signals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What more patterns/ important data points can I discern from this data, that might have utility ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one being, the RMS value of the signal. What else ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the recorded data &lt;a href=&quot;https://drive.google.com/file/d/0B9fbkDTtTJfSblh3b0c2OXBYYzA/view?usp=sharing&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; as a mat file, which can be loaded in matlab, the data in a variable called Track1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics of the problem: The data is collected from a shake table, a rigid platform, that oscillates back and forth, so the sharp edges in the original signal periodically, representing the point of return. &lt;/p&gt;&#xA;" OwnerUserId="1663" OwnerDisplayName="Chintan Pathak" LastActivityDate="2016-01-23T03:24:48.207" Title="How to analyse periodic data obtained from a vibration experiment?" Tags="|vibration|" AnswerCount="3" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3873" PostTypeId="2" ParentId="3872" CreationDate="2015-08-07T12:30:55.157" Score="3" Body="&lt;p&gt;This certainly isn't an answer, but more of a guidance on how to approach the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analysis that you choose to do depends &lt;em&gt;entirely&lt;/em&gt; on &lt;em&gt;why you are collecting the data in the first place.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the structure may have been designed and constructed to behave a certain way, and you are collecting data to determine whether it is behaving to spec.  You may be trying to track changes over time.  You may be collecting data under a protocol to determine under what types of stimuli your structure will fail.  You may be tracking data over time to determine when your part needs repair.  Each scenario, and dozens of others I can't even dream about, will determine how you need to analyze the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that when you do an experiment, you need to understand what question you will be asking of the data generated -- or the best use of your time is probably not to do the experiment in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a direct answer, I suggest dividing your data by \$\pi\$, which is as meaningful as any other suggestion sans some clue about why you are collecting the data.&lt;/p&gt;&#xA;" OwnerUserId="5293" OwnerDisplayName="Scott Seidman" LastActivityDate="2015-08-07T12:30:55.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3874" PostTypeId="2" ParentId="3872" CreationDate="2015-08-07T13:38:17.487" Score="0" Body="&lt;p&gt;I suggest a statistical approach, looking for any characteristics of the data values that are significant.  The characteristics seem to be:&#xA;absolute value&#xA;time&#xA;direction&#xA;rate of change&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From these you can establish other characteristics, such as period, frequency, repeat ability or variation, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other information that is available to you but not evident in the data would be things like, at what time did the structure collapse (if it did), what was the growth rate of cracks in the structure, did the structure change color, size, shape or other features during the vibration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statistical testing of the data displayed above plus other captured data may reveal significant points or relationships.&lt;/p&gt;&#xA;" OwnerDisplayName="Legendsk" LastActivityDate="2015-08-07T13:38:17.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3875" PostTypeId="1" CreationDate="2015-08-10T19:12:44.593" Score="2" ViewCount="291" Body="&lt;p&gt;I'm talking about a:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;smooth circular pipe without any valve at inlet or outlet&lt;/li&gt;&#xA;  &lt;li&gt;without pressure losses due to friction&lt;/li&gt;&#xA;  &lt;li&gt;subjected to a constant heat flux from outside on the entire pipe external surface which makes temperature rise from $T_i=973.15[K]$ to a $T_o=1073.15 [K]$&lt;/li&gt;&#xA;  &lt;li&gt;in which air flows at pressure of $p_i=10 [bar]$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this doubt was born studying the momentum balance:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\rho*v^2|_{out}-\rho*v^2|_{in}=p_{out}-p_{in}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;expressed using the perfect gas law $\rho=\dfrac{p}{R^**T}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and using the conservation law $v=\dfrac{\dot{m}}{\rho*A}$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is the cross sectional area of the pipe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing those in the balance and if I assume mass flow, composition, and cross sectional area constants, I get:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$\dfrac{\dot{m}^2*R^*}{A^2}*\left[\dfrac{T_{out}}{p_{out}}-\dfrac{T_{in}}{p_{in}}\right]=p_{out}-p_{in}$  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;because the &lt;em&gt;block&lt;/em&gt; outside of the parentheses is constant, this bring me to a fractional second order equation, in which $p_{out}$ is the only &lt;strong&gt;unknown&lt;/strong&gt;. The numerical solutions to this equation are (if I put the outside &lt;em&gt;block&lt;/em&gt; equals to 1 for simplicity):&lt;br&gt;&#xA;$1*\left[\dfrac{1073.15}{x}-\dfrac{973.15}{10^6}\right]=x-10^6$&lt;br&gt;&#xA;that I can rewrite as $\dfrac{1073.15}{x}-\dfrac{973.15}{10^6}=x-10^6$&lt;br&gt;&#xA;again as $\dfrac{1073.15}{x}-\dfrac{973.15}{10^6}-x+10^6=0$&lt;br&gt;&#xA;and definitively as $\dfrac{1073.15-\dfrac{973.15*x}{10^6}-x^2+10^6*x}{x}=0$&lt;br&gt;&#xA;denominator simply exclude $x=0$ as possible solution&#xA;while solving numerator equation bring me this two (and both numerically valid) solutions:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$x_1=-0.00107315$  &lt;/li&gt;&#xA;  &lt;li&gt;$x_2=1000000.0001$  &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;we can easily exclude first result because it has not a &lt;strong&gt;physical meaning&lt;/strong&gt; because can't exist negative pressure, so definitively $p_{out}=x_2=1000000.0001[Pa]=1.0000000001[bar] &amp;gt; p_{in}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result makes me &lt;strong&gt;weird&lt;/strong&gt; because it's the first time I heard about this pressure behaviour. I don't understand why!&lt;/p&gt;&#xA;" OwnerUserId="2157" LastEditorUserId="2157" LastEditDate="2015-08-12T20:20:19.613" LastActivityDate="2015-08-12T20:20:19.613" Title="Can pressure rise due to heat flux in a pipe?" Tags="|thermodynamics|pressure|heat-transfer|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3876" PostTypeId="2" ParentId="3834" CreationDate="2015-08-10T21:12:13.393" Score="1" Body="&lt;p&gt;With the structure (and masses) being distributed perfectly even on both sides of the servo, the servo will not require any additional force to rotate the structure since the mass on one side is always balanced out or negated by the mass on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will still have to overcome friction forces and depending on how you mount and use the servo (directly connected to the driving shaft of the servo or not), you may have to take into account whether your servo's motor shaft can handle such radial forces (and maybe axial forces too depending on your application).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, it is possible to rotate the structure with the servo.&lt;br&gt;&#xA;But in order to know how fast you can accelerate or decelerate the structure, you will have to use its moment of inertia.&lt;/p&gt;&#xA;" OwnerUserId="2430" LastActivityDate="2015-08-10T21:12:13.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3878" PostTypeId="2" ParentId="3868" CreationDate="2015-08-10T22:28:35.597" Score="3" Body="&lt;p&gt;If you are adding fluid at a constant rate a simple michelson interferometer seems like a possible solution. You should get a fringe every $\lambda/2$ m, so ~300 nm. As long as you can sample fast enough it is then just a case of counting the number of fringes to get the distance moved. The only difficulty I can see with this approach is low reflectance from you liquid surface, but that will be the same for any optical method and shouldn't be too much of a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultrasound will be difficult as you need to measure time of flight to the ns level to get that sort of resolution. It may still be possible using a lot of averaging, but I would think its not worth it.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-08-10T22:28:35.597" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3879" PostTypeId="2" ParentId="3875" CreationDate="2015-08-11T05:33:15.920" Score="1" Body="&lt;p&gt;I don't know what your x1 and x2 refer to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one-dimensional inviscid compressible flow with heat transfer you are referring to is called &quot;Rayleigh flow&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The static pressure will decrease for subsonic flow and increase for supersonic flow when heat is added. As heat is added, the Mach number will approach 1 and the static pressure will approach a value that is a function of the starting static pressure and Mach number, regardless of whether the flow is supersonic or subsonic when heating begins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total pressure will always decrease regardless of the starting Mach number. You can read about this in a compressible flow textbook. A popular one is by John Anderson.&lt;/p&gt;&#xA;" OwnerUserId="2594" LastActivityDate="2015-08-11T05:33:15.920" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3880" PostTypeId="2" ParentId="3836" CreationDate="2015-08-11T13:21:19.743" Score="0" Body="&lt;p&gt;&lt;strong&gt;tl;dr&lt;/strong&gt; National measurement institutes are responsible for ensuring the definition of SI units. They collaborate to ensure they get the same answer. Non-SI units are defined in terms of their SI equivalents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.bipm.org/en/about-us/&quot; rel=&quot;nofollow&quot;&gt;BIPM&lt;/a&gt; is responsible for determining and maintaining the standards for SI units in combination with national measurement institutes (NIST, NPL, PTB, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally, these were based on physical artifacts but nowadays they are mostly (except the kg) defined in terms of physical constants. Research on &lt;a href=&quot;http://www.nist.gov/pml/si-redef/kg_redefinition.cfm&quot; rel=&quot;nofollow&quot;&gt;redefining the kg&lt;/a&gt; is currently ongoing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each country has its own measurement laboratory which provides standards and services to help industry ensure that units are correct. They also regularly do international comparisons to ensure they all give the same results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-SI units are defined based on national and international laws and treaties. Historically, these were also physical artifacts. But now they are pretty much all defined in law in terms of their SI equivalent (either directly or indirectly). For example the inch is defined as exactly 25.4 mm and a US Gallon is 231 cubic inches.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-08-11T13:21:19.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3881" PostTypeId="1" AcceptedAnswerId="3882" CreationDate="2015-08-11T13:24:00.290" Score="2" ViewCount="4633" Body="&lt;p&gt;When an ideal gas with constant specific heat is throttled adiabatically, with negligible changes in kinetic and potential energies: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(a): ∆h = 0, ∆T = 0  &lt;/li&gt;&#xA;&lt;li&gt;(b): ∆h =0, ∆S &gt; 0  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The answer is supposed to be (b).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain why answer (a) is wrong and (b) is correct.&lt;/p&gt;&#xA;" OwnerUserId="21" LastEditorUserId="272" LastEditDate="2015-08-11T17:00:19.153" LastActivityDate="2015-08-12T06:34:32.553" Title="Adiabatic throttling of ideal gas" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3882" PostTypeId="2" ParentId="3881" CreationDate="2015-08-11T14:02:59.437" Score="3" Body="&lt;p&gt;You answer this by applying first and second laws of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First Law&#xA;$$Q - W = \triangle E_{sys}$$&#xA;since throttling process is adiabatic and there is no mechanical work applied on or extracted from the system then $Q$ and $W$ = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that leaves us with $\triangle E_{sys} = 0$, and for steady state system we can express $E$ as the sum of enthalpy, kinetic energy and potential energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E =  m(h + v^2/2 + gz)$$&#xA;and since we have negligible changes in kinetic and potential energies we finally get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\triangle h = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, as we know from second law of thermodynamics that for an adiabatic process &lt;a href=&quot;https://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Entropy&quot; rel=&quot;nofollow&quot;&gt;$\triangle S \geq 0$&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You should have a look on the throttling process on an h-s diagram of a reversed carnot cycle (refrigeration cycle).)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature behavior in throttling process is governed by the Joule-Thomson coefficient $\mu$&#xA;$$\mu = \left ( \frac{\partial T}{\partial P} \right )_h $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Joule-Thomson coefficient is a measure of the change in temperature at constant-enthalpy process, if it equals zero then temperature remains constant, if greater than zero temperature decreases and if less than zero temperature increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for an ideal gas the enthalpy is a function of only the temperature $h = h(T)$, and since enthalpy remains constant in throttling processes and $\mu = 0$ (You can check Cengel's Thermodynamics book for the proof), &lt;strong&gt;the temperature also remains constant.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so, all the choices are actually correct.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2015-08-12T06:34:32.553" LastActivityDate="2015-08-12T06:34:32.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3883" PostTypeId="1" CreationDate="2015-08-11T17:14:49.493" Score="0" ViewCount="13167" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LYHD8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LYHD8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone know how to solve for the OMC and MDD using mathematical solution? The book says its OMC is 13% and MDD of 2020. I tried using $a+bx+cx^3$ and getting its maxima but its far from 13%&lt;/p&gt;&#xA;" OwnerUserId="2599" LastEditorUserId="90" LastEditDate="2022-08-03T00:45:52.627" LastActivityDate="2022-08-03T00:45:52.627" Title="Optimum Moisture Content and Maximum Dry Unit Weight" Tags="|geotechnical-engineering|soil-mechanics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3886" PostTypeId="2" ParentId="3883" CreationDate="2015-08-11T18:50:27.470" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Proctor_compaction_test&quot; rel=&quot;nofollow noreferrer&quot;&gt;optimum moisture content&lt;/a&gt; is defined as the moisture content where you get maximum dry unit weight. So plot &lt;em&gt;moisture content&lt;/em&gt; ($w$) vs &lt;em&gt;dry density&lt;/em&gt; ($\rho_d$) and locate the maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without any interpolation (i.e. if we have to pick a value from your table) this occurs at $12.3 \%$ moisture content, and corresponds to a dry density of $2001.27 \frac {kg}{m^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we want to interpolate, you need to decide on a fit. The relationship could be ball-parked as quadratic, so fitting it (see image below) as such yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_d = -25.168w^2 + 633.08w - 2061.1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this for the maximum value of dry density results in $1919.67 \frac {kg}{m^3}$ at $12.57 \%$ moisture content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HK5Cj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HK5Cj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you have a measured point that is above your interpolated maximum. This means you could either measure more points (to reduce uncertainty), choose a different fit, etc. But that is the general idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have a curve fit, you can solve for dry density at any moisture content. And more importantly, you know where the optimum is, so if the field conditions are too dry or too wet prior to compaction, you can adjust accordingly if needed.&lt;/p&gt;&#xA;" OwnerUserId="34" LastEditorUserId="34" LastEditDate="2015-08-12T12:11:53.390" LastActivityDate="2015-08-12T12:11:53.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3887" PostTypeId="1" CreationDate="2015-08-11T19:25:09.590" Score="7" ViewCount="2830" Body="&lt;h2&gt;The quick:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've come up with a handful of possible options for how to structure the Datum Reference Frames (DRFs) and Feature Control Frames (FCFs) for this part.  I've narrowed it down to the two shown below, but I don't know which is better, or if there is maybe another way of doing this I'm overlooking. I believe both are compliant to &lt;a href=&quot;http://The%20acronyms%20are%20defined%20in%20ASME%20y14.5-2009%20;)%20FCF%20=%20feature%20control%20frame.%20DRF%20=%20Datum%20reference%20frame.%20ASME%20Y14.5-2009%20is%20a%20drafting%20standard,%20used%20in%20the%20U.S.%20and%20other%20countries%20to%20ensure%20prints%20are%20created%20and%20interpreted%20the%20same%20way%20across%20geographic%20and%20cultural%20divides.&quot; rel=&quot;noreferrer&quot;&gt;ASME Y14.5-2009&lt;/a&gt;, and I'm leaning towards option 1, but I wanted to get another set of eyes here  to make sure the intent is understandable and clearly communicated. ASME Y14.5-2009 is a drafting standard, used in the U.S. and other countries to ensure prints are created and interpreted the same way across geographic and cultural divides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sHsVS.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sHsVS.jpg&quot; alt=&quot;DRF construction options&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The dirty:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is a cast part.  It is cast to near net shape and then machined to final specifications.  I need to check both the relationship of the machined features to each other and some of the machined features to the casting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In checking the machined features to each other, I want to check the position of one group of features (hole pattern) to another group of features (different hole pattern).  For this, I've simply declared the first set as a datum feature, and created a compound true position FCF for the other set using the first set as the secondary datum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for checking the first set to the casting, I need to check each machined feature individually to it's cast member.  I'm struggling with how to apply a datum using the &quot;INDIVIDUALLY&quot; requirement (as defined in section 7.4.8, and figures 7-26 and 7-37 of ASME Y14.5-2009) to these holes for this FCF, while also maintaining the group datum for the other FCF. &lt;/p&gt;&#xA;" OwnerUserId="2598" LastEditorUserId="2246" LastEditDate="2015-08-11T21:00:03.480" LastActivityDate="2016-12-30T21:33:45.433" Title="Using a pattern of features as datum BOTH individually and as a group, for different FCFs" Tags="|design|drafting|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3889" PostTypeId="2" ParentId="3745" CreationDate="2015-08-11T21:29:25.160" Score="2" Body="&lt;p&gt;I've spent some time working out the geometry for this configuration and believe I've developed a set of cases that will work to determine the total effective throat of this connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is divided into three cases.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Case 1 - Undersized Fillet Weld $(L_2 &amp;lt; RO)$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For this case, the fillet weld is sized such that the vertical leg is less than the root opening $RO$ of the weld (see the figure below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SviQS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SviQS.png&quot; alt=&quot;Case 1&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values $\alpha_{RO}$ and $RO$ are found as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \alpha_{RO} = \arcsin \left( \frac{R_1 - E}{R_1} \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ RO = G + R_1 \left[ 1 - \cos(\alpha_{RO}) \right]. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing these, one can determine if $L_2 &amp;lt; RO$, in which case $E_T = E$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Case 2 - Medium Fillet Weld $ (RO &amp;lt; L_2 \leq R_1\tan (\alpha_{RO})) $&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In this case, the fillet weld is large enough that the toe of the vertical leg is past the height of the root opening, but the fillet isn't large enough that the total effective throat passes through part of the fillet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NPPoD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NPPoD.png&quot; alt=&quot;Case 2&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\alpha_{RO}$ and $RO$ already found, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_T = \sqrt{E^2 + (L_2 - RO)^2}. $$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Case 3 - Large Fillet Weld $ (L_2 &amp;gt; R_1\tan (\alpha_{RO})) $&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For Case 3, the fillet weld is large enough that the total effective throat passes through the body of the fillet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1wSuC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1wSuC.png&quot; alt=&quot;Case 3&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle of the fillet, $\alpha_F$ is found as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \alpha_F = \arctan \left( \frac{L_1}{L_2} \right). $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this, one can set up a system of equations to solve for $E_T$.  I've skipped this step for brevity.  It can be shown that, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  E_T = \frac{L_1 + E - RO \tan(\alpha_F)}{\sin(\alpha_{RO})\tan(\alpha_F) + \cos(\alpha_{RO})}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This methodology is now formalized in the article &quot;Measuring the Effective Throat of Groove Welds,&quot; which appears in the February 2017 edition of the AWS &lt;em&gt;Welding Journal&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2017-02-02T20:59:26.943" LastActivityDate="2017-02-02T20:59:26.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3892" PostTypeId="2" ParentId="3868" CreationDate="2015-08-12T00:43:55.203" Score="1" Body="&lt;p&gt;I suggest that you look at one of the follow three options.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Ultrasonic Sensor for Texas Instrument (Option 1)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;To measure liquid level from the bottom of tank one could use a piezo electric ceramic transducer combined with an Analog Front End (AFE) and a micro-controller to measure the liquid level. I believe with some additional work you can achieve your object. Check the references for more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hUq5N.jpg&quot; alt=&quot;Fluid Level Sensing&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Capacitive Fluid Level Sensors from Molex (Option 2)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;They have claimed that there system is capable of measuring very small changes. Molex still have to demonstrate there product to me. Either way it is over kill for my current project. If you happen to use them please come back and leave your feedback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JyUOK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JyUOK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;PSoC CapSense for Liquid Level Sensing for Cypress (Option 3)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This technology from Cypress is interesting. I have this as viable option for my current project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please refer to the references for additional infomation. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/2697/110&quot;&gt;Can I use a ultrasound sensor to measure water level?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?feature=player_embedded&amp;amp;v=HY_ahbqIdEM&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molex - Capacitive Fluid Level Sensors&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/mElhpwiBrUU&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cypress’s Automotive CapSense Liquid Level Sensing&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eetimes.com/document.asp?doc_id=1279778&quot; rel=&quot;nofollow noreferrer&quot;&gt;Use capacitive sensing to Implement reliable liquid level sensing—Part II&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-08-12T10:20:55.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3894" PostTypeId="2" ParentId="3661" CreationDate="2015-08-12T01:22:56.593" Score="2" Body="&lt;p&gt;AISC's magazine, Modern Steel Construction periodically publishes a guide to specifying grades of steel to help engineers stay aware of developments in the market. Their latest &lt;a href=&quot;http://msc.aisc.org/globalassets/modern-steel/steelwise/2015/steelwise.pdf&quot; rel=&quot;nofollow&quot;&gt;edition&lt;/a&gt;, from this February mentions A1085, but still suggests A500 Grade C as the standard. They recommend checking that A1085 is available and affordable in your area - it sounds like you already have. This may not be true in all regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than the maximum yield strength which isn't useful in your practice, there are a few other convenient features. Firstly, according to &lt;a href=&quot;http://msc.aisc.org/globalassets/modern-steel/steelwise/steelwise-september-2013.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; article, it has the same mechanical properties specified regardless of product form. By contrast, A500 Gr C tube has a different Fy for square/rectangular and round tubes. Also, A500 allows actual wall thickness to be reduced by a large variance. This was intended to allow for cheaper manufacture by less accurate methods, but with modern steel mills, tubes are consistently produced undersized. Because of this, calculations require reducing the wall thickness by 7% from nominal. While these issues are not major considerations in specifying steel, they will have some appeal in simplifying the design process and reducing opportunities for error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard also specifies minimum corner radii, which reduces the risk of corner cracks. Previously, AWS D1.1 and the AISC Manual have warned about the risk of corner cracks in square and rectangular A500 members subject to significant stress from welding or galvanizing. I'm not certain if this fear will go away, or just be easier to quantify if using the new standard since A500 is produced with fairly uniform radii already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main down side of specifying the A1085 tube is cost, as you point out. Additionally, you may not be able to count on availability right now if you are specifying projects outside of your specific area. One other down side if you do lighter work is that A1085 is not available with 1/8&quot; wall thickness like A500 is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your situation, these considerations may be a wash, meaning you should continue to specify A500 to reduce costs as long as it is still available. It seems probable that the distinct advantages for seismic design will lead to widespread use of this new tube standard. If so, it will eventually become the default nationally, and you may have to switch simply because A500 becomes less readily available. In the mean time, it might be appropriate to allow tube steel in your designs to conform with either spec, since the design values are fairly similar.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-08-12T01:22:56.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3895" PostTypeId="2" ParentId="3495" CreationDate="2015-08-12T03:19:44.727" Score="3" Body="&lt;p&gt;This is definitely not a simple repaving job. The second sentence gives it away: drainage system is being upgraded and curbs and drainage are being added. Converting a roadway from a flush shoulder with a swale (ditch) drainage system to one with curb and gutter is not a straightforward process. There are several considerations when installing new curbs on an existing roadway, and possible reasons for the reconstruction:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A drainage system will be required. Because a straight section of road slopes from the center to both sides, curb inlets will be needed on each side. A drainage system is typically connected at one or more points between sides of the roadway, because the water has to be channeled into a pond in order to treat it before sending into natural waterways. Placing large pipes across the roadway will inevitably necessitate the sort of reconstruction you see in the pictures.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Flush shoulder roadways have a different profile than those required for a curb and gutter roadway. With flush shoulder roads, water drains to the side anywhere and the swale has the proper slope to carry the water away. With curb, a gutter (or in your case the edge of the pavement adjacent to the curb) has to have a slope to carry the water to the nearest drainage inlet, which is placed at a low point. A gutter or edge of pavement has to match the slope of the rest of the roadway, so, especially in flat areas, the whole profile of the roadway needs to be adjusted to have what is referred to as a &quot;sawtooth&quot; where there are low points, high points, low points, and consistent, smooth slopes in between to direct the water to the drainage. This can sometimes be done by merely building up the pavement to create high points, but that may not be possible because it would reduce the clearance for vehicles under that pedestrian overpass. This could be the reason for the reconstruction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Future traffic considerations. Pavement designs take into account the predicted traffic volumes (especially the heavy vehicle loads) the road will experience in 20 years, the lifespan of the road. The current pavement design may be insufficient for this volume. Again, with that overpass, this may not be able to be resolved with merely making the pavement thicker.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Existing deficiencies. If there was rutting or potholes forming in that area caused by some weak soil beneath the roadway, the right way to resolve that is to reconstruct the roadway foundation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2606" LastActivityDate="2015-08-12T03:19:44.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3901" PostTypeId="2" ParentId="1742" CreationDate="2015-08-12T14:20:08.080" Score="4" Body="&lt;p&gt;The problem with quantifying impact force is the short duration of what is usually a 'peaky' pulse.   Bathroom scales are usually damped to give a steady reading of a static force and won't capture transient peaks.   Ballistic pendulums also have problems - some of which have been discussed.   However, for training young fighters, I'd suggest that you don't really need to quantify the force but, rather, compare their performance over time and let them compare themselves with other students.   To do this, all you need do is fill an open box (about 30 cm square) with a thick layer of modelling clay and fix the box to the wall.   When the student throws a punch, the clay will act as a witness material and deform plastically.   Just measure the maximum depth of the depression and record this on the student's personal record.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clay is, of course, resusable but you might have to experiment to get the right modelling clay:  Roma Plastilina was the preferred material for firearms ballistics but Plasticine or Play-Doh might be less painful to young karate-ka.&lt;/p&gt;&#xA;" OwnerUserId="2520" LastActivityDate="2015-08-12T14:20:08.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3902" PostTypeId="1" CreationDate="2015-08-12T14:26:28.197" Score="3" ViewCount="212" Body="&lt;p&gt;The EPA has come to our facility and has required us to raise 60 of our exhaust stacks to a given height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering with initial data how i can get &lt;strong&gt;final CFM.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example. We have a stack with a current height of 41.92'. Height required is 66.58' which is an increase of 24.66'. The stack system has been designed such that there is 36' of stack height that starts at a furnace exit (which is decoupled) and goes up to rooftop height and ties inline with an exhaust fan which in turn has additional stack height inline above it. We'll be replacing the 5.92' of stack above the fan with a 30.58' stack which will get us to our goal of 66.58'. I don't want to overdraft nor under draft the furnace with the stack height modification.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm not mistaken, as long as final CFM doesn't change or at least stays within a tolerance of initial CFM the additional stack height will not over-draft nor under-draft the furnace. &lt;a href=&quot;https://i.stack.imgur.com/Tf4qg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tf4qg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again all that is changing is stack height. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inner duct diameter is the same at 1.33FT. Temperature is 112 F. Initial SCFM is 2684 before modification. I believe the static pressure for the specific duct is .4inwc per 100ft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fan details: design CFM -  3570. Design Static Pressure - 2.5inwc. Fan RPM 1518 and designed max RPM is 2026.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need help finding a formula or something to get final CFM. &lt;/p&gt;&#xA;" OwnerUserId="2614" LastEditorUserId="272" LastEditDate="2015-08-12T14:31:02.627" LastActivityDate="2015-08-12T21:25:57.960" Title="Finding final CFM with a proposed stack height increase" Tags="|mechanical-engineering|fluid-mechanics|airflow|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3905" PostTypeId="2" ParentId="3875" CreationDate="2015-08-12T15:57:07.680" Score="0" Body="&lt;p&gt;I don't get your solutions to the equation, mine &lt;a href=&quot;http://www.wolframalpha.com/input/?i=1*%281073.15%2Fx+-+973.15%2F1e6%29+%3D+x-1e6&quot; rel=&quot;nofollow&quot;&gt;are:&lt;/a&gt;&#xA;$$&#xA;x_1=-0.00107315 \text{Pa}\\&#xA;x_2=1000000 \text{Pa}=10\text{bar}&#xA;$$&#xA;which means that the outlet pressure is the same as (or very close to) the inlet pressure as you would expect.&lt;/p&gt;&#xA;" OwnerUserId="2616" LastActivityDate="2015-08-12T15:57:07.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3906" PostTypeId="1" AcceptedAnswerId="3907" CreationDate="2015-08-12T16:45:16.400" Score="-3" ViewCount="1390" Body="&lt;p&gt;Is this how it works?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fingers generate moment.&lt;br&gt;&#xA;The load generates moment.&lt;br&gt;&#xA;Finger moment &gt; load moment, so the load is cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I use an equation to identify the most vulnerable point of the product?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More questions on this topic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the typical accelaration of scissors? $F=ma$, so that I can assume $F$? Other method of giving a logical assumption of F is also acceptable.&lt;/p&gt;&#xA;" OwnerUserId="2618" LastEditorUserId="2618" LastEditDate="2015-08-19T15:18:48.787" LastActivityDate="2015-08-19T15:18:48.787" Title="How to analyze the free body diagram of a pair of scissors?" Tags="|mechanical-engineering|statics|" AnswerCount="1" CommentCount="10" ClosedDate="2015-08-16T11:44:59.760" ContentLicense="CC BY-SA 3.0" />
  <row Id="3907" PostTypeId="2" ParentId="3906" CreationDate="2015-08-12T17:10:13.623" Score="5" Body="&lt;p&gt;Your general idea is correct. Your fingers apply a force which can be translated to a moment at the rotation axis. The reaction forces applied by the object to be cut must generate an equal  moment at the axis. If the necessary reaction forces are greater than the object's shear resistance forces, then the object is sheared through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, it is best to place the object as close to the axis as possible, so that the forces required to balance the moment are maximized. Too close, however (at which point the angle between the blades becomes greater than 90 degrees), and you run into the problem that the forces start becoming primarily horizontal (pushing the object out of the scissors) as opposed to vertical (cutting through the object).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because the blade's force is applied perpendicular to the blade, so the greater the angle between the blades, the greater the horizontal component of the force becomes. This component is useless since it does not serve to cut through the object, but instead simply tries to force the object out of the scissors. Only the vertical component actually shears the object. This is because the horizontal components of both blades point in the same direction, while the vertical components point in opposite directions.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2015-08-19T13:57:57.840" LastActivityDate="2015-08-19T13:57:57.840" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3908" PostTypeId="1" CreationDate="2015-08-12T18:03:19.947" Score="3" ViewCount="4879" Body="&lt;p&gt;I am trying to perform some experiments with a motor and I want to run it under realistic conditions (or at least not unloaded). Is there an easy way to load a motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple ideas I had include: attach water paddles to paddle water continuously, attach pulley to lift weight, clamping brake pads (getting desperate), ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working with a &lt;a href=&quot;http://www.surpluscenter.com/Electric-Motors/3-Phase-Motors/3-Phase-Motors-Base-Mount/1-3-HP-1710-RPM-230-460-3PH-MOTOR-TEFC-10-2844.axd&quot; rel=&quot;nofollow&quot;&gt;1/3 HP three-phase motor&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2619" LastEditorUserId="368" LastEditDate="2015-08-12T18:31:33.633" LastActivityDate="2015-08-21T18:24:19.647" Title="How can I apply a load to an electric motor to simulate realistic conditions for an experiment?" Tags="|mechanical-engineering|motors|gears|pulleys|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3909" PostTypeId="1" AcceptedAnswerId="3917" CreationDate="2015-08-12T18:52:57.830" Score="10" ViewCount="3134" Body="&lt;p&gt;Looking for some hints on how to create a &lt;strong&gt;rough estimate&lt;/strong&gt; for the following problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two steel gates with the same dimensions, same material - e.g. everything is the same.  The only difference is that the middle parts have different structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When applying some force to the top, the gate will start to be more and more deformed, and at some force the gate will touch the ground at the place where the blue arrow points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm looking for a &lt;em&gt;rough estimate&lt;/em&gt; of how much more&lt;/strong&gt; force is needed for the second gate - i.e. how much more &quot;sturdy&quot; is the second gate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't need any exact calculation, but probably will need some material data, so:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;common steel &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Closed_thin-walled_beam.png&quot; rel=&quot;noreferrer&quot;&gt;thin-walled beam&lt;/a&gt; (25mm x 25mm x 2mm wall thick)&lt;/li&gt;&#xA;&lt;li&gt;each joint point is welded, we can be simplify and assume that the welds are exactly as strong as the material itself&lt;/li&gt;&#xA;&lt;li&gt;the suspension points can hold infinite force&lt;/li&gt;&#xA;&lt;li&gt;and any other possible simplification - this problem isn't for any rocket-science but for solving an evening talk with a friend. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f2njv.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f2njv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2617" LastEditorUserId="272" LastEditDate="2015-08-12T19:47:32.027" LastActivityDate="2015-08-13T15:04:47.177" Title="Strength of a welded steel gate with vertical bars vs. crossed diagonal bars" Tags="|structural-engineering|beam|stresses|reinforcement|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3910" PostTypeId="2" ParentId="3909" CreationDate="2015-08-12T19:03:34.820" Score="8" Body="&lt;p&gt;While you've described your problem pretty well, I don't think you're going to find a satisfactory answer without having to run a fairly complex finite element analysis on both structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first gate structure will behave similarly to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Truss#Vierendeel_truss&quot;&gt;Vierendeel truss&lt;/a&gt; since you have all of the pieces essentially moment-connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second gate structure will likely fall somewhere between the Vierendeel and a traditional truss, though it's still, for the most part, moment connected with no real alignment of working points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, trusses are detailed such that their working points (i.e., the center of action of the axial force in the members) coincide on roughly the same point.  This is to reduce bending in any single member since the eccentricity is approximately zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second gate has some truss action due to the diamond-shaped section in the middle.  Unfortunately, since the working points of the diamond section don't meet up with the vertical/horizontal sections, you're losing some of the advantage of the truss action.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-08-12T19:03:34.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3911" PostTypeId="2" ParentId="3908" CreationDate="2015-08-12T19:42:51.457" Score="1" Body="&lt;p&gt;Attach a flat paddle across the axle that moves against the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I3LZc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I3LZc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;(image source: &lt;a href=&quot;http://woodgears.ca/dust/blower_design.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;woodgears.ca&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-08-12T19:42:51.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3912" PostTypeId="2" ParentId="3909" CreationDate="2015-08-12T20:23:41.863" Score="13" Body="&lt;p&gt;Assuming the joints are welded, for the top gate to deform as you draw it the vertical bars will have to bend into an &quot;S&quot; shape. The flexibility in bending will be proportional to the cube of the length, if everything else is the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stiffness of the three sections of the top gate will be proportional to $1/1^3 = 1$, $1/0.6^3 = 4.6$, and $1/0.4^3 = 15.6$. The total flexibility is dominated by the longest (middle) section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the bottom gate, the diagonal bars would be (to a first approximation) infinitely stiffer than the vertical bars since they carry shear in diagonal tension and compression, not in bending. The overall stiffness would be of the order of 4 or 5 times greater (based on the 4.6 above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could probably get away with less material in the diagonal bars (either thinner bars or fewer bars) but, a more detailed analysis is too much work to do by hand and for free! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't matter if the spacing of the diagonal bars matches the verticals, so long as the horizontal bars are strong enough to redistribute the load between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If stiffness is the only criterion, you might as well just have an outer rectangular frame and diagonal bracing, with no sections of &quot;vertical bars&quot; at all.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-08-12T20:23:41.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3913" PostTypeId="2" ParentId="3902" CreationDate="2015-08-12T21:25:57.960" Score="3" Body="&lt;p&gt;The resistance against the fan is influenced by the total resistance of the pipe down wind of of the fan (in additional to the resistance on the inlet side). The total resistance is calculated from the skin friction, diameter and length with equivalent lengths added for bends/turns, joints and the type of plenum used on the outlet. When you add length to this pipe you will increase the resistance on the fan and depending on the existing fan operating point (on the fan curve) it may reduce the CFM output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Options:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Hire a mechanical/ventilation engineer to ensure that those fans do not stall and destroy other equipment in your factory by not properly handling exhaust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Gather all the information you have available about the fans and the pressure on them, then hire a mechanical/ventilation engineer to do the work based on that data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Put the pipe on there and see what happens. You could increase the pipe size or reduce the skin friction so that it would be equivalent to the friction that the fan is experiencing pre-retrofit. You may be able to change the pitch of the fan blades so that its operating curve will be able to meet the new requirements. Removing bends within the system will also do wonders for any system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, I believe your best option is to bring in an expert that deals with this day in and day out. Its not one stack at a small operation you're dealing with, you have 60. What is the cost of hiring a Qualified &amp;amp; Licensed Engineer to do this work vs the down time incurred by shutting down one of the furnaces due to a calculation error.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2015-08-12T21:25:57.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3914" PostTypeId="2" ParentId="3868" CreationDate="2015-08-12T21:52:49.507" Score="2" Body="&lt;p&gt;(340.29(m/s))/(50 μm) = 6.8058 megahertz (or 146.933498 nanoseconds if you prefer) timing that tight is hard to measure on a micro controller, and requires a lot of knowledge to do on a micro processor, but certainly not out of the realm of possibility. That makes (ultra)sonics a possibility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Capacitive sensors are the other method for things like this, but you said &quot;no external body can touch the solution&quot; so if that means a metal plate (coated in plastic if you want) can't be built into the container it won't work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, if its water based what are the effects of the meniscus on your measurement?&lt;/p&gt;&#xA;" OwnerUserId="2624" LastActivityDate="2015-08-12T21:52:49.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3916" PostTypeId="1" CreationDate="2015-08-13T02:36:15.947" Score="1" ViewCount="198" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;br&gt;&#xA;Desktop computers and laptops can be configure to dual boot. Example: Ubuntu and Window on the same computer/laptop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; &#xA;Can a similar concept be applied to embedded system? If so this can greatly benefit electrical manufacturing testing. Electrical manufacturing testing depends on high throughput. So booting a complete for example Embedded Linux system is not efficient and almost never done in electrical manufacturing test environment. So with some type of electronic switch/signal (may be provided by the tester) is possible to load very fast a minimalist set of software that already program to the processor that will enable testing of the board. For example verify the external crystal (XT) is correct and functioning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per &lt;a href=&quot;https://stackoverflow.com/users/446477/ouflak&quot;&gt;@ouflak&lt;/a&gt; comment: I am aware of a particular NEC V850 micro-controller based application, where all micro-controllers were programmed using a gang programmer before been placed on the PCB board. During Electrical manufacturing test, the tester would force micro-controller to load very small program and communicate to the PCB board via the CAN bus to activate sections of the embedded system for testing purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly I would like to investigate how a beaglebone black like PCB running Android architecture/design can be modified to boot from a section of the flash for electrical manufacturing test only. There is no point loading and testing the graphic controller for &lt;a href=&quot;http://www.ti.com/lsds/ti/processors/sitara/arm_cortex-a8/am335x/overview.page&quot; rel=&quot;nofollow noreferrer&quot;&gt;TI Sitara&lt;/a&gt; as vendor TI as already done perform this function. All that is need is to test and verify the PCB manufacturing process.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-05-23T12:38:59.030" LastActivityDate="2024-01-20T07:21:27.163" Title="Can an embedded system be designed to support electrical manufacturing testing?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3917" PostTypeId="2" ParentId="3909" CreationDate="2015-08-13T02:47:14.420" Score="13" Body="&lt;p&gt;As grfrazee said, you won't know for sure until you do a finite element analysis. I was intrigued by this question as a colleague and I got into a discussion about this. While we both agreed the diagonal bracing would be better at resisting deflection, we wondered by what factor it would be better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We were really curious so we settled the debate and did a quick structural analysis on &lt;a href=&quot;http://skyciv.com/skyciv-structural-features&quot; rel=&quot;noreferrer&quot;&gt;SkyCiv Structural 3D&lt;/a&gt; (can try for free for one month if anyone is wondering). It took around an hour to set up both gates and analyze them mainly because we had to generate the node positions from scratch. Anyways here are the results of the linear static analysis which take into account the assumptions and simplifications you made. We applied a 5 kN POINT LOAD at both F1 and F2 and made each support a pin support at the locations you specified. Note that in the 3D colored results the deflection is 12X greater than the actual deflection of the gate in both scenarios - it is exaggerated so you can see the deflected shape of the gates.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Gate #1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$\text{y-deflection at the bottom-left of the gate} = 31.74\text{ mm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Max total deflection} = 32.10\text{ mm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r8i2b.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r8i2b.png&quot; alt=&quot;SkyCiv Structural 3D Deflection Result for Gate 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Gate #2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$\text{y-deflection at the bottom-left of the gate} = 7.84\text{ mm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{Max total deflection} = 7.55\text{ mm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K6ASN.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K6ASN.png&quot; alt=&quot;SkyCiv Structural 3D Deflection Result for Gate 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagonal bracing (Gate #2) is clearly the winner. So when both gates are subjected to the same load it looks like Gate #2 resists deflection better (i.e. is more stiff) by a factor of &lt;strong&gt;4.25&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more interesting points:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There's a pretty high bending stress at that top right support in both scenarios ~ 350 MPa.&lt;/li&gt;&#xA;&lt;li&gt;The analysis didn't take into account self-weight of the gates.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also let me add that there looks to be a scaling issue with the diagonal grid you have drawn, because when I modeled it I found that there were far less points than what was suggested by your diagram. I ensured that the parallel spacing between each rhombus was 300mm. This means the diagonal of each rhombus is roughly 424mm. Your gate is 3300mm in length so that means around 8 rhombi should fit across your gate in the x-direction - but you've drawn around 12. Just thought I'd let you know.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-08-13T05:21:06.827" LastActivityDate="2015-08-13T05:21:06.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3918" PostTypeId="1" CreationDate="2015-08-13T03:45:17.367" Score="3" ViewCount="11362" Body="&lt;p&gt;This may be a silly question, but I'm not a mechanical engineering. What's the best way to calculate the energy required to reduce the pressure of a gas stream below atmospheric pressure? Obviously you could work out the energy for isentropic expansion and apply an efficiency factor, but this seems like it may be inaccurate. I know my stream's mass flow rate, temperature and desired final pressure.&lt;/p&gt;&#xA;" OwnerUserId="2625" LastEditorUserId="272" LastEditDate="2015-08-13T11:37:15.870" LastActivityDate="2020-06-05T23:25:30.913" Title="Vacuum Pump Power Calculation" Tags="|pressure|pumps|vacuum|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3919" PostTypeId="2" ParentId="3918" CreationDate="2015-08-13T04:24:05.377" Score="4" Body="&lt;p&gt;Different techniques have been developed to evaluate the&#xA;efficiency of energy use by a vacuum pump. Most vacuum pump manufacturers catalog&#xA;their test results, including brake horsepower (actual hp) and cfm vs. vacuum level. Fairly&#xA;accurate evaluations of power needs can be made from such information sources.&#xA; For example, the relative efficiency of different pumps can be obtained by calculating the&#xA;cfm of free air removed per horsepower. Or input horsepower can be compared to the &quot;fluid&#xA;power horsepower&quot; delivered, which is proportional to the product of gauge vacuum and air&#xA;flow rate. All comparisons must be made at the same specific vacuum level, usually at 20&#xA;in. Hg or above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g. check this source &lt;a href=&quot;http://www.tuthillvacuumblower.com/dam/357.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.tuthillvacuumblower.com/dam/357.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2404" LastActivityDate="2015-08-13T04:24:05.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3920" PostTypeId="2" ParentId="3916" CreationDate="2015-08-13T07:52:29.480" Score="5" Body="&lt;p&gt;I have done something similar with a small 8-bit microcontroller in a medical device.  The factory had two pieces of firmware:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First, factory test firmware is downloaded.&lt;/strong&gt;  It has the algorithms for exercising the subassembly and talking to the factory test equipment.  Factory test are performed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second, field firmware is downloaded.&lt;/strong&gt;  Field firmware has its own field self-test (smaller in scope, doesn't use factory test equipment).  When the field self-test also passes, we conclude that the subassembly is good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allowed to remove the factory test code from the main firmware.   That made the field firmware easier to develop: I didn't have to worry about a remote possibility that some test mode would jump at the end user.  The memory footprint was also reduced.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2015-08-13T07:52:29.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3921" PostTypeId="1" CreationDate="2015-08-13T11:03:02.187" Score="2" ViewCount="104" Body="&lt;p&gt;I studied Medical Engineering and previously worked as a test engineer for flow measuring instruments (for three years now) and find myself in a dead-end job. There are virtually no matching jobs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I can program quite well, I can't get a job as a programmer, because the companies prefer &quot;full-fledged&quot; programmers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an engineering work which is a mixture of Programming and measuring technology | physics | electrical engineering | medical engineering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, PLC and automation technology you can not learn at home apparently. Therefore this is no choice for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an engineer I like logical systems. And I like to improve measurement systems or products. I like programming of course. Therefore automation technology sounds good. But it is difficult to enter this field. Maybe I should apply as an software engineer (C#) and demand less salary.&lt;/p&gt;&#xA;" OwnerUserId="2628" LastEditorDisplayName="user16" LastEditDate="2015-08-13T13:13:27.067" LastActivityDate="2015-08-13T13:13:27.067" Title="What are the career growth opportunities for an Engineer with Product Testing Experience?" Tags="|employment|" AnswerCount="2" CommentCount="6" ClosedDate="2015-08-13T13:22:52.800" ContentLicense="CC BY-SA 3.0" />
  <row Id="3922" PostTypeId="2" ParentId="3921" CreationDate="2015-08-13T11:30:02.883" Score="1" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; Thank you for updating the question. I will update the response appropriately in a few hour (After work) listen to this &lt;a href=&quot;http://money.cnn.com/2015/04/09/technology/google-people-laszlo-bock/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google doesn't care where you went to college&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When most employers seek new employee their are few key points employees pay close attention (I speak for US base companies). &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Attitude:&lt;/strong&gt; Do you have positive attitude?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;People Skills:&lt;/strong&gt; How well does the employee work with peers/co-works&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Problem Solving Skills:&lt;/strong&gt; When confronted with a problem, ability to resolve the problem?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Create Business Value:&lt;/strong&gt; Does the engineer understand the business value of the engineering problem and can the engineer solve the problem to bring economic value to the company?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you see all the above have nothing to do with engineering. My best advise is to figure out what you really makes you happy as an engineer (Update your post, so I can modify the answer appropriately), and real your current work experience to your passion. It is all about selling your skills and passion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an Engineer has acceptable qualifications and decent work experience then engineer is qualified for the job. Most import thing is when confronted with a engineering problem can the engineer solve the problem to bring economic value to the company. The cold hard truth is if the engineer can create economic value to the company the company will sustain, if the companies sustains, then engineer has more opportunity to follow the engineers passion. There is no magic in this formula.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://money.cnn.com/2015/04/09/technology/google-people-laszlo-bock/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google doesn't care where you went to college&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/119/in-engineering-fields-how-often-do-you-find-yourself-being-asked-to-be-a-sales&quot;&gt;In Engineering fields, how often do you find yourself being asked to be a “salesperson”, and what is your reaction?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-08-13T12:20:00.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3923" PostTypeId="2" ParentId="3374" CreationDate="2015-08-13T12:54:52.390" Score="0" Body="&lt;p&gt;Raise the ceiling level in one side of the room, likely away from the window, this would allow a heat biasing movement vertically which would induce a very small air flow.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2015-08-13T12:54:52.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3924" PostTypeId="2" ParentId="3921" CreationDate="2015-08-13T13:07:35.093" Score="3" Body="&lt;p&gt;First of all, no job is going to be perfect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second,  I've been working as an engineer since 1998 and spent all of that time as a test engineer, testing products. I've tested everything from early &quot;let's try it and see&quot; prototypes to testing medical devices during production, maintaining compliance with FDA regulations. I'd say the opportunities now are as good as I've seen them in the past 10 years. I get at least 2 or 3 calls a week from headhunters looking for engineers with a background in product development and testing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you have to understand what it is that you offer and sell employers on how your particular set of skills and experiences are a value to them, regardless of what industry you have experience in or what industry the position is in.     &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-08-13T13:07:35.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3926" PostTypeId="1" AcceptedAnswerId="3927" CreationDate="2015-08-13T13:21:31.570" Score="3" ViewCount="2618" Body="&lt;p&gt;I'm trying to crimp some pins onto 20 AWG wire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can either use a tool like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ihyNm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ihyNmt.jpg&quot; alt=&quot;specialty crimper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or I can just use pliers, which are obviously cheaper and more readily available. Why should I use the more expensive tool?&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorDisplayName="user16" LastEditDate="2015-08-13T14:16:13.800" LastActivityDate="2015-08-14T00:58:53.923" Title="What is the advantage of using a four-sided crimper vs. pliers?" Tags="|tools|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3927" PostTypeId="2" ParentId="3926" CreationDate="2015-08-13T14:14:41.340" Score="4" Body="&lt;p&gt;It's all about the amount surface area in contact between the two materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pair of pliers will create points of contact on two sides of the wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specialty crimper you linked will provide contact on four sides of the wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having four points of contact provides:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Better electrical conductance between the pins and the wires&lt;/li&gt;&#xA;&lt;li&gt;Better mechanical resistance (aka a friction fit) to pulling and the pins are more likely to stay attached.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-08-13T14:14:41.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3928" PostTypeId="1" AcceptedAnswerId="3938" CreationDate="2015-08-13T14:37:52.000" Score="6" ViewCount="3680" Body="&lt;p&gt;In my lab we use a number of materials which are &lt;a href=&quot;https://en.wikipedia.org/wiki/Hygroscopy&quot; rel=&quot;noreferrer&quot;&gt;hygroscopic&lt;/a&gt; and must therefore be stored in a dry environment. For that purpose we have numerous dry cabinets very similar to the one shown below (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00871AXHE&quot; rel=&quot;noreferrer&quot;&gt;Amazon link here&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These cabinets run on electricity, do not require  desiccant, and do not have a water outlet such as one finds with home dehumidifiers, yet they are able to maintain low relative humidity (RH) values of &amp;lt; 40%.  The temperature inside the cabinet is also the same as (or very close to) the temperature outside of the cabinet, and there are no other connections aside from the electrical plug.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How do these cabinets maintain such low humidity values?&lt;/li&gt;&#xA;&lt;li&gt;Assuming they are removing water from the air, where does the water go?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GK0BN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GK0BNm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2020-01-21T10:31:55.893" Title="How do dry storage cabinets work?" Tags="|hvac|dehumidification|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3930" PostTypeId="2" ParentId="2068" CreationDate="2015-08-13T15:11:56.497" Score="4" Body="&lt;p&gt;UPS has, or at least had when I was transit testing packaged products, a specific packaging test standard based on &lt;a href=&quot;http://www.ista.org/&quot; rel=&quot;nofollow&quot;&gt;ISTA&lt;/a&gt; standards and &lt;a href=&quot;http://www.astm.org/Standards/D4169.htm&quot; rel=&quot;nofollow&quot;&gt;ASTM D4169&lt;/a&gt;. There's also a MIL-STD, 810 something, but I don't remember the exact number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, for a parcel carrier, the test, regardless of standard, consists of environmental conditioning, a &quot;10 pt&quot; drop test and a vibration test. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The environmental conditioning typically represents the extremes of what would be expected. In our case, we conditioned products for 12h @ +150F and -30F with uncontrolled humidity. The referenced standards discuss other conditions such as a high humidity environment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The height of the drop is determined by the packaged weight. The product is dropped onto a rigid surface without inducing any vertical motion as follows: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The bottom vertex of the carton which includes the carton seam. &lt;/li&gt;&#xA;&lt;li&gt;Each of the three edges radiating from that vertex.  &lt;/li&gt;&#xA;&lt;li&gt;Flat on each of the 6 faces &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Following that, the product is subjected to random vibration (typically 0.52 gRMS) for 1 hour on each of the three axes with a vibration profile according to &lt;a href=&quot;http://www.astm.org/Standards/D4728.htm&quot; rel=&quot;nofollow&quot;&gt;ASTM D4728&lt;/a&gt;, ISTA or a profile developed from data collected by monitoring your particular distribution stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to what forces you design your product to withstand dependent on a lot of things, mainly what's the cost of product breakage? What are the specific stages of getting your product to the customer? How much does it weigh? and so on. How much engineering and money are you willing to put into the packaging to protect the product? Can you make the product itself robust to transportation, or is it delicate and requires significant packaging? Like most engineering undertakings, it's a tradeoff and there's no one good answer that applies to all products. I recommend looking at the referenced standards, and using those as a guide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get testing done without buying a test lab of equipment, there are plenty of labs across the country that will help you develop a test plan and perform the testing. I have personal experience working with &lt;a href=&quot;http://www.lansmont.com/&quot; rel=&quot;nofollow&quot;&gt;Lansmont&lt;/a&gt;, they make packaging test equipment, perform testing in house and provide consulting on packaging design. (&lt;strong&gt;Note:&lt;/strong&gt; My relationship with Lansmont is strictly as a prior customer, I have no personal, financial or other ties to them). They are great to work with. UPS also has a test lab and provides packaging design support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could go on and on about this stuff, but this is already long enough. If you have more questions, want further information or clarifications, I'll be glad to answer them.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2015-08-13T15:56:09.113" LastActivityDate="2015-08-13T15:56:09.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3931" PostTypeId="1" AcceptedAnswerId="3933" CreationDate="2015-08-13T16:13:38.813" Score="3" ViewCount="2462" Body="&lt;p&gt;We regularly bolt an angle iron against an existing building element (flange down), and use the flange as support for a new concrete slab. Is there any design approach to assist with the sizing / spacing of the chemical anchors as well as confirming the suitability of the angled section?  Epoxy anchors are one form of chemical anchor we use. There are other types of (chemical) bonding agents available like polyesters, vinyl esters as well as hybrid systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have tried various design software packages none of them is very helpful with this configuration. Therefore a manual approach is all that is left, it seems.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorDisplayName="user16" LastEditDate="2015-11-14T14:06:46.553" LastActivityDate="2015-11-14T14:06:46.553" Title="Angle Iron Cleat at Slab-Slab connection" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3933" PostTypeId="2" ParentId="3931" CreationDate="2015-08-13T16:44:18.567" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;...and use the flange as support for a new concrete slab&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By this, I assume you mean a shelf angle (or a connection angle) to support a concrete floor.  The image below is an example from the &lt;em&gt;PCI Design Handbook&lt;/em&gt;, 7th Edition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fXPY5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fXPY5.png&quot; alt=&quot;Shelf Angle&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section 6.6.5 of the &lt;em&gt;PCI Design Handbook&lt;/em&gt;, 7th Edition goes through a design example of an unstiffened connection angle.  You will likely find similar design examples in previous editions of that &lt;em&gt;Handbook&lt;/em&gt;, thought not necessarily with the same section number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.structuremag.org/wp-content/uploads/C-StrucDesign-Mcginley-May131.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; from &lt;a href=&quot;http://www.structuremag.org/?p=219&quot; rel=&quot;nofollow noreferrer&quot;&gt;STRUCTURE Magazine&lt;/a&gt; also goes through some design methodology for shelf angles.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-08-13T16:44:18.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3934" PostTypeId="2" ParentId="3926" CreationDate="2015-08-13T18:05:54.393" Score="2" Body="&lt;p&gt;The proper crimper will indent the cylindrical body of the contact, leaving the outside diameter unchanged.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pliers will simply squash the contact, making it oval, and larger than the original diameter in one direction.  This may prevent the contact from seating correctly (or at all) in the connector housing.  The distorted contact will probably make it impossible to use the contact removal tool to remove the contact from the connector.&lt;/p&gt;&#xA;" OwnerUserId="2638" LastEditorUserId="2639" LastEditDate="2015-08-14T00:58:53.923" LastActivityDate="2015-08-14T00:58:53.923" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="3935" PostTypeId="1" CreationDate="2015-08-13T21:21:56.483" Score="-3" ViewCount="175" Body="&lt;p&gt;How feasible (in practicing engineer sense) do you think is the general concept of &quot;computational engineering&quot;? That is, engineering design carried predominantly on computers, with as less physical hardware and &quot;hands-on&quot; as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sure CAD/CAE/etc. are standard tools, but their place as tools is currently lesser than what could be possible in software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does it hide something that someone coming from a pure software and mathematics background cannot see? What are the blind spots that you think software cannot handle? What can it handle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could computer simulation replace engineering &lt;em&gt;tacit knowledge&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Related:&#xA;&lt;a href=&quot;https://engineering.stackexchange.com/questions/207/how-do-engineers-really-use-numerical-simulation&quot;&gt;How do engineers really use numerical simulation?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2643" LastEditorUserId="2643" LastEditDate="2018-06-28T13:30:14.787" LastActivityDate="2018-06-28T13:30:14.787" Title="How feasible is &quot;computational engineering&quot;?" Tags="|computer-aided-design|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ClosedDate="2015-08-24T17:50:06.693" ContentLicense="CC BY-SA 4.0" />
  <row Id="3936" PostTypeId="2" ParentId="3935" CreationDate="2015-08-13T21:39:57.363" Score="1" Body="&lt;p&gt;At our current level of software technology and accessible computing power, computers can't do a whole design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computers can automate some parts of the design process that can be well specified in a formal sense.  For example, once you specify what pads have to be connected to what other pads on a PC board, minimum trace/width spacing and other rules, software can often do the actual routing of tracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even this is far from fire-and-forget with current technology, but the big issue is that this is only a small part of designing a electronic device.  Humans still have to decide what the device is supposed to do.  Nailing down the specs is a important part of designing something, and these kind of constraints are hard to codify in a machine-understandable way.  There are just too many possibilities such that software that understood all but dumbed down and contrived specifications is beyond our reach today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another part that would be very difficult for software to do is the actual creative engineering design part.  There are a vast number of possibilities, each with its own set of tradeoffs.  A computer could possibly verify that a particular design meets a particular set of specs (although, again, codifying those specs for anything but toy problems is too complex currently), but finding the optimal or reasonably optimal design is much much harder.  That's where the creativity, knowledge, and experience of the engineer comes in.  It will probably be quite a while yet before computers can mimic that for broad classes of real world problems.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-08-13T21:39:57.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3937" PostTypeId="2" ParentId="3916" CreationDate="2015-08-13T22:06:46.923" Score="2" Body="&lt;p&gt;What you ask is quite commonly done in various forms. For embedded microcontrollers, this usually takes one of two forms:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;There are two separate firmware builds, one for manufacturing test and one for normal operation.  The test process loads the first firmware into the device, does test, calibration, etc, then loads the second firmware into the device.  This may include customizing some of the values in the production firmware based on calibration measurements taken in the first phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Add manufacturing test features to the production firmware. This is more common in my experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often the device needs to communicate to a host computer in some way.  This usually means a command set is implemented. With this method, a few extra commands are added that are only used during manufacturing test.&#xA;If the end user never sends command directly (use a canned app on the host, for example), then you generally don't publish the protocol and you can add whatever commands you want. If the user communicates with the device directly, then the last step in manufacturing might be to disable the additional commands or features.  For example, I've used the trick a few times on serialized units where the extra features are only available when the serial number is the unprogrammed value of all 1s.  The last step in manufacturing is to allocate a new serial number and program it into the device if the device meets all measured specs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extra features don't have to be all firmware either, although you don't want to burden the production cost too much. In one case I implemented a command/response interface using the built-in UART, although the end unit is stand-alone with no communications interface. In this case, the UART signals were available via bare pads on the bottom of the board.  Pogo pins on the test fixture connected to them and allowed access to the command set. End users know nothing about this feature and only see two round pads on the bottom of the board.  In this case there are no security issues, so in the extremely unlikely case (hasn't happened as far that I know) a user monkeys around with it, they could end up &quot;breaking&quot; the device.  Of course that's no different than monkeying around with any other parts of the board you're not supposed to connect to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other cases, there are hidden secret ways to access &quot;test mode&quot;.  For example, the GPS unit in my car does strange things when certain buttons are held down during power up. I've created devices that required a certain unlikely button click sequence, usually within timing constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="282" LastEditDate="2015-08-14T02:32:27.840" LastActivityDate="2015-08-14T02:32:27.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3938" PostTypeId="2" ParentId="3928" CreationDate="2015-08-13T23:21:28.510" Score="3" Body="&lt;p&gt;Not to jump the gun from @hazzey, but one of the product description images describes how the cabinet works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y1xir.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y1xir.png&quot; alt=&quot;Cooling&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The English in the images is a little lacking (probably a translation from a Chinese manual or something), but it seems like the cabinet condenses the moisture inside, then freezes it on a collection board.  Once the space is dehumidified, it appears that the water melts and is wicked to the outside of the enclosure, where it then evaporates into the ambient air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, assuming my interpretation of the image is correct.  I'm pretty certain that by &lt;code&gt;the collecting board refreezes to water&lt;/code&gt; they mean &lt;code&gt;the ice on the collecting board remelts&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-08-13T23:21:28.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3939" PostTypeId="2" ParentId="3928" CreationDate="2015-08-14T01:07:04.383" Score="1" Body="&lt;p&gt;As I commented and @grfrazee answered, the particular dry cabinet claims to freeze the moisture.  I think that there is a general consideration that is being overlooked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the question, you compared the dry cabinet to a room dehumidifier, but a better comparison is probably a refrigerator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A room dehumidifier need to pass a large volume of air through it and take out a lot of water. This is the reason for the bucket or drain hose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A refrigerator (or dry cabinet) has a much smaller volume of air to work with. Just like a refrigerator doesn't have a drain (it does, but I'll get to that), a dry cabinet won't need a drain. A refrigerator (and I'm assuming that a dry cabinet is just a warmer version) has a small drain pan in its base. The condensation drips onto this and then evaporates normally. The amount of condensation is so low that it never over flows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the plumbing and controls will be slightly different, but the relative volume of treated air is similar.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-08-14T01:07:04.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3940" PostTypeId="1" CreationDate="2015-08-14T01:22:30.067" Score="-1" ViewCount="2646" Body="&lt;p&gt;From my point of view, it seems that a soldier armed with a &lt;strong&gt;full metal plate armor&lt;/strong&gt; was almost invulnerable at the time their opponents yielded swords, spears or bows. I understand that it couldn't be the case, but I'm not sure about the physics behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How could an archer beat this soldier? An arrow shooted with full power would have enough kinetic energy to penetrate it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How could a swordsman or a spearman beat this soldier? Would the impact of these weapons suffice?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Just out of curiosity, a modern gun would penetrate into the plates that easily? One headshot would do?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm thinking on a soldier wearing an armor like this one below, or even  bulkier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/3/34/Italian_-_Sallet_-_Walters_51580.jpg/378px-Italian_-_Sallet_-_Walters_51580.jpg&quot; alt=&quot;Italian suit of Armor, c.1450&quot;&gt;[1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Courtesy of Wikipedia&lt;/p&gt;&#xA;" OwnerUserId="2644" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.177" LastActivityDate="2015-08-14T06:15:43.123" Title="How was it possible to beat a soldier with a full plate armor?" Tags="|mechanical-engineering|materials|" AnswerCount="3" CommentCount="3" ClosedDate="2015-08-14T11:26:45.060" ContentLicense="CC BY-SA 3.0" />
  <row Id="3942" PostTypeId="2" ParentId="3940" CreationDate="2015-08-14T03:12:12.550" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;An arrow shooted with full power would have enough kinetic energy to penetrate it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heavy crossbows could penetrate even heavy plate armor.  Longbows could penetrate some lighter plate mail.  &lt;a href=&quot;http://indie-rpgs.com/archive/index.php?topic=10676.0&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt; states that the church banned crossbows in medieval battle since they were so lethally effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics behind it comes down to simple pressure.  You have an arrow or a bolt with a certain mass, traveling at a certain speed.  All of the kinetic energy of the projectile is concentrated on the razor tip.  Once the pressure at the tip exceeds the strength of the metal, you get penetration.  There is some balance of kinetic energy with the plate mail rebounding from the impact, but that might be a little beyond the scope of this discussion.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How could a swordsman or a spearman beat this soldier? Would the impact of these weapons suffice?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The weak point of any suit of armor is the joints.  This is where a knight was most vulnerable.  Lots of suits of armor had the joints reinforced with chainmail, leather, and little targa-like shields (like the circular things on your knight's armpit area).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A heavy enough swing could dent armor, which could then impede movement of that part of the body.  Do it enough times, and you effectively cripple your opponent or render him unable to fight back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may not be historically accurate, but if you want an example of how a spearman would beat a knight, watch the &lt;a href=&quot;https://www.youtube.com/watch?v=nqT22PlqnUA&quot; rel=&quot;nofollow&quot;&gt;Oberyn Martel vs. The Mountain&lt;/a&gt; duel from Game of Thrones.  The same would apply for a swordsman.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Just out of curiosity, a modern gun would penetrate into the plates that easily? One headshot would do?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Modern guns would have no issue with plate mail.  The advent of black powder weapons is the reason that plate mail was abandoned.  Have a read through &lt;a href=&quot;http://www.thehighroad.org/archive/index.php/t-454655.html&quot; rel=&quot;nofollow&quot;&gt;this thread&lt;/a&gt; for further information.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-08-14T03:12:12.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3943" PostTypeId="2" ParentId="3940" CreationDate="2015-08-14T04:49:18.790" Score="0" Body="&lt;p&gt;History for centuries past always depends on who writes it, and the Engineering behind the history is thus somewhat a matter of conjecture, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The English Longbow and &quot;Clothyard shaft&quot; are said by many to be the &quot;high tech&quot; weapons system that won them the Battle of Agincourt against the French. In simplistic terms, the longbow shafts could penetrate French plate armour far enough away to protect the front line of both English Noblemen, foot soldiers and pike bearers while the front line protected the bowmen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The detail is both far far more complex, and open to much discussion, but the summary is &quot;good enough&quot;. By itself the longbow almost certainly have been enough to swing the balance at Agincourt, but they were greatly helped by a complete lack of sensible behaviour by their opponents due to the concepts of Chivalry, societal position and, as often, the lure of wealth. Rather than me trying to explain how that makes sense - here's what some sources say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;From Wikipedia Longbow - see below:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Archery was described by contemporaries as ineffective against plate armour in the Battle of Neville's Cross (1346), the siege of Bergerac (1345), and the Battle of Poitiers (1356); such armour became available to European knights of fairly modest means by the late 14th century, though never to all soldiers in any army. Longbowmen were however effective at Poitiers, and this success stimulated changes in armour manufacture partly intended to make armoured men less vulnerable to archery. Nevertheless, at the battle of Agincourt in 1415 and for some decades thereafter, English longbowmen continued to be an effective battleﬁeld force.[34]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Battle_of_Agincourt&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Wikipedia - Battle of Agincourt&lt;/strong&gt;&lt;/a&gt; - a nicely detailed account with discussion of the uncertainties this far after the event and a good general impression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/English_longbow&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Wikipedia - English longbow&lt;/strong&gt;&lt;/a&gt; - good discussion good detail. Good section on armour penetration, modern simulations and pros and cons&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In an early modern test by Saxton Pope, a direct hit from a steel bodkin point penetrated Damascus mail armour.[28][29]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A 2006 test was made by Matheus Bane using a 75 lbf (330 N) draw (at 28&quot;) bow, shooting at 10 yards; according to Bane's calculations,&lt;br&gt;&#xA;  this would be approximately equivalent to a 110 lbf (490 N) bow at 250 yards.[30] &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Measured against a replica of the thinnest contemporary &quot;Jack coat&quot; armour, a 905 grain needle bodkin and a 935 grain curved broadhead penetrated over 3.5 inches (89 mm). (&quot;Jack coat&quot; armour could be up to twice as thick as the coat tested; in Bane's opinion such a thick coat would have stopped bodkin arrows but not the cutting force of broadhead arrows.)   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Against &quot;high quality riveted maille&quot;, the needle bodkin and curved broadhead penetrated 2.8&quot;.&lt;br&gt;&#xA;  Against a coat of plates, the needle bodkin achieved 0.3&quot; penetration.&lt;br&gt;&#xA;  The curved broadhead did not penetrate but caused 0.3&quot; of deformation of the metal.&lt;br&gt;&#xA;  Results against plate armour of &quot;minimum thickness&quot; (1.2mm) were similar to the coat of plates, in that the needle bodkin penetrated to a shallow depth, the other arrows not at all.&lt;br&gt;&#xA;  In Bane's view, the plate armour would have kept out all the arrows if thicker or worn with more padding.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Other modern tests described by Bane include those by Williams (which concluded that longbows could not penetrate mail, but in Bane's view did not use a realistic arrow tip), Robert Hardy's tests (which achieved broadly similar results to Bane), and a Primitive Archer test which demonstrated that a longbow could penetrate a plate armour breastplate. However, the Primitive Archer test used a 160 lbf (710 N) longbow at very short range, generating 160 joules (vs. 73 for Bane and 80 for Williams), so probably not representative of battles of the time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Telegraph &lt;a href=&quot;http://www.telegraph.co.uk/news/8648068/Battle-of-Agincourt-ten-reasons-why-the-French-lost.html&quot; rel=&quot;nofollow&quot;&gt;10 reasons why the French lost&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longbow Archers site - &lt;a href=&quot;http://www.longbow-archers.com/historyagincourt.html&quot; rel=&quot;nofollow&quot;&gt;Agincourt page&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You tube documentary 1h12m - looks good &#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=LWmM-jyE338&quot; rel=&quot;nofollow&quot;&gt;The Battle of Agincourt : The Bloodiest Battle of the Medieval Age &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6m video reenactment - &lt;a href=&quot;https://www.awesomestories.com/asset/view/English-Longbow-Battle-Scene-at-Agincourt-&quot; rel=&quot;nofollow&quot;&gt;Agincourt - Henry V version&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://kingjohnshouse.org.uk/the-longbow-and-its-impact-on-the-battle-of-agincourt/&quot; rel=&quot;nofollow&quot;&gt;Steven Whetan says ...&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the usual &lt;a href=&quot;https://www.google.co.nz/search?q=longbow+agincourt&amp;amp;biw=1422&amp;amp;bih=1041&amp;amp;tbm=isch&amp;amp;tbo=u&amp;amp;source=univ&amp;amp;sa=X&amp;amp;ved=0CFgQsARqFQoTCPKin7Xdp8cCFSHYpgod1fcLGQ&amp;amp;dpr=0.9#tbm=isch&amp;amp;q=%2Blongbow+%2Bagincourt&quot; rel=&quot;nofollow&quot;&gt;obligatory pictures&lt;/a&gt; most of which link to perhaps relevant pages. Agh. Wow. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-08-14T05:32:31.213" LastActivityDate="2015-08-14T05:32:31.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3944" PostTypeId="2" ParentId="3940" CreationDate="2015-08-14T06:15:43.123" Score="2" Body="&lt;p&gt;In response to plate armor, a variety of anti-armor weapons and techniques have emerged.  Different concepts were employed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hammers and maces that can give a concussion without penetrating the armor.&lt;/li&gt;&#xA;&lt;li&gt;Spikes that concentrate pressure at one point and thus pierce.&lt;/li&gt;&#xA;&lt;li&gt;Stabbing between the plates of the armor.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some weapons combined multiple approaches.  For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/War_hammer&quot; rel=&quot;nofollow&quot;&gt;war hammers&lt;/a&gt; often had a non-penetrating side for giving concussion and a spike for penetration.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2015-08-14T06:15:43.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3945" PostTypeId="1" AcceptedAnswerId="3947" CreationDate="2015-08-14T07:53:00.993" Score="1" ViewCount="64" Body="&lt;p&gt;I'm hoping to rebuild a joystick (sidewinder strategic commander), though the parts inside it are a bit too wobbly because they have been worn beyond tolerances. Due to this wobbling on one (rotation) axis, the sliding action is rough on the x/y plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any recommendation on how and what to use, if there is a compound I can apply to make up for what was lost through friction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I bring the plastic mechanisms back to spec? Also what lubricant is suggested afterwards?&lt;/p&gt;&#xA;" OwnerUserId="2648" LastEditorUserId="272" LastEditDate="2015-08-14T11:56:42.803" LastActivityDate="2015-08-14T12:49:16.440" Title="How to rebuild and lubricate worn plastic parts?" Tags="|plastic|tolerance|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3947" PostTypeId="2" ParentId="3945" CreationDate="2015-08-14T12:49:16.440" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Any recommendation on how and what to use, if there is a compound I can apply to make up for what was lost through friction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At first thought, I'm thinking that you scuff the surface of the plastic parts and try to rebuild the missing area with &lt;a href=&quot;http://www.jbweld.com/collections/epoxy-adhesives/products/j-b-weld-twin-tube&quot; rel=&quot;nofollow&quot;&gt;JB Weld&lt;/a&gt;.  After it cures, you can file away the unneeded areas to get the part back to &quot;original&quot; shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll want as smooth of a contact surface as you can get to reduce wear going forward, so don't forget to sand with increasingly fine grits of sandpaper.  I would probably start at 150 grit and work to at least 600 grit, maybe even 1000 grit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also what lubricant is suggested afterwards?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For lubricant, it really &lt;a href=&quot;http://machinedesign.com/mechanical-drives/lubrication-tips-plastic-gears-and-more&quot; rel=&quot;nofollow&quot;&gt;depends on the type of plastic&lt;/a&gt; your parts are made of.  The site I linked recommended a thicker synthetic lubricant (silicone-based) or mineral oil.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-08-14T12:49:16.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3948" PostTypeId="1" CreationDate="2015-08-14T13:34:02.797" Score="4" ViewCount="102" Body="&lt;p&gt;In my organization, every unit is built by hand, and every unit is tested. Every product has a written construction procedure and a written test procedure. We're in the process of updating and improving all those procedures. One potential problem we're trying to eliminate is when the checkout personnel skip a step during testing. If all you have is a list of fifty steps, and no step-by-step record of what you've already done, that's an easy thing to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One proposal is to require them to print a new copy of the procedure and check a box next to every step, every time a unit is checked. That's going to cost a lot of time and paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea is to digitize the whole thing. Have a server running somewhere that the checkout tech logs into, which will show them the relevant procedure for the unit they're testing. They check the boxes as they go through the procedure, and it all gets stored electronically. This has a higher initial overhead, but should save a lot of resources in the long term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best proposal I've heard so far is to just roll our own checkout database and interface, but I feel like that must be reinventing the wheel. Is there software to support this sort of thing already in existence? Are there other, better ideas? Is there an accepted industry standard for solving this problem?&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="122" LastEditDate="2015-08-14T16:51:11.503" LastActivityDate="2015-08-17T11:43:20.460" Title="How can I better structure documentation for a quality control process?" Tags="|quality-engineering|product-testing|documentation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3949" PostTypeId="1" AcceptedAnswerId="3950" CreationDate="2015-08-14T14:03:25.020" Score="3" ViewCount="727" Body="&lt;p&gt;I'm using a OnePlusOne and while using my sunglasses I can see the phone screen in portrait mode, however if I rotate the screen to landscape I can't see anything at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sunglasses are polarized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for this?&lt;/p&gt;&#xA;" OwnerUserId="2652" LastActivityDate="2015-08-14T14:17:37.753" Title="Why can't I see my phone screen in landscape with sunglasses?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2015-08-14T21:40:46.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="3950" PostTypeId="2" ParentId="3949" CreationDate="2015-08-14T14:17:37.753" Score="5" Body="&lt;p&gt;The screen of your phone uses a polarizing screens as well. When you hold you phone in portrait mode, the screens are oriented in the same direction so you can see the image. When you rotate it 90deg, the screens are not aligned and light cannot pass. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-08-14T14:17:37.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3951" PostTypeId="2" ParentId="3948" CreationDate="2015-08-14T17:50:39.200" Score="1" Body="&lt;p&gt;Have you looked at NI Test Stand? When I worked in the medical device industry that's what was used at every step along the way. The devices weren't built completely and then tested, but tested after each stage of production with a final programming and validation at the end. The devices were serialized by a ROM, which was tied to an engraved s/n on the exterior of the part. Each test stand, read the ROM and then added the results of each test to the record for each device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each test consisted of several steps and Test Stand allows requiring different criteria to be met before proceeding to the next step and the data for all tests are stored under each device's sn/ROM ID.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-08-14T17:50:39.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3952" PostTypeId="2" ParentId="3948" CreationDate="2015-08-14T18:36:20.313" Score="1" Body="&lt;p&gt;With out a cost benefit analysis it is hard to tell with method will bring the best financial value to your organization. I am assuming that the analysis suggest that using a test automation will bring the best value for your organization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base on the current industry standard test automation will bring you the best value in almost all cases. In one of my past experiences bar coding was used to serialize product. A manufacturing personal would scan the bar code and execute the automated test process. The product ID, test data, manufacturing personal ID (in some cases such as FDA), date, time etc were stored in server as suggested by your question. This enabled the quality organization to verify if all test and verification steps were executed prior to shipping the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These type of processors are followed in US automotive, and semiconductor industry. I believe similar processor are followed in Medical device manufacturing industry. Also some of the reputed semiconductor manufactures from the far east not only follow these practices but have taken to them to the next level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Product such as &lt;a href=&quot;http://www.ni.com/labview/&quot; rel=&quot;nofollow noreferrer&quot;&gt;LabView&lt;/a&gt; and &lt;a href=&quot;http://www.ni.com/lwcvi/&quot; rel=&quot;nofollow noreferrer&quot;&gt;LabWindows/CVI&lt;/a&gt; from &lt;a href=&quot;http://www.ni.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;National Instruments (NI)&lt;/a&gt; are the ones that I am most familiar. There is also &lt;a href=&quot;http://www.keysight.com/en/pc-2472896/benchvue-software?cc=US&amp;amp;lc=eng&quot; rel=&quot;nofollow noreferrer&quot;&gt;BenchVue&lt;/a&gt; from &lt;a href=&quot;http://www.keysight.com/main/home.jspx?cc=US&amp;amp;lc=eng&quot; rel=&quot;nofollow noreferrer&quot;&gt;Keysight Technologies&lt;/a&gt; with I believe originated from HP VEE. This product is evaluation from HP-VEE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you can use open source software such as Linux combined with a open source back-end server. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Most importantly&lt;/strong&gt; data collection enable the use statistical analysis tools such as &lt;a href=&quot;http://www.minitab.com/en-us/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Minitab&lt;/a&gt; to improve product quality, &lt;a href=&quot;https://en.wikipedia.org/wiki/Statistical_process_control&quot; rel=&quot;nofollow noreferrer&quot;&gt;statistical process controls (SPC)&lt;/a&gt;, trending, operation optimization, cost improvement and many more. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/17161/test-automation-framework-softwares-for-hardware-applications&quot;&gt;Test automation framework/software for hardware applications&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Agilent_VEE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Agilent VEE&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.hpl.hp.com/hpjournal/98may/may98a13.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Compiler for HP VEE&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2015-08-17T11:43:20.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3959" PostTypeId="1" AcceptedAnswerId="3960" CreationDate="2015-08-04T21:20:15.227" Score="12" ViewCount="26255" Body="&lt;p&gt;The TI MSP430F20XX series has a 12-bit internal ADC output, which is right-justified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference between a left-justified output and a right-justified output? What are their pros and cons?&lt;/p&gt;&#xA;" OwnerUserId="2666" OwnerDisplayName="prasad" LastEditorUserId="1586" LastEditDate="2015-08-20T17:31:30.033" LastActivityDate="2015-08-20T17:31:30.033" Title="What are left justified and right justified ADC results?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3960" PostTypeId="2" ParentId="3959" CreationDate="2015-08-04T21:47:38.767" Score="13" Body="&lt;p&gt;On this processor, the register that holds the conversion result is 16 bits wide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A right-justified result means that bits [(&lt;em&gt;N&lt;/em&gt;-1):0] (where &lt;em&gt;N&lt;/em&gt; is the number of bits of precision) of the register contain the ADC value and the most-significant bits of the register are set to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A left-justified result means that bits [15:(16-&lt;em&gt;N&lt;/em&gt;)] of the register hold the result, and bits [(15-&lt;em&gt;N&lt;/em&gt;):0] are set to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if your actual conversion result is 0x123, it would be read as 0x0123 if the register was right-justified, and as 0x1230 if it was left-justified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An advantage of left-justified results (on processors that support it) is that you can take just the most-significant byte of the register, giving you 8-bits of precision instead of the native precision. This can be useful if you don't need the extra precision, or have RAM constraints and want to store a large number of samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a right-justified value can be used directly without the scaling that a left-justified value would need. &lt;/p&gt;&#xA;" OwnerUserId="2660" OwnerDisplayName="Niall C." LastActivityDate="2015-08-04T21:52:48.327" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="3961" PostTypeId="1" AcceptedAnswerId="3962" CreationDate="2015-08-05T18:12:00.707" Score="10" ViewCount="231" Body="&lt;p&gt;I am writing a driver for the &lt;code&gt;Samsung K9WAG08U1D NAND flash&lt;/code&gt; chip. The specification of the memory chip mentions it has a page size of 2048 bytes (2kB). I am using a &lt;code&gt;TI MSP430F2619&lt;/code&gt; which has 4096 Bytes (4kB) of RAM. This means I need to allocate a 2k memory buffer just to write to flash. My application is a protocol converter and hence requires an additional buffer for handling to and fro transmission. &#xA;Please suggest me better approach to reduce the RAM requirement due to flash page size.&lt;/p&gt;&#xA;" OwnerUserId="2666" OwnerDisplayName="prasad" LastEditorUserId="1586" LastEditDate="2015-08-17T16:52:50.390" LastActivityDate="2015-08-17T16:52:50.390" Title="Saving RAM Memory when writing to 2K page size NAND flash" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3962" PostTypeId="2" ParentId="3961" CreationDate="2015-08-05T18:37:50.550" Score="7" Body="&lt;p&gt;You don't need to fill the page register all in one go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You begin a page write (i.e. the &quot;Page Program&quot; operation) by writing the Serial Data Input command (&lt;code&gt;0x80&lt;/code&gt;), the column address, and the row address.  Then you transfer the data to the page register (up to 2112 bytes).  This transfer can be broken up into chunks, with any delay between chunks you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have filled the page register, you begin the transfer from the page register to the array with the Page Program Confirm command (&lt;code&gt;0x10&lt;/code&gt;).&lt;/p&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastActivityDate="2015-08-05T19:11:42.750" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3963" PostTypeId="1" CreationDate="2015-08-06T04:46:03.270" Score="12" ViewCount="364" Body="&lt;p&gt;Why are typical baud rates not in multiples of 1000 instead of being in multiples of 300 i.e. 4800, 9600, 19200, 115200, etc...? Is there a mathematical or electrical reason behind this?&lt;/p&gt;&#xA;" OwnerUserId="2666" OwnerDisplayName="prasad" LastEditorUserId="1586" LastEditDate="2015-08-20T17:31:55.730" LastActivityDate="2015-08-20T17:31:55.730" Title="Why are typical UART baud rates not in multiples of 1000?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3964" PostTypeId="2" ParentId="3963" CreationDate="2015-08-06T07:28:09.277" Score="11" Body="&lt;p&gt;This goes way, way back to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bell_103_modem&quot;&gt;Bell 103 modem&lt;/a&gt; from 1962.  It operated at 300 bits per second for reasons deriving from the use of &lt;a href=&quot;https://en.wikipedia.org/wiki/Frequency-shift_keying&quot;&gt;audio frequency-shift keying&lt;/a&gt; to modulate data; the frequencies available were limited by the quality of the phone lines.  All common serial data rates since then are integer multiples of this rate.&lt;/p&gt;&#xA;" OwnerUserId="370" OwnerDisplayName="Mark" LastActivityDate="2015-08-06T07:28:09.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3965" PostTypeId="1" AcceptedAnswerId="3966" CreationDate="2015-07-30T10:33:42.867" Score="5" ViewCount="174" Body="&lt;p&gt;The development board for the touchscreen chip we got is too good ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chip provides the readouts over I2C, passes it to FPGA (I could provide the model if I find a magnifying glass... but I'd prefer a more generic answer anyway) and the FPGA passes the data over RS232 to an FDDI which connects over USB to the PC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except I don't want the signal over USB or RS232, I need it in I2C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a jumper that allows me to remove power from the FPGA, and there are convenient test points on the I2C bus which I can connect to the bus with my device, but the FPGA still remains connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of problems/interference can I expect from it - present on the bus, unpowered?&lt;/p&gt;&#xA;" OwnerUserId="203" OwnerDisplayName="SF." LastEditorUserId="1586" LastEditDate="2015-08-17T19:01:06.353" LastActivityDate="2015-08-17T19:01:06.353" Title="What is the behavior of unpowered FPGA pins on I2C bus?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3966" PostTypeId="2" ParentId="3965" CreationDate="2015-07-30T11:04:16.487" Score="6" Body="&lt;p&gt;It does, unfortunately depend a little on the FPGA itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But most cases that will probably not work, unfortunately. In many logic chips with programmable I/O type pins, the final stage looks a bit like this:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/x1GT6.png&quot; alt=&quot;schematic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;This is both a side-effect of the cheapest to make final stage and for protection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you turn off the power to that stage, the Vdd/Vcc voltage goes to 0V, so the top diode, D2, will conduct towards that. Since I2C is powered with resistors from the power rail in an open-collector type scheme, the diode will easily conduct those few mili-amps away, so damage is very unlikely, but the bus will never see a logic one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the FPGA is 1.8V or 2.7V or 3.3V itself, but 5V tolerant on the pins in question, the top diode will not be there and the problem may very well disappear, but other characteristics may still have similar influences. (JFETs that turn on because their biasing disappears at power-off, though highly unlikely, technically possible).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An option would be to break the traces and then (if you want the FPGA again later) to add decent analogue switches. Sometimes traces come along the board in such a way that a simple dual Analogue Switch can be dropped right across a gap you made with the scalpel. I've done that a couple of times. It's not something you'd design in, but for 400kHz I2C there are plenty chips that could do it and as an after-market mod, it's not even the ugliest you could think of.&lt;/p&gt;&#xA;" OwnerUserId="2683" OwnerDisplayName="Asmyldof" LastActivityDate="2015-07-30T11:04:16.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3967" PostTypeId="1" AcceptedAnswerId="3969" CreationDate="2015-07-29T10:38:09.310" Score="7" ViewCount="583" Body="&lt;p&gt;PSOC4 provides a quadrature decoder functionality but I was unable to find specifics on its performance in the datasheets. Sure I know I need to read and utilize the count until the 16 bits of its counter overflow, but that's some 32k per iteration of the main loop of my program. If a precise encoder is attached to a fast motor, that produces lots and lots of pulses in a very short time, and I'm worried the decoder module won't be able to keep up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does anyone know what is the maximum performance of the decoder block - the minimum timings of its input?&lt;/p&gt;&#xA;" OwnerUserId="203" OwnerDisplayName="SF." LastEditorUserId="1586" LastEditDate="2015-08-20T17:31:53.077" LastActivityDate="2015-08-20T17:31:53.077" Title="How fast movement (cycles/s) can I reliably read with PSoC4 quadrature decoder?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3968" PostTypeId="2" ParentId="3967" CreationDate="2015-07-29T21:07:15.520" Score="5" Body="&lt;p&gt;This doesn't get mentioned in some documentation for the PSoC quadrature decoder component, but in other versions (e.g. &lt;a href=&quot;http://www.cypress.com/file/135186/download&quot;&gt;here&lt;/a&gt;) it states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The clock input frequency should be greater than or equal to 10x the maximum A or B input frequency.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given that AFAICT you can clock the decoder at multi-MHz frequencies, I think it can likely keep up with most plausible motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; It seems the component I found the data for isn't quite the one the question was about so, whilst this answer may be useful to someone some day, it probably shouldn't be the highest voted nor the accepted one here. &lt;/p&gt;&#xA;" OwnerDisplayName="Simon Jenkins" LastActivityDate="2015-07-30T13:29:52.603" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="3969" PostTypeId="2" ParentId="3967" CreationDate="2015-07-30T10:39:37.447" Score="6" Body="&lt;h2&gt;Quadrature decoder&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Looking at the &lt;a href=&quot;http://www.cypress.com/file/131666/download&quot;&gt;PSoC 4 TCPWM datasheet&lt;/a&gt; the maximum operating frequency of the counter block is 48MHz, with a counter resolution of 21ns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that you have a high resolution &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotary_encoder#Incremental_rotary_encoder&quot;&gt;incremental rotary encoder&lt;/a&gt; with 10,000 counts per revolution, and spin the rotor at 15,000 rpm, that's 2,500,000 counts per second (2.5MHz), comfortably less than the 48MHz max. sampling rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sadly, it is not clear if the &lt;em&gt;PSoC 4 TCPWM&lt;/em&gt; suffers the same 10x sampling requirement as the &lt;a href=&quot;http://www.cypress.com/file/135186/download&quot;&gt;PSoC QuadDec&lt;/a&gt; mentioned in &lt;a href=&quot;http://embedded.stackexchange.com/a/85/186&quot;&gt;Simon Jenkins' answer&lt;/a&gt;. I suspect not, as the datasheet doesn't include the clock signals on the timing diagrams. This implies that &lt;em&gt;PSoC 4 TCPWM&lt;/em&gt; is edge driven (limited by the counter clock), while &lt;em&gt;PSoC QuadDec&lt;/em&gt; is driven by level at each clock edge and thus limited by the sampling at the clock frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the &lt;em&gt;PSoC 4 TCPWM&lt;/em&gt; does require over sampling, then the max 2.5MHz pulse train is still less than both 10x at 48MHz for the &lt;em&gt;PSoC 4 TCPWM&lt;/em&gt; and 28MHz for the &lt;em&gt;PSoC QuadDec&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only need 2000 PPR at 6000 RPM (as suggested in your comment), then you only need to support 200 KHz (2000*6000/60), so a 2MHz clock should be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Interrupt handling&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Another concern is whether the CPU will be able to cope with the interrupts this could generate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that you are running the quadrature right at the limit of the decoder block, i.e. 48MHz with a 16bit counter you will be getting 1465 overflow interrupts per second, so unless you can service your interrupt (including any context switching required) in less than 682 micro seconds, then you are going to struggle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, it looks like the the Cortex-M0 processors have very efficient ISR handling &lt;sub&gt;(See &lt;a href=&quot;http://community.arm.com/docs/DOC-2607&quot;&gt;A Beginner’s Guide on Interrupt Latency - and Interrupt Latency of the ARM® Cortex®-M processors&lt;/a&gt;)&lt;/sub&gt; so unless your ISR is very inefficient, it should be able to handle this.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Resolution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Incidentally, it looks like you will need to use X4 mode if you want to count individual quadrature state transitions. If you use X1 then you only see whole quadrature cycles (4 states), while if you use X2 then you only see half of the states.&lt;/p&gt;&#xA;" OwnerUserId="287" OwnerDisplayName="Mark Booth" LastActivityDate="2015-07-30T10:39:37.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3970" PostTypeId="1" CreationDate="2015-07-29T12:31:51.557" Score="6" ViewCount="189" Body="&lt;p&gt;I've already learned how to find which pins can be assigned to a functionality I just open the [arch]-pinfunc.h file and find the entries after &lt;code&gt;__&lt;/code&gt; for given functionality and then find the pin that provides it, left of the &lt;code&gt;__&lt;/code&gt; in the entry. Then I just write the right entry in .dts/.dtsi referencing the pin identity in a way specific to that entry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the worse problem is if the pin is already assigned - conflicting with my purpose - and I don't know what to. Not such a big problem if it's a default described in the datasheets, worse if it's an obscure helper functionality left over from development, and not even routed to the board so I won't find it in the schematics. The device tree for modern processors is extensive and long and normally I'd need to check every single entry (out of a couple hundred) to see if any uses the pin I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any easier/simpler way to find what functionality is currently assigned to given pin in .dts?&lt;/p&gt;&#xA;" OwnerUserId="203" OwnerDisplayName="SF." LastEditorUserId="1586" LastEditDate="2015-08-20T17:32:00.963" LastActivityDate="2015-08-20T17:32:00.963" Title="How to find who occupies pin X in Device Tree?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3971" PostTypeId="2" ParentId="3970" CreationDate="2015-07-30T23:00:50.850" Score="2" Body="&lt;p&gt;&lt;strong&gt;Possible Solution&lt;/strong&gt;&lt;br&gt;&#xA;The most simple way to find this is a simple text search through the *.dts file for the pin name. You may be able to look at some random pin in the *.dts file and compare it to the schematic to see the naming scheme to give you an idea of what it should look like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the the schematic has a GPIO Pin3 on Port A, you might see something like &quot;gpio_portA-3&quot; or some other permutation. Look for a pin in the *.dts file that also appears in the schematic to identify the naming pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have that, search for the pin you're actually interested in. If you don't find it right away, try different variations, and perhaps try regular expressions to match varying separating characters. If you find it, you should be able to see what the parent node is, which will tell you what the pin does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given &lt;a href=&quot;https://groups.google.com/forum/#!topic/beagleboard/AKHfQIuDdZI&quot; rel=&quot;nofollow&quot;&gt;this poor guy's adventures&lt;/a&gt; with reverse-engineering the BeagleBone Black's pinmux mappings, you should consider yourself lucky if a simple text search actually works for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caveat&lt;/strong&gt;&lt;br&gt;&#xA;Unfortunately, there's no guarantee that the pin you're interested in will even be present in the FDT. If a pin doesn't require direct interaction from the OS to work properly, it won't be there. Consider this example from &lt;a href=&quot;http://www.devicetree.org/Device_Tree_Usage&quot; rel=&quot;nofollow&quot;&gt;devicetree.org&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;/ {&#xA;    #address-cells = &amp;lt;1&amp;gt;;&#xA;    #size-cells = &amp;lt;1&amp;gt;;&#xA;&#xA;    ...&#xA;&#xA;    serial@101f0000 {&#xA;        compatible = &quot;arm,pl011&quot;;&#xA;        reg = &amp;lt;0x101f0000 0x1000 &amp;gt;;&#xA;    };&#xA;&#xA;    serial@101f2000 {&#xA;        compatible = &quot;arm,pl011&quot;;&#xA;        reg = &amp;lt;0x101f2000 0x1000 &amp;gt;;&#xA;    };&#xA;&#xA;    gpio@101f3000 {&#xA;        compatible = &quot;arm,pl061&quot;;&#xA;        reg = &amp;lt;0x101f3000 0x1000&#xA;               0x101f4000 0x0010&amp;gt;;&#xA;    };&#xA;&#xA;    interrupt-controller@10140000 {&#xA;        compatible = &quot;arm,pl190&quot;;&#xA;        reg = &amp;lt;0x10140000 0x1000 &amp;gt;;&#xA;    };&#xA;&#xA;    spi@10115000 {&#xA;        compatible = &quot;arm,pl022&quot;;&#xA;        reg = &amp;lt;0x10115000 0x1000 &amp;gt;;&#xA;    };&#xA;&#xA;    ...&#xA;&#xA;};&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This block describes a couple of memory-mapped devices, including serial, SPI, and interrupt controllers, and even GPIOs. All of those, with the possible exception of the interrupt controller, must use pins, yet they aren't specifically present here. This is because the manipulation of the pins is intended to be done by hardware, and the OS shouldn't have to know or care which pins are used. All the OS needs to know is how to interact with the controllers' registers, so the pin information may be left out of the FDT.&lt;/p&gt;&#xA;" OwnerUserId="2659" OwnerDisplayName="skrrgwasme" LastActivityDate="2015-07-30T23:00:50.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3972" PostTypeId="2" ParentId="3970" CreationDate="2015-07-31T08:35:38.260" Score="2" Body="&lt;p&gt;If your kernel is compiled with &lt;a href=&quot;https://www.kernel.org/doc/Documentation/filesystems/debugfs.txt&quot; rel=&quot;nofollow&quot;&gt;debugfs support&lt;/a&gt; (&lt;code&gt;CONFIG_DEBUG_FS&lt;/code&gt;), then often you can use debugfs to inspect pinctrl mappings at runtime.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$ mount -t debugfs none /sys/kernel/debug     # ensure debugfs is mounted&#xA;$ cat /sys/kernel/debug/pinctrl/pinctrl-maps  # prints info on pin mappings&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Not strictly related, but also useful:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;$ cat /sys/kernel/debug/gpio     # prints configuration and state of GPIO lines&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Debugging tools such as &lt;a href=&quot;http://www.lartmaker.nl/lartware/port/devmem2.c&quot; rel=&quot;nofollow&quot;&gt;devmem2&lt;/a&gt; are also very helpful as you can often use this to quickly inspect the value of the registers that control configuration of each pin.&lt;/p&gt;&#xA;" OwnerUserId="2941" OwnerDisplayName="Grodriguez" LastActivityDate="2015-07-31T08:40:52.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3973" PostTypeId="1" AcceptedAnswerId="3974" CreationDate="2015-07-28T22:52:34.127" Score="8" ViewCount="83" Body="&lt;p&gt;For example, if there is many floating point operations, it could &quot;rewire&quot; the area of some of its integer units to handle floating point instructions, and vice versa. Or if there isn't too many calculations, but there is many I/O need, it could parallelize its IO better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I know, the FPGAs load their HDL code on boot, but to me it seems not really impossible, to make it possible to reload differents  parts of a bigger HDL on the need, partially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a such FPGA already exist?&lt;/p&gt;&#xA;" OwnerUserId="9" OwnerDisplayName="peterh" LastEditorUserId="1586" LastEditDate="2015-08-20T17:51:25.650" LastActivityDate="2015-08-20T17:51:25.650" Title="Is there an FPGA which could &quot;re-wire&quot; itself on the need?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3974" PostTypeId="2" ParentId="3973" CreationDate="2015-07-29T02:27:22.067" Score="11" Body="&lt;p&gt;It seems that such (Xilinx) FPGAs already exist.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Reconfigurable_computing#Partial_re-configuration&quot;&gt;https://en.wikipedia.org/wiki/Reconfigurable_computing#Partial_re-configuration&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In partial reconfiguration, one part of the FPGA continues to run while another part is being reconfigured.&lt;/p&gt;&#xA;" OwnerDisplayName="Jacques" LastActivityDate="2015-07-29T02:27:22.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3978" PostTypeId="1" AcceptedAnswerId="3979" CreationDate="2015-07-30T16:40:59.113" Score="4" ViewCount="245" Body="&lt;p&gt;Atmel's ASF (Atmel Software Framework) is periodically updated by Atmel.  When an update is available, Atmel Studio notifies you and asks if you want to download the update.  If I install the update, this does not update any code in my existing projects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to determine what code changes have been made in the new ASF compared to what I previously used in a project.  And ideally I'd like to be able to cherry-pick the changes that I want to incorporate (and exclude those I don't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way I know to do this now is to create new project with the updated ASF (using the ASF Wizard to add all the drivers I used previously) and manually diff the files against my existing project.  Then manually copy/paste from the files I want to include.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I heard that Atmel uses an internal git repository for their ASF.  Is there a way to get at this directly and view the changes?  Or is there a better way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would also be nice to view the changes before installing the ASF update.&lt;/p&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastEditorUserId="2642" LastEditDate="2018-01-29T14:23:26.197" LastActivityDate="2018-01-29T14:23:26.197" Title="How can I determine the code changes in an Atmel ASF update?" Tags="|embedded-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3979" PostTypeId="2" ParentId="3978" CreationDate="2015-07-30T18:40:47.217" Score="4" Body="&lt;p&gt;I think what you're looking for is the &lt;a href=&quot;http://www.atmel.com/webdoc/asf/asf.VersionConversionWizard.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASF Conversion Wizard&lt;/a&gt;. According to the linked page, this wizard is supposed to be launched whenever the ASF version changes. If is not launching for you, you can open the ASF Wizard in your existing project, and under the Version tab, you can change the ASF version for your current project. Make sure the &quot;Trigger upgrade of modules&quot; box is checked:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1YGdJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1YGdJ.png&quot; alt=&quot;upgrade screenshot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image is from &lt;a href=&quot;http://www.atmel.com/webdoc/asf/asf.ModuleExplorer.VersionTab.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASF Wizard documentation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For viewing changes before you upgrade the ASF version, I think the release notes for a given ASF release are your only resource. They're admittedly not very complete, but I can't find any other documentation about them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of any way to cherry-pick different sections of an ASF release. I don't think this is a great idea anyway, unless you're willing to wade through a lot of dependencies. Not all components of the ASF are direct windows between your software and the hardware; some of it is middleware that other components of the same ASF release may rely on. By picking and choosing which aspects of the ASF to update, you risks losing APIs, structure definitions, and other code that you may need in a non-obvious way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last note: the only ASF git repository I can find is here: &lt;a href=&quot;https://spaces.atmel.com/gf/project/asf/scmgit/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://spaces.atmel.com/gf/project/asf/scmgit/&lt;/a&gt;. Unfortunately, it doesn't appear to have been updated since 2012. I can't find any more up-to-date repositories.&lt;/p&gt;&#xA;" OwnerUserId="2659" OwnerDisplayName="skrrgwasme" LastActivityDate="2015-07-30T18:40:47.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3980" PostTypeId="1" AcceptedAnswerId="3981" CreationDate="2015-07-31T13:43:05.453" Score="7" ViewCount="2557" Body="&lt;p&gt;In the Linux kernel's dtsi file (am335x-bone-common.dtsi) for the AM335x processor, there is the am33xx_pinmux tree, with values like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    i2c0_pins: pinmux_i2c0_pins {&#xA;        pinctrl-single,pins = &amp;lt;&#xA;            0x188 (PIN_INPUT_PULLUP | MUX_MODE0)    /* i2c0_sda.i2c0_sda */&#xA;            0x18c (PIN_INPUT_PULLUP | MUX_MODE0)    /* i2c0_scl.i2c0_scl */&#xA;        &amp;gt;;&#xA;    };&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How do I know what values to use for the &lt;code&gt;pinctrl-single,pins&lt;/code&gt; entry?&lt;/p&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastEditorUserId="1586" LastEditDate="2017-11-01T21:16:16.570" LastActivityDate="2017-11-01T21:16:16.570" Title="In a device tree for AM335x, how do I know what pinctrl values to use?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3981" PostTypeId="2" ParentId="3980" CreationDate="2015-07-31T13:43:05.453" Score="7" Body="&lt;p&gt;To update your .dts file (that includes the file am335x-bone-common.dtsi) for a pinmux assignment, consult the datasheet to determine which pin(s) are assigned to a function.  This is in section 4.3 (broken into subsections per module).  Clicking on the hyperlink for a pin (check your package type, ZCE BALL or ZCZ BALL) will take you to the corresponding entry in Table 4-1 (Ball Characteristics).  Use Table 4-1 to determine the correct mode to set for that pin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then consult the TRM, Table 9-7 (CONTROL_MODULE REGISTERS), on page 1363 for Rev K.  The table lists the registers for the default function (likely mode 0).  The pin-mux registers start at offset 0x800, and this offset must be subtracted to give the first value in the .dts file.  Consult section 9.2.2 (Pad Control Registers) for options for the second value in the .dts file (in addition to the mode setting).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, to assign pin C18 of the ZCZ package to SPI1 SCLK, click on the C18 link in table 4-53 of the datasheet (section 4.3.6.5).  This takes you to the &lt;code&gt;ECAP0_IN_PWM0_OUT&lt;/code&gt; pin in table 4-1.  Here you see that a mode value of 4 will select &lt;code&gt;spi1_sclk&lt;/code&gt; for that pin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then in the TRM, table 9-7 shows &lt;code&gt;conf_ecap0_in_pwm0_out&lt;/code&gt; registers start at address 0x964.  We must subtract 0x800, so the first value in the .dts is 0x164.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we want to select mode 4 (&lt;code&gt;MUXMODE=4&lt;/code&gt;).  If we also want to enable a pull-up (&lt;code&gt;PULLTYPESEL=1&lt;/code&gt;, &lt;code&gt;PULLUDEN=0&lt;/code&gt;), then the pad control field is 0x14.  The .dts entry would then be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;amp;am33xx_pinmux {&#xA;    spi1_pins_s0: spi1_pins_s0 {&#xA;        pinctrl-single,pins = &amp;lt;&#xA;            0x164 0x14 /* ecap0_in_pwm0_out.spi1_sclk */&#xA;        &amp;gt;;&#xA;    };&#xA;};&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastActivityDate="2015-07-31T13:43:05.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3982" PostTypeId="1" AcceptedAnswerId="3983" CreationDate="2015-08-03T14:48:51.097" Score="8" ViewCount="1124" Body="&lt;p&gt;I'm considering using the i.MX 7 processor for a new design.  The overall architecture appears to be similar to the i.MX 6, at least from comparing the block diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What mechanisms exist for communication between the Cortex-M4 core and the Cortex-A7 core(s)?  Does the stock Linux kernel provide any such mechanisms (which likely would assume a specific RTOS or firmware running on the M4), or is this custom firmware for both processors in Freescale's offerings?  The processors appear to share pretty much everything, so I would imagine you could implement a shared memory buffer with an atomic semaphore, or something along those lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, I would like to have the M4 firmware &quot;take over&quot; some peripherals while the A7 processor is in a low-power mode (i.e. manage periodic communication transfers or some data analysis until something &quot;interesting&quot; happens that requires the A7 to do something).  From the block diagrams and the i.MX 6 SoloX reference manual, it appears the hardware doesn't preclude this.&lt;/p&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastEditorUserId="2642" LastEditDate="2018-01-29T14:27:17.190" LastActivityDate="2021-10-21T23:05:25.787" Title="What IPC mechanisms exist for the i.MX 7 (or i.MX 6) multi-core processors?" Tags="|design|embedded-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3983" PostTypeId="2" ParentId="3982" CreationDate="2015-08-04T23:56:09.397" Score="6" Body="&lt;p&gt;The generic term for multi-core systems that run different kernels (or bare metal code) on different cores is Asymmetric multi-processing (&lt;strong&gt;AMP&lt;/strong&gt;). Other related  terms are &lt;a href=&quot;https://en.wikipedia.org/wiki/Heterogeneous_computing&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Heterogeneous Computing&lt;/strong&gt;&lt;/a&gt;, which refers to multiple cores of different types in the same system, and Multiprocessor System-on-Chip (&lt;a href=&quot;https://en.wikipedia.org/wiki/MPSoC&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;MPSoC&lt;/strong&gt;&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;Freescale happens to have an &lt;a href=&quot;http://www.freescale.com/files/32bit/doc/brochure/PWRARBYNDBITSRAS.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;overview document (PDF)&lt;/a&gt; discussing some of the terminology and design issues relating to such systems, which may be good background, though it is discussing a different OS (QNX) and product line (PowerPC). A frequently discussed free software combination is Linux on a processor an &lt;a href=&quot;http://www.freertos.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;FreeRTOS&lt;/a&gt; on the microcontroller. Several existing chips both from Freescale and other vendors have architectures similar to the i.MX7, such as TI's &lt;a href=&quot;https://en.wikipedia.org/wiki/OMAP#OMAP_4&quot; rel=&quot;nofollow noreferrer&quot;&gt;OMAP 4/5&lt;/a&gt;. There is also a lot of &lt;a href=&quot;http://blog.idv-tech.com/2014/02/26/zedboard-linux-freertos-amp-board-bringup-guide/&quot; rel=&quot;nofollow noreferrer&quot;&gt;discussion of and example code for&lt;/a&gt; AMP on Xilinx's dual-core Zynq ARM+FPGA chips.&lt;/p&gt;&#xA;&lt;p&gt;Since 2012 or so the mainline Linux kernel has had optional support for the &lt;a href=&quot;http://lxr.free-electrons.com/source/Documentation/remoteproc.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Remote Processor&lt;/a&gt; (remoteproc) and &lt;a href=&quot;http://lxr.free-electrons.com/source/Documentation/rpmsg.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Remote Processor Messaging&lt;/a&gt; (rpmsg) frameworks. remoteproc is a generic interface for the Linux kernel to control the &amp;quot;remote&amp;quot; core (power on/off, load firmware). rpmsg is the generic interface for sending messages between cores. The linked documentation pages detail their APIs and give more context. A developer would need to write a custom application-specific Linux kernel driver using this framework to expose I/O to userspace, and then need to write, eg, FreeRTOS firmware that &amp;quot;speaks&amp;quot; rpmsg. The rpmsg framework generally seems to use shared memory to actually implement communication between the cores.&lt;/p&gt;&#xA;&lt;p&gt;If one does not exist already, I assume Freescale will soon release documentation and examples specific to AMP with the i.MX7 and Linux. You might want to keep an eye on the &lt;a href=&quot;http://git.freescale.com/git/cgit.cgi/imx/linux-2.6-imx.git/refs/heads&quot; rel=&quot;nofollow noreferrer&quot;&gt;Freescale Linux git repositories&lt;/a&gt; as they implement i.MX7 support; these branches will presumably be merged into mainline in due time.&lt;/p&gt;&#xA;" OwnerUserId="35451" OwnerDisplayName="bnewbold" LastEditorUserId="35451" LastEditDate="2021-10-21T23:05:25.787" LastActivityDate="2021-10-21T23:05:25.787" CommentCount="2" CommunityOwnedDate="2021-10-21T23:05:25.787" ContentLicense="CC BY-SA 4.0" />
  <row Id="3984" PostTypeId="1" AcceptedAnswerId="3985" CreationDate="2015-07-29T14:56:20.197" Score="6" ViewCount="155" Body="&lt;p&gt;I have discovered a problem when using the SPI peripheral in slave mode on an Atmel SAM4S.  Occasionally words can become bit-shifted.  I can force this condition by enabling the peripheral some number of clocks after &lt;code&gt;NSS&lt;/code&gt; goes active.  I do this by waiting for NSSR, then using a delay loop to wait a specific time before I enable the peripheral.  I have confirmed on a scope that when the peripheral is enabled &lt;code&gt;n&lt;/code&gt; clocks after &lt;code&gt;NSS&lt;/code&gt; goes active, the words are shifted by &lt;code&gt;n&lt;/code&gt; bits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the peripheral gets in this state, the only way to recover is to perform a reset (i.e. set the &lt;code&gt;SWRST&lt;/code&gt; in the control register).  Disabling and re-enabling is not sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This issue has also been observed on a SAM9G45 MPU.&lt;/p&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastEditorUserId="1586" LastEditDate="2015-08-20T17:31:37.223" LastActivityDate="2015-08-20T17:31:37.223" Title="Why are words bit-shifted for an Atmel SAM4S SPI slave?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3985" PostTypeId="2" ParentId="3984" CreationDate="2015-07-29T14:56:20.197" Score="5" Body="&lt;p&gt;This is due to the design of their SPI peripheral.  In slave mode, the shift register is shifted each &lt;code&gt;SPCK&lt;/code&gt; (assuming &lt;code&gt;NSS&lt;/code&gt; is low and &lt;code&gt;SPIENS&lt;/code&gt; is set).  There is an internal counter (not shown in the block diagram) for the shift register, and when it reaches the word size, the contents are shifted to &lt;code&gt;SPI_RDR&lt;/code&gt; and from &lt;code&gt;SPI_TDR&lt;/code&gt;.  Thus if you have a word size of 8, and only 7 clocks are present for the first &lt;code&gt;NSS&lt;/code&gt;, on the next &lt;code&gt;NSS&lt;/code&gt;, the shift register contents will not be transferred until after the first clock cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &lt;code&gt;NSS&lt;/code&gt; goes high, this counter is not reset.  Similarly, setting &lt;code&gt;SPIDIS&lt;/code&gt; does not reset the counter.  Only &lt;code&gt;SWRST&lt;/code&gt; resets the counter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the method described to force this condition, glitches on &lt;code&gt;SPCK&lt;/code&gt; during &lt;code&gt;NSS&lt;/code&gt; (such as from EMI) can also result in a bit-shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/udIfG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/udIfG.png&quot; alt=&quot;SAM4S Slave Mode Block Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastActivityDate="2015-07-29T14:56:20.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3986" PostTypeId="1" AcceptedAnswerId="3987" CreationDate="2015-07-28T18:26:32.947" Score="13" ViewCount="157" Body="&lt;p&gt;For an embedded Linux system, if I have two or more network interfaces, how do I ensure that they always get the same interface names every boot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, I want, for example, eth0 to always map to one physical Ethernet port, eth1 to the next, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Linux &quot;distribution&quot; is home-grown, and I use devtmpfs for populating /dev. I use busybox for init (and most everything else), along with custom init scripts for system startup and shutdown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not need hotplug facilities of mdev or udev -- I'm referring to &quot;fixed&quot; Ethernet ports.&lt;/p&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastEditorUserId="2642" LastEditDate="2016-08-15T14:18:30.503" LastActivityDate="2016-08-15T14:18:30.503" Title="How do you ensure physical network interfaces always get the same interface name across reboots on an embedded Linux system?" Tags="|embedded-systems|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3987" PostTypeId="2" ParentId="3986" CreationDate="2015-07-28T18:26:32.947" Score="7" Body="&lt;p&gt;This works for me with Linux 3.9.0 on an x86_64 architecture.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/bin/sh&#xA;&#xA;# This assumes the interfaces come up with default names of eth*.&#xA;# The interface names may not be correct at this point, however.&#xA;# This is just a way to get the PCI addresses of all the active&#xA;# interfaces.&#xA;PCIADDRLIST=&#xA;for dir in /sys/class/net/eth* ; do&#xA;  [ -e $dir/device ] &amp;amp;&amp;amp; {&#xA;    PCIADDRLIST=&quot;`readlink -f $dir/device` ${PCIADDRLIST}&quot;&#xA;  }&#xA;done&#xA;&#xA;# Now assign the interface names from an ordered list that maps&#xA;# to the PCI addresses of each interface.&#xA;&#xA;# IFNAMES could come from some config file.  &quot;dummy&quot; is needed because of&#xA;# my limited tr- and awk-fu.&#xA;IFNAMES=&quot;eth0 eth1 eth2 dummy&quot;&#xA;&#xA;for dir in `echo ${PCIADDRLIST} | tr &quot; &quot; &quot;\n&quot; | sort` ; do&#xA;  [ -e $dir/net/*/address ] &amp;amp;&amp;amp; {&#xA;    MACADDR=`cat $dir/net/*/address`&#xA;    IFNAME=`echo $IFNAMES | awk '{print $1}'`&#xA;    IFNAMES=`echo $IFNAMES | awk '{ for (i=2; i&amp;lt;=NF; i++) printf &quot;%s &quot;, $i; }'`&#xA;    echo -n &quot;$IFNAME &quot;&#xA;    nameif $IFNAME mac=$MACADDR&#xA;  }&#xA;done&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastActivityDate="2015-07-28T18:26:32.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3988" PostTypeId="2" ParentId="3986" CreationDate="2015-08-12T23:43:15.477" Score="2" Body="&lt;p&gt;You mentioned that you do not &lt;strong&gt;need&lt;/strong&gt; udev but if you are not opposed to installing it then you may want to consider using &lt;a href=&quot;http://linux.dell.com/biosdevname/&quot; rel=&quot;nofollow&quot;&gt;biosdevname&lt;/a&gt;. There is a great &lt;a href=&quot;http://www.freedesktop.org/wiki/Software/systemd/PredictableNetworkInterfaceNames/&quot; rel=&quot;nofollow&quot;&gt;wiki page&lt;/a&gt; on &lt;a href=&quot;http://freedesktop.org&quot; rel=&quot;nofollow&quot;&gt;freedesktop.org&lt;/a&gt; that discusses this issue. As noted on that page, if you're using udev version 197 or higher then you don't even need biosdevname because udev will already do what you need.&lt;/p&gt;&#xA;" OwnerUserId="131" OwnerDisplayName="Olek Wojnar" LastActivityDate="2015-08-13T13:35:25.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3989" PostTypeId="1" AcceptedAnswerId="3990" CreationDate="2015-07-31T16:08:51.560" Score="8" ViewCount="173" Body="&lt;p&gt;I have an Atmel SAM4S MCU with a 12-bit internal ADC.  I would like to achieve 16-bit resolution at a sample rate of 500 Hz.  The input to the ADC is fairly low-frequency (about 20 Hz max frequency content), and low noise (less than one 12-bit lsb RMS).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I achieve 16-bit resolution with this processor's ADC?&lt;/p&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastEditorUserId="1586" LastEditDate="2015-08-20T17:31:40.287" LastActivityDate="2015-08-20T17:31:40.287" Title="How can I increase my processor's ADC resolution?" Tags="|electrical-engineering|embedded-systems|adc|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3990" PostTypeId="2" ParentId="3989" CreationDate="2015-07-31T17:14:47.303" Score="6" Body="&lt;p&gt;You can increase the effective resolution by intentionally oversampling the input signal. Here is brief summary of the idea from an application note on the topic provided by Atmel:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The theory behind ‘Oversampling and decimation’ is rather complex, but using the method is fairly easy. The technique requires a higher amount of samples. These extra samples can be achieved by oversampling the signal. For each additional bit of resolution, n, the signal must be oversampled four times. Which frequency to sample the input signal with, is given by Equation 3-1. To get the best possible representation of a analog input signal, it is necessary to oversample the signal this much, because a larger amount of samples will give a better representation of the input signal, when averaged.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The full application note can be obtained here: &lt;a href=&quot;http://www.atmel.com/images/doc8003.pdf&quot;&gt;http://www.atmel.com/images/doc8003.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your specific case, you would need to sample at about 8KHz to get the equivalent of 16 bits of resolution out of your 12-bit ADC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I arrived at this number:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;f_oversample_rate = f_current_sample_rate * (4 * (desired_bits - current_bits))&#xA;f_oversample_rate = 500Hz * (4 * (16 - 12))&#xA;f_oversample_rate = 500Hz * (4 * 4)&#xA;f_oversample_rate = 500Hz * 16&#xA;f_oversample_rate = 8KHz&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2659" OwnerDisplayName="skrrgwasme" LastActivityDate="2015-07-31T17:24:12.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3991" PostTypeId="1" AcceptedAnswerId="3992" CreationDate="2015-08-04T13:38:27.773" Score="11" ViewCount="1541" Body="&lt;p&gt;If I have a 10 MHz digital signal, how fast do I need to sample with a logic analyzer to avoid any bit errors?  Does the minimum sample rate increase if I want to attach a protocol decoder (e.g. SPI)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem&quot;&gt;Nyquist-Shannon sampling theorem&lt;/a&gt; and the fact that a 10 MHz square wave has significant energy at higher harmonics (30 MHz, 50 MHz, etc.).  However, a logic analyzer doesn't need to perfectly sample or reproduce a square wave to know if it is a logic 1 or 0.  I'm looking for what is practically necessary to avoid bit errors or protocol decoding errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the Saleae Logic Pro 16.&lt;/p&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastEditorUserId="1586" LastEditDate="2015-08-17T19:01:02.330" LastActivityDate="2015-08-17T19:01:02.330" Title="How fast should I sample with a logic analyzer?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3992" PostTypeId="2" ParentId="3991" CreationDate="2015-08-05T15:21:42.220" Score="7" Body="&lt;p&gt;The least you can do is given by &lt;a href=&quot;https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem&quot;&gt;Shannon-Nyquist theorem&lt;/a&gt;: to be able to read a signal, you shall sample at least twice its frequency. Here, you can sample at 20MS/s (Mega sample per second). Nevertheless, 2 points for one period is very few, and introduce much jitter on the rebuilt signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your logic analyzer won't share your signal clock. This is a common case in transmission, and I would advise you to do the same as most UART/SPI/I2C/... interface: take 3 points per state. This allow to filter EMC noise using a majority filter, and, when analyzing, allows to see significant skew between your signals. Please note that the clock of a 10MHz SPI bus goes both high and low every 100ns period, so you need to sample at 6 times your SPI frequency (60MS/s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;60MS/s is common for logical analyzers, and this will allow you to detect excessive skew (between your data and your clock) for most protocols (including SPI). Nonetheless, with a non-symmetric clock (with tHIGH != tLOW), you'll need to sample fast enough to sample the shortest part of the clock. If you clock is HIGH for 1µs and LOW for 9µs, you've got a 100kHz signal, but need 3MS/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer only considers the signal you want to analyze. For EMC issues, we usually use analog scopes with a sampling frequency 10 times higher than the highest frequency in the system, but it's a bit expensive, and off-topic, considering that the question is to decode an SPI bus.&lt;/p&gt;&#xA;" OwnerUserId="2664" OwnerDisplayName="Jacen" LastActivityDate="2015-08-05T15:21:42.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="3993" PostTypeId="1" AcceptedAnswerId="3994" CreationDate="2015-07-28T23:14:09.687" Score="13" ViewCount="353" Body="&lt;p&gt;I am thinking on analog circuits, whose circuit design could be read (or, at least, significantly altered) from software. Do they already exist?&lt;/p&gt;&#xA;" OwnerUserId="9" OwnerDisplayName="peterh" LastEditorUserId="1586" LastEditDate="2015-08-20T17:31:42.743" LastActivityDate="2015-08-20T17:31:42.743" Title="Do &quot;analog FPGAs&quot; exist?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="3994" PostTypeId="2" ParentId="3993" CreationDate="2015-07-29T00:56:32.823" Score="9" Body="&lt;p&gt;There are many chips that can be configured by choices made in software, which then have a perfectly analogue behaviour that gets altered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blocks that work like that go from Analogue Switch arrays and/or Multiplexers all the way past Programmable-Gain amplifiers through to complete integrated chips that have a combination of blocks, such as programmable gain, a whole bunch of analogue switches, analogue mixers and multiplexers and all such. Analog Devices are kings of doing things like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example a programmable gain amplifier may have only a discrete number of gains, 4, 16 or 32, or even 256 for a very high-end, luxury one, but it will gain any voltage on the input by that number, any real voltage that can be made by something, so they are truly analogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe there are even some actual FPGA-type devices, much like DSPICs, that include some of those kinds of blocks into the programmable gate array along with DACs (Digi-&gt;Analogue) and ADCs (Analogue-&gt;Digi), but no types spring to mind, so I could just be misremembering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sure many DSP System on Chip (SoC) devices made for old-school analogue video and/or multi-track audio manipulation and conversion would be a mix of such analogue trickery before the signal would be turned digital and processed further as a digital signal. Which then in some cases would be turned back to analogue and possibly again multiplexed or modulated into some kind of antenna signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No modern-to-future-day applications for such a complex SoC other than very specific Audio tools or metrology spring to mind right now, but it's not unlikely they still exist, since no one person will have heard of everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in more ways than one, the answer seems to be yes, they exist. Though of course not exactly like an FPGA, because a digital beast can never be fully compared to an analogue one.&lt;/p&gt;&#xA;" OwnerUserId="2683" OwnerDisplayName="Asmyldof" LastActivityDate="2015-07-29T00:56:32.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="3995" PostTypeId="2" ParentId="3993" CreationDate="2015-07-29T02:12:35.380" Score="11" Body="&lt;p&gt;Yes, it's called Field-programmable Analog Array (FPAA):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Field-programmable_analog_array&quot;&gt;https://en.wikipedia.org/wiki/Field-programmable_analog_array&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are exactly the analog version of the FPGAs. There are many different subtypes (current or voltage driven, discrete or continous time). They are built up from CABs (configurable analogous blocks), and from interconnects between them.&lt;/p&gt;&#xA;" OwnerDisplayName="Jacques" LastActivityDate="2015-08-01T19:58:32.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4000" PostTypeId="1" CreationDate="2015-08-07T20:25:01.297" Score="5" ViewCount="59" Body="&lt;p&gt;I am interested in a microcontroller for a low power application. I have been advise to look at MSP430 or Microchip PIC microcontrollers. I also wonder if &lt;a href=&quot;http://www.arm.com/products/processors/cortex-m/cortex-m0.php&quot; rel=&quot;nofollow&quot;&gt;ARM-Cortex-M0&lt;/a&gt; is a good choice too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a high level the system will have two analog sensors, few GPIO to control LED, and actuation mechanisms. The systems is intended to be powered with standard batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the critical parameters that warrant attention in researching for a suitable low power micro controllers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I posted this question on engineering SE and have not received any responses. I believe the question might be better suited for this forum.&lt;/p&gt;&#xA;" OwnerUserId="282" OwnerDisplayName="user277228" LastEditorUserId="1586" LastEditDate="2015-08-17T19:00:55.173" LastActivityDate="2015-08-17T19:00:55.173" Title="How to select a micro controller for a low power applications?" Tags="|electrical-engineering|embedded-systems|power|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4001" PostTypeId="2" ParentId="4000" CreationDate="2015-08-08T03:39:41.847" Score="8" Body="&lt;p&gt;Well, what are the use cases of your system? That's what will drive this decision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some considerations:&#xA;If the system spends 99% of its time sleeping then the key parameter is not going to be max current draw. It is actually the sleep current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest looking into CoreMark when comparing between controllers. CoreMark has a &lt;a href=&quot;https://www.eembc.org/ulpbench/&quot;&gt;power metric&lt;/a&gt; which is basically performance/mW. So if you wake up from sleep, how efficiently can you do what you need to do (perhaps log data) before going back to sleep. Unfortunately this list is pretty short right now, hopefully it will grow over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a data logger / metering application a key thing to consider is how much power your memory is going to take (in the uC or external), because that will be a major consumer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also consider various sleep modes. Does the controller have a sleep mode that supports internal ADC that can wake up the core and interrupt when a conversion is done if you need it to? Does it have an RTC that can run in sleep mode that can wake the controller up to poll for data periodically or at least maintain system time if you need it to? When the controller is in those sleep modes with select peripherals still operating what is the current draw? High end MSP430s like the 5 series support 5 sleep modes. The 1 series only has two I think. Do you even need all of that? If not, even a PIC10 will let you wake up from sleep on a pin transition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, many controllers will claim to be the best for low power, but your your exact application dictates which will be best suited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.eevblog.com/projects/ucurrent/&quot;&gt;Look into Dave Jone's uCurrent if you ever need to do really low current (nA) measurements.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="Nick" LastActivityDate="2015-08-11T22:07:27.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4007" PostTypeId="1" AcceptedAnswerId="4008" CreationDate="2015-08-07T00:45:24.977" Score="2" ViewCount="1705" Body="&lt;p&gt;I am currently working on project where the commands for a PC to the MSP430 will be communicated via UART. These commands need to be converted to binary to be able to written to the SPI bus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example 'a' from PC is translated to 0x61 on the microcontroller. What is really needed is hex 'a' Thus&lt;br&gt;&#xA;ASCII 'a' from PC = 0x61 =&gt; need to translate hex 0x0A  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question :&lt;/strong&gt; What is the most simplest way to convert ASCII to binary? I am using &quot;C&quot;. &lt;/p&gt;&#xA;" OwnerUserId="110" OwnerDisplayName="Mahendra Gunawardena" LastEditorUserId="1586" LastEditDate="2015-08-20T17:31:50.400" LastActivityDate="2015-08-20T17:31:50.400" Title="How to convert ASCII to binary to facilitate a UART to SPI bridge for an MSP430?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4008" PostTypeId="2" ParentId="4007" CreationDate="2015-08-07T13:43:00.970" Score="5" Body="&lt;p&gt;The input from the PC is a character, i.e. an ASCII-value. To understand how this is saved, have a look at the &lt;a href=&quot;http://www.asciitable.com/&quot; rel=&quot;noreferrer&quot;&gt;ASCII-table&lt;/a&gt;. The numbers '0' to '9' are represented by the ASCII-codes &lt;code&gt;0x30&lt;/code&gt; to &lt;code&gt;0x39&lt;/code&gt;. The capital letters 'A' to 'Z' are represented by &lt;code&gt;0x41&lt;/code&gt; to &lt;code&gt;0x5A&lt;/code&gt;, and the small letters 'a' to 'z' by &lt;code&gt;0x61&lt;/code&gt; to &lt;code&gt;0x7A&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could of course use a function like &lt;code&gt;sscanf&lt;/code&gt;, as suggested by @JohnO'M. Such functions (&lt;code&gt;printf&lt;/code&gt;, &lt;code&gt;scanf&lt;/code&gt;, ...) have many features but thus require a lot of power and memory&lt;sup&gt;[Citation needed]&lt;/sup&gt;, which we often don't want to spend on a microcontroller. A simple alternative would be to distinguish the 3 possible cases (0-9 or A-F or a-f) and subtract the correct number from the ASCII code to get to the result:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if( (s &amp;gt;= 0x30) &amp;amp;&amp;amp; (s &amp;lt;= 0x39)) {        // 0-9&#xA;    x = s - 0x30;&#xA;}&#xA;else {&#xA;    if( (s &amp;gt;= 0x41) &amp;amp;&amp;amp; (s &amp;lt;= 0x46)) {    // A-F&#xA;        x = s - 0x37;&#xA;    }&#xA;    else {&#xA;        if( (s &amp;gt;= 0x61) &amp;amp;&amp;amp; (s &amp;lt;= 0x66)) { // a-f&#xA;            x = s - 0x57;&#xA;        }&#xA;        else {                            // wrong input&#xA;            x = 0xFF;&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2669" OwnerDisplayName="hbaderts" LastActivityDate="2015-08-07T13:43:00.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4009" PostTypeId="1" AcceptedAnswerId="4010" CreationDate="2015-08-07T23:31:06.100" Score="6" ViewCount="201" Body="&lt;p&gt;I am currently working on develop an IOT architecture.  The sensing devices are to be integrated with microcontroller. The microcontroller will be integrated to some type of RF communication device.  As the community is well aware RF communication protocols that will fit the bill. Once in consideration are Bluetooth smart/BLE, uIP, zigbee and Zwave. I intend to use module primarily to address FCC certification issues. At this point Bluetooth smart/BLE and Zigbee are the prime candidates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; From an Embedded Systems design standpoint what are the pros and cons of using Bluetooth smart/BLE over Zigbee or vice versa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/UIP_(micro_IP)&quot; rel=&quot;nofollow&quot;&gt;uIP (micro IP)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" OwnerDisplayName="Mahendra Gunawardena" LastEditorUserId="8055" LastEditDate="2016-12-09T16:46:50.353" LastActivityDate="2018-05-22T16:44:03.267" Title="What are the pros and cons of using Zibgee over BLE or vice versa for IOT application?" Tags="|electrical-engineering|embedded-systems|iot|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4010" PostTypeId="2" ParentId="4009" CreationDate="2015-08-14T03:41:33.990" Score="2" Body="&lt;p&gt;Those two standards have numerous similarities and the differences really come down to flexibility versus simplicity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bluetooth_low_energy&quot; rel=&quot;nofollow&quot;&gt;Bluetooth&lt;/a&gt; is ubiquitous, feature-rich, and very mature. Newer cell phones will easily interface with your architecture as will a plethora of other devices. It is also relatively faster with a 1Mbit/s transmission data rate (about a quarter of that when you consider overhead). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/ZigBee&quot; rel=&quot;nofollow&quot;&gt;ZigBee&lt;/a&gt; is designed and optimized for an IOT world. Your devices will not have the flexibility of BLE but they will be less expensive and simpler. Don't underestimate simplicity, the simpler a device is the fewer insidious problems you'll have to troubleshoot as you go along! It is also relatively slower with a 250 kbit/s transmission data rate (still well within the rate requirement you mention in your comment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, ZigBee is the better choice unless you absolutely need a feature that only BLE can provide.&lt;/p&gt;&#xA;" OwnerUserId="131" OwnerDisplayName="Olek Wojnar" LastActivityDate="2015-08-14T03:41:33.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4011" PostTypeId="1" AcceptedAnswerId="4012" CreationDate="2015-08-03T23:35:45.480" Score="7" ViewCount="119" Body="&lt;p&gt;I need to configure the clock systems for a MSP430F5528 microcontroller. The current development board has a 24MHZ crystal attach to XT2 pins (XT2IN/XT2OUT). The current design will need to support the follow peripherals.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;UART&lt;/li&gt;&#xA;&lt;li&gt;ADC&lt;/li&gt;&#xA;&lt;li&gt;SPI&lt;/li&gt;&#xA;&lt;li&gt;Timers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Few GPIO's will be need for system power up. The system is will part of a low power application. The power budget has not yet been determined.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;About the MSP430 clock system&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MSP430 for the most part has three internal clocks&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;MCLK - Master Clock - CPU&lt;/li&gt;&#xA;&lt;li&gt;SMCLK - Sub Master Clock - fast peripherals &lt;/li&gt;&#xA;&lt;li&gt;ACLK - Auxiliary Clok - Slow peripherals&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wYV4t.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wYV4t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What needs to be considered when configuring the clock system base on the above criteria? &lt;/p&gt;&#xA;" OwnerUserId="110" OwnerDisplayName="Mahendra Gunawardena" LastEditorUserId="1586" LastEditDate="2015-08-20T17:31:23.250" LastActivityDate="2015-08-20T17:31:23.250" Title="What needs to be considered when configuring the clock system for a MSP430F5xxx micro-controller?" Tags="|electrical-engineering|embedded-systems|power|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4012" PostTypeId="2" ParentId="4011" CreationDate="2015-08-04T14:01:09.120" Score="4" Body="&lt;p&gt;It depends a bit on the quality you expect from some of the peripherals and what their qualities are at given clock speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will you want a very high speed SPI interface and can this specific chip support that on the Fast Auxiliary Clock? Or are you going to do 1MHz SPI at most? So also for all the other devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's of course a nice boasting thing to be able to say that everything runs on many-MHz, but if it's not needed it will create more power consumption, more heat and more programming effort for no result. It can be tempting to just do everything at maximum speed and highest accuracy, but in almost all cases a good systems design will start with indexing what are the minimum speed and accuracy for something to work and then taking the appropriate margins. (Caveat: If you may need a higher speed later, it may be advantageous to make the hardware such that scaling up is only a firmware issue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for activating the Crystal; I don't actually know this particular chip, but it seems compatible in build-up to the ones that I know that have a configurable clock source, that is selected at power-up, in which case setting the crystal as a source is a simple configuration and possibly the configuration of all the other clocks is too. If that isn't the case and you need to actively switch while already running the core, you need to make 100% sure the crystal is running before you switch over. The datasheet will make very clear mention of the procedures needed for that if this is the case. Usually it's something along the line of: Activate Clock --&gt; Wait for a bit in a register to be set --&gt; Switch over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to the ADC, that can become a very complicated case. If you need very low sampling jitter or highly accurate timing (for highly accurate control or very (very) high grade Audio) you will quickly get to the domain of using the Crystal as a clock source. Where you'll then also need to make some hardware considerations. Both in the Analogue domain and on the Crystal end, since crystal accuracy is influenced by the loading capacitance (which includes traces and pins).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But those applications are few and far between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases your timers and your UART will determine the clocks you need, since they are most sensitive to clock drift and/or offset. If you need a timer just for low-speed timing you are better off relying on a watch crystal, since both the crystal and the internal driving circuitry for such low frequency crystals are usually better suited to high accuracy and low drift over long time.&lt;/p&gt;&#xA;" OwnerUserId="2683" OwnerDisplayName="Asmyldof" LastActivityDate="2015-08-04T14:01:09.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4013" PostTypeId="1" AcceptedAnswerId="4014" CreationDate="2015-08-01T13:39:10.427" Score="12" ViewCount="19960" Body="&lt;p&gt;I have been working on and off on small embedded systems project on and off. Some of these projects used an ARM Cortex-M4 base processor. In the project folder there is a &lt;strong&gt;startup.s&lt;/strong&gt; file. Inside that file I noted the following two command lines.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;;******************************************************************************&#xA;;&#xA;; &amp;lt;o&amp;gt; Stack Size (in Bytes) &amp;lt;0x0-0xFFFFFFFF:8&amp;gt;&#xA;;&#xA;;******************************************************************************&#xA;Stack   EQU     0x00000400&#xA;&#xA;;******************************************************************************&#xA;;&#xA;; &amp;lt;o&amp;gt; Heap Size (in Bytes) &amp;lt;0x0-0xFFFFFFFF:8&amp;gt;&#xA;;&#xA;;******************************************************************************&#xA;Heap    EQU     0x00000000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How does one define the size of the &lt;strong&gt;heap&lt;/strong&gt; and &lt;strong&gt;stack&lt;/strong&gt; for a microcontroller? Is there any specific information in the datasheet to guide to arrive at the correct value? If so, what should one look for in the datasheet?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://community.arm.com/docs/DOC-9701&quot; rel=&quot;nofollow noreferrer&quot;&gt;Decoding the Startup file for ARM Cortex M4&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.lpcware.com/content/project/lightweight-ip-lwip-networking-stack/configuring-lwip/default-stack/heap-sizes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Default stack/heap sizes&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" OwnerDisplayName="Mahendra Gunawardena" LastEditorUserId="330" LastEditDate="2016-11-04T19:17:42.070" LastActivityDate="2016-11-04T19:17:42.070" Title="Defining the heap and stack size for an ARM Cortex-M4 microcontroller?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4014" PostTypeId="2" ParentId="4013" CreationDate="2015-08-01T14:06:14.877" Score="12" Body="&lt;p&gt;Stack and heap are software concepts, not hardware concepts. What the hardware provides is memory. Defining zones of memory, one of which is called “stack” and one of which is called “heap”, is a choice of your program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hardware does help with stacks. Most architectures have a dedicated register called the stack pointer. Its intended use is that when the program makes a function call, the function parameters and the return address are pushed to the stack, and they are popped when the function terminates and returns to its caller. Pushing onto the stack means writing at the address given by the stack pointer and decrementing the stack pointer accordingly (or incrementing, depending in which direction the stack grows). Popping means incrementing (or decrementing) the stack pointer; the return address is read from the address given by the stack pointer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some architectures (not ARM though) have a subroutine call instruction that combines a jump with writing to the address given by the stack pointer, and a subroutine return instruction that combines reading from the address given by the stack pointer and jumping to this address. On ARM, the address saving and restoring is done in the LR register, the call and return instructions do not use the stack pointer. There are however instructions to facilitate writing or reading multiple registers to the address given by the stack pointer, to push and pop function arguments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To choose the heap and stack size, the only relevant information from the hardware is how much total memory you have. You then make your choice depending on what you want to store in memory (allowing for code, static data, and other programs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A program would typically uses these constants to initialize some data in memory that will be used by the rest of the code, such as the address of the top of the stack, maybe a value somewhere to check for stack overflows, bounds for the heap allocator, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;http://community.arm.com/docs/DOC-9701&quot;&gt;the code you're looking at&lt;/a&gt;, the &lt;code&gt;Stack_Size&lt;/code&gt; constant is used to reserve a block of memory in the code area (via a &lt;code&gt;SPACE&lt;/code&gt; directive in ARM assembly). The upper address of this block is given the label &lt;code&gt;__initial_sp&lt;/code&gt;, and it is stored in the vector table (the processor uses this entry to set SP after a software reset) as well as exported for use in other source files. The &lt;code&gt;Heap_Size&lt;/code&gt; constant is similarly used to reserve a block of memory and labels to its boundaries (&lt;code&gt;__heap_base&lt;/code&gt; and &lt;code&gt;__heap_limit&lt;/code&gt;) are exported for use in other source files.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;; Amount of memory (in bytes) allocated for Stack&#xA;; Tailor this value to your application needs&#xA;; &amp;lt;h&amp;gt; Stack Configuration&#xA;;   &amp;lt;o&amp;gt; Stack Size (in Bytes) &amp;lt;0x0-0xFFFFFFFF:8&amp;gt;&#xA;; &amp;lt;/h&amp;gt;&#xA;&#xA;Stack_Size      EQU     0x00000400&#xA;&#xA;                AREA    STACK, NOINIT, READWRITE, ALIGN=3&#xA;Stack_Mem       SPACE   Stack_Size&#xA;__initial_sp&#xA;&#xA;&#xA;; &amp;lt;h&amp;gt; Heap Configuration&#xA;;   &amp;lt;o&amp;gt;  Heap Size (in Bytes) &amp;lt;0x0-0xFFFFFFFF:8&amp;gt;&#xA;; &amp;lt;/h&amp;gt;&#xA;&#xA;Heap_Size       EQU     0x00000200&#xA;&#xA;                AREA    HEAP, NOINIT, READWRITE, ALIGN=3&#xA;__heap_base&#xA;Heap_Mem        SPACE   Heap_Size&#xA;__heap_limit&#xA;&#xA;…&#xA;__Vectors       DCD     __initial_sp               ; Top of Stack&#xA;                DCD     Reset_Handler              ; Reset Handler&#xA;                DCD     NMI_Handler                ; NMI Handler&#xA;&#xA;…&#xA;&#xA;                 EXPORT  __initial_sp&#xA;                 EXPORT  __heap_base&#xA;                 EXPORT  __heap_limit&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="89" OwnerDisplayName="Gilles" LastActivityDate="2015-08-01T21:14:40.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4015" PostTypeId="2" ParentId="4013" CreationDate="2015-08-01T14:06:44.290" Score="7" Body="&lt;p&gt;The sizes of the stack and heap are defined by your application, not anywhere in the microcontroller's datasheet.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The stack&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The stack is used to store the values of local variables inside of functions, the previous values of CPU registers used for local variables (so they can be restored on exit from the function), the program address to return to when leaving those functions, plus some overhead for management of the stack itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When developing an embedded system, you estimate the maximum call depth you expect to have, add up the sizes of all the local variables in the functions in that hierarchy, then add some padding to allow for the overhead mentioned above, then add some more for any interrupts that might occur during execution of your program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative estimation method (where RAM isn't constrained) is to allocate far more stack space than you'll ever need, fill the stack with a sentinel value, then monitor how much you actually use during execution. I've seen debug versions of C language runtimes that will do this for you automatically. Then, when you've finished developing, you can reduce the stack size if you want.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The heap&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Calculating the size of the heap you need can be a trickier. The heap is used for dynamically allocated variables, so if you use &lt;code&gt;malloc()&lt;/code&gt; and &lt;code&gt;free()&lt;/code&gt; in a C language program, or &lt;code&gt;new&lt;/code&gt; and &lt;code&gt;delete&lt;/code&gt; in C++, that's where those variables live. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in C++ especially, there can be some hidden dynamic memory allocation going on. For example, if you have statically allocated objects, the language requires that their destructors be called when the program exits. I'm aware of at least one runtime where the addresses of the destructors are stored in a dynamically allocated linked list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to estimate the size of the heap you need, look at all the dynamic memory allocation in each path through your call tree, calculate the maximum and add some padding. The language runtime may provide diagnostics that you can use to monitor total heap usage, fragmentation, etc.&lt;/p&gt;&#xA;" OwnerUserId="2660" OwnerDisplayName="Niall C." LastActivityDate="2015-08-01T14:06:44.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4016" PostTypeId="2" ParentId="4013" CreationDate="2015-08-02T02:26:59.457" Score="5" Body="&lt;p&gt;In addition to the other answers, I'd like to add that when carving up RAM between stack and heap space, that you also need to consider the space for static non-constant data (e.g. file globals, function statics, and program-wide globals from a C perspective, and probably others for C++).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How stack/heap allocation works&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that the startup assembly file is one way of defining the region; the toolchain (both your build environment and run-time environment) mostly care about the symbols which define the start of stackspace (used to store the initial stack pointer in the Vector Table) and the start and end of heap space (used by the dynamic memory allocator, typically provided by your libc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In OP's example, only 2 symbols are defined, a size of stack at 1kiB and a size of heap at 0B. These values get used elsewhere to actually produce the stack and heap spaces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In @Gilles example, the sizes are defined &lt;em&gt;and used&lt;/em&gt; in the assembly file to set a stack space starting wherever and lasting the size, identified by the symbol Stack_Mem and sets a label __initial_sp at the end. Likewise for the heap, where the space is the symbol Heap_Mem (0.5kiB in size), but with labels at the beginning and end (__heap_base and __heap_limit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These get processed by the linker, which won't allocate anything within the stack space and heap space because that memory is occupied (by Stack_Mem and Heap_Mem symbols), but it can place those memories and all of the globals wherever it needs. The labels end up being symbols with no length at the given addresses. The __initial_sp is used directly for the vector table at link time, and the __heap_base and __heap_limit by your runtime code. The actual addresses of the symbols are assigned by the linker based on where it placed them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I aluded to above, these symbols don't actually have to come from a startup.s file. They can come from your linker configuration (Scatter Load file in Keil, linkerscript in GNU), and in those you can have finer grained control over placement. For example, you can force the stack to be at the beginning or end of RAM, or keep your globals away from the heap, or whatever you want. You can even specify that the HEAP or STACK just occupy whatever RAM is left over after globals are placed. NOTE though that you have to be careful that adding more static variables that your other memory will decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, each toolchain is different, and how to write the configuration file and what symbols your dynamic memory allocator will use will have to come from the documentation of your particular environment.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Stack Sizing&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As to how to determine stack size, many toolchains can give you a maximum stack depth by analyzing the function call trees of your program, IF you don't use recursion or function pointers. If you do use those, estimating a stack size and pre-filling it with cardinal values (perhaps via the entry function before main) and then checking after your program has run for a while where the max depth was (which is where the cardinal values end). If you have fully exercised your program to its limits, you'll know fairly accurately whether you can shrink the stack or, if your program crashes or no cardinal values are left, that you need to increase the stack and try again.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Heap Sizing&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Determining heap size is a bit more application dependant. If you only do dynamic allocation during startup, you can just add up the space required in your startup code (plus some overhead for memory management). If you have access to the source of your memory manager, you can know exactly what the overhead is, and possibly even write code to walk the memory to give you usage information. For applications that need dynamic runtime memory (e.g. allocating buffers for inbound ethernet frames) the best I can suggest is to carefully hone your stacksize and give the Heap everything that is left over after stack and statics.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Final note (RTOS)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;OP's question was tagged for bare-metal, but I want to add a note for RTOSes. Often (always?) each task/process/thread (I'll just write task here on out for simplicity) will be assigned a stack size when the task is created, an in addition to task stacks, there will likely be a small OS stack (used for interrupts and such)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task accounting structures and the stacks have to be allocated from somewhere, and this will often be from the overall heap space of your application. In these instances, your initial stack size often won't matter, because the OS will only use it during initialization. I have seen, for example, specifying ALL remaining space during linking be allocated to the HEAP and placing the initial stack pointer at the end of the heap to grow into the heap, knowing that the OS will allocate starting from the beginning of the heap and will allocate the OS stack just before abandoning the initial_sp stack. Then all of the space is used for allocating task stacks and other dynamically allocated memory.&lt;/p&gt;&#xA;" OwnerUserId="262" OwnerDisplayName="John O'M." LastActivityDate="2015-08-02T02:26:59.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4017" PostTypeId="1" CreationDate="2015-08-10T12:08:33.427" Score="4" ViewCount="305" Body="&lt;p&gt;MSP430F2619 has 256 bytes of information flash memory. This memory is again segmented in 4, 64 bytes segments. Segment A of this memory can't be erased during mass erase and production programming. I want to store device specific configuration data which is more than 105 bytes long. This I will store in Segment A and B. But how do I make sure that part in segment will remain even after mass erase or production programming. &lt;/p&gt;&#xA;" OwnerUserId="2666" OwnerDisplayName="prasad" LastEditorUserId="1586" LastEditDate="2015-08-17T19:00:51.687" LastActivityDate="2015-08-17T19:00:51.687" Title="Retain information memory on MSP430 in mass erase operation?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4018" PostTypeId="2" ParentId="4017" CreationDate="2015-08-11T07:42:40.317" Score="2" Body="&lt;p&gt;At runtime the LOCKA bit protects segment A from any kind of write or erase, but it also protects the &lt;em&gt;entire&lt;/em&gt; information memory from a mass erase operation. Its certainly possible to erase these areas during production programming but, depending on how you're doing the programming, you need to go more or less out of your way to do it. (It looks like for JTAG you'd need to explicitly clear LOCKA to do it, but the built in bootloader will clear it for you if you enter it from reset).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that segment A contains some device calibration data so if you need that, don't put your data over the top of it. &lt;/p&gt;&#xA;" OwnerDisplayName="Simon Jenkins" LastActivityDate="2015-08-11T07:42:40.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4019" PostTypeId="1" AcceptedAnswerId="4020" CreationDate="2015-07-31T15:28:14.623" Score="11" ViewCount="6022" Body="&lt;p&gt;A colleague mentioned that Thumb instructions were significantly slower than ARM instructions (this was for an AT91SAM7S32 processor).  Is this true?  What are the performance benefits of one instruction set over another?&lt;/p&gt;&#xA;" OwnerUserId="2642" OwnerDisplayName="Patrick" LastEditorDisplayName="user16" LastEditDate="2015-08-14T21:58:17.517" LastActivityDate="2015-08-14T21:58:17.517" Title="How do Thumb instructions compare to ARM instructions for performance?" Tags="|embedded-systems|computer-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4020" PostTypeId="2" ParentId="4019" CreationDate="2015-08-02T03:00:12.680" Score="11" Body="&lt;h2&gt;In General&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;THUMB instructions aren't to my knowledge intrinsically slower than ARM instructions, but rather are more limited in capability. If your code only needs the functionality of THUMB instructions, it will occupy less space than ARM, but would be the same number of instructions and, other things being equal, run at the same speed. If your code needs more functionality, than it would require more THUMB instructions than ARM instructions to execute and would take longer, again other things being equal (see below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THUMB is popular in microcontrollers because of the smaller size instructions for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Program space is often limited&lt;/li&gt;&#xA;&lt;li&gt;Many microcontrollers have 16-bit data buses to their internal flash&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the second reason, when your code doesn't require functionality from the ARM instruction set, THUMB code actually runs faster. This is because your instruction can be fetched in one I/O cycle from the flash instead of two. Depending on the speed of your flash interface, that second read can incur one or more wait-cycles per instruction where your CPU is simply stalled and cannot do anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This becomes less of an issue if you can copy your code to RAM before executing (which I've usually seen as 32-bit for recent ARM microcontrollers), where the only concern is code density. For that, many tools will try to find which representation is more efficient for a given function. If the compiler can produce THUMB code in fewer instructions it will, but if ARM results in fewer instructions you get ARM. This is the default mode for Keil, if I recall correctly.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Your specific chip&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For your particular chip (AT91SAM7S32) the documentation mentions that the flash controller has a prefetch buffer that can predict accesses to make things more efficient, which might improve ARM instruction execution. However, it also states that the prefetch is a &quot;dual 32-bit&quot; buffer that &quot;optimizes 16-bit accesses&quot; which is most suitable for &quot;running in Thumb mode&quot;, which seems to indicate that it isn't intended to speed up ARM instructions, but to allow your core to run faster in THUMB mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the diagrams, it looks like the flash on your chip actually has a 32bit data bus. The prefetcher seems to work by reading a whole 32bits, giving 16 to the CPU (in THUMB mode) and caching the whole 32bits. During the next cycle, when the CPU reads the second 16 bits, this time from cache, the flash controller is reading the next 32bits and caching it. In this way THUMB code can run without more than an initial wait even if the flash speed would be a bit slower than the CPU core speed. Section 19.2.2 &quot;Read Operations&quot; has more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your flash is a 32bit bus (as near as I can tell), if your CPU and Flash clocks are the same, THUMB will only give you code density over ARM. If you want your CPU core to run faster than Flash (and note, I didn't review all of the timing of this chip; I assume the CPU can run faster because they let you set wait states), than the prefetch gives a speed advantage to THUMB due to the reduction in actual flash accesses. However, that speed advantage is a per-instruction advantage. If the number of THUMB instructions vs ARM instructions is big enough, it will outweigh the per instruction speed resulting in ARM having a faster per-routine speed.&lt;/p&gt;&#xA;" OwnerUserId="262" OwnerDisplayName="John O'M." LastActivityDate="2015-08-02T03:00:12.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4021" PostTypeId="1" CreationDate="2015-08-14T01:20:24.487" Score="8" ViewCount="987" Body="&lt;p&gt;I work with a system in which the microcontroller is exposed to X-ray radiation from time to time. Exactly how much radiation is hard to say... we've never measured it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, we've seen a mysterious form of flash corruption on a small sample of our products in which random bits in code space are transitioning from 0 to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've previously attributed this to power supply issues and the controller not being reset quickly enough while data logging is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now we may be seeing it on a newer version of our product which has a higher end controller with better brown out protection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it time to consider X-rays as a possible source?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The X-rays would be released due to breaking current high currents with a vacuum interrupter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone ever heard of something like this before on nand-flash? Most of the articles I have seen have been with respect to Airport X-rays or &lt;a href=&quot;http://www.spansion.com/Support/Application%20Notes/Xray_Inspection_Test_Conditions_AN.pdf&quot; rel=&quot;nofollow&quot;&gt;X-ray inspection used in PCB FAB&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="Nick" LastEditorUserId="1586" LastEditDate="2015-08-17T19:00:41.500" LastActivityDate="2022-04-19T18:39:18.250" Title="Flash Corruption - Power Supply or X-ray Radiation?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4022" PostTypeId="2" ParentId="4021" CreationDate="2015-08-14T09:13:27.477" Score="6" Body="&lt;p&gt;There seem to be plenty of papers suggesting X-rays could be a real problem with NAND flash. I've not read it but the abstract for &lt;a href=&quot;https://ieeexplore.ieee.org/document/6616668&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; contains actual figures:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Only a few seconds of x-ray exposure corresponding to a total dose of merely 50 rad(Si) in a real-time x-ray source are required to induce errors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So if your exposure gets anywhere near this level then its not being paranoid to consider the possibility.&lt;/p&gt;&#xA;" OwnerDisplayName="Simon Jenkins" LastEditorUserId="6767" LastEditDate="2022-04-18T17:02:21.597" LastActivityDate="2022-04-18T17:02:21.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4023" PostTypeId="2" ParentId="4021" CreationDate="2015-08-14T10:07:42.657" Score="7" Body="&lt;p&gt;You should definitely consider X-rays as a possible source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An erased flash is full of 1s. Programming a flash is full of 0s. Over time, or under some stress, flash components may lose their gate charge, which leads to 0 to 1 transition. This is a well known behavior on NAND flash, but may also happen on NOR flash, even if those ones are much more robust by design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High energy radiations (X rays, gamma rays) are a well known source of stress for flash memories, and it is an important topic for aeronautic, aerospace and nuclear products. See &lt;a href=&quot;http://link.springer.com/chapter/10.1007/978-90-481-9431-5_19#page-1&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;https://books.google.fr/books?id=vaq11vKwo_kC&amp;amp;pg=PA538&amp;amp;lpg=PA538&amp;amp;dq=cosmic+rays+NAND&amp;amp;source=bl&amp;amp;ots=UJTQOljDtX&amp;amp;sig=TjNEFoweghgJHtiyFWnT85PNDKQ&amp;amp;hl=fr&amp;amp;sa=X&amp;amp;redir_esc=y#v=onepage&amp;amp;q=cosmic%20rays%20NAND&amp;amp;f=false&quot;&gt;there&lt;/a&gt; and &lt;a href=&quot;http://www.tanisys.com/effect-radiation-solid-state-drive-ssd/&quot;&gt;there&lt;/a&gt; my first google results for &quot;cosmic rays NAND&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the issue is a power cut during program or erase, it may probably occur at any moment during the operation, and you should have some parts with many 0 to 1 transitions, and some with very few. If you always have few bit errors, it's probably something else, such as EMC or radiations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yours products have to resist to this kind of stress, &lt;a href=&quot;https://en.wikipedia.org/wiki/Error_detection_and_correction#Error-correcting_code&quot;&gt;ECC&lt;/a&gt;, such as Hamming codes, are made for you. Nevertheless, error management in flash memory is a big topic.&lt;/p&gt;&#xA;" OwnerUserId="2664" OwnerDisplayName="Jacen" LastActivityDate="2015-08-14T10:07:42.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4029" PostTypeId="1" CreationDate="2015-07-29T01:57:17.740" Score="8" ViewCount="266" Body="&lt;p&gt;I've seen many microcontrollers use pull up or pull down resistors either as a setting for GPIO pins or as  part of the hardware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I know if I need a pull up or pull down resistor for a GPIO pin and if I do, how do I know which is right for my application? &lt;/p&gt;&#xA;" OwnerUserId="295" OwnerDisplayName="Dom" LastEditorUserId="1586" LastEditDate="2015-08-20T23:57:41.363" LastActivityDate="2015-08-20T23:57:41.363" Title="How do I know if I need a pull up or pull down resistor for a GPIO pin?" Tags="|electrical-engineering|embedded-systems|design|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4030" PostTypeId="2" ParentId="4029" CreationDate="2015-07-29T04:22:32.800" Score="7" Body="&lt;p&gt;You need a pull up or pull down resistor on a line of which the voltage level at least sometimes is &quot;unpredictable&quot; - for example, a floating input, but also an I2C line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the default of the line to be a digital 1, you use a pull up. If you want it to be a digital 0, you use a pull down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very typical example is a button. Suppose the software expects a pin to go down when a button is pressed. On the corresponding pin, you use a pull up and the button connects the pin with ground. &lt;/p&gt;&#xA;" OwnerDisplayName="user2662" LastActivityDate="2015-07-29T04:22:32.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4031" PostTypeId="2" ParentId="4029" CreationDate="2015-07-29T05:17:35.463" Score="3" Body="&lt;p&gt;In final layout it is also strongly recommended to pull every unused pin to the appropriate level. Do not leave them not connected as it can cause unpredictable behaviour at start time or during the running program.&lt;/p&gt;&#xA;" OwnerUserId="2674" OwnerDisplayName="Ruslan Gerasimov" LastActivityDate="2015-08-03T19:37:09.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4032" PostTypeId="2" ParentId="4029" CreationDate="2015-07-29T05:33:05.333" Score="0" Body="&lt;p&gt;Whether you pull it up, or down, is really up to you and your project needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned prior&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your code is looking to trigger on a falling edge or a &lt;code&gt;0&lt;/code&gt;, you would want to use a pull up to keep the level high when unused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your code is looking to trigger on a rising edge or a &lt;code&gt;1&lt;/code&gt;, you would want to use a pull down to keep the potential at 0, util your external source pulls it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pulling resistors should have a high ohmic value, enough to overcome the noise you need to eliminate, but not so strong that your external soucre can not overcome it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, by pulling down you will conserve a few mA of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say we have a 5k pull up @ 5v, that is &lt;code&gt;1mA&lt;/code&gt; per pin; when running off of batteries, every mA is precious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we pull down, we are generally just burning off the EMF noise, consuming a negligible number of mAs.&lt;/p&gt;&#xA;" OwnerUserId="204" OwnerDisplayName="Matt Clark" LastActivityDate="2015-07-29T05:33:05.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4033" PostTypeId="5" CreationDate="2015-08-14T19:34:18.550" Score="0" Body="&lt;p&gt;For digital systems such as computer or microcontrollers to manipulate an analog signal, the signal must first be converted to a digital one. An ADC is a device that can perform this transformation. The techniques used to accomplish this task can vary depending on the type of analog input signal and the desired digital signal format.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions should use this tag when they are about the usage or design of analog-to-digital converters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional Information:&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Analog-to-digital_converter&quot; rel=&quot;nofollow&quot;&gt;Wikipedia entry for Analog-to-Digital Converters&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2659" LastEditorUserId="2659" LastEditDate="2015-08-14T21:40:23.627" LastActivityDate="2015-08-14T21:40:23.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4034" PostTypeId="4" CreationDate="2015-08-14T19:34:18.550" Score="0" Body="Acronym for  &quot;Analog-to-Digital Converter&quot;. Refers to a device that can convert an analog input signal to a digital signal. Use this tag for questions about the usage or design of an ADC." OwnerUserId="2659" LastEditorUserId="2659" LastEditDate="2015-08-14T21:40:26.860" LastActivityDate="2015-08-14T21:40:26.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4035" PostTypeId="1" CreationDate="2015-08-14T21:54:19.550" Score="0" ViewCount="175" Body="&lt;p&gt;Totally numb question from non-engineer, so sorry about that. For my little home project I need to create a matrix of solenoids, so I need some kind of simulation. I wish to be able to create metal rod of specific volume and mass, and coil with specific number of turns, so being able to apply current source to coil, see rod movement, if it occurs, to be able to experiment with voltage, coil turns or rod mass. I have no experience with CAD software whatsoever I must say. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw some demo videos with solidworks where there was Newtonian Cradle simulation using motion analysis, is there the same 'simple' and 'intuitive' type of simulation for motion analysis with electromagnetic field? For example so I can create electromagnet, place some metal on it, then turn on current and it will attract (or repell) in live?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw such motion simulations in solidworks and inventor, also comsol multiphysics looks promising, can anyone suggest some?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thx.&lt;/p&gt;&#xA;" OwnerUserId="2665" LastActivityDate="2017-02-20T20:38:22.693" Title="Live motion simulation with respect to electromagnetic field software" Tags="|electrical-engineering|magnets|simulation|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4036" PostTypeId="5" CreationDate="2015-08-14T23:37:24.603" Score="0" Body="&lt;p&gt;C is a general-purpose, weakly typed programming language with static types. It's most common modern usage is in embedded systems and systems-level programming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tag should be used for questions about using C in engineering-related projects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More Information:&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/C_(programming_language)&quot; rel=&quot;nofollow&quot;&gt;Wikipedia Entry for the C language&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2659" LastEditorUserId="2659" LastEditDate="2015-08-15T20:53:31.887" LastActivityDate="2015-08-15T20:53:31.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4037" PostTypeId="4" CreationDate="2015-08-14T23:37:24.603" Score="0" Body="Refers to the C programming language. Use this tag for questions about using C in engineering projects." OwnerUserId="2659" LastEditorUserId="2659" LastEditDate="2015-08-15T20:53:36.247" LastActivityDate="2015-08-15T20:53:36.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4038" PostTypeId="2" ParentId="3935" CreationDate="2015-08-15T01:31:23.053" Score="1" Body="&lt;p&gt;One very significant thing that &quot;someone coming from a pure software and mathematics background&quot; might not &quot;see&quot; is the role that validation and verification play in the complete process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That can be more sophisticated than simply &quot;building a prototype to show that the computer-generated design actually works&quot;. A more effective way is to verify and validate the &lt;em&gt;techniques&lt;/em&gt; used in creating the design, so that, based on experiments, you know when the software has actually &lt;em&gt;designed&lt;/em&gt; something, and when it has only produced some pretty computer graphics.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can characterize that as redefining the purpose of engineering &quot;research and development&quot;, from &quot;figuring out how to fix stuff that doesn't work&quot; to &quot;figuring out how to acquire technology that &lt;em&gt;does&lt;/em&gt; work, within well-defined limitations&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That may be a long way from &quot;the singularity&quot;, but it can be orders of magnitude better and faster than what an experienced team of human designers can achieve on their own. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-08-15T01:31:23.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4039" PostTypeId="2" ParentId="3916" CreationDate="2015-08-15T03:03:40.687" Score="2" Body="&lt;p&gt;It is doable, but depends on having enough storage space to contain both the full image and the small test image. There is often an initial program that runs, often referred to as the bootloader. The bootloader's job is to set up and launch the main application, and often to provide a recovery path for updating firmware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would not be hard, and is done in some products, for the bootloader to check an I/O and use that to select the image to load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a real-world example: On my cell phone the is an initial part called &quot;fastboot&quot; that checks for a button (volume down I think; can't recall at the moment) where either it will load the main Android system if nothing is pressed, or a recovery image if it is pressed. The recovery image starts very fast, but has limited functionality. It can access the filesystem and it wouldn't be stretch to put some diagnostic test options in there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On other embedded systems, every byte of storage might count, and in those the designer has to weigh having diagnostics and recovery against fitting the required featureset at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: original answer had a comment asking for elaboration about detecting a button press or similar to select the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases the bootloader is written in code similar to the rest of the system (often C or assembly). As such, it has the same level of access to the hardware that the normal software has. In many cases, the button or similar is a simple I/O line that can be read as a direct GPIO on the processor, or via a simple bus read to an external GPIO Expander (e.g. SPI or I2C bus), either of which wouldn't occupy much space in the bootloader's image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For something that is explicitly designed for manufacture testing (as opposed to recovery or field testing) it would be less likely to be implemented on a user accessible button. A cheap but not easily accidentally tripped method would be a couple solder pads on the PCB that could be shorted together to invoke the test (ranging from using a simple screw driver to being implemented with a bed-of-nails style rig).&lt;/p&gt;&#xA;" OwnerUserId="262" LastEditorUserId="1586" LastEditDate="2015-09-02T23:28:32.863" LastActivityDate="2015-09-02T23:28:32.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4040" PostTypeId="1" CreationDate="2015-08-15T11:06:27.523" Score="3" ViewCount="2037" Body="&lt;p&gt;I use Labcenter Proteus Circuit simulation - you create circuit design, then apply current and see simulation in action - for example leds are blinking, or motor is rotating. If there is microcontroller - say AVR, you just load compiled hex file for it and see simulation live - again, if it was programmed to blink led, and there is one on pin, it will blink, or you can simulate button pressing and see what happens, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I know SolidWorks have SolidWorks Electric, SolidWorks Simulation, SolidWorks CircuitWorks (are those all?), is there any way to have such functionality in it? I mean, if I have added motor to project, then created circuit with avr microcontroller in SolidWorks electrical, is there a way to somehow combine those two, set .hex file for avr and run simulation to see how motor moves? Or Any alternative, how do engineers do such things?&lt;/p&gt;&#xA;" OwnerUserId="2665" LastEditorUserId="272" LastEditDate="2015-08-17T13:33:35.513" LastActivityDate="2016-04-19T17:54:43.123" Title="SolidWorks circuit simulation?" Tags="|simulation|circuits|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4043" PostTypeId="1" AcceptedAnswerId="4047" CreationDate="2015-08-16T10:44:26.503" Score="4" ViewCount="2375" Body="&lt;p&gt;I notice that quite a number of deep excavation projects adopt double I-section walers to form a box beam since standard rectangular hollow sections don't come with such large sizes. The I-sections are usually factory butt-welded together at the flanges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Thkkd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Thkkd.jpg&quot; alt=&quot;Braced Excavation: Struts and Walers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are such members spliced together? Unlike a single I-section, the inner flanges of the double I-section would not be accessible. There would also be problem connecting the webs due to the same accessibility issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the pictures I found on the internet, it isn't clear how this can be done to achieve effective moment transfer.&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2015-08-16T19:21:57.630" Title="How can double I-section walers be spliced to ensure effective moment transfer?" Tags="|structural-engineering|steel|geotechnical-engineering|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4044" PostTypeId="2" ParentId="4043" CreationDate="2015-08-16T12:06:00.910" Score="4" Body="&lt;p&gt;Double-I beams cannot be spliced like single-I beams for the reasons you mentioned. Indeed, they must be spliced like box beams, which suffer from the same problems (lack of access to internal face). To deal with this, it is common to weld plates at the ends of the beams to be spliced and then bolting the plates together, as seen below (&lt;a href=&quot;http://www.steelconstruction.info/Simple_connections#Column_splices&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kmYb1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kmYb1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-08-16T12:06:00.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4045" PostTypeId="1" CreationDate="2015-08-16T14:32:25.930" Score="2" ViewCount="1223" Body="&lt;p&gt;I have seen one video about fountains, and there are some kind of nozzles.  What type of nozzle is used in the fountain shown below? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CgFFM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CgFFM.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be seen in the photo, the fountain nozzle causes an oval shape to form from the water cascading back down.&lt;/p&gt;&#xA;" OwnerUserId="2678" LastEditorUserId="368" LastEditDate="2016-01-20T17:48:57.637" LastActivityDate="2021-08-20T15:44:02.427" Title="What type of fountain nozzle ejects water in an oval shape?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4046" PostTypeId="1" AcceptedAnswerId="4055" CreationDate="2015-08-16T15:19:12.180" Score="1" ViewCount="1418" Body="&lt;p&gt;I am trying to calculate the loads in the suspension components of a typical car double wishbone setup.  I have found &lt;a href=&quot;http://www.fsae.com/forums/showthread.php?11179-analysis-of-wishbones&quot; rel=&quot;nofollow&quot;&gt;a thread on the www.fsae.com forum&lt;/a&gt;, which seems to be exactly the kind of thing I am looking for. It involves considering the assembly as a 3D truss, and building up matrices, containing&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the unit vectors and cross products at the mounting points of the struts at the upright&lt;/li&gt;&#xA;&lt;li&gt;the forces and cross products at the tyre contact patch.  Which when the inverse of matrix 1 is multiplied by matrix 2, results in a third matrix containing the loads on the struts.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is, can someone give me the name of this method of truss analysis, and any textbooks / resources that specifically cover this method?&lt;/p&gt;&#xA;" OwnerUserId="2680" LastEditorUserId="272" LastEditDate="2015-08-17T11:51:52.663" LastActivityDate="2015-08-18T04:06:00.100" Title="Calculating automotive suspension wishbone loads by considering the structure as a 3D Truss" Tags="|mechanical-engineering|automotive-engineering|structures|statics|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4047" PostTypeId="2" ParentId="4043" CreationDate="2015-08-16T19:15:54.227" Score="5" Body="&lt;h3&gt;Bolts&lt;/h3&gt;&#xA;&lt;p&gt;Flanges can be added to the members as Wasabi mentioned in his answer. Another option would be to cut a small hole in the web of the members to allow for an iron worker to insert his hand to insert a bolt or hold a wrench. The reduction in strength from the access holes is minimal. This type of access is also used in the bolted connections of truss members. With this method, typical flat splice plates can be used.&lt;/p&gt;&#xA;&lt;h3&gt;Welds&lt;/h3&gt;&#xA;&lt;p&gt;Single sided complete joint penetration (CJP) welds are not a problem. They are done frequently. The joint will require a backer rod that will be left in place.&lt;/p&gt;&#xA;&lt;p&gt;In the case of these walers, fatigue won't be an issue, so partial joint penetration (PJP) welds would even work. As long as the splices were located in good locations, the full strength of the member wouldn't be required at the splice location.&lt;/p&gt;&#xA;&lt;p&gt;Weld examples from AWS:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SPxEG.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SPxEG.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/n2GfS.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n2GfS.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-08-16T19:21:57.630" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4051" PostTypeId="2" ParentId="2349" CreationDate="2015-08-17T06:19:51.020" Score="3" Body="&lt;p&gt;Petrol engines have a very poor efficiency at low load... no ifs or buts. The 34mpg efficiency is not quoted at low load.&lt;/p&gt;&#xA;&lt;p&gt;There is an ideal speed at which maximum mileage occurs... engine efficiency increases as load increases, however, air drag losses increase as the square of the speed.&lt;/p&gt;&#xA;&lt;p&gt;There are two major reasons for the poor efficiently at low load:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;At low load, a significant percentage of the work done is used to overcome engine friction.&lt;/li&gt;&#xA;&lt;li&gt;At low load, the effective compression ratio is very low, which in turn leads to a very low efficiency.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Power generated by an engine is the product of angular speed by the available torque at that angular speed. The function of the gearbox is merely to match the engine speed to the road speed.&lt;/p&gt;&#xA;&lt;p&gt;City driving mileage is lower than highway mileage mainly due to continuous braking which dissipates kinetic energy as heat, and idling at traffic lights or stop signs.  And accelerating away from those signs and lights.&lt;/p&gt;&#xA;" OwnerUserId="2684" LastEditorUserId="-1" LastEditDate="2022-02-01T22:01:26.723" LastActivityDate="2022-02-01T22:01:26.723" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="4053" PostTypeId="1" AcceptedAnswerId="4059" CreationDate="2015-08-17T12:21:36.487" Score="8" ViewCount="553" Body="&lt;p&gt;I live and work in the UK, a country known for relentless roadworks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last week I drove through roadworks on the M1 that lasted for 18 miles, necessitating road users driving at 50 mph. Of course, almost all of this impressive stretch was simply coned off, with active work only present on one or two small sites. Signs indicate that work will continue into next year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What justification is there for keeping such a large section of road cordoned off for improvements? Is it a safety or economical consideration? It would seem prudent to close off mile sections at a time to increase traffic flow and reduce the number of cones/signs/average speed cameras in use, but perhaps there is an underlying reason I am missing?&lt;/p&gt;&#xA;" OwnerUserId="2688" LastEditorUserId="368" LastEditDate="2015-08-17T21:15:38.577" LastActivityDate="2015-12-14T04:59:33.830" Title="What is the justification for working on such large stretches of road?" Tags="|civil-engineering|highway-engineering|construction-management|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4055" PostTypeId="2" ParentId="4046" CreationDate="2015-08-17T14:01:30.623" Score="2" Body="&lt;p&gt;Initially after reading your question I thought that it was using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Direct_stiffness_method&quot; rel=&quot;nofollow&quot;&gt;stiffness method&lt;/a&gt; but after seeing the image on page 7 of that forum post I don't think that's what it is. This is because the stiffness method relates forces to displacement using the members' stiffness properties. I don't see any of that discussed on the forum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, it just seems like a force balance in each of the XYZ axes and a moment balance about each of the XYZ axes giving rise to 6 equations. Depending on how your 3D truss is supported you may have 6 or more unknowns. Using matrices is just an efficient way to solve many simultaneous equations. I'm not sure there's a specific name for this given the context of your application. In terms of mathematics it is just linear algebra (i.e. solving a &lt;a href=&quot;https://en.wikipedia.org/wiki/System_of_linear_equations#Matrix_equation&quot; rel=&quot;nofollow&quot;&gt;system of linear equations&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps with more information about what you're trying to achieve (maybe a diagram of your suspension system) we can better guide you. The assembly can be modeled as a 3D Truss like you've stated. Normally you would come up with a Free Body Diagram (FBD) to figure out how the 3D Truss is supported and loaded. From there you would solve for the reaction forces where it is supported. This looks like where the 6x6 matrix comes into play - you'd use that to solve for the reaction forces of the suspension system on the chassis. Then you could use this to find the internal forces of the suspension members if that's what your ultimate goal is.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-08-18T04:06:00.100" LastActivityDate="2015-08-18T04:06:00.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4056" PostTypeId="1" CreationDate="2015-08-16T08:15:16.423" Score="3" ViewCount="642" Body="&lt;p&gt;I want to put LED lamps in water. Up until now I used epoxy glass for its isolation, but after one year it starts to become &lt;strong&gt;&lt;em&gt;opaque&lt;/em&gt;&lt;/strong&gt; and lose its clearness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes the epoxy glass to cloud and become opaque? How can I avoid this problem while keeping the lamps isolated from the water?&lt;/p&gt;&#xA;" OwnerUserId="2678" OwnerDisplayName="Soheil Paper" LastEditorUserId="33" LastEditDate="2016-01-26T03:48:27.370" LastActivityDate="2020-01-06T19:54:36.937" Title="Why does epoxy resin become opaque after long exposure to water and LED lamps?" Tags="|materials|waterproofing|epoxy|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4057" PostTypeId="2" ParentId="4056" CreationDate="2015-08-17T08:23:01.580" Score="3" Body="&lt;p&gt;I would guess that your current epoxy changes its optical properties because of ultraviolet (UV) light coming from the LEDs.  You may be able to find specialized UV-stable epoxies that don't suffer from UV.  You could rethink your mechanical design and use a glass (or UV-stable plastic, or sapphire) window in front of the LED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White LEDs are actually blue LEDs with additional phosphor (look at a spectrum below).  The tail of the blue part can extend far enough into the short wavelengths to reach UV.  If an LED is powerful and the polymer victim is close enough, then this residual UV can cause degradation over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/efXMV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/efXMV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(&lt;a href=&quot;http://www.ecse.rpi.edu/~schubert/Light-Emitting-Diodes-dot-org/chap21/chap21.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt; of the plot)&lt;/p&gt;&#xA;" OwnerUserId="63" OwnerDisplayName="Nick Alexeev" LastActivityDate="2015-08-17T08:23:01.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4058" PostTypeId="1" AcceptedAnswerId="4163" CreationDate="2015-08-17T18:21:48.513" Score="8" ViewCount="4902" Body="&lt;p&gt;The AISC 360-10 &lt;em&gt;Specification for Structural Steel Buildings&lt;/em&gt; gives provisions for calculating the maximum unbraced length of a compression flange that separates yielding moment from lateral torsional buckling (LTB).  This formula is (AISC 360-10, Eqn. F2-5):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L_p = 1.76r_y\sqrt{\frac{E}{F_y}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L_p =$ limiting length that separates full yield moment and LTB &lt;br&gt;&#xA;$r_y =$ radius of gyration about the $y$-axis &lt;br&gt;&#xA;$E =$ Young's modulus &lt;br&gt;&#xA;$F_y =$ yield strength of material &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that one is using regular structural steel, the Young's modulus of the material is assumed to be the same regardless of steel grade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation works out such that a steel with a &lt;em&gt;lower&lt;/em&gt; yield strength may actually be braced at a lesser interval than one with a &lt;em&gt;higher&lt;/em&gt; yield strength.  In other words, given the same beam size, the material with the higher yield strength buckles first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also found this to be applicable to design using the ASME &lt;em&gt;Boiler &amp;amp; Pressure Vessel&lt;/em&gt; code, specifically Division III, Subsection NF for supports.  With temperature effects on yield strength and Young's modulus taken into account, it's possible that a member at an elevated temperature may buckle at a longer length than one at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems counter-intuitive to me.  Why would a weaker material exhibit less LTB action with the same given length?&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-08-25T14:30:06.330" Title="Unbraced Length for Lateral Torsional Buckling vs. Yield Strength" Tags="|structural-engineering|steel|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4059" PostTypeId="2" ParentId="4053" CreationDate="2015-08-18T01:22:10.937" Score="13" Body="&lt;p&gt;I'll give you some considerations that you may not be aware of. Also, take the disclaimer that I'm in the US (but construction is construction, right?)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Frequently moving construction cones/barrels along the road is less efficient than setting them out once and only making minor adjustments to their location.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Depending on the type of work being done, some sections may require a certain length of time before the next bit of work is done. This is frequently the case with painting the lines. Painting specifications usually require that the concrete or asphalt sit for about a week before painting can begin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Not every crew can do all of the jobs. Sometimes scheduling certain crews to come do their part means that specific tasks may be held back while the workers are on a different job.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Keeping a large stretch of road closed allows for flexibility in where equipment and people can be moved after finishing their current job.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Shortening the construction zone may not have that great of an effect on the throughput of the road. If you take for granted that there will be a bottleneck somewhere, its exact location doesn't matter much. Also, once you are in a single line, the speed limit may be close to the optimum speed. Studies have shown that the highest volume of vehicles occur around 55 miles per hour.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is some safety benefit to long construction zones. The majority of accidents will occur at the beginning of the zone. Keeping workers away from those areas is safer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Probably the biggest reason when you combine all of the reasons above is that the contractor wants to define their area and their efficiency (read: money) is more important than the user's (you pay taxes no matter what).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Those are some reasons why long construction zones aren't entirely wasteful. That doesn't mean that there aren't ways that they can be improved, though. The problem is that there is limited benefit to the people controlling the zones to shortening the zone length.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-08-18T01:22:10.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4060" PostTypeId="2" ParentId="4045" CreationDate="2015-08-18T04:22:35.017" Score="1" Body="&lt;p&gt;This type of flow is called a water bell, or at least the laminar version of it is. &lt;a href=&quot;http://www.fontanafountains.com/products/spray-systems/fountain-nozzles-heads/water-bells&quot; rel=&quot;nofollow&quot;&gt;Here's the website of a manufacturer of these sorts of nozzles&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;a href=&quot;http://rspa.royalsocietypublishing.org/content/253/1274/289.short&quot; rel=&quot;nofollow&quot;&gt;G. I. Taylor wrote a paper on the various phenomena involved with water bells&lt;/a&gt;. (I have not read it, but someone reading this may find it to be interesting.)&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2015-08-18T04:22:35.017" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4061" PostTypeId="2" ParentId="4058" CreationDate="2015-08-18T05:12:29.423" Score="4" Body="&lt;p&gt;Slenderness ($\lambda = L/r$) is the ratio of a member's length to its smallest radius of gyration. It should make sense that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The less slender a member, the more its plastic strength needs to be considered rather than its Euler (buckling) strength.&lt;/li&gt;&#xA;&lt;li&gt;The more slender a member, the more its Euler (buckling) strength needs to be considered rather than its plastic strength.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In other words as slenderness increases there becomes a point where critical buckling stress becomes the limiting factor rather than plastic yield strength ($F_y$). The maximum allowed compressive strength is the &lt;strong&gt;minimum of yield strength and buckling strength&lt;/strong&gt;. This is illustrated in the diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h3yhK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h3yhK.jpg&quot; alt=&quot;Euler Strength and Buckling Strength&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \lambda = L_p/r_y = 1.76\sqrt{\frac{E}{F_y}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula you have provided has separated yielding moment from lateral torsional buckling (LTB) as you stated. This would be the slenderness point where the critical strength changes from the plastic strength to the Euler strength. If $F_y$ increases, then this point on the x-axis would move to the left. This means slenderness $\lambda$ would be smaller and hence the length of the member (or length between bracing points), $L$ should be smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vYDnF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vYDnF.png&quot; alt=&quot;Slenderness decreases if yield strength increases&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the formula it does seem counter-intuitive. But what you have to remember is that it's either going to fail due to plastic yielding or LTB. And so at higher yield strengths, the buckling strength falls below the yield strength at lower slenderness (smaller member length) than lower yield strengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-08-18T09:06:07.003" LastActivityDate="2015-08-18T09:06:07.003" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4062" PostTypeId="1" CreationDate="2015-08-18T13:12:45.370" Score="4" ViewCount="100" Body="&lt;p&gt;I have a collection of data obtained from calculations and I want to display it on a chart or whiz-wheel or similar, to be used &quot;in the field&quot; as a low-tech solution, since no computer will be available.  I have a single output variable calculated from three inputs.  What I want to avoid is making a giant book of tables or charts, and instead use something like the whiz-wheel that reloaders use for calculating propellant weight.  Are there any nifty graphical devices out there that can display three degrees of freedom on one chart?&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1586" LastEditDate="2015-08-20T16:58:35.530" LastActivityDate="2015-08-20T16:58:35.530" Title="How to display data with three degrees of freedom on a chart?" Tags="|statistics|data|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4063" PostTypeId="1" AcceptedAnswerId="4066" CreationDate="2015-08-18T13:24:24.987" Score="8" ViewCount="1419" Body="&lt;p&gt;The &lt;strong&gt;Darcy-Weisbach equation&lt;/strong&gt; is used to calculate the frictional pressure losses in pipes transporting incompressible fluids. This equation uses a dimensionless &lt;strong&gt;Darcy friction factor&lt;/strong&gt;, also known as the Moody factor, to account for the relative roughness of the pipe surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This empirical factor was experimentally determined by Moody and is normally read off of the &lt;a href=&quot;http://www.ndlr.ie/artefact/file/download.php?file=13347&amp;amp;view=2200&quot;&gt;Moody Chart&lt;/a&gt;. However I am implementing the pressure drop calculation in software, so I need a non-graphical way to find the Darcy friction factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equations for calculating the Darcy friction factor under laminar (Re &amp;lt; 2320) and turbulent (Re &gt; 4000) flow are readily available. But I haven't been able to find one that is valid for the transitional region which exists between laminar and turbulent flow (2320 &amp;lt; Re &amp;lt; 4000), also known as the 'critical zone'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that fluid flow is complex and difficult to predict in this region. However, &lt;strong&gt;is there a commonly used method that provides &lt;em&gt;reasonable estimates&lt;/em&gt; for the friction factor in this critical zone?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found a method described in a &lt;a href=&quot;http://www.chem.mtu.edu/~fmorriso/DataCorrelationForSmoothPipes2013.pdf&quot;&gt;student paper&lt;/a&gt;, but it hasn't been peer reviewed and is limited to smooth pipes only. I'm looking for something more tried and tested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If no formula is available, what approach do other engineers usually take to mitigate or solve this problem?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="2551" LastEditorUserId="14498" LastEditDate="2018-02-07T17:30:36.633" LastActivityDate="2018-02-07T17:30:36.633" Title="Estimating Darcy Friction Factors for Critical Flow" Tags="|fluid-mechanics|chemical-engineering|pipelines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4064" PostTypeId="1" AcceptedAnswerId="4103" CreationDate="2015-08-18T14:10:51.443" Score="4" ViewCount="905" Body="&lt;p&gt;I would like to build a &lt;em&gt;little water mill to place under the sink of the kitchen to produce electricity&lt;/em&gt;. Considering that the mill will work when the user opens the tap to wash a dish, using a dynamo and a little electric circuit (bridge rectifier and a low pass filter), I would be able to generate a continuous electric signal of about &lt;em&gt;6 V, 0.1 A&lt;/em&gt; for a couple of minutes. The idea is to charge a battery, typically an external one used to charge a smartphone, i.e. a battery with a capacity of &lt;em&gt;2500 mAh&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I am wondering if the repeated short sessions of charges will affect the battery in any way&lt;/strong&gt;. Obviously, the battery will need a couple of dish-washing sessions to be fully charged, but will these short sessions affect its efficiency, or even alter the charging time ?&lt;/p&gt;&#xA;" OwnerUserId="2693" LastEditorUserId="2246" LastEditDate="2015-08-18T22:14:52.067" LastActivityDate="2015-08-26T14:41:58.630" Title="What are the consequencies of intermittently charging a smartphone battery?" Tags="|electrical-engineering|battery|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4065" PostTypeId="1" CreationDate="2015-08-18T14:19:33.700" Score="10" ViewCount="660" Body="&lt;p&gt;I am working on the hydraulic seawater flow though a siphon spillway. My task is to determine at a location &lt;em&gt;in front&lt;/em&gt; of the spillway the flow velocity and profile when&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The flow enters the spillway&lt;/li&gt;&#xA;&lt;li&gt;The flow exits the spillway &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;See attached a drawing of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the basic formula for calculating the flow  &lt;em&gt;through&lt;/em&gt; the siphon is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta H = \xi \frac{v^2}{2g}&#xA;$$&#xA;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta H$ = pressure head &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\xi$ = total friction losses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v =flow velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;g = gravitational acceleration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is: what is the flow profile and velocity at the location in front of the spillway (see drawing) for a given pressure head, geometry and friction loss factor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that I am not looking for a full fledged CFD analysis at this stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help on how to tackle this problem is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2CzMN.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2CzMN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2695" LastEditorUserId="33" LastEditDate="2015-11-21T01:13:13.583" LastActivityDate="2018-09-07T14:00:30.147" Title="Hydraulic siphon spillway entry and exit velocity magnitude and profiles" Tags="|fluid-mechanics|hydraulics|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4066" PostTypeId="2" ParentId="4063" CreationDate="2015-08-18T15:31:26.100" Score="2" Body="&lt;p&gt;As an engineer, sometimes you need an answer, and it might not be the best answer, but you've got to get one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, there are two ways to handle this.  The first would be to &lt;strong&gt;extrapolate&lt;/strong&gt; the turbulent flow backwards to the end of the laminar flow.  Since the turbulent flow is always higher than the laminar flow, you'll get a higher than real friction flow in your piping, but since you're mainly using this to size your piping, pumps, etc, in the end you'll just wind up &lt;a href=&quot;https://en.wikipedia.org/wiki/Overengineering&quot; rel=&quot;nofollow&quot;&gt;overengineering&lt;/a&gt; the pumps, pipes, etc.  This isn't a bad thing, but it isn't desired.  However, this way you have an answer.  If your program states a warning when this happens &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;This is in the turbulent zone, so an accurate estimate is difficult. &#xA;  Instead a conservative amount is given.  A better design would be in&#xA;  either the laminar or turbulent zones&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And most engineers would be satisfied.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second way would be to require the user to be in either the laminar region or the turbulent region.  This is how most engineers handle this situation - they &lt;strong&gt;avoid the region altogether.&lt;/strong&gt;  This way, the system can be accurately and efficiently designed, to avoid overengineering.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-08-18T15:31:26.100" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4067" PostTypeId="2" ParentId="4062" CreationDate="2015-08-18T15:48:30.433" Score="3" Body="&lt;p&gt;Per @Dan's suggestion, I googled &quot;nomography&quot; and found an article titled &quot;The Lost Art of Nomography&quot; by Ron Doerfler, published in the UMAP Journal 30.4 (2009).  The article describes a compound-nomograph, which is pretty much what I need.  Here is a screen shot of such a device, taken from the article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PilKc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PilKc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image shows a nomograph for three independent variables, but the principle could be extended to any number of variables I believe.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-08-18T15:48:30.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4068" PostTypeId="2" ParentId="4064" CreationDate="2015-08-18T16:42:21.503" Score="4" Body="&lt;p&gt;I don't believe periodic charging would damage a smartphone, however it likely wouldn't charge it much either. A typical phone charger today outputs 1 Ampere of current at 5 Volts, and takes perhaps 2-4 hours to charge a phone. At 0.1 Ampere or a tenth of this current, it would take ten times as long to charge the phone. It would actually be longer since the phone is self-discharging during this time, and may not ever charge appreciably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the dynamo's output was rectified, filtered, &lt;em&gt;and regulated to 5.0v&lt;/em&gt;, it should work for demonstration purposes. It must be regulated however, a phone will never expect 6V and could be damaged. It should also be &quot;smooth&quot; DC or direct current, meaning there is little noise or variations in it's output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the dynamo can be expected to produce less than 5V, then a &quot;buck-boost&quot;, &quot;boost-buck&quot;, SEPIC, Zeta, etc. converter/regulator can all output a fixed voltage (5v) for an input which goes below and above the target. I don't recommend inexpensive eBay items, but something like this &lt;a href=&quot;http://www.ebay.com/itm/Digital-Display-Step-Up-Step-Down-DC-DC-Boost-Buck-Converter-Power-Module-/381299207276&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Step Up Step Down DC-DC Boost Buck Converter&quot;&lt;/a&gt; can convert a varying input voltage to a relatively stable output voltage. Even so, I'd recommend having it output 6V, and use a second &quot;buck&quot; regulator (with better &quot;line&quot; and &quot;load&quot; regulation and less noise) to produce a clean 5.00v for the phone. Of course, the more pieces that are added, the more efficiency is lost. The best solution would be to buy (or build!) a custom converter designed around the particular dynamo being used. Questions about such would be best directed to &lt;a href=&quot;https://electronics.stackexchange.com/&quot;&gt;electronics.stackexchange.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to recharge the phone from a kitchen sink tap or garden hose &lt;em&gt;quickly&lt;/em&gt;, consider a plumbing fitting which replaces the existing nozzle with an impeller under pressure from the water supply. The water &lt;em&gt;pressure&lt;/em&gt; can do much more work than the free-flowing &lt;em&gt;water mass and gravity&lt;/em&gt;, allowing the dynamo to spin faster and produce more current. Such topics as fluid pressure, work, fluid head etc, all fall under the category of &lt;a href=&quot;https://en.wikipedia.org/wiki/Fluid_dynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fluid Dynamics&lt;/a&gt; in physics.&lt;/p&gt;&#xA;" OwnerUserId="1673" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2015-08-18T16:42:21.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4070" PostTypeId="2" ParentId="100" CreationDate="2015-08-18T22:54:58.090" Score="4" Body="&lt;p&gt;It's not a perfect sphere but as close as I've gotten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suspend in heated jet, then drop.  Done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NwOLa.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NwOLa.gif&quot; alt=&quot;101&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to control the temperate carefully, too hot and it flies apart.&lt;/p&gt;&#xA;" OwnerUserId="2702" LastEditorUserId="2702" LastEditDate="2015-08-18T23:28:21.273" LastActivityDate="2015-08-18T23:28:21.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4072" PostTypeId="1" CreationDate="2015-08-19T08:44:24.530" Score="6" ViewCount="466" Body="&lt;p&gt;We have an electronic display product that will be installed in a street side application. The front part is made from metal and exposed to the elements so it will need to be painted to protect it from corrosion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what would be a good international standard to refer to when specifying the corrosion resistance of the paint and the final products?&lt;/p&gt;&#xA;" OwnerUserId="2705" LastEditorUserId="90" LastEditDate="2015-08-20T05:58:16.777" LastActivityDate="2015-08-20T05:58:16.777" Title="What is the relevant ISO or similar standard to refer to for painted metal corrosion testing?" Tags="|materials|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4074" PostTypeId="1" CreationDate="2015-08-19T11:37:01.040" Score="2" ViewCount="187" Body="&lt;p&gt;I wonder what is the best way to dry something that is quite thick. The gravity will attract the water down, but the sun will &quot;elevate&quot; the water. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;So I can choose to flip the object every 30 min so that each side gets the same amount of sun that will evaporate the water. But at the same time, the water that was slowly getting downward thanks to gravity will &quot;travel&quot; back to the other side because gravity is now attracting it to the other side. So, each time I turn the object the water inside the object will be kind of &quot;stuck in the thickness&quot;. &lt;/li&gt;&#xA;&lt;li&gt;Because of this gravity attraction, I can choose to only flip the object once every 2 hours or once every 6 hours. &lt;/li&gt;&#xA;&lt;li&gt;Capillarity should also be taken in account.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I wonder which would be the best interval between 2 flips based on these conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Let's the object is a latex mattress that was completely submerged in a swimming pool (20cm thick, 170 long and 110 wide)&lt;/li&gt;&#xA;&lt;li&gt;Let's say it will dry in a garden (not in contact with the grass), under sunny condition (about 25°C), with a humidity of about 65%, and low wind (let say between 11.9 and 19.7 km/h ―or 7.4–12.2 mph― : Level 3 on the Beaufort scale)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I also wonder which of these 3 factors, evaporation, gravity or capillarity would be the major one first at the beginning of the drying and then toward the end? &lt;/p&gt;&#xA;" OwnerUserId="2708" LastEditorUserId="2708" LastEditDate="2015-08-19T14:01:06.390" LastActivityDate="2016-02-17T02:12:24.117" Title="What is the best way to dry something that is quite thick?" Tags="|hydraulics|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4075" PostTypeId="2" ParentId="4072" CreationDate="2015-08-19T13:15:20.870" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;what would be a good international standard to refer to&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I claim to be no expert in international (or even American) standards, but a quick search on my company's database gives me some options that may be worth considering.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The front part is exposed to the elements so it will need to be painted to protect it from corrosion. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I assume you mean that the housing will be steel since you are concerned about corrosion.  Consider having a look at the following standards:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;ISO 12944-5, &lt;em&gt;Paints and varnishes — Corrosion protection of steel structures by protective paint systems — Part 5: Protective paint systems - Second Edition&lt;/em&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ASTM D1014, &lt;em&gt;Standard Practice for Conducting Exterior Exposure Tests of Paints and Coatings on Metal Substrates&lt;/em&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ASTM MNL20, &lt;em&gt;Corrosion Tests and Standards: Application and Interpretation - 2nd Edition&lt;/em&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;specifying the corrosion resistance of the paint and the final products?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You mention paint, but a galvanized coating may be fitting for your application as well.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;ISO 1461, &lt;em&gt;Hot dip galvanized coatings on fabricated iron and steel articles - Specifications and test methods - Third Edition&lt;/em&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ASTM A123, &lt;em&gt;Standard Specification for Zinc (Hot-Dip Galvanized) Coatings on Iron and Steel Products&lt;/em&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-08-19T13:29:02.473" LastActivityDate="2015-08-19T13:29:02.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4077" PostTypeId="1" AcceptedAnswerId="4079" CreationDate="2015-08-19T14:20:49.930" Score="3" ViewCount="5955" Body="&lt;p&gt;Does a standard formula exist for calculating the (maximal) deflection of a constrained rectangular plate that is loaded with a evenly distributed pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this &lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Useful tables of mechanics&lt;/a&gt;, however I'm not sure if it's correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wNnzz.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wNnzz.gif&quot; alt=&quot;http://www.roymech.co.uk/images/Plate_12.gif&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2509" LastActivityDate="2015-08-19T15:27:02.813" Title="Deflection of constrained rectangular plate" Tags="|mechanical-engineering|deformation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4078" PostTypeId="1" CreationDate="2015-08-19T14:28:27.930" Score="3" ViewCount="803" Body="&lt;p&gt;When considering long-term prestressed girder camber (for a bridge), what are the implications (if any) of long-term downward deflection on girder performance? Should prestressing be designed to ensure that long term camber is always upward?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using the PCI multiplier approach to calculating long term deflection.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Utah DOT Structures Design and Detailing Manual page 14-30&#xA;  &lt;a href=&quot;https://drive.google.com/file/d/0B8QRVMpaE6oYWTg3WU5ISDhkRGM/view&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://drive.google.com/file/d/0B8QRVMpaE6oYWTg3WU5ISDhkRGM/view&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yCY0D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yCY0D.png&quot; alt=&quot;Utah DOT Camber&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2015-08-26T15:48:05.137" LastActivityDate="2015-08-26T15:48:05.137" Title="Should long-term prestressed girder camber always be upward?" Tags="|structural-engineering|bridges|prestressed-concrete|deflection|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4079" PostTypeId="2" ParentId="4077" CreationDate="2015-08-19T14:35:04.930" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;deflection of a constrained rectangular plate&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;By this I assume you mean that all edges are fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My general go-to for these types of formulations is &lt;em&gt;Roark's Formulas for Stress and Strain&lt;/em&gt;, 7th Edition.  These formulations assume a flat plate with straight boundary conditions and constant thickness.  Also, Poisson's ratio for the material is assumed to be $\nu = 0.3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Table 11.4 of that text, Case No. 8a, Uniform load over entire plate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{max} = \frac{\alpha q b^4}{Et^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\alpha$ value is found by considering the dimensions of the plate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UMXEp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UMXEp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3YXMD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3YXMD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this formula is really only strictly valid for materials with $\nu = 0.3$.  If the material has a different Poisson's ratio, they may need to be adjusted.  &lt;em&gt;Roark's&lt;/em&gt; lists the following references next to this formulation that may be of some help if you have a differing $\nu$ value:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Timoshenko, S., and J. M. Lessells: &quot;Applied Elasticity,&quot; Westinghouse Technical Night School Press, 1925.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Evans, T. H.: Tables of Moments and Deflections for a Rectangular Plate Fixed at All Edges and Carrying a Uniformly Distributed Load, ASME J. Appl. Mech., vol. 6, no. 1, March 1939.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Timoshenko, S., and S. Woinowsky-Krieger; &quot;Theory of Plates and Shells,&quot; 2d ed., McGraw-Hill, 1959.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I do not have copies of these documents, so I cannot verify their utility for this problem.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-08-19T15:27:02.813" LastActivityDate="2015-08-19T15:27:02.813" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4080" PostTypeId="2" ParentId="4078" CreationDate="2015-08-19T14:55:06.613" Score="2" Body="&lt;p&gt;The questions in your title and in your body-text are slightly different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prestressed girder camber should usually be upwards. This is because one usually applies sufficient prestressing to cancel out (or overcome) all possible dead loads and some fraction of live loads. This means that without live loads the prestressing will probably be larger than the applied dead loads, even if one only considers the bending moment component and ignores the compression component of the prestress. The girder's total camber will therefore be upwards in the absence of live loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, it is theoretically possible (though a terrible idea, economically speaking) to prestress a beam such that the camber is downward. Just prestress the beam along its centroidal axis, such that the prestress overcomes the tensile stresses by simple compression (making sure not to rupture the beam by compression). In such a case the prestress would not generate an upwards camber and the girder's total camber would remain downwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, onto the second question: there is no problem in having long-term downwards camber. That just means that your prestress' bending moment component is smaller than the dead loads' total bending moment. So long as your compression component is taking up the slack in the long-term, there's no problem (assuming it's not so dramatic as to cross the acceptable deflection limits in the code, of course).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2015-08-24T19:08:49.303" LastActivityDate="2015-08-24T19:08:49.303" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4081" PostTypeId="1" AcceptedAnswerId="4083" CreationDate="2015-08-19T16:40:53.750" Score="6" ViewCount="667" Body="&lt;p&gt;I am designing a pattern for a sand casting. It has only one critical dimension which is a 4&quot; inner diameter cylinder. If I use a 4&quot; core, will the casting shrink away from the core? In other words will the inner diameter of the cylinder end up being 4.1&quot; or something like that? What kind of shrinkage can I expect for ductile iron?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cylinder needs to be a sliding fit on a 4&quot; diameter painted, mill finish steel column. The inner diameter of the cylinder should end up being no more than 4.05&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="33" LastEditDate="2016-01-27T02:24:04.337" LastActivityDate="2018-06-27T12:37:37.640" Title="How do the dimensions of a ductile iron casting change as it cools?" Tags="|manufacturing-engineering|casting|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4083" PostTypeId="2" ParentId="4081" CreationDate="2015-08-19T18:08:03.547" Score="7" Body="&lt;p&gt;Ductile iron is actually likely to expand during the casting process, as shown in the table found on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Casting_(metalworking)#Solidification_shrinkage&quot;&gt;Casting Wikipeda page&lt;/a&gt;, which is in part cited from &lt;a href=&quot;https://books.google.com/books?id=JVTJi30phCwC&amp;amp;pg=PA66&amp;amp;source=gbs_toc_r&amp;amp;cad=4#v=onepage&amp;amp;q&amp;amp;f=false&quot;&gt;this casting text&lt;/a&gt; by Stefanescu. However, as the tables show, the range of volume change is fairly large and not one-sided, which is to say the ductile iron may either expand or contract depending on other conditions of the casting process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, another concerning aspect is the range of precision that you're looking for in this casting. A tolerance of $^{+0.05&quot;}_{-0}$ is pretty tight for a ductile iron casting. The ISO standard that lists general tolerances for castings (ISO 8062-3) says that for a machine-molded sand casting of ductile iron, the tolerance range on a linear dimension of $100-160mm$ is at least $\pm 0.035&quot;$, and could be up to $\pm0.14&quot;$. For a hand-molded casting, that goes from $\pm0.098&quot;$ to $\pm0.24&quot;$. These are general tolerances, so you could look for tighter tolerances on your critical feature, but I would suggest (if it's possible in your process) to plan on machining out that 4&quot; diameter and set your core diameter up so that you are cleaning up the entire circumference of the bore. Otherwise, I think hitting the tolerance range you are looking for as-cast will be tricky.&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-08-19T18:08:03.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4084" PostTypeId="2" ParentId="4078" CreationDate="2015-08-19T21:05:10.497" Score="1" Body="&lt;p&gt;The PCI manual (3rd Edition, Section 8.7.1) states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is not recommended that prestressing levels be increased in order to reduce or eliminate long-term downward deflection that might be predicted if the multipliers in Table 8.7.1-1 are used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-08-19T21:05:10.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4085" PostTypeId="2" ParentId="3409" CreationDate="2015-08-20T00:26:15.783" Score="2" Body="&lt;p&gt;I'm sure there are specific strategies relating to pipe flanges, but for the broader issue of controlling that all of your bolts are tight enough, here is some background information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When tightening bolts in any clamping application, what's really important is the total clamping force. In any simple joint, and specifically when dealing with pipe flanges, the clamping force in the joint is the same thing as the tension in the bolt (also known as preload.) Torque is sometimes used as a surrogate for preload since the two are correlated, but for a number of reasons, it is not a very good tool for measuring clamping force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between torque and preload is based on some things that are fairly easy to quantify like the thread pitch of the thread, but also on things that are hard to quantify like the accuracy of the threadform, the quality of the coating, the cleanliness of the fasteners, etc. The amount of torque that is translated into tension is a relatively small percentage of the overall torque, with the remainder overcoming friction, often quoted as 10%. Because many of these factors are hard to quantify and hard to control without very detailed quality control, torque does not always offer an acceptably accurate way of predicting preload. (More on the torque-tension relationship: &lt;a href=&quot;http://www.portlandbolt.com/technical/faqs/tension-vs-torque-explained-sort-of/&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt; &lt;a href=&quot;http://www.assemblymag.com/articles/83789-fastening-the-truth-about-torque-and-tension&quot; rel=&quot;nofollow&quot;&gt;2&lt;/a&gt; &lt;a href=&quot;http://www.smartbolts.com/wp-content/uploads/2012/05/TheTorque-TensionRelationship.pdf&quot; rel=&quot;nofollow&quot;&gt;3&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To overcome this problem, when the amount of preload is critical, we measure the tension in the fastener directly to calibrate our tightening system. One way to do this is with a device colloquially known by it's brand name, &lt;a href=&quot;http://www.skidmore-wilhelm.com&quot; rel=&quot;nofollow&quot;&gt;Skidmore-Wilhelm&lt;/a&gt;. A bolt can be tightened around this device, and the tension is measured. In structural applications in the US, per the &lt;a href=&quot;http://www.boltcouncil.org/files/2014RCSCSpecification-withErrata.pdf&quot; rel=&quot;nofollow&quot;&gt;RCSC (PDF Warning)&lt;/a&gt; it would be typical to tighten 3 bolts out of each batch against the machine, using either a torque wrench, or by turning the nut a certain number of degrees past 'snug-tight.' If those 3 bolts fall into the acceptable tension range, then that torque or angle is used to tighten the actual production bolts until that batch of bolts runs out, or the next day, when the calibration procedure is performed again. Torque or angle are still being used as surrogates for pretension, but they are being calibrated for all of the unknown variables regularly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More recently, some methods for measuring the tension directly have emerged that are affordable enough that they can be permanently used for each bolt (whereas the Skidmore-Wilhelm costs thousands of dollars.) These include bolts with heads that change color when preloaded, washers that squirt out paint when preloaded, and bolts with a splined end that breaks off when preloaded. These methods are generally considered to be more reliable than conventional bolting by calibrated torque or turn-of-the-nut (the method that uses angles) but do increase materials costs, so they may or may not be appropriate for your application.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-08-20T00:26:15.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4086" PostTypeId="1" CreationDate="2015-08-20T09:37:57.970" Score="13" ViewCount="266" Body="&lt;p&gt;I am investigating the effects of bubble oscillations (calculated using the Rayleigh-Plesset equation for bubble dynamics) on the turbulence in two-phase flows. Since I am using cryogenic fluids, the thermal effects also play a role. I am trying to find out the connection between the oscillation frequencies and how they influence the turbulence dissipation rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that, when the frequencies are high, i.e., when the liquid-vapour interface oscillates at high frequencies, the turbulent kinetic energy in the flow increases. So, will the dissipation rate also increase proportionally to the turbulent kinetic energy? (epsilon proportional to k^1.5)&lt;/p&gt;&#xA;" OwnerUserId="2303" LastEditorUserId="272" LastEditDate="2015-08-20T11:57:38.030" LastActivityDate="2024-03-01T03:03:55.430" Title="How do the bubble oscillations in two-phase flows affect the dissipation rate of turbulent kinetic energy?" Tags="|fluid-mechanics|multiphase-flow|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4087" PostTypeId="2" ParentId="4045" CreationDate="2015-08-20T16:13:12.693" Score="2" Body="&lt;p&gt;The only way to achieve a curved water stream without an external force* is to change the direction it is being sprayed. The way the effect below is accomplished is that the nozzle rotates from a vertical to a horizontal angle, which causes the illusion that the stream is curving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kl9eB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kl9eB.jpg&quot; alt=&quot;curved water illusion&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oval shape is created by a trumpet-type nozzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UTiZg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UTiZg.jpg&quot; alt=&quot;trumpet fountain&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing the deflection and reducing pressure causes the flow to bulge outward and not go as high. The fountain then just oscillates between a &quot;high and narrow&quot; stream, and a &quot;medium height and wide&quot; stream. The resulting shape then also oscillates, making it look like an oval that collapses in on itself and then reemerges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sw58.com/enproduct/prod77.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; has some more sample pictures. Those are composed from multiple individual nozzles, but the same principle applies to e.g. trumpet nozzles, which create a 3-dimensional flow from one nozzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Gravity is one force, but will always accelerate downward causing a parabola shape. Wind is another, but again it acts in one direction and is not predictable/constant. &lt;a href=&quot;http://www.phikwadraat.nl/water_bells/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Surface tension can actually produce a spherical curve like that with a water bell&lt;/a&gt;, but requires laminar flow and the water in your case looks rather turbulent. Also it wouldn't change shape.&lt;/p&gt;&#xA;" OwnerUserId="269" LastEditorUserId="2246" LastEditDate="2015-08-20T16:37:36.903" LastActivityDate="2015-08-20T16:37:36.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4090" PostTypeId="1" AcceptedAnswerId="4092" CreationDate="2015-08-21T14:19:19.387" Score="1" ViewCount="3633" Body="&lt;p&gt;For a National Board Exam Review:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To what height will a barometer column rise if the atmospheric&#xA;  conditions are 13.9psia and 68F and the barometer fluid is mercury?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer 28.3in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have encountered questions with irrelevant variables; I assume temperature is irrelevant here; so I try;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ P = \rho gh }$$&#xA;$${ 13.9\frac{lb}{in^2} = (13.6)(62.4 \frac{lb}{ft^3})(\frac{1ft}{12in})^3 (32.2\frac{ft}{s^2})(\frac{12in}{ft})h }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im getting h = 0.073...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why isn't my solution with the right answer?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastActivityDate="2016-07-21T11:05:40.437" Title="Barometer Column with Mercury" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4091" PostTypeId="1" AcceptedAnswerId="4191" CreationDate="2015-08-21T16:07:18.440" Score="7" ViewCount="137" Body="&lt;p&gt;I have an (AC) power grid topology with generators, consumers, power lines, tranformers, etc. I also have a &quot;snapshot&quot; of the system's state, i.e., I know how much real power the generators feed, how much real power is used by the consumers, I know the reactive power values, and voltages. I also know the parameters of the power grid's infrastructure such as the maximum power of the transformers or the impedance of the power lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I am facing now is this: Suppose one generator feeds more real power, and one consumer uses more real power. How can I calculate what the load of the power grid's nodes is? I want to ascertain that, e.g., no transformer or power line is loaded beyond its limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I basically know that there is something called &quot;power flow analysis,&quot; which I've been reading up. The matrices, Newton-Raphson approach to solving the equations, etc., is not the problem: What I do &lt;em&gt;not&lt;/em&gt; know is how to apply the matrices to arrive at an answer to my question. I basically need help to formulate the equations: What are my input variables, what are my results?&lt;/p&gt;&#xA;" OwnerUserId="1775" LastActivityDate="2015-09-03T12:17:35.843" Title="How to apply power flow analysis?" Tags="|electrical-engineering|power|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4092" PostTypeId="2" ParentId="4090" CreationDate="2015-08-21T16:14:22.383" Score="4" Body="&lt;p&gt;You are making this too complicated.  You look up the density of mercury at 68&amp;deg;F (20&amp;deg;C) and find that it's 0.49057 pounds-mass/in&amp;sup3;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1 g, (13.9 pounds/in&amp;sup2;)/(0.49057 pounds/in&lt;sup&gt;3&lt;/sup&gt;) = 28.3 inches.  Yes, it's really that simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature is slightly relevant in that the density of mercury changes slightly with temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that pounds-force and pounds-mass come out to the same thing under 1 g acceleration, such as when sitting on the surface of the earth.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2016-07-21T11:05:40.437" LastActivityDate="2016-07-21T11:05:40.437" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4093" PostTypeId="1" AcceptedAnswerId="4101" CreationDate="2015-08-21T16:19:21.160" Score="6" ViewCount="3032" Body="&lt;p&gt;From my understanding, there are two uses of a gearing system: to change the speed of output rotation (trading it with torque), and to change the axis of rotation. Now, in a car, for example, it is necessary to have multiple available gear ratios, to allow for high torque and high acceleration when the car begins to move from stationary, and also high speed when the car is already on the move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I know that some systems still have gears when they only used a fixed gear ratio. For example, in my lab, we work with a robot arm, and my colleagues often talk about the gearing system in the arm. But the arm does not repeatedly change its gear ratio like in a car. So what are these gears actually doing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are to change the speed / torque of the output of the motor, and this is a fixed gear ratio, then why was the arm not designed with a different motor entirely -- one which provides the desired speed / torque properties? My intuition is perhaps that it is easier to mass produce motors that have high speed and low torque, and so it is more economical to buy one of these generic motors and attach it to a gearing system, rather than design a bespoke motor that has very high torque at its output by default....is this correct?&lt;/p&gt;&#xA;" OwnerUserId="2727" LastEditorUserId="2727" LastEditDate="2015-08-21T18:04:59.683" LastActivityDate="2015-08-21T21:39:36.390" Title="What is the use of gears in a device with a fixed gear ratio?" Tags="|mechanical-engineering|motors|gears|torque|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4094" PostTypeId="2" ParentId="4093" CreationDate="2015-08-21T17:24:14.307" Score="6" Body="&lt;p&gt;Electric motors typically only operate well at high speeds, and compared to something like an internal combustion engine, relatively low torque. Fixed gear ratios are used to cope with this inherent limitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, some designs will use a gear train so that they can design a point of failure in their system that is not the motor. If the system, robot arm in your example, uses a very expensive motor, you might want a cheaper component to fail and protect your motor in the event of an overload.&lt;/p&gt;&#xA;" OwnerUserId="2728" LastActivityDate="2015-08-21T17:24:14.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4096" PostTypeId="2" ParentId="3908" CreationDate="2015-08-21T17:33:14.097" Score="2" Body="&lt;p&gt;You could couple the motor to an electric generator equipped with braking resistors. This is how dynos driven by test vehicles are slowed down.&lt;/p&gt;&#xA;" OwnerUserId="2728" LastActivityDate="2015-08-21T17:33:14.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4097" PostTypeId="1" AcceptedAnswerId="4100" CreationDate="2015-08-21T17:33:55.650" Score="3" ViewCount="7601" Body="&lt;p&gt;I have always thought that lateral load is horizontal and not vertical in direction. For example, from this &lt;a href=&quot;http://web.mit.edu/course/4/4.441/1_lectures/1_lecture18/1_lecture18.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;MIT notes&lt;/a&gt; and many others:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The lateral load resisting systems for earthquake loads are similar to&#xA;  those for wind loads. Both are designed as if they are horizontally&#xA;  applied to the structural system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But the textbook I have (Mechanics of materials, 7th edition, Gere and Goodno, page 412) shows a vertical load on a beam and says it is lateral load. Here is the page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DZboI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DZboI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a convention for loading directions and the authors happens to use different one or should not Q above be called transverse load?&lt;/p&gt;&#xA;" OwnerUserId="2730" LastEditorUserId="272" LastEditDate="2015-08-21T19:29:03.720" LastActivityDate="2015-08-21T23:59:32.637" Title="Standard specification for direction of lateral load for beams" Tags="|structural-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4098" PostTypeId="2" ParentId="3785" CreationDate="2015-08-21T17:35:49.520" Score="0" Body="&lt;p&gt;Festo makes complete assemblies that fit your description. You can also buy the individual components from them, or from a company like McMaster Carr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;www.festo.com&lt;/p&gt;&#xA;&#xA;&lt;p&gt;www.mcmaster.com&lt;/p&gt;&#xA;" OwnerUserId="2728" LastActivityDate="2015-08-21T17:35:49.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4099" PostTypeId="2" ParentId="4097" CreationDate="2015-08-21T17:46:47.167" Score="4" Body="&lt;p&gt;The definition of &lt;a href=&quot;http://dictionary.reference.com/browse/lateral&quot; rel=&quot;nofollow&quot;&gt;lateral&lt;/a&gt; is: of or relating to the side; situated at, proceeding from, or directed to a side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In beam loading problems lateral refers to the part of the load that is not directed along the length (i.e. axis) of the beam while, axial refers to the load which is directed along the axis of the beam.  The lateral load can be horizontal, vertical, or somewhere in-between.  Since the weight of the beam is generally insignificant in beam loading problems, horizontal and vertical are indistinguishable.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-08-21T17:46:47.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4100" PostTypeId="2" ParentId="4097" CreationDate="2015-08-21T17:48:57.667" Score="4" Body="&lt;p&gt;The nomenclature adopted by the authors is unusual, yes. Usually that load $Q$ would be called a transverse load, as you've mentioned. That being said, there is nothing that distinguishes a lateral load from a transverse load other than traditional nomenclature. As @Chris mentions in &lt;a href=&quot;https://engineering.stackexchange.com/a/4099/272&quot;&gt;his answer&lt;/a&gt; (which I saw while writing this), lateral is a perfectly fine term, it's just... not as common in such a case.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-08-21T17:54:05.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4101" PostTypeId="2" ParentId="4093" CreationDate="2015-08-21T17:58:45.627" Score="3" Body="&lt;p&gt;There are various reasons why one might chose to use a geared motor with a fixed gear ratio for a robot arm over an ungeared motor:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;As Carlton commented on your question, it may be difficult to find an ungeared motor that meets the specific speed/torque requirements for a given application and a motor that can provide enough torque ungeared may be too large/massive for the robot arm (and may be too power hungry as well)&lt;/li&gt;&#xA;&lt;li&gt;From a controls perspective, assuming the motor is small and the gear ratio is used to increase torque/reduce speed, a large change in the angle of the motor would result in a small change in the angle of the robot arm; this would mean that you might be able to more accurately control the position of the arm by measuring and commanding the position of the motor (bigger angle = easier to measure)&lt;/li&gt;&#xA;&lt;li&gt;It is also entirely possible, that they happened to have a motor of a certain size available and decided to use it instead of procuring a different, perhaps more expensive, motor for the arm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm sure there are many other factors to consider, but, at a high level, this should give you an idea as to why the robot arm may not have been designed to use an ungeared motor.&lt;/p&gt;&#xA;" OwnerUserId="1061" LastActivityDate="2015-08-21T17:58:45.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4102" PostTypeId="2" ParentId="3908" CreationDate="2015-08-21T18:24:19.647" Score="0" Body="&lt;p&gt;You could use almost anything as a load. As has been mentioned before, a brake would work, so would a second motor, driven as a generator with a bank of power resistors as a load (They can get nice and toasty). A variable pitch propeller in air would work, varying the pitch would change the load. So would a centrifugal pump circulating a fluid, if you had a flow control valve opening and closing it would vary the load. The possibilities are pretty much wide open. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I think the easiest to implement would be using a second motor and a resistor bank, but that's mostly because that's what I did way back when I was in college. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-08-21T18:24:19.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4103" PostTypeId="2" ParentId="4064" CreationDate="2015-08-21T19:07:15.200" Score="4" Body="&lt;p&gt;The slower you charge a Lithium battery the better. They are not keen on very fast charge/discharge fluctuations.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;One reason is that faster charging also results in higher temperatures which stresses a battery and therefore reduces battery life.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another reason is that lithium plating might occur, where the transport rate of the Li-ions exceed the rate at which they can be 'stored' in the anode in a process called intercalation. Instead, the lithium deposits as Li metal onto the anode, which can cause short circuit and degrades capacity, since there are less lithium ions to store electrons.&#xA;The following image might visualize the storage process better:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tMr9w.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tMr9w.png&quot; alt=&quot;Li-ion intercalation&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;intercalation in cathode. The opposite happens in the anode(electrons are stored during charge and released during discharge)&lt;/em&gt;&#xA;&lt;a href=&quot;http://tymkrs.tumblr.com/post/7846476684/lithium-ion-battery-how-does-it-work&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;EDIT: To answer your question directly:no, short charge cycles do not influence battery life as long as you charge slowly and try not to stress it by making full charge swings from 0 to 100%. In fact, ideally, but maybe not practically, you should try to keep it near 50% at all times(at least, this is the rule of thumb used in EVs). &lt;/p&gt;&#xA;" OwnerUserId="2731" LastEditorUserId="2731" LastEditDate="2015-08-26T14:41:58.630" LastActivityDate="2015-08-26T14:41:58.630" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4104" PostTypeId="2" ParentId="4093" CreationDate="2015-08-21T19:59:13.770" Score="1" Body="&lt;p&gt;Gearing and belting etc. is used to change the speed and more importantly the torque of the input device (motor). Rotary loads are accelerated by torque to a commanded speed. You reduce the inertia of a moving load by deploying a gearbox or other speed reduction device. If for example you have a 5:1 reduction ratio, the motor torque is increased by a factor of 5 and the inertia is (1/5)**2 the non reduced value.&lt;/p&gt;&#xA;" OwnerUserId="2732" LastActivityDate="2015-08-21T19:59:13.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4105" PostTypeId="1" CreationDate="2015-08-21T20:46:27.973" Score="2" ViewCount="794" Body="&lt;p&gt;I have a couple of spur gears (0.4 mm module, ISO8/AGMA9 precision, 24 and 120 teeth). I want to make their backlash as small as possible. But also I don't want them to jam. Where can I get the normal clearance between their pitch surfaces (and the tolerance of the clearance)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there is some handbook to use?&lt;/p&gt;&#xA;" OwnerUserId="2734" LastEditorUserId="2734" LastEditDate="2015-08-21T20:56:29.413" LastActivityDate="2015-08-22T02:14:06.943" Title="Gear coupling and clearance between pitch surfaces" Tags="|gears|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4106" PostTypeId="2" ParentId="4093" CreationDate="2015-08-21T21:39:36.390" Score="6" Body="&lt;p&gt;A given mass of copper and iron can only produce so much torque : increasing torque further would require more current which means (a) the increased magnetic flux saturates the iron and (b) the I^2R losses in the copper overheat the motor. So more torque requires more iron and copper; i.e. a bigger, heavier, more expensive motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, because increased torque comes at the expense of increased I^2R losses, electric motors are most efficient when used at light torque loadings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no equivalent fundamental limitation on speed. Brushed motors have limits from brush wear, friction, heating, and erosion from sparking but newer motors (BLDC etc) have no such considerations up to the speed limits of their bearings, and material strength limits under centrifugal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to use a motor efficiently you will run it fast producing relatively little torque. Then it has torque in reserve when you need it for short term purposes (acceleration etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you need high torque and low speed, gearing a high speed motor will usually be more efficient, lower cost, lighter, using less material and thus probably cheaper than a direct drive motor used at low speed. &lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2015-08-21T21:39:36.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4107" PostTypeId="2" ParentId="4097" CreationDate="2015-08-21T23:59:32.637" Score="1" Body="&lt;p&gt;In the general case, you need to resolve the transverse load into &lt;em&gt;two&lt;/em&gt; components, in the planes of the two principal axes of the beam cross-section. Of course when beginning a study of combined loads, one of the two transverse components would probably be zero, which is apparently the case in the quote from the textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some application areas (e.g. the frames of buildings in civil engineering), the two transverse directions are almost always &quot;lateral&quot; and &quot;vertical&quot;, and those names for the directions relate to the origin of the loads - for example wind loads are &quot;lateral&quot; on the sides of the building, but snow loads are &quot;vertical&quot; on the roof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other applications, there are no obvious directions to describe the orientation of the principal axes of the beam, and the general term &quot;transverse load&quot; (or &quot;two transverse load components&quot;) would be more appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without more context from the book, it's not very clear what scenario is being described. If there is an implication that the X-Y plane in the diagrams is horizontal and the Z direction is vertical (if it is relevant at all), then I think using the term &quot;lateral&quot; is sensible.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-08-21T23:59:32.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4110" PostTypeId="2" ParentId="4105" CreationDate="2015-08-22T01:41:30.747" Score="1" Body="&lt;p&gt;The &lt;a href=&quot;http://diyhpl.us/~bryan/papers2/machinery_handbook/27_Gears_10.pdf&quot; rel=&quot;nofollow&quot;&gt;Machinery's Handbook&lt;/a&gt; is pretty amazing when it comes to this sort of thing. Go to pg2067 and that's the section on backlash. Depending on whether your spur gears are coarse or fine pitch the tolerance and &quot;recommended or proper&quot; amounts are shown. These are Table 1 (pg2068) and Table 3 (pg2072). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have to convert your module ($m$) to diametral pitch (inches) ($P$): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P = \dfrac{25.4}{m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your centre distance is of course: $\dfrac{mN_1}{2} + \dfrac{mN_2}{2} = \dfrac{(0.4)(24)}{2} + \dfrac{(0.4)(120)}{2} = 28.8 \text{ mm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also your module is quite small! You must be dealing with tiny gears - if not then I'd double check your calculations.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-08-22T02:14:06.943" LastActivityDate="2015-08-22T02:14:06.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4111" PostTypeId="1" AcceptedAnswerId="4114" CreationDate="2015-08-22T02:42:44.827" Score="2" ViewCount="38401" Body="&lt;p&gt;For a National Board Exam Review:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the Reynold's Number, $Re$ for water at 20 °C flowing in an open&#xA;  channel. The water flowing at a volumetric rate of 200 gal/sec. The&#xA;  channel has a height of 4 ft and a width of 8 ft. At this temperature,&#xA;  water has a kinetic viscosity of 1.104*10^-5 ft^2/s&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The given answer is 600,000. I have tried to solve the problem like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ D_{eq} = \frac{2*4*8}{4+8} = \frac{16}{3} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ V = \frac{Q}{A} = \frac{200 \frac{gal}{sec} * \frac{1ft^3}{7.48gal}}{4.8 ft}  = 5.570409982 \frac{ft}{s}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ Re = \frac{VD}{\mu_k} = \frac{ 5.570409982 \frac{ft}{s} \times \frac{16}{3} ft }{ 1.104\times 10^-5 \frac{ft^2}{sec} }  = 2691019.315 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I using the wrong equivalent diameter? Why am I not getting the correct answer?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="368" LastEditDate="2015-08-22T03:28:37.393" LastActivityDate="2024-01-05T09:29:16.963" Title="Reynolds Number for a Rectangular Duct" Tags="|fluid-mechanics|open-channel-flow|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4112" PostTypeId="1" CreationDate="2015-08-22T04:05:59.850" Score="2" ViewCount="348" Body="&lt;p&gt;I am using ANSYS Workbench 16.0, static structure analysis. An assembly is imported from NX 8.0.&#xA;In Design Modeler, I modified some parts and regrouped them, at last there are 129 bodies, and 9 parts are formed (one part contains all useless bodies and is suppressed). I renamed the bodies one by one (including suppressed ones). Everything looks fine.&#xA;But when I open Mechanical, I find one big body is missing, and many strange bodies appear randomly in 8 parts, which do not ever exist in Design Modeler. &#xA;I have renamed every single body that exists, but the unexpected bodies have raw names. &#xA;They are small parts, small faces or just empty objects with names like NX assembly and part names.&#xA;I've reset and refreshed the 'Model' step in static structure many times, but that doesn't help.&#xA;Please tell me how I can fix it?&lt;/p&gt;&#xA;" OwnerUserId="2740" LastActivityDate="2015-08-22T04:05:59.850" Title="unexpected bodies appear in ansys mechanical" Tags="|ansys|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4113" PostTypeId="1" AcceptedAnswerId="6034" CreationDate="2015-08-22T04:38:51.010" Score="4" ViewCount="2602" Body="&lt;p&gt;The &lt;a href=&quot;https://drive.google.com/file/d/0BwUq8KuaDLbTU2Z3YVd1cGo5YUY3WHlybUpISS0tU3pYNlEw/view?usp=sharing&quot; rel=&quot;nofollow&quot;&gt;Peng Robinson&lt;/a&gt; Equation of State gives a better approximation of the critical compressibility factor $Z_c$, where experimental data ranges between $0.24$ and $0.28$ (&lt;a href=&quot;http://books.google.com/books?id=2TpCxMgoxoQC&amp;amp;lpg=PA113&amp;amp;dq=peng%20robinson%20critical%20compressibility&amp;amp;pg=PA118#v=onepage&amp;amp;q=peng%20robinson%20critical%20compressibility&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;). Since the calculated value should be $0.307$ for any substance (&lt;a href=&quot;http://books.google.com/books?id=f-u-kPUWmrIC&amp;amp;lpg=PA13&amp;amp;ots=DbcNy90Tqi&amp;amp;dq=0.307%20critical%20compressibility&amp;amp;pg=PA13#v=onepage&amp;amp;q=0.307%20critical%20compressibility&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Source&lt;/a&gt;), I'd expect to obtain a rather close number by re-applying numerically the Equation of State, given a point $(T_c,P_{c})$ on the Clapeyron's Diagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I try to solve the Equation of State for $Z_c$ it gives $Z_c=0.321379$ (&lt;a href=&quot;https://drive.google.com/file/d/0BwUq8KuaDLbTV1BGZ2hnZ0ZMbGNEazcxYU9wUDFmR0xjOHNZ/view?usp=sharing&quot; rel=&quot;nofollow&quot;&gt;Mathematica&lt;/a&gt;), a solution with discrepancy of $4.6\%$ from the theoretical value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also when I use sightly different parameters for this EoS from &lt;a href=&quot;http://en.wikipedia.org/wiki/Equation_of_state&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;, the numerical result with the same code is $Z_c=0.311155$. &lt;/p&gt;&#xA;" OwnerUserId="2404" LastEditorUserId="2274" LastEditDate="2015-10-22T18:20:37.267" LastActivityDate="2017-04-05T18:50:11.420" Title="Discrepancy of calculated critical compressibility factor in Peng Robinson EoS" Tags="|thermodynamics|chemical-engineering|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4114" PostTypeId="2" ParentId="4111" CreationDate="2015-08-22T09:14:13.287" Score="4" Body="&lt;p&gt;As Dan pointed out you got an error in your calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you start by calculating the hydraulic diameter, which is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_h = \frac{4 \cdot A}{U}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ = cross sectional area of your duct&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U$ = &lt;strong&gt;wetted&lt;/strong&gt; perimeter of the duct&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This applies to any shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_h = \frac{4 \cdot 8 \cdot 4}{8 + 4 + 4 } = 8 \: \mathrm{ft}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your definition of the Reynolds number is correct&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Re = \frac{v\cdot L}{\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L$ = characteristic length, in the case of a duct it's the hydraulic diameter $d_h$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v$ = velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nu$ = kinematic viscosity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \frac{\dot{V}}{A} = \frac{200 \: \mathrm{\frac{gal}{s}} }{32 \: \mathrm{ft^2}} = \frac{26.74 \: \mathrm{\frac{ft^3}{s}}}{32 \: \mathrm{ft^2}} = 0.8355 \: \mathrm{\frac{ft}{s}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inserting yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Re = \frac{0.8355 \:\mathrm{\frac{ft}{s}}\cdot 8 \: \mathrm{ft}}{1.104 \cdot 10^{-5} \: \mathrm{\frac{ft^2}{s}}} \cong 600 000$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that the Reynoldsnumber is dimensionless, in your final calculation this might be true, but take a closer look at your second line, the units do not equal $\mathrm{\frac{ft}{s}}$.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2015-08-22T10:14:18.800" LastActivityDate="2015-08-22T10:14:18.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4117" PostTypeId="1" AcceptedAnswerId="4126" CreationDate="2015-08-22T12:58:26.757" Score="0" ViewCount="1110" Body="&lt;p&gt;Recently there has been road construction around on the main road leading into my office building. I observed the construction crew removing asphalt in the form of a trench as showed in pictures below. Later on they filled up with aggregate and poured a layer of black tar and filled with something. The trench has been filled and resurfaced in line with the existing road surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zKb29.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zKb29.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Iir9K.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iir9K.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attach is the before picture (Satellite) captured from google maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u5cQj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u5cQj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attach is the before picture (Street) captured from google maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w7cWX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w7cWX.jpg&quot; alt=&quot;Google Maps Street Image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; The rain water is being directed to a near by manhole. What is the engineering explanation for digging the trench later on was poured with liquid tar (asphalt crack filler)? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.gov.uk/government/uploads/system/uploads/attachment_data/file/4382/practicalguidetostreetworks.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Practical Guide to Street Works&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-08-23T14:14:48.317" LastActivityDate="2021-03-31T16:05:20.527" Title="What is the purpose of digging up trench and then filling with something similar to asphalt crack filler?" Tags="|civil-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4118" PostTypeId="1" AcceptedAnswerId="4122" CreationDate="2015-08-22T15:52:33.933" Score="0" ViewCount="755" Body="&lt;p&gt;I'm trying to understand the concept of mechanical resonance.&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mechanical_resonance&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Mechanical_resonance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find much references about how exactly is the phenomenon known. I.e. how was it discovered and how is it measured in contemporary engineering applications?&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2018-06-29T13:05:27.413" Title="How is Mechanical Resonance detected?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4119" PostTypeId="1" CreationDate="2015-08-22T18:47:58.930" Score="4" ViewCount="3574" Body="&lt;p&gt;For a small water pump like this one &lt;a href=&quot;http://www.ebay.de/itm/Hot-351PSI-DC-12V-High-Pressure-Water-Pump-4-3-L-Min-For-Marine-Boats-RV-Ger-/161376369321?hash=item2592c7faa9&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, how can I roughly estimate how much thrust it would deliver considering the maximal pressure (PSI) and the flow rate (L/min). As a nozzle let's assume it is just a tube with a diameter d. I am a little bit confused because pressure and flow rate probably depend on each other as well as on the nozzle diameter.&lt;/p&gt;&#xA;" OwnerUserId="1849" LastActivityDate="2015-08-23T01:16:17.487" Title="Maximal thrust from water pump" Tags="|pumps|thrust|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4122" PostTypeId="2" ParentId="4118" CreationDate="2015-08-22T23:43:02.263" Score="3" Body="&lt;p&gt;Mechanical resonance is all around us. Attach a weight to a string and swing it back and forth, you will easily recognize when you have hit the resonance frequency.  Bang a metal pipe and you will hear its resonance frequencies.   Find an opera singer and you will recognize the resonance of a crystal wine glass when it shatters.  Watch a &lt;a href=&quot;https://www.youtube.com/watch?v=j-zczJXSxnw&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tacoma_Narrows_Bridge_(1940)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tacoma Narrows Bridge&lt;/a&gt; collapse and you will easily recognize its aero-elastic resonance. I don't think that the discovery of these phenomena can be ascribed directly to any individual or group.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how it is measured in contemporary engineering applications, there are a number of different ways.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Accelerometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Accelerometers&lt;/a&gt; attached to mechanical structures are frequently used.  Higher frequency, lower amplitude resonances can be identified with &lt;a href=&quot;https://en.wikipedia.org/wiki/Piezoelectric_sensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;piezo-electric displacement sensors&lt;/a&gt;.  Often measurements can be made optically/visually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter what sensor is used, the analysis of the data is usually done in the frequency domain.  This is done by taking the time domain signal measured by your sensor and numerically calculating it's Fourier transform.  The resonances show up as peaks at specific frequencies, the height and width of which tells you about the characteristics of the resonance.  An example frequency domain signal from &lt;a href=&quot;http://cqgplus.com/2015/07/22/spectral-analysis-in-resonant-interferometry-following-the-traces-of-thermal-deformation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;some work I did in grad school&lt;/a&gt; measuring the resonances of &lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_cavity&quot; rel=&quot;nofollow noreferrer&quot;&gt;optical cavities&lt;/a&gt; is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://cqgplus.com/2015/07/22/spectral-analysis-in-resonant-interferometry-following-the-traces-of-thermal-deformation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DW8jF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2018-06-29T13:05:27.413" LastActivityDate="2018-06-29T13:05:27.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="4123" PostTypeId="2" ParentId="4119" CreationDate="2015-08-22T23:52:03.797" Score="1" Body="&lt;p&gt;This would be a very rough estimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = F v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you assume that the pumps efficiency $\eta$ is 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the power we assume 2.7 A x 12 V = 32.4 W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the flow rate you assume $4.3 \mathrm{\frac{L}{min}} = 7,16 \cdot 10^{-5} \mathrm{\frac{m^3}{s}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is due to the fact that flow rate and pressure are linked, as you already pointed out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B9Wof.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B9Wofm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see that the lower the pump head, for example the height the fluid has to overcome, the lower the flow rate. So I assumed the specified flow rate is specified for a 0 m head. Typically you would have diagrams that indicate at which pressure heads which flow rates could be achieved. In some / most cases the manufacturers of cheap pumps state $\dot{Q}$ for the lowest pressure head and $p$ for the lowest flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the diameter of 10 mm the equations yield&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$32,4 \: W = F \cdot \frac{\dot{Q}}{A} =  F \cdot \frac{7,16 \cdot 10^{-5} \mathrm{\frac{m^3}{s}}}{0,005^2 \pi \:\mathrm{m^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = 35,54 \: N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be the highest possible thrust that pump could achieve. Considering that all assumptions we made were to maximize the thrust it will be much lower in reality.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2015-08-22T23:52:03.797" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4124" PostTypeId="1" AcceptedAnswerId="5303" CreationDate="2015-08-22T23:52:56.703" Score="3" ViewCount="547" Body="&lt;p&gt;I am just posting here to verify that I am properly using a tool that is new to me, a micrometer. Here is what the precision label specifies, basically that it can measure up to an inch and down to 1/1000&quot; (a notch on the round scale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qKpGi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qKpGim.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to describe how I was converting the reading on the linear and on the round gauges and you tell me if I did it right.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IRQMy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IRQMym.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above picture, the linear scale is reading 2/10&quot; and the round scale is reading 15/1000&quot;.  So the total gauge I am measuring is &lt;code&gt;0.2&quot; + 0.015 = 0.215&quot;&lt;/code&gt;.  I am particularly curious to verify if the round gauge's each notch is indeed 1/1000&quot; and whether I am using the correct numerator in the fraction.&lt;/p&gt;&#xA;" OwnerUserId="2738" LastEditorUserId="1586" LastEditDate="2015-09-03T22:31:37.490" LastActivityDate="2015-09-04T20:41:44.167" Title="How to read micrometer?" Tags="|measurements|tools|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4125" PostTypeId="2" ParentId="4119" CreationDate="2015-08-23T00:13:31.757" Score="0" Body="&lt;p&gt;Firstly in the link you've provided it isn't clear whether the pressure is 351 psi or 35 psi. In the description it says 35 psi and in the images it looks like this too, but in the title it states 351 psi. I think it may be 35 psi as stated &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00DLKT4OO&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thrust force of something like a nozzle or rocket is derived from the fact that force is the change in momentum over time. The general thrust equation of a fluid stream jet is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \dot{m_e}V_e + (P_e - P_a)A_e$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{m_e} = \text{mass flow rate exiting the nozzle (kg/s)} \Rightarrow 4.3 \text{ L/min} = 0.07167 \text{ kg/s} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{V_e} = \text{velocity of stream exiting the nozzle (m/s)} = \text{??}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_e = \text{pressure of fluid exiting the nozzle (N/m$^2$ or Pa)} = 35 \text{ psi}= 241.3 \text { kPa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_a = \text{atmospheric pressure (N/m$^2$ or Pa)} = 101.3 \text{ kPa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_e = \text{area of outlet (m$^2$)} = \dfrac{\pi d^2}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to get a pretty accurate result, the only piece of information missing is the velocity of the exit stream, as well as the diameter of the nozzle, which you may be able to get from the manufacturer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance say $d = 10 \text{ mm}$ and $V_e = 3 \text{ m/s}$ then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \dot{m_e}V_e + (P_e - P_a)A_e$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = (0.07167)(3) + (241,300 - 101,300)\dfrac{\pi(0.01^2)}{4}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = 0.215 + 11$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = 11.215 \text{ N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see the first term $\dot{m_e}V_e$ looks like it contributes a negligible amount to the force - so we can probably ignore $V_e$ anyways as it is mostly the pressure difference between the nozzle outlet and atmosphere contributing to the force. So provided you can find $d$ of the nozzle outlet then you could get a pretty decent approximation!&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-08-23T00:41:55.570" LastActivityDate="2015-08-23T00:41:55.570" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4126" PostTypeId="2" ParentId="4117" CreationDate="2015-08-23T01:04:50.793" Score="1" Body="&lt;p&gt;Given the very little detail you've provided, it's not possible to be sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't think it's any more complicated than a simple resurfacing. The top layer has been planed off, and a new top layer laid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planing is done when the underlying structure is fine, but there are superficial problems: a maintenance regime of frequent planing and resurfacing can help reduce the whole-life maintenance bill, by fixing problems early before they propagate to the underlying structural layers (layers which are much more expensive to replace).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia currently says that this process is also called &quot;pavement milling&quot;, &quot;cold planing&quot;, &quot;asphalt milling&quot;, and &quot;profiling&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible reason (of several) why one might just plane the strip alongside the footway, is that traffic can cause indentations, bumps and grooves along the surface, and when that's along the edge of the road, that creates ponding of water after rainfall: instead of the water all draining off, some of it gets trapped in the dents and grooves. This creates a problem for pedestrians and any other kerbside activity getting splashed, and can decrease the lifetime of the surface.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2015-08-23T01:04:50.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4127" PostTypeId="2" ParentId="4119" CreationDate="2015-08-23T01:16:17.487" Score="0" Body="&lt;p&gt;This is best done by doing a momentum balance instead of trying to calculate pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a first order estimate, use the flow rate and the nozzle crossection area to compute the exit speed of the water.  This is the momentum of the ejected water per unit mass.  This divided by the time per ejection of each unit mass (which you get from the flow rate) is the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force = momentum / time = mass * speed / time&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-08-23T01:16:17.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4128" PostTypeId="1" CreationDate="2015-08-23T01:16:18.940" Score="1" ViewCount="181" Body="&lt;p&gt;I am looking for a formula that states the breaking point of a composite - How much pressure it can withstand before it 'snaps'. Specifically I am looking for a formula that states the materials stiffness. &lt;/p&gt;&#xA;" OwnerUserId="2736" LastEditorUserId="1832" LastEditDate="2015-08-25T13:28:26.380" LastActivityDate="2015-08-25T13:28:26.380" Title="Looking for a formula that states the stiffness of a composite" Tags="|mechanical-engineering|materials|structural-engineering|" AnswerCount="1" CommentCount="6" ClosedDate="2015-08-23T18:23:41.077" ContentLicense="CC BY-SA 3.0" />
  <row Id="4129" PostTypeId="1" AcceptedAnswerId="4132" CreationDate="2015-08-23T01:59:35.437" Score="0" ViewCount="2598" Body="&lt;p&gt;For a National Board Exam Review:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A floating cylinder 8cm in diameter and weighing 950 grams is placed in&#xA;  a cylindrical container 20cm in diameter and partially full of water.&#xA;  The increase in the depth of water in the container due to placing the&#xA;  float in it is:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer is 3cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ P_1 = \frac{0.950kg * 9.81 \frac{m}{s^2}}{ \frac{\pi}{4} * { (20cm * \frac{1m}{100cm}})^2 } = 296.649 \frac{N}{m^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ \gamma = 9.81 \frac{kN}{m^3} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ P_2 = P_{atm} = 101325 \frac{N}{m^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ P_1 + \gamma h = P_2 }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${h = 10.29m }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is my error? Am I using wrong approach? Wrong formula?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="2170" LastEditDate="2015-08-23T03:21:33.107" LastActivityDate="2015-08-23T03:45:57.310" Title="Height Increase by a Floating Cylinder Over a Larger Cylinder" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4130" PostTypeId="1" CreationDate="2015-08-23T03:40:48.623" Score="0" ViewCount="4211" Body="&lt;p&gt;For a National Board Exam Review:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the pressure 7000ft below the water surface of the ocean?&#xA;  Neglect compressibility.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer is 512000psf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ P_2 = P_1 + \gamma h }$$&#xA;$${ 14.7 \frac{lb_f}{in^2} * (\frac{ 12in }{ 1 ft})^2 = P_1 + (62.4 \frac{lb_f}{ft^3}) 7000ft }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get: 434683.2 psf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I using right formula? Is there better approach?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastActivityDate="2020-12-08T06:40:50.827" Title="Pressure beneath the Ocean with Neglect to Compressibility" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4131" PostTypeId="2" ParentId="4129" CreationDate="2015-08-23T03:41:17.030" Score="1" Body="&lt;p&gt;20cm diameter gives a cross sectional area of $\pi r^2$ or 314.15 cm2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;8cm diameter uses the same formula for a cross sectional area of 50.26 cm2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free space around the float in which the water may rise is the difference in cross sectional areas, 314.15-50.26 = 263.886 cm2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The float, uh, floats, so the water that is displaced must weigh the same as the float, or 950g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;950g / 1 g/cm3 = 950 cm3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;950cm3 / 263.886cm2 =  3.6cm.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-08-23T03:41:17.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4132" PostTypeId="2" ParentId="4129" CreationDate="2015-08-23T03:45:57.310" Score="0" Body="&lt;p&gt;The first part of your maths is OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_1$ = $296.649$ $Pa$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pressure at the base of a fluid is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P$ $=$ $\rho gh$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearrange the equation &amp;amp; $h$ $=$ $P/\rho g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using your numbers &amp;amp; the fact that the density of water is $1$ $g/cm^3$ or $1000$ $kg/m^3$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h$ $=$ $296.649/(1000)(9.81)$ = $0.0302$ $m$ = $3$ $cm$&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-08-23T03:45:57.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4133" PostTypeId="1" AcceptedAnswerId="4135" CreationDate="2015-08-23T08:47:52.650" Score="1" ViewCount="1382" Body="&lt;p&gt;I am trying to understand what the dynamic factor $K_v$ describes. I know that a low $K_v$ factor results in quieter and smoother work of the gear but what exactly does $K_v$ tell me?&lt;/p&gt;&#xA;" OwnerUserId="2749" LastEditorUserId="272" LastEditDate="2015-08-23T10:49:07.203" LastActivityDate="2018-05-30T10:26:00.947" Title="Gears - dynamic factor $K_v$" Tags="|mechanical-engineering|gears|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4135" PostTypeId="2" ParentId="4133" CreationDate="2015-08-23T11:06:39.500" Score="2" Body="&lt;p&gt;A &lt;a href=&quot;https://www.google.com/search?q=dynamic%20factor%20gear%20agma&quot; rel=&quot;nofollow&quot;&gt;google search&lt;/a&gt; for this actually returns pretty good information.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;The dynamic factor $K_v$ accounts for internally generated gear tooth loads which are induced by non-uniform meshing action (transmission error) of gear teeth.&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;It provides allowance for overload effects generated by a pair of meshing gears&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;It is used to account for the effect of tooth spacing and profile errors...&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt; It is a way to account for transmission errors. Transmission errors occur due to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Inaccuracies in the manufacture of the gear (gear teeth are never perfect).&lt;/li&gt;&#xA;&lt;li&gt;Meshing of the gear teeth during action (vibration due to tooth stiffness during meshing, tooth friction, tooth wear etc.).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The gear ratio will fluctuate during operation because of these transmission errors. Therefore, the gears will always experience slight accelerations/decelerations causing dynamic forces - hence it is called a 'dynamic factor'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a change to the way it was defined (reciprocated) by AGMA. The recent standard is that the factor should be above 1. Note that if $K_v = 1$ this represents perfectly accurate gearing. Under 1.1 is &quot;very accurate gearing&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-08-23T11:32:14.480" LastActivityDate="2015-08-23T11:32:14.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4136" PostTypeId="1" CreationDate="2015-08-23T13:54:01.503" Score="4" ViewCount="602" Body="&lt;p&gt;I want to know what bearings might allow a sphere (5cm diameter) to roll in-place most freely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest mechanism that matches what I might want to achieve is a &quot;ball transfer&quot;, I believe the term is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been unable to find:&#xA; - an explanation of the internal mechanisms they use&#xA; - if there are different implementations&#xA; - how freely the ball can roll, and in what motion relative to the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to have a mechanism that supports a ball the size and weight of a snooker ball, so it can be used as a trackball (to replace a PC mouse, or touchpad). An optical sensor from an optical mouse would be placed underneath, to detect the movement of the ball, so spinning on that point is not necessary, but the sphere should have the lowest possible friction when its surface sweeps over the sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason low friction is required, is it would allow the user to transverse a large screen distance by giving the ball momentum, and then stopping it would allow the user to resume precise placement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution I considered was to have extremely small castor-wheels, three of which would be placed around the base of the ball. But I have not found any such castors; their wheels would have to have a diameter of around 1cm or less. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea was to use a number of ball transfers to support the sphere itself, but again, ball transfers seem to be rather bulky, their balls having a diameter around 2-3cm, plus a rather large mechanism surrounding them. And I do not know how much friction they would still have, since they seem to be made to support far heavier loads (&gt;20kg) than that of the snooker ball (250g)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me any other solutions, and/or highlight pros/cons I need to be aware of?&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-01-13T21:43:15.290" Title="Bearing to allow a sphere to roll freely" Tags="|bearings|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4137" PostTypeId="2" ParentId="4128" CreationDate="2015-08-23T18:20:26.540" Score="1" Body="&lt;p&gt;I don't think there is a formula. For metals, at least, these values are all derived empirically via extensive testing, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Charpy_impact_test&quot; rel=&quot;nofollow&quot;&gt;the Charpy V-notch test&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Tensile_testing&quot; rel=&quot;nofollow&quot;&gt;tensile testing&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Fatigue_(material)&quot; rel=&quot;nofollow&quot;&gt;fatigue testing&lt;/a&gt;, etc. Typically you would find this information published in trade manuals or released by the manufacturer of the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If such a formula did exist, then couldn't you use it to solve for the &quot;ultimate&quot; composite material? &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-08-23T18:20:26.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4138" PostTypeId="1" CreationDate="2015-08-23T19:08:35.423" Score="3" ViewCount="869" Body="&lt;p&gt;I am investigating the use of stucco masonry versus Exterior Insulation and Finish Systems (EIFS) in the construction industry. I have no knowledge or training in this matter and my knowledge is limited to what I have read in the past few days.&lt;/p&gt;&#xA;&lt;h3&gt;Stucco&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R0n0f.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R0n0f.jpg&quot; alt=&quot;Stucco Cross Section&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As I understand it, both stucco and EIFS have a long history in the building construction industry. I have also read that there are two type of EIFS, PB (polymer-based) and PM (polymer modified) systems widely used in the building construction industry.&lt;/p&gt;&#xA;&lt;h3&gt;Exterior Insulation and Finish Systems (EIFS)&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zdpz7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zdpz7.jpg&quot; alt=&quot;Exterior Insulation and Finish System&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What are the moisture related issues in the of Stucco and EIFS with regards to mold buildup with respect to &lt;strong&gt;corrugated aluminum siding&lt;/strong&gt;? Once again my primary interest is with to respect to moisture and mold.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://diy.stackexchange.com/questions/9556/how-do-i-mitigate-moisture-problems-when-putting-up-drywall-over-stucco-with-tor&quot;&gt;How do I mitigate moisture problems when putting up drywall over stucco with torn tar paper?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Exterior_insulation_finishing_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Exterior insulation finishing system&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pieglobal.com/eifs-vs-stucco-whats-the-difference/&quot; rel=&quot;nofollow noreferrer&quot;&gt;EIFS vs. Stucco: What’s the Difference?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.firstext.com/stucco.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stucco System&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-08-24T22:28:37.850" Title="What are the moisture related issues in the use of Stucco and EIFS in the building construction Industry?" Tags="|civil-engineering|" AnswerCount="0" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4139" PostTypeId="1" CreationDate="2015-08-24T00:31:05.233" Score="4" ViewCount="328" Body="&lt;p&gt;We are working on a project requiring a temperature-controlled oven for melting lead, tin and other low-temperature metals.  The inner cylinder is aluminum.  The heat comes from Kanthal wire, for which we needed an electrical insulator capable of withstanding the heat.  The problem is, something must hold the fiberglass in place.  We just tried to melt the stuff end-to-end (does not hold) and with an overlap (gets blackened, shrinks, but does hold a tiny bit).  We have thought of a couple of alternative ideas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Add glass powder, sand, or broken glass to the area of the joint and have more glass available to melt&lt;/li&gt;&#xA;&lt;li&gt;Get some kind of glass fiber and &quot;sew&quot; the layers together somehow&lt;/li&gt;&#xA;&lt;li&gt;Some kind of chemical adhesive that can handle the same extremes of temperature&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The plan is to wrap a thin layer of fiberglass around the inner aluminum, then wrap with Kanthal wire.  Then wrap more fiberglass around that and cover with aluminum, and do enough layers that the outside is not hot to the touch.  An Arduino will control the circuit, and slowly bring the inside up to exactly the temperature we want using a thermocouple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we hold the fiberglass insulation in place without interfering with or shorting out the Kanthal wires that will go around the fiberglass?  &lt;/p&gt;&#xA;" OwnerUserId="2758" LastEditorUserId="368" LastEditDate="2015-08-24T15:56:33.107" LastActivityDate="2015-09-03T20:29:49.177" Title="How to connect butted or overlapping fiberglass insulation" Tags="|thermal-insulation|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4140" PostTypeId="1" AcceptedAnswerId="4167" CreationDate="2015-08-24T04:37:34.770" Score="3" ViewCount="14300" Body="&lt;p&gt;I made a test in Ansys but the result doesn't seem to be the final steady state. The two bodies didn't touch each other. How can I get the right result?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e809N.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e809Nm.jpg&quot; alt=&quot;manual contact&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/xQjmF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQjmFm.jpg&quot; alt=&quot;boundary condition&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZOgeL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZOgeLm.jpg&quot; alt=&quot;result&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2740" LastEditorUserId="272" LastEditDate="2015-08-24T15:28:58.383" LastActivityDate="2018-08-03T18:44:38.977" Title="How to simulate the contact between two distant parts in Ansys workbench？" Tags="|applied-mechanics|simulation|ansys|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4141" PostTypeId="1" AcceptedAnswerId="4147" CreationDate="2015-08-24T13:00:57.333" Score="3" ViewCount="518" Body="&lt;p&gt;The UK is about to start trials of &lt;a href=&quot;http://www.thetimes.co.uk/tto/life/motoring/article4524398.ece&quot; rel=&quot;nofollow&quot;&gt;wireless charging for moving cars&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of power transfer efficiency is likely to be achievable in such a system, compared to just plugging in the car directly?&lt;/p&gt;&#xA;" OwnerDisplayName="anon" LastEditorUserId="272" LastEditDate="2015-08-24T13:52:25.967" LastActivityDate="2015-08-24T19:28:53.900" Title="Efficiency of wireless charging for moving cars" Tags="|power-engineering|inductive-charging|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4143" PostTypeId="2" ParentId="4141" CreationDate="2015-08-24T13:56:21.127" Score="4" Body="&lt;p&gt;Based upon the link you provided, it doesn't sound like they know for certain what the end technologies will be and how they'll be tuned.  So the final answer regarding charging efficiency likely won't be known until the trial is completed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Inductive_charging&quot; rel=&quot;nofollow&quot;&gt;inductive charging&lt;/a&gt; cites a system that claims 86% efficiency for near field inductive charging for vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.wirelesspowerconsortium.com/technology/why-not-a-wire-the-case-for-wireless-power.html&quot; rel=&quot;nofollow&quot;&gt;Texas Instruments presentation&lt;/a&gt; cites an inductive charging system efficiency in the ballpark of 70%.  It also points out that depending upon how a wired charging solution is configured, a better configured inductive charging system may be more efficient for total system efficiency.&lt;sup&gt;1&lt;/sup&gt;  And it's worth noting that the voltage levels used in this study (5-7V) are significantly lower than what vehicle chargers use (120+V).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the biggest challenge in this trial will be the fact that the inductive charging is no longer &quot;near field.&quot;  That will likely have the greatest impact on overall system efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;&lt;em&gt;It's worth calling out the inherent bias from the piece as it's thinly veiled marketing material for newer generation TI chips.&lt;/em&gt;&lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-08-24T13:56:21.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4144" PostTypeId="2" ParentId="4118" CreationDate="2015-08-24T14:40:03.557" Score="1" Body="&lt;p&gt;I don't think you could point to any one person or group and say, &quot;They discovered mechanical resonance.&quot; It's something that's always been around and people have used it to their advantage for centuries, if not millennia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done more than a few vibration tests looking for the kind of resonances where the response is detrimental to whatever's being tested in a variety of machines or devices. A typical approach for a resonance search is to use some kind of measured, controlled input, such as a shaker table, and record the motion (usually with an accelerometer) of both the table and the DUT (device under test). The first step is usually a &quot;sine sweep&quot; where the shaker is operating over a range of frequencies, sweeping up and then down (or vice versa) and the data is examined for regions where the response is high relative to the excitation. In the end, what you're trying to identify are those frequencies where the DUT has a sensitivity that are also present in the environment which the device is supposed to operate.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-08-24T14:40:03.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4145" PostTypeId="2" ParentId="4141" CreationDate="2015-08-24T17:55:42.487" Score="-1" Body="&lt;p&gt;There's a name for &quot;inductive charging system coupled to a moving vehicle&quot;: a &#xA;&lt;strong&gt;Linear Induction Motor.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you tried running the car above a stationary magnetic field, it would just create drag and consume more of the battery's energy than it replenished. It would be like trying to use a battery to power a generator, then using the generator's output to try and charge the same battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you actively modulated the magnetic field so it appeared to be moving forward (relative to the car), you'd be better off at that point forgetting about the battery, disengaging the transmission (putting the car in neutral), and using it AS a linear induction motor to drag the car along.&lt;/p&gt;&#xA;" OwnerUserId="2763" LastActivityDate="2015-08-24T17:55:42.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4146" PostTypeId="1" CreationDate="2015-08-24T18:56:46.540" Score="5" ViewCount="197" Body="&lt;p&gt;I'm looking to redesign a seal my company currently uses. The original task was to re-source it from a lower cost supplier with the same design as is in current production. However, the supplier doesn't work with the type of material the seal is currently made of (rubber) so in the change of material, we may look at potential design changes. In the interest of cost savings, I'm looking to remove a thread feature from the seal and the mating part, and replace it with a snap fit feature. However, I'm concerned about the ability of the injection molding process to create the current design I'm considering. I'd normally go the supplier, but it's been two weeks since the last (unrelated) item I've sent to them and I haven't heard back, so I'm turning to Engineering.SE for a quicker turnaround time.&lt;/p&gt;&#xA;&lt;p&gt;A rough sketch of the design is below. The right edge is where the part is currently threaded, and that lip would interface with a similarly sized slot in the mating part, locating the seal properly and preventing it from pulling away (the part inside the smaller diameter lip must translate relative to the seal).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VaYU5m.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VaYU5m.png&quot; alt=&quot;Seal Sketch&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Potentially further complicating this is the fact that the outer diameter is currently reinforced with a steel ring that is fully enclosed in the rubber (plastic in the redesign.) Our customer will probably be looking to keep the strength and rigidity added by this.&lt;/p&gt;&#xA;&lt;p&gt;What I'm ultimately asking is whether it'll be possible to injection mold this part, particularly whether it's possible to construct a mold that can be removed from this part without destroying it. As a follow-up, if injection molding is entirely impossible, are there other plastic forming processes that could be used to bond and form around a steel ring and create such a part? Extrusion clearly won't work and I can't see vacuum forming working either, and the volume is too high for rapid prototyping to be feasible, but beyond that I don't know enough about the various options as far as plastic forming. (I haven't settled on a replacement material yet, but I'm guessing it'll be some sort of polyurethane.&lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-01-10T22:41:15.127" Title="Limits of Injection Molding Regarding Overhung Edges" Tags="|manufacturing-engineering|plastic|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4147" PostTypeId="2" ParentId="4141" CreationDate="2015-08-24T19:28:53.900" Score="4" Body="&lt;p&gt;This is not exactly a comprehensive answer, but I found an interesting article on &lt;a href=&quot;http://eandt.theiet.org/news/2014/apr/onroad-charging.cfm&quot; rel=&quot;nofollow&quot;&gt;The Institution of Engineering and Technology&lt;/a&gt;'s website. It is fairly interesting, if not incredibly technically detailed, and I encourage you to check it out for projects to learn more about. I'll talk about the most interesting parts here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has many interesting projects listed. Some are for bus routes, and rather than charging along the whole route, they use 'opportunistic charging' to charge at key places (presumably bus stops and traffic lights) to rapidly transfer energy. This is intended to minimize the disruption and possibly cost of construction (imagine having to dig up an entire busy road to add charging coils). This is called &quot;semi-dynamic charging&quot; by Transport Scotland, who are working on one such project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it sounds like in Korea, a more full route system has been operating.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[T]he Korea Advanced Institute of Science and Technology (KAIST) is running two online electric vehicle (OLEV) buses on a 12km continuous charging route in the city of Gumi. It claims 85 per cent maximum efficiency in power transfer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're question is focused on the pure efficiency of the charging capacity, but I think also interesting to consider (or possibly implied in the question) is for what cases these technological limitations make actual use a practical exercise. On that note, I will add that this technology may prove of more usefulness on highways, where range is of particular concern in the arena of electric vehicles, leading to projects going to great lengths to provide very close charging stations (25 to 50 miles) like &lt;a href=&quot;http://www.westcoastgreenhighway.com/electrichighways.htm&quot; rel=&quot;nofollow&quot;&gt;West Coast Green Highway&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2371" LastActivityDate="2015-08-24T19:28:53.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4149" PostTypeId="1" AcceptedAnswerId="4168" CreationDate="2015-08-24T21:36:03.453" Score="3" ViewCount="5647" Body="&lt;p&gt;&lt;a href=&quot;http://co.routt.co.us/building/handouts/notching.pdf&quot; rel=&quot;nofollow&quot;&gt;IBC 2308.8.2&lt;/a&gt; states a rule of not notching more than 1/6 the depth of a wooden joist (emphasis added):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Framing details.&lt;/strong&gt; Joists shall be supported laterally at the ends and at each support by solid blocking except where the ends of the&#xA;  joists are nailed to a header, band or rim joist or to an adjoining&#xA;  stud or by other means. Solid blocking shall not be less than 2 inches&#xA;  (51 mm) in thickness and the full depth of the joist. Notches on the&#xA;  ends of joists shall not exceed one-fourth the joist depth. Holes&#xA;  bored in joists shall not be within 2 inches (51 mm) of the top or&#xA;  bottom of the joist, and the diameter of any such hole shall not&#xA;  exceed one-third the depth of the joist. &lt;em&gt;Notches in the top or bottom&#xA;  of joists shall not exceed one-sixth the depth and shall not be&#xA;  located in the middle third of the span.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Would it be acceptable to notch a joist more than 1/6th of the depth if adding a steel plate or sistered wood beam (going the full length) or some other method (short of putting a column in the middle)? I need to cut about 2.5&quot; out of a 9.5&quot;-deep beam to get an HVAC duct through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The joist is between the basement and first floor. Above is, I believe, the living room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The joist spans 10'.&lt;/p&gt;&#xA;" OwnerUserId="2765" LastEditorUserId="368" LastEditDate="2015-08-25T15:49:14.793" LastActivityDate="2015-08-25T16:10:24.923" Title="Is it acceptable, with reinforcement, to notch a floor joist more than 1/6 of the depth?" Tags="|structural-engineering|beam|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4150" PostTypeId="2" ParentId="1864" CreationDate="2015-08-24T22:43:39.660" Score="2" Body="&lt;p&gt;In the United States, you can often use &lt;a href=&quot;http://nationalbridges.com/&quot; rel=&quot;nofollow&quot;&gt;nationalbridges.com&lt;/a&gt; and &lt;a href=&quot;http://uglybridges.com/&quot; rel=&quot;nofollow&quot;&gt;uglybridges.com&lt;/a&gt; to ascertain the current status of a bridge, along with some basic information about the owning agency/maintenance agency. Ordinary road/highway bridges are usually inspected on a biennial basis, and given a sufficiency rating out of 100. If a bridge is identified as having structural problems it will be listed as &quot;structurally deficient&quot;. Those with substandard geometry, safety railing, or insufficient vehicle capacity will be listed as &quot;functionally obsolete&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-08-24T22:43:39.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4151" PostTypeId="1" AcceptedAnswerId="4152" CreationDate="2015-08-24T22:47:30.640" Score="2" ViewCount="912" Body="&lt;p&gt;I have seen this somewhere in the wild, perhaps a space-efficient storage system in lieu of two shelving units. I'm trying to reproduce, and failing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine two stacks of trays which hold things that must not fall out, so the trays must remain horizontal. The trays are all connected to the same track, which loops around, so that the first stack is on &quot;the way down&quot; for the track, and the second stack on &quot;the way up&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: how are the trays connected to the track, so that they can move along the track eternally without ever tipping over?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sort of remember that there is a wheel on each tray that counteracts the rotation coming from the track, but I don't remember how.&lt;/p&gt;&#xA;" OwnerUserId="2766" LastEditorUserId="272" LastEditDate="2015-08-25T11:40:26.640" LastActivityDate="2015-08-25T11:40:26.640" Title="How to keep a tray horizontal when moving around a closed, vertical track?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4152" PostTypeId="2" ParentId="4151" CreationDate="2015-08-24T23:55:26.153" Score="3" Body="&lt;p&gt;Image in my mind: The track is on a wall, with the trays being perpendicular from the wall. The track is rectangular (with rounded corners obviously). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution 1:&lt;/strong&gt;&lt;br&gt;&#xA;If the things that the trays are holding are relatively light (to the tray/shelf), or if its feasible to perfectly balance the load on each tray, then you could just use gravity. You could attach the tray to the track carriage (for lack of a better word) via a bearing, and then just ensure that the CoG of the tray+contents is well below the pivot point (where the bearing is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution 2:&lt;/strong&gt;&lt;br&gt;&#xA;Have an additional track/slot that controls the angular position of the tray (by it's position, so it would not just 'follow' the other tracks). It would probably need to be a slot, and the 'track' would probably need to be redesigned as a slotted/indented track as well that was of greater depth than your angular-controlling slot so that the angular-control-slot could pass over the normal slots/track. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PYmQ0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PYmQ0.jpg&quot; alt=&quot;crude hand-drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be an idea to offset the arm by something other than the depicted offset (let's call it 90 degrees) as then the two tracks would overlap less. &lt;/p&gt;&#xA;" OwnerUserId="2768" LastEditorUserId="2768" LastEditDate="2015-08-25T00:22:10.337" LastActivityDate="2015-08-25T00:22:10.337" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4153" PostTypeId="1" AcceptedAnswerId="4154" CreationDate="2015-08-24T23:58:14.887" Score="1" ViewCount="867" Body="&lt;p&gt;For a National Board Exam Review: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A 29.53&quot; * 39.37&quot; pressure vessel contains ammonia with f = 0.041. Compute the minimum required discharge capacity of the relief device in $\mathrm{\frac{kg}{hr}}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution is shown as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = fDL = 0.041(\frac{29.53}{39.37})(\frac{39.37}{39.37}) = 0.03075 \mathrm{\frac{kg}{s}} = 110.71 \mathrm{\frac{kg}{hr}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't get is why DIVIDE? Isn't it already length and diameter?&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="368" LastEditDate="2015-08-25T15:34:07.260" LastActivityDate="2015-08-25T15:34:07.260" Title="Why divide the dimensions of a pressure vessel by 39.37 when calculating discharge capacity?" Tags="|fluid-mechanics|refrigeration|pressure-vessel|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4154" PostTypeId="2" ParentId="4153" CreationDate="2015-08-25T01:00:22.800" Score="3" Body="&lt;p&gt;This is because of units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Getting the right answer with your units and their approach&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the people have done in your question is converted everything to metres. The division of 39.93 is because that's how you convert inches into metres. L's value is actually 1m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1000mm}{25.4mm} = 39.37&#xA;$$&#xA;thus:&#xA;$$&#xA;39.37\frac{inches}{metre}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the equation that you're using is actually&#xA;$$&#xA;C\frac{kg}{sec} = f(metric) \cdot D(metres) \cdot L(metres)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The math&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.muellerindustries.com/uploads/pdf/Pressure%20Relief%20Valves_up.pdf&quot; rel=&quot;nofollow&quot;&gt;this report&lt;/a&gt; C is either in kg/min or lb/min, while D and L are in ft (or m). The value of f changes depending on whether you've been given the metric or imperial unit one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The valve you're dealing with is an R717 (look at the second page of the above report) and you've been given the metric value. This valve has an imperial F value of 0.5. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C \frac{lb}{min} = 0.5 \cdot \frac{29.54}{12} \cdot \frac{39.37}{12} = 4.03 \frac{lb}{min} &#xA;$$&#xA;&lt;em&gt;The #/12 is to get the value from inches to ft&lt;/em&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then converting to kg/hr:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C \frac{kg}{hr} = C \frac{lb}{min} \cdot \frac{60min}{hour} \cdot \frac{1kg}{2.2lb}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C \frac{kg}{hr} = 4.03 \frac{lb}{min} \cdot \frac{60min}{hour} \cdot \frac{1kg}{2.2lb} = 110.094 \frac{kg}{hr}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is a bit of an unfair question since whenever you're dealing with cross-unit-systems it's hard to keep track of what's going on. Did they at least give you the units of f so that you knew you had been supplied with the metric value?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2768" LastEditorUserId="2768" LastEditDate="2015-08-25T01:36:54.217" LastActivityDate="2015-08-25T01:36:54.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4155" PostTypeId="2" ParentId="2300" CreationDate="2015-08-25T01:09:44.783" Score="13" Body="&lt;p&gt;$Lb_m$ is not the base unit. The Slug is the base unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$32.2\ lb_m = 1\ slug$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert $1\ lb_m$ to $lb_f$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\ lb_m * \frac{1\ slug}{32.2\ lb_m} * 32.2 \frac{ft}{s^2} = 1\ lb_f$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $1\ lb_m$  will yield $1\ lb_f$ on Earth at STP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=4ePaKh9QyC8&quot;&gt;This video&lt;/a&gt; does an excellent job of explaining it.&lt;/p&gt;&#xA;" OwnerUserId="2769" LastEditorUserId="368" LastEditDate="2015-12-02T19:58:50.307" LastActivityDate="2015-12-02T19:58:50.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4156" PostTypeId="2" ParentId="4153" CreationDate="2015-08-25T01:37:15.927" Score="2" Body="&lt;p&gt;By dividing the diameter and length dimensions, which are in inches, by 39.37 you are converting them to metres. The problem you are trying to solve has inputs that are in both English and metric units, but the answer needs to be in metric/SI, so you need to convert everything to SI units prior/during calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 inch = 2.54 cm = 0.0254 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/0.0254 = 39.37&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was doing the calculation I would have multiplied by 0.0254 to convert inches to metres. It's just how I do things. The person who came up with the answer you're quoting divided by 39.37 instead.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-08-25T01:37:15.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4157" PostTypeId="2" ParentId="4140" CreationDate="2015-08-25T01:44:15.837" Score="1" Body="&lt;p&gt;Unless you're trying to use ANSYS to validate common-sense/newtonian physics then start the analysis with them already touching. &lt;/p&gt;&#xA;" OwnerUserId="2768" LastActivityDate="2015-08-25T01:44:15.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4158" PostTypeId="1" CreationDate="2015-08-25T07:12:42.227" Score="1" ViewCount="427" Body="&lt;p&gt;This model illustrates my question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Red is a motor&lt;/li&gt;&#xA;&lt;li&gt;Green is the drive shaft &lt;/li&gt;&#xA;&lt;li&gt;Orange are nylon straps &lt;/li&gt;&#xA;&lt;li&gt;Blue are spools for the straps&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I want to be able to drive multiple spools in multiple directions at the same time. They should be individually controllable and able to stay still, wind, or unwind. They may all move at the same speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess is that this has something to do with idler gears, planetary gears, and servo-controlled clutches. But I don't really understand how to employ these, or know whether I'm on the right track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate any help! I'm especially interested in solutions which are cost effective and robust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems common enough that there must be a solution already?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rewLd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rewLd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2773" LastActivityDate="2015-08-25T07:12:42.227" Title="Gearing for Multiple Forward / Reverse on Single Drive Shaft" Tags="|mechanical-engineering|motors|gears|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4159" PostTypeId="1" AcceptedAnswerId="4205" CreationDate="2015-08-25T08:03:14.777" Score="4" ViewCount="10642" Body="&lt;p&gt;What are the advantages and disadvantages of bevel gears?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I've already found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Advantages:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;possibility to change operation angle&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disadvantages:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;must be precisely mounted&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;shafts' bearings will be exposed to significant forces&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;limited gear ratio &amp;lt; 5&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do you know any other? Especially advantages.&lt;/p&gt;&#xA;" OwnerUserId="2749" LastActivityDate="2018-07-11T11:23:08.190" Title="Advantages and disadvantages of bevel gears" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4161" PostTypeId="2" ParentId="4136" CreationDate="2015-08-25T08:24:18.930" Score="3" Body="&lt;p&gt;There are a few patents for trackball that uses simple wheels. You can check them. An older one, is &lt;a href=&quot;http://www.google.com/patents/US4952919&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trackball mechanism US 4952919 A&lt;/a&gt;. It's bearing can be seen from this assembly drawing. &lt;a href=&quot;https://i.stack.imgur.com/oQ8cC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oQ8cC.png&quot; alt=&quot;Trackball mechanism US 4952919 A&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see it uses two wheels with friction padding that rotates with rotation of sphere that are used to discern x-y motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I don't understand your reason for less friction, if you're looking to reduce friction mounting sphere use ball transfer's design. You can see a cross section of SKF's ball transfer design below &lt;a href=&quot;http://www.skf.com/binary/30-97863/Ball-Transfer-units.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;from it's catalog.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QcqMj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QcqMj.png&quot; alt=&quot;SKF Balltransfer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a cheap and dirty solution, buy a balltransfer that has diameter 5cm ball, remove housing, drill a hole to the housing for sensor, put your sensor and assemble it. &lt;/p&gt;&#xA;" OwnerUserId="879" LastActivityDate="2015-08-25T08:24:18.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4162" PostTypeId="2" ParentId="2300" CreationDate="2015-08-25T11:18:59.847" Score="7" Body="&lt;p&gt;The textbook is incomplete.   Newton's Law is &lt;em&gt;usually&lt;/em&gt; written $F=ma$.   The SI unit of mass is the $kg$ and that of force is the $N$.   One of the advantages of SI is that it clarifies the distinction between mass and force (especially weight).   In the old British Imperial system there are several options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;we can measure mass in pounds_mass $lbm$;  the corresponding force&#xA;unit is the seldom-used poundal $pdl$.&lt;/li&gt;&#xA;&lt;li&gt;we can measure force in pounds_force $lbf$;  the corresponding mass&#xA;unit is the $slug$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, you'll often see $lbm$ and $lbf$ in the same document.   This is perfectly acceptable: it's equivalent to normalising Newton's Law with the gravitational acceleration to give $F=ma/g$.   It's the failure to state this that leads to the confusion.&lt;/p&gt;&#xA;" OwnerUserId="2520" LastActivityDate="2015-08-25T11:18:59.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4163" PostTypeId="2" ParentId="4058" CreationDate="2015-08-25T13:54:52.763" Score="4" Body="&lt;p&gt;As discussed in the earlier answer, if we look at the curve of moment capacity versus unbraced length we see three regions of behavior -- yielding, inelastic LTB, and elastic LTB (see Fig. C-F1.1 in the AISC &lt;em&gt;Steel Construction Manual&lt;/em&gt;). It is important to note that we only have inelastic LTB due to residual stresses. This is where that $0.7F_yS_x$ term is coming from (residual stresses are assumed to be $0.3F_y$). It is also important to note that the equation for critical stress at elastic LTB is in the form $\alpha + \sqrt{1+\beta}$ and is not a function of yield stress. &lt;em&gt;Alpha&lt;/em&gt; is a term for out-of-plane buckling of the compression flange and &lt;em&gt;beta&lt;/em&gt; is a term for torsional stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7aVZE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7aVZE.png&quot; alt=&quot;Moment Capacity&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, conceptually, we could look at a curve that disregards residual stresses - meaning we have just yielding and elastic LTB. When we increase $F_y$, the elastic LTB curve remains the same while $M_p$ increases. The result is that we transition to elastic LTB at a smaller unbraced length. One way to think about it is that with an increased $F_y$, it takes more force to yield the member, making it more likely that it will buckle before yielding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KWz9w.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KWz9w.png&quot; alt=&quot;Moment Comparison&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2015-08-25T14:30:06.330" LastActivityDate="2015-08-25T14:30:06.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4164" PostTypeId="1" CreationDate="2015-08-25T14:51:11.980" Score="1" ViewCount="113" Body="&lt;p&gt;A fan of Elon musk but have hard time understanding his engineering approach when he says use &quot;first principles&quot;. Could anyone elaborate on that ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we got a problem at hand what is the first principle way to solve the problem ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference&#xA;www.youtube.com/watch?v=NV3sBlRgzTI&lt;/p&gt;&#xA;" OwnerUserId="2777" LastEditorUserId="2777" LastEditDate="2015-08-26T10:30:30.707" LastActivityDate="2015-08-26T10:30:30.707" Title="What does it mean to design something by using &quot;first principles&quot;?" Tags="|product-engineering|" AnswerCount="0" CommentCount="1" ClosedDate="2015-08-25T15:27:37.963" ContentLicense="CC BY-SA 3.0" />
  <row Id="4165" PostTypeId="1" CreationDate="2015-08-25T15:22:28.050" Score="8" ViewCount="1899" Body="&lt;p&gt;I have a product of which we have shipped something like 500 units over the last five years. This product is not intended to be user serviceable; a failure of any component results in the unit being replaced. The majority of these units have never seen problems and are still working fine. Some have been damaged and come back for repairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about computing mean time to failure (MTTF)? Would I include only the units that have failed? Or would I also figure in all units that are currently operating? What about the fact that I only have the date of sale, not the date of installation? And that I don't know what fraction of the installed time the unit is running? Should I just make reasonable assumptions?&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="272" LastEditDate="2015-08-26T11:51:51.457" LastActivityDate="2015-09-02T22:58:42.700" Title="How do I properly compute mean time to failure?" Tags="|statistics|mathematics|failure-analysis|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4166" PostTypeId="1" CreationDate="2015-08-25T15:43:25.517" Score="3" ViewCount="828" Body="&lt;p&gt;I was wondering how you would &lt;strong&gt;control the flow of powder from a container&lt;/strong&gt; (with a small cylindrical dispenser at the bottom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The container is filled with powder. As the container drains, less pressure will be exerted on the bottom most powder, affecting the flow rate (at least, I think).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I select 32.5 grams and hit a button. What process would you use to deliver this measurement of powder from the nozzle?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Would a lasar work?&lt;/li&gt;&#xA;&lt;li&gt;Would there have to be a measurement of the current mass?&lt;/li&gt;&#xA;&lt;li&gt;Would you need to know density?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I also posted this on &lt;a href=&quot;http://www.quora.com/How-can-you-use-fuzzy-logic-to-control-the-flow-of-powder-from-a-container-in-varying-amounts?snids=1306788656%2C1306270469%2C1306270433%2C1306270405&amp;amp;nsrc=2&quot; rel=&quot;nofollow&quot;&gt;Quora&lt;/a&gt; and got a very nice introductory answer but something more specific could help me.  Also, I was given a link to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation&quot; rel=&quot;nofollow&quot;&gt;Weisbach equation&lt;/a&gt;, but the contents of the container will be a powder. Probably varying in their size and density as some powders will be finer than others. But in general, how would you solve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the pharmaceutical industry uses expensive equipment to achieve this, but I am not thinking industrial. They are filling tiny little pill capsules with exact proportions. The focus here is household use, with maybe 15–40 g servings&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am looking to do is input varying size of serving. 20 g, 22.5 g, 35.5 g serving. Optimally, the system will have an electronic motor component, CPU and a valve that opens/closes with use of the motor...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my part time jobs is at McDonalds. We have this cream/milk dispensing machine. You input double-double, and out comes 2 creams for the corresponding cup size. I wonder how those work? The paramount different in systems being those dispense liquids, and we're looking to dispense powder solids.&lt;/p&gt;&#xA;" OwnerUserId="2778" LastEditorUserId="368" LastEditDate="2015-08-25T21:06:15.133" LastActivityDate="2015-08-28T16:57:18.380" Title="Controlling the flow of powder from a container" Tags="|mechanical-engineering|materials|fluid-mechanics|pressure|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4167" PostTypeId="2" ParentId="4140" CreationDate="2015-08-25T15:58:08.803" Score="4" Body="&lt;p&gt;Okay so I ended up having a play in ANSYS Mechanical Workbench. I did a mock up of your problem. Here's how I've been able to successfully model it. Note that a lot of my decisions are based on the &lt;a href=&quot;http://www.ansys.com/staticassets/ANSYS/staticassets/resourcelibrary/presentation/2014-sd-mechanical-contact-best-practices.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Contact Best Practices Guide from ANSYS&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some very important points given your problem but also for contact modelling in general and a frictionless contact type:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As stated by Trevor there's no need for such a large gap initially - you don't need ANSYS to simulate that sort of behaviour because it should be obvious. So move the roller closer to the ring. I ended up moving the roller within 1mm of the ring. Just doing this on its own, however, isn't enough to get it to converge.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mesh $\rightarrow$ Advanced $\rightarrow$ Shape Checking $\rightarrow$ Aggressive Mechanical (This is generally a good idea for any contact modelling).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After selecting the contact type as frictionless look under Advanced $\rightarrow$ Formulation $\rightarrow$ Normal Lagrange. It has highest accuracy of all the methods, can handle large sliding and interference fit. If you use Augmented Lagrange with frictionless it's likely to diverge when the contact is over a small surface area like your problem. On the other hand if you were considering a bonded or no-separation contact type then Augmented Lagrange formulation seems to converge well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Frictionless Contact $\rightarrow$ Advanced $\rightarrow$ Interface Treatment $\rightarrow$ &lt;strong&gt;Adjust to Touch&lt;/strong&gt; (should be obvious why I chose this).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Frictionless Contact $\rightarrow$ Advanced $\rightarrow$ Pinball Region $\rightarrow$ Radius. Set the radius to be at least as large as the gap between the surfaces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Below was the deflection result (left) and Von-Mises Stress result (right). As you can see the stress distribution is &lt;strong&gt;not&lt;/strong&gt; great - around the contact area in the ring there are probably large stress concentrations. You might need a better mesh around the contact area as I spent hardly any time meshing. The main reason for this exercise was to prove that even though there is a gap between the contact surfaces there's appropriate controls in place to model it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1GSxM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1GSxM.png&quot; alt=&quot;ANSYS deflection and stress&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're expecting large deflections then you can turn ON &quot;Large Deflection&quot; in Analysis Settings $\rightarrow$ Solver Controls $\rightarrow$ Large Deflection. That might help it converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're wondering whether you could do this without moving the roller closer to the ring then you could probably play around with the Pinball Radius as well as another option called &quot;Predict For Impact&quot; in the Time Step Controls. I've never used this personally and I'm not sure what effect that would have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the time, read the &lt;a href=&quot;http://www.ansys.com/staticassets/ANSYS/staticassets/resourcelibrary/presentation/2014-sd-mechanical-contact-best-practices.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;best practices&lt;/a&gt; for contact modelling.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-08-25T16:04:17.303" LastActivityDate="2015-08-25T16:04:17.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4168" PostTypeId="2" ParentId="4149" CreationDate="2015-08-25T16:10:24.923" Score="3" Body="&lt;p&gt;&lt;strong&gt;The 1/6th rule&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IRC allows a notch 1/6th of the depth of a wooden joist (but not within the middle third) on the basis that the joist is sized sufficiently for the stress at midspan, and a 5/6th depth beam in the end thirds is hence sufficient for the lower stress experienced there. Hence a 1/6th depth notch will not affect the design capacity of the joist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Engineering&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In engineering terms it is eminently possible to reinforce a beam such that a deeper notch will not cause the beam to fail. A simple rule applicable in this case is that the maximum stress is proportional to the width of the beam times the depth cubed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if you wanted a beam that was 1/2 of the depth of the original beam, then in order to provide the same stress as the original beam would have experienced, you'd need a beam that was 8 times the width!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at your numbers, and trying to minimise the width as much as possible, I would say:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The IRC rules suggest you're allowed a beam depth of 5/6 of 9.5&quot; = 7.92&quot;.&lt;/li&gt;&#xA;&lt;li&gt;You want a beam that has a depth of 9.5&quot; - 2.5&quot; = 7&quot;&lt;/li&gt;&#xA;&lt;li&gt;Hence your depth is reducing by a factor of 7 / 7.92 = 0.884&lt;/li&gt;&#xA;&lt;li&gt;Hence your breadth needs to increase by a factor of 1 / (0.884^3) = 1.45&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So attaching a wooden joist to the side which is half as wide as your original beam should be sufficient. Alternatively, by using a steel plate (as you suggested) you would need much less material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are of course key points that I've missed here. You need to transfer the stress from the original beam into the attachment. You would hence need to design connections to transfer the stress, and over an appropriate length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The real world&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most jurisdictions it should be possible for engineering calculations to be carried out, proving that the reinforcement/modification is sufficient. Once the change is then approved by the local authority (on the basis of the calculations), work will then be able to go ahead. I would advise you to either contact a local engineer or contact your local building inspector.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Joist width vs IRC width&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned in one of your edits (since superseded I believe) that the joist was wider than the minimum required by IRC. Possibly this is because the joist is wider and/or longer than suitable for the minimum width. I would hence advise against thinking &quot;the joist is wider than the minimum width, therefore I can notch deeper&quot;.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-08-25T16:10:24.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4169" PostTypeId="2" ParentId="4166" CreationDate="2015-08-25T16:37:55.930" Score="4" Body="&lt;p&gt;Ok, stick with me. I am going to tell you how to do this at home on the cheap and easy, but first lets talk about a few industrial approaches. Because maybe you want to try for something more exciting! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several industrial approaches that are used for metering (this is the technical term you are looking for) solids out of container. They range from an &lt;a href=&quot;https://www.youtube.com/watch?v=TZ6S7FXAbUY&quot; rel=&quot;nofollow&quot;&gt;Auger or Screw Conveyor&lt;/a&gt; to my favorite the &lt;a href=&quot;https://www.youtube.com/watch?v=cLCjUXvLSsk&quot; rel=&quot;nofollow&quot;&gt;Shaking Screw Conveyor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be a fun project to build either of those at home, but what you are looking for is a far simpler batch style dispenser. This is accomplished with 2 large valves and a secondary metering vessel. At the bottom of your main storage container there is a valve A, which leads to the metering vessel; at the bottom of this vessel is valve B. To meter out exactly the right amount, size the metering vessel so that it holds the mass you want (i.e., 20 g). All you do is close valve B, open valve A, wait for the metering vessel to fill, close valve A and then open valve B. You will get exactly the same amount every time and you don't need to worry about the flow rate changing as the container empties. all that you need to do is be sure you size the valves large enough so that the solid can fall through them. If its a powder than a 1&quot; valve will work just fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a boring but exceptionally simple solution. Personally, I would try to build the shaking conveyor. &lt;/p&gt;&#xA;" OwnerUserId="2780" LastEditorUserId="368" LastEditDate="2015-08-28T16:57:18.380" LastActivityDate="2015-08-28T16:57:18.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4170" PostTypeId="1" CreationDate="2015-08-25T17:20:08.670" Score="6" ViewCount="1527" Body="&lt;p&gt;In an ideal diesel cycle, the cut-off ratio is set by the maximum gas temperature after combustion. That's all fine and good, but in a real-world engine what controls the cut-off ratio, and what are the limits to its flexibility? I feel like it must be the valve timing, but perhaps it is controlled by the fuel injection? Perhaps both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it common to have variable cut-off ratios based on ambient temperature to achieve higher power ratings at lower temperatures? Or is this not possible in a real diesel cycle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example: If an engine runs both during the summer (35 °C) and the winter (-10 °C) in the Midwest can its cut-off ratio be adjusted such that the final post-combustion gas temperature is the same in both cases leading to much more power in the winter than in the summer? &lt;/p&gt;&#xA;" OwnerUserId="2780" LastEditorUserId="2780" LastEditDate="2015-08-28T22:11:25.367" LastActivityDate="2017-03-22T14:33:03.150" Title="Limitations of the cut-off ratio in real-world diesel engines" Tags="|mechanical-engineering|engines|diesel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4172" PostTypeId="1" AcceptedAnswerId="8783" CreationDate="2015-08-25T20:58:12.193" Score="7" ViewCount="85" Body="&lt;p&gt;Is there a method to test for thermal oxide on a painted steel part after the paint has cured?  I'd rather the method be nondestructive but it only being destructive in the presence of thermal oxide is also acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically looking for laser oxide under 3mils of polyurethane, if that helps.&lt;/p&gt;&#xA;" OwnerUserId="2557" LastEditorUserId="33" LastEditDate="2015-08-26T02:55:20.583" LastActivityDate="2016-08-05T23:24:49.463" Title="What is a method to detect thermal oxide layer after steel is painted?" Tags="|manufacturing-engineering|product-testing|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4173" PostTypeId="2" ParentId="4165" CreationDate="2015-08-25T23:33:57.397" Score="4" Body="&lt;p&gt;If you don't have hard data, making assumptions (preferably &quot;reasonable&quot; ones) is the only option you have. (Maybe that's why engineers used to call their slide rules &quot;guessing sticks...&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't ignore the fact that most of the units have &lt;em&gt;not&lt;/em&gt; failed so far. A plausible approach to this would be to use the times-to-failure that you know, to fit the parameters of statistical model of the failure process. You also need to check that the predictions of the model are consistent with the raw data, before you use it to forecast anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A commonly used model in reliability engineering is the Weibull distribution, which can represent quite a wide range of different &quot;root causes&quot; of failure, and will automatically adjust to use the &quot;best&quot; shape of probability curve (within limits, of course) to match your real-world data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google will find plenty of hits for &quot;Weibull distribution tutorial&quot; etc, but if you are new to this it would be a good idea to get an overview of &quot;reliability engineering&quot; before you pitch into the details. A good place to start would be a professional engineering organization, for example the &lt;a href=&quot;http://asq.org/index.aspx&quot; rel=&quot;nofollow&quot;&gt;American Society for Quality (ASQ)&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most practical way to make the estimate would be to use some computer software rather than figuring out how to do the math by hand, but without more specifics of the problem, it's hard to recommend any particular package.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="1586" LastEditDate="2015-09-02T22:58:42.700" LastActivityDate="2015-09-02T22:58:42.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4175" PostTypeId="2" ParentId="2612" CreationDate="2015-08-26T01:28:16.907" Score="0" Body="&lt;p&gt;The &lt;code&gt;[MaxBotix HRLV MaxSonar-EZ][1]&lt;/code&gt; line (I'm looking at EZ4) claims to have a very steady analog output with 1mm accuracy, and has some supporting test documentation for their claims. About $40/sensor.&#xA;I'd like the I2C version in order to eliminate analog weirdnesses in the circuitry as a factor, but so far the I2C ones only go to 1CM accuracy.&lt;/p&gt;&#xA;" OwnerUserId="2784" LastActivityDate="2015-08-26T01:28:16.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4176" PostTypeId="2" ParentId="4165" CreationDate="2015-08-26T03:35:54.713" Score="5" Body="&lt;p&gt;First off, always remember that garbage in = garbage out; so if your data is garbage then your statistics will be garbage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation your optimal data would be something like Run Hours Until Failure and your entire dataset would have failed already.  With this in mind you may want to choose a conservative number from whatever statistic you calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you only have failure from sale date this may be skewed toward a higher MTTF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since not all of your product has failed yet you can look at a smaller subset of your population, say the 1st six months of production.  A higher percentage of these have failed most likely (since the product you sold last week shouldn't fail this week, hopefully).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your failure proportion is still too low then you may have to try to fit the data to a distrubution keeping in mind that you only have the low proportion of the distribution, i.e. you must extrapolate from the dataset to a fitted curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Weibull_distribution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weibull Distribution&lt;/a&gt; would work well here and is commonly used for MTTF data.  The idea here is to fit the proportion of your dataset which has failed to a corresponding proportion of a distribution.  If your proportion of products in your dataset which have failed was 48.66% then you would fit it to that probability on your hypothesized distribution as shown by the shaded area in the following image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z5Gze.jpg&quot; alt=&quot;PDF&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be rather intensive, however, for anything besides an exponential distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method of extrapolation is by  &lt;a href=&quot;http://reliawiki.org/index.php/Degradation_Data_Analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Degradation Analysis&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2557" LastActivityDate="2015-08-26T03:35:54.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4177" PostTypeId="2" ParentId="3776" CreationDate="2015-08-26T12:41:41.207" Score="0" Body="&lt;p&gt;In the UK roofs with access is 1.5 kN/m&lt;sup&gt;2&lt;/sup&gt;. But the roof may not have been designed for access. It is safe to say however that the roof fabric will have been designed for access during construction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not give an easily understood criterion such as no more than two people on the roof at any one time and to limit equipment carried to say 50 kg? At least this is clearer than an 'X' kN/m&lt;sup&gt;2&lt;/sup&gt; criterion to the layman.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were any long term works to be carried out or if there was any concern you should inspect from the underside, exposing structure from removal of any ceiling locally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside a report was done for Stanhope in the UK some years ago into insitutional over specifying of floor loads in the City of London (5 kN/m&lt;sup&gt;2&lt;/sup&gt; sometimes). One of the examples in the report showed a crowd standing in close formation with a quoted calculated load of 2.5 kN/m&lt;sup&gt;2&lt;/sup&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2786" LastEditorUserId="368" LastEditDate="2015-08-26T15:13:19.113" LastActivityDate="2015-08-26T15:13:19.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4178" PostTypeId="1" CreationDate="2015-08-26T13:44:31.220" Score="3" ViewCount="2177" Body="&lt;p&gt;I came up with the following design for a single wheel trailer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jRy8d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jRy8d.png&quot; alt=&quot;Single wheel trailer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NpfD8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NpfD8.png&quot; alt=&quot;Single wheel trailer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shock absorber is missing in this drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to mount the swingarm to the horizontal plate? It needs to pivot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about pillow blocks, but I'm not sure if they are suitable for this application. Design should be as simple as possible and easy to fabricate.&lt;/p&gt;&#xA;" OwnerUserId="2218" LastActivityDate="2017-12-04T03:39:37.737" Title="Designing the pivot point for a swingarm" Tags="|mechanical-engineering|design|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4179" PostTypeId="1" CreationDate="2015-08-26T14:29:07.657" Score="8" ViewCount="112" Body="&lt;p&gt;Assuming maximum efficiency is not the &lt;em&gt;primary&lt;/em&gt; concern (but net positive efficiency is), to what extent could a wind turbine operate in water and vice versa? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do the different turbine types fare in this respect? I.e., HAWT / VAWT (Savonius, Darrieus, Gorlov) for air turbines, reaction (Francis, Kaplan, Gorlov) / impulse (Turgo, Jonval) turbines (Pelton wheels are probably out) for water turbines?&lt;/p&gt;&#xA;" OwnerUserId="2788" LastEditorUserId="368" LastEditDate="2015-08-26T15:09:39.697" LastActivityDate="2015-08-26T15:39:46.467" Title="To what degree are wind and water turbines exchangeable?" Tags="|turbines|renewable-energy|wind-power|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4180" PostTypeId="2" ParentId="4179" CreationDate="2015-08-26T15:39:46.467" Score="5" Body="&lt;p&gt;Given that water is roughly 1000&amp;times; as dense as air, and incompressible besides, it should be fairly obvious: A wind turbine would be destroyed by the forces that water would put on it, and a water turbine wouldn't even notice the forces created by wind.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-08-26T15:39:46.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4182" PostTypeId="1" AcceptedAnswerId="4187" CreationDate="2015-08-26T17:21:01.837" Score="5" ViewCount="5477" Body="&lt;p&gt;For a given laminate made of composite material (composed of fibers oriented in a carefully designed sequence coated and embedded in a matrix material), where can I begin estimating the stiffness and/or strength of the material?  How can I know (within a beginning estimate tolerance) how to design a part out of composite materials?&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="272" LastEditDate="2015-08-26T20:55:16.033" LastActivityDate="2017-10-03T14:10:48.550" Title="How do I calculate an estimate for the properties of a composite material?" Tags="|mechanical-engineering|materials|composite|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4183" PostTypeId="2" ParentId="3043" CreationDate="2015-08-26T18:08:11.323" Score="1" Body="&lt;p&gt;This is a very interesting question. It might even be worth submitting to Modern Steel Construction's &lt;a href=&quot;http://msc.aisc.org/en/modernsteel/resources/steel-interchange/&quot; rel=&quot;nofollow&quot;&gt;Steel Interchange&lt;/a&gt;. Per the AISC 360-10 user note at the beginning of Chapter F, you are allowed to use the interaction equations in Sections H1 for beams in biaxial flexure without axial load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, Salmon &amp;amp; Johnson's &lt;em&gt;Steel Structures Design and Behavior&lt;/em&gt; 5th Edition, Section 7.11 suggests that the beam-column interaction equation is unconservative without axial load. The author suggests not relying on any plastic capacity. That is, for the yielding limit state, Mnx = FySx and Mny = FySy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on my Google searches, it appears that Australian standard AS4100 Section 8.3.4 &lt;a href=&quot;https://wiki.csiamerica.com/download/attachments/7636185/Steel+Structures+Design+Manual+to+AS+4100.pdf&quot; rel=&quot;nofollow&quot;&gt;uses this approach&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-08-26T18:08:11.323" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4186" PostTypeId="1" AcceptedAnswerId="4188" CreationDate="2015-08-26T18:19:54.513" Score="4" ViewCount="3363" Body="&lt;p&gt;I am trying to figure out if a motor I have is capable of spinning a wheel. Here is all of the information I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MOTOR: 190 g.cm min.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WHEEL: 26 cm diameter, weighs about 59 grams&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that I would have to calculate the linear speed of the wheel in order to convert it to the same units that the motor torque is listed in. However, I don't know if I am doing it correctly or what the next step would be to solve this. &lt;/p&gt;&#xA;" OwnerUserId="2164" LastEditorUserId="2164" LastEditDate="2015-08-28T03:04:11.953" LastActivityDate="2015-08-28T03:04:11.953" Title="How to calculate if a motor is capable for a load?" Tags="|mechanical-engineering|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4187" PostTypeId="2" ParentId="4182" CreationDate="2015-08-26T18:25:42.263" Score="5" Body="&lt;p&gt;There are quite a few ways to find out the stiffness and/or strength of a composite material, with a good understanding of the materials that make up the composite material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Pre-engineered composite materials&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several materials, prevalent mainly in the aerospace industry, are pre-engineered and pre-designed.  &lt;a href=&quot;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#safe=active&amp;amp;q=gy70+339+prepreg&quot; rel=&quot;nofollow noreferrer&quot;&gt;GY70/339&lt;/a&gt; is a common carbon fiber epoxy used extensively with highly tested properties.  The material is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pre-preg&quot; rel=&quot;nofollow noreferrer&quot;&gt;prepreg&lt;/a&gt;, which means the fibers are pre-oriented and the matrix to fiber ratio is carefully controlled in the factory selling the material.  The roll must be kept in careful thermal control to prevent the pre-impregnated roll from curing before desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Simple formula used for rough estimation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a far enough point, a very simple &lt;a href=&quot;https://en.wikipedia.org/wiki/Rule_of_mixtures&quot; rel=&quot;nofollow noreferrer&quot;&gt;rule-of-mixtures&lt;/a&gt; can be used for a lot of situations using composites.  This includes when one of the materials is even missing.  Unfortunately, the rule of mixtures can only figure out &lt;a href=&quot;https://en.wikipedia.org/wiki/Stiffness&quot; rel=&quot;nofollow noreferrer&quot;&gt;stiffness&lt;/a&gt;, not &lt;a href=&quot;https://en.wikipedia.org/wiki/Strength_of_materials&quot; rel=&quot;nofollow noreferrer&quot;&gt;strength&lt;/a&gt;.  Note the major differences in definition.  Most other methods for estimating materials can only measure stiffness, with strength being better defined as a maximum &lt;a href=&quot;https://en.wikipedia.org/wiki/Deformation_(mechanics)#Strain&quot; rel=&quot;nofollow noreferrer&quot;&gt;strain&lt;/a&gt;, then an actual strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. More complex empirical formula&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond the rule of mixtures, there are more and more complex formulas that can be utilized.  For example, the &lt;a href=&quot;http://www.doitpoms.ac.uk/tlplib/fibre_composites/stiffness.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Halpin-Tsai&lt;/a&gt; formula is utilized frequently.  Although semi-empirical, it is used in plenty of applications that need good rough estimates to investigate the plausibility of a design.  More complex methods can be seen by various sources, but usually this is considered acceptable with follow up material testing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. Highly complex formulas which still need testing&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;State-of-the-art is considered &lt;a href=&quot;https://www.library.ucdavis.edu/wp-content/uploads/2017/03/HDBK17-3F.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mil Handbook 17F, Section 3, Chapter 5&lt;/a&gt;, which starts on page 211 of the reference.  Section 5.2 gives estimates for each layer, than 5.3 combines each layer into an orientation based laminate using matrix methods.  These are complex but understandable formulas.  As you get further into the document, you will notice that while this system is considered good enough for people who can't afford testing, stress concentration factors, fatigue factors, etc are completely unknown - if you need the system to be more reliable testing will be required, or you can use heavy safety factors and use these equations (10:1 is frequently used in pressure vessel applications, with 5:1 allowable for emergency circumstances, such as seismic and wind loading).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this will be helpful to people asking &quot;where do I begin&quot; when it comes to composite material design.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2017-10-03T14:10:48.550" LastActivityDate="2017-10-03T14:10:48.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4188" PostTypeId="2" ParentId="4186" CreationDate="2015-08-26T21:14:44.927" Score="10" Body="&lt;p&gt;Check out &lt;a href=&quot;https://robotics.stackexchange.com/questions/7796/calculating-required-torque/7799#7799&quot;&gt;my answer to a similar question at the Robotics SE&lt;/a&gt;. You are asking only about a motor and a &lt;strong&gt;wheel&lt;/strong&gt;, and you do not give the weight of the vehicle (assuming there is one), so I cannot comment regarding vehicular motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, if you assume friction is negligible, the torque a motor outputs is used to accelerate a load. Then you have to ask, &quot;What is my desired acceleration?&quot; A [Honda Civic] can accelerate 0-60mph in about 9 seconds, while a [1968 VW bus] went the same 0-60mph in 37 seconds. Both vehicles eventually get to the same top speed, but higher torque values give you a higher acceleration rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate your own desired acceleration rate by giving a top speed you would like to reach and a time limit in which you would like to achieve that speed. So, for example, if you wanted to get to 10rpm in 4 seconds, you would need:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a = (\mbox{Top speed} - \mbox{Starting speed})/(\mbox{Time limit}) \\&#xA;a = (10\mbox{rpm} - 0\mbox{rpm})/(4\mbox{s}) \\&#xA;a = 2.5 \frac{\mbox{rpm}}{\mbox{s}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be used in an equation, you typically have to convert to radians per second, which is easy knowing that $2\pi \mbox{radians} = 1\mbox{revolution}$ and that $1\mbox{min} = 60\mbox{s}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mbox{rpm} = \frac{\mbox{rev}}{\mbox{min}} \\&#xA;$$&#xA;$$&#xA;a = 2.5\frac{\mbox{rpm}}{\mbox{s}} = 2.5\frac{\mbox{rev}}{\mbox{min s}}\\&#xA;$$&#xA;Convert to $\mbox{rad/s}^2$:&#xA;$$&#xA;a = 2.5\frac{\mbox{rev}}{\mbox{min s}} * \frac{1\mbox{min}}{60\mbox{s}} * \frac{2\pi\mbox{rad}}{1\mbox{rpm}} \\&#xA;\boxed{a = 0.2618 \mbox{rad/s}^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate your own target acceleration using the same method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have your target acceleration, you can use the torque formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau = I \alpha \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\tau$ is the motor torque &lt;strong&gt;in Nm&lt;/strong&gt;, $\alpha$ is the acceleration &lt;strong&gt;in $\mbox{rad/s}^2$&lt;/strong&gt;, and $I$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_moments_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;moment of inertia&lt;/a&gt; of your load &lt;strong&gt;in $\mbox{kg m}^2$&lt;/strong&gt;. For a uniform disc, the moment of inertia about the center is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;I = mr^2/2 \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the mass &lt;strong&gt;in kg&lt;/strong&gt; and $r$ is the wheel &lt;em&gt;radius&lt;/em&gt; &lt;strong&gt;in meters&lt;/strong&gt;. This is how you would solve for the required (minimum) torque for your target application; as the question has been asked you do not know the acceleration you can achieve, so you rearrange the torque equation to solve for it as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau = I \alpha \\&#xA;\frac{\tau}{I} = \alpha \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute in the definition for $I$ and you can determine your acceleration, &lt;strong&gt;after you convert to standard units&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau = 190 \mbox{g cm} \\&#xA;\tau = 190 \mbox{g cm} * \frac{1 \mbox{kg}}{1000 \mbox{g}} * \frac{1 \mbox{m}}{100 \mbox{cm}} \\&#xA;\tau = 0.019 \mbox{kg cm} \\&#xA;$$&#xA;Here the torque is tricky - to get to Nm from kg-m you need to multiply by the gravitational constant, $9.81 \mbox{m/s}^2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau = 0.019 \mbox{kg cm} * 9.81\mbox{m/s}^2 \\&#xA;\boxed{\tau = 0.1864 \mbox{Nm}}&#xA;$$&#xA;Divide the mass by 1000 to get to kg:&#xA;$$&#xA;m = 59 \mbox{g} * \frac{1 \mbox{kg}}{1000 \mbox{g}} \\&#xA;\boxed{m = 0.059 \mbox{kg}} \\&#xA;$$&#xA;And lastly divide the wheel diameter by 100 to get to meters &lt;strong&gt;and&lt;/strong&gt; by 2 to get to a radius:&#xA;$$&#xA;r = d/2 \\&#xA;r = (26 \mbox{cm} * \frac{1 \mbox{m}}{100 \mbox{cm}})/(2) \\&#xA;\boxed{r = 0.13 \mbox{m}} \\&#xA;$$&#xA;Now, plug those values into the rearranged torque equation:&#xA;$$ &#xA;\frac{\tau}{mr^2/2} = \alpha \\&#xA;\frac{0.1864 \mbox{Nm}}{(0.059 \mbox{kg} * 0.13^2 \mbox{m}^2)/2} = \alpha \\&#xA;\frac{0.1864 \mbox{Nm}}{0.0004986 \mbox{kg m}^2} = \alpha \\&#xA;\boxed{374 \mbox{rad/s}^2 = \alpha} \\&#xA;\alpha = 374 \mbox{rad/s}^2 * \frac{1 \mbox{rev}}{2\pi \mbox{rad}} * \frac{60 \mbox{s}}{1 \mbox{min}} \\&#xA;\boxed{\alpha = 3570 \mbox{rpm/s}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a lot, but it's not so unreasonable - &lt;a href=&quot;http://machinedesign.com/sensors/basics-motion-control-part-2&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article about motion control&lt;/a&gt; has the figure below: (&quot;back then&quot; - 2000 - people didn't put graphics on the page, it's linked as Figure 2)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kGbWV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kGbWV.jpg&quot; alt=&quot;Motor ramp&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can see the motor accelerates from stopped to 2000 rpm in 0.12 seconds, for an acceleration of 16,667 rpm/s. Your motor is 3,570 rpm/s for your given load. You probably won't see rates quite that high due to friction, but it's a good ballpark. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in conclusion, you haven't stated what &quot;adequate&quot; is, but whatever your case, yes this motor will probably be good enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS - again, &lt;a href=&quot;https://robotics.stackexchange.com/questions/7796/calculating-required-torque/7799#7799&quot;&gt;see my answer about climbing hills&lt;/a&gt; for more information on how ramps affect required torque. If you're building a vehicle. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:49:17.810" LastActivityDate="2015-08-26T21:14:44.927" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4189" PostTypeId="1" CreationDate="2015-08-27T00:51:51.850" Score="4" ViewCount="32593" Body="&lt;p&gt;What is the physical meaning behind the concept of shear lag in fibre reinforced composite structures or the concept in general for any structure?&lt;/p&gt;&#xA;" OwnerUserId="2799" LastEditorUserId="9" LastEditDate="2015-08-27T09:44:53.740" LastActivityDate="2015-08-27T17:34:15.180" Title="Physical meaning of shear lag" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="2" CommentCount="2" ClosedDate="2015-09-01T11:48:06.577" ContentLicense="CC BY-SA 3.0" />
  <row Id="4190" PostTypeId="2" ParentId="3649" CreationDate="2015-08-27T07:48:10.857" Score="1" Body="&lt;p&gt;&lt;strong&gt;Overview:&lt;/strong&gt;&lt;br&gt;&#xA;I wouldn't be worrying about clutch life. From my experience with FSAE events they're not long enough to actually wear down a clutch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However(!!):&lt;/strong&gt;&lt;br&gt;&#xA;One of the main reason racers actually blip the throttle/rev match is to avoid &lt;strong&gt;compression lockup&lt;/strong&gt;. This is when the engine braking caused by downshifting (and resultantly having to bring the engine up to a higher RPM to match the wheel speed) causes your wheels start rotating slower which causes the car to oversteer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;As per your other questions:&lt;/strong&gt;&lt;br&gt;&#xA;Allowing the engine to rev while the clutch is disengaged will not damage the clutch. Your proposed solution of allowing the engine to decelerate freely and only engaging the clutch at a preset level could very well work from a clutch-life point of view but it will probably result in slow downchanges with an unpredictable application of power/traction (suddenly the engine will be connected to the rear wheels when a split second before they had just been free-wheeling). This would be pretty disastrous/hard-to-compensate-for in corners.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternative solutions:&lt;/strong&gt;&lt;br&gt;&#xA;Usually when these types of systems (sequential dog-boxes with some type of smart shifting mechanism but no throttle control) are employed the driver will manually downshift (such as in V8 race cars in Australia) allowing them to rev match and whatnot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could look at installing a &lt;strong&gt;slipper clutch&lt;/strong&gt; on your engine/transmission which would then completely remove the need to rev-match at all and make the downshifting system just as easy as the upshifting one (without introducing additional clutch-wear or compression lockup). This would be by far the easiest solution (but unfortunately quite costly). &lt;/p&gt;&#xA;" OwnerUserId="2768" LastActivityDate="2015-08-27T07:48:10.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4191" PostTypeId="2" ParentId="4091" CreationDate="2015-08-27T12:15:09.547" Score="0" Body="&lt;p&gt;you can use this free software &lt;a href=&quot;http://www.interpss.org/&quot; rel=&quot;nofollow&quot;&gt;Internet technology based Power System Simulator (InterPSS)&lt;/a&gt; also there are a lot of better Commercial software.&lt;/p&gt;&#xA;" OwnerUserId="2800" LastEditorUserId="1586" LastEditDate="2015-09-03T12:17:35.843" LastActivityDate="2015-09-03T12:17:35.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4192" PostTypeId="2" ParentId="179" CreationDate="2015-08-27T13:19:36.580" Score="-2" Body="&lt;p&gt;Dynamics for engineers like myself, generally concerned with statics, can be fraught with easy to make errors, and misunderstandings. This formula is very useful for simply supported beams since it can be related quickly to the applied self weight loads and a proportion of live loading (generally 10%) without having to go into complications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also Cantilevers can use a similar constant (19.8 with udl, 15.8 with end point load). It all breaks down with continuous beams and frames. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I build in a natural frequency check with all beam designs to keep track of it. For timber structures for instance 8Hz is the target and for concrete floors/steel frames 4-6Hz - as a first pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also rough and ready methods for assessing dynamic responses around. I have to say dynamics still eludes and confuses me and always will! So I stay as simple as possible.&lt;/p&gt;&#xA;" OwnerUserId="2786" LastActivityDate="2015-08-27T13:19:36.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4193" PostTypeId="2" ParentId="4189" CreationDate="2015-08-27T13:51:25.197" Score="3" Body="&lt;p&gt;I can't speak to fiber-reinforced composites, but the general concept of shear lag is summed up in the AISC 360-10 &lt;em&gt;Specification&lt;/em&gt;, Commentary Section D3 as such:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shear lag is a concept used to account for uneven stress distribution in connected members where some but not all of their elements (flange, web,&#xA;  leg, etc.) are connected.  The reduction coefficient, $U$, is applied to the net area, $A_n$, of bolted members and to the gross area, $A_g$, of welded members. As the length of the connection, $l$, is increased, the shear lag effect diminishes. This concept is expressed empirically by the equation for $U$. Using this expression to compute the effective area, the estimated strength of some 1,000 bolted and riveted connection test specimens, with few exceptions, correlated with observed test results within a scatterband of ±10% (Munse and Chesson, 1963). Newer research provides further justification for the current provisions (Easterling and Gonzales, 1993).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, shear lag is a concept where the forces in a connection require a certain length to properly &quot;get out.&quot;  This is partially dependent on the relative stiffness of the connection vs. the direction of the application of the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider the figure below from the AISC 360-10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LsMIK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LsMIK.png&quot; alt=&quot;angle connection&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example on the left is more impacted by the effects of shear lag since the force in the connection has to transfer along the length $l$ of the weld. The $U$ value would likely be less than 1.0, indicating the influence of shear lag. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the middle example, the weld at the end of the angle is perpendicular to the force.  In this case, the $U$ value would likely be 1.0 (assuming the outstanding leg of the angle is also welded similarly) since the force is transferred uniformly to the perpendicular weld.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Table D3.1 of the AISC 360-10 tabulates the calculation of different $U$ values based on the connection geometry.  I have pasted a snippet of it below for a couple examples.  Your code of record may vary depending on your country and jurisdiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hWOB4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hWOB4.png&quot; alt=&quot;lag factors&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-08-27T17:34:15.180" LastActivityDate="2015-08-27T17:34:15.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4196" PostTypeId="2" ParentId="4189" CreationDate="2015-08-27T17:08:31.250" Score="12" Body="&lt;p&gt;Here is how I visualize what is physically happening (for steel sections):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cLMo9.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cLMo9.png&quot; alt=&quot;Shear Lag&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/g870P.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g870P.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you get closer to the connection to the gusset, all the force needs to be transferred through the bolts, so the stress has to flow toward the connected portion of the angle. In the limit, (L = 0) the net effective area would simply be the net area of the connected portion of the element.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-08-27T17:08:31.250" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4197" PostTypeId="1" AcceptedAnswerId="4198" CreationDate="2015-08-27T23:10:45.337" Score="3" ViewCount="1666" Body="&lt;p&gt;I just started mechanics of materials, and it seems that my knowledge from statics has gotten rusty over the summer. But there was a  new concept introduced in the first example problem today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a horizontal bar connected to a wall (fixed support). I understand that with a fixed support, there are horizontal and vertical reaction forces and a moment as well. However, when the professor cut the horizontal bar to calculate the internal loadings, he drew vectors for vertical and horizontal forces AND a moment. Since I didn't use this approach in statics, I don't get the reasoning behind adding a vector for the moment when the body is cross sectioned. Can someone explain to me why a moment is added at the newly cut end? &lt;/p&gt;&#xA;" OwnerUserId="2804" LastEditorUserId="368" LastEditDate="2015-08-28T22:53:28.370" LastActivityDate="2015-08-28T22:53:28.370" Title="Why is there a moment at the cross section end when you cut a body?" Tags="|mechanical-engineering|beam|statics|torque|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4198" PostTypeId="2" ParentId="4197" CreationDate="2015-08-28T00:15:40.623" Score="6" Body="&lt;p&gt;To put it simply your loads are causing the beam/bar to &lt;strong&gt;bend&lt;/strong&gt; right? So at the cut we need to consider the internal bending moment ($M$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like you said: the professor has cut the bar to calculate the INTERNAL forces experienced by the member. In any 2D problem (say in the x-y plane) the first thing that you learn is that whenever you take a cut there's three internal forces you need to consider:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Shear Force ($V$ or $Q$) perpendicular to the longitudinal axis of the member.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Axial Force ($N$) along the longitudinal axis of the member.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bending Moment ($M$) about the z-axis which is coming out of the plane/page.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We consider these three internal loads because we know that they can be experienced by the member along its length due to the reactions at the supports and the external loads. The one you seem confused about is the moment ($M$). Nearly all the time in statics your loads on the beam will cause &lt;strong&gt;bending&lt;/strong&gt; and this why we need to add a force vector for bending moment whenever we take a cut. Imagine taking a ruler, holding it at both ends and causing it to bend in the middle - a bending force is there that needs to be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just because we've taken a cut and considered that these three internal loads, it doesn't mean they ALWAYS exist. They MIGHT exist. For instance, if you take a cut of the beam/bar where shear force is 0, then bending moment is typically at its maximum. At the point where no internal shear force exists (i.e. it is switching direction), bending moment happens to be maximum. Another example is at free ends the bending moment is typically zero.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-08-28T00:48:06.077" LastActivityDate="2015-08-28T00:48:06.077" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4199" PostTypeId="2" ParentId="2905" CreationDate="2015-08-28T00:37:28.077" Score="6" Body="&lt;p&gt;You're looking for $\tau_{xz}$, which pertains to horizontal shear flow along the top flange. (Rather than the vertical shear flow considered in $\tau_{xy}$.) As Mark noted, the shear flow starts from the center of the flange and flows outward/down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2HHJI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2HHJI.png&quot; alt=&quot;Shear Flow&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulation for the shear stress calculation remains essentially the same. Only the area changes -- we're simply cutting out a different section, and because of the tube shape we have to make two cuts to remove our section. (Versus and I-shape where only one cut is required.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wrpdx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wrpdx.png&quot; alt=&quot;Section Cuts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://kisi.deu.edu.tr/mehmet.aktas/Dersnotlari/6.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;presentation&lt;/a&gt; has a pretty good explanation of shear stress.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="1832" LastEditDate="2015-08-28T02:20:25.773" LastActivityDate="2015-08-28T02:20:25.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4200" PostTypeId="1" AcceptedAnswerId="4204" CreationDate="2015-08-28T01:19:03.463" Score="3" ViewCount="279" Body="&lt;p&gt;I've been given the following to memorize; But my head is so crammed with many others that I need to find another way; Are these formulas Empirical? I was hoping that if I derive it from a much more primitive form; I maybe able to remember it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ (\frac{A}{F})_t = W_{ta} = \frac{Q_h \frac{kJ}{kg} }{3117} \frac{kg_{air}}{kg_{fuel}} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ (\frac{A}{F})_t = W_{ta} = \frac{Q_h \frac{kCal}{kg} }{745} \frac{kg_{air}}{kg_{fuel}} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ (\frac{A}{F})_t = W_{ta} = \frac{Q_h \frac{Btu}{lb_m} }{1340} \frac{kg_{air}}{kg_{fuel}} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know if I'm provided the answer to one of these; I can easily derive the others because I've memorized the conversions. What I don't get is how ${\frac{kJ}{kg}}$ got to ${ \frac{kg_{air}}{kg_{fuel}} }$. I cant figure out where the 3117 came from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${Q_h}$ is Higher Heating Value, ${ \frac{A}{F}}$ is Air to Fuel ratio&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorDisplayName="user16" LastEditDate="2015-08-29T00:45:58.320" LastActivityDate="2015-08-29T00:45:58.320" Title="Derivation for the Formulas of the Approximate Ultimate Analysis of Coal" Tags="|combustion|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4201" PostTypeId="2" ParentId="4197" CreationDate="2015-08-28T02:06:09.257" Score="2" Body="&lt;p&gt;It's useful to be clear about the fact that the 'cut' is an imaginary one.  Since the portion of the beam no longer attached to the wall (because of our imaginary cut) is not falling or rotating, it must be subjected to the same type of forces/torques as if it were attached directly to its own 'wall'.  The 'wall' in this case is actually the part of the beam on the other side of the imaginary cut though. &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-08-28T02:06:09.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4204" PostTypeId="2" ParentId="4200" CreationDate="2015-08-28T09:45:08.067" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are these formulas Empirical? I was hoping that if I derive it from a much more primitive form&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is &lt;strong&gt;&lt;em&gt;probably an approximation&lt;/em&gt;&lt;/strong&gt;, let's try to calculate stoichiometric air to fuel ratio and heat of combustion for liquid octane as an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ C_8H_{18} + \lambda(O_2 + 3.76N_2) \rightarrow 8CO_2 + 9H_2O + 3.76\lambda N_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we calculate heating value, Cengel (2005) defines the heating value as :&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the amount of heat released when a fuel is burned completely in a steady-flow process and the products are returned to the state of the reactants. In other words, the heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Making energy balance we can get $Q_h$ around $-47,891 \  \text{kJ/kg fuel}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, we calculate stoichiometric air to fuel $(\lambda = 8 + \frac{18}{4} = 12.5)$:&#xA;$$ (\frac{A}{F})_{stoich.} = \frac{12.5(32 + (3.76*28)}{(8*12) + 18} = 15.052 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in your formula:&#xA;$$ (\frac{A}{F})_{stoich.} = \frac{|Q_h|}{3117} = \frac{47891}{3117}= 15.36$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is slightly higher than the calculated theoretical value. Of course there is no direct relation between heating value of fuel and stoichiometric air to fuel ratio as there are many parameters that could affect that relation, you can derive the air to fuel ratio by making energy balance and substitute with elements' enthalpy and heat of formation but that won't be any easier than simply memorizing the formula you were given.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2015-08-28T09:45:08.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4205" PostTypeId="2" ParentId="4159" CreationDate="2015-08-28T13:03:40.213" Score="3" Body="&lt;p&gt;&lt;strong&gt;Advantage :-&lt;/strong&gt;&#xA;       Differing of the number of teeth (effectively diameter) on each wheel allows mechanical advantage to be changed. By increasing or decreasing the ratio of teeth between the drive and driven wheels one may change the ratio of rotations between the two, meaning that the rotational drive and torque of the second wheel can be changed in relation to the first, with speed increasing and torque decreasing, or speed decreasing and torque increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disadvantage :-&lt;/strong&gt;&#xA;            One wheel of such gear is designed to work with its complementary wheel and no other.&lt;/p&gt;&#xA;" OwnerUserId="2101" LastActivityDate="2015-08-28T13:03:40.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4206" PostTypeId="1" CreationDate="2015-08-28T13:44:31.100" Score="5" ViewCount="2387" Body="&lt;p&gt;I am attempting to figure out why a laser layout tool I have is drifting over time. It was designed with a small adjustment mechanism that relies on a spring to apply pressure to a fulcrum that rotates the entire laser. A screw is used to apply and release pressure on the spring thus moving the laser very small amounts left and right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am getting a drift in the lasers horizontal position of about 1/2inch at 100 feet over the course of several hours. I am doing this test at night and there is about a 15 degree Fahrenheit temperature from when I start the test to when it is complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a Sketch-Up model of the device and determined that 5 micron change in position (at the end of the fulcrum) is all it would take to move the laser 1/2 inch at 100ft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is it possible that my spring is expanding and contracting 5 microns with a 15 degree temperature change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an engineer and am looking for some speculation just to rule in or rule out the general possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spring looks similar to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9E39l.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9E39l.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter of the entire spring is about 1/4 inch. The diameter of the coiled metal wire is about 1/6 inch. The spring length is about 1/2&quot; uncompressed. It consists of about 8 loops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know the metal it is made of, speculation on different scenarios would be helpful as, again, I'm just trying to rule in or out the possibility.&lt;/p&gt;&#xA;" OwnerUserId="2812" LastEditorUserId="2246" LastEditDate="2015-08-28T15:14:21.513" LastActivityDate="2018-04-18T19:45:11.393" Title="Metal spring thermal expansion" Tags="|springs|metals|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4207" PostTypeId="1" CreationDate="2015-08-28T15:12:57.383" Score="0" ViewCount="1341" Body="&lt;p&gt;I want to break a blade as it attempts to cut into a piece of metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the metal in question was vibrating rapidly or with a significant amount of force, would the blade break in attempt to cut through the metal? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going on, would the same concept work on bolt-cutters, effectively breaking the blade on them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If metal will not break in this instance, but may bend, would this be an applicable concept if such a vibrating force were used on a bicycle U-Lock in order to combat bolt cutters? In essence, would it be an effective deterrent as far as bending the metal in the bolt cutter blades, rendering it useless against the U-Lock?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm tying to think about potential concepts that could be applied to bike locks in order to mitigate potential theft.&lt;/p&gt;&#xA;" OwnerUserId="2816" LastEditorDisplayName="user16" LastEditDate="2015-08-28T16:39:40.093" LastActivityDate="2015-09-15T17:23:44.230" Title="Can vibration break a blade?" Tags="|mechanical-engineering|materials|metallurgy|stresses|vibration|" AnswerCount="1" CommentCount="2" ClosedDate="2015-08-28T16:52:05.697" ContentLicense="CC BY-SA 3.0" />
  <row Id="4208" PostTypeId="2" ParentId="4207" CreationDate="2015-08-28T16:07:09.747" Score="3" Body="&lt;p&gt;Unless the blade is made of a ceramic material, there is very little chance of this working.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea you have in mind, I think, is something like the shattering of a wine glass by exciting one of it's resonance frequencies.  This works because the &lt;a href=&quot;https://en.wikipedia.org/wiki/Q_factor&quot; rel=&quot;nofollow&quot;&gt;$Q$ of the resonance&lt;/a&gt; (the ratio of amplitude of the motion to excitation amplitude) is very high so that a relatively small driving force creates vibrations which are large enough to drive the glass beyond its elastic limit at which point it shatters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will not work with a metal for two reasons.  First, metal is a relatively lossy material so that the $Q$ of its resonances are significantly lower than that of ceramics like glass (wine glass: $Q\sim10^6$, metal: $Q\sim10^3$).  The second, likely more important issue, is that metals don't usually break without being pushed significantly beyond their elastic limits, they just deform.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-09-15T17:23:44.230" LastActivityDate="2015-09-15T17:23:44.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4209" PostTypeId="1" AcceptedAnswerId="4210" CreationDate="2015-08-28T20:52:38.440" Score="20" ViewCount="7740" Body="&lt;p&gt;Why do we design cars to turn by turning the front wheel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't it better to keep all wheels straight and turn by rotating the wheels on one side of a car more then the wheels on the opposite side of the car, like tank tracks do? That way you can turn even if you are standing still. Why don't cars do this?&lt;/p&gt;&#xA;" OwnerUserId="2821" LastEditorUserId="90" LastEditDate="2015-08-29T01:02:44.080" LastActivityDate="2016-06-07T17:54:10.607" Title="Why do cars turn by turning the front wheel?" Tags="|automotive-engineering|wheels|" AnswerCount="7" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4210" PostTypeId="2" ParentId="4209" CreationDate="2015-08-28T21:12:37.247" Score="19" Body="&lt;p&gt;If you were going to turn left 90 degrees, without turning the wheels, then you wind up dragging the wheels sideways while you turn. &lt;a href=&quot;https://www.youtube.com/watch?v=9AfuD7xzsAU&amp;amp;t=16s&quot;&gt;16 seconds into this video shows exactly what I'm talking about&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So every time you try to back out of your driveway, or a parking spot, or turn into a parking spot, or turn anywhere for any reason, you're going to lay down rubber because the tires are rotating quickly while remaining nearly stationary as you turn. Again, look at the wheel slip in the video I linked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd be hard pressed to find &quot;four wheel differential steering&quot; in use anywhere for any reason. Tanks use it to reduce the ground pressure they exert so they don't sink in the dirt. Cars aren't going to sink into pavement so this isn't an issue. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-08-28T21:12:37.247" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4211" PostTypeId="2" ParentId="4170" CreationDate="2015-08-28T22:27:39.610" Score="2" Body="&lt;p&gt;OK, fair warning: I am answering my own question and am not a engines person. So this could be wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real limit in the engine is how hot certain parts can get without breaking. This temperature is related to the gas temperature after combustion via the cooling system and the cylinder design (convective heat transfer between the gas and the cylinder wall). This proportionality will vary from engine to engine, but for arguments sake lets just assume that all engines have some magic gas temperature limit that is governing the cut-off ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a real engine there is not a clear cut-off ratio because the compression, combustion, and expansion steps all bleed into each other. However, the final gas temperature still sets the volume at which combustion stops which sets a cut-off ratio analogy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that is governing this is not really the valve timing as the rpm of the engine is remaining the same and the timing during each revolution does not change. What does change is the Air to Fuel Ratio (AFR). This is controlled by the fuel injection system based on how much oxygen is coming into the engine and at what temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lowering the inlet temperature does two things: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It increases the mass flow of air through the engine.  &lt;/li&gt;&#xA;&lt;li&gt;It decreases the temperature of the air prior to combustion.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both of these things lead to a higher AFR at the same limiting gas temperature after combustion. The AFR is only controlled by the quantity of gas injected which is possible to dynamically control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my answer is yes, it can be adjusted on the fly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, that could be wrong. &lt;/p&gt;&#xA;" OwnerUserId="2780" LastActivityDate="2015-08-28T22:27:39.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4212" PostTypeId="2" ParentId="4209" CreationDate="2015-08-28T22:29:39.923" Score="7" Body="&lt;p&gt;While it might be possible to design a rear-wheel-drive car in such a fashion that the front wheels basically operated as casters and the steering was controlled by driving the rear wheels at different speeds (some automatic guided vehicles operate on this principle), such vehicles perform poorly when there is a loss of traction at speed.  If such a car were accelerating when the left rear wheel lost traction, the car would immediately start rotating to the left, and would likely travel some distance to the left before the driver could react to the situation.  Even if the car's control system used accelerometers and gyros to detect that condition faster and keep the car from going far off course, they couldn't react to the loss of traction until after it had already affected the vehicle's rotational velocity.  By contrast, holding the front wheels at a driver-controlled angle will keep the vehicle going straight even if one of the wheels loses traction.&lt;/p&gt;&#xA;" OwnerUserId="397" LastActivityDate="2015-08-28T22:29:39.923" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4213" PostTypeId="1" CreationDate="2015-08-28T23:00:27.320" Score="2" ViewCount="1069" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Question:&lt;br&gt;&#xA;  All other factors being equal, the diffusion rate of nickel is the highest in copper with ASTM grain size number of:&lt;br&gt;&#xA;  A. 4&lt;br&gt;&#xA;  B. 6&lt;br&gt;&#xA;  C. 8&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Honestly, the question does not appear to be very intuitive - I've searched through my notes and cannot seem to find any relatable formulae. I understand that the number of grains per square inch at 100X magnification is equatable to 2^(n-1), where n is the ASTM grain size number of the specimen.&lt;br&gt;&#xA;However, given the multiple-choice nature of this question and the apparent lack of information (I suppose the copper and nickel are pure, at room temperature, etc.), I must be overlooking something simple.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would anyone be willing to advise? &lt;/p&gt;&#xA;" OwnerUserId="2823" LastEditorDisplayName="user16" LastEditDate="2015-08-29T00:38:34.747" LastActivityDate="2015-10-30T23:38:43.873" Title="ASTM grain size: diffusion of nickel in copper" Tags="|materials|chemical-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4214" PostTypeId="2" ParentId="4209" CreationDate="2015-08-28T23:43:16.073" Score="7" Body="&lt;p&gt;The main reason is because when a car brakes/decelerates, the load is reduced on the rear wheels and increased on the front wheels. If your car was exclusively rear wheel steering you would lose steering in high speed braking. The converse does not occur because engine accelerations are typically much much less than braking decelerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forklifts are almost exclusively rear wheel steering for more maneuverability of a pallet in the front, but they are limited to low speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four wheel steering cars:&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Category:Vehicles_with_four-wheel_steering&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Category:Vehicles_with_four-wheel_steering&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of them are computer controlled and steer in opposite directions for low speed maneuverability and in the same direction for high speed strafing. None that I know of do caterpillar style rotation about the center axis.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2016-03-03T05:15:57.787" LastActivityDate="2016-03-03T05:15:57.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4215" PostTypeId="5" CreationDate="2015-08-29T00:46:32.500" Score="0" Body="" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-08-29T00:46:32.500" LastActivityDate="2015-08-29T00:46:32.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4216" PostTypeId="4" CreationDate="2015-08-29T00:46:32.500" Score="0" Body="DO NOT USE.  This is a meta tag and meta tags are actively discouraged within StackExchange." OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-08-29T00:46:32.500" LastActivityDate="2015-08-29T00:46:32.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4217" PostTypeId="1" CreationDate="2015-08-29T01:21:19.070" Score="1" ViewCount="321" Body="&lt;p&gt;With most microwaves I've used, I've noticed that once the timer goes off, there is a couple seconds where the &quot;microwave sound&quot; gets louder and it seems to be cooking more intensely before it actually shuts off. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="2825" LastEditorUserId="272" LastEditDate="2015-08-29T11:48:01.367" LastActivityDate="2015-08-29T13:40:08.043" Title="Why do microwaves get louder when the timer goes off?" Tags="|electrical-engineering|design|consumer-electronics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4221" PostTypeId="2" ParentId="4209" CreationDate="2015-08-29T09:07:12.640" Score="5" Body="&lt;p&gt;The &lt;a href=&quot;https://www.youtube.com/watch?v=3b5zH0G5pxE&quot;&gt;Jeep Hurricane&lt;/a&gt; has four-wheel steering that allows it to pivot 360° without moving. But it's very complex and expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would guess that skid-steer systems in general are heavier, bulkier, and more complex. And really, the only place they're particularly useful is when you need really tight steering. 99% of our driving has no use for really tight steering, so it doesn't make a lot of sense to design the vehicle for the other 1%.&lt;/p&gt;&#xA;" OwnerUserId="2835" LastActivityDate="2015-08-29T09:07:12.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4222" PostTypeId="2" ParentId="4206" CreationDate="2015-08-29T12:26:30.987" Score="6" Body="&lt;p&gt;Assuming the spring is not fully compressed/bottomed out, it doesn't really matter if the spring material is expanding or not because it's a spring - it will deform to fill the space between the coils. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of providing an increasing distance, it will increase the &lt;em&gt;force&lt;/em&gt; the spring exerts on your fulcrum. This shouldn't be an issue if you have the adjustment locked in with a set screw or similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could be happening is that the drift isn't due to the spring but some other part. &lt;a href=&quot;http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html&quot; rel=&quot;noreferrer&quot;&gt;This table&lt;/a&gt; gives some linear expansion coefficients. The metric side's units can be read as &quot;microns of expansion per meter of original length, per degree Kelvin.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you have an aluminum screw 1 meter long and you heat it 1K, it expands 22 microns. 0.5cm in original length, it expands 0.11 microns. If the temperature rise were 8.3K, about 15F, then a 0.5 cm screw would expand about 0.9 microns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say the effect you are seeing might be thermal expansion, but I would check other places for drift too, such as not locking down the setting after the adjustment is made. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2015-08-29T16:15:27.603" LastActivityDate="2015-08-29T16:15:27.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4223" PostTypeId="2" ParentId="4217" CreationDate="2015-08-29T13:40:08.043" Score="8" Body="&lt;p&gt;When the timer reaches zero, the microwave tube is turned off. Depending on your house wiring, this large reduction in load may allow the line voltage to rise a bit. The fan, which continues to run for a few seconds in order to get rid of the built-up heat, speeds up a bit as well.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-08-29T13:40:08.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4224" PostTypeId="1" CreationDate="2015-08-30T05:10:57.813" Score="2" ViewCount="978" Body="&lt;p&gt;If you force high-pressure gas through a packed-bed, the gas pressure drops. However, the ideal gas law states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \frac{RT}{v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so for an ideal gas, the pressure drop &lt;em&gt;must&lt;/em&gt; be associated with a property change of the gas itself as it moves through the packed bed. My question is: how does a typical gas moving through a typical packed bed adjust its pressure: is it primarily by a &lt;em&gt;decrease in temperature&lt;/em&gt; or by an &lt;em&gt;increase in specific volume&lt;/em&gt;? &lt;/p&gt;&#xA;" OwnerUserId="2625" LastActivityDate="2015-11-29T19:14:55.157" Title="Pressure drop across column in gas phase: what changes?" Tags="|pressure|gas|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4225" PostTypeId="1" AcceptedAnswerId="4236" CreationDate="2015-08-30T06:12:53.267" Score="0" ViewCount="610" Body="&lt;p&gt;For a National Board Exam Review:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A uniform area jet travels to $600 \frac{ft}{s}$ and a&#xA;  $100\frac{ft^3}{s}$. What is the horizontal force acts on the water jet&#xA;  if it undergoes a $180^\circ$ turn?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would normally provide an answer but even review instructor cannot answer question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${ F_x = \frac{ \gamma Q }{ g } \left(\nu_2 \cos(\theta) - \nu_1\right)}$$&#xA;$${ F_x = \frac{ 62.4 \frac{lb_m}{ft^3} ( 100 \frac{ft^3}{s} ) }{ 32.2 \frac{ft}{s^2} } \left(600 \frac{ft}{s^2} \cos(180^\circ)- 0\right) = -116273.2919\,lb_f }$$&#xA;Is the negative sign logical? All choices are from the range of ${1.35*10^5}$ to ${ 5.32*10^5 }$ lbf&lt;/p&gt;&#xA;" OwnerUserId="2170" LastEditorUserId="2844" LastEditDate="2015-08-30T17:37:48.817" LastActivityDate="2015-09-01T13:36:39.083" Title="The Horizontal Force of Water as it turns 180 degrees" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4226" PostTypeId="1" CreationDate="2015-08-30T10:53:32.250" Score="4" ViewCount="2615" Body="&lt;p&gt;I suppose this is a logical extension of my earlier question &lt;a href=&quot;https://engineering.stackexchange.com/questions/2398/contours-of-radial-flow-impeller-vane&quot;&gt;here&lt;/a&gt; - in fact, this is probably what I was meaning to ask in the first place. I posted as &quot;centrifugal pump&quot;, although the same concept can be observed in other (radial or axial) pumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting to it, pretty much any photo of a high-efficiency impeller you can find shows that the blade &quot;twists&quot;, or has three-dimensional curvatures that are much more complex than taking a flat blade profile and simply extruding it, i.e. this Francis turbine runner:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7cnYP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7cnYP.jpg&quot; alt=&quot;Francis turbine runner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First part of this question being, is this for the same reason that an airplane propeller twists - (forgive me, rather generalizing here) due to flow field conditions and the radial velocity effects? Are the same principles at work here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second part being, how does one go about constructing structures like this? There must be some math behind it that isn't completely CFD-based, as you can see the same ideas in the Rocketdyne F-1 turbopumps constructed in the '60s. However, no matter how many pump design sources I try to find, I can't find any that treat the impeller vanes as anything other than a two-dimensional chord on the impeller back shroud (see below image). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IzoAP.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IzoAP.gif&quot; alt=&quot;Impeller velocity triangles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in essence, how would I start to go from a basic pump design (inlet/outlet velocity triangles, simple impeller) to a more &quot;advanced&quot; or &quot;efficient&quot; design like an impeller of the above nature?&lt;/p&gt;&#xA;" OwnerUserId="1229" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2018-10-05T20:12:21.120" Title="&quot;Twist&quot; of centrifugal pump impeller" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4227" PostTypeId="1" AcceptedAnswerId="4239" CreationDate="2015-08-30T11:19:18.820" Score="1" ViewCount="125" Body="&lt;p&gt;I want to mount a heat source onto several heat pipes. The heat will be forwarded to coolers (four heat pipes, 200 mm) which are able to distribute up to 120 W without increasing the temperature of the device which should be cooled. The device itself will generate up to 100 W heat. Now I was wondering what would happen to the mounting point: Will the mounting point resp. the clamp used for fixing the heat source on the heat pipes also heat up? If yes, how much?&lt;/p&gt;&#xA;" OwnerUserId="2844" LastActivityDate="2015-09-30T13:29:07.580" Title="Mounting heat source onto heat pipes" Tags="|heat-transfer|" AnswerCount="1" CommentCount="6" ClosedDate="2015-10-01T20:44:07.230" ContentLicense="CC BY-SA 3.0" />
  <row Id="4228" PostTypeId="2" ParentId="4224" CreationDate="2015-08-30T11:49:16.023" Score="0" Body="&lt;p&gt;In packed beds that I've seen, the pressure loss across the bed is miniscule compared to the absolute pressure (on the order of 2%), and the length of the packed bed is sometimes 2 or 3 diameters (or more). Finally, the gas is usually undergoing a chemical reaction (which is generally exothermic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking over the net result of all of these effects:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Pressure drops by 2%, which could increase specific volume or temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Gas passes through long, uninsulated column which maintains gas temperature at or around original temperature, resulting in temperature to return to normal while maintaining density decrease&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Gas absorbs substantial heat from chemical interactions, which dominates the two effects above, resulting in temperature increase and an altogether different gas state which has reduced pressure (if pressure increases, the gas won't flow), with likely higher specific volume and higher temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only for effects 1 and 2, I'd argue that a packed bed results in a change in density, because the gradual temperature change is removed from the long travel length. However the final effect changes things to a chemical engineering problem. &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-08-30T11:49:16.023" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4230" PostTypeId="1" CreationDate="2015-08-30T12:45:35.750" Score="8" ViewCount="232" Body="&lt;p&gt;I was thinking about the relationship between water hammer and expansion waves. Pressure seems to drop across a surge wave travelling at the speed of sound in water hammer (after the fluid has been initially brought to rest by a compression). Pressure also drops across an expansion wave (e.g. a rarefaction fan). I know they are not the same, but my question is - &quot;is the wave travelling at the speed of sound in water hammer more like an expansion rather than a compression (or is it the other way round)?&quot;&lt;/p&gt;&#xA;" OwnerUserId="2689" LastActivityDate="2018-06-29T18:50:08.930" Title="Water hammer and expansion waves" Tags="|mechanical-engineering|pressure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4231" PostTypeId="2" ParentId="4165" CreationDate="2015-08-30T15:27:08.423" Score="2" Body="&lt;p&gt;The statistical tool &lt;a href=&quot;https://en.wikipedia.org/wiki/Weibull_distribution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weibull&lt;/a&gt; as suggested by the previous two responded is the tool of choice for &lt;a href=&quot;http://www.weibull.com/hotwire/issue94/relbasics94.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Mean Time To Failure&lt;/em&gt; (MTTF)&lt;/a&gt; calculations. Based on your comment as capture below, it appears that Weibull Analysis didn’t generate expected results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9ye03.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ye03.jpg&quot; alt=&quot;Comment from Stephen Collings&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most statisticians that I have worked with recommend a sample size of 30 for most statistical analysis. My suspicion is that the limited data size might not be helping the analysis. I suggest starting with a simple average and standard deviation calculation for time to failure based on the available data. You might have to make few reasonable assumptions when calculating time to failure base on your product. For example  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Assumption&lt;/strong&gt;: Time to failure (days) = Return date – Ship date&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With current technology and available data you might be able to refine your assumptions too.      &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Improved Assumption&lt;/strong&gt;:  Time to failure (days) = Customer product return&#xA;  ship date – Customer initial product receive date&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The point that I am making is good reasonable assumptions will help generate a good data set. Also in my experience basic average and standard deviation calculation will help gain good insight into the problem at hand.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other point to be aware is determine if the failures are due to &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Special cause&lt;/li&gt;&#xA;&lt;li&gt;Common cause&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Root cause analysis need to be performed on special cause failures and corrective action needs to be implemented. Common cause failures are just part of doing business in the specific industry and with the specific customer base. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this response finds a reasonable solution to the problem at hand. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Common_cause_and_special_cause_(statistics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Common cause and special cause (statistics)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.weibull.com/basics/lifedata.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Life Data Analysis (Weibull Analysis)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Reliability_engineering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reliability engineering&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.itl.nist.gov/div898/handbook/apr/apr.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Assessing Product Reliability&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-08-30T16:01:32.443" LastActivityDate="2015-08-30T16:01:32.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4232" PostTypeId="1" AcceptedAnswerId="4233" CreationDate="2015-08-30T17:31:08.970" Score="1" ViewCount="794" Body="&lt;p&gt;I'm an engineer in industrial design and despite that I studied mechanics, kinematics and dynamics, I got difficulties when I try to label or find components which could enable my designs have relative movements, i.e. one part rotates inside other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I can design most of them but I guess that some must be already created and would like to know how to search them, instead designing new ones or asking for help in each specific project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the basics relative movements and components that generates them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find a &lt;strong&gt;&lt;em&gt;&quot;movements library&quot;&lt;/em&gt;&lt;/strong&gt;, to be able to find what are the most common movements and what mechanisms or components are used to achieve them. For example, materials libraries are pretty common in many countries, in them you can find a huge list of materials, with their properties, application and providers.&lt;/p&gt;&#xA;" OwnerUserId="1262" LastEditorUserId="282" LastEditDate="2015-09-04T18:48:37.527" LastActivityDate="2015-09-04T18:48:37.527" Title="What are the basic mechanism, components or parts to generate relative movements?" Tags="|mechanical-engineering|design|kinematics|biomechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="4233" PostTypeId="2" ParentId="4232" CreationDate="2015-08-30T23:47:51.113" Score="3" Body="&lt;p&gt;There is a fascinating historical collection of mechanisms compiled in the late 19th century here: &lt;a href=&quot;http://507movements.com/about.html&quot; rel=&quot;nofollow&quot;&gt;Mechanical Movements&lt;/a&gt; and somebody is working through the task of making animations of them for the website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manufacturing technology has progressed since then, but the basic ideas are timeless. We used to have a printed copy of this book in our &quot;office library&quot; before the internet existed, as a source of design inspiration.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="1586" LastEditDate="2015-09-02T22:58:12.353" LastActivityDate="2015-09-02T22:58:12.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4234" PostTypeId="1" CreationDate="2015-08-31T08:44:57.910" Score="3" ViewCount="302" Body="&lt;p&gt;I have a Galaxy Note2 with a front glass layer that is broken, while the screen itself is ok, as well as the touch panel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand properly, the front glass has only a protective function and is not the same as a &quot;touch panel&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how technically does the glass transfer the touch actions to the touch panel? On the answer I will base the way of the glass replacement.&lt;/p&gt;&#xA;" OwnerUserId="2851" LastEditorUserId="90" LastEditDate="2015-08-31T11:07:01.323" LastActivityDate="2015-08-31T11:07:01.323" Title="How does the glass layer transfer the touch to the touch sensitive layer on Galaxy phones?" Tags="|sensors|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="4235" PostTypeId="2" ParentId="4234" CreationDate="2015-08-31T08:56:53.130" Score="3" Body="&lt;p&gt;Glass fronted touch screens use &lt;a href=&quot;https://en.wikipedia.org/wiki/Capacitive_sensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;capacitance&lt;/a&gt; to detect fingers. Multi touch will use projected capacitance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:TouchScreen_projective_capacitive.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E22E8.png&quot; alt=&quot;projected-capacitive touchscreen&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;image source: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:TouchScreen_projective_capacitive.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;however the capacitors are bonded to the glass panel. Replacing the screen will often be more expensive than just buying a new device and turning in the old one for recycling.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-08-31T08:56:53.130" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4236" PostTypeId="2" ParentId="4225" CreationDate="2015-08-31T11:02:46.160" Score="1" Body="&lt;p&gt;This can be solved easily by calculating momentum.  Momentum is mass x velocity.  Force is momentum / time.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-08-31T11:02:46.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4238" PostTypeId="1" AcceptedAnswerId="4240" CreationDate="2015-08-31T13:11:53.067" Score="6" ViewCount="3351" Body="&lt;p&gt;To pick a specific example, a flyover that goes over a ground-level road/rail need only be as tall so that the tallest vehicle travelling in the outermost lane can clear it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or another, an elevated road/metro rail that runs along a ground-level road need to be as tall as the tallest vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why are they made much taller than that?&lt;/p&gt;&#xA;" OwnerUserId="444" LastEditorUserId="1586" LastEditDate="2015-09-03T23:32:52.867" LastActivityDate="2016-06-25T04:11:27.043" Title="Why are flyovers/metro rail tracks much higher than traffic?" Tags="|civil-engineering|transportation|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="4239" PostTypeId="2" ParentId="4227" CreationDate="2015-08-31T13:28:12.957" Score="2" Body="&lt;p&gt;Well, how much the clamp heats up is determined by a great number of factors, chiefly (off the top of my head):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The properties of the pipe (material properties: density, material&#xA;type, size of contact area etc)&lt;/li&gt;&#xA;&lt;li&gt;How the clamp interfaces with the pipe (e.g. metal on metal, thermal&#xA;interface material between clamp and pipe, an insulator between clamp&#xA;and pipe). &lt;/li&gt;&#xA;&lt;li&gt;The properties of the clamp (material properties, initial temperature etc)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Regards the resp. abbeviation I think you're using it incorrectly. I'd say &quot;Will the mounting point w.r.t. the clamp.....&quot; where w.r.t. means with respect to.&lt;/p&gt;&#xA;" OwnerUserId="2174" LastActivityDate="2015-08-31T13:28:12.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4240" PostTypeId="2" ParentId="4238" CreationDate="2015-08-31T14:20:14.150" Score="18" Body="&lt;p&gt;If I'm understanding your question correctly, you're asking why a flyover has excess clearance beneath it for traffic below.  For example, if the tallest vehicle expected beneath the flyover is 12ft tall, the flyover may have 14ft clearance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many reasons for this, some of them more obvious than others.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You don't want to try to squeeze a 12ft tall vehicle through something 12ft exactly.  Vehicles have suspensions which, depending on the exact loading of the vehicle, may have a variance of a couple inches.  Thus, a lightly-loaded vehicle could be riding higher than the 12ft nominal clearance and hit the bottom of the flyover.&lt;/li&gt;&#xA;&lt;li&gt;Occasionally, heavy, oversized loads will need to be transported.  Having excess clearance is very beneficial for travel of such items, since their dimensions tend to be all over the place.&lt;/li&gt;&#xA;&lt;li&gt;Roads have to be resurfaced.  Often times, it is more economical to do a simple resurface where a new layer of asphalt is added on top of the existing layer.  This reduces the clearance between the road and the flyover.  Thus, building in some margin for the clearance is advantageous.&lt;/li&gt;&#xA;&lt;li&gt;If the road below is expected to be in use during construction of the flyover, the height of the flyover may need to be increased to provide clearance beneath any formwork (falsework) supporting construction of the flyover (thanks @CableStay).&lt;/li&gt;&#xA;&lt;li&gt;Bridges deflect under load.  Generally, the deflection is limited to a certain value that depends on the jurisdiction.  However, you don't want a deflected bridge to eat into the minimum clearance, hence some margin is built in.&lt;/li&gt;&#xA;&lt;li&gt;The height of the flyover could have nothing to do with required clearance.  Train tracks are designed to be as flat as possible and have no slopes where feasible.  Thus, if the train tracks are, say, 30ft above the road surface below, it would be silly to slope the train tracks down to make a minimum clearance of 15ft at the road level, just to slope the train tracks back up on the other side.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Where I went to middle school, there was a rail viaduct just down the road.  There were a couple times where a semi trailer struck the bottom of the viaduct due to inadequate clearance.  If I recall, this road had been recently resurfaced, so who knows if the clearance value had been updated afterward.  Just something to think about.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-10-07T14:55:28.113" LastActivityDate="2015-10-07T14:55:28.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4241" PostTypeId="2" ParentId="4238" CreationDate="2015-08-31T17:27:41.753" Score="4" Body="&lt;p&gt;In the United States the &lt;a href=&quot;https://law.resource.org/pub/us/cfr/manifest.us.html&quot; rel=&quot;nofollow&quot;&gt;public safety standards&lt;/a&gt;, designed by the &lt;a href=&quot;http://www.transportation.org/Pages/Default.aspx&quot; rel=&quot;nofollow&quot;&gt;American Association of State Highway and Transportation Officers&lt;/a&gt; (AASHTO), require that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Vertical clearance on State trunk highways and interstate systems in rural areas shall be at least 16 feet over the entire roadway width, to which an allowance should be added for resurfacing. On State trunk highways and interstate routes through urban areas a 16-foot clearance shall be provided except in highly developed areas. A 16 foot clearance should be provided in both rural and urban areas where such clearance is not unreasonably costly and where needed for defense requirements. Vertical clearance on all other highways shall be at least 14 feet over the entire roadway width to which an allowance should be added for resurfacing. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As others have pointed out, it is important to allow enough height for nonstandard loads.  In addition, ensuring that all of the bridges in the highway system conform to a certain level of requirements ensures that vehicles can be designed such that they can travel throughout the country without having to check the height of every flyover on their route.&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="964" LastEditDate="2015-09-12T20:18:57.087" LastActivityDate="2015-09-12T20:18:57.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4244" PostTypeId="1" CreationDate="2015-08-31T20:27:46.550" Score="10" ViewCount="599" Body="&lt;p&gt;I'm looking for a justification for the use of cotton duck bearing pads without a sole plate. I'm specifically considering their use with precast I-girders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have a two span bridge, simply supported for dead load, with expansion bearings at the abutments and fixed bearings at the piers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The configuration for the expansion bearings that I'm familiar with is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ebnrJ.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ebnrJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(bearing plate embedded in girder - welded to sole plate - which is attached to stainless steel sliding plate - that rests on a PTFE sheet - which is attached to the cotton duck pad)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The configuration that the Nebraska Department of Roads specifies for a &quot;fixed&quot; bearing is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ISbSJ.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ISbSJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(bearing plate embedded in girder - sitting on cotton duck pad)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How can this be justified as a fixed girder? Are there studies that support the reliability of counting on friction between the bearing and a steel plate?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If it's not truly a fixed girder, do I have a &quot;floating&quot; bridge? What design considerations/concerns does this raise? If you've encountered this consideration what is your design approach?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2016-12-30T18:26:57.243" Title="Detailing Fixed Cotton Duck Bridge Bearings" Tags="|bridges|bearings|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="4248" PostTypeId="2" ParentId="2015" CreationDate="2015-09-01T05:56:40.513" Score="0" Body="&lt;p&gt;Single start threads are commonly use since it has large mechanical advantage &amp;amp; also it has self locking property because in multistart thread,this property is lost&lt;/p&gt;&#xA;" OwnerUserId="2864" LastEditorUserId="33" LastEditDate="2015-09-10T15:51:42.843" LastActivityDate="2015-09-10T15:51:42.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4249" PostTypeId="2" ParentId="4225" CreationDate="2015-09-01T13:36:39.083" Score="0" Body="&lt;p&gt;Try considering the force required to stop the jet, then the same force again to accelerate the jet back to the same velocity, but in the opposite direction.&lt;/p&gt;&#xA;" OwnerUserId="2870" LastActivityDate="2015-09-01T13:36:39.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4251" PostTypeId="1" AcceptedAnswerId="4252" CreationDate="2015-09-01T20:43:17.637" Score="9" ViewCount="4977" Body="&lt;p&gt;I'm encasing the abutment ends of prestressed concrete bridge girders in concrete before the deck pour. (For lateral stability and to prevent the soil from spilling through.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Discussion of intermediate diaphragms for prestressed girders abounds, but I haven't found information on structural behavior of end diaphragms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For purposes of a consulting engineer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut / engineering judgment is telling me the effect is negligible, but I prefer to understand what's actually going on before making simplifying assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conceptually&lt;/strong&gt; (versus a girder without the diaphragms):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does this change deflections during the deck pour? What about long-term deflections?&lt;/li&gt;&#xA;&lt;li&gt;Does this change the live load moments &amp;amp;/or shears?&lt;/li&gt;&#xA;&lt;li&gt;Is there any logic by which the encased length should be a function of girder depth or span length?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I've put a sketch below since terminology and typical bridge configurations vary around the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LUcbf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LUcbf.png&quot; alt=&quot;Diaphragm&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-09-02T00:46:01.043" Title="What is the effect of end diaphragms on prestressed girder forces/deflections?" Tags="|bridges|prestressed-concrete|deflection|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4252" PostTypeId="2" ParentId="4251" CreationDate="2015-09-02T00:46:01.043" Score="3" Body="&lt;p&gt;I have no resources to point to in regards to this, but I'm fairly certain your gut-feeling is correct. In fact, end diaphragms exist in bridges primarily to hold the soil on the approach. They can also be useful for temporary lateral stability of the beams, but if that were the only reason, temporary metallic bracing would be much easier and cheaper. Lateral instability (buckling) in the final structure is usually not a controlling factor, especially given the (partial) bracing that is the slab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mid-span diaphragms help with transverse load distribution and stop the primary beams from &quot;opening&quot; along with the slab deflection. The same applies to some extent to end diaphragms, but to a much lesser extent since the load is transmitted almost immediately to the bearings. The bearings also help to impede the beams from &quot;opening&quot;. This is probably why you've found no resources in regards to end diaphragms: no one's really bothered to think too much about them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, answering your specific questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This would change deflections somewhat, since the end diaphragm is going to change load distributions from your slab to your beams. This will be most relevant if your bridge has a small distance between diaphragms (less than twice the distance between primary beams, according to classic slab methods such as Rüsch). If the diaphragms are more spread out, then they will have almost no influence in the load distribution and therefore will not influence deflections. This also applies to long-term deflections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long-term deflections are, however, influenced by one more factor, and that is the prestress' differed losses. Over time, the primary beams are going to try to shrink. This is due not only to the concrete's natural shrinkage, but also due to the prestress' compressive creep. If all the beams were exactly the same, the creep and shrinkage should progress similarly along all of them. In that case, the end diaphragm would not have an effect since all the beams would &quot;pull&quot; it the same amount, implying in a simple rigid-body translation, without any deformation on the diaphragm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That, however, would be in the perfect world, and that ain't ours. Creep and shrinkage are mysterious and unstable, with a lot of scatter. So even exactly similar beams would probably result in different creeps and shrinkages, meaning that the end-diaphragms would be deformed. The diaphragms deformations (which would appear as horizontal shear forces on the diaphragm) will create tensile forces in the primary beams over time and these tensile forces will influence the creep behavior of the beams over time, leading to a recursive effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the beams would never be exactly the same because the loading in each beam is different (perhaps the central beams have similar loads, but they will certainly not be similar to those at the transversal extremities of the bridge), which is sufficient to generate (slightly) different creep behaviors in the beams.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As mentioned in the beginning of item 1, this will change the stresses suffered in the beams since the load distribution from the slab to the beams is modified by the existence of the end diaphragms. If the diaphragms are sufficiently far apart, however, the effect is almost certainly negligible.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Here I have a bit less to go on. I can say, however, that at the firm where I work we usually embed the primary beam only a few centimeters (3-5 cm, usually) into the end diaphragms. This allows the end diaphragms' crack-control reinforcement to pass without much fuss. The primary beams are designed with all their reinforcement sticking out at both ends in order to anchor into the diaphragms. Obviously, if your beam is unusual or if is a deep beam, the behavior might be different.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-09-02T00:46:01.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4253" PostTypeId="1" CreationDate="2015-09-02T09:51:44.073" Score="4" ViewCount="134" Body="&lt;p&gt;When we have a part that tends to corrode; Is it possible to apply an electrode on it to sense the excess or deficit of electrons on it that make it prone to corrosion and accordingly send a current that will drive that excess or deficit of electrons away? So that it will protect the part from corrosion?&lt;/p&gt;&#xA;" OwnerUserId="2353" LastEditorUserId="272" LastEditDate="2015-09-02T12:26:26.760" LastActivityDate="2015-09-03T01:03:48.693" Title="Electrochemical protection for corrosion" Tags="|mechanical-engineering|automotive-engineering|corrosion|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4254" PostTypeId="1" AcceptedAnswerId="4256" CreationDate="2015-09-02T10:33:09.343" Score="2" ViewCount="986" Body="&lt;p&gt;I am taking organic solar cell course on Coursera, and I have a question about I-V curve for a certain solar cell. The question asks for the solar cell characteristics such $V_{oc}$, $P_{max}$ and $FF$. I get the $V_{oc}$ correctly but I couldn't calculate the other two correctly as they provide me with an excel sheet of the current and voltage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plotted the voltage on the x-axis and current,power on y-axis below. I need to get the correct $I_{max}$ and $V_{max}$ from the graph with (I-V) in blue and (P-V) in orange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the question from Coursera:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Download this dataset &lt;a href=&quot;https://i.stack.imgur.com/DJw1t.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;DataSheet&lt;/a&gt; (, decimal mark: dataset) and&#xA;  calculate the maximum power point (Pmax). Give answer in either W or&#xA;  mW, a valid form of answer would be '94' or '0.094'.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Hint: You may need to do an interpolation to get the correct answer,&#xA;  if the value you are looking for does not exist.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DJw1t.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DJw1t.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2876" LastEditorUserId="1586" LastEditDate="2015-09-08T16:59:20.133" LastActivityDate="2017-02-11T08:28:13.477" Title="How to get the maximum power point from I-V curve for Organic Photovoltaics (OPV)?" Tags="|chemical-engineering|renewable-energy|photovoltaics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4255" PostTypeId="1" CreationDate="2015-09-02T12:21:42.787" Score="8" ViewCount="2541" Body="&lt;p&gt;Have spent hours poring over this homework question but am having difficulty getting the right process going to solve the problem. I don't really know where to start, and I'd really be grateful if someone could show me the ropes (pun intended ^_^). There are two components to the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(a) What is the tension force $T$ in the cables at this instant (Newtons)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(b) What is the net horizontal thrust force required to produce the truck's motion? This includes drive force from the wheels, rolling resistance and air resistance.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WZXN0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WZXN0.jpg&quot; alt=&quot;The Problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My (flawed, obviously) attempt at A went something like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Draw FBD for block A&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Apply Newton's Second Law, where tension in cable is constant throughout its length $F_{y,A} = 2T-m_A\cdot g = m_A\cdot a_A$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Length of cord is constant, so $L = -2X_A+X_T + C$, and $-2a_A+a_T = 0$, so the acceleration of block A is $\frac{1}{2}a_T = 1.2\text{ m/s}^2$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rearranging equation from (2) provides $2T = m_A*g + \frac{1}{2}m_A\cdot a_T$, and then T = 456.5N.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Part B I struggle with altogether, because I don't have a clear process in my head to tackle the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit~ A written attempt at a solution - Not sure how to factor in the angled a(T) as proposed...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ou7xf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ou7xf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2877" LastEditorUserId="86" LastEditDate="2015-09-02T16:27:38.783" LastActivityDate="2015-12-14T20:49:26.130" Title="Dynamics Simple Pulley System - Stuck on the Process" Tags="|applied-mechanics|dynamics|pulleys|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="4256" PostTypeId="2" ParentId="4254" CreationDate="2015-09-02T12:22:22.073" Score="1" Body="&lt;p&gt;As you have the power-voltage curve, finding the maximum power point is easy: it's the turning point on the power-voltage curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the absolute values in your graph look completely wrong. If the graph is flipped upside down, and based at zero power, zero current and zero voltage, then the blue curve does indeed look like an I-V curve for a photovoltaic cell, and the orange curve looks like the corresponding power curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it looks like the shape of your curve is right (albeit upside down from the usual presentation), but the absolute values are wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to work out what the right absolute numbers are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that the lowest the actual voltage is going to be is zero, and the lowest observed value is currently marked as -10V. The curve is the right shape going left to right (it's upside down, but does not also need reflecting in a vertical axis), so the correction coefficient for voltage is going to be positive. You know that the lowest the actual current and the actual power can be is also zero, and their lowest observed values are marked as 0.1A and about 6.9W respectively - I say &lt;em&gt;lowest&lt;/em&gt;, because we need to flip the curve upside down. And because we do, you know the coefficient to correct them will be negative. And you know that $P_{real}=I_{real}V_{real}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the shape of the curves, it's safe to assume that current and voltage each need a linear transformation (i.e. $V_{real} = aV_{obs}+b$, where you need to find $a$ and $b$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives you a set of simultaneous equations you can solve, to transform observed voltage (current/power) [$V_{obs}, I_{obs}, P_{obs}$] into actual voltage (current/power) [$V_{real}, I_{real}, P_{real}$]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write out those simultaneous equations, and solve them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the recipe on how to solve it. You'll need to follow the recipe yourself, though: given that this is coursework, there's no use in me doing it for you. Be sure to go through the recipe yourself, and make sure you understand why each step exists - try to work out how I derived it.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-09-11T08:11:53.063" LastActivityDate="2015-09-11T08:11:53.063" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4258" PostTypeId="2" ParentId="2015" CreationDate="2015-09-02T13:49:00.813" Score="1" Body="&lt;p&gt;In Addition to the other answers, manufacturing of single-start threads is more simple thus easier to control from a quality perspective. Also, Multi-start threads are typically used for motion control not fastening(unless special cases as described above for specialty plastic fasteners and such)&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2015-09-02T13:49:00.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4260" PostTypeId="5" CreationDate="2015-09-02T17:36:36.893" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-09-02T17:36:36.893" LastActivityDate="2015-09-02T17:36:36.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4261" PostTypeId="4" CreationDate="2015-09-02T17:36:36.893" Score="0" Body="Use for questions about non-photovoltaic solar energy applications, such as passive solar heating or concentrated solar thermal power. For PV applications, use the photovoltaics tag instead." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-09-02T17:36:36.893" LastActivityDate="2015-09-02T17:36:36.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4263" PostTypeId="1" AcceptedAnswerId="5650" CreationDate="2015-09-02T19:13:50.997" Score="3" ViewCount="521" Body="&lt;p&gt;Two questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Has anyone had experience in retrofitting consumer (locally-available) square aluminum tube as telescoping structural support?&lt;/strong&gt;  Ergo; with 0.125&quot; wall thickness square tube at an outside dimension (OD) of 1.5&quot;, would minor superficial sanding and end beveling of an inner square tube with OD of 1.25&quot; enable a telescoping pair?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, &lt;strong&gt;could said pair be capable of implementing a gas spring internally for vertical assist?&lt;/strong&gt;  See image below, noted: gas spring improperly sketched upside down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loads and dimensions can be provided if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y23Hq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y23Hqm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDITION:&lt;/strong&gt;&#xA;Is there an implied ideal mount offset for a gas spring in this circumstance relative to the load distribution (please see 2nd attached image below)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nmvvO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nmvvO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2878" LastEditorUserId="2878" LastEditDate="2015-09-05T15:04:24.760" LastActivityDate="2015-10-03T15:01:49.433" Title="Assist for vertically telescoping support?" Tags="|mechanical-engineering|materials|tolerance|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4264" PostTypeId="2" ParentId="4253" CreationDate="2015-09-02T19:30:13.637" Score="1" Body="&lt;p&gt;What you're describing sounds like a galvanic cathode protection system.  In this system, a sacrificial anode is connected to the metal part to be protected.  The material in the anode is chosen such that it has a lower electropotential than the protected part so that it preferentially corrodes instead of the part.  The electrical connection between the part and the anode allows current to flow, which forces the part to become more cathode-like, which slows its rate of oxidation.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-09-02T19:30:13.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4265" PostTypeId="1" AcceptedAnswerId="4266" CreationDate="2015-09-02T19:31:02.100" Score="2" ViewCount="307" Body="&lt;p&gt;I am designing a block and tackle system to haul wood up a ramp onto my porch. The issue I am having is that the rope seems like it will rub against the top of the ramp and cause it to fray. How can I prevent this from happening?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PkHNo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PkHNo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2882" LastActivityDate="2015-09-02T20:35:42.867" Title="Block and Tackle with Ramp - stopping rope drag" Tags="|applied-mechanics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="4266" PostTypeId="2" ParentId="4265" CreationDate="2015-09-02T20:35:42.867" Score="3" Body="&lt;p&gt;Here is how I would solve this in true bodge-job fashion:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Materials:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2x bricks, the type with holes in&lt;/li&gt;&#xA;&lt;li&gt;1x short section of broom-handle&lt;/li&gt;&#xA;&lt;li&gt;1x short plastic pipe, or toilet-roll tube (bigger than broom handle)&lt;/li&gt;&#xA;&lt;li&gt;2x pencils, or bits of string, or something&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Assembly:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Put the pipe over the broom handle such that it can rotate freely&lt;/li&gt;&#xA;&lt;li&gt;Support the broom handle in the holes between the two bricks&lt;/li&gt;&#xA;&lt;li&gt;Optionally drill holes through the ends of the broom handle and put a pencil through it, to stop it sliding sideways and dropping out of the holes. Alternatively put a bit of string through it and tie to the brick, or anything really.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Usage:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Place the roller at the apex of the hill. The rope will roll over the roller then lift off as the truck gets to the top. The truck will drive over the top of the roller assembly as you pull it onto the flat section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jobsagoodun.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2015-09-02T20:35:42.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4268" PostTypeId="2" ParentId="4253" CreationDate="2015-09-03T00:59:29.960" Score="4" Body="&lt;p&gt;I believe what you are suggesting is impressed current corrosion protection.  It is used commonly in pipeline and underground storage tank applications. &lt;a href=&quot;http://www.pipingandcorrosion.com/how-cathodic-protection-works.html&quot; rel=&quot;nofollow&quot;&gt;Example&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ICCP is superior to sacrificial anode protection but is an active process while the latter is passive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not coat the substrate I would think it would be corroding at a fairly steady state; so you would need to supply the current continuously.  The case of when the automobile is not running would need to be addressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do coat the substrate then you would be adding a redundancy which drives up cost and you still have to account for a power source during automobile downtime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way it is important to note that they both require an anode.  If the anode in ICCP is less noble than steel then it will deteriorate.  If it is more noble then you would need constant current or galvanic attack on the substrate would occur.&lt;/p&gt;&#xA;" OwnerUserId="2557" LastActivityDate="2015-09-03T00:59:29.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4269" PostTypeId="2" ParentId="4253" CreationDate="2015-09-03T01:03:48.693" Score="3" Body="&lt;p&gt;There are two options: a passive method and an active method.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;Passive method&lt;/strong&gt; is to use a sacrificial zinc anode. The zinc corrodes instead of the metal that it is attached to. More information about the mechanics of the system can be found on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Galvanic_anode&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;Active method&lt;/strong&gt; sounds more like what you are asking for in your question. This method imposes a DC current on the structure to effectively drive the chemical reaction away from reaction with the metal that it is protecting. As above, more information can be found on &lt;a href=&quot;https://en.wikipedia.org/wiki/Cathodic_protection#Impressed_current_systems&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both of these methods provide constant protection and aren't turned on and off.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-09-03T01:03:48.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4270" PostTypeId="1" AcceptedAnswerId="4272" CreationDate="2015-09-03T07:33:44.673" Score="3" ViewCount="2725" Body="&lt;p&gt;&quot;Sketch a combinational circuit which takes as input, two 4-bit binary numbers,&#xA;A3A2A1A0 and B3B2B1B0, and which outputs the 8-bit product P7P6P5P4P3P2P1P0.&#xA;Assume that you have available sixteen 2-input AND gates and three 4-bit adders.&#xA;Please explain the reasoning behind your design.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a bit of fiddling I got close to the final circuit but I found that three 4-bit adders wouldn't be enough. To save time (and my complete lack of being able to explain anything) I found &lt;a href=&quot;http://www.adaelis.com/misc/temp/russian_mul/russian_mul.png&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;, which is almost exactly what I need except it used two 4-bit adders and one 6-bit adder. I want to know if it's even possible to construct a circuit that can multiply two 4-bit numbers using the gates specified above, and if so, how can you do it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="2888" LastActivityDate="2015-09-03T10:43:49.823" Title="How to create a logic circuit that multiplies two 4 bit numbers, using only 16 2-input AND gates and 3 4-bit adders?" Tags="|electrical-engineering|computer-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4271" PostTypeId="1" AcceptedAnswerId="5644" CreationDate="2015-09-03T09:56:25.713" Score="2" ViewCount="2619" Body="&lt;p&gt;I'm studying &quot;signals and systems&quot;. We can indentify two kinds of signals: power signal and energy signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definitions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy of a time-continous signal&#xA;$$ E_x = \int_{-\infty}^{+\infty}{|x(t)|}^2dt $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power of a time-continous signal&#xA;$$ P_x = \lim_{Z\rightarrow+\infty}{\frac{1}{2Z}\int_{-Z}^{+Z}{|x(t)|}^2dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An energy signal satisfies $0&amp;lt;E_x&amp;lt;+\infty$ and $P_x=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A power signal satisfies $0&amp;lt;P_x&amp;lt;+\infty$ and $E_x = +\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we have a signum function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$sgn(t)=\left\{\begin{matrix}&#xA;1 &amp;amp; if\;t \ge 0 \\ &#xA;-1 &amp;amp; if\;t &amp;lt; 0&#xA;\end{matrix}\right.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we calculate the formulas, then we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_x = 0 \;\; E_x=0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, as the title said: the signum function which kind of signal is?&lt;/p&gt;&#xA;" OwnerUserId="2891" LastEditorUserId="421" LastEditDate="2015-10-03T00:15:36.870" LastActivityDate="2015-10-03T00:15:36.870" Title="What kind of signal is the signum function? A power signal or an energy signal?" Tags="|telecommunication|power|energy|signal|systems-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4272" PostTypeId="2" ParentId="4270" CreationDate="2015-09-03T10:19:26.683" Score="0" Body="&lt;p&gt;The first step is ANDing all the combinations of inputs which takes all the 16 AND gates available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call those results $S_{00}$ $S_{01}$ $S_{02}$ $S_{03}$ $S_{10}$ etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is we arrange them in how we need to add them we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{matrix}&#xA;&amp;amp; &amp;amp; &amp;amp; S_{03}&amp;amp;S_{02}&amp;amp; S_{01} &amp;amp; S_{00} \\&#xA;&amp;amp; &amp;amp; S_{13}&amp;amp;S_{12}&amp;amp; S_{11} &amp;amp; S_{10} &amp;amp; \\&#xA;&amp;amp; S_{23}&amp;amp;S_{22}&amp;amp; S_{21} &amp;amp; S_{20} &amp;amp; &amp;amp; \\&#xA;S_{33}&amp;amp;S_{32}&amp;amp; S_{31} &amp;amp; S_{30} &amp;amp; &amp;amp; \\&#xA;\end{matrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S_{00}$ is the low bit of the output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you use 1 4-bit adder to add {0,$S_{03}$, $S_{02}$, $S_{01}$} and {$S_{13}$, $S_{12}$, $S_{11}$, $S_{10}$} Output is {$A_c$, $A_3$, $A_2$, $A_1$, $A_0$}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we substitute that result in the original table we see that this collapses the top 2 rows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{matrix}&#xA;&amp;amp; A_c &amp;amp; A_3 &amp;amp; A_2 &amp;amp; A_1 &amp;amp; A_0 &amp;amp; S_{00} \\&#xA;&amp;amp; S_{23}&amp;amp;S_{22}&amp;amp; S_{21} &amp;amp; S_{20} &amp;amp; &amp;amp; \\&#xA;S_{33}&amp;amp;S_{32}&amp;amp; S_{31} &amp;amp; S_{30} &amp;amp; &amp;amp; \\&#xA;\end{matrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_0$ is bit 1 of the output&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you use the second 4-bit adder to add {$A_c$, $A_3$, $A_2$, $A_1$} and {$S_{23}$, $S_{22}$, $S_{21}$, $S_{20}$} output is {$B_c$, $B_3$, $B_2$, $B_1$, $B_0$}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{matrix}&#xA;B_c &amp;amp; B_3 &amp;amp; B_2 &amp;amp; B_1 &amp;amp; B_0 &amp;amp; A_0 &amp;amp; S_{00} \\&#xA;S_{33}&amp;amp;S_{32}&amp;amp; S_{31} &amp;amp; S_{30} &amp;amp; &amp;amp; \\&#xA;\end{matrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B_0$ is bit 2 of the output&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you use the third 4-bit adder to add {$B_c$, $B_3$, $B_2$, $B_1$} and {$S_{33}$, $S_{32}$, $S_{31}$, $S_{30}$} output is {$C_c$, $C_3$, $C_2$, $C_1$, $C_0$}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The carry input on all the 4-bit adders is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total output is then: {$C_c$, $C_3$, $C_2$, $C_1$, $C_0$, $B_0$, $A_0$, $S_{00}$}&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-09-03T10:43:49.823" LastActivityDate="2015-09-03T10:43:49.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4273" PostTypeId="1" CreationDate="2015-09-03T10:20:34.620" Score="0" ViewCount="70" Body="&lt;p&gt;Good day! We (undergraduates w/o prior foundry experience) inspected a foundry for its health hazards, our problem is we are not permitted to be accompanied by a laboratory assistant, so we don't have any idea what the following parts are for, kindly help us, thank you for your help! :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)&lt;a href=&quot;https://i.stack.imgur.com/wvx0o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvx0o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GmVex.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GmVex.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2892" LastActivityDate="2015-09-03T13:46:26.720" Title="Identification of Foundry equipment" Tags="|mechanical-engineering|manufacturing-engineering|equipment-selection|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4274" PostTypeId="1" CreationDate="2015-09-03T10:58:15.387" Score="3" ViewCount="286" Body="&lt;p&gt;So I have to analyze this circuit, with inputs C0, A0, B0, A1, B1, A2 and B2, and outputs S0, S1, S2 and C3. It's been suggested I don't use a truth table to figure it out. I need to find out what the circuit does by deriving expressions for the outputs.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've identified the XOR gate, and the AND gate sections for the A,B pairs, are adders. I am yet to figure out what the logic block is. By putting in values A=101, B=111, C0=0, I have found out that this whole system is an adder. But I don't know what the logic block section is. That's what I've done so far.&lt;a href=&quot;https://i.stack.imgur.com/hJZWy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hJZWy.png&quot; alt=&quot;Inputs and Outputs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vDKAo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vDKAo.png&quot; alt=&quot;Internal structure of the logic block&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2893" LastEditorUserId="33" LastEditDate="2015-11-03T02:20:11.490" LastActivityDate="2015-11-03T02:20:11.490" Title="What does this Logic Circuit do?" Tags="|electrical-engineering|circuits|circuit-design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4275" PostTypeId="1" AcceptedAnswerId="4276" CreationDate="2015-09-03T11:09:09.517" Score="1" ViewCount="180" Body="&lt;p&gt;An input device sends a 20ms positive signal to perform an action. The signal is captured using an interrupt triggered on the rising edge of the positive signal. It appears that there might be some electrical noise on the signal line. I suspect this because the interrupt is getting trigger inadvertently. I currently don’t have access to an oscilloscope to verify the signal integrity.  I am using MSP430 microcontroller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;  What is the most effective/&lt;strong&gt;simple&lt;/strong&gt; way for the micro-controller to capture the 20ms positive signal with out modifying electrical hardware? &lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="1586" LastEditDate="2015-10-13T12:35:20.590" LastActivityDate="2017-02-18T22:09:31.490" Title="How can electrical noise be isolated in an Embedded System?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4276" PostTypeId="2" ParentId="4275" CreationDate="2015-09-03T11:17:49.630" Score="1" Body="&lt;p&gt;Add a &lt;a href=&quot;https://en.wikipedia.org/wiki/Low-pass_filter#Electronic_low-pass_filters&quot; rel=&quot;nofollow noreferrer&quot;&gt;low-pass filter&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be a resistor in series and capacitor in parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:RC_Divider.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i5N1U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't adjust the hardware then you will need to debounce the signal in software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That will mean taking the average of the last x samples and then seeing if it exceeds the trigger level&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-09-03T11:46:14.880" LastActivityDate="2015-09-03T11:46:14.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="4277" PostTypeId="1" CreationDate="2015-09-03T11:20:41.037" Score="4" ViewCount="119" Body="&lt;p&gt;&lt;a href=&quot;http://www.wver.de/index.php/abwassertechnik/109-abwassertechnik/klaeranlagenuebersicht/293-klaeranlage-heinsberg-kirchhoven&quot; rel=&quot;nofollow noreferrer&quot;&gt;Consider this overview picture of a WWTP&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KL7nB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KL7nBt.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the back, you see two oval tanks with bridges and walls, separating an inner from an outer area. These are aeration tanks.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a specific reason to build them like this, instead of circular, or is there a specific process that calls for this shape? I'm aware of aeration tanks with an aerated zone and an anoxic zone, built concentrically with circular tanks.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="368" LastEditDate="2016-01-19T18:44:21.253" LastActivityDate="2016-04-18T21:07:42.720" Title="Oval aeration tanks - what are they called, what are the key differences in operation to circular tanks?" Tags="|civil-engineering|environmental-engineering|waste-water-treatment|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4279" PostTypeId="1" CreationDate="2015-09-03T11:42:47.073" Score="7" ViewCount="148" Body="&lt;p&gt;If we have a tracking system placed in a tough desert environment with occasional sandstorms and temperatures that go as high as 50C, which actuator would be better: a hydraulic piston or an electric motor with a gearbox?&#xA;My main concern is performance and resistance to conditions&lt;/p&gt;&#xA;" OwnerUserId="2895" LastEditorUserId="2895" LastEditDate="2015-09-03T12:40:10.377" LastActivityDate="2015-09-15T04:04:44.470" Title="Which is more durable in sandy desert conditions: electric motors or hydraulic pistons?" Tags="|motors|hydraulics|actuator|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4280" PostTypeId="1" CreationDate="2015-09-03T12:33:05.150" Score="1" ViewCount="629" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z7WqN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z7WqN.png&quot; alt=&quot;Question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the formula given at the start of the question, I had to convert the numbers in part (a) to floating point notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answers for part (a) were:&#xA;(i) 1111.2^11 + 111&#xA;(ii) 1111.2^11 + 000&#xA;(iii) 0000.2^00 + 000&#xA;(iv) 1100.2^00 + 000&#xA;(v) 1010.2^11 + 101&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now i have no clue on what to write for 2(b), because I didn't really use a specific formula for finding out the answers for part (a).&#xA;Any help would be greatly&lt;/p&gt;&#xA;" OwnerUserId="2897" LastActivityDate="2015-10-03T14:32:42.370" Title="How to represent floating point numbers using logic gates?" Tags="|electrical-engineering|computer-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4281" PostTypeId="2" ParentId="4280" CreationDate="2015-09-03T12:43:27.800" Score="1" Body="&lt;p&gt;To get the exponent you need to get the index of the highest selected bit of the top 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the truth table with $B_6$ $B_5$ $B_4$ -&gt; exponent:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1** -&amp;gt; 11&#xA;01* -&amp;gt; 10&#xA;001 -&amp;gt; 01&#xA;000 -&amp;gt; 00&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the mantissa you shift the bits to the right $E$ bits. The underflow lands in T after a $4 - E$ shift to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in other words if $E = 3$ then $M$ will take $B_6$ $B_5$ $B_4$ $B_3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if $E = 2$ then $M$ will take $B_5$ $B_4$ $B_3$ $B_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and so one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T$ will take whatever bits M didn't take.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-09-03T13:15:00.723" LastActivityDate="2015-09-03T13:15:00.723" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4283" PostTypeId="2" ParentId="4273" CreationDate="2015-09-03T13:46:26.720" Score="1" Body="&lt;p&gt;The first picture looks like a top-loading electric kiln:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/W5A9N.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W5A9N.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A foundry might use one to bake refractory molds prior to casting.  Not sure what the second picture is.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-09-03T13:46:26.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4285" PostTypeId="2" ParentId="2149" CreationDate="2015-09-03T14:19:35.657" Score="0" Body="&lt;p&gt;Given the variables involved and the relatively low cost of polycarbonate tubing, I'd be tempted to just buy a length and test it.   Use a fairly chunky dancer to build in a factor of safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polycarbonate isn't brittle and so catastrophic failure is unlikely.   However, it scratches easily, which might degrade its light transmission over time.   Stress corrosion cracking can be a long-term problem but I don't imagine any corrosive materials would be involved in the application you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the top of the pole can't be built-in, think about guy-lines ('shrouds' in yacht-speak), possibly with a spreader to give more room for the dancer.&lt;/p&gt;&#xA;" OwnerUserId="2520" LastEditorUserId="2520" LastEditDate="2015-09-03T14:31:26.717" LastActivityDate="2015-09-03T14:31:26.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4286" PostTypeId="1" CreationDate="2015-09-03T15:19:35.133" Score="2" ViewCount="211" Body="&lt;p&gt;I'm getting oval balloons for my annotations instead of circular balloons even though I selected the circular setting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zLdbX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zLdbXm.png&quot; alt=&quot;Balloon Settings&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to edit/revert this annotation to become a normal circle?&lt;/p&gt;&#xA;" OwnerUserId="2901" LastEditorDisplayName="user16" LastEditDate="2015-09-03T18:46:49.103" LastActivityDate="2015-09-03T18:46:49.103" Title="How do I make my balloon annotations circular instead of oval-shaped?" Tags="|mechanical-engineering|solidworks|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="4289" PostTypeId="2" ParentId="4274" CreationDate="2015-09-03T17:58:12.270" Score="2" Body="&lt;p&gt;The circuit appears to be a ripple carry adder using full adders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When compared to this circuit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pFLks.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pFLksm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;a href=&quot;http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Comb/adder.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Image courtesy of this tutorial&lt;/em&gt;&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see how the X and Y inputs map to your A &amp;amp; B inputs.  The Z outputs map to your S outputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input and output names give you a hint as to the circuit function as well.  C0 implies an input carry bit from the previous adder.  C3 implies an output carry bit, with C1 and C2 being used inside of the chip.  S0 - S2 also imply a &quot;sum&quot; output of the input bits.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-09-03T17:58:12.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4290" PostTypeId="2" ParentId="2382" CreationDate="2015-09-03T18:54:00.060" Score="1" Body="&lt;p&gt;Almost everyone is partly right.   You can fail the ring-pull by simple overload in one 'cycle' or you can accumulate plastic strain in three or four cycles.   This wouldn't normally be considered even low-cycle fatigue but I don't know that there's a lower limit on how many cycles Paris' Law can be applied.&lt;/p&gt;&#xA;" OwnerUserId="2520" LastActivityDate="2015-09-03T18:54:00.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="4291" PostTypeId="2" ParentId="4139" CreationDate="2015-09-03T20:29:49.177" Score="2" Body="&lt;p&gt;First I recommend you use mineral wool or ceramic wool instead of fiberglass as they are rated for higher temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Material    Temperature&#xA;&lt;br&gt;Glass wool  230 - 260 °C&#xA;&lt;br&gt;Stone wool  700 - 850 °C&#xA;&lt;br&gt;Ceramic fiber wool  1200 °C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mineral_wool&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Mineral_wool&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, using insulation to separate the aluminum and resistance heating wire is not a reliable solution. You should use ceramic rods, ceramic insulator rigs or something similar to space the wire away from the aluminum. You may even consider not using aluminum and directly wrapping a ceramic container with the wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of options for securing the insulation around the outside. A thin sheet of aluminum or stainless would be a nice clean way to keep the insulation protected.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-09-03T20:29:49.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="4292" PostTypeId="2" ParentId="4263" CreationDate="2015-09-03T23:56:42.483" Score="1" Body="&lt;p&gt;I have experience with both of these but not for this purpose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Telescoping tubing usually has a thickness of .105&quot; .  Also alot of square tubing has flashing left on the inside.  Both of these may cause issues with fit up.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the gas springs, I have only seen them go up to 500# force.  Also since the gas is compressible the force is not constant across the length of travel.  I would definitely try to mechanically prevent the gas spring from bottoming out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Foremost the concern here is safety.  What happens when one or both of the springs fail?  Or if they are overloaded?  Are there any pinch points?  Is the operator in danger of 300# of force hitting the bottom of his chin when a pin is released?  These are issues to consider when evaluating alternatives.&lt;/p&gt;&#xA;" OwnerUserId="2557" LastActivityDate="2015-09-03T23:56:42.483" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5292" PostTypeId="2" ParentId="449" CreationDate="2015-09-04T11:21:14.263" Score="0" Body="&lt;p&gt;Meshing a body with 2D triangles or 3D tets can be done automatically but these elements don't give the best results:  quads and bricks are generally better.   However, meshing a body entirely with quads/bricks can't be done automatically and you have to manually partition it into blocks that can be automeshed.   This isn't trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a mesh well-suited to a thermal analysis isn't generally well-suited to, say, a vibration analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, running analyses with huge numbers of tiny elements isn't the problem it once was and so tailoring the mesh to the type of analysis is less important than it once was.   Also, the tet element designed by Burton and Clegg (&lt;em&gt;Tetrahedral Elements for Explicit Ballistics Simulations&lt;/em&gt;), seems to perform as well as a brick, so my first point might be less important than it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, automatic meshing has come a long way but is still the subject of much research.   Will it ever be fully automatic?   I'm inclined to doubt it.   Even with automatic remeshing of areas of high field gradients, I think a good initial choice of mesh will be useful.  &lt;/p&gt;&#xA;" OwnerUserId="2520" LastActivityDate="2015-09-04T11:21:14.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5294" PostTypeId="1" AcceptedAnswerId="5295" CreationDate="2015-09-04T12:45:05.283" Score="1" ViewCount="188" Body="&lt;p&gt;I am trying to identify the section designation so I can determine it's properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W5znc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W5znc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked under &quot;C 150 x 100 x 40&quot; but I can't find its properties or what it's correct designation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this section called? It's not just a &quot;channel&quot; is it?&lt;/p&gt;&#xA;" OwnerUserId="2895" LastEditorUserId="122" LastEditDate="2015-09-04T14:26:51.617" LastActivityDate="2019-09-18T01:07:18.810" Title="What is the name of this section?" Tags="|civil-engineering|beam|standards|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5295" PostTypeId="2" ParentId="5294" CreationDate="2015-09-04T13:52:22.720" Score="2" Body="&lt;p&gt;It appears to be either a C-Channel or a Box-Channel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the information gathered at this &lt;a href=&quot;http://www.johnsonrollforming.com/display.php/display/A2/category/3&quot; rel=&quot;nofollow noreferrer&quot;&gt;manufacturer's reference&lt;/a&gt;, it appears that you'll need to identify a few more characteristics of the beam before you can determine its properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sample image and expected specifications for box channel &lt;a href=&quot;http://www.johnsonrollforming.com/display.php/display/A2/category/3&quot; rel=&quot;nofollow noreferrer&quot;&gt;from that previous reference&lt;/a&gt;.  This would be a cold rolled section made of light gauge sheet metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K9x30.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K9x30t.jpg&quot; alt=&quot;box channel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;&lt;a href=&quot;https://i.stack.imgur.com/K9x30.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;click to expand&lt;/a&gt;&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also hot rolled C-channel, which has a different shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KRii1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KRii1t.png&quot; alt=&quot;steel c-channel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;a href=&quot;http://www.engineeringtoolbox.com/american-standard-steel-channels-d_1321.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;some sites&lt;/a&gt; will provide properties for standard sizes of that channel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to tell from the image, but it appears to be rectangular tubing, so the 150 is likely millimeters and is the channel height.  You'll either need to provide more information or perform additional research on your own in order to determine what the 100 and 40 values refer to.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-09-04T17:49:17.427" LastActivityDate="2015-09-04T17:49:17.427" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5297" PostTypeId="1" AcceptedAnswerId="5298" CreationDate="2015-09-04T15:21:11.230" Score="3" ViewCount="1885" Body="&lt;p&gt;Every morning when I get up, I use &lt;a href=&quot;http://www.mrcoffee.com/specialty/grinders/automatic-burr-mill-grinder-stainless-steel/BVMC-BMH23.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this coffee grinder&lt;/a&gt; to grind beans for my morning coffee. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/bX9v7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bX9v7t.png&quot; alt=&quot;grinder&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I remove and subsequently open the top of the bin, I often find that enough static electricity has built up in the grounds/bin that particles of coffee will literally fly out of the bin and onto my hand and the counter, making a mess. Coffee will also be stuck to the sides of the bin to varying degrees, making it necessary to rap the bin on the counter in order to dislodge the particles from the sides. Of course, more then fly out and increase the mess on my counter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made some observations of a few factors that seem to affect this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It happens to some degree with most beans, but I've noticed that the darker the roast, the greater the static charge. It's the worst with French Roast (my favorite), which is also oily since they add oil to the beans in the roasting process.&lt;/li&gt;&#xA;&lt;li&gt;The coarseness of the grind doesn't seem to matter except that a coarse grind produces heavier particles which don't tend to fly as far.&lt;/li&gt;&#xA;&lt;li&gt;Cleaning the bin had no observable effect on the static.&lt;/li&gt;&#xA;&lt;li&gt;Waiting a while after grinding doesn't seem to do much to alleviate the problem. I've ground the beans and waited 10-20 minutes before removing the bin and observed no apparent decrease in the amount of static.&lt;/li&gt;&#xA;&lt;li&gt;It is worse in winter when the air is dry, but I don't really let my house get too humid.&lt;/li&gt;&#xA;&lt;li&gt;Google reveals that this problem is not specific to any one brand or model of grinder, but seems to plague most burr grinders. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My questions are these:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the cause of this static charge in the grinding process?&lt;/li&gt;&#xA;&lt;li&gt;What modifications can the end-user make to either the grinder or the process to prevent static buildup?&lt;/li&gt;&#xA;&lt;li&gt;How could this be eliminated in the design itself?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="282" LastEditDate="2015-09-04T18:54:47.037" LastActivityDate="2023-03-06T23:11:57.380" Title="How to prevent static buildup in an electric burr mill coffee grinder?" Tags="|process-engineering|product-engineering|grinding|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5298" PostTypeId="2" ParentId="5297" CreationDate="2015-09-04T17:29:10.657" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the cause of this static charge in the grinding process?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My thought is that the coffee grinder is essentially acting like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Van_de_Graaff_generator&quot; rel=&quot;noreferrer&quot;&gt;Van de Graaff generator&lt;/a&gt;.  Since the coffee maker likely has a plastic frame, rests (most likely) on rubber feet, and the electrical wiring (most likely) is insulated from the grinding mechanism, you have the potential for charge to be stored in the grinder.  If I had to guess, there's probably some insulative gap in the grinder drive train between it and the motor that does not allow grounding of the charge.  This may be by design to avoid the potential for electric shock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, assuming you're using it on a countertop that is not metallic (i.e., nonconductive), even of there was a discharge path from the grinder through the body, it can't ground out anyway since the counter won't conduct the charge.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What modifications can the end-user make to either the grinder or the process to prevent static buildup?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I might try soldering a grounding wire to the outside of the burr cone.  This would have to be connected to a ground somehow.  For example, you could open up an electrical outlet (turn the power off to the outlet first!) and attach the grounding wire to the metal outlet box.  The outlet box should be connected to your house's main ground.  This will allow the charge to dissipate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this had a grounded plug (three-prong), I might just connect the grounding wire to the ground in the plug wire.  However, this is not the case for this particular coffee grinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, dryness tends to be conducive to static buildup.  For example, it's much easier to shock yourself on a doorknob in the middle of winter (very dry air) than in the summer (usually humid air, unless you live in a desert or something).  Wetting the beans a bit may help too.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How could this be eliminated in the design itself?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the grinder had more metallic parts, especially in the main frame and the hoppers, and these parts were all connected in such a way as to allow discharge to the ground, it would likely eliminate the static problem altogether.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-09-04T17:34:40.760" LastActivityDate="2015-09-04T17:34:40.760" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5299" PostTypeId="1" AcceptedAnswerId="5312" CreationDate="2015-09-04T18:19:43.767" Score="4" ViewCount="1923" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How is shelf deflection determined?  More specifically, how is the deflection determined when one or more edges of the plate have reinforcing edging applied to them?&lt;/p&gt;&#xA;&lt;p&gt;I have been able to find &lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;plate deflection equations&lt;/a&gt; in a number of places, but of the resources I have found, they are limited to plates without any edging at all.&lt;/p&gt;&#xA;&lt;p&gt;I have found a &lt;a href=&quot;http://www.woodbin.com/calcs/sagulator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;shelf sag calculator&lt;/a&gt; which happens to be well known and loved by woodworkers.  But their calculator only handles a single side of edging, and they don't provide any explanation of the formulas that are used.  And depending upon what deflections I find, I may need to be able to evaluate the deflection in a lattice structure with a supporting skin.&lt;/p&gt;&#xA;&lt;p&gt;I would assume that the edging can be treated as simple beams supporting the plate, but I don't know how to apply that support to the plate.&lt;/p&gt;&#xA;&lt;p&gt;Is there a standard set of equations and / or plate geometries that can be applied to estimate what the shelf deflection may end up being?  Or is this getting into the non-trivial end of analysis and design?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;  For a hobby project of mine, we need to build shelving to support a variety of items which cover a range of shapes and sizes.  Some items are large and comparatively lightweight (e.g. a student's desk).  Others are small, dense, and kind of heavy (e.g. concrete block).  And of course, there are some in the middle for both size and either side of the weight spectrum.&lt;/p&gt;&#xA;&lt;p&gt;We're planning on building several sets of shelving to accommodate the various sizes of things that need to be stored.  And while we do have to design for the unknowns of the future, we're still trying to optimize costs through the amount of material used.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-09-06T18:28:22.040" Title="How is shelf deflection determined?" Tags="|mechanical-engineering|beam|deflection|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5301" PostTypeId="1" CreationDate="2015-09-04T18:57:41.727" Score="2" ViewCount="540" Body="&lt;p&gt;I am looking for the technical term which describes the kind of slot one finds commonly in modern phones to hold SIM cards. More specifically, what does one call a slot that one pushes the SIM card into once and it is secured there, and then pushes on the SIM card once again so that it pops out and allows it to be removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no technical term, does somebody know where I may find information on how such a slot works so I may attempt to replicate it in a design?&lt;/p&gt;&#xA;" OwnerUserId="2914" LastEditorUserId="272" LastEditDate="2015-09-04T19:21:40.483" LastActivityDate="2015-09-05T18:39:31.180" Title="What is the Technical Term for this kind of slot?" Tags="|springs|terminology|nomenclature|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5302" PostTypeId="2" ParentId="5301" CreationDate="2015-09-04T20:20:09.020" Score="-1" Body="&lt;p&gt;What you are looking for is not a &quot;slot&quot;, since that means just a thin and wide hole.  In general these things are called &quot;connectors&quot;, but that is a very general term that covers just about anything that is intended to provide a removable electrical connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, look for something like &quot;SIM card connector&quot;.  Go to a site like Mouser and start drilling down.  When you don't know the right search term, it helps to come at it a few different ways.  For example, the brute force method would be to start at &quot;connector&quot;, then selectively click on more specific options as they are presented.  You do know some things about this connector, like it will be meant to be mounted on a PC board, as apposed to being bolted to a chassis or free-hanging from a bundle of wires.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-09-04T20:20:09.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5303" PostTypeId="2" ParentId="4124" CreationDate="2015-09-04T20:31:27.477" Score="1" Body="&lt;p&gt;These kind of micrometers have two scales, a coarse and a fine.  You add them together to get the final reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fine scale is etched around the part that turns.  With nothing in the microcometer and it closed all the way, the 0 on this scale should line up with the horizontal line on the fixed part to the left.  As you turn the handle to open the gap, you will see that the numbers that line up with the fixed line get larger.  These numbers are usually directly in the units the micrometer is intended to measure, which looks like mils (1/1000 inch) from your picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all works fine until the gap is opened so that the handle goes thru a full turn.  That's where the second more coarse scale comes in.  This scale is revealed as the gap is increased.  Usually there is one line per handle turn, and these are usually calibrated to come out to a convenient multiple.  For example, if each handle turn opens the gap 100 mils, then there is a line every 100 mils on the fixed coarse scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to know what the coarse scale multiplier is, but that's easy to see by rotating the handle one turn.  Let's say the coarse scale is marked every 100 mils and the handle in mils.  Then for example, if you can see 3 lines of the coarse scale revealed and the handle scale is at 37, then the measurement is 337 mils, or 0.337 inch.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorDisplayName="user16" LastEditDate="2015-09-04T20:41:44.167" LastActivityDate="2015-09-04T20:41:44.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5304" PostTypeId="2" ParentId="5301" CreationDate="2015-09-05T00:09:05.150" Score="-1" Body="&lt;p&gt;The word &quot;slot&quot; is confusing in your question and possibly not a big part of what you're after. I think what you are describing has two components. Firstly it is a &quot;latching&quot; mechanism (which is probably a more specific term than &quot;bistable&quot;), in that one press puts it into a certain state and a further press takes it back out of that state. Secondly it is a &quot;catch&quot; or &quot;retention&quot; mechanism, in that the card is retained when it is in one of the states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to replicate within a design it will very much depend on your specifics. If it is purely mechanical then having a sprung tab that drops into a hole when your item is depressed will retain it &quot;latched down&quot;, you will then need a method of releasing the tab when it is repressed and preventing it from relocating in the hole as it releases, perhaps by forcing the tab to follow a different route as it returns. Look at the simple mechanisms in various types of ballpoint pen for ways to achieve this. Retaining a card or similar whilst you're in the latched position could just be an extension of this idea, depending on what shape the card is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your design is electro-mechanical then you have several further options, perhaps look at the design of toasters for ideas around electromagnets and springs, or CD trays for motor-powered release mechanisms.&lt;/p&gt;&#xA;" OwnerUserId="157" LastActivityDate="2015-09-05T00:09:05.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5305" PostTypeId="2" ParentId="2190" CreationDate="2015-09-05T00:23:46.277" Score="1" Body="&lt;p&gt;I would guess that it's using a variant of an &lt;a href=&quot;http://eprints.lancs.ac.uk/20295/1/20295.pdf&quot; rel=&quot;nofollow&quot;&gt;anglepoise lamp type mechanism&lt;/a&gt;, which relies on a &lt;a href=&quot;https://en.wikipedia.org/wiki/Spring_(device)#Zero-length_springs&quot; rel=&quot;nofollow&quot;&gt;zero free-length spring&lt;/a&gt; and some geometry. By adjusting the offset of the spring from the lever arm different loads can be balanced. The clever part will be in making the whole assembly compact and invisible to the user.&lt;/p&gt;&#xA;" OwnerUserId="157" LastActivityDate="2015-09-05T00:23:46.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5307" PostTypeId="1" AcceptedAnswerId="5309" CreationDate="2015-09-05T00:53:52.910" Score="6" ViewCount="3330" Body="&lt;p&gt;In the rankine cycle, water is first pressurized and then heated to produce steam before it is used to turn a turbine.  As far as i know, this is what is considered the best way to extract mechanical work on the turbine, but i want to understand why this is so.  How is it that we can extract more work on a turbine from steam than from regular unheated, un pressurized water?&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2015-09-05T02:17:36.073" Title="Why use steam in the rankine cycle?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5308" PostTypeId="1" AcceptedAnswerId="5310" CreationDate="2015-09-05T01:11:32.173" Score="4" ViewCount="2297" Body="&lt;p&gt;In my admittedly naive understanding of the rankine cycle, steam is condensed back into water after passing through the turbine.  Why not reuse the steam instead?  It would seem to require less effort to reuse heated steam than to intentionally condense the water only to reheat it once again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure i'm misunderstanding something about the process.  Why do we have to condense the water at all?&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2015-09-05T02:27:38.547" Title="Why do we need to condense steam in the rankine cycle?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5309" PostTypeId="2" ParentId="5307" CreationDate="2015-09-05T02:11:01.900" Score="7" Body="&lt;p&gt;A good way to understand why the phase change is so important is by looking at the Pressure vs. Volume diagram for the Rankine cycle (path 1-2-3-4-1):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jhxl0.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jhxl0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the work output of any thermodynamic cycle is the integral of its P-V diagram (area under the curve).  If the water in the Rankine cycle were not heated to steam, its change in volume would be very small, both in the pumping/heating processes and in the expansion/cooling processes.  And, if its volume didn't change significantly, its P-V curve would have almost no area.  Thus, the cycle would produce much less work output than if the volume changed significantly and produced a P-V diagram with a large cyclic integral.  The dashed line (path 1-2-2'-3'-1) represents what the path might look like if plain water were used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many early heat engines used a gas-phase-only working fluid; it was not until Rankine or DeLaval came along (can't remember who), and discovered the benefits of using a phase change in the cycle.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-09-05T02:17:36.073" LastActivityDate="2015-09-05T02:17:36.073" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5310" PostTypeId="2" ParentId="5308" CreationDate="2015-09-05T02:27:38.547" Score="7" Body="&lt;p&gt;It's for almost the same reason as my answer to &lt;a href=&quot;https://engineering.stackexchange.com/questions/5307/why-use-steam-in-the-rankine-cycle&quot;&gt;your other question&lt;/a&gt; on the Rankine cycle:  The work output of the cycle increases as the integral (area) of the P-V diagram (or T-S) diagram is increased.  Condensing the steam to liquid water results in a huge change in specific volume, which makes the integral of the P-V diagram much larger than if no phase change occured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does seem like energy is being wasted in the condensation process, but think of how much energy it would take to compress the working fluid to the boiler inlet pressure if it were a gas.  (It would be much more energy than if we just had to pump a liquid to the same pressure).&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-09-05T02:27:38.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5311" PostTypeId="1" CreationDate="2015-09-05T12:22:18.573" Score="3" ViewCount="422" Body="&lt;p&gt;If a member is very slender then its length will be large compared to its cross-sectional area. I've heard and read that slender members can potentially cause the stiffness matrix of a beam/frame element to become ill-conditioned in FEM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How/why can a slender member cause the stiffness matrix to become ill-conditioned?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my own thoughts regarding this question, which may or may not be correct - I'd like some confirmation. A long slender member will have larger axial stiffness than flexural stiffness. In the stiffness matrix, the axial stiffness term has the length of the member ($L$) in the denominator. However, the flexural stiffness terms typically have $L^2$ and $L^3$ in their denominators. Therefore, given constant cross-sectional properties ($A$ and $I$), as $L$ grows the flexural stiffness terms approach zero much quicker than axial stiffness. Is this the reason for the potential for ill-conditioning? Or is there something I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-09-05T12:46:15.590" LastActivityDate="2015-09-06T01:40:08.093" Title="Stiffness Method FEM - Slender Members" Tags="|structural-engineering|finite-element-method|structural-analysis|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5312" PostTypeId="2" ParentId="5299" CreationDate="2015-09-05T14:34:32.150" Score="3" Body="&lt;p&gt;The exact deflection result is a function of the shelving's structural system. In the OP comments you mention that the two options would be either a wooden skin with metal edging or a wooden lattice skinned with wood.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's start with the wooden skin with metal edging. Structurally, this is equivalent to a slab with multiple supports. The sort of support depends on... the sort of support. If the shelf is fixed to a wall, then that shelf-face will have a fixed support (restricts vertical displacements and rotations). If the shelf is not fixed to a wall, then it must either rest on metal edging (which restricts the vertical displacements of the shelf but would allow for small rotations) which transmits the load to a column or there is no metal edging at all and the shelf itself needs to transmit the entirety of the load to the column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rectangular shelf, you will have up to four supports, one on each face. Considering only more plausible configurations, you have the following possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One face fixed on the wall*&lt;/li&gt;&#xA;&lt;li&gt;One face fixed on the wall, lateral faces with metal edging&lt;/li&gt;&#xA;&lt;li&gt;One face fixed on the wall, other faces with metal edging&lt;/li&gt;&#xA;&lt;li&gt;Two faces fixed on the wall (corner-shelf), other lateral face with metal edging&lt;/li&gt;&#xA;&lt;li&gt;Two faces fixed on the wall, other faces with metal edging&lt;/li&gt;&#xA;&lt;li&gt;All faces with metal edging&lt;/li&gt;&#xA;&lt;li&gt;All but one face with metal edging&lt;/li&gt;&#xA;&lt;li&gt;Both long faces with metal edging*&lt;/li&gt;&#xA;&lt;li&gt;Both short faces with metal edging*&lt;/li&gt;&#xA;&lt;li&gt;No edging, supported directly on columns.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As you can see, this is a long list and I've removed a few which I found to be implausible. For this reason, there is no simple equation that can be adopted. This is made worse by the fact that the behavior of slabs is defined by the following multivariate differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\partial^4w}{\partial x^4} + 2\dfrac{\partial^4w}{\partial x^2\partial y^2} + \dfrac{\partial^4w}{\partial y^4} = \dfrac{p}{D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D = \dfrac{Eh^3}{12\left(1-\nu^2\right)}$. This is a pain to solve, so no one does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ways of getting around these equations. I found &lt;a href=&quot;https://www.ideals.illinois.edu/bitstream/handle/2142/4273/engineeringexperv00000i00303.pdf?sequence=3&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; which shows a lot of the necessary math in each of the support configurations, but in practice one doesn't go to such lengths. I can't find any resources for the case of wooden slabs, but in reinforced concrete slabs there are different approaches obtained by smarter men that I (Marcus, Czerny, Nadai, etc) which result in dozens of tables of coefficients. These coefficients can then be used to figure out how much load is transmitted in the x-axis and how much in the y-axis. Knowing that, you can calculate the slab as two perpendicular beams, which should result in similar deflections. The biggest problem translating this to a wooden slab is wood's anisotropy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way of getting around that ugly differential and the one I recommend is a finite element model. This is also probably the only one that can handle the &quot;no edging, supported directly on columns&quot; case and it can take into consideration the flexibility of the supports (in the case of the metal edging). It may seem like overkill but it's the best way to get a good estimate of what your actual deflections will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that some of the options on the list are marked with an asterisk*. These cases are exceptions and can be quite trivially calculated, since they behave in a manner identical to that of beams. Also, long slabs (where one side is more than twice the other), can usually be assumed to be &quot;infinite&quot;, meaning that they too can be considered equal to beams in the perpendicular direction (if $x \geq 2y$, then the slab works like beams with span $L=y$). For a beam or slab fixed on a wall with all other faces free, the maximum deflection is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\delta = \dfrac{PL^3}{3EI} &amp;amp;&amp;amp; \delta = \dfrac{qL^4}{8EI}&#xA;\end{align}$$&#xA;for a concentrated load $P$ and a distributed load $q$, respectively, where $L$ is the slab's span from fixed to free face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slab supported at two opposing faces is equivalent to a simply supported beam of equal span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\delta = \dfrac{PL^3}{48EI} &amp;amp;&amp;amp; \delta = \dfrac{5qL^4}{584EI}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a concentrated load $P$ at midspan and a distributed load $q$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now onto the lattice method. If the lattice is very dense, it will behave very similarly to the slab above. If it is less dense, then the behavior may differ significantly. What the threshold is, I can't say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the lattice doesn't act like a slab, it will behave like a grid. Once again, this can't be trivially solved and the best solution is an analysis tool, though in this case the tool can be much simpler (grids can be solved analytically without the need of FEM).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-09-05T14:34:32.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5313" PostTypeId="2" ParentId="5301" CreationDate="2015-09-05T14:55:42.177" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO2&amp;amp;Sect2=HITOFF&amp;amp;u=%2Fnetahtml%2FPTO%2Fsearch-adv.htm&amp;amp;r=1&amp;amp;f=G&amp;amp;l=50&amp;amp;d=PTXT&amp;amp;p=1&amp;amp;S1=%22spring+card+adapter%22&amp;amp;OS=&quot; rel=&quot;nofollow&quot;&gt;One patent&lt;/a&gt; I found refers to this mechanism as the &quot;lock spring&quot; of the  &quot;card adapter.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;As shown in FIGS. 1 and 8, a lock spring is provided in the card adapter to prevent the inserted memory card M from falling out easily or moving out of place. When the memory card M is inserted into the card adapter, at first the shoulder portion M3 of the memory card M contacts the protruding portion 61 of the lock spring, and then the protruding portion engages the cut-out portion M4 formed on the side surface of the memory card M at a predetermined insert position. At this time, the memory card M is fixed within the card adapter. The memory card M can be removed from the card adapter simply by holding the end portion of the memory card M with fingers, etc. and pulling it out in the direction opposite from the inserting direction.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Patent Family ID:   39133830&lt;/p&gt;&#xA;" OwnerUserId="2920" LastEditorUserId="1832" LastEditDate="2015-09-05T16:02:56.903" LastActivityDate="2015-09-05T16:02:56.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5314" PostTypeId="1" CreationDate="2015-09-05T17:05:35.657" Score="3" ViewCount="1935" Body="&lt;p&gt;I am wondering whether and how superfluids (possibly at room-temperature, if that is possible at some point) can be of any practical use in mechanical or electrical engineering or other fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be delighted if you could provide me with some thoughts about hypothetical or even practical (at low temperatures) realizations of machines that use superfluids and might have a huge impact.&lt;/p&gt;&#xA;" OwnerUserId="2924" LastEditorUserId="33" LastEditDate="2015-09-10T23:03:12.000" LastActivityDate="2015-09-10T23:03:12.000" Title="Practical uses of superfluids" Tags="|mechanical-engineering|fluid-mechanics|friction|" AnswerCount="1" CommentCount="6" ClosedDate="2015-09-11T15:41:27.553" ContentLicense="CC BY-SA 3.0" />
  <row Id="5315" PostTypeId="2" ParentId="5301" CreationDate="2015-09-05T18:39:31.180" Score="5" Body="&lt;p&gt;In the datasheets, this type of memory card socket is called &quot;&lt;em&gt;Push in, Push out&lt;/em&gt;&quot;, or simply &quot;&lt;em&gt;Push-Push&lt;/em&gt;&quot;.  In contrast, the memory card socket that uses friction is called &quot;&lt;em&gt;Push in, Pull out&lt;/em&gt;&quot;.  Keep in mind that these datasheets are geared towards electrical engineers and industrial designers, who are not necessarily interested in the workings of the moving parts inside of the socket itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.digikey.com/product-search/en/connectors-interconnects/memory-connectors-pc-cards/1442985&quot;&gt;Here you can find datasheets&lt;/a&gt; for all kinds of memory card sockets (SIM, SD, microSD).&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2015-09-05T18:39:31.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5316" PostTypeId="2" ParentId="5311" CreationDate="2015-09-05T19:09:50.210" Score="0" Body="&lt;p&gt;Yes, you're right.   As you imply, in a long, slender beam the bending stiffnesses will be orders of magnitude less than the axial stiffness and mixing numbers that are orders of magnitude different in any numerical calculation can cause problems.&lt;/p&gt;&#xA;" OwnerUserId="2520" LastActivityDate="2015-09-05T19:09:50.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5317" PostTypeId="1" CreationDate="2015-09-05T19:19:03.143" Score="5" ViewCount="12639" Body="&lt;p&gt;In an 12v outdoor lighting circuit, is our electrical consumption (our electrical bill) determined by the 50W 12V bulb that is connected to the transformer, or the 300W rated output of the transformer?  I have equally received two answers; i.e.....(1) our electrical usage is based upon the 50W bulb, (2) our usage is determined by the transformer making available 300W, regardless how much of the 300W we use.  &lt;/p&gt;&#xA;" OwnerUserId="2925" LastActivityDate="2017-09-18T05:57:00.330" Title="12V outdoor lighting power consumption" Tags="|power-electronics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5318" PostTypeId="2" ParentId="5299" CreationDate="2015-09-05T19:28:04.413" Score="0" Body="&lt;p&gt;The first place I'd look, certainly for your flat plate design, is Roark's &lt;em&gt;Formulas for Stress and Strain&lt;/em&gt;.   This is a compendium of formulas build up over many years covering almost every structural element you'd ever want to design.   I've used it many times for commercial structural projects.   It includes formulas for the deflection of rectangular plates with various types of support, on any number of edges, and with various types of loading.   Making 'stiffest case' assumptions about the dimensions and the supports will let you calculate the 'stiffest case deflection'.   If that's too big, you'll have to rethink the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper copies of Roark aren't cheap so I'd try the library or a friendly structural engineer, who'll almost certainly know where to find one.   You might even find it on-line.   For a hobby project, Roark will be more than adequate.   When it runs out of steam, it's usually time to consider computational analysis which, unless you've the time and inclination to learn or money to pay for, is a substantial overhead.&lt;/p&gt;&#xA;" OwnerUserId="2520" LastEditorUserId="2520" LastEditDate="2015-09-06T10:41:58.883" LastActivityDate="2015-09-06T10:41:58.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5319" PostTypeId="2" ParentId="5317" CreationDate="2015-09-05T20:40:19.170" Score="4" Body="&lt;p&gt;The answer is #1, with a bit of an asterisk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the bulb pulls 50 watts as it's rated, the current (power divided by voltage) will be 4.17 amps. We &lt;a href=&quot;https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws&quot; rel=&quot;nofollow&quot;&gt;know&lt;/a&gt; that anywhere in a closed circuit, the current is the same, so the power supply is outputting 4.17 amps as well (assuming it is attached to only one light bulb.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if your power supply is outputting 12 volts (which it is because it's designed to regulate its output voltage) and it's only outputting 4.17 amps (which it is because of the resistance of the light bulb) we can calculate the power it is outputting as 4.17 amps times 12 volts and we end up back at 50 watts. The 300 watt rating on the power supply just means that that's what it can output, not that it always outputs that. If you attached 5 more light bulbs, all in parallel, then it would produce the whole 300 watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only trick is that we have checked how much power the power supply is outputting. Your electric bill is based on how much power you consume at line voltage which is the input to the power supply. Power supplies are not perfectly efficient (some energy gets lost as heat,) so your consumption will be a bit higher, but hopefully not a whole 300 watts. If we assume, conservatively, that your power supply is 50% efficient, you'd still only be drawing 100W at the input side, not 300. Also of note is the fact that if your power supply were outputting a whole 300 watts, the input would be even more than 300W.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-09-05T20:40:19.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5321" PostTypeId="2" ParentId="5311" CreationDate="2015-09-06T01:40:08.093" Score="2" Body="&lt;p&gt;You have the right idea, but it doesn't really make sense to say &quot;the ratio of bending to axial stiffness depends on $L^2$&quot; because $L$ isn't dimensionless. If you measured the length of same beam in mm and km you would reach two very different conclusions about the ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better version of the argument is to say that the axial stiffness is of order  to $A/L$ and the bending stiffness $I/L^3$. For a typical beam cross section, $I$ is of order $At^2$ where $t$ is the &quot;thickness&quot; or &quot;diameter&quot; of the beam. So the ratio of axial to bending stiffness is of the order of $t^2/L^2$ (or $A/L^2$ if you prefer that) which is a dimensionless quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some special situations this doesn't matter at all. For example, if the axis of the beam is aligned with one of the coordinate axes of the model, the axial and bending stiffness terms are not cross-coupled in any way, and there is nothing to degrade the answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, even with FE software that uses a simple-minded equation solver working in 64-bit arithmetic (double precision, 16 significant figures), there shouldn't be any serious problems with a ratio of the order of $10^6$, but if you exceed $10^8$ you are starting to live dangerously and at $10^{10}$, all bets are off unless you have some solid evidence that the results &lt;em&gt;are&lt;/em&gt; meaningful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have structural analysis FE code that works entirely in 32-bit arithmetic, my advice would be &quot;don't use it for anything except toy problems&quot; - but that advice may be over-pessimistic for some &lt;em&gt;non-structural&lt;/em&gt; applications of FE.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-09-06T01:40:08.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5322" PostTypeId="1" AcceptedAnswerId="5339" CreationDate="2015-09-06T09:01:27.770" Score="1" ViewCount="58" Body="&lt;p&gt;When a rectangular piece of plastic (say a credit card) experiences an increasing compression force, the plastic eventually buckles. Is it possible to find the equation of the plastic's cross section once it has buckled elastically? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have in mind is two fingers squeezing a plastic card. The fingers are $x$ cm apart, and the card itself has dimensions $l\times b$ cm and negligible thickness. Suppose $x \leq l$ (so that the card is compressed in a direction parallel to its original length $l$. Can we find an equation for the card's deflection $y$ in terms of $x,h,b$ and (I guess) the elastic modulus? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1YQ4v.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1YQ4v.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the &lt;a href=&quot;https://www.google.com.au/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0CB4QFjAAahUKEwiMoMTj4ePHAhXEN6YKHd5kAuQ&amp;amp;url=http%3A%2F%2Fweb.aeromech.usyd.edu.au%2FAMME2301%2FDocuments%2FChapter09.pdf&amp;amp;usg=AFQjCNE9E-ZyR6-WqQ7N_CkzOeYt6hDNJw&amp;amp;sig2=aLYWveV2N_tMTB0Fyl9ulQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler column buckling result&lt;/a&gt; work here? That is, can we expect a solution of the form $y=A\sin(\pi z/x)$? If yes, for small deflections is it safe to assume that the arc length of the curve is equal to $l$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm nervous about using this, because I'm assuming negligible thickness. Do I have to allow the card to have non-zero thickness? Can I then use this to find the value of $A$? &lt;/p&gt;&#xA;" OwnerUserId="2928" LastEditorUserId="2928" LastEditDate="2015-09-07T01:40:28.363" LastActivityDate="2015-09-07T14:23:31.987" Title="The Equation of a buckled card" Tags="|statics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5323" PostTypeId="1" CreationDate="2015-09-06T12:12:54.600" Score="6" ViewCount="309" Body="&lt;p&gt;Suppose that the system &#xA;$$x'(t)=Ax(t)+Bu(t)$$&#xA;is controllable in $\mathbb{R}^n$, where $A$ is $n \times n$, $B$ is $ n \times m$ and $u(t)$ is $m \times 1$&#xA;Show that the system&#xA;$$\left \{ \begin{array}{rclccc} x'(t) &amp;amp;=&amp;amp; Ax(t) &amp;amp;+&amp;amp; By(t)&amp;amp; \\ y'(t)&amp;amp;=&amp;amp;&amp;amp;&amp;amp;&amp;amp;u(t) \end{array} \right.$$&#xA;with the additional states $y\in\mathbb{R}^m$ and input $u\in\mathbb{R}^m$ is controllable in $\mathbb{R}^{n+m}$.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;My attempt&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If the first system is controllable, then the matrix&#xA;$$M = \begin{pmatrix} B &amp;amp; AB &amp;amp; A^2B &amp;amp;\ldots &amp;amp; A^{n-1}B \end{pmatrix}$$&#xA;has rank M. Let the second system&#xA;$$\begin{pmatrix}x' \\ y' \end{pmatrix} = \begin{pmatrix} A &amp;amp; B \\ 0 &amp;amp; 0\end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix}+\begin{pmatrix}0 \\ I \end{pmatrix}u(t)$$&#xA;Then, its Kalman matrix (aka controllability matrix) is&#xA;$$\bar{M} = \begin{pmatrix} B &amp;amp; AB &amp;amp; A^2B &amp;amp; \ldots &amp;amp; A^{n+m-1}B \\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \ldots &amp;amp; 0\end{pmatrix}$$&#xA;The matrix $A$ is $n \times n$ and, by Cayley-Hamilton theorem, &#xA;$$A^{n+l} = \sum_{k=0}^{n-1} \alpha_{k,l} A^{k}, l=0,1,...$$&#xA;I mean, $A^{n+l}$ is a linear combination of the first $n$ powers of $A$. This implies that we have $n$ linearly independent columns in $\bar{M}$ at the the terms until $A^{n-1}B$ (because the first system is controllable). But I don't see how to find the other $m$ columns. What am I doing wrong?&#xA;Probably my controlability matrix is wrong, since it only can have $n$ l.i rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could solve it for $y$ first and return to a system only in $x'$, substituting $u$ by some $\int u$. The Kalman matrix of first and second system would be the same and, by hypothesis, it would have $\text{rank }n$, correct? But why isn't my first approach correct? &lt;/p&gt;&#xA;" OwnerUserId="2929" LastEditorUserId="6790" LastEditDate="2020-05-04T11:28:40.340" LastActivityDate="2020-05-04T11:28:40.340" Title="Controllability of $x' = Ax + Bu(t)$ implies controllability of $\left \{ \begin{matrix} x' = Ax + By \\ y'=u(t) \end{matrix} \right.$" Tags="|control-theory|linear-control|linear-systems|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5324" PostTypeId="2" ParentId="5323" CreationDate="2015-09-06T13:14:59.773" Score="1" Body="&lt;p&gt;Couldn't you just define $y(t)$ and be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y'(t)=u(t) \\&#xA;y(t) = \int{u(t)}dt \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the system is controllable, as defined by the controllability matrix $M$ being full rank, it doesn't matter what the input signal is, does it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only issue I can think of is maybe if $u(t)$ is not able to be integrated, but that seems beyond the scope of the question. $M$ is defined only by $A$ and $B$, so as long as you can define $y(t)$ I don't think it matters what it is. At least, that would be my approach/defense. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-09-06T13:14:59.773" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5325" PostTypeId="1" AcceptedAnswerId="5326" CreationDate="2015-09-06T13:50:21.160" Score="17" ViewCount="6900" Body="&lt;p&gt;Something hit me during my last flight: our ears feel that the pressure around us changes quickly during liftoff and landing, they hurt more and more until we make them pop. However, the cabin must maintain a level of oxygen higher than outside the aircraft because it isn't dense enough at ~10km to breathe normally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean that the aircraft merely takes air from the outside, heats it up (it's  around -40°C outside at 10km) and adds oxygen to it before blowing it inside? How else? &lt;/p&gt;&#xA;" OwnerUserId="1326" LastEditorUserId="399" LastEditDate="2015-09-07T14:46:33.007" LastActivityDate="2015-09-07T14:46:33.007" Title="If aircraft are pressurised, why do our ears pop during liftoff and landing?" Tags="|pressure|aircraft-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5326" PostTypeId="2" ParentId="5325" CreationDate="2015-09-06T14:13:14.517" Score="19" Body="&lt;p&gt;Because the cabin isn't pressurized to sea level pressure instead it to about 8k ft equivalent. (while the plane is 4.5 times higher)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means there is less differential pressure than if the cabin was pressurized to sea level pressure. But it's still within the limit of what passengers feel comfortable with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://aviation.stackexchange.com/a/15331/609&quot;&gt;This answer&lt;/a&gt; on aviation.SE contains a plot of cabin pressure over time during a flight:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B94bv.jpg&quot; alt=&quot;Cabin pressure over time&quot;&gt;&#xA;&lt;br&gt;&lt;sup&gt;&lt;a href=&quot;http://www.meteormetrics.com/flightpressure.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The composition of air doesn't change with altitude just the pressure. This means that you only need to compress the air before blowing it in. &lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="-1" LastEditDate="2017-04-13T13:00:01.440" LastActivityDate="2015-09-06T14:13:14.517" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5327" PostTypeId="2" ParentId="5325" CreationDate="2015-09-06T14:18:55.113" Score="4" Body="&lt;p&gt;There's a difference between being pressurized to an absolute value and being pressurized to a relative value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if the cabin immediately pressurized to the same atmospheric pressure as your departure terminal, then your ears would not pop on ascent, but they would have to pop at some point on the descent unless the elevation of the landing terminal were the same as the departure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would also mean that the aircraft structure would have to be thick enough to withstand the full &lt;em&gt;differential&lt;/em&gt; pressure between up to sea level and cruising altitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead the cabin only pressurized to, say, 7psi (about 50kPa?) above whatever the outside pressure is, then there's no need for an elaborate destination pressure matching scheme and the aircraft walls can be made thinner, thus lighter and more fuel efficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative pressurization means the absolute pressure changes with altitude, so your ears pop.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-09-06T14:18:55.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5328" PostTypeId="2" ParentId="5299" CreationDate="2015-09-06T18:28:22.040" Score="1" Body="&lt;p&gt;To add to Wasabi's answer - especially dealing with wood's anisotropy, here are a few things to consider:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) IF wood is end grain, plywood, oriented strand board, etc then for all design purposes, it is isotropic - as the grain is going in the &quot;up&quot; direction - where no real bending occurs.  As a result, you can use the simpler equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Under most conditions, as shown by &lt;a href=&quot;http://link.springer.com/book/10.1007%2F0-306-48414-5&quot; rel=&quot;nofollow&quot;&gt;Dr. Jack Vinson&lt;/a&gt;, where your longest length is in the primary grain direction (that are difficult to go into without a lot of partial differential equations), and a non-linear term that &lt;em&gt;for the most part&lt;/em&gt; can be ignored, a shelf can be &quot;squished&quot; in the grain direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's say that your shelf is &quot;A&quot; wide and &quot;B&quot; long.  And the wood has &quot;a&quot; modulus in the &quot;A&quot; direction, and &quot;b&quot; modulus in the &quot;B&quot; direction.  If &quot;B&quot; is the grain direction, simply change the dimensions in the deflection calculator to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A&quot; wide by (a/b)*&quot;B&quot;, with modulus &quot;a&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculated deflection, for woodworking purposes, will be within an order of magnitude of the actual deflection.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-09-06T18:28:22.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5329" PostTypeId="2" ParentId="5317" CreationDate="2015-09-06T20:56:29.467" Score="4" Body="&lt;p&gt;A perfect ideal transformer converts some combination of Volts and Amps going into it, to some other combination of Volts and Amps coming out but a the same power.  Put another ways (Volts in)(Amps in) = (Volts out)(Amps out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By this logic, it is the power used by the bulb that matters.  Just because the transformer &lt;i&gt;could have&lt;/i&gt; supplied 300 W (and then drawn 300 W from the power line) is irrelevant.  All that means is that you could connect a brighter bulb or more bulbs that add up to 300 W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, back in reality, transformers aren't perfect.  OutPower = InPower x efficiency, where efficiency will always be somewhat less than 1.  As you might imagine, higher efficiency requires tricks that add to the cost of the transformer, so some compromise was inevitably made.  The efficiency might be 90%, for example, or maybe just 80%, or even lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to get some idea of this is to feel the transformer after it's been on a while.  If it's barely warmer than when off, its efficiency is good.  If it's considerably warmer than when off, then clearly it's dissipating power that is not getting delivered to the light.  Both the power to drive the light (50 W) and to heat the transformer are delivered by the power line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the light is off, the transformer is really just a inductor connected to the power line since its secondary is open-circuit and therefore has no current thru it and no relevance to the magnetic field in the transformer.  There will be current in the primary, which is ideally 90&amp;deg; out of phase with the voltage, so no net power is delivered to the transformer and you are not charged for this &quot;reactive power&quot;, as apposed to &quot;real power&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the wires the primary coil is made from don't have perfect 0 resistance.  Electrically, this looks like a resistor in series with the inductor.  The inductor still doesn't dissipate power, but the resistor does, and you do get billed for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So overall, the power drawn from the line by the transformer is &lt;i&gt;largely&lt;/i&gt; a function of how much power is drawn from its secondary, but there is always some loss which is made up for by drawing more power, and even with the light off there will be some power drawn.  The higher the transformer rating, the higher this no-load power will likely be.  Up to around 10% of the full power rating wouldn't be much of a surprise for such a transformer that was likely optimized for cost much more than efficiency or no-load power.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-09-06T20:56:29.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5330" PostTypeId="1" CreationDate="2015-09-07T00:16:01.063" Score="4" ViewCount="3085" Body="&lt;p&gt;I have two results that don't seem to match up and I can't figure out why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I built a hollow square section (HSS) using two different structural analysis softwares (they both agreed). However, they don't seem to agree with the AISC steel table despite the dimensions being identical. It seems to be the Torsion Constant (J) which is slightly off in the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could be the possible reason for this as I am totally stumped?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BPJgH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BPJgH.png&quot; alt=&quot;AISC Steel Table&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kWAjk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kWAjk.png&quot; alt=&quot;SkyCiv section builder result&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2936" LastEditorUserId="2936" LastEditDate="2015-09-07T03:01:57.063" LastActivityDate="2015-09-15T18:19:11.930" Title="What could be the reason for these sections not agreeing?" Tags="|structural-analysis|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5331" PostTypeId="1" CreationDate="2015-09-07T00:50:12.890" Score="0" ViewCount="322" Body="&lt;p&gt;I'm curious why IBM/Lenovo ThinkPad line is basically the only manufacturer/brand of mobility devices that includes circuitry and software that allows one to vastly prolong battery life of laptops used as stationary workstations by making sure to never charge the battery prior to it being discharged to below 96%, and having options to start charging only when the battery goes, say, below 46%, and stop charging at, say, 50% or 80%, vastly improving long-term battery health (compared to when the battery is always kept at 100%, and always topped up to 100% all the time, for months at a time, and somehow never happens to reach its advertised cycle count).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why the other laptops and mobile phones never seem to have any such functionality?  Is the circuitry behind such feature somehow intrinsically expensive and not worth it?  (Or is it just the case that stationary usage is not accounted for in the design of such other devices?)&lt;/p&gt;&#xA;" OwnerUserId="1451" LastEditorUserId="1586" LastEditDate="2015-09-09T00:40:01.243" LastActivityDate="2015-09-10T01:45:30.043" Title="Is battery-charge-stop-at-XX-percent circuitry not cheap?" Tags="|electrical-engineering|battery|circuits|cost-engineering|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ClosedDate="2015-09-08T14:00:08.393" ContentLicense="CC BY-SA 3.0" />
  <row Id="5332" PostTypeId="2" ParentId="5331" CreationDate="2015-09-07T01:20:56.577" Score="1" Body="&lt;p&gt;“Vastly prolong battery life” is a subjective statement.  A little bit internet research reveals (&lt;a href=&quot;http://www.laptopmag.com/articles/all-day-strong-longest-lasting-notebooks&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laptops with the Longest Battery Life&lt;/a&gt;) that a new IBM/Lenovo Thinkpad has an estimated battery life of 15+ hours while a new Toshiba Protégé has 15- hours of battery life. Based on the above link it is safe to assume that IBM/Lenovo Thinkpad have a superior battery management technology. If you dig deep into the feature function sets of these laptops you will find that each manufacturer offers a different value proposition. Manufacturers engineer devices in line with the value propositions. There could  many reasons behind other vendors not offering the same feature.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another factor could be intellectual property implications.  With the growth in mobile devices, power management and battery technology vendors patent innovations to achieve differentiation from their competition. Therefore investigating recent patent applications and filings is another avenue to explore if IBM/Lenovo or any of IBM/Lenovo business partners have patents that allow them a strategic advantage of the competition.  There a few reference that might assist in the investigation process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a battery power management technology aspect the cost of power management devices are extremely inexpensive. A quick price check of BQ240X battery charging IC from TI is listed at $0.85/1ku. There are other great power management IC vendors such as microchip and Fairchild to name a few. It is safe to assume that price might not be a compiling factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most obviously the fuel gauges the most common battery power management software option used by majority of the mobile device users. Looking deeper into a laptop you will find many advance battery management options. Below are the options that are available on my laptop. The reality is that consumers hardly use these option. Therefore, it is difficult to justify investing in engineering a feature function set that consumers don’t typically use, unless there is research to state otherwise. This should essentially answer the second part of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7lXjA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7lXjA.jpg&quot; alt=&quot;Windows 7 Battery Management&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary there are few enterprise mobile application where the battery charging feature function set are implemented as discussed in the question. So to answer your question, such feature function sets are engineered into devices that meet specific value propositions. Below is device that have the feature function set that are referenced in the question. The battery charger shut off when the battery is fully charged. Similarly the mobile device managers the battery charger per the specification. Both the battery charger, mobile device and battery has smarts to manage the charging. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H04fF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H04fF.jpg&quot; alt=&quot;Smart Charging Device&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vendors encounter many non-technical maneuvers such as intellectual property, business contracts, partnerships, regulations many other in implement battery technologies. It is highly like the reason is non-technical.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly it is worthwhile investigating different type of battery chemistry. Lithium Ion (LiIon) is the most popular and widely used in mobile devices, but Lithium Polymer (LiPo) is fast gaining recognition as a viable alternative. Battery vendors differentiate themselves by offering different value propositions to address battery life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some links to help get started with further research&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lsds/ti/power-management/battery-management-products-overview.page&quot; rel=&quot;nofollow noreferrer&quot;&gt;Power Management IC from Texas Instruments&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.google.com/patents/US20050200332&quot; rel=&quot;nofollow noreferrer&quot;&gt;Intelligent multiple battery charging station&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.freepatentsonline.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Free patent online&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://batteryuniversity.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battery University&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.microchip.com/stellent/groups/designcenter_sg/documents/market_communication/en027883.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Developing Affordable Mixed-Signal Power Systems for Battery Charger Applications&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.maccor.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battery and cell test equipment&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-09-10T01:45:30.043" LastActivityDate="2015-09-10T01:45:30.043" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5333" PostTypeId="1" CreationDate="2015-09-07T06:21:37.403" Score="1" ViewCount="45" Body="&lt;p&gt;I’m performing a sensitivity analysis for a model of a set of elementary chemical reactions. I’m considering the variation in species concentration ($c_{k}$) with respect to the forward reaction rate constants ($k_{f}$). Let’s consider for the moment that there is only one reaction (for simplicity). So, I have to consider the following sensitivity coefficient, which must be thermodynamic consistent (equilibrium constant $k_{eq}$ held constant) in order to define system properties (rate limiting step, etc):&#xA;$$\left ( \frac{\partial c_{k}}{\partial k_{f}} \right )_{k_{eq}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The species concentration follows a governing equation in time “t” of the form:&#xA;$$\frac{dc_{k}}{dt}=f\left ( c_{k}\left ( t,k_{f},k_{eq} \right ),k_{f},k_{eq} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can derive a differential equation for the sensitivity coefficients by taking partial derivatives with respect to the reaction constant $k_{f}$ (at $k_{eq}$ constant) on both sides of the above equation. This differential equation will have a jacobian term, which is no problem to calculate, and a term of the form:&#xA;$$\left ( \frac{\partial f}{\partial k_{f}} \right )_{k_{eq}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can calculate the above term by forward finite difference evaluating the reaction constant at an incremental value $k_{f}^{1}=k_{f}+\Delta k_{f}$, at constant $k_{eq}$. Here stems my concern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction constant actually depends on other basic parameters, pre exponential factor ($A_{f}$) and heat of adsorption of the participating species ($Q_{k}$), $k_{f}=k_{f}\left ( A_{f},Q_{k} \right )$. Both parameters have a defined range and since the derivative is taken at $k_{eq}$ held constant, both parameters set will vary. I’d like to verify that $k_{f}^{1}$ can actually be obtained from the range values of $A_{f}$ and $Q_{k}$. One way to do this is actually calculate the values $A_{f}$ and $Q_{k}$ that yield $k_{f}^{1}$, but this is too expensive because the numbers of parameters is too high and there is no need to calculate these new parameters. Is there any strategy to verify that a given $k_{f}^{1}$ belongs to the range specified for its basic parameters $A_{f}$ and $Q_{k}$?&lt;/p&gt;&#xA;" OwnerUserId="2919" LastActivityDate="2015-09-07T06:21:37.403" Title="Verify parameters variation in local sensitivity coefficients for microkinetic model" Tags="|thermodynamics|chemical-engineering|modeling|simulation|numerical-methods|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5334" PostTypeId="1" AcceptedAnswerId="5337" CreationDate="2015-09-07T07:13:54.313" Score="7" ViewCount="3415" Body="&lt;p&gt;I often see specifications for refurbishing works on gas pipelines calling for blanking spades, to separate the area worked on from the rest of the gas system. This is in addition to closing valves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling showed me some pictures of how blanking spades look like, but I don't yet understand the process of installing them. The most straightforward way would be to mount them between flanges, after a section of pipe has been valved off and flushed. How are blanking spades actually used?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorDisplayName="user16" LastEditDate="2015-09-07T15:28:37.910" LastActivityDate="2015-09-07T15:28:37.910" Title="How are blanking spades entered into gas pipelines?" Tags="|pipelines|safety|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5336" PostTypeId="2" ParentId="5331" CreationDate="2015-09-07T10:40:54.747" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;IBM/Lenovo ThinkPad line is basically the only manufacturer/brand of mobility devices that includes circuitry and software that allows one to vastly prolong battery life of laptops used as stationary workstations by making sure to never charge the battery prior to it being discharged to below 96%, and having options to start charging only when the battery goes, say, below 46%, and stop charging at, say, 50% or 80%, vastly improving long-term battery health.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Why is this apparently useful feature very uncommon? - Is the circuitry behind such feature somehow intrinsically expensive and not worth it? (Or is it just the case that stationary usage is not accounted for in the design of such other devices?)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no technical problem with doing this, and there are advantages in allowing partial state of charge, although not necessarily always as great as you may imagine. There are a number of factors that affect battery cycle life and manufacturers could choose other ones than you suggest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_{min}$ is already controlled by the control circuitry to a level that manufacturers consider safe and this probably varies somewhat between manufacturers. About 3V/cell is usual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know if patents cover any related aspects but this seems unlikely to have a blanket effect as partial charging in general has general prior art and LiIon technology is now old enough that early patents will have expired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major gain can be obtained by simply limiting $V_{max}$ from the usual 4.2V to 4.1V or 4.0V, with a significant gain in cycle life and loss in capacity per cycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A significant gain in cycle life can be had by increasing the current at which charging terminates. Usually LiIon / LiPo charge at 1C (mA = mAh) until Vmax is reached then hold voltage at $V_{max}$ and allow $I_{charge}$ to decrease under control of the battery chemistry. If $I_{charge}$ is allowed to decrease to about C/10 = 10% of max rate then the battery attains full capacity but the cell is pushed very near to its upper limit (close to the point where metallic Lithium may &quot;plate out&quot;) and the cycle life is decreased. By stopping charge at C/4 or even C/2 capacity is somewhat reduced but cycle life is increased. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Users can achieve their own improvements in cycle life in everyday use at the expense of available capacity.  When charging from &quot;empty&quot; at C/1 a LiIon battery will charge at CC (constant current) until $V_{max}$ is reached - usually at around 70% + of full capacity. If charging is terminated at this point (nop CV &quot;tail) then cycle life is increased significantly - maybe 2x to 3x. If charging from empty at C/1 this point is reached at about 0.7 hours = 42 minutes. So a say 40- minute charge from empty about reaches this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally reducing battery capacity per charge gives diminishing returns. eg if you use half capacity per charge cycle life would need to at least double before you got the same whole-of-lifetime capacity. ie N x 100% = 2N x 50%. In practice you may get say 3N cycles for C/2 capacity so a 50% plus gain in gross capacity over lifetime. &#xA;Reducing charge to 1/3 capacity may give 6 x as many cycles.&lt;br&gt;&#xA;But reducing charge to say 10% of capacity may give not too much more than 30 x as many cycles so still a not much more than 3x gain, if that.&lt;br&gt;&#xA;Those figures are 'out of my head' but I believe they are in of the generally correct order and will be enough affected by other factors that they can only be a guide.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium Ion batteries that must have extremely long cycle lives may be charged to about 2.8V/cell. This greatly reduces cell capacity (maybe to 30% of normal max) but allows maybe 8000 cycles. The Mars Opportunity Rover battery is run at this sort of level plus &quot;all the magic they can work&quot; to achieve this sort of result. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at www.batteryuniversity.com for more related information.&lt;br&gt;&#xA;These pages are relevant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.batteryuniversity.com/learn/article/charging_lithium_ion_batteries&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;BU-409: Charging Lithium-ion&lt;/strong&gt;&lt;/a&gt;&lt;br&gt;&#xA; &quot;Find out how to prolong battery life by using correct charge methods.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.batteryuniversity.com/learn/article/how_to_prolong_lithium_based_batteries&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;BU-808: How to Prolong Lithium-based Batteries&lt;/strong&gt;&lt;/a&gt;&lt;br&gt;&#xA;&quot;Discover what causes Li-ion to age and what the battery user can do to prolong its life&quot;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="1586" LastEditDate="2015-09-09T12:04:24.533" LastActivityDate="2015-09-09T12:04:24.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5337" PostTypeId="2" ParentId="5334" CreationDate="2015-09-07T13:52:05.587" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The most straightforward way would be to mount them between flanges, after a section of pipe has been valved off and flushed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is correct. Blanking spades, also known as slip plates, slip blinds or spectacle blinds, are usually slipped in between two flanges after a process line has been double blocked and bled. They are usually secured in place by the flange bolts and sealed by a gasket or O-ring, similar to how wafer pattern valves are installed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes dedicated 'blind flanges' are provided in the pipeline for this purpose, but often the inlet or outlet flanges on valves or equipment are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blanking plates are used to provide an additional layer of protection during maintenance work, guarding against leaking valves or negligent opening of an isolation valve. Other important considerations when using them are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;That they are installed in a sensible place, usually as close to the isolated equipment as feasible.&lt;/li&gt;&#xA;&lt;li&gt;That they are rated to withstand the upstream pressure.&lt;/li&gt;&#xA;&lt;li&gt;That they are removed before the pipeline is put back into service.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2551" LastActivityDate="2015-09-07T13:52:05.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5338" PostTypeId="2" ParentId="5334" CreationDate="2015-09-07T13:54:21.183" Score="1" Body="&lt;p&gt;You are correct, blanks are installed at a pipe flange where two runs connect. Typically you break the joint and pry the flange apart to install the blank, but this can cause problems if/when the prying action torques the next joint in the piping system and damages the gasket. I've heard of a person replacing every gasket in a large system because every time a joint was pried apart it ruined the adjacent gasket. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-09-07T13:54:21.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5339" PostTypeId="2" ParentId="5322" CreationDate="2015-09-07T14:23:31.987" Score="2" Body="&lt;p&gt;The axial compression of the member should be negligible compared to the sideways deflection. This means that it doesn't depend on the cross section of the card. This is &lt;em&gt;exactly&lt;/em&gt; Euler buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations for buckling are usually used to determine the load at which buckling will occur. This is where the cross section matters. Once buckling happens, the shape is only defined by the deflection. The length along the member doesn't change (appreciably).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this changes once enough force is applied to move into plastic behavior and hinges form.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-09-07T14:23:31.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5340" PostTypeId="1" CreationDate="2015-09-07T15:45:01.787" Score="8" ViewCount="7354" Body="&lt;p&gt;I have been wondering if there are &lt;strong&gt;certain media&lt;/strong&gt; (for instance glass, air, foam, etc.) &lt;strong&gt;which are peculiar given their chemical composition and/or structure allowing them to filter out mechanical waves&lt;/strong&gt; (examples include: sound, vibrations) of high or low frequencies. Do these exist? Could I please have some examples?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The less abstract example: I tend to notice the bass of the music booming over from the car next to me when I am stopped at a red light. I do not know if this is due simply to the amplitude of those low frequency components or whether there is mechanical filtering at play (suppressing the higher frequencies). This is the source of inspiration for asking my question.&lt;/p&gt;&#xA;" OwnerUserId="2945" LastEditorUserId="368" LastEditDate="2015-11-20T01:02:04.090" LastActivityDate="2015-11-20T01:02:04.090" Title="Is there a mechanical equivalent to a hi/low pass filter (as in electrical engineering)?" Tags="|mechanical-engineering|materials|vibration|audio-engineering|electronic-filters|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5341" PostTypeId="2" ParentId="5340" CreationDate="2015-09-07T16:24:22.013" Score="8" Body="&lt;p&gt;Yes, for sound specifically there is a term called &quot;acoustic impedance&quot; that, just like electrical impedance, is frequency dependent. Acoustic impedance results from the acoustic wave equation, which takes the same form as the electromagnetic wave equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any solid has an acoustic impedance just like every conductor has an electric impedance. Acoustic wave transmission depends on angle of incident and the frequency content, with the material impedance attenuating certain frequencies. Ducts act like waveguides, you can make acoustic resonators, etc. - there are a ton of parallels because the underlying equations are almost identical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like electrical impedance matching results in maximum electrical power transfer, acoustic impedance matching results in maximum acoustic power transfer. That's what ultra sound jelly is used for - impedance matching between the wand and skin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be used in reverse - intentional impedance mismatching results in reflected waves and very little acoustic power transmission. There is actually a line of &lt;a href=&quot;http://m.lowes.com/pd/QuietRock-Drywall-Panel/3069747&quot; rel=&quot;noreferrer&quot;&gt;drywall designed to exploit this&lt;/a&gt; to prevent/minimize the &quot;noisy neighbor.&quot;&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2015-09-08T11:36:45.317" LastActivityDate="2015-09-08T11:36:45.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5342" PostTypeId="2" ParentId="5340" CreationDate="2015-09-08T01:50:43.413" Score="9" Body="&lt;p&gt;In a more general sense, the electrical concepts of inductance, capacitance and resistance are equivalent to mass, spring constant and friction in the mechanical world. Voltage becomes force and current becomes velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the suspension in a vehicle is a carefully-tuned low-pass filter that uses the mass of the frame and body, the springs on the axles and the shock absorbers to &quot;block&quot; the high-frequency vibrations from the road from being coupled to the passengers and cargo inside.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-09-08T01:50:43.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5343" PostTypeId="2" ParentId="5330" CreationDate="2015-09-08T02:34:31.967" Score="3" Body="&lt;p&gt;Simple formulas for torsion constants of this type of hollow section are all approximate. The AISC tables may be using a different formula from the calculator and making some allowance for the radii at the corners of the section. Or, the AISC values might be derived from a detailed three-dimensional finite element analysis, which will converge to the &quot;correct&quot; torsion constant value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &quot;serious&quot; work, use the values from the data sheets, even though for your section the difference is too small to be important.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="2246" LastEditDate="2015-09-08T11:52:51.567" LastActivityDate="2015-09-08T11:52:51.567" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5344" PostTypeId="1" AcceptedAnswerId="28600" CreationDate="2015-09-08T03:39:59.077" Score="3" ViewCount="2642" Body="&lt;p&gt;I'm currently working on problems for my mechanics of material course, and I just learned how to calculate the shear stress on a bolt holding stacked plates. That whether it be a single or double shear, I have to account for every shear plane. However, what happens for the bearing stress? I know that it depends on the area of contact, which is diameter * thickness of the plate. If there were three plates (double shear) and I looked at only the portion of the bolt in each plate, would the bearing stress not differ? Do I have to look at the entire bolt at once to calculate the bearing stress?&lt;/p&gt;&#xA;" OwnerUserId="2804" LastActivityDate="2019-05-30T21:21:33.423" Title="Does the bearing stress vary along the bolt?" Tags="|stresses|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5345" PostTypeId="1" AcceptedAnswerId="5359" CreationDate="2015-09-08T06:48:31.593" Score="2" ViewCount="182" Body="&lt;p&gt;I have a system with a motor, some masses, damping and springs. One of the masses is loaded with an unknown external force F. I want to estimate the force F using data of the motor (mainly current). There are some unknown disturbances, like temperature and friction in the system. Can this be done or are there better ways to estimate the unknown force? There will be no feedback loop. The available data will look be a time series of the current acting on the motor.&lt;/p&gt;&#xA;" OwnerUserId="2948" LastEditorUserId="6790" LastEditDate="2020-05-04T08:02:51.813" LastActivityDate="2020-05-04T08:02:51.813" Title="Can a load be estimated using a Kalman filter?" Tags="|mechanical-engineering|control-engineering|sensors|kalman-filters|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5346" PostTypeId="1" CreationDate="2015-09-08T08:23:22.530" Score="10" ViewCount="1938" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://www.archdaily.com/395363/india-s-forgotten-stepwells&quot;&gt;an article&lt;/a&gt; that was linked on another site I frequent  on &quot;India's forgotten stepwells&quot;, which are basically elaborate wells with stairs to go down to the water level, sometimes 10 m or so below surface (maybe more, looking at the pictures). Some of these are whole temples underground, very impressive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was wondering how one would go about building such a thing, or rather how they would have gone about it 1000 years ago. Would you just dig a very big, deep hole, build a structure inside, then fill the sides (can't imagine this)? Is there a way to start at the surface, then work yourself down, basically building layers &lt;em&gt;under&lt;/em&gt; existing ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, how does/did one go about a brick well? Dig a deep hole, hope it doesn't collapse on top of you, start with brick once you hit the bottom during dry season? Or is there a way to work yourself down there? And does it make a difference nowadays; I mean are there new methods now with modern materials that let you do things they couldn't do 1000, 500 or 100 years ago?&lt;/p&gt;&#xA;" OwnerUserId="2949" LastEditorDisplayName="user16" LastEditDate="2015-09-09T01:12:13.673" LastActivityDate="2019-11-06T12:39:29.320" Title="How does one go about building an underground structure without concrete?" Tags="|civil-engineering|structural-engineering|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5347" PostTypeId="2" ParentId="5346" CreationDate="2015-09-08T13:09:57.717" Score="7" Body="&lt;p&gt;&lt;strong&gt;General:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumption that concrete was not available is probably correct but not certain. The Colosseum is largely concrete and the Pantheon (still) has the largest unreinforced concrete dome in the world. &lt;a href=&quot;https://en.wikipedia.org/wiki/Concrete&quot;&gt;Wikipedia&lt;/a&gt; advises that concrete of some form was known in 6000 BC and that eg Bedouins used it to built underground cisterns in desert areas from around 700 BC. But, I could find no clear indication that it was used in India until relatively modern times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Taj Mahal was built over ~ 20 years from 1632 and Saint Paul's Cathedral in London was built a few decades later. Both are (as far as my non civil engineering eye and brain can tell - having looked at both) approximately equivalent accomplishments. I'd have expected something like cement would be a given in both cases. Maybe not.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stepwells:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A web search showed (as you also will have discovered) that there is a vast amount ABOUT the wells available but almost none of it deals with construction methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better than most, but not really good enough, is Cyark's &lt;a href=&quot;http://www.cyark.org/projects/rani-ki-vav/in-depth&quot;&gt;&lt;strong&gt;Rani ki Vav&#xA;An Ancient, Royal Stepwell&lt;/strong&gt;&lt;/a&gt; which on that page gives an overview and on associated pages gives immense detail in cross section diagrams, photos, sketches and more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cyark.org/projects/rani-ki-vav&quot;&gt;&lt;strong&gt;3D interactive 'fly through'&lt;/strong&gt;&lt;/a&gt; - impressive and fun - shows you WHAT was achieved but does not answer the 'how'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://archive.cyark.org/rani-ki-vav-gallery-all#&quot;&gt;&lt;strong&gt;All resources&lt;/strong&gt;&lt;/a&gt; on one pahge, and you can subset them using the menu at top into 3D Point Clouds,  Drawings, Panoramas, Perspectives, Photographs and Videos. Some of the detailed material requires an account but there is much there that is accessoible. Hopefully this approaches your &quot;Stepwells 101&quot; level.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What appears to be &lt;strong&gt;very close to what you want&lt;/strong&gt; is described in a course outline, but without the detail in a school student's course outline. It's possible that the lecturer MAY make the course material avaialable. Course outline &lt;a href=&quot;http://www.tryscience.org/sites/default/files/form_function_and_spirituality_at_rani_ki_vav.pdf&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt; - more questions than answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tryscience.org/lp/form-function-and-spirituality-rani-ki-vav&quot;&gt;&lt;strong&gt;Form, Function, and Spirituality at Rani ki Vav&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They say - After completing this lesson, the student will be able to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;describe well engineering and functionality with regard to several different &#xA;well typologies.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;recall principles of weight bearing and the effect of water on various materials.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;explain the history and construction of Rani ki Vav,&lt;/strong&gt; the symbology of its iconography, and the source of its water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;relate the importance of water for survival to the prominent role water plays in religions across the globe.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;compare the ceremonial aspects of carving across sites around the world, illustrating that the amount of effort placed into the creation of buildings may indicate something beyond its functionality.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;reconstruct a scaled well or stepwell out of new material and engage the scientific method in testing its strengths and weaknesses.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rani ki vav:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://whc.unesco.org/en/list/922&quot;&gt;Unesco&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://asi.nic.in/asi_monu_tktd_gujarat_ranikivav.asp&quot;&gt;Archaeological survey of India&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Various:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimal stepwell mention, but looks useful - &lt;a href=&quot;http://www.infinityfoundation.com/hooja_book.htm&quot;&gt;Infinity Foundation sponsored new book project titled: &#xA;&quot;Channeling Nature: Hydraulics, Traditional Knowledge Systems, And Water Resource Management in India – A Historical Perspective&quot;&#xA;by Rima Hooja, PhD&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ancient-origins.net/history/beautiful-utilitarian-stepwells-india-are-danger-becoming-extinct-003770&quot;&gt;Ancient origins&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stepwell&quot;&gt;Wikipedia - Stepwell&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.indiasinvitation.com/stepwells_of_gujarat/&quot;&gt;&quot;India's invitation&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.britannica.com/technology/stepwell&quot;&gt;Britannica&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.archdaily.com/395363/india-s-forgotten-stepwells&quot;&gt;Your superb reference&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lowish:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dnaindia.com/lifestyle/report-step-well-architecture-a-classic-example-of-fusion-of-designs-in-gujarat-1474398&quot;&gt;Low&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sscnet.ucla.edu/southasia/Culture/Archit/Stepw.html&quot;&gt;UCLA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.atlasobscura.com/places/ancient-stepwells-india&quot;&gt;Atlas obscura&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Somewhat related only:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cyark.org/&quot;&gt;Cyark home page&lt;/a&gt; - just sit and watch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cyark.org/projects/monastery-of-geghard&quot;&gt;Geghard, Armenia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-09-08T13:48:21.317" LastActivityDate="2015-09-08T13:48:21.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5348" PostTypeId="1" CreationDate="2015-09-08T14:36:00.557" Score="0" ViewCount="397" Body="&lt;p&gt;I have a load of almost 15 kg. &#xA;Both the pulley diameter = 0.6m &#xA;Distance between the center of both the pulley is 0.8m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loads are permanently attached to the belt.&lt;a href=&quot;https://i.stack.imgur.com/MiiB9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MiiB9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i will use a HTD belt for this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me find the torque required for the motor?&#xA;Thinking of using a stepper or servo motor to rotate it in steps of 10cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i dont have any knowledge about the frictional losses&lt;/p&gt;&#xA;" OwnerUserId="2953" LastEditorUserId="1586" LastEditDate="2015-09-09T00:18:42.473" LastActivityDate="2015-09-09T00:18:42.473" Title="Torque requirement for the motor" Tags="|mechanical-engineering|torque|" AnswerCount="1" CommentCount="3" ClosedDate="2015-09-08T16:35:30.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="5349" PostTypeId="2" ParentId="5348" CreationDate="2015-09-08T15:56:41.527" Score="3" Body="&lt;p&gt;$$&#xA;\tau_{\mbox{required}} = \mbox{acceleration torque} + \mbox{friction losses} + \mbox{slope/hill climbing opposing force} \\&#xA;\tau_{\mbox{required}} = I\alpha + \mu F_{\mbox{normal}} r_{\mbox{effective}} + mg \sin{\theta} r_{\mbox{effective}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\tau_{\mbox{required}}$ is the required torque&lt;/li&gt;&#xA;&lt;li&gt;$I$ is the load inertia&lt;/li&gt;&#xA;&lt;li&gt;$\alpha$ is the desired peak acceleration&lt;/li&gt;&#xA;&lt;li&gt;$\mu$ is the cumulative coefficient of friction covering all friction losses in the drivetrain&lt;/li&gt;&#xA;&lt;li&gt;$F_{\mbox{normal}}$ is the normal force between the vehicle and the ground&lt;/li&gt;&#xA;&lt;li&gt;$r_{\mbox{effective}}$ is the effective hub radius&lt;/li&gt;&#xA;&lt;li&gt;$m$ is the mass of the vehicle&lt;/li&gt;&#xA;&lt;li&gt;$g$ is the gravitational constant&lt;/li&gt;&#xA;&lt;li&gt;$\theta$ is the slope of the ground you are on&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you can define all of the above terms, plug them in and you have an answer for your required torque. No need to update your question. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2015-09-08T19:32:31.160" LastActivityDate="2015-09-08T19:32:31.160" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5352" PostTypeId="2" ParentId="5331" CreationDate="2015-09-06T22:40:46.470" Score="1" Body="&lt;p&gt;All laptops (and especially cell phones) do &quot;fuel gauge&quot; of the battery.  All these mobile devices have a battery charger, which can be started or stopped on command.  I don't see why that would be a fundamentally expensive feature.  Some of them have software that allows user fine-grained control, and others use only on the functionality built into the OS.&lt;/p&gt;&#xA;" OwnerUserId="63" OwnerDisplayName="Nick Alexeev" LastActivityDate="2015-09-06T22:40:46.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5353" PostTypeId="2" ParentId="5314" CreationDate="2015-09-08T20:34:37.173" Score="1" Body="&lt;p&gt;Your comment, &quot;my interest is more whether we can make machines waste less energy due to a lower friction and how such things can get realized&quot; provoked me to respond:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On an initial thought, a fluid with no viscosity should be great because it means that there's no friction, right? I'm assuming this is your train of thought. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flaw here is that there's no viscous friction - lubricants &quot;want&quot; a low viscosity because viscous friction causes the lubricant to heat and thus draws power from the system, but at the same time a lubricant &lt;em&gt;needs&lt;/em&gt; viscosity because lubrication relies on &lt;a href=&quot;https://en.wikipedia.org/wiki/Fluid_bearing&quot; rel=&quot;nofollow&quot;&gt;thin films&lt;/a&gt; between the load-bearing surfaces to lubricate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The viscosity of the lubricant is what keeps the lubricant from spraying/rushing out from between the load-bearing surfaces. In light load applications the lubricant is typically a very light, low viscosity oil, like WD-40. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the design loads increase, the required viscosity of the lubricant increases. I don't know how much automotive experience you have, but there's a noticeable difference between engine oil and the gear oil in a manual transmission. As loading continues to increase &lt;a href=&quot;https://en.wikipedia.org/wiki/Grease_(lubricant)#Thickeners&quot; rel=&quot;nofollow&quot;&gt;thickeners&lt;/a&gt; begin to be added to the oil, producing grease. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, back to your question, if you tried using a zero-viscosity fluid as a lubricant you would find that it would shoot out from between the parts to be lubricated as soon as you applied any load because there is no viscosity present to slow the fluid from leaving the joint.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-09-08T20:34:37.173" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5354" PostTypeId="2" ParentId="5330" CreationDate="2015-09-09T01:03:06.237" Score="2" Body="&lt;p&gt;If you look at the &quot;book&quot; values and the &quot;calculated&quot; value of many shapes, you will often find that there are differences. This doesn't only happen with hollow sections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differences are usually small enough that it doesn't matter (and there is as much variation in the material properties).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @alephzero mentions, most of the variation is from the fillet or radius at the corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fun happens when a given shape changes properties between editions of the manual without changes in the dimensions. Sometimes this is errata, sometimes it is &quot;re-calculating&quot; the table properties to match the &quot;as-rolled&quot; properties. The rolling mill has tolerances, so it may just be that one side of the tolerance is more common than the other. This would skew the average properties.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-09-09T01:03:06.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5355" PostTypeId="1" AcceptedAnswerId="5356" CreationDate="2015-09-09T02:34:25.843" Score="-2" ViewCount="1338" Body="&lt;p&gt;I know lasers can cut or create holes on some hard materials such as diamond and metals. Is there any similar concept in tunnel engineering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what are the difficulties?&lt;/p&gt;&#xA;" OwnerUserId="2958" LastEditorUserId="272" LastEditDate="2015-09-09T11:27:40.627" LastActivityDate="2015-09-09T11:42:41.573" Title="Is it possible to use lasers to build tunnels instead of mechanical digging machines?" Tags="|materials|lasers|tunnels|" AnswerCount="2" CommentCount="4" ClosedDate="2015-09-09T11:52:13.223" ContentLicense="CC BY-SA 3.0" />
  <row Id="5356" PostTypeId="2" ParentId="5355" CreationDate="2015-09-09T07:36:46.643" Score="4" Body="&lt;p&gt;First is power those cutting lasers usually make relatively small cuts and take quite a bit of power and time. Scaling that up means waiting for a few decades to move a few meters if you can't get more power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second is conservation of mass. All that rock has to go &lt;em&gt;somewhere&lt;/em&gt;. If you use a laser it will evaporate. No-one wants to be breathing metal fumes in an enclosed space. They will also cool off again and deposit themselves everywhere.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-09-09T07:36:46.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5357" PostTypeId="1" AcceptedAnswerId="5362" CreationDate="2015-09-09T08:08:40.533" Score="4" ViewCount="1354" Body="&lt;h2&gt;Short form: How would one calculate the perfect shape of a range hood?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;See drawing below. Will the curve of the profile of the range hood be exponential? Quadratic? What if the range is enclosed from three sides that create a box open only from the front - what curve should one choose? Sigmoid? Cubic? Some sort of chain curve? How does one calculate the parameters of the curve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one select the minimal air flow necessary to suck the steam in?&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Long form:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I'll start with a drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/WgRDm.png&quot; width=&quot;200&quot; height=&quot;340&quot; alt=&quot;schematic drawing of range hood&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;EDIT&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I have added the sections &quot;Fan throughput&quot; and &quot;Transition from laminar flow to turbulent flow&quot;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Dimensions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assume the range is a 60x60 cm square that produces steam (or 60 cm diameter disc if that's easier). The range hood begins 70 cm above that and is 110 cm high, terminating in a 7cm diameter pipe (which I might have to replace for something wider, but that's what I'm working with right now). This pipe is 80 cm high and terminates in a fan. I assume having the fan in a pulling configuration is better than having it push into the pipe. In addition it'll be quieter (which I would prefer). The fan has a constant throughput and blows into open air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a wall behind the range. There may be walls on the sides but not the front which is where you access the pots. I assume that's better than just having the wall behind the range. As a bonus it will minimize splatters around the kitchen. The walls may also only go part way from the range hood towards the stove if that's optimal.&#xA;The hood may be wider than the range, but with side walls that might not be necessary. It may not protrude towards the cook.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fan throughput&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I have tried to calculate the necessary fan throughput by calculating the expansion rate of heated steam during normal use (one large and one medium heater turned on, a total of roughly $4000\mathrm{W}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure which of my simplifying assumptions are too strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the volume above the range (60 cm * 60 cm * 70 cm) is filled with steam at 100 degrees Celsius, normal atmospheric pressure, and is heated uniformly by a 4000W heating element with no heating loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, the excess volume would spill in all directions - to the top and to the sides. We want to suck enough air into the top that the whole excess is taken in, so that the sides have negative pressure, i.e. air is coming into the volume from the sides, not out of it, and only escaping through the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C8FLX.png&quot; width=&quot;300&quot; height=&quot;200&quot; alt=&quot;Steam volume&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us calculate the volume gained over one second:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;(starting temperature) $20 ^\circ \mathrm{C} = 293 \mathrm{K}$&lt;/li&gt;&#xA;&lt;li&gt;(specific heat of steam between $20$ and $100 ^\circ \mathrm{C}$) between $1.86$ and $1.89$ $\mathrm{J/(gK)}$, let's settle for roughly $c_P = 1.88 \mathrm{J/(gK)}$.&lt;/li&gt;&#xA;&lt;li&gt;(density of superheated steam at $100 ^\circ \mathrm{C}$) $\rho=597.42868305543 \mathrm{g/m^3}$&lt;/li&gt;&#xA;&lt;li&gt;(mass of steam) $m = \rho*0.6*0.6*0.7 \mathrm{m^3} = \rho * 0.252 \mathrm{m^3}$&lt;/li&gt;&#xA;&lt;li&gt;(power of heater) $H = 1000 \mathrm{W}$&lt;/li&gt;&#xA;&lt;li&gt;(energy increase) $\Delta U = H * 1\mathrm{s} = 1000 \mathrm{J}$&lt;/li&gt;&#xA;&lt;li&gt;(temperature increase) $\Delta T = \Delta U/(c_P*m)$&lt;/li&gt;&#xA;&lt;li&gt;(volume increase) $\Delta V=(nR/P) \Delta T$&lt;/li&gt;&#xA;&lt;li&gt;(gas constant) $R = 8.3144621 \mathrm{J/(K*mol)}$&lt;/li&gt;&#xA;&lt;li&gt;(pressure) $P = 101.3 \mathrm{kPa}$&lt;/li&gt;&#xA;&lt;li&gt;(amount of substance) $n = m/m_m \mathrm{mol}$&lt;/li&gt;&#xA;&lt;li&gt;(molar mass of water) $m_m = 18.02 \mathrm{g/mol}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Plugging everything in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\text{from 8:} \\&#xA;\Delta V &amp;amp; =(nR/P) \Delta T \\&#xA;\\&#xA;\text{from 7:} \\&#xA;\Delta V &amp;amp; =(nR/P) \Delta U/(c_P*m) \\&#xA;\\&#xA;\text{simplify:} \\&#xA;\Delta V &amp;amp; =nR \Delta U/(P*c_P*m) \\&#xA;\\&#xA;\text{from 11:} \\&#xA;\Delta V &amp;amp; =(m/m_m)R \Delta U/(P*c_P*m) \\&#xA;\\&#xA;\text{simplify:} \\&#xA;\Delta V &amp;amp; =R \Delta U/(P*c_P*m_m) \\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging everything into Google gives me the answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(((8.3144 joule) / (kelvin * mol)) * (1000 joule)) / (101.3 kilopascal * 1.88 * (joule / (gram * kelvin)) * 18.02 * (gram / mol)) =&#xA;0.00242275129 m^3&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $1000 \mathrm{W}$ gives us $0.002 \mathrm{m^3/s}$, and $4000\mathrm{W}$ gives us $0.008 \mathrm{m^3/s} = 16.95 \mathrm{CFM} = 8 \,  \mathrm{liters/s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the minimum according to the above model. However, the volume in question isn't made out of jelly, and so a lot more suction will be necessary to make sure steam doesn't escape out the sides. My gut feeling tells me around three times more, or &lt;strong&gt;24 liters per second&lt;/strong&gt;, will be adequate &lt;strong&gt;according to this model&lt;/strong&gt; (but below we reduce this to 5 liters per second in a realistic scenario).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can compute at what air throughput turbulent flow will begin for our range hood of 60x60cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dynamic viscosity of steam at 100 degrees celsius = 212 deg K can be read off for &lt;a href=&quot;http://www.engineeringtoolbox.com/steam-viscosity-d_770.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this table&lt;/a&gt; by using &lt;a href=&quot;http://www.engineeringtoolbox.com/air-altitude-pressure-d_462.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this table&lt;/a&gt; to get the absolute atmospheric pressure of $14.7 \, \mathrm{psia}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(dynamic viscosity of steam) $\mu = 0.013 \mathrm{centiPoise}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinematic viscosity of a fluid &lt;a href=&quot;http://www.engineeringtoolbox.com/dynamic-absolute-kinematic-viscosity-d_412.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;can be calculated&lt;/a&gt; using $\nu = \mu / \rho$ where $\rho$ is the density of steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(density of steam as above) $\rho=597.42868305543 \mathrm{g/m^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(kinematic viscosity of steam) $\nu = 2.18 * 10^{-5} \mathrm{m^2/s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to calculate the Reynolds number and keep it under $2300$ for it to indicate laminar flow. Above $4000$ it indicates turbulent flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Re = Q D_H / (\nu A)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ is the cross-sectional area of the range hood's opening: $A = 60\mathrm{cm} * 60\mathrm{cm} = 0.36 \mathrm{m^2}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q$ is the flow rate: $Q = 24 \, \mathrm{liters/s} = 0.024 \mathrm{m^3/s}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D_H$ is the hydraulic diameter for a square duct, which is just the length of the side: $D_H = 0.6 \mathrm{m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging everything in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;Re &amp;amp; = 0.024 * 0.6 / (2.18 * 10^{-5} * 0.36) \\&#xA;Re &amp;amp; = 183.48 \\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we are perfectly within laminar flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a circular pipe at $24 \, \mathrm{liters/s}$, we can calculate the minimum radius $r$ to be below turbulent flow thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;2300  &amp;amp; &amp;gt; 0.024 2r / (2.18 * 10^{-5} * \pi r^2) \\&#xA;2300 * 2.18 * 10^{-5} * \pi / (0.024 * 2) &amp;amp; &amp;gt; 1/r \\&#xA;(2300 * 2.18 * 10^{-5} * \pi / (0.024 * 2))^{-1} &amp;amp; &amp;lt; r \\&#xA;0.30 &amp;amp; &amp;lt; r \\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we would need roughly a half-meter pipe otherwise we are dealing with slightly turbulent flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the real world, we are dealing with air that only contains a portion of steam. The steam buildup will be an order of magnitude less than 8 liters per second (imagine how much water you'd have to put in a pot!). This makes me think that &lt;strong&gt;realistically we only need a fan rated about 5 liters per second&lt;/strong&gt;. On the other hand dry air is only two-thirds as viscous as steam, which will mean turbulence is easier. All in all, with those adjustments compared to our model we can multiply the $Re$ calculated above by 0.3, or in other words, we only need &lt;strong&gt;a pipe of 18 centimeters diameter&lt;/strong&gt; for perfectly laminar flow. A pipe of 7 centimeters diameter which I have will cause turbulent flow.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Transition from laminar flow to turbulent flow&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I understand because we are transitioning from laminar to turbulent flow, the correct profile curve will minimize pockets of air blocked by turbulence, which end up &quot;choking&quot; the flow, further reducing the laminar core and causing even more turbulence.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Range hood profile curve&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;My experience from acoustics, and horns specifically, gives me the hunch that if the range hood terminated in mid air, the perfect profile would be an exponential curve. However I appreciate that with a horn speaker the pressure changes on the order of 15 Hz or higher, while here we're purely dealing with constant pressure. Also in this case the range would not terminate in open space but in a confined box (cylinder) that's open on one side (say an arc of 90 degrees). Would the best curve then be a sigmoid? Some sort of chain curve? A cubic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How does one calculate the right kind of curve for the profile connecting the semi-closed box and the pipe inlet?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that the case where everything has a round cross-section is easier than when it's square, but can then be generalized using a little common sense.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Airflow&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I assume there is some minimal air flow that is necessary to suck in the steam. How does one calculate that? I guess this has to do with pressure building up where air flows from the kitchen to the range top; my guess is that you could calculate the pressure of the air flowing past the range, and if that pressure is high enough, the &lt;strong&gt;flow&lt;/strong&gt; will expand to cover the whole available space (i.e. all the way down to the range top). I understand that's a &lt;strong&gt;fluid&lt;/strong&gt; &lt;strong&gt;dynamics&lt;/strong&gt; question, but I know almost nothing on the subject.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Impedance&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Is there a good approach to this problem (the curve profile as well as the that uses the concept of impedance? I understand impedance based approaches only really deliver answers for AC signals, not DC, but perhaps one could calculate the perfect transformer as a limit as frequency tends to 0, or use an approximation (e.g. assume frequency is 0.1 Hz)?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;My range hood doesn't really work at all. That is, it's so inefficient that putting my hand just under the hood I don't feel any draft at all. Having looked at it, there are many, many things wrong with it that I won't bother to list - it's just bad, leaks air everywhere, etc. The market doesn't seem to provide any sensible solutions either. Not even the professional offerings seem to make sense from an aerodynamics perspective. That made me wonder what the perfect range hood &lt;em&gt;would&lt;/em&gt; be, and I don't know how to approach the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I have posted this on the Physics StackExchange previously, but it was pointed out to me that the question was off topic for that SE, and I was asked to move it to Engineering. It has been put on hold there.&lt;/p&gt;&#xA;" OwnerUserId="2960" LastEditorUserId="2960" LastEditDate="2015-09-15T21:36:12.420" LastActivityDate="2015-09-15T23:40:25.837" Title="Perfectly aerodynamic kitchen range hood" Tags="|fluid-mechanics|hvac|airflow|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="5358" PostTypeId="1" AcceptedAnswerId="5360" CreationDate="2015-09-09T08:09:39.283" Score="4" ViewCount="3232" Body="&lt;p&gt;I am trying to manipulate some data that were obtained from a three point load experiment on a porcine tibia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to obtain a stress-strain curve for the various specimens; however, the only data that was given from the Instron three-point loading set up was Load (N) vs Deformation/Extension (mm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;how do you convert the Load vs Deformation curve to a Stress-Strain curve&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="2940" LastEditorUserId="2246" LastEditDate="2015-09-09T22:08:15.593" LastActivityDate="2015-09-09T22:08:15.593" Title="Convert Load Deformation (Extension) Data to Stress-Strain Curve" Tags="|mechanical-engineering|materials|stresses|biomedical-engineering|biomechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5359" PostTypeId="2" ParentId="5345" CreationDate="2015-09-09T08:59:05.570" Score="1" Body="&lt;p&gt;First, start with a precise mathematical description of your system. Second, analyze it observability. If the system is observable, of course you can estimate the (augmented) state vector (along with the load as an unknown parameter). In practice, there are always some factors that are often a subject to experimental estimation rather than system identification. These are, for instance, the efficiency map of the motor, liquid friction coefficient (may be estimated from the step response), motor inertia etc. I do not know which exactly feedback loop you are taking about, but the mechanical torque in electrical machines can be estimated from the current, that is true.&lt;/p&gt;&#xA;" OwnerUserId="2961" LastActivityDate="2015-09-09T08:59:05.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5360" PostTypeId="2" ParentId="5358" CreationDate="2015-09-09T09:27:46.983" Score="5" Body="&lt;p&gt;It's fairly complicated. I found an article in The Bone Journal &lt;a href=&quot;http://www.thebonejournal.com/article/S8756-3282(98)00076-3/fulltext?refuid=S8756-3282(10)01400-6&amp;amp;refissn=8756-3282&quot;&gt;Comparison of three-point bending test and peripheral quantitative computed tomography analysis in the evaluation of the strength of mouse femur and tibia&lt;/a&gt; where they used tomogrphy to calculate inertia. However if you don't have CT you can try the hard way.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You'll have to calculate bending moment M. It should be simple.&lt;/li&gt;&#xA;&lt;li&gt;Locate Neutral Axis. You can assume it as a simple shape but probably you have a complicated shape and have to calculate it.&lt;/li&gt;&#xA;&lt;li&gt;Calculate Area Moment of Inertia. Again if you assume simple shape it's easy else complicated.&lt;/li&gt;&#xA;&lt;li&gt;Formula for bending stress is $\sigma = -\frac{M\cdot y }{I}$ which is called &lt;a href=&quot;http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/flexure-formula&quot;&gt;flexure formula&lt;/a&gt; M is moment, y is distance from neutral axis and I is area moment of inertia.&lt;/li&gt;&#xA;&lt;li&gt;Strain is $\varepsilon = \frac{y}{\rho }$ where $\rho$ is radius of curvature. A simple presantation is &lt;a href=&quot;https://www.google.com.tr/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CCEQFjAAahUKEwjM4Kqn0OnHAhXnpnIKHWGUBKU&amp;amp;url=http%3A%2F%2Fengr.bd.psu.edu%2Frxm61%2F213%2FCh6_sec3_4.ppt&amp;amp;usg=AFQjCNHJlOFr9JLexGpzpSwVC6aofMQOeg&amp;amp;sig2=_R_C6lMrIKcaiyVYC1beSw&quot;&gt;here (PDF)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://academic.uprm.edu/pcaceres/Courses/INME4011/MD-3A.pdf&quot;&gt;presentation (PDF)&lt;/a&gt; and &lt;a href=&quot;http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?topic=me&amp;amp;chap_sec=04.1&amp;amp;page=theory&quot;&gt;this course notes&lt;/a&gt; are also informative.&lt;/p&gt;&#xA;" OwnerUserId="879" LastActivityDate="2015-09-09T09:27:46.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5361" PostTypeId="2" ParentId="5355" CreationDate="2015-09-09T11:42:41.573" Score="1" Body="&lt;p&gt;Barring other difficulties, it is unclear what advantages laser cutting would have in such an application.  Lasers are &lt;a href=&quot;https://engineering.stackexchange.com/questions/2991/why-are-lasers-used-for-concentrated-light-applications-instead-of-incoherent-li&quot;&gt;useful in drilling/cutting of materials&lt;/a&gt; because of their high precision and ability to make very small cuts.  Digging a tunnel is not a precision activity though.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if a tunnel drilling machine were set up to use lasers, the cut between the bore and the tunnel wall would have a very small separation (~1 mm), and a massive amount of equipment would still be needed to remove the rest of the material from the bore.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-09-09T11:42:41.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5362" PostTypeId="2" ParentId="5357" CreationDate="2015-09-09T13:38:06.027" Score="3" Body="&lt;p&gt;Making a duct that flares along an exponential curve to 2 feet (60cm) in diameter would create a larger duct that won't solve your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be an expensive duct to make (think &lt;a href=&quot;http://www.all-brass-instruments.com/uploads/brass/20/200400529718-7.jpg&quot; rel=&quot;nofollow&quot;&gt;tuba bell&lt;/a&gt;), and if you focus on the flare instead of the overall dimensions, then the duct might come down too low and you won't have enough space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that a flared duct doesn't do anything to move air - that's all done with a fan. The duct also isn't going to generate noise - that's also all done by the fan. If you want a quiet, high-flow hood, get a quiet, high-flow fan. I searched online for &quot;low noise high flow kitchen hood fan&quot; and I found &lt;a href=&quot;http://www.broan.com/products/series/broan-evolution-qp4-series-1783e9c0-5b48-4785-b04b-d4d01fad493e&quot; rel=&quot;nofollow&quot;&gt;this fan&lt;/a&gt; that advertises &amp;lt;0.3 sones during normal operation, with a boost mode of 630 CFM. 630CFM means that all of the air in a 18x18 foot room gets removed every 4 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to me that you should ignore the hood shape and focus on the fan. At your desired air speeds, for your desired application, the shape of the hood is going to make virtually no difference at all. That is why there is &lt;a href=&quot;https://www.google.com/search?q=kitchen+hood&amp;amp;biw=1920&amp;amp;bih=955&amp;amp;tbm=isch&amp;amp;source=lnms&amp;amp;sa=X&quot; rel=&quot;nofollow&quot;&gt;such variety&lt;/a&gt; in kitchen hoods. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;:EDIT:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The same conclusion, this time with numbers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider evaluating the &lt;a href=&quot;https://en.wikipedia.org/wiki/Reynolds_number&quot; rel=&quot;nofollow&quot;&gt;Reynold's number&lt;/a&gt;, which &quot;is defined as the ratio of momentum forces to viscous forces and ... also used to characterize different flow regimes within a similar fluid, such as laminar or turbulent flow.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;a href=&quot;https://en.wikipedia.org/wiki/Reynolds_number#Flow_in_pipe&quot; rel=&quot;nofollow&quot;&gt;flow in a pipe&lt;/a&gt;, the Reynolds number can be calculated with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mbox{Re} = \frac{Q D_H}{\nu A} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Q$ is the volumetric flow rate in $m^3/s$, $D_H$ is the hydraulic diameter (which is just the diameter for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydraulic_diameter&quot; rel=&quot;nofollow&quot;&gt;circular tube&lt;/a&gt;), $\nu$ is the kinematic viscosity in $m^2/s$, $A$ is the cross-sectional area of the pipe in $m^2$, and $\mbox{Re}$ is the Reynolds number, a unitless number that indicates laminar flow for values under approximately 2300, turbulent flow for values over 4000, and transitional flow between the two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO, for air, here are the values I used in the equation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Going off the grid in your question (thanks for that!), if the bell end flare is 60cm wide at 6 grids wide, I estimated your duct to be about 1 grid wide, so a diameter about 10cm or 0.01m. &lt;/li&gt;&#xA;&lt;li&gt;This diameter gives the cross-sectional area $A = \pi r^2$, or $A = pi (0.05^2)m^2$, so $A = 0.00785m^2$. &lt;/li&gt;&#xA;&lt;li&gt;Your stated flow of $50 L/s = 0.050 m^3/s$. &lt;/li&gt;&#xA;&lt;li&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Viscosity#Air&quot; rel=&quot;nofollow&quot;&gt;kinematic viscosity&lt;/a&gt; of $\nu = 1.48 \times 10^{-5} m^2/s$. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That's it! So, plugging everything in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mbox{Re} = \frac{Q D_H}{\nu A} \\&#xA;\mbox{Re} = \frac{(0.050 m^3/s)(0.01m)}{(1.48 \times 10^{-5} m^2/s) (0.00785m^2)} \\&#xA;\mbox{Re} = \frac{5 \times 10^{-3}}{1.16 \times 10^{-7}} \\&#xA;$$&#xA;$$&#xA;\boxed{\mbox{Re} = 43,015} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, where a Reynolds number over 4,000 indicates turbulent flow, at your duct size and your flow rate, you are at 43,000, so an order of magnitude past the threshold for turbulent flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This demonstrates and hopefully reinforces my earlier statement, &quot;the shape of the hood is going to make virtually no difference at all,&quot; because again, at your flow rate, you're already in turbulent flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you can't do anything about the flow noise due to turbulence, the best thing you can do for noise reduction is to look at a fan that's designed for low noise emission. This is typically done by increasing the blade surface area while reducing the speed, because one of the largest factors for fan noise is the tip velocity on the blades. See &lt;a href=&quot;http://www.moorefans.com/pdfs/NOTES4_1.PDF&quot; rel=&quot;nofollow&quot;&gt;this guide&lt;/a&gt; for more information. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2015-09-15T23:40:25.837" LastActivityDate="2015-09-15T23:40:25.837" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="5364" PostTypeId="1" CreationDate="2015-09-10T05:39:37.080" Score="1" ViewCount="77" Body="&lt;p&gt;I'm currently building a work table that will have &quot;wings&quot; on each side that are fixed to the mid section of the table via a hinge.  These wings, in turn, have hinges that fix a support board to them which will rest against the mid section of the table.  I'm currently in the process of deciding where to place the support and am wondering if anyone has any engineering input as to an optimal placement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of how the rest of the table in constructed, I am limited as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$14'' \leq A \leq 19''$&lt;/li&gt;&#xA;&lt;li&gt;$B + C \leq 38''$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then there's also obviously the fact that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A^2 + B^2 = C^2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what the optimal setup is here.  The only scenario I've solved for is trying to make $A = B$ so that the angle $\theta$ between $A$ and $C$ is 45 degrees.&#xA;This implies $\sin(\theta) = B / C = 1 / \sqrt{2}$.  Plugging this into the equations I have above, I found that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A = B$ implies that $14'' \leq A = B &amp;lt;= \frac{38}{1 + \sqrt{2}} \approxeq 15.3$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are there any other more sophisticated approaches to this problem than the &quot;make $\theta=$ 45 degrees&quot; approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. - I studied Physics in college, so have a math / physics background, but know very little about applied engineering like this.  Also, let me know if you require / would like any additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/puSnw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/puSnw.jpg&quot; alt=&quot;Wing Up Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3trKA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3trKA.jpg&quot; alt=&quot;Wing Down Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2967" LastEditorUserId="2246" LastEditDate="2015-09-10T20:29:22.823" LastActivityDate="2019-12-28T18:28:27.177" Title="Optimizing support board placement on work table wing" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5365" PostTypeId="1" AcceptedAnswerId="5368" CreationDate="2015-09-10T09:37:59.240" Score="5" ViewCount="327" Body="&lt;p&gt;How can I compute the Frequency Response Function (FRF) if I use two three-axis MEMS accelerometer to measure the input excitation and output response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read from some books that the FRF can be computed as the ratio of response to excitation, e.g., $ H(s) = x(s)/F(s)$, where $s$ is the Laplace variable. or a similar way via Fourier transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, those books have not mentioned how to compute FRF when it comes to three-axis accelerometer. As there are three channels of data, &lt;em&gt;i.e.&lt;/em&gt;, X, Y, Z, should I compute the FRF for each channel separately, or I need to perform some kind of combination of these three axes before computing the FRF?&lt;/p&gt;&#xA;" OwnerUserId="2228" LastEditorUserId="41" LastEditDate="2015-09-10T20:28:54.303" LastActivityDate="2016-12-14T00:09:27.997" Title="How do I conduct an experimental modal analysis with a three-axis accelerometer?" Tags="|civil-engineering|structural-engineering|vibration|modal-analysis|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5367" PostTypeId="2" ParentId="5365" CreationDate="2015-09-10T12:28:11.267" Score="2" Body="&lt;p&gt;&lt;em&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt;&lt;/em&gt; Base on the question, I am suspecting you are looking for a theoretical explanation. Unfortunately I am  not in a position offer any insight in short order.&lt;/p&gt;&#xA;&lt;p&gt;But I can offer you following two ideas and source code to access data from three axis ADXL345 accelerometer to help you with your endeavors.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Three axis ADXL345 accelerometer implementation using a Embedded Linux platform&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=bd7ULY1frA4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Accelerometer Application on Embedded Linux&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Three axis ADXL345 accelerometer implementation using a TIVA ARM Cortex M4&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/6ZFy4zMGLgg&quot; rel=&quot;nofollow noreferrer&quot;&gt;TIVA - ADXL345 Accelerometer Demo Using I2C&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Your are welcome to fork the source code and use them as you wish.&lt;/p&gt;&#xA;&lt;p&gt;Also I suggest that you use a rate table to generate the input input excitation. Below is a suggestion.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W74aB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W74aB.jpg&quot; alt=&quot;3-axis Rate table&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Hope this post with combination of theoretical explanation will give you total solution.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://github.com/mahengunawardena/Tiva_I2C_Nokia_ADXL345&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tiva_I2C_Nokia_ADXL345 source code&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://github.com/mahengunawardena/BeagleboneBlack_ADXL345_GUI&quot; rel=&quot;nofollow noreferrer&quot;&gt;BeagleboneBlack_ADXL345_GUI source code&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.acutronic.com/us/products/3-axis/product-detail/view/3-axis-motion-simulator-serie-ac347-v.html?no_cache=1&amp;amp;nar=1&amp;amp;cHash=6281730b35d79e1ad086599273498a93&quot; rel=&quot;nofollow noreferrer&quot;&gt;3-Axis Motion Simulator Series AC347-V&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/3348/calculating-pitch-yaw-and-roll-from-mag-acc-and-gyro-data/3350#3350&quot;&gt;Calculating pitch, yaw, and roll from mag, acc, and gyro data&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-12-14T00:09:27.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5368" PostTypeId="2" ParentId="5365" CreationDate="2015-09-10T12:29:00.297" Score="3" Body="&lt;p&gt;The body of your question (the title is somewhat different) asks about computing the frequency response function (FRF) between a multi-input, multi-output (MIMO) system.  First off, the numerical transfer function/FRF between two time domain signals is usually calculated by calculating the cross power spectrum between the two signals and dividing it by the power spectrum of the input&#xA;$$&#xA;FRF_{A\rightarrow B}=\frac{S_{AB}(f)}{S_{AA}(f)}.&#xA;$$&#xA;National Instruments &lt;a href=&quot;http://www.engr.sjsu.edu/bjfurman/courses/ME120/me120pdf/FFT_tutorial_NI.pdf&quot; rel=&quot;nofollow&quot;&gt;Application Note 41&lt;/a&gt; has a helpful introduction to this type of analysis.  It is important to also calculate the coherence of the input and output signals so that you have an idea of how accurate the estimated transfer function is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question about how to deal with a MIMO system; usually people think about it as a matrix of transfer functions.  In your case there will be 9 independent transfer functions:&#xA;$$&#xA;\begin{pmatrix}&#xA; FRF_{x\rightarrow x} &amp;amp; FRF_{x\rightarrow y} &amp;amp; FRF_{x\rightarrow z} \\&#xA; FRF_{y\rightarrow x} &amp;amp; FRF_{y\rightarrow y} &amp;amp; FRF_{y\rightarrow z} \\&#xA; FRF_{z\rightarrow x} &amp;amp; FRF_{z\rightarrow y} &amp;amp; FRF_{z\rightarrow z}&#xA;\end{pmatrix}&#xA;$$&#xA;Understanding how any excitation shows up at the output is then just a matter of multiplying the x, y, and z inputs (expressed in the frequency domain) by the transfer function matrix.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-09-10T12:29:00.297" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5369" PostTypeId="2" ParentId="5340" CreationDate="2015-09-10T12:40:12.537" Score="5" Body="&lt;p&gt;There are analogs to filters in lots of mechanical systems. In fluid systems prone to pressure spikes perhaps because they use a fixed displacement pump, an accumulator will be used to filter out those spikes to prevent damaging other parts of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vibration isolator acts in much the same way. Sometimes they're used as is. Other times, for more critical applications, they're designed to filter specific frequencies, like the rotational speed of a motor or a pump, to filter those out and limit or prevent that energy from exciting the surrounding structure. That's done by controlling both the geometry and the material properties of the isolator. One of my first projects out of undergrad was designing just such a mount for a small pump motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another mechanical filter is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tuned_mass_damper&quot; rel=&quot;noreferrer&quot;&gt;tuned mass damper&lt;/a&gt;. Typically these are found inside skyscrapers and are intended to limit the movement of the building due to earthquakes. The same idea has been applied to &lt;a href=&quot;http://www.google.com/patents/US20120186611&quot; rel=&quot;noreferrer&quot;&gt;appliance motors&lt;/a&gt; as well, where a metal &quot;lollipop&quot; is cantilevered off the side of the motor. The resonant frequency of the mass and rod are matched to the operating frequency of the motor. The result is that when the motor turns, the energy at the motor's operating frequency goes into making the mass vibrate in the air instead of transmitting it to the structure of the appliance where it will be radiated as sound. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-09-10T12:40:12.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5371" PostTypeId="1" CreationDate="2015-09-10T14:05:42.777" Score="3" ViewCount="396" Body="&lt;p&gt;Looking at the turbocharger picture below there is a compressor and a turbine side.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you dont know the flow direction of the fluid, how can you distinguish which side is the compressor and which is the turbine? Normally for subsonic design this would mean increasing cross section area is a compressor and reverse for turbine. However, since the design is very different considering the shape of the blades between both sides there seems to be a better way to identify it. How?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why is on the left side of the pictures every second blade shorter? What effect does it prevent/provoke?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a special reason why the blades dont have a certain profile (for example NACA profiles) like in airplane turbines instead of (near) constant thickness? This part seems milled with a 5/6 axis machine, so price should not change with a profile shape design.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.glr.tu-darmstadt.de/media/glr/turbolader/DSC_6847_3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;turbine http://www.glr.tu-darmstadt.de/media/glr/turbolader/DSC_6847_3.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2971" LastActivityDate="2015-09-14T20:29:42.417" Title="Aerodynamic Turbocharger Design" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5372" PostTypeId="1" CreationDate="2015-09-10T15:02:47.997" Score="3" ViewCount="259" Body="&lt;p&gt;Measuring pressures have many different applications which require proper encapsulating of the sensors, but at the core there is a limited amount of techniques.&lt;/p&gt;&#xA;&lt;p&gt;To my understanding, the two leading technologies in high accuracy pressure sensing are:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Measuring resonant frequencies in e.g. quartz crystals&lt;/li&gt;&#xA;&lt;li&gt;Piezo-electric elements&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;And there are two leading technologies in pressure transfer from ambient pressure to the pressure sensing element (force collectors)&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Diaphragm&lt;/li&gt;&#xA;&lt;li&gt;Bourdon tube&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I found a easy to understand reference to precision and accuracy of pressure sensors &lt;a href=&quot;http://www.wika.us/upload/TA_0614_CEP_en_co_59242.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pressure sensor fundamentals: Interpreting accuracy and error&lt;/a&gt;.&#xA;Here they divide Accuracy into the following categories&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Nonlinearity&lt;/li&gt;&#xA;&lt;li&gt;Hysteresis&lt;/li&gt;&#xA;&lt;li&gt;Nonrepeatability&lt;/li&gt;&#xA;&lt;li&gt;Zero point error&lt;/li&gt;&#xA;&lt;li&gt;Span error&lt;/li&gt;&#xA;&lt;li&gt;Temperature impacts&lt;/li&gt;&#xA;&lt;li&gt;Long term drift&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Most high accuracy sensors promise an accuracy of 0.01% of full scale, but how does this distribute among the different types of errors?&lt;/p&gt;&#xA;&lt;p&gt;Are there any sensors promising significantly better accuracy?&lt;/p&gt;&#xA;&lt;p&gt;Is there some technologies that are better than others on e.g. hysteresis or temperature impacts?&lt;/p&gt;&#xA;&lt;p&gt;Has there been scientific comparisons between different sensors to measure these errors which constitute the total accuracy?&lt;/p&gt;&#xA;&lt;p&gt;What is the best way of going about to find the sensor best suited for my purpose?&lt;/p&gt;&#xA;" OwnerUserId="2972" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2022-08-11T07:03:17.323" Title="Pressure sensors with best precision" Tags="|pressure|sensors|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5373" PostTypeId="2" ParentId="2015" CreationDate="2015-09-10T15:27:40.533" Score="0" Body="&lt;p&gt;Advantages of double lead threading: Starting ease and eliminating cross-threading (i.e. happy customers). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common usages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any bottle cap or jaw lid&lt;/li&gt;&#xA;&lt;li&gt;Catsup bottles&lt;/li&gt;&#xA;&lt;li&gt;Tooth paste tubes&lt;/li&gt;&#xA;&lt;li&gt;Split bolts (Kearneys)&lt;/li&gt;&#xA;&lt;li&gt;Flash light end caps&lt;/li&gt;&#xA;&lt;li&gt;Any assembly not requiring high tension&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2974" LastEditorUserId="272" LastEditDate="2015-09-10T16:02:50.667" LastActivityDate="2015-09-10T16:02:50.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5374" PostTypeId="1" CreationDate="2015-09-10T15:40:08.860" Score="4" ViewCount="2240" Body="&lt;p&gt;With regard to Fan Sensors, there are two very common methods for detecting whether the fan is doing what it's supposed to or not.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Locked Rotor- Detects when the fan is working, or when it has stopped.&#xA;(on or off voltage signal)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tachometer- Detects the speed of the fan, to determine whether it is performing optimally.&#xA;(pulsed voltage signals)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My Question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q- What is the difference between using these two types of fan sensors in 'some' design?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional info:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know what this 'design' is, but I'm asking for the &lt;strong&gt;practical&lt;/strong&gt; differences if I picked either one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I understand that both setups use different voltage signals, and require different circuitry to operate. That is beyond what I think would be the &lt;strong&gt;practical&lt;/strong&gt; scope of this question.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2973" LastEditorUserId="272" LastEditDate="2015-09-10T16:03:16.363" LastActivityDate="2015-09-10T16:14:07.020" Title="Fan Sensors (Tachometer vs Locked Rotor)" Tags="|sensors|design|airflow|turbines|systems-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5375" PostTypeId="2" ParentId="5374" CreationDate="2015-09-10T16:14:07.020" Score="4" Body="&lt;p&gt;It's a trade-off between the simplicity of the sensor and how comprehensive the reported data is.  The locked rotor sensor only reports whether or not the motor is spinning, not how fast it's spinning or how much resistance is on the fan.  It is a very simple signal though, it's either on or off.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the tachometer signal is a bit more complex: the signal will need to be converted or interpreted in some way to extract the fan RPMs, but, you can use this data for much more than just indicating whether the fan is running or not.  You could compare the measured RPMs to a table of load vs. RPMs to estimate the load on the fan, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of which sensor to use should depend on what information you need of course.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-09-10T16:14:07.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5376" PostTypeId="2" ParentId="5344" CreationDate="2015-09-10T20:50:34.717" Score="5" Body="&lt;p&gt;Bolt shear is a limit state for the bolt, whereas bearing stress is more a limit state for the connected plates. Therefore, it makes sense to evaluate bearing stress on a plate-by-plate basis. When two or more plates are sharing the load, you can consider the total thickness in each loaded direction versus the total applied load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose that in principle, when two or more plates are sharing the load you could split the load between the plates based on relative bearing area and evaluate adequacy on a plate-by-plate basis. However, since the loaded 'width' is the same for all plates (i.e. the bolt diameter) this approach would give the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little qualitative thought experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e0wxn.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e0wxn.png&quot; alt=&quot;Bolt Bearing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are two shear planes and three bearing 'areas'&lt;/li&gt;&#xA;&lt;li&gt;The stress on the two shear planes is equal&lt;/li&gt;&#xA;&lt;li&gt;The $Plate_1$ load + $Plate_2$ load = $Plate_3$ load&lt;/li&gt;&#xA;&lt;li&gt;Load in $Plate_1$ &gt; load in $Plate_2$ (because $t_1$ &gt; $t_2$)&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bearing stress at $Plate_1$ = bearing stress at $Plate_2$ = $\frac{P}{bolt diameter(t1 + t2)}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If ($t_1$ + $t_2$) = $t_3$ then the bearing stress is uniform along the bolt length, otherwise it will vary&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2015-09-11T12:33:22.240" LastActivityDate="2015-09-11T12:33:22.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5377" PostTypeId="2" ParentId="5364" CreationDate="2015-09-11T02:33:59.630" Score="0" Body="&lt;p&gt;When looking at the piece from a force balancing analysis, the answer comes down to three cases: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the angle is very small, then the board is primarily horizontal and most of the vertical load will be multiplied by a factor of the secant of the small angle. A skinny member like that does not take compressive loads like this well (buckling). The shelf will not take much in the way of load. However, since the end of the support will likely be near the end of the board, the plate functions like a simply supported beam and has minimal deflection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the angle is very large, then the board is primarily vertical and takes most of the compression without the secant multiplier adding to much. However, due to geometry, the board will struggle to reach the end of the plate, possibly turning it into cantilever, leading to poor deflection control of the plate, but it can take large amounts of loading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally the middle case you described.  This can handle both cases well, deflection of the plate minimized and holding a reasonable amount of load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically in these cases a designer comes up with an angle and evaluates the loading the system can handle. Then, by comparing the performance of all members with regards to specified parameters (factor of safety on buckling, allowed deflection), the excess factors (high fos or low deflection) can sometimes be adjusted by changing the angle until the excess on each factor is the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this really depends on your specified parameters - what you want the system to do - high load carrying capability or minimal deflection.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-09-11T02:33:59.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5378" PostTypeId="2" ParentId="5364" CreationDate="2015-09-11T02:40:09.067" Score="2" Body="&lt;p&gt;As @hazzey mentioned in a comment under the OP, if nothing is going to break, then the choice is mostly arbitrary. So I'm going to assume that things do break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A structural element can basically collapse in three ways: excessive axial force (tension or compression), shear force or bending moment. Each of these is calculated differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simple things like a table, one can assume a simple calculation for the maximum supported tensile force &lt;span class=&quot;math-container&quot;&gt;$T_R$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$T_R = f_y\cdot A$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$f_y$&lt;/span&gt; is the allowable tensile stress and &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the total cross-section area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear force can also be calculated in a similar fashion:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q_R = f_q\cdot A$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$f_q$&lt;/span&gt; is the allowable shear stress. For steel one often uses &lt;span class=&quot;math-container&quot;&gt;$f_q = 0.6f_y$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bending moment has a series of complications due to lateral buckling. However, since that probably won't be the limiting factor in the design of your table, I'll skip over them as well. Therefore, we can take simply that the maximum stress at any point in the beam must be less than &lt;span class=&quot;math-container&quot;&gt;$f_y$&lt;/span&gt;. The applied stress can be obtained via&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \sigma = \dfrac{My}{12EI}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the bending moment, &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; is the height from the centroid of the desired point, &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is the material's elastic modulus, and &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt;, the cross-section's second moment of area (aka, moment of inertia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have left compression for last because it's the one with complications we can't avoid, since they will probably be the controlling factors. For short segments, the maximum allowable compressive force is equal to the maximum tensile force &lt;span class=&quot;math-container&quot;&gt;$T_R$&lt;/span&gt; presented above. That being said, due to buckling, this usually isn't the case. Buckling is a pain and has a bunch of complications in and of themselves, but for a ballpark value, we can use Euler's equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_E = \dfrac{\pi^2EI}{(KL)^2}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; is a factor depending on the element's support conditions. &lt;span class=&quot;math-container&quot;&gt;$P_E$&lt;/span&gt; is non-conservative, meaning the actual allowable compressive force will certainly be lower than this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also another way the table can fail and that is if it doesn't suit your needs. It's possible for the table to support the desired load but to do so by deforming completely, making it unappealing and unusable. Deflections, however, don't have simple equations and depend on the current conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now we need to find the stresses that occur in your table. This is its structural model (the partial dimensions are placeholders and change according to the selected angle and value of length &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;. The load is also a placeholder):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BHuiG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BHuiG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balancing act that is required is due to the fact that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if you increase the table span by moving the bracing closer to the base, you're going to increase the bending moment and shear force on the table. The deflections are also going to increase.&lt;/li&gt;&#xA;&lt;li&gt;if you lengthen the bracing (up to its maximum 19&quot;, for instance), you can reduce the stresses described above on the table, but lower the buckling load for the bracing.&lt;/li&gt;&#xA;&lt;li&gt;if you open up the bracing (increase its angle to the vertical), you can reduce the span of the table but you reduce the bracing's effectiveness.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The optimal solution needs to balance all of these considerations and it goes beyond the scope of this answer to actually obtain it (especially given that it depends on the materials and cross-sections). To perform the structural analyses, a 2D frame analysis tool like &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt; (free) can be used. That being said, 45 degrees is usually a good bet.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:28:27.177" LastActivityDate="2019-12-28T18:28:27.177" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="5379" PostTypeId="1" CreationDate="2015-09-11T03:53:11.913" Score="3" ViewCount="40" Body="&lt;p&gt;For a project I am working on I need to model the flow of argon in a long pipe. The flow is transient in nature since the tanks and regulator from which it enters cause the mass flow to change. Moreover, it reaches low temperatures due to the Joule-Thomson effect. Hence convective heat transfer from the pipe to the gas occurs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any recommended literature would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2914" LastEditorDisplayName="user16" LastEditDate="2015-09-11T20:05:55.753" LastActivityDate="2015-09-11T20:05:55.753" Title="Advice on Literature pertaining to transient pipe flow with heat transfer and friction" Tags="|mechanical-engineering|fluid-mechanics|heat-transfer|compressed-gases|" AnswerCount="0" CommentCount="3" ClosedDate="2015-09-13T12:21:49.150" ContentLicense="CC BY-SA 3.0" />
  <row Id="5380" PostTypeId="1" CreationDate="2015-09-11T05:41:04.057" Score="12" ViewCount="1225" Body="&lt;p&gt;Every time I search for &quot;aeronautical engineering&quot; on Google I see a few results about aerospace engineering. So they must be similar, but what is different about these careers?&lt;/p&gt;&#xA;" OwnerUserId="2977" LastEditorUserId="368" LastEditDate="2015-09-11T15:37:02.437" LastActivityDate="2016-02-22T04:07:00.857" Title="What is the difference between aeronautical and aerospace engineering?" Tags="|aerospace-engineering|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5383" PostTypeId="2" ParentId="5380" CreationDate="2015-09-11T12:17:47.423" Score="0" Body="&lt;p&gt;Technically, aeronautical engineers only deal with things inside Earth's atmosphere, while aerospace engineers may also deal with spaceflight. However, in practice the two terms are essentially interchangeable - you will want to look at each job offer / university course etc in detail to see what's being offered or required. &lt;/p&gt;&#xA;" OwnerUserId="2585" LastActivityDate="2015-09-11T12:17:47.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5384" PostTypeId="1" CreationDate="2015-09-11T12:29:10.627" Score="6" ViewCount="730" Body="&lt;p&gt;I am studying about wind turbine and turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some questions, but first, I explain some cases as follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low velocity air over an airfoil at a zero-incidence angle does not initiate flow separation. If the air velocity slowly increases, the flow separation starts from the blade trailing edge (TE). If the velocity or the incidence angle increases further, the separation point or stall point moves towards the blade leading edge (LE). If the stall point can be pushed towards the TE, the airfoil performance or the turbine blade performance increases as the attached flow transfers more energy to the blade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I want to know that how I can claim that efficiency of turbine increase up to given angle of attack?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The efficiency increase with increasing angle of attack while flow separation also increase with increasing angle of attack. I don't understand this case!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to above explanations, if angle of attack increases; separation point or stall point moves towards the blade LE and the turbine blade performance decreases as there is low attached flow. I know that when angle of attack increases, tangential force also increases and for this reason, the efficiency of turbine increases but I want to know about increasing efficiency from the point of flow separation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can these two things (increasing efficiency and increasing separation) be linked together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am grateful to all who guide me about these cases and if it is possible, please place useful links and papers related to these cases.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In fact, I am studying more about the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Wells_turbine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ws turbine&lt;/a&gt;. In my investigations, the air velocity(inlet velocity or axial  velocity) is constant and rotational velocity(r $\omega$) is changed and in other words, rotational speed is decreased in my calculations. this method is performed in the following paper. In this paper, There is the used equation for efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.4shared.com/office/IJ8vtfl-ce/A_comparison_of_computational_.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;A comparison of computational...&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\eta = \frac{T \cdot \omega}{\Delta p_\mathrm{t}\cdot Q}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T =$ shaft torque&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega =$ angular speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U = r \cdot  \omega =$ rotational speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta p_\mathrm{t} =$ total pressure drop across the turbine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = V_\mathrm{inlet} \cdot A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \pi\left(R_\mathrm{casing}^2-R_\mathrm{hub}^2\right) =$ through flow area&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{V_\mathrm{inlet}}{U} =$ flow coefficient&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, please see the following figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gK9DL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gK9DL.jpg&quot; alt=&quot;blade profile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, The  various  values  of  the  flow coefficient  were  achieved by  varying  the  rotational  speed  of  the rotor  at  a  constant  value  of  the  axial  velocity,  which  are  similar with  those  performed  by  Kim  et  al.  [1,2],  Takao  et  al.  &lt;a href=&quot;http://0up.ir/do.php?imgf=2_3753c.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;  and  Zahari et  al.  [4]. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.4shared.com/office/IJ8vtfl-ce/A_comparison_of_computational_.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;  Kim  TH,  Setoguchi  T,  Kinoue  Y,  Kaneko  K.  Effects  of  blade  geometry  on  performance  of  Wells  turbine  for  wave  power  conversion.  Journal  of  Thermal  Science&#xA;2001;10(4):293–300.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gK9DL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;  Kim  TH,  Setoguchi  T,  Kaneko  K,  Raghunathan  S.  Numerical  investigation  on  the&#xA;effect  of  blade  sweep  on  the  performance  of  Wells  turbine.  Renewable  Energy&#xA;2002;25:235–48.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://0up.ir/do.php?imgf=2_3753c.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;  Takao  M,  Setoguchi  T,  Kinoue  Y,  Kaneko  K.  Wells  turbine  with  end  plates  for wave  energy  conversion.  Ocean  Engineering  2007;34:1790–5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[4]  Zahari  Taha,  Sugiyono,  Sawada  T.  A  comparison  of  computational  and  experimental  results  of  Wells  turbine  performance  for  wave  energy  conversion. Applied  Ocean  Research  2010;32:83–90.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am grateful that guide me about these cases and discuss together about these cases.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2015-09-22T19:46:53.447" LastActivityDate="2015-09-22T19:46:53.447" Title="efficiency and flow separation" Tags="|mechanical-engineering|fluid-mechanics|airflow|turbines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5385" PostTypeId="2" ParentId="5371" CreationDate="2015-09-11T13:24:12.427" Score="1" Body="&lt;p&gt;&lt;strong&gt;Regarding point 1:&lt;/strong&gt;&#xA;To be honest I am not 100% sure. However, here are my thoughts:&#xA;As with all profiles the turning of the flow is easier when it is accompanied with a pressure gradient (from high to low pressure). Easier is quite an unspecific word lumping together a lot of undesirable aerodynamic features like boundary layer growth and or boundary layer separation. The gradient between high pressure and low pressure can be reduced by increasing blade length (reducing the pressure-gradient acting on the boundary layer).&#xA;Regarding your mention of subsonic flow and cross section: In the case of a transonic compressor the inflow is supersonic and the outflow is subsonic. Plus, the compressor flow is always from a lower to a higher radius (the turbine is the other way around). So both sides will feature a bigger area at the higher radius and a smaller area ant the smaller smaller radius.&#xA;So I guess we look at the compressor with the turbine in the background.&#xA;Because it features longer blades with reduced flow turning at the outer diameter and splitter blades (end-bend).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Regarding point 2:&lt;/strong&gt;&#xA;The design of a radial compressor is influenced by a lot of parameters one is manufacturability. I guess that the requested pressure rise of the compressor and the associated turning of the flow was not possible with just seven blades, however the machining and minimum blade thickness did not allow for 14 full blades. If you are able to provide a meridional view of pressure rise one would probably see that the first section of the compressor has a lower loading and the rear sections of the passage feature a higher loading because it can be distributed across a larger number of blades.&#xA;So in short the answer is: The design depends on the requirements in terms of manufacturability and thermodynamic behaviour (e.g. off-design characteristics). And what we see seems to be the optimal trade-off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Regarding point 3:&lt;/strong&gt;&#xA;You need to keep in mind that NACA profiles are based on constant free stream flow velocity. In an axial machine this assumption is somewhat valid because the difference in radii of inflow and outflow stream surface are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;small. In an radial machine the &lt;em&gt;free stream&lt;/em&gt; velocity changes due to the higher circumferential velocity (i.e. higher radius).&#xA;NACA profiles do not &lt;em&gt;account&lt;/em&gt; for centrifugal forces which play a huge role in radial machines.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-09-14T20:29:42.417" LastActivityDate="2015-09-14T20:29:42.417" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5386" PostTypeId="2" ParentId="5380" CreationDate="2015-09-11T13:42:20.433" Score="7" Body="&lt;p&gt;A space engineer deals with objects in space: from satellites, space stations, space vehicles and space probes to space debris. Space means high launch vibrations, vacuum, radiation concerns, high temperature gradients and extremes, complex trajectories, next to no maintenance (meaning: built in redundancy)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An aeronautical engineer deals with objects in the air, but not in space (space is commonly considered to start at 100km of altitude): mainly all types of airplanes and helicopters, but balloons and missiles fall under that category too. Aeronautics generally means cost effectiveness (mass produced), challenging mechanisms (in terms of performance and reliability over time, may mean redundancy), complex computational fluid dynamics, high stresses on the wings... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Aerospace&quot; is a mix of both. Since they are so different, I wouldn't say an aerospace engineer is worth a space engineer and an aeronautical engineer. Different fields, different skills, but such engineers particularly come in handy for the launchers, especially the new generation of renewable ones that fly (e.g. Virgin Galactic SpaceshipTwo, or Skylon). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: I'm a space engineer specialised in mechatronics. I have never worked in the aeronautics industry.&lt;/p&gt;&#xA;" OwnerUserId="1326" LastEditorUserId="1326" LastEditDate="2015-09-11T13:49:25.660" LastActivityDate="2015-09-11T13:49:25.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5387" PostTypeId="2" ParentId="5384" CreationDate="2015-09-11T14:03:19.797" Score="6" Body="&lt;p&gt;You are describing a very interesting engineering problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The requirement of a turbine is to convert fluid-energy into mechanical energy over a range conditions. Since the flow around the airfoils is governed by the (elliptic) Navier-Stockes-Equations, &lt;em&gt;an easy answer to your question is not possible in general&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only in certain edge cases it is possible to simplify the system in such a way that simple answers are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I explain what goes into some of those cases, I'd like clarify some words first. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;Efficiency&lt;/em&gt; is usually the ratio between the theoretical maximum and the actually achieved value. The above mentioned equation assumes that the energy extraction is only visible in a total pressure drop, $\Delta p_\mathrm{t}$. For low Mach-Number flows that is a reasonable assumption ($\mathrm{Ma}\ll 0.3$). If higher speeds are achieved a total ethalpy based definition of efficiency is needed.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Angle of Attack&lt;/em&gt; (AoA) describes the angle between the fluid velocity vector and a somewhat arbitrary line with respect to the airfoil. Since Wells turbines usually feature a symmetric profile the reference line is the symmetry-axis. A change in angle of attack can be achieved by three things: &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;rotating the blades &lt;/li&gt;&#xA;&lt;li&gt;change of rotational speed of the rotor&lt;/li&gt;&#xA;&lt;li&gt;change of windspeed.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As I said there is no easy answer to your question. To delve further into your question, we will tackle it in two steps. First, we have a look on the aerodynamic characteristics of an airfoil. Then, we will look at the turbomachinery part of the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1.&lt;/strong&gt; Wells Turbines often use symmetric &lt;a href=&quot;https://en.wikipedia.org/wiki/NACA_airfoil&quot; rel=&quot;nofollow noreferrer&quot;&gt;4 digit NACA profiles&lt;/a&gt;. The aerodynamic characteristics of such airfoils can easily be calculated using webtools. The following curves were created using &lt;a href=&quot;http://airfoiltools.com/airfoil/details?airfoil=naca0021-il#polars&quot; rel=&quot;nofollow noreferrer&quot;&gt;Air Foil Tools&lt;/a&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SFd2i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SFd2i.png&quot; alt=&quot;NACA0021-loverd&quot;&gt;&lt;/a&gt;&#xA;$c_l$ is the lift-coefficient which is defined as: $\frac{F_\mathrm{lift}}{\frac{1}{2}\rho u^2} = c_l$. The drag-coefficient ($c_d$) is defined in equal manner. The angle of attack is defined as &lt;em&gt;alpha&lt;/em&gt;. &lt;em&gt;Lift&lt;/em&gt; and &lt;em&gt;Drag&lt;/em&gt; both result from the aerodynamic forces acting on a profile (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Lift_(force)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; for additional illustrations and explanations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FhzMW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FhzMW.png&quot; alt=&quot;aeroforces&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of information contained in the three curves above. The two messages of interest here are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Within a certain range of AoA the lift changes proportional to the AoA.&lt;/li&gt;&#xA;&lt;li&gt;Within the same range of AoA the drag is considerable lower but increases over-proportional.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2.&lt;/strong&gt; A Wells Turbine consists (in its original form) a single blade row, which is rotating at a certain speed. The rotor-inflow-angle is a result of the vector addition of the axial inflow velocity and the rotational speed of the turbine rotor. The AoA is the difference of the inflow angle and the blade setting (i.e. angle between the symmetry-axis of the blade and a reference axis). The vector representation of the inflow is called a velocity triangle. This means for every pair of axial-velocity and circumferential-velocity the blade will have a different inflow-angle. And, because of the setting of the turbine blade it will have a different AoA.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When we define a simplified version of $\eta$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\eta = \frac{\text{WORK}}{\text{WORK }+\text{ LOSS}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see, that as long as the &lt;strong&gt;losses&lt;/strong&gt; are small and the change with AoA-variation is small (see loss-curves) the change in &lt;strong&gt;work&lt;/strong&gt; will dominate the equation. This results in a higher efficiency despite the increased losses because the work which can be extracted from the fluid increases faster than the loss increases. &lt;strong&gt;&lt;em&gt;But:&lt;/em&gt;&lt;/strong&gt; As it can be seen from the $c_l/c_d$-curve this is only true for a certain AoA-range. Outside of this range the separations will create over-proportional more loss. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are absolutely right to be confused. Because it seems to be strange that despite a larger separation (which is usually results in a loss) the efficiency actually increases. However, this is only true when the separations are small. Therefore correlation between separation-area and efficiency is not a causality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think of this problem in three stages. In the first stage we look at the blade profile. Here the &lt;strong&gt;losses&lt;/strong&gt; increase with higher AoA. In the second stage we also look at the &lt;strong&gt;lift&lt;/strong&gt;, and see that it also increases and increases faster than the losses. In the third stage we see from the definition of the &lt;strong&gt;efficiency&lt;/strong&gt; that the combination of &lt;em&gt;lift&lt;/em&gt; and &lt;em&gt;drag&lt;/em&gt; mathematically allow for the parallel increase of efficiency and losses if the extracted work increases sufficiently. But this statement is only true for a certain range of AoA. Outside of this range the &lt;em&gt;losses&lt;/em&gt; are bigger and grow faster and the efficiency will not increase anymore and even decrease.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-09-13T16:50:26.050" LastActivityDate="2015-09-13T16:50:26.050" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5388" PostTypeId="2" ParentId="2305" CreationDate="2015-09-11T16:49:32.400" Score="1" Body="&lt;p&gt;I think @regdoug is on the right track, but this is how I think of it:  The work done by the compressor is equal to the difference in enthalpy of the ammonia between the compressor inlet and outlet; $W_c=H_{out}-H_{in}$  On the following pressure-enthalpy diagram, points 2 and 3 are the states at the compressor inlet and outlet, respectively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fDbM1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fDbM1.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the fan speed in the condenser is increased, it will draw more heat out of the ammonia, which will cause the temperature and pressure of the ammonia in the condenser to drop.  Consequently, the enthalpy of the ammonia will also drop, as indicated by the dashed line in the pressure-enthalpy diagram (points 3' and 4').  Lower outlet enthalpy of the ammonia results in less work done by the compressor, according to the previous equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, the other points in the diagram are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-&gt;2 Evaporator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2-&gt;3 Compressor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3-&gt;4 Condenser&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4-&gt;1 Throttle&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-09-11T16:49:32.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5393" PostTypeId="1" CreationDate="2015-09-12T22:28:32.547" Score="8" ViewCount="7212" Body="&lt;p&gt;I'm simulating a turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The convergence of solution is monitored by checking the residuals of the numerically solved governing equations. &#xA;The &quot;viscous-spalarat-almaras&quot; model is used, and used air with constant density. Pressure-velocity coupled scheme and pressure based solver are selected.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the following figures, torque is increasing but the residual of continuity does not decrease and is almost constant. T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gK9DL.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gK9DL.jpg&quot; alt=&quot;Image 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://0up.ir/do.php?imgf=Capture_99287.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aSIXT.jpg&quot; alt=&quot;Image 2&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://0up.ir/do.php?imgf=Capture333.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zMnPg.jpg&quot; alt=&quot;Image 3&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to monitored torque, the solution is not converged.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2015-09-17T10:21:51.047" LastActivityDate="2015-09-17T10:21:51.047" Title="Residuals of continuity not converging" Tags="|fluid-mechanics|simulation|turbines|ansys|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5394" PostTypeId="2" ParentId="55" CreationDate="2015-09-13T03:46:55.093" Score="0" Body="&lt;p&gt;I have this 150 Watt panel, 12 volt(said to be). In testing it with two 100 Watt (equal too regular 100 watt old kind) bulbs used for lighting the room. This one panel can light up two other LED bulbs from the two on the ceiling. WOW, I thought that was impossible. That you only get 10% from the panel. Not so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0YCfX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0YCfXm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/unYNP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/unYNPm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two of these to power solar panel.&#xA;The LEDs look a little dimmer then normal. But the panel is only getting 11.61 Volts (about).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugged LEDs in a 12 volt power supply and there's a big difference in lighting strength.&lt;/p&gt;&#xA;" OwnerUserId="2988" LastEditorUserId="272" LastEditDate="2015-09-13T12:24:22.443" LastActivityDate="2015-09-13T12:24:22.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5396" PostTypeId="1" AcceptedAnswerId="5406" CreationDate="2015-09-13T14:35:18.607" Score="4" ViewCount="248" Body="&lt;p&gt;Let us assume you have a large forging press or drophammer, and drawings for a set of smaller parts that all need to be forged from the same material.  Would it be feasible to construct a forging die so that all the parts could be forged from a single billet and then separated as a machining step?  Or is this more expensive than making a forging die for each part and forging them separately from individual, smaller billets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of this question, you can assume that the parts all have a linear geometry -- say, if you wanted to forge a crankshaft and camshaft set for an engine in a single operation.&lt;/p&gt;&#xA;" OwnerUserId="964" LastEditorUserId="964" LastEditDate="2015-09-14T02:50:09.043" LastActivityDate="2015-09-14T16:09:40.013" Title="Is it feasible to forge multiple, smaller parts in the same forging operation?" Tags="|manufacturing-engineering|forging|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5397" PostTypeId="2" ParentId="5344" CreationDate="2015-09-13T16:03:00.737" Score="0" Body="&lt;p&gt;For a single bolt clamping one middle plate between two side plates (assume both of the same thickness but not necessarily the same as the middle plate), the joint can fail by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;shearing of the bolt, but it is double shear, so double the bolt&#xA;area.&lt;/li&gt;&#xA;&lt;li&gt;bearing (i.e crushing) failure of the hole in the middle plate alone, so bolt diameter x&#xA;plate thickness.&lt;/li&gt;&#xA;&lt;li&gt;bearing failure of the side plates but both have to&#xA;fail for the joint to fail, so double the bearing area of one hole.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The essential question you should ask is 'what area of material must fail for the whole joint to fail?'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must also check for the possibility of tensile failure of the plates:  if the holes are large compared to the plate width, this might be the failure mode.&lt;/p&gt;&#xA;" OwnerUserId="2520" LastEditorUserId="2520" LastEditDate="2015-09-13T16:08:47.237" LastActivityDate="2015-09-13T16:08:47.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5400" PostTypeId="2" ParentId="179" CreationDate="2015-09-14T07:47:03.873" Score="0" Body="&lt;p&gt;There is some more information on this in Ladislav Fryba's book &quot;Dynamics of Railway Bridges&quot; (1996). If you read chapter 4, you will see formula 4.53 on page 92:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_1 = 17.753 v_{st}^{-1/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $f_1$ being the first natural frequency in Hertz and $v_{st}$ the midspan deflection in mm. This is exacly the formula you are asking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation follows from the formula for the midspan deflection of a simply supported beam loaded by a uniformly distributed load &lt;em&gt;μg&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{st} = {5 \above 1pt 384} {\mu gl^4 \above 1pt EI} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is substituted in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_j = {\lambda_j^4 \above 1pt l^4} ({EI \above 1pt \mu})^{1/2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It yields $$ \lambda_1 = \pi $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting those equations into each other using g = 9.81 m/s^2 gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_1 = {\pi \above 1pt 2} ({5 \above 1pt 384} g)^{1/2}  v_{st}^{-1/2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerical evaluation of this equation yields the desired equation.&lt;/p&gt;&#xA;" OwnerUserId="2996" LastEditorUserId="2996" LastEditDate="2015-09-14T13:04:30.707" LastActivityDate="2015-09-14T13:04:30.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5401" PostTypeId="5" CreationDate="2015-09-14T08:26:01.563" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-09-14T08:26:01.563" LastActivityDate="2015-09-14T08:26:01.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5402" PostTypeId="4" CreationDate="2015-09-14T08:26:01.563" Score="0" Body="Computer hardware and software which forms a component of some larger system. They are expected to perform functions with relation time constrains, without human intervention. Embedded system can range from simple MP3 player to complex avionics system. " OwnerUserId="1116" LastEditorUserId="1116" LastEditDate="2015-09-14T12:26:21.127" LastActivityDate="2015-09-14T12:26:21.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5403" PostTypeId="1" CreationDate="2015-09-14T09:46:14.940" Score="1" ViewCount="1647" Body="&lt;p&gt;If I were to fully fill a sphere with water without any air trapped inside, how to find the pressure of the water at, say, the center of the sphere? &lt;/p&gt;&#xA;" OwnerUserId="2997" LastEditorUserId="272" LastEditDate="2015-09-14T13:22:49.640" LastActivityDate="2015-09-14T13:22:49.640" Title="Water pressure in a filled sphere" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5404" PostTypeId="2" ParentId="5403" CreationDate="2015-09-14T10:29:17.093" Score="7" Body="&lt;p&gt;$$p = p_0 + \rho g \frac{d_{sphere}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_0$ would be 1 Bar I assume, you are not specific under what conditions you fill the sphere and I assume the spheres border are made from a solid material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is just that the pressure is not dependent on the shape of the container. You just calculate the height of the water column for you desired point, which for the center would be $\frac{d_{sphere}}{2}$.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2015-09-14T10:29:17.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5405" PostTypeId="2" ParentId="5403" CreationDate="2015-09-14T11:15:34.127" Score="4" Body="&lt;p&gt;The pressure at the center would be the pressure at the top plus the extra due to the depth of one radius.  The &lt;i&gt;difference&lt;/i&gt; in pressure due to height is the same inside the sphere as water anywhere else in the same gravity (using the simplifying assumption that water is incompressible, which is quite valid for most human-scale circumstances on earth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty you have is deciding what the pressure is at the top of the sphere.  If there is a little hole at the top so that the pressure is atmospheric, then it's simple.  Otherwise, you have to know how much the sphere is pressurizing the water, which you haven't told us anything about.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-09-14T11:15:34.127" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5406" PostTypeId="2" ParentId="5396" CreationDate="2015-09-14T16:09:40.013" Score="5" Body="&lt;p&gt;I've previously worked in the forging industry and you can certainly forge multiple parts in a single stroke of a large press but there are problems.   The obvious ones are the size of the available press and the economics of the operation.   However, even from a metallurgical point of view, forging a complex part like a crankshaft in one stroke would induce more plastic deformation than the material could take, at least at room temperature.   The shaft would then have to be hot-forged to maintain the ductility of the material during deformation and the temperature required is critical.   I can't imagine that you could do a crankshaft and a camshaft at the same time.   And apart from that, complex parts are usually forged in stages, each stage taking the component closer to its final shape.   Again, I can't see this being done for a crankshaft and a camshaft in a single die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I could be out-of-date...&lt;/p&gt;&#xA;" OwnerUserId="2520" LastActivityDate="2015-09-14T16:09:40.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5407" PostTypeId="1" CreationDate="2015-09-14T17:25:07.933" Score="4" ViewCount="1459" Body="&lt;p&gt;What factors affect the penetration depth and weld deposition rate in arc welding?&#xA;How does polarity affect the penetration depth and weld deposition rate?&lt;br&gt;&#xA;My lecturer said that penetration depth is higher when using straight polarity when compared to reverse polarity.&#xA;But I've read in a textbook that it's the opposite. &lt;/p&gt;&#xA;" OwnerUserId="21" LastEditorUserId="21" LastEditDate="2015-09-15T00:36:43.257" LastActivityDate="2016-08-09T12:12:16.597" Title="Penetration depth and weld deposition rate in Welding" Tags="|mechanical-engineering|welding|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5408" PostTypeId="1" AcceptedAnswerId="5409" CreationDate="2015-09-14T18:46:12.063" Score="2" ViewCount="292" Body="&lt;p&gt;Just a short question: Is there any usefulness in doing block diagram reduction piecewise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I am asking is that I find it much (!) easier to just find the final $\frac{output}{input}$ transfer function using mathematics (the brute force method), but my professor does it stepwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am to do it stepwise, I still feel like I have to do it mathematically, rendering the stepwise reduction pretty useless as far as I can tell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this seems like a banal question, but I cannot find anything answering this question is my controls book.&lt;/p&gt;&#xA;" OwnerUserId="1721" LastActivityDate="2015-09-14T20:10:26.697" Title="Block Diagram Reduction: Is it necessary to do it stepwise?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5409" PostTypeId="2" ParentId="5408" CreationDate="2015-09-14T20:10:26.697" Score="3" Body="&lt;p&gt;When you are solving a problem in the real world, there are 2 main important requirements&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Get the correct answer. &lt;/li&gt;&#xA;&lt;li&gt;Solve it using sound methods. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There might be a half-dozen different ways to solve a given problem, so long as you arrive at the correct answer and don't pick a method involving rain dances or Satanic rituals, you are free to pick the method that you find easiest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the requirements in academia are often different. There is a good chance that your professor has a reason for using the method he does. If he asks you to solve a problem step-wise, you should do what he asks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's likely that, at a certain point, doing it directly rather than piecemeal becomes overly complex and involves massive equations. At this point, knowing how to do it step-wise becomes necessary, and your professor is preparing you for this. But the only way to know for sure is to ask him. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said, beyond this class, the only thing that will matter is getting the correct answer and being able to explain how you arrived at that answer, so if you get the same answer through both methods, both are equally valid. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-09-14T20:10:26.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5410" PostTypeId="1" AcceptedAnswerId="5441" CreationDate="2015-09-14T21:09:45.450" Score="6" ViewCount="614" Body="&lt;p&gt;I want to simulate a turbine in Fluent. For this reason, I studied some papers. In these papers, Turbulence is modeled using the realizable k–e model and Non-equilibrium wall functions are used in the near wall flow modeling. There are the following sentences in these papers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Boussinesq approach is applied to relate the Reynolds stresses to the mean velocity gradient.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The Reynolds stress is related to the mean velocity gradient by means of the Boussinesq approach. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the exact meaning of above sentences? And is there a direct option in Fluent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any adjustment that must be made in Fluent before starting the calculation for this approach?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="368" LastEditDate="2015-09-15T18:49:51.177" LastActivityDate="2020-08-18T08:44:51.240" Title="What is the Boussinesq approach?" Tags="|mechanical-engineering|fluid-mechanics|turbines|cfd|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5411" PostTypeId="1" AcceptedAnswerId="5426" CreationDate="2015-09-14T21:37:19.430" Score="4" ViewCount="6372" Body="&lt;p&gt;While studying some papers,  I didn't understand the following sentences:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;code&gt;deltP&lt;/code&gt; the pressure drop across the turbine, evaluated using the mass average values at the inlet and outlet sections of the computational domain, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the exact meaning of the &quot;mass average values&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And is there such a option of &quot;mass average values&quot; in Fluent?  I dont know that how to compute this parameter in Fluent.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="1293" LastEditDate="2015-09-15T16:24:00.730" LastActivityDate="2015-09-17T15:54:42.750" Title="Understanding pressure drop and mass average values" Tags="|mechanical-engineering|pressure|cfd|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5412" PostTypeId="2" ParentId="693" CreationDate="2015-09-14T21:49:29.390" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.automationdirect.com/static/specs/encstratusheatexchangers.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.automationdirect.com/static/specs/encstratusheatexchangers.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this document, heat load transfer through the walls of a metal enclosure is estimated in BTU/H as 1.25 x surface area (sq ft) x temperature differential (degrees F).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting to W/degC, you would get about .66 times the surface area of the box in square feet, or 7.1 times the surface area in square meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in degC/W, that's 1.52 ft^2, 218.9 in^2, or .14 m^2. Divide by the surface area to get the thermal resistance of the box.&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="18" LastEditDate="2015-09-15T12:00:20.543" LastActivityDate="2015-09-15T12:00:20.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5413" PostTypeId="2" ParentId="2172" CreationDate="2015-09-15T03:40:16.580" Score="2" Body="&lt;p&gt;I am unsure about the use of graphite or other materials for this purpose. However, kiln wash is a low cost solution that should give you the same quality as kiln shelf paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.bigceramicstore.com/info/ceramics/tips/tip96_kiln_wash.html&quot; rel=&quot;nofollow&quot;&gt;http://www.bigceramicstore.com/info/ceramics/tips/tip96_kiln_wash.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ceramicartsdaily.org/firing-techniques/soda-firing/the-many-layers-of-kiln-wash-how-to-find-the-best-kiln-wash-for-your-firing-temperature-and-methods/&quot; rel=&quot;nofollow&quot;&gt;http://ceramicartsdaily.org/firing-techniques/soda-firing/the-many-layers-of-kiln-wash-how-to-find-the-best-kiln-wash-for-your-firing-temperature-and-methods/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-09-15T03:40:16.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5414" PostTypeId="2" ParentId="663" CreationDate="2015-09-15T03:55:13.703" Score="1" Body="&lt;p&gt;I am not an expert in this field, but epoxy coating, a wired sacrificial anode (as opposed to being coated directly), or active imposed current may be good solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.anchorguard.com/reference_understand_how2.cfm&quot; rel=&quot;nofollow&quot;&gt;http://www.anchorguard.com/reference_understand_how2.cfm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-09-15T03:55:13.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5415" PostTypeId="2" ParentId="4279" CreationDate="2015-09-15T04:04:44.470" Score="2" Body="&lt;p&gt;The main concern in the design is the exposure of a high tolerance surface to the elements. Both systems can be made very robust. The hydraulic system would need a shroud to protect the rod from damage. Additional rod wiper rings would be a good idea too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electrical motor with gearbox would be very easy to protect and would be the most reliable if it is only transmitting rotation motion. If it is transmitting linear motion like a hydraulic piston the lead screw or ball screw will need to be protected in the same way as a hydraulic cylinder. It will require more dust protection to have the same reliability as a hydraulic cylinder.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-09-15T04:04:44.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5416" PostTypeId="2" ParentId="1789" CreationDate="2015-09-15T04:13:25.137" Score="2" Body="&lt;p&gt;It is a very broad question as there are many design considerations involved, but a good place to start is looking at BoPET(mylar) because it has a high strength to weight ratio and can be made very thin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clear mylar on the outside, possibly reflective mylar on the north, and black mylar on the inside. There are lots of other ways to optimize once you have clearer picture of the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/BoPET&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/BoPET&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-09-15T04:13:25.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5417" PostTypeId="2" ParentId="4074" CreationDate="2015-09-15T04:33:39.010" Score="2" Body="&lt;p&gt;Anything can dry with enough time and in the right conditions. The mattress left in 65% humidity will eventually be full of air at 65% humidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger question in most real world scenarios is time to dry; because like in this situation the object will mold within days and will take weeks or months to be low enough moisture not to mold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most economical solution would be vacuum bagging the entire mattress and pulling vacuum on it for probably a week. This would help inhibit mold growth (not guaranteeing that) and will accelerate the drying process. In an industrial setting a larger vacuum pump could be used and the drying process could occur much faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record, gravity has no influence on evaporation, they are separate areas of science. Gravity may make one side more wet by transporting via capillaries; which is perhaps a design consideration but is not related via science. The capillary structure, the material's gas permeability, temperature, pressure, and relative humidity are the primary factors that influence drying time.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2016-01-15T00:27:06.587" LastActivityDate="2016-01-15T00:27:06.587" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5418" PostTypeId="2" ParentId="3192" CreationDate="2015-09-15T04:43:22.880" Score="1" Body="&lt;p&gt;The vocabulary word you are looking for is &quot;Color Buffing&quot;; which is the final buffing process of polishing in a single direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=gyOodLNI4BQ&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=gyOodLNI4BQ&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.pjtool.com/buffing-polishing-terms.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.pjtool.com/buffing-polishing-terms.aspx&lt;/a&gt;&#xA;&lt;a href=&quot;http://expha.com/articles/13/&quot; rel=&quot;nofollow&quot;&gt;http://expha.com/articles/13/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-09-15T04:43:22.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5419" PostTypeId="1" CreationDate="2015-09-15T07:02:48.203" Score="4" ViewCount="49" Body="&lt;p&gt;A pump is connected to a glass tube to evacuate it. The reading shows 0.4 mbar when the pump is on because of a vacuum. As soon as the pump is turned off, the pressure in the tube becomes equal to atmospheric pressure.&lt;/p&gt;&#xA;" OwnerUserId="3004" LastEditorUserId="90" LastEditDate="2015-09-15T07:44:00.853" LastActivityDate="2015-09-15T07:44:00.853" Title="What is the reason for the instant loss of a vaccum through the sealed glass tube after shuting off a pump?" Tags="|pumps|vacuum|" AnswerCount="0" CommentCount="3" ClosedDate="2015-11-25T19:21:00.027" ContentLicense="CC BY-SA 3.0" />
  <row Id="5420" PostTypeId="2" ParentId="3192" CreationDate="2015-09-15T08:06:08.597" Score="0" Body="&lt;p&gt;I believe the &lt;a href=&quot;https://en.wikipedia.org/wiki/Random_orbital_sander&quot; rel=&quot;nofollow&quot;&gt;random orbital sander&lt;/a&gt; tool was invented exactly for this - it slides/vibrates while it spins &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This ensures that no single part of the abrasive material travels the same path twice during the same rotation. Because of this random sanding action, the tool does not leave swirl marks.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-09-15T08:06:08.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5421" PostTypeId="1" CreationDate="2015-09-15T11:40:17.213" Score="2" ViewCount="673" Body="&lt;p&gt;In a heating circuit, there's two (known to me) ways to decouple primary flow (through heat source) and the secondary flows (to consumers):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a hydraulic switch in the heating manifold&lt;/li&gt;&#xA;&lt;li&gt;complete hydraulic separation with a heat exchanger&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When is which solution more appropriate?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="90" LastEditDate="2015-09-15T12:37:49.597" LastActivityDate="2017-11-17T19:53:19.763" Title="Hydraulic switch vs heat exchanger" Tags="|heat-transfer|hvac|heating-systems|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5422" PostTypeId="1" AcceptedAnswerId="5447" CreationDate="2015-09-15T11:50:55.000" Score="2" ViewCount="79" Body="&lt;p&gt;I have gas pipeline with biogas, low pressure - approx 40mbar over atmosphere. I want to remove a small amount of condensate, inside a building, without creating an &lt;a href=&quot;http://www.engineeringtoolbox.com/hazardous-areas-classification-d_395.html&quot; rel=&quot;nofollow&quot;&gt;ex-zone&lt;/a&gt;. Small amount: I honestly don't know but it should be next to nothing. But I create a low point and want to be sure I don't block my pipeline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a valve that I can build into the pipeline at the low point, that allows me to ex-filtrate gas without creating an ex-zone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No need for an actual supplier, but some terms to aid my googling my writing of the RFQ would be great. Plus the technical info to decide if it's the right gadget for my problem!&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="272" LastEditDate="2015-10-16T12:27:18.683" LastActivityDate="2015-11-15T12:50:14.523" Title="Device to remove condensate, without creating a hazard (EX) zone" Tags="|mechanical-engineering|pipelines|safety|valves|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5423" PostTypeId="2" ParentId="5407" CreationDate="2015-09-15T12:38:49.567" Score="2" Body="&lt;p&gt;Quick answer, you are both right and wrong. Arc Welding has many types and for Shielded Metal Arc Welding (SMAW), Gas Metal Arc Welding (GMAW), Flux Cored Arc Welding (FCAW) and SAW processes electrode positive polarity increases depth but for Gas Tungsten Arc Welding (GTAW) electrode negative polarity increases penetration depth. I'm not at the office so can't give reference from books but &lt;a href=&quot;http://www.lincolnelectric.com/en-us/support/process-and-theory/Pages/variables-weld-penetration.aspx&quot; rel=&quot;nofollow&quot;&gt;here is a page from Lincoln Electric &lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="879" LastActivityDate="2015-09-15T12:38:49.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5424" PostTypeId="2" ParentId="5411" CreationDate="2015-09-15T12:51:56.897" Score="1" Body="&lt;p&gt;I think they are telling you that they are considering the bulk, or &lt;em&gt;macroscopic&lt;/em&gt;, properties of the fluid, and not the properties of the individual molecules. This is an important assumption in fluid mechanics, known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fluid_mechanics#/Continuum_hypothesis&quot; rel=&quot;nofollow&quot;&gt;Continuum Hypothesis&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the molecular level, tiny, rapid fluctuations in the fluid's physical properties such as velocity and density occur as the molecules move around, bumping into each other (e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Brownian_motion&quot; rel=&quot;nofollow&quot;&gt;Brownian motion&lt;/a&gt;). These fluctuations would be impossible to predict and model accurately, so fluid mechanics often treats fluids as continuous media with an average velocity, average viscosity, average pressure, etc, on the assumption that the sum of the fluctuations, across millions or billions of particles, cancel out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This greatly simplifies most fluid mechanics problems and is a valid assumption for most practical problems.&lt;/p&gt;&#xA;" OwnerUserId="2551" LastActivityDate="2015-09-15T12:51:56.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5425" PostTypeId="2" ParentId="5411" CreationDate="2015-09-15T14:03:03.227" Score="1" Body="&lt;p&gt;While I haven't heard this term specifically, it may be telling you to use &lt;em&gt;specific&lt;/em&gt; properties (those expressed in terms of mass) instead of non-mass-adjusted properties. For example, you would use the specific volume (${m^3}/{kg}$, the inverse of density) in calculating the pressure difference, because the actual volume has little meaning in a moving system. To me, this seems pretty basic, as I rarely used anything but specific values in my fluids and thermo classes, but I can see an academic paper wanting to be thorough in explaining their exact method. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-09-15T14:03:03.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5426" PostTypeId="2" ParentId="5411" CreationDate="2015-09-15T14:35:21.187" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What is the exact meaning of the &amp;quot;mass average values&amp;quot;?&#xA;And is there such a option of &amp;quot;mass average values&amp;quot; in Fluent? I don't know that how to compute this parameter in Fluent.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes, FLUENT has many surface integral reports like Area-weighted average and Mass average, quoting from FLUENT 6 user manual:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;&lt;strong&gt;Area-weighted average:&lt;/strong&gt;&lt;/em&gt; You can find the average value on a solid    surface, such as the average heat flux on a heated wall with&#xA;a    specified temperature.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;&lt;strong&gt;Mass average:&lt;/strong&gt;&lt;/em&gt; You can find the average value    on a surface in the flow, such as average enthalpy at a velocity inlet.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And this is an answer from &lt;a href=&quot;http://www.cfd-online.com/Forums/cfx/18663-mass-area-average.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;cfd-online&lt;/a&gt; about the difference between them:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The area average of a scalar is calculated by integrating the scalar times the area divided by the total area over the region. A mass averaged quantity is obtained by integrating the scalar time mass flow divided by total mass flow over the region.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\text{Area Average} = \frac{\int \phi \, dA}{A}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\text{Mass Average} = \frac{\int \phi \, dm^.}{m^.}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; is an arbitary scalar property of the flow.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-09-15T14:35:21.187" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5427" PostTypeId="2" ParentId="5411" CreationDate="2015-09-15T14:36:08.043" Score="1" Body="&lt;p&gt;If the analysis on the turbine is two- or three-dimensional, then the pressure can vary across the transverse and/or circumferential dimensions as well as between the entrance/exit.  Think about the air entering at the center of the inlet duct possibly having a different pressure than the air entering at the edge of the inlet duct.  In this case, you don't have just one pressure at the inlet and one at the outlet, you have a &lt;em&gt;pressure field&lt;/em&gt; at each.  So, the author(s) take an average of the pressure at the inlet and outlet, and use those averages to calculate the pressure difference.  The term &quot;mass-average&quot; in this context in analagous to a weighted-average.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-09-15T14:36:08.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5428" PostTypeId="2" ParentId="2121" CreationDate="2015-09-15T14:50:51.823" Score="1" Body="&lt;p&gt;I don't believe there is any fundamental difference between intentionally frozen and naturally frozen soils. Other than control of the temperature.&lt;/p&gt;&#xA;&lt;p&gt;I'm afraid the mechanics of soils (frozen or otherwise) is complex and depends on many factors such as the soil type, water content, grain-size distribution, compaction, etc. The soil properties used for design purposes are generally obtained through a combination of testing and empirically derived relationships. I don't think it would be appropriate for me to attempt to summarize the entire subject here. If you would like to know more details, this MIT OpenCourseware course may be helpful: &lt;a href=&quot;http://ocw.mit.edu/courses/civil-and-environmental-engineering/1-322-soil-behavior-spring-2005/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MIT - Soil Behavior&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;With regards to your question specifically about frozen soils. Every soil will have different properties when frozen depending on the unfrozen properties as well as the temperature, ice content, porosity, etc. Your intuition that the frozen soil will be stiffer than the unfrozen soil is correct. Frozen soil can be 10-100 times stiffer than the unfrozen soil &lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This book deals entirely with the design of frozen soils, including a chapter on Construction Ground Freezing, which was what you were inquiring about:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Frozen Ground Engineering (Second edition). Orlando B. Andersland and Branko Ladanyi. Chichester: John Wiley &amp;amp; Sons. 2004.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Yershov, E. D.. General Geocryology. Ed. Peter J. Williams. 1st ed. Cambridge: Cambridge University Press, 1998.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="-1" LastEditorDisplayName="user16" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-09-15T16:23:32.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5429" PostTypeId="1" CreationDate="2015-09-15T14:57:55.577" Score="4" ViewCount="469" Body="&lt;p&gt;So imagine we have a house with some isolation and some thermal inertia. This house also has a heating system that turns on (starts heating) when the observed temperature curve goes below some predefined temperature threshold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured that in the scenario where the heating system is off (the observed temperature is above the threshold the temperature decrease will give us information on the thermal isolation and inertia. We need of course to take in to account the external temperature, the wind, and the sun irradiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there was a way to dissociate the information of the thermal isolation and the information of the thermal inertia. Clearly, if we had information about the thermal power loss it would give an indication but unfortunately we don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data I have :  Observed inside temperature and desired inside temperature, outside temperature, wind, humidity from a weather station in the same city (so I guess only the external temperature can be used)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objective: Determine the conductivity and thermal mass of the house for sales purposes (sell more isolation, help them master energy efficient behavior  etc)&lt;/p&gt;&#xA;" OwnerUserId="3011" LastEditorUserId="1586" LastEditDate="2015-12-23T20:06:45.683" LastActivityDate="2015-12-23T20:06:45.683" Title="How to extract a building's heat isolation and thermal inertia based on temperature curves?" Tags="|thermodynamics|temperature|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5430" PostTypeId="2" ParentId="5410" CreationDate="2015-09-15T15:19:15.623" Score="5" Body="&lt;p&gt;To take into account the effect of turbulent fluctuations on a flow field, the Navier-Stokes equations are modified to include such effects. The obtained equations are called Reynolds-averaged Navier-Stokes (RANS) equations.&lt;/p&gt;&#xA;&lt;p&gt;As an example, the steady incompressible momentum equation can be written as follows (in &lt;a href=&quot;http://www.cfd-online.com/Wiki/Einstein_summation_convention&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein tensor notation)&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \rho \frac{\partial u_i}{\partial t} + \rho u_i \frac{\partial u_i}{\partial x_j} = -\frac{\partial p}{\partial x_i} + \frac{\partial t_{ij}}{\partial xj}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$t_{ij}$&lt;/span&gt; is the viscous stress tensor defined as:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ t_{ij} = 2\mu s_{ij}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; is viscosity and &lt;span class=&quot;math-container&quot;&gt;$s_{ij}$&lt;/span&gt; is the strain-rate tensor:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ s_{ij} = \frac{1}{2}(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i})$$&lt;/span&gt;&#xA;After time-averaging, one can obtain the following equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \rho \frac{\partial U_i}{\partial t} + \rho U_i \frac{\partial U_i}{\partial x_j} = -\frac{\partial P}{\partial x_i} + \frac{\partial }{\partial xj}(2\mu s_{ij} - \overline{\rho {u_j}'{u_i}'})$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The resulted term &lt;span class=&quot;math-container&quot;&gt;$- \overline{\rho {u_j}`{u_i}`}$&lt;/span&gt; is called the Reynolds-stress tensor (or turbulent shear stress), which is the basis of all turbulence models.&lt;/p&gt;&#xA;&lt;p&gt;The Boussinesq approximation is merely assuming that turbulent shear stress is analogous to viscous shear stress (by introducing a new term called eddy viscosity &lt;span class=&quot;math-container&quot;&gt;${\mu}_t$&lt;/span&gt;). Thus, one can write:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ - \overline{\rho {u_j}'{u_i}'} = {\mu}_t \frac{\partial U_i}{\partial x_j}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The approximation didn't solve the closure problem of turbulence, but it was used to come up with turbulent models that can model that eddy viscosity. By definition it's not a complete model, so there is no direct option for this in Fluent.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2020-08-18T08:44:51.240" LastActivityDate="2020-08-18T08:44:51.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="5431" PostTypeId="1" AcceptedAnswerId="5442" CreationDate="2015-09-15T16:06:10.750" Score="1" ViewCount="1632" Body="&lt;p&gt;Consider a centrifugal compressor. The compressor map usually plots the adiabatic head vs volumetric inlet flow for rated conditions of pressure, molecular weight and temperature at a given speed. But when pressure, temperature, molecular weight and speed are fixed, how can the inlet flow be made to vary ? Is it accomplished through the use of inlet vanes ?&lt;/p&gt;&#xA;" OwnerUserId="1181" LastEditorUserId="282" LastEditDate="2015-09-19T19:45:02.833" LastActivityDate="2015-09-19T19:45:02.833" Title="How can flow in a centrifugal compressor at fixed conditions vary?" Tags="|mechanical-engineering|compressors|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5432" PostTypeId="2" ParentId="2551" CreationDate="2015-09-15T16:52:27.117" Score="0" Body="&lt;p&gt;You could use electrically-initiated pyrotechnics like in &lt;a href=&quot;https://www.youtube.com/watch?v=MPOiDQ1-LuA&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt;.  They spontaneously ignite when electric current is applied, and immediately self-extinguish when current is removed.  The pyrotechnics would be on one side of a piston and the gas to be compressed would be on the other side.  When pressure falls below a certain threshold, you fire up the pyrotechnics for a bit until the desired pressure is reestablished, then turn them off.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While 10 GPa is rather high, realize that some military and research guns go as high as 1 GPa (granted, for only a few milliseconds at a time).  10 GPa is not unreachable, it would just require a vessel with very thick walls.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-09-15T16:52:27.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5433" PostTypeId="2" ParentId="5431" CreationDate="2015-09-15T17:56:38.497" Score="2" Body="&lt;p&gt;The generic equation for mass flow through a cross-section of area $A$, traveling at velocity $V$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot m = \rho \, V \, A&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ is the density of the fluid and $\dot m$ is the mass flow rate.  If the molecular weight, temperature, and pressure of the fluid are fixed, then density is fixed too.  That leaves $V$ and $A$ to play with.  Since $V$ will be largely determined by the compressor rotation speed, all you have left is to change the cross-sectional area of the inlet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the area for a particular flow rate is a bit more difficult in practice however, since the fluid velocity will want to increase as the area is increased.  Also, if the compressor is going to pump liquid, then there isn't much that can be done to alter the flow rate without incurring cavitation in the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is another alternative; you can feed back some of the outlet from the compressor to the inlet, so that the net outlet of the pump is reduced.  You could control the fraction of feed back (a.k.a. reflux) with a valve.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-09-15T18:05:52.957" LastActivityDate="2015-09-15T18:05:52.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5434" PostTypeId="2" ParentId="5330" CreationDate="2015-09-15T18:19:11.930" Score="2" Body="&lt;p&gt;The approximate equation for the torsional constant, $J$ of a thin-walled square tube is: $$J = tb^3$$ Where t is the wall thickness and b is the width between the wall centerlines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this equation you can verify that SkyCiv is ignoring the corner radius when calculating J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no simple closed-form solution for $J$ with rounded corners, but we &lt;em&gt;can&lt;/em&gt; observe that accounting for the corner radii increases $J$. Where the rectilinear corners produced stress concentrations, the curved corners produce a more 'uniform' torsional stress flow, and therefore a torsionally stiffer section. (As least, I think that's a valid mental model.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read the full AISC discussion of torsion in &lt;a href=&quot;http://academic.csuohio.edu/duffy_s/AISC_DG_9_Torsion.pdf&quot; rel=&quot;nofollow&quot;&gt;Design Guide 9&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://deepblue.lib.umich.edu/bitstream/handle/2027.42/4748/bab4733.0001.001.pdf?sequence=5&amp;amp;isAllowed=y&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; by Darwish and Johnston has a fairly extensive discussion of the effect of corner radii on torsional constants.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-09-15T18:19:11.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5435" PostTypeId="2" ParentId="4255" CreationDate="2015-09-15T19:10:02.400" Score="1" Body="&lt;p&gt;The cord from the truck to the first pulley is on a diagonal, so the length of that section does not increase linearly with the trucks motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L=-2X_A+\sqrt{{Y_t}^2+C^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $Y_t$ is the horizontal distance from the pulley to the truck, and C is the vertical distance. Defined by:$\frac{Y_t(t=0)}{\cos(\beta(t=0))}=\frac{C}{\sin(\beta(t=0))}=10m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating with respect to time yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=-2V_A+\frac{V_t\,Y_t}{\sqrt{Y_t^2+C^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And differentiating a second time yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=-2a_A+a_t\frac{C^2Y_t+{Y_t}^3}{({Y_t}^2+C^2)^\frac32}+\frac{C^2{V_t}^2}{({Y_t}^2+C^2)^\frac32}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there one can solve for the acceleration of the block, and plug in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y_t(t=0)\approx7.9m$$&#xA;$$C\approx6.1m$$&#xA;$$a_A(t=0)\approx3.2\frac{m}{s^2}$$&#xA;$$T(t=0)\approx540.6N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly there's a greater contribution to the acceleration from the angle of the cord changing due to the velocity (the ${V_t}^2$ term), then there is from the actual acceleration of the truck (the $a_t$ term). This makes sense if you think about how fast that truck is moving relative to the dimensions in the problem, and how relatively slowly the truck is accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once one has the tension, the total thrust of the truck (ie the wheel trust minus all losses other than tension) will just be the net force in the horizontal direction plus the horizontal tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{net}=\text{Thrust}-T\cos(\beta)=m_t\,a_t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Thrust}\approx 4825N$$&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-09-15T19:10:02.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5437" PostTypeId="1" CreationDate="2015-09-15T19:31:13.253" Score="1" ViewCount="3534" Body="&lt;p&gt;I am simulating an axial flow air turbine (the Wells turbine for extracting wave energy) using MRF. The problem is when i calculate the blade torque, it gives a value close to zero!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is converged well but the problem is with the moment report. my torque is very low . I tried different mesh and all give me almost the same number (which is very low compared to experiments). Is there any adjustment that must be made in fluent before starting the calculations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I divided the mesh into two domains. the first domain is the piping before and after the The turbine and the second domain contains the turbine blades. the domain containing the blades rotates at 2000 rpm while the other one is stationary. I meshed the geometry using Gambit. I assigned a negative rotational speed (2000) to the walls above the blades to maintain the turbine casing of the rotating domain stationary relative to the blades. the turbulence model is k-epsilon high Reynolds number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mentioned turbine have three major components(blade-hub-casing) that these components should be set in Fluent. I performed different meshes and Many attempts have been performed in order to construct a high quality mesh, especially at the blade tip clearance zone(please see second figure), but the results is not validated with experiments.I should validate three non-dimensional coefficients as follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TN5Fl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TN5Fl.jpg&quot; alt=&quot;enter link description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This turbine have tip gap(other its name is tip clearance), please see the following figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ekV0j.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ekV0j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I divided the mesh two domain and one material(material is air=default). This turbine have 8 blades((360/8)=45 deg=angle between blades). Due to the circumferential symmetry, only one eight of the annulus has been computed, &#xA;imposing periodic boundary conditions in the tangential direction. In my model, the computational domain has been restricted in the axial direction to four chord lengths upstream and eight chord lengths downstream of the blade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computational domain is as follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bUTf4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bUTf4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete image of turbine as follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e6y7d.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e6y7d.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I created  an interface between the rotating and non-rotating region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;boundary conditions and components and position of interfaces are as follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set up of Fluent is as follow:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I set hub (in the domain) to rotate with a rotational velocity of 0, in +y vector direction. also, I set the hub (in the frame) to rotate with an absolute velocity of 209.43(2000rpm), in the +y vector direction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Casing in stationary domain:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;velocity of zero,in +y vector direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;casing in rotary domain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rotational velocity of 0 ,in +y vector direction or a rotational velocity(209.43), to the motion of the turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for stability purposes it is better to set it to an rotational velocity of m/s, direction vector +y.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;inlet:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;absolute velocity of  m/s, normal to boundary.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;outlet:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;value of pressure is zero and I enabled option of radial equilibrium pressure distribution.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Operating pressure is default.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Periodic conditions set to +y vector direction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;*About calculation of tuque:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I computed tuque as follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Report&gt;Forces&gt;moments&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment center is 0,0,0; for the moment axis I choose the +y direction and wall zone is blade. Totally, apply print. And moment refers to 1 blade, because I simulate periodic: total moment in +y direction will be total moment for one blade multiplied by the number of blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According the following sentences in Ansys Fluent Help, Which do I choose? Absolute or fluid zone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O62qx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O62qx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Relative Specification indicates whether the velocities are absolute velocities (absolute) or velocities relative to the motion of each cell zone (Relative to Cell Zone). &lt;strong&gt;This selection is necessary only if your problem involves moving reference frames or sliding meshes.&lt;/strong&gt; If there is no zone motion, both options are equivalent.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do i need to adjust the reference values? &#xA;...........................................................&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thing that I should say is that, in order to judge the&#xA;convergence, torque coefficient generated by the rotor is monitored. when i saw torque coefficient is constant and it do not change, I stop calculation. Number of iterations is about 500. I think this number is very low because now that I see static pressure contour around of blade, it is not solved well. In fact, Static pressure contour is not good!!! I think that behavior of other quantities, such as the total&#xA;pressure at the inlet and outlet boundaries should be also monitored. Am I right?&#xA;I think that Streamlines and direction of velocity vector are correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see the following figures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Residuals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aSAfK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aSAfK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Streamlines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jcf9q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jcf9q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yE4Ev.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yE4Ev.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Velocity vectors on the blade:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X0rR6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X0rR6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8yYVg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8yYVg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Static pressure contour is not good!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pGqwC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pGqwC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oIBQH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oIBQH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image of turbine-please see axis of rotation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zPUFg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zPUFg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2015-09-26T04:29:34.950" LastActivityDate="2015-09-26T04:29:34.950" Title="Turbine simulation and Torque Calculation" Tags="|mechanical-engineering|fluid-mechanics|turbines|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5438" PostTypeId="2" ParentId="2551" CreationDate="2015-09-15T19:37:40.770" Score="2" Body="&lt;p&gt;Such high pressures can be achieved via &lt;a href=&quot;https://en.wikipedia.org/wiki/Diamond_anvil_cell&quot; rel=&quot;nofollow&quot;&gt;diamond anvil cells&lt;/a&gt; (DAC). The heart of these cells are two diamonds mounted on tungsten seats symmetrically in front of each other which are separated by metallic plate. Metallic plate has an aperture with loaded sample (gas, crystal, powder, etc.). These diamonds are then brought closer to each other by tightening bolts or by extension of piezo actuators. Pressures in such DACs can reach up to 200GPa or higher depending on the size of the diamond, it's quality, temperature and type of material being compressed. Amount of sample which can be compressed usually ranges from micrograms to milligrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would need a special chamber in order to load gas into DAC. In these chambers operator controls separation between diamonds in order to fill the aperture in metal gasket between diamonds with liquefied gas and enclose it safely for further compression&lt;/p&gt;&#xA;" OwnerUserId="2872" LastEditorUserId="2872" LastEditDate="2015-09-15T21:47:44.580" LastActivityDate="2015-09-15T21:47:44.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5440" PostTypeId="2" ParentId="1966" CreationDate="2015-09-15T21:27:50.187" Score="0" Body="&lt;p&gt;Thanks for answering the inquiry. And sorry for not following up. I have been working and this bookmark really sat all the time in a corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I came up with some what similar solution but more like a ready made solution with many associative problems. Because I had no better alternatives. Let me give a description.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The meter I had to choose has the flow range from 0-650 liters of diesel per hour. Plus it had to vibration shock resistant, magnetically sound working, water proof, must consider the counter precision after at least 0.1 million liter circulation, must be considerably small sized to fit the installation location.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Very few can manage those points in reasonable budget. the range that most closely works for me was $300-$600 USD meters, mostly produced in switzerland based company aquametro. Another company offered they can also manage the criteria.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;These meters involve simple structure of pd meters which you can produce for fur less and improvise your own customized meter. The biggest drawback was the precision after 100,000 liter circulation limit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As I was working with diesel flow, filter management was also important.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So I can say from the little search and discussion with professionals, that while choosing a meter, one must take consider the application criteria and the materials involved. For most cost economic decision and plausible application criteria, you might need to improvise over the existing products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And once you can figure out what is actually needed, you can mass manufacture it in China or anywhere you find it easy for around below 50 USD. But will need rigorous testing.  Meters are one of the oldest in the list. But for the leading manufacturers already worked on R&amp;amp;D for product development you must weight between the options. Must not compromise the quality or time/money.&lt;/p&gt;&#xA;" OwnerUserId="3019" LastEditorUserId="3022" LastEditDate="2015-09-16T00:30:05.203" LastActivityDate="2015-09-16T00:30:05.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5441" PostTypeId="2" ParentId="5410" CreationDate="2015-09-15T22:50:22.550" Score="2" Body="&lt;p&gt;The &lt;strong&gt;basic assumption&lt;/strong&gt; of the RANS approach (see the answer by @Algo) is that there is an overall steady solution to the problem. This overall steady solution features small scale oscillations &lt;em&gt;around&lt;/em&gt; the mean value. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In a first step the Navier Stokes equations (NS) are rewritten with this assumption &#xA;$u_i = \bar{u_i} + u_i'$&lt;/li&gt;&#xA;&lt;li&gt;The second step is to average the NS again &lt;/li&gt;&#xA;&lt;li&gt;By doing this only on term does not cancel out except: $\frac{\partial }{\partial x_j}(\overline{\rho {u_i}'{u_i}'})$ (this is a 3x3 matrix and it's called the Reynolds Stress Tensor)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since there is no analytical solution for the &lt;em&gt;mean of the product of the fluctuations&lt;/em&gt; &lt;strong&gt;an additional assumption&lt;/strong&gt; was needed to solve the equation system. &#xA;There were a lot of ideas on how to &lt;em&gt;model&lt;/em&gt; or &lt;em&gt;solve&lt;/em&gt; this problem. One turned out to be very useful because it predicted the reality very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is based on the observation that turbulence basically increases mixing of the fluid. To put it in other words: velocity gradients &lt;em&gt;mix&lt;/em&gt; out a lot faster in turbulent flows, as if the viscosity (momentum diffusion) is increased. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first step in the solution was to model the unknown term as a kind of additional viscosity term as shown in the answer from @Algo. Here the mean velocity gradient is known and only one unknown remains (after the RANS approach we were faced with a 3x3 matrix).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turbulence models are then used to calculate $\mu_t$. So the only adjustment you can make is to choose a turbulence model and in some cases tweak its parameters.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2015-09-15T22:50:22.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5442" PostTypeId="2" ParentId="5431" CreationDate="2015-09-15T23:04:13.523" Score="2" Body="&lt;p&gt;In general every compressor (axial and radial) will deliver a certain flow rate depending on the back-pressure down stream of the compressor. So by changing the pressure difference over the compressor: for example closing a throttle upstream or downstream of the compressor the flow rate will change accordingly. This can also be achieved by changing the &lt;em&gt;geometry&lt;/em&gt; of the compressor. One possibility are the mentioned variable inlet guide vanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a compressor map taken from &lt;a href=&quot;https://en.wikipedia.org/wiki/Centrifugal_compressor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;. As you can see (away from the surge line) every mass flow rate is connected to a pressure ratio. At constant rotor speed, the flow rate changes according to the pressure ratio which is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/stlhG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/stlhG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2015-09-15T23:04:13.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5443" PostTypeId="2" ParentId="5411" CreationDate="2015-09-15T23:24:03.197" Score="2" Body="&lt;p&gt;Averaging is a tricky business (see &lt;a href=&quot;https://www.researchgate.net/publication/245354712_Averaging_Nonuniform_Flow_for_a_Purpose&quot; rel=&quot;nofollow&quot;&gt;Cumsty 2006&lt;/a&gt; for an introduction). Basically the problem is, that our thermodynamics usually talk about &lt;strong&gt;a&lt;/strong&gt; pressure and &lt;strong&gt;a&lt;/strong&gt; temperature etc. But in real life the pressure and temperature are fields (i.e. varying in all three dimensions). So the question is: How to come up with &lt;strong&gt;one&lt;/strong&gt; representative value of (for example) the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the way the averaged value is used different averaging techniques are suited better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example think of the temperature of water in a big tank which is filled via two pipes from two different heat exchangers delivering two different water temperatures. Therefore not every water molecule has the same temperature. If you were supposed to calculate &lt;strong&gt;the&lt;/strong&gt; temperature in the tank you would need to come up with a way to calculate a representative temperature. How high would this mixed out temperature be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can of course just use the mean of the two pipes. But what if one pipe pumps more water into the tank. The solution to this particular problem is to weigh every pipe-flow with the mass-flow rate. This is called a mass-flow-average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb:&#xA;To validate numerical simulations the averaging method in CFD and experiment have to be the same. It might be a good idea to use area-averaging to compare basic values as temperature or pressure.&#xA;If the averaged values are used to do further calculations use mass-average for all except for static pressure which is always area-averaged. &lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-09-17T15:54:42.750" LastActivityDate="2015-09-17T15:54:42.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5445" PostTypeId="1" CreationDate="2015-09-16T07:21:47.133" Score="1" ViewCount="59" Body="&lt;p&gt;I want to create a closed container where I will put 12 temperature sensors for a project in my company. I'm a software engineer and I have to deal with this project so I need help because I'm not good at thermal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This container will look like a pan of 30 cm diameter. There will be one hot point at the bottom of the container to heat the inside of the container. What can I put in the container to have a homogeneous environment without overpressure; water, oil, sand or something else?&lt;/p&gt;&#xA;" OwnerUserId="3026" LastEditorUserId="90" LastEditDate="2015-09-16T08:46:45.893" LastActivityDate="2015-09-16T08:46:45.893" Title="Create a thermally homogeneous container" Tags="|fluid-mechanics|thermodynamics|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5446" PostTypeId="1" CreationDate="2015-09-16T10:51:49.037" Score="1" ViewCount="228" Body="&lt;p&gt;I have recently started research in mechanical engineering. Currently I am working in the field of &lt;strong&gt;friction stir welding&lt;/strong&gt;. During my study I came across a point where I have to find the &lt;strong&gt;mathematical relation between shear stress and temperature during plastic deformation of material&lt;/strong&gt;. I tried to find out about it on &lt;em&gt;google&lt;/em&gt; but did not get satisfactory answer. Can anyone please explain about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be specific, I am performing Friction stir welding on aluminum during which the temperature is going to be in the range 473-700 Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a difficult question for me hopefully not for others.&lt;/p&gt;&#xA;" OwnerUserId="3029" LastEditorUserId="2246" LastEditDate="2015-09-16T18:00:22.630" LastActivityDate="2015-09-16T18:00:22.630" Title="Mathematical relation between shear stress and temperature" Tags="|mechanical-engineering|heat-transfer|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5447" PostTypeId="2" ParentId="5422" CreationDate="2015-09-16T11:00:18.257" Score="1" Body="&lt;p&gt;There are &lt;a href=&quot;http://mankenberg.de/en/20/prgb:7/product:kondensatableiter-schwimmergesteuert/produkte.html&quot; rel=&quot;nofollow&quot;&gt;swimmer operated condensate traps&lt;/a&gt; (link contains cut view). The operating principle seems fairly simple, a swimmer in the pot opens a valve when enough condensate has accumulated and closes before the pot is empty. The specific product I linked has been accepted as tight enough for not creating an ex-zone, I'm sure there's others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I don't want to advertise for Mankenberg, it's just the first example I found.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-09-16T11:00:18.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5448" PostTypeId="1" CreationDate="2015-09-16T14:13:23.503" Score="3" ViewCount="4727" Body="&lt;p&gt;I'm studying a reactor, that can be imagined simply as a pipe in which a reaction advances.&lt;br&gt;&#xA;In this reactor, the inlet and outlet flow are constant as is the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that reaction is equi-molar no pressure change happens, otherwise in a reaction like steam reforming, where product moles are greater than reactant ones, I should expect an increase in pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$CH_4+H_2 O \rightleftharpoons CO+3H_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the pressure change due to change in the number of moles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: In a closed system (like a vessel) I'd use the ideal gas law: $pV=nRT$ ... but this is not the case!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;steady state grants me $\dot m|_{in} = \dot m|_{out}$&lt;br&gt;&#xA;I can express mass flow as $\dot m = \dot n \cdot MM$ where MM is molar mass&lt;br&gt;&#xA;and as  $\dot m = A \cdot u \cdot \rho$ where A is the cross sectional area (constant) u is mean speed and $\rho$ is density&lt;br&gt;&#xA;I express density with ideal gas law $\rho = {p \over {R \cdot T}} \cdot MM$&lt;br&gt;&#xA;finally I get $\dot n = A \cdot u \cdot {p \over {R \cdot T}}$&lt;br&gt;&#xA;now, since A,R,T are constants&#xA;taking the reaction abovementioned, if $\dot n$ rise up $u \cdot p$ have to follow this trend.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How, who, what, does make pressure constant while speed rise up ?&lt;/p&gt;&#xA;" OwnerUserId="2157" LastEditorUserId="2157" LastEditDate="2015-09-19T08:46:55.663" LastActivityDate="2015-09-22T17:24:14.217" Title="How can I calculate pressure change due to reaction?" Tags="|thermodynamics|chemical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5450" PostTypeId="2" ParentId="2641" CreationDate="2015-09-16T17:18:25.187" Score="3" Body="&lt;p&gt;The relative displacement tensor is the sum of the rotation tensor and the strain tensor. The strain tensor is always symmetric, while the rotation tensor is always antisymmetric. Once decomposed, the strain tensor is still a displacement tensor that has the rotation removed from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A displacement tensor can be defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P$ is vector field representing the displacement. If the material that started at point $\langle 2,4,8 \rangle$ ended at point $\langle 3,3,6 \rangle$ then $P(\langle 2,4,8 \rangle )=\langle 1,-1,-2 \rangle$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The displacement tensor $D$ is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D = \nabla P =&#xA;\begin{bmatrix}&#xA;\partial_x P_x &amp;amp; \partial_y P_x &amp;amp; \partial_z P_x\\&#xA;\partial_x P_y &amp;amp; \partial_y P_y &amp;amp; \partial_z P_y\\&#xA;\partial_x P_z &amp;amp; \partial_y P_z &amp;amp; \partial_z P_z&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the displacement tensor is constant, that implies the cube is deformed linearly, and thus P is a linear field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case:&#xA;$$P =&#xA;\begin{bmatrix}&#xA;0.2 \, x\\&#xA;0.8 y + 0.2 z\\&#xA;0.6 y + 0.4 z&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the original displacement tensor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D =&#xA;\begin{bmatrix}&#xA;0.2 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; 0.8 &amp;amp; 0.2\\&#xA;0 &amp;amp; 0.6 &amp;amp; 0.4&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have the displacement field we can find the new position $N$ as a function of original position $O$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N=O+P(O)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we take the corners of a unit cube we can calculate their new positions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\langle 0,0,0\rangle + P(\langle 0,0,0 \rangle) = \langle 0.0,0.0,0.0 \rangle\\&#xA;\langle 0,0,1\rangle + P(\langle 0,0,1 \rangle) = \langle 0.0,0.2,1.4 \rangle\\&#xA;\langle 0,1,0\rangle + P(\langle 0,1,0 \rangle) = \langle 0.0,1.4,0.6 \rangle\\&#xA;\langle 0,1,1\rangle + P(\langle 0,1,1 \rangle) = \langle 0.0,1.6,2.0 \rangle\\&#xA;\langle 1,0,0\rangle + P(\langle 1,0,0 \rangle) = \langle 1.2,0.0,0.0 \rangle\\&#xA;\langle 1,0,1\rangle + P(\langle 1,0,1 \rangle) = \langle 1.2,0.2,1.4 \rangle\\&#xA;\langle 1,1,0\rangle + P(\langle 1,1,0 \rangle) = \langle 1.2,1.4,0.6 \rangle\\&#xA;\langle 1,1,1\rangle + P(\langle 1,1,1 \rangle) = \langle 1.2,1.6,2.0 \rangle\\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;visually it would look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Wsudg.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wsudg.gif&quot; alt=&quot;Deformed Cube&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="1122" LastEditDate="2015-11-18T12:37:36.500" LastActivityDate="2015-11-18T12:37:36.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5451" PostTypeId="1" AcceptedAnswerId="5453" CreationDate="2015-09-16T22:53:40.500" Score="4" ViewCount="88" Body="&lt;p&gt;A motor gives its strength in N-cm or oz-in, but how do you know how much  you require? I know you can just calculate the minimum torque you need to move the load, but is there a typical amount over that minimum that is standard? Like 20% or reaching a certain acceleration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm using a motor for is my electrical engineering capstone project.  I need to rotate a mirror about the z axis at between 30/60 Hz. The mirror is the only moving component, so there aren't any other mechanical considerations besides that. I don't know the mirror size yet, as I figured I'd buy as good a motor as I can afford and pick a mirror that's small enough to work with it, but I want to know how to figure out approximately what a given torque will get me. I don't want to spend $50 on a motor with something like &quot;100 oz-in min holding torque&quot; and then get it and find out thats only strong enough for a 2&quot;x2&quot; mirror or something.&lt;/p&gt;&#xA;" OwnerUserId="3036" LastEditorUserId="1586" LastEditDate="2015-09-24T17:31:25.867" LastActivityDate="2015-09-24T17:31:25.867" Title="How much stronger than the bare minimum should a motor be?" Tags="|electrical-engineering|stepper-motor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5453" PostTypeId="2" ParentId="5451" CreationDate="2015-09-17T05:24:04.463" Score="2" Body="&lt;p&gt;Without more details about your design, the dynamics of your mechanism, or your calculations, its virtually impossible to address your question with any certainty.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, i offer some advice that may prove useful to you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If affordable, I would use double the minimum holding torque needed for the application because i usually find that my calculations don't usually account for all the true dynamics of the mechanism.  When starting out with an untested design, it's better to err on the side of caution.  Having too much torque is usually better than having too little.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on speed requirements, you may be even able to use gears to increase your torque while using a motor with less torque.  I may even recommend experimenting with gear ratios since its usually cheaper to buy a different gear than to buy a different motor.  Over time, you may find gears to be the better choice until you find the torque that meets your needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you're really confident that your calculations truly capture the physics, you might be able to get away with 20%-30% more than the minimum torque.  Ultimately, it all boils down to how well you can capture the physics of the mechanism.&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2015-09-17T05:24:04.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5455" PostTypeId="2" ParentId="5429" CreationDate="2015-09-17T07:50:08.710" Score="3" Body="&lt;p&gt;Yes, there are at least two ways to do it.&lt;/p&gt;&#xA;&lt;h3&gt;Static, deterministic analysis&lt;/h3&gt;&#xA;&lt;p&gt;One is to look at total energy in over a long time period, such that the internal temperature is the same at the start and the end. And if you take the time period to be long enough (30 days or more, for many buildings), then any changes in heat content of the thermal mass will be negligible when you divide out over the whole period. You can then integrate &lt;span class=&quot;math-container&quot;&gt;$T_{internal}-T_{external}$&lt;/span&gt; over time, to find the total number of Kelvin-seconds of temperature difference. And divide total energy put in by the heating system plus insolation, by your number of Kelvin seconds, to get the house's thermal conductivity in W/K. Once you've got that, you can then look at individual night's cooling curves to estimate the thermal mass of the house, e.g. by using a simple one-box simulation.&lt;/p&gt;&#xA;&lt;h3&gt;Bayesian, probabilistic time-series analysis&lt;/h3&gt;&#xA;&lt;p&gt;Another method, and one I've had a little involvement in myself, is the Bayesian approach, which makes the most of all the available data, and jointly estimates probability distributions for both parameters simultaneously. This method is harder to implement, but is a lot more powerful: you will probably only need a few days' worth of data; and you will get much more information out than you would from the static method.&lt;/p&gt;&#xA;&lt;p&gt;You write down the equation for the probability of observing the temperature series &lt;span class=&quot;math-container&quot;&gt;$T_{internal}$&lt;/span&gt; for arbitrary values of the house's thermal conductivity and thermal mass R, and given external temperature time series &lt;span class=&quot;math-container&quot;&gt;$T_{external}$&lt;/span&gt;; you add in prior distributions for your thermal mass, thermal conductivity, observation errors, and calculate a posterior distribution for thermal mass and thermal conductivity.&lt;/p&gt;&#xA;&lt;p&gt;The more information you've got, the more elaborate you can make the model. Add in time-series of heating-system energy use first, as that will be the most useful information. Then insolation, and wind speed and direction. Build up the complexity of your model one variable at a time, to ensure you've got sane, plausible results at each stage.&lt;/p&gt;&#xA;&lt;p&gt;For an example of this Bayesian approach for a single building element, see &lt;a href=&quot;http://dx.doi.org/10.1016/j.enbuild.2014.04.004&quot; rel=&quot;nofollow noreferrer&quot;&gt;&amp;quot;Inferring the thermal resistance and effective thermal mass of a wall using frequent temperature and heat flux measurements&amp;quot;&lt;/a&gt; by some of my colleagues. There will be more papers in this series - keep an eye out for new papers by the first two of those authors in particular.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-09-17T12:33:27.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5457" PostTypeId="1" CreationDate="2015-09-17T13:02:47.183" Score="1" ViewCount="133" Body="&lt;p&gt;In my position I'll frequently visit wastewater treatment plants, often when refurbishments work happen. What are best practices I should adopt to avoid accidents?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="104" LastEditDate="2015-09-19T18:48:25.580" LastActivityDate="2015-09-19T18:48:25.580" Title="What are best safety practices for visits to waste water treatment plants (WWTP)?" Tags="|safety|waste-water-treatment|" AnswerCount="1" CommentCount="9" ClosedDate="2015-09-17T13:58:45.147" ContentLicense="CC BY-SA 3.0" />
  <row Id="5458" PostTypeId="2" ParentId="5457" CreationDate="2015-09-17T13:02:47.183" Score="1" Body="&lt;p&gt;Some practices include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Announce visit to staff and inquire about potential hazards&lt;/li&gt;&#xA;&lt;li&gt;watch out for traffic &lt;/li&gt;&#xA;&lt;li&gt;watch out for machines that start on their own&lt;/li&gt;&#xA;&lt;li&gt;don't run, becareful around ladders etc, watch were to step, &lt;/li&gt;&#xA;&lt;li&gt;don't hang around under cranes, &lt;/li&gt;&#xA;&lt;li&gt;don't smoke in zones with explosion hazard or bring ignition sources &lt;/li&gt;&#xA;&lt;li&gt;don't touch sewage &lt;/li&gt;&#xA;&lt;li&gt;wash hands after visit &lt;/li&gt;&#xA;&lt;li&gt;only enter cellars or shafts when they are activly ventilated and/or after gas measurements have determined them to be safe &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-09-17T13:02:47.183" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5459" PostTypeId="2" ParentId="662" CreationDate="2015-09-17T14:15:20.173" Score="4" Body="&lt;p&gt;A good timeline and discussion of the events leading up to the disaster and the aftermath can be found &lt;a href=&quot;http://www.engineering.com/Library/ArticlesPage/tabid/85/ArticleID/175/Hyatt-Regency-Walkway-Collapse.aspx&quot; rel=&quot;nofollow&quot;&gt;here on Engineering.com&lt;/a&gt;. At the end of the article is a discussion of why Gillum (the engineering firm) was responsible and the ethics involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gillum attempted to shift/deflect blame by claiming that they never received the phone calls from Havens, the fabricator and that the project management denied them on-site inspection over cost concerns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite all of the other issues, the smoking gun is that two key drawings detailing the change from a one rod design to a two rod design, Shop Drawing 30 and Erection Drawing E3, were changed in January 1979. On February 16th 1979 Gillum received these two drawings, along with 40 others and returns them to Havens stamped as reviewed and approved. So not only was Gillum made aware of the proposals, they also asserted that they had reviewed and approved those changes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should also be noted that even the original one rod design would have been far short of the requirements of the Kansas City building code.   &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-09-17T14:15:20.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5460" PostTypeId="2" ParentId="5448" CreationDate="2015-09-17T15:24:42.200" Score="2" Body="&lt;p&gt;The change in pressure depends on the design of the reactor, specifically its boundary conditions like pressure and volumetric flow rate at the inlet and outlet.  Without any kind of compressor at the inlet, a reaction like the one you described will generally result in the reactor operating at the same pressure as the feed, but with the exit velocity and volumetric flow rate of the outlet stream being larger/faster than the feed.  If the reactants and products are all in the gas phase and are ideal gasses, and the temperature is held constant, you can calculate the change in inlet flow rate to outlet flow rate (volumetric) with this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta \dot V = \xi \left( \nu_p - \nu_r \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $\xi$ is the extent of reaction, and $\nu_p$, $\nu_r$ are the total stoichiometric numbers of the products and reactants.  Note that if $\nu_p=\nu_r$, the reaction is equimolar and the volumetric flow rate entering the reactor is the same as the flow rate exiting the reactor (which is intuitive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To better explain why the velocity increases while the pressure stays more or less the same, consider that you always need a negative pressure gradient through the reactor (assuming steady-state operation) so that the fluid will always flow from inlet to outlet.  That is, the outlet pressure, $P_{out}$, must always be less than the inlet pressure $P_{in}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing a mass balance on the reactor shows that the mass flow rate coming out must be exactly equal to the inlet flow rate (no gains/losses):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot m = \rho_{in} u_{in} A = \rho_{out} u_{out} A&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ is density, $u$ is velocity, and $A$ is the cross-sectional area of the reactor (constant).  We also have the ideal gas equation of state (assuming the reactants and products are in-fact ideal gasses):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P=\rho RT = \rho \frac{\bar R}{MW} T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P$ is pressure, $R$ is specific gas constant, $\bar R$ is universal gas constant, $MW$ is the average molecular weight, and $T$ is temperature.  We can solve the equation of state for $\rho$ and plug it in to the mass balance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{in} \, MW_{in} \, u_{in}  = P_{out} \, MW_{out} \, u_{out} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T$, $A$, and $R$ have canceled out.  If we assume that the pressure gradient is very small, $P_{in} \approx P_{out}$ (just enough to get a little bit of flow), then we can also cancel the pressure terms and rearrange:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{MW_{in}}{MW_{out}} = \frac{u_{out}}{u_{in}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it should be more apparant that, if the outlet average molecular weight is less than the inlet (i.e. products have a higher stoichiometric number than the reactants), then the outlet velocity must be higher to satisfy our mass balance and equation of state.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-09-22T17:24:14.217" LastActivityDate="2015-09-22T17:24:14.217" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5461" PostTypeId="1" CreationDate="2015-09-17T21:06:46.270" Score="3" ViewCount="2455" Body="&lt;p&gt;I want to make a precision syringe pump and need some O-rings for its piston. The cylinder is made of stainless steel (maybe later I'll order a glass one). The lower the friction ratio, the better the precision of the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What sort of rubber is the best in this case? BuNa? Viton? Silicone? Something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what should be used to keep it lubricated?&lt;/p&gt;&#xA;" OwnerUserId="2734" LastEditorUserId="282" LastEditDate="2015-09-19T19:44:40.580" LastActivityDate="2016-05-21T09:26:57.813" Title="Identifying O-rings with a low friction ratio" Tags="|mechanical-engineering|pumps|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5462" PostTypeId="1" AcceptedAnswerId="5464" CreationDate="2015-09-17T21:22:49.660" Score="8" ViewCount="431" Body="&lt;p&gt;I'm working on analysis of a bridge wall-pier for dead and live loads. When modeling the pier in finite element software (LARSA), I'm running up against several modeling questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is it valid to model the wall as a single column element? (Going to a plate mesh seems like overkill for a run-of-the-mill highway bridge.)&lt;/li&gt;&#xA;&lt;li&gt;How should the connectivity be modeled between the girders and the wall pier? Are rigid links appropriate? Joint constraints? Beam elements with really high moments of inertia?&lt;/li&gt;&#xA;&lt;li&gt;If I wanted to run modal analysis, would this change the modeling considerations?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9GXyA.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9GXyA.png&quot; alt=&quot;Wall Pier&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-09-18T00:41:02.590" Title="Simple Finite Element Modeling of Wall Piers" Tags="|structural-engineering|bridges|finite-element-method|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5463" PostTypeId="1" CreationDate="2015-09-17T23:39:20.590" Score="4" ViewCount="5637" Body="&lt;p&gt;In control theory, according to the definition of matched uncertainty, the uncertainty enters the system through the same channel as the control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the system:&#xA;$$\dot x = F(x)+G(x)u+Δ(x, t),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F$ and $G$ are known functions comprising the nominal system and $Δ$ is an uncertain function known only to lie within some bounds. For example, we may know a function $ρ(x)$ so that $| Δ(x, t) |≤ ρ(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we require that the uncertainty $Δ$ is of the form:&#xA;$$Δ(x, t) = G(x) · \barΔ(x, t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for some uncertain function $\barΔ$, this form is called the matching condition because it allows the system to be written as:&#xA;$$\dot x = F(x)+G(x)[u+ \barΔ(x, t)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where now the uncertainty $\barΔ$ is matched with the control $u$, that is, it enters the system through the same channel as the control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the physical examples for matched and unmatched uncertainties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Levine, William S., ed. The control handbook. CRC press, 1996.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3044" LastEditorUserId="3044" LastEditDate="2015-09-19T11:22:45.630" LastActivityDate="2018-04-23T06:44:06.813" Title="Matched and unmatched uncertainties" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5464" PostTypeId="2" ParentId="5462" CreationDate="2015-09-18T00:41:02.590" Score="6" Body="&lt;p&gt;I'll go through your questions one by one:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A simple model like you have shown is fine to get model loads. Once you get those loads, you can then use a more specific technique to design the pier. (strut and tie).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rigid links are ok. Just make sure that you are capturing the various induced moments. I usually have rigid links from the bearing locations to the pier nodes and rigid links from the center of girders to the bearings. That way you can see individual bearing axial loads and shears (and set moment/rotation releases properly).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All of the above should work for modal analysis. In modal analysis you will have to be extra careful that members are drawn along their centerlines so that you can apply mass accurately.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-09-18T00:41:02.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5465" PostTypeId="2" ParentId="4277" CreationDate="2015-09-18T01:41:59.113" Score="1" Body="&lt;p&gt;Those are called oxidation ditches.  They are aerated mechanically in the center where the bridge is located.  The length of the channel and the overall size of the process are functions of required detention time/space availability/and budget.&lt;/p&gt;&#xA;" OwnerUserId="3046" LastActivityDate="2015-09-18T01:41:59.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5466" PostTypeId="2" ParentId="1929" CreationDate="2015-09-18T03:14:37.993" Score="1" Body="&lt;p&gt;Diesel combustion does not involve a mixture. Fuel/air mixtures apply to the Otto cycle in gasoline engines. In diesels, air is compressed to top-dead-center where temperatures can reach 1000 degrees Fahrenheit, and then the fuel is injected and subsequently combusts, resulting in a power stroke.&lt;/p&gt;&#xA;" OwnerUserId="3047" LastActivityDate="2015-09-18T03:14:37.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5467" PostTypeId="1" AcceptedAnswerId="5475" CreationDate="2015-09-18T15:35:39.757" Score="5" ViewCount="732" Body="&lt;p&gt;I am currently tinkering with typewriters and two things striked me when I started to read about their maintenance:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They are supposed to run almost completely dry, except for a very tiny amount of lubricant at very specific points;&lt;/li&gt;&#xA;&lt;li&gt;Gummy or sticky gunk buildup - usually due to over-oiling + dust buildup + passage of time - is one of the main causes of typewriter malfunctioning.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As an avid cyclist I am, I have already noticed some oils have a tencency to get sticky as they dry out and mix with dust (e.g. gearbox oil), while others seem to be very stable and somewhat &quot;repel&quot; dust, mostly automatic transmission fluid - ATF, power steering fluid, or suspension oil. I believe these are somewhat intended to remain fluid and to actively repel dust, which seems to &quot;fall out of the way&quot; instead of get increasingly mixed with the oil and the working surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know a lot of typewriter folks use sewing machine oil (Singer, mostly), but I wonder if its characteristics are suitable for this need of not getting gunky as the years pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So que questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given the considerations above, would some type of oil or non-oily lubricant be more adequate than sewing-machine oil?&lt;/li&gt;&#xA;&lt;li&gt;Is the hydraulic fluids hypothesis a good one?&lt;/li&gt;&#xA;&lt;li&gt;Any other relevant consideration?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2015-09-19T17:47:59.363" Title="Which lubricant (oil or not) to use if I want to prevent sticky/gummy buildup?" Tags="|mechanical-engineering|tribology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5468" PostTypeId="1" CreationDate="2015-09-18T18:41:34.583" Score="1" ViewCount="1093" Body="&lt;p&gt;I have a 3 phase motor and 3 phase power supply (400V, 50Hz)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently this motor produces 1.5hp in torque and ~1750rpm. I need it to produce less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the best way to go about this, a rheostat or a VFD? The cheaper the better.&lt;/p&gt;&#xA;" OwnerUserId="3052" LastActivityDate="2015-09-21T12:21:20.783" Title="VFD/rheostat for motor speed control" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5469" PostTypeId="1" CreationDate="2015-09-19T03:34:38.550" Score="6" ViewCount="2353" Body="&lt;p&gt;We can get really humid days sometimes. Recently I was digging around in boxes and found some old mineral specimens, and I set out a few of the salt crystal rocks. I soon found that they had pools of water accumulating underneath them, which required some cleaning and attention. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After putting them on drinkable trays, I found that the surface tension was pulling the water off the bottom of the block and causing it to collect.  This got me thinking about passing the air in coming to this space through a baffle of salt blocks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good thing about salt blocks is that the absorbed moisture can be 'wicked' away. Whatever the solution, this needs to be a cycle that is ongoing using the material(s) installed. I am fine to imagining the installation of a plumbed drain or collection reservoir. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there another material or better technique for removing moisture from the air in this passive manner? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Passive meaning 'without an energized / motorized compressor to dehumidify'.  Mostly, the mechanism needs to be designed to leverage ambient energy or mechanics, and minimize maintenance cycles and materials handling.&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="104" LastEditDate="2015-09-21T03:57:16.823" LastActivityDate="2018-12-21T13:18:53.223" Title="Is there a more active passive technique for removing humidity from the air besides salt blocks?" Tags="|airflow|energy-efficiency|air-quality|dehumidification|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5470" PostTypeId="2" ParentId="5469" CreationDate="2015-09-19T07:34:36.490" Score="8" Body="&lt;p&gt;Silica gel is the standard for most purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the desiccant you find inside those paper sachets packed with new electronic equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It consists of tiny beads of silicon dioxide that are nanometres across (so that a little material has a lot of surface area), and although it's called a gel, the material itself is hard at room temperature and pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source material is very plentiful, and very cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in general separating water vapour from air is going to require work done: i.e. it will need continual energy inputs. In the case of a dessicant, you will need to apply dry heat to regenerate it from time to time.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-09-19T20:01:28.273" LastActivityDate="2015-09-19T20:01:28.273" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5471" PostTypeId="2" ParentId="5448" CreationDate="2015-09-19T09:23:40.207" Score="0" Body="&lt;p&gt;I think you're mixing things up. Quoting from this &lt;a href=&quot;http://www.chem.ox.ac.uk/vrchemistry/Conservation/page07.htm&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In chemical processes, the most important property to be conserved is the number of atoms of each kind that are present. Unlike nuclear processes, chemical reactions do not create or destroy atoms, or change one kind of atom into another. They only reshuffle the atoms that were originally present into different molecular combinations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meaning that mass is conserved whether number of moles is the same or not.&#xA;For a closed system (combustion reaction inside a piston cylinder), the rise in pressure is not dependent on whether the reaction is equimolar or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we used two different fuels, hydrogen and methane for example, stoichiometric hydrogen reaction is non-equimolar and methane is equimolar, However, both will cause an increase of pressure inside a piston cylinder, the rise in pressure is due to the dramatic increase in temperature of the exothermic reactions (with fixing the volume).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same thing for an open system, the result of the reaction will affect the temperature (enthalpy) of the stream due to the release of the energy stored in the fuel, while the pressure will stay the same since there is no reason for it to build up.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2015-09-19T09:30:49.117" LastActivityDate="2015-09-19T09:30:49.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5474" PostTypeId="1" CreationDate="2015-09-19T14:48:24.243" Score="6" ViewCount="800" Body="&lt;p&gt;A bolted end plate splice in a strut is designed to transmit compression by direct bearing of connecting cross-sections. If, after accounting for the bending moment due to strut action and additional moment due to moment amplification effects, the bolts themselves still experience no tension due to the large compressive force, is it possible to design the bolts to take only vertical shear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would the size of bolts affect the intended stiffness (underlined clause in red below) in the strut, assuming I provide the bare minimum to counter shear due to self-weight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UM3yl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UM3yl.png&quot; alt=&quot;clause 6.1.8.2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are answering with an example, you may use a 25m laced strut comprising of two numbers of UB914x305x287kg/m, taking a compression load of say 6000kN being spliced at a quarter span from each end support with 4 numbers of M20 bolts.&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2018-04-26T13:55:14.490" Title="How to interpret this BS 5950:2000 clause for splices in compression members?" Tags="|structural-engineering|steel|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5475" PostTypeId="2" ParentId="5467" CreationDate="2015-09-19T17:47:59.363" Score="3" Body="&lt;p&gt;The best option would probably be &quot;clock oil&quot; which is intended to be long lasting, non-acidic, and (importantly) non-spreading. If you apply it in small enough quantities, it will stay where you put it, and not move around through capillary action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sewing machine oil is probably a reasonable alternative, but the fact that it is formulated not to mark or damage any fabric or thread that it comes in contact with is probably irrelevant. Also it was not meant to be long-lasting. Singer used to recommend that their sewing machines should be oiled daily if they were in continuous use (i.e. in a factory.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clock repairers sometimes use transmission fluid to lubricate clock springs (if only because of the expense of using clock oil on big rusty springs in a low-quality clock!) but not for lubricating bearings and pivots.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-09-19T17:47:59.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5476" PostTypeId="2" ParentId="5469" CreationDate="2015-09-19T21:07:26.080" Score="1" Body="&lt;p&gt;How about this: You blow your humid air through the primary side of an air-air heat exchanger, here it is cooled a bit. Then this air passes through past a cold surface: either outside or underground, depending on oyour local climate. The air cools further and moisture condenses.&#xA;The cool air passes through the secondary side of our air-air heat exchanger, gets reheated (and so the relative humidity sinks) and cools the incoming air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would need energy input for the blower, which is small compared to a compresor in a heat pump. You would loose heat to the environment. The bigger the HX, the less heat you loose. actual design must take into account that water may condense in the HX. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;IIRC correctly, this was one trick used in an energy-optimized school, in this case they used an old subterreanean bunker nearby as the sink for the moisture.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-09-19T21:07:26.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5477" PostTypeId="1" CreationDate="2015-09-20T04:54:40.720" Score="4" ViewCount="3427" Body="&lt;p&gt;I have read on many sites that for lift to occur, the pressure below a wing must be higher than the pressure above it. This happens by somehow moving air faster over the top of the air foil so the pressure above the wing decreases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can't understand is: Why does the air above the wing have to move faster? Because on many sites it is stated that there is a misunderstanding that air moving upwards has to flow faster to reach the end at the same time with the air moving downwards. But that is wrong: &lt;a href=&quot;https://m.youtube.com/watch?v=6UlsArvbTeo&quot; rel=&quot;nofollow&quot;&gt;https://m.youtube.com/watch?v=6UlsArvbTeo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why else would the air upwards want to move faster? Why don't they both move at the same speed?&lt;/p&gt;&#xA;" OwnerUserId="3060" LastEditorUserId="90" LastEditDate="2015-09-20T14:20:49.733" LastActivityDate="2015-09-21T04:48:35.087" Title="How an airfoil works" Tags="|fluid-mechanics|pressure|aircraft-design|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5478" PostTypeId="2" ParentId="5477" CreationDate="2015-09-20T08:58:36.523" Score="3" Body="&lt;p&gt;You are partially right. The air does move faster on the top of the wing than at the bottom. This is true, however as you pointed out the reason for that is not that 2 air molecules enter the wing at the same time and due to the curvature the molecule on top has to travel faster before they meet again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what happens then? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sKzBF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sKzBF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the airplane starts moving a curl occurs at the end of the wing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/epaMS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/epaMS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to conservation of angular momentum curls always occur in pairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hb0Km.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hb0Km.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have air that is moved under the bottom of the wing and eventually with the air coming in the direction of the wing it moves over the wing again.&#xA;And this is the reason why the air moves faster on top of the wing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;References: Mostly memory and &lt;a href=&quot;http://www.luftfahrtarchiv.eu/index.php%3Foption%3Dcom_content%26view%3Darticle%26id%3D186:wie-fliegt-ein-flugzeug%26catid%3D39:grundkenntnisse%26Itemid%3D59&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2015-09-21T04:48:35.087" LastActivityDate="2015-09-21T04:48:35.087" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5479" PostTypeId="1" CreationDate="2015-09-20T09:33:27.760" Score="2" ViewCount="4371" Body="&lt;p&gt;I am currently preparing for an exam where on part is control theory. This is one task from a former exam where I don't know how to solve it (translated from German):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the z-transformed transfer function $G(z)$ of the time-discrete plant (&quot;Ersatzregelstrecke&quot;). The time-continous plant $G_s(s)$ is given. The sampling rate $T_A$ should be viewed as given.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$G_s(s) = \frac{1}{s^2 + 2s + 1}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Note: $$\mathcal{Z} \left \{\frac{a^2}{s \cdot (s+a)^2} \right\} = \frac{z}{z-1} - \frac{z}{z-e^{-aT}} - \frac{aTz \cdot e^{-aT}}{(z-e^{-aT})^2}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that I can simplify $G_s(s) = \frac{1}{(s+1)^2}$, but this is obviously not of the form $\frac{a^2}{s \cdot (s+a)^2}$...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate this?&lt;/p&gt;&#xA;" OwnerUserId="3061" LastEditorUserId="3061" LastEditDate="2015-09-20T09:42:24.887" LastActivityDate="2022-09-05T16:23:23.257" Title="How can I calculate the z-transformation of this transfer function of a time-discrete controller?" Tags="|control-theory|transfer-function|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5480" PostTypeId="1" AcceptedAnswerId="5503" CreationDate="2015-09-20T09:37:06.603" Score="1" ViewCount="393" Body="&lt;p&gt;For rectification I am given the following information&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_e$ = 0.95&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_f$ = 0.50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_b$ = 0.15&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v$ = 8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q$ = 1.17&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;**VLE Data**    &#xA;x       y&#xA;0.00    0.00&#xA;0.10    0.17&#xA;0.20    0.31&#xA;0.30    0.44&#xA;0.40    0.55&#xA;0.50    0.645&#xA;0.60    0.73&#xA;0.70    0.81&#xA;0.80    0.88&#xA;0.90    0.94&#xA;0.95    0.97&#xA;1.00    1.00&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The VLE data is derived from a given diagram. There may be some inaccuracies there. However drawing with pen and paper in the diagramm gives me the same error I'll explain later on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all this data I developed the McCabe Thiele Diagramm for this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting everything together in Excel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EPMrb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EPMrbm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grey dotted line indicates my $x_b$ and in orange I drew the theoretical stages. My Problem is: Usually you would come up with a whole number of stages, in this case I would get $9\frac{5}{7}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also an $E_k = 0.7$ given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n=\frac{n_{th}}{E_k} = 13.87$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Why do I not get whole numbers? I am under the impression that there always have to be whole numbers. However in Absorption we where supposed to give answers like $n = 1.2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought maybe the given mole fractions took $E_k$ into account and therefore I would get a whole number after dividing $n_{th}$ by 0.7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this just an error in my VLE Data and actually I would end up with n = 14? 13+1 to be accurate. It's hard to use the given diagramm since it's very small and I couldn't draw accurate enough in it. Also on what stage is the feed? I wouldn't get a whole number the either.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2015-09-22T07:01:58.573" Title="McCabe Thiele - number of stages" Tags="|thermodynamics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5481" PostTypeId="5" CreationDate="2015-09-20T09:47:35.520" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-09-20T09:47:35.520" LastActivityDate="2015-09-20T09:47:35.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5482" PostTypeId="4" CreationDate="2015-09-20T09:47:35.520" Score="0" Body="The transfer G(s) = X(s) / W(s) function describes the relationship between an input signal X(s) and an output signal W(s)." OwnerUserId="3061" LastEditorUserId="3061" LastEditDate="2015-09-20T16:43:56.383" LastActivityDate="2015-09-20T16:43:56.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5483" PostTypeId="2" ParentId="5479" CreationDate="2015-09-20T09:59:55.160" Score="2" Body="&lt;p&gt;In order to solve it, you have to divide $ G_s(s)$ by s first and then you have it in the form on which you can perform the Z-transform.&lt;/p&gt;&#xA;" OwnerUserId="3062" LastEditorUserId="3062" LastEditDate="2015-09-20T10:17:42.637" LastActivityDate="2015-09-20T10:17:42.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5484" PostTypeId="2" ParentId="5477" CreationDate="2015-09-20T11:12:05.903" Score="6" Body="&lt;p&gt;It is true that the air on the &quot;top&quot; (&lt;em&gt;suction side&lt;/em&gt;) is faster than on the &quot;bottom&quot; (&lt;em&gt;pressure side&lt;/em&gt;). The interesting thing is that the air on the suction side is so fast, it &lt;em&gt;overtakes&lt;/em&gt; the air on the pressure side (see illustration from &lt;a href=&quot;https://en.wikipedia.org/wiki/Lift_(force)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QUOFq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QUOFq.png&quot; alt=&quot;profile air flow from wikipedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to understand the velocity distribution around an air foil is not to look at it independent from the pressure distribution and curvature of the airfoil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by describing the flow from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stagnation_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;stagnation point&lt;/a&gt; this is the point at the very left in the figure where the free air flow hits the airfoil first (leading edge) and move along the suction side to the end of the air foil (trailing edge).&lt;br/&gt;&#xA;At the stagnation point we have a very high static pressure (imagine holding your hand out of the car window when driving really fast). Around the stagnation point the static air pressure is a lot higher than the air around it. From there on the flow is first accelerated (to the point of maximum profile thickness) and then decelerated (until it reaches the trailing edge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flow around and air foil is governed by a set of equations (i.e. momentum and energy conservations). These are the &lt;a href=&quot;https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Navier Stokes Equations&lt;/a&gt;. Without going into detail the equations are not linear. This means we end up with a chicken-and-egg problem. It means one cannot really answer your question in a step-by-step way. The answer to your question is actually (which is not really satisfactory): The flow field which develops around the airfoil is the only possible state this equation system allows for (conserving momentum and energy). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where as this sounds really discouraging there are some patterns which occur. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Whenever stream-lines are diverging the pressure has to be lower than the surrounding air.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whenever stream-lines are converging the pressure has to be higher than the surrounding air.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Let's look at an example:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/U9McQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U9McQ.jpg&quot; alt=&quot;http://www.rpi.edu/dept/chem-eng/WWW/faculty/plawsky&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flow which is approaching the cylinder (left to right) is converging in front of the stagnation point. The distance between the streamlines is getting smaller and the flow is moving around the cylinder. This happens because the high static pressure at the stagnation point pushes the air away from the cylinder and thereby compressing the flow.&lt;br/&gt;&#xA;When the flow has reached the top (or bottom) of the cylinder, the streamlines diverge. The distance between the stream lines increases because they flow towards the centre line. In order for the flow to flow &lt;em&gt;inwards&lt;/em&gt; the pressure has to be lower than the surrounding air. This lower pressure sucks the air inwards, you might also say the higher pressure of the surrounding air pushes the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's look at an airfoil again:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uYDL4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uYDL4.jpg&quot; alt=&quot;http://fearofflyingschool.com/&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;The stream lines (across the suction side) converge first and diverge than until they reach the trailing edge. This means the pressure at the upper side of the airfoil has to be lower. The opposite is true for the pressure (lower) side of the air foil. Here, the streamlines converge which means a higher pressure is present.&lt;br&gt;&#xA;When summing up all the pressure forces around the airfoil there is a net-upward force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that this all sounds kind of &lt;a href=&quot;http://www.urbandictionary.com/define.php?term=mambo%20jambo&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;mambo jambo&lt;/strong&gt;&lt;/a&gt;. The reason for it is, that &lt;strong&gt;the model&lt;/strong&gt; which is used in most of aerodynamics is &lt;a href=&quot;https://en.wikipedia.org/wiki/Continuum_mechanics&quot; rel=&quot;nofollow noreferrer&quot;&gt;continuum mechanics&lt;/a&gt;. So in a way you are asking why is the model behaving like this, where we actually told the model to behave in that way.&lt;br/&gt;&#xA;Our model of the flow is based on the observation of fluids. It is a macroscopic view of something a lot more complex. In principle we can explain everything based on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kinetic_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;kinetic theory of gases&lt;/a&gt; but this becomes impractical when one wants to calculate the lift of an airfoil.&lt;br/&gt;&#xA;However, think of gas as a bunch of air molecules which bounce around. They hit each other or surfaces. The rate with which they hit each other and surfaces is what is responsible for the &lt;em&gt;pressure&lt;/em&gt; we use in aerodynamics. This &lt;em&gt;bouncing&lt;/em&gt; is extremely high. Now think of the curvature of the suction side of the airfoil. It is curved downwards. When the air is flowing across the airfoils suction side &lt;em&gt;the room&lt;/em&gt; the air can fill out gets bigger. This means that the same amount of air-molecules have more space therefore the rate at which they hit will become lower (hence lower pressure), and since they have more room to move the bulk-velocity of all air-molecules will increase in this direction (which results in higher velocities). &lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-09-20T13:48:02.673" LastActivityDate="2015-09-20T13:48:02.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5485" PostTypeId="2" ParentId="5479" CreationDate="2015-09-20T12:00:45.507" Score="2" Body="&lt;p&gt;Observe that $G(s)$ is formed by $$\frac{1}{s^2} \to \frac{1}{(s+1)^2}$$ with the corresponding Z-transform/Laplace pairs $$\frac{1}{s^2} \to \frac{Tz}{(z-1)^2}$$ and $$\frac{1}{s+a} \to \frac{z}{z-e^{-aT}}$$&#xA;With this information you should be able to make the conversion adhering to all the rules of linearity, e.t.c, thats if you wanna find the representation from scratch.&#xA;Found this: A more elegant way is to go back to the time domain and compute the Z-transform sum. Or just use the tables that have become very popular &lt;a href=&quot;http://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html&quot; rel=&quot;nofollow&quot;&gt;Table of Laplace and Z Transforms&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="3065" LastEditorUserId="1586" LastEditDate="2015-09-28T18:00:21.787" LastActivityDate="2015-09-28T18:00:21.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5486" PostTypeId="1" CreationDate="2015-09-20T12:01:02.200" Score="2" ViewCount="164" Body="&lt;p&gt;I understand flaps increase the angle of attack so it increases lift, but how does it increase drag also? Do the lift and drag have a direct relationship? &lt;/p&gt;&#xA;" OwnerUserId="3063" LastEditorUserId="272" LastEditDate="2015-09-24T17:10:23.480" LastActivityDate="2015-09-24T17:10:23.480" Title="How do flaps work?" Tags="|mechanical-engineering|fluid-mechanics|aerospace-engineering|airflow|" AnswerCount="1" CommentCount="3" ClosedDate="2015-09-24T17:23:16.857" ContentLicense="CC BY-SA 3.0" />
  <row Id="5487" PostTypeId="2" ParentId="5477" CreationDate="2015-09-20T12:42:40.653" Score="1" Body="&lt;p&gt;The most common error in explanations of aerofoil lift is that it's due to the Bernoulli effect - but who told the air particles that separated at the leading edge that they had to meet up at the trailing edge?   The air flowing over the wing does move faster, but that's only part of the story.   The air that flows off the trailing edge (and there are no vortices at the trailing edge!) has a downwards component to its velocity.   For equilibrium, there must then be an upwards force on the wing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article 'airfoil' is instructive, as is: &#xA;&lt;a href=&quot;https://www.grc.nasa.gov/www/K-12/airplane/shape.html&quot; rel=&quot;nofollow&quot;&gt;https://www.grc.nasa.gov/www/K-12/airplane/shape.html&lt;/a&gt; .&lt;/p&gt;&#xA;" OwnerUserId="2520" LastActivityDate="2015-09-20T12:42:40.653" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5488" PostTypeId="2" ParentId="5486" CreationDate="2015-09-20T15:10:57.783" Score="3" Body="&lt;p&gt;As you say, flaps on aircraft wings increase lift at the expense of also increasing drag.  This also means that the minimum speed to get lift is lower.  These tradeoffs is why airplanes don't fly with flaps down all the time, only during takeoff and landing when the extended low speed range of lift is really useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a quick intuitive idea of why flaps also create more drag, put your hand out the window of a car traveling at highway speed.  Start with the hand flat and parallel to the ground.  In this configuration there will be no up/down force from the wind, just some backward push.  Now rotate your hand a little so that the leading edge is higher than the trailing edge.  You will feel the air pushing upwards, but there will also be more force pushing backwards.  In affect, the larger area presented perpendicular to the wind creates more drag.  If this area is oriented and shaped correctly, then it can also produce lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the opposite by tilting the leading edge of your hand down a bit.  You still get a bit more drag again due to the larger area facing the wind, but now the wind is also actively pushing down.  This is basically negative lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case that sounds silly, there are situations where this is useful too.  This is exactly what a &quot;spoiler&quot; does on a race car.  The wind is being used to provide a downward force (negative lift), which increases the sliding friction of the tires against the road, thereby allowing for more aggressive accelleration and cornering without slipping.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-09-20T15:10:57.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5490" PostTypeId="1" AcceptedAnswerId="5492" CreationDate="2015-09-21T11:22:14.297" Score="5" ViewCount="126" Body="&lt;p&gt;A journal I am reading about circular hydraulic jumps provides the equation and I quote:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; Depth after hydraulic jump, &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; depth before hydraulic jump, &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; velocity after hydraulic jump, &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; velocity before hydraulic jump, &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; radius of hydraulic jump.&lt;/p&gt;&#xA;&lt;p&gt;The condition for conservation of momentum can be written as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dp}{dt} = 2\pi R\rho HV^2 - 2\pi R\rho hv^2 = F_1-F_2$$&lt;/span&gt;&#xA;where&#xA;&lt;span class=&quot;math-container&quot;&gt;$$F_1 = 2\pi R \rho g \int_0^hx\cdot dx = \rho g\pi Rh^2$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$F_2 = 2\pi R \rho g \int_0^Hx\cdot dx = \rho g\pi RH^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I understand that units for &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\frac{dp}{dt}$&lt;/span&gt; are both &lt;span class=&quot;math-container&quot;&gt;$kg\ m\ s^{-2}$&lt;/span&gt;, and that they are equivalent.&lt;/p&gt;&#xA;&lt;p&gt;But I don't know how, in this example, &lt;span class=&quot;math-container&quot;&gt;$F_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_2$&lt;/span&gt; are derived i.e. where &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; and the integrals come from.  Could somebody please explain?&lt;/p&gt;&#xA;&lt;p&gt;Page 4 from: &lt;a href=&quot;http://adatbank.transindex.ro/vendeg/htmlk/pdf5017.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;On the circular hydraulic jump&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3013" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-09-26T21:15:44.187" Title="Equating $\frac{dp}{dt}$ and $F$ at a hydraulic jump?" Tags="|mechanical-engineering|fluid-mechanics|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5491" PostTypeId="2" ParentId="5468" CreationDate="2015-09-21T12:21:20.783" Score="1" Body="&lt;p&gt;1.5hp is 1.12kW, this is no torque, this is power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to decrease the output power of your motor. I assume that, for your load, the torque is a function of the rpm. With a &lt;a href=&quot;https://en.wikipedia.org/wiki/Synchronous_motor&quot; rel=&quot;nofollow&quot;&gt;synchronous 3 phase motor&lt;/a&gt;, the rpm are set by the power supply frequency, and this would prevent you to adjust output power. All you'll change is the system efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your current speed is 1750rpm, however, I assume that your motor is &lt;a href=&quot;https://en.wikipedia.org/wiki/Induction_motor&quot; rel=&quot;nofollow&quot;&gt;asynchronous&lt;/a&gt;. Therefore, you can decrease your effective speed by using a rheostat, as long as your &lt;a href=&quot;https://en.wikipedia.org/wiki/Induction_motor#Slip&quot; rel=&quot;nofollow&quot;&gt;slip&lt;/a&gt; is low enough to keep your motor running. Obviously, you'll still decrease your system efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, you should go for VFD: for a 1.12kW motor, this is cheap. Maybe not as cheap as a rheostat (it depends on your needs), but the price difference will probably be paid by the power saving, you'll avoid heat dissipation issues, and you may control your system more precisely than with rheostat.&lt;/p&gt;&#xA;" OwnerUserId="2664" LastActivityDate="2015-09-21T12:21:20.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5492" PostTypeId="2" ParentId="5490" CreationDate="2015-09-21T12:42:25.137" Score="3" Body="&lt;p&gt;$F=\frac{dp}{dt}$ is, in some sense, a more fundamental expression of Newton's law than $F=ma$ because $F=ma$ doesn't allow for situations with changing mass.  You can easily derive the second from from the former by using $p=mv$ which gives $F=m\frac{dv}{dt}=ma$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question seems to be more about how to derive the expressions for the force at the boundary of the hydraulic jump.  To do this, think about the hydraulic jump as a wall; on one side of the wall the height of the water is $h$ and on the other it is $H$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider an infinitesimal square in this wall at depth $x$; the pressure exerted on this square is given by &#xA;$$&#xA;P=\rho\ g\ x.&#xA;$$&#xA;The &lt;em&gt;force&lt;/em&gt; exerted on the square is calculated by multiplying by the surface area of the infinitesimal patch&#xA;$$&#xA;F_{inf}=\rho\ g\ x\ dx\ dy.&#xA;$$&#xA;Now you just have to sum up (with integration) all of the patches on the wall.  The y direction is easy&#xA;$$&#xA;\begin{align}&#xA;F_{tot}&amp;amp;=\rho\ g\int_{0}^{h}\left(\int_0^{2\pi R}dy\right) xdx\\&#xA;  &amp;amp;=\rho\ g\ 2\pi R\int_0^{h}x dx\\&#xA;  &amp;amp;=\pi R\ \rho\ g\ h^2&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nAAyS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nAAyS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-09-21T12:50:15.950" LastActivityDate="2015-09-21T12:50:15.950" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5493" PostTypeId="2" ParentId="5461" CreationDate="2015-09-21T13:24:30.107" Score="0" Body="&lt;p&gt;From experience designing and testing mechanical seals, I'll tell you straight up that it can vary a huge amount based mainly on several factors.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The squeeze on the o-ring&lt;/li&gt;&#xA;&lt;li&gt;The surface finish of the sliding surface&lt;/li&gt;&#xA;&lt;li&gt;O-ring material&lt;/li&gt;&#xA;&lt;li&gt;Lubrication&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At the sizes you're talking, squeeze on the o-ring would be my biggest concern. It will change due to variations in the o-rings themselves. Designing o-ring grooves properly, especially that small can be a challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With stainless steel, you'll want as smooth a surface finish as you can get. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For lubrication, it will depend on the application and what contaminants are tolerable, but for many applications, we use an inert white grease, &lt;a href=&quot;https://www.chemours.com/Lubricants/en_US/products/Products_Greases.html&quot; rel=&quot;nofollow&quot;&gt;Krytox&lt;/a&gt;. It's expensive but worth it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would start with a basic Viton o-ring lubricated with krytox grease, that is my go-to starting point. Unless there's some compelling reason, such as you can't tolerate any contamination whatsoever, then I'd consider a PTFE o-ring or a coated/impregnated o-ring but they get pricey fast too and the available sizes are limited. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For designing these kind of dynamic seals, I'd start with the &lt;a href=&quot;http://parker.com/portal/site/PARKER/menuitem.6c94058cc6466c6315731910237ad1ca/?vgnextoid=a141a35c7575e210VgnVCM10000048021dacRCRD&amp;amp;vgnextfmt=default&quot; rel=&quot;nofollow&quot;&gt;Parker O-ring handbook&lt;/a&gt; and design your cylinder/piston around a standard size using their recommendations. While you &lt;em&gt;can&lt;/em&gt; get custom o-rings made, it's almost never worth it unless your production volumes will be very, very large. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-09-21T13:24:30.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5494" PostTypeId="1" CreationDate="2015-09-21T14:45:18.367" Score="9" ViewCount="20921" Body="&lt;p&gt;I'm thinking about building another trailer. I've built lots of smaller trailers in the past, but this time I'd like to build myself a small tandem axle gooseneck rated for 7,500 lbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a certified welder, I have a Bachelor's degree in Physics, and I work as a software developer. I have the know how, but I would like some input on material selection.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Round Pipe&lt;/li&gt;&#xA;&lt;li&gt;Angle Iron&lt;/li&gt;&#xA;&lt;li&gt;I-Beam&lt;/li&gt;&#xA;&lt;li&gt;Rectangular Tubing&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am currently leaning towards the Rectangular Tubing as the best for support, but I cannot seem to find anything online that confirms this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After deciding on the best material for the trailer, where would I find good charts to tell me what size and thickness I should use? Obviously, I could go overkill, but I would like to build this one smarter rather than throwing as much iron at it that I have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any input? Is there a better group to post this in? I was looking for something along the line if Industrial Engineering, but this is all that pulled up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;br&gt;&#xA;I was trying to keep this a generic question where someone could tell me something like, &quot;Here is the formula we use, and this is how to use it...&quot; It looks like I won't get that, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My heaviest load would be a tractor with a front end loader and a brush cutter on the back with a total weight of 5500 to 6500 lbs. A tandem axle trailer with two (2) 3500-lbs axles can support this load fine. I have selected axles from &lt;a href=&quot;http://www.southwestwheel.com/store/showproduct.aspx?ProductID=3027&amp;amp;SEName=3500-lb-flexiride-torsion-axle-wbrakes&quot; rel=&quot;noreferrer&quot;&gt;Southwest Wheel's torsion axle with brakes&lt;/a&gt; (the front axle will have brakes, but not the rear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trailer length will be 18-foot, and have a gooseneck configuration (it distributes the weight better and pulls smoother than a bumper trailer). For calculations, I'm going to use 7500-lb capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking at the structural data for square tubing using a spec sheet &lt;a href=&quot;http://www.independencetube.com/downloads/Engineering_Brochure.pdf&quot; rel=&quot;noreferrer&quot;&gt;HERE&lt;/a&gt; (trying not to advertise another website, but that is where I see data). Page 21 shows data values for various sizes and thicknesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a line called Bending Factor. For an 18-foot trailer (18 x 12 = 216 inches), 3/8-inch thick 4x2 square tubing shows a Bending Factor of (x=1.03 , y=1.55).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was using &lt;a href=&quot;https://www.roguefab.com/calculator.php&quot; rel=&quot;noreferrer&quot;&gt;Rogue Fabrication's Calculator&lt;/a&gt; yesterday, where I entered the following values: Tube Shape=Square Tubing, Outside Diameter=4-in, Wall Thickness=0.1875-in, Material=&quot;Cheap seamed tube&quot;, Load=3800-lbs, Tube Length=216-in, and Hazard Factor=1, I got that my material is 1.22 times as strong as the loading conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I tried &lt;a href=&quot;https://www.easycalculation.com/engineering/mechanical/deflection-hollow-rectangular-beams.php&quot; rel=&quot;noreferrer&quot;&gt;EasyCalculation's Beam Deflection Calculator&lt;/a&gt;, with values of Length=216, Width=2, Height=4, Wall Thickness=0.1875, Force=3750. It shows a deflection of about 100 inches for 2 lengths of rectangular tubing. If I use 4 lengths, that drops the force down to 7500/4=1875 per beam, and deflection down to 50 inches. Those deflection values seem really high. That is more iron than most trailers have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The old tandem axle trailer I use now only has two (2) lengths of 4-inch angle iron (1/4-thick). It flexes a couple of inches, but not 50 inches. I must be missing something.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How do I calculate the amount of flex a 20-ft length of material would have?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If square tubing is not best, that's fine as long as you let me know what would be better and how you selected that configuration when you comment.&lt;/p&gt;&#xA;" OwnerUserId="3073" LastEditorUserId="3073" LastEditDate="2015-09-28T15:14:48.617" LastActivityDate="2023-01-29T15:41:31.010" Title="Steel selection for building a trailer" Tags="|mechanical-engineering|materials|structural-engineering|steel|" AnswerCount="7" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="5495" PostTypeId="2" ParentId="5474" CreationDate="2015-09-21T15:37:46.580" Score="2" Body="&lt;p&gt;You may already be aware of this, but BS 5950:2000 has been withdrawn and replaced by Eurocode 3 (BS EN 1993). The part you want is BS EN 1993-1-8: Design of joints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that EC3 also requires splice material be provided (and bolts to connect it) to carry at least 25% of the maximum compressive force in the column. So your bolts may need to carry more than just shear due to self weight. Also, depending on the purpose of this element you may need to consider possible tension in the structural integrity load case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the clause you highlighted. I believe it is referring to situations like the splice below. Note the locations of the bolts could easily result in a considerable reduction in bending stiffness in comparison to the sections themselves. See National Annex BS EN 1993-1-8 section 6.3 for guidance on calculating the rotational stiffness of a connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TrubW.jpg&quot; width=&quot;400&quot; height=&quot;200&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find the following document helpful as it provides design guidance for connections (including splices) together with worked examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.steelconstruction.info/The_Green_Books&quot; rel=&quot;nofollow noreferrer&quot;&gt;SCI P358 Joints in steel construction: Simple joints to Eurocode 3 (2014 reprint)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that there are standard splice details available which, if your situation allows it, are very much worth using.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2015-09-24T10:37:38.967" LastActivityDate="2015-09-24T10:37:38.967" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5496" PostTypeId="1" CreationDate="2015-09-21T17:21:42.243" Score="4" ViewCount="766" Body="&lt;p&gt;I am trying to find the formulas for making bends in sheet metal. I understand there are formulas for bend allowances. My issue is how do I determine where to bend if I want the inside dimension to be an exact number and if I want the outside to be an exact number. For both of the below questions the material is 6061 aluminum .080 thick with a .1 inside bend radius and 90 degree bends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example for the picture I included. If I wanted the piece to be 7&quot; on the inside and the sides to be .5&quot; I would do 7 + .5 + .5 + (2 x bend allowances) to get the total length of the plate I think? but how do I know where to put the bend lines to get 7&quot; exactly on the inside?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also on a totally separate plate how would I go about getting the total length of the unbent plate plus how would I know where to place the bends for an outside measurement of 7&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems like it might be so trivial but I just can't seem to grasp what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To clarify&lt;/strong&gt; it would be 7&quot; between verticals for the first one and 7&quot; on the outside of the verticals for the second one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not an exact part I need. It is an exercise in how to determine the lengths needed for what I am asking for. This information seems to be harder to find than gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I have read &lt;a href=&quot;https://engineering.stackexchange.com/questions/2589/how-do-i-size-metal-plates-to-get-the-correct-dimensions-after-folding&quot;&gt;How do I size metal plates to get the correct dimensions after folding?&lt;/a&gt;&#xA;and still don't understand. How does bend allowance and bend deduction help me. I am confused as ever and also extremely new at this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done a little bending on my brake and seem to have something so please clarify if you can. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a piece where I need the inside verticals to be exactly 7&quot; I drew a centerline and measured 3.5 + .05( half of radius) from center and drew a bend line. Did same on opposite side. It seemed like I got 7&quot; but it could be just luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also on a piece where I need exactly 7&quot; on the outside of the verticals I drew a centerline and did 3.5 + .05 - .080(thickness of the sheet metal) from center and seem to have gotten 7&quot; on the exterior verticals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these two formulas even remotely correct?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KDdMv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KDdMv.png&quot; alt=&quot;C shape&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3074" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-11-10T10:53:54.427" Title="Bend allowances when making a C-shape plate" Tags="|mechanical-engineering|manufacturing-engineering|autocad|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5499" PostTypeId="2" ParentId="5496" CreationDate="2015-09-21T19:04:15.897" Score="1" Body="&lt;p&gt;In Inventor, you can select the location (&quot;datum&quot;) you want to use for referencing where you intend the bend to be located. First, make a sheet metal part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yaUNf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yaUNf.png&quot; alt=&quot;Inventor 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, make the base face whatever you want the back of the part to be. Click the &quot;Flange&quot; button, then the edge you want to fold, and the dialog box will show the options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yTkhe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yTkhe.png&quot; alt=&quot;Inventor 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The options, from left to right, are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The edge you selected should be the back/outside of the flange&lt;/li&gt;&#xA;&lt;li&gt;The edge should be the start of the bend&lt;/li&gt;&#xA;&lt;li&gt;The edge you selected should be the front/inside of the flange&lt;/li&gt;&#xA;&lt;li&gt;The edge you selected should be the outermost projecting part of the bend. Note that if the bend is 90 degrees, this is the same as option 1.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In general the bend radius is determined by the sheet metal thickness; &lt;a href=&quot;http://www.custompartnet.com/wu/sheet-metal-forming&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt; has a lot of explanation about the bending process. At the end it says that &quot;the inside bend radius should equal at least the sheet thickness.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used Inventor and Solidworks to get parts manufactured and I've never had any issue with any of the default bend radii. Inventor has a free version available to students. If you're not using a CAD package to create the drawings I think you're probably wasting a lot of effort; any local university or even community college for that matter probably has a CAD lab you could use to get the drawings done. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-09-21T19:04:15.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5500" PostTypeId="2" ParentId="5469" CreationDate="2015-09-21T19:44:47.697" Score="1" Body="&lt;p&gt;I don't think you're going to find an answer that meets the intention of your question as I read it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes energy to convert water from a vapor to a liquid. Some materials (minerals, sugar, etc.) can absorb water vapor from the air because they are &lt;a href=&quot;https://en.wikipedia.org/wiki/Hygroscopy&quot; rel=&quot;nofollow noreferrer&quot;&gt;hygroscopic&lt;/a&gt;, but it also means that there is a finite limit to the amount of water they can absorb before they need to be dried or replaced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest you're going to get is the &lt;a href=&quot;https://engineering.stackexchange.com/a/5476/1633&quot;&gt;answer from mart&lt;/a&gt;, where you drop the air temperature, but this means that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You need a cave or other geothermal heat sink&lt;/li&gt;&#xA;&lt;li&gt;You need a blower to circulate cave-side air&lt;/li&gt;&#xA;&lt;li&gt;You need a blower to circulate room-side air&lt;/li&gt;&#xA;&lt;li&gt;You need a heater to reheat room-side air,&lt;/li&gt;&#xA;&lt;li&gt;Whatever condensate forms needs to be plumbed&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;and, most importantly,&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;You can only dehumidify to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dew_point&quot; rel=&quot;nofollow noreferrer&quot;&gt;dew point&lt;/a&gt; of the cave temperature. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Nothing will create a sustained dehumidification by itself. You need energy input. If such a material could absorb an infinite amount of humidity, then (1) why do we still have humidity (why isn't it all absorbed already), and (2), where would that infinite quantity of water go?&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-09-21T19:44:47.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5501" PostTypeId="1" CreationDate="2015-09-21T21:16:22.580" Score="1" ViewCount="106" Body="&lt;p&gt;Design a combinational circuit that multiplies two numbers together, and outputs the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest product will be 3 * 3 = 9.&#xA;Four bits to represent the product.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0: 00&#xA;1: 01&#xA;2: 10&#xA;3: 11&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need a 4 bit register for the outputs, right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I start my truth table? like this?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A B C D | Output | AB | CD&#xA;0 0 0 0   0         0    0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and so onto minterm 15?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will I form the expressions from the truth table?&#xA;I know how to solve with a K-map, but I'm having a hard time visualizing and building it.&lt;/p&gt;&#xA;" OwnerUserId="3077" LastEditorUserId="1586" LastEditDate="2015-09-24T00:24:57.393" LastActivityDate="2015-09-24T00:24:57.393" Title="Circuit that multiples two numbers together?" Tags="|electrical-engineering|circuit-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5502" PostTypeId="1" CreationDate="2015-09-21T23:51:19.793" Score="5" ViewCount="524" Body="&lt;p&gt;Whenever the question &quot;Why do airplanes fly?&quot; is asked (e.g. &#xA;&lt;a href=&quot;https://physics.stackexchange.com/questions/290/what-really-allows-airplanes-to-fly&quot;&gt;»What really allows airplanes to fly?«&lt;/a&gt;, &#xA;&lt;a href=&quot;https://engineering.stackexchange.com/questions/5477/how-an-airfoil-works&quot;&gt;»How an airfoil works?«&lt;/a&gt;) at a certain point the following is cited:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bernoulli's principle accounts for 20% of an airplane's lift…&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a very general formulation &lt;a href=&quot;https://en.wikipedia.org/wiki/Bernoulli%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bernoulli's Principle&lt;/a&gt; states&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_\mathrm{total} = p_\mathrm{dynamic} + p_\mathrm{static}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming low ( $M\ll0.3$ ) velocities this can be written as (resulting in an error of 5% at $M=0.3$ due to density change):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_\mathrm{total} = \frac{1}{2}\rho u^2 + p_\mathrm{static}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &lt;strong&gt;Bernoulli's Principle&lt;/strong&gt; is defined as that. Then ( in case of $M\ll0.3$ ) it &lt;strong&gt;does&lt;/strong&gt; account for &lt;strong&gt;100% of the lift&lt;/strong&gt;, since it describes how the static pressure difference between suction and pressure side results into drag and lift (neglecting skin friction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which additional assumptions have to be made to come up with a lift which is only 20% of the actual lift? Is this the lift one would calculate when the lift-force is based on the equal-transit-time-assumption? Or is this an error occurring when a delta wing also features vortex system so that the definition of the total pressure might be difficult?&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2017-08-18T05:41:18.483" LastActivityDate="2017-08-18T05:41:18.483" Title="What is &quot;Bernoulli-Lift&quot;, and why does it only account for 20% of an airplane's lift?" Tags="|aerospace-engineering|aerodynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5503" PostTypeId="2" ParentId="5480" CreationDate="2015-09-22T07:01:58.573" Score="2" Body="&lt;p&gt;Well a stage is an equilibrium stage ie. given some feed composition z, temperature and pressure, vapour of composition y goes up and liquid composition x goes down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you are calculating actual or theoretical stages, you will hardly ever get a whole number. If you look at the graph, last stage (where xb = 0.15), the 5/7 stage probably comes from the (Stage N-1 composition - xb)/(Stage N-1 composition - Stage N composition). You can verify this yourself. Looks about right from eyeballing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to see it is if your column had 10 theoretical stages, then xb will be around 0.12 (from the graph). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ek just accounts for the fact that in an actual column the stages will not be at equilibrium. Meaning on each stage n y will be less than what is predicted on the VLE curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude, you didn't do anything wrong. If this were an actual design and you come up with 13.87 actual stages required you would then select 14 stages. Then if you really want to meet xb=0.15 you can do that by playing with the bottom or top temperature.&lt;/p&gt;&#xA;" OwnerUserId="1103" LastActivityDate="2015-09-22T07:01:58.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5504" PostTypeId="1" CreationDate="2015-09-22T10:50:43.987" Score="1" ViewCount="888" Body="&lt;p&gt;I would like to know what properties would be necessary in a metal material used to construct an mining excavator bucket the brushes and pins for mining coal. I think that hardness and toughness would be necessary for the bucket however the brushes may need to be made of a softer material to save the pins and bucket from wear.&lt;/p&gt;&#xA;" OwnerUserId="3080" LastEditorUserId="1586" LastEditDate="2015-09-24T17:18:38.537" LastActivityDate="2015-09-24T17:18:38.537" Title="Material requirement for excavator bucket, brushes and pins" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="1" ClosedDate="2015-09-24T17:23:45.593" ContentLicense="CC BY-SA 3.0" />
  <row Id="5505" PostTypeId="2" ParentId="5501" CreationDate="2015-09-22T11:04:38.720" Score="1" Body="&lt;p&gt;You seem to be on the right track.  You have only 16 possible unique input states, represented as 4 bits.  I would start by writing out the truth table containing just the 4 input bits and the 4 output bits.  AB and CD may well end up as useful intermediate terms, but don't assume that until you see what the truth table looks like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding sub-expressions and how to combine them to make combinatorial logic implement a particular logic function is exactly what Karnough maps are intended for.  Your problem is small enough so that a K-map is tractable.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-09-22T11:04:38.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5507" PostTypeId="1" AcceptedAnswerId="5509" CreationDate="2015-09-22T12:05:07.420" Score="6" ViewCount="431" Body="&lt;p&gt;The following scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am tasked to design a 6.6m high reinforced concrete wall in an industrial building. We have a 10 ton Telehandler (telescopic loader) with 7m maximum reach, loading 1500kg loose material in a scoop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to cater for a &quot;runaway vehicle collision&quot; where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the vehicle reaches a maximum velocity of 2m/s before it collides with the wall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find out the worst case collision force on my wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually I would employ the design methodology for vehicle barriers and parapets in car parks in Eurocode 1 (EN 1991-1-1). But in this case the point of contact is significantly higher than the centroid of the colliding vehicle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that the front wheels of the loader leave the ground during the collision, thus there is some sort of energy dissipation there too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure that the long boom just frightens me and is enough of a distraction to stop me from thinking straight here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rk8ID.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rk8ID.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="1586" LastEditDate="2015-09-24T00:25:35.920" LastActivityDate="2015-09-24T00:25:35.920" Title="Telehandler wall collision forces" Tags="|civil-engineering|reinforced-concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5508" PostTypeId="2" ParentId="2832" CreationDate="2015-09-22T12:56:14.473" Score="1" Body="&lt;p&gt;It's absolutely possible and has been done for some time now in &lt;a href=&quot;http://www.coolsound.us/technology.html&quot; rel=&quot;nofollow&quot;&gt;thermoacoustic compressors&lt;/a&gt;. Originally they were developed for cryo-coolers to condense gases into liquids and that remains their principle application though there are companies working to bring this technology to a consumer level. These compressors have no, or at most one, moving parts (the sound source). They also have the benefit that they don't use any greenhouse gases.     &lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2015-09-22T16:14:31.983" LastActivityDate="2015-09-22T16:14:31.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5509" PostTypeId="2" ParentId="5507" CreationDate="2015-09-22T14:16:17.687" Score="4" Body="&lt;p&gt;In a situation like this you will need to resort to energy methods. Depending on how complicated you wanted to get you could do a dynamic calculation also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend looking at the guidance in Eurocode 1991-1-7 (Accidental Actions). In particular look at Annex C which describes dynamic design for impact. This essentially involves equating the kinetic energy of the object impacting the wall to the energy absorbed by deformation of the wall. There are a few simple formulas which can be used depending on how you classify the impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to consider a variety of loading cases for example: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Load at various heights - separate (or possibly simultaneous) impacts&#xA;of the loader and load. &lt;/li&gt;&#xA;&lt;li&gt;Load and loader all impacting as one 'lump'&lt;/li&gt;&#xA;&lt;li&gt;etc&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From my experience with impact calculations it is unusual for a structure to be able to withstand any significant impact through elastic deformations alone. In which case the structure will obviously be damaged. In this case you have a few options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Protect the structure from being hit in the first place&lt;/li&gt;&#xA;&lt;li&gt;Perform a plastic analysis and provide appropriate detailing so that the resulting plastic configuration can be achieved&lt;/li&gt;&#xA;&lt;li&gt;Allow the structure to be damaged in some controlled way (imagine a fuse or some kind of energy dissipating system)&lt;/li&gt;&#xA;&lt;li&gt;Accept that whatever is being hit will not exist after the impact and deal with the consequences (structural integrity)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In my view it is prudent to consider the consequences of your constraints being exceeded:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What if someone is driving too fast? &lt;/li&gt;&#xA;&lt;li&gt;What if someone is carrying a load which is too heavy?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To address your comment below. Clause 4.4 in EN 1991-1-7 states for accidental actions caused by forklift trucks:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The National Annex may give the value of the equivalent static design&#xA;  force F. &lt;strong&gt;It is recommended that the value of F is determined according&#xA;  to advanced impact design for soft impact in accordance with C.2.2&lt;/strong&gt;.&#xA;  Alternatively, it is recommended that F may be taken as 5 W, where W&#xA;  is the sum of the net weight and hoisting load of a loaded truck (see&#xA;  EN 1991-1,1 , Table 6.5), applied at a height of 0,75 m above floor&#xA;  level. However, higher or lower values may be more appropriate in some&#xA;  cases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So one way of approaching the problem would be to apply an equivalent single degree-of-freedom analysis: First obtain the maximum dynamic interaction force from Equation C.1 (substituting the stiffness of your wall for k as specified in clause C.2.2(1) ):&#xA;$$ F = v_r \sqrt{k m} $$&#xA;Where $v_r$ is the velocity at impact, k is the wall stiffness, and m is the mass of the impacting object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time duration of loading can be obtained from equation C.2:&#xA;$$ \Delta t = \sqrt{\frac{m}{k}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The static deflection can then be calculated by:&#xA;$$ \frac{F}{k} $$&#xA;A dynamic load factor assuming a rectangular pulse load (see Fig C.1) can  be calculated according to [1]:&#xA;$$ DLF = 1 - \cos{2\pi\frac{\Delta t}{T}} $$&#xA;Where T is the natural frequency of vibration of the wall. The dynamic displacement is then:&#xA;$$ \delta = DLF \frac{F}{k} $$&#xA;You will immediately notice that: &#xA;$$ 0 \leq DLF \leq 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then obtain the desired section forces/moments by using the dynamic displacement in the standard deflection formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of the wall stiffness k will depend on the impact location, construction of the wall, support conditions, if the wall remains elastic (if it is not elastic than the above analysis is no longer valid), etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book cited below is quite good for explaining how to perform an equivalent single-degree-of-freedom analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: Biggs, John M. &quot;Introduction to structural dynamics.&quot; (1964). (One of the best introductory books on structural dynamics in my opinion - despite its age)&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2015-09-22T21:53:40.897" LastActivityDate="2015-09-22T21:53:40.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5510" PostTypeId="2" ParentId="5504" CreationDate="2015-09-22T15:30:06.110" Score="2" Body="&lt;p&gt;As with all earth digging equipment, abrasion resistance is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ground engaging tools used to dig coal, particularly underground, the tools must not create sparks when they engage the ground or muck pile due to the potentially explosive environment from methane gas or coal dust.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-09-22T15:30:06.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5511" PostTypeId="2" ParentId="5469" CreationDate="2015-09-22T19:12:28.087" Score="2" Body="&lt;p&gt;Once again, forced to answer because not enough reputation to comment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several answers/comments have talked about requiring energy to reduce humidity.  @Chuck specifically states you need energy to convert from vapor to liquid.  In fact, the reverse is true.  When vapor condenses to liquid it gives up energy.  For water it's approximately 1000 BTU/pound of water.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sustainable system will require energy, either to regenerate the absorbant (salt, silica gel, other) or for a mechanical system such as a refrigeration cycle to make something cold enough to cause condensation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use an absorbent, then you could end up with an analog of an industrial absorption system.  In those, the working &quot;tower&quot; absorbs vapor from the gas (water from the air in this case) while an offline &quot;tower&quot; is regenerated by, for example, being heated with dry gas (air).  &lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2015-09-22T19:12:28.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5512" PostTypeId="1" CreationDate="2015-09-22T19:44:03.647" Score="9" ViewCount="2398" Body="&lt;p&gt;&lt;strong&gt;The Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where should a closure pour be located relative to centerline of girder and how should the closure pour width be detailed/designed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;And now, a little background:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When designing a steel girder bridge using phased construction with dead load deflections exceeding 2&quot;, my state (Nebraska, USA) requires a closure pour. No guidance is provided on how to locate or design the closure pour. As a point of reference, I'm dealing with a 100 ft span and a 44&quot; deep girder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done moderately extensive Googling and found varied suggestions. Align the edge of the pour with centerline of girder. Don't align the edge of pour with centerline of girder. 2' minimum width. 3' minimum width. And an interesting suggestion from &lt;a href=&quot;https://www.nevadadot.com/uploadedFiles/NDOT/About_NDOT/NDOT_Divisions/Engineering/Structures/Chapter16.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nevada DOT (p. 16-10)&lt;/a&gt; that: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The required width can be estimated by considering the&#xA;  closure pour to be a fixed-fixed beam and by limiting the stresses in the concrete to the&#xA;  cracking stress.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which sounds quite promising, though I haven't fully digested the particulars of execution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where my noggin really starts to hurt is when I start thinking about how my girder deflections will be affected by phasing and a closure pour. Say my girder has a tributary deck width of 96&quot; in the finished condition. But during phased construction, only the left 2/3 of this deck will be poured. Do people execute a series of analysis runs looking at all the intermediate conditions and the progressive deflections/stresses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tDxjv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tDxjv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2015-09-23T17:51:37.247" LastActivityDate="2016-07-06T05:26:12.683" Title="How should a bridge deck closure pour be positioned?" Tags="|structural-engineering|bridges|deflection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5514" PostTypeId="1" AcceptedAnswerId="5519" CreationDate="2015-09-23T03:59:12.370" Score="10" ViewCount="4875" Body="&lt;p&gt;I have a very basic understanding of how gears work, and while trying to learn more, I've run into a small point of confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a 2:1 gear ratio is achieved by having twice as many teeth on one gear than the other, then how are 250:1 ratios achieved? Certainly there aren't gears with 2500 teeth on one gear, and 10 on the other...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, here's a very tiny motor with a 250:1 gear ratio: &lt;a href=&quot;http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_253534_-1&quot; rel=&quot;noreferrer&quot;&gt;DC Motor with Gearhead&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="300" LastEditorUserId="1586" LastEditDate="2015-09-24T17:18:27.700" LastActivityDate="2016-02-17T15:54:04.537" Title="How are high gear ratios achieved in a small space?" Tags="|mechanical-engineering|gears|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5515" PostTypeId="1" CreationDate="2015-09-23T05:04:46.177" Score="3" ViewCount="183" Body="&lt;p&gt;What is &quot;hot-cure resin&quot;? Is it the resin that has been once melted and solidified again?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is used talking about the big glassy boxes shown in &lt;a href=&quot;http://www.artandsciencejournal.com/post/44699915623/sense-by-annie-cattrell-the-five-senses-become&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; image. Here is the context:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Working with neuroscientists Steve Smith and Mark Lythgoe, Cattrell&#xA;  captured fMRI digital data relayed while subjects were caught in the&#xA;  act of looking and listening, and then used RP to transform these&#xA;  isolated processes into computerised virtual models. Out of these she&#xA;  made waxy resin sculptures, embedding them in solid square&#xA;  ‘brain-boxes’ made of transparent hot-cure resin.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3091" LastEditorUserId="9" LastEditDate="2015-09-24T19:22:29.213" LastActivityDate="2015-09-24T19:22:29.213" Title="What kind of material is &quot;hot-cure resin&quot;?" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5516" PostTypeId="2" ParentId="5515" CreationDate="2015-09-23T05:15:54.157" Score="4" Body="&lt;p&gt;Sounds like a thermosetting polymer resin that cures (reacts to form crosslinks, making it solid) at temperatures above room temperature. Hence &quot;hot-cure.&quot; Though, in fairness, I've never heard that term. Curing is a thermally activated phenomenon, and, barring limiting kinetic effects, proceeds exponentially faster with linear increases in temperature. In that sense, all thermosetting resins are hot-cure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while resin can be used to refer to a thermoplastic melt, or a thermoset, it almost always refers to the latter. In that case, the resin can't be melted or re-melted once it has cured, because a thermosetting chemical reaction has taken place which can't be reversed by thermal action alone. Curing in the context of resin generally refers to that thermosetting chemical reaction.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-09-23T05:21:33.183" LastActivityDate="2015-09-23T05:21:33.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5517" PostTypeId="1" CreationDate="2015-09-23T08:33:38.063" Score="0" ViewCount="1173" Body="&lt;p&gt;This is a follow on to &lt;a href=&quot;https://engineering.stackexchange.com/questions/5334/how-are-blanking-spades-entered-into-gas-pipelines&quot;&gt;my previous question about blanking spades&lt;/a&gt;. There's a gas pipeline with a butterfly valve mounted between to flanges. I want to work on one side (call it the inert side) and need to close his side with a blanking spade to the other side (gas side). To enter the spade I have to close the valve and open a few bolts so I can push the spade in. But the moment I remove bolts, the force (What's Kraftschluss in english) between the valve and the flange on the gas side is weaker and I get a leak. At least that's what I think happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on pressure on the gas side, I either have a gas leak and a potential hazardous atmosphere in the working area, or  air infiltration into the pipe and a potential explosion hazard there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far all no butterfly valve I've seen could be mounted as a headpiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So either there's a trick to it or a short time hazardous atmosphere is accepted (and can be worked around with appropriate safety measures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the actual procedure to enter a spade into a pipeline in light of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I don't do or supervise these works. I'm just curios how the folks doing it (should) go about it. &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-09-23T12:37:16.573" Title="Exact procedure to mount a blanking spade near a butterfly valve" Tags="|pipelines|safety|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5518" PostTypeId="1" AcceptedAnswerId="5522" CreationDate="2015-09-23T12:04:21.123" Score="2" ViewCount="344" Body="&lt;p&gt;What would the ideal thickness of the dump trailer bed be, so it can withstand loading rock from a height of 1 meter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spacing of ribs is yet to be decided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is that the trailer be as light as possible, whilst being strong enough to withstand hauling rock every once in a while.&lt;/p&gt;&#xA;" OwnerUserId="3081" LastEditorUserId="9" LastEditDate="2015-09-24T19:22:20.780" LastActivityDate="2015-09-24T19:22:20.780" Title="What is the ideal metal sheet thickness used in a 1 tonne atv/tractor dump trailer?" Tags="|mechanical-engineering|metals|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5519" PostTypeId="2" ParentId="5514" CreationDate="2015-09-23T12:11:38.347" Score="8" Body="&lt;p&gt;One simple way is to do the gearing in a number of stages in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gear_train&quot; rel=&quot;nofollow noreferrer&quot;&gt;gear train&lt;/a&gt;.  Such a system uses gears which have two different gear sizes on the same wheel. In the example shown in the image below (from the &lt;a href=&quot;http://science.howstuffworks.com/transport/engines-equipment/gear-ratio3.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;How Stuff Works article on gear trains&lt;/a&gt;) each gear has a ratio of 2:1 such that the final ratio of the magenta gear to the blue gear is 8:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7kt9n.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7kt9n.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-09-30T12:05:40.157" LastActivityDate="2015-09-30T12:05:40.157" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5520" PostTypeId="2" ParentId="5517" CreationDate="2015-09-23T12:37:16.573" Score="1" Body="&lt;p&gt;You are correct, you can't disassemble a valve that's currently isolating system pressure. This means that you have to isolate the system (at least) one valve upstream of the valve you intend to blank. Note that, if your upstream piping branches between the intended valve and the next upstream valve, you need to shut the valve in &lt;em&gt;every&lt;/em&gt; upstream line. This could be a lot of valves and/or could significantly impact operations in other locations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, the considerate system designer will include a gasketed flange, with no valve installed, just downstream of an isolation valve for every major piece of equipment specifically for this reason - you shut the isolation valve, &lt;strong&gt;bleed the system&lt;/strong&gt;, then install the blank. The valve should, theoretically, be enough to hold back system pressure, but valves leak, so a &lt;strong&gt;gasketed&lt;/strong&gt; blank flange is generally the accepted method of working on a system that normally runs a hazardous fluid (acid, steam, flammable/toxic gas, etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are intending to take the equipment down for maintenance on a frequent/regular interval, sometimes spectacle flanges, seen below, are used. It's basically just a quick way to switch from an open pipe to an isolated pipe, but remember - always use new gaskets! A pipe without a seal doesn't!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0pJUi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0pJUi.jpg&quot; alt=&quot;Spectacle Flange&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-09-23T12:37:16.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5521" PostTypeId="1" CreationDate="2015-09-23T12:39:34.653" Score="1" ViewCount="327" Body="&lt;p&gt;I recently had the need to quickly build a fume extractor for soldering, and since at the place there was a ton of 80mm CPU fans lying around, I used those to build something like this (not to scale):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mL3sg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mL3sg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the components were (all airtight &quot;sealed&quot; with duct tape)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Intake of an old clothes dryer hose (100mm diameter, a bit more than 1m long when extended, fixed to some big metal lump to have it stay in place). This is positioned dynamically to suck in soldering fumes&lt;/li&gt;&#xA;&lt;li&gt;the hose itself&lt;/li&gt;&#xA;&lt;li&gt;A cut off pyramid shaped cardboard construction&lt;/li&gt;&#xA;&lt;li&gt;A 3x3 stack of 80mm CPU fans. Directly behind it, the air comes out (future enhancement might be some coal filter)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So the airflow is from 1 to 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This thing did its job, but not in any terrific way. So I was wondering if anyone with experience in fluid dynamics could tell me how to improve it, or more specifically which of my ideas to improve it might make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of this question, lets say the resources are limited to about the items described, reasonable additions that most people have in their workshop included (so e.g. hot glue is fine, but I am not looking for &quot;add a bigger fan&quot; kind of enhancements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The items are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The hose (2m of it, 100mm diameter)&lt;/li&gt;&#xA;&lt;li&gt;Enough cardboard&lt;/li&gt;&#xA;&lt;li&gt;10 fans, 80mm diameter, all assumed to be equal in performance&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My ideas now included:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;instead of making 3x3 fan configuration, I take two 2x2 configurations and put them &quot;in series&quot; (that is, a box like 4 to its right, with reasonable space in between, whatever that space needs to be)&lt;/li&gt;&#xA;&lt;li&gt;at point 3, add an additional 80mm fan &lt;strong&gt;or&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;at point 2, add an additional 80mm fan &lt;strong&gt;or&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;at point 1, add an additional 80mm fan (all assuming for that additional point I would only have one last fan left in case of a 3x3 configuration)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which of these ideas (or combinations of one additional fan and the other fan configurations) would noticeably improve the performance of the device (that is, move more air)? &lt;/p&gt;&#xA;" OwnerUserId="706" LastEditorUserId="1586" LastEditDate="2015-09-24T00:24:22.340" LastActivityDate="2015-09-24T00:24:22.340" Title="Best placing of fans for makeshift fume extractor" Tags="|mechanical-engineering|airflow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5522" PostTypeId="2" ParentId="5518" CreationDate="2015-09-23T12:46:43.780" Score="3" Body="&lt;p&gt;See &lt;a href=&quot;https://engineering.stackexchange.com/questions/5494/steel-selection-for-building-a-trailer&quot;&gt;this similar-ish question&lt;/a&gt;. The ideal thickness will be the thickness &gt;= the thickness required to keep stresses below the yield strength of &quot;the metal.&quot; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pick a metal, find its yield strength. &lt;/li&gt;&#xA;&lt;li&gt;Pick a factor of safety. Use this and the yield strength to calculate an allowable stress. &lt;/li&gt;&#xA;&lt;li&gt;Since you're dropping a load onto the trailer, find out spring/damping constants for the suspension as the suspension will affect peak loading.&lt;/li&gt;&#xA;&lt;li&gt;Pick a rib spacing. &lt;/li&gt;&#xA;&lt;li&gt;Draw a &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_and_moment_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;bending moment diagram&lt;/a&gt;, find the peak moment, then use that and your allowable stress to find your thickness.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that your scenario will be harder to design for (you should have a large factor of safety) because the rocks will not deform to create a true distributed load. Instead you'll have corners concentrating load in a (relatively) few number of places. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-09-23T12:46:43.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5523" PostTypeId="1" AcceptedAnswerId="5588" CreationDate="2015-09-23T13:11:17.030" Score="7" ViewCount="1790" Body="&lt;p&gt;I'm currently designing a small reinforced concrete bridge on private property. The project is in Scotland. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very small scale, a single lane, with a span of around 4.5m.  Service life is taken as 120 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Eurocode design for standard road bridges seems far too over-the-top (there will not be any lorries crossing this bridge for example), so I'm looking for a good load design precedent (with backup documentation). The largest vehicle that may foreseeably cross the bridge would be, say, a 30 tonne concrete mixer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's a good code that relates to, say, loading on temporary works or construction traffic loading then we feel that would be perfect for use on this bridge. Lest we over-engineer the thing and it costs a fortune... &lt;/p&gt;&#xA;" OwnerUserId="3093" LastEditorUserId="33" LastEditDate="2016-02-09T02:16:42.390" LastActivityDate="2016-02-09T02:16:42.390" Title="Looking for a loading code for use in the design of a small scale private bridge? (4.5m span, one lane)" Tags="|civil-engineering|structural-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5524" PostTypeId="2" ParentId="5521" CreationDate="2015-09-23T14:10:40.793" Score="1" Body="&lt;p&gt;If the fans have high air flow but low static pressure capabilities you would be better off with the 2x2 in series to overcome the flow restriction of the dryer hose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways to improve flow too...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pyramid looks like the picture and assuming your fans are mounted on point 4, and that point 4 is the intake, the cone of the pyramid needs to be much longer. If the fan is mounted right next to a 45deg bend, you are losing airflow just from that. The longer the cone, the less the angle the air needs to change, the better your flow will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also your using expanding clothes dryer hose that is likely ribbed on the inside, if possible run 2 hoses in parallel since I think that hose is restricting your flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the best flow, if you have a large amount of cardboard and tape available, just build a duct out of cardboard that keeps at least the same cross sectional area. &lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2015-09-23T14:10:40.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5525" PostTypeId="2" ParentId="5514" CreationDate="2015-09-23T14:17:31.600" Score="4" Body="&lt;p&gt;Using a worm drive you can have one tooth per rotation on one shaft and have that drive a gear with 250 teeth, for a 250:1 ratio in one stage.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2015-09-23T14:17:31.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5526" PostTypeId="2" ParentId="5523" CreationDate="2015-09-23T15:38:25.943" Score="4" Body="&lt;p&gt;I'm not particularly familiar with the Eurocode (I work in the US).  However, a 30-tonne concrete truck doesn't seem much smaller than what we call the AASHTO Design Truck in the US.  The Design Truck has a loading as shown below (ref. AASHTO &lt;em&gt;LRFD Bridge Specification,&lt;/em&gt; 2012).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GyzNA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GyzNA.png&quot; alt=&quot;design truck&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By my calculations, the metric equivalent of the Design Truck has 14.5-tonne axle loads (&lt;a href=&quot;https://www.google.com/#safe=off&amp;amp;q=32000+lbs+to+tonne&quot; rel=&quot;nofollow noreferrer&quot;&gt;32 kip is about 14.5 tonne&lt;/a&gt;).  AASHTO also uses the Design Tandem, which has a pair of 25-kip (&lt;a href=&quot;https://www.google.com/#safe=off&amp;amp;q=25000+lbs+to+tonne&quot; rel=&quot;nofollow noreferrer&quot;&gt;11.3-tonne&lt;/a&gt;) axles spaced 4ft (&lt;a href=&quot;https://www.google.com/#safe=off&amp;amp;q=4ft+to+m&quot; rel=&quot;nofollow noreferrer&quot;&gt;1.2 m&lt;/a&gt;) apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I'm sure you know, concrete is &lt;em&gt;heavy,&lt;/em&gt; and the trucks are heavier than one might expect given their footprint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that your standard lorrie has a similar loading to the Design Truck since they both are designed to carry shipping crates.  By this logic, designing for a 30-tonne concrete truck isn't really all that different than designing for a lorrie.  Depending on the wheel configuration of your concrete truck, its loading could actually be more critical than the lorrie.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, doing your bridge design for a lorrie load seems appropriate to me.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-09-23T19:30:52.223" LastActivityDate="2015-09-23T19:30:52.223" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5527" PostTypeId="1" CreationDate="2015-09-23T16:41:14.553" Score="1" ViewCount="118" Body="&lt;p&gt;Currently I am trying to rotate a piece of steel &gt;90 degrees without any power input. The piece of steel is attached on the wall. A servo is also attached on the steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question addresses turning the piece of steel if the servo fails to work. A battery is not supposed to be used. Generally like a servo may do it with power the rotation should be performed without power whilst in this case the rotation has to take place only once.&#xA;The trigger is a toggle switch that will drop,if pushed, a weight on the right side (see picture) in order to rotate the piece of steel. This rotation has only to be performed once. Doing this currently the required rotation &gt;90 degree does not take place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Where has the weight to tip the steel ideally so that the piece of steel can be moved/rotated with the least amount of force? I was thinking that the edge of on the right side of the steel would be most beneficial location to tip with the weight. Am I right?&lt;/li&gt;&#xA;&lt;li&gt;What weight must ‘the object’ be in order to move a 8cm x 2cm x 4cm piece of steel &gt;90 degrees. Hereby I have to know the required force on the steel in order to rotate it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Do I think too naively and have any mistake on my thoughts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oLCK1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oLCK1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3095" LastEditorUserId="3095" LastEditDate="2015-09-27T21:30:54.560" LastActivityDate="2015-09-27T21:30:54.560" Title="How can a piece of steel rotated without any power input?" Tags="|mechanical-engineering|control-engineering|kinematics|" AnswerCount="0" CommentCount="10" ClosedDate="2015-09-23T19:15:45.050" ContentLicense="CC BY-SA 3.0" />
  <row Id="5528" PostTypeId="1" AcceptedAnswerId="5530" CreationDate="2015-09-23T17:32:43.037" Score="6" ViewCount="435" Body="&lt;p&gt;According to &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0962373109&quot;&gt;Hoffman&lt;/a&gt;, the standard $k-\epsilon$ turbulence model incorporates a two-layer approach (inner region and outer region formulations to reperesent mixing length) when it comes to regions adjacent to wall surface $(y^+ \approx 30 - 50)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of the $k-\epsilon$ model is that a two additional equations are added to the system of flow governing equations, one equation for kinetic turbulence energy $k$ and the other is for the turbulence energy dissipation rate $\epsilon$, and the closure problem is solved by using dimensionless groups as boundary conditions such as turbulence intensity, turbulence length scale and hydraulic diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why the $k-\epsilon$ model uses zero equations models (algebraic models) for regions adjacent to the wall, what advantages do zero equations models have over $k-\epsilon$ if by definition zero equations models assume that the rate of production of turbulence and the rate of dissipation are approximately equal and they do not include the convection of turbulence?&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2015-09-24T12:52:50.950" Title="Why does the standard $k-\epsilon$ model use zero equation models?" Tags="|mechanical-engineering|fluid-mechanics|cfd|turbulence|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5529" PostTypeId="2" ParentId="5528" CreationDate="2015-09-23T18:17:43.960" Score="0" Body="&lt;p&gt;I know that from a CFD point of view (or any time you need a numerical solution to a system of ODEs/PDEs), zero-equation models have two major benefits: 1. they don't need to be integrated by the numerical algorithm, which reduces the computational complexity of the simulation and may improve solution convergence and numerical accuracy, and 2. they don't require any kind of state or history to be maintained (which reduces memory consumption).  The cost of this simplicity is, of course, accuracy, especially in regions of high turbulence, flow separation, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the problem, I usually start with k-e because it's simple and numerically stable, and only after I get a good initial solution and work out all the other bugs in the simulation do I experiment with more complex turbulence models.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-09-24T12:52:50.950" LastActivityDate="2015-09-24T12:52:50.950" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5530" PostTypeId="2" ParentId="5528" CreationDate="2015-09-23T19:26:55.650" Score="2" Body="&lt;p&gt;There are two ways of &quot;&lt;em&gt;wall treatment&lt;/em&gt;&quot; both feature two layers but at different $y^+$. The general idea is that the velocity gradients in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Boundary_layer&quot; rel=&quot;nofollow noreferrer&quot;&gt;boundary layer&lt;/a&gt; are so high that one would need a very high number of grid cells in order to resolve those gradients. In order to overcome that the flow close to the wall is modeled by one of the following algebraic models (see &lt;a href=&quot;http://www.cfd-online.com/Wiki/Two_equation_turbulence_models#Near-wall_treatments&quot; rel=&quot;nofollow noreferrer&quot;&gt;cfd-online&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Low Reynolds number treatment (LRN)&lt;/li&gt;&#xA;&lt;li&gt;High Reynolds number treatment (HRN)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both &lt;em&gt;treatments&lt;/em&gt; are based on the &lt;a href=&quot;http://www.cfd-online.com/Wiki/Law_of_the_wall&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Law of the Wall&lt;/strong&gt;&lt;/a&gt;. Numerous experiments have shown that the boundary layer of  a flat plate consists of two distinct regions. From the wall until $y^+ \approx 4$ it is called viscous sublayer with a linear velocity gradient and a logarithmic region from $y^+ \approx 30$. Both layers are joint by a region called buffer-layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one chooses a LRN approach the numerical grid needs to fully resolve the boundary layer in the log-region with (rule of thumb) the first grid-cell at $y^+ =1$. The viscous sub-layer is then modeled algebraically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case the HRN approach is chosen the first grid-cell should be in the log-region. Both layers viscous sublayer and log-layer are than modeled algebraically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SXibb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SXibb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br/&gt; &lt;em&gt;from Turbulence and Transport Phenomena by &lt;a href=&quot;https://www.researchgate.net/profile/Kemal_Hanjalic&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hanjalic&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Law_of_the_wall&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to your question has three parts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Due to the high velocity gradients in the boundary layer it is beneficial with respect to the computational costs to model the flow close to the wall.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The modeling of the boundary layer makes the simulations more robust.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The error introduced into the solution by algebraically modeling the boundary layer is generally small (obviously depending on the simulation itself). Your mentioned neglect of the convective transport of turbulent properties is correct (Which in certain cases will introduce an error).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-09-23T20:55:19.783" LastActivityDate="2015-09-23T20:55:19.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5531" PostTypeId="1" AcceptedAnswerId="5532" CreationDate="2015-09-23T21:22:32.013" Score="3" ViewCount="251" Body="&lt;p&gt;I would like to fold thin metal sheets to a particular angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: fold a 0.1mm thick steel strip of width 20mm and length 150mm to an angle of 45° along its length (10mm panels on each side of the fold).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to keep the radius of the fold as small as possible, ideally on the order of the thickness or smaller. I would also like to do this on a bench top. I plan to use a range of thicknesses, up to 1mm thickness. But I am flexible on this constraint (but need to use thicknesses &amp;lt;1mm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any tools/machines which can do this in a repeatable way (ideally with 1-2° variation or less)? Are there any standards for this type of forming?&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="1586" LastEditDate="2015-09-23T22:07:09.973" LastActivityDate="2015-09-24T02:20:33.850" Title="Folding of thin gauge metal sheets" Tags="|mechanical-engineering|metal-folding|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5532" PostTypeId="2" ParentId="5531" CreationDate="2015-09-23T21:32:32.547" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I would like to fold thin metal sheets to a particular angle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It sounds like you're looking for a &lt;a href=&quot;https://www.northerntool.com/shop/tools/product_7825_7825&quot; rel=&quot;nofollow noreferrer&quot;&gt;sheet metal brake&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B2zem.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B2zem.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure you can find less expensive models elsewhere.  Lots of places rent them too.&#xA;Heck, with the thicknesses you're using, you can make one with some hardwood and hinges.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-09-24T02:20:33.850" LastActivityDate="2015-09-24T02:20:33.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5534" PostTypeId="1" CreationDate="2015-09-24T08:00:13.707" Score="1" ViewCount="109" Body="&lt;p&gt;I'm finding difficult in implementing real time clock for pic16lf1902, wherein their are no serial protocols for communicating/interfacing with builtin LCD display. Is their a convenient way to solve the problem? For example using I2C software functions with DS1309 for LCD disply as used for PIC18 microcontrollers. &lt;/p&gt;&#xA;" OwnerUserId="3105" LastEditorUserId="1586" LastEditDate="2015-09-24T17:31:14.987" LastActivityDate="2015-09-25T11:25:13.200" Title="Real time clock for PIC16LF1902/03" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5535" PostTypeId="1" CreationDate="2015-09-24T10:17:58.657" Score="3" ViewCount="2447" Body="&lt;p&gt;Given the temperature $T_1$ and volume flow rate $\dot{V}$ at the start of the tube, which has a temperature of $T_0$ and an inner surface of $A_0$, what is the resulting temperature of the stream going out of the tube? ($T_2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed that the temperature $T_0$ is fixed, and the material is aluminium and the flow is laminar. Maybe i have to assume more?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it boils down to finding the heat transfer coefficient between the copper and the air stream - but this depends on the temperature difference, which changes as the air is heated up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here a small sketch of my problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7JclJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7JclJ.jpg&quot; alt=&quot;sketch of the problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="1293" LastEditDate="2015-09-24T11:14:13.157" LastActivityDate="2015-09-24T11:14:13.157" Title="How to calculate the resulting temperature for a stream of air through a hot tube?" Tags="|heat-transfer|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5536" PostTypeId="2" ParentId="5535" CreationDate="2015-09-24T10:53:58.957" Score="4" Body="&lt;p&gt;As long as you have enough data about the inlet conditions you don't need to assume whether the flow is laminar or turbulent (this is a very dangerous assumption), you can calculate Reynolds number to find that out:&#xA;$$ Re = \frac{\rho v D}{\mu} $$&#xA;where $\rho$ is the flow density, $v$ is the velocity, $D$ is the hydraulic diameter of the tube and $\mu$ is the dynamic viscosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow inside tubes is laminar for $Re &amp;lt; 2300$ and turbulent for $Re &amp;gt; 10,000$ (and transition in between), Assuming that you have a constant surface temperature tube $ T_s = \text{constant}$, making energy balance we have:&#xA;$$ Q^o = hA_s \theta_{lmtd} = m^o C_p \triangle T $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $A_s$ is the surface area $\pi D L$ and $\theta_{lmtd}$ is &lt;a href=&quot;https://en.wikipedia.org/wiki/Logarithmic_mean_temperature_difference&quot; rel=&quot;nofollow&quot;&gt;logarithmic mean temperature difference.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For laminar flow, Nusselt number can be calculated as (for fully developed flow and moderate temperature difference): $$ Nu = \frac{hD}{k} = 3.66$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For turbulent flow, you can use Gnielinski correlation, Dittus-Boelter equation or Sieder-Tate correlation. (This &lt;a href=&quot;https://en.wikipedia.org/wiki/Nusselt_number#Forced_convection_in_turbulent_pipe_flow&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; is very helpful).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After calculating $h$ from Nusselt number correlations (You might need to assume an exit temperature for the first iteration) substituting in energy balance equation, after some iterations the solution will converge and you'll finally obtain the exit temperature from the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can refer to the forced convection heat transfer chapter in this &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0072458933&quot; rel=&quot;nofollow&quot;&gt;book&lt;/a&gt; for more details and solved examples.)&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2015-09-24T10:53:58.957" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5537" PostTypeId="1" CreationDate="2015-09-24T12:52:46.023" Score="2" ViewCount="89" Body="&lt;p&gt;Consider 2 or 3 stirred tank reactors as a cascade. The sizes are different. A part of the throughput is recirculated from the last to the first. The mean hydraulic retention of course is straightforward to calculate. But what is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Residence_time_distribution&quot; rel=&quot;nofollow&quot;&gt;residence time distribution&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is a solved problem, but for some reason the solution is not in any textbook I've consulted so far. Only the classical case of &lt;a href=&quot;https://en.wikipedia.org/wiki/Continuous_stirred-tank_reactor&quot; rel=&quot;nofollow&quot;&gt;stirred tank reactor (STR) (all same size, no recirculation)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the effect of recirculation on the RT distribution? Logic says the distribution will be closer to a single vessel, but how to quantify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a significant effect on the HRT distribution of a STR cascade when the tanks have different sizes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated above, I'm sure this is a solved problem - I just haven't been able to find the solutions. &lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-09-24T12:52:46.023" Title="What is the retention time distribution in an STRcascade when the reactors are of uneven side and we have some recircultion?" Tags="|chemical-engineering|process-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5539" PostTypeId="1" AcceptedAnswerId="5633" CreationDate="2015-09-24T15:10:36.517" Score="2" ViewCount="334" Body="&lt;p&gt;I am intending to construct a small pressure vessel. It needs to be around 7 centimeters in length and width and around 1-2 centimeters in depth. It must be capable of being pressurised to 300 psi with sufficient space to contain a small amount of liquid (say 5cc) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Constraints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The body of the vessel must be made out of nickel-plated annealed brass.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The vessel will have a brass cap, held in place with silver solder. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The cap must be sufficiently thin to be pierced when it is inserted into a receptacle. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this design feasible?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3112" LastActivityDate="2015-10-02T13:07:44.440" Title="Is this pressure vessel feasible?" Tags="|mechanical-engineering|pressure|pressure-vessel|" AnswerCount="1" CommentCount="19" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5541" PostTypeId="1" CreationDate="2015-09-23T16:23:12.477" Score="2" ViewCount="2290" Body="&lt;p&gt;I am working on my BSc Electrical Engineering final-year project, and have been asked to do a literature review to include in the project report.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My project is an Arduino-based power strip which can be controlled wirelessly using an Android application, and has power consumption reporting capabilities (among other things). It doesn't involve reading scientific papers, journal articles and the like - rather it is simple, hands-on electronics. It can be constructed easily by following tutorials and Instructables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I chose it because it is in line with the whole 'home automation' theme, and will give me practical experience, not because I was looking to solve a scientific problem of some kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've read about literature reviews, it seems as if they only apply to research-based projects that &lt;em&gt;do&lt;/em&gt; solve a scientific problem. The student compares the previous work in that field, and expands on it in his project from a 'scientific' point of view. In my case, I just have a cool product that fits a particular market, but there is nothing scientifically new about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, what should I do differently compared to a 'normal' literature review?&lt;/p&gt;&#xA;" OwnerDisplayName="M. Hassaan" LastEditorUserId="1586" LastEditDate="2015-09-28T18:01:07.020" LastActivityDate="2015-09-28T18:01:07.020" Title="How to write a literature review for a DIY-style electronics project?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5542" PostTypeId="2" ParentId="5541" CreationDate="2015-09-23T16:57:16.270" Score="3" Body="&lt;p&gt;I advise you to discuss it with your professor/advisor, but since you asked...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just my perspective as a working engineer, not as an academic so take it for what it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What it sounds like you're really doing is product development. Rather than focus on academic research papers, your review should focus on: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Defining specifically the problem(s) you're solving and whose problems are they. (Market research) &lt;/li&gt;&#xA;&lt;li&gt;What needs do potential customers expect, want, desire to be met?&lt;/li&gt;&#xA;&lt;li&gt;What are the specific functional requirements of your system to meet those needs?&lt;/li&gt;&#xA;&lt;li&gt;How and how well do competing products meet those needs (even if you have to select multiple products to cover all of the functionality your product provides) &lt;/li&gt;&#xA;&lt;li&gt;Specifically, how will your product fulfill each of those requirements?&lt;/li&gt;&#xA;&lt;li&gt;What are the options for providing each function and why is the option you have selected the best? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another way to think about it is to think about your professor as a potential investor in your company that's going to make and sell this device. What information are you going to use to convince them that they should invest? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't address these things, then you're just cobbling together gadgets from the internet and not really engineering something.&lt;/p&gt;&#xA;" OwnerUserId="2061" OwnerDisplayName="DLS3141" LastActivityDate="2015-09-23T16:57:16.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5543" PostTypeId="2" ParentId="5541" CreationDate="2015-09-23T17:00:37.007" Score="4" Body="&lt;p&gt;The difference between research and product development, is research is done to understand some natural phenomenon, product development is done to create a practical device to solve a problem. In the product development context, it's usually called a &quot;market review,&quot; or similar. It's a review of similar devices on the market, their features, specs, etc. Normally, a large part of this work is not done by engineering. Ask your professor, but I'm guessing they don't want you just copying a commercially available product, and want you to be able to say, &quot;My design is an improvement on what's out there because it does ____.&quot;&lt;/p&gt;&#xA;" OwnerUserId="3114" OwnerDisplayName="Matt Young" LastActivityDate="2015-09-23T17:00:37.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5544" PostTypeId="1" CreationDate="2015-09-24T17:17:07.917" Score="5" ViewCount="318" Body="&lt;p&gt;If I have a water pump and a fast solenoid valve (can be open or closed) and turn on my pump at a constant voltage and apply pulse width modulation to my valve. Would the resulting thrust created at the water output be constant and can it be controlled nicely by the pulse width modulation process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the equations that could model the reaction of such a setup?&lt;/p&gt;&#xA;" OwnerUserId="1849" LastEditorUserId="282" LastEditDate="2015-09-26T21:15:54.563" LastActivityDate="2015-09-26T21:15:54.563" Title="PWM flow control and thrust" Tags="|electrical-engineering|pumps|thrust|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5545" PostTypeId="2" ParentId="5544" CreationDate="2015-09-24T18:07:17.707" Score="3" Body="&lt;p&gt;It's not clear, but I think you are asking whether you can control your solenoid valve with pulses instead of steady DC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, above some frequency, the mechanical system actuated by the solenoid won't &quot;see&quot; individual pulses, just the average.  Usually at a higher frequency, the solenoid itself will smooth out the pulses and maintain are more average steady current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrically, this is basically creating a switching power supply that controls the current thru the solenoid, with the solenoid being the inductor of the switching power supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is assuming your &quot;solenoid nozzle&quot; (do you really mean &quot;valve&quot;?) is intended for other than binary on/off operation, often called a &lt;i&gt;proportional valve&lt;/i&gt;.  Trying to drive a binary mechanism to in-between states may not end well.  Constantly banging it between on and off may be even worse.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-09-24T18:07:17.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5546" PostTypeId="5" CreationDate="2015-09-24T19:56:42.127" Score="0" Body="&lt;p&gt;The act of regulating the rate of fluid movement through flow conduits, or into and out of fluid containing vessels. Control action may or may not involve sensing of the flow rate and feedback of flow measurement, but in all cases involves some means of throttling the flow in either an on-off or proportional manner.&lt;/p&gt;&#xA;" OwnerUserId="820" LastEditorUserId="820" LastEditDate="2015-09-25T00:07:04.977" LastActivityDate="2015-09-25T00:07:04.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5547" PostTypeId="4" CreationDate="2015-09-24T19:56:42.127" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-09-24T19:56:42.127" LastActivityDate="2015-09-24T19:56:42.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5549" PostTypeId="1" CreationDate="2015-09-25T02:13:00.220" Score="0" ViewCount="585" Body="&lt;p&gt;Why are &lt;a href=&quot;http://fridgedimensions.com/refrigerators/reviews/all-refrigerator-best-in-refrigerator-efficiency/&quot; rel=&quot;nofollow&quot;&gt;freezerless refrigerators more efficient&lt;/a&gt; and just out of curiosity, about how much energy can I save per month per cubic feet by going all-refrigerator?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2015-09-25T02:43:09.897" LastActivityDate="2015-09-25T11:41:04.297" Title="Why are freezerless refrigerators more efficient?" Tags="|energy-efficiency|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5550" PostTypeId="2" ParentId="5549" CreationDate="2015-09-25T05:55:19.663" Score="1" Body="&lt;p&gt;Speaking about efficiencies in fridges is misleading. You typically use the concept &lt;a href=&quot;https://en.wikipedia.org/wiki/Coefficient_of_performance&quot; rel=&quot;nofollow&quot;&gt;Coefficient of Performance (COP)&lt;/a&gt;: The heat removed by the heat pump devided by the input in mechanical work. The higher you COP, the less energy input you need to remove a given amount of heat from your fridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The higher the temrpeature difference the heat pump has to bridge, the lower the COP. So if you have a freezer, you need to cool further below 0°C, this gives you a higher temp diff to the roomthe fridge is standing in. Hence a lower COP, and more energy needed. So far for thermodynamics. Other factors affecting the actual energy uptake are the insulation, how you operate your freezer, efficiency of the pump, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part of your question, I don't think a comparison between 'typical' freezers will help you much. Look up actual energy demands of actual freezers and make your comparison.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-09-25T05:55:19.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5551" PostTypeId="2" ParentId="5549" CreationDate="2015-09-25T08:07:46.537" Score="3" Body="&lt;p&gt;Heat flux is directly proportional to gradient of temperature - in other words, if you have an imperfect insulation wall, it will pass the more heat the higher the temperature difference between its two sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, your room temperature is 22°C. If the freezerless refrigerator internal temperature is 4°C, and the freezer temperature is -5°C, that's 18°C and 27°C difference between the inside and outside respectively. With the same insulation, the latter will leak heat inside faster than the former, necessitating the pump to run more frequently and remove it for a longer time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat flux is &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_conduction#Fourier.27s_law&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;linearly&lt;/em&gt; proportional&lt;/a&gt; to temperature gradient:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \overrightarrow{q}  = - k {\nabla} T $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining features - energy usage per joule of heat transmitted etc remain (almost) linear as well. Therefore the efficiency change between a refrigerator and a freezer of the same volume, size and construction is directly proportional to the difference between the two gradients. For the above example, 27/18 = 1.5, &lt;strong&gt;the freezer will use about 150% of similar refrigerator's energy&lt;/strong&gt;. But that largely depends on the thermostat settings, ambient temperature, construction, insulation used (the freezer is likely to have it thicker) and so on. If you add a joint refrigerator/freezer into the mix, this becomes quite tricky as the temperature gradient changes throughout the volume. Given a ballpark figure of freezer part occupying 20% of the volume, and assuming the changed geometry doesn't influence it, I can give you a rough estimate of &lt;strong&gt;10%&lt;/strong&gt; energy savings when using a pure refrigerator vs one of the same volume and parameters, but with a freezer part.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2015-09-25T11:41:04.297" LastActivityDate="2015-09-25T11:41:04.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5552" PostTypeId="2" ParentId="5534" CreationDate="2015-09-25T11:25:13.200" Score="1" Body="&lt;blockquote&gt;I'm finding difficult in implementing real time clock for pic16lf1902&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This PIC, as most 8 bit PICs, can drive a 32786 Hz low power &quot;watch&quot; crystal directly.  This peripheral is called the &lt;i&gt;Timer 1 oscillator&lt;/i&gt; because it directly drives timer 1.  Timer 1 and its oscillator can run during sleep, then cause the processor to wake periodically, like every second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You use the 1 second periodic wakeup to drive the real time clock logic.  Most of the time, you just increment the seconds counter and go back to sleep.  Since the processor is only run a few &amp;micro;s every second and sleeping the rest of the time, the average current draw is very low and appropriate for a real time clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum sleep current is 1 &amp;micro;A at 1.8 V and up to 85&amp;deg;C, with it usually being much less than that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;wherein their are no serial protocols for communicating/interfacing with builtin LCD display&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Huh?  What?  It doesn't come with a sensor for measuring the ripeness of tomatoes either, but neither has anything to do with the real time clock you are asking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And no, this PIC doesn't have a built-in LCD.  None of them do.  By what hallucination do you imagine a liquid crystal somehow &quot;integrated&quot; onto a silicon chip?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic problem appears to be that you don't understand what microcontrollers are.  Go back and learn some basics, and also learn to formulate and present questions clearly.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-09-25T11:25:13.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5553" PostTypeId="1" AcceptedAnswerId="5555" CreationDate="2015-09-25T14:34:11.360" Score="6" ViewCount="1096" Body="&lt;p&gt;We have some metal parts in our machines, which are anodised aluminium. Those machines are due to be shipped to the customer in the near future, but I have noticed some pitting or slight rust developing on some of the anodised aluminium surfaces. I have tried cleaning it with 70% IMS, but to no avail (the surface is clean, there is no dirt). This doesn't really affect the proper function of the machine, but for something that is supposed to be brand new and relatively expensive, I would like it to look better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a couple of photos to illustrate what I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LAufj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LAufjt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/jiCME.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jiCMEt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click for larger images&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody suggest a cleaning method for removing slight surface pitting/corrosion on anodized aluminum parts?&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2015-09-28T08:40:14.420" LastActivityDate="2015-09-28T08:40:14.420" Title="Cleaning anodised aluminium" Tags="|materials|chemical-engineering|surface-preparation|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5554" PostTypeId="2" ParentId="5553" CreationDate="2015-09-25T15:33:13.100" Score="3" Body="&lt;p&gt;You say you have, &quot;&lt;strong&gt;some&lt;/strong&gt; metal parts which are aluminum&quot; - would any of the other parts happen to be stainless steel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your parts could be experiencing &lt;a href=&quot;https://en.wikipedia.org/wiki/Galvanic_corrosion&quot; rel=&quot;nofollow noreferrer&quot;&gt;galvanic corrosion&lt;/a&gt;, where two dissimilar metals essentially form a battery in the presence of an electrolyte. The electrolyte could be moisture from condensation mixing with chemicals/salt in the air or residual chemicals on the part after the anodizing process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.fastenal.com/content/feds/pdf/Article%20-%20Corrosion.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;'s a PDF with more information on galvanic corrosion, and I've attached a picture below. The picture is of a stainless steel screw attached to aluminum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gKwE6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gKwE6.jpg&quot; alt=&quot;Galvanic Corrosion&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-09-25T15:33:13.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5555" PostTypeId="2" ParentId="5553" CreationDate="2015-09-25T16:00:34.020" Score="5" Body="&lt;p&gt;Chuck is right that there could be some galvanic action contributing to your issue. One other possibility is that your aluminum has been contaminated with steel particles at some point. Your second picture does seem to clearly show rust, which wouldn't be happening if the metal was all aluminum. It's less clear what the issue is in the first picture, but it could be smaller pieces of foreign material that interfered with the anodizing coating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, there are a few possible sources. The most likely is that the tooling used for the machining of your parts had been used on steel before. Since aluminum (at least the common alloys and tempers) is a very soft metal, it is easy to embed particles of other metals in the surface. For this reason, many shops will segregate cutters, grinding wheels, files, etc for ferrous and non-ferrous metals. Selection of the material used on the cutting tool, and speeds &amp;amp; feeds also have an impact on the risk of entrapping foreign material. Other possible sources are defects in the aluminum stock, contamination in the anodizing baths, or impact with a steel object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on your picture, I would guess that steel was embedded while the part was machined, and then the moisture in the anodizing process accelerated the rusting. This is problematic not just cosmetically, but also bad for the coating because as the rust progresses, it will expand and separate the anodizing from the base aluminum. The best way to solve this problem is probably to use dedicated tooling for these parts that is only used on non-ferrous materials, so there is minimal risk of getting steel into the aluminum in the first place. This segregation is especially important where aluminum is to be welded, because a small amount of steel in the weld pool can contaminate the weld and cause excessive porosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming this is the problem, cleaning it after the fact is fairly difficult. A chemical process won't work without damaging the whole surface, you have to actually remove the offending material or it will rust again. If it's loose enough and the anodizing layer is thin, you may be able to scrape it off with a knife or a needle file. Depending on the function of your part, you might be able to mill or face the surface down with clean tooling and re-anodize. One important thing is to take a big enough cut that you don't get any 'tearing' in the surface. Too thin a cut will tend to rub and push the material in deeper. If that's not possible, the next best thing would probably be to dig out that specific area by drilling or with a clean abrasive. Again, depending on the function of your part, you may then be stuck welding the spot full, re-surfacing, and re-anodizing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, you may decide that if the issue is only cosmetic, it's better to live with the stain than the cosmetic impact of a repair. Whatever you do, make sure you remove all of the steel, otherwise the rust stain will likely re-appear soon.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-09-25T16:37:53.580" LastActivityDate="2015-09-25T16:37:53.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5559" PostTypeId="1" CreationDate="2015-09-26T21:24:15.050" Score="3" ViewCount="1668" Body="&lt;p&gt;I want to ask about the principle or the idea that people who invented the geometric method of drawing a regular pentagon depended on. I know how to follow the steps described below to draw the pentagon but I want to know how these steps led us to a regular pentagon. &lt;strong&gt;They must have depended on a rule, law, idea or something so what is it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VrnA1.jpg&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5hpw9.jpg&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Jg7Nk.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I am talking about the geometric method explained in the pictures not the divider method.&lt;/p&gt;&#xA;" OwnerUserId="3134" LastEditorUserId="368" LastEditDate="2016-02-18T23:18:00.653" LastActivityDate="2023-04-14T16:51:30.660" Title="Principle behind the geometric method for constructing a regular pentagon" Tags="|design|drafting|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5560" PostTypeId="1" AcceptedAnswerId="5564" CreationDate="2015-09-26T22:26:33.333" Score="1" ViewCount="181" Body="&lt;p&gt;Say, in a firearm, after giving off a shot, if the barrel were to be closed off, does there exist a way to store some of it in an external pressure vessel, without requiring a mainly-external source of power and complex machinery?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forgive the possibly stupid question, I'm a bit of a layman as far as engineering goes.&lt;/p&gt;&#xA;" OwnerUserId="3136" LastActivityDate="2015-09-28T03:08:27.237" Title="Is it possible to store leftover pressure from burning a propellant in a pressure vessel?" Tags="|pressure|combustion|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5561" PostTypeId="2" ParentId="5559" CreationDate="2015-09-27T07:23:32.727" Score="1" Body="&lt;p&gt;An number of methods have been devised to construct pentagons, among them Richmond's method and Carlyle circles. Classical methods are based on mathematics: trigonometry and Pythagoras's theory of triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An animation of the method you are asking about is in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pentagon&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article of Pentagons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XACfI.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XACfIt.gif&quot; alt=&quot;pentagon animation&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click image to expand and see animation&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mathworld.wolfram.com/Pentagon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolfram&lt;/a&gt;, also gives a good description of the method,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Euclid showed how to inscribe a regular pentagon in a circle. Ptolemy also gave a ruler and compass construction for the pentagon in his epoch-making work The Almagest. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="90" LastEditorDisplayName="user16" LastEditDate="2015-09-27T16:11:19.413" LastActivityDate="2015-09-27T16:11:19.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5562" PostTypeId="2" ParentId="5559" CreationDate="2015-09-27T21:37:56.270" Score="4" Body="&lt;p&gt;There are many ways to do this, but they all depend on the same basic fact that the sines and cosines of the angles $36°$ and $72°$ involve $\sqrt 5$. For example $\cos 36° = (1 + \sqrt 5)/4$. Since $\sqrt 5 = \sqrt{1^2 + 2^2}$, These angles can be constructed by starting from a right-angled triangle with sides $1$, $2$, and hypotenuse $\sqrt 5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the OP's construction, is should be obvious how the $\sqrt 5$ appears in the length of AE and AB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euclid gave a (rather long winded) construction in &lt;a href=&quot;http://aleph0.clarku.edu/~djoyce/elements/bookIV/propIV11.html&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Elements&lt;/em&gt;, book IV, proposition 11&lt;/a&gt; and proved it was correct using only geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is usually easier to prove that &quot;quicker&quot; constructions like the OP's are correct using trig, rather than pure geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more general question of &quot;which regular polygons can be constructed using only ruler and compasses&quot; (i.e. no measuring instruments allowed) was answered by Gauss. Only 31 such polygons with an odd number of sides are known, the smallest ones having 3, 5, 15, 17, and 51 sides. The largest known one has 4,294,967,295 sides. It is an unsolved mathematical problem whether any more exist, and if so whether there are a finite or infinite number of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Constructible_polygon&quot; rel=&quot;nofollow&quot;&gt;Wikipedia: &lt;em&gt;Constructible polygon&lt;/em&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://oeis.org/A045544&quot; rel=&quot;nofollow&quot;&gt;OEIS: &lt;em&gt;Odd values of n for which a regular n-gon can be constructed by compass and straightedge&lt;/em&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://mathworld.wolfram.com/FermatPrime.html&quot; rel=&quot;nofollow&quot;&gt;Wolfram MathWorld: &lt;em&gt;Fermat Prime&lt;/em&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="683" LastEditorUserId="368" LastEditDate="2016-02-18T23:20:59.897" LastActivityDate="2016-02-18T23:20:59.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5563" PostTypeId="2" ParentId="5514" CreationDate="2015-09-28T02:52:28.110" Score="1" Body="&lt;p&gt;What determines the gear ratio is the ratio of input and output angular velocities. This is traditionally done by interlocking different sized circles together, but for high gear ratios other designs are often used because they can reduce the output angular velocity sufficiently without needing several stages of gears interacting. One of the most ingenious examples is the harmonic gear. There are plenty of videos of how they work on Youtube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=bzRh672peNk&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=bzRh672peNk&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3002" LastActivityDate="2015-09-28T02:52:28.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5564" PostTypeId="2" ParentId="5560" CreationDate="2015-09-28T03:01:43.907" Score="2" Body="&lt;p&gt;Absolutely.  There is a device called a bore evacuator which has been in service for a long time on large caliber guns.  Its purpose is to draw combustion gas out through the muzzle after the shot cycle is over, so that said gas (which is usually toxic) doesn't flow backwards out the breech and into the crew compartment (assuming the gun is mounted on a vehicle like a tank or self-propelled howitzer).  The picture below shows a bore evacuator on on an Abrams tank.  It is the bulge in the middle of the main gun barrel with the words &quot;99 Problems&quot; faintly visible on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ce71G.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ce71G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way it works is simple; after the bullet passes the bore evacuator, the high-pressure combustion gas behind the bullet is allowed to flow into the bore evacuator through small holes.  After the bullet exits the gun, the pressure in the tube begins to drop, eventually going below the pressure in the evacuator.  The gas then begins to flow out of the evacuator into the tube through holes that are slightly angled towards the muzzle.  The resulting discharge induces gas flow towards the muzzle and creates a slight vacuum at the breech.  When the breech is opened, hardly any combustion gas escapes into the crew compartment due to this forward flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bore evacuator stores the gas for only a few milliseconds, but it should not be difficult to store it longer (a few seconds or minutes) by using check valves.  In fact some bore evacuators in service do use check valves at the inlet to increase their effectiveness.  I don't know how much gas you need to store, but you may not need to close off the barrel completely; you may be able to harvest enough gas just during the normal shot cycle, before the bullet exits the gun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a German recoilless rifle called the &quot;Armbrust&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L5rAd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L5rAd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses captive pistons to completely hold the propellant combustion gas inside the weapon after firing.  It's a one-use weapon though, which may not work for what you're doing.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-09-28T03:08:27.237" LastActivityDate="2015-09-28T03:08:27.237" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5566" PostTypeId="1" AcceptedAnswerId="5569" CreationDate="2015-09-28T08:10:40.533" Score="9" ViewCount="13882" Body="&lt;p&gt;I've learned in school that pressure in water changes like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(h) = \rho g h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h$ is depth in meters, $\rho$ is density (e.g. 1000 $\frac{\text{kg}}{\text{m}^3}$ for water) and $g$ is gravitation acceleration ($\approx 9.81 \frac{\text{m}}{\text{s}^2}$) and $p$ is the pressure in Pascal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess there is no similar law for pressure in earth as it is to different, depending on where you are. But is there a rule of thumb? What do engineers who build tunnels / underground stations do?&lt;/p&gt;&#xA;" OwnerUserId="3061" LastEditorUserId="1586" LastEditDate="2015-09-28T18:00:27.077" LastActivityDate="2016-09-06T16:45:25.387" Title="How does pressure change with depth in earth?" Tags="|civil-engineering|pressure|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5567" PostTypeId="2" ParentId="5523" CreationDate="2015-09-28T12:44:25.293" Score="2" Body="&lt;p&gt;For any structure with a service life of 120 years I would always design it to the current, applicable design standard or code. You don't know what to expect during the service life. The owner could sell, a new settlement may develop, or some industry or mining activity may change the requirements of the bridge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless there is a clear sign on the bridge limiting the use, most engineers would look at the design code applicable during the time it was constructed to establish safe working loads. They would not be able to deduce that you have employed a &quot;simple design method&quot; and may make a bad judgement call.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing is, you already mentioned a 30 ton truck could possibly cross it. So are you be designing for it in any case? &lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2015-09-28T12:44:25.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5568" PostTypeId="2" ParentId="5294" CreationDate="2015-09-28T12:49:16.797" Score="4" Body="&lt;p&gt;In the Southern African context this would be defined as a &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;cold formed lipped channel&quot; section. I am not sure if it the same across the world though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one &lt;a href=&quot;http://www.macsteel.co.za/products/cold-formed-sections?b=55&quot; rel=&quot;nofollow&quot;&gt;supplier's link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="2637" LastEditDate="2015-10-01T06:02:06.777" LastActivityDate="2015-10-01T06:02:06.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5569" PostTypeId="2" ParentId="5566" CreationDate="2015-09-28T13:47:32.603" Score="10" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I guess there is no similar law for pressure in earth as it is to different, depending on where you are. But is there a rule of thumb? What do engineers who build tunnels / underground stations do?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I approach this question as an engineer who does a lot of work on buried pipes and occasionally has to qualify buried structures for nuclear power plants.  Also, for the sake of brevity, I assume you are talking about &lt;strong&gt;only&lt;/strong&gt; vertical loads on the structure (lateral loads are another complicated topic for foundation engineering).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soil can act similarly to fluid, depending on the soil type and even the type of structure that is being loaded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, flexible pipes such as PVC, HDPE, and steel can be assumed to be loaded by the soil prism directly above the pipe.  Piping is considered flexible if it can sustain a sizable deformation of its cross-section without rupturing.  Consider the image below from Moser &amp;amp; Folkman's &lt;em&gt;Buried Pipe Design&lt;/em&gt;, 3rd Edition (1):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9DGKj.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9DGKj.png&quot; alt=&quot;flexible pipe&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, since the pipe is considered more flexible than the soil, the pipe deforms under load such that no arching of the soil takes place.  As such, the load on the pipe is simply the soil density times the depth of soil, like in your example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things get more complicated for so-called rigid pipes, such as concrete pipe or transite (asbestos-cement) pipe.  In this case, the rigidity of the pipe is such that the soil on the sides of the pipe settle more than the diameter of the pipe itself and the pipe takes extra loading via soil arching.  Below I've pasted another image from Moser &amp;amp; Folkman (1) illustrating this phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nieoK.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nieoK.png&quot; alt=&quot;rigid pipe&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loading on the pipe depends on how it was buried (positive projection, trench, induced trench, etc.) and is really beyond the scope of this answer.  I've included a couple of references at the end of this answer for further reading.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For larger structures such as your tunnels or subway stations, determining the soil load is more complicated.  Are there adjacent structures applying load?  Has there been something done to stabilize the soil?  How are the different soil strata interacting, and how does the relative stiffness of each impact the total load?  If tunneling through rock, can the rock support itself without further reinforcement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these considerations and more that I can't think of at the moment come into play when determining the load on a buried structure.  &lt;strong&gt;There is no true rule of thumb when it comes to designing a buried structure&lt;/strong&gt; since there are so many considerations when it comes to the actual loading.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further Reading&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Moser, A.P. &amp;amp; Steven Folkman, &lt;em&gt;Buried Pipe Design&lt;/em&gt;, 3rd Edition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) &lt;a href=&quot;https://archive.org/details/TheTheoryOfLoadsOnPipesInDitchesAndTestsOfCementAndClayDrainTile&quot; rel=&quot;noreferrer&quot;&gt;Marston, A. &amp;amp; A. O. Anderson, &lt;em&gt;The Theory of Loads on Pipes in Ditches and Tests of Cement and Clay Drain Tile and Sewer Pipe&lt;/em&gt;, Feb. 1913.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Clarke, N.W.B., &lt;em&gt;Buried Pipelines: A Manual of Structural Design and Installation&lt;/em&gt;, 1968.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-09-28T14:53:42.253" LastActivityDate="2015-09-28T14:53:42.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5570" PostTypeId="1" AcceptedAnswerId="5575" CreationDate="2015-09-28T13:58:53.050" Score="51" ViewCount="10010" Body="&lt;p&gt;The recent &lt;a href=&quot;http://www.nature.com/news/the-science-behind-the-volkswagen-emissions-scandal-1.18426?WT.mc_id=SFB_NNEWS_1508_RHBox&quot;&gt;Volkswagen pollution test cheating scandal&lt;/a&gt; has shocked many people with its widespread extent as well as how it was hidden for so many years. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Last year, the International Council on Clean Transportation (ICCT) in&#xA;  Washington DC contracted scientists from the Center for Alternative&#xA;  Fuels Engines and Emissions at West Virginia University in Morgantown&#xA;  to test emissions from three light-duty diesel vehicles under&#xA;  more-realistic conditions than are possible in the lab. To do so, the&#xA;  scientists fitted cars with a portable emissions measurement system to&#xA;  gather a continuous stream of data over a variety of US road types.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The tests found that the levels of NOx emitted by a Volkswagen Jetta&#xA;  were 15–35 times greater than dictated by the US standard (31&#xA;  milligrams per kilometre), depending on road and driving conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How was this cheating program carried out, and why were the lab tests unable to detect the presence of this cheating program? &lt;/p&gt;&#xA;" OwnerUserId="164" LastEditorUserId="272" LastEditDate="2015-09-29T13:29:26.607" LastActivityDate="2015-12-02T19:55:37.277" Title="How was Volkswagen able to trick the lab pollution test?" Tags="|automotive-engineering|product-testing|air-quality|emissions|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5571" PostTypeId="2" ParentId="5566" CreationDate="2015-09-28T14:29:51.970" Score="8" Body="&lt;p&gt;As someone who has been involved with underground infrastructure to depths of at least 1400 metres, there are no rules of thumb. It all comes down to geology and the local conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soils behave differently to rock and sedimentary rock behave differently to igneous and metamorphised rock. Brittle rock behave differently to ductile rock. Brittle rock in the form of dikes and sills, can fail explosively when over stressed. Some mafic rock can exhibit a creep behaviour over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number, orientation and condition of rock discontinuities is a factor, as is the proximity of faults/shears. The condition of the faults and whether they are active is important as is the width of the fault or fault zone and whether the fault is smooth sided or infilled and if infilled what material fills the fault. Talc on faults only leads to problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The juxtaposition of brittle and ductile rock can induce localised stresses due each rock type behaving differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geotechnical holes can give information such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Core_recovery_parameters#Rock_quality_designation&quot;&gt;rock quality designation&lt;/a&gt; (RQD). Other drill holes in which three dimension stress cells have been placed can be over-cored so that the principle stresses for the rock mass in certain locations can be ascertained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At depth, lateral stresses can be higher than sub-vertical stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a tunnel or chamber is excavated underground, the stresses in the rock mass realign. If a system of closely spaced voids is introduced into the rock mass, zones of de-stressed rock can occur, where there rock is no longer under the influence of the virgin rock stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other situations, the lack of confinement introduced when a tunnel or chamber has been excavated can cause the walls of the void to contract; in some cases 50 mm or more.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-09-28T15:24:34.270" LastActivityDate="2015-09-28T15:24:34.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5574" PostTypeId="1" CreationDate="2015-09-28T15:44:26.793" Score="5" ViewCount="246" Body="&lt;p&gt;I want to implement a piece of glass to set various pieces of steel on and take picture through it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pieces of steel weigh anywhere between 20-150 lbs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of the glass is so pictures can be taken of both sides of the piece at the same time to speed up production instead of having to flip it over by hand.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm no expert in this, but I am assuming there are many different compositions of glass as there is steel.  According to Wikipedia, steel has a hardness of 4-4.5 and glass has a hardness of 5.5.  I don't know how true this holds, which is why I'm here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what this boils down to is I'm looking for a sheet of glass that will with stand weights over 150 lbs and won't be scratched by constant pieces of steel being laid on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question could be as easy as &quot;All glass will work&quot; but I've seen glass on cars that have been scratched and gouged by other parts, so I wasn't sure how true the Mohs scale I saw was. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem I have thought about is which is better, Anti-Glare or Anti-Reflection.  I would like to try and keep a high quality picture through the glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;Here is a picture as requested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PE130.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PE130.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3150" LastEditorUserId="3150" LastEditDate="2015-09-28T16:43:19.270" LastActivityDate="2015-10-16T03:44:51.073" Title="Scratchless Glass against steel" Tags="|materials|" AnswerCount="3" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5575" PostTypeId="2" ParentId="5570" CreationDate="2015-09-28T17:40:25.593" Score="62" Body="&lt;p&gt;After &lt;a href=&quot;http://www.theicct.org/use-emissions-testing-light-duty-diesel-vehicles-us&quot; rel=&quot;noreferrer&quot;&gt;the 2014 ICCT report&lt;/a&gt; revealed that these light-duty passenger diesel vehicles were emitting too much NOx and US regulators confronted VW about the results, VW did some testing and proposed a voluntary software recall to recalibrate the various emissions control devices on the affected vehicles. From &lt;a href=&quot;http://www.theicct.org/use-emissions-testing-light-duty-diesel-vehicles-us&quot; rel=&quot;noreferrer&quot;&gt;the California Air Resources Board's (CARB's) in-use compliance letter to Volkwagen AG, 2015-09-18&lt;/a&gt;, following that recall:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To have a more controlled evaluation of the high NOx observed over the&#xA;  road, CARB developed a special dynamometer cycle which consisted of&#xA;  driving the Phase 2 portion of the FTP repeatedly. This special cycle&#xA;  revealed that VW's recall calibration did increase Diesel Exhaust&#xA;  Fluid (DEF) dosing upon initial startup; however, dosing was not&#xA;  sufficient to keep NOx emission levels from rising throughout the&#xA;  cycle. This resulted in uncontrolled NOx emissions despite the SCR&#xA;  reaching sufficient operating temperatures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What's significant here is that CARB &lt;em&gt;changed the test cycle&lt;/em&gt; for this additional testing. &lt;a href=&quot;http://www3.epa.gov/nvfel/testing/dynamometer.htm&quot; rel=&quot;noreferrer&quot;&gt;The EPA's standard test cycles are publicly available&lt;/a&gt; and consist of set patterns of acceleration and braking, each intended to simulate a real trip* characterized by a certain type of driving (urban highway, rural up- and downhill, and so on). Here's a visualization of Federal Test Procedure 75 (FTP-75):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www3.epa.gov/nvfel/methods/quickdds.htm&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e05ck.gif&quot; alt=&quot;&quot; title=&quot;Click through for more EPA test cycle charts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manufacturers typically know the parameters of the test cycles to which their vehicles will be subjected in a given market. They then design their vehicles to pass the test. It's the regulator's responsibility to make sure the testing provides a reasonably accurate picture of how the vehicle will perform during its lifetime. This transparency should benefit everyone involved by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;giving the manufacturers a clear path to compliance and helping them to plan R&amp;amp;D and control costs;&lt;/li&gt;&#xA;&lt;li&gt;reassuring legislators and the public that the regulator is effectively enforcing the laws that protect public health and the environment;&lt;/li&gt;&#xA;&lt;li&gt;allowing the regulator to perform controlled testing on a limited number of vehicles and extrapolate that to the in-use fleet, rather than testing every vehicle (not economically feasible).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, making the details of test cycles known also has the potential to allow manufacturers to circumvent testing in a way that gives them a business advantage without actually providing the mandated emissions reduction. This is sometimes called &quot;cycle beating&quot; and it's nothing new in the automotive industry.** What may be new here is how VW went about beating the cycle. Continued from the compliance letter, emphasis mine:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;CARB shared its test results with VW on July 7, 2015. CARB also shared&#xA;  its results with the EPA. Several technical meetings with VW followed&#xA;  where VW disclosed that Gen1, Gen2 and the 2015 model-year improved&#xA;  SCR vehicle (known as the Gen3) had a second calibration intended to&#xA;  run only during certification testing. During a meeting on September&#xA;  3, 2015, &lt;strong&gt;VW admitted to CARB and EPA staff that these vehicles were&#xA;  designed and manufactured with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Defeat_device&quot; rel=&quot;noreferrer&quot;&gt;defeat device&lt;/a&gt; to bypass, defeat, or&#xA;  render inoperative elements of the vehicles' emission control&#xA;  system.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That goes way beyond a failure of due diligence. You can imagine a situation where an automaker is &quot;teaching to the test,&quot; i.e., designing their emissions system with a strict focus on what they have to do to pass testing, and just fails to realize or discover some flaw that will result in non-compliant levels of emissions on average. You might call that shoddy or incompetent work; you could even say the regulator's at fault there, for not having a better suite of testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, VW designed the software that controls the vehicle's emissions system to cheat the test. It's very much like hiring someone to write an essay or sit an exam for you, while you take the credit; the vehicle system that EPA and CARB were testing was literally not the same system that consumers were purchasing and driving, making the test meaningless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The International Business Times, referencing &lt;a href=&quot;http://www.theicct.org/john-german&quot; rel=&quot;noreferrer&quot;&gt;ICCT Senior Fellow John German&lt;/a&gt; as the source, &lt;a href=&quot;http://www.ibtimes.com/volkswagen-emissions-scandal-what-defeat-device-how-does-it-work-why-cant-you-see-it-2112350&quot; rel=&quot;noreferrer&quot;&gt;reported earlier this week&lt;/a&gt; that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It’s not yet known exactly what Volkswagen was measuring. German said&#xA;  that it will probably be months before we find out what exactly the&#xA;  company was doing. The EPA explained that the &quot;device&quot; can't be turned&#xA;  off by the user, and Volkswagen will now be required to fix the issue&#xA;  without the owner incurring charges.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While we don't know exactly how VW did it, it's clear that the software they wrote for these vehicles contained some sort of algorithm(s) to detect, based on data from the comprehensive suite of sensors included in modern vehicles' &lt;a href=&quot;https://en.wikipedia.org/wiki/On-board_diagnostics&quot; rel=&quot;noreferrer&quot;&gt;OBD&lt;/a&gt; systems and their knowledge of the parameters of US regulatory test cycles, when the vehicle was and was not being run through one of these test cycles. &lt;a href=&quot;https://www.washingtonpost.com/news/the-switch/wp/2015/09/22/the-tech-behind-how-volkswagen-tricked-emissions-tests/&quot; rel=&quot;noreferrer&quot;&gt;This Washington Post article&lt;/a&gt; is the best lay summary I've read so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.wired.com/2015/09/vw-fool-epa-couldnt-trick-chemistry/&quot; rel=&quot;noreferrer&quot;&gt;Some reports suggest that all they did was detect whether the steering wheel was moving&lt;/a&gt;, because when hooked up in the lab there are no turns—but since the CARB letter suggests that a &quot;special dynamometer cycle&quot; was able to get the control unit to switch out of the special &quot;test&quot; mode, I expect the actual algorithm was a bit more sophisticated than that. &lt;a href=&quot;http://www3.epa.gov/otaq/cert/documents/vw-nov-caa-09-18-15.pdf&quot; rel=&quot;noreferrer&quot;&gt;The EPA's 2015-09-18 Notice of Violation to VW&lt;/a&gt; makes this specific allegation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &quot;switch&quot; senses whether the vehicle is being tested or not based&#xA;  on various inputs including the position of the steering wheel,&#xA;  vehicle speed, the duration of the engine's operation, and barometric&#xA;  pressure. These inputs precisely track the parameters of the federal&#xA;  test procedure used for emission testing for EPA certification&#xA;  purposes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As for why the lab tests didn't detect the presence of this cheating program, they're just not designed to catch deliberate, sophisticated fraud. These programs are expensive, they're funded by taxpayer dollars and the regulations they enforce makes cars more expensive, meaning taxpayers who own vehicles get hit with costs from multiple directions. Auto manufacturers are large and have a lot to lose by cheating the system—the financial impact of this scandal on VW was already in the billions of dollars even before the news broke and sent their stock down 40% (tens of billions in market capitalization) and before any fines have been announced (also potentially in the tens of billions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Costs are a huge, enormous, cannot-be-overstated issue in the world of environmental (and public health) regulation. Cost, not science, is nearly always what kills advancements in this area. So we trust that auto makers will make a reasonable effort to comply and we pay for testing that basically amounts to: Put the car on rollers; &quot;drive&quot; it through a test cycle with a bag around its tailpipe; see what's in the bag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of testing, as long as the test cycles are known in advance, is just not capable of detecting fraud within the control software. In fact, considering the level of sophistication possible in modern software algorithms, even hiding some or all of the details of the test cycle might not be sufficient to defeat this approach entirely, as long as the manufacturer is determined enough and the testing takes place in artificially controlled lab conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the ICCT group did was to use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Portable_emissions_measurement_system&quot; rel=&quot;noreferrer&quot;&gt;PEMS&lt;/a&gt;*** to measure tailpipe emissions under real-world conditions, while driving on actual roads with actual traffic patterns that VW's software didn't recognize as corresponding to emissions testing. This allowed them to gather more accurate data about the real performance of the vehicle emissions systems—but at a cost of $50,000 for only three vehicles and with a very limited scope of what they were looking for.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;More Reading&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.npr.org/2015/09/24/443053672/how-a-little-lab-in-west-virginia-caught-volkswagens-big-cheat&quot; rel=&quot;noreferrer&quot;&gt;NPR Morning Edition: How a Little Lab in West Virginia Caught Volkswagen's Big Cheat&lt;/a&gt; (includes brief interviews with some of the scientists involved)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.wired.com/2015/09/vw-owners-arent-going-like-fixes-diesels/&quot; rel=&quot;noreferrer&quot;&gt;Wired: VW Owners Aren't Going to Like the Fixes for their Diesels&lt;/a&gt; (why cheat in the first place?)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.explainthatstuff.com/how-dynamometers-work.html&quot; rel=&quot;noreferrer&quot;&gt;Explain that Stuff: How do dynamometers work?&lt;/a&gt; (see &quot;chassis dynamometers&quot; at the bottom)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://media.vw.com/release/1075/&quot; rel=&quot;noreferrer&quot;&gt;Statement by the Supervisory Board of Volkswagen AG, Sep 25, 2015&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www3.epa.gov/otaq/cert/violations.htm&quot; rel=&quot;noreferrer&quot;&gt;US EPA: Notices of Violations, Light Cars &amp;amp; Trucks&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* &lt;a href=&quot;http://www.arb.ca.gov/research/resnotes/notes/96-11.htm&quot; rel=&quot;noreferrer&quot;&gt;Research Note 96-11: Driving Patterns and Emissions: A New Testing Cycle (CARB 1996)&lt;/a&gt; provides a brief description of how a test cycle is developed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** For example, see &lt;a href=&quot;http://www.transportenvironment.org/sites/te/files/media/T%26E%2098-3_0.pdf&quot; rel=&quot;noreferrer&quot;&gt;Cycle-Beating and the EU Test Cycle for Cars (Kågeson 1998)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*** Say it just like it sounds—&quot;pems&quot;—or call it a lab-in-a-box, but don't confuse it with &lt;a href=&quot;http://robotics.eecs.berkeley.edu/~varaiya/papers_ps.dir/PeMS_TRB2002.pdf&quot; rel=&quot;noreferrer&quot;&gt;PeMS&lt;/a&gt;, &lt;a href=&quot;http://www.monsol.com/home/pems-predictive-emissions-monitoring-systems/&quot; rel=&quot;noreferrer&quot;&gt;PEMS&lt;/a&gt;, &lt;a href=&quot;http://www.thermoscientific.com/content/tfs/en/product/particulate-matter-continuous-emissions-monitoring-system-pm-cems.html&quot; rel=&quot;noreferrer&quot;&gt;PM CEMS&lt;/a&gt;...&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-09-30T16:13:43.637" LastActivityDate="2015-09-30T16:13:43.637" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5576" PostTypeId="1" AcceptedAnswerId="5699" CreationDate="2015-09-28T19:33:40.583" Score="3" ViewCount="420" Body="&lt;p&gt;Take a hollow aluminium cylinder with outer radius $r$ and length $h$, capped with two circular endcaps.  How thick does the aluminium have to be, that is what is the inner diameter of the cylinder, to withstand 1 atm on the outside and 0 mbar on the inside, without crumpling?&lt;/p&gt;&#xA;" OwnerUserId="3153" LastEditorUserId="1122" LastEditDate="2015-10-08T20:55:51.100" LastActivityDate="2015-10-08T20:55:51.100" Title="How thick does an aluminium vacuum cylinder have to be?" Tags="|materials|metals|geometry|vacuum|buckling|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5577" PostTypeId="2" ParentId="5574" CreationDate="2015-09-28T21:33:30.170" Score="4" Body="&lt;p&gt;Consider using tempered glass. Because it has a compressive residual stress at the surface, it will take greater force to scratch it, all else equal. The downside is that a deep enough scratch will shatter the entire plate. However, producing such a scratch may be quite challenging depending on the exact pressure involved, which would depend specifically on the steel part and its weight, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally dislike using the Mohs hardness scale as it is a relative scale, intended only for minerals, which relies on the qualitative measure of &quot;A readily scratches B and B does not readily scratch A, so Mohs( A ) &gt; Mohs( B )&quot;, but it is the only data commercial producers supply. They often claim values of 8 or 9, but those values are not likely accurate. A more likely range is 6-7, putting it above most steels in the 4-5 range, and typical window glass in the 5-6 range. Unfortunately, it is challenging to find Vickers or Brinell hardness measurement (or any quantitative values really) for glass. You may have to contact a supplier to get a sample, then do your own testing (or farm it out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure your steel parts are free of grit, sand, etc. as those are going to cause by far more damage than the steel. And of course, like anything subject to wear conditions, you will have to replace the glass periodically if it experiences frequent, continuous use. It is unclear how long it will take before your ability to image the underside suffers. If you are working with very high-strength steels, wear will occur more rapidly.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-09-28T21:41:38.747" LastActivityDate="2015-09-28T21:41:38.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5578" PostTypeId="2" ParentId="100" CreationDate="2015-09-28T21:38:56.473" Score="0" Body="&lt;p&gt;After the outside of a Prince Rupert drop solidifies, it will quickly contract.  During this process, if there is nowhere for the glass inside to go this will cause the outside to be under significant tension, virtually guaranteeing that it will crack (crackle-glass is formed by quenching an entire glass piece briefly; the exterior layer will crack immediately, but if all the cracked glass pieces are in contact with glass that is still molten the overall piece will remain intact).  While it's possible to cool glass slowly enough to prevent cracking, reducing the peak tensile load sufficiently to prevent cracking will also reduce the amount by which such a load can be shifted toward being compressive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This difficulty may be overcome by lowering the glass relatively slowly into the water (the tail is still attached to the rod from which it came).  Doing this will mean that while part of the outside of the glass has solidified and is contracting, the liquid glass in the middle will, during most of this contraction, have a continuous path of liquid glass which extends out of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point the glass entering the water will be so thin that it's no longer possible for liquid glass to flow through the center, but by the time that happens the larger portions of the glass will have contracted almost as much as they're going to, so the amount of liquid glass that would still need to be displaced to avoid creating tension will be pretty small, and so the amount of tension created by an inability to displace any more liquid glass from the interior will be likewise small.  If the region of the glass which is thick enough to allow liquid flow through the center overlaps the region that's thin enough to avoid breaking when it cools, the drop may be cooled to room temperature without premature failure.  A uniform spherical blob, however, would have nowhere to displace the interior liquid to avoid having the pressure of that liquid fracture the exterior.&lt;/p&gt;&#xA;" OwnerUserId="397" LastActivityDate="2015-09-28T21:38:56.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5579" PostTypeId="2" ParentId="5574" CreationDate="2015-09-28T22:00:42.070" Score="4" Body="&lt;p&gt;If hard transparent plates like synthetic sapphire glass plates (Mohs 9) are not available or too expensive, I would cover the glass with some type of cheap quickly replaced transparent material like a automated roll of vinyl sheeting. &lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2015-09-28T22:00:42.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5581" PostTypeId="2" ParentId="666" CreationDate="2015-09-29T02:56:30.213" Score="0" Body="&lt;p&gt;Have you looked at the design that a Parr reactor uses? It would require a bit of machining (mostly lathe work), if that is an option to you. Basically, there are two C-shaped pieces of metal that fit around one end of the vessel (forming a circle). The bottom goes under a lip at the top of the vessel, and the top goes over a lip on the vessel cap. Bolts thread into the top of the C and press down on a metal ring that sits on top of the lip of the vessel cap (the ring is just there to protect the lip of the vessel cap, as the bolts will mar the surface they press against - if the surface gets really bad, you can just replace the ring instead of the entire cap). There is a ring cut out of the bottom of the cap, where a teflon ring can be inserted, and it is pressed into the top of the vessel (where there are a couple of lines cut into the metal to form a better seal). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative with less (or no) lathing would be to have a lip on the top that you drill holes in, matching up with holes that you tap on the lip of the vessel. The lips could be metal rings that you weld on. The downside of this would be that it would be harder to find a way to put a sealing ring in and get a good seal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will probably be able to understand this better after looking at some of the pictures of Parr reactors online, then reading through again a time or two. Some decent pictures are found &lt;a href=&quot;http://i.ytimg.com/vi/HSxrWYfLNyI/0.jpg&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://hamiltoninstrument.com/image/cache/data/reactors/Reactor%20Head-500x500.jpg&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3163" LastEditorUserId="272" LastEditDate="2015-09-29T11:38:25.930" LastActivityDate="2015-09-29T11:38:25.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5583" PostTypeId="1" AcceptedAnswerId="5797" CreationDate="2015-09-29T03:05:37.480" Score="4" ViewCount="752" Body="&lt;p&gt;I´m developing a tool to generate an &lt;a href=&quot;http://www.openfoam.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;openFoam&lt;/a&gt; wind tunnel case of a wind turbine.  I already have the blade geometry constructed, but I'm really new to CFD and parametric meshing.  I found some articles where they mention a C topology to mesh the airfoil and also some software (&lt;a href=&quot;http://bladedesigner.sourceforge.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;bladedesigner&lt;/a&gt;) that mentions a topology named O10H, but none of these has a reference about how to implement these topologies and if there are others.  They also do not specify what the best use cases for each one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me to get started making an efficient mesh for my tool? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; At the end I want to have a domain like the picture below, with other blade orientation, but first I need to define the mesh topology around the blade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dv73A.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dv73A.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My example Blade:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0kNCh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0kNCh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the mesh topologies that I found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bladedesigner O10H topology&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B37LM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B37LMm.png&quot; alt=&quot;Bladedesigner O10H&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Various C topologies&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RW352.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RW352m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/el5Mr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/el5Mr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Xwv8o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xwv8o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dFhvz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dFhvz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2211" LastEditorUserId="2211" LastEditDate="2015-09-30T13:30:14.423" LastActivityDate="2019-11-21T18:55:13.090" Title="Efficient 3D mesh topology for CFD simulation of turbine blade" Tags="|airflow|cfd|wind-power|meshing|wind-tunnels|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5585" PostTypeId="1" CreationDate="2015-09-29T05:36:46.700" Score="2" ViewCount="74" Body="&lt;p&gt;In order to run some numerical simulations, I would like to know some rough values (estimates) of the velocity at which the crude oil flows through the rock or soil matrix from which it is extracted. I found lots of data about production rates (bpd), which is the main parameter of interest from the engineering point of view, but so far, nothing about this technical aspect. One last thing, I am an outsider (civil engineer and applied mathematician), not an specialist.    &lt;/p&gt;&#xA;" OwnerUserId="3166" LastEditorUserId="90" LastEditDate="2015-09-29T08:59:48.157" LastActivityDate="2016-11-29T12:26:38.917" Title="Crude Oil Velocity Flow Within Rock-Soil Matrix" Tags="|petroleum-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5588" PostTypeId="2" ParentId="5523" CreationDate="2015-09-29T11:50:39.520" Score="6" Body="&lt;p&gt;I'm going to take a slightly unconventional approach, and suggest you design to an &lt;em&gt;assessment&lt;/em&gt; standard; namely BD21/01&lt;a href=&quot;http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol3/section4/bd2101.pdf&quot;&gt; [pdf link]&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasons are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It's a UK standard, and you're in the UK&lt;/li&gt;&#xA;&lt;li&gt;It provides loading details for vehicle classes which are lighter than the full 40/44 tonnes.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The rest of this answer probably won't make much sense to those not familiar with BD21. I thought it useful to include this advice, but I don't think a full explanation of all terms and loading to BD21 is appropriate here.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that seeing as it's a private bridge, the surfacing maintenance will be minimal, and such the design should be for &lt;strong&gt;Poor&lt;/strong&gt; surfacing. Clearly the traffic flow is &lt;strong&gt;Low&lt;/strong&gt;, so you're looking at Figure 5.4 for your K factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an absolute minimum, I think your bridge needs to be sufficient for all fire engines (i.e. group 1 FE loading to BD21). Beyond that, you've suggested a concrete mixer truck might need to use it for access. A quick bit of googling and I've found &lt;a href=&quot;http://www.hanson.co.uk/en/technical-information/truck-information&quot;&gt;two&lt;/a&gt; different &lt;a href=&quot;http://www.smithsconcrete.co.uk/truck.htm&quot;&gt;links&lt;/a&gt; which suggest a concrete truck might be 26 tonnes or 32-33 tonnes. The difference between 26 tonnes and 40 tonnes loading for a 5m span is a K factor of 0.85 vs 0.87. Given the small difference, and that you wanted to allow the larger concrete trucks on your bridge, I think you should design for the 40 tonnes loading.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-09-29T11:50:39.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5590" PostTypeId="1" CreationDate="2015-09-29T13:21:48.387" Score="5" ViewCount="134" Body="&lt;p&gt;As we all know, Volkswagen is in the middle of a scandal centered on its use of special software to lower tailpipe emissions during standard emission testing protocols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is why this software needed to be switched off in other circumstances. Why can't the car always operate in low-emission mode? Is there some negative effect on performance, component life?&lt;/p&gt;&#xA;" OwnerUserId="3174" LastEditorUserId="272" LastEditDate="2015-09-29T13:29:04.220" LastActivityDate="2015-09-29T17:05:30.203" Title="Why did Volkswagen use special software ONLY during emissions testing?" Tags="|automotive-engineering|air-quality|emissions|" AnswerCount="0" CommentCount="5" ClosedDate="2015-11-17T12:28:48.727" ContentLicense="CC BY-SA 3.0" />
  <row Id="5591" PostTypeId="1" AcceptedAnswerId="5594" CreationDate="2015-09-29T13:25:58.347" Score="1" ViewCount="1266" Body="&lt;p&gt;My understanding so far is that a jump occurs when the Froude Number $\left(Fr=\frac{v}{\sqrt{2gh}}\right)$ of the radial flow falls with a decreasing velocity due to friction and viscosity; and the jump occurs at the critical point where $Fr=1$.  However, I also learnt that the height of the jump is dependant on the difference in $Fr$ before and after the jump which contradicts the former theory where there is no difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody please explain and help correct my understanding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://thatsmaths.com/2014/01/09/white-holes-in-the-kitchen-sink/&quot; rel=&quot;nofollow&quot;&gt;http://thatsmaths.com/2014/01/09/white-holes-in-the-kitchen-sink/&lt;/a&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hydraulic_jump#Tabular_summary_of_the_analytic_conclusions&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Hydraulic_jump#Tabular_summary_of_the_analytic_conclusions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3013" LastEditorUserId="3013" LastEditDate="2015-09-29T15:28:26.577" LastActivityDate="2017-08-28T15:15:36.040" Title="What causes circular hydraulic jumps in kitchen sinks?" Tags="|mechanical-engineering|fluid-mechanics|hydraulics|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5593" PostTypeId="1" CreationDate="2015-09-29T14:52:15.730" Score="-1" ViewCount="502" Body="&lt;p&gt;I am building a sumo-bot and our competitors have thin sticky tires, while we have wider and less sticky tires. The diameter is the same, and the gearbox/motor is the same. Who will win?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Sticky tires: &lt;a href=&quot;https://www.pololu.com/product/694&quot; rel=&quot;nofollow&quot;&gt;https://www.pololu.com/product/694&lt;/a&gt; &amp;amp; wide tires: &lt;a href=&quot;https://www.pololu.com/product/62&quot; rel=&quot;nofollow&quot;&gt;https://www.pololu.com/product/62&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3181" LastEditorDisplayName="user16" LastEditDate="2015-09-29T15:14:11.397" LastActivityDate="2015-09-29T18:31:25.457" Title="Sumo-bot Wheel Help - Sticky tires or Wide &amp; less friction tires?" Tags="|friction|" AnswerCount="1" CommentCount="4" ClosedDate="2015-09-30T02:54:24.853" ContentLicense="CC BY-SA 3.0" />
  <row Id="5594" PostTypeId="2" ParentId="5591" CreationDate="2015-09-29T15:31:44.277" Score="2" Body="&lt;p&gt;Below is a diagram of hydraulic jump that I found on Google Images that shows all the key elements.  In &quot;Region 1&quot;, $Fr &amp;gt; 1.0$ and in &quot;Region 2&quot;, $Fr &amp;lt; 1.0$.  In between these two regions, the Froude number is transitioning from one value to the other.  If the upstream Froude number is increased, the ratio of fluid height before and after the jump will increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, this diagram is for one-dimensional flow; The sink problem is a little different because it's two-dimensional; the cross-sectional area of the flow is increasing as the flow moves outward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4RdTi.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4RdTi.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found this image, which shows another important feature: the flow leading up to the jump is steadily decreasing in velocity and increasing in height.  These two events are what drive down the Froude number to the point that it falls below 1.0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rutsU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rutsU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how does the jump start in the first place?  Look at the second image; even before we get to the jump, the height of the fluid downstream is slightly &lt;em&gt;higher&lt;/em&gt; than the fluid upstream.  This causes the fluid to want to flow upstream, against the flow.  More specifically, there is a surface wave trying to flow upstream.  It can't flow upstream, however, because its velocity is slower than the fluid velocity.  Eventually though, the fluid slows enough (due to friction) such that it approaches the speed of the surface wave.  At this point the wave is not carried away by the flow any more, it stays put and pushes back against the flow.  This causes a build-up of fluid until finally the kinetic energy of the incoming fluid balances the potential energy of the fluid after the jump.  At this point an equilibrium is reached.  We can use Bernoulli's equation to analyze this equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{v^2}{2} + gz + \frac{p}{\rho} = constant$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is a form of energy balance.  The three terms in the equation represent kinetic, gravitational potential, and pressure (potential) energy (respectively).  Ignoring losses due to friction, their sum is constant across the hydraulic jump.  The flow entering the jump has a lot of kinetic energy, but not much potential energy.  The flow leaving the jump however, has less kinetic energy but a lot more potential energy because its higher than the upstream flow.  The pressure of the fluid probably doesn't change much, so we can just ignore the pressure term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equating the sums of energies of the inlet and outlet flows gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{v_{in}^2}{2} + gz_{in} = \frac{v_{out}^2}{2} + gz_{out}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a nice derivation of the momentum equation for this case in the Wikipedia article on hydraulic jump.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-09-29T16:53:05.753" LastActivityDate="2015-09-29T16:53:05.753" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5596" PostTypeId="2" ParentId="5593" CreationDate="2015-09-29T18:31:25.457" Score="3" Body="&lt;p&gt;Assuming &lt;em&gt;everything&lt;/em&gt; about the vehicles is the same - &lt;strong&gt;mass especially&lt;/strong&gt;, but also shape, center of gravity, etc., such that the entire problem boils down to tire grip, you will lose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://physics.stackexchange.com/a/154485&quot;&gt;this answer&lt;/a&gt; for more information, but essentially, &lt;em&gt;assuming your vehicle weight is low enough that you're not going to deform the competition surface&lt;/em&gt;, static friction is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F_{\mbox{friction}} = \mu_{\mbox{static}} P_{\mbox{tire}} A&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mu_{\mbox{static}}$ is the static coefficient of friction, $P_{\mbox{tire}}$ is the pressure the tires exert on the ground, and $A$ is the contact surface area. However, because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_{\mbox{tire}} = \frac{F_{\mbox{normal}}}{A}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F_{\mbox{normal}}$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Normal_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;normal force&lt;/a&gt; of the vehicle, you wind up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F_{\mbox{friction}} = \mu_{\mbox{static}} \frac{F_{\mbox{normal}}}{A} A&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F_{\mbox{friction}} = \mu_{\mbox{static}} F_{\mbox{normal}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, again, assuming &lt;em&gt;everything&lt;/em&gt; else about the two robots is the same, and that you're not operating on a surface like mud or something else that will appreciatively deform under the tires, then the traction/friction force will be higher in the vehicle with a higher coefficient of friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, your tires should start slipping before theirs do, and since &lt;a href=&quot;https://physics.stackexchange.com/a/569&quot;&gt;static friction is greater than dynamic friction&lt;/a&gt;, once your tires start to slip they get a significant &quot;pushing&quot; advantage over you. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2015-09-29T18:31:25.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5598" PostTypeId="2" ParentId="5494" CreationDate="2015-09-29T20:03:48.283" Score="9" Body="&lt;h1&gt;Here's the formula(s) we use&lt;/h1&gt;&#xA;&#xA;&lt;h3&gt;Beam Bending (&lt;a href=&quot;https://en.wikipedia.org/wiki/Deflection_(engineering)&quot;&gt;available on Wikipedia&lt;/a&gt;)&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$EI\frac{d^4\,\delta y}{d\,x^4}=q(x)$$&#xA;$$I=\int (y-\bar y)^2 dA$$&#xA;$$\bar y= \frac1{A}\int y \;dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_{max} = y_{max}E\frac{d^2\,\delta y}{d\,x^2}_{max}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $A$ is the cross-sectional area of the beam, $y$ is the position along the axis in the direction of the beam loading, $\delta y$ is the deflection in the direction of loading, $E$ is the modulus of elasticity (search matweb to get a value for your material), and finally $q(x)$ is the load per distance function.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Here's how to use them&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;For a rectangular tube with height $H$, width $W$ and thickness $t$ we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar y=0$$&#xA;$$I=W\int_{-\frac{H}2}^{\frac{H}2}y^2\;dy-(W-2t)\int_{-\frac{H}2+t}^{\frac{H}2-t}y^2\;dy=\frac{H^3W-(H-2t)^3(W-2t)}{12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $H=4\,in\,,\quad W=2\,in\,,\quad t=.1875\,in$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I\approx4.2\,in^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there's beam loading, this is likely where you ran into difficulty. First lets take a look at a cantilever beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here there are just two points that are loaded, the support and the tip. Think of a diving board scenario. We'll say the support is at $x=0$ and the load $F$ is at $x=L$&#xA;$$q(x)= -\delta(x)F+\delta(x-L)F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EI\frac{d^4\,\delta y}{d\,x^4}=q(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EI\frac{d^3\,\delta y}{d\,x^3}=\int_{0^-}^xq(x) dx = F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically saying that there's a constant shearing stress in the beam the whole way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EI\frac{d^2\,\delta y}{d\,x^2}=\int F dx +C=F(x-L)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This expression is for the bending moment in the beam. We know that the free end must have a bending moment of zero so we set the integration constant to accommodate that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d\,\delta y}{d\,x}=\frac1{EI}\int F(x-L) dx +C=\frac{F}{EI}(\frac12 x^2-Lx)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This represents the slope of the deflected beam. Here we know that the slope must be zero at the support so we've set the integration constant accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta y=\frac{F}{EI}\int \frac12 x^2-Lx \; dx +C=\frac{F}{EI}(\frac16 x^3-\frac{L}2 x^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we know the deflection is zero at the support so we set eh integration constant accordingly. Now if we just want to look at the deflection at the end we plug in $x=L$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta y=-\frac{FL^3}{3EI} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This corresponds to the equation on last website in your post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.matweb.com/search/DataSheet.aspx?MatGUID=f7666326ceb3482f87a9f41ace1d1fb0&quot;&gt;matweb&lt;/a&gt; for medium alloy steel we have $E=30\,000\,ksi$ So plugging in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta y= -\frac{3.750\,klb \, (216 in)^3}{3\,30000\,ksi\,4.2\,in^4}\approx -100\,in$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly what the online calculator produced. However, if you tried to load a beam like this it would permanently deform. An 18 foot lever arm is really long and will bend the snot out of a 4in thin wall beam with only moderate difficulty. The issue is that a trailer is not a cantilever beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so let's take a look at a more reasonable loading scenario. Let's model the axles as a point loads located $40\,in$ and $80\,in$ from the end of the trailer, the $7500 lbf$ load as distributed over the rearward $18\,ft$ and the gooseneck support an additional $5ft$ in front of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now some of our loads aren't known yet, but we can figure out some of them in the process. Some of them we can't though, so let's add an additional constraint. The weight distribution will be split between the axles according to the variable $\alpha$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{axles}=F_{rear} \frac1{\alpha}=F_{front}\frac1{(1-\alpha)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q(x)=-\frac{F}{L}H(L-x)+F_{axels}(\alpha\delta(x-x_{rear})+(1-\alpha)\delta(x-x_{front})+(F-F_{axels})\delta(x-x_{goose})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EI\frac{d^3\,\delta y}{d\,x^3}=\begin{cases} &#xA;      -\frac{F}{L}x &amp;amp; x\leq x_{rear} \\&#xA;      -\frac{F}{L}x+F_{axels}\alpha &amp;amp; x_{rear} \lt x\leq x_{front} \\&#xA;      -\frac{F}{L}x+F_{axels} &amp;amp; x_{front} \lt x\leq L \\&#xA;      F_{axels}-F &amp;amp; L \leq x &#xA;   \end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then integrating again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EI\frac{d^2\,\delta y}{d\,x^2}=\begin{cases} &#xA;      -\frac{F}{2L}x^2 &amp;amp; x\leq x_{rear} \\&#xA;      -\frac{F}{2L}x^2+F_{axels}\alpha (x-x_{rear}) &amp;amp; x_{rear} \leq x\leq x_{front} \\&#xA;      -\frac{F}{2L}x^2+F_{axels} (x-(1-\alpha)x_{front}-\alpha x_{rear}) &amp;amp; x_{front} \leq x\leq L \\&#xA;      (F_{axels}-F) (x-x_{goose}) &amp;amp; L \leq x &#xA;   \end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this bending moment must be continuous and both ends must be zero as there is no bending moment applied at the ends (they are free to rotate) This leads to an additional constraint that can be used to find $F_{axles}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{axels}=F\frac{x_{goose}-\frac{L}2}{x_{goose}-(1-\alpha)x_{front}-\alpha x_{rear}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to keep the expressions shorter lets leave $F_{axels}$ in the expressions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the slope will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d\,\delta y}{d\,x}=\frac1{EI}\begin{cases} &#xA;      -\frac{F}{6L}x^3+C_1 &amp;amp; x\leq x_{rear} \\&#xA;      -\frac{F}{6L}x^3+F_{axels}\alpha \frac12 ( x-x_{rear})^2+C_1 &amp;amp; x_{rear} \leq x\leq x_{front} \\&#xA;      -\frac{F}{6L}x^3+F_{axels}(\alpha \frac12 (x-x_{rear})^2+(1-\alpha)\frac12(x-x_{front})^2)+C_1 &amp;amp; x_{front} \leq x\leq L \\&#xA;      (F_{axels}-F)\frac12 (x-x_{goose})^2 +C_2 &amp;amp; L \leq x &#xA;   \end{cases}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And at this point I moved over to a numerical solution. I integrated again and found values for all the constants such that both the slope and displacement were continuous and the displacement at the goose and the rear axle were zero. The resulting deflection had a maximum at about 2 inches. But I used the full load and I should have used half the load giving 1 inch. That sounds about right to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the peak bending moment is $9\, kN \,m$ which when multiplied by half our height and divided by our area moment give a peak stress of $38 ksi$ this is about 13% the yield strength of the medium alloy steel on matweb. You mihgt hink this would be sufficient then however, this is only for a static trailer, not one moving and bumping around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acceleration forces on a trailer could easily triple the load over short periods. Additionally, the bumps in the road will cycle the loading making it not the yield strength you want to look at but the fatigue strength at the appropriate number of cycles you'd like the trailer to last. The fatigue strength may be as low as 10% of the yield strength, so I would want a minimum load factor of about 30 (3/10%), then add a factor of safety of 2 and your beams need to be about 60 times stronger than would be required to just meet your yield stress in a static load scenario. In short, I would go with bigger beams.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-09-29T20:03:48.283" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5601" PostTypeId="1" AcceptedAnswerId="5602" CreationDate="2015-09-30T02:11:15.903" Score="6" ViewCount="816" Body="&lt;p&gt;I am printing a wheel with a 2.5mm hole in the middle to accept a shaft of a motor with a set screw. The problem I am having is that the wheel is always wobbly when spinning (sometimes more, sometimes less without changing any parameters). Is there anything I can do to make this print as perfect as possible and prevent the wheel from wobbling?&lt;/p&gt;&#xA;" OwnerUserId="2164" LastEditorUserId="41" LastEditDate="2015-11-16T15:11:05.313" LastActivityDate="2016-11-10T17:33:17.213" Title="How do you make a 3D print with an accurately aligned bore hole?" Tags="|mechanical-engineering|3d-printing|prototyping|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5602" PostTypeId="2" ParentId="5601" CreationDate="2015-09-30T03:08:53.290" Score="10" Body="&lt;p&gt;3D printing bores/holes is inherently and wildly inaccurate. You can continuously tweak the model, material, and print configurations to get better results, but for best results, in my experience, redraw the hole to a slightly smaller size than your target and reprint. Then, after printing, use a drill to get the size and geometry more precise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your case, re-design the hole in your 3D model to 2mm-2.2mm and use a 2.5mm drill after it's printed. Make sure with your infill amount that after drilling you still have a solid hole surface(infill is not exposed) Also, if you have a drill-press, use that for better drilling results.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2015-09-30T19:58:20.800" LastActivityDate="2015-09-30T19:58:20.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5603" PostTypeId="1" AcceptedAnswerId="5614" CreationDate="2015-09-30T10:04:35.117" Score="3" ViewCount="436" Body="&lt;p&gt;Instead of using an alternator to charge the battery, and  a starter to start the engine, why not just put one electric motor which is always connected with the engine? When we want to start the motor, we supply electricity to it; and when the engine is started, it spins the motor to generate electricity and charge the battery.&lt;/p&gt;&#xA;" OwnerUserId="3063" LastEditorUserId="498" LastEditDate="2017-09-14T15:28:13.720" LastActivityDate="2017-09-14T15:28:13.720" Title="Why not use just one electric motor in a car?" Tags="|mechanical-engineering|electrical-engineering|electric-vehicles|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5604" PostTypeId="1" CreationDate="2015-09-30T10:52:56.813" Score="1" ViewCount="1394" Body="&lt;p&gt;I have PCHR trace files of the above mentioned server . But unfortunately I couldn’t find an appropriate decoding software to read those files. I tried manual through a c program but it didn’t work well.&#xA;files are of the format &lt;code&gt;.log&lt;/code&gt; for example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;BSC0007_[PCHR]02Log20150928102948_20150928103252.log&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the header of this HEX files contains this when it decoded to UNICODE&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;.........................&#xA;.........................&#xA;.........................&#xA;.........................&#xA;....TT__TOOLMBSC.........&#xA;.V900R015....C00SPH588...&#xA;..4........Ü_.ß..MBSC CHR&#xA;.........................&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;CHR: (Call History Record) Detailed info for the abnormal call.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PCHR: (Performance CHR) Limited info for all calls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can anyone help me with these&lt;/p&gt;&#xA;" OwnerUserId="3205" LastEditorUserId="3205" LastEditDate="2015-10-09T04:12:18.697" LastActivityDate="2015-10-09T04:12:18.697" Title="seeking help for decoding a call log / PCHR trace files (Call logs)of Huawei V900 R015 machine(.log files)" Tags="|telecommunication|" AnswerCount="0" CommentCount="8" ClosedDate="2015-09-30T14:18:31.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="5606" PostTypeId="2" ParentId="5514" CreationDate="2015-09-30T12:47:58.007" Score="4" Body="&lt;p&gt;Planetary Gear Sets can provide a high reduction in a small space. Also, they can be connected in a similar fashion to gear train to further increase the reduction. Also, one benefit is the output shaft is co-axial with the input shaft.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2015-09-30T12:47:58.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5607" PostTypeId="2" ParentId="5494" CreationDate="2015-09-30T12:59:20.073" Score="4" Body="&lt;p&gt;Here is some additional information and lengthy discussion on trailer design criteria. There is even a white paper in the thread on loading and safety factors that should be used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.eng-tips.com/viewthread.cfm?qid=261920&quot; rel=&quot;nofollow&quot;&gt;Loads for Trailer Design&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of other threads on that site providing good information regarding trailer design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, I'd start my structure design with some type of rectangular steel section in mind for a trailer. They are regularly available and &quot;easy&quot; to work with(cut, drill, weld, mount other components, etc). &lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="1586" LastEditDate="2015-10-18T14:28:00.317" LastActivityDate="2015-10-18T14:28:00.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5610" PostTypeId="2" ParentId="5570" CreationDate="2015-09-30T16:18:38.187" Score="5" Body="&lt;p&gt;Details have emerged in the German press that indicate this was known about in 2007 (&lt;a href=&quot;http://blog.caranddriver.com/report-bosch-warned-vw-about-diesel-emissions-cheating-in-2007/&quot; rel=&quot;nofollow&quot;&gt;from German newspaper Bild am Sonntag, as translated by Automotive News&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to the report, Bosch claims to have supplied diesel&#xA;  engine-management software to Volkswagen under the impression that it&#xA;  would be used only in vehicle testing. That software, which was able&#xA;  to activate emissions-control devices when a testing environment was&#xA;  detected and deactivate them during normal driving, somehow ended up&#xA;  in production vehicles. According to Bild am Sonntag, Bosch wrote to&#xA;  Volkswagen in 2007 warning the automaker that using this software in&#xA;  publicly sold vehicles was illegal.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In a statement released last week, Bosch revealed that it supplied&#xA;  common-rail fuel-injection systems, as well as supply and dosing&#xA;  modules for exhaust-gas treatment, on the Volkswagen and Audi models&#xA;  at the center of the growing emissions-cheating scandal. “As is usual&#xA;  in the automotive supply industry, Bosch supplies these components to&#xA;  the automaker’s specifications,” the statement reads. “How these&#xA;  components are calibrated and integrated into complete vehicle systems&#xA;  is the responsibility of each automaker.”&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Automotive News reports that the crisis began when&#xA;  then–VW brand chief Wolfgang Bernhard and engineer Rudolf Krebs began&#xA;  developing a new diesel engine for the U.S. market in 2005. Bernhard&#xA;  and Krebs realized that an AdBlue urea exhaust-treatment system would&#xA;  be needed to meet U.S. emissions standards, at an estimated cost of&#xA;  $335 per vehicle. Reportedly, VW finance heads determined this cost&#xA;  was too high, as a company-wide cost-cutting exercise was then&#xA;  underway.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand most, if not all, of these vehicles have an electronic anti-skid braking system and/or an electronic stability control for roll, pitch and yaw. As the tests are done on a &quot;rolling road,&quot; which only uses the driven axle, the sensors for these must be disabled, as the undriven wheels are stationary.&lt;/p&gt;&#xA;" OwnerUserId="3214" LastEditorUserId="368" LastEditDate="2015-12-02T19:55:37.277" LastActivityDate="2015-12-02T19:55:37.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5613" PostTypeId="2" ParentId="3609" CreationDate="2015-09-30T17:28:49.120" Score="1" Body="&lt;p&gt;Expanding Dan's answer (after some research), The main reasons for the so-called instability in porous media that is subjected to a heat flux (e.g. Solar irradiance) with air as a working medium are the behavior of dynamic viscosity of air with temperature (Viscosity increases as temperature increases) and the bell-shaped distribution of concentrated incident solar flux on the porous receiver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many experiments mentioned in the same paper suggested a relationship between incident flux and velocity of air, If a local high flux spot was formed in a porous medium mass flow rate is decreased and temperature is increased (high local dynamic viscosity leading to a lower local flow rate) and vice versa for local low flux spots, which means that the porous medium temperature can exceed the design temperature leading to failure even if the outlet temperature is low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be shown by integrating the Forchheimer extension of Darcy's law using a correlation for viscosity behavior with temperature and ideal gas law, the following curve relating the quadratic pressure difference and outlet temperature can be obtained:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GGdoJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GGdoJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown that at specific incident fluxes at the same quadratic pressure difference there could exist two different outlet temperatures (hence the instability).&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2015-09-30T17:28:49.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5614" PostTypeId="2" ParentId="5603" CreationDate="2015-09-30T20:31:27.330" Score="5" Body="&lt;p&gt;One problem with generating electrical energy from a car engine is that the RPMs are constantly changing, and consequently the power output of an electrical generator would vary with engine speed.  An alternator is very good at responding to these variations; the current in the rotor can be quickly adjusted to change the power output at a given RPM.  I don't know how you would do the same with an electric motor as a generator, but I imagine that it would involve rejecting some of the surplus energy as waste, which would be inefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is that starting an engine and using it to generate electrical energy each require controlling very different amounts of electrical power.  When starting the engine, a high amount of electrical power is needed, but only for a few seconds.  When using the engine to &lt;em&gt;generate&lt;/em&gt; power, typically only a small amount of power is needed, but it has to be generated continuously, as long as the engine is running.  This means that the starter motor has to be built to withstand much higher current than the alternator, and the alternator has to be built to operate much longer than the starter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all comes down to the fact that the starter and the generator each have different design requirements.  You could use one device that does the job of both, but it might be something of a &quot;jack of all trades, master of none&quot;.  The best solution for the time being seems to be the use of a separate starter and generator.  That may change as technology advances though.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-09-30T20:31:27.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5615" PostTypeId="2" ParentId="5463" CreationDate="2015-09-30T22:24:24.163" Score="1" Body="&lt;p&gt;One answer can be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Magnetic suspension systems, in general, suffer from two principal&#xA;  components of uncertainty. The first of these are the parameter&#xA;  variations, where the electromagnetic coil characteristics change due&#xA;  to resistance heating, and the coefficients of the – curve drift with&#xA;  temperature. The second important source of uncertainty is the&#xA;  external interaction forces on the suspended body, which are typically&#xA;  unknown. Robustness to parameter variations relaxes the accuracy&#xA;  constraints on the force–current–airgap characterization of the&#xA;  electromagnets, while resistance to external forces determines the&#xA;  dynamic stiffness of the suspension. If the dynamical equations for&#xA;  the voltage–current characteristics of the power supply are not&#xA;  modeled in the suspension equations, then the parameter variations and&#xA;  the external disturbance force terms appear in the same level of&#xA;  differentiation as the system input, which happens to be the coil&#xA;  current in this case. Uncertainties that occur in the same order of&#xA;  differentiation as the control inputs are termed as matched&#xA;  uncertainties.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The matching conditions are typically restrictive for arbitrary&#xA;  nonlinear systems and, if the voltage–current dynamics of the power&#xA;  supply are modeled in the suspension equations, then the uncertainty&#xA;  terms appear in levels of differentiation other than that of the&#xA;  voltage input to the system and are, therefore, termed as unmatched&#xA;  uncertainties. Modeling of the voltage– current characteristics and&#xA;  consequent compensation of the resulting unmatched uncertainties is&#xA;  essential to an efficient and cost-effective design of the magnetic&#xA;  suspension system. By ignoring the dynamics of the voltage–current&#xA;  characteristics, the controller may demand arbitrarily high current&#xA;  slewing rates which, in turn, lead to an oversized power supply&#xA;  system. Therefore, compensation of unmatched uncertainties in&#xA;  nonlinear control system design is an important step in the context of&#xA;  magnetic suspension systems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mittal, Samir, and Chia-Hsiang Menq. &quot;Precision motion control of a magnetic suspension actuator using a robust nonlinear compensation scheme.&quot; Mechatronics, IEEE/ASME Transactions on 2.4 (1997): 268-280.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3044" LastActivityDate="2015-09-30T22:24:24.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5616" PostTypeId="2" ParentId="4213" CreationDate="2015-09-30T22:52:28.923" Score="2" Body="&lt;p&gt;Since this homework is likely long-past turned in, a direct answer probably can't hurt. Sorry if this doesn't help you with your homework, but hopefully it will still provide useful knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ASTM grain number is inversely proportional to grain size, so that the largest grain size number corresponds with the smallest grains. Assuming the grain size distribution is approximately uniform, the grain boundary surface area per unit volume of material increases with decreasing grain size. Thus higher ASTM grain size number corresponds with higher specific grain boundary area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grain boundaries are less densely packed with atoms, as they can be thought of as organized collections of dislocations. Diffusion of atoms in materials occurs most rapidly where atomic packing is lowest, because there is a lower energy potential to be overcome by the diffusing atoms. Diffusion rate is exponentially dependent on the additive inverse of the energy barrier (think Arrhenius). Thus diffusion occurs more rapidly at grain boundaries than in grain volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall effect is that higher ASTM number gives more specific grain surface area gives more regions of rapid diffusion. Diffusion therefore occurs most rapidly in a material with the highest ASTM grain number, all else equal. The correct answer is thus (C. 8).&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2015-09-30T22:52:28.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5621" PostTypeId="1" AcceptedAnswerId="5636" CreationDate="2015-10-01T11:48:03.677" Score="2" ViewCount="2095" Body="&lt;p&gt;I am in the process of helping my son design a sort of model of the &lt;a href=&quot;http://www.bloodhoundssc.com/project/car&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bloodhound SSC car&lt;/a&gt; for his school project, using balsa wood, model aircraft wheels, etc... It doesn't have to a like of like, but vaguely resemblant so that you can tell what it is just by looking at it. It would also be nice if the car could move and I was thinking of using on this small CO2 gas canisters as a means of propulsion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OVvrz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OVvrz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get some which are threaded, as above, or some which are just plain. What I haven't been able to find is a way to quickly release the pressure from the canister to propel the car. &lt;strong&gt;Are there any small on/off manual valves that would fit this type of gas canister?&lt;/strong&gt; I haven't been able to find any, the only ones apparently in existence are for fitting onto bike tyres to (re)inflate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only method I have come across is people using some sort of puncturing device (e.g. with a nail) that is generally sprung-loaded to pierce the gas canister when launching the model car. This puncturing device can be made part of the car, but more generally tends to be part of the &quot;launching pad&quot; as it adds weight and unnecessary complexity to the car. I would like to avoid this if I can.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-11-01T15:52:50.907" Title="Valve for mini CO2 gas canister?" Tags="|valves|propulsion|compressed-gases|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5622" PostTypeId="1" CreationDate="2015-10-01T17:27:51.960" Score="6" ViewCount="14362" Body="&lt;p&gt;I have a structural analysis question.  I want to know if the hydrostatic pressure in a water tank is a live load or a dead load and why?&lt;/p&gt;&#xA;" OwnerUserId="3059" LastEditorDisplayName="user16" LastEditDate="2015-10-01T18:23:23.400" LastActivityDate="2015-10-06T11:14:32.437" Title="Is hydrostatic pressure a dead load or a live load?" Tags="|civil-engineering|structures|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5623" PostTypeId="2" ParentId="5622" CreationDate="2015-10-01T18:10:01.257" Score="4" Body="&lt;p&gt;This will likely depend on your jurisdiction, but here is my interpretation based on US codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dead loads are typically considered well-known and well-defined.  For example, if you have a W8x13 beam, you know that that piece of steel is 13 pounds per linear foot.  This is why the LRFD load factor for dead load is lower than other loads - there is less inherent variability in the loading.  This is unlike a live load, where the occupancy of a building isn't well-known and is really just an educated guess based on the space's function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrostatic loads are very easy to calculate and, more importantly, very easy to calculate to a high degree of accuracy.  For anything that acts like a fluid, the pressure at the bottom is simply the density of the fluid times the height of the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ASCE 7-05, Chapter 3 is titled &lt;em&gt;Dead Loads, Soil Loads, and Hydrostatic Loads&lt;/em&gt;.  Though it doesn't explicitly include hydrostatic pressures on tanks, it does discuss lateral pressures due to soil and groundwater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, the hydrostatic pressure on a tank wall is therefore a Dead Load.  However, check with the authority having jurisdiction or your code of record to be certain.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-10-01T18:10:01.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5624" PostTypeId="2" ParentId="5622" CreationDate="2015-10-01T20:57:50.523" Score="2" Body="&lt;p&gt;It may depend on your location, but in the USA most codes eventually reference back to ASCE 07 &quot;Minimum design loads for buildings and other structures.&quot; In there, one load type is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;H = load due to lateral earth pressure, ground water pressure, or pressure of bulk material.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The load is then included the load combinations in the same place as live load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be argued that in your case, the load is easy to calculate accurately. This could make it seem like it should be factored as a dead load. &lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-10-01T20:57:50.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5625" PostTypeId="1" AcceptedAnswerId="5626" CreationDate="2015-10-01T23:21:58.187" Score="2" ViewCount="368" Body="&lt;p&gt;So, I must do the following conversion from my engineering textbook: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.155 \frac{m^2 * ^{\circ} C}{W}$ to $\frac{^{\circ}F * ft^2 * hr}{Btu}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can get as  far as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.155 \frac{m^2 * ^{\circ} C}{W} * (\frac{1 ft}{0.3048 m})^{2} * \frac{1 W}{3.4123 \frac{Btu}{h}} = .39108 \frac{ft^2 * ^{\circ}C *hr}{Btu}$, but I can't figure out how to do the final conversion properly, which involves converting that from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$.39108 \frac{ft^2 * ^{\circ}C *hr}{Btu}$ to $\frac{^{\circ}F * ft^2 * hr}{Btu}$. &lt;/p&gt;&#xA;" OwnerUserId="3226" LastEditorUserId="3226" LastEditDate="2015-10-01T23:41:12.927" LastActivityDate="2015-10-02T03:13:12.507" Title="Convert 1 one compound unit to another involving temperature" Tags="|heat-transfer|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5626" PostTypeId="2" ParentId="5625" CreationDate="2015-10-02T03:13:12.507" Score="3" Body="&lt;p&gt;I suspect you're converting a value of &lt;a href=&quot;https://en.wikipedia.org/wiki/R-value_(insulation)&quot; rel=&quot;nofollow&quot;&gt;thermal resistance&lt;/a&gt;, which means that the temperature units represent a temperature &lt;em&gt;difference&lt;/em&gt;, rather than specific values of temperature.  In this case, the constant 32 drops out of the conversion and you would just just multiply by $\frac{1.8^\circ F}{^\circ C}$.  Also, I agree with idkfa that your math is wrong in the conversion of power and area.  I get 0.489.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-10-02T03:13:12.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5627" PostTypeId="1" AcceptedAnswerId="5652" CreationDate="2015-10-02T08:49:10.083" Score="3" ViewCount="24178" Body="&lt;p&gt;How can I choose between Implicit or Explicit Analysis in ABAQUS for a simple rectangular composite structure with 2 layers of unidirectional lamina? I am trying to do a linear-static analysis to capture the tensile strain for the specimen. &lt;/p&gt;&#xA;" OwnerUserId="2799" LastEditorUserId="7461" LastEditDate="2017-05-15T16:07:34.110" LastActivityDate="2022-11-08T12:12:22.773" Title="What is the difference between implicit and explicit analysis in ABAQUS?" Tags="|finite-element-method|abaqus|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5628" PostTypeId="2" ParentId="5622" CreationDate="2015-10-02T09:38:03.863" Score="6" Body="&lt;p&gt;Eurocode looks into this with very much detail, see the &quot;Loading Standards&quot; list, attached. You can see that there is a document dedicated to actions in silos and tanks. This deals with hydrostatic forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as a rule of thumb you have two types of loads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Self Weight (Dead Load): Loads that are fixed and do not change during the lifespan of the structure. The load factors only accommodate variances in quality of workmanship and materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imposed Loads (Live Loads): loads which may vary during the lifespan of the structure. This means anything which may change. This would include your hydrostatic pressure, since the water level may change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KfQRm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KfQRm.png&quot; alt=&quot;Eurocode &amp;quot;Loading Standards&amp;quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2015-10-02T09:38:03.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5629" PostTypeId="1" AcceptedAnswerId="5631" CreationDate="2015-10-02T09:54:56.790" Score="8" ViewCount="675" Body="&lt;p&gt;From time to time I am tasked with additions or alterations to old buildings or structures. I would need to establish the &quot;as built&quot; structure and from that analysis see how the amended/new structure suits the existing structural components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge lies in the assumptions one is forced to make. Over the years things change, so do the material properties and standard member sizes. For example, I have come across a bombed bridge that had to be rehabilitated in Angola. The bridge was built by the Portuguese government some time before the Angolan war. The bridge was constructed from large steel I-beams as main bearers. We could not establish the type of steel used for the I-beams, and the sectional properties we established by measuring the beams with a tape measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do engineers determine the material properties of members of old or historic structures?&lt;/strong&gt; I am not aware of any document or database that captures all steel (or rebar) manufacturers of the world with their respective product lists and material properties over time, or what design standards were historically applicable at various times in various regions of the world.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="368" LastEditDate="2015-10-02T23:52:02.207" LastActivityDate="2015-10-06T12:59:01.683" Title="Determining the properties of members of historic structures" Tags="|materials|civil-engineering|structural-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5631" PostTypeId="2" ParentId="5629" CreationDate="2015-10-02T10:50:54.907" Score="7" Body="&lt;p&gt;I have never encountered a resource which compiled a worldwide database of historic structural data. But individual countries do have some resources. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the limited work I have done with historic structures (in the UK) I have used these resources:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.thenbs.com/PublicationIndex/DocumentSummary.aspx?PubID=67&amp;amp;DocID=261469&quot;&gt;Historical structural steelwork handbook&lt;/a&gt; contains material and section properties for UK and European cast iron, wrought iron and steel sections since the mid 19th century. It also includes references to historic building codes as well as loading and other design information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.concretebookshop.com/tr70-historical-approaches-to-the-design-of-concrete-buildings-and-structures-1869-p.asp&quot;&gt;TR70 Historical approaches to the design of concrete buildings and structures&lt;/a&gt; which contains historic reinforcement and concrete mix properties as well as historical design guidance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, additional research is usually required. National libraries often carry historic copies of building regulations and other design guidance. Institutions like the Institution of Structural Engineers and the Institution of Civil engineers in the UK often have historical construction information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine that outside of Western Europe and North America finding similar guidance would be very challenging or impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper may be helpful to get started doing historical research: &lt;a href=&quot;http://www.icevirtuallibrary.com/doi/abs/10.1680/cien.2006.159.1.42&quot;&gt;Chrimes, M. Historical research: a guide for civil engineers, Civil Engineering, Proceedings of the Institution of Civil Engineers, Volume 159, Issue 1, February 2006, pp. 42–47.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2015-10-02T10:50:54.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5632" PostTypeId="1" CreationDate="2015-10-02T11:08:58.213" Score="1" ViewCount="84" Body="&lt;p&gt;I'd like to have electronic control of the open/close timing of a source of air at the bottom of a tank of liquid such that I have control of when a single bubble is formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experimentation would be required but I expect the open period to be at &lt;em&gt;most&lt;/em&gt; 250ms - the more scope for lower period options the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process of bubble forming would be repeated over and over again - with a 50% duty-cycle the frequency of new bubble formation would be a minimum of 2Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say 5~10Hz would be ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple bubbles will need to be formed in close vicinity (~10mm) so the device would either have an appropriate form-factor, or some kind of tubing system would be put in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will have close to arbitrary control over the control signals, but I expect some form of digital control per 'channel', preferably parallel and &lt;em&gt;direct&lt;/em&gt;, i.e. &lt;code&gt;0=closed&lt;/code&gt;, &lt;code&gt;1=open&lt;/code&gt;, as opposed to some kind of interpreted data stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I understand that the dynamics of &lt;em&gt;bubbles&lt;/em&gt; will be all over the place, in terms of control I'd expect low latency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question: what kind of devices am I looking at here?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there off the shelf components that can do this?  Cost is a concern at some level but is not necessarily the overriding requirement.&lt;/p&gt;&#xA;" OwnerUserId="3231" LastEditorDisplayName="user16" LastEditDate="2015-10-02T14:23:37.400" LastActivityDate="2015-10-16T02:53:32.933" Title="How to control the timed release of small bubbles in a liquid?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5633" PostTypeId="2" ParentId="5539" CreationDate="2015-10-02T13:07:44.440" Score="2" Body="&lt;p&gt;Sure. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0015V45F4&quot; rel=&quot;nofollow&quot;&gt;CO2 cartridges&lt;/a&gt; are made of &lt;a href=&quot;https://www.reddit.com/r/Blacksmith/comments/119rd7/co2_cartridges_as_a_metal_source/&quot; rel=&quot;nofollow&quot;&gt;steel&lt;/a&gt; and hold &lt;a href=&quot;https://www.physicsforums.com/threads/12g-co2-cartridge-energy-in-j-please-help.686293/&quot; rel=&quot;nofollow&quot;&gt;850 psi&lt;/a&gt;. You want to use brass, which has &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultimate_tensile_strength&quot; rel=&quot;nofollow&quot;&gt;80% of the yield strength of steel&lt;/a&gt; (200MPa vs 250MPa) to hold 35% of that pressure. Not to mention the fact that you specify a 7x7x&quot;1-2&quot;cm vessel, which is 49-98cc, to hold 5cc of liquid, so you're left with up to 93cc of brass to use as the wall thickness, where again, the same dimensions as a CO2 cartridge should suffice due to the difference between yield strength and desired pressure limits. The use of silver solder won't matter because you can use whatever surface area you need to provide enough contact area to provide a solid joint - see the comment above about having plenty of excess volume. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-10-02T13:07:44.440" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5634" PostTypeId="2" ParentId="5629" CreationDate="2015-10-02T13:51:55.830" Score="7" Body="&lt;p&gt;My knowledge of historical work only extends as far as the borders of the United States, unfortunately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for US work, the American Institute of Steel Construction (AISC) does publish a &lt;a href=&quot;http://www.aisc.org/content.aspx?id=2868&quot;&gt;historical shapes database&lt;/a&gt; showing the different sizes used over the past 80 years or so and some of the manufacturers that made steel members.  If you can take measurements of the beam's dimensions, it's possible that you can back-figure what size it's supposed to be.  Given that, if the shape isn't made anymore, you might be able to determine around when it was made and correlate that to the common material specifications of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, if you don't have access to the design drawings or specifications used to construct the bridge and you absolutely have to know the material type, I don't see how you can get around taking a sample and having it tested in a lab.  For something as critical to human safety as a bridge, I personally think it's a justifiable cost, especially considering some of the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_bridge_failures#2000.E2.80.93present&quot;&gt;bridge failures that have occurred&lt;/a&gt; in the last decade or so.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="122" LastEditDate="2015-10-02T15:59:12.790" LastActivityDate="2015-10-02T15:59:12.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5635" PostTypeId="1" AcceptedAnswerId="5637" CreationDate="2015-10-02T15:13:41.620" Score="2" ViewCount="392" Body="&lt;p&gt;In the context of Bernoulli, does &quot;static pressure&quot; typically refer to $p$, or to $p + \rho gz$?&lt;/p&gt;&#xA;" OwnerUserId="3232" LastActivityDate="2018-10-13T15:18:05.407" Title="Is gravity included in &quot;static pressure&quot;?" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5636" PostTypeId="2" ParentId="5621" CreationDate="2015-10-02T15:34:49.137" Score="1" Body="&lt;p&gt;In the end, I decided to use an &lt;a href=&quot;http://www.estesrockets.com/rockets/engines/mini/1502-1-4a3-3t&quot; rel=&quot;nofollow noreferrer&quot;&gt;Estes 1/4A3-3T rocket engine&lt;/a&gt; (the smallest one they do):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xoy8X.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xoy8X.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll see how it goes!!&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-10-02T15:34:49.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5637" PostTypeId="2" ParentId="5635" CreationDate="2015-10-02T17:00:53.310" Score="4" Body="&lt;p&gt;$p_0$ is the static pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho g z$ is the hydrostatic pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{2} \rho v^2$ is the dynamic pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(z) = p_0 + \rho g z + \frac{1}{2} \rho v^2$&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2015-10-02T17:13:21.883" LastActivityDate="2015-10-02T17:13:21.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5638" PostTypeId="1" CreationDate="2015-10-02T17:19:51.017" Score="3" ViewCount="4506" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IW5j0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IW5j0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to calculate the force sustained on the three labelled members of this frame (Members 1, 2 &amp;amp; 3). The frame will experience a uniform pressure of 1000 kPa in the area enclosed by the members. It will be screwed into walls on all 4 sides. You can imagine it as being installed in a 4m by 4m shaft. And the goal is to brace a door in the shaft, which might experience wind pressures of a 1000 kPa. Once I have found the force I will calculate the deflection in order to select the correct steel members for the frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those curious, I just made up the numbers and the scenario but I am working on a similar situation and it’s been many years since I studied statics. Any help or guidance would be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="3233" LastEditorUserId="2246" LastEditDate="2015-10-03T21:41:25.783" LastActivityDate="2015-10-06T16:46:01.053" Title="How to calculate the forces on members of a frame, due to a distributed pressure load?" Tags="|structural-engineering|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5639" PostTypeId="2" ParentId="2948" CreationDate="2015-10-02T17:22:28.683" Score="1" Body="&lt;p&gt;It looks like your top points are not constrained in the X-Z plane like you've claimed. This would allow each top point to swing back and forth towards the and away from the center of the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the original model, perhaps your bending stiffness is to low. If there is no bending stiffness that would also explain the problem. Even if in reality there was no bending stiffness and the mesh deforms just find without coming apart, numerically the solution could be unstable. Imagine two stiff links with a hinge, one end anchored, and pulling on the other. Obviously the solution is that the links will be pulled straight, but numerically, the center joint could bend slightly back and forth and the restoring force would be tiny, so it would probably not converge.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-10-02T17:22:28.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5640" PostTypeId="1" AcceptedAnswerId="5641" CreationDate="2015-10-02T17:30:03.270" Score="4" ViewCount="198" Body="&lt;p&gt;I'm designing a crucible / extruder, and right now it has a .6mm hole to extrude from. Will this work? A chart with various metals listed would be nice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pictures:&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click to expand&lt;/em&gt;&lt;/sub&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UPzKX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UPzKXt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/qdmj3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qdmj3t.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UYLOJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UYLOJt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3235" LastEditorUserId="282" LastEditDate="2015-11-17T01:09:40.860" LastActivityDate="2015-11-17T01:09:40.860" Title="How small can you get a stream of molten steel to be at low pressure?" Tags="|mechanical-engineering|thermodynamics|metallurgy|metals|experimental-physics|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5641" PostTypeId="2" ParentId="5640" CreationDate="2015-10-02T18:01:51.830" Score="4" Body="&lt;p&gt;Making a nozzle with that long and thin of a hole is not feasible, but you could make a larger hole behind it and just have the last 3mm be 0.6mm in diameter. This would be feasible in a ceramic, but due to cost, I might make the nozzle into an insert that goes into a larger piece made of cheaper material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things to be concerned with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Surface_tension#The_breakup_of_streams_into_drops&quot; rel=&quot;nofollow&quot;&gt;Stream breakup due to surface tension.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Flow velocity/pressure&lt;/li&gt;&#xA;&lt;li&gt;Heat transfer&lt;/li&gt;&#xA;&lt;li&gt;Manufacturability &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Each of those could be their own question, answerable by people with different expertise. So I'd recommend asking them separately.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="1633" LastEditDate="2015-10-02T18:51:34.500" LastActivityDate="2015-10-02T18:51:34.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5642" PostTypeId="2" ParentId="2807" CreationDate="2015-10-02T18:39:02.950" Score="2" Body="&lt;p&gt;sys·tem     /ˈsistəm/       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;noun: a set of connected things or parts forming a complex whole, in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a set of things working together as parts of a mechanism or an interconnecting network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plural noun: systems&#xA;&quot;the state railroad system&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;synonyms: structure, organization, arrangement, complex, network; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;informal: setup &quot;a system of canals&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;en·gi·neer  /ˌenjəˈnir/     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;noun: engineer; plural noun: engineers&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a person who designs, builds, or maintains engines, machines, or public works.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;synonyms: originator, deviser, designer, architect, inventor, developer, creator; mastermind &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;the prime engineer of the approach&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;verb: engineer; 3rd person present: engineers; past tense: engineered; past participle: engineered; gerund or present participle: engineering&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;design and build (a machine or structure).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&quot;the men who engineered the tunnel&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, a Systems Engineer is a person who designs, builds, or maintains a set of connected things or parts forming a complex whole. Doesn't matter if that complex whole is a rocket ship, an airplane, a computer network, or a waste treatment facility. The Systems Engineer is the guy that makes things work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the classic &quot;glass half full, glass half empty&quot; test for optimism/pessimism. Any engineer can tell you the problem isn't half full or half empty. The problem is that the volume of the glass is twice a much as it needs to be to efficiently contain the liquid. The Systems Engineer is the guy who comes up with a way to successfully reduce the volume of the glass so it works properly, with maximum efficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the liquid is still in it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And while people (customers) continue to drink out of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is why &quot;Freakin' Awesome&quot; should be a perfectly acceptable job title. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;LEM&lt;/p&gt;&#xA;" OwnerUserId="3237" LastActivityDate="2015-10-02T18:39:02.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5644" PostTypeId="2" ParentId="4271" CreationDate="2015-10-02T18:44:47.967" Score="2" Body="&lt;p&gt;First point is that signum is not continuous with the jump at $t=0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second because of the absolute value in the integral it has the same $E$ and $P$ as a constant signal $1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that $E_{sgn}=+\infty$ and $P_{sgn}=1$&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-10-02T18:44:47.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5645" PostTypeId="2" ParentId="5638" CreationDate="2015-10-02T19:31:27.707" Score="5" Body="&lt;p&gt;The force on the perimeter will depend on the boundary conditions at the edges of each plate.  Determining the precise distribution of load at each point of the supported edges of the panels is going to be a headache if done by hand.  I would determine the maximum force per unit length of the edges and use that as a uniform linear force on the supporting members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I would approach this is to use formulations from &lt;em&gt;Roark's Formulas for Stress and Strain&lt;/em&gt;, 7th Edition, Table 11.4:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Edges Simply-Supported&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mfnc5.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mfnc5.png&quot; alt=&quot;simply supported&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Edges Fixed&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yAXGN.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yAXGN.png&quot; alt=&quot;fixed&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will want to solve for $R_{max}$.  For the Edges Fixed case, I believe $R_{max}$ is the same as for the simply-supported case.  Also, in the Fixed case, you will have to include the edge moments into your design.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Otherwise&lt;/strong&gt;, if you're just looking for an approximate method to determine the force on the supporting members, first determine the total force on a panel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = (1000 kPa)A_{panel}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, take that force and divide it by the perimeter of the panel ($P_{panel}$) to determine your uniform load per unit length:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w = F/P_{panel}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give you a load analogous to the $R_{max}$ determined if you follow &lt;em&gt;Roark's&lt;/em&gt; above, though this will be an average instead of the peak lineal load.  It will under-estimate the loads on the long edges and overestimate the forces on the short edges.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-10-02T19:52:19.283" LastActivityDate="2015-10-02T19:52:19.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5647" PostTypeId="2" ParentId="5635" CreationDate="2015-10-03T10:00:33.810" Score="2" Body="&lt;p&gt;Without any further information on where you want to apply &quot;Bernoulli&quot; it is hard to answer the question. But in general the static pressure $p$ does not include $\rho g z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets recall again &lt;a href=&quot;https://en.wikipedia.org/wiki/Bernoulli%27s_principle&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;Bernoulli's principle&lt;/strong&gt;&lt;/a&gt; which states:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Along&lt;/strong&gt; a &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Streamlines,_streaklines,_and_pathlines&quot; rel=&quot;nofollow&quot;&gt;streamline&lt;/a&gt;&lt;/strong&gt; $\frac{v^2}{2} + gz + \frac{p}{\rho}$ is &lt;strong&gt;constant&lt;/strong&gt;. Here $g$ is the gravity which is the force field acting on the fluid. This equation/model is true for small fluid speeds (with respect to its speed of sound, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Mach_number&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Mach Number&lt;/em&gt;&lt;/a&gt; for further information on compressibility effects). If one multiplies the equation with $\rho$ all parts of the equation have the &lt;a href=&quot;https://en.wikipedia.org/wiki/Dimensional_analysis&quot; rel=&quot;nofollow&quot;&gt;dimension&lt;/a&gt; of a pressure and the &lt;em&gt;contstant&lt;/em&gt; is referred to as total pressure, $p_\mathrm{t}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho\frac{v^2}{2} + \rho gz + p = $ Total Pressure $ = p_\mathrm{t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if one assumes an isentropic process, it is possible to convert one form of pressure into another form of pressure (always along the streamline).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the times $\Delta \rho$ and $\Delta z$ are small along the streamline which is investigated that it can be neglected. Then Bernoulli's principle can be simplified to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho\frac{v^2}{2} + p = p_\mathrm{t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The static pressure, $p$, in this equation therefore does not &lt;em&gt;include&lt;/em&gt; the &lt;em&gt;gravity&lt;/em&gt;. A drastic change of the altitude along the streamline needs to be addressed by including an additional term in the equation.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2015-10-03T10:00:33.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5650" PostTypeId="2" ParentId="4263" CreationDate="2015-10-03T15:01:49.433" Score="2" Body="&lt;p&gt;Aluminium rubbing on aluminum generally will wear very rapidly. You generally want some sort of plastic bushing to reduce friction and wear. (Like with luggage handles there's just a little piece of plastic between the tubes) you should also have a bushing for the other end of the joint (the one inside that's not visible.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should do a beam analysis to check that the smaller beam won't buckle right at the joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the piston shouldn't matter very much, but the minor benefit is probably worth the small effort to offset it.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-10-03T15:01:49.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5652" PostTypeId="2" ParentId="5627" CreationDate="2015-10-03T21:31:53.707" Score="9" Body="&lt;p&gt;The basic difference between implicit and explicit dynamics solutions is that an explicit solution takes account of the finite propagation speed (at the speed of sound) of dynamic effects through the material. To do that, you need a mesh which is fine enough to represent the spatial effects (e.g. a &quot;stress wave&quot; propagating through the structure), and time-steps of the same order of magnitude as the transit time of sound waves from each element to its neighbours. If the time steps exceed that size, the response will usually be unstable and the analysis will fail after a few time steps. The time step size is limited by the &lt;em&gt;smallest&lt;/em&gt; element in the model, not by the average size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Implicit solution methods smear out those local effects. The propagation of dynamic effects around the structure is controlled by the inertia (mass) of the structure, not by local speed of sound. You can think of an implicit solution method as assuming the speed of sound is infinite, or (equivalently) that any applied load affects all of the structure instantaneously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mesh for an implicit solution only needs to be fine enough to capture the overall deformation of the structure, and the time steps only need to be small enough to capture the frequency spectrum of the response that you are interested in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you need the fine detail from an explicit analysis, an implicit solution will usually run much quicker - possibly several orders of magnitude quicker. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if you were modelling the dynamics of a road vehicle for typical driving conditions, you are probably mainly interested in the low-frequency response to bumps in the road etc, and the propagation of stresss and strains from one end of the vehicle to the other as &quot;stress waves&quot; with a timescale of the order of 1ms is irrelevant. But if you want to model the deformation of the same vehicle in a crash, capturing the local high-speed behaviour of the structure as it deforms &lt;em&gt;is&lt;/em&gt; important, and you need to do an explicit analysis. Typically, the time step size for explicit dynamics is of the order of a microsecond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A static analysis is &quot;implicit&quot; by definition - the static response ignores any transient behaviour that occurs while the loads are being applied to the structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Abaqus there is some commonality between the Explicit and Implicit (a.k.a. Standard) analysis procedures. For example, you might want to do a dynamics analysis where the loads are suddenly removed from a pre-loaded structure. In that case, you could first do an (implicit) stress analysis, and use the steady-state results as the initial conditions for an (explicit) dynamics analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, none of the above mentioned linear or non-linear behaviour - both explicit and implicit methods can be either linear or nonlinear. But in &quot;real world&quot; applications, there are usually quicker ways to model the high speed &lt;em&gt;linear&lt;/em&gt; dynamics response of a structure, so the models analysed with Abaqus Explicit are usually nonlinear.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-10-03T21:31:53.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5653" PostTypeId="1" CreationDate="2015-10-03T22:40:33.830" Score="10" ViewCount="1105" Body="&lt;p&gt;I am struggling a bit with an external diffusion problem.  I am trying to calculate the concentration at the surface (as well as surface reaction rate) and would like some help or guidance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I have thus far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction taking place, is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5C%5C2%20H_2%20S%20&amp;plus;%20SO_2%20%5Crightarrow%200.5%20S_6%20&amp;plus;%202%20H_2%20O%5C%5C%20%5Cdownarrow%20%5Cdownarrow%20%5Cdownarrow%20%5Cdownarrow%20%5C%5C%202A&amp;plus;B%20%5Crightarrow%200.5C&amp;plus;2D&quot; alt=&quot;D&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the concentration of B at the surface of a spherical catalyst particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flux:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5C%5CN_%7BB%2Cr%7D%3D-cD_%7BB%2C%5Ctext%7Bmix%7D%7D%5Cfrac%7B%5Cpartial%20y_B%7D%7B%5Cpartial%20r%7D%20&amp;plus;%20y_b%20%28N_%7BA%2Cr%7D&amp;plus;N_%7BB%2Cr%7D&amp;plus;N_%7BC%2Cr%7D&amp;plus;N_%7BD%2Cr%7D%29%5C%5C%20%5C%5C%5C%5C%20%5Cfrac%7BN_%7BB%2Cr%7D%7D%7BN_%7BA%2Cr%7D%7D%3D%5Cfrac%7B-1%7D%7B-2%7D%5CRightarrow%20N_%7BA%2Cr%7D%3D2N_%7BB%2Cr%7D%5C%5C%20%5Cfrac%7BN_%7BB%2Cr%7D%7D%7BN_%7BC%2Cr%7D%7D%3D%5Cfrac%7B-1%7D%7B0.5%7D%5CRightarrow%20N_%7BC%2Cr%7D%3D-0.5N_%7BB%2Cr%7D%5C%5C%20%5Cfrac%7BN_%7BB%2Cr%7D%7D%7BN_%7BD%2Cr%7D%7D%3D%5Cfrac%7B-1%7D%7B2%7D%5CRightarrow%20N_%7BD%2Cr%7D%3D-2N_%7BB%2Cr%7D%5C%5C%20%5C%5C%5C%5C%20%5Ctherefore%20N_%7BB%2Cr%7D%3D%5Cfrac%7B2cD_%7BB%2C%5Ctext%7Bmix%7D%7D%7D%7By_B-2%7D%5Cfrac%7B%5Cpartial%20y_B%7D%7B%5Cpartial%20r%7D&quot; alt=&quot;D&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, from the diffusion equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7Br%7D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20r%7D%28r%5E2%20N_%7BB%2Cr%7D%29%3DR_B&quot; alt=&quot;D&quot;&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;R_A&lt;/code&gt; can be approximated by the first order reaction rate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5C%5CR_A%5Capprox-2%20K_1%20P_%7BSO_2%7D%5E%7B0.5%7D%5C%5C%5C%5C%20%5Ctext%7Bfrom%20stoichiometry%2C%20%7DR_B%3D0.5%20R_A%5C%5C%5Ctherefore%20R_B%3D%20-K_1%20P_%7BB%7D%5E%7B0.5%7D&quot; alt=&quot;D&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7Br%7D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20r%7D%20%5Cleft%20%28%20r%5E2%5Ctimes%5Cfrac%7B2cD_%7BB%2C%5Ctext%7Bmix%7D%7D%7D%7By_B-2%7D%5Cfrac%7B%5Cpartial%20y_B%7D%7B%5Cpartial%20r%7D%20%5Cright%20%29%20%3D-2K_1%28y_B%20P%29%5E%7B0.5%7D&quot; alt=&quot;D&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(just ignore the &quot;&lt;code&gt;2&lt;/code&gt;&quot; after the &lt;code&gt;=&lt;/code&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the boundary conditions that I think I should use, are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5C%5Cy_B%28r%3D0%29%3Dy_%7BB%2C%7D_%7B%5Ctext%7Bsurf%7D%7D%5C%5C%20y_B%28%5Cdelta%29%3Dy_%7BB%2C%7D_%7B%5Ctext%7Bbulk%7D%7D&quot; alt=&quot;d&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;, at all times, I already have the values of all components' bulk concentrations, and I also have values for &lt;code&gt;D_i,j&lt;/code&gt; and &lt;code&gt;D_i,mix&lt;/code&gt; for all &lt;code&gt;i&lt;/code&gt;,&lt;code&gt;j&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are my boundary conditions chosen correctly for solving for the surface concentration of B (i.e. c_B or y_B or P_B, which are all related)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need surface values for calculation of the effectiveness factor. I can use any way to calculate surface values with the values that I already have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I chose r to be any point in the radial direction, even &quot;past&quot; the sphere (when going from r=0, the centre), delta = the thickness of the boundary layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit 2:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems I may have over-complicated it. Based on &lt;a href=&quot;https://www.youtube.com/watch?v=FKO4K-KPlb4&amp;amp;list=PL61BFC1C064B40049&amp;amp;index=59&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;this&quot;&gt;this&lt;/a&gt; video, the control volume considered is only the gas part - the boundary layer. This is correct, since the reaction is assumed to only take place on the catalyst surface and not in the gas phase itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case, $R_B=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\therefore \large{ \frac{\partial }{\partial r}\left ( r^2 \frac{2cD_{B,\text{mix}}}{y_B-2} \frac{\partial y_B}{\partial r} \right)=0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at $y_B(0)=y_{B,\text{surf}}$ and $y_B(\delta)=y_{B,\text{bulk}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;!! Ahh, I've just realized a mistake in my boundary conditions. At $r=0$, we are at the center of the sphere, so that boundary condition is incorrect. !!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's try again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $y_B(r=r_{sphere})=y_{B,\text{surf}}$ and $y_B(\delta)=y_{B,\text{bulk}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pastebin.com/CQwYYyJj&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;From Matlab&quot;&gt;From Matlab&lt;/a&gt;: $\large{y_B= 2+{\left (y_{B,\text{bulk}}-2  \right )} \left ( \frac{y_{B,\text{surf}}-2}{y_{B,\text{bulk}}-2} \right )^{\left (\frac{r_{\text{sphere}}\left ( \delta -r \right )}{r\left ( \delta -r_{\text{sphere}} \right )}  \right )} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what? How do I get the surface concentration values? Since I do not know the thickness of the boundary layer ($\delta$)?&lt;/p&gt;&#xA;" OwnerUserId="3245" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.193" LastActivityDate="2018-09-13T04:05:58.807" Title="External diffusion: calculation of surface concentration" Tags="|heat-transfer|chemical-engineering|process-engineering|numerical-methods|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5654" PostTypeId="1" CreationDate="2015-10-03T23:03:35.823" Score="0" ViewCount="70" Body="&lt;p&gt;I recently bought a servo whose torque is 2.2kg/cm, if i wanted to rotate (using the servo) an item that weighs 6.8KG (15lbs) at a length of 0.4572M (18 inches) 60 degrees would it be possible?&lt;/p&gt;&#xA;" OwnerUserId="3068" LastActivityDate="2015-10-04T03:38:37.543" Title="Servo Torque question" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5655" PostTypeId="2" ParentId="5653" CreationDate="2015-10-04T03:19:11.900" Score="-1" Body="&lt;p&gt;If we can assume the sphere has a radius $r_0$, and that $\delta$ is the thickness of the boundary layer surrounding the sphere, then the boundary conditions I would use are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_B(r= r_0 + \delta)=y_{B,bulk}$$&#xA;$$\frac{\partial y_B}{\partial r} \bigg|_{r=0}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one (Dirichlet boundary condition) is what you have already.  The second one (Neumann boundary condition) is due to the symmetry of the spherical particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the diffusion through the boundary layer will be a separate equation from the diffusion through the sphere.  You'll need to set some kind of continuity condition so that the two solutions produce the same value of $y_B$ where they intersect at the surface of the sphere.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-10-04T03:49:52.240" LastActivityDate="2015-10-04T03:49:52.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5656" PostTypeId="2" ParentId="5654" CreationDate="2015-10-04T03:38:37.543" Score="3" Body="&lt;p&gt;I think the units for the servo torque should be $kg \cdot cm$.  Anyway, the torque required to rotate the mass if it is sticking 90 degrees out to the side is $(6.8\,kg)(0.4572 \,m)(9.8 \frac{m}{s^2}) = 30.5\, N \cdot m$.  The torque of your servo is $(2.2\,kg\cdot cm)(\frac{1m}{100cm})(9.8 \frac{m}{s^2})=0.216\,N\cdot m$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the servo is drastically undersized for this application.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-10-04T03:38:37.543" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5659" PostTypeId="1" CreationDate="2015-10-04T15:36:07.450" Score="2" ViewCount="61" Body="&lt;p&gt;Small refrigerator compressors in the US generally all have exactly the same design and I am trying to understand their starting mechanism better. How exactly does it work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I generally understand is that the motor is a squirrel cage motor which has two sets of windings in a &quot;star delta&quot; configuration. On the outside of the motor is a small black box with a two holes that connect to two pins coming out of the compressor called the &quot;positive temperature coefficient relay&quot;. One pin is the hot for the &quot;star&quot; windings and the other is for the &quot;delta&quot; windings. When the compressor starts the PTC directs house current to the first pin very briefly, then due to the wire getting hot, quickly switches over the current over to the other pin (the &quot;run&quot; pin). Is this basically correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, my main question is what kind of star delta motor is being used? I believe there are 5 different types: Normal, Enhanced, Combined, Partially Wound, and Uninterrupted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are two sets of windings in star delta motor, that would seem to increase the weight of the motor a lot, but makes it cheap to start it, since you only need a very cheap, small relay to do it. Is that the tradeoff being made from a design perspective? I assume that in operation only the run windings are used, so the start windings essentially are dead weight at that point, is that right?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2015-10-04T15:36:07.450" Title="How are small refrigerator compressors started?" Tags="|motors|compressors|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5663" PostTypeId="1" AcceptedAnswerId="7199" CreationDate="2015-10-05T01:49:44.163" Score="2" ViewCount="245" Body="&lt;p&gt;I've heard the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mark_48_torpedo&quot; rel=&quot;nofollow&quot;&gt;Mk. 48 torpedo&lt;/a&gt; is wire-guided, why not make it wire powered? The range would only be limited by the length of the wire. I'm wondering why this is not used and what problems it would have.&lt;/p&gt;&#xA;" OwnerUserId="863" LastEditorUserId="33" LastEditDate="2016-01-31T02:28:00.307" LastActivityDate="2016-01-31T14:40:37.570" Title="Is it possible to power a torpedo via a wire?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5665" PostTypeId="1" AcceptedAnswerId="5666" CreationDate="2015-10-05T10:57:10.960" Score="10" ViewCount="802" Body="&lt;p&gt;I am having trouble to calculate the differential equations of a simplified loading bridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is build as shown in the picture below (just a sketch):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YEj4q.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YEj4q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use the Newton approach, I am getting the following equations by neglecting friction, air resistance and changes in the length of rope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;m_k \ddot{x}_{k} = F_{A} + F_{S} \sin(\varphi) \\&#xA;m_G \ddot{x}_{G} = -F_{S} \sin(\varphi) \\&#xA;m_G \ddot{z}_{G} = m_{G} g - F_{S} \cos(\varphi)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I look at the kinematic relationships from the gripper (the circle with the weight $m_G$) I get the following equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_{G} = x_{k} + l \sin(\varphi) \\&#xA;z_{G} = l \cos(\varphi)\\&#xA;\varphi = \omega t = \dot{\varphi} t&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the weights $m_k$ and $m_G$ and the length $l$ but the values are not important right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to have two differential equations at the end. &#xA;One equation shall show the relationship between the driving force $F_A$ and the path of the trolley $x_k$ (with derivations)&#xA;The other equation shall show the relationship between driving force $F_A$ and angle of the rope $\varphi_G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that I want to make the transfer functions (Laplace transformation etc.) but that is not the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I can not seem to find those equations. My best approach so far looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;m_{k} \ddot{x}_{k} = F_{A} + F_{S} \sin(\varphi)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that means if&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;m_G \ddot{x}_{G} = -F_{S} \sin(\varphi) \\&#xA;F_{S} \sin(\varphi) = -m_{G} \ddot{x}_{G} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;m_{k} \ddot{x}_{k} = F_{A} - m_{G} \ddot{x}_{G} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if I derive $x_{G}$ like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_{G} = x_{k} + l \sin(\varphi) \\&#xA;\dot{x}_{G} = \dot{x}_{k} + l \dot{\varphi} \cos(\varphi) \\&#xA;\ddot{x}_{G} = \ddot{x}_{k} + l \left[ \ddot{\varphi} \cos(\varphi) - \dot{\varphi}^{2} \sin(\varphi) \right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am actually getting stuck here because I can not find a way to eliminate $\varphi$ from the equations. The addition theorems are not helping me at all (or I'm using them correctly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have an idea of how I should continue at this point?&#xA;I hope I don't need a complete solution. I am actually more interested at doing this myself and hope to get a push towards the right direction.&lt;/p&gt;&#xA;" OwnerUserId="3258" LastEditorUserId="1832" LastEditDate="2016-01-16T22:04:01.397" LastActivityDate="2016-04-29T10:23:21.007" Title="Differential equations of a (simplified) loading bridge" Tags="|mechanical-engineering|control-engineering|modeling|mathematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5666" PostTypeId="2" ParentId="5665" CreationDate="2015-10-05T11:31:14.220" Score="5" Body="&lt;p&gt;My guess is that you probably need another differential equation for the angular movement, that will involve the inertia, such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_G l^2 \ddot{\varphi} = m_G g l \sin(\varphi)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ddot{\varphi} = \frac{g}{l} \sin(\varphi)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then maybe use the small angles approximation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sin(\varphi) \simeq \varphi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverted_pendulum&quot; rel=&quot;noreferrer&quot;&gt;inverted pendulum&lt;/a&gt; example.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2015-10-05T11:31:14.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5667" PostTypeId="1" AcceptedAnswerId="5668" CreationDate="2015-10-05T13:26:22.700" Score="4" ViewCount="4803" Body="&lt;p&gt;I understand the difference between the two in terms of how they function; one will release all of it's energy in small pulses with the other creates a constant stream of radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is for materials processing such as marking or cutting.  Why would I use one over the other?&lt;/p&gt;&#xA;" OwnerUserId="3260" LastEditorUserId="272" LastEditDate="2015-10-05T14:33:18.960" LastActivityDate="2015-10-05T14:33:18.960" Title="Difference between continuous wave laser and a pulsed laser" Tags="|materials|optics|machining|lasers|cutting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5668" PostTypeId="2" ParentId="5667" CreationDate="2015-10-05T14:32:27.687" Score="4" Body="&lt;p&gt;Pulsed lasers release their energy in very short pulses which can have incredibly high &lt;strong&gt;peak&lt;/strong&gt; powers.  A run-of-the-mill nanosecond laser will have a peak power in the multi-kilowatt range while a femtosecond laser can easily reach into the megawatt range.  In contrast, CW lasers generally do not reach power levels in excess of a few hundred Watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When these pulsed lasers are focused to a small spot size, the intensity is generally high enough to significantly alter the molecular state of the material through non-thermal processes such as ablation.  These non-thermal processes are useful for creating very small features because they have a smaller &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat-affected_zone&quot; rel=&quot;nofollow&quot;&gt;'heat affected zone'&lt;/a&gt;.  The very high peak powers can also make use of nonlinear absorption in materials which would otherwise be transparent (and therefore not able to be processed by that wavelength).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, the energy from a tightly focused CW laser is deposited over long enough timescales that the material simply heats up and melts.  This can be more useful for processes such as welding.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-10-05T14:32:27.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5669" PostTypeId="1" CreationDate="2015-10-05T15:03:11.460" Score="7" ViewCount="579" Body="&lt;p&gt;I have a question regarding bit error rate (BER) performance over the Rayleigh channel. I observed BER performance over the AWGN channel using Simulink in MATLAB. I used to change the &quot;SNR&quot; parameter in the AWGN block to get the BER.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if I want to edit the block to get the BER but in Rayleigh fading channel (I want for every SNR, to get the BER). There is a Rayleigh fading channel block but I don't know how to use it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to observe the BER performance of MPSK and MQAM over Rayleigh fading channel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure below shows my blocks for observing BER in fading channel, the parameters in the fading block and my result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In AWGN I generated binary bits, then modulated them by using the BPSK. After that I added the noise and then I demodulated it and finally I got the BER for every SNR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I tried to edit it so I added the Multipath Rayleigh fading channel block to observe the BER in fading channel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lj9FM.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lj9FM.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the Doppler spectrum type, there are several types: Jakes, Flat, Gaussian, Rounded, Restricted Jakes, Asymmetric Jakes, Bi-Gaussian, Bell and &quot;specify as dialog parameter&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &quot;Result&quot; figure, the simulated one is the blue one (wrong) and the theoretical one (red) I got by using the &quot;berfading&quot; command, which is the correct one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, my result in not correct. Are my blocks correct? Are there any parameters I should change, especially in the fading block, to get the correct result?&lt;/p&gt;&#xA;" OwnerUserId="3251" LastEditorUserId="368" LastEditDate="2015-10-06T23:03:13.517" LastActivityDate="2015-10-06T23:03:13.517" Title="BER peroformance of MPSK and MQAM in Rayleigh fading channel" Tags="|electrical-engineering|telecommunication|signal|wireless-communication|simulink|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5670" PostTypeId="1" CreationDate="2015-10-05T16:13:59.480" Score="5" ViewCount="993" Body="&lt;p&gt;First condition assuming perfect rolling, no slip, from what I understand is that the rolling resistance coefficient can be calculated as the displacement arm of the normal force the tire &quot;feels&quot;, in which should be equal to the force about its center times the radius of the tire (essentially equating two moments?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am attempting to do a finite element analysis on a tire, solving for rolling resistance for the tire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach is to take the sum of the shear in the contact patch, multiply it by contact area to result in total tractive force (Fundamentals of Vehicle Dynamics). If this force is then subtracted by static friction, shouldn't this result in total rolling resistance force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if my conceptual understanding is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are industrial ways of getting rolling resistance of tires?&lt;/p&gt;&#xA;" OwnerUserId="2883" LastActivityDate="2015-10-05T20:16:47.693" Title="How do I calculate rolling resistance for tire on pavement?" Tags="|mechanical-engineering|automotive-engineering|measurements|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5672" PostTypeId="2" ParentId="5670" CreationDate="2015-10-05T20:16:47.693" Score="3" Body="&lt;p&gt;Find the moment of inertia of the tire, either experimentally by spinning it with a motor or with a &lt;a href=&quot;http://www.mathworks.com/company/newsletters/articles/improving-mass-moment-of-inertia-measurements.html?refresh=true&quot; rel=&quot;nofollow&quot;&gt;torsional pendulum&lt;/a&gt;, or by measuring it with CAD models (warning: your properties need to be dead on for this - watch out for belting skewing your numbers). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know what it takes to get the tire to accelerate, try &lt;a href=&quot;http://www.wired.com/2014/07/a-rolling-object-accelerating-down-an-incline/&quot; rel=&quot;nofollow&quot;&gt;putting it on a ramp&lt;/a&gt; and timing the results. Gravity should accelerate the object, but rolling resistance would dissipate energy. At low speeds wind resistance should be negligible. There should then be a difference between finishing times of the ideal calculation (which you could solve with energy) and the real trial, where the difference would be due to the rolling resistance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your comment on static friction, I would just point out that static friction is what generates the torque that rotates the tire, it should not contribute any losses to the rotation (remember that the normal force is $g\cos{\theta}$). Remember that work (energy) is &lt;a href=&quot;https://en.wikipedia.org/wiki/Work_(physics)#Work_and_energy&quot; rel=&quot;nofollow&quot;&gt;force times displacement&lt;/a&gt;, and so, if the tire does not slip, the static friction force doesn't act on any distance, so no energy is dissipated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rolling resistance causes energy loss because of the energy it takes to deform the tire and/or road. The bulge in the bottom of a tire is always at the bottom because, as the wheel turns, the tire is constantly flexing into and then out of that shape. Natural damping in the tire means that it takes more energy to bend it into shape than you get out by relaxing the tire. This is where the energy is lost for rolling resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more about rolling resistance at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rolling_resistance&quot; rel=&quot;nofollow&quot;&gt;Wikipedia entry&lt;/a&gt;, where it states &lt;a href=&quot;https://en.wikipedia.org/wiki/SAE_J1269&quot; rel=&quot;nofollow&quot;&gt;SAE J1269&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/SAE_J2452&quot; rel=&quot;nofollow&quot;&gt;SAE J2452&lt;/a&gt; are SAE-defined test procedures to measure rolling resistances of tires.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-10-05T20:16:47.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5673" PostTypeId="2" ParentId="5638" CreationDate="2015-10-05T20:40:13.840" Score="2" Body="&lt;p&gt;Adding to @grfrazee's answer, there is an intermediate method, less rigorous than Rourke but more precise than simply dividing by the perimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the trapezoidal method, where one divides the load being applied to each supporting member. The only relevant information is the boundary conditions of the plate: whether it is free, pinned or fixed. Usually, slabs are said to fix each other, so if you have a beam supporting two slabs, one to either side, then both slabs are considered fixed on that face (the only exception is if one of the slabs is a cantilever, in which case the cantilevered slab is considered fixed but the other is pinned, since a cantilevered slab won't resist rotations due to loading on the other slab). On the other hand, if a beam is supporting only one slab, then that slab-face is considered pinned. If the edge of a slab is not supported by a beam, then it is considered free, of course. This method doesn't work for slabs supported directly by columns. It also assumes the slab is under uniform loads which cover the entire slab and is not strictly valid for other loading conditions (concentrated loads or variable loading within a slab).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if two adjacent and perpendicular faces of a slab are equally supported (both are pinned or both are fixed), then one draws a 45° line starting from their shared vertex. If, however, one face is pinned and the other is fixed, then the line must be 60° (from the fixed face). These lines describe the load division to each supporting member. (After a quick googling, I've discovered some people simply consider 45° in all cases. Do as you wish, but I think giving more load to the stiffer member makes sense. That being said, I can't prove it, nor can I demonstrate why 60° is a good angle, I simply learned this in college).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extending these lines until they fully divide the slab area, you now know the fraction of load that will go to each face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be easily understood with an example such as this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T49Kk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T49Kk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have three slabs A, B and C. Slab A is supported by beams B2.1, B3.1 and B4.1 and has a free end (the dotted line). Slab B is supported by beams B2.2, B3.2, B4.1 and B5.1. Slab C is supported by beams B1, B2.2, B4.2 and B5.2. In this example, lets just look at the load distribution of Slabs A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slab A is free on one face, pinned on faces B2.1 and B3.1 and fixed on B4.1. This is because the stiffness of slab B on the other side of B4.1 impedes (or at least limits) the rotation around face B4.1, while the torsional rigidity of beams B2.1 and B3.1 is considered insufficient to influence the slab's rotation. Given the rules described above, lines are drawn from the vertices between the faces. In this case, since B4.1 is fixed and B2.1 and B3.1 are pinned, the lines must be at 60° from B4.1. The remaining load is distributed between the pinned faces. The load distribution is therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5v6uX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5v6uX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same reasons as described above, Slab B is fixed on faces B2.2 and B4.1 (due to the slabs on the other side of the beams) and pinned on faces B3.2 and B5.1. Since both B2.2 and B4.1 are fixed, the line between them is at 45°. Since both B3.2 and B5.1 are pinned, the line between them is also 45°. Since B4.1 and B2.2 are fixed and B3.2 and B5.1 are pinned, the lines between them are at 60° from the fixed faces. The load distribution is therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HTWdZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HTWdZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one can calculate how much load goes to each beam by multiplying the &quot;tributary area&quot; of the beam by the applied load. In the case of beam B3, for example, the applied load is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ q_{B_3} = \begin{cases}&#xA;\dfrac{A_3 \cdot q_A}{L_{B_{3.1}}} \text{ along B3.1} \\&#xA;\dfrac{B_3 \cdot q_B}{L_{B_{3.2}}} \text{ along B3.2}&#xA;\end{cases}$$&#xA;where $q_{B_3}$ is the uniform load on beam B3, $A_3$ and $B_3$ are the tributary areas found in the figures and $q_A$ and $q_B$ are the loads applied in each slab (in force-per-area units). The loading of beam B3 is therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OVrKf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OVrKf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of beam B4, the loads would be&#xA;$$ q_{B_4} = \begin{cases}&#xA;\dfrac{A_4 \cdot q_A + B_4 \cdot q_B}{L_{B_{4.1}}} \text{ along B4.1} \\&#xA;\dfrac{C_4 \cdot q_C}{L_{B_{4.1}}} \text{ along B4.2}&#xA;\end{cases}$$&#xA;(C4 is the tributary area of beam B4 in slab C, which hasn't been drawn). Along B4.1 one must obviously add the loads applied by both slabs supported by the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations are somewhat simplified, since they consider the loading along the beam to be uniform, which isn't really the case. The loading should actually be trapezoidal according to the tributary area. To calculate this more precise form, the loading on the beam is found by&#xA;$$q(x) = p \cdot h(x)$$&#xA;where $q(x)$ is the value of the distributed load at point $x$, $p$ is the area-load on the slab and $h(x)$ is the height of the tributary area perpendicular to the beam's axis (displayed as $h_{3.1}(x)$ in the second figure above). The loading therefore becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1QTbP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1QTbP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2015-10-06T16:46:01.053" LastActivityDate="2015-10-06T16:46:01.053" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5674" PostTypeId="2" ParentId="5622" CreationDate="2015-10-05T20:51:45.503" Score="3" Body="&lt;p&gt;@grfrazee says it's a dead load, @NamSandStorm and @hazzey say it's a live load, and now I'm going to stop half way between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on what your tank is doing:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it a frequently flushed tank, which empties and fills frequently? Then that's a live load.&lt;/li&gt;&#xA;&lt;li&gt;Is it a storage tank, whose fluid levels rarely ever change? Then that's a dead load.&lt;/li&gt;&#xA;&lt;li&gt;Is it something in between, where, for example, there's a minimal level that will never (or rarely) be crossed, but the level itself may vary above that level? Then consider the minimal level a dead load and the variance a live load. The inverse  case is also possible: if you have the storage tank above but you want to take into consideration the possible effects of the rare moments where it's flushed? Then consider its standard level as a dead load and apply a negative load as a live load.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-10-05T20:51:45.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5677" PostTypeId="2" ParentId="5622" CreationDate="2015-10-06T11:14:32.437" Score="0" Body="&lt;p&gt;From a practical perspective, when in doubt (and if it does not lead to an overly conservative design) look at both cases, and after analysis choose the worst case scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas the codes are the guide (and there appears to be some contradiction between US and European codes from the information provided), in any case the final result must be a structure that can serve it's design purpose.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastActivityDate="2015-10-06T11:14:32.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5678" PostTypeId="2" ParentId="5629" CreationDate="2015-10-06T12:59:01.683" Score="5" Body="&lt;p&gt;When it comes to concrete and steel, there were few standards for materials and design prior to about 1920, and you may not find any prior to 1905, at least that is my experience.  Reinforced concrete technology was largely proprietary prior to about 1910 - as such the first multi-story reinforced concrete building in the UK (Weaver's Warehouse 1897 to 1906 I believe) was built under license.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Civil Engineer's Handbook in the UK can be a good guide for what was going on in the early days (UK and Commonwealth), and the Institution of Civil Engineers library has older copies of this.  For example, the water to cement ratio was found in the 1923 version of the Handbook in the UK for the first time (known as the Abrams Ratio and was expressed as volume of cement to volume of water rather than today's w/c m/m ratio).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design of the Weaver's Warehouse indicated that the use of reinforcement at that time resulted in loads being supported by an 'arching' effect rather than the 'ductility' design approach of present.  This means that older structures could be at greater risk of brittle failure modes.  I suspect that this is because there was little knowledge of embedment length, etc prior to the 1930's, and steel was often 'fixed' at the ends.  The proprietary construction process of that project required that the slabs/beams be load tested as the structure was built.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From memory the first codes for steel and cement (eg BS 12) in the UK were released in the early 1900's, around about 1906 or so, I don't have the exact dates. Cements were far more coarsely ground in those days, and chemically very much different.  There was little C3A and gypsum added, for example, prior to about 1920, as it was introduced at a later date when manufacturers found it could reduce the energy and cost of production.  Furthermore, because little was known about the bonding of reinforcing steel and embedment length, it was often smooth, not deformed, prior to 1920.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other side of the Atlantic, the American Concrete Institute was founded around 1904, and the first building regulations were adopted around 1910 with a number of standards in place around 1912.  The Portland Cement Assocation, with Duff Abrams later as its first Chief Engineer, was founded around 1917.  Abrams was to originate the ACI concrete mix design method, the concept of water to cement ratio and reinforcing steel bonding capacity in the US.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have found in the US and Europe on structures dating back to about 1910 (but mostly during the 1920's) is that you can find just about anything, and testing is strongly advised. It is not unusual to find concrete typically designed and produced on a loose volume basis (ie by scoop or shovel 1:2:4 and no water to cement ratio).  One project we looked at, an older bridge foundation and deck, had concrete mixture composition that varied from one end of the bridge to the other: the aggregates were a rounded river gravel on one end and the other end had a crushed granite, and they blended together as they reached the centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another significant issue is the foundation design and construction.  You can find anything.  I have seen structures carrying load successfully for 50 years or more but with timber piles that were located only half under the cap (ie extending beyond the edges).  So again there is no substitute for investigation and/or testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For concrete there is an international standard BS EN 13791 that includes statistical strength assessment for structures where no history of the structure is known. This can be of use for comparing the concrete to modern structural codes.  But if there is to be a change of loading condition then there is no substitute for identifying cover depths and steel/section geometry using non-destructive tests.  And of course reinforcing steel corrosion can be a significant factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For structural steel a critical issue is often the bolts/welds as well as the sections. The assessment is very much of an art here, and may require radiographic/ultrasonic/torque testing if the situation demands it. Very old steel structures are often riveted which introduces further complexity.  As far as I am aware, there aren't any historical guides on bolts and rivets.  Previous posts have given some references that could be of use for the sections.  If in doubt about a steel structure, you can obtain coupons from the sections and have them tested for tensile capacity and ductility.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastActivityDate="2015-10-06T12:59:01.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5682" PostTypeId="1" CreationDate="2015-10-07T12:34:53.900" Score="0" ViewCount="242" Body="&lt;p&gt;Recently I was experimenting with a metal piece, and I came across a challenge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small piece of thin copper metal (5cm x 3cm x 0.2 cm) has to be pushed forward and to be rotated. The problem is how to get a force behind the metal piece and have a force to rotate it. The metal piece must go from left side to the right side of the picture2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial idea was to bind the metal piece with a metal bar and to rotate it so that metal will rotate accordingly. However the bar is too thick. I need to use  a thin, ideally &amp;lt;5mm, string, metal piece or anything else to control (to move forward and to rotate) the metal piece (see picture below). Using a 0.2 cm copper wire  after tying a knot with the metal piece I failed to be able to rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you give me some hints what kind of thin string or bar or anything else I may use to accomplish a forward movement through the gap of 0.5 cm between object 1 and 2 (see picture 2) and the rotation of the metal piece afterwards? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture 1:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hKJs6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hKJs6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture 2:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rwico.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rwico.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3095" LastEditorUserId="3095" LastEditDate="2015-10-09T23:07:59.883" LastActivityDate="2015-10-10T12:38:22.380" Title="How to move a thin metal piece with a tiny string or anything else?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="12" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5684" PostTypeId="2" ParentId="5682" CreationDate="2015-10-07T17:46:14.350" Score="2" Body="&lt;p&gt;Unless I am missing some major constraint, you have answered your own question just by stating it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You have a gap that is 0.5cm = 5mm. (This is the upper limit.)&lt;/li&gt;&#xA;&lt;li&gt;You tried a wire that was 0.2mm. (This didn't work.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You therefore need something between those two sizes. Either make something that size or get some wire that is that size. Literally the first chart that a Google search brought me was &lt;a href=&quot;http://www.wiringfordcc.com/wire_sizes.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. It shows many different sizes of wire within your range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, all of this seems very simple, which leads me once again to think that there are constraints that are missing from your question.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-10-07T17:46:14.350" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5685" PostTypeId="1" AcceptedAnswerId="5698" CreationDate="2015-10-07T22:32:50.567" Score="7" ViewCount="8427" Body="&lt;p&gt;The Reynolds Number is defined as $Re = \frac{\rho v L}{\mu}$, where $\rho$ is the density, $v$ is the fluid velocity, $L$ is the characteristic length, and $\mu$ is the dynamic viscosity.  By definition, inviscid flow implies that $\mu=0$.  By this formula, this would make the reynolds number infinite and thus turbulent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to have laminar inviscid flow?  Must inviscid flow always (necessarily) be turbulent?&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2015-10-11T06:41:39.833" Title="Is inviscid flow necessarily turbulent?" Tags="|fluid-mechanics|turbulence|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5686" PostTypeId="1" AcceptedAnswerId="5688" CreationDate="2015-10-08T00:25:27.780" Score="5" ViewCount="168" Body="&lt;p&gt;I am going to machine something out of a small 40mm long x 8mm diam tungsten rod.  The majority will be hollowed out using a larger bit, but at the very tip, I need a small hole, the smaller the better between .05mm and .6mm.  I understand an L/D for that small hole should be less than 60%?  Did I get that right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be done with a standard micro carbide bit?  I guess the length of the hole can be adjusted to make this work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of the nozzle will be to pour out accurately micro amounts of melted steel.  My parts include the ability to push steel wire at the rate of 1/27 mm^3 / step (of a stepper motor), so as you can see a tiny amount!  But if not possible, I can bump that number up some.&lt;/p&gt;&#xA;" OwnerUserId="3235" LastEditorDisplayName="user16" LastEditDate="2015-10-08T00:54:18.857" LastActivityDate="2015-10-08T04:11:00.143" Title="What's the smallest pin-like hole I can machine into a piece of tungsten using a metal lathe and common carbide tools?" Tags="|machining|microfluidics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5687" PostTypeId="2" ParentId="5685" CreationDate="2015-10-08T00:38:33.580" Score="1" Body="&lt;p&gt;&lt;strong&gt;The simple answer&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reynolds number can be more simply defined as the ratio of Inertial forces to Viscous forces. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;viscous &gt;&gt; inertial = Laminar&lt;/li&gt;&#xA;&lt;li&gt;viscous &amp;lt;&amp;lt; inertial = Turbulent&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is why when speed/flow rate increases fluids go towards/become turbulent as you're increasing the inertia of the fluid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this logic if you had inviscid (no viscous forces) flow then the inertial forces would have to dominate which should lead to Turbulent flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The fringe case&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're asking if it's possible as a 'fringe case' well... then we're getting to the boundary between engineering and science. Inviscid flow in itself is just an assumption to simplify the NS equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK In real life there's nothing that is truly inviscid, just things that are &lt;em&gt;almost&lt;/em&gt; inviscid. Thus as soon as you have some incredibly-close-to-zero viscosity value, then you &lt;em&gt;should&lt;/em&gt; have some equally incredibly-close-to-zero flow value that allows laminar flow. &lt;/p&gt;&#xA;" OwnerUserId="2768" LastEditorUserId="2768" LastEditDate="2015-10-08T00:49:32.413" LastActivityDate="2015-10-08T00:49:32.413" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5688" PostTypeId="2" ParentId="5686" CreationDate="2015-10-08T01:35:46.263" Score="3" Body="&lt;p&gt;&lt;strong&gt;Reposting my comment for clarity:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Before we get into making this: have you confirmed that a wall thickness of ~0.3mm around the hole is going to be thick/strong enough to resist the pressure required to push the molten steel through your nozzle/hole in the first place? That's very thin.... I think you might have to look at a higher ratio than 60%&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to mention the difficulty in actually performing a drill operation with that much precision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just saw your other &lt;a href=&quot;https://engineering.stackexchange.com/questions/5640/how-small-can-you-get-a-stream-of-molten-steel-to-be-at-low-pressure&quot; title=&quot;question&quot;&gt;question&lt;/a&gt;.  As per the recommendation on the accepted answer a 3mm deep hole is probably much more reasonable than the 0.3mm one that you're proposing (due to the pressure forces that the steel will be at blowing out the tip of your nozzle). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would give a L/D ratio of 5 (500%) which off intuition sounds hard to machine. I can't comment on whether it would work with a bit sticking 3mm out of the chuck but if it doesn't then you could always drill the hole in 1mm (or smaller) segments. What I mean by this is extend the bit out of the chuck by 1mm and drill into the piece by that amount. Then change the bit to extend by 2mm out of the chuck and drill it again. Repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the deeper drills the bit would be supported by the surrounding tungsten, and on the earlier passes the bit would extend out only a short amount and thus be unlikely to break due to lateral load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Realistic&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be best to contact the machine shop where you plan to get this made. If you're making it yourself I would have a play around with a small amount of tungsten bar and see what's possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit in response to comments&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had this sort of thing in my mind:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/b2f07.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b2f07.jpg&quot; alt=&quot;Diagram of nozzle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However now I realise that it would be more sensible/easier to just leave the drill profile in the nozzle:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/EQEjy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EQEjy.jpg&quot; alt=&quot;Realistic diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which negates my concerns. &lt;/p&gt;&#xA;" OwnerUserId="2768" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-10-08T04:11:00.143" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5689" PostTypeId="1" AcceptedAnswerId="5695" CreationDate="2015-10-08T03:46:57.153" Score="4" ViewCount="554" Body="&lt;p&gt;They've printed jet engines with it that work after polishing.  So how is the accuracy measured and how accurate is the DMLS process?  I can't find the required info anywhere.  &lt;/p&gt;&#xA;" OwnerUserId="3235" LastEditorUserId="272" LastEditDate="2015-10-08T14:15:28.650" LastActivityDate="2015-10-08T15:29:09.173" Title="What is the accuracy of direct metal laser sintering (DMLS)?" Tags="|machining|manufacturing-engineering|lasers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5690" PostTypeId="1" CreationDate="2015-10-08T08:11:30.290" Score="3" ViewCount="542" Body="&lt;p&gt;Biogas and sewage (typically around 60% CH&lt;sub&gt;4&lt;/sub&gt;, rest CO&lt;sub&gt;2&lt;/sub&gt;) gas is typically treated in activated carbon (ac) filters to remove H&lt;sub&gt;2&lt;/sub&gt;S and siloxanes. The ac filters need a small amount of oxygen (&gt;0,4%) to work. Additionally, the moisture content needs to be controlled. This is where the confusion starts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is logical that the gas may not be saturated with water, as condensation in the filter would cause ac particles to clog and would lose filter surface. But I've seen different statements about the optimum moisture content: One source claims &lt;a href=&quot;http://www.tll.de/imperia/ft_biog/tbi50314.pdf&quot; rel=&quot;nofollow&quot;&gt;50% relative moisture is optimal&lt;/a&gt;, others go for drying the gas as much as economically sensible (10-30% rel. moisture). So far I've found no reason given for why water is even beneficial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the role of water in gas cleaning by ac, and how much is actually optimal?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="368" LastEditDate="2015-11-20T00:28:52.903" LastActivityDate="2015-11-20T00:28:52.903" Title="Gas cleaning in activated carbon filters - How much moisture is needed and why?" Tags="|process-engineering|environmental-engineering|biogas|fluid-filtration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5691" PostTypeId="1" CreationDate="2015-10-08T10:51:55.570" Score="3" ViewCount="230" Body="&lt;p&gt;My question is what are the ideas behind the complicated historical water clocks. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Water_clock&quot; rel=&quot;nofollow&quot;&gt;wikipedia&lt;/a&gt;, modern versions of the historical water clock rely on the principle of the syphon. So I ask how to translate it to the mechanical language. In particular, as is evident from my question, I'm very interested in knowing the inner-workings of &lt;a href=&quot;https://en.wikipedia.org/wiki/Ismail_Al-Jazari#Castle_clock&quot; rel=&quot;nofollow&quot;&gt;al-jazary's &quot;castle clock&quot;&lt;/a&gt;, which is an early example of programmable analog computer.&lt;/p&gt;&#xA;" OwnerUserId="3283" LastEditorUserId="1293" LastEditDate="2015-10-09T14:28:29.813" LastActivityDate="2015-10-09T14:28:29.813" Title="How did Al-jazary's &quot;castle clock&quot; work?" Tags="|mechanical-engineering|engineering-history|timing|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5693" PostTypeId="2" ParentId="5691" CreationDate="2015-10-08T12:02:01.257" Score="4" Body="&lt;p&gt;There is an excellent description of the castle clock in &lt;a href=&quot;http://www.muslimheritage.com/article/al-jazari%E2%80%99s-castle-water-clock-analysis-its-components-and-functioning&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; on muslimheritage.com.  It describes the full workings of the clock, including all of the mechanisms for things such as doors opening and closing and for the sun moving across the clock.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most interesting part, for me at least, is how he used water to make a relatively accurate timing device.  As with all timing devices, the key is to find a physical process with a very consistent rate.  Al-jazary used the outflow of water through a small hole to achieve this.  The difficulty of using the outflow of water from a reservoir is that; as the height of the water decreases, the pressure at the bottom changes, and the rate of outflow decreases.  Al-jazary had a rather ingenious solution though.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H8MHW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H8MHWm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image above the main reservoir would be filled with water, and, as the water was released, the main float would sink and transmit the motion to the clock.  In order to deal with the problem of decreasing pressure (and outflow rate), Al-jazary added another smaller reservoir at the output of the main reservoir.  The small reservoir had a floating stopper at the top; as the smaller reservoir drained the stopper would release more water, and as it filled the stopper would release less water (an example of a mechanical feedback loop).  This maintained the pressure in the secondary reservoir at a constant level.  The flow regulator could then be adjusted to give a precise rate of water release.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h6WlQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h6WlQm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-10-08T12:02:01.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5695" PostTypeId="2" ParentId="5689" CreationDate="2015-10-08T15:29:09.173" Score="3" Body="&lt;p&gt;Accuracy will vary a lot depending on what you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the simplest level you are limited by the layer thickness, which is in the $20-100~\mu m$ range. This places a limit on the accuracy of the dimensions that can be achieved in the z direction. Theoretically you can achieve very high accuracy in x and y, but in reality you are limited by the roughness of the the part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the real problem with most of these additive technologies is that the resulting surfaces are very rough, hence why polishing is needed to finish the surface. I don't know for DMLS specifically, but for related methods, such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Selective_laser_melting&quot; rel=&quot;nofollow&quot;&gt;SLM&lt;/a&gt;, $30~\mu m$ roughness would not be bad. Generally, the roughness is highest where the part is angled with respect to the layers as you then can get additional effects from going over steps.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-10-08T15:29:09.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5696" PostTypeId="1" CreationDate="2015-10-08T18:07:14.247" Score="6" ViewCount="61" Body="&lt;p&gt;I'm considering putting a circuit board in a ventilated NEMA 4 enclosure outdoors. I'm a little concerned about humidity and condensation inside the enclosure. Do I need to conformal coat my circuit boards, or take any other precautions?&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2015-10-09T13:43:22.817" Title="If my electronics are outdoors in a ventillated NEMA 4 enclosure, should I conformal coat my circuit boards?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5697" PostTypeId="2" ParentId="5690" CreationDate="2015-10-08T18:24:17.410" Score="2" Body="&lt;p&gt;In terms of bed capacity, dry gas—0% moisture, or as close as you can get—is generally optimal. Section 12.4 of &lt;a href=&quot;https://www.waveland.com/browse.php?t=380&amp;amp;pgtitle=Air%20Pollution%20Control%20by%20Cooper-Alley&quot; rel=&quot;nofollow&quot;&gt;Air Pollution Control (Cooper &amp;amp; Alley, 4th ed.)&lt;/a&gt; characterizes the impact of several factors on the theoretical equilibrium saturation capacity of a fixed-bed adsorption system:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Thickness of the adsorption zone (in which the sorbent is not fully saturated);&lt;/li&gt;&#xA;&lt;li&gt;Deviation from the isotherm (heat wave);&lt;/li&gt;&#xA;&lt;li&gt;Displacement of adsorbate by sorbed water vapor;&lt;/li&gt;&#xA;&lt;li&gt;Sustained loss of capacity due to residual moisture in the bed after regeneration.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It seems that the combined effect on bed capacity of the latter two (moisture-related) factors is about as influential as the combined effect of the former two (non-moisture-related) factors &lt;em&gt;if&lt;/em&gt; the system is run to complete saturation (which would be abnormal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, it really depends on how long much demand is placed on the system between regeneration cycles. The text implies that a typical capacity factor might be around 30%. Notably, the effect of the moisture-related factors is proportionally greater near the bed inlet. (This makes sense when you consider that the first factor, adsorption zone thickness, is only relevant as the leading edge of the adsorption zone approaches the outlet of the bed.) So it seems generally reasonable to expect that high moisture content in the gas might reduce the effective capacity of the bed by as much as half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, there's a trade-off between the cost of increasing the bed size and the cost of dehumidifying the influent gas stream, and as the gas gets drier and drier it's more costly to remove that last bit of moisture (diminishing returns). So while &lt;strong&gt;ideally your filtration system operates at 0% humidity, the optimum level in practical terms depends on the relative cost of dehumidification and increased bed size&lt;/strong&gt;. Selecta may simply have decided 50% is good enough, from a cost perspective, to sell in the market for which that presentation was created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see a method of regeneration indicated in the slides you linked but it seems reasonable to assume pressure swing adsorption (PSA) for biogas enhancement (see, e.g., &lt;a href=&quot;http://www.adnett.org/dl_gas_treatment.pdf&quot; rel=&quot;nofollow&quot;&gt;Anaerobic Digestion of Agro-Industrial Wastes, Schomaker et al&lt;/a&gt;). Other regeneration methods, if used, might have different implications for residual moisture (but also cost). Something to think about.&lt;/p&gt;&#xA;" OwnerUserId="368" LastActivityDate="2015-10-08T18:24:17.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5698" PostTypeId="2" ParentId="5685" CreationDate="2015-10-08T20:08:02.037" Score="4" Body="&lt;p&gt;No, inviscid flows are not necessarily turbulent. If there is nothing to &quot;trip&quot; the turbulence, then the flow will remain laminar. Features which could trip the turbulence include vibration, small temperature fluctuations, any geometric imperfections, velocity field imperfections, and other similar things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Potential_flow&quot; rel=&quot;nofollow&quot;&gt;potential flow&lt;/a&gt; is a type of inviscid flow. Potential flow solutions are laminar solutions to the Navier-Stokes equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As another example, if care is taken to avoid vibration and other flow imperfections, it seems you can generate laminar pipe flows at arbitrarily high Reynolds numbers. &lt;a href=&quot;http://www.annualreviews.org/doi/abs/10.1146/annurev-fluid-122109-160652&quot; rel=&quot;nofollow&quot;&gt;Laminar pipe flows have been obtained at Reynolds numbers of about 100,000 under these conditions&lt;/a&gt;, which is far higher than the typical transition Reynolds number of about 2,000. From what I understand, there is no indication that 100,000 is any hard limit; you probably could get higher with better experimental setups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How I think about it is this: Viscosity helps damp out flow imperfections, allowing laminar flows to occur with more imperfections. If you have a truly inviscid flow, it needs to be perfect to not trigger instabilities which lead to turbulence. If you were able to obtain an inviscid flow, you should expect turbulence due to the imperfections inherent in reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer a question you put in a comment in the other answer, yes, I do believe using a turbulence model for inviscid flow is prudent. For RANS, the Reynolds stress still will exist if the flow is inviscid, and for LES the same is true for the residual stress.&lt;/p&gt;&#xA;" OwnerUserId="49" LastEditorUserId="49" LastEditDate="2015-10-09T00:16:22.413" LastActivityDate="2015-10-09T00:16:22.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5699" PostTypeId="2" ParentId="5576" CreationDate="2015-10-08T20:35:02.420" Score="3" Body="&lt;p&gt;I found this formula &lt;a href=&quot;http://www.eng-tips.com/viewthread.cfm?qid=91966&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{crit}=\frac{2\,E\,t}{D}\left(\frac1{(n^2-1)\left(1+\left(\frac{2\,n\,L}{\pi\,D}\right)^2\right)^2}+\frac{t^2}{3(1-\nu^2)D^2}\left(n^2-1+\frac{2\,n^2-1-\nu}{\left(\frac{2\,n\,L}{\pi\,D}\right)^2-1}\right)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you should set $p_{crit}$ to about 3 atm to be safe, then find $E$ and $\nu$ for your aluminum, plug those in along with your diameter, and length. Then plug in a few integer values of $n&amp;gt;1$ to find a thickness $t$ that is strong enough for all values of $n$.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-10-08T20:35:02.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5700" PostTypeId="1" CreationDate="2015-10-08T22:17:52.907" Score="1" ViewCount="182" Body="&lt;p&gt;I am supporting 25,448 pounds over a series of cables, spaced 20' apart, over a span of approximately 424', left to right.  The load is mostly in an area of the middle 10' of the span.  I am using synthetic rope/cable with 68,500# capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to design a plate to secure the cable on both sides.  The plate will be attached to a curbing that is concrete roughly 24&quot; thick.  How can I determine how thick my plate should be and what anchors I should use? I am planning to use 24&quot; x 24&quot; plates with an anchor in each corner.&lt;/p&gt;&#xA;" OwnerUserId="3290" LastEditorUserId="368" LastEditDate="2015-10-08T22:36:52.727" LastActivityDate="2016-02-07T01:28:46.830" Title="How do I determine plate thickness and anchor size for a 10-ton span supported by synthetic rope?" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5701" PostTypeId="1" AcceptedAnswerId="5702" CreationDate="2015-10-08T23:38:25.087" Score="6" ViewCount="847" Body="&lt;p&gt;I have a 10 mm depth x 5 cm length 99% Tungsten rod.  How can I machine the described hole so that instead of having a bevel from the tip of the main drill-hole bit, it has a perfect 90 degree corner all around as if the tip of the drill bit were like a flat router bit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is not feasible, I can think about a design that doesn't use this larger hole.  For this design, the inside of the large hole will be filled with an AlO3 ceramic tube that snuggly surrounds 1mm D of round stainless steel wire that flows through the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wGS3p.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wGS3p.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3235" LastEditorUserId="368" LastEditDate="2015-10-09T15:46:18.587" LastActivityDate="2016-01-30T00:25:40.773" Title="How can I finish a round hole so it has 90 degree corners (flat bottom)?" Tags="|mechanical-engineering|machining|metals|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5702" PostTypeId="2" ParentId="5701" CreationDate="2015-10-08T23:47:34.737" Score="7" Body="&lt;p&gt;You are likely referring to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Counterbore&quot; rel=&quot;nofollow&quot;&gt;counterbore hole&lt;/a&gt;, which may be drilled using a bit of the same name. Alternately, if you can't find a bit of the appropriate dimensions, it may be possible to remove the material using an end-mill tool in a CNC mill of smaller diameter than the hole (say 3 mm to the hole's 6 mm), and having the CNC machine mill a series of circles until the hole is the appropriate shape and the appropriate material is removed. From what I understand, using an end mill of the same diameter as the hole may cause it to get stuck and fracture in the titanium, which would result in a wasted component and a wasted tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Ethan48 helpfully added that a CNC machine is not required, and a conventional milling machine may be used in conjunction with a rotary table. Align the axis of the hole with the axis of the table, and have the axis of the bit away from the table axis the appropriate amount. The table would then have the same effect as the CNC machine by rotating the part through the small circles.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-10-09T16:43:14.060" LastActivityDate="2015-10-09T16:43:14.060" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5704" PostTypeId="2" ParentId="5682" CreationDate="2015-10-09T09:46:02.487" Score="0" Body="&lt;p&gt;Something like that should do it, probably better to be made of metal (brass?) given the thin walls. Just have to make sure it won't damage the part it is pushing (may need to replace the chamfer by a filet to round it off and specify a &quot;smooth&quot; finish for the surface in contact with the pushed part):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rAODt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rAODt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4.5mm diameter should just about fit in the 5mm gap and the 2.5mm slot should accommodate the 2mm thick plate. I have added a chamfer to help guide the metal plate in the pusher. You will need to get this machined by a machinist on a lathe or similar.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2015-10-10T12:38:22.380" LastActivityDate="2015-10-10T12:38:22.380" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5705" PostTypeId="2" ParentId="5696" CreationDate="2015-10-09T13:43:22.817" Score="4" Body="&lt;p&gt;Requirements depend on environment,but, yes, conformal coating is, at a minimum, a really really good idea. Apart from condensation and misc stuff that manages to get in, ants can wreak havoc, depending on ventillation holes other denizens may also do their bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a 'quick and easy' coating material that is a lot better than nothing at all then clear polyurethane &quot;varnish:&quot; spray is useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many commercial products offered for this purpose. Dow Corning make some good ones but there are many others as well. (My only association with Dow Corning is as a satisfied customer). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a useful &lt;a href=&quot;http://www.dowcorning.com/content/etronics/etronicscoat/etronics_cc_tutorial.asp&quot; rel=&quot;nofollow&quot;&gt;tutorial page&lt;/a&gt; from Dow Corning. While it obviously steers you towards their products it also provides good general advice. &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2015-10-09T13:43:22.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5706" PostTypeId="1" AcceptedAnswerId="5707" CreationDate="2015-10-09T14:10:41.717" Score="5" ViewCount="173" Body="&lt;p&gt;A question asked about the &lt;strong&gt;necessary curing pressures for a composite resin in the autoclave&lt;/strong&gt;. Looking through the corresponding &lt;a href=&quot;http://hexcel.com/Resources/DataSheets/Prepreg-Data-Sheets/F155_us.pdf&quot; rel=&quot;nofollow&quot;&gt;datasheet&lt;/a&gt; (HexPly F155), I'm confused:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Cure Procedure&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Apply vacuum of 22 inches (74 kPa) Hg minimum.&lt;/li&gt;&#xA;  &lt;li&gt;Apply 85 + 15 – 0 psig (586 + 103 – 0 kPa) pressure for laminates.&lt;/li&gt;&#xA;  &lt;li&gt;Apply 45 + 15 – 0 psig (310 + 103 – 0 kPa) pressure for sandwich.&lt;/li&gt;&#xA;  &lt;li&gt;Vent vacuum bag to atmosphere when pressure reaches 20 psi (138 kPa).&lt;/li&gt;&#xA;  &lt;li&gt;During cool-down when the part temperature falls below 140°F (60°C), pressure can be released and the test panel removed from the&#xA;  autoclave and debagged.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first step is clear, but not the next steps, for instance &lt;code&gt;85 + 15 – 0 psig (586 + 103 – 0 kPa)&lt;/code&gt; is causing me a headache and I can't make sense of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd guess that one is the external positive pressure and the other is negative vacuum pressure under the sheet, but I couldn't for instance have a 103 kPa vacuum.&lt;/p&gt;&#xA;" OwnerUserId="782" LastEditorUserId="782" LastEditDate="2015-10-09T16:14:49.880" LastActivityDate="2015-10-09T16:14:49.880" Title="How do I interpret the curing pressure of a composite from this data sheet?" Tags="|composite|composite-resin|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5707" PostTypeId="2" ParentId="5706" CreationDate="2015-10-09T14:43:57.473" Score="7" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Cure Procedure&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Apply vacuum of 22 inches (74 kPa) Hg minimum.&lt;/li&gt;&#xA;&lt;li&gt;Apply 85 + 15 – 0 psig (586 + 103 – 0 kPa) pressure for laminates.&lt;/li&gt;&#xA;&lt;li&gt;Apply 45 + 15 – 0 psig (310 + 103 – 0 kPa) pressure for sandwich.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I believe Steps #2 and #3 are providing you a tolerance for the applied pressure.&lt;/p&gt;&#xA;&lt;p&gt;So, for Step #2, it's telling you to apply 85 psig.  You can go +15 psig higher (for a total of 100 psig) but should not go lower than 85 psig.  Normally, this is written as &lt;span class=&quot;math-container&quot;&gt;$85^{+15}_{-0}$&lt;/span&gt; psig or 85 +15/-0 psig.&lt;/p&gt;&#xA;&lt;p&gt;For Step #3, apply 45 psig, with an upward tolerance of +15 psig (total 60 psig) but go no lower than 45 psig total pressure.  Normally, this is written as &lt;span class=&quot;math-container&quot;&gt;$45^{+15}_{-0}$&lt;/span&gt; psig or 45 +15/-0 psig.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-10-09T15:25:41.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5708" PostTypeId="1" CreationDate="2015-10-09T14:58:17.613" Score="3" ViewCount="138" Body="&lt;p&gt;I have some transfer functions and I was wondering if it would be possible to approximate this to an electric circuit equivalent. For example, given the Nyquist plot below, is it acceptable to hypothesise that a circuit with RC components can represent this Nyquist plot since we know that the Nyquist plot of a low pass (RC) filter is circular such as the one below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/70K1d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/70K1d.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="610" LastEditorUserId="90" LastEditDate="2015-10-13T12:47:17.787" LastActivityDate="2015-10-13T12:47:17.787" Title="Is there a way to obtain a circuit of some sort given a Nyquist plot?" Tags="|electrical-engineering|control-engineering|control-theory|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5709" PostTypeId="1" AcceptedAnswerId="5752" CreationDate="2015-10-09T18:12:35.390" Score="0" ViewCount="87" Body="&lt;p&gt;I'm trying to figure out the theoretical calculations for a Stirling Engine system so I can determine optimal piston size, optimal piston displacement and number of pistons. I want to calculate which configuration has the most amount to torque to turn something else, and the highest amount of RPM's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have managed to achieve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Internal energy = Thermal difference + Work&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ΔU = Q + W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Work = Rotational Kinetic Energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W = \dfrac{1}{2}Iw^2$$&#xA;(3) Assuming that Inertia is standard wheel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \dfrac{1}{2}Mr^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Equation I have so far&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1}{4}Mr^2w^2 = ΔU - Q$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am unsure on how I can relate the equation to the pistons. According to HyperPhysics &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Internal energy is defined as the energy associated with the random,&#xA;  disordered motion of molecules.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That sounds like pressure to me, and I know that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pressure = Force / Area&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I think I need a better way to express the physics of pistons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Q component of internal energy is also difficult for me to find, because a heat source (like a candle flame), will be creating a heat differential on the piston. This heat differential in Joules, will be a little difficult to find as well.&lt;/p&gt;&#xA;" OwnerUserId="3305" LastEditorUserId="1832" LastEditDate="2015-10-09T22:22:35.810" LastActivityDate="2015-10-12T22:39:24.120" Title="Find optimal configuration for Stirling Engine" Tags="|automotive-engineering|experimental-physics|pistons|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5712" PostTypeId="1" CreationDate="2015-10-09T21:22:51.640" Score="5" ViewCount="507" Body="&lt;p&gt;A product can be UL recognized or UL listed. I understand that there is some difference between these two things, but I have never had a full and clear explanation of the difference. What is, in fact, the difference?&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2015-10-10T04:19:12.893" Title="What is the difference between being UL recognized and being UL listed?" Tags="|safety|ul|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5713" PostTypeId="1" CreationDate="2015-10-09T22:26:00.267" Score="15" ViewCount="57367" Body="&lt;p&gt;I understand that the reynolds number is given by the expression $Re=\frac{\rho v L}{\mu}$, where $\rho$ is the density, $v$ is the fluid velocity and $\mu$ is the dynamic viscosity.  For any given fluid dynamics problem, $\rho$, $v$, and $\mu$ are trivially given.  But what exactly is the characteristic length $L$?  How exactly do I calculate it?  What can I use from a given problem to determine the characteristic length automatically?&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2015-10-15T11:58:59.830" Title="How to determine the characteristic length in reynolds number calculations in general?" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5714" PostTypeId="2" ParentId="5713" CreationDate="2015-10-09T23:05:11.323" Score="2" Body="&lt;p&gt;There are three main ways to determine which groups of terms (more general than just length or time scales) are relevant. The first is by math, which could involve solving a problem or an analogous or appropriate problem analytically and seeing which terms appear and making selections which simplify things as appropriate (more on this below). The second approach is by trial and error, more or less. The third is by precedent, usually when someone else in the past has already done some sort of the previously mentioned analysis in this problem or related ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of ways to do theoretical analysis, but one useful one in engineering is non-dimensionalizing governing equations. Sometimes, the characteristic length is obvious, as is the case in a pipe flow. But other times, &lt;em&gt;there are no obvious characteristic lengths&lt;/em&gt;, as is the case in free shear flows, or a boundary layer. In these cases, you can make the characteristic length a free variable, and &lt;em&gt;choose one which simplifies the problem&lt;/em&gt;. &lt;a href=&quot;http://www.math.illinois.edu/~rdeville/teaching/558/nondimensionalization.pdf&quot; rel=&quot;nofollow&quot;&gt;Here are some good notes on non-dimensionalization&lt;/a&gt;, which have the following suggestions for finding characteristic time and length scales:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;(always) Make as many nondimensional constants equal to one as possible.&lt;/li&gt;&#xA;  &lt;li&gt;(usually) Make the constants that appear in the initial or boundary conditions equal to one.&lt;/li&gt;&#xA;  &lt;li&gt;(usually) If there is a nondimensional constant that, if we were to set it equal to zero, would simplify the problem significantly, allow it to remain free and then see when we can make it small.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The other main approach is to solve a problem entirely and see which groups of terms appear. Generally the relevant length is obvious if you are grabbing the term from this type of theoretical analysis, though this sort of analysis is often easier said than done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do you figure out a good length if you don't have a theoretical analysis to go off of? Often, it doesn't matter too much which length you pick. Some people seem to think this is confusing, because they were taught that turbulence transition occurs at $Re$ of 2300 (for a pipe), or 500,000 (for a flat plate). Recognize that in the pipe case, it doesn't matter if you pick the diameter or radius. That just scales the critical Reynolds number by a factor of two. &lt;em&gt;What does matter is making sure that any criteria you use are consistent with the definition of the Reynolds number you use, and the problem you are studying&lt;/em&gt;. It's tradition that dictates that we use the diameter for pipe flows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to be general, analysis or experimentation could suggest another number, say the Biot number, which also has a &quot;characteristic length&quot; in it. The procedures in this case are identical to that already mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes you can make a heuristic analysis to determine the relevant length. In the Biot number example, this characteristic length is usually given as the volume of an object divided by its surface area, because this makes sense for heat transfer problems. (Larger volume = slower heat transfer to center and larger surface area = faster heat transfer to center.) But I suppose it's possible to derive this from certain approximations. You can make a similar argument justifying the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydraulic_diameter&quot; rel=&quot;nofollow&quot;&gt;hydraulic diameter&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="49" LastEditorUserId="49" LastEditDate="2015-10-11T03:09:23.970" LastActivityDate="2015-10-11T03:09:23.970" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5715" PostTypeId="1" AcceptedAnswerId="5873" CreationDate="2015-10-10T00:39:46.823" Score="0" ViewCount="191" Body="&lt;p&gt;I have a 15 pound weight that I use to crush 8.4 oz red bull cans. A metal barbell with a handle and 2 ends. Not 100% flat ends, but close to flat. About 12 inches from end to end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hold the barbell by one end and lower it to crush a standing redbull can. sometimes the can crushes completely, other times it gets crooked. There doesn't seem to be much predictability in regards the various methods of striking it. Only that multiple small strikes allows it to be mended and crushed by hand also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is - how can I predictable crush the can with the least effort? I think my strikes are consistent but the cans are in need of some pre-crushing to create fault lines were it will crash like a controlled demolition, or at least that is my current theory.&lt;/p&gt;&#xA;" OwnerUserId="3308" LastEditorUserId="33" LastEditDate="2015-10-10T12:57:22.453" LastActivityDate="2015-10-22T00:04:54.797" Title="Crushing drink cans" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5716" PostTypeId="2" ParentId="5566" CreationDate="2015-10-10T00:51:01.447" Score="4" Body="&lt;p&gt;Your question is specific to the pressure change with depth in the earth. When that earth consists of soils, the lateral and vertical pressures can be calculated in a number of different ways, depending upon whether your soil is sand, or clay and whether there is ground water present.  It can be quite a complex matter, as the following illustrates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ratio of Horizontal to Vertical Pressure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, in excavations, under backfilled conditions, and under foundations, horizontal pressure and vertical pressure are not considered to be equivalent, and are dependent upon soil-structure interaction, in terms of active, passive and at rest conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Active conditions are where the structure is moving away from the soil (decreasing pressure on the structure).  Passive conditions occur where the structure is moving towards the soil (increasing pressure on the structure) and at rest is where the soil has reached its natural state. You can imagine that all three of these conditions could be observed in a retaining structure, as it may rotate or deform during its lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, most theories will provide coefficients that can be used to calculate the ratio of horizontal to vertical pressure based upon the state of the soil/structure interaction and the properties of the soils.  Some are based upon Poisson's Ratio.  I have even used a temperature-based Poisson's ratio to conduct an elastic analysis of horizontal and vertical pressures in bituminous pavement structures using the Boussinesq equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Effective Stress&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where groundwater is present the pressure is expressed in terms of &lt;a href=&quot;https://en.wikipedia.org/wiki/Effective_stress&quot; rel=&quot;nofollow&quot;&gt;effective stress&lt;/a&gt;, that is the difference between the total stress and the pore water pressure.  This is tricky to understand but has to do with soil buoyancy and other factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider a point of interest 10 m below the ground surface, and uniform sands that have a natural density of 1300 kg/m3, the total stress at the 10 m depth of interest would be 130 kPa. Now consider that the free surface of the groundwater table is at a constant depth of 2 m and assume the density of water to be 1000 kg/m3.  The pore pressure at the depth of 10 m would be based upon an 8 m column of water, so that the pore pressure would be 80 kPa at the depth of interest. Thus the effective stress at 10 m becomes 130 kPa - 80 kPa = 50 kPa.  This is a very simplified expression as there can be lots of other factors, for example fluctuations in water level, so-called 'quicksand' conditions and for retaining structures such things as drainage, among many other considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sands (Cohesionless Soils)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sandy (cohesionless) soils, &lt;a href=&quot;https://en.wikipedia.org/wiki/Rankine_theory&quot; rel=&quot;nofollow&quot;&gt;Rankine Theory&lt;/a&gt; (elasticity) is often applied.  For this the angle of shear resistance of the soil (friction angle) and the angle of inclination of the excavation/retaining structure becomes critical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friction angle of sandy soil is best measured in the laboratory, but it is also considered roughly equivalent to the natural angle of repose of the loose, dry material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clays (Frictionless Soils)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For soils with a cohesive element, such as clays and clay silt combinations, Coulombs (Wedge) Theory (plasticity) is commonly applied. Under this analysis, the soil is imagined as a wedge (free-body) behind the structure, and as the solution is non-determinate, the a variety of potential failure surfaces is tried until the solution converges on a maximum soil pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Soils with Friction and Cohesion&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coluomb's Theory can be used on soils that exhibit both friction and cohesion.  Rankine's method is not suitable for cohesive soils. However, determining the ratio of horizontal to vertical stress may require further analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often the ratio can be established by determining the states of stress as represented by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Mohr%27s_circle&quot; rel=&quot;nofollow&quot;&gt;Mohr's Circle&lt;/a&gt;.  These properties are often measured by &lt;a href=&quot;https://en.wikipedia.org/wiki/Triaxial_shear_test&quot; rel=&quot;nofollow&quot;&gt;Triaxal Shear Tests&lt;/a&gt; where a column of soil is tested in the laboratory under a range of confining pressures.  This can establish the cohesive strength and friction angle of the material and the ratio of horizontal to vertical stress according to depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;General Elastic Theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other theoretical methods that are often used to compute the horizontal and vertical pressures beneath a point of a foundation.  Commonly two methods are applied: 1) Westergaard Theory and 2) Boussinesq Theory.  The ratio of horizontal to vertical pressure at some point beneath the surface is largely a function of the estimated value of &lt;a href=&quot;https://en.wikipedia.org/wiki/Poisson%27s_ratio&quot; rel=&quot;nofollow&quot;&gt;Poisson's Ratio&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Westergaard Theory is elastic theory applied to layered media. This is the case in most conditions typically found in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boussinesq Theory is elastic theory applied to a homogeneous elastic half space.  Whereas this may not be so applicable to all soils, it does find frequent application under simplifying assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Closure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a taste of the more common analysis techniques that are used to evaluate earth pressures in excavations, under foundations and behind retaining structures. There are others, for example Log Spiral Analysis for braced excavations, which is frequently used.  While the theories can be complex, when one considers the great difficulty in establishing the true composition of subsurface soil conditions (ie the existence of layers, layer thicknesses and the variability of the properties of the soils), it becomes clear that pressure/stress analysis requires a great deal of experience and skill.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastActivityDate="2015-10-10T00:51:01.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5717" PostTypeId="2" ParentId="5715" CreationDate="2015-10-10T01:45:32.323" Score="1" Body="&lt;p&gt;Three things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, you're assuming every can is made identically to each other. There may be slight differences in how some cans are made which alters the strength and crushing characteristics of the cans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, to increase the reliability of crushing results you need the same perfect alignment between can and the crushing weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly, the action of the crushing weight, how it strikes each can needs to be exactly the same when crushing cans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the impression your system for crushing the cans is manual. Such a system can introduce all types of random fluctuations which results in different crushing outcomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A well designed and made automated system will have fewer random fluctuations and produce more consistent results.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-10-10T01:45:32.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5718" PostTypeId="2" ParentId="5712" CreationDate="2015-10-10T04:19:12.893" Score="6" Body="&lt;p&gt;Products can be given a UL Listing Mark or the can be classified as a UL Recognized component.  The simplest way to think about the difference between these two is that a UL Listing Mark is meant for products that are &quot;complete products.&quot;  Things like appliances or equipment that you could, for example, buy in a store.  A UL Recognized Component, on the other hand, is something that is meant for use in some end product.  They usually aren't complete products in themselves.  And they usually have conditions of acceptability that dictate how they can be used in end products.  Examples of UL recognized components might be a plastic material or an electric motor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I'm an appliance manufacturer.  And let's say I am designing a new washing machine.  If I want my washing machine to be UL listed, I can use UL Recognized Components in my design and this will prevent UL from having to evaluate the safety and construction of the components of my washing machine.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symbols also look different.  The UL Listing mark is a &quot;UL&quot; inside a circle, with the word &quot;LISTED&quot; underneath it.  The UL Recognized Component mark looks like a backwards &quot;R&quot; followed by a &quot;U&quot;.&lt;/p&gt;&#xA;" OwnerUserId="198" LastActivityDate="2015-10-10T04:19:12.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5719" PostTypeId="1" AcceptedAnswerId="5721" CreationDate="2015-10-10T05:58:34.710" Score="1" ViewCount="2038" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s9LVf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s9LVf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi, I'm having some problems with this truss question that I have in my homework. So far this is my solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I worked out my solutions to the reaction forces but don't know if they're correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is the find the internal forces ZB&#xA;YB&#xA;SU&#xA;SV&#xA;TV&#xA;RT&#xA;RS&#xA;QS&#xA;JL&#xA;JK&#xA;NP&#xA;MP&#xA;HO&#xA;HM&#xA;IO&#xA;MO&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easier way to find all these? e.g. are some of them the same?. Wouldn't using either method of joints or sections be too tedious here?.&#xA; Thanks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[![enter image description here][2]][2]&lt;/p&gt;&#xA;" OwnerUserId="2994" LastEditorUserId="2994" LastEditDate="2015-10-13T14:05:39.500" LastActivityDate="2015-10-13T14:05:39.500" Title="Loaded Truss problem" Tags="|civil-engineering|structures|statics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5720" PostTypeId="1" AcceptedAnswerId="5729" CreationDate="2015-10-10T06:47:34.927" Score="1" ViewCount="887" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SkVHG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SkVHG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm supposed to be finding the reaction forces first but from the FBD of the entire truss where am I supposed to start. If I take the moment at A then I get 3 unknowns at Bx Dy and Dx. If I get the sum of forces in x direction equals to 0 then I get 3 unknowns again. Same thing with the Y direction, I get two unknowns. &lt;/p&gt;&#xA;" OwnerUserId="3309" LastEditorUserId="33" LastEditDate="2015-11-07T14:57:17.003" LastActivityDate="2015-11-07T14:57:17.003" Title="Frame/truss problem with sliding joint" Tags="|civil-engineering|structural-engineering|statics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5721" PostTypeId="2" ParentId="5719" CreationDate="2015-10-10T10:36:12.527" Score="5" Body="&lt;p&gt;It may be worth reconsidering your deduced zero force members. But for this question it is not really necessary to determine which are zero force members in advance as this will come out in the analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not important if the truss is determinate or not in order to get the member forces your are interested in (highlighted in red). Approach this question using the &lt;a href=&quot;http://www.learnengineering.org/2013/08/structural-analysis-method-of-sections.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;method of sections&lt;/a&gt;. I have marked sensible section cuts in green. You can get the member forces of JL and JK by considering the equilibrium of joint J. It is worth taking note that the member forces requested hints at the appropriate solution method.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/PdcO1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PdcO1.jpg&quot; alt=&quot;Figure1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in order to get the member forces RT, RS, and QS make cut 3:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6srGl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6srGl.jpg&quot; alt=&quot;Figure2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that since you have the reaction forces at A already the member forces RT, RS, QS can be solved by equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting member forces for NP, MP, and MO is a little bit more tricky, but once you have member force for MH (by making cut 4) you can make the cut marked in blue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a bit tedious using method of sections, but there isn't a faster method that I am aware of. Even modelling this in a structural analysis software package would probably take longer ...&lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2015-10-10T10:36:12.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5722" PostTypeId="2" ParentId="5715" CreationDate="2015-10-10T13:08:52.480" Score="2" Body="&lt;p&gt;To answer your first question, no, you can't predict exactly how the can will crush. There are too many variables in the manufacturing of the can and the crushing force. This is typical in most buckling problems. There are many &quot;modes&quot; (shapes) that the structure can fail in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty of how the can can crush can be reduced by adding a failure point (dent) in the can. This will concentrate the force on that point and almost certainly guarantee where the the can will fail. It will also greatly reduce the force required to crush the can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing that will help is to make something that will hold the base of the can in one position. A slightly larger can that is cut down to only cover the base will keep the base from moving.  This will keep the can from sliding to the side while force is being applied to it. A typical cause of incompletely crushed cans is that they are pushed out from under the weight because the top doesn't stay parallel to the base.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-10-10T13:08:52.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5723" PostTypeId="1" AcceptedAnswerId="5725" CreationDate="2015-10-10T14:00:46.633" Score="3" ViewCount="1634" Body="&lt;p&gt;I have an application (a tube with parts in it) which needs a certain static pressure in it in order to contain a sustainable air flow. Unfortunately the fan I wanted to use in a push-configuration is not able to create this pressure. Thus I thought that I could install another fan with a higher cfm at the end of the tube (in pull-configuration), and thereby increasing the static pressure? Would that be useful, or is there another way to increase the static pressure in the tube?&#xA;Schematic is:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nqQbo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nqQbo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither F1 or F2 are able to provide the necessary static pressure alone.&lt;/p&gt;&#xA;" OwnerUserId="2844" LastEditorUserId="2844" LastEditDate="2015-10-10T15:11:22.287" LastActivityDate="2020-02-23T19:06:32.240" Title="Increase static pressure" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5724" PostTypeId="2" ParentId="5346" CreationDate="2015-10-10T15:17:47.333" Score="2" Body="&lt;p&gt;Considering your question about ancient wells, as Russell McMahon has indicated, there does not appear to be much information on ancient construction techniques of this type.  So lets conduct a thought experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, modern drilled caisson foundations, say, for example, 3 m diameter shafts to depths of 30 m, may often pass through water bearing strata to reach sound bearing strata, that is either rock or highly consolidated, dense soils.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically this is accomplished by advancing a steel casing as the drilling head extends to greater depths. Typically the shaft is filled with a dense drilling mud (sodium bentonite clay slurry) to counterbalance hydrostatic pressures (to stop water from blowing soil back in through the base of the excavation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Returning to antiquity, consider that knowledge of the application of the wheel, lever and arch were well known at that time, and arch construction typically employed braced timber forms that required precision of design and fabrication (this is the entire basis of the load capacity and performance of an arch - its geometrical precision).  It is not beyond the imagination to expect that the interlocking stone or masonry shaft could be constructed, one course at the time, as the shaft was dug by hand.  Borrowing ideas from arch construction (applied horizontally rather than vertically), timber bracing/forms could be used to ensure the precision and integrity of the shaft until completed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the shaft neared its destination (water bearing strata), if necessary rubble or gravel (or masonry elements for that matter) could be used as ballast to counterbalance the effects of hydrostatic pressure on soils at the bottom.  This ballast could be removed or left behind, as the situation required, upon completion.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastEditorUserId="3266" LastEditDate="2015-10-10T15:29:14.977" LastActivityDate="2015-10-10T15:29:14.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5725" PostTypeId="2" ParentId="5723" CreationDate="2015-10-10T16:33:34.047" Score="4" Body="&lt;p&gt;In order to get the air through the pipe, the pressure rise of the first fan has to be so high, that the pressure-loss within the tube is compensated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your first fan (F1) is not powerful enough you can add a second fan (F2) to rise the pressure at the end of the pipe to ambient pressure, or use a more powerful fan (F3) instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6o0Wn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6o0Wn.png&quot; alt=&quot;setch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sketch shows the static pressure on the y-axis and the tube at the x-axis&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2015-10-10T16:33:34.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5726" PostTypeId="2" ParentId="2516" CreationDate="2015-10-10T17:12:01.567" Score="6" Body="&lt;p&gt;From what I read, you are looking at the pressure the sand between the sheet piling exerts on them.  In this case, I see two possibilities: (1) &lt;strong&gt;log-spiral analysis&lt;/strong&gt; or (2) elastic analysis of &lt;strong&gt;Boussinesq&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Log Spiral Analysis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The log spiral analysis assumes that soil pressure is mobilised by a soil mass that follows the shape of a log spiral curve.  This is commonly used for braced trench excavations, and the curve of the mass must intersect the surface at the perpendicular. The analysis is non-determinate, so a trial and error graphical (scaled) method is recommended, but we have worked out a computer based algorithm that does this trial and error process computationally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case though, in your trial and error analysis, you can consider that the curve must be forced to occur within the geometrical limits of the distance between piled walls.  So it could represent a realistic condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Log spiral is suggested as applicable to all passive soil retention problems.  I think this assumption would be applicable to your situation, but this is something that should be verified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Boussinesq Elasticity Theory&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boussinesq theory can be used to look at lateral (and vertical) pressure problems where deformation does not occur.  In your case deformation will likely occur, but assuming that it cannot will produce higher stresses/pressures than are expected (there is no relaxation under the theory) so it will be a conservative result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there is the assumption of an elastic half space within Boussinesq theory. As your system is restricted by hydrostatic pressure, it could be considered to &lt;em&gt;behave&lt;/em&gt; as an elastic half space.  But more information would be required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other Considerations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very good, comprehensive, but dated information source is the &lt;strong&gt;Steel Sheet Piling Design Manual (1984)&lt;/strong&gt;.  Cellular cofferdams and pressure analysis is included, however, and a copy can be viewed at scribd.com &lt;a href=&quot;http://www.scribd.com/doc/27226946/Steel-Sheet-Piling-Design-Manual&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the photo provided there is no doubt going to be construction traffic travelling along the region between the piles. I have used Boussinesq (as modified) specifically for this purpose on previous projects, to ensure that the structure can withstand these loadings. This is another very important issue to be studied - it will require the analysis of the specific equipment, track patterns and loadings - essentially the equipment manufacturers data.  Your analysis should also be closely coordinated with the construction programme, to include the numbers and likely configurations of equipment that will be used.  Not an easy task.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Schematic Of Suggested Analysis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure below, the suggested approach is shown.  Of course all of the conditions are not known, for example the locations of sea/river bed, the hydrostatic conditions between the sheet piled retaining elements, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pCsll.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pCsll.png&quot; alt=&quot;combination boussinnesq and log spiral&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construction loadings at the top of the section can be modelled using the track patterns/footprints and associated loadings.  Boussinesq theory is used to compute the lateral stresses at the retaining structure as illustrated by the yellow and green stress envelopes, and these can be superimposed to accommodate any surface loading configuration that is desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The log spiral analysis, however, is an iterative process, where the origin of the curve, at point &lt;strong&gt;O&lt;/strong&gt; must be perturbed such that the curve always intersects point &lt;strong&gt;A&lt;/strong&gt; at right angles and also intersects point &lt;strong&gt;C&lt;/strong&gt; at the base of the excavation. This yields a series of soil envelopes within &lt;strong&gt;ABC&lt;/strong&gt; that reach a maximum value as illustrated by the curve and points above point &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this considers a curved failure surface.  The assumption of passive conditions is difficult to assess, but near to corners of the cofferdam the &lt;em&gt;box&lt;/em&gt; effect should provide sufficient rigidity.  Towards the centre of the sides of the &lt;em&gt;box&lt;/em&gt; this assumption needs further examination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traditional way to carry out the log spiral analysis is graphically. That is to construct a log spiral template to scale according to a scale drawing and shift it around the drawing under the constraints of points &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;C&lt;/strong&gt;.  The area of &lt;strong&gt;ABC&lt;/strong&gt; is calculated for each trial until a clear maximum is reached.  However we have developed an algorithm that will carry this out computationally, so no graphical analysis is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending upon your geometry, you may not encounter a maximum, instead you may be limited by point &lt;strong&gt;D&lt;/strong&gt;.  In this case the envelope defined by &lt;strong&gt;DBC&lt;/strong&gt; would be the value of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most difficult aspects of such an analysis will be to establish the &lt;em&gt;worst case&lt;/em&gt; base condition.  Careful consideration will be needed to determine which events could coincide, in terms of equipment configurations, fluctuations in water levels and other issues, such as potential de-watering risks.  A risk-based approach may be advised, that warrants more than the traditional factor of safety methods.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastEditorUserId="3266" LastEditDate="2015-10-16T15:05:30.177" LastActivityDate="2015-10-16T15:05:30.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5727" PostTypeId="1" CreationDate="2015-10-10T17:14:31.220" Score="6" ViewCount="156" Body="&lt;p&gt;Does limit switch &lt;strong&gt;LS1&lt;/strong&gt; remain pressed on until the piston moves to &lt;strong&gt;LS2&lt;/strong&gt;?&lt;a href=&quot;https://i.stack.imgur.com/XcidL.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XcidL.jpg&quot; alt=&quot;Limit switches are shown as 1 and 2 &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3311" LastEditorUserId="122" LastEditDate="2015-10-11T23:28:13.860" LastActivityDate="2015-10-11T23:28:13.860" Title="Schematic Reading - Fluid Control Limit Switch Question" Tags="|mechanical-engineering|sensors|pistons|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5728" PostTypeId="1" AcceptedAnswerId="5745" CreationDate="2015-10-11T02:26:05.770" Score="4" ViewCount="342" Body="&lt;p&gt;Assume I put an RF antenna in front of an object (typically the face of someone sitting in a chair, so able to move about +/- 10cm in all directions) at a typical distance of 1m. If I consider the system antenna/face as a resonant cavity, I'd get a fundamental resonant frequency of about 150MHz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a general physics education but no experience at all with RF antennas and resonators, so I have no orders of magnitude in mind. I realize such a 'resonant' cavity would be extremely poor, but how poor? My question is: is it conceivable to have enough resonance with basic antenna and processing electronics to be able to measure the mean distance to the face with a precision of say 1cm (i.e. measure the mean resonance frequency with about 1.5MHz precision), or is it an obvious and complete no go?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were conceivable, how 'clean' would the background need to be? Assuming there could be many unconstrained background objects (typically at a distance twice larger) all in a small room, could I just pick up the main resonance frequency around 150MHz for my measurement or would all the reflections / interferences with the other objects + room walls mess the signals completely up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I'm not expecting a detailed solution to my problem (although I'll happily take it if it comes :) I'm more looking for an &quot;expert's feeling&quot; like &quot;yes, people have done that for decades&quot;, &quot;maybe, you'd have to try&quot; or &quot;No way dude, you're completely dreaming&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, if anyone sees any other (as cheap as possible, and most importantly automated and non intrusive) way to get such a measurement (+/- 1cm at 1m distance) please let me know.&lt;/p&gt;&#xA;" OwnerUserId="3316" LastEditorUserId="1586" LastEditDate="2015-10-13T12:35:08.100" LastActivityDate="2015-10-13T12:35:08.100" Title="Measuring the distance to an object with RF resonance" Tags="|electrical-engineering|rf-electronics|distance-measurement|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5729" PostTypeId="2" ParentId="5720" CreationDate="2015-10-11T02:27:54.880" Score="4" Body="&lt;p&gt;You seem to be treating all the connections between members as fully fixed but rather they are all hinges with the exception of joint E. The contact at E can be treated like a roller, thus there's a force normal to member DE as shown below which I have called $F_E$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O0yFc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O0yFc.png&quot; alt=&quot;Member DE&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the sum of moments about D:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum M_D = 0: F_E(5) = 650(7.5)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\therefore F_E = 975 \text{ N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the sum of forces in the direction normal to the member:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum F_n = 0: F_E = 650 + F_D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_D = F_E - 650$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_D = 975 - 650 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\therefore F_D = 325 \text{ N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can split $F_E$ into its x and y components:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{E,x} = F_E\text{cos}(30) = 975\text{cos}(30) = 844.4 \text{ N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{E,y} = F_E\text{sin}(30) = 975\text{sin}(30) = 487.5 \text{ N}$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's look at the rest of the system including the resultant force of $F_E$ at E - remember to switch the directions of $F_E$ as this is how it acts on member AE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4FjLw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4FjLw.png&quot; alt=&quot;The rest of the system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't do the rest but you should be able to do some force balances in the x and y-axes and then a moment balance about A to get the reactions: $A_x, A_y \text{ &amp;amp; } B_x $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use &lt;a href=&quot;https://skyciv.com/3d-structural-analysis-software&quot; rel=&quot;nofollow noreferrer&quot;&gt;SkyCiv Structural 3D&lt;/a&gt; to check your solution for this as shown below but you need to be careful so that you model the connections properly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EG2CL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EG2CL.png&quot; alt=&quot;SkyCiv 3D Structural Analysis Software&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-10-11T07:46:39.660" LastActivityDate="2015-10-11T07:46:39.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5731" PostTypeId="2" ParentId="5713" CreationDate="2015-10-11T04:13:26.347" Score="4" Body="&lt;p&gt;This is a practical, empirical question, not a theoretical one that can be &quot;solved&quot; by mathematics. One way to answer it is to start from what Reynolds number means physically: it represents the ratio between &quot;typical&quot; inertia forces and viscous forces in the flow field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you look at a typical flow pattern, and choose the best length measurement to represent that ratio of forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in flow through a circular pipe, the viscous (shear) forces depend on the velocity profile from the axis of the pipe to the walls. If the velocity along the axis of the pipe remains the same, doubling the radius will (roughly) halve the rate of shear between the axis and the walls (where the velocity is zero). So the radius, or the diameter, are a good choices for the characteristic length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously Re will be different (by a factor of 2) if you choose the radius or the diameter, so in practice everybody makes the same choice and everybody uses the same critical value of Re for the transition from laminar to turbulent flow. From a practical engineering point of view, the size of a pipe is specified by its diameter since that is what is easy to measure, so you might as well use the diameter for Re also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a pipe that is approximately circular, you might decide (by a similar sort of physical argument) that the &lt;em&gt;circumference&lt;/em&gt; of the pipe is really the most important length, and therefore compare the results with circular pipes by using an &quot;equivalent diameter&quot; defined as (circumference / pi).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the &lt;em&gt;length&lt;/em&gt; of the pipe doesn't have much influence on the fluid flow pattern, so for most purposes that would be a poor choice of characteristic length for Re. But if you are considering flow in a very short &quot;pipe&quot; where the length is much less than the diameter, the length might be the best number to use as the parameter describing the flow.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-10-11T04:13:26.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5732" PostTypeId="1" CreationDate="2015-10-11T09:22:18.937" Score="4" ViewCount="2362" Body="&lt;p&gt;A wide variety of trusses exist for bridges and roofs. There are many different permutations when it comes to arranging the internal truss members. Depending on the direction, a diagonal member may be subjected to either compression or tension. Using this outline of a truss top chord and bottom chord, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MpeX5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MpeX5.png&quot; alt=&quot;truss outline&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one goes about searching for the most efficient arrangement of internal web members within the truss, without going through a tedious trial-and-error process?&lt;/p&gt;&#xA;" OwnerUserId="620" LastActivityDate="2017-01-25T12:27:54.437" Title="Arranging members within a truss efficiently" Tags="|structural-engineering|steel|structural-analysis|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5735" PostTypeId="1" AcceptedAnswerId="5737" CreationDate="2015-10-08T05:21:18.700" Score="1" ViewCount="1694" Body="&lt;p&gt;I'm thinking about using the &lt;a href=&quot;https://www.bopla.de/en/enclosure-technology/product/ultramas/ultramas-um-309-l-enclosures/um-32009.html&quot; rel=&quot;nofollow&quot;&gt;Bopla UM 32009 housing&lt;/a&gt; for my product. It has little standoffs for attaching PCBs. These standoffs are made of plastic, and seem to be drilled (or molded?) to a diameter of 2.0 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions: What type of screw would be appropriate for this hole diameter? A regular M2 machine screw won't work well. An M2.5 might work, but would probably deform the standoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An M2.2 self tapping screw might work I suppose, but I think the hole is a bit large for this (recommended is something like 1.8 mm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some kind of imperial screw that fits well into a 2.0 mm plastic hole?&lt;/p&gt;&#xA;" OwnerUserId="3324" OwnerDisplayName="avl_sweden" LastEditorUserId="368" LastEditDate="2015-10-12T16:18:28.457" LastActivityDate="2015-10-12T16:18:28.457" Title="What screw for 2.0 mm plastic hole in housing" Tags="|plastic|fasteners|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5736" PostTypeId="2" ParentId="5732" CreationDate="2015-10-11T19:46:36.243" Score="3" Body="&lt;p&gt;You say your members will be either in tension or compression. This is not entirely correct. The members supporting your UDL's will be in bending. If you design the entire structure as a pinned frame structure you will end up with the remaining members being either in tension or compression. In that case all your nodes have to be pinned to avoid moment transfer across the nodes. If you want to avoid designing for bending (except self weight of the member), your 'base' nodes would be all your corners around the perimeter of your frame as well as locations where the supports and any point loads are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would then have to look at the sectional properties of your members to ensure you don't end up with slender members for the compression members. This is the one of the reasons you might need an iterative (trial and error) process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would start by dividing the structure into triangles. Something like the image below. I think that these members would be the bare minimum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EtqCW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EtqCW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="2637" LastEditDate="2015-10-12T05:36:47.977" LastActivityDate="2015-10-12T05:36:47.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5737" PostTypeId="2" ParentId="5735" CreationDate="2015-10-11T19:56:05.027" Score="2" Body="&lt;p&gt;Typing &quot;screws&quot; in the search field of the Bopla website finds &lt;a href=&quot;https://www.bopla.de/en/service/technical-information/plastic/general-information-on-plastic-enclosures/screws-and-attachments-for-plastic-enclosures.html&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the screw you need is has a shank diameter of 1.7–1.8 mm and a thread diameter of 2.2 mm. The head is rounded.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="368" LastEditDate="2015-10-12T16:14:39.260" LastActivityDate="2015-10-12T16:14:39.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5738" PostTypeId="1" CreationDate="2015-10-11T21:17:32.443" Score="3" ViewCount="41" Body="&lt;p&gt;Cathode ray tubes (CRT) are analog addressing displays because you can light a point on its surface applying a continuous electric or magnetic field to deflect electrons beam. Virtually, there is no limit on the resolution that CRT can reach. On the other hand, TFT-LCD, for example, are digital displays because you can only light discrete points of its surface. Moreover, you are limited to a fixed set of points, so resolution is limited and fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any technologies (old and new) that allows an analog addressing of a flat panel display with no fixed resolution? For example, a thin film of electroluminescent material between two continuous (versus discrete addressing lines in common displays) electrodes that allows addressing any point of its surface with analog signal without the need of an intermediate analog-digital conversion.&lt;/p&gt;&#xA;" OwnerUserId="1616" LastEditorUserId="1586" LastEditDate="2015-10-13T12:33:13.383" LastActivityDate="2015-10-13T12:33:13.383" Title="Analog addressing flat panel display" Tags="|electrical-engineering|consumer-electronics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5739" PostTypeId="1" CreationDate="2015-10-12T01:51:04.827" Score="4" ViewCount="84" Body="&lt;p&gt;This question relates to the &lt;a href=&quot;https://en.wikipedia.org/wiki/ATEX_directive&quot; rel=&quot;nofollow&quot;&gt;ATEX directive&lt;/a&gt;, which is the European regulation governing the design of products to be used in explosive atmospheres. In particular, I am asking about the part relating to non-electrical (i.e. mechanical) devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have experience in applying non-electrical ATEX to products containing springs? If so, do you know of any good guidelines as to how to conduct the assessment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that the main potential ignition source would be a mechanical spark, which could conceivably be generated if the spring was to fail. So, the question boils down to: how to assess whether this is a genuine concern or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible option would be to assume as a worst-case that all the stored energy in the spring is converted into heat in the spark, and compare that to the minimum energy needed to ignite the explosive mixture; however, my gut feeling is that would be very conservative.&lt;/p&gt;&#xA;" OwnerUserId="3307" LastEditorUserId="272" LastEditDate="2015-10-12T12:28:51.377" LastActivityDate="2015-10-12T12:28:51.377" Title="How does non-electrical ATEX apply to springs?" Tags="|mechanical-engineering|springs|mechanical-failure|regulations|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5740" PostTypeId="2" ParentId="5739" CreationDate="2015-10-12T05:48:18.567" Score="5" Body="&lt;p&gt;This is not an authoritative answer as I don't know the ATEX regulation that well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have it however right that the ability to form a spark is relevant here. So your next step would be to look at the materials you use. Steel gives sparks on occassion. Most ATEX mechanical tools (hammers, wrenches) are from Bronze or Beryllium, so I would research in that direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another avenue of research: Some pneumatic ATEX tools do use steel components, here the moving parts are in an oil bath.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-10-12T05:48:18.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5741" PostTypeId="1" CreationDate="2015-10-12T08:14:52.017" Score="4" ViewCount="216" Body="&lt;p&gt;I have a stainless steel (316) 2B finish strip.  I tried buffing the material for a shiny surface but it takes lot of time to finish my production.  So, I want to know if there is any chemical available or any more simple process which I can do in-house to get a shiny surface on SS316 material?&lt;/p&gt;&#xA;" OwnerUserId="3329" LastEditorUserId="368" LastEditDate="2015-10-12T16:10:23.517" LastActivityDate="2015-10-12T16:11:35.693" Title="How can I get a quick 2B shine on grade 316 stainless steel?" Tags="|materials|steel|surface-preparation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5743" PostTypeId="2" ParentId="5728" CreationDate="2015-10-12T10:43:32.323" Score="2" Body="&lt;p&gt;I'd put this in the not possible/very difficult category. Simply because if you are dealing with RF your wavelength is going to be on the order of 10cm - 1m+ which is going to place a limit on the accuracy you can achieve with any simple method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be interest if thinks it can be done and how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For other methods I think &lt;a href=&quot;https://en.wikipedia.org/wiki/Computer_stereo_vision&quot; rel=&quot;nofollow&quot;&gt;stereo vision&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Structured-light_3D_scanner&quot; rel=&quot;nofollow&quot;&gt;structured illumination&lt;/a&gt; are probably the easiest/most commonly used for that sort of thing. &lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-10-12T10:43:32.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5744" PostTypeId="1" CreationDate="2015-10-12T11:33:14.483" Score="2" ViewCount="1181" Body="&lt;p&gt;I'm looking at a project where stationary CHP engines will be replaced, leaving as much of the infrastructure in place as possible. Current exhaust velocity in the chimney is ~42m/s and it will be a bit higher (~47m/s) with the new engines. I've seen 'around 40 m/s' as a typical exhaust velocity to design the chimney for, but I've also seen far smaller values and none of this with a source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a 'good' design value for the exhaust velocity? Please provide a source/and or reasoning!&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-10-06T01:58:34.457" Title="Flow velocity in exhaust chimney - typical design value" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5745" PostTypeId="2" ParentId="5728" CreationDate="2015-10-12T12:18:21.717" Score="2" Body="&lt;p&gt;I do not believe this will be an effective scheme.  The &lt;a href=&quot;https://en.wikipedia.org/wiki/Q_factor&quot; rel=&quot;nofollow&quot;&gt;$Q$ of a resonator&lt;/a&gt;, a number which describes the strength of the resonant peak, &lt;a href=&quot;https://www.rp-photonics.com/q_factor.html&quot; rel=&quot;nofollow&quot;&gt;is given by&lt;/a&gt; &#xA;$$ &#xA;Q=\frac{f_0}{\delta f}\approx f_0\frac{2\pi\ t_{RT}}{\ell},&#xA;$$&#xA;where $f_0$ is the resonant frequency, $\delta f$ is the FWHM of the resonance, $t_{RT}$ is the round-trip time, and $\ell$ is the round-trip loss.  The approximation is for $\ell\ll1$, but it actually overestimates the $Q$ value for $\ell\sim1$ so it will be useful for this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest problem for your application are the losses.  I don't know exactly how much power a human face will reflect back towards the RF antenna, but I would think 5% is a relatively high value.  Putting $\ell=0.95$ in the above equation gives $Q\sim3,$ which is a very low value.  Good resonators have $Q$ values between 100 and 10,000.  Using the other definition of the $Q$ parameter given above you can see that the width of the resonant peak will be &#xA;$$&#xA;\delta f=\frac{f_0}{Q}\simeq50\ \text{MHz}.&#xA;$$&lt;br&gt;&#xA;So, the accuracy you can expect in the measurement is on the order of 50 MHz.  You may be able to do somewhat better with a really clean measurement, but the accuracy won't be anywhere near 1.5 MHz. &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-10-12T13:57:40.953" LastActivityDate="2015-10-12T13:57:40.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5746" PostTypeId="2" ParentId="5738" CreationDate="2015-10-12T12:21:42.843" Score="2" Body="&lt;p&gt;There are too many wrong assumptions in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, there are certainly theoretical limits on the resolution CRTs can reach.  The electron beam is never a perfect point when it hits the screen.  At the lowest level, electrons are probability functions.  However, well above that, you can't get them to line up perfectly in a straight line, even at the macro level.  Then the phosphor has some thickness, and will bleed out the spot on its own a little too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention that TFT-LCD displays have pre-determined finite pixels.  That is true.  However, this is a pointless off-topic comment since you then ask about flat panel displays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CRTs can and have been made flat, so that's one example of analog addressable flat panel displays.  There are also displays made by deflecting thin laser beams with mirrors, which take analog deflection signals at their lowest levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course any discrete display can be wrapped with circuitry to take analog signals as inputs.  This is done routinely by LCD monitors that plug into a SVGA output of a computer.  SVGA carries analog signals, yet the LCD monitors are inherently discrete, at least in position.  By converting a analog signal to digital, you can drive a discrete output device with a analog signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern computer displays are driven from interfaces, like HDMI for example, that are digital to avoid this extra conversion step.  That is more efficient, less prone to noise, and eliminates the need to try to sync up to the pixel clock (since the pixel clock is provided directly).&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-10-12T12:21:42.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5747" PostTypeId="2" ParentId="5741" CreationDate="2015-10-12T14:39:48.287" Score="1" Body="&lt;p&gt;The most economic thing to do is to use raw material that already has the finish you want. Stainless is most commonly available in a 'Brushed' #4 finish, or a 'mirror' #8 finish, or a number of bead blasted options, although there are a number of other choices like swirl patterns. These will be readily available in both 304 and 316, less common for the more exotic alloys. Special considerations are needed for handling, processing and cutting these materials. Usually they come with a protective film on the finished side, but it's still possible to damage the finish. Care should be taken not to roll the material across a table with grinder grit on it, for example. When waterjet cutting, you'll typically get better results cutting the material 'upside down' with the jet entering the unfinished side and exiting the finished side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you must finish it yourself, it is indeed a time consuming process. I'm not aware of any chemical equivalent, but getting your process optimized is important. If you just start by buffing, progress will be very slow. The process should start by obtaining a flat surface (through machining or scraping) then sanding or grinding with progressively finer grits then finally cut and color buffing. As long as your parts are flat and relatively large, there is a machine known by its brand name &quot;Timesaver&quot; which will do most or all of these steps, but they are big and expensive. Depending on your location, you may be able to find someone who has one and get them to run your part through.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="368" LastEditDate="2015-10-12T16:11:35.693" LastActivityDate="2015-10-12T16:11:35.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5748" PostTypeId="1" CreationDate="2015-10-12T15:08:35.613" Score="5" ViewCount="832" Body="&lt;p&gt;I studied a manual of Ansys CFX about turbulence models and I came across the term  &quot;non equilibrium boundary layer&quot;.  What does this term mean?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="272" LastEditDate="2015-10-12T15:58:21.333" LastActivityDate="2016-02-08T18:38:25.550" Title="What is a non-equilibrium boundary layer?" Tags="|mechanical-engineering|fluid-mechanics|turbulence|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5749" PostTypeId="1" AcceptedAnswerId="5754" CreationDate="2015-10-12T15:48:11.197" Score="8" ViewCount="1524" Body="&lt;p&gt;I live in Melbourne, Australia and a lot of the creeks here have been diverted through underground storm water pipes. I guessed that this is because land is highly sought after for development, and a lot of these creeks have been built over, however there are also a lot of cases where there is nothing built over the top of them at all. They simply carry the normal water flow through a small concrete pipe, leaving an empty valley on the surface which would only carry flood waters during heavy rainfall. What's the reason for going to the cost and effort of doing this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a safety concern having creeks in residential areas? Do creeks have some effect on the land which would affect nearby buildings? Is it just for the convenience of being able to cross them without a bridge?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Example&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Dandenong Creek in Bayswater area @ -37.836524 lat, 145.254465 long (water flow goes from the right to the left of the image):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BDt0d.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BDt0d.jpg&quot; alt=&quot;Dandenong Creek&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also seen a creek which briefly goes both above and below ground. The rushing water of the underground section could be seen through an overflow while the topside creek was mostly just stagnant water. It seemed completely pointless and made no sense to me.&lt;/p&gt;&#xA;" OwnerUserId="3332" LastEditorUserId="33" LastEditDate="2015-10-13T01:15:02.323" LastActivityDate="2015-10-27T08:39:14.480" Title="Why are creeks sometimes piped underground in urban areas?" Tags="|civil-engineering|water-resources|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5750" PostTypeId="1" AcceptedAnswerId="25183" CreationDate="2015-10-12T19:08:39.697" Score="5" ViewCount="4844" Body="&lt;p&gt;I'm running a simple tensile test simulation for a single element. Under Results I made a Chart object with Stress vs Strain, which can be manually exported to a csv file by right clicking on the table. The trouble arises from the parametrization. I've parametrized temperature in Workbench so that the same simulation is run once for each temperature. However, the results are not saved for each substep for each design point, and it would be tedious to have to open Mechanical, re-solve, and export the Chart as many times as I have temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally there would be a macro that runs every time the simulation is solved in Mechanical, exporting the Chart's table to a csv (appending it if possible so that the data for all temperatures is in the same file).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to do this? I would imagine it could be achieved with JavaScript in Mechanical (for which I cannot find any documentation). &lt;/p&gt;&#xA;" OwnerUserId="3334" LastEditorUserId="282" LastEditDate="2016-03-02T00:13:21.090" LastActivityDate="2018-12-19T20:56:49.717" Title="ANSYS Workbench/Mechanical: Automatically export chart?" Tags="|mechanical-engineering|simulation|ansys|finite-element-method|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5752" PostTypeId="2" ParentId="5709" CreationDate="2015-10-12T22:39:24.120" Score="1" Body="&lt;p&gt;To &quot;optimize&quot; requires a desired outcome. Sterling engines, like most mechanical/thermodynamic systems are highly complex. You could optimize for efficiency, cost, power-to-weight, and many others. For learning the theory before you get to that stage I would recommend the wiki article, then do web searches based on your new vocabulary after that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stirling_engine#Theory&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Stirling_engine#Theory&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-12T22:39:24.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5753" PostTypeId="1" AcceptedAnswerId="5762" CreationDate="2015-10-12T23:46:41.957" Score="5" ViewCount="1739" Body="&lt;p&gt;This is a basic control theory question, since Control Theory is a part of applied mathematics but also of engineering I was unsure whether to ask this here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the transfer function of a system is $G(s)=1/(s^2+3s+2)$, consider the design of a PI closed loop control system with unit feedback using proportional gain $k_p$ and integral gain $k_i$, both of which are positive. Determine the range of gain for which the closed loop system is stable.&#xA;What I did was this, I went the Routh-Hurwitz way: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2tb8f.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2tb8f.jpg&quot; alt=&quot;My solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After completing the Routh table, I went ahead and reasoned a little about what conditions need to be met in order to avoid sign changes on the main column, however it seems that these conditions are never met! I get that $k_i$ should be less than zero when the problem clearly specifies it will always be positive.&#xA;Have I done something wrong? Is my reasoning right? Is the answer &quot;The controller is never stable&quot;?&lt;/p&gt;&#xA;" OwnerUserId="3336" LastEditorUserId="90" LastEditDate="2015-10-13T12:43:54.960" LastActivityDate="2015-10-13T16:12:30.543" Title="Determine the range of values for a PI controller" Tags="|control-engineering|control-theory|transfer-function|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5754" PostTypeId="2" ParentId="5749" CreationDate="2015-10-13T01:13:19.550" Score="7" Body="&lt;p&gt;In more developed areas (cities) burying streams is popular because it allows for more land to be developed without the fear of flooding. Also, streams in these areas tend to become very blighted and void of life. At that point you might as well hide them from sight.&lt;/p&gt;&#xA;&lt;p&gt;The area that you show doesn't seem to be one of these areas. There are a few reasons that I can think of that might lead to the situation that you are describing.&lt;/p&gt;&#xA;&lt;h3&gt;Swampy Area&lt;/h3&gt;&#xA;&lt;p&gt;The area seems to be fairly flat. In its natural state, it may have been a swamp. Providing a buried pipe provides a freely draining area what will keep the water flowing in times of low flow. This will keep the area from being swampy and the negative aspects of that: mosquitoes and smell.&lt;/p&gt;&#xA;&lt;p&gt;You can see from the satellite images that there are some drainage structures that will help surface water to flow into the buried pipe.&lt;/p&gt;&#xA;&lt;h3&gt;Park Areas&lt;/h3&gt;&#xA;&lt;p&gt;The grass above the stream can now be used as a park without worrying about the dangers of children playing around a stream. This makes the area dual purpose without completely reducing the flood control capacity of the area. In times of heavy rain, the area above the pipes can still be used to to retain excess water.&lt;/p&gt;&#xA;&lt;h3&gt;Slowly Fixing Past Mistakes&lt;/h3&gt;&#xA;&lt;p&gt;It could also be that at one time the stream was completely buried. The local authority might be slowly exposing the stream to restore the natural condition. Without knowing the timeline of the area, it it hard to tell if this is a possibility.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-10-13T01:13:19.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5755" PostTypeId="1" AcceptedAnswerId="5756" CreationDate="2015-10-12T00:18:41.260" Score="4" ViewCount="420" Body="&lt;p&gt;I'm wondering what the technical name is for this kind of stepper-motorized arrangement (a positioning system? an XY table?)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JLtsL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JLtsL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it typically used for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to have two steppers as opposed to generic DC motors, so it's presumably for precision work, and I'm trying to see if it would be appropriate for my (experimental biology-related) application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unstable kind of two-axis arrangement I suppose would make it too wobbly for something like CNC-machining, but perhaps might work well for scientific experiments.&lt;/p&gt;&#xA;" OwnerUserId="3573" OwnerDisplayName="OrCa" LastEditorUserId="1586" LastEditDate="2015-10-13T12:47:53.397" LastActivityDate="2018-08-14T05:14:17.433" Title="What is this kind of X-Y motorized system?" Tags="|mechanical-engineering|stepper-motor|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5756" PostTypeId="2" ParentId="5755" CreationDate="2015-10-12T01:25:33.783" Score="11" Body="&lt;p&gt;This would definitely fit under any of the following names:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;XY Table&lt;/li&gt;&#xA;&lt;li&gt;2D Positioner&lt;/li&gt;&#xA;&lt;li&gt;XY Stage&lt;/li&gt;&#xA;&lt;li&gt;XY Positioner&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1358" OwnerDisplayName="Michael Karas" LastActivityDate="2015-10-12T01:25:33.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5757" PostTypeId="2" ParentId="5755" CreationDate="2015-10-12T01:29:17.737" Score="8" Body="&lt;p&gt;Well &lt;a href=&quot;http://www.manualpositioning.com/html_products/XY_Motorized_Linear_Stage-134.html&quot; rel=&quot;noreferrer&quot;&gt;that part's manufacturer &lt;strong&gt;calls it&lt;/a&gt; a &quot;Precision XY Motorized Linear Stage&quot;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The terminology is not hard and fast, though.  I've also heard those called &quot;XY tables&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally recommend that we only call them &quot;tables&quot; if they either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Have a relatively large platform, EG:&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/WYt6c.png&quot; alt=&quot;&amp;quot;large&amp;quot; table&quot;&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Or the piggyback actuator is adequately supported. EG:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QHIWW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QHIWW.png&quot; alt=&quot;supported table&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3339" OwnerDisplayName="Brock Adams" LastActivityDate="2015-10-12T05:40:18.183" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5758" PostTypeId="1" CreationDate="2015-10-13T01:37:52.977" Score="5" ViewCount="132" Body="&lt;p&gt;I have a project where the contractor is using ASTM F1852 Tension Control (&lt;a href=&quot;http://www.appliedbolting.com/twist-off-bolts-bolting-method.html&quot;&gt;twist-off&lt;/a&gt;) bolts. The project is outdoors and uses weathering steel and weathering fasteners. The main members are hollow steel sections that are utilizing bolted splices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the contractor is tensioning the splice bolts, he is letting the twisted off portion of the bolt (spline) fall into the the hollow sections. You can hear the metal pieces ping all the way down the tower legs. There are no access holes to be able to retrieve these pieces. They are permanently in the hollow steel members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have used these types of bolts in other structures, but this is the first time that I have run into the situation where the splines are left in place. I have not seen an part of a specification that specifically speaks to this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Will this cause any future issues for the structure?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possible issue that I can see is that in the future, the collection of splines will rust together and clog the weep holes in the base of the members.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-10-13T16:32:52.327" Title="What could be the downsides of leaving twist-off bolt splines in the interior of exterior columns?" Tags="|civil-engineering|structural-engineering|bolting|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5759" PostTypeId="1" CreationDate="2015-10-13T03:58:26.190" Score="1" ViewCount="175" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/692jd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/692jd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is taken from Versteeg and Malalasekhara CFD book.&#xA;δx, δy, and δz are given as the dimensions of the sides of the fluid element under consideration along the three axes respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I've read the same mass flow rate derivation in Anderson's Fundamentals of Aerodynamics and understood it perfectly. &lt;a href=&quot;https://i.stack.imgur.com/vNxpI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vNxpI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3337" LastEditorUserId="1586" LastEditDate="2015-10-13T12:33:36.703" LastActivityDate="2015-10-13T19:50:48.780" Title="Can somebody explain how this equation(based on conservation of mass of a fluid) came about?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5761" PostTypeId="2" ParentId="5759" CreationDate="2015-10-13T09:18:00.727" Score="2" Body="&lt;p&gt;So $\rho\delta x\delta y \delta z$ is just the mass of a cube of size $\delta x\delta y \delta z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $\frac{\partial}{\partial t}\rho\delta x\delta y \delta z$ is the change in mass over time of this cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expending out this equation we get: $$\frac{\partial}{\partial t}\rho\delta x\delta y \delta z = \frac{\partial\rho}{\partial t} \delta x\delta y \delta z + \rho \frac{\partial \delta x}{\partial t} \delta y \delta z  + ...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then conclude that $\frac{\partial \delta x}{\partial t} = 0$ as your fluid element is fixed in size. Using the same argument for $y$ and $z$ you get the equation given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your next step will be to find the net mass flow through the fluid element, which is $-\nabla \rho \textbf{v} \delta x\delta y \delta z$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then equating the two terms to get the conservation of mass.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-10-13T09:18:00.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5762" PostTypeId="2" ParentId="5753" CreationDate="2015-10-13T14:26:22.290" Score="5" Body="&lt;p&gt;There is a mistake in your expressions. The coefficient of $s$ is $2+k_p$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions are: $$\frac{1}{3} \left(3 \left(k_p+2\right)-k_i\right)&amp;gt;0$$ $$k_i&amp;gt;0$$ $$k_p&amp;gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simplifies to: $$k_p&amp;gt;0$$ $$0&amp;lt;k_i&amp;lt;3 k_p+6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the transfer function is $$ \frac{1}{s^2+3 s + c} $$ where c is some positive constant, then the conditions simplify to: $$k_p&amp;gt;0$$ $$0&amp;lt;k_i&amp;lt;3 ( k_p+c)$$&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2015-10-13T16:12:30.543" LastActivityDate="2015-10-13T16:12:30.543" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5763" PostTypeId="1" AcceptedAnswerId="11995" CreationDate="2015-10-13T14:32:24.623" Score="5" ViewCount="704" Body="&lt;p&gt;I have raw data obtained from EMED barefoot scan containing a matrix of pressure sensors over about 70 frames. This totals 70 matrices that record a snapshot of the pressure over the duration of a person's natural walk over the pressure plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to ask if anyone knows the algorithm that is typically used to determine the Center of Pressure Line (indicated by the path on the heatmap below). One thing I have tried is to consider each row as an array and find the index of the maximum pressure across the row, and create a line by matching up all the points that correspond to these indices. My approach fails to produce a smoothed line (even when using Gaussian smoothing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bnjqi.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bnjqi.gif&quot; alt=&quot;center of pressure line across foot heatmap&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3343" LastEditorUserId="3343" LastEditDate="2015-10-13T16:14:39.703" LastActivityDate="2016-10-24T15:01:26.147" Title="Center of Pressure Line from Barefoot Scan (EMED)" Tags="|biomechanics|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5764" PostTypeId="2" ParentId="5758" CreationDate="2015-10-13T15:54:09.303" Score="2" Body="&lt;p&gt;One aspect I would be concerned with is if any utilities have been run yet.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Galvanized parts sitting on soon-to-be pressurized copper tubing could be a recipe for disaster in the making.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Galvanized parts striking electrical components likely won't be an issue as the electrical connections should be strong enough to not be affected.  However, this seems to be an invitation for Murphy's Law to rear its head.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the comments, debris accumulating at the base of the main members could cause a host of issues from preferential corroding to blocking the weep holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another consideration is with future construction.  If future work requires drilling up through the base of those main members, additional debris could fall down during the drilling process.  Likewise, that debris could present additional scraping hazards when pulling new utilities through that hole.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-10-13T15:54:09.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5765" PostTypeId="2" ParentId="5763" CreationDate="2015-10-13T16:29:22.407" Score="0" Body="&lt;p&gt;The way you mathematically describe your data should depend on how you plan to use the data. Just any smooth equation may not represent something meaningful. I am not familiar with practices in biomechanics but center of pressure is typically understood as a point, not a line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the simplest way I can see to process this data would be to rotate the plot 90 degrees (switch x and y). Then separate your data into the 6 levels or however many you have. The first level gets one data point, second level gets two data points and so on (assuming your scale is linear). Then you can model it with standard polynomial regression. This is kind of a crude way to go about it; depending on your implementation you can likely modify the standard polynomial regression to account for the &quot;weight&quot; of a data point instead making multiple points form single points.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2246" LastEditDate="2015-10-13T17:05:36.387" LastActivityDate="2015-10-13T17:05:36.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5766" PostTypeId="2" ParentId="5758" CreationDate="2015-10-13T16:32:52.327" Score="2" Body="&lt;p&gt;If they were coated bolts, I would worry that you are adding more uncoated material inside to corrode (the shear plane becomes exposed metal with reduced cathodic protection) but since they are weathering steel, there isn't a big difference. One potential issue would be that they prevent water from evaporating at the bottom of the tube, even if the weep hole is still functioning. Liquid water will escape through the weep hole, but the inside of the tube and especially the top of the base plate will be damp. As I understand it, weathering steel needs wet and dry cycles to build and maintain its corrosion resistant layer, so enough airflow for the inside to dry out will be important. Depending on your climate, I don't know if a pile of the splines will limit airflow and prevent the dry half of the cycle from every happening. This could plausibly lead to accelerated corrosion at a pretty critical part of the column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, I don't believe that the RCSC addresses the issue, but you might be able to stretch the AISC Code of Standard Practice to require it as part of final cleaning in 7.18. More realistically, cleaning of any materials not required for the construction is probably in a specific contract. I'm a little unclear how the splines are getting stuck on the inside, since the shear wrench has to touch the spline, so there should be plenty of access to control them. Also, if there is not access to the spline side, how is the inspector verifying that the end is properly sheared? Maybe your situation is in a tapered column or something, but I'm having trouble visualizing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, I don't think I'd worry much about it, but if you are worried it might be easier to add a few extra weep holes that to try to get them to remove the splines.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2015-10-13T16:32:52.327" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5767" PostTypeId="2" ParentId="5759" CreationDate="2015-10-13T19:50:48.780" Score="0" Body="&lt;p&gt;The confusion grounds in a different frame of reference (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_and_Eulerian_specification_of_the_flow_field&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; on Lagrangian and Eulerian description).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first equation describes the rate of change of a fluid element. &lt;br/&gt; &lt;em&gt;One might think of it as &quot;sitting on the fluid element&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equation describes the rate of change within a system. &lt;br/&gt; &lt;em&gt;Here one is an observer, looking at multiple fluid elements flying through the observed system.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2015-10-13T19:50:48.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5768" PostTypeId="1" CreationDate="2015-10-13T21:52:11.687" Score="1" ViewCount="62" Body="&lt;p&gt;I have to find a solution to uniform values on 3-axis accelerometers on different android devices. I mean, a device with a specific accelerometer, under the same acceleration, give me values for 3 axis that are different from the values of another device with different accelerometer. I need some kind of normalization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to calculate sensitivity for different accelerometers (reading values with smartphone under +1g and -1g and calculating half of the difference for each axis) and I thought calculate the difference of sensitivity between two accelerometers (supposing to take one of these as the target for my normalization) and sum or subtract it to each output value of each other accelerometer. Is it the right way??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I do otherwise? Is possible to solve my problem?&lt;/p&gt;&#xA;" OwnerUserId="3346" LastActivityDate="2015-10-13T22:31:47.760" Title="Uniform values from different MEMS 3-axis accelerometers" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5769" PostTypeId="2" ParentId="5768" CreationDate="2015-10-13T22:12:24.923" Score="1" Body="&lt;p&gt;This is actually very easy to solve. I will state some assumptions, but even if these are not true you can still solve the problem with a slightly more complex solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumption 1:&lt;/strong&gt; The smallest negative value and largest positive value on any axis will have approximately the same magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumption 2:&lt;/strong&gt; The 3 axes on a single device are calibrated against each other such that the maximum absolute value on all three axes is around the same magnitude.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Solution&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;When your app starts, start recording the maximum value ever seen on any axis. This would be something similar to &lt;code&gt;normalizer = max(normalizer, abs(x), abs(y), abs(z))&lt;/code&gt; on each new reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then when using the readings, normalize them by dividing by the normalizer value. This will give you values between -1 and +1 on all devices, regardless of their native range.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note: If assumption 2 is not correct, you will need to use separate normalizers for each axis. If assumption 1 is incorrect (highly unlikely) you would need separate normalizers for negative and positive readings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="95" LastEditorUserId="95" LastEditDate="2015-10-13T22:31:47.760" LastActivityDate="2015-10-13T22:31:47.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5771" PostTypeId="1" AcceptedAnswerId="5773" CreationDate="2015-10-14T02:35:34.203" Score="3" ViewCount="204" Body="&lt;p&gt;I'm having problems understanding the worked answer for this problem: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/29pZi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/29pZi.png&quot; alt=&quot;Problem definition&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's part of the worked answer:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Rxy6q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rxy6q.png&quot; alt=&quot;Part of worked answer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand why $\delta_C = \dfrac{4}{9} \delta_A$ . Can you please explain that part of it?&lt;/p&gt;&#xA;" OwnerUserId="1330" LastEditorUserId="255" LastEditDate="2015-10-14T16:05:38.490" LastActivityDate="2015-10-25T10:07:06.263" Title="Virtual Work - Determining Relationships in deflections" Tags="|mechanical-engineering|civil-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5773" PostTypeId="2" ParentId="5771" CreationDate="2015-10-14T09:49:49.937" Score="5" Body="&lt;p&gt;This is just similar triangles. The corresponding sides of the two triangles are in proportion. You basically equate each triangle's base-to-height ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\text{Base of Triangle 1}}{\text{Height of Triangle 1}} = \dfrac{\text{Base of Triangle 2}}{\text{Height of Triangle 2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{\text{AB}}{\delta_A} = \dfrac{\text{BC}}{\delta_C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta_C = \dfrac{\text{BC}}{\text{AB}} \delta_A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta_C = \dfrac{0.04}{0.09} \delta_A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta_C = \dfrac{4}{9} \delta_A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think back to your geometry days. In this case the two triangles in question are similar because the three pairs of corresponding angles are equal (AAA). Therefore the corresponding sides are all in proportion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same can be done for the other two triangles drawn in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastActivityDate="2015-10-14T09:49:49.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5774" PostTypeId="2" ParentId="5771" CreationDate="2015-10-14T10:17:59.407" Score="0" Body="&lt;p&gt;Between ABC there is a single member. Hence the angle between AB and BC is always 90 degrees. Hence if AB rotates by theta, then BC also rotates by theta. As the length AB is 90mm, and the length AC is 40mm, then (by similar triangles):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta_C = \dfrac{40}{90} \delta_A$&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-10-14T10:17:59.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5775" PostTypeId="1" CreationDate="2015-10-14T13:47:37.320" Score="1" ViewCount="177" Body="&lt;p&gt;I got a question thats been bugging me for a while:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that you're given a straight square beam (say one meter long), and you attach three strain gauges on each face. One at 0 degrees, one at 45 degrees and one at 90 degrees. All of these gauges are flat against the beam, so they measure in plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is: given these strains, could you reconstruct the geometry?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Assumptions:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The material is linear&lt;/li&gt;&#xA;&lt;li&gt;The deformations are small&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've been trying it for a while, and I can always resolve for the compression and curvature, but the rotation seems to elude me. For simple cases (like pure bending) it  works fine, the sensor at 45 degrees are linearly correlated to the bending. However, when shear is present, these seems not to be the case anymore. As of yet, I can't find a way to correct for this. &#xA;Anyone got an idea? This problem has probably already been solved by someone else, but I can't seem to find a good reference to this (for the 3D case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: images included&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Front view of a small section of beam. The red images indicate strain gauges. As can be seen, only strains in plane can be measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YTzQZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YTzQZ.png&quot; alt=&quot;Front view of a small section of beam. The red lines indicate strain gauges. As you can see, only the strains in plane can be measured&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isometric view of a small section of beam, showing two strain gauges as well as the axis. The torsion mentioned would work on the y axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PbGkL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PbGkL.png&quot; alt=&quot;Isometric view of a small section of beam, showing two strain gauges as well as the axis.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3300" LastEditorUserId="3300" LastEditDate="2015-10-14T14:34:59.250" LastActivityDate="2015-11-13T20:44:18.723" Title="From strain to displacement in 3d" Tags="|structural-engineering|sensors|deformation|structural-analysis|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5776" PostTypeId="1" CreationDate="2015-10-14T14:33:41.110" Score="2" ViewCount="149" Body="&lt;p&gt;I am simulating an known as , using CFX as a solver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This turbine have 8 blades((360/8)=45 deg=pitch angle). Due to the circumferential symmetry, only one eight of the annulus has been computed, imposing periodic boundary conditions (periodic rotational) in the tangential direction. The computational domain has been restricted in the axial direction to four chord lengths upstream and six chord lengths downstream of the blade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computational domain is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, these are set with specified e. Reference pressure is 1 atm. At the inlet I input a normal speed and at the outlet I have Static Pressure SST as turbulence model setting advanced turbulence controls such as curvature correction and production limiter. Frozen rotor interfaces connect the domains. Rotation domain rotates wit and No-slip boundary conditions are imposed on blade and hub and casing in stationary and rotation domain. and also Rotation domain is set as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the values of production limiter and curvature correction and conservation target are default values and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is as follows, there is an error in end of set up in CFX.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Set up of passages in 360 is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am grateful that guide me about above cases.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2015-10-19T10:43:31.623" LastActivityDate="2015-10-19T10:43:31.623" Title="Problem in an axial turbine" Tags="|mechanical-engineering|fluid-mechanics|turbines|ansys|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5778" PostTypeId="1" AcceptedAnswerId="5828" CreationDate="2015-10-14T18:34:18.793" Score="8" ViewCount="373" Body="&lt;p&gt;Turbulent flow involves extremely small scale fluctuations which are usually too small to resolve by using the Navier-Stokes equations directly.  To overcome this, we model each transient variable $\Phi$ (e.g. velocity, pressure, etc...) as the sum of a time average quantity $\bar{\Phi}$ and some small fluctuation.  By introducing both of these new variables into the navier stokes equations, we obtain more unknowns than equations.  In order to obtain a unique solution, we need to introduce a turbulence model which consists of at least one additional equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of different turbulence models developed from the Boussinesq Hypothesis, which roughly says that the Reynolds stresses are proportional to the mean rate of deformation by way of the so-called &quot;turbulent viscosity&quot;.  The different turbulence models vary by how they estimate the &lt;em&gt;turbulent viscosity&lt;/em&gt;.  Models of this type include, but are not limited to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$k-\epsilon$ models&lt;/li&gt;&#xA;&lt;li&gt;$k-\omega$ models&lt;/li&gt;&#xA;&lt;li&gt;Mixing Length models&lt;/li&gt;&#xA;&lt;li&gt;Spalart-Allmaras models&lt;/li&gt;&#xA;&lt;li&gt;Reynolds Stress models&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With such a wide selection of turbulence models, it can be difficult to assess which model is appropriate for each modeling situation.  I understand that with infinitely many potential simulation needs, there is no one-size-fits-all answer.   But, are there any general guidelines on how to select an appropriate turbulence model?  What factors should I consider when selecting a model?  What kinds of problems are known to be effective for the models listed above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I'm mostly interested in the time-averaged turbulence models, I also welcome responses for spatially-averaged models (e.g. Large Eddy Simulation models).&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="368" LastEditDate="2015-10-15T00:07:38.843" LastActivityDate="2017-02-22T20:30:24.757" Title="How do I select an appropriate turbulence model?" Tags="|fluid-mechanics|turbulence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5780" PostTypeId="2" ParentId="5775" CreationDate="2015-10-14T19:46:53.597" Score="1" Body="&lt;p&gt;&lt;strong&gt;TL;DR You don't have enough measurements&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Took a while, but it can be proven that (using only the strains on the sides), that there is no way to calculate the resulting shape. The reason for this is the (nonlinear) influence of shear and the lack of constraints.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way around (from displacement to strain) is possible, which is why programs like ANSYS can find a result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, in &lt;em&gt;most&lt;/em&gt; cases, the shear is negligible, in which case you can describe the shape as a combination of Torsion, Compression and Curvature (both x and y) from the given measurements.&lt;/p&gt;&#xA;" OwnerUserId="3300" LastActivityDate="2015-10-14T19:46:53.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5781" PostTypeId="1" CreationDate="2015-10-14T23:36:20.157" Score="6" ViewCount="943" Body="&lt;p&gt;I am working on designing and creating a solar powered car for the &lt;a href=&quot;http://www.solarcarchallenge.org/challenge/&quot; rel=&quot;noreferrer&quot;&gt;Solar Car Challenge&lt;/a&gt;. The car will have to race from Dallas, TX to Minneapolis, MN in 5 days (~5 hours of driving a day) along back roads. My team is still in the planning/concept generation phase and I'm in charge of designing the solar array. To begin designing, I need to first know how much power the car will consume per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My textbook gives the power consumption of a car as $$ P_{w} = F_{foward}v = (F_{roll}+F_{air})v  $$ where $F_{roll} = \mu_{r}mg$ and $F_{air} = (1/2) C_{d}A\rho v^2 $. I'm assuming a rolling resistance coefficient of 0.017, a drag coefficient of 42, and the estimated mass and frontal area of the car are 275 kg and 2.45 m$^2$, respectively. I thought a good way to estimate the minimum required speed $v$ would be to get the elevation of the route (Google Maps with &quot;avoid highways and toll roads&quot; selected) at various points and plug it into the formula above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;a href=&quot;http://www.doogal.co.uk/RouteElevation.php&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; elevation finder, I was able to get the elevation at a certain distance for ~27,000 points and export the data into Excel. If I understand correctly, $V(t)$ would be given by $$ V(t) = \frac{\left(\sum _{n=1}^{27298}\sqrt{dx^2+dy^2}\:\right)}{dt} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $dx$ is a change in horizontal distance, $dy$ is a change in elevation, and $dt$ is change in time. I have the data and I have the equation, but I'm not sure how to convert this into something that Excel can use to get my speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://onedrive.live.com/redir?page=view&amp;amp;resid=A918FADDFF95F58F!4468&amp;amp;authkey=!AJNjcENeOFqDTyE&quot; rel=&quot;noreferrer&quot;&gt;Here&lt;/a&gt; is the spreadsheet of all the data points.&lt;/p&gt;&#xA;" OwnerUserId="3364" LastEditorUserId="368" LastEditDate="2015-10-15T17:11:34.500" LastActivityDate="2015-10-16T16:31:29.503" Title="How can I calculate the rate of power consumption of a solar car along a known route with changes in elevation?" Tags="|automotive-engineering|mathematics|data|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5782" PostTypeId="1" AcceptedAnswerId="5787" CreationDate="2015-10-15T01:53:19.380" Score="10" ViewCount="641" Body="&lt;p&gt;Weathering steel is supposed to form a protective patina when exposed to wetting and drying cycles. This protective patina prevents more corrosion from occurring. This is what differentiates it from regular (carbon) steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In a situation where the environment does not allow the patina to form, does weathering steel perform any differently than carbon steel?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking that there might be situations where weathering steel without a patina might actually corrode faster (perform worse) than carbon steel. The alloy differences may cause the change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My situation is a steel deck with a layer of aggregate on top. The aggregate will likely keep the top of the deck from ever completely drying or at least greatly slow the process. This will keep a patina from forming. Various people that I have talked with at the client have said that they would rather have carbon steel instead of weathering steel. They seem to think that it will perform better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the &lt;em&gt;real answer&lt;/em&gt; is to provide a layer of protection (waterproofing, paint, etc.) over the steel, but this is not an option. I wanted to provide my exact situation in case someone requests specifics.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2016-02-09T02:16:02.637" Title="Does weathering steel corrode the same as carbon steel when it doesn't develop a protective patina?" Tags="|steel|corrosion|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5783" PostTypeId="2" ParentId="5713" CreationDate="2015-10-15T02:51:53.517" Score="8" Body="&lt;p&gt;I would like to approach this question from a mathematical perspective which can be fruitful as discussed in some of the comments and answers. The given answers are useful, however i would like to add:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In general the smallest available length scale is the characteristic length scale.&lt;/li&gt;&#xA;&lt;li&gt;Sometimes (e.g. in dynamic systems) there is no fixed length scale to choose as a characteristic length scale. In such cases often a dynamic length scale can be found.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;strong&gt;Characteristic length scales:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;TL;DWTR:&lt;/strong&gt;&lt;/em&gt; for &lt;span class=&quot;math-container&quot;&gt;$R/L\ll1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the characteristic length scale; for &lt;span class=&quot;math-container&quot;&gt;$R/L\gg1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the characteristic length scale. This implies that the smaller length scale is (usually) the characteristic length scale.&lt;/p&gt;&#xA;&lt;p&gt;Consider the pipe flow case discussed in the other answers; there is the radius &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; but also the length &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; of the pipe. Usually we take the pipe diameter to be the characteristic length scale but is this always the case? Well, lets look at this from a mathematical perspective; let's define the dimensionless coordinates:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\bar{x}=\frac{x}{L} \quad \bar{y}=\frac{y}{R} \quad \bar{u}=\frac{u}{U} \quad \bar{v}=\frac{v}{V} \quad \bar{p}=\frac{p}{\rho U^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here, &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; are &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; coordinate and velocity scales but not necessarily their characteristic scales. Note that the choice of the pressure scale &lt;span class=&quot;math-container&quot;&gt;$P=\rho U^2$&lt;/span&gt; is only valid for &lt;span class=&quot;math-container&quot;&gt;$\mathrm{Re}\gg1$&lt;/span&gt;. The case &lt;span class=&quot;math-container&quot;&gt;$\mathrm{Re}\ll1$&lt;/span&gt; requires a rescaling.&lt;/p&gt;&#xA;&lt;p&gt;Transforming the continuity equation to dimensionless quantities:&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\boldsymbol{\nabla}\cdot\boldsymbol{u}=0 \rightarrow \partial_{\bar{x}}\bar{u}+\partial_{\bar{y}}\bar{v}=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;which can only be the case when we assume &lt;span class=&quot;math-container&quot;&gt;$\frac{U}{V}\frac{R}{L}\sim1$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\frac{V}{U}\sim\frac{R}{L}$&lt;/span&gt;. Knowing this, the Reynolds number may be redefined:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathrm{Re}=\frac{UR}{\nu}=\frac{U}{V}\frac{R}{L}\frac{VL}{\nu}=\frac{VL}{\nu}=\hat{\mathrm{Re}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Similarly, let's transform the Navier-Stokes equations (&lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-component only to keep it short):&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\boldsymbol{u}\cdot\boldsymbol{\nabla u}=-\frac{1}{\rho}\boldsymbol{\nabla}p+\nu\triangle\boldsymbol{u}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\bar{u}\partial_{\bar{x}}\bar{u}+\bar{v}\partial_{\bar{y}}\bar{u}=-\partial_{\bar{x}}\bar{p}+\frac{1}{\mathrm{Re}}\left[\frac{R}{L}\partial_{\bar{x}}^{2}\bar{u}+\frac{L}{R}\partial_{\bar{y}}^{2}\bar{u}\right]$$&lt;/span&gt;&#xA;We see here the Reynolds number occuring naturally as part of the scaling process. However, depending on the geometric ratio &lt;span class=&quot;math-container&quot;&gt;$R/L$&lt;/span&gt;, the equations may require rescaling. Consider the two cases:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;strong&gt;The pipe radius is much smaller than the pipe length (i.e. &lt;span class=&quot;math-container&quot;&gt;$R/L\ll1$&lt;/span&gt;):&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The transformed equation then reads:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\bar{u}\partial_{\bar{x}}\bar{u}+\bar{v}\partial_{\bar{y}}\bar{u}=-\partial_{\bar{x}}\bar{p}+\frac{1}{\mathrm{Re}}\frac{L}{R}\partial_{\bar{y}}^{2}\bar{u}$$&lt;/span&gt;&#xA;Here we have a problem because the term &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{\mathrm{Re}}\frac{L}{R}$&lt;/span&gt; could be very large and a properly scaled equation only has coefficients &lt;span class=&quot;math-container&quot;&gt;$O(1)$&lt;/span&gt; or smaller. So we require a rescaling of the &lt;span class=&quot;math-container&quot;&gt;$\bar{x}$&lt;/span&gt; coordinate, &lt;span class=&quot;math-container&quot;&gt;$\bar{v}$&lt;/span&gt; velocity and &lt;span class=&quot;math-container&quot;&gt;$\bar{p}$&lt;/span&gt; pressure:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\hat{x}=\bar{x}\left(\frac{R}{L}\right)^{\alpha}\quad\hat{v}=\bar{v}\left(\frac{R}{L}\right)^{-\alpha}\quad\hat{p}=\bar{p}\left(\frac{R}{L}\right)^{\beta}$$&lt;/span&gt;&#xA;This choice of rescaled quantities ensures that the continuity equation remains of the form:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\partial_{\hat{x}}\bar{u}+\partial_{\bar{y}}\hat{v}=0$$&lt;/span&gt;&#xA;The Navier-Stokes equations in terms of the rescaled quantities yields:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\bar{u}\partial_{\hat{x}}\bar{u}+\hat{v}\partial_{\bar{y}}\bar{u}=-\partial_{\hat{x}}\hat{p}+\frac{1}{\mathrm{Re}}\partial_{\bar{y}}^{2}\bar{u}$$&lt;/span&gt;&#xA;which is properly scaled with coefficients of &lt;span class=&quot;math-container&quot;&gt;$O(1)$&lt;/span&gt; or smaller when we take the values &lt;span class=&quot;math-container&quot;&gt;$\alpha=-1,\,\beta=0$&lt;/span&gt;. This indicates the pressure scale didn't need any rescaling but the length and velocities scales have been redefined:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\hat{x}=\bar{x}\frac{L}{R}=\frac{x}{R}\quad\hat{v}=\bar{v}\frac{R}{L}=\bar{v}\frac{V}{U}=\frac{v}{U}\quad\hat{p}=\bar{p}=\frac{p}{\rho U^{2}}$$&lt;/span&gt;&#xA;and we see that the characteristic length and velocity scale for respectively &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; isn't &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; as assumed at the beginning but &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;&lt;strong&gt;The pipe radius is much larger than the pipe length (i.e. &lt;span class=&quot;math-container&quot;&gt;$R/L\gg1$&lt;/span&gt;)&lt;/strong&gt;&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;p&gt;The transformed equation then reads:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\bar{u}\partial_{\bar{x}}\bar{u}+\bar{v}\partial_{\bar{y}}\bar{u}=-\partial_{\bar{x}}\bar{p}+\frac{1}{\mathrm{Re}}\frac{R}{L}\partial_{\bar{x}}^{2}\bar{u}$$&lt;/span&gt;&#xA;Likewise to the previous case, &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{\mathrm{Re}}\frac{R}{L}$&lt;/span&gt; could be very large and requires a rescaling. Except this time we require a rescaling of the &lt;span class=&quot;math-container&quot;&gt;$\bar{y}$&lt;/span&gt; coordinate, &lt;span class=&quot;math-container&quot;&gt;$\bar{u}$&lt;/span&gt; velocity and &lt;span class=&quot;math-container&quot;&gt;$\bar{p}$&lt;/span&gt; pressure:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\hat{y}=\bar{y}\left(\frac{R}{L}\right)^{\alpha}=\frac{y}{L}\quad\hat{u}=\bar{u}\left(\frac{R}{L}\right)^{-\alpha}\quad\hat{p}=\bar{p}\left(\frac{R}{L}\right)^{\beta}$$&lt;/span&gt;&#xA;This choice of rescaled quantities again ensures that the continuity equation remains of the form:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\partial_{\bar{x}}\hat{u}+\partial_{\hat{y}}\bar{v}=0$$&lt;/span&gt;&#xA;The Navier-Stokes equations in terms of the rescaled quantities yields:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\hat{u}\partial_{\bar{x}}\hat{u}+\bar{v}\partial_{\hat{y}}\hat{u}=-\partial_{\bar{x}}\hat{p}+\frac{1}{\mathrm{\hat{\mathrm{Re}}}}\partial_{\bar{x}}^{2}\hat{u}$$&lt;/span&gt;&#xA;which is properly scaled with coefficients of &lt;span class=&quot;math-container&quot;&gt;$O(1)$&lt;/span&gt; or smaller when we take the values &lt;span class=&quot;math-container&quot;&gt;$\alpha=1\,\beta=-2$&lt;/span&gt;. This indicates the length, velocities and pressure scales have been redefined:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\hat{y}=\bar{y}\frac{R}{L}=\frac{y}{L}\quad\hat{u}=\bar{u}\frac{L}{R}=\bar{u}\frac{U}{V}=\frac{u}{V}\quad\hat{p}=\bar{p}\left(\frac{L}{R}\right)^{2}=\bar{p}\left(\frac{U}{V}\right)^{2}=\frac{p}{\rho V^{2}}$$&lt;/span&gt;&#xA;and we see that the characteristic length, velocity and pressure scales for respectively &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; isn't &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\rho U^{2}$&lt;/span&gt; as assumed at the beginning but &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\rho V^{2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;In case you had forgotten the point of this all: for &lt;span class=&quot;math-container&quot;&gt;$R/L\ll1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the characteristic length scale; for &lt;span class=&quot;math-container&quot;&gt;$R/L\gg1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the characteristic length scale. This implies that the smaller length scale is (usually) the characteristic length scale.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Dynamic length scales:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Consider diffusion of a species into semi-infinite domain. As it is infinite in one direction, it does not have a fixed length scale. Instead a length scale is established by the 'boundary layer' slowly penetrating into the domain. This 'penetration length' as the characteristic length scale is sometimes called is given as:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\delta\left(t\right) = \sqrt{\pi D t}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is the diffusion coefficient and &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; is the time. As seen, there is no length scale &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; involved as it is determined completely by the diffusion dynamics of the system. For an example of such a system see my answer to &lt;a href=&quot;https://physics.stackexchange.com/questions/203018/show-that-the-boundary-layers-diffuse-out-from-the-plate-with-speed-sqrt-frac/203293#203293&quot;&gt;this&lt;/a&gt; question.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-10-15T11:58:59.830" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5784" PostTypeId="1" CreationDate="2015-10-15T08:41:24.247" Score="12" ViewCount="4925" Body="&lt;p&gt;The &lt;a href=&quot;http://turkstream.info/project/technology/&quot;&gt;TurkStream project site&lt;/a&gt; claims the gas pipeline will use pipes &lt;em&gt;made from 39 millimeters of high-quality carbon manganese steel&lt;/em&gt; with additional plastic coatings on both sides. Yes, they claim 39 millimeters (roughly 1,54 inches) of steel walls - that's a whole lot of steel! For comparison, the &lt;a href=&quot;https://en.wikipedia.org/wiki/M41_Walker_Bulldog&quot;&gt;M41 Walker Bulldog&lt;/a&gt; light tank had only 38 mm of armor plating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is such wall thickness realistic for an industrial gas pipeline? Why would you need that much steel for a gas pipeline?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="368" LastEditDate="2015-10-15T17:36:21.683" LastActivityDate="2015-10-16T11:58:31.950" Title="How thick is it reasonable for the walls of a steel gas pipeline to be?" Tags="|mechanical-engineering|civil-engineering|steel|pipelines|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5785" PostTypeId="2" ParentId="5784" CreationDate="2015-10-15T10:24:22.523" Score="19" Body="&lt;p&gt;It seems realistic to me. This is an undersea pipeline at depths of over 2 km. The pressure would be considerable at those depths (on the order of 20 MPa or 200 atmospheres). The pipe would need to be thick enough to withstand these (very high) pressures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.europipe.com/121-1-Nord-Stream.html&quot; rel=&quot;noreferrer&quot;&gt;Nord Stream&lt;/a&gt; pipeline under the Baltic sea is at depths of up to 210m and uses 1220 mm diameter pipe with a wall thickness ranging from 26.8 to 41 mm. The TurkStream site specifies depths more than 10 times this, but with a pipe diameter of 813 mm. Without doing any calculations that seams reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Undersea pipelines are known to have wall thicknesses of &lt;a href=&quot;https://en.wikipedia.org/wiki/Submarine_pipeline#Submarine_pipeline_characteristics&quot; rel=&quot;noreferrer&quot;&gt;up to 75mm&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2015-10-15T20:50:29.037" LastActivityDate="2015-10-15T20:50:29.037" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5786" PostTypeId="2" ParentId="5784" CreationDate="2015-10-15T12:50:57.513" Score="13" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is such wall thickness realistic for an industrial gas pipeline?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking through ASME B36.10M-2004, &lt;em&gt;Welded and Seamless Wrought Pipe&lt;/em&gt;, there are plenty of pipes that have thicknesses in that range.  For example, 16&quot; diameter, Schedule 160 pipe is 1.594&quot; thick:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hU2Sl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hU2Sl.png&quot; alt=&quot;ASME Table 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This translates to about 40.5mm thickness.  As another extreme example, 24&quot; diameter, Schedule 160 pipe is a whopping 2.344&quot; (59.5mm) thick!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @mg4w says, the pressures at play for some of these pipelines can be enormous, so the pipe wall needs to be commensurately thick to withstand those pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that per Section 1 of ASME B36.10M, it &quot;covers the standardization of dimensions of welded and seamless wrought steel pipe for high or low temperatures and pressures.&quot;  That's not to say that pipe cannot be made with thicknesses outside of the ranges in the standard, which is likely the case for some of these heavy industrial applications.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-10-15T12:50:57.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5787" PostTypeId="2" ParentId="5782" CreationDate="2015-10-15T13:06:38.243" Score="7" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;In a situation where the environment does not allow the patina to form, does weathering steel perform any differently than carbon steel?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Weathering steel is specifically designed to form a protective coating (i.e., patina) of rust that prevents the material underneath from corroding.&lt;/p&gt;&#xA;&lt;p&gt;Referring to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Weathering_steel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; (1) on the topic,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Using weathering steel in construction presents several challenges. Ensuring that weld-points weather at the same rate as the other materials may require special welding techniques or material. Weathering steel is not rustproof in itself. If water is allowed to accumulate in pockets, those areas will experience higher corrosion rates, so provision for drainage must be made. Weathering steel is sensitive to humid subtropical climates. In such environments, &lt;strong&gt;it is possible that the protective patina may not stabilize but instead continue to corrode&lt;/strong&gt;. For example, the former Omni Coliseum, built in 1972 in Atlanta, never stopped rusting, and eventually large holes appeared in the structure. This was a major factor in the decision to demolish it just 25 years after construction. The same thing can happen in environments laden with sea salt. Hawaii's Aloha Stadium, built in 1975, is one example of this.&lt;/p&gt;&#xA;&lt;p&gt;The rate at which some weathering steels form the desired patina varies strongly with the presence of atmospheric pollutants which catalyze corrosion. While the process is generally successful in large urban centers, the weathering rate is much slower in more rural environments.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As you say, it would seem that your case, where the steel cannot fully dry, the protective patina cannot form.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;My situation is a steel deck with a layer of aggregate on top. The aggregate will likely keep the top of the deck from ever completely drying or at least greatly slow the process.  This will keep a patina from forming. Various people that I have talked with at the client have said that they would rather have carbon steel instead of weathering steel. They seem to think that it will perform better.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However, given the conditions you've specified, &lt;strong&gt;I cannot see how regular, uncoated carbon steel would perform any better than weathering steel in these conditions&lt;/strong&gt;.  I would expect that both will corrode approximately at the same rate in these conditions.  My assumption is backed up by the information available on &lt;a href=&quot;http://www.steelconstruction.info/Weathering_steel&quot; rel=&quot;nofollow noreferrer&quot;&gt;SteelConstruction.info&lt;/a&gt; (2):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Alternate wet/dry cycles are required for the adherent ‘patina’ to form. Where this cannot occur, due to continuously wet or damp conditions, a corrosion rate similar to that of ordinary structural steel must be expected. Examples include weathering steel elements submerged in water, buried in soil or covered by vegetation. If weathering steel is used in such cases, it should be painted and the paint should extend above the level of the water, soil or vegetation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Regular carbon steel will be more cost effective than weathering steel, so I assume this is where your client's concern lies.&lt;/p&gt;&#xA;&lt;p&gt;Personally, I would go with galvanized steel for the decking, or put some sort of &lt;a href=&quot;http://www.sumtercoatings.com/epoxy-primers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;epoxy primer&lt;/a&gt; over the steel (I assume stainless steel is out of the question).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;1.) &lt;a href=&quot;https://en.wikipedia.org/wiki/Weathering_steel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia - Weathering Steel&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;2.) &lt;a href=&quot;http://www.steelconstruction.info/Weathering_steel&quot; rel=&quot;nofollow noreferrer&quot;&gt;SteelConstruction.info - Weathering Steel&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-02-09T02:16:02.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5789" PostTypeId="1" CreationDate="2015-10-15T17:06:06.203" Score="7" ViewCount="396" Body="&lt;p&gt;A lot of production equipment requires periodic calibration as part the documentation that the equipment is under control. The companies selling the equipment knows this and sometimes lowers the price of the equipment and offer an expensive service agreement (better to sell a long time service than a product once).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wouldn't it be valid to use redundancy and drop calibration - what are the chances that both drift or breaks at the same time (AND that the process comes out of control). If one goes out then it just time to order a calibration..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example could be monitoring systems like temperature loggers. Having two loggers (perhaps from different manufactures) and just drop calibrating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do anyone have some arguments why this would &lt;strong&gt;NOT&lt;/strong&gt; be such a good idea? &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Update&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Thank you very much for the big interest in the topic!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I changed the header to use the term &lt;strong&gt;&lt;em&gt;periodic calibration&lt;/em&gt;&lt;/strong&gt; instead of just &lt;em&gt;calibration&lt;/em&gt; - in order to be more clear. I would also like to elaborate here a bit more on what I mean by &lt;em&gt;periodic calibration&lt;/em&gt; and &lt;em&gt;proven state of control&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In EU GMP guidelines for &lt;strong&gt;process equipment&lt;/strong&gt; the following is stated: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5.3 Calibration&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5.30 Control, weighing, measuring, monitoring and test equipment that is critical for assuring the quality of intermediates or APIs should be calibrated according to written procedures and an established schedule. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5.31 Equipment calibrations should be performed using standards traceable to certified standards, if existing.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5.32 Records of these calibrations should be maintained.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5.33 The current calibration status of critical equipment should be known and verifiable.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5.34 Instruments that do not meet calibration criteria should not be used.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;5.35 Deviations from approved standards of calibration on critical instruments should be investigated to determine if these could have&#xA;  had an impact on the quality of the intermediate(s) or API(s)&#xA;  manufactured using this equipment since the last successful&#xA;  calibration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ec.europa.eu/health/files/eudralex/vol-4/pdfs-en/2005_10_03_gmp-partii-activesubstance_en.pdf&quot; rel=&quot;nofollow&quot; title=&quot;EU Guidelines to GMP vol. 4 part II&quot;&gt;EU Guidelines to GMP vol. 4 part II&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This industry commonly uses two calibrations to prove/document that the equipment has been under &lt;strong&gt;control&lt;/strong&gt; the entire production period (the period between &lt;em&gt;before/start&lt;/em&gt; and &lt;em&gt;after/end&lt;/em&gt; calibrations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My point is that it seems unlikely that both devices measures wrongly AND that the process is out of control (unless the root cause affects both the measurements and condition). The idea of redundancy instead of calibration is derived from the principle of separating the control and monitoring system - since a drift would not be caught and the process could then go out of control/specs.  &lt;/p&gt;&#xA;" OwnerUserId="3372" LastEditorUserId="3372" LastEditDate="2016-02-23T06:40:10.250" LastActivityDate="2016-02-23T06:40:10.250" Title="Can redundancy be used instead of periodic calibration?" Tags="|measurements|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5790" PostTypeId="2" ParentId="5789" CreationDate="2015-10-15T17:49:20.670" Score="10" Body="&lt;p&gt;There are a few basic issues. The premise of this idea is that the errors of each measurement device will balance out, so with enough devices, the average value will stay roughly accurate. If not, at least you would be able to notice the difference between the two devices and know when to re-calibrate. For some types of measurements, this is roughly true, but for others it is not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first big problem is that some types of measuring devices will always tend to drift in the same direction. For example, if your device compares a force against a spring, that spring will get weaker over time as it is fatigued, never stronger, so your measured values will all drift in the same direction. Here, averaging the results from multiple devices wouldn't help, and you wouldn't even notice the error as the springs will decay at roughly the same rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other fundamental problem is that even for a measurement where errors were bilateral and random, it would take a lot of devices to be statistically confident that your average was good. Just using two or three devices - all errors could well be in the same direction, so your average value would still be wrong. You would need a lot of measurement devices to get any precision in your average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to these two basic principles, there are a bunch of practical reasons to use calibrated measurement devices when values matter. One is that handling the specimen to measure it a number of times would take more time, machinery, and/or handling which increases costs and risk of damage. Another is that for most work governed by a standard, units will be defined with respect to an absolute value (usually a physical constant, but sometimes an artifact) and if your measurements don't relate to that value, you can't be sure that others will agree with them. For some critical work, a traceable lineage to a governmentally established constant is required behind each measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all of these reasons, it is almost always better to use one instrument that is periodically calibrated instead of a bunch of instruments with unknown accuracy and trying to get a statistical average.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Based on your edit I think I understand that you don't want to avoid calibration altogether, just to reduce the frequency and reduce the risk that an instrument drifting out of spec (but within it's calibration period) is going to lead to a failure of your final product. This is a good point - there is a risk analysis to be done of process failures based on the frequency of instrument failures, the frequency of control process failures, and the cost of a defective product. Based on those factors, you could add two instruments, maybe only one of which is in its designated calibration cycle, with the second just acting as a trouble monitor. If the two instruments disagree, you won't be certain which is right, but you'll at least know that something is going wrong and to check calibration. There's nothing wrong with this idea, and it will probably help detect some sudden failures, although the point that common mode failures won't be detected still stands. On the other hand, it's probably not a big economic impact to keep both instruments within their calibration period, (ie. don't reduce the calibration interval) and then you could have more confidence in your results.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2015-10-23T14:28:43.247" LastActivityDate="2015-10-23T14:28:43.247" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5791" PostTypeId="2" ParentId="5789" CreationDate="2015-10-15T19:32:54.847" Score="1" Body="&lt;p&gt;It's always a trade-off. Usually between economics and accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two temperature loggers, and one drops out of calibration, but you don't know which one, you now have much larger uncertainties on your temperature observations. And you've got twice the bill for recalibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nice thing about calibrated kit, is that this, in most cases, gives you a pretty reasonable idea of the distribution of errors on your observations, and that makes it much easier to account for them. Whereas having N bits of uncalibrated kit give you a lot of uncertainty about what the actual distribution of errors is, which makes it much harder to manage them and account for them. So you've got a system where you've paid extra for redundancy, but have not compensated for the reduced accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have to calibrate each time one of your recorders disagrees with another, you could end up doing more than twice the recalibrations you do currently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the two recorders drift off calibration, but in sync with each other, then you'll miss a critical recalibration. And understanding the distribution of such common-mode failures is really difficult; so once again, you won't have a known error distribution that you could measure and manage; instead you'll have an unknown error distribution, and that will increase your costs.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-10-15T20:11:20.917" LastActivityDate="2015-10-15T20:11:20.917" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5792" PostTypeId="2" ParentId="5781" CreationDate="2015-10-15T19:42:30.073" Score="1" Body="&lt;p&gt;I think it's going to be too complicated to calculate from the basics, as you are attempting. You can't assume a constant speed; rolling resistance will depend on the nature of the road surface, the gradient, and how wet the road surface is' and air resistance will depend on wind speed and direction. The uncertainties in your calculation will swamp the realities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, get a test vehicle out on the road ASAP and monitor its energy consumption in minute detail. Use a GPS logger, altimeter, accelerometer, and detailed engine measurements, to calculate how the car's power consumption varies with speed, gradient, road surface condition, and vehicle mass. Then use that empirical data for your estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't think it will make that much difference, because your answer is going to be the same, whatever the numbers say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vehicle will need to be as light and as aerodynamic as possible, with as much of the upward-facing surface covered in high-efficiency low-weight PV as is possible without compromising vehicle safety. (NB, that should be on the surface of the car, not some PV array roof lifted above the car - think of the aerodynamics!). So you should be thinking about a custom array of cells tiled over the car body, not off-the-shelf PV panels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's because insolation at its best is only about $1000W/m^2$, and that's at mid-day. And you won't get PV that's much better than 22% efficient. You're just not going to have much surface area to play with, and you're going to have to squeeze every last joule out.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2015-10-15T19:42:30.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5793" PostTypeId="2" ParentId="5781" CreationDate="2015-10-15T20:17:02.920" Score="1" Body="&lt;p&gt;Not sure if this approach is in the format you need, but it might help.  The work required to go uphill by a height $y$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W=mgy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is mass and $g$ is gravitational acceleration.  Assuming mass and gravity are constant, we can differentiate this equation with respect to time to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{W}=mg \frac{dy}{dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can also use the chain rule to rewrite it as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{W}=mg \frac{dy}{dx} \frac{dx}{dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which $x$ is the horizontal distance traveled.  Then, $\frac{dx}{dt}$ is just velocity, so the equation reduces to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{W}=mgv \frac{dy}{dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You already have the data for $\frac{dy}{dx}$, so you can factor out the velocity and insert this equation into your first equation for total power consumption:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_w = (F_{air} + F_{roll} + mg \frac{dy}{dx} ) \cdot v $$&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-10-15T20:17:02.920" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5794" PostTypeId="2" ParentId="5789" CreationDate="2015-10-15T22:23:42.000" Score="2" Body="&lt;p&gt;Having &lt;em&gt;two&lt;/em&gt; measuring devices usually a bad idea for avoiding &lt;em&gt;recalibration&lt;/em&gt;. If the devices give different results, you don't know which is right until you recalibrate, or do some fault-finding. Of course two different readings tell you &lt;em&gt;something&lt;/em&gt; is wrong with the measurement system, which is more information than just one wrong reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have &lt;em&gt;three&lt;/em&gt; or more devices, it is often sensible to use &quot;majority voting&quot; to decide that one device is giving bad data and should be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As other answers have said, you also need to consider &quot;common-mode&quot; causes of errors which affect every device the same way - not only &quot;uncontrollable&quot; things like environmental conditions, but also issues like using a common electrical power supply, which might go out of specification without any indication and affect the output of all the devices.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2015-10-15T22:23:42.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5795" PostTypeId="2" ParentId="5781" CreationDate="2015-10-16T01:27:48.873" Score="1" Body="&lt;p&gt;The equation you included in your question only considers horizontal distances (x and y). You need to use slope distances (three dimensional distances: x, y and z) as these will be the true distances travelled by the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the distances in your data spreadsheet are &lt;strong&gt;slope lengths&lt;/strong&gt;, sum all the lengths to give the total slope distance between the start and the finish. Divide this by the total amount of time you expect the vehicle to be driven. This will give you the average speed required for every stage of the competition, under ideal conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the data in the spreadsheet calculate the gradient of each section of road, noting sections of steep gradient. This will probably be the most important information you can derive from the data in your spreadsheet. Try to find continuous sections of road where the gradient is approximately constant and adjust the average speed required for such sections based on road conditions and the road gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When determining the total driving time, make allowances for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Driver exchange – I'm assuming each driver will drive no more than 2&#xA;to 2.5 hours before being replaced by a fresh driver.&lt;/li&gt;&#xA;&lt;li&gt;Scheduled maintenance during driving time, if any.&lt;/li&gt;&#xA;&lt;li&gt;Unscheduled maintenance/repairs.&lt;/li&gt;&#xA;&lt;li&gt;Gradient of the road for each leg/stage and its impact on power&#xA;requirements.&lt;/li&gt;&#xA;&lt;li&gt;Rest periods, if any&lt;/li&gt;&#xA;&lt;li&gt;The affect of weather and the environment: cloud obscuring the Sun,&#xA;rain, wind, heat, dust.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other things you'll need to consider are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The affect of wind on the performance of the vehicle, with headwinds&#xA;and cross winds impeding travel while tail winds would assist travel.&lt;/li&gt;&#xA;&lt;li&gt;The affect of wind on the fatigue of the driver.&lt;/li&gt;&#xA;&lt;li&gt;Condition of the road – potholes, rough patches, etc.&lt;/li&gt;&#xA;&lt;li&gt;Radius of curvature of curves and how they may impact on speed&lt;/li&gt;&#xA;&lt;li&gt;The affect of rail road crossings, such as roughness and the loss of kinetic energy and speed in crossing the tracks. Also, having to&#xA;wait for trains.&lt;/li&gt;&#xA;&lt;li&gt;The possibility of collision with wild or farm animals.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-10-16T01:27:48.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5796" PostTypeId="2" ParentId="5632" CreationDate="2015-10-16T02:53:32.933" Score="1" Body="&lt;p&gt;Yes they do make pneumatic valves that meet your criteria. These ones definitely exceed your criteria, but will come with a large price tag. I recommend searching vendors for &quot;high speed pneumatic valve&quot;.&#xA;&lt;a href=&quot;http://ph.parker.com/us/12051/en/pulse-valves-miniature-high-speed-high-vacuum-dispense-valve&quot; rel=&quot;nofollow&quot;&gt;http://ph.parker.com/us/12051/en/pulse-valves-miniature-high-speed-high-vacuum-dispense-valve&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.festo.com/net/SupportPortal/Files/10047/MH234-PSI-US.PDF&quot; rel=&quot;nofollow&quot;&gt;https://www.festo.com/net/SupportPortal/Files/10047/MH234-PSI-US.PDF&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very possible that standard direct acting pneumatic valves would meet your criteria; they will just not publish it because it is something they dont consider an important feature or dont want to be held to. You could get something like these and test to see if they meet your spec:&#xA;&lt;a href=&quot;http://www.zoro.com/parker-solenoid-air-control-valve-18-in-24vdc-b310bb549c/i/G2536274/&quot; rel=&quot;nofollow&quot;&gt;http://www.zoro.com/parker-solenoid-air-control-valve-18-in-24vdc-b310bb549c/i/G2536274/&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.omega.com/pptst/V60_SERIES.html?pn=OM-AIR-V60P6DDA-XP0200&quot; rel=&quot;nofollow&quot;&gt;http://www.omega.com/pptst/V60_SERIES.html?pn=OM-AIR-V60P6DDA-XP0200&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note; If this is in water, you can increase bubble uniformity by adding an inorganic salt like NaCl or MgCl to increase the surface tension.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-16T02:53:32.933" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5797" PostTypeId="2" ParentId="5583" CreationDate="2015-10-16T03:24:58.750" Score="1" Body="&lt;p&gt;I modeled a vertical axis turbine back in 2010 in a two dimensional rotating reference frame in OpenFOAM with good results. I am not up to date with the new tools available, but you want to make sure that your geometry is in fact rotating according to the model (perhaps your periodic boundaries are handling this for you).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used gmsh (&lt;a href=&quot;http://geuz.org/gmsh/&quot; rel=&quot;nofollow&quot;&gt;http://geuz.org/gmsh/&lt;/a&gt;) for my mesh generation. You want to make sure your boundary layer on the blade surface is hexahedral and the remaining volume can be filled with tetrahedrals as they are very forgiving of complex geometery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would highly recommend reposing your question on cfd-online. Lots of experts on there with specific OpenFOAM experience.&#xA;&lt;a href=&quot;http://www.cfd-online.com/Forums/openfoam/&quot; rel=&quot;nofollow&quot;&gt;http://www.cfd-online.com/Forums/openfoam/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-16T03:24:58.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5798" PostTypeId="1" CreationDate="2015-10-16T03:44:28.040" Score="5" ViewCount="827" Body="&lt;p&gt;I've been reading the Feynman lectures, and in &lt;a href=&quot;http://feynmanlectures.caltech.edu/II_09.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this lecture about lightning,&lt;/a&gt; he describes how the Earth is negatively charged, and how there is thus an electric field of about &lt;span class=&quot;math-container&quot;&gt;$100\ \tfrac{\text{V}}{\text{m}}$&lt;/span&gt; near the surface of the earth. &lt;a href=&quot;https://i.stack.imgur.com/EH5Sz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EH5Sz.png&quot; alt=&quot;Here was the diagram to explain&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;To me, it seems like &lt;span class=&quot;math-container&quot;&gt;$100\ \tfrac{\text{V}}{\text{m}}$&lt;/span&gt; could provide a lot of energy, a lot of small appliances run on &lt;span class=&quot;math-container&quot;&gt;$9\ \text{V}$&lt;/span&gt; batteries, and the power point provides &lt;span class=&quot;math-container&quot;&gt;$240\ \text{V}$&lt;/span&gt;, which would be the potential difference between &lt;span class=&quot;math-container&quot;&gt;$2.4\ \text{m}$&lt;/span&gt; of air, not all that much. There are probably problems in attempting to harness the electric field around the earth, but if I were to design a simple machine to harness the energy, how would I do so?&lt;/p&gt;&#xA;" OwnerUserId="3376" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2023-12-28T07:56:58.067" Title="How could we possibly harness the energy from the Earth's negative charge?" Tags="|electrical-engineering|energy|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5799" PostTypeId="2" ParentId="5574" CreationDate="2015-10-16T03:44:51.073" Score="0" Body="&lt;p&gt;Tempered glass and clean parts will help as starrise has mentioned. It will scratch over time however, so you may however want to consider a different, more reliable process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase the amount of light in the photography area and use a quality digital camera, you should be able to take a quality photo of the part even when it is not perfectly still; such as in a persons hands or attached to a gantry. If straps or hands are in the way of the photo, you could use an electromagnetic lift attachment for taking the bottom shot.&#xA;&lt;a href=&quot;http://www.zoro.com/mag-mate-lifting-magnet-250-lb-cap-5-in-oal-pnl0250/i/G3212553/&quot; rel=&quot;nofollow&quot;&gt;http://www.zoro.com/mag-mate-lifting-magnet-250-lb-cap-5-in-oal-pnl0250/i/G3212553/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-16T03:44:51.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5800" PostTypeId="2" ParentId="5784" CreationDate="2015-10-16T03:50:32.997" Score="0" Body="&lt;p&gt;Because of work conditions of pipe (fluids or gas, lifespan, pressure inside the pipe...), and I think that was calculated by design engineers and the result accordance with application code/standard. For the tank you should not compare like this because work condition and application code/standard not the same. When repairing or maintaining the pipe, it is much more difficult.&lt;/p&gt;&#xA;" OwnerUserId="3375" LastEditorUserId="3375" LastEditDate="2015-10-16T03:57:40.407" LastActivityDate="2015-10-16T03:57:40.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5801" PostTypeId="2" ParentId="5784" CreationDate="2015-10-16T04:06:42.563" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why would you need that much steel for a gas pipeline?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Remember that the force on the inside of a pipe is proportional to pressure*diameter. So a large pipe needs to be much thicker than a small pipe for the same pressure. &lt;/p&gt;&#xA;" OwnerUserId="3377" LastEditorUserId="272" LastEditDate="2015-10-16T11:58:31.950" LastActivityDate="2015-10-16T11:58:31.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5802" PostTypeId="1" CreationDate="2015-10-16T06:01:09.067" Score="3" ViewCount="365" Body="&lt;p&gt;Which type of turbine (Pelton, Kaplan, Francis, etc.) would be suitable and deliver the best efficiency when driven by a high velocity forced air jet of 125-200 mph through a nozzle of 2-2.5 inches diameter? The desired outcome is maximum torque in the 300-400 rpm range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The turbine should preferably be enclosed.&lt;/p&gt;&#xA;" OwnerUserId="3378" LastEditorUserId="3378" LastEditDate="2015-10-19T04:35:15.417" LastActivityDate="2023-12-13T17:00:20.877" Title="Which type of Turbine (Pelton, Francis, etc.) is most suitable to be driven by high velocity forced-air jet?" Tags="|mechanical-engineering|fluid-mechanics|turbines|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5803" PostTypeId="2" ParentId="5502" CreationDate="2015-10-16T07:10:53.123" Score="2" Body="&lt;p&gt;The Bernoulli-Newton lift debate has been around for a long time, and now has many other theories in the mix. This hyperphysics link sums it up pretty well; &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/fluids/airfoil.html&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/fluids/airfoil.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of confusing explanations out there on both sides of the fence. Like the bogus “20%” claim you have come across ;-). It is a more complex topic than most people credit it and I can not say I am truly happy with any of the &quot;book&quot; definitions I have found over the years. I agree with your mathematical evaluation of Bernoulli's principle explaining 100% of the lift, which the equation does, but it does not explain the misconceptions surrounding the debate. It is a bit like arguing &quot;is the world described by a Polar or Cartesian coordinate system?&quot;, because it is the same world regardless of which way you choose to describe it. I have spent a fair amount of time studding this topic and will take a shot at my own (perhaps more layman's terms) explanation. It sounds like you already have a firm understanding, but this may help you to explain it to others:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lift is the process by which an airfoil traveling through a fluid at a velocity, transforms a small forward force into a large perpendicular force. Think of a wing/airfoil like a gear ratio between forward force and downward force. Different wings/airfoils have different gear ratios(lift coefficients). Conservation of momentum (per Newton) tells us that if we move more air down at a lower velocity, we can lift much more efficiently. You can do this math in your head, because momentum has a linear velocity term (p=mv), and kinetic energy has a squared velocity term(ke=1/2mv^2). Thus, you can see that as the velocity of the downward air increases, the power required grows exponentially (1-&gt;4-&gt;9-&gt;16-&gt;25) while lifting force only grows linearly (1-&gt;2-&gt;3-&gt;4-&gt;5).  This is why a Cessna 172 can fly, but it can not fly vertical. The lift produced by the wings(low velocity, more air) is substantially more than the lift produced by the propeller(high velocity, less air), even though both scenarios have the same power input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This insight is highly useful for explaining lift all by itself. Whatever magic occurs to make x many kg of air move down at v velocity is outside of our concern and we can still explain and evaluate lift effectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we dive deeper into our understanding however, like when trying to understand why airfoils stop working at high angles of attack, we need new explanations to model reality. With our previous understanding we would have expected a nice trigonometric gear ratio change based on our angle of attack, but since airfoils do not behave that way in practice we need to understand them deeper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bernoulli's principle explains why the laminar flow of air over the top of the wing is critical to lift. If the angle of attack is too large or if the laminar airflow on the top of the wing is disrupted; turbulence will be generated and the airfoil will stall. In a stall, some amount of air is still being deflected downward by the bottom of the airfoil; even in the absence of negative pressure on the top of the wing. In this instance there is “lift” being generated in a “purely Newton” fashion, but the glide ratio will be that of a triangular shaped rock ;-). There is so little downward force relative to a large forward force that it is not considered lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Newton perspective of lift is correct in the sense that the wing/airfoil, as a complete system, produces lift based on a gear ratio of that system. It however does not explain why we can achieve such a high gear ratio(lift coefficient) in practice. To fulfill this need we use Bernoulli's principle along with many other fluid dynamics principles to help explain the complex physics occurring. Other mathematics and theorems can be used to explain the physics in more useful and/or in more confusing ways ;-). For example I find the Kutta-Joukowski Lift Theorem to be highly irritating and unintuitive. That said, I'm sure it has saved people from some very nasty differential equations over the years ;-).&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-16T07:10:53.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5804" PostTypeId="1" AcceptedAnswerId="5805" CreationDate="2015-10-16T08:26:13.677" Score="1" ViewCount="820" Body="&lt;p&gt;What is the difference between &lt;em&gt;Transport Phenomena&lt;/em&gt; first edition (1979) and second edition (2002)?&lt;br&gt;&#xA;Which one do you recommend to be bought, taking into account that I am a student and that the first one is cheaper?&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2015-10-16T10:26:40.550" Title="&quot;Transport Phenomena&quot; (Bird, Stewart, Lightfoot) difference between edition 1 and 2" Tags="|fluid-mechanics|heat-transfer|chemical-engineering|" AnswerCount="1" CommentCount="6" ClosedDate="2015-10-16T22:38:09.153" ContentLicense="CC BY-SA 3.0" />
  <row Id="5805" PostTypeId="2" ParentId="5804" CreationDate="2015-10-16T10:26:40.550" Score="2" Body="&lt;p&gt;Taken from the &lt;a href=&quot;http://eu.wiley.com/WileyCDA/WileyTitle/productCd-0470115394.html&quot; rel=&quot;nofollow&quot;&gt;Wiley&lt;/a&gt; page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;About the Revised 2nd Edition: Since the appearance of the second&#xA;  edition in 2002, the authors and numerous readers have found a number&#xA;  of errors--some major and some minor. In the Revised 2nd Edition the&#xA;  authors have endeavored to correct these errors. A new ISBN has been&#xA;  assigned to the Revised 2nd Edition in order to more easily identify&#xA;  the most correct version.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;New content:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Revised to include more extensive reference to applications of material covered and the addition of appendices.&lt;/li&gt;&#xA;  &lt;li&gt;Expanded coverage of: transport properties in 2-phase systems, boundary-layer theory, heat/mass transfer coefficients, dimensional&#xA;  analysis and scaling&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Concerning your issues with the English language used, your comment exhibits a good grasp of the English language and I would argue that a good understanding of Calculus is much more valuable than being a native English speaker for reading this book. Besides, many canonical literature is written in English and it would be good to get some practice in reading and understanding such texts. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2015-10-16T10:26:40.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5806" PostTypeId="2" ParentId="5789" CreationDate="2015-10-16T11:22:41.083" Score="0" Body="&lt;p&gt;More independent measurements with the same accuracy only change the probability profile of the error, not the worst case error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example is rolling two dice.  You can consider each die as providing a average value of 3.5&amp;plusmn;2.5.  Any one roll of any one die can be anywhere in that range with equal probability.  However, the average of two dice has a triangular profile.  Six different possible combinations result in 3.5, whereas only one combination each results in 1 or 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rolling more dice narrows the probability of the average around 3.5, but the probability of 1 or 6 never goes to 0.  The worst case is still the same as before.  No amount of extra dice can change the worst case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since most of the time for production measurements, we care about the worst case error, multiple measurements doesn't help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, note that the better average case of the multiple dice only works because each measurement (each die) is independent of the others.  The same measurement process, even performed independently by different instruments, may have some common sources of error.  For example, temperature, atmospheric pressure, age, etc, can all affect different instruments that measure the same thing in the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's rare that you get truly independent measurements, and then you only get a better average answer, not a better worst case anyway.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-10-16T11:22:41.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5807" PostTypeId="1" CreationDate="2015-10-16T15:19:09.807" Score="2" ViewCount="65" Body="&lt;p&gt;One of the processes in a current product design catalyzes the conversion of atmospheric oxygen into ozone: &#xA;$$&#xA;3\text{O}_2+\text{energy}\rightarrow 2\text{O}_3&#xA;$$&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ozone&quot; rel=&quot;nofollow&quot;&gt;Ozone&lt;/a&gt; is highly reactive and, over time, will degrade the components in the sealed enclosure.  One technique which I have read about for removing the ozone is to actively cycle the air in the enclosure through a catalyst such as silver oxide ($\text{Ag}_2\text{O}$) or manganese oxide ($\text{MnO}$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the enclosed volume is 1-2 liters, is it necessary to actively cycle the air, or can the catalyst simply be placed in the enclosure? Are there any other techniques for ozone removal which I should consider?&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-10-20T01:26:35.813" Title="How can I remove Ozone from a sealed enclosure" Tags="|chemical-engineering|product-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5808" PostTypeId="1" CreationDate="2015-10-16T16:17:56.270" Score="5" ViewCount="10812" Body="&lt;p&gt;Given this type of arrangement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EKCuB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EKCuB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;- image by &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Simple_beam_with_center_load.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hermanoere&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the beam is of rectangular cross-section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically thinking of a wooden beam where I can look up some material characteristics (e.g. modulus of elasticity) from places such as &lt;a href=&quot;http://www.woodworkweb.com/woodwork-topics/wood/146-wood-strengths.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I only know&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Force&lt;/li&gt;&#xA;&lt;li&gt;Length between supports&lt;/li&gt;&#xA;&lt;li&gt;Width and thickness(height) of beam&lt;/li&gt;&#xA;&lt;li&gt;Modulus of elasticity of beam material&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Deflection_(engineering)#Center-loaded_simple_beams&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; suggests I also need to know &quot;Area moment of inertia of cross section&quot; but I have no idea how to obtain this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Do I have enough information to calculate deflection?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Footnote: I ultimately want to know how thick the beam should be so that&#xA;it won't break if I sit on it - but to avoid asking unacceptably many &#xA;questions at once, I have broken my problem up into simpler and more general questions.&lt;/p&gt;&#xA;" OwnerUserId="1683" LastEditorUserId="1586" LastEditDate="2015-10-18T14:27:37.187" LastActivityDate="2017-01-03T17:09:08.543" Title="How to calculate deflection of a simple beam with a load in the center?" Tags="|structures|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5809" PostTypeId="2" ParentId="5781" CreationDate="2015-10-16T16:31:29.503" Score="2" Body="&lt;p&gt;Having done this before when I was in college on the solar car team, designing and building cars for Sunrayce '97 and '99, I'll tell you that you're on the right track, but you need to start with a simpler dataset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with analyzing something like a basic grade, something that you can check using hand calculations (they still teach that stuff, right?) then extend it to larger and larger data sets. Similarly you'll need to add terms for efficiencies (Array, battery, driveline, drive) building a more and more complex model as you go. You'll also have to allow for a certain amount of power recovery via regenerative braking and the possibility of coasting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start, I noticed that you only have one velocity, $v$, which I take to be the velocity relative to the ground. You can start there, but the aerodynamic terms are relative to the speed at which the air flows over the body of the car. Your model will work, but only for perfectly still air.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that your car's $C_d$ is much lower than 0.42, I recall a target for our team's '99 car of 0.17 in the software analysis (it will be higher in reality, but if it's 0.42, you're doing it wrong.) You should also develop drag coefficients for off axis flows since in the real world the flow over the vehicle body will almost never be on axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your model develops, you'll have to refine it using results of your testing program, which, in part, should serve the needs of developing the inputs for your model based on actual data. If you can, do some wind tunnel modeling, preferably full-scale to get a handles on the aerodynamic loads and how they vary. At the speeds you need to go to be competitive, the bulk of your power will be spent fighting aerodynamic resistance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing in school that's quite as satisfying as seeing your car performing as your model predicts, singing down the road at the speed limit on nothing but array power. Your model will never be perfect, but you can get close.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-10-16T16:31:29.503" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5810" PostTypeId="2" ParentId="5808" CreationDate="2015-10-16T16:58:03.643" Score="4" Body="&lt;p&gt;Yes, you likely have enough information in order to solve your equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_area_moments_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article listing various area moment of inertias&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would use the rectangular area equation for your bench.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a filled rectangular area with a base width of b and height h&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9WEx1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9WEx1.png&quot; alt=&quot;filled rectangle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which has an area moment of inertia of: $I = b*h^3/12$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In order to apply that, take a reasonable amount of deflection (say $1/4^&quot;$ or so) and work the equations backwards in order to determine the minimum area moment of inertia that's required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, one of the two dimensions of your material is likely fixed.  So you can determine the remaining dimension based upon the area moment of inertia equation.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-10-16T17:08:32.223" LastActivityDate="2015-10-16T17:08:32.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5811" PostTypeId="2" ParentId="5808" CreationDate="2015-10-16T17:56:56.147" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am specifically thinking of a wooden beam...&#xA;  Wikipedia suggests I also need to know &quot;Area moment of inertia of cross section&quot; but I have no idea how to obtain this.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're using standard wood dimensional lumber sizes, the American Wood Council (AWC) publishes a table of different sizes vs. section properties in the &lt;a href=&quot;http://www.awc.org/pdf/codes-standards/publications/nds/AWC-NDS2015-Supplement-ViewOnly-1411.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;2015 NDS Supplement&lt;/a&gt;, Section 3.1.  For example, consider the screenshot below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dH7YD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dH7YD.png&quot; alt=&quot;NDS Values&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were using a 6x12 beam, its moment of inertia, $I_x$, is 697.1 in&lt;sup&gt;4&lt;/sup&gt; about its major axis.  This is one part of the measurement of the beam's &lt;em&gt;stiffness&lt;/em&gt;, or its resistance to deflection - the other part being its &lt;em&gt;modulus of elasticity&lt;/em&gt;, as you've already identified.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you don't already have the formulation, the deflection of a simply-supported beam with a point load at midspan is given by the following (per the AISC 14th Ed. &lt;em&gt;Steel Construction Manual&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BzzxW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BzzxW.png&quot; alt=&quot;deflection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are solving for the variable $\Delta_{max}$.  As you've stated, you have all of the parts needed to calculate deflection except for $I$.  @GlenH7's answer already gives you the formulation to determine $I$ if you have a shape not covered by the NDS Supplement, so I won't repeat that information here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the NDS contains a code for designing wood structures.  This, or another similar code, is usually the code of record for wooden structures designed in the United States.  There are additional provisions to consider if you are doing a &quot;proper&quot; deflection analysis per the code that are beyond the scope of this Question, but I get the feeling that you don't really need to go that in-depth.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, this type of scenario is very easily worked out by trial and error if all you're trying to do is see if a wooden beam is OK to sit on.  And given how cheap dimensional lumber is (in the United States, at least), I would just err on the side of caution and get a bigger size than you think you need and it will &lt;em&gt;probably&lt;/em&gt; work.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-12-02T13:55:06.063" LastActivityDate="2015-12-02T13:55:06.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5812" PostTypeId="1" CreationDate="2015-10-16T18:25:19.943" Score="4" ViewCount="179" Body="&lt;p&gt;My company has over 100k communication devices deployed. We collect failure information from the devices. We have two type of data failure count per hour for each device and we use that to calculate failure percentage. We are developing a pro active monitoring system which will help us better maintain the system. One of the challenge is to prioritize devices using failure count or failure percentage. Is there any statistical or mathematical theory which will help us create a priority list using both count and percentage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 1: All the devices are not the same. But for this process we will consider to be same. To have a good system performance we have to prioritize our troubleshooting.&lt;/p&gt;&#xA;" OwnerUserId="3317" LastEditorUserId="3317" LastEditDate="2015-10-18T01:21:18.083" LastActivityDate="2015-10-22T17:08:27.410" Title="Statistical approach to defect detection and prioritization" Tags="|statistics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5813" PostTypeId="1" AcceptedAnswerId="5821" CreationDate="2015-10-16T19:28:13.597" Score="9" ViewCount="2919" Body="&lt;p&gt;In Rankine cycle there's the boiler where water gets boiled into overheated steam. On input there's a pump that delivers more water, and on output there's a turbine that picks up the energy of the compressed steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bhQ5I.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bhQ5I.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steam pressure is roughly identical against the turbine and the pump; the pressure of the boiler section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the steam power the turbine instead of backing up and forcing the pump to turn backwards? - well, this one is simple, power delivered to the pump. But then how comes the turbine produces more power than the pump takes? The pump, after all, must overcome the same pressure that propels the turbine and deliver the same amount of water that is being ejected as steam. I'm missing some significant element of the device. What is it?&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2016-11-26T10:34:47.767" Title="How in Rankine cycle the turbine generates more power than the pump takes?" Tags="|thermodynamics|steam|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5814" PostTypeId="2" ParentId="5813" CreationDate="2015-10-16T20:15:34.263" Score="0" Body="&lt;p&gt;There is a pressure differential across the turbine which prevents the pressure from backing up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mpoweruk.com/images/rankine_pv.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mpoweruk.com/images/rankine_pv.gif&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/1VacK.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1VacK.gif&quot; alt=&quot;Somebody&amp;#39;s Ideal Rankine Cycle P-vs-vol&quot;&gt;&lt;/a&gt;&#xA;On the above chart between point 2 and point three the fluid passes through the turbine which expands while it pushes the turbine and reduces pressure. If you ignored the shaft and rotation of the turbine that point could almost be replaced by a nozzle with the same effect on the cycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your description it sounds like you're describing a pumped storage system or you are ignoring the heating/combustion stage (point 4 to point 1 on my chart). This stage is what increases the pressure of the system beyond the pressure that the pump creates. &lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2015-10-16T20:15:34.263" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="5815" PostTypeId="2" ParentId="3356" CreationDate="2015-10-16T20:35:00.203" Score="7" Body="&lt;p&gt;If you are really considering catastrophic failure, what you are asking actually has several parts.  The first is the coatings aspect, the second is the metal structure, the third is the connections, the fourth is the cables, the fifth would be the substructure and the sixth is the foundation.  A failure of any one of these systems (with the exception of coatings) could lead to the type of catastrophic failure you are considering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Golden Gate Bridge was constructed in the mid to late 1930s (around 1933 to 1938 I believe, see &lt;a href=&quot;http://www.baybridgeinfo.org/timeline&quot; rel=&quot;nofollow noreferrer&quot;&gt;timeline&lt;/a&gt;) and I will come back to this in later paragraphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to note that the loading conditions currently on the bridge in terms of axle loads and traffic volumes are perhaps magnitudes higher than the original designer could have envisaged.  These loads will have a significant impact upon the risk of catastrophic failure, and for this reason there has likely been significant structural strengthening carried out in the Golden Gate Bridge over the years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that I noted on the timeline was that there already has been a catastrophic failure of the upper and lower decks due to a 7.1 earthquake on October 17, 1989.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EboUt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EboUt.png&quot; alt=&quot;Golden Gate Catastrophic Failure 1989&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Coatings&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking coatings are designed for a given service life.  Provided that they are applied with adequate workmanship and quality, for a given material in a given environment the performance is fairly predictable; ie they will lose their protective capability over time due to oxidation, temperature, solar radiation and abrasion effects. Many designs are geared towards a certain loss in terms of mils/yr, so that for a 10 year service interval there will be a certain dry film thickness of coating specified.  &lt;a href=&quot;http://corrosion.ksc.nasa.gov/pubs/nasa-std-5008b.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA&lt;/a&gt;  Kennedy Space Center has detailed documentation on these issues.  Once this protective measure is breached then corrosion of the underlying steel may take place.  So it is important that coatings are renewed at regular intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Structural Steel&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structural steel generally has a well defined atmospheric corrosion rate for a given environment and this is based upon it's proximity to the sea.  Studies done by NASA in Florida on metal coupons from seafront to a distance inland about 30 km indicate that the danger zone is within roughly 1000 m of the sea. After that the atmospheric corrosion rates dropped off significantly.  But the Golden Gate Bridge is clearly within a danger zone.  From &lt;a href=&quot;http://corrosion.ksc.nasa.gov/atmos.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA&lt;/a&gt; data, Pont Reyes, California would have an expected atmospheric corrosion rate of about 20 mils/yr.  Certainly Caltrans will have similar research data.  If left unprotected by coatings, and if this number were valid for your immediate environment, then you could take the thickness of the structural section and reduce exposed surfaces by 20 mils/yr to roughly estimate what would happen to an unprotected structure with time.  Of course the NASA studies are using modern grades of steel, the Golden Gate steel was produced in the 1930s, so there could be a greater annual loss. This is atmospheric corrosion; the structural steel that is within range of the seawater spray and splash zone will actually experience a more severe corrosion condition, and protective measures there will likely include sacrificial anode systems in addition to coatings.  That is, an anode material such as zinc is electrically connected to the structural steel in seawater spray exposure zones because it is more susceptible to corrosion.  It is sacrificed in order to direct corrosion activity away from the structural steel.  These also must be replaced at intervals.  There are also impressed current systems too, these are called cathodic protection and they work in a similar fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what does this mean for the structure? A structural analysis would be required to determine where the critical loading point occurs; ie where the force of a fully loaded bridge would exceed the ultimate loading capacity of the corroded, diminished sections.  When structural elements oxidize, the oxidized portion of the section can no longer support load, and the remaining sound steel under the corrosion product is the issue of importance.  But you can see that if coatings/protective measures are not reapplied when required then the  corrosion rate of structural steel could endanger the structural capacity of the bridge very quickly, perhaps within a few years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Connections&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structural connections are very sensitive to fatigue and this also exacerbates their sensitivity to corrosion.  In other words elements of higher stress cycling are also more susceptible to corrosion.  The two factors are quite significant in determining lifespan.  For this reason, and I do not have any specific knowledge of this matter for the Golden Gate Bridge, I suspect that the connections in the bridge today have been essentially entirely replaced, perhaps several times at high stress locations, since initial construction.  These are the most critical part of the structural steel framework for the bridge, from a load and lifespan perspective.  If they fail, then catastrophic failure follows.  It is also possible that additional protective measures are taken for the connections in terms of additional coating thickness or specialized treatments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cables&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Golden Gate Bridge is a suspension bridge so the cables are an integral part of the load capacity of the structure.  These cables normally would have a protective sleeve around them to ensure that they are isolated from the environment, and cable sleeves today are typically filled with protective lubricants. I do not have any knowledge specific to the Golden Gate Bridge but I would expect that the design does include protective sleeves, at least for the main cables.  Corrosion of cables can very quickly lead to very serious structural problems and for that reason they are inspected frequently.  One of the problems with sleeved cables is that the cable itself is hidden from view, unlike other elements of the bridge, so the inspection would require non-destructive techniques that may include ultrasonic and radiographic testing.  I am not aware of any loss per year allowances for cables, if they show any signs of corrosion then I would expect they will be scheduled for replacement or repair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Substructure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concrete substructure (foundation piers) elements would take the brunt of the most severe exposure condition, that is sea spray and splash zone.  Wetting and drying conditions in the sea environment are the most corrosive.  Here any exposed concrete may degrade at the surface due to salt scaling effects and this, along with carbonation (ie conversion of calcium hydroxide to calcium carbonate in the concrete) can reduce the effective protection of reinforcing steel in the concrete by reducing the thickness of the protective concrete cover.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are generally well known formulas for concrete in sea splash and spray zones for time to corrosion of reinforcing steel.  Pioneers in this field (among many others  since) were Beaton and Stratfull (&lt;a href=&quot;http://www.dot.ca.gov/research/researchreports/1961-1963/62-17.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1962&lt;/a&gt;) and Stratfull (&lt;a href=&quot;http://www.dot.ca.gov/research/researchreports/1964-1965/64-06.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1964&lt;/a&gt; and &lt;a href=&quot;http://www.dot.ca.gov/research/researchreports/1966-1967/67-09.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1967&lt;/a&gt;) of Caltrans. More recently Collepardi and Clear have introduced improvements (Collepardi introduced diffusion theory in the early 1970s and Clear's work in the mid 1970s was a refinement of the Beaton and Stratfull analysis).  These formulas are used at the time of design and can be used throughout the life of the structure for estimating remaining life.  This has to do with the concentration of chlorides with depth in the concrete, that are carried into the concrete by seawater and increase in concentration with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason protective coatings and other systems such as cathodic protection are used to prevent reinforcing steel corrosion.  I would suspect that one or more of these treatments are in place at the Golden Gate Bridge, and I would also expect that portions of the concrete and reinforcement have been replaced in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, if the corrosion were to be left untreated in the substructure then there could be a very serious catastrophic failure.  If left completely untreated, the concrete of the type used on the Golden Gate Bridge in the 1930's would probably reach a critical stage within 25 to 30 years.  The age of the bridge, being some 75 to 80 years of age means that this substructure concrete has likely had substantial repair, replacement and protective treatment by now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Foundations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concrete foundations, if permanently below the low tide line, generally are not at high risk, but can present problems.  It is not commonly known, but the concrete of foundations of the Golden Gate Bridge contain Pozzolans, which was an advanced technique for its time.  Pozzolans are usually added for improved durability, but they can play a major role in reducing temperature related distortions that can lead to cracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/povJu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/povJu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact one of the more tragic bridge collapses in history occurred for this very reason during May 1980.  The &lt;a href=&quot;http://www.metrojacksonville.com/article/2013-sep-sunshine-skyway-bridge-disaster-may-9-1980&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sunshine Skyway Bridge&lt;/a&gt; in Tampa collapsed killing 35 persons.  The engineer's investigation concluded that the cause of the collapse was directly attributable to cracks in the below water pier foundations due to the high temperature of the concrete at the time of construction.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastEditorUserId="3266" LastEditDate="2015-10-20T08:19:24.970" LastActivityDate="2015-10-20T08:19:24.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5818" PostTypeId="1" AcceptedAnswerId="5823" CreationDate="2015-10-17T13:06:39.367" Score="6" ViewCount="145" Body="&lt;p&gt;Just something I was wondering about when I've seen &lt;a href=&quot;https://i.stack.imgur.com/xVr4T.jpg&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; on reddit today. Unlike in firearms where the pressure propels the projectile, in a railgun, wouldn't the pressure ahead/behind of the projectile slow it down instead? If so, would it be by a significant amount?&lt;/p&gt;&#xA;" OwnerUserId="3136" LastEditorUserId="1560" LastEditDate="2015-10-21T12:53:21.833" LastActivityDate="2015-10-21T12:53:21.833" Title="Does negative pressure inside the barrel affect low-powered railguns?" Tags="|mechanical-engineering|pressure|firearms|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5820" PostTypeId="2" ParentId="5818" CreationDate="2015-10-17T19:18:14.867" Score="3" Body="&lt;p&gt;Yes a pressure differential would act to slow down the projectile. But I see no reason why the chamber could not be vented behind the projectile, in which case this would greatly reduce any resistance due to a pressure differential. There is also no reason the projectile has to seal the barrel as in a traditional gun. So pressure could equalize by passing around the projectile also. How significant any slowing of the projectile would be would depend heavily on the construction of the gun and this would need to be considered in the design. &lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2015-10-17T19:18:14.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5821" PostTypeId="2" ParentId="5813" CreationDate="2015-10-17T22:54:47.360" Score="6" Body="&lt;p&gt;Pressure is force/area, if the area offered by the pump impeller is smaller than the area against wich the steam must force its way out of the boiler, and both the turbine and the pump are linked, then, the same pressure will result in a smaller force on the pump than on the turbine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets suppose that the pump is a piston pump, and on the steam side there is a piston engine (for simplicity). In a certain part of the cycle, both the pump piston and the engine piston valves are open towards the boiler (the pump is feeding water to the boiler and the engine is taking steam from the boiler).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;face&quot; of the pump piston has, lets say, a surface area of 10cm², while the steam engine piston has a surface area of 100cm². Lets suppose the pressure on the boiler is 200kPascal. This means that the pump piston will need to force its way against 200000N/M² * 0.001M² = 200N of force. While this pressure on the steam engine produces 200000N/M² * 0.1M² = 20kN. It's clear that in a direct linkage between the pump and the engine, the steam engine will produce much more force than the pump needs to feed water inside the boiler against the pressure gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparision, lets suppose that instead of pumping water into the boiler, the pump takes steam and pumps it into the boiler. If the pump piston is smaller (meaning a smaller stroke or bore), the amount of water mass that gets out of the boiler via the engine would be bigger than the amount of mass that gets back into the boiler via the pump. If both where equal the engine would not generate any power. While if the pump were able to put more steam into the boiler than it were to get out, you would have an above unit power source - i.e. defying the laws of thermodynamics. Soon the boiler would run out of mass to heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, as what gets pumped inside the boiler is liquid water, and water as a liquid has a much higher density than steam (meaning the same volume holds more mass as liquid than as gas - besides the fact that liquids have a fixed volume while the gas can expand), when that small volume is pumped into the boiler as water, the thermal energy added to the system will expand this mass into a steam that has a lot of pressure and much less density, so to get the same mass of water (as steam) out of the boiler via the engine, you will need a much bigger swept volume than the one you used to put that water inside, resulting in a difference in forces exerted by this gas over the piston of the pump (via the incoming water column) versus the engine piston, deciding the direction towards wich the system will move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Got the idea ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the ability of the water to change phase and occupy a bigger volume than the fixed volume liquid results in a net power gain into this system. From where comes the energy to cause the phase change, which then causes thisincrease in pressure? It's from the heat source. Where goes the energy rejected? It goes out of the condenser, where steam becomes liquid, losing volume, to be injected again into the boiler as a smaller and fixed volume liquid, and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key idea here is pressure equals force divided by area.&lt;/p&gt;&#xA;" OwnerUserId="1087" LastEditorUserId="-1" LastEditDate="2016-11-26T10:34:47.767" LastActivityDate="2016-11-26T10:34:47.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5822" PostTypeId="2" ParentId="5813" CreationDate="2015-10-18T00:30:32.133" Score="4" Body="&lt;p&gt;The main reason is that it takes much less energy to compress a liquid than a gas by the same pressure difference. The pump takes a little bit of energy to compress the water, but a huge amount of energy is released when the steam expands in the turbine.  This is why a phase change is used in the Rankine and related cycles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is that the pump doesn't need to over-match the &lt;em&gt;power&lt;/em&gt; of the turbine, as you said - it only needs to over-match the &lt;em&gt;pressure&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the difference in energy between gas and liquid compression when looking at an enthalpy-vs-pressure table.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-10-18T00:35:39.170" LastActivityDate="2015-10-18T00:35:39.170" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5823" PostTypeId="2" ParentId="5818" CreationDate="2015-10-18T00:52:47.853" Score="3" Body="&lt;p&gt;For both conventional guns and rail guns, the resistance from pressure ahead of the projectile is usually negligible. Venting probably won't help much either if the projectile gets near the speed of sound because the air ahead of the projectile won't have time to move out of the way. You could fire in an evacuated chamber to completely avoid the air resistance, but the gains probably wouldn't justify the added complexity.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-10-18T00:52:47.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5825" PostTypeId="1" CreationDate="2015-10-18T06:12:22.187" Score="1" ViewCount="143" Body="&lt;p&gt;Is there any equation to calculate the switching transient of a power Transformer? How much it will be for a 30MVA transformer (33/11kV) at no load?&lt;/p&gt;&#xA;" OwnerUserId="3400" LastEditorUserId="1586" LastEditDate="2015-10-18T14:27:47.740" LastActivityDate="2015-10-18T14:27:47.740" Title="Switching transient of a transformer on no load" Tags="|electrical-engineering|power-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5826" PostTypeId="1" CreationDate="2015-10-18T13:34:00.777" Score="4" ViewCount="4176" Body="&lt;p&gt;Can we use Grade 410 Stainless Steel for replacement of Grade 420 Stainless Steel? The application is for engineering components which have torsional and fatigue load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the substitution is that 420 is not commonly available in the Indian market. Can we use 410 stainless as a replacement for 420 stainless? Does it has any major differences with regard to chemical and mechanical properties? &lt;/p&gt;&#xA;" OwnerUserId="3404" LastEditorUserId="33" LastEditDate="2015-10-18T18:13:25.180" LastActivityDate="2018-06-06T00:22:31.977" Title="Use of 410 Stainless Steel for replacement of 420 Stainless Steel" Tags="|materials|steel|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5828" PostTypeId="2" ParentId="5778" CreationDate="2015-10-18T13:57:00.153" Score="3" Body="&lt;p&gt;Disclaimer: As you requested &lt;em&gt;general guidelines&lt;/em&gt; here they are. After having a converged solution, you need to investigate it thoroughly and look for clues that the choice of the turbulence model altered the flow in an un-physical way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Rules of Thumb for Turbulence Models&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The more additional transport equations you solve:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the more possibilities exists to make errors (e.g. wrong assumptions)&lt;/li&gt;&#xA;&lt;li&gt;the better the capabilities of the simulation are to capture the physics involved&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Usually &lt;em&gt;mixing length&lt;/em&gt; models have ONE additional transport equation, k−ϵ and k−ω have TWO and Reynolds Stress models have SEVEN. &lt;br/&gt;&#xA;The Spalart-Allmaras model is a highly tuned model and has only one additional transport equation. It was tuned for flight regimens (Reynolds  and Mach number) of airplanes. The application in other fields might result in bigger errors.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;k−ϵ models&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The k-ϵ-Model is commonly optimised for external flow (around bodies)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;k−ω models&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The k−ω-Model is commonly optimised for internal flows (inside pipes, close to walls)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;Mixing Length models&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Are very very basic and will only produce meaningful results when the turbulence modelling was not necessary in the first place.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;Spalart-Allmaras models&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Are very robust, fast and widely tested. Additional testing is required when not applied in their actual scope (flow around airplanes)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;&lt;blockquote&gt;&#xA;  &lt;p&gt;Reynolds Stress models&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Cutting Edge Models need a lot of computational power and are not as robust.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="3583" LastEditDate="2017-02-22T20:30:24.757" LastActivityDate="2017-02-22T20:30:24.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5829" PostTypeId="2" ParentId="5826" CreationDate="2015-10-18T18:26:12.217" Score="5" Body="&lt;p&gt;In general, I would be hesitant to switch the grade of stainless steel used. You are asking to switch from a more unique grade to a common grade. There was probably a good reason why the grade 420 was chosen for the original design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that in mind, the differences between the two grade seem to mainly be that 420 is harder than 410 (see these references: &lt;a href=&quot;http://www.pennstainless.com/stainless-grades/400-series-stainless/410-stainless-steel/&quot;&gt;410&lt;/a&gt; and &lt;a href=&quot;http://www.pennstainless.com/stainless-grades/400-series-stainless/420-stainless-steel/&quot;&gt;420&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what you have described of the requirements (e.g. fatigue is an issue), it seems strange that grade 420 was originally specified. Typically a harder material is more brittle and has a lower fatigue strength. Using that one factor, it would seem that 410 would be an OK substitution. The original designer should have known this though. There is probably a good reason why grade 420 was chosen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't ask the designer, it would not be a good idea to substitute a cheaper material.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-10-18T18:26:12.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5830" PostTypeId="2" ParentId="5789" CreationDate="2015-10-18T20:48:53.413" Score="4" Body="&lt;p&gt;The topic is – as easily seen from length and number of answers – not easy. This also means it does highly depend on every specific case to answer this topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part I:&lt;/strong&gt;&lt;br&gt;&#xA;If we define that the process-parameter ( $p$ ) hast to be monitored. The measurement ( $m$ ) of the process parameter always contains an uncertainty ( $u$ ). So lets state the following model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m = p + u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of every calibration is to specify the uncertainty. Strictly spoken the uncertainty can only be specified for the calibration which was already done. However, a very reasonable assumption is, that the uncertainty will not change dramatically. But, it will change over time. So the model will look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m = p + u(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your proposed redundancy has different effects on the uncertainty, depending on the way redundancy is achieved. One way could be to measure the same process parameter with one additional sensor of the same type (1). Another way would be to either use a different sensor (2) or even observe a different process parameter with a different sensor (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just using the same sensor to observe the same parameter will most likely have a similar degradation of the calibration quality. Because the cause of the degradation is not changed, so both sensor signals have possibly a higher uncertainty over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two other options have a potential to come up with an uncertainty-model which would not make the calibration un-necessary but might increase re-calibration periods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part II:&lt;/strong&gt;&lt;br&gt;&#xA;In any case there is a significant &lt;strong&gt;difference between error and uncertainty&lt;/strong&gt;. If it would be possible to &lt;em&gt;know&lt;/em&gt; the error of every measurement one could easily compensate for that. &lt;br&gt;&#xA;But the only thing a calibration can provide is the uncertainty of a measurement. Or put it the other way around a calibration will provide you with an estimate how certain one can be that in (usually) 95% of all measurements the returned value will be in the specified interval around the true value.&#xA;The following figure shows the upper and lower bounds of the uncertainty assuming a very simple degradation model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zj1wr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zj1wr.png&quot; alt=&quot;uncertainty bounds&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you want to keep your process parameter at a level of &quot;1&quot; and your uncertainty has an initial value of +/-10% (which is huge but nicer to illustrate). The reason you have to control parameter in the first place is because you need to prove that your process (or product) has a specific quality. As it can be easily be seen without re-calibration (bounds m+ and m-, dotted) after some imaginary time the uncertainty has steadily increased. By re-calibrating in the middle of the interval you reduce the uncertainty (m+r and m-r, solid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your description I assume that you need to report or save the calibration before and after every period together with the measurement-logs of the process to prove your quality of the whole production cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is absolutely understandable to reduce the effort and associated costs of recalibration. The only way to do it is do gather more knowledge of this specific process. &lt;br&gt;&#xA;Right now the brute-force approach is to calibrate two times and assume the uncertainty in between those calibrations was within in the two calibrations.&lt;br&gt;&#xA;If it would be possible to obtain a better model for the uncertainty degradation the number of calibrations could be reduced. However, in order to have a better model it is necessary to combine a larger number of calibrations. One possible way might be to use all the calibrations and recalibration available. It might be possible to develop a better understanding of the uncertainty and thereby develop a better model which would extend the calibration intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example given the sensors would degrade in the manner depicted in the figure above, a solution could be to change the control parameters of the production system control over time. But the way to do this or if this is even possible highly depends on your process which we don't know yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, even though it is counterintuitive having redundant sensors does not necessarily decrease uncertainty. &lt;br&gt;&#xA;By adding an additional sensor of the same type to the system you basically double the number of measurements keeping the same uncertainty. The only way a redundant sensor reduces uncertainty is when the first sensor measurement was not representative of the system and did not only have uncertainty from the sensor but additional fluctuations from the system itself. Think of a measurement like shooting at a target. Depending on your skills you will have a certain probability of hitting the bulls eye. That means by shooting more often you only increase of hitting the bulls eye but you never decrease the spread of your shots. Every shot has the same probability of hitting the bulls eye. Likewise measuring more often increases the chance of measuring the &quot;right&quot; value, but this is not of interest here. It is of interest how certain you can be that every measured value is within a certain &quot;interval around&quot; the right value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summing it up:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Adding sensors to a measurement does not necessarily decrease the uncertainty of the combined measurement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Having two sensors of the same type will not change the degradation rate of the combined measurement value compared to a single sensor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the possibility is big that one sensor would return erroneous values during the production run, than having redundant sensors makes a lot of sense. &lt;strong&gt;But&lt;/strong&gt;, this has nothing to do with the uncertainty of the values. In this case the post-calibration would reveal that the process was not properly monitored due to sensor failure so there is not way to assure the quality of the production run at all. In case of sensor failure the other sensor would have kicked in and provided (at best) the same quality as if one sensor would have survived the whole production run.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The only way to reduce the number of calibrations is to collect all calibrations from the past and try to build a sensor-degradation-model from that which might show that the current number of calibrations is unnecessarily high.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By utilising different sensor types or measurement of process parameters it might be possible to come up with a uncertainty model which degrades slower and is more robust with respect to sensor failure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2015-10-19T11:04:08.827" LastActivityDate="2015-10-19T11:04:08.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5832" PostTypeId="1" AcceptedAnswerId="5847" CreationDate="2015-10-18T23:23:20.613" Score="1" ViewCount="220" Body="&lt;p&gt;What does trip angle mean in the context of &lt;a href=&quot;http://www.pqcontrols.com/m410.php&quot; rel=&quot;nofollow&quot;&gt;slope  / level sensors / inclinometers&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="3412" LastEditorUserId="33" LastEditDate="2015-10-20T01:33:53.173" LastActivityDate="2015-10-20T01:33:53.173" Title="What is the meaning of a &quot;trip angle&quot; in a level sensor?" Tags="|sensors|terminology|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5833" PostTypeId="1" CreationDate="2015-10-19T08:38:41.287" Score="-3" ViewCount="207" Body="&lt;p&gt;Combustible dust is known as an explosion hazard, especially in large factories etc. They can create powerful explosions and bring down entire buildings if it is disturbed by a large enough vibration and then sparked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My theory is that by drying out and grinding up trash, especially if done at the nano level for further efficiency, and with proper filtration and machine lubrication the energy could be harnessed, and the fuel doesn't need anything but a spark to ignite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trash is readily available and even what some would consider in the way considering it just sits in land fills emitting poisonous gas and possibly radiation in later stages. Note: This paragraph is obsolete and trash is already incinerated regularly as a means of energy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Try this at home. Very slowly and incrementally pour some coffee creamer and light the trail. Anything is highly flammable and creates a somewhat explosive reaction, however small, it will engulf the entire stream vertically upward, creating a way to not only burn the fuel but the air around it. I would compare that to the electron duplicating characteristics of a mirror.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also general nanotechnology is in the experimental stages. To draw an unrelated idea, the friction reduction capabilities of dry lithium come from something called a basal plane, in crystal structures consisting of edges and even gaseous or liquid bearing cavities. A good example would be the bottom of a soda can. In a small puddle it closes the cavity and creates buoyancy allowing it to glide like a hovercraft. You could even use the nanodust theoretically in a synthetic nano crystal, giving you a low friction, leak free means of lubrication with no grease buildup, breakdown or dust problems.&lt;/p&gt;&#xA;" OwnerUserId="3416" LastEditorUserId="3416" LastEditDate="2015-10-19T16:33:58.787" LastActivityDate="2015-10-19T17:49:33.673" Title="Why isn't trash ground into dust and used as fuel?" Tags="|energy-efficiency|energy|" AnswerCount="1" CommentCount="3" ClosedDate="2015-10-19T17:50:02.117" ContentLicense="CC BY-SA 3.0" />
  <row Id="5834" PostTypeId="1" CreationDate="2015-10-19T10:42:18.530" Score="2" ViewCount="187" Body="&lt;p&gt;I have to mount a glass fibre woven mat onto a stainless steel foil and using staples seems to be the only way to do it. The environment is one where a large variety of solvent vapors will be encountered, and as far as I know no adhesive could survive all of them: alkanes, chlorinated hydrocarbons, alcohols, ketones, ethers...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A regular paper stapler works fine, except for the staples which tend to be galvanized steel at best. Stainless steel staples would seem to be an option, but they don't seem to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone can suggest a better approach?&lt;/p&gt;&#xA;" OwnerUserId="3417" LastEditorUserId="33" LastEditDate="2015-10-20T00:52:35.040" LastActivityDate="2015-10-20T09:31:34.890" Title="Alternatives to stainless steel staples" Tags="|steel|fasteners|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5835" PostTypeId="1" CreationDate="2015-10-19T11:13:09.160" Score="3" ViewCount="457" Body="&lt;p&gt;I'm very much new to pneumatics and I was wondering if there was such a thing as a reversible pneumatic actuator? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm designing a system that has to be lightweight, pneumatic and use a mechanical input to store energy in a pneumatic manner. The system must then be able to translate this energy back to kinetic energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my research I have found that there are no pneumatic motors that can be back-driven to run as a compressor, and the addition of a compressor adds weight to the system so I'm trying to avoid this. Essentially, I was wondering if I used a generator/motor to convert the mechanical input to electrical energy, would an actuator have the ability to convert electrical energy into a source of potential energy?&lt;/p&gt;&#xA;" OwnerUserId="3269" LastEditorDisplayName="user16" LastEditDate="2015-10-19T17:47:17.047" LastActivityDate="2015-10-21T11:45:40.923" Title="Pneumatic Actuators - System Reversibility. Is it possible to use an actuator to store energy?" Tags="|mechanical-engineering|motors|compressed-air|energy-storage|actuator|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5837" PostTypeId="2" ParentId="5833" CreationDate="2015-10-19T11:51:19.627" Score="5" Body="&lt;p&gt;Combustible dust &lt;strong&gt;&lt;em&gt;is&lt;/em&gt;&lt;/strong&gt; used as a source of fuel, just not exactly in the scenario that you envisioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coal-fired power plants first grind the chunks of coal into powder that's small enough to be injected into the burners.  There's a considerable amount of &lt;strike&gt;alchemy&lt;/strike&gt; chemistry involved based upon the composition of the coal.  Grinding the coal prior to burning provides a more consistent size and a greater surface area which results in a more easily regulated and controlled combustion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trash is sometimes used as a fuel source, but not as frequently as it once was.  The biggest challenge here is controlling the amount of pollutants that go out as part of the exhaust.  To my knowledge, an additional challenge in burning trash is a relatively low energy yield combined with a comparatively higher pollutant profile.  More pollutants mean greater amounts of scrubbing and therefore a lower yield from the power plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding grinding to trash to nano scale, I suspect that doing so is well past the point of reasonable returns.  It will take more energy to process to this level than what you'll see in return post-combustion.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-10-19T17:49:33.673" LastActivityDate="2015-10-19T17:49:33.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5838" PostTypeId="1" AcceptedAnswerId="5844" CreationDate="2015-10-19T12:54:08.443" Score="6" ViewCount="1029" Body="&lt;p&gt;I'm making a remote starter for a car using a Raspberry Pi.  The problem is that the car has a manual transmission.  The driver needs to remember to place the gear shift in neutral when parking, otherwise all kinds of bad stuff could happen when the car is started in-gear.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What possible solutions do I have for making an interlock that prevents the car from starting in gear?  It's a 2001 Honda Accord, if that helps.  I have seen transmissions with sensors (switches) on every gear, but this transmission does not have those.  I thought of monitoring the speedometer sensor, but I think that info would come too late.  I'm thinking I might have to put one or more switches on the shift lever to detect its position, but this sounds like a lot of work.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2023-02-11T21:01:06.080" Title="Remote start on a car with manual transmission" Tags="|electrical-engineering|automotive-engineering|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5839" PostTypeId="2" ParentId="5835" CreationDate="2015-10-19T13:54:24.690" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gae1W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gae1W.png&quot; alt=&quot;Air Storage&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A,B,C, and D are normally closed solenoid valves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The valves with the arrows above them are pneumatic check valves, such as reed valves, with the arrow indicating flow direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, with no power, pushing the piston in and out of the air cylinder causes an inlet check valve to open and a check valve to the flask to open and you act to pressurize the flask. This is really a mechanical analog of an H-bridge used in rectifying AC power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To actuate, operate A&amp;amp;B or C&amp;amp;D as a pair; one valve (A or D) aligns high pressure (flask) air to the piston while the other valve (B or C) aligns the exhaust port back to the exhaust/inlet cap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that it will get harder to manually/externally stroke the air cylinder as pressure builds in the flask, and that your actuation ability is limited by whatever pressure is in the flask. You can cap pressure in the flask with a relief valve set to your desired pressure. I would definitely use a second relief valve on any pressure vessel for safety, one that isn't designed to operate regularly (i.e., at a higher pressure than your target flask pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;:EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically yes, the exhaust port in a pneumatic system discharges to atmosphere (through a &lt;a href=&quot;https://www.google.com/search?q=pneumatic+muffler&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;muffler&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're familiar with electronics, consider the following analogy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QpASi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QpASi.png&quot; alt=&quot;Diode check valve analogy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A check valve and a diode can both be &quot;forward biased&quot; and allow &quot;flow&quot;, or &quot;reverse biased&quot; and not allow &quot;flow&quot;. In valves, &quot;flow&quot; is volumetric flow; in diodes, &quot;flow&quot; is electron flow (current). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both instances, there's some very small &quot;bias&quot; required to put the device in operation; in a diode this is a bias voltage (~0.6V) required to allow current to flow; in a check valve this is essentially a &quot;bias pressure&quot; (very small), which is required to hold open the valve internals to allow fluid flow across the valve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, diodes and check valves, if you apply &quot;pressure&quot; (literal pressure or voltage) the wrong way, you are &quot;reverse biasing&quot; the device and you will get no flow across it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, keeping that in mind, I made several copies of the schematic with high and low pressures drawing in orange/blue, respectively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Afjrd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Afjrd.png&quot; alt=&quot;Flow paths&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the (numbers) in the center of the image. Paths (1) and (2) show how pressures exist in the system when you are &lt;em&gt;driving&lt;/em&gt; the cylinder in or out; (3) and (4) show how pressures exist when you are &lt;em&gt;actuating&lt;/em&gt; the cylinder in or out. Now I'll walk you through each flow path. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.automationdirect.com/adc/Overview/Catalog/Pneumatic_Components/Pneumatic_Valves_-a-_Accessories/General_Purpose_Solenoid_Directional_Control_Air_Valves&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pneumatic solenoid valves&lt;/a&gt; A,B,C,D are &quot;normally closed&quot; and are not energized, so they do not allow flow.&lt;/li&gt;&#xA;&lt;li&gt;Because the piston is being driven in, the top side of the &lt;a href=&quot;https://www.youtube.com/watch?v=TZ5_8H4Bkho&quot; rel=&quot;nofollow noreferrer&quot;&gt;cylinder disc&lt;/a&gt; is trying to &quot;suck in&quot; air (at a low pressure), while the bottom side is trying to &quot;push&quot; air (at a higher pressure). &lt;/li&gt;&#xA;&lt;li&gt;The pressures in the piston ports forward bias the top and bottom check valves and reverse bias the middle two check valves. This lines up exhaust/inlet air to the low pressure side of the piston and aligns the flask to the high pressure side of the piston. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(2):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The same general steps as with (1), but in reverse. Here the middle 2 check valves are forward biased and the top and bottom check valves are reverse biased. &lt;/li&gt;&#xA;&lt;li&gt;Flow is still lined up between the low pressure side of the piston and exhaust/inlet are and between the high pressure side and the flask. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(3):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In this scenario the cylinder is being actuated. Solenoid valves A&amp;amp;B are energized, lining up flask air to one side of the piston and exhaust/inlet air to the other side of the piston. &lt;/li&gt;&#xA;&lt;li&gt;This differential pressure across the cylinder disc acts to pull the rod in. &lt;/li&gt;&#xA;&lt;li&gt;Note here that the check valves are all either reverse biased or have the &lt;em&gt;same&lt;/em&gt; pressure on both sides. In scenarios (1) and (2), the pressure in the flask is always $P_{\mbox{flask}} = P_{\mbox{Piston}} - P_{\mbox{bias}}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(4):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Again, the same as (3) but in reverse. Solenoid valves C&amp;amp;D are energized, lining up the high/low pressure sides appropriately. &lt;/li&gt;&#xA;&lt;li&gt;The differential pressure across the cylinder acts to push the rod out. &lt;/li&gt;&#xA;&lt;li&gt;Again, the check valves are either reverse biased or not biased at all. There is no flow across any of the check valves. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2015-10-21T11:45:40.923" LastActivityDate="2015-10-21T11:45:40.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5840" PostTypeId="1" AcceptedAnswerId="5842" CreationDate="2015-10-19T14:59:08.427" Score="3" ViewCount="1691" Body="&lt;p&gt;What's the mechanism behind disposable syringes jamming? How do you prevent this from happening or reverse the effect? They are a lot cheaper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that once the syringe is pressed all the way in, the plunger sticks and can not be pulled back out.&lt;/p&gt;&#xA;" OwnerDisplayName="user3421" LastEditorUserId="33" LastEditDate="2015-10-20T01:02:20.840" LastActivityDate="2015-10-20T01:02:20.840" Title="What is the purpose and cause of disposable syringes being single use only?" Tags="|mechanical-engineering|medical-devices|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5841" PostTypeId="2" ParentId="2285" CreationDate="2015-10-19T16:47:05.987" Score="4" Body="&lt;p&gt;SF. explained nicely &lt;em&gt;how&lt;/em&gt; the rifle achieves this behavior, but the question remains as to &lt;em&gt;why&lt;/em&gt; you would design a rifle to do this.  When firing an automatic weapon, the accuracy decreases very quickly (after a few rounds are fired) because the recoil causes the weapon to jump around.  What the AN-94 does is delay the recoil of the first round from reaching the shooter's body until just after the second round is fired.  This makes it possible to get two rounds on-target, both with first-round accuracy (or nearly so).  There are other automatic weapons out there that do this; the generic term is &quot;hyperburst&quot; I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the recoil has to eventually be absorbed by the shooter, so you can only get a few rounds off with this method.  After that you need to reduce the rate of fire to some sustainable level so that the shooter can control the weapon.  That's why the AN-94 reduces its rate of fire after two rounds.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-10-19T16:47:05.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5842" PostTypeId="2" ParentId="5840" CreationDate="2015-10-19T17:38:37.100" Score="7" Body="&lt;p&gt;There's a story on them &lt;a href=&quot;http://www.pbs.org/newshour/rundown/self-destructing-syringes-force-safer-injection-practices/&quot; rel=&quot;noreferrer&quot;&gt;at PBS&lt;/a&gt;. From the article:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A nurse injects a patient with a syringe of antibiotics, reloads and moves on to the next patient in line. The syringe isn’t sterilized, the needle is not replaced, and the patient is at risk of contracting a disease from the very shot that is supposed to cure him.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;About 40 percent of all injections are given with unsterilized, reused syringes and needles, reports the World Health Organization. An estimated 1.3 million deaths — and 21.7 million new Hepatitis B infections — occur each year as a result of the unsafe practice.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The devices, called auto-disable syringes, jam up after one injection and can’t be reused. The casing is engineered to break if someone tries to forcefully refill it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Regular disposable syringes can also make their way back into the marketplace after being used. People will take syringes from a health clinic’s trash, wash them and repackage them for resale.&lt;/p&gt;&#xA;&lt;p&gt;“You can find in a market a perfectly well packed syringe, then you look closely and you can still see some blood in the syringe from the last person,” he said.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The article continues with all the reasons you can think of for not wanting to re-use a syringe.&lt;/p&gt;&#xA;&lt;p&gt;The syringe appears to be invented by a company called &lt;a href=&quot;http://www.starsyringe.com/product-k1.htm&quot; rel=&quot;noreferrer&quot;&gt;&amp;quot;Star Syringe&amp;quot;&lt;/a&gt;, who has licensed their intellectual property to 13 manufacturers. From their website:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A small ring etched on the inside of the barrel allows the specifically adapted plunger to move in one direction and not the other. After one complete injection is given the plunger will automatically lock in place and break if forced, thus rendering the syringe useless.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The following graphic is from &lt;a href=&quot;http://www.zdnet.com/article/this-syringe-will-now-self-destruct/&quot; rel=&quot;noreferrer&quot;&gt;ZDNet&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8DJPM.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8DJPM.png&quot; alt=&quot;Auto-disable syringe&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Finally, there's a video on YouTube that &lt;a href=&quot;https://www.youtube.com/watch?&amp;amp;v=jmsyuPUS-EM&quot; rel=&quot;noreferrer&quot;&gt;shows the operation of the syringe&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-10-19T19:42:16.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5843" PostTypeId="2" ParentId="5835" CreationDate="2015-10-19T18:36:48.700" Score="1" Body="&lt;p&gt;A Variable Displacement Axial Piston Motor could be used is this manner. It also has the advantage of also being an adjustable positive displacement pump. So you could have it set at maximum displacement while charging the system, and then set to zero displacement while waiting to discharge the system (no valves), and at low displacement wile discharging.&lt;/p&gt;&#xA;" OwnerUserId="3014" LastEditorUserId="3014" LastEditDate="2015-10-20T03:16:43.757" LastActivityDate="2015-10-20T03:16:43.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5844" PostTypeId="2" ParentId="5838" CreationDate="2015-10-19T20:39:22.210" Score="4" Body="&lt;p&gt;I can't think of a more elegant way than adding a sensor or sensors either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An inductive sensor may be a good option for a switch if you have enough room below and the stick has metal in it. If not, you could epoxy a magnet to the hidden part of the shifting stick and use a hall effect sensor (or reed switch but I don't recommend it). A roller style limit switch may also be an option, but I would recommend going non-contact if possible.&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B008MU1GEY&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/gp/product/B008MU1GEY&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.mouser.com/ProductDetail/Littelfuse/55100-3H-02-A/?qs=nyo4TFax6Nff4PypTg%2FOjg%3D%3D&quot; rel=&quot;nofollow&quot;&gt;http://www.mouser.com/ProductDetail/Littelfuse/55100-3H-02-A/?qs=nyo4TFax6Nff4PypTg%2FOjg%3D%3D&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to include the parking break in the interlock system. It would be a little insurance if the sensor fails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An accelerometer would also be some good insurance. If it feels a kick when it starts the engine it immediately stops and prevents further input until the car is manually started. Even though this response will be very fast, the car could still creep forward 6 inches or more so it would still be a safety issue and couldn't be the primary interlock system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as previously mentioned in the comments you will want to make sure that if you are bypassing the clutch interlock you make sure that it is still enforced when starting the car manually.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-19T20:39:22.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5845" PostTypeId="2" ParentId="5802" CreationDate="2015-10-20T01:06:24.093" Score="2" Body="&lt;p&gt;Pelton, Kaplan, and Francis turbines are designed for non-compressible fluids (aka water not air). While they will certainly spin, they would be very inefficient and not what you are after. You will want to use a pneumatic motor or exhaust turbine like Dave Tweed recommended in the comments. Read the last paragraph of the function section for reference:&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pelton_wheel#Function&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Pelton_wheel#Function&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pneumatic is generally not a good way to transfer power. Most farm implements use PTO mechanical rotation or hydraulic power. Hydraulics have many benefits over pneumatic; efficiency, noise, power to weight and more reliable in dirty environments. The only benefits of pneumatics that I see are no return loop plumbing and less risk of oil contaminating product (still some risk as most pneumatic systems need oil). I would even consider generating electrical power off the PTO and using electric motors before considering pneumatic.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-20T01:06:24.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5846" PostTypeId="2" ParentId="5807" CreationDate="2015-10-20T01:26:35.813" Score="1" Body="&lt;p&gt;My first thought would be to fill and purge the sealed container with an inert gas like nitrogen or argon. Without oxygen present, no ozone will be created. Additional details regarding the origin of the ozone may be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CFC molecules break down ozone, but are consequently regulated because of it.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Chlorofluorocarbon#Regulation&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Chlorofluorocarbon#Regulation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-20T01:26:35.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5847" PostTypeId="2" ParentId="5832" CreationDate="2015-10-20T01:31:12.343" Score="1" Body="&lt;p&gt;From the link that you provided, it would seem that trip angle is a preset angle at which the sensor will &quot;trip&quot; or &quot;switch on&quot;. The results of this can depend on purpose of the tilt meter. It can be wired to sound an alarm, turn on a light, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this &lt;a href=&quot;http://www.spectronsensors.com/appsheets/SAN-218-2109.pdf&quot; rel=&quot;nofollow&quot;&gt;spec sheet&lt;/a&gt; on another tile switch, the trip angle can be adjustable.  The spec sheet also gives an example circuit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another &lt;a href=&quot;http://www.pqcontrols.com/m420.php&quot; rel=&quot;nofollow&quot;&gt;sensor on the site&lt;/a&gt; that you linked gives possible uses.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-10-20T01:31:12.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5848" PostTypeId="2" ParentId="5798" CreationDate="2015-10-20T02:01:12.567" Score="5" Body="&lt;p&gt;I think Failmond's first post does a solid job with the analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a voltage potential across a resistor(the air). You can not remove air from the equation as it is what is sourcing and sinking the voltage potential in question. The maximum energy you can harness is what it is already sinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of power harnessable is indeed very low, but if I were to build  a theoretical device to harness this power it would involve high altitude balloon system with lots of surface area. High altitude would mean high potential and higher the surface area the higher the current. It would be tethered with a light weight insulated conductor. The conductor could be very small gauge wire because of the very very small current, but the insulation would be very thick and have a high dielectric strength. Once the power made it to the ground it would connect to one leg of a specialized high voltage dcdc converter involving a spark gap and a transformer to convert the power down low enough to be useful. The other leg would be connected to an array of small copper wires going across the ground in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, its possible, but at the end of the day you would produce a continuous 100 watts of power and have incurred millions/billions of dollars in expense. You have also created a really good lightning rod which would destroy the system unless even more expensive measures were taken. On a similar note, you can not harness power from lightning because it will destroy any reasonably sized system. And an unreasonably sized system, sized to handle 99% of lightning strikes without failure, would be many many many times over non-cost effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its not reasonable to harness this knowledge for power production. However, in the future it could possibly be used for sensors much like an electroreceptive fish.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Electroreception#Electrolocation&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Electroreception#Electrolocation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-20T02:01:12.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5850" PostTypeId="2" ParentId="3301" CreationDate="2015-10-20T03:06:54.120" Score="1" Body="&lt;p&gt;In addition to the other answers, Search google for &quot;threaded tube bung&quot; or &quot;threaded bung&quot; you can find all sorts of sizes/styles and threads. They are typically used in conjunction with tube-type fabrication/weldments. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2015-10-20T03:06:54.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5851" PostTypeId="1" CreationDate="2015-10-20T03:33:49.607" Score="2" ViewCount="295" Body="&lt;p&gt;Looking into purchasing a drone. One has a dedicated 5-channel multispectral camera (Micasense RedEdge) the other has a modified Sony RGB camera that resolves the NIR spectrum. Can someone sum up the pros and cons of each of the above systems? &#xA;Can one filter spectral data taken with a modified camera ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the manufacturer's description of the modified camera:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We specifically modify our cameras internally and externally to ensure we have technically appropriate 3-band data for either&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;NIR vegetation mapping - To use NIR mapping effectively, one&#xA;processes the imagery via algorithms such as NDVI (Normalised&#xA;Difference Vegetation Index), which is a ratio of the NIR band and&#xA;other visible bands.&lt;/li&gt;&#xA;&lt;li&gt;Natural colour mapping (RGB)&lt;/li&gt;&#xA;&lt;li&gt;Underwater mapping - designed specifically for water penetration&#xA;and underwater bathymetric mapping&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All with the SAME modified camera, but simply screwing the appropriate filter onto the lens (we supply the 3 special filters after the camera modification).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, it suggests that the &quot;3 bands&quot; are not collected simultaneously as one needs to change filters. Am I correct?&lt;/p&gt;&#xA;" OwnerUserId="3425" LastEditorUserId="282" LastEditDate="2015-10-31T16:37:32.023" LastActivityDate="2015-10-31T16:37:32.023" Title="UAV 5-Channel Multispectral camera vs. modified RGB camera that resolves NIR" Tags="|aerospace-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5853" PostTypeId="1" CreationDate="2015-10-20T10:02:27.103" Score="3" ViewCount="2714" Body="&lt;p&gt;I have done some research and asked in a few forums for gold plating on anodized aluminum  and stainless steel but haven't yet found the answer to my questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the chemicals for pre-processing the surface and the suitable gold solution for these tasks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stainless steel I believe they use Wood's nickel for pre-processing. Is it the only or the easiest method? How about anodized aluminum?&lt;/p&gt;&#xA;" OwnerUserId="3430" LastEditorUserId="1586" LastEditDate="2015-10-23T06:15:05.430" LastActivityDate="2015-10-23T06:15:05.430" Title="How to do gold plating on Anodized aluminum and stainless steel?" Tags="|steel|chemical-engineering|aluminum|surface-preparation|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5855" PostTypeId="1" CreationDate="2015-10-20T11:48:53.277" Score="7" ViewCount="3394" Body="&lt;p&gt;I'm thinking of installing a rather large aquarium on the 4th floor of a residential high-rise which has a total of 27 floors. I plan to place it directly against a concrete column with a diameter of around 4 feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the main issues are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Overloading the slab such that it falls through the slab.&lt;/li&gt;&#xA;&lt;li&gt;Not overloading the floor to the extent that it falls through the floor but overloading the column.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Who do I need to talk to in order to find out if this is feasible before the project actually starts? The building manager, a civil engineer, or....?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load is spread over about 8.5 ft². Are high-rise buildings designed to support live loads of this type and magnitude?&lt;/p&gt;&#xA;" OwnerUserId="3432" LastEditorUserId="368" LastEditDate="2015-10-20T19:31:24.993" LastActivityDate="2015-10-20T19:49:41.993" Title="How significant is a 8,500 lb concentrated load in a large (25+ floors) residential high-rise?" Tags="|civil-engineering|structural-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5857" PostTypeId="2" ParentId="5855" CreationDate="2015-10-20T13:03:08.847" Score="9" Body="&lt;p&gt;The answer is that you will have to have an engineer look at it if you want an &amp;quot;official&amp;quot; answer.  The details below may help to show you whether or not it will be feasible.&lt;/p&gt;&#xA;&lt;h3&gt;Load&lt;/h3&gt;&#xA;&lt;p&gt;The load that you gave is &lt;span class=&quot;math-container&quot;&gt;$8,500\ \mathrm{lb}/8.5\ \mathrm{ft^2}=1000\ \mathrm{lb/ft^2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Code&lt;/h3&gt;&#xA;&lt;p&gt;Floors are designed for typical loads. In addition to the weight of the floor itself (dead Load), the floor is designed for a live load that takes the form of a uniform pressure. The exact amount used depends on the the expected use of the floor. You can see the variations from the table on page two of &lt;a href=&quot;http://www2.iccsafe.org/states/Florida2001/FL_Building1/PDFs/Chapter%2016_Structural%20Loads.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this document&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Typical apartment (residential) floor loads are &lt;span class=&quot;math-container&quot;&gt;$40\ \mathrm{lb/ft^2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h3&gt;Your Situation&lt;/h3&gt;&#xA;&lt;p&gt;As you can see from simply comparing the loads, you will be locally applying more load that the floor was designed for.&lt;/p&gt;&#xA;&lt;p&gt;As you have noted, there are some mitigating factors it your situation.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You are placing the load near a column.&lt;/li&gt;&#xA;&lt;li&gt;You aren't applying this same load to the entire floor.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-10-20T19:49:41.993" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5858" PostTypeId="1" CreationDate="2015-10-20T15:05:22.050" Score="2" ViewCount="345" Body="&lt;p&gt;Is it possible to have a helical gear and a worm gear with the teeth angled the same amount, so that the helical gear can drive the worm gear, unlike a spur gear.&lt;/p&gt;&#xA;&lt;p&gt;Can the helical gear also move the worm gear along the axle like a gear rack. until the worm gear is stopped and then it resumes driving the worm gear normally.&lt;/p&gt;&#xA;&lt;p&gt;This is to make an OR gate, I can show an example using Lego Technic gears here:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XoeQj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XoeQj.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/jreF1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jreF1.gif&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3435" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-10-20T16:39:09.357" Title="Helical - Worm Drive" Tags="|gears|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5860" PostTypeId="1" AcceptedAnswerId="5861" CreationDate="2015-10-20T21:03:42.257" Score="2" ViewCount="128" Body="&lt;p&gt;Here's my situation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dMIvO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dMIvO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to drop a payload via parachute with some accuracy.  I'm worried that in the above setup, the line will interfere with parachute deployment or - if not deployment - operation during flight (imagine wind pushing the chute into the line).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't tested this yet but plan to.  Does anyone know how the payload to anchor connection might be engineered to ensure full deployment throughout the descent? &lt;/p&gt;&#xA;" OwnerUserId="3443" LastEditorUserId="33" LastEditDate="2015-10-20T21:15:19.007" LastActivityDate="2015-10-21T10:07:56.027" Title="Guiding a parachute via anchor: Possible?" Tags="|mechanical-engineering|aerodynamics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5861" PostTypeId="2" ParentId="5860" CreationDate="2015-10-20T21:43:06.040" Score="3" Body="&lt;p&gt;As the diagram shows you may be able to use a standoff arm to maintain separation between the payload and the guide line. However friction between the standoff and the guide line may cause the two to tangle or jam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given no knowledge of the payload layout or size. You could run the guide line down the center line of the paracute, not sure if it would be better to continue down center of payload or route the line around the perimeter.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2015-10-20T21:43:06.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5863" PostTypeId="1" CreationDate="2015-10-21T05:11:17.743" Score="0" ViewCount="218" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Case 1&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The development of Case 1 of this problem in which the axes of the prisms intersect at right angles, is shown in Diagram 1. Note that interedge lines are required for the development of both solids, their positions being found by projectors drawn from the different views of the drawing to the full view of the end.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Case 2&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Case 2 of the problem in which the axes intersect at 45°, is developed in a similar manner. The projections for this are shown on the plate at Fig 1(a). In order to avoid confusion of lines, projectors are sometimesdrawn as shown on this plate, that is, only rheir starting points are indicated, as between the two views of the octagonal prism in this drawing. First, reproduce the projections. Next, develop the strechouts for the solids on the lines MN and M'N', and draw edge and interedge lines through their respective points. The positions of the interedge lines are found by projecting the point A in the plan across to the full view, as shown in (d) afterwards locating the points x and y at x' and y' in (d'); then they are projected to the elevation and carried to the development in the usual manner, the resulting figures at (b) and (c) completing the development of the solids.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What I need to know is, should both drawings be drawn in an elevation and plan way? And where exactly is the problem in both cases that should be developed?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TOBWp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TOBWp.jpg&quot; alt=&quot;Diagram 1&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/qlkDa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qlkDa.jpg&quot; alt=&quot;Diagram 2&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3418" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-07-05T16:37:06.067" Title="Two intersecting prisms (typed and with diagrams)" Tags="|design|drafting|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5864" PostTypeId="2" ParentId="5860" CreationDate="2015-10-21T10:07:56.027" Score="2" Body="&lt;p&gt;The chute should probably be shaped to pull itself away from the line like a kite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be simpler however to use the line itself as the brake instead of using a parachute.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-10-21T10:07:56.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5865" PostTypeId="1" CreationDate="2015-10-21T11:16:39.260" Score="5" ViewCount="295" Body="&lt;p&gt;Before widespread use of digital computers lofting was done using a thin flexible strip, called spline, held in place by a series of weights. This technique is a type of interpolation (curve exactly passes over sample points) using a physical analog model (minimal potential energy of a thin flexible beam).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WE9vw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WE9vw.jpg&quot; alt=&quot;Analog lofting spline&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the advent of digital computers and CAD tools, engineers developed a digital version of lofting splines, but instead of using a similar interpolation technique they chose an approximation technique by means of distant control points called &lt;a href=&quot;https://en.wikipedia.org/wiki/Non-uniform_rational_B-spline&quot; rel=&quot;nofollow noreferrer&quot;&gt;NURBS&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RmXpi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RmXpi.png&quot; alt=&quot;Digital lofting spline (NURBS)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an approximation approach you need to perform an iterative process to manually fit a curve to the desired path. In the field of graphic design this may not be a drawback, but in the field of engineering, where accuracy is a maximum, an interpolating approach is much more useful because you can easily interpolate smooth surfaces from a few sample points (analogous to use a lofting spline or a set of French curves). Why was this decision made?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I clarify that this is not a question about a particular software, but on the history of the development of CAD tools.&lt;/p&gt;&#xA;" OwnerUserId="1616" LastEditorUserId="6767" LastEditDate="2018-12-29T08:54:16.780" LastActivityDate="2018-12-29T17:42:12.723" Title="Why choose approximation over interpolation to implement digital lofting?" Tags="|mathematics|software|drafting|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="5866" PostTypeId="2" ParentId="5461" CreationDate="2015-10-21T14:55:04.440" Score="2" Body="&lt;p&gt;You may want to consider using &lt;a href=&quot;http://www.mnrubber.com/Design_Guide/6-3.html&quot; rel=&quot;nofollow&quot;&gt;quad-rings&lt;/a&gt; (sometimes called X profile o-rings) instead of o-rings.  I worked and a mechanical engineer for a packaging machinery company that created a lot of piston filler machines with stainless steel barrels and Delrin pistons.  The seals were always quad rings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;McMaster Carr has a &lt;a href=&quot;http://www.mcmaster.com/#o-rings&quot; rel=&quot;nofollow&quot;&gt;wide variety&lt;/a&gt; of o-rings available, including X profile o-rings with sizes down to 3mm ID x 6mm OD.&lt;/p&gt;&#xA;" OwnerUserId="3440" LastActivityDate="2015-10-21T14:55:04.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5867" PostTypeId="2" ParentId="5812" CreationDate="2015-10-21T15:14:04.167" Score="1" Body="&lt;p&gt;What you are considering, I think, is a system reliability problem.  From your description my impression is that these devices may or may not be connected, that they form a network and that you are looking to improve the reliability of operation of the entire system.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;System Reliability&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The solution you are seeking is complicated by the fact that the reliability of one part of your system may affect the reliability of another route within the system.  So, in addition to identifying the reliability of individual components, you will need to identify how the network is connected; where events are mutually exclusive, where the communication flow is sequential, and where it may be in parallel.  Of course if you have control of this network structure, ie you can redirect the routing of data within the system, then you may be able to optimize the network, which is one of the goals of your opening post, I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next issue of great importance is to establish whether the probability of failure of a device is independent of all other devices.  You may decide that the reliability of a device is affected by the devices around, or connected to it.  If so, this will affect the mathematics of your analysis considerably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all events are independent, then you can apply classical probability laws to your calculation.  However, if they are not independent, then you will need to apply conditional probability, or &lt;strong&gt;Bayes' Theorem&lt;/strong&gt;.  Using a computer, it is feasible apply brute-force Monte Carlo methods by assigning random variables (probability mass or density functions) to each element to model the behaviour. A widely used domain specific language for this type of analysis is the &lt;a href=&quot;https://www.r-project.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;R Language&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, take the case illustrated below that is borrowed from &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471040657&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hahn and Shapiro, Statistical Models in Engineering&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y7dBw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y7dBw.png&quot; alt=&quot;System Analysis&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We shall examine three stages in series ($S_1$, $S_2$, $S_3$) that determine the reliability of the whole system ($S$).  If we constructed a Venn diagramme, we would be looking for the area of the diagramme where the three overlap - their intersection.  We define these stages to behave as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Stage 1&lt;/strong&gt;:  Events &lt;strong&gt;are not mutually exclusive but are independent&lt;/strong&gt;.  Survival depends upon the survival of A and/or B.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Stage 2&lt;/strong&gt;:  Survival depends upon the survival of C only.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Stage 3&lt;/strong&gt;:  Events &lt;strong&gt;are not mutually exclusive but are independent&lt;/strong&gt;.  Survival depends upon the survival of D and/or E and/or F.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Where events are operating in parallel, and they &lt;strong&gt;are not mutually exclusive but are independent&lt;/strong&gt;, then the collective survival is computed according to:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Pr = \bigcup Pr_i = 1 - \prod (1 - Pr_i) $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the diagramme and equations above, we now can determine the overall system reliability:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Pr_S = Pr_1 \times Pr_2 \times Pr_3 $$&#xA;$$ Pr_1 = 1 - ( 1 - 0.9 ) \times ( 1 - 0.8 ) = 0.98 $$&#xA;$$ Pr_2 = Pr_C = 0.95 $$&#xA;$$ Pr_3 = 1 - ( 1 - 0.9 ) \times ( 1 - 0.9 ) \times ( 1 - 0.5 ) = 0.995 $$&#xA;$$ Pr_S = 0.98 \times 0.95 \times 0.995 = 0.926 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the probability of survival for the above system is 0.926.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where elements of the analysis &lt;strong&gt;are not independent&lt;/strong&gt;, then we enter the realms of conditional probability and &lt;strong&gt;Bayes' Theorem&lt;/strong&gt; must be applied.  For example if the reliability of B depends upon the reliability of A, then the probability of event B given event A is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Pr(B \mid A) = \frac{Pr(AB)}{Pr(A)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be generalized to produce Bayes' Theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Pr(A_i \mid B) = Pr(A_i) \times \frac{Pr(B \mid A_i)}{\sum Pr(B \mid A_i) \times Pr(A_i)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the $rhs$ term $Pr(A_i)$ is often identified as the &lt;strong&gt;prior probability&lt;/strong&gt; and the remaining fractional term is the factor by which the prior probability is updated as more experimental data is obtained.  For example, you could assign a prior probability to a device from a certain manufacturer based upon your belief or experience, until more data is acquired.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Monte Carlo&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The above analysis yields a single answer for the entire system.  You may prefer instead to develop a range of possible outcomes as a probability density function.  If this is desired, then instead of applying single reliability values for the analysis (as was done in the example), random variables expressed as density (continuous) or mass (discrete) functions would be assigned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The random variables are perturbed by running a large number of computer simulations with pseudo random number generators.  Each of the simulations is captured and the collection of results is used to develop an overall probability density function for the system reliability.  With this you can better engineer your system in that you would then be able to look at variance as a key factor, something that is not possible to do under the previous scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The selection of the correct representation for the random variable of interest is of great importance.  The commonly used normal (Gaussian) distribution may not be appropriate in that it permits negative values as outcomes.  Instead the variables should comply with the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^{\infty} f(x)dx $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For time to failure analyses, an exponential distribution is often used.  However, if a general frequency of failure is of interest, independent of time, then it may be possible to move between a normal distribution and log-normal distribution, as this permits compliance with the above form.  Of course the appropriate selection depends upon the nature of the events or processes within the system to be simulated.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Constraint Propagation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Another approach that may be beneficial is to model the system by considering each device (or subnet of devices) as a constraint.  This is an inference net and the goal is to propagate values through interconnected logic constraints that obey the laws of probability.  Instead of yielding a  single value, as offered by the first scenario, or yielding a distribution of possible outcomes, as in the second scenario, all inputs and outputs are intervals over the range 0 to 1. This will result in an interval bounding the expected reliability, for example [0.92,1.0].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technique is formally identified as &lt;strong&gt;constraint propagation&lt;/strong&gt; by Winston and Horn and was originated by &lt;em&gt;J. R. Quinlan, INFERNO: a cautious approach to uncertain inference, Comput. J. 26 (1983) 255-269&lt;/em&gt;.  If you are considering applying this technique then I would strongly advise reading Quinlan's paper, who was a researcher at the RAND Institute at the time this was published.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excellent review is found within Chapter 23 of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201083191&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lisp 3rd Edtion, Winston and Horn&lt;/a&gt; and Chapter 3 of &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0201533774&quot; rel=&quot;nofollow noreferrer&quot;&gt;Artificial Intelligence, Winston&lt;/a&gt;.  Within &lt;em&gt;Winston and Horn&lt;/em&gt; you will find source code that will permit you to prototype your network rather quickly, according to the probability model you are using. If you are not familiar with Lisp you can download cross-platform versions of &lt;a href=&quot;http://trac.clozure.com/ccl/wiki/ReleaseNotes/1.10&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clozure Common Lisp&lt;/a&gt; or &lt;a href=&quot;http://sourceforge.net/projects/clisp/files/&quot; rel=&quot;nofollow noreferrer&quot;&gt;CLISP&lt;/a&gt; at no cost.  This technique is ideally suited for functional programming languages and I have noticed &lt;a href=&quot;https://wiki.haskell.org/Sudoku#Constraint_Propagation_.28a_la_Norvig.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Haskell&lt;/a&gt; is used for this type of analysis too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The benefit of using constraint propagation is that you can work with uncertainly as to your assessment of the risk of failure.  This is more of a &lt;em&gt;plausibility&lt;/em&gt; approach rather than strict probability.  Each logic gate is assigned an upper and lower probability of reliability, within the range 0 to 1.  The spread between these two bounds is a measure of your certainty as to the reliability of the element of interest. These values then propagate throughout the network in accordance with the laws of probability (not simple arithmetic) and this allows you to immediately consider the impact of a sudden change of a specific node on the reliability of the entire system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The illustration below provides a schematic of an example set of constraint boxes, each with inputs, $L_i$ and $U_i$, and outputs, $O_i$.  All of the values in the network must be contained within the interval [0,1], and the $Pr_i$ values must be within the interval [$L_i$,$U_i$].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QTZDa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QTZDa.png&quot; alt=&quot;Constraint Propagation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system initializes with all values as $L_i = 0.0$ and $U_i = 1.0$.  As the nodal values are each updated, the system automatically propagates new values to all nodes in accordance with the laws of probability.  The &lt;strong&gt;Or&lt;/strong&gt; constraints consider survival if either of the inputs ($L_i$ or $U_i$) so indicate, the &lt;strong&gt;And&lt;/strong&gt; constraints consider survival only when both values so indicate (more conservative).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end result, in this case $O_C$, will yield an interval that will indicate your lower end reliability; for instance [0.92, 1.0].  If the upper end reliability were to drop below 1.0, it could portend serious network reliability problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works forward and backwards, so it is possible to consider &lt;em&gt;what-if&lt;/em&gt; scenarios, ie, what would be the effect on the entire system if a node were replaced, for example.  It should also be a possible to produce colour-coded output of the network to quickly illustrate where problems are emerging, perhaps in real time, although I have not used it for this purpose.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Closure&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;For a network with 100,000 nodes, consider breaking the network down into blocks of contiguous, like units. Model each of these in your system first, then develop beyond to look at the whole.  No doubt you will discover potential benefits very quickly under this process.      &lt;/p&gt;&#xA;" OwnerUserId="3266" LastEditorUserId="3266" LastEditDate="2015-10-22T17:08:27.410" LastActivityDate="2015-10-22T17:08:27.410" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5868" PostTypeId="1" CreationDate="2015-10-21T15:53:07.550" Score="3" ViewCount="1947" Body="&lt;p&gt;I have a binary mixture of water and methane. The system is at 2000kPa and I am trying to find the saturation temperature of the water in the mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know Tsat is a function of partial pressure, but cant find a formula. I do not know the ratio of water to methane in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do know that the system is currently at 100C. If I assumed that 100C is the Tsat for the water, is there a way to calculate the ratio of water to methane?&lt;/p&gt;&#xA;" OwnerUserId="3448" LastEditorDisplayName="user16" LastEditDate="2015-10-21T15:59:45.487" LastActivityDate="2015-10-30T19:18:29.100" Title="Saturation Temperature of Binary Mixture" Tags="|thermodynamics|chemical-engineering|temperature|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5869" PostTypeId="1" CreationDate="2015-10-21T18:55:31.343" Score="4" ViewCount="109" Body="&lt;p&gt;I designed from scratch an analog PID controller to mantain a temperature selected in the range of 60 to 80ºC with a room temp say = 20ºC.  The &quot;plant&quot; is a simple 4R7 5 W resistor heated by a PWM-controlled current (100% to 0% range).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any given setpoint, say 67ºC, I know that room temp is lower, so, if heat is not supplied, the cold source (room) will be pulling the temperature continuously down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: where should I &quot;center&quot; my PWM duty cycle excursion?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If $V_{err}$ = 0, DC = 50%? I heat the resistor with more or less current, pivoting on $V_{err}$ =0&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For $V_{err}$ =0, DC = 0% (minimum or null current)?  I heat the resistor with a varying current only if Verr is positive&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Option 2) seems somewhat reasonable to me, but, at the same time, I see that I am just controlling the heating side of this. Do I leave the room alone doing the cooling with no current supplied by me?&lt;/p&gt;&#xA;" OwnerUserId="3454" LastEditorUserId="272" LastEditDate="2015-10-21T20:00:12.240" LastActivityDate="2016-04-19T14:28:23.287" Title="Analog PID controller - settings wrt $V_{err}$ - setpoint always higher than room temp" Tags="|control-engineering|control-theory|temperature|heating-systems|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5870" PostTypeId="2" ParentId="5868" CreationDate="2015-10-21T19:28:21.617" Score="0" Body="&lt;p&gt;Water is a compressed liquid at 2000 kPa and 100C, and the methane is well above it's critical temperature, so it's essentially all gas.  From this, I would think that the liquid phase is almost entirely water, and the gas phase is almost entirely methane.  So, the presence of methane probably has very little effect on the saturation temperature of the water.  Tsat for the water is probably close to what it would be for pure water, 212C at 2000 kPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My source for water saturation temperature is &lt;a href=&quot;http://www.tlv.com/global/TI/calculator/steam-table-pressure.html&quot; rel=&quot;nofollow&quot;&gt;TLV&lt;/a&gt;, and source for methane properties is &lt;a href=&quot;http://www.engineeringtoolbox.com/methane-d_1420.html&quot; rel=&quot;nofollow&quot;&gt;Engineering Toolbox&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-10-21T19:28:21.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5871" PostTypeId="2" ParentId="5869" CreationDate="2015-10-21T23:25:06.537" Score="0" Body="&lt;p&gt;More details about the design of your PID controller would be helpful. Is this a purchased unit or one you programmed yourself on specific hardware? Assuming your PID controller is functioning correctly and that your temperature input it calibrated; it will eventually converge. Tuning and triggering based on PWM instead of discrete on/off will just affect the time it takes to converge. I am not familiar with your Verr term; if this is your feedback error; zero error should equal zero output once the PID stabilizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like your heating source is under powered for the task at hand. This could be why your setpoint is always higher than room temp. Your controlled input(your resistive heater) must be greater than your loss at the target temperature(maybe 100W at 67C). Im not sure what &quot;room&quot; constitutes in your question, but in a small 20ft x 20ft insulated room, a 5 watt resistor can be on 24 hours a day and make little difference in the temperature. If you instead hook up a relay (may require modifying your algorithm to be discrete) and a 500+ watt heater, the PID now has excess heating capacity and can do its job by regulating the duty cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PWM is probably overkill for a simple application like room heating; as it typically does not have lots of dynamic loading. Once your heating source is sized correctly you can use power mosfets to accomplish this if necessary. If you attempt this on a forced convection heating device, make sure to leave the fan motor on non-PWM power.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-21T23:25:06.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5872" PostTypeId="2" ParentId="5853" CreationDate="2015-10-21T23:48:53.477" Score="2" Body="&lt;p&gt;Both materials you have selected are not good candidates for electroplating because they both protect themselves already with an oxide coating. &quot;anodized&quot; is referring to an intentional increased amount of surface oxide to protect the metal. This oxide must be removed, so you really want to ask &quot;how to gold plate aluminum&quot; not anodized aluminum. In order to plate them you have to go through an acid pickling process that removes the oxide, then nickel plating(has better adhesion), then gold plating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an expert and there are may steps involved especially for high quality finishes. I recommend asking your question &lt;a href=&quot;http://forum.caswellplating.com/forum/electroplating-questions&quot; rel=&quot;nofollow&quot;&gt;here (Caswell &gt; Electroplating Questions)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="1586" LastEditDate="2015-10-23T04:05:50.210" LastActivityDate="2015-10-23T04:05:50.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5873" PostTypeId="2" ParentId="5715" CreationDate="2015-10-22T00:04:54.797" Score="0" Body="&lt;p&gt;There was a very slight wobble of the weight as it went from &quot;standing&quot; to &quot;dropping&quot;. The standing position had no wobble, but it created a wobble when released. The solution was hold the weight freely and to control the wobble, so when allowed to drop by gravity, no additional forces occur in transition. Can crushes are then relatively predictable. Only it hurts my back a little from the extra forces and time.&lt;/p&gt;&#xA;" OwnerUserId="3308" LastActivityDate="2015-10-22T00:04:54.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5874" PostTypeId="2" ParentId="5865" CreationDate="2015-10-22T02:01:28.107" Score="2" Body="&lt;p&gt;I think your history isn't quite right here. The first widely-used CAD systems used Bezier curves, where the end points of each section &lt;em&gt;do&lt;/em&gt; lie on the curve. For cubic Bezier curves, the intermediate points also lie on the &lt;em&gt;tangent lines&lt;/em&gt; at the end of each segment. This corresponds quite closely to the constraints that are implied by using a physical drafting spline as in your image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bezier curves are still used - for example to define the character shapes (glyphs) in digital fonts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step was the realization that there are other mathematical ways to express the &lt;em&gt;same&lt;/em&gt; curve shapes as Bezier curves, which are more convenient for some purposes. A commonly used alternative is B-splines, which can represent everything that Bezier curves can do, plus the capability of having &lt;em&gt;discontinuities&lt;/em&gt; of slope (i.e. &quot;kinks&quot;) along the length of the curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is valuable in CAD software because there is no difference in the math required to handle a polygon made of straight line segments, or a smooth curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But B-splines were not the complete solution to the problem of computer geometry, because they can only represent &lt;em&gt;approximations&lt;/em&gt; to circles, and engineers just love circles!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nurbs (Non-uniform Rational B-splines) were the next level of generalization. The key feature of Nurbs is that they can represent curves such as circles and ellipses &lt;em&gt;exactly.&lt;/em&gt; The math required to work with Nurbs is more complicated than for B-splines, but you can write &quot;real-world&quot; CAD software that uses Nurbs to define &lt;em&gt;every&lt;/em&gt; geometrical object. Since there are now well-established software libraries to do the math that underlines Nurbs geometry, you don't need to understand the math in detail to be able to &lt;em&gt;use&lt;/em&gt; it in developing new CAD application software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;International standards for defining and transferring CAD geometry between different computer systems, like IGES and STEP, use Nurbs geometry as their basic mathematical foundation.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/B-spline&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/B-spline&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/B%C3%A9zier_curve&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/B%C3%A9zier_curve&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Non-uniform_rational_B-spline&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Non-uniform_rational_B-spline&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Historical timeline:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bezier curves (or their equivalent, Bernstein polynomials): first defined in 1912, first use in computer graphics in the 1960s.&lt;/li&gt;&#xA;&lt;li&gt;B-splines: The math was actually discovered in the 19th century, but efficient computer algorithms date from the 1970s.&lt;/li&gt;&#xA;&lt;li&gt;Nurbs - first computer implementation in 1989, first practical CAD systems in the 1990s.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other types of splines have been proposed, including &quot;minimum-energy splines&quot; which model the behaviour of an &lt;em&gt;idealized&lt;/em&gt; physical spline as illustrated by the OP), &quot;minimum-curvature-variation splines&quot;, B-splines with an extra &quot;tension&quot; parameter to allow the user to control the amount of curvature at different parts of the curve, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, there is no universally accepted objective criterion for judging the &quot;best&quot; spline. This &lt;a href=&quot;http://levien.com/phd/LevienSequinCAD09_014.pdf&quot; rel=&quot;nofollow&quot;&gt;http://levien.com/phd/LevienSequinCAD09_014.pdf&lt;/a&gt; discusses some of the issues, and includes the results of a survey in which the minimum-energy spline fit was &lt;em&gt;not&lt;/em&gt; subjectively judged to be the &quot;best&quot; fitted curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since Nurbs or B-splines can approximate &lt;em&gt;any&lt;/em&gt; spline shape to any required degree of accuracy, the debate about what is the &quot;best&quot; shape is perhaps not of critical importance - you can learn how to use whichever type of spline your CAD system provides to produce the result that you (personally and subjectively) like best. An important practical issue is the &quot;stability&quot; of the spline fit, i.e. whether a small changes in the control points can produce undesirable large and non-local changes in the spline shape.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2015-10-22T17:22:20.540" LastActivityDate="2015-10-22T17:22:20.540" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5875" PostTypeId="1" CreationDate="2015-10-22T06:12:30.477" Score="1" ViewCount="357" Body="&lt;p&gt;A naive question, yes, I know, but it's something that has plagued me for a long time. Whether by secreting a substance, controlled vibrations, metamaterials, or powered systems, is it possible to make something that would otherwise sink, not sink, preferably to an extent where (possibly with extra wide soles) boots may allow walking on water, even if only for a short period of time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a hypothetical contemplation, so anything goes, including things that react explosively or otherwise violently with water, high-tech materials, experimental tech, etc.&lt;/p&gt;&#xA;" OwnerUserId="3136" LastEditorUserId="3136" LastEditDate="2015-10-22T09:27:52.677" LastActivityDate="2015-10-22T11:56:26.557" Title="Is it possible to engineer water-walking boots?" Tags="|fluid-mechanics|control-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5876" PostTypeId="2" ParentId="5875" CreationDate="2015-10-22T11:56:26.557" Score="8" Body="&lt;p&gt;There are some possibilities, but none of them sound practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, let's take a look at ordinary displacement.  Let's say the boots need to hold up 200 pounds, so 200 pints of water need to be displaced, or 25 gallons.  That's 12.5 gallons per boot just to hold you up with both feet in the water.  It's 25 gallons per boot for &quot;walking&quot; where all the weight must be supported by a single foot regularly.  25 gallons is about 95 liters, which is the volume of a cube about 460 mm on a side.  No matter how you shape that volume, you're not going to be &quot;walking&quot; with something that big attached to your shoes, let alone somehow be able to balance and not fall over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given enough power, you can in theory pump water down to create enough reaction force to hold you up.  Again though, this won't leave you with anything like a &quot;walking&quot; motion to move forwards.  You've probably seen pictures of people hovering over water with a sort of &quot;jet pack&quot; that squirts water down at high volume and high speed.  These obviously work, but arranging that at your feet will be tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason &quot;walking&quot; just won't work on water is that walking relies on friction with the ground to move laterally and to maintain balance.  Think how differently you walk just on ice, and that motion is still relying on the little friction between the bottoms of your shoes and the ice.  With water there will be no static friction at all, only viscous friction.  The human control system for walking just won't work with no static friction, and balancing will be very very difficult at best, given that the force holding you up is well below your center of mass, and you can't rely on the usual friction for sideways forces to dynamically correct for the inevitable imbalances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few more possible techniques I can think of for creating upwards force on special boots atop water, but these are pointless since anything like walking won't be possible.  With the right control system and steerable force, you could stay upright and propel yourself in the desired direction, but this is not &quot;walking&quot; by any reasonable interpretation.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-10-22T11:56:26.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5877" PostTypeId="2" ParentId="5869" CreationDate="2015-10-22T12:12:21.097" Score="1" Body="&lt;p&gt;You seem to be ignoring the I term, since you ask about &quot;centering&quot;.  You can think of the I term as automatically providing the centering value, although thinking of &quot;centering&quot; will probably lead to other misconceptions in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also say this is all done with analog electronics.  Carefully store your circuit in some archival packaging, and maybe you can find a museum that wants it.  Then go back and do it right with a microcontroller.  That way you can tweak the parameters much more simply, see what's going on and log the internal state of the PID controller more easily, and possibly apply non-linear techniques as appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That all said, have you looked at the really simple and obvious thermostat control scheme.  That will have some output ripple, but since you've given no spec on accuracy, we don't know at this point that it's not sufficient.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-10-22T12:12:21.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5878" PostTypeId="1" AcceptedAnswerId="5879" CreationDate="2015-10-22T12:34:54.370" Score="4" ViewCount="2762" Body="&lt;p&gt;I have a small vessel with a rectangular cross-section which needs to be ventilated at a relatively slow rate, ~$3\ \tfrac{\text{L}}{\text{hr}}$.  The connections to the vessel will be tubes with an approximately 0.5&quot; diameter, but the cross-section of the vessel is 4&quot; x 1&quot;.  The diagram below gives a rough idea of the cross-section (viewed from above) with the height of the chamber being about 1&quot;.  The part circled in orange is the topic of this question. The flow will be forced in from the left and will exit passively from the right (or vice versa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I couple the 0.5&quot; tube to the chamber such that the flow in the chamber is uniform across the entire cross-section for the entire length of the duct?  In addition, how can I couple the two in the smallest amount of space possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JkH0K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JkH0K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-10-22T15:49:26.743" LastActivityDate="2015-10-22T20:11:22.960" Title="How can I create uniform laminar flow in a short duct?" Tags="|fluid-mechanics|airflow|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5879" PostTypeId="2" ParentId="5878" CreationDate="2015-10-22T19:43:04.800" Score="7" Body="&lt;p&gt;Quick and dirty CFD simulation of your problem using ANSYS Fluent 14.5:  I used a 2D duct, 8&quot; x 4&quot; with a 45-degree angle going from the inlet pipes to the main chamber.  Assuming 3 liters per hour flow through a half-inch diameter pipe gave me an inlet velocity of 6.6 mm/s.  Air enters from the left and exits through the right.  Inlet and outlet were set to a constant pressure of 101 kPa.  I used the realizable $k-\epsilon$ turbulence model with standard wall functions.  Computed velocity contours are below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mhs9o.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mhs9o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that the flow becomes very uniform by the time the angled parts meet the main duct.  It may be hard to read off the image, but the difference in velocity between points A and B is about 1 mm/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I re-ran the problem with the inlet going directly into the main duct - no expander section.  Velocity contours are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yQcwB.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yQcwB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Velocity at point A is nearly zero (a dead-spot).  Velocity at B is about 4 mm/s.  The flow through the duct becomes uniform after about 1 inch into the duct,  where I'm defining uniformity as when the velocity at the center of the duct is within 1 mm/s of the velocity near the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most interesting thing about the second case is the formation of dead spots in the corners.  The flow is fairly uniform, but the dead spots may be bad if you need good ventilation there.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-10-22T20:11:22.960" LastActivityDate="2015-10-22T20:11:22.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5880" PostTypeId="1" AcceptedAnswerId="5889" CreationDate="2015-10-22T20:53:36.683" Score="0" ViewCount="174" Body="&lt;p&gt;Let's say I have a bar attached to the rim of a rotating wheel, &lt;a href=&quot;https://i.stack.imgur.com/Zdg7G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zdg7G.png&quot; alt=&quot;sketch&quot;&gt;&lt;/a&gt;, and it's in a vacuum so we can ignore drag. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wheel is rotated, does the bar experience a lateral force (perpendicular to the bar)? What if the rotation speeds up impulsively: will the bar experience a lateral force then?&lt;/p&gt;&#xA;" OwnerUserId="3468" LastEditorUserId="3468" LastEditDate="2015-10-23T12:50:02.147" LastActivityDate="2015-10-23T13:45:41.840" Title="Does a rotating bar experience lateral force?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5881" PostTypeId="1" AcceptedAnswerId="5882" CreationDate="2015-10-22T20:57:29.887" Score="5" ViewCount="2463" Body="&lt;p&gt;More specifically, I'd like to know if unmanned rockets need to be specially designed to withstand the heat generated from air friction and shock waves during launch.  I found the diagram below on the NASA website which shows some fairly high stagnation temperatures, but the problem is that I don't know how fast a typical rocket travels on its way to, say, low earth orbit.  I also don't know what the velocity profile looks like, i.e. Mach number vs. altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the information I've found only talks about orbital velocity and reentry temperatures, neither of which is very useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qBriJ.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qBriJ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2023-11-24T22:43:58.857" Title="How hot does a rocket nose cone get during launch?" Tags="|aerospace-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5882" PostTypeId="2" ParentId="5881" CreationDate="2015-10-22T21:13:28.383" Score="3" Body="&lt;p&gt;Judging from this profile of a Delta IV launch, it doesn't appear like it gets very warm at all, since it maintains a speed of less than 3.5 km/s until it is mostly out of the &quot;thick&quot; atmosphere (~90 km). A few calculations show stagnation temperatures as below. Though these temperatures are very high, they are not achieved until the higher altitudes, where the density is very low, so the surface temperature will not likely equal the temperature in the shock layer (or boundary layer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{0,11km} \approx 265 K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{0,45km} \approx 2470K $ (but massive error due to perfect gas/isentropic assumption due to Mach number of ~6)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{0,90km} \approx 6300K$ (again, same argument as for 45 km)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nasaspaceflight.com/wp-content/uploads/2010/11/A58.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oMfJO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3015" LastEditorUserId="272" LastEditDate="2015-10-22T21:34:34.707" LastActivityDate="2015-10-22T21:34:34.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5883" PostTypeId="1" AcceptedAnswerId="5886" CreationDate="2015-10-22T23:42:29.920" Score="2" ViewCount="166" Body="&lt;p&gt;Please note, this is not a shopping question; this is a &lt;em&gt;what type of component&lt;/em&gt; works best for this scenario question.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have an induction heater circuit and a 5A max. / 24V wall wart, and I'd like to switch off / on the output of the wall wart so that I can control the power output of the induction heater.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About 7 years ago I found something by the name of SSR or SCR (or something) on digikey and that cost a reasonable amount (&amp;lt; 10 USD), and now I can't find it any longer.  The SSR's on there that can handle that much power cost around 50 USD, which is too much to design with.  The frequency can be low (5-100 Hz) range, since it &lt;em&gt;is&lt;/em&gt; a relatively slow heating process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plan to switch the relay on/off with a raspberry pi, and will probably need a transistor in between.  I don't want a mechanical relay since the thing will probably break before even making real use of the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The component I found prior was upright flat-rectangular with 4 thru-hole terminals coming out of the bottom smaller side.  I found some that look like that but like I said they were in the 50 USD range.  Also, that component if I recall correctly was for switching on/off AC for a different type of heater, so maybe &lt;em&gt;that's&lt;/em&gt; why it was cheaper?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What types of alternative components should I be considering based upon these requirements?&lt;/p&gt;&#xA;" OwnerUserId="3235" LastEditorDisplayName="user16" LastEditDate="2015-10-23T21:17:29.853" LastActivityDate="2015-10-23T21:47:10.923" Title="Identifying affordable alternatives for low-frequency high-current DC-DC solid state relays" Tags="|electrical-engineering|power-electronics|consumer-electronics|" AnswerCount="2" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5884" PostTypeId="1" CreationDate="2015-10-22T23:49:54.863" Score="3" ViewCount="1776" Body="&lt;p&gt;I am examining a high rise building project of 30 floors with a podium underneath and 2 basements. The structure in the tower and the podium varies between flat slabs, steel structure for wide spans, a huge shear wall around the tower's core, etc. The basement space has become inefficient as a parking area so I'd like to transfer this whole structure to an 8x8 or 16x16 grid of columns by using the transfer floor/slab. How can I estimate the required thickness of the transfer floor?&lt;/p&gt;&#xA;" OwnerUserId="3470" LastEditorUserId="255" LastEditDate="2015-12-16T09:29:31.780" LastActivityDate="2015-12-16T09:29:31.780" Title="How to estimate the required thickness of a transfer floor" Tags="|civil-engineering|structures|building-design|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5885" PostTypeId="1" AcceptedAnswerId="5887" CreationDate="2015-10-23T01:59:50.227" Score="4" ViewCount="466" Body="&lt;p&gt;I am in a first-year engineering design group where we are in the process of designing a cooling vest for motorcyclists for use in the city (1-2h length). Water will circulate through the vest, absorbing body heat, and then that heat will be dumped into a heat sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We initially decided on using Peltier chips for our heat sink, but when we brought this up with our instructor he was unenthusiastic (they are highly inefficient apparently). Then we decided to look into short-acting (1-2h) endothermic chemical reactions, but here ran into information overload.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of practical options are there for such a heat sink?&lt;/p&gt;&#xA;" OwnerDisplayName="user3471" LastEditorUserId="272" LastEditDate="2015-10-23T11:57:31.723" LastActivityDate="2015-10-23T11:58:47.887" Title="What are heat sink options for a circulating-water cooling vest?" Tags="|heat-transfer|design|cooling|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5886" PostTypeId="2" ParentId="5883" CreationDate="2015-10-23T05:38:09.493" Score="2" Body="&lt;p&gt;SSR or Solid State Relay is what you are after. You should be able to trigger this one right off the 3 volt rpi gpio. &lt;strike&gt;&lt;a href=&quot;http://www.aliexpress.com/item/wholesale-Industrial-FOTEK-Solid-State-Relay-SSR-40A-with-Protective-Flag-SSR-40DA-40A-DC-control/32372533315.html&quot; rel=&quot;nofollow&quot;&gt;FOTEK SSR-40DA&lt;/a&gt;&lt;/strike&gt;. Here is a DC one &lt;a href=&quot;http://www.digikey.com/product-detail/en/CPC1709J/CLA280-ND/1277135&quot; rel=&quot;nofollow&quot;&gt;CPC1709J&lt;/a&gt;. Use a voltage divider or zener diode so you dont exceed the 1.2v input. You will likely need a heat sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to consider using a mosfet instead and running higher frequency.  Higher frequency means you will transfer more power for a given coil size. Also reliability and efficiency will likely be higher than an SSR pushed to 100Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one switches low side (which is safe in your 24v application), but it is nice because it triggers right off of the 3 volt rpi gpio without a driver.&lt;a href=&quot;https://www.digikey.com/product-detail/en/RFD3055LE/RFD3055LE-ND/458758&quot; rel=&quot;nofollow&quot;&gt;RFD3055LE 60V 11A N-Channel Mosfet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a really nice mosfet wiring tutorial. &lt;a href=&quot;http://bildr.org/2012/03/rfp30n06le-arduino/&quot; rel=&quot;nofollow&quot;&gt;bildr.org mosfet tutorial&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2015-10-23T21:47:10.923" LastActivityDate="2015-10-23T21:47:10.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5887" PostTypeId="2" ParentId="5885" CreationDate="2015-10-23T06:03:03.357" Score="5" Body="&lt;p&gt;What you first need to establish is your temperatures. Body temp min and max; ambient min and max. Select the worst case. Then take some guesses as to how much heat you have to pump. Is the vest insulated on the outside or does it have to cool the ambient as well? I would blind guess that you will need about 200 watts of cooling by the time you figure your losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peltier modules are inefficient, but have a much nicer form factor than an R-134A compressor (like what is sitting under your fridge). You will have to weigh the befits of each. With the low temperature differential it will likely turn out that Peltier modules are actually more efficient than a prebuilt compressor designed for a high temperature differential. Also keep in mind that if you pay top dollar you can get a lot more efficient peltier modules than the $5 Chinese ones. Here are some nice reasonably priced ones with good specifications;  &lt;a href=&quot;http://tetech.com/peltier-thermoelectric-cooler-modules/high-performance/&quot; rel=&quot;nofollow&quot;&gt;tetech.com&lt;/a&gt;. From the COP graph in the specifications pdf you can calculate very precisely how efficient a module will be for a given temperature differential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im not sure about your application, but remember that endothermic reactions are either not sustainable (throw them away when you are done) or require recharging. Recharging requires cooling; so you are just delaying the need to have a heat pump somewhere else in the process.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="272" LastEditDate="2015-10-23T11:58:47.887" LastActivityDate="2015-10-23T11:58:47.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5888" PostTypeId="1" CreationDate="2015-10-23T09:19:53.363" Score="2" ViewCount="166" Body="&lt;p&gt;What is the general approach for calculating the heat transfer through a body? In this special case I want to calculate the $\Delta T$ (i.e. the temperature difference between the side and the top/bottom) for a given $P_{heat}$ through the following cylinder:&lt;a href=&quot;https://i.stack.imgur.com/cCcso.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cCcso.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat power is applied on the side of the outer cylinder (equally distributed), and is distributed both on the upper side and on the lower side of the cylinder. Without the trench in the cylinder I know the formula, but how does it change with the trench? (The trench is not going through the whole cylinder, only through half of it resulting in a common base).&#xA;The formula for the whole cylinder is with $S_m$ the average way from side to the base, $A_Q$ the area covered by the heating source on the side, $A_B$ the area on the bottom of the cylinder and $\lambda_{Al}$ the heat transfer capacity:&#xA;$$\dot{Q}=\frac{\lambda_{Al}\cdot(A_B-A_Q)}{S_m\cdot\ln\left(\frac{A_B}{A_Q}\right)}\cdot \Delta T$$&#xA;Btw., is that formula correct, or are there other ways to calculate that?&lt;/p&gt;&#xA;" OwnerUserId="2844" LastEditorUserId="2844" LastEditDate="2015-10-23T11:46:58.103" LastActivityDate="2015-10-23T11:46:58.103" Title="Calculate heat transfer for a partly hollow cylinder?" Tags="|heat-transfer|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="5889" PostTypeId="2" ParentId="5880" CreationDate="2015-10-23T10:46:57.830" Score="4" Body="&lt;p&gt;In a vacuum acceleration is all towards the centre of rotation. Acceleration and force are vectors in the same direction, hence with no lateral acceleration there is no lateral force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the real world it's not in a vacuum and hence there is an additional force acting on the bar, caused by drag. This drag is lateral to the bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An increase/decrease in rotational speed (assuming the impulse for this change of speed comes from a torque at the centre of rotation) would cause an angular acceleration, i.e. the acceleration would not be towards the centre of rotation. As previously mentioned, acceleration and force are vectors in the same direction, hence the angular acceleration (which is lateral to the bar if taken over an infinitely small time period) would result in a lateral force on the bar.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2015-10-23T13:45:41.840" LastActivityDate="2015-10-23T13:45:41.840" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5890" PostTypeId="2" ParentId="5883" CreationDate="2015-10-23T12:22:47.580" Score="0" Body="&lt;p&gt;If you want to use an SSR, you'll need to get one designed for switching DC. Like &lt;a href=&quot;http://www.aliexpress.com/item/solid-state-relay-SSR-25DD-25A-actually-3-32V-DC-TO-5-110V-DC-SSR-25DD/1154754963.html?spm=2114.031010208.3.41.qum1ti&amp;amp;ws_ab_test=searchweb201556_2_71_72_73_74_75,searchweb201527_4,searchweb201560_9&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; Unlike mechanical relays which can switch AC or DC, SSR's are designed for switching either AC or DC loads.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-10-23T12:22:47.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5891" PostTypeId="2" ParentId="3883" CreationDate="2015-10-23T12:39:39.273" Score="1" Body="&lt;p&gt;One of the things that struck me about this question is that the course booklet states the maximum value is higher than the results of the test, that is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Maximum from test:     2001.27 kg/m3&#xA;Maximum from booklet:  2020    kg/m3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Aside from the fact that it is hard to see where this value has come from, if tests are failed on this basis, the Contractor may experience delays and this in turn may lead to a claim on the project - or at the very least potentially embarrassing arguments.  When a higher value is used for acceptance of the Contractor's work than was actually measured in the laboratory, there must be a valid reason for it.  So let's put this under scrutiny, by examining the data with commonly accepted curve fitting methods.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This post contains the code and data behind this procedure (for those who may find it useful), but others may want to skip to the images and conclusions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Data File&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Using &lt;strong&gt;GnuPlot&lt;/strong&gt;, we look at: (1) linear regression using quadratic forms and (2) interpolation between points using cubic splines.  The data file for this process, &lt;strong&gt;points.txt&lt;/strong&gt;, follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;X    Y         X    Y&#xA;8.4  1447.38   8.4  1447.38&#xA;10.2 1808.53   10.2 1808.53&#xA;12.3 2001.27   12.3 2001.27&#xA;14.6 1683.60   16.8 1524.81&#xA;16.8 1524.81&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Columns 1 and 2 are the data given.  In Columns 3 and 4 we eliminate point 4 as an outlier (it's not an outlier, but for the sake of comparison, let's consider it).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;GnuPlot Source File&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The gnuplot script, &lt;strong&gt;density.gpl&lt;/strong&gt; follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# gnuplot 4.6 patchlevel 4&#xA;&#xA;# quadratic fit&#xA;f0(x) = a0*x**2+b0*x+c0&#xA;f1(x) = a1*x**2+b1*x+c1&#xA;&#xA;# set formats&#xA;set output 'density.png'&#xA;set xlabel 'Moisture, %'&#xA;set ylabel 'Dry Density, kg/m3'&#xA;set yrange [1400:2200]&#xA;set key inside horizontal top right&#xA;set terminal pngcairo dashed # permits dashed lines&#xA;set terminal pngcairo size 640,480 enhanced font 'Verdana,10'&#xA;&#xA;# pt 1 +, pt 2 X, pt 3 *, pt 4 box, pt 5 filled box, pt 6 circle, pt 7 filled circle&#xA;# lt 1 solid, 2 ---, 3 ..., 4 -.-, 5 -..-, 6 solid, 7 - - -  &#xA;&#xA;# color styles (for display)&#xA;set style line 1 lt 1 lc rgb 'black'  lw 2 pt 7 ps 1.2 &#xA;set style line 2 lt 1 lc rgb 'red'    lw 2 pt 2 ps 1.2&#xA;set style line 3 lt 1 lc rgb 'blue'   lw 2 pt 4 ps 1.2&#xA;set style line 4 lt 1 lc rgb 'green'  lw 2 pt 6 ps 1.2&#xA;set style line 5 lt 1 lc rgb 'yellow' lw 2 pt 2 ps 1.2&#xA;&#xA;# black styles (for printing)&#xA;set style line 10 lt 1 lc rgb 'black' lw 2 pt 7 ps 1.2 &#xA;set style line 20 lt 2 lc rgb 'black' lw 2 pt 2 ps 1.2&#xA;set style line 30 lt 4 lc rgb 'black' lw 2 pt 4 ps 1.2&#xA;set style line 40 lt 5 lc rgb 'black' lw 2 pt 6 ps 1.2&#xA;set style line 50 lt 7 lc rgb 'black' lw 2 pt 2 ps 1.2&#xA;&#xA;# fit data to forms&#xA;fit f0(x) 'points.txt' using 1:2 via a0,b0,c0&#xA;fit f1(x) 'points.txt' using 3:4 via a1,b1,c1&#xA;&#xA;# plot the results&#xA;plot f0(x) with lines ls 2 smooth bezier title 'Quadratic', \&#xA;     f1(x) with lines ls 3 smooth bezier title 'Quadratic w/o 4', \&#xA;     'points.txt' using 3:4 with lines ls 4 title 'Cubic Spline w/o 4' smooth csplines, \&#xA;     'points.txt' using 1:2 with lines ls 1 title 'Cubic Spline' smooth csplines, \&#xA;     'points.txt' using 1:2 with points ls 1 notitle&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h1&gt;Moisture-Density Curves&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Next we run gnuplot from the command line (Linux Debian OS), view with the &lt;strong&gt;ImageMagick&lt;/strong&gt; programme &lt;strong&gt;display&lt;/strong&gt; and capture all output to &lt;strong&gt;stats.txt&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;gnuplot density.gpl &amp;amp;&amp;gt; stats.txt &amp;amp;&amp;amp; display density.png &amp;amp;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The resulting set of moisture-density curves follows :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RqN2a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RqN2a.png&quot; alt=&quot;Moisture-Density Curves&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Moisture-Density Statistics&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;From &lt;strong&gt;stats.txt&lt;/strong&gt;, the linear regression of all points (columns 1 and 2) yields:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;degrees of freedom    (FIT_NDF)                        : 2&#xA;rms of residuals      (FIT_STDFIT) = sqrt(WSSR/ndf)    : 121.785&#xA;variance of residuals (reduced chisquare) = WSSR/ndf   : 14831.5&#xA;&#xA;Final set of parameters            Asymptotic Standard Error&#xA;=======================            ==========================&#xA;&#xA;a0              = -25.1676         +/- 7.493        (29.77%)&#xA;b0              = 633.079          +/- 189.8        (29.98%)&#xA;c0              = -2061.07         +/- 1147         (55.64%)&#xA;&#xA;correlation matrix of the fit parameters:&#xA;&#xA;                a0     b0     c0     &#xA;a0              1.000 &#xA;b0             -0.995  1.000 &#xA;c0              0.979 -0.994  1.000 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These coefficients are identical to those of CoryKramer's post.  Now the best fit for Columns 3 and 4 (with the so-called outlier, point 4, discarded) becomes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;degrees of freedom    (FIT_NDF)                        : 1&#xA;rms of residuals      (FIT_STDFIT) = sqrt(WSSR/ndf)    : 4.71227&#xA;variance of residuals (reduced chisquare) = WSSR/ndf   : 22.2055&#xA;&#xA;Final set of parameters            Asymptotic Standard Error&#xA;=======================            ==========================&#xA;&#xA;a1              = -29.4084         +/- 0.3123       (1.062%)&#xA;b1              = 750.537          +/- 8.016        (1.068%)&#xA;c1              = -2783.63         +/- 48.58        (1.745%)&#xA;&#xA;correlation matrix of the fit parameters:&#xA;&#xA;                a1     b1     c1     &#xA;a1              1.000 &#xA;b1             -0.996  1.000 &#xA;c1              0.982 -0.995  1.000 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Just as an aside, there is little difference in the correlation coefficients between the two sets of data, but there is a very large difference in the standard error of the coefficients.  This could mean either: that there is in fact an outlier, that there are too few data points or that the mathematical model is unsatisfactory.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Conclusion&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;It is clear that the stated maximum value of &lt;code&gt;2020 kg/m3&lt;/code&gt; is not supported by the data.  From inspection of the Moisture-Density plots, the maximum values from the various curve fitting methods do not exceed the measured test value, only the position of the optimum moisture content changes.  It is also clear that if a dispute were to emerge on a construction project, the Contractor would have a legitimate case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, which curve is the most appropriate?  Well the Proctor test method puts a great deal of reliability on each data point, in that the error of the test method should be minimal.  For this reason only 5 test values are required.  Therefore, it is logical to interpolate between these points, so my vote would be to use the cubic spline approach, including all points, rather than use the linear regression model.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastActivityDate="2015-10-23T12:39:39.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5895" PostTypeId="1" CreationDate="2015-10-23T16:13:19.593" Score="5" ViewCount="81" Body="&lt;p&gt;I am trying to calculate large volumetric strains but I am somewhat confused whether Green strain is the right thing to be using or whether I should use good old-fashioned engineering strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work I am doing is looking at strain of a deformable pipe. I am measuring longitudinal ($L_s$), circumferential ($C_s$) and volumetric ($V_s$) strain. As the strains are large and I want to avoid errors due to rotations, I have opted to use the Green-Lagrange strain tensor for the $L_s$ and $C_s$ in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;	\frac{d\mathbf{x}^2-d\mathbf{X}^2}{2d\mathbf{X}^2}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $X$ represents the undeformed configuration and $x$ the deformed configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that I should be using normal engineering strain for the volume but I just want some clarification.&lt;/p&gt;&#xA;" OwnerUserId="3476" LastEditorUserId="1832" LastEditDate="2015-10-23T18:01:15.653" LastActivityDate="2015-10-23T18:01:15.653" Title="Are Green strain calculations apt for volumetric strains?" Tags="|deformation|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5897" PostTypeId="2" ParentId="5884" CreationDate="2015-10-23T19:57:55.800" Score="2" Body="&lt;p&gt;This is a common issue, especially considering the &lt;a href=&quot;http://www.telegraph.co.uk/news/newstopics/howaboutthat/10635036/UKs-most-expensive-parking-space-sold-for-400000.html&quot; rel=&quot;nofollow&quot;&gt;value of a parking space&lt;/a&gt; in cities like London. In principle, columns can be moved around to a certain extent using a transfer structure in order to improve the parking layout. However, as the commentators have pointed out, this depends on many factors and it is impossible to estimate the required thickness here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reasonable solution is to work closely with the structural engineer to reach an acceptable column configuration. Better yet, if you have a good engineer, ask &lt;em&gt;them&lt;/em&gt; to work out a configuration of columns/parking spaces which maximises the number of parking spaces. You may be surprised what they are able to come up with.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2015-10-23T20:03:46.307" LastActivityDate="2015-10-23T20:03:46.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5900" PostTypeId="1" CreationDate="2015-10-24T04:58:47.113" Score="3" ViewCount="105" Body="&lt;p&gt;Why is the jacket or sheath of fibre optic endoscopes always black?&#xA;Is it just a convention, or is there a more logical explanation for this?&lt;/p&gt;&#xA;" OwnerUserId="3489" LastEditorUserId="90" LastEditDate="2015-10-24T06:59:36.850" LastActivityDate="2016-02-06T23:50:52.210" Title="Why is the jacket or sheath of fibre optic endoscopes always black?" Tags="|optics|biomedical-engineering|medical-devices|fiber-optics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5901" PostTypeId="1" CreationDate="2015-10-24T11:57:44.660" Score="5" ViewCount="1442" Body="&lt;p&gt;Why is it that as the frequency approaches a pole or zero of a system, the slope of the magnitude of the transfer function changes by 20dB/decade?&#xA;Whenever I see this stated when reading about transfer functions, it's never explained. What causes this property of transfer functions?&lt;/p&gt;&#xA;" OwnerUserId="3491" LastActivityDate="2015-10-26T12:41:18.097" Title="Why do the poles and zeroes of a transfer function cause a slope change in frequency response?" Tags="|electrical-engineering|control-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5902" PostTypeId="1" CreationDate="2015-10-24T17:10:52.317" Score="2" ViewCount="3359" Body="&lt;p&gt;How can I calculate the strain in a simply supported beam from having the maximum bending moment (25kNm) and the maximum bending stress (45MPa)?&lt;/p&gt;&#xA;" OwnerUserId="3494" LastEditorUserId="9" LastEditDate="2015-10-24T21:21:52.940" LastActivityDate="2015-10-24T21:22:00.357" Title="How to calculate strain from maximum bending moment and stress?" Tags="|statics|stresses|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5903" PostTypeId="2" ParentId="5900" CreationDate="2015-10-24T17:28:13.347" Score="3" Body="&lt;p&gt;I really have no idea, but I did come up with the following possibilities.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;To minimize reflections from the instrument itself.  Light reflecting off the endoscope could interfere with the doctor's view.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because little (if any) body tissue inside the body is black, thus making it easier to distinguish the instrument from body parts.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because that's the natural color of the plastic including additives.  Adding dyes could be undesirable if they leech over time, and since it's not a typical consumer item there's no need to make it colorful and flashy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To shield the fibers from outside light, thus minimizing interference.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-10-24T17:57:15.927" LastActivityDate="2015-10-24T17:57:15.927" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5904" PostTypeId="2" ParentId="5901" CreationDate="2015-10-24T17:40:41.570" Score="3" Body="&lt;p&gt;Have you read about roll-off? Wikipedia has a great explanation of this topic. Basically when you have a pole or zero of a system, you are representing, in this case, the transfer function of a circuit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/math/d/e/0/de0b638a870ba64ff1205d841d7ebf7a.png&quot; alt=&quot;Ec1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/math/8/f/5/8f5b79d7f856dfca8dc0b8dcf064308b.png&quot; alt=&quot;Ec2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To express the magnitude in decibels of the system on a frequency point you use this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/math/0/e/b/0ebd5dc4966155306ab7c76f69db3fc4.png&quot; alt=&quot;Ec3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or expressed as a loss:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/math/6/e/4/6e4773f85468bb0e419a55562bcd6a27.png&quot; alt=&quot;Ec4&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At frequencies well above ω=1, this simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/math/b/9/e/b9e380bcc443e174e369f9dd73b7ef25.png&quot; alt=&quot;Ec5&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you want to know the difference in decibels between to frequencies, you may have to use the same equation but this time using (ω2/ω1) in the logarithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/math/b/c/f/bcf95d998c2fa2e1b7c6d39b0fab2bbe.png&quot; alt=&quot;Ec6&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the difference here is a decade, or a frequency ten times ω1, the logarithm will be always equal to 1. Then you multiply this result with 20 because of the fifth equation, then you have: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/math/4/b/c/4bc7f6ac6b6951cabbd7718e1857008b.png&quot; alt=&quot;Ec7&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's it! Maybe you really want to know another reason besides the one I gave to you, but it is explained mathematically. &lt;/p&gt;&#xA;" OwnerUserId="3495" LastActivityDate="2015-10-24T17:40:41.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5905" PostTypeId="1" AcceptedAnswerId="5915" CreationDate="2015-10-24T18:17:12.103" Score="2" ViewCount="5107" Body="&lt;p&gt;We are given a standard diagram of a rotating cylinder with the parabolic shape in the rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the distance between the vertex of parabola to the top of the parabola.&lt;/li&gt;&#xA;&lt;li&gt;the original height of the water. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have a really general question regarding this problem. From where the vertex of the parabola begins, up to the top, the pressure distribution varies with radius? And from the vertex of the parabola to the bottom of the cylinder, it only varies with height, since radius is fixed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: can I split the two different areas and integrate $P = \dfrac{\rho\omega^2}{2} -\gamma z$ twice? &#xA;Once while holding the radius constant but the height a variable, and another while holding the height constant but integrating with respect to the radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm planning on using $\text{d}F= P\cdot\text{d}A$ when integrating each one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A link to an image that looks similar to the one I have : &lt;a href=&quot;http://demonstrations.wolfram.com/FluidRotatingInACylinder/&quot; rel=&quot;nofollow&quot;&gt;http://demonstrations.wolfram.com/FluidRotatingInACylinder/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3496" LastEditorUserId="2819" LastEditDate="2016-03-15T05:33:20.647" LastActivityDate="2016-03-15T14:10:01.257" Title="Force exerted on wall of a rotating cylinder filled with fluid" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5906" PostTypeId="2" ParentId="3301" CreationDate="2015-10-24T19:40:06.230" Score="1" Body="&lt;p&gt;It looks like a known problem.  Some people use similar for potentiometers on older gear when cases were thicker and penny pinching was not essential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a part that might fit and may be a stock item somewhere.  I searched for &lt;em&gt;potentiometer nut thick panel&lt;/em&gt; in Google image search. I have seen similar made of brass with round rim with two slots for a hollow wrench.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;em&gt;The Caption for this picture was &lt;strong&gt;Note unusual &quot;deep threaded&quot; nut!&lt;/strong&gt; so it is unusual even when encountered in the field.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.abalonevintage.com/1960s_Gibson_EMS-1235_guitar_pot_shaft.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.abalonevintage.com/1960s_Gibson_EMS-1235_guitar_pot_shaft.jpg&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would try and find a suitable clinch nut from some vendor, check the pictures in this image search for most likely candidate.  Thread pitch may limit choices. Some types do not have the thread all the way through even if the hole does go as they are riveted into place. The collapsing types have the benefit of usually a thin wall design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=clinch+nut&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0CAcQ_AUoAWoVChMI1LHj5-3byAIVygcsCh0zhgRL&amp;amp;biw=1400&amp;amp;bih=778&quot; rel=&quot;nofollow&quot;&gt;https://www.google.com/search?q=clinch+nut&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0CAcQ_AUoAWoVChMI1LHj5-3byAIVygcsCh0zhgRL&amp;amp;biw=1400&amp;amp;bih=778&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some similar items are called a furniture connector in some places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ap-magazine.com/upload/fastener/Product/Image1/Screw35439m95925.jpg&quot; rel=&quot;nofollow&quot;&gt;http://www.ap-magazine.com/upload/fastener/Product/Image1/Screw35439m95925.jpg&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://i01.i.aliimg.com/photo/v0/908973835/Custom_stainless_steel_carbon_steel_M6_M8.jpg&quot; rel=&quot;nofollow&quot;&gt;http://i01.i.aliimg.com/photo/v0/908973835/Custom_stainless_steel_carbon_steel_M6_M8.jpg&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="3497" LastEditorUserId="3497" LastEditDate="2015-10-24T19:51:02.603" LastActivityDate="2015-10-24T19:51:02.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5907" PostTypeId="2" ParentId="5902" CreationDate="2015-10-24T20:00:50.383" Score="3" Body="&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory#Relation_between_curvature_and_beam_deflection&quot; rel=&quot;nofollow&quot;&gt;Wikipedia's page on Euler-Bernoulli Beam Theory&lt;/a&gt;:&#xA;$$&#xA;\sigma_x = -zE\cfrac{\mathrm{d}^2w}{\mathrm{d}x^2}\,,~~&#xA; M = -EI\cfrac{\mathrm{d}^2w}{\mathrm{d}x^2} \,,~~&#xA;\varepsilon_{x} = -z\cfrac{\mathrm{d}^2w}{\mathrm{d}x^2} \,.&#xA;$$&#xA;Therefore,&#xA;$$&#xA;  \sigma_x = E\,\varepsilon_x ~,~~ M = \frac{EI}{z}\,\varepsilon_x \,.&#xA;$$&#xA;You can eliminate the ratio&#xA;$$&#xA;  \frac{I}{z} = \frac{M}{\sigma_x} \,.&#xA;$$&#xA;But you will still need to know $E$ to find the strain.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="1832" LastEditDate="2015-10-24T21:22:00.357" LastActivityDate="2015-10-24T21:22:00.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5908" PostTypeId="1" CreationDate="2015-10-25T08:01:14.093" Score="5" ViewCount="328" Body="&lt;p&gt;I am simulating a turbine and I have some question about standard wall function in FLUENT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard wall function requires $y^+$ value between 30 to 60 on the blade. but the following quote states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When using wall function models, the $y^+$ value should ideally be above&#xA;  15 to avoid erroneous modelling in the buffer layer and the laminar&#xA;  sub-layer &lt;a href=&quot;http://www.computationalfluiddynamics.com.au/turbulence-part-3-selection-of-wall-functions-and-y-to-best-capture-the-turbulent-boundary-layer/#comment-7471&quot; rel=&quot;noreferrer&quot;&gt;[Source]&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why $y^+$ value should ideally be above 15 by using wall functions?&#xA;Can massive separation be modeled with $y^+$ value between 30 to 60 on the blade? because it is mentioned according to FLUENT help:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Traditionally, there are two approaches to modeling the near-wall&#xA;  region. In one approach, the viscosity-affected inner region (viscous&#xA;  sublayer and buffer layer) is not resolved. Instead, semi-empirical&#xA;  formulas called &quot;wall functions'' are used to bridge the&#xA;  viscosity-affected region between the wall and the fully-turbulent&#xA;  region. The use of wall functions obviates the need to modify the&#xA;  turbulence models to account for the presence of the wall.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, I think that there is a velocity profile in viscosity-affected inner region as a wall function that it can almost model boundary layer and massive separation. For example, wall functions with 30&#xA;&#xA;&lt;p&gt;I studied &lt;a href=&quot;https://www.sharcnet.ca/Software/Fluent6/html/ug/node512.htm&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; but I don't understand the following sentences:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The logarithmic law for mean velocity is known to be valid for . In&#xA;  FLUENT, the log-law is employed when . It should be noted that, in&#xA;  FLUENT, the laws-of-the-wall for mean velocity and temperature are&#xA;  based on the wall unit, $y^*$ , rather than $y^+$. These quantities are&#xA;  approximately equal in equilibrium turbulent boundary layers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is difference between $y^*$ and $y^+$? What is the exact meaning of &lt;em&gt;&quot;equilibrium turbulent boundary layers&quot;&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, my question is that when wall function approach is used, viscosity-affected inner region (viscous sublayer and buffer layer) is computed by using semi-empirical formulas with good accuracy. Is this statement correct?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="1293" LastEditDate="2015-10-29T10:35:24.403" LastActivityDate="2024-03-27T22:06:25.767" Title="Wall functions and separation in turbine blade simulation" Tags="|mechanical-engineering|fluid-mechanics|turbulence|cfd|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5909" PostTypeId="1" AcceptedAnswerId="5922" CreationDate="2015-10-25T09:08:25.237" Score="4" ViewCount="120" Body="&lt;p&gt;Sorry if this is the wrong forum, I'm not an engineer but I'd need a pro to answer my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Short version:&lt;/strong&gt;&#xA;Are there any tests or studies, if a carpet in an apartement can reduce noise from the apartment below. I'm searching for numbers rather than oppinions. The noise comes in all frequencies and shapes, from trampling to yelling, machine noise etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there aren't any tests, please give me your oppinion, if you're an experienced professional in this field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long version:&lt;/strong&gt;&#xA;I need to find a way to reduce noise in my apartment. I'm living in the topmost floor, so only the noise from below is the problem. Apparently our landlord put laminate in all apartements, without any form of insulation. My neighbors aren't the type of people that seem to be (neither could be made) aware, that noise exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After browsing the web for months, trying out various earplugs for &gt;300$, hanging thick curtains, rearanging my apartement etc. I'm just searching for something more. If it's 5% less noise - fine. If it doesn't help at all I don't want to spend time and money on it. &#xA;I know, that I propably should better find a new place to live or make my neighbors be more silent, but this is not a perfect world, things cannot be done in the best possible way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure why Mythbusters never tested this stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BcsHw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BcsHw.jpg&quot; alt=&quot;Soundproof bed&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;RED&lt;/strong&gt; = Insulation (Foam, ..., whatever)&#xA;&lt;strong&gt;White&lt;/strong&gt; = Heavy wood&#xA;&lt;strong&gt;Blue&lt;/strong&gt; = Inner shell, also heavy wood&#xA;What if I layed carpet and built something like this? I set up directional lamps for every wall and the ceiling. Sure, light is not sound, but it bounces too, right? Or will that totaly not work? Maybe I worsen it by that, like I build a trumpet..&lt;/p&gt;&#xA;" OwnerUserId="3502" LastEditorUserId="3502" LastEditDate="2015-10-25T11:57:39.770" LastActivityDate="2015-10-26T15:27:26.820" Title="Carpet for noise reduction (tests / studies)" Tags="|architecture|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5910" PostTypeId="1" AcceptedAnswerId="5919" CreationDate="2015-10-25T22:37:41.507" Score="11" ViewCount="5481" Body="&lt;p&gt;In the buckling of columns we know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \dfrac{n^2\pi^2EI}{L^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest value of P occurs when $n=1$ which gives a simple buckling shape (one wave):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{cr} = \dfrac{\pi^2EI}{L^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However for $n &amp;gt; 1$, as shown below the buckling shape is more complex and has many waves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x2Mm9.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x2Mm9.jpg&quot; alt=&quot;Buckling shapes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is do the buckling mode shapes for $n &amp;gt; 1$ ever occur in reality? If the column begins to buckle as per the shape for $n = 1$ then wouldn't it just continue to buckle like this until failure? How would the other buckling modes ever occur?&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="33" LastEditDate="2015-10-25T23:36:34.827" LastActivityDate="2015-10-26T16:16:41.110" Title="Buckling: Do buckling mode shapes of n &gt; 1 occur in reality?" Tags="|structural-engineering|beam|columns|buckling|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5911" PostTypeId="1" CreationDate="2015-10-26T00:18:35.750" Score="6" ViewCount="2797" Body="&lt;p&gt;If I pour a sample of aggregate into a container full of water and measure the displacement of the water due to the sample, does this give me Bulk Volume or Solid Volume?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what is the difference?&lt;/p&gt;&#xA;" OwnerUserId="3507" LastEditorUserId="1832" LastEditDate="2015-10-26T00:40:41.080" LastActivityDate="2015-10-26T15:06:12.573" Title="Bulk Volume vs. Solid Volume" Tags="|civil-engineering|construction-management|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5912" PostTypeId="2" ParentId="5910" CreationDate="2015-10-26T00:40:02.990" Score="8" Body="&lt;p&gt;If you are looking at the column as being supported at the ends, you are correct that the &lt;strong&gt;n=1&lt;/strong&gt; mode gives the lowest buckling load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other modes (n=2,3,...) aren't useless though. Long columns are often braced at regular intervals to reduce the unbraced length of the column. For a given length of column, these braces &lt;em&gt;force&lt;/em&gt; the column to buckle under a different mode (n=2,3,...) with the corresponding increase in buckling load.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-10-26T00:40:02.990" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="5913" PostTypeId="1" CreationDate="2015-10-26T01:18:27.403" Score="3" ViewCount="3236" Body="&lt;p&gt;So I was told that $V_x$ is equal to 10V.&#xA;I tried reasoning out why $V_x$ was 10V and this is what I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-You can treat the Vx as an open circuit which means current does not flow through the resistor (meaning it's a short circuit) and thus from KVL we get $V_x$ = 10V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am still not convinced that $V_x$= 10V. &#xA;A better explanation would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vojlP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vojlP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3508" LastEditorUserId="110" LastEditDate="2015-10-27T07:29:46.047" LastActivityDate="2015-10-28T02:44:06.607" Title="How do you find &quot;this&quot; Vx in the given circuit?" Tags="|electrical-engineering|circuits|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5914" PostTypeId="2" ParentId="5911" CreationDate="2015-10-26T02:08:06.360" Score="7" Body="&lt;p&gt;Bulk volume is the total volume of void and solid material. Solid volume is the volume occupied by the solid material only, i.e. not including porosity. Phrased mathematically: $V_{bulk}=V_{void}+V_{solid}$. If an aggregate is poured into a container full of water, the displaced water reflects the solid volume only, since the space between the aggregate particles is filled with water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out in the comments, for sufficiently fine aggregate or porous media, if air becomes trapped then the volume of liquid displaced may reflect a value between solid volume and bulk volume. Take care with how you measure. &lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-10-26T15:06:12.573" LastActivityDate="2015-10-26T15:06:12.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5915" PostTypeId="2" ParentId="5905" CreationDate="2015-10-26T02:37:21.890" Score="0" Body="&lt;p&gt;You don't need to split up the fluid domain into two parts.  You can use the following equation for the pressure in a rotating fluid which I got from &lt;a href=&quot;http://www.mne.psu.edu/cimbala/Learning/Fluid/Rigid_body/rigid_body.htm&quot; rel=&quot;nofollow&quot;&gt;this PSU website:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=p_a-\rho g z+ \frac{\rho r^2 \omega^2}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p \equiv$ pressure at location $(r,z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_a \equiv$ external (e.g. atmospheric) pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho \equiv$ fluid density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g \equiv$ gravity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$z \equiv$ height above the origin of the parabola (negative for the fluid below the parabola)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r \equiv $ radial distance from the axis of rotation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega \equiv$ angular velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you're only concerned with the pressure at the wall, $r$ is a constant and the only variable remaining is $z$.  Just integrate this equation with respect to area over the height of the fluid (like you mentioned already) and that should give you total force on the container walls.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-10-26T02:37:21.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5916" PostTypeId="1" CreationDate="2015-10-26T04:17:28.057" Score="1" ViewCount="908" Body="&lt;p&gt;Background Information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My design team is trying to create a small RC car that can climb a 30 degrees incline that is 6inches high. We have two motors on a 3 wheels vehicle with the motors powering the 2 back wheels. The wheels are about 20cm length and 18cm width (15cm without) with the tires. The tires we currently have are made of plastic. The track is painted on wood so is pretty slippery. 4 double A batteries for each wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Our vehicle is unable to make it up the incline.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some ways to make the vehicle go up the incline? &lt;/p&gt;&#xA;" OwnerUserId="3511" LastEditorUserId="272" LastEditDate="2015-10-26T11:55:44.013" LastActivityDate="2015-10-27T17:30:52.740" Title="Design of Wheels/Tires for a RC Car" Tags="|design|friction|wheels|" AnswerCount="1" CommentCount="4" ClosedDate="2015-10-27T19:58:30.497" ContentLicense="CC BY-SA 3.0" />
  <row Id="5917" PostTypeId="2" ParentId="5913" CreationDate="2015-10-26T06:43:08.110" Score="-1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wlP1c.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wlP1c.jpg&quot; alt=&quot;Thevenin Circuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When calculating $V_{th}$ , $V_{AB}$ is open circuited as in the diagram. As a result there is no current flow across $\frac{5}{11} \Omega$.&#xA;Thus   $V_x = 10V$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thévenin's theorem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/E38u3pLHSOc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thevenin's Equivalent&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-10-27T07:24:43.700" LastActivityDate="2015-10-27T07:24:43.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5918" PostTypeId="2" ParentId="5913" CreationDate="2015-10-26T11:57:15.090" Score="2" Body="&lt;p&gt;You are right in that $V_x$ can't be 10 V.  Fortunately, 10V for $V_x$ simplifies the circuit considerably, making it particularly easy to analyze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $V_x$ is the same voltage as the supply on the left, both ends of the 5/11 &amp;Omega; resistor are at the same voltage.  That means the current thru it is zero.  That means the current thru anything directly in series with is must also be zero.  Therefore, the current thru the 5/11 &amp;Omega; resistor, the 10 V power supply, and the 1 &amp;Omega; resistor at right are all 0.  The voltage across any resistor with 0 current thru it must also be 0.  We therefore know that point A is at 0 V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the loop with the 2 V supply and the two 1 &amp;Omega; resistors in isolation, we can easily see that 1 A must be circulating, as indicated by your curved arrow.  1 A thru a 1 &amp;Omega; resistor causes a 1 V drop, so the right end of this loop must be 1 V higher than the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting this back into the rest of the circuit that is known to carry 0 current just doesn't work.  According to previous logic, point A must be at 0 V, but adding the +2 V of the supply and -1 V of the resistor to the 10 V at $V_x$ yields 11 V at A, which is a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly something is wrong somewhere.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="282" LastEditDate="2015-10-28T02:44:06.607" LastActivityDate="2015-10-28T02:44:06.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5919" PostTypeId="2" ParentId="5910" CreationDate="2015-10-26T12:20:43.247" Score="7" Body="&lt;p&gt;Whether or not buckling modes with $n&amp;gt;1$ exists depends on how you look at the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @hazzey notes in his answer, columns with bracing may display buckling modes with $n&amp;gt;1$. These buckling modes, however, are simply equivalent to the $n=1$ modes of the individual segments that compose the column. To be clear, this doesn't mean that the segments behave independently (you'll never have two consecutive unbraced lengths buckling to the same side), only that any $n&amp;gt;1$ mode can be composed by a series of continuous $n=1$ modes for the unbraced lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you have a column with a single bracing which buckles, do you consider that an $n&amp;gt;1$ mode for the entire column or an $n=1$ mode for each of the unbraced lengths? Both? Your call.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6AJff.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6AJff.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To paraphrase @starrise's comment on @hazzey's answer, this can be demonstrated by looking at the buckling equation:&#xA;\begin{align}&#xA;P &amp;amp;= \left(\dfrac{n}{L}\right)^2\pi^2EI \\&#xA;P_{column,\,n=2} &amp;amp;= \left(\dfrac{2}{L}\right)^2\pi^2EI \\&#xA;P_{segment,\,n=1} &amp;amp;= \left(\dfrac{1}{\frac{L}{2}}\right)^2\pi^2EI = \left(\dfrac{2}{L}\right)^2\pi^2EI \\&#xA;\therefore P_{column,\,n=2} &amp;amp;= P_{segment,\,n=1}&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2015-10-26T16:16:41.110" LastActivityDate="2015-10-26T16:16:41.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5920" PostTypeId="2" ParentId="5901" CreationDate="2015-10-26T12:41:18.097" Score="2" Body="&lt;p&gt;This is just a result of the definition of poles and zeros of a transfer function.  A transfer function in the Laplace domain &lt;a href=&quot;http://web.mit.edu/2.14/www/Handouts/PoleZero.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;is written as&lt;/a&gt;&#xA;$$&#xA;H(s)=\frac{b_ms^m+...+b_2s^2+b_1s+b_0}{a_ns^n+...+a_2s^2+a_1s+a_0}&#xA;$$&#xA;By factoring the polynomials (which is often not easily achieved by hand) you can rewrite this as&#xA;$$&#xA;H(s)=\frac{(s-z_1)(s-z_2)...(s-z_m)}{(s-p_1)(s-p_2)...(s-p_n)}.&#xA;$$&#xA;This form is what defines the poles and zeros; the $p_i$ are the poles and the $z_i$ are the zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how this leads to the 20 dB per decade rule, lets consider a concrete example:&#xA;$$&#xA;H(s)=\frac{(s-1000)}{(s-10)(s-100000)}.&#xA;$$&#xA;I've intentionally spread the poles and zeros very far out to illustrate the point.  When $s\ll 10$ (i.e. $s$ is much smaller than all of the poles and zeros), then this reduces approximately to $H(s)\approx\frac{1000}{10*100000}$ and the transfer function is flat.  When $s\sim100$ (i.e. much larger than the first pole but smaller than the zero) this becomes approximately $H(s)\approx\frac{1000}{s*100000}$ and the transfer function is falling at 20 dB per decade.  Upon crossing the zero, at say $s\sim10000$, the transfer function becomes approximately $H(s)\approx\frac{s}{s*100000}=\frac{1}{100000}$ and the transfer function is again flat.  Finally for $s\gg100000$ the transfer function becomes approximately $H(s)\approx\frac{s}{s*s}=\frac{1}{s}$, and it is falling at 20 dB per decade again.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honestly, the best way to get a feel for this rule is to simply plot a number of transfer functions.  There is an online tool &lt;a href=&quot;http://www.onmyphd.com/?p=bode.plot.online.generator&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; which allows you to type in an arbitrary number of poles and zeros to see how the Bode plot looks.  The plot below shows its output for the example I described above except that it sets the DC gain to 0 dB.  The 'asymptotic plot' is the approximation that I've described above, and the 'real plot' is the actual Bode plot of the transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BzNgD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BzNgD.png&quot; alt=&quot;magnitude&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lpoCs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lpoCs.png&quot; alt=&quot;phase&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-10-26T12:41:18.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5921" PostTypeId="1" CreationDate="2015-10-26T15:08:51.520" Score="7" ViewCount="3806" Body="&lt;p&gt;I have been working on creating a 1-D model of a material with a layer of heat resistant material stuck to it. For this I have been trying to use the implicit finite-differential method. The equations are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{E}_{in} - \dot{E}_{out} = \dot{E}_{st}&#xA;$$&#xA;or&#xA;$$&#xA;\frac{1}{\alpha} \frac{dT}{dt}= \frac{d^2T}{dx^2}&#xA;$$&#xA;Where $\alpha = \frac{k}{\rho C_p} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if there is only one material, an internal node equation would look like (I am assuming all flow in into the node for sign convention, n is the node location, p is the time step, and x is the thickness of a node):&#xA;$$&#xA;\frac{k A}{\Delta x}(T^{p+1}_{n+1}-T^{p+1}_{n}) + \frac{k A}{\Delta x}(T^{p+1}_{n-1}-T^{p+1}_{n}) = \frac{C_p \rho  A \Delta x }{\Delta t}(T^{p+1}_{n}-T^{p}_{n})&#xA;$$&#xA;or&#xA;$$&#xA;(1+2*Fo)T^{p+1}_n - Fo(T^{p+1}_{n-1}+T^{p+1}_{n+1}) = T^p_n&#xA;$$&#xA;Where, $Fo = \frac{\alpha \Delta t}{(\Delta x)^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface node equation looks like:&#xA;$$&#xA;{h A}(T_\infty-T^{p+1}_{n}) + \frac{k*A}{\Delta x}(T^{p+1}_{n-1}-T^{p+1}_{n}) = \frac{C_p \rho A \Delta x }{2 \Delta t}(T^{p+1}_{n}-T^{p}_{n})&#xA;$$&#xA;or&#xA;$$&#xA;(1+2Fo + 2FoBi)T^{p+1}_n-2Fo(T^{p+1}_{n-1}+BiT_\infty) = T^p_n&#xA;$$&#xA;Where, $Bi = \frac{h\Delta x}{k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to work pretty good for me when I use just one material. I give it some initial conditions and run it to steady state then check it against a simple steady state resistance model and there is a match. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I begin to bump into issues when I put in an internal material boundary. This is the equation I was using to simulate it (This calculation is for a node in material A adjacent to a node in material B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{(k_b +k_a )A}{\Delta x_b+\Delta x_a}(T^{p+1}_{n+1}-T^{p+1}_{n}) + \frac{k_a A}{\Delta x_a}(T^{p+1}_{n-1}-T^{p+1}_{n}) = \frac{C_{pa} \rho_a A \Delta x_a }{\Delta t}(T^{p+1}_{n}-T^{p}_{n})&#xA;$$&#xA;or&#xA;$$&#xA;(1+ Fo + Fo_{ab})T^{p+1}_n-FoT^{p+1}_{n-1}-Fo_{ab}T^{p+1}_{n+1} = T^p_n&#xA;$$&#xA;Where, $Fo_{ab} = \frac{\alpha_{ab} \Delta t}{\Delta x_a(\Delta x_b+\Delta x_a)}$ and $\alpha_{ab} = \frac{k_b+k_a}{\rho_a C_{pa}} $ Note: if the materials are the same $Fo_{ab} = Fo$ this equation becomes a same as the boundary equation above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this boundary added I again created a steady-state resistive model to check against and found about a 10% difference, which was not expected. (note:updated to add in an averaging of k values to remove this difference, not sure if its right, but it fixed the error)&#xA;Below is a diagram showing the two model setups:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3UjDV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3UjDV.jpg&quot; alt=&quot;model diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the test case, I used the arbitrary constants:&#xA;$$\Delta t = 0.15$$&#xA;$$Nodes_a = Nodes_b = 5$$&#xA;Note, number of nodes is different than in the picture above (4).&#xA;$$\Delta x_a = \Delta x_b = Length/(Nodes-0.5) = 10/4.5 = 2.2222$$&#xA;$$h_a = h_b = 1$$&#xA;$$\rho_a = \rho_b = 1$$&#xA;$$C_{pa} = C_{pb} = 1$$&#xA;$$k_a = 5$$&#xA;$$k_b = 10$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I be getting this 10% difference between the two models? (updated to fix difference, still not sure if the new equation is correct though)&#xA;Is this equation I am using for the material boundary correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw, many of the equations for surface nodes and boundary nodes came from the book: &quot;Fundamentals of Heat and Mass Transfer&quot; by Incropera and DeWitt.5th edition, section 5.9.2.&lt;/p&gt;&#xA;" OwnerUserId="3518" LastEditorUserId="282" LastEditDate="2015-10-28T02:44:02.420" LastActivityDate="2022-08-01T17:16:08.520" Title="Modeling Transient Heat Transfer between two 1-D materials" Tags="|mechanical-engineering|thermodynamics|heat-transfer|finite-element-method|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5922" PostTypeId="2" ParentId="5909" CreationDate="2015-10-26T15:27:26.820" Score="4" Body="&lt;p&gt;I have a background in acoustics, noise control and vibrations. I haven't done much in the way of residential applications though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honestly, your best bet is to lay down heavy carpet with an underlayment designed to maximize sound absorption, like &lt;a href=&quot;http://www.soundisolationcompany.com/solutions-products/soundproof-floors/carpet-floors/privacy-premium-carpet-underlay/&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;. That doesn't mean that your problem will go away, or even be affected significantly.Since you're hearing sounds of all frequencies, in all likelihood there are &quot;flanking paths&quot; that allow sound from your downstairs neighbors into your apartment. These are typically things like ducts, pipes, walls and other vertical passages inherent in a multistory building. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason there aren't a lot of quantifiable measures and material ratings for sound reduction is that the testing required to give good results adds to the product cost. Most people are simply looking for a rug or carpet that will make their room look great. They aren't willing to pay a lot more for a material that's been tested. In any case, too much of a material's effectiveness in a given application depends on other factors that are beyond the control of the manufacturer. Just to give you an idea, consider what happens when sound waves hit a material surface. The energy from the sound can do three things. It can be reflected, it can be transmitted through or it can be absorbed by the material. What happens depends not just on the material, but also a great deal on the frequency of the sound, the angle of incidence and what's behind the material (i.e. the structure of the building). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's entirely possible to design and construct buildings that minimize noise pollution between rooms/units, but for pre-existing structures, it can be all but impossible&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-10-26T15:27:26.820" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5923" PostTypeId="1" AcceptedAnswerId="5935" CreationDate="2015-10-26T15:57:06.453" Score="5" ViewCount="1354" Body="&lt;p&gt;I'm trying to determine a level of vibration testing I should perform on my product. I understand electrical stress/life testing and I'm looking for a mechanical equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a DC motor running in my application that generates a certain level of vibration. I'm planning to do some environmental testing to my electronics and one of the tests is vibration, so I plan to subject the electronics to the level of vibration above what the motor produces, my thinking being that a higher level may be equivalent to electrical life testing. My product needs to last ten years so I'd like to know that the vibration I see now won't cause problems in several years time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I can put an accelerometer on my electronics and measure ~2 g. To simulate over the product's lifetime should I be applying a higher level of vibration, say 3-5 g?&lt;/p&gt;&#xA;" OwnerUserId="3519" LastEditorUserId="368" LastEditDate="2017-12-14T18:12:15.093" LastActivityDate="2017-12-14T18:12:15.093" Title="What level of vibration testing should I perform on my product?" Tags="|mechanical-engineering|vibration|product-testing|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5924" PostTypeId="2" ParentId="5913" CreationDate="2015-10-26T16:19:48.380" Score="2" Body="&lt;p&gt;You start by making A-B an open circuit.  This implies that the current entering or leaving node A from the load is 0 Amps.  This implies that the current in the two 1 $\Omega$ resistors have the same current flowing through them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The open circuit across A-B also implies that 0 current is flowing into or out of Node B.  This means 0 current is flowing through $\frac{5}{11}\Omega$ resistor, so there is 0 voltage drop across that resistor, so $V_x$ is also 10 V.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves you with 10 V at $V_x$ and $0.2V_x = 2 V$ across the voltage controlled voltage source.  This means the current flowing in the loop with the two 1 $\Omega$ resistors is $\frac{2 V}{2\Omega} = 1 A$.  This also means the voltage between the voltage controlled voltage source and the bottom 1 $\Omega$ resistor is $12 V$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this information you can calculate the voltage drop across the lower 1 $\Omega$ resistor to be $1V$, which gives you a $V_{th}$ of $11 V$.&lt;/p&gt;&#xA;" OwnerUserId="198" LastActivityDate="2015-10-26T16:19:48.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5925" PostTypeId="2" ParentId="5921" CreationDate="2015-10-26T16:26:18.453" Score="10" Body="&lt;p&gt;I am guessing, you are solving the following system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_t T = \alpha_i \partial_x^2 T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\alpha_i=\kappa_i/\rho_i c_{p,i}$ for material $i$, subject to the conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T\left(x,0\right)=T_\infty\quad T\left(0,t\right) = T_h \quad -\kappa\partial_xT\left(L,t\right)=h\left[T\left(L,t\right)-T_\infty\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can approach simulating different materials in the following way:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Define an interface at a node at $x_b$; make $\alpha_i$ a function of $x$, have $\alpha=\alpha_1$ in region $0 \le x \le x_b$ and $\alpha=\alpha_2$ in region $x_b \le x \le L$.&lt;/li&gt;&#xA;&lt;li&gt;Impose thermal equilibrium and heat flux continuity at the interface; i.e.:&#xA;$$T\left(x_b^-\right)=T\left(x_b^+\right)\quad -\kappa_1\partial_xT\left(x_b^-\right)=-\kappa_2\partial_xT\left(x_b^+\right)$$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would make the node corresponding to $x=0$ equal to $T=T_h$ and the node corresponding to $x=L$ equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\kappa_2\frac{T\left(L^-,t\right)-T\left(L,t\right)}{\Delta x}=h\left[T\left(L,t\right)-T_\infty\right]$$&#xA;$$T\left(L,t\right)=\frac{T\left(L^-,t\right)-\beta T_\infty}{1-\beta} \quad  \beta=\frac{h\Delta x}{\kappa_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $T\left(L^-,t\right)$ is the node to the left of $x=L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution based on Finite Volume&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to take another look at this and came up with a solution based on Finite Volume which I think works better. Coincidentally, a 1D FV solution is the same as a 1D FD solution except the grid is staggered. Let's start simple and add complexity after everything is working correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some important assumptions i have made in the analysis:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nodes are monospaced at $\Delta x=L/N$ where $L$ is the length of the domain and $N$ the number of nodes; this simplifies the analysis considerably. &lt;/li&gt;&#xA;&lt;li&gt;The boundaries and interface are located exactly halfway between nodes, this allows a second order accurate approximation of temperatures and gradients at those boundaries.&lt;/li&gt;&#xA;&lt;li&gt;I do not bother with the time integration of these systems, i leave that up to a relevant ODE solver which can handle linear systems (see usage of &lt;code&gt;odeint&lt;/code&gt; solver provided by Scipy in the Python code appended below). A good solver has adaptive time stepping allowing me to avoid any issues with numerical instability related to the CFL criterium.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We start with simple diffusion between two boundaries defined by the system:&#xA;$$\partial_{t}T=\alpha\partial_{x}^{2}T$$&#xA;$$T\left(x,0\right)=T_{\infty}\quad T\left(0,t\right)=T_{h}\quad T\left(L,t\right)=T_{\infty}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In FV we define an average temperature over the control volume $dV=Adx$:&#xA;$$\bar{T}_{i}=\frac{\int_{V}TdV}{\int_{V}dV}=\frac{1}{\Delta x}\int_{i-\frac{1}{2}\Delta x}^{i+\frac{1}{2}\Delta x}Tdx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then integrate our PDE:&#xA;$$\frac{1}{\Delta x}\int_{i-\frac{1}{2}\Delta x}^{i+\frac{1}{2}\Delta x}\partial_{t}Tdx=\frac{\alpha}{\Delta x}\int_{i-\frac{1}{2}\Delta x}^{i+\frac{1}{2}\Delta x}\partial_{x}^{2}Tdx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_{t}\bar{T}_{i}=\frac{\alpha}{\Delta x}\left[\left.\partial_{x}T\right|_{i-\frac{1}{2}\Delta x}-\left.\partial_{x}T\right|_{i+\frac{1}{2}\Delta x}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_{t}\bar{T}_{i}=\frac{\alpha}{\Delta x^{2}}\left[\bar{T}_{i-\Delta x}-2\bar{T}_{i}+\bar{T}_{i+\Delta x}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $0 \le i \le N-1$ are the inclusive nodes, i.e. the nodes lie within the domain, according to this schematic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/avE07.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/avE07.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conversion from node position to spatial position is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$i=N\frac{x}{L}-\frac{1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this staggered grid, the boundaries $x=0$ and $x=L$ are located halfway between two nodes. However, if we imagine a ghost node at $i=-1$ and $i=N$, we can define the boundary temperatures as:&#xA;$$T_{h}=\frac{\bar{T}_{-1}+\bar{T}_{0}}{2}\rightarrow\bar{T}_{-1}=2T_{h}-\bar{T}_{0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{\infty}=\frac{\bar{T}_{N-1}+\bar{T}_{N}}{2}\rightarrow\bar{T}_{N}=2T_{\infty}-\bar{T}_{N-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This then yields the following ODE's for $i=0$ and $i=N-1$:&#xA;$$\partial_{t}\bar{T}_{0}	=	\frac{\alpha}{\Delta x^{2}}\left[\bar{T}_{-1}-2\bar{T}_{0}+\bar{T}_{1}\right]&#xA;	=	\frac{\alpha}{\Delta x^{2}}\left[2T_{h}-3\bar{T}_{0}+\bar{T}_{1}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_{t}\bar{T}_{N-1}	=	\frac{\alpha}{\Delta x^{2}}\left[\bar{T}_{N-2}-2\bar{T}_{N-1}+\bar{T}_{N}\right]&#xA;	=	\frac{\alpha}{\Delta x^{2}}\left[\bar{T}_{N-2}-3\bar{T}_{N-1}+2T_{\infty}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we are in a position to define our linear system. If we define a temperature vector:&#xA;$$\bar{\boldsymbol{T}}=\begin{bmatrix}\bar{T}_{0} &amp;amp; \bar{T}_{1} &amp;amp; \cdots &amp;amp; \bar{T}_{N-2} &amp;amp; \bar{T}_{N-1}\end{bmatrix}^{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the ODE system can be written as:&#xA;$$\partial_{t}\bar{\boldsymbol{T}}=\omega\left(\boldsymbol{A}\cdot\bar{\boldsymbol{T}}+\boldsymbol{b}\right) \quad \omega = \frac{\alpha}{\Delta x^{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&#xA;$$\boldsymbol{A}=\begin{bmatrix}-3 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA;1 &amp;amp; -2 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; \ddots &amp;amp; \ddots &amp;amp; \ddots &amp;amp; 0\\&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; -2 &amp;amp; 1\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; -3&#xA;\end{bmatrix}\quad \boldsymbol{b}=\begin{bmatrix}2T_{h}\\&#xA;0\\&#xA;0\\&#xA;0\\&#xA;2T_{\infty}&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this system for $T_h=1$, $T_{\infty}=0$, $L=1$, gives the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip1in$ &lt;a href=&quot;https://i.stack.imgur.com/GMRqO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GMRqO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the temperature profiles are given at different times and the markers are the analytical solution given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T=T_h-\left(T_h-T_{\infty}\right)\frac{x}{L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Adding a convective boundary condition&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's change the boundary condition at $x=L$ to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-k\partial_{x}T\left(L,t\right)=h\left(T\left(L,t\right)-T_{\infty}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We discretize this condition:&#xA;$$-k\left(\frac{\bar{T}_{N}-\bar{T}_{N-1}}{\Delta x}\right)=h\left(\frac{\bar{T}_{N-1}+\bar{T}_{N}}{2}-T_{\infty}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and rearrange for the ghost node temperature at $i=N$:&#xA;$$\bar{T}_{N}=\frac{\mathrm{Nu}_{x}}{1+\frac{1}{2}\mathrm{Nu}_{x}}T_{\infty}+\frac{1-\frac{1}{2}\mathrm{Nu}_{x}}{1+\frac{1}{2}\mathrm{Nu}_{x}}\bar{T}_{N-1}\quad\mathrm{Nu}_{x}=\frac{h\Delta x}{k}$$&#xA;which is now a function of a grid Nusselt number which specifies the relative importance of convective to conductive heat transfer. If $\mathrm{Nu}_{x}\ll1$ then convection is negligible compared to conduction and vice versa if $\mathrm{Nu}_{x}\gg1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that if we take the limits:&#xA;$$\lim_{\mathrm{Nu}_{x}\rightarrow0}\bar{T}_{N}=\bar{T}_{N-1}\quad\lim_{\mathrm{Nu}_{x}\rightarrow\infty}\bar{T}_{N}=2T_{\infty}-\bar{T}_{N-1}$$&#xA;we retrieve a no flux condition as $\mathrm{Nu}_{x}\rightarrow0$ and the previously found constant temperature condition as $\mathrm{Nu}_{x}\rightarrow\infty$. This gives a good indication of the correctness of the implementation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To incorporate this condition we modify the ODE to read:&#xA;$$\partial_{t}\bar{T}_{N-1}	=	\frac{\alpha}{\Delta x^{2}}\left[\bar{T}_{N-2}-2\bar{T}_{N-1}+\bar{T}_{N}\right]&#xA;	=	\frac{\alpha}{\Delta x^{2}}\left[\bar{T}_{N-2}-\left(2-\frac{1-\frac{1}{2}\mathrm{Nu}_{x}}{1+\frac{1}{2}\mathrm{Nu}_{x}}\right)\bar{T}_{N-1}+\frac{\mathrm{Nu}_{x}}{1+\frac{1}{2}\mathrm{Nu}_{x}}T_{\infty}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which modifies $A$ and $b$ to:&#xA;$$\boldsymbol{A}=\begin{bmatrix}-3 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA;1 &amp;amp; -2 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0\\&#xA;0 &amp;amp; \ddots &amp;amp; \ddots &amp;amp; \ddots &amp;amp; 0\\&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; -2 &amp;amp; 1\\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; -\left(2-\frac{1-\frac{1}{2}\mathrm{Nu}_{x}}{1+\frac{1}{2}\mathrm{Nu}_{x}}\right)&#xA;\end{bmatrix}\quad \boldsymbol{b}=\begin{bmatrix}2T_{h}\\&#xA;0\\&#xA;0\\&#xA;0\\&#xA;\frac{\mathrm{Nu}_{x}}{1+\frac{1}{2}\mathrm{Nu}_{x}}T_{\infty}&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain the temperature at $x=L$, we have:&#xA;$$T_{L}=\frac{\bar{T}_{N-1}+\bar{T}_{N}}{2}=\frac{\frac{1}{2}\mathrm{Nu}_{x}}{1+\frac{1}{2}\mathrm{Nu}_{x}}T_{\infty}+\frac{1}{1+\frac{1}{2}\mathrm{Nu}_{x}}\bar{T}_{N-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result for the same system as above with $\mathrm{Nu}=N\mathrm{Nu}_{x}=1$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip1in$ &lt;a href=&quot;https://i.stack.imgur.com/oDYP2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oDYP2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analytical solution is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T=T_h-\left(T_h-T_{\infty}\right)\frac{\mathrm{Nu}}{1+\mathrm{Nu}}\frac{x}{L}$$&#xA;$$\mathrm{Nu}=\frac{hL}{k}=\mathrm{Nu}_{x}\frac{L}{\Delta x}=N\mathrm{Nu}_{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Simulating a composite material&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simulate a composite material, we define an interface located at $x=x_b$ located at node $i=i_b$ where the region $1$ ($0\le x\lt x_b$) is characterized by $\alpha_1$ and region $2$ ($x_b\lt x\le L$) is characterized by $\alpha_2$. At the interface, there must be thermal equilibrium and thermal flux continuity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{1}\left(x_{b},t\right)=T_{2}\left(x_{b},t\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-k_{1}\partial_{x}T_{1}\left(x_{b},t\right)=-k_{2}\partial_{x}T_{2}\left(x_{b},t\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $i_b^-$ and $i_b^+$ are the nodes $i$ closest to the interface in region $1$ and $2$ respectively; i.e. $i_b^-=\mathrm{floor}(i_b)$ and $i_b^+=\mathrm{ceil}(i_b)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We discretize the equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\bar{T}_{i_{b}^{-},1}+\bar{T}_{i_{b}^{-}+\Delta x,1}}{2}=\frac{\bar{T}_{i_{b}^{+}-\Delta x,2}+\bar{T}_{i_{b}^{+},2}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-k_{1}\frac{\bar{T}_{i_{b}^{-}+\Delta x,1}-\bar{T}_{i_{b}^{-},1}}{\Delta x}=-k_{2}\frac{\bar{T}_{i_{b}^{+},2}-\bar{T}_{i_{b}^{+}-\Delta x,2}}{\Delta x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and solve for the ghost nodes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{T}_{i_{b}^{-}+\Delta x,1}=\frac{\kappa-1}{\kappa+1}\bar{T}_{i_{b}^{-},1}+\frac{2}{\kappa+1}\bar{T}_{i_{b}^{+},2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{T}_{i_{b}^{+}-\Delta x,2}=\frac{2\kappa}{\kappa+1}\bar{T}_{i_{b}^{-},1}-\frac{\kappa-1}{\kappa+1}\bar{T}_{i_{b}^{+},2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\kappa = k_1/k_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ODEs closest to the interface node are changed to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_{t}\bar{T}_{i_{b}^{-}}=\frac{\alpha_{1}}{\Delta x^{2}}\left[\bar{T}_{i_{b}^{-}-\Delta x}-\left(2-\frac{\kappa-1}{\kappa+1}\right)\bar{T}_{i_{b}^{-}}+\frac{2}{\kappa+1}\bar{T}_{i_{b}^{-}+\Delta x}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_{t}\bar{T}_{i_{b}^{+}}=\frac{\alpha_{2}}{\Delta x^{2}}\left[\frac{2\kappa}{\kappa+1}\bar{T}_{i_{b}^{+}-\Delta x}-\left(2+\frac{\kappa-1}{\kappa+1}\right)\bar{T}_{i_{b}^{+}}+\bar{T}_{i_{b}^{+}+\Delta x}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which modifies the linear system to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\partial_{t}\bar{\boldsymbol{T}}=\boldsymbol{\omega}\cdot\left(\boldsymbol{A}\cdot\bar{\boldsymbol{T}}+\boldsymbol{b}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\boldsymbol{A}=\begin{bmatrix}-3 &amp;amp; 1\\&#xA;1 &amp;amp; -2 &amp;amp; 1\\&#xA; &amp;amp; \ddots &amp;amp; \ddots &amp;amp; \ddots\\&#xA; &amp;amp;  &amp;amp; 1 &amp;amp; -2 &amp;amp; 1\\&#xA; &amp;amp;  &amp;amp;  &amp;amp; 1 &amp;amp; -\left(2-\gamma\right) &amp;amp; \frac{2}{\kappa+1}\\&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \frac{2\kappa}{\kappa+1} &amp;amp; -\left(2+\gamma\right) &amp;amp; 1\\&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; 1 &amp;amp; -2 &amp;amp; 1\\&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \ddots &amp;amp; \ddots &amp;amp; \ddots\\&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; 1 &amp;amp; -2 &amp;amp; 1\\&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; 1 &amp;amp; -\left(2-\delta\right)&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gamma=\frac{\kappa-1}{\kappa+1}\quad\delta=\frac{1-\frac{1}{2}\mathrm{Nu}_{x}}{1+\frac{1}{2}\mathrm{Nu}_{x}}&#xA; $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b=\begin{bmatrix}2T_{h}\\&#xA;\\&#xA;\\&#xA;\\&#xA;\\&#xA;\\&#xA;\\&#xA;\\&#xA;\\&#xA;\frac{\mathrm{Nu}_{x}}{1+\frac{1}{2}\mathrm{Nu}_{x}}T_{\infty}&#xA;\end{bmatrix}\quad\boldsymbol{\omega}=\frac{1}{\Delta x^{2}}\begin{bmatrix}\alpha_{1}\\&#xA; &amp;amp; \ddots\\&#xA; &amp;amp;  &amp;amp; \ddots\\&#xA; &amp;amp;  &amp;amp;  &amp;amp; \ddots\\&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \alpha_{1}\\&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \alpha_{2}\\&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \ddots\\&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \ddots\\&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \ddots\\&#xA; &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp;  &amp;amp; \alpha_{2}&#xA;\end{bmatrix}&#xA; $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I have defined a new matrix $\boldsymbol{\omega}$ which takes care of different values of $\alpha$ at different nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain the temperature at $x=x_b$, we have:&#xA;$$T_{x_{b},1}=\frac{\bar{T}_{i_{b}^{-},1}+\bar{T}_{i_{b}^{-}+\Delta x,1}}{2}=\frac{\kappa}{\kappa+1}\bar{T}_{i_{b}^{-},1}+\frac{1}{\kappa+1}\bar{T}_{i_{b}^{+},2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{x_{b},2}=\frac{\bar{T}_{i_{b}^{+}-\Delta x,2}+\bar{T}_{i_{b}^{+},2}}{2}=\frac{\kappa}{\kappa+1}\bar{T}_{i_{b}^{-},1}+\frac{1}{\kappa+1}\bar{T}_{i_{b}^{+},2}&#xA; $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which shows $T_{x_{b},1}=T_{x_{b},2}$ as imposed by the boundary condition at the interface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this system, with similar parameters as above and $\alpha_1=k_1=10$ and $\alpha2=k_2=1$ yields this nice result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hskip1in$ &lt;a href=&quot;https://i.stack.imgur.com/x7wTI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x7wTI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analytical solution is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{1}=T_{h}+\left(T_{\infty}-T_{h}\right)\frac{1}{\kappa}\frac{\Gamma\Pi}{1+\Pi}\frac{x}{L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{2}=T_{h}+\left(T_{\infty}-T_{h}\right)\frac{1+\Gamma\Pi\frac{x}{L}}{1+\Pi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma=\frac{\mathrm{Nu}}{1+\mathrm{Nu}}\quad\Pi=\frac{1}{\Gamma}\frac{L}{x_{b}}\frac{\kappa}{1-\kappa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Python code:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following Python code is used to generate the results.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import math&#xA;import numpy as np&#xA;import scipy.integrate as spi&#xA;import matplotlib.pyplot as plt&#xA;&#xA;from __future__ import division, print_function&#xA;&#xA;l, xb = 1, 0.5&#xA;a2, alpha = 1, 10 #alpha = a1/a2&#xA;k2, kappa = 1, 10 #kappa = k1/k2&#xA;Th, Tinf = 1, 0&#xA;Nu = 1 #= hL/k2&#xA;tp = l**2/a2&#xA;&#xA;n = 100&#xA;dx = l/n&#xA;Nux = Nu/n&#xA;&#xA;ib = n*xb/l-0.5&#xA;ibm = int(math.floor(ib))&#xA;ibp = int(math.ceil(ib))&#xA;&#xA;A = np.zeros((n,n))&#xA;b = np.zeros((n,))&#xA;omega = np.zeros((n,n))&#xA;&#xA;A[0,0], A[0,1], b[0], omega[0,0] = -3, 1, 2*Th, alpha*a2/dx**2&#xA;for i in xrange(1,ibm):&#xA;    A[i,i-1], A[i,i], A[i,i+1], omega[i,i] = 1, -2, 1, alpha*a2/dx**2&#xA;A[ibm,ibm-1], A[ibm,ibm], A[ibm,ibm+1], omega[ibm,ibm] = 1, -(2-(kappa-1)/(kappa+1)), 2/(kappa+1), alpha*a2/dx**2&#xA;A[ibp,ibp-1], A[ibp,ibp], A[ibp,ibp+1], omega[ibp,ibp] = 2*kappa/(kappa+1), -(2+(kappa-1)/(kappa+1)), 1, a2/dx**2&#xA;for i in xrange(ibp+1,n-1):&#xA;    A[i,i-1], A[i,i], A[i,i+1], omega[i,i] = 1, -2, 1, a2/dx**2&#xA;A[n-1,n-2], A[n-1,n-1], b[n-1], omega[n-1,n-1] = 1, -(2-(1-Nux/2)/(1+Nux/2)), Nux/(1+Nux/2)*Tinf, a2/dx**2&#xA;&#xA;def diff_eq(T,t):&#xA;    return np.dot(omega, np.dot(A,T) + b)&#xA;&#xA;i = np.linspace(0,n-1,n)&#xA;x = l/n*(i+1/2)&#xA;t = tp*np.logspace(-4,1,6)&#xA;T0 = Tinf*np.ones((n,))&#xA;&#xA;sol = spi.odeint(diff_eq, T0, t)&#xA;&#xA;## boundary condition at x=0&#xA;x = np.insert(x, 0, 0)&#xA;sol = np.insert(sol, 0, Th, axis=1)&#xA;## boundary condition at x=l&#xA;x = np.insert(x, n+1, l)&#xA;Tl = (Nux/2*Tinf + sol[:,-1])/(1+Nux/2)&#xA;sol = np.insert(sol, n+1, Tl, axis=1)&#xA;&#xA;gamma = Nu/(1+Nu)&#xA;pi = l/xb/gamma*kappa/(1-kappa)&#xA;ana1 = Th+(Tinf-Th)*(gamma*pi/kappa*x/l)/(1+pi)&#xA;ana2 = Th+(Tinf-Th)*(1+gamma*pi*x/l)/(1+pi)&#xA;ana = np.append(ana1[x&amp;lt;xb],ana2[x&amp;gt;=xb])&#xA;&#xA;plt.plot(x, np.transpose(sol), '-',&#xA;         x[::n//10], ana[::n//10], 'o')&#xA;plt.xlabel('Spatial coordinate, x')&#xA;plt.ylabel('Temperature coordinate, T')&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second order approximations on boundaries&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to quickly show that these approximations are indeed second-order accurate. To do this we take a Taylor expansion about the relevant boundary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T\left(x_{b}+\frac{1}{2}\Delta x,t\right)=T\left(x_{b},t\right)+\partial_{x}T\left(x_{b}\right)\cdot\frac{1}{2}\Delta x+\frac{1}{2}\partial_{x}^{2}T\left(x_{b}\right)\cdot\left(\frac{1}{2}\Delta x\right)^{2}+O\left(\Delta x^{3}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T\left(x_{b}-\frac{1}{2}\Delta x,t\right)=T\left(x_{b},t\right)-\partial_{x}T\left(x_{b}\right)\cdot\frac{1}{2}\Delta x+\frac{1}{2}\partial_{x}^{2}T\left(x_{b}\right)\cdot\left(\frac{1}{2}\Delta x\right)^{2}+O\left(\Delta x^{3}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we add these equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T\left(x_{b}+\frac{1}{2}\Delta x,t\right)+T\left(x_{b}-\frac{1}{2}\Delta x,t\right)=2T\left(x_{b},t\right)+O\left(\Delta x^{2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and rearrange we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T\left(x_{b},t\right)=\frac{T\left(x_{b}+\frac{1}{2}\Delta x,t\right)+T\left(x_{b}-\frac{1}{2}\Delta x,t\right)}{2}+O\left(\Delta x^{2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is our approximation for the temperature at a boundary shown to be indeed second-order in $\Delta x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we subtract the two Taylor expansion:&#xA;$$T\left(x_{b}+\frac{1}{2}\Delta x,t\right)-T\left(x_{b}-\frac{1}{2}\Delta x,t\right)=\partial_{x}T\left(x_{b}\right)\Delta x+O\left(\Delta x^{3}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and rearrange, we get:&#xA;$$\partial_{x}T\left(x_{b}\right)=\frac{T\left(x_{b}+\frac{1}{2}\Delta x,t\right)-T\left(x_{b}-\frac{1}{2}\Delta x,t\right)}{\Delta x}+O\left(\Delta x^{2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the approximation for the temperature gradient at a boundary, again shown to be second-order accurate in $\Delta x$.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastEditorUserId="3100" LastEditDate="2015-12-16T05:52:29.530" LastActivityDate="2015-12-16T05:52:29.530" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="5926" PostTypeId="2" ParentId="5749" CreationDate="2015-10-26T16:38:06.890" Score="5" Body="&lt;p&gt;&lt;strong&gt;What's the reason for going to the cost and effort of doing this?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many places, placing a stream into a culvert can be cheaper in the long run. Streams (particularly fast flowing ones) erode the land and may require erosion protection added in the future. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An advantage is reduced land use: Land that does not have a stream running through it can be used for other purposes, although the culvert will need to, obviously, be assessed before plomping (technical term..) a multi-storey car park on top of it! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it a safety concern having creeks in residential areas?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Highly dependent on the area in question and the nature of the creek or stream. Naturally, a stream is a hazard to people of all ages, particularly kids though. But so is a culvert if the access to it is not cordoned off. As answered previously, an open grassy pitch is more favourable than a flowing stream. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Do creeks have some effect on the land which would affect nearby buildings?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Creeks have a tendency to meander in the long term (decades to centuries), and deeply scour the nearby banks in the short term (years). Scoured banks can become steep and rocky, and even collapse. If there are buildings located nearby, then this can cause structural damage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is it just for the convenience of being able to cross them without a bridge?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an engineer with Highway experience you'd be amazed at how many culverts manage water flow run under many of our roads. In Scotland, it's not unusual for there to be a 600mm diameter culvert every few hundred metres or so on many trunk roads and motorways helping ease water through them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the time, these are never seen or acknowledged as they can be down steep road verges. This is because these roads tend to be built very straight (much like railways), opposed to older roads that simply follow ancient paths and travelling routes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A final note:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Culverts can be used to change the velocity of flowing water. For example if a stream has a narrow channel, the water will flow quickly causing it to quickly erode the bank. You can slow the water velocity in an area by passing it through a wide culvert. With a wider area of flow, the water will travel more slowly. This can cause advantages down stream. Similarly, the same effect can be caused upstream by forcing the water through a narrower passage. This will cause the water to back-up up stream reducing it's speed. &lt;/p&gt;&#xA;" OwnerUserId="3093" LastEditorUserId="3093" LastEditDate="2015-10-27T08:39:14.480" LastActivityDate="2015-10-27T08:39:14.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5927" PostTypeId="1" AcceptedAnswerId="5928" CreationDate="2015-10-26T16:38:07.030" Score="1" ViewCount="5023" Body="&lt;p&gt;I'm not sure how to calculate force $P$ when the moment is a given with the force $P$ at an angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also having trouble breaking the moment into the $x$ and $y$ components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SXy9z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SXy9z.jpg&quot; alt=&quot;force P causes a moment of 500 N-m about point A. Determine force P&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3522" LastEditorUserId="1832" LastEditDate="2015-10-26T18:26:21.603" LastActivityDate="2015-10-26T18:26:21.603" Title="Calculate the force when given the moment" Tags="|mechanical-engineering|torque|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ClosedDate="2015-10-26T18:26:57.690" ContentLicense="CC BY-SA 3.0" />
  <row Id="5928" PostTypeId="2" ParentId="5927" CreationDate="2015-10-26T17:46:01.040" Score="1" Body="&lt;p&gt;$$&#xA;\tau = 500 \mbox{Nm} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau = r \times F \\&#xA;\tau = |r| |F| \sin{(\theta)} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the lower left angle is $43^{\circ}$, then the top angle from the bar $r$ to vertical is then $180 - 90 - 43 = \boxed{47^{\circ}}$. This means that $\theta$, the angle between the bar $r$ and the force $P$ is $47^{\circ} + 15^{\circ} = \boxed{\theta = 62^{\circ}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau = |r| |F| \sin{(\theta)} \\&#xA;|F| = \frac{\tau}{r \sin{(\theta)}} \\&#xA;|F| = \frac{ 500\mbox{Nm} }{(1.8 \mbox{m} ) \sin{(62^{\circ})}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\boxed{F = 314.6 \mbox{N}} \\&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-10-26T17:46:01.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5931" PostTypeId="2" ParentId="5923" CreationDate="2015-10-27T01:37:52.483" Score="2" Body="&lt;p&gt;Since you are planning on testing you are already ahead of a lot of companies. Trevor makes some good points in the comments. Failure from vibration is certainly non-linear, but it typically errors in favor of failure. A good initial test would be to take it to the extreme; put 10g or 20g accelerations on it at various frequencies. If it survives a week of that, it is a very high certainty that a 2g vibration will have an infinite fatigue limit. If it doesnt survive, you can dial it back until it does, you now have a metric for improving the vibration isolation or testing higher quality components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are published standards for this, but official standards have to be purchase. Its a corrupt system in my opinion, but you can find older versions with some creative web searching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;br&gt;&#xA;&lt;a href=&quot;http://media.digikey.com/pdf/data%20sheets/varitronix%20pdfs/typical%20vi%20or%20vim%20spec.pdf&quot; rel=&quot;nofollow&quot;&gt;LCD tested per IEC 68-2-6&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.eleq.com/lmbinaries/eleq_vstr_typical_current_transformer_set.pdf&quot; rel=&quot;nofollow&quot;&gt;Transformer tested per IEC 68-2-6&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A list of a companies testing abilities &lt;a href=&quot;http://www.environlab.com/pdf/Scope-Expansion-Announcement.pdf&quot; rel=&quot;nofollow&quot;&gt;(full list)&lt;/a&gt;:&lt;br&gt;&#xA;Sinusoidal Vibration CEI / IEC 68-2-6,&#xA;Edition 6, 1995 &lt;br&gt;&#xA;Combined&#xA;Cold/Vibration&#xA;(Sinusoidal)&#xA;CEI / IEC 68-2-50, Edition 1, 1983&lt;br&gt;&#xA;Combined Dry&#xA;Heat/Vibration&#xA;(Sinusoidal)&#xA;CEI / IEC 68-2-51, Edition 1, 1983&lt;br&gt;&#xA;Combined&#xA;Temperature (Cold&#xA;and Dry&#xA;Heat)/Vibration&#xA;(Sinusoidal)&#xA;CEI / IEC 68-2-53, Edition 1, 1984&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-27T01:37:52.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5932" PostTypeId="1" CreationDate="2015-10-27T01:49:53.717" Score="4" ViewCount="157" Body="&lt;p&gt;I've currently got a mechanism that is powered via a shaft connected to a ducted fan. The ducted fan is powered by an existing air flow through the duct. I am looking to re-design the fan to increase torque/performance. The fan will be injection molded and is about 2&quot; is diameter. With no-load, the current fan spins at 10k RPM. I need to make measurements under load, but it's significantly slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the parameters of the fan blades be adjusted to maximize the output torque?&lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="368" LastEditDate="2015-10-29T21:29:03.963" LastActivityDate="2015-10-29T21:29:03.963" Title="Inline ducted fan/air turbine blade design to maximize output torque" Tags="|mechanical-engineering|fluid-mechanics|airflow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5933" PostTypeId="2" ParentId="389" CreationDate="2015-10-27T02:09:28.703" Score="1" Body="&lt;p&gt;The use of any idea/technology is dependent on its cost effectiveness. Efficiency is just one part of the analysis. If the modules were free and never went bad, they would certainly be in use. However, high temperature modules are expensive for their power produced; making them currently not cost effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EnergyNumbers does a good job of explaining the thermodynamics involved. Because of Carnot you cant harvest significant amount of power in this fashion. This is also why money for peltier modules is better spent on increased engine efficiency/performance instead. However, high temperature waste heat will be there regardless and when thermoelectric becomes cost effective, you will see it in use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite what people may advertise, no idea/technology is &quot;green&quot; until it can pull its own weight in a cost benefit analysis. You have to look at the whole picture; how many dollars and raw materials did it consume to provide this &quot;green&quot; energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would like to discover it for yourself, here are some high temperature modules you can price and evaluate. Remember the watts listed is the consumed power not the generated power. &lt;a href=&quot;http://tetech.com/peltier-thermoelectric-cooler-modules/high-temperature/&quot; rel=&quot;nofollow&quot;&gt;tetech.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-27T02:09:28.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5935" PostTypeId="2" ParentId="5923" CreationDate="2015-10-27T17:08:34.967" Score="3" Body="&lt;p&gt;It's not just about selecting a &lt;strong&gt;level&lt;/strong&gt; of vibration. There's a lot more that goes into developing a vibration test program that does what you're talking about. There's probably not any one vibration test that will address everything you're looking to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you need to understand the vibration environment that your product will see. Are you only interested in the operating environment? What about the vibration during manufacturing, transport and installation. If it's portable, you should consider that as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, what kind of thing are you designing? A mass produced near-commodity product, say something along the lines of a cheap clock radio? Or is it a unique high cost item, like a satellite that can't be damaged in testing? Or something in the middle of that spectrum? How much are you willing to invest in testing? How much are your customers willing to pay for the resulting reliability? At a certain point, it's better to simply replace/repair a product than it is to design the thing that will last forever. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When developing a vibration test program, intensity is only one component of the input, frequency is another. The rotating frequency of your motor will likely dominate, but there will be other factors as well and if you're serious about this, it will pay off to develop a random vibration profile based on your device's vibration environment. Does the operating vibration of your device excite any of the resonances of your subcomponents/systems? I've seen transformers and large caps pull away from pcb's because the vibration the system experienced was at the component's resonant frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many degrees of freedom are significant? In other words, is the vibration mainly up/down, right/left, fore/aft? What about pitch, roll yaw?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To deliver a product out of the gate with high reliability, you're talking about developing a HALT/HASS (Highly Accelerated Life Test/ Highly Accelerated Stress Screening) test program for your product. You'll have to break a few test samples, but if that kind of reliability is what you're after, HALT/HASS will get you closer than just picking a few specs from the internet and testing to those. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the plan goes something like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Identify frequencies of interest (Operating frequencies, component resonances, environmental inputs etc.) You can do this experimentally (sine sweep, resonance search), analytically or in some combination. &lt;/li&gt;&#xA;&lt;li&gt;Collect vibration data from the device in operation for each axis you want to test and for each, compute a PSD (Power Spectral Density) which tells you how much energy is distributed over any bandwidth. That level represents one standard deviation away from the mean acceleration of zero and the integral of the PSD plot is the overall RMS acceleration. &lt;/li&gt;&#xA;&lt;li&gt;Start shaking the DUT (device under test) while it's operating at increasing levels of vibration until something breaks. Find and fix that failure and repeat until something else fails and so on. Each time something fails, use a time varying stress model, typically some sort of cumulative damage model, to estimate the B1 life (the time at which reliability = 99% at 100% stress) Typically, this data analysis would be done with software like ALTA. &lt;/li&gt;&#xA;&lt;li&gt;If the B1 life isn't enough at that point, fix the issue and retest. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Typically, multiple units will be tested at once until they all fail or the B1 life prediction becomes adequate. Fix the failures that showed up during testing and re-test. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2015-10-27T17:08:34.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5936" PostTypeId="2" ParentId="5916" CreationDate="2015-10-27T17:30:52.740" Score="1" Body="&lt;p&gt;I'm going out on a limb and assuming that the reason the car doesn't make it up the incline is because the wheels are too slippery.  Therefore, you'll want to increase the coefficient of friction between the wheels and the ramp.  You can do this many different ways, but in general you need to add some material to your wheels that's softer/stickier than the current plastic.  The simplest way would be to wrap a few rubber bands around the wheels.  You could also cut some rubber sheeting (available at Home Depot, etc.) to fit around the wheel and glue it on.  The possibilities are endless.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-10-27T17:30:52.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5937" PostTypeId="1" CreationDate="2015-10-27T23:50:21.047" Score="0" ViewCount="73" Body="&lt;p&gt;&lt;strong&gt;Requirements:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I need to produce $20\ \mathrm{W}\,\mathrm{hr}^{-1}$ by burning wood.  &lt;/li&gt;&#xA;&lt;li&gt;The specific energy of wood is $21\ \mathrm{MJ}\,\mathrm{kg}$  &lt;/li&gt;&#xA;&lt;li&gt;Find the required mass of wood to produce $20\ \mathrm{W}\,\mathrm{hr}^{-1}$  &lt;/li&gt;&#xA;&lt;li&gt;Assume ideal conditions  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I tried:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=E/T$ where $P = 20\ \mathrm{W}\,\mathrm{hr}^{-1}$ and $T = 1\ \mathrm{hr} = 3600\ \mathrm{s}$, which yields $E=72\ \mathrm{kJ}$ when rearranged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I know that wood has a specific energy of $21\ \mathrm{MJ}\,\mathrm{kg}^{-1}$, thus specific energy is $E/m$ where $E=72\ \mathrm{kJ}$ as found above, which yields $21\times 10^6 = 72\times10^3/m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $m$ gives 3.4 milligrams of wood. I appear to have missed something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What have I missed?&lt;/p&gt;&#xA;" OwnerUserId="3538" LastEditorUserId="282" LastEditDate="2015-10-31T16:37:19.633" LastActivityDate="2015-10-31T16:37:19.633" Title="Required mass of wood to produce 20 Watts/Hour" Tags="|energy|power|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5938" PostTypeId="2" ParentId="5937" CreationDate="2015-10-28T00:20:55.920" Score="1" Body="&lt;p&gt;There it is! The left-hand-side of the specific energy equation is $21\times 10^6\ \mathrm{J}\:\mathrm{kg}^{-1}$, but the final units you've written are $\mathrm{mg}$. However, the $\mathrm{kg}$ would carry through since no conversion is noted, so that $m=3.4\times 10^{-3}\ \mathrm{kg}$, which is $3.4\ \mathrm{g}$, a more reasonable answer for perfectly efficient combustion of organic material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in the comments, one of the desired values is called power, but has units of $\mathrm{W}\,\mathrm{hr}^{-1}$, which is rate of change of power. My answer is predicated on the assumption the desired value is $20\ \mathrm{W}$.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-10-29T14:00:04.887" LastActivityDate="2015-10-29T14:00:04.887" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5939" PostTypeId="2" ParentId="5932" CreationDate="2015-10-28T00:35:46.940" Score="1" Body="&lt;p&gt;Fan/turbine design is a highly involved process that lots of very smart people dedicate their careers to. If this is for an experiment, certainly try building one, but if this is for an application/product with efficiency requirements, I would recommend sourcing a pre-built one spec'd to the proper torque and rpm required by the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can probably gain a lot of understanding in this by optimizing the fan/turbine you already have. A rule of thumb we used in water turbines was maximum power will be delivered at approximately half of the freewheeling RPM. This is probably different for gas turbines, but might be a good place to start. If it is currently loaded at 1000 RPM or less you are likely only receiving a fraction of the power your current turbine is capable of producing. Remember that you are not trying to maximize output torque(typically max at 0rpm) you are trying to maximize output power (RPM x Torque). It may be as simple as changing the gear ratio to your load or optimizing your load for your turbines maximum power RPM.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-28T00:35:46.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5940" PostTypeId="1" AcceptedAnswerId="5943" CreationDate="2015-10-28T10:23:35.860" Score="0" ViewCount="27" Body="&lt;p&gt;I want to know if the sensor nodes in any cluster in a wireless sensor network may detect events in a different ratio i.e.some nodes could detect events more others in the same cluster, so is power consumption in these nodes will be higher than other nodes? &#xA;i want to know if there is an application in which this assumption or question is possible with take in our concentration that we can applied clustering to increase the lifetime of this network.&lt;/p&gt;&#xA;" OwnerUserId="3540" LastEditorUserId="3540" LastEditDate="2015-10-28T12:22:58.480" LastActivityDate="2015-10-28T13:36:37.597" Title="Could a one sensor node in a cluster sense a data more the others?" Tags="|telecommunication|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5941" PostTypeId="1" CreationDate="2015-10-28T12:20:40.357" Score="3" ViewCount="1138" Body="&lt;p&gt;I am considering some empirical signal propagation models (like Hata, Ericsson models) to calculate 4 parameters: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Path loss&lt;/li&gt;&#xA;&lt;li&gt;Received Signal Strength (RSS)&lt;/li&gt;&#xA;&lt;li&gt;signal-to-interference-plus-noise ratio&lt;/li&gt;&#xA;&lt;li&gt;Throughput&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Path loss is okay as each propagation model has its own formula, but how can the other three parameters to be calculated?&lt;/p&gt;&#xA;" OwnerUserId="3541" LastEditorUserId="33" LastEditDate="2015-11-17T02:12:36.803" LastActivityDate="2015-11-17T02:12:36.803" Title="signal-to-interference-plus-noise ratio (SINR) and others calculation" Tags="|electrical-engineering|telecommunication|wireless-communication|signal|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5943" PostTypeId="2" ParentId="5940" CreationDate="2015-10-28T13:36:37.597" Score="0" Body="&lt;p&gt;The function of a node and how it interacts on the mesh network (assuming a mesh-connected node) are two different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such nodes almost certainly will have a microcontroller in them that performs the node function, like measuring temperature periodically or whatever.  This micro then also usually decides when it should attempt to communicate on the mesh network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the micro can take many measurements per mesh connection, or connect many times without taking any measurements.  The two are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the mesh is only powered up at regular intervals, then a node can't report any data no matter what activity it might detect.  It has to wait until the next mesh interval.  Until then, it has no way to pass on any of its data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some nodes might consume more power if whatever they are connected to is more active.  Such activity could cause a node to wake up and do something, then go back to low power state until the next event.  Other types of nodes just sample something periodically, so wouldn't take more power.  It's all up to whatever policies are encoded in the firmware of the micro.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-10-28T13:36:37.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5944" PostTypeId="1" CreationDate="2015-10-28T18:16:20.310" Score="4" ViewCount="18139" Body="&lt;p&gt;Does any one here know of an authoritative source (ASME, ISA, NFPA, OSHA, etc..) that gives a definition for the following in one of their standards or codes and the number for that standard or code:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;LO (Valve Locked Open)&lt;/li&gt;&#xA;&lt;li&gt;LC (valve Locked Closed)&lt;/li&gt;&#xA;&lt;li&gt;CSO (Valve Car Sealed Open)&lt;/li&gt;&#xA;&lt;li&gt;CSC (Valve Car Sealed Closed)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for an explanation, I'm looking for a citable code.&lt;/p&gt;&#xA;" OwnerUserId="3014" LastEditorUserId="41" LastEditDate="2015-10-28T18:41:26.820" LastActivityDate="2023-09-15T03:14:56.540" Title="LO/LC, CSO/CSC Valve Definition" Tags="|mechanical-engineering|valves|standards|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5945" PostTypeId="1" CreationDate="2015-10-28T19:35:49.420" Score="4" ViewCount="1743" Body="&lt;p&gt;I have a simple linear translation stage that is driven by a screw, and I would like to attach a stepper motor to the screw drive.  The axis of the screw and the axis of the motor are approximately colinear.  The issue is that, as the screw turns it recedes away from (or advances towards) the motor.  The spacing between the motor shaft and the screw drive can change by as much as 10 mm.  A diagram of the problem and an example of the type of stage are shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I couple the stepper motor to the screw drive in a way that allows for this expansion/contraction of the space between the two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xgYs4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xgYs4.png&quot; alt=&quot;The problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0nkmh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0nkmhm.jpg&quot; alt=&quot;Translation stage&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-10-28T20:39:23.077" LastActivityDate="2015-10-28T23:07:37.773" Title="How to couple a motor to a screw drive?" Tags="|mechanical-engineering|motors|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="5947" PostTypeId="2" ParentId="5601" CreationDate="2015-10-28T22:10:11.573" Score="0" Body="&lt;p&gt;Here is my solution for those with similar questions - cut out a small portion of the part (in the model) and print that. Replace that part with a piece of wood that fits the cut out, and drill into the wood instead. Attach them together with either screws or glue (don't use CA with ABS).&lt;/p&gt;&#xA;" OwnerUserId="2164" LastActivityDate="2015-10-28T22:10:11.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5948" PostTypeId="1" CreationDate="2015-10-28T22:18:59.693" Score="2" ViewCount="81" Body="&lt;p&gt;I have a motor whose specifications say that its torque is 190g*cm. If I have the formula for the mass of the wheel vs radius, is there any way I can find the radius to get the best linear speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula that relates mass and radius is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m=1.14r-0.948$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the m is mass in grams and r is radius in cm&lt;/p&gt;&#xA;" OwnerUserId="2164" LastEditorUserId="2164" LastEditDate="2015-11-03T22:48:25.150" LastActivityDate="2015-11-03T22:48:25.150" Title="How do I calculate the best mass for the highest linear speed?" Tags="|mechanical-engineering|motors|torque|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5949" PostTypeId="1" CreationDate="2015-10-28T22:30:12.813" Score="4" ViewCount="144" Body="&lt;p&gt;I studied a thesis about a axial turbine and but i didn't understood the bold part of following sentences. What is the exact meaning of &quot;dot product of the pressure and viscous force coefficients on each face with the specified force vector&quot;? This thesis is performed by using Fluent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*The forces coefficients  (CA and CT) on blade surface were calculated by summing the dot product of the pressure and viscous force coefficients on each face with the specified force vector.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="272" LastEditDate="2015-10-29T13:49:13.187" LastActivityDate="2015-10-29T13:49:13.187" Title="What is the exact meaning of &quot;dot product of the pressure and viscous force coefficients on each face with the specified force vector&quot;?" Tags="|mechanical-engineering|fluid-mechanics|turbines|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5950" PostTypeId="1" CreationDate="2015-10-28T22:41:37.783" Score="7" ViewCount="14851" Body="&lt;p&gt;Often it's desirable to reduce the turbulence level in an internal flow &lt;em&gt;without changing the Reynolds number or fluid&lt;/em&gt;. This could be measured by the &lt;a href=&quot;http://www.cfd-online.com/Wiki/Turbulence_intensity&quot; rel=&quot;noreferrer&quot;&gt;turbulence intensity&lt;/a&gt;, though other measures might be better in different circumstances. I haven't seen a good compilation of these techniques, and I believe such a compilation could be very useful to engineers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are the ways I'm aware of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Flow_straightener&quot; rel=&quot;noreferrer&quot;&gt;Flow straighteners&lt;/a&gt; reduce turbulence and swirl, but have an associated pressure drop. You can also get similar effects with meshes, porous media, and fibers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's commonly believed that turbulence transition begins in pipe flows at a Reynold numbers of about $2300$ in general, but laminar flows at far higher Reynolds numbers can be obtained with better quality experimental setups as discussed in &lt;a href=&quot;http://www.annualreviews.org/doi/abs/10.1146/annurev-fluid-122109-160652&quot; rel=&quot;noreferrer&quot;&gt;this review paper on turbulence transition in pipe flows&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Experimental evidence suggests that the laminar state can be achieved in pipe flows over a wide range of $Re$ with the record standing at $Re = 100,000$ by Pfenniger (1961). Reynolds himself managed to achieve $Re&#xA;= 13,000$, and Ekman (1911) later improved on this to $∼50,000$ using Reynolds’ original apparatus. There have been many others (Barnes &amp;amp; Coker 1905, Draad et al. 1998, Gilbrech &amp;amp; Hale 1965, Goldstein 1965, Hof et al. 2003, Nishi et al. 2008, Paterson &amp;amp; Abernathy 1972, Schiller 1921, Wygnanski &amp;amp; Champagne 1973) who have achieved Hagen-Poiseuille flow over a wide range of $Re$ greater than 20,000, which is the upper value in Figure 2. Achieving laminar flows at high values of $Re$ is an indication of the quality of an experimental facility and gives some confidence that the observations will not be contaminated by extraneous background disturbances such as entrance flow effects, convection, and geometrical irregularities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From what I understand, in addition to the effects listed above, surface roughness and vibration can also trigger turbulence, so you can delay the development of turbulence with smoother surfaces and/or vibration damping in some cases. The standard transition criteria appears to be correct, but for systems that don't take special precautions to eliminate disturbances.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Relaminarization&lt;/em&gt; is when a turbulent flow becomes laminar, and is a stricter version of what I'm after. I'm interested in reducing the turbulence level. The flow could still remain turbulent, just less so. Here are a few ways to do this discussed by this &lt;a href=&quot;http://users.ictp.it/~krs/pdf/1979_003.pdf&quot; rel=&quot;noreferrer&quot;&gt;review article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;dissipation of turbulence (could happen from enlargement in a pipe)&lt;/li&gt;&#xA;&lt;li&gt;curved or rotating flows (could introduce swirl, though)&lt;/li&gt;&#xA;&lt;li&gt;acceleration (could happen from a contraction in a pipe)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are a number of papers on suppression of turbulence via sound emission or vibration. I don't know much about this, but I'm aware that it exists. If anyone knows more about this, or knows a good place to learn more, I'm interested.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you can change the fluid entirely, that could be a good option to reduce turbulence by reducing the Reynolds number. There also are a variety of additives (mainly for water?) which can suppress turbulence (polymers, fibers, viscous liquids, etc.), but that's an entire other subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;other&lt;/em&gt; methods exist?&lt;/p&gt;&#xA;" OwnerUserId="49" LastEditorUserId="49" LastEditDate="2015-10-29T20:25:54.793" LastActivityDate="2018-04-28T16:52:51.650" Title="Ways to reduce turbulence in internal flows" Tags="|fluid-mechanics|turbulence|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5951" PostTypeId="2" ParentId="5945" CreationDate="2015-10-28T23:07:37.773" Score="2" Body="&lt;p&gt;All you need is a sliding base that can keep the motor from counter-rotating. You could use a &lt;a href=&quot;http://www.mcmaster.com/#linear-motion-carriages/=zkhnf2&quot; rel=&quot;nofollow&quot;&gt;linear carriage&lt;/a&gt; or a shaft and a &lt;a href=&quot;http://www.mcmaster.com/#standard-linear-bearings/=zkhneq&quot; rel=&quot;nofollow&quot;&gt;linear bearing&lt;/a&gt; to create the sliding platform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You attach the fixed end or shaft to the screw drive housing, either directly or via an intermediary like screwing both to lumber, and then you mount the motor on the moving platform. Couple the motor and screw drive together with any &lt;a href=&quot;http://www.mcmaster.com/#shaft-couplings/=zkhq4m&quot; rel=&quot;nofollow&quot;&gt;ordinary coupling mechanism&lt;/a&gt;; flanges, flexible couplings, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget end stops and making the wiring long enough. First prototype phase is typically where you break all your expensive equipment. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-10-28T23:07:37.773" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="5952" PostTypeId="2" ParentId="5900" CreationDate="2015-10-28T23:44:48.323" Score="0" Body="&lt;p&gt;To prevent ambient light from &quot;leaking&quot; into the fiber optic cables. It's mainly a redundancy (extra layer of protection)&lt;/p&gt;&#xA;" OwnerUserId="3549" LastActivityDate="2015-10-28T23:44:48.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5953" PostTypeId="2" ParentId="3623" CreationDate="2015-10-29T01:26:07.233" Score="2" Body="&lt;p&gt;Some IR sensors just receive. In this case they distinguish only by difference in infrared emission from the environment. These sensors could not see a hand if it was inside an oven mitt for example or if the hand was the same temperature as the surface, but in most cases it would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a photo gate setup you could have an IR led constantly on to illuminate the surface. It would work as long as the reflectiveity of your surface was always different from a hand (not sure if we are talking raw carbon fiber here or shiny epoxy coated carbon fiber). Dirt on a shiny surface would cause problems as would external heat sources like sunlight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An infrared proximity sensor uses triangulation and will be able to distinguish your targets more reliably, but will still have sunlight issues. &lt;a href=&quot;http://www.societyofrobots.com/member_tutorials/book/export/html/71&quot; rel=&quot;nofollow&quot;&gt;More info on IR sensors&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will want to look at an ultrasonic sensor. An ultrasonic sensor could easily handle the task and works better in dirty and sunny environments. A capacitive sensor would also work, but only at short range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sensor Examples:&lt;br&gt;&#xA;&lt;a href=&quot;https://www.sparkfun.com/search/results?term=ultrasonic%20proximity%20sensor&quot; rel=&quot;nofollow&quot;&gt;Ultrasonic and infrared proximity sensors&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.automationdirect.com/adc/Overview/Catalog/Sensors_-z-_Encoders/Ultrasonic_Proximity_Sensors/18mm_Round,_400mm_Maximum_Sensing_Distance_(UK_Series)&quot; rel=&quot;nofollow&quot;&gt;Ultrasonic proximity sensors&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.omega.com/pptst/E2K-X.html?pn=E2K-X4ME1&quot; rel=&quot;nofollow&quot;&gt;Capacitive proximity sensor&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-29T01:26:07.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5955" PostTypeId="1" CreationDate="2015-10-29T04:28:59.623" Score="3" ViewCount="733" Body="&lt;p&gt;I'm currently developing a treadmill for clinic gait research. If I were to build a flat treadmill with no moving surfaces what materials or coatings would be worth exploring as walking surface. Ideally, the surface would be durable, safe to touch with bare skin, capable of supporting a healthy adult, and low cost. The user would most likely use the treadmill wearing socks, made of either cotton, wool, spandex, or nylon. Delrin has been explored but is not adequate and produces a residual friction load on the user, producing artifacts in gait. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't necessarily have a suggestion for a material, please suggest a way of calculating the equivalent friction coefficient between a surface and a fabric. This method could be experimental or based on theoretical methods using a materials database. &lt;/p&gt;&#xA;" OwnerUserId="3551" LastActivityDate="2019-07-12T03:19:28.863" Title="Suggestions for low friction surface materials for a &quot;beltless static treadmill&quot;" Tags="|mechanical-engineering|materials|design|friction|biomechanics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5956" PostTypeId="2" ParentId="5763" CreationDate="2015-10-29T04:43:50.110" Score="0" Body="&lt;p&gt;You might try looking at the k-means function in opencv for Python or C. Additionally, the k-means function in Matlab may be useful as well. To further clarify, I believe OP would describe the center line as a neutral axis nearly parallel to the sagittal plane.&lt;/p&gt;&#xA;" OwnerUserId="3551" LastActivityDate="2015-10-29T04:43:50.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5957" PostTypeId="2" ParentId="5949" CreationDate="2015-10-29T10:02:05.333" Score="1" Body="&lt;p&gt;I am unsure of the context of you question but perhaps this will give some insight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a pressure and viscous stress contribution to the force on a boundary. This force can be calculated from the Navier-Stokes equations by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\int_{V}\left[\boldsymbol{\nabla}p+\mu\triangle\boldsymbol{v}\right]dV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewriting this slightly to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\int_{V}\boldsymbol{\nabla}\cdot\left[p\boldsymbol{I}+\mu\boldsymbol{\nabla}\boldsymbol{v}\right]dV$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can transform this integral to a surface integral using the divergence theorem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\int_{S}\left[p\boldsymbol{I}+\mu\boldsymbol{\nabla}\boldsymbol{v}\right]\cdot\boldsymbol{n}dS$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the unit normal vector to the boundary.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2015-10-29T10:02:05.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5958" PostTypeId="2" ParentId="5941" CreationDate="2015-10-29T12:07:21.517" Score="3" Body="&lt;p&gt;First, you should link in a reference to the Hata and Ericsson models. I haven't heard of them, and a quick search on Google didn't give anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Path loss and received signal strength are typically calculated using the Friis Transmission Equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Friis_transmission_equation&quot; rel=&quot;nofollow&quot;&gt;http://en.wikipedia.org/wiki/Friis_transmission_equation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To use this equation you must know the distance between the antennas, frequency, the gain of both the transmitter and receiver antennas, and the transmitter power. Then you can calculate the received signal strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you asked about the SNR (signal to noise ratio). In order to calculate the noise power, you have to know the bandwidth of the receiver. The noise power at the receiver can be calculated as kTB, where k is the Boltzman's constant, B is the receiver's bandwidth in Hz, and T is the temperature in degrees Kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simplified version of this is simply -174 dBm + 10*log(B)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The -174 dBm is kTb at room temperature in 1 Hz of bandwidth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also asked about calculating interference. In general this cannot be calculated. If you are expecting in band interference, it must be measured at the receiver site. Since all countries tightly regulate electrical emissions, and if you are operating in a controlled band (controlled by the FCC or similar agency), then your inband interference should be minimal. If you are operaing on one of the ISM bands, then you have to expect almost anything in terms of interference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, you asked about throughput. I am guessing that you mean throughput in terms of bits per second. This is a system level question that I can't possibly answer. It depends entirely on the bandwidth available, and parameters such as modulation type, and the SNR requirements for the receiver.&lt;/p&gt;&#xA;" OwnerUserId="2351" LastActivityDate="2015-10-29T12:07:21.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5960" PostTypeId="1" CreationDate="2015-10-29T15:30:07.150" Score="-1" ViewCount="134" Body="&lt;p&gt;For a 2-bit machine, there are 4 bit patterns, each bit pattern represents a unique integer value in base 10 according to different methods: unsigned, signed magnitude, 1's complement, 2's complement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write down all the possible bit patterns (in increasing order) for a 4-bit machine, and for each bit pattern, write down the unique integer value in base 10 according to unsigned, signed magnitude, 1's complement, 2's complement methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RdvCP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RdvCP.png&quot; alt=&quot;My answer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please check if this is right?&lt;/p&gt;&#xA;" OwnerUserId="3561" LastActivityDate="2015-10-29T16:56:05.700" Title="Need help with bit patterns" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2016-01-29T18:40:35.213" ContentLicense="CC BY-SA 3.0" />
  <row Id="5961" PostTypeId="2" ParentId="5960" CreationDate="2015-10-29T16:56:05.700" Score="2" Body="&lt;p&gt;It looks like your answers for the various interpretations are correct for what you have currently. However, you're missing 7 bit patterns. You still need $1001, 1010, 1011,$ etc. In 4-bits, the highest bit pattern should be $1111$, not $1000$. In base 10, this would be like asking for all numbers that can be expressed in 2 digits and then stopping at 10. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2015-10-29T16:56:05.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5962" PostTypeId="1" AcceptedAnswerId="5965" CreationDate="2015-10-29T18:18:57.320" Score="12" ViewCount="474" Body="&lt;p&gt;Hooke's Law defines a linear-elastic relationship between stress and strain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma = E\epsilon $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel behaves very much like a linear-elastic material, following Hooke's Law closely. It does, however, display non-elastic behaviors such as relaxation. Relaxation is the behavior wherein a member under constant strain displays variable (and reducing) stress over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is relaxation plastic? If the relaxed member were released, how would it behave? Would it follow a path defined by its elastic modulus? If this is the case, then it will end with a plastic deformation, no? After all, when stressed, the member will have reached $\left(\sigma_1, \epsilon_1\right)$. After relaxation, it will reach $\left(\sigma_2, \epsilon_1\right)$. Once released, it would have to reach $\sigma =0$, which implies in $\epsilon = \epsilon_1 - \dfrac{\sigma_2}{E}$ and since $\sigma_2 &amp;lt;\sigma_1$, that implies in a nonzero $\epsilon$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there some other behavior? Does the elastic modulus change to allow for a return without plastic deformations? &lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-10-29T23:28:41.497" Title="Is steel relaxation plastic?" Tags="|materials|steel|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="5965" PostTypeId="2" ParentId="5962" CreationDate="2015-10-29T23:23:35.580" Score="4" Body="&lt;p&gt;In short, yes, relaxation should probably be considered plastic deformation, as plastic strain is defined as &lt;strong&gt;non-recoverable deformation&lt;/strong&gt; when applied stresses are removed.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Definitional Explanation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Assume you have a sample of some material, in this case steel, and want to apply a load for an extended period of time, sufficiently long for noticeable relaxation to occur. The load is not sufficient to leave the elastic regime. Just after the load is applied but before relaxation begins, the strain in the material due to the load is $\varepsilon_{0}$. If the load is immediately removed, again before relaxation proceeds, all $\varepsilon_{0}$ is recovered and the steel material returns to its original shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead the material experiences the load sufficiently long so that relaxation proceeds, and the load is removed, then $|\varepsilon_{1}|&amp;lt;|\varepsilon_{0}|$ is recovered. As a result, not all of the strain is recovered. Therefore, there must have been &lt;strong&gt;non-recoverable strain&lt;/strong&gt; of $|\varepsilon_{0}-\varepsilon_{1}|&amp;gt;0$ due to relaxation. Therefore, by definition, relaxation is plastic deformation.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Thermodynamic &amp;amp; Kinetic Explanation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;If the definitional explanation is insufficient, we can also look at this from a thermodynamic and kinetic point of view. Suppose for the moment the steel is instead a single crystal of pure iron. Elastic strain stores energy in the crystal lattice. Because the energy is higher than its rest state, there is free energy available to do work, and thus a driving force for reorganization of atoms in the crystal lattice. There are also point defects in the lattice in the form of vacancies, or missing atoms. Random fluctuations cause neighboring atoms to fill the vacancies, which results in the vacancies moving around the lattice. The vacancies provide a means for reorganization of the atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if the strain is not isotropic (i.e. is not purely hydrostatic), then the lattice strain field makes vacancies slightly larger in tensile-strain directions than in compressive-strain directions. As a result, the energy barrier to moving in the tensile directions will be lower than in the compressive directions. Think of the atoms being squeezed out from between their compressive-direction neighbors along the tensile directions. There will thus be a net flow of atoms in the crystal, with atoms tending to move from directions of high compression to directions of high tension. The overall, long term effect is to extend the crystal in the directions of tension and shorten the crystal in the directions of compression, causing a &lt;strong&gt;non-recoverable deformation&lt;/strong&gt;. The same effects occur with multiple grains, except the mechanics are complicated by the presence of grain boundaries and varying crystal orientations. The same effects also occur with the presence of interstitial atoms like carbon, and they probably have negligible effect on vacancy motion as they don't get in the way (though I am not 100% sure of this part, see &lt;strong&gt;note&lt;/strong&gt; below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is a most-likely theory based on theories of vacancy flow and grain boundary migration due to thermal stresses (e.g. creep and grain-growth) and from dislocation motion, which have been observed directly. The described behavior for relaxation, however, has not been observed directly to the best of my knowledge (i.e. with a tunneling electron microscope).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Note&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;*Interstitial atoms will have lower energy in interstitial sites aligned with the tensile directions, as those sites are slightly increased in volume. This is related to anelastic strain and martensite formation, but may or may not have an impact on relaxation. However it is worth noting that purely axial strain may induce anisotropic properties in the steel.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-10-29T23:28:41.497" LastActivityDate="2015-10-29T23:28:41.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5966" PostTypeId="2" ParentId="5955" CreationDate="2015-10-30T00:32:26.097" Score="7" Body="&lt;p&gt;While it is an interesting idea, I don't think a &quot;static treadmill&quot; is going to give you results similar to walking. A PFTE(teflon) surface would be really low friction, but I think there are some bigger issues you may need to address first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Walking is essentially a repeated process of balancing, tipping, and re-balancing. The algorithms your brain uses to do this process will not be the same as the algorithms it will develop to slide your feet &lt;em&gt;as if&lt;/em&gt; you were walking. The physics is different and your brain will accommodate the new physics. So while I'm sure one could become quite proficient at doing it, I don't think that it will model actual walking/running very well. Let me give you some examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When your body is in the re-balancing stage it is providing a net amount of force to move the mass of your body forward. The movement of your body relative to the surface provides a somewhat constant velocity for your legs to rely on and manage. On your static treadmill you legs will be permitted to travel at any velocity they choose and may even change that velocity during a stride without consequence.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Putting the person in a harness artificially changes their weight; so now the person's stride will change because the legs do not have to support as much weight during the walking process.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The low friction surface will not provide the stability needed for a normal stride. You can walk on ice, but your stride is much different and purposeful than when you walk normally.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastEditorDisplayName="user748" LastEditDate="2016-10-02T04:42:20.390" LastActivityDate="2016-10-02T04:42:20.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5967" PostTypeId="1" AcceptedAnswerId="9088" CreationDate="2015-10-30T01:00:59.637" Score="2" ViewCount="171" Body="&lt;p&gt;By definition, prestressing is a self-balanced load. This is because it is in fact merely an applied state of internal stresses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can also be demonstrated mathematically (in this case for a simple parabolic cable) via Lin's load-balancing method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a cable which follows the layout given by&#xA;$$ y(x) = \dfrac{4e}{L^2}x^2 $$&#xA;where $e$ is the vertical distance between the maximum and minimum points of the parabola, $L$ is the span of the cable, and $x=y=0$ is located at the minimum point of the layout, the equivalent load is &#xA;$$ q = \dfrac{8Pe}{L^2}$$&#xA;where $P$ is the prestress force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the total vertical load given by the distributed load is $\dfrac{8Pe}{L}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This force is countered by the vertical forces applied at the anchors. These are equal to $P\sin(\theta) \approx P\theta$ and $y'(x) \approx \theta$, so the forces are each equal to $Py'\left(\frac{L}{2}\right) = \dfrac{4Pe}{L}$ and combined are equal (and opposite) to the total distributed load. So the total global load applied is null. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what if we consider friction losses between the cable and the duct in a post-tensioned beam? Then the combined forces at the extremities remain equal to $\dfrac{8Pe}{L}$, but the total distributed load will be reduced by the losses, implying in a non-balanced load, which is impossible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be adjusted by the anchorage slip losses, but what if we assume a hypothetical anchor which doesn't slip? Is this impossible and slippage is necessary to make the load balanced? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how are the remaining losses (elastic deformation, creep, shrinkage and relaxation) affected? &lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2015-10-30T13:34:28.733" LastActivityDate="2019-12-28T18:22:08.407" Title="How is prestressing with losses self-balanced?" Tags="|civil-engineering|structural-engineering|concrete|prestressed-concrete|" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="5968" PostTypeId="2" ParentId="5950" CreationDate="2015-10-30T01:15:09.143" Score="2" Body="&lt;p&gt;I think you have a pretty comprehensive list. Some things could be more general such as flow straighteners could also include airfoils for correcting flow around obtuse geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other geometry which may fall under your scope can isolate turbulent areas of flow from laminar areas of flow. Like in a &lt;a href=&quot;http://images2.labplanet.com/365-240-ffffff/opplanet-vwr-flow-indicator-ball-pk1-k538-vwr.jpg&quot; rel=&quot;nofollow&quot;&gt;spinning ball flow indicator &lt;/a&gt;for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the primary physical ways I see that turbulence is reduced. Some of your examples fall into more than one category, but I think looking at it from the physics side will make a lot more sense and give you the ability to derive more ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turbulence reduction by:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Changing the Reynolds number; fluid properties, velocity, scale.&#xA;&lt;li&gt;Changing the geometry; shallow diverging walls, airfoils, flow straighteners, flow isolators&#xA;&lt;li&gt;Eddy size reduction; vibration or dither for example could encourage small eddies to form instead of large eddies, airfoils, flow straighteners&#xA;&lt;li&gt;Dissipation of turbulence to heat (energy removal); flow straighteners, porous media, fibers or polymers suspended in the fluid.&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-10-30T01:15:09.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5969" PostTypeId="1" CreationDate="2015-10-30T02:20:41.220" Score="3" ViewCount="502" Body="&lt;p&gt;I know that radiant barriers (i.e., reflective films applied over or under insulated spaces) don't technically have R-values (because they primarily reflect heat, instead of primarily slowing convection and conduction as does traditional insulation), but they do have &lt;em&gt;some&lt;/em&gt; insulating effect, which could be quantified as equivalent to an R-value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., if I put a radiant barrier under R-19 insulation, what equivalent insulation will that space experience?  Holding conditions constant, an HVAC system will see a reduced load after the addition of the radiant barrier.  One could find the pure insulation level that matches that load reduction &amp;ndash; perhaps it's R-30? &amp;ndash; and based on that say the radiant barrier is &quot;functionally equivalent to R-11.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Amendment:&lt;/strong&gt; Yes, there are good theoretical reasons why radiant barriers are not assigned an R-value.  However, as one user notes, &lt;em&gt;we only ever talk about &quot;R&quot;&lt;/em&gt;.  For example, &lt;a href=&quot;http://www.naima.org/insulation-knowledge-base/residential-home-insulation/how-much-insulation-should-be-installed.html&quot; rel=&quot;nofollow&quot;&gt;where I live the standard for home insulation is R-30&lt;/a&gt;.  Can I reliably achieve that through a &lt;em&gt;combination&lt;/em&gt; of blanket insulation and a radiant barrier?  MIMA, hardly an objective trade organization, &lt;a href=&quot;http://www.naima.org/publications/RP051.PDF&quot; rel=&quot;nofollow&quot;&gt;did a test in Florida 30 years ago&lt;/a&gt; which, while highlighting shortcomings of radiant barriers, did note reasonably consistent heat flow reduction from properly installed barriers.  I would assume that with the benefit of 30 years of knowledge and increased emphasis on insulation we can say &lt;em&gt;something&lt;/em&gt; useful about the functional value of adding a radiant barrier to an insulation stack.&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="77" LastEditDate="2015-10-30T14:24:54.417" LastActivityDate="2022-12-10T02:39:22.857" Title="Effective R-value of radiant barriers" Tags="|hvac|thermal-insulation|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5970" PostTypeId="2" ParentId="5969" CreationDate="2015-10-30T03:28:41.917" Score="5" Body="&lt;p&gt;Insulating against &lt;a href=&quot;https://en.wikipedia.org/wiki/R-value_(insulation)#Relationships&quot; rel=&quot;nofollow&quot;&gt;thermal conduction&lt;/a&gt; can be represented as an R-value because it is a linear calculation. I can add two or more R-values together and I will see a linear benefit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Insulating against &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_radiation#Radiative_heat_transfer&quot; rel=&quot;nofollow&quot;&gt;thermal radiation&lt;/a&gt; is more complex: You can't add two radiation barriers and get one that is twice as good. Insulation is based on the temperature of the barrier itself and the temperature of the environment radiating to it (and occurs on both sides of the barrier). It gets even more complicated when it is placed between insulation or when there are materials against it on either side. Now only the radiation that permeates the surrounding material is being reflected. The absorption of the material is doubled because the radiation is absorbed again on the way back out. Thus radiant barriers are always more effective when at least one side of them faces an air gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though it can't be represented by an R-value, you can reasonably calculate its effectiveness when against an air gap for a specific temperature differential. Just use the equations in the thermal radiation link above.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="77" LastEditDate="2015-10-30T08:57:36.713" LastActivityDate="2015-10-30T08:57:36.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5971" PostTypeId="2" ParentId="5969" CreationDate="2015-10-30T06:58:36.173" Score="1" Body="&lt;p&gt;No, the radiant effect cannot be quantified as an equivalent R-value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't work like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're trying to co-opt a measurement that it's completely unsuited for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ericnutsch has explained one of the main reasons why not, and I'll repeat it: R-values sum; radiant barriers do not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R-values measure thermal insulation against heat-transfer by conduction. Radiant barriers offer little to no thermal insulation against heat-transfer by conduction. That's why the whole multi-foil thing failed.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if I put a radiant barrier under R-19 insulation, what equivalent insulation will that space experience? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Do it and find out. But remember that you'll have to account for lots of factors in different ways to if you were using better insulation, such as the emissivity of stuff around it. They are not swappable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Holding conditions constant&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They never are. If you are interested in a particular real-world situation, do a real-world test.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2015-10-30T06:58:36.173" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5972" PostTypeId="2" ParentId="5950" CreationDate="2015-10-30T07:04:39.343" Score="2" Body="&lt;p&gt;Dolphin skin seems to have properties that limit drag, also by reducing turbulence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the late 1950s the aerodynamicist Max Kramer claimed that a dolphin ensured a low level of friction drag by maintaining the laminar flow over most parts of its body. The dolphin’s skin having an unusually ordered inner structure was considered to be a natural compliant wall effectively reducing the flow disturbances in the boundary layer&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://biomimetic.pbworks.com/f/Dolphin+skin+as+a+naturalPavlov.pdf&quot; rel=&quot;nofollow&quot;&gt;Pavlov: Dolphin skin as a natural anisotropic&#xA;compliant wall&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for explaining how and why that works, someone with more understanding of fluid dynamics than me has to read the paper. I'd suspect that some anisotropic properties in the skin mean that movement orthogonal to the flow gets absorbed, the energy dissipated. But that's pure conjecture at this point.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="255" LastEditDate="2015-10-30T16:25:16.107" LastActivityDate="2015-10-30T16:25:16.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5973" PostTypeId="1" AcceptedAnswerId="5978" CreationDate="2015-10-30T07:16:32.320" Score="2" ViewCount="117" Body="&lt;p&gt;for example :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sender:&lt;/strong&gt; A cellular signal tower(Base Station(BS)) sends data(voice, GPRS, VOIP) to the cellphone. The sender will obviously have the carrier frequency and more power(voltage+current) to generate the carrier signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Receiver:&lt;/strong&gt; A cell phone which receives the data signal (voice, GPRS, VOIP) from the BS. This requires less energy because the cell phone is receiving data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know, when the receiver sends data(voice,GPRS,VOIP) to the BS, how does it do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the cell phone generate it's own carrier signal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, then how can a 3.7 volt battery be available for 8-12 hours?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The receiver's signal goes on to same carrier signal which is generated by the BS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, How? Does the carrier signal use refraction/reflation properties to send data back to BS?&lt;/p&gt;&#xA;" OwnerUserId="3565" LastEditorUserId="110" LastEditDate="2015-10-30T10:05:36.953" LastActivityDate="2015-10-30T17:24:13.907" Title="How does a carrier signal work from a receiver's perspective?" Tags="|electrical-engineering|telecommunication|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5978" PostTypeId="2" ParentId="5973" CreationDate="2015-10-30T17:24:13.907" Score="0" Body="&lt;p&gt;When a cell phone is sending, it generates its own carrier.  The power for doing this comes from the cell phone's battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Batteries in a cell phone may last 12 hours, but that is not while transmitting.  Depending on how &quot;off&quot; the cell phone is, it is only occasionally listening to the cell tower, and even less often sending anything to it.  When not moving, it may send nothing at all for long periods of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The battery wouldn't last 12 hours with you continuously talking on the cell phone.  That requires frequent transmitting, which uses a lot more power than just sitting there, even in standby mode listening for others to call it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transmitting power also depends on distance to the cell tower.  Actually the phone doesn't measure distance, but the cell tower tells it how well the message was received.  The phone cranks up the power as needed achieve whatever it considers a acceptable error rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite all that, the predominant energy drain on the battery can be illuminating the screen.  If you want long life per charge, don't make the phone power up the screen.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-10-30T17:24:13.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5979" PostTypeId="2" ParentId="5868" CreationDate="2015-10-30T19:18:29.100" Score="1" Body="&lt;p&gt;For the ratio of methane and water use henry's law:&#xA;From NIST site: constant for methane 0.0014  mol/kg bar at 298K.&#xA;At 2000 kpa = 20 bar,&#xA;Concentration in water is 0.028 mol/kg at 298 K. At your temp of 373K, you should use the temperature correction equation to get the correct constant. This equation is on the same NIST page for methane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the question of the influence on boiling point: It sounds like you are thinking of the Clausius-Clapeyron equation to estimate boiling point from vapor pressure and heat of vaporization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My opinion is that a better thermodynamic answer would be to say that there would simply be less dissolved methane at the boiling point of water at 2 mPa and 485 K. So, like the other comments said, the saturation point is assumed as 212 degC with less dissolved gas than at 100 degC. &lt;/p&gt;&#xA;" OwnerUserId="3569" LastActivityDate="2015-10-30T19:18:29.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5982" PostTypeId="1" AcceptedAnswerId="5985" CreationDate="2015-10-30T22:36:52.547" Score="3" ViewCount="1233" Body="&lt;p&gt;I'm in school for electronics engineering.  Would someone lay out the high level steps in becoming a professional engineer (PE) in the state of California?&lt;/p&gt;&#xA;" OwnerUserId="3575" LastEditorDisplayName="user16" LastEditDate="2015-10-31T14:59:06.887" LastActivityDate="2015-11-17T23:57:04.853" Title="What is the sequence of exams to become a Professional Engineer in California?" Tags="|electrical-engineering|licensure|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5983" PostTypeId="1" AcceptedAnswerId="5986" CreationDate="2015-10-31T02:53:00.950" Score="2" ViewCount="1491" Body="&lt;p&gt;How do I know when to use an upward or a downward curve when I'm drawing a bending moment diagram for a uniformly-distributed load (UDL) and what's the most reliable way of identifying the maximum bending moment?&lt;/p&gt;&#xA;" OwnerUserId="3577" LastEditorUserId="2246" LastEditDate="2015-10-31T15:21:29.920" LastActivityDate="2015-10-31T15:21:29.920" Title="Drawing Bending Moment Diagrams" Tags="|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5984" PostTypeId="1" AcceptedAnswerId="5987" CreationDate="2015-10-31T06:35:28.450" Score="9" ViewCount="292" Body="&lt;p&gt;In addition to power and performance, one of the reasons VW cheated their emissions results was to improve the fuel economy of their cars. Why exactly does this happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm familiar with cars and how cars work so feel free to get technical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a follow-up question: Where is the &quot;sweet spot&quot; between burning more gas per mile but having that burned gas release fewer emissions, and burning less gas per mile but releasing more emissions per gallon burned?&lt;/p&gt;&#xA;" OwnerUserId="3579" LastActivityDate="2016-03-01T03:03:36.023" Title="Why does reducing the amount of emissions control improve the fuel economy of a car?" Tags="|mechanical-engineering|automotive-engineering|environmental-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5985" PostTypeId="2" ParentId="5982" CreationDate="2015-10-31T10:23:32.370" Score="5" Body="&lt;p&gt;There are slight variations in requirements between individual states. Please check with the respective State Board of Professional Engineering for those exceptions. At a very high level steps are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Obtain a Bachelors in Electrical Engineering (BSEE) from an ABET accredited engineering program&lt;/li&gt;&#xA;&lt;li&gt;Successfully complete the Fundamentals in Engineering exam to obtain Engineering in Training (EIT) status&lt;/li&gt;&#xA;&lt;li&gt;Gain professional experience (typically four years minimum)&lt;/li&gt;&#xA;&lt;li&gt;Learn about California state's licensure requirements&lt;/li&gt;&#xA;&lt;li&gt;Successfully complete the Professional Engineer exam&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The above is also detailed by the National Society of Professional Engineers (NSPE):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;How To Get Licensed (NSPE)&lt;/strong&gt;   &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you're a graduate from an engineering program approved by your&#xA;  state's licensure board, you can become classified as an &quot;engineer&#xA;  intern&quot; or &quot;engineer-in-training&quot; by successfully completing the&#xA;  Fundamentals of Engineering (FE) exam. Achieving EI or EIT status&#xA;  signals that you have mastered the fundamental requirements and taken&#xA;  the first step toward earning your PE licensure.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All states require that candidates complete four years of &lt;a href=&quot;http://www.nspe.org/resources/licensure/resources/demonstrating-qualifying-engineering-experience-licensure&quot; rel=&quot;nofollow noreferrer&quot;&gt;qualifying&#xA;  engineering experience&lt;/a&gt;, typically under the supervision of a&#xA;  professional engineer. ...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Each state (as well as the District of Columbia and all U.S.&#xA;  territories) has its own &lt;a href=&quot;http://www.nspe.org/resources/licensure/licensing-boards&quot; rel=&quot;nofollow noreferrer&quot;&gt;licensure board&lt;/a&gt;, administering its own&#xA;  exam and required qualifications. To find out more about your state's&#xA;  licensure requirements, visit the Web site of your state licensing&#xA;  board state-by-state summary.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In most cases, the final step in attaining licensure is successfully&#xA;  completing the Principles and Practice of Engineering (PE) exam in&#xA;  your state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/2658/do-references-for-the-california-pe-application-have-to-be-california-pes&quot;&gt;Do references for the California PE application have to be California PE's?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nspe.org/resources/licensure/how-get-licensed&quot; rel=&quot;nofollow noreferrer&quot;&gt;How To Get Licensed? (National society of Professional Engineers)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-11-17T23:57:04.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5986" PostTypeId="2" ParentId="5983" CreationDate="2015-10-31T10:33:45.940" Score="4" Body="&lt;p&gt;The direction a bending moment diagram is drawn depends on the sign convention being used. The most common sign convention is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ag9He.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ag9He.gif&quot; alt=&quot;PositiveSignConvention&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is good practice to indicate the sign convention used whenever you draw a BMD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we are interested in the sign of the bending moment for a UDL we can take a cut such as:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hWNCF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hWNCF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which case we can see from the sum of moments at the cut:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M = Rx - w\frac{x^2}{2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And whatever the sign found from this is the sign for the BMD. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simple structures you can obtain the sign of the bending moment by thinking of the deformed shape. If the deformed shape resembles that shown in the sign convention above then the sign of the BMD will be positive (tension on bottom). So for a simply supported beam with a UDL acting downwards the bending moment would be positive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be aware that sometimes a BMD is drawn with the positive moment axis pointing downwards, this is called drawing 'on the tension side' and is common in concrete design as it indicates where reinforcement should be placed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fastest, most reliable way of finding the maximum bending moment, especially for simple cases, would be to use established beam design formulas such as: &lt;a href=&quot;http://www.awc.org/codes-standards/publications/da6&quot; rel=&quot;nofollow noreferrer&quot;&gt;DA 6 - Beam Design Formulas with Shear and Moment Diagrams&lt;/a&gt;. If you knew where the location of maximum bending moment was, you could take a cut at that location and calculate it directly as shown above. Otherwise, you will need to resort to integrating the shear force (either directly, or graphically) to obtain the BMD and then locating the maximum that way.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2015-10-31T11:40:46.413" LastActivityDate="2015-10-31T11:40:46.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5987" PostTypeId="2" ParentId="5984" CreationDate="2015-10-31T11:58:58.273" Score="9" Body="&lt;p&gt;Not all of the pollutants come directly from the fuel. Some (e.g. NOx) come from the air itself, based on the combustion temperature in the cylinder. Running a cylinder at a higher temperature (by leaning the mixture) results in both greater efficiency and greater NOx emissions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, running the cylinder at a lower temperature reduces NOx emissions, but at some point, CO and HC emissions go up because of incomplete combusion (which is why catalytic converters are used).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;sweet spot&quot; you ask about in your followup question is very much dependent on the specific design of the engine, and it is, of course, the target of ongoing (and proprietary) research.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2015-10-31T13:47:29.340" LastActivityDate="2015-10-31T13:47:29.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5988" PostTypeId="1" AcceptedAnswerId="5991" CreationDate="2015-10-31T15:49:06.700" Score="4" ViewCount="1532" Body="&lt;p&gt;&lt;strong&gt;Regarding micro inverters for solar power systems:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there technically a difference between an inverter classified as a &quot;micro inverter&quot; and and inverter classified as a &quot;pure sine wave&quot; inverter?  Or is it simply a matter of using extra jargon to describe the size of the inverter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem I'm trying to solve:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to purchase a micro inverter based solar system, where each panel has it's own dedicated inverter.  Enphase seems to have coined the term &quot;micro inverter&quot; for this purpose, but the reseller doesn't offer a micro inverter that is big enough for a 315 watt panel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm wondering if I can purchase a pure sine wave inverter rated at 400 watts, for each 315 watt panel.  Would those wire up the same way as the &quot;micro inverter&quot; solution?&lt;/p&gt;&#xA;" OwnerUserId="3581" LastActivityDate="2015-11-02T07:29:52.437" Title="&quot;Micro Inverters&quot; vs &quot;Pure Sine Wave&quot; for Solar Systems" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5989" PostTypeId="1" CreationDate="2015-10-31T18:43:49.410" Score="3" ViewCount="191" Body="&lt;p&gt;I have a type of pressure gauge (material in contact with medium: 1.4542 (17-4 PH / 630)²) that I would like to use for sulphur hexaflouride (SF&lt;sub&gt;6&lt;/sub&gt;) at room temperature and pressures of typically 6 bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the manufacturer is not really interested to support me with information if this sensor is compatible with SF&lt;sub&gt;6&lt;/sub&gt;, either they tell me to find out myself or just not to use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wonder - how would you approach this issue, what limiting factors other than the material in contact with the medium and the max pressure could there be?&lt;/p&gt;&#xA;" OwnerUserId="3582" LastEditorUserId="33" LastEditDate="2015-10-31T21:09:58.913" LastActivityDate="2015-10-31T22:09:16.643" Title="Determining compatibility of pressure gauge with sulphur hexaflouride (SF6)" Tags="|materials|pressure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5990" PostTypeId="2" ParentId="5989" CreationDate="2015-10-31T19:43:02.917" Score="2" Body="&lt;p&gt;In terms of chemical compatibility SF6 is a relatively inert gas, it will not react with most metals, plastics, glasses. In fact there used to be a loudspeaker company in Canada that used the gas as a high voltage insulator in their electrostatic speakers, the gas being in constant contact with a thin mylar film used for the speaker diaphragm (around .25 mil thick). So to really get an engineering solution I would just ask the manufacturer of your sensor to give you information about what materials the sensor is made of and contact the makers of the SF6 gas to make sure there would be no harmful reaction between the sensor and the gas.&lt;/p&gt;&#xA;" OwnerUserId="628" LastEditorUserId="628" LastEditDate="2015-10-31T22:09:16.643" LastActivityDate="2015-10-31T22:09:16.643" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5991" PostTypeId="2" ParentId="5988" CreationDate="2015-10-31T21:46:03.243" Score="2" Body="&lt;p&gt;&quot;Micro&quot; and &quot;pure sine wave&quot; are orthagonal, even if &quot;micro&quot; had a real definition.  &quot;Micro&quot; is just a relatively content-free marketing term.  It says that the inverter is intended for a small array or single panel, but of course without a spec it doesn't mean anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything connected back to the power line should be, and probably must be due to local laws, a sine wave inverter.  &quot;Pure&quot; again is a silly marketing term.  Surely the inverter will produce some harmonics.  I guess the term is trying to say that the level of harmonics is low enough that the power company won't care.  That's probably true, but you need to check that yourself since either you or the licensed electrician that installs it will be on the hook, not the sales guy who's telling you it's &quot;pure&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ignore micro, nano, or turbo, but make sure the output is within requirements in your jurisdiction.  Otherwise, make sure the inverter will handle the volts, amps, and watts you will give it.  If you have a 315 W panel, then a 350 or 400 W inverter sounds about right.  A little more than the minimum is good in that it provides some margin.  A lot more makes it too expensive, and it may not even run on a small power input.  Too little is obviously bad.  It either won't work or something will get fried.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-10-31T21:46:03.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5992" PostTypeId="1" CreationDate="2015-10-31T23:43:47.183" Score="1" ViewCount="1049" Body="&lt;p&gt;How do spray nozzles work? How do they convert liquid to aerosol? Is it possible to print tiny spray nozzles from metal powder?&lt;/p&gt;&#xA;" OwnerUserId="597" LastEditorUserId="421" LastEditDate="2015-11-01T01:11:50.467" LastActivityDate="2015-11-01T21:16:50.567" Title="How to design spray nozzle (like in pulverizer) for 3d print?" Tags="|liquid|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5993" PostTypeId="2" ParentId="5944" CreationDate="2015-11-01T02:34:46.717" Score="2" Body="&lt;p&gt;That would be Process Industry Practices, Practice Ref. PIC001. Issued 7/15/1998.&#xA;Appendix A1 contains exactly what you listed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I just came across this online: &lt;a href=&quot;http://www.cheme.utm.my/staff/arshad/images/lecture/mkh1223/pid_sample.pdf&quot; rel=&quot;nofollow&quot;&gt;PIC001 Sample, Not for Commercial Use&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3569" LastEditorUserId="3569" LastEditDate="2015-11-17T21:13:45.817" LastActivityDate="2015-11-17T21:13:45.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="5995" PostTypeId="1" AcceptedAnswerId="6022" CreationDate="2015-11-01T11:30:45.920" Score="4" ViewCount="614" Body="&lt;p&gt;I have the same problem as described in this question:&lt;br&gt;&#xA;&lt;a href=&quot;https://engineering.stackexchange.com/q/5535/2844&quot;&gt;How to calculate the resulting temperature for a stream of air through a hot tube?&lt;/a&gt;, but I want to increase the transferred heat by adding fins onto the inner side of the tube, parallel to the flowing air. Is that useful? If yes, how does the formula mentioned in the answer (&lt;a href=&quot;https://engineering.stackexchange.com/a/5536/2844&quot;&gt;https://engineering.stackexchange.com/a/5536/2844&lt;/a&gt;) change?&lt;br&gt;&#xA;Furthermore, how do I get from the temperature difference to the transferred heat power?&lt;/p&gt;&#xA;" OwnerUserId="2844" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-02-19T18:31:13.177" Title="Calculate thermal flow for air flowing through a tube, with fins" Tags="|heat-transfer|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="5996" PostTypeId="1" CreationDate="2015-11-01T13:34:02.920" Score="3" ViewCount="1588" Body="&lt;p&gt;Moran and Shapiro  Fundamentals of Engineering Thermodynamics (p147) has a worked example:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Steam at a pressure of $15$ bars and a temperature of $320^{\circ}C$ is contained in a large tank. Connected to the tank through a valve is a turbine followed by a small initially evacuated vessel with a volume of $0.6m^3$. The valve is opened and the vessel; fills with steam until the pressure is $15$ bars and the temperature $400^{\circ}C$. The valve is then closed. The filling process takes place adiabatically and kinetc and potential energy effects are negligible. Determine the amount of work developed by the turbine in kJ ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The solution is then given but in a comment at the end they add:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the turbine were removed and steam allowed to flow adiabatically into the smaller vessel without doing work, the final steam temperature in the vessel would be $477^{\circ}C$, as may be verified&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How do you find the temperature without the turbine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They start by observing that with their assumptions the energy rate balance reduces to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dU_{cv}}{dt} = -\dot{W} + \dot{m_i}h_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and integrating this and re-arranging they have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ W_{cv} = h_i\Delta m_{cv} - \Delta U_{cv}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the smaller vessel is initially evacuated the changes are just the final values which they indicate with the subscript &lt;code&gt;2&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;W_{cv}&amp;amp;=&amp;amp;  h_i\Delta m_{cv} - \Delta U_{cv}\\&#xA;  &amp;amp;=&amp;amp; h_i m_2 - m_2 u_2\\&#xA;  &amp;amp;=&amp;amp; m_2(h_i - u_2)\\&#xA;  &amp;amp;=&amp;amp; 386.6 kJ&#xA;\end{eqnarray*}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;steamtables(400C, 15bar, v) = $0.203\,m^3/kg$  (use for mass)&lt;br&gt;&#xA;  steamtables(400C, 15bar, $u_2$) = $2951.3\,kJ/kg$&lt;br&gt;&#xA;  steamtables(320C, 15bar, $h_i$) = $3081.9\,kJ/kg$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the temperature of the steam in the second question&lt;/p&gt;&#xA;" OwnerUserId="3590" LastEditorUserId="3590" LastEditDate="2015-11-11T12:22:21.247" LastActivityDate="2015-11-11T12:22:21.247" Title="Temperature when steam expands into a vacuum" Tags="|thermodynamics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="5997" PostTypeId="1" CreationDate="2015-11-01T14:57:47.007" Score="1" ViewCount="170" Body="&lt;p&gt;I have purchased a fuel cell, and the supply pressure must not exceed 0.55 bar. My question is how can I supply the hydrogen at this pressure given that it is below atmospheric pressure.&lt;/p&gt;&#xA;" OwnerUserId="3591" LastEditorUserId="1832" LastEditDate="2015-11-01T19:59:54.140" LastActivityDate="2015-11-02T02:53:38.720" Title="How to supply hydrogen to a fuel cell at a pressure lower than the atmospheric pressure?" Tags="|pressure|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="5998" PostTypeId="1" CreationDate="2015-11-01T16:20:16.130" Score="5" ViewCount="468" Body="&lt;p&gt;I studied a paper about simulation of turbines using Fluent that included the following sentence, which I don't understand:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All  converged solutions  had  a  mass imbalance  of less  than &#xA;  0.3%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What does the mass imbalance of converged solutions refer to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I ensure this condition is met in Fluent?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="282" LastEditDate="2015-11-07T21:18:24.487" LastActivityDate="2015-11-07T21:18:24.487" Title="How can I set an upper limit on mass imbalance in Fluent?" Tags="|mechanical-engineering|cfd|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="5999" PostTypeId="2" ParentId="5997" CreationDate="2015-11-01T16:35:04.603" Score="1" Body="&lt;p&gt;The fuelcellstore.com website offers a pressure reducing regulator that supplies the 0.55 bar. It does not say if this is a gauge or atmospheric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The regulator uses a fitting for silicone tubing and the site suggests using silicone tubing. The tubing they suggest using is not a vacuum rated tubing (usually wire/polyester reinforced silicone) but rather tubing with a max pressure rating of 10 psi. (Again, they fail to specify psig rather than psia). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel safe assuming gauge pressures for these (i.e. 0.55 barg = 1.55 bar absolute at sea level) but suggest you continue researching this since Hydrogen can be dangerous. &lt;/p&gt;&#xA;" OwnerUserId="3569" LastEditorUserId="3569" LastEditDate="2015-11-01T18:19:26.497" LastActivityDate="2015-11-01T18:19:26.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6001" PostTypeId="2" ParentId="5998" CreationDate="2015-11-01T17:46:06.390" Score="5" Body="&lt;p&gt;Mass imbalance is a measure of the error in the solution.  You would typically set the limit for this in the &quot;Solution Controls&quot; section of the inputs (if using the GUI).  For mass imbalance, set the tolerance for residuals on the continuity equation.  The solver will keep iterating until it finds a solution that satisfies all your residual tolerances, or until it reaches the limit of iterations that you set.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-11-01T17:46:06.390" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6002" PostTypeId="1" CreationDate="2015-11-01T21:16:38.293" Score="1" ViewCount="411" Body="&lt;p&gt;I am simulating an axial turbine by using Fluent. The convergence of solution is monitored by checking the residuals of the numerically solved governing equations. Moreover, in order to judge the convergence, the behaviour of other quantities, such as the total pressure at the inlet and outlet boundaries, and torque coefficient generated by the rotor, are also monitored. Here, the default convergence criterion of each residual is reduced in order to allow the monitored quantities to stagnate at consistent values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, according the following figures, Is the convergence of solution ok? and particularly about residuals? As you see in sixth figure, residuals increase and decrease!!! I want to know that this event is logical or not. I already thought that residuals should decrease decidedly because of convergence. But in this simulation, residuals increases and decreases!! What is reason for this event?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iElIS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iElIS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eXeyr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eXeyr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TPyUt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TPyUt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i8G1I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i8G1I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zkUZ7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zkUZ7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fIX1K.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fIX1K.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FqFKx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FqFKx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NGq7g.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NGq7g.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YmlFN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YmlFN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2015-11-05T21:45:57.733" LastActivityDate="2015-11-07T12:07:35.953" Title="Is the convergence of solution ok in Fluent?" Tags="|fluid-mechanics|turbines|cfd|" AnswerCount="0" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6003" PostTypeId="2" ParentId="5992" CreationDate="2015-11-01T21:16:50.567" Score="4" Body="&lt;p&gt;Nozzle design can be fairly complex, but basically the nozzle speeds up the velocity of the fluid to the point where inertial forces exceed surface tension forces, and the fluid breaks up into drops.  The &lt;em&gt;Weber number&lt;/em&gt;, $We = \frac{\rho u^2 l}{\sigma}$ is a convenient parameter to do a basic analysis on a nozzle flow.  In this equation, $\rho$ is the fluid density, $u$ is its velocity, $l$ is a characteristic length (usually droplet diameter), and $\sigma$ is the fluid surface tension.  When $We &amp;gt; 1$, the fluid is likely to break up into smaller droplets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the Weber number equation to calculate how fast your fluid must be traveling to break up into the target droplet diameter.  This velocity, combined with the flow rate of the fluid through the nozzle, can be used to estimate the orifice area: $A = \frac{\dot V}{u}$ (area is volumetric flow rate divided by velocity).  Compare the orifice diameter to the resolution of your printer; if the diameter is too small, then you might have to drill the orifice separately, after the nozzle is printed.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-11-01T21:16:50.567" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6004" PostTypeId="1" CreationDate="2015-11-01T22:31:01.587" Score="2" ViewCount="35" Body="&lt;p&gt;Is there correlation between friction of boundary layer(or inertia and viscous forces in boundary layer) and values of inlet and outlet pressure in an axial turbine? For example, when friction of boundary layer decrease or boundary layer is not captured well, how to change values of inlet and outlet pressure in a turbine?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="1293" LastEditDate="2015-11-02T16:40:07.180" LastActivityDate="2015-11-02T16:40:07.180" Title="Is there correlation between friction of boundary layer and values of inlet and outlet pressure?" Tags="|fluid-mechanics|turbines|cfd|" AnswerCount="0" CommentCount="1" ClosedDate="2016-03-04T18:53:56.457" ContentLicense="CC BY-SA 3.0" />
  <row Id="6005" PostTypeId="1" CreationDate="2015-11-01T23:17:32.263" Score="1" ViewCount="74" Body="&lt;p&gt;I am simulating an axial turbine, I tried different mesh topologies and all give me almost the same number. In fact, Boundary layer is not captured well, Reynolds number is equal to $4.4 \times 10^5$ and according to mentioned formula in the following link, I computed $y_1$ (First cell height) by using $y^+ = 30$ and $y^+ = 60$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the two cases($y^+=30$ and $y^+=60$), the boundary layer is not captured well (Why?). In the other words, according to the following figure, there is no adjacent wall. And also, I studied the following link:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, performance parameters are not validated well. In fact, there is a difference with the experimental value. This difference is not big but it is not good! I think that it is difficult to capture BL by using low Re number turbulence model (for example,  $k-\epsilon$ turbulence model).&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2015-11-13T13:38:05.250" LastActivityDate="2015-11-13T13:38:05.250" Title="Boundary layer is not captured well in axial turbine blade simulation" Tags="|mechanical-engineering|fluid-mechanics|turbines|cfd|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6006" PostTypeId="1" AcceptedAnswerId="6029" CreationDate="2015-10-31T22:20:01.323" Score="6" ViewCount="3485" Body="&lt;p&gt;&lt;strong&gt;TL;DR: how can I calculate the disturbance at each joint due to coupling forces in a two-link planar robot manipulator actuated by two independent DC motors?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm studying control theory and trying to work through a simple example using a two-link planar robot manipulator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hfUPe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfUPe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to simulate PID control of the planar two-link manipulator, where each joint is actuated by an independent DC motor. The input is two continuous signals $$\theta_1(t), \theta_2(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which represent the desired angle for each joint at time &lt;em&gt;t&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following this paper: &lt;a href=&quot;http://www.scirp.org/journal/PaperInformation.aspx?PaperID=22044&quot; rel=&quot;nofollow noreferrer&quot;&gt;Modeling a Controller for an Articulated Robotic Arm&lt;/a&gt;, I can obtain an expression for the voltage I need to apply to each motor to achieve a desired angle. The paper also describes the PID controller necessary to maintain the desired angle given an error signal $$e(t) = \theta_{desired}(t)-\theta_{actual}(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since each motor is controlled independently, coupling effects among joints due to varying configurations during motion are treated as disturbance inputs. &lt;strong&gt;My question is: how can I model this coupling effect in order to &quot;simulate&quot; the error signal &lt;em&gt;e(t)&lt;/em&gt; for system under ideal conditions?&lt;/strong&gt; By ideal conditions I mean that the disturbance due to coupling among joints accounts for 100% of the error signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current thought is to use an expression for the dynamics of the two-link planar manipulator, as per the following book: &lt;a href=&quot;http://www.cds.caltech.edu/~murray/books/MLS/pdf/mls94-complete.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Mathematical Introduction to Robotic Manipulation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FPqIJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FPqIJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way, at time &lt;em&gt;t&lt;/em&gt;, we determine the voltage to achieve the desired angles for each independent motor, then plug that into our DC motor model to obtain the generated torques: $$\tau_1, \tau_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then plug these torques into the dynamics to get the actual angles, taking into account the coupling forces, and then use the actual angles compared to the desired angles in order to generate the error signal that feeds into the PID control loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this approach make sense? If not, where have I gone wrong and how can I simulate the error signal due to coupling forces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; one commenter points out PID control may not be optimal for this problem. If this is the case, what alternative control strategies should I use?&lt;/p&gt;&#xA;" OwnerUserId="3594" OwnerDisplayName="louism" LastEditorUserId="3594" LastEditDate="2015-11-02T03:00:18.960" LastActivityDate="2016-07-22T01:35:48.043" Title="Feedback control of two-link planar manipulator" Tags="|control-engineering|control-theory|dynamics|kinematics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6007" PostTypeId="2" ParentId="5997" CreationDate="2015-11-02T02:53:38.720" Score="1" Body="&lt;p&gt;The rating is for gauge pressure. So as Brandon Morris mentioned; it will be 0.55 bar as read on a gauge and 1.55 bar absolute (read from vacuum). If it was below atmospheric pressure it would not be feasible. They are basically limiting the pressure because the structure of the fuel cell and its membranes can not take a large differential pressure without damage. You will just need to source a regulator that delivers this low of pressure.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-02T02:53:38.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6009" PostTypeId="2" ParentId="5988" CreationDate="2015-11-02T07:29:52.437" Score="0" Body="&lt;p&gt;Stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't just attach &lt;em&gt;any&lt;/em&gt; inverter on a PV panel or array, even if it does have the right power rating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your inverter needs to be a PV inverter specificaly. It has to be designed to do max-power-point tracking. That's because the panel doesn't have any specific potential difference: that has to be imposed on it by the rest of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you're connecting to the grid, the inverter may have to meet specific local regulations (e.g. G83 in Great Britain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole idea of micro-inverters is that you have one per panel. And any micro inverter will be a sine-wave generator. The need for &quot;purity&quot; of that sine wave, and the ability to synchronise phases with the grid, as well as behaving properly upon grid disconnection, is what goes into things like G83 certification.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2015-11-02T07:29:52.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6011" PostTypeId="2" ParentId="5996" CreationDate="2015-11-02T08:47:24.040" Score="1" Body="&lt;p&gt;For the second question we have only the third value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Carlton gives hints. The work is set to zero so the equations become:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{eqnarray*}&#xA;0 &amp;amp;=&amp;amp;  h_i\Delta m_{cv} - \Delta U_{cv}\\&#xA;  &amp;amp;=&amp;amp; h_i m_2 - m_2 u_2\\&#xA;  h_i &amp;amp;=&amp;amp; u_2\\&#xA;\end{eqnarray*} &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;state of the steam is determined by any two properties&lt;/em&gt; and we have the pressure and the internal energy. So we just need go to the steam tables to get the answer ... unfortunatlely my free steam tables at &lt;a href=&quot;http://www.spiraxsarco.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.spiraxsarco.com/&lt;/a&gt; do not provide this combination and the other place does but wants $30.00 to tell me the answer. But my printed tables tell me that the answer is about 480C. &lt;/p&gt;&#xA;" OwnerUserId="3590" LastActivityDate="2015-11-02T08:47:24.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6012" PostTypeId="1" AcceptedAnswerId="7164" CreationDate="2015-11-02T09:45:26.133" Score="2" ViewCount="1344" Body="&lt;p&gt;For a project using rotating voice coils motor in order to balance a system I need to determine the damping coefficient of these voice coils in order to insert it in a ADAMS model and analyse its performances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the oscillation data obtained during a physical test on a mock-up, I used the Matlab tool &lt;a href=&quot;http://www.vibrationdata.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;vibration data&lt;/a&gt; in order to find the damping ratio of these voices coils. To find this value the vibrationdata tool makes a FFT and then a &quot;Half-Power Bandwidth Curve fit&quot; method...&#xA;At the end I got : damping ratio = 0.061, as well as the natural frequency $(f_n)$ 3.1Hz and the Q factor 5.8&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I was not sure about either to insert a damping coefficient or damping ratio value in my ADAMS model, I sent an email to MSC and they answered me &quot;In Adams, bushing required the damping coefficient values, in the units as:&#xA;- newton-sec/meter&#xA;- newton-meter-sec/deg&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I have to calculate the damping_coefficient using the damping_ratio, natural frequency...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obtained using the expression : $$c = 2\,m\omega_n \times \text{(damping ratio)}$$&#xA;with $\omega_n = 2\pi f_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get a certain value but I am totally not sure about the use of the mass $m$ in this expression as I want to get the damping coefficient of a rotating system... &#xA;Can I directly replace this mass by the mass inertia coefficient around which the system rotate ($I_{yy}$ for example from the Matrix)  ?&#xA;Or what else should I do ?&#xA;I would like to specify a damping coefficient in Translation in my ADAMS model...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KLflm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KLflm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On this awful drawing I just tried to sketch the two rotating voice coils in blacks with the magnets (right and left) rotating around the axis y... I would like to know what should I consider in term of mass or inertia to find the damping coefficient in the expression above.&lt;/p&gt;&#xA;" OwnerUserId="2486" LastEditorUserId="1293" LastEditDate="2015-11-02T16:44:51.460" LastActivityDate="2016-01-29T15:36:24.567" Title="How to determine the damping coefficient?" Tags="|mechanical-engineering|vibration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6013" PostTypeId="1" AcceptedAnswerId="6018" CreationDate="2015-11-02T11:46:32.560" Score="0" ViewCount="120" Body="&lt;p&gt;Most/all nuclear fission reactors require cooling.  Due to radioactive decay heat, this is true even when the reactor is switched off.  Hence, there are requirements for nuclear plants to have power at all times: for example, in the United States, they must  have backup battery power for 4–8 hours (&lt;a href=&quot;http://www.pennlive.com/midstate/index.ssf/2011/03/nuclear_power_plants_in_us_vul.html&quot; rel=&quot;nofollow&quot;&gt;example source&lt;/a&gt;), with the assumption that power is subsequently restored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, power is not quickly restored to a nuclear reactor and there is a prolonged &lt;a href=&quot;https://en.wikipedia.org/wiki/Loss-of-coolant_accident&quot; rel=&quot;nofollow&quot;&gt;loss of cooling&lt;/a&gt; event, what is the risk of large-scale radioactive contamination of the environment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purpose of this question, assume nuclear reactor models that are currently common in the United States or western Europe.&lt;/p&gt;&#xA;" OwnerUserId="3599" LastEditorUserId="3599" LastEditDate="2015-11-02T13:48:15.457" LastActivityDate="2016-04-21T04:29:11.887" Title="When backup cooling fails in reactors without passive safety, what is the risk of radioactive contamination?" Tags="|safety|nuclear-technology|nuclear-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6014" PostTypeId="2" ParentId="5982" CreationDate="2015-11-02T17:12:57.013" Score="2" Body="&lt;p&gt;California has a few notable exceptions - they try to favor graduates from the in-state schools, but are very picky on structural engineers.  Here they are, in no particular order:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You do not need to have 4 years of experience if you are a resident and graduated from a California School.  2 years is acceptable.&lt;/li&gt;&#xA;&lt;li&gt;For all licenses except Civil, working in industry (v. studying under a guiding engineer) is considered acceptable experience, so long as the test score is high enough&lt;/li&gt;&#xA;&lt;li&gt;The ethics examination needs to be completed by hand and submitted.  You do not know if you passed until you get your information back.&lt;/li&gt;&#xA;&lt;li&gt;Feedback isn't done by e-mail, it's done by postcards.  Always include 2 self addressed post cards in your packet when you mail it in.&lt;/li&gt;&#xA;&lt;li&gt;Reviewers are assigned based upon last name.  Redoing the packet will mean that it will get reviewed again by the same reviewer.&lt;/li&gt;&#xA;&lt;li&gt;Electrical, Mechanical, and Civil have some interesting rules on personal recommendations.  Civil can only have other Civil.  Mechanical can have Civil or Mechanical.  Electrical can only have Civil or Electrical.  Anything else (besides structural!) can have any recommendation desired.&lt;/li&gt;&#xA;&lt;li&gt;To gain a structural license, you have to have a Civil license.  Then another 4 years of additional experience studying as a licensed Civil engineer under a structural engineer.  Then you have to get Structural Engineer recommendations, then take the 2 part structural examination.  It's very complicated for structural engineers.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-11-02T17:12:57.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6015" PostTypeId="1" CreationDate="2015-11-02T18:11:09.677" Score="3" ViewCount="25716" Body="&lt;p&gt;Dyna Beads are a brand of small ceramic beads that are claimed to have a balancing effect on vehicle tires. Here is &lt;a href=&quot;https://www.youtube.com/watch?v=eq263AYgyYg&quot; rel=&quot;nofollow&quot;&gt;a video showing them in action&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do these work? What are the physical principles involved?&lt;/p&gt;&#xA;" OwnerUserId="3602" LastEditorUserId="368" LastEditDate="2015-11-06T20:52:34.793" LastActivityDate="2018-09-22T13:39:35.327" Title="How do tire-balancing beads work?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6016" PostTypeId="1" AcceptedAnswerId="6019" CreationDate="2015-11-02T20:45:50.640" Score="2" ViewCount="6372" Body="&lt;p&gt;I know that the neutral axis occurs where the sum of stresses (bending + axial) is equal to 0. I have found the stresses and drawn a diagram to represent forces. How do I determine the neutral axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wwGxq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wwGxq.png&quot; alt=&quot;Question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bm4Dj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bm4Dj.jpg&quot; alt=&quot;What ive done so far&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3577" LastEditorUserId="33" LastEditDate="2015-11-03T02:11:36.867" LastActivityDate="2015-11-03T02:41:37.363" Title="Finding Neutral Axis of a simple shape undergoing axial load" Tags="|mechanical-engineering|structural-engineering|structures|applied-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6017" PostTypeId="1" AcceptedAnswerId="6028" CreationDate="2015-11-02T23:07:32.820" Score="5" ViewCount="154" Body="&lt;p&gt;I can't find anything on this. Every source I can find on UV cured polymers describe the process and how they're cured using UV radiation but nothing on what the resistance to UV is once it has cured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically I'm looking at the Stratasys PolyJet (&lt;a href=&quot;http://www.stratasys.com/materials/material-safety-data-sheets/polyjet&quot; rel=&quot;nofollow&quot;&gt;http://www.stratasys.com/materials/material-safety-data-sheets/polyjet&lt;/a&gt;) line of resins which are supposed to be very similar to PMMA in mechanical properties but I can't find anything on whether this extends to UV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone with more knowledge on polymers can help me out that would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3605" LastEditorUserId="90" LastEditDate="2015-11-03T00:15:32.480" LastActivityDate="2015-11-03T13:24:03.833" Title="UV resistance of UV cured photopolymers" Tags="|materials|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6018" PostTypeId="2" ParentId="6013" CreationDate="2015-11-03T01:41:50.410" Score="2" Body="&lt;p&gt;There are different types of nuclear reactors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most commons ones are pressure water reactors and boiling water reactors. The &lt;a href=&quot;http://www.world-nuclear.org/info/Nuclear-Fuel-Cycle/Power-Reactors/Nuclear-Power-Reactors/&quot; rel=&quot;nofollow&quot;&gt;World Nuclear Association&lt;/a&gt; has published the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pressure Water Reactors (PWR)&lt;/strong&gt; are currently used in: the US, France, Japan, Russia &amp;amp; China. The number of such reactors is 277.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 80 &lt;strong&gt;Boiling Water Reactors (BWR)&lt;/strong&gt;, predominantly in the US, Japan and Sweden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both PWR and BWR type reactors use enriched uranium for fuel and ordinary water as the coolant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, there are 49 &lt;strong&gt;Pressurised HEAVY Water Reactors (PHWR)&lt;/strong&gt; used in Canada and India. Unlike PWRs and BWRs, these reactors use natural uranium as their fuel and heavy water as the coolant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other types of reactors that have been used are Gas-Cooled Reactors, of which 15 where constructed in the UK. These reactors used natural and enriched uranium for fuel and carbon dioxide as the coolant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Russians have also built 15 Light Water Graphite Reactors which use enriched uranium for fuel and ordinary water as a coolant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Germans and Russians have built Fast Breeder Reactors which use plutonium and uranium as fuel and liquid sodium as the coolant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stanford University provides additional information about &lt;a href=&quot;http://large.stanford.edu/courses/2013/ph241/kallman1/docs/nuclear_reactors.pdf&quot; rel=&quot;nofollow&quot;&gt;reactor types&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing to appreciate is the &lt;strong&gt;fuel assemblies&lt;/strong&gt; used in nuclear reactors. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_fuel&quot; rel=&quot;nofollow&quot;&gt;fuel&lt;/a&gt; for PWR, BWR and PHWR reactors (the most common currently used reactors) is uranium oxide powder that has been formed into small cylindrical pellets. The pellets are then placed into rods. The rods are then assembled into bundles and bundles are brought into close proximity to other bundles in the core of the reactors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fission within the reactor core is via neutron interaction between the pellets in the rods, the rods in the bundles and between the bundles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To effectively stop fission and heat production, the bundles would have to be separated to prevent neutron interaction between bundles, rods within bundles would have to be separated to prevent interaction between rods and pellets removed from the rods and separated to stop interaction between pellets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is impractical for general operations and is only undertaken when a core has to be shut down so spent fuel can be replaced. Its why moderators are used and why coolant must be continually supplied to the reactor core to control neutrons and the rate of fission and heat generation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If reactors are deprived of a continual supply of coolant for a prolonged period the core will overheat. Depending on the duration of coolant deprivation, the reactor structure may be damaged and environmental contamination may eventuate as happened at &lt;a href=&quot;https://en.wikipedia.org/wiki/Chernobyl_disaster&quot; rel=&quot;nofollow&quot;&gt;Chernobyl&lt;/a&gt; in 1986 and &lt;a href=&quot;https://en.wikipedia.org/wiki/Fukushima_Daiichi_nuclear_disaster&quot; rel=&quot;nofollow&quot;&gt;Fukushima&lt;/a&gt; in 2011.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are extreme cases and it is why the restoration of cooling, as quickly as possible, was a very high priority at the stricken plant at Fukushima, even though they were unsuccessful due to the extensive damage at the plant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical factors in whether there will be radioactive contamination resulting from coolant deprivation of a reactor core is the design and construction of the plant and the duration of coolant deprivation. Deprive a nuclear reactor of coolant long enough and contamination will occur. Nuclear reactions produce a lot of heat.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-11-03T01:41:50.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6019" PostTypeId="2" ParentId="6016" CreationDate="2015-11-03T02:41:37.363" Score="2" Body="&lt;p&gt;In order to calculate the neutral axis under this definition (see my comment under the OP), you need to figure out the stress distribution for each of the applied loads and then superpose them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for this we need to calculate the properties:&#xA;$$\begin{align}&#xA;A &amp;amp;= 80\cdot80 - 60\cdot60 = 2800 \\&#xA;I &amp;amp;= \dfrac{80\cdot80^3 - 60\cdot60^3}{12} = 2333333&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, under the axial force, the stress distribution is uniform and equal to (positive for tension)&#xA;$$\sigma = \dfrac{N}{A} = \dfrac{72000}{2800} = +25.71$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, under the moment, the stress distribution is linear, and the maximum stresses are equal to&#xA;$$\sigma = \dfrac{My}{I} = \dfrac{(72000\cdot(60+40))\cdot(\pm40)}{2333333} = \pm123.43$$&#xA;where the stresses on the left side are negative (compression) and on the right side are positive (tension). The stress profile due to bending can therefore be described by the equation&#xA;$$\sigma = -123.43 + \dfrac{2\cdot123.43}{80}y$$&#xA;where $y=0$ is on the left side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, adding the effect of the axial load's stress, this equation becomes&#xA;$$\sigma = -123.43 +25.71 + \dfrac{2\cdot123.43}{80}y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to find the neutral axis, you just need to find the zero of this equation:&#xA;$$\begin{gather}&#xA;\sigma = 0 = -97.72 + \dfrac{2\cdot123.43}{80}y \\&#xA;y = \dfrac{97.72}{\left(\dfrac{2\cdot123.43}{80}\right)} = 31.7&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there you have it, the neutral line is 31.7 mm from the left side.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-11-03T02:41:37.363" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6020" PostTypeId="1" AcceptedAnswerId="6021" CreationDate="2015-11-03T02:57:00.033" Score="15" ViewCount="13800" Body="&lt;p&gt;Surprisingly this hasn't been asked before, so I must be missing something simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use engineering stress and engineering strain in this eq. Stress = (Young's modulus) × (strain). This eq. is used in analysis of bending beams, twisting shafts and in buckling. So the final equation of bending $(\frac{M}{I} = \frac{\sigma}{y})$ and torsion $(\frac{T}{I} = \frac{\tau}{r})$ will give us value of engineering stress but not the value of stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why are we considering engineering stress instead of true stress while we know it will not give correct value of stress?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some things I read are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Difficult to measure.&lt;/li&gt;&#xA;&lt;li&gt;Not that much of a difference and we can just apply a Factor of Safety.&lt;/li&gt;&#xA;&lt;li&gt;&quot;We don't consider materials to change their cross-sectional area after loading, since we design to have no plastic deformation the elastic region is most important, therefore what happens after the proportional limit is not important&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Firstly, 1 and 2 are not real reasons for me. Number 3 seems plausible since we always design in the elastic region, but is this it? &lt;strong&gt;Does engineering strain even give valid information after the proportional limit?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3609" LastEditorUserId="2246" LastEditDate="2015-11-03T14:02:32.043" LastActivityDate="2015-11-03T14:58:18.543" Title="Why do we even use engineering stress?" Tags="|materials|structural-engineering|stresses|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6021" PostTypeId="2" ParentId="6020" CreationDate="2015-11-03T04:25:10.287" Score="14" Body="&lt;p&gt;&lt;strong&gt;We use engineering strain even though it is not the &quot;correct&quot; value because in most cases, specifically in the elastic regime, engineering strain differs negligibly from true strain.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For linear elastic, Hookean materials, it is generally the case strain at the elastic limit is very small. Even the strongest steels, for example, have an upper limit when cold worked of about $\sigma_{\textrm{el}}=1\times 10^{9}\ \textrm{Pa}$. The modulus of steel is approximately $E=200\times 10^{9}\ \textrm{Pa}$. Thus $\varepsilon_{\textrm{el}}=0.005=0.5\%$ for the strongest steels. So at the onset of plastic deformation, engineering strain is $0.5\%$. Many useful elastic materials have much lower engineering strain at their elastic limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an isotropic, Hookean elastic solid, the following is true&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varepsilon_{x_{1}} = \frac{1}{E}\left[\sigma_{x_{1}}-\nu\left(\sigma_{x_{2}}+\sigma_{x_{3}}\right)\right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;without loss of generality in choice of $x_{i}$. So in uniaxial tension at the elastic limit, $\sigma_{x_{2}}=\sigma_{x_{3}}=0$ assuming the material is free to contract. Thus $\varepsilon_{x_{2}}=\varepsilon_{x_{3}}=-\frac{\sigma_{\textrm{el}}\nu}{E}=-\nu\varepsilon_{\textrm{el}}$. Since the Poisson ratio $\nu$ is approximately 0.3 for steels in the elastic regime, the cross-sectional linear compressive strain is $0.0015$. The cross-sectional area at the elastic limit is thus $(1-0.0015)^{2}A_{0}$, or very close to $0.997$ times the original area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the true strain is $\frac{1}{0.997}$ times larger than engineering strain at the elastic limit, or about $1.003$ times, or about $0.3\%$ larger. Keep in mind this is at the elastic limit of an exceptionally strong linearly elastic material, and so is a reasonably conservative estimate of the difference between true strain and engineering strain in the elastic regime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the above analysis is reasonably useful for linearly elastic, Hookean solids, it doesn't hold quite as well for polymers and biological materials. Such materials are usually viscoelastic (or another class of material entirely), and thus obey different rules in their behavior. True strain also diverges quite wildly from engineering strain in the plastic regime, as evidenced in the following plot (found &lt;a href=&quot;https://www1.us.elsevierhealth.com/books.elsevier/companionsites/JenkinsKhanna/mmd/yieldpoint/yieldpointa.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www1.us.elsevierhealth.com/books.elsevier/companionsites/JenkinsKhanna/mmd/yieldpoint/Image200.gif&quot; alt=&quot;True stress true strain plot&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your points:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Measuring changes in cross-sectional area during deformation &lt;em&gt;is&lt;/em&gt; difficult. It requires careful placement of calibrated instrumentation on precisely machined test samples. One could use &lt;a href=&quot;https://en.wikipedia.org/wiki/Strain_gauge&quot; rel=&quot;noreferrer&quot;&gt;strain gages&lt;/a&gt; placed on the sides of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tensile_testing#Tensile_specimen&quot; rel=&quot;noreferrer&quot;&gt;tensile bar&lt;/a&gt; to measure lateral strain in uniaxial tension and compression in &lt;a href=&quot;https://en.wikipedia.org/wiki/Tensile_testing#Equipment&quot; rel=&quot;noreferrer&quot;&gt;tensile testing equipment&lt;/a&gt;. Obtaining statistically meaningful results takes many samples, as well as significant time, effort and cost.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There &lt;em&gt;is&lt;/em&gt; little difference. I hope I have adequately explained above how small the difference is: I calculated approximately $0.3\%$ difference in a conservative case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The idea that we can ignore anything beyond the end of the elastic regime, or that we always design for the elastic regime, is not true. Plastic deformation can often be worth studying. Modeling continuous shape-forming processes such as rolling, drawing, extruding, etc. requires a deep understanding of the mechanics of plastic deformation to perform successfully, and to that end true stress and true strain are invaluable. Specifically for wire drawing, see (this &lt;strong&gt;&lt;a href=&quot;http://www.antaac.org.mx/assets/10.-theory-of-wiredrawing.pdf&quot; rel=&quot;noreferrer&quot;&gt;pdf&lt;/a&gt;&lt;/strong&gt;) and find equation 7. Plastic deformation is also useful for modeling materials that must permanently deform in some expected use cases, such as car body panels and frame components during a collision. The plastic deformation is useful because it absorbs kinetic energy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I apologize, I didn't actually answer the question for stress. However, it should be fairly clear that the same points apply to stress as apply to strain given their linear relationship in the elastic regime. Again, in the plastic regime, there can be large variations.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.247" LastActivityDate="2015-11-03T14:58:18.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6022" PostTypeId="2" ParentId="5995" CreationDate="2015-11-03T06:36:39.050" Score="4" Body="&lt;p&gt;Internally finned tubes are not that common because of the high cost of manufacture. There aren't many sources for this kind of heat transfer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ruor.uottawa.ca/bitstream/10393/10870/1/EC52380.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; paper provides a detailed description of numerical methods used to calculate heat transfer in a finned tube with fully developed laminar flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It evaluates fins with the following measurement limits: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of Fins:$$8\le M\le32$$&#xA;Fin Length (ft):$$0.2\le\ell\le0.9$$&#xA;Fin angle:$$1.5°\le\beta\le3°$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fTZRk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fTZRk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do have laminar flow, you can use the following chart to estimate the Nusselt number for a certain fin design:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F7UTE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F7UTE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the Nusselt number to calculate the heat transfer coefficient, h:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Nu= \frac {hd}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optimally, you would use the hydraulic diameter rather than the nominal tube diameter. &lt;a href=&quot;http://buet.ac.bd/me/icme/icme2001/cdfiles/Papers/Thermal/29_FINAL_TE-71(159-163).pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; uses a method of determining hydraulic diameter of an internally finned tube using the ratio of cross sectional area of flow to the wetted perimeter. This area would be irritating to calculate for fins and I don't think it would affect the result any more than the other assumptions we have made. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also of course, use the thermal conductivity, k, value for your fluid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly use this equation to get your heat transfer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=hA_s\theta_{lmtd}=mC_p\Delta{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.banglajol.info/index.php/JME/article/download/3473/2912&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; paper analyzes T shaped internal fins and gives a good analysis on how fins affect friction factor and a comparison of smooth tube vs finned. &lt;/p&gt;&#xA;" OwnerUserId="3569" LastEditorUserId="3569" LastEditDate="2015-11-04T01:15:15.923" LastActivityDate="2015-11-04T01:15:15.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6023" PostTypeId="1" CreationDate="2015-11-03T07:20:58.217" Score="1" ViewCount="2242" Body="&lt;p&gt;If water at the same temperature is heated with the same heat power source in two different ambient temperatures (one ambient is low and the other ambient is high) then which water sample will reach the equilibrium state of temperature more quickly, or both will take same time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is related, if we just change the ambient temp. and start heating the object then how will it affect to thermal resistance and thermal time constant?&lt;/p&gt;&#xA;" OwnerUserId="3610" LastEditorUserId="90" LastEditDate="2015-11-03T08:09:35.937" LastActivityDate="2015-11-03T13:14:18.760" Title="thermal resistance and thermal time constant with ambient temperature" Tags="|thermodynamics|heat-transfer|temperature|thermal-insulation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6024" PostTypeId="2" ParentId="6023" CreationDate="2015-11-03T07:53:08.040" Score="1" Body="&lt;p&gt;The time constant can be defined as a function of density ($\rho$), volume ($V$), heat capacity ($C_p$), surface area ($A_s$), and heat transfer coefficient ($h$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau = \frac{\rho\ V\ C_p}{h\ A_s}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat transfer coefficient is a function of thermal conductivity and length. &#xA;$$&#xA;  h=\frac{k}{x}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As temperature of water increases, so does its thermal conductivity. This increases the heat transfer coefficient, which decreases the time constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A higher temperature starting point would mean faster heat transfer. &lt;/p&gt;&#xA;" OwnerUserId="3569" LastEditorUserId="272" LastEditDate="2015-11-03T13:14:18.760" LastActivityDate="2015-11-03T13:14:18.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6025" PostTypeId="1" CreationDate="2015-11-03T08:44:39.570" Score="3" ViewCount="198" Body="&lt;p&gt;I have purchased a wind turbine with a power of 300 W, the start up wind speed is 2.5 m/s, rated wind speed is 11 m/s, and the survival wind speed is 45 m/s, and what I'm going to do is to make an educational modeling for how to generate electrical power using a wind turbine, I mean that I will provide the wind using an axial fan, &lt;strong&gt;My question is, how can I predict the power of the fan and its max flow rate required to make the turbine rotate at its maximum power?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3591" LastEditorUserId="3591" LastEditDate="2015-11-03T13:32:19.087" LastActivityDate="2015-11-03T18:45:08.447" Title="What is the power and max flow rate required to rotate a 300 W wind turbine?" Tags="|fluid-mechanics|airflow|renewable-energy|wind-power|wind-tunnels|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6027" PostTypeId="2" ParentId="6020" CreationDate="2015-11-03T10:24:49.933" Score="10" Body="&lt;p&gt;Adding to @starrise's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to your dismissal of reasons 1 and 2, you are forgetting to consider the cost-benefit analysis regarding them. As @starrise showed in their answer, the difference is usually not substantial (though other materials will usually have larger differences).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the materials always show intrinsic variation of their values. Steel's modulus of elasticity has a range defined in different articles  from $\pm6\%$ &lt;a href=&quot;http://www.jcss.byg.dtu.dk/~/media/Subsites/jcss/english/publications/probabilistic_model_code/steelpr.ashx?la=da&quot;&gt;[a]&lt;/a&gt; to $\pm15\%$ &lt;a href=&quot;http://www.assakkaf.com/Papers/Journals/Uncertainties_in_Material_Strength_Geometric_and_Load_Variables.pdf&quot;&gt;[b]&lt;/a&gt; (for the 95% confidence interval).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what's the point of considering the true strain in everyday engineering practice if all the other properties (including yield strength and cross-section dimensions) will have random fluctuations that are all but certain to drown out the &quot;error&quot; due to the use of engineering strain instead of true strain?&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2015-11-03T10:30:00.373" LastActivityDate="2015-11-03T10:30:00.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6028" PostTypeId="2" ParentId="6017" CreationDate="2015-11-03T13:24:03.833" Score="2" Body="&lt;p&gt;I can provide an anecdotal answer.  We use UV cured adhesives to bond pieces inside of our UV lasers.  These lasers produce intense UV light, some of which is scattered around inside of the laser head.  The exposure to those specific UV wavelengths is much more intense than the exposure of a piece of material laying out in the sun.  Bond failure, however, is not an issue for us.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-11-03T13:24:03.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6029" PostTypeId="2" ParentId="6006" CreationDate="2015-11-03T15:31:40.120" Score="7" Body="&lt;p&gt;Convert the nonlinear model to state-space form, $x'=f(x,u)$, and linearize it to get a linear state-space or transfer function representation. You can use this to design a PID controller and simulate it that together with the original nonlinear model to see how the controller performs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted, the performance of the linear controller will most likely deteriorate far from the operating point. You can design a tracking controller based on feedback linearization. This will give you the control action based on the current configuration and desired trajectory. One downside is that it will not be that robust if there are large variations in the parameters (masses, lengths). The theory for this is in the book by Isidori (&lt;a href=&quot;http://www.springer.com/us/book/9783540199168&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this will be tedious algebraic manipulation. You can find several worked examples of these computations being performed using Mathematica. (&lt;a href=&quot;http://wolfram.com/xid/0puxqmekk3-zuv5k2&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt;, &lt;a href=&quot;http://wolfram.com/xid/0b6h7b4gu9ny0lgnhqytaw-huhgr6&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt;, &lt;a href=&quot;http://wolfram.com/xid/0bsw81boyzl1mwhqd5-3fl5mk&quot; rel=&quot;noreferrer&quot;&gt;link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;Here is my attempt with some arbitrary numerical values and sinusoid reference inputs. I have also made the assumption that the masses are concentrated at the end of the links. (If you also want to include the inertia and assume that the mass is uniformly distributed, just update the Lagrangian suitably.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HKDd8.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HKDd8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2015-11-03T19:36:31.170" LastActivityDate="2015-11-03T19:36:31.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6030" PostTypeId="2" ParentId="6025" CreationDate="2015-11-03T17:28:02.553" Score="4" Body="&lt;p&gt;You need to push air through the turbine at 11 m/s, or just above. That's what the rated windspeed means: it's the lowest windspeed at which the turbine hits rated power, at standard atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of air you need is determined by the area of the rotor. Let's say it's $\alpha m^2$. So you know you need to push through $11\alpha m^3/s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assuming you have some shroud or something so that &lt;strong&gt;all&lt;/strong&gt; the air pushed by the fan passes through the turbine rotor, then that's the rate of air that the fan moves. Then you need the specific fan power, to know how much power you'll draw - let's say it's $\phi kJ/m^3$. Multiply that by the air volume, to get the power requirement of the fan. In the above case, that would be $11\alpha\phi kW$. You'll want to add a bit of contingency to that, to account for things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, your required power in is going to be your power out, divided by your turbine efficiency, and then divided by your fan efficiency. &lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-11-03T18:45:08.447" LastActivityDate="2015-11-03T18:45:08.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6031" PostTypeId="1" AcceptedAnswerId="6602" CreationDate="2015-11-03T23:03:22.640" Score="12" ViewCount="9537" Body="&lt;p&gt;Magnetic locks seem to be fairly common on the doors of modern buildings, but I question their reliability when compared to a traditional, mechanical lock. I realize that the strength that it would take to actually pry a door away from a good electromagnetic lock would certainly be enough to damage the door or frame, just as it would be with a mechanical lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had an electromagnet strong enough, however, wouldn't it be possible to negate a magnetic lock by fastening your oppositely-polarized electromagnet on the outside of the door opposite the magnet on the inside so that, when activated, the outer magnet would overpower the attraction between the inner magnet and the magnetic material on the door, effectively opening the locked door?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, I feel this would be a pretty big disadvantage when compared to a traditional mechanical lock. Not that a powerful and portable electromagnet is necessarily a readily available and affordable thing, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the practicality of actually owning and operating such a powerful electromagnet, is this even a plausible thing to do?&lt;/p&gt;&#xA;" OwnerUserId="3620" LastEditorUserId="368" LastEditDate="2015-11-05T16:58:39.190" LastActivityDate="2020-11-09T14:35:19.117" Title="Can a magnetic lock be negated or overcome by a stronger electromagnet?" Tags="|mechanical-engineering|electrical-engineering|magnets|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6033" PostTypeId="1" CreationDate="2015-11-04T12:20:03.823" Score="1" ViewCount="67" Body="&lt;p&gt;Consider a &lt;a href=&quot;http://www.erv-gmbh.at/en/flanges-collars-butt-ends/167-v-boerdel-din-2642&quot; rel=&quot;nofollow&quot;&gt;collar&lt;/a&gt; (V-Bördel) and a &lt;a href=&quot;http://www.erv-gmbh.at/en/flanges-collars-butt-ends/178-glatte-bunde-en-1092-1-typ-32&quot; rel=&quot;nofollow&quot;&gt;flat collar&lt;/a&gt; (glatter Bund) for a flanged connection, in conjunction with a loose flange.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are reasons to use one over the other? I want to know because a tender specified a flat collar but it seems (I'm not on site, just helping with clerical work an spotting the discrepancy) the constructor installed collars. Most likely they simply got a better deal on the one type. Is there an important functional difference?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2015-11-04T12:20:03.823" Title="Piping construction, flanged connection - Difference between Collar and Flat collar" Tags="|mechanical-engineering|piping|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6034" PostTypeId="2" ParentId="4113" CreationDate="2015-11-04T12:39:13.833" Score="3" Body="&lt;p&gt;The original &lt;a href=&quot;http://dns2.asia.edu.tw/~ysho/YSHO-English/2000%20Engineering/PDF/Ind%20Eng%20Che%20Fun15,%2059.pdf&quot; rel=&quot;nofollow&quot;&gt;Peng Robinson (1976)&lt;/a&gt; paper has the equation rearranged in cubic form with respect to Z. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z^3-(1-B)Z^2+(A-3B^2-2B)Z-(AB-B^2-B^3)=0$$&#xA;$$A=\frac{aP}{R^2T^2}$$&#xA;$$B=\frac{bP}{RT}$$&#xA;$$Z=\frac{Pv}{RT}$$&#xA;$$a(T_c)=0.45724\frac{R^2T_c^2}{P_c}$$&#xA;$$b(T_c)=0.07780\frac{RT_c}{P_c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the critical point, A and B are 0.45724 and 0.07780, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the cubic equation is &lt;a href=&quot;http://www.wolframalpha.com/input/?i=z%5E3-%281-.0778%29*z%5E2%2B%28.45724-3*.0778%5E2-2*.0778%29*z-%28.45724*.0778-.0778%5E2-.0778%5E3%29%3D0&quot; rel=&quot;nofollow&quot;&gt;solved&lt;/a&gt; for its three roots at the critical point, it yields one real root and two imaginary roots. &#xA;$$Z_c=0.321379$$&#xA;$$Z_c=0.30041-0.01199i$$&#xA;$$Z_c=0.30041+0.01199i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average of these three roots is:&#xA;$$Z_c=0.30739967$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the critical point, there is only one phase in the system. However, the cubic equation is judged based on how good it is at finding the three roots corresponding to each phase in the system. I assume the standard practice in comparing cubic equations of state is to take the average of all three roots whether they are real or not. &lt;/p&gt;&#xA;" OwnerUserId="3569" LastActivityDate="2015-11-04T12:39:13.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6035" PostTypeId="1" AcceptedAnswerId="6036" CreationDate="2015-11-04T14:16:09.203" Score="5" ViewCount="160" Body="&lt;p&gt;I look at PID controler and do not understand derivatve term. I know it's used in order to not let the signal to overshoot the set value. BUT when the sigal is arising, its derivative is positive. So when we add the positive derivative term, the signal gets even higher, so its derivative becomes even higher!&#xA;I think that if the signal is rising too fast it should be slowed down by derivative term, which means that derivative term should produce NEGATIVE value to DECREASE the signal that controlls the object. How does it work in reality? Is there any good simulation online of how PID works? I've seen some simulations, but they weren't very informative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more question: I want to check if I understand the integral term properly. Let's assume our PV is arising and getting closer and closer to the SP. During that time integral part of the PID output is GROWING all the time, until the PV reaches SP, than the integral part has already grown so much that the PV overshoots the SP. So PV rises above SP and the integral part becomes smaller and smaller (but still bigger than zero), and finally it becomes so small, that it gets negative smaller than zero). That casues the PV to start decreasing. It gets smaller and smaller and again, integral part is so big (on the minus part this time) that the PV goes down too much and gets smaller than the SP.&#xA;And that's why integral part causes oscillations. Was my example right? Is that how it really works?&lt;/p&gt;&#xA;" OwnerUserId="3627" LastEditorUserId="3627" LastEditDate="2015-11-04T14:21:34.267" LastActivityDate="2020-05-10T20:47:32.973" Title="How does the DERIVATIVE term slow down the signal?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6036" PostTypeId="2" ParentId="6035" CreationDate="2015-11-04T15:28:46.243" Score="2" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ e = r - y$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$e \rightarrow$&lt;/span&gt; error&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$r \rightarrow$&lt;/span&gt; reference input&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$y \rightarrow$&lt;/span&gt; actual output of the plant&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; is increasing (&lt;span class=&quot;math-container&quot;&gt;$\left[\frac{d e}{d t}\right]&amp;gt;0$&lt;/span&gt;) this means that &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; is becoming smaller and smaller compared to &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;. So the input needs to be increased to increase &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your analysis of the integral term is right.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="18238" LastEditDate="2020-05-10T20:47:32.973" LastActivityDate="2020-05-10T20:47:32.973" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6037" PostTypeId="1" CreationDate="2015-11-04T19:53:57.167" Score="2" ViewCount="413" Body="&lt;p&gt;I am simulating an axial turbine. Under relaxation factors are default.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Residuals are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vkvAZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vkvAZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P9HcT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P9HcT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CvqEf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CvqEf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also, mass flow rate, torque (in direction of Y) and inlet and outlet total pressure are as follows(in order to judge on convergence):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque (in direction of Y):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k2JTo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k2JTo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g2lHI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g2lHI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inlet and outlet total pressure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/80WwU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/80WwU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B41Jk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B41Jk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass flow rate in inlet and outlet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q65gL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q65gL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have some question about Residuals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see in first figure, residuals decrease until about iteration of 1500 and then after iteration of 1500, residuals increase. I want to know that this event is logical or not. I already thought that residuals should decrease decidedly because of convergence. But in this simulation, residuals increases and decreases!! What is reason for this event?&#xA;According above figures, Is solution converged?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="33" LastEditDate="2015-11-05T02:39:19.220" LastActivityDate="2015-11-05T02:39:19.220" Title="How to determing convergence in a CFD model" Tags="|mechanical-engineering|fluid-mechanics|cfd|" AnswerCount="0" CommentCount="2" FavoriteCount="0" ClosedDate="2015-11-11T22:08:44.047" ContentLicense="CC BY-SA 3.0" />
  <row Id="6038" PostTypeId="1" CreationDate="2015-11-04T20:27:11.587" Score="2" ViewCount="477" Body="&lt;p&gt;How does Microsoft project an image so close to the eye yet understandable by the brain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the display transparent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This link has good information but is speculative.&#xA;&lt;a href=&quot;https://www.reddit.com/r/HoloLens/comments/2v0nac/hololens_display_tech_and_what_to_expect_from_it/&quot; rel=&quot;nofollow&quot;&gt;https://www.reddit.com/r/HoloLens/comments/2v0nac/hololens_display_tech_and_what_to_expect_from_it/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What materials are used?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a list of patents see this: &lt;a href=&quot;http://vrwiki.wikispaces.com/Microsoft+HoloLens&quot; rel=&quot;nofollow&quot;&gt;http://vrwiki.wikispaces.com/Microsoft+HoloLens&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3629" LastEditorUserId="3629" LastEditDate="2015-11-05T19:13:40.013" LastActivityDate="2015-12-10T22:13:26.870" Title="What optics and display are used in Microsoft HoloLens?" Tags="|design|optics|computer-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6039" PostTypeId="1" CreationDate="2015-11-04T20:33:32.607" Score="2" ViewCount="357" Body="&lt;p&gt;I want to monitor how much water we're pumping out of our shallow well every day (residential). We want to figure out how many gallons per day we're using. Doesn't have to be crazy accurate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First thought was hooking up some kind of inline flow meter to a computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I thought, hey, since the pump always outputs the same amount of water when it's running, wouldn't monitoring the amount of time the pump is running be fairly accurate as well without needing an inline flow meter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking a relay or something similar that's triggered when the pump turns on. What would be a simple way to hook this up to a computer? A usb trigger thingy? (don't know if exists)&#xA;Or an arduino? &lt;/p&gt;&#xA;" OwnerUserId="2080" LastEditorUserId="2080" LastEditDate="2015-11-04T23:47:41.177" LastActivityDate="2018-02-26T09:39:59.277" Title="Monitoring water flow from a pump" Tags="|pumps|water-resources|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6040" PostTypeId="2" ParentId="2470" CreationDate="2015-11-04T20:51:07.790" Score="-1" Body="&lt;p&gt;You might want to use (if you can afford it) a material known as aerogel. It is known to be one of the best insulators known . The material has some drawbacks though, it is very brittle so you would have to build a support matrix to use it for your application and the material has a slight bluish tinge to it, don't know if that would be a problem for you.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2015-11-04T20:51:07.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6041" PostTypeId="1" AcceptedAnswerId="6043" CreationDate="2015-11-04T22:44:48.653" Score="3" ViewCount="363" Body="&lt;p&gt;I'm trying to solve the question attached below (please note the handwritten values are ones that I calculated/not part of question) I am assuming that to find the load $q$ I need to get the bending stress (which from my calculation was $45\text{ N/mm}^2$) and substitute into equation of Bending stress $= \dfrac{My}{I}$ to solve for M which should be equated to $\dfrac{ql^2}{4}$ then find $l$, however I think I messed up in the first part of the question because I'm not taking the area of part A into consideration, can you please advise me on how to approach depth-varying beams?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FrDpv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FrDpv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3577" LastEditorUserId="33" LastEditDate="2015-11-05T02:29:54.277" LastActivityDate="2015-11-05T02:29:54.277" Title="Calculating stresses of section with varying depth" Tags="|mechanical-engineering|structural-engineering|structures|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6043" PostTypeId="2" ParentId="6041" CreationDate="2015-11-05T01:17:58.963" Score="4" Body="&lt;p&gt;As always, the first thing you need to do is calculate your properties.&#xA;\begin{align}&#xA;A_A &amp;amp;= 150\cdot10 + 6\cdot(200-10) = 2640\text{ mm}^2 \\&#xA;\overline{y}_A &amp;amp;= \dfrac{150\cdot10\cdot195 + 6\cdot190\cdot95}{2640} = 151.8\text{ mm} \\&#xA;I_A &amp;amp;= \dfrac{150\cdot10^3}{12}+150\cdot10\cdot(151.8-195)^2+\dfrac{6\cdot190^3}{12}+6\cdot190\cdot(151.8-95)^2 \\&#xA;I_A &amp;amp;=9919274\text{ mm}^4 \\&#xA;A_B &amp;amp;= 150\cdot10 + 6\cdot(100-10) = 2040\text{ mm}^2 \\&#xA;\overline{y}_B &amp;amp;= \dfrac{150\cdot10\cdot95 + 6\cdot90\cdot45}{2040} = 81.76\text{ mm} \\&#xA;I_B &amp;amp;= \dfrac{150\cdot10^3}{12}+150\cdot10\cdot(81.76-95)^2+\dfrac{6\cdot90^3}{12}+6\cdot90\cdot(81.76-45)^2 \\&#xA;I_B &amp;amp;=1369647\text{ mm}^4&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you need to observe the three sources of stress in your structure:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The axial load at point B;&lt;/li&gt;&#xA;&lt;li&gt;The moment due to the eccentricity of the load at point B;&lt;/li&gt;&#xA;&lt;li&gt;The moment due to the load $q$ to be determined.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, what you need to do is calculate the stresses due to the known sources (#1 and #2) to see what slack you have for the load $q$. Strictly speaking you should check points A and B, but I'll only do A here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the axial load at point B is equal to $70\cdot2040=142800\text{ N}$. This immediately gives us a uniform stress of&#xA;$$\sigma_1 = \dfrac{142800}{2640} = 54.09\text{ N/mm}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the resultant force of the load at point B is located at the centroid of the section at point B, which is different from the centroid at point A. This causes a moment, which is equal to&#xA;\begin{align}&#xA;M &amp;amp;= 142800\left((200-151.8)-(100-81.76)\right) = 4278288\text{ Nmm} \\&#xA;\therefore \sigma_2 &amp;amp;= \dfrac{My}{I} = \dfrac{4278288\cdot(200-151.8)}{9919274} =20.79\text{ N/mm}^2&#xA;\end{align}&#xA;Note the stress is being calculated in the top fiber, since that is where the tensile stresses occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, just these loads give you already $\sigma = 54.09+20.79=74.88\text{ N/mm}^2$, meaning you are allowed an increment of $\sigma_3 = 115-74.88 = 40.12\text{ N/mm}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that you need to back-calculate the load via stress, so:&#xA;\begin{align}&#xA;M &amp;amp;= \dfrac{qL^2}{2} \\&#xA;\sigma_3 &amp;amp;= \dfrac{My}{I} \\&#xA;\therefore \sigma_3 &amp;amp;= \dfrac{qL^2y}{2I} \\&#xA;\therefore q &amp;amp;= \dfrac{2I\sigma_3}{yL^2} = \dfrac{2\cdot9919274\cdot40.12}{(200-151.8)\cdot2500^2} = 2.64\text{ N/mm}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may have bundled the calculations, but that's the general gist of it.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-11-05T01:17:58.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6044" PostTypeId="2" ParentId="6039" CreationDate="2015-11-05T02:07:25.443" Score="1" Body="&lt;p&gt;Typically a well pump fills a pressure tank. By knowing the max and min fill levels of the pressure tank and how often it cycles, you would roughly know consumption. The measurement would be off by the amount used when the pump was running, but it would get you close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To measure the difference between min and max fill, turn off all water use in the house except for a bath tub faucet or a place you can fill a five gallon bucket. Turn on the water until the pump starts then turn the water off. Wait until the pump is finished then begin filling and emptying a 5 gallon bucket until the pump starts again. Counting the number of buckets gives you the volume difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also try to determine the flow rate of your pump, but since the flow rate is pressure dependent, it may give you a less accurate answer unless you put a pressure transducer in the line and calibrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a variety of ways you could count the pump start up/ durration. It really depends on your skill set as to which one is easier. An arduino, raspberry pi, or PC with a &lt;a href=&quot;https://labjack.com/&quot; rel=&quot;nofollow&quot;&gt;daq&lt;/a&gt; or &lt;a href=&quot;http://makeymakey.com/&quot; rel=&quot;nofollow&quot;&gt;other inputs&lt;/a&gt; would all work well. A relay to the pressure switch would work, but a little dangerous and unnecessary. A &lt;a href=&quot;https://www.sparkfun.com/products/11005&quot; rel=&quot;nofollow&quot;&gt;split core current sensor&lt;/a&gt; is a nice solution that can also tell you when there are issues with your pump. You could also do it simply with a well placed microphone. You could use a &lt;a href=&quot;https://pypi.python.org/pypi/soundmeter&quot; rel=&quot;nofollow&quot;&gt;python library&lt;/a&gt; on the PC with standard audio input or a &lt;a href=&quot;https://learn.sparkfun.com/tutorials/sound-detector-hookup-guide&quot; rel=&quot;nofollow&quot;&gt;pre-made board&lt;/a&gt; for arduino or rpi.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-05T02:07:25.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6045" PostTypeId="1" CreationDate="2015-11-05T02:12:46.517" Score="2" ViewCount="614" Body="&lt;p&gt;I expected to find more documentation on analytical solutions to determine the contact force between bodies, where deformations (&lt;span class=&quot;math-container&quot;&gt;$\delta = \delta_{1} + \delta_{2}$&lt;/span&gt;) and impact time (&lt;span class=&quot;math-container&quot;&gt;$t = t_{f} - t_{i}$&lt;/span&gt;) are unknown.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The time interval begins at contact between bodies (with &lt;span class=&quot;math-container&quot;&gt;$m_{1}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$v_{1,i}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$m_{2}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$v_{2,i}$&lt;/span&gt;) and ends at the instant &lt;span class=&quot;math-container&quot;&gt;$v_{f} = v_{1,f} = v_{2,f}$&lt;/span&gt;.  Without deformation, &lt;span class=&quot;math-container&quot;&gt;$d = 0$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$d = \frac {v'-v}{2}*t$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathrm {dP} = F\mathrm {dt}$&lt;/span&gt; require the time interval &lt;span class=&quot;math-container&quot;&gt;$t = 0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F \to \infty$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Avoiding deformation analysis, is it possible to apply &lt;span class=&quot;math-container&quot;&gt;$\lim_{t \to 0}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\lim_{d \to 0}$&lt;/span&gt; to calculate a finite, average contact force?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;If not, how is deformation (&lt;span class=&quot;math-container&quot;&gt;$\delta = \delta_{1} + \delta_{2}$&lt;/span&gt;) or time duration (&lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;) calculated?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;While accurate solutions are readily obtained by Finite Element Methods (continuum mechanics), as shown, &lt;strong&gt;do analytical solutions exist?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Uxehc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uxehc.jpg&quot; alt=&quot;FEA&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Many engineering texts do contain partial solutions (for axial, bending, and torsional impact), where deformation is calculated for the impacted body only- &lt;strong&gt;they do not account for deformation in both bodies&lt;/strong&gt;.  This results is a conservative estimate of contact force and stress.&#xA;Contact forces and stresses are well documented (ex. &lt;a href=&quot;https://en.wikipedia.org/wiki/Contact_mechanics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Contact Mechanics&lt;/a&gt;); however, they require that contact forces are known.  Please explain how these might be introduced to solve for deformation in both bodies.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;My incomplete solution follows.  I am interested in critiques and/or practical, analytical approaches to this type of problem.&lt;/p&gt;&#xA;&lt;p&gt;Conservation of Momentum: &lt;span class=&quot;math-container&quot;&gt;$$m_{1}(v_{f} - v_{1,i}) = m_{2}(v_{f} - v_{2,i}) \; \Rightarrow \; \therefore v_{f} = \frac {m_{1}v_{1,i}-m_{2}v_{2,i}}{m_{1}-m_{2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Without a known force, deformation is solved with respect to energy.  For the ideal case, bodies are linear elastic (with stiffness &lt;span class=&quot;math-container&quot;&gt;$k = \frac{EA}{L}$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is the Modulus of Elasticity, &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is contact area, and &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the length perpendicular to contact area).  Certainly, these large assumptions result in error; regardless, they allow for an estimate.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta E_{Kinetic} = E_{Deform}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\frac {1}{2}m_{1}v_{1,i}^2 + \frac {1}{2}m_{2}v_{2,i}^2]_\text{i} - [\frac {1}{2}(m_{1} + m_{2})v_{f}^2]_\text{f} = \frac {1}{2}k_{1}\delta_{1} + \frac {1}{2}k_{2}\delta_{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I am unable to complete the system of equations- I know that both bodies:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Deform during the same time interval&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Have equal contact forces/surface stresses&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;strong&gt;Can this problem be solved analytically?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2538" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-11-17T18:30:28.367" Title="Impact force with unknown deformations or time duration?" Tags="|stresses|applied-mechanics|dynamics|deformation|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6047" PostTypeId="1" AcceptedAnswerId="6058" CreationDate="2015-11-05T09:04:40.973" Score="5" ViewCount="353" Body="&lt;p&gt;What is the difference between an automatic transmission (AT) and an automated manual transmission (AMT)?&lt;/p&gt;&#xA;" OwnerUserId="3633" LastActivityDate="2015-11-07T20:56:40.263" Title="What is the difference between an automatic transmission and an automated manual transmission?" Tags="|mechanical-engineering|automotive-engineering|power-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6048" PostTypeId="1" AcceptedAnswerId="7867" CreationDate="2015-11-05T09:20:01.100" Score="5" ViewCount="3432" Body="&lt;p&gt;Figure 2 in &lt;a href=&quot;http://uk.mathworks.com/help/simulink/examples/modeling-an-automatic-transmission-controller.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this tutorial&lt;/a&gt; represents the shift map/schedule of an automatic transmission. How is this map constructed? I want to know how to design a shift map myself for another transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shift map is below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vFfNqm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vFfNqm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3633" LastEditorUserId="33" LastEditDate="2015-11-07T14:43:26.717" LastActivityDate="2016-03-11T20:48:35.690" Title="How is the shift map of an automatic transmission constructed?" Tags="|mechanical-engineering|automotive-engineering|power-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6050" PostTypeId="2" ParentId="6048" CreationDate="2015-11-05T10:30:59.047" Score="-1" Body="&lt;p&gt;It is written approximately at the end of the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot; The engine torque map, and torque converter characteristics used in the simulations are shown in Figure 10 and Figure 11.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I guess, the map/schedule is built from the graphs.&lt;/p&gt;&#xA;" OwnerUserId="3636" LastActivityDate="2015-11-05T10:30:59.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6051" PostTypeId="1" CreationDate="2015-11-05T11:55:24.913" Score="6" ViewCount="9317" Body="&lt;p&gt;I've been having trouble doing this school assignment. My professor has not taught the class how to read control block diagrams so I assumed that we were supposed to research on it. I've watched a few YouTube tutorials online and have been able to grasp the rough concept of it. But I'm stuck halfway through. Shown below is the question itself and my work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wQJxH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wQJxH.jpg&quot; alt=&quot;Question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V1yqn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V1yqn.jpg&quot; alt=&quot;Working&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried expanding everything out but I'm not able to get it into the form of the transfer function that the question is asking for.&lt;/p&gt;&#xA;" OwnerUserId="3637" LastEditorUserId="272" LastEditDate="2015-11-06T21:27:12.880" LastActivityDate="2020-05-08T06:51:38.417" Title="Calculating a transfer function from a block diagram" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6052" PostTypeId="2" ParentId="6051" CreationDate="2015-11-05T13:06:02.843" Score="1" Body="&lt;p&gt;You are almost there.  Your last line in the second section, $D(s)=F(s)-X(s)$, is pointless; that information is already included in the above lines.  Instead, you need to close the loop by making the replacement $E(s)\rightarrow \theta_r(s)-\theta(s)$ and then solve for the two transfer functions you are asked for.  The final equation is&#xA;$$&#xA;\theta(s)=B(s)\left\{D(s)+A(s)\left[\theta_r(s)+\theta(s)\right]\right\},&#xA;$$&#xA;which you can solve for both of the requested transfer functions.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2015-11-05T13:06:02.843" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6053" PostTypeId="1" CreationDate="2015-11-05T18:55:03.103" Score="2" ViewCount="8169" Body="&lt;p&gt;I was looking over practice problems for an upcoming exam and I am struggling with root locus. I understand how to plot the basic root locus but for some of the aspects when gain is asked to be calculated for plots I'm stumped. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Z80D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Z80D.png&quot; alt=&quot;Practice problem from exam study guide&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if anyone knew a good step by step method to solving something like this?&lt;/p&gt;&#xA;" OwnerUserId="3640" LastActivityDate="2023-01-21T17:19:02.250" Title="Determining Gain from a root locus plot" Tags="|control-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6055" PostTypeId="1" CreationDate="2015-11-05T20:39:58.133" Score="4" ViewCount="424" Body="&lt;p&gt;I'm making weldment drawings that involve a long W-Beam with a number of flanges welded to the web. The problem is that because the weldment is so large, lines become difficult distinguish from each other when the weldment is scaled to fit the page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to be able to do is scale the X- and Y-axes of the view independently so lines become clear while still fitting everything on the page, like SDS/2 does. Solidworks does not appear to have this feature.&lt;/p&gt;&#xA;" OwnerUserId="3644" LastEditorUserId="33" LastEditDate="2015-11-25T19:35:00.157" LastActivityDate="2015-11-25T19:35:00.157" Title="Can drawing views in Solidworks be scaled without keeping aspect ratio?" Tags="|solidworks|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6056" PostTypeId="1" AcceptedAnswerId="6057" CreationDate="2015-11-05T21:51:20.493" Score="2" ViewCount="604" Body="&lt;p&gt;I saw a &lt;a href=&quot;http://www.cnn.com/2015/11/05/travel/macau-studio-city-wheel/index.html&quot; rel=&quot;nofollow&quot;&gt;story on CNN today&lt;/a&gt; about a figure-eight ferris wheel being built in Macau.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious what path the cabins take, and how this occurs. Do they actually follow a figure eight pattern? Do they move from one wheel to the other somehow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story has a picture with the caption&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The cabins don't actually cross in the middle. They follow a gourd-shaped track. Each ride takes around 15 minutes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand what that means. A &quot;gourd-shaped&quot; track?&lt;/p&gt;&#xA;" OwnerUserId="3645" LastActivityDate="2015-11-05T22:23:51.733" Title="How does the figure eight ferris wheel work?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6057" PostTypeId="2" ParentId="6056" CreationDate="2015-11-05T22:04:20.570" Score="10" Body="&lt;p&gt;It looks like the cars don't actually move in a figure eight, though it is designed to look like they do from afar. In the image below you can see that the cars travel around an hourglass-shaped (or gourd-shaped) track which is behind the outer yellow steel structure that looks like a figure eight.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.travelandleisure.com/articles/studio-city-macau-ferris-wheel&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R5fuF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They describe the track as gourd-shaped because some gourds, such as the one shown below, describe the shape of the track very well even though others do not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://blogs.ntm.org/rachel-chapman/2012/07/19/an-old-nahuatl-tale/gourd/&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h2WNLm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-11-05T22:23:51.733" LastActivityDate="2015-11-05T22:23:51.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6058" PostTypeId="2" ParentId="6047" CreationDate="2015-11-05T22:10:48.253" Score="5" Body="&lt;p&gt;In answering this question I'm assuming you already know the difference between an Automatic Transmission (AT) and a Manual Transmission (MT).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Conventional Automatic Transmission (AT)&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Minimal to no power interruption&lt;/li&gt;&#xA;&lt;li&gt;Automatic moving-off&lt;/li&gt;&#xA;&lt;li&gt;Various gear ratios&lt;/li&gt;&#xA;&lt;li&gt;Efficiency: 90-95%&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Automated Manual Transmission (AMT)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Functionality is the same as a Z-speed manual transmission in constant-mesh or synchromesh but the process of shifting gears and moving off are automated.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Power interruption like a manual transmission (MT)&lt;/li&gt;&#xA;&lt;li&gt;Automatic moving-off&lt;/li&gt;&#xA;&lt;li&gt;Efficiency: 92-97% (pretty much the same as a MT)&lt;/li&gt;&#xA;&lt;li&gt;Reputation for poor shift quality and reliability&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are more - I wasn't sure if you were looking for functional differences or characteristics. See the reference I've provided for more info - it is a fantastic textbook.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Automotive Transmissions: Fundamentals, Selection, Design and Application (Naunheimer et al.)&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2015-11-07T20:56:40.263" LastActivityDate="2015-11-07T20:56:40.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6059" PostTypeId="2" ParentId="6015" CreationDate="2015-11-06T05:33:06.680" Score="6" Body="&lt;p&gt;Short of a mathematical proof, the concept provided in the &lt;a href=&quot;http://www.innovativebalancing.com/HowItWorks.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;manufacture's description&lt;/a&gt; is valid- &lt;strong&gt;the beads dynamically distribute mass to align the center of mass with the geometric center of rotation.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The following time steps are listed in the above link.  I have tried to clarify the physics of each time step.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Tire at Rest:&lt;/strong&gt;  The beads rest on the tire floor due to gravity.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Tire in Motion:&lt;/strong&gt;  The beads distribute uniformly by friction as the tire begins to rotate, where they are held in place by centrifugal force (&lt;span class=&quot;math-container&quot;&gt;$F=m \frac {v^2}{r}$&lt;/span&gt;), acting perpendicular to the tire wall.  Note that beads would remain in this state for a perfectly balanced tire.  &lt;em&gt;For completeness, gravity still acts on the beads but is small relative to centrifugal forces.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Heavy Spots in Tire:&lt;/strong&gt;  In this time step it is important to remember that the vehicle's suspension allows the wheel assembly to move vertically- Upward motion is resisted by  the car's suspension (spring) while downward motion is assisted by the suspension (spring) and resisted by tire pressure against the roadbed.  As the 'heavy spots' in the tire are rotated at higher velocities, their centrifugal (inertial) forces physically move the tire up and down- poorly balanced tires can literally cause '&lt;em&gt;wheel hop&lt;/em&gt;'!  As the tire moves (up and down), the beads, with their own masses resisting motion, do not move rigidly with the tire's translation; they move relative to the tire.  Note that without vertical movements, only centrifugal forces act on the beads and they maintain their new position on the tire wall.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Imagine a bead when the tire moves up (ie. the 'heavy spot' on top)-  &lt;em&gt;Reversed when the tire moves down (ie. 'heavy spot' on bottom)&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At the top:  As the tire moves up, the bead does not.  No longer guided by the tire wall it maintains its tangential velocity until it reestablishes contact at a new tire wall location, further from the imbalance.&lt;/li&gt;&#xA;&lt;li&gt;At the bottom: As the tire moves up, the bead is lifted with the tire and does not change its location in the tire wall.&lt;/li&gt;&#xA;&lt;li&gt;On a side between top and bottom:  As the tire moves up, the bead rolls down the tire, changing its relative location in the tire wall further from the imbalance.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Reduced Tire Oscillations:&lt;/strong&gt;  Each oscillation (tire movement, up and down) moves the beads progressively further from the imbalance ('heavy spot'), reducing the imbalance. Therefore, the tire becomes more balanced each oscillation until the tire is balanced.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;No Tire Vibration:&lt;/strong&gt;  The beads are held by centrifugal force in their balanced state.  Because no imbalance exists, there are no vertical movements of the tire to disrupt their positions.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;However, there are practical considerations worth mentioning.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Because the beads are free-floating (unlike mounted weights), accelerations create transient imbalance.  For example, until the beads are distributed, their mass actively contributes to  wheel imbalance.  Similarly, road bumps, braking or driving accelerations will disrupt the equilibrium state.&lt;/li&gt;&#xA;&lt;li&gt;Because the mass is distributed furthest from the center of rotation, more force is required to accelerate the wheel assembly, resulting in decreased fuel efficiency. This effect is magnified by larger diameter tires.  Additionally, more mass is required to balance the larger tire mass.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Note: Tire balancing machines do not validate this method because they have a fixed axis, unlike the axis on a vehicle that moves.  Without movement, the beads gather at the 'heavy spot'.  To accurately measure the effect, the machine must allow movement.&lt;/p&gt;&#xA;&lt;p&gt;While tire balancing beads are a valid method to correct imbalance, their effectiveness depends on speed, tire size, and driving accelerations.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Also note that steel BBs, sand, water, or similar may also be used to dynamically distribute mass, but beware of abrasive or chemical effects (that may damage the tire).  There is also a liquid that solidifies in its balanced state, &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B004HL5TPO&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;liquid tire balance&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastEditorUserId="-1" LastEditorDisplayName="user16" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-11-22T00:47:27.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6061" PostTypeId="1" CreationDate="2015-11-06T14:37:13.633" Score="9" ViewCount="1981" Body="&lt;p&gt;I understand that the most important parameters for a filter are basically pressure drop and thickness, both of which affect efficiency. However why does thickness cause a change in efficiency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm particularly interested in the use of nanofibers as filters, and why decreased diameter increases efficiency. &lt;/p&gt;&#xA;" OwnerUserId="3419" LastEditorUserId="368" LastEditDate="2015-11-20T00:27:36.567" LastActivityDate="2015-11-20T00:27:36.567" Title="Why does filter efficiency decrease with thickness?" Tags="|mechanical-engineering|fluid-mechanics|fluid-filtration|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6062" PostTypeId="2" ParentId="6061" CreationDate="2015-11-06T15:34:25.847" Score="7" Body="&lt;p&gt;More stuff in the way of the flow causes more back-pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a thicker filter as being two filters in series.  For example, if a filter develops 0.1 PSI across it when 100 CFM air flow is going thru it, then two filters in the same 100 CFM air flow must together drop 0.2 PSI across them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what exactly your criteria for efficiency are, more filters in series may decrease efficiency.  If the first filter blocks all the particles you care about, then extra filters only cause more back pressure without any better filtering.  For example, consider two window screens on the same window.  The outer one already blocks all mosquitos, so the inner one doesn't make the overall screen perform better.  It does however make it more difficult to see thru the overall screen.  If you consider efficiency as mosquitos blocked per light blocked, then two screens is clearly less efficient than one screen.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-06T15:34:25.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6064" PostTypeId="1" CreationDate="2015-11-06T17:29:20.527" Score="4" ViewCount="1753" Body="&lt;p&gt;I'm confused about some concepts concerning interaction diagrams for reinforced concrete columns. Hopefully somebody will be able to answer these:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Point with zero tensile strain, but before uniform compression (no moment)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the calculation for this point, I see that the distance $c$ (distance to neutral axis from extreme fiber) is assumed to be the height of the cross section. I understand that the strains are all compressive which means that no tensile zone will exist below the neutral axis, but why wouldn't it be assumed that $c = d$ (distance from extreme fiber to last layer of steel). Since the analysis of columns with axial + moment is essentially an extension of beam analysis, why would the concept of effective depth not be used? In other words, I think the strain distribution for this point should act from the extreme fiber to the last layer of steel.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Stresses in the steel layers are computed based on&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;a) An assumed value for $c$ (for the zero tensile strain point with moment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) A calculated value for $c$ (for the balanced point and below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this value of $c$ when the stress in the steel layers is computed it can come out to be less than, equal to, or greater than the yield stress for the corresponding reinforcement. I don't understand how this can happen. In beam design a strain distribution is assumed (under, over, or balanced), and forces are computed based on the stress in the steel. The assumption you make determines whether you initially know the stress in the steel or if it has to be computed. Either way since the steel is assumed to be elastic-perfectly plastic, the stress in the steel is never going to be greater than the yield stress. Based on this concept, how can the stress in the steel come out to be greater than the yield stress when forming the interaction diagram. From what I can tell if the stress in the steel comes out to be greater than the yield stress, the stress in the steel is set equal to the yield stress for the purpose of computing forces. Can someone explain this iterative process for computing the steel stresses?&lt;/p&gt;&#xA;" OwnerUserId="3653" LastEditorUserId="1832" LastEditDate="2015-11-06T17:49:34.190" LastActivityDate="2016-10-18T00:26:57.600" Title="Reinforced Concrete Columns - Interaction Diagram Question" Tags="|reinforced-concrete|columns|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6065" PostTypeId="2" ParentId="6053" CreationDate="2015-11-06T18:13:30.080" Score="2" Body="&lt;p&gt;First determine the damping ratio $\zeta$ and natural frequency $\omega$ of the closed loop poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general characteristic equation is $s^2+2 \zeta  s \omega +\omega ^2$. For the desired pole locations the characteristic equation is $(s+10 -8.83 i) (s+10 +8.83 i)$. Equate the coefficients and solve for $\zeta$ and $\omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now draw lines from the origin to the desired closed-loop poles at $-\zeta  \omega \pm i \sqrt{1-\zeta ^2} \omega $. The lines must intersect with the root-locus plot to get a feasible $K$. The $K$ value at which the intersection occurs is the value you are looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You did not provide the transfer function, so I made one up. Your analysis will be something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nt7wj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nt7wj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution according to your professor's logic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic equation is $$1 + K G(s) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies $$K = \left| \frac{1}{G(s)}\right|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the figure we know that $G(s)$ has two finite poles and one finite zero. So write $$G(s) = \frac{s-z}{(s-p_1)(s-p_2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed loop pole needs to be at $s = -10 +8.83 i$ (the conjugate will give the same result). Thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K = \left|\frac{\left(-10+8.83 i-p_1\right) \left(-10+8.83 i-p_2\right)}{-10+8.83i-z}\right|$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you again need to read off the values of the open-loop zero and open-loop poles from the figure to get $z$, $p_1$, and $p_2$.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2015-11-06T21:58:15.520" LastActivityDate="2015-11-06T21:58:15.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6066" PostTypeId="2" ParentId="6051" CreationDate="2015-11-06T18:37:23.203" Score="1" Body="&lt;p&gt;This is the equation you want to solve for $\theta$.&#xA;$$\theta =\frac{\left(\theta _r-\theta \right) \left(K_d s+\frac{K_i}{s}+K_p\right)+D}{s^2-\alpha }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is $$\theta =\frac{s^2 K_d \theta _r+D s+K_i \theta _r+s K_p \theta _r}{s^2 K_d+K_i+s K_p+s^3-\alpha  s}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get $G_r$, assume $D=0$. You can do this because the system is linear and superposition holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_r=\frac{\theta}{\theta_r}=\frac{s^2 K_d+K_i+s K_p}{s^2 K_d+K_i+s K_p+s^3-\alpha  s}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, to get $G_D$ assume $\theta_r=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_D=\frac{\theta}{D}=\frac{s}{s^2 K_d+K_i+s K_p+s^3-\alpha  s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The systems are third order.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2015-11-06T18:37:23.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6069" PostTypeId="1" AcceptedAnswerId="6070" CreationDate="2015-11-07T16:31:29.847" Score="3" ViewCount="2001" Body="&lt;p&gt;I would like to know what mechanical parts can be used to convert a rotation movement from an axis to another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: You have the shaft of a motor rotating vertically, but you need to rotate a part horizontally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already found the worm drive assembly but I would like to know if an &quot;exhaustive&quot; list exists somewhere with pro/con for each.&lt;/p&gt;&#xA;" OwnerUserId="3664" LastEditorUserId="282" LastEditDate="2015-11-07T19:08:45.653" LastActivityDate="2015-11-07T22:15:12.683" Title="Transform/change/convert rotation axis" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6070" PostTypeId="2" ParentId="6069" CreationDate="2015-11-07T16:38:22.793" Score="6" Body="&lt;p&gt;The go-to solution is &lt;a href=&quot;https://en.wikipedia.org/wiki/Bevel_gear&quot;&gt;bevel gears&lt;/a&gt;. They are used when the shafts would intersect and allow the same ratios as normal gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also make one of the gears a &lt;a href=&quot;https://en.wikipedia.org/wiki/Crown_gear&quot;&gt;crown gear&lt;/a&gt; so the other gear can be a normal spur gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Universal_joint&quot;&gt;universal joint&lt;/a&gt; is also an option if the angle isn't too big.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-11-07T22:15:12.683" LastActivityDate="2015-11-07T22:15:12.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6071" PostTypeId="1" CreationDate="2015-11-07T20:15:14.020" Score="2" ViewCount="5848" Body="&lt;p&gt;I am simulating a turbine that has curves. Mesh is structure completely. But there is high skewness in tip clearance and some parts of around of the blade:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K2xa3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K2xa3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I use SIMPLEC Method, one factor (Skewness correction) appear under the SIMPLEC method in ANSYS Fluents method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How I must adjust this factor? I dont know which number must be set? 1&amp;lt; or 1 , 2, 3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used K-e turbulence model and when I set Skewness correction to 5 or 20, the following error is appeared:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Error: Divergence detected in AMG solver: k&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Error: Divergence detected in AMG solver: k Error Object: #f&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does this happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest a way to reduce the skewness in ansys fluent and how can I check Skewness in Fluent?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="9" LastEditDate="2015-11-07T22:43:46.767" LastActivityDate="2018-07-23T13:57:32.323" Title="How do I set skewness correction in fluent?" Tags="|mechanical-engineering|fluid-mechanics|turbines|cfd|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6072" PostTypeId="2" ParentId="3324" CreationDate="2015-11-08T07:07:28.900" Score="-1" Body="&lt;p&gt;The explanation is incorrect. The laser is never more than half &quot;in the pit&quot;. The reflection is land or 1/2 pit plus 1/2 non pit.&lt;/p&gt;&#xA;" OwnerUserId="3667" LastActivityDate="2015-11-08T07:07:28.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6073" PostTypeId="2" ParentId="3726" CreationDate="2015-11-08T08:41:40.847" Score="3" Body="&lt;p&gt;Algo has given you a good answer about the detail. There is a much broader, more widely-applicable, kind of answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting any form of energy to heat introduces huge inefficiencies. It is the lowest-quality form of energy. So if you want to do any work with that energy at all, it is best to do it before it gets turned into heat, whenever possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once it's in the form of heat, then the quality of the energy (what we call its &lt;em&gt;exergy&lt;/em&gt;, that is it's ability to do work) depends on the difference between its temperature, and the temperature of the cold reservoir you're using as your heat sink. That's typically, but not always, ambient temperature. If you want to do any work with that heat, then you do it when the difference in temperatures is highest, and when the cold-reservoir temperature is lowest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you wouldn't go from heat to chemical energy to kinetic energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever you do an energy-conversion, you lose some ability to do work - i.e. you lose some exergy. When you do non-heat to heat conversion, you lose a &lt;em&gt;lot&lt;/em&gt; of exergy. So round-trip processes, where you go from, say, chemical energy (methane) to heat and then back again, are really inefficient, and you'll get much less out than you put back in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, you wouldn't use the Sabatier process to power a vehicle, because that would involve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;chemical energy -[combustion]-&gt; heat -[Sabatier]-&gt;&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;chemical energy -[combustion]-&gt; heat -[pistons]-&gt; kinetic energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it really does not matter &lt;em&gt;which&lt;/em&gt; conversion process is in that box that's currently marked &quot;Sabatier&quot;: this convoluted series of conversions would not make sense, whatever that process is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, you'd just go straight from chemical energy to heat to kinetic energy, which is what combustion engines do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Going from methane to heat to methane is going to be very inefficient. You'd only do it in really exceptional circumstances.&lt;/strong&gt; I can't think of any right now, but I'm sure someone could contrive a corner-case where it made sense; e.g. some peculiar circumstance where you were able to move high-grade heat but could not move methane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it would be pointless to put a Sabatier (or similar) process in a methane-powered car. If you needed higher efficiency, the thing to do is to invest in vehicle efficiency: lower speeds, cleaner-burn higher-temperature engines, lighter vehicle weight, more aerodynamic profile, higher-efficiency transmission, kinetic-energy recovery systems, and so on. What you &lt;em&gt;wouldn't&lt;/em&gt; do is add weight through adding inefficient unnecessary kit just to turn your heat back into fuel: just burn less fuel in the first place.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-11-08T12:17:45.510" LastActivityDate="2015-11-08T12:17:45.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6074" PostTypeId="1" CreationDate="2015-11-08T09:51:33.200" Score="2" ViewCount="859" Body="&lt;p&gt;I can't find any information about this on Google, so I thought I'd ask it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a LCD Screen-like panel which uses a mirror-like liquid crystal so that I can make a pane of glass into a mirror electronically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;pixels&quot; do not have to be as small as a normal LCD display,&#xA;in my case I want &quot;bands&quot; of pixels that have a dimension of 30x2 centimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's not really a display, it's more like a dynamic mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this technology exist? I remember seeing something like this a couple of years ago but I can't remember where.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my prototype I'd have a big glass panel resembling a window, and using for example an Arduino I'd like to switch some segments of the panel on or off, and those segment will reflect light with a mirror-grade finish.&lt;/p&gt;&#xA;" OwnerUserId="1299" LastActivityDate="2018-02-13T10:04:58.023" Title="Mirror-like reflective liquid crystal display panel" Tags="|electrical-engineering|liquid|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6077" PostTypeId="2" ParentId="6071" CreationDate="2015-11-09T06:21:56.510" Score="1" Body="&lt;p&gt;First try setting skewness correction to 0, and see if Fluent produces a converged solution.  If not, then you have other problems that need to be addressed first.  If it converges, then increase skewness correction to 1 or 2 and retry.  To address the error regarding divergence in the AMG solver, try reducing your pressure correction under-relaxation factor to less than 1.0, like 0.8 or 0.7.  This will require more iterations but will be more tolerant of skewed cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry I don't have any advice based on hard theory to give you.  I usually just end up playing around with different settings in Fluent until I get it to work.  If you have a tech support contract with ANSYS, their support staff are very knowledgeable and usually can spot problems in a Fluent model pretty quickly.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-11-09T06:21:56.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6078" PostTypeId="1" CreationDate="2015-11-09T09:22:13.580" Score="2" ViewCount="376" Body="&lt;p&gt;I have some gcode created with dmap2gcode. This and Meshcam are the only software I could find that could would take a large image and create the necessary gcode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I would like to try cutting this on a different machine. This machine however is only connected to a SolidCam PC and cant be messed with. I know that it can produce Heidenhain gcode files which seem to be the closest format to what is used in the .ngc file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there software which could translate the EMC gcode to Heidenhain gcode? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample .ngc&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;( Code generated by dmap2gcode-0.07.py widget )&#xA;( by Scorch - 2014 )&#xA;(Settings used in dmap2gcode when this file was created)&#xA;(=========================================================)&#xA;(dmap2gcode_set show_axis  1 )&#xA;(dmap2gcode_set invert     0 )&#xA;(dmap2gcode_set normalize  0 )&#xA;(dmap2gcode_set cuttop     1 )&#xA;(dmap2gcode_set cutperim     1 )&#xA;(dmap2gcode_set disable_arcs 0 )&#xA;(dmap2gcode_set yscale     127 )&#xA;(dmap2gcode_set toptol     -0.127 )&#xA;(dmap2gcode_set vangle     45 )&#xA;(dmap2gcode_set stepover   1.02 )&#xA;(dmap2gcode_set plFEED     254 )&#xA;(dmap2gcode_set z_safe      6.35 )&#xA;(dmap2gcode_set z_cut       -12.7 )&#xA;(dmap2gcode_set diatool    6.35 )&#xA;(dmap2gcode_set origin     Default )&#xA;(dmap2gcode_set tool       Ball )&#xA;(dmap2gcode_set units      mm )&#xA;(dmap2gcode_set plunge     simple )&#xA;(dmap2gcode_set feed       381 )&#xA;(dmap2gcode_set lace       None )&#xA;(dmap2gcode_set cangle     45.0 )&#xA;(dmap2gcode_set tolerance  0.0254 )&#xA;(dmap2gcode_set splitstep  0 )&#xA;(dmap2gcode_set gpre       &quot;G17 G90 G64 P0.001 M3 S3000&quot; )&#xA;(dmap2gcode_set gpost      &quot;M5|M2&quot; )&#xA;(dmap2gcode_set scanpat    &quot;Rows&quot; )&#xA;(dmap2gcode_set scandir    &quot;Alternating&quot; )&#xA;(dmap2gcode_set imagefile  x )&#xA;(dmap2gcode_set ROUGH_TOOL     Ball )&#xA;(dmap2gcode_set ROUGH_DIA      6.35 )&#xA;(dmap2gcode_set ROUGH_V_ANGLE  45 )&#xA;(dmap2gcode_set ROUGH_R_FEED   381 )&#xA;(dmap2gcode_set ROUGH_P_FEED   254 )&#xA;(dmap2gcode_set ROUGH_STEPOVER 1.02 )&#xA;(dmap2gcode_set ROUGH_DEPTH_PP 2.54 )&#xA;(dmap2gcode_set ROUGH_OFFSET   0.508 )&#xA;(dmap2gcode_set ROUGH_SCANPAT  &quot;Rows&quot; )&#xA;(dmap2gcode_set ROUGH_SCANDIR  &quot;Alternating&quot; )&#xA;(=========================================================)&#xA;G17 G90 G64 P0.001 M3 S3000&#xA;G21&#xA;G91.1&#xA;G0 Z6.3500&#xA;F381.0000&#xA;G18&#xA;F254.0000&#xA;G0 X254.0000 Y127.0000&#xA;F381.0000&#xA;G1 Z-11.7156&#xA; X248.1755 Z-11.7342&#xA;G2 X247.6719 Y127.0000 Z-11.7835 I0.0438 K-3.0451&#xA;G1 X245.5041 Z-11.7844&#xA; X244.5407 Z-11.8400&#xA; X242.7233 Z-11.8420&#xA;G2 X242.2197 Y127.0000 Z-11.7970 I-0.5310 K-3.1026&#xA;G1 X233.1107 Z-11.7978&#xA; X231.2495 Z-11.7528&#xA;G2 X230.7240 Y127.0000 Z-11.7038 I-0.5473 K-3.0262&#xA;G1 X222.0310 Z-11.7541&#xA; X220.6734 Z-11.7722&#xA; X218.9655 Z-11.9030&#xA; X218.1772 Z-11.9114&#xA;G2 X217.5860 Y127.0000 Z-11.9992 I0.2959 K-4.0213&#xA;G1 X215.1993 Z-12.1024&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;a sample from the .H file:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 BEGIN PGM PLATTENP1 MM&#xA;1 BLK FORM 0.1 Z X+0.000 Y-500.000 Z-42.000&#xA;2 BLK FORM 0.2 X+500.000 Y+0.000 Z+0.000&#xA;3 ; TEIL-NAME    : PVC_UNTEN - KOPIE&#xA;4 ; PROGRAMM-NR. : 1&#xA;5 ; DATUM - ZEIT : 30-OCT-2015 - 11:13:10&#xA;6 ; ERSTELLT MIT SOLIDCAM&#xA;7 ; PP-STAND 2.0 VOM 28.10.2015&#xA;8 * VERWENDETE WERKZEUGE:&#xA;9 ; T19 SCHAFTFRÄSER D19.746&#xA;10 ; T12 SCHAFTFRÄSER D5.968&#xA;11 * WERKZEUGE/JOBS:&#xA;12 TOOL CALL 19 Z S7000&#xA;13 *     HSR-R-CONT-FERTIGMODELL&#xA;14 CYCL DEF 32.0 TOLERANZ&#xA;15 CYCL DEF 32.1 T0.05&#xA;16 CYCL DEF 32.2 HSC-MODE:0 TA3&#xA;17 L X+142.093 Y-142.093 R0 FMAX M3&#xA;18 L Z+16.000 FMAX&#xA;19 L Z+0.490 FMAX&#xA;20 CC X+140.696 Y-140.696&#xA;21 CP IPA+45.000 IZ-0.044 DR+ F3000&#xA;22 CC X+140.696 Y-140.696&#xA;23 CP IPA+90.000 IZ-0.087 DR+&#xA;24 CC X+140.696 Y-140.696&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3683" LastEditorUserId="33" LastEditDate="2015-11-10T13:46:00.320" LastActivityDate="2017-10-31T01:31:29.273" Title="Is conversion from EMC (*.ngc) to heidenhain (*.h) CNC code possible?" Tags="|computer-aided-design|cnc|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6079" PostTypeId="2" ParentId="6078" CreationDate="2015-11-09T12:59:39.107" Score="1" Body="&lt;p&gt;Without knowing much more, your best bet would be to switch your Heidenhain controller from Conversational Mode to G-Code mode(this is a setting in the controller) This way you can run g-code on the Heidenhain. You may want to more closely look at that code and make sure you clear all modal codes/offsets and run the program in single block mode if you're going to try it out.  Good luck!&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2015-11-09T12:59:39.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6080" PostTypeId="2" ParentId="6074" CreationDate="2015-11-09T13:59:59.973" Score="0" Body="&lt;p&gt;The closest thing I can think of to this would be a &lt;a href=&quot;http://www.ti.com/lsds/ti/analog/dlp/how-dlp-works.page&quot; rel=&quot;nofollow&quot;&gt;digital micromirror device&lt;/a&gt;, like the kind that are used in digital light projectors. See the video &lt;a href=&quot;http://bcove.me/eu6n0s0q&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, explanation of operation is about 2:10 in the video. I don't think you're going to find an LCD mirror that's see-through when not in use (the application I think you're asking for), but I could be wrong. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-11-09T13:59:59.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6081" PostTypeId="2" ParentId="6074" CreationDate="2015-11-09T16:44:12.830" Score="0" Body="&lt;p&gt;As a materials person, one approach would be a material that can change phases with electronic stimuli such that one phase has no electronic band gap (a conductor) and the other phase has a band gap in the visible range (a semi-conductor). I am unaware of and such material, and should such a material exist it would probably be seeing extensive use in Dubai and Las Vegas if it were economical. The biggest challenge in designing such a material for production is ensuring the phase change material is perfectly flat as though polished in the reflective state, and not too rough in the transparent state, e.g. to have a negligible volume change associated with the phase change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest existing materials I am aware of do not have image reflection, but instead become opaque, and there have been several approaches, including &lt;a href=&quot;https://en.wikipedia.org/wiki/Smart_glass#Suspended_particle_devices_.28SPDs.29&quot; rel=&quot;nofollow&quot;&gt;suspended particle&lt;/a&gt;, electrochromic, etc (see the link). Such a material does what I've explained except instead of changing band-gaps with phase, simply causes changing organization of ionized or ionizable materials. The issue with modifying this to become reflective instead of merely opaque is again related to roughness. The material must reflect nearly all of the incident light, and must do so with a smooth, as-if-polished surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way most LCDs work is by electrically activated, nematic (threadlike) liquid phase crystals which twist into and out plane with the panel, blocking light as required. None of the material properties change as a result of the process, so the LCD crystals would need to be always reflective, which might create an unusual and distracting appearance when the pixels are &quot;off&quot;, though it would act nicely as a mirror when on. How to create a reflective nematic liquid-phase crystal is an interesting question in itself. Additionally, to have a polished mirror effect, the crystals would need to line up perfectly, or the image would be distorted in a distracting way. As it stands with displays, if the crystals are a bit unaligned the display still colors the light passing through effectively.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2015-11-09T16:44:12.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6083" PostTypeId="1" CreationDate="2015-11-10T00:26:40.087" Score="6" ViewCount="8452" Body="&lt;p&gt;As part of an ongoing Raspberry Pi project, I want to automatically open and close my car windows depending on the interior temperature and whether or not it's raining.  Currently, when I manually hold the window switch down after the window is completely raised or lowered, the lights (headlights, interior lights) dim slightly, which indicates to me that current is still flowing to the window motor.  I'm worried about overloading the motors with my automated system, since I was just planning to blindly apply current to the motor for a fixed period of time.  So, my question really is, does the motor rely on the user releasing the switch after the window opens/closes, is it just designed to handle small periods of deadlock, or is there some more elegant protection going on in there?&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="90" LastEditDate="2015-11-10T03:09:14.407" LastActivityDate="2024-01-24T00:43:53.143" Title="Do car power windows have limit switches?" Tags="|electrical-engineering|automotive-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6084" PostTypeId="5" CreationDate="2015-11-10T03:53:50.477" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-11-10T03:53:50.477" LastActivityDate="2015-11-10T03:53:50.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6085" PostTypeId="4" CreationDate="2015-11-10T03:53:50.477" Score="0" Body="The physics of buildings, specifically thermal conduction, convection, &amp; radiation; heating and cooling; ventilation; transmission and management of water vapour &amp; pollutants" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-11-10T14:11:58.137" LastActivityDate="2015-11-10T14:11:58.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6087" PostTypeId="1" CreationDate="2015-11-10T05:16:07.407" Score="4" ViewCount="138" Body="&lt;p&gt;I have read a specific term to describe the process when water pools behind a restriction such as a bridge during a flood. I cannot for the life of me remember what the word was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term specifically refers to the process of the ponding of water upstream of a restriction during a flood creating an area where the water pools higher and velocity is slower. Essentially the flood version of localised water detention due to a outlet restriction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could help me out with the exact terminology that would be great. Have been searching all over to no avail.&lt;/p&gt;&#xA;" OwnerUserId="3696" LastEditorUserId="33" LastEditDate="2015-11-11T00:37:28.167" LastActivityDate="2015-11-11T00:37:28.167" Title="Terminology for ponding behind a bridge during a flood" Tags="|civil-engineering|hydrology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6089" PostTypeId="1" AcceptedAnswerId="6111" CreationDate="2015-11-10T08:04:23.520" Score="5" ViewCount="382" Body="&lt;p&gt;I'm trying to find a material suitable to be printed and with the properties of whiteboard, easy to clean and reusable. I'm having difficulties to find alternative materials, or providers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know &lt;a href=&quot;http://www.magicwhiteboard.co.uk/&quot; rel=&quot;nofollow&quot;&gt;magic whiteboard&lt;/a&gt; is an alternative but it isn't reusable, only last two times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I say printable I mean than I can print on it, lines or other shapes and images like on paper and I want this to be permanent. I didn't mean that it needs to be printed in a 3D printer, sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I'm looking for simple whiteboard, but I need to print on it and make it lightweight and handy (mobile phone size). In addition I would like to be able to use different and thinner pens on it.&lt;/p&gt;&#xA;" OwnerUserId="1262" LastEditorUserId="1262" LastEditDate="2015-11-11T22:29:21.243" LastActivityDate="2019-07-29T16:30:38.547" Title="Printable and reusable whiteboard material" Tags="|materials|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6090" PostTypeId="1" CreationDate="2015-11-10T08:53:00.350" Score="5" ViewCount="197" Body="&lt;p&gt;Here's a zoomed piece of &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f4/ACE_Altamont_Pass.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this image from Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wUDUr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wUDUr.jpg&quot; alt=&quot;bridge tower close-up&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a close-up of a tower holding a railroad bridge. A and B are two of the four columns in the tower's corners. I assume that they are placed symmetrically, so arrow B points onto the typical &quot;outerwards&quot; surface and arrow A points onto typical &quot;inwards&quot; surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that where arrow B points the surface is solid and where arrow A points there's a lattice instead. Note the numerous rivets alongside each beam - the beams look like they are box girders assembled such that each of them has three solid surfaces and one lattice surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the reasoning for having a lattice on one of the surfaces only?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="374" LastEditDate="2015-11-11T07:40:12.723" LastActivityDate="2015-11-11T07:40:12.723" Title="Why is this box girder made of lattice on one side only?" Tags="|civil-engineering|structural-engineering|steel|beam|bridges|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6091" PostTypeId="1" CreationDate="2015-11-10T10:29:47.950" Score="0" ViewCount="6918" Body="&lt;p&gt;I am simulating a axial turbine and number of cell are about 3000000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under relaxation factors are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure: 0.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;momentum:0.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turbulent kinetic energy: 0.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turbulent dissipation rate: 0.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Velocity formulation is relative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turbulence model is K-E relizable. The convergence of solution is monitored by checking the residuals of the numerically solved governing equations. Moreover, in order to judge the convergence, the behaviour of other quantities, such as the total pressure at the inlet and outlet boundaries, and torque coefficient generated by the rotor, are also monitored. Here, the default convergence criterion of each residual is reduced in order to allow the monitored quantities to stagnate at consistent values. It seems that convergence is ok but there is reverse flow!! why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;residuals :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FWP2o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FWP2o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;torque coefficient:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O0MQ2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O0MQ2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;total pressure at the inlet and outlet boundaries:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VK9Cs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VK9Cs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass flow rate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LfJ1z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LfJ1z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I continued this run and is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RgRqc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RgRqc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bw0HT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bw0HT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CrQ8l.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CrQ8l.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FktGk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FktGk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Residuals are almost constant. What is the reason? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is convergence OK?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am grateful that guide me about this case. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2015-11-11T07:36:41.490" LastActivityDate="2016-03-30T10:12:21.150" Title="Why is there reverse flow in Fluent?" Tags="|mechanical-engineering|fluid-mechanics|turbines|cfd|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6092" PostTypeId="2" ParentId="6083" CreationDate="2015-11-10T12:28:51.670" Score="6" Body="&lt;p&gt;People do dumb things, and car companies know this.  They try to make basic user-abusable things like this idiot proof, all though this is difficult since some idiots can be quite clever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The few electric car electric window systems I've looked at don't have limit switches.  They do have a controller for each motor that senses the current.  When the window reaches a mechanical limit, the motor slows down and the current goes up.  The controller keeps the voltage applied for maybe 1/2 second, then shuts off the motor.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-10T12:28:51.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6093" PostTypeId="2" ParentId="6090" CreationDate="2015-11-10T13:58:05.160" Score="4" Body="&lt;p&gt;I can see a few reasons why three of the sides would be solid and one side open. They are listed in roughly order of decreasing importance.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Because of how the tower legs (columns) slope, the laced (lattice) side is the least exposed to the weather. Providing a solid surface on the three other sides limits the amount of rain and snow that will travel down the legs. This will help protect the steel and rivets from corrosion. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The laced side prevents the entire leg from being closed and allows any moisture to dry without pooling.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another minor reason may be that keeping one side mostly open allows for easier access to actually construct the legs. Riveting take a lot of manual work, so being able to place hands and tools inside the member eases construction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-11-10T13:58:05.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6094" PostTypeId="2" ParentId="6087" CreationDate="2015-11-10T14:12:25.767" Score="2" Body="&lt;p&gt;I believe that the phrase that you are looking for is either &lt;strong&gt;backwater&lt;/strong&gt; or &lt;strong&gt;headwater&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iGiB3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iGiB3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sub&gt;Image from from &lt;a href=&quot;http://epg.modot.org/index.php?title=748.4_Headwater_and_Backwater&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://epg.modot.org/index.php?title=748.4_Headwater_and_Backwater&lt;/a&gt; &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-11-10T14:12:25.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6095" PostTypeId="2" ParentId="6091" CreationDate="2015-11-10T14:37:42.397" Score="1" Body="&lt;p&gt;Reverse flow is usually not a big deal, there's even an option in the text user interface to suppress the warning.  If you're using higher order discretization in your solver, you may want to instead try a first order approach first.  Get that to converge, then run a few iterations with second order and see if you still get reversed flow.  The solution obtained from the first-order method will provide good initial conditions for the second-order method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The y-axis on your mass flow rate is hard to see, but it looks like the convergence is hovering around zero.  All the values on the y-axis look to be zero, which means that graph is zoomed in quite far.  Based on your residuals plot, I think your solution is pretty well converged at this point.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-11-10T14:37:42.397" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6096" PostTypeId="1" CreationDate="2015-11-10T15:01:24.767" Score="4" ViewCount="105" Body="&lt;p&gt;I know that the effectiveness of metals used for electromagnetic shields heavily depends on the used frequency or frequency range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not find a resource in the web (tables or graphs) that give me the information which material (especially ferromagnetic vs. diamagnetic) should be selected when shielding a component from a given frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some information about this in the web? Which is the most effective metal for a frequency of 125kHz?&lt;/p&gt;&#xA;" OwnerUserId="3699" LastEditorUserId="272" LastEditDate="2015-11-10T21:03:44.977" LastActivityDate="2016-06-19T16:44:41.367" Title="Frequency dependence of electromagnetic shield metal" Tags="|electrical-engineering|metals|rf-electronics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6097" PostTypeId="1" AcceptedAnswerId="6098" CreationDate="2015-11-10T17:55:19.880" Score="6" ViewCount="523" Body="&lt;p&gt;If we have a simple adiabatic cylinder with a piston on top and we heat the gas from the inside, the gas starts to expand and move the piston up. This process is isobaric; but if the pressure on the piston doesn't change, how does it move up?&lt;/p&gt;&#xA;" OwnerUserId="1613" LastEditorUserId="282" LastEditDate="2015-11-14T18:14:58.667" LastActivityDate="2015-11-14T18:14:58.667" Title="How does expanding gas in isobaric process move a piston?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6098" PostTypeId="2" ParentId="6097" CreationDate="2015-11-10T18:48:07.250" Score="9" Body="&lt;p&gt;In reality, such a process is not &lt;em&gt;truly&lt;/em&gt; isobaric; a small rise in temperature will cause the gas to build up an infinitesimally small amount of pressure, which moves the piston up an infinitesimally small distance, which restores the pressure to its original value.  If you assume that the piston can react quickly to the heat being applied, then it behaves nearly isobarically.  Or, in mathematical terms, if $\Delta T$ is the rise in temperature from one state to the next, the process becomes isobaric as ${\Delta T \to 0}$.  This is also referred to as a quasi-static process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Textbooks frequently make simplifications like this because they don't expect students to ask such questions!&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-11-10T23:54:59.860" LastActivityDate="2015-11-10T23:54:59.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6099" PostTypeId="2" ParentId="6096" CreationDate="2015-11-10T22:00:34.433" Score="2" Body="&lt;p&gt;To the best of my knowledge, it's only a matter of &lt;a href=&quot;https://en.wikipedia.org/wiki/Skin_effect&quot; rel=&quot;nofollow&quot;&gt;skin effect&lt;/a&gt;. Your shield should be larger than the skin effect of your electromagnetic field, to be able to fully reflect it (minus some losses). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are just 2 rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For any conductive material, the higher frequency you go, the thicker the skin effect is.&lt;/li&gt;&#xA;&lt;li&gt;For any conductive material, the less resistive it is,  the thicker the skin effect is.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Wikipedia's page linked above seems to indicate that &lt;a href=&quot;https://en.wikipedia.org/wiki/Permalloy&quot; rel=&quot;nofollow&quot;&gt;Permalloy&lt;/a&gt; is good for shielding, but Permalloy's page indicates that its mechanical characteristics make it inappropriate for elaborated shapes. Copper seems to stay a good choice.&lt;/p&gt;&#xA;" OwnerUserId="2664" LastEditorUserId="2664" LastEditDate="2015-11-10T23:28:59.177" LastActivityDate="2015-11-10T23:28:59.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6101" PostTypeId="1" CreationDate="2015-11-11T01:36:35.277" Score="1" ViewCount="151" Body="&lt;p&gt;What would happen if I tempered aluminum of aeronautical use? Being more specific, an alloy of 96% of aluminum and 4% of copper, without modifications and 2% of impurities. (Al 2024)&lt;/p&gt;&#xA;" OwnerUserId="3704" LastEditorUserId="33" LastEditDate="2015-11-11T01:43:03.477" LastActivityDate="2022-02-23T04:08:52.227" Title="What would happen if I tempered aluminum of aeronautical use (Al 2024)?" Tags="|metallurgy|aluminum|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6103" PostTypeId="1" AcceptedAnswerId="6106" CreationDate="2015-11-11T09:35:06.280" Score="3" ViewCount="84" Body="&lt;p&gt;I have a school case study to prepare a (business) plan for office building construction with at least one level of underground parking. I have it pretty much done, but I still need to clarify one thing (teacher is not available):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) What is implied by &quot;Underground water level varies from 5-6m&quot; for the construction? Currently I ignore it. Does it affect the underground parking if I consider only 1 level? Does it have (any/significant) impact on building costs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) What would be depth of typical base/foundation for an office complex building (3 stories, 50k sqm)?  Does the underground water interfere in any way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tried googling for relation of underwater level and foundations, but couldn't really find anything. &lt;/p&gt;&#xA;" OwnerUserId="3709" LastEditorUserId="2246" LastEditDate="2015-11-11T16:44:02.710" LastActivityDate="2015-11-11T17:16:00.317" Title="Office Complex Foundations/Base Depth and Undeground Water" Tags="|building-design|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6106" PostTypeId="2" ParentId="6103" CreationDate="2015-11-11T16:55:33.757" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;(1) What is implied by &quot;Underground water level varies from 5-6m&quot; for the construction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Typically this means that the groundwater table starts between 5m and 6m below grade.  In other words, if you dig over 5m into the soil, you will likely hit the water table and have water in your excavation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construction sites use water mitigation strategies to deal with groundwater, which can be anything from using sump pumps to putting a slurry wall around the excavation.  It all depends on the in situ condition and cost.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(2) What would be depth of typical base/foundation for an office complex building (3 stories, 50k sqm)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without knowing anything about the subsurface condition and what kind of equipment will be housed in the building, this question is really impossible to answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could have bedrock 3m down.  The soil could be cruddy organics or sand, which would require drilling piles down to a more stable soil strata.  Your office building could house a heavy piece of equipment, requiring a special foundation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, not knowing the footprint of the building makes it difficult to guess at a foundation design.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the underground water interfere in any way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Groundwater pressure can make a building buoyant, which needs to be considered in design of the foundation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will likely have to include some kind of foundation drainage system into your building design to shed the groundwater away.  Also, a sump pump system in the foundation will likely be needed, similar to what is done in residential basements.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-11-11T17:16:00.317" LastActivityDate="2015-11-11T17:16:00.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6108" PostTypeId="1" AcceptedAnswerId="6109" CreationDate="2015-11-12T00:17:49.883" Score="12" ViewCount="3314" Body="&lt;p&gt;For example i have an exhaust pipe with very hot air from burning biofuel, and i have a turbine at the end of the exhaust that rotates and generates electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it more efficient to use the heat to boil water then use the steam produced to rotate the turbines? As in why is more electricity generated by using steam to rotate the turbines as opposed to not using steam when burning the same amount of biofuel?&lt;/p&gt;&#xA;" OwnerUserId="3538" LastEditorUserId="3538" LastEditDate="2015-11-12T19:56:38.177" LastActivityDate="2017-05-30T07:20:15.247" Title="Why is using steam to rotate a turbine more efficient?" Tags="|electrical-engineering|turbines|steam|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6109" PostTypeId="2" ParentId="6108" CreationDate="2015-11-12T01:56:53.243" Score="12" Body="&lt;p&gt;The main reason is that a turbine requires a pressure drop to extract energy from the working fluid.  The drop in temperature that is observed in a turbine is a result of the expansion of the fluid; the turbine doesn't have a way to extract the heat energy directly from the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total work done by the fluid is typically expressed as a change in enthalpy, which is the sum of internal energy (heat) and work done by expansion (pressure drop): $\Delta H = \Delta U + \Delta (PV)$.  If the exhaust pressure of your combustor is not much higher than ambient pressure, then there won't be much of a pressure drop across the turbine and hence not much work will be done by the gas.  The gas will exit the turbine at a relatively high temperature, indicating that it still has a lot of energy that wasn't extracted by the turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to capturing this wasted energy is to instead take some of that heat energy and convert it to pressure energy by boiling water - now you have a high-pressure working fluid that's much more useful for driving a turbine.  The turbine is now able to extract much more of the original heat energy in the form of pressure, hence higher efficiency.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-11-12T13:14:01.080" LastActivityDate="2015-11-12T13:14:01.080" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6111" PostTypeId="2" ParentId="6089" CreationDate="2015-11-12T07:34:06.850" Score="5" Body="&lt;p&gt;Get a print on whatever material is easily available from local print houses. Either rigid or poster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Install a glass screen in front of it, and you have your easily cleanable surface.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2015-11-12T07:34:06.850" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6114" PostTypeId="1" CreationDate="2015-11-12T12:22:10.640" Score="3" ViewCount="268" Body="&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt; We are in the process of developing a sensing application. We are currently using an Evaluation board from a vendor for initial development work that has a MSP430 microcontroller. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first version (000) of the board had 240 ABMM &lt;strong&gt;I3C&lt;/strong&gt; Ceramic surface mount microprocessor crystal from Abracon on the board. This evaluation board with custom software works fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second version (001) of the board has a 240 ABMM &lt;strong&gt;F4C&lt;/strong&gt; crystal. The second board with the custom software doesn’t work at all. Although the crystal is from the same vendor the crystals appear to be slightly different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Picture of Crystal from the working EVM board&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3ldSZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3ldSZt.jpg&quot; alt=&quot;Crystal I3C&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Picture of Crystal from the defective EVM board&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/djA7H.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/djA7Ht.jpg&quot; alt=&quot;Crystal F4C&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the schematic for the crystal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AZSyf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AZSyf.jpg&quot; alt=&quot;Schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What is the difference between the two crystals? (&lt;strong&gt;I3C&lt;/strong&gt; vs &lt;strong&gt;F4C&lt;/strong&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt;: I'm wondering if one of the components is not a genuine part and is a knockoff.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.abracon.com/Resonators/ABMM.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Datasheet for Ceramic surface mount microprocessor CRYSTAL 240 ABMM&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="368" LastEditDate="2015-11-18T18:13:45.313" LastActivityDate="2015-11-18T18:13:45.313" Title="What is the difference between I3C and F4C microprocessor crystals?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6115" PostTypeId="2" ParentId="6108" CreationDate="2015-11-12T13:49:42.730" Score="4" Body="&lt;p&gt;Heating water to make steam is not necessarily more efficient, but a lot more practical.  What you describe is how internal combustion engines work, for example, so it's a valid concept.  However, they do this in bursts and use liquid and carefully engineered fuel, which makes the implementation more practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a continuous system as you describe, the fuel is burned at high pressure.  Consider the mechanical difficulty of adding more fuel into the system while sealing against that pressure.  You also have to get the unburnt waste out somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While basic physics does not prevent what you describe, practical engineering does.  It's simpler to burn the fuel at ambient pressure, and use the heat to make high pressure inside a specially designed pressure vessel.  Put another way, it's a lot easier to get heat across a pressure seal than solids with somewhat unpredictable shapes and sizes.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-12T13:49:42.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6116" PostTypeId="2" ParentId="6114" CreationDate="2015-11-12T14:08:44.137" Score="3" Body="&lt;p&gt;Are you sure those are really genuine Abracon crystals?  The markings and packaging look Abracon-like, but the details don't match the datasheet you linked to.  Abracon may use that case in different series, so possibly there is a different datasheet that actually applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without the datasheet you don't know if it's a date code, tolerance indicator,  tells you the required load capacitance, or something else.  The &quot;240&quot; probably indicates 24 MHz, but that's just a guess, and is contrary to what the datasheet says.  Send the picture to Abracon tech support.  They can either tell you what the markings mean, or who the counterfeiter is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second not working, you've only said the whole thing doesn't work.  There is a lot more to a micro running than the crystal.  Check with a scope to see if the crystal is oscillating.  Make sure to use the highest impedance probe setting, and probe the output of the micro (into the crystal).  I have seen even 10x scope probes kill the oscillation if placed on the output of the crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the crystal is not oscillating, that doesn't mean the crystal is broken.  Something else, like a bad solder joint, may be keeping the micro from turning on the crystal driver.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-12T14:08:44.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6117" PostTypeId="2" ParentId="6089" CreationDate="2015-11-12T14:13:47.277" Score="2" Body="&lt;p&gt;I would think that the plastic (vinyl?) that print shops use for banners could be used for this purpose. We know that it is able to be printed on. The only unknown (to me) is how easily it can be erased. My impression is that any slick and flat surface should work for dry erase markers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point in its favor is that it is durable. Being hung up outside for long periods of time has to be a more harsh use than rubbing on it with an eraser.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-11-12T14:13:47.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6118" PostTypeId="1" CreationDate="2015-11-11T15:06:23.240" Score="1" ViewCount="1339" Body="&lt;p&gt;I'm considering attaching a stainless steel bar on a sailboat masthead.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial thought was to use a bar with a cross section of 3/8x2&quot; and 1/4x2&quot;.  The bar will be supported for 3&quot; of its 5&quot; length.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the breaking strength if a load is applied 1&quot; past the support?&lt;br&gt;&#xA;What kind of safety margin would I have if the downward load is 2500 lbs?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would I be better off using rectangular channel 1&quot; x 2&quot; with a 1/8&quot; or 1/4&quot; wall thickness?&lt;/p&gt;&#xA;" OwnerDisplayName="dave abrams" LastEditorDisplayName="user16" LastEditDate="2015-11-12T16:08:29.637" LastActivityDate="2015-11-13T22:10:25.340" Title="Breaking strength of stainless steel bar" Tags="|mechanical-engineering|stresses|deflection|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6119" PostTypeId="2" ParentId="6108" CreationDate="2015-11-12T16:17:43.023" Score="4" Body="&lt;p&gt;You are almost describing a gas turbine engine.  These are used to generate electrical power, and also to power aircraft.  But, in a gas turbine the output of the combustor is at high pressure, and that is used to turn a turbine.  And, that is a different combustion cycle from a steam cycle.&lt;/p&gt;&#xA;" OwnerUserId="2066" LastActivityDate="2015-11-12T16:17:43.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6120" PostTypeId="2" ParentId="6118" CreationDate="2015-11-12T20:06:47.197" Score="4" Body="&lt;p&gt;There are a number of unknowns in the problem statement:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The geometry of the bar&lt;/li&gt;&#xA;&lt;li&gt;The location of the supports &lt;/li&gt;&#xA;&lt;li&gt;Whether the beam is simply supported or clamped at these supports&lt;/li&gt;&#xA;&lt;li&gt;Whether the load is a point load or distributed&lt;/li&gt;&#xA;&lt;li&gt;Whether the load is in lbf or lbm&lt;/li&gt;&#xA;&lt;li&gt;What type of stainless steel is being used, etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, a rough estimate can be found if we make a number of assumptions. Assume that the beam is simply supported with asymmetric point load as shown in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory#Three-point_bending]&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page on beam bending&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum moment is &#xA;$$&#xA;  M_{\max} = \frac{P\,a\,b}{L}&#xA;$$&#xA;where $P$ is the applied load, $a,b$ are the distances of the load point from the supports, and $L$ is the distance between the supports.  For this situation the maximum stress is&#xA;$$&#xA;  \sigma_{\max} = \frac{M\,h}{I} = \frac{P\,a\,b\,h}{L\,I}&#xA;$$&#xA;where $h$ is the height of the beam cross-section and $I$ is the area moment of inertia. For a beam with a square cross-section,&#xA;$$&#xA;  I = \frac{1}{3} w h^3&#xA;$$&#xA;where $w$ is the width of the beam cross-section. So&#xA;$$&#xA;  \sigma_\max = \frac{3\,P\,a\,b\,h}{L\,w\,h^3} = \frac{3\,P\,a\,b}{L\,w\,h^2}&#xA;$$&#xA;Plugging in some numbers, $P$ = 2500 lbf, $a$ = 2 in, $b$ = 1 in, $h$ = 2 in, $L$ = 3 in, $w$ = 3/8 in, we get&#xA;$$&#xA; \sigma_\max = \frac{3 \times 2500 \times 2 \times 1 }{3 \times 3/8 \times 2^2} = 3333\,\, \text{lbf/sq.in.} \approx 3500\,\text{psi}&#xA;$$&#xA;Note that if the load is in lbm (or more accurately in &quot;slugs&quot;), then &#xA;$$&#xA;  P = 2500 \times 32 &#xA;$$&#xA;and we have&#xA;$$&#xA;  \sigma_\max \approx 106000 \,\text{psi}&#xA;$$&#xA;Now we have to compare this stress with the yield stress of stainless steels.  If you look at the &lt;a href=&quot;http://www.matweb.com/search/DataSheet.aspx?MatGUID=dc78ba5a80e347d182805eeb0a50e28c&amp;amp;ckck=1&quot; rel=&quot;nofollow&quot;&gt;MatWeb page on stainless steels&lt;/a&gt; you will find that the yield strength can vary between 24,700 psi and 160,000 psi. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have to pick your material such that even if there are microscopic cracks in the material (and cyclic loading condition), the beam will not fail by direct fracture or fatigue fracture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increase in stress due to the presence of a small flaw can be estimated using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Stress_intensity_factor&quot; rel=&quot;nofollow&quot;&gt;Stress intensity factor&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that you have a tiny edge crack in the beam.  We can approximate the stress near the crack as uniform and uniaxial.  This allows us to use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stress_intensity_factor#Edge_crack_in_a_plate_under_uniaxial_stress&quot; rel=&quot;nofollow&quot;&gt;stress intensity factor for an edge crack&lt;/a&gt; for a rough estimate of the stress intensity factor:&#xA;$$&#xA;   K_{\rm I} = \sigma_0\sqrt{\pi c}\left[1.12 - 0.23\left(\frac{c}{h}\right) + 10.6\left(\frac{c}{h}\right)^2&#xA;    - 21.7\left(\frac{c}{h}\right)^3 + 30.4\left(\frac{c}{h}\right)^4\right] \,.&#xA;$$&#xA;where $\sigma_0$ is a uniform applied stress, $c$ is the crack length and $h$ is the height of the beam cross-section.  If the crack length is $c$ = 0.01 in, $h$ = 2 in, and $\sigma_0$ = 3500 psi, we get&#xA;$$&#xA;  K_{\rm I} = 695 \,\,\rm{psi.in}^{1/2}&#xA;$$&#xA;The stress near the crack can be found using the relation&#xA;$$&#xA;  \sigma = \frac{K_{\rm I}}{\sqrt{2\pi \,r}}&#xA;$$&#xA;but we assume that there exists a material property called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fracture_toughness&quot; rel=&quot;nofollow&quot;&gt;fracture toughness&lt;/a&gt; ($K_{\rm{Ic}}$) that allows us to compare $K_{\rm{I}}$ with experimental data.  For some stainless steels, $K_{\rm{Ic}}$ = 100,000 psi-in$^{1/2}$ (which is much larger than our calculated value of $K_{\rm I}$).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fracture toughness values can be found in the ASM Metals Handbooks and a large list of publications can be found in the review articles by Hudson, CM and Seward, SK: &lt;em&gt;A Compendium of sources of fracture toughness and fatigue-crack growth data for metallic alloys&lt;/em&gt;, &lt;strong&gt;International Journal of Fracture&lt;/strong&gt;, vol. 14, R151-R184, 1978; vol. 20, R58-R117, 1982; vol. 39, R43-R63, 1989 vol. 48, R19-R43, 1991.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="3488" LastEditDate="2015-11-13T22:10:25.340" LastActivityDate="2015-11-13T22:10:25.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6121" PostTypeId="1" AcceptedAnswerId="6123" CreationDate="2015-11-12T20:31:05.983" Score="4" ViewCount="2251" Body="&lt;p&gt;If biofuel is burnt, for example wood, and the stove I am building is made of metal, what is the most appropriate metal material I should use to avoid excessive corrosion in the exhaust pipe of the stove?&lt;/p&gt;&#xA;" OwnerUserId="3538" LastEditorUserId="33" LastEditDate="2015-11-13T02:28:16.793" LastActivityDate="2015-12-10T22:01:34.043" Title="What metal material to use for stove exhaust pipe?" Tags="|materials|combustion|corrosion|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6122" PostTypeId="2" ParentId="6121" CreationDate="2015-11-12T21:30:56.967" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;what is the most appropriate metal material i should use to avoid excessive corrosion in the exhaust pipe of the stove?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're looking for corrosion protection, stainless steel is hard to beat.  However, its corrosion resistance comes at a price.  You will likely find that a stainless exhaust pipe will be &lt;strong&gt;MUCH&lt;/strong&gt; more expensive than a regular mild steel pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stoves have been made for hundreds of years using much cheaper plain mild steel, wrought iron, or cast iron for fittings and piping.  There's no reason that these should not work for really any kind of biofuel.  If you want to help mitigate corrosion (on the outside of the pipe), you can paint it with a heat-resistant paint.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-11-12T21:30:56.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6123" PostTypeId="2" ParentId="6121" CreationDate="2015-11-12T21:37:28.300" Score="9" Body="&lt;p&gt;&lt;strong&gt;It depends on flue gas temperature.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;High temperature $(&amp;gt;250\ \textrm{C})$: use galvanized steel.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Low temperature $(&amp;lt;250\ \textrm{C})$, use stainless steel or PVC as appropriate.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;See below for an explanation.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flues conveying higher temperature flue gas are typically made from galvanized, rolled, plain steel, as sulfuric acid condensation doesn't occur, so acid damage risk is low. The galvanization also protects plain steel from atmospheric humidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flues conveying lower temperature flue gas, such as in a high-efficiency furnace, are typically made from PVC or stainless steel depending on the expected operating temperatures. If the temperature is near or above the glass-transition temperature of the PVC used (approximately $80\ \textrm{C}$), then stainless steel should be used instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The breakpoint between higher and lower temperatures here is the point at which condensation of sulfuric acid occurs. The reasoning is that combustion fuel contains non-negligible quantities of sulfur, which combines with oxygen and water to form sulfuric acid. Note that water is a byproduct of combustion of organic materials, so ambient humidity is irrelevant as to whether the process can occur, it only affects the temperature at which it occurs. See &lt;a href=&quot;https://chemengineering.wikispaces.com/Acid+dewpoint&quot;&gt;this link&lt;/a&gt; for more details. The image below (from the link) can be used to estimate the sulfuric acid dew point from the gaseous sulfur trioxide concentration and the water concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://chemengineering.wikispaces.com/file/view/Sulfuric_Acid_Dew_Point.png/242362135/Sulfuric_Acid_Dew_Point.png&quot; alt=&quot;chart for estimation of sulfuric acid dewpoint&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based only on the provided chart, a safe minimum dew-point is likely $250\ \textrm{C}$. I chose this temperature as sulfur trioxide and water both appear to have a logarithmic relationship with dew point, and we don't know the concentration of either, so must assume both are high to be safe. If your flue gases are below this temperature, or the flue inner wall is below this temperature, then you should use at least something as corrosion resistant as stainless steel. Zinc (galvanization) reacts readily with sulfuric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Engineering Note:&lt;/strong&gt; Chlorine compounds, including hydrochloric acid, may also be necessary to examine depending on the specifics of the fuel composition. I have not considered these, but stainless steel and PVC should both be resistant to these compounds. Zinc reacts readily with hydrochloric acid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SAFETY NOTE:&lt;/strong&gt; It is important to be safe when dealing with flue gases, as carbon monoxide poisoning is a serious danger. The gas is colorless and odorless, and you will not notice symptoms until you are already severely poisoned. Most commercially available CO alarms are of questionable quality. Be sure to have a licensed, qualified installer help with any installation involving flue gas venting. See &lt;a href=&quot;http://www3.epa.gov/airquality/carbonmonoxide/&quot;&gt;this EPA link&lt;/a&gt; for more information on carbon monoxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a qualified HVAC engineer. Only a materials engineer. Please consult a qualified HVAC engineer, or at least licensed, qualified HVAC contractor, before attempting to install a home-built system. For your safety.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2015-11-12T21:37:28.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6128" PostTypeId="1" AcceptedAnswerId="6130" CreationDate="2015-11-13T15:52:31.983" Score="2" ViewCount="3973" Body="&lt;p&gt;Lavoisier's principle of mass conservation states that the mass of a system must remain constant over time. But, as far as I know, there's no similar principle for density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is it possible to develop an alloy as dense as gold but made of constituents less dense than gold?&lt;/p&gt;&#xA;" OwnerUserId="3734" LastActivityDate="2015-11-13T18:10:39.060" Title="Alloys with same density as gold" Tags="|materials|alloys|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6129" PostTypeId="1" CreationDate="2015-11-13T16:40:47.573" Score="4" ViewCount="948" Body="&lt;p&gt;Consider 1D heat conduction on a rod of length L.  Assume the rod has a known density $\rho$, specific heat $c_p$ , and thermal conductivity $k$, all of which are constant with respect to temperature. Suppose that the initial temperature of the rod is constant $T_0$.  I want apply a constant heat flux $Q$ to the rod at one end, with the goal of heating this end of the rod to a desired temperature $T_{desired}$.  Assume the other end of the rod has an insulated (zero heat flux) boundary condition.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without explicitly solving the heat equation $\rho c_pu_t=ku_{xx}$, is there a simpler formula I can use to estimate how much time it takes for the heated end of the rod to rise by a a factor of $\Delta T = T_{desired}-T_0$?  Is there a formula that only requires knowing $Q,\Delta T, L,\rho, c_p,$ and $k$ to estimate this time?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2015-11-13T16:46:26.190" LastActivityDate="2015-11-14T13:23:01.057" Title="Estimating the time required to heat a rod by a certain $\Delta T$" Tags="|heat-transfer|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6130" PostTypeId="2" ParentId="6128" CreationDate="2015-11-13T17:58:31.287" Score="5" Body="&lt;p&gt;Unfortunately, your best bet is abusing the existence of isotopes of tungsten to produce &quot;heavy&quot; tungsten. A potential, but unexplored, alternative is to alloy tungsten with an interstitial atom other than carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no way to make tungsten look or behave like gold beyond superficially, i.e. spraypainting it or electroplating it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No other element or mixture of elements can make it to the density of gold without requiring denser constituents.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Explanation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There are only a finite number of elements to play with, unfortunately, so options are very limited. I am going to assume when you say &quot;less dense than gold&quot; you mean without using any elements that have a density higher than gold at standard temperature and pressure, and in their as-typically-processed, pure states. To have an alloy with the same density of gold but be made of constituents that all have lower density than gold would require very special and unusual properties of those materials. Density of single crystals is controlled by four factors: crystal structure, atomic weight, interatomic spacing, and quantity of point defects.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Crystal Structure&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Changing the crystal structure of an elemental metal can change its density. The change in density is usually very small, on the order of $1\%$, so we would need to start with either tungsten (W, $19.25\ \textrm{g/cc}$) or uranium (U, $19.05\ \textrm{g/cc}$). Other appropriate elements are simply too far below the density of gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately tungsten occupies only the body-centered cubic (BCC) structure at all temperatures and pressures when solid (&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/2/22/Phase_diagram_of_tungsten_(1975).png&quot;&gt;phase diagram&lt;/a&gt;). Uranium has an orthorhombic structure at room temperature, and changes first to tetragonal around $950\ \textrm{K}$, then to BCC at about $1050\ \textrm{K}$ (&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/2/2c/Phase_diagram_of_uranium_(1975).png&quot;&gt;phase diagram&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that as the temperature increases, density decreases for two reasons: an increase in vacancy density, and an increase in interatomic spacing due to higher average atomic kinetic energy. It is unlikely uranium will have density $19.3\ \textrm{g/cc}$ at any temperature at atmospheric pressure (we'll get to elevated pressure momentarily). At any rate, you'd have a hard time fooling anyone with a radioactive slab of glowing hot metal. Or at the least you'd probably have folks in black suits knocking at your door.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Atomic Weight&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We can change atomic weight by adding or removing neutrons from existing elements to create isotopes. How you would go about mass-producing specific isotopes is beyond the scope of this answer. The stable isotopes of Tungsten are &lt;a href=&quot;https://en.wikipedia.org/wiki/Isotopes_of_tungsten&quot;&gt;listed here&lt;/a&gt; and have 182 to 186 nucleons, weighing between $181.95$ and $185.95\ \textrm{g/mol}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atomic weight of Tungsten is $183.84\ \textrm{g/mol}$, and the value is an average of the isotopic content of typical samples of Tungsten containing the various stable isotopes. To increase the density of tungsten to that of gold, an increase of $0.25\%$ is required, so increasing the atomic weight to $184.3\ \textrm{g/mol}$ should be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uranium has no stable isotopes, but it does have two that last for billions of years, specifically 235 and 238. Unfortunately those are practically the only isotopes in existence, and virtually all of it is 238. The next isotope up, 239, only has a half life of 30 minutes, so by the time you made it it would already be decaying. Even still, that isn't enough of an increase in atomic weight to push it up to the density of gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that, even if you went this route, neutron bombardment creates byproducts like hydrogen and helium which create density-reducing defects such as vacancies and voids, so you would have to reprocess your material after you make it to remove the defects. Alternately, you could process via atomic centrifuge set up to process tungsten, instead of uranium, and isolate the isotopes, then recombine them to the appropriate average atomic weight.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Point Defects&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;More vacancies would only reduce the density, which means we would need to start with something heavier than gold, so that won't work. More substitutional atoms would also generally require at least one denser component as density virtually always follows a linear rule of mixtures in alloys, so that won't work either. Because of the linear rule of mixtures issue, virtually all multi-metal alloys are out, and only pure elements have any real chance of approaching the density of gold without using elements that have higher density than gold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last point defect to consider is interstitial defects, where atoms small enough to fit between the metal atoms are used as alloying elements. Interstitials generally increase the density, as with carbon in steel. Unfortunately, tungsten does not dissolve carbon at standard temperature and pressure, as shown in the &lt;a href=&quot;http://patentimages.storage.googleapis.com/EP1606426B1/imgf0001.png&quot;&gt;phase diagram&lt;/a&gt;, so that idea is out for carbon. I am unable to find a phase diagram for nitrogen, so it is unclear whether it would be possible to produce a tungsten-interstitial alloy. If it is up to the requisite $0.25\%$ atomic percent, then this method may also work.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Interatomic Spacing&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Aside changing the magnitude of the electromagnetic force constant, the only way to change interatomic spacing is via the application of stress. In this case, we would want hydrostatic compressive stress. Again, we only need an increase of $0.25\%$ for tungsten, so a decrease in lattice constants of one-third that or $0.083\%$ would suffice. That is a strain of $0.00083$. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Tungsten&quot;&gt;modulus of tungsten&lt;/a&gt; is $411\ \textrm{GPa}$, so the required hydrostatic compressive stress is $341\ \textrm{MPa}$, or a pressure of close to $50,000\ \textrm{psi}$. Fooling anyone by this method would require some interesting apparatus.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-11-13T18:10:39.060" LastActivityDate="2015-11-13T18:10:39.060" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6131" PostTypeId="2" ParentId="5955" CreationDate="2015-11-13T19:12:11.027" Score="3" Body="&lt;p&gt;If you have to select a material for this application, I would use ultra high molecular weight polyethylene (UHMW). It has a very low coefficient of friction, very high wear resistance, and is a lot cheaper than Teflon.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2015-11-13T19:12:11.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6134" PostTypeId="1" AcceptedAnswerId="6139" CreationDate="2015-11-14T04:49:34.043" Score="6" ViewCount="6730" Body="&lt;p&gt;The question has been surfacing periodically in the media and books over the past 100 years or so. I realize we're talking about a lot of material and work hours to begin with; the subsequent maintenance would be a nightmare. There are no power stations, nor radio towers between the continents; there is no viable way (or is there?) to build in vents along the way; the bottom of the ocean is uneven, and there's a mountain ridge in the middle; and who knows what problems would be encountered if the engineers decided to actually drill under the ocean floor all the way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pumping the air out might turn out to be extremely expensive. Leaving the air in might create problems for the trains: that's a lot of air in a tight space to push through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How feasible is it? Or is it even technically possible to have a tunnel under the Atlantic? &lt;/p&gt;&#xA;" OwnerUserId="3740" LastEditorUserId="282" LastEditDate="2015-11-14T18:15:08.783" LastActivityDate="2018-06-23T13:02:56.373" Title="Is a Transatlantic tunnel technically possible today? How feasible would it be?" Tags="|civil-engineering|tunnels|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6135" PostTypeId="1" CreationDate="2015-11-14T05:26:05.363" Score="1" ViewCount="518" Body="&lt;p&gt;I am installing a retaining wall in front of an existing bridge abutment that is supported by a spread footing. Because of the geometry of the site, ground anchors are required to help support the wall.  These ground anchors will be be bored under the spread footing of the abutment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YXgW1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YXgW1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bored diameter of the ground anchors is approximately 6&quot;. These holes will be filled with grout after the steel anchors have been installed. The distance from the bottom of the spread footing to the anchor can be as little as a few feet vertically. This has raised a few concerns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest concern seems to be causing the bridge abutment to settle while the borehole is advanced. A secondary concern is the bored hole collapsing from the high soil pressure before it can be filled with grout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are these concerns reasonable and what methods can be used to mitigate the risks?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these boreholes were typical drilled shafts, I would say that the holes should be cased while being drilled or slurry should be used. Neither of these options seems to be a common practice for ground anchors. This can't be a unique situation though.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-11-15T00:13:05.947" Title="What are the risks and possible mitigations of installing ground anchors under spread footings?" Tags="|civil-engineering|geotechnical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6136" PostTypeId="1" CreationDate="2015-11-14T07:51:16.000" Score="1" ViewCount="119" Body="&lt;p&gt;Is thermal conductivity of nano sized particles and bulk particles the same? If not, what's the difference and is there any relation between thermal conductivity and and particle size?&lt;/p&gt;&#xA;" OwnerUserId="3741" LastEditorUserId="1293" LastEditDate="2015-11-14T14:28:43.037" LastActivityDate="2015-11-14T14:28:43.037" Title="Thermal conductivity relationship with particle size in nanofluids" Tags="|fluid-mechanics|heat-transfer|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6137" PostTypeId="2" ParentId="6135" CreationDate="2015-11-14T08:28:43.423" Score="1" Body="&lt;p&gt;Your concerns are reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention soil, not rock. I'm assuming the end of the anchors will not be embedded in rock, even as a point anchor. I get the impression the grouted anchors will be totally encapsulated in soil (no rock anywhere) and they will act as a dowel and not a tensioned blot or cable. Dowels offer passive support, whereas tensioned anchors are an active means of support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Placing grouted anchors as you mention will stiffen the ground beneath the bridge footing and cause a realignment of stresses underneath the footing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that soil will be drilled into and the proximity of the footing, drilling techniques with minimal vibration would be best so as to minimize any affect on the bridge footing - something like auger drilling would be good if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit - Pre-injecting Soil with Chemical Grout&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following on from our discussion in the comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that casing the holes is not an option due to the requirement for cement grout binding with the soil. Pre-injecting the soil with a chemical grout prior to drilling anchor holes would bind the soil reducing the possibility of the anchor holes collapsing after being drilled.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2015-11-15T00:13:05.947" LastActivityDate="2015-11-15T00:13:05.947" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6138" PostTypeId="2" ParentId="6136" CreationDate="2015-11-14T09:24:51.640" Score="3" Body="&lt;p&gt;You should add more details to your question but since you added the fluid mechanics tag I'll assume you mean nanofluids. Yes, of course there is a difference between thermal conductivity of solid particles (nano) and base fluid. The addition of nano-sized particles to the main fluid has been found to enhance the thermophysical properties of the base fluid (the main concept of nanofluids) [a].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was found that the thermal conductivity of the nanofluid depends on many factors [b]:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Thermal conductivity of the base fluid and nanoparticles $k_f$ and $k_p$&lt;/li&gt;&#xA;&lt;li&gt;Volume fraction $\phi$&lt;/li&gt;&#xA;&lt;li&gt;Surface area (shape of the nanoparticles suspended in the&#xA;base fluid)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The effective thermal conductivity $k_{eff}$ of the nanofluid can be calculated using Maxwell and Hamilton and Crosser equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Maxwell equation&lt;/strong&gt; (Valid for a fluid having a lower thermal conductivity than solid nanoparticles, and you can see that the effective thermal conductivity depends only on both $k$ values of fluid and solid nanoparticles and volume fraction):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k_{eff} = k_f (1 + \frac{3(k_p - k_f)\phi}{(k_p - k_f)-(k_p - k_f)\phi})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hamilton and Crosser equation&lt;/strong&gt;:&#xA;$$k_{eff} = k_f (\frac{k_p + k_f (n-1) + (n-1)(k_p - k_f)\phi}{k_p + k_f (n-1)- (k_p - k_f)\phi})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is valid for non-spherical particles, and $n$ is the shape factor that is obtained experimentally for different materials, you may refer to &lt;a href=&quot;http://www.researchgate.net/publication/224453999_Stochastic_thermal_transport_of_nanoparticle_suspensions&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; for more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[a] Vincenzo Bianco, Oronzio Manca, Sergio Nardini and K Vafai -Heat Transfer Enhancement With Nanofluids (2015).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[b] D. C. Hernandez Aita - Design and Optimization of Volumetric Solar Receivers based on Nanoparticles with Supercritical CO2 (2014).&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2015-11-14T09:24:51.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6139" PostTypeId="2" ParentId="6134" CreationDate="2015-11-14T09:37:38.883" Score="9" Body="&lt;p&gt;I'm assuming you are asking about a tunnel between North America and Europe.&lt;/p&gt;&#xA;&lt;p&gt;There are two types of tunnel that &amp;quot;might&amp;quot; be possible: excavating a tunnel through the rock beneath the sea bed or sinking prefabricated sections of tunnel to the sea floor, joining them up and sealing them against water inflow.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Problem #1 - Depth of the North Atlantic Ocean&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The depth of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Atlantic_Ocean&quot; rel=&quot;noreferrer&quot;&gt;Atlantic Ocean&lt;/a&gt; is between 3700 and 5500 metres. Such depths are going to make it extremely difficult to sink the tunnel sections,  assemble them, ensure they can survive the water pressure and be water tight.&lt;/p&gt;&#xA;&lt;p&gt;Additionally, such deep water will increase the ground stresses in the rock in the sea bed, which will be problematic for an excavated tunnel.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Problem #2 - Geological Reconnaissance&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;For an excavated tunnel, the sea bed will need to drilled along the proposed path of the tunnel at closely spaced intervals so core sample can be taken, so that geologists can determine what type are rock or soil (sediments) are there and ascertain material properties for the rock. They will also need to know what geological structural discontinuities are present, such as faults, where they are located, how they are orientated and how adverse they could be.&lt;/p&gt;&#xA;&lt;p&gt;Drilling in waters 5500 metres presents a whole lot of challenges, one of them being the drill string would be unsupported for the first 5500 metres of its length.&lt;/p&gt;&#xA;&lt;p&gt;Geological reconnaissance would also be required for a sunk tunnel to ensure the sea bed could support the weight of the tunnel sections. Having sections of tunnel subside into soft sediments is something that would need to be prevented.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Problem #3 - Ventilation&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Ventilation shafts rising though 5500 m of water would be problematic in terms of construction, support and possible impact by whales, ships or submarines. They would also need to be water tight to prevent water inundation of the tunnel.&lt;/p&gt;&#xA;&lt;p&gt;The only way to ventilate such a tunnel would be via ventilation tunnels parallel to the main tunnel. These could double-up as service tunnels for inspections, maintenance and rescue, should there be a fire in the main tunnel.&lt;/p&gt;&#xA;&lt;p&gt;For reasons of safety, ventilation tunnels would need to be duplicated, at least.&lt;/p&gt;&#xA;&lt;p&gt;Ventilation fans would need to be placed at the ends of the vent tunnels. They would need to be large and they would consume a lot of electricity.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Problem #4 - Traversing an Active Volcanic Zone&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;There is an entity called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mid-Atlantic_Ridge&quot; rel=&quot;noreferrer&quot;&gt;Mid Atlantic Ridge&lt;/a&gt;. This is an active volcanic zone that is pushing apart the continental plates that separate the Americas from Europe and Africa. This ridge is visible above the water on Iceland.&lt;/p&gt;&#xA;&lt;p&gt;Putting a tunnel on or through the Mid Atlantic Ridge may not be possible. If it were possible, the tunnel would require regular maintenance to remedy damage done by the Ridge.&lt;/p&gt;&#xA;&lt;p&gt;Putting ventilation tunnels through the Mid Atlantic Ridge would be give additional ventilation problems.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;&lt;strong&gt;Overall&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;The distance between Canada and Ireland is close enough to 3000 km.&lt;/p&gt;&#xA;&lt;p&gt;The cost of construction a trans Atlantic tunnel would be hideously expensive!&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-11-14T09:37:38.883" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6140" PostTypeId="2" ParentId="6129" CreationDate="2015-11-14T13:23:01.057" Score="1" Body="&lt;p&gt;Well no not to any desired temperature, but you can estimate within engineering accuracy what the temperature rise will be by dimensional analysis. It is not an exact solution but will be accurate within an order of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this problem it is quite trivial but for more exotic equations (perhaps with heat transfer to the surroundings, or 2D etc.) dimensional analysis is a powerful tool which can provide insight into the dynamics of an equations which may otherwise be unsolvable (e.g. due to non-linearity, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, lets state the system we are dealing with:&#xA;$$\partial_{t}T=a\partial_{x}^{2}T\quad T\left(x,0\right)=T_{0}$$&#xA;with thermal diffusion coefficient $a=k/\rho c_p$ and boundary conditions:&#xA;$$-k\partial_{x}T\left(0,t\right)=Q\quad\partial_{x}T\left(L,t\right)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the dimensional analysis we will determine characteristic time ($\Delta t$), length ($\Delta x$) and temperature ($\Delta T$) scales which will result in the above equations to only contain terms of $O(1)$ and smaller when they are non-dimensionalized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We define the relevant dimensionless variables:&#xA;$$\theta=\frac{T-T_{0}}{\Delta T}\quad\eta=\frac{x}{\Delta x}\quad\tau=\frac{t}{\Delta t}&#xA; $$&#xA;Substituting these dimensionless variables in the equations yields:&#xA;$$\partial_{\tau}\theta=\frac{a\Delta t}{\Delta x^{2}}\partial_{\eta}^{2}\theta\quad\theta\left(\eta,0\right)=0$$&#xA;$$-\partial_{\eta}\theta\left(0,\tau\right)=\frac{Q\Delta x}{k\Delta T}\quad\partial_{\eta}\theta\left(\frac{L}{\Delta x},\tau\right)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain equations where all terms are $O(1)$ or less we define:&#xA;$$\Delta x=L\quad\Delta t=\frac{L^{2}}{a}\quad\Delta T=\frac{QL}{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which result in the equations becoming:&#xA;$$\partial_{\tau}\theta=\partial_{\eta}^{2}\theta\quad\theta\left(\eta,0\right)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\partial_{\eta}\theta\left(0,\tau\right)=1\quad\partial_{\eta}\theta\left(1,\tau\right)=0$$&#xA;which are indeed $O(1)$ or smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having determined the characteristic scales of the system, it follows that it take approximately time $\Delta t = L^2/a$ to increase the temperature by $\Delta T = \frac{QL}{k}$. Hopefully, you find this useful.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2015-11-14T13:23:01.057" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6141" PostTypeId="1" CreationDate="2015-11-14T17:51:45.377" Score="0" ViewCount="395" Body="&lt;p&gt;This question is about engineering drawing and representing a 3D shape on a 2D paper, so there are &quot;four&quot; angles of projections which we can use to draw, but I'm not sure about that, and I don't understand the meaning of &quot;first angle&quot; and &quot;second angle&quot;. What does the &quot;angle&quot; mean in the first place? What are the quadrants which they are in? Please explain these concepts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The angle&lt;/li&gt;&#xA;&lt;li&gt;The quadrants&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please include a detailed answer for how many angles are there, and all the possible manners for drawing an orthogonal projection. and do not refer me to external books or links unless you specify a specific portion of them.&lt;/p&gt;&#xA;" OwnerUserId="3745" LastEditorUserId="106" LastEditDate="2015-11-14T19:42:25.937" LastActivityDate="2015-11-14T19:44:12.270" Title="Where is the &quot;angle&quot; in the first, second, etc. angle projections?" Tags="|drafting|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6143" PostTypeId="1" CreationDate="2015-11-14T17:57:39.270" Score="4" ViewCount="999" Body="&lt;p&gt;I am building a very simple turbine that runs on steam. I want to make it as simple as possible, thus avoiding impulse and reaction turbines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to design the turbine to fit in an exhaust pipe (about 15cm diameter) so that steam would rotate it, and generate electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many blades must the turbine have? I know wind turbines are most efficient with 3 blades, does this also apply to simple steam turbines?&lt;/p&gt;&#xA;" OwnerUserId="3538" LastActivityDate="2015-12-15T21:07:18.483" Title="How many blades for a steam turbine?" Tags="|turbines|steam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6144" PostTypeId="1" AcceptedAnswerId="6165" CreationDate="2015-11-14T17:42:32.860" Score="5" ViewCount="4542" Body="&lt;p&gt;I'm trying to understand and identify the equations to use in defining the relationship between wind velocity, turbine rotor diameter, and power output for a wind turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify my question, let's use the following assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Average wind speed of 8 m/s.  &lt;/li&gt;&#xA;&lt;li&gt;Rotor length of 40m for an overall diameter of 80m.&lt;/li&gt;&#xA;&lt;li&gt;Ideal conditions, so we can neglect direction of wind with respect to the positioning of the turbine.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3757" OwnerDisplayName="Dwiparna Datta" LastEditorUserId="163" LastEditDate="2015-11-14T19:48:32.520" LastActivityDate="2021-08-20T10:37:58.713" Title="Relationship between wind velocity, turbine rotor diameter, and power output" Tags="|energy|renewable-energy|power|wind-power|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6145" PostTypeId="2" ParentId="6141" CreationDate="2015-11-14T19:37:59.827" Score="2" Body="&lt;p&gt;Take some object, and place it so that your line of sight is perpendicular (technically, normal) to the front of the object. You see the front of the object only - essentially, a two-dimensional view. If you were to superimpose a grid on the object, it would look like a Cartesian plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/1/1a/Cartesian_coordinates_2D.svg/518px-Cartesian_coordinates_2D.svg.png&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/1/1a/Cartesian_coordinates_2D.svg/518px-Cartesian_coordinates_2D.svg.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the four quadrants, which you might have heard of in geometry. They begin with quadrant I, in the upper right-hand corner, and then move counterclockwise. Quadrant II is in the upper left-hand corner, quadrant III is in the lower left-hand corner, and quadrant IV is in the lower left-hand corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can say that each quadrant has an angle associated with it: the angle between the two perpendicular planes (for they have depth, not just length) that bound the quadrant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a &lt;strong&gt;first-angle projection&lt;/strong&gt;, the horizontal plane from the first quadrant rotates down, extending the first angle to 180 degrees. You then put the view from the horizontal place of the object below the view from the vertical plane of the object. In a &lt;strong&gt;third-angle projection&lt;/strong&gt;, the horizontal plane from the third quadrant rotates up, extending the third angle to 180 degrees. You then put the view from the horizontal place of the object above the view from the vertical plane of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to say, this is the first time I've heard the term &quot;second-angle projection&quot;. Europe (and Asia, I believe) tend to use first-angle projection, while third-angle projection is common in the United States. Second-angle projection is seldom used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.freewebs.com/fadillah218/ORTHOGRAPHIC%20PROJECTION.pdf&quot; rel=&quot;nofollow&quot;&gt;This pdf&lt;/a&gt; has a pretty good (and in-depth) explanation of the idea of projections, and the ones I gave you an overview of.&lt;/p&gt;&#xA;" OwnerUserId="106" LastEditorUserId="106" LastEditDate="2015-11-14T19:44:12.270" LastActivityDate="2015-11-14T19:44:12.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6147" PostTypeId="2" ParentId="6144" CreationDate="2015-11-14T19:43:11.267" Score="9" Body="&lt;p&gt;It's not a simple relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let's deal with the kinetic energy of the wind passing through the rotor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of air passing through it in one second, $m$, is equal to the the density of the air ($\rho$), times the surface area of the rotor (${\pi}r^2$), times the velocity of the air ($v$). i.e. ${\rho}{\pi}r^2v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of that air is just $\frac{1}{2}mv^2$ which is $\frac{1}{2}{\rho}{\pi}r^2v^3$ in each second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, an idealised wind turbine would capture as high a proportion of that energy as it possibly could: that proportion is given by the Betz limit, which is $\frac{16}{27}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the real world, we don't achieve perfection, so at best the real proportion is a bit lower than that. And also, in the real world, we have to make trade-offs based on cost. And that means that at higher speeds, which happen less often, although we could capture much more power, it's just not worth the extra expense of uprating all of the electronics and the connection to the grid, for those times, because it wouldn't represent much extra energy over the lifetime of the turbine, but would be a lot of extra cost. So the blades get feathered at higher speeds - at any speed above what we define as the &lt;em&gt;rated wind speed&lt;/em&gt; - and the power generated is capped at the &lt;em&gt;rated power&lt;/em&gt;. And there gets to be some rare wind speed where, in the interests of a long and healthy lifetime for the turbine, it's most economic to shut it down completely until wind speeds drop again: that's the &lt;em&gt;cut-out wind speed&lt;/em&gt;. And at very low wind speeds, there's virtually no energy in the wind - nothing worth harvesting, and not enough to get the blades moving. So below a certain speed - the &lt;em&gt;cut-in wind speed&lt;/em&gt; - no power is generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this gives us a curve of how generated power varies with windspeed that looks like this: (&lt;a href=&quot;http://www.elm.eru.rl.ac.uk/ins4.html&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CCuDl.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CCuDl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2015-11-14T20:07:42.853" LastActivityDate="2015-11-14T20:07:42.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6148" PostTypeId="1" AcceptedAnswerId="6149" CreationDate="2015-11-14T22:39:47.933" Score="3" ViewCount="297" Body="&lt;p&gt;I've recently designed what is essentially an aluminium box. I've done the entire design process of CAD software SolidWorks and then done some simulations of loading of the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through iterations I've figured out the minimum thickness of the aluminium (AL6061 T6) box material to be quite small - 0.25 mm. This seemed almost too small to me so I looked up kitchen 'tin' foil for comparison. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Aluminium_foil&quot; rel=&quot;nofollow&quot;&gt;Wikipedia page&lt;/a&gt; puts tin foil thickness at between 0.2 mm and 0.006 mm. That means that the upper end of the tin foil is easily comparable with the thickness of my box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that tin foil is probably not AL6061 and it's probably not tempered at all but this has thrown me into quite a bit of confusion as to whether or not I can trust my simulation results (I've not given any specific details of my simulation since I don't want to turn this into a homework style question - but I'm happy to share if it helps).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is; can I trust the results at AL6061 T6 is significantly stronger than aluminium foil (it would help if I can find what aluminium the foil is made of)?&lt;/p&gt;&#xA;" OwnerUserId="3748" LastEditorUserId="86" LastEditDate="2015-11-14T23:20:27.237" LastActivityDate="2015-11-15T05:36:51.837" Title="Intuition vs software results" Tags="|design|computer-aided-design|aluminum|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6149" PostTypeId="2" ParentId="6148" CreationDate="2015-11-14T23:07:48.557" Score="7" Body="&lt;p&gt;Most soda cans in the US are on the order of 0.1 mm thick according to the &lt;a href=&quot;http://www.airliners.net/aviation-forums/non_aviation/read.main/977127/&quot;&gt;sources I've seen&lt;/a&gt;. So 0.25 mm seems plausible to me, assuming that you made no errors in your calculations.&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2015-11-14T23:07:48.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6150" PostTypeId="1" AcceptedAnswerId="6151" CreationDate="2015-11-15T00:42:01.457" Score="2" ViewCount="86" Body="&lt;p&gt;I need a small air compressor for a robotics application. I currently have one of those 12V tire inflators (200 psi, 1 CFM or so) that is sized well for this purpose (sample image below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the future I would like to replace this with a lighter compressor with similar specifications (for weight savings). Most of the weight of these compressors is a large DC brushed motor (the whole big silver cylinder in the photo below), while the actual compression module is relatively small. These motors are roughly 350W peak (30A, 12V from spec sheet). Similar DC brushless motors (by wattage) are &amp;lt;150 grams. Certainly they might be at a higher RPM, but even with gears this would still be a lot smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason that these huge (3-4 kg) brushed motors couldn't be replaced by tiny brushless motors? If so, why can't I find such devices anywhere? It wouldn't be cost effective for these cheapo inflators (you need more complex controls for brushless), but it seems like there would be enough demand for lightweight compressors for someone to manufacture a few.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sj1eM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sj1eMt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3749" LastEditorDisplayName="user16" LastEditDate="2015-11-15T01:19:08.323" LastActivityDate="2015-11-15T01:19:08.323" Title="Lightweight Air Compressor" Tags="|mechanical-engineering|compressors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6151" PostTypeId="2" ParentId="6150" CreationDate="2015-11-15T01:18:14.047" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If so, why can't I find such devices anywhere?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a word - &lt;strong&gt;&lt;em&gt;Cost&lt;/em&gt;&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running some quick searches for 300W DC motors, the brushless motors were typically 3x - 5x more expensive than their brushed counterparts.  And as you pointed out, brushless motors require additional and more expensive controls than a comparable brushed motor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth pointing out that small compressors like the one you asked about are commodity grade products.  Profit margins on them are slim and purchase decisions are frequently driven by cost alone.  It is the rare consumer who would consider paying a significantly higher amount of money for a comparatively smaller benefit of decreased weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I think you're overestimating the demand for lighter weight compressors.  Most places that I'm aware of that are using compressors purchase them infrequently, and they tend to be fixed installations where the weight of the compressor doesn't matter that much.  And for the cases where weight and portability is a concern, cost becomes an over-riding factor.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That said, would you be able to replace the existing motor with a brushless motor?  Presuming you have the ability to create the additional controls and gearing to go along with using the brushless motor, I don't see why not.  And for what I suspect is going to be a relatively low utilization rate, you're likely going to see an equivalent lifespan of the product.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2015-11-15T01:18:14.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6152" PostTypeId="1" AcceptedAnswerId="6153" CreationDate="2015-11-15T01:56:27.710" Score="0" ViewCount="70" Body="&lt;p&gt;Do we have the technology to illuminate a small region on the dark side of the Earth with an artificial satellite in orbit around the Earth? By illuminate I mean bright enough to light objects for regular humans nearby to see better than they could have otherwise. Assume low Earth orbit and no moon to make it easier.&lt;/p&gt;&#xA;" OwnerUserId="3751" LastActivityDate="2015-11-15T02:30:16.380" Title="Illuminate the Earth with an artificial satellite?" Tags="|aerospace-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6153" PostTypeId="2" ParentId="6152" CreationDate="2015-11-15T02:30:16.380" Score="3" Body="&lt;p&gt;Sure. Even a cubesat could do this (if only for a few seconds). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put a relatively powerful laser (maybe 100W?) on a satellite, focus and aim it at a square meter or so of ground, and light it up. 100 km (LEO) is short enough that a good laser can focus down to a pretty small land area, meaning a smallish laser can focus all of its power on a small area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 100W laser is about as bright as 10 100W incandescent light bulbs, and only weights a few kilograms. Add some solar panels and batteries, and your satellite can light up your 1 square meter of dirt for a bit each orbit.&lt;/p&gt;&#xA;" OwnerUserId="3749" LastActivityDate="2015-11-15T02:30:16.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6154" PostTypeId="2" ParentId="6148" CreationDate="2015-11-15T04:04:04.963" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The Wikipedia page puts tin foil thickness at between 0.2 mm and 0.006 mm. That means that the upper end of the tin foil is easily comparable with the thickness of my box.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ooook. Let's look at this, from an engineering perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, scales. The smallest measurement from Wikipedia vs the largest is a 33x factor difference. This is perhaps hard to conceptualize in mm, but imagine a 1m stick. Now imagine 33 of them. This is the difference between those measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you need to be &lt;em&gt;very&lt;/em&gt; careful looking at those ranges and saying &quot;oh that means my measurement is about tin foil.&quot; Understanding orders of magnitude is a very important engineering principle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So my question is; can I trust the results at AL6061 T6 is significantly stronger than aluminium foil (it would help if I can find what aluminium the foil is made of)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is good to question unintuitive software results. But be careful to not combine different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may (as in this case) attempt to compare things which are not good to compare. Experience helps with this.&lt;/p&gt;&#xA;" OwnerUserId="4" LastActivityDate="2015-11-15T04:04:04.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6156" PostTypeId="1" CreationDate="2015-11-15T11:01:32.923" Score="1" ViewCount="129" Body="&lt;p&gt;I'm building a rotary inverted pendulum to demonstrate reinforcement learning principles, however I have little experience with mechatronics. I will be building something similar to &lt;a href=&quot;https://www.youtube.com/watch?v=Cd96_8iwP6s&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However instead of plastic I will be using 10 mm aluminum tubes. The arm (horizontal) will be about 150mm long and the pendulum (vertical) will be about 150mm to 200mm with a little weight at the end (a few hundered grams).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have started my construction but the problem I have now is choosing a motor. I started looking through sparkfun and some chinese stores for motors until I realised that I have no idea what type or how powerful of a motor I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to avoid the complex calculations, I was hoping someone could give me a ball park estimation of the torque I would need, and the best type of motor for this application (stepper or a classic DC motor) considering my setup?&lt;/p&gt;&#xA;" OwnerUserId="3755" LastEditorUserId="1293" LastEditDate="2015-11-15T11:04:34.860" LastActivityDate="2015-11-15T11:04:34.860" Title="Motor for inverted rotary pendulum" Tags="|mechanical-engineering|motors|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6157" PostTypeId="1" AcceptedAnswerId="6164" CreationDate="2015-11-15T12:16:15.840" Score="1" ViewCount="944" Body="&lt;p&gt;Is there any emitter/receiver (passive or active) systems that allow distance estimation in the range of 1~2 meters with an accuracy of few centimetres and don't need to have a direct line of sight as optical systems mostly do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about using such sensors as nodes to put on a suit that a human can wear. As a result, a graph can be obtained where each node represents one of those sensors and each edge describes the distance from this node to the other nodes. Using graph optimization techniques, the graph would converge to the correct joint position of the human. Such a system could replace expensive motion capture methods using calibrated external cameras.&lt;/p&gt;&#xA;" OwnerUserId="1849" LastEditorUserId="282" LastEditDate="2015-11-17T01:43:09.690" LastActivityDate="2015-11-17T23:45:48.473" Title="Short range distance estimation systems" Tags="|electrical-engineering|sensors|distance-measurement|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6158" PostTypeId="1" AcceptedAnswerId="6161" CreationDate="2015-11-15T12:25:42.243" Score="1" ViewCount="771" Body="&lt;p&gt;I know definition but how would you explain to complete layik ( as me ) what is MEMS and how it is used.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples: Silicon nanopore membranes (SNM) have been prototyped from silicon substrates by an innovative process based on MEMS technology.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://whatis.techtarget.com/definition/micro-electromechanical-systems-MEMS&quot; rel=&quot;nofollow&quot;&gt;micro-electromechanical systems (MEMS)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3756" LastEditorUserId="1586" LastEditDate="2015-11-17T00:52:23.877" LastActivityDate="2015-12-12T06:15:37.457" Title="What is micro-electromechanical systems (MEMS)?" Tags="|mechanical-engineering|applied-mechanics|power-electronics|systems-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6160" PostTypeId="2" ParentId="6158" CreationDate="2015-11-15T14:34:05.097" Score="3" Body="&lt;p&gt;MEMS is a semiconductor based process that can create miniature structures in silicon. When these structures (sensors) are subjected to mechanical stresses (pressure, acceleration, yaw etc) they are able generate electrical signals. These signals are conditioned using an &lt;strong&gt;A&lt;/strong&gt;pplication &lt;strong&gt;S&lt;/strong&gt;pecific &lt;strong&gt;I&lt;/strong&gt;ntegrated &lt;strong&gt;C&lt;/strong&gt;ircuit (ASIC) to create a sensor system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some of images representing manufacture of a &lt;strong&gt;MEMS differential pressure sensor&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X4C29.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X4C29.jpg&quot; alt=&quot;MEMS pressure sensor silicon stack up&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above silicon structure electrically is represented as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2f3X8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2f3X8.jpg&quot; alt=&quot;MEMS pressure sensor electrical representation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qZeiS.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qZeiS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The piezoresistive sensing elements are structured to form a wheatstone bridge. When pressure is applied on a thin diaphragm the mechanical stresses are converted to an electrical signal using a wheatstone bridge resistor structure. This signal then processed using an ASIC to create a stable and predictable electrical signal. The electrical output is used by other device such as a micro-controllers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zOUei.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zOUei.gif&quot; alt=&quot;MEMS + ASIC + micro controller&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.maximintegrated.com/en/design/reference-design-center/ref-circuits/5319.html/tb_tab1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Liquid-level measurement system uses a compensated silicon pressure sensor and precision Delta-Sigma ADCS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://compliantmechanisms.byu.edu/content/introduction-microelectromechanical-systems-mems&quot; rel=&quot;nofollow noreferrer&quot;&gt;Introduction to Microelectromechanical Systems (MEMS)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/5367/110&quot;&gt;How do I conduct an experimental modal analysis with a three-axis accelerometer?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/1749/110&quot;&gt;Ah, strong am I with the Force, but how do I measure it?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-12-12T06:15:37.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6161" PostTypeId="2" ParentId="6158" CreationDate="2015-11-15T14:39:43.903" Score="5" Body="&lt;p&gt;One way to think of MEMS is regular mechanical systems, but at a very small scale.  Often these tiny systems are fabricated using technology developed for making silicon electronic chips, like nanometer-scale photolithography and etching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, describing MEMS as just downscaled regular mechanical systems is doing the concept injustice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various physical properties scale differently with linear dimension.  For example, simply scaling a ant to the size of a elephant wouldn't work.  The mass the legs have to support is proportional to the volume (cube of the linear scale), but the strength of the legs only proportional to their crossectional area (square of linear scale).  Such a system can't be scaled up linearly and still function as the original.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of different scaling of different physical properties to the tiny dimensions of MEMS, some things we do routinely at the human scale are impossible, but new things are possible that would be impractical at human scale.  For example, rotary electric motors with conventional windings would be extremely difficult in MEMS.  However, at such short distances, the electrostatic force can be harnesses much more easily, and MEMS motors using it are quite possible and efficient.  At nanometer distances, just a few volts can cause quite significant forces relative to the much much lower mass (remember, mass scales with the cube of linear dimension) of the parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is just one example, but illustrates how you enter a different world at micrometer or nanometer dimensions.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-15T14:39:43.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6162" PostTypeId="1" AcceptedAnswerId="6168" CreationDate="2015-11-15T15:11:56.637" Score="3" ViewCount="285" Body="&lt;p&gt;I am studying embedded systems and microcontrollers and now I have to face the interrupts. I understood the main concepts around them and how to implement them inside a C code, but I want to know how they work at hardware level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this scheme on the internet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kvTLq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kvTLq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I cannot understand it at full. Can you explain me a bit? Don't worry to use technical terms, I think that I will understand them. However I prefer to search some terms that I don't know rather than limiting your knowledge while answering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice engineering at everyone!&lt;/p&gt;&#xA;" OwnerUserId="3762" LastEditorUserId="1586" LastEditDate="2015-11-17T01:43:17.093" LastActivityDate="2015-11-17T01:43:17.093" Title="Hardware behind interrupts" Tags="|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6163" PostTypeId="2" ParentId="6162" CreationDate="2015-11-15T16:24:34.720" Score="2" Body="&lt;p&gt;Each time the processor is about to go to the next instruction it will check the interrupt mask to see if an interrupt has triggered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so it will push the current program counter to the stack and jump to the position in the matching position in the vector. Changing to kernel mode and disabling interrupts in the process.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2015-11-15T17:07:20.007" LastActivityDate="2015-11-15T17:07:20.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6164" PostTypeId="2" ParentId="6157" CreationDate="2015-11-15T18:52:27.863" Score="1" Body="&lt;p&gt;I think you have approached your problem from the wrong direction. There are currently no sensors that measure distance between each other regardless of material in the path. It would be possible to make a device with optical or ultrasonic &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_of_flight&quot; rel=&quot;nofollow&quot;&gt;time of flight&lt;/a&gt; technology, but that would be a project all by itself. These sensors are very expensive and need calibrated for a fixed environment. It would require lots of development to remove all the variables from a biomechanics type application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you look at currently available sensing technology and look how you can apply those sensors in new ways. Optical is likely the future in this field. Computer software can be written to splice multiple images and reverse render the position and texture of objects. If this still does not solve the problem, you could look at using wavelengths outside the visible spectrum such as near infrared.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-15T18:52:27.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6165" PostTypeId="2" ParentId="6144" CreationDate="2015-11-15T19:32:26.943" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Betz%27s_law&quot; rel=&quot;nofollow&quot;&gt;Betz limit&lt;/a&gt; will give you a good theoretical maximum from your wind velocity and swept area. Your real world turbine will fall somewhere below this number based on its efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For measuring power production over a period of time, you need wind data for that particular area. Wind data can be iterated and summed for a particular turbine, but generally the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rayleigh_distribution&quot; rel=&quot;nofollow&quot;&gt;Rayleigh distribution&lt;/a&gt; is used for a close approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like EnergyNumbers mentioned, it is not economically feasible to collect power from very high speed wind that only occurs once in a while, so turbines have braking and feathering systems to protect themselves in a power off state during these situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://img.bhs4.com/01/7/017b203448efc4b4f63e85957c56b03c439b1219_large.jpg&quot; rel=&quot;nofollow&quot;&gt;Skystream3.7 power production&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.wind-power-program.com/small_turbines.htm&quot; rel=&quot;nofollow&quot;&gt;Power production evaluation of small wind turbines&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-15T19:32:26.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6166" PostTypeId="2" ParentId="6143" CreationDate="2015-11-15T20:29:11.760" Score="2" Body="&lt;p&gt;The reason wind turbines use three blades is because it reduces the load fluctuations (fatigue) on the turbine shaft an bearings. Wind at high altitude is faster than the wind at low altitude. A two bladed rotor would be the most economical if it were not for the largest horizontal load being on the upper blade while simultaneously the lowest horizontal load being on the bottom blade. Three blades more evenly distributes this load, allowing the shaft and bearings to be smaller which makes it the most economical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Turbine&quot; and &quot;simple&quot; are kindof oxymorons ;-)  Impulse and reaction turbines are the simplest &lt;em&gt;efficient&lt;/em&gt; turbine designs. It may initially seem easier to install an axial turbine, but there is significant amount of complex engineering involved. It would be pretty easy to extract 10% of the available power by guessing at the turbine design, but to achieve higher and higher efficiency a larger and larger amount of engineering is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic variable would be the number of blades like you mentioned. More blades generally means higher pressure differential and higher torque. There are many many more variables involved however, such as desired output rpm, flow velocity, blade shape, angle of attack, change of that angle of attack as it approaches the tip, inlet and outlet diameter, additional layers of blades as the steam expands, etc.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-15T20:29:11.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6167" PostTypeId="2" ParentId="6061" CreationDate="2015-11-15T23:35:15.467" Score="2" Body="&lt;p&gt;Filters can be evaluated by a few different parameters. &quot;efficiency&quot; is a broad term that doesnt lend to a better understanding in this application. Here are some better metrics to consider:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Maximum flowrate (same as pressure drop or efficency when considering a power loss from the pressure drop. Proportionally increases with filter surface area.)&lt;/li&gt;&#xA;&lt;li&gt;Maximum pressure (before rupture)&lt;/li&gt;&#xA;&lt;li&gt;Material (chemical resistance)&lt;/li&gt;&#xA;&lt;li&gt;Storage (how much contamination the filter can hold; how quickly pressure drop increases)&lt;/li&gt;&#xA;&lt;li&gt;Maximum diameter of particles that can flow through. (ex. 1 micron, 10 micron, &lt;a href=&quot;http://www.showmegold.org/news/Mesh.htm&quot; rel=&quot;nofollow&quot;&gt;400 mesh&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Other technology/principles utilized (vortex, oil wetting, diatomaceous earth filter aid, self cleaning, gravity settling, scrubbing, etc)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In a &lt;a href=&quot;http://www.filtersource.com/c49/c9/string-wound-cartridge-filters-c34.html&quot; rel=&quot;nofollow&quot;&gt;cotton(or other fiber) wound filter cartridge&lt;/a&gt; there is additional thickness beyond what is necessary to archive the 1 micron maximum particle rating. This additional media increases the storage of the filter; giving it open surface area even as the filter is contaminated. If there was a single 1 micron membrane on the outer surface it would quickly plug with even the slightest contamination. So while the thicker filter may initially have a larger pressure drop for a given flow and surface area, it will last much longer than a thin filter that would plug quickly and have an even larger pressure drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nano fibers are attractive for filtering because the bulk ratio of empty space to fibers is much greater for a desired particle size rating. The larger amount of empty space both reduces flow resistance(pressure drop) and increases the storage capacity of the filter. It may be easier to visualize at the extremes; you could make a 400 micron filter with 1 inch diameter machined rods placed side by side, but the max flow rate and storage would be very low compared to the same surface area of cotton fabric composed of small fibers.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-15T23:35:15.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6168" PostTypeId="2" ParentId="6162" CreationDate="2015-11-16T00:28:59.553" Score="3" Body="&lt;p&gt;A interrupt is when the processor goes and executes some pre-determined code based on some event.  This is done so that, with the cooperation of the specially executed code, the processor can return to executing the original code in such a way that its operation is not affected other than some additional time taken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all it is, but there's a lot behind this relatively short definition.  For example, the event is usually some hardware completing a task (so that the software now needs to &quot;service&quot; it) or a external event, but can also be caused internally in the CPU.  Examples of the latter in some processors are when a unrecognized opcode is encountered for execution, a divide by 0, attempt to reference non-existant memory, or even a condition deliberately set by the software.  There is great variation between processors in the details of how interrupts are handled.  Saying something as simplistic as the program counter is pushed onto the stack and the interrupt code is executed is missing many important issues and is downright wrong for some processors.  Not all processors even have a stack, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the interrupt envolves interrupting the existing code to execute something else, and the processor must be able to resume the existing code, some state has to be saved as part of the interrupt mechanism.  At the very least, this requires saving the PC (program counter).  The mechanism for storing the PC and then later restoring it to resume execution of the original code is usually done similarly to how subroutines are handled on that processor.  For example, the PC could be stored in a special place, special register, pushed on a stack, etc.  All these and more have been done historically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it must be possible for interrupts to be transparent to the interrupted code (other than a delay between two instructions), this usually means some other state must be saved and restored on entry and exit of the interrupt routine.  What state needs to be restored, and whether hardware mechanisms are provided to do this or facilitate this is highly dependent on the processor architecture.  For example, old style PICs save the PC on the hardware stack for you, and the rest is up to the interrupt code.  Anything it uses must be saved first so that it can be restored later.  For example, on these PICs you can't do much without using the W register, so just about any useful interrupt routine would need to save the W register early on and then restore is shortly before exiting the interrupt code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More high end processors might provide a whole set of &quot;shadow&quot; registers just for use during interrupts.  The hardware switches to using these alternate registers automatically upon interrupt and switches back on return from interrupt.  In this case, the interrupt code can just use registers normally without worrying about trashing them from the interrupted code's point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many many processor architectures out there, and the details of how interrupts are handled can vary significantly between architectures.  Some processors have a single location execution goes on interrupt.  Others have a different location for each posssible interrupt condition.  On processors with virtual memory or protection mechanism to handle hostile processes, it gets more complicated.  Multiple threads of execution makes things complicated (which process' context does the interrupt code run with, for example?).  Then there are interrupt priorities, or not.  You generally don't want the same interrupt condition to be able to interrupt the interrupt code that is trying to handle it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many many details and many many schemes have been devised to implement interrupts, but they all conform to the definition in the first paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to learn how interrupts work is to understand them on one specific machine.  For starters, pick a simple one, like a original PIC 16.  Once you understand that, it will be easier to understand the extra wrinkles added by higher end processors.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2015-11-16T00:37:50.510" LastActivityDate="2015-11-16T00:37:50.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6169" PostTypeId="1" CreationDate="2015-11-16T01:14:42.690" Score="1" ViewCount="66" Body="&lt;p&gt;I am making a reprap rostock (delta style printer) and I am trying to use as many parts I already have as possible. The stepper I have are 7.2 degrees per step. Even with microstepping, would this resolution get me high quality prints?&lt;/p&gt;&#xA;" OwnerUserId="2164" LastActivityDate="2015-11-16T01:14:42.690" Title="What is the maximum degrees per step of a stepper motor for a 3D printer?" Tags="|stepper-motor|3d-printing|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6170" PostTypeId="1" AcceptedAnswerId="6171" CreationDate="2015-11-16T02:37:53.643" Score="6" ViewCount="406" Body="&lt;p&gt;I want to convert rotary motion (the orange propeller) onto a square path (the dark gray path). Is this a valid way to go about it? Would it be possible to use a square peg instead of a round one (the red circle) in order to keep the peg level as it travels? If not how could I keep it level? I'm a noob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wwoX2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wwoX2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3Apv8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Apv8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LNMMa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LNMMa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GTvkA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GTvkA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; 2015/11/16 - Posted another option as an answer below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zfV8s.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zfV8s.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3765" LastEditorUserId="3765" LastEditDate="2015-11-16T17:06:27.493" LastActivityDate="2015-11-16T17:09:14.177" Title="Is this Square Track Cam valid?" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6171" PostTypeId="2" ParentId="6170" CreationDate="2015-11-16T02:52:10.687" Score="4" Body="&lt;p&gt;There is an existing method using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Reuleaux_triangle#Mechanism_design&quot; rel=&quot;nofollow noreferrer&quot;&gt;reuleaux triangle&lt;/a&gt;. It was used for moving film in a stepwise motion in projectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Luch2_greifer.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/evYKq.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2015-11-16T02:52:10.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6172" PostTypeId="2" ParentId="6170" CreationDate="2015-11-16T09:38:19.957" Score="4" Body="&lt;p&gt;If you're going with square peg and don't strictly restrict its rotation (e.g. with mechanism like in Ratchet freak's answer), you WILL run into problems with technical tolerances - it will be catching on the corners and with any other shape than round you will have this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qgBJh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qgBJh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your original image mechanism can work okay, but modifying the peg to be square will break it.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-11-16T09:38:19.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6173" PostTypeId="1" CreationDate="2015-11-16T10:58:12.323" Score="5" ViewCount="6993" Body="&lt;p&gt;Here's a &lt;a href=&quot;https://www.youtube.com/watch?v=jDErZ0TR30w&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube video&lt;/a&gt; with a concept of a bridge which is currently under construction in Saint Petersburg. The bridge largest span is cable stayed and designed to drive traffic over the Neva river fairway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest span rests on two pylons placed symmetrically such that they are inclined towards the fairway axis. Something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sr484.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sr484.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This differs much from &quot;usual&quot; design where pylons are built upright. They may have &quot;A&quot; shape but still don't have inclination along the road axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual strategy of building cable stayed bridges is that you first build the pylons and temporary supports, then assemble the deck on those supports, then mount the cables, then remove the temporary supports. Maybe the pylons become evenly loaded once construction is complete but clearly building those inclined pylons of dozens tons of reinforced concrete poses a challenge - both the pylons and their foundations need to support the extra loads that appear simply because the pylons are inclined. Building an upright pylon looks much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like this design just asks for extra problems and doesn't provide any benefit compared to pylons built upright.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why design a bridge with pylons inclined towards the river fairway instead of upright pylons?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="255" LastEditDate="2015-11-16T16:44:33.980" LastActivityDate="2016-03-15T23:38:21.733" Title="Why design a cable stayed bridge with pylons inclined towards the obstacle being spanned?" Tags="|civil-engineering|structural-engineering|bridges|reinforced-concrete|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6174" PostTypeId="1" AcceptedAnswerId="6224" CreationDate="2015-11-16T12:10:04.283" Score="6" ViewCount="13919" Body="&lt;p&gt;I need to design the roof structure for a sports facility. I have no information with regard to HVAC design at this stage. Also, the HVAC may change in future. Thus I don't know what load to apply to the structure. I intend to give the HVAC and electrical engineers a weight envelope in which they can operate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standardised list available which would give an indication of what loads I can assume for this?&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="1832" LastEditDate="2015-11-18T12:26:17.880" LastActivityDate="2018-10-02T14:06:54.003" Title="Dead Load / Self Weight for ceilings &amp; HVAC" Tags="|structural-engineering|hvac|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6175" PostTypeId="1" CreationDate="2015-11-16T13:39:46.370" Score="13" ViewCount="4742" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Concentrated_solar_power&quot;&gt;Concentrated solar power&lt;/a&gt; looks on the surface to be an excellent source of grid power. It's sustainable, has no obvious waste products to deal with either during construction or operation, and with thermal storage it can generate power at night or on a cloudy day. It seems to address every problem with fossil fuel, nuclear, PV, or wind generation. Is there some disadvantage to this that I'm not seeing? Shouldn't we be building these plants as fast as possible?&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="368" LastEditDate="2015-11-19T17:31:45.377" LastActivityDate="2017-04-21T00:45:33.623" Title="What are the disadvantages of concentrated solar thermal power?" Tags="|renewable-energy|solar-energy|electrical-grid|" AnswerCount="7" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6176" PostTypeId="2" ParentId="6175" CreationDate="2015-11-16T14:25:10.990" Score="15" Body="&lt;p&gt;There are no silver bullets when it comes to decarbonising the grid. That's good news: it means you can stop looking for one, and accept that every form of generation has its pros and cons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/a/78848/4066&quot;&gt;Concentrated solar power [CSP] relies on direct sunlight&lt;/a&gt;. Not just ambient daylight, but lots and lots of &lt;em&gt;direct&lt;/em&gt; sunlight all year round. Whereas ordinary photovoltaics [PV] will work pretty much anywhere in the world, from the tropics to the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CSP doesn't work at all scales. Prototypes are still determining what's the most economical. PV works at pretty much any scale, from watts to gigawatts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CSP is still in its infancy. There are maybe a couple of thousand grid-scale prototypes in the world, ~ 4-5 GW total capacity, and the ones with in-built  storage for 24/7 operation are even rarer, because the economics stink. We've little idea on lifetime performance, optimal design, optimal maintenance regimes, and so on. Whereas PV is a commodity item manufactured in vast quantities - we're nearing a billion PV panels in the world now, with about 200GW capacity - fifty times as much capacity, and five or six orders of magnitude more repeatable units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CSP is complex and time-consuming to design, install and commission. PV is very fast and very easy to design, install and commission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Storage is, in most of the world, not a high-value commodity. So having in-built storage isn't - in and of itself - an inherently valuable thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know what its lifetime cost per unit of electricity supplied will be. We've only a few young prototypes to go on. That kind of uncertainty is a disincentive to investors.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2015-11-16T14:56:14.920" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6178" PostTypeId="2" ParentId="6173" CreationDate="2015-11-16T16:29:17.900" Score="6" Body="&lt;p&gt;&lt;strong&gt;Why build pylons which are inclined &lt;em&gt;away from&lt;/em&gt; the obstacle being spanned?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as being aesthetically interesting, this can also be structurally efficient. Personally I love the many bridges by &lt;a href=&quot;https://en.wikipedia.org/wiki/Santiago_Calatrava&quot;&gt;Calatrava&lt;/a&gt; using this concept, in particular the &lt;a href=&quot;https://en.wikipedia.org/wiki/Puente_del_Alamillo&quot;&gt;Puente del Alamillo&lt;/a&gt;. The pylon is actually fully in compression under dead load: the cable stress and the self-weight of the pylon resolve into a force directly down the axis of the pylon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why build pylons which are inclined &lt;em&gt;towards&lt;/em&gt; the obstacle being spanned?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the main answer is aesthetics. &quot;Art&quot; taking pride of place, increasing costs. Because here we have the self-weight of the pylon and the cable stress acting together, both bending the pylon in the same direction. To counteract this you will typically use a lot of prestress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so I &lt;em&gt;can&lt;/em&gt; see that it could make engineering sense if, due to your site constraints, you had a short main span and long backspans; because then the backspan cable force (acting against pylon self-weight) could be greater than the main span cable force. But that would be very unusual and isn't the case in the picture you've provided.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;A note on construction of inclined pylons&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your &quot;usual strategy&quot; is, in fact, not common for larger cable-stayed bridges. Far more common is building a section of the pylon and a section of the deck, joining them with a cable, and then repeating. Following this method, the out of balance load of the self-weight on an inclined pylon is much reduced.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-11-16T16:29:17.900" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6179" PostTypeId="5" CreationDate="2015-11-16T16:43:41.287" Score="0" Body="&lt;p&gt;Highway engineering covers the design of highways, in particular:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;layout (vertical and horizontal)&lt;/li&gt;&#xA;&lt;li&gt;safety&lt;/li&gt;&#xA;&lt;li&gt;surfacing design&lt;/li&gt;&#xA;&lt;li&gt;drainage&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="255" LastEditorUserId="33" LastEditDate="2016-02-23T03:07:48.140" LastActivityDate="2016-02-23T03:07:48.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6180" PostTypeId="4" CreationDate="2015-11-16T16:43:41.287" Score="0" Body="Highway engineering covers the design of highways, in particular: layout (vertical and horizontal), surfacing design and drainage" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2015-11-16T17:33:00.570" LastActivityDate="2015-11-16T17:33:00.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6181" PostTypeId="2" ParentId="6170" CreationDate="2015-11-16T16:48:43.337" Score="1" Body="&lt;p&gt;The reuleaux triangle is a great solution but I would prefer something that has fewer parts and is more compact...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another option. The idea is to sandwich a bent peg between offset tracks. So the sandwich goes like this: Bottom Track, peg, Propeller, Top Track (Transparent). And Yes, I would round out the track corners more than depicted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SruXM.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SruXM.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3765" LastEditorUserId="3765" LastEditDate="2015-11-16T17:09:14.177" LastActivityDate="2015-11-16T17:09:14.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6182" PostTypeId="2" ParentId="6175" CreationDate="2015-11-16T18:08:59.980" Score="9" Body="&lt;p&gt;&lt;strong&gt;In the UK&lt;/strong&gt;, the sun does not shine every day.   We also need the most power in winter when the sun shines the least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore every solar power station must be backed up by anther power station that can work when the sun is not shining.    Concentrated solar power allows short term storage of the steam, so at least copes with 5 minutes of cloud cover, it does not help much with the “winter problem”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However in an area when the peak power need is on hot sunny days due to AC solar power become a much better option.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then need to ask if Concentrated Solar Power (CSP) is worth the effort compared to photovoltaics (PV).  PV is a commodity and is getting cheaper every year, so all CSP can offer over PV is the short term storage of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there was great investment in CSP in the correct areas of the world, with many designer being build and perfected, it may then prove to be very good.   However I expect that we will never know, as the case is hard to make for the investment, given the PV is well understood and AC can be turned off for a few minutes if a cloud stops the PV working.    (Storing “coldness” in a building is a method of effective energy storage.) &lt;/p&gt;&#xA;" OwnerUserId="1226" LastActivityDate="2015-11-16T18:08:59.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6183" PostTypeId="2" ParentId="6175" CreationDate="2015-11-16T20:17:46.687" Score="10" Body="&lt;p&gt;&lt;em&gt;Everything&lt;/em&gt; we do as humans has consequences of some sort. Although not a &lt;em&gt;huge&lt;/em&gt; downside (in my opinion), critics have often cited the plant's supposed adverse effects on wildlife:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Concentrated_solar_power#Effect_on_wildlife&quot;&gt;Wikipedia Article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It has been noted that insects can be attracted to the bright light caused by concentrated solar technology, and as a result birds that hunt them can be killed (burned) if the birds fly near the point where light is being focused onto. This can also affect raptors who hunt the birds. Federal wildlife officials have begun calling these 'eco-friendly' power towers &quot;mega traps&quot; for wildlife.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.natureworldnews.com/articles/12918/20150223/solar-farm-set-hundreds-birds-ablaze.htm&quot;&gt;Nature World News&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Unfortunately, about two hours into the test, engineers and biologists on site started noticing &quot;streamers&quot; - trails of smoke and steam caused by birds flying directly into the field of solar radiation. What moisture was on them instantly vaporized, and some instantly burst into flames - at least, until they began to frantically flap away. An estimated 130 birds were injured or killed during the test.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The veracity of these claims is in question however (from the Wikipedia article):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, the story about the Ivanpah Solar Power Facility was exaggerated, numbering the deaths in many tens of thousands, spreading alarm about concentrated solar power (CSP) plants, which was not grounded in facts, but on one opponent's speculation. According to rigorous reporting, in over six months, actually only 133 singed birds were counted. And not only are the number of bird deaths far lower than the hundreds of millions to billions that die annually from collisions with windows, vehicles, and power lines., but, by focusing no more than 4 mirrors on any one place in the air during standby, at Crescent Dunes Solar Energy Project, in 3 months, the death rate dropped to zero fatalities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3776" LastActivityDate="2015-11-16T20:17:46.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6184" PostTypeId="2" ParentId="6114" CreationDate="2015-11-16T23:12:18.103" Score="3" Body="&lt;ul&gt;&#xA;&lt;li&gt;240 ABMM I3C is 2013/September date code&lt;/li&gt;&#xA;&lt;li&gt;240 ABMM F4C is 2014/June date code&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/alDsp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/alDsp.jpg&quot; alt=&quot;Abracon ABMM date code&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details can be found at &lt;a href=&quot;http://www.beckwithelectronics.com/abracon/mark.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Beckwith Electronics/Abracon Corporation&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also try re-flowing the solder around the crystal. Also based on the image it appears that the crystal is some distance away from the micro-controller. It is best to place the crystal and the capacitors as close to the micro-controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also look like the load and shunt capacitance are don't match the crystal specification too.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2015-11-18T17:52:58.830" LastActivityDate="2015-11-18T17:52:58.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6185" PostTypeId="2" ParentId="6175" CreationDate="2015-11-17T01:33:39.063" Score="3" Body="&lt;p&gt;Concentrated solar power plants are best located in isolated areas that receive a lot of sunshine all year round, which basically means arid or semi-arid regions. Most of these regions don't have large populations so long power transmission lines will be needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To increase performance and efficiency, each CSP plant will need to be uniquely designed and constructed for the path and elevation of the sun at each location where a CSP plant will be established; particularly for the sun tracking system that will be needed to continuously focus the light from a moving sun onto one location. The path of the sun varies throughout the day and from day to day - low elevations in winter, high elevations in summer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power output will be optimal when the sun is at it's highest point in the sky and functioning well 2 to 3 hours either side of that. During early mornings and late afternoons the CSP will not receive much usable sunlight.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-11-17T01:33:39.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6186" PostTypeId="1" CreationDate="2015-11-17T02:00:46.077" Score="11" ViewCount="6296" Body="&lt;p&gt;I usually design precast concrete members for the common concrete strength of f'&lt;sub&gt;c&lt;/sub&gt;=4,000 psi at 28 days. Occasionally, a mix design will be submitted that has a very high strength. In one case, the mix that was submitted had a history of cylinder breaks above 7,000 psi. This is 175% of the specified strength!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally in engineering extra strength is good, but this seems excessive. Steel grades typically specify a strength range (min &lt;strong&gt;and&lt;/strong&gt; max), but I have never seen a maximum compressive strength in a concrete specification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Should I reject these very-high strength concrete mixes?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What could be the complications of too high of a compressive strength?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are only two possible areas of concerns that I can think of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Modulus of Elasticity is directly related to strength. A high strength may cause the concrete to be more prone to cracking (serviceability concern).&lt;/li&gt;&#xA;&lt;li&gt;Un-designed high strength may be detrimental in seismic design where you want members to fail in certain locations and in certain modes.&#xA;Note, my situations typically include moments (either beams or beam-columns). There likely wouldn't be any concern in a purely compression member.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reason for submitting such a high strength mix seems to be that the precaster wants to get the piece out of the form quickly, so they add lots of cement to insure that the strength requirement is met.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2018-11-25T02:06:45.730" Title="Should I reject a concrete mix design for having too high of a compressive strength?" Tags="|civil-engineering|structural-engineering|reinforced-concrete|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6187" PostTypeId="2" ParentId="6157" CreationDate="2015-11-17T02:02:31.987" Score="0" Body="&lt;p&gt;Ultrasonic sensing is can be used for distance measurement. Ultrasonic proximity sensing technology is commonly used in automotive as part of &lt;a href=&quot;https://en.wikipedia.org/wiki/Advanced_driver_assistance_systems&quot; rel=&quot;nofollow noreferrer&quot;&gt;Advanced Driver Assistance Systems&lt;/a&gt;. The basic theory behind this technology is &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_of_flight&quot; rel=&quot;nofollow noreferrer&quot;&gt;Time-of- Flight&lt;/a&gt; (TOF). Below few images to help visualize the idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xGM0A.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xGM0A.jpg&quot; alt=&quot;Park Assist-1&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BO8P4.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BO8P4.gif&quot; alt=&quot;Park Assist-2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To achieve this you can use &lt;a href=&quot;http://www.ti.com/product/pga450-q1&quot; rel=&quot;nofollow noreferrer&quot;&gt;PGA450 Ultrasonic Analog Front End&lt;/a&gt; (AFE) from Texas Instruments. As I understand this sensor is capable of distances up to 7M.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The PGA450-Q1 device can measure distances ranging from less than 1 meter up to 7 meters, at a resolution of 1 cm depending on the transducer-transformer sensor pair used in the system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Below is a diagram describing the concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fc7RS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fc7RS.jpg&quot; alt=&quot;PGA450 AFE&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you might want to check out &lt;a href=&quot;http://www.ti.com/product/TDC1000-Q1/datasheet&quot; rel=&quot;nofollow noreferrer&quot;&gt;TDC1000 Ultrasonic AFE&lt;/a&gt;. I am much more familiar with this part. But I have not used this part for proximity sensing. Once again below is a diagram describing the application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1rGvC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1rGvC.jpg&quot; alt=&quot;TDC1000 AFE&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also TDC1000 has a configuration where the emitter and receiver are seperate that might be interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5OY8N.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5OY8N.jpg&quot; alt=&quot;TDC1000 AFE Separate emitter and receiver&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some references that you might find interesting&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.freescale.com/applications/automotive/adas:ADAS&quot; rel=&quot;nofollow noreferrer&quot;&gt;Advanced Driver Assistance Systems (ADAS)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/2176/is-the-reliability-of-automotive-sensor-technology-impeding-the-success-of-self&quot;&gt;Is the reliability of automotive sensor technology impeding the success of self driving car?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lsds/ti/sensors/ultrasonic-sensing-applications.page#proximity-sensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proximity sensing&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/tool/tida-00151&quot; rel=&quot;nofollow noreferrer&quot;&gt;Automotive Ultrasonic Sensor Interface for Park Assist or Blind Spot Detection Systems&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=9&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CE0QtwIwCGoVChMIq9Di2K2WyQIVhGs-Ch3QhwPb&amp;amp;url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DrjEO4PsTnP4&amp;amp;usg=AFQjCNFccjB5OWF4_n_Ilz4HJjbe_uHfmw&amp;amp;sig2=pJojaKTNTbfsEzLIVa0fww&amp;amp;bvm=bv.107467506,d.cWw&quot; rel=&quot;nofollow noreferrer&quot;&gt;PGA450-Q1 for Ultrasonic Sensing&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.engineering.com/Library/ArticlesPage/tabid/85/ArticleID/8307/Ultrasonic-Sensor-SOC-can-Sense-Objects-and-Distance-for-Automotive-and-Industrial-Designs.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ultrasonic Sensor SOC can Sense Objects and Distance for Automotive and Industrial Designs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/2697/110&quot;&gt;Can I use a ultrasound sensor to measure water level?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0CB0QFjAAahUKEwi76_DN0JjJAhXCPT4KHSQkDuw&amp;amp;url=http%3A%2F%2Fwww.ti.com%2Flit%2Fsnaa220&amp;amp;usg=AFQjCNHiIW0c4gybaffO0MlWBYrpGYK47g&amp;amp;sig2=4jIjgrHtE75F1TqW7ReYNw&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ultrasonic Sensing Basics for Liquid Level Sensing, Flow Sensing, and Fluid Identification Applications&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-11-17T23:45:48.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6188" PostTypeId="2" ParentId="6175" CreationDate="2015-11-17T02:54:23.027" Score="-1" Body="&lt;p&gt;I would like to point out some &quot;additional&quot; information. I think that these CSP might also be used to create Hydrogen from electrolysis. Thus increasing the offer which might be important for development of the hydrogen fuel technology in cars. Because, only recently, I read that these cars provide much greater range than electric cars but the density of &quot;hydrogen stations&quot; is much to low for this technology to evolve. Thus, the CSP might have additional benefits for which they weren't even intended. It would have another positive effect, people don't have to think about how to store the &quot;electric energy&quot; because it was converted to usable &quot;chemical energy&quot; (although storing hydrogen isn't too easy by itself).&lt;/p&gt;&#xA;" OwnerUserId="3780" LastEditorUserId="3782" LastEditDate="2015-11-17T05:55:36.110" LastActivityDate="2015-11-17T05:55:36.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6189" PostTypeId="1" CreationDate="2015-11-17T03:58:09.210" Score="1" ViewCount="337" Body="&lt;p&gt;I'm designing a machine to test a static object (I'll refer to it as A)'s resistance to deflection due to impact force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/glNCN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/glNCN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This design uses a pneumatic(probably wont be using hydraulic) cylinder to apply the impact force to the surface of A. (However, if I were to use the pneumatic cylinder alone to apply the force, I would have to position the surface right at the end of the piston stroke. If I allowed the piston stroke to be slightly more than the distance between the two to completely apply the force on the surface(overexerting the cylinder), I will damage the cylinder over time and reduce its lifespan.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, I added an extended head to the piston end with the use of springs and linear guides(shafts) within it. In my head, I picture it as the head exerting the impact force(F) first at a pretty high speed and then when the head stops at the surface of A the springs absorb the remaining of the piston stroke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have some concerns too so I'm hoping some of you more experienced and knowledgable people out there can help.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Since the speed is moving at a pretty fast speed(completes the stroke(completely extends) in 1 second), would the head be able to totally transfer the force F from the cylinder piston to the surface of A or would the springs somewhat absorb the force and lower the resultant output force? How should I select springs for this purpose?&lt;/li&gt;&#xA;&lt;li&gt;To determine the force of the cylinder, use Force = mass * acceleration. The thing is, I have the equivalent load mass(60kg) but I don't have the acceleration. I only have the distance(stroke length = 300mm) and maximum piston velocity(300mm/s). Since the velocity is pretty fast and stroke length is not that long, how should I calculate the acceleration? Can I assume that the acceleration will be constant since the piston will extend completely pretty fast?&lt;/li&gt;&#xA;&lt;li&gt;The cylinder applies a force equivalent to a 60kg load. How should I be calculating the output? In terms of force, F=ma or kinetic energy, 1/2*m(v^2)? Since there will be spring energy in play as well(?)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3781" LastActivityDate="2016-01-18T11:54:03.217" Title="How do the springs really work in my design?" Tags="|mechanical-engineering|design|springs|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6190" PostTypeId="1" AcceptedAnswerId="6214" CreationDate="2015-11-17T04:19:21.503" Score="5" ViewCount="264" Body="&lt;p&gt;Does the speed of the water flowing radially outwards from a water jet impacting a sink depend on how far the tap is opened?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment, I understand that the maximum velocity of the water in the vertical jet is constant regardless of how far the tap is open (as it is accelerated by gravity, and disregarding air-resistance), but the volume flow rate is greater the further the tap is opened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that a higher volume flow rate results in a greater depth around the bottom of the jet which reduces the effect of friction between the water and the sink as the contact surface of the water has reduced, which would decelerate the water at a lower rate than if it were shallower, but am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain precisely &lt;em&gt;how&lt;/em&gt; opening a tap affects the radial speed of the water around the bottom of the jet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should have clarified that, for my question, the area is variable.&#xA;What I had in mind was more like a &lt;em&gt;variably sized circular hole beneath a constant-height water tank&lt;/em&gt;.  I used the word &quot;tap&quot; for simplicity, but I realise now this is a bad substitute, and has caused more confusion than intended.&lt;/p&gt;&#xA;" OwnerUserId="3013" LastEditorUserId="3013" LastEditDate="2015-11-19T01:20:39.193" LastActivityDate="2015-11-19T01:20:39.193" Title="Does opening a tap in a sink affect the radial speed of the water around the jet?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6191" PostTypeId="2" ParentId="6186" CreationDate="2015-11-17T04:22:49.687" Score="4" Body="&lt;p&gt;&lt;strong&gt;Reliability of Data&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you able to repeat the test to confirm the 7ksi break? Did you have a second 28 day cylinder or 7 day breaks which were also higher than typical?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How sure are you that the cylinder was formed properly and followed ASTM?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Application Specific Requirements&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the specifications from the design engineer call for a max and min ksi or only a minimum strength?   I have projects which utilize cement in flexible applications and require max and min range to firstly provide strength and secondly fail in a planned motion to prevent major cracking (paved soil cement).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you spoken to the design engineer about the results to see if he/she has any hesitations about high strength concrete for the application?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Customer Satisfaction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher strength concrete costs more than a leaner product, the customer or your company are not getting their money's worth.. or the concrete supplier is ripping themselves off.  However, these would not necessarily be reasons to reject the concrete just something to keep in mind depending on who you serve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Less Generally&lt;/em&gt;&lt;/strong&gt;&#xA;After writing all of that and looking at your profile it seems like you may be the design engineer, doh.  I would most likely look at the pros and cons of why you didn't select the higher strength concrete to begin with and throw out the reasons that involve cost.   Will the higher strength reduce work-ability of the structure, more apt to chip than absorb when manhandled on site.  Is there a strength to thickness ratio that needs to be maintained that would mean the higher strength would require a thicker or thinner design that it no longer meets. Will the higher strength change curing time, increase reactivity of the cement which would change your production rate coming out of the precast plant?&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2015-11-17T04:22:49.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6192" PostTypeId="2" ParentId="6189" CreationDate="2015-11-17T10:39:53.557" Score="1" Body="&lt;p&gt;It seems to me you've made quite a mess of what is being exerted, measured and absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface A receives impact of certain force exerted by the inertia of the head plus (low, probably negligible) tensile strength of the expanded springs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's definitely not proportional to acceleration of the piston - the acceleration, and the force exerted by the piston against the head is; but that is before the contact and far, far less than the impact force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual force is proportional to DECELERATION of the head, as it hits against the surface, and to the mass of the head. Since the deceleration upon impact lasts milliseconds and speed goes to zero from a considerable value over such a short time, the force exerted is much higher than subsequent force from the spring pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here begins in that the force is highly non-linear. As the surfaces touch and begin the compression, the force is much lower than when the depression/deflection of A (and the head) reaches maximum. And even then it changes as the pressure wave only begins travel through the medium, and pressure (=force) drops despite the object still yielding to the impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could calculate the &lt;em&gt;average&lt;/em&gt; force during the impact by measuring the time of the impact (e.g. with high-fps camera), but in reality it's not the force you should be measuring...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you should be measuring is the impact energy. And that's pretty easy: $ {1 \over 2} m v^2 $ with mass of the impact head for &lt;em&gt;m&lt;/em&gt;, and its speed right before the impact as &lt;em&gt;v&lt;/em&gt;. This is the primary, most informative piece of data about the impact. Speed is secondary, as different materials react differently to varied speed impact of the same energy (e.g. metal dents/bends/rips at low-velocity, but &lt;em&gt;splashes&lt;/em&gt; at extreme speeds). If the behavior doesn't vary much within reasonable speed range, it's omittable. Another piece of data would be the impact surface; a piercing point will behave differently than a blunt ball or a flat block. You might want pressure from a &lt;em&gt;static&lt;/em&gt; force. But force alone, in impact measurement isn't all that helpful.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-11-17T10:39:53.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6194" PostTypeId="1" CreationDate="2015-11-17T17:46:11.960" Score="0" ViewCount="1004" Body="&lt;p&gt;I studied many papers and I face with concept of  numerical diffusion,but I don't understand it. Can someone explain briefly about numerical diffusion?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorDisplayName="user16" LastEditDate="2015-11-17T17:50:12.700" LastActivityDate="2015-11-18T08:44:20.493" Title="What is the numerical diffusion?" Tags="|mechanical-engineering|cfd|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6195" PostTypeId="2" ParentId="6194" CreationDate="2015-11-17T17:55:45.523" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Numerical_diffusion&quot; rel=&quot;nofollow&quot;&gt;Numerical diffusion&lt;/a&gt; is an artifact of solving finite differences numerically (typically on a computer). Error in discretization, formulation of the finite difference equations, or just machine precision can all cause an increase in (thermal/momentum) diffusivity of the material being simulated, causing the results to be (sometimes wildly) inaccurate.&lt;/p&gt;&#xA;" OwnerUserId="3015" LastActivityDate="2015-11-17T17:55:45.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6196" PostTypeId="1" AcceptedAnswerId="6201" CreationDate="2015-11-17T19:11:38.180" Score="5" ViewCount="113" Body="&lt;p&gt;I want to know why ceramic or disk capacitors are not rated directly like electrolyte ones? Is it just a convention or some other reason is behind this? Like, the ceramic ones are rated as 762k. then the value of the capacitance is 7600pF. (7-1st significant digit, 6-2nd significant digit and 2=multiplying factor(100))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But an electrolyte one is directly printed like:- 3400uF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QEOvY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QEOvY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Nn8U9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nn8U9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3757" LastEditorUserId="3757" LastEditDate="2015-11-17T21:17:31.330" LastActivityDate="2019-10-31T00:11:18.790" Title="Why Ceramic Capacitors Are Not Rated Directly Like Electrolytic Ones?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6197" PostTypeId="2" ParentId="6175" CreationDate="2015-11-17T19:24:29.543" Score="1" Body="&lt;p&gt;Some of these answers already touched on the environmental impact of CSP, I would like to add &lt;a href=&quot;http://www.researchgate.net/profile/Rebecca_Hernandez2/publication/257200435_Environmental_impacts_of_utility-scale_solar_energy/links/02e7e5249f14ea3528000000.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; as a source. It reiterates on the points made by others who answered the OP's question and pointed out the issues with biodiversity (specifically bird species) being affected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, nobody pointed out the water use required for the cooling system which is also important. As time goes on we are going to need to be more careful with our water use, as we are already in need of a &quot;Blue Revolution&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that CSP is an awesome supplement to PV systems and other types of alternative energy. It does make solar capable of achieving a more stable grid on industrial-scale installations. However we have to be careful of how we build it and how much we build it to minimize environmental costs. &lt;/p&gt;&#xA;" OwnerUserId="3792" LastActivityDate="2015-11-17T19:24:29.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6199" PostTypeId="1" CreationDate="2015-11-17T20:15:17.593" Score="5" ViewCount="296" Body="&lt;p&gt;Why is the heat addition process in otto cycle assumed constant volume while in diesel cycle it is considered constant pressure?&lt;/p&gt;&#xA;" OwnerUserId="3794" LastEditorUserId="1560" LastEditDate="2015-11-18T01:35:54.023" LastActivityDate="2015-11-18T01:35:54.023" Title="Diesel Cycle Vs. Otto cycle" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6200" PostTypeId="1" CreationDate="2015-11-17T22:40:57.773" Score="1" ViewCount="112" Body="&lt;p&gt;&lt;strong&gt;Context:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to create a mathematical model for centrifugal pumps. This model should describes how the operational variables of centrifugal pumps $(Q, \Delta P, ...)$ related to it's design parameters $(Stages, Dimensions, ...)$ and it's use-case $(\rho, \mu, ....)$ .The intention behind this job is to develop an educational simulation environment. Anyway, after spending reasonable time reading+hacking various references I didn't find an &lt;em&gt;idiomatic, integrated, already provided model&lt;/em&gt; to use. And I decided to assemble it using different parts from different sources. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I think that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One important part of the model is to write $f(\eta,Q,...)=0$ or finding relation between &lt;strong&gt;Efficiency&lt;/strong&gt;, &lt;strong&gt;Flow&lt;/strong&gt; + &lt;strong&gt;design parameters&lt;/strong&gt; on &lt;em&gt;allowable&lt;/em&gt;, &lt;em&gt;preferred&lt;/em&gt; or at list &lt;em&gt;rated&lt;/em&gt; flow range.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposal:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To overcome the above bulleted item of my modeling problem, I picked mandatory conditions of an acceptable &lt;strong&gt;Ecopump&lt;/strong&gt; project's centrifugal pump. Reference: &lt;a href=&quot;http://europump.net/news/22/44/HANDBOOK-ASSESSING-THE-ENERGY-EFFICIENCY-OF-PUMPS-AND-PUMP-UNITS&quot; rel=&quot;nofollow&quot;&gt;Assessing the Energy&#xA;Efficiency of Pumps and Pump Units&lt;/a&gt; By: &lt;em&gt;Bernd Stoffel&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Ecopump&lt;/strong&gt; is the name of the successful flagship project that &lt;a href=&quot;http://www.europump.org/&quot; rel=&quot;nofollow&quot;&gt;Europump&lt;/a&gt;&#xA;  launched in 2004 to bring the pump industry’s commitment to energy&#xA;  savings to the fore.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Following equations from the referred handbook, describes minimum $Efficiency$ required at these three different flow rates: 1-$BEP$, 2-part-load$(PL)$ and 3-over-load$(OL)$ , in terms of design parameters:    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\eta_{BEP,min\, requ} = -11.48*(\log(n_s))^2-0.85*(\log(Q_{BEP}))^2-&#xA;0.38*\log(n_s)*\log(Q_{BEP})+88.59*\log(n_s)+13.46*\log(Q_{BEP})\qquad (7.4)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\eta_{PL,min\, requ} = 0.947*\eta_{BEP,min\, requ}\qquad (7.5)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\eta_{OL,min\, requ} = 0.985*\eta_{BEP,min\, requ}\qquad (7.6)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$#Where:$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_{PL} = 0.75*Q_{BEP}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_{OL} = 1.1*Q_{BEP}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the above conditions are three inequality rule that define a floor for a real pump $\eta$ value, I make an important suggestion here about a conceptual minimum Ecopump &lt;em&gt;which exactly overlays those minimum requirements&lt;/em&gt; so $\eta_{real}==\eta_{min}$ ,and using those three points for $\eta$ it is possible to interpolate a second order polynomial and find $\eta(Q)$ for that conceptual minimum Ecopump with specific design parameters.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect that this model along with its numerical results could help instructors to cover following topics in a classic pump course:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How efficiency of a pump is related to other parameters.&lt;/li&gt;&#xA;&lt;li&gt;How much a pump efficiency should be, to fulfill mandatory items of standards. &lt;/li&gt;&#xA;&lt;li&gt;How does $\eta$ change related to $Q$  &lt;/li&gt;&#xA;&lt;li&gt;How to deal with modern standards about pump efficiency    &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Part One:&lt;/strong&gt;&lt;br&gt;&#xA;Am I on the right way? Or is(are) there any better way(s) to achieve the mentioned four goals.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also:&lt;br&gt;&#xA;The original $(7.4)$ equation have an additional $-C$ constant with the following description:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The constant C(in [%]) depends on &lt;strong&gt;MEI&lt;/strong&gt; and additionally on the type and nominal speed of the size.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Part Two:&lt;/strong&gt;&lt;br&gt;&#xA;From the handbook it is not clear for me how to calculate this constant, but for my approximate educational model I simply put it aside, Is this approximation acceptable? How much it affects my model? How to substitute more accurate value for $C$? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any idiomatic points to create a better model are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2404" LastEditorUserId="2404" LastEditDate="2015-11-22T07:37:05.483" LastActivityDate="2015-11-22T07:37:05.483" Title="Efficiency model for centrifugal pumps, using relations from Europump's Ecopump project" Tags="|mechanical-engineering|pumps|modeling|process-engineering|education|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6201" PostTypeId="2" ParentId="6196" CreationDate="2015-11-17T23:07:38.620" Score="3" Body="&lt;p&gt;Capacitors are basically rated the same.  The two minimum mandatory ratings are the capacitance and the maximum voltage.  This applies to electrolytics, ceramics, and others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that it gets into details like ESR (equivalent series resistance), maximum ripple current, temperature range, leakage current or resistance, upper frequency, and more.  These parameters are all really limits of how the specific capacitor varies from ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different technologies have different tradeoffs.  Electrolytics usually have good energy density per volume and per price, but low lifetime and high ESR.  Ceramics have low ESR, higher temperature tolerance, and can therefore tolerate higher ripple currents.  They can exhibit piezo effects and the capacitance my drop significantly at higher voltages.  Again, everything is a tradeoff.  The reason we have so many different types of capacitors is because different parameters matter more in different applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the printing, many electronic parts are marked with a floating point scheme, especially when there is little room.  The &quot;104&quot; on the cap at right means 10 x 10&lt;sup&gt;4&lt;/sup&gt; pF, or 100 nF.  The cap at left is physically much bigger, so there is more room to print a more elaborate label.  SMD caps usually have no label at all.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-17T23:07:38.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6202" PostTypeId="2" ParentId="6199" CreationDate="2015-11-18T00:42:35.593" Score="2" Body="&lt;p&gt;In an Otto cycle, fuel is introduced prior to compression and ignited quickly when the piston is at or near top dead center.  This produces a sharp rise in pressure because combustion happens quickly compared to how fast the piston is moving.  In the Diesel cycle, fuel begins to enter near the end of the compression stroke, and continues past top dead center, into the expansion stroke.  For this reason, the pressure is kept more-or-less constant because fuel is being added as the volume is increasing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=15&amp;amp;ved=0CFYQFjAOahUKEwjcicaM3pjJAhVIRyYKHTzZDPA&amp;amp;url=https%3A%2F%2Fwww.ideals.illinois.edu%2Fbitstream%2Fhandle%2F2142%2F4436%2Fengineeringexperv00000i00160.pdf&amp;amp;usg=AFQjCNHb5CqxWMgesGq8fD6VuJU5vQVHpQ&amp;amp;sig2=3G18fbqmrqdiDHZm0s1NjQ&quot; rel=&quot;nofollow&quot;&gt;This paper &lt;/a&gt; is a bit dated, but well written and contains a brief explanation for the constant-pressure assumption on page 16.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-11-18T00:42:35.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6203" PostTypeId="1" AcceptedAnswerId="6217" CreationDate="2015-11-18T04:19:27.050" Score="7" ViewCount="1154" Body="&lt;p&gt;I was a doing a placement at a company, and they were designing a new heat pump. I was told the more compact you can make a heat-pump, the better it is for efficiency, and they mentioned something about this was because the power density was increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get there are benefits from using less material and making something lighter which would improve efficiency in that regard, but I don't get the benefit of a higher power density in a heat pump. I understand that in a heat engine like a combustion engine you want a high compression ratio so you'd want to make the thing as compact as possible, but why is a better power density also good for a heat-pump? Is it this just the thing with all types of machinery which have some sort of volume which is compressed? If so Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I didn't really want to say COP but rather efficiency to keep the question more general. These guys seemed pretty certain that having a small high rpm heat pump gave improved performance, whilst I assume they meant COP could be something else they are looking at. Maybe a better way to ask this is: Are there any theoretical benefits to the COP of a heat pump by making it smaller?&lt;/p&gt;&#xA;" OwnerUserId="3797" LastEditorUserId="3797" LastEditDate="2015-11-21T13:54:14.723" LastActivityDate="2015-11-21T13:54:14.723" Title="Are smaller heat-pumps more efficient?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6204" PostTypeId="1" CreationDate="2015-11-18T04:40:05.060" Score="1" ViewCount="544" Body="&lt;p&gt;Are there any free online databases where I can access data for the wear rates of materials? I tried using CES EduPack, but I'm having no luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe other free and open databases are also existing on the net? Where and how could I find the wear rate in them?&lt;/p&gt;&#xA;" OwnerUserId="3798" LastEditorUserId="9" LastEditDate="2016-01-28T11:49:46.723" LastActivityDate="2016-01-28T11:49:46.723" Title="How to find the wear rate of materials in the net?" Tags="|mechanical-engineering|materials|data|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6209" PostTypeId="2" ParentId="6173" CreationDate="2015-11-18T08:27:16.383" Score="4" Body="&lt;p&gt;As AndyT says - it seems it's aesthetics in most cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the above as the answer, the following is essentially comment, but seemed worth posting as it gives a lot of detail of what goes into a design and how there are issues that are not apparent to &quot;outsiders&quot; but which can be important. eg in this example the span is very slightly curved resulting in a 66 mm deflection in the roadway (under 3 inches) but it caused extra difficulty. And the effort needed to construct this bridge and the tight tolerances are not at all obvious from inspection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seemed worthwhile as while the original question asks about just the angled towers this shows how many other unseen factors can affect the design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cmnzl.co.nz/assets/sm/3513/61/0010-E78Jackson.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;ORMISTON ROAD CABLE STAYED BRIDGE&lt;/strong&gt;&lt;/a&gt;&lt;br&gt;&#xA;This paper provides an overview to selected construction challenges encountered&#xA;during the building of the Ormiston Road cable stayed bridge, an iconic cable stayed&#xA;composite steel and concrete bridge constructed in the Sir Barry Curtis Park,&#xA;Manukau City, Auckland. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand the description correctly, one tower is in considerable compression and the other in tension. This is a small road bridge but various constraints make technical difficulties greater than in some much larger bridges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some key comments - the whole paper is worth reading.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The construction of the cable stay bridge was technically very complex due to the&#xA;asymmetric geometry and very tight tolerances specified. The bridge deck is on a&#xA;radius of approx 37kms, which sounds very flat but results in variations in levels due&#xA;to curvature of 66mm along the length of the bridge. The 45.5m pylons are made up&#xA;of a 28m section of reinforced concrete tapered from 1.8m diameter at the base to&#xA;1.3m diameter at the top, with a 5.5m high structural steel box to provide anchorage&#xA;for the stay cables and topped with a 12m lattice spire made of stainless steel and&#xA;glass. To further complicate matters both pylons are inclined back longitudinally at 15&#xA;degrees and angled together at 5 degrees and were not self supporting.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There was&#xA;very little tolerance in ensuring the stay cables were correctly aligned between the&#xA;pylon and the deck anchorages. The angular rotational tolerance of 0.25 degrees&#xA;commonly specified for cable stayed bridges &lt;strong&gt;required the positional tolerance of the&#xA;stay anchorages to be within 3mm.&lt;/strong&gt; With this level of accuracy much of the&#xA;construction effort and risk mitigation was focussed on survey integrity and&#xA;conservation of construction tolerances. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The concrete pylons are angled in two directions providing a dynamic element to the&#xA;bridge. They are also positioned closer to the western abutment than the eastern&#xA;meaning the back span is considerably shorter than the fore span. &lt;strong&gt;This asymmetry&#xA;generates considerable uplift on the western abutment which is resisted with deep&#xA;tension piles.&lt;/strong&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Normal Drossbach ducting could not be used as the tendon sheathing after research&#xA;showed that Drossbach could collapse at about 12m head of concrete. 100NB steel&#xA;pressure pipe was used as an alternative, which could cope with the high hy&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tendons were assembled on the ground prior to lifting and placing inside the piles&#xA;which already had the reinforcing cage installed. It took a synchronised effort of 3&#xA;cranes using 6 snatch blocks and an excavator to successfully lift the 45 m long&#xA;flexible tendons from horizontal to vertical, without kinking the tendon, so that they&#xA;could be lowered into the pile casing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The pile tendons pass through the western abutment and terminate in the deck. This&#xA;meant that the tendons could not be stressed and grouted until the deck had been&#xA;poured, some 9 months later. &lt;strong&gt;As a temporary measure to prevent corrosion of the&#xA;strand, **a sodium hydroxide solution was introduced to the pile tendons to create a&#xA;protective alkali environment.&lt;/strong&gt; Regular pH testing was used to monitor and maintain&#xA;alkalinity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;While the bridge span is short at 70 m, the effective tributary load area for the cables&#xA;was of similar magnitude to a much larger cable stayed bridge because of the large&#xA;deck width and resulted in similarly sized cable stays. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2VdxW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2VdxW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bbrpy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bbrpy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Footbridge with &quot;leaning tower&quot; at Brown Owl (hoo?) in New Zealand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/84TqG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/84TqG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.nz/maps/@-41.1043367,175.0915203,294m/data=!3m1!1e3&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Location on Google maps&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2015-11-18T10:33:48.007" LastActivityDate="2015-11-18T10:33:48.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6212" PostTypeId="1" AcceptedAnswerId="6218" CreationDate="2015-11-18T13:38:15.003" Score="3" ViewCount="727" Body="&lt;p&gt;I'm an undergraduate student of Electrical Power Engineering currently in 3rd semester, one of the courses we  study is Digital Logic Design(dld) and I have found much interest in it but I am unclear whether the subject would have application in our field i.e. power engineering or not....Could anyone guide me that how should I study dld with what things in mind related to future perspective?&lt;/p&gt;&#xA;" OwnerUserId="1422" LastEditorUserId="1422" LastEditDate="2016-04-21T16:19:52.873" LastActivityDate="2022-04-18T17:02:17.117" Title="Relation of Digital Logic Design to Electrical Power Engineering?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6213" PostTypeId="1" CreationDate="2015-11-18T19:10:27.853" Score="5" ViewCount="805" Body="&lt;p&gt;Is there a general way to convert rotational energy from a gear or something into lateral motion toward or away from the gear? I could use a piston type construction, but I only want the lateral motion to be in effect when the rotation just begins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance I have a motor that turns 180 degrees each direction at an equal time interval. When it is rotating clockwise, I need a &quot;foot&quot; on the end to move toward the pivot by about 1/4 the total length of the leg (about 4 centimeters) and stay there until the counterclockwise motion begins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When rotating counter clockwise, foot is extended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When rotating clockwise, foot is drawn in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Edit:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a drawing. It may not be any help at all, but I will try to explain it as best I can with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/txx2i.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/txx2i.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In figure one the leg is at it's full extension forward and placed on the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In figure two the leg pivots backward to push the body forward. this happens when anything walks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure three is where I'm having the problem: the foot needs to lift off the ground so it can begin it's forward motion to take another step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In figure four, the foot is at full forward extension and is ready to be put back on the ground for the second step, which is continued back at step 1, 2, 3, 4 and repeat repeat ect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any living thing that walks follows these steps in order to move. I would like to replicate it in my robot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE;&lt;/strong&gt; I have not drawn any of the leg, since the solution to this problem will likely determine what it looks like. sorry if this doesn't help at all, but I think it must be required for there to be an image for this site or something like that. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have considered both the answers for a similar &lt;a href=&quot;https://robotics.stackexchange.com/questions/8501/lifting-robotic-leg-with-only-one-servo&quot;&gt;question&lt;/a&gt; on robotics stack exchange, and most of them will not work because they require continuous motion whereas I only have 180* cycles. The accepted answer ended up not working because it interfered with the foot design needing to move straight up and down and because there was no-way to stiffen the gears of the mechanism reasonably without creating too much strain for the motor when it reached full extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea I've fiddled with is an equal gear ratio essentially always keeping the foot level and being triggered by extraneous motion upon the top gear, but it failed for the same reasons as above.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastEditorUserId="-1" LastEditDate="2017-04-13T12:49:17.810" LastActivityDate="2019-02-07T07:01:30.290" Title="How to transfer rotational energy into lateral energy only when rotating one way?" Tags="|mechanical-engineering|gears|applied-mechanics|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6214" PostTypeId="2" ParentId="6190" CreationDate="2015-11-18T20:21:52.507" Score="2" Body="&lt;p&gt;The speed at which water leaves the tap is not constant, as it has a finite area, which remains roughly constant, so increased flow does indeed increase the velocity. ($\dot m= \rho\, VA$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we can still ask the question of how the flow profile of water flowing outward from a jet is dependant on flow rate when jet velocity is held constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let's calculate the reynolds number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Re=\frac{\rho V D}{\mu} \approx \frac{1000\frac{kg}{m^3}\,2\frac{m}{s}1\,cm}{900 Pa\,\mu s}\approx 20000$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the viscous effects are going to be tiny during the transition, which means we can use bernoulli's equation. The pressure will be at atmospheric, before and after the transition, and the height will be the same, so the velocity will be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the spray will just get thicker, but remain at the same velocity when the flow rate is increased. Of course, far from the jet the viscous effects will begin to matter more as the fan thickness decreases, and the reynolds number drops. Then indeed the thicker flow will remain at high speed for longer.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-11-18T20:21:52.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6215" PostTypeId="1" AcceptedAnswerId="6628" CreationDate="2015-11-18T20:28:32.723" Score="2" ViewCount="325" Body="&lt;p&gt;I have a problem in understanding of gears (spur gears). I designed some spur gears but I don't understand how the modulus works. I saw that this modulus is some standard thing and it's calculated by diametral pitch. Do I need to make this diametral pitch from standards, or could I make whatever I want? How does this work?&lt;/p&gt;&#xA;" OwnerUserId="3807" LastEditorUserId="9" LastEditDate="2015-11-18T22:57:53.853" LastActivityDate="2017-04-27T15:43:06.050" Title="Spur gears design" Tags="|gears|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6216" PostTypeId="1" CreationDate="2015-11-18T22:31:41.613" Score="2" ViewCount="1458" Body="&lt;p&gt;I have a stream ($10,000$ $m^3/d$) of inert gas and steam. The temperature of the stream is $195$ $^oC$, and the absolute pressure is $1455$ $kPa$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find the amount of steam that can be present in the system. I can assume $100\%$ quality saturated steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would I be correct in approximating the steam fraction to be: &#xA;$P_{sat@T}$ / $P_{stream}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $195$ $^oC$, the $P_{sat}$ of steam is $1398$ $kPa$. Therefore the fraction of steam is:&#xA;$1398/1455$ $=$ $96\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can then find the volume of steam: $0.96*10,000$ $=$ $9,600$ $m^3/d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my reasoning correct. Any other ways to calculate this question?&lt;/p&gt;&#xA;" OwnerUserId="3448" LastEditorUserId="90" LastEditDate="2015-11-19T00:14:57.970" LastActivityDate="2015-11-19T00:42:35.900" Title="Calculating the Fraction of Steam in a Mixture" Tags="|thermodynamics|pressure|temperature|steam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6217" PostTypeId="2" ParentId="6203" CreationDate="2015-11-18T22:34:05.337" Score="3" Body="&lt;p&gt;As far as I know the COP of a heat pump doesnt vary greatly with the size of the unit only with the technology used, quality of components, and the correspondingly price. As a general engineering rule, its easier to make something cheaper/more effective when you have less constraints. So having no footprint constraint would generally make the unit cheaper/higher COP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not work in HVAC, but generally a large building will invest in two large heat pump air handlers integrated into one system. They have two for redundancy purposes and no more than two because maintenance increases with increased number of units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reliability of small units were very high, they could potentially be a better solution because of the elimination of the large ducting required. With current technology however, small units increase facility maintenance. Another aspect is control. One or two large units is easy to control. Many small units, all with different set points and pressure differentials can cause a lot of maintenance headaches.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-18T22:34:05.337" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6218" PostTypeId="2" ParentId="6212" CreationDate="2015-11-19T00:10:12.820" Score="1" Body="&lt;p&gt;Below is a job description for a Senior Electrical Engineer position at Princeton power systems to help you with future perspective. &lt;a href=&quot;https://web.archive.org/web/20220331101227/https://princetonpower.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Princeton Power System&lt;/a&gt; makes power inverters for the Renewable Energy market. As you would see few key words that address the relationship between Digital Logic Design and Electrical Power Engineering.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;FPGA – Field Programmable Gate Array&lt;/li&gt;&#xA;&lt;li&gt;Power Electronics&lt;/li&gt;&#xA;&lt;li&gt;Microprocessors&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Senior Electrical Engineer&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Job Description:&lt;/em&gt; You will be a key member of the Product Development&#xA;team developing state of the art Power Conversion products for use in&#xA;Alternative Energy, Renewable Energy, Energy Storage, Micro-Grid, and&#xA;Backup Power applications. Tasks will include design of power&#xA;conversion systems, design of electrical control circuitry,&#xA;development of control algorithms and software, performing&#xA;system-level analysis and design, printed circuit board design, and&#xA;supporting fabrication and testing of components and systems.&lt;/p&gt;&#xA;&lt;p&gt;Responsibilities:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Control system and auxiliary electronics circuits design&lt;/li&gt;&#xA;&lt;li&gt;Overall system architecture and schematic design&lt;/li&gt;&#xA;&lt;li&gt;Printed circuit board design and design supervision&lt;/li&gt;&#xA;&lt;li&gt;Simulation or analysis of electronic circuits ranging from small electronics to high-power electronics&lt;/li&gt;&#xA;&lt;li&gt;Software development for &lt;strong&gt;microprocessor&lt;/strong&gt; and &lt;strong&gt;FPGA&lt;/strong&gt;-based control&#xA;systems and PC-based graphical user interfaces&lt;/li&gt;&#xA;&lt;li&gt;Electrical system test procedure development, testing, and test reporting&lt;/li&gt;&#xA;&lt;li&gt;Analysis of system design reliability and design for reliability&lt;/li&gt;&#xA;&lt;li&gt;Generation of electrical schematics and manufacturing-related documentation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Required:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At least a Bachelor’s degree in Engineering or other pertinent field.&lt;/li&gt;&#xA;&lt;li&gt;Detailed understanding of modern electrical engineering principles and practice&lt;/li&gt;&#xA;&lt;li&gt;Experience with printed circuit board design&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Ideal:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Experience with high-&lt;strong&gt;power electronics&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Software development experience&lt;/li&gt;&#xA;&lt;li&gt;Experience with system simulation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Most of &lt;a href=&quot;https://en.wikipedia.org/wiki/SCADA&quot; rel=&quot;nofollow noreferrer&quot;&gt;Supervisory control and data acquisition (SCADA)&lt;/a&gt; have microprocessor and/or FPGA, thus the close relationship between Digital Logic Design and Electrical Power Engineering.&lt;/p&gt;&#xA;&lt;p&gt;Hope this helps with some ideas related to future perspective and the relationship between Digital Logic Design and Electrical Power Engineering. Investigating Renewable Energy Industry and power electronics related supervisory control system might be a good start.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://ieeexplore.ieee.org/document/1569247&quot; rel=&quot;nofollow noreferrer&quot;&gt;FPGA implementation of a fastflex supervisory control and data acquisition (SCADA) remote telemetry unit&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://ieeexplore.ieee.org/document/5735058&quot; rel=&quot;nofollow noreferrer&quot;&gt;RTU hardware design for SCADA systems using FPGA&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="6767" LastEditDate="2022-04-18T17:02:17.117" LastActivityDate="2022-04-18T17:02:17.117" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6219" PostTypeId="2" ParentId="6216" CreationDate="2015-11-19T00:42:35.900" Score="1" Body="&lt;p&gt;If the saturation pressure at $195^\circ C$ is $1398 kPa$, and you're above this pressure, technically you have compressed liquid water.  Never the less, we'll assume you have 100% quality saturated steam for a moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Gibbs phase rule says that $f=C-P+2$, where $f$ is the number of degrees of freedom, $C$ is the number of components in the system (two in your case, inert gas and steam), and $P$ is the number of phases (one in your case, everything is vapor phase).  This means you have 3 total degrees of freedom.  However, you only specified two (pressure, temperature), so the third (mass fraction of steam or inert gas) remains unspecified.  Unfortunately, you cannot determine the steam content with the given information alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you were hinting at the concept of partial pressures with the calculation you showed.  Unfortunately, you don't know the partial pressure or the mass fraction of the steam, so this doesn't help much either.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-11-19T00:42:35.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6220" PostTypeId="2" ParentId="6174" CreationDate="2015-11-19T01:00:02.983" Score="-1" Body="&lt;p&gt;In my experience, appealing to codes that may or may not have been intended for your specific application is a costly business practice. There is no guarantee a load rating will be sufficient for the variables involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were in your situation, I would contact the intended HVAC company or another prominent HVAC company in you area. Take one or two of the engineers out to lunch and ask them to review your design as a colleague. They will immediately be able to see the red flags, be able to give you a model number for the largest possible air handler for that size building, know the capability of cranes in your area, and will be 100% familiar with the codes in your area. They will be glad to help because in a couple months they will need to ask you a roof structure question for one of their unrelated projects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you do find the exact code specification you are looking for I would encourage you to reach out for those kind of connections to experts in your community.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-19T01:00:02.983" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6221" PostTypeId="1" AcceptedAnswerId="6274" CreationDate="2015-11-19T01:07:00.053" Score="3" ViewCount="10028" Body="&lt;p&gt;Are there any established solutions for truning a gear (or wheel) half-circle in one direction, back in the opposite direction, and repeating? Mechanism must be driven by a motor and continuous rotary motion. Compact and simple mechanisms preferred. Below are two possible solutions I dreamed up. Which one is better and why? Or what should I do instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea I had was to use a motorized disk (orange) with a slotted shaft(yellow) to create reciprocating motion. Then by extending the shaft and connecting it to a second disk (Red) I could achieve it, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EgflS.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EgflS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another Idea I had was to use a full-square Scotch Yoke (Yellow) with gear teeth on the bottom. Properly sizing the connecting gear would allow for 180 degrees reciprocating motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mDWE7.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mDWE7.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DelpQ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DelpQ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3765" LastEditorUserId="3765" LastEditDate="2015-11-19T11:54:51.013" LastActivityDate="2019-03-24T09:13:41.523" Title="Rotary Motion to 180 Degrees Reciprocating Mechanism" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6223" PostTypeId="1" AcceptedAnswerId="6237" CreationDate="2015-11-19T10:51:00.163" Score="5" ViewCount="1728" Body="&lt;p&gt;I'm well aware that there are many different factors to be considered, such as ease of installation/maintenance, conversion from AC to DC, environment, distance, speed, frequency of turns, and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting all of those aside for now, suppose we build a 10-mile long demonstration track with some hills and turns along the way and featuring both a third rail and an overhead cable. Suppose further that we have two trains (one equipped with a pantograph, the other with contact shoes). They have the exact same weight and are programmed to accelerate and decelerate at the same rate, and run at the same speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one will have consumed more electricity at the end of the ten-mile run?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus Question: A third rail is a lot wider and thicker than an overhead wire. Is more electricity wasted that way? &lt;/p&gt;&#xA;" OwnerUserId="3740" LastEditorUserId="3740" LastEditDate="2015-11-19T14:06:40.007" LastActivityDate="2015-11-20T14:36:58.250" Title="Third Rail vs Overhead Cable: which is more efficient in terms of electricity consumption?" Tags="|electrical-engineering|energy-efficiency|power|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6224" PostTypeId="2" ParentId="6174" CreationDate="2015-11-19T15:24:00.120" Score="7" Body="&lt;p&gt;I don't think what you are looking for really exists, in terms of a standardized list of HVAC/services loadings. Building codes don't usually specify dead loads, probably because they are, in principal, known. Typically only imposed (live) loads are specified. From the buildings I have worked on the services loadings were essentially copied from similar buildings/agreed with the HVAC contractor. There are some generic guidelines available. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.amazon.co.uk/Structural-Engineers-Pocket-Book-Eurocodes/dp/0080971210&quot; rel=&quot;nofollow&quot;&gt;The Structural Engineers Pocketbook&lt;/a&gt; uses 0.15 kPa for ceiling and services&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Arup Structural Scheme Design Guide has 0.25 kPa nominal and 0.4 kPa for HVAC (I only have an old version so don't know if that has changed...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.steelconstruction.info/Design_codes_and_standards&quot; rel=&quot;nofollow&quot;&gt;Steel Construction.info&lt;/a&gt; site (which is a pretty good resource) has 0.25 kPa as being a typical services loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ascelibrary.org/doi/book/10.1061/9780784412916&quot; rel=&quot;nofollow&quot;&gt;ASCE 7-10&lt;/a&gt; in Table C3-1 lists a dead load of 4 psf as a minimum 'mechanical duct allowance'. Which is likely where the 4 psf (0.19 kPa) mentioned in the comments originally came from and is consistent with the above references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.designingbuildings.co.uk/wiki/Structural_systems_for_offices#Standard_allowances_for_a_dead_load&quot; rel=&quot;nofollow&quot;&gt;Designing Buildings Wiki&lt;/a&gt; has 0.85 kPa for Raised floors, ceiling and building services equipment combined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hard to say if these would be applicable to your situation. Also &#xA;you would need to deal with heavy equipment (condensers, fans, heaters, etc) which are not considered in these loadings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HVAC contractors tend to want to put holes all over your structure as well, so it would probably save time and money to consult with them early in the design and sort out what they need.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2015-11-19T17:56:43.650" LastActivityDate="2015-11-19T17:56:43.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6225" PostTypeId="1" CreationDate="2015-11-19T16:03:37.463" Score="1" ViewCount="66" Body="&lt;p&gt;I need a relation between Biological oxygen demand and depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ultimate [BOD] is known, and I need to calculate a [BOD] at any depth of water tank; temperature is constant. The tank is two meters deep and the incubation lasts seven days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimate biological oxygen demand is 0.05 mg/L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need constant-rate equations for deoxygenation and biological oxygen demand.&lt;/p&gt;&#xA;" OwnerUserId="3811" LastEditorUserId="1832" LastEditDate="2015-11-19T22:07:13.787" LastActivityDate="2015-11-19T22:07:13.787" Title="What is the relation between the Biological oxygen demand and the depth of water tank?" Tags="|civil-engineering|environmental-engineering|waste-water-treatment|" AnswerCount="0" CommentCount="3" ClosedDate="2015-11-19T22:28:15.743" ContentLicense="CC BY-SA 3.0" />
  <row Id="6227" PostTypeId="1" CreationDate="2015-11-19T17:16:18.530" Score="2" ViewCount="695" Body="&lt;p&gt;I have used the &lt;em&gt;Simulation Data Inspector&lt;/em&gt; in Simulink to log a dataset of two time varying signals.&#xA;The only option I see concerning exporting is to .mat or to workspace, which basically produce the same type of output:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simulink.sdi.Dataset&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no way of opening this file or dataset, and I need to export it to Excel or to a simple text file with this format:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Time  Var1  Var2&#xA;0  53  34&#xA;1  47  78&#xA;2  67  24&#xA;3  54  87&#xA;4  45  54&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(tab separated values).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to use the logged data in LaTeX for a report, but as of now I am stuck with a 1x1 Dataset in the Workspace which I cannot use or read. &#xA;I have checked the MATLAB manual online, but the export-function they refer to does not work with this simulink.sdi.dataset:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt; export(voutyngve2,'XLSfile','data.xls')&#xA;Undefined function 'export' for input arguments of type 'Simulink.sdi.Dataset'.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can someone please give me a tip?&lt;/p&gt;&#xA;" OwnerUserId="3814" LastActivityDate="2016-08-25T11:36:33.200" Title="Export simulink sdi dataset from MATLAB 2015b" Tags="|matlab|simulink|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6228" PostTypeId="1" AcceptedAnswerId="6229" CreationDate="2015-11-19T19:15:26.153" Score="4" ViewCount="148" Body="&lt;p&gt;I need to spin a threaded rod to raise a picture on a frame. The weight of the picture + the parts that it hangs on will be around 6 pounds. Two motors are going to be used to lift it via threaded rods (like in 3d printer platforms) but I need them to lift about 8&quot; in less than 4 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unfamiliar with the differences between a DC and stepper motor and my worry is that when the picture is coming back down, a DC motor could overrun or would it not have enough power to strip the rod or nut (even over time)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a larger stepper, say like a 128oz/in and an M8x1.25 rod (.31&quot; with 20 threads/in) is it reasonable for me to use steppers or should I go with DC and rely on timing to stop the lowering motion? &lt;/p&gt;&#xA;" OwnerUserId="3817" LastEditorUserId="421" LastEditDate="2015-11-19T19:38:52.070" LastActivityDate="2015-11-19T20:08:04.160" Title="High RPM - Stepper vs Brushless Motor" Tags="|mechanical-engineering|motors|stepper-motor|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6229" PostTypeId="2" ParentId="6228" CreationDate="2015-11-19T20:08:04.160" Score="3" Body="&lt;p&gt;Direct answers to question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Whether a motor could overrun depends on motor power, momentum of the load, and resistance of the system. This could be a problem even with switches, but can be overcome with careful design.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Steppers are overkill if you need to move between two points without stopping between them. See below.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What you need is a linear actuation system, translating rotational motion from a motor into linear motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Threaded rod can be used in principle, but it is not designed for linear actuation. It is designed for static structural applications. Threaded rod has sharp threads which can suffer damage and can wear quickly. The products of wear, and damaged (especially folded-over) threads can cause the actuator to seize during operation. Instead I recommend using something like an acme screw with appropriate nuts. The threads are broader and blunted, and the steels used are typically much harder than those of threaded rod with a smoother surface, all contributing to lower wear rates and less risk of damage, reducing the risk of seizing. Threaded rod also has finer pitch, resulting in less distance per rotation, and is less mechanically efficient than acme, lead, or ball screws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice between a DC brushless motor and a stepper motor should be based on repeatability. A stepper is designed for ensuring that X number of electric power pulses translates into exactly Y distance, every time, within some tolerance, without any feedback mechanism. A brushless motor can not achieve the same repeatability without a tight feedback mechanism. If you don't need repeatability then a brushless motor is cheaper and easier to build working electronic systems for as it doesn't require pulse train generation, and operates with continuous DC power. I am assuming that you intend to raise/lower something between two positions and not stop anywhere in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on momentum, braking is also a concern to ensure you don't overrun the limit switches or damage anything when stopping. You have a 6 pound mass, but you don't give a speed. If the speed is significant, then braking could be an issue. Otherwise, the limit switches could swap the DC motor from the power source to a bank of resistors tied back into the motor: a braking loop. The short-circuit from the motor to itself provides braking, and the resistors dump the energy as heat outside the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have your scenario correct, then a brushless motor directly connected to an acme screw (or similar) with limit switches at each end, with a braking loop as required, is a simple design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proper sizing and quantitative design of each component of the overall system could each be its own question, if you need more information.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2015-11-19T20:08:04.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6230" PostTypeId="1" AcceptedAnswerId="6231" CreationDate="2015-11-19T23:41:08.030" Score="18" ViewCount="52632" Body="&lt;p&gt;A wifi user is in a different room than the router. The computer is having a hard time connecting and receiving the wifi signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the wifi signal from the router to the computer be improved by opening a door to the room where the computer is?&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2018-01-07T10:43:30.377" LastActivityDate="2020-04-03T06:19:51.563" Title="Can wifi signal reception be improved by opening a door?" Tags="|signal|wifi|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6231" PostTypeId="2" ParentId="6230" CreationDate="2015-11-20T00:42:04.807" Score="27" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the wifi signal from the router to the computer be improved by opening a door to the room where the computer is?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe, but probably not to a noticeable degree.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All structures, including doors, impede the wireless signal from the router by some amount.  Generally, the amount of impedance added by the door is a negligible amount and would not be sufficient to noticeably improve the quality of the signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, differing types of door construction have differing impacts on the signal.  A hollow core, wooden door won't impede the signal all that much at 4 dB.  A solid core, wooden door will present more impedance at 6 dB. A steel door would provides the most at 11 dB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a computer right at the edge of receiving a reliable signal and being blocked by a steel door, and there is an otherwise direct line of site to the router except for the door, then it might be possible to improve signal reception by opening the door.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More than likely though, there are other structural elements that are having a more significant impact on the quality of the wireless signal. Especially since it's unlikely that there is a direct line of sight between the computer and the wireless router.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Because we like numbers, I dug a bit further to identify some common sources of attenuation.  Source data is courtesy of &lt;a href=&quot;https://web.archive.org/web/20101204021026/http://www.3com.com/other/pdfs/products/en_US/101900.pdf&quot; rel=&quot;noreferrer&quot;&gt;3COM and the Internet Archive&lt;/a&gt; and was also linked from &lt;a href=&quot;https://web.archive.org/web/20110311182958/http://wireless.navas.us/wiki/Wi-Fi&quot; rel=&quot;noreferrer&quot;&gt;Navas.us and the Internet Archive&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth pointing out that they don't provide any attenuation measurements for differing floor types.  That's likely due to the wide variety of framing and construction materials that can reasonably be seen for floor construction.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Building Material                   2.4 GHz Attenuation&#xA;Solid Wood Door 1.75&quot;                        6 dB&#xA;Hollow Wood Door 1.75&quot;                       4 dB&#xA;Interior Office Door w/Window 1.75&quot;/0.5&quot;     4 dB&#xA;Steel Fire/Exit Door 1.75&quot;                  13 dB&#xA;Steel Fire/Exit Door 2.5&quot;                   19 dB&#xA;Steel Rollup Door 1.5&quot;                      11 dB&#xA;Brick 3.5&quot;                                   6 dB&#xA;Concrete Wall 18&quot;                           18 dB&#xA;Cubical Wall (Fabric) 2.25&quot;                 18 dB&#xA;Exterior Concrete Wall 27&quot;                  53 dB&#xA;Glass Divider 0.5&quot;                          12 dB&#xA;Interior Hollow Wall 4&quot;                      5 dB&#xA;Interior Hollow Wall 6&quot;                      9 dB&#xA;Interior Solid Wall 5&quot;                      14 dB&#xA;Marble 2&quot;                                    6 dB&#xA;Bullet-Proof Glass 1&quot;                       10 dB&#xA;Exterior Double Pane Coated Glass 1&quot;        13 dB&#xA;Exterior Single Pane Window 0.5&quot;             7 dB&#xA;Interior Office Window 1&quot;                    3 dB&#xA;Safety Glass-Wire 0.25&quot;                      3 dB&#xA;Safety Glass-Wire 1.0&quot;                      13 dB&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;n.b. I used Internet archive links as I wasn't able to find current links for either source.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://superuser.com/q/298290&quot;&gt;This SuperUser question&lt;/a&gt; goes into some detail about understanding the power output from wireless routers.  There's quite a bit of variability involved, so I won't attempt to summarize it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dataloggerinc.com/content/resources/white_papers/332/the_basics_of_signal_attenuation/&quot; rel=&quot;noreferrer&quot;&gt;And this vendor link&lt;/a&gt; goes into the calculations involved in trying to more quantitatively identify the effects of signal attenuation.  And beyond an obligatory note that decibels (dB) are not directly additive or subtractive in the mathematical sense, this calculation is also a bit involved so I'm not attempting to summarize it here either.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorUserId="-1" LastEditDate="2017-03-20T10:04:44.983" LastActivityDate="2015-11-23T01:58:51.617" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6232" PostTypeId="1" CreationDate="2015-11-20T03:43:55.593" Score="2" ViewCount="101" Body="&lt;p&gt;The power went out in my block just now, so I got in my car and drove around and found a utility truck next to a pole that is at the junction between the block circuit and an adjoining circuit (which was on). At that area are two large buildings which used to be industrial buildings fed by three phase power, but the buildings were recently converted to be residential. The pole looked fine. It had some stuff running down the side of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The guy was in his cherry picker basket, but only about 10 feet up the pole, well below the wires. He was messing with something for a minute or two, then he made a motion with his hand and there was a blue flash and the whole neighborhood came back on, street lights, houses, everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think he must have caused the outage because I got there in my car in only 2 or 3 minutes after the lights went out, so either it was the fastest power company response time in history, or he deliberately cut off the whole neighborhood for some reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could he have been doing?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="3820" LastEditDate="2015-11-20T09:19:42.567" LastActivityDate="2015-11-20T13:10:22.093" Title="Short power outage, deliberate?" Tags="|electrical-engineering|energy|power|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6233" PostTypeId="2" ParentId="6230" CreationDate="2015-11-20T06:14:45.093" Score="8" Body="&lt;p&gt;This is engineering.  Test it.  It's not like it takes a lot of money or time to do experimentally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as what the theory says, it will always have a slight effect.  The magnitude of this effect depends on the nature of the location, the door, and the wall.  More metallic/heavy objects block the signal, so if it's a weak signal and a bad environment, it could make a difference.&lt;/p&gt;&#xA;" OwnerUserId="3823" LastActivityDate="2015-11-20T06:14:45.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6234" PostTypeId="2" ParentId="6232" CreationDate="2015-11-20T13:10:22.093" Score="1" Body="&lt;p&gt;We can only speculate (so this question may get closed), but maybe something accidentally happened while work was going on, and this caused a fuse to blow.  The arc you saw was power being switched back on after a new fuse was installed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another less likely possibility is that they had to make some changes to the connection and shut off a switch temporarily.  They probably knew it would only take a few minutes and just did it instead of going thru the time consuming process of letting everyone know.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-20T13:10:22.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6235" PostTypeId="2" ParentId="4065" CreationDate="2015-11-20T13:36:40.543" Score="0" Body="&lt;p&gt;As a simplification I would assume that the flow going into the pipe flowed mostly uniformly towards the pipe from all directions, meaning that the flow velocity would drop off rapidly and there would be very little energy loss associated with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, I would assume that the water exiting the pipe would pretty much keep going straight with the same velocity, the kinetic energy in the flow would then be dissipated through viscous and turbulent effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this flow profile would be affected by any objects (such as turbines) placed in the way, and at low flow velocities the jet would diffuse over a significantly shorter distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for a more accurate analysis I think you need to look into jet dynamics, or just use a simulation, though jet simulations are sensitive to choice of turbulent model so make sure to validate your model.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-11-20T13:36:40.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6237" PostTypeId="2" ParentId="6223" CreationDate="2015-11-20T14:08:52.970" Score="4" Body="&lt;p&gt;Your &quot;Bonus question&quot; holds the answer here.  Assuming that the two trains are perfectly identical, and that the power consumption is identical (which, if you have the same weight, motor, and size, it will be), you are down to transmission loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resistance of a conductor is R=L*Rho/A.&#xA;L=length&#xA;Rho=resistivity (material property)&#xA;A=cross sectional area&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that L is the same for both systems, it comes down to cross-sectional area.  The rail is larger, so less power is lost in transferring the power to the train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P=I^2*R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P=power, I=current, R=resistance (from above)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the act of transferring current to the train, power is lost.  And more is lost in the lines/rails than in actually transferring to the train, and because the rails have lower resistance, they waste less power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Typically, when overhead lines are used to power trains, it is due to safety concerns.  If people are around, and could potentially walk on the rails, like an above-ground, street level (&quot;trolley&quot;) train, it makes a lot of sense to not electrocute them when they step on the rails.&lt;/p&gt;&#xA;" OwnerUserId="3823" LastEditorUserId="3823" LastEditDate="2015-11-20T14:36:58.250" LastActivityDate="2015-11-20T14:36:58.250" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6238" PostTypeId="1" AcceptedAnswerId="6266" CreationDate="2015-11-20T15:14:08.613" Score="4" ViewCount="136" Body="&lt;p&gt;I have a rotary pump and want to gain knowledge about its characteristics. I have an uncalibrated flow meter. At best I want to get a flowrate/pressure-diagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand, I need to analyse the flow sensor in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things I have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Voltage and current measurement equipment&lt;/li&gt;&#xA;&lt;li&gt;Counter and/or microcontroller for evaluating rotary sensor pulses&lt;/li&gt;&#xA;&lt;li&gt;oscilloscope&lt;/li&gt;&#xA;&lt;li&gt;vast set of tubing and fittings which can be connected to the pump and the flow sensor.&lt;/li&gt;&#xA;&lt;li&gt;graduated beaker&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Things I don't have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pressure sensors&lt;/li&gt;&#xA;&lt;li&gt;calibrated sensors&lt;/li&gt;&#xA;&lt;li&gt;calibrated pumps.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can I gain pump characteristics without usage of pressure sensors?&lt;/li&gt;&#xA;&lt;li&gt;How can I calibrate the flow sensor without a calibrated source of flow rate?&lt;/li&gt;&#xA;&lt;li&gt;What is a setup for getting the full Q/P-Diagram without?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3786" LastEditorUserId="1586" LastEditDate="2015-11-25T21:37:24.553" LastActivityDate="2015-11-25T21:37:24.553" Title="Method for determining pump characteristics" Tags="|mechanical-engineering|pumps|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6239" PostTypeId="1" AcceptedAnswerId="6241" CreationDate="2015-11-20T16:27:47.437" Score="7" ViewCount="3358" Body="&lt;p&gt;My father-in-law would like to install a hinged stair system in his garage that allows him comfortable access to his attic space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He wants the stairs to swing up flush with the ceiling when not in use, and then to easily swing down when he needs it, and plans on using a set of pulley and weights as a counterbalance to make raising and lowering the stairs easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking of the stairs as a &quot;lever&quot;, where the lever itself is the &quot;load&quot; (i.e. the load is distributed).  If he places the ropes in the middle of the stair system, then obviously the &quot;effort&quot; is in the middle, while the fulcrum will always be at an extreme end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My confusion is, since the load is distributed evenly, is this a second class lever or a third class lever?   Really I don't care what class it is, but it would be nice to understand how to roughly calculate the weight required to lift the stair system.  :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously we can just install it and keep adding weight to see what works in order to solve the problem, but I'd like to rough out some calculations first to see if there are any gotchas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this &lt;a href=&quot;https://engineering.stackexchange.com/questions/2114/how-to-calculate-lever-force-when-lever-has-uniformed-distributed-load&quot;&gt;link&lt;/a&gt; to a related Question, but this situation is just different enough that I can't tell if it solves my problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints on what the equation(s) would look like to calculate the required weight (in pounds) to lift the stair system that weighs, say, $L$ pounds, and the ropes are $d$ distance from the hinge?&lt;/p&gt;&#xA;" OwnerUserId="3830" LastEditorUserId="2246" LastEditDate="2015-11-20T17:03:12.380" LastActivityDate="2020-05-31T12:40:15.490" Title="Calculate effort to raise hinged lever with distributed load" Tags="|applied-mechanics|statics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6241" PostTypeId="2" ParentId="6239" CreationDate="2015-11-20T17:01:36.467" Score="4" Body="&lt;p&gt;The answer for this will depend on the location of where you attach your counterweight rope, but the generic solution is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you have a staircase of weight $W$.  It can be assumed that this weight is centered at the midpoint of the staircase, or $\frac{L}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, suppose you have a countering force of $F$ supplied by your counterweight, at a location of $d$ from the hinge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zgqIh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zgqIh.png&quot; alt=&quot;FBD&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simple balance of moments about the hinge of the staircase, where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{WL}{2} = Fd$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should know everything in the equation except for $F$.  Solving for $F$, you find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = \frac{WL}{2d}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see that the closer you attach the counterweight rope to the hinge, the larger the countering force you will need.  Just think of a teeter-totter on a playground - the heavier kid always wins.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It should be evident, but make sure that you use consistent units when solving this equality (i.e., don't mix feet and inches, etc.).  Also, you will want your counterweight slightly heavier than what's required so that it keeps the staircase closed without any outside effort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for a cheap and effective counterweight, drywall screws with sand filling in between works quite well.  A concrete block works fine too but requires more preparation and mess.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2015-11-20T17:13:34.073" LastActivityDate="2015-11-20T17:13:34.073" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6242" PostTypeId="1" CreationDate="2015-11-20T23:27:02.740" Score="2" ViewCount="5446" Body="&lt;p&gt;I want to construct a freestanding pull-up bar like the one described &lt;a href=&quot;http://samueltoepke.com/?p=33&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I was wondering what would be the best angle for a diagonal support considering that the things I care are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;That it doesn't fall over when doing a muscle-up or swinging from the bar.&lt;/li&gt;&#xA;&lt;li&gt;That it uses the least amount of floor space, i.e. a diagonal support attached at a height of 4 feet at 45 degrees needs a base of 4 feet but a diagonal support at the same height but 30 degrees from the column that holds the bar (or 60 degrees form the base) only needs a base of 2.3 feet. By base, I mean the horizontal beam that closes the triangle between the vertical column and the diagonal support. So I prefer a base of 2.3 feet because it uses less floor space.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So I'm trying to develop an equation with variable angle and height to minimize base length but that it can support the forces for muscle-ups or swinging from the bar. I weight between 130-140 pound and I don't foresee anyone over 200 pound using it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been a long time since I've done any static/dynamic analysis, so I don't really know where to start.&lt;/p&gt;&#xA;" OwnerUserId="3838" LastEditorUserId="1832" LastEditDate="2015-11-21T00:24:14.577" LastActivityDate="2015-11-22T23:50:31.793" Title="Best angle and length for a diagonal wood support for a pull-up bar" Tags="|structural-engineering|structures|statics|dynamics|structural-analysis|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6243" PostTypeId="2" ParentId="6242" CreationDate="2015-11-21T00:33:29.017" Score="1" Body="&lt;p&gt;In any product design, there are lots of variables. The angle (or more specifically base width in this case) determines how eccentric the load can be and will not make a huge difference on the total vertical load as that is limited by vertical 2x4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of product design involves prototyping and measuring that prototype, then using engineering to optimize what you have measured. The trade offs in this application are basically &quot;how stable do you want it to be&quot; vs a smaller foot print. The easiest way to optimize this would be to make a foldable version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the linked design will work well. One thing I would change is to make sure the horizontal supports(where your feet will be) are against the floor; in their current state they are a trip hazard.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-21T00:33:29.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6247" PostTypeId="1" CreationDate="2015-11-21T19:33:09.430" Score="3" ViewCount="3395" Body="&lt;p&gt;When bolting something like a flange or a dead/blind end to a port on a pressure vessel (lets say rated to 2 ATM), is there a good rule of thumb on how much of the threading should be engaged into the tapped hole in the vessel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an ISO standard or something and if so can someone provide a reference to the standard?&lt;/p&gt;&#xA;" OwnerUserId="3848" LastActivityDate="2015-11-23T15:16:07.357" Title="Rule for bolt thread engagement" Tags="|mechanical-engineering|fasteners|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6248" PostTypeId="2" ParentId="6247" CreationDate="2015-11-21T19:42:24.717" Score="3" Body="&lt;p&gt;I have always used the rule of thumb that the bolt has to be engaged at least one bolt diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This partly comes from looking at structural nuts. Structural nuts develop the full strength of the bolt in tension. These nuts typically engage about one bolt diameter.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2015-11-21T19:42:24.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6251" PostTypeId="1" CreationDate="2015-11-22T06:08:40.997" Score="1" ViewCount="173" Body="&lt;p&gt;Does rock expansion play a role in increased recovery of oil when heat is supplied to the reservoir at the hands of steam or hot water ?&lt;/p&gt;&#xA;" OwnerUserId="3854" LastActivityDate="2016-10-30T10:50:23.817" Title="Oil recovery by thermal expansion of rock" Tags="|fluid-mechanics|petroleum-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6252" PostTypeId="2" ParentId="6251" CreationDate="2015-11-22T07:10:17.643" Score="2" Body="&lt;p&gt;Any rock expansion would hinder oil recovery due to a reduction in void space either between grains or in fissures. For oil to flow more readily it needs connected void spaces in which to collect and then flow through. The more tightly packed a rock is the less the ability of oil to flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oil flows well in porous media like sand and sandstone due to the void/pore spaces. Shale doesn't have void spaces like sandstone and getting oil out of shale is very difficult. Previously shale had to be mined, crushed and heated on the surface to extract the oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the advent of shale fracking, shale is fractured in-situ, underground. The introduced voids allow oil to collect and flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heating the oil makes it less viscous which increases its ability to flow more easily. This is why oil is heat via steam underground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To test this, put some butter in a pan and try to make it flow. It won't because it's solid. Heat the butter and see how it flows as more easily as it gets hotter.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-11-22T07:10:17.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6253" PostTypeId="2" ParentId="6242" CreationDate="2015-11-22T07:43:35.510" Score="0" Body="&lt;p&gt;I think @ericnutsch's design analysis is slightly oversimplified for your purpose. Plus, you both omitted a key factor in the successful construction of your design — your &lt;strong&gt;&lt;em&gt;fastening scheme&lt;/em&gt;&lt;/strong&gt;!&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Design Analysis&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;As ericnutsch correctly points out:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Your design is about making tradeoff decisions&lt;/li&gt;&#xA;&lt;li&gt;One tradeoff is: base width vs eccentric load carrying capacity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, I disagree and doubt you will want to go with a foldable version. There are too many barriers to that solution from a design, construction and usage standpoint. I think you can construct a perfectly acceptable and functional static (non-folding) solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His analysis, however, fails to take into account a key design variable you have at your disposal — the &lt;strong&gt;&lt;em&gt;diagonal angle&lt;/em&gt;&lt;/strong&gt; of the support. In other words, how far up the vertical column to connect the diagonal strut. Given that you want to make this in one shot and don't want to make a career out of it, you pretty much want to get this pretty close to correct on the first try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tradeoffs, therefore, are as follows. As the height of the connection point increases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You use more construction material (i.e., wood in this case)&lt;/li&gt;&#xA;&lt;li&gt;Your horizontal stabilization decreases (beyond a certain point of optimization, let's call this the &lt;strong&gt;&lt;em&gt;sweet spot&lt;/em&gt;&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here's where it gets tricky. The sweet spot will vary depending on the amount of eccentricity of the load. In other words, the sweet spot will equate the angle of the truss with the angle of the horizontal component of the load it's carrying. This is obviously true in the case when the horizontal load is zero. Increasing the eccentricity angle will increase the angle of the sweet spot. Which is, as previously mentioned, variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you will have to make your best guess based on how you plan to use the thing as to what the optimal angle is. In this case, I would say the 60 degree angle as you have shown in your design drawings is close enough.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Fastening Scheme&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Simply put. You want the highest fastener tension you can get without compromising the structural integrity of the wood. Therefore, I would do the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Make the largest possible counterbores in the wood.&lt;/li&gt;&#xA;&lt;li&gt;Use washers (that fit the counterbores) to distribute the load normal to the fastened surfaces.&lt;/li&gt;&#xA;&lt;li&gt;Use the longest possible wrench (to give you maximum leverage for torque) when tightening your bolts.&lt;/li&gt;&#xA;&lt;li&gt;Consider trying to find some washers with teeth in them (or leaf spring) to grip the wood so they don't spin radially (about their center axis).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2015-11-22T23:50:31.793" LastActivityDate="2015-11-22T23:50:31.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6254" PostTypeId="1" CreationDate="2015-11-22T12:38:26.527" Score="1" ViewCount="649" Body="&lt;p&gt;Is there a formula, which would allow to estimate water turbine rotor mass based on head and water flow, or based on the final energy output provided by the turbine? I am interested in rough estimate, not a precise formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The range of parameters I am interested in is head from 0 m to 500 m, and flow from $0\;\mathrm{m}^3/\mathrm{s}$ to $1000\;\mathrm{m}^3/\mathrm{s}$. I do not know if there is very significant difference in rotor mass between the Kaplan and the Francis turbine, but ideally I am interested in rotor mass of both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I am interested in scaling of the turbine size in general. (How big would be the turbine of 10 kW power planet? And how big for 100 MW power plant?) Turbine mass is as good parameter to get idea about the scaling as rotor diameter or other turbine size-related parameter that is used in praxis. &lt;/p&gt;&#xA;" OwnerUserId="3858" LastEditorUserId="3858" LastEditDate="2015-11-29T22:01:52.603" LastActivityDate="2015-11-30T18:41:28.947" Title="Calculating water turbine mass based on head and flow" Tags="|turbines|" AnswerCount="2" CommentCount="9" FavoriteCount="0" ClosedDate="2015-11-30T18:55:12.630" ContentLicense="CC BY-SA 3.0" />
  <row Id="6255" PostTypeId="1" CreationDate="2015-11-22T18:00:25.447" Score="1" ViewCount="100" Body="&lt;p&gt;This &lt;a href=&quot;https://www.washingtonpost.com/national/national-security/supercomputers-offer-tools-for-nuclear-testing--and-solving-nuclear-mysteries/2011/10/03/gIQAjnngdM_story.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Washington Post&lt;/a&gt; news article states that with the advent of computer simulation of nuclear tests, live tests are no longer needed.&lt;/p&gt;&#xA;&lt;p&gt;Generally speaking there are 3 aspects of an Explosion resulting from a nuclear weapon test that are tested:&lt;/p&gt;&#xA;&lt;p&gt;(a) &lt;em&gt;effectiveness&lt;/em&gt;, (b) &lt;em&gt;yield&lt;/em&gt;, and (c) &lt;em&gt;explosive capability&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Russia  carried out its last nuclear test in 1990, UK in 1991,the US in 1992,  France and China in 1996, India &amp;amp; Pakistan in 1998.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_nuclear_weapons_tests#Soviet_Union&quot; rel=&quot;nofollow noreferrer&quot;&gt;List Of Nuclear Weapon Tests&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Therefore, does this mean that all aspects of a Nuclear Explosion can be simulated using Super computers or is there any &lt;strong&gt;aspect&lt;/strong&gt; of a &lt;strong&gt;Nuclear Explosion&lt;/strong&gt; that cannot be simulated by Super Computers?&lt;/p&gt;&#xA;" OwnerUserId="3842" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-11-23T20:32:13.843" Title="Is there any aspect of an Explosion resulting from a nuclear weapon test that cannot be simulated using super computers?" Tags="|simulation|computer-aided-design|nuclear-technology|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6256" PostTypeId="1" CreationDate="2015-11-22T18:18:55.980" Score="5" ViewCount="191" Body="&lt;p&gt;I would like to create an interface that increases the frequency out of the wall, 60 hz, to 24khz. I have chosen 24khz as the operating frequency of an induction heater I am planning on building and I would like some advice as to how I should go about doing that. Here is a block diagram of what I am attempting to construct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hTfE0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hTfE0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the power ratio to remain as near to unity as possible through each block so I must also take into account the impedance and heat loss through each step as I design them. If anyone can assist me on how I should design the frequency multiplier circuit, I would be very grateful.&lt;/p&gt;&#xA;" OwnerUserId="3859" LastEditorUserId="3859" LastEditDate="2015-11-23T00:01:03.627" LastActivityDate="2015-11-24T13:40:39.483" Title="Convert 60hz 120Vrms to 24khz 120Vrms" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6257" PostTypeId="1" AcceptedAnswerId="6259" CreationDate="2015-11-23T01:15:17.043" Score="13" ViewCount="4022" Body="&lt;p&gt;I've see the same lids every day but I never really thought about their construction. There is an indent in a &amp;quot;Solo Travel Lid&amp;quot; for coffee cups that is just above the hole that you drink the coffee from.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cHn26.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cHn26.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can see the crescent shaped indent in the image. What is it for? Does it somehow increase fluid flow? Is it just a good spot for your upper lip? If so, why is there not a matching one for the nose? I tried googling this but I got no answer.&lt;/p&gt;&#xA;" OwnerUserId="3806" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-12-01T23:56:56.467" Title="From an engineering standpoint, what is the purpose of the indent in a coffee lid?" Tags="|design|applied-mechanics|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6259" PostTypeId="2" ParentId="6257" CreationDate="2015-11-23T05:50:47.110" Score="25" Body="&lt;p&gt;From the &quot;Solo Traveler&quot; &lt;a href=&quot;http://www.google.com/patents/US4589569&quot; rel=&quot;nofollow noreferrer&quot;&gt;patent&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A more specific object of the present invention is to provide a lid having an opening formed therethrough to enable drinking, and having a recess formed in the lid adjacent the opening to accommodate the upper lip of one drinking from the cup&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is also illustrated in the patent diagrams.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/K5c8M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K5c8M.png&quot; alt=&quot;Patent Image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across this &lt;a href=&quot;http://www.theatlantic.com/technology/archive/2011/05/the-rise-of-the-plastic-disposable-coffee-cup-lid/238573/&quot; rel=&quot;nofollow noreferrer&quot;&gt;interesting article&lt;/a&gt; on the history, use, design, and styles of disposable lids.  The Solo Traveler is prominently featured for its excellent aesthetics and design .  The article reiterates that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the Solo Traveler lid was designed to accommodate the nose and lip of a drinker. In accomplishing this design goal, the necessary height of the lid made it useful for foam-topped gourmet coffees.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additionally, when upright, the recess acts as a reservoir (complete with drainhole) to catch drips/spills/splashes that might otherwise run down the cup and create the infamous coffee stain ring.&lt;/p&gt;&#xA;" OwnerUserId="2538" LastEditorUserId="2538" LastEditDate="2015-11-23T06:40:41.060" LastActivityDate="2015-11-23T06:40:41.060" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6261" PostTypeId="2" ParentId="6256" CreationDate="2015-11-23T12:06:44.073" Score="5" Body="&lt;p&gt;There is no handy frequency multiplier block like you show in your diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the best way to do this is to convert the incoming AC power to DC, then chop that up to drive the work coil.  If the work coil is fully insulated, then it doesn't need to be isolated, which saves having to use a transformer and the associated cost, size, and efficiency loss.  If you do need isolation, then have the chopper drive the transformer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One wrinkle you shouldn't ignore at the power levels you are probably intending is the power factor this unit presents to the AC line.  The AC to DC conversion should include &lt;i&gt;power factor correction&lt;/i&gt;.  This is required in some jurisdictions above a certain power level.  In the US, it's not so much a requirement as that large customers get billed more for low power factors.  In the EU, there are outright regulations for minimum power factor depending on power draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any competent power supply you get at such power levels will do power factor correction anyway.  That industry is highly competitive, and you won't be able to beat buying a off the shelf supply unless you have very unusual requirements or your volumes are very high.  There is a lot of engineering that goes into a good AC to DC power supply, and it's somewhat of a specialty of its own.  It often doesn't make sense to design your own power supply.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Added&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You say you want the output to be a specific frequency.  That doesn't change anything.  You control the oscillator or chopper that makes this frequency from DC.  You apparently want to exploit resonance, which is a reasonable idea.  So make sure that the system resonates at a reasonable frequency, and that the oscillator is tuned to that frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this has anything to do with the 60 Hz power frequency.  That's way too low for resonance with reasonable size parts.  Using individual 60 Hz cycles to &quot;ping&quot; the resonant circuit at 60 or 120 Hz makes no sense.  The Q would have to be very high to sustain 24 kHz ringing for 200 cycles between 120 Hz pings, which isn't practical when you want to take significant power from it.  You also would have to artificially create the 24 kHz content to ring the resonant circuit with anyway.  You might as well do this every cycle with a 24 kHz oscillator, then once or twice each power line cycle.  That concept makes no sense.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2015-11-24T13:40:39.483" LastActivityDate="2015-11-24T13:40:39.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6263" PostTypeId="2" ParentId="6247" CreationDate="2015-11-23T15:16:07.357" Score="8" Body="&lt;p&gt;When dealing with pressure vessels, you should not rely on rules of thumb.  You should rely instead on the &lt;a href=&quot;https://www.asme.org/shop/standards/new-releases/boiler-pressure-vessel-code-2013&quot;&gt;ASME Boiler and Pressure Vessel Code&lt;/a&gt; or whatever code is required by your local enforcement or regulatory body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are far too many variables for a one-size-fits-all answer.  The design pressure/temperature, materials of construction, flange configuration, gasket material, etc all play a role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bolts are not typically used for these types of connections.  Instead you would install a threaded stud in the vessel and then use a washer and nut to hold the flange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section VIII Division 1 UG-43 states that a tapped hole for a stud shall engage the stud for a length not less than the larger of $d_s$ (the nominal stud diameter) or $$0.75d_s \times \Big( \frac {maximum \space allowable \space stress \space value \space of \space stud \space material \space at \space design \space temp}{maximum \space allowable \space stress \space value \space of \space tapped\space material \space at \space design \space temp} \Big)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;except that the thread engagement need not exceed 1.5$d_s$.&lt;/p&gt;&#xA;" OwnerUserId="3440" LastActivityDate="2015-11-23T15:16:07.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6264" PostTypeId="2" ParentId="6255" CreationDate="2015-11-23T16:22:15.360" Score="4" Body="&lt;p&gt;The type of machine that runs the simulation (e.g. super computer, slide rule, pen and paper, etc.) is less important than the physical and mathematical models that describe the process being simulated.  Throwing more computing resources at a simulation whose fundamental equations are flawed will not improve the accuracy of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, there are very likely some aspects of a nuclear explosion that currently cannot be simulated (at least not accurately) with a super computer, or any computer for that matter.  But, the reason is not the lack of computational power, it's a lack of complete understanding of the fundamental physics.  Without periodic testing to verify and improve the models, we don't know how valid the simulation results are.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-11-23T16:22:15.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6265" PostTypeId="2" ParentId="3431" CreationDate="2015-11-23T16:38:06.803" Score="3" Body="&lt;p&gt;After some more thought on this, I'm realizing that the Bond number is more of a measure of the ratio of &lt;em&gt;body&lt;/em&gt; forces to surface tension.  Since I am looking at acceleration by way of &lt;em&gt;surface&lt;/em&gt; forces, I'm not sure the traditional Bond number is applicable here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Mark's answer regarding dimensional analysis still seems valid, so I think the best thing for me to do is keep my modified version of the equation, but not call it the Bond number anymore; just call it the ratio of surface forces to surface tension, if there isn't already a number with that definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some searching, I have found the Euler number (ratio of pressure force to inertial force), and the Weber number (ratio of inertial force to surface tension).  The product of these two numbers effectively gives me what I want (ratio of surface force due to pressure to surface tension):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Eu = \frac{\Delta p}{\rho v^2}&#xA;$$&#xA;$$&#xA;We = \frac{\rho v^2 l}{\sigma}&#xA;$$&#xA;$$&#xA;Eu\cdot We=\frac{\Delta p\cdot l}{\sigma}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2015-11-23T16:52:25.683" LastActivityDate="2015-11-23T16:52:25.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6266" PostTypeId="2" ParentId="6238" CreationDate="2015-11-23T17:07:00.250" Score="2" Body="&lt;p&gt;I'm sure there's a better way, but this is how I would do it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the oscilloscope have a function generator? If so, make it pulse 100 times per second or so, and attach the counter/microcontroller to it's output to provide you with a basic timer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using the scope-counter timer, you can record how long it takes to fill the graduated cylinder. Divide your known volume by the time taken to fill the cylinder to get the flow rate. This value can be used to calibrate your flow meter. Try again at another flow rate to check if it's linear or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now find the Q/P diagram, using the fittings. Connect 'a bunch' of fittings together. Estimate the expected pressure drop across them, using some reference text (e.g. Crane Fluid Co's TP-410, or your favorite fluid flow textbook). Then start the pump and monitor the flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connect a bunch more fittings to the pump discharge, run the pump, noting down the flowrate and estimate the pressure drop. Do this a bunch of times, then plot Q/P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the pressure drop across the fittings is so great that there's no flow, the estimated pressure drop is equal to the pump maximum head.&lt;/p&gt;&#xA;" OwnerUserId="2551" LastActivityDate="2015-11-23T17:07:00.250" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6267" PostTypeId="2" ParentId="6238" CreationDate="2015-11-23T18:06:43.153" Score="3" Body="&lt;p&gt;I am working under the assumption that we are talking about a small pump with a flow rate appropriate to measure using a graduated cylender.  It's a bit tricky with the tools described but I would make a vertical column using your vast set of tubing as a home made pressure sensor.  Also you can use a stop watch and your graduated cylender to calibrate your flow sensor.  Use the cross sectional area and mark the pressure required to push the fluid to various heights.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Run the pump and see how high it can push the column.  This will give the maximum pressure.  Turn off the pump, cut the tubing at a known height and measure flow rate using your freshly calibrated flow sensor.  Calculate your pressure output using head and friction losses at the known flow rate.  Repeat for various heights of column to create the various data points on your Q/P curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large pumps this will be impractical.&lt;/p&gt;&#xA;" OwnerUserId="869" LastActivityDate="2015-11-23T18:06:43.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6268" PostTypeId="1" AcceptedAnswerId="6269" CreationDate="2015-11-23T18:07:37.427" Score="4" ViewCount="438" Body="&lt;p&gt;When you compress a bicycle pump and seal the opening, the pump will automatically return to it's initial position when you release it. I suppose this is how air suspension works as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/durma.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/durma.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you have a internal combustion engine without any gas and turn it over - for instance with a kickstarter, until the piston reaches the very top on the compression cycle - why doesn't it return to the bottom by itself by the air compression alone? Is there anything blocking this, or is the air simply not powerful enough to turn the engine over?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See comments from &lt;a href=&quot;https://physics.stackexchange.com/questions/220153/turning-an-engine-over-with-engine-compression&quot;&gt;Physics StackExchange&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3873" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2015-11-23T21:10:17.390" Title="Turning an engine over with engine compression" Tags="|pressure|energy|compressed-air|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6269" PostTypeId="2" ParentId="6268" CreationDate="2015-11-23T18:26:49.117" Score="4" Body="&lt;p&gt;In theory, it should perform as you say. This is how &lt;a href=&quot;https://en.wikipedia.org/wiki/Aircraft_engine_starting#Hand_starting.2Fpropeller_swinging&quot; rel=&quot;nofollow&quot;&gt;prop starting&lt;/a&gt; an aircraft engine works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An internal combustion engine, unlike your bike pump, is designed for high speed, high power, and high temperatures. These design considerations give it higher internal friction and leakage (valves and cylinder seals) when doing a slow and cold propping operation. This would be a reason why some engines may not respond as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pipercubforum.com/handprop.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.pipercubforum.com/handprop.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorDisplayName="user16" LastEditDate="2015-11-23T18:53:42.913" LastActivityDate="2015-11-23T18:53:42.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6270" PostTypeId="2" ParentId="6255" CreationDate="2015-11-23T20:32:13.843" Score="3" Body="&lt;p&gt;The accuracy of a simulation depends more on how well your equations and assumptions model reality, than the resolution of the simulation (computing power, as Carlton has stated). More computing power does let you have your solution in much less time however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A model, by definition, is not the real thing. No model can be trusted unless it is tested in the real world. This is why we still have wind tunnels even though we extensively use computation fluid dynamics(CFD) for flow modeling. Its just numbers until you have validated the model. Once the model is validated you can extend it or extrapolate it to predict conditions that you can not test. For example testing a small bomb to predict the outcome of a large bomb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case future prediction may very well be restricted to modeling only; leveraging past tests. Nuclear bomb testing today has a way of impeding world peace (oh, we arent building nuclear weapons, we are just testing them)  ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been lots of test in the past to validate the models they are likely using now. So for the United States at least, I wager their models are highly accurate, even including weather and GIS data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just skimmed the article, but they refer to the model as a &quot;tool&quot;, not a replacement for testing. Dont let yourself reverse causation and correlation in your thinking. They dont only model with super computers because &quot;there is no reason to test&quot;. They only model with super computers because there is no way any (&lt;em&gt;sane&lt;/em&gt;) government will allow them to test.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-23T20:32:13.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6271" PostTypeId="2" ParentId="6268" CreationDate="2015-11-23T21:10:17.390" Score="6" Body="&lt;p&gt;There is another reason for this in multi-cylinder engines.  While one cylinder is compressing the air, another is decompressing it.  The net result is that the torque as a function of position gets smoother as the number of cylinders is increased.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-23T21:10:17.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6272" PostTypeId="2" ParentId="3041" CreationDate="2015-11-23T22:43:43.563" Score="5" Body="&lt;p&gt;The bricks themselves should be fine, at least for one cycle, IF they are heated reasonably slowly. Plenty of people have put bricks in barbecues and bonfires. Remember they were already fired in a kiln when they were made, to fuse the clay particles together so that they wouldn't turn into mud as soon as it rained. The only exception is if the bricks are wet. In this case the water inside will turn to steam. Pressure will build up and they will fracture and crumble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would be better off with proper firebricks. These, in addition to being designed for high temperature, have a higher air content / lower density which improves their insulating properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firebricks are best stacked loose, or with the proper cement. An individual brick will not suffer much differential thermal expansion, but a large structure almost certainly will, so I'd expect your cement / mortar to crack pretty quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, I agree with ratchetfreak. Engineering bricks for structure, firebricks for high temperatures and to insulate and protect the structure.&lt;/p&gt;&#xA;" OwnerUserId="1575" LastActivityDate="2015-11-23T22:43:43.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6273" PostTypeId="2" ParentId="6254" CreationDate="2015-11-24T01:48:13.140" Score="2" Body="&lt;p&gt;I am not sure why you would want rotor mass as it much more material dependent than design or power dependent. I have seen kaplan turbines built from both plastic, brass, and steel. Obviously the mass would differ greatly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you are making one out of gold, I imagine that the mass is not end goal. What are you trying to do with this mass value? It is likely going to differ turbine for turbine because mass is something that does not need optimized in this design process. If you do indeed need the mass, the best approach would be to contact manufactures of turbines for the flow range your are interested in and ask for the shipping weight of a replacement rotor.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-24T01:48:13.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6274" PostTypeId="2" ParentId="6221" CreationDate="2015-11-24T02:32:34.827" Score="0" Body="&lt;p&gt;Selection of a mechanism highly depends on the end application. Here are some ways to evaluate a mechanism vs your application:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Input (360 deg rotation in this case)&lt;/li&gt;&#xA;&lt;li&gt;Output (180 deg rotation motion in this case)&lt;/li&gt;&#xA;&lt;li&gt;Relationship between input and output (mechanical advantage, velocity, and acceleration for a given position). (linear, parabolic, sinusoidal, other crazy polar math, etc)&lt;/li&gt;&#xA;&lt;li&gt;Number of moving parts&lt;/li&gt;&#xA;&lt;li&gt;Number of rolling surfaces (rollers, gears)(expensive)&lt;/li&gt;&#xA;&lt;li&gt;Number of sliding surfaces (pins, slots)(reduced life)&lt;/li&gt;&#xA;&lt;li&gt;Number of rotation points (bearing locations)&lt;/li&gt;&#xA;&lt;li&gt;Can the mechanism be statically and dynamically balanced (for high speed)&lt;/li&gt;&#xA;&lt;li&gt;What is the footprint of the mechanism&lt;/li&gt;&#xA;&lt;li&gt;What loads does the mechanism have to endure (material required, fatigue stress, etc)&lt;/li&gt;&#xA;&lt;li&gt;Backlash (new and worn)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You may be able to use &lt;a href=&quot;https://en.wikipedia.org/wiki/Four-bar_linkage#Classification&quot; rel=&quot;nofollow&quot;&gt;crank-rocker four bar linkage&lt;/a&gt;. I haven't sketched it out, but 180 degrees is likely its theoretical maximum. A spring at either end of the range would help avoid top dead center, but power transfer at those high angles would be difficult. In either example you could replace the orange pin connection with a four-bar. That would eliminate one pin and slot giving you a longer service life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four bar linkages are attractive because they employ only low cost rotation connections that have long service lives. Rack and pinion is well suited for when low backlash linear output is required, but can be expensive. Slots and linear bearings are less desirable because of service life, but sometimes reduce the number of components or provide the desired output.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-24T02:32:34.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6275" PostTypeId="1" CreationDate="2015-11-24T05:02:39.487" Score="0" ViewCount="958" Body="&lt;p&gt;I'm wondering just how cold it would have to be before electric power can no longer flow through overhead power lines. I know power lines generate a certain amount of heat just as a side effect from electric current, but I don't know if sucking off this heat will make them fail, or if very cold air will make them fail for some other reason. &lt;strong&gt;I'm looking for the reason and the approximate air temperature where failure will occur.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also transformers and metal pylons to worry about. Presumably wooden telephone poles would be fine, but the big metal pylons in long-distance lines might suffer too much thermal contraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in &lt;strong&gt;overhead power lines&lt;/strong&gt; only, which are exposed directly to air. Transformers and the metal pylons count too. I'm &lt;strong&gt;not&lt;/strong&gt; interested in underground lines. I'm &lt;strong&gt;not&lt;/strong&gt; interested in power plants themselves, which of course have their own vulnerabilities. Just the transmission infrastructure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm not interested in stormy weather either. I don't care about snow or ice accumulating on the cables, or rain short-circuiting something or wind blowing something down. All I care about is the cold ambient air affecting power transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how cold does it have to be before overhead power lines fail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; For all I know there is no lower limit. Very low temperatures might actually be a benefit if they safeguard against overheating.&lt;/p&gt;&#xA;" OwnerUserId="3510" LastEditorUserId="3510" LastEditDate="2015-11-24T09:16:34.707" LastActivityDate="2015-11-24T10:30:53.163" Title="At what cold ambient air temperature will power lines fail?" Tags="|electrical-engineering|temperature|infrastructure|electrical-grid|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6276" PostTypeId="1" CreationDate="2015-11-24T09:10:18.343" Score="4" ViewCount="313" Body="&lt;p&gt;Consider the system with unitary negative feedback such that the open loop transfer function is&#xA;$$G(s) = \frac{as + 1}{s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Determine the value of $a$ such that the phase margin is 45º&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Determine the stationary state error for the unitary ramp input&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) For $a &amp;gt; 0$, what is the gain margin&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt:&lt;/strong&gt; My doubt is about the item (c). That what I've made:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$GM = \frac{1}{|G(j\omega)|}$ for $\omega$ such that $\text{phase}(G(j\omega)) = -180º$. In this case&#xA;$$\text{phase}(G(j\omega)) = -180º \iff \text{arctan}(a \omega) - 180º = -180º \iff \text{arctan}(a \omega) = 0 \iff \omega = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for $\omega \to 0$, $|G(j\omega)| \to \infty \Rightarrow GM \rightarrow 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2929" LastActivityDate="2016-01-26T17:12:45.377" Title="Negative feedback Gain and Phase Margin" Tags="|control-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6278" PostTypeId="2" ParentId="6275" CreationDate="2015-11-24T10:30:53.163" Score="3" Body="&lt;p&gt;Here are some pylons in Norilsk, Northern Siberia, where &lt;a href=&quot;http://www.themoscowtimes.com/business/article/tmt/194983.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;temperatures get down towards -50&amp;deg;C&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Oyr5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Oyr5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(&lt;a href=&quot;http://www.telegraph.co.uk/sponsored/rbth/features/9718598/norilsk-nickel-mine-visit.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's all a question of design and engineering: pylon spacing, the range of max and min ambient temperatures, and the range of currents carried on the cables - there will be whole design manuals that cover this, so a full answer would be way too long for this format. The short answer to your question is &quot;it depends&quot;. Overhead electricity lines do exist in some of the coldest permanently-inhabited places on Earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So low ambient temperatures do not in and of themselves cause power lines to fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Poor engineering, inadequate maintenance, or conditions that are well outside the design range will cause them to fail.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2015-11-24T10:30:53.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6280" PostTypeId="2" ParentId="6251" CreationDate="2015-11-24T17:50:02.303" Score="0" Body="&lt;p&gt;The thermal expansion of a rock grain is severely limited by the overburden stress applied due to the overlaying strata. Since this force is usually quite massive, the amount of expansion per degree of temperature change is not significant. In addition, it takes a huge amount of energy to effectively heat a reservoir rock mass since it is dissipated into the fluid in the pore space, effecting a viscosity change. Thermal expansion of the fluid is a factor however. When the fluid expands it increases in volume and thereby enhances the production pressure gradient.&lt;/p&gt;&#xA;" OwnerUserId="3887" LastActivityDate="2015-11-24T17:50:02.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6281" PostTypeId="2" ParentId="2349" CreationDate="2015-11-24T18:02:50.357" Score="1" Body="&lt;p&gt;That chart is suspect.  If you look at the Web page it was taken from, it is clear that the author of that page does not understand the basic terms that he is using, like &quot;load&quot;, &quot;power&quot;, &quot;torque&quot;, etc.  The chart may be correct if we interpret &quot;load&quot; as &quot;torque&quot;, as it is well known that engine efficiency improves as torque increases and spin decreases (hence, we have &quot;overdrive&quot;).&lt;/p&gt;&#xA;" OwnerUserId="3888" LastActivityDate="2015-11-24T18:02:50.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6283" PostTypeId="1" CreationDate="2015-11-24T22:52:49.917" Score="3" ViewCount="178" Body="&lt;p&gt;I am an electrical engineer that took two semesters of Thermodynamics about 4 decades ago.  I am aware of the theory behind the Carnot cycle and such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in a condo in the Boston area that is all-electric (except for a fireplace I am not using and has its flue closed up).  Recently, I replaced the windowbox A/C with a windowbox A/C and heat pump.  I have been using the heat pump solely to heat the apartment while the outside temps have been above 40 F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other heat is simple resistance heating, which I will call &quot;100% efficient&quot; even though the electricity was generated at a plant somewhere with a heat engine of far less efficiency.  I look at the heat pump as being a way to recover some of the heat lost in that inefficiency because, theoretically, it is better than 100% efficient because the heat coming out of it is the sum of the electric power running it plus whatever energy taken from the outside air.  I also know that this efficiency decreases as the outside temperature falls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now as winter is starting to close in, outside temps will become consistently below freezing.  At what point is it not worth it to run the windowbox heat pump?  Just asking if some HVAC sorta people know a rule of thumb regarding this.&lt;/p&gt;&#xA;" OwnerUserId="3891" LastEditorUserId="33" LastEditDate="2015-11-25T13:29:33.950" LastActivityDate="2015-11-25T13:37:53.533" Title="At what (outside) temperatures is it efficient to run my windowbox heat pump?" Tags="|thermodynamics|hvac|heating-systems|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6284" PostTypeId="1" AcceptedAnswerId="6286" CreationDate="2015-11-25T00:16:19.183" Score="6" ViewCount="1572" Body="&lt;p&gt;I am trying to bore out a red brass pipe from 3/8&quot; to 10mm using my drill press and a standard twist HSS bit.  Running the drill at 600rpm I can't get more than half an inch in before the drill starts binding.  I tried running first with cutting oil, and then, thinking maybe the oil wasn't sticky/viscous enough, tried with Permatex silver, which only got me a few mm further before binding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that I need some sort of metal-cutting bit that has a narrower shank so that there isn't friction except where it's actively cutting.  If so, what are such bits known as?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other practice I might be missing to make this work?&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2015-11-26T08:06:10.587" Title="How to drill larger bore in pipe?" Tags="|mechanical-engineering|machining|drilling|brass|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6285" PostTypeId="2" ParentId="6283" CreationDate="2015-11-25T01:34:05.853" Score="2" Body="&lt;p&gt;Recently I reviewed the technical specifications for a number of heat pumps (reverse cycle air conditioners) with a view to recommending one for purchase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest outside temperature for a heat pump (reverse cycle air conditioner) to operate varies according to the manufacturer and the unit. According to the manufacturer's technical specifications, in heating mode, some units I reviewed were capable of operating at outside temperatures of $-15$ $^oC$ ($5$ $^oF$) and other had their lowest outside temperatures at $-10$ $^oC$ ($14$ $^oF$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lowest outside operating temperature for your unit should be stated in the operating manual, which you should have received. If you don't have one, look at the technical specifications for your unit on the manufacturer's website.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-11-25T01:34:05.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6286" PostTypeId="2" ParentId="6284" CreationDate="2015-11-25T02:07:27.500" Score="6" Body="&lt;p&gt;What you are looking for is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Reamer&quot;&gt;reamer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some 10mm reamers:&lt;BR&gt;&#xA;&lt;a href=&quot;http://www.aliexpress.com/item/Free-shipping-cnc-nc-2015-Reamer-10-0-X30X75L-High-speed-reamer-high-degree-of-finish/32244988192.html&quot;&gt;http://www.aliexpress.com/item/Free-shipping-cnc-nc-2015-Reamer-10-0-X30X75L-High-speed-reamer-high-degree-of-finish/32244988192.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aliexpress.com/item/Square-End-10mm-Cutting-Diameter-6-Flutes-HSS-Hand-Reamer-Milling-Cutter/32240625696.html&quot;&gt;http://www.aliexpress.com/item/Square-End-10mm-Cutting-Diameter-6-Flutes-HSS-Hand-Reamer-Milling-Cutter/32240625696.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aliexpress.com/item/Free-Shipping-New-1-pcs-Solid-Carbide-Rotary-File-Burr-Cylindrical-half-circular-nose-10mm-W1018M06/650695215.html&quot;&gt;http://www.aliexpress.com/item/Free-Shipping-New-1-pcs-Solid-Carbide-Rotary-File-Burr-Cylindrical-half-circular-nose-10mm-W1018M06/650695215.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-11-25T02:07:27.500" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6289" PostTypeId="1" CreationDate="2015-11-25T06:08:05.530" Score="1" ViewCount="624" Body="&lt;p&gt;I have this system below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kNn90.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kNn90.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the schematic of a linear accelerometer moving horizontally, where $m$ is the total mass of the slide, $b$ denotes the viscous damping, and $k$ represents the spring constant. The relative position between the moving mass and the case is measured by a linear variable differential transformer (LVDT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I derive the dynamic equation describing the relationship between $V$, the voltage output of the LVDT, and $x_1$ , the external position, and show that $V$ indeed can be used to measure the external acceleration? I'm supposed to also state any assumptions made and any possible problems and remedies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the mass spring damper system has the equation $mx_1'' + bx_1' + kx = 0$, and the moving core that is pulled by the mass spring damper system will induce a voltage in the LVDT. If I was given a transfer function for the LVDT, $G$, then I have $V = Gx_1$. But I don't know how to link the 2 concepts together to derive the relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the assumptions, I would say the damping force is constant, i.e. the case is smooth. I'm wondering how the mass hitting the stopper will affect the system. Should it be made of rubber to absorb the impact and to minimize the shock wave?&lt;/p&gt;&#xA;" OwnerUserId="3894" LastEditorUserId="421" LastEditDate="2015-11-25T08:11:06.860" LastActivityDate="2015-11-25T08:11:06.860" Title="Deriving relationship between LVDT and mass spring damper" Tags="|mechanical-engineering|sensors|springs|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6290" PostTypeId="2" ParentId="2265" CreationDate="2015-11-25T06:55:11.633" Score="12" Body="&lt;p&gt;The motor tends to be a cheap Synchronous AC motor. The design uses the shift in AC polarity (Going from Positive to Negative phases and back) to create a magnetic field in a coil, which interacts with a multi-poled permanent magnet. As the magnetic polarity shifts in coil, the magnet moves accordingly (opposites attract). Once it is moving, it's easier for the magnetic poles to attract. The permanent magnet is attached to a shaft, which goes through multiple gears to reduce rotations and increase the torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Djk5x.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Djk5x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the middle of the motor is a plate. Underneath it is a coil in plastic bobbin. Now notice the hole marked 1. It has fins. Some come from the bottom of the motor housing, some from the plate that's hiding the coil. That plate will take the magnetic field from the top of the coil and pass it to the fins connected to it. The bottom housing will take the magnetic field from the bottom of the coil and pass it to the fins connected to it. &lt;strong&gt;These alternating fins create the Stators of the synchronous motor.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coil and fins can be seen in this video:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=CzhcJDqQ_h0&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=CzhcJDqQ_h0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;There are two reasons for the motor to change direction.&lt;/em&gt;&lt;/strong&gt; The first, is that the motor is cheap and nothing was added to force it to go in one direction. Typically more expensive motors, one of the gears will have a stop notch, that will prevent it from going backwards. It would stall the motor for one half of the AC phase, then continue the way it should, if it starts the wrong way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more relevant reasons is twofold. One, the fins that make the stators of the motor are not equal size. This is to prevent the motor from getting stuck, moving back and force from equal torque. (If you push a car one way, and then push it back the other way the same exact force and distance, the car will never move from that spot, just rock back and forth gently). Since the permanent magnet can stop between these uneven sized fins, next time it starts, it will get pulled one way or the other. And since the motor could start anywhere on the AC phase, so depending on how the magnet is facing compared to the stator magnetic field, it could be pulled one way or the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;TLDR cheap motor with no directional stopping gear, loose tolerances, uneven fin/stator sizing, and uncertain AC Positive or Negative starting phase leads to the motor randomly starting in either direction&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3895" LastActivityDate="2015-11-25T06:55:11.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6291" PostTypeId="1" CreationDate="2015-11-25T08:25:29.060" Score="2" ViewCount="133" Body="&lt;p&gt;I am trying to do PID control for my servo electronic motor system. This system uses SSI encoder output for feedback mechanism. I will calculate PID parameters according to ziegler nichols tuning methods. However I must know transfer function of this system while determining PID parameters. So how can I extract transfer function of my system? Is there any way by using MATLAB/SIMULINK or Labview? I think I need a methodology for this? I do not know how can I do. Can anyone explain clearly.&lt;/p&gt;&#xA;" OwnerUserId="3896" LastEditorUserId="1586" LastEditDate="2015-11-25T21:37:21.263" LastActivityDate="2015-11-25T21:37:21.263" Title="How can I extract transfer function an unknown nonlinear system?" Tags="|mechanical-engineering|dynamics|transfer-function|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6292" PostTypeId="2" ParentId="6291" CreationDate="2015-11-25T10:29:32.140" Score="4" Body="&lt;p&gt;As I remember (learned it 15 yrs ago) You need to have an experimental data, which will help you to find &quot;linear&quot; sections and the their limits (with taking into account the things like hysteresis), so you will be able to find transfer function for each band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. Why do you think that the servo motor system is nonlinear? Please describe the the mechanism to which motor is connected? What parameter you want to control?&lt;/p&gt;&#xA;" OwnerUserId="3897" LastEditorUserId="3897" LastEditDate="2015-11-25T10:36:54.630" LastActivityDate="2015-11-25T10:36:54.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6293" PostTypeId="1" CreationDate="2015-11-25T11:30:12.900" Score="8" ViewCount="6346" Body="&lt;p&gt;&lt;strong&gt;Why can we neglect the inertial (but not viscous) term in Navier-Stokes at low flow and high viscosity?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete Navier-Stokes: $\rho \frac{D\vec{v}}{Dt}=\rho g - \nabla P+ \mu \nabla ^2 \vec{v}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inertial term:&#xA;$\frac{D\vec{v}}{Dt}=&#xA;\frac{\partial\vec{v}}{\partial t}+&#xA;\frac{\partial\vec{v}}{\partial x}v_x+&#xA;\frac{\partial\vec{v}}{\partial y}v_y+&#xA;\frac{\partial\vec{v}}{\partial z}v_z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as we assume a stationary flow and low rate: $&#xA;\frac{\partial\vec{v}}{\partial t}=0, &#xA;\frac{\partial\vec{v}}{\partial x}\approx0,&#xA;\frac{\partial\vec{v}}{\partial y}\approx0,&#xA;\frac{\partial\vec{v}}{\partial z}\approx0$. And so it follows that the inertial term can be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in my material it is also stated that the $\mu \nabla ^2 \vec{v}$ will be the dominating term during these circumstances. Why will it not be so that $\nabla ^2 \vec{v} \rightarrow&#xA;\frac{\partial^2\vec{v}}{\partial x^2}\approx0,&#xA;\frac{\partial^2\vec{v}}{\partial y^2}\approx0,&#xA;\frac{\partial^2\vec{v}}{\partial z^2}\approx0&#xA;\rightarrow \mu \nabla ^2 \vec{v} \approx 0$ as well?&lt;/p&gt;&#xA;" OwnerUserId="3898" LastEditorUserId="282" LastEditDate="2015-12-08T09:14:12.030" LastActivityDate="2015-12-08T09:14:12.030" Title="Why can we neglect the inertial (but not viscous) term in Navier-Stokes at low flow and high viscosity?" Tags="|fluid-mechanics|chemical-engineering|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6294" PostTypeId="1" CreationDate="2015-11-25T12:03:27.233" Score="-1" ViewCount="227" Body="&lt;p&gt;Usual cameras can capture the visible spectrum, and some others the infra-red spectrum. But the world is so much than that, isn't it? I wonder why cameras that could see a larger electro-magnetic spectrum does not exist, let's say from gamma to radio waves. And I am talking about an image representation, not just about detecting intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine. Select a specific spectrum and the camera could show you what it sees in false colours. That could be very interesting to see heat losses, Wifi or cellular emissions, astronomy etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it is a technical problem, that it would be needed to have lenses and sensors for each sub-spectrum. But I believe it is possible for some ones, and that some telescopes do it. Am I wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what spectrum cannot we build lenses and sensors for the moment? How hard would it be to have only one device that could capture all that?&lt;/p&gt;&#xA;" OwnerUserId="3899" LastEditorUserId="3899" LastEditDate="2015-11-26T15:30:11.123" LastActivityDate="2015-11-26T15:30:11.123" Title="Would it be hard to create a camera that could see the whole electro-magnetic spectrum?" Tags="|sensors|measurements|optics|" AnswerCount="3" CommentCount="2" ClosedDate="2015-11-25T17:22:55.837" ContentLicense="CC BY-SA 3.0" />
  <row Id="6295" PostTypeId="2" ParentId="6294" CreationDate="2015-11-25T12:41:00.660" Score="2" Body="&lt;blockquote&gt;Would it be hard to create a camera that could see the whole electro-magnetic spectrum?&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no one technology that can &quot;see&quot; all wavelengths, and then the concept of &quot;whole spectrum&quot; is totally silly anyway since its bounds are infinite.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-25T12:41:00.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6296" PostTypeId="1" CreationDate="2015-11-25T13:24:58.687" Score="8" ViewCount="2064" Body="&lt;p&gt;When an infrastructure project is complete, the professional team has to prepare drawings representing the infrastructure, as it was completed in the end. All on-site changes and deviations from designs are incorporated in these drawings. This is very important particularly to services which are difficult to locate once they have been covered, like pipelines and under ground power lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past we would mark/label these drawings &quot;&lt;strong&gt;as built&lt;/strong&gt;&quot;. Now we have been informed that we should change that to &quot;&lt;strong&gt;as recorded&lt;/strong&gt;&quot;, because &quot;as built&quot; implies that the drawings are a perfect representation of what in fact has been built. By using the phrase &quot;as recorded&quot;, any liability / responsibility for incorrect drawings becomes unclear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the proper term to use?  And to what extent is the person preparing the &quot;as built&quot; drawings responsible for inaccuracies?&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorDisplayName="user16" LastEditDate="2015-11-25T16:50:16.220" LastActivityDate="2021-07-09T09:53:32.643" Title="As built vs as recorded" Tags="|civil-engineering|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6297" PostTypeId="2" ParentId="6283" CreationDate="2015-11-25T13:37:53.533" Score="1" Body="&lt;p&gt;The operating performance of air-source heat pumps (ASHP) will vary by model: mostly, but not entirely, driven by the choice of working fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put an individual plug-in electricity meter on the ASHP power supply: you might be surprised. There are two things that can bite you. One is that some ASHPs have a built-in resistance heater that they will switch to when the source temperature gets low. The other is that they will periodically run a defrost cycle too, which will be both energy-hungry and noisy (some defrost at 72dB).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll get higher efficiency from your ASHP if you can lower the temperature that it delivers heat to you: turning it down a couple of degrees can make a big difference. But watch out for the dew point temperature in your room: running your room cooler raises the risk of mould, and the consequential risks to health and property.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2015-11-25T13:37:53.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6298" PostTypeId="1" CreationDate="2015-11-25T14:01:22.363" Score="8" ViewCount="782" Body="&lt;p&gt;The &lt;a href=&quot;http://www.weldguru.com/nonferrous-metals.html#tin&quot;&gt;site&lt;/a&gt; states that Tin cannot be welded, but after I made some research, I discovered that tin, when alloyed with copper or used as a protective coating for steel, can be welded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why pure Tin isn't weldable? What are its properties that made it non-weldable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no prior background to welding and is just starting to study the metallic materials that is used in engineering. I also tried to research answers to this question first before I posted it here, but I can't find anything, if possible could you also include a link/book reference so that I can read it.&lt;/p&gt;&#xA;" OwnerUserId="2892" LastEditorUserId="1293" LastEditDate="2015-11-25T17:39:51.347" LastActivityDate="2016-02-04T17:50:38.853" Title="Weldability of Pure Tin" Tags="|materials|metals|welding|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6299" PostTypeId="2" ParentId="6296" CreationDate="2015-11-25T15:55:08.570" Score="6" Body="&lt;p&gt;It's an interesting proposal. Certainly &quot;as builts&quot; are not always accurate to exactly how it was built. I'm not certain that &quot;as recorded&quot; is a great phrase - the drawing &lt;em&gt;is&lt;/em&gt; the record, supposedly of how it was built. Saying &quot;this is a record of how it was recorded&quot; is tautological.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to who has the liability - I actually don't know! I've only been involved in &quot;as built&quot; drawings once, on a project where the contractor &quot;self certificated&quot;, i.e. the contractor was responsible for checking that what he built was right. But my team (the consultants) prepared the &quot;as built&quot; drawings. I &lt;em&gt;believe&lt;/em&gt; (from a vague memory) that we were employed by the contractor to produce &quot;as builts&quot; separate from our main contract with them. Essentially they instructed us exactly what to put on the &quot;as built&quot; drawings. Therefore the responsibility should lie with them, but our team's initials were on the drawings; which would therefore mean we took responsibility for them. But it was the contractor who submitted them to the client, which obviously means he had responsibility to the client for their accuracy. Whether the contractor would then have any comeback on us... I don't know, I'd have to ask a lawyer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, I can't see that relabelling them as &quot;as recorded&quot; would absolve anyone of any legal responsibility.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-11-25T15:55:08.570" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6300" PostTypeId="2" ParentId="6055" CreationDate="2015-11-25T16:09:03.783" Score="3" Body="&lt;p&gt;No this is not possible, as it would distort the part. You may be able to achieve an acceptable view with the following techniques:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Broken Views- These would allow you to cut out the long sections between the flanges allowing one to visualize the whole beam.&lt;/li&gt;&#xA;&lt;li&gt;Hiding lines - with a zoomed out view you can hide the lines that are blending together to only show the outer edges of the w. This allows you to show the entire beam to scale.&lt;/li&gt;&#xA;&lt;li&gt;An angled view- If you turn off perspective and view the beam from the end, then rotate by a small angle you can effectively get a changed aspect ratio, but it will be obvious that it is an angled view.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2015-11-25T16:09:03.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6301" PostTypeId="2" ParentId="6294" CreationDate="2015-11-25T16:14:09.103" Score="6" Body="&lt;p&gt;Look at it this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple compass will detect from zero to about  0.1 Hz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your AM radio detects around a MHz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your cellphone around a few GHz or so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radar systems anywhere from 100kHz  to  1 THz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special materials can handle parts of the range from 10 microns to 0.1 micron (which includes the visible)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special crystals &amp;amp; other things can detect hard X-rays, gamma rays, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image &quot;fusion&quot; is a hot and growing field, which involves combining images from different wavelength ranges to provide more information/discrimination about objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THere's no reason, or possibility, of having a single tool respond to all those wavelength ranges.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2015-11-25T16:14:09.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6302" PostTypeId="2" ParentId="6294" CreationDate="2015-11-25T17:10:25.417" Score="2" Body="&lt;p&gt;First, I like your question. You might ponder what you imply when you want to &quot;see&quot; the common parts of the electromagnetic spectrum (e.g. visible light, radio waves, x-rays etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a camera that can &quot;see&quot;  a broader spectrum, you are asking for a camera that can transform a broad spectrum of wavelengths into visible light (since we can only use visible light to interpret what the camera is picking up). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of this logic: I create a camera that is sensative to visible and infrared radiation. The camera transforms this combined spectrum range into visible light. Problem: if I am seeing the full visible light spectrum, where do I have any unique wavelengths of visible light left to represent infrared wavelengths? Therefore, you have some limitation on overlapping spectrum ranges with visible light. Maybe you'd see greater intensity or brightness in some colors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are other ways to analyze electromagnetic radiation, like in terms of a spectrogram, we can tell which frequencies are more &quot;present&quot; in a signal. But analyzing electramag radiation in this manner for a large spectrum has some harsh limitations: 1) you'd need a lot of different antennas (one size does not fit all) 2) an insanely fast sampling frequency for higher frequency radiation 3) a large bottle of aspirin for the headache you'd get trying to process all that information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it would be possible for a hypothetical biological eyeball to have sensitivity to a larger band of the spectrum. You'd need a brain that could process the information. It is possible, since some creatures like &lt;a href=&quot;http://www.webexhibits.org/causesofcolor/17C.html&quot; rel=&quot;nofollow&quot;&gt;Butterflies&lt;/a&gt; can make good use of ultraviolet radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I think most engineers on this website have less experience designing eyeballs (excluding some ninja bio-engineers if they exist).&lt;/p&gt;&#xA;" OwnerUserId="1174" LastActivityDate="2015-11-25T17:10:25.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6303" PostTypeId="2" ParentId="6296" CreationDate="2015-11-25T19:08:10.613" Score="11" Body="&lt;p&gt;In aerospace and defence projects, &quot;as built&quot; is the used term and the correctness of the as-built data is the responsibility of the whole company (contractor/subcontractor). The as-built data is certificated by the quality department, not by engineers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technical details of as-design, as-built, as-maintained data sets can be found in the Configuration Management literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For civil engineering and architects, here's an excerpt from the American Institute of Architects Best Practices document (BP.10.10.04): &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Risk management advisors recommend that architecture and engineering professionals use the term “record drawings,” saying there is no such thing as an “as-built” drawing or document (Schinnerer, 2002). Because portions of record drawings are based on information provided by outside sources, architects cannot be expected to verify the information (Schinnerer, 2002).&#xA;  Victor O. Schinnerer &amp;amp; Co. recommends including a provision in the professional services contract that acknowledges this understanding. For examples of this language, see Best Practice, “Certifying As-Built or &#xA;  Record Drawings.”&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2015-11-25T19:08:10.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6304" PostTypeId="1" AcceptedAnswerId="6313" CreationDate="2015-11-25T21:15:02.287" Score="3" ViewCount="37" Body="&lt;p&gt;My goal is to have an array of LEDs arranged in such a way that I can light the view of my action cam. My action cam has a horizontal field of view of 175°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first idea was to determine the angle of light emission of a given LED and stack them next to each other, each time being rotated by their angle of light emission around each LEDs vertical axis. So if I have an angle of light emission of 15°, I need equally spaced 175 / 15 = 12 LEDs. To make mounting easier, I could create an element that is shaped in a circular way, so I don't have to calculate each rotation but only watch out for the same distances between the LEDs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I would do the same for the vertical axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the best approach or is there a better approach?&lt;/p&gt;&#xA;" OwnerUserId="3903" LastEditorUserId="3903" LastEditDate="2015-11-25T21:38:37.127" LastActivityDate="2015-11-26T03:47:53.337" Title="Equal illumination of an actioncam's field of view" Tags="|mechanical-engineering|electrical-engineering|design|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6305" PostTypeId="2" ParentId="6293" CreationDate="2015-11-25T22:35:37.847" Score="4" Body="&lt;p&gt;Usually what is implied by low flowrate and high viscosity is that we are dealing with a so-called low Reynolds number flow. The Reynolds number is dimensionless number which is the ratio of inertial forces ($\rho U U$) and viscous force ($\mu U/L$):&#xA;$$\mathrm{Re}=\frac{\rho U U}{\mu U/L}=\frac{\rho U L}{\mu}$$&#xA;For low $\mathrm{Re}$ viscous forces dominate (laminar regime) and for high $\mathrm{Re}$ inertial forces dominate (turbulent regime). Dimensionless numbers like $\mathrm{Re}$ show up naturally through a process known as 'scaling' in which the equations are made non-dimensional; through this process it is then possible to say which terms are negligible based on the values of relevant dimensionless numbers. For more information check my answer to &lt;a href=&quot;https://physics.stackexchange.com/questions/197848/why-is-the-reynolds-number-the-way-it-is-why-is-its-order-the-way-it-is/197852#197852&quot;&gt;this&lt;/a&gt; question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, saying 'low flow and high viscosity' is not sufficient to say we are dealing with a low $\mathrm{Re}$ flow because it also depends on the length scale $L$ (usually a pipe diameter, etc) and the density $\rho$ (of air or water), but it is usually implied that is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now saying for a low flowrate that $\partial_\beta u_\alpha \approx 0$ is incorrect; what you probably mean is that $u_\beta\partial_\beta u_\alpha \ll \mu\partial_\beta^2u_\alpha$. This justifies the simplification of the equations by saying that $u_\beta\partial_\beta u_\alpha\approx 0$ which physically means that inertial terms are completely negligible compared to viscous terms. $u_\beta\partial_\beta u_\alpha\approx 0$ does not imply $\partial_\beta u_\alpha \approx 0$ rather the low flowrate implies $u_\beta\approx0$ while $\partial_\beta u_\alpha$ can be significant. Consider the order-of-magnitude estimation of $\partial_\beta u_\alpha \sim U/L$; for small values of $L$ (contributes to low $\mathrm{Re}$) it may be much larger than order $O(U)$. A similar order-of-magnitude analysis of the viscous terms $\partial_\beta^2 u_\alpha \sim U/L^2$ shows that these will be even more significant. Hence, the reason why the inertial terms are negligible but the viscous terms are not.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2015-11-25T22:35:37.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6307" PostTypeId="1" CreationDate="2015-11-26T02:14:07.830" Score="0" ViewCount="62" Body="&lt;p&gt;Is it theoretically possible and/or economically viable to construct batteries capable of supplying an entire city with electricity for a day or two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the big problems with power plants of any kind is waste motion. The turbines keep churning, the head of steam continues to form, during the off-hours. If we could somehow store all that energy for later use, we would need a lot fewer power plants, and would be able to spare a lot of resources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anyone working on this, or is it just not worth it? Would batteries of this kind do more harm than good?  &lt;/p&gt;&#xA;" OwnerUserId="3740" LastActivityDate="2015-11-26T02:38:30.887" Title="Are super-batteries theoretically possible?" Tags="|battery|" AnswerCount="1" CommentCount="3" ClosedDate="2015-11-26T08:13:34.940" ContentLicense="CC BY-SA 3.0" />
  <row Id="6308" PostTypeId="2" ParentId="6307" CreationDate="2015-11-26T02:19:44.667" Score="1" Body="&lt;p&gt;Yes, it's theoretically possible. But it wouldn't necessarily be any more efficient than the current power plant method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The battery is just the vehicle to store the potential energy. But the energy being stored has to be created in the first place before it can be stored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same reason why electric cars are not necessarily &quot;green&quot; in all cases. Yes, the electric power comes from a charging station. But the charging station has to get that energy from somewhere. And if the power comes from a power plant converting hydrocarbon fuel stores, well then there is no reduction in overall hydrocarbon emissions from using the electric car. The emission points have merely been redistributed.&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2015-11-26T02:38:30.887" LastActivityDate="2015-11-26T02:38:30.887" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6310" PostTypeId="1" CreationDate="2015-11-26T02:41:06.330" Score="1" ViewCount="1082" Body="&lt;p&gt;I want to grind flexible polyurethane foam to a size of about 50-100 microns.&#xA;The feedstock is shredded 1.5 lb/ft^3 polyurethane foam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried a commercial meat grinder, which works but is very slow, and &#xA;I tried a hammermill, which works, especially if it's mixed with ice, but nothing's really been effective.&lt;/p&gt;&#xA;" OwnerUserId="3909" LastEditorUserId="33" LastEditDate="2016-01-25T14:14:01.780" LastActivityDate="2016-12-21T19:20:25.177" Title="What is an effective method of turning polyurethane foam into a powder?" Tags="|grinding|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6311" PostTypeId="2" ParentId="6310" CreationDate="2015-11-26T02:45:23.237" Score="0" Body="&lt;p&gt;Have you considered adjusting the nozzle on a pressure washer to create a coaxial (zero degree) exit angle and using it like a fluid powered knife?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/MATIA-X59ro?t=3m23s&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/MATIA-X59ro?t=3m23s&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2015-11-26T02:57:51.523" LastActivityDate="2015-11-26T02:57:51.523" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6313" PostTypeId="2" ParentId="6304" CreationDate="2015-11-26T03:47:53.337" Score="2" Body="&lt;p&gt;Instead of generating a perfect set of lights, which may be expensive, it should be possible to correct some uneven illumination via software post-processing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution is to generate a static, white-field exposure of the space you wish to image. Essentially, drape the area you wish to image with something as close to perfectly white as possible, and then capture an image of the area with the same illumination setup you will use on whatever you wish to capture. The drape should be white and matte, i.e. not have specular (mirror-like) reflection, but instead have diffuse reflection, like paper or a cloth. Once you've captured the static, white-field image, you have captured all of the uneven-ness of your illumination setup. You can then subtract the uneven-ness from any future images to correct for uneven illumination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A drawback to this method is that it works best with flat objects, as the illumination uneven-ness may change if a 3D object is inserted into the frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mathworks.com/help/images/examples/correcting-nonuniform-illumination.html&quot; rel=&quot;nofollow&quot;&gt;This Matlab tutorial&lt;/a&gt; gives a general idea of one way of correcting uneven illumination. Their method involves determining background illumination via software (which you can also do). Since you have access to your own hardware, you may also capture the background illumination as discussed above, and follow the remainder of the procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, non-uniformities in illumination should still be minimized so as to ensure you capture as much information as possible through a wide-spread histogram distribution. If the illumination non-uniformities dominate the overall image lightness, then you aren't capturing as much lightness information from the objects in the frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photography.SE may also be a useful source of information.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2015-11-26T03:47:53.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6314" PostTypeId="2" ParentId="6284" CreationDate="2015-11-26T08:06:10.587" Score="1" Body="&lt;p&gt;Reamers mentioned in an ericnutsch's answer are good solution (remember that these reamers are designed to increase diameter of existing aperture up to 0,5mm), but may be the better solution is to use senker.&#xA;But seems your problem is not in using wrong reamer type.&#xA;I would suggest you to check the straightness of your bit and drill. May be a drill or a bit have some wobbling due to overheating or overloading in past, also your drill motor must have enough power (rpm is not a major parameter). Lack of power may cause the same effect with using senker or drill bit.&lt;/p&gt;&#xA;" OwnerUserId="3897" LastActivityDate="2015-11-26T08:06:10.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6315" PostTypeId="2" ParentId="6293" CreationDate="2015-11-26T09:28:35.240" Score="2" Body="&lt;p&gt;Since the final term is proportional to $\nabla ^2 \vec{v}$, and not $|\vec{v}|$, the term can be large even if the velocity magnitude is small. Consider the simple case of no-slip laminar flow in an x-oriented pipe. This is a unidirectional flow, so we can discard $v$ and $w$ and focus on $u$. We'll take as given that the flow speed is small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just because $u$ is generally small, however, doesn't mean we can conclude that $\nabla u$ is also generally small. In fact, the narrower the pipe, the larger the magnitude of $\nabla u$. Looking at the gradient field of the horizontal velocity, we see it tends to point inwards, towards the center of the pipe, where the velocity is at a maximum. This means we have an negative divergence, for which the magnitude is dependent on the &lt;em&gt;sharpness of the variation&lt;/em&gt; in the velocity, and not its overall magnitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $\nabla ^2 \vec{v} = \{-|\nabla ^2 u|, 0, 0\}$ is not insignificant, which of course is just Navier Stokes accurately predicting the tendency of viscous forces to cause a pressure drop in steady state, horizontal pipe flow (or you could call it the necessity of a pressure drop to force flow against viscous forces, as you please). Cancel out the gravitational and time varying terms and see for yourself.&lt;/p&gt;&#xA;" OwnerUserId="2578" LastEditorUserId="2578" LastEditDate="2015-11-26T09:41:34.817" LastActivityDate="2015-11-26T09:41:34.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6316" PostTypeId="1" CreationDate="2015-11-26T09:46:20.947" Score="0" ViewCount="134" Body="&lt;p&gt;How does one protect electronics against extreme accelerations/impacts? I mean accelerations of at least 2000g, possibly an order of magnitude more; essentially &quot;payload&quot; of an artillery shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circuitry would consist of some radio receiver, a controller of an electric primer, and some kind of power source to keep both running for a couple minutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Size and weight aren't much of a problem (&amp;lt;1m, &amp;lt;1ton) but the acceleration would be sustained over about 1s of time, not just a momentary impact, also the initial jerk ($\dot{a}$) would be an extreme spike.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2015-11-26T14:41:07.360" Title="How are electronics hardened against extreme accelerations?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6317" PostTypeId="1" CreationDate="2015-11-26T13:51:47.557" Score="7" ViewCount="262" Body="&lt;p&gt;What are the downsides for galvanized steel pipes + press fittings vs. welded steel pipes in a heating application? Diameter would be DN25/1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: The customer specifies only steel, only welded connections for heating pipes starting at this size. The contractor wants to use press fittings, presumably because this way the installation will be easier. The rooms the installation will take place are rather full. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there good technical reasons to avoid galvanized steel pipes + press fittings? Corrosion, longevity, others? &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="1293" LastEditDate="2015-11-26T16:24:43.230" LastActivityDate="2018-06-11T15:23:05.810" Title="Heating circuit: Galvanized steel with press fitting vs welded steel pipes" Tags="|mechanical-engineering|pipelines|heating-systems|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6318" PostTypeId="2" ParentId="6316" CreationDate="2015-11-26T14:41:07.360" Score="1" Body="&lt;p&gt;There are many considerations, some of which are secret or proprietary to those that figured them out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major issue is how to connect the electronic parts and hold them in place.  Individual transistor, resistor, and even ICs are small and light and fully encapsulated in epoxy or some other material.  Keeping these parts in place and properly connected to each other is a tough problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any good direct answers, but I do remember reading about how the first proximity fuses for anti-aircraft shells were developed during WWII.  They made it work back then with vacuum tubes, so it should be possible today with smaller and lighter semiconductors.  If I remember right, one of the breakthroughs was silver in the solder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go read up on the history of proximity fuses in WWII.  It's a interesting bit of history by itself, and you should be able to find pointers to how the electronics was made to survive, which might lead to pointers to more modern research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course one has to wonder why someone would be asking this here.  If you're professionally designing electronics for artillery shells, you should have access to the classified research.  If you don't have access to this information, then it's hard to see a legitimate reason you should be privy to the information.  I wouldn't tell you more even if I did know it.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-26T14:41:07.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6322" PostTypeId="1" CreationDate="2015-11-27T02:50:13.237" Score="1" ViewCount="1194" Body="&lt;p&gt;I am given the rated apparent power, frequency, rated speed, and rated voltage of a 3-phase wye-connected synchronous generator. Also, I know that the generator delivers the rated apparent power at leading power factor. Given the armature resistance Ra, Xd, and Xq, and the power angle, how can I solve for the power factor? Note that I am not given the real power.&lt;/p&gt;&#xA;" OwnerUserId="3922" LastActivityDate="2015-12-01T07:12:44.963" Title="synchronous generator power factor calculation with missing information" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6323" PostTypeId="1" AcceptedAnswerId="6324" CreationDate="2015-11-27T06:01:27.993" Score="6" ViewCount="281" Body="&lt;p&gt;I posted this question on &lt;a href=&quot;https://electronics.stackexchange.com/questions/202839/three-heaters-on-single-pcb&quot;&gt;electronics SE&lt;/a&gt;, but was suggested that I post it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I basically having three heaters (serpentine pattern) on a PCB (fiberglass board), as the following image shows:&lt;a href=&quot;https://i.stack.imgur.com/OiMs1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OiMs1.jpg&quot; alt=&quot;PCB Heater&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ones at the end would be heated to 95°C, the ones next to it at about 50-55°C, while the middle one to about 72°C. They are spaced 0.3mm apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is because the 55°C heater is between heaters with higher temperatures, the heat gets transferred onto that area and goes well beyond 60°C, which I'm not able to control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried removing the area between the heaters to see if it would help somehow, but the problem persists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I do to &quot;confine&quot; the temperature to the heater area? &lt;/p&gt;&#xA;" OwnerUserId="3924" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2015-11-27T08:32:05.943" Title="Having three heaters on single PCB" Tags="|electrical-engineering|materials|heat-transfer|temperature|power|" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6324" PostTypeId="2" ParentId="6323" CreationDate="2015-11-27T07:49:35.023" Score="4" Body="&lt;p&gt;Increasing the insulation between the heat sources there is quite difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, you can sink the heat away:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Add a full copper fill on the backside of the PCB and connect that into a heat sink that is kept cool externally.&lt;/li&gt;&#xA;&lt;li&gt;Add copper rectangles between the separate heating elements.&lt;/li&gt;&#xA;&lt;li&gt;Connect the rectangles to back side with plenty of vias.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This should keep the areas between heaters from heating too much. If you need accurate temperature in the in-between areas also, it may be necessary to adjust the heat sink cooling based on feedback also.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2015-11-27T07:49:35.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6325" PostTypeId="1" CreationDate="2015-11-27T12:46:59.990" Score="1" ViewCount="16259" Body="&lt;p&gt;I've had an experience of having outside air heavily polluted by wildfire smoke, and now I'm reading up on the wildfire happening in Southeast Asia. When the air outside got too polluted to breathe, I covered a plain old fan with a wet cloth and closed all windows. That helped me filter out the air somewhat, but I would like to know if there are more effective measures to take in such situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cdc.gov/features/wildfires/&quot; rel=&quot;nofollow&quot;&gt;The CDC's site&lt;/a&gt; says you should use air conditioner's air filtering function if you can't relocate. Let's say you can't, what can you do to make the air in your home breathable?&lt;/p&gt;&#xA;" OwnerUserId="14" LastEditorUserId="368" LastEditDate="2015-11-30T17:58:39.597" LastActivityDate="2019-01-23T13:02:01.217" Title="How to filter outside air during wildfire to safe breathable level?" Tags="|combustion|air-quality|fluid-filtration|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6326" PostTypeId="1" AcceptedAnswerId="6327" CreationDate="2015-11-27T13:28:46.553" Score="7" ViewCount="1964" Body="&lt;p&gt;This might be a stupid question, but I'm new to mechanics, so please bear with me. Suppose I have a compound gearing system defined in the image below (not to scale). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZAqiM.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZAqiM.jpg&quot; alt=&quot;compound gear system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green gear is the driving gear, and the blue gear is a compound gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have a driving torque of 10kg.cm, and the ratio is 1:6, this means the output torque is 60kg.cm, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At which point is this torque increase? Is it on the outer edge of the smaller gear (A), or larger gear (B)?&lt;/p&gt;&#xA;" OwnerUserId="391" LastActivityDate="2019-03-12T23:52:54.743" Title="Compound gear system ratio" Tags="|mechanical-engineering|gears|applied-mechanics|torque|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6327" PostTypeId="2" ParentId="6326" CreationDate="2015-11-27T14:30:28.853" Score="4" Body="&lt;p&gt;First, kg-cm is not a unit of torque.  On the surface of the earth, your 10 kg-cm would actually be 98 N-cm, or 0.98 Nm.  If these gears are on a satellite in space, then kg-cm for torque is completely nonsensical.  For sake of example, I'll assume you are relating mass to force by gravity here on the surface of the earth.  Please be more careful with units next time.  There is no place in engineering for sloppiness with units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the 0.98 Nm applied to the left gear would be seen as 6 times that, or 5.9 Nm at the right gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This torque exists everywhere on the right gear.  Torque is not position-dependent.  You can think of torque as the position-independent way of expressing a rotational force.  To find the force at a particular point, divide the torque by the radius at that point.  For example, at A the force is (5.9 Nm)/(0.01 m) = 580 N.  At B the force is (5.8 Nm)/(0.06 m) = 98 N.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-27T14:30:28.853" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6328" PostTypeId="2" ParentId="6325" CreationDate="2015-11-27T14:52:14.680" Score="3" Body="&lt;p&gt;There are are three independent problems with air containing &quot;smoke&quot; from a fire:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Particles&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Poisonous or unhealthy combustion gasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Oxygen depletion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you're really close to the fire, #1 is the biggest problem.  As you already found, plain old mechanical filtration is the simplest way to address this.  There are commercial air filters for HVAC systems, but those are usually more designed to filter out dust, not the much smaller smoke particles.  There are filters for that too.  Just about any pleated paper filter will likely be good enough.  Examples of common ones include the air filter for the engine in your car, and the cabin air filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best answer is therefore to go get commercial filters of the right kind and force air thru them somehow.  The &quot;somehow&quot; is the difficult part.  Ordinary window fans and the like are optimized for moving large volumes of air around, but generally can't handle much pressure drop.  The type of filter that will remove smoke particles will cause enough of a pressure drop that a typical &quot;room fan&quot; would probably move very little air thru the filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better alternative, keeping to reasonably available parts, is to use a &quot;exhaust fan&quot;.  These are meant to move air thru a pipe a ways, and are meant to handle a little back pressure.  You would have to rig up your own jig to have such a fan suck in outside air thru a filter, then exhaust it into the house.  The air flow will be low, so you have to be careful to keep the house otherwise shut as tightly as possible.  You want the only air input to the house to come thru the filter.  It will then find its way out thru the inevitable cracks on its own.  You want the flow to be enough so that air only exits thru these cracks, even when the wind is blowing.  That will be hard to do with most houses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single exhaust fan, even if the filter doesn't decrease its air flow much, won't likely be enough for a whole house.  If you have to rig your own filter, it gets even worse.  Paper towels over a window screen will probably work well enough to filter smoke particles, but the performance of such a filter will be poor compared to commercial pleated paper filters designed for the purpose.  You will need much more filter area to compensate for the increased back pressure.  &lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-27T14:52:14.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6329" PostTypeId="1" CreationDate="2015-11-27T15:19:25.197" Score="2" ViewCount="39" Body="&lt;p&gt;We've all read articles on rising sea levels and the impact it will have on coastal cities in the coming decades. There are &lt;a href=&quot;http://ss2.climatecentral.org/#8/42.332/-71.020?show=satellite&amp;amp;projections=0-RCP85-SLR&amp;amp;level=5&amp;amp;unit=feet&amp;amp;pois=hide&quot; rel=&quot;nofollow&quot;&gt;even maps&lt;/a&gt; that illustrate how varying levels of rising seas will impact the dry land in and around these cities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, aside from deep carbon cuts that can help postpone or reverse rising sea level trends, what sorts of engineering proposals could these cities undertake to maintain their current areas of dry land? Whether its constructing massive gates around the shores of the city limits or something else entirely, could we avoid large portions of these urban areas going underwater in the future?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm gonna go ahead and say spare no expense when it comes to strategies.&lt;/p&gt;&#xA;" OwnerUserId="3929" LastActivityDate="2015-11-27T15:19:25.197" Title="In what ways could coastal cities like New York/Boston/Miami mitigate future sea level rises and maintain current dry land area?" Tags="|mechanical-engineering|geotechnical-engineering|environmental-engineering|" AnswerCount="0" CommentCount="3" ClosedDate="2015-11-27T18:34:15.480" ContentLicense="CC BY-SA 3.0" />
  <row Id="6330" PostTypeId="2" ParentId="6322" CreationDate="2015-11-27T15:32:40.497" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zRPwi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zRPwi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Here is a generic approach for finding parameters from given ones in such situations. Starting from the equivalent circuit where $R_a$ is the armature resistance and $jX_{ar}$ is the armature reactance, $V_{T}$ is the rated terminal voltage and $E_f$  is the excitation voltage.&#xA;Corresponding parameters are:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;rated apparent power : &lt;em&gt;Use this to calculate $I_a$&lt;/em&gt;  &lt;/li&gt;&#xA;&lt;li&gt;frequency : &lt;em&gt;see &lt;strong&gt;Xd&lt;/strong&gt; and &lt;strong&gt;Xq&lt;/em&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;rated speed : -  &lt;/li&gt;&#xA;&lt;li&gt;rated voltage : $V_T$  &lt;/li&gt;&#xA;&lt;li&gt;Ra : $R_a$  &lt;/li&gt;&#xA;&lt;li&gt;Xd, and Xq : &lt;em&gt;Use these to calculate armature reactance (depending on the type of generator [salient pole or cylindrical])&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;and the power angle : $\delta$&lt;/li&gt;&#xA;&lt;li&gt;power factor : $\cos \theta$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3930" LastEditorUserId="-1" LastEditDate="2015-11-28T19:29:17.643" LastActivityDate="2015-11-28T19:29:17.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6331" PostTypeId="2" ParentId="6276" CreationDate="2015-11-27T15:42:18.167" Score="1" Body="&lt;p&gt;&lt;strong&gt;a)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two poles at the origin contribute to a phase of -180${}^{\circ}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zero at $-\frac{1}{a}$ contributes a phase ranging from 0 to 90${}^{\circ}$, with a phase 45${}^{\circ}$ at the corner frequency $\frac{1}{a}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus at the corner frequency the phase is -135${}^{\circ}$. The phase margin can be 45${}^{\circ}$ if at this frequency the magnitude is 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left|\frac{\frac{i a}{a}+1}{\left(\frac{i}{a}\right)^2}\right| = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sqrt{2} a^2=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\frac{1}{\sqrt[4]{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;b)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steady-state error for a ramp input is zero because it is a type 2 system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;c)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gain margin is infinite. This is because the Bode phase plot is always above -180${}^{\circ}$ for $a&amp;gt;0$.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2015-11-27T15:42:18.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6333" PostTypeId="1" CreationDate="2015-11-28T01:07:11.483" Score="1" ViewCount="245" Body="&lt;p&gt;In my controls class, we're covering phase and gain margins. We're asked to increase the phase margin of a given plant to greater than 45 degrees using a lag compensator. Initially, the phase margin of the open loop is -13 degrees, as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZqHft.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZqHft.png&quot; alt=&quot;Open loop plant, gain and phase margins&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that a lag compensator &lt;em&gt;decreases&lt;/em&gt; phase angle around a given frequency. This is the filter, before and after I apply it to the plant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r114M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r114M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now I have a phase margin of -45.7 degrees. Does this satisfy the original criteria of greater than 45 degrees phase margin?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the phase margin is a measure of safety, because crossing -180 degrees will effectively turn your negative feedback into positive feedback, making the system unstable. This tells me I should use only positive feedback, because the feedback will always be 180 degrees out of phase. But, it's not clear to me whether I was supposed to increase this phase margin somehow using only lag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, here are the compensated and uncompensated closed loop systems:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ur3dO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ur3dO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it interesting that the maximum amplitudes of either of these functions also roughly correspond to their -270 degree crossings. To calculate these, I'm using MATLAB's &lt;code&gt;feedback()&lt;/code&gt;, which uses negative feedback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clarification on whatever it is I'm doing would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="3934" LastActivityDate="2015-11-28T01:07:11.483" Title="Details of the phase margin" Tags="|control-engineering|control-theory|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6334" PostTypeId="1" CreationDate="2015-11-26T06:34:00.717" Score="1" ViewCount="20" Body="&lt;p&gt;(This is a follow-up to the previous &lt;a href=&quot;https://electronics.stackexchange.com/questions/199140/pcb-heater-design-issues&quot;&gt;question&lt;/a&gt; I posted.) I'm basically trying to control three different tempeartures on a single PCB. It looks like the following:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/P8zpH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P8zpH.jpg&quot; alt=&quot;PCB Heater&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The middle pattern heats up to about 75 deg. Celsius, while the far-right and far-left ones heats up to 95 deg. Celsius. The ones between them are supposed to be controlled to 55 deg. Celsius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I'm facing problems. I think, because of the relatively high temperatures of the surrounding areas, I'm not able to control the 55 deg. heater, and the temperature keeps rising. (I'm using thermocouples to measure the temperature)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a solution, I thought of removing the area between the heaters, so that the heat gets dissipated into the air. However, I'm not entirely sure of that theory, and needed some opinions about that. Also, the design cannot be changed at this point, so the spacing between them is pretty much a constant.&lt;/p&gt;&#xA;" OwnerUserId="3924" OwnerDisplayName="Koushik S" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2015-11-28T01:10:54.820" Title="Three heaters on single PCB" Tags="|temperature|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2015-11-28T14:18:21.610" ContentLicense="CC BY-SA 3.0" />
  <row Id="6335" PostTypeId="2" ParentId="6323" CreationDate="2015-11-26T06:51:28.910" Score="3" Body="&lt;p&gt;Well, air has &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_thermal_conductivities&quot; rel=&quot;nofollow&quot;&gt;worse thermal conductivity than practically any solid&lt;/a&gt;, so chances are that a hole will decrease the heat transmission. There are however solids with thermal conductivity very close to that of air, so if that board was made (on purpose) from one of those... you won't see much improvement for your hassle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since you mentioned fiberglass while I typing this (doesn't quite look a common FR4 I have to say), then you might hope for halving of the conductivity given that Wikipedia give 0.024 Wm&lt;sup&gt;-1&lt;/sup&gt;K&lt;sup&gt;-1&lt;/sup&gt; for air and 0.045 for fiberglass. I suspect with fiberglass though there can be a lot of variation between between various flavors because it's porous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.ewh.ieee.org/soc/cpmt/presentations/cpmt0412b.pdf&quot; rel=&quot;nofollow&quot;&gt;this presentation&lt;/a&gt; thermal conductivity of FR4 is a lot better at 0.25 W/m·K (5x) than that of pure fibeglass given by Wikipedia. (And actually that presentation agrees with &lt;a href=&quot;https://en.wikipedia.org/wiki/FR-4#Properties&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/FR-4#Properties&lt;/a&gt; but on the through-plane value). In-plane value is even greater there at 0.81 W/m·K to 1.059 W/m·K. I'm guessing the epoxy binder in FR4 contributes a lot to that. So on conducted heat alone you could hope for 10x (or even 40x) improvement in thermal isolation a with hole then. But then there's also radiated heat and depending how the board is oriented convection may contribute too. There are some examples/caculations on the web for radiative heat transfer between some simple geometries,  e.g. &lt;a href=&quot;http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node137.html#SECTION07541000000000000000&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; but I couldn't find something for yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is &lt;a href=&quot;https://www.mentor.com/pcb/hyperlynx/thermal/&quot; rel=&quot;nofollow&quot;&gt;software that can do this sort of complex thermal simulation&lt;/a&gt;, alas I don't know any free&amp;amp;good such, and it depends of course on having good material data. Actually, of the freebie stuff &lt;a href=&quot;http://poweresim.com/newhelp/EN/thermal.jsp&quot; rel=&quot;nofollow&quot;&gt;http://poweresim.com/newhelp/EN/thermal.jsp&lt;/a&gt; might help to some extent, but it seems it can only do some chunky blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, one thing you could do to estimate this is to position a termocouple in air at the same distance form the hot traces as the cold[er] traces are. And then measure what this registers. I suspect this would be a resonable approximation for what a hole could hope to achieve (ignoring he self-heating of the cold[er] traces).&lt;/p&gt;&#xA;" OwnerUserId="480" OwnerDisplayName="Respawned Fluff" LastActivityDate="2015-11-26T07:48:30.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6336" PostTypeId="1" CreationDate="2015-11-28T01:43:25.563" Score="2" ViewCount="268" Body="&lt;p&gt;As I was reorganizing my air tool arsenal, a theoretical question popped in my mind I'd like to know the answer to. In designing air flow between an air compressor and a tool (such as a nailer or an impact wrench), when the tool is not taking air from the compressor, is there a mechanism to stop backflow of air from the tool back to the compressor?  Are those mechanisms in the tool, the compressor, or both? Meaning, if the compressor suddenly emptied out, would air leave the tool to fill the compressor or is there a backflow stopper of some sort?&lt;/p&gt;&#xA;" OwnerUserId="2738" LastActivityDate="2015-11-28T19:56:29.307" Title="Air tool to compressor backflow" Tags="|airflow|compressed-air|compressors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6338" PostTypeId="1" CreationDate="2015-11-28T01:52:53.217" Score="6" ViewCount="8234" Body="&lt;p&gt;I am well familiar with the concept of &lt;code&gt;gasket&lt;/code&gt;. From my non-mechanical engineer perspective, I can understand how a standard rubber gasket works in various plumbing and automotive applications (such as O rings etc). However, working on both auto engines and residential plumbing over years I have noticed some fittings forming a sealed metal-to-metal connection without using a rubber gaskets. Some examples are galvanized or copper unions that look like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g471Mm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g471Mm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;In metal-to-metal connections like this that require a high pressure seal, what insures that small cracks in the connection (that a rubber O ring would be conforming to due to its soft texture) do not leak?&lt;/p&gt;&#xA;" OwnerUserId="2738" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-08-28T20:07:57.943" Title="How metal fittings seal connection without rubber gaskets?" Tags="|pressure|airflow|metals|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6340" PostTypeId="2" ParentId="6338" CreationDate="2015-11-28T04:04:15.570" Score="7" Body="&lt;p&gt;The common pipe threads that are used in buildings for water and gas are tapered threads.  The thread is cut on the cone rather than cylinder.  In the US, these threads are called National Pipe Thread Tapered (or NPTT&lt;sup&gt;1&lt;/sup&gt;, or simply NPT).  They don't seal metal-to-metal.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Thread_seal_tape&quot;&gt;Thread seal tape&lt;/a&gt; is wound onto the male thread before it's screwed in.  When the tapered threads bottom-out, they compress the tape, and it fills the gaps.  Unlike a gasket or O-ring, the thread seal tape is good for only one mate.  If the thread needs to be unscrewed, the old tape is cleaned out, and fresh tape is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; There is also National Pipe Thread Straight (NPTS), which uses a flat gasket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; There is a dozen major families of pipe threads.  Some of them have a polymer gasket or O-ring, others don't.  The trade-offs are between cost, ease of cutting thread along with other installation procedures, pressure and aggressiveness of the fluid in the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;3&lt;/sup&gt; There are pipe connection systems that seal metal cone to metal cone.  Swagelok, for example.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2015-11-28T04:04:15.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6342" PostTypeId="2" ParentId="6336" CreationDate="2015-11-28T19:56:29.307" Score="1" Body="&lt;p&gt;I doubt there is a backflow stopper, and I've never personally seen one on such a system.  It doesn't make sense because:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;The tool isn't going to generate higher pressure than what the compressor is feeding it, so there is no danger from a little bit of air going back into the compressor tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;There is very little air stored in a tool, so basically nothing to block against.  Any air in the tool is just a side affect of the tubing and the like having some volume.  The volume of air in the tool is tiny compared to the compressor tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Since there is no problem to solve, any cost spent on such a valve would only make the tool more expensive, larger, and heavier without any benefit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, putting a backflow prevention valve on tools would be silly.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-28T19:56:29.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6343" PostTypeId="1" AcceptedAnswerId="6344" CreationDate="2015-11-28T23:05:42.363" Score="0" ViewCount="254" Body="&lt;p&gt;Since all it really is is a heating element, what happens if I use a 240v/50Hz clothes iron with a 110v/60Hz supply?&lt;/p&gt;&#xA;" OwnerUserId="3942" LastEditorUserId="3942" LastEditDate="2015-11-30T02:12:55.130" LastActivityDate="2015-11-30T02:12:55.130" Title="What happens if I use a 240v clothes iron in USA?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6344" PostTypeId="2" ParentId="6343" CreationDate="2015-11-28T23:13:41.613" Score="7" Body="&lt;p&gt;In general, anything with a heating element will produce only 1/4 the heat on 1/2 the voltage, since power is proportional to voltage squared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, you'll probably be disappointed with the results.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-11-28T23:13:41.613" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6347" PostTypeId="2" ParentId="6326" CreationDate="2015-11-29T15:30:37.837" Score="1" Body="&lt;p&gt;Yes the torque on the driver(green or smaller pinion)would be present on the driven(blue or larger gear) multiplied by the gear ratio i.e. 6 times as much. This torque would be constant over the entire cross section of the driven gear. The rotational force would vary along the cross section inversely with the radius. In contrast the force remains constant across the interface or mesh of the gears( neglecting friction)&lt;/p&gt;&#xA;" OwnerUserId="3946" LastActivityDate="2015-11-29T15:30:37.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6348" PostTypeId="2" ParentId="6338" CreationDate="2015-11-29T19:07:48.590" Score="5" Body="&lt;p&gt;In addition to tapered threads for creating seals, there are Compression Fittings which use a matching male and female mating low angle taper to create a sealing surface. When properly torqued, they rely on elastic deformation to &quot;squish&quot; the mating surfaces together to create a seal. These are re-usable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crush washers are used in a similar fashion to create a seal. Instead of a taper, a softer metallic washer(usually aluminum or copper) is used between the male and female threads. When correctly torqued, the crush washer will plastically deform to create a seal between the fittings. When the connection needs to be serviced, usually the crush washers are replaced or annealed before re-assembly. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2015-11-29T19:07:48.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6349" PostTypeId="2" ParentId="4224" CreationDate="2015-11-29T19:14:55.157" Score="2" Body="&lt;p&gt;Frictional Heat!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simple terms, the pressure drop that is observed is a result of the friction between the moving fluid, and the stationary walls and column packing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This friction reduces the pressure of the fluid, and in turn heats up both the fluid, and the packing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This topic is discussed at length in Moody's treatise on the subject:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moody, L.F. (1944). &quot;Friction Factors for Pipe Flow&quot;. Transactions of the ASME 66 (8): 671–684.&lt;/p&gt;&#xA;" OwnerUserId="2551" LastActivityDate="2015-11-29T19:14:55.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6350" PostTypeId="1" AcceptedAnswerId="6358" CreationDate="2015-11-29T22:19:11.990" Score="3" ViewCount="98" Body="&lt;p&gt;Given this circuit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aQ9eD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aQ9eD.jpg&quot; alt=&quot;Question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to deduce the relationship between the output voltage &lt;span class=&quot;math-container&quot;&gt;$U_A(t)$&lt;/span&gt; an d the input voltage &lt;span class=&quot;math-container&quot;&gt;$U_E(t)$&lt;/span&gt; using Kirchhoff's current law. We're allowed to assume that &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$L $&lt;/span&gt;and &lt;span class=&quot;math-container&quot;&gt;$C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_2$&lt;/span&gt; are given to us.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution so far:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ U_E = U_R + U_{C1} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ U_{C1}=U_L+U_{C2} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$U_A = U_{C2} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \rightarrow  U_E = U_R+U_L+U_A$$&lt;/span&gt;&#xA;At point &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;, i want to use Kirchhoff's current law:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$I_R=I_{C1}+ I_L= I_{C1}+ I_{C2} $$&lt;/span&gt;&#xA;Is my last step correct? How to continue?&lt;/p&gt;&#xA;" OwnerUserId="3952" LastEditorUserId="6767" LastEditDate="2018-12-30T23:32:19.087" LastActivityDate="2018-12-30T23:32:19.087" Title="connectedness between the output voltage and input voltage" Tags="|electrical-engineering|control-engineering|modeling|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6353" PostTypeId="1" CreationDate="2015-11-30T01:12:36.320" Score="4" ViewCount="3231" Body="&lt;p&gt;I am trying to derive the maximum beam slope formula $\theta = \dfrac{wL^3}{24EI}$ for this cantilever beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g6qqc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g6qqc.png&quot; alt=&quot;cantilever beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What method can I use for this derivation? So far I can only find simply supported beams and derivations for the $\theta_{max}$ formula at the free end (e.g., &quot;slope at free end&quot; column of &lt;a href=&quot;http://1.bp.blogspot.com/_aDnUYI_aMy4/TR4fAHUroxI/AAAAAAAAADY/IpSVcm4-mto/s1600/Beam+deflections.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;this table&lt;/a&gt;). I tried doing it by hand, but I cannot seem to get to the correct answer.&lt;/p&gt;&#xA;" OwnerUserId="3956" LastEditorUserId="368" LastEditDate="2015-12-17T23:22:08.673" LastActivityDate="2015-12-17T23:22:08.673" Title="Derivation of beam slope for a cantilever beam" Tags="|mechanical-engineering|beam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6354" PostTypeId="2" ParentId="6353" CreationDate="2015-11-30T02:11:09.450" Score="9" Body="&lt;p&gt;The way to derive the deflection is via the relation $w = EI\dfrac{d^4\delta}{d^4x}$, which means that the deflection is the fourth integral of the loading (and rotation can be taken as the derivative of the deflection and therefore equal to the third integral of the loading).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's begin with the loading:&#xA;$$w = w_{max}\dfrac{x}{L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first integral of the load is the shear force:&#xA;$$Q = \int w = w_{max}\dfrac{x^2}{2L} + C_1$$&#xA;we know that in this case $C_1 = Q(0) = 0$ since it is the free end of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second integral of the load is the bending moment:&#xA;$$M = \int Q = w_{max}\dfrac{x^3}{6L} + C_2$$&#xA;we know that in this case $C_2 = M(0) = 0$ as well since it is the free end of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third integral of the load is the beam's rotation (times its stiffness):&#xA;$$\theta = \dfrac{1}{EI}\int M = \dfrac{1}{EI}\left(w_{max}\dfrac{x^4}{24L} + C_3\right)$$&#xA;we know that in this case $\theta(L) = 0$ since it is the fixed end of the beam, therefore&#xA;\begin{align}&#xA;\theta(L) = 0 &amp;amp;= w_{max}\dfrac{L^3}{24} + C_3 \\&#xA;\therefore C_3 &amp;amp;= -w_{max}\dfrac{L^3}{24} \\&#xA;\therefore \theta &amp;amp;= \dfrac{w_{max}}{EI}\left(\dfrac{x^4}{24L} - \dfrac{L^3}{24}\right) \\&#xA;\therefore \theta(0) &amp;amp;= -\dfrac{w_{max}L^3}{24EI}&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-11-30T02:11:09.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6356" PostTypeId="2" ParentId="6350" CreationDate="2015-11-30T13:19:40.823" Score="1" Body="&lt;p&gt;One approach to solving this problem is to consider each part as a resistor with complex impedance.  Now you can solve as you would if each part was just a resistor.  You do the same series, parallel, or voltage divider arithmetic, except that it will be performed on complex numbers instead of the simple real numbers if the parts were all true resistors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to think of R and C1 as forming a single-pole low pass filter, and L and C2 a two-pole low pass filter.  You can get the Bode plot from inspection and a few seconds with a calculator.  The tricky part with this method will be finding exact values near the rolloff frequencies when the impedances of the two filters interact.  If you mostly need the response some distance below or above the rolloff frequencies, then this approach is simpler.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-11-30T13:19:40.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6358" PostTypeId="2" ParentId="6350" CreationDate="2015-11-30T15:14:53.223" Score="4" Body="&lt;p&gt;You can write two ODEs describing the relationship, and that would be in the time-domain. Alternatively you could describe things in the frequency domain and get the relationship as a transfer function. I will show the latter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Kirchhoff's current law at the two nodes &lt;em&gt;K&lt;/em&gt; and &lt;em&gt;A&lt;/em&gt;, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{U_E-U_K}{R}=\frac{U_K-U_A}{L s}+C_1 s U_K$$&#xA;$$\frac{U_K-U_A}{L s}=C_2 s U_A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two equations can be solved for the two unknowns $U_A$ and $U_K$. We are only interested in the former.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{U_A}{U_E}=\frac{1}{C_1 C_2 L R s^3+C_2 L s^2+C_1 R s+C_2 R s+1}$$&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2015-11-30T15:14:53.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6361" PostTypeId="1" AcceptedAnswerId="6367" CreationDate="2015-11-30T16:06:43.817" Score="12" ViewCount="505" Body="&lt;p&gt;All the electrical vehicles I've ever ride make that humming sound (either cars or trains).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly causes that sound? Is it drag force in the gears? Is it a fan?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it because of the coils (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Coil_noise&quot;&gt;coil noise&lt;/a&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="3914" LastEditorUserId="33" LastEditDate="2016-01-27T02:51:49.217" LastActivityDate="2016-01-27T02:51:49.217" Title="What causes the sound coming from electrical vehicles?" Tags="|mechanical-engineering|electrical-engineering|automotive-engineering|power-electronics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6362" PostTypeId="2" ParentId="6325" CreationDate="2015-11-30T17:35:37.827" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Centrifugal_fan&quot; rel=&quot;nofollow&quot;&gt;Centrifugal fans&lt;/a&gt; and long, cylindrical filters are the ideal situation to filter super small particles at a measurable flow rate.  The idea is simple, you have a long cylinder wrapped with filter material, exposed to the environment.  Then, you pull air from the middle of the cylinder.  Voila!  A lot of filtered air with plenty of surface area for removal.  These are typically called &lt;strong&gt;&lt;em&gt;Candle Filters&lt;/em&gt;&lt;/strong&gt;.  Pictures can be found &lt;a href=&quot;http://www.cecofilters.com/Documents/Organics_Catalog.pdf&quot; rel=&quot;nofollow&quot;&gt;in this catalog&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Careful engineering of the filter elements is required however - you don't want the sides to cave in, or the bottom to open out.  With a good design, the material can be washable, and inserting a spray cleaner into a cylinder is pretty simple.  To accomplish both at the same time, in industrial applications, the filter elements are inserted into a large pressure vessel that blows the fouled gas stream across the filter elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you need some filtration from the store, get an in-room filter with an &lt;a href=&quot;https://en.wikipedia.org/wiki/Electrostatic_precipitator&quot; rel=&quot;nofollow&quot;&gt;Electrostatic Precipitator&lt;/a&gt;.  This is a simple, low volume system that should suit your home needs reasonably.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-11-30T17:35:37.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6363" PostTypeId="2" ParentId="6254" CreationDate="2015-11-30T18:41:28.947" Score="2" Body="&lt;p&gt;For world building purposes, quick interpolation would be sufficient.  Fortunately, I have a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0072871024&quot; rel=&quot;nofollow&quot;&gt;textbook&lt;/a&gt; with some data points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 125 MW (HP/LP) two stage turbine has 26&quot; long blades, with a 14' x 14' exhaust piping while it's 400 MW cousin has 33.5&quot; long blades, with an 18' x 20' exhaust piping.  A 1000 MW (HP/LP/LP) three stage turbine has 43&quot; long blades - exhaust piping is not given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these are sub-critical (not &lt;a href=&quot;https://en.wikipedia.org/wiki/Supercritical_steam_generator&quot; rel=&quot;nofollow&quot;&gt;super-critical&lt;/a&gt;), so the head can be expected to be well under the 3,200 psi range - 2,400 psi is a typical value for most sub-critical systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, without a condenser, a turbine is useless - unless you are talking gas turbines.  For the 125MW, expect a 36'-6&quot; x 24'-4&quot; x 23'-7&quot; condenser with 48&quot; water inlet piping.  400MW is 43'-6&quot; x 38'-0&quot; x 34'-3&quot;, with 78&quot; inlet water piping.  I'd imagine 2 - 500 MW condensers would be used on a 1000 MW turbine, which are 44'-6&quot;, x 41'-0&quot; x 36'-1&quot; each and have 84&quot; piping, with 18'-0&quot; x 20'-0&quot; exhaust openings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While these don't give you mass, these do give you dimensions as you were looking for.  Hopefully this helps.  &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-11-30T18:41:28.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6367" PostTypeId="2" ParentId="6361" CreationDate="2015-12-01T00:49:27.313" Score="9" Body="&lt;p&gt;Correct, coil noise is generated in the windings of the motor because of Pulse Width Modulation (&lt;a href=&quot;https://en.wikipedia.org/wiki/Pulse-width_modulation&quot;&gt;PWM&lt;/a&gt;) driving current. Different frequencies of modulation cause corresponding sound frequencies (may create half or double frequency oscillations as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To change the speed of an electric motor, PWM is required to regulate voltage in DC motors and generate variable frequency wave forms in Variable Frequency Drives (&lt;a href=&quot;https://en.wikipedia.org/wiki/Variable-frequency_drive&quot;&gt;VFD&lt;/a&gt;) for AC motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Brushless_DC_electric_motor&quot;&gt;brushless DC motor&lt;/a&gt; is essentially a 3 phase AC motor coupled with an application specific VFD. These are common in electric remote controlled aircraft and often play tones for a power up self check with nothing but the motor acting as a speaker. There are many examples online of people trying this DIY; for example, &lt;a href=&quot;https://www.youtube.com/watch?v=kyV6eopQBRM&quot;&gt;Brushless motor music (YouTube)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="368" LastEditDate="2015-12-17T23:02:29.853" LastActivityDate="2015-12-17T23:02:29.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6368" PostTypeId="2" ParentId="6325" CreationDate="2015-12-01T01:33:59.953" Score="4" Body="&lt;p&gt;The short answer is that if the air in or around your home is not safe and breathable, you need to &lt;strong&gt;leave the area and possibly seek medical attention&lt;/strong&gt;. You're extremely unlikely to be able to MacGyver your way out of a truly life-threatening air quality event caused by a wildfire (or any comparable industrial accident). Follow the advice laid out in the CDC article you linked. There is no &lt;em&gt;One Weird Trick that Firefighters Hate&lt;/em&gt; method for avoiding relocation/evacuation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you belong to a sensitive group—elderly, asthmatic, pregnant, etc.—there may be shelter space available at a local hospital or other public, professionally-ventilated building during extreme air quality events. Watch the news, listen to the radio, contact your local government office, etc. to find out about this sort of resource.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Understanding short-term ambient smoke exposure&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's reasonable to be concerned about exposure to smoke at levels that are irritating or unpleasant, even if not immediately life-threatening. What most people want to know is, what are the long-term harms of breathing this smoke? How can I stay comfortable in my home while also protecting my health?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.arb.ca.gov/carpa/toolkit/data-to-mes/wildfire-smoke-guide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;California Air Resources Board's wildfire smoke guide for public officials&lt;/a&gt; summarizes the issue this way (emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One concern that may be raised by members of the general public is&#xA;  whether they run an increased risk of cancer or of other chronic&#xA;  health conditions (e.g. heart disease) from short-term exposure to&#xA;  wildfire smoke. People exposed to toxic air pollutants at sufficient&#xA;  concentrations and durations may have slightly increased risks of&#xA;  cancer or of experiencing other chronic health problems. However, &lt;strong&gt;in&#xA;  general, the long-term risks from short-term smoke exposures are quite&#xA;  low.&lt;/strong&gt; Short-term elevated exposures to wildfire carcinogens are also&#xA;  small relative to total lifetime exposures to carcinogens in diesel&#xA;  exhaust and other combustion sources. Epidemiological studies have&#xA;  shown that urban firefighters exposed to smoke over an entire working&#xA;  lifetime have about a threefold increased risk of developing lung&#xA;  cancer (Hansen 1990). This provides some perspective on the magnitude&#xA;  of potential risks from short-term wildfire events.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Short-term&quot; in the context of a wildfire is usually on the order of a few days. Though the fires themselves can burn for much longer—&lt;a href=&quot;https://en.wikipedia.org/wiki/Rim_Fire&quot; rel=&quot;nofollow noreferrer&quot;&gt;one recent wildfire in California burned for over a year&lt;/a&gt;—wind and weather patterns blow the smoke in different directions at different times, so that it's very rare for any particular area to be smoked in for more than a few weeks, unless it's surrounded by or right next to the fire (in which case I would expect the area to have been officially evacuated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being smoked in can be very unpleasant (I know from experience) but unless you have a pre-existing respiratory condition it's unlikely to pose a real danger to the average adult. When the smoke gets so thick that it poses an acute risk to the average person, your only reliable solutions are to leave the area or find shelter in a building that has a well-engineered ventilation system. Local agencies may direct sensitive groups to official shelter areas, often shopping malls, as their systems are already designed to supply comfortable, conditioned air to a great many people. Even if not officially designated as a shelter, an large, air-conditioned commercial building (&lt;em&gt;not&lt;/em&gt; an apartment building) is almost always going to be better-equipped for this purpose than your home.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Keeping smoke out of your home&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you're very concerned about your exposure to smoke and your ability to relocate during extreme air quality events, hire a licensed and bonded HVAC contractor who specializes in high-efficiency central heating and air systems.  You need a tight building envelope and a positive pressure ventilation (PPV) system if your goal is to prevent smoke infiltration, plus better-than-consumer-grade filtration to effectively remove the particulates that do make it into the home (e.g., on your clothing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This will be an expensive retrofit&lt;/strong&gt;, generally more so the older the home is (newer homes are built with generally tighter envelopes). I'm not sure PPV is even feasible in the typical single-family home and, if it is, it will cost an arm and a leg—I mentioned it half as a joke (think &lt;a href=&quot;https://i.stack.imgur.com/YIQHU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;space suits and airlocks&lt;/a&gt;). Realistically, a professional whole-house job is probably not an option, and you might be thinking—well, I'll just do it myself on the cheap. Not so fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/6328/368&quot;&gt;Olin's answer&lt;/a&gt; suggests that sealing your house well enough to make a DIY fan/filter arrangement workable is &quot;hard to do&quot; and while that is correct, it's also a bit of an understatement. You could do more harm than good by blowing in outside air that's not filtered properly—or, for that matter, by backdrafting the exhaust from a gas appliance into your home. Improperly ventilated gas appliances do kill people every year in the U.S.; while it's &lt;a href=&quot;http://www.resnet.us/blog/wp-content/uploads/2014/03/Singer_Rapp_RESNET_CombustionSafety_2014_0226_rev1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;statistically less likely than being hit by lightning&lt;/a&gt;, a bad DIY job in a house with no CO detector is a great way to increase your odds of winning this fabulous prize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you want to do, if this will be a DIY project, is &lt;strong&gt;focus on simply sealing your building envelope&lt;/strong&gt;—check and replace weatherstripping and door sweeps, feel for drafts around windows, doors and vents. You'll still want to hire an HVAC company to come do a home energy audit/weatherization, but this will be relatively inexpensive. They'll check floor registers, bathroom fans, range hoods and many other potential draft sources and seal any they find as best they can—few homeowners have the experience and equipment required to do a competent job at this themselves. The more you prevent infiltration of smoke while your doors and windows are closed, the less demand you place on any consumer-grade indoor filter you buy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, go out and buy &lt;a href=&quot;https://en.wikipedia.org/wiki/HEPA&quot; rel=&quot;nofollow noreferrer&quot;&gt;HEPA&lt;/a&gt; filtration units sufficient to cover your living area. Specifically, you're looking for a device that sits on the floor or on a shelf that recirculates the air in the room through a removable filter. Don't be fooled by &quot;artificial cheese flavor product&quot; style marketing tricks; many cheaper units will say &quot;HEPA&quot; on the box and make claims that sound impressive but will not actually be tested and proven to meet this (or any) specific standard:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;HEPA filters, as defined by the DOE standard adopted by most American&#xA;  industries, remove at least 99.97% of airborne particles 0.3&#xA;  micrometers (µm) in diameter.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The diameter part is important; &lt;a href=&quot;http://www.airnow.gov/index.cfm?action=aqibasics.particle&quot; rel=&quot;nofollow noreferrer&quot;&gt;not all particle pollution is created equal&lt;/a&gt; and there are any number of ways to filter large particles. What would you think if I sold you a filter that was guaranteed to remove 99.97% of airborne particles 100 µm in diameter? You might not realize I'm selling you a cotton shirt, and the particles it removes are the size of sand on a beach. Look for products that have been &lt;a href=&quot;https://www.ashrae.org/File%20Library/docLib/Technology/FAQs2014/TC-02-04-FAQ-68.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;tested and certified to deliver real HEPA filtration (PDF)&lt;/a&gt;. They cost more, because testing isn't free—but cheaper models may or may not protect you at all from &lt;a href=&quot;http://jaha.ahajournals.org/content/4/7/e001653.abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;the smaller particles that we believe cause the most harm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer to the manufacturer's specifications to see how much square footage the unit covers and keep in mind that closed doors, small hallways and stairways all tend to inhibit airflow, so a smaller filter in each living space is often more effective than a larger filter in a central area. Follow the recommended filter replacement schedule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avoid &quot;ionic&quot; type units, &lt;a href=&quot;http://www.arb.ca.gov/research/indoor/o3g-list.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;particularly the ones in this list&lt;/a&gt;; some of them &lt;a href=&quot;http://www.arb.ca.gov/research/indoor/aircleaners/certified.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;at least won't poison you&lt;/a&gt; but I'm not confident they offer any real filtration benefit. In an extreme situation where you've exhausted all other options and can't leave the home, I guess bring all the filter units you have into one room, shut the door and turn them on—but I make no guarantees it will protect your health, especially if you fail to follow the safety recommendations of organizations like the CDC. (At the &lt;a href=&quot;https://en.wikipedia.org/wiki/Survivalism&quot; rel=&quot;nofollow noreferrer&quot;&gt;prepper&lt;/a&gt; level of paranoia, you could go shopping for &lt;a href=&quot;http://www.cdc.gov/niosh/npptl/topics/respirators/disp_part/default.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;an expensive respirator&lt;/a&gt; with a dedicated air supply but this seems a bit like bringing your own personal parachute on an airplane—if you expect to need it, you probably shouldn't make the trip at all.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have central air, there should also be one or more filters on your return air ducting. These are sometimes only designed to trap larger particles but you should learn about your system and the filters it accepts, and perform regular maintenance to keep them in working order. If you heat your home with a fireplace or an older wood stove or you are a smoker, you have much more pressing indoor air quality concerns than ambient wildfire smoke; deal with those first.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2017-09-11T17:36:36.640" LastActivityDate="2017-09-11T17:36:36.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6370" PostTypeId="1" CreationDate="2015-12-01T04:47:53.603" Score="3" ViewCount="662" Body="&lt;p&gt;I would like to know what I am doing wrong for this PID controller design problem. I have all the parameters calculated ($\zeta$,$\omega_n$,$\sigma$...) and I have my target point set up. But now I need to find the zeros for the PID controller. I am have the equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum \text{angles from zeros} - \sum \text{angles from poles} = (2k-1)\times 180 $$&lt;br&gt;&#xA;where k = #poles - # zeros of the open loop system....I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case with the PID controller and the Plant function I have 4 poles and 2 zeros (unknown).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So using geometry I find the angles of all the the open loop transfer function and then I am given the the hint &lt;strong&gt;&quot;either set both zeros at the same location or set one close to the origin and calculate the location of the other&lt;/strong&gt; I chose the latter. So I picked a point close to the origin found its angle and then found the angle of the new zero by subtracting the angle of the poles from the angle of the zero and then subtract that from 180 to get the angle for the new zero. The root locus for this doesn't even come close to my desired point so I assume I am doing this completely wrong.&lt;/p&gt;&#xA;" OwnerUserId="3640" LastEditorUserId="1293" LastEditDate="2015-12-01T06:24:26.883" LastActivityDate="2015-12-01T06:24:26.883" Title="Finding the zeros for a PID controller" Tags="|control-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6371" PostTypeId="2" ParentId="6322" CreationDate="2015-12-01T07:12:44.963" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gUgDI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gUgDI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For salient pole generators, it is merely an extension of Anonymous's answer. The armature current needs only to be broken down into its perpendicular components and applied correspondingly to the respective inductances.&lt;/p&gt;&#xA;" OwnerUserId="3975" LastActivityDate="2015-12-01T07:12:44.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6372" PostTypeId="2" ParentId="6186" CreationDate="2015-12-01T18:04:38.080" Score="2" Body="&lt;p&gt;From a legal liability stand point, you as the design engineer would have to take responsibility for the higher strength concrete.  In this sense, I would delay approval until new calculations are run to establish the higher strength concrete will meet the design criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my concerns, in decreasing order of importance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) The higher strength concrete undoubtedly has a &lt;strong&gt;&lt;em&gt;higher density&lt;/em&gt;&lt;/strong&gt; and therefore more &lt;strong&gt;&lt;em&gt;dead weight&lt;/em&gt;&lt;/strong&gt;, due to cement being 3x the density of water.  This will certainly affect the seismic calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Will this product be &lt;em&gt;capable of meeting 7 ksi, but only rejected if it tests less than 4 ksi?&lt;/em&gt;  In this case, then you don't have 7 ksi concrete, you have 4 ksi concrete with the density of 7 ksi concrete - a real headache for seismic loading.  If they submit 7 ksi concrete, they need to show that they are producing 7 ksi concrete, not just meeting the original specification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Steel work (admittedly not my area of expertise) - will placement of the steel be as consistent in the 7 ksi concrete as in the 4 ksi?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) Due to the increased amount of cement, will there be any additional problems with environmental exposure, etc?  How does the reliability fare?&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-01T18:04:38.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6373" PostTypeId="1" CreationDate="2015-12-01T21:50:53.150" Score="7" ViewCount="4227" Body="&lt;p&gt;I come from a mathematics background and I'm familiar with laplace transforms in-as-far-as they provide an algebraic method to solve ordinary differential equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when I read about transfer functions, I can clearly see the one-to-one correspondence between the poles of a transfer function and the eigenvalues used to solve the homogeneous ODE.  But I'm rather confused as to why the zeros of the transfer function are of particular interest to control engineers.  If we're ultimately interested in stability of the system, why even bother with the zeros of the transfer function?  The poles seem to tell you everything you really need to know.  Does the location of a transfer function's zero with respect to its poles matter in some way?  More importantly, how does their existence (or lack there of) affect how we design a suitable controller?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2015-12-01T22:08:58.937" LastActivityDate="2015-12-08T13:36:32.340" Title="What do the zeros of the transfer function tell you about a system?" Tags="|control-engineering|transfer-function|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6374" PostTypeId="2" ParentId="6373" CreationDate="2015-12-01T22:53:01.067" Score="2" Body="&lt;p&gt;You can think of poles as low pass filters and zeros as high pass filters.  They are in some way two sides of the same coin.  You can't decide stability solely by looking at poles, any more than solely looking at zeros.  For one thing, a zero can be used to offset the affects of a pole, or the other way around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each has the opposite affect on the response to frequency and phase shift.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Added&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I see there is some discussion about the concept of poles can be used to offset the effects of zeros.  As Willpower points out, zeros don't directly cancel poles, and I fixed my statement above to not say so.  However, some combinations of poles and zeros can, over some range, offset the affects of other poles and zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the transfer function of old vinyl records.  The information in the grooves was deliberately high-pass filtered, then the inverse of this filter applied in the playback circuit to ideally get a flat frequency response from original signal to final reproduced signal.  There were various poles and zeros in the system along the way, some unavoidable due to the process, others deliberately introduce, and other poles and zeros applied to offset them.  In this case, the ideal net result was a flat transfer function over the audio range.  For more information on this, look up something called the RIAA equalization curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would get a very wrong idea of the overall system by looking only at the poles and not the zeros, or just the poles and zeros in just the playback amp, for example.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2015-12-02T13:41:34.333" LastActivityDate="2015-12-02T13:41:34.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6375" PostTypeId="1" AcceptedAnswerId="6380" CreationDate="2015-12-01T23:07:27.200" Score="9" ViewCount="6870" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bIFBO.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bIFBO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a cantilever shaft 6inches in length supported on one end using a ball bearing. I am having problems of shaft wobbling in the bearing. I cannot have support on the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I help this situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shaft is made of aluminum, rotating at around 120 rpm. &lt;/p&gt;&#xA;" OwnerUserId="3179" LastEditorUserId="41" LastEditDate="2015-12-03T16:54:31.577" LastActivityDate="2020-05-31T15:36:09.517" Title="How can the wobble of a rotating cantilever shaft be reduced?" Tags="|mechanical-engineering|bearings|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6376" PostTypeId="2" ParentId="6375" CreationDate="2015-12-02T00:05:59.210" Score="3" Body="&lt;p&gt;Your shaft is a beam.  You must have something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X========X========&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the X's are bearings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things you can do to improve the situation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Slower rate of rotation&lt;/li&gt;&#xA;&lt;li&gt;Use stiffer material for the shaft&lt;/li&gt;&#xA;&lt;li&gt;Use bearings that control the orientation of the shaft, not just it's&lt;br&gt;&#xA;location (e.g. don't use bearings embedded in a sphere)&lt;/li&gt;&#xA;&lt;li&gt;Increase the diameter of the shaft&lt;/li&gt;&#xA;&lt;li&gt;Make the stick out distance shorter&lt;/li&gt;&#xA;&lt;li&gt;make the portion of the shaft between the bearings larger in diameter&lt;/li&gt;&#xA;&lt;li&gt;Change the distance between the bearings. Exactly 'to what' depends&#xA;on more information than you've given.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The phenomenon you're dealing with is 'shaft whirl' - you can read about it in a Millwright's book.&lt;/p&gt;&#xA;" OwnerUserId="3909" LastEditorUserId="90" LastEditDate="2015-12-02T01:31:12.823" LastActivityDate="2015-12-02T01:31:12.823" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6377" PostTypeId="2" ParentId="3013" CreationDate="2015-12-02T00:09:58.047" Score="-2" Body="&lt;p&gt;bolt cutters are the normal tool used by self store places and such to remove locks.&lt;/p&gt;&#xA;" OwnerUserId="3909" LastActivityDate="2015-12-02T00:09:58.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6378" PostTypeId="2" ParentId="6373" CreationDate="2015-12-02T00:42:32.970" Score="5" Body="&lt;p&gt;Zeros are &lt;em&gt;very&lt;/em&gt; important when you are looking at how to control a system. A good way to look at this is to examine a &lt;a href=&quot;https://en.wikipedia.org/wiki/Root_locus&quot; rel=&quot;nofollow noreferrer&quot;&gt;root locus&lt;/a&gt; plot. There are several rules on how to sketch a root locus, but the basic idea is to first plot the poles and zeros of the open loop system on the real-imaginary plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take this system as an example (code for matlab):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G = tf([1 -1],[1 2 4]);&#xA;pzplot(G)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can see in this pole zero plot, there is a stable complex conjugate pair of poles and a zero in the right half plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CBYj7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CBYj7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You already know that the poles are the eigenvalues of the system, and become the defining factors of the system's behavior (you can have distinct real, repeated real, or complex conjugate pairs). But what the root locus does is it asks: &quot;What will happen to my system's poles if I try to apply feedback with a certain gain?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the rules of sketching a root locus, you'll get this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JAKUn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JAKUn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this shows is where the poles of the system are when you increase the feedback gain. What do you notice about one of the poles when the gain is 10? The system has an unstable pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moral of the story: when you want to apply feedback control, as you increase the gain, &lt;em&gt;the closed loop poles will start at the open loop poles for low gain and then move towards and terminate at the open loop zeros as gain increases&lt;/em&gt;. Therefore a controls engineer cares deeply about the zeros of a system. This is true for SISO systems, life gets more complicated for MIMO systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please beware of what Olin said about pole-zero cancellation. This is only &quot;true&quot; in mathematic land. &lt;a href=&quot;http://ctms.engin.umich.edu/CTMS/index.php?aux=Extras_PZ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pole-zero cancellation&lt;/a&gt; is not really possible.&lt;/p&gt;&#xA;" OwnerUserId="1174" LastEditorUserId="1174" LastEditDate="2015-12-08T13:36:32.340" LastActivityDate="2015-12-08T13:36:32.340" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6379" PostTypeId="1" AcceptedAnswerId="6389" CreationDate="2015-12-02T00:51:38.413" Score="6" ViewCount="5212" Body="&lt;p&gt;I am designing a box and lid (for a home-made micro computer). Using Autodesk Inventor (Student Version), I've created my box which looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I plan on 3D printing (ABS Plastic) the box and lid in the end.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CdyHb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CdyHb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I am going to create a rectangular lid to encapsulate the contents. I know accuracy will be important to the printing, so I want to make sure I have the sizing right. If it helps you answer my question, the dimensions of the top of the box are 3in by 4in and I want my lid and its outer lip to be around 0.1in thick. I would like the seal to be pretty tight (won't fall off easily) but if it's too small I risk breaking the ABS Plastic when putting on or taking off the lid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, however, is what should the difference in size between the outer rim of the box's top and the inner rim of the lid be?&lt;/p&gt;&#xA;" OwnerUserId="3981" LastEditorUserId="191" LastEditDate="2015-12-02T11:30:21.827" LastActivityDate="2015-12-03T01:19:40.010" Title="What should the size difference be between a lid and a container?" Tags="|design|plastic|3d-printing|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6380" PostTypeId="2" ParentId="6375" CreationDate="2015-12-02T02:04:03.627" Score="14" Body="&lt;p&gt;Rotating shafts are subject to deflections, even without applied (transverse) loads.  To reduce excessive shaft deflection:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Avoid operating speeds near the &lt;a href=&quot;https://en.wikipedia.org/wiki/Critical_speed&quot; rel=&quot;nofollow noreferrer&quot;&gt;critical shaft speed&lt;/a&gt;, where the shaft vibrates at its natural frequencies.&lt;/li&gt;&#xA;&lt;li&gt;Ensure the shaft is concentric, where its center of mass rotates about its geometric center.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assuming the shaft is concentric and does not operate at the critical shaft speed, deflections are caused by applied loads but depend on geometry and material.&lt;/strong&gt;  From your diagram, construct a Free Body Diagram and system of equations to solve for the reaction moment at the bearing(s).&#xA;$$\sum F: \qquad 0 = R_{bearing} - (F_{pulley}+F_{distrib})$$&#xA;$$\sum M_{@bearing}: \qquad 0 = M_{bearing} + F_{pulley}d_{pulley} - F_{distrib}d_{distrib}$$&#xA;&lt;sup&gt;&lt;em&gt;Note that $F_{distrib}$ is the &lt;a href=&quot;https://web.iit.edu/sites/web/files/departments/academic-affairs/Academic%20Resource%20Center/pdfs/Distributed_Loading.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;resultant force of the distributed load&lt;/a&gt;.&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://bendingmomentdiagram.com/free-calculator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;shear and bending moment diagrams&lt;/a&gt;, the bearing(s) must resist the reaction moment, $M_{bearing}$, required by static equilbrium (or the shaft is free to rotate about its transverse axes).  Therefore:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Decrease force by reducing shaft speed&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Select bearing(s) rated for the reaction moment, reaction forces, and operating speeds.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;strong&gt;Modify bearing type or configuration:&lt;/strong&gt;&#xA;  &#xA;  &lt;p&gt;A single ball bearing is not designed to resist bending moments.  Instead, consider using wide needle bearings, double bearings, or (preferably) two separated bearings.  Multiple bearings resolve the reaction moment to reaction forces applied at a distance ($M = Fd$), where greater distance between bearings reduce $R_{bearing}$.  These design options are illustrated below.&#xA;  &lt;img src=&quot;https://machinedesign.com/site-files/machinedesign.com/files/archive/motionsystemdesign.com/images/Ball-screw-end-mounted.jpg&quot; alt=&quot;Bearings&quot;&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From the &lt;a href=&quot;http://www.engineersedge.com/beam_calc_menu.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;deflection equations&lt;/a&gt;, the following design parameters further reduce shaft deflections (note that these parameters are only applicable if the bearing(s) resist the bending moments):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Increase shaft diameter&lt;/li&gt;&#xA;&lt;li&gt;Select a stiffer material&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Used in combination, these parameters are used to design shafts according to design requirements (geometry, material, loads, allowable deflections).&lt;/p&gt;&#xA;" OwnerUserId="2538" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.267" LastActivityDate="2015-12-04T17:58:25.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6381" PostTypeId="1" AcceptedAnswerId="6456" CreationDate="2015-12-02T02:29:42.313" Score="6" ViewCount="1965" Body="&lt;p&gt;In the book, Design of Liquid-Propellant Rocket Engines, there is a practice problem written as such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Determine the design values of $(C_*), (C_f), and  (I_s)_t$ for the engine thrust chambers of the stages of the hypothetical Alpha vehicle, with the following assumed design parameters.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am given the fuel, the fuel mixture ratio, the nozzle stagnation pressure $ (P_c)_n $, the specific heat ratio, and the nozzle expansion ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can calculate $(C_*)$ fairly easy, the problem is calculating $(C_f)$. The book says that you can derive a theoretical vacuum $(C_f)$ using a figure. However when I look at the figure its listing, It does not give me a clear path to derive the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the equations I need to calculate $(C_f)$ or the theoretical vacuum $(C_f)$ ( I have the equation used to convert the theoretical vacuum into the actual thrust coefficient. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the pages I'm talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yCgRJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yCgRJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KEMZg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KEMZg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8Y4vM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Y4vM.jpg&quot; alt=&quot;Fig 1.11 is the reference figure&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig 1.17 is the referenced figure, however it does not show the equation it uses to graph its chart. What would be the equation needed to graph the chart used in fig 1.17&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KAhv0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KAhv0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3976" LastEditorUserId="3976" LastEditDate="2015-12-07T14:03:58.720" LastActivityDate="2015-12-08T15:39:21.617" Title="What is the equation to calculate the Thrust coefficient within a set of parameters?" Tags="|aerospace-engineering|thrust|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6384" PostTypeId="1" AcceptedAnswerId="6387" CreationDate="2015-12-02T09:22:45.243" Score="3" ViewCount="5693" Body="&lt;p&gt;What determines the maximum torque of a DC motor?&#xA;How is this related to the equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=k\omega$$&#xA;$$T=kI$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$E$ is electromagnetic force / voltage (V)&lt;/li&gt;&#xA;&lt;li&gt;$\omega$ is rotational speed (rad/s)&lt;/li&gt;&#xA;&lt;li&gt;$T$ is torque (Nm)&lt;/li&gt;&#xA;&lt;li&gt;$I$ is current (A)&lt;/li&gt;&#xA;&lt;li&gt;$\phi$ is flux (Wb)&lt;/li&gt;&#xA;&lt;li&gt;$k$ is the motor constant (V/(rad/s)=Nm/A) such that $k=K\phi$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3633" LastEditorUserId="368" LastEditDate="2015-12-17T23:14:53.030" LastActivityDate="2015-12-17T23:14:53.030" Title="What determines the maximum torque of a DC motor?" Tags="|electrical-engineering|motors|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6385" PostTypeId="1" AcceptedAnswerId="7207" CreationDate="2015-12-02T10:26:17.377" Score="2" ViewCount="2771" Body="&lt;p&gt;Assume you have a system model that is controlled by a PID controller (feedback). You want to add feedforward to improve the characteristics of the controlled system. How can you design the transfer function of the feedforward?&#xA;(Cf. &lt;a href=&quot;http://engineeronadisk.com/book_modeling/images/feedback4.gif&quot; rel=&quot;nofollow&quot;&gt;this image&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On slide 18 of &lt;a href=&quot;http://research.che.tamu.edu/groups/Seminario/ProcessControl-Fall2012/Handouts/Chap_15_Marlin_2002.pdf&quot; rel=&quot;nofollow&quot;&gt;this presentation&lt;/a&gt;, it is suggested to use a gain and a lead-lag filter:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;K*(T1*s+1)/(T2*s+1).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3633" LastEditorUserId="3633" LastEditDate="2015-12-03T09:49:44.113" LastActivityDate="2018-05-24T19:52:22.260" Title="How to improve a PID control loop by adding feedforward?" Tags="|control-engineering|feedback-loop|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6386" PostTypeId="1" CreationDate="2015-12-02T12:08:37.587" Score="4" ViewCount="76" Body="&lt;p&gt;Assume that you have an axial fan placed in open air with no duct. How can I go about calculating the air speed at a distance x from the fan?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fan is about 1 m above the ground in an open space. I want to use the fan to generate power from a wind turbine, so I need to know what the air speed will be at the turbine.&lt;/p&gt;&#xA;" OwnerUserId="3989" LastEditorUserId="368" LastEditDate="2015-12-17T23:17:00.317" LastActivityDate="2015-12-17T23:17:00.317" Title="How to size a fan for open air application?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6387" PostTypeId="2" ParentId="6384" CreationDate="2015-12-02T12:18:31.627" Score="3" Body="&lt;p&gt;Torque of a motor is directly proportional to current.  Therefore, the maximum torque a motor can produce is limited by the maximum current it can handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum steady state current a motor can handle is limited by the heat it can safely dissipate, which is proportional to the square of the current.  This means you have to be careful with that spec.  Just 41% more current causes twice the heating, and twice internal temperature rise above ambient, which is likely to cause damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to exceed the maximum allowed steady current for short periods of time.  Some motors may actually come with such specifications.  To really push a motor to its limits without damage requires tracking the assumed temperature.  You usually model the temperature as slowly decaying to ambient with a time constant you usually have to guess at or measure.  This could be multiple minutes.  You can get some idea of this by knowing the maximum allowed steady state current and making a assumption about internal temperature above ambient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of system lets you do some short pulses of somewhat higher current, which you then pay for in a lower allowed maximum current for some time afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally you don't want to exceed the maximum allowed steady current by 2x, especially for a brushed motor where there are other limits on the current.  Even for a brushless DC motor with good temperature modeling, I'd consider 3-4x a upper limit on short bursts of current intended for extra torque.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-12-02T12:18:31.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6388" PostTypeId="1" AcceptedAnswerId="6390" CreationDate="2015-12-02T15:33:39.007" Score="9" ViewCount="204" Body="&lt;p&gt;I am looking to test the &lt;a href=&quot;https://www.tekscan.com/products-solutions/force-sensors/a401&quot; rel=&quot;nofollow noreferrer&quot;&gt;FlexiForce&lt;/a&gt; sensor in measure force, however for accurate readings the load needs to be disturbed evenly over the sensing area. As such I am having &quot;discs&quot; made to cover the sensing area. For my application the thinner the material of the disc the better (1 mm or less in thickness). Which of the following materials will be most suitable to manufacture the disc for the stiffest/strongest possible part?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Brass &amp;amp; Copper&lt;/li&gt;&#xA;&lt;li&gt;Aluminium&lt;/li&gt;&#xA;&lt;li&gt;Stainless Steel&lt;/li&gt;&#xA;&lt;li&gt;Mild/Carbon Steel&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The disc will be 25.4 mm in diameter with a thickness of 1 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an illustration of what I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/83hze.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/83hze.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3270" LastEditorUserId="368" LastEditDate="2015-12-17T23:18:49.047" LastActivityDate="2019-07-07T18:25:40.863" Title="Material selection for weight scale" Tags="|materials|metals|manufacturing-engineering|deformation|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6389" PostTypeId="2" ParentId="6379" CreationDate="2015-12-02T16:09:31.107" Score="6" Body="&lt;p&gt;To make the lid fit on reliably with as tight as possible, the only difference you need in the box and lid dimensions is the resolution of the 3-D printer. You need to account for manufacturing inaccuracies and inconsistencies. Rapid prototyping is very reliable and not subject to the same issues as machining like tool wear and fixturing issues, but no part will ever be 100% to the design. This sort of information should be available from whoever is printing the parts for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't want a full perimeter interference fit, as the corners of the box will be pretty stiff and won't compress easily, and a full perimeter interference isn't necessary to lock it on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you should look at some method of securing the lid besides it being a snug fit. There are a couple simple ways to do this, especially when 3-D printing the parts. Some kind of &lt;a href=&quot;https://en.wikipedia.org/wiki/Snap-fit&quot; rel=&quot;noreferrer&quot;&gt;snap fit&lt;/a&gt; will prevent you from adding extra parts to the design, but will add extra stress to the printed parts. This shouldn't be a big deal, because you don't need a lot of interference, but you may want to give yourself some extra strength and bump up to 1/4&quot; thick walls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is adding holes to the corners of the case to allow the lid to be screwed on. If you make the holes about the size of the shank of the screw, they should automatically thread the hole, and you won't need to tighten it too much to keep the top closed if it's just sitting on your desk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get even more complex, you could probably even make a hinged top, buy a standard door hinge at your nearest hardware store, and use a snap fit on the other side to lock it down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Play around with some of these ideas in Inventor and see what looks good. 3-D printing is a powerful tool that allows you to create some complex geometry, and while a couple of rectangles is really easy, if you experiment with the CAD software to see what you can design, the 3-D printer can probably do a lot of that stuff without much extra effort. &lt;/p&gt;&#xA;" OwnerUserId="41" LastEditorUserId="368" LastEditDate="2015-12-02T19:12:42.097" LastActivityDate="2015-12-02T19:12:42.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6390" PostTypeId="2" ParentId="6388" CreationDate="2015-12-02T16:23:15.013" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Which of the following materials will be most suitable to manufacture the disc for the stiffest/strongest possible part.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems that you need a material that will transform a non-uniform load into a more-or-less uniform load by providing consistent bearing.  As such, you're looking at something like a plate on an elastic foundation, albeit on a much smaller scale.  This type of problem can be quite complex due to the interplay between the stiffness of the plate and the foundation and the loading on the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only variables you seem to have locked down are the 1 mm thickness and 25.4 mm diameter of the disc.  Strength is not an issue here (assuming you have a ductile material); stiffness is the only real requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are truly looking to get the stiffest material for your disc, the carbon steel you have listed is probably your winner, with stainless a possibility depending on the alloy.  Assuming all the discs have the same thickness/diameter, the only thing that will then affect stiffness is the material's &lt;a href=&quot;https://en.wikipedia.org/wiki/Young&amp;#39;s_modulus&quot; rel=&quot;nofollow&quot;&gt;modulus of elasticity&lt;/a&gt;.  The greater the modulus, the stiffer the disc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tungsten_carbide&quot; rel=&quot;nofollow&quot;&gt;Tungsten carbide&lt;/a&gt; might be worth looking into for your disc as well.  It's incredibly stiff (twice that of steel) and not terribly expensive.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="368" LastEditDate="2015-12-17T23:19:35.997" LastActivityDate="2015-12-17T23:19:35.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6391" PostTypeId="2" ParentId="2673" CreationDate="2015-12-02T16:23:28.730" Score="2" Body="&lt;p&gt;It all depends on the sterilization level you want to achieve. As you say, you would need to know the death kinetics parameters, or equivalently the log-decimal reduction time (D) and how it changes with temperature (Z). As sterilization is a problem that takes place almost everywhere in food industry, a standard sterilisation cycle lasts for 30 min. at 121 degC at the cold spot of your reactor. This ensures that the spores of many &lt;em&gt;Bacillus&lt;/em&gt;, &lt;em&gt;Clostridium&lt;/em&gt; and other thermophilic microorganisms cannot survive.&lt;br&gt;&#xA;For a 600 m3 reactor the easiest is to have a temperature probe inside of the reactor.&#xA;As how much steam? That depends on how well isolated your reactor is. &#xA;The best is to make a feedback loop with setpoint 121 degC that would send steam whenever temperature starts to drop. Otherwise, you can try to do the same manually.  &lt;/p&gt;&#xA;" OwnerUserId="3991" LastActivityDate="2015-12-02T16:23:28.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6392" PostTypeId="1" AcceptedAnswerId="6416" CreationDate="2015-12-02T16:42:57.263" Score="6" ViewCount="155" Body="&lt;p&gt;In commercial fire sprinkler systems or large farm irrigation systems, is the thrust produced by the water a major design consideration? At what stage in the design process is this taken into account?&lt;/p&gt;&#xA;" OwnerUserId="3992" LastEditorUserId="368" LastEditDate="2015-12-17T22:59:27.007" LastActivityDate="2015-12-17T22:59:27.007" Title="Thrust consideration in sprinkler and irrigation systems" Tags="|systems-design|thrust|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6393" PostTypeId="1" CreationDate="2015-12-02T18:34:08.017" Score="3" ViewCount="192" Body="&lt;p&gt;What is the terminology for having different/spaced-out on and off set points for a particular controlled process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a system does not have the set points set appropriately, it could exhibit something like contact bounce.&lt;br&gt;&#xA;This is not limited to any particular kind of system. It can be for a electro-mechanical system with physical contacts and switches, or it can be for a software system. &quot;Debouncing&quot; can be done in a variety of ways, such as physical distances, time delays, introducing some kind of hysteresis behavior, etc.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;E.g.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A thermostat controls the AC in a house. It is &lt;strong&gt;set to turn on at 90 degrees and turn off at 70 degrees&lt;/strong&gt;. When the thermostat reads 90 degrees, it turns the AC on, and when the thermostat reads 70 degrees, it turns the AC off.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Bad Examples&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If this system did not have the correct set points, it would exhibit something akin to &lt;strong&gt;contact bounce&lt;/strong&gt;:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the &lt;strong&gt;thermostat is set to switch on/off at 80 degrees&lt;/strong&gt;, when the thermostat reads 80 degrees, it turns the AC on, but could bounce back and forth between on and off while the temperature varies slightly from 79-81 degrees.&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;Similarly, if the &lt;strong&gt;thermostat is set to switch on at 80 degrees and off at 79 degrees&lt;/strong&gt;, when the thermostat reads 80 degrees, it will turn the AC on, and as the AC cools the temperature down from 80 to 79, it will turn the AC off. This will occur relatively rapidly but not as rapidly as the example given above.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2430" LastEditorUserId="368" LastEditDate="2015-12-17T22:56:45.343" LastActivityDate="2015-12-17T22:56:45.343" Title="Terminology for setting Set Points to avoid &quot;Contact Bounce&quot;-like characteristics" Tags="|control-engineering|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6395" PostTypeId="2" ParentId="6373" CreationDate="2015-12-02T22:07:07.093" Score="2" Body="&lt;p&gt;I would refer you to &lt;a href=&quot;http://www.researchgate.net/publication/245320516_Poles_and_Zeros_of_Linear_Multivariable_Systems_A_Survey_of_the_Algebraic_Geometric_and_Complex_Variable_Theory&quot; rel=&quot;nofollow&quot;&gt;&quot;Poles and Zeros of Linear Multivariable Systems: A Survey of the Algebraic, Geometric and Complex Variable Theory&quot;, A. G. J. Macfarlane, N. Karcanias&lt;/a&gt; for a very detailed analysis of zeros.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In picturesque terms, poles can be thought of as system resonances coupled to input and output, and zeros as associated with anti-resonances at which propagation through the system is blocked.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For example, sending an exponential input $e^{3 t}$ through the system $\frac{s-3}{s+1}$ results in the output $e^{-t}$. The system blocked an input that was blowing up, and the output decays to zero!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there are many notions of zeros: transmission, decoupling, invariant, and system. These definitions work not just for scalar systems, but for multivariable systems as well.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="368" LastEditDate="2015-12-03T01:00:49.267" LastActivityDate="2015-12-03T01:00:49.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6396" PostTypeId="1" CreationDate="2015-12-02T22:58:36.610" Score="1" ViewCount="235" Body="&lt;p&gt;I've looked everywhere for this seemingly simple question, but I cannot for the life of me find out how to easily cut a specific amount of holes in a plate of metal with my cnc. Am I going to have to make each individual hole, or is there a program that automatically generates an svg, g-code, stl, etc. for evenly spaced holes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm attempting to cut a grid pattern using a Shapeoko and a Grbl Arduino shield.&lt;/p&gt;&#xA;" OwnerUserId="3997" LastEditorUserId="1586" LastEditDate="2015-12-23T20:06:49.023" LastActivityDate="2019-10-25T06:40:06.297" Title="How to cut evenly spaced uniform holes in a plate of metal?" Tags="|machining|cnc|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6397" PostTypeId="1" CreationDate="2015-12-03T00:56:44.353" Score="2" ViewCount="79" Body="&lt;p&gt;My question is naive but I'm having a hard time finding a precise yes/no answer when I ask google. The general advice seems to be &quot;it depends&quot;, but are there voltage levels where it really does not depend? My intuition is that there are some voltages (eg. 5v) that are so low that it does not matter how many amps the power source can supply because the human body will give enough resistance, even in worst case scenarios, so that current will be limited to safe levels. Is this intuition correct or am I looking at this the wrong way?&lt;/p&gt;&#xA;" OwnerUserId="3998" LastActivityDate="2015-12-03T12:26:18.677" Title="Is there such a thing as a voltage level that is safe for humans in all circumstances?" Tags="|safety|" AnswerCount="1" CommentCount="4" ClosedDate="2015-12-03T12:48:39.757" ContentLicense="CC BY-SA 3.0" />
  <row Id="6398" PostTypeId="2" ParentId="6379" CreationDate="2015-12-03T01:19:40.010" Score="3" Body="&lt;p&gt;You seem to want an a slight interference fit so the top will snugly seat into the housing. Depending on your wall thickness and size you could probably oversize the &quot;male&quot; portion of the cap like .002-.003&quot; compared to the &quot;female&quot; dimension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, and understanding that 3D printing generally cannot hold those tolerances I would try an alternative fastening method. Unless you have some time and material to play with to keep tweaking the 3D file for shrinkage and reprint until it fits how you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly consider a snap fit like so(upper right and lower left examples):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/whOTe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/whOTe.jpg&quot; alt=&quot;snap fit examples&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could probably get away with only one &quot;tapered beam&quot; and have the top rotate downward along the edge opposite of the snap fit downward and snap into the case. Think tv remote battery cover. You would have better allowances for fit and would be a nice detail for your project. Make sure you have a detail to be able to remove the top like a little slit for a screwdriver or coin to pry it open. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could always use screws as well&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2015-12-03T01:19:40.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6399" PostTypeId="1" AcceptedAnswerId="6407" CreationDate="2015-12-03T07:04:30.133" Score="6" ViewCount="2238" Body="&lt;p&gt;I am modelling my structure using FEM, and my structure contains line elements (1D) such as beams and columns and area elements (2D) such as walls and columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that, how can I proceed to do modelling, and how can I interpret the results for stress and strain, especially at the places where 1D elements meet the 2D elements? Take for example, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A beam's Center of Gravity might not be located within the slab, but in reality the beam and slab are still touching each other. In other words the beam is offset from the slab&lt;/li&gt;&#xA;&lt;li&gt;A wall's center of gravity might not be located within the slab,  but in reality the wall and slab are still touching each other. In other words the wall is offset from the slab&lt;/li&gt;&#xA;&lt;li&gt;A beam and a column are supposed to be connected together, except that now the beam is offset in left and column in right. If going by line element, they are no longer touching, although in reality, they still are.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is this a solved problem? &lt;strong&gt;Any research literature that I can refer to, or any software packages that you can recommend&lt;/strong&gt; that handle this gracefully?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="3353" LastEditDate="2015-12-03T08:55:54.100" LastActivityDate="2015-12-03T17:49:25.867" Title="How should one model a structural model with both line elements and area elements?" Tags="|structural-engineering|finite-element-method|structural-analysis|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6400" PostTypeId="2" ParentId="6399" CreationDate="2015-12-03T07:40:55.923" Score="0" Body="&lt;p&gt;This mostly depends on your modelling software. You would definitely have nodes at your supports as well as at locations where you have point loads. For line loads you could create a mesh which has nodes exactly where features are. The line loads would then start/end at these nodes. Alternatively you can create a fairly dense mesh (250 mm x 250 mm aperture). On this you can distribute the line loads, adjusted as an area load per 1/16 sqm.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2015-12-03T07:40:55.923" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6401" PostTypeId="2" ParentId="6399" CreationDate="2015-12-03T08:12:15.843" Score="2" Body="&lt;p&gt;A good FEM software package will allow the centroid of the geometric property to be offset from the mesh location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when modelling a beam and slab you may have the mesh at the top of beam/underside of slab level. The line elements then connect to the surface elements at the nodes (which should be fairly regular). The beam geometry will then have an offset of the distance from top of beam to centroid of beam (half the beam's height if it's symmetrical), while the slab will have an offset of half the slab depth.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2015-12-03T08:12:15.843" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6403" PostTypeId="2" ParentId="6397" CreationDate="2015-12-03T12:26:18.677" Score="1" Body="&lt;p&gt;If there is a universally safe voltage level, it is so low to be of little practical value, probably in the 10s of mV range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with your spec is that it's current that causes the problems, not voltage.  Given a particular voltage, the current depends on the resistance, which can vary greatly between different points of the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside, we are largely bags of saltwater, which has low resistivity.  Our skin, in contrast, has high resistance, although that varies greatly depending on wetness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For most &quot;ordinary&quot; cases, 5 V is so low compared to our skin resistance that you can't even feel it.  However, if I dipped two fingers in salt water and then touched them across a 5 V supply, I probably would feel it.  Take a 9 V battery and touch it briefly to your tongue.  You will definitely feel that, and it will be rather unpleasant.  If you're clever, you can kill yourself with a 12 V car battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, that's all been outside the body.  If you connect directly to points inside the body, then not only will the current for the same voltage be much higher, but depending on where you connect, the current tolerance can be much lower.  I expect 100 &amp;micro;A thru the heart muscle in the right place could cause serious problems.  I don't know how much voltage that would take, but probably much less than 5 V.  Even a few 100 mV between selected points in the brain can very likely mess up critical systems.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2015-12-03T12:26:18.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6405" PostTypeId="2" ParentId="6399" CreationDate="2015-12-03T16:01:28.587" Score="0" Body="&lt;p&gt;In general, for beams, I look not at the stresses, but at the bending moments (which they separate for the beams v. the panels).  Then use s = Mc/I.  Then I can offset for the beam off the panel stress.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-03T16:01:28.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6406" PostTypeId="1" AcceptedAnswerId="6488" CreationDate="2015-12-03T17:13:29.183" Score="7" ViewCount="4473" Body="&lt;p&gt;While this question pertains to smartphone , it is being asked from purely an electrical engineering view (My electrical knowledge is very basic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to understand if charging a cell phone and using it is recommended or not, I came across this &lt;a href=&quot;http://batteryuniversity.com/learn/article/charging_lithium_ion_batteries&quot; rel=&quot;nofollow noreferrer&quot;&gt;Charging Li Ion battery&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting extracts (emphasis mine)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;portable devices sit in a charge cradle in the on position. The current drawn through the device is called the &lt;strong&gt;parasitic load&lt;/strong&gt; and can distort the charge cycle. Battery manufacturers advise against parasitic loads while charging because it induces mini-cycles, but this cannot always be avoided....&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A portable device should be turned off during charge. This allows the battery to reach the set voltage threshold and current saturation point unhindered. A &lt;strong&gt;parasitic load confuses the charger&lt;/strong&gt; by depressing the battery voltage and preventing the current in the saturation stage to drop low by drawing a leakage current. A battery may be fully charged, but the prevailing conditions will prompt a continued charge, causing stress.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wanted to verify this by testing to the extent possible with limited means. I ran two tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For measurement of voltage/Current used an app &lt;a href=&quot;https://play.google.com/store/apps/details?id=ccc71.at.free&quot; rel=&quot;nofollow noreferrer&quot;&gt;3C Toolbox&lt;/a&gt;, which gives measurement at every 1% change in both tabular and graph form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First test was normal charging in air plane mode. Had to keep the phone on to run the app&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second test, was to charge the phone while simulating usage. Instead of actual usage, kept screen display on at 50% brightness to simulate usage load, since this would give a steady load and had read elsewhere that this would approximate a 2G data download load&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to compare with the graph quoted in the linked article pasted below&#xA;&lt;a href=&quot;https://i.stack.imgur.com/imosa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/imosa.jpg&quot; alt=&quot;Battery University Graph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Test results for voltage and current are pasted below. The first graph on the left side is of first test and next is of second test (pale gold line at bottom shows screen on status)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Charging Time&lt;/strong&gt;.  Identical in both cases -121 minutes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Voltage&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bvodj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bvodj.jpg&quot; alt=&quot;Voltage Compared&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Current&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/k7CpY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k7CpY.jpg&quot; alt=&quot;current compared&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Temperature&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2Dkc7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Dkc7.jpg&quot; alt=&quot;temperature compared&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit Notes&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have edited the question, making it more focused to &lt;strong&gt;validate&lt;/strong&gt; my interpretations of the results of test. In this context :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) I fully agree that this is not a rigorous scientific experiment &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b). Battery University talks of effects observed while charging with device on. Here the comparison from my tests, is not between  charging while device  switched &quot;off&quot; and &quot;on&quot;. It is a comparison between charging while the device is on with &quot; less &quot; load  and &quot; more&quot; load (device on with screen &quot;off &quot; and device on with screen &quot;on at 50% brightness&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having made these points, I believe that these tests offer &lt;strong&gt;indicative&lt;/strong&gt; trends and is reasonable to compare with Battery University and draw conclusions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inferences&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Voltage&lt;/strong&gt;.  This offers the most striking contrast.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;a). While the time taken to reach maximum voltage and voltage levels are nearly same, the voltage graph shows steep ramp up compared to the first. This means that the topping up of charge is gradual in the first case, since more time is spent at various voltages. This translates to longer &lt;strong&gt;battery life  per cycle&lt;/strong&gt; (assuming equal discharge conditions applied to both cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b). Battery University says &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A parasitic load confuses the charger by depressing the battery voltage...., &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;this can be observed in the second case &lt;strong&gt;more prominently&lt;/strong&gt; and confirms the behaviour&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Current&lt;/strong&gt;.   &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;a). The current graph in the second graph, exhibits a pronounced  &lt;strong&gt;saw tooth&lt;/strong&gt; shape before it drops off. This corresponds to &lt;strong&gt;mini cycles&lt;/strong&gt; caused by &lt;strong&gt;parasitic load&lt;/strong&gt; as mentioned by Battery University&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The current drawn through the device is called the parasitic load and can distort the charge cycle. Battery manufacturers advise against parasitic loads while charging because it induces mini-cycles, ....&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;b).  Further, Battery University statement below cannot be established, since in both cases, the battery is being charged while in &quot; on&quot; condition and measured drop is nearly the same when you compare the maximum and minimum&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A parasitic load confuses the charger by depressing the battery voltage and preventing the current in the saturation stage to drop low by drawing a leakage current. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Temperature&lt;/strong&gt;.  The  battery temperature in second case is marginally higher by 2°C  and while it is not significant, it can be argued that it is not &lt;strong&gt;desirable for battery&lt;/strong&gt;, since increased internal temperature while charging leads to more internal stress.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Charging Time&lt;/strong&gt;. Same in both cases and the charger having a limited capability to supply voltage or current,&#xA;Would imply that there has to be a trade-off between quality of charging with more load, as is evident from the graphs above&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inferences above are in line with Battery University and consequently &lt;strong&gt;one can conclude that charging battery, while device is switched &quot;on &quot;or &quot;in use&quot; is NOT a good practice&lt;/strong&gt;. This is further supplemented by &lt;a href=&quot;http://ww1.microchip.com/downloads/en/AppNotes/01149c.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt; best practices&lt;/a&gt; shared and amplified by @ericnutsch (thanks)   &lt;a href=&quot;https://engineering.stackexchange.com/questions/6406/charging-a-smartphone-battery-while-using-it#comment11477_6409&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Seeking anwers for confirmation or negation of conclusion highlighted above (backed by technical sources and citations). Additional inputs on charging methodology implemented by major OEM cell manufacturers  to  cater for  &quot;charge while in use&quot;, would be appreciated&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please ask if any other information can be provided apart from above and cell phone is Honor 6, with 3100 mAh battery.OEM wall charger of 2A used&lt;/p&gt;&#xA;" OwnerUserId="3945" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-11-20T16:53:28.760" Title="Charging a smartphone battery while using it" Tags="|electrical-engineering|battery|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6407" PostTypeId="2" ParentId="6399" CreationDate="2015-12-03T17:49:25.867" Score="8" Body="&lt;p&gt;Say we have a slab supported on a beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lVpmp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lVpmp.png&quot; alt=&quot;Beam composite with slab&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The centroid of the slab and the centroid of the beam are not coincident. Fortunately, in FEM software packages the geometric centroid of the element can be offset from the nodes that define the element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sketch below shows a case where the shells have been offset such that the nodes are at the bottom face and the beams have been offset such that the nodes are at the top face. This way, the section properties accurately reflect the true geometry but the shell and beam elements are able to share nodes (which is essential).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GcpNx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GcpNx.png&quot; alt=&quot;Extruded 3D view of beam with slab&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for implementation. Most FEM packages should be able to handle this. I'm familiar with LARSA so my notes below are specific to that software package, but SAP, RISA, MIDAS, LUSAS, ABAQUS, etc. etc. etc. should have equivalent capabilities. Furthermore, there's likely quite a few different ways of solving this problem. What follows is just one of them -- just because it seems like some pictures might help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a wireframe and extruded view of a quick beam-slab model I put together. The plate elements and beam elements are shown at their centroids. Note that the plate and beam elements share nodes but their centroids are offset from one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IoY2e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IoY2e.png&quot; alt=&quot;Wireframe and Extruded Views&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, this offset was accomplished in two steps. The beam itself was defined using Section Composer with a geometry that places the node at the top face of the beam. Then, member end offsets were used to offset the top of the beam half the plate thickness downward. The plate elements are defined using nodes at the plate centroid so these two steps shift the beam geometry down so that the top of the beam is 'in contact' with the bottom of the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4PcPy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4PcPy.png&quot; alt=&quot;Member End Offsets&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, look for something like &quot;member end offsets&quot; in your FEM package.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-12-03T17:49:25.867" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6408" PostTypeId="2" ParentId="6393" CreationDate="2015-12-03T18:56:37.960" Score="1" Body="&lt;p&gt;&quot;Hysteresis&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your mechanical thermostat air-con example the bimetallic strip is designed to toggle closed when set point is exceeded. The toggle action gives a quick closing of the contacts and minimises arcing. The toggle action requires the bimetallic strip to reduce in temperature by a degree or two before switching back on again. (20 degrees in your example would be too much. You want the hysteresis to be small enough that you don't experience much fluctuation in temperature but not so small that it cycles the air-con too frequently.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a software system the hysteresis can be more easily adjusted. The code for such a device might look like this pseudo-code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;setpoint = 80&#xA;hysteresis = 0.8&#xA;if actual &amp;gt;= setpoint: output on                 // start cooling&#xA;if actual &amp;lt; (setpoint - hysteresis): output off  // stop cooling&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2015-12-03T23:11:20.870" LastActivityDate="2015-12-03T23:11:20.870" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6409" PostTypeId="2" ParentId="6406" CreationDate="2015-12-04T01:53:57.600" Score="4" Body="&lt;p&gt;I am not familiar enough with your tests and your data to draw any conclusions from it. Perhaps if you infer your own conclusions citing specific parts of your data I would be able to follow better and see if your conclusions are sound. I am not sure if the data you are seeing is significant or just noise in the testing process. I am also not an EE so EEs feel free to correct me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically a modern battery charging/power system has functionality to charge the battery, pull from the battery, and draw from line power directly when available(5v usb in this case). This allows the battery to charge according to what is best for its chemistry. No discharge on the battery occurs until it is disconnected from line power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This link describes it much better than I can: &lt;a href=&quot;http://ww1.microchip.com/downloads/en/AppNotes/01149c.pdf&quot; rel=&quot;nofollow&quot;&gt;Designing A Li-Ion Battery Charger and Load Sharing System With Microchip’s Stand-Alone Li-Ion Battery Charge Management Controller&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, it does not harm the battery to charge and use it simultaneously. I believe that this is a common perception because the phone may operate much warmer than it usually does. This however, is due to the fact that both the battery and electronics are giving of peak amounts of heat; not because damage is being done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Older lead acid systems like on older cars (not sure about the new ones) don't do this, and just put the battery right on the bus. Cars don't have to hold a very close voltage tolerance, and lead acid is very forgiving so the system works fine. They could however increase battery life if they managed it in the controlled way as mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;br&gt;&#xA;I have re-read your question and Mahendra Gunawardena's answer. The battery charging schematic in the link I provided, and cited in Mahendra's answer is evidence that the best practice is to use a charging chip that correctly handles use and charging simultaneously. Companies like Samsung, HTC, Motorola, Apple, and Google, meet or exceed best practice. I can not provide you an electrical schematic of one of these phones, but by logical deduction I can assure you that if a superior solution exists, they are using it. &lt;strong&gt;Consequently, I reject your hypothesis that &quot;charging [a cell phone] battery, while [the] device is switched &quot;on &quot;or &quot;in use&quot; is NOT a good practice.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I think I have found your misunderstanding on the voltage: In your voltage graphs, you arrived at 4335mV and 4250mV. The starting points were 3930mV and 3981mV respectively. Note that your axis is not the same on each graph and that &lt;strong&gt;the slope of each voltage graph is basically the same&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While your battery university link does have a nice collection of resources; their information is not up to date. If the battery and load are on the same bus, yes what they state is true and it will harm the battery over time, but modern electronics do not do it this way anymore; as seen in the microchip electrical schematic. Also if it was occurring, the small amount of parasitic load from the phone being on in airplane mode would have nearly the same negative effect as a larger load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To dive into the voltage a bit further; there is no harm in the final charge voltages being slightly different either. It is also likely that the chip controlling the change targeted a different final voltage based on temperature compensation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To trust this data you would have to preform this test tens of times and analyse your data statistically. Then you would need to graph it vs other variables like temperature to check if the chip was compensating or not. End of the day, your measurement here is lost in the noise of many other factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current graph is inconclusive because we have no idea if that means load draw(just the phone electronics) or line draw (battery and load). We also dont know how they are measuring this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already reviewed that the temperature would be increased because the usage also produces heat that adds to the total. This heat is not an issue for the small differences seen; like your 2degC for example.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="282" LastEditDate="2015-12-08T09:13:43.223" LastActivityDate="2015-12-08T09:13:43.223" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6410" PostTypeId="1" AcceptedAnswerId="6426" CreationDate="2015-12-04T02:24:26.047" Score="2" ViewCount="80" Body="&lt;p&gt;I'm trying to measure water elevation as a function of distance in a water tank, viewed through the glass wall, as shown here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jWsei.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jWsei.jpg&quot; alt=&quot;tank&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The height variations that's of interest to me is very small. Would anyone know of a lens that can be placed in front of a camera to stretch the image vertically? This will help enhance the vertical resolution of the photo. Other suggestions are also welcome.&lt;/p&gt;&#xA;" OwnerUserId="4010" LastActivityDate="2015-12-04T20:39:25.253" Title="Lens to &quot;stretch&quot; or enhance vertical resolution in a photo" Tags="|measurements|optics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6411" PostTypeId="1" CreationDate="2015-12-04T02:26:29.427" Score="4" ViewCount="209" Body="&lt;p&gt;I found this article on &lt;a href=&quot;http://voxel.at/voxel/blog/macro-photography-m12-usb-camera-modules&quot; rel=&quot;nofollow&quot;&gt;hacking macro lenses&lt;/a&gt;. The author used a lens extension tube to allow closer focusing. IE The lens's Minimum Object Distance (MOD) was significantly reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find an affordable lens that suits my requirements, so I would also like to extend or shorten an existing lens to achieve my goals. Basically I am trying to hack a close-focus machine vision industrial camera from an &lt;a href=&quot;http://www.uctronics.com/ov5647-camera-board-w-m12x0-5-mount-lens-fully-compatible-with-raspberry-pi.html&quot; rel=&quot;nofollow&quot;&gt;S-Mount Raspberry Pi camera&lt;/a&gt;. The problem I'm running into is that compatible lenses have too high MODs. &lt;strong&gt;Given the constraints below, how do I solve the variables?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Constants:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Object Size: 4&quot; x 5&quot;&lt;/li&gt;&#xA;&lt;li&gt;Object Distance: 4.5&quot; (Object to Lens)&lt;/li&gt;&#xA;&lt;li&gt;Camera Sensor: 0.25&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;The Object should be in focus and snug to the edges of the captured image&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Avoid fish-eye or wide angle lenses, if possible&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Variables:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Focal Length (Lens to principal focus)&lt;/li&gt;&#xA;&lt;li&gt;Image Distance / Back Focal Length (Lens to sensor)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/class/phscilab/imagei.html&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; which discusses Optics and the Lens Equation. I could simply solve the lens equation as follows: 1/115 mm + 1/x = 1/6 mm. In this case, the image distance would be 6.3303 mm. Is it really that simple though?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can use any standard lens (3 mm, 4 mm, 6 mm, 8 mm, etc). My concern is that extension rings result in image magnificaion which would cut-off the image. I could plug my numbers into the magnification equation. and then extend my object size to include a 20% buffer, for example. But making that adjustment changes the figures for the lens equation. The two equations are intrinsically linked and I'm not sure how to solve them inclusively.&lt;/p&gt;&#xA;" OwnerUserId="3765" LastEditorUserId="368" LastEditDate="2015-12-17T22:48:39.027" LastActivityDate="2015-12-17T22:48:39.027" Title="Modifying a fixed camera lens for close focus" Tags="|optics|mathematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6412" PostTypeId="2" ParentId="6410" CreationDate="2015-12-04T04:20:18.737" Score="1" Body="&lt;p&gt;Assuming you need to gather displacement data from every position along the horizontal dimension of the tank simultaneously, you may be looking for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Cylindrical_lens&quot; rel=&quot;nofollow&quot;&gt;cylindrical lens&lt;/a&gt;, which focuses light onto a line, as noted in the link. With a proper lens setup, you should be able to change the aspect ratio of the field of view and capture greater resolution along the vertical without affecting the horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be worth putting a uniform, high-contrast background behind the tank and using plenty of illumination of both the water and background to assist with accurate measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate method to consider is &lt;a href=&quot;https://en.wikipedia.org/wiki/Interferometry&quot; rel=&quot;nofollow&quot;&gt;interferometry&lt;/a&gt;, and it may offer greater accuracy than imaging devices. From the link (emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Interferometers are widely used in science and industry for the measurement of &lt;strong&gt;small displacements&lt;/strong&gt;, refractive index changes and surface irregularities.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am unsure how to set up an interferometer to measure displacements in water height, however.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2015-12-04T04:20:18.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6413" PostTypeId="1" CreationDate="2015-12-04T05:21:21.133" Score="2" ViewCount="6473" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fbTjs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fbTjs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above mechanism it is required to find the generalized coordinates to write equation of motion of the MDOF system. I just want to ask how do we actually approach such a problem? How to identify which kinematic degrees of freedom are relevant and independent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer given in this particular case is 4. Two rotation of the pulleys, and two vertical motions of the masses B and C. Now how to visualize this? How do I methodically come to this conclusion? Again, is there any other combination possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All pulleys are massless and frictionless.&lt;/p&gt;&#xA;" OwnerUserId="4011" LastEditorUserId="33" LastEditDate="2015-12-05T13:08:07.523" LastActivityDate="2015-12-05T13:08:07.523" Title="How do you find the kinematic degrees of freedom of a mechanism?" Tags="|applied-mechanics|dynamics|kinematics|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6414" PostTypeId="1" AcceptedAnswerId="6446" CreationDate="2015-12-04T06:56:49.583" Score="4" ViewCount="2255" Body="&lt;p&gt;I need a 8 m x 4 m x 0.6 m RC plinth (mechanical requirement) which needs to be built onto loose gravel. Obviously I am going to over excavate, back fill with selected material and bed my plinth on that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plinth accommodates 2 x 65 m&lt;sup&gt;3&lt;/sup&gt; water tanks, placed symmetrically.  The tanks are $\phi$3 m, 9 m high and weigh 796 kN when full. The wind load on each creates a 206 kNm moment and horizontal force of 38 kN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to design the plinth for bending and check the soil for shear failure. I assume the max. bearing capacity of the soil to be 150 kPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to simulate this using FEM. How do I address the support conditions. I understand that I should use springs as supports, but how would I get the equivalent spring constants if my spring is soil?&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="1832" LastEditDate="2015-12-07T10:50:54.110" LastActivityDate="2015-12-17T22:52:31.983" Title="How to model elastic support in FEM?" Tags="|structural-engineering|finite-element-method|foundations|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6416" PostTypeId="2" ParentId="6392" CreationDate="2015-12-04T07:34:19.430" Score="3" Body="&lt;p&gt;Product design often involves juggling many parameters at the same time, prioritizing them, then testing a design. In sprinklers themselves, the forces associated with the thrust are much lower that the forces the sprinkler experiences during assembly or from internal pressure; so typically they are not a design priority.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For irrigation sprinklers the thrust is not a big consideration in the structural design, because the flow, and resultant thrust is comparatively small. It is however important for &lt;a href=&quot;https://en.wikipedia.org/wiki/Impact_sprinkler&quot; rel=&quot;nofollow&quot;&gt;impact sprinklers&lt;/a&gt;, as it is their method of operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fire sprinklers themselves don't generate a lot of force, but there is a lot of dynamic force on the &lt;a href=&quot;http://wermac.org/steel/pipesupports.html&quot; rel=&quot;nofollow&quot;&gt;pipe supports&lt;/a&gt; on large diameter pipes that turn corners to deliver water to the sprinklers. Fluids can exert very large forces without being a free jet. Here is a big nasty document on &lt;a href=&quot;http://www.wermac.org/pdf/piping_engineering.pdf&quot; rel=&quot;nofollow&quot;&gt;pipe engineering&lt;/a&gt; that might make your thermo fluids homework look nicer ;-)&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-12-04T07:34:19.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6417" PostTypeId="2" ParentId="6386" CreationDate="2015-12-04T07:59:05.787" Score="1" Body="&lt;p&gt;I think that an Inlet/Outlet disk in free air is probably general enough that some smart dead person solved it theoretically. However, mortals use &lt;a href=&quot;http://www.openfoam.com/&quot; rel=&quot;nofollow&quot;&gt;OpenFOAM&lt;/a&gt; or other CFD software to solve problems like these ;-).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your application, that amount of work is probably not justified. Basic rule is that a larger diameter fan will allow you to have a larger separation distance between the two (for visual appeal) and still maintain reasonable efficiency. By reasonable I mean 10% of available wind energy as opposed to 1% or less. I assume you are turning a wind turbine for demonstration purposes; or we have an entirely different discussion on our hands ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a shot in the dark, I would estimate that with a wind turbine and fan of equal diameter, you would have the following &quot;wind coupling&quot; efficiency (not considering blade design, turbine rpm, or motor/generator efficiency).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Close as possible: 50%&lt;br&gt;&#xA;1 Diameter away: 10%&lt;br&gt;&#xA;2 Diameters away: 2%&lt;br&gt;&#xA;3 Diameters away: 1%&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-12-04T07:59:05.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6418" PostTypeId="2" ParentId="6411" CreationDate="2015-12-04T10:18:21.943" Score="3" Body="&lt;p&gt;I think you are misinterpreting what your variables are. The focal length is fixed for a given lens and cannot be changed. What you can change is the image to lens distance. Combined these will determine what point is in focus using the lens equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data sheet for the camera you give, gives the focal distance as 6.68 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plugging this into $$\frac{1}{u}+\frac{1}{v} = \frac{1}{f} $$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{6.68} - \frac{1}{114.3} = \frac{1}{u} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ u = 7.1 mm $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that this is not a simple thin lens, so the lens formula will not be exact, and you to not know what the current lens to image distance is, you might be able to get a rough estimate by finding where the image is currently in focus. In general, as the focus distance moves further away the image to lens distance will approach the focal plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best bet is probably a bit of trial and error you want to try and raise the lens by just under 1 mm from the image plane. I'm not sure how easy this is to do accurately and would probably just play with it until you get close to what you want.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-12-04T10:18:21.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6422" PostTypeId="2" ParentId="6406" CreationDate="2015-12-04T11:56:21.483" Score="-4" Body="&lt;p&gt;&lt;strong&gt;mobile technology,&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.&lt;strong&gt;By pass mode&lt;/strong&gt; is not available, &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Your battery have cycle, Charging and Discharging, So need to chargeable FULL, In emergency use &lt;strong&gt;Battery Bank&lt;/strong&gt;, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Don't charge cont, Full Battery 95%, 85%,92%, Wait after dry to minmum 30 to 20%, your &lt;strong&gt;battery life&lt;/strong&gt; will be more ( Practical Hence proved ) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In case you are using in battery charging condition, Profit for mobile    and Battery companies ( point 3.), &#xA; 4.1 &lt;strong&gt;&lt;em&gt;OEM please refer about OEM&lt;/em&gt;&lt;/strong&gt;, They are not to be R&amp;amp;D / Own brand manufactures. The are branding company that all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I never recommend to charge within use.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4013" LastActivityDate="2015-12-04T11:56:21.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6423" PostTypeId="2" ParentId="6414" CreationDate="2015-12-04T12:31:18.557" Score="4" Body="&lt;p&gt;What you need is the modulus of subgrade reaction (MSR) of the soil. This is a measure of soil deflection under a given pressure, so the unit is in (for example) kPa/m, or equivalently, as I'm more accustomed to see it, kN/m&lt;sup&gt;3&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MSR is obtained via plate-bearing tests, but there are also quite a few different tables that give some &quot;typical values&quot; for different soil compositions (obtained via plate-bearing tests). That being said, these tables can present huge ranges for a given soil type. I've found some tables &lt;a href=&quot;http://www.ce.memphis.edu/3137/Powerpoint%20Handouts/7%20-%20Subgrade%20Characterization.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (page 37), &lt;a href=&quot;http://www.sefindia.org/forum/download.php?id=1209&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (Table C.3 on pages 1397-1398) and &lt;a href=&quot;https://books.google.com.br/books?id=YfFP2ejTujoC&amp;amp;pg=PA16&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (Table 1.2 page 16).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically this does not mean, however, that you can simply get the MSR and multiply it by the tributary area of each element since MSR isn't a soil property but a shorthand for the interaction between the soil and the foundation (different foundation sizes and shapes imply in different MSR values). There are equations you can use to find the corrected MSR, references to which can be found &lt;a href=&quot;https://books.google.com.br/books?id=YfFP2ejTujoC&amp;amp;pg=PA128&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (section 5.2.2 starting on page 128).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, I have seen some people (including books) simply adopt the default MSR, so do that at your own risk. Foundations aren't my specialty, so I can't even say how significant the correction equation is.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-12-04T12:31:18.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6424" PostTypeId="1" AcceptedAnswerId="6425" CreationDate="2015-12-04T15:12:14.390" Score="0" ViewCount="10281" Body="&lt;p&gt;Below is a picture of two supporting posts in my basement. The posts are directly underneath the kitchen/dining room.  Would it be feasible to remove one or both of the posts? What would the cost roughly be to make this change? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I know that I should have a structural engineer take a look before actually making any changes. Also, I live in the mid-west. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p9iqJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p9iqJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4016" LastActivityDate="2015-12-04T15:49:54.720" Title="Can I remove one or both of these support posts in my basement?" Tags="|structural-engineering|beam|" AnswerCount="1" CommentCount="4" ClosedDate="2015-12-04T17:26:19.580" ContentLicense="CC BY-SA 3.0" />
  <row Id="6425" PostTypeId="2" ParentId="6424" CreationDate="2015-12-04T15:49:54.720" Score="12" Body="&lt;p&gt;Whenever one speaks about removing a structural element (column or beam), the initial hypothesis must always be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NO! GOD NO! WHAT ARE YOU DOING!&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The second hypothesis must always be:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No. No. No. No.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That being said, it can be done. But it is not trivial and requires a serious investigation by a structural engineer. This is a &quot;last resort&quot; issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those columns are transferring a load from the beam which can clearly be seen on the ceiling to the foundation. If you remove the columns, then the beam will have to be able to take that load and transfer it across the room to the columns in the walls. The forces the beam will have to withstand will be multiplied many times (possibly 45 times, according to a simplified model).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with all certainty, if you simply remove the columns, the beam will collapse and probably take the rest of your house (this is the basement after all) and you with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you'd need to effectively redesign the structural layout of your basement. Without columns, you'll need to replace the current beam. That's basically a given. Then you'll have two choices:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Place a much stronger beam in its place, capable of withstanding the increased forces. This almost certainly implies in an increase in beam height (and therefore reduction of usable height under the beam) since beam strength is proportional to the cube of the height.&lt;/li&gt;&#xA;&lt;li&gt;Place a grid of multiple beams (parallel and perpendicular to the current beam) on the ceiling, so that each beam absorbs a very small load and is capable of transferring it across the room. This has the additional issue that you'd need to know where the columns (or structural walls) in your wall are: if you don't have many places to rest your beams, your grid won't be efficient.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are then a bunch of issues of how to make the beams and the ceiling slab work in tandem to improve the beams' efficiency and to certify that the slab isn't &quot;floating&quot; over the beam, how to tie your new beams to the columns (if the beams are of concrete), limiting deflections and vibrations so that your dinning room doesn't sink or feel like a rock and roll concert when your cat walks over it, how to safely remove the existing columns and beam prior to doing your new layout, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, you can do that. But are you really, really, &lt;em&gt;REALLY&lt;/em&gt;, &lt;strong&gt;&lt;em&gt;REALLY&lt;/em&gt;&lt;/strong&gt; sure you want to?&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2015-12-04T15:49:54.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6426" PostTypeId="2" ParentId="6410" CreationDate="2015-12-04T18:30:59.217" Score="2" Body="&lt;p&gt;How about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/boZoJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/boZoJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Photo 1. Inclined tank level scale.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inclined scale means that graduations can be spread more widely for ease of reading. For example, if 1 mm resolution is required and the line rotated by 84 degrees the 1 mm marks stretch out to 10 mm making it much more easy to read and interpolate. cos(84°) = 0.1. This effectively stretches the vertical resolution as required in the question but without the use of a lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not obvious from the question whether an automated vision system is to be used but either way this should make reading the level somewhat easier.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2015-12-04T20:39:25.253" LastActivityDate="2015-12-04T20:39:25.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6427" PostTypeId="2" ParentId="6413" CreationDate="2015-12-04T18:40:54.730" Score="4" Body="&lt;p&gt;It's a question of how many independent variables you need in order to write out the equations of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's walk through it step by step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by just considering Pulley 1 and Mass A. If they're joined by a massless inextensible cable (as shown in your figure) then the displacement of Mass A can be determined as a function of the rotation and radius of Pulley 1. So we only need 1 degree of freedom - the rotation of Pulley 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we added a spring in between Pulley 1 and Mass A then we would no longer be able to directly relate the position of Mass A with the rotation of Pulley 1. Now we need 2 degrees of freedom - the rotation of Pulley 1 and the displacement of Mass A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now add Pulley 2 into the mix. If Pulley 1 were connected to Pulley 2 by a massless inextensible cable, then we could define the rotation of Pulley 2 as a function of the rotation of Pulley 1. However, we've got a spring in there so we need to know the rotation of both pulleys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing with this logic and looking at Mass B and Mass C -- if either one were connected to Pulley 2 using a massless inextensible cable, we could define its position as a function of the rotation of Pulley 2. However, we've got springs attached to both masses, so we need independent variables to describe their positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then Mass A and Mass C need to be connected to the ground via springs to keep the whole thing from unraveling and dropping all the weights on the floor. (I think)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, we need four degrees of freedom -- rotation of Pulley 1, rotation of Pulley 2, displacement of Mass B, and displacement of Mass C. The displacement of Mass A is a function of the rotation of Pulley 1.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-12-04T18:40:54.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6428" PostTypeId="1" CreationDate="2015-12-04T20:24:27.267" Score="1" ViewCount="1919" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9H7DN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9H7DN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem was assigned in my Structural Analysis class, and I am having some difficulty with it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor said the problem is &quot;very easy&quot; if we just think about what the fixed end moment on a cantilever is. However, I don't see how this is relevant considering the member ACE is not a cantilever... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other way I see to do this is using the moment distribution method, but I'm uncertain as to what happens with the internal hinges; we've never had a problem like this in class, and there is no relevant example in my textbook. Any suggestions? &lt;/p&gt;&#xA;" OwnerUserId="4018" LastEditorUserId="33" LastEditDate="2015-12-06T13:36:36.807" LastActivityDate="2015-12-06T13:36:36.807" Title="Beam Reaction Problem With Hinged Joints" Tags="|structural-engineering|beam|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6429" PostTypeId="2" ParentId="6413" CreationDate="2015-12-04T21:48:12.503" Score="3" Body="&lt;p&gt;I often approach this type of problem by identifying &lt;a href=&quot;http://www.engineering.union.edu/~tchakoa/mer312/Lectures/Lecture%202.pdf&quot; rel=&quot;nofollow&quot;&gt;constraints&lt;/a&gt;.  Each independent constraint limits the number of degrees of freedom, required to completely define the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$DOF = DOF_{system} - \sum \text{Contraints}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;In 3D systems&lt;/strong&gt;, each body has 6 degrees of freedom:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Translations in $(x, y, z) \rightarrow 3 \;\text{DOF}$&lt;/li&gt;&#xA;  &lt;li&gt;Rotations $(\theta_{x}, \theta_{y}, \theta_{z}) \rightarrow 3 \;\text{DOF}$&#xA;  $$\therefore DOF_{system} = 6(N_{bodies})$$&#xA;  &#xA;  &lt;hr&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;In 2D systems&lt;/strong&gt;, each body has 3 degrees of freedom:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Translation in $(x, y) \rightarrow 2 \;\text{DOF}$&lt;/li&gt;&#xA;  &lt;li&gt;Rotation $(\theta) \rightarrow 1 \;\text{DOF}$&#xA;  $$\therefore DOF_{system} = 3(N_{bodies})$$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the 2D system shown (with 5 bodies), calculate the degrees of freedom for the unconstrained system.&#xA;$$\therefore DOF_{system} = 15$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, determine the imposed constraints on each body:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The pulleys are constrained by $(x, y)$- each pulley removes $2 \;\text{DOF}$ from the system.  &lt;/li&gt;&#xA;&lt;li&gt;The masses are constrained (though not explicitly) by $(x, \theta)$- each mass removes $2 \;\text{DOF}$ from the system.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Lastly, recognize that body A and its adjacent pulley are not independent- they are kinematically linked by $y=r \theta$, where both move (according to the systems &lt;a href=&quot;http://www.aerostudents.com/files/vibrations/multipleDegreeOfFreedomSystems.pdf&quot; rel=&quot;nofollow&quot;&gt;differential equations&lt;/a&gt;).  This additional constraint to the system requires that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$DOF = 15 - \sum [2(2 \;\text{DOF})_{pulleys} + 3(2 \;\text{DOF})_{masses} + (1 \;\text{DOF})]$$&#xA;$$= 4 \;\text{DOF}$$&lt;/p&gt;&#xA;" OwnerUserId="2538" LastEditorUserId="2538" LastEditDate="2015-12-04T23:35:41.130" LastActivityDate="2015-12-04T23:35:41.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6433" PostTypeId="1" CreationDate="2015-12-05T08:34:16.103" Score="5" ViewCount="104" Body="&lt;p&gt;Imagine the following initial condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two completely miscible fluids &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt; with different dynamic viscosities $\eta_A$ and $\eta_B$ are seperated in a container with fixed Volume $V_0$ and length $L$. The temperature $T_e$ is constant across the container, the time is $t_0=0$. The mass fraction $w_B(x)$ at this point in time is a step function (or a convenient approximation of it), as shown in the sketch below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With those definitions in mind, what is the time $t_m$ at which the mass fraction $w_B$ is within $\pm \delta \%$ of its equilibrium value across the whole container?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cf27D.png&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Initial condition of two miscible fluids in a container.&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cf27D.png&quot; alt=&quot;&quot; title=&quot;Initial condition of two miscible fluids in a container.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing which came to mind was the diffusion equation:&#xA;$$\frac{\partial\phi(\mathbf{r},t)}{\partial t} = \nabla \cdot \big[ D(\phi,\mathbf{r}) \ \nabla\phi(\mathbf{r},t) \big]$$&#xA;From there, I probably 'just' need an approximation for $D(\phi(\mathbf{r},t))$ and solve the differential equation. Is this the right direction? What approximations would be appropriate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Stoke-Einstein-Equation enables me to calculate $D$:&#xA;$$ D = \frac{k_\mathrm{B} \cdot T}{6 \pi \cdot \eta \cdot r}$$&#xA;but I don't know the radius $r$. I just dont know how to move on.&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="368" LastEditDate="2015-12-16T00:32:40.300" LastActivityDate="2016-03-10T23:05:19.813" Title="How long does it take for two initially seperate fluids to reach a certain homogenity across their container?" Tags="|fluid-mechanics|diffusion|" AnswerCount="1" CommentCount="12" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6434" PostTypeId="2" ParentId="6428" CreationDate="2015-12-05T19:49:25.147" Score="2" Body="&lt;p&gt;Remember that hinges cannot transfer moment - only shear. Try drawing three separate free body diagrams (member ACD, member DE, and member EFG) knowing that the only reactions at the ends of member DE can be shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free body diagram that you come up with for ACD should look much more manageable than the initial problem - a end loaded cantilever with redundant roller support.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-12-05T19:49:25.147" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6435" PostTypeId="1" AcceptedAnswerId="6436" CreationDate="2015-12-06T03:14:43.013" Score="21" ViewCount="2222" Body="&lt;p&gt;I know that digging tunnels is always &lt;em&gt;much&lt;/em&gt; more costly than building ways or train above ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't the Channel Tunnel start around the coastline? Why does it have an around 10 km long portion under land on the British side?&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditDate="2016-02-01T09:57:27.163" LastActivityDate="2019-06-03T01:15:15.613" Title="Why does the Channel Tunnel enter the ground around 10 km from the coast?" Tags="|civil-engineering|geotechnical-engineering|rail|tunnels|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6436" PostTypeId="2" ParentId="6435" CreationDate="2015-12-06T09:00:44.307" Score="33" Body="&lt;p&gt;The picture, below, of the exaggerated long section of the Channel Tunnel was taken from &lt;a href=&quot;https://en.wikipedia.org/wiki/Channel_Tunnel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MFNuz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MFNuz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Full-sized image &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Channel_Tunnel_geological_profile_1.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the limiting factors for the Channel Tunnel are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Railways don't like steep gradients&lt;/li&gt;&#xA;&lt;li&gt;The tunnels comprising the Channel Tunnel were excavated using tunnel&#xA;boring machines (TBMs). Like railways, they cannot tolerate steep&#xA;gradients.&lt;/li&gt;&#xA;&lt;li&gt;The tunnel was excavated in chalk marl (green coloured material in&#xA;the picture). This was due to its depth (not being too shallow and&#xA;not being too deep) and its ability to be easily dug but also it&#xA;would cause major support issues for the tunnels.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you look at the long section of the tunnel there is some high ground on both coasts. The width of the high ground on the French side is about 2 to 3 km, whereas on the English side the width about 7 to 8 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reasons of maintaining a comfortable gradient for the TBMs and the rail line and to position the tunnels in the lower part of the chalk marl combined with the width of the high ground on the coast and where the chalk marl is located within the high ground and because the chalk marl is inclined on the English side of the tunnel, the portal for the tunnel on the  English side had to be positioned some 9 to 10 km from the coast.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="9" LastEditDate="2019-06-03T01:15:15.613" LastActivityDate="2019-06-03T01:15:15.613" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="6439" PostTypeId="2" ParentId="6406" CreationDate="2015-12-06T21:05:30.373" Score="4" Body="&lt;p&gt;First of all it is highly unlikely that the one would find public information of major OEM cell phone manufactures battery charging methodology back by technical source and citations as this would be a breach of confidentiality.  Having read your post and references I can safely assume that you have some understanding of rechargeable batteries and related charging methodologies.  Also source referenced in your post are very credible.&lt;/p&gt;&#xA;&lt;p&gt;Also in modern rechargeable battery systems don't directly charge a rechargeable batteries as per the reference in the post. Custom silicon IC do charge batteries as per the diagram in the above post.&lt;/p&gt;&#xA;&lt;p&gt;As per your own reference a battery charging system can be designed such that use of the mobile device and battery changing can safely occur simultaneously.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Knvc3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Knvc3.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;&lt;em&gt;This image was extracted from Microchip Application Note link cited both by @beeshyams and @ericnutsch.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;Reading through your post I believe you have done a good top down analysis of the workings of rechargeable batteries and battery charging system. Thus I have described below a bottom up engineering design approach for a rechargeable battery system for mobile devices used in enterprise and medical application. It is my belief that this would help you in understand of rechargeable battery systems.&lt;/p&gt;&#xA;&lt;p&gt;In both the cases the customer/consumer was seeking a mobile device that would guarantee more than 8 hours of usage without recharging. One application was primary at room temperature but the other application had a wider temperature requirements. Also note that this process is fairly dated and some of the applications has been already been &lt;a href=&quot;https://en.wiktionary.org/wiki/sunset#Verb&quot; rel=&quot;nofollow noreferrer&quot;&gt;sunsetted&lt;/a&gt;.  The high level steps were&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Select a rechargeable battery vendor&lt;/li&gt;&#xA;&lt;li&gt;Determine the appropriate battery charging circuit&lt;/li&gt;&#xA;&lt;li&gt;Design and Test both hardware and software for the mobile devices per the customer requirements&lt;/li&gt;&#xA;&lt;li&gt;Furnish parametric data to validate customer/consumer requirements&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h3&gt;Select a rechargeable battery vendor&lt;/h3&gt;&#xA;&lt;p&gt;Many vendors were evaluated including vendors such as Maxwell, Sanyo and Sony.  Although all vendors offered similar batteries at high level, parametrically particularly at extreme temperatures some venders had superior performances. Test systems from &lt;a href=&quot;http://www.maccor.com/Products/HevPhevEvTester.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maccor&lt;/a&gt; and &lt;a href=&quot;http://www.cadex.com/en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cadex&lt;/a&gt; were used for battery parametric testing.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oeedj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oeedj.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Determine the appropriate battery charging circuit&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ti.com/lsds/ti/power-management/battery-charger-ic-products.page&quot; rel=&quot;nofollow noreferrer&quot;&gt;Texas Instrument&lt;/a&gt; and &lt;a href=&quot;http://www.microchip.com/pagehandler/en-us/technology/batterymanagement/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Microchip&lt;/a&gt; both offers very good battery charging silicon and support. I am sure other vendors offer superior battery changing silicon unfortunately I don’t have much experience with other vendors&lt;/p&gt;&#xA;&lt;h3&gt;Design and Test both hardware and software for the mobile devices per the customer requirements&lt;/h3&gt;&#xA;&lt;p&gt;This was very interesting and challenging phase, because generating, storing and displaying parametric data as per the above images on the question was very difficult particularly for extreme temperatures.&lt;/p&gt;&#xA;&lt;h3&gt;Furnish parametric data to validate customer/consumer requirements&lt;/h3&gt;&#xA;&lt;p&gt;I suggest you refer to Maccor, Cadex, &lt;a href=&quot;http://biz.maxell.com/en/product_rechargeable/?pci=6&amp;amp;pn=rb0003&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maxell&lt;/a&gt;  and &lt;a href=&quot;http://www.sonyenergy-devices.co.jp/en/products/corporation.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sony&lt;/a&gt; information device level parametric data. Also I suggest you take a look at &lt;a href=&quot;http://www.prba.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Portable Rechargeable Battery Association (PRBA)&lt;/a&gt; and  International lithium battery regulations where there is a better chance of finding public information.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lit/slyt420&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battery Management Solutions&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/5331/is-battery-charge-stop-at-xx-percent-circuitry-not-cheap&quot;&gt;Is battery-charge-stop-at-XX-percent circuitry not cheap?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/3632/what-is-the-minimum-current-required-to-charge-a-smartphone-battery&quot;&gt;What is the minimum current required to charge a smartphone battery?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=NqNI75LtDho&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battery charging, management with the Fuel Tank BoosterPack&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-12-08T00:41:56.873" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="6441" PostTypeId="1" AcceptedAnswerId="6458" CreationDate="2015-12-07T03:09:39.203" Score="3" ViewCount="2356" Body="&lt;p&gt;I'm designing a little tester for a project which uses a pneumatic cylinder. The pneumatic cylinder's purpose is to apply an impact on a hard surface to test for deflection of the surface. When I looked around catalogs, the pneumatic cylinders have a theoretical force output, for example 3,016 N at 0.6 MPa with an extension speed of 0.8 m/s and a stroke length of 300 mm (preliminary selection although I'm not too sure).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How do I calculate the impact energy upon collision with the hard surface, with this information I currently have? There is going to be an amount of impact energy that I need to achieve.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I understand that when it comes to energy, the formula to use is $\frac{1}{2} mv^2$. What should the mass be? Is it the mass of the cylinder or whatever is placed on the piston rod end (I intend to put some kind of bumper at the end)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are &quot;impulse forces&quot; something I should be looking at?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I can't move on to select and appropriately sized pneumatic cylinder for my tester until I understand how to calculate the force output and impact energy.&lt;/p&gt;&#xA;" OwnerUserId="3781" LastEditorUserId="368" LastEditDate="2015-12-16T00:27:07.990" LastActivityDate="2016-01-17T05:44:16.100" Title="Pneumatic cylinder force output and impact energy" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6442" PostTypeId="1" CreationDate="2015-12-07T04:48:59.327" Score="1" ViewCount="3927" Body="&lt;p&gt;Assuming that I have a water source ($Q$) that can flow into 3 drains ( $Q_1$, $Q_2$, $Q_3$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ye6Xw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ye6Xw.png&quot; alt=&quot;Single-inflow, multiple-outflow drain&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that we know that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The drain geometry is predetermined—we know the $x,y,z$ coordinates of the start and end node of the river/drain and we know the cross sectional information and length of the river/drain&lt;/li&gt;&#xA;&lt;li&gt;The velocity is approximated by &lt;a href=&quot;https://en.wikipedia.org/wiki/Manning_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;manning formula&lt;/a&gt;, $$V=\frac{k}{n}R_h^\frac{2}{3} S^\frac{1}{2}$$&lt;/li&gt;&#xA;&lt;li&gt;The flow is incompressible flow, so $Q=VA$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$R_h$ is the hydraulic radius of the river/drain&lt;/li&gt;&#xA;&lt;li&gt;$S$ is the slope&lt;/li&gt;&#xA;&lt;li&gt;$k$ conversion between the SI and English units&lt;/li&gt;&#xA;&lt;li&gt;$n$ the manning coefficient&lt;/li&gt;&#xA;&lt;li&gt;$A$ is the cross section area of the river/drain&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For one, I know that the water flow volume must be conserved&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q=Q_1+Q_2+Q_3+...$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't know what are the other factors that could help us determine $Q_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit: After some research, I think that I need use &lt;a href=&quot;https://en.wikipedia.org/wiki/Bernoulli%27s_principle#Incompressible_flow_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bernoulli equation&lt;/a&gt; to include headloss in this calculation, but I have no idea how to do it, how to proceed?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="3353" LastEditDate="2016-03-11T06:42:00.177" LastActivityDate="2016-03-11T06:42:00.177" Title="How to calculate how much flow there is from a single source to multiple drains" Tags="|fluid-mechanics|open-channel-flow|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6443" PostTypeId="1" CreationDate="2015-12-07T08:33:14.227" Score="2" ViewCount="714" Body="&lt;p&gt;I have to build an &lt;a href=&quot;https://www.youtube.com/watch?v=vw-QKToDMVY&quot; rel=&quot;nofollow&quot;&gt;egg tray&lt;/a&gt;. The material used for such trays is 2 mm stainless steel welding rods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I queried nearby workshops they said a TIG welder is needed to weld such material and it is very difficult to weld such thin rods: they bend or lose temper. I don't have a TIG welder. Is there a cost-effective and simple way to do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found &lt;a href=&quot;http://www.bssa.org.uk/topics.php?article=150&quot; rel=&quot;nofollow&quot;&gt;an article about soldering stainless steel&lt;/a&gt;. I have a soldering iron. Can stainless steel be soldered with the soldering iron used for circuit soldering? What type of flux and lead would I need? Will it produce an effective and permanent bond?&lt;/p&gt;&#xA;" OwnerUserId="4040" LastEditorUserId="368" LastEditDate="2015-12-16T00:14:40.873" LastActivityDate="2015-12-16T00:15:08.447" Title="How to join 2 mm stainless steel welding rods without a TIG welder" Tags="|structural-engineering|welding|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6444" PostTypeId="1" AcceptedAnswerId="6445" CreationDate="2015-12-07T12:02:59.790" Score="-2" ViewCount="141" Body="&lt;p&gt;The high water has some energy E = mgh. In the basic school, we were told that hydropower plants extract the energy of &quot;falling water&quot;. Now, &lt;a href=&quot;https://physics.stackexchange.com/questions/221243&quot;&gt;that banned principle&lt;/a&gt; says that if your water falls upon still turbines, you will loose energy. The water will turn into a foam, vapor and heat up until it slows down to the speed of rotating turbine. At this point, half of the &quot;falling energy&quot; is lost. This means that speed of turbines must match the speed of the water stream. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the waterfall has the highest speed at the basement and, in order to capture all the energy, you need to stop the water completely because water escaping at speed v, carries energy $E = mv^2/2$, where mass m grows linearly with time (your losses accumulate over time). These losses are proportional to water density m and speed v. Which means we loose more energy with each time unit if outbound stream speed v is higher. The only way to achieve 100% efficiency is to stop the water completely, to v=0. However, this implies no flow and no electricity produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these considerations visited me when I did some switching power supplies. Switching power supply achieves 100% efficiency by operating in one of 2 modes: full power, when water input is 100% open to accelerate the turbine and itself, and zero power, when fresh water input is closed and previously accelerated water altogether with massive turbine slow down by the energy consumer (assume a spring that we charge against its force -- it will slow down the stream with turbine). In electronics, the role of moving water and rotor is played by constant inductor, whose inductance is constant, unlike the water, which would arrive all the time and accumulate at v whenever you open the switch. I therefore would stop the water completely in the lowland, before releasing it there, achieving 100% extraction. But, it seems that in real hydro plants and wind mills, turbine rotates constantly, without any switching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, what theoretical efficiencies can wind and hydro turbines achieve?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4041" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2015-12-16T00:10:55.847" Title="What theoretical efficiencies can wind / hydro turbines achieve?" Tags="|energy-efficiency|renewable-energy|power-engineering|wind-power|" AnswerCount="1" CommentCount="2" ClosedDate="2015-12-07T14:44:40.447" ContentLicense="CC BY-SA 3.0" />
  <row Id="6445" PostTypeId="2" ParentId="6444" CreationDate="2015-12-07T12:07:39.380" Score="3" Body="&lt;p&gt;The theoretical maximum efficiency of a wind turbine is the Betz Limit, which is $\frac{16}{27}$ ( ~ 59.3%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result &lt;a href=&quot;http://www.reuk.co.uk/Betz-Limit.htm&quot; rel=&quot;nofollow&quot;&gt;dates back to 1919&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Betz limit is, as you note, speed invariant. To get an intuition as to why, consider the system speeded up or slowed down. This changes the fluid's velocity, but does not change any of the issues around the dispersion of the fluid after the rotor. This means that the limit is driven not by absolutely velocity, but rather the ratio of the speed after the rotor, to the speed before it. That ratio is 1:3 - the fluid loses two-thirds of its speed in the perfect rotor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's different in hydro where you've got a vertical drop. There, the turbine can take almost all of the speed out of the water. That's because there's a reservoir both before and after the turbine, and a vertical drop after the turbine. That means that gravity can do the work of dispersing the water after a turbine, and there's enough volume of space immediately behind the rotor for the water to travel into slowly. And as water is much denser, and much less compressible, than air, it displaces it easily. So we get real-world hydro efficiencies of ~90%.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="368" LastEditDate="2015-12-16T00:10:55.847" LastActivityDate="2015-12-16T00:10:55.847" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6446" PostTypeId="2" ParentId="6414" CreationDate="2015-12-07T12:39:34.013" Score="3" Body="&lt;p&gt;Before jumping into elastic support conditions, you have to realise that it's &lt;strong&gt;not&lt;/strong&gt; a ground beam.  For three reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;With a minor axis width to depth ratio of 6ish, it's not far away from a square.  If you stick a wide support somewhere near the ends of the section, the load will simply arch over the supports in pure compression.  You'd almost be able to build it in brickwork!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ground beams by definition span between supports.  Your plinth is continuously supported by the fill.  It's a raft if anything.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, how will you apply the moment?  Water does't transfer bending moment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The approximations suggested by Wasabi of the ground elasticity will totally negate your efforts at an accurate assessment of stresses within both the supports and the plinth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're going to use an FEA analysis, you'll be looking at modelling the plinth as several layers of concrete cubes, sat on several more layers of cubes of fill, all within a surrounding matrix of cubes representing the gravel.  If you're designing a skyscraper have at it, otherwise the soil investigations and modelling effort required would make this one of the most over designed structures I've ever seen.  Do the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Treat the plinth as infinitely stiff&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Check that the allowable bearing pressure isn't exceeded by either the dead weigh with a uniform ground pressure, or a triangular ground pressure due to the moment when the tanks are empty, or both.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Design the plinth for bending and shear due to the same load patterns&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ignore the lateral 38.  The plinth won't slide due to self weight and friction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You could do this in an afternoon after getting back from the pub.  Flippancy aside, I've done similar (your approach) for a multimillion pound project.  Don't.  It's a pig.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="2947" LastEditDate="2015-12-17T22:52:31.983" LastActivityDate="2015-12-17T22:52:31.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6447" PostTypeId="1" CreationDate="2015-12-07T12:48:00.697" Score="5" ViewCount="1337" Body="&lt;p&gt;I have a closed cylinder with a gas inside at 3,000 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the temperature profile (over time) through the cylinder wall, something like figure (b) shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tufts.edu/as/tampl/en43/lecture_notes/ch4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MysRj.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, however, $T_\infty$ does not stay fixed at 3,000 K over time but changes (cooling) due to heat transfer through the cylinder wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I was not able to find any solved problem like that in textbooks because any existing solved problem considers the temperature inside the bulk of gas as a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I model this problem?&lt;/p&gt;&#xA;" OwnerUserId="4043" LastEditorUserId="368" LastEditDate="2015-12-16T00:05:08.357" LastActivityDate="2015-12-17T16:38:10.953" Title="How to model heat transfer through a pressure vessel wall if the reservoir temperature is not constant" Tags="|mechanical-engineering|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6448" PostTypeId="2" ParentId="6443" CreationDate="2015-12-07T13:47:16.073" Score="3" Body="&lt;p&gt;I think brazing would be correct for your application. Unless, the entire tray system needs to be stainless steel for hygienic and/or anti-corrosive reasons. Brazing is like soldering, you likely may be able to find a smaller handheld torch and the correct brazing rod for less than the cost of a TIG welding service. Also, it will require some skill and practice to get it right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, depending on your budget, you may want to look into a spot welder. That might do the trick for you. Or, speak with a sheet metal fabrication shop and ask them if they can spot weld your trays. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="368" LastEditDate="2015-12-16T00:15:08.447" LastActivityDate="2015-12-16T00:15:08.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6449" PostTypeId="1" AcceptedAnswerId="6465" CreationDate="2015-12-07T14:26:56.197" Score="5" ViewCount="422" Body="&lt;p&gt;I'm studying the design of storm sewers and I had two questions relating to air gaps in storm pipes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;According to sources I've come across, flow through storm pipes&#xA;require an air gap so as to maintain open channel flow. Do I need to&#xA;maintain a specific amount of air gap? Or will any amount of air gap do?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, does it &lt;em&gt;really&lt;/em&gt; matter whether or not I have an air gap? Do we&#xA;maintain an air gap so as to prevent a partial vacuum and self siphonage within the pipes? In which case why would it matter? Furthermore, since&#xA;storm sewer manholes (inlet pits) are open to the air, wouldn't that alleviate  any pressure differentials?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3832" LastActivityDate="2015-12-08T17:34:58.093" Title="How much air gap is required in a storm pipe?" Tags="|civil-engineering|hydrology|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6450" PostTypeId="1" CreationDate="2015-12-07T14:49:15.400" Score="10" ViewCount="5508" Body="&lt;p&gt;A cylindrical or a spherical pressure vessel, having inner diameter $d$ &amp;amp; thickness of wall $t$, is called thin if $\large \frac{t}{d}&amp;lt;\frac{1}{20}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bullet$ Thin cylindrical pressure vessel &lt;/p&gt;&#xA;&#xA;&lt;p&gt;hoop or circumferential stress $ \sigma_h=\large\frac{pd}{2t}$  &amp;amp; longitudinal stress $ \sigma_l=\large\frac{pd}{4t}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bullet$ Thin spherical pressure vessel &lt;/p&gt;&#xA;&#xA;&lt;p&gt;hoop or circumferential stress $ \sigma_h=\large\frac{pd}{4t}$  &amp;amp; longitudinal stress $ \sigma_l=\large\frac{pd}{4t}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both the cases we consider only two stresses i.e. the system of plane stresses neglecting the radial stress $(\sigma_r)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is why do we neglect the radial stress for thin pressure vessels? &#xA;Can it be explained by Lame's equation used for thick pressure vessels:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_h, \sigma_r=\frac{B}{x^2}\pm A\ \text{?}$$  &lt;/p&gt;&#xA;" OwnerUserId="4042" LastEditorUserId="4042" LastEditDate="2015-12-09T05:19:26.090" LastActivityDate="2016-05-21T22:56:50.073" Title="Why do we neglect the radial stress in case of a thin pressure vessel?" Tags="|mechanical-engineering|applied-mechanics|pressure-vessel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6454" PostTypeId="2" ParentId="6450" CreationDate="2015-12-07T16:51:10.297" Score="10" Body="&lt;p&gt;I've been doing large pressure vessel design for a while, and ultimately it comes down to what is negligible v. what is not.  Especially in the realm of atmospheric storage vessels, scrubbers, or vessels within 10 atmospheres of pressure, there isn't a lot of radial stress compared to the &lt;em&gt;other conservative&lt;/em&gt; factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the realistic engineering world, by the time you deal with the following uncertainties (all within tolerance):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Vessel is never round (typical tolerance is 0.5%)&lt;/li&gt;&#xA;&lt;li&gt;Vessel is never the correct diameter (typical tolerance is 1%)&lt;/li&gt;&#xA;&lt;li&gt;Environmental loading is only defined by a once in 50 year hazard (while vessel life is typically 20 years)&lt;/li&gt;&#xA;&lt;li&gt;Environmental loads are defined by 3-7 empirical factors, with errors in the 1%-15%, typically conservative.&lt;/li&gt;&#xA;&lt;li&gt;Buckling loads typically dominate on any vessel undergoing vacuum or handling atmospheric pressure - this is due to environmental axial buckling (i.e. the vessel is bending under the environmental loading, and may buckle at the inside throat).&lt;/li&gt;&#xA;&lt;li&gt;Factors of safety greater than necessary to ensure conservative design.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By the time the radial stress is added in and evaluated, the change in Von Mises stress is significantly lower in magnitude than these additional factors.  In addition, by eliminating the radial stress, a linear relationship can be utilized between pressure, diameter, and thickness - allowing for simple, easy estimating solutions for use in deriving prices of vessels.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-12-07T17:17:33.860" LastActivityDate="2015-12-07T17:17:33.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6456" PostTypeId="2" ParentId="6381" CreationDate="2015-12-07T18:43:37.627" Score="1" Body="&lt;p&gt;So I figured it out, I first took this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(C_F) = (CT_F) - (E)(P_a)/(P_C)_N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I isolated $(CT_f)$ and substituted $(C_F)$ equation resulting in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C = \sqrt{\frac{2y^2}{y-1)}*\frac{2}{y+1}^{y+1/y-1}*(1-\frac{P_e}{P_C}^{\frac{y-1}{y}})+e(\frac{P_e - P_a}{P_C})} + e\frac{P_e}{P_C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I solved for P_e using this equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = \frac{(\frac{2}{y+1})^\frac{1}{y-1}*(\frac{(P_C)_n}{P_e})^\frac{1}{y}}{\sqrt{\frac{y+1}{y-1}*(1-\frac{P_e}{(P_C)_N})^(\frac{y-1}{y})}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value $(P_a)$ is ambient pressure, so I will plugin in 14.6959488 in the place of $(P_a)$. 14.6959488 is the pressure in psi at ground level ( I hate imperial but the book's units are imperial ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give me a value for $(CT_F)$ which I can then plug back into the original equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(C_F) = (CT_F) - (E)(P_a)/(P_C)_N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which will give me the value of $(C_f)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;only took me two weeks.&lt;/p&gt;&#xA;" OwnerUserId="3976" LastEditorUserId="3976" LastEditDate="2015-12-08T15:39:21.617" LastActivityDate="2015-12-08T15:39:21.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6457" PostTypeId="2" ParentId="659" CreationDate="2015-12-07T22:53:50.400" Score="3" Body="&lt;p&gt;A little late to the game. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A method recommended in B&amp;amp;W's &lt;strong&gt;Steam, Its Generation and Use&lt;/strong&gt;, is to drill small, closely set holes in the metal substrate, insert the individual wires into the holes, and peen them in place.  This avoids the issues others have raised of welding affecting the properties.  It provides intimate contact with the measured surface.  It does not introduce any additional materials.  It will serve for any temperatures that the substrate and thermocouple materials themselves can withstand.  It should be moderately robust mechanically.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net couple is the same as that of the standard wire to wire joint, but is made of each of the wires to the substrate, taken in series.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At my suggestion, a client used this method successfully in a fired boiler.  &lt;/p&gt;&#xA;" OwnerUserId="998" LastActivityDate="2015-12-07T22:53:50.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6458" PostTypeId="2" ParentId="6441" CreationDate="2015-12-07T23:42:07.417" Score="1" Body="&lt;p&gt;It's not an easy problem, there are a few things going on at the same time here.  Most obviously, the piston has kinetic energy from being accelerated through the bore, which gets transferred to the test item upon impact.  This can be calculated from $E_{KE}=\frac{1}{2}mv^2$, as you stated.  What's not as obvious is the additional energy transferred to the surface by the piston if it's still being pushed by compressed air during the impact.  The total energy delivered by the piston can be hard to calculate, so you might be better off trying to measure it instead.  Or, maybe use calculations to get in the ball park then confirm with experiment.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could launch an object vertically with the piston and measure it's maximum height, or use a pendulum.  The change in potential energy of either will be the energy transferred by the piston.  Use the formula $E=mg\Delta h$ to calculate the delivered energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, depending on your application, the time interval over which the impact occurs could be important.  Materials behave differently under different rates of strain; armor is tested differently than structural steel, for example.  You'll want to verify that the piston impacts the surface at a speed representative of your final application.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-12-07T23:42:07.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6460" PostTypeId="1" AcceptedAnswerId="6461" CreationDate="2015-12-08T04:51:30.537" Score="3" ViewCount="124" Body="&lt;p&gt;I have two columns of water at different (but constant) temperatures. If I calibrate a pressure transducer using the first column at differing heights, when I measure the second column how do I account for the change in temperature (and water density)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to measure pressure in megapascals (MPa) so for the calibration I currently plot pressure transducer mV output at different column heights against the calculated megapascal for the corresponding height and temperature of water using the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pressure = height (m) x water density (kg/m&lt;sup&gt;3&lt;/sup&gt;) x gravitational constant (m/s&lt;sup&gt;2&lt;/sup&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the density of the second column is different so this value needs to be adjusted.&lt;/p&gt;&#xA;" OwnerUserId="4049" LastEditorUserId="368" LastEditDate="2015-12-15T00:32:43.617" LastActivityDate="2015-12-18T19:20:17.930" Title="Calibrating and measuring pressure in Megapascals using a pressure transducer" Tags="|fluid-mechanics|pressure|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6461" PostTypeId="2" ParentId="6460" CreationDate="2015-12-08T13:57:58.833" Score="2" Body="&lt;p&gt;You can either compensate for it in your measurement or neglect it and lump it in with the other error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First though, you need to decide if the error would even be worth correcting. At 4 °C, the density of water is 1000 kg/m^3 and 958 kg/m^3 at 100 °C. That's a maximum of 4.2% at those temp extremes. Over small variations in temperature, correcting for it is like picking the fly poop out of the pepper... it's insignificant and pales in comparison to the other sources of error like your instrumentation. The real question is: Can your application tolerate that error? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do decide to compensate for it, you need to measure temperature (and that has it's own errors) and in your equation make density a function of temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ p = h_{col} * \rho (T_{water}) * g $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess though, based on experience is that once you look into it, the variation in density due to expected water temperature changes is going to be negligible compared to things like manually reading a height scale unless you're doing some crazy stuff.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="368" LastEditDate="2015-12-15T00:33:24.633" LastActivityDate="2015-12-15T00:33:24.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6463" PostTypeId="1" AcceptedAnswerId="6476" CreationDate="2015-12-08T14:52:14.787" Score="6" ViewCount="831" Body="&lt;p&gt;Normal sealings and gaskets are made from NBR for use in conjunction with aqueous solutions of ethylene glycol which are common coolants for various purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However to keep sulphur out of a cooling system and the surrounding area I want to change NBR sealings against silicone sealings. I've read some vague information about reduced stability of these sealings in glycol containg coolants. But what is the detailed behaviour? Will it intumesce, brittle or deteriorate otherwise?&lt;/p&gt;&#xA;" OwnerUserId="3786" LastEditorUserId="282" LastEditDate="2015-12-15T09:12:38.077" LastActivityDate="2020-05-14T17:27:08.700" Title="What happens to silicone gaskets (siloxane rubber) in presence of ethylene glycol?" Tags="|materials|cooling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6465" PostTypeId="2" ParentId="6449" CreationDate="2015-12-08T16:06:19.033" Score="5" Body="&lt;p&gt;You're right, if your pipe is completely full then you don't have open channel flow. In other words, you can no longer use &lt;a href=&quot;https://en.wikipedia.org/wiki/Manning_formula&quot;&gt;Manning's equation&lt;/a&gt; to model the flow in that pipe. You would have to use either &lt;a href=&quot;https://en.wikipedia.org/wiki/Hazen%E2%80%93Williams_equation&quot;&gt;Hazen-Williams&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation&quot;&gt;Darcy-Weisbach&lt;/a&gt; as you can no longer ignore the pressure term in Bernoulli's equation. As the depth of the water approaches full pipe depth, Manning's is going to become increasingly unreliable. So &lt;strong&gt;any&lt;/strong&gt; air gap, while theoretically would relieve pressure, you can't rely on Manning's alone to model this. Furthermore, even if you model a pipe using pressure equations, models do not reflect the reality of how the pipe was installed anyways. So you shouldn't just say, yes I've provided 10 cm of air gap, therefore the pipe is not pressurized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, you don't &lt;strong&gt;have&lt;/strong&gt; to ensure open channel flow in a pipe, you could design a system to function near full pipe for a 100-year storm. But as you mentioned in the second part of your question, any nearby manholes or inlets will surcharge and overflow, thus alleviating your pressure differentials (and flooding your site). If there is nothing to alleviate the pressure, the overflow can literally carry away manhole lids or blow them off in spectacular fashion in extreme cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider that if your designed pipe is at capacity the amount of additional flow before you see flooding will be very small. Lastly, pressurized pipes are going to have additional stresses on them that will limit their durability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have to consider whatever regulations that you are bound by (minimum pipe size, for example) and economical considerations. You don't want to install a pipe that never exceeds 50% capacity, but I always consider that if you're digging the trench no matter what, it's better to be safe and go with a larger pipe to save yourself a headache later when you don't have enough capacity and you cause flood damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gold standard for modeling all of these situations easily is &lt;a href=&quot;https://en.wikipedia.org/wiki/Storm_Water_Management_Model&quot;&gt;SWMM&lt;/a&gt; which exists as the computational engine in various design softwares. The free &lt;a href=&quot;http://www.epa.gov/water-research/storm-water-management-model-swmm&quot;&gt;EPASWMM&lt;/a&gt; is a good place to start if you are researching modeling pipes.&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2015-12-08T16:06:19.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6466" PostTypeId="2" ParentId="6449" CreationDate="2015-12-08T17:34:58.093" Score="3" Body="&lt;p&gt;There are additional features of the air gap besides pressure relief (Especially in small domestic pipes).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Maintenance clean out access&lt;/li&gt;&#xA;&lt;li&gt;View port for viewing if flow is blocked&lt;/li&gt;&#xA;&lt;li&gt;Alleviating a small amount of BOD from system&lt;/li&gt;&#xA;&lt;li&gt;Reducing odor problems with adequate air flow&lt;/li&gt;&#xA;&lt;li&gt;Preventing stagnant water formation while not storming.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of these should be considered in addition to modeling - otherwise you might eliminate some of the standard &quot;features&quot; normally seen.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-08T17:34:58.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6467" PostTypeId="1" CreationDate="2015-12-08T17:49:43.313" Score="14" ViewCount="6718" Body="&lt;p&gt;I am a homeowner who likes to save money by performing &lt;a href=&quot;https://en.wikipedia.org/wiki/Do_it_yourself&quot;&gt;DIY work&lt;/a&gt;. When possible, I prefer to serve as my own &lt;a href=&quot;https://en.wikipedia.org/wiki/General_contractor&quot;&gt;general contractor&lt;/a&gt; to reduce the cost of a project. However, when I talk with professionals about my plans and ask for advice, they often tell me that I need to consult with a professional engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want my home to be safe but hiring an engineer sounds like an expense I can't afford. How can I tell when it's really necessary to consult an engineer before performing work on my home? Why can't I just hire a licensed contractor instead?&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2015-12-08T17:59:37.160" LastActivityDate="2018-09-24T18:09:38.320" Title="When do I need to consult a professional engineer during a home renovation?" Tags="|home-improvement|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6469" PostTypeId="2" ParentId="6467" CreationDate="2015-12-08T19:54:40.017" Score="6" Body="&lt;p&gt;I can't answer for every home, but I do know California Board of Professional Engineer's regulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The board dictates for &lt;a href=&quot;http://bpelsg.ca.gov/pubs/building_design_auth.pdf&quot;&gt;design&lt;/a&gt;, that anyone can design an improvement of a two story home with a basement.  However, this comes with a two-fold liability for the design:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The design must meet with pre-engineered building designs in California Building Code:  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If any portion of the structures described above does not meet the conventional woodframe requirements described in Title 24 of the California Code of&#xA;  Regulations or in the building codes of the local jurisdiction, then the building official having jurisdiction shall require the plans, calculations, and&#xA;  specifications for that portion of the structure to be prepared and signed and sealed by a licensed engineer or a licensed architect.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;If there are any problems, the liability goes directly to the designer.  &lt;a href=&quot;http://bpelsg.ca.gov/laws/pe_act.pdf&quot;&gt;Per California Professional Engineering Code 6737.3&lt;/a&gt;:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This section shall not prohibit a licensed contractor, while engaged in the business of&#xA;  contracting for the installation of electrical or mechanical systems or facilities, from designing&#xA;  those systems or facilities in accordance with applicable construction codes and standards for&#xA;  work to be performed and supervised by that contractor within the classification for which his or&#xA;  her license is issued, or from preparing electrical or mechanical shop or field drawings for work&#xA;  which he or she has contracted to perform. Nothing in this section is intended to imply that a&#xA;  licensed contractor may design work which is to be installed by another person. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As such, if you want someone else to design something new, and have the liability for designing it, they need to be a Professional Engineer.  If you want to modify it yourself, you can - so long as you follow the pre-engineered plans.  But the liability falls upon yourself.  No contractor will design it for you to do it yourself - that's the law.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-08T19:54:40.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6471" PostTypeId="2" ParentId="489" CreationDate="2015-12-08T22:20:05.397" Score="3" Body="&lt;p&gt;Carbides of heavy metals are always good candidates. Tungsten carbide has 15,63 g·cm−3. The other parameters are also &quot;impressive&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="3786" LastActivityDate="2015-12-08T22:20:05.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6474" PostTypeId="2" ParentId="6467" CreationDate="2015-12-09T04:25:41.577" Score="3" Body="&lt;p&gt;To add to &lt;a href=&quot;https://engineering.stackexchange.com/a/6469/368&quot;&gt;Mark's answer&lt;/a&gt;; most houses are built based on state building code, not designed by a &lt;a href=&quot;http://www.ipels.idaho.gov/forms_pubs/se_faq.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Structural PE&lt;/a&gt; (structural is a defined specialization in many states). A Structural PE is expensive and usually only comes into play when you want to do something tricky that is outside of what the code allows. The code can only cover so many cases. For structural, material, and load exceptions, you need a Structural PE. And for spacing, layout, and use exceptions you need an Architect. Many firms will have both on staff. An experienced general contractor is good at staying within what the code supports because it is much cheaper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are doing a major remodel it is often good to go to an engineer or architect, because they do this for a living and will catch the things you might not think of like toilet spacing or egress requirements. Its much cheaper to get those things correct upfront as opposed to when you go to install things or when your house is inspected before a sale. Also, if you are getting a loan for a remodel a bank will require a drawing of the proposed layout and sometimes require that drawing to be from an architect or engineer.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-12-09T16:46:32.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6475" PostTypeId="2" ParentId="6467" CreationDate="2015-12-09T04:39:19.437" Score="14" Body="&lt;p&gt;Here's a simple analogy&lt;sup&gt;1, 1a&lt;/sup&gt; - if you're planning on modifying the skeleton of your house, then you ought to consult with a licensed Engineer before committing to any work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if your renovation is altering one of the house's systems, such as the plumbing, electrical, or heating, then you can most likely rely upon the contractor performing the work and you do not need to consult with a licensed Engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;So what's the skeleton of a house?&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;It's the framework of the foundation, walls, floors, and ceilings of the building.  More accurately, it is the structural components that keep your dwelling standing as well as held together.  Also keep in mind that significant amounts of nearby earthwork can affect the structure of the house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5skQN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5skQNt.jpg&quot; alt=&quot;House structure&quot;&gt;&lt;/a&gt; &#xA;&lt;a href=&quot;https://i.stack.imgur.com/shf9E.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/shf9Et.jpg&quot; alt=&quot;Foundation&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on link for larger image&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;So what counts as a modification then?  It's obviously more than drilling a hole into a wall in order to hang a picture, right?&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;A structural modification means moving or removing a piece that was originally part of the dwelling's skeleton.  For example, there might be metal posts holding up beams running the length of your basement or garage.  Those posts are helping transfer forces from one area of the house into a lower area.  Moving those posts would affect the skeleton of the house.  Another example would be the joists between the floors of the house.  Those joists help distribute the load within each floor of the house.  Significant modifications to them could impact the structural integrity of the house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qalvF.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qalvFt.jpg&quot; alt=&quot;Load bearing wall&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on link for larger image&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Okay, so I can't touch anything associated with the foundation or framing of the house, right?&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Well, probably not.  But that's Murphy's Law speaking.  In other words, the thing you want to change is likely to be a critical part of your house's structural integrity.  But there are non-load-bearing walls within a house.  Believe it or not, those walls can be knocked out with almost zero impact to the structural integrity of the house. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Wait, what? So I can rip out the wall I want?&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Slow down again.  Yes, there are &lt;em&gt;some&lt;/em&gt; walls that can safely be ripped out.  There are (many) other walls where you'd be foolish to do so.  How do you tell the difference?&lt;sup&gt;2&lt;/sup&gt;  Hire a structural Engineer and have a proper analysis done on your dwelling and the proposed change.  Reasonably speaking, the original construction of the house wouldn't have put the majority of the walls in place unless they were necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Blah, blah, blah.  You're just trying to perpetuate a racket and drive up Engineering services revenue.&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Please keep in mind that &lt;a href=&quot;http://www.nspe.org/resources/ethics/code-ethics&quot; rel=&quot;noreferrer&quot;&gt;Professional Engineers have an ethical obligation to uphold individual and societal safety&lt;/a&gt; over any considerations for profit.  The reality is that making changes to the structural elements of your house can result in catastrophic damage to your house if incorrectly performed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most good contractors will avoid any potentially catastrophic project from a healthy fear of the liability associated with the project.  That's a key warning sign that you should heed.  That apparently expensive consult with a licensed Engineer is nothing compared to the potential expenses you would be facing to fix an errant and ill advised change to the structure of your dwelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Ah, c'mon.  It's just a simple, stupid post that's in my way.  Moving it won't affect anything.  Right?&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Quite the contrary.  Shifting where loads are transferred can increase the amount of applied force by an exponential rate.  Said another way, &quot;simple&quot; shifts in things can create really big problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Okay, wait.  Wow! this is apparently very complicated stuff.  What should I do?&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Structural analysis &lt;strong&gt;&lt;em&gt;is&lt;/em&gt;&lt;/strong&gt; complicated, and there are a number of factors involved that have to be considered.  Generally speaking, the best thing you can do is to consult with a licensed Engineer and let them know what it is you &lt;em&gt;want&lt;/em&gt; to try and accomplish.  Also let them know what you're willing to compromise upon, so they can help balance between project expenses and requirements.  Their job is to help you identify the right balancing point between your dream project and the cold, harsh reality of physics.  As an added bonus, they have likely already helped many other clients address similar projects.  And they will likely be able to point out additional considerations that you would not have considered.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Footnotes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1:&lt;/sup&gt; Please note, this answer is intentionally written in a conservative perspective.  Why? Because it's the prudent approach to take when the consequences of a mistake could be catastrophic or just exceptionally expensive to repair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1a:&lt;/sup&gt; Also note, there are many variables at play and the range of considerations can be quite wide.  There are modifications that should be answered with &quot;Are you crazy?!&quot; as well as modifications where the answer is &quot;Sure, that shouldn't be an issue.&quot;  This answer is meant to provide a starting point as a simple &quot;rule of thumb.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2:&lt;/sup&gt; Well scoped questions could potentially be asked here on &lt;a href=&quot;http://engineering.stackexchange.com&quot;&gt;Engineering&lt;/a&gt; or on &lt;a href=&quot;http://diy.stackexchange.com&quot;&gt;DIY / Home Improvement&lt;/a&gt;.  Asking about how to identify non-load-bearing walls, or asking about what building codes allow for modifications to joists ought to be on-topic for one or both of those sites.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-12-09T19:40:00.733" LastActivityDate="2015-12-09T19:40:00.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6476" PostTypeId="2" ParentId="6463" CreationDate="2015-12-09T04:41:08.513" Score="3" Body="&lt;p&gt;Based on the available data it looks like it will perform well: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.efunda.com/designstandards/oring/oring_chemical.cfm?SM=none&amp;amp;SC=Ethylene%20Glycol&quot; rel=&quot;nofollow&quot;&gt;efunda.com -  (4) Good, both for static and dynamic seals&lt;/a&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pspglobal.com/fluid-compatibility/chemicals-e.html&quot; rel=&quot;nofollow&quot;&gt;pspglobal.com -  1 = Recommended&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Temperature and additives may change this, however.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="368" LastEditDate="2015-12-15T00:31:47.493" LastActivityDate="2015-12-15T00:31:47.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6477" PostTypeId="2" ParentId="6442" CreationDate="2015-12-09T05:26:16.940" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Volumetric_flow_rate&quot; rel=&quot;nofollow&quot;&gt;Volumetric flowrate&lt;/a&gt; is conserved; giving the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q.soruce = Q.drain1 + Q.drain2 + Q.drain3 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to determine how much flow goes where you have to determine the flow resistance of each. To reduce complexity (especially the first time around) you should hold the upstream pressure constant (this is actually the pressure in the center of your fork. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P = 700 kPa &lt;a href=&quot;https://en.wikipedia.org/wiki/Pressure_measurement#Absolute.2C_gauge_and_differential_pressures_-_zero_reference&quot; rel=&quot;nofollow&quot;&gt;Gauge&lt;/a&gt; (above atmospheric)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many factors that affect flow resistance. The primary factors are pipe diameter and length. Obviously the most simple case is all three resistances being equal and all 3 pressure drops being equal, so consequently all flows would also be equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve something more complicated we first need to know the pressures at all 3 outlets. If it is to atmosphere the gauge pressure is 0kPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once pressure drops are established we need to calculate resistance either theoretically for simple geometry or empirically for complex geometry. There is also the option of Computational Fluid Dynamics for theoretically solving complex geometry. Some good places to start are &lt;a href=&quot;http://www.engineeringtoolbox.com/major-loss-ducts-tubes-d_459.html&quot; rel=&quot;nofollow&quot;&gt;major losses&lt;/a&gt; and &lt;a href=&quot;http://www.engineeringtoolbox.com/orifice-nozzle-venturi-d_590.html&quot; rel=&quot;nofollow&quot;&gt;nozzle pressure drop&lt;/a&gt;. If you are dealing with low pressure or big changes in height you will also have to account for gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have all the outlet resistances and all the outlet pressure drops, you can solve for each outlet flow; then sum them to get the total flow. You can then remove one variable and solve for it under different conditions or divide the flows over the total flow to get ratios.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="368" LastEditDate="2015-12-16T00:23:20.487" LastActivityDate="2015-12-16T00:23:20.487" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6478" PostTypeId="1" CreationDate="2015-12-09T11:29:41.790" Score="1" ViewCount="2142" Body="&lt;p&gt;I am a student in a Drafting 1 class and we are going over all of the section views (full, half, revolved, offset, etc.) and also all about the cutting planes and different ways for that and how to do them in Autocad. Anyways I'm reading from the book and I still don't understand it very well.  Could someone here explain it well?&lt;/p&gt;&#xA;" OwnerUserId="4069" LastEditorUserId="1832" LastEditDate="2015-12-09T12:18:46.747" LastActivityDate="2015-12-09T19:23:22.400" Title="What are sectional views and cutting plane lines?" Tags="|drafting|autocad|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6480" PostTypeId="1" CreationDate="2015-12-09T14:08:53.660" Score="6" ViewCount="752" Body="&lt;p&gt;We have a boiler exhaust stack with a base dimension of 120&quot; ID and an overall height of 84'.  21' up, the stack begins a slow transition to 7'6&quot; ID.  The transition takes approximately 35' to reach its smallest diameter in the stack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In designing the new stack I was wondering if I could just make a straight stack or if the transition is actually performing a function.  Why would a boiler exhaust stack be tapered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existing stack is manufactured of Alloy 5052-F and I want to make the new one of CS construction.&lt;/p&gt;&#xA;" OwnerUserId="4071" LastEditorUserId="368" LastEditDate="2015-12-11T19:27:36.137" LastActivityDate="2015-12-11T19:27:36.137" Title="Why do boiler exhaust stacks use tapered transitions?" Tags="|mechanical-engineering|hvac|boilers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6481" PostTypeId="1" AcceptedAnswerId="6487" CreationDate="2015-12-09T14:35:09.580" Score="6" ViewCount="322" Body="&lt;p&gt;I want to contain a very small PCB (roughly a square centimetre) in a casing. For demonstration purposes I want to 3D print a couple of those casings as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious method available would be using ABS for the printed casings. However, nylon appears to be smoother and tougher. Those traits are often a plus during demonstrations. However, since the PCB contains a S band antenna I'm slightly worried about the RF properties of the material used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any downsides to switching from ABS to nylon that I'm not seeing?  I'm especially interested in the RF characteristics in 2-4 GHz range. Cost is not an issue.&lt;/p&gt;&#xA;" OwnerUserId="1070" LastEditorUserId="33" LastEditDate="2016-01-27T02:44:59.417" LastActivityDate="2016-01-27T02:44:59.417" Title="Are there any differences between using nylon or ABS for RF electronic cases?" Tags="|electrical-engineering|materials|rf-electronics|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6482" PostTypeId="2" ParentId="6478" CreationDate="2015-12-09T19:23:22.400" Score="3" Body="&lt;p&gt;Imagine you wanted to show a drawing of a car. You can show the &lt;em&gt;exterior&lt;/em&gt; features just fine, but how do you show the interior features? If, for instance, you wanted to show where the steering wheel was with respect to the driver's seat, how could you do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of different ways to do it, but the most straightforward is with a &lt;em&gt;section view&lt;/em&gt;. In a section view, you take an imaginary saw and &lt;em&gt;cut&lt;/em&gt; the object along a &lt;em&gt;section plane&lt;/em&gt;. Now you have a cut-open car! Now, if you were to look at the side of the car, the passenger's door, passenger seat, ... the entire passenger's half of the car is gone! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all that is left is the &lt;em&gt;driver's&lt;/em&gt; side of the car, and in looking at it in profile with the passenger's half removed you can now clearly see where the steering wheel lies with respect to the driver's seat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put together a crude drawing for you to follow along with below. Note the arrows on the section view - the dashed line shows the person viewing the drawing where the object was cut, and then the arrows show which half you're going to look at. If the arrows pointed down instead of up then you would be looking at the passenger's half of the (American) car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4hNej.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4hNej.png&quot; alt=&quot;Section view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, notice that hatching /////// or \\\ or XXXXXX that exists in the section view. This shows you which surfaces were cut. Surfaces that are not hatched were not cut. So, in the example above, you could infer that the back seat is maybe a bench seat that spans the width of the car, where the driver's seat is a bucket seat that is not attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're kind of awkward to look at in the beginning, but once you get the concept (cut the object open, then look at the cut surface), then it becomes a lot easier. Section views can show lots of things, but internal features that are otherwise hard to display would be the most common reason I would use a section view. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-12-09T19:23:22.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6483" PostTypeId="2" ParentId="6480" CreationDate="2015-12-09T19:30:19.073" Score="6" Body="&lt;p&gt;The taper in the stack performs two functions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;A tapered section is a lot more structurally resistant to wind loads&lt;/strong&gt;  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This seems counter intuitive at first - after all, a tapered cantilever beam with a fixed load at the end deflects more than a straight beam.  However, there is less surface area at the top of the stack for winds to push. Since, in the US, ASCE 07-10 and ASME STS-1 utilize a &lt;a href=&quot;https://en.wikipedia.org/wiki/Wind_gradient#Engineering&quot;&gt;wind gradient&lt;/a&gt;, the loads at the top of the stack are significantly larger than the loads at the base.  Due to reduction in surface area, this translates to less loading for the base to counteract, and overall reduces the stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical stack design of this construction is to size the top for &lt;em&gt;optimal velocity&lt;/em&gt;, and make that the wind design. Then factor in several methods for making the stack - a linear reduction is a simple fabrication, and tapering down thickness between sections (since you don't need to resist as much loading near the top as the bottom) makes for material cost reduction that are well worth the engineering effort for the manufacturer.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Outlet Velocity Increase&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By reducing the cross sectional area of the flow area at the outlet, additional velocity will increase the plume height and result in &lt;a href=&quot;http://yosemite1.epa.gov/oaqps/eogtrain.nsf/fabbfcfe2fc93dac85256afe00483cc4/c9862a32b0eb4f9885256b6d0064ce2b/$FILE/Lesson%206.pdf&quot;&gt;better dispersion&lt;/a&gt;, which avoids downwash effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are additional thermodynamic implications - for example preventing sulfuric acid condensation.  These are likely more a function of whatever acid brick you will use to line the steel stack and the thermal reduction (I should mention that certain types of phenolic resin coating can act as a great acid brick - it's worth investigating).&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-09T19:30:19.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6485" PostTypeId="2" ParentId="6442" CreationDate="2015-12-09T23:02:00.767" Score="2" Body="&lt;p&gt;This is essentially explaining how to solve parallel flow problems using the fundamentals on fluid dynamics - let me know if you need additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assign some variables.  Let's let:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Q1, Q2, Q3, and Q be the flow rates in each of the pipes (Q is the overall flow).  Note Q = V * A, where V1, V2, and V3 are the fluid velocities. A1, A2, A3 being the channel area used.&lt;/li&gt;&#xA;&lt;li&gt;Z1, Z2, Z3 and Z are the heights above sea level of each of the outlets (Z is the cross point).&lt;/li&gt;&#xA;&lt;li&gt;Let's assume you are working with manning formula - in this case, convert the Zs into different slopes - S1, S2, and S3.  If your drains slope changes along the way, you can model it using an average slope, or you can model using several pieces of pipe in parallel.  The average slope is easier, and generally accurate.&lt;/li&gt;&#xA;&lt;li&gt;While X, Y, and Z are nice - let's use L1, L2, and L3 for the lengths of each pipe.&lt;/li&gt;&#xA;&lt;li&gt;Finally, we need the energy in each of these points.  Note that power is used because for a &lt;em&gt;steady state&lt;/em&gt; assumption, the energy would be infinite.  So, Power = Flow Rate * Pressure (&lt;a href=&quot;http://www.wolframalpha.com/input/?i=1%20psi%20*%201%20gpm&quot; rel=&quot;nofollow&quot;&gt;Check out the units!&lt;/a&gt;).  Here, we use P1, P2, P3 and P for the powers at each ending - P Being the power at the cross point.  For the power consumed by friction along the length, let's use Pf1, Pf2, and Pf3.&lt;/li&gt;&#xA;&lt;li&gt;Rho is the density of the fluid.  D is hydraulic diameter (D1, D2, D3 for consistency).  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now the total power at the cross point must be the total power - expended via friction along the pipes or delivered at the outlet.  That's just &lt;a href=&quot;https://en.wikipedia.org/wiki/Conservation_of_energy&quot; rel=&quot;nofollow&quot;&gt;conservation of energy&lt;/a&gt;.  For a bunch of reasons though, power is not in units of Watts like it should be for fluid dynamics.  It's in units of length - the height of a reservoir of energy.  It's called head.  I'm explaining this in terms of power, and it's valid - but it's not historically accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power expended by the pipes: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pf1 = rho*Q1*f*(L1/D1)*(Q1/A1)^2/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f is of course the &lt;a href=&quot;https://en.wikipedia.org/wiki/Darcy_friction_factor_formulae&quot; rel=&quot;nofollow&quot;&gt;Darcy Friction Factor&lt;/a&gt;.  Of course, to solve for f, you need to obtain Q.  But Q is unknown!  So, we guess at an f, check the problem via iteration, solve for Q, then check that the guess for f was valid.  If we're close, move on - otherwise guess at a new f and keep going. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this friction formula in mind, then the power consumed by each of the channels is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P1 = (Z-Z1) * rho * Q1 + Q1^3*rho/(2*g*A1) + Pf1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, at last we have the equations needed to solve this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1 + Q2 + Q3 = Q&#xA;P1 + P2 + P3 = P&#xA;Q1 = A1*(k/n)*D1^(2/3)&lt;em&gt;S1^(1/2)&#xA;Q2 = A2&lt;/em&gt;(k/n)*D2^(2/3)&lt;em&gt;S2^(1/2)&#xA;Q3 = A3&lt;/em&gt;(k/n)*D3^(2/3)*S3^(1/2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this, we have five equations and 3 unknowns - the various Q1, Q2, and Q3.  The simplest solution is to iterate through and find 3 Qs to solve all 5 equations, disregarding the most empirical assumptions first (in this case, I'd disregard the f's - followed by the Manning equations.  Energy and Mass conservation is top priority).&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-09T23:02:00.767" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6486" PostTypeId="1" AcceptedAnswerId="16400" CreationDate="2015-12-10T00:35:23.007" Score="4" ViewCount="114" Body="&lt;p&gt;How far must the surface temperature of a UL-listed combustion appliance rise to prohibit the manufacturer from specifying it for a zero clearance to combustibles installation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I looked in UL 127 but was unable to find a straightforward answer there.)&lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2017-07-23T01:57:27.090" Title="What surface temperature (rise) is required to (dis)qualify a combustion appliance from obtaining a &quot;zero clearance to combustibles&quot; rating?" Tags="|combustion|heating-systems|ul|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6487" PostTypeId="2" ParentId="6481" CreationDate="2015-12-10T03:01:58.627" Score="3" Body="&lt;p&gt;I am not sure if this will help, but when I have used any plastic in antenna designs, or RF power amplifiers, I would put them in a microwave oven and see if they heated up. Most would stay cold, but every now and then some would get warm, and sometimes I could hear some audible arcing. My only experience is at Ham Radio frequencies but I think it may still be a valid test for what you are doing. After all, a microwave oven works at 2.450 GHz. This has always worked for me to see if any insulating material would react to RF.&lt;/p&gt;&#xA;" OwnerUserId="4063" LastEditorUserId="368" LastEditDate="2015-12-15T00:20:09.933" LastActivityDate="2015-12-15T00:20:09.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6488" PostTypeId="2" ParentId="6406" CreationDate="2015-12-10T10:12:06.797" Score="1" Body="&lt;p&gt;Based on the answers here it shows that interpretation of my results was wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was independently validated and supported by Bruce Huang of Battery University. Main points from him were&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Parasitic Load would not delay current drop indefinitely, it does do momentarily.&lt;/li&gt;&#xA;&lt;li&gt;Keeping the screen on at 50% load does not create a significant load to create noticeable parasitic load&lt;/li&gt;&#xA;&lt;li&gt;Charging time may vary depending on the load&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Full text of his reply and my answer based on his and inputs obtained from &#xA;&lt;a href=&quot;https://android.stackexchange.com/a/131169/131553&quot;&gt;this answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would still hope to obtain some inputs on OEM implementation of charging&lt;/p&gt;&#xA;" OwnerUserId="3945" LastEditorUserId="1832" LastEditDate="2017-11-20T16:53:28.760" LastActivityDate="2017-11-20T16:53:28.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6489" PostTypeId="2" ParentId="6486" CreationDate="2015-12-10T12:51:29.253" Score="1" Body="&lt;p&gt;I'd have posted this as a comment, but too much text.  I found a &lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=6&amp;amp;ved=0ahUKEwjbk-yHqdHJAhWIWSYKHQCWAEMQFgg7MAU&amp;amp;url=http%3A%2F%2Fsweets.construction.com%2Fswts_content_files%2F5624%2F743355.doc&amp;amp;usg=AFQjCNFhdH5haYeaYXIkk5_-NsxZPcKbxw&amp;amp;sig2=xsz4wYxFKwuTdi7VaLpnBg&quot; rel=&quot;nofollow&quot;&gt;doc online&lt;/a&gt; whose preamble listed all the following specifications as being related to zero-clearance design:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The following published specifications, standards, or tests that apply to flexible, fire rated grease dust wrap in this section:&#xA;&#xA;A.  NFPA 96 (All Editions, including 2004, 2008, 2011), NFPA 101, 90A&#xA;B.  International Code Council Evaluation Service (ICCES)&#xA;C.  1997 ICBO Uniform Mechanical Code (ICBO UMC)&#xA;D.  1997 Uniform Building Code (UBC)&#xA;E.  2006, 2009 and 2012 International Mechanical Code (IMC) &#xA;F.  2006 and 2009 IAPMO Uniform Mechanical Code (IAPMO UMC)&#xA;G.  2010 National Building Code of Canada&#xA;H.  ASTM E2336 Internal Fire Test, Zero Clearance to Combustibles, Grease Duct Enclosure&#xA;I.  CAN/ULC-S144-09 and -12 Standard Method of Fire Resistance Test – Grease Duct Assemblies.&#xA;J.  ASTM E-119 Engulfment Fire Test for 2 Hour Grease Duct Enclosure&#xA;K.  ASTM E-119 Fire Wall Test, 2 Hour Rating&#xA;L.  ASTM E-84 Standard Test Method for Surface Burning Characteristics of Building Materials&#xA;M.  ASTM E-814, UL/ULC 723 Standard Test Method for Fire Tests of Through-Penetration Fire Stops&#xA;N.  ASTM E-136 Standard Test Method for Noncombustibility&#xA;O.  ASTM E-518 Standard Test Method for Thermal Resistance (Durability)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;An ad (PDF) for some ductwork quotes the following specs that their product meets:&#xA;SETTING THE STANDARD…&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ZeroClear™ meets the following stringent codes and standards:&#xA;• UL1978 – Standard for Grease Ducts – 2000° test&#xA;• UL2221 – Standard for Tests of Fire Resistive Grease&#xA;Duct Enclosure Assemblies&#xA;• ASTM E2336 – Standard Test Methods for Fire Resistive&#xA;Grease Duct Enclosure Systems – 1 hour (max) rating with ZeroClear&#xA;– 2 hour (max) rating with ZeroClear Plus&#xA;• NFPA96 – Standard for Ventilation Control and Fire Protection&#xA;of Commercial Cooking Operations&#xA;• International Mechanical Code&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How one tracks down all those specs is another problem :-(&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2015-12-10T12:51:29.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6490" PostTypeId="1" CreationDate="2015-12-10T14:39:23.453" Score="4" ViewCount="2671" Body="&lt;p&gt;&lt;strong&gt;In brief: between a standard quick-release skewer and a bolt-on skewer, which will provide greater clamping force?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;More detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern bicycle wheels are held in place by clamping skewers that run through hollow axles.  The typical &quot;quick release&quot; design has a built-in lever on a cam that you close with you hand to secure the wheel, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IH0eL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IH0eL.jpg&quot; alt=&quot;QR skewer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you prefer not to have the quick-release mechanism on your hub for whatever reason, it's possible to retrofit your axles with &quot;bolt-on&quot; skewers which forego the cam and lever for simpler hex heads:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ldZnb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ldZnb.jpg&quot; alt=&quot;Bolted skewer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The torque spec for bolt-on skewers is generally 7 N-m, while Shimano recommends 5.0 - 7.5 N-m tightening torque on their quick-release skewers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: How would these torque values applied through their respective systems (7.5 N-m through a cam versus 7 N-m through a threaded fastener) compare?  Which would likely result in a higher clamping force (if we're assuming proper lubrication with no contamination, etc.)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, people seem to believe the bolt-on design clamps down better, but I've seen some people insist the opposite is true.  Conventional wisdom in the bicycle world is notoriously dodgy, so I was hoping to get some expert insight.&lt;/p&gt;&#xA;" OwnerUserId="4085" LastEditorUserId="4085" LastEditDate="2015-12-10T17:06:24.587" LastActivityDate="2020-12-09T01:24:31.800" Title="Clamping force of a bicycle quick release skewer" Tags="|mechanical-engineering|bicycles|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6491" PostTypeId="2" ParentId="6490" CreationDate="2015-12-10T15:56:02.667" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My question: between these two designs, which could I expect to offer stronger clamping force?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The only truly measurable quantity here is the torque of the bolt-on skewer, which you have stated is 7 N-m.  Using this, it's possible to determine the clamping force of the skewer, but it's not really relevant right now (and we'd need to know a lot more about the skewer's properties).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're correct that the whole &lt;code&gt;&quot;leave a bit of a dent in the palm of your hand.&quot;&lt;/code&gt; thing is dodgy.  As far as I know, there is no ASTM standard test for hand denting (though this is tongue-in-cheek, I &lt;em&gt;really&lt;/em&gt; hope someone can prove me wrong, just because of how ridiculous that would be).  Personally, I like to torque the quick-release skewers on my bike so that I can barely get them closed, and I like to think I have pretty strong hands.  I've never stopped to look at the dent in my hand afterward, and my bike doesn't seem to notice a difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the variability inherent with this sort of rule-of-thumb approach, &lt;strong&gt;I don't think there's a way to truly answer this question.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2015-12-10T15:56:02.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6494" PostTypeId="1" AcceptedAnswerId="6504" CreationDate="2015-12-10T17:40:55.017" Score="4" ViewCount="1254" Body="&lt;p&gt;I have to calculate the uncertainty of measurement for an LVDT Linear displacement transducer used for measuring length. Uncertainty calculations are ruled by ISO 17025 but there are no (at least haven't found) specific guidelines for the mathematical procedure of uncertainty for LVDT sensors. There are many guides for more common measuring instruments like head micrometers and calipers though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any official document to which I can refer to for the requirements and maths for calculating the uncertainty of measurement for this kind of sensors?&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2015-12-11T10:13:23.453" Title="Uncertainty Calculation of LVDT displacement transducers for length measurement" Tags="|sensors|measurements|mathematics|standards|metrology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6497" PostTypeId="1" CreationDate="2015-12-10T19:56:17.010" Score="0" ViewCount="1187" Body="&lt;p&gt;Are there any materials, such as a type of tempered glass, that can not be drilled or penetrated by any means (steel drills, diamond drills, saws, lasers, ablation with focused ion beam machines, etc) without some sort of dramatic evidence of such drilling (cracks throughout the entire piece, or a dramatic change of color of an area covering several square centimeters, etc)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, are there materials that can not accept holes over a certain size, but that will accept smaller holes?  For instance, maybe there is a material that can accept 50 micron holes but not 1mm holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machinist may or may not have access to a full-fledged laboratory with million-dollar equipment, including things like focused-ion beam machines.&#xA;If there is a material that resists attacks using inexpensive machines (under $100,000) but not things like focused ion beam machines, that would still be interesting to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to build a tamper-evident container about the size of a brick, perhaps 120mm x 120mm x 180mm.  That rules out some materials, such as nitrogen triiodide, which can not be used to make containers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that there are existing ways of building these containers using electronic sensors connected to something, such as a random arrangement of printed wires encased in epoxy, or some sort of mechanical trigger.  I seek to develop a more sensitive scheme that is harder to bypass by having the container itself be made out of a material that amplifies the effect of an intrusion attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty familiar with tamper-evident seals, and find they all are either pretty easily defeatable (i.e. in a lab setting; perhaps they are good enough for particular field applications), or require detailed non-automated inspection to determine tampering (such as requiring the detailed comparison of before &amp;amp; after photos, as I hear is done using random solder and wire-brush marks on some nuclear program auditing seals).  I am also familiar with some micro-level tamper-evident schemes, such as putting an additional patterned metal layer on a chip, with capacitive sensors on the layer below to sense tampering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize these sorts of clever electronic or mechanical schemes can provide some degree of tamper evidence.  Lately I've become curious if these traditional approaches can be improved upon by deploying a special material that is inherently tamper evident.&lt;/p&gt;&#xA;" OwnerUserId="4092" LastEditorUserId="4092" LastEditDate="2015-12-12T01:48:39.513" LastActivityDate="2023-05-18T19:09:20.807" Title="Are there any materials that can not be drilled or sawed?" Tags="|materials|machining|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6498" PostTypeId="2" ParentId="6121" CreationDate="2015-12-10T22:01:34.043" Score="3" Body="&lt;p&gt;Just to add some unique engineering solutions that are used, depending upon applications:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Brick works wonders - it resists the acid in the smoke well and can withstand the heat.  Not metal, but well worth it if the application is right.&lt;/li&gt;&#xA;&lt;li&gt;For power plants, typical material use is acid brick lined steel.&lt;/li&gt;&#xA;&lt;li&gt;For industry, such as fume exhaust of potential violently reactive chemicals (not just limited to combustion), typically Fiberglass Ducting is utilized, or fluoropolymer coated steel ducting.  Both have excellent corrosion resistance, and the flouropolymer coating can withstand 150C.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-10T22:01:34.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6499" PostTypeId="1" AcceptedAnswerId="6501" CreationDate="2015-12-11T00:37:31.507" Score="4" ViewCount="94" Body="&lt;p&gt;I have a 5&quot; x 4&quot; x 3/8&quot; plastic panel (wall thickness of 1/8&quot;, ribbed). I need to slide the panel over a 12&quot; plastic track that is U-shaped (like a skateboard ramp). This motion could be repeated 2 million+ times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CkuPE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CkuPE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The panel is rigid and maintains 2 contact edges when traversing the curved surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: Should I be worried about wearing-out the plastic components at this scale?  What options are there to protect this mechanism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;plain plastic on plastic rails / tracks / sliding pegs&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ebay.com/itm/10PCS-MR74-open-4-7-2-mm-Miniature-Bearings-ball-Mini-bearing-4X7X2mm-/400958903666?hash=item5d5b035572:g:yBQAAOSwHnFVrLig&quot; rel=&quot;nofollow noreferrer&quot;&gt;mini ball bearings&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on using cheap bearings for this?&lt;/p&gt;&#xA;" OwnerUserId="3765" LastEditorDisplayName="user16" LastEditDate="2015-12-11T16:46:17.557" LastActivityDate="2015-12-11T18:41:48.970" Title="Are Rolling-Element Bearings Practical for Light Loads?" Tags="|mechanical-engineering|materials|mechanical-failure|bearings|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6500" PostTypeId="2" ParentId="6497" CreationDate="2015-12-11T00:57:39.397" Score="4" Body="&lt;p&gt;You may want to reword your question. What you are looking for is tamper evident materials or &lt;a href=&quot;https://en.wikipedia.org/wiki/Tamper-evident_technology&quot; rel=&quot;nofollow&quot;&gt;tamper evident technology&lt;/a&gt;. Searching for those may give you some better results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be aware that anyone with enough time and money can defeat tamper evident materials/designs. I am not sure what you are building, but you may want to consider a non-passive system such as a computer based machine lockout or internet based web reporting of the violation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tempered glass is a good one. It is manufactured with built in stresses. If you attempt to drill or cut it, the whole sheet will explode into small peices. There may be an exotic way to cut it, but like you mentioned it would not be cost effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example would be tempering metal or quenching it. A material that was easily machined before can be made very hard and difficult to machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other non-reversible processes that may be of interest too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Shrink-fitting&quot; rel=&quot;nofollow&quot;&gt;Shrink fit assembly&lt;/a&gt; involves heating the large part or cooling the small part and easily fitting them together. Once they are the same temperature it is almost impossible to cleanly remove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Epoxy potted circuit boards are difficult to tamper with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fiber filled composites and gummy metals paired with very hard metals or ceramics are very hard to saw because the saw tends to bite in and break teeth quickly. Slow working abrasives are usually used for those purposes.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="368" LastEditDate="2015-12-11T18:22:34.063" LastActivityDate="2015-12-11T18:22:34.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6501" PostTypeId="2" ParentId="6499" CreationDate="2015-12-11T03:03:48.887" Score="3" Body="&lt;p&gt;Without knowing additional details, my attempt at a catch-all answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt; attempt to minimize wear. Millions of cycles is a lot for even low-load contact surfaces. Use components with high hardness and smooth materials. PTFE or UHMWPE backed by metal for full-contact bearings, and AISI 52100 or AISI 440C for ball bearings. There may be an issue with contact angles at the curves.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Theory&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The primary issue with millions of cycles of amply supported contacting surfaces is almost certainly going to be wear. A simple model for wear is given by &lt;a href=&quot;http://www.tribology-abc.com/abc/wear.htm&quot; rel=&quot;nofollow&quot;&gt;the Archard equation&lt;/a&gt; - (&lt;a href=&quot;https://en.wikipedia.org/wiki/Archard_equation&quot; rel=&quot;nofollow&quot;&gt;Wiki&lt;/a&gt;), which holds that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Q = k\:p\:s&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Q$ is the wear volume, $K$ is a dimensionless constant, $W$ the normal load, $L$ the sliding distance, and $H$ the hardness of the softest contacting material. The Wiki link separates hardness $H$ from $k$. The idea behind the model is that contacting surfaces don't meet uniformly, but instead at randomly distributed, microscopic height differences called asperities. As a result, the actual contact area is much smaller than the apparent macroscopic contact area, and consequently the contact loads are orders of magnitude larger than the apparent macroscopic load. Thus, the size, distribution, ductility, and hardness of the asperities plays a role, as do lubrication, mean height difference (i.e. smoothness), and the presence of external detritus. The constant $k$ depends on all of these parameters, and is typically thus typically empirically determined. However, knowing the proportionality relationships is useful in making first-passes at material selection. Specifically: increasing hardness reduces asperity formation rate, reducing wear; and decreasing surface roughness decreases number of asperities, reducing wear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore to minimize wear, a smooth, hard material, with an appropriate lubricant should be used. It would also be helpful to minimize external detritus (e.g. dirt, dust) entering the system. You can possibly go either route you describe in the question, depending on the specifics of your setup. I am &lt;strong&gt;assuming&lt;/strong&gt; you require motion along a specific path with minimal deviation, like a train on rails.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;First-pass Material Selection&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;For plastic-on-plastic with no balls, it would probably help to use parts formed from a hard metal or ceramic and coated with either polyfluorocarbons (PTFE, e.g Teflon) or ultra-high-molecular-weight polyethylene (UHMWPE). The trick here is that your setup requires changing direction smoothly. Getting custom made, non-linear polymer rails and bearings may be more costly than linear alternatives. Additionally, there is the problem of changing contact angles: the bearings would have to be shaped like the inner surface of a toroid, or like a saddle. If using a more typical cylindrical bearing, when the bearing meets the curve, only the leading and trailing edges will be in contact, which will cause damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ball-bearings, there would need to be a special track that solves the issues of leading/trailing edge contact. Ball bearings may be made of metal or ceramic, though ceramics like silicon nitride ($\textrm{Si}_{3} \textrm{N}_{4}$) may be overkill for a low-temperature application as they are more expensive than their metal counterparts. Typically chrome steel (usually near AISI 52100) or stainless steel (usually near AISI 440C) are used as they readily form martensite (a hard, metastable steel phase) during processing and can thus be made quite hard and wear resistant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the weight truly is so low, adding any bearings at all will dramatically increase the weight. Instead of having bearings, is it possible to shape the panel to fit into a suitable-width groove dug into the track surface? Then make both out of teflon or UHMWPE, and back the track with brass or steel. A thin layer of dense lubricant could even make the panel float slightly, which would reduce wear to virtually zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the edges are in contact, as noted in the comments, then a track could still work, but would require rounded panel contact edges. The lubricant &quot;float trick&quot; would be messy and probably wouldn't cause flotation when rounding corners.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorDisplayName="user16" LastEditDate="2015-12-11T18:41:48.970" LastActivityDate="2015-12-11T18:41:48.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6502" PostTypeId="1" AcceptedAnswerId="6503" CreationDate="2015-12-11T06:58:38.283" Score="0" ViewCount="467" Body="&lt;p&gt;What equations are needed to describe the forces and properties of the kind of mechanisms used in a machine that harnesses the gravitational potential energy of a suspended weight to rotate the shaft of a permanent magnet DC generator (i.e. the &lt;a href=&quot;http://gravitylight.org&quot; rel=&quot;nofollow&quot;&gt;gravity light&lt;/a&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, these machines aim to slow the descent of a suspended weight and convert that slow downward motion into high rpm rotary motion that is used to do work. As far as I can tell, this slowdown can be achieved using some kind of governor (in the case of a gravity powered clock air resistance from some spinning baffles is utilized), or by the torque required to turn the shaft on the generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the relationship between the mass of the weight, the torque required to turn the generator, and/or any gearing that might be helpful be described mathematically?&lt;/p&gt;&#xA;" OwnerUserId="4095" LastEditorUserId="368" LastEditDate="2015-12-11T18:46:08.323" LastActivityDate="2015-12-11T18:46:08.323" Title="Equations relevant to gravity powered clocks and machines such as the gravity light" Tags="|mechanical-engineering|gears|energy-efficiency|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6503" PostTypeId="2" ParentId="6502" CreationDate="2015-12-11T09:51:46.550" Score="2" Body="&lt;p&gt;Taking the force exerted by the mass and converting it to rotational force or torque will involve the following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = g*(mass) where g = 9.81 and&lt;br&gt;&#xA;T = r*F where r is the radius of the pulley from which the suspended mass is hanging. This is the torque that turns the generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When other factors such as gears are included the gear ratios need to be multiplied with the torque. Example equations can be seen below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gear 1 radius is 20&lt;br&gt;&#xA;gear 2 radius is 10&lt;br&gt;&#xA;if the Torque in the shaft of gear 1 is 5 N.m then the torque in gear shaft 2 is:&lt;br&gt;&#xA;T2 = 10/20*T1 = 2.5 N.m.&lt;br&gt;&#xA;Gear calculations can become more complicated with modulus and gear teeth etc being included but this is the basics of it.&lt;/p&gt;&#xA;" OwnerUserId="4096" LastEditorUserId="368" LastEditDate="2015-12-11T18:45:06.780" LastActivityDate="2015-12-11T18:45:06.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6504" PostTypeId="2" ParentId="6494" CreationDate="2015-12-11T10:13:23.453" Score="4" Body="&lt;p&gt;For the Maths of calculating uncertainty the standard document is the &lt;a href=&quot;http://www.bipm.org/en/publications/guides/gum.html&quot; rel=&quot;nofollow&quot;&gt;GUM&lt;/a&gt;. Which describes all the maths but can be somewhat unclear if you don't already have some idea how it is supposed to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your current level of expertise there are several good introductions to uncertainty calculations. I would recommend &lt;a href=&quot;http://www.npl.co.uk/publications/guides/a-beginners-guide-to-uncertainty-of-measurement&quot; rel=&quot;nofollow&quot;&gt;A Beginner's Guide to Uncertainty of Measurement&lt;/a&gt;, which is freely downloadable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any uncertainty calculation you need a model of your output given the inputs. For your case a very simple model would be purely linear, with distance proportional to voltage measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$L=\alpha V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this simple case, for fixed $\alpha$, the uncertainty in your distance is $\alpha u_v$ but more generally for a function $f$ with various inputs $x_i$.&#xA;$$u^2=\sum_i (\frac{df}{dx_i})^2 u_{x_i}^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your uncertainties in V (or your general $x_i$) can be determined either using prior knowledge or by repeated measurement (at a given distance or similar).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calibration of your system is necessary to determine any coefficients in your model, such as $\alpha$, and to help inform the choice of model itself. The necessary calibration will vary in a case by case basis, but for your setup I guess you will want to measure a series of known distances and plot a calibration curve of distance against voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to note that as you have uncertainty in your voltage and in the distance of your standard lengths used for calibration, as well as other non-linearities no there will inevitably be some uncertainty in your calculation of $\alpha$, unlike in my simple example. This will add some additional uncertainty proportional to the voltage to your measurement.&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2015-12-11T10:13:23.453" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6505" PostTypeId="1" CreationDate="2015-12-11T14:39:02.890" Score="2" ViewCount="63" Body="&lt;p&gt;As part of a study at work, I have to calculate the cutting speed and feed rates for cutting plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found a formula for cutting speed calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v={{C_V*K}\over{T^m+t^{x_v}+s^{y_v}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$C_V$ is a coefficient that is dependent on processing conditions;&lt;/li&gt;&#xA;&lt;li&gt;$m$ is the specific retention figure;&lt;/li&gt;&#xA;&lt;li&gt;$x_v,\ y_v$ are coefficients relative to the cutting depth and feed;&lt;/li&gt;&#xA;&lt;li&gt;$K$ is a correction coefficient;&lt;/li&gt;&#xA;&lt;li&gt;$T$ is sharpness resistance; and&lt;/li&gt;&#xA;&lt;li&gt;$s$ is the feed rate.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;variable descriptions may not be completely correct as they were translated from Russian&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For cutting metal there are lots of tables and sources for finding the coefficients but I can't find any for cutting plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the cutting speed for plastic? Is there a different formula that doesn't involve these coefficients, or another method I can use to identify the coefficients?&lt;/p&gt;&#xA;" OwnerUserId="4101" LastEditorUserId="368" LastEditDate="2015-12-11T19:22:55.317" LastActivityDate="2015-12-11T19:22:55.317" Title="Identifying coefficients to determine cutting speed for PA6 plastic" Tags="|mechanical-engineering|machining|plastic|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6507" PostTypeId="2" ParentId="2166" CreationDate="2015-12-11T18:55:41.177" Score="1" Body="&lt;p&gt;Centrifugal pumps typically go by &lt;a href=&quot;https://en.wikipedia.org/wiki/Nominal_Pipe_Size&quot; rel=&quot;nofollow&quot;&gt;nominal pipe size&lt;/a&gt; for the inlet and outlet connections.  The first number is typically the inlet, and the second is the outlet.  Normally the outlet is smaller (which is why you'll see 10&quot;x8&quot; centrifugal pumps but never 8&quot;x10&quot;).  This is partly due to mechanical efficiency and partly to get a justifiably higher pressure. However in small sizes such as this one, both connections are the same.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-11T18:55:41.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6508" PostTypeId="1" AcceptedAnswerId="6533" CreationDate="2015-12-11T20:44:47.790" Score="10" ViewCount="1214" Body="&lt;p&gt;I was searching for information about pressure-volume processes when I read, in a heat engineering book, &quot;The process is polytropic with $PV^n = \text{constant}$.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I know that a real life process follows this model? Are all processes related to this heat engineering, pressure-volume, etc., in fact polytropic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if they aren't then how can I tell when that model is valid?&lt;/p&gt;&#xA;" OwnerUserId="4110" LastEditorUserId="368" LastEditDate="2015-12-15T00:08:15.653" LastActivityDate="2015-12-15T00:08:15.653" Title="Are all pressure-volume processes polytropic?" Tags="|thermodynamics|pressure|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6509" PostTypeId="1" AcceptedAnswerId="6510" CreationDate="2015-12-11T22:02:13.497" Score="2" ViewCount="94" Body="&lt;p&gt;I have a quadrotor aircraft (4 motors/propellers around a center housing) with some electronics in an electronics housing. Looking to cool the electronics with propeller downwash. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RDKZW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RDKZW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a vent on the side of the electronics housing and one on the bottom. I assume the pressure on bottom P0 is atmospheric. Trying to determine which way the air will flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would pressure P1 be higher or lower than atmospheric? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The propellers do work on the fluid and increase the dynamic pressure but there is also air at high velocities flowing past the side vents. Having trouble consolidating propeller dynamics with Bernoulli's principle.&lt;/p&gt;&#xA;" OwnerUserId="4112" LastActivityDate="2015-12-12T00:33:47.067" Title="Using propeller downwash for electronics cooling" Tags="|fluid-mechanics|heat-transfer|aircraft-design|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6510" PostTypeId="2" ParentId="6509" CreationDate="2015-12-12T00:33:47.067" Score="1" Body="&lt;p&gt;It depends on the velocity across P1 and the pressure differential across your blades. It would be hard to determine without testing or using computation fluid dynamics. That is probably overkill and its much easier to just change the geometry to do what you need. Hazzey mentioned a scoop in the comments; that would work well but it would also send rain and dust through. You can also do it the opposite way by putting down facing outlet to generate negative pressure. This would allow you to have a large area on the bottom vent and be able to filter the intake air.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-12-12T00:33:47.067" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6511" PostTypeId="1" AcceptedAnswerId="6512" CreationDate="2015-12-12T17:04:15.657" Score="2" ViewCount="3451" Body="&lt;p&gt;I have a book that claims:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A metal will have more of a tendency to dissolve another metal of higher valency than one of a lower valency.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I interpret this as saying that a metal with higher valency will dissolve more easily into a host metal than it would if it had a lower valency. For example, $\textrm{Fe}^{3+}$ as impurity atoms and $\textrm{Na}^{+}$ as host will work better than vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then in the summary of the chapter it says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A valence that is the same as or less than the host material will have more appreciable solubility.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which of the two statements is correct?&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="368" LastEditDate="2015-12-15T00:02:45.063" LastActivityDate="2015-12-15T00:02:45.063" Title="How does the relative valency of the impurity and host metals affect solubility in a substitutional solution?" Tags="|materials|metallurgy|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6512" PostTypeId="2" ParentId="6511" CreationDate="2015-12-12T17:26:55.907" Score="2" Body="&lt;h2&gt;Summary&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Your book is correct. The same valence (+3/+3 solvent/solute) will be most likely to have large solubility, while higher valences (+3/+4) will be more likely to have high solubility than lower valences (+3/+2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting them in order as an example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Solvent / Solute&#xA;&#xA;    +3  /  +3       Highest Likelihood of Solubility&#xA;    +3  /  +4       &#xA;    +3  /  +2       Lowest Likelihood of Solubility&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The Hume-Rothery rules for solubility are necessary, but not sufficient, conditions for complete solid-solution solubility. That means if any of the rules are violated, complete solubility does not occur. However, even if all of the rules are met, complete solubility is not guaranteed. Mixing &lt;em&gt;always&lt;/em&gt; occurs due to entropy, though possibly in infinitesimally small ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For posterity, the rules are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Difference of less than 15% in the ratio of the radii of the solute and the solvent, so that&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left( \frac{\left|r_{solute} - r_{solvent}\right|}{r_{solvent}} \right)\leq 0.15&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;There must be the same or similar crystal structures between the solute and solvent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Complete solubility occurs when the solvent and solute have the same valency. &lt;strong&gt;A metal will be more likely to dissolve a solute of higher valency than a solute of lower valency.&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The solvent and solute should have similar electronegativity. Intermetallics tend to form when the difference is large.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The specific part you are interested in is in &lt;strong&gt;bold&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These rules, more or less stated the same way, can be found from several sources, including &lt;a href=&quot;https://en.wikipedia.org/wiki/Hume-Rothery_rules&quot; rel=&quot;nofollow&quot;&gt;The Wiki&lt;/a&gt; and &lt;a href=&quot;http://www.msm.cam.ac.uk/phase-trans/2004/titanium/hume.rothery.html&quot; rel=&quot;nofollow&quot;&gt;The University of Cambridge Department of Material Science and Metallurgy&lt;/a&gt;. Note that the Cambridge link states the &lt;em&gt;opposite&lt;/em&gt; of the bolded portion of rule 3. The Cambridge link is possibly incorrect, but it is difficult to tell based on the literature whether or not rule 3 is even useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thorough review of the literature surrounding the rules can be found &lt;a href=&quot;http://eprints.soton.ac.uk/172201/&quot; rel=&quot;nofollow&quot;&gt;at this link&lt;/a&gt;. The paper is unfortunately downloadable in Word Document form, but seems to be clean. A brief summary of the relevant part is that the relative valence factor rule (rule 3), specifically the bolded part, seems to only be valid when monovalent Cu, Ag, and Au are alloyed with B-subgroup elements of higher valence (using the Old IUPAC nomenclature, so the right-hand-block of the periodic table). The explanation involves Fermi surface and Brillouin zone interactions in the B-subgroup elements. What I take away from the review is that this particular part of rule 3 is not terribly useful in practice.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-12-14T20:51:47.790" LastActivityDate="2015-12-14T20:51:47.790" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6516" PostTypeId="1" AcceptedAnswerId="6560" CreationDate="2015-12-13T09:07:18.393" Score="1" ViewCount="3849" Body="&lt;p&gt;A wooden police baton has to be hung on the back of an officer and released only when s/he wants it to be released. I thought about &quot;push to stow, push to release&quot;. This is a thought experiment, I am not a manufacturer...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The baton is smooth without grooves so alternative solution of releasing by rotation (like BNC plugs, or push/push pen) won't work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I find a simple design of such contraption or what terms would I use to describe this?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is a rough model of a four bar (ABCFD) which push point E toward the object to hold (LMNOQ)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object, represent here as rectangle LMNOQ, moves on a fixed bar (HL).&#xA;When pushed up (represented by piston A1 expanding), point E moves towards the edge NQ. The angle FDB (in the locked position) is &gt; 180 , so it should be locked, as with locking pliers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the open position, LMNOQ is lower, and point E is far from it.&#xA;(this model is only for showing the idea. I am sure it should be modified to work in the real world)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HOW can a SECOND push unlock point D?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;img src=&quot;https://i.stack.imgur.com/7iQf6.png&quot; alt=&quot;partial clamping mechanism&quot;&gt;]&#xA;[&lt;img src=&quot;https://i.stack.imgur.com/d8aoC.jpg&quot; alt=&quot;baton is held&quot;&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GAp3W.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GAp3W.jpg&quot; alt=&quot;baton released&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4120" LastEditorUserId="33" LastEditDate="2015-12-17T02:08:56.130" LastActivityDate="2015-12-17T09:02:28.423" Title="Mechanism for a mechanical latch/clamp" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6517" PostTypeId="1" CreationDate="2015-12-13T11:51:40.943" Score="10" ViewCount="159" Body="&lt;p&gt;I'm interested in controlled oxidation of nickel in a methane reformer with diameter ~3 m and height ~1.5 m. In this reactor, CH&lt;sub&gt;4&lt;/sub&gt; is converted into H&lt;sub&gt;2&lt;/sub&gt; and CO, and CO into H&lt;sub&gt;2&lt;/sub&gt; and CO&lt;sub&gt;2&lt;/sub&gt; but for this problem, we assume that the methane and water stopped flowing in the reactor and that all these substances are not present in it during the oxidation process. So the idea for the nickel oxidation is to let some air flow in the reactor by controlling the following three parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The percentage of oxygen in the air I let flow in the reactor.&lt;/li&gt;&#xA;&lt;li&gt;Its flow rate.&lt;/li&gt;&#xA;&lt;li&gt;The temperature of the air injected&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are the realistic values that these three parameters can take?&lt;/strong&gt;  I've searched the internet but I could not find any information about how the air generation step (that's how it's called) works and what exactly is producing the air flow (turbine?). &lt;/p&gt;&#xA;" OwnerUserId="4121" LastEditorUserId="368" LastEditDate="2015-12-17T23:40:52.420" LastActivityDate="2016-12-06T19:35:41.690" Title="How do I choose operating parameters for input air to a methane reformer?" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6518" PostTypeId="1" CreationDate="2015-12-13T16:59:12.793" Score="5" ViewCount="916" Body="&lt;p&gt;I am looking to purchase some new Vernier calipers, but I am overwhelmed by the amount of specs which are being offered.  I need help in understanding how to evaluate them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not want a dial or digital set of calipers, but rather am interested in any additional qualities such as the accuracy that these may offer which Vernier calipers can not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accuracy is the most important quality that I am looking for.  For the sake of the question, assume that I have budget around US $2,000.  What aspects of the Vernier calipers should I be looking at in order to have a high degree of accuracy?  What else should I be examining with the calipers as part of this purchase? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some specific sub-questions that I haven't been able to figure out.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Graduation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;A common amount is 0.02 mm, does this mean we can advance each measurement by a minimum of 0.02 mm? Therefore cannot get any closer a measurement from any 1 point on the scale?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Parallax-Free&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I have read somewhere online that parallax-free aids the readability of calipers, thus allowing for less human error - If this is true, does this parallax-free attribute come at a cost? Do we lose some accuracy? (I think I read parallax-free Vernier calipers do not have a separate upper movable ruler like some of the others.)&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Metric&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I'm having a hard time identifying what's standard.  What are common units offered?&lt;/p&gt;&#xA;" OwnerUserId="4123" LastEditorUserId="368" LastEditDate="2015-12-14T16:00:54.307" LastActivityDate="2016-03-01T13:46:56.047" Title="How to evaluate new Vernier calipers?" Tags="|measurements|equipment-selection|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6519" PostTypeId="2" ParentId="6516" CreationDate="2015-12-13T22:19:59.770" Score="0" Body="&lt;p&gt;From the limited information given in your question, it sounds like you might want a mechanism similar in operating principal to what is called a &quot;push to open&quot; drawer slide. On these drawers the draw doesn't open when you pull on it, you have to push it past a detent point and then it opens by itself basically. Hope this helps !&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2015-12-13T22:19:59.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6520" PostTypeId="2" ParentId="6518" CreationDate="2015-12-14T03:59:22.500" Score="3" Body="&lt;p&gt;If your budget is correct at US $2000, you have very limited choice.  You'll either have to opt for a 1000+ mm long caliper, or an antique one, probably used by da Vinci himself.  The most expensive Mitutoyo vernier caliper they sell of the standard 150 mm length is less than UK £200.  The accuracy is quoted at +/- 0.05 mm.  I'm not sure if you can go better with a vernier. 0.02 is common with dial calipers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your specifications and focus on accuracy seem at odds with my experience.  If accuracy is primary, then go digital (to 0.01 mm for £500).  They're easier to use, zero out, and do relative measurements.  They use batteries of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to metric/Imperial, you get both but metric is more common in Europe whilst Imperial is more common in the Colonies.  I don't think  that you can have both on the same instrument like a rule.  Digital can though.  If you're one of those people who do engineering in 'ths, then there exist fractional display digital calipers that can read for example, 1 3/16th inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also please do not confuse accuracy with precision.  Precision, like resolution, is the degree of repeatability of readings.  You can easily have a very accurate 0.05 mm dial caliper that is better than a crappy 0.01 mm digital one from the DIY store.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="2947" LastEditDate="2015-12-15T00:26:39.057" LastActivityDate="2015-12-15T00:26:39.057" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6521" PostTypeId="2" ParentId="5346" CreationDate="2015-12-14T04:35:04.897" Score="5" Body="&lt;p&gt;Sinking a brick shaft is fairly easy technically, but the skills may have been lost a few years back.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You dig a hole.&lt;/li&gt;&#xA;&lt;li&gt;Line the hole with mortared bricks from the inside.&lt;/li&gt;&#xA;&lt;li&gt;Drive pins into the sides below the brickwork so that it corbels over the pins and is supported.&lt;/li&gt;&#xA;&lt;li&gt;Dig out under the brickwork, extending the hole downwards somewhat.&lt;/li&gt;&#xA;&lt;li&gt;Build more lining.&lt;/li&gt;&#xA;&lt;li&gt;Remove pins and repeat ...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It's how mine shafts going down 1000 feet were sunk.  I don't know your well's ground conditions, but I can speculate that if they where conducive to some self support, the brickwork could be replaced with un mortared stonework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like this, but here soft ground conditions (Bolton, England) means that an iron ring is used instead of pins. And the ring is supported from the surface:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KXVMG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXVMG.jpg&quot; alt=&quot;shaft&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems we've forgotten so much stuff these days and have to rely on complex technology or a Dremel.  Shame really.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="2947" LastEditDate="2019-11-06T12:39:29.320" LastActivityDate="2019-11-06T12:39:29.320" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="6522" PostTypeId="2" ParentId="4053" CreationDate="2015-12-14T04:59:33.830" Score="-3" Body="&lt;p&gt;There is some merit in the argument that one mile on one mile off construction would be dangerous.  Works vehicles would have to enter and exit every mile from live traffic.  They'd go from 50mph to some stupid speed like 2mph and back every mile go get along the works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would seriously question though whether it's cheaper per se.  It's &lt;strong&gt;easier.&lt;/strong&gt;  A lot of road jobs are rolled up into large multi year contracts /partnerships where a safety case has to be made.  As part of the UK's war on the motorist, Health &amp;amp; Safety plays a disproportionate role in evaluation of long term contract /partnership bids.  It can thus be used as a lever against the pencil heads to extract more money from them.  It's really a license to print money as in the UK it's politically incorrect to challenge Health &amp;amp; Safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're not the only one to complain about those works.  They've been criticised in the House.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2015-12-14T04:59:33.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6523" PostTypeId="1" CreationDate="2015-12-14T11:08:56.697" Score="5" ViewCount="181" Body="&lt;p&gt;Consider the following DC motor parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rotor resistance, $R\ [\Omega]$&lt;/li&gt;&#xA;&lt;li&gt;Motor constant, $k_t\ [\mathrm{V/rpm}]$&lt;/li&gt;&#xA;&lt;li&gt;Inertia, $I\ [\mathrm{kg\cdot m^2]}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How do they scale with motor size? Do $R$ and $k_t$ depend on $I$?&lt;/p&gt;&#xA;" OwnerUserId="3633" LastEditorUserId="368" LastEditDate="2015-12-14T16:35:46.700" LastActivityDate="2021-02-05T22:04:05.123" Title="How do DC motor parameters scale with motor size?" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6524" PostTypeId="1" CreationDate="2015-12-14T13:06:46.103" Score="2" ViewCount="1246" Body="&lt;p&gt;I'm working on a lifting system that uses a ball screw and linear shaft as guide and I would like to determine which shaft diameter would be enough for my application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is a sketch of the system that uses 3 linear shafts and 3 linear bearings. The grey transparent huge block weighs 40 kg and its dimensions (which I guess are important to determine radial loads?) are 90 cm (width) x 70 cm (depth). How can I determine (approximately) the diameter required for the 3 shafts and also determine whether 2 shafts would be enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CNdfm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CNdfm.png&quot; alt=&quot;sketch of the system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading this guide:  &lt;a href=&quot;http://www.nookindustries.com/LinearLibraryItem/Linear_Components_Design_Considerations&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nookindustries.com/LinearLibraryItem/Linear_Components_Design_Considerations&lt;/a&gt; I found that using a diameter of 1/2&quot; with a length of 35 cm for the shaft would create a deflection that produces a misalignement angle around 0.25° (which is acceptable if I refer to what is written on that guide). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am really not convinced that a single shaft with such a small diameter could support that 40kg load.&lt;/p&gt;&#xA;" OwnerUserId="3664" LastEditorUserId="33" LastEditDate="2016-02-09T20:29:20.817" LastActivityDate="2016-02-09T20:37:47.297" Title="How to choose linear shaft diameter for a lifting system?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6526" PostTypeId="1" AcceptedAnswerId="6543" CreationDate="2015-12-14T13:24:30.367" Score="10" ViewCount="2058" Body="&lt;p&gt;For simplicity we're going to ignore aerodynamics and vehicle mass for a moment...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are small (around 1 litre) car engines designed to tolerate a higher duty cycle than large (2+ litre) car engines? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cars are usually all travelling at roughly the same speed which means the engines will be putting out proportionally similar amount of power, however a small engine would be operating at a greater proportion of its output than a large one to maintain the same speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some background to this, I have a 1.1 l car and I spend quite a lot of time with my foot completely flat on the floor, especially going uphill on a high speed road (60 or 70 mph limit) to keep up with the traffic. I think my record's somewhere around 2 minutes on full throttle.&lt;/p&gt;&#xA;" OwnerUserId="4132" LastEditorUserId="2819" LastEditDate="2015-12-16T05:30:50.937" LastActivityDate="2020-04-10T01:42:19.417" Title="Are small car engines designed to tolerate a higher duty cycle than large car engines?" Tags="|automotive-engineering|engines|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6528" PostTypeId="2" ParentId="6526" CreationDate="2015-12-14T14:27:28.770" Score="8" Body="&lt;p&gt;The primary issue (ignoring secondary issues like internal wear) is whether or not the engine can get rid of the waste heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that all engines for a given fuel have roughly the same overall thermal efficiency, for a given power output, a certain amount of waste heat needs to be dissipated regardless of the physical size of the engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat is dissipated through the radiator, and if it is effective, then for any given power level, the coolant temperature will be stable. If the radiator is being driven beyond its capacity, the coolant temperature will keep rising.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the bottom line is this: When operating at full throttle, does the coolant temperature stabilize at a safe level? If not, you should consider getting a different car for that kind of driving.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2015-12-14T17:16:00.333" LastActivityDate="2015-12-14T17:16:00.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6529" PostTypeId="1" CreationDate="2015-12-14T14:35:12.887" Score="10" ViewCount="672" Body="&lt;p&gt;In many books the permittivity of a material is only listed as dielectric constant for one or a couple of frequencies. 1 kHz is used often (for example in the Plastics Technology Handbook, 4th ed, ISBN-13: 978-0-8493-7039-7), but I'm looking for the permittivity for a frequency at least 2000 times higher (2-4 GHz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does one calculate the permittivity of a material for a higher frequency? Is the dielectric constant still useful here?&lt;/p&gt;&#xA;" OwnerUserId="1070" LastEditorUserId="282" LastEditDate="2015-12-15T06:13:24.503" LastActivityDate="2017-02-22T18:21:36.270" Title="How to calculate permittivity for higher frequencies?" Tags="|electrical-engineering|materials|rf-electronics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6531" PostTypeId="1" CreationDate="2015-12-14T18:50:25.560" Score="6" ViewCount="1812" Body="&lt;p&gt;I've been working on the design of a reinforced concrete wall pier for a three-span bridge. I roughed out the design based on a number of example plan sets and now that I'm taking a crack at the actual design calculations, I'm questioning the logic of the example plans. (Plus, I just dislike &quot;designing&quot; on the basis of &quot;Well, we've been doing it this way and nothing's fallen down yet!&quot;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The design is thus:&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UgNaW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UgNaW.png&quot; alt=&quot;Wall Pier Layout&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My questions are:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Why would the piles be stopped short of the cap?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;With the piles stopped short of the cap, what is the load path from the girders to the piles for vertical and lateral loads?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm also confounded by the fact that I'm typically seeing US #6 bars for the (a) bars (which seems large given that the cap will not see much flexure) and US #5 bars for the (c) bars (which seems small given that it looks like these bars would be necessary to transfer lateral loads from the cap into the wall).&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="368" LastEditDate="2015-12-14T23:51:06.403" LastActivityDate="2019-01-03T14:38:44.410" Title="How are loads transferred to bearing piles in a wall pier?" Tags="|structural-engineering|bridges|reinforced-concrete|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6533" PostTypeId="2" ParentId="6508" CreationDate="2015-12-14T21:29:42.090" Score="6" Body="&lt;p&gt;No, all thermodynamic processes are not polytropic. As an example, consider a real engine cycle. Any undergraduate engineering thermodynamics text should have a plot showing the true P-V diagram. You also can see &lt;a href=&quot;http://4mechtech.blogspot.com/2013/12/indicator-diagram-or-p-v-diagram-actual_9.html&quot; rel=&quot;noreferrer&quot;&gt;examples&lt;/a&gt; online. Clearly parts of this cycle can not be modeled with a polytropic process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A polytropic process is an approximation, much like a linear regression. It is correct to the extent by which the real and approximate curves are similar, however you define &quot;similar&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ideal gases, sometimes with constant specific heats, you can derive certain polytropic relations, e.g. for an &lt;a href=&quot;https://en.wikipedia.org/wiki/Isothermal_process&quot; rel=&quot;noreferrer&quot;&gt;isothermal process&lt;/a&gt; of an ideal gas you get $P V =$ constant, for an adiabatic process with no entropy generation you get the &lt;a href=&quot;https://en.wikipedia.org/wiki/Isentropic_process&quot; rel=&quot;noreferrer&quot;&gt;isentropic process&lt;/a&gt; relations, etc. These are polytropic processes, and their use often follows the validity of their assumptions. That is, if the gas is ideal and the temperature is constant, the (polytropic) isothermal process is a good model. It should be clear that for other equations of state you will not necessarily get a polytropic process keeping the other assumptions the same.&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2015-12-14T21:29:42.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6535" PostTypeId="2" ParentId="486" CreationDate="2015-12-15T14:19:56.573" Score="4" Body="&lt;p&gt;For high accuracy over long distances it's typical to use general surveying techniques.  You use a total station (&lt;a href=&quot;https://www.sccssurvey.co.uk/surveying-equipment-instruments/total-stations.html&quot; rel=&quot;nofollow&quot;&gt;like this&lt;/a&gt;) which will get you 1.5mm accuracy in a single shot.  They're no laser tape measure. Repeated set ups /readings with some statistical corrections should get you easily below 1mm.  Note too that these are immune to thermal expansion of any tape, but of course the subject will be.  This too can be allowed for though with come mathematical compensation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good thing is that you don't have to buy a total station.    You can rent it for a few days. Better still (contingent on budget) is to get a survey guy in to do it for you.  This may be the way to go considering the experience required to use this kind of kit. It'll take a beginner a day just to stand it up accurately over a mark.  I think that the accuracy you're asking for can be achieved at reasonable cost.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2015-12-15T14:19:56.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6536" PostTypeId="2" ParentId="6467" CreationDate="2015-12-15T16:22:14.627" Score="1" Body="&lt;p&gt;Why can't I just hire a licensed contractor instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, because a contractor does not have the education and is not required or expected by law to know what he is doing or telling you and/or to protect your best interests in doing so.  A design professional (Professional Engineer or Registered Architect) does all of the above.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when the stuff hits the fan and your DIY work collapses or starts to leak, a contractor is not responsible nor expected by law for the design of the work or for contract specs and you will not be able to recover your defective DIY work.  Assuming your contractor does not disappear, the contractor's attorney will simply state that his proposal simply listed the work that you wanted him to do and he has no idea how a house works or what the design is all about (he doesn't have enough education remember?).  And the contractor wins every time because that's the law.  Don't let his tongue impress you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even if his proposal states that the work shall comply with all building codes, if the work doesn't pass inspection, a contractor is allowed to say I didn't know this code and I will have to charge you extra change order because by law a contractor is not expected to know the code (he doesn't have enough education remember?).  Don't let his shiny red pick-up truck impress you (or his attorney's Mercedes).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even if he gives you a labor and materials guarantee, if its defective design or a code violation, case law says a contractor is not expected to know the job is in violation or defective work if its simply something you asked or directed him to do as the owner and general contractor on the job.  By the way, most roofing contractors for example have a statement that if the roof develops cracks more then 1/4 inch, the roofing guarantee is VOID my friend.  Don't let the contractor's attorney's Mercedes impress you.&lt;/p&gt;&#xA;" OwnerUserId="4140" LastEditorUserId="368" LastEditDate="2015-12-15T18:02:54.900" LastActivityDate="2015-12-15T18:02:54.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6537" PostTypeId="1" CreationDate="2015-12-15T16:56:05.153" Score="5" ViewCount="1845" Body="&lt;p&gt;I am looking at a company-written standard for cast parts. The castings are blanks for driveshaft yokes, similar to what is pictured below. The standard defines general tolerances for the dimensions on the part, and it classifies them into two categories: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Outside dimensions and center-to-center distances&lt;/li&gt;&#xA;&lt;li&gt;Inside dimensions (core dimensions) &lt;sup&gt;1&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There is some confusion between us and our supplier as to what constitutes an inside dimension. Obviously, anything that requires a core (such as pilot holes for the yoke ear holes seen below) will be an inside dimension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when our internal inspection requirements were laid out, things such as the distance from inside edge to inside edge of the ears were determined to be inside dimensions, while the outside edge measurement on the ears is an outside dimensions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lT7Zdm.jpg&quot; alt=&quot;Yoke&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, is this terminology common to castings (or even forgings) in general? I have looked at ISO standards regarding general tolerances, but they make no reference to it. If it is common, are inside dimensions strictly related to features requiring cores? Should other dimensions be considered &quot;inside&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, the tolerances for outside dimensions are asymmetrical, with the high end being larger than the low end (e.g. +1.0/-0.5), and the inside tolerances are symmetrical to the same absolute value as the low end of the outside dimensions (e.g. +/- 0.5 to match the first example). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1. This is verbatim how it is listed in the standard.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2018-01-22T23:09:52.120" Title="What are 'inside' and 'outside' dimensions in reference to castings?" Tags="|mechanical-engineering|manufacturing-engineering|casting|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6539" PostTypeId="1" CreationDate="2015-12-15T17:29:30.340" Score="4" ViewCount="821" Body="&lt;p&gt;I'm using hard drive platters to create a Tesla Turbine. I'd like to get as much efficiency out of it as possible. That being said, for safety concerns, I'd need to know a generic range at which they'd shatter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's no range that anyone has found, then how could I (safely) test what speed they shatter at?&lt;/p&gt;&#xA;" OwnerUserId="4142" LastActivityDate="2019-04-01T14:57:54.013" Title="At How Many RPMs Would Aluminum HDD Platters Shatter?" Tags="|materials|thermodynamics|pressure|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6540" PostTypeId="2" ParentId="6539" CreationDate="2015-12-15T19:03:43.930" Score="3" Body="&lt;p&gt;I think it's going to depend upon the individual platters and what type of drive you sourced them from.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technology for hard disk drives (HDD) that can easily handle 5,400 rpm is comparatively old and very easily sourced. Slightly newer technology allows for drives that can handle 7,200 rpm.  Keep in mind that those are the internal speeds of the drives themselves during regular operations, and not necessarily their upper rotational speed limit.  And there are newer drives that can handle up to 15,000 rpm with claims of some companies researching 20,000 rpm drives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, you'd be wise to take a conservative value.  &lt;a href=&quot;https://www.youtube.com/results?search_query=cd+shattering&quot; rel=&quot;nofollow&quot;&gt;YouTube has a number of videos of CD's shattering&lt;/a&gt; when running at excessive rpm, and there's a particularly awesome one &lt;a href=&quot;https://www.youtube.com/watch?v=zs7x1Hu29Wc&quot; rel=&quot;nofollow&quot;&gt;captured at an extremely high fps rate&lt;/a&gt; of a CD shattering when the disc was spun at 23,000 rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it may be possible to run older 5,400 rpm style platters at much higher speeds, as demonstrated in &lt;a href=&quot;https://www.youtube.com/watch?v=3lP_nK_Tqp8&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; that claims to run the platters at 15,000 rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prudent approach would be to test some platters at your preferred target rate along with a very substantial safety margin.  This &lt;a href=&quot;http://www.instructables.com/id/Build-a-15,000-rpm-Tesla-Turbine-using-hard-drive-/&quot; rel=&quot;nofollow&quot;&gt;instructable tutorial&lt;/a&gt; shows how to build a Tesla Turbine and run the platters at 15,000 rpm.  It would also seem prudent to build some form of protective barrier around the turbine should any of the parts shatter.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2015-12-15T19:14:20.540" LastActivityDate="2015-12-15T19:14:20.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6541" PostTypeId="1" CreationDate="2015-12-15T23:31:43.030" Score="4" ViewCount="295" Body="&lt;p&gt;I need to use a custom part inside high vacuum (cryo vacuum), for example as a shadow mask. If PLA/ABS are not volatile, I could just print the part I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, are ABS or PLA too volatile to be used in cryo vacuum chambers?&lt;/p&gt;&#xA;" OwnerUserId="602" LastEditorUserId="421" LastEditDate="2015-12-16T04:26:05.777" LastActivityDate="2017-10-26T15:55:01.393" Title="Are ABS/PLA too volatile to be used in cryo vacuum?" Tags="|materials|3d-printing|vacuum|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6542" PostTypeId="2" ParentId="6541" CreationDate="2015-12-16T04:23:08.603" Score="3" Body="&lt;p&gt;&lt;strong&gt;It is unlikely polymers will be an issue, but the additives used may pose problems. Polymers should have negligible volatility due to their long chain nature. Polymer &lt;em&gt;additives&lt;/em&gt; may still pose an issue. You may need to test any particular filament material to ensure it meets your needs.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Polymer Chain Mobility&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Polymers do not behave like small molecules in terms of mobility. Whereas some small molecules are relatively free to leave the bulk at any temperature due to the nature of neighbor-neighbor bonding, the polymer molecules of interest have purely covalent bonding along their backbones. For a molecule to leave the bulk and enter a gas phase, not only does the molecule have to mechanically disentangle from its neighbors, the entire chain must simultaneously defeat secondary bonding along its entire length, which can be many thousands of atoms long. While it is possible for chains to disentangle, that only occurs above the glass transition temperature, or $T_g$, of the polymer in question. Above the $T_g$ is when &lt;em&gt;reptation&lt;/em&gt; - relative, cooperative chain motion - becomes possible. Reptation time is inversely proportional to temperature, as noted in &lt;a href=&quot;https://en.wikipedia.org/wiki/Reptation&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, polymers should have negligible volatility, especially below their glass transition temperatures. The lowest $T_g$ is of the butadiene rubber in the ABS copolymer (acrylonitrile-butadiene-styrene), at about $-90\ \textrm{°C}$. The styrene has a $T_g$ of about $105\ \textrm{°C}$. Because ABS is a block or graft copolymer, meaning the different components reside on the same chain, the molecules are effectively &quot;locked in&quot; below the $T_g$ of the styrene component. PLA, or poly(lactic acid), typically has a $T_g$ of about $60\ \textrm{°C}$. If your cryo chamber is below these temperatures then the molecules have virtually no freedom to move relative to one another, let alone leave the bulk. Note that this is not a guarantee due to the stochastic nature of kinetics, and the dearth of studies on polymer volatility.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Additives&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What you more likely need to be worried about are polymer modifiers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;diluents: decreases melt viscosity, reduces $T_g$&lt;/li&gt;&#xA;&lt;li&gt;compatibilizers: allows better polymer blending&lt;/li&gt;&#xA;&lt;li&gt;trace solvents: remaining from production, or water from the atmosphere&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;all of which are small molecules. Some of these are almost certainly present in printed polymers due to the special requirements of printing. Diluents are likely in some filament-printer systems to achieve optimal high-resolution flow. Compatibilizers are unlikely in &quot;pure&quot; ABS and PLA, but are possible depending on the specifics of the filament material. Trace solvents are possible especially if the filament is of low quality. Additionally, some forms of PLA are mildly hydrophilic, and may absorb water. Drying printed PLA may be useful.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="368" LastEditDate="2015-12-16T17:41:45.993" LastActivityDate="2015-12-16T17:41:45.993" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6543" PostTypeId="2" ParentId="6526" CreationDate="2015-12-16T05:27:08.717" Score="5" Body="&lt;p&gt;Most (modern) small and large car engines are designed for 100% duty cycle. This means that at 100% rated power(gas pedal all the way down) the engine can run continuously.  Heat dissipation is the limiting factor like Dave Tweed stated. Cars that are not designed to continuously dissipate 100% of the heat generated at max power require the driver to watch the temperature gauge to limit the power use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern engines do not have this problem because the engine is &lt;a href=&quot;https://en.wikipedia.org/wiki/Governor_(device)&quot; rel=&quot;noreferrer&quot;&gt;governed&lt;/a&gt; (speed regulated) below the cooling capacity of the radiator. Most modern engines use electric fans on the radiators that are independent of engine rpm; greatly increasing continuous cooling capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Older cars and &quot;high performance&quot; cars may have power that exceeds the cooling capacity. Any engine that has had the maximum engine speed regulation removed or any engine that can be &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Redline&quot; rel=&quot;noreferrer&quot;&gt;red lined&lt;/a&gt;&quot; can also overheat. An engine boosting system such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrous_oxide_engine&quot; rel=&quot;noreferrer&quot;&gt;nitrous oxide&lt;/a&gt; also exceeds the cooling capacity and thus must be used intermittently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will often see both large and small cars pulled over for overheating along a steep hill on a hot day. In this instance the &quot;duty cycle&quot; under these operating conditions was not continuous (100%). However, duty cycle is typically not used to describe this behavior, because it is a design expectation that it can operate continuously. The engine was simply operating outside of its designed range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Duty cycle is not influenced by the size of the engine, but rather, duty cycle is a design parameter when designing an engine system. Most cars would be designed for continuous duty, while race cars would be designed for intermittent.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-12-16T05:27:08.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6545" PostTypeId="1" AcceptedAnswerId="6548" CreationDate="2015-12-16T08:02:52.703" Score="6" ViewCount="5536" Body="&lt;p&gt;I am working on a simple program that would help welder to count how much ferrite, austenite etc. there is in weld. I am focusing on stainless steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have decided to use 3 graphs: Schaeffler, De Long and WRC. What I don't know is for which materials can I use each of these graphs, to get reliable results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe, that in each of those charts I can use only Cr-Ni steels. Am I right? Also, standard in my country (PL) that concerns stainless steels splits them into groups like: ferritic steel, austenitic steel, ferritic-austenitic steel, martensitic steel and heatproof steel (hope I translated it right). So which group of steels can I use in three diagrams I mentioned?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt;&#xA;&lt;strong&gt;Which groups of steel can be used in Schaeffler, De Long, WRC diagram to get phase composition of weld?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4084" LastEditorUserId="368" LastEditDate="2015-12-16T17:38:07.400" LastActivityDate="2015-12-17T00:22:51.683" Title="Schaeffler, De Long, and WRC welding diagrams - which steels can be used?" Tags="|steel|metallurgy|welding|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6546" PostTypeId="1" CreationDate="2015-12-16T14:25:49.047" Score="7" ViewCount="645" Body="&lt;p&gt;I am working on a system that consists of a hanging-pendulum thrust balance with two pivot points, one in the front and one in the back There is a heavy weight applied in the front. I can change the position of the center of gravity by applying counterweights on the rear part. See attached image.&lt;a href=&quot;https://i.stack.imgur.com/UFi4a.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UFi4a.png&quot; alt=&quot;Sketch of my setup. The center of mass is drawn red.&quot;&gt;&lt;/a&gt;. The position of the center of gravity is known as well as the moments of inertia with respect the pivot points&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to simplify this system to an equivalent physical pendulum with one pivot point. However, because I have two pivot points, I have two moments of inertia, one with respect to each one of the pivots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the beginning I thought about taking the mean average of both, and then calculate the period of a physical pendulum or compound pendulum, just as &lt;a href=&quot;https://www.boundless.com/physics/textbooks/boundless-physics-textbook/waves-and-vibrations-15/periodic-motion-123/the-physical-pendulum-432-1488/&quot; rel=&quot;noreferrer&quot;&gt;Hyperphysics&lt;/a&gt; suggests: $T=2\pi\sqrt{\frac{I_{y}}{mgLcm}}$, where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$I_{y}$ is the average moment of inertia,&lt;/li&gt;&#xA;&lt;li&gt;$m$ is the mass of the system,&lt;/li&gt;&#xA;&lt;li&gt;$L_{cm}$ is the distance pivot-center of mass.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but this just does not seem right. Can somebody give me a hint on how to do this? Better, is it even reasonable to do such simplification? Shall I use a matrix notation? My purpose is to study how the thrust balance will respond when a force is applied upon it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have analyzed the oscillations of the system for several counterweights. From the study of those oscillations I can obtain the damping coefficient $\zeta$ and the natural frequency of the undamped system $\omega_{n}$, but I still need $I_{y}$ to complete the equation $\omega_{n}=\sqrt{\frac{k}{I_{y}}}$.&lt;/p&gt;&#xA;" OwnerUserId="4152" LastEditorUserId="368" LastEditDate="2015-12-16T17:37:02.927" LastActivityDate="2022-12-11T03:04:25.557" Title="What is the equivalent moment of inertia of a two-pivot pendulum?" Tags="|mechanical-engineering|dynamics|thrust|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6547" PostTypeId="1" CreationDate="2015-12-16T14:46:29.420" Score="6" ViewCount="233" Body="&lt;p&gt;This is the situation - I have three heat sources, called CHP, burner 1 and burner 2 on the same circuit. All three feed to the same heating manifold, with various heat demands. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preferably I want to heat with the CHP; if that is not sufficient, additionally with burner 1; and only use burner 2 if that is still insufficient. The CHP is always running and not controlled for temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this scheme is that the CHP burns (free) sewage gas and earns money (electricity), burner 1 burns sewage gas and burner 2 uses (expensive) natural gas. Burner 2 is installed in case all other heat providers fail or the sewage gas runs out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my idea for a control scheme, where the actual temperatures are more examples than set in stone:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In normal operation, the return flow is 75 °C or higher&lt;/li&gt;&#xA;&lt;li&gt;If return temp. &amp;lt; 73 °C, burner 1 is activated and heat controlled to 73 °C return&lt;/li&gt;&#xA;&lt;li&gt;If return temp. &amp;lt; 71 °C, burner 2 is activated and heat controlled to 71 °C return&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the last case, burner 1 should run at nominal power and burner 2 at enough power to raise return temp. to 71 °C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will I run into problems with two burners on the same circuit controlling for different temperatures?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="368" LastEditDate="2015-12-16T17:32:32.857" LastActivityDate="2015-12-28T20:42:28.933" Title="Is it a problem to have two heat sources controlled for different temperatures on the same heating circuit?" Tags="|control-engineering|heating-systems|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6548" PostTypeId="2" ParentId="6545" CreationDate="2015-12-16T18:21:11.683" Score="6" Body="&lt;p&gt;&lt;strong&gt;You are not exactly right. The purpose of Cr and Ni in stainless steel, besides the stainless part, is to tailor the microstructure. Cr promotes ferrite, Ni promotes austenite. Other elements have similar effects and must be taken into consideration. Beware of carbide formation changing properties and reducing weldability.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The three different graphs plot three different sets of equivalents. Schaeffler accounts for Ni/C/Mn vs Cr/Mo/Si/Nb. De Long accounts for nitrogen as an additional Ni equivalent. The WRC diagram accounts for Ni/C/N/Cu, and Cr/Mo/Nb. The effectiveness ratios are different among the different graphs. What happens &lt;em&gt;in between&lt;/em&gt; these different equivalents is &lt;em&gt;not&lt;/em&gt; clear, so be careful modeling combinations that don't meet the requirements of any one graph individually. If you have both Cu and Si, for example, none of the graphs can identify what microstructures will occur. Your program can make interpolations, but you should warn any users you have done so and the results may not be correct.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Microstructure Stabilizing Elements&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Each of the diagrams has two axes, and points in the diagram area correspond to expected microstructures. Each axis is associated with either &lt;strong&gt;chromium equivalent&lt;/strong&gt; or &lt;strong&gt;nickel equivalent&lt;/strong&gt;, typically with the former on the horizontal axis and the latter on the vertical axis. In steels, chromium is a &lt;strong&gt;ferrite stabilizer&lt;/strong&gt;, meaning that steels rich in chromium will tend to form ferritic microstructures. In contrast, nickel is an &lt;strong&gt;austenite stabilizer&lt;/strong&gt;, and steels rich in nickel will tend to form austenite. Naturally, all three diagrams reflect this, with upper-left regions having high Ni equivalent, low Cr equivalent, and high austenite, and bottom-right regions having the opposite. See the Schaeffler diagram below (the other two have similar overall appearance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l3Ndz.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l3Ndz.gif&quot; alt=&quot;Schaeffler Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind an equivalent is that it provides an estimate of the effect of different alloying elements on stabilization of ferrite or austenite. Some elements are &quot;better&quot; at stabilizing their respective microstructure than others. For example, carbon stabilizes austenite 30 times more effectively than nickel, while manganese is about half as effective as nickel. Of course if you want austenite, you can't simply load up the steel with carbon or it will become extremely brittle and hard to work with. The fact that different elements have different efficiencies is what allows engineers to tailor steel properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted on the axes in this particular diagram, the Cr and Ni equivalents include more than just those two elements. Specifically, carbon and manganese are austenite stabilizers, while molybdenum, silicon, and niobium are ferrite stabilizers. A more complete (but probably not exhaustive) list of each type:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ferrite Stabilizers: Cr, Si, Mo, W, Al, Ti, Nb&lt;/li&gt;&#xA;&lt;li&gt;Austenite Stabilizers: Ni, C, Mn, N&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In addition to the ferrite and austenite stabilizers, some elements also form carbides and nitrides, specifically&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Carbide Formers: Cr, W, Mo, V, Ti, Nb, Ta, Zr&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which can increase strength, but reduce ductility. They can also reduce fatigue properties by providing crack nucleation sites, and can alter creep properties. Carbide formers can also reduce weldability as they tend to form at microstructure boundaries when solidified rapidly and then allowed to cool slowly. Brittle microstructure boundaries, as with carbides, tend to cause excessive brittleness in the weld and greatly reduce impact toughness.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why Do Stabilizing Elements Work?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Low-carbon steel goes through three phases as it solidifies and cools, starting with ferrite $\delta$-iron, austenite $\gamma$-iron, and then ferrite $\alpha$-iron. In the Fe-Cr phase diagram shown below, note that the upper and lower ferrite regions at the pure iron vertical axis eventually merge with increasing Cr, moving right on the diagram. This indicates that Cr &lt;em&gt;stabilizes&lt;/em&gt;, in a thermodynamic sense, the ferrite phase: it is more energetically favorable with high Cr content. The reverse effect occurs with increasing Ni, so that the austenite region spreads outward, eliminating both ferrite regions, as seen in the Fe-Ni phase diagram. Ni makes austenite more energetically favorable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lJnbK.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lJnbK.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K5pok.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K5pok.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://steel.keytometals.com/articles/art50.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt; for a good starting point for more information.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How Should I Use the Diagrams?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The Schaeffler diagram takes into account the following equivalents:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\textrm{Ni eq}=\textrm{Ni}+30\times \textrm{C}+0.5\times\textrm{Mn}$&lt;/li&gt;&#xA;&lt;li&gt;$\textrm{Cr eq}=\textrm{Cr}+\textrm{Mo}+1.5\times \textrm{Si}+0.5\times \textrm{Nb}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The De Long diagram has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\textrm{Ni eq}=\textrm{Ni}+30\times \textrm{C}+0.5\times\textrm{Mn}+30\times\textrm{N}$&lt;/li&gt;&#xA;&lt;li&gt;$\textrm{Cr eq}=\textrm{Cr}+\textrm{Mo}+1.5\times \textrm{Si}+0.5\times \textrm{Nb}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The WRC-1992 stainless steel diagram has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\textrm{Ni eq}=\textrm{Ni}+35\times\textrm{C}+30\times\textrm{N}+0.25\times\textrm{Cu}$&lt;/li&gt;&#xA;&lt;li&gt;$\textrm{Cr eq}=\textrm{Cr}+\textrm{Mo}+0.7\times\textrm{Nb}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sources for &lt;a href=&quot;http://www.bssa.org.uk/topics.php?article=121&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schaeffler and De Long&lt;/a&gt; and &lt;a href=&quot;http://www.migweld.de/english/service/welding-stainless-steels/wrc-diagram-for-standard-analysis.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;WRC-1992&lt;/a&gt;. The diagrams are only strictly valid when only the relevant equivalent stabilizing elements are present in significant quantities, and no other stabilizers are present. If only Cr and Ni are present, any of the diagrams should be accurate, though you may want to verify that they indicate the same microstructures given the same values of Cr and Ni. In contrast, if both Cu and Si are present in a material, none of the diagrams should be expected to produce accurate results. It is possible to interpolate between the diagrams, but such a model would be a new model, and should be validated experimentally before being asserted as useful or accurate. At the very least, users should be warned that such interpolation is occurring.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-12-17T00:22:51.683" LastActivityDate="2015-12-17T00:22:51.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6550" PostTypeId="1" CreationDate="2015-12-16T19:16:54.503" Score="4" ViewCount="1650" Body="&lt;p&gt;Within MATLAB Community, I found an example of a simple quadcopter model control built with Simulik and SimMechanics, The controllers handle position and orientation control. I want to extend this controller and add a speed controller (Quadcopter Vx,Vy,Vz), and keep the position control as well. What should be my starting point? I believe I can't use two PID controllers for the same input since they'll interfere with each other, thus one PID will oscillate. The current controller is given below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oJlWc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oJlWc.png&quot; alt=&quot;Controlling 4 motor of quad with PID and saturation logic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would you suggest I should start with? I intend to use MATLAB/Simulink. Thanks for the information in advance.&#xA;Note: I have every sensor I need on the model to measure errors.&lt;/p&gt;&#xA;" OwnerUserId="4157" LastActivityDate="2020-03-06T12:52:30.987" Title="Speed Control of Quadcopter along with Position Control Using PID with Simulink" Tags="|control-engineering|simulink|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6554" PostTypeId="2" ParentId="6550" CreationDate="2015-12-17T01:51:32.813" Score="4" Body="&lt;p&gt;As you mention, PID control is for Single Input, Single Output (SISO) control. If you want Multi Input, Single Output (MISO) or Multi Input, Multi Output (MIMO) control, you need to look at other methods of control, such as Linear Quadratic Regulators (LQR) or state space control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MATLAB has a control system toolbox that will do both, and every quality control text will cover both in explicit detail. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-12-17T01:51:32.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6557" PostTypeId="2" ParentId="6547" CreationDate="2015-12-17T06:52:28.643" Score="2" Body="&lt;p&gt;Your control logic appears to be sound. And with that logic, they could each be on separate simple thermostats if necessary. Just make sure all of your temperature sensors are sensing at the same location; or lots of craziness will happen. If you would like us to review your hydraulic plumbing you will need to attach a plumbing diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important thing to consider is that your thermostats will have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Deadband#Thermostats&quot; rel=&quot;nofollow&quot;&gt;dead band&lt;/a&gt;. More expensive thermostats may have an adjustable dead band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dead band is the difference between when a controller starts and stops. So in your scenario burner 1 will start at 73 °C and will turn off at (possibly) 74 °C. This gives you a dead band of 1 °C. Depending on the thermal mass of your system, 1 °C may mean a cycle time of 1 hour or 30 seconds. If it is the latter; your themostat, relays and other hardware will wear out quickly so you will want to increase the amount of thermal mass or dead band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overlapping them would not necessarily be a bad thing; and would help the system quickly return to the 75 °C set point after a large load. Your choice here depends on how critical the 75 °C target is and how expensive fuel is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Burner 1 on when between 73 °C and 75 °C&lt;br&gt;&#xA;Burner 2 on when between 71 °C and 75 °C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will also want to add some sort of notification system like an led to let you know when Burner 2 is on and that it is costing you money. Thermostats, relays, and temperature sensors do go bad. It would be nice to know if the system is acting up before the bill at the end of the month.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="368" LastEditDate="2015-12-17T18:48:21.067" LastActivityDate="2015-12-17T18:48:21.067" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6558" PostTypeId="1" AcceptedAnswerId="6563" CreationDate="2015-12-17T08:04:17.747" Score="4" ViewCount="82" Body="&lt;p&gt;For the last hour I have been trying to understand what $\dfrac{\partial p}{\partial s}ds$ is in Euler's equation but I have a problem. You can see this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7scvT.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7scvT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5zEOd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5zEOd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that &#xA;$p\ dA = \text{Force}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$γ ds dA = mg$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but what is $\left(p + \dfrac{\partial p}{\partial s} dA\right)$?&lt;/p&gt;&#xA;" OwnerUserId="4162" LastEditorUserId="1832" LastEditDate="2015-12-17T09:55:26.270" LastActivityDate="2015-12-23T17:11:57.997" Title="∂p / ∂s ds Component in Euler's Fluids Equation" Tags="|mechanical-engineering|fluid-mechanics|civil-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6560" PostTypeId="2" ParentId="6516" CreationDate="2015-12-17T09:02:28.423" Score="0" Body="&lt;p&gt;The mechanism you're looking for is named &lt;strong&gt;push-push latch&lt;/strong&gt;. They are used in a lot of places and there are various patents for that. You can check &lt;a href=&quot;https://www.google.com/patents/EP1596030A2?cl=un&quot; rel=&quot;nofollow noreferrer&quot;&gt;this patent&lt;/a&gt; granted at 2005. It is for a small holder but it can be modified for bigger projects. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/yWADU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yWADU.png&quot; alt=&quot;Figure from Patent EP1596030 A2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if I were you I'd desing a plastic holder such as the stress holder at &lt;a href=&quot;http://kinetxbatons.com/product/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kinetxbatons&lt;/a&gt; picture given below. Moving parts and mechanisms are prone to errors and dirt is a real problem. You can check snap-fit design books I've linked in another &lt;a href=&quot;https://engineering.stackexchange.com/questions/2933/snap-fit-design/3341#3341&quot;&gt;answer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lboZ2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lboZ2.png&quot; alt=&quot;Kinetxbatons stress holder&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="879" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-12-17T09:02:28.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6563" PostTypeId="2" ParentId="6558" CreationDate="2015-12-17T13:33:38.760" Score="3" Body="&lt;p&gt;$\frac{\partial p}{\partial s}$ is the pressure gradient through the control volume.  Multiplying it by $ds$, the length of the control volume, gives the change in pressure from one side of the control volume to the other.  It is essentially a first-order Taylor-series expansion of the function $p(s)$.  This method is used a lot in fluid dynamics derivations; the most thorough explanation I've seen is in &quot;Computational Fluid Dynamics&quot; by John Anderson.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-12-17T13:33:38.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6564" PostTypeId="1" CreationDate="2015-12-17T13:33:51.200" Score="0" ViewCount="399" Body="&lt;p&gt;I have a rod heater mounted in a brass rod and I need to coat the rod with a solid thermal insulator. I have the following options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;PTFE&lt;/li&gt;&#xA;&lt;li&gt;Teflon&lt;/li&gt;&#xA;&lt;li&gt;Polyurethane foam&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, I doubt that the foam would be suitable when it comes to be in direct contact with the brass rod. How can I choose the right insulator for this application?&lt;/p&gt;&#xA;" OwnerUserId="3591" LastEditorUserId="368" LastEditDate="2015-12-17T17:02:40.180" LastActivityDate="2015-12-17T17:02:50.273" Title="How to choose an insulating material for a heated metal rod" Tags="|thermodynamics|heat-transfer|thermal-insulation|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6565" PostTypeId="2" ParentId="6564" CreationDate="2015-12-17T13:42:17.780" Score="1" Body="&lt;p&gt;I would say fiber glass would suffice.  It's non-combustable and a good insulator.  You can use regular house insulation fiber glass or special shething thats available online.  Depends partly on your engineering constraints.&lt;/p&gt;&#xA;" OwnerUserId="4167" LastActivityDate="2015-12-17T13:42:17.780" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6567" PostTypeId="1" CreationDate="2015-12-17T15:04:12.057" Score="1" ViewCount="1616" Body="&lt;p&gt;I am trying to find Poisson's Ratio, given the following information:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A cylindrical specimen of some metal alloy 52 mm in diameter is&#xA;  stressed elastically in tension. A force of 43,354 N produces a&#xA;  reduction in specimen diameter of 0.0031 mm. The elastic modulus of&#xA;  the material is 821 GPa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I calculated the lateral strain as $0.0031\ \mathrm{mm}/52\ \mathrm{mm} = 5.96\times 10^{-5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $\sigma = F/A$, I calculated the axial strain as the stress divided by the elastic modulus. Since $E = 821\ \mathrm{GPa}$ and the stress is $\sigma = 43,354\ \mathrm{N}/(\pi(\frac{52}{2}\ \mathrm{mm})^2)= 0.02041\ \mathrm{GPa}$, the axial strain is $\epsilon_a = 2.486 \times 10^{-5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing the lateral strain by the axial I get the answer $\nu = -2.397$, which seems too big (in module) to be right. I know that the algebra is correct, so the problem must be somewhere else. But where?&lt;/p&gt;&#xA;" OwnerUserId="4168" LastEditorUserId="368" LastEditDate="2015-12-17T21:49:14.037" LastActivityDate="2016-03-01T14:09:27.773" Title="Poisson's ratio" Tags="|materials|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6570" PostTypeId="2" ParentId="6447" CreationDate="2015-12-17T16:38:10.953" Score="4" Body="&lt;p&gt;I don't know the whole solution, but I can show you a way you can probably go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @nluigi pointed out in his comment, the governing equations of your problem are given in &#xA;&lt;a href=&quot;https://engineering.stackexchange.com/questions/5921/modeling-transient-heat-transfer-between-two-1-d-materials/5925&quot;&gt;Modeling Transient Heat Transfer between two 1-D materials&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with temperature $T$, thermal diffusivity $\alpha=\frac{k}{\rho \, c_P}$, time $t$, space $x$, thermal conductivity &#xA;$k$, density $\rho$ and specific heat capacity $c_P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental solution of this differential equation is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;H(x,t) = \frac{1}{(4 \pi \alpha t)^{\frac{n}{2}}} \, \exp \left( - \frac{|x|^2}{4 \alpha t} \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the dimension of your space coordinate $x$ and $|x|^2 = \sum_{i=1}^n x_i^2$ is the square of the &#xA;Euklid norm (with $x_i$ being the coordinates of $x$ in a cartesian coordinate system). Please note that $x$ may be &#xA;a scalar value for one-dimensional problems ($n=1$), a two-dimensional vector for two-dimensional problems ($n=2$) &#xA;or a three-dimensional vector for three-dimensional problems ($n=3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example for a one-dimensional problem is heat transfer through a wall. Your problem is (approximately) &#xA;two-dimensional. A three dimensional problem would be a hot or cold sphere within a large room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate a solution for a specific problem, you need the initial conditions, given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T(x,t=0) = T_0(x)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and calculate the convolution between the fundamental solution and the initial conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T(x,t) = (H * T_0) (x,t) = \int\limits_{{\mathbb R}^n} H(x-y,t) \, T_0(y) \, \mathrm{d}y&#xA;\\&#xA;T(x,t) = \frac{1}{(4 \pi \alpha t)^{\frac{n}{2}}} \int\limits_{{\mathbb R}^n} \exp \left( - \frac{|x-y|^2}{4 \alpha &#xA;t} \right) T_0(y) \, \mathrm{d}y&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I understand your problem correctly, you can use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T_0(x) = \left\{ \begin{array}{ll} T_I &amp;amp; \mbox{for } |x| &amp;lt; x_B \\ 0 &amp;amp; \mbox{otherwise} \end{array}\right.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $T_I$ being your inner temperature at the beginning and $x_B$ being the coordinate of your wall. Note that since the differential equation is linear, you can add any baseline temperature to the solution as well as multiply the solution with any scale. Thus for the sake of mathematical solution, you could calculate with $T_I=1$ and scale the result. However, I'm an engineer and it doesn't hurt much to just drag $T_I$ along.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently had the one-dimensional problem, and this one can still be handled with some effort:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T(x,t) = \frac{1}{\sqrt{4 \pi \alpha t}} \int\limits_{-\infty}^{+\infty} \exp \left( - \frac{|x-y|^2}{4 \alpha t} &#xA;\right) T_0(y) \, \mathrm{d}y&#xA;\\&#xA;= \frac{T_I}{\sqrt{4 \pi \alpha t}} \int\limits_{-x_B}^{+x_B} \exp \left( - \frac{|x-y|^2}{4 \alpha t} &#xA;\right) \, \mathrm{d}y&#xA;\\&#xA;= \frac{T_I}{2} \left( \mathrm{erf} \left( \frac{x+x_B}{\sqrt{4 \alpha t}} \right) -&#xA;\mathrm{erf} \left( \frac{x-x_B}{\sqrt{4 \alpha t}} \right) \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int\limits_0^x \exp\left( -t^2 \right) \, \mathrm{d}t$ being the error function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\alpha=1$, $T_I=1$, $x_B=1$ the following graph shows the solution for some $t$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kmSGQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kmSGQ.png&quot; alt=&quot;1d-solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for higher dimensions of the initial conditions given above, the equation needs to be transformed into spherical coordinates first. For two dimensions you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T(x,t) = \frac{1}{4 \pi \alpha t} \int\limits_{{\mathbb R}^2} \exp \left( - \frac{|x-y|^2}{4 \alpha t} &#xA;\right) T_0(y) \, \mathrm{d}y&#xA;\\&#xA;\Rightarrow\quad&#xA;T(x,t) = \frac{T_I}{4 \pi \alpha t} \int\limits_{0}^{r_B} \left( \int\limits_0^{2\pi} \exp \left( - \frac{|x-y|^2}{4 \alpha t} \right) \, \mathrm{d}\varphi(y) \right) r(y) \, \mathrm{d}r(y)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;giving (at least me) considerable headaches when trying to evaluate. Probably somebody out there knows what&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\int\limits_0^{2\pi} \exp \left( - |x-y|^2 \right) \, \mathrm{d}\varphi(y)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;evaluates to, but I currently don't. Probably there also is a much cleaner approach to evaluate the convolution integral in two dimensions.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-12-17T16:38:10.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6574" PostTypeId="2" ParentId="6467" CreationDate="2015-12-18T02:28:55.280" Score="1" Body="&lt;h3&gt;First you have to decide how far down the rabbit hole to go...&lt;/h3&gt;&#xA;&lt;p&gt;As with all things, the answer is going to be complicated and unique to every project. The easy way is to stop now and hire a Contractor. The Contractor is supposed to know what they are doing and know when an additional professional (i.e. Engineer) is needed.&lt;/p&gt;&#xA;&lt;p&gt;This reduces your future liability for doing it wrong (i.e. places the liability on the Contractor) and allows you to get the project done without lots of research.&lt;/p&gt;&#xA;&lt;h3&gt;No turning back...&lt;/h3&gt;&#xA;&lt;p&gt;If you get to the point of your DIY project that you are asking yourself this question, you have likely already gone too far. Also, you likely don't want to ask anyone else (or pay for anyone else).&lt;/p&gt;&#xA;&lt;p&gt;There are lots of rules-of-thumb or general guidelines that could be followed, but there are always exceptions (and ways to convince yourself that your project is special). The rules-of-thumb are all based on something. That something is usually...&lt;/p&gt;&#xA;&lt;h3&gt;Building Codes!&lt;/h3&gt;&#xA;&lt;p&gt;If you are in the US, get acquainted with the building codes that govern your structure. (Other countries should have similar codes.)&lt;/p&gt;&#xA;&lt;p&gt;For a house, the two that you have to worry about are typically:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://publicecodes.cyberregs.com/icod/irc/2012/&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Residential Code&lt;/a&gt; and&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nfpa.org/codes-and-standards/document-information-pages?mode=code&amp;amp;code=70&quot; rel=&quot;nofollow noreferrer&quot;&gt;National Electrical Code&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Both of these are available online as linked above. If your local jurisdiction requires a special year, check here: &lt;a href=&quot;https://law.resource.org/pub/us/code/safety.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Resource.org&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Note, all of these links are legal. Don't go searching the shady parts of the internet.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;h3&gt;To the Books!&lt;/h3&gt;&#xA;&lt;p&gt;Read the section of the above codes that apply to your project. Try not to get distracted by the vast amount of other things in there though. You can lose hours of your life by finding interesting things that you never knew were specified by code.&lt;/p&gt;&#xA;&lt;p&gt;If you have found the section that applies to your project and it all made sense, congratulations, you should have an answer. The answer may be specified by the code, in which case you now know how to proceed. If it isn't covered by the code (or you don't understand), then you need to hire someone.&lt;/p&gt;&#xA;&lt;p&gt;It is that simple.&lt;/p&gt;&#xA;&lt;p&gt;Oh wait, there is something more. You might not have a choice...&lt;/p&gt;&#xA;&lt;h3&gt;Permits!&lt;/h3&gt;&#xA;&lt;p&gt;You were going to get the permit that is required by your local city/county right? &lt;strong&gt;RIGHT?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The kind (and some are helpful, some are not) people in the permit office will definitely have opinions about what you will need to hire someone to do. Their opinions also come backed with fines if you don't comply.&lt;/p&gt;&#xA;&lt;h3&gt;California Guide&lt;/h3&gt;&#xA;&lt;p&gt;California has a guide that covers most of these issues: &lt;a href=&quot;http://www.bpelsg.ca.gov/pubs/consumer_guide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Guide For Consumers&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-02-01T04:10:15.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6575" PostTypeId="2" ParentId="5494" CreationDate="2015-12-18T02:39:13.200" Score="1" Body="&lt;p&gt;The easy answer is don't design - cheat.  Go and look for a trailer similar to what you're after.  Photograph it and measure all the bits. (Don't act  like you're trying to nick it).  Similar sections will do, but I'd err on larger sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your issues will be:-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Welding the joints.  I'm not sure what type of welding you're certified in, but fillet welding 10mm steel is not the same as tacking on car wings.  &lt;/li&gt;&#xA;&lt;li&gt;Brakes.  You need to make sure that they work.  How will you test them?  Just the fact that the trailer doesn't roll down your drive doesn't mean they're working.&lt;/li&gt;&#xA;&lt;li&gt;In England, if you took this onto a public highway, it would need a test certificate.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I hate the whole Health &amp;amp; Stupidity nonsense, but &lt;strong&gt;do not underestimate the responsibility you will be assuming if you drive this down a road at speed.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2015-12-18T02:39:13.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6576" PostTypeId="1" AcceptedAnswerId="6630" CreationDate="2015-12-18T05:08:20.240" Score="10" ViewCount="19540" Body="&lt;p&gt;A simple thermostat will turn on at one temperature and off at a higher temperature. This keeps the thermostat from cycling on and off too quickly. The difference between these values is sometimes called &lt;a href=&quot;https://en.wikipedia.org/wiki/Deadband#Thermostats&quot; rel=&quot;nofollow noreferrer&quot;&gt;dead band&lt;/a&gt; and sometimes called &lt;a href=&quot;https://en.wikipedia.org/wiki/Hysteresis#Control_systems&quot; rel=&quot;nofollow noreferrer&quot;&gt;hysteresis&lt;/a&gt;. Both words apply to more complex processes like hydraulic valves and PID controllers, but what I am curious about is which one is better used to describe a simple thermostat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What term(s) do you use in this context? Please cite sources if you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some background, this question came up as a result of this &lt;a href=&quot;https://engineering.stackexchange.com/questions/6547/is-it-a-problem-to-have-two-heat-sources-controlled-for-different-temperatures-o&quot;&gt;Engineering SE question&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2020-08-19T02:09:23.277" Title="What do you call the difference between the on and off temperatures in a simple thermostat?" Tags="|control-engineering|terminology|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6579" PostTypeId="1" CreationDate="2015-12-18T09:29:22.843" Score="1" ViewCount="58" Body="&lt;p&gt;I have one question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Fluent, should be generated same mesh in rotational faces for creating periodic rotational conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way that unmatched meshes are applied for creating periodic rotational conditions?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="1586" LastEditDate="2016-02-18T19:50:07.497" LastActivityDate="2016-02-18T19:50:07.497" Title="periodic rotational conditions in Fluent" Tags="|fluid-mechanics|cfd|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6580" PostTypeId="2" ParentId="6460" CreationDate="2015-12-18T10:05:33.477" Score="-1" Body="&lt;p&gt;you can olso put temperature gauge for each column vessel with Instrument calibrated accepted because ; 1# the volume of column vessel &amp;amp; level of water are not the same in between this is normal condition (ambient) event not pressurized yet, temperature and pressure of each vessel not the same due to gravity same case&#xA;Hydrostatic test for under water pipe line offshore to Onshore. due to sea water temperature are not stabile, I agree with the 1st answer&lt;/p&gt;&#xA;" OwnerUserId="4178" LastActivityDate="2015-12-18T10:05:33.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6581" PostTypeId="2" ParentId="6031" CreationDate="2015-12-18T10:23:25.393" Score="3" Body="&lt;p&gt;Yes you can negate or deguass an electronic magnet actively. The older door electromagnetic locks uses permanent magnets and electromagnets hybrid called bonded magnets. The magnet is locked when the power is turned off allowing the permanent magnet to stick. When electricity is applied it creates a field that to temporarily neutralize the permanent magnet allowing it to release so when the power goes out the default is locked like in an insane-asylum . While the door is propped open some one can easily depolarize or degauss the permanent magnet side by walking by and holding a much stronger magnet north to north or south to south quickly and permanently weakening the door's permanent magnet strength.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2015-12-20T20:04:28.193" LastActivityDate="2015-12-20T20:04:28.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6582" PostTypeId="2" ParentId="6576" CreationDate="2015-12-18T12:25:43.930" Score="7" Body="&lt;p&gt;I'd say &quot;deadband&quot; is the more end-user term, and &quot;hysteresis&quot; more of a engineering term.  Engineers will understand deadband in this context, but Joe Sixpack or even a electrician installing a thermostat may not understand what &quot;hysteresis&quot; means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I disagree that these words apply to more complex processes.  What is going on in a simple thermostat is exactly hysteresis, which is fine to express as deadband to laymen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are overthinking this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more precise meaning of deadband is actually different from hysteresis.  A deadband is a region where the system doesn't respond to changes.  This does occur in hysteresis, but the general case does not require hysteresis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose you have a proportional temperature controller instead of on/off like what the thermostat is doing.  Let's say it has a 1&amp;deg; deadband and the temperature is slowly rising from 65&amp;deg; to 70&amp;deg;.  This will cause a decreasing control response.  If the temperature now drops, the control response won't change until it gets to 69&amp;deg;, then will increase with decreasing temperature again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a deadband is usually undesirable, so systems are specified not to exceed a particular deadband.  This is common in mechanical systems due to the combination of looseness of fit and static friction.  It is often called the &quot;backlash&quot; in gear systems.  Electronic systems can have a deadband caused by two diode drops, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary, it is OK to call the difference between the two hysteresis thresholds a deadband, but not all deadbands are due to hysteresis.  In common usage, this is rather a fine distinction and I wouldn't count on people realizing this.  Usually what is meant by the deadband in a system is clear enough from context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a engineering setting, use &quot;hysteresis&quot; when it really is hysteresis.  In a lay setting, use &quot;deadband&quot; and don't expect anyone to understand nuances of hysteresis versus some other cause of a deadband.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2015-12-29T00:11:41.260" LastActivityDate="2015-12-29T00:11:41.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6583" PostTypeId="2" ParentId="6576" CreationDate="2015-12-18T12:29:47.380" Score="0" Body="&lt;p&gt;The &quot;dead band&quot; is the range where nothing happens. &amp;lt;--&lt;em&gt;updated for acccuracy thanks to Ericnutsch's comment&lt;/em&gt;&#xA;A simple temperature control system  shuts down in the dead band, as heating only activates at  $T_{low}$ and stops at $T_{high}$, while the cooling portion should activate only at $T_{high} + x$ and deactivate at $T_{low}+y$ .&#xA;There are two dead bands $T_{low}$ to $T_{low}+y$ and  $T_{high} to $T_{high} + x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Hysteresis&quot; is the term applied to any system whose state is path-dependent. You can have a control system, or for that matter, a mechanical system in general without a dead band but with hysteresis.  Here's a sample curve from &lt;a href=&quot;http://www.uk.tdk-lambda.com/technical-data/glossary-terms.aspx?i=H&quot; rel=&quot;nofollow noreferrer&quot;&gt;tdk-lambda&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/5JbWs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5JbWs.png&quot; alt=&quot;Magnetic hysteresis&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="615" LastEditDate="2015-12-29T11:25:47.267" LastActivityDate="2015-12-29T11:25:47.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6586" PostTypeId="1" CreationDate="2015-12-18T15:56:15.717" Score="7" ViewCount="3926" Body="&lt;p&gt;I'm reading an introduction to structural engineering book and there's a fundamental concept I'm trying to relate to real life:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In construction, we regularly bend iron rebar into a variety of shapes and rebar stays locked in whatever bendy shape we give it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, if the rebar holds its deformed position that means it's behaving plastically and &lt;em&gt;not&lt;/em&gt; elastically. When a material transitions from elastic behavior to plastic behavior, doesn't that mean it has yielded? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it has indeed yielded, shouldn't bent rebar exhibit vastly reduced bearing capacity, particularly about the areas it has been bent? &lt;/p&gt;&#xA;" OwnerUserId="3832" LastActivityDate="2015-12-20T00:14:28.467" Title="Material Properties: Does bending reduce iron rebar's strength?" Tags="|materials|structural-engineering|reinforced-concrete|reinforcement|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6587" PostTypeId="1" CreationDate="2015-12-18T16:37:16.373" Score="-1" ViewCount="380" Body="&lt;p&gt;After finish designing a product and having some &lt;code&gt;STL&lt;/code&gt; files, I would like to make some movie of it, and animate it .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example same as Apple does in its marketing videos, where you see the product zoomed in and camera covering it from many directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;code&gt;Solidworks&lt;/code&gt; can't do that, is there other software that can do that ?&lt;/p&gt;&#xA;" OwnerUserId="4182" LastEditorDisplayName="user16" LastEditDate="2015-12-18T17:03:02.497" LastActivityDate="2015-12-18T17:03:02.497" Title="Create a movie from STL file?" Tags="|computer-aided-design|solidworks|" AnswerCount="1" CommentCount="1" ClosedDate="2015-12-18T17:03:15.380" ContentLicense="CC BY-SA 3.0" />
  <row Id="6588" PostTypeId="2" ParentId="6587" CreationDate="2015-12-18T16:54:47.863" Score="2" Body="&lt;p&gt;Almost any 3D CAD programs can make movies.  How detailed you want the movies, various textures (i.e. &quot;shiny metal&quot;), etc... is where the devil resides.  As long as your software will render the material properly to your satisfaction, and you are proficient in the software, it should make a movie and you should use it.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-18T16:54:47.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6590" PostTypeId="2" ParentId="6586" CreationDate="2015-12-18T17:39:13.930" Score="6" Body="&lt;p&gt;It depends what you mean by bearing capacity. Plastic deformation increases yield strength at the expense of ductility. Also possibly toughness and fatigue durability if strained too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metals virtually always exhibit increased yield strength when strained plastically below their strain recovery temperatures. The phenomenon is referred to as &lt;strong&gt;strain hardening&lt;/strong&gt;, and is important in a number of product and processing applications as another tool for tailoring material properties. The increase yield strength comes with a number of drawbacks, the most readily apparent of which is reduced elongation at failure. There is a tradeoff between yield strength amd ductility. Additionally, too much plastic strain with no recovery or annealing can result in the formation of voids, microcracks, and other microscopic, volumetric defects. Volumetric defects provide crack nucleation sites when dynamically loaded and accelerate failure in dynamic fatigue, as well as creating stress concentration sites which can reduce durability and toughness eve after annealing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below shows the effects of cold work on strength, hardness, and ductility. Note that as cold work increases to the right, strength and hardness increase, while ductility decreases. While the plot is for a specific metal, the curve shapes apply broadly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZFqCh.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZFqCh.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2015-12-18T18:14:52.273" LastActivityDate="2015-12-18T18:14:52.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6592" PostTypeId="2" ParentId="6460" CreationDate="2015-12-18T19:20:17.930" Score="0" Body="&lt;p&gt;The pressure will be the same regardless of the temperature.  So, the two columns will have the correct pressure recording.  However, the height of the column will vary.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. - Say you have your pressure transmitter reading for calibration a 10C column (999.7 kg/m^3).  The results are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.5m - 4.9 kPa - 7 mV; 1m - 9.8 kPa - 10 mV; 2m - 19.6 kPa - 13 mV;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you measure the second column at 70C (977.8 kg/m^3), and it reads 7mV.  Then the column will still be 4.9kPa at the base, but the new height will be 51.1 cm, not the original 50 cm.  The pressure will still be 4.9kPa though.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-18T19:20:17.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6596" PostTypeId="5" CreationDate="2015-12-19T09:42:10.240" Score="0" Body="&lt;p&gt;A magnet (from Greek μαγνήτις λίθος magnḗtis líthos, &quot;Magnesian stone&quot;) is a material or object that produces a magnetic field. This magnetic field is invisible but is responsible for the most notable property of a magnet: a force that pulls on other ferromagnetic materials, such as iron, and attracts or repels other magnets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A permanent magnet is an object made from a material that is magnetized and creates its own persistent magnetic field. An everyday example is a refrigerator magnet used to hold notes on a refrigerator door. Materials that can be magnetized, which are also the ones that are strongly attracted to a magnet, are called ferromagnetic (or ferrimagnetic). These include iron, nickel, cobalt, some alloys of rare earth metals, and some naturally occurring minerals such as lodestone. Although ferromagnetic (and ferrimagnetic) materials are the only ones attracted to a magnet strongly enough to be commonly considered magnetic, all other substances respond weakly to a magnetic field, by one of several other types of magnetism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ferromagnetic materials can be divided into magnetically &quot;soft&quot; materials like annealed iron, which can be magnetized but do not tend to stay magnetized, and magnetically &quot;hard&quot; materials, which do. Permanent magnets are made from &quot;hard&quot; ferromagnetic materials such as alnico and ferrite that are subjected to special processing in a powerful magnetic field during manufacture, to align their internal microcrystalline structure, making them very hard[citation needed] to demagnetize. To demagnetize a saturated magnet, a certain magnetic field must be applied, and this threshold depends on coercivity of the respective material. &quot;Hard&quot; materials have high coercivity, whereas &quot;soft&quot; materials have low coercivity.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2015-12-21T15:29:43.430" LastActivityDate="2015-12-21T15:29:43.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6597" PostTypeId="4" CreationDate="2015-12-19T09:42:10.240" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-12-19T09:42:10.240" LastActivityDate="2015-12-19T09:42:10.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6600" PostTypeId="2" ParentId="6586" CreationDate="2015-12-20T00:14:28.467" Score="1" Body="&lt;p&gt;Bringing rebar  to full plastic behaviour and then releasing doesn't make it any less strong.  It may make it stronger through work hardening.  If you don't repeat this too often, you won't introduce fatigue.  You might recognise &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/6/61/Rock_plasticity_compression_plain.svg/300px-Rock_plasticity_compression_plain.svg.png&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or similar diagrams from your book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The allowable stresses for rebar allow for them to be fully bent to various predetermined or custom shapes.  They're made and bent to quite rigorous standards.  Just don't re-bend one from shape code 00 to code 75 too often.  Har har.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2015-12-20T00:14:28.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6601" PostTypeId="1" AcceptedAnswerId="6609" CreationDate="2015-12-20T19:25:05.857" Score="8" ViewCount="11588" Body="&lt;p&gt;I have calculated the shear flows and center of an L shaped profile. I assume they are correctly calculated (I followed the same steps as during the lectures for an I shaped profile). However, I have some problems with the interpretation of the results. The results are shown in my simple sketch.&lt;/p&gt;&#xA;&lt;p&gt;The height of the profile is 1.5 times the base (but this actually doesn't matter) and thickness is very small compared to other dimensions. Given this, I calculated that the shear center is about 0.49xH to the left of the center of gravity (c.g.) to counter act the torsion of the shear flow.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1jmFQ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1jmFQ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h1&gt;My confusion regarding the results:&lt;/h1&gt;&#xA;&lt;p&gt;The net result of the vertical shear flow is equal to the vertical force V. The net result of torsion of the shear flows is equal to the torsion moment of the vertical force (0.49xHxV).&#xA;&lt;strong&gt;But there is no horizontal force to counter act the horizontal shear flows. How is this possible?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I am confused because in the lectures we have only seen shear flows in symmetric profiles (like C an I).&lt;/p&gt;&#xA;" OwnerUserId="4201" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-01-06T02:05:51.187" Title="Interpretation of shear flow/shear center of L-profile" Tags="|mechanical-engineering|stresses|structural-analysis|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6602" PostTypeId="2" ParentId="6031" CreationDate="2015-12-20T23:48:57.767" Score="9" Body="&lt;p&gt;I think it may be possible but more difficult than it first appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vYOMn.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vYOMn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know this, but I suspect that the three poles are arranged as south-north-south (or vice-versa) and that the coil is wound, pushed into the black slots and potted in position. Once the (electro) magnet hits the keeper the magnetic circuit is closed. As anyone who has played with a horseshoe magnet will know, opening the closed loop is very difficult. Interfering with this closed circuit from outside would be very difficult due to the relatively low permeability of the air relative to the iron. Iron is about 5,000 times more permeable than air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a3rde.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a3rde.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we can see that with the lock open the exposed faces are poles of the electro-magnet. Note also that the magnetic path is twice as wide in the centre pole as on the upper and lower poles so that flux density is fairly constant. Once the lock closes the flux forms a loop through the iron core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think magnetic cancellation is unlikely to be practical.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2015-12-21T22:25:27.670" LastActivityDate="2015-12-21T22:25:27.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6604" PostTypeId="1" CreationDate="2015-12-21T08:37:24.873" Score="1" ViewCount="321" Body="&lt;p&gt;I have a Downham Mini Borer (similar/identical to an Elliot) that has no mechanical stop on how far the Z-axis can be wound in the upward direction. This means it's possible to wind it up against some internal, physical limit. That's the situation mine is now in and it doesn't want to wind back the other way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine the hand-wheel has a worm gear, which turns a wheel around a nut on the vertical threaded bar in the pic below. At the hidden end of the threaded bar, that thread must end, either as an abrupt stop, or as a gradual transition back to the full diameter of the round bar, and it's that that the nut is jammed against. I've tried getting at that nut by removing the automatic feed selector, but the casting that accepts the drive doesn't want to separate from the main head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QRmkq.jpg&quot; alt=&quot;Downham&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have any experience of this phenomena on this or a similar borer/mill?&lt;/p&gt;&#xA;" OwnerUserId="4206" LastActivityDate="2016-07-08T20:49:09.123" Title="Z-axis of Downham (Elliot) Mini Borer is jammed. How to recover?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6605" PostTypeId="1" CreationDate="2015-12-21T10:40:07.570" Score="-1" ViewCount="92" Body="&lt;p&gt;Im looking at the possibility of creating an object with a counterweight within itself. The object for this purpose would be a cylinder with a longitudinal central axis and the cylinder itself should be counteracted by an offset weight inside the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if this is theoretically possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see diagram for a clearer explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hTWAg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hTWAg.png&quot; alt=&quot;diagram 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to create a large area of inbalance on the periphery of the cylinder.&lt;/p&gt;&#xA;" OwnerUserId="4186" LastActivityDate="2015-12-21T16:30:12.710" Title="A counterweight with one weight inside the other" Tags="|design|" AnswerCount="1" CommentCount="2" ClosedDate="2015-12-21T21:52:17.220" ContentLicense="CC BY-SA 3.0" />
  <row Id="6606" PostTypeId="1" CreationDate="2015-12-21T12:24:47.397" Score="1" ViewCount="962" Body="&lt;p&gt;I have a general question without many specific parameters, but I hope someone will be able to point me in the right direction - at least what actuator best serves my purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on a project where I will be moving a 6 kg object in an oscillating manner up and down, from ideally about 0.1 Hz to 20 Hz at different times. The max vertical motion I would hope to be 30 cm or so. I will be controlling the speed with a microcontroller via serial. It would be ideal to have some control of vertical range of motion, though this is not necessary if it proves to be too difficult. Ideally it would also be able to brake quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not quite sure how to go about this - my first thought was a servo, but I assumed that it would be difficult to get it to move up and down as quickly as I would like, and perhaps the startup torque would be a problem on the up stroke. It would otherwise be ideal as I could control the angle, and therefore the vertical range. Perhaps there is a very fast and high torque servo that can handle this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then thought of using an industrial sewing machine motor with a clutch to vary the speed, but all of these seem to use a mechanical clutch which would be burdensome to control (would have to use a separate motor to change the speed via a lever). Also, I would have to convert the rotation to linear motion, and I'd rather avoid making more complex parts if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also thought of a linear actuator, but these seem to be rather slow, and pneumatics are too imprecise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea as to what kind of actuator to use? I'm aware I am slightly vague in my usage - this is an art project so there is some flexibility as to how it turns out, other than the general parameters I've mentioned. Any suggestions would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4209" LastEditorUserId="1832" LastEditDate="2015-12-21T15:34:08.497" LastActivityDate="2015-12-22T19:26:03.710" Title="Actuator recommendation for oscillating motion of a 6 kg object" Tags="|motors|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6607" PostTypeId="2" ParentId="6601" CreationDate="2015-12-21T14:24:11.250" Score="2" Body="&lt;p&gt;You are right that for vertical applied load, the shear flow in the horizontal direction must add up to zero to satisfy equilibrium between the internal stress and the externally applied load. Therefore the shear flow diagram which is drawn cannot be right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key thing is that shear flow (like bending stress) must be calculated with respect to the principal axis directions of the cross section. Unlike I beams, the principal axes of an angle section do not line up wit the horizontal and vertical directions. The principal axes are rotated slightly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is necessary to calculate the section properties about the principal (rotated) axes. Then consider your applied load as a vector sum of two forces, one acting in each of the principal directions. And then calculate the shear flow separately for the forces acting about the two axes, and add together the results. It's a bit long winded unfortunately!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a simplified method of doing the calculation as well, which uses only the x and y coordinates of the section (not the rotated coordinates of the principal axes). This saves on a lot of trig! You can find an explanation of this here:&#xA;&lt;a href=&quot;http://aerostudents.com/files/aircraftStructuralAnalysis/bendingShearAndTorsion.pdf&quot; rel=&quot;nofollow&quot;&gt;bending stress and shear flow in thin walled sections&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4212" LastEditorUserId="4212" LastEditDate="2016-01-06T02:05:51.187" LastActivityDate="2016-01-06T02:05:51.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6608" PostTypeId="2" ParentId="6605" CreationDate="2015-12-21T16:21:49.297" Score="2" Body="&lt;p&gt;Yes, it's possible. Any cylinder whose center of gravity is not in the geometric center has a large area of inbalance on its periphery. Such a cylinder is only stable when the center of gravity is in the lowest possible position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily obtain such a cylinder by glueing something heavy enough onto the interior wall of a hollow cylinder.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="4122" LastEditDate="2015-12-21T16:30:12.710" LastActivityDate="2015-12-21T16:30:12.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6609" PostTypeId="2" ParentId="6601" CreationDate="2015-12-21T18:42:56.717" Score="7" Body="&lt;p&gt;Firstly, the shear center is the point at which an applied load produces no torsion on the section. For a singly or doubly symmetric section, the shear center will lie on the axis of symmetry. For the unequal leg angle shown here, we can say by inspection that the shear center must be at the intersection of the two legs because the shear flow will produce no moment about this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AMuf5.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AMuf5.png&quot; alt=&quot;Shear Center of Unequal Leg Angle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general equation for shear flow in an unsymmetric section is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q=\frac{I_xV_x - I_{xy}V_y}{I_xI_y - I_{xy}^2}\sum xA  - \frac{I_yV_y - I_{xy}V_x}{I_xI_y - I_{xy}^2}\sum yA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the section is symmetric, the $I_{xy}$ term is equal to zero. For a symmetric section subjected only to vertical shear $V_y$ this equation reduces to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q = \frac{V_y}{I_x}\sum yA = \frac{VQ}{I}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for an unsymmetric section we need the full generalized equation. The result is something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WvK54.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WvK54.png&quot; alt=&quot;Shear Flow Diagrams for Unequal Leg Angle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can see that if we integrate the shear force over the length of the horizontal leg, the result will be zero. Integrating over the vertical leg will equal the applied shear force $V_y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a fairly good discussion/example of this &lt;a href=&quot;http://www.ae.msstate.edu/tupas/exp/A14.8_ex1.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://books.google.com/books?id=D2JtIxV-WQMC&amp;amp;pg=PA471&amp;amp;lpg=PA471&amp;amp;dq=shear%20flow%20in%20unsymmetric&amp;amp;source=bl&amp;amp;ots=yoJj8yy1Yo&amp;amp;sig=cOa6atR4Kb7Yfa5EMZTpbNdrXvI&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjQoKSvv-3JAhWHlh4KHRqtCYs4FBDoAQgjMAI#v=onepage&amp;amp;q=shear%20flow%20in%20unsymmetric&amp;amp;f=false&quot; rel=&quot;noreferrer&quot;&gt;here (page 474)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-12-21T18:42:56.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6610" PostTypeId="2" ParentId="6567" CreationDate="2015-12-21T19:40:30.293" Score="4" Body="&lt;p&gt;A couple of notable points:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Metals rarely exceed a few hundred GPa Modulus - see the &lt;a href=&quot;https://en.wikipedia.org/wiki/Young%27s_modulus#Approximate_values&quot; rel=&quot;nofollow&quot;&gt;wikipedia page&lt;/a&gt; for several listed values.  My hunch is this is a textbook error, and that a different modulus would give the correct answer.  For example, 82.1 GPa would lead to an axial strain of 2.486×10−4, leading to a Poisson's ratio of 0.2397, which makes sense.  The 82.1 GPa is in between Aluminum and Brass on the table - a realistic value for a metal.&lt;/li&gt;&#xA;&lt;li&gt;Some objects really do have a super high, or even negative modulus - but these are usually composite materials.  The textbook lists a range between 0 - 1/2 as acceptable values for Poisson's ratios of engineering, non-composite materials (See &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B008BWTPVY&quot; rel=&quot;nofollow&quot;&gt;Mechanics of Materials, Section 2.13&lt;/a&gt;).  This makes 0.2397 a better answer.&lt;/li&gt;&#xA;&lt;li&gt;All scientific values should have a decimal point.  A given value of 821 GPa does not list an accuracy.  821. (or 821.0) GPa states that the accuracy is +/- 0.5 GPa by having a decimal point.  This is part of &lt;a href=&quot;https://en.wikipedia.org/wiki/Significant_figures&quot; rel=&quot;nofollow&quot;&gt;significant figures&lt;/a&gt;.  This highlights that the 821 GPa is a textbook error on an additional level.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-21T19:40:30.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6611" PostTypeId="2" ParentId="6606" CreationDate="2015-12-21T20:11:54.230" Score="4" Body="&lt;p&gt;Vertical position is:&#xA;$$&#xA;y = A \sin{(\omega t)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your frequency is 20Hz, your angular frequency is $2\pi f$, or 125.66 rad/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your max vertical motion is 30cm, then you could say it's 15cm +/- 15cm, so your amplitude is 15cm = 0.15m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your position is given by:&#xA;$$&#xA;y = 0.15 * \sin{(125.66 t)} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, first derivative, your speed is given by (chain rule):&#xA;$$ &#xA;\array{&#xA;dy/dt = &amp;amp; 0.15* (\cos{(125.66 t)} * 125.66) \\&#xA;dy/dt = &amp;amp; 18.85 \cos{(125.66 t)}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, second derivative, do the chain rule again:&#xA;$$&#xA;\array{&#xA;d^2y/dt^2 = &amp;amp; 18.85 * (-\sin{(125.66t)} * 125.66) \\&#xA;d^2y/dt^2 = &amp;amp;-2368.6 \sin{(125.66t)}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line above is the second derivative of position, a.k.a. acceleration. As $F = ma$, and the max value of $-\sin{(\mbox{stuff})} = 1$, this gives you a peak acceleration of $2368.6 m/s^2$. If gravitational acceleration is $9.81 m/s^2$ (it is), then you are looking at a peak acceleration of $2368.6/9.81 = 241\mbox{g's}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to the $m/s^2$ definition of your peak acceleration, you're looking at a force of $F = (2368.6 m/s^2)(6kg)$, or 14,211 Newtons, or 3,194 lbf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you are interested in the power required to do this, that's force times speed. Because you get peak force when speed is zero and vice versa, let's just multiply the equations together and work from there, as peak power is going to happen somewhere in between peak force and peak speed. And again, since these are sine and cosine, we can drop the negative because it's absolute values we're concerned with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\array{&#xA;P = &amp;amp; F*(dy/dt) \\&#xA;P = &amp;amp; (6*2368.6 \sin{(125.66t)} N)(18.85 \cos{(125.66 t)} m/s) \\&#xA;}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearrange, and then use the double-angle formula $\sin{(2\theta)} = 2\cos{\theta}\sin{\theta}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\array{&#xA;P = &amp;amp; 267,894 \sin{(125.66t)}\cos{(125.66t)} \\&#xA;P = &amp;amp; 133,947 \sin{(2 * (125.66 t)} (Nm/s)\\&#xA;}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And again, here the peak value of $\sin{(\mbox{stuff})} = 1$, so the peak power you're going to require is $P = 133,947 Nm/s$, or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\boxed{P = 134 \mbox{kW}}\\$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think of the peak power as $P_p = V_p * I_p$, then you could consider RMS power as $P_{\mbox{rms}} = \frac{V_p}{\sqrt{2}}\frac{I_p}{\sqrt{2}}$, or $P_{\mbox{rms}} = P_p/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, going this route, if your home electrical service is 120VAC, then you're looking at drawing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;I = P/V = (134,000/2)/120\\&#xA;I = 558A&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put this into perspective, your home probably has &quot;100 amp service&quot;, meaning that if you added all of the circuits in your house together you could only get 100A combined - you're trying to draw all the power available to a block. The breakers in your house are probably rated for 15A each, though, so this would blow the circuit well before you ever got close to what you're asking for. Also the motor would burn up. And the clutch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--EDIT--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After getting back to my computer, I set up a spreadsheet with the above equations to run through some different scenarios. I found a math error this way - I had left off mass in the power equation $P = Fv$, or $P = mav$. What I calculated was just $P = av$, which is incorrect. It threw my numbers off by a factor of , which was 6 in this case, so where I thought you were drawing 93A at 120VAC you are actually drawing 558A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, if you lower your frequency from 20Hz to 5Hz, then your power requirement drops from 67kW, or 90HP - about the output power of an economy class sedan, to 1kW, or about 1.4HP. This would have you drawing about 8.7A at 120VAC, which is entirely within the realm of possibilities for actual implementation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with a reasonable set of specifications, you can go hunting for applicable actuators. At this point, your &quot;general question without many specific parameters&quot; is too vague to give much guidance beyond some broad categories. For instance, I would hope you have considered how the following plays into your decision:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Cost&lt;/li&gt;&#xA;&lt;li&gt;Positional accuracy&lt;/li&gt;&#xA;&lt;li&gt;Complexity&lt;/li&gt;&#xA;&lt;li&gt;Noise&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Those are the things I would consider, in approximately that order of priority. Once you set a budget for what you are (and are not) willing to spend, then that should weed out a lot of actuators. Beyond that, your positional accuracy should then seal the deal. You said you didn't want to use pneumatics because &quot;they are too imprecise&quot;, but you have not stated how precise you want your positioning to be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have actually used &lt;a href=&quot;https://www.youtube.com/watch?v=CyFDYbK1ebc&quot; rel=&quot;nofollow&quot;&gt;this controller&lt;/a&gt; before for some relatively fine position control, but that controller is about \$900, and you still need to buy the cylinder (probably about \$100) and a suitable air compressor (maybe \$150-\$200), and pneumatic hose and fittings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use hydraulics, but that requires a hydraulic power unit (~\$400), a position regulator (~\$100), and then the cylinder (~\$100) and the hydraulic hose and fittings, which are likely to be considerably more expensive than pneumatic hose and pneumatic fittings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use electric linear actuators, but those are typically much slower. At 5Hz, for a +/- 15cm stroke, you're looking at a top speed of 4.7 m/s. &lt;a href=&quot;https://www.firgelliauto.com/products/high-speed-actuator&quot; rel=&quot;nofollow&quot;&gt;This &quot;high speed&quot; linear actuator&lt;/a&gt; moves at 4.5 inches per second, or about 0.11m/s, which would correspond to a frequency of about 0.1Hz for you - one cycle every ten seconds instead of five cycles every second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the above are linear actuators. You can go with a motor (probably would be my choice), but as you mentioned this will require a cam/lever/crank system to convert the rotary motion to linear motion. Electric motors rated for 1.5HP or 1kW should be pretty easy to find, I would guess for about \$150 or so, but you'd still need a motor controller, which could easily run \$250-\$500, and then, if you care about position, you would need a position encoder for the motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It all gets pretty costly pretty quickly, which I know is a frustrating answer, but that's engineering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to recap: first, set a budget. As in most things, this should significantly narrow the scope of what's feasible for you.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2015-12-22T14:26:28.857" LastActivityDate="2015-12-22T14:26:28.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6612" PostTypeId="1" CreationDate="2015-12-22T01:04:57.190" Score="1" ViewCount="39" Body="&lt;p&gt;To what degree is a Car's crash worthiness dependent on max speed? If a car could only reach a max speed of 45 mph, would it require less protective measures than your average car that goes 60+ easily? I am interested particularly in the US and the World Forum for Harmonization of Vehicle Regulations. &lt;/p&gt;&#xA;" OwnerUserId="4214" LastActivityDate="2015-12-22T17:42:11.420" Title="Do Car Safety (Colision) Regulations Depend on Max Speed?" Tags="|regulations|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6613" PostTypeId="1" AcceptedAnswerId="6614" CreationDate="2015-12-22T08:54:11.327" Score="5" ViewCount="941" Body="&lt;p&gt;For torsional constant, analytical formulas are available for some standard steel sections as shown in this &lt;a href=&quot;https://en.wikipedia.org/wiki/Torsion_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about arbitrary shapes? How can I obtain the formulas for torsional constants? If no such formulas are available, are there any first principles that I can use to deduce the approximate formulas or FEM calculation for torsional constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: a related question about the FEM formulation of torsional constant is being asked &lt;a href=&quot;https://engineering.stackexchange.com/questions/6622/how-do-i-use-fem-to-derive-the-torsional-constant-of-an-arbitrarily-shape&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-12-23T07:39:12.723" Title="First principle derivation for torsional constant" Tags="|materials|structural-engineering|steel|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6614" PostTypeId="2" ParentId="6613" CreationDate="2015-12-22T09:15:15.277" Score="4" Body="&lt;p&gt;For arbitrary thick-walled shapes, there are no generic formulas available. You have to calculate the torsion constant using numerical methods, e.g. FEM or BEM.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2015-12-22T09:15:15.277" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6618" PostTypeId="2" ParentId="6606" CreationDate="2015-12-22T16:59:45.233" Score="0" Body="&lt;p&gt;For the lower frequency, the easiest setup I can think of would be a motor (connected via a &lt;a href=&quot;https://en.wikipedia.org/wiki/Variable-frequency_drive&quot; rel=&quot;nofollow&quot;&gt;variable frequency drive&lt;/a&gt;) to the object, through a gearbox to reduce the frequency (common VFDs will leave your motor between 0-60 Hz).  This then turns a large &lt;a href=&quot;https://en.wikipedia.org/wiki/Crank_(mechanism)&quot; rel=&quot;nofollow&quot;&gt;crank&lt;/a&gt;.  Connect the bottom end of your object via spring to the ground to ensure stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So long as your wheel is 30cm in diameter, your object will displace the same fixed 30cm each time.  The gear box will make sure you have enough force to lift the 6 kg object by increasing the torque.  Based on Chuck's Dynamic analysis at 2 Hz, you will need to be able to have (6kg * g + 142 N) * 15 cm = 30 N * m of torque.  Your typical motor will have way less than this.  But, your gearbox should be a 30:1 reducer to make your object oscillate at 2 Hz.  In this case, you only need 1 N * m.  A &lt;a href=&quot;http://www.automationdirect.com/static/specs/ironhorsesprs.pdf&quot; rel=&quot;nofollow&quot;&gt;reliable supplier of motors&lt;/a&gt; has a 1/3 hp motor that can handle this loading easily.  I'd also recommend investing in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Flywheel&quot; rel=&quot;nofollow&quot;&gt;large flywheel&lt;/a&gt; to ensure things run smoothly.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2015-12-22T19:26:03.710" LastActivityDate="2015-12-22T19:26:03.710" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6619" PostTypeId="2" ParentId="6612" CreationDate="2015-12-22T17:42:11.420" Score="3" Body="&lt;p&gt;You might look up NCAP and ASIL-related documentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in general,  all vehicles of a given class are required to meet safety standards at a given speed.   It's rather doubtful that any vehicle will be approved as a personal passenger vehicle if it cannot attain highway speeds;  similarly it's unlikely that anyone would bother to require crashworthiness at impact speeds of, say 250 km/h . (for one thing, it's essentially impossible to keep g-forces below lethal levels in the kind of stopping distances involved)&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2015-12-22T17:42:11.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6621" PostTypeId="1" AcceptedAnswerId="6631" CreationDate="2015-12-22T22:58:38.233" Score="3" ViewCount="111" Body="&lt;p&gt;Could someone please explain me why the velocity of point P and P' aren't the same? Appearently it is because the angular velocity of point A and B aren't the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GH4r6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GH4r6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are only given that the angular velocity of A is 10rad/s clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? I would think that, as these two elements are connected , they would all rotate with the same angular velocity. If A rotates with a speed of 10rad/s it would be logical for me that point B just follows and rotates with the same angular velocity.&lt;/p&gt;&#xA;" OwnerUserId="4218" LastActivityDate="2015-12-23T23:13:13.400" Title="Why do 2 points on a mechanism not have the same angular velocities?" Tags="|kinematics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6622" PostTypeId="1" AcceptedAnswerId="6638" CreationDate="2015-12-22T23:53:56.660" Score="8" ViewCount="1214" Body="&lt;p&gt;In &lt;a href=&quot;https://engineering.stackexchange.com/q/6613/3353&quot;&gt;this question&lt;/a&gt; I ask about how to perform a first-principle derivation of the torsional constant of a section. It appears that there is no such analytical derivation for torsional constant, so my question therefore becomes: what about FEM derivation for torsional constant of arbitrary shapes section? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I am not interesting in just using a FEM package without actually understanding the basic principle. I want to be able to derive the FEM formulation from first principles.&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2022-12-06T09:48:11.317" Title="How do I use FEM to derive the torsional constant of an arbitrary shape?" Tags="|modeling|finite-element-method|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6624" PostTypeId="1" AcceptedAnswerId="21876" CreationDate="2015-12-23T09:17:42.740" Score="10" ViewCount="662" Body="&lt;p&gt;I'd like to compute the response to a step function of a electrical/thermal system. Generally I can &quot;easily&quot; compute the transfer function $H$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H(\omega) = \frac{V_{out}(\omega)}{V_{in}(\omega)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Fourier transform ($\mathcal{F}$) of the Heaviside function is (computed with WA):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathcal{F}(\theta(t)) = V_{in}(\omega) = \sqrt{\frac{\pi}{2}}\delta(\omega)+\frac{i}{\sqrt{2\pi}\omega}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, noting $\mathcal{IF}$ the Inverse Fourier transform:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{out}(t) = \mathcal{IF} \left\{ \left( \sqrt{\frac{\pi}{2}}\delta(\omega)+\frac{i}{\sqrt{2\pi}\omega} \right) H(\omega) \right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check my math I tried to compute the response for a simple RC system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6H6lP.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6H6lP.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should get the well known charge of the capacitor. The transfer function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H(\omega) = \frac{1}{1+i\omega R C}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computing the Inverse Fourier transform ($\mathcal{IF}$) with WA ($R=C=1$) I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CKnbV.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CKnbV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be correct if we were going backward in time :/. So the question is... What am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did the same using Laplace Transforms and everything works fine... But I don't understand why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I don't want another method, I just want to understand what's wrong in my approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. the reason why I am using WA is that for my more complicated system I need to compute the Fourier transforms using WA.&lt;/p&gt;&#xA;" OwnerUserId="2437" LastEditorUserId="2437" LastEditDate="2015-12-23T18:01:28.360" LastActivityDate="2018-12-20T05:03:10.640" Title="Response of a system to a step function (Heaviside function)" Tags="|electrical-engineering|mathematics|signal|transfer-function|signal-processing|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6625" PostTypeId="1" AcceptedAnswerId="6651" CreationDate="2015-12-23T10:15:03.643" Score="5" ViewCount="625" Body="&lt;p&gt;I ve tried to solve this problem in so many ways but still didn't manage to do it...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y9G91.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y9G91m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the correct way to solve it please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This arm of this mechanism has a length of 0,2m. The piston has an angular velocity of 2000 tours/min. What would be the velocity of point D for an angle theta of 60 degrees?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that what I am missing is the angle formed by the arm and the line, which is 50mm long. Example like here (different &#xA;exercise):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l0kNH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l0kNHm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to look for this angle beta which could help me solve the problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;expected answer:2,88m/s&lt;/p&gt;&#xA;" OwnerUserId="4218" LastEditorDisplayName="user16" LastEditDate="2015-12-23T16:25:29.133" LastActivityDate="2015-12-26T00:38:08.930" Title="How to calculate the velocity of a piston-like linkage?" Tags="|mechanical-engineering|kinematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6626" PostTypeId="2" ParentId="6576" CreationDate="2015-12-23T14:03:14.763" Score="0" Body="&lt;p&gt;I think the &lt;a href=&quot;https://en.wikipedia.org/wiki/Deadband&quot; rel=&quot;nofollow&quot;&gt;Wikipedia Deadband article&lt;/a&gt; explains it fairly well.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Hysteresis Vs. Deadband&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A deadband (sometimes called a neutral zone or dead zone) is an interval of a signal domain or band where no action occurs (the system is 'dead' - i.e. the output is zero)  ...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Deadband is different from hysteresis. With hysteresis there is no&#xA;  dead zone, and so the output is always in one direction or&#xA;  another. Devices with hysteresis have memory, in&#xA;  that previous system states dictate future states. Examples of devices with hysteresis are single-mode&#xA;  thermostats and smoke alarms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Examples of deadband I've come across:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Air handling unit damper actuator. Deadband prevented actuator moving until output changed by, say, 5%. If damper is at 32% then it won't move until the setpoint goes to 27 or 37% and then it will move to that position. This is to prevent wear and tear on the mechanisms.&lt;/li&gt;&#xA;&lt;li&gt;Data collection where data points are logged 'on-change'. For digital I/O this is straight-forward. For an analog reading which is constantly changing (think two or three decimal places on a temperature) setting a deadband of, say, 0.5 degrees dramatically reduces the amount of data-logging (but at the expense of fine-grained trending).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hysteresis examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The room thermostat is the simplest and most well known.&lt;/li&gt;&#xA;&lt;li&gt;A two level tank fill system is another. The pump is switched on when the low level sensor detects tank empty. Pump turns off when the high-level sensor detects tank full. The hysteresis is the difference in level between the two switches.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2015-12-23T22:03:49.300" LastActivityDate="2015-12-23T22:03:49.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6627" PostTypeId="2" ParentId="6558" CreationDate="2015-12-23T15:10:51.137" Score="0" Body="&lt;p&gt;Whenever a fluid flows in a pipe, it flows due to difference in pressure. Due to this difference in pressure, there is a pressure gradient which is $\dfrac{\partial p}{\partial s}$. $\dfrac{\partial p}{\partial s}\ \text{d}s$ is the change in pressure over the length $\text{d}s$. $p+\dfrac{\partial p}{\partial s}\ \text{d}s$ is the pressure at the exit in terms of pressure function $p$ at inlet + change in pressure over length $\text{d}s$.&#xA;So, $\left(p+\dfrac{\partial p}{\partial s}*ds\right)dA$ is also the force like $p\cdot \text{d}A$ with pressure $p$ at the exit equals $p+\dfrac{\partial p}{\partial s}\ \text{d}s$.Pressure variation in fluid is given by Tailor's series expansion with respect to length $s$ about $p$. $p(s + \Delta s) = p(s) + \dfrac{\partial p}{\partial s}\ \Delta s + O(\Delta s2)$. $O(\Delta s2)$ is the higher order term which is neglected.&lt;/p&gt;&#xA;" OwnerUserId="4224" LastEditorUserId="1832" LastEditDate="2015-12-23T17:11:57.997" LastActivityDate="2015-12-23T17:11:57.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6628" PostTypeId="2" ParentId="6215" CreationDate="2015-12-23T21:26:28.157" Score="4" Body="&lt;p&gt;Whenever you choose to design standard gears, you typically have two main parameters: the number of teeth $N$ and the module, or modulus $m$. The module can generally be interpreted as the distance between the gear teeth. In fact, the module is defined as the circular pitch $p$ (the circumferential distance between the same point on two neighbouring teeth) divided by pi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the parameter 'module' is popular is because it allows you to work with rational numbers. You can calculate the gear's pitch diameter from the circular pitch $d=Np/\pi$, but this means either the pitch diameter or the circular pitch is an irrational number. It is not useful or practical to have either a standard set of values for gears (circular pitch) or the distance between gear shafts (from pitch diameters) be based on irrational numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead, the module is preferred, and used in sets of standardised values (i.e. you are more likely find off the shelf gears of module like 4mm or 5mm than something like 4.123mm). You can calculate pitch diameter from the module via $d = Nm$ and all the variables can be rational numbers. For example, a gear of 16 teeth and module of 2.5mm turns out to have a pitch diameter of 40mm. All nice round numbers, easy to work with without having to bring out a calculator too often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because module can be interpreted as the distance between gear teeth, it is also important to note that only gears with the same module can mesh together. Also, in general, gears with a large module can withstand higher bending stresses at the root of teeth.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2015-12-24T10:08:22.513" LastActivityDate="2015-12-24T10:08:22.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6629" PostTypeId="2" ParentId="6625" CreationDate="2015-12-23T22:07:48.180" Score="4" Body="&lt;p&gt;The angle $\beta$ can be obtained from the geometry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0.2*1000 \sin (\beta )+50 \sin (60 {}^{\circ})=150 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives $\beta = 0.5627 rad$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the angular velocity of the arm relative to B is $w$, the velocity of the point D can be computed as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_D=\frac{2000}{60} (2 \pi ) \{0,0,1\}\times \frac{50}{1000} \{-\sin (60&#xA;   {}^{\circ}),\cos (60 {}^{\circ}),0\}+\{0,0,w\}\times 0.2 \{-\sin (\beta&#xA;   ),-\cos (\beta ),0\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_D= \{0.169161 w-5.23599,-0.106699 w-9.069,0.\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $x$ component of the velocity at D is zero, so setting the first element to zero we get $w = 30.9527 $. Substitute this in the second element to get the vertical velocity as $-12.3716$. It is moving downwards at 12.3716 m/s.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2015-12-23T22:07:48.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6630" PostTypeId="2" ParentId="6576" CreationDate="2015-12-23T22:55:38.310" Score="5" Body="&lt;p&gt;Set point and trigger point are two terms that I am most familiar with respect to temperature controls. My experiences is limited to &lt;a href=&quot;http://thermotron.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thermotron&lt;/a&gt; and &lt;a href=&quot;http://www.espec.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ESPEC&lt;/a&gt; environmental temperature chambers. I believe searching online resources section of both these organization's might lead to better focused definitions. Below is short list of terms defined in a Watlow controller.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ov9dv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ov9dv.jpg&quot; alt=&quot;Temperature Controller terms&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;In addition Watlow PID Controller user guide defines the following.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Positive Dead band&lt;/li&gt;&#xA;&lt;li&gt;Zero Dead band&lt;/li&gt;&#xA;&lt;li&gt;Negative Dead band&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Below are excerpts from the user guide.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RdjcA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RdjcA.jpg&quot; alt=&quot;Positive Dead band&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2BfMs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2BfMs.jpg&quot; alt=&quot;Zero Dead band&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kIzvb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kIzvb.jpg&quot; alt=&quot;Negative Dead band&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;On-Off Controller&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k5aIw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k5aIw.jpg&quot; alt=&quot;On-Off Controller&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;There is more interesting infomation with regard the on/off temperatures in the User Guide.&lt;/p&gt;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0ahUKEwjM2JyYifPJAhVMFj4KHR7dBpYQFggnMAA&amp;amp;url=https%3A%2F%2Fwww.watlow.com%2Fdownloads%2Fen%2Fmanuals%2Fpmpid.pdf&amp;amp;usg=AFQjCNHZMcbg6j9z6d1xjOZxF_pfX6hycQ&amp;amp;sig2=WI6OksHIXt84X3aLL714Lw&quot; rel=&quot;nofollow noreferrer&quot;&gt;User's Guide EZ-ZONE® PM PID Controller Models&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.omega.com/prodinfo/temperaturecontrollers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;On/Off Control&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.omega.com/temperature/Z/pdf/z110-114.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Introduction To Temperature Controllers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-12-24T07:28:09.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6631" PostTypeId="2" ParentId="6621" CreationDate="2015-12-23T23:13:13.400" Score="3" Body="&lt;p&gt;The velocity of P' is perpendicular to AD, and the velocity of P is perpendicular to BP. So clearly the directions are not the same. Thus there is relative velocity between P and P' as shown in the diagram. Now, if gyroscopes are mounted at D, P', or any point on AD, except A, they will all measure the same angular velocity. A gyroscope mounted at P will measure the same value as one mounted anywhere between B and P, excluding B.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2015-12-23T23:13:13.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6632" PostTypeId="2" ParentId="6547" CreationDate="2015-12-24T08:02:24.087" Score="0" Body="&lt;p&gt;&amp;quot;Will run into problems&amp;quot; is a loaded question. If you properly design logic you should not encounter problems. But there are various types of temperature controllers in market that can help you avoid problems. Although not directly applicable, I am mostly familiar with &lt;a href=&quot;http://www.watlow.com/products/controllers/index.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Watlow temperature controllers&lt;/a&gt; and some insights into &lt;a href=&quot;http://www.omega.com/prodinfo/temperaturecontrollers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;omega temperature controllers&lt;/a&gt; which are used in environmental chambers which perform similar functions. I suggest taking a look at these type of controllers because they offer out of the box solutions. Each vendor has different functions, features, and definitions but at a high level the following are the most common:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;On-Off controllers&lt;/li&gt;&#xA;&lt;li&gt;Proportional Controllers&lt;/li&gt;&#xA;&lt;li&gt;Proportional plus integral controllers&lt;/li&gt;&#xA;&lt;li&gt;Proportional plus integral plus derivative (PID) controllers&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Bellow are some excerpts from Watlow user guide that might be of interest&lt;/p&gt;&#xA;&lt;h3&gt;On-Off controllers&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O0k0S.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O0k0S.jpg&quot; alt=&quot;On-Off Controller&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Proportional Controllers&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wZJVh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wZJVh.jpg&quot; alt=&quot;Proportional Controllers&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Proportional plus integral (PI) controllers&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sn8jI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sn8jI.jpg&quot; alt=&quot;Proportional plus integral controllers&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Proportional plus integral plus derivative (PID) controllers&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vDq9O.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vDq9O.jpg&quot; alt=&quot;Proportional plus integral plus derivative controllers&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Using functions and features available in these controller you will able avoid common problems.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0ahUKEwjM2JyYifPJAhVMFj4KHR7dBpYQFggnMAA&amp;amp;url=https%3A%2F%2Fwww.watlow.com%2Fdownloads%2Fen%2Fmanuals%2Fpmpid.pdf&amp;amp;usg=AFQjCNHZMcbg6j9z6d1xjOZxF_pfX6hycQ&amp;amp;sig2=WI6OksHIXt84X3aLL714Lw&quot; rel=&quot;nofollow noreferrer&quot;&gt;User's Guide EZ-ZONE® PM PID Controller Models&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.omega.com/temperature/Z/pdf/z110-114.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Introduction To Temperature Controllers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/6630/110&quot;&gt;What do you call the difference between the on and off temperatures in a simple thermostat?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2015-12-28T20:42:28.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6635" PostTypeId="5" CreationDate="2015-12-24T15:50:46.337" Score="0" Body="&lt;p&gt;The Biomedical engineering tag covers the design concepts to medicine and biology for healthcare purposes.It combines the design and problem solving skills of engineering with medical and biological sciences to advance health care treatment, including diagnosis, monitoring, and therapy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use this tag when someone who has trained as a biomedical engineer is likely to be able to answer/ask your question. As with most disciplines, biomedical engineering encompasses a lot of specialisms and it may be necessary to further sub-categorise your question using additional tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Biomedical Engineering Sub-disciplines&lt;/strong&gt;                                            &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biomedical engineering is a broad field.If your question is specific to any of the following disciplines, consider using the more specific tag instead of or in addition to this one:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Biosystems engineering&lt;/li&gt;&#xA;&lt;li&gt;Biotecnology&lt;/li&gt;&#xA;&lt;li&gt;Biotechnical engineering&lt;/li&gt;&#xA;&lt;li&gt;Biomolecular engineering&lt;/li&gt;&#xA;&lt;li&gt;Bioresource engineering&lt;/li&gt;&#xA;&lt;li&gt;Bioprocess engineering&lt;/li&gt;&#xA;&lt;li&gt;Cellular engineering&lt;/li&gt;&#xA;&lt;li&gt;Genetic engineering&lt;/li&gt;&#xA;&lt;li&gt;Health and Safety engineering&lt;/li&gt;&#xA;&lt;li&gt;Food and Biological Process Engineering&lt;/li&gt;&#xA;&lt;li&gt;Biochemical engineering&lt;/li&gt;&#xA;&lt;li&gt;Clinical engineering&lt;/li&gt;&#xA;&lt;li&gt;Medical device / Medical equipment/ Health technology&lt;/li&gt;&#xA;&lt;li&gt;Micro and Nanotechnology&lt;/li&gt;&#xA;&lt;li&gt;Neural engineering&lt;/li&gt;&#xA;&lt;li&gt;Pharmaceutical engineering&lt;/li&gt;&#xA;&lt;li&gt;Rehabilitation engineering&lt;/li&gt;&#xA;&lt;li&gt;Systems Physiology&lt;/li&gt;&#xA;&lt;li&gt;Tissue engineering&lt;/li&gt;&#xA;&lt;li&gt;Electrical engineering&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Biomedical Engineering and other fields&lt;/strong&gt;                                        &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This field seeks to close the gap between engineering and medicine.And for this biomedical engineering covers a vast area of knowledge, it overlaps with many basic science and engineering disciplines with varying amounts.These are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Physics&lt;/li&gt;&#xA;&lt;li&gt;Chemistry&lt;/li&gt;&#xA;&lt;li&gt;Materials Science&lt;/li&gt;&#xA;&lt;li&gt;Electrical and Electronics Engineering&lt;/li&gt;&#xA;&lt;li&gt;Mechanical engineering&lt;/li&gt;&#xA;&lt;li&gt;Systems engineering&lt;/li&gt;&#xA;&lt;li&gt;Nanoengineering&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4231" LastEditorUserId="4231" LastEditDate="2015-12-24T22:38:53.687" LastActivityDate="2015-12-24T22:38:53.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6636" PostTypeId="4" CreationDate="2015-12-24T15:50:46.337" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2015-12-24T15:50:46.337" LastActivityDate="2015-12-24T15:50:46.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6638" PostTypeId="2" ParentId="6622" CreationDate="2015-12-24T18:32:08.213" Score="4" Body="&lt;p&gt;You can find an implementation of a finite element used in computation of arbitrary shape section torsional constant here:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://projects.ce.berkeley.edu/feap/elmt11.f&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://projects.ce.berkeley.edu/feap/elmt11.f&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I don't see any reference in the implementation other than a comment in the source code: &amp;quot;St. Venant torsion problem solution by warping and Prandtl stress function.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;I don't see any reference on their website either, but hope the source would give you an idea.&#xA;&lt;a href=&quot;http://www.ce.berkeley.edu/projects/feap/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ce.berkeley.edu/projects/feap/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4234" LastEditorUserId="36066" LastEditDate="2022-12-06T08:22:56.190" LastActivityDate="2022-12-06T08:22:56.190" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6639" PostTypeId="1" AcceptedAnswerId="6730" CreationDate="2015-12-24T21:48:10.713" Score="1" ViewCount="577" Body="&lt;p&gt;I am in the process of selecting a linear rail system for a 5ft x 10ft cnc laser cutter. &lt;a href=&quot;http://openbuildspartstore.com/delrin-mini-v-wheel-kit/&quot; rel=&quot;nofollow&quot;&gt;Delrin polymer rollers&lt;/a&gt; on 2 inch square extruded aluminum tube is an economically attractive solution. I have had success with them on a 3d printer, but I am not sure how well they will handle higher speeds (2m/s) and 12hr per day operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The environment is reasonably dust free, but not a clean room. I suspect the erosion of the Delrin over time may cause tolerance issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some successful implementations of delrin or other plastic rollers in an commercial/industrial setting? Will the plastic rollers typically outlive the bearings?&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-01T23:59:28.313" Title="What is the longevity of Delrin rollers on aluminum linear rail?" Tags="|mechanical-engineering|failure-analysis|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6642" PostTypeId="1" AcceptedAnswerId="6648" CreationDate="2015-12-25T09:07:29.317" Score="2" ViewCount="94" Body="&lt;p&gt;I would like to design a long, uniform magnetic field using Hemholtz coils but am not sure how far I can separate them before the field lines diverge and become two loops that go around each coil rather than one which goes through both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have formulae for finding the magnetic flux at different points but I am not sure how to use them to determine this distance I am trying to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have experience in this area who could offer some advice?&lt;/p&gt;&#xA;" OwnerUserId="4189" LastEditorUserId="282" LastEditDate="2015-12-28T01:54:45.510" LastActivityDate="2015-12-28T01:54:45.510" Title="How to calculate the maximum spacing between Hemholtz coils?" Tags="|mechanical-engineering|electrical-engineering|magnets|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6645" PostTypeId="1" CreationDate="2015-12-25T15:14:02.213" Score="4" ViewCount="193" Body="&lt;p&gt;Some cell phone cases have a waterproof rating of IP-68, often specifying that they can tolerate immersion to 6 feet for 30 minutes or 3 feet for 60 minutes.  My understanding, however, is that a good seal (like an o-ring) would fail near a certain water pressure without much time dependence.  So, why the strange time dependence?  This makes me wonder about diffusion of water across a poor seal being a factor, but I am guessing instead that these numbers are simply conservative times used in final manufacturer QC checks.  Anyway, is there a theoretical predictive model to determine the expected &quot;waterproof time vs depth&quot; scaling?&lt;/p&gt;&#xA;" OwnerUserId="2132" LastActivityDate="2016-08-22T22:12:12.137" Title="Waterproof Case Time Limits" Tags="|waterproofing|diffusion|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6646" PostTypeId="1" CreationDate="2015-12-25T16:57:26.453" Score="1" ViewCount="1978" Body="&lt;p&gt;After solving my problem in APDL, and started doing some post-processing, I want to find how much the deformed shape has rotated (i.e. the maximum angle the furthest node has made comparing with the original unreformed shape), I'm doing 2D analysis, is this exactly what is shown as, z-component of rotation, or there is other thing I need to know?  and in which units, deg or rad? &lt;/p&gt;&#xA;" OwnerUserId="4245" LastActivityDate="2022-11-21T20:32:57.213" Title="How to find the angle of deformation in Ansys APDL?" Tags="|mechanical-engineering|modeling|finite-element-method|ansys|structural-analysis|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6648" PostTypeId="2" ParentId="6642" CreationDate="2015-12-25T22:13:50.813" Score="2" Body="&lt;p&gt;Short answer: Not very far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had this question one or two years ago. At that time I did a simple calculation of $B$ along the symmetry line for Helmholtz coils with non-standard distance, i.e. the position of the two coils being different from $+R/2$ and $-R/2$ ($R$ being the radius of the two coils). The homogeneity of $B$ decreases quite quickly when deviating from the standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia gives you the equation. For a single conductor loop, according to Biot-Savart,&#xA;$$&#xA;B(x)=\frac{\mu_0 I}{2}\cdot \frac{R^2}{(R^2+x^2)^{\frac 32}}&#xA;$$&#xA;with $B$ being the magnetic flux density (more precisely its vector component along the symmetry axis of the loop), $\mu_0$ the vacuum permeability, $I$ the current along the conductor loop, $R$ the radius of the conductor loop and $x$ the distance to the center of the loop along the symmetry axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following plot shows $B(x-d)+B(x+d)$ for some $d \geq R/2$ for $\frac{\mu_0 I}{2} = R = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PjU4W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PjU4W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, $B$ drops significantly in the center between the two coils once deviating more than 20% from $d=R/2$. You can calculate for yourself how much deviation you wish to accept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be reminded that this is just an evaluation of the magnetic flux density along the central symmetry line of a coil pair. If you are interested in the full 2D field, things become more complicated. But this simple evaluation gives you some idea what happens when you move the two coils apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a longer homogeneous field you need more coils.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="4122" LastEditDate="2015-12-25T22:19:09.377" LastActivityDate="2015-12-25T22:19:09.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6651" PostTypeId="2" ParentId="6625" CreationDate="2015-12-26T00:38:08.930" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;First you build a kinematic chain between &lt;em&gt;A&lt;/em&gt;, &lt;em&gt;B&lt;/em&gt; and &lt;em&gt;D&lt;/em&gt; with radius $r$ between &lt;em&gt;AB&lt;/em&gt; and the length $\ell$ between &lt;em&gt;BD&lt;/em&gt;. The orientation angles from vertical are $\theta$ and $\phi$ respectively.&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$(x_B,y_B) = (x_A,y_A) + (-r \sin\theta,r \cos \theta)$&lt;/li&gt;&#xA;&lt;li&gt;$(x_D,y_D) = (x_B,y_B) + (-\ell \sin\phi, - \ell \cos \phi)$&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;You find the angle $\phi$ from the constraint that $x_A-x_D = r \sin\theta + \ell \sin \phi = x_{AD}$ $$\sin \phi = \frac{x_{AD}}{\ell}-\frac{r}{\ell} \sin \theta $$&lt;/li&gt;&#xA;&lt;li&gt;Next you differentiate with respect to time using the chain rule to get the velocities &#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$(\dot{x}_B,\dot{y}_B) = (-r \dot{\theta} \cos\theta, -r \dot{\theta}\sin\theta)$&lt;/li&gt;&#xA;&lt;li&gt;$(\dot{x}_D,\dot{y}_D) = (\dot{x}_B,\dot{y}_B) + (-\ell \dot{\phi} \cos\phi, \ell \dot{\phi} \sin \phi)$&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;You find the rotational velocity of the connecting rod from the constraint $\dot{x}_D=-r\dot{\theta}\cos\theta -\ell \dot{\phi} \cos \phi=0$ $$ \dot{\phi} = - \frac{r \dot{\theta} \cos\theta}{\ell \cos{\phi}}$$&lt;/li&gt;&#xA;&lt;li&gt;The collar speed is&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$ v = \dot{y}_D = \ell \dot{\phi} \sin \phi - r \dot{\theta} \sin\theta = -\frac{r \sin(\theta+\phi)}{\cos\phi} \dot{\theta} $$ where $\dot{\theta}$ is the rotation of the disk in radians per second.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2015-12-26T00:38:08.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6653" PostTypeId="1" CreationDate="2015-12-26T05:54:38.963" Score="0" ViewCount="3116" Body="&lt;p&gt;I am need to match &lt;a href=&quot;https://www.matbase.com/material-categories/metals/ferrous-metals/steel/low-alloy/low-carbon/material-properties-c15-low-carbon-steel.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;DIN C15 steel&lt;/a&gt; with an Indian steel but I don't know the equivalent grade in the Indian standard. For example, DIN C40 steel is known as EN8 under the IS standard; similarly, what is the IS grade for C15 steel?&lt;/p&gt;&#xA;" OwnerUserId="3404" LastEditorUserId="368" LastEditDate="2016-12-28T00:19:45.183" LastActivityDate="2016-12-28T00:27:28.213" Title="C15 Equivalent material in IS standard" Tags="|mechanical-engineering|steel|standards|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6654" PostTypeId="1" CreationDate="2015-12-26T08:18:23.797" Score="1" ViewCount="54581" Body="&lt;p&gt;They differ in frequency of about 6000 Mhz. What are the practical differences of these devices. What are the ups and downs vs the other?&lt;/p&gt;&#xA;" OwnerUserId="4251" LastEditorDisplayName="user16" LastEditDate="2015-12-26T15:24:10.650" LastActivityDate="2020-01-06T22:41:51.480" Title="What are the key differences between the X-Band and the S-Band radars?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6655" PostTypeId="2" ParentId="6654" CreationDate="2015-12-26T11:08:26.900" Score="4" Body="&lt;p&gt;Different wavelengths of electromagnetic radiation have different penetrating and reflection properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wavelength produced by S Band radars is 8 to 15 cm (2 to 4 GHz) and the wavelength produced by X Band &lt;a href=&quot;http://www.everythingweather.com/weather-radar/bands.shtml&quot; rel=&quot;noreferrer&quot;&gt;radars&lt;/a&gt; is 2.5 to 4 cm (8 to 12 GHz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the smaller wavelengths produced by X Band radars they use small antennas. Such antennas can easily fit onto smaller boat, consequently X Band radar is used for marine applications.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Because of the smaller wavelength, the X Band radar is more sensitive and can detect smaller particles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Meteorological X Band radars are used to study &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;cloud development because they can detect the tiny water particles and also used to detect light precipitation such as snow. X band radars also attenuate very easily, so they are used for only very short range weather observation&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Most major airplanes are equipped with an X band radar to pick up turbulence and other weather phenomenon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The radio waves produced by S Band radars are not easily attenuated, this gives them greater penetrating power and they are used for &quot;near and far weather observations&quot;. Their penetration capabilities allow them to see through heavy weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of S Band antennas can exceed 7.6 m (25 ft) in size.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-12-26T11:08:26.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6656" PostTypeId="2" ParentId="6645" CreationDate="2015-12-26T12:17:16.027" Score="1" Body="&lt;p&gt;As the comments suggest, it's generally something the manufacturer is comfortable standing behind &amp;amp; waranteeing.  While you're correct that a seal &lt;em&gt;should&lt;/em&gt; either hold or breach,  in practice a &quot;soft&quot; seal such as an o-ring can be subject to creep under pressure.  Given enough time, that deformation will lead to a leak.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW, my advice would be only to wear cheap watches you don't mind losing if you need to keep a watch on while diving; if you're a serious scuba fellow then buy a watch (expensive) rated as immersible at depth indefinitely.&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="368" LastEditDate="2016-02-24T19:33:23.943" LastActivityDate="2016-02-24T19:33:23.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6659" PostTypeId="1" AcceptedAnswerId="6673" CreationDate="2015-12-27T01:57:30.460" Score="4" ViewCount="154" Body="&lt;p&gt;I would like to essentially flatten or smooth the surface of a viscous liquid after dropping an object into it, without actually touching the liquid. For example, if I drop a rock into a tub of oil, I need the oil to move into the void and return to having a smooth surface as fast as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that I need to create some sort of force that acts against the liquid surface from above, or a force from the sides that acts as a kind of 'wiper' to smooth it out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked into using ferrofluids and electromagnets on the sides or bottom of the liquid vat/container, however many demonstrations I have seen result in a 'spiky' surfaces when magnets are brought near the liquid due to non-uniform field lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also looked into trying to suspend diamagnetic nanomaterials in the liquid which could be repelled by a magnetic field from above the vat, however I'm not sure about whether diamagnetic materials exist which would could be used like ferrofluids in the sense of staying suspended in the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My job depends on being able to do this, so if anyone has any ideas I'd really appreciate hearing them.&lt;/p&gt;&#xA;" OwnerUserId="4189" LastEditorUserId="282" LastEditDate="2015-12-28T01:55:08.050" LastActivityDate="2015-12-28T14:55:33.230" Title="How to 'smooth/flatten' the surface of a disturbed liquid without contacting it?" Tags="|fluid-mechanics|magnets|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6661" PostTypeId="1" CreationDate="2015-12-27T03:16:06.903" Score="4" ViewCount="395" Body="&lt;p&gt;I am doing a project on a generator. Instead of using an alternator I am using a dynamo. I then amplify the voltage produced using a DC to DC voltage booster. However, can I get a higher AC voltage by connecting it directly to a DC to AC inverter? I am still green in this topic, I hope you guys could explain it clearly and perhaps you could give some helpful resources.&lt;/p&gt;&#xA;" OwnerUserId="4256" LastEditorUserId="1832" LastEditDate="2015-12-27T04:43:37.237" LastActivityDate="2018-07-09T17:14:22.303" Title="Can a DC voltage booster be connected directly to a DC to AC inverter?" Tags="|electrical-engineering|power-electronics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6663" PostTypeId="2" ParentId="6661" CreationDate="2015-12-27T06:08:22.517" Score="1" Body="&lt;p&gt;The short answer is yes a DC to DC booster can be connected to DC - AC power inverter. But the key is to define the electrical parameters for your project, and then research for components at will meet you design specification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters there are wide ranged of DC to AC power inverter in the market. Below are two such power inverter example with different VDC inputs and VAC outputs.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;12VDC to 110VAC power inverter used in North America automotive aftermarket&lt;/li&gt;&#xA;&lt;li&gt;415-600 VDC to 480 VAC power inverter used in renewable energy industry&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;12VDC to 110VAC Power Inverter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f8id3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f8id3.jpg&quot; alt=&quot;Automotive Power Inverters&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J2dDg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J2dDg.jpg&quot; alt=&quot;Automotive Power Inverter Specification&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;415-600 VDC to 480 VAC power inverter&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tuuHK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tuuHK.jpg&quot; alt=&quot;Renewable Energy Power Inverter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wXQWN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wXQWN.jpg&quot; alt=&quot;Renewable Energy Power Inverter specifications&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your end goal the generator will have to boost the DC voltage to match the power inverter requirements. Other factors to consider wattage of the power inverter, and efficiency to name a few.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this will get you started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Referances:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.smps.us/power-inverter.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Your guide to DC to AC POWER INVERTERS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.princetonpower.com/products/battery-integrated-inverter-3-port.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battery Integrated Inverter (3 Port) — BIGI-250 from Princeton Power Systems&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://accessories.us.dell.com/sna/productdetail.aspx?c=us&amp;amp;l=en&amp;amp;s=&amp;amp;cs=04&amp;amp;sku=A0145295&amp;amp;dgc=ST&amp;amp;cid=293344&amp;amp;lid=5616479&amp;amp;acd=12309152537461010&amp;amp;ven1=sE1inYhPj&amp;amp;ven2=,&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ultra Compact 150 Watt Power Inverter/Charger&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2015-12-27T12:07:57.017" LastActivityDate="2015-12-27T12:07:57.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6664" PostTypeId="1" CreationDate="2015-12-27T06:14:27.823" Score="1" ViewCount="189" Body="&lt;p&gt;I think there is a mechanism that switches off the transmitter in order to turn on the receiver in the antenna. Do you know the technical terms by which this is done?&lt;/p&gt;&#xA;" OwnerUserId="4251" LastEditorUserId="33" LastEditDate="2015-12-27T13:49:32.147" LastActivityDate="2016-01-04T12:25:48.290" Title="What mechanism in the radar that enables the switch between transmission and reception?" Tags="|electrical-engineering|marine-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6667" PostTypeId="1" AcceptedAnswerId="6736" CreationDate="2015-12-27T15:29:02.230" Score="4" ViewCount="2683" Body="&lt;p&gt;I am familiar with the basic concepts of fits and tolerances, and I am aware of the various online calculators for determining them. However, I don't understand how to apply those numbers to my actual CAD design for the parts that I want to fit together. None of the calculators seem to account for the actual manufacturing tolerances of the shop I'm working with (in this case +/- 0.1 mm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to be more specific, let's say I have a shaft and hole and I want to work with a nominal size of 6 mm. I want a &quot;locational clearance&quot; fit between them. The calculators tell me the maximum and minimum hole sizes, etc., but I don't know how to gather from this information what I should make the nominal sizes of the hole and shaft in my CAD model. And how can a calculator tell me that anyway, if it doesn't know what size my manufacturing tolerance is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody elucidate?&lt;/p&gt;&#xA;" OwnerUserId="4264" LastEditorUserId="1832" LastEditDate="2015-12-27T15:47:02.067" LastActivityDate="2016-01-02T10:01:09.650" Title="How do you apply fits and tolerances when designing hole and shaft?" Tags="|mechanical-engineering|machining|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6670" PostTypeId="1" AcceptedAnswerId="6686" CreationDate="2015-12-28T00:20:40.777" Score="7" ViewCount="1821" Body="&lt;p&gt;I noticed many kitchen utensils or stand mixer attachments are made with injected aluminum. Aluminum (~\$20/kg) is usually more expensive than stainless steel (~\$10/kg) and is not compatible with dishwashers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I have a peeler made with anodized aluminum and a KitchenAid attachment with some aluminum parts that I cannot put in my dishwasher. I don't understand why engineers are still making such utensils in this material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good reason for this?&lt;/p&gt;&#xA;" OwnerUserId="2070" LastEditorUserId="33" LastEditDate="2016-01-05T02:07:38.870" LastActivityDate="2017-05-05T12:07:23.853" Title="Why do manufacturers still make aluminum kitchen utensils?" Tags="|steel|aluminum|" AnswerCount="7" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6671" PostTypeId="2" ParentId="6670" CreationDate="2015-12-28T02:18:51.147" Score="3" Body="&lt;p&gt;I'm guessing that shipping cost could be a reason. For a pallet (or several) of utensils, shaving a few percent weight off of each one could amount to a significant savings in shipping cost, especially if the trade route is between China and Europe or North/South America.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2015-12-28T02:18:51.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6672" PostTypeId="2" ParentId="6670" CreationDate="2015-12-28T02:37:32.950" Score="4" Body="&lt;p&gt;Re-tooling and re-equipping a factory to use a different metal can be very expensive. If the required rate of return from re-tooling/re-equipping cannot be achieved be it simply isn't done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The production rate from some utensil manufacturers may be low and the profit margins small resulting in no major new investing in changing to other metals and re-equipping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineering is more than just science, design, materials and efficiencies, economics and profit also plays a significant role.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2015-12-28T02:37:32.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6673" PostTypeId="2" ParentId="6659" CreationDate="2015-12-28T04:25:46.717" Score="4" Body="&lt;p&gt;Some quick dimensional analysis for a Newtonian fluid suggests that the settling time scales as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = \sqrt{\frac{L}{g}} f\left(\frac{\nu}{\sqrt{g L^3}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\tau$ is the time it takes for the liquid surface to settle, $L$ is some characteristic length of the dropped object (say, the diameter of a ball), $\nu$ is the kinematic viscosity, and $g$ is the acceleration due to gravity. I expect the function $f$ to increase as the parameter inside it increases, which makes sense. The higher the viscosity, the slower the settling, and the stronger the gravity, the faster the settling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, influencing these parameters in some way could help. Unfortunately, you have no control over these parameters aside from the viscosity, which you might influence via heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An oscillation applied to the container could help smooth things out. The speed by which this occurs surely is strongly dependent on the viscosity for a Newtonian fluid. For a non-Newtonian fluid (say, peanut butter), I'm not sure this will necessarily work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also imagine an air jet applied parallel to the surface could help, depending on the viscosity, especially if its direction is varied. Targeted jets striking at an angle to the surface might help too, but this could require a complicated control system or manual human intervention to do the targeting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suction applied directly above the hole may help, again, depending on the viscosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(After writing this, I noticed Chuck said some similar things in the comments. He also recommended a centrifuge, which I didn't think of.)&lt;/p&gt;&#xA;" OwnerUserId="49" LastEditorUserId="49" LastEditDate="2015-12-28T14:55:33.230" LastActivityDate="2015-12-28T14:55:33.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6674" PostTypeId="1" CreationDate="2015-12-28T08:54:28.677" Score="3" ViewCount="561" Body="&lt;p&gt;I would like to make/obtain a micro water dripping pump, as a component of an item I am designing, to irrigate a plant with a water drop every few seconds in a closed loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &quot;micro&quot; I mean few cm³ in size. I imagine this should not be bigger than a walnut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I insist on a closed loop, meaning the system isn't connected to a water input like water tap or such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What technology is available? I would prefer a system with no or little electrical input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would such pump come ready-made or would I have to make it myself from scratch?&lt;/p&gt;&#xA;" OwnerUserId="4271" LastEditorUserId="4271" LastEditDate="2015-12-28T09:07:01.357" LastActivityDate="2024-01-09T04:18:56.787" Title="How to make a micro water dripping pump with little or no electrical input?" Tags="|pumps|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6676" PostTypeId="1" CreationDate="2015-12-28T09:37:19.640" Score="1" ViewCount="885" Body="&lt;p&gt;What is the generic name of the electronic hand-held device that measures the relative height of two different points (altitude, etc.)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, suppose I'm climbing a tree. I turn on the device, put it in my pocket, and once I've climbed 15 feet from the ground, the device should indicate &quot;15 feet&quot; on its display. When I climb further up to 20 feet, the device should indicate &quot;20 feet&quot; on its display. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used to measure different levels using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dumpy_level&quot; rel=&quot;nofollow&quot;&gt;Dumpy level&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Total_station&quot; rel=&quot;nofollow&quot;&gt;total station&lt;/a&gt; &quot;distant measurer-laser.&quot; I need a different kind of device that can take levels more quickly but I don't know what this device is called or how to select a particular model. In addition to the name of the device, what properties of the device should I be looking at in order to take measurements more quickly?&lt;/p&gt;&#xA;" OwnerUserId="4273" LastEditorUserId="33" LastEditDate="2015-12-30T23:41:07.380" LastActivityDate="2015-12-30T23:41:07.380" Title="What is the name for a digital height measuring device?" Tags="|civil-engineering|metrology|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6677" PostTypeId="2" ParentId="6674" CreationDate="2015-12-28T13:52:21.523" Score="5" Body="&lt;p&gt;Despite the terminology differences I believe this is what you are looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.treegator.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.treegator.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ready-made pump is called gravity. The pressure of gravity pushes down on the closed bladder and exerts light pressure to the water that is higher than the pressure outside the pores. This small pressure differential pushes the water in the bladder through the pores at a slow rate to provide drip irrigation over an extended period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This meets all of your requirements listed.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="368" LastEditDate="2015-12-28T17:14:23.473" LastActivityDate="2015-12-28T17:14:23.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6679" PostTypeId="2" ParentId="6676" CreationDate="2015-12-28T14:12:19.797" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/6676/a-height-measure-digital-device-or-leveler#comment12153_6676&quot;&gt;MWijnand is correct&lt;/a&gt; that what you are looking for is an altimeter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you will not find anything that could provide you with accuracy anywhere near the scale of a correctly used total station. From my experience you would be lucky to find something which would be able to find precision down to the 5' of elevation and accuracy with this device would take more time than setting up a total station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To provide accurate differential altitude measurements you would need two of the same altimeters, both calibrated and showing exact altitude while placed beside each other. One would remain on the bottom of your measurement while the other would travel to the top. Once reaching the height you would like to measure they should remain still for 5-10 minutes to normalize followed by a reading. With my work we typically take 3 cycles of readings and average (or throw out a really bad reading). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for needing two altimeters is because they measure altitude based on barimetric pressure. The pressure they are reading would change any time the weather also changes (wind changed direction to a degree would cause a flux in the pressure across the sensor).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Simplest solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Continue to use the total station or carry a heavy plumb bob up the tree with you.&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2015-12-29T00:37:17.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6682" PostTypeId="1" AcceptedAnswerId="6684" CreationDate="2015-12-28T18:59:59.827" Score="7" ViewCount="23131" Body="&lt;p&gt;I have a pump that I want to use to pump a solvent. It is hard black plastic so I'm pretty confident it is either ABS or polypropylene (it wasn't expensive enough to be some more exotic plastic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solvent will rapidly dissolve ABS but it will work great if its PP. Anyone have a strategy to tell which it is without destroying it if it happens to be ABS?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2020-08-22T17:02:58.523" Title="How to determine whether plastic is ABS or Polypropylene?" Tags="|plastic|" AnswerCount="5" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6684" PostTypeId="2" ParentId="6682" CreationDate="2015-12-28T20:42:48.290" Score="8" Body="&lt;p&gt;This is a great question.  Typically the engineering method to identify plastics is with a &lt;a href=&quot;http://www.boedeker.com/burntest.htm&quot;&gt;Burn Test&lt;/a&gt;.  As fun as it sounds, it is destructive.  So the next place to go is to try to test specific gravity if possible.  ABS sinks in water (SG = 1.06), Polypropylene would float (SG = 0.946).&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2015-12-28T20:42:48.290" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6685" PostTypeId="1" CreationDate="2015-12-28T21:09:25.520" Score="1" ViewCount="90" Body="&lt;p&gt;Suppose we have an AC motor which needs to lower an object at a set speed. We use a PID loop to determine and correct positional errors. Presumably the PID loop will keep the current low as most of the work will be done by gravity. By lowering or stopping the current, will the motor be able to maintain the set speed, or will gravity take over and descend the object at its own will?&lt;/p&gt;&#xA;" OwnerUserId="4281" LastActivityDate="2015-12-29T15:45:15.013" Title="Will AC Motor Maintain Speed of Descending Object" Tags="|motors|current|ac|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6686" PostTypeId="2" ParentId="6670" CreationDate="2015-12-28T21:56:53.963" Score="13" Body="&lt;p&gt;Yes, steel (even some stainless) may be cheaper than aluminum, but the material cost of an item is seldom the majority of the total cost, especially a small item such as a potato peeler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making aluminum parts with complex, curved shapes can be fairly easily done by casting. Aluminum pours at around 1500 °F, which can be achieved with low-cost furnaces and contained by low-cost crucibles. The molds can be easily made, and you'll find lots of instructions for doing so, e.g. using &lt;a href=&quot;http://3dtopo.com/lostPLA/&quot;&gt;lost-plastic casting from 3D-printed templates done in the home&lt;/a&gt;. Aluminum can also be easily machined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stainless steel, however, has a much higher pour temperature (above 2500 °F), which makes melting it, keeping it uncontaminated, and casting it &lt;a href=&quot;http://www.artmetal.com/blog/abominog/2009/01/casting_stainless_steel&quot;&gt;much more difficult&lt;/a&gt;. This is why you'll seldom see a consumer item made from cast stainless; if it involves stainless it will generally be stamped or machined, either of which seriously boosts the price of the item. In a market where price is king, and speed to market is queen, the fact that the results aren't strictly dishwasher safe is a minor consideration.&lt;/p&gt;&#xA;" OwnerUserId="3941" LastEditorUserId="368" LastEditDate="2015-12-29T23:49:58.347" LastActivityDate="2015-12-29T23:49:58.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6687" PostTypeId="2" ParentId="6670" CreationDate="2015-12-28T22:07:51.963" Score="6" Body="&lt;p&gt;I looked up 6000 grade aluminum to be &lt;a href=&quot;http://www.metalprices.com/metal/aluminum/aluminum-6061-extrusion-billet-price&quot;&gt;about \$2/kg&lt;/a&gt;, while 304 stainless steel is about &lt;a href=&quot;http://www.metalprices.com/metal/stainless-steel/stainless-steel-flat-rolled-coil-304&quot;&gt;\$3.40/kg&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means stainless steel is about 70% more expensive than aluminum, which is consistent with my experience in doing mechanical design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where your numbers came from that aluminum is 4x more expensive than stainless steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have a KitchenAid stand mixer, and I put the whisk attachment in the dishwasher and it came out oxidized (black-ish). I was pretty pissed too to find out that I spent $300 for a thing that needs to be hand washed. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-12-28T22:07:51.963" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6688" PostTypeId="2" ParentId="6682" CreationDate="2015-12-28T22:23:28.823" Score="1" Body="&lt;p&gt;It all depends what are the tools you dispose.&lt;br&gt;&#xA;Infrared analysis will tell you easily whether it is ABS or PP. If the material is too thick for a transmission measurement, you can always use ATR (Attenuated total reflectance), which allows you to measure some microns inside of the solid. Raman would do the trick as well.  For the three methods, the CH and CH2 group spectra will only appear in PP. On the contrary, ABS will show aromatic rings, which are active at different wavebands which allow to see the difference.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermal analysis (e.g. differential scanning calorimetry) would also be useful; it is destructive but you only need a tiny piece. As ABS and PP have different glass transition temperature and melting point, it is possible to distinguish them.   &lt;/p&gt;&#xA;" OwnerUserId="3991" LastEditorUserId="3991" LastEditDate="2015-12-29T15:02:25.977" LastActivityDate="2015-12-29T15:02:25.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6690" PostTypeId="1" CreationDate="2015-12-28T23:17:54.183" Score="8" ViewCount="160" Body="&lt;p&gt;I am trying to establish an appropriate distance between a spectrograph output window and a CMOS sensor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spectrograph creates a focused image on the CMOS sensor but I am trying to restrict the light as the initial pixels are capturing light of higher intensity from 0th order wavelengths. Post processing is possible, but I would rather physically restrict the wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to block out some of the pixels in the border. Therefore, I need to determine at what distance the blocking window needs to be placed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the image below will illustrate my problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dTRpc.jpg&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Problem illustration&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dTRpc.jpg&quot; alt=&quot;A 16.64 by 14.04 mm sCMOS sensor with 6.5 micron pixels is positioned at a distance of 43.35 mm from a spectrograph exit window. A square blocking window of unknown dimensions is placed between the sensor and exit window at an unknown distance from the sensor. All components are parallel.&quot; title=&quot;Problem illustration&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The setup is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Motor operated scanning mirror&lt;/li&gt;&#xA;&lt;li&gt;Headwall VNIR Spectrograph&lt;/li&gt;&#xA;&lt;li&gt;PCO Edge VNIR 2560 x 2160 pixels sCMOS sensor&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am trying to &quot;see&quot; 2560 spatial pixels and 500th-1600th spectral pixels. So, I am thinking the horizontal window dimensions would remain the same (at 16.64 mm), but the vertical dimension is unknown. For example, vertical dimension would be 7.15 mm if the window is placed EXACTLY on the sensor. I am trying to establish how far to place the window. I can place the window anywhere between 5 and 10 mm from the sCMOS sensor.&lt;/p&gt;&#xA;" OwnerUserId="4284" OwnerDisplayName="Akhil" LastEditorUserId="368" LastEditDate="2016-12-28T18:11:03.750" LastActivityDate="2024-03-22T18:05:20.637" Title="How can I estimate blocking window distance in an sCMOS sensor setup?" Tags="|electrical-engineering|sensors|optics|cmos|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6691" PostTypeId="2" ParentId="6690" CreationDate="2015-12-29T00:21:43.743" Score="0" Body="&lt;p&gt;I would place the window directly on the sensor, if at all possible.&lt;/p&gt;&#xA;" OwnerUserId="2639" OwnerDisplayName="Peter Bennett" LastActivityDate="2015-12-29T00:21:43.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6692" PostTypeId="2" ParentId="6685" CreationDate="2015-12-29T07:37:48.727" Score="1" Body="&lt;p&gt;You will need &lt;a href=&quot;https://en.wikipedia.org/wiki/Variable-frequency_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;variable frequency drive (VFD)&lt;/a&gt; with &quot;&lt;a href=&quot;https://www.yaskawa.com/pycprd/articles/detail?a=88161&quot; rel=&quot;nofollow noreferrer&quot;&gt;regenerative braking&lt;/a&gt;&quot; or a &quot;braking resistor&quot;(the cheaper option) for this application. &quot;Braking&quot; is not necessarily a standard feature (especially on small VFDs) so you will have to specify it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VuKXK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VuKXK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common configuration is to have the VFD output a frequency proportional an input(0-10v or 4-20mA). They can also be controlled via digital protocols(ethernet, modbus, canbus) and many even have PID controllers built into them. For your application all you should need to do is put the drive in constant velocity mode and send it a speed input or manually set that speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency is what controls the speed of the 3 phase AC motor(single phase will not work). When an external load like gravity exceeds the target speed the VFD will divert the energy to resistors or put it back on to the grid depending on your configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, this &lt;a href=&quot;http://www.vfds.com/5hp-7.5hp-230v-saftronics-vfd-c1020051&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saftronics C102005-1&lt;/a&gt; specifies that it has a &quot;Built-in braking transistor&quot;. Others may say &quot;Built in dynamic braking&quot; or the like. The resistor will usually be external. You will have to read the user manual or contact the manufacturer to determine the required power capacity and ohms of the resistor bank.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2015-12-29T15:45:15.013" LastActivityDate="2015-12-29T15:45:15.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6693" PostTypeId="2" ParentId="6682" CreationDate="2015-12-29T07:57:49.297" Score="5" Body="&lt;p&gt;There is often a non-critical location on the plastic extrusion that you can test some of the solvent on. Rubbing a little with a paper towel you can see discoloring and if any of the plastic is removed. Acetone will detect ABS really well if your specific solvent is more slow acting. The blemish is usually just a dull area in the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that even though the plastic housing may survive the solvent, your bearings and seals may not. Another consideration is flammability and the likeliness of a failed pump catching fire. You may want to do a search for pumps specifically designed for your application or highly chemically inert pumps like &lt;a href=&quot;http://www.zoro.com/shurflo-pump-diaphragm-115-vac-8000-811-288/i/G0496711/&quot;&gt;santoprene diaphragm pumps&lt;/a&gt; or silicone tube &lt;a href=&quot;https://en.wikipedia.org/wiki/Peristaltic_pump&quot;&gt;peristaltic pumps&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-12-29T07:57:49.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6694" PostTypeId="2" ParentId="6674" CreationDate="2015-12-29T08:24:37.933" Score="1" Body="&lt;p&gt;I would initially assume that by &quot;closed loop&quot; you mean like in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Terrarium&quot; rel=&quot;nofollow&quot;&gt;Terrarium&lt;/a&gt;. The water travels via evaporation and condensation and the energy input is the light that heats the plant and the walls of the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the environment is not closed off in this fashion you will have to produce water by condensing it from the air (like rain). Otherwise any container will eventually run out. This could be done with thermometric modules but would use a lot of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you meant &quot;closed loop&quot; in terms of pumping from a reservoir for a limited period of time; there are lots of small pumps available. All of them require a small amount of electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some I found in a quick search:&#xA;&lt;br&gt;&lt;a href=&quot;https://www.adafruit.com/products/1150&quot; rel=&quot;nofollow&quot;&gt;Peristaltic Pump&lt;/a&gt;&#xA;&lt;br&gt;&lt;a href=&quot;http://www.aliexpress.com/item/Brand-New-Solar-Power-Fountain-Pool-Water-Pump-Garden-Plants-Watering-Kit-Solar-Powered-Pump-with/32457721287.html&quot; rel=&quot;nofollow&quot;&gt;Solar Pump&lt;/a&gt;&#xA;&lt;br&gt;&lt;a href=&quot;http://www.aliexpress.com/item/U119-Free-Shipping-1-2A-DC-12V-Micro-Submersible-Motor-Water-Pump-14L-Min-840L-H/32440657187.html&quot; rel=&quot;nofollow&quot;&gt;12v Submersible&lt;/a&gt;&#xA;&lt;br&gt;&lt;a href=&quot;http://www.aliexpress.com/item/Free-Shippiing-5PCS-New-Mini-Micro-Water-Priming-Gear-Pump-DC-1-5V-12V-RS-360SH/32252758490.html&quot; rel=&quot;nofollow&quot;&gt;12v Gear Pump&lt;/a&gt;&#xA;&lt;br&gt;&lt;a href=&quot;http://www.ebay.com/itm/High-Quality-Miniature-liquid-water-pump-100-150ml-min-24V-3W-24VDC-3Watt-A-/201475966744&quot; rel=&quot;nofollow&quot;&gt;Diaphram Pump&lt;/a&gt;&#xA;&lt;br&gt;&lt;a href=&quot;http://www.ebay.com/itm/HP104-Miniature-Inline-Only-Centrifugal-Micro-Water-Pump-6-VDC-1-5-Watt-5-GP-/231587767607&quot; rel=&quot;nofollow&quot;&gt;6v Centrifugal Pump&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-12-29T08:24:37.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6695" PostTypeId="2" ParentId="2333" CreationDate="2015-12-29T11:30:35.890" Score="2" Body="&lt;p&gt;After some more searching, I think that the above material is a carbon-based 'forged composite' &lt;a href=&quot;https://www.google.co.uk/search?q=forged+composite&amp;amp;biw=1841&amp;amp;bih=898&amp;amp;site=webhp&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;sqi=2&amp;amp;ved=0ahUKEwiJ6Z_Q-oDKAhWFOBoKHeihCCcQ_AUIBygC&quot; rel=&quot;nofollow&quot;&gt;as shown in this Google image search query&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was apparently developed by Lamborghini and Callaway Golf, who both use it in their products, and is composed of carbon strands in a resin matrix. It is injected or laid up as a malleable paste which is then presumably cured to produce the final product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantages over regular carbon cloth are presumably lower layup costs (this stuff can be injected and doesn't need labour-intensive hand layup) and easier production of small and detailed pieces that are hard to form with cloth. The main disadvantage is presumably a loss of specific stiffness due to having no control over fibre orientation in the final product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you're really interested, &lt;a href=&quot;http://www.quantumcomposites.com/pdf/papers/2011-ASC-montreal-forged-suspens.pdf&quot; rel=&quot;nofollow&quot;&gt;here's a good paper from the guys who developed it about using it in the new Lamborghini&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1161" LastEditorUserId="368" LastEditDate="2015-12-29T23:29:08.150" LastActivityDate="2015-12-29T23:29:08.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6696" PostTypeId="1" AcceptedAnswerId="6704" CreationDate="2015-12-29T13:04:38.090" Score="7" ViewCount="1220" Body="&lt;p&gt;How can I control a Four-Bar Infinity Coupler such that it does not accidentally become a Paralellogram Linkage? It seems to me that, when the bars are all horizontal, gravity could change the movement of the green driven linkage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=NZZtXUcRZVs&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=NZZtXUcRZVs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eJLCW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eJLCW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WhECH.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WhECH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3765" LastActivityDate="2019-05-15T11:19:18.637" Title="Design of a Four-Bar Infinity Coupler Linkage" Tags="|mechanical-engineering|design|linkage|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6697" PostTypeId="2" ParentId="6696" CreationDate="2015-12-29T16:14:00.303" Score="3" Body="&lt;p&gt;Correct, the linkage will randomly try to switch modes at the two top dead center locations. I have actually built one in the past and attempted to power only one of the rotational bars and expected the other to stay in its mode from flywheel inertia on that leg. However, the the linkage would still randomly switch modes and abruptly (sometimes catastrophically) change rotational direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some possible solutions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Control the speed of both rotational links by coupling them with a timing belt or gears. This will only work for parallel mode; not for the lemniscate mode as skibulk has corrected.&lt;/li&gt;&#xA;&lt;li&gt;Control the speed of both rotational links with electronic motor control. Constant speed for parallel mode, variable speed for lemniscate mode. The control is mainly important near top dead center; which may simplify things.&lt;/li&gt;&#xA;&lt;li&gt;Have a second identical linkage with the rotational bars coupled at 90 degrees out of phase. This also only works for parallel mode as skibulk has corrected. This is how old steam engines coupled power to additional wheels (could not find a good image unfortunately). Its a design aspect that you don't really respect until you have confronted the problem on your own :-)&lt;/li&gt;&#xA;&lt;li&gt;Another option is to bound the linkage such that it will never reach the modes. This is obviously of no value if you want to do full rotation, but this is how a lot of vehicle suspension works (&lt;a href=&quot;https://en.wikipedia.org/wiki/Watt%27s_linkage&quot; rel=&quot;nofollow&quot;&gt;Watt's linkage&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;With some additional work you may be able to maintain the lemniscate mode with just one motor controlled input. My linkage would randomly switch modes because I was attempting to transfer constant torque which is not possible. With enough inertia on the driven leg and if the control system controls the speed such that there is no force transfer through the floating link near top dead center; the system should maintain the lemniscate mode. If friction is roughly equal on each, removing torque inputs and outputs should allow it to coast though top dead center. If input or outputs are not removable/controllable it gets more complicated.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2015-12-30T01:04:37.383" LastActivityDate="2015-12-30T01:04:37.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6698" PostTypeId="2" ParentId="2333" CreationDate="2015-12-29T16:53:56.587" Score="0" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Forged_composite&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt; the composite is a novel construction using carbon fibers interleaved or layered with graphite sheets, all mixed with resin paste. The mixture is molded and cured simultaneously. Essentially compression molding but with a different raw material. It is not clear how the raw material is produced initially, and it is likely all we can do is speculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the graphite sheets were aligned with the surface it might give the appearance you see, since each would be slightly out of alignment with the surface by different amounts, and reflect light at slightly different angles. The two photos show that effect when compared to each other. To me the surface doesn't look like fibers, but like fragments, though we would need a closeup to be sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, it could just be a thin layer of randomly oriented chopped carbon material at the surface for show. Whether it looks good or not is another question.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2015-12-29T16:53:56.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6700" PostTypeId="1" CreationDate="2015-12-29T18:46:09.900" Score="2" ViewCount="101" Body="&lt;p&gt;I need to design a controllable Bernoulli ball airflow levitation setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, the ball will be about 80 mm in diameter, and I need to be able to operate the height in range from zero to one meter from the air outlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What equations can be used to compute height vs airflow and outlet geometry?&lt;/p&gt;&#xA;" OwnerUserId="4294" LastEditorUserId="33" LastEditDate="2016-01-05T01:54:31.377" LastActivityDate="2016-01-05T01:54:31.377" Title="What equations can be used to determine the height of Bernoulli ball levitation?" Tags="|fluid-mechanics|airflow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6701" PostTypeId="1" CreationDate="2015-12-29T19:20:52.383" Score="3" ViewCount="844" Body="&lt;p&gt;I want to test the frequency response of stringed musical instruments . My test set-up includes a piezo film transducer which is just taped onto the instrument and fed a signal which then vibrates the instrument. The resulting sound is picked up by a calibrated microphone and sent to a spectrum analyzer.  The strings are damped with memory foam so their vibrations are damped. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if it matters what kind of test signal I send to piezo element. Should I use single frequencies, a sweep or &quot;chirp&quot; signal , or a broadband noise signal such as pink or white noise. &lt;/p&gt;&#xA;" OwnerUserId="628" LastEditorUserId="368" LastEditDate="2015-12-29T23:42:10.093" LastActivityDate="2015-12-30T12:41:14.603" Title="What type of test signal is appropriate for frequency response analysis of string instruments?" Tags="|sensors|audio-engineering|frequency-response|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6702" PostTypeId="2" ParentId="6701" CreationDate="2015-12-29T19:41:40.410" Score="3" Body="&lt;p&gt;First, can you really actuate white noise? If yes, White noise would probably show many natural frequencies and harmonics. However, make sure the actuation bandwidth is sufficient for your frequency range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the piezo can be adjusted to your frequencies of interest, a more deterministic approach would be to do a frequency sweep. For that, start from the lowest frequency and increase the frequency towards the highest frequency that the piezo permits. This would give the complete frequency response for the instrument (its strings, structure, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would be able to capture the frequencies of the strings (all of them), the instrument structure, and maybe even the objects in the room, so a non-echo room would be an advantage.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastEditorUserId="1097" LastEditDate="2015-12-29T20:30:36.673" LastActivityDate="2015-12-29T20:30:36.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6703" PostTypeId="2" ParentId="6701" CreationDate="2015-12-29T21:27:19.817" Score="1" Body="&lt;p&gt;It is helpful to be clear about what you are trying to measure.  The frequency response (aka transfer function) of the string will be given by&#xA;$$&#xA;T(f)=\frac{O(f)}{I(f)},&#xA;$$&#xA;where $T(f)$ is the frequency response, $I(f)$ is the input signal, and $O(f)$ is the output signal and all three are a function of frequency.  In order to have a meaningful frequency response function, it is important that $I(f)$ and $O(f)$ have meaningful units, which is achieved through proper calibration of your drive and sensor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One difficulty in your situation is that you don't have a good measurement of $I(f)$, so you have to make some assumptions about it's responsivity to your electrical drive.  Since audio frequencies are much lower than typical piezo resonances, you can probably assume that the frequency response of the piezo element has a flat amplitude, but the phase may start to roll off at higher audio frequencies.   If your goal is simply to measure the amplitude of the $T(f)$ and ignore the phase, then you should be safe simply driving the piezo with white noise and taking the $O(f)$ to be the amplitude of the frequency response function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a perfect world where both your sensor and transducer have no noise and no saturation, then a quick white noise measurement is enough to fully measure the frequency response of any system.  In the real world, you have to be careful of the following (I'm sure I'm missing some):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Saturation of the peizo&lt;/li&gt;&#xA;&lt;li&gt;Saturation of the microphone&lt;/li&gt;&#xA;&lt;li&gt;Saturation of the electronics driving the piezo&lt;/li&gt;&#xA;&lt;li&gt;Saturation of the electronics reading the microphone&lt;/li&gt;&#xA;&lt;li&gt;Producing a sound which is below the background noise&lt;/li&gt;&#xA;&lt;li&gt;Producing a sound which is below the noise level of the microphone.&lt;/li&gt;&#xA;&lt;li&gt;Nonlinearities in the instrument causing response at frequencies other than the driving frequency.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some example reasons that you might want to use a signal other than white noise:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Measuring at discrete frequencies (swept sine) allows you to make higher SNR measurements at the specific frequencies which you are interested in because you can integrate for longer.&lt;/li&gt;&#xA;&lt;li&gt;If the response is very low at some of the frequencies you are interested in, then your sensor may not pick up the sound.  In this case, you may want to drive harder in certain frequency bands (pink noise).  &lt;/li&gt;&#xA;&lt;li&gt;On the other hand, if the response is very strong at certain frequencies, it may drive your sensor into the nonlinear regime which will destroy the measurement.  In this case you may want to drive less in certain frequency bands (pink noise).  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2015-12-30T12:41:14.603" LastActivityDate="2015-12-30T12:41:14.603" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6704" PostTypeId="2" ParentId="6696" CreationDate="2015-12-29T21:59:32.507" Score="3" Body="&lt;p&gt;Gearing is complicated by the fact that the driver bar moves at a constant speed while the driven bar changes speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this gearing Solution work? The gear should only engage when the center bar is fully extended to the right or fully extended to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/99uRy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/99uRy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=I4V3NqwZG0o&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=I4V3NqwZG0o&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=aPUcdGnf2uk&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=aPUcdGnf2uk&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=c9jSFZZX6uA&amp;amp;feature=youtu.be&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=c9jSFZZX6uA&amp;amp;feature=youtu.be&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=cw0Wco_O600&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=cw0Wco_O600&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3765" LastEditorUserId="3765" LastEditDate="2015-12-30T12:24:17.747" LastActivityDate="2015-12-30T12:24:17.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6705" PostTypeId="1" AcceptedAnswerId="6721" CreationDate="2015-12-30T02:35:30.890" Score="6" ViewCount="4622" Body="&lt;p&gt;For me, it's a bit hard to imagine where shear center can be, and what purpose it serves. &lt;a href=&quot;http://www.civilengineeringterms.com/mechanics-of-solids-2/shear-center/&quot;&gt;The definition&lt;/a&gt; I found online is very vague:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Shear center is defined as the point on the beam section where load is&#xA;  applied and no twisting is produced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given the general profile of a section, is there a proper mathematical derivation/formula that shows how the shear center is defined/derived?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="33" LastEditDate="2016-01-01T01:36:21.477" LastActivityDate="2022-07-13T22:55:59.670" Title="What is the mathematical derivation for shear center of a beam?" Tags="|mechanical-engineering|structural-engineering|beam|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6706" PostTypeId="2" ParentId="6676" CreationDate="2015-12-30T05:51:09.820" Score="1" Body="&lt;p&gt;GPS would be the best solution provided the desired measurement location has access to multiple satellites (not in a cave, steel building, deep canyon, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like Air mentioned, smartphones have built in GPS modules that will work well in many situations. This &lt;a href=&quot;http://gpsworld.com/accuracy-in-the-palm-of-your-hand/&quot; rel=&quot;nofollow&quot;&gt;gps world article&lt;/a&gt; estimates consumer grade smartphone resolution to be 2 to 10 meter depending on signal quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turn-key unit with 1cm accuracy: &lt;a href=&quot;http://www.tigersupplies.com/Products/Ashtech-ProMark-120-Receiver-with-L1-GPSGLONASS__ASH990664-03.aspx&quot; rel=&quot;nofollow&quot;&gt;Ashtech ProMark 120&lt;/a&gt;&lt;br&gt;&#xA;A good overview of diy modules: &lt;a href=&quot;https://www.sparkfun.com/tutorials/169&quot; rel=&quot;nofollow&quot;&gt;Sparkfun GPS Module Comparison&lt;/a&gt;&lt;br&gt;&#xA;A $900 1cm resolution module: &lt;a href=&quot;http://hackaday.com/2013/08/05/centimeter-level-precision-gps-for-500/&quot; rel=&quot;nofollow&quot;&gt;Piksi GPS receiver featured on Hackaday&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2015-12-30T05:51:09.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6707" PostTypeId="2" ParentId="6676" CreationDate="2015-12-30T08:11:43.723" Score="1" Body="&lt;p&gt;Have you considered using a laser level device?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen estate agents (realtors) use &lt;a href=&quot;https://www.sccssurvey.co.uk/surveying-equipment-instruments/leica-disto-laser-measurers.html&quot; rel=&quot;nofollow&quot;&gt;these&lt;/a&gt; for measuring the size of rooms. You could point one at the ground and it would give you your elevation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantages are relative cheapness and reliability, the main disadvantage is it requires line-of-sight.&lt;/p&gt;&#xA;" OwnerUserId="2551" LastActivityDate="2015-12-30T08:11:43.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6708" PostTypeId="1" CreationDate="2015-12-30T09:33:47.400" Score="0" ViewCount="691" Body="&lt;p&gt;As in the picture below, there is a static force $A$ being applied at 45 degrees to a rigid rod attached to a rigid plate. The plate is supported by two rods of length $B$ and diameter $D$, each of circular cross-section, which are fixed at their ends opposite the plate. The force is applied halfway between the two rods. The weight of the objects are negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Od3Jk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Od3Jk.jpg&quot; alt=&quot;Diagram.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fGuU8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;See here&lt;/a&gt; for a more precise diagram, but note that there should also be a bending moment at the point where the load is applied due to the parallel-axis theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective is to determine the deflection in each of the two rods. Assuming no other parts in the picture other than the two rods deflect, how do I determine the deflection of the two steel rods?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I think a better way to visualise it is that it is kind of like a cantilever; where the two fixed ends are like attached to a wall. This was how I saw the problem.&lt;/p&gt;&#xA;" OwnerUserId="3781" LastEditorUserId="1832" LastEditDate="2015-12-31T01:24:40.207" LastActivityDate="2019-12-28T18:28:05.877" Title="Deflection of parallel rods with force at an angle" Tags="|structural-engineering|applied-mechanics|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6709" PostTypeId="2" ParentId="6708" CreationDate="2015-12-30T12:55:36.087" Score="0" Body="&lt;p&gt;How accurate are you trying to be? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the roller is ideal, it won't allow a torque, so resolve the impact force to x/y component forces at the center of the roller. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the base of the black roller pedestal is fixed (if the legs are going to flex, then it's not!), then the x/y forces at the top of the black pedestal will resolve to an x/y force and a moment at the base of the pedestal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the steel plate doesn't flex (if it's real, it will!), then it will transmit the forces and moment to the tops of the rods. You can then look up the formulas for deflection for the loads and sum the results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But again, if the legs flex, the black roller pedestal will move. You are calling the force an &quot;impact&quot;, which implies that it is a short-term, dynamic load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An impact has an impulse profile, which shows force plotted over time. The area under the curve (integral) gives the impulse, which is typically constant for a given impact regardless of the &quot;springiness&quot; of the collision. A springy surface increases the impulse duration, reducing the peak force. This is why hitting an airbag is better than hitting a wall - lower peak force, even though both will bring your face to a stop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this is to say that dynamic situations are extremely difficult to evaluate. These are typically done with numerical simulations, because &lt;em&gt;dynamic&lt;/em&gt; deflection reduces the force transmitted &lt;em&gt;at that particular instance.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerical simulations are used because you wind up with iterations around &quot;loops&quot; like the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Assume nothing deflects, find max loading. &lt;/li&gt;&#xA;&lt;li&gt;Use that loading to deform the structural members.&lt;/li&gt;&#xA;&lt;li&gt;With the members deformed, the applied force is lower (because force times distance, energy, &quot;drains&quot; kinetic energy from the impact source)&lt;/li&gt;&#xA;&lt;li&gt;Run the process again with the lower applied force.&lt;/li&gt;&#xA;&lt;li&gt;Get new deflections, those get a new impact force, repeat, repeat, repeat. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Eventually this should stabilize at a particular deflection, but now that the structure has been deflected in some period of time there is an acceleration in the structure, which adds a force at each member that deflected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your question were about deflection under static load then I would say it's probably a suitable final exam question for a deformable bodies course (5th semester mechanical/civil engineering course). As it stands, this is more along the lines of a finite element analysis class, which would probably be a 4th year undergrad/graduate course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is to say, if you are looking for accurate results, you're not going to find them here. You need to know more information about the source of the impact, the shape, size, material (including grade) of all of the structural members, a finite element analysis software package, and probably at least one class on how to use the software. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2015-12-30T13:04:53.550" LastActivityDate="2015-12-30T13:04:53.550" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6710" PostTypeId="1" CreationDate="2015-12-30T14:34:49.760" Score="1" ViewCount="239" Body="&lt;p&gt;I have a bit of an ambitious project on hand and was looking for some advice/ideas.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My team and I are working on a project that involves retrofitting a commercial grade freezer (model MasterBilt IHC-48) into an environment chamber that would be used for HASS testing. HASS testing involves perform a rapid temperature cycling test and vibration/shock test, which would be used to stress electronic products.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example this rapid temperature cycling might involve going from +&lt;strong&gt;65C to -10C in 2 minutes&lt;/strong&gt; and vice versa.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The medium inside the chamber would be air.  The dimensions inside the freezer are 27&quot;x58&quot;x46&quot; (0.686m X 1.47m X 1.17m).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gray area for me is how to achieve this &lt;strong&gt;rapid cooling&lt;/strong&gt;.  What I am looking for are ideas/suggestions on how to retrofit this freezer to achieve this rapid cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas/comments are welcomed.&lt;/p&gt;&#xA;" OwnerUserId="4300" LastEditorUserId="4300" LastEditDate="2015-12-30T18:14:17.497" LastActivityDate="2016-01-01T08:27:17.610" Title="Rapid Cooling System for Commerical Freezer" Tags="|thermodynamics|design|temperature|cooling|" AnswerCount="1" CommentCount="12" ClosedDate="2016-01-02T15:18:55.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="6712" PostTypeId="1" AcceptedAnswerId="6713" CreationDate="2015-12-30T17:06:34.867" Score="2" ViewCount="4034" Body="&lt;p&gt;Are there any established (compact and simple) solutions for turning two 5mm rods/shafts side-by-side together by driving a singe motor? The torque is about 5kg-cm. The one I'm thinking of is the coupling rod:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JFAVT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JFAVT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there disadvantages to this solution compared to others, e.g., gearing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the coupling rod is a good solution, where could I get my hands on those &quot;tabs&quot; where the coupling rod connects? Is the coupling rod supposed to be connected using some small shoulder bolts and nuts?&lt;/p&gt;&#xA;" OwnerUserId="4303" LastEditorUserId="4303" LastEditDate="2015-12-30T17:42:08.413" LastActivityDate="2016-01-02T21:34:21.737" Title="Rotating two shafts together - coupling rod?" Tags="|mechanical-engineering|motors|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6713" PostTypeId="2" ParentId="6712" CreationDate="2015-12-30T17:34:18.443" Score="7" Body="&lt;p&gt;The usual solution is to use gears, possibly with a toothed belt or chain to connect them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single connecting rod like you show is not sufficient &amp;mdash; there's an ambiguity when the rod is at the extreme left or right position, allowing the driven shaft to turn in either direction. A train locomotive uses &lt;em&gt;two&lt;/em&gt; connecting rods, 90&amp;deg; out of phase on either end of the axles, in order to resolve the ambiguity. Whenever one of the rods is at one of its ambiguous positions, the other rod is not, and that's what makes sure the driven shaft turns in the correct direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Jackshaft_(locomotive)&quot;&gt;This Wikipedia article&lt;/a&gt; talks about the various means of coupling parallel shafts with connecting rods. &lt;a href=&quot;https://en.wikipedia.org/wiki/Jackshaft_(locomotive)#/media/File:PRR_DD1_running_gear.jpg&quot;&gt;This picture&lt;/a&gt; from that article is particularly good, because you can clearly see that the counterweights on either side of the motors are 90&amp;deg; out of phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Jackshaft_(locomotive)#/media/File:PRR_DD1_running_gear.jpg&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/b/bc/PRR_DD1_running_gear.jpg/1024px-PRR_DD1_running_gear.jpg&quot; alt=&quot;photograph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2015-12-30T18:26:45.060" LastActivityDate="2015-12-30T18:26:45.060" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6714" PostTypeId="2" ParentId="6710" CreationDate="2015-12-30T17:46:49.730" Score="2" Body="&lt;p&gt;Cooling a little over a cubic meter of air by 75K in 2 minutes requires on the order of 1 kW of average continuous heat transfer. It might be simpler to consider having two reservoirs of air, one hot and one cold, and using fans to transfer air from one reservoir or the other to the test chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuous operation spreads the workload out over a longer period of time, requiring a lower-powered chiller. Rapid temperature changes would depend only on how fast you can move the air.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2015-12-30T17:46:49.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6715" PostTypeId="1" CreationDate="2015-12-30T20:02:20.010" Score="3" ViewCount="294" Body="&lt;p&gt;I am trying to build a linear actuator for which I am trying to select a appropriate motor. I need to move about 1 kg of weight at about 1.2 m/s using a stepper motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What specifications of the motor should I be looking at? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know power, RPM and torque all play a big role, but not quite sure how to determine the motor based on the combination of those three. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, torque happens when force is applied at an angle. RPM is just rotations per minute. The pulley's diameter will also change things. Wouldn't the belt rotate faster if I increase the diameter of the pulley since it covers more area of the pulley or is it the other way around? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a bicycle for instance, the chain seems to move faster on a smaller gear which have large diameter but it goes slower. How will this affect the rpm and torque?&lt;/p&gt;&#xA;" OwnerUserId="4304" LastEditorDisplayName="user16" LastEditDate="2016-01-02T15:20:53.040" LastActivityDate="2016-01-02T15:20:53.040" Title="Motor selection" Tags="|mechanical-engineering|power|" AnswerCount="2" CommentCount="0" ClosedDate="2016-01-03T16:04:01.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="6716" PostTypeId="2" ParentId="6715" CreationDate="2015-12-30T20:32:14.843" Score="2" Body="&lt;p&gt;First you need set your self tolerances, how critical is speed? If you want exactly 1.2m/s with +-0.5m/s? Also you will have to define a safety factor, so you generally want to increase the motor specs by 50% always better to have extra power to play with. Gears work in ratios, Bigger the gear in diameter with appropriate number of teeth will give you more torque smaller the gear higher rpm lower torque simple relation. &lt;/p&gt;&#xA;" OwnerUserId="4305" LastActivityDate="2015-12-30T20:32:14.843" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6718" PostTypeId="2" ParentId="6708" CreationDate="2015-12-31T01:15:09.383" Score="2" Body="&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Looking back at this answer I realize that the entire method is flawed. The final answers for &lt;span class=&quot;math-container&quot;&gt;$M_B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B_y$&lt;/span&gt;, however, are correct. However, I believe this cannot be proven without FEA software or the direct stiffness method, which would be too lengthy and complex to present here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;This can be solved once you know your structural analysis. You'll need to solve the entire structure for its internal forces and then you can calculate each of the rods for their deflections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take this as your structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ukMaS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ukMaS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can immediately realize that the vertical component of the applied load won't affect anything in the rods other than their axial loads, so we are going to sideline it for now and pretend we only have a horizontal load. We'll remember it later on. This only works because the horizontal bar is defined as rigid in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the shear force along each of the two rods will be constant and equal to &lt;span class=&quot;math-container&quot;&gt;$B_x = C_x = -\frac{1}{2}F_x$&lt;/span&gt;. Also, the moment due to the applied force will be resisted by the fixed ends but also by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Couple_(mechanics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;force couple&lt;/a&gt; of the axial forces on the rods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total moment around point C is equal to (observing the fact that &lt;span class=&quot;math-container&quot;&gt;$M_B = M_C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B_y = -C_y$&lt;/span&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sum M_C = -F_x \cdot (a+b) - 2 \cdot B_y \cdot c + 2 \cdot M_B = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to find &lt;span class=&quot;math-container&quot;&gt;$F_{y,B}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$M_B$&lt;/span&gt; to solve the entire structure. To do this, we need to list what we know.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The shear force along the rods is equal to &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}F_x$&lt;/span&gt;. This means that the total change in moment along the rods will be equal to &lt;span class=&quot;math-container&quot;&gt;$b \cdot \frac{1}{2}F_x$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The axial forces in the rods (&lt;span class=&quot;math-container&quot;&gt;$B_y$&lt;/span&gt;) will turn to shear forces in the horizontal bar, meaning there will be a change in moment along the horizontal bar from point D to A equal to &lt;span class=&quot;math-container&quot;&gt;$B_y \cdot c$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The bending moment around point A for the vertical rod is equal to &lt;span class=&quot;math-container&quot;&gt;$a \cdot F_x$&lt;/span&gt;. Since point A is in the middle of the horizontal bar and the axial forces in the rods are equal in magnitude, this bending moment will be equally split between both sides of the horizontal beam.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So now we know that the bending moment diagram from point B to point A will have to go from &lt;span class=&quot;math-container&quot;&gt;$M_B$&lt;/span&gt; at B to &lt;span class=&quot;math-container&quot;&gt;$M_B - b \cdot \frac{1}{2}F_x$&lt;/span&gt; at D to &lt;span class=&quot;math-container&quot;&gt;$-\frac{1}{2}a \cdot F_x$&lt;/span&gt; at A. Since we know that the change in moment from D to A is equal to &lt;span class=&quot;math-container&quot;&gt;$B_y \cdot c$&lt;/span&gt;, we can conclude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_B - b \cdot \frac{1}{2}F_x + B_y \cdot c = -\frac{1}{2}a \cdot F_x$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this and the sum of moments around C,&lt;/strike&gt; we can solve for &lt;span class=&quot;math-container&quot;&gt;$M_B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B_y$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;M_B &amp;amp;= \dfrac{F_xb}{4} \\&#xA;B_y &amp;amp;= -\dfrac{F_x(2a+b)}{4c}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We must then just remember the vertical component of the applied load and add it to &lt;span class=&quot;math-container&quot;&gt;$B_y$&lt;/span&gt; , which therefore becomes &lt;span class=&quot;math-container&quot;&gt;$-\dfrac{F_y}{2}-\dfrac{F_x(2a+b)}{4c}$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$C_y$&lt;/span&gt;, meanwhile, becomes &lt;span class=&quot;math-container&quot;&gt;$-\dfrac{F_y}{2}+\dfrac{F_x(2a+b)}{4c}$&lt;/span&gt;. Once you know the internal forces in each of the rods, you can then calculate the deformations due to these. Since you don't define which deflection component you're after, I'll leave that part to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make it perfectly clear, you need to calculate the deformations under this condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RxsK1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RxsK1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All figures created with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:28:05.877" LastActivityDate="2019-12-28T18:28:05.877" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="6720" PostTypeId="1" CreationDate="2015-12-31T17:12:31.867" Score="3" ViewCount="166" Body="&lt;p&gt;I have a IES luminaire file of a lamp and some polar candela distribution plots of lenses like:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/98A8l.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/98A8l.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I create a new luminaire file combining the lamp with every lens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could manually sample the plots in order to retrieve the information from the plot but I haven't been able to fully understand how to weight the candelas values from the IES using the lens plots. I assume the plots are representing the iluminare of an ideal candela through a lens, but I'm not sure. Even if I were correct, there's various distributions (ways to weight each candela value of an IES) that for an ideal candela will output that plot.&lt;/p&gt;&#xA;" OwnerUserId="4310" LastEditorUserId="368" LastEditDate="2015-12-31T17:23:16.043" LastActivityDate="2015-12-31T17:23:16.043" Title="Generating a IES file from an illuminaire and a lens" Tags="|optics|simulation|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6721" PostTypeId="2" ParentId="6705" CreationDate="2015-12-31T17:50:33.160" Score="8" Body="&lt;p&gt;&lt;strong&gt;SHEAR CENTER&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why do we care?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because we need to know if the beam is subjected to torsion in addition to flexure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point in the cross section (or outside the cross section) where we can apply load and produce beam &lt;em&gt;bending without twisting&lt;/em&gt;. Loads applied anywhere other than the shear center will produce both bending (moment) and twisting (torsion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally when we think about a beam in bending, we only think about, well...the bending. But depending on how that load is applied to the cross section (i.e. is the load through the shear center), there may also be twisting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/omkyc.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/omkyc.png&quot; alt=&quot;I-Beam Deflections&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are shear center locations for some typical structural sections (from &lt;a href=&quot;http://academic.csuohio.edu/duffy_s/AISC_DG_9_Torsion.pdf&quot; rel=&quot;noreferrer&quot;&gt;AISC Design Guide 9&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W5MI2.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W5MI2.png&quot; alt=&quot;Structural Section Shear Centers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for doubly symmetric sections, the shear center is coincident with the centroid. In other words, the shear center is at the intersection of the axes of symmetry. In singly symmetric sections, the shear center lies somewhere along the axis of symmetry. Lucky for us, in typical civil engineering applications we very often deal with symmetric sections. (Aero folks get saddled with the harder math.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Intuitive Approach to Estimating Shear Center&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters, you can probably look at the two I-beam load cases above and intuitively sense that applying a load offset from the centerline will produce beam twisting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more quantifiable procedure is to visualize the shear flows and estimate the point about which the net moment produced by the shear flow will be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EPRKG.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EPRKG.png&quot; alt=&quot;Shear Flows&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the Mathematically Inclined&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I could go bonanzas with the LaTex (and I may do so a bit later), but since my lunch break is coming to an end, I'll refer you to &lt;a href=&quot;http://aerostudents.com/files/aircraftStructuralAnalysis/bendingShearAndTorsion.pdf&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; explanation.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2015-12-31T17:50:33.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6722" PostTypeId="2" ParentId="6715" CreationDate="2015-12-31T23:04:12.780" Score="1" Body="&lt;p&gt;I feel like I say this a lot, but it seems to be a commonly overlooked design spec. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque is a force applied (orthogonally) at some distance, but torque is &lt;em&gt;also&lt;/em&gt; moment of inertia times angular acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is your arrangement? You have a mass and a speed spec, but how long should it take to &lt;em&gt;accelerate&lt;/em&gt; to that speed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are trying to move a mass orthogonal to the gravity vector, then theoretically (no friction), it doesn't take any power to move it at any speed because power is work per time, and work is force times distance, and &lt;em&gt;there is no force&lt;/em&gt; (other than friction/air resistance) &lt;em&gt;required to hold an object at a constant speed!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real life friction and wind resistance matter, but those are probably negligible at your desired speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So again, to spec your motor, you need to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Determine &lt;em&gt;how quickly&lt;/em&gt; you want to get to your rated speed, then&lt;/li&gt;&#xA;&lt;li&gt;Choose a mechanism to convert rotary to linear motion, then &lt;/li&gt;&#xA;&lt;li&gt;Find the moment of inertia &lt;em&gt;of the entire system&lt;/em&gt; at the motor shaft, then&lt;/li&gt;&#xA;&lt;li&gt;Multiply the moment of inertia by your desired acceleration to get your &quot;rated torque&quot;, then&lt;/li&gt;&#xA;&lt;li&gt;Multiply your linear speed by your gearing/belting/reduction ratio to get your &quot;rated rpm&quot;, then&lt;/li&gt;&#xA;&lt;li&gt;Multiply your torque by your rpm (in rad/s) to get your rated &lt;em&gt;motor (output) power&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would look for a motor at least 10% higher power than whatever you calculate because of efficiency loss in your rotary-linear conversion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can play around with gearing, but keep in mind you need to update moment of inertia (torque) and rpm when you do. Eventually you should reach a design that an off-the-shelf motor meets or exceeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have questions about how to calculate a moment of inertia, that's a good question and you should ask it. As it stands, I hope this explains how to go about choosing motor torque, rpm, and power. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2015-12-31T23:04:12.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6723" PostTypeId="1" AcceptedAnswerId="6875" CreationDate="2016-01-01T05:57:11.507" Score="4" ViewCount="682" Body="&lt;p&gt;I am studying the performance of volumetric solar receivers (specifically ceramic foams) versus the change of many parameters: solar flux, surface temperature, porosity, pore diameter, flow pressure and flow velocity. I have seen two ways in the literature to model porous media:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Drawing a complex 3D structure that is analogous to the porous zone.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;While this might be a very-problem-specific solution, but turbulence modelling in that case is not a problem.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Averaging the governing partial differential equations to account for the nature of porous media and introducing new terms to the equations (porosity $\epsilon$ and superficial velocity).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And this is the approach I am taking which is used in porous models in many CFD solvers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, while reading about the treatment of turbulence in porous media in the &lt;a href=&quot;http://aerojet.engr.ucdavis.edu/fluenthelp/html/ug/node274.htm&quot; rel=&quot;nofollow&quot;&gt;FLUENT user guide&lt;/a&gt; I found that turbulence is not exactly modelled:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;turbulence in the medium is treated as though the solid medium has no effect on the turbulence generation or dissipation rates. This assumption may be reasonable if the medium's permeability is quite large and the geometric scale of the medium does not interact with the scale of the turbulent eddies. In other instances, however, you may want to suppress the effect of turbulence in the medium.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I find &lt;em&gt;&quot;medium's permeability is quite large&quot;&lt;/em&gt; quite ambiguous (What values are considered &lt;em&gt;large&lt;/em&gt;?) and don't know whether my model will be badly affected by this treatment or not and to what extent? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So has anyone came through this before? or am I overcomplicating things for a parametric study?&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2016-01-04T07:34:35.840" LastActivityDate="2016-01-12T06:32:22.570" Title="How safe is this assumption for turbulence modelling in porous media?" Tags="|cfd|turbulence|porous-medium|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6724" PostTypeId="1" AcceptedAnswerId="6727" CreationDate="2016-01-01T09:35:52.363" Score="1" ViewCount="249" Body="&lt;p&gt;I have a setup in which a bolt threaded on two ends with an unthreaded shank in the middle will be rotated by the unthreaded shank area. That is, the bolt's axis of rotation is roughly the red cross:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pu9wQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pu9wQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if there's a small part/connector that I could fit over the unthreaded shank so that a shaft could be connected to the connector such that the bolt can be rotated. Perhaps, something like below where the top is a hex/square hole and the bottom is a threaded hole?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qgatq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qgatq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, are there low-profile setups where a bolt can be rotated around the centre?&lt;/p&gt;&#xA;" OwnerUserId="4301" LastEditorUserId="4301" LastEditDate="2016-01-01T09:50:18.040" LastActivityDate="2017-08-14T14:10:06.203" Title="Rotating a bolt around the centre" Tags="|mechanical-engineering|bolting|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6725" PostTypeId="1" CreationDate="2016-01-01T10:27:10.733" Score="1" ViewCount="1004" Body="&lt;p&gt;This is slightly similar to &lt;a href=&quot;https://engineering.stackexchange.com/q/6708/1832&quot;&gt;a question I previously asked&lt;/a&gt; but now I decided to simplify the question and model it similar to a design I have in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sketched out a slightly simplified model. It is a cantilever rod with two supporting rods of slightly larger diameter, connected by a plate. (Ignore the weight of the individual parts. For illustration purposes, I draw this to resemble a cantilever beam from the side view but in actual design, this drawing is actually the top view.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dEosz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dEosz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My aim is to calculate the deflection/bending in the rods (more importantly, the two side/supporting rods) due to the force applied. The plate can be considered rigid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll break down the question into two parts, so that I can understand the concept better:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If the three rods were of the same length (the middle rod is not extended and force is applied at the end of the three rods), how do I calculate the forces acting on each rod and then the deflection?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;With the middle rod being longer than the supporting two rods. How should I approach calculating the deflection in each of the rods?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3781" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-01-05T22:46:29.463" Title="Cantilever with three parallel rods" Tags="|applied-mechanics|structural-analysis|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6726" PostTypeId="1" CreationDate="2016-01-01T13:02:19.860" Score="0" ViewCount="194" Body="&lt;p&gt;In a book, There is chapter of &quot;Centre of Gravity&quot; before &quot;Moment of Inertia&quot; and in the chapter Centre of Gravity, the list of some centroid and it's location of C.G (i.e x̅ and y̅) is given. You can refer similar &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_centroids&quot; rel=&quot;nofollow&quot;&gt;table&lt;/a&gt; from wikipedia. And with the help of it, the examples of finding the location of C.G is explained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I want to prove the location of C.G for &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_centroids&quot; rel=&quot;nofollow&quot;&gt;that&lt;/a&gt; centroids. For example, I want to prove that y̅=4r/3π for the semicircular. How can I prove it?&lt;/p&gt;&#xA;" OwnerUserId="4005" LastActivityDate="2016-01-01T16:52:32.527" Title="How can I prove the location of centre of gravity of centroid?" Tags="|mechanical-engineering|civil-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6727" PostTypeId="2" ParentId="6724" CreationDate="2016-01-01T15:52:25.153" Score="1" Body="&lt;p&gt;Clamp collars can attach to the bolt and shank:&#xA;&lt;a href=&quot;http://www.mcmaster.com/#shaft-collars/=10hrt5k&quot; rel=&quot;nofollow&quot;&gt;http://www.mcmaster.com/#shaft-collars/=10hrt5k&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Holes can be tapped in the sides of the clamp collars and then can be bolted together by making a bracket as needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clamping handers are also sold that bolt together, but these are in limited sizes: &lt;a href=&quot;http://www.mcmaster.com/#routing-collars/=10hrtms&quot; rel=&quot;nofollow&quot;&gt;http://www.mcmaster.com/#routing-collars/=10hrtms&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-01T15:52:25.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6728" PostTypeId="2" ParentId="6725" CreationDate="2016-01-01T16:10:01.447" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Draw the system in 3D CAD software SolidWorks (30 day free trial is available), and use the built in FEA (finite element analysis) to calculate deflections.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want an approximate hand calculation, there may be a beam case of fixed one end, and support in the middle. There are similar beam formulas at:&#xA;&lt;a href=&quot;http://www.engineersedge.com/beam_calc_menu.shtml&quot; rel=&quot;nofollow&quot;&gt;http://www.engineersedge.com/beam_calc_menu.shtml&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The load at the support in the middle could then be used to get the approximate deflection in the other two beams.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-01T16:10:01.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6729" PostTypeId="2" ParentId="6667" CreationDate="2016-01-01T16:28:18.510" Score="1" Body="&lt;p&gt;Use nominal hole sizes in CAD for both the shaft and hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll also commonly used shaft and hole tolerance using English units because in the U.S. English sizes are still more common for industrial materials. Standard cold drawn 1/4 shafting has a tolerance of +0/-.002: &lt;a href=&quot;http://www.mcmaster.com/#standard-steel-rods/=10hs6l4&quot; rel=&quot;nofollow&quot;&gt;http://www.mcmaster.com/#standard-steel-rods/=10hs6l4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For easy assembly, the hole tolerance should be .250 +.005/-.000. These can be converting into nominal metric values by 25.4.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-01T16:28:18.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6730" PostTypeId="2" ParentId="6639" CreationDate="2016-01-01T16:40:43.060" Score="0" Body="&lt;p&gt;Plastic roller are good for lower load, speed, and accuracy applications, and also FDA requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel v roller and rails may work better for your application. These are available at: &lt;a href=&quot;http://www.mcmaster.com/#v-groove-rails/=10hsfa7&quot; rel=&quot;nofollow&quot;&gt;http://www.mcmaster.com/#v-groove-rails/=10hsfa7&lt;/a&gt; and &lt;a href=&quot;http://www.mcmaster.com/#6318k61/=10hsgx6&quot; rel=&quot;nofollow&quot;&gt;http://www.mcmaster.com/#6318k61/=10hsgx6&lt;/a&gt; and also from the Dual V company.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-01T16:40:43.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6731" PostTypeId="2" ParentId="6726" CreationDate="2016-01-01T16:52:32.527" Score="4" Body="&lt;p&gt;For a planar surface, the centroid is defined by the equation&#xA;$$\overline{y} =\dfrac{\int y\ \text{d}A}{\int\ \text{d}A} = \dfrac{\int y\ \text{d}A}{A}$$&#xA;where $y$ is the distance of each point from the origin and $A$ is the surface's area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, the center of gravity/mass is defined by &#xA;$$\text{C.G.}=\dfrac{\int y\rho(y)\ \text{d}A}{\int\ \rho(y)\text{d}A} = \dfrac{\int y\rho(y)\ \text{d}A}{M}$$&#xA;where $\rho(y)$ is the density profile of the surface and $M$ is the total mass of the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the case of a uniformly-dense surface, $\rho(y) = \rho$ and $M = A\rho$, which means that&#xA;$$\text{C.G.} = \dfrac{\int y\rho\ \text{d}A}{A\rho} = \dfrac{\int y\ \text{d}A}{A} = \overline{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for uniformly-dense surfaces (or volumes), the centroid and the center of gravity are the same. For non-uniformly-dense surfaces, they are not.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-01-01T16:52:32.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6732" PostTypeId="1" CreationDate="2016-01-01T19:11:36.723" Score="2" ViewCount="77" Body="&lt;p&gt;I am simulating an axial turbine. In this turbine, there are two domains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first domain is the piping before and after the blade and the second domain contains the turbine blades. The domain containing the blades rotates at 2000 rpm=209.43 rad/s while the other one is stationary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7LPos.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7LPos.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There three pair of rotational periodic faces in numerical model( one is for inlet stationary domain and other one is outlet stationary domain and other one is for rotation domain). In fact, there are three pairs faces parallel to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I validated torque of turbine with experimental values but pressure drop at inlet and outlet of turbine doesn't validate with experimental value correctly. In fact, pressure drop is law and it should be higher. How do i solve this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am grateful that guide me about this case.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2016-01-03T10:36:13.533" LastActivityDate="2016-01-03T10:36:13.533" Title="problem in the calculation of pressure drop for an axial turbine" Tags="|mechanical-engineering|fluid-mechanics|turbines|cfd|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6735" PostTypeId="1" AcceptedAnswerId="6738" CreationDate="2016-01-02T03:27:18.340" Score="0" ViewCount="669" Body="&lt;p&gt;Suppose there is an accelerating mass along a straight line, and you would want to convert that linear motion into rotary motion. Is it possible to convert that accelerating linear motion into constant speed rotary motion? Could you suggest mechanisms that I can use?&lt;/p&gt;&#xA;" OwnerUserId="4318" LastEditorUserId="1832" LastEditDate="2016-01-02T03:33:22.437" LastActivityDate="2016-01-02T11:44:55.140" Title="Converting accelerating linear motion into constant rotational motion" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="1" CommentCount="6" ClosedDate="2016-01-02T15:21:54.690" ContentLicense="CC BY-SA 3.0" />
  <row Id="6736" PostTypeId="2" ParentId="6667" CreationDate="2016-01-02T10:01:09.650" Score="2" Body="&lt;p&gt;I think you may be getting a little confused here. There is a fit tolerance, which determines how easy it is to insert and rotate a pin in a joint, and then there is machining tolerance, which is how accurately a shop can match your nominal values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your shop can't meet your printed fit tolerances, you're using the wrong shop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, if you must go with a particular vendor (3d printing, for example), and you have no other choice, then make your nominal value the mean of your MIN and MAX values and then change your tolerance values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for an easy example, say your hole should be 6mm +/- 0.05mm. Find the MIN (6-0.05 = 5.95mm) and the MAX (6+0.05 = 6.05mm), and average them - 6mm. Tolerances remain unchanged and this was a trivial problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider though 6mm +0.1mm -0mm. Now do the same, MIN=6mm, MAX=6.1mm, &lt;em&gt;new nominal value of 6.05mm&lt;/em&gt;, and new tolerance of +/- 0.05mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's functionally the same spec, but the &quot;new&quot; format would get you some dirty looks at a machine shop. However, now you can compare your needs: +/-0.05mm, to the shop capability: +/-0.1mm, and can see that they cannot meet your tolerancing needs and are thus unlikely to produce a satisfactory product. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-01-02T10:01:09.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6737" PostTypeId="2" ParentId="6700" CreationDate="2016-01-02T10:15:45.970" Score="2" Body="&lt;p&gt;I would try using the drag equation to determine the force on the ball. You have the capability to vary air speed, which would change the drag. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the hard part you would have trying to do this would be to characterize the air velocity as a function of distance from the source and angle from the axis of the source. If you assumed the air velocity was constant along the axis, then congratulations - you made a blowgun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you would have so much difficulty that you would be better if trying to determine the air velocity - height relationship empirically and then trying to fit a curve to that data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could even take those empirical results, use the drag equation, and see if that will give you wind speed by distance. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-01-02T10:15:45.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6738" PostTypeId="2" ParentId="6735" CreationDate="2016-01-02T11:44:55.140" Score="2" Body="&lt;p&gt;Do you really mean&lt;br&gt;&#xA;ACCELERATING linear  to&#xA;CONSTANT rotary motion?  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A variable ratio drive as in the DAF cone drive now used in a number of Japanese gearboxes would work.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Provided the accelerating linear motion is repeatable and predictable, below one turn this can be seen as a &quot;cam&quot;.&lt;br&gt;&#xA;The linear motion is coupled to the cam surface where cam radius from centre decreases in inverse ratio to the linear velocity.&lt;br&gt;&#xA;For more than one turn the cam becomes a conic surface with again a radius proportional to inverse of velocity as the linear drive accelerates. This becomes impractical for large variations of linear velocity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Add electronics in the middle and it becomes an alternator-motor set.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="399" LastActivityDate="2016-01-02T11:44:55.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6739" PostTypeId="1" CreationDate="2016-01-02T16:58:15.593" Score="0" ViewCount="682" Body="&lt;p&gt;I'm hoping I'm asking this in the right place. I'm trying to determine how I can improve the water flow within a project I'm working on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BKAQ1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BKAQ1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flow starts out at the large water pump on the bottom left, flows through a radiator and into a water to air intercooler. Cooling the water to air intercooler should warm the water which will then flow to a tee where a part of the water will flow through normally and a small portion of the water gets sucked in by a smaller pump, pushed through an ESC and electric RC motor and continues back to the original pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The large pump flows 500GPH&#xA;the small pump flows 90 GPH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Diameter of the large pump and hoses is 3/4&quot; &#xA;and the diameter for the smaller pump is 5/16&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water should be at 80*F (Or ambient temps) exiting the radiator and probably 170* after the intercooler, ESC, and Motor. (This are estimates as I don't really know the temps post radiator)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a small radiator I could place between the small pump and esc to cool the water more if needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut feeling is that the Tee's with the reducers are a poor choice thus why I'm asking how I could improve this system.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorDisplayName="user16" LastEditDate="2016-01-02T23:43:58.130" LastActivityDate="2016-04-04T04:15:51.643" Title="How can I improve water flow within this diagram?" Tags="|fluid-mechanics|liquid|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6740" PostTypeId="1" CreationDate="2016-01-02T18:23:08.507" Score="2" ViewCount="691" Body="&lt;p&gt;I have a fan-like setup where a 4 mm-thick rod is attached to a small motor-driven shaft (also 4 mm thick) such that the rod turns about its centre as the shaft turns:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lky6c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lky6c.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a closing mechanism for a catflap. It normally can rotate in both directions, but when late at night, it should lock the outward direction but allow the inward direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flap's right rail slides over the rod so when the rod rotates, the flap opens up. The flap's other rail slides over a similar rod, but is connected to a bearing instead. Here's a rough sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IjfdS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IjfdS.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I block the shaft from rotating further clockwise (or the rod, for that matter) regardless of the current position? I'm hoping to have the shaft/rod blocked (using another motor) so that it won't rotate by +5 degrees or so even under some 7 kg-cm torque.&lt;/p&gt;&#xA;" OwnerUserId="4301" LastEditorUserId="368" LastEditDate="2016-01-04T23:01:47.447" LastActivityDate="2016-01-04T23:01:47.447" Title="Locking the shaft of a small fan-like setup" Tags="|mechanical-engineering|torque|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6741" PostTypeId="2" ParentId="6723" CreationDate="2016-01-02T20:31:07.610" Score="4" Body="&lt;p&gt;It is just saying that since there is no turbulence or dissipation being computed; your solution will be less accurate for lower and lower permeability. As I read it; eddy currents in this model will cross the medium. For a thin wire fence this would match reality; for a furnace filter it would not because the air is laminar as it exits the filter. The extent to which this assumption affects your solution depends on your geometry and how much the turbulence affects your measured results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you post on the &lt;a href=&quot;http://www.cfd-online.com/Forums/fluent/&quot; rel=&quot;noreferrer&quot;&gt;CFD-Online Fluent Forum&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some existing forums that may be of use: &lt;a href=&quot;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=porous%20fluent%20site:%20www.cfd-online.com%2F%20site:www.cfd-online.com&quot; rel=&quot;noreferrer&quot;&gt;Porous Fluent CFD-Online search&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to consider an opensource solver for this situation as opposed to closed source fluent that has undesirable assumptions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.openfoam.com/&quot; rel=&quot;noreferrer&quot;&gt;OpenFOAM Website&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.cfd-online.com/Forums/openfoam-solving/135962-porous-media.html&quot; rel=&quot;noreferrer&quot;&gt;A Porous Media in OpenFOAM Forum&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.tfd.chalmers.se/~hani/kurser/OS_CFD_2008/HaukurElvarHafsteinsson/haukurReport.pdf&quot; rel=&quot;noreferrer&quot;&gt;Porous Media in OpenFOAM&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://web.stanford.edu/~csoulain/OF_Training/OPENFOAM_TRAINING_PART_1v5.pdf&quot; rel=&quot;noreferrer&quot;&gt;Porous Media and Heat Transfer In OpenFOAM&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-01-02T20:31:07.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6742" PostTypeId="2" ParentId="6740" CreationDate="2016-01-02T20:52:19.350" Score="1" Body="&lt;p&gt;One method to control a shaft this way would be to use an &lt;a href=&quot;http://www.tinyclutch.com/magnetic-clutches/&quot; rel=&quot;nofollow&quot;&gt;electric clutch&lt;/a&gt; to connect to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Sprag_clutch&quot; rel=&quot;nofollow&quot;&gt;Sprag Clutch&lt;/a&gt;, sometimes called a &quot;&lt;a href=&quot;http://www.vxb.com/page/bearings/CTGY/One-Way-Bearings&quot; rel=&quot;nofollow&quot;&gt;one-way bearing&lt;/a&gt;&quot;. However, this would be a little expensive for your task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method that may be more appropriate for you application is to use a plain hinge at the top (a piano hinge would work well), and a &lt;a href=&quot;https://learn.adafruit.com/secret-knock-activated-drawer-lock/installation&quot; rel=&quot;nofollow&quot;&gt;solenoid spring latch&lt;/a&gt; near the middle or bottom (chose location based on toque requirements). There is a &lt;a href=&quot;http://www.aliexpress.com/wholesale?&amp;amp;SearchText=solenoid%20locks&quot; rel=&quot;nofollow&quot;&gt;wide variety&lt;/a&gt; of different types. You could also construct your own from a plain solenoid if necessary. Just have your timer/control engage the solenoid when you want the one-way operation to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots more options if the design is changed to sense the inbound cat and control the system accordingly.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2016-01-02T22:16:44.417" LastActivityDate="2016-01-02T22:16:44.417" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6743" PostTypeId="2" ParentId="6712" CreationDate="2016-01-02T21:34:21.737" Score="0" Body="&lt;p&gt;This &lt;a href=&quot;https://engineering.stackexchange.com/questions/6696/design-of-a-four-bar-infinity-coupler-linkage/6697#6697&quot;&gt;Engineering SE Question&lt;/a&gt; explains why a single parallel link will not work. When the link is at &lt;a href=&quot;https://en.wikipedia.org/wiki/Dead_centre_(engineering)&quot; rel=&quot;nofollow noreferrer&quot;&gt;top dead center&lt;/a&gt; there is nothing keeping the wheels in time. The mechanism can switch to a lemniscate mode which will make the wheels rotate in opposite directions. To maintain a parallel mode with just links you have to use a second link 90 degrees out of phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the following diagram there are two rotating shafts connected by a blue link and a red link. The two links are 90 degrees out of phase. The red link in its current position can not transmit any torque; this is called  &lt;a href=&quot;https://en.wikipedia.org/wiki/Dead_centre_(engineering)&quot; rel=&quot;nofollow noreferrer&quot;&gt;top dead center&lt;/a&gt;. If only the red link was here; the wheels would be permitted to go in opposite directions. With the addition of the blue link however, the shafts must continue to rotate in the same direction. Correspondingly, when the shafts rotate 90 degrees, the red link transmits the load while the blue link passes top dead center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZCHGJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZCHGJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steam engines use links 90 degrees out of phase to couple the two wheels. It can not be seen from one side of the engine though. The blue link is on one side of the engine, and the red is on the other. You can see this in Dave Tweed's answer by looking at the counterweight positions on the engine at the right (there are two separate steam engines in the photo).&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-01-02T21:34:21.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6744" PostTypeId="1" AcceptedAnswerId="6757" CreationDate="2016-01-03T01:19:18.717" Score="4" ViewCount="95" Body="&lt;p&gt;Recently I watched &lt;a href=&quot;http://www.imdb.com/title/tt3659388/&quot; rel=&quot;nofollow&quot;&gt;The Martian&lt;/a&gt; and the space vehicle &lt;em&gt;Hermes&lt;/em&gt; in the movie had a rotating platform. I wonder if that would significantly cause the vehicle to be inclined in direction? Does a spinning platform/living space on a space vehicle, used to provide artificial gravity, create momentum that inclines the trajectory of that vehicle?&lt;/p&gt;&#xA;" OwnerUserId="4331" LastEditorUserId="368" LastEditDate="2016-01-04T22:59:02.527" LastActivityDate="2016-01-04T22:59:37.387" Title="Does a rotating platform on a space vehicle affect the vehicle's trajectory?" Tags="|mechanical-engineering|aerospace-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6745" PostTypeId="1" CreationDate="2016-01-03T05:48:09.547" Score="1" ViewCount="2365" Body="&lt;p&gt;I want to determine how much force in Newtons will be need to bend a solid cylinder of magnesium.  The cylinder is (x) cm in diameter,  and it is (l) cm long.  The density of normal magnesium alloy is 1.8 grams per cubic cm. The yield strength is 130 MPa, the tensile strength is 220 MPa, the Young's modulus is 45 GPa, and the melting temperature is around 615 degrees Celsius.  All the information about magnesium alloy can be found &lt;a href=&quot;http://mg.tripod.com/asm_prop.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the amount of force needed to bend a magnesium cylinder with variable diameter and length, or the minimum force needed to break the magnesium cylinder by bending? If I had a general formula, I could then use Newton's law to find how much mass can be put on the cylinder in order to bend it or break it by bending, based on the gravity of the planet Earth. &lt;/p&gt;&#xA;" OwnerUserId="4333" LastEditorUserId="368" LastEditDate="2016-01-04T22:55:12.927" LastActivityDate="2016-02-03T23:04:43.660" Title="How can I find the amount of force needed to bend and break a magnesium cylinder?" Tags="|mechanical-engineering|metals|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6749" PostTypeId="2" ParentId="590" CreationDate="2016-01-03T17:23:32.303" Score="2" Body="&lt;p&gt;This is a rather complex question as there are multiple variables and many proposed designs for thorium fuel cycles to consider as well; but it seems your primary interest is in whether or not leaving the Pa-233 in solution will adversely affect the thorium nuclear cycle to the point where it makes more sense to remove this isotope to be reintroduced after it beta decays to our much needed U-233 fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer this question briefly let's first assume a thermal reactor (as in the neutrons are well moderated and have energies ideal for fission of U-233). Next let's make an assumption about composition with 98% Th-232, 1% Pa-233 and 1% U-233.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross sections of each of these isotopes (how 'big' they are to a thermal neutron) are approximately: Th-232, 7.37 barns for absorption; Pa-233, 40 barns for absorption; U-233, 529 barns for fission. If you do not know what a 'barn' is, basically it is nothing more than describing the 2D size of the target nuclei as far as having an interaction with the incoming neutron. 1 barn = 10&lt;sup&gt;-24&lt;/sup&gt; cm&lt;sup&gt;2&lt;/sup&gt; and was named such because on atomic scales, as the old saying goes, &quot;...is as big as a barn.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This information can be used to derive the average distance a neutron will travel before it has a 'collision/interaction' with one of these atoms (also &#xA;known as the transport mean free path). The function is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l=\frac{1}{\sigma N-\frac{2}{3A}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$l$ is the transport mean free path;&lt;/li&gt;&#xA;&lt;li&gt;$\sigma$ is the isotope cross section;&lt;/li&gt;&#xA;&lt;li&gt;$N$ is the number density of the isotope;&lt;/li&gt;&#xA;&lt;li&gt;$A$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass_number&quot; rel=&quot;nofollow&quot;&gt;number of protons and neutrons in the isotope&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since they are all very similar in number of protons and neutrons we can eliminate the $\frac{2}{3A}$ term. Also, this function is primarily used for scattering and calculating energy loss of a neutron through a given depth of material but it works just as well for absorption leaving us with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$l \sim \frac{1}{\sigma N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula gives the average (ish) distance a neutron will travel through a material before having an interaction with an atom (absorption, fission, scattering, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With some quick number crunching (skipping the exact number densities and going with the % of compositions) we can easily see the average distance traveled by the neutron is over an order of magnitude shorter for the U-233 and Th-232 vs. the Pa-233 isotope so its effects on the 'efficiency' of this reactor would be negligible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does Pa-233 formation affect the reactor efficiency? &lt;strong&gt;Yes.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Is it critical to remove Pa-233 to have a viable thorium fuel cycle? &lt;strong&gt;No.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Does the geometry of the reactor effect efficiency? Yes, but that is a whole other question. ;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="553" LastEditorUserId="553" LastEditDate="2016-01-04T21:48:56.170" LastActivityDate="2016-01-04T21:48:56.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6750" PostTypeId="1" CreationDate="2016-01-03T20:04:42.293" Score="-1" ViewCount="102" Body="&lt;p&gt;As part of my final year project, I am designing a racehorse simulator. I want to use 2 hydraulic actuators - one at the front and one at the back to simulate jumping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any suggestions on what I need to make this happen and how to control it? I have no experience using hydraulics. The shaded area under the simulator is where I intend to attach them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ihnQd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ihnQd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3113" LastEditorDisplayName="user16" LastEditDate="2016-01-03T21:01:06.460" LastActivityDate="2016-01-03T21:53:17.647" Title="Hydraulic cylinders" Tags="|control-engineering|hydraulics|" AnswerCount="2" CommentCount="1" ClosedDate="2016-01-04T02:09:45.680" ContentLicense="CC BY-SA 3.0" />
  <row Id="6751" PostTypeId="2" ParentId="6750" CreationDate="2016-01-03T20:27:40.300" Score="1" Body="&lt;p&gt;Hydraulics normally run very high pressures and produce very large forces and are normally limited to heavy industries; use pneumatics instead if you want a fluid solution. At the minimum, air cylinder, values, fittings, and airline tubing is needed. If a solenoid instead of manually controlled valve is used, a control system such as programmable logic controller is needed.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-03T20:27:40.300" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6752" PostTypeId="2" ParentId="6750" CreationDate="2016-01-03T21:47:25.707" Score="3" Body="&lt;p&gt;Hydraulic (or pneumatic) cylinders are merely a means to provide a linear force between two points. You select the cylinder based on how much force, how much distance and how much speed you need for the relative motion of those two points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is up to you, the designer, to constrain that force through additional mechanical linkages in order to get the specific motion you desire. For example, if you study a typical backhoe digger, you'll see that each cylinder forms one side of a triangle. The other sides are rigid links that control how the changing length of the cylinder produces the desired motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f6/JCB_3CX_Backhoe_loader.jpg&quot; alt=&quot;photo of backhoe&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step for your simulator is to figure out exactly what movements you want to produce, including a consideration of the accelerations you intend to use, then design the linkages that will allow those movements while not allowing unwanted movements. Only then can you consider how to power those movements with cylinders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your question stands, this is a very broad topic &amp;mdash; probably too broad for StackExchange's Q and A format.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2016-01-03T21:53:17.647" LastActivityDate="2016-01-03T21:53:17.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6754" PostTypeId="2" ParentId="6745" CreationDate="2016-01-03T21:54:23.967" Score="1" Body="&lt;p&gt;The simple answer: Use &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow&quot;&gt;classical beam theory&lt;/a&gt; to calculate the load distribution of your cylinder with your loads applied. Then compare against the strength of your material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more complex answer: &lt;a href=&quot;https://en.wikipedia.org/wiki/Strength_of_materials&quot; rel=&quot;nofollow&quot;&gt;Strength of materials&lt;/a&gt; is quite a large subject on its own, and providing a satisfactory answer to your question requires significant more information on your application. Why do you want to predict the behavior? Is it just for the sake of prediction? Do you want to prevent the cylinder from permanent deformation or breaking? Or do you want to deliberately deform and break the cylinder? Do you need to consider safety factors? How exactly is the normal force applied? How exactly is the cylinder mounted? How well do you know your material properties?&lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="368" LastEditDate="2016-01-04T22:56:37.867" LastActivityDate="2016-01-04T22:56:37.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6755" PostTypeId="1" CreationDate="2016-01-03T23:05:30.920" Score="0" ViewCount="55" Body="&lt;p&gt;I am looking to learn more about how engineers efficiently search for solutions to problems, particularly mechanical engineers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computer scientists sometimes speak of &quot;search space reduction&quot; when they are trying to make an algorithm more tractable. For example, if a search space is small, trial-and-error or &quot;brute force&quot; may be a very efficient method to implement because it is easy to execute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do engineers have a similar concept? &lt;/p&gt;&#xA;" OwnerUserId="648" LastActivityDate="2016-01-04T22:50:42.730" Title="Is there an equivalent concept in engineering to CS's &quot;search space reduction&quot;?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="6" ClosedDate="2016-01-04T05:41:19.537" ContentLicense="CC BY-SA 3.0" />
  <row Id="6756" PostTypeId="1" AcceptedAnswerId="6760" CreationDate="2016-01-04T01:28:09.690" Score="6" ViewCount="16049" Body="&lt;p&gt;I've been designing an electric skateboard which is driven by only one of the rear wheels (for the moment). As I was picking out motors for the board, I soon realized I would need to find a balance between torque needed to climb hills, and RPM to get a good top speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm searching for the best KV (RPM/v) motor/gearing ratio combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all calculations I've assumed the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Total mass (board + rider) = 85 kg&lt;/li&gt;&#xA;&lt;li&gt;Voltage = 12 Cells * 3.2V (LiFePo4) = 38.4 V&lt;/li&gt;&#xA;&lt;li&gt;Max Speed = 40 km/h = 11.11 m/s&lt;/li&gt;&#xA;&lt;li&gt;Hill Grade = 15%&lt;/li&gt;&#xA;&lt;li&gt;Coefficient of friction between rubber and asphalt: 0.65&lt;/li&gt;&#xA;&lt;li&gt;Diameter of wheel = 80 mm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I tried assuming maximum kinetic energy as being equal to maximum gravitational potential energy at the top of the hill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_{k,max} = \frac{1}{2}mv_{max}^2 = E_g = mgh_{max}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then solving for $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd take that and use $y/x = 0.2$, solve for $x$, then use some trigonometry to setup a Newton's Second Law problem and solve for force needed to overcome friction and gravity (Equilibrium problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using $T = F \times r$, find torque needed at the wheels, assume some gearing ratio that keeps that and the $V_{max}$. Then solve for RPM and divide by total voltage for motor KV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My mechanics are rather rusty and I could use some help.&lt;/p&gt;&#xA;" OwnerUserId="4346" LastEditorUserId="421" LastEditDate="2016-01-04T02:08:11.697" LastActivityDate="2020-03-06T14:12:18.287" Title="Calculating torque needed to climb a graded hill" Tags="|electrical-engineering|motors|applied-mechanics|energy|kinematics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6757" PostTypeId="2" ParentId="6744" CreationDate="2016-01-04T01:52:50.837" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum&quot; rel=&quot;nofollow&quot;&gt;Conservation of angular momentum&lt;/a&gt; creates a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gyroscope&quot; rel=&quot;nofollow&quot;&gt;gyroscopic force&lt;/a&gt; when the angle of the space craft is changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trajectory however is independent of orientation. You could change the velocity of your spacecraft (speed and direction) without changing the orientation (angle relative to the universe). This could be done with multiple rocket engines firing at different amplitudes to get the desired thrust vector or with a rocket engine that could be re-positioned about the crafts center of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If changing the orientation of the craft is important for engine positioning, solar, or other; 2 living spaces could be used rotating in opposite directions. The gyroscopic force would still put a large load on bearings however, but slow changes in orientation would be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option would be to have large control gyroscopes that have rotational inertia at the same order of magnitude as the rotating living space. These could be rotated to counteract the living space angular momentum and control the angle of the craft.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="368" LastEditDate="2016-01-04T22:59:37.387" LastActivityDate="2016-01-04T22:59:37.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6758" PostTypeId="2" ParentId="6755" CreationDate="2016-01-04T02:17:17.227" Score="1" Body="&lt;p&gt;&quot;Trial and error&quot; and &quot;brute force&quot; are terms I frequently use in engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thomas Edison is famous for his &quot;brute force&quot; invention of a long life incandescent light bulb. Other incandescent light bulbs had been created before him, but burnt out quickly. He was reasonably certain a long lasting bulb could be created if he had the right filament material; so his company proceeded to test every combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However most of the time engineers use physics and industry knowledge to more efficiently solve problems. The Hoover Dam was engineered based on physical principles and past experimentation so that it only needed built one time.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="368" LastEditDate="2016-01-04T22:50:42.730" LastActivityDate="2016-01-04T22:50:42.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6759" PostTypeId="2" ParentId="6755" CreationDate="2016-01-04T02:18:59.653" Score="2" Body="&lt;p&gt;I think the analogy does not fit, because of a confusion between levels of abstraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you need to attach a plastic bar and an aluminium rod. Solving this engineering problem involves deciding whether to use a mechanical fastener, or a joint, or an adhesive, or whatever else; or experimenting with a few different methods until you find something that works. This is a &lt;em&gt;design&lt;/em&gt; level issue. Influencing factors might be things like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What loads that will be applied?&lt;/li&gt;&#xA;&lt;li&gt;Is it visible and does it need to look neat?&lt;/li&gt;&#xA;&lt;li&gt;What environmental conditions will it be subject to? (underwater or in air, temperatures, pressures, etc.).&lt;/li&gt;&#xA;&lt;li&gt;What wear will it take and how long does it need to last?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now imagine you need to find all 5-digit integer palindromes that are a multiple of an input integer &lt;code&gt;&amp;lt;n&amp;gt;&lt;/code&gt;. Solving this computer science problem involves choosing a programming language/platform, selecting an algorithm, and writing some code; or experimenting with a few different methods until you find something that works. This is also a &lt;em&gt;design&lt;/em&gt; level issue. Influencing factors might be things like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How often will be it called and how fast does it need to perform?&lt;/li&gt;&#xA;&lt;li&gt;Is this a sub-routine of a larger program or a stand-alone program?&lt;/li&gt;&#xA;&lt;li&gt;How will it be called? (Function, command line, HTTP request, etc.)&lt;/li&gt;&#xA;&lt;li&gt;What sort of output should it produce - who are the consumers?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Reducing the search space in computer science is at a different level of abstraction to the above design tasks. Say for example you write the program in Python and you run it as a stand-alone program from the command line so that a human can view the output. A crude implementation might be sufficient:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/python&#xA;&#xA;def isPalindrome(n):&#xA;    return str(n) == str(n)[::-1]&#xA;&#xA;def isMultiple(n, d):&#xA;    return (n % d) == 0&#xA;&#xA;for i in xrange(10000,99999+1):&#xA;    if isMultiple(i, 17) and isPalindrome(i):&#xA;        print i;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This was very quick to write, and there is no need to optimise or improve it because doing so would be a waste of time considering how it is intended to be used. The analogous solution in the engineering problem is to use a glue-gun or some tape because the plastic bar is only there to stop the sliding aluminium rod form moving around during transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what if those design requirements changed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the palindrome generator needs to be called many times per second from an app with a large user-base over an HTTP connection then I may consider a different approach re. platform, language, interface, etc. but I would also certainly consider reducing the search space of the algorithm. Instead of a loop over all integers with two tests, I'd only consider multiples of &lt;code&gt;&amp;lt;n&amp;gt;&lt;/code&gt; to start with by starting at the first one and adding &lt;code&gt;n&lt;/code&gt; each time. This is an optimisation or &lt;em&gt;performance&lt;/em&gt; level issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if the plastic bar had to attach to the aluminium rod in order to prevent a user-facing keypad panel from deflecting during general use over a 5 year lifetime then I may consider a different connection type, such as a simple shaped joint with adhesive, and I would certainly consider the equivalent &lt;em&gt;performance&lt;/em&gt; level issue of what adhesive to use. I would look up adhesive strengths for both material types and effective operating conditions depending on what environment it would be used in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you see, the analogy doesn't really work. Reducing the search space is highly specific to both the task at hand and how it's implemented. It's about trivial rejection, to eliminate (or not consider in the first place) unsuitable candidates. CS: Reduce the total number of tests that need to be performed; EN: Eliminate unsuitable adhesives. Both highly specific optimisations of the task at hand. What you're talking about is a general purpose way to reduce the &quot;search space&quot; of possible &lt;em&gt;design&lt;/em&gt; solutions. Which doesn't really make sense.&lt;/p&gt;&#xA;" OwnerUserId="95" LastEditorUserId="368" LastEditDate="2016-01-04T22:50:04.067" LastActivityDate="2016-01-04T22:50:04.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6760" PostTypeId="2" ParentId="6756" CreationDate="2016-01-04T05:29:10.320" Score="13" Body="&lt;p&gt;I think you're overcomplicating things. To push 85 kg up a 15% slope against gravity of 9.8 m/s&lt;sup&gt;2&lt;/sup&gt; requires a force of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sin (\arctan (0.15) ) = 0.1483 \approx 0.15  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = 85 kg \cdot 9.8 \frac{m}{s^2} \cdot 0.15 = 125 N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an 80 mm wheel, this requires a torque of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = 125 N \cdot 0.04 m = 5 Nm$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this at a forward speed of 11.11 m/s requires a fair amount of power:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = Force \cdot Velocity = 125 N \cdot 11.11 \frac{m}{s} = 1400 W$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is about 36 A from your 38.4 V battery pack, and your wheels will be turning at&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{11.11 \frac{m}{s}}{\pi\cdot 0.08 \frac{m}{rev}} \cdot 60 \frac{sec}{min} = 2600 RPM$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does any of this help you move forward?&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="4346" LastEditDate="2016-01-04T20:57:26.740" LastActivityDate="2016-01-04T20:57:26.740" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="6761" PostTypeId="2" ParentId="6664" CreationDate="2016-01-04T05:54:38.773" Score="1" Body="&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Duplexer&quot; rel=&quot;nofollow&quot;&gt;this wikipedia article&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A duplexer is an electronic device that allows bi-directional (duplex) communication over a single path. In radar and radio communications systems, it isolates the receiver from the transmitter while permitting them to share a common antenna. Most radio repeater systems include a duplexer. Duplexers can be based on frequency (often a waveguide filter), polarization (such as an orthomode transducer), or timing (as is typical in radar).[1]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4349" LastActivityDate="2016-01-04T05:54:38.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6763" PostTypeId="1" CreationDate="2015-12-31T22:59:23.020" Score="1" ViewCount="3379" Body="&lt;p&gt;I could use a hand choosing the mechanism for joining the ends of 2 flat metal bars (each around 3/4&quot; wide and 1/8&quot; thick) so that they pivot freely but without play.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm currently planning is to drill a 1/8&quot; hole in one and a 1/4&quot; hole in the other, put a 1/8&quot; diameter rivet through the bar with the 1/8&quot; hole, then a thin plastic (Teflon maybe) washer, put a Teflon plain sleeve bearing (looks like a tiny 1/8&quot; long pipe with a .126&quot; I.D. and a 1/4&quot; O.D.) into the 1/4&quot; hole, pass the rivet through that, then through another thin plastic washer, a thin metal washer, and then close the rivet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's important to avoid metal-on-metal contact, but I'm not certain about that.  Is there a simpler design that I could use?&lt;/p&gt;&#xA;" OwnerUserId="4410" OwnerDisplayName="uncaged" LastEditorUserId="33" LastEditDate="2016-01-05T02:12:23.047" LastActivityDate="2016-04-04T04:02:19.077" Title="What is a good method of creating a pivot point between flat metal bars?" Tags="|robotics|joining|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6764" PostTypeId="2" ParentId="6763" CreationDate="2016-01-01T18:22:36.737" Score="1" Body="&lt;p&gt;Thrust bearings and thrust washers are made for this exact purpose.   Your design intent is pretty close to that of a thrust bearing but more complex.  &lt;/p&gt;&#xA;" OwnerDisplayName="SteveO" LastActivityDate="2016-01-01T18:22:36.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6765" PostTypeId="2" ParentId="6664" CreationDate="2016-01-04T12:25:48.290" Score="1" Body="&lt;p&gt;The issue isn't necessarily &quot;switching&quot; between reception and transmission, but making sure the receiver doesn't get damaged by the high voltage put onto the antenna by the transmitter.  Since the higher level system knows when it's transmitting and when its receiving, it simply ignores the receiver during transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the problem of a receiver.  The received RF signals may only be &amp;micro;V, but the transmitter can put 10s or even 100s of volts onto the same node.  Note that this problem has been around for a long time, and is not unique to radar.  Two-way radios and &quot;walkie talkies&quot; have this problem, as do cell phones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is generally some kind of switching and/or clamping the signal in the receiver.  In a old walkie talkie I had, the push to talk switch physically switched the antenna between the receiver and transmitter, and the speaker between power amp output and microphone amp input.  Radar systems need to be much faster, so the switch is electronic.  There will also be protection circuitry on the input of the receiver to keep it from getting fried by the inevitable RF that makes it thru the switch even when off.  Another issue is recovery time for the receiver.  Even if no damage occurs, it may take a little while for the receiver to be able to pick up &amp;micro;V signals again after having been subjected to volts.  This all has to be taken into account in the design of the receiver, and the &quot;switch&quot; mechanism may &quot;reach into&quot; the receiver a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At short wavelengths like some radar, it can be possible to use the same pipe bi-directionally.  This usually done with waveguides, so is limited to wavelengths short enough to allow the waveguide to be practical in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no single universal answer to this, but hopefully the above gives you general ideas.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-01-04T12:25:48.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6766" PostTypeId="2" ParentId="6531" CreationDate="2016-01-04T12:32:52.547" Score="1" Body="&lt;p&gt;The piles are not stopping short of the cap.  Of course this depends on your definition.  The wall and top beam form the pile cap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load path is whatever you want it to be.  That's the fun of engineering.  Think about strut and tie models, or truss models.  Follow the loads and make sure that the materials are capable of resisting the stresses at each level.  If you can establish a load path then you will have a stable structure.  Now reinforced concrete is much smarter than all of us, and it will find the most efficient path, but it is up to you to give it at least one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top transfer beam requires at least minimum steel requirements.  Looking at the model in a simple way, one could treat the top beam as a two-span continuous beam between the 3 taller piles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertical bars in the wall also need at least minimum steel reinforcement. Close spacing of smaller bars offer better crack control than large bars spaced further apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two outer piles are battered to offer lateral support to the pile cap.&#xA;I am assuming that for the other direction, the bending capacity of the piles are sufficient for lateral support, this would need to be determined and checked and additional means provided if required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top of the piles may have bearing plates welded to them as well to establish load transfer to the piles. &lt;/p&gt;&#xA;" OwnerUserId="4337" LastEditorUserId="4337" LastEditDate="2016-01-04T23:00:32.763" LastActivityDate="2016-01-04T23:00:32.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6768" PostTypeId="2" ParentId="6670" CreationDate="2016-01-04T13:27:34.160" Score="3" Body="&lt;p&gt;Having spent several years in product development at Kitchen-Aid's parent company, I can guarantee that if there were a less expensive way to provide the same function to the consumer, that is how it would be done. They have whole teams of engineers tasked only with cost reduction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that items are marked as &quot;hand wash only&quot; often have more to do with the detrimental effect of the dishwasher environment on components of the attachment other than the base metal.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2016-01-04T13:27:34.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6769" PostTypeId="1" CreationDate="2016-01-04T14:21:26.773" Score="0" ViewCount="116" Body="&lt;p&gt;I need to configure my code to make retransmission in slotted Aloha in case of collision, how to choose the timer ? &lt;/p&gt;&#xA;" OwnerUserId="3540" LastActivityDate="2016-01-04T14:26:45.443" Title="how to calculate the time for retransmission in slotted Aloha" Tags="|telecommunication|wireless-communication|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6770" PostTypeId="2" ParentId="6769" CreationDate="2016-01-04T14:26:45.443" Score="1" Body="&lt;p&gt;Keep in mind that the other guy (who you collided with) will also try to retransmit. So you need a random factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can start by waiting 1 to 16 slots and if you collide again (6.25% chance both parties picked the same slot) then you wait for 1 to 32 slots and so on.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2016-01-04T14:26:45.443" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6771" PostTypeId="1" CreationDate="2016-01-04T14:44:42.477" Score="1" ViewCount="2426" Body="&lt;p&gt;I have a heater rod inside a brass rod, which is wrapped by PTFE insulation. The dimensions of the assembly are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Heater diameter = 9.5 mm&lt;/li&gt;&#xA;&lt;li&gt;Brass rod diameter = 25 mm&lt;/li&gt;&#xA;&lt;li&gt;Insulation diameter  = 100 mm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When the heater heats the assembly, will the assembly get tighter or looser? The final temperature that the brass rod would reach is 100 °C and the initial temperature is 20 °C.&lt;/p&gt;&#xA;" OwnerUserId="3591" LastEditorUserId="368" LastEditDate="2016-01-04T22:29:29.910" LastActivityDate="2016-01-04T22:32:10.473" Title="Determination of thermal expansion of a hollow cylinder" Tags="|thermodynamics|heat-transfer|thermal-insulation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6772" PostTypeId="2" ParentId="6771" CreationDate="2016-01-04T16:50:22.837" Score="2" Body="&lt;p&gt;Plastics have a thermal expansion rate of approximately 5 to 10 times that of metals. The PTFE will expand the most and the brass expansion is about 50% more than a steel heater, so the assembly will get looser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html&quot; rel=&quot;nofollow&quot;&gt;Here is a link to a chart of linear expansion coefficients.&lt;/a&gt; For a 9.5 mm part with 80 °C temperature change, the steel extends .0082 mm, brass .0142 mm and PTFE .0851 mm.             &lt;/p&gt;&#xA;" OwnerUserId="4311" LastEditorUserId="368" LastEditDate="2016-01-04T22:32:10.473" LastActivityDate="2016-01-04T22:32:10.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6773" PostTypeId="1" CreationDate="2016-01-04T17:00:46.507" Score="3" ViewCount="203" Body="&lt;p&gt;Suppose I have a block of ice resting on an insulated surface in a closed room.  Suppose the room also contains air at a temperature above the ice's melting point.  I want to simulate the transient behavior of the ice melting and flowing (due to gravity) onto the insulated surface over time.  Particularly, I'm interested in tracking the evolution of the interfaces between each phase (solid, liquid, and gas) over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure that the heat equation and navier stokes equations will be involved, but I'm not sure how to adjust them to account for the existence of three phases (solid ice, liquid water, and air) and how to model the transition between phases (e.g. equation of state).  What are the complete set of equations that I need to model this situation?  More importantly, what is the best approach to handle the evolution of the interfaces over time?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2016-01-04T17:46:27.827" LastActivityDate="2016-02-04T17:15:53.220" Title="What system of pde's models ice melt flow?" Tags="|fluid-mechanics|thermodynamics|heat-transfer|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6774" PostTypeId="2" ParentId="6763" CreationDate="2016-01-04T17:20:31.173" Score="1" Body="&lt;p&gt;Use a 1/4&quot; dia x 3/16&quot; long shoulder screw with 1/16&quot; thick washer. Also use nut and washer on the thread. You can get the screw at: &lt;a href=&quot;http://www.mcmaster.com/#90298a530/=10jcicb&quot; rel=&quot;nofollow&quot;&gt;http://www.mcmaster.com/#90298a530/=10jcicb&lt;/a&gt;&#xA;A steel shoulder screw will be much stronger than a rivet and teflon (plastic) bearings. Steel on steel is commonly used in pivots from door hinges to motorcycle chains.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastEditorUserId="4311" LastEditDate="2016-01-04T18:01:00.570" LastActivityDate="2016-01-04T18:01:00.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6775" PostTypeId="2" ParentId="2234" CreationDate="2016-01-04T17:48:16.393" Score="1" Body="&lt;p&gt;There are many companies that sell gearmotors, and their specifications will give output torque for input power. A few gearmotors link are below, and you can compare their ratings to your calculations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.baldor.com/brands/baldor-reliance/products/motors/gearmotors/ac-gearmotors&quot; rel=&quot;nofollow&quot;&gt;http://www.baldor.com/brands/baldor-reliance/products/motors/gearmotors/ac-gearmotors&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.mcmaster.com/#gearmotors/=10jct48&quot; rel=&quot;nofollow&quot;&gt;http://www.mcmaster.com/#gearmotors/=10jct48&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.globalindustrial.com/p/motors/Motors-Gearmotors/dc-right-angle-motors/gearmotor-dc-right-angle-90rpm-cm34d17nz35a?infoParam.campaignId=T9F&amp;amp;gclid=Cj0KEQiAzai0BRCs2Yydo8yptuIBEiQAN3_lFq10KrIML5XDwgHC3HtQlVn68AaBDrbCJwKGaV-TxBYaArPz8P8HAQ&quot; rel=&quot;nofollow&quot;&gt;http://www.globalindustrial.com/p/motors/Motors-Gearmotors/dc-right-angle-motors/gearmotor-dc-right-angle-90rpm-cm34d17nz35a?infoParam.campaignId=T9F&amp;amp;gclid=Cj0KEQiAzai0BRCs2Yydo8yptuIBEiQAN3_lFq10KrIML5XDwgHC3HtQlVn68AaBDrbCJwKGaV-TxBYaArPz8P8HAQ&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-04T17:48:16.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6776" PostTypeId="1" AcceptedAnswerId="6778" CreationDate="2016-01-04T18:18:02.730" Score="6" ViewCount="1163" Body="&lt;p&gt;We are testing a new ultrasound product using a Texas Instruments PGA450 automotive transducer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three of us are sharing a lab and we need some kind of ultrasound-blocking partition walls around each product to prevent interference. Think cubicle walls that absorb ultrasound.&lt;/p&gt;&#xA;" OwnerUserId="2169" LastEditorUserId="368" LastEditDate="2016-01-04T22:25:59.820" LastActivityDate="2017-05-06T13:27:18.743" Title="How can I block ultrasound from an automotive transducer at about 8'?" Tags="|mechanical-engineering|electrical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6777" PostTypeId="2" ParentId="6531" CreationDate="2016-01-04T18:30:33.683" Score="2" Body="&lt;p&gt;As @Frank mentions in his answer, the piles aren't stopping short of the cap since the piles are more than sufficiently embedded in the wall. This therefore means that the forces go through the topping beam, down the wall and then into the piles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, if you're going to make the wall axial-load-bearing, I don't see the purpose for raising the piles so far into the wall. Steel piles usually only need around 1-2 ft of embedment length into the cap for effective load transfer. Here's a diagram provided by a Brazilian pile supplier (Gerdau-AçoMinas):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YyJQh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YyJQh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very few codes actually define minimal embedment lengths, but most seem to hover around 1-2 ft:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.udot.utah.gov/main/uconowner.gf?n=1838506422268074&quot; rel=&quot;nofollow noreferrer&quot;&gt;This thesis&lt;/a&gt; suggests (for cylindrical steel piles) 1 ft with reinforcement or 2 ft without.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://scholarsarchive.byu.edu/cgi/viewcontent.cgi?article=3627&amp;amp;context=etd&quot; rel=&quot;nofollow noreferrer&quot;&gt;This thesis&lt;/a&gt; has experimental results pointing towards 1.5 pile diameters.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://wisdotresearch.wi.gov/wp-content/uploads/tsrpilefootingfixity.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This American state survey&lt;/a&gt; lists multiple different state's suggestions and most are up to 2 ft (only NE, NJ, ND suggest 3 ft).&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.dtic.mil/dtic/tr/fulltext/u2/a138862.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This report&lt;/a&gt; concludes that embedment should be equal to two pile diameters or heights.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, all of these assume that the pile is embedding into an actual cap, which must be sufficiently wide in both horizontal directions. For an example, here's another diagram from the same supplier (basically, there must be 25 cm of concrete around the pile):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5Eyy5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Eyy5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if your wall isn't thick enough to count as a standard cap, I'd guess the extended embedment length allows for the forces to be transmitted to the pile at lower stresses along the wall. No idea how to quantify that, though. Personally I'd rather just thicken the foot of the wall and leave the piles at the minimal embedment length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load paths are simple enough. The vertical loads obviously just go right through to the piles (I'm assuming the girders lie directly above the piles). The horizontal loads go from the girders to the bearing pads to the topping beam and then, via &lt;em&gt;shear friction&lt;/em&gt;, go to the wall (as you mention, the (c) bars would be responsible for this) and then are transferred to the piles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the reinforcements, it's hard to tell without dimensions and loading: are they close to the minimal reinforcement? Is the anchorage length of the (c) bars sufficient?&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-01-04T18:30:33.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6778" PostTypeId="2" ParentId="6776" CreationDate="2016-01-04T18:57:40.077" Score="4" Body="&lt;p&gt;Typical towels do a pretty good job of not echoing ultrasound around too much, and also attenuate ultrasound going thru them.  Ordinary sheetrock walls are really bad because they are great mirrors for ultrasound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are special materials for absorbing sounds at various frequencies, but hanging a bunch of large beach towels from the ceiling around each work area will likely be cheaper and just as effective.  Two or three towels an inch or so apart should help a lot.  You will still either have to put sound absorbent material on the ceiling, or make sure the towels extend to the ceiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining path will then be by bouncing off the floor.  That's not so simple.  Some &quot;area rugs&quot; should help, but some of them reflect ultrasound much better than others.  Ideally you want something that has roughness extending over &amp;frac12; wavelength or so, but that will be hard to find.  Beach towels on the floor will work to attenuate ultrasound, but won't last long as they aren't designed to take that sort of abuse.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2016-01-04T22:24:38.550" LastActivityDate="2016-01-04T22:24:38.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6779" PostTypeId="1" CreationDate="2016-01-04T19:58:41.703" Score="-1" ViewCount="249" Body="&lt;p&gt;What kind of formulas/equations are commonly used to implement Stewart Platforms in electronics and mechanics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a co-ordinate system, how would you determine the position of each actuator, etc?&lt;/p&gt;&#xA;" OwnerUserId="4281" LastActivityDate="2016-01-05T12:15:08.200" Title="Stewart Platform Formulas" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2016-01-05T19:09:58.520" ContentLicense="CC BY-SA 3.0" />
  <row Id="6780" PostTypeId="1" AcceptedAnswerId="6781" CreationDate="2016-01-04T20:39:11.923" Score="0" ViewCount="8659" Body="&lt;p&gt;I have a DC motor with a shaft with a diameter of x. It would produce some torque T because torque is proportional to radius. If I stuck a pulley on the shaft of radius 2x, then I would produce twice as much torque, right? And then if I attach this to a gear with a ratio of 2:1, I would again double the torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image to clarify:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ho88X.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This bugs me because it means the larger a pulley I can fit on a shaft the more torque I get. What am I missing? Are the RPMs reduced?&lt;/p&gt;&#xA;" OwnerUserId="4346" LastEditorUserId="368" LastEditDate="2016-01-04T22:19:51.137" LastActivityDate="2016-01-05T17:36:33.397" Title="What is the relationship between drive shaft diameter and torque output?" Tags="|applied-mechanics|gears|torque|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6781" PostTypeId="2" ParentId="6780" CreationDate="2016-01-04T21:10:00.120" Score="6" Body="&lt;p&gt;No.  Your basic problem is that &lt;i&gt;&quot;torque is proportional to radius&quot;&lt;/i&gt; is wrong.  Torque is the expression of normalized rotational force independent of radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if a motor can deliver 10 Nm torque, then a pulley with 1 m radius can produce 10 N of force, at 500 mm radius 20 N, at 2 m radius 5 N, etc.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-01-04T21:10:00.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6782" PostTypeId="2" ParentId="5494" CreationDate="2016-01-04T21:22:18.980" Score="2" Body="&lt;p&gt;For the structure of a trailer rectangular section tube is likely to be the most efficient compromise between strength stiffness and ease of design and manufacture. Round tube is a bit stronger weight for weight but much more difficult to assemble and join accurately, simply because rectangular tube has convenient flat surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already mentioned thing like this aren't designed by calculus in the real world and by far your best bet is to copy an existing design as failures in this sort of application tend to occur when you get unexpected load concentrations rather than considering the design as an approximated beam so unless you have access to FEA software then paper calculations are a bit pointless. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-04T21:22:18.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6783" PostTypeId="1" AcceptedAnswerId="6787" CreationDate="2016-01-04T21:27:18.527" Score="-1" ViewCount="173" Body="&lt;p&gt;I'm having trouble identifying the diameter of the reinforced steel bars in the structural steel design for a slab and a beam of a building complex. You can see it &#xA;&lt;a href=&quot;https://www.dropbox.com/s/xszs9wurf8gmk65/POS%20300%2C400%2C500%2C600%2C700%2C800%20%281%29.pdf?dl=0&quot; rel=&quot;nofollow&quot;&gt;here (PDF)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm told there's column design in this file, but I only see beam design. Is the diameter number of the steel before or after the diameter sign? Why are there numbers like 12/20 after the diameter sign? Is there any column design here at all?&lt;/p&gt;&#xA;" OwnerUserId="4356" LastEditorUserId="368" LastEditDate="2016-01-04T22:14:16.540" LastActivityDate="2016-01-04T23:53:06.830" Title="Identifying the diameter of reinforced steel bars in a column design schematic" Tags="|civil-engineering|structural-engineering|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6784" PostTypeId="2" ParentId="2478" CreationDate="2016-01-04T21:56:09.647" Score="1" Body="&lt;p&gt;While it's not fundamentally impossible there in no real reason to do it as in most circumstances it will be more cost effective to either make the component in stainless steel throughout or to plate it with some other metal such as chromium or nickel. The problem is that stainless steel are quite complex alloys and plating processes generally work on a molecular level so there is a big challenge in plating the right proportion of the constituent elements onto the mild steel surface in the right structure and proportions as opposed to depositing a layer of just one elemental metal.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-04T21:56:09.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6785" PostTypeId="2" ParentId="6780" CreationDate="2016-01-04T22:30:21.227" Score="1" Body="&lt;p&gt;It is more useful to say that torque is proportional to the ratio of the radii of a pair of pulleys or gears. Also RPM is reduced or increased in the same ratio so a small diameter gear driving a larger one increases torque but reduces RPM at the output shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., if you had a motor producing 100 Nm of torque at 100 RPM connected to a pulley of diameter 100 mm to an output shaft via a pulley with diameter 200 mm, the torque on the output shaft would be 200 Nm but the RPM would be 50 RPM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way, if you ignore friction, gearing systems allow you to have more torque and less RPM or less torque and more RPM but you can't gain any power (in fact you always lose a bit).&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="368" LastEditDate="2016-01-05T17:36:33.397" LastActivityDate="2016-01-05T17:36:33.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6786" PostTypeId="2" ParentId="6725" CreationDate="2016-01-04T23:26:10.330" Score="1" Body="&lt;p&gt;Have you considered modelling this in some software? You can get a pretty good idea of what is occurring without the hassle of wondering if your hand calculations are correct. I set this up in about 10 minutes (but I'm experienced in the software). I did a vertical load of 10 kN where you had your force and I did a plate thickness of 15 mm and a width of 100 mm (probably a little too excessive). You might also want to consider self-weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/76v61.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/76v61.png&quot; alt=&quot;SkyCiv Structural 3D Analysis Software&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used &lt;a href=&quot;https://skyciv.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;SkyCiv Structural 3D&lt;/a&gt; for this analysis. It is affordable but powerful online analysis software. You can sign up for a free 30 day trial. You can get started right away because it's all online. If you sign up let me know your username and I'll send you the file that I used to do the analysis and you can make the appropriate adjustments to suit your problem.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="368" LastEditDate="2016-01-05T22:46:29.463" LastActivityDate="2016-01-05T22:46:29.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6787" PostTypeId="2" ParentId="6783" CreationDate="2016-01-04T23:53:06.830" Score="0" Body="&lt;p&gt;The column design schematics are &quot;POS S1&quot; and &quot;POS S2&quot;, on the bottom center-right of the page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nomenclature here has some peculiarities I'm unsure about, so someone else might come up with a better answer, but here goes... You'll see that there are basically two ways to describe the rebar:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Longitudinal bars tend to be in the form &quot;&amp;#10112; $4\ \phi 16\ l=391\ (2)$&quot;: This translates to 4 rebars identified by &amp;#10112; elsewhere which have a 16 mm diameter and are each 391 cm long. I'm not sure what the (2) means.&lt;/li&gt;&#xA;&lt;li&gt;Transversal bars are described similarly (when their dimensions are shown), but they are then often repeated in this drawing in a manner I don't quite understand: &quot;&amp;#10116; $18\ \phi 8/10/20$&quot;. I don't understand what the $/10/20$ is supposed to mean. That being said, when their positions along the column are shown as &amp;#10116; $14\ \phi 8/20$, that just means that there are 14 cages made of 8 mm rebar distributed with a 20-cm spacing.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-01-04T23:53:06.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6788" PostTypeId="1" AcceptedAnswerId="6796" CreationDate="2016-01-05T04:38:02.783" Score="4" ViewCount="89" Body="&lt;p&gt;The interior girder moment formula for one lane loaded for the AASHTO LRFD method is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;mg^{SI}_{moment}&amp;amp;=\left(1.75+\frac{S}{3.6}\right)\left(\frac{1}{L}\right)^{0.35}\left(\frac{1}{N_c}\right)^{0.45} \\&#xA;&amp;amp;= \left(1.75+\frac{13}{3.6}\right)\left(\frac{1}{100}\right)^{0.35}\left(\frac{1}{3}\right)^{0.45} \\&#xA;&amp;amp;= 0.65\ \mathrm{lane/web}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this formula derived? I have not been able to find the original research paper.&lt;/p&gt;&#xA;" OwnerUserId="252" LastEditorUserId="368" LastEditDate="2016-01-05T16:55:53.983" LastActivityDate="2016-01-05T16:56:45.863" Title="Derivation of the AASHTO formula of interior girder moment" Tags="|civil-engineering|bridges|aashto|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6789" PostTypeId="1" CreationDate="2016-01-05T07:33:17.260" Score="2" ViewCount="1793" Body="&lt;p&gt;I am currently designing a winch for a submersible vehicle and need to implement a self reversing screw shaft in it for ensuring proper spooling of the tether cable. It's also known as diamond screw shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any mechanical design handbook or guidelines from where i can get the design details for design of such a screw shaft ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To know what is a self reversing screw, please refer to this &lt;a href=&quot;http://www.abssac.co.uk/p/Self+reversing+screws/8/#.VotuS_n5i01&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4361" LastActivityDate="2016-01-05T07:33:17.260" Title="Design of a Self Reversing Screw" Tags="|mechanical-engineering|design|" AnswerCount="0" CommentCount="2" ClosedDate="2016-01-05T16:47:09.320" ContentLicense="CC BY-SA 3.0" />
  <row Id="6790" PostTypeId="1" AcceptedAnswerId="6797" CreationDate="2016-01-05T09:11:59.517" Score="2" ViewCount="295" Body="&lt;p&gt;During the '80s and the '90s, a lot of media cases were made of thermoformed PVC. I believe this trend was created by Disney and Blackbourn with their VHS cases. Then with DVDs, industrial standards for media casing have changed in favor of injection moulded PP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the latest Neo Geo AES cases are made of PP plastic but early Neo Geo AES cases were completely different. Internal trays were clearly made through thermoforming, as they have strong light reflection properties, but I'm not sure how external cases were made. They give the impression of being made of leather but are certainly some kind of plastic imitation. How were the external cases processed? Is it also PVC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JRj0A.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JRj0A.jpg&quot; alt=&quot;Alpha Mission complete set: case, manual and cartridge&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4362" LastEditorUserId="368" LastEditDate="2016-01-05T16:40:14.910" LastActivityDate="2016-01-05T16:54:16.590" Title="How can I process plastic to give the impression of a leather finish?" Tags="|materials|process-engineering|plastic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6791" PostTypeId="2" ParentId="6725" CreationDate="2016-01-05T09:58:36.210" Score="1" Body="&lt;p&gt;The other answers suggest the use of FEA software. However this answer will instead focus on an analytical approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the main difficulties of the problem you are try to solve is that it is a &lt;em&gt;statically indeterminate&lt;/em&gt; problem. In other words, the structure is more constrained than simpler structures (such as a simple cantilever) such that &lt;em&gt;equilibrium&lt;/em&gt; of forces and moments is not enough to solve the structure. This is because, when using equilibrium only, there are more unknown variables than equations. More information is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to solve a statically indeterminate problem, &lt;em&gt;compatibility&lt;/em&gt; must also be enforced, i.e. the structure needs to make sense geometrically. This will make sense later. To enforce compatibility, a method called the &lt;em&gt;Force Method&lt;/em&gt; will be employed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, we need to convert the statically indeterminate problem into a statically determinate problem by making cuts at arbitrary and appropriate locations, simplifying the geometry. So this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/OvC0t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;becomes this:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/PcLuk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, by making these cuts, it makes it seem as if the structure does not interact with each other at the sites of the cuts, like the way cutting a rope means one cut end no longer interacts with the other cut end, allowing the rope to be pulled apart. However, this is not the case, and so we need to add forces and bending moments at the sites of the cuts we have made to account for the interaction of the structure at these regions. In the example of a rope, this is akin to adding tension forces at the cut ends. These 'cut' forces are internal forces, and so, at every cut, the forces and bending moments must be equal and opposite. So the forces and moments to add back are:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/yb3C8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this moment in time, the values of these forces to add back are unknown variables. So, let us add up all the forces up (the externally applied force in our newly formed statically determinate structure + the cut forces) to obtain three cantilevers:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/vwd2R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you can now easily determine the deflections of these three cantilevers individually, like you would for any other simple cantilever, in terms of $F$, plus the unknowns, $P_1, P_2, M_1, M_2$. The problem now is, how do you solve for the four unknowns? We need four equations after we have exhausted our use of &lt;em&gt;equilibrium&lt;/em&gt;! This is the part where you actually enforce &lt;em&gt;compatibility&lt;/em&gt;, by using 'compatibility conditions'. In this case, these are the equation forms of saying &quot;we need each of the cantilevers to bend such that the ends are still the right distance apart from each other&quot;. Because you have a rigid plate connecting the three ends, you know the distance between the ends will not change. So we obtain two 'compatibility conditions':&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta_{1} = \delta_{2}$$&#xA;$$\delta_{2} = \delta_{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\delta$ is the deflection of each of the cantilevers (1,2,3) at the point the cantilever is connected at the plate. Also, note that the cantilevers must have the same rotation at the point connecting the plate, otherwise the rigid plate cannot be straight:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Zv0NB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This provides the remaining two required compatibility conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_{1} = \theta_{2}$$&#xA;$$\theta_{2} = \theta_{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\theta$ is the rotation of each cantilever at the point connecting it to the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you easily determine an expression for the deflections of the system as desired!&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2016-01-05T10:04:32.450" LastActivityDate="2016-01-05T10:04:32.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6792" PostTypeId="2" ParentId="6670" CreationDate="2016-01-05T10:05:53.057" Score="2" Body="&lt;p&gt;Beside very cheap products, the only aluminum parts I have found in my kitchen tools are either injected or molded. Why this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the fusion point of aluminum is much lower than steel, plain steel molds can be easily used. The opposite is not true and for steel casting, it requires disposable sand molds. Thus, the process of steel casting is more complex and way more expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realized that it is not easy to find stainless steel injection molded parts apart from very big stuff such as car rims or machine frames. Almost all steel parts are either machined, laminated, forged or heat pressed, but never injection molded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So engineers will use stainless steel or plastic for any part that can be simply laminated, forged, heat pressed or injected. However, for more complex parts the only remaining choices are: injection molded or CNC machined. The latter is too expensive for high volume production for both steel and aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel injection modeling is neither a solution because of the process complexity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;KitchenAid is probably focused on cost reduction and will choose the most affordable. Using aluminum is the best compromise in terms of cost because aluminum casting is not much expensive and offers good mechanical performances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the same part in stainless steel would make the part five to ten times more expensive. It is worth it, just because you can safely put it in the dishwasher? Certainly not...&lt;/p&gt;&#xA;" OwnerUserId="2070" LastActivityDate="2016-01-05T10:05:53.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6793" PostTypeId="2" ParentId="6779" CreationDate="2016-01-05T12:15:08.200" Score="4" Body="&lt;p&gt;A Stewart platform is a type of parallel manipulator with (usually) 6 prismatic actuators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For controlling the platform, you would firstly need the inverse kinematics equations for this type of manipulator. These would give you the actuator states needed to place the end effector, i.e. the platform in our case, in a certain position. &lt;a href=&quot;http://www.techfak.uni-bielefeld.de/~fszufnar/publications/Szufnarowski2013.pdf&quot; rel=&quot;nofollow&quot;&gt;This paper&lt;/a&gt; might be worth looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how much experience you have in robotics, but I would also recommend Introduction to Robotics: Analysis, Control, Applications by S. B. Niku if you can get your hands on it.&lt;/p&gt;&#xA;" OwnerUserId="4216" LastActivityDate="2016-01-05T12:15:08.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6794" PostTypeId="1" CreationDate="2016-01-05T13:09:16.457" Score="4" ViewCount="299" Body="&lt;p&gt;Does anyone know if typical/common worms and worm gears self-locking (plastic or metal)? Are the self-locking ones specially designed (with specific angles) in order to be self-locking? By self-locking, I mean that the worm gear can't turn the worm but the worm can turn the worm gear. Are they considered &quot;special&quot; worm drives, and thus cost more to make (and thus, less commonly used)?&lt;/p&gt;&#xA;" OwnerUserId="4301" LastEditorUserId="4301" LastEditDate="2016-01-05T13:19:55.750" LastActivityDate="2016-01-05T14:09:01.330" Title="Are typical worms and worm gears self-locking?" Tags="|mechanical-engineering|motors|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6795" PostTypeId="2" ParentId="6794" CreationDate="2016-01-05T14:09:01.330" Score="5" Body="&lt;p&gt;The vast majority of worm gears are self locking naturally, it wouldn't be considered a special feature, just something you should confirm in your final design. Generally any single start worm screw will be self locking, there are multi-start worm screws that may not be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are doing something where backdriving would be hazardous (not just risk minor damage to machinery) you should probably have an independent brake and not rely on the gear mechanism to hold a sustained load.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-01-05T14:09:01.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6796" PostTypeId="2" ParentId="6788" CreationDate="2016-01-05T15:15:30.293" Score="5" Body="&lt;p&gt;The live load distribution formulas in the AASHTO LRFD Bridge Design Specifications cannot be derived. As I understand it, they are based on calibration to extensive finite element modeling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.trb.org/Main/Public/Blurbs/159050.aspx&quot;&gt;This is probably the NCHRP report you're looking for.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="368" LastEditDate="2016-01-05T16:56:45.863" LastActivityDate="2016-01-05T16:56:45.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6797" PostTypeId="2" ParentId="6790" CreationDate="2016-01-05T16:54:16.590" Score="1" Body="&lt;p&gt;For injection molding this is typically done by etching or burning(EDM) a pattern onto the surfaces of the injection mold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;check out this PDF for more detailed information on this:&#xA;&lt;a href=&quot;http://eschmanntextures.com/fileadmin/user_upload/flyer_neu/texturesguide_eng.pdf&quot; rel=&quot;nofollow&quot;&gt;http://eschmanntextures.com/fileadmin/user_upload/flyer_neu/texturesguide_eng.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Search around for injection mold surface texture guide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty of resources out there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;re: Is it PVC? Probably not, but you can do a burn test or send it to a lab for material testing.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-01-05T16:54:16.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6798" PostTypeId="1" CreationDate="2016-01-05T18:45:45.880" Score="4" ViewCount="123" Body="&lt;p&gt;In the article &lt;a href=&quot;http://www.wseas.us/e-library/conferences/2009/lalaguna/MANOCO/MANOCO17.pdf&quot; rel=&quot;nofollow&quot;&gt;Controller Design using Bifurcation Map for Aircraft Spin Recovery&lt;/a&gt;, the sliding surface derivative is obtained using a numerical method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steady space system is defined by the following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\underline{x}}=A(\underline{x},t)+B(\underline{x},t)\underline{U}, \\&#xA;\underline{x}=[V,\alpha,\beta,p,q,r,\phi,\theta], \\&#xA;s=\left[\begin{array}{c}&#xA;\dfrac{d}{dt}(\phi-\phi_d)+\lambda_1(\phi-\phi_d) \\&#xA;\dfrac{d}{dt}(\alpha-\alpha_d)+\lambda_2(\alpha-\alpha_d) \\&#xA;\dfrac{d}{dt}(\beta-\beta_d)+\lambda_3(\beta-\beta_d) \\&#xA;\end{array}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda_1$, $\lambda_2$ and $\lambda_3$ are positive real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The control law from reaching law in Eq. (6) can be derived&#xA;  as:&#xA;  $$\dot{s}=\dfrac{\partial{s}}{\partial{x}}\dot{x}=\dfrac{\partial{s}}{\partial{x}}(A+Bu)=-Qsgn(s)+Kf(s).\tag{8}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By solving the above equation (8) for u (3 x 1), one gets expressions&#xA;  for u as:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$u=-\left(\dfrac{\partial{s}}{\partial{x}}B\right)^{-1}\left[\dfrac{\partial{s}}{\partial{x}}A+Qsgn(S)+Kf(s)\right].\tag{9}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The matrix $\frac{\partial{s}}{\partial{x}}$ is computed numerically.&#xA;  The control inputs computed are subject to position and rate&#xA;  constraints on control surface deflections as listed in the Table 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which numerical method is used to compute $\dfrac{\partial s}{\partial x}$ in $u$?And also how would be the exact form of matrix of $\dfrac{\partial s}{\partial x}$ (element of $\dfrac{\partial s}{\partial x}$ in columns and raws)?&lt;/p&gt;&#xA;" OwnerUserId="4368" LastEditorUserId="68" LastEditDate="2016-01-08T17:21:43.417" LastActivityDate="2016-01-08T17:21:43.417" Title="Sliding mode surface derivative using numerical method" Tags="|control-engineering|aerospace-engineering|numerical-methods|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6802" PostTypeId="1" CreationDate="2016-01-05T23:09:44.553" Score="3" ViewCount="890" Body="&lt;p&gt;In pre-cast bridges, one often lays beams over the columns and then places thin pre-cast &quot;pre-slabs&quot; over the transversal spans between the beams. These slabs already contain all the requisite positive transversal (between-beams) steel reinforcement. All other reinforcements (positive longitudinal, negative transversal and longitudinal) are then placed and &lt;em&gt;In-situ&lt;/em&gt; concrete is poured over the &quot;pre-slabs&quot; up until the requisite thickness. Steel framing contained within the &quot;pre-slabs&quot; certifies that the pre-cast and &lt;em&gt;in situ&lt;/em&gt; concrete work together under ultimate loads. Here's an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5SBqg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5SBqg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how does one calculate the slab's positive transversal reinforcement? The steel must be stressed so as to create a &lt;a href=&quot;https://en.wikipedia.org/wiki/Couple_(mechanics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;force couple&lt;/a&gt; with the resultant of the concrete's compression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the &quot;pre-slab&quot; needs to be able to withstand a load equal to its own self-weight and that of the wet &lt;em&gt;in situ&lt;/em&gt; concrete. If this were calculated by traditional methods, a certain steel area $A_{s,1}$ would be obtained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additional loads (paving, moving loads, etc) are then applied as well once the &lt;em&gt;in situ&lt;/em&gt; concrete has hardened. These loads are therefore resisted by the force couple composed of compression of the &lt;em&gt;in situ&lt;/em&gt; concrete and tension of the reinforcing steel. If this were also calculated by traditional means, a steel area $A_{s,2}$ would be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, is the resultant necessary steel equal to $A_{s,total} = A_{s,1}+A_{s,2}$? I was initially concerned that there might be an interference in the compression regions, but given that the pre-slabs are usually less than half the total slab thickness (in the example above, a 7-cm pre-slab supports 13 cm of &lt;em&gt;in situ&lt;/em&gt; concrete, for a total slab thickness of 20 cm), I don't believe that'd be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, I'm not entirely certain of this conclusion and would like confirmation or correction on this matter. This summation of areas means that all of the steel would be stressed below their ultimate capacity in the first stage (such that $A_{s,total}\sigma = A_{s,1}f_{yd}$) and then would reach the ultimate stress under final loads (such that $A_{s,total}\left(f_{yd} - \sigma\right) = A_{s,2}f_{yd}$). Is this fine or is there some behavior I'm not taking into consideration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, then there's the case where the compression zones of the different stages interfere one with the other, but I'll leave that for a future question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; Here's an example of a pre-slab blueprint, including the steel reinforcement. This case is for the pre-slab including a cantilever, which explains the gap: that's where the beam's transversal steel joins with the &lt;em&gt;in situ&lt;/em&gt; concrete. The units are centimeters. The &quot;criss-cross&quot; in the cross-section is a steel truss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HKgTP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HKgTP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.planasfrance.fr/produit/predalle-en-beton/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This site&lt;/a&gt; describes pre-slabs thusly (according to my rough French translation): &quot;The pre-slab is layer 6 to 8 cm thick, creating a sort of base, whereupon once installed on the site, one must add a compression layer.&quot; It also has the following picture of a real pre-slab, where the trusses responsible for certifying that the pre-cast and &lt;em&gt;in situ&lt;/em&gt; concrete layers work in unison are clearly visible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G3Hbx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G3Hbx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-02-03T12:31:57.287" LastActivityDate="2016-02-03T12:31:57.287" Title="Multi-stage/multi-section reinforced concrete slab design" Tags="|civil-engineering|structural-engineering|reinforced-concrete|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6804" PostTypeId="1" CreationDate="2016-01-06T11:17:00.167" Score="2" ViewCount="131" Body="&lt;p&gt;I have a car moving in straight line with certain time-varying acceleration and there is another car moving in a curvilinear way along some curved path with some time-varying acceleration. The inertial reference frame is say fixed at some point in space. The relative velocity and acceleration at any instant is calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the relative velocity of the car moving in curved path from the point of view of the car moving in straightline, different from the relative velocity of the car moving in straightline from the point of view of the car moving in the curved path?&#xA;I know that in mathematical formulation this will involve the instantaneous angular velocity of the car following the curved path. But that is not very intuitive to me. Is there any way I can understand this situation clearly and more intuitively,i.e., Visualize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ev2tL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ev2tL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P2XfG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P2XfG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4011" LastActivityDate="2016-01-06T15:20:28.383" Title="Simple rotational dynamics" Tags="|mechanical-engineering|dynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6807" PostTypeId="2" ParentId="6804" CreationDate="2016-01-06T15:20:28.383" Score="3" Body="&lt;p&gt;Get a friend, and stand 15 feet (5m) from your friend. Your friend holds out their arm and points at you, then you do the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you try turning in a circle. Your friend's shoulder doesn't move because your relative linear velocity is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your shoulder has to move to keep pointing at your friend, though, because &lt;em&gt;from your point of view&lt;/em&gt;, your friend is running circles around you. Your friend &lt;em&gt;does&lt;/em&gt; have a relative linear velocity, and so you have to constantly adjust to keep pointing at them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same reason why it looks like the sun translates across the sky. The sun is (relatively) stationary, but the Earth rotates. Since you are on the Earth, your point of view is that the Earth is stationary and the sun is translating - this is why the original view was that the Earth was the center of the solar system. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-01-06T15:20:28.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6809" PostTypeId="1" CreationDate="2016-01-07T04:56:54.327" Score="1" ViewCount="154" Body="&lt;p&gt;I have a project for which I need a pneumatic cylinder to impart an impact force in such a way that work-done/energy is 4.5 Joules, and the input pressure is 10 bar.  How can I solve this so that I can derive bore &amp;amp; stroke of piston?&lt;/p&gt;&#xA;" OwnerUserId="4389" LastEditorUserId="272" LastEditDate="2016-01-07T13:45:40.140" LastActivityDate="2016-01-07T14:33:48.507" Title="Calculate pneumatic cylinder dimensions from work done" Tags="|fluid-mechanics|pressure|gas|engines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6810" PostTypeId="1" AcceptedAnswerId="6819" CreationDate="2016-01-07T05:50:11.737" Score="1" ViewCount="107" Body="&lt;p&gt;I work in a lab, and I've been trying to replace a lost AC-DC power adapter. I got one recently that has the exact specs that I need, which are apparently pretty rare (15V, 400 mA). However, when I test the power adapter with a voltmeter, it's actually outputting ~22V, despite what it says on the plug. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when I plug in the power adapter, the camera &quot;turns on&quot; because the green LED on it comes on. However, when I try to run the camera software, it behaves exactly as it does if I have the camera unplugged or turned off. So the software is not &quot;seeing&quot; it, even though it's plugged in, and plugged into the  firewire port on the PC (yes, it uses firewire... ugh). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now I don't know why the camera is not actually working. My hypothesis -- is it possible that the camera is not really working because the power adapter is bad, and that the specs are too far off from what it wants? Even though the LED light turns on, maybe the voltage/current is close enough to make something work, but not close enough for it to actually function properly?? Is this stupid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just not sure if this is a software or hardware problem... &lt;/p&gt;&#xA;" OwnerUserId="4390" LastActivityDate="2016-01-07T14:42:49.433" Title="Is it possible that an electronic device &quot;half-works&quot; when the power adapter supplies too much voltage?" Tags="|electrical-engineering|consumer-electronics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6811" PostTypeId="1" AcceptedAnswerId="6834" CreationDate="2016-01-07T07:18:38.747" Score="5" ViewCount="278" Body="&lt;p&gt;I recently bought my first lathe (A circa 50s PREMO, Australian built with tapered roller bearings) and the spindle has a little bit of wiggle parallel to the axis of the spindle. I decided to pull the spindle out to take a look at 'er, but I ran into a spot of bother with the gear that mates with the reverse tumbler. (And just to be clear, I'm absolutely certain the gear needs to come off before the spindle will come out, it's bigger than the ID of the cone pulley)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may seem silly, but I can't work out how that thing is supposed to come off. The gear sits on top of a threaded portion of the spindle (And indeed before getting to the gear I had already removed two nuts from that end of the spindle), so that was my first thing to try. Grabbed it with a chain wrench, gave it a fairly good torque, nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next thing I noticed was that there was a keyway cut in the gear, so I took a look in there. Sure enough, the threads extend under the gear, but curiously there doesn't seem to be a thread cut on the inside of the gear. I would have expected to see the tell-tale profile of the internal thread, but nothing; smooth. My next thought was that it was simple pressed on with a key, but I couldn't see a key, and in fact I'm able to push an allen key down the keyway the full thickness of the gear, there's nothing there. In addition to the lack of a key, there doesn't seem to be a matching keyway on the spindle for the gear to key into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What troubles me is that if there's a key and I try to unscrew it, I'll break something, and if it's threaded and I try to press it off, I'll break something. Where should I go from here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oLlwN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oLlwN.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DkFdS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DkFdS.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2227" LastEditorUserId="2227" LastEditDate="2016-01-07T10:39:40.263" LastActivityDate="2016-01-08T02:07:44.523" Title="How can I tell if a gear is pressed or threaded?" Tags="|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6812" PostTypeId="1" AcceptedAnswerId="6820" CreationDate="2016-01-07T10:15:23.173" Score="4" ViewCount="10796" Body="&lt;p&gt;Suppose a 'D' shaped shaft with a 4 mm diameter like this type&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nOVs5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nOVs5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is fitted inside a module 1 (plastic) gear with a round hole with the same diameter. If the gear has a set screw, will this still be a good fit? Will all the force be transferred to the set screw or will it be distributed onto the plastic surface as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should only round shafts fit inside gears with a round hole? Would connecting D shaft to a round hole gear be less preferable than connecting a D shaft to a D hole - even with a set screw? Yes, there's a set screw, but the flat part of the shaft will have a smaller contact surface, right?&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="272" LastEditDate="2016-01-07T13:00:59.080" LastActivityDate="2016-01-07T14:56:01.810" Title="Using a 'D' shaped shaft with a set screw" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6813" PostTypeId="2" ParentId="6810" CreationDate="2016-01-07T11:33:02.433" Score="3" Body="&lt;p&gt;Modern power adapters tend to be electronically regulated and not vary much in output voltage with load changes. Older ones that used iron core transformers were often less stable in output voltage with Vout rising to a substantially higher than specified value on no load. If you are 'lucky' the camera load may have reduced the voltage enough to prevent damage and/or the camera designers may have had their wits about them - but no guaranteed on either count, alas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voltage supplied should ideally be close to correct - usually 14V to 16V or so would be OK.  22V is higher than may be safe and may have caused damage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current rating of the adapter needs to be equal or higher than what is specified. Higher will almost always do no harm if voltage is OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other thing to check is polarity. Most adapters have +ve on the sleeve and -Ve on the center pin but a small number have the opposite polarity. It is almost always essential to use the correct polarity. A very very few devices* do not care but this is very rare. Reversed polarity at best will not work and at worst may destroy the equipment and/or the adapter.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Such as ones that I build :-).&lt;br&gt;&#xA;Within reason, plug something in and it will work :-).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="399" LastActivityDate="2016-01-07T11:33:02.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6816" PostTypeId="1" CreationDate="2016-01-07T13:19:11.087" Score="-1" ViewCount="284" Body="&lt;p&gt;Assume we have a platform fixed to a pivot point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0pBlI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0pBlI.png&quot; alt=&quot;Forgive the crude image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use 2 rods, either side of the pivot point, but at different distances from it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to lift ROD 1 by a certain amount, how much would ROD 2 have to lower?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other information is needed in solving this problem?&lt;/p&gt;&#xA;" OwnerUserId="4281" LastActivityDate="2016-01-07T15:14:29.747" Title="Pivot Point Equations" Tags="|mechanical-engineering|measurements|" AnswerCount="1" CommentCount="3" ClosedDate="2016-01-07T20:28:56.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="6817" PostTypeId="2" ParentId="6809" CreationDate="2016-01-07T13:42:25.327" Score="2" Body="&lt;p&gt;It is possible to calculate this under the assumption that all of the energy from the expansion goes into mechanical energy instead of being lost to heat transfer through the walls.  In this case the process is called &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html&quot; rel=&quot;nofollow&quot;&gt;adiabatic&lt;/a&gt;, and the work done is given by &#xA;$$&#xA;W = P_0V_0^\gamma\frac{V_f^{1-\gamma}-V_0^{1-\gamma}}{1-\gamma},&#xA;$$&#xA;where &#xA;$$&#xA;\gamma=\frac{C_p}{C_v}&#xA;$$&#xA;is the ratio of the &lt;a href=&quot;https://www.ohio.edu/mechanical/thermo/property_tables/air/air_Cp_Cv.html&quot; rel=&quot;nofollow&quot;&gt;specific heat capacities of the gas&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cross-sectional area (bore) is given by $A$ and the initial and final lengths of the piston (stroke) are $\ell_0$ and $\ell_f$, then the volume can be written as $V_0=A\ell_0$ and $V_f=A\ell_f$.  Plugging this in gives&#xA;$$&#xA;W=P_0A\frac{\ell_0 - \frac{\ell_0^\gamma}{\ell_f^{\gamma-1}}}{\gamma-1}.&#xA;$$&#xA;If the gas you are using is air at room temperature, then gamma is given by&#xA;$$&#xA;\gamma=\frac{1.005}{0.718}=1.400.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2016-01-07T14:33:48.507" LastActivityDate="2016-01-07T14:33:48.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6819" PostTypeId="2" ParentId="6810" CreationDate="2016-01-07T14:42:49.433" Score="1" Body="&lt;p&gt;What Russell said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the reason the LED lights but the device no longer appears to work may be because the LED is just wired to indicate power, and the rest of the device is now blown out due to the much higher than specified voltage you applied to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power LEDs are sometimes just a LED and resistor connected in series across the input power.  These LEDs don't need to be particularly bright, so are often run at less than their maximum current rating.  Many common indicator LEDs are rated for 20 mA, but are still plenty bright enough at 10 mA to tell you the power is on.  Figuring the green LED drops 2.1 V and is just connected to the supply with a resistor, your 22 V as apposed to 15 V would result in 54% more current thru the LED.  If it was only run at half its maximum current normally, this wouldn't even be pushing it.  Even if the LED was set to run at its maximum rating normally, 54% overcurrent for a typical indicator LED won't blow it out.  It will be brighter than normal, and it will have a shorter lifetime than rated, but would most likely work fine for days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circuitry in the rest of the device apparently wasn't so forgiving, and now it's just a hunk of junk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next time get a &lt;i&gt;regulated&lt;/i&gt; supply of the right voltage.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-01-07T14:42:49.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6820" PostTypeId="2" ParentId="6812" CreationDate="2016-01-07T14:56:01.810" Score="3" Body="&lt;h1&gt;Quick Answer&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can fit D shaped shaft into round hole if diameters are OK.&lt;/li&gt;&#xA;&lt;li&gt;D shaft to D hole without a setscrew will be preferable to D shaft to round hole with setscrew if there is no axial movement. Best is D shaft to D hole with setscrew(s). However you can't always find D holes because they're more costly to manufacture.&lt;/li&gt;&#xA;&lt;li&gt;Shear forces act on setscrew if holding torque isn't exceeded. You can check holding torque below.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Setscrew holding power&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Setscrews' holding power is proportional to its diameter you can check &lt;strong&gt;Marks’ Standard Handbook for Mechanical Engineers, Eleventh Edition, Table 8.2.18&lt;/strong&gt;. gives holding power for Cup-Point Setscrews. Maximum load you can get from a setscrew from that table is 7000lb axial holding power for 1 in setscrew. The torque is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{F_{a}}{2}D_{s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F_{a}$ is holding power and $D_{s}$ is shaft diameter in inches. Sorry I don't have metric at hand. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;D shaft&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Advantage of D shaft is it prevents shaft from spinning inside the gear hub. So it is preffered in gears.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;An example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you choose No0 screw (1.524mm diameter), axial holding power is 50 lb (222N) so your holding torque is 444 Nmm. If you exceed this torque setscrew will fail but because of D shaft it will hold till material failure (on plastic shaft probably).&lt;/p&gt;&#xA;" OwnerUserId="879" LastActivityDate="2016-01-07T14:56:01.810" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6821" PostTypeId="2" ParentId="6816" CreationDate="2016-01-07T15:14:29.747" Score="1" Body="&lt;p&gt;This problem can be simply understood via &lt;a href=&quot;https://en.wikipedia.org/wiki/Similarity_(geometry)&quot; rel=&quot;nofollow&quot;&gt;similar triangles&lt;/a&gt;. When you raise one rod, it creates a triangle between its original point (where it connects with the platform), its final position and the fulcrum. The other rod creates a &lt;em&gt;similar triangle&lt;/em&gt;. You can therefore find the relation between how much each rod moves via the difference in scale of the triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one triangle is $X$ times larger than the other, then that triangle will also move $X$ times more than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it's almost trivial with a casual glance at the link above, I'll leave it to you to find the difference in scale of the triangles.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-01-07T15:14:29.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6822" PostTypeId="1" AcceptedAnswerId="6828" CreationDate="2016-01-07T15:28:19.507" Score="1" ViewCount="643" Body="&lt;p&gt;I am trying to find a method of calculating the minimum shear area of the following rectangular thread form. It is a 4.125&quot; square thread, 8 T.P.I., straight, right-hand thread. The thread is loaded in the axial direction, thus I would like to calculate the minimum thread shear strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/28GtV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/28GtV.jpg&quot; alt=&quot;Thread&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="738" LastEditorUserId="738" LastEditDate="2016-01-07T15:42:39.673" LastActivityDate="2016-01-07T20:21:46.767" Title="How to find the minimum shear area of a rectangular thread" Tags="|mechanical-engineering|threads|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6823" PostTypeId="1" CreationDate="2016-01-07T16:55:29.697" Score="5" ViewCount="473" Body="&lt;p&gt;I am trying to make a 3D model.  It consists of a cylinder resting on an extruded ellipse as you can see in the picture below.  I now want to join the top edge of the ellipse with the top edge of the circle.  In a sense, I want there to be a chamfer between the side surface of the ellipse-prism and the top surface of the cylinder.  However, I don't even know what tool to use.  Is it a chamfer?  Is it called something else?  The problem is that the distance from the cylinder to the edge of the ellipse changes depending on whether you are looking at the north/south/east/west sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using Autodesk's free 123D design software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kqOQG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kqOQG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4403" LastEditorUserId="1290" LastEditDate="2016-01-08T08:37:25.340" LastActivityDate="2016-01-12T17:15:09.080" Title="How do you create a chamfer in 3D CAD that joins the edges of two different shapes?" Tags="|mechanical-engineering|computer-aided-design|3d-printing|solidworks|autocad|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6824" PostTypeId="2" ParentId="6798" CreationDate="2016-01-07T17:25:50.957" Score="3" Body="&lt;p&gt;It's hard to say exactly which method they would use with certainty, but it's pretty reasonable to assume that they'd specify the method if it was anything special.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They easiest way to do this would be with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow&quot;&gt;finite difference scheme&lt;/a&gt;. In this you approximate a derivative of a function by computing it at two locations that are a small, but known distance apart in your state space. The derivative (slope) is then the 'rise' (difference between the two values of the function) divided by the 'run' (the distance in state space) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The matrix in question is composed of the derivatives of each term of $s$ with respect to each variable in $x$. Treating $s$ as a function of $x$ you can readily form approximations of the derivatives as follows:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial s}{\partial \alpha} \approx \frac{s(V,\alpha +\Delta\alpha,\beta,p,q,r,\phi,\theta) - s(V,\alpha -\Delta\alpha,\beta,p,q,r,\phi,\theta)}{2\Delta\alpha} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I've introduced $\Delta\alpha$ as a small perturbation to give to nearby but different values at which to evaluate $s$. This will give will approximate s at $x = (V,\alpha,\beta,p,q,r,\phi,\theta)$ (between the two locations) and the error of the approximation will be proportional to ${\Delta\alpha}^2$.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="4412" LastEditDate="2016-01-08T17:20:23.273" LastActivityDate="2016-01-08T17:20:23.273" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6826" PostTypeId="1" CreationDate="2016-01-07T18:54:38.603" Score="1" ViewCount="36" Body="&lt;p&gt;I am simulating a turbine and I have some doubts about the correctness of my approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this turbine, there are three volumes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kOKuO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kOKuO.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two volumes are the piping before and after the blade and the other volume contains the turbine blades. The domain containing the blades rotates at 2000 rpm = 209.43 rad/s while the other two volumes are stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I modeled this turbine in Gambit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, are there any problems? Gambit is weak for modeling this stage.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2016-05-20T06:34:07.600" LastActivityDate="2016-05-20T06:34:07.600" Title="How to set up the boundary between volumes in a simulated turbine" Tags="|mechanical-engineering|fluid-mechanics|turbines|cfd|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6827" PostTypeId="2" ParentId="6823" CreationDate="2016-01-07T19:13:41.917" Score="5" Body="&lt;p&gt;This type of feature is typically called a &quot;loft&quot;, but I would caution you that this should be used only if needed. Lofted features will cost you a lot of money if you try to get them machined. If you're going with 3d printing it doesn't matter too much (complexity is free). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, here's some pictures of before:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WzXOj.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WzXOj.png&quot; alt=&quot;Loft Input&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And after a &quot;loft&quot; command:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3kKJ9.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3kKJ9.png&quot; alt=&quot;Loft Output&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you do a loft command, it's useful to give &quot;guidelines&quot; that to help the program determine what path the loft feature should take to get from profile 1 (the ellipse) to profile 2 (the circle). Here I just added a straight line from the circle to the ellipse on the long (XY) and short (ZY) planes of the ellipse. If you wanted the loft to have a concave/convex (&quot;bubbled&quot;) appearance, you would do it by altering the guide lines in this step. Note that I colored the guidelines RED here because I thought Solidworks' gray was a little hard to see. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ci0ek.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ci0ek.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, use the loft command. Assign the circle and ellipse as the &quot;Profiles&quot; and the guidelines as the guidelines, then accept the output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ADerI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ADerI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is similar in Autodesk Inventor, which I could show if you wanted, but I don't have (have never used) 123D, so I can't speak to whether or not that package has this tool.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-01-07T19:13:41.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6828" PostTypeId="2" ParentId="6822" CreationDate="2016-01-07T20:21:46.767" Score="1" Body="&lt;p&gt;The area equation of the thread is PI * D * nut length * .060/(.060*.065)&#xA;PI = 3.14, D is 4.125-2*.050, nut length not given, probably 1&quot; to 4&quot;, .060/(.060*.065) ratio of thread to machined out area&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-07T20:21:46.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6829" PostTypeId="2" ParentId="625" CreationDate="2016-01-07T20:52:50.570" Score="1" Body="&lt;p&gt;In short, whether there are long-term savings to be made by using super hydrophobic surfaces (SHS) on ships is still an open question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been many claims, but the benefits have yet to be shown on real ships. There are many factors that come into play, such as durability of the surface, its cost, whether the surface must be cleaned in special ways etc. These will require testing over long periods of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some discussion of the potential of SHS in the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triantafyllou, Michael,&#xA;&quot;Science and Technology Challenges and Potential Game-Changing Opportunities &quot;,&#xA;Special Report 306: Naval engineering in the 21st century the science and technology foundation for future naval fleets,&#xA;&lt;a href=&quot;https://seagrant.mit.edu/publications/MITSG_10-36.pdf&quot; rel=&quot;nofollow&quot;&gt;https://seagrant.mit.edu/publications/MITSG_10-36.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A recent book also discusses SHS. See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perlin, Marc and Ceccio, Steven,&#xA;&quot;Mitigation of Hydrodynamic Resistance: Methods to Reduce Hydrodynamic Drag&quot;,&#xA;2015, 150 pp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be aware that there are quite a few people making unsubstantiated claims for drag reduction.&#xA;My advice is to treat any claims of large performance gains with suspicion, and to keep an eye on developments &#xA;reported by the International Towing Tank Conference (ITTC).&#xA;&lt;a href=&quot;http://ittc.info/index.php&quot; rel=&quot;nofollow&quot;&gt;http://ittc.info/index.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm sorry, but I can't help with your question about how SHS can be used to effect on aircraft.)&lt;/p&gt;&#xA;" OwnerUserId="1908" LastEditorUserId="1908" LastEditDate="2016-01-08T03:57:25.627" LastActivityDate="2016-01-08T03:57:25.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6830" PostTypeId="1" CreationDate="2016-01-07T20:54:49.840" Score="9" ViewCount="4274" Body="&lt;p&gt;I am master student and doing an assignment of Finite element method. In the instruction I could not understand the derivation of the weak form, which should not be difficult. I'm sorry for posting this easy and most likely not helpful question for other people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the derivation is about the weak form of the integral formulation of 4th ODE. It is a simple beam deformation in the interval 0 and L.&#xA;$$ \int_0^L\frac{d^2 w}{d x^2}EI\frac{d^2 \hat{u}}{d x^2}dx = ...$$&#xA;$$ \left( \left. w EI \frac{d^3 \hat{u}}{d x^3} \right|_{x=0} \right. &#xA;-\left( \left. \frac{d w}{dx} EI \frac{d^2 \hat{u}}{d x^2} \right|_{x=0} \right. &#xA;-\left( \left.  w EI \frac{d^3 \hat{u}}{d x^3} \right|_{x=L} \right. &#xA;+\left( \left. \frac{d w}{dx} EI \frac{d^2 \hat{u}}{d x^2} \right|_{x=L} \right. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about partial integration&#xA;$$\int u(x) v'(x) \, dx = u(x) v(x) - \int v(x) \, u'(x)  dx $$&#xA;then I ended up&#xA;$$\frac{d^2u}{dx^2}\frac{dw}{dx} - \int \frac{dw}{dx}\frac{d^3x}{dx^3}d x$$&#xA;if I continue the partial integration to the 2nd term then derivative will be 4th order...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I get the following?&#xA;$$ \left( \left. w EI \frac{d^3 \hat{u}}{d x^3} \right|_{x=0} \right. &#xA;-\left( \left. \frac{d w}{dx} EI \frac{d^2 \hat{u}}{d x^2} \right|_{x=0} \right. &#xA;-\left( \left.  w EI \frac{d^3 \hat{u}}{d x^3} \right|_{x=L} \right. &#xA;+\left( \left. \frac{d w}{dx} EI \frac{d^2 \hat{u}}{d x^2} \right|_{x=L} \right. $$&lt;/p&gt;&#xA;" OwnerUserId="4405" LastEditorUserId="19906" LastEditDate="2020-12-04T12:31:11.673" LastActivityDate="2020-12-04T12:31:11.673" Title="Derivation of the weak form for the euler-bernoulli beam equations" Tags="|beam|finite-element-method|mathematics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="6832" PostTypeId="1" AcceptedAnswerId="8395" CreationDate="2016-01-07T21:28:34.023" Score="5" ViewCount="2008" Body="&lt;p&gt;I am trying to calculate the power distribution of a planetary gear system, specifically a Wolfrom system like this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8kRDA.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8kRDA.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find the output power, I need to find the torque for the rings 5 and 6. Torque and angular velocity of the sun 1 is known. Also I have all the other angular velocities. I am trying to figure out a methodology to design the free body diagram of this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NVzkC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NVzkC.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the positive sign is clockwise, how can I add the torque T5 and T6 to this free body diagram?&lt;/p&gt;&#xA;" OwnerUserId="4388" LastEditorUserId="368" LastEditDate="2016-01-08T00:12:13.237" LastActivityDate="2016-04-12T12:20:11.130" Title="Determine the sign of torques for a planetary gear system" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6834" PostTypeId="2" ParentId="6811" CreationDate="2016-01-08T02:07:44.523" Score="6" Body="&lt;p&gt;I decided that because of the lack of threads, it was more likely to be pressed on. So I whipped up a little shopmade gear puller, gave a little tug, and off it came. As it turns out, the gear and the spacer that I'd thought to be two pieces was actually one piece, and the key was under the spacer rather than the gear. Mystery solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lisxR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lisxR.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Pzrln.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pzrln.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2227" LastActivityDate="2016-01-08T02:07:44.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6835" PostTypeId="1" CreationDate="2016-01-08T05:16:51.743" Score="1" ViewCount="3824" Body="&lt;p&gt;I would like to control the locking of a gear, using a solenoid.  My original idea was to simply activate the solenoid to pull back the lock (a tooth that sticks into the gear), but that means that I'd need to keep applying current to the solenoid as long as I want the gear unlocked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm thinking it would be better to have a mechanism between the solenoid and the gear, so that a pulse to the solenoid would unlock the gear, and another pulse would lock it.  Kind of like the mechanism in a click type pen.  What would be a good choice of mechanism to accomplish this, considering how solenoids have a very short stroke distance?&lt;/p&gt;&#xA;" OwnerUserId="4410" LastEditorUserId="272" LastEditDate="2016-01-08T12:53:55.613" LastActivityDate="2016-01-08T12:53:55.613" Title="Gear Locking Mechanism" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6836" PostTypeId="2" ParentId="6835" CreationDate="2016-01-08T06:53:23.103" Score="3" Body="&lt;p&gt;I wouldn't suggest putting objects into gear teeth if the object engages the teeth while a motor is powering the gear, the gear will likely be damaged. If there is compressed air available, I would suggest a solenoid valve operated air cylinder pushing a brake pad against a rotating shaft. Also, electric brakes are available: &lt;a href=&quot;http://www.mcmaster.com/#brakes/=10l6j16&quot; rel=&quot;nofollow&quot;&gt;http://www.mcmaster.com/#brakes/=10l6j16&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-08T06:53:23.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6837" PostTypeId="2" ParentId="6835" CreationDate="2016-01-08T11:39:14.177" Score="1" Body="&lt;p&gt;Your toggle design has the disadvantage that the same input from the controller both locks and unlocks the gear. This can be problematic if the state of the lock is not what the controller thinks it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead using an actuator that remains in position you can move it forward to lock and backwards to unlock. Whether the locking mechanism is a brake pad or a pawl that meshes with the gear is irrelevant to this.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2016-01-08T11:39:14.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6838" PostTypeId="1" CreationDate="2016-01-08T12:06:36.843" Score="3" ViewCount="239" Body="&lt;p&gt;From &lt;a href=&quot;https://www.quora.com/Bend-a-beam-heavily-Now-freeze-that-geometrical-state-and-get-rid-of-any-tensions-Apply-the-opposite-force-Does-it-have-the-same-horizontal-shape-of-the-beginning&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, I read that if you have a straight beam, bend it with force at one end, and then create a beam with the resulting curvature, the equal and opposite force &lt;em&gt;won't&lt;/em&gt; make a straight beam.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the shape of a curved beam that &lt;em&gt;will&lt;/em&gt; become straight given a point force on the end? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assume the cross section is rectangular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either a closed solution or a way to approximate the shape will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to design a spring for a 3d printer.  For technical reasons, printing a 3d helix is difficult.  A design that has a constant cross section in one axis is best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One design that I've seen is an &lt;a href=&quot;http://d2q1ebiag300ih.cloudfront.net/uploads/image/file/5254/GM-S-2.jpg&quot; rel=&quot;nofollow&quot;&gt;zigzag-shaped spring&lt;/a&gt;.  It seemed to be, however, that the stress isn't evenly distributed, with more strain on the corners and less on the straight parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought was to make an &lt;a href=&quot;http://www.clker.com/cliparts/P/P/y/a/2/j/plain-football-outline-md.png&quot; rel=&quot;nofollow&quot;&gt;eye-shape&lt;/a&gt; with pointed corners that compresses to two flat lines.  My intuition is that, if the shape compresses to a flat line then it's spreading the tensile forces evenly and won't break at a weak point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the link above, I read that I can't assume a constant $I$ in the Euler-Bernouli beam equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\left(EI{\frac {\mathrm {d} ^{2}w}{\mathrm {d} x^{2}}}\right)=q $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://courses.washington.edu/me354a/Curved%20Beams.pdf&quot; rel=&quot;nofollow&quot;&gt;This PDF&lt;/a&gt; claims that the neutral axis of a curved beam isn't the same as in a straight beam (where it is just the centeroid).  There is a formula for distance of the neutral axis from the inside of the curve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \text{Height of Neutral Axis} = {\frac h {\ln(r_o / r_i)}} - r_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the radius of a function $y$ at point $x$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R =\left| \frac { \left(1 + y'^{\,2}\right)^{3/2}}{y''}\right|,&#xA;\qquad\mbox{where}\quad&#xA;y' = \frac{dy}{dx},\quad y'' = \frac{d^2y}{dx^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With those, I hoped to find the shape that, at rest, looks like an eye and when squeezed becomes two flat lines.&lt;/p&gt;&#xA;" OwnerUserId="4416" LastEditorUserId="1832" LastEditDate="2016-01-08T19:53:36.247" LastActivityDate="2016-01-08T22:31:33.033" Title="Design a beam that is straightened with force on one end" Tags="|mechanical-engineering|beam|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="6839" PostTypeId="1" CreationDate="2016-01-08T17:11:22.187" Score="1" ViewCount="258" Body="&lt;p&gt;I have been interested in the prospect of leveraging &lt;a href=&quot;https://en.wikipedia.org/wiki/Van_der_Waals_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;Van der Waals force&lt;/a&gt; in an adhesive material for several years now.  The most notable example I have seen thus far is the highly publicized &lt;a href=&quot;https://geckskin.umass.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;GeckSkin&lt;/a&gt; product resulting from biomimicry of Gecko locomotion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some key features of Nano-hair structure include:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;anisotropic attachment&lt;/li&gt;&#xA;  &lt;li&gt;high pulloff to preload ratio&lt;/li&gt;&#xA;  &lt;li&gt;low detachment force&lt;/li&gt;&#xA;  &lt;li&gt;material independence / van der Waals adhesion&lt;/li&gt;&#xA;  &lt;li&gt;self-cleaning&lt;/li&gt;&#xA;  &lt;li&gt;anti-self matting&lt;/li&gt;&#xA;  &lt;li&gt;non-sticky default state.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is obvious far reaching impacts of such materials in everything from robotics to home decor and without a supply (I have already contacted the team involved with little feedback on estimated time to market) I have been investigating the prospect of good old fashioned DIY.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my pursuit I have come across a simple method of &lt;a href=&quot;http://www.nisenet.org/catalog/programs/biomimicry_synthetic_gecko_tape_through_nanomolding&quot; rel=&quot;nofollow noreferrer&quot;&gt;NanoMolding synthetic gecko tape&lt;/a&gt;.  The key ingredient of which is the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00C10VPCY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Millipore Isopore Polycarbonate Filter Membrane (Product# TMTP04700)&lt;/a&gt; which conveniently enough is sold on Amazon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information with respect to robotics can be found &lt;a href=&quot;https://robotics.eecs.berkeley.edu/~ronf/Gecko/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically one could use the membrane to create a negative mold from which a more permanent reusable mold (possibly a Resin?) could then be used for future casts.  I figure resolution may be somewhat of an issue in this case, however, so this leads me to my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;TL;DR&lt;/em&gt;&lt;/strong&gt;&#xA;I would like to know if there are other materials with similar surface properties that may be good candidates to replace the membrane.  Something more readily available, reusable, lower cost, natural or man made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Bonus&lt;/em&gt;&lt;/strong&gt;  I am also open to suggestion on alternative methods for creating the &lt;a href=&quot;https://en.wikipedia.org/wiki/Synthetic_setae&quot; rel=&quot;nofollow noreferrer&quot;&gt;synthetic setae&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; I had a hard time finding a place for this question as it is applicable to so many fields and there currently is not a dedicated &lt;strong&gt;&lt;em&gt;&lt;a href=&quot;https://meta.stackexchange.com/questions/265751/stack-exchange-site-for-material-science&quot;&gt;Material Science&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt; StackExchange community. &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="4418" LastEditorUserId="-1" LastEditDate="2017-03-20T10:32:31.157" LastActivityDate="2016-01-08T19:02:25.640" Title="Alternative substrate for nanomolding synthetic &quot;gecko tape&quot;" Tags="|materials|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6840" PostTypeId="1" CreationDate="2016-01-08T17:58:01.103" Score="1" ViewCount="1042" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RlkMR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RlkMR.jpg&quot; alt=&quot;original question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem 3/155 The light rod is pivoted at O and carries the 2kg and 4kg particles. If the rod is released from rest at $\theta=60°$ and swings in the vertical plane, calculate (a) the velocity $v$ of the 2-kg particle just before it hits the spring in the dashed position and (b) the maximum compression $x$ of the spring. Assume that $x$ is small so that the position of the rod when the spring is compressed is essentially horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part I am most confused about it part a) as I am not sure how to come up with the energy equation using angular motion instead of linear motion. Trying to solve the question I assumed I had to use $\frac12I\omega^2$ instead of $\frac12mv^2$.The height for GPE of mass a and mass b I am also unsure how to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to part a, I assumed the energy equation: $GPE_a=KE_b-GPE_b$ where GPE is gravitational potential energy and KE is kinetic energy. The kinetic energy of A = 0 since it is initially at rest, and GPE of mass b is negative due to loss of GPE where I took the vertical plane as the datum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may have the wrong assumptions and I am open to other answers people may have.&lt;/p&gt;&#xA;" OwnerUserId="4419" LastEditorUserId="4419" LastEditDate="2016-01-08T18:55:25.533" LastActivityDate="2016-04-10T16:15:33.237" Title="Need advice answering this kinetics of particles energy method dynamics problem" Tags="|mechanical-engineering|dynamics|energy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6841" PostTypeId="2" ParentId="6838" CreationDate="2016-01-08T19:27:12.800" Score="-1" Body="&lt;p&gt;$y=\dfrac{Wx^2(3l-x)}{6EI}$ is the equation for deflection of a beam supported at one end and load at the other end (&lt;a href=&quot;http://www.engineersedge.com/beam_bending/beam_bending9.htm&quot; rel=&quot;nofollow&quot;&gt;reference&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E$ is modulus of elasticity. For steels it's 30x10^6 psi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I$ (moment of inertia) is $\dfrac{bh^3}{12}$ for rectangular sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just design the spring going up by the amount of calculated deflection. &lt;/p&gt;&#xA;" OwnerUserId="4311" LastEditorUserId="1832" LastEditDate="2016-01-08T19:43:48.440" LastActivityDate="2016-01-08T19:43:48.440" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6844" PostTypeId="2" ParentId="6326" CreationDate="2016-01-08T23:00:07.707" Score="0" Body="&lt;p&gt;I think that the most useful answer to your question is that, in practice we are most concerned about the relative torque between two linked shafts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way the forces on gears are only of interest to people who design gears and if you are designing a gearbox it's the ration of torque/RPM which is of primary interest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example shaft Shaft B will see 6x the torque of shaft A and rotate at 1/6 RPM. At this point the diameter of the smaller gear on shaft B is irrelevant until we know what it is coupled to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that you are correct in saying that the ratio of torque between shaft A and shaft B is 1:6 and the ratio of angular velocity(RPM) is 6:1 ie shaft B has 6 time more torque and rotates at 1/6th of the speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For it to be  true compound gear train you also need a shaft C coupled t the smaller gear on shaft B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the gear on shaft C was 6cm in diameter then you would have a final drive ratio of 1:(6x6) ie 1:36 &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-08T23:00:07.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6845" PostTypeId="1" CreationDate="2016-01-09T02:30:15.900" Score="1" ViewCount="199" Body="&lt;p&gt;I'm looking to electroplate brass rings with gems glued on top of them (though I believe this is then considered electroforming as the copper builds on the glue or epoxy and around the ring and gem?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to know is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What amp and volt size rectifier would be best? Some say 3amp is enough for small scale Jewelry others insist on going up to 25 to 30amp. I can get a 30amp 20volt online for relatively cheap - &lt;a href=&quot;http://m.aliexpress.com/item/1919812106.html&quot; rel=&quot;nofollow&quot;&gt;http://m.aliexpress.com/item/1919812106.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(They have other smaller rectifiers, all have US plugs but do not state the plug outlet voltage. I live in Australia and need to be sure I can use an adaptor (not a converter) to plug into a 220v outlet).&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;If I use brass rings do I need to paint them first with a copper conductive paint? What if I want to eventually do silver plating? Would brass be okay to silver plate? I know I would need a different bath for that though, I'm just thinking of possibilities for the future. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Instead of using a copper conductive paint could I instead make my bath with copper sulphate, sulphuric acid and distiled water mixed together? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is the copper sulphate I am currently using for the bath: &lt;a href=&quot;http://www.bunnings.com.au/manutec-500g-copper-sulphate-soluble_p2961523&quot; rel=&quot;nofollow&quot;&gt;http://www.bunnings.com.au/manutec-500g-copper-sulphate-soluble_p2961523&lt;/a&gt;&#xA;Is this pure enough? For my anode can I use a 99% pure copper sheet to will only 100% suffice? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lastly, do I need to agitate the bath continually? I wouldn't mind a slightly organic look and have read that without agitation this will happen but I do not know to what extent.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have experimented with copper plating a silver plated ring but the copper did not build up and part of the ring was eaten way at. I used the copper sulphate and boiling water for the bath, a copper pipe for the anode, and hooked the leads up to a 6 volt battery which attached to the anode and my cathode was a copper wire which wrapped around the ring. Not sure where I went wrong if it was the wrong type of copper sulphate, if I need conductive paint, a rectifier instead of a battery or what. Any assistance anyone could provide would be so very greatly appreciated. I've done so much research but am still at a bit of a loss as to what I need to improve upon to make the process work. I'm very keen to start plating my own jewelry. &lt;/p&gt;&#xA;" OwnerUserId="4424" LastEditorUserId="272" LastEditDate="2016-01-11T13:44:58.270" LastActivityDate="2016-01-11T13:44:58.270" Title="Electroplating details for a beginner" Tags="|brass|" AnswerCount="1" CommentCount="1" ClosedDate="2016-01-12T16:36:13.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="6846" PostTypeId="2" ParentId="6845" CreationDate="2016-01-09T03:57:52.663" Score="2" Body="&lt;p&gt;My first suggestion would be to forget about voltage, current is the important thing. If you can, obtain a current controlled power supply, that way you can finely tune exactly the current that you want, rather than trying to balance the resistance of an electrolyte solution against a fixed voltage supply. A cheap way to emulate this could perhaps be to add a multimeter and variable resistor in series with your plating setup. To do this, you'll need a power supply rated for more current than you need, and a high-ish voltage, perhaps 30-40, or even 50 volts. Then you can watch the current measurement, and tune it with the variable resistor. As for numbers, different settings will give you different results, so I think you should experiment. Perhaps you could make up some test coupons, plate them all at different currents, and decide what you like. Also, the absolute current setting isn't what counts, it's the current per surface area, something else to keep in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might experience poor adhesion of copper to brass, brass doesn't seem to like to stick very well. A common technique is to first plate with nickel, then plate with copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purity of your components is probably fine, but purer is likely always better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Agitation is again something that you should experiment with, and decide what you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the sounds of it, I would guess that you accidentally made the ring the anode and the copper pipe the cathode; you must make sure that the positive terminal of your power supply is connected to the anode. It's also possible some kind of coating on the ring may have inhibited the plating in some way, but I wouldn't have expected that to result in corrosion of the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things I know nothing about are electroplating non-conductive materials (glue, epoxy, etc) and electroplating of non-metallic materials (conductive paint).&lt;/p&gt;&#xA;" OwnerUserId="2227" LastActivityDate="2016-01-09T03:57:52.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6847" PostTypeId="2" ParentId="5461" CreationDate="2016-01-09T11:20:58.910" Score="1" Body="&lt;p&gt;To my answer of metal o ring...&#xA;Non-vented rings (OI &amp;amp; OE) are designed for lower pressure ranges. The perfectly closed O Ring design avoids ingress of the media into the seal. Moderate load, moderate pressure capability.&#xA;Internally (OVI) or externally (OVE) vented O Rings are designed for higher pressure. Besides the own seating load the seal is also energised by the system pressure.&#xA;Spring-energised metal O Ring for Internal pressure (OSI) and (OSE) for external pressure are designed for extreme low leakage at moderate pressures.&#xA;Gas filled metal O Ring for high temperature and moderate pressure applications. OGI for internal pressure and OGE for external pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details plz visit :&#xA;&lt;a href=&quot;http://www.htmsfutom.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.htmsfutom.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4427" LastActivityDate="2016-01-09T11:20:58.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6849" PostTypeId="1" AcceptedAnswerId="6858" CreationDate="2016-01-09T14:55:44.470" Score="1" ViewCount="706" Body="&lt;p&gt;I was studying the design of joints and came to knuckle joints. Generally  compressive strength is greater than tensile strength, but  &lt;strong&gt;considering knuckle joints compressive strength and tensile strength are considered equal&lt;/strong&gt; (as mentioned below). What are the reasons for assuming equal strengths?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BJ2Ek.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BJ2Ek.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.co.in/books?id=YaYQAgAAQBAJ&amp;amp;lpg=PA126&amp;amp;ots=0xUWYjKLSK&amp;amp;dq=WHY%20KNUCKLE%20joint%20crushing%20stress%20is%20equal%20to%20its%20tensile%20stress&amp;amp;pg=PA126#v=onepage&amp;amp;q=WHY%20KNUCKLE%20joint%20crushing%20stress%20is%20equal%20to%20its%20tensile%20stress&amp;amp;f=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a link to the page, shown above, in Google Books.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2116" LastEditorUserId="2116" LastEditDate="2016-01-10T04:06:31.450" LastActivityDate="2016-01-10T09:02:57.350" Title="Knuckle Joint: compressive strength equal to tensile strength" Tags="|design|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6854" PostTypeId="2" ParentId="6849" CreationDate="2016-01-09T21:08:38.573" Score="0" Body="&lt;p&gt;The ratio of compressive to tensile strength is dependent upon the steel. A36 structural steel has higher tension yield than compressive yield, 36kpsi vs 22kpsi, reference: &lt;a href=&quot;http://www.matweb.com/search/datasheet.aspx?matguid=d1844977c5c8440cb9a3a967f8909c3a&amp;amp;ckck=1&quot; rel=&quot;nofollow&quot;&gt;http://www.matweb.com/search/datasheet.aspx?matguid=d1844977c5c8440cb9a3a967f8909c3a&amp;amp;ckck=1&lt;/a&gt;&#xA;A very brittle tool steel will have a much higher compressive strength than tensile strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the cold drawn and stainless steels used commonly in machine components, usually only yield strength is published. These steels probably have yield and compressive strengths in the same range. The following test report (from 1942) that tested both compressive and tensile strength of stainless steels showed comparable results for compression and yield: &lt;a href=&quot;http://nvlpubs.nist.gov/nistpubs/jres/28/jresv28n4p499_A1b.pdf&quot; rel=&quot;nofollow&quot;&gt;http://nvlpubs.nist.gov/nistpubs/jres/28/jresv28n4p499_A1b.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-09T21:08:38.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6855" PostTypeId="1" AcceptedAnswerId="6857" CreationDate="2016-01-10T00:57:39.170" Score="3" ViewCount="332" Body="&lt;p&gt;I see that POM gears are often not preferred in high torque settings - but how about 0.4Nm? Is that amount of torque too much for module 1 POM bevel gears? Say they are 16-tooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is 0.4Nm considered moderately high torque?&lt;/p&gt;&#xA;" OwnerUserId="4301" LastActivityDate="2016-01-10T04:38:32.350" Title="Nylon gears for 0.4Nm torque" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6857" PostTypeId="2" ParentId="6855" CreationDate="2016-01-10T04:38:32.350" Score="3" Body="&lt;p&gt;module 1 16T bevel POM (also called acetal or Delrin) gears are rated at 0.2Nm at &lt;a href=&quot;http://www.huco.com/products.asp?p=true&amp;amp;cat=285&quot; rel=&quot;nofollow&quot;&gt;http://www.huco.com/products.asp?p=true&amp;amp;cat=285&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The webpage also list larger module gears torque ratings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.4 Nm is considered low torque in industrial machinery, but could be considered high torque in other fields.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-10T04:38:32.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6858" PostTypeId="2" ParentId="6849" CreationDate="2016-01-10T07:14:21.367" Score="2" Body="&lt;p&gt;Assuming they're equal, it allows you to use the more commonly available tensile strength data, while also getting a conservative solution. If it turns out they're not equal, then your design will still work correctly but might not make optimal use of the materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming uniaxial tensile and compressive strengths are equal is a common practice in engineering and is built into the popular von Mises and Tresca yield criteria which are used for all kinds of structures. If you want to take advantage of higher compressive strength, you can use less common failure criteria like Mohr-Coulomb.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="90" LastEditDate="2016-01-10T09:02:57.350" LastActivityDate="2016-01-10T09:02:57.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6860" PostTypeId="2" ParentId="6604" CreationDate="2016-01-10T16:16:45.717" Score="1" Body="&lt;p&gt;In this instance, the issue was as predicted - the wheel/nut driven by the worm connected to the hand wheel had jammed against an internal stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution I went with begins with removing the conventional nut at the top of the vertical screw pictured. This allows the plate at the top end of the large diameter section to lift off the end of the screw (taking the chuck etc. with it). It only needs to be lifted far enough to expose the top end of the screw. At this point I found a suitably sized nut, drilled it to half an inch and cut/filed a slot in it to accept the Woodruff key. It can then be placed on the end of the screw and used to turn it (clockwise, viewed from above) to release the internal jam. It did not require much torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: There's a near-horizontal linkage that transfers load from an internal counter-weight to make raising the head easier. That needs to be removed to allow the quill to rise higher than normal, but watch out for the tension in the chain when doing that. Take steps to constrain the linkage.&lt;/p&gt;&#xA;" OwnerUserId="4206" LastActivityDate="2016-01-10T16:16:45.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6862" PostTypeId="1" AcceptedAnswerId="6863" CreationDate="2016-01-11T00:56:35.047" Score="8" ViewCount="3546" Body="&lt;p&gt;I came across this question of &amp;quot;can an isothermal process also be adiabatic?&amp;quot;, and at first thought that yes, why not? If the temperature is constant and there is no heat loss to the environment, it seems perfectly possible.&lt;/p&gt;&#xA;&lt;p&gt;Now, I don't have too much experience in this field, but I came across a lot of vague statements that this is not possible at all.&lt;/p&gt;&#xA;&lt;p&gt;Could someone with a bit more expertise give me a clear answer?&lt;/p&gt;&#xA;" OwnerUserId="4442" LastEditorUserId="9448" LastEditDate="2022-09-28T06:53:22.667" LastActivityDate="2022-09-28T06:53:22.667" Title="Can an isothermal process also be adiabatic?" Tags="|thermodynamics|heat-transfer|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="6863" PostTypeId="2" ParentId="6862" CreationDate="2016-01-11T02:29:05.513" Score="7" Body="&lt;p&gt;If a process is both isothermal and adiabatic, it is implied that the work done on the system is being stored somewhere other than the internal energy of the working fluid.  (Or conversely, if the system is &lt;em&gt;doing&lt;/em&gt; work, the energy is coming from somewhere other than internal energy.)&lt;/p&gt;&#xA;&lt;p&gt;The classic example of such a process is &lt;a href=&quot;https://en.wikipedia.org/wiki/Joule_expansion&quot; rel=&quot;nofollow noreferrer&quot;&gt;free expansion&lt;/a&gt; of an ideal gas (also called Joule expansion), though in this case there is no work being done at all.&lt;/p&gt;&#xA;&lt;p&gt;Another example I can think of is a well-insulated vertical piston-cylinder filled with a saturated mixture of gas and liquid. The piston is frictionless and is allowed to move freely up and down. If a stirring rod is immersed in the liquid and spun, it will do work on the liquid, thus raising its enthalpy. Normally, this would result in a temperature increase, however since the mixture is saturated and held at a constant pressure (via the piston), its temperature must also be constant as required by the Gibbs phase rule. The added energy from stirring is instead used to raise the piston. Basically, the shaft work passes through the fluid and into the piston in the form of increased potential energy; the only change in the working fluid is the proportions of gas and liquid.&lt;/p&gt;&#xA;&lt;p&gt;A similar setup that is both adiabatic and isothermal might also be possible using a chemical reaction instead of a saturated mixture. Consider the same piston-cylinder as before, but this time it's filled with two gasses in chemical equilibrium with each other; i.e. a reaction like &lt;span class=&quot;math-container&quot;&gt;$A{\rightleftharpoons}B$&lt;/span&gt; where the reaction is exothermic in one direction and endothermic in the other. If the piston is used to compress the gas, it will heat up, and the increased pressure will cause the reaction to seek a new equilibrium. If the new equilibrium is in the endothermic direction, it would reduce the temperature and possibly cancel out the increase from the compression. The work on the piston would go into the chemical potential of the mixture without changing its temperature. I don't know of any reaction like this, but I also can't think of any reason why it wouldn't be possible. Perhaps someone with more knowledge in chemistry can check me on this.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2020-08-18T05:18:01.330" LastActivityDate="2020-08-18T05:18:01.330" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="6864" PostTypeId="1" CreationDate="2016-01-11T10:23:17.290" Score="2" ViewCount="154" Body="&lt;p&gt;I am trying to determine under what conditions falling hail will dent automobile panels. For example, is it determined by the size of the hail, the speed of impact, force, energy, etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are many variables that must be taken into account to get to an answer, such as the material strength, angle of impact, thickness of material, alloy compound, age of car, etc. but surely with the global automobile market worth over $9 trillion there must be some useful rule or heuristic for designing these panels to resist hail damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read &lt;a href=&quot;http://www.collectionscanada.gc.ca/obj/s4/f2/dsk3/OWTU/TC-OWTU-22.pdf&quot; rel=&quot;nofollow&quot;&gt;The Numerical Prediction of Panel Dent Resistance Incorporating Panel Forming Strains&lt;/a&gt; but it does not take hail as a main force into account. I am not an engineer so I don't know how to get a figure from this paper.&lt;/p&gt;&#xA;" OwnerUserId="4444" LastEditorUserId="368" LastEditDate="2016-02-03T16:33:05.250" LastActivityDate="2016-02-03T19:46:59.790" Title="How big does hail need to be to damage auto body panels?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6865" PostTypeId="1" CreationDate="2016-01-11T11:39:52.587" Score="4" ViewCount="691" Body="&lt;p&gt;To avoid the inconvenience of buying compressed &quot;air&quot; cleaners and scavenging quarters for refilling tires, I'd like to get a pump that can do 72 PSI (vapor pressure of difluoroethane minus atmosphere pressure) at ~2 liters/minute for dusting, and 35 PSI @ ~40 L/minute for filling tires (I calculated to pressurize a 33L tire @30 PSI to 35 PSI, 10.4L of uncompressed air is needed, and I'd like to do this in 15s). The dilemma is that I can't seem to find any portable pump that can operate at both these 2 points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have one of those 12V portable pumps that can go up to 250 PSI, but the time it takes to fill a single tire is painfully slow (and loud). The problem is that at a low pressure of like 30 PSI, the flow is pretty much the same (I measured about 6 L/minute with a balloon) as at a higher pressure like 80 PSI. In other words, the motor isn't delivering it's full power at low pressures. This is just like the &lt;a href=&quot;https://mechanics.stackexchange.com/questions/5574/why-do-engines-have-so-much-unused-horsepower/&quot;&gt;limited power band on diesel and gasoline engines&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there pumps that have a wide power band that would meet my requirement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or do pumps in general have limited operating points? Here's the impression I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(top - high pressure, low flow)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;piston&lt;/li&gt;&#xA;&lt;li&gt;diaphragm&lt;/li&gt;&#xA;&lt;li&gt;centrifugal pumps&lt;/li&gt;&#xA;&lt;li&gt;axial pumps / propellers&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(bottom - low pressure, high flow)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea is to make a piston pump with a large bore volume coupled to an electric motor with a constant power curve (torque = power / RPM). That way, it can operate at both high flow, low pressure (high RPM, low torque on the motor), and low flow, high pressure. But I don't think a constant power motor exist, so you'll end up needing a transmission, which will make it complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other ways?&lt;/p&gt;&#xA;" OwnerUserId="4443" LastEditorUserId="-1" LastEditDate="2017-04-13T12:47:21.447" LastActivityDate="2017-03-24T11:37:53.107" Title="Are there pumps that can vary their pressure and flow rate, while maintaining the same power?" Tags="|pumps|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="6866" PostTypeId="1" AcceptedAnswerId="6869" CreationDate="2016-01-11T13:53:43.720" Score="5" ViewCount="603" Body="&lt;p&gt;When preparing load calculations for a slab, the quick method is to define the tributary areas of a slab for the beams/columns below. These are generally taken as lines at mid-span of the slab panels and, depending on the support conditions, at the edges at angles of 30 &amp;amp; 60 or 45 &amp;amp; 45 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do the 60/30 and 45/45 angle ratios come from? Example distributions are shown in the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lK6UC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lK6UC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that in reality the ratios depend on the stiffness and end conditions of the various structural members, I also take it that the tributary areas are somewhat affected by the load they are carrying. Would excessive loading on one panel, compared to the adjacent panel affect the areas?&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="2246" LastEditDate="2016-01-11T16:43:13.617" LastActivityDate="2018-01-09T08:43:03.613" Title="Where do the tributary angles/ratios for tributary areas come from?" Tags="|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6868" PostTypeId="2" ParentId="6840" CreationDate="2016-01-11T14:38:40.503" Score="1" Body="&lt;p&gt;Initial energy (all potential):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_1=g l_1 m_1 \sin (\theta )-g l_2 m_2 \sin (\theta ) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy just before impact (all kinetic, and neglecting the mass of the rod):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_2=\frac{1}{2} m_1 \left(l_1 \omega \right){}^2+\frac{1}{2} m_2&#xA;   \left(l_2 \omega \right){}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy when the spring is completely compressed (all potential, and since $x$ is small neglect the the potential energy from the masses):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_3=\frac{k x^2}{2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\omega$ is the angular velocity. Solve for $\omega$ from the equation $E_1==E_2$ to get $\omega= 2.58002$. Then the velocity of $m_2$ just before impact can be computed as $v= l_2\omega= 1.16101$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compression x can be computed using $E_1==E_3$ and the final equation. Thus $x=\sqrt{\frac{2 E_1}{k}}=0.012$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Numeric values used $m_1=4,m_2=2,l_1=\frac{300}{1000},l_2=\frac{450}{1000},\theta =60 {}^{\circ},g=9.8,k=35000$)&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-01-11T14:38:40.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6869" PostTypeId="2" ParentId="6866" CreationDate="2016-01-11T18:05:17.707" Score="4" Body="&lt;p&gt;I wouldn't be surprised if these tributary angles are merely a rule-of-thumb obtained over time. (Correct me if I'm wrong!). While it makes obvious sense that, in a slab with neighboring fixed and pinned edges, the fixed edge will support a larger proportion of the load, I can't point to any derivation of the 30/60° division. It might have something to do with plastic hinges, which take a very similar shape as the tributary boundaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, different loads in different panels shouldn't affect the tributary areas due to the superposition principle. Assuming this principle is valid, the result of loading slab $A$ with $Q_A$ and slab $B$ with $Q_B$ should be equal to loading only slab $A$ and then adding the results due to loading only slab $B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously different from saying that loading slab $B$ won't affect slab $A$: any load applied to one slab will always generate stresses in neighboring, joined slabs (a fact usually ignored). It merely means that loading one slab won't affect the calculated tributary areas for another.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-01-11T18:05:17.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6870" PostTypeId="1" CreationDate="2016-01-11T20:20:04.887" Score="1" ViewCount="36" Body="&lt;p&gt;In the YouTube video &lt;a href=&quot;https://www.youtube.com/watch?v=q0erDDuPO7w&quot; rel=&quot;nofollow&quot;&gt;Spherical 4-Bar Infinity Mechanism&lt;/a&gt; by Thang010146, Dead Points are not compensated for by mechanical means. How can &lt;strong&gt;Autodesk Inventor Constraints&lt;/strong&gt; be applied to such a mechanism to ensure that it does not change &quot;modes&quot; into a &lt;a href=&quot;https://www.youtube.com/watch?v=XQCVg_iXV7U&quot; rel=&quot;nofollow&quot;&gt;Spherical 4-Bar Paralellogram Mechanism&lt;/a&gt;? I'm &lt;strong&gt;not&lt;/strong&gt; looking for a mechanical solution. I'm trying to learn the constraint logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.dropbox.com/s/h5zx74ltkz0v20c/Spherical-Linkage.zip?dl=0&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is my own, similar assembly. One of the bars keeps switching direction on me.&lt;/p&gt;&#xA;" OwnerUserId="3765" LastActivityDate="2016-01-11T20:20:04.887" Title="Autodesk Inventor Constraints for Clockwise / Counterclockwise / Dead Points" Tags="|modeling|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6871" PostTypeId="1" CreationDate="2016-01-11T21:51:50.147" Score="7" ViewCount="2753" Body="&lt;h2&gt;Statement&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In a crank system, as described in the picture, we know the values of: the width of the connecting rod ($m$), the width of the crank ($b$), the torque in the crank ($M$) and the angular velocity $\omega$ which we apply to this torque. We want to know the result force ($F_2$) of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iSPri.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iSPri.png&quot; alt=&quot;Crank&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note: the diagram is missing $\gamma$, which is the angle between $F_1$ and $F$&lt;/strong&gt;. Also, $F_1$ and $F$ are poorly drawn; they should verify $(1)$ below. It is just an error in the diagram, not in the concept of the question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;My attempt was the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By the diagram, we can see:&#xA;   \begin{align}&#xA; F_1 &amp;amp;= F \cos \gamma&amp;amp;&amp;amp;(1) \\&#xA; \frac{b}{\sin \alpha} &amp;amp;= \frac{m}{\sin \beta} &amp;amp;&amp;amp;(2)\\&#xA; \alpha + \beta + \gamma + 90 &amp;amp;= 180 &amp;amp;&amp;amp;(3)\\&#xA; F_2 &amp;amp;= F_1 \cos \beta &amp;amp;&amp;amp;(4)&#xA; \end{align}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Puting $(1)$ in $(4)$: \begin{align} F_2 = F \cos \gamma \cos \beta &amp;amp;&amp;amp; (5)&#xA; \end{align}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;First of all, we need a relation between $\cos \beta$ and $\alpha$.&#xA;  The only one is $(2)$, where we can get $\cos \beta$ in function of&#xA;  $\alpha$ by isolating $\cos \beta$ after having applied $\sin \beta =&#xA; \sqrt{1-\cos^2 \beta}$, so:&#xA;   \begin{align}&#xA; \frac{b}{\sin \alpha} &amp;amp;= \frac{m}{\sin \beta} \nonumber\\&#xA; \sin \beta &amp;amp;= \frac{m \cdot \sin \alpha}{b}\\&#xA; \sqrt{1-\cos^2 \beta} &amp;amp;= \frac{m \cdot \sin \alpha}{b}\nonumber\\&#xA; 1 - \cos^2 \beta &amp;amp;= \frac{m^2 \cdot \sin^2 \alpha}{b^2}\nonumber\\&#xA; \cos^2 \beta &amp;amp;= \frac{b^2 - m^2\cdot \sin^2\alpha}{b^2}\nonumber\\&#xA; \cos\beta &amp;amp;=  \sqrt{\frac{b^2 - m^2\cdot \sin^2\alpha}{b^2}}\nonumber\\&#xA; \cos \beta &amp;amp;= \frac{\sqrt{b^2 - m^2\cdot \sin^2\alpha}}{b}\nonumber&#xA; \end{align}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Puting this in $(5)$: \begin{align} F_2 = F \cos \gamma \cdot&#xA; \frac{\sqrt{b^2 - m^2\cdot \sin^2\alpha}}{b} &amp;amp;&amp;amp; (7) \end{align}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now, we need a relation between $\gamma$ and $\alpha$. From $(3)$, we've&#xA;  got: \begin{align} \gamma = 90 - \alpha - \beta &amp;amp;&amp;amp; (8) \end{align}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;However, We want $\gamma$ only in function of $\alpha$, without $\beta$. So we go to $(2)$ again to calculate $\beta$ in function of $\alpha$. By&#xA;  applying $\arcsin$, we've got: \begin{align}\beta = \arcsin&#xA; \left(\frac{m \cdot \sin \alpha}{b}\right) &amp;amp;&amp;amp; (9)\end{align}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;So, putting it into $(8)$: \begin{align*} \gamma = 90 - \alpha -&#xA; \arcsin \left(\frac{m \cdot \sin \alpha}{b}\right) \end{align*}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And putting it inside $(7)$: \begin{align*} F_2 = F \cos  \left[90 -&#xA; \alpha - \arcsin \left(\frac{m \cdot \sin \alpha}{b}\right)\right] &#xA; \cdot \frac{\sqrt{b^2 - m^2\cdot \sin^2\alpha}}{b} \end{align*}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;We can even go further. We know that $\cos(90^\circ - x) = \sin x$.&#xA;  So: \begin{align} F_2 = F \sin  \left[\alpha + \arcsin \left(\frac{m&#xA; \cdot \sin \alpha}{b}\right)\right]  \cdot \frac{\sqrt{b^2 - m^2\cdot&#xA; \sin^2\alpha}}{b} \end{align}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Simplifying with Wolfram Alpha:  \begin{align*} F_2 &amp;amp;= F \left(&#xA; \sin(\alpha) \sqrt{1-\frac{m^2}{b^2}\sin^2(\alpha)} + \frac{m}{b}&#xA; \sin{\alpha } \cos { \alpha} \right) \cdot \frac{\sqrt{b^2 - m^2\cdot&#xA; \sin^2\alpha}}{b}\\  F_2 &amp;amp;= F \cdot \sin\alpha \cdot \frac{ m\cos&#xA; \alpha \sqrt{b^2 - m^2 \sin^2\alpha} + b^2 - m^2 \sin^2 \alpha }{b^2}&#xA; \end{align*}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Finally, if we want to express this identity in function of $\omega$,&#xA;  $t$ and $M$ we just have to substitute $\alpha$ by $\omega t$ and $F$&#xA;  by $M/m$, so:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;\begin{align*}  \boxed{F_2 = \frac{M}{m} \cdot \sin(\omega t) \cdot&#xA; \frac{ m\cos(\omega t) \sqrt{b^2 - m^2 \sin^2(\omega t)} + b^2 - m^2&#xA; \sin^2 (\omega t) }{b^2}} \end{align*}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Is this solution correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;PS&lt;/em&gt;: my level of physics studies is quite low, so, please, restrict to this level of difficulty.&lt;/p&gt;&#xA;" OwnerUserId="4452" LastEditorUserId="1832" LastEditDate="2016-01-11T22:56:19.997" LastActivityDate="2016-01-17T02:00:47.613" Title="Force in a crank system" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6872" PostTypeId="1" AcceptedAnswerId="6874" CreationDate="2016-01-12T01:04:44.097" Score="2" ViewCount="1077" Body="&lt;p&gt;I'd like to design some things with bolts, fan belts, levers, motors, etc.  So I'd add a bolt here, drill a hole there and maybe take it apart and rearrange.  I'd like to do it in 3D and found SketchUp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be used to design, create and print out dimensioned components for manufacture and assembly? I'm not knocking it, but is Sketchup anything other than a really good conceptual visualization tool?  &lt;/p&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="33" LastEditDate="2016-01-15T01:39:43.330" LastActivityDate="2016-01-15T01:39:43.330" Title="Can Sketchup produce drawings that are adequate to manufacture components?" Tags="|computer-aided-design|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6873" PostTypeId="1" CreationDate="2016-01-12T01:11:49.993" Score="1" ViewCount="4131" Body="&lt;p&gt;Can a worm gear be used for linear motion as well as for rotation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific. Suppose I want the worm gear to rotate as well as move forward, is it possible to achieve that ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ur4Pl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ur4Plm.png&quot; alt=&quot;worm gear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4454" LastEditorUserId="272" LastEditDate="2016-01-12T12:48:18.527" LastActivityDate="2022-11-02T08:24:05.283" Title="Worm Gear and Worm Shaft for 2 axis" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6874" PostTypeId="2" ParentId="6872" CreationDate="2016-01-12T04:50:35.607" Score="1" Body="&lt;p&gt;If Sketchup cannot produce dimensioned drawings of parts and assemblies, then it is just conceptual. Solidworks has by far the most market share of PC-based 3D CAD for machinery, and has a one-month evaluation. AutoDesk Inventor and Solidedge are similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If 2D is okay, you can get a free AutoCAD clone &lt;a href=&quot;http://www.3ds.com/products-services/draftsight-cad-software/free-download/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.plm.automation.siemens.com/en_us/products/solid-edge/free2d/index.shtml&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt;'s another free 2D product with the same functionality as AutoCAD, but it was designed for Windows and is easier to use if not familiar to AutoCAD command line or DOS commands. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These companies are competitors to AutoDesk, and they give away 2D software so AutoDesk makes less money with AutoCAD.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastEditorUserId="1832" LastEditDate="2016-01-12T13:06:51.050" LastActivityDate="2016-01-12T13:06:51.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6875" PostTypeId="2" ParentId="6723" CreationDate="2016-01-12T06:21:59.910" Score="1" Body="&lt;p&gt;Short answer: The FLUENT approach is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many information are lost due to the space-averaging process - over a representative elementary volume - of the governing equations (which is the essence of every porous model that tries to avoid the complexity of the real geometry of the porous media), So any turbulence model is not to &lt;em&gt;&quot;reproduce the fine structure dynamics of the flow but to take into account information embedded in smaller scale for large scale modelization.&quot;&lt;/em&gt;[1]. The FLUENT approach simply ignores this fact or even assumes that there is no turbulence energy dissipation or generation due to the porous zone (similar approach was taken by Antohe and Lage to develop a new turbulence model that lead to trivial solutions ($k = 0$ or $\epsilon = 0$)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that STAR-CCM+ code follows a similar way in turbulence treatment:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The effect of a porous region on turbulent flow depends on the internal&#xA;  structure of the porous medium. Where turbulence is present, the&#xA;  turbulence scales are determined from the geometric structure of the porous&#xA;  medium. As it is not possible for STAR-CCM+ to predict the turbulence&#xA;  scales directly, you specify the appropriate values on the porous region.&#xA;  Turbulence quantities in fluid leaving the porous region are constructed&#xA;  from the user-defined values; &lt;strong&gt;they are not transported from the upstream&#xA;  side of the porous region.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, I think if you are pragmatic enough to have a model that can sacrifice the accuracy of results of the turbulence model you use, these assumptions are the best you can get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] $k–\epsilon$ Macro-scale modeling of turbulence based on a two scale analysis in porous media - Francois Pinson, Olivier Gregoire, and Olivier Simonin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] A new turbulence model for porous media flows. Part I: Constitutive equations and model closure - Federico E. Teruel and Rizwan-uddin.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2016-01-12T06:32:22.570" LastActivityDate="2016-01-12T06:32:22.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6876" PostTypeId="2" ParentId="6865" CreationDate="2016-01-12T07:48:51.423" Score="1" Body="&lt;p&gt;There are many cases in engineering where one device does not economically cover two situations. A sports car can go 250mph and a pickup can tow a very large trailer, but a vehicle that can do both would not be economical to own or drive. You would be better off buying one of each; even though at first glance the situation seems wasteful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of your applications will fall in the diaphragm or piston pump region. And sure, your could put a continuously variable transmission on and air compressor, but it would be complicated and very expensive. For your situation there are much easier/cheaper ways to increase portability and performance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have access to a 120v outlet or a 1500W inverter, the cost to cover both situations is minimal. This low-cost oil-less &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00K34UZBW&quot; rel=&quot;nofollow&quot;&gt;DEWALT Model DWFP55126&lt;/a&gt; compressor will easily handle both of your applications. It has a maximum delivery of 2.6 SCFM (74L/minute) at 90psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This smaller pancake &lt;a href=&quot;http://www.homedepot.com/p/HDX-2-Gal-Pancake-Air-Compressor-0210284C/205467817&quot; rel=&quot;nofollow&quot;&gt;HDX Model 0210284C&lt;/a&gt; compressor would only require a 400W inverter. It max flowrate is only 0.7 SCFM (20L/min), but it will continue filling the storage tank while you are getting ready for the next tire, so the effective delivery for an intermittent task like filling tires will be greater than the listed amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The noise level and longevity of the pump will depend on the quality of pump purchased and the rpm it operates at.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-01-12T07:48:51.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6877" PostTypeId="2" ParentId="5923" CreationDate="2016-01-12T08:59:30.013" Score="-1" Body="&lt;p&gt;Q- The aim of this research study is mainly based on the comparison of measured frequency response function results for the threshold with and without damage, and comparing them with the theoretical model to update the modal parameters.&#xA;The study will be performed on isotropic beam, to study the vibration characteristics. First Step, the beam will be tested without damage, and the frequency response function will be measured, then the natural frequencies and mode shapes, can be evaluated. To ensure these values, analytical model based on modal analysis approach will be utilized to predict the structural dynamic response of the beam and comparing it with the experimentally FRF. Second Step, cut in the beam, will be induced and getting again the FRF experimentally which will be used as input data to the analytical model to get the modal parameters. Last Step, the results from the previous steps will be treated to update the modal parameters (mass and stiffness matrix).&lt;/p&gt;&#xA;" OwnerUserId="4460" LastActivityDate="2016-01-12T08:59:30.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6878" PostTypeId="2" ParentId="6873" CreationDate="2016-01-12T13:01:38.530" Score="1" Body="&lt;p&gt;Interestingly, this is the opposite of the desired behavior of zoom lenses.  Optics design houses go to some lengths with cams and retainers to allow the lens to extend without rotation.  You might try looking up old (1970s) zoom lens mechanical patents to see if there's anything there you could use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as to pure worm gears, since the worm only rotates if it's constrained from translating, you'd need some &quot;secondary&quot; worm that translated the &quot;primary&quot; worm at half the drive gear's linear rate.  Even then I would be concerned that the tiniest error in step-down rate would cause catastrophic lockup.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-01-12T13:01:38.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6879" PostTypeId="2" ParentId="6830" CreationDate="2016-01-12T15:25:43.147" Score="5" Body="&lt;p&gt;It's easier to understand this identity if you start with the partial differential equation for the Euler-bernoulli beam deflection equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^2}{dx^2}\left[ EI \frac{d^2u}{dx^2}\right] = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and work your way down to the weak form.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply both sides of the equation by an arbitrary test function $w$.  Then apply integration by parts (only once) over the domain [0,L].  You will obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w\frac{d}{dx}\left[ EI \frac{d^2u}{dx^2}\right]\bigg|_0^L - \int_0^L \left[ EI \frac{d^2u}{dx^2}\right] \frac{dw}{dx}= 0.$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, apply integration by parts again, we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w\frac{d}{dx}\left[ EI \frac{d^2u}{dx^2}\right]\bigg|_0^L - \frac{dw}{dx}EI \frac{d^2u}{dx^2}\bigg|_0^L + \int_0^L \frac{d^2w}{dx^2}EI \frac{d^2u}{dx^2}=0$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging the terms evaluated at 0 and L to the right hand side, you will obtain the equation you state.&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2016-01-12T15:25:43.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6880" PostTypeId="2" ParentId="6872" CreationDate="2016-01-12T16:09:40.330" Score="3" Body="&lt;p&gt;I've used Sketchup for the last 6 years as my go-to 3D engineering program.  When combined in the Pro version with Layout, it can be fully dimensioned, rendered, and turned into full scale 2D production drawings that the shop can build from.  In the trial version, without the Layout program, it is only conceptual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said - Sketchup is intended for architects.  It's built with a bunch of tools for them in mind - including an absolute up and down.  As such, I'd only use Sketchup for items on the scale of houses.  In my line of work, we build giant ducting systems and process vessels - Sketchup is perfect.  &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2016-01-12T16:09:40.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6882" PostTypeId="1" CreationDate="2016-01-12T18:09:51.707" Score="1" ViewCount="60" Body="&lt;p&gt;Is there any suggestion on sliding surface state  derivative matrix (ds/dx) format?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ds/dx = [ds/beta,ds/dalpha,ds/dp,ds/dq,ds/dr,ds/dphi;&#xA;         ds/beta,ds/dalpha,ds/dp,ds/dq,ds/dr,ds/dphi;&#xA;         ds/beta,ds/dalpha,ds/dp,ds/dq,ds/dr,ds/dphi];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ds/dx = [0,0,ds/dp,0,0,ds/dphi;&#xA;         ds/beta,0,0,0,ds/dr,0;&#xA;         0,ds/dalpha,0,ds/dq,0,0];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;problem described below &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the article &lt;a href=&quot;http://www.wseas.us/e-library/conferences/2009/lalaguna/MANOCO/MANOCO17.pdf&quot; rel=&quot;nofollow&quot;&gt;Controller Design using Bifurcation Map for Aircraft Spin Recovery&lt;/a&gt;, the sliding surface derivative is obtained using a numerical method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steady space system is defined by the following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\underline{x}}=A(\underline{x},t)+B(\underline{x},t)\underline{U}, \\&#xA;\underline{x}=[V,\alpha,\beta,p,q,r,\phi,\theta], \\&#xA;s=\left[\begin{array}{c}&#xA;\dfrac{d}{dt}(\phi-\phi_d)+\lambda_1(\phi-\phi_d) \\&#xA;\dfrac{d}{dt}(\alpha-\alpha_d)+\lambda_2(\alpha-\alpha_d) \\&#xA;\dfrac{d}{dt}(\beta-\beta_d)+\lambda_3(\beta-\beta_d) \\&#xA;\end{array}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\lambda_1$, $\lambda_2$ and $\lambda_3$ are positive real numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The control law from reaching law in Eq. (6) can be derived&#xA;  as:&#xA;  $$\dot{s}=\dfrac{\partial{s}}{\partial{x}}\dot{x}=\dfrac{\partial{s}}{\partial{x}}(A+Bu)=-Qsgn(s)+Kf(s).\tag{8}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By solving the above equation (8) for u (3 x 1), one gets expressions&#xA;  for u as:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$u=-\left(\dfrac{\partial{s}}{\partial{x}}B\right)^{-1}\left[\dfrac{\partial{s}}{\partial{x}}A+Qsgn(S)+Kf(s)\right].\tag{9}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The matrix $\frac{\partial{s}}{\partial{x}}$ is computed numerically.&#xA;  The control inputs computed are subject to position and rate&#xA;  constraints on control surface deflections as listed in the Table 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How would be the exact form of matrix of $\dfrac{\partial s}{\partial x}$ (element of $\dfrac{\partial s}{\partial x}$ in columns and raws)?&lt;/p&gt;&#xA;" OwnerUserId="4412" LastEditorUserId="4412" LastEditDate="2016-01-13T07:07:23.897" LastActivityDate="2016-01-13T07:07:23.897" Title="Sliding mode surface derivative exact matrix format (ds/dx)" Tags="|control-engineering|aerospace-engineering|numerical-methods|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6883" PostTypeId="2" ParentId="2992" CreationDate="2016-01-12T20:21:50.423" Score="2" Body="&lt;p&gt;I would just like to add, in addition to the other answers regarding the EM spectrum, that anything that follows the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wave_equation&quot; rel=&quot;nofollow&quot;&gt;wave equation&lt;/a&gt; can be focused because all of the principles of constructive and deconstructive superposition apply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things outside the EM spectrum that follow the wave formula include pressure waves (when focused, commonly referred to as &lt;a href=&quot;https://en.wikipedia.org/wiki/Sonar&quot; rel=&quot;nofollow&quot;&gt;sonar&lt;/a&gt;), and even matter waves, such as focused ocean waves, called &lt;a href=&quot;https://en.wikipedia.org/wiki/Rogue_wave&quot; rel=&quot;nofollow&quot;&gt;rogue waves&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-01-12T20:21:50.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6884" PostTypeId="2" ParentId="2911" CreationDate="2016-01-12T21:29:25.437" Score="2" Body="&lt;p&gt;Strictly efficiency is a ratio of the energy you put into a system compared to the useful work you get out so for example an electric motor that was 90% efficient might draw 100W of electrical power and produce 90W of mechanical power at the shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawback with using efficiency as a measure for vehicles is that it doesn't directly tell you want you need to know since it is, by definition a ratio of power and doesn't in itself tell you the amount of energy required to travel a particular distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual amount of energy used by a car to travel a particular distance is determined not just by the efficiency of its engine but also its mass, aerodynamic drag (plus other resistive forces) and any other power used by sub-systems (eg air conditioning, lights etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind you can have a heavy car which is very efficient (in the sense that it converts a lot of the energy from burning fuel into mechanical power) but still has poor fuel economy because it needs more energy to accelerate its mass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So overall quoting mpg or litres per km gives you more useful information about how much fuel a vehicle actually uses. Just saying that a vehicle is xxx% efficient doesn't actually tell you much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you could quote the power use by a vehicle in standard conditions in kW but that's still less useful as fuel is generally sold by volume and distance is a better approximation of the energy requirements of a given journey than the time taken. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-12T21:29:25.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6885" PostTypeId="2" ParentId="2349" CreationDate="2016-01-12T22:53:50.567" Score="2" Body="&lt;p&gt;Bear in in mind that the chart shown relates to the thermal efficiency of the engine alone and doesn't consider a moving car as a complete system and also that the vertical axis appears to be labeled as a ratio rather than a percentage efficiency so 0.2 on the graph is equivalent to 20% &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a car as a whole, engine load is not independent of speed as drag forces tend to increase with the square of speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason that IC engines have lower efficiency at lower loads is that a lot of the energy losses are fairly constant, regardless of the power being produced; these include the power required for things like air induction, pumping fuel, coolant and lubricants and powering electrical systems. So at lower power output these losses represent a much greater proportion of the overall power produced by the engine. For example, say that an engine has a constant requirement of 1 kW for all the ancillaries. If it is producing an output power of 1 kW then these losses are half of the total power being produced, whereas when the engine is developing 50 kW than they are just 2% of the total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a larger engine with higher power output these (approximately) constant losses will tend to be larger than in a smaller capacity engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the efficiency of the engine doesn't relate directly to mpg. An efficient engine in a heavy car could still have higher fuel consumption than a relatively inefficient engine in a light car. To put it another way, mpg is related to how much energy you need to travel a certain distance but efficiency measures how much of that energy is wasted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram 'load' is the torque produced by the engine. For the purposes of testing it will be connected to some adjustable load (e.g., generator and a bank of resistors). The power produced is this torque multiplied by the angular velocity (RPM). Note that all engines have a practical limit on RPM due to the acceleration forces on moving parts (especially valves and connecting rods) so the output power is a function of both the load and the RPM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way, 25% load is not the same as 25% of peak power. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="368" LastEditDate="2016-02-10T17:20:32.630" LastActivityDate="2016-02-10T17:20:32.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6886" PostTypeId="2" ParentId="3502" CreationDate="2016-01-12T23:25:37.253" Score="1" Body="&lt;p&gt;If you are pumping using suction, the flow is driven by the pressure difference between atmospheric pressure and the vacuum. So if atmospheric pressure is 1 bar then 1 bar is the maximum pressure difference you can create, regardless of the pump design or power as you can't have negative pressure (in the sense of less pressure than a vacuum). So in this case the flowrate is quickly limited by the size of the orifice you are pumping through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get higher pumping pressures you need to be directly pressuring the fluid you are trying to pump&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-12T23:25:37.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6887" PostTypeId="2" ParentId="3736" CreationDate="2016-01-12T23:46:55.850" Score="1" Body="&lt;p&gt;To give you a very simplified overview of the process. If we assume a simple carbon steel eg a typical spring steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important thing to understand is that steel has a critical temperature at which its crystal structure changes this is around 800 deg C but varies for different alloys and roughly corresponds to a change from a red to a yellow glow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steel is then cooled from this temperature at different rates to get different results. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Annealing : by cooling very slowly eg by packing in insulating material or in a temperature controlled oven the steel becomes softer and more ductile. This process also relives work hardening and residual stress. &lt;/li&gt;&#xA;&lt;li&gt;Hardening : if the steel is cooled rapidly eg by quenching in oil it takes on a structure with high hardness and high tensile strength, however in this condition it is very brittle and usually requires tempering to be useful. &lt;/li&gt;&#xA;&lt;li&gt;Tempering : involves heating to a specific temperature (usually in the range 170-300 deg C) this reduces brittleness and improves toughness, precisely controlling this temperature is used to produce an optimum balance of hardness and toughness for different applications. &#xA;-Normalising : the steel is heated to critical temperature and allowed to cool in air, this neither hardens nor fully anneals the steel but refines the grain structure, improving both strength and toughness. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Carburising may apply to a number of processes from producing steel from raw materials to surface hardening processes like case hardening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes : accurate temperature control is extremely important in all aspects of heat treating and in most cases it is important to research the precise methods and specific temperatures needed for a given grade of steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general steels with higher carbon content will be harder and have higher tensile strength after heat treating but may be less tough than medium carbon steels. Steel with less than around 0.4% carbon content will not harden significantly. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-12T23:46:55.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6888" PostTypeId="1" AcceptedAnswerId="6889" CreationDate="2016-01-13T01:35:41.500" Score="4" ViewCount="1448" Body="&lt;p&gt;For a 3D printing application, I need to have a spinning disc(that I will rotate via geared stepper motors) that will wobble as little as possible. By wobble, I mean that it will only rotate around it's axis and not shift side to side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to build a 3d printer that uses polar coordinates rather than traditional Cartesian coordinates. Instead of using leadscrews, I want to instead rotate two discs. Disc 1 is mounted to disc 2 which is mounted to an immobile surface. If I can precisely rotate both, I should be able to 3d print. So the question is how can I mount them such that they don't require much torque to spin and also spin accurately around their axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have ordered some lazy Susan type bearings from McMasterCar, but they wobble too much under use. I've heard that there are axial and thrust bearings and that the ones I ordered are thrust. Should I use a combination of the two (and how would you suggest mounting it?) or should I use the diagonal type that has both combined in one assembly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disc is 8 inches in diameter and I would like to keep the assembly thin. The disc is spinning parallel to the ground; if I were to mount the bearing underneath, it would be in compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sketch showing one of the axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vm9RQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vm9RQm.jpg&quot; alt=&quot;close up&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sketch shows an overview of the whole system:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HIQ87.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HIQ87m.jpg&quot; alt=&quot;system overview&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4466" LastEditorDisplayName="user16" LastEditDate="2016-02-05T21:23:41.820" LastActivityDate="2016-02-05T21:23:41.820" Title="How to secure a rotating disc so that it has no wobble" Tags="|mechanical-engineering|3d-printing|bearings|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6889" PostTypeId="2" ParentId="6888" CreationDate="2016-01-13T01:50:13.157" Score="1" Body="&lt;p&gt;The bearings required will depend on the loads and rotational speeds involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the situation you describe you will probably have a combination of axial and thrust loads ie the thrust load is the weight of the disk and axial loads come from torque on the drive shaft from the drive torque and unevenly distributed loads on the disk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to consider how accurately your assembly will be aligned, different types of bearings have different tolerances for misalignment. For a driven shaft you generally need a pair of bearings per shaft so that it is supported at two points to minimise bending loads on the bearings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this sort of project often the most convenient solution is to use 'pillow block' type bearings which are an assembly which includes the bearing and a housing with mounting holes as a complete assembly and you can just mount the whole thing to a suitable point in your structure  without having to design and build a housing for the bearings from scratch. Again you need to make some sort of estimate of the radial and axial loads involved and choose a bearing to suit from the manufacturers specifications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional thought&lt;/strong&gt; : the situation you describe is broadly similar to the hub on a rear wheel drive car so it might help to look at the way that hubs are designed. It may even be that an off-the-shelf hub unit from a lightweight vehicle/robot  (or perhaps an industrial conveyor)  would be the easiest solution rather than designing and making something from scratch &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-05T20:35:25.983" LastActivityDate="2016-02-05T20:35:25.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6890" PostTypeId="2" ParentId="3736" CreationDate="2016-01-13T02:09:52.630" Score="1" Body="&lt;h2&gt;Work Hardening&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Metals deform plastically by means of shear between atomic layers. However, shearing an entire plane of atoms simultaneously takes proprotionally enormous stress. Instead, a half-plane of atoms perpendicular to both of the shear direction and shear plane gets pushed about one atom radius away from the shear plane, forming a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dislocation&quot; rel=&quot;nofollow&quot;&gt;&lt;strong&gt;dislocation&lt;/strong&gt;&lt;/a&gt; at the shear plane. The dislocation is easily moved by application of shear stress in a manner likened to that of an inch-worm. As dislocations move, they get caught on other defects (especially other dislocations, grain boundaries, and hard inclusions) and require greater stress to keep them moving. While the full atomic level process that causes work hardening is not fully understood (though there are theories such as dislocation pinning and &lt;a href=&quot;https://en.wikipedia.org/wiki/Frank-Read_Source&quot; rel=&quot;nofollow&quot;&gt;Frank-Read sources&lt;/a&gt;), dislocations increase in density with increasing plastic deformation. As they increase in density, they tend to run into each other more, requiring ever greater stress to continue deformation, resulting in strain hardening. Consequently, with increasing plastic deformation, or work, ductility decreases and yield strength increases.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Annealing&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Dislocations deform crystals locally producing strain fields, which store energy. As a result, dislocations increase the &lt;a href=&quot;https://en.wikipedia.org/wiki/Free_energy&quot; rel=&quot;nofollow&quot;&gt;free energy&lt;/a&gt; available in a crystal. Free energy drives spontaneous processes, which in this case is the reduction of strain energy by removal of dislocations. Metal crystals also have &lt;a href=&quot;https://en.wikipedia.org/wiki/Vacancy_defect&quot; rel=&quot;nofollow&quot;&gt;vacancies&lt;/a&gt;, or missing atoms, which can interact with dislocations to move them to the surface of the crystal where they are readily annihilated. Dislocations may also annihilate each other if two opposing (up vs. down) half planes meet. Increasing temperatures causes dislocation removal processes to occur exponentially more rapidly, eventually occuring within minutes or even seconds depending on the metal and the section thickness. As a result of dislocation density reduction, the properties of the metal change opposite that of work hardening: yield strength decreases while ductility increases.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Tempering&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Tempering is a mostly unique process in that it is almost exclusively used with respect to martensitic steel. The term is also used with other materials such as glass and some high-performance polymers. If a hardenable steel is quenched rapidly enough the microstructure that forms contains &lt;a href=&quot;https://en.wikipedia.org/wiki/Martensite&quot; rel=&quot;nofollow&quot;&gt;martensite&lt;/a&gt;. Martensite forms as a result of carbon solubility in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Austenite&quot; rel=&quot;nofollow&quot;&gt;austenite&lt;/a&gt; phase of steel at elevated temperatures. Ferrite has very low carbon solubility, so as temperature decreases carbon proceeds to diffuse out of the iron-rich phase. If the cooling rate is slow, the carbon is able to diffuse out and form &lt;a href=&quot;https://en.wikipedia.org/wiki/Cementite&quot; rel=&quot;nofollow&quot;&gt;cementite&lt;/a&gt;. If the cooling rate is sufficiently rapid, the diffusion process is too slow and the austenite transforms to martensite instead of ferrite, trapping carbon in the martensite crystals. Ferrite has a BCC lattice, while martensite has a strained BCT lattice due to the trapped carbon atoms. The sudden volume change of the transformation, creates internal stresses in the grains causing rapid, intra-grain work hardening. As a result, the strength of the martensitic steel is high and the ductility low, making it brittle. Tempering is effectively an annealing process, reducing the effects of the transformation-induced work hardening, and restoring some ductility. Interestingly, a carefully crafted tempering schedule can restore significant ductility without significantly reducing strength. The restored ductility is due to the reduction in dislocation density, while the maintained strength is due to the carbon-strained BCT lattice. It takes more stress to move a dislocation through a strained lattice. If the tempering goes on too long or at a too-high temperature, the carbon will diffuse out of the martensite, forming ferrite and pearlite, and return strength and ductility to that of a non-martensitic steel.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Quenching&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Quenching is useful for producing martensite, as noted, but may also be used to produce other microstructural components including &lt;a href=&quot;https://en.wikipedia.org/wiki/Bainite&quot; rel=&quot;nofollow&quot;&gt;bainite&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Pearlite&quot; rel=&quot;nofollow&quot;&gt;pearlite&lt;/a&gt;, and combinations of all of the above, according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Isothermal_transformation_diagram&quot; rel=&quot;nofollow&quot;&gt;TTT diagrams&lt;/a&gt;, also known as isothermal transformation diagrams. It is possible to schedule heat treatments to take advantage of quenching by rapidly reducing the temperature of a steel component to a specific temperature and holding it there. Such a quench and hold treatment allows one particular microstructure, as noted on the TTT, to form instead of the others. Naturally if such a temperature is higher than the boiling point of water, another medium such as heated oil must be used. By quenching, atomic mobility is decreased exponentially, preventing or slowing diffusion-based processes. The formation of pearlite and bainite are both diffusion based, while martensite is diffusionless. Specifically, martensite requires there to be minimal carbon diffusion.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Carburizing&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Carburizing is a diffusion-based process by which carbon is implanted in a metal at elevated temperatures. By now you can probably guess that elevated temperatures make the process go exponentially faster. The process can also be controlled based on the external carbon source concentration, and the amount of time spent processing. A carbon gradient forms in the metal, with the highest concentrations at the surface and decreasing inward. As noted in the tempering section with martensite, higher carbon content distorts the iron lattice, increasing internal stress, hindering dislocation mobility and increasing strength. Higher strengths are also associated with higher hardness and ultimately higher wear resistance. Note however that martensite does not form as a result of carburization, instead the existing BCC lattice becomes highly strained. Parts such as gears are often formed from a relatively low carbon steel and then carburized. This gives a shell of hard, wear resistant steel surrounding a ductile, damage-tolerant core which limits through-cracks. That way the surface may wear and chip, but the entire gear won't crack. A related process is nitriding, using nitrogen instead of carbon. Some metals exhibit extreme brittleness with low concentrations of one or both materials, so not all metals may be strengthened this way.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Final Notes&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;While most of the above has been applied to steel, these concepts may typically be broadly applied. Work hardening and annealing apply to all sufficiently impure metals at and above room temperature, with a few exceptions (gallium and mercury spring to mind). &quot;Pure&quot; metals also exhibit unusually high ductility compared with metals containing even slight (&gt;0.1%) impurities, as do single crystals. Six-nines fine gold can be worked endlessly at room temperature as the dislocations are annihilated as fast as you can create them. Pure iron anneals itself over the course of a few minutes at room temperature. However even A 0.01% carbon impurity is enough to raise the one-hour-per-inch annealing temperature to hundreds of Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tempering generally only applies to martensitic steels in the realm of metals, as far as I am aware. Some nickel alloys are capable of forming a diffusionless martensitic phase, and if they were widely used commercially I imagine tempering would be applicable to them as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quenching is applicable to any metal that requires rapid cooling as part of a heat-treatment schedule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carburizing applies to any metal for which it may be desirable to implant carbon atoms in the crystal lattice. Aluminum and gold are not good candidates for carburizing as they have virtually no solubility. Nitriding is the same process as carburizing but with nitrogen. Other non-metal elemental gases are generally deleterious for all properties, especially oxygen and hydrogen. Hydrogen makes virtually every metal extremely brittle, and oxygen oxidizes metal.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-01-13T02:09:52.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6891" PostTypeId="1" CreationDate="2016-01-13T03:34:12.947" Score="1" ViewCount="203" Body="&lt;p&gt;A previous question here was relating wind deflection to drag creation in vehicle solar panel installations. In a residential installation drag is not the issue, but uplift is a critical concern.  Since the efficiency of solar panels is affected by temps there will need to be sufficient spacing above the roof. I'd like to know whether a solid or permeable windbreak could provide useful protection for the panels against damage from hurricane force winds or would it instead increase uplift?&lt;/p&gt;&#xA;" OwnerUserId="4467" LastActivityDate="2016-01-13T11:58:47.223" Title="Wind protection for solar installation" Tags="|airflow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6895" PostTypeId="2" ParentId="6891" CreationDate="2016-01-13T11:58:47.223" Score="1" Body="&lt;p&gt;Regulatory requirements will usually stipulate a mounting that will meet uplift conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uplift is very approximately proportional to velocity-cubed normal to panel (eg under it facing up - so slower than through flow velocity BUT may be close. Halving peak velocity reduces max uplift by factor of V^3 = 8 times !  So a permeable windbreak can help muchly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: All advice worth what you paid for it (or less).&lt;br&gt;&#xA;Check regulations.&lt;br&gt;&#xA;Ask an engineer qualified to certify installations in your area.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drag ~~~~= 0.5 x density x Area x Drag_Coefficient x Velocity^3  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density at sea level = 1.2 kg/m^3.&lt;br&gt;&#xA;For A in m^2, V in m/S, g = 9.8 ~= 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force in kg ~= 0.06 x A x V^3&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2016-01-13T11:58:47.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6897" PostTypeId="2" ParentId="4136" CreationDate="2016-01-13T18:59:03.483" Score="1" Body="&lt;p&gt;For something like this with very low loads it might be that something as simple as a teflon ring underneath  would be the easiest solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something else to consider in terms of the hi speed vs high precision problem ie the ability to both traverse a large screen and make very precise movements can also be addressed by having a variable dpi sensor which is switched by a button, this is now quite a common feature on gaming mice. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-13T18:59:03.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6898" PostTypeId="1" CreationDate="2016-01-13T18:26:47.107" Score="0" ViewCount="318" Body="&lt;p&gt;Which U.S. states offer Professional Engineer (PE) licensure for software engineering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my research, few of the state licensing boards listed on the &lt;a href=&quot;http://ncees.org/licensing-boards/&quot; rel=&quot;nofollow&quot;&gt;NCEES site&lt;/a&gt; include the Software Engineering discipline.  Most focus on engineering disciplines such as Civil, Chemical, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I checked with NCEES, they recommended contacting each state &lt;strong&gt;individually&lt;/strong&gt;!  Yuck. &lt;/p&gt;&#xA;" OwnerDisplayName="Lawrence P. Kelley" LastEditorUserId="368" LastEditDate="2016-01-13T20:23:54.633" LastActivityDate="2016-01-13T20:23:54.633" Title="Which states offer PE licensure for Software Engineering?" Tags="|licensure|software|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="6899" PostTypeId="1" CreationDate="2016-01-13T20:30:13.383" Score="2" ViewCount="320" Body="&lt;p&gt;I am working on ''Active noise cancellation for air vent'' project. I am using &lt;strong&gt;NI myRIO1900&lt;/strong&gt; it has &lt;strong&gt;FPGA module(Xylinx Z-7010)&lt;/strong&gt; , Audio Technica ATR 3350 omni directional microphones and Tevion Loudspeakers. I have done the programming using LabVIEW . I am using &lt;strong&gt;sinusoidal noise source of (200-500 Hz)&lt;/strong&gt; as test noise signal to simulate the noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have implemented &lt;strong&gt;LMS(least mean squares)&lt;/strong&gt; algorithm on LabVIEW, as a preliminary results I am getting around 8dB noise reduction but these results are pretty random in a sense only at certain volume levels( if I adjust very carefully), the noise cancellation is happening and the variation in step size is not giving any decent change ( again I have to make volume level changes to see some results).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over all, I am getting some results but I don't have control over the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my set up &#xA;&lt;a href=&quot;https://i.stack.imgur.com/vGsF7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vGsF7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;[the pipe length is 1 meter]&#xA;I know I have to consider secondary path modelling and feedback modelling. I have actually completed the coding of FBFxLMS(Feedback Filtered x LMS Algorithm) also but I don't have any reasonable results with this also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read in 2-3 places that without proper acoustical design it is not possible to reduce the noise effectively. Quoting from &lt;a href=&quot;http://www.ti.com/lit/an/spra042/spra042.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper page 5&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;''If the acoustical design of the system is not optimized,the digital&#xA;  controller may not be able to attenuate the undesired noise&#xA;  adequately.''  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am using this paper as my major reference.&lt;br&gt;&#xA;The only point I am aware of is using low frequency noises (&amp;lt;= 500Hz) to consider the sound wave propagation as plane acoustic wave. I am completely unaware of any other details on how to better my acoustical design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I am wondering how to improve the acoustical design of my setup for effective noise cancellation?&lt;/strong&gt;&lt;br&gt;&#xA;What are the problems using just cardboard material to simulate the environment of air vent ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me if some one has experience with ANC or Acoustical design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;br&gt;&#xA;&lt;strong&gt;I am just using this model as prototype to demonstrate effective and controlled Active Noise cancellation.I don't have any practical model to compare to(right now!).&lt;/strong&gt;&lt;br&gt;&#xA;I am looking for some useful references or previous works dealing with acoustical design of Active Noise Cancellation(ANC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit2:&lt;br&gt;&#xA;More details:&lt;br&gt;&#xA;1.In the picture the Anti noise speaker has just been inserted without proper sealing.&lt;/p&gt;&#xA;" OwnerUserId="4480" LastEditorUserId="4480" LastEditDate="2016-01-14T17:21:45.050" LastActivityDate="2016-01-14T17:21:45.050" Title="Acoustical Design considerations for Active Noise Cancellation Project for Air vent" Tags="|audio-engineering|signal-processing|" AnswerCount="2" CommentCount="8" ClosedDate="2016-01-14T15:25:25.603" ContentLicense="CC BY-SA 3.0" />
  <row Id="6900" PostTypeId="2" ParentId="4136" CreationDate="2016-01-13T21:32:00.963" Score="3" Body="&lt;p&gt;To answer the question on what would allow it to roll &quot;most freely&quot;, an air bearing would be my recommendation, as it would let the ball spin in place for long durations. Air bearing bands around the ball would not interfere with the optical tracking sensor, and can handle hand pressure on the ball with little to no added resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spherical air bearings are also a commonly used air bearing type, so it should not be too difficult to find an existing 5cm one that would fit you needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this would be one of the more expensive solution to the problem, and you need a compressed air supply... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another non-contact bearing would be magnetic levitation, but that would be even more complicated and would likely heat up the sphere considerably.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastEditorUserId="1459" LastEditDate="2016-01-13T21:43:15.290" LastActivityDate="2016-01-13T21:43:15.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6901" PostTypeId="2" ParentId="6899" CreationDate="2016-01-13T21:40:02.993" Score="2" Body="&lt;p&gt;I don't think there's enough room in a comment to add all of this, so that's why I'm posting this as an answer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What do you mean when you say, &quot;I have implemented LMS on Labview&quot;? What is LMS?&lt;/li&gt;&#xA;&lt;li&gt;What FPGA board are you using? Is this an off-the-shelf part or some home brew hardware?&lt;/li&gt;&#xA;&lt;li&gt;You're not happy with your test results - what are they? Can you post before and after spectral results?&lt;/li&gt;&#xA;&lt;li&gt;How are you simulating noise?&lt;/li&gt;&#xA;&lt;li&gt;How are you measuring audio output? All I see in your image is two speakers and a cardboard duct. &lt;/li&gt;&#xA;&lt;li&gt;What is the purpose of the duct? Are you trying to simulate a larger duct system? (Note that ducting has its own set of acoustic properties)&lt;/li&gt;&#xA;&lt;li&gt;How do the dimensions of the duct compare with the simulated duct system? (Assuming this is the reason for the duct - why you're doing this at all to begin with still isn't clear) (Dimensions meaning length, width, height, number of bends, etc.)&lt;/li&gt;&#xA;&lt;li&gt;How are the speakers inserted into the duct? Is it just a hole they are resting against? Are the holes sealed?&lt;/li&gt;&#xA;&lt;li&gt;Have you taken into account the fact that butting the speaker face against the duct forms a &lt;a href=&quot;https://en.wikipedia.org/wiki/Helmholtz_resonance&quot; rel=&quot;nofollow&quot;&gt;Helmholtz resonator&lt;/a&gt; with the hole and the speaker body? You may be better off using just the speaker cone instead of the entire speaker body. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In general, looking at your setup, I would bet that you don't have the microphone co-located with the noise canceling speaker, so you are not acting on the actual magnitude and phase of noise &lt;em&gt;at the canceling speaker&lt;/em&gt; to actually achieve cancellation correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I would point out, as I did above, that your &quot;duct&quot; is very poorly sealed and that fact, combined with the material (cardboard) you've chosen to use, will mean that the chamber doesn't resonate at the frequency you want it to, &lt;em&gt;even if&lt;/em&gt; it accurately represented your final system, which I sincerely doubt it does. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I keep mentioning how your test duct compares with the actual duct because longer duct lengths add acoustic impedance and so even if your &quot;scale&quot; test works well it may still not work correctly when installed in the system. Scale testing, and getting scale factors setup correctly, is a science in its own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What resources have you been using so far in working on this project? &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-01-13T21:40:02.993" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6903" PostTypeId="1" CreationDate="2016-01-13T23:51:37.563" Score="3" ViewCount="311" Body="&lt;p&gt;I am trying to implement a simple single-phase fully-controlled thyristor converter using Simulink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YZKth.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YZKth.jpg&quot; alt=&quot;Single-phase fully-controlled thyristor converter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As load I have placed just an inductor, the firing angle is set to a=60 deg., f=50 Hz and the rms of the voltage source is 230 V. The output voltage is the one on the picture below,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xLod9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xLod9.jpg&quot; alt=&quot;Output voltage over time&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any idea how can I fix the part I have circled? The waveform is supposed to be repetitive, so there should be a negative part as in the next half periods.&lt;/p&gt;&#xA;" OwnerUserId="4484" LastEditorUserId="33" LastEditDate="2016-01-14T03:29:43.083" LastActivityDate="2016-01-14T03:29:43.083" Title="Single-phase fully-controlled thyristor converter simulation using simulink" Tags="|simulation|simulink|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6904" PostTypeId="2" ParentId="2832" CreationDate="2016-01-14T03:58:08.760" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Knudsen_pump&quot; rel=&quot;nofollow&quot;&gt;Knudsen pump&lt;/a&gt; has zero moving parts and is based on thermal diffusion (gas flows from low to high temperature end of a tube). The back pressure the flow can resist is termed the thermomolecular pressure difference and is a function of the ratio between the gas mean free path and the dimensions of the tube walls - &lt;a href=&quot;http://phys.org/news/2008-11-gas-minerals.html&quot; rel=&quot;nofollow&quot;&gt;modern advancements&lt;/a&gt; in the concept have used various materials such as zeolites consisting of nano-scale pores to improve this ratio.&lt;/p&gt;&#xA;" OwnerUserId="2201" LastActivityDate="2016-01-14T03:58:08.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6907" PostTypeId="1" CreationDate="2016-01-14T13:17:04.980" Score="8" ViewCount="277" Body="&lt;p&gt;I am interested in making a very small scraper blade that moves back and forth along a submerged surface, however I need to make it as hydrodynamic as possible so I disturb the liquid as little as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand I can model different things such as wings in ANSYS, however most of the tutorials only focus on those that have one direction of travel. I need to design the most efficient shape that will be moved in both directions in a liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are a lot of variables to deal with, but I need some sort of starting point on how to approach the problem. Can anyone offer any suggestions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: included a picture to explain a bit better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OAY39.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OAY39.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4189" LastEditorUserId="4189" LastEditDate="2016-01-16T17:17:44.993" LastActivityDate="2019-08-26T17:02:25.273" Title="How to design the most hydro/aerodynamic shape for a 'vehicle' that goes back and forth (using ANSYS)?" Tags="|fluid-mechanics|ansys|aerodynamics|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6908" PostTypeId="2" ParentId="6899" CreationDate="2016-01-14T13:51:19.700" Score="2" Body="&lt;p&gt;The first big problem is with your speaker placement; the 'anti-noise' speaker should be as close to the 'noise' speaker as possible.  Another big problem is with the speakers you are using; they are way to small for an open-volume system.  Even with both of these issues fixed, I doubt you will be able to get significant noise cancellation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple explanation of noise cancellation that you can read on the side of a box of noise cancellation headphones is very oversimplified.  What noise cancellation headphones actually do is not to add 'anti-noise'; rather they try to maintain a fixed pressure inside of the &lt;strong&gt;closed&lt;/strong&gt; system of the headphone in the presence of pressure disturbances from outside the system.  Even in this small closed system, &lt;a href=&quot;https://travel.stackexchange.com/a/25909/9872&quot;&gt;they aren't very effective&lt;/a&gt; once the wavelength of the sound gets close to the size of the closed volume (1 kHz = 34 cm is roughly the upper end for active noise cancellation in headphones).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are going to have any success with your system, you are going to need to target frequencies which are even lower than the frequencies you are currently trying; I'd say you should start at 70 Hz.  At these frequencies, your speakers don't have the ability to move enough air to create effective sound for either disturbance or cancellation; you need something more like a subwoofer with a driver size of at least 6 inches.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="-1" LastEditDate="2017-04-13T12:52:14.487" LastActivityDate="2016-01-14T13:51:19.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6909" PostTypeId="1" AcceptedAnswerId="6910" CreationDate="2016-01-14T14:05:17.647" Score="3" ViewCount="83" Body="&lt;p&gt;We are designing a diagnostic electronic tool for a vibrating table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our PCB there are a gyroscope, an accelerometer and a bluetooth module.&#xA;Our PCB is attached to a vibrating table that works at 12 Hz with 10 mm peak-to-peak amplitude. It has to work for at least three years &quot;always on&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we are choosing the power connector, what are the technical elements we have to check before buying one? Are there special connectors for that specific use? &quot;Vibrating-environment connectors&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are going to pot the pcb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our fear is that with that amplitude and with that frequency a normal soldered connector would fail and break itself.&lt;/p&gt;&#xA;" OwnerUserId="4490" LastActivityDate="2016-11-05T17:30:26.753" Title="How to choose an electronic connector/cable that works in a vibrating environment?" Tags="|mechanical-engineering|power-electronics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6910" PostTypeId="2" ParentId="6909" CreationDate="2016-01-14T14:45:19.647" Score="1" Body="&lt;p&gt;Use some type of strain relief on the power cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clamp it down half an inch away from the connection point. That reduces the amount of flex on power connector.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2016-01-14T14:45:19.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6911" PostTypeId="1" AcceptedAnswerId="6912" CreationDate="2016-01-14T15:32:34.497" Score="3" ViewCount="162" Body="&lt;p&gt;I am having hard time to fully undestand what those formula's actually mean. I know the meaning of each element:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I : moment of inertia&lt;/li&gt;&#xA;&lt;li&gt;H : angular momentum &lt;/li&gt;&#xA;&lt;li&gt;$\alpha$: angular acceleration&lt;/li&gt;&#xA;&lt;li&gt;$\omega$: angular velocity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$I_{G}\alpha = \dot{\vec{H}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_{G} \dot{\vec{\omega}}= \dot{\vec{H}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain the meaning and/or give and example/context to clarify things?&lt;/p&gt;&#xA;" OwnerUserId="4218" LastEditorUserId="4218" LastEditDate="2016-01-14T15:37:59.553" LastActivityDate="2020-08-08T08:55:20.490" Title="How to interpete the formula: $I_{G}\alpha = \dot{\vec{H}}$?" Tags="|mechanical-engineering|dynamics|kinematics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6912" PostTypeId="2" ParentId="6911" CreationDate="2016-01-14T21:02:27.603" Score="3" Body="&lt;p&gt;Perhaps it is helpful to compare the equation to its linear counterpart.&lt;br&gt;&#xA;$$&#xA;\begin{align}&#xA;\vec{F}&amp;amp;=\dot{\vec{p}}=m\vec{a}=m\dot{\vec{v}}\\&#xA;\vec{\tau}&amp;amp;=\dot{\vec{H}}=I\vec{\alpha}=I\dot{\vec{\omega}}&#xA;\end{align}&#xA;$$&#xA;This linear formula says that the force on an object is equal the the rate of change of the linear momentum which is equal to the mass times the acceleration which is equal to the mass times the rate of change of the velocity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In rotational dynamics, torque takes the place of force, the moment of inertia takes the place of mass, angular momentum takes the place of linear momentum, and angular velocity takes the place of linear velocity.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2016-01-14T21:02:27.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6914" PostTypeId="1" AcceptedAnswerId="6919" CreationDate="2016-01-15T02:13:14.547" Score="2" ViewCount="418" Body="&lt;p&gt;I have a setup where I want to move a knife like blade through a body of liquid inside a container and I want to control the blade from outside the container. Apart from using magnets to move the blade, is there any way to form a dynamic seal around the blade as it moves along the container length? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of using two long lengths of rubber that sandwiched the 1mm blade, but I am not sure how this would hold once some pressure was added to the container, especially if the blade had an oddly shaped cross section (like sharp curves that would be hard to seal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a fused silica layer at the top of the container, which is in contact with the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tPT0U.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tPT0U.jpg&quot; alt=&quot;Vat with scraper blade&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am wondering if there are any seals anyone knows of that could be used in this scenario?  I'm thinking of possibly using ferrofluid as a seal at this stage.&lt;/p&gt;&#xA;" OwnerUserId="4189" LastEditorDisplayName="user16" LastEditDate="2016-01-17T15:04:32.850" LastActivityDate="2016-01-18T07:53:37.900" Title="How to form a seal around a sharp moving blade?" Tags="|mechanical-engineering|fluid-mechanics|magnets|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6915" PostTypeId="2" ParentId="6914" CreationDate="2016-01-15T04:27:47.627" Score="3" Body="&lt;p&gt;Directly Coupled Rodless Cylinders used sealing technology that holds up to over 100 psi of air, see &lt;a href=&quot;https://www.smc.eu/portal_ssl/WebContent/local/UK/Pneu_Book/Rodless_Cylinders.jsp?box=box_basket_v2.jsp&quot; rel=&quot;nofollow&quot;&gt;Rodless Cylinders&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4311" LastEditorUserId="282" LastEditDate="2016-01-16T19:39:27.263" LastActivityDate="2016-01-16T19:39:27.263" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6916" PostTypeId="1" AcceptedAnswerId="7030" CreationDate="2016-01-15T04:57:22.833" Score="2" ViewCount="6066" Body="&lt;p&gt;I have an application where I need two motors to drive one shaft via the use of belt drives but I need help on how to arrange such a system. The system constraints are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;motor type - induction motors $(48\ \textrm{V}, 750\ \textrm{W})$&lt;/li&gt;&#xA;&lt;li&gt;both motors will have pulleys of same dimensions&lt;/li&gt;&#xA;&lt;li&gt;both motors will drive single o/p shaft with bigger pulley attached to it&lt;a href=&quot;https://i.stack.imgur.com/U0uyA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U0uyA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please see the image of my current work, and please let me know if with this arrangement I can have more output speed and torque. If you have any other ideas, please let me know.&lt;/p&gt;&#xA;" OwnerUserId="4496" LastEditorUserId="421" LastEditDate="2016-01-18T06:06:07.903" LastActivityDate="2023-11-30T15:16:44.630" Title="Using two motors and belt drives to drive one shaft" Tags="|mechanical-engineering|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6917" PostTypeId="1" CreationDate="2016-01-15T10:44:29.430" Score="2" ViewCount="1138" Body="&lt;p&gt;I'm writing a patent application and my device incorporates a similar design as what I've seen used on Dovetale plates. However, instead of a preset hole for positioning a bolt to mount to, a grooved slot is machined into the part so the bolt can be positioned at any point along the slide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a canonical technical name for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iwaXb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iwaXb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2260" LastEditorUserId="1560" LastEditDate="2016-01-15T16:34:29.067" LastActivityDate="2016-01-16T00:43:40.367" Title="How would you refer to a bolt mounting slide as opposed to a simple hole?" Tags="|mechanical-engineering|bolting|terminology|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6918" PostTypeId="2" ParentId="6917" CreationDate="2016-01-15T13:43:45.297" Score="2" Body="&lt;p&gt;Simply describe its function. For example &quot;plate A is fixed to plate B with a bolted connection....prior to tightening the fastener (x), a slotted hole (xx) in plate A allows for relative adjustment of the plates fig X. Your artwork will compliment your descriptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a rough draft, but just describe it's function. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-01-15T13:43:45.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6919" PostTypeId="2" ParentId="6914" CreationDate="2016-01-15T13:45:21.277" Score="2" Body="&lt;p&gt;Depending on the axes the blade shall be allowed to move, i.e. the degrees of freedom of your blade, you should try to seal the axes separately. A solution would also depend on anything besides the blade being allowed inside the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best solution (seen from sealing) would be if you can live with additional rods inside your container. Then you could construct a system of rods or levers to move the blade around and just seal the rods. There are lots of sealing solutions for moving rods around that hold up to a few thousand bar against liquids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's not possible and the blade's cross section is your only concern, you could add a mechanical adapter to your blade with the blade's cross section inside and some more easy to seal cross section outside as in the following sketch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qWZKQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qWZKQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is not enough, you are looking for something like a linear moving seal. Depending on the pressure difference you want to seal you might go with a system of plates that has tongues and grooves which are made with a transition fit or slight interference fit, e.g. made from teflon or some liquid-repellent material.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2016-01-15T13:45:21.277" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="6920" PostTypeId="1" CreationDate="2016-01-15T13:45:48.707" Score="3" ViewCount="145" Body="&lt;p&gt;I'm looking at a &lt;a href=&quot;https://www.ifixit.com/Teardown/MasterLock+1500EDBX+Teardown/14148&quot; rel=&quot;nofollow noreferrer&quot;&gt;teardown of a Masterlock&lt;/a&gt;. The guts look like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P0UJN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P0UJN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's strange to me is the black, plastic pinion that appears to be smooth and a strangely shaped cam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GXp4L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GXp4L.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the cam magnified:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ULt02.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ULt02.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motor doesn't seem to be encoded in anyway to prevent stall and the PCB only has one microswitch, but it's for the spring-loaded, shuttle-like metal piece rather than the cam. So I wonder how the motor knows when to stop without stalling. There doesn't seem to be a current shunt on the PCB either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My take is that the square groove of the cam (top facing) faces down in the housing. The round groove on the right of the square groove is where the plastic pinion contacts. That leaves the two not-so-visible grooves facing the mat (ie those facing away from the housing). Does anyone know how the cam is turned? The bottom groove of the cam appears round, so it could probably grip on the black plastic &quot;pinion&quot;. But then, if cam shaft is the axis of rotation, that round face can't be the contact face with the pinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how how the shaft is rotated or how the motor is encoded such that it won't stall?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll also magnify the black plastic pinion here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KoXWL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KoXWL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="4302" LastEditDate="2016-01-15T16:03:20.863" LastActivityDate="2016-01-15T16:03:20.863" Title="A round, plastic pinion in a lock?" Tags="|mechanical-engineering|gears|motors|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="6921" PostTypeId="1" CreationDate="2016-01-15T16:08:12.563" Score="1" ViewCount="1041" Body="&lt;p&gt;Assume that we have the differential equation of a mass-spring-damper model as follows: &#xA;$$&#xA;m\frac{d^2y}{dt^2}+c\frac{dy}{dt}+ky(t)=F(t)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How it could be implemented in MATLAB to do the following steps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, convert the differential equation to a difference equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, finding the discrete-time transfer function of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, finding the regression equation of this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally deriving Recursive Least Squares (RLS) and Ordinary Least Squares (OLS) to estimate the free parameters of the system ($m$,$c$, and $k$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information on this system is available at &lt;a href=&quot;http://www.mathworks.com/help/ident/gs/about-system-identification.html#bshf2u3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mass-Spring-Damper System Excited by Force F(t)&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is greatly appreciated&lt;/p&gt;&#xA;" OwnerUserId="4503" LastEditorUserId="8055" LastEditDate="2017-05-16T16:28:49.157" LastActivityDate="2017-05-16T16:28:49.157" Title="How to estimate a mass-spring-damper parameters in MATLAB using RLS and OLS?" Tags="|mechanical-engineering|control-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6922" PostTypeId="2" ParentId="6917" CreationDate="2016-01-15T17:42:53.330" Score="1" Body="&lt;p&gt;A 'slot' is probably the most concise way to describe this type of hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To describe it precisely you could say 'a slot with ends radius R and distance x between centres'. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-15T17:42:53.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6923" PostTypeId="1" CreationDate="2015-11-28T18:33:59.553" Score="4" ViewCount="69" Body="&lt;p&gt;I have a metal lathe with a broken part. The part is not available, so I have to machine one up from scratch.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this part is the lower half of the crossfeed assembly on the turret cutter, it has to be accurate, which is no problem for straightness and tolerance.  However, as ways and gibs have to be able to hold lubricant (so the parts don't just slide on each other and wear out), there must be a standard for surface finish that's preferable for this type of assembly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't seem to find a clear specification for what to use, so what would be an appropriate finish type for those conditions?&lt;/p&gt;&#xA;" OwnerUserId="6915" OwnerDisplayName="personal privacy advocate" LastEditorUserId="33" LastEditDate="2016-02-15T20:30:35.250" LastActivityDate="2016-05-15T22:20:12.757" Title="Appropriate finish type for a new lathe crossfeed?" Tags="|machining|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6924" PostTypeId="2" ParentId="6923" CreationDate="2016-01-15T18:11:15.227" Score="1" Body="&lt;p&gt;The traditional way to achieve this is to machine your part to the correct geometric tolerance and assemble it with a layer of marking blue applied to the mating surfaces. Work the part backwards and forwards a few time and then disassemble then inspect and remove any high spots with a scraper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative would be to use fine abrasive paste and lap the surfaces to their final fit/finish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you say you don't necessarily want a mirror polish surface as this won't hold oil well but similarly you don't want any ridges or ripples which will impair accuracy and accelerate wear so the goal is a flat surface with scratches in it rather than the troughs/ridges from as as-machined surface. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-15T18:11:15.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6925" PostTypeId="1" AcceptedAnswerId="7154" CreationDate="2016-01-15T18:24:20.900" Score="6" ViewCount="144" Body="&lt;p&gt;According to this &lt;a href=&quot;http://home.mit.bme.hu/~simon/publications/Sensys_2004_FloodingTSP.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;, which deals with time synchronization in wireless networks, this kind of delay occurs, however I do not understand its explanation. The paper says the reciever (radio) has to determine the bit offset of the message from a known synchronization byte. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition from the paper is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Byte Alignment Time&lt;/em&gt; —the delay incurred because of the different byte alignment of the sender and receiver. This time is deterministic and can be computed on the receiver side from the bit offset and the speed of the radio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If somebody familiar with this phenomenon, can you please explain it to me?&lt;/p&gt;&#xA;" OwnerUserId="4504" LastEditorUserId="282" LastEditDate="2016-01-16T18:51:36.550" LastActivityDate="2016-01-29T10:09:12.197" Title="What is Byte Alignment Time in radio communication?" Tags="|electrical-engineering|terminology|wireless-communication|rf-electronics|digital-communication|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6926" PostTypeId="2" ParentId="352" CreationDate="2016-01-15T18:50:43.520" Score="1" Body="&lt;p&gt;The wattage of an incandescent bulb is simply the electrical power that it dissipates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often the limiting factor is that this type of bulb dissipates a lot of this energy as heat so one of the most common reasons for limiting the wattage is that you have something like a fabric or paper lampshade which would be damaged or might even catch fire is a high output bulb was fitted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is in exterior lights where the fixture is usually sealed and so cannot dissipate heat by convection.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-15T18:50:43.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6927" PostTypeId="2" ParentId="3080" CreationDate="2016-01-15T19:26:05.783" Score="7" Body="&lt;p&gt;There are a few  principal advantages to having more bolts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is that the loads are more evenly distributed, especially when the rigidity of the fixture itself is a bit marginal and when it is important to ensure that no separation occurs eg in flange joints of high pressure fluid systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly having more  bolts for the same nominal load means smaller hole diameter and so small flange diameter which may be useful when you need to package things as tightly as possible (eg in automotive engine applications.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thirdly more fasteners can improve redundancy ie if you have 4 bolts and 1 is below spec or improperly assembled then you lose 25% of the design strength if you have 10 bolts and one is wrong then you only lose 10%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other side of the coin is that using a lot of small fasteners to support a load much larger than their individual capacity can cause cascade failures if you get unforeseen loading conditions and you can get joints 'unzipping'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand there may be instances where a large number of fasteners complicate assembly and maintenance, especially if access is limited and it is likely that the the fasteners might corrode or otherwise jam or seize. Similarly smaller diameter fasteners may have a smaller acceptable torque range than larger ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth bearing in mind that the best practice in designing bolted fixtures is for the bolts to work by clamping two surfaces together so that shear forces are resisted by friction between the mating surfaces rather than being carried directly by the bolts. Similarly bolts usually require moderate clearance in no threaded holes and so aren't suitable as the sole means of providing accurate alignment between two parts. Where this is required it's usual to have something like a stud or notch arrangement to provide a positive means of alignment. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-11-09T14:01:00.403" LastActivityDate="2018-11-09T14:01:00.403" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="6928" PostTypeId="2" ParentId="6917" CreationDate="2016-01-15T19:44:34.787" Score="2" Body="&lt;p&gt;Generically it would just be called a 'slot' or formally a 'slotted hole.' Your specific example shows a bigger slot going part of the depth and a smaller slot machine all the way though. This would specifically be called a 'counterbore slot' and is a fairly common detail which allows a slotted hole that can be used with a socket head screw with a flush head-side surface. It's called a counterbore slot because a counterbored hole is a hole with a recess that allows the head of a screw (other than a flat head screw for which the term is a countersink) to sit flush or recessed inside the parent metal.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-01-15T19:44:34.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6929" PostTypeId="2" ParentId="6917" CreationDate="2016-01-16T00:43:40.367" Score="0" Body="&lt;p&gt;In my design community, we call that a spotfaced slot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Spotface&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Spotface&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4495" LastActivityDate="2016-01-16T00:43:40.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6933" PostTypeId="1" AcceptedAnswerId="6934" CreationDate="2016-01-16T13:46:38.757" Score="1" ViewCount="693" Body="&lt;p&gt;I have interested in walking bicycle. For the concept is just bicycle that you don't cycling pedal but stepping or striding to drive wheel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which leads me to think that is it better if we use linear or ellipse motion of leg to transmit to bicycle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest mechanic I could think is piston crankshaft like a car. It convert linear motion to rotation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How disadvantage of the mechanic compare to cycling motion?&lt;/p&gt;&#xA;" OwnerUserId="1606" LastActivityDate="2016-01-16T17:07:58.697" Title="Why there is no piston style pedal bicycle?" Tags="|power|pistons|bicycles|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6934" PostTypeId="2" ParentId="6933" CreationDate="2016-01-16T17:07:58.697" Score="5" Body="&lt;p&gt;The biggest problem with a &quot;two cylinder&quot; crankshaft is that there is a dead spot in the torque curve whenever either &quot;piston&quot; is at TDC or BDC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have your feet directly on the pedals of a conventional circular crank, you actually have the ability to apply force over greater than a 180&amp;deg; arc &amp;mdash; by a combination of flexing your ankles and pushing forward/backward with your thigh muscles &amp;mdash; which eliminates the dead spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to use a mechanical connecting rod with a crankshaft, you need some other way to eliminate the dead spot. This typically involves using more than two pistons, or by incorporating a large flywheel into the mechanism. None of this is practical or desirable on a bicycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen toy vehicles for small children that use such a mechanism. but invariably, they need to put their feet down on the ground from time to time to push themselves out of the dead spot.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-01-16T17:07:58.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6935" PostTypeId="1" AcceptedAnswerId="6936" CreationDate="2016-01-16T17:39:12.263" Score="1" ViewCount="52" Body="&lt;p&gt;How to find out experimentally to measure the vibrational frequency of some type of spring without electronic or electrical instrument?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am just wondering about the advance stage of the discovery of the theory of vibrations it became in the 19th century&lt;/p&gt;&#xA;" OwnerUserId="252" LastActivityDate="2016-01-17T21:16:49.667" Title="How to find out experimentally without electronic or electrical instrument to measure the vibrational frequency of some type of spring" Tags="|mechanical-engineering|experimental-physics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6936" PostTypeId="2" ParentId="6935" CreationDate="2016-01-16T18:27:18.337" Score="2" Body="&lt;p&gt;A fairly straightforward way to do this is to have the sample spring connected via a mechanical linkage to a device which traces a line on a moving spool or drum using ink on paper or similar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These sorts of paper trace devices were used until relatively recently for all kinds of devices to measure some variable over time, barometers like this are very common. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you know the speed that the paper spool is moving in the y axis and the relationship between the extension of the spring and the x axis you can get pretty accurate displacement vs time recordings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the technology to do this is very similar to that of clocks and watches. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-01-17T06:26:44.660" LastActivityDate="2016-01-17T06:26:44.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6937" PostTypeId="1" AcceptedAnswerId="6938" CreationDate="2016-01-16T18:59:40.263" Score="1" ViewCount="80" Body="&lt;p&gt;How could I design a paper spool or drum that is moving at constant speed without electrical system? What mechanisms are available?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that some paper tracing system was being used in 19th century to track vibration frequency.&lt;/p&gt;&#xA;" OwnerUserId="252" LastEditorUserId="33" LastEditDate="2016-01-17T03:26:26.810" LastActivityDate="2016-01-17T06:43:50.237" Title="What mechanisms can be used to move a drum at constant speed without an electrical system?" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6938" PostTypeId="2" ParentId="6937" CreationDate="2016-01-16T19:35:17.220" Score="4" Body="&lt;p&gt;There are a few ways to go about this. One easily available solution is simply to adapt the mechanism from a mechanical clock or watch. This will give a regular series of pulses rather than a true constant speed but in many cases this will not be a problem and may even be an advantage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach is to use a regulator or governor to achieve a constant speed, either directly or via a mechanical control system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On way to achieve this, for example is to have a heavy weight turning a pulley with a large reduction ratio and coupled to some sort of damper (like a fan). In this case acceleration due to gravity and aerodynamic drag on the fan will quickly reach a stable equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another method is to have an approximately constant speed drive and a mechanism which regularly punches or otherwise marks the paper roll so you have a regular 'tick' as a baseline for comparison. For example a pendulum is a simple and effective way of achieving a regular and repeatable time interval. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-01-16T19:51:38.090" LastActivityDate="2016-01-16T19:51:38.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6940" PostTypeId="2" ParentId="5665" CreationDate="2016-01-16T21:34:31.900" Score="6" Body="&lt;h2&gt;Kinematics and dynamics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/94sUM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/94sUM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are the steps to solve problems of this nature.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Analize the kinematics of the system.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $_{o}\vec{r}_{OR}$ + $_{o}\vec{r}_{RP}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $_{o}\vec{r}_{OR}$ + $R(\varphi) _{B}\vec{r}_{RP}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $\big(x_{k}î + 0j + 0k \big)$ + $\big(\sin(\varphi)l î + 0j + \cos(\varphi)lk\big)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $_{o}\vec{r}_{OP}$ = $\big[\big(x_{k}+\sin(\varphi)l\big)î + 0j + \big(\cos(\varphi)l\big)k\big]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note: $R(\varphi)$ is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotation_matrix&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotation matrix&lt;/a&gt; and $x_{G} = x_{k}+\sin(\varphi)l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the time derivatives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $\dot{x_{G}}$ = $\dot{x_{k}}+\cos(\varphi)\dot{\varphi}l$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $\ddot{x_{G}}$ = $\ddot{x_{k}} + l\cos(\varphi)\ddot{\varphi} - l\sin(\varphi)\dot{\varphi}^{2}$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Use Newton's equation:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $m_{k}\ddot{x_{k}} = F_{A} - m_{G}\ddot{x_{G}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute $x_{G}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $m_{k}\ddot{x_{k}} = F_{A} - m_{G}\big(\ddot{x_{k}} + l\cos(\varphi)\ddot{\varphi} - l\sin(\varphi)\dot{\varphi}^{2}\big)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $\big(m_{k}+m_{G}\big)\ddot{x_{k}} + m_{G}\big(l\cos(\varphi)\ddot{\varphi}\big) - m_{G}\big(l\sin(\varphi)\dot{\varphi}^{2}\big)= F_{A}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the z axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $F_{Z}$ = $m_{G}g-l\big(\cos(\varphi)\dot{\varphi}^{2}+\sin(\varphi)\ddot{\varphi}\big)$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Use Newton's second law for rotation:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $I\ddot{\varphi}$ = $F_{Z}l\sin(\varphi)-\big(m_{G}\ddot{x_{G}}\big)l\cos(\varphi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{Z}l\sin(\varphi) = m_{G}gl\sin(\varphi)-l^{2}\big(\cos(\varphi)\sin(\varphi)\dot{\varphi}^{2}+\sin(\varphi)^{2}\ddot{\varphi}\big)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\big(m_{G}\ddot{x_{G}}\big)l\cos(\varphi) = m_{G}\big(l^{2}\cos(\varphi)^{2}\ddot{\varphi}\big) - m_{G}\big(l^{2}\cos(\varphi)\sin(\varphi)\dot{\varphi}^{2}\big)+m_{G}\ddot{x_{K}}l\cos(\varphi)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using trigonometry identities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{5.em}$ $\big(I+m_{G}l^{2}\big)\ddot{\varphi}$ = $m_{G}gl\sin(\varphi)-m_{k}l\cos(\varphi)\ddot{x_{k}}$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Done! Now you can rest... &#xA;$\ddot\smile$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4499" LastEditorUserId="5136" LastEditDate="2016-04-29T10:23:21.007" LastActivityDate="2016-04-29T10:23:21.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6941" PostTypeId="2" ParentId="6921" CreationDate="2016-01-16T23:07:48.207" Score="1" Body="&lt;p&gt;&lt;strong&gt;Matlab with some tips&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good method is using the Billinear transform (&lt;a href=&quot;https://en.wikipedia.org/wiki/Bilinear_transform&quot; rel=&quot;nofollow&quot;&gt;Tustin's&#xA;   method&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = tf('s')&#xA;Hc = 1/(m*s^2+c*s+k)   % Tf continuous-time&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Observation.: First declare your variables: m, c and k.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matlab provides some functions that help with the transformation. The function &lt;a href=&quot;http://www.mathworks.com/help/control/ref/c2d.html&quot; rel=&quot;nofollow&quot;&gt;c2d()&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ts = 0.1;&#xA;Hd = c2d(Hc,Ts,'tustin')&#xA;step(Hc,'--',Hd,'-')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where Ts is the sample time.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;To find the difference equation, apply the inverse Z transform after using tusting method in your equation. Remember that you need to use Laplace transform first!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\hspace{2.5em}$ $Y(s)\big(ms^{2}+cs+k\big) = \mathcal{L}\{F(t)\}$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can find good info about OLS &lt;a href=&quot;http://www.karenkopecky.net/Teaching/eco613614/Matlab%20Resources/OLS.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4499" LastActivityDate="2016-01-16T23:07:48.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6942" PostTypeId="2" ParentId="6871" CreationDate="2016-01-17T02:00:47.613" Score="3" Body="&lt;p&gt;I would solve this by conservation of power. Linear power is force times speed; rotational power is torque times angular velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're trying to find linear force, you know torque and angular velocity, so the only thing you need to find is linear velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortunately, you know angular velocity and the radius of the crank, so the linear velocity is just $ m\cos{\omega t}$, where you have chosen m to be a length and not mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, putting it all together:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F = \frac{\tau \omega}{m \cos{\omega t}}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-01-17T02:00:47.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6944" PostTypeId="2" ParentId="6441" CreationDate="2016-01-17T05:44:16.100" Score="0" Body="&lt;p&gt;In general pneumatics are quite precise when it comes to force but modeling energy in this context brings in a lot of additional unknowns which make be difficult to measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common method of impact testing is to use an impactor fixed to a pendulum as it's fairly easy to determine the kinetic energy at the lowest point in the swing directly from the initial gravitational potential energy and this gives a very repeatable test. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also important to bear in mind that, even when you measure kinetic energy that doesn't really give you a general material property which you can extrapolate from in the same way as something like tensile strength as strain rate is usually a very important factor in impact behavior so the priority is usually to simulate the real world conditions you are interested in as closely as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to use a pneumatic system for some operational reason then the most useful approach may be to treat it as an air spring. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-17T05:44:16.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6945" PostTypeId="2" ParentId="6937" CreationDate="2016-01-17T06:43:50.237" Score="0" Body="&lt;p&gt;If you are trying to simulate 19th century technology, based on your other question, you could achieve this to a pretty good precision with a calibrated metronome and a skilled operator turning a crank. That could easily provide enough precision for your vibration measurement experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things don't need to be any more complicated than they have to.&lt;/p&gt;&#xA;" OwnerUserId="554" LastActivityDate="2016-01-17T06:43:50.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6946" PostTypeId="1" CreationDate="2016-01-17T07:20:42.203" Score="1" ViewCount="147" Body="&lt;p&gt;I would like to display an image onto a screen using a digital light projector and somehow record the amount of light the screen is receiving from each pixel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have thought about using a visible light filter over an array of very small photovolatic cells and recording the voltage output by each cell, but the logistics of the wiring would be immense, not to mention I can't seem to find cells small enough.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have searched for 'UV screens' and many other terms but I can't seem to find anything. There are many results for UV diodes but they are too large to get the resolution I need. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know if such screens exist?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let me explain a common problem where this is an issue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This frequently occurs with 3D printers that use ultraviolet-cured resin:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/EKX49.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EKX49.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory is: projector displays black&amp;amp;white (binary, not grayscale; uniform regions of 100% brightness and 0% brightness) image onto the bottom of a transparent container of the resin, curing one &quot;layer&quot; of the item, then the layer is lifted (using a mechanical actuator), the space fills with more uncured resin, and next layer is displayed and cured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality the projector's light is non-uniform - some areas brighter, other dimmer. This depends on a lot of hard to determine factors, like light source construction, lens alignment errors etc. As effect, rhe resin will cure to a different degree in different areas of the image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Independently, the projector is capable of displaying full grayscale (or even color) images (for no other reason that that's what's cheapest due to demand and economy of scale). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if you know the brightness layout of the projected image, you can compensate in software - adjust the projected image of item layer, by dimming areas that were brightest; display a grayscale image instead of 1bpp, that results in actual uniform exposure of the &quot;screen&quot; surface, and uniform curing of the resin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in order to generate these adjustments, you must obtain the individual map of brightness of the projected image, as generated with this specific projector. With visible light, that wouldn't be very hard, just take a photo of the screen displaying a white sheet using a normal camera. This being ultraviolet though makes it quite tricky.&lt;/p&gt;&#xA;" OwnerUserId="4189" LastEditorUserId="203" LastEditDate="2017-08-30T11:17:06.330" LastActivityDate="2022-04-25T11:04:00.377" Title="How to digitalize a UV image (create an array of the light output at the pixel level)?" Tags="|optics|photovoltaics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6950" PostTypeId="2" ParentId="6946" CreationDate="2016-01-17T14:06:08.217" Score="-2" Body="&lt;p&gt;This is a bit like a perpetual motion machine.  You have a digital image.  You project that image onto a screen.  Then you want to capture that image from the screen to make a digital image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious answer is that this process is silly.  Just use your original digital image.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-01-17T14:06:08.217" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6951" PostTypeId="1" AcceptedAnswerId="6952" CreationDate="2016-01-17T15:52:00.113" Score="5" ViewCount="1676" Body="&lt;p&gt;What options do I have for creating a resistance to a mechanical movement apart from simple friction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume I have a lever, which will result in either circular or linear motion.  I want to be able to apply a resistance to this motion, so that the user will have to apply extra effort to move the lever in either direction.  I could use a spring, which would resist motion in one direction, but this would then result in a restoring force when the initial displacement force was removed.  I want the extra force to act in both directions, so that the lever is &quot;heavy&quot; to move, but will stay in its new position when displaced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the force to be able to be calibrated and remain constant, so I don't want to have something like a brake pad, which will wear, or just a nut to do the joint up tight, as this will be too difficult to achieve a calibrated force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something that could be done with a spring or pneumatic air muscle or by using hydraulics?&lt;/p&gt;&#xA;" OwnerUserId="4526" LastEditorUserId="4526" LastEditDate="2016-01-20T16:54:40.893" LastActivityDate="2018-06-08T10:48:35.123" Title="How do I add a resistance to mechanical movement in both directions, apart from friction?" Tags="|mechanical-engineering|hydraulics|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6952" PostTypeId="2" ParentId="6951" CreationDate="2016-01-17T16:21:11.200" Score="8" Body="&lt;p&gt;The two basic mechanical options are either a viscous damper or inertia or indeed a combination or both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of the two a damper is probably the closest to what you want as it will provide a force which opposes the direction of movement which is proportional to the speed of movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding mass (perhaps as a counterweight and pulley system) won't on it's own do exactly what you describe as it resists acceleration but it may be that a small adjustable  mass will help you to fine tune the feel. The downside of a mass damper system on its own is that it won't provide a positive 'lock' and the position could creep under accidental pressure or vibration so it's more useful if the lever is for constant adjustment rather than being set to one position and left. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are quite a lot of off the shelf dampers available in liner and rotary formats and many are adjustable. You also have the option of using cams or bar linkages to have variable or progressive resistance across the travel of the lever. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more sophisticated option is to use an active system with something like a solenoid or pneumatic actuator which provides active force feedback and even maintains an exact position below a minimum force input. In this case you could design an electric control system to provide whatever force feedback profile you wanted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, a friction brake is the simplest way to achieve exactly what you want as it provides a constant resistive force. To be honest, the wear rate of the friction material in this context is not likely to be significant, especially considering that hydraulic and pneumatic systems require a certain amount of maintenance in themselves.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="368" LastEditDate="2016-01-20T16:49:22.857" LastActivityDate="2016-01-20T16:49:22.857" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6953" PostTypeId="1" AcceptedAnswerId="6959" CreationDate="2016-01-17T17:35:35.417" Score="2" ViewCount="288" Body="&lt;p&gt;I have an issue with simulink, basically it's to do with a second order system, well two first order systems in series. As I understand it as you increase the damping factor (above 1), the system should respond slower and be more sluggish. The damping factor = (tau1+tau2)/(2root(tau1*tau2)). So, looking at this system with damping factor 1(with it's response):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1LL8e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1LL8e.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this system with damping factor 1.19 and it's response:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/u9eZZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u9eZZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What gives?! The system with the damping factor 1 hasn't even reached steady state by 250 secs while the system with damping factor 1.19 has had a faster response, why is this happening?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading.&lt;/p&gt;&#xA;" OwnerUserId="4530" LastEditorUserId="4530" LastEditDate="2016-01-17T18:09:00.260" LastActivityDate="2016-01-18T13:17:59.820" Title="Problem with process controls/simulink?" Tags="|electrical-engineering|control-engineering|chemical-engineering|process-engineering|simulink|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6954" PostTypeId="2" ParentId="6951" CreationDate="2016-01-17T21:07:07.180" Score="1" Body="&lt;p&gt;Dampers provides resistive force as a function of velocity, while friction is more constant. You can buy an off the shelf damper: look at the compression and extension section in the following link for dampers that work in both directions: &lt;a href=&quot;http://www.mcmaster.com/#adjustable-dampers/=10q42q4&quot; rel=&quot;nofollow&quot;&gt;http://www.mcmaster.com/#adjustable-dampers/=10q42q4&lt;/a&gt;&#xA;If you are looking for something less expensive, get an used pneumatic cylinder with ports on both ends, and run a small tube from the port on one end to the other port. If you need more damping, add a manual flow control to the tube, or fill the cylinder with oil.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastEditorUserId="4311" LastEditDate="2016-01-18T01:02:43.503" LastActivityDate="2016-01-18T01:02:43.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6955" PostTypeId="2" ParentId="6935" CreationDate="2016-01-17T21:16:49.667" Score="1" Body="&lt;p&gt;If the frequency is high enough, its sound can be compared to a tuning fork, ref &lt;a href=&quot;https://en.wikipedia.org/wiki/Tuning_fork&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Tuning_fork&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-17T21:16:49.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6956" PostTypeId="2" ParentId="3080" CreationDate="2016-01-17T21:51:44.577" Score="0" Body="&lt;p&gt;In the machinery industry the holo-krome bolt selectors are used regularly to get tightening torque and resultant bolt tensions: the green one is for inch and blue is for metric: &lt;a href=&quot;https://www.google.com/#q=holo-krome+screw+selector+cards&quot; rel=&quot;nofollow&quot;&gt;https://www.google.com/#q=holo-krome+screw+selector+cards&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If feasible, the miniature sizes such as 2.5mm are avoided. Simplified, bolt strength is proportional to cross section area, or PI*R^2. In comparing a 6mm to a 2.5mm, the correspond radii are 3 and 1.25, The corresponding strength ratios are approximately 3^2 and 1.25^2, or 9 to 1.56, or a ratio of 5.8. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bigger screws may need blue locktite to keep from becoming loose due to vibration: the length of the screw under tension generally needs to be 4x the diameter to be considered vibration resistant.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-17T21:51:44.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6957" PostTypeId="1" CreationDate="2016-01-17T22:06:12.823" Score="1" ViewCount="536" Body="&lt;p&gt;Consider a solid torus, which is a circular disk with diameter d with the disk's centre swept around a circle of diameter d, such that there is an open centre also of diameter d.  This torus is standing on its side so as to achieve its maximum height.  How it is supported so that it does not fall to either side is not an issue for this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given, arbitrary material, how can I calculate the maximum size that this solid torus could be before collapsing under its own weight in 1g?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For purposes of this question, consider the ground to be of infinite strength.&lt;/p&gt;&#xA;" OwnerUserId="4494" LastEditorUserId="4494" LastEditDate="2016-01-18T01:26:37.560" LastActivityDate="2016-01-20T08:47:19.123" Title="How to calculate the maximum size of a torus constructed in a given material" Tags="|structural-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6958" PostTypeId="1" CreationDate="2016-01-17T22:52:50.203" Score="2" ViewCount="3632" Body="&lt;p&gt;Why does a radial engine need a master rod? Why can't slave rods be the only rods? Would &lt;a href=&quot;https://www.youtube.com/watch?v=w27heVUxDQ8&quot; rel=&quot;nofollow&quot;&gt;this spider bearing&lt;/a&gt; (around 1:05 in the video) work for an IC engine, why or why not?&lt;/p&gt;&#xA;" OwnerUserId="4532" LastEditorUserId="272" LastEditDate="2016-01-18T13:03:50.140" LastActivityDate="2020-03-22T10:42:25.633" Title="Why does a radial engine need a master rod?" Tags="|mechanical-engineering|engines|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6959" PostTypeId="2" ParentId="6953" CreationDate="2016-01-17T23:02:55.830" Score="2" Body="&lt;p&gt;&lt;strong&gt;Time Constant and systems&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;A second order LTI system in Laplace domain:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\hspace{2.5em}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$H(s) = \frac{{\omega_{n}}^{2}}{s^{2}+\zeta\omega s+{\omega_{n}}^{2}}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The solution is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\hspace{2.5em}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$h(t) = \frac{{\omega_{n}}}{\sqrt{1-\zeta^{2}}}e^{-\zeta {\omega_{n}} t}sin({\omega_{n}} \sqrt{1-\zeta^{2}}t)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that the time constant depends on the product of the damping and the frequency!&lt;/p&gt;&#xA;&lt;p&gt;The denominator is called chacateristic equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\hspace{2.5em}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$s^{2}+\zeta{\omega_{n}} s+{{\omega_{n}}}^{2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\hspace{2.5em}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$r_{1,2} = \frac{-\zeta{\omega_{n}}\pm \omega_{n}\sqrt{1-\zeta^{2}}}{2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We have three forms for the solution:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MnHt3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MnHt3.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Overdamped: &lt;span class=&quot;math-container&quot;&gt;$r_{1} \neq r_{2}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\in$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\Re$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Critically damped: &lt;span class=&quot;math-container&quot;&gt;$r_{1} = r_{2}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\in$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\Re$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Underdamped: &lt;span class=&quot;math-container&quot;&gt;$ r_{1} = {r_{2}}^{*}$&lt;/span&gt;. Where &lt;span class=&quot;math-container&quot;&gt;${r_{2}}^{*}$&lt;/span&gt; is the complex conjugate of &lt;span class=&quot;math-container&quot;&gt;$r_{1}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In the s plane, should look like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AzznH.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AzznH.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The figure above show us a complex response (the conjugate is implicated). Note that &lt;span class=&quot;math-container&quot;&gt;$-\zeta\omega_{n}$&lt;/span&gt; its in the real part of the solution! So, is responsible for the time response and, the imaginary part, &lt;span class=&quot;math-container&quot;&gt;$\omega_{n}\sqrt{1-\zeta^{2}}$&lt;/span&gt; is responsible for the oscilation.&lt;/p&gt;&#xA;&lt;p&gt;Note that the real part is the exponential term in the solution!&lt;/p&gt;&#xA;" OwnerUserId="4499" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-01-18T13:17:59.820" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="6960" PostTypeId="2" ParentId="6957" CreationDate="2016-01-18T00:16:11.077" Score="3" Body="&lt;p&gt;I think there is no such limit. The only possible factor which comes to mind is the stress at the foot of the torus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you momentarily consider the structure to be infinitely rigid, then the region of contact of the torus with the ground will be equal to an infinitesimal point, leading to infinite stresses on the foot of the torus (and the ground beneath it). If you consider deformations, then the torus will no longer be a perfect torus since it will deform at the foot to cover an area such that the stress is lower than the material's yield strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a perfectly elasto-plastic material, this basically allows the torus to deform as much as necessary, so it doesn't offer a limit either. Assuming any other stress-strain relationships, it then becomes pretty much impossible to answer generically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Internal forces within the torus don't worry me too much. Looking at the top span of the torus, there will be bending there, sure, but the maximum bending moment will be a function of the square of the span, which you have defined as equal to twice the diameter (considering the span equal to the distance between the center points of the different sides of the torus). Bending, however, is resisted by a cross-section's second moment of area, which is a function of the fourth power of the diameter. This means that if you double your diameter, you will quadruple (4x) your bending moment, but your second moment of area will increase sixteen-fold. So, actually, the larger your diameter, the farther from collapse you get. This doesn't even take into consideration the arching behavior of the torus, which will drastically reduce your bending moment and transform it into compressive forces. One could then bring up the possibility of buckling, but given how non-slender this torus would be, that also won't be an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom half of the torus works similarly: the bending moment shouldn't be of great concern for the same reasons as described above. The arching behavior in this case will actually generate tensile forces, but these will be overcome by the compressive forces due to the torus' own weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I don't see where one could find a maximum size for a given material. In fact, the material you need to worry about isn't the torus', but the ground's. You can make a torus out of whatever material you want, but if you put it on weak clay, it will be gobbled up. If you make a Death-Star-sized torus and put it on stone, it will probably crush the stone and be gobbled up, too.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-01-18T00:16:11.077" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="6961" PostTypeId="2" ParentId="6951" CreationDate="2016-01-18T02:49:00.593" Score="6" Body="&lt;p&gt;Elaborating on &lt;a href=&quot;https://engineering.stackexchange.com/questions/6951/how-do-i-add-a-resistance-to-mechanical-movement-in-both-directions-apart-from#comment12733_6951&quot;&gt;CarlWitthoft's comment&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_damping&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetic damping&lt;/a&gt;, sometimes called eddy-current damping, is a standard way of introducing damping in metallic systems.  The principle is fairly simple; as a magnet moves past a conducting material like a metal, it induces circular currents (eddy currents) in the material.  The resistivity of the conductor converts this current flow into heat, taking energy out of the system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most dramatic example of this is dropping a strong magnet down a copper tube (cool videos &lt;a href=&quot;https://www.youtube.com/watch?v=5BeFoz3Ypo4&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;https://www.youtube.com/watch?v=zCUSXoL4SDE&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, and &lt;a href=&quot;https://www.youtube.com/watch?v=E97CYWlALEs&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-01-20T16:50:05.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6962" PostTypeId="2" ParentId="6951" CreationDate="2016-01-18T07:27:11.630" Score="1" Body="&lt;p&gt;Your question immediately brings dampers into mind, but I think you might still be happier with friction. It depends on whether you want the resistance to increase with speed of movement (damper) or to remain constant (friction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep the friction steady over the lifespan of the device, you can use a spring to push a brake pad against the rotating surface. If they are both made of e.g. steel, wear should be quite slow. As the surface wears, the spring will keep the force constant even if the dimensions change a millimeter or two.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-01-18T07:27:11.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6963" PostTypeId="2" ParentId="6914" CreationDate="2016-01-18T07:53:37.900" Score="1" Body="&lt;p&gt;One option is not to try to design a leak-proof seal, but instead just limit the leaks to acceptable levels and make maintenance possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use rubber seals like you planned to.&lt;/li&gt;&#xA;&lt;li&gt;Add a trough that will catch any leaking resin. Preferrably made of a material such as teflon or silicone that the resin will not stick to.&lt;/li&gt;&#xA;&lt;li&gt;Arrange so that the resin in the trough will eventually harden, making it easy to clean off.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-01-18T07:53:37.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6964" PostTypeId="1" AcceptedAnswerId="6971" CreationDate="2016-01-18T10:30:20.017" Score="3" ViewCount="259" Body="&lt;p&gt;First of all, I would like to know about the service life of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Flashback_arrestor&quot; rel=&quot;nofollow&quot;&gt;flashback arrestor&lt;/a&gt;. In my laboratory, I got three flashback arrestors {WITTgas RF53N(1/4”NPT-F), Brass} from the ex-students who used those items in 2012 and already finished their graduation. I joined the laboratory in 2015. In this circumstance, I want to know if there is any possible way to check those flashback arrestor. All the flashback arrestors they used are for the premixed combustion in a Constant Volume Combustion Chamber. My purpose is also same. But I want to know first whether those flashback arrestors are dead or alive. &lt;/p&gt;&#xA;" OwnerUserId="4439" LastEditorUserId="272" LastEditDate="2016-01-18T13:05:15.467" LastActivityDate="2016-01-18T16:00:57.520" Title="Service Life of a Flashback Arrestor" Tags="|piping|safety|gas|valves|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6965" PostTypeId="2" ParentId="6189" CreationDate="2016-01-18T11:54:03.217" Score="1" Body="&lt;p&gt;In measuring response to impact you really want to be using a system where the kinetic energy of the impactor is well known and very repeatable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also important to realise that in most cases impact testing will be destructive, or at least semi-destructive as it's only in what would normally be considered failure conditions (eg plastic deformation or fracture etc) that there is a meaningful difference between an 'impact' and a static force, which is a different situation again from measuring dynamic frequency response.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most straightforward way to give an impactor a consistent kinetic energy is to have it fall a measured height under gravity, either dropping straight down (for a vertical impact) or as a pendulum arrangement (for horizontal impact). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are circumstances where pneumatic actuators might be used (for example in ballistics testing) but these are generally used more like a gun to fire a free projectile at a target than as conventional linear actuators and in this case you would be considering the air pressure used as the energy source. In this sort of situation it is also common to use high speed photography to make accurate measurements of velocity and displacement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with using a hydraulic actuator is that there is no straightforward way to accurately measure its velocity at the point of impact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue is that impact testing tends to be very context specific as there are several possible different modes of failure which depend on several interrelated factors. For example even when the impactor has a known kinetic energy you can get different results between a high mass at low speed (eg a a big hammer) and a low mass at high speed (eg a bullet). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with this in mind it is almost always most useful for impact testing to reflect as closely as possible the real world situation that you are trying to model. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-18T11:54:03.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6966" PostTypeId="2" ParentId="2307" CreationDate="2016-01-18T12:14:16.547" Score="1" Body="&lt;p&gt;On of the problems with vacuum balloons is that the potential gains over using hydrogen or helium are actually pretty small, something like 10%. On the other hand the difference is stress between containing a gat at or around ambient pressure and containing a vacuum safely are orders of magnitude greater. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is magnified by the fact that a gas filled balloon can afford to be a bit leaky as the pressure difference between inside and outside is small gas loss will be slow unless you have a really big hole. In the case of a vacuum container however any leak will result in rapid loss of vacuum and may even cause a catastrophic failure of the structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net result of this is that whatever exotic materials you can develop you will still be better off using them as the skin of a gas filled balloon as the fundamental performance trade off still applies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TO put it another way, while something like graphene might make a vacuum&#xA; balloon possible the chances are that a helium/hydrogen balloon will still be better. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-18T12:14:16.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6967" PostTypeId="1" CreationDate="2016-01-18T12:30:17.250" Score="3" ViewCount="88" Body="&lt;p&gt;Suppose we craft a reinforced concrete pylon for a cable-stayed bridge. It's very tall and rather thin. No one builds falsework for the entire pylon - it'd be very impractical. Instead, a part of falsework is built for the lowest part of the pylon, filled with concrete, then some time later that falsework is removed and the next part is built for the higher part of the pylon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the pylon in not really a single piece of concrete, it a combination of pieces seamlessly united together because the concrete surface of the earlier part is kept clean until falsework is moved up and so new portion of concrete just sticks to it pretty well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How &quot;monolitic&quot; is such monolitic concrete? How much does it differ from concrete put into falsework size of the entire structure?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="2246" LastEditDate="2016-01-18T15:41:53.973" LastActivityDate="2016-01-18T15:41:53.973" Title="How monolitic is concrete when falsework is being crafted for parts of future structure only?" Tags="|materials|concrete|reinforced-concrete|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6968" PostTypeId="2" ParentId="6957" CreationDate="2016-01-18T12:46:33.980" Score="2" Body="&lt;p&gt;There is a limit on the size of an object made of a specific material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple reason is that for an arbitrary shape,(assuming that all material properties remain the same) its volume (and hence mass) increases with the cube of the scale factor but it's cross section at any point increases with the square. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you consider a cube of side length 1m and mass 1kg (volume 1m^3 and cross sectional area 1) The compressive stress at the base of the cube is its weight divided by area of the base ie 10N/1m^2 = 10Pa &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you double the length of the side, now we have a volume of 8m^3 and a base area of 4m^2 so the new stress is 80/4 = 20Pa &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double it again (side = 4m) volume = 64m^3  mass = 64kg weight = 640N base area = 16m^2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so stress at base = 640/16 = 40Pa &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So...the stress due to the object's own weight is proportional to the scale factor so you reach a scale where this is greater than the yield stress of the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put very simply the weight of an object increases with the cube of its scale and its cross section only increases with the square. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the maximum size object you can make from any given material is determined by its yield stress. This is the fundamental reason why it is not currently possible to build a 'space elevator'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of actually calculating the maximum size of a torus, the easiest way would be to use FEA software. Modeling a torus of unit radius should be easy enough then if you find the maximum stress as a fraction of UTS it should be a straightforward extrapolation. In fact many CAD packages will allow you to make a parametric model which you can link to a spreadsheet to get a general solution for any proportions and material properties. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-01-19T19:16:33.547" LastActivityDate="2016-01-19T19:16:33.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6969" PostTypeId="1" AcceptedAnswerId="6986" CreationDate="2016-01-18T14:28:21.247" Score="10" ViewCount="16144" Body="&lt;p&gt;Lets say you have a measurement device of which you do not know the accuracy, and a reference measurement device. Both measure a variable $x$. The range of interest is $x_0&amp;lt;x &amp;lt; x_1$. How do you determine the accuracy of the unknown device in this range?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My course of action would be to gather values for both devices from $x_0$ to $x_1$ and building a distribution of &lt;a href=&quot;https://en.wikipedia.org/wiki/Errors_and_residuals&quot;&gt;errors&lt;/a&gt;. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Accuracy_and_precision#ISO_Definition_.28ISO_5725.29&quot;&gt;accuracy&lt;/a&gt; could then be the error span, $\pm3\sigma$ or something similar - is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The reference measurement device is calibrated and has virtually no error&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="46" LastEditorUserId="46" LastEditDate="2016-01-18T18:52:01.460" LastActivityDate="2016-01-19T15:09:03.940" Title="How do you determine the accuracy of a measurement device?" Tags="|measurements|statistics|metrology|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6970" PostTypeId="2" ParentId="6967" CreationDate="2016-01-18T15:22:51.057" Score="3" Body="&lt;p&gt;In a continuous structure such as a beam or a column (under the expected loads for such elements), the structure is effectively indistinguishable from a truly monolithic one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has to do with how the elements are poured, with joints perpendicular (or at least not parallel) to the load direction. Columns are poured so that the joints are horizontal, while the load is primarily vertical. Whatever horizontal loads there may be (due to wind, for example) are usually relevant for bending. But in bending the column suffers tension, which the protruding rebar with sufficient anchorage length in both the new and old concrete absorbs, and compression, which is perfectly transferred from one concrete surface to another since the joint is perpendicular to the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very long beams requiring multiple pours are set with joints which are inclined at shallow angles in order to avoid creating a preferential plane for shear forces. One shouldn't make the joints horizontal, though, since that risks the different segments (more like layers) working independently instead of as a monolithic whole. Therefore, the same reasoning given above for the column applies to the beam as well: it suffers mostly bending, where the resultant tension is absorbed by the appropriately-anchored rebar and the compression is resisted by simple transfer between the concrete surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, pre-cast beams are often placed over consoles as in the figure below. Consoles, unlike cantilevers, suffer almost exclusively shear forces and must therefore have &quot;shear friction&quot; reinforcements. The book &lt;a href=&quot;https://www.scribd.com/doc/229329770/16/SHEAR-FRICTION-DESIGN-METHOD&quot; rel=&quot;nofollow noreferrer&quot;&gt;Design of Concrete Structures&lt;/a&gt; (as well as many others) has a description of the method in section 4.9 (pg 160). In new structures these consoles (and the column core between them, of course) should be poured all at once to avoid any joints. That being said, this isn't always possible. If you are reinforcing an existing structure, you might need to create such a console on a column which doesn't have one. You therefore have to deal with a joint between the new concrete of your console and the old concrete of the existing structure. This is an obvious problem since it creates a preferential shearing plane. The shear friction method considers such cases with a friction coefficient $\mu$ which is a function of the type of shearing plane (fully monolithic; new concrete on old, roughened concrete; new concrete on old, non-roughened concrete; steel on concrete). The book linked above has the appropriate coefficients listed on pg 161.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vUSx8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vUSx8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sub&gt;&lt;sup&gt;&lt;a href=&quot;http://www.oberbeton.com.ua/en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/sup&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that, yes, pour-joints may have an influence on the behavior of the structure if they are parallel (or close enough) to the direction of the load. If a beam were poured with vertical joints, it would create a preferential shear plane which would be quite tricky to reinforce. If a partially-loaded column (where the axial load is concentrated in a sub-region of the column's head) were absurdly poured with vertical joints, &lt;a href=&quot;https://en.wikipedia.org/wiki/Saint-Venant%27s_Principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saint-Venant's Principle&lt;/a&gt; may not really apply and the load may not be adequately distributed to segments of the column not directly under the loaded region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the specific case of the pylon mentioned in your question, I'm unsure since I've never designed one myself. As a first guess I'd consider it worrisome to create a horizontal joint near one of the cable anchors since that region must suffer significant shear forces which would be near-parallel enough to make me nervous about the joint behaving as a shearing plane. My completely ignorant, zero-experience approach would be to plan the joints so as to maximize the smallest distance between a joint and a cable. Also, maybe make the joints inclined so as to increase the angle between the cables and the joint (I don't even know if that's plausible in this case). Maybe both.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-01-18T15:22:51.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6971" PostTypeId="2" ParentId="6964" CreationDate="2016-01-18T16:00:57.520" Score="3" Body="&lt;p&gt;The &lt;a href=&quot;http://2.wittgas.com/filepool/Datenblaetter/usa/armaturen/flashback_arrestor_rf53n_dn_nsk_nu_u_usa.pdf&quot; rel=&quot;nofollow&quot;&gt;manufacturer says this&lt;/a&gt; about maintanence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;annual testing of the non-return valve, body leak tightness&#xA;  and flow capacity is recommended &lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;WITT is happy to supply special test equipment&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Flashback Arrestors are only to be serviced by the&#xA;  manufacturer; the dirt filter may be replaced by competent&#xA;  staff&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given that you inherit this equipment and don't know how it was treated I would do the recommended testing, or have it done. I would likewise replace the dirt filter.&lt;br&gt;&#xA;Contact Witt (or their sales rep/service partner) about these steps to get an idea if you can do this yourself, have it done by a technician in house or need external help.  &lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-01-18T16:00:57.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6976" PostTypeId="2" ParentId="6916" CreationDate="2016-01-18T22:13:04.543" Score="0" Body="&lt;p&gt;Two issues I see: The belt tensioner(s) should be placed on the &quot;slack&quot; portion of the belt, not between the motor and the load - similar to a bicycle chain derailleur. The tensioner rollers are relatively small and may be smaller than the minimum backbend diameter specification of the belt.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-18T22:13:04.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6979" PostTypeId="2" ParentId="6969" CreationDate="2016-01-19T05:50:28.080" Score="3" Body="&lt;p&gt;The only way to determine the accuracy to which any measuring device provides measurements is to calibrate it against a device of known accuracy and known errors for measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You technique is partially correct; don't just do the error measurement for the limits of the device as one population or sample bin. This is because measurement errors are not always uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for readings between 0 &amp;amp; 1, the error might be -0.2 and for readings between 2 &amp;amp; 3 the error might be +0.6. Your testing must be done in ranges or bands, irrespective of whether the units are mm (for rulers), m/s (for anemometers or speedometers) or Pa (for barometers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each range/band you determine the error for that range/band and then apply that error to the measurement taken from the device that needed calibrating.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-01-19T05:50:28.080" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6980" PostTypeId="1" CreationDate="2016-01-19T09:55:45.900" Score="1" ViewCount="45" Body="&lt;p&gt;For a project we need to repurpose 2 buried PE pipes, by connecting them and closing off segments. My question is about how this closing off can be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some details:&lt;br&gt;&#xA;The lines are roughly parallel, DN 600 and DN 350. The connecting piece will be in DN 350. After the connection the pipework will look like an &quot;H&quot;, then we close of one branch in each pipe to get a &quot;U&quot;.&lt;br&gt;&#xA;The medium is air at low pressure. To do the works, the pipes will be dug up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An obvious solution would be to cut the old pipes and weld or flange the new pipe with a reduction and 2 arches on. I want to know if there's an alternative way, where I use weld on stubs and just close the branches of the old pipes I don't need. How would I close the branches?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit to add: After sleeping over this, we will very likely have the old branches excavated and disposed so the question is moot (for now).&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-01-20T06:42:48.357" LastActivityDate="2016-01-20T06:42:48.357" Title="How to close a segment of a buried PE pipe" Tags="|civil-engineering|pipelines|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6981" PostTypeId="1" AcceptedAnswerId="6997" CreationDate="2016-01-19T10:11:22.373" Score="2" ViewCount="375" Body="&lt;p&gt;What kind of furnace is used to create alloys for aluminum die casting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have only been able to find information about how to create alloys or how to process them afterwards in the die casting process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insight on which kind of furnace you use to create the initial alloy would be highly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="4543" LastEditorUserId="272" LastEditDate="2016-01-19T13:00:20.197" LastActivityDate="2024-02-19T08:10:20.833" Title="Furnace for aluminum alloy creation" Tags="|aluminum|alloys|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6986" PostTypeId="2" ParentId="6969" CreationDate="2016-01-19T13:22:59.330" Score="2" Body="&lt;p&gt;Your approach is broadly correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are only interested in the accuracy of your system you probably want to use something like the maximum error. Your accuracy is then +/- Max error with the assumption that real errors are uniformly distributed within this range (a uniform distribution will often be an overestimation but is a simple option when no better information is available).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this approach will often produce large errors due to systematic effects which can easily be corrected by fitting a curve (normally linear) through the plot of measured and true values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should correct for the bias in your instrument and you can then calculate the uncertainty based on the standard deviation of the residuals. The total uncertainty is normally a multiple of $\sigma$, the choice is fairly arbitary, so you should state the multiple (k value), or the associate coverage factor.. You should also state what distribution you are assuming as this will effect what multiple gives a specific coverage. E.g. For a Gaussian 95 % coverage k~2, but for a uniform distribution 95 % coverage k~1.68&lt;/p&gt;&#xA;" OwnerUserId="153" LastActivityDate="2016-01-19T13:22:59.330" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="6987" PostTypeId="1" AcceptedAnswerId="6994" CreationDate="2016-01-19T13:54:09.730" Score="8" ViewCount="16818" Body="&lt;p&gt;I'm working on an application in which sparks (specifically from static discharge) present a severe safety hazard.  One component in question is a piece of PVC water pipe.  It has been proposed to &quot;ground&quot; the pipe to minimize the risk of static shocks between humans and the pipe, and between other components and the pipe.  I am skeptical as to whether this component needs to be grounded at all; it's not conductive, so my intuition is that it isn't able to cause a shock.  Normally I would just go along with this, however the proposed methods of grounding involve expensive adhesives and coatings which also have a long lead time for ordering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can a PVC pipe cause a static discharge? If so, what industry practices are out there to mitigate this phenomena?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pipe is being used in a test fixture for testing energetic materials, i.e. combustible solids. The energetic material is loaded into a length of the pipe, along with some sensors and other components. The material is ignited, after which the pipe is destroyed. The static spark concern is only during the loading process, when the energetic material could be ignited prematurely. The pipe is not used in its normal manner, like to transfer particles or fluids.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="368" LastEditDate="2016-01-19T19:03:03.627" LastActivityDate="2020-12-11T21:16:30.713" Title="Will PVC build up an electric charge and cause a spark?" Tags="|electrical-engineering|safety|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6988" PostTypeId="2" ParentId="6969" CreationDate="2016-01-19T15:09:03.940" Score="1" Body="&lt;p&gt;I was on a team of quality engineers (but not one of the experts), and they had a visual where they used a 2d plot where X axis was first measurement and Y was second measurement of the same observable feature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They would repeat the measure/remeasure and create what they called a &quot;sausage chart&quot;.  They would eliminate the outlying 2% of samples and draw a &quot;sausage&quot; around the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could visually see the quality of the measurement system by observing how close the data points fell to the 45deg angle line.&lt;/p&gt;&#xA;" OwnerUserId="4495" LastActivityDate="2016-01-19T15:09:03.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6990" PostTypeId="2" ParentId="6957" CreationDate="2016-01-19T16:40:34.020" Score="6" Body="&lt;p&gt;Gravitation will create a stress distribution inside any body. For homogeneous and isotropic materials there is two locations to care about: maximum tensile stress and maximum compressive stress. This is because the behavior of most materials differ when subjected to tensile versus compressive loads. (Ductile materials can be torn apart under tensile load but not under pure compression load; brittle materials usually have a higher compressive strength than tensile strength.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to continuum mechanics, the stresses depend on the geometry, the density and the gravitational acceleration only. For a body at rest, conservation of momentum evaluates to&#xA;$$&#xA;\rho f_i + \frac{\partial \sigma_{ij}}{\partial x_j} = 0&#xA;$$&#xA;where $\rho$ is the material's density, the $f_i$ are the internal forces ($f_i=-g\delta_{i3}$ when the gravitation is $g$ and is acting against negative 3 direction) and the $\sigma_{ij}$ are the stresses. (I'm using Einstein summation convention, so take the sum over $j$ in the $\sigma$-term in the above equation.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done a simple FEM calculations of your problem. Just to make sure, here is the geometry I calculated (dimensions in m):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZpD3C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZpD3C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the torus is not perfect, there is a small flat area at the bottom I had to introduce to retain the torus in the simulation (produced by cutting 1 m off the perfect torus). This introduces a small perturbation at the bottom, but you need a finite area to support the torus, otherwise stresses will become infinite at the bottom (see also below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum compressive stress is at the bottom (where the torus is touching ground) as already stated by the other answers. This is given by the minimum principle stress as shown in the next image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OjzTg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OjzTg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture shows the surface distribution of the minimum principle stress. Note that I have cut the scale at -50 MPa. The maximum compressive stress is at the bottom, and the best way to calculate that is (as given by the other answers) to divide mass by area. (This is the reason you need a finite area to support the torus.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the support area is large, the maximum compressive stress may also be achieved at the inner diameter. This maximum compressive stress for the geometry shown is at 32 MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum tensile stress is at the lowest point of the inner cirle. This is given by the maximum principle stress as shown in the following picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iO13X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iO13X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this geometry, the maximum tensile stress calculates to approx. 28 MPa. The density I used for the calculations is 2.41 g/cm³ (concrete). The gravitational acceleration is 9.81 m/s² in the simulation. The total mass of the object is 11.9e+9 kg (11.9 million metric tons), its volume 4.93e+6 m³.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For many materials, the compressive stresses at the bottom will most likely be the limiting design factor. However for extremely brittle materials, the tensile stresses may be the limit. E.g. when using concrete you will need to add steel to areas with tensile stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this still does not really answer your questions, since you have asked for an arbitrary material. As the others have pointed out, the stresses scale with geometry and density, i.e. doubling the size or doubling the density doubles the maximum stress. Now you have anything you need to calculate the stresses. To evaluate the maximum size compare stresses to strengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Before actually building, think about safety factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPS: This stress distribution is also quite intuitive if you think a little about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I updated the density just now – I had a slightly wrong number in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for ductile materials this is still not the whole answer, since you can allow the material to yield in some areas. The material will then plastically deform in those areas. This is fine, as long as the non-yielding structure can still carry all loads. This approach can be formalized. The associated term in German is 'plastische Stützzahl' which translates to 'plastic support number', but I don't know the correct English translation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the yield limit of the material need not be the ultimate limit, depending on your material. But it certainly is a good first estimate.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="4122" LastEditDate="2016-01-20T08:47:19.123" LastActivityDate="2016-01-20T08:47:19.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6991" PostTypeId="1" CreationDate="2016-01-19T16:54:16.863" Score="3" ViewCount="185" Body="&lt;p&gt;I need to replace a solenoid valve, and I need help in identifying the specifications that I should check for an equivalent replacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a picture of the valve itself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tefKH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tefKHt.jpg&quot; alt=&quot;solenoid valve&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click for larger image&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it matters, the valve is used to control water flow with a whole house humidifier.  The required inlet and outlet pipe sizes are understandably constrained by the existing piping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.achrnews.com/articles/92540-understanding-solenoid-valves&quot; rel=&quot;nofollow noreferrer&quot;&gt;This site&lt;/a&gt; provides a good overview of the terminology involved, but also mentions a few terms that aren't listed on the label.  Their review is echoed by &lt;a href=&quot;http://controltrends.org/by-industry/commercial-hvac/actuators-and-valves/01/solenoid-valve-basics/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;, which covers many of the same terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My online searching has yielded a couple of suggested equivalent parts, as well as a different version of the value offered by the OEM of the humidifier.  Some of the suggested equivalent parts are roughly half the price of the OEM replacement part.  But I would like to make sure that I'm truly buying an equivalent part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What specifications do I need to verify before purchasing a replacement solenoid valve?&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorUserId="368" LastEditDate="2016-01-19T19:00:11.833" LastActivityDate="2016-01-20T15:56:21.820" Title="How to evaluate solenoid valve specifications" Tags="|electrical-engineering|valves|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6992" PostTypeId="1" AcceptedAnswerId="6993" CreationDate="2016-01-19T17:44:37.480" Score="-1" ViewCount="141" Body="&lt;p&gt;I am trying to solve the following problem. If someone can just point me in the right direction it would be appreciated.  I don't have the textbook and online resources are proving to be absolutely worthless.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A piston-cylinder device containing a fluid is fitted with a paddle&#xA;  wheel stirring device operated by the fall of an external weight of&#xA;  mass 47 kg . As the mass drops by a height of 6.1 m , the paddle wheel&#xA;  makes 10800 revolutions. Meanwhile the free moving piston&#xA;  (frictionless and weightless) of 0.59 m diameter moves out by a&#xA;  distance of 0.70 m&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;What is the net work for the system if the outside pressure is 101&#xA;  kPa?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4552" LastEditorUserId="368" LastEditDate="2016-01-19T18:56:31.173" LastActivityDate="2016-01-19T18:56:31.173" Title="Find net work on the system" Tags="|thermodynamics|" AnswerCount="1" CommentCount="2" ClosedDate="2016-01-19T18:57:06.147" ContentLicense="CC BY-SA 3.0" />
  <row Id="6993" PostTypeId="2" ParentId="6992" CreationDate="2016-01-19T18:18:45.117" Score="1" Body="&lt;p&gt;The work that the system does on the atmosphere by pushing the piston outward is just $$W=\int{PdV}$$&#xA;From your problem statement, it looks like you have a constant pressure of 101 kPa and a change in volume of $\Delta V=\frac{\pi}{4}(0.59\ \mathrm{m})^2(0.7\ \mathrm{m})=0.19\ \mathrm{m^3}$.  The work done by the system is then $W=(101\,\mathrm{kPa})(0.19\ \mathrm{m^3})=19.3\,\mathrm{kJ}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, the energy added to the system by the falling weight is only $\Delta E=mg \Delta h=(47\ \mathrm{kg})(9.81\ \mathrm{\frac{m}{s^2}})(6.1\ \mathrm{m})=2.8\ \mathrm{kJ}$, so either the numbers are wrong or there was some other energy input besides the falling weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.chem.purdue.edu/gchelp/howtosolveit/Thermodynamics/ExpansionWork.html&quot; rel=&quot;nofollow&quot;&gt;This link on expansion work pretty much sums it up.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="368" LastEditDate="2016-01-19T18:55:45.190" LastActivityDate="2016-01-19T18:55:45.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6994" PostTypeId="2" ParentId="6987" CreationDate="2016-01-19T18:58:27.373" Score="6" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I am skeptical as to whether this component needs to be grounded at all; it's not conductive, so my intuition is that it isn't able to cause a shock.&lt;/p&gt;&#xA;&lt;p&gt;My question is basically (1) can a PVC pipe cause a static discharge&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Don't confuse &lt;em&gt;conductivity&lt;/em&gt; with &lt;em&gt;ability to hold charge&lt;/em&gt;.  If you've ever rubbed a balloon against your hair to make it statically cling to the balloon, you've witnessed a non-conductive material holding a charge.&lt;/p&gt;&#xA;&lt;p&gt;This problem is fairly common for &lt;a href=&quot;https://woodworking.stackexchange.com/questions/1489/is-pvc-pipe-safe-to-use-for-a-dust-collection-system/&quot;&gt;woodworkers who set up dust collection&lt;/a&gt; in their workshops.  PVC piping is often used as a carrier for sawdust back to a central dust collection tank.  The (usually) dry sawdust flowing rapidly past the PVC causes charge to build up, similar to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Van_de_Graaff_generator&quot; rel=&quot;noreferrer&quot;&gt;Van de Graaff&lt;/a&gt; generator.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;(2) if yes, what industry practices are out there to mitigate this phenomena?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The easiest way to &lt;a href=&quot;http://www.leevalley.com/en/shopping/Instructions.aspx?p=44424&quot; rel=&quot;noreferrer&quot;&gt;ground a PVC pipe&lt;/a&gt; is to pull a length of bare copper wire through the entire pipe length (on the &lt;em&gt;inside&lt;/em&gt; of the pipe) and connect it to a building ground.  This will not ground out the entire pipe since portions of the pipe away from the grounding wire (i.e., on the other side of the pipe cross-section) will be electrically insulated by the PVC itself, but it will mitigate the &amp;quot;danger&amp;quot; of a larger static discharge.&lt;/p&gt;&#xA;&lt;p&gt;The most effective way to ground a PVC pipe would be to line the inside of the pipe with some sort of wire mesh, but that is both impractical and fairly hard to achieve.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-01-19T19:03:58.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="6995" PostTypeId="1" CreationDate="2016-01-19T19:30:09.347" Score="4" ViewCount="1249" Body="&lt;p&gt;My state (USA - Nebraska) uses &quot;NU&quot; type prestressed girders (though the question should equally apply to all geometric variants of prestressed concrete &quot;I&quot; girders).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ICtIw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ICtIw.png&quot; alt=&quot;NU Girder Strand Layout&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 6 &quot;rows&quot; available for the prestressing strand. Moving from the bottom row upward, a maximum of 18, 18, 12, 6, 2, and 2 strands can be accommodated in each row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that theoretically the most efficient design will result when the rows are filled from the bottom up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Virtually all the plans I've seen fully fill that bottom row. However, I consistently run into cases where I'm &lt;strong&gt;not&lt;/strong&gt; inclined to fully fill the bottom row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1.&lt;/strong&gt; I need less than 18 strands total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 2.&lt;/strong&gt; I need 18 or more strands total, but the principle of &quot;fully fill the bottom row first&quot; requires debonding to satisfy the stress limits at transfer. Whereas, if I strategically partially fill several rows I can avoid debonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of one of my strategic layouts is below (shown at midspan; the orange strands are harped at 0.4L and 0.6L):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1gbEH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1gbEH.png&quot; alt=&quot;Example Strand Layout&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question initially arose because the standard drawings require that specific strands in the bottom row be extended into the concrete diaphragms at supports (see the top figure). If the bottom row isn't full, how will the contractor decide where to place the strands and which (if any) strands should extend into the diaphragms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More generally, is there any problem with partially filling rows?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also interested in rules-of-thumb for strand layout.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2016-01-19T23:53:44.307" Title="Strand layout in prestressed concrete girders: is a partially filled bottom row acceptable/advisable?" Tags="|structural-engineering|bridges|prestressed-concrete|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="6996" PostTypeId="1" CreationDate="2016-01-19T19:45:21.547" Score="8" ViewCount="9427" Body="&lt;p&gt;According to the United States &lt;a href=&quot;https://www.fhwa.dot.gov/bridge/material.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;2014 National Bridge Inventory&lt;/a&gt; the number of concrete bridges far outstrips the number of steel bridges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reinforced Concrete: 253,336&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prestressed Concrete: 148,333&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TOTAL CONCRETE: 401,669&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;versus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TOTAL STEEL: 181,095&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would concrete bridges outnumber steel bridges?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Basis for Categorizing By Material Type&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The concrete and steel totals are based on the &quot;Main Structure Type&quot; inventory code from the &lt;a href=&quot;https://www.fhwa.dot.gov/bridge/mtguide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;NBI Recording and Coding Guide&lt;/a&gt;. This is a three digit code defined as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J1dWl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J1dWl.png&quot; alt=&quot;NBI material code&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mZwZ5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mZwZ5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Motivation for Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In the state where I work we often design steel and concrete &quot;alternates&quot; for each bridge. However, we consistently see the concrete alternate (usually prestressed) selected instead of the steel. I'm trying to understand why this would be since steel seems to offer quite a few benefits...quick construction in the field, predictable material properties, often lighter than concrete girders, damage is usually readily inspectable, etc. etc. I've always chalked it up to some sort of inertial regional preference (when I was in Chicago we saw tons of steel, but in California and Nebraska it's mostly been concrete) and thought I'd see if there were some more substantive reasons.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2016-01-20T17:11:33.327" LastActivityDate="2016-12-10T08:08:30.180" Title="Why are concrete bridges more prevalent than steel bridges in the United States?" Tags="|structural-engineering|bridges|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="6997" PostTypeId="2" ParentId="6981" CreationDate="2016-01-19T20:03:19.660" Score="2" Body="&lt;p&gt;Normally the production of specific alloys is a fairly specialised process and they would be bought in as ingots or other stock types from commercial manufacturer rather than being made as part of a manufacturing process. This is as much about the quality control and analysis required to produce a given alloy to a specification as the actual mechanics of melting and handling it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When aluminium is refined from bauxite it is reduced by electrolysis, therefore aluminium refining plants are generally sited close to an economical source of electricity like a hydro-electric power station. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have ingots of metallic aluminium it is relatively easy to melt and both gas fired and induction furnaces are commonly used. Aluminium is usually melted in a refractory ceramic crucible although steel crucibles can be used in some circumstances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to produce a specific alloy you first need to know the composition of the metal you are starting with and then add the appropriate alloying elements in the correct proportions. Depending on the process flux may also be added to aid in separating out oxides and exclude atmospheric gasses (although aluminium is self-fluxing to a certain extent, in the  sense that is quickly forms a tough oxide layer on the surface of the molten metal which is effective in excluding the atmosphere ), it may also be necessary to add material to scavenge unwanted impurities and (especially in the case of aluminium) to remove any dissolved gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminium furnaces can be very simple indeed although, for obvious reasons the degree of sophistication will depend on the scale of production and the quality required from the finished product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One special case is where magnesium is a significant alloying element, in this case the very high reactivity of magnesium requires special safety precautions as molten magnesium can burn uncontrollably in air and as such it is not generally wise to melt high magnesium alloys in an open furnace. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-16T20:26:17.180" LastActivityDate="2016-02-16T20:26:17.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6998" PostTypeId="2" ParentId="6996" CreationDate="2016-01-19T20:07:16.857" Score="0" Body="&lt;p&gt;Steel will have lesser life span than concrete used in a bridge in water. Therefore concrete is a preferred choice.&lt;/p&gt;&#xA;" OwnerUserId="4554" LastActivityDate="2016-01-19T20:07:16.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="6999" PostTypeId="1" CreationDate="2016-01-19T20:31:25.720" Score="4" ViewCount="66" Body="&lt;p&gt;I am trying to come up with an analytical way to determine the optimal gland size that an elastomer o-ring will sit in. The amount of free space in the gland is determined by the thickness of two backup rings that surround it. Thus, the thicker the back-ups, the less free space to allow for thermal expansion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first step was to determine the internal stress on the o-ring due to temperature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coefficient of Thermal Expansion, $\alpha = 1.6\times10^{-4}\ \mathrm{\frac{in}{in-°F}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modulus, $E = 1900\ \mathrm{\frac{lbf}{in^2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature, $\Delta T = 400 \mathrm{°F}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stress: $$\begin{align}&#xA;\sigma=E\epsilon=E\alpha\Delta T &amp;amp;= (1900\ \mathrm{\frac{lbf}{in^2}})(1.6\times10^{-4}\ \mathrm{\frac{in}{in-°F}})(400\ \mathrm{°F}) \\&#xA;&amp;amp;=122\ \mathrm{\frac{lbf}{in^2}} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This where I am stuck relating this back to the dimensions of the gland. How can I move forward in order to relate this to the amount of free space in the gland? Assume worst case when the o-ring is fully constrained with no room for expansion.&lt;/p&gt;&#xA;" OwnerUserId="738" LastEditorUserId="368" LastEditDate="2016-01-19T22:16:57.617" LastActivityDate="2016-01-19T22:16:57.617" Title="Relating internal stress of an o-ring due to thermal expansion to amount of free volume in a gland" Tags="|elastic-modulus|seals|thermal-expansion|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7000" PostTypeId="2" ParentId="2096" CreationDate="2016-01-19T21:41:03.990" Score="2" Body="&lt;p&gt;In general you would raise mating surfaces in order to make them easier to machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having the surface raised above the surrounding material means that you don't have to worry so much about tools colliding with edges of the feature and it's fairly intuitive that skimming a raised surface is a simpler operation than machining a recessed pocket and it minimises the amount of material that you need to remove. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly if you are machining a raised surface with a bit of 'meat' to it you can concentrate on surface finish and flatness without too much concern about the precise depth of cut and you have some material to work with if you need to rework a part during original manufacture or in-service maintenance/repair. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A raised surface also means that the overall dimensions of mating parts can be less precise as slightly oversized parts will just overlap rather than interfering with anything. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-19T21:41:03.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7001" PostTypeId="2" ParentId="6995" CreationDate="2016-01-19T21:55:40.783" Score="3" Body="&lt;h3&gt;Strands in Upper Rows Versus Debonded Strands&lt;/h3&gt;&#xA;&lt;p&gt;You have the right idea that placing strands in upper rows is better than debonding strands. Debonded strands are more time consuming (e.g. cost more) than bonded strands. Also, placing a strand or two in the very top row usually helps to reduce the number of debonded strands needed. It is relatively easy to check if adding those strands will help.&lt;/p&gt;&#xA;&lt;h3&gt;Partially Filling Rows&lt;/h3&gt;&#xA;&lt;p&gt;As long as the strands are symmetric about the centerline, there is no problem with partially filling rows. If you think about it, that is exactly what is being done at the top of the beam in your first example (only four strands at top).&lt;/p&gt;&#xA;&lt;h3&gt;Continuous Strands&lt;/h3&gt;&#xA;&lt;p&gt;I have never seen strands being continued through the ends like you show, so I can't make a comment on those. I'm not entirely sure what the purpose is supposed to be.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-01-19T21:55:40.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7002" PostTypeId="2" ParentId="6996" CreationDate="2016-01-19T22:35:20.480" Score="11" Body="&lt;p&gt;I live in Brazil, so the relevance of my answer to your US experience is somewhat questionable. This is exacerbated by the fact that Brazil is almost 100% concrete, with very few steel or wooden structures (including buildings). That being said, some of these concepts are location-agnostic and others may apply to the US as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, as mentioned by others in the comments, concrete requires basically no maintenance for the duration of its service life. Steel, on the other hand, rusts and therefore must be maintained (or, well, should be, anyway...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Brazil at least, concrete is far cheaper than steel, which in infrastructure projects is usually the defining factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to the above, (non-prestressed) concrete requires less-skilled workers. Concrete is basically just &quot;put up the formwork, place the reinforcement and pour the concrete.&quot; Steel requires expert welders and careful torquing of screws. In Brazil the lack of steel-skilled workers is such that a welder easily earns more than an engineer. Weld-checkers swim in cash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bridges are often built far from urban centers, meaning your choice of constructor may be limited, reducing the quality of your workers. This relates to the above. Your supply lines may also be problematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concrete is more flexible, allowing for the easy treatment of curved beams, including box-girders which excel at withstanding torsion. It also trivially allows for variable-cross-section elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For long spans, prestressed concrete is usually the most efficient choice.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-01-19T22:35:20.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7003" PostTypeId="2" ParentId="6995" CreationDate="2016-01-19T23:53:44.307" Score="3" Body="&lt;p&gt;@hazzey has answered your main question quite effectively, so I'll just add two comments.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;While I agree that partially filling rows is perfectly fine, I would just like to add that it may be better to distribute the strands along their rows instead of bunching them all near the centerline (but keep the strands of different rows aligned) as in the figure below. Notice that the orange strands remained at the centerline since I assume they need to harp into the web. You could also bring the outer strands in by one space and then redistribute the strands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H8je9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H8je9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distributing your strands horizontally along the beam will reduce splitting-stresses at the anchors. It also improves the quality of the concrete bond, but as of a certain distance between strands this loses relevance. That being said, I assume that the actual drawings will display the distance between the strands.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;While I've never worked with such &quot;standardized&quot; cross-sections, I've never encountered a rule-of-thumb other than &quot;use the cheapest solution to achieve your objective.&quot; Cheapest usually means using the least possible total steel area (and therefore force). That implies in jamming as many strands as possible as low as possible (at the midspan, at least) in order to increase the efficiency of your prestress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, you currently have sixteen strands (6 + 6 + 4). I assume not, but could this be reduced to fourteen (or fewer) on the bottom row only? What about sixteen (or more, so long as the total area of steel is lower) strands of a lesser diameter? Less force but with a greater lever arm. This would be a better solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't reduce your area of steel, then you only have two choices: keep your sixteen strands on the bottom and debond; or use your current layout. Debonding strands is more expensive and therefore goes against the &quot;cheapest solution&quot; rule, so your layout is the correct choice.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-01-19T23:53:44.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7006" PostTypeId="2" ParentId="6996" CreationDate="2016-01-20T02:58:37.987" Score="8" Body="&lt;h3&gt;Initial Cost&lt;/h3&gt;&#xA;&lt;p&gt;When span lengths allow, concrete spans are cheaper than steel spans. This is partially because of material costs and partial because of labor (fabrication) costs.&lt;/p&gt;&#xA;&lt;p&gt;There are a few situations where concrete is not feasible. These situations could be:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Very long span lengths.&lt;/li&gt;&#xA;&lt;li&gt;Skewed spans.&lt;/li&gt;&#xA;&lt;li&gt;Shallow structure depths.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;In these situations, steel might be used even if it is more expensive.&lt;/p&gt;&#xA;&lt;p&gt;Also, concrete spans are usually composed of typical cross sections. Once the forms for these precast, prestressed beams are made, the labor costs are greatly reduced. Every steel girder has the same amount of labor whether it has been previously done or not.&lt;/p&gt;&#xA;&lt;h3&gt;Long-term Maintenance&lt;/h3&gt;&#xA;&lt;p&gt;Concrete spans have reduced maintenance as compared to steel spans. Steel spans typically require periodic painting and cleaning. This isn't a requirement for concrete spans. Also, concrete does not require the same level of inspection as steel spans do. There is no need to check for fatigue cracks in concrete spans.&lt;/p&gt;&#xA;&lt;h3&gt;Added Together&lt;/h3&gt;&#xA;&lt;p&gt;When both the initial cost and and long-term costs are added together, concrete is typically cheaper. This is why there are more concrete bridges.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-01-20T17:33:27.197" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7011" PostTypeId="1" CreationDate="2016-01-20T12:54:28.130" Score="6" ViewCount="135" Body="&lt;p&gt;I'm interested in some theory &lt;em&gt;Eurocode 1: Actions on structures - Part 2: Traffic loads on bridges&lt;/em&gt;. How were the limits chosen (e.g. maximum deck acceleration = 3.50 m/s&lt;sup&gt;2&lt;/sup&gt;)? Why is the lower limit of $n_0$ governed by dynamic impact criteria and given by the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_0 = \begin{cases}&#xA;80/L, &amp;amp;4m&amp;lt;L&amp;lt;20m \\[2ex]&#xA;23,58L^{-0,592}, &amp;amp;20m&amp;lt;L&amp;lt;100m&#xA;\end{cases}$$&lt;/p&gt;&#xA;" OwnerUserId="4564" LastEditorUserId="368" LastEditDate="2016-01-20T17:44:01.570" LastActivityDate="2016-01-20T17:44:01.570" Title="What is the theory behind Eurocode 1 - Part 2: Traffic loads on bridges?" Tags="|civil-engineering|bridges|eurocodes|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7012" PostTypeId="2" ParentId="2220" CreationDate="2016-01-20T15:09:01.387" Score="1" Body="&lt;p&gt;The  issue described here sounds like galling. This is where the mating threads adhere to each other and tear chunks out of the surface which then wedge into the  thread locking it completely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symptoms are that the nut will go on fine for a number of turns and very suddenly seize in both directions, at this point its usually impossible to remove without destroying the threads and you often see quite long sections of one thread embedded in the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some steels are more susceptible than others, with stainless being a particular offender and the finish of the thread is also a factor is is the use of impact wrenches etc. So a change in steel specification could indicate that the new steel is more susceptible to galling or that it has different machining properties which affect the surface finish. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-20T15:09:01.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7013" PostTypeId="2" ParentId="6991" CreationDate="2016-01-20T15:39:21.067" Score="2" Body="&lt;p&gt;You need verify that your replacement has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1/16&quot; orifice(commonly available)&lt;/li&gt;&#xA;&lt;li&gt;Now, what are the size of the theads on the inlet and outlet(1/8&quot; npt?) it might be stamped somewhere on the valve body. Also, unscrew the 90deg adapter and check. &lt;/li&gt;&#xA;&lt;li&gt;24/60 would indicate 24volts AC at 60Hz.(not very common) If you can check the two lines connecting the solenoid with a multimeter you can verify this yourself.&lt;/li&gt;&#xA;&lt;li&gt;the wattage of the valve: 6 watts?(hard to see)&lt;/li&gt;&#xA;&lt;li&gt;it looks like a standard brass housing&lt;/li&gt;&#xA;&lt;li&gt;Normally Open or Normally Closed? (NO, NC) When NOT energized, is there flow through the valve?&lt;/li&gt;&#xA;&lt;li&gt;Is it suitable/designed for your application? Water.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once you can answer all of these questions, you can easily find a replacement valve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, check ebay/google/amazon by description of part/machine/Serial Numbers on valve and/or by valve specifications. You'll likely find what you need there as well. I did a quick search and found several likely replacements.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2016-01-20T15:56:21.820" LastActivityDate="2016-01-20T15:56:21.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7014" PostTypeId="1" CreationDate="2016-01-20T15:53:30.137" Score="1" ViewCount="2514" Body="&lt;p&gt;As per title, I would like the equation to calculate the effective failure rate of a system or branch of parallel/redundant components whose failure rates follow a Weibull distribution. I found what I need for components whose failure rate is described by an exponential distribution at pag. 394 (table 6.2.1-3, equation 1) of the book &quot;system reliability toolkit&quot; that can be found &lt;a href=&quot;https://books.google.co.uk/books?id=--jRl7hQ4uUC&amp;amp;printsec=frontcover#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&#xA;I would like to find the analogous for Weibull distribution. &lt;/p&gt;&#xA;" OwnerUserId="4566" LastEditorUserId="1586" LastEditDate="2016-03-22T18:00:36.200" LastActivityDate="2016-03-22T18:00:36.200" Title="Formula to calculate the total failure rate of a system composed by parallel components whose failure rate is described by Weibull distributions?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7015" PostTypeId="1" AcceptedAnswerId="13761" CreationDate="2016-01-20T16:47:56.673" Score="5" ViewCount="2287" Body="&lt;p&gt;While I've designed tons of plastics for corrosive service, the fasteners to hold the plastics together are inevitably metallic.  There have been a lot of ways to protect them from rusting (due to exposure to things like &lt;em&gt;33% HCl, 50% H2SO4, 50% NaOH, and 15% NaOCl&lt;/em&gt;, for example) that I've seen, but here is a new one I haven't heard of until today - &lt;strong&gt;Alloy 20.&lt;/strong&gt;  It's a nickel alloy apparently and seems to hold up pretty well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my concerns about Alloy 20 is whether the parts are easy to obtain and available across the world? How does it compare with Grade 5 Titanium, Hastelloy C-276, and 316 SST as a fastener in plastics?&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="368" LastEditDate="2016-01-20T17:02:00.697" LastActivityDate="2017-02-14T15:28:14.697" Title="How does Alloy 20 compare with Ti 6-4, C-276 and 316 stainless for corrosive service?" Tags="|mechanical-engineering|materials|chemical-engineering|corrosion|fasteners|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7016" PostTypeId="1" CreationDate="2016-01-20T17:07:07.510" Score="3" ViewCount="62" Body="&lt;p&gt;This is related to construction of a bipedal robot. I have a doubt regarding the stages involved in walking of a biped. In the one leg support phase, when the body moves in a circular path about one leg, who is providing the force? How can the joint at the knee or hip provide the force for the body to rise and move forward?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tktOg.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tktOg.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4569" LastEditorUserId="368" LastEditDate="2016-01-20T17:36:08.100" LastActivityDate="2016-01-20T21:29:41.843" Title="Which joint is providing the force to a bipedal robot to move the body forward?" Tags="|dynamics|robotics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7018" PostTypeId="2" ParentId="7016" CreationDate="2016-01-20T21:29:41.843" Score="3" Body="&lt;p&gt;Bipedal walking has been &lt;a href=&quot;http://blogs.discovermagazine.com/inkfish/2014/09/26/walking-really-is-just-falling-and-catching-yourself/#.Vp_5cEorKUk&quot; rel=&quot;nofollow&quot;&gt;aptly compared to the art of controlled falling&lt;/a&gt;.  And once we consider that perspective, it's a little easier to describe what's taking place when someone is walking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, the following stages occur during a single step:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Subject's stepping foot is lifted up through a combination of the quadriceps and gluteus muscles.&lt;/li&gt;&#xA;&lt;li&gt;The subject's center of gravity is shifted forward, and the subject starts to fall forward.  This is where forward motion from walking comes from.&lt;/li&gt;&#xA;&lt;li&gt;Subject places stepping foot on the ground and stops the falling motion.&lt;/li&gt;&#xA;&lt;li&gt;Various minor adjustments are made to assure both forward and lateral stability.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From there, the process repeats for the next step(s).&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2016-01-20T21:29:41.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7019" PostTypeId="1" AcceptedAnswerId="7022" CreationDate="2016-01-20T22:11:39.683" Score="4" ViewCount="6225" Body="&lt;p&gt;I am practicing engineering drafting and sometimes I am near on finishing my work but suddenly I make a mistake. Is it possible to remove the ink? If yes, how? Where can I buy such tools? I am using a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B001U3RCGQ&quot; rel=&quot;nofollow&quot;&gt;Staedtler's Mars Matic 0.5mm&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3703" LastEditorUserId="2246" LastEditDate="2016-01-21T17:29:46.550" LastActivityDate="2016-08-19T21:42:18.570" Title="Is it possible to remove Staedtler ink from a paper?" Tags="|drafting|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7020" PostTypeId="2" ParentId="7019" CreationDate="2016-01-20T22:21:06.320" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to remove the ink?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A quick &lt;a href=&quot;https://www.google.com/search?q=erase%20mars%20matic&amp;amp;ie=utf-8&amp;amp;oe=utf-8#safe=off&amp;amp;q=mars%20matic%20erase%20ink&quot;&gt;Google search&lt;/a&gt; isn't turning up anything, as I more or less expected.  Ink isn't meant to be erased, so I'm going to hazard that there is no product that will erase the ink for this pen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, the only &quot;erasable&quot; ink pens I see are the cheap, consumer-grade pens that use a more standard-grade ink (erasable is a relative term - I've never seen an erasable pen where you can totally remove the ink from the paper).  Since the Staedtler pens are more professional-grade and use higher-quality inks, they are made to be indelible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;sometimes I am near on finishing my work but suddenly I make a mistake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The tongue-in-cheek response to this would be &quot;stop making mistakes.&quot;  I know that's not really possible given that we're all human.  However, when it comes time to inking, you should really be focused on tracing lines that you already know are correct.  Inking is not the time to make changes.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2016-01-20T22:21:06.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7021" PostTypeId="1" AcceptedAnswerId="7036" CreationDate="2016-01-20T22:25:29.150" Score="2" ViewCount="10226" Body="&lt;p&gt;I'm curious about pumping a fluid through two different tubes (coming from two different pumps) into a Y connector. The third side of the connector runs into a collector without any resistance (other than that of the length of tubing).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If each pump runs at a different rate, say one is twice as strong as the other, do I run any risk of any backing up into the weaker pump? What if one is three times stronger than the other, etc.?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the third side of my Y connector (the side the fluid leaves through) now pump with the equivalent of pump a + pump b total power? Are they averaged together? Is it diminished? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does the type of fluid matter for parts 1 and 2 above?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This question assumes that all tubing is the same thickness, but answers regarding different thicknesses could also be interesting (and worth extra credit!)&lt;/p&gt;&#xA;" OwnerUserId="4573" LastEditorUserId="368" LastEditDate="2016-01-21T00:14:32.057" LastActivityDate="2016-01-21T20:09:13.730" Title="How do pumped flows combine in a Y connector?" Tags="|fluid-mechanics|pumps|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7022" PostTypeId="2" ParentId="7019" CreationDate="2016-01-20T23:20:32.633" Score="9" Body="&lt;p&gt;To remove such ink it needs to scraped off, usually with a scalpel blade. This works well on thicker and more studier media such as film or vellum, but it will always be obvious where the erasure was made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the amount of ink that needs to be removed and the robustness of the media, sometimes an ink erasure can work (such as a Staedtler blue ink eraser, &lt;a href=&quot;http://www.staedtler.com/en/products/pencils-accessories/erasers/mars-plastic-combi-526-508-premium-quality-combi-eraser/&quot;&gt;Mars Plastic Combo&lt;/a&gt;), but again it will be obvious where the erasure was made. This can work on paper, but only with thicker robust paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to present a drawing of professional standard, redraw the entire drawing.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-01-20T23:20:32.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7023" PostTypeId="1" AcceptedAnswerId="7028" CreationDate="2016-01-21T00:01:33.367" Score="0" ViewCount="2896" Body="&lt;p&gt;I accidentally dropped my &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B001U3RCGQ&quot; rel=&quot;nofollow&quot;&gt;Mars Matic 0.5mm&lt;/a&gt; and noticed that maybe I broke the needle. I was having a hard time writing with it so I tried to clean it, and while cleaning I accidentally bent the needle part. After all the trouble that I experienced I manage to straighten the needle very carefully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now it's working fine but I need to shake it rigorously. I am planning on buying a new one because of the hassle of using it but there is also a side of me that says I should not buy a new one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other way that I can fix my pen? It's slightly bent now but by shaking it rigorously it can still write.&lt;/p&gt;&#xA;" OwnerUserId="3703" LastEditorUserId="368" LastEditDate="2016-01-21T00:17:11.680" LastActivityDate="2016-01-21T03:15:32.297" Title="How do I fix a technical drawing pen with a bent needle?" Tags="|drafting|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7024" PostTypeId="2" ParentId="7021" CreationDate="2016-01-21T00:14:24.247" Score="3" Body="&lt;p&gt;Pumps output based on a &lt;a href=&quot;http://www.pentairpool.com/images/product_images/wholegood/WhisperFlo/WhisperFloCurve_new.gif&quot; rel=&quot;nofollow&quot;&gt;pump curve graph&lt;/a&gt; that has been determined empirically by the manufacturer for that specific pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question, &quot;Twice as strong as the other&quot;, is not meaningful because it is a complex relationship of output pressure(head) and output flow. If you have the pump curve for both pumps and the &lt;a href=&quot;http://www.engineeringtoolbox.com/total-pressure-loss-ducts-pipes-d_625.html&quot; rel=&quot;nofollow&quot;&gt;minor and major losses&lt;/a&gt; you can solve for a flow direction, total flow, pressures, etc.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-01-21T00:14:24.247" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7025" PostTypeId="5" CreationDate="2016-01-21T00:20:36.787" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-01-21T00:20:36.787" LastActivityDate="2016-01-21T00:20:36.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7026" PostTypeId="4" CreationDate="2016-01-21T00:20:36.787" Score="0" Body="Questions about tools and techniques for technical drawing. If your problem is mainly with a particular software package, consider using a tag for that software package instead." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-01-21T00:20:36.787" LastActivityDate="2016-01-21T00:20:36.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7027" PostTypeId="1" CreationDate="2016-01-21T01:28:48.693" Score="3" ViewCount="254" Body="&lt;p&gt;I'm developing a code to simulate thermal performance of automotive radiators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My results are 40% under-predicted. But when I multiply them to 1.6, they match completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the graph that compares the results of program and experimental work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o5kEp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o5kEp.jpg&quot; alt=&quot;Comparing experimental and numerical results&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be noted, this number (1.6) differs when we use the other data sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Procedure of the program is as following :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;given air velocity, coolant flow rate, inlet air and coolant temperatures.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;assume outlet air and water temperatures (it should be a logical assumption).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;calculate air and fluid properties (using interpolation in existing tables based on mean temperatures)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;calculate heat transfer coefficient of water according to flow regime (using Petukhov or Gnielinski equations)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;calculate Colburn j factor (using Chang and Wang empirical correlation)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;calculate the total thermal resistance between air and coolant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;calculate NTU and effectiveness&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate heat rejection&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate outlet temps and compare them with assumed temps. If it doesn't meet the specified error, go to step 3 and repeat calculations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;clc;clear all&#xA;% _____Jung Experiment_______&#xA;% Air Velocity m/s&#xA;    Va=4;&#xA;water_Flowrate=2e-3/3;% m3/s (40 L/min)&#xA;% Top tank and inlet air temperatures (deg C)&#xA;twih=85; t_ambient=25;&#xA;% Specifying the j-factor correlation (Chang &amp;amp; Wang)&#xA;cor='CW';&#xA;% Entering the Radiator characteristics&#xA;cas='C:\Users\Solid\Desktop\jung.xlsx';&#xA;&#xA;% ___ Calculation ___&#xA;ITD=twih- t_ambient;&#xA;% Export Geometrical details from jung.xlsx file and calculate the needed&#xA;% Areas like AT(total heat transfer area etc).&#xA;[R,L,Afrt,Afrf,Ac,Afront,sigma,Af,Aw,AT...&#xA;    Lp,Fp, Fl, Ll, alpha, Fd, Tp, deltaf, Nlb,Td,LH,af,Dh,NT,kf...&#xA;    RH,RW,RD,N,Nct,Np,Nf,Yl,Dm,D_major,W,tt]=Area_Calc(cas);&#xA;% Radiator tubes are flat. So we need hydraulic diameter(dw). &#xA;[dw,A_cross,Perimeter]=hydro2(Dm,L);&#xA;% ___ Pre_Procssing ___ &#xA;t_top_tank=twih;&#xA;twi= t_top_tank;&#xA;tai = t_ambient;&#xA;%  defining an error;&#xA;error=.01;&#xA;% Assuming outlet temperatures of water(Two) and Air(Tao) logically.&#xA;Tao=50; Two=75; &#xA;Y=1; &#xA;% defining a counter;&#xA;ll=0;&#xA;        while Y&#xA;            Q=0;&#xA;        ll=ll+1;&#xA;%         At the end of the loop, calculated temperatures will be         compared&#xA;%         with the tepms of the past iteration.So we save the tepms of     the&#xA;% past iteration here.  &#xA;        Two1=Two; Tao1=Tao;&#xA;% Calculating the properties like viscosity(uw(coolant),ua(air)),Prandtel&#xA;% Number (Pra(air) ,Prw(coolant)) etc.&#xA;        [ uw,ua,rhow,rhoa,cw,ca,kw,ka,Prw,Pra,tam ]=...&#xA;            phyprop2(twi ,tai,Two,Tao);&#xA;%       Air side Reynolds number. ReLp is the reynolds number based on&#xA;% Louver pitch. &#xA;        ReLp=rhoa*Va*Lp*1e-3/(ua);&#xA;        ReDh=rhoa*Va*Dh*1e-3/(ua);&#xA;%         Air mass flowrate&#xA;mair=Ac*rhoa*Va;&#xA;&#xA;% --- Coolant side Reynolds number and mass flowrate ---&#xA;[Rew,mwater]=hydro(dw,A_cross,water_Flowrate,rhow,uw,...&#xA;    NT,'I');&#xA;% coolant heat transfer coff.&#xA;[hw]=hc2(Yl,dw,Rew,kw,Prw,1);&#xA;% Air heat transfer coff. and friction factor&#xA;% (friction factor(f) will be calculated but won't used in this program )&#xA;[ha,jcor,f]=ha2...&#xA;    (Lp,Fp,Fl, Ll, alpha,Fd, Tp, deltaf, Nlb,Td,LH,Dh,ua,rhoa,ca,Pra,...&#xA;    Va,cor,ReLp,ReDh,Dm,tt);&#xA; % ----------- UA , NTU--------&#xA; l=Fl*1e-3/2;&#xA;M=sqrt(2*ha/(kf*deltaf*1e-3));&#xA;% fin and Surface efficiency&#xA;nf=tanh(M*l)/(M*l);&#xA;no=1-Af*(1-nf)/(AT); Q=0;&#xA;% ----------Capacity Rate-----------&#xA;Ca=mair*ca; Cw=mwater*cw;&#xA;&#xA;if min(Ca,Cw)==Ca&#xA;Cmin=Ca;&#xA;Cmax=Cw;&#xA;Q=1;&#xA;else&#xA;Cmin=Cw;&#xA;Cmax=Ca;&#xA;end&#xA;Cs=Cmin/Cmax;&#xA;% Calculation of overall heat transfer coefficient using thermal&#xA;% resistances between air and water(fouling and wall resistances are&#xA;% negligible)&#xA;UA=(( 1/(hw*Aw)+1/(no*ha*AT) ) )^-1;&#xA;NTU=UA/Cmin;&#xA;% -------------  E - NTU (calculation of effectiveness _epe_ using         existed &#xA;% correlations) -----------------&#xA;epe=eps1(Q,N,Cs,NTU,Cmin,Ca);&#xA;% Air and water outlet temps&#xA;tao=tai+epe*Cmin*(twi-tai)/Ca;&#xA;two=twi-epe*Cmin*(twi-tai)/Cw;&#xA;&#xA;Tao=tao;&#xA;Two=two;&#xA;% do these temps meet the errors ?&#xA;Y=abs(Tao-Tao1)/Tao&amp;gt;error2 ||...&#xA;    abs(Two-Two1)/Two&amp;gt;error2;&#xA;        end&#xA;%         Radiator Heat Rejection&#xA;        q=epe*Cmin*ITD;&#xA;        disp(q)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What's wrong?&lt;/p&gt;&#xA;" OwnerUserId="4574" LastEditorUserId="1832" LastEditDate="2018-06-06T21:18:37.757" LastActivityDate="2018-06-06T21:18:37.757" Title="Automotive Radiator Simulation" Tags="|fluid-mechanics|automotive-engineering|heat-transfer|simulation|heat-exchanger|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="7028" PostTypeId="2" ParentId="7023" CreationDate="2016-01-21T03:15:32.297" Score="2" Body="&lt;p&gt;You can buy a replacement nib. The following picture is from eBay. I just wanted a picture to show you that it is possible to remove the existing nib and replace it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest you make enquires with pen supplier, or search the internet for a replacement nib.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/peN6c.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/peN6c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-01-21T03:15:32.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7029" PostTypeId="2" ParentId="7019" CreationDate="2016-01-21T05:02:05.340" Score="2" Body="&lt;p&gt;I have used sand paper on drawings before, with varied results. Experiment with a mix of grits on a scrap piece, and see if it will work with your paper. If no sand paper, try nail files. If nothing, redraw.  Also Frixion makes magical erasable pens. &lt;/p&gt;&#xA;" OwnerUserId="4577" LastActivityDate="2016-01-21T05:02:05.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7030" PostTypeId="2" ParentId="6916" CreationDate="2016-01-21T05:36:36.313" Score="2" Body="&lt;p&gt;I would suggest something closer to the following:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/B54bV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B54bV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also when controlling the system, make sure the speed control is on only one of the motor shafts and you drive both motors with the same output from the speed control loop.   In other words don't try to control the speed of both motors with separate motor controllers.&lt;/p&gt;&#xA;" OwnerUserId="4556" LastActivityDate="2016-01-21T05:36:36.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7032" PostTypeId="2" ParentId="7027" CreationDate="2016-01-21T10:29:34.757" Score="2" Body="&lt;p&gt;Either your definition of &quot;completely&quot; is a bit unusual, or there's something wrong with those charts, because in those charts, 1.6 x the model does not completely match &lt;em&gt;any&lt;/em&gt; of the six observations you have (assuming the point values are at the centre of each shape).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way: your model outputs don't match your observations, with or without a 1.6 fudge factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's some combination of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;your assumptions don't match reality;&lt;/li&gt;&#xA;&lt;li&gt;your inputs don't match reality;&lt;/li&gt;&#xA;&lt;li&gt;your model doesn't match reality;&lt;/li&gt;&#xA;&lt;li&gt;your observations don't match reality;&lt;/li&gt;&#xA;&lt;li&gt;there are things outside the system boundary of your model that are crucial to the results, that you are not accounting for (this is kind of a subset of the first bullet point).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, you need to go through, step-by-step, verifying or correcting each of those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a first guess: have you actually calibrated your model at any point: that is, taken a set of inputs that you can be absolutely sure of, and a set of observations that you can be absolutely sure of, and got a match. Or is what you're presenting to us, actually an uncalibrated model - in which case, why should it look like reality? The second-best way to tackle this: does your model reproduce the results of some other, calibrated model?&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2016-01-21T10:29:34.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7033" PostTypeId="1" CreationDate="2016-01-21T12:20:10.597" Score="5" ViewCount="2778" Body="&lt;p&gt;How a Material &lt;a href=&quot;https://en.wikipedia.org/wiki/Fatigue&quot; rel=&quot;nofollow noreferrer&quot;&gt;fatigues&lt;/a&gt; can be characterized by a Wöhler curve where each datapoint corresponds to a failure with the stress on the y-Axis and the number of cycles on the x-axis. An example is the following Wöhler curve: &lt;a href=&quot;https://i.stack.imgur.com/MtGwP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MtGwP.jpg&quot; alt=&quot;Example Wöhler curve.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does this curve change with the frequency with which a cycle count is reached?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="2015" LastEditDate="2016-11-02T20:26:52.957" LastActivityDate="2016-11-02T23:58:43.933" Title="How does the S-N curve (or Wöhler curve) of a material change with frequency?" Tags="|mechanical-engineering|structural-engineering|materials|metallurgy|fatigue|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7034" PostTypeId="2" ParentId="7014" CreationDate="2016-01-21T12:35:57.657" Score="1" Body="&lt;p&gt;RAC used to make their docs available for free; not so much any more.  Here's a link to a previous edition: &lt;a href=&quot;http://reliabilityanalytics.com/Rome_Laboratory_Reliability_Engineers_Toolkit.pdf&quot; rel=&quot;nofollow&quot;&gt;ReliabilityToolkit1993&lt;/a&gt;. Take a look around page 90 for some derivations of reliability equations.   Put simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) serial components&lt;/strong&gt; In this setup, if any single component fails, the system fails.  THerefore, the reliability is the product of the individual item's reliabilities.  E.g., if you have three items with predicted failure rates of 99%, 95%, and 99%,  the system value is 93.11 % &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) parallel components&lt;/strong&gt;  In the simplest case where you need one out of N working, calculate the probablity that all of them will fail and subtract from 1 to get the net reliability.  For the same three items above, then,  $P_f  = 1 - (1-.99)*(1-.95)*(1-.99) = 9.99995e-01 $.   Now, if you need to have, say  N out of M identical parallel items running, you have to write out a slightly longer equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nearly any system can be organized as a collection of series and parallel component operations, so the above 2 rules handle most situations.  There are additional tweaks for spare components which are left running ('fully parallel') or spares which are left off until needed, presumably increasing their lifetime after being turned on, etc.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your specific question, pretty much just place the Weibull function for each component where the exponentional function is in the document you referenced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe there are some good handbooks available for download somewhere at MIL DITC , NASA, etc.  as well, so a little websearching may be helpful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ETA: there's a treasure trove of calculators at &lt;a href=&quot;http://reliabilityanalyticstoolkit.appspot.com/&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt;, along with detailed equations and explanations of each one.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-01-21T12:35:57.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7036" PostTypeId="2" ParentId="7021" CreationDate="2016-01-21T14:23:35.493" Score="4" Body="&lt;p&gt;&lt;strong&gt;General rule when dealing with fluids&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluids will travel the path of least resistance. Treat them as dumb and lazy and you'll understand the process better; fluids will only travel where you make them and if you allow them to take a shorter path they will take it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt; If the smaller pump puts less pressure than the third side orifice then the small pump will likely cavitate and back flow.  The relative pressure of the two pumps will play into where they operate on the pump curve however if they both push higher pressure than the orifice there will not be back flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2&lt;/strong&gt; The resulting pressure of the orifice is a complicated answer solved by Bernoulli's Equation (&lt;a href=&quot;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=bernoulli%20equation&quot;&gt;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=bernoulli%20equation&lt;/a&gt;). In real life the texture, size, fullness, route and shape of the pipe opening play major roles in determining if the pressure is higher, lower or nil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3&lt;/strong&gt; They type of fluid does not change how physics work, unless you're talking about pumping different fluids in each pump and making the Y into a mixing chamber (asking for trouble). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do run into an issue that the pressure differential in the system is too great and the smaller pump would back flow are you able to create an air gap in the (liquid) system? Rather than directly piping the two pumps into each other have them pour into a common container which gravity discharges through the final orifice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4 Bonus based on comment&lt;/strong&gt; The same answers would apply to liquids or gasses. you would still need the same type of information.  In a previous life, I designed piping systems which covered hundreds of miles and thousands of junctions in each system. There were 3-8 fans passing 600k-1,000k cfm into different points in the system. We had to determine with a fairly high reliability (90%+-) what the pressure and air flow at each junction would be or faced shutdown by regulators. &lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2016-01-21T14:23:35.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7038" PostTypeId="1" AcceptedAnswerId="7054" CreationDate="2016-01-21T17:45:38.973" Score="4" ViewCount="541" Body="&lt;p&gt;I am interesting in replacing a solid strut with tubing, but I do not want to destructively test the strut to determine its yield point. I do not want to make assumptions about the yield point based on material, because heat treatment and other factors can affect the yield point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I cantilever suspend the strut and record its deflection due to a weight placed on the end, to infer the strength of the strut?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words lets say the strut deflects 5 inches when it is cantilevered and a weight is placed on its end. The tubing under test is placed the same way and only deflects 4 inches. Can I assume that the tubing will be an adequate replacement for the strut (ignoring factors like fatique, age microcracking, etc etc)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can do math that would suggest the replacement tubing will be significant lighter and stronger than the bar strut, but I am trying to find a way to actually physically test it.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="33" LastEditDate="2016-01-24T02:46:03.427" LastActivityDate="2016-01-24T02:58:52.967" Title="Can the strength of a strut be determined by deflections?" Tags="|mechanical-engineering|structural-engineering|stresses|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7039" PostTypeId="2" ParentId="7038" CreationDate="2016-01-21T20:08:41.037" Score="2" Body="&lt;p&gt;The deflection will tell you the rigidity of the material but this doesn't vary much for steels and so it doesn't tell you anything about the yield stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's fairly unlikely that something like a solid strut has been heat treated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also given that it's solid it's unlikely to be anything other than mild steel, because as you have identified, if the designer was sufficiently concerned about its strength to use a high strength steel then they would also have used tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What size of tube would be an equivalent replacement depends a lot on loading. For something mostly under bending stress a tube will save a lot of weight, for something in pure tension there is no real advantage to using tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is a bit of a generalisation but if it is an application sufficiently critical that a solid, high strength steel strut is required then I would leave it alone if I were you.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-01-21T20:41:18.390" LastActivityDate="2016-01-21T20:41:18.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7040" PostTypeId="2" ParentId="7021" CreationDate="2016-01-21T20:09:13.730" Score="2" Body="&lt;h2&gt;Pumps are complicated&lt;/h2&gt;&#xA;&lt;p&gt;Many questions regarding fluid machinery start with these two assumptions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A pump or gravity is needed to make the fluid go faster&lt;/li&gt;&#xA;&lt;li&gt;High pressure means faster flow&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;All of these are correct in principle, but none of them are accurate.  The first is inaccurate.  Due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Conservation_of_mass&quot; rel=&quot;nofollow noreferrer&quot;&gt;conservation of mass&lt;/a&gt;, fluid flowing into a pipe on one end will be equal to the flow out the other end.  While the pump will make the fluid flow &lt;em&gt;faster&lt;/em&gt;, the reality is the pump imparts energy on the flow, until the added pressure loss from the higher flow rate is equal to the added energy.&lt;/p&gt;&#xA;&lt;p&gt;Because of how complicated the energy loss in a pipe can become (&lt;a href=&quot;https://en.wikipedia.org/wiki/Reynolds_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reynolds numbers&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Darcy_friction_factor_formulae&quot; rel=&quot;nofollow noreferrer&quot;&gt;Darcy friction factor&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydraulic_diameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydraulic diameter&lt;/a&gt;, etc.), since most engineering problems deal with water and piping, &lt;a href=&quot;https://en.wikipedia.org/wiki/Hazen%E2%80%93Williams_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hazen Williams equations&lt;/a&gt; are typically used.  In addition to simplifying the pipe, we simplify the pump.  Pumps have a &lt;a href=&quot;https://www.youtube.com/watch?v=fiJeOLkLV5I&quot; rel=&quot;nofollow noreferrer&quot;&gt;curve&lt;/a&gt;, which is always convex.  By taking the max flow rate (at no pressure), and the maximum pressure (at no flow), a &lt;a href=&quot;https://en.wikipedia.org/wiki/Linear_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;linear equation&lt;/a&gt; can be developed - just use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Linear_equation#Intercept_form&quot; rel=&quot;nofollow noreferrer&quot;&gt;intercept form&lt;/a&gt;.  This simplifies the pump.  Now, a set of equations can be developed called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hardy_Cross_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hardy Cross Method&lt;/a&gt;.  My &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/0135259738&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;textbook&lt;/a&gt; shows a typical setup:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EXLHh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EXLHh.jpg&quot; alt=&quot;Hardy Cross Setup&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;A source of unknown pressure is feeding the system 3 cfs of liquid - via valves, pumps, or who knows what.  The exits have 1 and 2 cfs - matching the conservation of mass.  Again, pressure at these points is unknown, but not unlike pressure loss across a resistor, the energy difference between each point means that energy is conserved across the loop.  This is why Hardy-Cross Method can be seen to be the fluid version of &lt;a href=&quot;https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kirchoff's Voltage Law&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As can be seen, the top line (only 6&amp;quot; and a full 3000' + 4000') will get minimal flow, while the bottom line, which is shorter and has larger diameter pipes, should be expected to hold most of the flow.  When adding various pumps, it can be seen that even with sufficiently large pumps, circulation starts to become inevitable:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ruN7Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ruN7Z.jpg&quot; alt=&quot;HardyCross Examples&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;The Y-Connector&lt;/h3&gt;&#xA;&lt;p&gt;Your Y-Connector method is interesting because you now have an unknown flow rate - but a known pressure.  In this case, you need to use the air in the atmosphere to connect the pressures in a loop (See for example, &lt;a href=&quot;http://www.innovyze.com/bookstore/cwbsa/Chap5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;page 15&lt;/a&gt; of this pdf, showing how this is done.  Also, don't forget to add in all of the &lt;a href=&quot;http://www.engineeringtoolbox.com/minor-loss-coefficients-pipes-d_626.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;minor loss coefficients&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;However, all of this is a model.&lt;/strong&gt;  To prevent backflow, &lt;em&gt;&lt;strong&gt;always add a check valve to your pumps.&lt;/strong&gt;&lt;/em&gt;  Because you never know what could happen, and backflow can be very expensive.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-01-21T20:09:13.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7041" PostTypeId="2" ParentId="7038" CreationDate="2016-01-21T21:33:41.533" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can I cantilever suspend the strut and record its deflection due to a weight placed on the end, to infer the strength of the strut?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Steels all have (more or less) the same modulus of elasticity.  Therefore, two different steel samples of the same cross-section but with differing yield strengths will have essentially the same deflection when loaded.  So, no, you can't use deflection to infer the material strength of the steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is that struts &lt;em&gt;typically&lt;/em&gt; only resist compressive or tensile forces.  Thus, the cross-sectional &lt;a href=&quot;https://en.wikipedia.org/wiki/Area&quot; rel=&quot;nofollow&quot;&gt;area&lt;/a&gt; of the strut is what matters (not so much its &lt;a href=&quot;https://en.wikipedia.org/wiki/Section_modulus&quot; rel=&quot;nofollow&quot;&gt;section modulus&lt;/a&gt;).  We don't know what your layout looks like, so I can't say for sure what function yours serves.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I do not want to make assumptions about the yield point based on material, because heat treatment and other factors can affect the yield point.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't say what this strut is for or what machine, structure, etc. it will be mounted on.  However, most low and mild carbon steels are not heat treated to achieve their structural properties (stainless is a different matter).  A safe assumption for a lower-end yield point for mild steel is probably 30 ksi, which I base on ASTM A36 (36 ksi yield) and ASTM A1011 (33 ksi), which are both common steels in the industry.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In other words lets say the strut deflects 5 inches when it is cantilevered and a weight is placed on its end. The tubing under test is placed the same way and only deflects 4 inches. Can I assume that the tubing will be an adequate replacement for the strut (ignoring factors like fatigue, age microcracking, etc etc)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We have established that steel has a typical lower-end yield around 30 ksi and that two similar cross-sections will have the same deflection, even if their yield strengths differ.  If your new tubing member deflects less than the existing strut, you can say that the tube is stiffer than the strut, but you cannot truly say it is stronger since you don't know the yield strength of the existing strut's metal.  You could always err on the conservative side and assume a higher yield strength for the strut (say 100 ksi, as a very high end value) and back-figure the needed tube size based on a lower yield stress for it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: this should not be construed as any sort of approved engineering advice.  It is merely my take on the matter and should be accorded as such.  Do not modify a structure if you are afraid doing so will have the potential to cause injury or death.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2016-01-21T21:33:41.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7043" PostTypeId="1" AcceptedAnswerId="7044" CreationDate="2016-01-22T00:51:53.827" Score="0" ViewCount="135" Body="&lt;p&gt;I'm interested in learning what I might call &quot;advanced statics&quot; but am not sure what concepts this would include in engineering parlance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it actually be called &quot;matrix structural analysis&quot;?&lt;/p&gt;&#xA;" OwnerUserId="252" LastEditorUserId="2434" LastEditDate="2016-01-22T16:22:05.593" LastActivityDate="2016-02-03T16:07:49.900" Title="What engineering topic builds upon the concepts of basic statics?" Tags="|civil-engineering|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7044" PostTypeId="2" ParentId="7043" CreationDate="2016-01-22T02:26:29.827" Score="3" Body="&lt;p&gt;The field of statics is based on analyzing finitely many (solid) members and discrete forces acting in equilibrium.  It is a purely algebraic method of analysis (no calculus is necessary).  To move to topics more advanced than this while still assuming equilibrium, you have to start eroding away at the &quot;discreteness&quot; assumption.  In other words, you have you start considering infinitesimal contributions summed over an area or a volume (i.e. you really need vector calculus to do this for generalized 3d problems).  Continuum Solid Mechanics, particularly steady state linear elasticity, is pretty much the next step up from statics if you're analyzing solids in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is somewhat of a middle ground between the discrete nature of statics and continuum solid mechanics.  The  Finite Element Method makes additional assumptions to interpolate a continuous problem into discretely many pieces and resolves the degrees of freedom as a system of equations. But I would consider it more as a method to solve partial differential equations than an &quot;advanced form of statics&quot;, because FEM is more general and applies not only to solid mechanics, but also fluid mechanics, heat transfer, electromagentics, and pretty much anything that can be described by a PDE. &lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2016-02-03T16:07:49.900" LastActivityDate="2016-02-03T16:07:49.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7045" PostTypeId="1" CreationDate="2016-01-22T06:46:01.160" Score="1" ViewCount="149" Body="&lt;p&gt;Imagine the setup in this image:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/EUOgv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EUOgv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a platform resting on a fulcrum. On both sides, at the same distance from the fulcrum, we have 2 motors which control the height on either side of the platform by rotating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the left hand motor were rotated by +500 encoder counts, can we assume that right hand motor would need to rotate by -500 encoder counts to compensate for the change in height?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, how could I calculate the required change in encoder counts?&lt;/p&gt;&#xA;" OwnerUserId="4281" LastActivityDate="2016-01-22T12:18:13.067" Title="Fulcrum Movement" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7046" PostTypeId="2" ParentId="7045" CreationDate="2016-01-22T12:18:13.067" Score="2" Body="&lt;p&gt;If the two motor systems are identical other than being mirror images of each other, then yes, each motor will rotate the same amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not true if the angles are a little out of phase, the lengths of the rods are different, or the positions relative to the fulcrum are different.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-01-22T12:18:13.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7047" PostTypeId="1" CreationDate="2016-01-22T14:08:51.413" Score="0" ViewCount="813" Body="&lt;p&gt;What options are available to create a vacuum using hydraulics?  I am working on constructing a small planter which needs a vacuum to operate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planters operate best at a pressure of 10 PSI.  I think flow is around 30-50 gallon per minute range per row unit. This design has 2 row units.&lt;/p&gt;&#xA;" OwnerUserId="3390" LastEditorUserId="3390" LastEditDate="2016-01-23T23:50:07.920" LastActivityDate="2016-01-23T23:50:07.920" Title="What options are available to create a vacuum with hydraulics?" Tags="|hydraulics|vacuum|" AnswerCount="1" CommentCount="5" ClosedDate="2016-01-23T00:13:02.907" ContentLicense="CC BY-SA 3.0" />
  <row Id="7048" PostTypeId="2" ParentId="7047" CreationDate="2016-01-22T14:38:31.473" Score="1" Body="&lt;p&gt;There are devices called aspirators which use a stream of water to create a vacuum by the venturi effect. They basically consist of a T connector with the suction inlet at right angles to the the water flow. Across the joint the water flow is constricted, which accelerates it, dropping the pressure and creating suction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vacuum pressure generated is fairly modest and they use quite a lot of water but they are extremely simple. But if you need the vacuum to do a significant amount of work this may not be adequate. However is you can include a source of compressed air in your system then much higher pressures are possible using the same principal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is practical really depends on the context, for example whether you want discrete pulses eg to pick something up with a suction cup or a continuous suction more like a vacuum cleaner. In many cases a self contained electric vacuum pump will be the most convenient solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously there is ultimately the possibility to operate a vacuum pump or piston via a hydraulic motor or actuator. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-22T14:38:31.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7049" PostTypeId="2" ParentId="2124" CreationDate="2016-01-22T15:17:18.167" Score="2" Body="&lt;p&gt;If you count gasoline as &quot;gas&quot;, many car manufacturers are already testing homogeneous charge compression ignition engines. It is where they use the compression stroke to ignite the fuel mixture in the cylinder. The fuel is not direct injected, it is premixed in the intake charge before it makes it into the cylinder, so diesel type direct injectors are not necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same technology could be used for other gaseous fuels such as compressed natural gas.&#xA;Such as the experimental compression ignition natural gas engine made by toyota, referenced here &lt;a href=&quot;http://papers.sae.org/2007-01-0176/&quot; rel=&quot;nofollow&quot;&gt;http://papers.sae.org/2007-01-0176/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So such things do exist, though uncommon.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastEditorUserId="1459" LastEditDate="2016-01-22T18:22:51.087" LastActivityDate="2016-01-22T18:22:51.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7051" PostTypeId="1" CreationDate="2016-01-22T17:09:40.840" Score="3" ViewCount="922" Body="&lt;p&gt;I've been watching &lt;a href=&quot;https://www.youtube.com/watch?v=i_M4gtNkNHQ&quot; rel=&quot;nofollow&quot;&gt;a clip&lt;/a&gt; of a robot dragging a heavy load behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering, in a fairly high level, &lt;strong&gt;how does the robot maintain its balance&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I know that with a Segway, there is a gyroscope which senses in which direction the Segway is tilting, to allow the wheels to move accordingly to maintain balance. Is it the same principle with stabilising a robot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also seen that some robots use accelerometers for stabilisation. Do robots need both an accelerometer &lt;em&gt;and&lt;/em&gt; a gyroscope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there also be some kind of proportional–integral–derivative (PID) controller, such that when the robot notices that it is tipping backwards, the applied extra forward force does not cause the robot to overshoot?&lt;/p&gt;&#xA;" OwnerUserId="2727" LastEditorUserId="2246" LastEditDate="2016-01-22T21:17:03.420" LastActivityDate="2016-02-03T16:10:29.950" Title="How do robots maintain stability?" Tags="|mechanical-engineering|control-engineering|sensors|robotics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7052" PostTypeId="2" ParentId="7051" CreationDate="2016-01-22T19:43:39.133" Score="2" Body="&lt;p&gt;Clearly there are a lot of different potential solutions to this depending on the application and design approach. Also, as is clear from the video, bipedal robots are still a long way from achieving the level of elegance in walking that most humans manage with little conscious effort. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the most important comparison is with humans, it's worth looking at how humans maintain balance. This can involve a lot of senses working in concert. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key organ is the set of tubes in the inner ear, which is essentially a set of accelerometers. Humans also have an innate sense of where our limbs are in space in addition to a fairly refined sense of both external pressure and exactly how much force our muscles are providing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly sight is also very important, not just for sensing obstacles but in providing a reference point for speed, position and orientation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another very important factor is that the human body has a lot of degrees of freedom and hundreds of different muscles that can act as springs and dampers as well as actuators. Indeed, this is one of the big challenges in humanoid robot walking. Although it's not to difficult to assemble the main joints (e.g. hips, knees and ankles) and provide them with pairs of pneumatic pistons, in real humans the whole body is involved in maintaining balance with constant, subtle and dynamic adjustments in stance to adjust the centre of mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the actual mechanism of the control system it's fairly clear that a human brain is a very different thing from conventional software control. One of the big challenges is that, unlike something like steering a car where a certain control input gives a known response, with walking the whole geometry of the system is constantly changing and the input required might be the sum of a large number of different muscles, indeed there might be a large number of possible inputs to achieve the same output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind there are a number of approaches to control. The first is what you might call a 'brute force approach' where you have many sensors and a lot of computing power to model the whole system as a series of free body diagrams and try to model the whole system in sufficient detail that PID control can be effective. But as already mentioned the obstacle here is that the response required for a given correction may be ambiguous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach is to come up with a algorithm that responds to certain conditions in certain ways with a more limited degree of feedback-based control, i.e. you have a predetermined 'standard' gait and make corrections if something goes wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third and potentially more powerful method is to use an evolutionary learning type approach where you give the software access to the sensors and actuators and establish some criteria for 'success' and just let it get on with it by trial and error. In this case the software develops control systems pretty much at random and tries them out, deleting the versions which have low success scores and randomly combining the ones that work better. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="2246" LastEditDate="2016-01-22T21:21:45.610" LastActivityDate="2016-01-22T21:21:45.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7054" PostTypeId="2" ParentId="7038" CreationDate="2016-01-23T03:04:20.717" Score="4" Body="&lt;p&gt;The other answers say that this isn't possible in the way that you want to do it, but they don't provide the equations that show why.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Deflection of a Cantilever&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Deflection_%28engineering%29#End-loaded_cantilever_beams&quot; rel=&quot;nofollow&quot;&gt;equation&lt;/a&gt; for the deflection of a cantilever beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta = \frac{FL^3}{3EI} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where: $F=Force \\&#xA;L= \text{Length of cantilever} \\&#xA;E=\text{Modulus of Elasticity} \\&#xA;I=\text{Moment of Inertial}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, none of these components have anything to do with strength. The only variable that is affected by the material is $\text{E}$. This is basically constant for steels.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Column Strength&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You didn't necessarily ask about the material strength of your tube though. You asked about replacing a strut. A strut is a column. The compressive capacity of a column is ultimately controlled by &lt;a href=&quot;https://en.wikipedia.org/wiki/Buckling#Columns&quot; rel=&quot;nofollow&quot;&gt;Euler Buckling&lt;/a&gt; (hint: it rhymes with &quot;oiler&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\frac{\pi^2EI}{(KL)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where: $\text{K is dependent on the end connections} \\&#xA;\text{(well, see above, everything else is the same)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also shows that buckling capacity isn't dependent on material strength. Assuming that the original solid bar is the same material as your replacement tube, there is only one parameter to compare $\text{I}$, (&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_moments_of_inertia#Moments_of_inertia&quot; rel=&quot;nofollow&quot;&gt;Moment of Inertia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the diameter of your original solid bar and use that to determine the the required diameter and thickness of the wall so that the moments of inertia match.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;What if the materials are not the same?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;It the materials are not the same, then the &lt;strong&gt;E&lt;/strong&gt; value will change. Everything else in the equations stays the same, so &lt;strong&gt;EI&lt;/strong&gt; will need to stay the same. The moment of inertia may go up or down depending on how the modulus of elasticity changes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ends up meaning that a stiffer strut will have a greater &lt;strong&gt;EI&lt;/strong&gt;. This also contributes to a higher buckling force. If buckling is the controlling failure mode, then a strut that deflects less will be stronger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Insert standard warnings about there being lots of other factors that could affect this. Don't use this for things related to human life or anything that you care about.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2016-01-24T02:58:52.967" LastActivityDate="2016-01-24T02:58:52.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7055" PostTypeId="2" ParentId="3872" CreationDate="2016-01-23T03:24:48.207" Score="0" Body="&lt;p&gt;Yes. It is important to know what your goal is in analyzing vibration data.  You mentioned monitoring a structure but then mention RMS.  Structural vibration and rotating vibration are two very different things.  There are also many reasons for studying vibration responses and properties, I.e. Condition monitoring, fault detection, tolerances, resonance frequencies etc.  I understand you are trying to learn, but you do need some context for what you are trying to learn.  And I'm not reading that in your question description.&lt;/p&gt;&#xA;" OwnerUserId="4603" LastActivityDate="2016-01-23T03:24:48.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7056" PostTypeId="1" CreationDate="2016-01-23T10:09:43.973" Score="1" ViewCount="320" Body="&lt;p&gt;I am looking to find a seal like the one in the image that seals around the outside of a rectangular object and allows it to move, but when I search &quot;square seal&quot; I just get round o-rings that have a square cross section. Can anyone tell me what I need to search for to find such a seal as the one in the picture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sb9Hz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sb9Hz.jpg&quot; alt=&quot;Square rubber seal&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is to be used so that the glass in the picture below can be moved forward and back through the container holding a liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MMYYY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MMYYY.jpg&quot; alt=&quot;Container assembly&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4189" LastActivityDate="2016-01-23T10:09:43.973" Title="Seals that form around a moving (linear) rectangular object" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7058" PostTypeId="1" CreationDate="2016-01-23T18:10:17.910" Score="2" ViewCount="100" Body="&lt;p&gt;The electromagnetic field can be thought of as a superposition of light waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am attempting a physical Fourier Transform - i.e. take simple plane waves of the electromagnetic field (a.k.a. light) and recreate basic fields by superposing the plane waves correctly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My background is in theoretical physics and so I have little to no knowledge of applications. I know that the field of a rotating electric dipole becomes the simple plane wave I am looking for at large enough distances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So my question is&lt;/em&gt;: How do I actually create a rotating dipole field, or do you have a better idea that suits the application? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not wish to use LED's due it's quantum nature and I wish to stay within the fuzziness of classical mechanics.&lt;/p&gt;&#xA;" OwnerUserId="4612" LastEditorDisplayName="user16" LastEditDate="2016-01-26T14:45:28.147" LastActivityDate="2016-01-26T23:52:47.910" Title="How to create rotating, artificial dipoles?" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7059" PostTypeId="1" AcceptedAnswerId="49949" CreationDate="2016-01-23T20:34:36.600" Score="3" ViewCount="357" Body="&lt;p&gt;I'm well aware of Helmut Hoeltzer's &quot;Mischgerat&quot; analog computer, and I read that it not only served as a kind of &quot;fly by wire&quot; computer on board of the missile, but also as a simulator on the ground that calculated V2 trajectories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I didn't find any clue about how it did it -  how Hoeltzer calculated optimal trajectory to hit the target (cities). I think the trajectory had to calculated accurately because only by following an accurate trajectory a ballistic missile could hit a city sized target.&lt;/p&gt;&#xA;" OwnerUserId="3284" LastEditorUserId="90" LastEditDate="2016-01-24T00:43:02.200" LastActivityDate="2022-03-17T19:44:15.353" Title="How was the trajectory of the V2 ballistic missile (during boost phase) calculated?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7060" PostTypeId="1" AcceptedAnswerId="7065" CreationDate="2016-01-23T21:36:50.767" Score="5" ViewCount="204" Body="&lt;p&gt;recently I bought 3 solar panels rated at 5V 200 mA each. I want to use them to charge a 5V battery bank to charge a phone. Thinking about the proper way to put them, I thought i can connect all in parallel to get maximum current, but realized that if the sun light was a little weak it will no generate full 5v thus preventing charging. So I decided to put 2 in parallel to give the equivalent of one 5V solar panel, connected in series with the 3rd panel to give the equivalent of 10v. sacrificing a little current to get higher voltage, to allow the charging to happen on a wider range of sun light power. The following picture shows the wiring and the schematic I intend to replicate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4oFHJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4oFHJt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lHrv9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lHrv9t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that I have an equivalent of 10v, 400mA solar panel. I used a 7805 voltage regulator to cut down the excess to 5v. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IbnN3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IbnN3t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final step, I added a standard diode to prevent the panels from leaking the battery in the shade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and now measuring:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A0RIw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A0RIwt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I didn't think about this before putting the diode, but is it okay to put the blocking diode on the ground wire? because I know some applications do not use the ground except for safety (i.e. 3 phase system). it would be helpful also to avoid the 0.7v drop across the diode before the regulator.&lt;/li&gt;&#xA;&lt;li&gt;according to previous calculations, I'm supposed to get a maximum of 10v output before regulation, and considering that the sun was pretty shinny today, why was the reading I got not more then 6v? I have measured across each panel seperatly and got around 5.5V, are the connections right?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1736" LastEditorDisplayName="user16" LastEditDate="2016-01-25T19:37:59.243" LastActivityDate="2023-12-21T21:15:38.310" Title="Did I wire these solar panels correctly?" Tags="|electrical-engineering|photovoltaics|solar-energy|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7061" PostTypeId="1" CreationDate="2016-01-23T22:31:30.627" Score="6" ViewCount="21729" Body="&lt;p&gt;Why is that the shape of the tension test specimen a dogbone-like shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it is so that the deformation is confined to the narrow center region and to reduce the likelihood of fracture to occur at the ends of the specimen. But why is this so important? Why can't a cylinder shaped specimen that has a uniform radius all along the specimen be used? Why does it, for example, matter if it fractures near the end (near the region where you clasped the specimen in the machine)?&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="368" LastEditDate="2016-01-25T19:24:05.257" LastActivityDate="2022-11-19T04:33:50.657" Title="Why use a 'dogbone' shape for tensile testing specimens?" Tags="|mechanical-engineering|metallurgy|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7062" PostTypeId="2" ParentId="7061" CreationDate="2016-01-23T23:07:27.187" Score="10" Body="&lt;p&gt;You half-answered your own question. Preventing failure in the grips is important. Additionally, grips of tensile testing machines have teeth to achieve a sufficiently strong grip that can withstand the forces required to deform the sample longitudinally. The teeth typically cause plastic deformation of the gripped portion of the sample. The plastic deformation at the grips may change the material properties, and definitely changes the sample geometry. The sample geometry changes in such a way as to create stress concentrations in the gripped region. Both changes would contribute to an improper measurement if failure occurred at the grips in a cylindrical or bar specimen without enlarged ends.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-01-23T23:07:27.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7063" PostTypeId="2" ParentId="7061" CreationDate="2016-01-24T02:40:21.900" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/7062/33&quot;&gt;Starrise&lt;/a&gt; gave a good explanation of the reasons why a dog-bone shape is important to the tension portion of a tension test, but there is another property that is typically measured at the same time: &lt;strong&gt;Elongation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elongation is measured by placing a gauge on the reduced section. It is important that the elongation occurs in this area so that the measurements are accurate. If the movement of the machine grips is used, then any plastic deformation of the grips will add to the elongation measured.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-01-24T02:40:21.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7064" PostTypeId="2" ParentId="7060" CreationDate="2016-01-24T15:01:59.103" Score="2" Body="&lt;p&gt;That's a mess, and your labels about what is + and what is - out of each panel seem inconsistant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the most effective use of the panels, wire them all in series.  That will put out around 15 V under full sun.  Now use a buck converter to make a regulated 5 V from that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many buck converter chips available off the shelf at these low voltages.  Finding one with a built-in switch and synchronous rectification shouldn't be too hard.  You only have to supply the inductor, a few caps, and maybe a charge pump diode depending on the chip you chose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other advantage of a switching power supply is that you probably won't need heat sinking.  The resulting overall circuit will be smaller, cheaper, and give better performance than just throwing a 7805 regulator at it.  At 10 V in and 5 V 1 A out, the 7805 will dissipate 5 W.  That's way beyond what it can do without a heat sink.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-01-24T15:01:59.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7065" PostTypeId="2" ParentId="7060" CreationDate="2016-01-24T15:44:54.967" Score="4" Body="&lt;p&gt;Putting a single panel in series with two other panels that are in parallel does not accomplish what you think it does. The overall current of such a setup is limited by the single panel to 200 mA, so the three panels will not produce any more power than you'd get by just putting two panels in series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single solar cell can be thought of as a current source in parallel with a silicon diode. The current source is driven by the incoming light. The diode &quot;shorts out&quot; the current source, which is why the voltage across a single cell can never be more than about 0.65 V, the forward drop of a silicon diode. A 5V panel is approximately 10 such cells connected in series. The current through all of the cells will be limited by the cell that is receiving the least amount of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the blocking diode in your diagram is pointing the wrong way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your panels are rated at 5V, and your &quot;battery bank&quot; requires 5V to charge, then you don't need to do anything more than put all three panels in parallel and hook them directly to the battery. Forget about blocking diodes.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-01-24T15:44:54.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7066" PostTypeId="1" AcceptedAnswerId="7804" CreationDate="2016-01-24T17:16:11.273" Score="3" ViewCount="1367" Body="&lt;p&gt;It sounds logical untill I tried some thought experiments. So I must be doing something wrong. Or I am missing a fundamental aspect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I thought of was: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-lets say you have an orthogonal parallelepiped which is pulled in tension by a stress-strain test. (in the long axis or z-direction).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the lateral sides are 5 mm and 4 mm and the axial length is 7 mm.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;-It has been elongated by the test untill the specimen had an axial length of 11 mm and hypothetically the 4 mm lateral side did not change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the original volume was 4x5x7= 140 mm^3 and since there is no net volume change the final length of the lateral side that did change has to be: &#xA;140= a x 11 x 4 -&gt; a= 35/11 mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now if you calculate the poisson's ratio: v= - Ex/Ez&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&gt; Ex = (35/11 -5)/5= -4/11 and Ez= (11-7)/7= 4/7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v= 4/11 : 4/7 = 7/11 = 0.64 &gt; 0.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How I understood it: There is no volume change, so when you elongate one direction the other two will compensate this by compressing. And this poisson ratio of 1/2 is in the extreme case that one direction does not compress and so the other side has to take all the compression. Like in the situation that you elongate the axial side to twice it's orginal length, then in order to have same volume the two other directions have to become each 1/4 of their orginal length or one direction has to become 1/2 it's orginal length. This will indeed then give poisson ratio of 1/4 and 1/2. So I don't get why with other numerical examples I don't get this same 1/4 and 1/2 ratio.&lt;/p&gt;&#xA;" OwnerUserId="4117" LastActivityDate="2016-03-07T13:32:01.230" Title="Why is the Poisson's ratio for isotropic 1/4 and the maximum value 1/2 (when there is no net volume change)?" Tags="|mechanical-engineering|materials|metallurgy|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7067" PostTypeId="2" ParentId="7051" CreationDate="2016-01-24T20:46:02.170" Score="4" Body="&lt;p&gt;PID, LQR, and LQG feedback control laws are only small pieces of the solution. Alone, these methods generally cannot solve the problem of balanced bipedal walking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why is because these aforementioned control laws are designed to send the dynamics to a single, stable equilibrium point.  If our goal is only to keep the robot balanced while standing, the aforementioned control laws suffice.  Standing upright consists of a single point in the robot's dynamical configuration space and these control laws can make that point stable with respect to small perturbations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stable walking, on the other hand, consists of the dynamics following a periodic trajectory in the configuration space (also known as a stable limit cycle in ordinary differential equation terminology).  Designing control laws to ensure this is extremely difficult due to the extreme non-linearities associated with the dynamics.  Often, this cannot be done analytically and is too complex (or risky) to accomplish by mere trial and error.   Instead, this is often done numerically through a process called trajectory optimization, where an &quot;ideal path&quot; of the robot is prescribed and a numerical method determines the best approximation to this path that the robot can achieve, where the criterion for &quot;best&quot; is determined by a specially chosen cost function which accounts for both the ideal path and the limitations of the robot.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, the solution involves non-constant gains on the actuators.  This is something that linear feedback laws like PID cannot achieve.  PID's come into play when trying to achieve a particular action with a particular actuator, but is far from sufficient when tackling the overall problem of dynamic balance while walking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This discussion, of course, also assumes that you have a &quot;good model&quot; of the robot's dynamics to begin with.  That, my friend, is an entirely different struggle in and of itself.&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2016-02-03T16:10:29.950" LastActivityDate="2016-02-03T16:10:29.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7068" PostTypeId="2" ParentId="7059" CreationDate="2016-01-24T21:10:52.270" Score="1" Body="&lt;p&gt;I suspect you've seen this already, but there's a book &quot;Analog Computing&quot; by Bernd Ulmann which goes into some detail on the V2 guidance package.  It's mostly about the hardware, not the software or governing equations, but it's an interesting read none the less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to read through the technical publications of the time to see what the state of the art in trajectory computation was.  That would at least give you some idea what trajectory models Hoelzer &lt;em&gt;might&lt;/em&gt; have used.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-01-24T21:10:52.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7069" PostTypeId="2" ParentId="2419" CreationDate="2016-01-24T23:29:15.730" Score="1" Body="&lt;p&gt;It depends what you mean by 'best' &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a homogeneous material like steel a sphere will (in theory) be the lightest structure which will contain a given volume at a given pressure. However spheres do have certain disadvantages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-They don't stack well, so if you are distributing industrial gases, for example,  they are difficult to store and transport and take up a lot of space compared to their useful volume. &#xA;- They are difficult to manufacture, especially in composites. &#xA;- Spheres are often not a convenient shape to handle, especially in applications like breathing apparatus and fire extinguishers where this is very important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice the most useful shape for a pressure vessel is often a cylinder with hemispherical or ellipsoidal ends, often with one end concave to allow it to stand on its end. This is not quite as efficient in term of the minimum thickness of material required for the walls but this is often outweighed by the practical considerations already noted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most general engineering textbooks will have a section on algebraic formulae for calculating the stresses in various different types of pressure vessel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that composite pressure vessels are increasingly common. These are often made by winding a resin impregnated yarn around a former and as well as cylinders toroidal vessels can be made in this way, which come close to being as weight efficient as spheres but are generally more convenient. With composite materials it is also possible to orientate the yarn so as to have different strength in different directions, enhancing material efficiency, which is easier to do with a cylinder or torus than a sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scale matters as well so for very large scale bulk storage (eg LPG container ships) a lot of the problems with constructing spheres may become less relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of internal vs external pressure the most important difference is that external pressure will impose compressive forces on the vessel so some geometries may be subject to buckling in addition to simple stresses  but for common types of pressure vessel this won't make a huge difference in general design. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-01-24T23:42:50.317" LastActivityDate="2016-01-24T23:42:50.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7070" PostTypeId="2" ParentId="7051" CreationDate="2016-01-25T02:55:10.420" Score="2" Body="&lt;p&gt;The robot likely has a body accelerometer &amp;amp; integrates it to use a velocity tracking controller with some sort of explicit or implicit proportional &amp;amp; integral control for the body.&lt;br&gt;&#xA;- This would keep the robot moving at the same speed as the weight it is dragging goes up.&lt;br&gt;&#xA;The balance of the robot body can be measured 2 ways:&lt;br&gt;&#xA;1) the most straightforward is to use the geometry of the robot &amp;amp; measurements at each joint&#xA;2) or a Kalman filter could be used with a gyroscope  for the high frequency portion of motion and the joint measurements for the very low frequency portion.&#xA;Another option would be to have a force transducer that measures the pull from the basket it is dragging and use this to generate a counteracting torque on the leg joints. &lt;/p&gt;&#xA;" OwnerUserId="4556" LastActivityDate="2016-01-25T02:55:10.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7071" PostTypeId="1" AcceptedAnswerId="7074" CreationDate="2016-01-25T05:59:33.450" Score="0" ViewCount="71" Body="&lt;p&gt;I would like to create beam models having various cross sections on ProE tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I create a beam model like below by only extrusion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NEwmF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NEwmF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain what is the best way to model various cross sections of beams?&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="4624" LastEditDate="2016-01-25T06:05:28.067" LastActivityDate="2016-01-25T07:13:21.723" Title="beam cross section modeling by extrusion on proe" Tags="|mechanical-engineering|structural-engineering|modeling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7072" PostTypeId="1" AcceptedAnswerId="7083" CreationDate="2016-01-25T06:25:35.537" Score="4" ViewCount="25617" Body="&lt;p&gt;I've heard that materials that need to slide against each other should be of different materials. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="4410" LastEditorUserId="33" LastEditDate="2016-01-25T14:12:27.963" LastActivityDate="2018-10-22T10:53:03.283" Title="Why use bushings of different material than what they contact?" Tags="|materials|friction|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7073" PostTypeId="2" ParentId="7072" CreationDate="2016-01-25T07:06:36.853" Score="4" Body="&lt;p&gt;I haven't heard this bit of advice and as with most things in engineering it really comes down to WHICH materials and in what CONTEXT. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of one main reason why you might want to make them out of different materials: Differing hardness. What this means is that only one of the materials should 'wear' and then you could design it so that this part could be changed out easily. For example: if you had something following a guide, you wouldn't want to have to replace the whole guide when it started to wear, so if you make the thing that follows the guide softer then it will wear and you can just replace that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yes. Without more information I can't really give any more advice. That statement is not universally true though. &lt;/p&gt;&#xA;" OwnerUserId="2768" LastActivityDate="2016-01-25T07:06:36.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7074" PostTypeId="2" ParentId="7071" CreationDate="2016-01-25T07:13:21.723" Score="1" Body="&lt;p&gt;If pro-e is anything like solidworks then you need to draw a 2D version of the extrusion pattern. In your case, one large square with the 4 smaller squares within it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then select the area you wish to extrude and extrude it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend looking up some basic tutorials for ProE as I'm sure this would be covered in the early stages. &lt;/p&gt;&#xA;" OwnerUserId="2768" LastActivityDate="2016-01-25T07:13:21.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7076" PostTypeId="2" ParentId="7019" CreationDate="2016-01-25T13:30:14.277" Score="-1" Body="&lt;p&gt;Not tested yet, but try using fabric whitener such as Bayclin.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastEditorUserId="2246" LastEditDate="2016-01-25T14:05:07.287" LastActivityDate="2016-01-25T14:05:07.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7077" PostTypeId="2" ParentId="6916" CreationDate="2016-01-25T13:40:59.480" Score="0" Body="&lt;p&gt;My 2 cents:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Two motors for one shaft is not recommended because one motor will have more load (no motor can be exactly the same, even if same type and size) including induction motors.&lt;/li&gt;&#xA;&lt;li&gt;Induction motor speed depends on electrical power frequency (and number of motor poles). But you can install bigger motor pulley &amp;amp; smaller shaft pulley - but this will make your torque lower.&lt;/li&gt;&#xA;&lt;li&gt;More torque is certainly, since you use 2 motors. You can get higher torque by using smaller motor pulley &amp;amp; bigger shaft pulley - but this will make your speed lower.&lt;/li&gt;&#xA;&lt;li&gt;Be careful that belt drives damaged when installed on very small pulley. Dented belt or notched v-belt can be used for small pulleys.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4630" LastEditorUserId="33" LastEditDate="2016-01-25T14:07:02.000" LastActivityDate="2016-01-25T14:07:02.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7078" PostTypeId="2" ParentId="2908" CreationDate="2016-01-25T13:45:14.570" Score="0" Body="&lt;p&gt;I recommend alumina balls (not steel alloys though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used ball mills in 5 ton capacity and pot mill in lab (250 grams) using alumina balls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think alumina balls are inert (already oxidized) so that even though they may introduce some impurity, I don't believe this will affect your alloy chemically like Fe and Cr will. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With alumina balls, the rate of wear is very low.&#xA;Everyday use requires only a regular top-up of balls every 3 months. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alumina balls also can be used for cryogenic temperature (CMIIW). &#xA;Something to keep in mind though: pot mill temperature will be very hot (almost boiling) after milling, so it may prove difficult to maintain cryogenic. &lt;/p&gt;&#xA;" OwnerUserId="4630" LastEditorUserId="2598" LastEditDate="2016-04-27T20:26:31.290" LastActivityDate="2016-04-27T20:26:31.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7080" PostTypeId="1" AcceptedAnswerId="7081" CreationDate="2016-01-25T14:32:16.590" Score="46" ViewCount="8297" Body="&lt;h1&gt;Background&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In the world of &lt;a href=&quot;https://www.youtube.com/watch?v=kklBFX62F8U&quot; rel=&quot;noreferrer&quot; title=&quot;2012 Olympics - Elite Men&amp;#39;s Main Event&quot;&gt;bicycle motocross, also known as BMX racing&lt;/a&gt;, gearing is a hotly-debated topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the bikes are all single-speed, gear ratio is a fixed number defined as &lt;code&gt;chainwheel / cog&lt;/code&gt; (front gear divided by rear gear). Altering your gear ratio is understood as an immediately-noticeable tradeoff between acceleration and top-end speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a series of common gear ratios:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;╔════════════╦═════╦════════╗&#xA;║ Chainwheel ║ Cog ║ Ratio  ║&#xA;╠════════════╬═════╬════════╣&#xA;║         43 ║  16 ║ 2.6875 ║&#xA;║         41 ║  15 ║ 2.7333 ║&#xA;║         44 ║  16 ║ 2.75   ║&#xA;╚════════════╩═════╩════════╝&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In 2012, a company called &lt;a href=&quot;http://rennendesigngroup.com&quot; rel=&quot;noreferrer&quot;&gt;Rennen Design Group&lt;/a&gt; created a supposed breakthrough innovation called &lt;a href=&quot;http://rennendesigngroup.com/decimal.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;decimal gearing&quot;&lt;/a&gt;. The claim is that through manipulations of tooth profile and ring diameter, in-between gear ratios can be created - for example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;╔════════════╦═════╦════════╗&#xA;║ Chainwheel ║ Cog ║ Ratio  ║&#xA;╠════════════╬═════╬════════╣&#xA;║ 43         ║  16 ║ 2.6875 ║&#xA;║ 45.7       ║  17 ║ 2.6882 ║&#xA;║ 37.7       ║  14 ║ 2.6929 ║&#xA;║ 43.1       ║  16 ║ 2.6938 ║&#xA;║ 41         ║  15 ║ 2.7333 ║&#xA;║ 41.1       ║  15 ║ 2.74   ║&#xA;║ 52.2       ║  19 ║ 2.7473 ║&#xA;║ 44         ║  16 ║ 2.75   ║&#xA;║ 44.2       ║  16 ║ 2.7625 ║&#xA;╚════════════╩═════╩════════╝&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Table is not exhaustive.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For example - a 44.2 tooth gear actually only has 44 teeth, but the tooth spacing, tooth profile, and chainwheel diameter is supposed to have been manipulated to create a larger gear.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the world of BMX racing, the existence of in-between gear ratios like this is a Really Big Deal. Since the man behind Rennen has a Master's from MIT - and since most BMXers would rather hit jumps than do math or measure things - nobody has really ever checked up on whether or not this is valid. Some questions were asked a long time ago in the dusty corners of a BMX forum, but the testing methods didn't properly control for all variables and the thread descended into a bunch of name-calling and ad-hominem attacks.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The Actual Question&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this physically possible?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand &quot;gear ratio&quot; to be defined as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a given gear ratio &lt;code&gt;x / y&lt;/code&gt;, one rotation of the gear with &lt;code&gt;x&lt;/code&gt; teeth will result in &lt;code&gt;x / y&lt;/code&gt; rotations of the gear with &lt;code&gt;y&lt;/code&gt; teeth.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a gear ratio of 44/16, one full rotation of the 44 tooth gear (chainwheel) should result in 2.75 rotations of the 16 tooth gear (cog).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a &quot;decimal ratio&quot; of 44.2/16, one full rotation of the 44.2 tooth gear (which again - only has 44 teeth) is supposed to result in 2.7625 rotations of the 16 tooth gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My biggest reservation is the fact that a chain-driven drivetrain is a TIMED DRIVETRAIN.&lt;/strong&gt; No matter how big or small you make the teeth on the chainwheel, if they fit the chain, they're only going to push as many links through per rotation as the chainwheel has teeth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a true 44.2 tooth chainwheel, one would expect that 442 links get pushed through over 10 full rotations of the chainwheel - but that's not the case. Only 440 links will ever get pushed through to the cog because only 44 links get pushed through per full rotation of the chainwheel. I actually spent my whole afternoon yesterday taking video and counting links and measuring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not a scientist. My &lt;a href=&quot;http://www.gbcscrusaders.com/main-page.html&quot; rel=&quot;noreferrer&quot; title=&quot;my graduating class had ten people in it&quot;&gt;high school&lt;/a&gt; didn't even offer a physics course. I'm just a racer that trains really hard and knows how to do basic math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were a belt-driven system, I would completely understand how a manipulation of the chainwheel diameter would change the effective ratio - but it's not. It's a timed drivetrain, limited by the physical dimensions of the chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have several hundred dollars and months of training and metrics invested in these stupid chainwheels. If someone could confirm or deny my theories, I would really appreciate it. I just want some closure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a photo of the teeth from a 41 tooth chainwheel on top of the teeth from a 41.2 tooth chainwheel - both are Rennen gears:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j4TBD.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j4TBDt.jpg&quot; alt=&quot;41t and 41.2t teeth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a 41t on top of a 41.2t:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UE790.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UE790t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the 41.2t on top of the 41t, from behind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q1LRl.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q1LRlt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4629" LastEditorUserId="4629" LastEditDate="2016-01-26T18:45:12.837" LastActivityDate="2016-01-30T14:26:59.967" Title="Is it possible for a bicycle chainwheel to have a fractional number of teeth?" Tags="|mechanical-engineering|gears|measurements|" AnswerCount="9" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="7081" PostTypeId="2" ParentId="7080" CreationDate="2016-01-25T15:33:50.077" Score="19" Body="&lt;p&gt;I suspect that the answer to this is that, ultimately the gear ratio comes from the ratio of diameters of the gears rather than the number of teeth, although in most circumstances practicality dictates that they are proportional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have a 10 tooth cog and a 40 tooth chainwheel. It's fairly simple to imagine that you could remove every other tooth from the 40 tooth wheel while keeping the diameter the same and maintain exactly the same gear ratio. Similarly you could have a completely gearless wheel (putting aside issues of slippage) driving a chain which drove a geared cog. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this is mind, my suspicion would be that what they have done is increased the gear wheel diameter slightly but kept the same number of teeth by spacing them out slightly more and that there is enough tolerance in the whole system to get away with it. It may even be that the tooth spacings are not identical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that every so often a chain link is skipped and also that only a small proportion of teeth are directly engaged with the chain at any one time &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as there is a possible geometry which avoids a collision between teeth and chain pins I think it should work. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;More thoughts&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;It has also occurred to me that it is possible that the gear wheel itself is slightly elliptical; i.e., it has the same circumference as a conventional gear but is slightly squashed in one axis. I initially dismissed this as it wouldn't give a constant ration, but it also occurred to me that the torque input to a bike crank isn't completely constant anyway so it may not matter and &lt;a href=&quot;http://www.cyclingweekly.co.uk/news/latest-news/osymetric-chainrings-do-they-work-28044&quot; rel=&quot;nofollow&quot;&gt;apparently eliptical or otherwise non-circular chainrings are a thing&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've mentioned a few time in comments that the relationship between number of teeth and gear ratio is not absolute if the tooth pitch is not consistent. While it is usually desirable for meshing gears/chains to keep a constant pitch and profile there is no fundamental principle which ties N teeth to a particular diameter and it is the ratio of diameters that ultimately determines gear ratios, as this is defined by the relationship between moments, torque and angular velocity. A trivial proof of this is that you can remove an arbitrary number of teeth from a given gear and it will still provide the same torque ratio (as long as it does not slip).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind it seems reasonable that you could achieve a small fraction ratio change by increasing the effective diameter and retaining the same number of teeth as long as the tolerances in the system allowed it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate this consider a gear of large diameter D1 with one tooth meshing with a cog of diameter D2 via a chain. In this case it is self evident that this system behaves as a zero slip pulley and  you can change the size of D1 to get any ratio (D1/D2) you want as long as the single tooth is engaged in the chain (say between 12 o'clock and 6 o'clock).&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-01-28T13:37:07.093" LastActivityDate="2016-01-28T13:37:07.093" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="7082" PostTypeId="2" ParentId="7080" CreationDate="2016-01-25T16:59:45.770" Score="5" Body="&lt;p&gt;My interpretation of the closeup images such as provided by kivetros is that the teeth themselves are wider or narrower at the top, and that the rate at which the teeth widen down to the &quot;bottom&quot; of each interval controls how far towards the axis the chaing links can drop.  So what you get is teeth which maintain the link spacing but which allow the chain to drop to a specific radius on the chainwheel.  In this way the link spacing is maintained so as not to destroy the chain, but the actual operational diameter is controlled to the equivalent of a fractional tooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes, I guess, that all &quot;normal&quot; chainwheels have a standard tooth-depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.geartechnology.com/issues/0812x/profile-shift.pdf&quot; rel=&quot;nofollow&quot;&gt;This article provides&lt;/a&gt; more detail.  &lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorDisplayName="user16" LastEditDate="2016-01-27T14:31:02.373" LastActivityDate="2016-01-27T14:31:02.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7083" PostTypeId="2" ParentId="7072" CreationDate="2016-01-25T17:01:44.790" Score="7" Body="&lt;p&gt;In addition to &lt;a href=&quot;https://engineering.stackexchange.com/a/7073/368&quot;&gt;ashgetstazered's answer&lt;/a&gt;, some materials (Aluminum, I'm looking at you!) will gall badly when meshed against themselves, while moving reasonably smoothly against a different material (e.g. bronze bushing). So if you have a good reason to produce one part out of a given material, you then choose the mating part's material to minimize galling or wear.&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-01-25T19:13:28.783" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7084" PostTypeId="2" ParentId="7080" CreationDate="2016-01-25T17:12:24.190" Score="3" Body="&lt;p&gt;@Chris Johns' answer pretty much nails it, though I will add my two cents as well.  Apparently, in the world of BMX racing the difference between a 44 tooth and a 45 tooth chain wheel is large enough that there's a demand for intermediate ratios.&lt;br&gt;&#xA;Short of making a new chain with different link spacing, I think the only way to do this is by taking advantage of the mechanical tolerances in link spacing to create a tooth pattern that doesn't line up perfectly with the links.  I would guess that Rennen's teeth are spaced a little farther apart than normal (to increase the sprocket diameter), but the ability of the chain and its links to stretch a little bit allows this.  About half the teeth on the sprocket must be engaged at any time, so you spread the mismatch between tooth and link spacing out over 22 teeth or so.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2016-01-25T18:41:57.620" LastActivityDate="2016-01-25T18:41:57.620" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7085" PostTypeId="2" ParentId="7080" CreationDate="2016-01-25T17:16:42.573" Score="29" Body="&lt;h3&gt;Gear ratio&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Roller_chain&quot;&gt;Roller Chain&lt;/a&gt; drive system, and as such it's a timed system.  The ratio between the two connected pieces depends &lt;em&gt;entirely&lt;/em&gt; on the number of teeth on either end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you altered the diameter of the root of each tooth so the chain can sit higher or lower, at the end of 1,000 revolutions the chain will have moved the same amount on gears with the same number of teeth as the altered gears, irrespective of the root diameter, or tooth profile.  As long as the chain is engaged and doesn't skip links, then the gear ratio is the same as any other gear with the same number of teeth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take your &quot;normal&quot; 41t gear and rotate it 10 times, it will move the same amount of chain (to within tolerances for the gears and chain) as the modified 41t gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If tolerances allow the chain to sit higher or lower in the teeth and &quot;modify&quot; the diameter, the reality is that the same amount of chain will be moved, but you'll experience more noise as the chain is rattling (loose) or creaking (tight) around the modified gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the &lt;em&gt;gear ratio&lt;/em&gt; between the pedals and the wheel does not change.  If so, what does change?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Torque ratio&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;If the tooth profile is modified so it's a little loose, then the chain will start to disengage under tension, and pull itself up out of the tooth a little as it approaches the top of the gear on the way forward to the pedals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the torque transfer to the gear is actually a little further away from the center of the gear, which provides more torque given the same chain tension as a comparable gear which is fully engaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does decrease how well the chain engages in the gear, and increases both chain and gear wear.  This might be an acceptable tradeoff if the torque is increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that these gears are used at relatively low speeds, and torque matters more than the gear ratio.  In fact, for mountain biking, you might say the torque ratio is more important than the gear ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the gear is designed to push the chain away from the center of the gear as it approaches the top, and thus increases the torque ratio, then this claim is possible, and the effect should be measurable.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Testing&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Put a 200 lb scale on the pedal and tie a very heavy weight to string wound around the rim of the wheel.  You should find that one ring produces a higher or lower torque transfer.  Note that if you do this with lighter weights, the tension on the chain might not be sufficient to pull the chain up the slopes designed into the gear that pull the chain away from the center of the gear.&lt;/p&gt;&#xA;" OwnerUserId="784" LastEditorUserId="368" LastEditDate="2016-01-25T19:04:02.760" LastActivityDate="2016-01-25T19:04:02.760" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="7086" PostTypeId="2" ParentId="7072" CreationDate="2016-01-25T17:31:53.693" Score="11" Body="&lt;p&gt;As already mentioned in many contexts you will make one part essentially consumable in order to reduce wear on an expensive or difficult-to-replace part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly if you have a situation like a shaft mounted in a cast iron casing (eg an engine block) it makes sense to protect both, relatively complex and expensive parts with a cheap, expendable and easily replaceable intermediate e it a bearing or bushing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason is that a lot of materials that have good friction properties may not be suitable for both parts; e.g., sintered bronze makes good bushings but you wouldn't usually want to make a shaft out of it. A similar situation is white metal bearings, which are cast in place and a relatively low tech way to get a high tolerance fit. Similarly, bushings are generally well-supported so it's often acceptable to favour good friction characteristics over strength in a way that you couldn't with a shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another consideration is galling. This is effectively cold welding of touching surfaces under load, which ends up tearing chunks off one or both surfaces, causing deep surface damage at a much faster rate than straightforward wear. Some metals or combinations of metals are particularly vulnerable to this; e.g., stainless steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also manufacturing considerations. It's generally easier to make an accurately round shaft than a hole, so bushing materials will often be relatively easy to machine or at least tolerant of size variation. For example, if you fit a steel shaft in an undersized nylon bushing, it will fairly quickly wear itself in to a reasonable bearing fit without harming either part, whereas a steel shaft in an undersized steel bushing would be in risk of scoring or seizing up altogether. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are equally circumstances where similar materials are fine, especially in contexts where overall wear needs to be minimised or where it's possible to provide good lubrication. In a fluid dynamic or hydrostatic bearing, the bearing surfaces should never actually touch as they are held apart by pressurised lubricant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific materials matter as well for example aluminium-on-aluminium sliding would be much less desirable than cast iron on cast iron. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-07-15T19:14:31.000" LastActivityDate="2016-07-15T19:14:31.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7087" PostTypeId="2" ParentId="7080" CreationDate="2016-01-25T18:13:17.783" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Altering your gear ratio is understood as an immediately-noticeable tradeoff between acceleration and top-end speed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In general&lt;/em&gt; this is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as I said in a comment, you're conflating two things under the convenient heading of the gear ratio.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;top-end speed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is limited by development: when you spin out at your maximum rpm (of the cranks), the tooth ratio - coupled with wheel+tyre circumference - tells you exactly how fast you're going.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at least at low speeds,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;acceleration&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is limited by how much force you can put through a small number of teeth over less than a whole revolution. This is limited by the mechanical advantage of the crank-&gt;chainring-&gt;sprocket-&gt;wheel/tyre system. You have 4 different effective radii affecting the leverage in there, and if the crank length alters it (which it does), I don't see why chainring radius shouldn't as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have no idea whether allowing tooth count and radius to diverge like this actually works, works well, or has any other disadvantages. It seems like it would be a pretty small effect if it does work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be fair, it's also misleading to give fractional tooth counts, because it encourages this conflation: that 44.2 is as you say exactly like a 44-tooth ring for top speed purposes, but gives (or implies) slightly more leverage for acceleration. Your measurement of development (chain links moved per revolution) doesn't address this.&lt;/p&gt;&#xA;" OwnerUserId="4635" LastEditorUserId="4635" LastEditDate="2016-01-25T18:25:11.630" LastActivityDate="2016-01-25T18:25:11.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7088" PostTypeId="2" ParentId="7080" CreationDate="2016-01-25T18:23:00.733" Score="2" Body="&lt;p&gt;By making the overall gear slightly smaller while keeping the same number of teeth, they give you a little more torque, with the same ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this looks a little dangerous, as if the space between the teeth is shorter than the distance between the links, then only the last tooth that connected to a link is really connected. It also mean that that tooth starts pulling a little earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, by making the gear larger, only the first tooth that connects really pulls, the torque is slightly lower, and the chain kind of falls on the next tooth when the connected link leaves the gear.&lt;/p&gt;&#xA;" OwnerUserId="4639" LastActivityDate="2016-01-25T18:23:00.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7089" PostTypeId="2" ParentId="121" CreationDate="2016-01-25T18:26:48.027" Score="1" Body="&lt;p&gt;In general there will be an optimal cutting speed and feedrate for any given combination of tool material and the material being machined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is often drive by the heat heat generated by cutting but factors such as toughness, ductility and tool geometry will also have an effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recommended spindle speeds for drills, mills etc are really just translating the linear cutting speed into a rotational format which is more convenient to work with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality there are all sorts of compromises involved and quoted figures are often an average value for general use. For example you might use a low cutting speed and heavy fed rate to get the fastest rate of material removal but a higher speed with lighter feed for a better final finish. Similarly recommended tools speed are often a compromise between material removal rate and tool wear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific machine characteristics also matter a lot for example a basic pillar drill with an AC motor and pulley drive may simply not have enough torque to run a large hole saw at the optimum speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall it isn't too difficult to tell when a speed is 'right' for a particular job and my experience is that the tables are very useful to get it pretty much right in most circumstances but there is quite a wide margin to adjust them to fit the needs of a particular job. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-25T18:26:48.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7090" PostTypeId="2" ParentId="7080" CreationDate="2016-01-25T18:27:14.867" Score="2" Body="&lt;p&gt;A very interesting question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think what has been done here is commonly known as a Gear &lt;strong&gt;Profile Shift&lt;/strong&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This procedure changes the tooth profile and the working pitch diameter of a gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.geartechnology.com/issues/0812x/profile-shift.pdf&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; provides information about the profile shift, and &lt;a href=&quot;http://khkgears.net/wp-content/uploads/2015/10/gear_guide.pdf&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt; is a technical reference on gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt; Forewarning, the gear guide is 13 MB in size. &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something else:&#xA;It sounds like you are doing BMX racing on a very professional level,&#xA;so I assume you already did that - still, I have to ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you were testing the new decimal chainwheels, did you replace all other transmission components (rear hub cog and chain) with new parts as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because a used bike chain becomes permanently stretched, and therefore has a larger pitch, so it won't fit a new chainwheel well. &lt;/p&gt;&#xA;" OwnerUserId="4637" LastEditorUserId="4637" LastEditDate="2016-01-30T14:26:59.967" LastActivityDate="2016-01-30T14:26:59.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7091" PostTypeId="2" ParentId="2042" CreationDate="2016-01-25T18:44:10.270" Score="1" Body="&lt;p&gt;The problem is that there is no absolute material property which generally describes material toughness in the same way that, for example, yield stress describes strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the comments charpy testing allows meaningful comparison between bulk materials but is less useful in terms of actual calculations of real world performance without so fairly thorough modelling based on the specific application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crucial issue is that toughness may be very dependant on specific geometries and indeed surface finish and is significantly affected by local stress raisers and strain rate. There is also a very big difference in whether the loading is distributed or acts at a point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This inherent ambiguity in toughness as  a general property is one of the reasons why cars are required to pass physical impact testing for safety certification but not for things like the strength of suspension parts. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-25T18:44:10.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7093" PostTypeId="1" CreationDate="2016-01-25T23:15:14.397" Score="1" ViewCount="167" Body="&lt;p&gt;I made a waterfall hoping to use water as a piston for suction. I'm wondering if some of you could help explain why it's not working the way I was expecting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the basic design: I have a 4 inch by 2.5 foot tube with a 5/8 inch opening on each side. The top has got a hose going to a drain of a 40 gallon tank and falls free at the bottom to fall in to the 40 gallon tank. My thoughts on this was that the tube filled with water would act as a hydraulic piston, though it's sucking air backwards in to the drain of the suction tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2I8cH.jpg&quot; alt=&quot;basic design&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4648" LastEditorUserId="33" LastEditDate="2016-01-26T01:47:09.810" LastActivityDate="2016-01-26T01:47:09.810" Title="Help with hydraulic design" Tags="|hydraulics|siphon|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7094" PostTypeId="2" ParentId="7093" CreationDate="2016-01-25T23:38:33.053" Score="5" Body="&lt;p&gt;Siphoning works when water goes from a higher point to a lower point, such as from an aquarium on a stand into a bucket on the floor next to the stand. Siphoning only works when the start point is higher than the end point because there is a net decrease in potential energy of the water from start to end, so the system is reducing its total useful energy. In your case, the start point is the same as the end point. To see this, pick any point in your cycle and note that water starting at that point must eventually return to that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we select the lowest point on the tube, then water going up the tube would gain potential energy proportional to the height difference of the lowest point of the tube and the highest point of the tube, and lose exactly the same amount of potential energy on the return trip from the highest point to the lowest point of the tube. The net potential energy change is thus zero. There are also frictional losses in the tube and other components,  and to overcome these frictional losses requires useful energy. Thus the total energy required to move the water around the cycle is greater than zero. Energy can neither be created nor destroyed, and useful energy can not be collected from a uniform temperature reservoir (i.e. the atmosphere), therefore, you must add useful energy in some form to move water around the cycle. Generally this would be accomplished by a powered pump of some sort.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-01-25T23:38:33.053" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7099" PostTypeId="2" ParentId="7080" CreationDate="2016-01-26T11:00:24.113" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible for a bicycle chainwheel to have a fractional number of teeth?  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;TLDR: Yes, but they are not suitable for BMX.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding chainwheels &lt;strong&gt;can&lt;/strong&gt; provide continuously variable transmission.  They have been re-invented several times over the years.  A recent example is the Wavetrans Bicycle Transmission.  The ratio is adjusted by moving segments of gears radially on an carrier.  Non-integer effective ratios occur when the disengaged sector has a fractional effective length. A chain tensioner is needed to take up the slack as the ratio is varied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system has not been a commercial success over the years as derallieur and hub gears have proved more effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A review is:&#xA;&lt;a href=&quot;http://bikeretrogrouch.blogspot.co.uk/2016/01/new-is-old-again-expanding-chainring.html&quot; rel=&quot;nofollow&quot;&gt;http://bikeretrogrouch.blogspot.co.uk/2016/01/new-is-old-again-expanding-chainring.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thought: If gearing is critical for the discipline and cost is unimportant then having a range of rear wheels/tyres with slightly varying diameter will provide intermediate gear progression, as will slightly varying the tyre pressure.&lt;/p&gt;&#xA;" OwnerUserId="4656" LastActivityDate="2016-01-26T11:00:24.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7100" PostTypeId="1" CreationDate="2016-01-26T11:39:48.250" Score="5" ViewCount="961" Body="&lt;p&gt;So I would like to ask you about a problem I encountered.&#xA;The case study goes like this: I want to study species distribution inside a packed bed reactor, at first at steady state but eventually what interests me is the dynamic behavior of the reactor. The reactants are gases. For the smimle steady state case the mass balance of each species if defined by:&lt;br&gt;&#xA;$$u_s\frac{\partial C_j}{\partial z}=\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{k}$$&#xA;where $u_s$ is the superficial velocity (m/s), $C_j$ the $j$ species concentration (mol/m^3), $z$ the axial direction, $\rho_b$ the catalyst density in the reactor ($\text{kg}_\text{cat}/\text{m}^3$), $\nu_{j,k}$ the $j$ species stoichiometric index in $k$ reaction and $R_k$ the $k$th reaction rate ($\text{mol}/(\text{kg}_\text{cat}*\text{s})$).&#xA;Reaction rates are known functions of concentrations, temperature and pressure and superficial velocity is given by:&#xA;$$u_s=\frac{Q}{A}=\frac{\sum_{j=1}^{N_c}F_j \frac{Rg T}{P}}{A}$$&#xA;where $Q$ is the volumetric flowrate ($\text{m}^3/\text{s}$), $A$ the cross-section area, $F_j$ the $j$th molar flowrate ($\text{mol}/\text{s}$), $Rg$ universal gas constant, $T$ is the temperature, $P$ is the pressure and $N_c$ is the number of components. While, when dealing with gases it is not correct to assume that velocity does not change along the reactor because of mole changes, in this case study I found (in terms of mass balance between reactors inlet and outlet) that the error produced is insignificant.&lt;br&gt;&#xA;Now the easiest way to solve this is by discretizing the differential equation via finite difference scheme and you get a set of algebraic equations like:&lt;br&gt;&#xA;$$u_s\frac{(C_{i,j}-C_{i-1,j})}{\delta z}=\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{i,k}$$&#xA;where $i$ is the spatial node number (not sure that's the correct terminology).&lt;br&gt;&#xA;For the case where I want to study the dynamic behavior of the system the equation would be like:&lt;br&gt;&#xA;$$\frac{\partial C_{i,j}}{\partial t}=-u_s\frac{(C_{i,j}-C_{i-1,j})}{\delta z}+\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{i,k}$$&#xA;or discretized:&#xA;$$\frac{(C_{t,i,j}-C_{t-1,i,j})}{\delta t}=-u_s\frac{(C_{t,i,j}-C_{t,i-1,j})}{\delta z}+\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{t,i,k}$$&#xA;&lt;strong&gt;Please correct me if I am wrong&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOW, here is where my problem starts. The assumption of constant velocity (calculated at the inlet of the reactor) is not safe anymore. Explanation:&lt;br&gt;&#xA;1. Velocity contant both in space and in time (calculated by $F$ at inlet of reactor at $t=0$): Not good. Gives large deviations in mass balance when inlet flowrate changes (checked it when the system reached steady state).&lt;br&gt;&#xA;2. Velocity constant in space but not time (calculates by $F$ at inlet of reactor at each $t$): Good in terms of mass balance but not good in terms of dynamic behavior because at each time interval the velocity might be correct BUT changes for the whole length of the reactor thus does not take into account the residence time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are my choices now? At first I thought to calculate the velocity in space like:&#xA;$$u_{s,i}=\frac{Q_i}{A}=\frac{\sum_{j=1}^{N_c}F_{i,j} \frac{Rg T}{P}}{A}$$&#xA;but $F_j=u_s C_j A\Rightarrow u_s=f(u_s)$.&lt;br&gt;&#xA;And also the mass balance equation that i showed earlier does not apply. It must be like:&#xA;$$\frac{\partial (u_sC_j)}{\partial z}=\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{k}$$&#xA;and discretized&#xA;$$\frac{((u_sC_j)_i-(u_sC_j)_{i-1})}{\delta z}=\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{i,k}$$&#xA;or&#xA;$$u_{s,i}\frac{(C_{i,j}-C_{i-1,j})}{\delta z}&#xA;+C_{i,j}\frac{(u_{s,i}-u_{s,i-1})}{\delta z}&#xA;=\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{i,k}$$&#xA;In the first case i can solve it but what happens when it gets dynamic? I have more variables than equations as $(u_sC_j)_{t,i}$ is one variable and $(C_j)_{t,i}$ is another. Is there a way to overcome this problem?&lt;br&gt;&#xA;In the second case I am almost sure that this system of equations does not have one unique solution. Thus my solver keeps giving me wrong (in terms of realistic) solutions (without reaction both velocity and concentration changes in an arithmetically correct way but no reaction should give no change).&lt;br&gt;&#xA;Is this statement true or false? Should this last equation along with inlet boundary conditions give me one and only one solution?&lt;br&gt;&#xA;Also should I solve along with the mass balance another equation like continuity equation? (I guess along with an equation of state, like $PV=nRT$). If so, I will also face the problem with unique solution because my equations will be:&#xA;$$u_{s,i}\frac{(C_{i,j}-C_{i-1,j})}{\delta z}&#xA;+C_{i,j}\frac{(u_{s,i}-u_{s,i-1})}{\delta z}&#xA;=\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{i,k}$$&#xA;and&#xA;$$u_{s,i}\frac{(\rho_i-\rho_{i-1})}{\delta z}&#xA;+\rho_i\frac{(u_{s,i}-u_{s,i-1})}{\delta z}=0$$&#xA;and&#xA;$$\rho_i=MW_{mean}\frac{P}{Rg T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New Comments referring to progress made on this problem&lt;/strong&gt;&lt;br&gt;&#xA;So, today I tried to solve the dynamic mass balance equation (DMBE) thinking that each quantity inside a differential should be treated like one. Explanation: in the DMBE we have actually 2 unknown variables, concentration $C$ and velocity $u$. One way to solve it is, as mentioned by nluigi bellow, to solve it along with an equation to calculate velocity (will test it tomorrow and get back to you).&lt;br&gt;&#xA;The way I tried (&lt;em&gt;and I think I made it&lt;/em&gt;) is to think my unknown variable as concentration $C$ and an other quantity like $G=C*u$ (molar flux). In this case the discretized equation do have one unique solution, given initial (in time) and boundary (in space) conditions. But still I need one more equation. The one that I thought is shown below along with the mass balance:&#xA;$$\frac{\partial C_i}{\partial t}=-\frac{\partial G_i}{\partial z}+\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{i,k}$$&#xA;$$C_i=\frac{G_i}{\sum_{j=1}^{N_{comp}} G_i}\frac{P}{RgT}$$&#xA;Now, the solution of these equations seems to give perfect results both in terms of mass balance as well as in terms of dynamic behavior, but I would like someone to make a comment regarding the validity of this set of equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;br&gt;&#xA;Dear all,&lt;br&gt;&#xA;After a lot of tests on solving the last system of equations (last two in this post) I am able to say that the solution does not give good results in terms of dynamic behavior. The truth is that my hopes where really high for this one but they didn't pay off. I may be doing an error during modelling and simulation which may result to this &lt;em&gt;problematic&lt;/em&gt; behavior.. but I dont know. I also tested what nluigi mentioned in his answer ($(u \rho A)_{in}=(u \rho A)_{out}$ for each volume) but still the dynamic behavior is not good enough. You see, when I make a step change on one of the inlet component's flowrate, velocity changes at the same timestep all over the reactor's length resulting to an error on outlet molar flowrate calculation $(F_i=C_iuA)$. Here is a figure that might help you understand what I am saying, look closer at $t=5s$ which if the time when the step change is made on one of the inlet flowrates.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KRDqx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KRDqx.gif&quot; alt=&quot;Superficial velocity versus reactor&amp;#39;s length AND versus time&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steady state $(t&amp;gt;&amp;gt;time of step change)$ values are correct but the dynamic of the system is not what one would expect. Actually the problem I am experiencing is (in my opinion) attributed to the fact that a term like $\frac{\partial u}{\partial t}$ does NOT appear in the equation where velocity is calculated.&lt;br&gt;&#xA;For this reason I decided to solve the mass balance equation&#xA;$$\frac{\partial C_{i,j}}{\partial t}=-u_i\frac{(C_{i,j}-C_{i-1,j})}{\Delta z}+\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{i,k}$$&#xA;along with momentum balance:&#xA;$$\rho _i\frac{\partial u_i}{\partial t}+&#xA;\rho _iu_i\frac{u_i - u_{i-1}}{\Delta z}+=0$$&#xA;and an equation to calculate density:&#xA;$$\rho_i=\sum_{j=1}^{N_{comp}}C_{i,j}mw_j$$&#xA;Until now, I thing that the results produced this way are looking good but.. let's say that I hope I will not need to update this question again... The next figure shows results of inlet (&lt;em&gt;blue line&lt;/em&gt;, at reactor's inlet) and outlet (&lt;em&gt;red line&lt;/em&gt;, at reactor's outlet) velocity versus time while tree step changes are imposed at one of the inlet flowrates. As you can see residence time is taken into account and the calculated (this one you can't see but I can assure you that it does) outlet flowrates are smooth and hopefully correct. (Do not mind about the different scale. the last results where for different flowrates)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pykww.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pykww.jpg&quot; alt=&quot;Velocity VS Time&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the question (one might ask): I really need a comment on the &lt;em&gt;correctness&lt;/em&gt; and &lt;em&gt;validity&lt;/em&gt; of this set of equations.&lt;/p&gt;&#xA;" OwnerUserId="4655" LastEditorUserId="4655" LastEditDate="2016-02-11T11:02:50.323" LastActivityDate="2016-08-31T18:19:47.650" Title="Dynamic mass balance in plug flow reactor" Tags="|fluid-mechanics|chemical-engineering|modeling|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7101" PostTypeId="1" AcceptedAnswerId="7102" CreationDate="2016-01-26T13:47:00.400" Score="1" ViewCount="284" Body="&lt;p&gt;Does the material with which I make the mold for carbon fiber parts affect the finish the part will get? If it does, I would like to understand how a given material may affect the finish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing that is confusing me is the exact definition of &lt;code&gt;finish&lt;/code&gt;. Does it refer only to the smoothness of the surface of the final part, or does it also refer to gloss and polish?&lt;/p&gt;&#xA;" OwnerUserId="4659" LastEditorDisplayName="user16" LastEditDate="2016-01-26T14:36:38.747" LastActivityDate="2016-06-16T19:26:15.297" Title="Impact of mold material in carbon fiber part finish" Tags="|materials|composite|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7102" PostTypeId="2" ParentId="7101" CreationDate="2016-01-26T14:40:46.217" Score="2" Body="&lt;p&gt;The finish of the mould will certainly have an effect on the finished part. How much of an effect depends on the exact process, for example if you use a gel coat then that will pick up the surface texture of the mould pretty closely. Without a gelcoat the ratio of fibre to resin will determine how much the weave of the fabric comes through. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case it's not so much the mould material as such as the finish of the mould surface and of course any release agent used will have an effect as well. It's certainly possible to get a glossy part straight out of the mould as long as t is prepared properly. In practice you would balance the time and expense you put into maintaining the mould with the amount of time spent on finishing the part after it comes out of the mould. eg if you are making dozens of parts it might well be worth getting the mould as close to perfect as you can to save time spent finishing each part. If you're just making a one-off you might not worry so much about the mould finish and sort out the finish on the part itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another consideration is that for high performance laminated parts you may find that the optimum resin to fibre ratio is just enough to wet the fibres and you get the weave of the mat showing through regardless of the mould finish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of definitions finish can refer to overall surface smoothness as well as gloss as these are really just two aspects of the same thing. &#xA;This is a significantly different thing from  the geometric accuracy of the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarity low volume moulding processes like silicone, plaster etc tend to be more likely to pick up small surface defects than an CNC machined steel mould. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-16T19:26:15.297" LastActivityDate="2016-06-16T19:26:15.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7107" PostTypeId="1" AcceptedAnswerId="7108" CreationDate="2016-01-26T20:39:10.750" Score="0" ViewCount="39" Body="&lt;p&gt;I have heard that error correcting codes are used to store and retrieve data in hard-drives but are they used in the processors in common laptops? Is the information encoded before it is processed?&lt;/p&gt;&#xA;" OwnerUserId="4671" LastActivityDate="2016-01-26T23:50:04.613" Title="Error correcting codes in processors" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7108" PostTypeId="2" ParentId="7107" CreationDate="2016-01-26T22:20:21.623" Score="2" Body="&lt;p&gt;Error correction means extra bits are stored along with the data.  Think of it as a advanced checksum.  A checksum can detect when the data is corrupt.  A error correcting &quot;checksum&quot; allows recovering some limited numbers of errors, usually a single bit error in the chunk of data being protected by the checksum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this isn't free, it is done only when the benefits are important.  In some cases it's cheaper to allow for a higher native bit error rate, but add bits to get the effective overall bit error rate back down via error correcting codes.  This is often done with dynamic RAM modules that plug into ordinary PC motherboards.  Whether laptops use them or not is best determined by reading the documentation for individual laptops.  Magnetic disks is another common application.  This time it's because their native bit error rate is unacceptably low for the densities and read/write speed people demand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error correcting bits are usually determined and checked with dedicated combinatorial logic running in parallel with fetches and stores.  That way this process largely doesn't get in the way of memory operations, and is largely transparent to the entity that is reading/writing data from/to the memory.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2016-01-26T23:50:04.613" LastActivityDate="2016-01-26T23:50:04.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7109" PostTypeId="2" ParentId="2353" CreationDate="2016-01-26T23:24:51.563" Score="0" Body="&lt;p&gt;One of the most important consideration is what sort of production volume you are looking at. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example injection moulding is frequently used for high volume plastic components as cost per unit is low but setup costs are high. At the other end of the scale something like vacuum forming is good for making small volumes as the tooling can be made relatively easily but it does have limitation on the forms that you can use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty with high volume moulding processes is that you need quite a detailed understanding of the manufacturing process to optimise a design and there may be very specific requirements for things like draft angles, fillet radii,  acceptable changes of section, allowance for shrinkage etc. etc. Not to mention implications for the design of the tooling which may not be at all obvious from the end product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good option for box-like shapes is to fabricate it using already available moldings or extrusions for example you may be able to find a rectangular section extrusion which can be adapted with relatively simple modifications like cutting to length, drilling holes and adding fill-in plates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest I would be quite surprised if there isn't an existing plastic box somewhere which is close enough to your requirements to avoid having to manufacture it from scratch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point here is that the specific manufacturing process selected has a profound influence on the suitability of a given design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that the best approach is to focus on defining what properties, performance  and standards the enclosure needs to achieve, such as environmental resistance, internal dimensions and any requirement for mounting points etc etc and then approach a manufacturer to discuss the best way to achieve these objectives. &lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="4357" LastEditorUserId="7194" LastEditDate="2016-07-15T19:09:59.240" LastActivityDate="2016-07-15T19:09:59.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7111" PostTypeId="1" AcceptedAnswerId="7171" CreationDate="2016-01-27T02:25:29.153" Score="0" ViewCount="56" Body="&lt;p&gt;When a fluid, like air, is moving at some velocity and &quot;hits&quot; or collides with a surface:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the surface is at an angle, the fluid will tend to be propelled in a similar direction (anything else?).  &lt;/li&gt;&#xA;&lt;li&gt;If the surface is perpendicular, the fluid will tend to kind of splash out in all directions (anything else?).  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For generally turbulent, laminar, or mixed flow conditions?&#xA;&lt;em&gt;The surface has mass and is relatively stationary for the purposes of this question.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each of those scenarios, what are the (qualitative) local effects of physical properties like pressure, density, etc. for both turbulent or laminar flow conditions?  Does the air-like fluid compress into a denser form and the local pressure increase where the fluid is coming in contact with the surface?&lt;/p&gt;&#xA;" OwnerUserId="2602" LastEditorUserId="1832" LastEditDate="2016-01-28T11:11:20.867" LastActivityDate="2016-01-29T19:13:18.490" Title="Air colliding with surface, local effects?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7112" PostTypeId="1" CreationDate="2016-01-27T07:16:53.187" Score="0" ViewCount="127" Body="&lt;p&gt;I'm just starting to dive into digital logic and I'm stuck on making a circuit that multiplies two 2-bit numbers. I partially understand that you need to use adders but I just cant fully wrap my head around the concept. I have made my truth table and then proceeded with the K-maps, but that's where I'm stuck. Any help would be much appreciated!  &lt;/p&gt;&#xA;" OwnerUserId="4678" LastEditorUserId="4678" LastEditDate="2016-01-27T07:28:18.320" LastActivityDate="2016-01-27T18:01:26.590" Title="How would I go about designing a logic circuit that multiplies two 2-bit numbers?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7114" PostTypeId="2" ParentId="7100" CreationDate="2016-01-27T10:43:17.323" Score="1" Body="&lt;p&gt;A couple of points:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Have you considered the stability of your discretization? Specifically are you sure that: $$C=\frac{u\Delta t}{\Delta x}&amp;lt;1$$ This may be difficult in the case $u$ is variable but perhaps you can make an estimate of its average value?&lt;/li&gt;&#xA;&lt;li&gt;Can you calculate the new velocity based on the mass balance? i.e. the total mass at any point in the reactor must be equal: $$\rho A u = \rho_{in} A_{in} u_{in} = \rho_{out} A_{out} u_{out}$$ You could calculate at each point in the reactor: $$u= \frac{\rho_{in} A_{in} u_{in}}{\rho A}$$ where $\rho$ is the total density at that point in the reactor determined from $$\rho = \sum_i^N c_i m_i$$ where $c_i$ is the species concentration and $m_i$ is the molecular mass.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2016-01-27T10:43:17.323" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7116" PostTypeId="2" ParentId="7080" CreationDate="2016-01-27T13:16:20.683" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A &quot;44.2 tooth&quot; gear still can only push 44 links of chain per revolution, which pushes a 16 tooth cog around 2.75 times. There's no getting around that.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that is the question that has everyone's head smoking. So I made a visualization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have small gear with 40px diameter and 4 cogs and a large gear with 80px diameter and 8 cogs, respectively. The chain is indicated by red dots and fits perfectly onto the gears. As one would expect, chain links are picked up at the top of gear and released at the bottom.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BIoqS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BIoqS.png&quot; alt=&quot;4-to-8 ratio&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second picture, I increased the diameter of the larger gear to 90px, but it still has only 8 cogs (I call it the 8.9999/4 ratio). As you can see, I had to move the gears closer together to adjust for the larger circumference (with a less significant increase, as from 41 to 41.2, the chain might only sit slightly tighter).&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XXaec.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XXaec.png&quot; alt=&quot;4-to-8.999 ratio&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does increased radius mean that the smaller gear moves faster? I'd say: No.&lt;br&gt;&#xA;As you can see in the second picture, the chain links only grip at the very bottom of the gear. The cogs move faster than the chain links, which means the gear subtly skids under the chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other answers suggested that this setup might lead to a higher torque. However, I don't think so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, increasing the size of the larger gear should decrease the torque.&#xA;Secondly, according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Work_%28physics%29#Torque_and_rotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; work is $W = \tau\phi$. Thus, as long as the angular speed of the chain ($\phi$) does not change, you will get the same torque($\tau$) for your work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this setup, the chain only grips at the top of the chainwheel (larger gear); I've visually verified this. The other rollers float between the teeth as you continue on down the gear.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we look closely at your pictures, we can see that the cogs are very close together so I suspect the chain links fit perfectly in-between. Towards the top, the cogs become slimmer. This might cause the tight chain links to slide upwards and become more loose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another illustration (again, I greatly exaggerated the effect):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tI4FY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tI4FY.png&quot; alt=&quot;More Sprocketes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4673" LastEditorUserId="4673" LastEditDate="2016-01-30T12:23:06.247" LastActivityDate="2016-01-30T12:23:06.247" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7119" PostTypeId="1" AcceptedAnswerId="7126" CreationDate="2016-01-27T16:36:41.783" Score="2" ViewCount="1544" Body="&lt;p&gt;Let's say I have a pump pushing fluid through a 250 ft pipe, 1.5'' diameter, at a rate of 20 gallons per minute. I would like to calculate the delta in energy required to pump this fluid with and without a restriction at the end of the 250 ft line. The restriction results in 0.01 psi pressure drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What equation is needed to calculate this energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on comments below, I was able to figure this out. First, I need to convert my $\Delta P$ to pressure head using the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta P = \rho_{fluid} * g * h$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g =\text{ acceleration due to gravity}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h =\text{ pressure head}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula itself is a simplified form of what can be found &lt;a href=&quot;http://www.engineeringtoolbox.com/static-pressure-head-d_610.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_2 - P_1 = \gamma (h_2 - h_1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\gamma = \rho_{fluid} * g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us all the ammunition we need with the help of the pump power equation found &lt;a href=&quot;http://www.engineeringtoolbox.com/pumps-power-d_505.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{\text{pump_power}} = q * \rho_{fluid} * g * h$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q =\text{ fluid flow, volume per time}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you plug in our $\Delta P$ formula from above (you have to rearrange to get the h on it's own), you get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{\text{pump_power}} = q*\rho_{fluid}*g*\frac{\Delta P}{\rho_{fluid} * g}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which, with handy cancellations, leads to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{\text{pump_power}} = q * \Delta P$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nice! Dimensionnally, if your flow is in gallon per minute (which it was for me) you'll want to convert to cubic inches per second - this way, if your $\Delta P$ is in psi (which again, mine was) some of your inches will cancel out. You'll then have to convert BACK to feet from inches, but then you can easily go from $\frac{\text{ft-lb}}{s}$ to horsepower or kilowatts or whatever you want.&lt;/p&gt;&#xA;" OwnerUserId="4694" LastEditorUserId="4694" LastEditDate="2016-01-28T15:32:14.230" LastActivityDate="2016-01-28T15:32:14.230" Title="How do I calculate the additional energy required to pump fluid through a restriction?" Tags="|energy|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7120" PostTypeId="2" ParentId="7112" CreationDate="2016-01-27T18:01:26.590" Score="4" Body="&lt;p&gt;Since there are only 16 possible inputs to this multiplier and its output is only 4 bits, it can be easily implemented as a lookup.  In case you think that sounds silly, consider that many FPGAs actually implement soft logic this way.  All you need is a memory with 16 words of 4 bits each.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-01-27T18:01:26.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7121" PostTypeId="2" ParentId="3115" CreationDate="2016-01-28T02:37:28.880" Score="2" Body="&lt;h3&gt;Standard&lt;/h3&gt;&#xA;&lt;p&gt;The official description of the test can be found in one of the following depending on your location of interest:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://shop.bsigroup.com/ProductDetail/?pid=000000000030087025&quot; rel=&quot;nofollow noreferrer&quot;&gt;BS 1377 Part 4&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.astm.org/Standards/D6951.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASTM D 6951&lt;/a&gt; or &lt;a href=&quot;http://www.astm.org/Standards/D3441.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASTM D 3441&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;These should have all of the answers that you could possibly want. I'll answer your questions directly though.&lt;/p&gt;&#xA;&lt;h3&gt;Questions&lt;/h3&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a standard number of times that the hammer must be dropped?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Typically the test is run continuously from the surface of the ground to a depth of interest. It could also be run from the bottom of a drilled hole if the soil layers above that point are not of interest.&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;What is recorded after an xx number of hammer drops?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The depth of penetration is recorded after each drop. The test is continuous from beginning to end. The penetration depth should be continuously increasing as the test progresses. That is unless refusal is reached for some reason.&lt;/p&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;What happens to that data? What does it correlate with to give us the strength of the ground? What is that strength measured in?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Yes, there are various conversion equations to convert the number into other values. This &lt;a href=&quot;http://docs.lib.purdue.edu/cgi/viewcontent.cgi?article=1544&amp;amp;context=jtrp&quot; rel=&quot;nofollow noreferrer&quot;&gt;research report&lt;/a&gt; shows some comparisons.&lt;/p&gt;&#xA;&lt;p&gt;Some typical comparisons are California Bearing Ratio (&lt;a href=&quot;https://en.wikipedia.org/wiki/California_bearing_ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;CBR&lt;/a&gt;) and density. These comparisons are based off of best fits of experimental data and are dependent on soil type.&lt;/p&gt;&#xA;&lt;p&gt;In general think of the Dynamic Cone Penetration Test as a cheap and quick check of subgrade strength.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-01-28T02:37:28.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7122" PostTypeId="2" ParentId="6204" CreationDate="2016-01-28T06:54:05.673" Score="1" Body="&lt;p&gt;Maybe you should check out &lt;a href=&quot;http://www.tribology-abc.com/abc/adhesion.htm&quot; rel=&quot;nofollow&quot;&gt;tribology&lt;/a&gt;. There is such thing as specific wear rate, k, for example, which is given for a pair of materials in dynamic contact. Say steel over steel has very low coefficient, even better than PTFE over steel, for example.&lt;/p&gt;&#xA;" OwnerUserId="4710" LastEditorUserId="1832" LastEditDate="2016-01-28T11:09:48.283" LastActivityDate="2016-01-28T11:09:48.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7124" PostTypeId="1" CreationDate="2016-01-28T07:14:03.310" Score="4" ViewCount="161" Body="&lt;p&gt;I have been studying the &lt;a href=&quot;http://www.ulb.ac.be/scmero/documents/these/Hanieh03.pdf&quot;&gt;sky hook damping system&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On page 14 of paper linked above, I found that the maximum transmissibility of sky hook system can be calculated by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{max} = \dfrac{1}{2\xi}$$&#xA;where $\xi$ is the damping ratio and $T_{max}$ is the maximum transmissibility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condition for maximum transmissibility was $\xi&amp;gt;0$, and $\dfrac{\omega_d}{\omega_n}=1$ in the following transmissibility equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T = \sqrt{\dfrac{1+\left(2\xi\dfrac{\omega_d}{\omega_n}\right)^2}{\left(1-\left(\dfrac{\omega_d}{\omega_n}\right)^2\right)^2+\left(2\xi\dfrac{\omega_d}{\omega_n}\right)^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, when I simply substitute $\dfrac{\omega_d}{\omega_n}=1$, I get the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{max} = \dfrac{\sqrt{1+\left(2\xi\right)^2}}{2\xi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is wrong with this equation? Could anyone explain how to correctly induce the maximum transmissibility of skyhook system?&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="1832" LastEditDate="2016-01-28T11:21:28.870" LastActivityDate="2016-01-28T18:50:41.273" Title="Maximum transmissibility of sky hook damper" Tags="|mechanical-engineering|structural-engineering|control-engineering|vibration|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7126" PostTypeId="2" ParentId="7119" CreationDate="2016-01-28T09:56:15.797" Score="3" Body="&lt;p&gt;The pump power equation can be used to calculate the energy required in both scenarios. Convert the additional $\Delta P$ to head, then calculate the difference in required hydraulic pump power. Both of these equations are on Engineering Toolbox. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/pump-head-pressure-d_663.html&quot; rel=&quot;nofollow&quot;&gt;Converting head (ft or m) to pressure (psi or bar, kg/cm2) and vice versa&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/pumps-power-d_505.html&quot; rel=&quot;nofollow&quot;&gt;Calculate pump hydraulic and shaft power&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the question in your comment about the energy balance equation, Bernoulli's Equation is an energy balance for fluids. &lt;/p&gt;&#xA;" OwnerUserId="3569" LastEditorUserId="3569" LastEditDate="2016-01-28T10:10:42.203" LastActivityDate="2016-01-28T10:10:42.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7127" PostTypeId="1" CreationDate="2016-01-28T09:56:45.943" Score="2" ViewCount="1189" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Fiberscope&quot; rel=&quot;nofollow noreferrer&quot;&gt;fiberscope&lt;/a&gt; is the snakelike, fiber-optic camera that doctors use during an endoscopy. The tips of these scopes can articulate: you can control the rate at which they bend in one or more directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YqlZv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YqlZv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the tip on one end of this fiberscope can bend in various directions and at various angles. You control this via the controller on the other end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this done? I'm assuming it involves a sort of spring/flexible tube and some sort of cable system that controls the rate at which the spring/tube bends, but I can't imagine how this would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; This image from &lt;a href=&quot;http://annals.org/article.aspx?articleid=706038&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; refers to a (or possibly four) &quot;tip bending control wire(s)&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lk4Sl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lk4Sl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://biology-forums.com/index.php?action=gallery;sa=view;id=11818&quot; rel=&quot;nofollow noreferrer&quot;&gt;here's another one&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the question is: what do these wires do? Do they bend the tip by contracting/expanding? Is there some sort of mechanical device in the &quot;bendy&quot; part that is controlled by these wires?&lt;/p&gt;&#xA;" OwnerUserId="4714" LastEditorUserId="4714" LastEditDate="2016-01-28T15:15:32.410" LastActivityDate="2016-01-28T17:45:55.777" Title="How does the articulation system on a fiberscope work?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7129" PostTypeId="1" CreationDate="2016-01-26T13:01:45.550" Score="2" ViewCount="90" Body="&lt;p&gt;I would like some help defining the specification of my motor, as I have little experience in this field. I've tried researching through Google, but the information is insufficient or unrelated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projects involves horizontal movement of a platform with a load of up to &lt;strong&gt;10kg&lt;/strong&gt; along at a speed of &lt;strong&gt;10mm/sec&lt;/strong&gt;. As you can see this is a large load and very low speed. The movement has to be pretty smooth. I'm unsure how to specify this when searching for motors (I think it's referred to as jitter). I have yet to decide on the sliding mechanism if it will be using wheels or just gliding on a cylindrical rail, but I'll be trying to minimise the coefficient of friction to reduce the motors force needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was wondering if anyone can help me identify the key properties of a motor so it provides very smooth movement?  Or if people have experience designing supporting components to the system and why I should focus on that aspect instead of the motor.&lt;/p&gt;&#xA;" OwnerUserId="13665" OwnerDisplayName="Kiran" LastEditorUserId="33" LastEditDate="2016-01-28T14:33:17.433" LastActivityDate="2016-01-28T20:51:19.050" Title="What are the key properties to consider when choosing a DC electrical motor for linear movement?" Tags="|mechanical-engineering|motors|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7130" PostTypeId="2" ParentId="7129" CreationDate="2016-01-26T13:29:46.580" Score="0" Body="&lt;p&gt;Not sure where you are starting, this question may be deleted as too broad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Start by understanding the physics involved - forces, energies etc. Look at hyperphysics.org.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Get some practical advice by looking to see how others do it in your area, on Google etc. Types of rails, friction, drive mechanisms etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Now you need to control the movement - from a switch to a highly complex arrangement. This depends on what you need to do , which you will have learned from the physics above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Go to your friendly retail motor supplier, hardware, electronics store and speak to them - they are amazingly knowledgable, helpful and informed and will help you if you have done your homework.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SE is also a great place to bring questions you have researched thoroughly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you haven't researched, you WILL get downvoted and perhaps even banned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Research Papers and Journals? How many Millennia do you have?&lt;/p&gt;&#xA;" OwnerDisplayName="ChrisR" LastActivityDate="2016-01-26T13:29:46.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7131" PostTypeId="2" ParentId="7129" CreationDate="2016-01-26T13:33:05.417" Score="0" Body="&lt;p&gt;If you use a worm gear arrangement, you could use almost any gear that was fast enough, given the constraints of your speed vs. torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know the torque requirements, of course, but a reducing gear could make many worm gear or pulley arrangement work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of jitter, a worm or rack and pinion arrangement can be very stable. &lt;/p&gt;&#xA;" OwnerDisplayName="jdv" LastActivityDate="2016-01-26T13:33:05.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7132" PostTypeId="2" ParentId="7129" CreationDate="2016-01-26T14:49:19.910" Score="0" Body="&lt;p&gt;I would look into a threaded rod with a nut fixed to the platform: that way, you can have a motor that spins very fast and gives you smooth, linear movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll often have problems getting a motor to give you very smooth movement since most controllers give you a jerking motion at lower speeds. If you add in any sort of gearing (such as an ACME rod) so that the motor can spin at higher speeds with less resistance, you will get better results. It won't help you with jerking, but you can ensure a consistent speed using an encoder (rotary on your motor/transmission or linear on your platform) to give your controller feedback, so that it can adjust its output based on if it's moving too fast or too slow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: you'll never get smooth movement unless you use a fast motor geared down to be slow, and the only real way to guarantee a consistent speed is to measure it.&lt;/p&gt;&#xA;" OwnerUserId="4771" OwnerDisplayName="Trevor Gross" LastActivityDate="2016-01-26T14:49:19.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7133" PostTypeId="1" AcceptedAnswerId="7144" CreationDate="2016-01-28T15:13:30.727" Score="4" ViewCount="1699" Body="&lt;p&gt;I have been reading a lot on &lt;a href=&quot;https://en.wikipedia.org/wiki/Flexure_bearing&quot; rel=&quot;nofollow&quot;&gt;flexure bearings&lt;/a&gt;, but it's still not really clear what they are. From everything I've seen, it seems like a flexure bearing is just a spring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give me a simple explanation of what they are and their use?&lt;/p&gt;&#xA;" OwnerUserId="4725" LastEditorUserId="33" LastEditDate="2016-01-29T14:06:41.203" LastActivityDate="2016-01-29T14:06:41.203" Title="What is a flexure bearing?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7134" PostTypeId="1" CreationDate="2016-01-28T15:34:19.037" Score="5" ViewCount="1049" Body="&lt;p&gt;What is the property of not shattering, in for example a &lt;a href=&quot;https://en.wikipedia.org/wiki/Brisance&quot; rel=&quot;nofollow&quot;&gt;brisant explosion&lt;/a&gt;, called? And what kind of steel would I want that has this property?&lt;/p&gt;&#xA;" OwnerUserId="4726" LastEditorUserId="33" LastEditDate="2016-01-29T01:45:14.780" LastActivityDate="2016-01-29T01:45:14.780" Title="What do you call the steel property that reduces the likelihood of shattering?" Tags="|materials|steel|terminology|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7135" PostTypeId="2" ParentId="7134" CreationDate="2016-01-28T15:56:07.607" Score="1" Body="&lt;p&gt;This would be the opposite of brittleness, 'toughness' seems like the right term but 'ductility' might also apply.&lt;/p&gt;&#xA;" OwnerUserId="4132" LastActivityDate="2016-01-28T15:56:07.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7136" PostTypeId="2" ParentId="7134" CreationDate="2016-01-28T16:36:57.990" Score="0" Body="&lt;p&gt;I believe @squigbobble is on the right track.  The issue is related to crack propagation, namely the energy to propagate the crack in the material.  The reason ductility plays a big roll here is that ductility &quot;dulls&quot; the tip of this crack requiring greater energy to continue opening it.  This slows the rate of crack propagation significantly.&lt;/p&gt;&#xA;" OwnerUserId="4697" LastEditorUserId="2246" LastEditDate="2016-01-28T16:41:33.880" LastActivityDate="2016-01-28T16:41:33.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7137" PostTypeId="2" ParentId="7134" CreationDate="2016-01-28T16:41:27.887" Score="9" Body="&lt;p&gt;There are three (or four) possible terms for what you are asking: &lt;a href=&quot;https://en.wikipedia.org/wiki/Resilience_(materials_science)&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;resilience&lt;/em&gt;&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Ductility&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;ductility&lt;/em&gt;&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Toughness&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;toughness&lt;/em&gt;&lt;/a&gt; (a steel that is &lt;em&gt;resilient&lt;/em&gt;, &lt;em&gt;ductile&lt;/em&gt; and/or &lt;em&gt;tough&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A material's resilience describes its ability to absorb energy by deforming elastically. Its ductility (or malleability, there is a slight difference between the terms, but its not relevant here) represents how much energy it can absorb via plastic deformations. Its toughness encompasses both elastic and plastic deformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These terms are actually quantifiable and are equal to the area beneath the stress-strain diagram. Resilience equals the area under the elastic regime, ductility under the plastic regime, and toughness under the entire curve (and is therefore equal to the sum of the material's resilience and ductility).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pGuip.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pGuip.png&quot; alt=&quot;stress-strain diagram representing resilience and ductility&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the second part of your question was asking for specific steels, that's out-of-scope for this site, given how different steels are available in different countries.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-01-28T16:41:27.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7138" PostTypeId="2" ParentId="7127" CreationDate="2016-01-28T17:45:55.777" Score="3" Body="&lt;p&gt;The principal is fairly simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the diagram there are 4 wires or cables around the circumference of the sleeve. They are constrained so that they can slide freely along its length but are otherwise fixed in the 12, 3, 6 and 9 o'clock positions. The ends of the wire are attached to the head end but free at the control end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To bend the wire in one direction you pull one wire, this effectively shortens the length of wire inside the sleeve and so tends to want to form the inside of a radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also think about it by a reverse analogy. If you bend a flexible tube the inside of the bend gets shorter and the outside gets longer here you are just pulling on one side to create the bend. You can create abend in any direction by pulling adjacent wires different amounts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You the 'joint' is created by having most of the sleeve incompressible along it's length (eg like a tightly coiled spring) except for where you want the joint (eg by opening up the coils of the spring). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is broadly the way that the flexing of human fingers work. Most of the hand muscles are actually in the forearm and are connected to the fingers by long tendons. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-28T17:45:55.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7139" PostTypeId="1" CreationDate="2016-01-28T18:38:30.037" Score="4" ViewCount="2030" Body="&lt;p&gt;I have a question about the methods to decide whether a beam/frame is determinate or indeterminate. Typically you can count the number of reaction forces and compare them to the number of elements to determine this. The typical equation for a determinate beam is $r = 3N$, where, $r$ is the number of reaction forces and $N$, the number of elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is where does this equation come from? I understand that mathematically you have 3 available equations so you can only have 3 unknowns (without an internal hinge or something else). However, I am trying to grasp what this means physically? Is there no physical interpretation of this equation to be had, and it's all mathematical? Do the 3 available equations come from an assumption in elasticity?&lt;/p&gt;&#xA;" OwnerUserId="3653" LastEditorUserId="1832" LastEditDate="2016-01-29T14:12:44.497" LastActivityDate="2017-04-20T13:57:54.423" Title="Statics and Structural Analysis: Determinacy in beams/frames" Tags="|structural-engineering|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7140" PostTypeId="2" ParentId="7124" CreationDate="2016-01-28T18:50:41.273" Score="1" Body="&lt;p&gt;This isn't exactly my field, but I might have an idea. Sometimes people simplify equations and throw away terms that will be insignificant in value or close to zero. It's possible that a series expansion was performed and only the first term (which would be 1/2ξ) is considered.&lt;/p&gt;&#xA;" OwnerUserId="3653" LastActivityDate="2016-01-28T18:50:41.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7141" PostTypeId="1" AcceptedAnswerId="7145" CreationDate="2016-01-28T19:13:25.423" Score="8" ViewCount="4205" Body="&lt;p&gt;When building things like floors to walk on, we use materials like 2&quot;x10&quot; wood joists or steel I-beams. I'm primarily interested in knowing why a 2&quot;x10&quot; piece of wood resists bending downward when you have it &quot;standing up&quot; (ie, the 2&quot; side facing the sky) as opposed to when you have it &quot;laying down&quot; (ie, the 10&quot; side facing the sky).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems intuitive to me based on experience, but I can't actually explain why we orient floor joists the way we do even though it seems obvious that will lead to a stronger floor. I imagine whatever the reason is, it also explains why I-beams have such a thin webbing between the two sides.&lt;/p&gt;&#xA;" OwnerUserId="4728" LastEditorUserId="255" LastEditDate="2016-01-29T12:07:39.463" LastActivityDate="2023-01-20T01:12:08.053" Title="Why does the orientation of a floor joist affect its strength?" Tags="|civil-engineering|structural-engineering|" AnswerCount="7" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7142" PostTypeId="2" ParentId="7141" CreationDate="2016-01-28T19:27:19.107" Score="5" Body="&lt;p&gt;The web or wide part of the 2x10 is the part that resists bending. A wider 2x? or web will resist bending under heavier loads. The force being transferred through the web means that more material needs to stretch or compress for it to fail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine bending a 2x10 in plane with the 10 inch part. The bottom part will be stretching along the length of the joist and the top will be compressing. Having more material to resist those forces makes a stronger joist (or I-beam). Using a 2x10 with the web face-up would have very little material resisting the stretching or compressing which would lead to failure much more easily.&lt;/p&gt;&#xA;" OwnerUserId="2733" LastActivityDate="2016-01-28T19:27:19.107" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7143" PostTypeId="2" ParentId="7141" CreationDate="2016-01-28T19:48:46.427" Score="4" Body="&lt;p&gt;In the context of engineering calculations this is down to a property called 'second moment of area', the way that this is defined is a bit obtuse but is describing something which is reasonably intuitive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider a joist, supported at both ends with a weight at a point in the middle it will (obviously) bend. This bending force is trying to stretch the bottom edge and compress the top edge so logically there is a line along the middle where it is neither bending nor compressing and similarly the greatest stress is at the top and bottom edge, decreasing towards the middle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do the full analysis it turns out that the stiffness of the beam is directly proportional to its width but also to the cube of it's depth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in terms of bending the distance between centreline and edge of a beam (in the direction of load) is much more important than just its cross-sectional area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By extension the material near the centreline is contributing much less to the stiffness than material near the edge and something like an I-beam is just a way of dispensing with much of this material which is adding mass but not contributing much to stiffness. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-28T19:48:46.427" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7144" PostTypeId="2" ParentId="7133" CreationDate="2016-01-28T20:34:04.767" Score="4" Body="&lt;p&gt;From the Wikipedia article you linked,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A flexure bearing is a bearing which allows motion by bending a load element.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it's something that provides support, but also allows motion by &lt;em&gt;bending&lt;/em&gt; a component (as opposed to sliding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several examples on the Wikipedia page, but I would add (in addition to those examples) that probably the most common flexure bearing would be the vibration mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is typically a piece of rubber that is designed to provide support, and it allows some relative motion by bending. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have vibration mounts holding the engine and transmission in my car, holding the ball joints, sway bars, and other suspension components. They're in the blender in my kitchen, in my washing machine, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could consider any rubber foot on a component designed to prevent rattle/hum (refrigerators, etc.) as vibration mounts and thus flexure bearings. They all provide support and allow motion by bending. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-01-28T20:34:04.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7145" PostTypeId="2" ParentId="7141" CreationDate="2016-01-28T20:44:12.890" Score="7" Body="&lt;p&gt;Get a gallon of milk, and shake it up and down by moving your wrist only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, hold your forearm out and shake it by moving your elbow only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, hold your entire arm out and shake it by moving your shoulder only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass didn't change, but the force it took to move it goes up (a lot!) because it is farther from you. This phenomenon, when referring to rotating a mass (as in rotating about your wrist, elbow, or shoulder) is called the &quot;moment of inertia&quot; or, &lt;em&gt;specifically&lt;/em&gt;, the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Moment_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass moment of inertia&lt;/a&gt;.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, it becomes &quot;harder&quot; to apply the same force to &lt;em&gt;anything&lt;/em&gt; the farther from the force origin you get. In instances other than &lt;em&gt;mass&lt;/em&gt; specifically, as in I-beams or wood beams, the property of interest is called the &quot;area moment of inertia&quot; or, to add to the confusion, also the &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Second_moment_of_area&quot; rel=&quot;nofollow noreferrer&quot;&gt;second moment of area&lt;/a&gt;.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, the more material (cross-sectional area) you can put away from the force, the less of an impact that force has on the material. It's very similar to how your shoulder has to work so hard to shake the milk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also consider that, for a given arc of motion, the material spanned by that arc increases linearly with distance, and so you could also consider that your force is getting applied to more material (though this isn't quite technically correct), and because it's getting applied to more material the effective force is diminished. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;:EDIT:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I made a picture (not animated, sorry) that shows a traversed arc, and how it remains constant if the traversed &lt;em&gt;angle&lt;/em&gt; decreases linearly with increasing &lt;em&gt;distance&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of this graphic is to demonstrate strain. If your shoulder did the same work as your wrist, then the gallon of milk would go up and down the same &lt;em&gt;linear distance&lt;/em&gt;, but it would not traverse the same &lt;em&gt;angle&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KWrSV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KWrSV.png&quot; alt=&quot;Bending and Deflection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if you attempt to bend something, as in a floor joist, the change in length (strain) in the material may be the same, but if you can put the material farther from the bend axis then that strain corresponds to a smaller deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try a numeric example. Consider a very short, very wide rectangular beam whose cross-sectional width is 10m and cross-sectional height is 0.1m. This beam is representative of a slice, but I'm using it for this example to hopefully drive the point across. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area moment of inertia of the beam about &lt;em&gt;its&lt;/em&gt; x-axis ($I_x$) is &lt;a href=&quot;https://en.wikipedia.org/wiki/Second_moment_of_area#Examples&quot; rel=&quot;nofollow noreferrer&quot;&gt;given by&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;I_x = \frac{bh^3}{12} \\&#xA;I_x = \frac{(10)(0.1^3)}{12}\\&#xA;I_x = 0.000833 \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area moment of inertia for the beam is very small about the x-axis - it is very easy to &quot;twist&quot; about the x-axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Second_moment_of_area#Parallel_axis_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;parallel axis theorem&lt;/a&gt;, let's set that up at a distance of 1m from the bending axis. This means that the area moment of inertia about the &lt;em&gt;bending axis&lt;/em&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;I_x = I_{x'} + Ad_y^2 \\&#xA;I_x = 0.000833 + (1)(1^2) \\&#xA;I_x = 1.000833\\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see, the area moment of inertia for bending is about 1 for this example, where the distance of the beam from the bending axis is 1. Now, if the beam is moved out to a distance of 2m from the bending axis, the parallel axis theorem gives the new area moment of inertia to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;I_x = I_{x'} + Ad_y^2 \\&#xA;I_x = 0.000833 + (1)(2^2) \\&#xA;I_x = 4.000833 \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when the distance is doubled, the area moment of inertia &lt;em&gt;approximately&lt;/em&gt; quadruples. Now, consider the bending stress:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sigma = \frac{My}{I} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bending moment, times the distance from the bending axis, divided by the area moment of inertia. Now, for the first example, the beam was at 1m, and the area moment of inertia was &lt;em&gt;approximately&lt;/em&gt; 1. This means that the stress works out to &lt;em&gt;approximately&lt;/em&gt; $\sigma = \frac{M(1)}{(1)}$, or $M$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the beam is moved out to 2m, the moment of inertia became &lt;em&gt;approximately&lt;/em&gt; 4, but now the distance $y$ to get to that beam is $2$, so the bending stress becomes &lt;em&gt;approximately&lt;/em&gt; $\sigma = \frac{M(2)}{4}$, or $\frac{1}{2}M$. Putting the beam out at twice the distance caused the bending stress to halve. Strain is linearly related to stress by Young's modulus (assuming elastic deformation), so this means that the strain, or elongation, is also halved by moving the beam twice as far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this was an illustrative example where I picked the beam size to drive the area moment of inertia of the beam itself to a negligible value such that the parallel axis theorem was the dominating factor. This was done to reinforce the &lt;em&gt;concept&lt;/em&gt; that material placed farther from the bending axis results in less stress (and thus less deformation/bending) at the edge of the material.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2016-01-29T14:59:51.417" LastActivityDate="2016-01-29T14:59:51.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7146" PostTypeId="2" ParentId="7129" CreationDate="2016-01-28T20:51:19.050" Score="3" Body="&lt;p&gt;You need more specifications to decide what you need. Power is force times speed, but force is mass times &lt;strong&gt;acceleration&lt;/strong&gt;. You want to move at a constant speed, so there is no force. If you assume a frictionless surface, then any actuator could be used because the required power is zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to choose how quickly you want to get from zero to top speed to determine an acceleration, multiply that by your load inertia (if you're going through gears/belts/other reductions) to get a force, then multiply that force by your top speed to get a peak power. Add a margin for friction, but I think you'll find that actuators come in certain &quot;bins&quot; where one class of actuators will be insufficient or marginal and the next class up will be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't say how far you need to move your platform, which would be another spec in finding a linear actuator or in designing an actuated platform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest considering looking at an electric linear actuator and putting your platform on wheels or &lt;a href=&quot;http://www.mcmaster.com/#linear-bearings/=10vro8a&quot; rel=&quot;nofollow&quot;&gt;carriages&lt;/a&gt;. Electric linear actuators are relatively cheap, easy to power and control, and can have modest ranges of motion (up to a meter would probably be feasible). Beyond that, you may want to consider some kind of pulley or cart system with a motor and encoder, but again, you'll still need to perform the calculations above - peak force (or torque if you're going with pulleys), and peak power from the force (or torque) and linear (or angular) velocity. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-01-28T20:51:19.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7148" PostTypeId="1" AcceptedAnswerId="7152" CreationDate="2016-01-29T01:46:30.760" Score="5" ViewCount="579" Body="&lt;p&gt;I'm a beginner at autodesk inventor.&lt;br&gt;&#xA;For some instances when I click a region to extrude it doesn't come alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The screenshot below illustrates something even stranger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should I do to make this not to happen?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Yg1xp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yg1xp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4735" LastEditorUserId="4428" LastEditDate="2016-01-30T19:35:55.333" LastActivityDate="2016-01-30T19:35:55.333" Title="How do I extrude a region in autodesk inventor?" Tags="|autodesk-inventor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7151" PostTypeId="1" CreationDate="2016-01-29T05:15:39.690" Score="3" ViewCount="4104" Body="&lt;p&gt;Assume that a person is touching a bare wire when the person is connected directly to the ground system of a home electrical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a hot wire from, say, a washing machine, suddenly shorts to its chassis, doesn't that route to ground, and isn't our human suddenly now part of a deadly circuit involving the hot wire?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question came about from hearing about people who do this on purpose. Many people now practice &quot;earthing&quot;, physically linking themselves to the earth ground plug of their home's electrical system.  They go as far as sleeping on mesh wire sheets plugged into ground to protect from stray EMF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is not to be a commentary on the earthing theories.&lt;/p&gt;&#xA;" OwnerUserId="4737" LastEditorUserId="61" LastEditDate="2016-01-30T21:03:52.043" LastActivityDate="2018-09-01T22:39:59.253" Title="Is there a danger to a human that is linked to ground wire of home electrical system?" Tags="|electrical-engineering|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7152" PostTypeId="2" ParentId="7148" CreationDate="2016-01-29T08:18:12.237" Score="9" Body="&lt;p&gt;In a sketch, you cannot have two lines that intersect themselves. You have to split them with the Split tool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your bottom line for example, must be divided into 10 parts.&lt;/p&gt;&#xA;" OwnerUserId="4428" LastActivityDate="2016-01-29T08:18:12.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7154" PostTypeId="2" ParentId="6925" CreationDate="2016-01-29T10:09:12.197" Score="5" Body="&lt;p&gt;This happens on a bit level of the receiver electronics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every byte you have to receive 8 bits. At this point in communication you only decoded the signal (eg. converted the electrical power or voltage) levels to bits. You receive bits in the buffer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are constantly filling the buffer (on a bit level), so you can check the buffer to see if you received anything. First you receive a lot of 000000, but then at one point message arrives, and bits in the buffer start to have different values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you are lucky first bit of the message arrives on let's say 8th or 16th place in the buffer (or 0th). This are all the bits that also start a logical representation of a byte. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not that lucky, you just pushed in a 0 on 8th place of a buffer, and you would start filling the message in on the 9th place, which would in mean first seven bits from the original first byte would be received as first byte. And a second byte would be received as a last bit of the original first byte and first 7 bits of the original second byte and so on. This would mess things up good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to be able to slide in some zeros in to the buffer in order to check up with the start filling the message in on the &quot;start of byte position&quot; of the buffer. But since you know how many zeros you had to fill in, you know how much delay this added for certain message (if you know the frequency of transmission). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;Just to make things more exact:&#xA;you push the preamble in the buffer at random bytes, this is how you know you are getting something. sync is then used to sync the trasmission (eg. fill in the zerros so you start on right bit position).&lt;/p&gt;&#xA;" OwnerUserId="4682" LastActivityDate="2016-01-29T10:09:12.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7155" PostTypeId="2" ParentId="7141" CreationDate="2016-01-29T10:28:23.127" Score="4" Body="&lt;p&gt;The first thing to take into account is how stress builds in a material. This is rather intuitive and something we all know, but takes a bit of thinking to back it up mathematically. It's quite easy to get sucked into the world of 'Second moment of area' which is incredibly important for calculations, but not strictly necessary for understanding how it works (I hear screams of &quot;heresy&quot;..).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, when a beam bends, the bottom fibres are stretched and pulled apart (&lt;strong&gt;tension&lt;/strong&gt;), while the top fibres crush together (&lt;strong&gt;compression&lt;/strong&gt;). The beam will fail (e.g break) when either the top fibres reach their &lt;em&gt;maximum compressive strength&lt;/em&gt;, or the bottom fibres reach their &lt;em&gt;maximum tensile strength&lt;/em&gt;. Simple really! See my vquick sketch below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OgRv2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OgRv2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the implications of this is that at some point between the bottom and top fibres, there is a point where there is no stress (where it transitions from tension to compression). The maximum bending moment (engineer talk, not strictly important for understanding!) that a beam can take is limited by the compressive or tensile strength of a beam, and so working backwards we can work out the difference between a beam that is standing upright and one lying flat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done a quick hand calculation with a few assumptions (for simplicity I've assumed the beam will fail in compression and has a compressive strength of 10N/mm^2 and changed the beam from 2&quot;x10&quot; to 20mm x 100mm). Starting with this, you can work out the 'area' of the stress triangle. You then multiply this by the area of the beam that this stress triangle occurs over. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AxDqc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AxDqc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the 'taller' of the two beams has a larger stress triangle and thus can resist more bending! This is because &lt;strong&gt;more of the beam&lt;/strong&gt; is located &lt;strong&gt;further away&lt;/strong&gt; from the neutral axis (the point where the beam is neither in compression or tension)! Basically if you were to slowly increase the load, the top fibre of the taller beam would take longer to reach the killer 10N/mm^2 limit that I've given it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this is clear and simple, it's a little irritating to explain what is a relatively simple concept because there's so much ground to cover. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I plan on re doing those sketches/calcs soon, hopefully with some colour. I quickly drew them up at work so didn't put much effort into them! &lt;/p&gt;&#xA;" OwnerUserId="3093" LastEditorUserId="3093" LastEditDate="2016-01-29T14:14:00.430" LastActivityDate="2016-01-29T14:14:00.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7157" PostTypeId="2" ParentId="7151" CreationDate="2016-01-29T11:38:32.567" Score="4" Body="&lt;p&gt;No, the whole point of the earth in an electrical system is to provide a safe direct route away from the rest of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there were any risk to someone connected to earth, there would also be a risk to every other device connected to the same earth. That would defeat the purpose of the earth. When a device develops a fault resulting in power going to earth, then as long as the circuit is designed properly, almost all the current will go through no other device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People who practice earthing their own bodies out of some peculiar ideas about health benefits may well have problems, but these are the remit of the &lt;a href=&quot;https://cogsci.stackexchange.com/&quot;&gt;Cognitive Sciences Stack Exchange&lt;/a&gt; and/or problems of science education, not problems from faulty electrical devices.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="-1" LastEditDate="2017-04-13T12:56:17.013" LastActivityDate="2016-01-29T11:38:32.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7159" PostTypeId="2" ParentId="7151" CreationDate="2016-01-29T12:29:00.567" Score="4" Body="&lt;p&gt;It &lt;i&gt;should&lt;/i&gt; be safe.  The safety standards already assume a user is connected to ground, since that's the worst case in terms of probability of electrocution.  You are supposed to be two independent failures away from getting zapped, already assuming you are connected to ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For equipment with a conductive shell, that shell is tied to ground.  To get zapped, line has to get shorted to the shell, and the shell connection to ground has to be broken.  If the first doesn't happen, then there is nothing &quot;hot&quot; to connect yourself to.  If the second doesn't happen, then the breaker will trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For insulated equipment, the device often has no ground connection at all.  However, things inside are &quot;double insulated&quot;.  Wires are insulated on their own, but then shell then provides another layer of insulation between hot and user.  Any conductive parts then also have minimum clearance and creepage requirements from anything hot, and units are generally &quot;hi-pot&quot; tested and tested for leakage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though you are supposed to be safe when grounded, it does remove one extra layer of protection.  Supposed the unlikely failures do happen so that the case of a unit gets connected to line power.  If you're grounded, then there is a path thru you, and you get zapped when you touch the case.  If you weren't connected to anything else, you might not even notice you were touching something with live power on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why there are special rules in areas where people are particularly likely to be connected to ground via low resistance, such as in a bathtub or around any sink.  In some jurisdictions, outlets in such areas must have ground fault interrupters.  That's a special kind of circuit breaker that not only trips due to excessive current, but also when there is a imbalance of current going out one of the AC leads but not coming back on the other.  The missing current must be going from the line side to ground via some separate and unintended path, which causes a ground fault interruptor to shut off the power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary, while you are supposed to be safe, you are one level closer to a the right failures causing you harm, although the probability of such a set of failures is low.  If you plan to deliberately ground yourself regularly, it might be a good idea to replace all your circuit breakers with ground fault interruptor types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course grounding yourself to somehow protect yourself against EM waves is silly, and can have the opposite affect in many circumstances, but that's another discussion.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-01-29T12:29:00.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7160" PostTypeId="2" ParentId="7141" CreationDate="2016-01-29T13:56:13.347" Score="2" Body="&lt;p&gt;All structural materials collapse when they suffer excessive deformations and stresses. If you stretch an elastic too far, it'll snap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joists are considered beam elements. This means that the primary internal force they need to withstand is bending moment. This bending moment is resisted by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Couple_(mechanics)&quot; rel=&quot;nofollow&quot;&gt;force couple&lt;/a&gt; within the joist. The moment causes a rotation of the joist: the top of the joist will be compressed and the bottom will be stretched. This means that the force couple will be composed of the total compressive force on the top half of the joist and the total tensile force on the bottom half (which must be equal under simple bending).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the moment due to a force couple is equal to the product of one of the forces and the distance between them. So, for a joist placed in the strong (&quot;tall&quot;) as opposed to the weak (&quot;wide&quot;) configuration, the distance will be greater, meaning that the requisite forces to withstand the bending moment will be reduced. Since, at the end of the day, what causes the element to collapse is excessive stresses, you want to minimize the total internal compressive/tensile forces (and therefore stresses) and therefore always put elements in their &quot;strong&quot; configuration.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-01-29T13:56:13.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7162" PostTypeId="2" ParentId="7139" CreationDate="2016-01-29T14:39:54.757" Score="1" Body="&lt;p&gt;Determinancy in beams/frames is much like that in any other system of equations. If you can set up a set of equations for your beam/frame where the number of equations (what you know) is equal to the number of variables (unknowns) then your system is determinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a mismatch between number of equations and number of variables you have an over-defined (or under-defined) system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps, I'm afraid I'm not familiar with the $r=3N$ equation you mentioned.&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2016-01-29T14:39:54.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7163" PostTypeId="2" ParentId="7139" CreationDate="2016-01-29T14:40:16.693" Score="3" Body="&lt;p&gt;Whether a structure is determinate or indeterminate is just a description of how easy or hard it is to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 2D planar structure such as a beam or frame, there are only three global equations at your disposal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= 0 \\&#xA;\sum F_y &amp;amp;= 0 \\&#xA;\sum M_A &amp;amp;= 0&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are that the sum of horizontal and vertical forces are null, as is the sum of moments around an arbitrary point $A$. Now, if you have only three reactions, then you only have three unknowns. Since the number of unknowns is equal to the number of equations, then you can &quot;trivially&quot; solve for these reactions with just the equations of static equilibrium above. Therefore, statically-determinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you have more than three reactions, then you have more variables than equations. You can't solve the solution using just the static equilibrium equations. Therefore, statically indeterminate or hyperstatic. For these structures, you need to add more equations which define the compatibility of deformations along your structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a continuous beam under a uniform load $q$ with three supports: one at each end and one somewhere along the span (not at mid-span, which would allow cheating), as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1eGeZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1eGeZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much vertical force goes to each support? You have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= A_x + B_x + C_x = 0 \\&#xA;\sum F_y &amp;amp;= A_y + B_y + C_y - q(L_1+L_2) = 0 \\&#xA;\sum M_A &amp;amp;= B_yL_1 + C_y(L_1+L_2) - \frac{1}{2}q(L_1+L_2)^2 = 0&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can immediately afirm that $A_x = B_x = C_x = 0$. But now we have two equations for three unknowns ($A_y$, $B_y$ and $C_y$). We can use the $\sum F_y$ equation to find $A_y = q(L_1+L_2) - B_y - C_y$. But then we throw that into the $\sum M_A$ equation and get two variables we can't solve for. We're stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would need to observe that the rotation of segment $\overline{AB}$ at point $B$ is equal to the rotation of segment $\overline{BC}$ at point $B$. Armed with this information, we can create a new equation which describes the compatibility of deformations at point $B$. With this new equation, we now have as many equations as we have unknowns and can therefore solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we didn't have one of those supports (either $A$ or $B$ or $C$), then this problem would be trivial to solve. That is all that is meant by statically determinate or indeterminate. Is it determinate? Then it's simple to solve. Is it indeterminate? Then it's a headache.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the structure were three-dimensional, then there are six equations of static equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= 0 \\&#xA;\sum F_y &amp;amp;= 0 \\&#xA;\sum F_z &amp;amp;= 0 \\&#xA;\sum M_{A,x} &amp;amp;= 0 \\&#xA;\sum M_{A,y} &amp;amp;= 0 \\&#xA;\sum M_{A,z} &amp;amp;= 0 \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, sum of forces in $x$, $y$ and $z$ are null and moments around the $x$, $y$ and $z$ axes for an arbitrary point $A$ are null. So in this case, a statically determinate structure has equation $r = 6N$.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-04-20T13:57:54.423" LastActivityDate="2017-04-20T13:57:54.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7164" PostTypeId="2" ParentId="6012" CreationDate="2016-01-29T15:36:24.567" Score="0" Body="&lt;p&gt;You need to find the effective mass of a rotating rod at the point where the damping force is applied. This equals $$m_{eff} = \frac{I_{pivot}}{\ell^2} $$ where $I_{pivot}$ is the MMOI about the pivot, and $\ell$ is the distance where the force is applied to the pivot.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2016-01-29T15:36:24.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7165" PostTypeId="2" ParentId="3348" CreationDate="2016-01-29T15:39:55.150" Score="13" Body="&lt;p&gt;So my longer answer below assumes that the board will undergo acceleration and during this time you still need to be able to measure your pitch, roll and yaw within a short amount of time. If the board will be stationary for all measurements then Mahendra Gunawardena's answer will work perfectly for you. If this is going into a device like a segway or model plane or multirotor or anything that moves around, you may want to keep reading. This post is about how to use all three sensors though a method called sensor fusion. Sensor fusion allows you to get the strengths of each sensor and minimize the effects of each sensors weaknesses.&lt;/p&gt;&#xA;&lt;h1&gt;Sensor characteristics and background&lt;/h1&gt;&#xA;&lt;p&gt;First understand that an accelerometer measure all forces being applied to it, not just the force of gravity. So in a perfect world with the accelerometer in a stationary position without any vibrations you could perfectly determine which way is up using some basic trigonometry as shown by Mahendra Gunawardena's answer. However since an accelerometer will pick up all forces, any vibrations will result in noise. It should also be noted that if the board is accelerating you can not just use simple trigonometry as the force the accelerometer is reporting is not only the earths force of gravity but also the force that is causing you to accelerate.&lt;/p&gt;&#xA;&lt;p&gt;A magnetometer is more straightforward then an accelerometer. Movement will not cause problems with it but things like iron and other magnets will end up effecting your output. If the sources causing this interference are constant its not to hard to deal with but if these sources are not constant it will create tons of noise that is problematic to remove.&lt;/p&gt;&#xA;&lt;p&gt;Of the three sensors, the gyroscope is arguable the most reliable and they are normally very very good at measuring rotational speed. It is not affected by things like iron sources and accelerations have basically no impact on their ability to measure rotational speed. They do a very good job of reporting the speed at which the device is turning at, however since you are looking for an absolute angle you have to integrate the speed to get position. Doing this will add the error of the last measurement to the error of the new measurements since integration is basically a sum of values over a range, even if the error for one measurement is only 0.01 degrees per second off, in 100 measurements, your position can be off by 1 degree, by 1000 measurements, you can by off by 10 degrees. If you are taking hundreds of measurements a second, you can see this causes problems. This is commonly called gyro drift.&lt;/p&gt;&#xA;&lt;h1&gt;Sensor fusion&lt;/h1&gt;&#xA;&lt;p&gt;Now the beauty of having all of these sensors work together is that you can use the information from the accelerometer and magnetometer to cancel out gyro drift. This ends up allowing you to giving you the accuracy and speed of the gyro without the fatal flaw of gyro drift.&lt;/p&gt;&#xA;&lt;p&gt;Combining the data from these three sensors can be done in more then one way, I'll talk about using a complementary filter because its far simpler then a kalman filter and kalman filters will eat up much more resources on embedded systems. Often times a complementary filter is good enough, simpler to implement(assuming your not using a pre-built library) and lets you process the data faster.&lt;/p&gt;&#xA;&lt;p&gt;Now onto the process. The first steps you need to do is to integrate the gyroscope output to convert the angular speed into angular position. You will also most likely have to apply a low pass filter on the accelerometer and magnetometer to deal with noise in the output. A simple FIR filter like the one shown below works here. With some trigonometry you can find the pitch and roll with the accelerometer and the yaw with the magnetometer.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;filteredData = (1-weight)*filteredData + weight*newData&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;The weight is just a constant that can be adjusted depending on how much noise you have to deal with, the higher the noise is the smaller the weight value will be. Now combining the data from the sensors can be done by the following line of code.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;fusedData = (1-weight)*gyroData + weight*accelMagData&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;It should be noted that the data is a vector of the pitch, roll and yaw. You can just use three variables to do this as well instead of arrays if you want. For this calculation the gyro provides a position in degrees in pitch, roll and yaw, the magnetometer provides an angle for yaw while the accelerometer provides its own numbers for pitch and roll.&lt;/p&gt;&#xA;&lt;p&gt;If you still want more information you can google &amp;quot;sensor fusion with complementary filter&amp;quot; there are plenty of articles about this.&lt;/p&gt;&#xA;" OwnerUserId="4747" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-01-29T15:39:55.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7167" PostTypeId="1" CreationDate="2016-01-29T18:41:55.840" Score="4" ViewCount="790" Body="&lt;p&gt;I need to have an electrical probe machined; but I am unsure how to specify my requirements.  I am an electrical engineer with only moderate mechanical knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probe will be machined from 0.125&quot; diameter stainless steel round stock. The last 0.25 inch before the point necks down to a smaller diameter - maybe 0.031&quot; - 0.062&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is in how to specify the pointed end of the probe. I want it to be sharp enough to dig in a little to the circuit under test but not so sharp that the point is fragile.  Also, I don't want to specify overly-tight tolerances and push up the cost. I am assuming a Swiss screw type machine will be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is &quot;sharpness&quot;? Is it the angle leading to the point or is it the diameter at the very end?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial stab at it would be something like:&#xA;&quot;a cone with 60 degree angle sides&quot;.  But this seems to imply an infinitely small point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I specify a flat end of a certain diameter? (i.e. a conical frustum instead of a cone?)  &lt;/p&gt;&#xA;" OwnerUserId="4751" LastActivityDate="2016-01-29T21:37:34.670" Title="How to specify and tolerance a &quot;sharp&quot; point?" Tags="|mechanical-engineering|machining|tolerance|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7168" PostTypeId="2" ParentId="2353" CreationDate="2016-01-29T18:59:21.107" Score="0" Body="&lt;p&gt;The &quot;extrusions&quot; on the side that you ask about are there to hold a circuit board.  The board slides in vertically between the ridges. There are several locations to provide for multiple boards or a board in different positions. &lt;/p&gt;&#xA;" OwnerUserId="4751" LastActivityDate="2016-01-29T18:59:21.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7169" PostTypeId="2" ParentId="7167" CreationDate="2016-01-29T19:00:39.207" Score="1" Body="&lt;p&gt;Effective sharpness depends on both the angle of the cone and the geometry of the very edge/point. In this case it sounds like you aren't aiming for very deep penetration into the surface so the tip of the point itself is likely to be the key factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to consider the ways that a sharp point wears. In general this can either be that material from the point is worn away, chipped or that the point becomes bent. This is a combination of the mechanical properties of the steel and the geometry of the point. Most sharp tools are heat treated and the precise temper is a crucial variable in tandem with the geometry in determining performance as creating a close approximation of a geometrically 'ideal' point or edge relies on controlling a combination of abrasive wear and plastic deformation which is not well catered for by conventional tolerancing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice specifying the manufacturing process is probably more meaningful than geometric tolerances in the context of sharpness. Normally sharp edges are finished b y abrasive grinding and/or lapping and it may be the case that a manual finishing process makes the most sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the scale and overall geometry you are talking about there is an obvious parallel with sharpening the tungsten electrodes used in TIG welding and it may even be worth considering these as an easily available source of hard, small diameter rod, High speed steel and tungsten carbide are also possible contenders. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would probably also be worth considering having the larger diameter part as a separate holder with a collet or similar to hold a replaceable probe tip. This arrangement would certainly be easier to manufacture than turning a small diameter neck in stainless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example you might specify that the point is sharpened to say 60 degrees and finished on a 1200 grit abrasive wheel. Also bear in mind that if you just specify that you want the tip sharp then sharpening things is something that any machinist should immediately grasp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also be that you can find an off the shelf source of points as there are quite a few applications where these are used, carbide tipped scribers immediately spring to mind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also as general advice if you are designing something out of your specific expertise and experience it is always worth talking to whoever is going to make it and telling them what you actually want the part to achieve. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-01-29T19:12:39.993" LastActivityDate="2016-01-29T19:12:39.993" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7170" PostTypeId="1" CreationDate="2016-01-29T19:01:56.810" Score="1" ViewCount="320" Body="&lt;p&gt;Is there a mathematical method or reference source/handbook for computing the compressive forces necessary for thread rolling? I realize I could use FEM, but that is a lot of work, so I am hoping to avoid it if at all possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am designing a frame that will hold two parallel radial rolls and gearing to drive the rolls into the workpiece while rotating and I need to know the approximate forces involved so I can (1) dimension the frame to be strong and rigid enough, and (2) accommodate appropriate gearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables are the material being rolled (typically annealed alloy steel or stainless steel), the diameter of the material (expected to be between 1/4&quot; and 3/4&quot;) and the thread pitch. The type of thread is Acme.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not asking for an answer that computes the compressive force. I am looking for an answer that tells me how to do the computation myself, or at least explains the theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I can go to LMT Fette and they can just tell me off the bat, but I would rather not be going hat-in-hand to vendors every time I need to solve a problem.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="368" LastEditDate="2016-03-30T17:32:14.720" LastActivityDate="2017-02-24T00:00:55.003" Title="Compute compressive forces necessary for thread rolling" Tags="|mechanical-engineering|threads|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7171" PostTypeId="2" ParentId="7111" CreationDate="2016-01-29T19:13:18.490" Score="3" Body="&lt;p&gt;When the velocity of the air is less than about Mach 0.3, there is hardly any change in pressure or density as it impacts the surface.  For flows greater than $M=0.3$, you can calculate the stagnation pressure, density, and temperature with the following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{p_0}{p}=\left( 1+ \frac{\gamma - 1}{2}\cdot M^2 \right)^\frac{\gamma}{\gamma -1}&#xA;$$&#xA;$$&#xA;\frac{\rho_0}{\rho}=\left( 1+ \frac{\gamma - 1}{2}\cdot M^2 \right)^\frac{1}{\gamma -1}&#xA;$$&#xA;$$&#xA;\frac{T_0}{T}=1+ \frac{\gamma - 1}{2}\cdot M^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $M$ is the mach number and $\gamma$ is the heat capacity ratio.  The subscript $0$ is the stagnation value and no subscript is the free-stream value.  The fluid &quot;stagnates&quot; at the point of impact with the surface, at which point its kinetic energy transforms into thermal and pressure energy, hence the increase in pressure, density, temperature, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond the stagnation zone, its not so easy to calculate the flow field.  I ran a test case in ANSYS Fluent with an 80mm pipe shooting air at a solid wall 100mm away at a velocity of 10 m/s.  The Reynolds number is about 28,000 for these conditions.  I used the Reynolds stress model to calculate turbulence.  The pressure and density fields are mostly constant as should be expected for this velocity.  The velocity field is plotted in the image below.  You can see the air make several vortexes along the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MoLCT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MoLCT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this gives you some things to think about for your problem.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-01-29T19:13:18.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7172" PostTypeId="1" CreationDate="2016-01-29T19:15:27.463" Score="1" ViewCount="668" Body="&lt;p&gt;My question is pretty simple. The transfer function of an LTI system seems rather limited in use since the initial conditions must be zero. What are some practical engineering uses of it, and examples? I understand it's mathematical use, but fail to see where it's utility would really come in handy except in very simple mathematical models. &lt;/p&gt;&#xA;" OwnerUserId="4752" LastActivityDate="2016-01-30T07:57:19.190" Title="Practical examples of LTI transfer function" Tags="|control-engineering|transfer-function|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7173" PostTypeId="2" ParentId="7172" CreationDate="2016-01-29T20:11:00.367" Score="3" Body="&lt;p&gt;Why would a linear, time-invariant system require initial conditions to be zero? This is completely incorrect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A linear, time-invariant system is any system that is &lt;strong&gt;linear&lt;/strong&gt; (no state terms multiplying one another or themselves) and &lt;strong&gt;time-invariant&lt;/strong&gt;, meaning that the coefficients don't change with respect to time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple system would be an RLC circuit. Resistance in a circuit doesn't affect its capacitance or inductance, and the same is true for the other terms. The circuit is linear, and, provided you're not actively tuning the circuit (changing any of those values), the system is time-invariant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An RLC circuit &lt;em&gt;can&lt;/em&gt; be tuned by variable capacitors or variable inductors, and such a tunable circuit is commonly referred to as a &quot;radio&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR, you could have an RLC circuit that features a voltage proportional to, say, a shaft spinning. If you called the voltage &quot;back EMF&quot;, then this RLC circuit is referred to as a &quot;motor&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motor circuit is probably one of the most-used LTI systems, at least for instruction. You can do all kinds of control theory with motors (see also: robots), but heat transfer, fluid dynamics... a lot of systems can be adequately represented by LTI systems. Those systems that don't exactly meet the LTI criteria can typically be linearized and used as an LTI system. See also: Inverted pendulum control, a.k.a. the Segway or Hoverboard. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-01-29T20:11:00.367" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7174" PostTypeId="2" ParentId="7167" CreationDate="2016-01-29T21:24:39.673" Score="1" Body="&lt;p&gt;Comments:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Do not specify manufacturing methods; what method is used is best decided by the shop depending on what tools they have and what they want to do with the job; the number of parts ordered will have a big effect on their approach. In general, you should never try to tell a machine shop how to do their job, because they know MUCH more than you do about what is practical and efficient.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In this case the dimensions are non-critical, so do not specify them. The tapered end of the probe is probably fine 0.25±0.05&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the point, give them a sample of the material it needs to penetrate: &quot;Sufficiently sharp to penetrate sample material annexed.&quot; They will figure out what needs to be done based on that.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For this type of job, I would expect a swaging machine to be used, assuming the parts are being made in quantity. Of course, if you go to a shop that does not have a swaging machine, then they will use some other approach, or they will buy a swaging machine if the order is big enough. There is also a special kind of machine called a &quot;pointing machine&quot; which would be another possibility.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2016-01-29T21:37:34.670" LastActivityDate="2016-01-29T21:37:34.670" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7175" PostTypeId="2" ParentId="7170" CreationDate="2016-01-29T22:02:45.597" Score="2" Body="&lt;p&gt;If you have a vendor you work with, or even if you don't work with them but you're aware of their company, then I would ask them directly for the formulas/references you should use to spec the housing for their equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from working with sales application engineers that they are very eager to help you design your system if it means you're (telling them you're) buying their equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a degreed mechanical engineer, this is the route I would take. Ask the people that know. If they seem unwilling to help, politely inform them that you'll check with other manufacturers - suddenly people are tripping over themselves to help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On their &lt;a href=&quot;http://www.lmtfettetools.com/&quot; rel=&quot;nofollow&quot;&gt;home page&lt;/a&gt;, LMT Fette has a big graphic in the middle that says, &quot;Have a thread rolling application question? Ask our Experts here. &lt;a href=&quot;http://www.lmtfettetools.com/contact.html&quot; rel=&quot;nofollow&quot;&gt;Click here&lt;/a&gt;&quot; Sales application engineers exist to help you - use them!&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="368" LastEditDate="2016-03-30T17:29:22.430" LastActivityDate="2016-03-30T17:29:22.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7176" PostTypeId="2" ParentId="7151" CreationDate="2016-01-29T22:16:09.180" Score="-3" Body="&lt;p&gt;As long as the resistance of the ground path is less than the resistance of the body of the person touching the wire, the electricity will flow through the ground path. This will always be the case for any normal electrical ground.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2016-01-29T22:16:09.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7177" PostTypeId="2" ParentId="5701" CreationDate="2016-01-29T23:08:46.910" Score="0" Body="&lt;p&gt;Wow, tungsten, huh? Whenever I see exotic materials, usually a design mistake is being made. The other answer by starrise is wrong for various reasons. Counterbores are normally only used for making sockets, not boring blind holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you actually want to do this, the right strategy would be to set up the work piece in a four-jaw chuck on a lathe and rough the bore with a carbide jobbing drill mounted in the footstock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your next step will be to dress a cylindrical form in Aluminum Oxide J 40-60 grit using a rotary dresser. Mount the form in the tailstock of the lathe and then finish the hole. Clean the lathe carefully, because now it will have grit all over it. If your bore needs to have a good finish (not apparent that it does, given what you have written so far), you can repeat the process with a 200 grit form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next you need to get an extension mount for your tailstock and mount a smaller diameter carbide drill to rough the small bore. Once again use a grinding form to finish the small bore. Depending on the tolerance of the small bore you will probably need to redress the form because it will wear down as it grinds out the bore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously everything needs to be concentric all the time, so you have to be checking runout carefully on every step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this work can be also be done in a drill press (or mill) if you make a suitable workholding fixture. Using a drill press or mill your maximum precision will be 0.005&quot; and 0.002&quot; respectively. A lathe will be 0.0002&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2016-01-30T00:25:40.773" LastActivityDate="2016-01-30T00:25:40.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7179" PostTypeId="1" AcceptedAnswerId="7180" CreationDate="2016-01-30T01:27:48.890" Score="3" ViewCount="4796" Body="&lt;p&gt;I am prototyping a nested waveguide made of four cylindrical sections.  Each piece is turned on a lathe from Delrin (acetal) rod.  Each piece fits inside the next larger one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no moving parts.  It is an antenna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:  how much larger must the ID of a cylinder be, than the OD of the one which fits inside it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my drawing, I have indicated that they be 10 thousandths of an inch larger, but I do not know if this is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to be able to disassemble the cylinders during experiment, to take measurements, and then put them back together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each cylinder is to be electroformed with nickel on the outside, so the thickness of the metal layer will also be taken into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/snB0y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/snB0ym.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an updated drawing with tolerances added and revised diameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AWiiQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AWiiQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4755" LastEditorUserId="41" LastEditDate="2016-02-02T17:06:03.383" LastActivityDate="2016-02-02T17:06:03.383" Title="How much larger must the ID of a cylinder be than the OD for a Slip Fit? Material: Delrin (Acetal)" Tags="|mechanical-engineering|machining|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7180" PostTypeId="2" ParentId="7179" CreationDate="2016-01-30T03:48:24.523" Score="4" Body="&lt;p&gt;0.01&quot; is probably a decent place to start as a slip fit, though you can certainly go tighter. I might even question if you actually want a slip fit for this application. You say it's for an antenna, and if you want it to extend and stay extended on its own like other collapsible antennae, a slip fit probably won't accomplish that. You'll need something that creates some friction but not so much that it can't be overcome by a person pulling on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, what's potentially more important than the actual dimension is the tolerance. Those nominal dimensions will not be what you actually get, especially when you call out things down to the thousandth of an inch. For a slip fit, how I would actually dimension it is to make the nominal dimensions the same for the inner and outer, and then create a plus tolerance (e.g. +0.005&quot;/-0) on the OD and a minus tolerance on the ID. Using 5 thousandths on each end will get you a 10 thousandths slip at most. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you need to know that the place you're sourcing these parts from can match that tolerance. 0.005&quot; isn't that tough for a good machine shop, though I'm not sure how workable that plastic is, I'm more used to dealing with steels and cast irons. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2016-01-30T03:48:24.523" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7182" PostTypeId="2" ParentId="7172" CreationDate="2016-01-30T07:36:14.850" Score="2" Body="&lt;p&gt;Imagine hitting a pendulum by a hammer with same force and same direction. The pendulum's response will be always same, yesterday, today, tommorow, and 1 year after. It means, between your impact force by hammer and pendulum's respone, there will be an unchanged relation independent on time. Here, you can model this system as LTI system. Hammer impact and pendulum's response will be system input and output, respectively. And, the relation between input and output will be a transfer function.&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="4624" LastEditDate="2016-01-30T07:57:19.190" LastActivityDate="2016-01-30T07:57:19.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7183" PostTypeId="1" AcceptedAnswerId="7197" CreationDate="2016-01-30T08:24:05.997" Score="0" ViewCount="73" Body="&lt;p&gt;I have two bevel small (POM) gears that I'd like to test. I'm hoping to test their strength and their maximum torque. One is 15-tooth and is around 16MM wide and the other is 45-tooth and is around 46MM wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of a good, simple way for building a rig that hold these gears (and their 6MM shafts) in place in order for the test to take place? Would it be a good idea to just use 3 timbre boards and screw them together? Something like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GoAJN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GoAJN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the boards are held by screws. Would this suffice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="4302" LastEditDate="2016-01-30T08:36:09.167" LastActivityDate="2016-01-31T14:19:47.713" Title="Gear testing rig" Tags="|mechanical-engineering|gears|torque|product-testing|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7188" PostTypeId="2" ParentId="2190" CreationDate="2016-01-30T12:39:56.813" Score="-2" Body="&lt;p&gt;The only &quot;spring&quot; i know with constant force is the magspring, used in several industrial applications to overcome gravity.&lt;/p&gt;&#xA;" OwnerUserId="4763" LastActivityDate="2016-01-30T12:39:56.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7189" PostTypeId="1" AcceptedAnswerId="7204" CreationDate="2016-01-30T13:59:48.527" Score="1" ViewCount="166" Body="&lt;p&gt;I'm going to simulate an &lt;strong&gt;automotive radiator&lt;/strong&gt; . The method called $\varepsilon$- NTU , suggests following steps to calculate the heat rejection,$Q$, of a heat exchanger:&#xA;&lt;br/&gt;1-calculate $U$(overall heat transfer coefficient)&#xA;&lt;br/&gt;2- calculate $NTU$&#xA;&lt;br/&gt;3- calculate $\varepsilon$ and finally calculate $Q(W)$.&#xA;&lt;br/&gt;My question is about calculation of $U$ in automotive radiators. We can neglect fouling and wall resistances and write:&#xA;$$&#xA; \frac{1}{UA} = \frac{1}{\eta_aA_ah_a}+\frac{1}{A_wh_w}&#xA;$$&#xA;$U$: Overall heat transfer coefficient. &lt;br/&gt;$A$ : Air side or waterside area. &lt;br/&gt;$A_a$ : Air side heat transfer area&lt;br/&gt;$A_w$: Waterside heat transfer Area &lt;br/&gt;$h$ : heat transfer coefficient&lt;br/&gt; $\eta_a$ air side surface efficiency.&lt;br/&gt;I want to use &lt;em&gt;Shanoun &amp;amp; Webb&lt;/em&gt; formula to calculate the first term of the above equation(Air thermal resistance). They propose following equation for &lt;strong&gt;automotive radiators&lt;/strong&gt;:  $$ (\eta Ah)_a=(\eta_f Ah)_{l} +(\eta_f Ah)_{S1}+(\eta_f Ah)_{S2}+h_eA_e$$  You can find regions named &lt;em&gt;louvered(l)&lt;/em&gt;,&lt;em&gt;S1&lt;/em&gt;,&lt;em&gt;S2&lt;/em&gt; and &lt;em&gt;e&lt;/em&gt; in following image: &lt;a href=&quot;https://i.stack.imgur.com/BbHPm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BbHPm.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;  This is image of a fin between &lt;em&gt;2&lt;/em&gt; flat tubes.&lt;br/&gt;&lt;strong&gt;Question&lt;/strong&gt;:  &lt;em&gt;Shanoun and Webb&lt;/em&gt; just enter areas relating to one fin in their equation. So when we want to calculate $UA$, we account total surface of water$A_w$ but just a small surface of air side$A_a$. It makes the answer wrong. How can we use this formula to calculate $UA$ ?&lt;/p&gt;&#xA;" OwnerUserId="4574" LastEditorUserId="33" LastEditDate="2016-01-30T19:50:44.953" LastActivityDate="2016-01-31T21:25:04.930" Title="Simulation of Automotive Radiators" Tags="|fluid-mechanics|automotive-engineering|simulation|cooling|heat-exchanger|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7190" PostTypeId="1" CreationDate="2016-01-30T19:57:01.367" Score="2" ViewCount="76" Body="&lt;p&gt;I am not entirely sure if this is the correct forum to ask this question so please refer me on if there is a better arena. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am creating a small weather station and I wish to measure soil temperature. I am aware that digging a hole and then burying the sensor will alter the aeration of the soil and therefore give imprecise readings. I have therefore decided to integrate the sensor into a notch in a steel tube and then drive this tube into the ground. This means that there will be minimal disruption of the soil composition and the sensor will touch the soil both inside the tube and outside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concern is that the steel pole will retain heat differently to the soil (both from any showing to the sun above ground and more generally below the ground) and in doing so skew the temperature readings. Can you recommend a coating that I could put on the metal to reduce its conductivity/radiative properties and therefore minimise its impacts on the sensor?&lt;/p&gt;&#xA;" OwnerUserId="4768" LastActivityDate="2016-01-30T20:16:20.133" Title="What paint/spray on material should I use to insulate a metal pole?" Tags="|sensors|metals|soil|thermal-insulation|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7191" PostTypeId="2" ParentId="7190" CreationDate="2016-01-30T20:16:20.133" Score="1" Body="&lt;p&gt;Instead of trying to add insulating coating to the metal tube, I would try to get rid of the tube that protrudes above ground.  The tube would serve as an introduces for the sensor.  The tip with the sensor could be designed such that it stays on the tube as when you push, but separates from the tube when you pull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or try to interrupt the thermal conduction path to the sensor.  Insert a plastic section somewhere along the length of the tube.  The plastic part doesn't have to be thick.  It could be thinner than the tube diameter.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2016-01-30T20:16:20.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7192" PostTypeId="2" ParentId="7151" CreationDate="2016-01-30T21:02:52.080" Score="2" Body="&lt;p&gt;I'm no electrical engineer and my last training in safety around voltages is ages back. I will answer the question according to the wording it had in the first versions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Many people now practice &quot;earthing&quot;, physically linking themselves to the earth ground plug of their home's electrical system. They go as far as sleeping on mesh wire sheets plugged into ground to protect from stray EMF. [...] If a hot wire from, say, a washing machine, suddenly shorts to its chassis, doesn't that route to ground, and isn't our happy &quot;earther&quot; suddenly now part of a deadly circuit involving the hot wire?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, yes. As pointed out in &lt;a href=&quot;https://engineering.stackexchange.com/a/7159/61&quot;&gt;Olins answer&lt;/a&gt;, all or most devices with a conducting chassis have this grounded for safety reasons. In these cases, it &lt;em&gt;should&lt;/em&gt; make little difference if the person is grounded.However ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at another scenario: Your alarm clock is (not your cell phone but) an older model with an AC connection. You managed to damage the cable near where it enters the (plastic) chassis and touch the live wire when fumbling for the thing one early morning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1) You are not grounded in any special way, but lying on a cotton sheet on some mattress on a wooden bed. Resistance for electric current to the ground will be very high. Electrically, you are - grossly simplified - now a &lt;a href=&quot;https://en.wikipedia.org/wiki/Capacitor&quot; rel=&quot;nofollow noreferrer&quot;&gt;capacitor&lt;/a&gt; connected to an alternating current via a 1k$\Omega$ resistor (your arm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2) Same as before but you lie on a grounded mesh. This time you are a 1k$\Omega$ resistor between the live wire and ground. Far more current will flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High resistance towards grounding plays a minor role in safety around eletricity. The normal case is that a device is built safely as described. For work on electrical equipment, there's 5 basic rules: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;switch the power supply to the device you work on off&lt;/li&gt;&#xA;&lt;li&gt;secure it against switching on, &lt;/li&gt;&#xA;&lt;li&gt;confirm by measurement that there's indeed no voltage&lt;/li&gt;&#xA;&lt;li&gt;first ground then short circuit the live wires so if the device is switched on the circuit breaker will engage &lt;/li&gt;&#xA;&lt;li&gt;cover nearby devices, if still connected to power supply&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So you consider how not to touch a life wire in the first place, not about where the current will go when you do or what the resistance is. The only instance I know of where this was given serious consideration is safety shoes, these need a high resistance according to the relevant industry codes.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-01-30T21:02:52.080" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7193" PostTypeId="1" CreationDate="2016-01-30T21:49:53.107" Score="2" ViewCount="1130" Body="&lt;p&gt;How scalable is the wave &lt;a href=&quot;http://www.gizmag.com/wave-disc-generator-combustion-engine/19394/&quot; rel=&quot;nofollow&quot;&gt;disk generator engine&lt;/a&gt;?  Are there fundamental limits to how large or small the engine can be and if so, what are they?  I'm not looking for answers addressing microscale issues, but rather, I want to know if the fluid dynamics inside the engine require a certain size range in order for the combustion to work properly.&lt;/p&gt;&#xA;" OwnerUserId="398" LastActivityDate="2016-02-10T16:45:56.173" Title="How Scalable is the Wave Disk Generator Engine" Tags="|mechanical-engineering|automotive-engineering|engines|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7197" PostTypeId="2" ParentId="7183" CreationDate="2016-01-31T14:19:47.713" Score="3" Body="&lt;p&gt;Torque ratings for bevel POM (also called acetal or Delrin) gears are at &lt;a href=&quot;http://www.huco.com/products.asp?p=true&amp;amp;cat=285&quot; rel=&quot;nofollow&quot;&gt;http://www.huco.com/products.asp?p=true&amp;amp;cat=285&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't find timbre boards on google, I assume you meant timber wood boards. I don't think it is feasible with wood boards screwed together to get the necessary accuracy to holding the center lines of the gears to each other for valid test results. A precision machined gearbox would be ideal, but the cost would probably be prohibited. For practical purposes I think it is best to go with manufacture's torque ratings.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-01-31T14:19:47.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7198" PostTypeId="2" ParentId="2602" CreationDate="2016-01-31T14:25:59.427" Score="0" Body="&lt;p&gt;All the answers so far have taken the single-process deliberately planned approach.  The way this problem is solved in many real life applications is to use distributed processes each attempting to locally optimize but competing with each other.  As a result, the optimum distribution falls out almost as a bi-product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, raise the price until demand is reduced to equal the supply.  Even this system can be more distributed by using a method known as &quot;auction&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put yet another way, instead of the communist solution you seem to be asking about, you should also consider the capitalist solution.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-01-31T14:25:59.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7199" PostTypeId="2" ParentId="5663" CreationDate="2016-01-31T14:40:37.570" Score="4" Body="&lt;p&gt;The main reason is that the power requirement for running the torpedo is vastly larger than that for merely sending control signals to it.  This means &lt;i&gt;much&lt;/i&gt; thicker wire would be needed to actually power the torpedo over the wire.  Even if cost is not a consideration, a usefully long spool of such thick wire would be very large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can reduce the size of the wire by increasing the voltage, which thereby reduces the current at the same power.  However, that then increases the demands on insulation, which takes significant space.  Thin insulation, like enamel coated, can't stand off much voltage.  To make things even worse, this all has to work in the a conductive medium (seawater).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how much power a torpedo requires, but probably a few kW at least.  Let's say 5 kW just to pick something (if anyone knows the real number, please tell us).  That could be 10 A at 500 V, 50 A at 100 V, etc.  Even considering the heat sinking provided by the seawater, 50 A still requires a substantial wire, and 100 V insulation is going to double the thickness, which quadruples the volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to stop here since figuring out the volume of wire, volume and weight of copper, power loss, and voltage drop requires knowing too many specifics that I can only guess at.  However, it should still be obvious that this is going to be a lot worse than the torpedo spooling out double-stranded #30 wire with thin insulation behind it.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-01-31T14:40:37.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7204" PostTypeId="2" ParentId="7189" CreationDate="2016-01-31T21:25:04.930" Score="0" Body="&lt;p&gt;I should multiply the No. of fins to the right-hand-side statement. So we have:&#xA;$$(\eta hA)_a=No.of fins \times( (\eta_fhA)_l+(\eta_fhA)_{S1}+(\eta_fhA)_{S2}+h_eA_e)$$&lt;/p&gt;&#xA;" OwnerUserId="4574" LastActivityDate="2016-01-31T21:25:04.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7205" PostTypeId="2" ParentId="6639" CreationDate="2016-01-31T23:28:07.020" Score="0" Body="&lt;p&gt;Given that you're using extruded aluminium rails polymer rollers are likely to be the best solution as anything harder will likely make the wear rate of the aluminium more of an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest using extruded aluminium for a machine tool is always going to be a cheapest rather than best solution compared to ground steel  or cast iron.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-01-31T23:28:07.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7206" PostTypeId="1" CreationDate="2016-02-01T00:16:50.460" Score="0" ViewCount="238" Body="&lt;p&gt;Predict the diffusion coefficient for nitrogen through a mixture containing 50% hydrogen, 30% ammonia, 15% nitrogen, and 5% water at 25°C and 1 atm. This was my thought process on which equation to start with but do I assume I'm dealing with water and ammonia in a gas phase? Would this even be a correct equation to use?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions: Stagnant mixture&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D_{AM} = \frac {1-x_A}{\sum_{i=1}^n \frac{x_i}{D_{Ai}}} = \frac{1-x_A}{\frac{x_B}{D_{AB}}+\frac{x_C}{D_{AC}}+\frac{x_D}{D_{AD}}}$$&#xA;$A=\text{N}_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B=\text{H}_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C=\text{NH}_3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D=\text{H}_2\text{O}$&lt;/p&gt;&#xA;" OwnerUserId="4785" LastEditorUserId="1832" LastEditDate="2016-02-01T13:02:11.117" LastActivityDate="2016-04-01T14:08:28.760" Title="Mass Transfer Question" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7207" PostTypeId="2" ParentId="6385" CreationDate="2016-02-01T05:11:38.577" Score="3" Body="&lt;p&gt;The idea of using a feedforward component in a control system is to provide near &lt;strong&gt;instant action&lt;/strong&gt; of the plant output to the input command. So a suitable choice of transfer function which connects the command to the plant input directly is an inverse model of the plant. The feedback control can then serve to 'clean up' any error in the feedforward component perhaps due to modeling error or drift in the parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually the feedforward DC gain used is less than or equal to 1.0.&lt;/p&gt;&#xA;" OwnerUserId="820" LastEditorUserId="820" LastEditDate="2018-05-24T19:52:22.260" LastActivityDate="2018-05-24T19:52:22.260" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7208" PostTypeId="1" AcceptedAnswerId="7238" CreationDate="2016-02-01T07:25:25.200" Score="8" ViewCount="6862" Body="&lt;p&gt;I've read that it's not a good idea to rub two surfaces of the same material against each other, e.g., plastic against plastic, metal against metal. Plastic against metal is preferred. Why is this the case? Is it because that the coefficient of friction is higher for same materials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.tribology-abc.com/abc/cof.htm&quot;&gt;this site&lt;/a&gt;, steel-steel gives 0.5, plastic-plastic gives 0.3, but plastic-metal gives 0.25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose two rollers rotate and rub against each other. Will it be better for one to be made of metal and one of plastic?&lt;/p&gt;&#xA;" OwnerUserId="4302" LastActivityDate="2016-05-05T23:49:18.593" Title="Not a good idea to rub two of the same material against each other?" Tags="|mechanical-engineering|friction|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7209" PostTypeId="1" AcceptedAnswerId="16189" CreationDate="2016-02-01T09:38:51.200" Score="0" ViewCount="192" Body="&lt;p&gt;Can a screw such as the one below be used as an air engine in pressure range between 700 kPa and 2500 kPa &lt;br&gt;&#xA;on other words Can I reverse compression process with a screw to work as air engine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eC33J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eC33J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4790" LastEditorUserId="4790" LastEditDate="2016-02-02T20:31:36.673" LastActivityDate="2017-11-06T22:03:56.360" Title="Using compression screw as air engine" Tags="|mechanical-engineering|compressed-air|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7210" PostTypeId="1" CreationDate="2016-02-01T10:37:28.920" Score="7" ViewCount="485" Body="&lt;p&gt;I am interested in the kinetic energy T and the potential energy U of a three dimensional Timoshenko beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 2D Timoshenko beam the energies read:&#xA;$$ T = \dfrac{1}{2} \int\limits_o^l \left( \rho A \left( \dfrac{\partial w}{\partial t} \right)^2 + \rho I \left( \dfrac{\partial \Phi}{\partial t} \right)^2 \right) dx$$&#xA;$$ U = \dfrac{1}{2} \int\limits_o^l \left( EI \left( \dfrac{\partial \Phi}{\partial x} \right)^2 + kAG \left( \dfrac{\partial w}{\partial x} - \Phi \right)^2 \right) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in these formulas $w$ and $\Phi$ are just defined in the x-y-Plane. How can I derive the energies for bending in both direction (x-y-Plane and y-z-Plane) and torsion (around x-axis)?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: I figured out a part of the question by myself: Because Timoshenko beams are a linear theory, one can simply add the energies of two deformations and get the energies of the superposition of the deformations. Therefore:&#xA;$$ T = \dfrac{1}{2} \int\limits_o^l \left( \rho A \left( \dfrac{\partial v}{\partial t} \right)^2 + \rho A \left( \dfrac{\partial w}{\partial t} \right)^2 + \rho I \left( \dfrac{\partial \theta}{\partial t} \right)^2 + \rho I \left( \dfrac{\partial \Phi}{\partial t} \right)^2 \right) dx$$&#xA;$$ U = \dfrac{1}{2} \int\limits_o^l \left( EI \left( \dfrac{\partial \Phi}{\partial x} \right)^2 + EI \left( \dfrac{\partial \theta}{\partial x} \right)^2 + kAG \left( \dfrac{\partial v}{\partial x} - \theta \right)^2 + kAG \left( \dfrac{\partial w}{\partial x} - \Phi \right)^2 \right) dx$$&#xA;Where $w: \mathbb{R} \times \mathbb{R}^{+} \rightarrow \mathbb{R}$ is the deflection in the x-y-plane and $\Phi: \mathbb{R} \times \mathbb{R}^{+} \rightarrow \mathbb{R}$ is the slope in the x-y-plane. Respectively $v$ and $\theta$ for the y-z-plane. &lt;strong&gt;Can someone please confirm this?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still the energies for torsion are missing. Can the theory of Saint-Venant applied here?&lt;/p&gt;&#xA;" OwnerUserId="4792" LastEditorUserId="4792" LastEditDate="2016-02-02T08:46:14.550" LastActivityDate="2016-08-05T12:19:43.640" Title="Energy of a 3D Timoshenko beam" Tags="|mechanical-engineering|structural-engineering|beam|energy|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7212" PostTypeId="2" ParentId="7193" CreationDate="2016-02-01T13:06:36.313" Score="3" Body="&lt;p&gt;&quot;Let me say this about that&quot;  (extra points to those who recognize the originator of that phrase).&lt;br&gt;&#xA;First:  yeah, they've invented a cool-looking device.  But, they appear to be getting nowhere so far as either advanced funding or a fieldable model produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second:  I have seen at least 5 or 6 &quot;groundbreaking new powerful/efficient/wunderdevice&quot; internal combustion engine designs pop up over the last 30 years (not even counting the Wankel rotary).  Not one has been able to show both minimal life testing and improved efficiency over modern piston-crankshaft engines.   I see no reason to expect this design to do so either.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further, quite specific, counterarguments, take a look at  &lt;a href=&quot;http://newenergyandfuel.com/http:/newenergyandfuel/com/2013/09/04/arpa-e-backing-the-msu-wave-disk-engine/&quot; rel=&quot;nofollow&quot;&gt;this review&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-02-01T13:06:36.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7213" PostTypeId="2" ParentId="7206" CreationDate="2016-02-01T13:12:16.783" Score="1" Body="&lt;p&gt;One thing to keep in mind for mass transfer is to stay consistent when using x's and y's. You identified that this must be in the gas phase at STP, yet the equation you used has x's which represent the mole fraction of components in liquid phase. Mixing up these can get you in trouble later on with multiphase systems. Here is the equation for the effective diffusion coefficient of gas A with respect to the total gas mixture. Further explanation can be found &lt;a href=&quot;http://pubs.acs.org/doi/abs/10.1021/ie50483a022&quot; rel=&quot;nofollow&quot;&gt;here.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D'_{A} = \frac {1-y_A}{\sum_{i=1}^n \frac{y_i}{D_{Ai}}} = \frac{1-y_A}{\frac{y_B}{D_{AB}}+\frac{y_C}{D_{AC}}+\frac{y_D}{D_{AD}}}$$&#xA;$A=\text{N}_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$B=\text{H}_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C=\text{NH}_3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D=\text{H}_2\text{O}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your textbook might be a little different, but I don't like adding subscripts to denote mixtures. I noticed your equation has $D_{AM}$. I changed this to D prime of component A, $D'_A$. &lt;/p&gt;&#xA;" OwnerUserId="3569" LastActivityDate="2016-02-01T13:12:16.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7214" PostTypeId="1" CreationDate="2016-02-01T14:44:14.817" Score="5" ViewCount="414" Body="&lt;p&gt;&lt;img src=&quot;https://i1108.photobucket.com/albums/h407/robertbranham/Oops2%20cm_zpsfeu02sgh.jpg&quot; alt=&quot;Table Sketch&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am building a table with four legs which must support a dead weight of 250 kg, and possible peak dynamic load of 1000 kg (A 200 kg person climbs onto and off of it).  The legs come straight down from the box, rigidly fixed to an aluminum tubing frame.  They flare outward and end in free-rolling wheels.  Dimensions are as shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The application is very weight-sensitive.  Would legs made from 6061-T6 aluminum round tubing, 1.0&quot; OD x &quot;.093 wall be sufficient or should I bite the weight bullet and go with steel tubing?&lt;/p&gt;&#xA;" OwnerUserId="4796" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.287" LastActivityDate="2016-03-04T03:55:11.617" Title="Cantilever Table Legs" Tags="|materials|aluminum|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7215" PostTypeId="1" CreationDate="2016-02-01T15:43:08.183" Score="4" ViewCount="372" Body="&lt;p&gt;The ASTM dicates a plain rectangular shape.  I understand that the material necks pretty quickly but I frequently get breaks in the grips, particularly with films of thickness less than 0.002.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dog bone samples seem as though they would help, so why not use those?&lt;/p&gt;&#xA;" OwnerUserId="4697" LastActivityDate="2016-02-01T20:34:24.780" Title="Why does tensile testing in thin films (ASTM D882) require a non-dogboned sample?" Tags="|materials|product-testing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7216" PostTypeId="1" CreationDate="2016-02-01T15:57:17.113" Score="4" ViewCount="8760" Body="&lt;p&gt;This one's a bit of a doozy, so &lt;strong&gt;please feel free to comment on any one single part or parts to this problem.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The setup is simple--I have a container (an empty bottle) which I fill with a certain pressure of compressed air through a car tire valve assembly attached to the top of the bottle. On the bottom is an orifice made crudely by forcing a heated paperclip through the bottom of the bottle, small enough such that a thumbtack can fit snugly through it, effectively sealing the hole. The whole thing is attached to a small toy car; Removing the thumbtack releases a stream of air that acts as a thrust force. I'm running this trial with a 2 L bottle at room temperature, and for incremental pressures of 10 psi from 30 psi to 60 psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an assignment, I have to determine a few things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. THE NATURE OF THE FLOW:&lt;/strong&gt; My understanding is that adiabatic flow is different from isothermal flow is different from compressible/non-compressible flow, etc. Unfortunately, this is literally the first time that I've heard any of these terms, and while I researched and understand their definitions I cannot tell which of the following my setup is. Would anyone with a more professional of experience know which my setup might be, and why? This affects which of the equations I can use to find...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. MASS FLOW RATE &amp;amp; THRUST FORCE:&lt;/strong&gt; I did see this link on calculating the flow rate of air through a pressurized hole: &lt;a href=&quot;https://physics.stackexchange.com/questions/131032/calculate-flow-rate-of-air-through-a-pressurized-hole&quot;&gt;https://physics.stackexchange.com/questions/131032/calculate-flow-rate-of-air-through-a-pressurized-hole&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my understanding is that the Bernoulli equation described in the link above does not apply because if the ratio of outside pressure to bottle pressure is lower than 0.528, then the flow is choked. This, however, opened up another can of worms, in that I had trouble finding an equation that does work concisely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found quite a few, but I am the world's worst researcher and am probably missing something very helpful. &lt;strong&gt;I need to know the BEST one given my setup.&lt;/strong&gt; The ideal scenario would be a time-dependent one; I found &lt;a href=&quot;http://www.osti.gov/scitech/servlets/purl/786430&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; website (the equation is towards the bottom of the page) that split the mass flow rate into a time dependent equation that had a time-dependent pressure equation and time-dependent temperature equation as part of its components. I tried plugging values (pressure of 30 psi) into the pressure equation and received some pressure marginally different from 30 psi for a time of 10 seconds into the trial. This seemed to conflict with the fact that a lot of air sounded like it was coming out of the bottle in the first 10 seconds when I ran the trial. OR, if this equation is absolutely right, am I doing something wrong with plugging in my variables or failing to account for a certain variable or effect with my experimental setup? For plugging in, I used the temperature in absolute kelvin, the gas constant R = 287.058 J mol^-1 K^-1, gamma = 1.4, orifice area in m^2, volume of bottle in m^3, and pressure in psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Given the nature of my setup, what specific mass flow rate equation would apply best, and why? THERE ARE A LOT, and trying to look for the right one myself has gotten me nowhere the past few days. Note that my setup involves something that is time-dependent, as pressure, mass, velocity of the gas being expelled are changing over time as air is expelled.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. VELOCITY OF GAS THROUGH ORIFICE:&lt;/strong&gt; Again, I've had trouble researching this. I know that I can relate this exit velocity to mass flow rate, since:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass flow rate = (gas density)*(Area of orifice) *(exit velocity), so if I solve for the former three I can easily determine exit velocity. This would be made much easier, however, with an equation describing the escape velocity. This is, again, better as time-dependent and pressure-dependent, as changing pressure means changing gas density and changing exit velocity.&lt;/p&gt;&#xA;" OwnerUserId="4798" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2016-02-03T13:26:54.260" Title="Determining mass flow rate and exit velocity of compressible flow" Tags="|fluid-mechanics|compressible-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7218" PostTypeId="1" AcceptedAnswerId="7223" CreationDate="2016-02-01T18:08:36.813" Score="6" ViewCount="116" Body="&lt;p&gt;I'm looking at something called at a Steadicam that stabilizes cameras. It looks like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hj29C.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hj29C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the weights at the bottom are used for dampening/stabilizing. If the arm holding the weights were shorter/longer, would the stabilizing effect be affected? Would the centre of gravity change if the arm lengths were different (and thus affecting stability)?&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="33" LastEditDate="2016-02-02T04:06:52.120" LastActivityDate="2016-02-08T21:59:58.707" Title="How does the location of a mass affect its dampening performance?" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7219" PostTypeId="2" ParentId="7216" CreationDate="2016-02-01T19:05:53.920" Score="2" Body="&lt;p&gt;You are correct that the Bernoulli equation will not work here because you're in the compressible flow regime.  As you already learned, the flow through the orifice will become sonic when the pressure in the bottle exceeds about 30 psi.  What you have here is probably mostly adiabatic expansion of the air in the bottle (no heat transfer), however the problem will be easier to solve if you assume isothermal (constant temperature) expansion instead.  If isothermal doesn't produce accurate results, I can elaborate on how to do adiabatic instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for calculating the velocity of the gas:  I usually use the choked flow equation &lt;a href=&quot;https://en.wikipedia.org/wiki/Choked_flow&quot; rel=&quot;nofollow&quot;&gt;found on Wikipedia&lt;/a&gt; for this type of problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot m=C_dA\sqrt{k \rho_0 P_0 \left( \frac{2}{k+1} \right)^\left( \frac{k+1}{k-1} \right)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\dot m$ is the mass flow rate, $C_d$ is the discharge coefficient, $A$ is the cross-sectional area of the hole, $k$ is the heat capacity ratio (1.4 for air), $\rho_0$ is the upstream density and $P_0$ is the upstream pressure.  Pressure is known and density can be calculated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\rho_0 = \frac{P_0}{RT}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R$ is the &lt;em&gt;specific&lt;/em&gt; gas constant (not to be confused with the universal gas constant) and is equal to $287 \frac{J}{kg \cdot K}$ for air.  You can probably estimate the hole diameter by measuring the diameter of the thumb tack that fits in it.  That just leaves $C_d$ - from experience, a value of 0.7 should work for this problem.  You can adjust as necessary as you collect more empirical data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sub-sonic flow, I use the following equation, which I got from &lt;a href=&quot;http://oai.dtic.mil/oai/oai?verb=getRecord&amp;amp;metadataPrefix=html&amp;amp;identifier=AD0692866&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot m = A \cdot \sqrt{\frac{2g \gamma}{R(\gamma -1)}} \frac{P_0}{\sqrt{T_0}} \frac{\left[ \left( \frac{P_0}{P} \right)^\frac{\gamma-1}{\gamma}-1 \right]^\frac{1}{2}}{\left( \frac{P_0}{P} \right)^\frac{\gamma +1}{2\gamma}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_0$ and $T_0$ are the upstream pressure and temperature, $P$ is downstream pressure, and $g$ is the gravitational constant.  I used $\gamma$ for heat capacity ratio instead of $k$ in this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mentioned in your question, the pressure in the bottle is decreasing with time, which changes the rate of mass flow.  What you need to do is solve the choked flow equation over time.  It is a non-linear ordinary differential equation, but is fairly easy to solve numerically using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler_method&quot; rel=&quot;nofollow&quot;&gt;forward-Euler method&lt;/a&gt; and a spreadsheet program.  The general solution method is to calculate the pressure, then the rate of mass transfer, then total mass remaining for several discrete time steps.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First calculate the initial mass of air in the bottle using the ideal gas equation of state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;m=\frac{PV}{RT}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R$ is again the specific gas constant here.  You know the initial pressure; plug it in to the choked flow equation to get the rate of mass loss.  Now multiply this rate by a small time step, 0.1 second might be a good initial guess.  This gives you the approximate mass of air that escaped over the first 0.1 second after the stopper was removed.  Now subtract this mass off the initial mass that you calculated already.  Using the new mass, recalculate the pressure, recalculate the rate of mass transfer, calculate the new mass, and repeat until the pressure drops below 30 psi.  Below 30 psi, you're not at choked flow any more, so you should switch to the equation for subsonic flow.  Repeat the above until the pressure drops to atmospheric, or the length of time covers the time interval of interest.  This may sound complicated, but it's actually fairly easy to implement with a spreadsheet.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2016-02-03T13:05:24.940" LastActivityDate="2016-02-03T13:05:24.940" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7220" PostTypeId="2" ParentId="7215" CreationDate="2016-02-01T20:34:24.780" Score="2" Body="&lt;p&gt;ASTM D882 is for plastic films.  For thins films of many other materials, dog-bone specimens are more commonly used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deformation of dog-bone specimens is not homogeneous through the length of the full specimen.  That means that the cross-head displacement of the testing machine is not necessarily equal to the displacement of the gage region. As a result, you will need some sort of extensometer to measure the change in length of the specimen which makes the test a bit more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, it can be argued that the deformation is not homogeneous in a rectangular specimen either - which is why you run into breaks at the grips.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2016-02-01T20:34:24.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7221" PostTypeId="1" AcceptedAnswerId="7225" CreationDate="2016-02-02T02:35:47.273" Score="3" ViewCount="250" Body="&lt;p&gt;So I'm a high school student doing some competition (a wallet-sized one, unfortunately) and I'm supposed to make an electric car. Think an RC-car, (about that size) without the remote control. I want to make one from scratch, so I decided on using an arduino for a micro controller. Unfortunately, I don't really have an adviser that's particularly knowledgeable about this stuff, so I decided to ask here. (brace yourself for quite a few possibly dumb and vague questions. I do understand this question will likely be edited/closed for being too vague, but if, by any chance, someone has answers, I'd also like explanations -- mostly because I want to learn something.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've decided on using an aluminum chassis and soft wheels, since the competition wants a car that can go as fast as possible. I've chosen soft wheels because that should give me higher friction, and an aluminum chassis because I want to reduce weight. Now, here comes the first problem. The competition specifies a maximum of a 9v battery to power everything, including the arduino and the motors. Now, of course, we only need to go about 10 meters (I just want sub 3s times). My original design involved having 4 separate dc brushed/brushless motors, and then connecting each dc motor directly to the wheel. The dc motors could be held on with brackets, and then connected to an Arduino connected to a MOSFET, at which point I could then use a motor encoder to see how far to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now have a couple of questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Should I try having a 9v to 12v converter? or a 9v to 24v converter? That might cut the current significantly, but I'd also be able to use a much more powerful motor. The problem I have with this is that it might be adding a lot of potential weight -- as I understand it, I can't convert directly from DC 9v to something else.&lt;/li&gt;&#xA;&lt;li&gt;I've decided on having a rear wheel drive rather than a 4wd -- but is it worth getting a gearbox and an axle and just having one motor?&lt;/li&gt;&#xA;&lt;li&gt;Should I use a gearbox at all? Or should I try to change the speed based on the size of the wheel I use?&lt;/li&gt;&#xA;&lt;li&gt;Would I try to maximize RPM or torque by getting a geared-down brushless?&lt;/li&gt;&#xA;&lt;li&gt;If I have a 9v battery, can I have 2 9v motors in parallel? Or is that unsafe because 9v is only nominal?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If anyone has tips on how to improve the question, I'd be glad to incorporate them into my edits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;some info, by the way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;looking for sub3 second times&lt;/li&gt;&#xA;&lt;li&gt;using banebots 2-7/8 inch diameter wheels&lt;/li&gt;&#xA;&lt;li&gt;hard floor&lt;/li&gt;&#xA;&lt;li&gt;will need to back up after 9m&lt;/li&gt;&#xA;&lt;li&gt;only going straight, doesn't need to be able to turn&lt;/li&gt;&#xA;&lt;li&gt;thinking of a PID loop with a physical rotary quadrature encoder if one axle. thoughts?&lt;/li&gt;&#xA;&lt;li&gt;the car will be ~1kg and I won't be carrying anything else.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4805" LastEditorUserId="368" LastEditDate="2016-05-17T14:44:19.563" LastActivityDate="2016-05-17T14:44:19.563" Title="A couple of questions about making an electric hobby car from scratch" Tags="|mechanical-engineering|motors|electric-vehicles|" AnswerCount="1" CommentCount="1" ClosedDate="2016-02-02T14:47:55.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="7222" PostTypeId="1" AcceptedAnswerId="7484" CreationDate="2016-02-02T05:52:46.400" Score="7" ViewCount="1284" Body="&lt;p&gt;I am faced with a problem where I have to design a reinforced masonry beam for biaxial bending. The governing code is ACI 530-11. I cannot find a provision in this Code for biaxial bending. The only part that addresses this issue is in Section 2.2.3.1 which states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The unity formula can be extended when biaxial bending is present by replacing the bending stress quotients with the quotients of the calculated bending stress over the allowable bending stress for both axes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately this section deals with unreinforced masonry. It is very strange to me that reinforced masonry is not addressed.  Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4808" LastEditorUserId="33" LastEditDate="2016-02-17T20:07:19.337" LastActivityDate="2016-05-02T11:15:04.887" Title="How is biaxial bending considered for reinforced masonry design?" Tags="|structural-engineering|structures|building-design|masonry|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7223" PostTypeId="2" ParentId="7218" CreationDate="2016-02-02T09:18:09.593" Score="1" Body="&lt;p&gt;The device &lt;em&gt;seems&lt;/em&gt; to work like this: The weight attached to the arm always pulls down. You tilt the camera right, the weight moves left and up so it pulls against the tilt. Same for forward/backward tilt etc. The red knobs seem to be for adjusting where exactly the camera sits relative to the weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now to your question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the arm holding the weights were shorter/longer, would the stabilizing effect be affected? Would the centre of gravity change if the arm lengths were different (and thus affecting stability)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The  weight needs to be under the center of gravity of the grip+camera, else the stabilization would not work. The stabilization effect will be greater when the weight is lower (along a vertical line) as the weight will have a longer lever to pull back against tilt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my description is unclear, do a home experiment: Take a hammer at the grip so that the hammer head is on the side of your little finger and downward, see how easy or hard it is to tilt the hammer depending on where you grip the shaft. You may also want to walk around holding your camera and something about as heavy as the steadicam in one hand, to see if it's comfortable for you.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-02-02T09:18:09.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7225" PostTypeId="2" ParentId="7221" CreationDate="2016-02-02T11:29:41.997" Score="3" Body="&lt;p&gt;Answers to your specific questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No, voltage converters won't help, at best they will waste 10% of your power. Think about this : they involve transformers. two different windings on a magnetic core to translate one voltage to another. That might help you understand that any given motor (which is just a winding on a magnetic core to translate voltage into motion) can be tuned to run off any voltage by changing the winding. You'll see the same RC car motors with 13,15,17 turns to tune performance.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Instead, learn the relationship between voltage and speed (Kv, rpm/volt, the relationship is simply Speed = Kv * voltage) and current and torque (Kt, Newton-metres/amp). &#xA;A high speed motor (high Kv) gives little torque, and a low speed motor gives higher torque. So with direct driven wheels you want a LOW speed motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a mass (1kg), a distance (10m), and a time (3s). If you've done basic high school physics you know how to get speed and acceleration from distance and time. Work out 2 cases : linear acceleration from 0 to twice the average speed, and covering the distance at average speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll recommend you stick to metric, it makes some of the math easier. If your motor Kv is specified in rpm/volt, you have to convert that to radians/second/volt. In metric units, the torque relationship Kt = 1/Kv, so torque is Kt * current, or 1/Kv * current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about electric power, P = V * A and mechanical power = Nm/second = torque * rotation rate in radians/second, and the law of conservation of energy, you'll see why Kt = 1/Kv and understand how (high speed, low torque) and (low speed, high torque) can provide the same power.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now acceleration and mass give you force, and force * wheel radius gives you the torque required for acceleration.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also multiply mass * gravity(9.8N/kg) * coefficient of friction to get the force required to push the car at constant speed, and convert that to a torque. (You don't know the coefficient of friction : try 0.2 for a perfectly smooth floor and 0.6 for carpet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you know the torque required just for acceleration, and for travelling across the floor. The actual torque you need is the sum of these. (adding a safety factor like another 25% isn't a bad idea).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the speeds you need, in m/s, and a specific wheel radius, divide speed by radius to get rotational rate in radians/second. (Convert that to RPM if you want).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now that you know the speed (rad/sec or rpm) and torque (Nm or Newton metres) you can go back to selecting a motor. Can you find a motor with the right Kv, where 9V * Kv is anywhere close to your desired rotation speed? What current do you need to achieve the torque? If it's over 0.5A you'd better forget that little 9V battery, and that's being optimistic...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing not, and that's where you need a gearbox to translate from (high speed, low torque) to (low speed, high torque). So if your motor speeds are 10x too high, use a 10:1 gearbox, multiply your required speed by 10, and divide the required torque by 10. Can you meet your speed and current requirements now?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this may look daunting but each bit is relatively simple maths. If there are specific sticking points, think about them for a bit, and ask if you need to.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Having laid that groundwork, the rest of the questions should be easy...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you have 4 motors, each delivers 1/4 the torque and consumes 1/4 the current, so there's not much to choose between 1 motor and 4 in overall efficiency, but if one small motor can't supply the torque you need, 2 or 4 might work.&lt;/li&gt;&#xA;&lt;li&gt;The huge chunk of maths above will tell you if you need a gearbox. You probably do.&lt;/li&gt;&#xA;&lt;li&gt;You don't want to maximise RPM OR torque, you want just enough of both to cover the requirements and a bit (50% or so) to spare if possible. (I haven't mentioned inefficiencies like friction that eat away at performance little by little. This is long enough without that!). Brushless motors aren't a whole lot better at delivering torque, where they win is by running much faster, so you'd need to gear them down more. Probably not worth it here.&lt;/li&gt;&#xA;&lt;li&gt;You can run motors in parallel, but remember their currents add up, and your little battery can't supply much current. (&lt;a href=&quot;http://www.farnell.com/datasheets/2006032.pdf&quot; rel=&quot;nofollow&quot;&gt;Good batteries have datasheets&lt;/a&gt; which show how much current they can supply, for how long. In this case, 0.5A for much less than an hour, down to 4.8V)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="435" LastEditorUserId="435" LastEditDate="2016-02-02T12:02:46.110" LastActivityDate="2016-02-02T12:02:46.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7226" PostTypeId="1" CreationDate="2016-02-02T11:46:21.710" Score="0" ViewCount="483" Body="&lt;p&gt;I am using a nichrome wire for my project.&lt;br&gt;&#xA;Q.1. I want to know which source should I use as my voltage AC or DC? I know that the average power dissipated in a resistor in one cycle(in AC source) is equal to the average power dissipated in DC source. But I want to know which would be efficient for me ... I want to use this to cut fabric.&lt;br&gt;&#xA;Q.2 First of all the power dissipated in the circuit will be in the wire(which will act as a resistor). And also the wire will radiate heat energy(Stefans law depending upon emmisivity). So the temperature of the wire will increase for some time and then become constant. &lt;br&gt;&#xA;&lt;strong&gt;I want to know if there is energy dissipating in any other form?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 1: I figured out a little bit. Now please tell me if there is something to be added in place of something or not! &lt;strong&gt;But the problem is that e i.e. emissivity is varying. Then how to use the below formula. We are trying to find temperature but a term in the formula of temperature is itself dependent on temperature :(&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V^2/R=e\sigma AT^4 + mS + Something$$&lt;/p&gt;&#xA;" OwnerUserId="4776" LastEditorUserId="4776" LastEditDate="2016-02-04T12:15:40.813" LastActivityDate="2016-04-04T13:09:33.763" Title="Energy transfer in a hot wire cutter?" Tags="|thermodynamics|heating-systems|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7228" PostTypeId="1" CreationDate="2016-02-02T16:43:57.793" Score="4" ViewCount="4422" Body="&lt;p&gt;I'm making a dust extractor for a desktop &lt;a href=&quot;http://carbide3d.com/nomad/&quot; rel=&quot;nofollow noreferrer&quot;&gt;CNC milling machine&lt;/a&gt;.  My design process was very crude: I looked at some DIY dust-separating cyclones, and it seemed like they just work without any advanced calculations, so I visually copied some of those designs at a smaller scale, sat a large PC cooling fan over the top, and put the whole thing in a coffee can:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Oj5Yb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oj5Yb.jpg&quot; alt=&quot;Fan in a Can version 2&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Vv9sf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vv9sf.jpg&quot; alt=&quot;Fan in a Can v2 cutaway&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;My reasoning was that (1) the 18W fan blows out a lot of air, (2) the only place for that air to come from is through the hole at the top of the cone, and (3) the only way air can get into the sealed cone / dust receptacle (the bottom can) is through the tube at the side.  So, it would have to suck a reasonable amount of air through that red pipe.&lt;/p&gt;&#xA;&lt;p&gt;It does work to some extent (&lt;a href=&quot;https://vimeo.com/153922186&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt;), but not as well as I hoped.  You can see that it barely picks up sawdust through a 1-inch hose, and ideally I'd like it to draw in dust at a range of 20-50mm or so-- I can't position the inlet any closer than that to the point where the dust is created. Plus, I'd like to fit a dust filter over the exhaust hole, which will reduce the air flow even more.&lt;/p&gt;&#xA;&lt;p&gt;My question is basically &amp;quot;how can I make it suck harder?&amp;quot;.  But more specifically, how do I start to analyse what's happening, and what determines the amount of air pulled in?&lt;/p&gt;&#xA;&lt;p&gt;Most Google references are either way over my head, or talk very specifically about pumps in volute casings; obviously I removed the volute casing from my fan in order to fit it in the enclosure.  If I understand correctly, the point of the volute is to make the fast narrow stream exiting the fan into a slower, wider stream at higher pressure (why?), and I guess that something similar happens in my can:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WjatS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WjatS.jpg&quot; alt=&quot;Stages of air flow&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;A) incoming air spirals down; dust is thrown out and drops through bottom of cone&#xA;B) clean air spirals up through fan inlet&#xA;C) fast air exits fan circumference&#xA;D) ...and expands into volume below, losing speed and increasing pressure&#xA;E) higher-pressure air blown out into atmosphere&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Stage C/D is the part I'm most uncertain about.  The rest of the design comes from recipes, but I've changed the way air exits the fan and would like to understand how this might affect things.&lt;/p&gt;&#xA;&lt;h3&gt;Edit 4 Feb 2016:&lt;/h3&gt;&#xA;&lt;p&gt;Per @ericnutsch's suggestion, I made rough measurements of the air flowing into stages A and C, and got a figure of about 0.009 m&lt;sup&gt;3&lt;/sup&gt;min&lt;sup&gt;-1&lt;/sup&gt; at both points.  I also found the manufacturer's data for the fan, which suggests I should be able to do better-- the stated max is 1.09 m&lt;sup&gt;3&lt;/sup&gt;min&lt;sup&gt;-1&lt;/sup&gt;.  For comparison, my crappy handheld vacuum scores around 0.024 m&lt;sup&gt;3&lt;/sup&gt;min&lt;sup&gt;-1&lt;/sup&gt;.  I've ordered a pressure sensor to investigate further...&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JXY7X.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JXY7X.jpg&quot; alt=&quot;AVC BA10033B12U fan spec&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4815" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-05-03T00:55:55.057" Title="How to increase the air flow rate through a dust-separating cyclone?" Tags="|pumps|airflow|turbomachinery|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7229" PostTypeId="2" ParentId="7141" CreationDate="2016-02-02T19:13:25.140" Score="0" Body="&lt;p&gt;A beam reacts to bending moment by actually bending a bit in the direction of moment. This bending creates strain in cross section of the beam, tension at lower face, compression at top face. Strain is directly proportional to stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's for simplicity assume we have bend a beam that has 1 cm&lt;sup&gt;2&lt;/sup&gt; of material on top and bottom and a web that is 10 cm wide and the max stress this material is capable of resisting is 1 kg/cm&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maximum moment this beam can handle is 1 kg x 10/2 at the top and 1kg x 10/2 at the bottom, put together it can handle 1 kg x 10 cm = 10 kgcm momentum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we widen the beam so that the web is 20 cm wide but same amount of material the new beam again has 1 cm&lt;sup&gt;2&lt;/sup&gt; material on top and 1 cm&lt;sup&gt;2&lt;/sup&gt; material on the bottom. But this beam can handle 1 kg x 20 cm = 20 kgcm momentum. Turning the joist upside-up will effectively increase its web width!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2016-02-02T19:38:33.410" LastActivityDate="2016-02-02T19:38:33.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7230" PostTypeId="1" AcceptedAnswerId="7231" CreationDate="2016-02-02T20:33:49.490" Score="4" ViewCount="3055" Body="&lt;p&gt;I see that most worm gears are self-locking but I'm looking to make my own non-self-locking POM worm gear. It'll only be of module 1. Does anyone know what needs to be changed in order to make a self-locking worm gear non-locking? Is it just the angle of the threads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many motorised locks tend to have a worm gear on the motor shaft though - like &lt;a href=&quot;https://i.stack.imgur.com/DWkYT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;. The lock is designed to be manually turned as well. Does that mean the friction caused by manual turning will be high?&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="4302" LastEditDate="2016-02-03T16:27:55.760" LastActivityDate="2016-02-03T19:21:49.470" Title="Designing non-self-locking worm gear" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7231" PostTypeId="2" ParentId="7230" CreationDate="2016-02-02T22:00:23.390" Score="4" Body="&lt;p&gt;Whether or not a worm gear is self locking depends on the lead angle and the coefficient of friction between worm and wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will self lock when the coefficient of friction is greater than the arctan of the lead angle (from memory plus quick google so not guaranteed but if you are designing it yourself I'm assuming that you can take the time to look it up in Machinery's Handbook). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this is not an absolute as varying conditions and vibration can change this in practice. Also for high ratios it is likely to be self locking whether you like it or not. Similarly it is not a sharp cutoff and backdriving a worm gear will result in very high friction so there are very limited circumstances where it is something that you would want. It is certainly not a practical way of getting an equivalent drive in forward and reverse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the lock, I can't see exactly how it works but bear in mind that the whole point of a lock is to couple or decouple a mechanical linkage when you insert a key. In this case I suspect the key turns the cylinder and the motor turns the whole barrel. In this case the worm gear must be self locking or it wouldn't be a lock.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-03T19:21:49.470" LastActivityDate="2016-02-03T19:21:49.470" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7232" PostTypeId="2" ParentId="7226" CreationDate="2016-02-02T23:32:26.463" Score="1" Body="&lt;p&gt;Most of the cutting wire's heat loss will be through conduction: into the air, into the mounting hardware, and especially into the material being cut.  Melting the material is an endothermic process, meaning it will lower the temperature of your wire more than heat diffusion alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these heat losses (plus the radiated heat) will increase with the temperature of the wire, though it won't be a simple linear relationship. The wire's temperature will rise to an equilibrium value where the rate of heat loss is equal to the power supplied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the wire is a resistor, the power $P$ will be determined by $P=V^2/R$, where $V$ is the DC voltage or RMS AC voltage, and $R$ is the resistance of the circuit.  Assuming the cutting wire has a much higher resistance than the wires supplying it-- which it should-- then $R$ is just the resistance of the cutting wire, and is inversely proportional to its cross-sectional area.  Putting that all together, loosely speaking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{eq}\propto{P}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V=IR$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=VI=\frac{V^2}{R}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R\propto \frac{1}{d^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_{eq}\propto {V^2}{d^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $T_{eq}$ is the equilibrium temperature, $d$ is the wire diameter, and there is some constant of proportionality that would be too hard to work out. In practice you just pick a thin wire and experiment to find the lowest voltage that gets it hot enough to cut your material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't make any difference whether you use AC or DC, except that of course the specific choice of power supply affects the overall efficiency.  But just to make this explicit, &lt;strong&gt;do not run mains AC through your cutting wire!&lt;/strong&gt;  You don't need anything like that much voltage or current, and having a bare wire connected to a live wall outlet is extremely dangerous.  And it'd melt your cutting wire anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, the reason some hot wire cutters use AC is mainly that something like a 30V AC power supply is simpler and cheaper than a comparable DC supply, though I suspect this was more an issue in the olden days.&lt;/p&gt;&#xA;" OwnerUserId="4815" LastActivityDate="2016-02-02T23:32:26.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7233" PostTypeId="2" ParentId="6802" CreationDate="2016-02-02T23:54:29.203" Score="1" Body="&lt;p&gt;I'm inferring that you're in the UK and this is a real design rather than some theoretic statics exercise.  I've not come across pre-slabs before.  They're not permanent form work are they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately your  solution is somewhat trivial  :-(   Top slab design can quite involved, and the load analysis certainly is.  There's Eurocode 2, BS5400 and it's offshoot, BD44.  These will give you the detailed processes to go through.  If these pre-slabs are meant to be structural I suggest referring to the codes for how they are to be incorporated into the load bearing design.  Those bureaucrats love to include this kind of stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget that the top surface of the slab passing over the beams might be in compression, so you'll need bottom steel to resist the moments.  In this situation a pre-slab would have no effect and have to be ignored.  This all depends on the type of analysis you perform, and agreement with the client for your said methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaking from ignorance vis a vis pre-slabs, I'm still surprised that you only need to consider the wet weight of concrete for it's steel design.  This is negligible.  You also need to consider construction loads, and I would have expected that there would be a standard loading applied to allow for this.  Some paddy is going to want to drive a cherry picker onto the un poured deck.  Remember that bridges that fall down, usually fall down during construction when loads are in flux and difficult to quantify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bending is only one design criterion affecting the amount of steel.  There will also be flexural shear, punching shear, cracking, fire resistance and minimum bar spacings.  It might take you a week to design the slab, excluding the load analysis.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2016-02-02T23:54:29.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7235" PostTypeId="2" ParentId="7230" CreationDate="2016-02-03T00:45:15.840" Score="3" Body="&lt;p&gt;Like Chris Johns mentioned, locking is dependent on lead angle and a coefficient of friction; specifically a coefficient of static friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a worm gear design to back drive you can increase the lead angle or decrease the friction coefficient, both of which increases the ratio of torque to friction. It will not be efficient, and even though it is spinning it may not transfer much load before locking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many &quot;locking&quot; worm drives can back drive when they are already in motion when a force was applied. This is because the dynamic friction coefficient is less than the static friction coefficient. This is why many safety dependent systems also have brakes since the &quot;locking&quot; characteristic of the worm drive is not a guarantee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to run a worm drive the other direction with reasonable efficiency you may want to look at a &lt;a href=&quot;https://www.youtube.com/watch?v=hvz9A0c_HeQ&quot; rel=&quot;nofollow&quot;&gt;ball wormdrive&lt;/a&gt; similar to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Ball_screw&quot; rel=&quot;nofollow&quot;&gt;ball screw&lt;/a&gt; design. The ball bearings have rolling resistance which is much lower than sliding friction. Probably too expensive for most projects though.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-02-03T00:45:15.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7236" PostTypeId="2" ParentId="7228" CreationDate="2016-02-03T01:02:11.650" Score="1" Body="&lt;p&gt;If your cyclone design is correctly separating particles without a major pressure drop; it is working satisfactory and its influence on the flow rate is minimal. If you increase the capacity of your fan and separation starts failing, you may have to modify your design or scale it up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not have a high enough flow rate, you need to get a larger fan. In addition to this you need to make sure you land at a good location on that fans &quot;fan curve&quot;. A fan curve is the relationship between pressure and flowrate. Many fans are rated in maximum pressure (suction is different, but related if you cant get that data) and maximum flow rate. Maximum flow rate occurs at zero pressure drop and maximum pressure occurs at zero flow rate. So in your design you will need to do some testing to make sure the fan you select lands somewhere nicely in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some fans that may work well for the application:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.zoro.com/dayton-psc-blowers/g/00011669/&quot; rel=&quot;nofollow&quot;&gt;Dayton Blowers&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.zoro.com/search?q=motorized%20impeller&quot; rel=&quot;nofollow&quot;&gt;Motorized Impellers&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-02-03T01:02:11.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7237" PostTypeId="2" ParentId="7214" CreationDate="2016-02-03T03:23:30.583" Score="2" Body="&lt;p&gt;Steel or aluminum can be made equally strong. Generally aluminum can more easily be designed to have a higher strength to weight ratio because it has the a larger &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_area_moments_of_inertia&quot; rel=&quot;nofollow&quot;&gt;area moment of inertia&lt;/a&gt; for a given mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like Olin Lanthrop mentioned in the comments, the weakest link will be your connection between the table top and the legs. You can't count on the friction of the surface to hold the legs from moving outward(even in directions the wheels may not roll). If you add cross bracing (steel cables, aluminum rod, etc) your design will be lighter overall. Even a single 1/8&quot; rod between wheels on the bottom would increase the strength considerably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary failure modes that will need to be designed for are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Buckling of the legs, hard to calculate that shape without FEA. If design fails in testing; increase wall thickness or add cross bracing (think triangles).&lt;/li&gt;&#xA;&lt;li&gt;Joint failure between the legs and the table top. Make sure that this joint exceeds the strength of the legs and table top; especially if you go without cross bracing.&lt;/li&gt;&#xA;&lt;li&gt;Buckling of the table top. Probably the easiest to calculate. Make sure to include the moments added by the legs that are cantilevered.&lt;/li&gt;&#xA;&lt;li&gt;Deflection beyond limits. You may find that even though a given design is strong enough it may not hold the tolerances you wish. Under load this table will bow in the middle. The permissible deflection for a given load may require thicker members or more structure.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-02-03T03:23:30.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7238" PostTypeId="2" ParentId="7208" CreationDate="2016-02-03T04:15:37.170" Score="5" Body="&lt;p&gt;Friction is probably one of the primary reasons; as friction also correlates to wear. Friction is much more complicated than the equation we learned back in physics. Published friction coefficients are very much ball park figures because the friction greatly depends on the micro surface texture. This &lt;a href=&quot;http://www.school-for-champions.com/science/friction_causes.htm#.VrF3GGUrLCI&quot; rel=&quot;nofollow&quot;&gt;causes of friction link&lt;/a&gt; does a good job explaining some of the mechanics involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the main factors influencing lower friction for dislike materials is that there is a lower likeliness that there will be adhesion between them (atomically and macroscopically). A materials professor I had once lectured that if you were to polish two pure copper surfaces perfectly smooth in a vacuum and touch the surfaces together they would immediately become one piece of metal (&lt;a href=&quot;https://en.wikipedia.org/wiki/Cold_welding&quot; rel=&quot;nofollow&quot;&gt;cold welding&lt;/a&gt;). Similar adhesion occurs for plastic; perhaps even more so in the standard environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another consideration is that plastic on plastic can not dissipate heat nearly as fast as metal on plastic, which could lead to additional adhesion wear or failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes the erosion of softer material is used to &quot;dry&quot; lubricate a surface such as oil impregnated brass bushings or graphite motor brushes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the harder material can stay smooth and true, reducing the friction and wear of the system in the long term. And like welf mentioned in the comments, the softer surface is usually a consumable (bushing, brush, etc) that is more easily replaceable.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2016-05-05T23:49:18.593" LastActivityDate="2016-05-05T23:49:18.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7239" PostTypeId="1" CreationDate="2016-02-03T05:56:18.997" Score="2" ViewCount="1253" Body="&lt;p&gt;I am considering moving from abroad to the US, and hoping to obtain an engineering license for employment. I have a bachelors and graduate degree in physics, but I have heard that many states require a bachelors in an ABET accredited program to become licensed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which states do NOT require an engineering bachelors? If they don't require this, do they have some other requirement instead?&lt;/p&gt;&#xA;" OwnerUserId="4819" LastActivityDate="2018-06-04T09:31:24.240" Title="Which US states require a bachelors in Engineering to become licensed?" Tags="|licensure|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7240" PostTypeId="1" CreationDate="2016-02-03T06:56:11.733" Score="1" ViewCount="326" Body="&lt;p&gt;I'm looking at a P&amp;amp;ID of a heating circuit.  The burner/boiler is non-modulating. Red is flow/hot, blue is return/cold. AFAIK the pump is fixed speed (no &lt;a href=&quot;https://en.wikipedia.org/wiki/Variable-frequency_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;FI&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WvKCg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WvKCg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand the control logic, the temperature sensor on the return line (or rather the control circuit it's a part of) will open the three way valve and mix hot water into the return, and close when the return flow temp is warm enough. The boiler delivers a constant $\Delta T$, controlling for return temperature ensures that the flow temperature is correct. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this control scheme likely?&lt;/li&gt;&#xA;&lt;li&gt;Why not control for flow temp. directly?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-02-03T10:42:27.540" LastActivityDate="2016-02-03T11:55:44.830" Title="Heating circuit - help me understand control scheme of return temperature" Tags="|control-engineering|heating-systems|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7242" PostTypeId="2" ParentId="7240" CreationDate="2016-02-03T11:02:49.253" Score="1" Body="&lt;p&gt;Typically a three way valve is used in these systems to maintain minimum flow through the boiler. There are many ways of configuring such a system but the objective is to maintain flow through the boiler rather than allowing it to cook itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the return temperature to modulate a three way valve to either direct more flow to the load (on high return temperature) or more flow through the boiler (on low return temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the boiler will also need to be switched off on high inlet temperature to prevent damage.&lt;/p&gt;&#xA;" OwnerUserId="4210" LastEditorUserId="4210" LastEditDate="2016-02-03T11:55:44.830" LastActivityDate="2016-02-03T11:55:44.830" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7244" PostTypeId="2" ParentId="7239" CreationDate="2016-02-03T11:13:52.640" Score="2" Body="&lt;p&gt;As you have noted each state in the USA has different requirements. The general requirements are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Obtain a Bachelor's in Engineering (BSE) from an ABET accredited engineering program. The disciplines that I am familiar are Electrical, Civil, and Mechanical. I suggest you review the references below for more detailed information. &lt;/li&gt;&#xA;&lt;li&gt;Successfully complete the Fundamentals in Engineering exam to obtain Engineering in Training (EIT) status.&lt;/li&gt;&#xA;&lt;li&gt;Gain professional experience (typically four years minimum).&lt;/li&gt;&#xA;&lt;li&gt;Successfully complete the Principles and Practice of Engineering (PE) exam.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Considering that you have bachelor's and graduate degrees in physics, you have a few options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Enroll in a ABET accredited engineering program to obtain a Bachelor's in Engineering. With your physics background you are most likely to get a significant amount of credit. It is likely that you will be able to obtain a Bachelor's in Engineering in about 2 year depending on the University.&lt;/li&gt;&#xA;&lt;li&gt;Submit your credential to the licensing authority; hopefully they will accept your credentials and allow you sit for the EIT exam. You will have to do more research on this topic. I suggest that you review the &lt;a href=&quot;http://www.ieagreements.org/accords/washington/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Washington Accord&lt;/strong&gt;&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;If you are already a licensed, practicing engineer, submit those credentials to the appropriate licensing authority. They may accept your current credentials. I have submitted my USA credentials to &lt;a href=&quot;https://www.engineersaustralia.org.au/national-engineering-register&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineers Australia&lt;/a&gt; and obtained registration in Australia for an Australia-based engineering project. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;References&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ieagreements.org/accords/washington/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Washington Accord&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Regulation_and_licensure_in_engineering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Regulation and licensure in engineering&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/5985/110&quot;&gt;What is the sequence of exams to become a Professional Engineer in California?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nspe.org/resources/licensure/how-get-licensed&quot; rel=&quot;nofollow noreferrer&quot;&gt;How To Get Licensed (National society of Professional Engineers)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-06-04T09:31:24.240" LastActivityDate="2018-06-04T09:31:24.240" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7245" PostTypeId="1" AcceptedAnswerId="7248" CreationDate="2016-02-03T13:24:48.743" Score="20" ViewCount="33865" Body="&lt;p&gt;When iron is melted, I guess it has to be transported and contained. I think the container in which it is has to be able to withstand higher temperatures than what you want to melt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.engineeringtoolbox.com/melting-temperature-metals-d_860.html&quot;&gt;this webiste&lt;/a&gt;, &quot;Iron, Wrought&quot; has a melting temperature of 1482 - 1593&amp;nbsp;°C. There are a couple of other metals which have higher melting points (e.g. Wolfram (tungsten) with over 3400&amp;nbsp;°C), but all I can think of are much more expensive. So what material is the oven / &quot;bottle&quot; / &quot;basin&quot; (or however you call it) made of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Side question: Iron has been melted for quite a while now. I guess this has changed over the years. Of which materials was it before?)&lt;/p&gt;&#xA;" OwnerUserId="3061" LastEditorUserId="33" LastEditDate="2016-02-03T20:36:39.643" LastActivityDate="2019-02-08T16:22:34.857" Title="What material is used to hold molten iron in a furnace?" Tags="|metallurgy|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7246" PostTypeId="2" ParentId="7141" CreationDate="2016-02-03T13:48:28.417" Score="0" Body="&lt;p&gt;Why do we orient floor joists in with the longest side vertically? as in the shape of the letter &quot;I&quot; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AS mentioned in other posts, the name of the theory is &quot;mass moment of inertia&quot; of which there are whole textbooks written to explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, quite simply, the formula for calculated this moment of inertia (for a rectangle like a 2x10 joist) is (bh^3)/12 or in English &quot;B&quot; x &quot;H Cubed&quot; over 12. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;when the joist is oriented this way, its longer side (10&quot;) is oriented as the &quot;H&quot; (and the base is 2). Therefore when oriented this way, the wood resists loads much heavier than if it was &quot;laid flat&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to bend a plastic ruler different ways. There is one way which will be very difficult. This is explained by moment of inertia!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I beams are so slim in the middle is that they are able to retain their strength in this orientation, but to save on weight, they are minimal in design (because the weight of the beam also adds to the forces that need to be calculated) So why would you make the beam any heavier than it needs to be if it is  just as strong without the extra weight?&lt;/p&gt;&#xA;" OwnerUserId="4824" LastActivityDate="2016-02-03T13:48:28.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7247" PostTypeId="2" ParentId="7245" CreationDate="2016-02-03T14:08:31.320" Score="5" Body="&lt;p&gt;As &lt;a href=&quot;https://engineering.stackexchange.com/users/435/brian-drummond&quot;&gt;Brian Drummond&lt;/a&gt; noted, the &quot;basin&quot; is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Crucible&quot; rel=&quot;nofollow noreferrer&quot;&gt;crucible&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A crucible is a container that can withstand very high temperatures and is used for metal, glass, and pigment production as well as a number of modern laboratory processes. While crucibles historically were usually made from clay, they can be made from any material that withstands temperatures high enough to melt or otherwise alter its contents.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A detailed answer to the question can be found in the linked Wikipedia article. The short answers are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Iron age: clay&lt;/li&gt;&#xA;&lt;li&gt;Medieval period: introduction of new tempering material for the ceramic crucibles (&lt;a href=&quot;https://en.wikipedia.org/wiki/Mullite&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mullit&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Post Medieval: graphite&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3061" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-02-03T14:08:31.320" CommentCount="0" CommunityOwnedDate="2016-02-03T14:08:31.320" ContentLicense="CC BY-SA 3.0" />
  <row Id="7248" PostTypeId="2" ParentId="7245" CreationDate="2016-02-03T15:29:01.217" Score="22" Body="&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Crucibles are lined with refractory materials. Steel processing makes use of graphite or a combination of chromite and magnesite for direct contact with the melt. Cast iron processing often uses engineered clays, also known as alumina-magnesia-silica mixtures. Graphite is harder to form than clay-type refractories. To be suitable as a refractory, a material must meet a number of property requirements to be both economical and safe.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Refractory Materials&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;As you noted, iron has a high-end melting point of about 1,540 °C on the far left side of the $\textrm{Fe-C}$ phase diagram below, in the form of pure iron. There are two categories of materials with higher melting points, but only a few of those materials are both economical and safe. Generally, any material with a melting point high enough to withstand the melting points of commercially used metals such as iron, copper, and aluminum are called &lt;strong&gt;refractory materials&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bBlX0.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bBlX0.jpg&quot; alt=&quot;Fe-C phase diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source: &lt;a href=&quot;http://ispatguru.com/the-iron-carbon-phase-diagram/&quot; rel=&quot;noreferrer&quot;&gt;ispatguru.com&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Refractory Metals (Not Useful for Foundries)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The first category of high melting point materials, of which you noted one material, are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Refractory_metals&quot; rel=&quot;noreferrer&quot;&gt;refractory metals&lt;/a&gt;. Note that these are not generally referred to as refractories or refractory materials in the foundry industry. They consist of niobium, molybdenum, tungsten, tantalum, and rhenium, (Nb, Mo, W, Ta, Re) and have melting points ranging from about 2,500 °C to 3,500 °C. While the melting points are high enough and they have sufficient strength as structural materials, and some impact toughness to boot, there are a number of factors limiting their use.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;High reactivity with oxygen&lt;/li&gt;&#xA;&lt;li&gt;High reactivity with other metals&lt;/li&gt;&#xA;&lt;li&gt;High cost per weight&lt;/li&gt;&#xA;&lt;li&gt;High density&lt;/li&gt;&#xA;&lt;li&gt;High heat capacity&lt;/li&gt;&#xA;&lt;li&gt;High heat conduction&lt;/li&gt;&#xA;&lt;li&gt;Difficult to shape (requires either carefully controlled melting in a vacuum or &lt;a href=&quot;https://en.wikipedia.org/wiki/Powder_metallurgy&quot; rel=&quot;noreferrer&quot;&gt;powder metallurgy&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Refractory Ceramics (Useful for Foundries)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The second category of refractory materials are based on a variety of ceramics, and are called &lt;strong&gt;refractory ceramics&lt;/strong&gt;, or more commonly just &lt;strong&gt;refractories&lt;/strong&gt;. However, not just any ceramic is suitable. Ideally the ceramic would have extremely high atomic bond strength, or higher affinity for oxygen than the metal being melted. These would make the material relatively inert with respect to the molten metal. Such a ceramic must also be easily formable, have low heat capacity and heat conduction, and should be reasonably inexpensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphite is a reasonable choice for direct contact with steel and aluminum as the carbon-carbon bond strength is very high and it has a sufficiently high melting temperature, higher than its decomposition temperature in atmosphere. Graphite is somewhat more expensive to form than alternatives, though the crucibles tend to last longer. Graphite crucibles are high strength, though as with all ceramics low impact toughness. It has low density and lower heat capacity and heat conductivity than the refractory metals. Magnesite $\left(\textrm{MgCO}_3\right)$ and chromite $\left(\textrm{FeCr}_2\textrm{O}_4\right)$ are also common steel refractories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other choices are those systems below the $\textrm{Fe} + \textrm{O}_2 \rightleftharpoons \textrm{FeO}_2$ line on the Ellingham diagram below.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Chromia $\left(\textrm{Cr}_2\textrm{O}_3\right)$ can be used for some materials, but has high density and high heat capacity, as well as somewhat higher cost.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Silica $\left(\textrm{SiO}_2\right)$ is suitable for metals and alloys with lower melting temperatures, but has low thermal shock resistance. Pure silica (fused silica) has much higher thermal shock resistance but is very expensive and difficult to form. It is used in telescope mirrors.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Alumina $\left(\textrm{Al}_2\textrm{O}3\right)$ and magnesia $\left(\textrm{MgO}\right)$ are commonly used in cast-iron applications, where graphite is preferred for low alloy steel.  Engineered clays, which are effectively specific mixtures of alumina, magnesia, and silica, are also frequently used for large-scale cast iron applications as they are very inexpensive and very easy to form in-place in 100 ton and larger applications. Additionally, cast iron has a lower melting point than steel (see the vertical line at about 4.3% carbon on the $\textrm{Fe-C}$ phase diagram and follow it up to the liquid region) and thus somewhat less strict requirements on refractory durability and reactivity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Generally, lime $\left(\textrm{CaO}\right)$ is not used for structural materials as it is too brittle and tends to turn to powder quickly. It is however sometimes used as a binder additive, but the industry is moving away from this as calcium attacks other refractories reducing durability. See the Ellingham diagram below: lime is lowest on the diagram.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Titania $\left(\textrm{TiO}2\right)$ and manganese oxide $\left(\textrm{MnO})\right)$ are not generally used, though I do not know why; likely some combination of heat capacity and mechanical properties.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Ellingham Diagram (Selecting Stable Refractories)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The way to read an Ellingham Diagram, for our purposes, is that moving up on the graph means a decreasing affinity for oxygen, while moving down means an increasing affinity. The diagonal lines with chemical equations indicate the standard free energy of that reaction (vertical axis) at the given temperature (horizontal axis). If, at a given temperature, one reaction line is above another, the higher reaction will proceed toward pure metal plus oxygen (chemical reduction) while the lower reaction will proceed toward metal oxide (chemical oxidation). Therefore, refractory materials with higher affinity for oxygen than the molten metal will be chemically stable during melting. Note that additional diagrams exist or can be made for non-oxide materials using thermodynamic principles and some experimentation, and are a harder to come by on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nhr08.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nhr08.jpg&quot; alt=&quot;Ellingham Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source: &lt;a href=&quot;http://www.doitpoms.ac.uk/tlplib/ellingham_diagrams/ellingham.php&quot; rel=&quot;noreferrer&quot;&gt;Cambridge Ellingham Diagram Tutorial&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-02-05T00:49:55.633" LastActivityDate="2016-02-05T00:49:55.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7250" PostTypeId="2" ParentId="7226" CreationDate="2016-02-03T16:40:25.577" Score="0" Body="&lt;p&gt;The power a resistor will dissipate is the same for a particular DC voltage applied to it, or AC of the same RMS voltage.  The only difference between these AC and DC cases is that the power comes in clumps over time with AC, while continuous with DC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the AC is in the form of a sine, then the power is also a sine but at twice the frequency.  For example, if you drive your nichrome wire from a transformer output running from 60 Hz power, then the power into the wire will vary from 0 to maximum at 120 Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AC at the same RMS will have the same average power, so the question is at what frequency does the non-continuous power of AC matter.  To answer that, you have to look at the thermal time constant of the wire.  AT 120 Hz, you get a pulse of heat every 8.3 ms.  That is almost certainly a lot shorter than the time the wire can meaningfully change temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that old LEBs (light emitting bulbs) used filaments considerably finer than your cutting wire.  The light output is a non-linear function of temperature, such that a little more temperature makes disproportionately more light.  Even so, LEBs weren't generally considered to flicker at the line frequency.  For example, you could swing a bulb around at the end of a cord and not notice flicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since LEBs exhibited little flicker at 50 and 60 Hz, your much thicker wire with longer thermal time constant isn't going to have meaningful temperature ripple either.  For your purposes, 50 and 60 Hz AC is continuous in heating your wire, and you only see the average over a longer period than a AC cycle or two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your purposes, 12 V DC and 12 V AC at 50 or 60 Hz are equivalent.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-02-03T16:40:25.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7251" PostTypeId="2" ParentId="6773" CreationDate="2016-02-03T16:53:21.677" Score="4" Body="&lt;p&gt;If you have only one substance (e.g., only water/ice) and neglect the fluid motion, this is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Stefan_problem&quot; rel=&quot;nofollow&quot;&gt;Stefan problem&lt;/a&gt;, and the basic equations are known. Wikipedia provides them in 1D, although they appear to be normalized (unit latent heat and conductivity, for example). &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S001793100400290X&quot; rel=&quot;nofollow&quot;&gt;Here's a paper with more general conditions&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For higher dimensions and including fluid motion, you'll still need the Navier-Stokes equations and thermal energy equation, along with some interface conditions that simply generalize those discussed for the 1D Stefan problem. If you want the solid phase to move as well, you should include the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cauchy_momentum_equation&quot; rel=&quot;nofollow&quot;&gt;Cauchy equation&lt;/a&gt;. (Unfortunately, due to time constraints I am not able to generalize the interface conditions at this time, but I am sure this answer will help you get started.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tracking the interfaces is still a research problem. Common numerical methods include the &lt;a href=&quot;https://en.wikipedia.org/wiki/Volume_of_fluid_method&quot; rel=&quot;nofollow&quot;&gt;volume of fluid method&lt;/a&gt; or the &lt;a href=&quot;https://en.wikipedia.org/wiki/Level_set_method&quot; rel=&quot;nofollow&quot;&gt;level set method&lt;/a&gt;. While I have taken a CFD course which covered the basics of these two methods, I am not really qualified to make a particular recommendation. I get the impression that which method you choose depends on what you value, e.g., some methods may conserve mass, while others might not but have other attractive properties.&lt;/p&gt;&#xA;" OwnerUserId="49" LastActivityDate="2016-02-03T16:53:21.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7252" PostTypeId="1" CreationDate="2016-02-03T17:00:34.910" Score="6" ViewCount="155" Body="&lt;p&gt;Building codes such as BS5950 provide clear guidelines on how to consider P-Δ effect due to sway of a vertically loaded portal frame or single storey building, causing additional moment in the columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should this sway frame effect be extended to the horizontal members of the braced frame structure in an excavation if the walls are not so stiff (such as an interlocking pipe pile wall system in water) and can move longitudinally (in-plane)? If so, how would the effects of P-delta (I am only interested in the big delta) be applied to the waling-strut combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate if answerers can point to relevant literature to support the answer.&lt;/p&gt;&#xA;" OwnerUserId="620" LastEditorUserId="620" LastEditDate="2016-02-04T01:03:37.870" LastActivityDate="2018-04-26T13:36:24.280" Title="Should P-Δ effect be considered for struts of a braced excavation?" Tags="|steel|geotechnical-engineering|structural-analysis|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7253" PostTypeId="2" ParentId="7245" CreationDate="2016-02-03T17:29:19.550" Score="11" Body="&lt;p&gt;Molten ferrous metals are often handled in steel ladles with a refractory lining. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's only since about the 1860s that any ferrous metals other than cast iron (which has a significantly lower melting point than steel) were handled in a molten state in any sort of quantity. Before that, steel production generally involved carburisation of iron or decarburisation of cast iron in a furnace and wrought iron is not a castable material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically, wrought iron was produced in bloomery furnaces. These are essentially stacks of alternating iron ore and charcoal sealed with a layer of clay on the outside which are allowed to burn over a long period with an air draught coming in through a hole near the bottom. This process produces a spongy mass of metallic iron mixed with silicate slag. The mass would be repeatedly hammered while hot (but not molten) to remove porosity and create an approximately homogeneous ingot, albeit with fine laminations of silica slag—this is 'wrought iron.' The laminar structure contributes significantly to the mechanical properties of wrought iron.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later industrial processes such as 'puddling' decarburised cast iron by stirring it with long iron rods on a bed of sand with indirect heat in a reverberatory furnace. Bloomery furnaces are able to reduce the iron oxides in the ore to produce metal but aren't hot enough to melt it in bulk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cast iron is produced in a 'cupola furnace,' historically built from brick although modern ones tend to be steel with a refractory lining. Charges of iron ore and charcoal (or coke) are fed into the top of the stack and molten metal accumulates in a well a the the bottom where it can be 'tapped' by punching through a clay plug. In iron smelting (from ore) these furnaces would normally be tapped straight into sand ingot moulds producing 'pig iron' which would either be remelted into cast iron components or further processed to produce wrought iron or steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cupola furnaces introduce a lot of carbon into the iron (around 5%) which reduces its melting point to a temperature where it is practical to cast and as such cast iron can be produced with forced air (as opposed to pure oxygen) and at temperatures within the scope of simple refractory materials like fireclay which, not being very mechanically strong are usually used as a liner for the actual structure of the furnace/ladles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could just about get away with using unlined steel ladles for cast iron but the lining greatly prolongs their life and reduces the rate of heat loss from the metal between the furnace and the moulds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The furnaces used for both smelting iron ore into pig iron and remelting ast iron are essentially similar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8lX3m.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8lX3m.jpg&quot; alt=&quot;Steel ladle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H0oCp.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H0oCp.jpg&quot; alt=&quot;Steel ladle cross section&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="368" LastEditDate="2016-02-04T18:20:24.467" LastActivityDate="2016-02-04T18:20:24.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7254" PostTypeId="1" CreationDate="2016-02-03T20:50:04.237" Score="0" ViewCount="618" Body="&lt;p&gt;I have a working &lt;a href=&quot;https://github.com/ChristopheJacquet/PiFmRds&quot; rel=&quot;nofollow&quot;&gt;FM transmitter&lt;/a&gt; on a raspberry pi2. I'm running my FM raido at 103.3MHz, and the antenna that I connected to the GPIO4 is a 70cm copper wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;More data:&lt;/strong&gt;(checked with a &lt;a href=&quot;http://media.onsugar.com/files/2011/01/01/5/89/898472/c6ad4c6e7afa83f7_digital_multimeter.jpg&quot; rel=&quot;nofollow&quot;&gt;multimeter&lt;/a&gt; between GPIO4 and pin6-ground)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;DC voltage: 1.2V&lt;/li&gt;&#xA;&lt;li&gt;AC voltage: 1.8V&lt;/li&gt;&#xA;&lt;li&gt;Current: 18.4mA&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update:&lt;/strong&gt; I manually checked the distance by tunning cell phone fm to 103.3MHz. it was around 20 meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.How far is my FM transmitter signal going?(Is there a formula?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.How do i increase the transmitting distance?(amplify the signal)&lt;/p&gt;&#xA;" OwnerUserId="3565" LastEditorUserId="3565" LastEditDate="2016-02-04T09:12:06.043" LastActivityDate="2016-02-04T13:38:01.127" Title="How far can my FM transmitter signal reach?" Tags="|electrical-engineering|telecommunication|rf-electronics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7255" PostTypeId="2" ParentId="7226" CreationDate="2016-02-04T00:48:50.797" Score="1" Body="&lt;p&gt;AC or DC is fine, however do not exceed 48 volts. This is the known safe limit to ensure that you will not be injured even in wet conditions. This is most easily done by purchasing a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B015G8E7JU&quot; rel=&quot;nofollow&quot;&gt;sealed dc power supply&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat dissipation by the wire changes primarily based on the energy input and the speed at which it is pulled through the material (phase change conduction). Find your nichrome wire resistance per foot in &lt;a href=&quot;http://jacobs-online.biz/nichrome_wire.htm&quot; rel=&quot;nofollow&quot;&gt;this table&lt;/a&gt; and do some &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmlaw.html&quot; rel=&quot;nofollow&quot;&gt;V=IR Ohms Law&lt;/a&gt; calculations for getting in the ball park, but I recommend getting a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B007V1B0W8&quot; rel=&quot;nofollow&quot;&gt;PWM LED dimmer&lt;/a&gt; so you can easily adjust the power going to the wire as needed. Note that the linked dimmer is only rated for 12volts 8Amps, but that should be more than enough. This will be very handy especially if you frequently change the wire gauge, length of wire, or material being cut.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-02-04T00:48:50.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7257" PostTypeId="1" CreationDate="2016-02-04T06:17:48.280" Score="0" ViewCount="94" Body="&lt;p&gt;I am a beginner in ProE and simulation. I would like to apply force in ProE simulation like below picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jnmXJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jnmXJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain how to put the payload on a remote point?&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="1832" LastEditDate="2016-02-04T09:20:23.400" LastActivityDate="2016-02-04T09:20:23.400" Title="How can I apply remote force in Pro/E" Tags="|mechanical-engineering|structural-engineering|modeling|simulation|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7258" PostTypeId="2" ParentId="7254" CreationDate="2016-02-04T07:30:03.910" Score="1" Body="&lt;p&gt;The main factor limiting FM signal reach is often the height of transmitter and receiver antennas over ground, not the power. The Earth is curved and antennas must be in (or very close to) the zone of direct visibility (you should be able to see one antenna from the top of another).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer waves can bend over the Earth curvature, and short waves (the source talks about 3MHz and somewhat above) can reflect from the ionosphere to increase the distance. However 103.3 MHz you talk about is too high to benefit significantly from any of these two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More can be found &lt;a href=&quot;http://penerbit.uthm.edu.my/pdf/syarahan_perdana_siri_1.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that Earth curvature is not a problem, the distance depends on both the power of transmitter and sensitivity of the receiver (and sensitivity of the receiving part largely depends on antenna construction and quality). Hence it is also difficult to say without knowing these details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if there is any material like building walls between transmitter and receiver, the limiting factor most likely will be the thickness and properties of this material (wood? bricks? iron?), not the distance.&lt;/p&gt;&#xA;" OwnerUserId="4835" LastEditorUserId="4835" LastEditDate="2016-02-04T13:38:01.127" LastActivityDate="2016-02-04T13:38:01.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7259" PostTypeId="1" CreationDate="2016-02-04T08:25:08.927" Score="2" ViewCount="1702" Body="&lt;p&gt;I have a furnace for aluminum recycling wherein dross formation is a major issue. It reduces the yield by about 10-12%.&lt;/p&gt;&#xA;&lt;p&gt;I am looking for a change in process parameters or other solutions through which I can reduce the dross formation.&lt;/p&gt;&#xA;" OwnerUserId="3179" LastEditorUserId="90" LastEditDate="2021-12-25T17:26:34.693" LastActivityDate="2021-12-25T17:26:34.693" Title="How can dross/slag formation be reduced during aluminum recycling?" Tags="|materials|metallurgy|aluminum|recycling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="7260" PostTypeId="2" ParentId="7259" CreationDate="2016-02-04T11:08:11.690" Score="5" Body="&lt;p&gt;The dross is mostly composed of aluminium oxide which is already present on the surface of the metal before it even goes into the furnace, as the aluminium melts the oxides separate and float to the top. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recycled aluminium will tend to be made up of relatively thin material with a high surface area to volume ratio and so a relatively high proportion of surface oxide compared to ingots, especially if the material has been anodised. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if the oxide layer is 4 micrometers thick then on a 5kg ingot this is a negligible amount but if you are melting 0.1mm thick foil then a the oxide layer on both sides is getting on for 10% of the total thickness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a lot of the dross is simply separating out what is already present in the starting material which you can do very little about .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can minimise any additional oxide formation during melting by controlling the atmosphere in the furnace to reduce oxygen availability although how practical this is depends on the setup being operated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Separating the material by thickness and adding the thinner stuff to already molten metal would also help a bit.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-04T13:58:05.133" LastActivityDate="2016-02-04T13:58:05.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7261" PostTypeId="1" CreationDate="2016-02-04T12:25:21.217" Score="0" ViewCount="702" Body="&lt;p&gt;I need to create a transparent box that can stay warm inside while in cold temperatures, below 0 degrees, for as long as possible. Glass and plastic are transparent but afaik they don't insulate that well. Is there something else that I could use?&lt;/p&gt;&#xA;" OwnerUserId="4844" LastEditorUserId="33" LastEditDate="2016-02-04T13:49:59.963" LastActivityDate="2016-02-04T13:49:59.963" Title="What is a transparent material that insulates well?" Tags="|heat-transfer|temperature|plastic|thermal-insulation|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ClosedDate="2016-02-04T16:49:52.720" ContentLicense="CC BY-SA 3.0" />
  <row Id="7262" PostTypeId="2" ParentId="7261" CreationDate="2016-02-04T12:35:16.493" Score="2" Body="&lt;p&gt;To borrow from a certain Brit physician, &quot;I think it's a bit more complicated than that.&quot; For example, you could build/buy a double-walled box with a vacuum between the layers.  That's esssentially what a thermos bottle is;  there are varieties of drinkware that are built this way.  But if you were to build a double-walled glass enclosure that's not evacuated (i.e. easier to build), then fluid dynamics comes into play. While dry air is a pretty good insulator, heat transfer depends both on the thickness of the layer and maintaining a static environment.  Too thin, and not much insulating power.  But, as manufacturers of storm windows know, if the spacing exceeds about a cm or so, the air starts to form a vortex, moving up on the warm side and down on the cold side. This leads to a lot of conductive heat transfer at the glass-air interface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you really absolutely positively have to make the entire thing transparent, consider leaving a &quot;peep hole&quot; but covering the rest with a reflective metallic surface.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-02-04T12:35:16.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7263" PostTypeId="1" CreationDate="2016-02-04T15:12:47.993" Score="0" ViewCount="125" Body="&lt;p&gt;I understand that insulation, absorption and diffusion in acoustics are different concepts. Notwithstanding, would it be correct to assume they may be part of the same process when all are used to achieve 'sound proofing' and 'acoustic treatment' in a certain scenario?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it differently, is it correct to say that by treating the sound within a room (with proper absorption and diffusion techniques) we can limit (up to a certain extent) the pressure of certain sound waves that will exit the room?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, I would think that by treating a room with high absorption coefficients, a certain amount of sound energy will be transformed into heat energy (for instance using mineral wool). Shouldn't this mean that less energy will 'get out' of the room? Also, shouldn't this mean that we could potentially plan a room's construction with both sound proofing and acoustic treatment at the same time, without them being completely separate activities at the design stage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, in the aforementioned example could we not say sound insulation, diffusion and absorption may be part of the same process instead of being treated separately. When a room's structure (both inner and outer) is defined can't we limit sound transmission to a certain degree by adding non parallel walls for instance, thus decreasing vibrations?&lt;/p&gt;&#xA;" OwnerUserId="4845" LastEditorUserId="4845" LastEditDate="2016-02-04T17:56:08.917" LastActivityDate="2016-02-04T21:07:24.047" Title="Sound insulation, absorption and diffusion. Can they be part of the same process when wanting to both soundproof and acoustically treat a room?" Tags="|materials|audio-engineering|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7266" PostTypeId="2" ParentId="6773" CreationDate="2016-02-04T17:15:53.220" Score="0" Body="&lt;p&gt;You may want to consider analyzing this problem in two parts; first model the melting of the ice with no fluid flow (the Stefan problem that Ben Trettel identified), &lt;em&gt;then&lt;/em&gt; model the fluid flow separately using the Navier-Stokes equations with the ice melting results as a boundary condition for the inlet flow of water.  I'm assuming that you're only interested in the flow of water over the insulated surface, and not the flow down the sides of the ice cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ice melting and the thin-film flow of the water are two specialized types of problems, and trying to capture both of them in the same analysis might be difficult.  You'd have to choose a mesh structure, solution methods, time step, etc. that works for both problems.  By separating into two analyses, you get to optimize the approach for each one.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-02-04T17:15:53.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7267" PostTypeId="2" ParentId="6298" CreationDate="2016-02-04T17:50:38.853" Score="3" Body="&lt;p&gt;In general when we talk about welding metals the process involves melting and re-fusing two separate pieces in order to join them. This has the potential to create a joint at least as strong as the parent metal and so it is most relevant to high strength materials like steel and aluminium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of lower strength, soft materials like pure tin weldability becomes a bit of a moot point because there is no real reason to do it. This is also partly to do with the distinction in terminology between welding, brazing and soldering. In brazing and soldering a filler metal is used with a lower melting point than the parent metal so there is no melting of the parent metal and fusion/adhesion between the filler and parent occurs only at the surface with little or no penetration or mixing of the to into the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fusion welding relies on being ably to melt a small, controlled area of metal bridging the joint. With lower melting point metals this becomes increasingly difficult to do without melting the whole thing and brazing or soldering are much more practical options. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that it is very difficult to think of an application where you would want to fabricate anything out of pure tin. Bearing in mind that it has a melting point of only 232 degC and is extremely soft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tin/copper alloys ie. bronzes are an entirely different matter and have material properties closer to mild steel and a melting point of around 900 degC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that tin is often a major constituent of solder (especially as a replacement for lead solders). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This illustrates that specific alloys can have vastly different properties to their pure metal constituents and there are relatively few engineering applications for pure metals apart from coatings/foils, the only major exception being copper, gold and silver which are used for their electrical and thermal properties. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-04T17:50:38.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7269" PostTypeId="2" ParentId="2307" CreationDate="2016-02-04T18:36:29.330" Score="3" Body="&lt;p&gt;I think it is (theoretically) possible, but it didn't work when I tried. My main problem was that I didn't have a good way of stopping the required high charge from ionizing the surrounding air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I accidentally posted a duplicate question on the &lt;a href=&quot;https://physics.stackexchange.com/questions/201987/levitating-capacitor&quot;&gt;Physics Forum&lt;/a&gt;...&lt;/p&gt;&#xA;" OwnerUserId="4847" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2016-02-04T18:36:29.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7270" PostTypeId="2" ParentId="7263" CreationDate="2016-02-04T18:39:57.557" Score="1" Body="&lt;p&gt;This is a very good question and is in fact subject to in-depth research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factors one has to take into account involve differential acoustability (a fairly newly constructed term in order to address the various changes that occur when synergistic processes in acoustics take place).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to consider the net rate of absorption, diffusion and reflection per square metre (this is a definition similar to intensity). One then would have to treat the average density of a room as roughly uniform, and then partition it by assigning values from a continuum ranging from 0 to &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(Net rate) * (Average power of sound output) * (differential acoustability) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This then will allow one to get a spatial-temporal function of density-like quantities to be Fourier-transformed in order to get the average rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a fairly comprehensive analysis on the matter published here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.multi-science.co.uk/ba-room_acoustics.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.multi-science.co.uk/ba-room_acoustics.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4848" LastEditorDisplayName="user748" LastEditDate="2016-02-04T20:02:55.910" LastActivityDate="2016-02-04T20:02:55.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7272" PostTypeId="1" AcceptedAnswerId="7274" CreationDate="2016-02-04T20:53:40.973" Score="1" ViewCount="75" Body="&lt;p&gt;I am looking into making a strong electro magnet.&#xA;I want it to be able to hold around 100 pounds of weight or more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00GXTPBRY&quot; rel=&quot;nofollow&quot;&gt;Would a magnetic core like this work at all?&lt;/a&gt; &#xA;I'm planning on wrapping wire around one of these 'C' shaped ferrite cores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much current should I expect to be using?&#xA;How many turns or windings around the core should I expect?&#xA;Is 22 gauge copper solid core wire recommended?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rough estimate would help very much. &#xA;Also If there is any formula or information you could add so that I could start to answer some of these questions I would greatly appreciate it. &lt;/p&gt;&#xA;" OwnerUserId="4853" LastActivityDate="2016-02-04T21:57:54.327" Title="Making a strong electro magnet" Tags="|power-electronics|magnets|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7273" PostTypeId="2" ParentId="7263" CreationDate="2016-02-04T21:07:24.047" Score="1" Body="&lt;p&gt;Acoustic impedance is very similar (mathematically the same as) to electrical impedance. When an electrical wave hits something with infinite impedance (like unconnected wires), electricity doesn't &quot;leak&quot; out of the wire. Instead, the wave is &lt;em&gt;reflected&lt;/em&gt; back at the source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intermediate impedances allow some of the wave to pass while reflecting the rest back at the source. Only when the impedance of the load matches that of the source will there be no reflected waves. This is called &quot;impedance matching&quot; and is studied a lot in power systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impedance matching is the reason why doctors use ultrasound gel - without a significant amount of acoustic energy is lost at the probe/air interface and at the air/skin interface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Acoustic&lt;/em&gt; insulation doesn't really use the same transfer mechanism that &lt;em&gt;thermal&lt;/em&gt; insulation uses. Acoustic signals are pressure &lt;em&gt;waves&lt;/em&gt; and act like such. Material thickness doesn't make much difference in attenuating sound, but the material &lt;em&gt;impedance&lt;/em&gt; does; this is how things like &lt;a href=&quot;http://www.certainteed.com/products/gypsum/board/acoustics/340906&quot; rel=&quot;nofollow&quot;&gt;acoustic wallboard&lt;/a&gt; works - not by making the drywall &lt;em&gt;thicker&lt;/em&gt; but by &lt;em&gt;layering different materials&lt;/em&gt;. Each layer represents a new acoustic interface, which means a new opportunity to &lt;em&gt;reflect&lt;/em&gt; sound waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this is to say that you can't modify an interior chamber's acoustic properties without affecting how that chamber passes (or doesn't pass) acoustic waves to its surroundings. Churches and concert halls will typically have a very long reverb time because their (stone, typically) walls are very high impedance, which means that acoustic sources will produce higher sound pressure levels. &lt;a href=&quot;http://www.me.psu.edu/lamancusa/me458/8_rooms.pdf&quot; rel=&quot;nofollow&quot;&gt;Read more&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-02-04T21:07:24.047" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7274" PostTypeId="2" ParentId="7272" CreationDate="2016-02-04T21:57:54.327" Score="1" Body="&lt;p&gt;The basic principle is that the force exerted by an electromagnet (solenoid) is proportional to the number of turns in the coil and the current passing through the coil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gauge of wire required will be determined by the current. The size of the core doesn't affect the strength of  the magnet directly but the maximum field strength per unit area is limited so, in practice it does need to be scaled with the force required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing to use as a core would be soft iron or steel, it's dimensions are determined by its cross sectional area and the length required to physically fit the core on it. The ferrite core you linked to probably isn't large enough and in any case ferrite is quite brittle and probably won't stand up to being used to lift significant weights. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://www.daycounter.com/Calculators/Magnets/Solenoid-Force-Calculator.phtml&quot; rel=&quot;nofollow&quot;&gt;calculator&lt;/a&gt; should help &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also note that large electromagnets are potentially dangerous as they can cause injury if you trap your fingers with them and you are going to be dealing with fairly significant currents not to mention the fact that electromagnetic lifting devices do not fail safe and lifting 100lbs needs to be treated with caution&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-04T21:57:54.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7275" PostTypeId="1" AcceptedAnswerId="7379" CreationDate="2016-02-05T01:08:42.373" Score="4" ViewCount="172" Body="&lt;p&gt;I am trying to work out how a liquid inside a container will be affected by the back and forth sliding of a Teflon coated plate such as the one in the picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eBZT7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eBZT7.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to work out the shear rate of a fluid between two plates moving at different velocities using the v/h equation, but it doesn't take into account things like the walls on my container, or the fact that the plates may be hydrophobic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to move the plate (which is in contact with the liquid) back and forth without the liquid moving. So my question revolves around firstly whether it is even possible to move the screen back and forth without significantly displacing a certain liquid it is in contact with underneath, and secondly how I would go about calculating to what extent the liquid would be disturbed.  &lt;/p&gt;&#xA;" OwnerUserId="4189" LastEditorUserId="4189" LastEditDate="2016-02-07T10:30:10.837" LastActivityDate="2016-02-11T15:24:25.287" Title="Shear rate of a fluid in a container with a sliding PTFE top plate. Will the liquid move with the plate?" Tags="|fluid-mechanics|microfluidics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7276" PostTypeId="1" CreationDate="2016-02-05T02:11:19.883" Score="1" ViewCount="1810" Body="&lt;p&gt;House remodeling.   I need to enlarge (from 7 to 10-12 ft) and shift (about 5ft sideways) an existing header where one passes between rooms.  On one side, the header has 2x8 ceiling joists nailed to its face; cathedral ceiling 2x12 rafters, on the other side, are nailed to the top of it.  The existing header is a flitch beam, steel sandwiched between two 2x10s. I have computed (pretty conservatively) the load on the new header at about 900 pounds per linear foot (total) and about 575 plf of live load.  There are several ways to proceed, and I can't decide among:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Sister a 2x10 piece of LVL (engineered wood) to the exposed face (the cathedral side) of the existing header.  Unfortunately, the original carpentry was a little sloppy, and the existing header isn't directly on top of the girder which supports the floor which supports the wall with the header).  Unfortunately, it's off to the side of the face where I'd add the LVL, so it'd be unacceptably off center after I did that.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rip out the existing header and replace it with a 2-ply LVL beam (3.5&quot; x 9.5&quot;).  This is &quot;best&quot; in the sense it's very strong (so can make the opening 12ft wide instead of 10ft).  But a gigantic PITA: have to build temporary supporting walls on both sides of existing header, sawzall the thing out of there, etc.  Oh, did I mention that almost every piece of Romex in the house runs right along the edge of this header ?   At least I can do my own electrical work ...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Like #2, except after all the sheetrock is off, try to remove the flitch plate from the existing header, and the 1/2&quot; plywood from the part that wasn't formerly over the opening, and then insert a new fitch plate of the proper length.  Probably a non-starter too, especially if they used glue when they built the original flitch beam.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Put some sort of piece of structural steel under the existing header.  I'm currently favoring this option.   Not sure what to use - angle, square tube, channel ?   Strength shouldn't be an issue.   Deflection definitely is - there's no framing or door, but there is sheetrock; I've been told up to 1/2&quot; might be ok.  That means I need a section with a moment of inertia of about 10 in^4 (if I'm computing correctly).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thoughts on what I should do?&lt;/p&gt;&#xA;" OwnerUserId="4856" LastEditorUserId="1832" LastEditDate="2016-02-05T09:22:33.717" LastActivityDate="2016-02-05T09:22:33.717" Title="best way to reinforce existing structural header?" Tags="|mechanical-engineering|structural-engineering|structures|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7277" PostTypeId="1" CreationDate="2016-02-05T04:38:56.700" Score="1" ViewCount="1441" Body="&lt;p&gt;The formula for the area moment of inertia for a rectangular cross-section is b * h^3 / 12.  But it looks like when computing the value for angle-iron in this table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineersedge.com/standard_material/Steel_angle_properties.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.engineersedge.com/standard_material/Steel_angle_properties.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... they have assumed that the piece is bent into a flat plate, so that both sides are aligned to resist the bending force.  Of course, this is completely unrealistic.   Only one face is typically resisting the bending force (the vertical face in most applications) and the other face simple keeps the vertical face from buckling.   As I missing something here ?&lt;/p&gt;&#xA;" OwnerUserId="4856" LastEditorUserId="33" LastEditDate="2016-02-05T14:31:11.090" LastActivityDate="2016-02-05T14:31:11.090" Title="How is the area moment of inertia calculated for angle iron?" Tags="|mechanical-engineering|structural-engineering|steel|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7278" PostTypeId="1" CreationDate="2016-02-05T04:49:38.610" Score="1" ViewCount="2633" Body="&lt;p&gt;I really don't know why one person should create cavalier and the other one should create cabinet when it comes to oblique projection, are there any reasons why one should use cavalier instead of cabinet and vice versa?&lt;/p&gt;&#xA;" OwnerUserId="3703" LastActivityDate="2018-12-01T14:57:22.287" Title="What's the purpose of cavalier and cabinet in oblique projection?" Tags="|drafting|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7279" PostTypeId="1" CreationDate="2016-02-05T06:35:13.827" Score="4" ViewCount="61" Body="&lt;p&gt;The idea, as I understand it, is to transfer fragile material (glass bulbs in this case, I believe) from one moving &lt;em&gt;something&lt;/em&gt; to another, or over to a platform. This is taking place in the 1920's, I think. Anyway, here's the quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, I suffered myself to be taken to Battersea; and there, sure&#xA;  enough, I found a workshop, duly labelled as the premises of The New&#xA;  Transport Company, Limited, and spacious enough to accommodate a&#xA;  double railway line with a platform.  The affair was unquestionably&#xA;  real, so far.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The platform was not provided with a station: its sole equipment was a&#xA;  table with a row of buttons on it for making electrical contacts. Each&#xA;  line of railway had on it a truck with a steel lid. The practical part&#xA;  of the proceedings began by placing an armchair on the lid of one of&#xA;  the trucks and seating me in it.  A brimming glass of water was then&#xA;  set at my feet.  I could not imagine what I was expected to do with&#xA;  the water or what was going to happen; and there was a suggestion of&#xA;  electrocution about the chair which made me nervous.  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Gattie then sat down majestically at the table on the platform with&#xA;  his hand hovering over the buttons.  Intimating that the miracle would&#xA;  take place when my truck passed the other truck, he asked me to choose&#xA;  whether it should occur at the first passage or later, and to dictate&#xA;  the order in which it should be repeated.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I was by that time incapable of choosing; so I said the sooner the&#xA;  better; and the two trucks started.  When the other truck had passed&#xA;  mine I found myself magically sitting on it, chair and all, with the&#xA;  glass of water unspilled at my feet.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Does &quot;a double railway line&quot; mean two tracks running parallel to each other? Are they linked? Is a switch involved?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are the trucks - trucks (i.e. bogies) that look like this:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G59Vym.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G59Vym.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or are they flatbed freight cars that look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cmg6Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cmg6Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Most important&lt;/strong&gt;: how was the author (G.B. Shaw) transferred (along with the glass of water) over to the other - moving - whatever it was?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3740" LastActivityDate="2016-02-05T06:35:13.827" Title="Could someone explain to me the process described in this quote from G.B. Shaw (fragile load transfer system of some sort)?" Tags="|rail|" AnswerCount="0" CommentCount="10" ClosedDate="2017-05-31T11:47:08.993" ContentLicense="CC BY-SA 3.0" />
  <row Id="7280" PostTypeId="1" AcceptedAnswerId="7282" CreationDate="2016-02-05T07:46:49.950" Score="3" ViewCount="71" Body="&lt;p&gt;I'm hoping to design a simple gearbox for a 1-seater car for kids. So far I'm only using discs of paper I cut out and rotating them by hand. I wonder what the canonical approach is to prototyping/experimenting with gearboxes. I guess the prototype won't need to be big, but just 10CM or so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should it be done virtually in a CAD software instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm very new so any tip will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="4302" LastEditDate="2016-02-05T07:53:27.553" LastActivityDate="2016-02-05T11:33:41.703" Title="Canonical approach to experimenting with a gearbox design" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7281" PostTypeId="2" ParentId="7228" CreationDate="2016-02-05T09:04:16.167" Score="1" Body="&lt;p&gt;&lt;strong&gt;TL, DR&lt;/strong&gt; You need a stronger and different fan. Try a vac.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What you need to understand about fans&lt;/strong&gt;&lt;br&gt;&#xA;Fans are characterised by their power and their flow-pressure curve. Power is pressure times volume flow. Actualy, that's wrong but it's still useful. You fan delivers a certain and has to overcome a pressure difference. Less pressure - more flow and vice versa. Every system has a pressure drop that rises with flow. The flow will be determined by where the curve of the fan and the system curve meet. &lt;a href=&quot;http://www.engproguides.com/fancurve.html&quot; rel=&quot;nofollow&quot;&gt;Learn more here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your system curve (that you don't know) and fan curve meet at 0,009m³/min and 50mm H2O pressure. Take the diagram, draw a straight line through this point and the origin. Pretty steep. The actual sytem curve is somewhere to the left of this line. This tells us that you need a fan that delives a stronger pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to do is IMO attach a vac. cleaner to suction side of your cyclone and try this at different settings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What you need to understand about cyclones&lt;/strong&gt;&lt;br&gt;&#xA;There's a relationship between how effective a cyclone is and the pressure drop, with more pressure drop meaning more effective separation. Of course cyclones can be designed more or less efficient for a certain duty but with a given cyclone this relationship holds. Pressure drop of the cyclone depends on flow:&lt;br&gt;&#xA;More flow -&gt; more speed in cyclone -&gt; stronger centrifugal force to overcome friction (also more pressure drop)&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-02-05T09:04:16.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7282" PostTypeId="2" ParentId="7280" CreationDate="2016-02-05T09:17:23.293" Score="2" Body="&lt;p&gt;In designing a gearbox the first step is to work out the ratios you need in the context of the torques and rotational speeds you need, bearing in mind that other parts of the drive train such as the differential and wheel diameter will also effect the final drive ratio.  This will allow you to select the most appropriate overall design and specify parts which can meet these requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These basic parameters are simple enough to calculate in themselves but need to be considered in the context of the whole design as they depend on the overall performance requirements for the car. In particular the range of gear ratios required will depend on the mass of the car and the performance of the engine/motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For something like this designing gears from scratch is probably not something you want to do so this will likely be a case of sourcing component eg. gears, shafts bearings etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to start thinking about how the whole thing will be packaged which also needs to consider how the gearbox casing relates to other parts of the system. Here a lot depends on what you want to achieve and there are many potential different solutions. I'm assuming that a car for kids will be relatively low speed and power and so potential solutions could include chain and pulley drives as well as gears. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also well worth considering whether there is some existing off the shelf component which can be adapted to meet your needs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To expand on my reply to the first comment. A gearbox is a case where you really need to get quite a lot of the design parameters calculated on paper before you think about prototyping. They key parameters like the ratios and loads on individual parts are largely defined by external factors such as engine characteristics and desired vehicle performance so at the very least you need to give yourself a reasonable starting point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These parameters are reasonably easy to calculate and don't require FEA or calculus, especially if you are using off the shelf components. It's also entirely reasonable to do most of the design of a gearbox 'on paper' and a working prototype would normally only be needed for things like reliability testing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prototyping is most useful for working out fine detail like how the gearbox housing is physically connected to other parts of the vehicle in terms of its size, shape and where mounting points need to go and whether it interferes with other parts. But you can't even begin to do this until you have worked out the basic parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These should be based on your best estimate of things like vehicle weight and engine performance as well as desired top speed and acceleration. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-05T11:33:41.703" LastActivityDate="2016-02-05T11:33:41.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7283" PostTypeId="2" ParentId="4178" CreationDate="2016-02-05T09:33:45.477" Score="0" Body="&lt;p&gt;Rubber or plastic bushes are the conventional solution for suspension mountings. They cope well with shock loading and vibration, help to isolate vibration from  the wheels and are reasonable tolerant of slight misalignment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually mounted as a press fit in a simple tubular housing with a bolt providing the connection to the suspension arm. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-05T09:33:45.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7284" PostTypeId="1" CreationDate="2016-02-05T09:56:08.763" Score="1" ViewCount="142" Body="&lt;p&gt;I wonder if there's a way to rotate a small crank by turning a large wheel. Here's a diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D6RZb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D6RZb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to design the slot on the wheel for the crank pin? A standard, straight radial slot won't do since the diameter of the small wheel is less than the radius of the big wheel. Is there some some mechanism for achieving this?&lt;/p&gt;&#xA;" OwnerUserId="4303" LastEditorUserId="1586" LastEditDate="2016-03-22T18:01:17.643" LastActivityDate="2016-03-22T18:01:17.643" Title="Rotating a small crank with a large wheel" Tags="|mechanical-engineering|gears|" AnswerCount="0" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7286" PostTypeId="2" ParentId="3488" CreationDate="2016-02-05T12:17:42.690" Score="1" Body="&lt;p&gt;There appears to be a big fundamental problem with this design in that the trailer is not articulated. While this will indeed make it easier to reverse, in some circumstances it is potentially dangerous as whenever you turn a corner the rear of the trailer will swing out into the adjacent lane or footpath. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that this wouldn't actually even  be considered a trailer under EU regulations and in any case they are required to be connected to the towing bracket using an approved ball hitch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the issue that this arrangement effectively couples the mass of the trailer to the rear of the car in the yaw plane  which is going to have a very significant effect on its handling in addition to the effective increase in the rear overhang and is likely to be considered a modification to the vehicle itself in the EU and as such would require individual type approval. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-05T12:39:23.703" LastActivityDate="2016-02-05T12:39:23.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7287" PostTypeId="1" AcceptedAnswerId="7294" CreationDate="2016-02-05T12:34:16.900" Score="0" ViewCount="51" Body="&lt;p&gt;I'm studying how industrial plastic film pallet wrappers work. The wrappers apply a certain amount of pre-stretch to the plastic film (by means of different rolls) before applying it to the object that has to be wrapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the object to be wrapped isn't spherical, i.e. it has edges, then the plastic film is pulled more when it reaches an edge, and so there is an variation in how much the plastic stretches.  I believe that this variation in stretching leads to the plastic film breaking when the pallet is rotating very quickly and subsequently has high pull on the film.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there techniques that make it possible to even out these variations and to make the tension the film is under during application more uniform?&lt;/p&gt;&#xA;" OwnerUserId="4857" LastEditorDisplayName="user16" LastEditDate="2016-02-05T22:54:21.167" LastActivityDate="2016-02-05T22:54:21.167" Title="Plastic wrapper stretch" Tags="|plastic|" AnswerCount="1" CommentCount="3" ClosedDate="2016-02-06T00:17:43.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="7289" PostTypeId="1" CreationDate="2016-02-05T15:13:00.513" Score="2" ViewCount="134" Body="&lt;p&gt;Other than higher creep and relaxation moduli, why are polymerics rarely used for large scale structures, architectural, oil rigs, pipes, or other? It seems like, although they have weaker mechanical properties, they have better corrosion resistance etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are polymerics just cheaper?&lt;/p&gt;&#xA;" OwnerUserId="3419" LastEditorUserId="33" LastEditDate="2016-02-05T21:10:01.793" LastActivityDate="2017-10-28T03:25:58.313" Title="Why don't we often use polymerics for large scale structures?" Tags="|mechanical-engineering|materials|civil-engineering|structural-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7290" PostTypeId="2" ParentId="7289" CreationDate="2016-02-05T16:49:09.813" Score="4" Body="&lt;p&gt;Polymers are in fact used - wood is a very common construction material (although perhaps one should consider it a composite?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really have a straightforward answer, but polymers (= plastics) are also prone to degradation caused by the environment: UV light is bad for them, water might cause them to swell and go soft, they are prone to creep. Also, as you say, they are simply weak (lower Young's modulus). Their range of operating temperatures are very bad too: you can subject metals to fairly wide range of low and high temperatures without changing their properties very much, but this span is much narrower for polymers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like wood, they are easy to ignite (unless flame retardants are used which on the other hand they often are) but unlike wood, many plastics lose their capacity to carry loads at higher temperatures (just 100-200 °C). This also happens with metals, but they yield at much higher temperatures than polymers (plastics).&lt;/p&gt;&#xA;" OwnerUserId="3690" LastEditorUserId="1832" LastEditDate="2016-02-05T22:57:08.693" LastActivityDate="2016-02-05T22:57:08.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7291" PostTypeId="1" CreationDate="2016-02-05T17:01:29.943" Score="4" ViewCount="641" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Opposed-piston_engine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Opposed cylinder design&lt;/a&gt; for gasoline engines is rather uncommon, though several have been made in the past. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expansion forces in an opposed cylinder engine are balanced. For ever force there must be an equal and opposite force. Thus V shaped cylindar arrangements create forces that cause the engine block to vibrate. This seems like waste of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Napier_Deltic&quot; rel=&quot;nofollow noreferrer&quot;&gt;Napier Deltic&lt;/a&gt; engines, a type of opposed cylindar engine,  are described as &quot;high strung&quot; and require more maintenance. I would expect the same of any opposed cylinder design due to the use of gears instead of a single crankshaft. Ignoring the maintainence issue, is the opposed cylindar design more efficient at burning fuel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sF3Qk.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sF3Qk.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4214" LastActivityDate="2016-10-15T17:01:23.700" Title="How efficient is the opposed cylinder combustion engine design?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7292" PostTypeId="1" CreationDate="2016-02-05T18:38:05.453" Score="2" ViewCount="1350" Body="&lt;p&gt;I have a fairly tall bridge abutment (about 12 ft.). Vertical loads are supported by bearing piles and lateral loads from the retained soil are supported by a sheet pile wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I consider just the cantilevered sheet pile wall for the condition where scour has occurred (i.e. maximum exposed height of sheet pile), the sheet pile alone is insufficient to carry the lateral load from retained soil - the deflection at top of abutment cap will exceed the allowable 1/2&quot; for a &lt;a href=&quot;http://www.skylinesteel.com/File%20Library/Datasheets/PZ-PS.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PZ22 sheet pile section&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, I've been told to add tie rods to carry some of the lateral load back to a &quot;dead man&quot; / anchor block. Because I already have a grade beam as part of my abutment system, I'm using that as the anchor block...bad idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Clarifying &quot;Grade Beam&quot; terminology -- at the end of the bridge deck, there is a separate 20' slab of concrete called the &quot;approach slab&quot; spanning between the abutment and the grade beam. This helps prevent localized settlement at the beginning of bridge, which produces that &quot;kathunk-kathunk&quot; experience as you drive across bridges. The grade beam is just there to support the approach slab (almost like a mini bridge span).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zySdq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zySdq.png&quot; alt=&quot;Abutment Section&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I go about calculating the load in each of the tie rods?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I completely ignore the contribution of the sheet pile and just assume all the lateral load is transferred through the tie rods? Do I consider the contribution of both the sheet pile and tie rod using strain compatibility?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I put a sketch of the entire abutment system below - in the final configuration it's a pretty rigid system. The abutment / wing / grade beam are tied together with rebar and everything is supported on piles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UkyEL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UkyEL.png&quot; alt=&quot;Abutment System&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, it seems like the center tie rods will carry more load than the exterior rods but I'm not sure how to check this mathematically... though I suspect I'm eventually going to end up with some plate approximation and a massive plug-and-chug from &lt;a href=&quot;http://materiales.azc.uam.mx/gjl/Clases/MA10_I/Roark&amp;#39;s%20formulas%20for%20stress%20and%20strain.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roark's Formulas for Stress and Strain&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2016-02-05T18:38:05.453" Title="Calculating Tie Rod Force in Bridge Abutment System" Tags="|structural-engineering|bridges|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7293" PostTypeId="2" ParentId="2083" CreationDate="2016-02-05T21:15:58.170" Score="1" Body="&lt;p&gt;Structural steel are not usually intended to be heat-treated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat treatment involves heating the steel beyond a critical temperature (typically around 850 °C) and then cooling it rapidly to change its crystal structure. This results in a 'stressed' structure which tends to increase tensile strength and hardness at the expense of toughness and ductility. The tempering process involves reheating hardened steel to a very specific temperature (typically in the rang 180-300 °C) to partially reverse this process with the specific tempering temperature allowing fine control over the balance between hardness and toughness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if the steel has not been quench-hardened in the first place then heating and slow-cooling will not have any great effect on its mechanical properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put this another way tempering and annealing reverse the effect of quench hardening but won't soften steel which hasn't been hardened. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally structural steels are designed to have good tensile strength and ductility in a non-heat-treated state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly there is little point in hardening rivets for either hot- or cold-forming so once they cool they just go back to the state they were in before they were heated.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="1832" LastEditDate="2016-02-05T22:32:11.520" LastActivityDate="2016-02-05T22:32:11.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7294" PostTypeId="2" ParentId="7287" CreationDate="2016-02-05T22:42:57.770" Score="0" Body="&lt;p&gt;One way used to obtain more uniform loading when wrapping around square corners is to have the film run through a spring loaded dancer bar. When a corner pulls the film faster, the dancer bar rotates. Also, the dancer bar is connected to rotary potentiometer, which is wired to a DC SCR control board and speeds up the DC motor. This method provides control at a relatively lower control cost. A servo, encoder, and sensor system could provide very precise control for using high pre-stretch ratios, but may not be implemented in commercial systems as stretch wrappers are relatively lower cost machinery.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2016-02-05T22:42:57.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7296" PostTypeId="1" AcceptedAnswerId="7549" CreationDate="2016-02-06T08:06:01.243" Score="6" ViewCount="258" Body="&lt;p&gt;I'm hoping to make a small mechanical demo (~10cm tall) that models an orbit that is similar to the Earth-Moon orbit - that is, only one side of the Moon faces Earth. I'm trying to keep it simple, so here's my idea so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QtN7S.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QtN7S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the blue and yellow represent Earth and Moon, respectively. These are meant to represent a different planet and its moon though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orbit of the Planet (blue) and its Moon (yellow) has the following differences:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Planet is eccentric&lt;/li&gt;&#xA;&lt;li&gt;Only one face of the Planet and see its Moon, i.e. the horizontal bar on each disc should remain parallel. The bars are for explanatory purposes only.&lt;/li&gt;&#xA;&lt;li&gt;Planet's rotation axis may shift (vertically along the green bar), but the Moon's orbital axis is fixed, i.e. Moon should be always in contact with the outer ring housing&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is: how should the two discs be connected in order to to have the above properties? I've thought about using springs, but I think it'd be quite flimsy though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know about orreries, but typically they don't self-adjust orbits though. Also, I was hoping for a simpler mechanism given that there doesn't need to be empty space between the two discs. I'm only hoping to model the motion on just one plane.&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="33" LastEditDate="2016-02-21T21:20:10.830" LastActivityDate="2016-02-22T01:38:30.853" Title="How would you mechanically model an Earth-Moon like orbit?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7297" PostTypeId="1" AcceptedAnswerId="7298" CreationDate="2016-02-06T11:50:53.497" Score="9" ViewCount="21585" Body="&lt;p&gt;I was studying the designation of names to refrigerants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is the basic formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R - (m - 1) ( n + 1) ( o )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;br&gt;&#xA;m = number of carbon atoms in the refrigerant&lt;br&gt;&#xA;n = number of hydrogen atoms in the refrigerant&lt;br&gt;&#xA;o = number of fluorine atoms in the refrigerant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So R-134a has:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 + 1 = 2 carbon atoms&lt;br&gt;&#xA;3 - 1 = 2 hydrogen atoms&lt;br&gt;&#xA;4 fluorine atoms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What does the 'a' at the end mean?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2116" LastEditorUserId="33" LastEditDate="2016-02-08T14:44:31.910" LastActivityDate="2021-04-28T02:07:41.860" Title="What does &quot;a&quot; mean at the end of a refrigerant's designation (R-134a)?" Tags="|chemical-engineering|refrigeration|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7298" PostTypeId="2" ParentId="7297" CreationDate="2016-02-06T13:02:10.783" Score="13" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;R134a..Here &amp;quot;a&amp;quot; is used to denote that it is an isomer. R134 and R134a have same chemical formula and atomic weight but different chemical structures.&lt;/p&gt;&#xA;&lt;p&gt;R134 has NBP of  about -19 C whereas R134a has a NBP of about -26C.&lt;/p&gt;&#xA;&lt;p&gt;And dont use capital 'A' in R134a …. &amp;quot;A&amp;quot; denotes that the refrigerant is non -azeotropic.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.researchgate.net/post/What_does_A_in_R-134_A_refrigerant_stand_for&quot; rel=&quot;noreferrer&quot;&gt;ResearchGate&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3569" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-02-10T23:27:38.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7301" PostTypeId="1" AcceptedAnswerId="7302" CreationDate="2016-02-06T15:00:27.497" Score="2" ViewCount="56" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;: I'm a statistician working with a design of experiments example regarding plasma etching of circuits, and unfortunately, the author does not define what is meant by the units used for the etch rate ($\mathring{A}$/min). A google search has proved fruitless due to the specialized symbol and the fact that other texts also assume you already know what the $\mathring{A}$ symbol means (which is probably true for those who work in this field, but not for the rest of us). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What does the $\mathring{A}$ unit mean in a plasma etching experiment, for &quot;Etch rate ($\mathring{A}$/min).&quot;&lt;/p&gt;&#xA;" OwnerUserId="4868" LastEditorUserId="33" LastEditDate="2016-02-11T02:26:14.443" LastActivityDate="2016-02-11T02:26:14.443" Title="Units for plasma etching of circuits: A (with degree symbol)/min" Tags="|circuits|metrology|nomenclature|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7302" PostTypeId="2" ParentId="7301" CreationDate="2016-02-06T15:16:16.473" Score="5" Body="&lt;p&gt;Anisotropic etching is a linear process, like welding, so presumably the units would be distance per time. Unlike welding it is at a small scale. See &lt;a href=&quot;https://www.slideshare.net/mobile/minh65/plasma-etching&quot;&gt;this slideshare&lt;/a&gt;. In this case units are &lt;strong&gt;Angstroms per minute&lt;/strong&gt;. One Angstrom is &lt;strong&gt;$10^{-10}\ \textrm{m}$&lt;/strong&gt;, indicating the process is on an atomic scale. Since it appears to be used for integrated circuit fabrication, those units are sensible depending on the scale.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-02-06T15:16:16.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7303" PostTypeId="1" CreationDate="2016-02-06T16:37:14.253" Score="-1" ViewCount="332" Body="&lt;p&gt;I have an old Pratt &amp;amp; Whitney lathe and I need a drill chuck for the&#xA;spindle sleeve. Does anyone know what kind of cone I need? It's definitely not a morse cone, since I already tested it. Since I can only post two images, here are the dimensions of the cone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;smaller diameter: 9.2mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;larger diameter: 12.2mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;length: 57.8mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might be some error in measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/duhRM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/duhRM.jpg&quot; alt=&quot;Lathe&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4goZS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4goZS.jpg&quot; alt=&quot;Conus&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4871" LastActivityDate="2018-06-11T03:38:40.857" Title="Pratt &amp; Whitney lathe drill chuck" Tags="|drilling|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7304" PostTypeId="1" CreationDate="2016-02-06T17:48:03.660" Score="3" ViewCount="328" Body="&lt;p&gt;I'm working on paper cut mechanism, I use gears to move blade to cut paper. I would like to get knowledge about the technique of seling gear shafts to a metal frame. It would help me in designing elements in fusion 360, the limitations and sizes that I can use. I found information about gears and bending metal sheet fabrications enough but I failed to find any info about gear shift fixing to metal like it was done in my pictures provided below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPD: To due to my poor english my question was not clear enough I had to provide more detailed pictures and edit my initial post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm making case (frame made from metal sheets, bending fabrication) to hold gears, but some places are narrow and tiny, I'm not sure will be possible to seal pins in those places. If I get more info how the shafts are assembled into a metal cases I could design it in right way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nXnZv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nXnZv.jpg&quot; alt=&quot;Sealed shafts in real&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Il3WM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Il3WM.jpg&quot; alt=&quot;Shafts in Fusion 360&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4872" LastEditorUserId="4872" LastEditDate="2016-02-23T20:04:11.380" LastActivityDate="2016-09-22T18:31:16.533" Title="What is technology used to seal a gear shaft to metal frame?" Tags="|gears|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7305" PostTypeId="1" CreationDate="2016-02-06T20:07:07.743" Score="0" ViewCount="66" Body="&lt;p&gt;Don't know about you but here in Spain, in the practice lessons we use some software related to the field (for example, ADAMs for Multibody Systems in MechE, right now). Then when the course ends, we don't mention / use the software until you start working in a company and you might don't remember anything at all or you just don't know enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know what can I do in order &lt;strong&gt;not to miss this knowledge and really master all this software&lt;/strong&gt; we are supposed to know. Do we really need to master all the software or a few are enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practical example: I used MATLAB in few practices, but if I had to use MATLAB in a company I don't know how I should proceed. &lt;/p&gt;&#xA;" OwnerUserId="4246" LastEditorUserId="2918" LastEditDate="2016-02-09T08:04:49.843" LastActivityDate="2016-02-09T08:04:49.843" Title="Continuation of learning CAD/CAM/CAE packages" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ClosedDate="2016-02-07T17:02:53.847" ContentLicense="CC BY-SA 3.0" />
  <row Id="7306" PostTypeId="2" ParentId="7305" CreationDate="2016-02-06T22:01:43.747" Score="1" Body="&lt;p&gt;There are a few aspects to this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly it probably isn't practical to get hold of all of the software you might encounter in a future job because it is likely to be expensive and you have no way of knowing what version your future employer might be using. Many companies also use specialist or proprietary software that you may not even know about. It is also not entirely unknown for universities to use obscure or generally awful software for their own internal reasons. I was taught CAD on a dismal platform which was totally unsuited to general engineering because my university was sponsored by the company which made it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second issue is finding a way to practice which will actually be useful. Mechanical Engineering courses tend to be fairly broad  so it makes a certain amount of sense to pick an area that you are particularly interested in and make a project out of it. Similarly any practice you do will have much more value getting you a job if it is easy for you to demonstrate evidence of it. For example saying that you built a robot over the summer as a fun project to improve your skills will carry a lot more weight in an interview than saying that you spent 2 hours a day memorising Autocad keyboard shortcuts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example something like designing and building a working model aircraft would be good as you have something tangible to show for your efforts and it will demonstrate and ability to tackle real world problems rather than just abstract exercises. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing is that an employer probably won't expect a recent engineering graduate to be a technical expert the moment they walk in. They will be more interested in your ability to learn and adapt and have a sound understanding of basic principles rather than lot of memorised information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the most useful area to focus on is CAD as this is pretty widely applicable to any field. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-07T00:40:30.923" LastActivityDate="2016-02-07T00:40:30.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7307" PostTypeId="2" ParentId="7303" CreationDate="2016-02-06T22:24:53.673" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.lathes.co.uk/pratt&amp;amp;whitney/&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; Suggests that M1 tapers are standard but there are examples with Jarno tapers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your measurements seem closest to Jarno #4 but don't match exactly &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://littlemachineshop.com/reference/tapers.php&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; has tables of the common taper sizes so you may be able to deduce it from there. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-06T22:24:53.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7309" PostTypeId="2" ParentId="5900" CreationDate="2016-02-06T23:50:52.210" Score="0" Body="&lt;p&gt;Industrial polymers are often black by default because carbon black pigment is added to stabilise them against environmental degradation, especially the effects of UV light. This is certainly the reason why tyres are almost always black. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And black goes with everything :P&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-06T23:50:52.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7312" PostTypeId="2" ParentId="2876" CreationDate="2016-02-07T01:06:02.150" Score="0" Body="&lt;p&gt;A properly specified weld will as reliable as any other type of joint and castings are subject the their own set of potential defects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of performance (in the absence of a specific context) there won't be much to choose between a ductile iron casting and a steel fabrication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small numbers fabrication will almost certainly be cheaper and quicker. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that your design as you have presented it could be made as is by a fabrication shop in a few hours but would need to be modified somewhat for casting to accommodate draft angles and fillets. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-07T01:06:02.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7315" PostTypeId="2" ParentId="5512" CreationDate="2016-02-07T03:07:57.790" Score="1" Body="&lt;p&gt;This isn't going to be a very authoritative answer, but its the best I've got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can give you no evidence to why this is good practice, but at my previous job where I used to work with such deck closure pours, we had a standard detail, with the pour width equal to 15 cm (approx. 6&quot;) for slabs usually around 22 cm (approx 8.5&quot;). Where that value came from or what benefits or drawbacks it has, however, I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the location of the pour, we didn't really think too much about it. We'd never put it glued to a beam, though: that just makes your life more complicated since the load profiles for the &quot;Phase 2&quot; beams will be radically different (and can cause your &quot;Phase 2&quot; to be slightly tilted, since one beam will deflect more than the other). We tried to have a somewhat symmetric &quot;Phase 2&quot; structure, even if that implied in a slightly larger final span between the &quot;Phase 1&quot; and &quot;Phase 2&quot; beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about where to put the pour, however, I'd wager a good position is around one fifth of the total span between the &quot;Phase 1&quot; and &quot;Phase 2&quot; beams. I come to this considering that slab as a fixed-fixed beam (which implies I'm considering the slab to be one-way). Such a beam has, under uniform load, zero bending moment approximately one-fifth of the span away from the supports (more precisely, $\dfrac{1}{6}\left(3-\sqrt{3}\right) \approx 21.13 \%$). This should reduce the stress directly over this spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t3lSD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t3lSD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option would be to see where the lanes of the final bridge will be and put the pour as close to the center of one of the lanes as possible: this should effectively eliminate the risk of having heavy vehicles driving with wheels directly over the pour (motorcycles probably aren't much of an issue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding how deflections are taken into account when doing the pour, that's actually a really good question to which I don't have much of an answer. It's worth pointing out, however, that at such a moment your bridge will be under a significantly reduced load. If the bridge is entirely brand new, it will probably be completely closed and therefore only under its own dead-load (plus some load from the work being done on the bridge, but that's minimal). This means that, should the tributary areas of load going to each of the beams in &quot;Phase 1&quot; and &quot;Phase 2&quot; be somewhat similar, the phases should actually deflect similarly as well, in which case you won't have much of a problem. If you're reinforcing an existing bridge, however, you may well have to leave at least one lane open in the &quot;Phase 1&quot; segment. This will lead to variable deflections between the different phases (not to mention vibrations which I've heard make these pours very problematic), but I honestly don't really know what to do about it. &lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-02-07T03:07:57.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7316" PostTypeId="1" AcceptedAnswerId="7317" CreationDate="2016-02-07T06:47:32.460" Score="-1" ViewCount="2201" Body="&lt;p&gt;The problem in my text (Thermodynamics An Engineering Approach, Cengel &amp;amp; Boles, 2nd Ed.) is stated as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force, which can be expressed as $F_b=\rho_{air} g V_{balloon}$, will push the balloon upward. If the balloon has a diameter of $30ft$ and carries two people, $140 lb_m$ each, determine the acceleration of the balloon when it is first released. Assume the density of air is $\rho_{air}=0.0724 \frac{lb_m}{ft^3}$, and neglect the weight of the ropes and the cage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: $45.1 \frac{ft}{s^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can easily find a solution in S.I. units. It is the U.S. Customary units that I am having difficulty with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given:&#xA;$$&#xA;r = 15ft \\&#xA;m_{people} = 2 \cdot 140 lb_m \\&#xA;\rho_{air}=0.0724 \frac{lb_m}{ft^3} \\&#xA;$$&#xA;Let us begin by finding $V$ and $F_b$:&#xA;$$&#xA;V_{balloon} = \frac{4}{3}\pi r^3 =  \frac{4}{3}\pi (15ft)^3 = 14137 ft^3 \\&#xA;F_b=\rho_{air} g V_{balloon} = \frac{0.0724\frac{lb_m}{ft^3}}{32.174\frac{lb_m}{lb_f}\frac{ft}{s^2}}\cdot32.174\frac{ft}{s^2} \cdot 14137 ft^3 = 1023 lb_f\\&#xA;$$&#xA;Next let us find all masses:&#xA;$$&#xA;m_{people} = 2 \cdot 140lb_m = 280 lb_m = \frac{280lb_m}{32.174\frac{ft}{s^2}} = 8.703slugs \\ (Q1: \: how\: do\: I\: make\: the\: units\: work\: out?) \\&#xA;m_{He} = \frac{\rho_{air} V_{balloon}}{7} = \frac{0.0724\frac{lb_m}{ft^3}}{7 \cdot 32.174\frac{lb_m}{lb_f}\frac{ft}{s^2}} \cdot 14137 ft^3 = 4.545 slugs \\&#xA;(Q2: \: how\: do\: I\: make\: the\: units\: work\: out?) \\&#xA;m_{total} = m_{people} + m_{He} = (8.703 + 4.545) slugs = 13.25 slugs$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us determine the $F_{net}$ and finally the acceleration by Newton's Second Law:&#xA;$$&#xA;F_{total} = m_{total}g = 13.25 slugs \cdot 32.174\frac{ft}{s^2} = 426lb_f\\&#xA;(Q3: \: how\: do\: I\: make\: the\: units\: work\: out?) \\&#xA;F_{net} = F_b - F_{total} = (1023-426) lb_f = 596.7 lb_f \\&#xA;a = \frac{F_{net}}{m_{total}} = \frac{596.7 lb_f}{13.25 slugs} = 45.1 \frac{ft}{s^2} \\&#xA;(Q4: \: how\: do\: I\: make\: the\: units\: work\: out?) \\&#xA;$$&#xA;I was able to duplicate the solution but I had to fudge on the units because I am not sure how to convert between $lb_f$ to $lb_m$ to $slugs$. Please advise by using the very same calculations in this problem. Please select any of Q1 to Q4 to demonstrate how to use unit conversion factors in U.S. customary units.&lt;/p&gt;&#xA;" OwnerUserId="622" LastEditorUserId="622" LastEditDate="2016-02-07T19:05:01.617" LastActivityDate="2016-02-07T19:05:01.617" Title="Find Acceleration Of A Buoyancy Force Of A Balloon In U.S. Customary Units (1-49E)" Tags="|fluid-mechanics|thermodynamics|statics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7317" PostTypeId="2" ParentId="7316" CreationDate="2016-02-07T14:35:08.853" Score="0" Body="&lt;p&gt;If it helps, convert the input values to whatever units you are comfortable with, then convert the answer to whatever units the question is asking for.  If the question doesn't specify units for the answer, and this isn't implied from discussions in class, then answer in any reasonable units you like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I said &lt;i&gt;reasonable&lt;/i&gt; units, since deliberately tweaking the professor's or TA's nose is not a good idea.  He might be fine with meters/second instead of feet/second or miles/hour, but furlongs/fortnight is just deliberate provocation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, pounds-force, pounds-mass, and slugs are all very common units that are well defined out there, so answering that part of your question is pointless.  Go look them up.  Whether you like them or not, they are units you should at least know how to use, perhaps just looking up the conversion constants when you need them.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-02-07T14:35:08.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7318" PostTypeId="1" AcceptedAnswerId="7460" CreationDate="2016-02-08T00:04:47.807" Score="1" ViewCount="187" Body="&lt;p&gt;I'm looking for a linear actuator that can lift a microphone of about 700 grams. It's for a goose-neck microphone (together with shock-mount) that will pop out of a small pulpit. Only the top part is bent so it will stick out a small part when at the lowest position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.alibaba.com/trade/search?fsb=y&amp;amp;IndexArea=product_en&amp;amp;CatId=&amp;amp;SearchText=electric%20linear%20actuator&quot; rel=&quot;nofollow&quot;&gt;electrical actuators&lt;/a&gt; I have found so far are made for heavy lifting and are far to slow (and noisy). I need to be able to adjust the height quickly when a different speaker comes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm looking for would be more like a printer head carriage in an ink-jet printer. It does not have to be that fast, 10cm/second would be good enough with a stroke of +/-45cm. It would be great if it could also send feedback on its position (or if it could be driven to a specific point like a servo).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anything like that exist off the shelf or would I have to custom-build the whole thing? I'm planning to use an &lt;a href=&quot;http://www.netduino.com/netduinoplus2/specs.htm&quot; rel=&quot;nofollow&quot;&gt;Netduino+&lt;/a&gt; (with a motor shield) to drive it.&lt;/p&gt;&#xA;" OwnerUserId="4878" LastEditorUserId="272" LastEditDate="2016-02-08T12:58:02.860" LastActivityDate="2016-02-16T10:56:31.727" Title="Fast and quitet linear actuator for lifting less than a kilo" Tags="|mechanical-engineering|electrical-engineering|motors|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7319" PostTypeId="1" CreationDate="2016-02-08T01:54:41.087" Score="1" ViewCount="36" Body="&lt;p&gt;Like the title states, I am looking for a way to approximate the torque vs rotor speed curve of an AC induction motor using the standard specs given in a specs sheet:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;LRA&lt;/li&gt;&#xA;&lt;li&gt;Pull-up torque&lt;/li&gt;&#xA;&lt;li&gt;break down torque&lt;/li&gt;&#xA;&lt;li&gt;rated torque&lt;/li&gt;&#xA;&lt;li&gt;torque = 0 at synchronous speed&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am trying to approximate with least squares. I don't know what two parameter equation would be the best way to go with.&lt;/p&gt;&#xA;" OwnerUserId="4882" LastEditorUserId="1586" LastEditDate="2016-02-18T18:42:49.657" LastActivityDate="2016-02-18T18:42:49.657" Title="two parameter equation to approximate torque v.s rotor speed curve of AC induction motors" Tags="|motors|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7320" PostTypeId="1" CreationDate="2016-02-08T06:22:12.743" Score="1" ViewCount="64" Body="&lt;p&gt;In the interest of saving space, I'd like to have a stool that can double as a table. I'd like the height to be adjustable from 16 inches to 32 inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't find any pneumatic/spring piston with such a long travel (10&quot; at most). It seems the problem with long travel pistons is that the amount of weight that can be held when fully extended is low due to piston bore volume being so big, allowing the air to be easily compressed. Likewise, an  unreasonably large force will be needed to completely compress the spring when you want to retract the piston all the way. While it would be nice to have an extra tall stool to sit on, most of the time it will only be used as a table when extended that high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want this to be another shopping list question, so please also discuss any principles and about why this is or isn't practical, or propose alternate solutions&lt;/p&gt;&#xA;" OwnerUserId="4443" LastEditorUserId="33" LastEditDate="2016-02-08T14:17:36.657" LastActivityDate="2016-02-08T14:17:36.657" Title="Are there design reasons why there aren't pneumatic pistons for chairs with a stroke &gt;= 400 mm?" Tags="|springs|pistons|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7321" PostTypeId="1" CreationDate="2016-02-08T07:56:33.150" Score="4" ViewCount="2990" Body="&lt;p&gt;I'm writing a script to extract all the dimensions from a 2D drawing in NX. In doing so, I have been able to get the dimensions and tolerances on all sheets, along with the zone numbers. This has been fairly easy. However, I'm running into a brick wall when it comes to the other &quot;dimensions&quot; or drawing specifications. I don't know how to loop through the properties of the Surface Finish or the parallelism annotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my code currently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;' VBA Script to extract all the dimensions in an NX part file to an MS Excel Spreadsheet.&#xA;' The details extracted are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;' Sheet No. | Zone | Balloon No. | Description\Measurement Direction | Nominal Value | &#xA;' Tolerance Type | Upper Tolerance | Lower Tolerance | Maximum Value | Minimum Value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option Strict Off&#xA;Imports System&#xA;Imports System.IO&#xA;Imports NXOpen&#xA;Imports NXOpen.UF&#xA;Imports NXOpen.UI&#xA;Imports NXOpen.Annotations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Module ExportDimensionsAndAnnotations&#xA;    'Declare global variables&#xA;    Dim theSession as Session = Session.GetSession()&#xA;    Dim theUfSession As UFSession = UFSession.GetUFSession&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sub Main()&#xA;    ' First read the active part file from NX.&#xA;    ' Then, loop through the sheets.&#xA;    ' Then, loop through the dimensions in the sheet&#xA;    ' Grab the zone, text, tolerance type, tolerance and the measurement direction for each.&#xA;    ' Grab the nearest annotation in that zone.&#xA;    ' Build an array or a dictionary (similar to JSON or Python's Dictionary class)&#xA;    ' Then, loop through the text in the document, and do the same steps, except for those that are inapplicable, of course.&#xA;    ' Ask for a savefile location.&#xA;    ' Create an Excel file in that location&#xA;    ' Open the Excel file.&#xA;    ' Write array to the excel file.&#xA;&#xA;    If IsNothing(theSession.Parts.Work) Then 'Error handling&#xA;        MsgBox(&quot;This code requires an active part. Please open a drawing first and then try again.&quot;)&#xA;        Exit Sub&#xA;    End If &#xA;    Dim workPart as Part&#xA;    workPart = theSession.Parts.Work&#xA;&#xA;    Dim theUISession as UI = UI.getUI&#xA;    Dim saveFileName as String&#xA;    Dim saveDialog as New System.Windows.Forms.SaveFileDialog&#xA;    Dim objExcel as Object&#xA;    Dim objWorkbook as Object&#xA;    Dim excelFileExists as Boolean = False&#xA;    Dim objWorksheet as Object&#xA;    Dim colSheetNumber as Integer = 1&#xA;    Dim colZone as Integer = 2&#xA;    Dim colBalloon as Integer = 3&#xA;    Dim colDescrMatlDir as Integer = 4&#xA;    Dim colNomValue as Integer = 5&#xA;    Dim colTolerType as Integer = 6&#xA;    Dim colTolerUpper as Integer = 7&#xA;    Dim colTolerLower as Integer = 8&#xA;    Dim colMaxValue as Integer = 9&#xA;    Dim colMinValue as Integer = 10&#xA;    Dim colComment as Integer = 11&#xA;&#xA;    Dim valueSheetNumber as String&#xA;    Dim valueZone as String&#xA;    Dim valueBalloon as String&#xA;    Dim valueDescrMatlDir as String&#xA;    Dim valueNomValue as String&#xA;    Dim valueTolerType as String&#xA;    Dim valueTolerUpper as String&#xA;    Dim valueTolerLower as String&#xA;    Dim valueMaxValue as String&#xA;    Dim valueMinValue as String&#xA;    Dim valueComment as String&#xA;&#xA;    Dim rowNumber as Integer = 1&#xA;&#xA;    With saveDialog&#xA;        .DefaultExt = &quot;xlsx&quot;&#xA;        .FileName = &quot;Exported Data&quot;&#xA;        .Filter = &quot;MS Excel Spreadsheets (*.xlsx)|*.xlsx|All Files (*.*)|*.*&quot;&#xA;        .FilterIndex = 1&#xA;        .OverwritePrompt = True&#xA;        .Title = &quot;Select a file where you'd like to save the exported data.&quot;&#xA;    End With&#xA;    saveDialog.ShowDialog()&#xA;    saveFileName = saveDialog.FileName&#xA;    If saveFileName =&quot;Exported Data&quot; Then&#xA;        MsgBox(&quot;You failed to select a save file. Exitting the macro.&quot;)&#xA;        Exit Sub&#xA;    End If&#xA;    objExcel = CreateObject(&quot;Excel.Application&quot;)&#xA;    if objExcel is Nothing Then&#xA;        theUISession.NXMessageBox.Show(&quot;Error&quot;, theUISession.NXMessageBox.DialogType.Error, &quot;Could not start Excel, journal exiting&quot;)&#xA;        Exit Sub&#xA;    End If&#xA;    If File.Exists(saveFileName) Then&#xA;        excelFileExists = True&#xA;        objWorkbook = objExcel.Workbooks.Open(saveFileName)&#xA;        objWorksheet = objWorkbook.Sheets.Add&#xA;    Else&#xA;        objWorkbook = objExcel.Workbooks.Add&#xA;        objWorkbook.saveAs(saveFileName)&#xA;        objWorksheet = objWorkbook.Sheets(1)&#xA;    End If&#xA;&#xA;    objWorksheet.cells(rowNumber, colSheetNumber).Value = &quot;Sheet Number&quot;&#xA;    objWorksheet.cells(rowNumber, colZone).Value = &quot;Drawing Zone&quot;&#xA;    objWorksheet.cells(rowNumber, colBalloon).Value = &quot;Balloon Number&quot;&#xA;    objWorksheet.cells(rowNumber, colDescrMatlDir).Value = &quot;Description Text \ Measurement Direction&quot;&#xA;    objWorksheet.cells(rowNumber, colNomValue).Value = &quot;Nominal Value&quot;&#xA;    objWorksheet.cells(rowNumber, colTolerType).Value = &quot;Tolerance Type&quot;&#xA;    objWorksheet.cells(rowNumber, colTolerUpper).Value = &quot;Upper Tolerance&quot;&#xA;    objWorksheet.cells(rowNumber, colTolerLower).Value = &quot;Lower Tolerance&quot;&#xA;    objWorksheet.cells(rowNumber, colMaxValue).Value = &quot;Maximum Value&quot;&#xA;    objWorksheet.cells(rowNumber, colMinValue).Value = &quot;Minimum Value&quot;&#xA;&#xA;    For Each tempDim as Annotations.Dimension in workPart.Dimensions&#xA;        rowNumber = rowNumber + 1&#xA;&#xA;        Dim theSheet as Drawings.DrawingSheet = AskDrawingSheet(tempDim)&#xA;        valueSheetNumber = getSheetNumber(theSheet).ToString&#xA;        valueZone = getZone(theSheet, tempDim)&#xA;        valueBalloon = &quot;&quot;&#xA;        valueNomValue = tempDim.ComputedSize&#xA;&#xA;        valueDescrMatlDir = tempDim.GetType.ToString&#xA;        ' Clean up the material direction.&#xA;        valueDescrMatlDir = Replace(valueDescrMatlDir, &quot;NXOpen.Annotations.&quot;,&quot;&quot;)&#xA;&#xA;        Select Case valueDescrMatlDir&#xA;            Case &quot;DiameterDimension&quot;&#xA;                valueDescrMatlDir = &quot;ø&quot;&#xA;            Case &quot;CylindricalDimension&quot;&#xA;                valueDescrMatlDir = &quot;ø&quot;&#xA;            Case &quot;MinorAngularDimension&quot;&#xA;                valueDescrMatlDir = &quot;∠&quot;&#xA;            Case &quot;HorizontalDimension&quot;&#xA;                valueDescrMatlDir = &quot;Length&quot;&#xA;        End Select&#xA;&#xA;        valueTolerLower = &quot;-&quot;&#xA;        valueTolerUpper = &quot;-&quot;&#xA;        valueMaxValue = &quot;-&quot;&#xA;        valueMinValue = &quot;-&quot;&#xA;        valueComment = &quot;-&quot;&#xA;&#xA;        valueTolerType = tempDim.ToleranceType.ToString&#xA;        Select Case valueTolerType&#xA;            Case Is = &quot;UnilateralBelow&quot;&#xA;                valueTolerLower = tempDim.LowerMetricToleranceValue.ToString&#xA;                valueMaxValue = tempDim.ComputedSize&#xA;                valueMinValue = tempDim.ComputedSize + tempDim.LowerMetricToleranceValue&#xA;            Case Is = &quot;UnilateralAbove&quot;&#xA;                valueTolerUpper = tempDim.UpperMetricToleranceValue.ToString&#xA;                valueMaxValue = tempDim.ComputedSize + tempDim.UpperMetricToleranceValue&#xA;                valueMinValue = tempDim.ComputedSize&#xA;            Case Is = &quot;LimitsAndFits&quot;&#xA;                valueTolerLower = tempDim.LimitFitDeviation&#xA;                valueTolerUpper = tempDim.LimitFitGrade&#xA;                valueMaxValue = &quot;As per standards&quot;&#xA;                valueMinValue = &quot;As per standards&quot;&#xA;            Case Is = &quot;BilateralOneLine&quot;&#xA;                valueTolerUpper = tempDim.UpperMetricToleranceValue.ToString&#xA;                valueTolerLower = (-1*tempDim.UpperMetricToleranceValue).ToString&#xA;                valueMaxValue = tempDim.ComputedSize + tempDim.UpperMetricToleranceValue&#xA;                valueMinValue = tempDim.ComputedSize - tempDim.UpperMetricToleranceValue&#xA;            Case Is = &quot;BilateralTwoLines&quot;&#xA;                valueTolerLower = tempDim.LowerMetricToleranceValue.ToString&#xA;                valueTolerUpper = tempDim.UpperMetricToleranceValue.ToString&#xA;                valueMaxValue = tempDim.ComputedSize + tempDim.UpperMetricToleranceValue&#xA;                valueMinValue = tempDim.ComputedSize + tempDim.LowerMetricToleranceValue&#xA;            Case Is = &quot;None&quot;&#xA;                valueTolerUpper = &quot;NA&quot;&#xA;                valueTolerLower = &quot;NA&quot;&#xA;                valueMaxValue = tempDim.ComputedSize&#xA;                valueMinValue = tempDim.ComputedSize&#xA;            Case Is = &quot;Basic&quot;&#xA;                valueTolerUpper = &quot;NA&quot;&#xA;                valueTolerLower = &quot;NA&quot;&#xA;                valueMaxValue = tempDim.ComputedSize&#xA;                valueMinValue = tempDim.ComputedSize&#xA;            Case Is = &quot;LimitOneLine&quot;&#xA;                valueTolerUpper = tempDim.UpperMetricToleranceValue.ToString&#xA;                valueTolerLower = tempDim.LowerMetricToleranceValue.ToString&#xA;                valueMaxValue = tempDim.ComputedSize + tempDim.UpperMetricToleranceValue&#xA;                valueMinValue = tempDim.ComputedSize + tempDim.LowerMetricToleranceValue&#xA;            Case Else&#xA;                valueComment = &quot;Error, &quot; &amp;amp; valueTolerType &amp;amp; &quot; is a tolerance type that hasn't been accounted for in the code. Please modify the script.&quot;&#xA;        End Select&#xA;&#xA;&#xA;        objWorksheet.cells(rowNumber, colSheetNumber).Value = valueSheetNumber&#xA;        objWorksheet.cells(rowNumber, colZone).Value = valueZone&#xA;        objWorksheet.cells(rowNumber, colBalloon).Value = valueBalloon&#xA;        objWorksheet.cells(rowNumber, colDescrMatlDir).Value = valueDescrMatlDir&#xA;        objWorksheet.cells(rowNumber, colNomValue).Value = valueNomValue&#xA;        objWorksheet.cells(rowNumber, colTolerType).Value = valueTolerType&#xA;        objWorksheet.cells(rowNumber, colTolerUpper).Value = valueTolerUpper&#xA;        objWorksheet.cells(rowNumber, colTolerLower).Value = valueTolerLower&#xA;        objWorksheet.cells(rowNumber, colMaxValue).Value = valueMaxValue&#xA;        objWorksheet.cells(rowNumber, colMinValue).Value = valueMinValue&#xA;        objWorksheet.cells(rowNumber, colComment).Value = valueComment&#xA;    Next&#xA;&#xA;    for each annote as Annotation in workPart.Annotations&#xA;&#xA;    Next&#xA;&#xA;    objWorkbook.save()&#xA;    objWorkbook.close()&#xA;    objExcel.Quit()&#xA;    objWorksheet = Nothing&#xA;    objWorkbook = Nothing&#xA;    objExcel = Nothing&#xA;    MsgBox(&quot;Completed the extraction successfully! Check &quot; &amp;amp; saveFileName &amp;amp; &quot; for the data.&quot;)&#xA;&#xA;End Sub&#xA;&#xA;Function getSheetNumber(ByVal theSheet as Drawings.DrawingSheet) As String&#xA;    ' This function returns the number of a sheet given the DrawingSheet object.&#xA;    Dim sheetNum as Integer&#xA;    Dim theSheetBuilder as Drawings.DrawingSheetBuilder&#xA;    theSheetBuilder = theSession.Parts.Work.DrawingSheets.DrawingSheetBuilder(theSheet)&#xA;    sheetNum = theSheetBuilder.Number&#xA;    theSheetBuilder.Destroy()&#xA;    Return sheetNum.ToString&#xA;End Function&#xA;&#xA;&#xA;Function AskDrawingSheet(ByVal theObject As TaggedObject) As Drawings.DrawingSheet&#xA;&#xA;    Dim theView As View = TryCast(theObject, View)&#xA;    If Not theView Is Nothing Then&#xA;        Dim sheetTag As Tag = Nothing&#xA;        Try&#xA;            theUfSession.Draw.AskDrawingOfView(theView.Tag, sheetTag)&#xA;            Return Utilities.NXObjectManager.Get(sheetTag) ' the drawing it is on&#xA;        Catch ex As NXException&#xA;            Return Nothing  ' it is a model view&#xA;        End Try&#xA;    End If&#xA;&#xA;    Dim viewName As String = Nothing&#xA;    Dim status As Integer = Nothing&#xA;    Try&#xA;        theUfSession.View.AskViewDependentStatus(theObject.Tag, status, viewName)&#xA;    Catch ex As NXException&#xA;        Return Nothing&#xA;    End Try&#xA;    If status = 0 Then Return Nothing ' it is a model mode object&#xA;&#xA;    Dim viewTag As Tag = Nothing&#xA;    theUfSession.View.AskTagOfViewName(viewName, viewTag)&#xA;    Dim viewType As Integer = Nothing&#xA;    Dim viewSubtype As Integer = Nothing&#xA;    theUfSession.View.AskType(viewTag, viewType, viewSubtype)&#xA;    If viewType = 0 Then Return Nothing ' it is view dependent in a modeling view&#xA;&#xA;    Dim drawingTag As Tag = Nothing&#xA;    theUfSession.Draw.AskDrawingOfView(viewTag, drawingTag)&#xA;    Return Utilities.NXObjectManager.Get(drawingTag)  ' the drawing it is on!&#xA;&#xA;End Function&#xA;&#xA;Function getZone(ByVal theSheet As Drawings.DrawingSheet, ByVal theDimension As Annotations.Dimension) As String&#xA;    Dim borderBuilder As Drawings.BordersAndZonesBuilder&#xA;    borderBuilder = theSession.Parts.Work.Drafting.BordersAndZonesObjects.CreateBordersAndZonesBuilder(theSheet.BordersAndZones)&#xA;&#xA;    Dim numHorizontalZones As Integer = (theSheet.Length - borderBuilder.LeftMargin - borderBuilder.RightMargin) / borderBuilder.HorizontalSize&#xA;    Dim numVerticalZones As Integer = (theSheet.Height - borderBuilder.BottomMargin - borderBuilder.TopMargin) / borderBuilder.VerticalSize&#xA;&#xA;    'calculate zone wrt bottom left of drawing (ZoneOrigin.BottomLeft)&#xA;    Dim Hcell As Double = (theDimension.AnnotationOrigin.X - borderBuilder.LeftMargin) / borderBuilder.HorizontalSize&#xA;    Dim Vcell As Double = (theDimension.AnnotationOrigin.Y - borderBuilder.BottomMargin) / borderBuilder.VerticalSize&#xA;&#xA;    Hcell = Math.Ceiling(Hcell)&#xA;    Vcell = Math.Ceiling(Vcell)&#xA;&#xA;    Dim theZoneOrigin As Drawings.BordersAndZonesBuilder.ZoneOrigin = borderBuilder.Origin&#xA;    borderBuilder.Destroy()&#xA;&#xA;    Dim verticalLetterNum As Integer&#xA;    Dim verticalLetter As Char&#xA;&#xA;    Dim horizontalNum As Integer&#xA;&#xA;    If theZoneOrigin = Drawings.BordersAndZonesBuilder.ZoneOrigin.BottomLeft Or theZoneOrigin = Drawings.BordersAndZonesBuilder.ZoneOrigin.TopLeft Then&#xA;        horizontalNum = Hcell&#xA;    Else&#xA;        'origin on right side&#xA;        horizontalNum = numHorizontalZones - Hcell + 1&#xA;    End If&#xA;&#xA;    If theZoneOrigin = Drawings.BordersAndZonesBuilder.ZoneOrigin.BottomLeft Or theZoneOrigin = Drawings.BordersAndZonesBuilder.ZoneOrigin.BottomRight Then&#xA;        verticalLetterNum = Asc(&quot;A&quot;) + Vcell - 1&#xA;        verticalLetter = Chr(verticalLetterNum)&#xA;&#xA;    Else&#xA;        'origin on the top&#xA;        verticalLetterNum = Asc(&quot;A&quot;) + numVerticalZones - Vcell&#xA;        verticalLetter = Chr(verticalLetterNum)&#xA;&#xA;    End If&#xA;    Return verticalLetter &amp;amp; horizontalNum.ToString&#xA;End Function&#xA;&#xA;Public Function GetUnloadOption(ByVal dummy As String) As Integer &#xA;    'Unloads the image immediately after execution within NX&#xA;    GetUnloadOption = NXOpen.Session.LibraryUnloadOption.Immediately&#xA;    '-------------------------------&#xA;End Function&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;End Module&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone help me out here?&lt;/p&gt;&#xA;" OwnerUserId="4885" LastEditorUserId="315" LastEditDate="2016-12-02T16:32:36.270" LastActivityDate="2016-12-02T16:32:36.270" Title="How do I loop through annotations in NX using the NXOpen API in VB?" Tags="|mechanical-engineering|design|cad|software|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7322" PostTypeId="2" ParentId="7297" CreationDate="2016-02-08T15:39:10.323" Score="2" Body="&lt;p&gt;If a refrigerant contains 'a' at its end it is the isomer of that refrigerant.&#xA;Example 'R123' is 'C2H2F3Cl2' and 'R123a' is the isomer of the same molecule.&lt;/p&gt;&#xA;" OwnerUserId="4889" LastActivityDate="2016-02-08T15:39:10.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7323" PostTypeId="2" ParentId="3078" CreationDate="2016-02-08T18:29:33.093" Score="1" Body="&lt;p&gt;To find a ready-made device that utilizes a six-bar mechanism, you can dig through some of the classical resources below. Instead of pointing out a few devices, I think it would be better to point you to these databases.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;507 Mechanical Movements: Mechanisms &amp;amp; Devices&lt;/strong&gt;. &lt;a href=&quot;http://www.amazon.in/507-Mechanical-Movements-Mechanisms-Devices/dp/1603863117&quot; rel=&quot;nofollow&quot;&gt;(amazon link)&lt;/a&gt;. This book is also available for free as an animated database.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mechanisms and Mechanical Devices Sourcebook&lt;/strong&gt;. &lt;a href=&quot;http://www.amazon.in/Mechanisms-Mechanical-Devices-Sourcebook-Edition/dp/0071704426&quot; rel=&quot;nofollow&quot;&gt;(amazon link)&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have used most of them frequently while attending a course on mechanisms design and found all of them equally useful.&lt;/p&gt;&#xA;" OwnerUserId="2918" LastActivityDate="2016-02-08T18:29:33.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7324" PostTypeId="2" ParentId="5748" CreationDate="2016-02-08T18:38:25.550" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For the equilibrium boundary layer, the local skin friction coefficient is independent of two parameters, both the &lt;strong&gt;streamwise distance&lt;/strong&gt; and the &lt;strong&gt;Reynolds number&lt;/strong&gt;, based on the momentum thickness, and the boundary layer thickness is proportional to the streamwise distance. On the other hand, for the non-equilibrium boundary layer, the local skin friction coefficient depends on the above two parameters. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is from &lt;a href=&quot;http://link.springer.com/chapter/10.1007%2F978-1-4020-6472-2_30&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: This seems self explanatory. Please comment if you think an explanation is required. I have highlighted the parameters in &lt;strong&gt;bold&lt;/strong&gt; myself.&lt;/p&gt;&#xA;" OwnerUserId="2918" LastActivityDate="2016-02-08T18:38:25.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7325" PostTypeId="1" CreationDate="2016-02-08T19:13:27.573" Score="2" ViewCount="76" Body="&lt;p&gt;I am looking into any requirements to obtain a CE mark that require control logic to be positive. In other words, if an electronic circuit has logic that either resets or turns the circuit off, is there a requirement that the control line activates the reset/shutdown by a low (ground) to high (such as TTL) transition or state change?&lt;/p&gt;&#xA;" OwnerUserId="4891" LastActivityDate="2016-05-10T23:39:14.133" Title="Is there a CE mark requirement for positive control logic?" Tags="|electrical-engineering|control-engineering|control-theory|regulations|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7326" PostTypeId="2" ParentId="449" CreationDate="2016-02-08T19:19:18.167" Score="6" Body="&lt;p&gt;While the other guys explained the theoretical framework behind meshing, the practice is markedly different and it is not at all automatic in industries where quality of mesh is of utmost importance given that finite element analysis results cover a great deal of the product development process.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Let's first understand how meshing is done:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Meshing for structural domains are of three types: 1D meshing, 2D meshing, and 3D meshing based on the type of elements used for meshing.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;1D meshing: line element&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2D meshing: quad/tria element&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;3D meshing: hexa(brick)/penta/tetra elements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which mesh to use i.e. 1D, 2D, or 3D is primarily dependent on the &lt;em&gt;computational accuracy required, computational cost (time required to solve the problem), and aspect ratio of the domain&lt;/em&gt;. The highest aspect ratio should be more than 10 (as a thumbs rule in general) to neglect a dimension and go for a low-dimension mesh. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me explain.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A domain that is 100X50X80 has all comparable dimension and the highest aspect ratio is 100/50=3. Therefore, 3D elements will be used to mesh that part.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A domain that is 100X50X8 has one dimension negligible and the highest aspect ratio is 100/8=12. Therefore, 2D elements will be used. A sheet metal part is a perfect example of this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A domain that is 100X5X8 has two dimensions negligible and the highest aspect ratio is 100/5=20. Therefore, 1D elements will be used. A truss assembly serves as an example.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once you decide the type of elements to be used, element quality comes into picture. &lt;strong&gt;To maintain quality, meshing must be done manually&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All meshing software comes with an automesh option, which works only with mappable parts and straight faces/blocks. Most of the explanations in other answers (esp. @Wes's answer) are related to what is done in the background for automesh to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea then, is to &lt;strong&gt;divide your domain into multiple patches and automesh them patch by patch and continuously ensuring connection between the patches&lt;/strong&gt;. Ensuring connectivity is mostly automatic based on a tolerance based check. 1D meshing is easier in these aspects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing is to maintain mesh flow and symmetry. Mesh flow indicates the transformation of element sizes. When you have to represent a complex feature, element size will change from bigger to smaller. This shouldn't happen in a flash and gradual change of size is to be maintained. Also, symmetrical parts should have symmetrical mesh to maintain integrity of results from FEA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the above points will help in maintaining mesh quality. However, meshing software usually have a provision to check the mesh quality using a few parameters which can be adjusted as per one's requirement. A final check on quality and connectivity is essential to ensure quality results from FEA.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Some qualities expected from a good mesh:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;from 1D mesh&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No issue with connectivity of nodes&lt;/li&gt;&#xA;&lt;li&gt;No duplicate elements&lt;/li&gt;&#xA;&lt;li&gt;Maintain minimum and maximum length&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;from 2D/3D mesh&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Less than 5 degree warpage angle {calculated by splitting a quad into two trias and finding the angle between the two planes which the trias form}&lt;/li&gt;&#xA;&lt;li&gt;Aspect ratio less than 5 {dividing the maximum length side of an element by the minimum length side of the element.}&lt;/li&gt;&#xA;&lt;li&gt;Skew angle more than 60 degree {the minimum angle between the vector from each node to the opposing mid-side and the vector between the two adjacent mid-sides at each node of the element. Ninety degrees minus the minimum angle found is reported.}&lt;/li&gt;&#xA;&lt;li&gt;Jacobian more than 0.7 {The Jacobian ratio is a measure of the deviation of a given element from an ideally shaped element. The Jacobian value ranges&#xA;from -1.0 to 1.0, where 1.0 represents a perfectly shaped element. The ideal shape for an element depends on the element type.}&lt;/li&gt;&#xA;&lt;li&gt;Tria elements with angle between 20 and 120 degree&lt;/li&gt;&#xA;&lt;li&gt;Quad elements with angle between 45 and 135 degree&lt;/li&gt;&#xA;&lt;li&gt;Maintain minimum and maximum length&lt;/li&gt;&#xA;&lt;li&gt;Element connectivity&lt;/li&gt;&#xA;&lt;li&gt;Less than 10% tria elements in 2D mesh&lt;/li&gt;&#xA;&lt;li&gt;2D element normals oriented in the same direction for a particular parts.&lt;/li&gt;&#xA;&lt;li&gt;Tet collapse for tetra elements {Defined as the distance of a node from the opposite face divided by the area of the face multiplied by 1.24}&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;from all mesh&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Numbering the nodes and elements properly in defined ranges&lt;/li&gt;&#xA;&lt;li&gt;Minimal deviation from geometry and deviation supported by sound engineering judgement.&lt;/li&gt;&#xA;&lt;li&gt;Special connections between different types (1D/2D/3D) of elements properly defined.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, all these quality parameters can vary depending on the type of analysis, accuracy required, company guidelines, and computational cost.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why these stuffs aren't automated:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Finite element analysis requires a correct mesh to give correct results. This correctness can't be defined with a few parameters and even then, they will be contradictory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again for different types of analyses, mesh quality definition may be different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Material, geometric, and contact non-linearity complicate the requirements further while defining a good mesh. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One initial roadblock I have observed using automesh feature is the incorrect representation of geometry to maintain quality of the mesh in other aspects. Both of them are important. Also, representation of geometry can be simplified with good engineering judgements which is hard to automate as it varies case by case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, Hypermesh is a very popular commercial meshing package from Altair Engineering which has a Batchmesher application that does the meshing for you. However, it fails to maintain proper geometry deviations and connections between elements for complex parts.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;tl;dr:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is how meshing is done professionally&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Decide what kind of mesh to be used&lt;/li&gt;&#xA;&lt;li&gt;Mesh the parts patch by patch and ensure proper connections&lt;/li&gt;&#xA;&lt;li&gt;Maintain mesh flow and symmetry&lt;/li&gt;&#xA;&lt;li&gt;Do all quality checks and ensure quality&lt;/li&gt;&#xA;&lt;li&gt;Ensure proper connectivity&lt;/li&gt;&#xA;&lt;li&gt;Check geometry deviations and finite element mass&lt;/li&gt;&#xA;&lt;li&gt;Deliver the model to analysts who may again re-mesh certain areas depending on the analysis requirements.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS:&lt;/strong&gt; I am new to this forum and this is one of my first few answers that I have put a lot of effort. I would really appreciate if I get some feedback. I have a few Quora answers on meshing and FEA where these points are explained in some detail with graphics. &lt;a href=&quot;https://practicalfea.quora.com/&quot; rel=&quot;noreferrer&quot;&gt;&lt;strong&gt;[Practical Finite Element Analysis]&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2918" LastEditorUserId="2918" LastEditDate="2016-02-09T19:21:41.497" LastActivityDate="2016-02-09T19:21:41.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7327" PostTypeId="2" ParentId="7325" CreationDate="2016-02-08T19:52:27.563" Score="1" Body="&lt;p&gt;I don't know for sure, but I doubt that would be the case because all of the E-Stop circuits I have worked with signal shutdown on the transition from high to low; this ensures shutdown even if the E-Stop wiring gets disconnected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how would any operator ON-OFF switches work? Would they be &quot;hot&quot; the entire time the equipment was powered off, such that the high-low transition or low state signaled power-on, just so the high state or low-high transition could signal power-off?&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-02-08T19:52:27.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7328" PostTypeId="1" AcceptedAnswerId="13451" CreationDate="2016-02-08T20:50:21.843" Score="11" ViewCount="19417" Body="&lt;h3&gt;Crushing Question&lt;/h3&gt;&#xA;&lt;p&gt;While reading a material specification for crushed aggregate, the text mentions both &amp;quot;crushed stone&amp;quot; and &amp;quot;crushed gravel&amp;quot; are acceptable. These materials sound very similar, but they are mentioned individually. I assume that this means that they are not the same.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the difference?&lt;/li&gt;&#xA;&lt;li&gt;How can I tell the difference if I see them on-site?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h3&gt;Example&lt;/h3&gt;&#xA;&lt;p&gt;This &lt;a href=&quot;https://www.michigan.gov/documents/aero/P-209_282160_7.pdf&quot; rel=&quot;noreferrer&quot;&gt;specification&lt;/a&gt; from Michigan is an example:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Aggregates shall consist of clean, sound, durable particles of crushed stone, crushed gravel, or crushed slag...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-02-09T19:17:49.550" Title="What is the difference between &quot;Crushed Stone&quot; and &quot;Crushed Gravel&quot; aggregate?" Tags="|civil-engineering|geotechnical-engineering|" AnswerCount="4" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7332" PostTypeId="2" ParentId="7218" CreationDate="2016-02-08T21:59:58.707" Score="0" Body="&lt;p&gt;I think the most simple version would place the CG of the whole system right at the handle! this would eliminate any need for active electronic or mechanical damping.&#xA;So the range of the adjustable counter weight lever would allow for working with different shapes and masses of camcorders.&#xA;you would adjust to system by holding it and moving fast to the sides and back and forth and observing any rotation or sway. then by fine tuning the distance of the counter-weight you minimize the imbalance till you're satisfied with calibration. I guess the knobs are to adjust the grip for intentional tilt and sway to capture dynamic sports scenes!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-02-08T21:59:58.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7336" PostTypeId="1" CreationDate="2016-02-09T09:09:02.633" Score="2" ViewCount="248" Body="&lt;p&gt;It's known that the positive displacement pumps are self-priming, however can I operate the pump without water in its casing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I can use the positive displacement pumps as a priming pump for the centrifugal pump, however will I need to fill its casing with water before the priming process?&lt;/p&gt;&#xA;" OwnerUserId="3591" LastEditorUserId="33" LastEditDate="2016-02-09T13:00:28.307" LastActivityDate="2016-08-04T04:59:53.067" Title="Positive displacement pumps" Tags="|fluid-mechanics|pumps|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7338" PostTypeId="2" ParentId="7336" CreationDate="2016-02-09T10:52:17.860" Score="2" Body="&lt;p&gt;Yes, even self-priming pumps generally require water in their casing to pump. 'Self-priming' generally just means that they are capable of pumping a mixture of air/water, not that they can pull a vacuum &lt;em&gt;and&lt;/em&gt; pump the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some self priming pumps are available with an integrated vacuum pump, however these are special purpose and not generally cost-effective for most applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason you couldn't use a small header tank to provide prime, instead of the positive displacement pump? E.g:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1YsTH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1YsTH.png&quot; alt=&quot;Header Tank&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2551" LastEditorUserId="2551" LastEditDate="2016-02-09T10:57:34.137" LastActivityDate="2016-02-09T10:57:34.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7343" PostTypeId="1" CreationDate="2016-02-09T15:18:26.673" Score="7" ViewCount="413" Body="&lt;p&gt;Supposedly a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tesla_valve&quot; rel=&quot;nofollow&quot;&gt;Tesla valve&lt;/a&gt; will impede fluid or air flow in one direction whilst allowing flow to move freely through the valve entering from the opposite direction, without using any moving parts. Also the alleged effectiveness of the valve is based on its length, with a longer configuration being more efficient. &#xA;Does anyone know if such a valve is scalable or does the properties of the valve diminish with greater size?&lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="272" LastEditDate="2016-02-09T17:56:10.553" LastActivityDate="2016-02-09T18:26:05.483" Title="Is it possible to replicate a Tesla valve in large scale?" Tags="|mechanical-engineering|fluid-mechanics|airflow|valves|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7344" PostTypeId="1" AcceptedAnswerId="7345" CreationDate="2016-02-09T16:13:55.363" Score="14" ViewCount="15183" Body="&lt;p&gt;I've read that Mach 0.3 is pretty much the upper limit for treating air as an incompressible fluid.  The sources I've read seem to treat this as a given, without proof or justification.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this the limit?  Is there a mathematical justification for this?  Also, does this limit only apply to air? If not, then what does the limit depend on?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2016-02-09T16:24:15.167" LastActivityDate="2019-03-05T12:40:03.327" Title="Why is Mach 0.3 the threshold separating compressible and incompressible flow?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7345" PostTypeId="2" ParentId="7344" CreationDate="2016-02-09T16:29:31.983" Score="12" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Compressible_flow&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia gives the reason&lt;/a&gt; for Mach 0.3 as due to the fact that this achieves ~5% change in density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a &lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/machrole.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA page&lt;/a&gt; that describes (analytically!) the relationship. I cited the source, but I'll reproduce the work here for posterity, in the event their links change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with conservation of momentum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;(\rho V) dV = -dp \\&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the fluid density, &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; is the velocity, and &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the pressure. for isentropic flow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ &#xA;\frac{dp}{p} = \gamma \frac{d\rho}{\rho} \\&#xA;dp = \left( \frac{\gamma p}{\rho} \right) d\rho \\&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; is the specific heat ratio. The ideal gas law gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;p = \rho R T \\&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the specific gas constant and &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the absolute temperature. So, substituting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;dp = \gamma R T d\rho &#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of sound can be calculated by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\gamma R T = a^2 \\&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the speed of sound, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;dp = a^2 d\rho \\&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the expression above into the conservation of momentum equation gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;(\rho V)dV = -a^2 d\rho \\&#xA;-\left(\frac{V^2}{a^2}\right)dV/V = d\rho/\rho \\&#xA;-M^2 dV/V = d\rho/\rho \\&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the Mach number. &lt;em&gt;This&lt;/em&gt; gives a Mach number of 0.3 to be approximately a 5% change in density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a note, this is based on the Mach number, which in turn is dependent on the speed of sound in the gas, so it's automatically adjusted on a per-gas basis.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2019-03-05T12:40:03.327" LastActivityDate="2019-03-05T12:40:03.327" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="7346" PostTypeId="2" ParentId="7343" CreationDate="2016-02-09T18:26:05.483" Score="3" Body="&lt;p&gt;This device works on momentum effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the fluid had no momentum, it could just follow the channels wherever they go, and the resistance would be the same in both directions.  In the context of the fluid, the is what happens when the viscocity is high and the speed low relative to the size of the channels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, this only works over a range of Reynolds numbers.  For example, if the channels were large, the fluid was molasses, and the speeds slow, you wouldn't see much difference in pressure between directions.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-02-09T18:26:05.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7347" PostTypeId="2" ParentId="6524" CreationDate="2016-02-09T19:20:05.920" Score="2" Body="&lt;p&gt;If I interpret this correctly this is a broadly similar arrangement to a screw jack with additional guide rods (green) . If you are talking about lifting the load with a single screw (and no other support/guides) then the situation is a little bit different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really you have two separate considerations. First is the force required to lift the load itself and secondly is the requirement to keep it flat (ie. avoid bending forces one the screw) . In an ideal case these are the same and you just have a single lifting screw directly below the centre of mass. In practice there will always be some bending moment on the screw from small variations in the load distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in practice you have the 'active' part of the system which is lifting the load and the 'passive' part which is constraining it to move only in the vertical axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like bending moments from any misalignments and variations in load distribution are going to be more of a factor than the axial loads on the shafts in this case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice this sort of condition tends to be empirical rather than something which ends itself to straightforward analysis so I would say that going with supplier tables for load is a sensible way to look at the problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the load on the shaft 40kg for a 12.5mm thread diameter seems perfectly reasonable even for a relatively low strength thread. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important thing is how well the liner bearings take bending stresses from any misalignment of the load which is down to their fit and stiffness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally I would use 4 guide rods with the screw in the centre so that they are all seeing equal loading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that my practical experience says that the arrangement is absolutely fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to support the load with a single screw and nothing else then I would say go to maybe 25-30mm diameter to be absolutely safe and reliable ( &lt;a href=&quot;http://www.tiranti.co.uk/EdgeImpactShop/product.php?Product=949&amp;amp;Content=Sketch%20Stand%20Studio%20Equipment%20%26%20Armatures%20Modelling%20Stands%20%20A%20light%2C%20yet%20stable%2C%20modelling%20stand%2C%20made%20with%20strong%20tubular%20aluminium%20legs%2C%20which%20can%20be%20folded%20for%20easy%20transport%20and%20storage.%20The%20stand&quot; rel=&quot;nofollow&quot;&gt;almost identical arrangements&lt;/a&gt; are used by stone carvers and sculptors for adjustable height work stands) &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-09T20:37:47.297" LastActivityDate="2016-02-09T20:37:47.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7348" PostTypeId="1" CreationDate="2016-02-09T20:46:54.440" Score="11" ViewCount="666" Body="&lt;p&gt;In the recent 6.4 earthquake that hit Taiwan, it was seen in a newswire photograph of rubble that metal cans were embedded in the concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tDN3n.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tDN3n.jpg&quot; alt=&quot;cans in concrete&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quote from an official regarding this practice:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For such purposes in construction, it was not illegal prior to September 1999, but since then styrofoam and formwork boards have been used instead&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;suspect&lt;/em&gt; that the reason to use it is that the cans and the associated empty space is cheaper than actually having solid concrete (and may be partly necessary to reduce the weight of the concrete and thus associated stresses with it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the 'Styrofoam and formwork boards' being the acceptable standard now makes me wonder how different that actually is from empty cans.&lt;/p&gt;&#xA;" OwnerDisplayName="user348" LastEditorUserId="33" LastEditDate="2016-02-10T20:12:23.650" LastActivityDate="2016-04-29T20:46:55.523" Title="How different is using styrofoam rather than empty cans when pouring concrete?" Tags="|civil-engineering|structural-engineering|structures|concrete|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7349" PostTypeId="2" ParentId="7275" CreationDate="2016-02-09T21:41:29.533" Score="4" Body="&lt;p&gt;In order for the fluid to not be affected at all by the movement of the plate it would require either zero friction between the plate and the liquid or for the fluid to have zero viscosity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact that the mechanics of moving the plate will inevitably involve some flexing, however small, of the tank and plate and transfer of vibrations into the tank by whatever actuator mechanism you use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also external factors like convection in the fluid and environmental vibrations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So zero perturbation is not going to be possible and you need to decide what would be an acceptable level. As well as the friction properties of the surface and the fluid in question the rate of movement of the plate and its scale will be crucial factors. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-11T12:39:48.510" LastActivityDate="2016-02-11T12:39:48.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7350" PostTypeId="2" ParentId="6433" CreationDate="2016-02-09T22:08:44.140" Score="3" Body="&lt;p&gt;This is a difficult question to answer in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some point to consider are : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Differences in density will cause one fluid to tend to sink under the other, this will me faster or slower depending on their relative densities. This immediately introduces some energy into the system. BUT their viscosities will also affect how that affects mixing. For example if you had water and ethanol, both fairy dense, low viscosity  and miscible you can imagine that the you will have transient turbulence as one flows below the other which will speed up mixing a lot. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Convection : any small differences in temperature/pressure  between the fluids will create currents and pressure waves which may speed up mixing. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In some ways this is beyond the scope of an engineering question because the approximations required for an 'ideal' answer are completely unrealistic. To me this is  a purely mathematica exercise rather than engineering per se. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of ideal gasses you are probably bes off looking at the average time it take for a gas molecule to cross from one side to the other so the average velocity component in the x axis as a function of temperature would give you some approximation. Although the conditions for two liquids (frequent collisions) compared to two low pressure gasses (few collisions may give different results) &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-09T22:22:20.377" LastActivityDate="2016-02-09T22:22:20.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7353" PostTypeId="2" ParentId="7348" CreationDate="2016-02-09T23:14:16.900" Score="9" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I &lt;em&gt;suspect&lt;/em&gt; that the reason to use it is that the cans and the associated empty space is cheaper than actually having solid concrete (and may be partly necessary to reduce the weight of the concrete and thus associated stresses with it).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You're probably (mostly) correct.  For beams, reducing the weight of the beam itself will reduce stresses in the beam and the members supporting it, as well as reducing the material cost of construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for columns, reducing weight is not necessarily a good thing.  Columns are influenced by what's called &lt;em&gt;P-M behavior,&lt;/em&gt; which is the combination of axial load and bending moments on a column.  In general, if one increases the axial load on a concrete column, one also increases the bending capacity of that column (until one of these parameters hits a critical value).  The image below shows a simplified P-M curve, and you can see that increasing axial load increases moment capacity up to about midway up the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZjrGx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZjrGx.gif&quot; alt=&quot;P-M&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;(&lt;a href=&quot;https://3.bp.blogspot.com/_Z_G2vpC6FXg/TM2efDx7Q7I/AAAAAAAAAUo/ZckDij_6Qrs/s1600/dtcolumn.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove part of a column's cross-section, you remove it's axial capacity, which is not a good thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you'll notice that there are little layers of concrete between the cans.  Ideally, one would want these larger so that the &quot;webs&quot; of concrete can transfer force between the layers.  Having essentially two separate layers of concrete ~3&quot; thick isn't strong at all.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, the 'Styrofoam and foamwork boards' being the acceptable standard now makes me wonder how different that actually is from empty cans.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In practice, there's probably not a huge difference since both introduce a void in the concrete.  Styrofoam might be a bit better since it won't deform as much as a tin can and thus hold up to the hydrostatic pressures of concrete (during a pour) better.  Also, it's probably easier to get styrofoam that is manufactured to a controlled process than it is to get consistent tin cans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you end up with is something like a waffle slab, just that the holes in the waffle are interior to the slab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D8XTI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D8XTI.jpg&quot; alt=&quot;waffle slab&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;(&lt;a href=&quot;http://civilengineersforum.com/waffle-slab/&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an aside, the engineer in me really recoils at the thought of doing this to a concrete slab.  If I saw a contractor doing that to one of my designs, we would have problems.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2016-04-29T20:46:55.523" LastActivityDate="2016-04-29T20:46:55.523" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7354" PostTypeId="1" AcceptedAnswerId="7358" CreationDate="2016-02-10T00:06:52.717" Score="4" ViewCount="2538" Body="&lt;p&gt;I get that you normalize stress to engineering stress, so that it is independent of the cross sectional area. This means you can calculate (roughly) the stress needed to elongate the specimen to the degree you want. But why normalize the strain aswell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have thought that the orginal length of the specimen does not matter. Lets say you want to make the specimen 2 mm bigger, then I would have thought that you would need the same engineering stress no matter how long the specimen is. Thus that elongation is only dependant on cross sectional area and load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example you have a specimen, with orginal length, of 60 mm and 100 mm and then you want (l-lo) to be 2 mm. I would have thought that it would take same amount of engineering stress. And that a 2% strain would lead to different stresses needed, since in one this means an elongation of 1.2 mm and with the other an elongation of 2 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that the length of a specimen also determines how 'easily' a specimen elongates?&#xA;That it more difficult to elongate a 100 mm to 102 mm than a 60 to 62 mm. Because the tensile stress-strain curve implies that to elongate a 60 mm specimen to 61.2 takes same amount of stress as elongating a specimen to 62 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the length of the specimen relates to how easily you can elongate it, does this then also explain why the gauge length should be specified when giving the percent elongation (%EL) when talking about ductility? Because most of the plastic deformation is at the neck, so not uniform and thus independant of the length of the specimen orginally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So before necking the elongation is dependant of orginal length and engineering stress and after necking is independant of orginal length?&lt;/p&gt;&#xA;" OwnerUserId="4117" LastActivityDate="2016-02-10T05:27:52.090" Title="Why is strain also normalized to the parameters of engineering strain?" Tags="|materials|metallurgy|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7355" PostTypeId="1" CreationDate="2016-02-10T01:50:42.707" Score="5" ViewCount="248" Body="&lt;p&gt;I'm learning about how processors send an address to ram through the address bus, and receive and send data from ram through the data bus. In the videos I have watched, the processor was an 8 bit processor that had 8 wires coming out for the data bus, and 8 wires coming out for the address bus, where each wire carried one bit of information, either off or on. That got me wondering. Does a 32 bit processor basically work the same way, but it can send more bits through the address/data bus at one time. What about a 64 bit processor. I haven't learned about registers yet, but I'm getting there. Does it have something to do with the size of the registry? I understand the parts of a processor, but I don't understand what people mean when they classify a processor by its bit capacity.&lt;/p&gt;&#xA;" OwnerUserId="4912" LastEditorUserId="1586" LastEditDate="2016-02-11T01:52:35.440" LastActivityDate="2016-02-11T01:52:35.440" Title="What is the significance of a processors bit capacity?" Tags="|electrical-engineering|computer-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7356" PostTypeId="2" ParentId="7354" CreationDate="2016-02-10T03:09:02.887" Score="2" Body="&lt;p&gt;Indeed, deformation is a function of an element's length. The longer the element, the lower the stress required to obtain an equal deformation. This is because stress is proportional to strain (adimensional, such as 2%), not to deformation (such as 2 mm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you have two objects: a crystaline structure composed of nine iron atoms ($\alpha$ allotrope), and a twelve-$\alpha$-iron-crystal-long strand. Both can be seen below. You apply the same tensile force $F$ to each of the objects in the horizontal direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cAusNm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cAusNm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5INL3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5INL3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what does it mean when an element is stretched? It means that all of their constituent particles and their chemical bonds are stretched. The single crystal deforms by some very tiny value $\delta$. The &quot;chain&quot; however, transmits this same force $F$ along each of its constituent blocks, such that each of them also deform by the same value $\delta$ as the single crystal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you applied a force $F$ to each of the objects. The single crystal deformed such that its length is now equal to $\ell + \delta$. The chain, however, deformed such that it is now $L + 12\delta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so you see that for the same force (or stress), two objects of equal cross-section but different lengths present different elongations (in linear units). However, they do present the same deformations $\epsilon$ (adimensional), since $\epsilon = \dfrac{\delta}{\ell} = \dfrac{12\delta}{L} = \dfrac{12\delta}{12\ell}$.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-02-10T03:09:02.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7357" PostTypeId="2" ParentId="7355" CreationDate="2016-02-10T03:29:34.197" Score="2" Body="&lt;p&gt;A 64 bit processor has a datapath and memory address width of 64 bit. The integer size is also 64 bits. In addition the registers, address bus and data bus is 64 bits wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise in a 32 bit processor the datapath, memory address width, integer size, register size, etc are 32 bits wide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the free online &lt;a href=&quot;https://www.edx.org/course/embedded-systems-shape-world-utaustinx-ut-6-03x&quot; rel=&quot;nofollow&quot;&gt;embedded systems course&lt;/a&gt; from University of Texas on &lt;a href=&quot;http://www.edx.org/&quot; rel=&quot;nofollow&quot;&gt;edx.org&lt;/a&gt; that might interest you. This topic is covered and there are like minded student you can bounce back questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/64-bit_computing&quot; rel=&quot;nofollow&quot;&gt;64-bit computing&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/32-bit&quot; rel=&quot;nofollow&quot;&gt;32-bit computing&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.edx.org/course/embedded-systems-shape-world-utaustinx-ut-6-03x&quot; rel=&quot;nofollow&quot;&gt;Embedded Systems - Shape The World&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="1586" LastEditDate="2016-02-11T01:03:51.850" LastActivityDate="2016-02-11T01:03:51.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7358" PostTypeId="2" ParentId="7354" CreationDate="2016-02-10T05:06:00.123" Score="5" Body="&lt;p&gt;From what I understand of the question, there are three interlinked concepts that are causing confusion, each of which could be a separate question.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;1. Is axial stress dependent on length of a specimen?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Axial &lt;strong&gt;stress is independent&lt;/strong&gt; of length. @Wasabi's answer goes into more detail, but the core concept is that a stress of $\sigma$ on a specimen of length $L$ has the same effect on a specimen &lt;em&gt;per unit length&lt;/em&gt; as a stress of $\sigma$ on a specimen of length $2L$, or indeed of any length.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;2. Is axial strain dependent on length of a specimen?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Axial &lt;strong&gt;strain is independent&lt;/strong&gt; of length. This follows from the answer to (1.) above, and from Hooke's law, specifically that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sigma = E \varepsilon&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that if $\sigma$ is independent of length, so must be $\varepsilon$ since they are linearly related.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;3. Is axial deformation dependent on length of a specimen?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Axial &lt;strong&gt;deformation is dependent&lt;/strong&gt; on length. It also depends on cross-sectional area (assuming constant cross-section) and tensile modulus. This follows from a quick and dirty derivation of the axial deformation formula. Consider a constant-cross-section tensile member of area $A$, length $L$, experiencing a uniform load $P$, with modulus $E$. The member will experience a deformation of $\delta$. Note that by definition $\varepsilon = \delta / L$ and $\sigma = P / A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Hooke's law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sigma = E \varepsilon&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and substituting our notes gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{P}{A} = E \frac{\delta}{L} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which after some rearranging yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\delta = \frac{PL}{AE}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the equation for axial deformation of constant-cross-section tensile members. As should be obvious from the equation, axial deformation depends on length, as well as area.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How Can I Think About This Intuitively?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Elasticity is just like springs, and indeed Hooke's law is also used for linear springs. Now consider two springs such that one has twice the number of coils as the other, but the rest of their parameters are the same. Assume they are also weightless. If you held just the end of each spring, hooked identical weights to the other ends and let them uncoil toward the floor, the spring with twice the number of coils would extend twice as far. As a result, the change in length per per unit original length, equivalent to the strain, would be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same is true of two tension rods, one twice as long, experiencing the same stress with the same cross-sectional area. The one that is twice as long would have double the deformation, but they would have the same strain.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How Can I Use This Information?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As an example, consider a tensile rod. Suppose you must minimize axial deformation. Then you would choose a material with high tensile modulus so that stress has less effect on strain, and thus less effect on deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose instead you have a family of tensile rods all made of the same material. If each must not deform more than 1 mm, then the longer rods must have a larger cross-sectional area to make up for the additional &quot;slack&quot; from their added length.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What About Plastic Deformation?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The above three questions only apply in the elastic regime. In the plastic regime, volume is practically conserved in most materials, which means that any instantaneous extension of a tensile specimen must cause an instantaneous reduction of area such that the volume gained by increasing length is equal to the volume lost by reducing the cross-section area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is an instability inherent to tensile plastic deformation called necking, which you noted. Namely, if one length-wise segment of a specimen is slightly narrower than the rest of the specimen, stresses will be concentrated at the narrow region and it will reach the elastic limit slightly before the rest of the specimen. Plastic deformation will thus begin at the slightly narrower section, causing a greater decrease in area than the rest of the specimen, further concentrating stresses, which causes further and greater deformation. The specimen will tend to fail wherever it is narrowest, and will do so unstably in a smooth, constant-cross-section specimen. That is, it is impossible to predict where failure will occur without intentionally notching the specimen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware! Necking does not necessarily proceed the same way in polymers. In polymers, necking is associated with polymer chain alignment, which locally strengthens the material so much it prevents further deformation there, unlike metals. Instead, the neighboring regions begin to neck causing them to strengthen, and so on, until the entire specimen becomes aligned.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-02-10T05:27:52.090" LastActivityDate="2016-02-10T05:27:52.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7359" PostTypeId="2" ParentId="7355" CreationDate="2016-02-10T08:39:40.177" Score="3" Body="&lt;p&gt;There are several components in a microprocessor that may all have their individual size: data bus, address bus, registers, and ALU (arithmetic logic unit). Modern CPUs also have caches, memory management units, floating point and single instruction multiple data (SIMD) units. (Also nowadays you have multiple CPU cores in one CPU.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All those units may have their own bit size. E.g. the 68000 had a 16 bit data bus, 24 bit address bus, 32 bit registers and a 16 bit ALU (and none of the modern stuff).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically, the processor classification in bits (8, 16, 32, 64 and so on) gives the size of the ALU. This means that an 8 bit processor can handle 8 bit of data in one cycle, a 16 bit processor 16 bit data and so on. This does not mean that they can't handle larger data, it just takes longer. E.g. the 68000 required two cycles to add two 32 bit values.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2016-02-10T08:39:40.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7364" PostTypeId="1" AcceptedAnswerId="7366" CreationDate="2016-02-10T14:46:07.643" Score="3" ViewCount="9564" Body="&lt;p&gt;When calculating the reactions of a simply supported beam under any type of loading we assume that the moment about the reaction points is &lt;strong&gt;zero&lt;/strong&gt;. What if our assumption is wrong and in actual practice there is a rotational effect about that point? Why do we make such inappropriate assumption?&lt;/p&gt;&#xA;" OwnerUserId="4889" LastEditorUserId="368" LastEditDate="2016-02-10T19:15:43.047" LastActivityDate="2016-02-10T19:35:05.593" Title="Why do we assume the moment is zero when analyzing a simply supported beam?" Tags="|mechanical-engineering|beam|statics|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7365" PostTypeId="2" ParentId="7348" CreationDate="2016-02-10T14:49:38.297" Score="3" Body="&lt;p&gt;The inclusion of cans was almost certainly to reduce the volume of (expensive) concrete and had nothing to do with any weight savings by including voids. This type of practice has been seen throughout the world when building standards are lax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quote about this type of construction not being illegal before 1999 may be due to confusion on the part of the journalist. It could be a case where any building standards were not enforced before 1999. If the new (1999) standard allowed for Styrofoam voids, it doesn't mean that the concrete was designed to use them. It would just be allowable if engineered that way from the beginning.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2016-02-10T14:49:38.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7366" PostTypeId="2" ParentId="7364" CreationDate="2016-02-10T15:04:02.707" Score="-1" Body="&lt;p&gt;I got it as we can observe there will be no rotation because of the supports so that is the reason we take moment about that point as zero&lt;/p&gt;&#xA;" OwnerUserId="4889" LastActivityDate="2016-02-10T15:04:02.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7367" PostTypeId="2" ParentId="7364" CreationDate="2016-02-10T16:23:49.650" Score="8" Body="&lt;p&gt;This rule is typically applied when studying statics. Static means that your structure or object does not move. If the moments didn't all add up to zero, that would mean there was a net force action on the object, which would cause it to accelerate and move. Since it is static and not moving, we know that whatever forces are acting on the object, they must add up to zero. If you were studying something that moves like a see-saw then the moments might not add up to zero and there is potential for the object to move, but if you're studying a beam in a building you can be sure that the moments add to zero unless the building is falling down!&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-02-10T16:23:49.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7368" PostTypeId="1" AcceptedAnswerId="7369" CreationDate="2016-02-10T16:28:45.660" Score="3" ViewCount="11447" Body="&lt;p&gt;In the diagram below, the compression member is $AC$. What is its unbraced length for determining buckling capacity? This also matters because certain codes put limits on the maximum value of $\frac{KL}{r}$ for bracing members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ntDtA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ntDtA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the conservative answer is to consider the entire length, AC. This can be too conservative when lots of braces are used (total weight of steel increases). It can also be conservative if the actual compressive force is low enough that a member size is increased solely to meet the $\frac{KL}{r}$ requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can justify to myself that the tension member braces the compression member in one direction (in the plane of the page), but can the tension member be considered to provide bracing in the orthogonal direction (in and out of the page)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two reasons why I think this might be possible:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The tension brace provides resistance solely from being present. (i.e. something is better than nothing)&lt;/li&gt;&#xA;&lt;li&gt;The other brace is in tension so it will provide additional restraint similar to a bow string.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note: Some codes have requirements that a brace be able to withstand 5% of the axial force in the member being braced. I would think that this would be the lower limit for restraint coming from the tension member.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="2246" LastEditDate="2016-02-10T18:57:00.137" LastActivityDate="2016-02-27T17:04:46.660" Title="What is the unbraced length of the compression portion of an X brace?" Tags="|structural-engineering|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7369" PostTypeId="2" ParentId="7368" CreationDate="2016-02-10T18:56:01.383" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is its unbraced length for determining buckling capacity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;AISC defines $L$ as the &lt;strong&gt;laterally&lt;/strong&gt; unbraced length of the member (emphasis mine) (AISC 14th Ed. &lt;em&gt;Steel Construction Manual&lt;/em&gt;, Part 16, Section E2).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I can justify to myself that the tension member braces the compression member in one direction (in the plane of the page)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would agree.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;but can the tension member be considered to provide bracing in the orthogonal direction (in and out of the page)?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Note: Some codes have requirements that a brace be able to withstand 5% of the axial force in the member being braced. I would think that this would be the lower limit for restraint coming from the tension member.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Simply put, I don't think you'll be able to argue that the tension member provides enough out-of-plane resistance to bending of the compression member since the tension member would resist the out-of-plane load via bending, which is a relatively &quot;soft&quot; support condition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This also matters because certain codes put limits on the maximum value of $\frac{KL}{r}$ for bracing members.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Further discussion from the Commentary Section E2 says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H7u2X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H7u2X.png&quot; alt=&quot;Commentary&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's generally a good idea to stay under 200 for your slenderness ratio, though the Commentary discussion appears to give you an &quot;out&quot; if you think you need it.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2016-02-10T18:56:01.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7370" PostTypeId="1" AcceptedAnswerId="7372" CreationDate="2016-02-10T19:07:51.610" Score="2" ViewCount="75" Body="&lt;p&gt;I'm trying to design a small metal telescopic track (6cm x 2cm when extended) where small objects may be arbitrarily positioned inside:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GNTMw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GNTMw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above example, the green block jams the tracks as the bottom cannot move upward. Does anyone know of a better design such that the tracks won't be jammed? I've thought about putting a sleeve on the inside, but if it was 6cm long and covers the entire extended length, it'd overhang when retracted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4303" LastActivityDate="2016-02-11T21:02:28.607" Title="Avoiding jams in telescopic tracks" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7371" PostTypeId="2" ParentId="7364" CreationDate="2016-02-10T19:26:26.233" Score="2" Body="&lt;p&gt;This is something of a nonsensical question, as zero moment is not an assumption here but part of the &lt;strong&gt;definition&lt;/strong&gt; of this class of beam problem. A &quot;simply supported&quot; beam is any beam that can be represented as supported at one end by a hinge/pin and at the other by a roller (or, sometimes another hinge/pin). These are, again, &lt;em&gt;classes&lt;/em&gt; of supports that &lt;em&gt;represent&lt;/em&gt; real-world objects; in particular, they represent real-world objects that are not capable of transferring a moment to the member they support. &lt;strong&gt;By definition&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Rotational effect&quot; is not a meaningful term. The requirement that moment is zero means that the ends of the beam are free to rotate. However, if you meant &lt;em&gt;what if there is a nonzero moment&lt;/em&gt; transferred at the support(s) of an actual beam, the answer is simply that the type of analysis used for the &quot;simply supported&quot; class of problem would no longer apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you chose to represent a real beam as simply supported in your analysis and then found that there was actually a moment being transferred at one or both ends, the only bad assumption would be that this real beam could accurately be represented as a simply supported beam. You would have to classify it as something else. Not every real-world beam can be adequately analyzed using some common framework like &quot;simply supported&quot; or &quot;fixed-fixed&quot; or &quot;cantilever&quot;—some problems are more complex, and don't have a name. Some, we don't have an analytical solution for at all.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-02-10T19:35:05.593" LastActivityDate="2016-02-10T19:35:05.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7372" PostTypeId="2" ParentId="7370" CreationDate="2016-02-10T19:49:03.283" Score="2" Body="&lt;p&gt;Here are a few basic solutions you could try:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Fist, you could put a pin or bump sticking out from the center of the inside channel so that before the channel bumps the edge of the green block (wedging it) the pin bumps the center of the green block (un-wedging it.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You could try making the inside of the track something smooth so the green block is less likely to get wedged. For example a thin layer of smooth plastic on the inside of the bigger channel would help the block slide more easily.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You could install an accordion-folding guard that keeps objects away from the sides of the large channel when the small channel is extended. It would require you to make the large channel a little longer, but not twice as long like you would have to do with your original solution.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-02-10T19:49:03.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7373" PostTypeId="1" AcceptedAnswerId="7450" CreationDate="2016-02-11T03:05:28.320" Score="1" ViewCount="488" Body="&lt;p&gt;I am a draftsman. I make project of mechanical pieces and machinery. Some pieces works like knifes or sleeve, so to reduce the friction we design the ideal material to manufacture this pieces. The steel Vc 130 or 131 are options of choices. What is the usage of steel grades VC-130 and VC-131? What the difference in applications between them, on tools and gears projects?&lt;/p&gt;&#xA;" OwnerUserId="4913" LastEditorUserId="4913" LastEditDate="2016-02-11T17:37:19.263" LastActivityDate="2016-02-15T23:14:08.423" Title="Difference between VC-130 and VC-131 steel" Tags="|mechanical-engineering|steel|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7374" PostTypeId="1" AcceptedAnswerId="7380" CreationDate="2016-02-11T06:28:32.960" Score="2" ViewCount="1275" Body="&lt;p&gt;I need to import an assembly from CATIA into SimMechanics but in CATIA, the .stl file format is available only for part files and not for assembly files. How can I do this? Is it possible to convert between the .CATProduct and .stl file formats?&lt;/p&gt;&#xA;" OwnerUserId="3003" LastEditorUserId="368" LastEditDate="2016-02-11T17:05:14.497" LastActivityDate="2016-02-11T17:06:03.667" Title="How to import a CATIA assembly into SimMechanics" Tags="|matlab|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7375" PostTypeId="1" AcceptedAnswerId="7381" CreationDate="2016-02-11T06:38:11.087" Score="6" ViewCount="6514" Body="&lt;p&gt;What is the down-stand beam called at an edge of a surface bed or raft foundation? It is not exactly a grade beam or stem wall. It is more like a toe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DtqXP.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DtqXP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="2246" LastEditDate="2016-02-11T14:27:50.363" LastActivityDate="2021-12-11T13:41:33.993" Title="What is the down-stand beam called at an edge of a surface bed or raft foundation?" Tags="|structural-engineering|foundations|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7376" PostTypeId="2" ParentId="7348" CreationDate="2016-02-11T07:56:22.877" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jZ1bC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jZ1bC.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one can see, an organised, designed way of construction does work!&#xA;Styrofoam blocks have been used in the past. Controlling uplift is the big challenge. This has been addressed by the various manufacturers of the void formers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/46XoJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/46XoJ.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most notable building is the Square at the Frankfurt airport which is above the railway station and houses a Hilton.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Creating voids in slabs using voidformers (cans, or in this case plastic balls) is an accepted way of construction and fullfills Eurocode requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The benefits:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;reduction in self weight &lt;/li&gt;&#xA;&lt;li&gt;reduction in member sizes due to less load being carried&lt;/li&gt;&#xA;&lt;li&gt;floor-to-floor height of buildings is reduced because beams are smaller&lt;/li&gt;&#xA;&lt;li&gt;foundations can be smaller&lt;/li&gt;&#xA;&lt;li&gt;cost savings due to the above&lt;/li&gt;&#xA;&lt;li&gt;CO2 footprint of the building is less due to less material being used and the CO2 emissions during production of cement are reduced.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;More detail can be found at www.cobiax.com.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="368" LastEditDate="2016-02-11T17:01:11.420" LastActivityDate="2016-02-11T17:01:11.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7377" PostTypeId="1" CreationDate="2016-02-11T08:39:51.930" Score="3" ViewCount="105" Body="&lt;p&gt;I am trying to model a prototype resembling a section of vein channels. I am wondering if there are some basic rules for such fluid channels, for instance when a vein splits in two capillaries, is there some quantity conserved (area of the section or else)? &#xA;I observed that with my current design (which is done by hand just imitating pictures I saw) the fluid I pump in tends to flow very fast in some channels and very slow in others. I'd want some sort of common front to advance.&#xA;I am a mathematician and have ZERO education on fluid dynamics or else :) any help is much appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I saw the comment answer and I apologise because I am not sure I will be able to formulate my question in a decent way for engineers :). I am printing my model in rigid plastic and the application is scanning its section in the X-ray CT lab. Here's a couple of pics that I hope will clarify my position:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/5TVYL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5TVYL.png&quot; alt=&quot;My current prototype&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FindS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FindS.png&quot; alt=&quot;I need to scan the section&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to have a viscous fluid (honey+iodine) put in the top &quot;funnel&quot; and start measuring in the CT lab, observing it flowing into the bottom channels.&#xA;Observe that the channels have a rectangular section. &lt;/p&gt;&#xA;" OwnerUserId="4929" LastEditorUserId="4929" LastEditDate="2016-02-12T09:01:53.050" LastActivityDate="2016-02-12T13:53:57.850" Title="Are there basic rules to properly design vein channels?" Tags="|fluid-mechanics|microfluidics|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7378" PostTypeId="2" ParentId="7370" CreationDate="2016-02-11T13:43:26.633" Score="1" Body="&lt;p&gt;Add a third track that fits inside the moving track, but is fixed to the outer track. That way, objects are always moving from a narrow track to a wider track, regardless of the position of the moving track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or just make the outer track the one that moves in the first place.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-02-11T13:43:26.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7379" PostTypeId="2" ParentId="7275" CreationDate="2016-02-11T13:47:16.853" Score="4" Body="&lt;p&gt;As an addition to @ChrisJohns answer: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you want to limit the effect of the plate oscillations on the motion of the liquid as much as possible you need to make the time scale for viscous diffusion much larger than that of the plate oscillations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the plate oscillates with an angular frequency $\omega$, the liquid has a kinematic viscosity $\nu$ and the length scale is the height of the box $H$ then you want the ratio:&#xA;$$\epsilon=\frac{\nu}{\omega H^2}\ll1$$&#xA;here $\frac{H^2}{\nu}$ is the viscous time scale and $\frac{1}{\omega}$ is the oscillation time scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by @ChrisJohns, this requires very low viscosities or large angular frequencies. However, there will alway be some pertubation of the liquid by the plate but $\epsilon\ll1$ should limit it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take note, that this assumes that the height of the box is the characteristic length scale; this assumes the absence of any interference by walls. I would redesign the box to have a height much smaller than any of the other two dimensions if possible but still such that $\epsilon\ll1$.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastEditorUserId="3100" LastEditDate="2016-02-11T15:24:25.287" LastActivityDate="2016-02-11T15:24:25.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7380" PostTypeId="2" ParentId="7374" CreationDate="2016-02-11T14:03:28.023" Score="4" Body="&lt;p&gt;&lt;strong&gt;Short answer&lt;/strong&gt;: you can't directly import CATIA assemblies in SimMechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long answer&lt;/strong&gt;: SimMechanics supports import of CAD models (assemblies or parts) from Pro/Engineer, SolidWorks or Autodesk Inventor via SimMechanics Link (see &lt;a href=&quot;http://uk.mathworks.com/products/simmechanics/features.html#importing-cad-models&quot; rel=&quot;nofollow&quot;&gt;Importing CAD Models&lt;/a&gt; on the MathWorks website). The &lt;a href=&quot;http://uk.mathworks.com/products/simmechanics/download_smxmlschema.html&quot; rel=&quot;nofollow&quot;&gt;SimMechanics Import XML Schema&lt;/a&gt; should allow import from any CAD system, but you need to be able to export to XML from your CAD system, and the XML must follow the SimMechanics Import XML Schema. For you, that probably means having to write a custom interface layer in CATIA to export to the correct XML file schema.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="368" LastEditDate="2016-02-11T17:06:03.667" LastActivityDate="2016-02-11T17:06:03.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7381" PostTypeId="2" ParentId="7375" CreationDate="2016-02-11T14:11:02.040" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the down stand beam called at an edge of a surface bed or raft foundation? It is not exactly a grade beam or stem wall. It is more like a toe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is probably a regional nomenclature thing, since I'm seeing quite a few ways for naming this.  If you call it a grade beam, I don't think anyone will take it amiss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples below of different ways of naming it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Grade Beam&lt;br&gt;&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GQF9M.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GQF9M.jpg&quot; alt=&quot;grade beam&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;(&lt;a href=&quot;http://makatimidrisecondo.weebly.com/mat-foundations.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Exterior Beam&lt;br&gt;&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Jcje7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jcje7.png&quot; alt=&quot;exterior beam&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;(&lt;a href=&quot;http://www.jwkhomeinspections.com/blog/joe-keresztury-san-antonio-home-inspector-new-home-slab-foundation.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stem Wall or Thickened-Edge Slab&lt;br&gt;&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ndOIA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ndOIA.jpg&quot; alt=&quot;stem wall&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;(&lt;a href=&quot;https://www.nachi.org/structural-design-foundations-home-inspector.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;source)&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Beam Strip&lt;br&gt;&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Aa5bE.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Aa5bE.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;(&lt;a href=&quot;http://www.abuildersengineer.com/2013_01_01_archive.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2016-02-11T14:11:02.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7382" PostTypeId="2" ParentId="6865" CreationDate="2016-02-11T16:40:21.077" Score="0" Body="&lt;p&gt;It is possible to use a venturi nozzle to create a low pressure/high volume airflow from a high pressure/low volume one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence these are very similar to the way that a venturi gas burner works from a pressurised gas tank. The way it works is that high pressure gas is expanded through a small diameter nozzle, accelerating it and causing a consequent pressure drop which sucks more air into the stream through holes in the back of a large diameter shroud. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is &lt;a href=&quot;http://www.lasertools.co.uk/product/5539&quot; rel=&quot;nofollow&quot;&gt;one example&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-11T16:40:21.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7383" PostTypeId="2" ParentId="7370" CreationDate="2016-02-11T16:47:13.637" Score="1" Body="&lt;p&gt;If you beveled the edges of the end of the inner track so that they form a ramp rather than a square shoulder that would help a lot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if the two tracks are not a very close fit and the objects in question could jam in the gap some sort of flexible baffle may help. In both cases the idea is to create a smooth ramp transition between the two rail elements so there are no sharp corners for things to snag on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly having non-parallel sides (some approximation of a 'V' section) is also likely to help.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-11T21:02:28.607" LastActivityDate="2016-02-11T21:02:28.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7384" PostTypeId="1" CreationDate="2016-02-11T17:07:23.987" Score="3" ViewCount="94" Body="&lt;p&gt;I'm doing my senior design class and I have design a chemical plant that produces phosphoric acid by reacting phosphate rock with sulfuric acid. The phosphate rock has a lot of impurities, some of which I know how to separate but there are some impurities that I'm completely lost about. The impurities are: $\text{Al}_2\text{O}_3$, $\text{Fe}_2\text{O}_3$, $\text{MgO}$, $\text{Na}_2\text{O}$, $\text{K}_2\text{O}$. How would these react in a mixture of water, sulfuric acid and phosphoric acid and how would I purify them out of my crude phosphoric acid? Or at least, what's a reasonable assumption to make to make my plant design a little straightforward.&lt;/p&gt;&#xA;" OwnerUserId="4935" LastEditorUserId="1832" LastEditDate="2016-02-11T17:16:55.660" LastActivityDate="2016-02-11T17:16:55.660" Title="Purification of wet-process phosphoric acid" Tags="|control-engineering|chemical-engineering|process-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7385" PostTypeId="1" CreationDate="2016-02-11T18:07:21.967" Score="3" ViewCount="49" Body="&lt;p&gt;I have a balun of rated voltage 20V; however, my transmitted voltage is approximately 1V. Would the balun operate under this condition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the data sheet of the balun only the rated voltage and the max voltage (20V and 120V, respectively) are given.&lt;/p&gt;&#xA;" OwnerUserId="4937" LastActivityDate="2016-02-12T20:13:05.320" Title="Can a balun work below rated voltage?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7387" PostTypeId="1" AcceptedAnswerId="7422" CreationDate="2016-02-11T20:53:05.213" Score="0" ViewCount="544" Body="&lt;p&gt;Consider two streams of air flowing through the nozzles to an unlimited volume with atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flow is the same, but the cross section areas of the circular nozzles and the speeds differ, so the flow through the smaller nozzle is proportionally faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, the stream which is faster at the nozzle would transfer the air farther.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any empirical rule which can be used to measure air stream dissipation rates given the initial speed and nozzle cross section area?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of Lattice-Bolzmann numerical methods, and can run the experiments too, but as i am new to air dynamics, may be I am overlooking something obvious.&lt;/p&gt;&#xA;" OwnerUserId="4294" LastEditorUserId="4294" LastEditDate="2016-02-12T13:46:34.327" LastActivityDate="2019-03-06T05:20:35.303" Title="Air flow dissipation rate" Tags="|airflow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7388" PostTypeId="1" AcceptedAnswerId="7419" CreationDate="2016-02-12T04:03:54.420" Score="-1" ViewCount="1917" Body="&lt;p&gt;I am considering a two wheeler petrol engine. How is the mass of air entering the intake manifold calculated? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that I have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Manifold Absolute Pressure sensor&lt;/li&gt;&#xA;&lt;li&gt;Intake Air Tmperature sensor&lt;/li&gt;&#xA;&lt;li&gt;Engine Temperature sensor&lt;/li&gt;&#xA;&lt;li&gt;Throttle Position sensor&lt;/li&gt;&#xA;&lt;li&gt;Engine rpm sensor &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please help me understand how to estimate the amount of air entering the engine for combustion.&lt;/p&gt;&#xA;" OwnerUserId="4942" LastEditorUserId="33" LastEditDate="2016-02-12T14:00:19.907" LastActivityDate="2016-02-13T20:59:07.057" Title="How is the mass of air in the intake manifold calculated in an electronically fuel injected vehicle?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7389" PostTypeId="1" AcceptedAnswerId="7392" CreationDate="2016-02-12T06:25:46.467" Score="2" ViewCount="261" Body="&lt;p&gt;Similarly, does there exist an index with the maximum building depths and heights in the United States?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this map of soil types in California by UC Davis, but I don't think this directly translates into maximal building depths: &lt;a href=&quot;http://casoilresource.lawr.ucdavis.edu/gmap/&quot; rel=&quot;nofollow&quot;&gt;http://casoilresource.lawr.ucdavis.edu/gmap/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4940" LastActivityDate="2016-02-12T10:46:22.220" Title="What practical geographical and engineering constraints limit the depth to which you can build underground?" Tags="|civil-engineering|structural-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7390" PostTypeId="1" AcceptedAnswerId="7393" CreationDate="2016-02-12T06:39:03.733" Score="-4" ViewCount="1007" Body="&lt;p&gt;I have been investigating pros and cons USB Type-C. Obviously their are many advantages in innovation, so are the disadvantages. On such disadvantage is &lt;a href=&quot;http://money.cnn.com/2016/02/05/technology/usb-c-cords/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Those new USB-C cords can fry your laptop&quot;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/05CeZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/05CeZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Advantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Good bye to proprietary laptop power brick.&lt;/li&gt;&#xA;&lt;li&gt;Support both USB 2.0 and USB 3.0&lt;/li&gt;&#xA;&lt;li&gt;Higher data rate of 5 Gbps with USB 3.0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Disadvantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Poor design could fry your expensive laptop.&lt;/li&gt;&#xA;&lt;li&gt;Shorter cable length. USB 3.0 cable length is 10 feet, where is USB 2.0 is 16 feet.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: What are other disadvantages that need to be considered when recommending USB Type-C?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/USB#USB_Type-C&quot; rel=&quot;nofollow noreferrer&quot;&gt;USB&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/USB_Type-C&quot; rel=&quot;nofollow noreferrer&quot;&gt;USB Type-C&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://giuseppemelle.wordpress.com/2010/06/06/usb-3-0-the-pros-and-cons/&quot; rel=&quot;nofollow noreferrer&quot;&gt;USB 3.0 – The Pros and Cons&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.laptopmag.com/articles/usb-type-c-faq&quot; rel=&quot;nofollow noreferrer&quot;&gt;USB Type-C FAQ: Everything You Need to Know&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.usb.org/developers/presentations/pres1108/SuperSpeed_USB_DevCon_Cable_and_Connnector_Ling.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;SuperSpeed USB Developers Conference - USB.org&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.chassis-plans.com/should-usb-3-0-be-used-instead-of-usb-2-0-in-your-next-computer/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Should USB 3.0 Be Used Instead of USB 2.0 in Your Next Computer?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2016-02-12T13:10:11.133" Title="What are the documented pros and cons of USB Type-C?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2016-02-18T02:44:18.273" ContentLicense="CC BY-SA 3.0" />
  <row Id="7391" PostTypeId="2" ParentId="7377" CreationDate="2016-02-12T07:18:44.673" Score="2" Body="&lt;p&gt;If I understand your question correctly, you are designing and building a small network of pipes or the like that involves different diameters. You have little experience with this kind of design work. You are looking for rough heuristics to help you get started. I assume pipe flow, not open channels.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... when a vein splits in two capillaries, is there some quantity conserved (area of the section or else)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is actually a good idea. In a first approximation, if you conserve area flow velocity will be equal in all branches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd advise you however to look at and understand the &lt;a href=&quot;https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation&quot; rel=&quot;nofollow&quot;&gt;Darcy-Weisbach equation&lt;/a&gt;. Looks daunting, but its probably less work to understand than to build endless prototypes. The upshot is that specific pressure loss (pressure / length) depends strongly on flow velocity, but not only - so our first approximation is wrong but maybe useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing you need to understand is &lt;a href=&quot;https://en.wikipedia.org/wiki/Pipe_network_analysis&quot; rel=&quot;nofollow&quot;&gt;pipe network analysis&lt;/a&gt;. Then you can build a simulated model of your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give that you say &quot;ZERO education on fluid dynamics&quot; I'll throw in a link on how to &lt;a href=&quot;http://www.engineeringtoolbox.com/pump-system-curves-d_635.html&quot; rel=&quot;nofollow&quot;&gt;read pump and system curves&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="33" LastEditDate="2016-02-12T13:53:57.850" LastActivityDate="2016-02-12T13:53:57.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7392" PostTypeId="2" ParentId="7389" CreationDate="2016-02-12T10:46:22.220" Score="7" Body="&lt;p&gt;The deepest human made excavation is the &lt;a href=&quot;https://en.wikipedia.org/wiki/TauTona_Mine&quot; rel=&quot;noreferrer&quot;&gt;Tau Tona Mine&lt;/a&gt; in South Africa, which is 3900 metres deep. This is accessed by two vertical shafts, each 2000 metres long. At the bottom of the first shaft there is a short tunnel, at the end of which is the top of the second shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The depth limit on the shafts is due to the capabilities of the winding ropes: due to the weight of the ropes,  the strength of the steel used, the unavoidable manufacturing flaws in the ropes and the fatigue induced in the ropes over time as the ropes are wound and unwound from the drum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to dig an excavation for a building we will need to get the unwanted material out of the excavation so we will need a winder and a kibble (the bucket used to hoist the material out of the hole). With current technology, this limit is 2000 metres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other important factors which limit the depth to which we can dig and construct anything underground are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ground stresses - if they're too large the excavation for your construction will collapse or be squeezed so its cross-sectional area is reduced.&lt;/li&gt;&#xA;&lt;li&gt;Geothermal heat - if the temperature is too hot, people can't work in the excavation for the construction&lt;/li&gt;&#xA;&lt;li&gt;Competency of the material being dug - the strength of the material being dug will determine the maximum dimension of the walls of the excavation and of the overlying roof, if applicable.&lt;/li&gt;&#xA;&lt;li&gt;The material being dug - dig in rock and you can go down to 2000&#xA;metres; dig in sand and you won't get too far, but if you freeze the sand so it can be dug like rock you can dig deeper into it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other factors include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The proximity of geological structures, such as faults and shears;&#xA;how active such structures are and the extent of the fractured zone&#xA;of rock surrounding the structure.&lt;/li&gt;&#xA;&lt;li&gt;The availability of engineering construction materials such as steel,&#xA;concrete, reinforcing bars.&lt;/li&gt;&#xA;&lt;li&gt;How the walls of the excavation will be supported during and after&#xA;construction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-02-12T10:46:22.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7393" PostTypeId="2" ParentId="7390" CreationDate="2016-02-12T13:10:11.133" Score="1" Body="&lt;p&gt;No -- any cable built &lt;strong&gt;wrong&lt;/strong&gt; can fry your laptop.  The kerfluffle in the blogosphere is about off-brand cables which were incorrectly wired.  The whole point of USB-C is that the connector itself is orientation-agnostic and it (the &lt;em&gt;connector&lt;/em&gt;) cannot be inserted in a way that sends power to the wrong lead.  Only a miswired cable can do that.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-02-12T13:10:11.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7394" PostTypeId="1" AcceptedAnswerId="7398" CreationDate="2016-02-12T15:18:07.183" Score="20" ViewCount="12176" Body="&lt;p&gt;I have heard a couple of times that an operating nuclear power plant which was shut down (non-emergency; e.g. for a regular check) needs over 24 hours (up to 72 hours?) to get up running again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does it take that long?&lt;/p&gt;&#xA;" OwnerUserId="3061" LastEditorUserId="2246" LastEditDate="2016-02-12T17:02:26.937" LastActivityDate="2017-02-13T22:03:52.110" Title="Why does it take so long to restart a nuclear power plant?" Tags="|power|nuclear-technology|electrical-grid|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7395" PostTypeId="2" ParentId="2832" CreationDate="2016-02-12T16:19:53.613" Score="5" Body="&lt;p&gt;Yes. An amazing device called a tromp or &lt;a href=&quot;https://en.wikipedia.org/wiki/Trompe&quot; rel=&quot;nofollow noreferrer&quot;&gt;trompe&lt;/a&gt;. The flow of water over and down a funnel with a straw or tube elevated above the water level. The flowing water drags the surrounding air along with it, pulling the air through the straw and oxygenating the water with small air bubbles. The water travels through the tube under the river or stream it's placed in and as it moves horizontally through the tube the air bubbles escape into one or two air tanks connected to the pipe..this compresses the air. As long as water is flowing, even quite slowly, the air will compress in the tanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About 100 years ago a large scale mining operation in Canada used a tromp to power all the pneumatic drills etc. Why it's not being used today is puzzling?&lt;/p&gt;&#xA;" OwnerUserId="4906" LastEditorUserId="1290" LastEditDate="2017-08-04T16:37:48.460" LastActivityDate="2017-08-04T16:37:48.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7396" PostTypeId="2" ParentId="7394" CreationDate="2016-02-12T16:25:12.457" Score="7" Body="&lt;p&gt;The answer really boils down to two factors: safety and testing.  I'm going to give a generic summary of these two things below, but the real answer is quite complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crux of nuclear plant operation revolves around nuclear safety.  I'm not talking about personal safety, which is the purview of the Occupational Safety and Health Administration (OSHA), though that does have some factor.  More, this is general safety to the public against a radiologic event.  Nuclear plants are designed such that the risk of such an event is minimized as much as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a plant powers up, it undergoes different &lt;a href=&quot;http://www.nuclearelectricalengineer.com/nuclear-power-plant-modes-explained-here/&quot;&gt;Modes&lt;/a&gt;.  Each Mode has its own set of testing and acceptance criteria that must be met before the plant can be further elevated in Mode.  There are a lot of systems, and these things take time.  Systems critical to nuclear safety especially have a large amount of scrutiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nuclear plant will only become fully operational once all systems pass their tests and the plant is safe to run.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2016-02-12T16:25:12.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7398" PostTypeId="2" ParentId="7394" CreationDate="2016-02-12T16:49:28.670" Score="37" Body="&lt;p&gt;When a reactor is shut down the core produces much less heat, but they &lt;em&gt;do&lt;/em&gt; still produce heat through a mechanism known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Decay_heat&quot;&gt;decay heat&lt;/a&gt;. The fact that the core is producing less heat means that the coolant temperature is going to drop, but how far that temperature drops depends on the decay heat generation rate. This in turn is based on operating history, or the power at which the plant was operating before shutdown. This can be large for commercial plants, because they typically operate at or very near capacity and the power companies bring coal or natural gas plants up and down to modulate grid capacity. Decay heat after a day is about half a percent of the power history, which, for a 500 MW plant operating at capacity, means that the &lt;em&gt;decay heat&lt;/em&gt; could be 2.5 MW. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if there's a brief shutdown, the decay heat generation rate is so high that the primary plant stays hot and thus they can normally start up pretty &quot;quickly&quot;. I say &quot;quickly&quot; because, while the &lt;em&gt;primary&lt;/em&gt; (radioactive side) of the plant may still be hot, the &lt;em&gt;secondary&lt;/em&gt; steam plant will likely have gotten cold. For secondary plant startups, one of the big concerns is moisture formation in the piping. This happens when steam touches the (relatively) cold pipe. Moisture in the steam plant can cause all kinds of terrible things, but primarily the damage comes from &lt;a href=&quot;https://www.youtube.com/watch?v=PVLFJefTjkQ&quot;&gt;water hammer&lt;/a&gt; in the piping and moisture impingement of the turbine blades. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record: I know this because I was a Navy nuke. In my stint in the Navy, the most terrifying thing I every witnessed on the ship was a steam pipe, maybe 18 inches in diameter, literally jumping 2-3 inches with every water hammer blow, knowing that if the pipe failed that everyone in the engineroom would probably be cooked alive. Keep in mind that, in the video linked above, the steam is likely at or just above atmospheric pressure and very low flow and it still sounds like someone's beating on that radiator with a hammer. That pipe is probably an inch or less in diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condensate that forms when the steam touches the piping gets &quot;entrained&quot; in the steam flow through the pipe. The steam pushes this plug of water at very high speed, like a hammer (hence &quot;water hammer&quot;), breaking turbine blades and damaging piping and especially piping joints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are devices called &quot;moisture traps&quot; or &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Steam_trap&quot;&gt;steam traps&lt;/a&gt;&quot; that remove moisture from the system during normal operation, but the volume of condensate formed on cold plant startup is so much that the moisture traps can't keep up. This, combined with the danger presented by water hammer and moisture impingement in the turbine means that steam is admitted to the steam plant very, very, very slowly. The plant operators have to periodically go around to manually-operated steam traps to &quot;&lt;a href=&quot;https://www.youtube.com/watch?v=3Ide-3wpe4E&quot;&gt;blow down&lt;/a&gt;&quot; the condensate. (Note: the steam plant in that video is horrendous and I wouldn't work there, but the growling sound it makes when the condensate clears and steam starts to exit is exactly how I remember it sounding)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to summarize to now: the &quot;quick&quot; (24-hour) startup is typically limited by moisture generation in the secondary steam plant, caused by steam contacting cold pipes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;primary&lt;/em&gt; plant start has the potential to take much, much longer. Most (all?) reactors in the US are &lt;a href=&quot;https://en.wikipedia.org/wiki/Pressurized_water_reactor&quot;&gt;pressurized water reactors&lt;/a&gt;. This means that, despite being at 2-3 times (or more!) the temperature at which water normally boils, there is enough pressure in the primary plant to keep the water in its liquid form. This is a &lt;strong&gt;lot&lt;/strong&gt; of pressure, and the piping in the primary plant has very thick walls to withstand that pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thick walls means that there is the potential for the &lt;em&gt;inside&lt;/em&gt; of the pipe to be &quot;hot&quot; while the &lt;em&gt;outside&lt;/em&gt; of the pipe is &quot;cold&quot;. These are relative terms; everything is hot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warming up the primary plant is a chicken-and-egg problem. The primary concern here is ensuring that no steam ever forms in the reactor. Steam is actually a pretty good insulator, meaning that, if it ever &lt;em&gt;did&lt;/em&gt; form in the reactor, suddenly there would be nothing to cool the fuel, so it would get very very hot very quickly (read: melt). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you have to keep the system pressurized high enough that steam doesn't form in the reactor. &lt;em&gt;BUT&lt;/em&gt;, if you were to put that much pressure on the piping while it was cold, it would fracture, via a mechanism called, &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fracture#Brittle_fracture&quot;&gt;brittle fracture&lt;/a&gt;&quot;. This is a sudden and catastrophic failure that can be avoided if the piping is heated to the point that it has some ductility. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you need to heat the piping up, but you can't get it so hot that it boils. So you heat it a little, then increase the pressure a little, then heat, pressurize, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically there are pauses known as &quot;soaks&quot;, which give the metal in the piping time to equalize temperature. This prevents internal stresses from building up because the inside of the pipe is &quot;hot&quot; and the outside is &quot;cold&quot;. The soaks normally take a large portion to a majority of the startup time - soaks are generally 12-24 hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you heat up to a soak point, then typically pressurize to an intermediate pressure, heat to another soak point, then increase pressure to a higher intermediate pressure, then heat and pressurize together. All of this is done to stay under fracture limits known as the &quot;brittle fracture prevention limit&quot;, which again, is to ensure that the temperature-pressure the piping is subjected to is such that the pipes don't fall off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, once you have warmed up the primary plant, then you can begin to bring the secondary plant online, so it's usually 2 days for the primary and then another day for the secondary - this is the 72 hour startup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned, decay heat keeps the primary plant hot for a long time (up to maybe a month), so unless you're in an extended outage you can usually start up pretty &quot;quick&quot;, where again &quot;quick&quot; is about 24 hours. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2016-02-12T19:47:13.090" LastActivityDate="2016-02-12T19:47:13.090" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7401" PostTypeId="2" ParentId="7385" CreationDate="2016-02-12T20:13:05.320" Score="1" Body="&lt;p&gt;I don't see why not.  It would help to know your application, as I designed baluns in a very specific case (shield current suppression).  If you are using a simple centre-tapped transformer as your balun, then of course you can use it under the rated voltage - it won't arc and the magnetic current would be less.&lt;/p&gt;&#xA;" OwnerUserId="4403" LastActivityDate="2016-02-12T20:13:05.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7402" PostTypeId="2" ParentId="2349" CreationDate="2016-02-12T20:24:37.293" Score="3" Body="&lt;p&gt;I can give you an analogy: For a DC electric motor, efficiency is highest at a point LESS THAN its maximum rated power.  &lt;a href=&quot;https://i.stack.imgur.com/NnK4R.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NnK4R.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loss mechanisms are friction from fast rotation, and magnetic core losses (current going through the iron alloy) as a torque-independent parameter. If it weren't for magnetic losses, the efficiency would be maximum at torque=0. So understand your own loss mechanisms of the motor and see if the same type of curve can apply.&lt;/p&gt;&#xA;" OwnerUserId="4403" LastActivityDate="2016-02-12T20:24:37.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7403" PostTypeId="1" CreationDate="2016-02-12T22:04:34.040" Score="3" ViewCount="7281" Body="&lt;h3&gt;Background&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I have always assumed (or heard) that adhesives require a minimum thickness of adhesive to function properly. One example of this is not clamping two pieces of wood together so tightly that all of the glue squeezes out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumption seems to be backed up by the following articles on minimum adhesive bond line:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;From &lt;a href=&quot;http://www.masterbond.com/techtips/understanding-bond-line-thickness&quot; rel=&quot;nofollow&quot;&gt;MasterBond&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bond line thickness is one of the important factors to be considered in designing a bond joint. Most commonly used bond line thicknesses in a bond joint range from about 0.002 to 0.007 inches.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From &lt;a href=&quot;http://www.epotek.com/site/files/Techtips/pdfs/tip6.pdf&quot; rel=&quot;nofollow&quot;&gt;Epotek&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Minimum bond line thickness often receives special attention in part specifications but, is often overlooked and seldom controlled or monitored during process and manufacturing engineering. The risks of inadequate bond line could be low strength...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Why this might not be the case&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I ask about this because recently a machine shop that was fabricating a part for my design asked about adding adhesive to the connection of two components where no adhesive was specified. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The part is a press (interference) fit between two steel components.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine shop was offering to use a product that specifically states that it is used in interference fit situations. The idea of a minimum bond line seems to directly contradict the use of an adhesive in an interference fit. This is a &lt;a href=&quot;http://na.henkel-adhesives.com/us/content_data/341573_Design_World_22014_Reprint.pdf&quot; rel=&quot;nofollow&quot;&gt;white paper&lt;/a&gt; from the supplier that also talks about the use:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Retaining compounds are used ... for interference fitted joints: • added to existing interference fit designs to increase joint strength and reliability; ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So which is it?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Do adhesives have a minimum thickness (bond line)?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Do adhesives in an interference fit have a use?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastActivityDate="2017-12-20T20:46:01.127" Title="Do adhesives have a minimum thickness (bond line) to be useful?" Tags="|adhesive|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7404" PostTypeId="2" ParentId="7403" CreationDate="2016-02-12T22:55:01.843" Score="1" Body="&lt;p&gt;Many adhesives work best when you have the minimum thickness of adhesive you can get while still effectively wetting the joint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is particularly the case for unfilled resin based adhesives which tend to be somewhat brittle. In this case a thinner bond thickness results in a more flexible joint with less risk of cracking. However different adhesives do vary widely in their cap filling performance. In this case the minimum will often be specified to ensure that the joint is properly wetted and the maximum is the point at which mechanical strength starts to suffer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adhesives are applicable to certain types of interference fit, particularly in fitting bearings. Often this will be used as an alternative to a very tight pressed or shrunk fit. Similarly thread locking compounds for threaded fasteners fall into the same general category and are generally used to provide a seal and resist vibration rather than carry the principal load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface finish and porosity will also be important as this will cause local variations in the gap between abutting surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that some applications rely on capillary action to completely fill a joint, in which case gap thickness is critical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength of adhesive joints is usually also strongly dependant on the direction of load and here it is also consider the mechanical strength of the adhesive itself compared to the adhesion strength to the substrate. For example adhesive joints to wod will often fail by tearing off the wood surface before the adhesive itself fails in which case a thicker bond line would tend to be preferable to ensure good wetting and penetration. In metal to metal joints it is more commons for the adhesive itself to fail so a thinner line may help to minimise strains (especially in the 'peel' mode). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said all of that it is hard to generalise too much as 'adhesive' covers a huge range of materials from low viscosity cyanoacrylates to reinforced resins to rubber based materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often adhesives intended for larger gasps will have a higher viscosity, lower hardness and will often be a liquid filled with a powder of fiber reinforcement (JB weld is a good example). By extension consider materials like epoxy putty or fibreglass (GRP) where the material itself can form adhesive bonds to other surfaces but there isn't really a meaningful bond line thickness as such. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-12T23:12:54.697" LastActivityDate="2016-02-12T23:12:54.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7405" PostTypeId="2" ParentId="7394" CreationDate="2016-02-12T23:40:37.507" Score="13" Body="&lt;p&gt;Xenon is a result of the nuclear reaction and a neutron poison.  If you don't wait for the xenon to decay, it eats up too many neutrons and you can't go critical.  They always say &quot;there are not enough rods to pull.&quot;  If you have a nice new reactive core, you can get started sooner.  If the core is old, you will have to wait a long time before enough xenon (and other poisons) decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plant I used to work at cost about a million dollars a day for an outage.  Believe me, if they could start any sooner, they would.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2016-02-12T23:40:37.507" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7406" PostTypeId="1" AcceptedAnswerId="7420" CreationDate="2016-02-13T00:19:47.997" Score="5" ViewCount="3082" Body="&lt;p&gt;I have an abrasive cutoff saw that I use to cut metal, it runs at 3800RPM. You cannot cut aluminum with these saws, so I bought a carbide blade rated for 3800RPM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went to install it, but after looking in the manual I don't think it is such a good idea. It says &quot;No toothed blades&quot;. A regular saw runs around 1200RPM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't I put toothed blades in my cutoff saw? The blade is rated for that speed, would there be a problem?&lt;/p&gt;&#xA;" OwnerUserId="4921" LastEditorUserId="33" LastEditDate="2016-02-14T02:57:19.700" LastActivityDate="2022-12-23T22:20:31.570" Title="What prevents using a toothed blade with a cutoff saw designed for abrasive discs?" Tags="|machining|tools|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7407" PostTypeId="1" CreationDate="2016-02-13T04:11:54.403" Score="2" ViewCount="9959" Body="&lt;p&gt;If we place a obstacle in the fluid flow, fluid will be circulating behind the obstacle. What is this called?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be a wake or recirculation region. What is the difference?&lt;/p&gt;&#xA;" OwnerUserId="4955" LastEditorUserId="33" LastEditDate="2016-02-14T03:03:00.193" LastActivityDate="2016-02-14T03:03:00.193" Title="What is the difference between wake and recirculation region?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7408" PostTypeId="1" CreationDate="2016-02-13T06:20:28.890" Score="88" ViewCount="40160" Body="&lt;p&gt;Glass is fragile and impractical to transport, install and repair. Even worse, glass kills and hurts people when it breaks. Falling to the streets like guillotines during earthquakes and bomb raids. During wars people put tape on their windows to prevent shattering. When that meteor exploded over Chelyabinsk, people got hurt by standing inside of a window watching the sky when the shock wave hit them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are perfectly transparent plastics, for example the PET material used to make coca cola bottles. Why aren't windows made out of that instead of glass (fragile ceramics)? It seems to be much cheaper, safer and more practical to handle. Is there any advantage at all to make windows out of glass? Is this a billion dollar business idea, and if so, why haven't anyone realized it yet?&lt;/p&gt;&#xA;" OwnerUserId="4958" LastEditorUserId="33" LastEditDate="2016-02-17T15:48:20.553" LastActivityDate="2022-10-26T09:00:45.670" Title="Why do glass windows still exist? (Why haven't they been replaced by plastics?)" Tags="|materials|chemical-engineering|plastic|safety|" AnswerCount="9" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="7409" PostTypeId="2" ParentId="7408" CreationDate="2016-02-13T08:27:43.623" Score="10" Body="&lt;p&gt;After some research,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we are using glass windows as it was used traditionally. &lt;strong&gt;Actually plastic windows exists&lt;/strong&gt;. Though it may take time for most of us to adapt to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the reasons to consider glass windows are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Up to 80% of all recycled glass can be reclaimed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Recycled glass uses 40% less energy than manufacturing new glass.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Recycling doesn’t compromise glass’ quality or structure and no toxins are produced in its recycling. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.copywriterskitchen.com/2009/02/27/are-you-afraid-of-your-plastic-food-containers/&quot; rel=&quot;noreferrer&quot;&gt;Reference&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dir.indiamart.com/impcat/plastic-windows.html&quot; rel=&quot;noreferrer&quot;&gt;Plastic windows availablity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2116" LastEditorUserId="4705" LastEditDate="2016-02-15T17:17:30.800" LastActivityDate="2016-02-15T17:17:30.800" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7411" PostTypeId="2" ParentId="7408" CreationDate="2016-02-13T10:13:04.423" Score="31" Body="&lt;p&gt;What are the requirements for window glass?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Resistance to moisture&lt;/li&gt;&#xA;&lt;li&gt;Resistance to UV radiation&lt;/li&gt;&#xA;&lt;li&gt;Resistance to cleaning agents&lt;/li&gt;&#xA;&lt;li&gt;Very high transmission in the visible spectrum&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;(There are many more, from an engineering point of view)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Most transparent plastics are not resistant enough. &lt;a href=&quot;https://en.wikipedia.org/wiki/Poly%28methyl_methacrylate%29&quot;&gt;Plexiglas&lt;/a&gt; is one example which fulfills the requirements enough to be used that way. See this &lt;a href=&quot;http://www.ledil.com/sites/default/files/Documents/Technical/Material/PLEXIGLAS_Garantie.pdf&quot;&gt;brochure from evonic&lt;/a&gt; where they guarantee you 30 years without noticeable yellowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest problem with PMMA for household use is its &quot;softness&quot; (more prone to scratches) and mediocre resistance to cleaning agents as compared to glass. Now, there are many different flavours of PMMA, and you can coat a Plexiglas sheet to make it more durable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PMMA has many advantages over &lt;a href=&quot;https://en.wikipedia.org/wiki/Soda-lime_glass&quot;&gt;soda-lime glass&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Higher transmission in the visible spectrum&lt;/li&gt;&#xA;&lt;li&gt;Lower density&lt;/li&gt;&#xA;&lt;li&gt;Easier processing&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is at the moment probably just not cost effective enough to make PMMA as resilient as glass for use as window glass (for household use!, PMMA &lt;em&gt;is&lt;/em&gt; used in many different application as a better alternative to the classical inorganic glass).  This may change in the near future, as material science and the transparent plastics industry progresses.&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="4357" LastEditDate="2016-02-13T20:39:36.027" LastActivityDate="2016-02-13T20:39:36.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7412" PostTypeId="2" ParentId="7408" CreationDate="2016-02-13T11:27:52.353" Score="117" Body="&lt;p&gt;There are two main reasons why glass is still preferred over say PMMA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is durability. As long as it isn't broken, the glass in a window can easily last for hundreds of years in good condition. In particular it is a lot more resistant to scratches than comparable plastics and isn't really subject to much in the way of environmental degradation. Windows are very prone to getting scratched when they are washed as they accumulate small particles of grit on their surface which gets rubbed around the surface during cleaning. Even with scratch resistant coatings no transparent plastics get anywhere near the hardness of glass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most glasses are also much more resistant to environmental degradation from sunlight and various chemicals in the environment. Even the most resistant plastics start to discolour and become brittle over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second factor is stiffness. Glass has a much higher Young's Modulus than PMMA. In bottles etc which are stiffened by their shape this doesn't matter much but, as windows tend to be large, flat, thin panels stiffness is a big issue, affecting their ability to be sealed into their frames and their optical properties. So a plastic window would need to be substantially thicker than a glass one to have the same stiffness with consequences for optical quality and cost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may also be issues with gas permeability in the context of double glazed windows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition many of the safety concern raised in the question are addressed by laminated and tempered glass. Tempered glass is heat treated to control internal stresses, making it significantly stronger than float glass with the additional benefit that if it does break the entire plate fractures into small granules rather than sharp shards. Laminated glass consists of alternating layer of glass and a polymer film, producing a composite sheet with very high strength and toughness, potentially to the point where it can be usefully bullet resistant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another aspect of this is that side and rear windows in vehicles are often required to be made from tempered glass for safety reasons as they can be safely broken to allow access and extraction of passengers in an accident if doors are jammed or inaccessible. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-01-24T16:11:52.940" LastActivityDate="2018-01-24T16:11:52.940" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7413" PostTypeId="1" CreationDate="2016-02-13T11:29:52.543" Score="0" ViewCount="5673" Body="&lt;p&gt;Flywheel is used to store energy for the non power producing strokes of an IC(internal combustion) engine,  do we use a fly wheel in an motorcycle and if yes what is it's size and where is it placed?&lt;/p&gt;&#xA;" OwnerUserId="4889" LastActivityDate="2016-02-14T07:00:36.790" Title="Flywheel in a motorcycle" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7414" PostTypeId="1" CreationDate="2016-02-13T14:12:27.727" Score="1" ViewCount="2591" Body="&lt;p&gt;&lt;img src=&quot;https://d2vlcm61l7u1fs.cloudfront.net/media/a36/a36bb994-cc56-49ad-98d2-60ca096d0d54/phpREiYcV.png&quot; alt=&quot;picture of the question and the dimensions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the exercise that I just can't seem to figure out. The answer should be 71.9 Nm with coordinate angles 44.2, 131 and 103.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the couple moments caused by the forces $F_1$, $F_2$, and $F_3$ are -20j and 24i and -16k respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't seem to figure out the moment caused by the two $F_4$ forces. First I thought it only caused a moment around x-axis, but this is wrong. Then I tried $M = r \cdot F$. But I just can't express $r$ (from tip of $-F_4$ to tip of $F_4$) in a vector. &#xA;So can anyone help me with finding the coordinates of the ends of the bar where the forces $F_4$ work on, so that I can make a position vector out of it?&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.297" LastActivityDate="2018-11-25T15:44:27.327" Title="Determining the magnitude and coordinate angles of a resultant couple" Tags="|statics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7416" PostTypeId="2" ParentId="7413" CreationDate="2016-02-13T19:03:10.873" Score="1" Body="&lt;p&gt;It's the same as on any IC engine &amp;mdash; the flywheel is mounted directly on the crankshaft. Why would it be anywhere else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;YouTube: &lt;a href=&quot;https://www.youtube.com/watch?v=qgU2mtZ2axM&quot; rel=&quot;nofollow&quot;&gt;HOW TO REMOVE A MOTORCYCLE FLYWHEEL&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-02-13T19:03:10.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7417" PostTypeId="2" ParentId="7407" CreationDate="2016-02-13T19:56:40.627" Score="1" Body="&lt;p&gt;Generally the term 'wake' is used to describe any downstream disturbance to flow caused by an obstacle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically the wake will expand outwards in an approximate 'V' shape until it eventually dissipates into the general 'noise' of the flow whereas recirculation occurs is a reasonably well defined region immediately downstream of the obstacle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recirculation is a specific condition where the flow separates from the body of the obstacle. This creates a low pressure area immediately downstream which sucks fluid back into this region from the main flow with the net result of creating a circulating vortex or pair of vortices. This condition creates marked increase in drag compared to situations where the flow remains attached to the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If flow separation is likely it is often better to deliberately induce turbulence early rather than have an abrupt separation of laminar flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a major and complex subject in fluid mechanics and the above should be considered an introduction rather than a comprehensive explanation. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-13T20:01:48.690" LastActivityDate="2016-02-13T20:01:48.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7418" PostTypeId="2" ParentId="3468" CreationDate="2016-02-13T20:06:53.887" Score="1" Body="&lt;p&gt;A useful rule of thumb for plausibility checking of gas balloon lift is that helium and hydrogen provide roughly 1kg of lift per cubic metre. This is close enough for back of the envelope sums and to check that you haven't dropped a decimal point in more detailed calculations. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-13T20:06:53.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7419" PostTypeId="2" ParentId="7388" CreationDate="2016-02-13T20:59:07.057" Score="1" Body="&lt;p&gt;An IC engine is (for these purposes) a pump. It moves a certain volume of gas through on each cycle (one revolution for a 2-stroke, two for a 4-stroke). That amount is reported as the displacement of the engine (110 cc) in your case. Then you need to determine the mass of that air in that volume.You can arrive at that using the ideal gas law and the molecular weight of the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ PV=nRT $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Values in order: Pressure($Pa$), Volume (displacement in your case$m^{3}$), number of moles, the ideal Gas constant (8.314 $J\cdot K^{-1} \cdot mol^{-1}$) and temperature (of the air $K$). Solve for $n$ and multiply by the molecular weight (~.0288 $kg \cdot mol^{-1}$ for air). That'll be the mass of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is method ignores the volume taken up by the fuel itself, pressure loss between the intake manifold and cylinder etc.. It's a pretty good estimate, but won't be accurate enough to make the engine run acceptably well. You'll need to apply an empirical coefficient to tune it to the right value to control your fuel injection. That coefficient will not be constant over all RPM and other operating conditions. The good news is that the coefficient will be close to unity and won't change (significantly) over the lifetime of the engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That tuning process can be involved. Ideally, you'd have an air-fuel ($\lambda$ sensor on the exhaust to measure how far from stoich the mix is. Tuning with an exhaust gas temp sensor is also possible (never had to do that myself), and may be cheaper. The last option would be to tune the engine on a dynamometer (measures power), although tuning to produce maximum power will usually produce a slightly fuel rich mixture. The $\lambda$ sensors are the best way to go and can be integrated into your controller to make the system more robust, but you could borrow one temporarily to tune and run the engine without it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I've specified everything in base SI units. Other units will work too, but make sure that you get the value of $R$ right and use an absolute temperature (Kelvin or Rankine, not Celsius or Fahrenheit)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Engine temp and TP are not strictly necessary for the calculation, but TP is often used to tweak fuel delivery during changes in speed/TP.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="68" LastActivityDate="2016-02-13T20:59:07.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7420" PostTypeId="2" ParentId="7406" CreationDate="2016-02-13T21:14:37.820" Score="3" Body="&lt;p&gt;I would certainly hesitate to install a toothed blade on a saw designed for abrasive disks. My biggest concern would be that the dynamic loading is likely to be significantly different and as these things are dangerous enough to begin with it seems and unnecessary risk that some unforeseen set of circumstances could cause a catastrophic problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were you I would get a basic cutoff or mitre saw designed for toothed blades. Multi-material saws are now readily available and woodworking saws work fine for aluminium and there aren't expensive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that it is not really good practice to use he same machine for cutting steel and aluminium as a mixture of steel and aluminium swarf is extremely flammable ie &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermite&quot; rel=&quot;nofollow&quot;&gt;thermite&lt;/a&gt; especially when the steel cutting process is abrasive. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-13T21:14:37.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7421" PostTypeId="1" CreationDate="2016-02-13T21:41:19.107" Score="0" ViewCount="24" Body="&lt;p&gt;I am trying to obtain solubility data for a polymer in sc-CO2 and the data is reported at each T and P as $$w \ \ \times \ 10^6$$ Does anyone know what this means?&lt;/p&gt;&#xA;" OwnerUserId="4977" LastActivityDate="2016-02-13T21:47:19.500" Title="Solubility Data" Tags="|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7422" PostTypeId="2" ParentId="7387" CreationDate="2016-02-13T21:42:03.003" Score="1" Body="&lt;p&gt;In a gas flow into an open volume there will be a fairly well defined point (at some distance form the nozzle) at which the flow transitions from laminar to turbulent and this is usually where most of the energy dissipation occurs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can  be seem in the real world in the smoke pattern from a recently extinguished candle wick. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common formal definition of this is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Reynolds_number&quot; rel=&quot;nofollow&quot;&gt;Reynolds Number&lt;/a&gt; which gives a reasonable approximation of whether flow conditions are dominated by viscous or inertial forces. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-13T21:42:03.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7423" PostTypeId="2" ParentId="7421" CreationDate="2016-02-13T21:47:19.500" Score="1" Body="&lt;p&gt;It looks like w is the inverse of parts per million ie if w = 1 the solubility is 1 part per million &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-13T21:47:19.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7425" PostTypeId="1" AcceptedAnswerId="7434" CreationDate="2016-02-13T23:35:37.203" Score="5" ViewCount="1377" Body="&lt;p&gt;I was looking at hydrogen as alternative fuel and came across &lt;a href=&quot;http://www.autoblog.com/2009/08/20/greenlings-why-choose-a-fuel-cell-or-an-internal-combustion-eng/&quot; rel=&quot;nofollow&quot;&gt;an article&lt;/a&gt; stating that although $\textrm{H}_2$ produces more energy than conventional fuel, it produces less horsepower in internal combustion engines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that more energy ($143\ \textrm{MJ/kg}$ for $\textrm{H}_2$ vs $46.4\ \textrm{MJ/kg}$ for gasoline) would mean more heat, more temperature, more pressure and thus more force and horsepower. Why, then, does $\textrm{H}_2$ produce less horsepower than gasoline in internal combustion engines?&lt;/p&gt;&#xA;" OwnerUserId="4979" LastEditorUserId="368" LastEditDate="2016-04-11T22:09:34.173" LastActivityDate="2016-04-11T22:11:40.290" Title="Why does hydrogen fuel produce less horsepower than gasoline in internal combustion engines?" Tags="|mechanical-engineering|combustion|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7426" PostTypeId="1" CreationDate="2016-02-14T00:12:06.243" Score="1" ViewCount="463" Body="&lt;p&gt;I wish to determine the required torque of a 55rpm gear drive motor assembly to drive a fishing reel with a 4.6:1 turn ratio that will yield 27&quot;of line retrieval per turn or 1 crank revolution.  Lift weight minimum 1 lb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to direct couple a 12vdc 55rpm right angle gear drive rated @ 8kg-cm to a fishing reel crank shaft that has a 1:3.4 ratio turning a 4 cm spool. Line is retrieval is 12&quot; per crank revolution. How much can it lift before stall. I'm looking for a continuous lift of at least 1 lb.&lt;/p&gt;&#xA;" OwnerUserId="4980" LastEditorUserId="41" LastEditDate="2016-02-15T14:15:05.490" LastActivityDate="2017-11-04T02:27:40.077" Title="torque requirement" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7427" PostTypeId="2" ParentId="7413" CreationDate="2016-02-14T06:48:13.560" Score="2" Body="&lt;pre&gt;&lt;code&gt;do we use a fly wheel in an motorcycle&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Yes we do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to store the energy generated in the power stroke which is supplied that energy to other strokes and the traction effort of motorcycle&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;where is it placed?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The location of flywheel is directly next to the engine shaft (crack shaft)directly next to flywheel is a clutch with a spring arrangements and a diaphragm clutch (its motorcycle). The following image will give you an idea how that arrangement is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XQki9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XQki9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here is an illustrative image of the ECG (engine clutch gearbox)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/igWzb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/igWzb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2116" LastEditorUserId="2116" LastEditDate="2016-02-14T07:00:36.790" LastActivityDate="2016-02-14T07:00:36.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7428" PostTypeId="1" CreationDate="2016-02-14T07:58:35.013" Score="6" ViewCount="197" Body="&lt;p&gt;Now, I've been conducting beam simulation by using Pro/E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to change payload(force) of beam step by step, and then plot the relation between deflection and payload(load range : 50kg to 500kg, step size : 25kg).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, this task is too much time consuming, so that I would like to use a programming skill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the first time, I checked matlab-Pro/E interface, and I realized I can do something with SimMechanics tool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, unfortunately, my computer doesn't have the SimMachanics license.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I've been looking for other ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My computer has labview and MS-office as a programming language, as well as matlab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I programmably change payload by using these programming languages without additional charged tool like SimMechanics? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is possible, Could anyone give me something like source code, documents or helpful site addresses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Pro/E provides desired function, it would be great. In that case, I don't need to make a program.&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="1586" LastEditDate="2016-02-18T19:15:27.057" LastActivityDate="2016-06-19T07:28:49.560" Title="Programmable parameter change of Pro/E simulation by using Excel VBA , Matlab, Labview or Pro/E own function" Tags="|modeling|simulation|matlab|labview|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7429" PostTypeId="1" AcceptedAnswerId="7433" CreationDate="2016-02-14T09:14:21.797" Score="1" ViewCount="65" Body="&lt;p&gt;There seems to be no noiseless thermic motor for cars or motorcycles on the market.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a technical obstacle at making such a product ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to isolate/eliminate noises from the explosion process and from the moving parts ?&lt;/p&gt;&#xA;" OwnerUserId="4986" LastActivityDate="2016-02-15T13:04:14.603" Title="Technical obstacles in making a noiseless thermic motor?" Tags="|motors|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ClosedDate="2016-02-15T13:11:11.773" ContentLicense="CC BY-SA 3.0" />
  <row Id="7430" PostTypeId="2" ParentId="7408" CreationDate="2016-02-14T09:24:46.473" Score="15" Body="&lt;p&gt;I actually did make windows out of plexiglas for an outbuilding. I could work them to the shape I wanted and they were lightweight.  However, I discovered that they indeed scratch easily: just trying to get the protective paper off I both scraped and melted the plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want &lt;em&gt;hard&lt;/em&gt; windows for houses and primary buildings.  So what you describe would be addressed by laminates, as used in automobiles.  I'm sure you can buy that. But most people find having shutters to be cheaper and satisfactory if severe weather may cause breakage, along with breakage warranty and insurance, or the cheapness of replacing &lt;em&gt;plain&lt;/em&gt; glass panes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are regulations for requiring &lt;em&gt;tempered&lt;/em&gt; glass for some windows that a child may fall into. Tempered glass in general addresses the dangers you raise. I don't know about regulations for glass clad skyscrapers where earthquakes are prone, but I'll bet they don't fall even if shaken because they are hung and can move independantly.&lt;/p&gt;&#xA;" OwnerUserId="4705" LastActivityDate="2016-02-14T09:24:46.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7431" PostTypeId="1" CreationDate="2016-02-14T11:24:37.743" Score="1" ViewCount="974" Body="&lt;p&gt;I saw in this post on &lt;a href=&quot;http://www.smartstuffs.com/2016/02/14/self-driving-cars-roads-soon/&quot; rel=&quot;nofollow&quot;&gt;Self driving cars&lt;/a&gt; that said the cars developed by Google for the Google Self Driving Car Project, have sensors that can detect objects like humans, plastic bags within a range of two football fields. In my question, I have said &quot;200 metres&quot; which is rough figure because I think a football field would be anywhere between 90 metres to 120 metres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of sensors are capable of detecting objects in such a long range? If such sensors are available, are they used in any other devices or vehicles today?&lt;/p&gt;&#xA;" OwnerUserId="4990" LastEditorUserId="4991" LastEditDate="2016-02-14T12:46:12.203" LastActivityDate="2016-02-16T05:18:18.237" Title="Can sensors detect objects 200 metres away?" Tags="|electrical-engineering|sensors|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7432" PostTypeId="2" ParentId="7431" CreationDate="2016-02-14T12:26:39.073" Score="3" Body="&lt;p&gt;Yes, these kind of sensors exists, they're called radars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Military grade naval radars can detect a coin from tens of kilometers away, so that kind of performance (~1m² objects detected from ~100 meters) is not surprising. The keys are frequency (the higher the frequency, higher the accuracy) and signal processing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have doubts about the idea of dozens of these radars co-existing and overlapping each other's frequencies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : Also, it is possible to use high definition cameras and have a computer analyzing the images to detect moving objects,but it is more expensive (in terms of money and calculation capacity needed) than a radar-based technology&lt;/p&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="4991" LastEditDate="2016-02-14T12:50:27.033" LastActivityDate="2016-02-14T12:50:27.033" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7433" PostTypeId="2" ParentId="7429" CreationDate="2016-02-14T12:37:51.203" Score="4" Body="&lt;p&gt;Considering the nature of what an explosion is, I doubt we can just suppress the sound at it's source. The only way is to use a different motor technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, car manufacturers tend to isolate the motor compartment from the outside with basic noise-suppressing technologies which are for example : covering the compartment with absorbing materials like foams or using bushed fixations (&lt;a href=&quot;https://en.wikipedia.org/wiki/Bushing_%28isolator%29&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). Most highly priced cars are well isolated, for example I know that you barely hear the sound of the motor in recent BMW cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture of some bmw compartment. You can see that when the cover is closed, a plastic joint isolates the cover from the chassis.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bFyPA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bFyPA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="4946" LastEditDate="2016-02-15T13:04:14.603" LastActivityDate="2016-02-15T13:04:14.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7434" PostTypeId="2" ParentId="7425" CreationDate="2016-02-14T13:03:58.410" Score="5" Body="&lt;p&gt;One consideration is the relative densities of hydrocarbon fuels compared to hydrogen. If we assume that the fuels are completely vapourised in the combustion chamber then the mass of fuel depends on the volume, temperature and pressure. So energy density per kg doesn't necessarily tell you much about how many molecules of fuel you can get in one charge or indeed the chemical energy available per molecule. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also issues in IC engines relating to the combustion properties of hydrogen. One particular problem is that hydrogen will detonate at a broad range of air/fuel ratios which has knock on effects for ignition timing and compression ratios which in turn affect how well the chemical energy available from combustion can be converted into useful mechanical power. Premature detonation under compression (especially in hot cylinders)  is a particular problem with hydrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen is often associated with low compression ratio engines like Wankel (and other rotary) engines for exactly this reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen also burns hot compared to other fuel gases, so even when overall energies are comparable hydrogen engines can produce localised hot spots in the combustion chamber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gasoline is a mixture of fairly short chain hydrocarbons which can be approximated to C8 but the specific mixture of different molecules has a significant effect on its combustion behavior and there is quite a lot of scope for adjusting a particular fuel formulation to get the best performance. This is much more difficult with hydrogen. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="368" LastEditDate="2016-04-11T22:11:40.290" LastActivityDate="2016-04-11T22:11:40.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7435" PostTypeId="2" ParentId="7408" CreationDate="2016-02-14T13:26:53.543" Score="47" Body="&lt;p&gt;Here is a plastic (most likely PMMA) window, in a boat, after only 37 years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jubox.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jubox.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the obvious scratches, the outer surface has developed a cloudiness : possibly from degradation due to UV light, and (towards the LH end) you can see a cubelike pattern of stress cracks, rather like a toughened glass windscreen after a pebble hit it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You really can't tell what you're looking at through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this application, a PMMA window is still the best solution, rather than the difficulty of making a glass window fit the curve of the hull. But it shows the limitations compared to glass, which remains usable for centuries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as replacing glass windows, the cost of a specialist will be high for any type of repair in any technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But absent custom curves, the raw material (glass) is cheap : about £1/sq foot, much cheaper than perspex or polycarbonate, and much easier and faster to cut. (Watching a professional exploit the special fracture properties of glass is impressive. Contrast with the difficulty of cutting perspex or polycarbonate!)&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="435" LastEditDate="2016-02-14T13:34:19.213" LastActivityDate="2016-02-14T13:34:19.213" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7437" PostTypeId="2" ParentId="7414" CreationDate="2016-02-15T00:30:56.757" Score="2" Body="&lt;p&gt;I am going to first make an assumption that contradicts the figure, namely that the top tube is level. Then the moment from $F_4$ is along the vector v.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v=\left\{\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the magnitude is  $$ 150 \frac{3}{10} \cos (30 {}^{\circ})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the total moment can be computed. $$M= 100 \frac{2}{10} \{0,-1,0\}+120 \frac{2}{10} \{1,0,0\}+80 \frac{2}{10} \{0,0,-1\}+150 \frac{3}{10} \cos (30{}^{\circ}) v$$&#xA;$$M =\left\{{51.5568, -47.5568, -16}\right\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the magnitude of M is $71.9426 Nm$, and the coordinate angles of its direction are $$ \left\{\cos ^{-1}\left(\frac{51.5568}{71.9426}\right),\cos&#xA;   ^{-1}\left(-\frac{47.5568}{71.9426}\right),\cos&#xA;   ^{-1}\left(-\frac{16}{71.9426}\right)\right\}$$&#xA;$$ = \{44.2225{}^{\circ}, 131.379{}^{\circ}, 102.85{}^{\circ}\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, the result agrees with the answer you expect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the top tube is not level then the computations are not so straightforward. I show the computations that were done for the general case in Mathematica. The result is different (but when $h=0$ you get the earlier result.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7T4jQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7T4jQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2016-02-15T17:07:47.210" LastActivityDate="2016-02-15T17:07:47.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7438" PostTypeId="1" CreationDate="2016-02-11T12:56:49.460" Score="2" ViewCount="227" Body="&lt;p&gt;My question is about simultaneous heat and mass transfer in rotating cylinder system. System can be detailed as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A rotating cylinder contains wet porous substance in a lump shape. &lt;/li&gt;&#xA;&lt;li&gt;Air stream enters the cylinder axially and can exit it by perforations on back surface of cylinder or radially by perforations on its curved surface.&lt;/li&gt;&#xA;&lt;li&gt;Entering air flow is slightly higher temperature than porous medium and carries moisture from surface of porous medium.&lt;/li&gt;&#xA;&lt;li&gt;RPM of cylinder is about +/- 25 for most of 1 hrs' run time. But for 5 minutes in middle of runtime gets as high as 1400 RPM.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In low RPM period, I have modeled the system's heat and mass transfer w.r.t. NTU analogy. Equations are roughly as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Re = \frac{ \dot{m_{air}} D_{drum}}{\mu m_{porous}}  $,&#xA;$Pr = \frac{Cp_{air} \mu}{k_{air}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$j = a Re^{-b}$, &#xA;$ NTU = \frac{j}{Pr^{\frac{2}{3}}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h_{porous \, air} A_{cylinder} = (NTU)\times \dot{m_{air}} Cp_{air}  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_{air\, to\, porous } = h_{porous \, air} A_{cylinder} \times (T_{air} - T_{porous})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass transfer is characterized based on NTU too. Driving force is partial pressure difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main source of heat transfer is temperature difference b/w inflowing air and porous substance, $ T_{air} = T_{porous} + 45$ at start. For mass transfer $(X_{saturation at porous surface} - X_air)$, mole-fraction difference between input air water content and saturated layer at porous substance's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I want to know, how would heat transfer conditions change under very high RPM. It will be a completely mixed system with more resistance to axial flow than before. So would NTU analogy still be used under such conditions? If not, how can I modify or replace it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple figures for system:- &#xA;&lt;a href=&quot;https://i.stack.imgur.com/PPFMc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PPFMc.png&quot; alt=&quot;System at low RPM&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FjSod.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FjSod.png&quot; alt=&quot;System at high RPM&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions are welcome.&lt;/p&gt;&#xA;" OwnerUserId="4949" OwnerDisplayName="user1768201" LastEditorUserId="33" LastEditDate="2016-02-23T13:55:17.550" LastActivityDate="2016-02-23T13:55:17.550" Title="Modeling heat and mass transfer modes inside rapidly rotating cylinder" Tags="|thermodynamics|heat-transfer|modeling|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7441" PostTypeId="1" AcceptedAnswerId="7443" CreationDate="2016-02-15T09:32:36.263" Score="2" ViewCount="5483" Body="&lt;p&gt;An example is &lt;a href=&quot;http://www.omega.com/pptst/PX409-485_PXM409-485.html&quot; rel=&quot;nofollow&quot;&gt;this pressure transmitter&lt;/a&gt; which has a BSL accuracy of 0.08%. I know the term &quot;Best fit straight line&quot; - BFSL. Is BSL just a short hand for this (&quot;Best straight line&quot;)?&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="46" LastEditDate="2016-02-15T11:29:40.813" LastActivityDate="2016-02-15T12:33:31.710" Title="What does &quot;BSL Accuracy&quot; mean?" Tags="|sensors|metrology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7442" PostTypeId="2" ParentId="7431" CreationDate="2016-02-15T10:09:20.167" Score="1" Body="&lt;p&gt;LIDAR is another possible candidate. This works on a similar principal to radar but uses laser light rather than radio waves for scanning. It is already widely use for things like high resolution mapping and creating digital models of complex objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also has a few potential advantages over radar, including higher resolution and the fact that it is no impinging on an already crowded RF spectrum, similarly light doesn't pose the sort of electromagnetic interference problems associated with radar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice it is likely that a practical automated car would use a variety of different sensors for different tasks. Perhaps forward looking Lidar for detailed obstacle detection, video for all round situational awareness and ultrasound for low speed maneuvering (eg parking).  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-15T10:09:20.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7443" PostTypeId="2" ParentId="7441" CreationDate="2016-02-15T10:46:07.850" Score="2" Body="&lt;p&gt;BFSL (Best Fit Straight Line) is the same as BSL (Best Straight Line). It is calculated from the calibration's extracted data, where the manufacturer test the sensor for several known input values (in this case, pressures). Once these calibration points are measured, they are approximated with a straight line (least-square method) and the BSL value is an error value that indicates how close the measured points are to this straight line.&lt;/p&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="46" LastEditDate="2016-02-15T12:33:31.710" LastActivityDate="2016-02-15T12:33:31.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7446" PostTypeId="1" CreationDate="2016-02-15T14:35:26.520" Score="4" ViewCount="906" Body="&lt;p&gt;I have a cylinder with a shaft in its center that I need to rotate at 120 RPM on 180°, for costing reasons, I would like, if possible, to reuse the same DC motor that I already use for another system. That motor has a speed of 6000 RPM so I need a reduction ratio of 1:50.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a lot of space under the cylinder, so I decided to move that motor to make it parallel to the cylinder (and then to the cylinder shaft), this also means that I won't be able to direct drive the cylinder shaft and I will need to use at least a pair of pulleys (with belt :)) or gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what would be the best way to reduce the speed of the motor adding the least backlash ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I see these 5 solutions on which I would like to get external reviews on them : &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Only pulleys: since I'm using GT pulleys and the maximum reduction ratio with 2 pulley is 1:4, I need at least 3 pairs of pulleys to get 1:50.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planetary gearbox with 1:50 ratio + a pair of pulleys or gears having the same teeth count (just because the shaft are parallels) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Planetary gearbox with a ratio smaller than needed + a pair of pulley or gears that reduce again the speed appropriatelly to get the 1:50 at the end. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A pair of gears giving me the 1:50 ratio directly (if I find a gear with enough teeth..)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Replacing the motor by a servo which I do not know why are larger but shorter than the motors I found, and so it could fit under the cylinder.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3664" LastActivityDate="2019-05-02T01:02:50.580" Title="comparing backlash of different mechanisms" Tags="|gears|pulleys|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7448" PostTypeId="2" ParentId="7446" CreationDate="2016-02-15T22:03:19.657" Score="1" Body="&lt;p&gt;It sounds like, short of replacing the motor options (2) or (3) would be the most practical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compound sets of gears or pulleys is going to end up quite complex and by the time you've packaged 3 sets of shafts and bearings you are eating into any cost savings from keeping the motor and adding a lot of complexity to your design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of option 2 vs option 3 it really just depends on what parts are easiest/cheapest to source. DC motors with integrated planetary gearboxes are very common so it really comes down to whether it is more convenient to mount your existing motor with a separate gearbox or just get a complete new unit. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-15T22:03:19.657" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7449" PostTypeId="2" ParentId="7408" CreationDate="2016-02-15T22:08:51.440" Score="4" Body="&lt;p&gt;Hardness aside, weathering effects (from e.g. rainfall, snow, hail, temperature and moisture variations etcetera) and UV degradation are of course important to consider for windows. In general, glass is much more resistant to such effects compared to plastics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also something called creep, which is essentially time-dependent deformation during long times (months or years). You might have seen it if you've ever left something heavy on a cardboard box for a long time. Plastics creep pretty easily. I don't know if this actually would be a major problem for a regular window perpendicular to the ground, but it might not be good for a window that will carry static loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plastics will also whiten due to deformation. They might also discolor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, these properties are modifiable (to some extent) through various additives.&lt;/p&gt;&#xA;" OwnerUserId="5027" LastActivityDate="2016-02-15T22:08:51.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7450" PostTypeId="2" ParentId="7373" CreationDate="2016-02-15T23:14:08.423" Score="3" Body="&lt;p&gt;VC-130 and VC-131 are names from the Brazilian standard.  A comparison can be found at &lt;a href=&quot;http://www.joinville.ifsc.edu.br/%7Epaulosergio/Ciencia_dos_Materiais/Classifica%C3%A7%C3%A3o%20dos%20a%C3%A7os.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Paulo Sergio's website&lt;/a&gt;.  Table 4 suggests that VC-130 is equivalent to AISI-D3 and VC-131 is equivalent to AISI-D6.  However, these stainless steels are low sulfur  (see the &lt;a href=&quot;http://www.villaresmetals.com.br/pt/Produtos/Acos-Ferramenta/Trabalho-a-frio/VC131&quot; rel=&quot;nofollow noreferrer&quot;&gt;Villares VC-131 page&lt;/a&gt;). I think the &amp;quot;Tool &amp;amp; Die Steels&amp;quot; site has the composition wrong.&lt;/p&gt;&#xA;&lt;p&gt;The main difference between the two steels is the Tungsten content.  Tungsten tends to reduce pitting corrosion (see the &lt;a href=&quot;http://www.outokumpu.com/en/products-properties/more-stainless/the-effects-of-alloying-elements%E2%80%8B/Pages/default.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Outokompu page&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-02-15T23:14:08.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7451" PostTypeId="1" AcceptedAnswerId="7455" CreationDate="2016-02-16T00:33:18.647" Score="3" ViewCount="1551" Body="&lt;p&gt;I'm unsure if this is the correct community, so feel free to move it if you feel it's in the wrong area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to develop a water-tight/sealed box for a backup flash drive. The idea is to make something that is weatherproof enough so I can keep the box outside or in my basement for years, and if I ever need to retrieve the data, I can cut the box open and retrieve the USB stick. This would need to completely resist water/moisture and withstand temperature changes between say -40 C to 40 C without it breaking down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the concept I have so far: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;First layer: Put USB stick into air-tight plastic container. Could be something cheap like a sealable zip-lock bag to protect from:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Second layer: fill a small box - not much larger than the stick - with a polymer sealant/caulk that will eventually solidify, fitting the form of the box.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the best material can I use for layer #2?&lt;/p&gt;&#xA;" OwnerUserId="5028" LastEditorUserId="272" LastEditDate="2016-02-16T15:15:07.277" LastActivityDate="2018-09-28T09:30:15.363" Title="Best way to make a sealed container for a long term storage of a USB drive?" Tags="|materials|seals|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7452" PostTypeId="1" AcceptedAnswerId="7454" CreationDate="2016-02-16T00:38:33.303" Score="5" ViewCount="8443" Body="&lt;p&gt;I'd like to make my own structure that will hold roughly 50-100lbs of equipment. Since I need this to be portable, I feel like aluminum tubing would be the most practical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My structure will basically be a 2.5'-10' cube. I want to get 2.5' tube sections that I can connect very sleekly (word?) and make a cube that is 2.5' at its smallerst and 10' at its largest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a cue from a photography backdrop, they have 1&quot; diameter tubing that sort of fits into itself. The tube is not flared, but it's pinched so it's the same size as the inner diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for this &quot;pinch&quot; I'm talking about? And if so, is there a tool I can buy to do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ww71m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5029" LastEditorUserId="5029" LastEditDate="2016-02-16T01:29:00.767" LastActivityDate="2020-04-17T18:09:01.497" Title="What is the opposite of flaring a tube (pinching?) and is there a tool to do this with aluminum?" Tags="|aluminum|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7453" PostTypeId="2" ParentId="7408" CreationDate="2016-02-16T01:31:13.170" Score="7" Body="&lt;p&gt;Another issue is that the coefficient of thermal expansion of many plastics is higher than that of window glass.  Frames for plastic windows would have to be able to accommodate more expansion / contraction than those for glass windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.eplastics.com/Plastic/plastics_library/Coefficients-of-Thermal-Expansions-of-Plexiglass&quot;&gt;http://www.eplastics.com/Plastic/plastics_library/Coefficients-of-Thermal-Expansions-of-Plexiglass&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sdplastics.com/acryliteliterature/1121DFFPhysicalProperties%5B1%5D.pdf&quot;&gt;http://www.sdplastics.com/acryliteliterature/1121DFFPhysicalProperties%5B1%5D.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5030" LastActivityDate="2016-02-16T01:31:13.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7454" PostTypeId="2" ParentId="7452" CreationDate="2016-02-16T01:37:06.837" Score="6" Body="&lt;h2&gt;Summary&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The general term for changing tube and pipe diameter is &lt;strong&gt;swaging&lt;/strong&gt;. The tools you are looking for exist but vary widely in price and complexity. It may be more straightforward to find a local metal supplier who can do the swaging for you based on your specifications.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Flaring vs. Swaging&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I am unsure if there is a specific name for the process which is the opposite of flaring, however flaring is not quite what you are looking for. The term flaring is generally applied to the process of making the end of a tube or pipe or tube a conical shape of larger diameter than the original tube or pipe, especially for special plumbing fittings called &lt;a href=&quot;https://www.homefixitparts.com/FittingSelectionGuide.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;flare fittings&lt;/a&gt; and is shown below. The image shown is from the link provided. The flared end of a copper tube is shown at the marker (C) in the image, while the nut marked (B) goes over the left-hand threads on the male fitting, sealing the connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jEx4o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jEx4o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general term for continuously deforming hollow cylinders by mechanical action to change their diameters is &lt;strong&gt;swaging&lt;/strong&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Swaging&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;). Swaging is similar to drawing and extrusion. Drawing and extrusion are terms generally applied to solid, constant-cross-section bodies such as rods, while swaging is generally applied to tube and pipe. Additionally, the swaging process is usually applied only to a portion of the length of a tube or pipe for some engineered purpose, as you are seeking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A diagram demonstrating the process of swaging, which is done by drawing or pressing a tube or pipe through a die, is shown below. The pipe is moving from the left side to the right side of the image, through the yellow-colored die, and its outside diameter is being decreased as a result. The image is from &lt;a href=&quot;http://www.felss.com/en/processes-machines/rotary-swaging/about-rotary-swaging/rotary-swaging-overview/&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.felss.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RcI2j.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RcI2j.jpg&quot; alt=&quot;Swaging diagram.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Options&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Benchtop swaging equipment does exist, such as at this link from &lt;a href=&quot;http://www.mscdirect.com/product/details/42098178&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.mscdirect.com&lt;/a&gt; , though depending on what you want to do the price may vary and end up significantly lower, such as at this link from &lt;a href=&quot;http://www.directindustry.com/prod/op/product-17876-434326.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.directindustry.com&lt;/a&gt;. No affiliation or endorsements are implied with these links and they are intended for illustrative purposes only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to find a local metal part supplier who has all of the tools and order the pieces custom-made to your specifications. The materials and labor to shape the materials will probably end up being less expensive and higher quality than with a novice DIY approach if this is a one-off or proof-of-concept. Additionally, you have experts who know how to cold work metal and may be able to provide minor additional advice.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-02-16T01:37:06.837" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7455" PostTypeId="2" ParentId="7451" CreationDate="2016-02-16T01:48:26.420" Score="5" Body="&lt;p&gt;If you have access to a vacuum pack sealer, a vacuum sealed pack for the fist layer would be better option than a zip lock bag. Whatever you use, a first layer will be needed to protect the contacts of the USB stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second layer, as I understand it, you are wanting to make an encasing weather proof 'brick' around a sealed USB stick. I would suggest clear &lt;a href=&quot;https://en.wikipedia.org/wiki/Silicone&quot; rel=&quot;noreferrer&quot;&gt;silicone&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being clear you'll be able to see the USB stick inside. Some of the relevant properties of silicone are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Low toxicity&lt;/li&gt;&#xA;&lt;li&gt;Low chemical reactivity&lt;/li&gt;&#xA;&lt;li&gt;Thermally stable in a temperature range of $-100\ ^\rm{o}C$ to $250\&#xA;   ^\rm{o}C$&lt;/li&gt;&#xA;&lt;li&gt;It repels water &amp;amp; is used to create water tight seals&lt;/li&gt;&#xA;&lt;li&gt;It does not support microbial growth&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you need to retrieve the USB stick the silicone can be easily cut with a sharp knife.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-02-16T01:48:26.420" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7456" PostTypeId="2" ParentId="7431" CreationDate="2016-02-16T05:18:18.237" Score="2" Body="&lt;p&gt;The sensors which the article you linked are referring to are &lt;a href=&quot;https://en.wikipedia.org/wiki/Lidar&quot; rel=&quot;nofollow&quot;&gt;LIDARs&lt;/a&gt;. A LIDAR is a laser used in conjunction with an optical scanning system. By using sophisticated software and a LIDAR it is possible to do what is called &quot;scene generation&quot; which means creating a 3D model of the terrain (and objects) in front of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LIDARS ranges are only limited by the power of the laser and resolution of the scanner. In theory you could map the surface of the moon with a LIDAR if you had a powerful enough laser. A scene generation system with a range of 200 yards (&quot;two football fields&quot;) is feasible and many such systems exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such systems are not commonly used, even in autonomous systems like UAVs, for several reasons: they are expensive, fragile, and suffer from motion effects. For example, for aerial UAVs one of the big problems is what is called &quot;canyoning&quot;. As the vehicle traverses an area the parts of the scene to the forward right and left change very rapidly due to angular effects. This makes it very hard to compute the scene fast enough, except for things that are right in front of you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main use of 3D LIDARs today is to scan stationary objects like buildings and vehicles so they can be easily entered into a computer system as 3D models.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2016-02-16T05:18:18.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7457" PostTypeId="1" AcceptedAnswerId="7463" CreationDate="2016-02-16T08:47:22.857" Score="1" ViewCount="75" Body="&lt;p&gt;I recently got a gizmo installed at home that samples my electricity usage once every 10 seconds. I've become a little transfixed watching the graph update as various appliances in my home turn on and off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, turning on lights results in a small and simple bump up a couple of hundred watts. Turning on the microwave a huge 1.2KW jump. The washing machine usage moves up and down depending on where it is in the cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started wondering whether there'd be some ways to automatically work out what devices are running by what you might call their 'power signature'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially what I want is an algorithm that watches what is effectively a 0.1Hz signal representing usage in watts, and can say 'Looks like your washing machine was on between 09:00 and 11:00 today.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complication of course is that multiple appliances can be running. As a human you can sometimes resolve that (E.g. This shape looks like the washing machine was on, but someone also turned on the microwave in the middle).&lt;/p&gt;&#xA;" OwnerUserId="5036" LastActivityDate="2016-02-16T13:50:53.993" Title="Are there any methods to work out what appliances are running in my home from an aggregate power consumption sampled 6 times a minute?" Tags="|power|energy-efficiency|signal-processing|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7458" PostTypeId="1" CreationDate="2016-02-16T08:58:43.883" Score="2" ViewCount="1712" Body="&lt;p&gt;Why Heat Exchangers are preferred to work in Constant Pressure Process than Constant Volume Process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mostly, all problems are solved in Constant Pressure Process&lt;/p&gt;&#xA;" OwnerUserId="5037" LastActivityDate="2016-02-16T18:13:12.470" Title="Why Heat Exchangers are preferred to work in Constant Pressure Process than Constant Volume Process?" Tags="|mechanical-engineering|thermodynamics|thermal-insulation|thermal-expansion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7460" PostTypeId="2" ParentId="7318" CreationDate="2016-02-16T10:56:31.727" Score="2" Body="&lt;p&gt;If you're not set on the motor shield, you could use a stepper motor with a pulley instead, plus some form of linear guide (eg. an Igus slide, if you don't want it just dangling). Small Nema 17 size stepper motors easily generate enough torque to lift 700g, and you can use &lt;a href=&quot;http://reprap.org/wiki/StepStick&quot; rel=&quot;nofollow&quot;&gt;StepStick&lt;/a&gt;-style drivers, which can be driven easily from an Arduino.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need it to be very quiet, the &lt;a href=&quot;http://www.watterott.com/en/SilentStepStick&quot; rel=&quot;nofollow&quot;&gt;SilentStepStick&lt;/a&gt; is a driver which is completely inaudible, while many other stepper drivers cause audible hum/buzz in the motors.&lt;/p&gt;&#xA;" OwnerUserId="5046" LastActivityDate="2016-02-16T10:56:31.727" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7462" PostTypeId="2" ParentId="7451" CreationDate="2016-02-16T13:14:29.383" Score="2" Body="&lt;p&gt;1) You should not use a regular USB stick. It simply has too much exposed circuitry inside that can go bad. You should rather use a memory card or a dedicated, weatherproof stick like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Corsair Flash Voyager&lt;/li&gt;&#xA;&lt;li&gt;LaCie RuggedKey &lt;/li&gt;&#xA;&lt;li&gt;Patriot Supersonic Boost XT &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;2) Rather than trying to create a container around the thumbdrive you should use a case that's already designed to do that. One of best performers are aluminium containers with rubber or silicone seal that is compressed when the box is closed. This compression gives much better performance than using silicone caulk or glue. Quick search revealed they're easily found: &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0028CITY2&quot; rel=&quot;nofollow&quot;&gt;Aluminium airtight storage cylinder&lt;/a&gt;. I have no experience with this particular product, but it does everything I wrote: 2-piece body with o-ring inbetween.&lt;/p&gt;&#xA;" OwnerUserId="1931" LastActivityDate="2016-02-16T13:14:29.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7463" PostTypeId="2" ParentId="7457" CreationDate="2016-02-16T13:50:53.993" Score="3" Body="&lt;p&gt;The word that will unlock your searching is &lt;strong&gt;disaggregation&lt;/strong&gt; - there's quite a lot of research going on right now into energy disaggregation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These generally involve some kind of learning algorithm, and the energy-disaggregation field is quite young, so you won't find a reliable off-the-shelf package.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you will find current research which will help you design your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Daniel A Kelly (aka Jack Kelly, and currently of Imperial College London) has worked in this field - check out his papers; and Clym Stock-Williams of Uniper (formerly part of E.On) recently reported at a Data Science for Energy conference organised by the Alan Turing Institute (Edinburgh, Jan 2016) that data at 0.1 Hz was as good as data at 1 Hz for disaggregation - the more frequent 1-second data did not lead to a more skillful model (YMMV).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, do a literature search (Google Scholar or similar) for energy disaggregation or electricity disaggregation, and you'll find new papers every month or so, in this fast-evolving field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for a single dwelling, where you know what's going on at any one moment, the best way to do this would be to train your own model, as you'll be able to get very good data on what is being used, when: even as far as listening for when the fridge and freezer compressors turn on and off.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2016-02-16T13:50:53.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7464" PostTypeId="2" ParentId="7451" CreationDate="2016-02-16T15:02:25.243" Score="4" Body="&lt;p&gt;Others have given you an idea of how to seal the data storage device, but I'd like to make you aware of another issue.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;USB sticks (especially cheap ones) aren't designed for data archival.  The memory storage mechanism degrades over time from many different causes; entropy, cosmic rays, and radioactive decay come to mind.  It is entirely possible that you open up your excellently-designed, sealed box after a few years only to find that the memory is unreadable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few relatively cheap ways to increase your chances of recovering the data:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Buy a high quality flash drive with &lt;a href=&quot;https://en.wikipedia.org/wiki/USB_flash_drive#Longevity&quot; rel=&quot;nofollow&quot;&gt;SLC-based memory&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Buy multiple flash drives and write the same data to each&lt;/li&gt;&#xA;&lt;li&gt;Buy a high capacity flash drive and write many copies of the same data&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="272" LastActivityDate="2016-02-16T15:02:25.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7467" PostTypeId="2" ParentId="7458" CreationDate="2016-02-16T18:13:12.470" Score="3" Body="&lt;p&gt;I'm assuming that you want to know why we use the heat capacity at constant pressure, $c_p$, instead of at constant volume, $c_v$, in analyzing heat exchangers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason is that the pressure drop through most heat exchangers will be minimal, so a constant-pressure assumption works well.  This is especially true for liquids, whose thermodynamic properties are not very sensitive to changes in pressure.  You will of course need to verify and justify this assumption for any heat exchanger analysis that you do.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-02-16T18:13:12.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7468" PostTypeId="2" ParentId="7328" CreationDate="2016-02-16T19:40:09.743" Score="-5" Body="&lt;p&gt;I think crushed gravel means crushed bricks and roof tiles (tennis court). Crushed stone could be crushed natural shingle, for example from a river.&lt;/p&gt;&#xA;" OwnerUserId="5057" LastEditorUserId="4357" LastEditDate="2016-02-16T21:42:10.663" LastActivityDate="2016-02-16T21:42:10.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7469" PostTypeId="2" ParentId="7451" CreationDate="2016-02-16T20:09:20.227" Score="3" Body="&lt;p&gt;I'm not sure that silicon caulk is the idea material for this for two reasons. Firstly it is very viscous and forms a skin fairly quickly so it is not that easy to ensure that a big blob of it around something will actually make a very good seal and thick sections take very long time to cure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly the curing process releases acetic acid which has the potential to corrode the very thing you are trying to protect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it was me it would either use something like a preserving jar or get a short length of PVC pipe and bond an end cap on each end if you want it tamper proof or use threaded pipe fittings to do the same thing if you want easy access, just seal he threads with PTFE tape or sealing compound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, whatever solution you go for I would, at the very least add a couple of sachets of desiccant granules to the package to remove any moisture from the air inside, or if you wan to be really thorough you can get oxygen scavenging inserts for archival purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have mentioned a flash drive is probably a better choice for archiving than USB. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-16T20:09:20.227" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7470" PostTypeId="2" ParentId="7452" CreationDate="2016-02-16T20:22:42.063" Score="3" Body="&lt;p&gt;For a one-off structure swaging or flaring tube is probably a bit of a long way round to do it. There are actually quite a few off the shelf solutions for joining tube, &lt;a href=&quot;http://www.rosscastors.co.uk/tube-inserts-fittings/connect-a-tube.html&quot; rel=&quot;nofollow&quot;&gt;an example at random&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something else to consider is that, for something which is disassembled and moved about a lot, if your system relies on one tube fitting inside another then and damage to the ends of the tube will mean that they no longer fit properly and the method of fit will have  significant effect on the stiffness of the whole structure. With this in mind any system which either puts a permanent  rigid insert in the end of the tube or clamps on the outside is likely to be preferable in terms of both durability and stiffness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth saying that any operation which involves ductile forming of metals tends to be a bit fiddly as the parameters depend a lot on the &lt;em&gt;precise&lt;/em&gt; material properties of the raw materials which can vary a bit from batch to batch. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-16T21:39:08.223" LastActivityDate="2016-02-16T21:39:08.223" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7471" PostTypeId="1" CreationDate="2016-02-16T22:56:11.127" Score="4" ViewCount="153" Body="&lt;p&gt;We have:&#xA;$$ s(t) = 4u(t)−u(t−1)−2u(t−2)−2u(t−3.5)+2u(t−4) $$&#xA;where $u$ is the Heaviside function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had to draw the graph and find the minimum sampling period so we wont have loss of signal. I have drawn the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t=kT$&#xA;where $T$ is the sampling period.So $T=t/k$ so how do I find T?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nYpE4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nYpE4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5062" LastEditorUserId="1586" LastEditDate="2016-02-18T18:42:33.020" LastActivityDate="2016-02-18T18:42:33.020" Title="Finding the sampling period to avoid loss of signal" Tags="|electrical-engineering|signal|signal-processing|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7473" PostTypeId="2" ParentId="4074" CreationDate="2016-02-17T02:12:24.117" Score="5" Body="&lt;p&gt;Solar heating would not &quot;elevate&quot; moisture in the object, but instead evaporates whatever is on the sunny side, allowing moisture from the bottom to soak back up by capillary action.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best bet (if you can) would be to set the object on the end with the least surface area, such that gravity pulls the water out the bottom faster.  Setting on its end forces the contained water (not supported by capillary action) to pool deeper, which gives the water at the bottom more pressure to flow out.  Setting it on its narrow end also allows more area to be exposed for evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some period of time (based on the material and size), the primary method of losing water will switch from gravity flow to evaporation.  At this point, it would be best to flip the object to the other end as frequently as possible, since the evaporation will happen faster where the object has more moisture, so you want it as uniform as possible on the portion with the large surface area.  Here, mounting it on the narrow end helps again, since then it takes the remaining water longer for the moisture to slowly flow from one end to the other by gravity.&lt;/p&gt;&#xA;" OwnerUserId="5067" LastActivityDate="2016-02-17T02:12:24.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7475" PostTypeId="1" CreationDate="2016-02-17T06:39:41.497" Score="1" ViewCount="1010" Body="&lt;p&gt;I am trying to come up with a better way to adjust the angle between the two steel pieces in the picture, and lock it in place. At the moment I just have a nut and bolt that I loosen and then tighten, but it is very cumbersome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TJ9Ns.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TJ9Ns.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be perfect is some sort of geared dial that I could turn which maintains the force between the plates so I just turn it to whatever angle I need without loosening and tightening again. I'm just not sure what exactly I would search for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone have any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4189" LastActivityDate="2016-02-18T02:03:22.980" Title="Ideas for a simple angle adjustment/lock for a steel plate arm" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7476" PostTypeId="1" CreationDate="2016-02-17T07:39:17.587" Score="2" ViewCount="28" Body="&lt;p&gt;I have a cylinder with water, containing an arbitrarily shaped (i.e. a free surface), but still a primarily cylindrical object. The setup is subjected to transient heat loading (dT +20 -10C at room temperature), for a period of 6 months. I have a dataset regarding the temperature for the whole period and I am interested in simulating the temperature within the object. Now obviously running a coherent simulation with such a detailed dataset is beyond the computational capacity of mine, and I was wondering if there is a way to tweak heat transfer parameters and/or the setup to get my answers relatively accurately.&lt;/p&gt;&#xA;" OwnerUserId="5073" LastActivityDate="2016-02-17T07:39:17.587" Title="Solidworks Long Term Transient Heat Transfer" Tags="|heat-transfer|solidworks|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7477" PostTypeId="1" CreationDate="2016-02-17T08:17:52.130" Score="2" ViewCount="724" Body="&lt;p&gt;I want to use a mix of polyurethane and acetone to coat some polyester fabric. I have the choice between using a PU glue or a PU foam but because polyurethane isn't &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_polyurethane_applications#/media/File:Purgrid.png&quot; rel=&quot;nofollow&quot;&gt;a single material&lt;/a&gt; with a fixed composition but rather a range of chemicals sharing similar chemistry, I wonder which one would be better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the lightest coating possible it should be water proof as well as UV resistant one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here are my questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;● Should I go with a PU glue or a foam?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;● Will the UV resistant property of the foam would still work once it is dissolved in acetone?&lt;/p&gt;&#xA;" OwnerUserId="2708" LastEditorUserId="421" LastEditDate="2016-02-17T16:01:00.957" LastActivityDate="2018-03-03T22:35:45.530" Title="Polyurethane coating on fabric" Tags="|materials|chemical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7479" PostTypeId="2" ParentId="7471" CreationDate="2016-02-17T13:13:24.307" Score="4" Body="&lt;p&gt;The smallest feature that you need to capture is the dip between 3.5 and 4.  If you use a sampling period which is greater than 0.5, then you may end up not capturing this feature.  The sampling period must therefore be $T\leq 0.5$ in order not to lose any features of the signal.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2016-02-17T13:13:24.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7481" PostTypeId="2" ParentId="7471" CreationDate="2016-02-17T16:18:06.963" Score="2" Body="&lt;p&gt;To prevent loss of signal you need to sample at twice the highest frequency presented in the signal. This is also known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Nyquist_rate&quot; rel=&quot;nofollow&quot;&gt;Nyquist rate&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that time &lt;strong&gt;&lt;em&gt;t&lt;/em&gt;&lt;/strong&gt; is in seconds, the smallest time period is about 0.5s or the highest time frequency is 2Hz. So if sampled between 2Hz and 4Hz the processed signal will be subjected to what is known as aliasing. If the signal is sampled above 4Hz, there will no data loss. Below are few references that gives better detailed explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/Fy9dJgGCWZI&quot; rel=&quot;nofollow&quot;&gt;Sampling Rate, Nyquist Frequency, and Aliasing (data acquisition effects)&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=vJ8V8ipSZ50&amp;amp;feature=youtu.be&quot; rel=&quot;nofollow&quot;&gt;Sampling, Aliasing and Nyquist&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Nyquist_frequency&quot; rel=&quot;nofollow&quot;&gt;Nyquist frequency&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="282" LastEditDate="2016-02-18T00:17:27.640" LastActivityDate="2016-02-18T00:17:27.640" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7482" PostTypeId="1" CreationDate="2016-02-17T16:25:31.053" Score="2" ViewCount="1643" Body="&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I notice that there is quite a bit of similarity between the standard heat equation and the energy equation (from the navier stokes equations).  The heat equation is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial\left(\rho h\right)}{\partial t} + \nabla\cdot\left(u\rho h \right) - \nabla\cdot\left(k\nabla T \right)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\rho,h,u,k,T,$ are the density, total specific enthalpy, fluid velocity, thermal conductivity, and temperature, respectively.  Note that for many applications, it suffices to write the total specific enthalpy as $h=c_pT$, where $c_p$ is the constant pressure specific heat capacity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the energy equation (from the navier stokes system) is given as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial\left(\rho E\right)}{\partial t}+\nabla\cdot\left(u\rho E\right)-\nabla\cdot(k\nabla T)+\nabla\cdot(\sigma u)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E,\sigma$ are the total energy and stress tensor, respectively.  Here, the total energy E is the sum of internal and kinetic energy $E=e+K$, with $=e=c_vT$ and $K\frac{1}{2}|u|^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since both equations are derived from the conservation of energy, it is no surprise that both equations share the following terms: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Rate of change in Enthalpy (time derivative term) &lt;/li&gt;&#xA;&lt;li&gt;Advection of internal heat (divergence term)&lt;/li&gt;&#xA;&lt;li&gt;Heat diffusion (laplacian term)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The energy equation has additional terms which are not found in the heat equation.  These additional terms are specific to compressible fluid flow and account for&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Rate of change in &amp;amp; advection of kinetic energy (K)&lt;/li&gt;&#xA;&lt;li&gt;Stress contributions to energy ($\sigma$)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In many situations, the kinetic energy and stress contributions are very small when the fluid flow is incompressible, and thus can be neglected.  This results in the energy equation and heat equation being almost identical except for one subtle difference: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A subtle difference&lt;/strong&gt;&lt;br&gt;&#xA;The heat equation is formulated with the constant pressure specific heat capacity $c_p$ while the energy equation uses the constant volume heat capacity $c_v$.  Yet, I assume that they are derived from the same principle of conservation of energy.  If so, I would expect the two expressions to use the same specific heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This becomes more important when dealing with different types of fluid flows.  I have seen models for incompressible flow where the energy equation is simply formulated in terms of the standard heat equation (as above) with the constant pressure heat capacity $c_p$. Since the temperature does not affect incompressible fluid flow, one can think of this as a &quot;scalar transport&quot; which depends on the resolved fluid velocity $u$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Application: Solidification &amp;amp; Melting&lt;/strong&gt;&lt;br&gt;&#xA;In my particular case, I want to model phase change (solidification and melting).  For incompressible flow, it suffices to rewrite the total specific enthalpy as the sum of internal and latent heat ($h=c_pT+h_L$) and substitute h in the equation.  For compressible flows, the equation is written in terms of internal energy $c_vT$, not enthalpy.  So, I'm not 100% sure that it would suffice to simply add the latent heat term as if it were a part of the total internal energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to know for sure is if I understand why the incompressible energy equation is written in terms of $c_p$ while the compressible energy equation is written in terms of $c_v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the energy equation for incompressible flow is formulated in terms of constant pressure heat capacity $c_p$, why isn't the energy equation compressible flow also formulated in terms of constant pressure heat capacity $c_p$.  Why is the heat capacity different for incompressible and compressible flow?  What accounts for the difference?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, for my specific application, how do these differences affect how to add latent heat effects for solidification and melting of a compressible fluid?   Can I simply add the latent heat to internal energy term $e$?  Or do I need to completely reformulate the compressible energy equation in terms of enthalpy in order to add latent heat effects?&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2016-02-17T17:50:14.840" Title="Why use different specific heat capacities in the energy equations for incompressible and compressible fluid flow?" Tags="|fluid-mechanics|thermodynamics|heat-transfer|energy|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7483" PostTypeId="2" ParentId="7482" CreationDate="2016-02-17T17:50:14.840" Score="1" Body="&lt;p&gt;Regardless of context, when you do an energy balance on a thermodynamic system, you're usually interested in (among other things) the change in internal energy of the system and the work done on or by the system by compressing or expanding it.  Enthalpy $(dh=c_p dT)$ is more-or-less just a convenience function for the combination of internal energy and work: $dh=du+p \cdot dV$.  Any derivation that uses enthalpy (or constant-pressure heat capacity), can also be performed using constant-volume heat capacity instead.  The only difference should be whether or not expansion/compression work appears explicitly in the solution or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say that your system is incompressible, you're basically saying that any changes in volume are negligible, i.e. $dV \approx 0$.  The $p \cdot dV$ term then disappears from the definition of enthalpy, so $dh \approx du$.  That is, for incompressible materials, there isn't much difference between the heat capacity at constant pressure and at constant volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your question on where to add the latent heat, I don't have much experience with solidification and melting, but it sounds like a good idea to lump them together in the total energy term.  You'll have another degree of freedom that you need to track: the fraction of material that is liquid vs. that which is solid.  You could do something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;E_{tot} = \alpha \cdot u_l + (\alpha -1) u_s +KE+PE+...&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E_{tot}$ is the total energy, $KE$ and $PE$ are kinetic and potential energy, $\alpha$ is the fraction of material in the liquid phase, and $u_l$ and $u_s$ are the internal energy of the liquid and solid phases.  As the material melts, $\alpha$ increases towards 1.0, and the change in energy is captured in the first two terms.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-02-17T17:50:14.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7484" PostTypeId="2" ParentId="7222" CreationDate="2016-02-17T20:06:14.377" Score="5" Body="&lt;h3&gt;Unity Equation&lt;/h3&gt;&#xA;&lt;p&gt;The Unity Equation is a very standard method of analyzing a section under combined loads. Even though the masonry code doesn't specifically call it out in the reinforced pages, it would be hard to argue that it wasn't a reasonable assumption.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{f_b^1}{F_b^1}+\frac{f_b^2}{F_b^2}\le1$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The equation could even be &lt;span class=&quot;math-container&quot;&gt;$\le\frac{4}{3}$&lt;/span&gt; if stress increases were acceptable.&lt;/p&gt;&#xA;&lt;h3&gt;Code&lt;/h3&gt;&#xA;&lt;p&gt;I agree that I couldn't find any other reference to biaxial bending in the masonry code. I also didn't find any more clear discussion in the masonry design handbooks that I had available.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-05-02T11:15:04.887" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7485" PostTypeId="1" CreationDate="2016-02-17T22:21:53.867" Score="-2" ViewCount="1223" Body="&lt;p&gt;Could someone please explain why, when there is no secondary load, the applied voltage of the primary roughly equals the back EMF of the primary? Since the primary coil is an inductor I wondered exactly how the back EMF is always there and always cancelling the applied voltage when there is no secondary load. Take an inductor in a DC circuit for example: the back EMF of the inductor only cancels the applied voltage (source voltage) at $t=0$. I know transformers use AC but how can it continually match the applied EMF when there is no secondary load?&lt;/p&gt;&#xA;" OwnerUserId="5063" LastEditorUserId="1586" LastEditDate="2016-02-18T18:48:58.357" LastActivityDate="2016-02-18T22:56:08.667" Title="Transformers with no load" Tags="|electrical-engineering|power-electronics|circuits|inductive-charging|" AnswerCount="1" CommentCount="2" ClosedDate="2016-02-19T01:05:32.610" ContentLicense="CC BY-SA 3.0" />
  <row Id="7486" PostTypeId="2" ParentId="7475" CreationDate="2016-02-18T02:03:22.980" Score="1" Body="&lt;p&gt;I would suggest extending the fixed bar enough so that a hole could be tapped, then get a &lt;a href=&quot;http://www.mcmaster.com/#91404A505&quot; rel=&quot;nofollow&quot;&gt;wing head bolt&lt;/a&gt;&#xA;and thread it through, holding the bar in place. You are going to have to tighten something if you want to restrict movement. &lt;/p&gt;&#xA;" OwnerUserId="5091" LastActivityDate="2016-02-18T02:03:22.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7487" PostTypeId="2" ParentId="7328" CreationDate="2016-02-18T05:29:34.080" Score="1" Body="&lt;p&gt;Stone is a general term and there is no real size limit on that and more like everyday term.&#xA;However, gravel is an engineering term and in geotechnical engineering depending on the governing body it is generally between 2 and 40mm. USCS, AASHTO and a couple of others have their own sizes which are fairly close to this range. &lt;/p&gt;&#xA;" OwnerUserId="5092" LastEditorUserId="5092" LastEditDate="2016-02-18T05:36:32.393" LastActivityDate="2016-02-18T05:36:32.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7488" PostTypeId="2" ParentId="7252" CreationDate="2016-02-18T05:54:53.663" Score="1" Body="&lt;p&gt;Generally, in geotechnical engineering you won't allow much movement, especially for excavation support. Nevertheless, you will always have small movements but this is usually too small to consider P-Δ effect and safety factors account for it. Most software automatically do this. If you really want to be conservative allow $0.02H$, where $H$ is your height of excavation as movement and calculate P-Δ.&lt;/p&gt;&#xA;" OwnerUserId="5092" LastEditorUserId="1832" LastEditDate="2016-02-18T10:10:46.237" LastActivityDate="2016-02-18T10:10:46.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7490" PostTypeId="2" ParentId="7406" CreationDate="2016-02-18T06:03:23.733" Score="2" Body="&lt;p&gt;Generally a company is not going to recommend you do something outside the scope of the intended product design because that would invoke liability. There are lots of variables like blade mounting tolerances, bearing loading, shroud design, chip removal, intended cutting material, motor hp and rpm as you mentioned. And because of these variables, someone like myself won't be able to tell you if it is safe or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, cutting aluminum with a high speed carbide blade is common industry practice. They are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Cold_saw&quot; rel=&quot;nofollow&quot;&gt;cold saws&lt;/a&gt; in contrast to &quot;hot&quot; abrasive saws. They typically employ coolant and the cut speed is controlled by computer to ensure maximum safe cut speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your situation, a regular &lt;a href=&quot;http://www.homedepot.com/p/Ryobi-13-Amp-7-1-4-in-Circular-Saw-CSB125/205216317&quot; rel=&quot;nofollow&quot;&gt;wood circular saw&lt;/a&gt; or &lt;a href=&quot;http://www.homedepot.com/p/Ryobi-14-Amp-10-in-Compound-Miter-Saw-in-Green-TS1345L/205105634&quot; rel=&quot;nofollow&quot;&gt;wood miter saw&lt;/a&gt; outfitted with a &lt;a href=&quot;http://www.homedepot.com/p/Diablo-10-in-x-84-Tooth-Laminate-Non-Ferrous-Metal-Cutting-Saw-Blade-D1084L/100660707&quot; rel=&quot;nofollow&quot;&gt;high-tooth-count carbide blade&lt;/a&gt; will cut aluminum profiles just great without any coolant.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-02-18T06:03:23.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7491" PostTypeId="2" ParentId="7403" CreationDate="2016-02-18T07:25:29.020" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://books.google.com/books?id=D9Eq9WOKiPwC&amp;amp;pg=PA133&amp;amp;lpg=PA133&amp;amp;dq=minimum%20bond%20line%20and%20polymerization&amp;amp;source=bl&amp;amp;ots=IxFt-n_pPi&amp;amp;sig=IN4IgHtPCjKUXvoMEzhdSdHfeyg&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwiy8f6K1IDLAhVHyGMKHeSQABgQ6AEIPjAF#v=onepage&amp;amp;q=minimum%20bond%20line%20and%20polymerization&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Polymers Physical Properties&lt;/a&gt; states &quot;Joints made with smaller bond lines are epoxy-starved and have lower strengths.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a chemist, but there is some physical amount of space required for the polymerization process. It may be that in confined spaces the polymer chains are more parallel to the bond line instead of randomized (might be a good question for Chemistry SE). In my internet research many companies state that they use a filler with a constant particle size to control the minimum bond thickness (glass micro beads in your masterbond link).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interference fit adhesives such as thread locking compounds dont have the bond strengths that other &quot;minimum bond line thickness&quot; epoxies and adhesives do and they are not exposed to tension, only shear. Perhaps they too would be stronger at a specific thickness, but in this case bond strength is less important than increasing the friction between the two tightly mechanically fit components. Natures thread lock, rust, does a pretty good job of it.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-02-18T07:25:29.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7492" PostTypeId="1" CreationDate="2016-02-18T07:59:34.997" Score="4" ViewCount="9477" Body="&lt;p&gt;If you have a graph with a tensile stress-strain curve of a typical metal or alloy and you superimpose a compressive stress-strain curve on it (from same metal or alloy), how would this look like if you had to sketch it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that when you talking about engineering stress and strain, you will get a typical metal stress-strain curve for tension and then in comparison the compressive one follows same path for the elastic region and then after that rises above the tensile one. &#xA;And you will notice three differences:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;compressive one will have no necking and thus no maximum/tensile point.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;it will fracture differently and with way higher strain&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;like mentioned before; F needed for compression is higher than for tension in the plastic region because of the increasing area and so engineering stress will be higher as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now how would tensile look like versus compression if we talking about real stress and strain? Especially regarding point 3, will compression still rise above tension curve in the plastic region?&#xA;So in other words will delta F be higher than delta area in compression? Will the extra rise in Force for compression with respect to tension be compensated by the fact that for compression the area increases and for tension it decreases and thus true stress will be equal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it not much to ask, could someone sketch a true stress and strain curve with both tension and compression for a typical metal?&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="1586" LastEditDate="2016-02-18T18:42:12.407" LastActivityDate="2017-02-14T22:54:12.847" Title="How does a graph with tensile stress-strain curve and compressive stress-strain curve superimposed look like involving true stress and true strain?" Tags="|mechanical-engineering|materials|metallurgy|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7493" PostTypeId="2" ParentId="2176" CreationDate="2016-02-18T08:04:37.243" Score="2" Body="&lt;p&gt;Current sensor and computing technology permits a vehicle to be operated statistically safer by a computer than by a human.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Statistically&quot; is the key word here. Every day there are situations where someone will die in a car accident regardless of the decision making involved. Neither humans nor computers can see the future. The difference between the two is the human is socially permitted to make mistakes while a computer is not. If we truly want safety we should be building track based transportation systems where physics, not computers, ensure that collisions do not occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your Toyota reference for example stated &quot;Do not rely exclusively on rear cross traffic alert or blind spot monitoring systems.&quot; because the driver is the permissible liability; the scapegoat so to speak. No company is going to take responsibility where it doesn't have to; especially when a human still has partial control of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liability will be the largest obstacle to autonomous vehicles. &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_motor_vehicle_deaths_in_U.S._by_year&quot; rel=&quot;nofollow&quot;&gt;92 people died every day from car crashes in the US in 2012&lt;/a&gt;, but the blame was easily socially distributed to those individuals involved. If google or some other large corporation sells an autonomous car to everyone in the US and it only killed 1 person per day, people would be outraged, sue, and shut the whole operation down.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-02-18T08:04:37.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7496" PostTypeId="2" ParentId="7485" CreationDate="2016-02-18T13:52:05.810" Score="1" Body="&lt;p&gt;The specific question is too confused to be answerable, so I'll try to explain some basics of transformers in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A transformer with open secondary is just a inductor.  The concept of back EMF makes no sense here.  What will happen is the current will lag the voltage by 90&amp;deg; in the ideal case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power delivered to the inductor is current times voltage.  Note that this averages to 0 over any whole cycle when the current and voltage are sines with 90&amp;deg; phase shift between them.  Energy is put into the inductor during half the cycle, then given back during the other half.  The average power transfer is 0.  The energy that is absorbed and then given back is temporarily stored in the magnetic field of the transformer core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, the inductor will not be purely inductive, but will also appear to have some resistance in series with its inductance.  That resistance is the DC resistance of the wire the primary coil is made from.  That resistance will dissipate power proportional to the square of the current thru it.  In a &quot;good&quot; transformer or inductor, this resistance is small compared to the impedance due to the inductance over the frequency range the device is intended to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the voltage and current, the current will lag the voltage a little less than 90&amp;deg; due to this resistance.  The average of the voltage x current over a whole cycle is now a little positive, which is the energy dissipated by the resistance each cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting even closer to reality, there will also be some loss in storing then retrieving magnetic energy to/from the transformer core each cycle.  This will also appear as a resistive component electrically.  The resistance will appear in parallel with the primary inductance.  Both together are in series with the DC resistance of the coil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a load is connected to the secondary, it looks a lot like the core gets more lossy to the circuit driving the primary.  The primary looks more resistive, which accounts for the higher power it now draws at the same voltage.  Like the core loss resistance, this additional resistance appears in parallel with the inductance of the primary, from the point of view of the circuit driving the primary.  In a ideal transformer with ideal load, the secondary removes energy from the magnetic core exactly as the primary tries to put it there.  The result is the primary current is now in phase with its voltage, and therefore looks purely resistive.  This of course never happens exactly because there are always inevitable losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ideal case, a transformer primary looks like whatever impedance is connected to the secondary, divided by the square of the turns ratio.  For example, let's say a transformer has a 1:3 turns ratio.  You put 12 VAC into the primary, and ideally get 36 V out of the secondary.  If there is 10 &amp;Omega; on the secondary, then it is delivering (36V)&lt;sup&gt;2&lt;/sup&gt;/10&amp;Omega; = 130 W.  This has to go into the primary, which means the primary current is 130W/12V = 10.8 A, which means the primary looks like a 12V/10.8A = 1.11 &amp;Omega; resistor.  Note that 1.11&amp;Omega;/10&amp;Omega; = 1/9, which is the square of the turns ratio.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2016-02-18T22:56:08.667" LastActivityDate="2016-02-18T22:56:08.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7498" PostTypeId="1" CreationDate="2016-02-18T14:36:11.443" Score="1" ViewCount="1079" Body="&lt;p&gt;When building a trapezium dome, is it better to construct the dome row by row or by first making a vertical arch?  Is some form of bracing required?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dz7QA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dz7QAm.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5100" OwnerDisplayName="David Millette" LastEditorUserId="5100" LastEditDate="2016-02-18T19:38:52.050" LastActivityDate="2016-02-19T04:31:32.967" Title="Which is the best way to construct a trapeziun dome with a 76' diameter" Tags="|structural-engineering|structures|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7499" PostTypeId="1" AcceptedAnswerId="7504" CreationDate="2016-02-18T15:54:37.907" Score="0" ViewCount="1210" Body="&lt;p&gt;I have made Image which is self explanatory. It has the question&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KzIPd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KzIPd.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Suppose I have a metal plate that is travelling at V m/s.&lt;br /&gt;&#xA;A stationary body acts as a shearing body. When the metal plate contacts the still body, it starts to shear off. The plane in which it will be sheared is shown in diagram (red line). So I want to ask how much force will get applied for the shearing.&lt;/p&gt;&#xA;&lt;p&gt;This is just an idea that I'm working through. I haven't seen this in any book that I have read (I haven't yet seen various books yet). This problem came in my mind yesterday while I was dreaming.&lt;/p&gt;&#xA;&lt;p&gt;Also I am assuming that the acceleration of the moving plate will be equal to change in velocity (after shearing the plate stops), so a = (0 - V) / 1 (unit time)&lt;/p&gt;&#xA;&lt;p&gt;So question is&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;how much mass should be considered for the equation F = ma ?&lt;/p&gt;&#xA;&lt;p&gt;basically I want to calculate &lt;strong&gt;force required&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If am wrong in some calculation then correct me.&lt;/p&gt;&#xA;" OwnerUserId="2116" LastEditorUserId="-1" LastEditorDisplayName="user16" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-02-19T20:22:10.173" Title="shear force calculation with velocity and mass" Tags="|design|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7500" PostTypeId="1" AcceptedAnswerId="7501" CreationDate="2016-02-18T16:43:51.280" Score="4" ViewCount="553" Body="&lt;p&gt;I'm following &lt;a href=&quot;http://openmicros.org/index.php/articles/94-ciseco-product-documentation/raspberry-pi/217-getting-started-with-raspberry-pi-gpio-and-python&quot; rel=&quot;nofollow&quot;&gt;this tutorial&lt;/a&gt; on how to control Raspberry Pi GPIO pins with Python scripts.  Instead of connecting the test circuit (LED + resistor) between the ground and the GPIO pin however, they connect it between the +3v pin and GPIO (see section 4, last paragraph).  The author says this is to have the current come from the 3v pin rather than the GPIO pin.  However, from my limited understanding of electricity, the magnitude of the current through the GPIO pin should be the same regardless of how the polarity has been set, the only difference would be its direction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I wrong about the current?  Is it actually better to draw power from the 3v pin instead of the GPIO pin?&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1586" LastEditDate="2016-03-01T23:32:03.937" LastActivityDate="2017-05-09T01:18:18.127" Title="What is the advantage of using GPIO pins as sinks instead of sources?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7501" PostTypeId="2" ParentId="7500" CreationDate="2016-02-18T18:13:55.897" Score="2" Body="&lt;p&gt;I think the reason is if the GPIO is used the micro is starved of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the method described 3.3V is the power source. So the micro is not starved of power. If the 3.3V was provided by the GPIO pin, the micro is sourcing the power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am generally used to seeing a transistor used to switch on/off elements like a LED. Below is an example schematic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xx9IL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xx9IL.jpg&quot; alt=&quot;LED Schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further details can be found in &lt;a href=&quot;http://www.physics.udel.edu/~nowak/phys645/The_bipolar_transistor.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;The bipolar transistor&lt;/a&gt; article.&lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="282" LastEditDate="2016-03-02T00:17:43.737" LastActivityDate="2016-03-02T00:17:43.737" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7502" PostTypeId="1" AcceptedAnswerId="7516" CreationDate="2016-02-18T22:03:55.527" Score="5" ViewCount="399" Body="&lt;p&gt;I am trying to decipher some as-built drawings for an old wood frame building. The purpose is to generate a preliminary load takeoff. I cannot figure out what the following is saying: &lt;a href=&quot;https://i.stack.imgur.com/JRjSv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JRjSv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like &quot;3 Ply Built up roof &quot; but the rest makes no sense. Any seasoned structural engineers can tell me what this says?&lt;/p&gt;&#xA;" OwnerUserId="4808" LastActivityDate="2016-04-05T18:45:08.193" Title="Wood roof specs on engineering drawing" Tags="|structural-engineering|drafting|architecture|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7503" PostTypeId="1" CreationDate="2016-02-18T23:54:06.950" Score="5" ViewCount="1399" Body="&lt;p&gt;I recently got a laser that is water cooled via a couple of clear plastic tubes (about .5 cm diameter).  I have heard it can happen that algae or other undesirable inhabitants can get into such tubes and cause problems.  I have also heard that wrapping the tubes in something dark to keep light out can help with this issue.  However, I have not heard of a good material for wrapping the tubes with, and I am looking for suggestions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally I would like to be able to slide a sheath of some opaque material over the tubes.  Something like heat shrink (unshrunk) might work.  I have also seen braided nylon sheaths for covering cables, but they don't look very opaque.  What's the best option?&lt;/p&gt;&#xA;" OwnerUserId="5103" LastEditorUserId="2819" LastEditDate="2016-02-19T02:48:38.637" LastActivityDate="2019-01-05T07:21:23.770" Title="How to prevent algae in water cooled laser tubes?" Tags="|design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7504" PostTypeId="2" ParentId="7499" CreationDate="2016-02-19T01:04:49.717" Score="3" Body="&lt;p&gt;It seems like what you're really after is the force, not the mass. Now, if you're assuming that the plate will shear off, then the easiest way of calculating the force isn't through Newton's laws, but by calculating the force necessary to shear off the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplistically, that can be found by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F = A\tau $$&#xA;where $A$ is the transversal cross-section of the plate along the shearing plane (assuming thin-walled theory applies here) and $\tau$ is the shear strength of the chosen material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't consider dynamic effects (buckling, warping, possibility of the plate simply bouncing off instead of shearing), but that's impossible to do with the information presented.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-02-19T01:04:49.717" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7505" PostTypeId="2" ParentId="7503" CreationDate="2016-02-19T03:02:47.280" Score="8" Body="&lt;p&gt;I use &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0038QHXPS&quot;&gt;polypropylene glycol&lt;/a&gt; in my CO2 laser tube under florescent lights and have not had any problems. It also comes with the added benefits of freeze protection and corrosion protection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would still like to block out the light you can use &lt;a href=&quot;http://www.usplastic.com/catalog/item.aspx?itemid=23758&amp;amp;catid=707&quot;&gt;black hdpe tubing&lt;/a&gt; for plumbing and helical wrap the tube with &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00004Z4DU&quot;&gt;blue masking tape&lt;/a&gt;. You could also spray paint the tube with black spray paint and remove it with acetone in the future if necessary. The most maintainable would be to put it inside an enclosure; which would have the added benefits of structurally protecting it and keeping users safe from high voltage.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-02-19T03:02:47.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7506" PostTypeId="2" ParentId="7498" CreationDate="2016-02-19T04:31:32.967" Score="1" Body="&lt;p&gt;By constructing the dome row by row you have the opportunity to attach the bottom row to the ground so it becomes secure prior to attaching the rest of the dome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the dome is constructed by joining vertical segments, the dome can't be securely attached to the ground until it has been constructed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, an internal scaffold may be needed to be erected prior to constructing the dome. The scaffold would act as a temporary support during construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If constructing by rows, the upper rows will not be self supporting, due to the shallow angles of the upper rows of the dome, until the dome is complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If constructing by vertical strips, the dome won't become stable until most of it has been completed.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-02-19T04:31:32.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7509" PostTypeId="2" ParentId="325" CreationDate="2016-02-19T11:07:06.103" Score="2" Body="&lt;p&gt;PWM is more efficient than potentiometer, but on the other hand sometimes the flicker can be problematic. For example for filming video or when objects are moving fast, such as car headlights (PWM is prohibited for headlights in some countries).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use a switching mode power supply (SMPS), which basically means adding an inductor and capacitor as a filter to the PWM signal to get a DC voltage / current going to the LED.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-02-19T11:07:06.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7510" PostTypeId="1" AcceptedAnswerId="7513" CreationDate="2016-02-19T12:31:09.313" Score="1" ViewCount="134" Body="&lt;p&gt;I have to write a scientific paper where I refer to a CAD drawing. Now, I'm a bit confused, because I have to refer to the length &quot;parameters&quot; on that drawing (as opposed to the angle &quot;parameters&quot;), e.g. the segment lengths &lt;em&gt;l1&lt;/em&gt;, &lt;em&gt;l2&lt;/em&gt;, ... Should I refer to these length parameters as &quot;length measurements&quot; or as &quot;length measures&quot;? I used both as a synonym so far, but I'm not sure if that is accurate.&lt;/p&gt;&#xA;" OwnerUserId="1925" LastActivityDate="2016-02-19T18:02:45.373" Title="'Length measure' vs. 'length measurement'" Tags="|mechanical-engineering|computer-aided-design|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7511" PostTypeId="2" ParentId="7510" CreationDate="2016-02-19T12:51:10.310" Score="4" Body="&lt;p&gt;The usual practice is to refer to measurements on a drawing as 'dimensions'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example you might say 'dimensions L1, L2 and L3 are determined by calculation'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term dimensions also implies that you are talking about nominal values whereas 'measurements' implies real world data. The difference between dimensions and measurements is defined by tolerances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a particular need to be unambiguous you can talk about linear, radial, angular or diametric dimensions. But dimension is the generic terms for any measurement on a drawing. In most cases this will be entirely clear from the units used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also best to be careful using terms like 'length' generically as they can be misleading. For example the diameter of a hole is a linear dimension but it is not a length. Length implies that you are referring on a straight line on the drawing as opposed to say, the distance between two parallel lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general terminology is : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Length : distance between two points forming a straight line&lt;/li&gt;&#xA;&lt;li&gt;Radius : used for incomplete circular arcs eg fillets or bends &lt;/li&gt;&#xA;&lt;li&gt;Diameter : used for complete circles eg holes and shafts &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In CAD the term 'parameters' is often used to refer specifically to dimension values in a model which are variable and assigned values from some external table, formula, function etc. Parametric values will also often be related to one another eg L2 is defined as being half L1. For example you might design a generic part which is produced in several variations so you can produce a drawing for each version simply by referencing a table of dimensions rather than manually redimensioning the model for every version required. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-19T13:09:39.037" LastActivityDate="2016-02-19T13:09:39.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7512" PostTypeId="2" ParentId="7503" CreationDate="2016-02-19T13:03:11.057" Score="3" Body="&lt;p&gt;I work at a laser manufacturer, and we ship all of our water-cooled lasers with a polypropylene glycol/water mixture, and we recommend that it be replaced on a yearly basis.  The coolant tubes are translucent so the fluid is exposed to room lights.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2016-02-19T13:03:11.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7513" PostTypeId="2" ParentId="7510" CreationDate="2016-02-19T13:24:54.557" Score="0" Body="&lt;p&gt;I might be misreading your question. As I understand it, your question is more about English language usage of engineering terminology than about an engineering issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, a measurement is a reading of a quantity. In your situation it is length, for an electrical engineer it might be volts or amps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, my interpretation of a measure is an action undertaken, such as: &quot;the air strike in Foobar province was a measure against counter insurgency&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your paper, I would suggest you use &lt;strong&gt;length measurements&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-02-19T13:24:54.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7516" PostTypeId="2" ParentId="7502" CreationDate="2016-02-19T15:19:15.453" Score="7" Body="&lt;p&gt;I am a practicing structural engineer with plenty of wood experience. That said, this is a bit of a tricky note because I believe it was probably written by an architect on a set of architectural drawings (as opposed to by a structural engineer on a set of structural drawings). Therefore, I think the note includes information about both the structural makeup of the roof diaphragm as well as the architectural components of the roofing system. That said, I can at least get you started. I am assuming that you are in the US, but if you're in another country, the abbreviations may be a bit different. I am less sure about the second line, but am confident about the meaning of the other 4 lines.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The first line reads, &lt;code&gt;3 PLY BU ROOF&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;I agree that it is communicating something related to the (mostly) architectural composition of a built up roof. It likely means that the roof is a 3 ply &lt;a href=&quot;http://construction.about.com/od/Finishes/a/Types-Of-Built-Up-Roofing-And-Its-Benefits.htm&quot; rel=&quot;noreferrer&quot;&gt;built up roof&lt;/a&gt;. Depending on your intent here, if it's structural for example, this doesn't have too much importance to you.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second line, &lt;code&gt;o/ 3/4&amp;quot; O.F. PLYWOOD INO.&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;This line is a description, of some sort, of the plywood. &lt;code&gt;o/&lt;/code&gt; is, I think, an abbreviation for the word &amp;quot;over.&amp;quot; So the objects on this line are under the 3 ply built up roof. &lt;code&gt;3/4&amp;quot;&lt;/code&gt; is describing the plywood thickness, &lt;code&gt;OF&lt;/code&gt; is an old way to describe &amp;quot;oriented fiber&amp;quot; material, this is similar to oriented strand board (OSB) that you see at the hardware store today. This is a little confusing, since plywood is not oriented fiber board. It could have another meaning. &lt;code&gt;OF&lt;/code&gt; could also mean &amp;quot;outside face&amp;quot; and this may be an instruction about how to apply a coating related to the built up roof to the outside face of the plywood. Also, since the plywood is given a span rating (see next point), no plywood thickness note is required so the &lt;code&gt;3/4&amp;quot;&lt;/code&gt; may not be referring to the plywood at all.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The third and fourth line, &lt;code&gt;32/16 (UNBLKD) W/ 8d @ 6&amp;quot; EN &amp;amp; RO&lt;/code&gt;&lt;/p&gt;&#xA;&lt;p&gt;These lines are decribing the strucutral requirements of the plywood and associated fasteners.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;32/16&lt;/code&gt; is a &amp;quot;span rating&amp;quot; for the plywood. From the &lt;a href=&quot;http://www.apawood.org/apa-trademark&quot; rel=&quot;noreferrer&quot;&gt;APA&lt;/a&gt;,&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Two numbers separated by a slash. The left-hand number is the maximum recommended center-to-center spacing for supports in inches when the panel is used for roof sheathing with long dimensions across supports. The right-hand number is the maximum center-to-center spacing of supports in inches when the panel is used for subflooring with the long dimension across supports.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;(UNBLKD)&lt;/code&gt; is instructing that the edges of the plywood do not need to be &lt;a href=&quot;https://en.wikipedia.org/wiki/Blocking_(construction)&quot; rel=&quot;noreferrer&quot;&gt;blocked&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;W/&lt;/code&gt; is an abbreviation for the word &amp;quot;with&amp;quot;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;8d @ 6&amp;quot;&lt;/code&gt; is instructing the use of 8d nail at a spacing of 6&amp;quot; on center.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;code&gt;EN &amp;amp; RO&lt;/code&gt; is instructing that the previously mentioned nailing is applied as &lt;strong&gt;E&lt;/strong&gt;dge &lt;strong&gt;N&lt;/strong&gt;ailing and at &lt;strong&gt;R&lt;/strong&gt;ough &lt;strong&gt;O&lt;/strong&gt;penings. This means that the 6&amp;quot; oc nailing is applicable at the endges of the roof and around any openings.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The fifth line, &lt;code&gt;12&amp;quot; FN&lt;/code&gt; is instructing a 12&amp;quot; on center &amp;quot;field nailing.&amp;quot; This means that the plywood should be nailed to all the rafters/joists in areas that are not diaphragm edges at a 12 oc spacing.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Good luck and please comment when you figure it out. I've passed your image around the office and it's generated quite the lively discussion!&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-04-05T18:45:08.193" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7517" PostTypeId="2" ParentId="7368" CreationDate="2016-02-19T15:35:33.710" Score="1" Body="&lt;p&gt;The second edition of the AISC Sesimic Design Manual (SDM) contains many wonderfully worked out example problems. I highly recommend that you purchase this resource if you are designing braced frames in a professional setting. It can be purchased from the AISC, and if you are ever planning to sit and take the SE exam, you'll need to buy it anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 5.2.1 in the SDM is the complete design of the diagonal brace in an ordinary concentric moment frame. In this example, while discussing the effective length of the brace, it states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By inspection the laterally braced length of the diagonal brace in the in-plane direction is half the overall length. For buckling out-of-plane, &lt;strong&gt;if both of the diagonals are continuous for their full length and are connected at the intersection point, then the effective length factor, K, is 0.5&lt;/strong&gt; (El-Tayem and Goel, 1986; Picard and Beaulieu, 1987).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Emphasis mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in an effective length for &lt;em&gt;both&lt;/em&gt; the in-plane and out-of-plane bending of the brace to be 0.5 * L.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2016-02-27T17:04:46.660" LastActivityDate="2016-02-27T17:04:46.660" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7518" PostTypeId="2" ParentId="7510" CreationDate="2016-02-19T18:02:45.373" Score="2" Body="&lt;p&gt;The conventional term would be a &lt;em&gt;linear dimension&lt;/em&gt; if you just want to specify that your measurement is of a length. There are also &lt;em&gt;angular dimensions&lt;/em&gt; for angles, and more complicated dimensional features like tolerances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're discussing the CAD drawing in terms of the CAD program (not just to describe the object it shows) and you want to identify your parameters you would want to distinguish between a &lt;em&gt;driving dimension&lt;/em&gt; and a &lt;em&gt;driven dimension&lt;/em&gt;. A driving dimension is a length, angle or other number that you enter which is driving your model. A driven dimension is a length that you are showing in your drawing, but which is automatically determined based on the driving dimensions you provide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that specific CAD packages have more complicated names to describe how a dimension is created (eg. Autocad distinguishes between linear and aligned dimensions to describe how it sets the axis of the dimension) but in the context of the final drawing, most dimensions still fit into these two categories.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-02-19T18:02:45.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7519" PostTypeId="1" CreationDate="2016-02-19T18:10:38.070" Score="3" ViewCount="249" Body="&lt;p&gt;I'm calculating the combined measurement uncertainty u(c) of a measurement for a length displacement sensor. Amongst other factors, one individual uncertainty affecting the error of measurement is the uncertainty of the gage block I'm using. In the calibration certificate of the gage block the uncertainty of measurement is expressed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$U(E)=0.08\ \mathrm{µm}+1.2\ \mathrm{L}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the coverage factor $k=2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nominal thickness of the gage block is 10 mm and usually uncertainty is expressed as just  ±0.08 µm. In this case, what does $L$ stand for?&lt;/p&gt;&#xA;" OwnerUserId="4088" LastEditorUserId="4088" LastEditDate="2016-02-19T22:53:26.890" LastActivityDate="2016-02-19T22:53:26.890" Title="What does L mean in Uncertainty Measurement Result U(E)=0.08+1.2L ;k=2?" Tags="|measurements|" AnswerCount="1" CommentCount="1" ClosedDate="2016-02-22T14:03:06.547" ContentLicense="CC BY-SA 3.0" />
  <row Id="7520" PostTypeId="1" CreationDate="2016-02-19T18:22:44.843" Score="13" ViewCount="3771" Body="&lt;p&gt;&lt;strong&gt;Are there any materials known to be sound-insulating as well as thermally-conductive?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cost is not an overriding concern, but preferably costing &quot;reasonable amounts&quot; and much less than national defense spending levels.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is a thought experiment that I'm pondering regarding the idea of a computer case which conducts heat while isolating or dampening all of the sound inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's multiple holes in the idea when fully formed regarding financially reasonable, allowing air-flow, etc, but to keep this question narrow I want to just dismiss all of those concerns for the thought experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recognize conducting one type of energy while insulating against another is not a straight-forward thing.  The closest material that comes to mind is water, but that's not completely accurate because it does conduct kinetic energy (i.e. hydraulics).  It is better described as acting as a sort of low-pass filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if no materials come to mind, I'd be interested in hearing of something that could &lt;em&gt;filter&lt;/em&gt; sound very effectively while conducting heat as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to put the qualifier that the material is a solid, but I'd be interested really in any material with these properties just because I believe (perhaps wrongly?) that there are likely a very small number of them. Again, don't think in terms of the goal concept I mentioned regarding a computer case, this is just one step towards that thought experiment (which is likely not reasonable, or else products would be on the market right now that provide such).&lt;/p&gt;&#xA;" OwnerUserId="280" LastEditorDisplayName="user16" LastEditDate="2016-02-19T18:25:24.197" LastActivityDate="2016-03-25T09:34:29.563" Title="Are there any materials known to be sound-insulating but thermally-conductive?" Tags="|materials|thermal-conduction|sound-isolation|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7521" PostTypeId="2" ParentId="7520" CreationDate="2016-02-19T19:08:32.243" Score="9" Body="&lt;p&gt;I can think of a few possible ways to approach this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the first thing to look at is the design of the case itself. They generally tend to be made up of thin, flat panels riveted to a frame. In this sort of situation you can get a lot of benefit from the use of a relatively small amount of sound deadening material. This approach is widely used on automotive body panels and has the advantage that a fairly small strip can reduce the ability of the panel to transmit sound significantly and can be as simple as bitumen impregnated felt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also relatively thin films or coatings can have a significant effect on acoustic response without providing much thermal insulation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly there is probably a lot of potential in increasing the damping of the whole structure of the case, for example by using rubber gaskets or spacers at metal to metal joints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In more general terms it is often useful to think not so much in terms of sound insulation as a general material property but to look at the way that the structure as a whole responds to the vibration frequencies that it is subject to and 'tune' it to respond in the way that you want by adjusting its mass, stiffness and damping so that its resonant frequency is as far away as possible from the excitation frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So overall I would advise looking at the dynamic behaviour of the case as a whole rather than just  a specific material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that there are a few materials which are worth considering. Aluminium and copper (especially in high purity) have excellent thermal conductivity and are fairly 'soggy' in their mechanical properties, also cast iron has particularly good vibration damping characteristics as well as reasonable thermal conductivity (and I would love to see a cast iron PC case). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also be worth considering composites. Carbon fibre has the potential for good thermal conductivity and it may be worth investigating aluminium filled resins and  these may give a bit more flexibility in tuning the structure of teh case to get the best acoustic response. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible approach might be to sandwich a gel or powder between two sheets of aluminium or copper. There have also been experiments in completely submerging a computer system in oil which would provide both acoustic damping and convective cooling. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-19T19:23:50.020" LastActivityDate="2016-02-19T19:23:50.020" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7522" PostTypeId="2" ParentId="7520" CreationDate="2016-02-19T20:35:16.090" Score="1" Body="&lt;p&gt;Since nearly all the sound (aside from speakers themselves) originates in cooling fans, it's going to be really tough to block sound while allowing exhaust airflow. You'll need a different, expensive, cooling system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the next noisiest is the hard drive, which if well heat-sunk could be covered locally with sound-absorbing material (or switch to all-SSD at some expense).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sum: you  are far better off eliminating sources of sound than trying after the fact to ameliorate the problem.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-02-19T20:35:16.090" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7523" PostTypeId="2" ParentId="7519" CreationDate="2016-02-19T20:44:14.020" Score="4" Body="&lt;p&gt;Found it at &lt;a href=&quot;http://www.nist.gov/calibrations/upload/mono180.pdf&quot; rel=&quot;nofollow&quot;&gt;NIST&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are four defined tolerance grades in ISO 3650; 00, 0, 1 and 2.&#xA;  The algorithm for the length tolerances are shown in table 2.3, and&#xA;  there are rules for rounding stated to derive the tables included in&#xA;  the standard. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\text{Table 2.3} \\&#xA;\begin{array}{cc}&#xA;\text{Grade} &amp;amp; \text{Deviation from Nominal} \unicode{0x000A} \text{Length (µm)} \\&#xA;00 &amp;amp; (0.05 + 0.0001L) \\ &#xA;0 &amp;amp; (0.10 + 0.0002L) \\&#xA;1 &amp;amp; (0.20 + 0.0004L) \\&#xA;2 &amp;amp; (0.40 + 0.0008L) \\&#xA;\end{array}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Where L is the block nominal length in millimeters.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The ISO standard does not have an added tolerance for measurement&#xA;  uncertainty; however, the ISO tolerances are comparable to those of&#xA;  the ANSI specification when the additional ANSI tolerance for&#xA;  measurement uncertainty is added to the tolerances of Table 2.1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="615" LastEditorUserId="368" LastEditDate="2016-02-19T22:34:22.213" LastActivityDate="2016-02-19T22:34:22.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7524" PostTypeId="1" AcceptedAnswerId="7530" CreationDate="2016-02-20T02:19:39.323" Score="4" ViewCount="2907" Body="&lt;p&gt;In my stem class we are going to have a competition where we have to build a toothpick bridge that can support as must weight as possible, while using only toothpicks and glue. These are the rules:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each toothpick costs \$10,000.&lt;/li&gt;&#xA;&lt;li&gt;Each end of a toothpick glued to another end of a toothpick costs \$10,000.&lt;/li&gt;&#xA;&lt;li&gt;The bridge must span at least 1 foot. &lt;/li&gt;&#xA;&lt;li&gt;We can't build beams down to the floor for support.&lt;/li&gt;&#xA;&lt;li&gt;The base of the bridge, the part cars would drive on if it were a real bridge, can't have any large gaps in it, although I'm not entirely how much this matters.&lt;/li&gt;&#xA;&lt;li&gt;We will have a time limit when building it, but I don't know what it will be. We can bring a hairdryer to help the glue dry faster.&lt;/li&gt;&#xA;&lt;li&gt;It must hold a lot of weight, and be cheap. I don't know how much each of those factors matter, but I'd imagine the weight it can hold is much more important.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My teacher wants  us to do lots of research before the project and have a clear plan, so I have been looking into what kind of bridge to build. It seems like the obvious choice is a beam bridge, but I could be wrong. I found &lt;a href=&quot;http://www.warwickallen.com/bridges/BeamBridges.htm&quot; rel=&quot;nofollow&quot;&gt;this site&lt;/a&gt; talking about how a truss can help a beam bridge hold more weight, but I don't quite understand how it helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a beam bridge is my best option, would a truss significantly help at this scale? And if so, which kind of truss? The warren looks more efficient as far as costs are concerned, but I don't know if the pratt or any other kind has a major advantage. Any advice on this would be greatly appreciated, thanks!&lt;/p&gt;&#xA;" OwnerUserId="4676" LastEditorUserId="255" LastEditDate="2016-04-27T16:26:08.873" LastActivityDate="2016-04-27T20:21:48.573" Title="Most efficient way to build a toothpick bridge" Tags="|structural-engineering|civil-engineering|bridges|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7525" PostTypeId="2" ParentId="7520" CreationDate="2016-02-20T02:26:35.220" Score="1" Body="&lt;p&gt;If the hard disk drive was replaced by with a solid state drive and the heat sink for the CPU chip was made bigger and externalized it would be possible to have a silent computer which cooled directly to the atmosphere or could be cooled by other means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, a heat sink is attached to a CPU chip and all this is contained within the computer case, whether it be a laptop or a desktop. To cool the heat sink fans are placed in the computer case to provide airflow over the heat sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one on the larger panels of the computer case acted as a heat sink it could be connected to the CPU chip via thermal conductors and the heat produced by the CPU chip drawn away and radiated to the atmosphere or cooled by other means, such as a water jacket. This would eliminate the need for noisy cooling fans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, replacing noise producing hard disk drive with silent solid state drives would eliminate the other main source of noise produced by computers.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-02-20T02:26:35.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7526" PostTypeId="1" AcceptedAnswerId="7544" CreationDate="2016-02-20T03:42:11.050" Score="8" ViewCount="11792" Body="&lt;p&gt;First of all, I'm interested in train &lt;strong&gt;track&lt;/strong&gt; only, not the rolling stock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are tracks built to cope with really cold weather? An example might be some place in Canada or Siberia. Ice would accumulate all the time, changing the shape of the rail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the ballast or foundation cope with the thawing tundra in summer? What about the really nasty stuff like blizzards dumping snow over the track, or sleet, or freezing rain?&lt;/p&gt;&#xA;" OwnerUserId="3510" LastEditorUserId="33" LastEditDate="2018-05-08T02:14:25.440" LastActivityDate="2022-12-23T20:36:28.810" Title="How do train tracks handle really cold weather?" Tags="|materials|temperature|rail|transportation|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7527" PostTypeId="2" ParentId="7520" CreationDate="2016-02-20T05:37:13.643" Score="2" Body="&lt;p&gt;The first couple of things that come to mind are metallic 'wools' (ex. steel wool/stainless steel 'scrub pads'), or carbon foams.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Metal wools will provide minimal thermal insulation (metal strands conduct, air gaps insulate, so it balances out mostly), but a large amount of vibration dampening 'limp mass.'&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Carbon foams/aerogels/what not: aerogels are &lt;em&gt;very&lt;/em&gt; effective sound dampening materials, but are also quite insulative. Carbon-based foams (i.e. make an organic gel/foam, then carburize in a reducing-atmosphere oven) could be easily made with a larger pore size using freeze-drying processes to help lower the thermal insulation, without reducing the acoustic properties too much.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An afterthought: You could also use a vinyl/carbon-fiber type composite (maybe filled with alumina for enhanced thermal conductivity) to make a somewhat flexible (albeit a bit pricey) panel for constructing your computer box. The panel flexibility should work well for damping many of the vibrations &amp;amp; absorbing the energy (esp. by not 'buzzing' as much where panels meet, or where noisy fans/HDDs are mounted to panels).&lt;/p&gt;&#xA;" OwnerUserId="5123" LastEditorUserId="350" LastEditDate="2016-02-21T18:56:23.763" LastActivityDate="2016-02-21T18:56:23.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7528" PostTypeId="1" AcceptedAnswerId="7529" CreationDate="2016-02-20T09:00:15.823" Score="3" ViewCount="320" Body="&lt;p&gt;I just watched this video: &lt;a href=&quot;https://www.facebook.com/HigherPerspective/videos/1162007547164896/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.facebook.com/HigherPerspective/videos/1162007547164896/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what I see. If the spinner's magnetic, the bowl is made of iron and that the wood piece is magnetic too. Shouldn't the iron bowl mess the magnetic field? Can someone explain how this works?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another similar image from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vatCu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vatCu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5126" LastEditorUserId="33" LastEditDate="2016-02-21T19:15:59.943" LastActivityDate="2016-02-21T19:15:59.943" Title="How does a Levitron work?" Tags="|magnets|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7529" PostTypeId="2" ParentId="7528" CreationDate="2016-02-20T10:36:22.173" Score="2" Body="&lt;p&gt;This is an example of spin stabilised magnetic levitation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's really just two magnets repelling each other, one in the wooden base, one in the spinning top; with opposite poles facing each other. This system of two magnets  isn't stable on its own but the fact that the top is spinning provides gyroscopic stabilization. The bowl, bubbles, smoke and bits of wood are just cosmetic and have nothing to do with how it actually works.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-21T01:57:29.070" LastActivityDate="2016-02-21T01:57:29.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7530" PostTypeId="2" ParentId="7524" CreationDate="2016-02-20T11:44:41.480" Score="5" Body="&lt;p&gt;As you are spanning 10 feet and the length of the toothpicks is small in comparison you have quite a lot of options in terms of the design of the structure you use. In fact one approach would be to design the bridge as if it was a monolithic material and construct that as a mesh of toothpicks as if it wa a finite element analysis model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point you will want to refine your specification to prioritise between cost and load bearing capacity. In this case cost and weight are pretty much the same considerations you have a fixed price for materials (and presumably the design and construction are 'free'). As strong as possible and as cheap as possible is an open ended problem and at some point you need to pick a specific target to aim for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the materials you have to work with a lot depends on the properties of the glue joints. In particular, how flexible are they and how strong are they relatively in compression and tension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought would be a girder arch bridge as this gives you an efficient overall structure which can be constructed from relatively short members. One example is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ponte_dell%27Accademia&quot; rel=&quot;nofollow&quot;&gt;Ponte dell'Accademia&lt;/a&gt; in venice and looking at the early industrial and pre-industrial wooden bridges is likely to be a useful line of research. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another key consideration is how your 'foundations' are situated you will notice that building from bank to bank across a river required different designs to crossing a gorge or valley. Similarly if you don't need to have a completely flat road deck you have a lot more options. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update :&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having clarified that the span is 1 foot not 10 foot (which I probably should have questioned in the first place). Then a box girder construction does become quite a sensible approach. In terms of detailed design it is really a case of optimisation and as you have no control over the basic materials (toothpicks and glue) then it is a question of generating a design which meets the strength requirements using the fewest toothpicks. By extension this means looking at how the design will be evaluated, for example will the 'best' bridge be the one which holds the most weight or the cheapest one which holds a specific minimum weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of real world engineering ultimately comes down to this ability to properly evaluate and define a problem and find a good solution ie the ability to determine &lt;em&gt;exactly&lt;/em&gt; what problem you are trying to solve and how success is determined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example saying that it must carry 20lbs of weight and be as cheap as possible is a well defined problem as is cost no more than $500,000 and carry as much weight as possible. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-20T17:28:27.480" LastActivityDate="2016-02-20T17:28:27.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7531" PostTypeId="1" CreationDate="2016-02-20T14:19:06.627" Score="5" ViewCount="321" Body="&lt;p&gt;I have a small idea/project I am working on in which I want to use a water fiber optic sensor that can detect whether the water stream is broken and if so shut off the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having looked into fiber optic sensor systems in general, all I have managed to work out is I would need an intrinsic sensing region system.  I do not know what type of Optical Modulation mechanism I would need so have hit a wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any guidance into which of these systems, Intensity modulated, Phase modulated, Wavelength modulated or Polarization modulated, could be useful would be massively appreciated as I'd love to be able to look into this further.&lt;/p&gt;&#xA;" OwnerUserId="1149" LastEditorUserId="33" LastEditDate="2016-02-21T19:37:08.350" LastActivityDate="2016-02-22T08:08:25.443" Title="Could a water stream be used as a fiber optic sensor and detect when the stream is broken?" Tags="|control-engineering|flow-control|fiber-optics|" AnswerCount="3" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7532" PostTypeId="1" AcceptedAnswerId="7533" CreationDate="2016-02-20T19:44:45.290" Score="1" ViewCount="99" Body="&lt;p&gt;On my laptop, I set a variable that makes it so the line tool defaults to Cartesian coordinates. On my desktop, I can't seem to find the variable name. I have been searching for hours.&lt;/p&gt;&#xA;" OwnerUserId="4855" LastEditorUserId="33" LastEditDate="2016-02-21T19:23:49.953" LastActivityDate="2016-02-21T19:23:49.953" Title="How do I set the default coordinate system for the line tool in Autocad?" Tags="|autocad|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7533" PostTypeId="2" ParentId="7532" CreationDate="2016-02-20T22:15:44.680" Score="1" Body="&lt;p&gt;The variable is DYNPIFORMAT.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Controls whether pointer input uses polar or Cartesian format for coordinates.&lt;/p&gt;&#xA;&lt;p&gt;This setting applies only to a second or next point.&lt;/p&gt;&#xA;&lt;p&gt;0 Polar&lt;/p&gt;&#xA;&lt;p&gt;1 Cartesian&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;(From: &lt;a href=&quot;https://knowledge.autodesk.com/support/autocad-lt-for-mac/learn-explore/caas/CloudHelp/cloudhelp/2015/ENU/AutoCAD-LT-MAC/files/GUID-03D36D36-D734-43C3-BA4D-EA68C573F73C-htm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://knowledge.autodesk.com/support/autocad-lt-for-mac/learn-explore/caas/CloudHelp/cloudhelp/2015/ENU/AutoCAD-LT-MAC/files/GUID-03D36D36-D734-43C3-BA4D-EA68C573F73C-htm.html&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="4855" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-02-20T22:15:44.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7534" PostTypeId="1" CreationDate="2016-02-20T22:23:49.353" Score="3" ViewCount="656" Body="&lt;p&gt;In my course of dynamics of structures, I am struggling with some basic principles. As a practice, I am plotting (or attempting to plot) the curves that are shown in the course book, with the help of Matlab. For example, I want to plot the phase angle theta as a function of beta, the dimensionless frequency, for a damped SDOF. The code I have is the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%Stiffness of the spring in [N/m]&#xA;k = 1;&#xA;%Mass in [kg]&#xA;m = 1;&#xA;%Damping constant in [Ns/m]&#xA;c = 0.1;&#xA;%Initial displacement in [m]&#xA;u_0 = 0;&#xA;%Initial velocity in [m/s]&#xA;v_0 = 0;&#xA;%% &#xA;%Resonance frequency in [rad/s]&#xA;omega_res = sqrt(k/m);&#xA;%Resonance frequency in [Hz]&#xA;f_res = (omega_res)/(2*pi);&#xA;%Eigenperiod in [s]&#xA;T_res = 1/f_res;&#xA;%% &#xA;%Critical damping in [Ns/m]&#xA;c_c = 2*sqrt(k*m);&#xA;%Damping ratio in [/]&#xA;xi = c/c_c;&#xA;%% &#xA;%Damped eigenfrequency in [rad/s]&#xA;omega_d = omega_res * sqrt(1-xi^2);&#xA;%Damped eigenfrequency for strongly damped system in [rad/s]&#xA;omega_hat = omega_res * sqrt(xi^2-1);&#xA;%% &#xA;%Timevector&#xA;t = 0:0.1:100;&#xA;%Betavector&#xA;beta_v = 0:0.001:4;&#xA;%D as a function of beta&#xA;D = 1./sqrt((1-beta_v.^2).^2+(2*beta_v.*xi).^2);&#xA;&#xA;plot(beta_v,D)&#xA;axis([0 4 0 10])&#xA;xlabel('Dimensionless frequency [-]')&#xA;ylabel('Dynamic amplification factor [-]')&#xA;&#xA;%theta as a function of beta&#xA;theta = atan((2*beta_v*xi)./(1-beta_v.^2));&#xA;&#xA;figure&#xA;plot(beta_v,angle(theta),'m')&#xA;axis([0 4 0 pi])&#xA;grid on&#xA;xlabel('Dimensionless frequency [-]')&#xA;ylabel('Phase angle [rad]')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But plotting gives me the curve of an undamped SDOF. The curve I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pONPd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pONPd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot I am supposed to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1tOLv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1tOLv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a clue what is wrong in my approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second struggle occurs when you want to plot the the response for a particular beta. The curve I achieved was not what I expected. My code is the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%Response&#xA;beta_a = 0.2;&#xA;D_a = 1./sqrt((1-beta_a.^2).^2+(2*beta_a.*xi).^2);&#xA;theta_a = atan((2*beta_a*xi)./(1-beta_a.^2));&#xA;p_hat = 1;&#xA;omega = beta_a.*omega_res;&#xA;u_p = p_hat/k.*D_a.*exp(i.*(omega.*t-theta_a));&#xA;B = u_0 - p_hat./k.*D_a.*exp(-i.*theta_a);&#xA;A = (v_0 + xi*omega_res*B - p_hat/k*D_a.*i*omega*exp(-i.*theta_a))/omega_d;&#xA;u_h = exp(-xi.*omega_res.*t).*(A.*sin(omega_d.*t) + B.*cos(omega_d.*t));&#xA;u = u_p + u_h;&#xA;figure&#xA;plot(t,real(u))&#xA;grid on&#xA;xlabel('Time [s]')&#xA;ylabel('Displacement [m]')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me with this line of thought? What's wrong?&lt;/p&gt;&#xA;" OwnerUserId="5132" LastEditorUserId="5132" LastEditDate="2016-02-24T18:10:25.703" LastActivityDate="2016-02-24T18:10:25.703" Title="How do I create a graph of the phase angle and response of a harmonic excitation in Matlab?" Tags="|civil-engineering|structures|dynamics|matlab|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7535" PostTypeId="1" CreationDate="2016-02-20T23:00:34.210" Score="3" ViewCount="1347" Body="&lt;p&gt;I have two files open in AutoCAD. One uses horizontal text in radius annotation (see the first picture). The second uses a straight line in radius annotation (see the second picture). I cannot find any difference in the settings of the two files. How do I get the second file's annotation like the first?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture 1:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JJNzF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JJNzF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture 2:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CAnUe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CAnUe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4855" LastEditorUserId="33" LastEditDate="2016-02-21T19:18:35.600" LastActivityDate="2016-09-04T12:16:01.400" Title="How do I change the text alignment in a radius annotation with AutoCAD?" Tags="|autocad|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7536" PostTypeId="1" AcceptedAnswerId="7537" CreationDate="2016-02-21T00:46:55.283" Score="24" ViewCount="6211" Body="&lt;p&gt;I'm referring to the process shown in this video:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/Quyr5R1Rbfw?t=20&quot; rel=&quot;noreferrer&quot;&gt;https://youtu.be/Quyr5R1Rbfw?t=20&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or this image from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U2vdC.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U2vdC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In it, a large warship is launched into the water by essentially dropping it sideways down some ramps and off of a pier.  The ship rolls hard to one side, and then oscillates back to the other, making the process seem like a fairly risky one.  For instance, if it rolls back towards the pier too aggressively it might strike the structure and cause damage both to the pier and the shiny new warship.  Or if it rolls too far on the initial drop, the ship might capsize.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, what are the advantages of launching a large ship sideways like this, as opposed to, say, dropping or gently lowering it vertically into the water?&lt;/p&gt;&#xA;" OwnerUserId="5133" LastEditorUserId="33" LastEditDate="2016-02-21T20:10:21.470" LastActivityDate="2016-02-22T00:36:30.187" Title="Why would you launch a large ship by dropping it sideways?" Tags="|mechanical-engineering|structural-engineering|safety|naval-engineering|ships|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7537" PostTypeId="2" ParentId="7536" CreationDate="2016-02-21T01:24:27.897" Score="24" Body="&lt;p&gt;A specific reason for doing this is simply when there isn't enough room to do a bow or stern first launch. This is often the case when a ship or boat is built in a yard on a river or canal either because the hull is especially long or the channel it is being launched into is narrow. There is also the consideration that a sideways launch can be done from any quay that will take the weight but a bow first launch requires a specially constructed slipway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that if you go in sideways the load is distributed along the whole length of the hull whereas bow first has the potential to put a big bending load on the hull. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of risk that sort of roll should pose no problem at all for an ocean going ship and it is likely to be fairly heavily ballasted to compensate as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small boats can be craned into the water but with larger tonnage hulls this just isn't practical due to their sheer weight. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="5123" LastEditDate="2016-02-22T00:36:30.187" LastActivityDate="2016-02-22T00:36:30.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7538" PostTypeId="1" CreationDate="2016-02-21T05:04:38.750" Score="0" ViewCount="1534" Body="&lt;p&gt;Looking to drive a series of belts for lateral translation using a series of pallet bands (92&quot; rubber bands about 1&quot; wide). Approximately 16 bands would be driven by a singular band as illustrated (roughly) below. I am hoping to use a geared up stepper motor with planetary gearbox that will respond to sound levels to effect the speed of translation. I am wondering if anyone has any advice about using rubber bands as belts in a pulley driven system, ideal arrangements for transfer of power, pulley types etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FLX9J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FLX9J.jpg&quot; alt=&quot;upper zig zaging band for power distribution from stepper motor, lower bands only need to slowly tanslate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hTnaZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hTnaZ.jpg&quot; alt=&quot; &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5096" LastEditorUserId="1832" LastEditDate="2016-02-21T11:45:47.653" LastActivityDate="2016-02-21T18:10:32.870" Title="Rubber Band for Belt in Belt Drive system" Tags="|mechanical-engineering|pulleys|building-physics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7539" PostTypeId="1" CreationDate="2016-02-21T07:19:56.530" Score="2" ViewCount="9407" Body="&lt;p&gt;How do I find the entropies for T1 = -10 deg C and T2 = 30 deg C, respectively using the R-134a refrigerant chart (SI):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wvQCS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvQCS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1363" LastEditorUserId="33" LastEditDate="2016-02-21T19:07:44.390" LastActivityDate="2016-02-21T19:13:49.590" Title="How to find the entropy for a given temperature using P-H chart for a refrigerant?" Tags="|thermodynamics|refrigeration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7540" PostTypeId="2" ParentId="7539" CreationDate="2016-02-21T08:03:31.793" Score="3" Body="&lt;p&gt;Instead of giving you fish I will teach you how to catch it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't a particular entropy at a given temperature. Considering the wet region there should be given a dryness fraction value for the entropy. With dryness fraction value and temperature combined, you can calculate the entropy of the refrigerant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$entropy (s)  = sf + x * sfg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;br&gt;&#xA;$\text{sf = the entropy of saturated liquid R-134a}$&#xA;$\text{x = the dryness fraction (between 0 and 1)}$&#xA;$\text{sfg = the entropy change during the evaporation process}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the image below. The sf value can be taken from the chart. See the left value. It's between 1 and .9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now see the (sfg + sf) value on the right. When you equate it with a certain dryness fraction x you will get entropy at that particular point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rQfyi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rQfyi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2116" LastEditorUserId="33" LastEditDate="2016-02-21T19:13:49.590" LastActivityDate="2016-02-21T19:13:49.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7541" PostTypeId="2" ParentId="7536" CreationDate="2016-02-21T11:48:20.797" Score="3" Body="&lt;p&gt;End to end, it would bend and snap in half under its own weight.&#xA;I know it's just a movie, but watch &quot;Titanic&quot; for a visual.&#xA;The support required otherwise would be huge and expensive.&#xA;A boat/ship will naturally rock side to side without breaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impact with the water will act as a brake and keep it from rocking too far.&#xA;The weight of the ship will cause it to sink more and more, adding to friction with the water. The water &lt;strong&gt;does not&lt;/strong&gt; want to compress downward so it surges &lt;strong&gt;upward&lt;/strong&gt; into a wall. Gravity pulls the wall back down, even though only part of it is still pressing backwards on the ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physics will also fight it rocking back against the dock, inertia and gravity.&#xA;Think about a bolt in your ceiling attached to a weight, like a bowling ball with a rope. Pull it back from the center where it hangs, to your face, and let it go. It swings out away from you, and then swings back towards your face.&#xA;It will not swing back far enough to hit you because gravity is pulling it down, and friction with the air slows it down. (Water has even more friction than air!) Internal friction of the individual strands of the rope against each other also comes into play, but does not illustrate the point enough to mention in great detail.&lt;/p&gt;&#xA;" OwnerUserId="5134" LastEditorUserId="5136" LastEditDate="2016-02-21T19:05:26.767" LastActivityDate="2016-02-21T19:05:26.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7542" PostTypeId="2" ParentId="7538" CreationDate="2016-02-21T18:10:32.870" Score="1" Body="&lt;p&gt;I would seriously consider using separate motors for each element. A drive train as long as that has a very high chance of causing a lot of practical problems with friction, slip and backlash as well as the difficulties of keeping everything aligned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chances are that a single motor would need to be significantly more powerful than the sum of individual ones due to the additional resistance of the shared drive system and the additional complexity is likely to wipe out any cost savings of one large over several small motors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also be concerned about hanging the mounting point directly on a rubber belt of such a large span, even a modest load is going to cause significant sag of the belt which will further exacerbate problems with friction and alignment. This should be resolvable by using rails to support the weight of the moving elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If having a common drive is essential then I suggest that using a horizontal shaft driving each belt with bevel gears or a similar right angle drive system is likely to be less trouble. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-21T18:10:32.870" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7543" PostTypeId="2" ParentId="7531" CreationDate="2016-02-21T19:35:28.263" Score="0" Body="&lt;h3&gt;Is it possible?&lt;/h3&gt;&#xA;&lt;p&gt;Yes, it is certainly possible. It may even be easier than you are assuming.&lt;/p&gt;&#xA;&lt;h3&gt;Water stream light pipes&lt;/h3&gt;&#xA;&lt;p&gt;Using a water stream as a light pipe is a somewhat common science experiment. You can look up various videos and how-tos, but the general principal is that the light will follow the water stream as it flows.&lt;/p&gt;&#xA;&lt;p&gt;An &lt;a href=&quot;https://www.highlightskids.com/science-experiment/make-light-pipe&quot; rel=&quot;nofollow noreferrer&quot;&gt;example project&lt;/a&gt; shows this image of the light in the stream:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JUQCGm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JUQCGm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Once you have the light going through the stream, all you need is a detection method which may be a simple as a light sensor. Depending on the amount of ambient light, special methods might be required to filter out unwanted light. The modulation might not be required.&lt;/p&gt;&#xA;&lt;p&gt;I'm not sure how long of a length this will work for though.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-02-21T19:35:28.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7544" PostTypeId="2" ParentId="7526" CreationDate="2016-02-21T20:03:36.250" Score="15" Body="&lt;h2&gt;Precipitation&lt;/h2&gt;&#xA;&lt;h3&gt;Snow&lt;/h3&gt;&#xA;&lt;p&gt;Snow can be a problem for running trains, but it really doesn't affect the rail/ballast. Just like on highways, the snow needs to be moved away, but it doesn't have many other effects.&lt;/p&gt;&#xA;&lt;p&gt;Trains are used to plow through small amounts (Wikipedia):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WZOeA.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WZOeA.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;and large amounts (Wikipedia):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N4gD4.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N4gD4.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Ice&lt;/h3&gt;&#xA;&lt;p&gt;Ice could cause more of an issue due to adding thickness to the rails. This is mostly handled by making the trains travel slower. The weight of the wheels on the rail is enough to crush through any ice that has accumulated, i.e. trains are really heavy. As long as the train doesn't move too fast, it isn't in danger of derailing.&lt;/p&gt;&#xA;&lt;h2&gt;Track Components&lt;/h2&gt;&#xA;&lt;h3&gt;Switches&lt;/h3&gt;&#xA;&lt;p&gt;Switches need to move in order to operate. Ice and snow accumulations can cause issues if they keep the components from moving. It is for this reason that most switches have &lt;em&gt;switch heaters&lt;/em&gt; installed. These are gas powered heaters that blow hot air in the appropriate locations to melt any snow or ice.&lt;/p&gt;&#xA;&lt;h3&gt;Rail&lt;/h3&gt;&#xA;&lt;p&gt;Like mentioned above, the rails aren't really affected by ice or snow. They are just chunks of metal. The biggest issue that rails face is the &lt;em&gt;cold&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Rails are restrained by rail anchors from expanding and contracting due to temperature changes. If the weather gets cold enough, the rail won't be able to contract enough and the stress will be so great that the rail will break. This isn't as much of an issue as you might think though. Many sections of railroad use the individual rails to transmit signal controls. They are set up to failsafe, so that a broken rail is the same as a broken connection. This will show a stop signal.&lt;/p&gt;&#xA;&lt;h3&gt;Ties&lt;/h3&gt;&#xA;&lt;p&gt;Ties are hunks of wood. They aren't really affected.&lt;/p&gt;&#xA;&lt;h3&gt;Ballast and subgrade&lt;/h3&gt;&#xA;&lt;p&gt;As you mentioned, freeze and thaw will cause issues due to volume changes. These effects can be foreseen and dealt with.&lt;/p&gt;&#xA;&lt;p&gt;The ballast section should be designed to drain water. If there is no water, there is less chance of volume changes. The same can be said for the subgrade. Drainage is key.&lt;/p&gt;&#xA;&lt;p&gt;Apart from a good design, freezing and thawing happens every year. It is very predictable. During the time periods where these effects are likely to cause issues, the track can be inspected regularly to keep it in geometric tolerance.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-02-21T20:03:36.250" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7545" PostTypeId="2" ParentId="7534" CreationDate="2016-02-21T21:00:17.747" Score="4" Body="&lt;p&gt;With your code, and with $c = 0.3$, I get the following results:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gM0UG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gM0UG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yFEtq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yFEtq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/AelDV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AelDV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amplitude and phase look OK.  But the displacement does not show any damping.  To see why read on below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure about your notation.  So the equations below may differ from those in your textbook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ODE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ODE you are trying to solve is&#xA;$$&#xA;  \ddot{u} + 2\xi\omega_{\text{res}}\dot{u} + \omega_{\text{res}}^2 u = \frac{\hat{p}}{m}\text{Re}\left[\exp(i\omega t)\right]&#xA;$$&#xA;where&#xA;$$&#xA; \xi = \frac{c}{c_c} ,~~ c_c = 2 m \omega_{\text{res}}, ~~ \omega_{\text{res}} = \sqrt{\frac{k}{m}}&#xA;$$&#xA;and $c$ is the damping factor, $k$ is the stiffness, $m$ is the mass, $\omega$ is the forcing frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Homogeneous solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The homogeneous solution is&#xA;$$&#xA;  u_h = A\,\exp(-\xi\omega_{\text{res}} t)\sin(\omega_d t + \phi_d)&#xA;$$&#xA;where $\phi_d$ is a phase and &#xA;$$&#xA;  \omega_d = \omega_{\text{res}}\sqrt{1 - \xi^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Notice that the exponential decay term occurs only in the homogeneous solution.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Particular solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular solution is&#xA;$$&#xA;  u_p = \frac{\hat{p}}{k} \frac{\text{Re}\left[\exp[i(\omega t - \phi)]\right]}{\sqrt{\left[1 - \beta_a)^2\right]^2 + \left[2\xi\beta_a\right]^2}}&#xA;$$&#xA;where &#xA;$$&#xA;  \beta_a = \frac{\omega}{ \omega_{\text{res}}}, ~~\tan\phi = \frac{2\xi\beta_a}{1 - \beta_a^2} \,.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your solution the homogeneous term is zero and therefore there is no decay term in your solution.  Otherwise your implementation looks correct.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="3488" LastEditDate="2016-02-23T20:53:48.720" LastActivityDate="2016-02-23T20:53:48.720" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7547" PostTypeId="2" ParentId="7531" CreationDate="2016-02-22T00:20:56.423" Score="2" Body="&lt;p&gt;As long as you are not sending data, any light input in one end should be detectable with a &lt;a href=&quot;http://www.dickblick.com/products/grafix-clear-lay-assorted-colors/&quot; rel=&quot;nofollow noreferrer&quot;&gt;color filter&lt;/a&gt; covered &lt;a href=&quot;https://www.sparkfun.com/products/9088&quot; rel=&quot;nofollow noreferrer&quot;&gt;light sensor&lt;/a&gt; at the other end. Example: green film for a green led/laser. Light at the sensor means the beam is not blocked, no light means it is blocked. No fancy modulation should be necessary unless you are trying to filter out similar background colors from the sensor. A simple on-off pulse at 1000Hz would do well for this purpose if background light noise is a problem. If lots of these are used in close proximity you may want to pick a different frequency for each. For a small beam a LED should be fine, but depending on water impurities, diameter of the stream, and length of the stream a laser may be required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is stream diameter and bend radius. There is probably some complex math involving a lot of variables we don't know, but basically as your stream(fiber) gets larger diameter or the bend radius gets smaller, more and more light escapes the stream(fiber). Here is a good &lt;a href=&quot;https://www.linkedin.com/pulse/bend-radius-really-concern-olivia-ou&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; on it. In the image below you can see how light escapes the &quot;macrobend&quot;. You will want to reduce this by decreasing the stream diameter or increasing your bend radius so more light makes it to your receiver.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cIWri.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cIWri.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://www.sparkfun.com/products/241&quot; rel=&quot;nofollow noreferrer&quot;&gt;940nm infrared receiver and transmitter&lt;/a&gt; may also be an option. This has the added benefit of not being visible, but will potentially have less range because water is more absorptive at that frequency. If needed they do make powerful IR laser emitters such as this &lt;a href=&quot;http://www.aliexpress.com/item/808nm-500mW-5-6mm-Infrared-IR-Laser-Diode-Specially-for-Producing-Green-Lasers/1917646581.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;808nm IR laser diode&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will want to choose light source that is in the visible or near infrared regions. Other frequencies are absorbed much more quickly. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;full wiki article&lt;/a&gt; and graph below.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aWcEd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aWcEd.jpg&quot; alt=&quot;Water Electromagnetic Absorption Graph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-02-22T00:20:56.423" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7548" PostTypeId="2" ParentId="2908" CreationDate="2016-02-22T00:56:25.183" Score="1" Body="&lt;p&gt;If cost isn't an extreme object, then hou might have lowest contamination by using a Co+WC makeup for your balls, and use a cryoplastic cup to avoic abraiding between balls &amp;amp; cup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The WC is extremely thermally stable &amp;amp; has an incredibly high surface hardness (not to mention density). As long as your Co binder could hold up to the stresses (or you could find another suitable binder), you should enjoy much lower contamination, along with accelerated milling times/efficiency due to the increased density &amp;amp; hardness of your milling balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WC is readily available in powder form, ready to be formed &amp;amp; sintered using any machinery capable of heating &amp;amp; injecting molten Co (or Ni) to 'wet' it. Ofc, if the Co proves untenable due to contamination or processing costs, you could always use a low-outgas, cryo-rated epoxy to wet/bind the WC powder (might work better anywise, as I think about it more).&lt;/p&gt;&#xA;" OwnerUserId="5123" LastActivityDate="2016-02-22T00:56:25.183" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7549" PostTypeId="2" ParentId="7296" CreationDate="2016-02-22T01:38:30.853" Score="0" Body="&lt;p&gt;I started these as comments but they're way too long for comments. Basically, please clarify what you want. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The orbit of the Planet (blue) and its Moon (yellow) has the following differences: Planet is eccentric.&quot; Why would the eccentricity of the planet's orbit affect its moon? You don't have a sun in that model, so really I don't understand what the planet's orbit has to do with any of this.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Only one face of the Planet [can] see its Moon.&quot; This isn't how our moon works. With our moon, only one side of it can see the planet. The way you have it described, it would be like if only the Western hemisphere could see the moon, and nobody in the Eastern hemisphere would have ever seen it. Also, the way you describe the motion (the bars are always parallel) is like both scenarios put together - the moon can only be seen from one hemisphere and that moon only has one face that can be observed. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Planet's rotation axis may shift&quot; Here's where you really lost me. Maybe it's because &lt;em&gt;your drawing isn't labeled&lt;/em&gt; (!), but you state you are, &quot;only hoping to model the motion on just one plane.&quot; As I mention above, there is no sun, so I'm assuming the spinning disc that is your planet it spinning on its axis. If it is shown as spinning on its axis, then you can't move the planet to adjust the tilt of that axis and keep it planar. It looks like you might mean to do this by shifting the point of rotation, but all that would do is cause the entire planet to rotate eccentrically.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please clarify your question by:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Labeling your drawing&lt;/li&gt;&#xA;&lt;li&gt;Add several drawings showing &quot;snapshots&quot; of the motion you would like to demonstrate. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also bear in mind that, to show rotation in a plane, the axis of rotation has to be orthogonal to the plane. As soon as you change the angle of the axis of rotation you &lt;em&gt;cannot&lt;/em&gt; have a planar display. You could have a &lt;em&gt;projection&lt;/em&gt;, but I'm not sure that you could depict a 2D projection of an off-axis sphere rotating with a &quot;simple&quot; setup. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-02-22T01:38:30.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7551" PostTypeId="1" CreationDate="2016-02-22T05:13:33.327" Score="4" ViewCount="170" Body="&lt;p&gt;I'm looking for a small ring that allows a thin cable to pass through but can also lock to close that opening, like on bicycle brake lever screws (shown below). However, I'm having trouble finding the part because I don't know what it's called.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wguRU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wguRU.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do they have an official/technical name?&lt;/p&gt;&#xA;" OwnerUserId="5146" LastEditorUserId="368" LastEditDate="2016-02-22T18:48:35.903" LastActivityDate="2016-02-22T18:48:35.903" Title="What is a slotted open ring with a thread called (like on bicycle brake levers)?" Tags="|mechanical-engineering|terminology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7552" PostTypeId="1" CreationDate="2016-02-22T07:46:17.083" Score="3" ViewCount="427" Body="&lt;p&gt;How to read the dimension labels of this IC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ibWka.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ibWka.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5147" LastEditorUserId="33" LastEditDate="2016-02-23T14:21:54.157" LastActivityDate="2016-02-23T14:21:54.157" Title="How do I read the component dimension labels given in a schematic?" Tags="|mechanical-engineering|electrical-engineering|data|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7553" PostTypeId="2" ParentId="7531" CreationDate="2016-02-22T08:08:25.443" Score="1" Body="&lt;p&gt;Fiber optics depend on the surface reflecting light by total internal reflection. Therefore the properties of the outer surface are crucial, and reflection depends on what your understanding of a &quot;stream&quot; is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the water is in direct contact with anything solid (or liquid) (most likely anything but air), it will not work. Total internal reflection depends on the outer material being transparent (for the wavelength considered) with the refractive index being smaller than that for water (1.333).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your surface is not smooth (having air bubbles or the like), you will most likely get a significant intensity drop of transmitted light. Detection of a broken stream in such cases will depend on the power of your light source, the sensitivity of your detector and the length of the stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the surface is smooth, it should work. However, my everyday life experience with water tells me that freestream liquid jets tend to become unstable and separate into drops. I presume there are a lot of publications out there on this phenomenon. This is something that would need further investigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a curly surface might cause troubles. I don't know if there are any investigations on this, but you should have a look into this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your comment on turbulence: Turbulence alone does not influence transmisison of light in water. However, it influences the outer water surface of a free stream. Nice pictures of this process are here: &lt;a href=&quot;http://dx.doi.org/10.1016/j.ijmultiphaseflow.2013.11.013&quot; rel=&quot;nofollow&quot;&gt;http://dx.doi.org/10.1016/j.ijmultiphaseflow.2013.11.013&lt;/a&gt; (open access article). They show that turbulent free streams are unstable and disperse into small drops.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2016-02-22T08:08:25.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7554" PostTypeId="2" ParentId="7552" CreationDate="2016-02-22T08:43:10.433" Score="8" Body="&lt;p&gt;The figures underneath the solid lines are in millimeters, the ones above the line are in inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The X.XXX - X.XXX is a range, to account for variations in manufacturing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example, the width shown on the plan view of the chip (top) is given as 0.220 - 0.310 inches. This means that the actual value could be anything in that range. Some chips might be 0.23 ins wide, some might be 0.25 ins wide, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, there are ± signs after some numbers. For example, the pin spacing on the bottom right elevation is given as 0.100 ± 0.010 ins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the actual value could be anything in the range between 0.1 - 0.01 ins, and 0.1 + 0.01 ins, i.e. 0.099 - 0.101 ins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="2551" LastActivityDate="2016-02-22T08:43:10.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7555" PostTypeId="1" AcceptedAnswerId="7556" CreationDate="2016-02-22T08:52:02.870" Score="0" ViewCount="100" Body="&lt;p&gt;Recently I have been researching torque and horsepower.  I know that the lower the car gear, the more the torque.  However, whenever I see specifications of a car, it says : &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;X lb-ft of torque at Y rpm.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, shouldn't they specify the gear that was measured in? Or is this the torque the engine makes, measured before the gearbox? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a little confused, I would be happy if you could help me with some explanations.&lt;/p&gt;&#xA;" OwnerUserId="5149" LastEditorUserId="272" LastEditDate="2016-02-22T15:08:17.543" LastActivityDate="2016-02-22T15:08:17.543" Title="Torque in different car gears?" Tags="|torque|engines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7556" PostTypeId="2" ParentId="7555" CreationDate="2016-02-22T09:03:24.887" Score="2" Body="&lt;p&gt;The torque indicated by the car manufacturer is usually measured at the engine's output, without use of the gearbox.&#xA;However it is still a good indicator of the car's global performance, but only if you pay attention to which rpm provides the higher torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example : 100 Nm @3000 rpm is a better performance than 100 Nm @7000 rpm since the first one is available in &quot;normal&quot; driving while the second is available pushing the engine very high in rpm which only occurs in automobile sports or &quot;unlawful&quot; driving.&lt;/p&gt;&#xA;" OwnerUserId="4946" LastActivityDate="2016-02-22T09:03:24.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7558" PostTypeId="1" AcceptedAnswerId="7563" CreationDate="2016-02-22T11:57:04.473" Score="3" ViewCount="194" Body="&lt;p&gt;I'm trying to find the canonical method for transferring heat from a 70–80 °C source to heat up some air in a box to 45–50 °C. My understanding of heat pipes is that the liquid inside has to evaporate in order for heat transfer to occur. Many heat pipes use water, but water doesn't boil at 50 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the canonical/best way for transferring heat at 70 °C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a simple sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Phuab.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Phuabt.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the black rod is the proposed heat pipe insulated with some rubber (green).&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="368" LastEditDate="2016-02-22T18:45:15.730" LastActivityDate="2016-02-22T18:45:47.983" Title="Transferring heat using a pipe at temperatures below the boiling point of water" Tags="|thermodynamics|heat-transfer|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7559" PostTypeId="2" ParentId="7558" CreationDate="2016-02-22T12:33:39.470" Score="1" Body="&lt;p&gt;Like you suggest, a heat pipe is a good solution, and most of them do not use water but application-specific heat-transfer fluids (&lt;a href=&quot;https://www.thermalfluidscentral.org/encyclopedia/index.php/Working_Fluids_and_Temperature_Ranges_of_Heat_Pipes&quot; rel=&quot;nofollow&quot;&gt;some examples&lt;/a&gt;) that can evaporate at temperatures below 100 °C. Those are used in processor cooling in computers for example.&lt;/p&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="368" LastEditDate="2016-02-22T18:45:47.983" LastActivityDate="2016-02-22T18:45:47.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7560" PostTypeId="1" AcceptedAnswerId="7625" CreationDate="2016-02-22T13:20:31.930" Score="2" ViewCount="9599" Body="&lt;p&gt;I am trying to solve the problem below without using moments. To find the resultant force is simply a  matter of adding the force components together, but I cannot find any method of finding a location of the resultant without using moments. Is there any other way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z8PKg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z8PKg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems is from Hartog's &lt;em&gt;Mechanics&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This page explains how to solve it using moments: &lt;a href=&quot;http://www.leancrew.com/mechanics/problem006.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.leancrew.com/mechanics/problem006.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1721" LastEditorUserId="33" LastEditDate="2016-02-26T01:05:12.053" LastActivityDate="2016-02-26T14:31:30.173" Title="Find location of resultant of forces without using moments" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7561" PostTypeId="2" ParentId="7558" CreationDate="2016-02-22T13:25:05.130" Score="2" Body="&lt;p&gt;Yes, many heat pipes use water as a working fluid but the pipe is evacuated of air before the water is introduced so they will work fine at 50 °C.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was first presented with a heat pipe, I went into the bathroom and put it under the hot water to feel it conduct the heat.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastEditorUserId="368" LastEditDate="2016-02-22T18:45:41.530" LastActivityDate="2016-02-22T18:45:41.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7562" PostTypeId="2" ParentId="7551" CreationDate="2016-02-22T13:50:21.313" Score="7" Body="&lt;p&gt;The term appears to be &quot;&lt;a href=&quot;https://www.google.se/search?q=adjusting%20barrel&amp;amp;tbm=isch&quot;&gt;Adjusting Barrel&lt;/a&gt;&quot; or &quot;Barrel Adjuster&quot;. &lt;/p&gt;&#xA;" OwnerUserId="4716" LastActivityDate="2016-02-22T13:50:21.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7563" PostTypeId="2" ParentId="7558" CreationDate="2016-02-22T13:50:31.813" Score="4" Body="&lt;p&gt;The advantage of a heat pipe is that it is a passive way of getting a high rate of heat transfer between two points using a single, compact component. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a heat pipe is only one part of a heat transfer system and you also need heat exchangers of some sort at the source and sink to effectively  transfer heat from one medium  to another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common way to transfer heat to air is to use a radiator (they actually work by convection not radiation bu that's what they are called). Essentially you pump hot water through a network of pipes with a large surface area exposed to the air. Air has a low coefficient of thermal conductivity so the surface area has a big impact on its performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation you describe is broadly similar to a domestic central heating system or water cooling in an engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you use a heat pipe or pumped coolant is largely down to what is most convenient for your particular installation. Heat pipes tend to be more attractive for small compact systems whereas pumps will tend to give more performance in large scale and more demanding applications and also give more flexibility in the relative positions of the source and sink as you can make the pipe runs as long as you want as long as it is well insulated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pumps also have the advantage that you can use the flow-rate to control the rate of heat transfer. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-22T13:50:31.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7564" PostTypeId="1" AcceptedAnswerId="7565" CreationDate="2016-02-22T15:20:08.283" Score="0" ViewCount="98" Body="&lt;p&gt;I am thinking of playing around with food in partial vacuum conditions (ie water boiling at room temperature). But, as it is more a curiosity, I'd rather not spend too much money right now. That inspired me to be a little creative with stuff I already have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a pressure canner (heavy duty aluminum), an AC vacuum and sheets of 3/16&quot; acrylic. The acrylic would be nice to use as I can keep an eye on what is happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think the acrylic is thick enough to withstand the pressure, so I was thinking of ways to combine them somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a proposed solution. For the first sheet on the low pressure side, drill some holes in the sheet to reduce the pressure, but make the holes small enough that it is not able to come to an equilibrium while the pump is in operation. Then another sheet is placed on top of it with a gasket to keep them separate. This next sheet would have smaller/fewer holes in order to .. In theory, this could continue indefinitely, then an unperforated sheet at the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an illustration (note, I missed gaskets around the vacuum outlet):&lt;a href=&quot;https://i.stack.imgur.com/ZfzMS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZfzMS.png&quot; alt=&quot;vacuum semiperforated seal&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a=pressure canner &lt;/li&gt;&#xA;&lt;li&gt;b=sheets of acrylic &lt;/li&gt;&#xA;&lt;li&gt;c=gaskets &lt;/li&gt;&#xA;&lt;li&gt;d=vacuum outlet&lt;/li&gt;&#xA;&lt;li&gt;e=larger holes&lt;/li&gt;&#xA;&lt;li&gt;f=smaller holes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am not particularly sure I want to do this, but I am curious if the general concept is sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possibility is that it could work, but the holes would be smaller than I could reasonably do in my home shop. Acrylic isn't great for drilling.&lt;/p&gt;&#xA;" OwnerUserId="5153" LastEditorUserId="1832" LastEditDate="2016-02-22T16:30:52.007" LastActivityDate="2016-02-22T22:51:19.347" Title="Spreading vacuum load using stacked perforated sheets" Tags="|vacuum|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7565" PostTypeId="2" ParentId="7564" CreationDate="2016-02-22T22:51:19.347" Score="2" Body="&lt;p&gt;No, this idea won't work. Except for the start-up transient, there will be no flow of air through all of those holes, and the top (unperforated) plate will still be experiencing the full force of the pressure difference between outside and inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better to just stack up N sheets with no holes and no space between them.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-02-22T22:51:19.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7567" PostTypeId="1" AcceptedAnswerId="7575" CreationDate="2016-02-23T01:03:06.067" Score="9" ViewCount="920" Body="&lt;p&gt;The power generated by a wind turbine is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Power} = \frac{1}{2}C\rho AV^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\rho = \text{Air density}$&lt;/li&gt;&#xA;&lt;li&gt;$C = \text{Coefficient of performance}$&lt;/li&gt;&#xA;&lt;li&gt;$A = \text{Frontal area}$&lt;/li&gt;&#xA;&lt;li&gt;$V = \text{Velocity of the wind}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In other words, the power is proportional to the square of the length of the blades and the cube of the velocity of the wind. As we know, the velocities of winds are high at high altitudes. So instead of building many smaller wind turbines, why can't we just build a giant wind turbine that is, say, 1000 m tall? That may be an engineering challenge at first, but won't that be more economical in the end? After all, the Burj Khalifa in Dubai has been 828 m tall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't we build three instead of one pillars to support such a structure? Why can't we build one at the sea?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Wind_turbine#Records&quot; rel=&quot;noreferrer&quot;&gt;The Vestas V164 has a rated capacity of 8.0 MW, has an overall height&#xA;  of 220 m (722 ft), a diameter of 164 m (538 ft), is for offshore use,&#xA;  and is the world's largest-capacity wind turbine since its&#xA;  introduction in 2014.&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5159" LastEditorUserId="368" LastEditDate="2016-02-23T17:06:33.287" LastActivityDate="2021-10-21T22:23:58.003" Title="Why don't we just build a giant wind turbine?" Tags="|electrical-engineering|civil-engineering|structural-engineering|renewable-energy|wind-power|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7568" PostTypeId="1" AcceptedAnswerId="8171" CreationDate="2016-02-23T01:51:29.803" Score="12" ViewCount="5226" Body="&lt;p&gt;How can large bridges, with spans on the order of 1 km, be made resistant to earthquakes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm no expert on quakes, but there are at least two kinds of shaking: lateral and vertical. The vertical shaking in particular really worries me. I don't know how any kind of shock absorbtion can be built into a massive, tall structure like a building or bridge tower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I was originally going to ask about suspension bridges, but then I read that suspension bridges are not great for heavy trains. The background of this question lies in my exploring the idea of a Bering Bridge (from Alaska to Siberia, crossing the Bering Strait), which would be primarily a train bridge. And it would need to handle freight trains, the heaviest of all sometimes exceeding 100 tons per car.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I will just ask the question about large bridges in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the largest quake in Alaska was a 9.4 Richter Scale in 1964, hitting Fairbanks. (I don't know how to convert to Moment-Magnitude scale). Is it possible to build large bridges that won't collapse from that? Ideally, we want the bridge to not collapse during a quake even if it's under full load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I know it's not cost effective to build such a bridge. For one thing, far east Siberia has no rail network (or much civilization in general). There was a tunnel project proposed which I think they said was cheaper than a bridge, which I couldn't understand because digging through tough rock seems a lot harder than driving a pile in 50-m deep water. The tunnel project was suspended IIRC, no surprise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, I'm just exploring if such a bridge is technically possible and can be quake-resistant. If you need something closer to reality, I suppose we can look at the Golden Gate Bridge in San Francisco. I saw &lt;a href=&quot;http://kalw.org/post/what-could-collapse-golden-gate-bridge#stream/0&quot;&gt;here&lt;/a&gt; they were working to make it safe in quakes up to 8.3, but it didn't go into the details. And keep in mind the Golden Gate Bridge does not handle freight trains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, is it possible to quake-proof or seriously resist quakes in large bridges loaded with a freight train? The bridge would not necessarily have to remain completely undamaged. I just don't want it to snap and drop the train in the ocean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible?&lt;/p&gt;&#xA;" OwnerUserId="3510" LastEditorUserId="33" LastEditDate="2016-02-23T02:14:19.447" LastActivityDate="2016-05-21T13:34:02.087" Title="How can large bridges be made resistant to earthquakes?" Tags="|structural-engineering|bridges|seismic|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7569" PostTypeId="2" ParentId="1977" CreationDate="2016-02-23T02:59:00.143" Score="5" Body="&lt;p&gt;If one side of foundation has soil packed more loosely than the other side, the foundation may cause the ground to sink, and building may crumble. Even if it has a steel-reinforced concrete foundation, the building may tip sideways, causing uneven floors, or wood to shift and split, even if concrete remains in same shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my area we have a lot of caves, some hidden, that collapse under weight and cause sinkholes. Large buildings (Like a Tesco) have cracks in the walls EVEN AFTER the soil was repacked, and is still re-settling and causing NEW cracks 17 years later. One of the oldest houses in town has the foundation 17 feet deep, so that it hits bedrock. (5.1816 meters)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repacking the soil with power-hammers, or using a DIFFERENT filler material that is less prone to settle over time, is also a common practice. As is digging deeper just to pour a thicker foundation (or a wider one with a smaller footprint on the surface, like a submerged pyramid).&lt;/p&gt;&#xA;" OwnerUserId="5134" LastActivityDate="2016-02-23T02:59:00.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7570" PostTypeId="5" CreationDate="2016-02-23T03:31:33.550" Score="0" Body="&lt;p&gt;This tag is automatically assigned to migrated questions. This tag should never be applied by a person.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2016-02-23T03:31:33.550" LastActivityDate="2016-02-23T03:31:33.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7571" PostTypeId="4" CreationDate="2016-02-23T03:31:33.550" Score="0" Body="Do not use this tag." OwnerUserId="33" LastEditorUserId="33" LastEditDate="2016-02-23T03:31:33.550" LastActivityDate="2016-02-23T03:31:33.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7573" PostTypeId="2" ParentId="7567" CreationDate="2016-02-23T08:46:26.160" Score="5" Body="&lt;p&gt;The cost of a wind turbine is not proportional to its size - like everything that needs to be manufactured, the cost exponentially increases as manufacturing becomes more complex, beyond what has been done. In terms of real estate on the ground, building a giant wind turbine may be more efficient (maybe not in fact, because stability issues may require the base to be much wider to the point two turbines could have fit), but it is by no means economical. Why build the highest skycraper ever built (if not for the fame) if you have the space for many houses? Not to mention the scary hazard of having a multiton generator plus hundreds of metres of rotating death threatening to kill anyone within a 1.5km radius if things go wrong. I also suspect the shaft will have too high an inertia to spin at a sensible rate when the wind is not constant: the voltage generated is proportional to the shaft speed.&#xA;Finally, I nearly forgot: will you be the very lucky man climbing the 1km ladder to maintain the generator? Okay this is more of a joke rather than a real issue, since the wind turbine would be so big a lift could be built inside it, but that's one more thing that you do not need in reasonably sized wind turbines. &lt;/p&gt;&#xA;" OwnerUserId="1326" LastEditorUserId="1326" LastEditDate="2016-02-23T10:58:06.407" LastActivityDate="2016-02-23T10:58:06.407" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7574" PostTypeId="1" CreationDate="2016-02-23T11:50:02.027" Score="4" ViewCount="486" Body="&lt;p&gt;I am taking a phase sensitive measurement of an AC magnetic field using an SR830 Lock-In Amplifier and a magnetic sensor. I know I have to average over a number of measurement cycles to get an accurate reading. The amount of time I should average is determined by the measurement frequency, the time constant on the lock-in amplifier, the filter slope, the noise measured by the sensor and the required accuracy. See the following diagram I have pulled out of the SR830 manual (T = time constant, ENBW (Equiv. noise bandwidth). Wait time = min. time that should be allowed before averaging starts):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CnSvc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CnSvc.png&quot; alt=&quot;T = time constant, ENBW (Equiv. noise bandwidth). Wait time = min. time that should be allowed before averaging starts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have measured the spectral noise density of my sensor using the lock-in too:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o2enj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o2enj.png&quot; alt=&quot;noise density of sensor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now say I want to take a measurement that would have an expected output of 1μV at 10~Hz with an accuracy of 1%. How many cycles/how long would I have to average? Could you help to explain how you figured this out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will expand if the question is unclear.&lt;/p&gt;&#xA;" OwnerUserId="5161" LastEditorUserId="5161" LastEditDate="2016-02-23T14:30:00.983" LastActivityDate="2016-02-24T13:51:50.453" Title="How can I determine how many cycles to average for when taking a measurement using Lock-in Amp if I know sensor spectral noise density?" Tags="|mechanical-engineering|electrical-engineering|sensors|measurements|experimental-physics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7575" PostTypeId="2" ParentId="7567" CreationDate="2016-02-23T12:59:39.513" Score="15" Body="&lt;p&gt;Not everything scales linearly.  In particular, the cross-sectional area of supports required scales faster than height of a structure, all else held constant.  This explains why ants have tiny thin legs compared to elephants.  An ant linearly scaled up to elephant size would not be able to stand, or would snap its legs trying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same thing happens to wind turbines.  You get some advantages to making them bigger, as you mention, but you are also ignoring the disadvantages.  Not only must the structural support for a large turbine be disproportionately larger than a smaller one, there is also more wind loading, and that loading is higher up.  That puts disproportionately more torque on the mounting that has to be countered somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are manufacturing and maintenance issues.  Building 500 m blades will be difficult, especially considering they would need to be assembled in the field where it will be more costly and more difficult to do well than in a controlled environment in a factory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wind turbines have gotten very large in recent years, for the reasons you mention.  Material and manufacturing advances may allows them to get even larger and still make economic sense, but due to the non-linear nature of how various things scale, there will always be some finite sweet spot.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2016-02-23T17:07:31.070" LastActivityDate="2016-02-23T17:07:31.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7576" PostTypeId="1" CreationDate="2016-02-23T13:52:32.600" Score="1" ViewCount="244" Body="&lt;p&gt;I have a fluid flowing through an annular finned tube. The heat transfer (cooling of the fluid in the pipe) is done by natural convection. I am curious about the relationship with how the mass flow rate of the fluid in the pipe affect the amount of heat transfer which can be achieved.&lt;/p&gt;&#xA;" OwnerUserId="5163" LastActivityDate="2016-03-26T18:15:50.750" Title="How does the mass flow rate through an annular fin tube affect the heat transfer Q?" Tags="|fluid-mechanics|heat-transfer|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7577" PostTypeId="2" ParentId="2245" CreationDate="2016-02-23T14:29:09.577" Score="5" Body="&lt;p&gt;The Inertial response for a generator is characterised by its Inertia Constant, H, with units of seconds, defined as (&lt;a href=&quot;http://orca.cf.ac.uk/35740/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Samarakoon&lt;/a&gt;, p40):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;the ratio of kinetic energy stored at synchronous speed &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; to the&#xA;generator kVA or MVA rating, &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ H = \frac{0.5J\omega^2}{S}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;An equivalent Inertia Constant for an entire system can be estimated: (&lt;a href=&quot;http://wie.sagepub.com/content/32/6/573.full.pdf+html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ekanayake, Jenkins, Strbac&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;H_{equivalent} = \sum_{gens} H_{gen}/S_{gen}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;A value for the GB system (in 2008) was estimated at 9s (by &lt;a href=&quot;http://orca.cf.ac.uk/35740/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Samarakoon&lt;/a&gt;), projected to drop as far as 3s in 2020 with a high wind penetration.&lt;/p&gt;&#xA;&lt;p&gt;When modelling Inertial response (more commonly referred to as frequency response), a power system can be simplified to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Transfer_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;transfer function&lt;/a&gt; (&lt;a href=&quot;http://wie.sagepub.com/content/32/6/573.full.pdf+html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ekanayake, Jenkins, Strbac&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{1}{2H_{equivalent}s +D}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is known as the Damping Coefficient - the term encapsulates response from frequency responsive demand (&lt;a href=&quot;http://ieeexplore.ieee.org/xpl/abstractAuthors.jsp?arnumber=6397579&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mu,Wu,Ekanayake,Jenkins,Jia&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;An available proxy for the Inertia Constant is the Primary&lt;sup&gt;1&lt;/sup&gt; Frequency Control characteristic required by each system operator (MW/Hz). These are compared for 8 different systems by &lt;a href=&quot;http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=4077135&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rebours et al&lt;/a&gt;; ranging from 20570MW/Hz for UCTE (Union for the Co-ordination of Transmission of Electricity - European synchronous system) to approximately 600MW/Hz for Belgium.&lt;/p&gt;&#xA;&lt;p&gt;As lower inertia generators (e.g. wind) displace higher inertia generators (i.e. steam), the inertia constant tends to fall. This means that, to maintain overall stability, generators must react more quickly to sudden changes generation or changes in demand. This is often cited as a limiting factor in the connection of wind, especially to smaller &amp;quot;island&amp;quot; networks (e.g. &lt;a href=&quot;http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=1525120&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lalor, Mullane, O'Malley&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;&lt;sup&gt;1 - &lt;em&gt;Note: primary/secondary/tertiary response/reserve are defined in different ways on different power systems, as noted by &lt;a href=&quot;http://ieeexplore.ieee.org/xpl/abstractAuthors.jsp?arnumber=6397579&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rebours&lt;/a&gt;.&lt;/em&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="4476" LastEditorUserId="4476" LastEditDate="2021-01-08T17:58:58.323" LastActivityDate="2021-01-08T17:58:58.323" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="7578" PostTypeId="2" ParentId="7304" CreationDate="2016-02-23T20:54:15.800" Score="1" Body="&lt;p&gt;In that sort of sheet metal assembly the shafts will often be a press fit to their housing  with self bearing nylon gears ie the gears rotate on the shaft rather than being fixed to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will usually provide an adequate seal in itself as they will either be self lubricating (nylon or PTFE) or use a fairly viscous grease or a dry lubricant  which needs much less sealing than eg pressurised oil. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-23T20:54:15.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7579" PostTypeId="1" CreationDate="2016-02-24T00:02:10.710" Score="1" ViewCount="764" Body="&lt;p&gt;I would like to ask a specific conceptual question which bothers me for quite some time! First of all i do know the difference in between reversible and irreversible processes. What is thought in Thermodynamics courses is the maximum work can be obtained via reversible processes. Now please consider the following example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air (assume as ideal gas) at 5 bar and 298.15K (25℃) is expanded to 1 bar and 298.15K by a &lt;strong&gt;&lt;em&gt;mechanically reversible processes&lt;/em&gt;&lt;/strong&gt;: Heating at constant pressure followed by cooling at constant volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one considers the corresponding PV diagram , the work is calculated as the area under the curve which is obviously larger than the reversible isothermal expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my questions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How many different reversible paths can be drawn in between two different states at the same temperature (there can be infinite number of irreversible paths)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How can heating an ideal gas at constant pressure and cooling at constant volume be a reversible process (these are not adiabatic or isothermal)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Is it possible to say that : There can be many different reversible paths between two specified states, all of which will give larger work than corresponding irreversible paths but also vary in between themselves so that it is not possible to state which reversible path will give the highest work before specifying the path itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all in advance for your sincere help and answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: From MIT thermodynamics course notes: the reversible one produces the maximum work of all possible processes between two states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so there should not be more than one reversible process between two states. Then how can the process path given in the question be reversible as we can already perform the same change with isothermal expansion.&lt;/p&gt;&#xA;" OwnerUserId="5169" LastEditorUserId="5169" LastEditDate="2016-02-24T09:26:45.577" LastActivityDate="2016-04-03T14:51:40.130" Title="Reversible &amp; Irreversible Processes" Tags="|thermodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7585" PostTypeId="2" ParentId="2279" CreationDate="2016-02-24T04:51:07.143" Score="0" Body="&lt;p&gt;You can use this formula to calculate it based on the x, horizontal distance and z vertical distance.&#xA;Simply put z=0 and increment it say by 0.1 in excel while x starts from B/2.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MF5tQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MF5tQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formula&#xA;&lt;a href=&quot;https://i.stack.imgur.com/srx4T.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/srx4T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sample calculations&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oJq2J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oJq2J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OZfDB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OZfDB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5092" LastActivityDate="2016-02-24T04:51:07.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7586" PostTypeId="1" CreationDate="2016-02-24T12:16:37.843" Score="11" ViewCount="17715" Body="&lt;p&gt;Why do we use multiple reinforcing bars instead of one with a big diameter in reinforced concrete?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does increasing the number of reinforcing bars means that the reinforced concrete will have a higher tensile strength?&lt;/p&gt;&#xA;" OwnerUserId="5179" LastEditorUserId="33" LastEditDate="2016-02-24T13:49:02.833" LastActivityDate="2016-12-14T19:33:17.477" Title="Why do we use multiple reinforcing bars instead of one bar with a big diameter in reinforced concrete?" Tags="|civil-engineering|steel|reinforced-concrete|" AnswerCount="7" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7587" PostTypeId="2" ParentId="7586" CreationDate="2016-02-24T12:50:00.713" Score="10" Body="&lt;p&gt;The main purpose of rebar is to improve the tensile strength of concrete and in practice most of these loads come from bending rather than pure tension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a beam in subject to bending forces the greatest stress are at the edges and faces of the beam  so just having one big bar running down the centre wouldn't do very much as this part of the structure sees very little load until it starts to fail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having smaller diameter rebar distributed throughout the structure also distributes the load from the concrete to the steel more effectively as there is a greater area of contact for adhesion between the two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice the size and placement of the steel reinforcement will be determined by the expected loading on the structure and is a compromise between strength, weight, cost and the practicalities of assembling the steelwork during construction. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-24T12:50:00.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7588" PostTypeId="2" ParentId="7574" CreationDate="2016-02-24T13:46:36.547" Score="0" Body="&lt;p&gt;In the absence of an analytic answer (which I'm sure exists somewhere),  you can try running some post-test analyisis.  Take a very long sample, then calculate means and standard deviations when chopping the sample into a collection of subsamples.  By changing the subsample length, you can see at what sample length the variance of the calculated mean and the variance of the calculated standard deviation  are small enough for your desired results.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., collect 1e6 samples.   Run the statistical analysis on 1e4 subsamples of length 100,  on 1e3 samples of length 1000, etc.  See how the data work out.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-02-24T13:46:36.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7589" PostTypeId="2" ParentId="7574" CreationDate="2016-02-24T13:51:50.453" Score="1" Body="&lt;p&gt;I'll start by telling you the complete way to calculate the noise in the measurement, and then I'll tell you how to make a quicker estimate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lock-in Amplifiers:&lt;/strong&gt; First of all, it is helpful to explain how a lock-in amplifier works.  It works by taking the input signal and demodulating it against the internal (or externally supplied) sine wave reference.  If your input signal is $V_i = A_i\cos(\omega_i t+\phi)$ and the internal reference is $V_r=\cos(\omega_r t)$, then the demodulated signal is given by &#xA;$$&#xA;\begin{align}&#xA;V_d &amp;amp;= \cos(\omega_r t)\ A_i\cos(\omega_i t + \phi) \\&#xA;&amp;amp;= \frac{A_i}{2}\left[\cos((\omega_i-\omega_r)t - \phi) + \cos((\omega_i + \omega_r)t + \phi) \right]&#xA;\end{align}&#xA;$$&#xA;Usually $\omega_i\approx\omega_r$ because the lock-in's frequency reference is also used to drive the system being measured.  In this case the first term is near DC while the second term has twice the frequency of the driving signal.  After the demodulation, the lock-in amplifier low-pass filters the signal to get rid of the second term leaving only &#xA;$$&#xA;V_d=\frac{A_i}{2}\cos((\omega_i-\omega_r)t - \phi).&#xA;$$&#xA;The bandwidth of this low pass filter is what sets the frequencies you are able to see as well as the noise in the measurement.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Full Noise Calculation:&lt;/strong&gt; The RMS noise is usually calculated from the PSD (PSD = ASD$^2$) with an extension of &lt;a href=&quot;https://en.wikipedia.org/wiki/Parseval%27s_theorem&quot; rel=&quot;nofollow&quot;&gt;Parsaval's theorem&lt;/a&gt; which says that the RMS variance of a time domain signal is given by&#xA;$$&#xA;\sigma^2=\int_0^\infty PSD(f)df&#xA;$$&#xA;In the case of a lock-in amplifier you have to take the low-pass filter into account by multiplying by the &lt;em&gt;shifted, double sided&lt;/em&gt; filter. Once you have the variance, the standard deviation is just the square root of the variance, and the noise in your measurement after averaging over many cycles is given by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_error&quot; rel=&quot;nofollow&quot;&gt;standard error of the mean&lt;/a&gt;&#xA;$$&#xA;SE = \frac{\sigma}{\sqrt{N}}&#xA;$$&#xA;where $N$ is the number of cycles that you average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Quick Estimate:&lt;/strong&gt; Now for what you are probably most interested in.  You can make a quick estimate of the noise at a given frequency by simply taking the value of your measured noise at that frequency and multiplying by the square root of twice the bandwidth:&#xA;$$&#xA;\sigma(f)\simeq ASD(F)\times\sqrt{2\ BW}.&#xA;$$&#xA;The noise is you measurement after averaging over many cycles is then given by &#xA;$$&#xA;SE(f)=\frac{\sigma(f)}{\sqrt{N}}.&#xA;$$&#xA;So, for your example of 10 Hz, assuming that the bandwidth of your lock-in filter is set to 1 Hz, the noise would be approximately &#xA;$$&#xA;\sigma(10\ Hz) = 600 \frac{nV}{\sqrt{Hz}}\ \sqrt{2\ Hz}=850\ nV.&#xA;$$&#xA;If you average for 100 cycles, then the noise comes down to&#xA;$$&#xA;SE(10\ Hz) = \frac{850\ nV}{\sqrt{100}}=85 nV.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2016-02-24T13:51:50.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7590" PostTypeId="2" ParentId="7586" CreationDate="2016-02-24T14:26:04.167" Score="2" Body="&lt;p&gt;From a purely conceptual standpoint, one large bar of the same area as multiple smaller bars provides the same moment capacity for a concrete beam.  This is assuming that the center of the bars are all at the same depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distribution of the bars (multiple smaller bars) help to limit cracks by spreading the tension force through a greater width of concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple smaller bars also help when looking at the interaction at the interface of the concrete and reinforcing steel. A single large bar has less surface area than multiple smaller bars. This means that for a given load, the stress between the surface of the bar and the concrete is greater in the single bar case. This has applications to the development length of the reinforcing.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2016-02-24T14:26:04.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7591" PostTypeId="1" CreationDate="2016-02-24T14:53:55.913" Score="10" ViewCount="2734" Body="&lt;p&gt;A &lt;strong&gt;MIMO&lt;/strong&gt; system with 2 input and 2 output &lt;strong&gt;decoupling&lt;/strong&gt; method to a &lt;strong&gt;SISO&lt;/strong&gt; system is described in many articles and books. How about &lt;strong&gt;m*n size transfer functions&lt;/strong&gt; systems? How can we generalize the method for example to 3*3 or 3*7 MIMO systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a 2*2 MIMO system description:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XnYRY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XnYRY.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;with $\mathrm{D_{11}(s)=D_{22}(s)=1}$ to the form&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\mathrm{D(s)}=\begin{bmatrix} D_{11}(s) &amp;amp; D_{12}(s) \\ D_{21}(s) &amp;amp;&#xA;&amp;gt; D_{22}(s) \\ \end{bmatrix}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Here we specify a decoupled response and the decoupler with the&#xA;  structure in Equation&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$G_p(s)D(s)=\begin{bmatrix} G_{11}(s) &amp;amp; 0 \\0 &amp;amp; G_{22}(s)&#xA;&amp;gt; \end{bmatrix} \\ \begin{bmatrix} G_{11}(s) &amp;amp; G_{12}(s) \\ G_{21}(s) &amp;amp;&#xA;&amp;gt; G_{22}(s) \end{bmatrix} \begin{bmatrix} 1 &amp;amp; D_{12}(s) \\ D_{21}(s) &amp;amp; 1&#xA;&amp;gt; \end{bmatrix}&#xA;&amp;gt; = \begin{bmatrix} G^*_{11}(s) &amp;amp; 0 \\ 0 &amp;amp; G^*_{22}(s) \end{bmatrix}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;And we can solve four equations in four unknowns to find&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$D_{12}(s) = -\frac{G_{12}(s)}{G_{11}(s)} \\ D_{21}(s) =&#xA;&amp;gt; -\frac{G_{21}(s)}{G_{22}(s)} \\ G_{l1}(s) = G_{11}(s) = -\frac{G_{12}(s)G_{21}(s)}{G_{22}(s)} \\ G_{l2}(s) = G_{22}(s) = -\frac{G_{21}(s)G_{12}(s)}{G_{11}(s)} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4368" LastEditorUserId="368" LastEditDate="2017-04-13T19:21:46.850" LastActivityDate="2017-04-13T19:21:46.850" Title="MIMO (multi input - multi output) system decoupling method" Tags="|control-engineering|aerospace-engineering|systems-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7592" PostTypeId="2" ParentId="7586" CreationDate="2016-02-24T15:26:58.023" Score="17" Body="&lt;p&gt;There are a few reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm firstly going to assume you're talking about replacing a bunch of small rebars by a single reasonably-sized one: i.e. instead of $15\phi8$ (7.54 cm&lt;sup&gt;2&lt;/sup&gt;), using $1\phi32$ (8.04 cm&lt;sup&gt;2&lt;/sup&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason is to improve ease of constructibility. Reinforced concrete beams also have transversal reinforcement, and it's very common to place rebars at the four corners of the transversal reinforcement to tie everything together into a self-supporting cage (this also inhibits crushing of concrete at the corners of the transversal reinforcement). But that just means you should place a lower boundary of 2 rebar in each face (upper and lower). But why not adopt $4\phi16$ (8.04 cm&lt;sup&gt;2&lt;/sup&gt;) instead of $15\phi8$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple small rebar improve the behavior of the concrete-steel interface. The total surface area to transmit the tensile forces from the steel to the concrete at the anchorages is greater, reducing the tensile stress and therefore shortening the anchorage and lap splice lengths. This also reduces the cracks that will develop in the concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, one can usually also be more efficient with smaller rebar: if what you actually need is 7.5 cm&lt;sup&gt;2&lt;/sup&gt;, you'll need either $15\phi8$ (7.54 cm&lt;sup&gt;2&lt;/sup&gt;), $10\phi10$ (7.85 cm&lt;sup&gt;2&lt;/sup&gt;), $7\phi12.5$ (8.59 cm&lt;sup&gt;2&lt;/sup&gt;), $4\phi16$ (8.04 cm&lt;sup&gt;2&lt;/sup&gt;), $3\phi20$ (9.42 cm&lt;sup&gt;2&lt;/sup&gt;), $2\phi25$ (9.82 cm&lt;sup&gt;2&lt;/sup&gt;), or $1\phi32$ (8.04 cm&lt;sup&gt;2&lt;/sup&gt;). $\phi8$ gives the most efficient use of steel, putting as little extra as possible. Obviously, however, this isn't always the case. It just often is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm going to assume you're talking about using one ridiculously large steel bar: i.e. instead of $20\phi20$ (68.2 cm&lt;sup&gt;2&lt;/sup&gt;), using $1\phi94$. Some of the points raised here are still valid even if you're talking about being reasonable, but their significance is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason is cost. This concept implies in making custom-diameter rebar for each beam (otherwise, all you're saying is that current rebar is too small), which would impede use of scale to reduce costs. A factory can reduce the unit-cost of a $\phi10$ bar by making millions of them. If each beam in each job site will require a custom diameter, such cost-savings aren't possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a single $\phi94$ rebar would weigh almost 54 kg/m, meaning you'd need to use a crane to put it into position, while $20\phi20$ can be easily placed by hand by a single worker, one at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well, at the extremities of your beam the steel will probably need to be bent. The larger the diameter of the rebar, the larger its bending radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then there's this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6PdUz.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6PdUz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the center of gravity of your steel will be farther from the face of the concrete, it will also be less efficient: you'll need more steel to achieve the same strength. This also applies for the first part of this answer, where we were talking about reasonable sizes, but then the difference is obviously much smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why not use a sheet of steel then, instead of one large circular rebar? Hell, since it would no longer have the spaces between the rebar, you could choose a sheet which is thinner than the multiple rebar's diameters, and which is therefore more efficient as well! But then your surface area would be reduced, increasing the transfer stresses and therefore the anchorage and lap splice lengths. Also, a sheet is a two-dimensional element, so there may be other transversal behaviors which may lead to problems. Also, how would you pour the concrete for the bottom face of the beam?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;At the end of the day, the best rule of thumb (or, well, the one I try to use) is to do your best to fit everything into as few layers of rebar as possible, but filling those layers as much as possible (while still leaving comfortable space for proper concrete pours, including the necessary space for the vibrator). The only exception is the last layer (farthest from the relevant face of the beam), which can be left emptier (but preferably with a number of bars that's allows it to follow a symmetric pattern of the previous layers). Well, balance that with being efficient with the steel: if this requires using large rebar and results in an adopted area of steel much larger than if another layer were added with smaller rebar, then maybe another layer is best.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-02-24T15:33:16.177" LastActivityDate="2016-02-24T15:33:16.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7593" PostTypeId="1" AcceptedAnswerId="7594" CreationDate="2016-02-24T16:50:51.693" Score="1" ViewCount="1263" Body="&lt;p&gt;I am modelling a railway track in ANSYS where the different soil layers have different material properties (including different damping ratios).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I know what frequency of vibrations to expect in the track, I can model the damping ratio using beta-damping in the soil elements only. I do not want to use alpha-damping, since it is defined globally in ANSYS and therefore would apply to all other materials for which I do not want to include damping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no problems in the results, but I need to justify why it is okay not to include the alpha-coefficient. I have found a &lt;a href=&quot;http://designspace.com/staticassets/ANSYS/staticassets/resourcelibrary/confpaper/2002-Int-ANSYS-Conf-197.PDF&quot; rel=&quot;nofollow&quot;&gt;report online&lt;/a&gt; that says &quot;In many practical structural problems, the $\alpha$ damping (or mass damping) which represents friction damping may be ignored ($\alpha=0$).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I need either a published paper or relevant textbook, or a detailed justification why beta-damping is a good approach, and I have been unable to find it so far.&lt;/p&gt;&#xA;" OwnerUserId="5183" LastEditorUserId="1832" LastEditDate="2016-02-24T18:57:12.473" LastActivityDate="2016-02-24T19:16:16.210" Title="Rayleigh damping in Finite Element Models using beta-coefficient only" Tags="|ansys|finite-element-method|" AnswerCount="1" CommentCount="2" ClosedDate="2016-02-25T14:48:51.540" ContentLicense="CC BY-SA 3.0" />
  <row Id="7594" PostTypeId="2" ParentId="7593" CreationDate="2016-02-24T19:16:16.210" Score="3" Body="&lt;p&gt;The report in your link explains it briefly in the sentence after your quote. The beta term models &quot;structural&quot; or &quot;hysteretic&quot; damping, which is described earlier in the report.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key fact about hysteretic damping is that, for a fixed amplitude of motion, it dissipates the same amount of &lt;em&gt;energy&lt;/em&gt; in each cycle of the motion, &lt;em&gt;independent of the frequency&lt;/em&gt;. That is often a good model for &quot;internal energy dissipation&quot; within a structure. I'm not a civil engineer but I could imagine it is a plausible model for the ballast or earth under a rail track squidging about as it dissipates energy. It is certainly a good model for the (small) amount of energy dissipated by the metal rails themselves as they vibrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alpha term doesn't have an obvious physical interpretation, but it can be mathematically useful. It has the same mathematical property as beta, namely that the mode shapes of the damped system are identical to the undamped. (In general that is not the case, and the damped modes can have different phases in different parts of the structure, i.e. they can look more like &quot;travelling waves&quot; than the undamped vibration modes). Using a combination of alpha and beta, you can get a fairly constant level of equivalent modal damping factors within a narrow band of frequencies, while the damping factors for modes with both higher and lower frequencies are more heavily damped. (That can be nice if you are modelling a system which is not fixed to anything, because the alpha term damps out any rigid body motion which you are probably not interested in, but the beta term does not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I don't have a reference for this (I've known it for too many decades already!) but a search for &quot;hysteretic damping&quot; should turn up something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind that modelling damping is rarely an &quot;exact science&quot;. The best source of advice would be &quot;best practice&quot; in your particular application - either specialist textbooks, or relevant papers in journals. You are probably not the first person who wants to model this!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-02-24T19:16:16.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7596" PostTypeId="1" CreationDate="2016-02-24T23:10:19.360" Score="6" ViewCount="15451" Body="&lt;p&gt;I need to figure out what type of steering does &lt;a href=&quot;http://www.himoto.cz/rc-modely-aut/himoto-buggy-zmoto-z-3-4x4-rtr-1-10--2-4ghz--4/&quot; rel=&quot;noreferrer&quot;&gt;this 1:10 Himoto Buggy Zmoto Z-3&lt;/a&gt; RC car have in order to be able to calculate the steering angle from servo position for a robotics project. Is it the same mechanism as &lt;a href=&quot;https://en.wikipedia.org/wiki/Steering#Rack_and_pinion.2C_recirculating_ball.2C_worm_and_sector&quot; rel=&quot;noreferrer&quot;&gt;Rack-And-Pinion steering&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess the steering angle change will not be constant (steering won't be linear) from servo since the rotational movement of servo is translated in the steering mechanism? I mean, whether one degree from 0 to +1 on servo will translate to the same steering angle change as for example step from +45 to +46 on servo. If not, what is the non-linear relationship there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are pictures of the steering mechanism (more at the link above):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/e0JdI.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e0JdI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/AEF57.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AEF57.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/FRlaj.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FRlaj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="414" LastActivityDate="2017-11-29T15:28:41.417" Title="RC car steering mechanism type" Tags="|mechanical-engineering|automotive-engineering|wheels|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7597" PostTypeId="2" ParentId="7596" CreationDate="2016-02-24T23:34:24.937" Score="6" Body="&lt;p&gt;The key feature is that the steering has &lt;a href=&quot;https://en.wikipedia.org/wiki/Ackermann_steering_geometry&quot; rel=&quot;nofollow&quot;&gt;Ackerman geometry&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the steering arms coming off the back of the hubs are connected to the steering linkage at a point inboard and behind the pivot point of the hub. If  you draw a line through these two pivot points for both wheels they will meet at or near the centre of the rear axle (although the exact location may vary to produce specific handling characteristics). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of this geometry is that, when  turning the inside wheel will turn through a greater angle than the outside, reflecting the fact that the inside wheel follows a smaller turn radius. In a basic Ackerman geometry lines extending the axis of the front stub axles will meet at the centre of the turning radius, although again this may vary to suit particular handling characteristics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the steering is controlled by a bell crank linkage rather than rack and pinion, so  you may have to make some measurements to work out the precise relationship between servo position and steering input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice there are also a lot of other related factors which affect the relationship between steering input and the actual behaviour of the vehicle as the steering geometry is also affected by suspension movement and the tracking, caster and camber angles of the wheels as well as the dynamic behaviour of the vehicle as a whole. This is particularly the case with something like an RC car which tends to have low mass and high torque so there may well be significant steering effects from factors like  bump steer and roll steer, how the differential distributes torque to the rear wheels (ie the amount of slip in the differential and rear wheels) and the dynamic behaviour of the front tyres. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-25T18:56:08.490" LastActivityDate="2016-02-25T18:56:08.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7598" PostTypeId="1" AcceptedAnswerId="23989" CreationDate="2016-02-25T00:53:13.350" Score="4" ViewCount="586" Body="&lt;p&gt;Individual-head (i.e. not deluge) sprinkler systems come in three flavors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Wet-pipe systems have water pressure at the head at all times -- they are the quickest to respond, but can't be subject to freezing conditions and are too risky for water-sensitive spaces&lt;/li&gt;&#xA;&lt;li&gt;Dry-pipe systems have compressed air in the pipe between the head and the dry-pipe valve -- when the head fuses, the dry-pipe valve is opened by the pressure change and delivers water to the system.  This can be used when a freeze risk exists, but is slower to respond.&lt;/li&gt;&#xA;&lt;li&gt;Preaction systems are used to provide external confirmation (from say an alarm system) before the sprinklers activate -- the alarm system must open the preaction valve before water can flow onto the fire.  This is used when accidental discharges must be controlled at all costs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This raises the question: under NFPA 13/13R/13D, can individual fire zones be protected by a dry-pipe or preaction system while the remainder of the building has wet-pipe sprinkler protection? For instance, the attic, garage, and unfinished basement in a home could be dry-pipe protected to avoid freeze-ups while the finished spaces are protected by a conventional wet-pipe system.&lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2018-10-01T00:50:23.430" Title="Can a dry-pipe or preaction valve be used to create a branch off a wet pipe sprinkler system?" Tags="|safety|building-design|fire-sprinkler|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7599" PostTypeId="2" ParentId="6173" CreationDate="2016-02-25T01:56:34.253" Score="3" Body="&lt;p&gt;I believe there is a sensible engineering reason no one has pointed out yet.  In the picture in the original question the central span appears to be slightly longer than twice length of each outer cable-supported span.  This implies a greater load from each half of the central span than from each outer cable-supported span.  Furthermore, the cables from strictly vertical towers would have to become more shallow to reach the greater distance to the center of the central span which would further increase the tension necessary to support the same partial vertical load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would result in unbalanced tension on vertical towers and tend to pull them inward and distort the bridge.  Having the towers lean outward--and/or be pulled outward by additional tension into the ground support--may be one way of accommodating the imbalance (such as in the asymmetrical example in the answer by @RussellMcMahon), but it may be that the level of tension required becomes impractical for the load and span distance required and given the supporting structure on the riverbed for the bridge in the question.  In any case, it would certainly seem to require more structural support--and thus expense--to have the towers lean outward against even greater tension to support an even further reach to the center of the long central span.  (This may be why conventional wisdom was having trouble coming up with a workable and affordable design, if that was true of this case.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, it appears that by having the towers tilt inward the cables are able to maintain a more balanced profile with less stress being added to the design just to balance it out.  The tops of the towers are near each midpoint between the center of the central span and the outer reach of each outer cable-supported span, so the cables under the greatest tension (and having the greatest horizontal component) are the most symmetrical... to then keep the lateral forces on each tower balanced out.  It's more like the base of vertical towers were simply slid farther apart while keeping the tops fixed, which means the structure and expense are more like that for a shorter central span using symmetrical cables from vertical towers rather than the accelerating expense for the longer spanned distance with conventional designs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact distance between cable mounts on the span may not be exactly the same for the central span and outer spans, and may vary slightly across each span to slightly vary the load each supports as the mounting point on the tower becomes farther from the center between the pair of partial loads.  Each incrementally closer pair of cables can then be placed so as to balance out the lateral tension at the tower and keep the load on the tower directed along its axis of compression strength.  The engineering math to work out the optimal placements is beyond me.  It's possible that the cable load spacing is uniform after all; it just doesn't necessarily have to be with this approach.&lt;/p&gt;&#xA;" OwnerUserId="5190" LastEditorUserId="5190" LastEditDate="2016-03-15T23:38:21.733" LastActivityDate="2016-03-15T23:38:21.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7600" PostTypeId="1" AcceptedAnswerId="7602" CreationDate="2016-02-25T03:36:22.737" Score="4" ViewCount="420" Body="&lt;p&gt;See here are the shapes which are made on the suction head of compressor, &lt;a href=&quot;https://i.stack.imgur.com/s9dUq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s9dUqm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are these, and why are they on the suction head pipe? &lt;/p&gt;&#xA;" OwnerUserId="5191" LastEditorUserId="33" LastEditDate="2016-02-25T14:20:40.220" LastActivityDate="2024-01-03T03:43:11.500" Title="What are these bent shapes on this elbow in a compressor suction pipe?" Tags="|mechanical-engineering|fluid-mechanics|piping|compressors|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7601" PostTypeId="1" AcceptedAnswerId="7612" CreationDate="2016-02-25T07:38:41.467" Score="0" ViewCount="375" Body="&lt;p&gt;I open the door for 30s of a 140L capacity 80W (240VAC, 0.8A) barfridge with a 20L freezer compartment (partition open to main compartment at one end) at a temperature of -5℃ with an internal main compartment temperature at 2℃, with initially 30L of 50-50 glass and plastic containers container mainly water inside, in a room at an ambient temperature of 41℃ and 85% humidity (Sydney today), to &lt;strong&gt;take out&lt;/strong&gt; a 6-pack of medicinal beer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably, the cold, denser air from inside the fridge flows out,&#xA;   and is replaced by the hot, lighter air from the room.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;How much of the fridge and freezer compartment air is exchanged in that time period?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;How long does it take the exchanged air in the fridge to equilibrate&#xA;to its initial temperature of 2℃? What equation describes this? Is there an initial temperature rise?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;How many grams of ice will precipitate from the exchanged air into&#xA;the freezer compartment?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After @Chris Johns' answer, I rolled the barfridge out and actually looked at the manufacturer's plate, and then updated my question from my original eyeball estimates.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Changed 40L to &lt;strong&gt;140L&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Added power info: &lt;strong&gt;&quot;80W (240VAC, 0.8A)&quot;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my calculations, $ P = V­_{rms} * I_{rms} $ = 240VAC * 0.8A = 192 W,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so efficiency,  $ \eta = P_{out} / P_{in} \%% $ $ = {80W \over 192W} = 41.7\%% $&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Changed freezer compartment capacity eyeball estimate from 10L to &lt;strong&gt;20L&lt;/strong&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The refrigerant is $CCl_2F_2-R12$, 85g.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Clarified the internal contents of the fridge as containing a total of &lt;strong&gt;30L&lt;/strong&gt; of plastic and glass containers (50-50) containing mostly water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Clarified the temperature of the freezer compartment at &lt;strong&gt;-5℃&lt;/strong&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I presume you can calculate the coefficient of performance (CoP) from this data?&lt;/p&gt;&#xA;" OwnerUserId="5099" LastEditorUserId="5099" LastEditDate="2016-02-29T04:32:27.780" LastActivityDate="2016-02-29T04:32:27.780" Title="Refrigerator temperature - compartment gas volume exchange, time to equilibrate?" Tags="|thermodynamics|heat-transfer|airflow|cfd|gas|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7602" PostTypeId="2" ParentId="7600" CreationDate="2016-02-25T07:53:27.203" Score="3" Body="&lt;p&gt;Notice that the &quot;shapes&quot; are in a right angled bend where the flow changes directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes vanes are placed in right angled bends to reduce shock losses by reducing the turbulence of the flow as it goes through the bend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;shapes&quot; the picture appear to be on the outside of the bend, in which case they have been placed there to reinforce the bend to make it stronger.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-02-25T07:53:27.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7603" PostTypeId="1" CreationDate="2016-02-25T08:15:02.403" Score="1" ViewCount="151" Body="&lt;p&gt;I'm trying to model a blast in autodyn using axysimmetric wedge formulation in order to remap a 3d problem later. But I don't know how to calculate the radius of the explosive to match the given mass, or how to see the mass of constituent of multibody Euler part. Does anyone have any suggestions on this account?&lt;/p&gt;&#xA;" OwnerUserId="5193" LastEditorUserId="1832" LastEditDate="2016-02-25T10:34:35.290" LastActivityDate="2016-02-25T10:34:35.290" Title="Is there any way to see mass of a part in autodyn?" Tags="|ansys|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7604" PostTypeId="1" AcceptedAnswerId="7611" CreationDate="2016-02-25T08:32:59.877" Score="-1" ViewCount="1867" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6bpGs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6bpGs.jpg&quot; alt=&quot;determine the length of b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is 8 kN*m clockwise.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the question, but with me 6 kN/m is 4 kN/m and 2 kN/m is 2.5 kN/m. And the length is not 4 but 9 metres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not yet at the chapter of equations of equilibrium, plus it says the couple moment is not 0, so I assume it just about the loadings and not including the reactive forces and reactive moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I call the top one F1 and the bottom loading F2 I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_1+F_2=0= -4b\dfrac{1}{2}+2.5(b+a)\dfrac{1}{2}$$&#xA;solving this i got $a=0.6b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment around the free end of the bar (not $A$, but opposite side) due to the loadings is (counter clockwise positive):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M= -F_1\cdot\text{center of triangle} + F_2\cdot\text{center of triangle}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you can replace the loading with force $F_1$ and force $F_2$ and its line of action is through the center of the triangle area ($\dfrac{1}{3}\text{base}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M= -8= -4b\dfrac{1}{2}\cdot\dfrac{1}{3}b+2.5(b+a)\dfrac{1}{2}\cdot\dfrac{1}{3}(b+a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what am I doing wrong?&#xA;Because $M$ can never be negative with me, plus the answer should be $b=5.625$ and a= $1.539$.&#xA;But this to me makes no sense, because then $F_1+F_2\neq0$.&#xA;And if I should take the reactive forces into account at $A$ then you can never have still a moment, because then it is not static anymore.&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="4117" LastEditDate="2016-02-25T11:24:38.243" LastActivityDate="2016-02-25T11:34:41.697" Title="Reduction of a simple distributed loading" Tags="|statics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7605" PostTypeId="1" AcceptedAnswerId="15738" CreationDate="2016-02-25T08:36:38.920" Score="4" ViewCount="1487" Body="&lt;p&gt;I am working on modeling of air-heater component. A simplified representation of system is as follows:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/NIKAS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NIKAS.png&quot; alt=&quot;Heater duct block diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cold air flow enters steel duct at one end. Inside duct is Calrod heating element shaped as ellipse. Follwing heat transfer occur simulteneously:-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Air gains heat from surface of heated calord. &lt;/li&gt;&#xA;&lt;li&gt;Air also exchanges heat with duct inner surface. &lt;/li&gt;&#xA;&lt;li&gt;Calrod element radiates heat to duct inner surface. &lt;/li&gt;&#xA;&lt;li&gt;Duct loses heat by radiation to surroundings as well as convection.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;System is solved by taking energy balance equations on Air mass, Duct mass and Heater element mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is in regards to convection between air and duct. In model, I have considered whole assembly to be made of 5 parts along duct length. In each part respective thermal balance on air, duct and heated element mass is taken. In each part, I consider air to be lump-sum mass. Entering air gains heat from calrod and gains some temperature ( which would in real world be average temperature). Problem occurs when that temperature is used for convection transfer with duct. In actual process, air flow is fast enough (0.013 $\frac{m^3}{sec}$) so that air that actually is in contact with heater element doesn't reach duct surface. So air that exchanges heat with duct inner surface is at lower temperature than average air's temperature. This gives incorrect duct temperature predictions. How can I properly set $\Delta T$ for convection heat exchange with duct?    &lt;/p&gt;&#xA;" OwnerUserId="4949" LastEditorUserId="272" LastEditDate="2016-02-25T12:48:24.863" LastActivityDate="2017-06-07T03:30:42.613" Title="Heat Transfer coefficient calculation for hot air flow inside rectangular duct" Tags="|heat-transfer|modeling|convection|radiation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7607" PostTypeId="1" CreationDate="2016-02-25T09:48:57.477" Score="3" ViewCount="89" Body="&lt;p&gt;My company is working on a machine vision application.  One of our requirements is that the camera peer through a frame (like a picture frame - see &lt;a href=&quot;https://drive.google.com/open?id=0B6TLk-4SL4TeQVVUOXRvOEFkMTg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;  ), where the top and bottom border are etched with machine vision targets.   The  frame is mounted on the gray rectangle (one side of a box) in the photo.  The frame is 1/2&quot; aluminum and about 14&quot; long and between 3&quot; and 8&quot; high.  The  targets are laser etched.  The top and bottom targets are parallel.&#xA;For various reasons,  we'd like to vary the gap between the targets.  Our original thought is that we'd need to create a difference size frame (see picture below - red frame)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OLHJz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OLHJzm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of our engineers suggested that instead of building one frame for each size, we could make a vertical slot on either size of  the box and position two rows of targets (actually two long pieces of aluminum) accordingly, somehow fastening them in the slot. We could use a metal spacer, measured to a particular gap to insure that the top and bottom row of targets are correctly oriented. Correct orientation means that the targets rows are parallel and are spaced to a known distance.  My question is as follows.  Is it possible to achieve high tolerances (e.g. 1/100mm) as it relates to orientation using the second approach?  How would we do it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer:  I'm not a mechanical engineer.  Apologies if I'm not including important information (or if I've included irrelevant information).&lt;/p&gt;&#xA;" OwnerUserId="5196" LastEditorUserId="272" LastEditDate="2016-02-25T12:45:21.507" LastActivityDate="2016-04-25T23:34:01.230" Title="Placing machine visions targets with high accuracy" Tags="|mechanical-engineering|machining|aluminum|tolerance|product-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7608" PostTypeId="2" ParentId="7600" CreationDate="2016-02-25T10:19:34.373" Score="3" Body="&lt;p&gt;I suspect that the crescent shaped sections penetrate all the way through the pipe and are &lt;a href=&quot;http://www.dobyverrolec.com/turning-vane-system.html&quot; rel=&quot;nofollow&quot;&gt;turning vanes&lt;/a&gt; to help direct the flow around the sharp corner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pipe of this size it is much easier to fabricate a mitered corner than a radiused bend and the vanes are intended to make the change in flow direction less abrupt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case holes have been cut in the pipe, the vanes (probably cut from sections of pipe) pushed through and then welded on from   the outside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case they are pretty crude and all have the same tight radius so I'm not entirely convinced that they are doing much good. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-25T10:19:34.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7611" PostTypeId="2" ParentId="7604" CreationDate="2016-02-25T11:34:41.697" Score="1" Body="&lt;p&gt;Defining $F_1$ as the triangular load, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_1 = -\dfrac{4b}{2} = -2b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Defining $F_2$ as the uniform load, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_2 = 2.5(a+b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you stated, $F_1+F_2 = 0 \therefore b = -5a$. This result may seem odd, given that it's negative, but it's correct. After all, the resultant force of a triangular load is half of what it would be if the load were uniform and equal to the maximum value. So, for a triangular load to have the same resultant force as a uniform load while both occupy the same length ($a=0$), the triangular load must have a maximum value which is double that of the uniform load. If the triangular load's maximum value is greater than double, then it can occupy a shorter length while having the same resultant force ($a&amp;gt;0$, that's the case displayed in the original question with 6 and 2 kN/m). However, if the triangular load's maximum value is less than double the uniform load (as is your case), then the triangular load must occupy a greater length to have the same resultant force ($a&amp;lt;0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the moment due to a force couple is $M = F \times D$, where $D$ is the distance between the forces in the couple. Now, the centers of action of $F_1$ and $F_2$ are (from the free end):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;D_{F_1} &amp;amp;= \dfrac{b}{3} \\&#xA;D_{F_2} &amp;amp;= \dfrac{a+b}{2}&#xA;\end{align}$$&#xA;Therefore $D = \left|\dfrac{a+b}{2} - \dfrac{b}{3}\right| = \left|\dfrac{3a+b}{6}\right| = \left|\dfrac{-2a}{6}\right| = \dfrac{|a|}{3}$. Thus, $M = -2b \times \dfrac{|a|}{3} = 10a \times \dfrac{|a|}{3} = -8 \therefore a = -\sqrt{2.4} \therefore b = 5\sqrt{2.4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Checking our work:&#xA;$$\begin{align}&#xA;F_1 &amp;amp;= -10\sqrt{2.4} \\&#xA;F_2 &amp;amp;= 2.5(-\sqrt{2.4}+5\sqrt{2.4}) = 10\sqrt{2.4} \\&#xA;&amp;amp;\therefore F_1 + F_2 = 0 \text{ OK!}\\&#xA;M_A &amp;amp;= -10\sqrt{2.4}\left(4-\dfrac{5\sqrt{2.4}}{3}\right) + 10\sqrt{2.4}\left(4-\dfrac{4\sqrt{2.4}}{2}\right) = 10\sqrt{2.4}\left(\dfrac{5\sqrt{2.4}}{3}-\dfrac{4\sqrt{2.4}}{2}\right) \\&#xA;&amp;amp;= 24\left(\dfrac{5}{3}-2\right) = -\dfrac{24}{3} = -8\text{ OK!}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-02-25T11:34:41.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7612" PostTypeId="2" ParentId="7601" CreationDate="2016-02-25T12:00:05.257" Score="1" Body="&lt;p&gt;If you open the door of a small upright fridge for 30 seconds it's reasonable to assume that virtually all of the cold air inside will be replaced with warm air from outside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The missing information in this question is the power and coefficient of performance of the fridge, this tells you the rate at which it is able to move heat from inside to outside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a first approximation you are looking at the heat exchange required to cool 40L of air from 41 deg C to 2 deg C. This is simply the specific heat capacity of air multiplied by the mass of air multiplied by the temperature difference. And works out at around 40 Joules so basically very little. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more detailed approach would consider the change in enthalpy of condensing out and freezing the water vapour present and also the heat transfer from the fridge body and contents during the time that the door is open, but again we are talking about quite small  numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will put much more demand on the fridge when you put in cans of beer which are at ambient room temperature as the heat capacity of the liquid is much greater than that of the air that it replaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg the heat capacity of 1 litre of water is around 4200 J/C so a 40 deg C temperature difference would require 168 KJ of heat transfer to achieve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say the fridge has a power consumption of 100W and a CoP of 3 then is can transfer 300W of heat and so would take at least 10 minutes to cool 1 litre of water, although this figure will be further limited by the heat transfer rate between the water container and the air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Just opening the door for a short and taking something out is fairly minor in terms of load on the fridge. &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The minimum time (and energy requirement) to cool an object from its current temperature to fridge temp and return to equilibrium depends on the performance of the fridge, the initial temperature difference and heat capacity of the object being cooled.&lt;/li&gt;&#xA;&lt;li&gt;The actual time taken will depend on the rate of heat transfer between the cold air in the fridge and the object. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-25T12:42:49.997" LastActivityDate="2016-02-25T12:42:49.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7613" PostTypeId="2" ParentId="7607" CreationDate="2016-02-25T13:10:21.773" Score="1" Body="&lt;p&gt;0.01 of a mm is pretty tight by any standards, for example a typical jobing lathe might be graduated in divisions of 0.025mm. So 0.01mm is not impossible but certainly not trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This level of accuracy is also getting into the realms where thermal expansion and cleanliness of surfaces become significant factors. To put this into perspective a 200mm length of aluminium will expand by around 0.05mm between 10 and 30 deg C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sensible approach would be to build the structure upwards from a very flat base, something like a calibrated surface plate using precisely machined spacers, probably cylindrical columns which you would need to be made to length for each gap spacing you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative approach would be to use parallel high accuracy &lt;a href=&quot;https://en.wikipedia.org/wiki/Ball_screw&quot; rel=&quot;nofollow&quot;&gt;ball screws&lt;/a&gt; to adjust the gap, similar to the way the saddle on a lathe travels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method would be rather more complex than using calibrated spacers but gives you a continuous range of adjustment which allows you to manually compensate for things like thermal expansion and wear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also question whether aluminium is the best material for this level of accuracy. Something like &lt;a href=&quot;http://www.westyorkssteel.com/tool-steel/gauge-plate/&quot; rel=&quot;nofollow&quot;&gt;precision ground tool steel&lt;/a&gt; would give much better stiffness and dimensional accuracy as an off the shelf product. Steel is typically 3 times as stiff as aluminium alloys and typically has much better hardness and machinability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of practical implementation a good machinist should be able to provide you with a work around solution if you can tell them exactly what you need. A general purpose and 'foolproof' solution will probably require a bit more investment in design and manufacture. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-25T22:40:57.823" LastActivityDate="2016-02-25T22:40:57.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7614" PostTypeId="2" ParentId="7304" CreationDate="2016-02-25T14:06:09.423" Score="0" Body="&lt;p&gt;In addition to chris johns answer of self lubricated plastic gears (nylon, acetal, ptfe, etc) and possibly used with a heavier grease, you won't need a sealed housing... You really need to look at those shorter cantilevered shafts. They need to be secured sufficiently.. Maybe spot welded or press fit, if your housing wall thickness is enough. By looks, it doesn't seem thick enough to be reliable for a press fit. You can locally add a bearing to those locations or possibly design custom shafts that are bolted onto the housing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the circlips, you can use those to locate the gears horizontally along the shafts. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-02-25T14:06:09.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7615" PostTypeId="1" CreationDate="2016-02-25T15:01:23.653" Score="2" ViewCount="986" Body="&lt;p&gt;I need to know how I can attach a rotary encoder to something string/wire that I can pull and it retracts automatically &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea was to take apart a retractable dog lead leash and affix it to it somehow, but I'm not seeing how I could make that work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've included a picture of the dog leash(right) and the rotary endoer(left).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only goal here is to measure distance/velocity of a retractable string/wire. I'm not tied down to using a dog leash or rotary encoder, open to any suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other solutions that are not very expensive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CJQ66.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CJQ66.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5200" LastEditorDisplayName="user16" LastEditDate="2016-02-26T12:54:56.173" LastActivityDate="2016-02-26T12:55:06.940" Title="How can I attach a rotary encoder to a retractable dog lead?" Tags="|mechanical-engineering|electrical-engineering|prototyping|" AnswerCount="4" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="7616" PostTypeId="1" CreationDate="2016-02-25T15:14:25.807" Score="0" ViewCount="223" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fQw1v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fQw1v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not yet at the chapter of equations of equilibrium, plus it says the couple moment is not 0, so I assume it just about the loadings and not including the reactive forces and reactive moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I call the top one F1 and the bottom loading F2 I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_1+F_2=0= -4b\dfrac{1}{2}+2.5(b+a)\dfrac{1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solving this i got $a=0.6b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment around the free end of the bar (not $A$, but opposite side) due to the loadings is (counter clockwise positive):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M= -F_1\cdot\text{center of triangle} + F_2\cdot\text{center of triangle}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you can replace the loading with force $F_1$ and force $F_2$ and its line of action is through the center of the triangle area ($\dfrac{1}{3}\text{base}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M= -8= -4b\dfrac{1}{2}\cdot\dfrac{1}{3}b+2.5(b+a)\dfrac{1}{2}\cdot\dfrac{1}{3}(b+a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what am I doing wrong?&#xA;Because M can never be negative with me, plus the answer should be $b= 5.625$ and $a= 1.539$.&#xA;But this to me makes no sense, because then $F_1+F_2\neq0$.&#xA;And if I should take the reactive forces into account at $A$ then you can never have still a moment, because then it is not static anymore.&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="1832" LastEditDate="2016-02-25T15:26:43.473" LastActivityDate="2016-02-25T15:41:26.353" Title="Reduction of a simple distributed loading (correct question)" Tags="|statics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7617" PostTypeId="2" ParentId="7616" CreationDate="2016-02-25T15:41:26.353" Score="1" Body="&lt;p&gt;The answer can be found with &lt;a href=&quot;https://engineering.stackexchange.com/a/7611/1832&quot;&gt;the same process used in the previous question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Defining $F_1$ as the downwards load, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_1 = -\dfrac{4b}{2} = -2b$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Defining $F_2$ as the upwards load, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_2 = \dfrac{2.5(a+b)}{2} = 1.25(a+b)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you stated, $F_1+F_2 = 0 \therefore b = \dfrac{5}{3}a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the moment due to a force couple is $M = F \times D$, where $D$ is the distance between the forces in the couple. Now, the centers of action of $F_1$ and $F_2$ are (from the free end):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;D_{F_1} &amp;amp;= \dfrac{b}{3} \\&#xA;D_{F_2} &amp;amp;= \dfrac{a+b}{3}&#xA;\end{align}$$&#xA;Therefore $D = \dfrac{a+b}{3} - \dfrac{b}{3} = \dfrac{a}{3}$. Thus, $M = 2b \times \dfrac{a}{3} = \dfrac{10}{3}a \times \dfrac{a}{3} = 8 \therefore a = \sqrt{7.2} \therefore b = \dfrac{5}{3}\sqrt{7.2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Checking our work:&#xA;$$\begin{align}&#xA;F_1 &amp;amp;= -\dfrac{10}{3}\sqrt{7.2} \\&#xA;F_2 &amp;amp;= 1.25(\sqrt{7.2}+\dfrac{5}{3}\sqrt{7.2}) = \dfrac{10}{3}\sqrt{7.2} \\&#xA;&amp;amp;\therefore F_1 + F_2 = 0 \text{ OK!}\\&#xA;M &amp;amp;= \dfrac{10}{3}\sqrt{7.2}\cdot\dfrac{\sqrt{7.2}}{3} = \dfrac{72}{9} = 8\text{ OK!}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-02-25T15:41:26.353" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7618" PostTypeId="2" ParentId="7598" CreationDate="2016-02-25T16:12:37.520" Score="1" Body="&lt;p&gt;From a risk management standpoint, dry pipe or preaction systems should only be used when antifreeze wont work (E.g. too cold) and when heat tracing is unreliable or impractical (E.g. power outages). &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5.4.2* Piping in areas that cannot be maintained reliably above 40°F (4°C) shall be protected by use of one of the following methods:&#xA;  (1)*Antifreeze system using a listed antifreeze solution in accordance with NFPA 13&#xA;  (2) Dry pipe system&#xA;  (3) Preaction system&#xA;  (4) Listed dry pendent, dry upright, or dry sidewall sprinklers extended from pipe in heated areas&#xA;  (5) Heat tracing in accordance with 6.7.2.2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3569" LastActivityDate="2016-02-25T16:12:37.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7619" PostTypeId="2" ParentId="7560" CreationDate="2016-02-25T16:28:26.750" Score="2" Body="&lt;p&gt;To solve it without moments, use the parallelogram of forces to replace two forces with their single resultant force &lt;strong&gt;which acts through the point where the two force vectors intersect&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the resultant of the two 300 forces is a force of $300\sqrt2$ at an angle of 45 degrees, &lt;strong&gt;acting through the point 3 inches from the corner of the structure&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat the process for other pairs of forces, till only one force is left, and that gives the magnitude, direction, &lt;strong&gt;and line of action&lt;/strong&gt; of the resultant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hartog is a fairly old book (but still a good book!). Before computers, the quickest way to solve this sort of problem was to make a scale drawing of the force parallelograms instead of doing arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-02-25T16:28:26.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7620" PostTypeId="5" CreationDate="2016-02-25T17:06:58.987" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-02-25T17:06:58.987" LastActivityDate="2016-02-25T17:06:58.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7621" PostTypeId="4" CreationDate="2016-02-25T17:06:58.987" Score="0" Body="questions about computational fluid dynamics" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-02-25T17:06:58.987" LastActivityDate="2016-02-25T17:06:58.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7622" PostTypeId="1" CreationDate="2016-02-25T17:38:33.293" Score="2" ViewCount="168" Body="&lt;p&gt;Consider a control problem of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d \vec{x}}{dt} = F(\vec{x}, \vec{u})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\vec{u}$ is the control inputs and $\vec{x}$ is the state variables, and all we want to do is drive the system from $\vec{x}_0$ to $\vec{x}_1$ in some given amount of time $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is known about how many solutions for $\vec{u}$ exist for such a problem?&lt;/li&gt;&#xA;&lt;li&gt;How do engineers address/interpret the existence of multiple solutions? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5203" LastEditorUserId="1832" LastEditDate="2016-07-17T03:26:07.733" LastActivityDate="2016-07-17T04:16:33.373" Title="Multiple solutions in optimal control" Tags="|control-engineering|control-theory|mathematics|optimal-control|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7623" PostTypeId="2" ParentId="7615" CreationDate="2016-02-25T18:31:26.930" Score="5" Body="&lt;p&gt;Instead of a shaft driven rotary encoder, you might be able to build an &lt;a href=&quot;http://heliosoph.mit-links.info/make-your-own-rotary-encoder/&quot; rel=&quot;nofollow noreferrer&quot;&gt;optical encoder&lt;/a&gt; that uses optical sensors to read a pattern printed on a disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple encoder uses white/black alternating sectors so it can only tell you the speed of the disk, not the direction or position:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U7eR2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U7eR2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image credit: &lt;a href=&quot;http://heliosoph.mit-links.info/make-your-own-rotary-encoder/&quot; rel=&quot;nofollow noreferrer&quot;&gt;heliosoph - Make your own rotary encoder&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More complex sensors can use multiple optical sensors and use encoding stripes on the disk so it can determine position and direction. There are &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotary_encoder#Ways_of_encoding_shaft_position&quot; rel=&quot;nofollow noreferrer&quot;&gt;multiple ways to do the encoding.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's simple binary encoding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9tFL6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9tFL6.png&quot; alt=&quot;Binary encoded disk&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image credit: &lt;a href=&quot;https://commons.wikimedia.org/w/index.php?curid=1429319&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cburnett - Image:Encoder Disc (3-Bit).svg, CC BY-SA 3.0&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this has the drawback that if the disk stops between sectors, the bit pattern is ambiguous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotary_encoder#Gray_encoding&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gray Encoding&lt;/a&gt;, any two adjacent codes differ by only one bit position, which eliminates that ambiguity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UyB77.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UyB77.png&quot; alt=&quot;Gray Encoded disk&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image credit: &lt;a href=&quot;https://commons.wikimedia.org/w/index.php?curid=801435&quot; rel=&quot;nofollow noreferrer&quot;&gt;jjbeard - Replacement for Image:Encoder-disc.png, Public Domain&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="440" LastEditorUserId="440" LastEditDate="2016-02-25T18:37:38.023" LastActivityDate="2016-02-25T18:37:38.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7624" PostTypeId="2" ParentId="7615" CreationDate="2016-02-25T18:36:40.270" Score="0" Body="&lt;p&gt;Depending on the degree of accuracy you want it may be acceptable to pinch the cord between two rubber wheels and fix the encoder to one of the axles.  A bit like the &lt;a href=&quot;http://www.mig-welding.co.uk/wire-feed.htm&quot; rel=&quot;nofollow&quot;&gt;wire feed on a MIG welder&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-25T18:36:40.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7625" PostTypeId="2" ParentId="7560" CreationDate="2016-02-25T22:19:57.120" Score="3" Body="&lt;h2&gt;Method to Solve the Problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This problem requires only 2 things:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;one must remember that a force applied to a rigid body may be considered to act on the body anywhere along its &lt;a href=&quot;https://en.wikipedia.org/wiki/Line_of_action&quot; rel=&quot;nofollow noreferrer&quot;&gt;line of action&lt;/a&gt;, and  &lt;/li&gt;&#xA;&lt;li&gt;the application of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallelogram_of_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;parallelogram of forces (PoF) method&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To solve the problem, simply find the resultant force of two of the forces by applying the parallelogram of forces method. Proceed by finding the point of intersection of this resultant force's line of action and the line of action of one of the remaining original forces. Again apply the parallelogram of forces method. You will now have the resultant force from 3 of the original forces as well as its location. Proceed in this manner until you are left with a single resultant force and no original forces.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Solving the Problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Let's go ahead and perform this analysis on the problem given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by considering the 500 unit force applied to the upper left corner of the plate and the 500 unit force acting at the midpoint of the left edge of the plate. By inspection we see that the intersection point of their lines of action is at the midpoint of the left edge of the plate. Using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Law_of_cosines&quot; rel=&quot;nofollow noreferrer&quot;&gt;Law of Cosines (LoC)&lt;/a&gt; we can find the length (and angle) of the resultant of these two forces,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;c^2 = a^2 + b^2 -2ab \cdot cos (\gamma)&#xA;$$&#xA;$$&#xA;c^2 = 500^2 + 500^2 - 2 \cdot 500\cdot500\cdot cos(150^\circ)&#xA;$$&#xA;$$&#xA;c = 966&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3LyLlm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3LyLlm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We proceed in the same manner. I did this manually (mathematically via LoC) but I'm too lazy to type it all out. I did, however, check my work by doing it graphically in AutoCAD. Here are some pictures of each step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find intersection of lines of action for resultant and top 500 unit load:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7KELBm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7KELBm.png&quot; alt=&quot;Find intersection of lines of action for resultant and top 500 unit load&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply PoF method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/THG6pm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/THG6pm.png&quot; alt=&quot;Apply PoF method&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find intersection of lines of action for resultant and bottom 300 unit load:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sFykHm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sFykHm.png&quot; alt=&quot;Find intersection of lines of action for resultant and bottom 300 unit load&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply PoF method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TsY9Vm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TsY9Vm.png&quot; alt=&quot;Apply PoF method&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find intersection of lines of action for resultant and right 300 unit load:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d2OtPm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d2OtPm.png&quot; alt=&quot;Find intersection of lines of action for resultant and right 300 unit load&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply PoF method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BXcBom.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BXcBom.png&quot; alt=&quot;Apply PoF method&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final Result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U07AEm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U07AEm.png&quot; alt=&quot;Final Result&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there you are, how to find the magnitude, location, and angle of the resultant resulting from multiple force vectors applied to a rigid body.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Graphic Statics&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The above are a few of the basic principles of a pretty cool but obscure method of static analysis called &lt;em&gt;Graphic Statics&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to know more, you should check out the book &lt;em&gt;Shaping Structures | Statics&lt;/em&gt;, by Zalewski &amp;amp; Allen. As, alpehzero alluded to, graphic statics used to be an extraordinarily handy tool to solve otherwise tricky analysis problems prior to the advent and subsequent proliferation of computers. Note that research into graphic statics methods, particularly with respect to optimizing truss deisgn, &lt;a href=&quot;https://architecture.mit.edu/sites/architecture.mit.edu/files/attachments/lecture/StructuralOptUsingGraphicStatics_r.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;is still being done today&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2016-02-26T14:31:30.173" LastActivityDate="2016-02-26T14:31:30.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7629" PostTypeId="1" CreationDate="2016-02-25T14:44:33.487" Score="0" ViewCount="39" Body="&lt;p&gt;I need to know how I can attach a rotary encoder to something string/wire that I can pull and it retracts automatically &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea was to take apart a retractable dog lead leash and affix it to it somehow, but I'm not seeing how I could make that work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've included a picture of the dog leash(right) and the rotary endoer(left).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other solutions that are not very expensive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;The only goal here is to measure distance/velocity of a retractable string/wire. I'm not tied down to using a dog leash or rotary encoder, open to any suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CJQ66.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CJQ66.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5200" OwnerDisplayName="James Wierzba" LastActivityDate="2016-02-26T01:14:08.543" Title="How can I attach a rotary encoder to a retractable string/wire?" Tags="|prototyping|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2016-02-26T12:55:34.317" ContentLicense="CC BY-SA 3.0" />
  <row Id="7630" PostTypeId="2" ParentId="7615" CreationDate="2016-02-25T16:03:55.003" Score="0" Body="&lt;p&gt;If you're wanting to attach a retractable string to a knob, then go kick that old broken lawnmower/generator/weedeater/whatever in your garage one last time (just for the gratification of it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take out a screwdriver/saw and cut off the pull-start mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside, you have a spool of string, a retracting spring (which you &lt;strong&gt;&lt;em&gt;DO NOT&lt;/em&gt;&lt;/strong&gt; want to have &quot;pop out,&quot; as it takes many minutes &amp;amp; a few nicks/cuts to rewind &amp;amp; re-seat), and 1-2 &quot;little levers&quot; mounted so they &quot;stick out&quot; into the center when the string is pulled, latching onto ths shaft of the (now removed) motor that the pull-starter &lt;em&gt;used&lt;/em&gt; to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now just make sure to put a cap on your rotary encoder that the &quot;little levers&quot; can engage, then slap her together &amp;amp; start tugging ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. This question actually &lt;em&gt;would be&lt;/em&gt; a bit more on-topic on &lt;a href=&quot;http://engineering.stackexchange.com&quot;&gt;http://engineering.stackexchange.com&lt;/a&gt; )&lt;/p&gt;&#xA;" OwnerUserId="5123" OwnerDisplayName="Robherc KV5ROB" LastActivityDate="2016-02-25T18:46:06.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7631" PostTypeId="2" ParentId="7615" CreationDate="2016-02-25T18:01:17.620" Score="1" Body="&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hryxh.png&quot; alt=&quot;schematic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. This illustration clearly demonstrates the problem of making mechanical sketches with a schematic editor.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mount the dog-lead on a base board with a sturdy capstan and connect the encoder to the capstan with a flexible link.&lt;/li&gt;&#xA;&lt;li&gt;Wrap the dog-lead one or more turns around the capstan to give a good grip.&lt;/li&gt;&#xA;&lt;li&gt;Make a stop so that the rewinding stops before the wire end runs off the capstan.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" OwnerDisplayName="transistor" LastActivityDate="2016-02-25T18:01:17.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7633" PostTypeId="1" AcceptedAnswerId="7641" CreationDate="2016-02-26T05:35:26.377" Score="5" ViewCount="4086" Body="&lt;p&gt;A textbook is attempting to convince me that the following figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ShjuW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ShjuW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Represents the generic zones for a Case B wind loading including roof overhangs based on the ASCE 7-10 Chapter 28, Part 2 (Enclosed Simple Diaphragm Low-Rise Buildings) procedure. This procedure addresses roof overhangs via footnote 8 on Figure 28.6-1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Where zone E or G falls on a roof overhang on the windward side of the building, use EOH and GOH f`or the pressure on the horizontal projection of the overhang.  Overhangs on the leeward and side edges shall have the basic zone pressure applied.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I absolutely do not agree that the figure is a correct application of the Chapter 28, Part 2 wind loading provisions for Case B wind direction. Please see below for ASCE 7-10 Figure 28.6-1 Case B zoning. You may observe that zones H, E , A and C correspond, while zones G and F have been flipped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1CC2J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1CC2J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if the textbook is not applying the wind provisions properly, or if I am missing something in the code? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming ASCE 7-10 Figure 28.6-1 is correct and footnote 8 applies, I believe the following (edited) figure shows the correct distribution.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/J9AGB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J9AGB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4808" LastEditorUserId="7083" LastEditDate="2017-10-14T23:17:12.177" LastActivityDate="2017-10-14T23:17:12.177" Title="ASCE 7-10 Wind Loading with roof overhangs" Tags="|civil-engineering|structural-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7636" PostTypeId="1" AcceptedAnswerId="7671" CreationDate="2016-02-26T12:32:43.340" Score="1" ViewCount="829" Body="&lt;p&gt;What is the meaning of &lt;em&gt;clutch fill&lt;/em&gt;?&#xA;See for example &lt;a href=&quot;http://www.dcsc.tudelft.nl/~bdeschutter/private_20100705_acc_2010/data/papers/0918.pdf&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3633" LastEditorUserId="272" LastEditDate="2016-02-26T19:47:06.993" LastActivityDate="2016-02-29T15:40:20.803" Title="What is the meaning of 'clutch fill'?" Tags="|mechanical-engineering|fluid-mechanics|automotive-engineering|terminology|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7637" PostTypeId="1" AcceptedAnswerId="7640" CreationDate="2016-02-26T12:51:41.597" Score="2" ViewCount="427" Body="&lt;p&gt;Please take a look at following figure:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/S2cBw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S2cBw.png&quot; alt=&quot;Rectangular narrow channel with heat transfer by convection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;System considered is a narrow rectangular channel of 2.5 mm depth (W x L is 60 mm x 150 mm). Since channel has low depth, it can be considered as mini-channel. Bottom surface is hot and at $~ 100\,^o \text{C}$, top surface is initially at ambient $20\,^o \text{C}$. Air flow enters channel at $20\,^o \text{C}$. &#xA;I want to know how for such narrow channel heat transfer is defined for &lt;strong&gt;air&lt;/strong&gt; as working fluid. Specifically:&#xA;$$ Nu = f(Re, Pr)$$&#xA;a correlation of Nusselt number as function of Reynolds and Prandtl number. Re range is in 500 - 2000. &#xA;Any suggestion of literature reference are welcome.&lt;/p&gt;&#xA;" OwnerUserId="4949" LastEditorUserId="1293" LastEditDate="2016-02-27T09:55:20.437" LastActivityDate="2016-02-27T09:55:20.437" Title="Heat transfer characteristics in narrow rectangular channel" Tags="|heat-transfer|convection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7638" PostTypeId="2" ParentId="7636" CreationDate="2016-02-26T12:55:36.070" Score="1" Body="&lt;p&gt;It sounds like they are talking about engaging the clutch to the point where the friction surfaces are touching and slipping somewhat but without transmitting any significant torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current technology of automatic or semi automatic transmissions uses a pair of wet plate clutches which are better able to tolerate some slip as part of normal gear changes than conventional dry ones. Wet clutches have a smaller coefficient of friction but larger surface area (achieved with stacked plates) and larger pressures so it is practical to take them to a point where the plates are in contact but little or no torque is transmitted without causing excessive wear or heating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This presetting allows the clutch pressure to be ramped up as desired to accomplish a smooth transfer between the two. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-02-26T12:55:36.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7640" PostTypeId="2" ParentId="7637" CreationDate="2016-02-26T15:13:45.363" Score="3" Body="&lt;p&gt;From &quot;Heat Transfer, 2nd Edition&quot; by A.F. Mills, equation 4.51 gives a formula for the average Nusselt number for flow between two parallel plates.  This might be appropriate for your duct because you can probably ignore the sides and just treat it like two parallel plates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\overline{Nu} = 7.54 + \frac{0.03\frac{D_H}{L}Re\cdot Pr}{1+0.016 \left[ \frac{D_H}{L}Re\cdot Pr\right]^\frac{2}{3}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, $D_H$ is the hydraulic diameter, which is just twice the plate spacing for your case.  $L$ is the plate length, and $Nu$, $Re$, and $Pr$ are the Nusselt, Reynolds, and Prandtl numbers.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-02-26T15:13:45.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7641" PostTypeId="2" ParentId="7633" CreationDate="2016-02-26T17:02:26.767" Score="6" Body="&lt;p&gt;I believe you are correct, the diagram in your book is incorrect in the way it has labeled Zones F and G. This is a bit of a stretch, but I think this may be related to/as a result of an error in Figure G6-5 of a publication released by ASCE called, &lt;em&gt;Wind Loads, Guide to the Wind Loads Provisions of ASCE 7-10&lt;/em&gt; by Mehta &amp;amp; Coulbourne and published by ASCE Press that I came across last month. This publication is an ASCE officially endorsed &quot;How To&quot; guide for ASCE7-10 wind provisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure G6-5 in this publication (below) is an attempt to replicate the image shown in Figure 28.6-1 of ASCE7-10 but was pretty horribly botched. You'll notice the same mistake with Zones F and G for Case B (called &quot;Longitudinal in my book's figure) as well as some general drafting errors with the linework and arrow hatches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q526P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q526P.png&quot; alt=&quot;Figure G6-5 from Guide to the Wind Loads Provisions of ASCE 7-10 by Mehta &amp;amp; Coulbourne&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There doesn't seem to be a published errata for this publication and when I attempted to contact ASCE about the figure last month I didn't get a response. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that point, I also think that there are at least three other errors in your figure as well. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Note 8 of Figure 28.6-1 of ASCE7-10 states (emphasis mine),&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Where zone E or G falls on a roof overhang on the windward side of the building, use E&lt;sub&gt;OH&lt;/sub&gt; and G&lt;sub&gt;OH&lt;/sub&gt; for the pressure on the horizontal projection of the overhang. &lt;strong&gt;Overhangs on the leeward and side edges shall have the basic zones pressure applied.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore, I think that the pressures on the overhangs on the side edges should be E and G instead of E&lt;sub&gt;OH&lt;/sub&gt; and G&lt;sub&gt;OH&lt;/sub&gt; as shown in your figure. The application of E&lt;sub&gt;OH&lt;/sub&gt; and G&lt;sub&gt;OH&lt;/sub&gt; on the windward side in your figure is correct.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is also an error with the boundary line between E&lt;sub&gt;OH&lt;/sub&gt; and G&lt;sub&gt;OH&lt;/sub&gt; on the windward side. The boundary between E&lt;sub&gt;OH&lt;/sub&gt; and G&lt;sub&gt;OH&lt;/sub&gt; should project parallel to the axis of the building.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NttQem.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NttQem.png&quot; alt=&quot;Correct E&amp;lt;sub&amp;gt;OH&amp;lt;/sub&amp;gt; and G&amp;lt;sub&amp;gt;OH&amp;lt;/sub&amp;gt; boundary&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;There is nothing in Figure 28.6-1 of ASCE7-10 that suggests the length of Zone E on the gable side of a structure is equal to $ 2a$ as implied in your figure. I believe that Zone E (and G) should extend all the way to the ridge as shown in Figure 28.6-1 of ASCE7-10. The width of Zone E is equal to $2a$ &lt;strong&gt;only on the eave side&lt;/strong&gt; of the structure.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2016-02-27T13:52:03.860" LastActivityDate="2016-02-27T13:52:03.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7642" PostTypeId="2" ParentId="6958" CreationDate="2016-02-26T18:46:22.833" Score="-1" Body="&lt;p&gt;The geometry of the radial engine prevents all the rods from being directly connected to the crankshaft.  All of the cylinders are in a single plane, thus the connecting rods are also in that same plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Radial_engine&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Radial_engine&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-02-26T18:46:22.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7644" PostTypeId="1" AcceptedAnswerId="7645" CreationDate="2016-02-26T20:12:05.647" Score="3" ViewCount="6356" Body="&lt;p&gt;I'm attempting to find the equations of motion (and eventually transfer functions) for a mass-spring-damper system, but one that is slightly different from your generic damped system example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below I've given a picture of essentially what the system looks like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a large box with mass m and spring k inside of it. This box is damped by damper c to the left-hand wall of the room. I apply a force F(t) to the side of the box, causing both the box and perhaps the mass inside to move. x is measured from the left-hand side of the room to the center of mass, so x will increase &lt;em&gt;both&lt;/em&gt; when the box moves and when the mass inside the box moves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I imagine happens is that for low frequency forcing functions, the spring is essentially nonexistent and the equation of motion looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F(t) = m\ddot{x} + c\dot{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the forcing function frequency approaches the resonance frequency of the system, not only will the box move to the right, but the mass will vigorously experience motion &lt;em&gt;inside&lt;/em&gt; of the  box. I'm not quite sure what the equation of motion looks like for this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for one differential equation that captures both of these - low frequency and high frequency forcing functions. At the end of the day I really want a transfer function to use in Simulink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{w_n^2}{s^2 + 2{w_n}{\zeta}s + {w_n^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This transfer function doesn't seem to model what I want, even though it's the generic second order system transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated. ^^&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WmVnu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WmVnu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mq14A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mq14A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5218" LastEditorUserId="5218" LastEditDate="2016-02-26T21:31:25.237" LastActivityDate="2016-02-26T21:51:02.563" Title="Mass-Spring-Damper Dynamic System" Tags="|mechanical-engineering|electrical-engineering|dynamics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7645" PostTypeId="2" ParentId="7644" CreationDate="2016-02-26T21:51:02.563" Score="3" Body="&lt;p&gt;Call the displacement of the box (where it is attached to the damper and spring) y(t).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force of the damper is $$-c\cdot \dot{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force of the spring is $$k\cdot(x - y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y(t) is the point at which the three forces on it balance to zero:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F - c\cdot \dot{y} + k\cdot(x - y) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x(t) depends only on the spring force and the mass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k\cdot(x - y) + m\cdot \ddot{x} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combine and solve!&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-02-26T21:51:02.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7647" PostTypeId="2" ParentId="1753" CreationDate="2016-02-27T03:06:32.303" Score="4" Body="&lt;p&gt;Water main failures are greater in the winter months than in the spring or summer. As a water utility professional in northern Illinois we have around 120 failures each year with 20 + years of data. Each and every year we have 90 of the 120 between December and March (Fact). The question that indeed is hard to define is why? Here are the facts I see year after year and I base my opinion on.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each year as the frost gains depth and speed 1 foot to 3 feet, between December 15th and January 7th we will have 25 main breaks.&lt;/li&gt;&#xA;&lt;li&gt;From the 7th of January till the 21st of Feb. frost stays at 3 feet we will have 40 main breaks.&lt;/li&gt;&#xA;&lt;li&gt;From Feb 21st till March 20th as the frost is coming out quickly 3 feet till 1 foot or less we will get 25 main breaks. &lt;/li&gt;&#xA;&lt;li&gt;Many of the older pipes are made from cast iron that is very rigged and brittle. its like trying to bend a candy cane. 95 percent of failures are on this type of pipe.&lt;/li&gt;&#xA;&lt;li&gt;The newer pipe is made from ductile iron pipe that is less rigged, allowing it to flex and bend with the frost speed and depth both in and out.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As far as frozen water mains the only two I have known of were both shallow 4 foot or less, a dead end run, and very few service lines on them 1 or two.&#xA;I could go on and on about the subject but be it known I have been watching frost, cold weather, age of pipe, and material composition for 24 years and the whole water main failure is still a great mystery many days. &#xA;Water Utility Superintendent Northern Illinois.  &lt;/p&gt;&#xA;" OwnerUserId="5221" LastEditorUserId="33" LastEditDate="2016-02-27T14:06:31.230" LastActivityDate="2016-02-27T14:06:31.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7648" PostTypeId="1" AcceptedAnswerId="7956" CreationDate="2016-02-27T05:32:40.703" Score="5" ViewCount="1318" Body="&lt;p&gt;I am working on a flywheel currently, and I was wondering what is the force needed to recenter the flywheel when it is spinning and it is nudged slightly off axis? Is it related to gyroscopic precession?&lt;/p&gt;&#xA;" OwnerUserId="3335" LastEditorUserId="1586" LastEditDate="2016-03-01T23:28:23.353" LastActivityDate="2016-03-15T14:07:24.480" Title="Calculate Forces on a Flywheel" Tags="|mechanical-engineering|flywheels|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7649" PostTypeId="2" ParentId="6958" CreationDate="2016-02-27T13:58:06.337" Score="1" Body="&lt;p&gt;Having a master rod means that the bottom bearings of the connecting rods follow a fixed path throughout the cycle. If they are all attached to the crankshaft via a spider bearing it  the spider bearing itself has an extra degree of freedom to rotate around the crank bearing, having a master rod constraints this rotation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the con rod bottom bearings are centred at some radius from the centre of crank bearing they will exert a torque when the line of action of the cylinder is not on a direct line through the centre of both. The master rod resists this torque which is ultimately transmitted to the cylinder preventing the con rod bottom bearings from rotating around the centre of the crank bearing relative to the centreline of the master connecting rod. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the geometry this could potentially lead to collisions between the con rods and cylinder walls and is likely to waste power by making the cylinders fight against each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A master rod would not be required if the geometry of the engine is such that the torques on the con rod bottom bearings balance out naturally, this will depend on overall geometry, crank throw, stroke, number of cylinders etc. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-27T14:05:21.357" LastActivityDate="2016-02-27T14:05:21.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7653" PostTypeId="2" ParentId="6622" CreationDate="2016-02-28T03:52:35.260" Score="3" Body="&lt;p&gt;This is a problem which is usually solved in books on elasticity theory.  The underlying math is based on the solution to the Laplace PDE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do a Google search for Larry J. Segerlind's book &quot;Applied Finite Element Analysis Second Edition&quot;, he lays out the math behind calculating J using the Prandtl stress function method (see Chap. 8 'Torsion of Non-circular Sections').  There is another method which is derived using a warping function, and the math is similar.  The derivation using either method requires some sophisticated math, so be prepared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't find the Segerlind book, there are others on programming the finite element method which use this same problem as an illustration.  It's also used in the more advanced Boundary Element Method as well.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-02-28T03:52:35.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7656" PostTypeId="1" AcceptedAnswerId="7657" CreationDate="2016-02-28T16:28:45.943" Score="1" ViewCount="4687" Body="&lt;p&gt;We call air which is pressurised as &quot;compressed air&quot;. But we don't call pressurised liquid as compressed. I know that liquids are incompressible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;why particular word &quot;compressed&quot; is the created, we could call pressurised air instead of compressed air?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there any significant difference between them (aside from the words), if yes then explain?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2116" LastEditorUserId="1832" LastEditDate="2016-02-28T17:30:23.253" LastActivityDate="2016-02-29T17:56:37.747" Title="Basic difference between &quot;compress&quot; and &quot;pressurise&quot;" Tags="|pressure|compressed-air|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7657" PostTypeId="2" ParentId="7656" CreationDate="2016-02-28T17:36:13.930" Score="3" Body="&lt;p&gt;As your question implies 'pressurise' can apply to increasing the pressure of any fluid whereas 'compression' indicates a change in volume. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To illustrate; a typical industrial gas cylinder will be filled to around 200 bar and might have a volume of 100l at 1 bar that gas would occupy 20000l. Often this is as much about storing and transporting produced gasses in a convenient package (eg argon for welding) as it is using the stored pressure to do work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An incompressible fluid on the other hand occupies almost exactly the same volume (disregarding changes of state) regardless of pressure so 1kg of water has a volume of 1l at 1 bar or 100 bar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly using the term 'compressor' for a category of machine distinguishes it within the more general category of pumps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not unusual to use terms like 'pressurised gas' but, in very general terms, we talk about compressed air simply because it is produced by a compressor. The term also implies that it is available at (more or less) ambient temperature as opposed to air which has been pressurised by heating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly without any other context 'compressed air' tends to imply high pressure air used to operate air tools or similar pneumatic machinery or plant. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-29T17:56:37.747" LastActivityDate="2016-02-29T17:56:37.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7658" PostTypeId="1" CreationDate="2016-02-28T19:09:57.320" Score="2" ViewCount="342" Body="&lt;p&gt;I have a vertical exhaust pipe with an air inlet at the bottom. Biofuel is burnt also at the bottom of the exhaust pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means air is drafted from the bottom of the pipe through the inlet, burning the biofuel, then out from the top of the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine the air speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I use Bernoulli's equation? &lt;/p&gt;&#xA;" OwnerUserId="3538" LastEditorUserId="33" LastEditDate="2016-02-29T13:56:19.537" LastActivityDate="2016-06-28T16:02:11.447" Title="How can I determine air speed in an exhaust pipe?" Tags="|fluid-mechanics|thermodynamics|airflow|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7660" PostTypeId="1" CreationDate="2016-02-29T01:12:36.147" Score="2" ViewCount="96" Body="&lt;p&gt;I need to set up a system with a DC motor that will cause it to run for a small period of time, then stop, and then be able to repeat this until disconnected from its power source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas of how I could achieve this would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="5244" LastEditorUserId="1832" LastEditDate="2016-02-29T01:50:58.643" LastActivityDate="2016-03-01T23:21:07.510" Title="Setting up a &quot;pulsing&quot; motor?" Tags="|electrical-engineering|motors|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7661" PostTypeId="2" ParentId="7660" CreationDate="2016-02-29T01:42:34.943" Score="4" Body="&lt;p&gt;There are 3 obvious solutions &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;mechanical : have a second, continuously running motor connected to either a cam or a rotary contact switch which makes/breaks the power circuit. This sort of approach is simple and intuitive  and tends to be good for prototyping but may be expensive and less reliable for production versions. It is analogous to the valves in a 4 stroke IC engine. &lt;/li&gt;&#xA;&lt;li&gt;analogue electronics : a timer circuit can be constructed using basic components like capacitors, diodes and transistors to drive a power relay/transistor. There are lots of possible ways to achieve this but a web search for 'timer circuit' will yield lots of results and a bit of research should give you a design that you can assemble without needing a lot of knowledge of electronics.  &lt;/li&gt;&#xA;&lt;li&gt;Digital : a chip based system will give you the most flexibility to adjust the timings and set different on/off periods either via software or by potentiometer input. This may be the most versatile solution but requires a bit more knowledge to implement and may be overkill if you just want a simple on/off timer. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There isn't much point in giving too much detail here as your best bet is to have a look a the various options in detail for yourself and see what make the most sense for your particular application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible complication is that motors are inductive loads so rapidly switching them can potentially produce power spikes in your circuit how much of an issue this is depends on the type and power of the motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further option is to use a stepper motor and digital control this allows you to use digital logic to turn the motor by a defined number of steps (angular increments), this may be preferable if you are looking for high angular precision rather than purely time based inputs. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="1586" LastEditDate="2016-03-01T22:51:22.853" LastActivityDate="2016-03-01T22:51:22.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7662" PostTypeId="1" AcceptedAnswerId="7673" CreationDate="2016-02-29T04:40:15.863" Score="6" ViewCount="770" Body="&lt;p&gt;Building codes such as the Uniform Building Code (UBC) or International Building Code (IBC) require the determination of &lt;a href=&quot;http://timber.ce.wsu.edu/supplements/seismic/eslf.htm&quot; rel=&quot;nofollow&quot;&gt;Equivalent Static Lateral Forces&lt;/a&gt; in order to compute $P-\Delta$ effect and to perform checks on the structural design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all these overturning moment computations and structural checks are done (in the building code) assuming a &lt;a href=&quot;http://www.scc.kit.edu/scc/sw/msc/Nas120/appendix_f_final.pdf&quot; rel=&quot;nofollow&quot;&gt;lumped mass model&lt;/a&gt; (a cantilever column). But what if the structure is irregular? How can we compute the  Equivalent Static Lateral Forces in an irregular building/structure discretized by finite element? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These quantities have clear physical meaning in a cantilever column, but may not even necessarily have clear meaning/clear equivalent in an irregular building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for the FEM way of computing equivalent static forces that act on global level. Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To illustrate why I need FEM version of Equivalent Static Load: &lt;strike&gt;consider that the overturning moment is defined as $M_o=F_xh$, where $F_x$ is the lateral load at &lt;em&gt;floor&lt;/em&gt; level, and $h$ is the distance between the top floor to the point where overturning moment is calculated.&lt;/strike&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://site.iugaza.edu.ps/sshihada/files/2012/02/Handout-11-12.pdf&quot; rel=&quot;nofollow&quot;&gt;consider $P-\Delta$ checking ratio&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{P_x\Delta}{V_xh_s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_x$ is the total unfactored gravity load at and above level $x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta$ is the seismic story drift by seismic forces&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_x$ seismic shear between level $x$ and $x-1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h_s$ sotry height below level $x$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Consider also accidental torsional effects ( &lt;a href=&quot;https://drive.google.com/file/d/1XZcPZpa_ACp-Fw9ROSnKY2iJDgHe-FBY9L8rrGCTqgPe_Cy65Evs40bMHFoPR40mHK6n4mN4_qEsYFhz/view&quot; rel=&quot;nofollow&quot;&gt;EC8 example, section 2.5.3, pp 39&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_t=F_xe_y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_x$ is the equivalent lateral static force in $x$ direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_y$ is the accidental eccentricities&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take note that all these involved quantities are evaluated &lt;strong&gt;&lt;em&gt;on floor basis&lt;/em&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we do FEM seismic dynamic analysis, then all I get is the forces for each node, not at each floor. So we need to be able to resolve FEM seismic nodal forces into floor forces. &lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="3353" LastEditDate="2016-03-03T08:04:00.487" LastActivityDate="2016-03-03T08:04:00.487" Title="How to derive Equivalent Static Load for irregular structure modeled with FEM?" Tags="|structural-engineering|finite-element-method|structural-analysis|seismic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7664" PostTypeId="1" AcceptedAnswerId="7669" CreationDate="2016-02-29T09:19:07.290" Score="2" ViewCount="210" Body="&lt;p&gt;I am interested in  structural stability of earthen bunds. They are susceptible to being eroded by water. Let us assume I have a rice field that is surrounded on all four sides by earthen bunds. I have seen in my travels two geometries (shape of bunds). Some are rectangular and others are curved. Is there an objective way to analyze(mathematically I mean) which geometry would be more resistant to flooding i.e. rectangular(or square) vs. curved ? Water can hit the bund vertically from above(during precipitation) and/or horizontally (water from surrounding fields can flow towards a bund)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of a rice field surrounded by bunds - &lt;a href=&quot;http://www.rkmp.co.in/content/bund-preparation-and-maintenance-for-water-management&quot; rel=&quot;nofollow&quot;&gt;Earthen bunds&lt;/a&gt;. &#xA;This one is a bit curved but I have seen rectangular bunds as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume I have flat surface area. &lt;/p&gt;&#xA;" OwnerDisplayName="user5251" LastEditorDisplayName="user5251" LastEditDate="2016-03-01T08:31:27.787" LastActivityDate="2016-03-01T08:31:27.787" Title="Is a particular bund/embankment geometry more stable than others?" Tags="|structural-engineering|geotechnical-engineering|structural-analysis|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7665" PostTypeId="2" ParentId="7658" CreationDate="2016-02-29T09:39:01.663" Score="1" Body="&lt;p&gt;I would suggest measuring the flow directly using a flowmeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are generally two types of flowmeter: mass flow, and volumetric flowmeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you are only interested in the velocity, you probably don't want a thermal mass or Coriolis flowmeter. Suitable volumetric flow meter technologies, in ascending order of price and accuracy, are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Variable area/rotometer&lt;/li&gt;&#xA;&lt;li&gt;Turbine&lt;/li&gt;&#xA;&lt;li&gt;Vortex&lt;/li&gt;&#xA;&lt;li&gt;Ultrasonic&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2551" LastActivityDate="2016-02-29T09:39:01.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7668" PostTypeId="1" CreationDate="2016-02-29T12:57:02.697" Score="13" ViewCount="2314" Body="&lt;p&gt;I haven't found a text that explains the PID controller in simple words. I know the theory: that it calculates the derivative and the proportional gain and the integral etc, but I need to know in &lt;em&gt;reality&lt;/em&gt; what is the output of each function and each combination of functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, starting with the proportional: it sends an input that is proportional to the error recorded. So if the error is 5 V, does it do $\frac{1}{2}\cdot5\text{ V}$ to diminish it? or $\frac{1}{5}\cdot5\text{ V}$? or $-\frac{1}{5}\cdot5\text{ V}$ or what? I don't understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the derivative, it monitors the derivative across a specific time? And then does what? Also, what if there is a noise/disturbance at the begining, so the PID controller will not have normal use rates of change to compare? Same with integral. Can you point me to a good resource or explain me please? &lt;/p&gt;&#xA;" OwnerUserId="4846" LastEditorUserId="1832" LastEditDate="2016-03-01T23:25:12.640" LastActivityDate="2016-12-24T00:53:28.420" Title="PID controller explanation" Tags="|control-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7669" PostTypeId="2" ParentId="7664" CreationDate="2016-02-29T13:23:22.107" Score="1" Body="&lt;p&gt;Bund stability will be affected by a number a factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;materials used to construct the bund&lt;/li&gt;&#xA;&lt;li&gt;compaction of the materials used&lt;/li&gt;&#xA;&lt;li&gt;the height of the bund&lt;/li&gt;&#xA;&lt;li&gt;the width of the bund&lt;/li&gt;&#xA;&lt;li&gt;shear strength of the material used to construct the bund&lt;/li&gt;&#xA;&lt;li&gt;the degree water saturation and depth of saturation&lt;/li&gt;&#xA;&lt;li&gt;whether any part of the bund is allowed to dry out completely&lt;/li&gt;&#xA;&lt;li&gt;the slope angle of the walls of the bund&lt;/li&gt;&#xA;&lt;li&gt;whether the bund has been protected by facing materials like hard&#xA;stones that will absorb the energy of water hitting the bund&lt;/li&gt;&#xA;&lt;li&gt;the force of the water hitting the bund&lt;/li&gt;&#xA;&lt;li&gt;water drainage through the bund, if applicable&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, there are no simple formulae that can be used to analyze slope or bund &lt;a href=&quot;https://en.wikipedia.org/wiki/Slope_stability_analysis&quot; rel=&quot;nofollow&quot;&gt;stability&lt;/a&gt;. The Swedish slip circle method or similar slicing methods would be the most appropriate techniques as these methods are used for slopes composed of soils and I'm assuming the bunds you are referring to are made of soil or a combination of soil and stones.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-02-29T13:23:22.107" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7670" PostTypeId="2" ParentId="7668" CreationDate="2016-02-29T14:11:16.713" Score="4" Body="&lt;p&gt;PID controllers use tuning parameters to adjust the response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the equation for PID control:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RP56K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RP56K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three K-subscript terms are the tuning parameters, and there's one for each term of the PID controller output: proportional, integral, and differential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, with an error of +5V and a Kp of 0.3, the output would be 1.5V. Likewise for the integral and differential terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, these parameters are determined experimentally. The &lt;a href=&quot;http://staff.guilan.ac.ir/staff/users/chaibakhsh/fckeditor_repo/file/documents/Optimum%20Settings%20for%20Automatic%20Controllers%20(Ziegler%20and%20Nichols,%201942).pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ziegler-Nichols&lt;/a&gt; (pdf) tuning method is a simple heuristic method that used to be very popular in industry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These days, most off-the-shelf PID controllers and PLC functions have built-in tuning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="2551" LastActivityDate="2016-02-29T14:11:16.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7671" PostTypeId="2" ParentId="7636" CreationDate="2016-02-29T15:40:20.803" Score="3" Body="&lt;p&gt;The paper you referenced defines what the term means. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To ensure precise synchronization [when changing the drive path from one clutch to another one], before clutch engagement, it is&#xA;  necessary to actuate the oncoming clutch to a position where the&#xA;  clutch packs are in contact. This process is called clutch fill...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-02-29T15:40:20.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7672" PostTypeId="1" AcceptedAnswerId="7678" CreationDate="2016-02-29T15:48:24.223" Score="2" ViewCount="2001" Body="&lt;p&gt;Is there a good way to estimate the power consumption of an oven/heating device given the volume of the oven? I'm trying to do really, really dirty/quick math - I know that the average oven uses probably between 2-4000 Watts, but I'd be looking for a similar estimate with a volume of about 1.3 cubic &lt;em&gt;inches,&lt;/em&gt; so it would be far, far less than a normal oven. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm under the assumptions that heating up and insulating a smaller space would be significantly easier (better than just the linear ratio in size) than a regular oven. Is there a way to figure out how much power input would be needed to heat up that much volume in ~20 minutes (or any time)? I'm not sure of all of the factors that would need to come into play (I'm sure insulation/heat retention is important, but I don't know a typical number for that). I'm not sure if convection or conduction would be better, but presumably the one that would use less power? The maximum temperature would be ~400F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a project I just thought of, which is why everything is somewhat vague. I'm basically just looking for the worst case ceiling, and the best case ideal to have something to start from. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="5257" LastActivityDate="2016-02-29T20:11:59.353" Title="How to estimate power consumption for an oven given volume?" Tags="|mechanical-engineering|heat-transfer|chemical-engineering|power|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7673" PostTypeId="2" ParentId="7662" CreationDate="2016-02-29T15:52:08.120" Score="7" Body="&lt;p&gt;The short answer to your question,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can we compute the Equivalent Static Lateral Forces, the overturning moment and other quantities in an irregular building/structural discretized by finite element?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is, simply, you can't. ASCE7-10 speaks directly to this in Section 12.6 when it describes the conditions under which Equivalent Lateral Force (ELF) can and cannot be used. If your structure is located in an area with moderate to high seismic activity (SDC C-F), you cannot use ELF if your structure has &quot;structural irregularities.&quot; See Table 12.6-1 in ASCE7-10 for details, I've simplified a bit. There are certain structural irregularities that do not preclude ELF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there &lt;em&gt;are&lt;/em&gt; a couple different ways you can go about calculating equivalent static seismic forces for highly irregular structures using FEM. These forces are not directly analogous to the forces you'd get using ELF method from ASCE 7, but they are forces obtained via rational analysis and commonly used/accepted in order to simplify the seismic analysis of irregular structures. They are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Perform a (fixed base) time history analysis on your structure. Use the results to generate acceleration response spectra at each level of your structure. Use the accelerations from all of your RS to design the structure. (Most conservative) &lt;/li&gt;&#xA;&lt;li&gt;Perform a (fixed base) time history analysis on your structure. Pull the peak acceleration in all 3 directions across all your time steps for each node. Use these accelerations to design the structure. &lt;/li&gt;&#xA;&lt;li&gt;Perform a (fixed base) time history analysis on your structure. Use the accelerations at each time step to design your structure (at each time step). The most conservative design controls. (Least conservative)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Time History to RS Method (#1)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This method is the most conservative of the three. The general procedure would be to apply the peak acceleration from your response spectra to the structural elements at each level. In my industry, when using this method, we even multiply the peak acceleration by 1.5 to ensure that we envelope the actual acceleration including any multi-modal effects. This is generally only done for designing components within the structure or modifications to an old structure when time &amp;amp; budget do not allow a more accurate (\$$$$) analysis. It is considered less conservative for component design because you, theoretically, know the natural frequency of the piece of equipment you are working with, and therefore can pull a more accurate acceleration off the response spectra.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Max Time History Acceleration Method (#2)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This method is fairly straightforward. After performing your time history analysis, you can extract acceleration time histories for every node in your FE model. For each node in your model, pull the highest acceleration in the X, Y, and Z directions. Multiply these accelerations with your tributary nodal mass to obtain an equivalent static seismic nodal force in all three directions. This is still considered (highly) conservative because you are pulling accelerations from different time steps and combining them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several variations of this method related to exactly which accelerations/time steps you design to. The tricky part is in convincing yourself and/or the body having jurisdiction over your design that you are enveloping reality.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Time step Time History Acceleration Method (#3)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This method is similar to #2 except you will design your structure/run your analysis for the seismic accelerations at each time step. At each time step, for each node in your model, pull the acceleration in the X, Y, and Z directions. Multiply these accelerations with your tributary nodal mass to obtain the equivalent static seismic nodal force in all three directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is the least conservative but, by far, most computationally expensive approach and is mostly unrealistic. Your time histories will likely have upwards of 10,000-100,000 time steps. It is incredibly expensive to chug through the analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've personally used all 3 methods at one time or another in my career. #2 seems to hit the sweet spot with respect to the accuracy:expense ratio. &lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2016-03-01T00:53:08.707" LastActivityDate="2016-03-01T00:53:08.707" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="7674" PostTypeId="1" CreationDate="2016-02-29T15:58:29.670" Score="4" ViewCount="7682" Body="&lt;p&gt;How many power strokes occur for every complete rotation in a regular 3-cylinder engine? Or in other words, every how many degrees does a new power stroke takes place? Does anyone know please?&lt;/p&gt;&#xA;" OwnerUserId="4846" LastEditorUserId="272" LastEditDate="2016-02-29T16:41:46.907" LastActivityDate="2021-02-10T17:25:07.393" Title="How many power strokes per rotation in a 3 cylinder engine?" Tags="|automotive-engineering|engines|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7675" PostTypeId="2" ParentId="7674" CreationDate="2016-02-29T16:16:38.390" Score="8" Body="&lt;p&gt;Each cylinder produces a power stroke for every (2-stroke engine) or every other (4-stroke engine) rotation of the crankshaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3-cylinder, 4-stroke engine will produce 3 power strokes for every 2 rotations (720&amp;deg;), or one for every 720&amp;deg;/3 = 240&amp;deg; of rotation.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-02-29T16:16:38.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7676" PostTypeId="2" ParentId="7674" CreationDate="2016-02-29T16:33:04.193" Score="5" Body="&lt;p&gt;A stroke consists of either one expansion, or one contraction. As such it corresponds to half of a rotation of the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a three cylinder engine, each cylinder will stroke with 2 strokes per revolution, so that's 6 strokes per revolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in a four stroke engine, it takes four strokes for each cylinder to complete one cycle, and only one of those strokes is the power stroke. Thus, there are 1.5 power strokes per revolution, or a power stroke every 240 degrees.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="272" LastEditDate="2016-02-29T21:16:44.490" LastActivityDate="2016-02-29T21:16:44.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7677" PostTypeId="2" ParentId="7672" CreationDate="2016-02-29T18:59:16.140" Score="1" Body="&lt;p&gt;The volume, in itself, doesn't make much of a difference as this just determines the mass of air that you need to heat and air has quite a small specific heat capacity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the volume is directly related to the external surface area, which will have a significant effect. Assuming that the walls of your oven are reasonably comparable to those of your base for comparison in terms of heat transfer coefficient then the relative power requirements will be in the same ratio as surface area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if both ovens are cubes and your reference oven is a 10 inch cube it will have a surface area of 10x10 inches per side (with six sides) so 600 square inches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 1 inch cube will have a surface area of 6 square inches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the ratio of surface areas is 600/6 = 100 so a reasonable approximation would be that the 1 inch oven requires 1/100th of the power to maintain a stable temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as a first approximation you are looking at tens of watts rather than thousands. This is about the same order of magnitude as a candle and so seems fairly reasonable to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that for such a small volume it is fairly easy to be a bit more generous with the insulation than with a larger oven. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms time to warm up the key factor is the thermal mass of the oven structure and insulation which is proportional to volume ie a cubic relationship so you get even more benefit from reducing the size. Here I'm ignoring the scaling effects of the thickness of the insulation to have comparable heat flux per surface area but in terms of the practicalities of building the thing this is fairly trivial for a general case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple way to make a very efficient oven on the scale of a few inches is simply to carve it out of soft fire bricks and, if necessary line it with a short length of stainless steel box section. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-02-29T19:23:20.857" LastActivityDate="2016-02-29T19:23:20.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7678" PostTypeId="2" ParentId="7672" CreationDate="2016-02-29T20:11:59.353" Score="2" Body="&lt;p&gt;If you can make a guess at the interior and exterior surface temperatures, then you can use the following simplified steady-state equation for conduction through an arbitrary shape.  It is from A.F. Mills, &quot;Heat Transfer 2nd Edition&quot;, equation 3.32:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{Q} = k \Delta T \sum_i S_i&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\dot Q$ is the rate of heat transfer, $k$ is the thermal conductivity of the material, $\Delta T$ is the temperature difference across the wall of the material, and $S_i$ is a &quot;shape factor&quot; for surface $i$, which is a function of the size and shape of the material through which the heat is flowing.  If your oven will be a cube with interior width $W$ and uniform insulation thickness, $L$, on all sides, then it can be decomposed into two basic shapes; 6 walls and 12 edges.  I'm ignoring the corners because the amount of heat lost there will be small.  According to table 3.2 from the same book, the shape factors for these surfaces will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sum_i S_i = 6S_{wall} + 12S_{edge} = 6\frac{W^2}{L} + 12(0.54W)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I'll assume the interior of your oven will be a cube, ~1 inch (0.0254 meters) on all interior dimensions, with insulation thickness of ~1 inch.  So, $W=L=0.0254m$.  The total shape factor will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;S_{tot} = 0.152+0.165 = 0.317m&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the interior temperature is 400F (477K) and the outer surface is at 120F (322K), then the temperature difference will be $477K-322K = 155K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming fiberglass insulation, like the kind on a common kitchen stove, the thermal conductivity will be $\approx 0.11 \mathrm{\frac{W}{m\cdot K}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting it all together:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{Q} = (0.11\mathrm{\frac{W}{m\cdot K}})(155K)(0.317m) = 5.4 W&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, after the oven is heated up and running at steady state, you'll need about 6 watts to keep it at operating temperature.  Like I said, this solution depends on knowledge of the surface temperature, so if it's way off from my guess of 120F, you'll need to re-do the calculation.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-02-29T20:11:59.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7679" PostTypeId="2" ParentId="7668" CreationDate="2016-02-29T22:01:47.803" Score="9" Body="&lt;p&gt;In intuitive terms I have found the following explanation to be useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of argument lets say that our system is filling a bucket with a hole in it water from a tap. We measure the depth of water  in the bucket and control the flow rate of water  via a tap. We want to fill the bucket as quickly as possible but don't want it to overflow.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;proportional&lt;/strong&gt; element is a linear measure, in this case the height of the water in the bucket this is a useful measure of how full the buck is at a given time but it tells us nothing about how quickly it is filling so by the time we notice it is full it may be too late to turn off the tap or if we fill it too slowly water will leak through the hole faster than it fills and it never quite gets full. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On paper this sound s like it should be enough on it's own and in some cases it is, however it break down when the system itself is inherently unstable (like an inverted pendulum or a fighter jet) and the lag between measuring the error and the input taking effect is slow compared to the rate at which external noise induced perturbations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;derivative&lt;/strong&gt; element is the rate of change of the water level. This is especially useful when we want to to fill the bucket as quickly as possible eg we might open the tap as far as it will go at the start to fill it quickly but close it off a bit once the level gets near the top so we can be a bit more precise and not over fill it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;integral&lt;/strong&gt; element is the total volume of water added the bucket. If the bucket has straight sides this doesn't matter much as it fills at a rate proportional  to the flow of water BUT if the bucket has tapered or curved sides then the volume of water in it starts to have an effect to the rate at which the water level changes. More generally because this is an integral it accumulates over time so applies a greater response if the P and D elements aren't correcting enough eg by maintaining the bucket at half full. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at this is that the integral is a measure of the cumulative error over time and is effectively a check on how effective the control strategy is at achieving the intended result and is able to modify the input depending on how the system really behaves over a period of time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the P (proportional) element is proportional to the variable that you want to control (like a simple thermostat) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the D (derivative) element is proportional to the rate of change of that variable &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the  (integral) element is perhaps the most difficult to understand but relates to the &lt;em&gt;quantity&lt;/em&gt; that your P parameter is measuring typically this will be a cumulative quantity like volume, mass, charge, energy etc. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-12-24T00:53:28.420" LastActivityDate="2016-12-24T00:53:28.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7680" PostTypeId="1" CreationDate="2016-02-29T22:35:37.117" Score="7" ViewCount="145" Body="&lt;p&gt;I have a system that can be modeled with the following picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yxpbx.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yxpbx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a mass $m$ connected to a spring $k$ and a dashpot $d$. These are both connected to another dashpot $c$. A force $F(t)$ is applied at the junction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some effort tackling a system of linear ODEs, I've found a transfer function that describes this system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{X(s)}{F(s)} = \frac{ds + k}{(mc + md)s^3 + (mk + cd)s^2 + (ck)s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm almost certain this is correct considering I've checked over the math multiple times (and the units check out, which is always a plus).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system has certain parameters: $m = 1$, $Q = 20$, $\omega = 50$ Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this we can calculate the following parameters: $k = m(2\pi\omega) = 98696$ N/m, $d = \frac{sqrt(mk)}{Q} = 15.7$ Ns/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I estimate $c = 4$ Ns/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given these parameters, I expect the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Impulse response that asymptotically approaches a value.&lt;/li&gt;&#xA;&lt;li&gt;Step response that grows without bound.&lt;/li&gt;&#xA;&lt;li&gt;Bode plot that has a peak at around 50 Hz.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first two are true, but this is what my Bode plot &lt;em&gt;actually&lt;/em&gt; looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L1Goy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L1Goy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I expect a peak at 50 Hz is because when $F(t)$ has a frequency near $\omega$, the amplitude of motion of the mass should rapidly increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something that I'm modeling incorrectly here? Or is there some major misconception about how I think this system operates?&lt;/p&gt;&#xA;" OwnerUserId="5218" LastActivityDate="2016-03-01T17:13:21.350" Title="Unexpected Results from my Transfer Function" Tags="|control-engineering|dynamics|transfer-function|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7681" PostTypeId="2" ParentId="7660" CreationDate="2016-03-01T00:51:08.740" Score="1" Body="&lt;p&gt;One possible option is to use a micro-controller with an &lt;a href=&quot;https://en.wikipedia.org/wiki/H_bridge&quot; rel=&quot;nofollow noreferrer&quot;&gt;H-bridge&lt;/a&gt; driver. The micro controller can command when to switch the motor on and off. Also, using this scheme the motor can be switched on and off as well as rotated clockwise (CW) or counter-clockwise (CCW). Also most H-bridge drivers can command to shut off power, which can be significant in an underwater application. With a micro controller you could setup a timer to periodically repeat the process. Also most H-Bridge drivers have built in circuitry to handle back EMF or power spikes. I think a DC motor might be the best choice per your comments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/549/motor-driver-board-limited-current/555#555&quot;&gt;Motor Driver board: limited current&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sparkfun.com/datasheets/Components/General/L298N.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Full-Bridge Motor Driver Dual - L298N&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.sparkfun.com/datasheets/Robotics/TB6612FNG.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dual TB6612FNG (1A)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.microchip.com/stellent/groups/SiteComm_sg/documents/DeviceDoc/en543041.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brushed DC Motor Basics (pdf)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=0-YXBxzKx7g&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brushed DC Motor Basics Part 1 of 2 (Webinar)&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=v4l0TnCGoSc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brushed DC Motor Basics Part 2 of 2 (Webinar)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-03-01T23:21:07.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7683" PostTypeId="2" ParentId="7680" CreationDate="2016-03-01T02:18:12.330" Score="3" Body="&lt;p&gt;Checking the units is an excellent way to double check your work; kudos for doing so.  However, the next step in checking to see if your results make sense is to check limits.  In your case, you can use physical intuition to identify how the system would act at very low frequencies, and how it would act without any damping.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At very low frequencies ($s\rightarrow0$) the responsiveness of the system should be to move in phase with an amplitude transfer function that goes flat at DC ($\frac{X(s)}{Y(s)}\rightarrow1$).  Right now you have an infinite responsiveness at DC; applying a slight force moves the mass by an infinite amount.  Another way to see this is by considering the system without any damping ($c\rightarrow0$, $d\rightarrow0$) in which case your TF would become &#xA;$$&#xA;\frac{X(s)}{F(s)}\rightarrow\frac{1}{ms^2},&#xA;$$&#xA;when it should look more like the undamped harmonic oscillator TF which is given by&#xA;$$&#xA;\frac{k}{ms^2 + k}.&#xA;$$&#xA;So, I don't know exactly where you've gone wrong, but you are missing a constant term in the denominator of your TF.&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2016-03-01T13:07:13.460" LastActivityDate="2016-03-01T13:07:13.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7684" PostTypeId="2" ParentId="6083" CreationDate="2016-03-01T02:46:36.150" Score="2" Body="&lt;p&gt;Window motors used to have a circuit breaker to avoid high stall current. Now many cars control the motors with a microcontroller, similar to a Pi or Arduino. If the motor stops ( sudden rise in current) the program stops the motor, sometimes reversing it assuming it has caught a body part. The voltage across a low resistence is the current signal. It will monitor current to learn &quot;normal&quot; values which change with wear and sticking adding load to the motor. This is either for safety or to make the purchase and especially the repair more profitable.&#xA;Part of the learning includes running the window all the way up and down to store the running and stall currents.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2016-03-01T02:46:36.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7685" PostTypeId="2" ParentId="5838" CreationDate="2016-03-01T02:52:55.043" Score="1" Body="&lt;p&gt;Your car has a safety switch to avoid starting in gear with the clutch engaged. Use it. Running the start signal through the park brake switch is also a good idea since it will be on if parked in neutral, and should always be used.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2016-03-01T02:52:55.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7686" PostTypeId="2" ParentId="5984" CreationDate="2016-03-01T03:03:36.023" Score="1" Body="&lt;p&gt;Because most controls reduce efficiency, since they are aimed at protecting the catalyst - not reducing pollution. An air/fuel ratio of 14.7/1 keeps the catalyst happy. Engines can and do run cleaner and better without the catalyst, but the EPA is not only decades behind chemistry - it is not concerned with the environment.&lt;/p&gt;&#xA;" OwnerUserId="5265" LastActivityDate="2016-03-01T03:03:36.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7687" PostTypeId="2" ParentId="7668" CreationDate="2016-03-01T03:39:58.940" Score="15" Body="&lt;p&gt;A PID function that most people use every day is the hand-eye coordination to steer a car or bicycle. Your eyes are the input, the angle of the steering wheel/handle bars is the output. The set point is usually the center of your lane (until a deer hops out or a dog chases you).&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;Your mind has to &lt;strong&gt;constantly consider 3 different factors&lt;/strong&gt; when preforming this task. The importance it places on each factor is based on past experience, which is called &quot;tuning&quot; in the PID world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proportional:&lt;/strong&gt; &lt;em&gt;&quot;I am a long ways from the center of the lane, I should turn back that direction.&quot;&lt;/em&gt; &lt;br&gt; &#xA;Naturally, if I am further away I want turn sharper than if I am very close. This will allow me to get back to the center of my lane in a timely manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Derivative:&lt;/strong&gt; &lt;em&gt;&quot;I better not just yank the wheel/ handle bars over that direction or I will over-correct, roll, and crash.&quot;&lt;/em&gt; &lt;br&gt;&#xA;You may be in the gutter, but your driving experience teaches you that if you turn sharply things will change very quickly and you need to reduce how sharp you turn to keep from over shooting your set-point and entering on-coming traffic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Integral:&lt;/strong&gt; &lt;em&gt;&quot;The wind keeps pushing me to the edge of the road, and I have to turn into it to stay on coarse&quot;&lt;/em&gt;&lt;br&gt;&#xA;You are pretty close to the center of your lane, but not quite where you want to be. Proportional is small because you are really close and Derivative is small because you are not changing very fast. Integral is the term that steps in and says &quot;Hey now, I know we are not off by much but we have been off for a pretty long time; how about we turn into the wind so we can hold our set point.&quot;&#xA;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PID's aren't perfect, and your steering abilities are actually quite a bit better than a standard PID. You are smart enough to realize that when the wind disappears (for some unknown reason) you zero out your integral term and don't wander into opposing traffic while waiting for the wind to come back. Humans also self tune during operation by considering other inputs like accelerations and physics, while most machines/computers are not currently capable of this. &lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-01T03:39:58.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7690" PostTypeId="1" CreationDate="2016-03-01T06:19:21.563" Score="5" ViewCount="1372" Body="&lt;p&gt;Intuitively, building can be represented as &lt;a href=&quot;http://www.scc.kit.edu/scc/sw/msc/Nas120/appendix_f_final.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a lumped mass model&lt;/a&gt; ( the taller the merrier!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SauGW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SauGW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a building is a lumped mass cantilever column also makes the global stability checks easier because building codes often assume building structure as a cantilever column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if my building structure is discretized and response analyzed in terms of FEM ( with column/beam as line element, and slab/wall as area element), then is it possible that, from the stiffness matrix and everything I have in FEM, I can still approximate the structure as a lumped mass model? If yes, how to do it? How to calculate the errors involved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A practical application of doing this is &lt;a href=&quot;https://engineering.stackexchange.com/q/7662/3353&quot;&gt;in this question&lt;/a&gt;. &lt;strong&gt;But for the purpose of this question, let's assume that I do it for academic purpose and don't really want to care about the application.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-10-21T20:48:18.463" Title="How can we derived the equivalent Lumped Mass Model for irregular building and how to calculate the errors?" Tags="|finite-element-method|structural-analysis|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="7691" PostTypeId="2" ParentId="6670" CreationDate="2016-03-01T10:23:19.050" Score="1" Body="&lt;p&gt;The aluminium has different temper, for the utensil usually use 1 series aluminum alloy which is much softer, so it's easier to cast. And aluminum  is light, also aluminum can be recycled which is environmentally friendly. Most of all, for business, profits are very important. &lt;/p&gt;&#xA;" OwnerUserId="5273" LastEditorDisplayName="user16" LastEditDate="2017-05-05T12:07:23.853" LastActivityDate="2017-05-05T12:07:23.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7694" PostTypeId="1" AcceptedAnswerId="7700" CreationDate="2016-03-01T10:48:56.533" Score="0" ViewCount="210" Body="&lt;p&gt;On January 1991, the esteemed Swedish professor of optics filed a patent titled &quot;&lt;a href=&quot;http://www.google.com/patents/WO1996036893A1?cl=en&quot; rel=&quot;nofollow&quot;&gt;Coherence filter&lt;/a&gt;&quot;, that blocks coherent light regardless of laser light wavelength using destructive interference and lets through ordinary incoherent light. Sometime in 1999, several University College of London professors published &quot;Detection of coherent light in an incoherent background&quot;which uses both optical and digital signal processing with an interferometer to measure the self-coherence function of the incident radiation to produce an interferogram which is subsequently narrow band filtered to yield a sinc envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if it is possible to make a tunable coherence filter with 1.8 nanometer bandwidth for visible light which lets through ordinary incoherent light with less than 5db loss. The reason I chose a bandwidth of 1.8 nanometer is because Edmund Optics manufactures part 87293 which is a spectrophotometer with that wavelength resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In certain applications, sensing , filtering and protection from laser radiation is required for protection of eyes.&lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="2651" LastEditDate="2016-03-04T01:37:40.040" LastActivityDate="2016-03-04T01:37:40.040" Title="Is it possible to make a tunable coherent light filter with 1.8 nanometer bandwidth for visible light?" Tags="|optics|lasers|signal-processing|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7695" PostTypeId="2" ParentId="6518" CreationDate="2016-03-01T13:46:56.047" Score="0" Body="&lt;p&gt;If you are looking for smaller Least Count, it would come at the cost of overall length i.e. the least measurable dimension getting smaller, the span of the instrument would also decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graduation is the size of the indicator, the line markings on the scale and not the measurable quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vernier Callipers with both the units, centimetre and inches in one instrument are available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parallax error is a human error and to lessen the effect some opticals aids may be provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A digital Vernier Callipers would solve your problem as the result would be displayed and hence the graduation and the parallax errors are removed. Also facility of changing units is also available in digital Vernier Callipers.&#xA;So instead of looking for a Vernier Callipers with better Graduations, reduced Parallax error and specific units, a digital Vernier Callipers would be more handy.&lt;/p&gt;&#xA;" OwnerUserId="5276" LastActivityDate="2016-03-01T13:46:56.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7696" PostTypeId="2" ParentId="229" CreationDate="2016-03-01T13:53:51.210" Score="-1" Body="&lt;p&gt;Pressure is force applied per unit area. It arises due to external forces on the surface of an object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When external forces are applied, in order to avoid deformation internal forces are generated which are called Stresses. Both pressure and stress have the same unit. &lt;/p&gt;&#xA;" OwnerUserId="5276" LastActivityDate="2016-03-01T13:53:51.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7698" PostTypeId="2" ParentId="6567" CreationDate="2016-03-01T14:09:27.773" Score="1" Body="&lt;p&gt;The method you used can be modified a bit, by using True stresses and True strains instead of engineering stresses and strains. &#xA;Poisson's ratio is based on conservation of volume of an object.&#xA;Engineering stresses and strains do not conserve volume whereas True stresses and True strains conserve volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation between True stress and True strain is different from the engineering quantities. Using that relation may solve your problem.&lt;/p&gt;&#xA;" OwnerUserId="5276" LastActivityDate="2016-03-01T14:09:27.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7699" PostTypeId="1" CreationDate="2016-03-01T15:17:46.020" Score="0" ViewCount="424" Body="&lt;p&gt;I was wondering what the most non-sound-conductive materials are, especially where those materials are abutting materials that are sound conductive.  For example, imagine a long bar or block, and sound is being conducted through one half, but the sound needs to be killed so that it doesn't flow through the other half.  What materials would be best for the latter?&lt;/p&gt;&#xA;" OwnerDisplayName="user5278" LastActivityDate="2017-10-28T03:14:47.407" Title="Most non-sound-conductive materials, especially when abutting conductive material" Tags="|materials|sound-isolation|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7700" PostTypeId="2" ParentId="7694" CreationDate="2016-03-01T15:48:11.563" Score="2" Body="&lt;p&gt;Since you mentioned &quot;pilot,&quot; I'm going to guess you're looking for a way to block laser pointers from distracting airline pilots.  It really should be sufficient to use the same thing used in laboratories, i.e. goggles which physically attenuate at and near the laser wavelength.   Despite that patent's claims, I'm skeptical that their grating system will operate over the angular range (field of view) needed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can post the actual requirements you're trying to meet, we might be able to suggest other alternatives.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-03-01T15:48:11.563" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7701" PostTypeId="2" ParentId="7680" CreationDate="2016-03-01T17:13:21.350" Score="1" Body="&lt;p&gt;The transfer function appears correct, except that the x and y deviations from the unstreched locations and not as depicted in the figure. Otherwise, when $y=x$ then there is no space for the spring $k$ and damper $d$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This needs to be equated to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1}{c m+d m}\frac{d s+k}{s}\frac{1}{s^2+\frac{s w}{Q}+w^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing terms we get the following two equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ w^2=\frac{c k}{c m+d m} $$ $$\frac{w}{Q}=\frac{c d+k m}{c m+d m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 6 variables ($m$, $k$, $d$, $c$, $Q$, $w$) of which 3 are known ($m$, $Q$, $w$). This leaves us with 2 equations and three unknowns, and thus there are multiple solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing I noticed is that $c$ and $d$ have opposite signs. This is a bit odd because it means that $c$ or $d$ must have a negative sign! (The following is a screen shot of the computations in Mathematica.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2D0HK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2D0HK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $k &amp;lt;&amp;lt; d$ then the zero will almost cancel out the pole at the origin. This was the additional condition I used to arrive at the following values &#xA;$$ c=\frac{5 \pi  \left(1+\sqrt{16000000000 \pi ^2-1599}\right)}{2-20000000 \pi&#xA;   ^2}$$ $$d=\frac{1+\sqrt{16000000000 \pi ^2-1599}}{4000000 \pi } $$ $$k=\frac{1}{1000} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5eNSB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5eNSB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then this is the Bode plot with the peak as expected at ~$314 rad/s$ or $50 Hz$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZgSGu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZgSGu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-03-01T17:13:21.350" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="7702" PostTypeId="1" CreationDate="2016-03-01T17:21:54.200" Score="2" ViewCount="2500" Body="&lt;p&gt;I am trying to make an advanced car physics simulation for a game. What I need to do now is to calculate engine RPM in quite an accurate way. As the timesteps between calculations are very small (60 frames per second or 0.166667 ms) I can afford to use old frame data in a new frame so that's not a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to do is to calculate engine RPM. The values that I can access are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Throttle $f_\mathrm{throttle}$ (a scalable &lt;code&gt;float&lt;/code&gt; value between &lt;code&gt;0.0&lt;/code&gt; and &lt;code&gt;1.0&lt;/code&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Clutch $f_\mathrm{clutch}$ (just like throttle - a scalable &lt;code&gt;float&lt;/code&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Engine torque output $T_\mathrm{engine}$ based on torque curve (torque curve is lineary interpolated curve between RPM and Torque axis using RPM as input which returns torque as &lt;code&gt;float&lt;/code&gt; output).&lt;/li&gt;&#xA;&lt;li&gt;Wheel angular velocity $\omega_\mathrm{wheels}$ (a &lt;code&gt;float&lt;/code&gt; value in &lt;code&gt;rad/s&lt;/code&gt;, for drive wheels it is $T / I$, for free-rolling wheels it is $V/r$).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I need to calculate the engine rotation speed in revolutions per minute.&lt;/p&gt;&#xA;" OwnerUserId="5281" LastEditorUserId="2819" LastEditDate="2016-03-03T04:27:09.380" LastActivityDate="2016-03-03T04:27:09.380" Title="Calculating Car Engine RPM for Game" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7703" PostTypeId="1" CreationDate="2016-03-01T17:45:03.340" Score="1" ViewCount="67" Body="&lt;p&gt;I want to know what the &lt;a href=&quot;http://tienda.bricogeek.com/sensores/403-sensor-de-fuerza-resistivo-circular.html&quot; rel=&quot;nofollow&quot;&gt;circular force sensor&lt;/a&gt; outputs when it is connected to the breadboard (with resistor). Does it output force measured in grams? When I slightly touch the sensor with a finger, the output value varies around 500-600.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question: Do I need to somehow calibrate the output values in order to use them for other measurements?&lt;/p&gt;&#xA;" OwnerUserId="5283" LastEditorUserId="272" LastEditDate="2016-03-02T13:04:58.740" LastActivityDate="2016-03-02T13:04:58.740" Title="Resistive force sensor: some basic questions about its functionality" Tags="|electrical-engineering|pressure|sensors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7704" PostTypeId="1" CreationDate="2016-03-01T19:53:02.410" Score="3" ViewCount="653" Body="&lt;p&gt;Will 25% ammonia vapor balance water vapor between 350°C - 550°C corrode 304 SS? If you know the answer how did you find it what was your source or equation or book?  Will it cause pitting or stress corrosion cracking as it moves through a 304SS pipe?&lt;/p&gt;&#xA;" OwnerUserId="5282" LastActivityDate="2016-03-02T15:01:31.430" Title="Will ammonia vapor corrode 304SS?" Tags="|chemical-engineering|corrosion|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7705" PostTypeId="1" CreationDate="2016-03-01T12:31:05.143" Score="2" ViewCount="711" Body="&lt;p&gt;What is the process called when you upscale your proof of concept to a full size prototype? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a traffic light circuit using LEDs to fill size lights and power supply? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is if the term exists or if it is still prototyping as mentioned in the comments. &lt;/p&gt;&#xA;" OwnerDisplayName="Terry" LastActivityDate="2016-03-01T21:45:21.690" Title="What is the process called for up-sizing a proof-of-concept prototype to full-scale?" Tags="|prototyping|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7706" PostTypeId="2" ParentId="7705" CreationDate="2016-03-01T13:01:36.900" Score="0" Body="&lt;p&gt;Assuming you're not satisfied with using the generalized &quot;Prototyping&quot; term, you could use &quot;Full Scale Prototyping,&quot; &quot;Full Sized Prototyping,&quot; or &quot;Prototype Up-Scaling&quot;/&quot;Prototype Up-Sizing.&quot;&lt;/p&gt;&#xA;" OwnerUserId="5123" OwnerDisplayName="Robherc KV5ROB" LastEditorUserId="368" LastEditDate="2016-03-01T21:45:21.690" LastActivityDate="2016-03-01T21:45:21.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7707" PostTypeId="2" ParentId="7705" CreationDate="2016-03-01T13:07:13.473" Score="3" Body="&lt;p&gt;In &lt;strong&gt;&lt;em&gt;my&lt;/em&gt;&lt;/strong&gt; experience (your mileage may vary) these are the sorts of processes and terms used in product development:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;breadboard&lt;/strong&gt; is your proof-of-concept work; can be messy, often by taking something that's existing and adding/subtracting/hacking/cutting/jumpering/bludgeoning it into something to validate your idea&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Prototypes&lt;/strong&gt; are your pre-production products - hardware specifically built to achieve your idea, often with copious amounts of rework and abuse layered on. Sometimes you have more than one prototype build depending on the complexity of the product&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Pre-production&lt;/strong&gt; or &lt;strong&gt;pilot&lt;/strong&gt; units come after the prototypes are cleaned up and validated against spec - generally they are functionally complete and (almost) ready for mass production&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Mass production&lt;/strong&gt; units are fully validated, accepted and ready to crank out en masse (until the factory fires a 'stop ship' at you).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The road from breadboard to prototype to finished product in the context of product is called &lt;strong&gt;NPI&lt;/strong&gt; or &lt;strong&gt;New Product Introduction&lt;/strong&gt; - this encompasses the entire development path from idea to release.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The initial phase is referred to as the &lt;strong&gt;design phase&lt;/strong&gt; - the design is thought up, calculations are done, the breadboard is built, debugged and abused to give you enough information to try and build what you really need.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The prototype phase is referred to &lt;strong&gt;EVT&lt;/strong&gt; or &lt;strong&gt;engineering validation testing&lt;/strong&gt; - protos are built and the design team validates the functionality of the prototype(s) against specification, and releases changes as required to achieve compliance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, to try and answer your question, in my experience the &lt;strong&gt;process&lt;/strong&gt; that gets you from breadboard to prototype is a mix of debugging and preliminary EVT - you take your 'something', test it against whatever requirements you have (and whatever requirements your breadboard is capable of withstanding), adjust your hardware/design/requirements accordingly, generate and release your product documentation (BOM, schematic, mechanical drawings, firmware, etc.) and built prototypes.&lt;/p&gt;&#xA;" OwnerUserId="5285" OwnerDisplayName="Adam Lawrence" LastActivityDate="2016-03-01T13:07:13.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7708" PostTypeId="2" ParentId="7705" CreationDate="2016-03-01T13:16:26.027" Score="2" Body="&lt;p&gt;It's still a prototype, but its a prototype for a different purpose.  People use different languages to refer to this.  Personally, I call it &quot;iterative prototyping&quot;.  Others might call it moving from a &quot;works like&quot; prototype to a &quot;looks like/works like&quot; prototype, e.g., &lt;a href=&quot;http://umaine.edu/amc/2013/08/07/what-is-a-prototype&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://umaine.edu/amc/2013/08/07/what-is-a-prototype&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of what you call it, you should at all times remain aware of where that prototype lies in your process.  My favorite question for students during a design process involving prototyping is &quot;why are you building this prototype?&quot;.  Obviously, you're building a prototype because you're asking it a question.  If you knew the answer to the question you're trying to address with the prototype, then there wouldn't be a need to build it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I may wax a bit philosophical, going from small size to large size sometimes  shows an idea may not scale well.  My absolute favorite example comes from the story of Diognetus in Vitruvius' Ten Books on Architecture.  Diognetus was fired from his retainer for the city of Rhodes, because an upstart proposed a scale model for the purpose of defense against siege engines, nabbing them by crane.  When attacked by the Helepolis, the siege engine to end all siege engines, the &quot;breaker of cities&quot;, he couldn't build it, because it didn't scale well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They managed to rehire Diognetus, at his own price! (&lt;a href=&quot;http://lexundria.com/vitr/10.16/gw&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://lexundria.com/vitr/10.16/gw&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He at first refused to listen to their entreaties; but when&#xA;  afterwards the comely virgins and youths, accompanied by the priests,&#xA;  came to solicit his aid, he consented, on condition that if he&#xA;  succeeded in taking the machine, it should be his own property. This&#xA;  being agreed to, he ordered a hole to be made in that part of the wall&#xA;  opposite to the machine, and gave general as well as particular&#xA;  notices to the inhabitants, to throw on the other side of the hole,&#xA;  through channels made for the purpose, all the water, filth, and mud,&#xA;  that could be procured. These being, during the night, discharged&#xA;  through the hole in great abundance, on the following day, when the&#xA;  helepolis was advanced towards the wall, it sunk in the quagmire thus&#xA;  created: and Demetrius, finding himself overreached by the sagacity of&#xA;  Diognetus, drew off his army.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i6iZ2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i6iZ2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moral of the story is hold out for your price!&lt;/p&gt;&#xA;" OwnerUserId="5293" OwnerDisplayName="Scott Seidman" LastActivityDate="2016-03-01T13:16:26.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7709" PostTypeId="1" AcceptedAnswerId="7713" CreationDate="2016-03-01T20:26:43.833" Score="3" ViewCount="504" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/7680/unexpected-results-from-my-transfer-function&quot;&gt;I recently asked a question about odd results from a transfer function&lt;/a&gt;, and while looking over my work, I conjectured that maybe I'm having a more fundamental problem. Since this is a design problem I don't expect anyone to give a full-fledged answer, but any hints or prodding in the right direction would be appreciated. Here goes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'm looking to model a one-axis gimbal in Simulink, which requires a transfer function to plug into my simulation. The gimbal has the following properties that I need to capture in a transfer function:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Pushing the gimbal (giving it an impulse $\delta$) will move it some &lt;em&gt;finite distance&lt;/em&gt; $x$. This is equivalent to having an impulse response that converges to a finite value.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The gimbal has a resonant frequency of $\omega$. Since I will be using active PID control on it in Simulink, if I attempt to drive the gimbal with a forcing frequency near $\omega$, I should expect high amplitude oscillations in the position $x$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These two responses combined will produce a very realistic expectation: pushing on the gimbal will move it (though damping will slow it down to a halt if the force is removed), but if you get near the resonance frequency of the structure, you'll excite the first elastic mode and have large-amplitude oscillations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a picture of what I &lt;em&gt;thought&lt;/em&gt; the system should look like, approximated with springs and dampers (because finding the transfer functions of these elements is easy to compute):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wojVY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wojVY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My rationale is as follows: imagine just the mass $m$ and the damper $c$. This will produce the first response required, having the impulse response converge to a finite non-zero value. To force the resonancy condition, I should attach to the mass a spring-and-dashpot so that I can use the simple equation of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{x} + 2\zeta\omega\dot{x} + \omega^2x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you look at the link I've provided, I'm having a whole host of problems reproducing the two required conditions in my model. I think something is wrong with how I've modeled the system, but I don't quite know what's going wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated. Further, if anyone knows of papers that describe modeling of one-axis gimbal systems (I've found &lt;em&gt;plenty&lt;/em&gt; on two-axis gimbal systems, but those are way too complicated for what I'm trying to accomplish here), that would also be helpful.&lt;/p&gt;&#xA;" OwnerUserId="5218" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-03-02T15:03:34.987" Title="Approximating a Gimbal using Springs and Dampers" Tags="|mechanical-engineering|dynamics|systems-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7710" PostTypeId="2" ParentId="7699" CreationDate="2016-03-01T23:01:36.960" Score="1" Body="&lt;p&gt;To dampen vibrations the easy way, you need lots of mass and some kind elastomer/damper to convert vibrations to heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of making this as a single piece would be casting a soft elastomer in a mold filled with lead shot.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2016-03-01T23:01:36.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7711" PostTypeId="2" ParentId="5750" CreationDate="2016-03-02T07:10:02.557" Score="1" Body="&lt;p&gt;There is an option to retain all the solutions. Look in the parameter table for a check box. After you have it checked for all design points, run the simulation and it will create multiple project files of the form:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;name_dp0&#xA;name_dp1&#xA;name_dp2&#xA;name_dp3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;open each one to get the results. Or, find all the &lt;code&gt;.mechdb&lt;/code&gt; files, rename them as &lt;code&gt;.mechdat&lt;/code&gt; and import them into a single workbench project file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if you are savvy (which I am not) you can use &lt;code&gt;IronPython&lt;/code&gt; scripting to drive the export of the csv files from each project.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2016-03-02T07:10:02.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7712" PostTypeId="2" ParentId="7703" CreationDate="2016-03-02T13:03:20.873" Score="1" Body="&lt;p&gt;There is a link to the company's &lt;a href=&quot;https://www.sparkfun.com/datasheets/Sensors/Pressure/fsrguide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;integration guide&lt;/a&gt; on the page you linked which is very explanatory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These sensors show a decrease in resistance with increasing applied force.  I have no idea what you mean by &quot;connected to the breadboard (with resistor)&quot;, but the most simple way to read this device would be to connect it to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Multimeter&quot; rel=&quot;nofollow noreferrer&quot;&gt;multimeter&lt;/a&gt; set to read resistance.  Once you know the resistance you can calculate the approximate force from the curve on page 5 of the &lt;a href=&quot;https://www.sparkfun.com/datasheets/Sensors/Pressure/fsrguide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;integration guide&lt;/a&gt; (shown below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you are reading 500 Ohms of resistance across the force sensor, then the approximate mass on the sensor (force $\neq$ mass, argh) is 4 kg.  In order to get a more accurate measurement though, you will need to calibrate the sensor with a range of known masses since every sensor produced by the company will be slightly different.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sparkfun.com/datasheets/Sensors/Pressure/fsrguide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qeEX8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2016-03-02T13:03:20.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7713" PostTypeId="2" ParentId="7709" CreationDate="2016-03-02T14:52:48.093" Score="3" Body="&lt;p&gt;In my &quot;day job&quot; I do a lot of (mostly nonlinear!) dynamics modelling and simulation of mechanical systems. One thing that always raises a &quot;red flag&quot; for me is degrees of freedom with no mass, especially if they are at places where  somebody is trying to apply a load to the model, as you have done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that in the physical structure there is a rod of shaft connecting the &quot;payload&quot; (i.e. the mass $m$) to the &quot;motor&quot; (which applies the force $F$ ). The shaft will have a torsional stiffness (modelled by your $k$) but it also has some mass. The rotating part of the motor also has mass - possibly more than the connecting shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model should be capable of representing the basic behaviour of the system if you remove all the damping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With no damping, your model has no &quot;resonant frequency&quot; at all. It is just a disconnected mass, with a massless spring attached to it. Trying to create a non-zero-frequency vibration mode from nowhere by adding the dampers isn't likely to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you add a second mass at the point where the force is applied, representing the real mass of the shaft and the rotating part of the motor, then (again ignoring the dampers) you now have a two-degree-of-freedom system with two modes. One mode is a zero-frequency mode corresponding to free rotation at constant angular velocity, the other is your &quot;resonance&quot; with a non-zero frequency $\omega$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two modes now represent the two &quot;interesting&quot; things that physical structure does, and two dampers will act to control those two modes. It's a reasonable guess that $c$ will mainly damp out the rigid body rotation, and $d$ will mainly damp out the unwanted vibration in the drive shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The $m$ and $k$ in your model would create a vibration mode if you applied a fixed-amplitude &lt;strong&gt;displacement&lt;/strong&gt; to the end of the spring, instead of a &lt;strong&gt;force&lt;/strong&gt;. That would be equivalent to mounting $m$ and $k$ on a vibration testing table and shaking the end of the spring. But if you try to use the model in that way, the damper $c$ is irrelevant, because it has a prescribed displacement at both ends (one end fixed, the other end moving). But the way I interpret your diagram and description, $F$ is meant to represent a force, not a displacement.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-03-02T15:03:34.987" LastActivityDate="2016-03-02T15:03:34.987" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7714" PostTypeId="1" CreationDate="2016-03-02T14:53:11.330" Score="2" ViewCount="1556" Body="&lt;p&gt;You might see in &lt;a href=&quot;http://billionsinchange.com/solutions/limitless-energy&quot; rel=&quot;nofollow&quot;&gt;this short, 2-minute video&lt;/a&gt; that billionaire Manoj Bhargava talks about creating cables from graphene to conduct heat from the underground sections of the earth to the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as per the properties of graphene, it is only a good conductor if the number of layers is one or very few. Once you start stacking it up, it is just normal graphite and it's conductivity gets reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's proposed in the video is a cable made from graphene. My question is how is it possible to make a graphene rope without stacking up graphene layers (and thus ending up with graphite) or without converting it into nanotubes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to know why stacking up graphene reduces it's conductivity? Why don't the layers behave independent of each other?&lt;/p&gt;&#xA;" OwnerUserId="5294" LastEditorUserId="1832" LastEditDate="2016-03-02T15:05:05.403" LastActivityDate="2019-12-22T19:01:00.147" Title="How can one make graphene cables?" Tags="|thermal-conduction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7715" PostTypeId="2" ParentId="7704" CreationDate="2016-03-02T15:01:31.430" Score="2" Body="&lt;p&gt;Following Paper gives Corrosion rate for 304 SS in Ammonia + Water Vapor environment:- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Louis Caruso, Harold Mitchels,&quot;Resistance of copper-nickel alloys to ammonia corrosion in simulated steam condenser environments&quot; in &lt;em&gt;Proceedings of American Power&lt;/em&gt;, vol. 4, 1980, pp. 319-323.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this literature does not consider corrosion at high temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally SS wont corrode at Room temperature. So you may have to look for reference that considers both high temp and ammonia. Following might help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nickelinstitute.org/TechnicalLiterature/AISI/9013_StainlessSteelsinAmmoniaProduction.aspx&quot; rel=&quot;nofollow&quot;&gt;https://www.nickelinstitute.org/TechnicalLiterature/AISI/9013_StainlessSteelsinAmmoniaProduction.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have not read it completely but it discusses stainless steels from process equipment in ammonia production plants.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following references should help too (though you have to go through them to check their usefulness and these references won't be available freely ):-&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;MTI, Materials Selector for for Hazardous Chemicals, MS-6: Ammonia and Caustic Soda&lt;/li&gt;&#xA;&lt;li&gt;NACE Conference Paper 03523 Stress Corrosion Cracking of Austenitic Stainless Steel in Fertilizer Plants&lt;/li&gt;&#xA;&lt;li&gt;NACE Conference Paper 02439 External SCC Phenomena of Austenitic Stainless Steel in Urea Plants&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope this helps,&lt;/p&gt;&#xA;" OwnerUserId="4949" LastActivityDate="2016-03-02T15:01:31.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7716" PostTypeId="1" CreationDate="2016-03-02T15:48:46.333" Score="6" ViewCount="218" Body="&lt;p&gt;A well-known method to measure sheet resistance is the van der Pauw method. Looking at &lt;a href=&quot;https://en.wikipedia.org/wiki/Van_der_Pauw_method&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; they state that the cloverleaf configuration is the most advised. Two questions naturally arise:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why is this the best configuration?&lt;/li&gt;&#xA;&lt;li&gt;Why is it a cloverleaf and not a fully &quot;filled&quot; circle?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2437" LastEditorUserId="1832" LastEditDate="2016-03-02T18:00:29.490" LastActivityDate="2022-12-24T16:45:08.750" Title="Thin film sheet-resistance (van der Pauw method)" Tags="|electrical-engineering|systems-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7718" PostTypeId="1" AcceptedAnswerId="7719" CreationDate="2016-03-02T19:11:53.217" Score="8" ViewCount="7367" Body="&lt;p&gt;I was wondering about the main difference between using heat pipes and a fluid-based cooling system. In both cases the heat is transported away from the heat-generating area to a place where the heat is dissipated. What is now the advantage of using a fluid instead of heat pipes? Is it the lack of heat transfer points (a fluid has no heat exchange points, except of the radiator and thermal pad, while heat pipes have to be connected)?&lt;br&gt;&#xA;Thus, when is it more beneficial to use a fluid cooling system instead of heat pipes?&lt;/p&gt;&#xA;" OwnerUserId="2844" LastActivityDate="2018-09-08T22:46:21.997" Title="Heat pipes vs fluid cooling" Tags="|heat-transfer|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7719" PostTypeId="2" ParentId="7718" CreationDate="2016-03-02T21:19:10.370" Score="7" Body="&lt;p&gt;Heat pipes are very useful when you have limited space to work with or you want an entirely sealed and self contained unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting example of this is that heat pipes are used to cool the legs of oil pipelines which run through permafrost areas, the cooling is required to stop the heat generate by friction in the pipe and solar heating from conducting through the support legs and melting the ground under them. Here the advantage is that they are small, simple and require no power supply and little maintenance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly heat pipes are now a pretty standard component in CPU coolers. For laptops they may be the only way to package adequate cooling for modern processors and even for desktops they make it more convenient to design and package high capacity coolers by making it possible to move the heat-sink away from the processor itself for greater size and better airflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that water cooling for high performance desktop PCs is becoming increasingly mainstream so there is a certain amount of overlap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid cooling becomes more attractive as the heat loads get larger and you get beyond the limitations of needing to miniaturise pumps.  It also has the advantage that it is reasonably simple to design a custom system using fluid cooling in terms of piping routes and radiator and fan capacities using fairly standard off the shelf components. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that if you have large and geometrically complex components with large volumes it may be easier to have channels running through their existing internal structure than to try to find surfaces to attach heat pipes and you can cool a large and complex structure with a centralised pump and radiator eg an engine block. Similarly once you have the basic elements of a liquid cooled system it is very scalable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage of fluid systems is that they work across a large range of temperatures, eg water cooling will work between 0C and 100C and that range can be further extended with additives and it is usually fairly easy to control the system to get a target temperature at various points in the system by the simple expedient of controlling the fluid flow-rate. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-09-08T22:46:21.997" LastActivityDate="2018-09-08T22:46:21.997" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="7720" PostTypeId="1" CreationDate="2016-03-02T21:38:49.863" Score="5" ViewCount="3967" Body="&lt;p&gt;I was told that the proportional gain alone cannot the drive the error signal to zero in a feedback loop. Why is that? If there is a tiny error, there will be a tiny or larger (depending on the number of proportional gain) correction, so why won't the error signal reach zero? When the error tends to zero, the correction will tend to zero as well. But just before error becomes zero, there WILL BE corrective action, so why it cannot achieve the desired output?&lt;/p&gt;&#xA;" OwnerUserId="4846" LastEditorUserId="272" LastEditDate="2016-03-03T19:11:09.850" LastActivityDate="2016-03-03T20:08:12.020" Title="Why can't proportional gain alone, reduce the error to zero?" Tags="|electrical-engineering|control-theory|feedback-loop|pid-control|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7721" PostTypeId="1" CreationDate="2016-03-02T22:33:47.840" Score="6" ViewCount="334" Body="&lt;p&gt;I have been sent a clip of a drawing with the following annotations on it, but I've never seen them before and can't figure out what they mean. Is there a standard interpretation of these callouts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9y3Bj.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9y3Bj.png&quot; alt=&quot;Example 1&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2q5j0.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2q5j0.png&quot; alt=&quot;Example 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially I thought that they might be some metric notation for I beams with the height shown in mm, but weld callouts on the drawing use fractional inches and american electrode designations. Furthermore the members they refer to are rectangular and square tubes, not I shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some ambiguity if the material in question is steel or aluminum (an email said aluminum but the electrodes for welding are steel.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will obviously also contact the person who provided the drawing, but they got it third hand so it may take a long time to get to the source.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-03-02T22:33:47.840" Title="I-beam annotation for tube on drawing" Tags="|drafting|standards|documentation|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7722" PostTypeId="2" ParentId="7720" CreationDate="2016-03-02T23:16:36.013" Score="1" Body="&lt;p&gt;Because by definition, to have a non-zero output, you must have a non-zero error. This means that the output cannot match the setpoint perfectly. Only by adding an integrator can you drive the error to zero while maintaining a non-zero output.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-03-02T23:16:36.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7723" PostTypeId="1" CreationDate="2016-03-03T00:22:33.423" Score="2" ViewCount="154" Body="&lt;p&gt;Below is a picture 4-wheel vehicle steering linkage. I figured that in order to steer correctly $(W+D)tan\theta_1=Dtan\theta_2$ where $\theta_1$ is the angle of the outside front wheel, $\theta_2$ is the angle of the inside front wheel, $W$ is the distance between the two front wheels, and $D$ is the radius of curvature of the arc the rear inside wheel follows. The front wheels are the ones at the bottom of the picture. How can I figure out what lengths of the segments of the linkage will satisfy the equation for all values of $\theta_1$, $\theta_2$, and $D$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jEkaT.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jEkaT.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5299" LastEditorUserId="5299" LastEditDate="2016-03-03T01:30:44.407" LastActivityDate="2016-03-03T14:56:21.757" Title="Lengths of Connectors in Steering Linkage" Tags="|automotive-engineering|linkage|geometry|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7724" PostTypeId="2" ParentId="7723" CreationDate="2016-03-03T01:03:50.960" Score="2" Body="&lt;p&gt;To find the lengths of the components of the linkage mechanism you are going to have to use trigonometry. I'd envisage doing so for the case when both $\small\sf\theta_1$ and $\small\sf\theta_2$ are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the lengths of the components of the linkage on each side of the vehicle are the same do the calculations for one side of the vehicle and use the overall distance of $\small{W/2}$. For that however, you will need to know the angles between the linkage components for when $\small\sf\theta_1$ and $\small\sf\theta_2$ are zero.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2016-03-03T13:22:39.090" LastActivityDate="2016-03-03T13:22:39.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7725" PostTypeId="2" ParentId="7720" CreationDate="2016-03-03T02:14:59.803" Score="0" Body="&lt;p&gt;Consider what happens if the desired response is not constant - for example a vehicle doing a controlled acceleration or braking (i.e. not just the &quot;floor the gas pedal till you hit the speed limit&quot; style of driving!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way for proportional gain to make a correction is to lag behind the desired response. To reduce the errors, You want something in the controller that measures the way things are &lt;em&gt;changing&lt;/em&gt; with time.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-03-03T02:14:59.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7726" PostTypeId="1" AcceptedAnswerId="7727" CreationDate="2016-03-03T03:14:46.340" Score="6" ViewCount="1953" Body="&lt;p&gt;I have the following to wheels that rotate on the same axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fscJy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fscJy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that they are stuck together, if I rotate the green one the red one rotates at the same speed and when I stop the green one the red one stops. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm looking for the name of the mechanism that makes it so that when I stop rotating the green wheel and I hold it in place, the red one keeps turning.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm building a flywheel (represented here by the red wheel) and the green wheel here is connected with a belt to a crank. I want to make it so that the flywheel keeps turning even when the crank is fully stopped.&lt;/p&gt;&#xA;" OwnerUserId="5300" LastEditorUserId="272" LastEditDate="2016-03-03T12:46:13.357" LastActivityDate="2016-03-08T12:07:26.553" Title="What do you call a mechanism where rotation is transfered in one direction only?" Tags="|mechanical-engineering|terminology|flywheels|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7727" PostTypeId="2" ParentId="7726" CreationDate="2016-03-03T03:27:04.260" Score="5" Body="&lt;p&gt;It's called a &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Freewheel&quot;&gt;freewheel&lt;/a&gt;&quot; or &quot;overrunning clutch&quot;.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-03-03T03:27:04.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7728" PostTypeId="1" AcceptedAnswerId="7742" CreationDate="2016-02-25T14:35:57.193" Score="2" ViewCount="906" Body="&lt;h3&gt;Bulding&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I live in a new (~10 years old) 2 story duplex home in the Eastern United States.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a completely interior bathroom on the first floor. It is between 2 other (exterior) rooms, the garage, and the firewall. It has a shower (plastic covering, not tile) but no tub.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also have an unfinished basement. The foundation is concrete slab with CMU walls, which extend about 18 inches above ground level. There is a small (I think it's called a &quot;port&quot;) window in one corner. At the opposite corner is a steel door that opens to a poured concrete stairwell topped with Bilco doors. It's an open space, I'd estimate a little under 1500ft&lt;sup&gt;2&lt;/sup&gt;. We use about 2/3 of the space for general storage (decorations, unused furniture, food, the usual stuff) and 1/3 for my woodshop (hand tools, power tools, lumber, etc).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Risk&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://www.tornadohistoryproject.com&quot; rel=&quot;nofollow&quot;&gt;http://www.tornadohistoryproject.com&lt;/a&gt;, my County has seen 31 recorded tornadoes since 1950, with the following breakdown by scale:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;| Fujita Scale | Number |&#xA;| ------------ | ------ |&#xA;| 3            | 1      |&#xA;| 2            | 11     |&#xA;| 1            | 13     |&#xA;| 0            | 6      |&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(the Tornado History Project gets it's data directly from NOAA but uses the term &quot;Fujita Scale&quot; consistently - I don't know whether that means all their data after 2007 is actually scaled on EF or not - they don't seem to specify).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Advice&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The general advice is always &quot;get to a basement or an internal room on the lowest floor.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense. Roofs and exterior walls are for obvious reasons the first surfaces to be damaged and destroyed. Upper floors are more susceptible to structural damage. Direct wind and debris from tornadoes also is much less likely to extend below ground level, making basements in general a good place to be. In a lot of cases (as in mine) the basement is used as storage for  food and/or provisions - helpful in the event of being trapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also clear why fully interior rooms are an acceptable alternative - they put more structure between you and the storm. Since wind-induced structural failures happen from the outside in, they're also the rooms most likely to be left standing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Questions&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Are there any other factors to consider?&lt;/strong&gt; What are some other risk/benefits that I've not considered?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Speaking in the most general sense, if you have both options, &lt;strong&gt;does one tend to be ubiquitously safer&lt;/strong&gt;, overall?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Speaking specifically to my home, given that A) the historical risk of major structural failure is low and B) the basement is wide open and filled with sharp would-be missiles, &lt;strong&gt;which would you recommend for my family&lt;/strong&gt;, and why?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5329" OwnerDisplayName="Chris Tonkinson" LastEditorUserId="33" LastEditDate="2016-03-03T14:16:22.103" LastActivityDate="2016-03-03T14:29:31.323" Title="Tornado Safety: Basement or first floor interior room?" Tags="|structural-engineering|safety|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7729" PostTypeId="2" ParentId="7702" CreationDate="2016-03-03T04:24:12.303" Score="3" Body="&lt;p&gt;There are many variables, but much of your system can be measured empirically with a cars internal sensors and OBD-II system (or possibly manufactures data, but I wouldn't hold my breath). For RPM there are two scenarios you would have to independently model; &lt;strong&gt;clutch engaged&lt;/strong&gt; and &lt;strong&gt;clutch disengaged&lt;/strong&gt;.&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Clutch engaged:&lt;/strong&gt; As hazzey mentioned in the comments, the engine will turn at a gear ratio of the wheel speed. Take your RPM and put it into the torque curve(for your given throttle position) to give you torque. Torque times your gear ratio and tire radius gives you force. Then solve F=MA with the mass of the vehicle to give you acceleration. Then use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Equations_of_motion#Uniform_acceleration&quot; rel=&quot;nofollow&quot;&gt;kinematic equation v=a*t+v0&lt;/a&gt; to solve for your new velocity based on your current velocity and time step. Which then is used to calculate your new RPM... and so goes the iteration. Similarly you could solve it with kinetic energy equations instead.&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Clutch disengaged:&lt;/strong&gt; This case can be modeled by considering energy input and internal friction of the engine. When you stop pressing on the gas, even without the wheels on the ground, the system will slow down. You could measure this for a particular car empirically by graphing RPMs when throttle input is removed. The only difference between the wheels off the ground and the clutch disengaged is the amount of inertia or kinetic energy that is stored in the spinning system and additional friction imposed by the other spinning components. This could also be determined empirically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take RPM to the torque curve(for your given throttle position) which gives you torque.&lt;br&gt;&#xA;Torque * RPM = power input to the free spinning engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power input = internal friction + kinetic energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Internal friction is calculated from an empirical curve based on RPM. Then solve for kinetic energy input. Then use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Moment_of_inertia&quot; rel=&quot;nofollow&quot;&gt;moment of inertia&lt;/a&gt; (found empirically for a specific engine) to solve for your new RPM... put that back in again and iterate.&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;Note that the engine friction is still obviously present in the clutch engaged model as well, but is probably insignificant compared to other forces and coefficients (drag for example). Its usually best to get the model running in a simple mode first. You can always increase the accuracy of your model later if necessary. Every where I say &quot;empirically&quot; can be replaced by a fudge factor(an assumed/guessed coefficient) for initial model testing. For a highly accurate model however you will need highly accurate data.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-03T04:24:12.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7730" PostTypeId="2" ParentId="7726" CreationDate="2016-03-03T04:42:42.863" Score="5" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Freewheel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Freewheel Clutch&lt;/a&gt; or&lt;br&gt;&#xA;&lt;a href=&quot;http://www.bocabearings.com/bearing-inventory/One-Way-Bearings/19134/3x6-5x6-HF0306KF&quot; rel=&quot;nofollow noreferrer&quot;&gt;One-Way-Bearing&lt;/a&gt; or&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sprag_clutch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sprag Clutch:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fk9ij.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fk9ij.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;More information:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.gmnbt.com/free-wheel-clutches.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Freewheel Clutch Types&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.gmnbt.com/sprag-clutches.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sprag Clutch Information&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.konaflex.fi/datafiles/userfiles/onet/fi/attachments/Freewheel_Clutch.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cross+Morse Freewheel Clutch Catalog&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-03T04:42:42.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7731" PostTypeId="2" ParentId="7720" CreationDate="2016-03-03T05:58:19.443" Score="0" Body="&lt;p&gt;&lt;em&gt;//begin edit//&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading &lt;a href=&quot;https://engineering.stackexchange.com/a/7740/2819&quot;&gt;Chris Mueller's answer&lt;/a&gt; I have a better understanding for the misconceptions associated with this question (my past self included).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When a positive controller output is required to hold the systems current state; it will never reach zero error&lt;/strong&gt; because the controller needs the error term to compute this positive output as shown in Chris' proof. This includes this assumption of the process side of the loop being; y(s)= P * u(s). This positive output is required for most real world systems and is referred to as &lt;a href=&quot;https://en.wikipedia.org/wiki/PID_controller#Steady-state_error&quot; rel=&quot;nofollow noreferrer&quot;&gt;steady state error&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;strong&gt;when controller output is not required to hold the systems current state; it is possible to reach zero error&lt;/strong&gt; because the controller only needs to output the change required. For positioning with a simple hydraulic cylinder controlled with a proportional flow valve for example, no output is required to hold position. This would also apply in other special cases like heating an adiabatic (perfectly insulated) container. The process side of the loop in these cases would perhaps be; y(s)= P + u(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in a &quot;state-holding system&quot; only... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;//end edit//&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;...at infinity, yes, you will reach your setpoint. However, most of us don't have until infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good analogy is; you have a destination you would like to go to, like the grocery store for example (this is your setpoint). You decide to walk half of your current distance to the grocery store each day (this is overdamped, underdamped has a whole different set of problems)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Day one you are half way, &quot;great progress!&quot;. Day two you are 3/4 of the way, &quot;not bad&quot;. Day three you are 7/8 of the way, &quot;thought I would be there by now&quot;. Day four you are 15/16 of the way, &quot;I sure am hungry&quot;. Day five, your are 31/32 of the way, &quot;I give up&quot;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GjHFR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GjHFR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on what you consider &quot;at the setpoint&quot; and what you consider &quot;too much time&quot;. In this scenario we died of starvation in the parking lot of the grocery store.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;//begin edit//&lt;/em&gt;&#xA;Lets say a company you work for makes $100 a second when the internal temperature of an adiabatic (perfectly insulated) container is brought to within 0.1°C the set point. &#xA;&lt;em&gt;//end edit//&lt;/em&gt;&#xA;We could wait an hour for a proportional term to get us there and lose all that money, or we use a PI that could get us there in 2 minutes, or we could nicely tune a PID to get us there in 10 seconds.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-03-03T20:08:12.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7732" PostTypeId="1" AcceptedAnswerId="7735" CreationDate="2016-03-03T06:17:12.893" Score="1" ViewCount="1133" Body="&lt;p&gt;How strong is the adhesion between Ultra High Molecular Weight Polyethylene(UHMWPE) Fiber and Epoxy Resin normally? If we take the UHMWPE fiber sheet as it is without subjecting it to any chemical or plasma treatment (for improved surface bonding between UHMWPE fiber and Epoxy resin), can a laminate be formed through compression molding? Will there be any issues or  loss of mechanical properties like impact resistance for UHMWPE?&lt;/p&gt;&#xA;" OwnerUserId="4775" LastEditorUserId="33" LastEditDate="2016-03-03T14:08:38.723" LastActivityDate="2016-03-03T14:08:38.723" Title="How strong is the adhesion between UHMWPE Fiber and Epoxy Resin?" Tags="|materials|composite|composite-resin|adhesive|epoxy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7733" PostTypeId="1" AcceptedAnswerId="7739" CreationDate="2016-03-03T06:21:07.887" Score="1" ViewCount="5391" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/34p8p.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/34p8p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the horizontal and vertical components of the reactions at the supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the horizontal force for the pin by adding the forces to 0 and I know I can find the vertical reaction of the pin by summing the forces in the y direction. But I get stuck on what do do to find the vertical reaction at the roller support which I need to calculate the vertical force at the pin. If I did a moment calculation at the pin do I count the 10kN force at the top?, don't you only count the stuff that's only in a straight line from the place you're calculating the moment from?.&lt;/p&gt;&#xA;" OwnerUserId="5303" LastActivityDate="2018-11-02T00:59:01.573" Title="Horizontal and vertical forces on supports" Tags="|civil-engineering|statics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7734" PostTypeId="2" ParentId="7718" CreationDate="2016-03-03T06:38:37.593" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_pipe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heat pipes&lt;/a&gt; use a fluid that phase changes at a temperature between the two temperatures it is transporting between. It evaporates at one end, then condenses at the other, the fluid then runs via gravity or wicked via a membrane to the other end where it evaporates again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are low maintence as there are no moving parts, and when they are well designed, they can be quite effective. They are very effectivly used in evacuated tube solar thermal systems that actually just use water and propylene glycol as the working fluid for that temperature range.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/G37at.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G37at.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The down side to heat pipes is that they only work well around that designed phase change temperature. Above that, all the working fluid is gas and an insulator, below that and most of the working fluid is liquid that doesn't move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Pumped&quot; or &quot;Active&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Water_cooling&quot; rel=&quot;nofollow noreferrer&quot;&gt;liquid cooling&lt;/a&gt; relies on the high heat capacity of a fluid (usually water or water with glycol). These systems have moving parts, but they can transfer a lot of thermal energy over a wider range of temperatures. Also for a given footprint (like under the hood of your car), active liquid cooling can transfer a lot more heat than an equivalently sized heat pipe system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently it is usually cheaper to use heat pipes in small or very repeatable applications, and active fluid cooling for large amounts of heat transfer and more custom situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even evacuated tube solar collectors still use a pumped fluid system to transfer the heat to the hot water tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dk0ns.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dk0ns.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-03T06:38:37.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7735" PostTypeId="2" ParentId="7732" CreationDate="2016-03-03T06:48:14.243" Score="2" Body="&lt;p&gt;I am not a chemist or material scientist, but it is my experience that reliable adhesion to a smooth Polyethylene surface is nearly impossible even with &quot;plastic&quot; rated epoxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However where it is a fiber sheet, the epoxy, even with zero surface adhesion, will mechanically hold each fiber it encapsulates. I can not speak to what mechanical properties will result as it would be highly dependent on the fiber geometry. Some online research and testing is probably your best route.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-03T06:48:14.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7736" PostTypeId="1" AcceptedAnswerId="29194" CreationDate="2016-03-03T10:07:57.973" Score="12" ViewCount="3386" Body="&lt;p&gt;Is there some sort of equivalent standardized schematical (topological) diagram -such as Piping and Instrumentation Diagram, Electrical Schematics, Process Flow Diagrams, etc.- for the realm of mechanical engineering? I just need to draw the overall architecture that would describe the main components of the system and how they are connected, without going into too much detail. I have seen some diagrams but they seem to be ad hoc, as they have no fixed resemblance to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O93J7.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O93J7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: I'm looking for something in this complexity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there is more than one way to do this but it would be nice to be more universally understood. Looked into OMG SysML but that didt at first glance seem to have what i was looking for.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="5400" LastEditDate="2016-03-28T06:37:08.157" LastActivityDate="2019-07-16T09:45:22.277" Title="Schematic Diagrams for Mechanical Systems" Tags="|mechanical-engineering|drafting|standards|diagram|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7737" PostTypeId="2" ParentId="7733" CreationDate="2016-03-03T10:21:46.210" Score="0" Body="&lt;p&gt;Take&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sum of horizontal forces: horizontal component of the 10 kN force and one horizontal reaction force.&lt;/li&gt;&#xA;&lt;li&gt;Sum of vertical forces: vertical component of the 10 kN force and two vertical reaction forces.&lt;/li&gt;&#xA;&lt;li&gt;Sum of moments around left support pin (dashed lines in following picture). There is a contribution by the 10 kN force (project it perpendicular to the connection line between its point of action and the left support pin) and a contribution by the vertical reaction force in the rolling support (right).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7NpQU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7NpQU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3633" LastEditorUserId="3633" LastEditDate="2016-03-03T12:16:02.130" LastActivityDate="2016-03-03T12:16:02.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7738" PostTypeId="1" AcceptedAnswerId="7744" CreationDate="2016-03-03T10:23:55.467" Score="1" ViewCount="1545" Body="&lt;p&gt;I'm trying to analyze a simple shaft supported by two bearings and subject to a single point load. But I keep coming across assumptions that only consider the vertical reactions and neglect the moment reactions of the bearings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://faculty.ccri.edu/rpreliasco/website/engr%202050_files/hw9.pdf&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; states this assumption in examples: 5-73 and 5-86. I've tried to find an explanation for this but keep finding examples that just state the assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm at a loss not having my textbooks with me at the moment but from looking at examples I've noticed in some cases that the bearings take the form of a simple pin support hence the bending moment is zero at those points. See &lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Drive/Shaft_design.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for an example (although no assumption stated).&lt;/p&gt;&#xA;" OwnerUserId="5308" LastEditorUserId="5308" LastEditDate="2016-03-03T11:48:03.577" LastActivityDate="2016-03-03T14:39:54.863" Title="Bearing Alignment Moment Reactions" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7739" PostTypeId="2" ParentId="7733" CreationDate="2016-03-03T12:53:35.943" Score="2" Body="&lt;p&gt;You don't always have to use moments to calculate reactions. In this situation it's possible just by balancing horizontal and vertical forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have one force at an angle. With all our reactions at right angles, you should convert this force into it's two resultants (7.1kN in both the vertical and horizontal direction). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is only a single horizontal reaction so this must solely resist the 7.1kN horizontal load. Therefore this reaction is 7.1kN. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other force is 7.1kN upwards. You can proportionally divide this between the two supports, as we know that the closer support will take more of the load. The proportion would simply be 3/4 to the pin and 1/4 to the roller as we know the distances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing 7.1kN into these proportions means the rollers vertical reaction is 1.76kN downwards and the pin is 5.33kN downwards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a final check, taking moments around any point will give a result of 0 which is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MBza9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MBza9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3093" LastEditorUserId="3093" LastEditDate="2016-03-04T09:24:20.177" LastActivityDate="2016-03-04T09:24:20.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7740" PostTypeId="2" ParentId="7720" CreationDate="2016-03-03T13:13:49.093" Score="5" Body="&lt;p&gt;To understand why proportional gain won't drive the error point to zero, it is best just to look at the math.  Consider the PID loop shown in the image below.  The loop algebra in the $s$ domain comes out to&#xA;$$&#xA;\begin{align}&#xA;  e(s)&amp;amp;=r(s)-y(s)\\&#xA;  y(s)&amp;amp;=P\ u(s)\\&#xA;  u(s)&amp;amp;=\left(k_p+\frac{k_i}{s}+k_ds\right)e(s),&#xA;\end{align}&#xA;$$&#xA;where I have used $P$ to represent the plant/process.  Putting all of these together gives&#xA;$$&#xA;e(s)=\frac{r(s)}{1+P\left(k_p+\frac{k_i}{s}+k_ds\right)}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we now turn off the integral and derivative terms by setting $k_i=k_d=0$, then this becomes&#xA;$$&#xA;e(s)=\frac{r(s)}{1+Pk_p}.&#xA;$$&#xA;The only way to make this zero is to have &lt;strong&gt;infinite proportional gain&lt;/strong&gt;, i.e. $k_p=\infty$, regardless of the frequency/timescale unless the plant itself has a term which becomes infinite at low frequency.  If we add the integrator back in by only setting $k_d=0$, then the error point becomes&#xA;$$&#xA;e(s)=\frac{r(s)}{1+Pk_p+\frac{Pk_i}{s}}.&#xA;$$&#xA;At low frequencies $s\rightarrow0$ which makes the last term in the denominator become infinite which makes the error point go to zero.  So, we can see that &lt;strong&gt;an integrator drives the error point to zero at low frequencies&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:PID_en_updated_feedback.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DioUS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2016-03-03T18:35:03.820" LastActivityDate="2016-03-03T18:35:03.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7741" PostTypeId="2" ParentId="7723" CreationDate="2016-03-03T13:15:50.943" Score="1" Body="&lt;p&gt;This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Ackermann_steering_geometry&quot; rel=&quot;nofollow&quot;&gt;Ackerman Steering&lt;/a&gt;. In an ideal case a line drawn through the wheel pivot point and steering arm pivot point will cross the centre point of the rear axle when the steering is centered. In practice other dynamic effects like roll etc may mean that the best position for this point will vary a bit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the steering arms are the right angle to the hub you can still have toe in or out by having adjustable length track rods. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="86" LastEditDate="2016-03-03T14:56:21.757" LastActivityDate="2016-03-03T14:56:21.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7742" PostTypeId="2" ParentId="7728" CreationDate="2016-03-03T14:29:31.323" Score="4" Body="&lt;p&gt;The Wikipedia page on &lt;a href=&quot;https://en.wikipedia.org/wiki/Tornado_intensity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tornado Intensity&lt;/a&gt; has example pictures of damage.&lt;/p&gt;&#xA;&lt;p&gt;Your location has mostly EF1 and EF2 tornados historically (I'm assuming EF scale).&lt;/p&gt;&#xA;&lt;h3&gt;EF1&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XipYam.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XipYam.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;EF2&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VyuIAm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VyuIAm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;EF3&lt;/h3&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/14cIHm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/14cIHm.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Result&lt;/h3&gt;&#xA;&lt;p&gt;Just looking at those pictures, I wouldn't want to be above ground if the basement is an option. You don't want to make a decision based on how strong you think the tornado is. Just go to the basement if it is possible.&lt;/p&gt;&#xA;&lt;p&gt;It isn't likely that the wind will be strong enough to throw your tools around in the basement. And if it is, I certainly wouldn't want to be above ground.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-03-03T14:29:31.323" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7744" PostTypeId="2" ParentId="7738" CreationDate="2016-03-03T14:39:54.863" Score="1" Body="&lt;p&gt;Many bearings (single row ball bearings for example) are meant to allow slight misalignment of angle without significant counteracting torque. This allows them to be used in situations where there are two bearings supporting a shaft without having to have the alignment between the two bearing supports be extremely critical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that unless there is significant bending in the shaft, the torque applied by the bearing would be negligible compared to the side or axial loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many bearings that are meant to handle torque (double row angled ball bearings for example) are in reality composed of two (or more) bearings that are really using radial loads to apply the torque, so even these bearings could be modeled without using bending moments.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2016-03-03T14:39:54.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7746" PostTypeId="1" AcceptedAnswerId="7753" CreationDate="2016-03-03T15:42:45.900" Score="2" ViewCount="1758" Body="&lt;p&gt;I have this figure for my engineering drawing I am currently taking this semester I am quite baffled about this $1\dfrac{1}{4}$ radius circle with $1\dfrac{1}{4}$ Drill hole. How do I do that? as a reference I have this picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qeP7L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qeP7L.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5312" LastActivityDate="2017-06-05T23:51:23.180" Title="What is the Radius of 1 and 1/4 inch drill on a 1 and 1/4 Circle" Tags="|drafting|autocad|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7747" PostTypeId="1" CreationDate="2016-03-03T17:40:15.317" Score="2" ViewCount="18377" Body="&lt;p&gt;How can i calculate the weight bearing capacity of a section of aluminium box section? Known variables of section: Dimension, thickness, distance from load and supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pictorial example might help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2lfS4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2lfS4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if there's anything else needed to solve this problem&lt;/p&gt;&#xA;" OwnerUserId="4281" LastEditorUserId="4281" LastEditDate="2016-03-03T17:51:49.883" LastActivityDate="2016-04-02T19:32:54.817" Title="Calculate Weight Bearing Capacity" Tags="|aluminum|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ClosedDate="2018-05-11T11:35:38.143" ContentLicense="CC BY-SA 3.0" />
  <row Id="7748" PostTypeId="2" ParentId="7747" CreationDate="2016-03-03T18:21:48.813" Score="2" Body="&lt;p&gt;The key issue here is deciding how you want to define 'weight bearing capacity'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you load a beam like this two important things happen firstly the beam deflects under the load and at the same time the stress in the beam increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By convention the 'strength' of a structure is the stress it can withstand before it is permanently deformed (ie does not return to its original shape when the load is removed). This occurs when the maximum stress in the structure exceeds the Yield Stress of the material. For ductile materials like aluminium and steel, this will not necessarily break the beam but it may cause it to fold up and collapse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second consideration is deflection. In some cases deflection may be unacceptable in design terms before yield stress is reached. For example if you want to support a pendulum from your beam with the bottom 10mm above the floor and the beam deflects 11mm you have a problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To actually calculate stress and deflection some &lt;a href=&quot;http://www.engineersedge.com/beam_calc_menu.shtml&quot; rel=&quot;nofollow&quot;&gt;standard formulas&lt;/a&gt; can be used for simple cases. Your example is closest to &lt;a href=&quot;http://www.engineersedge.com/beam_bending/beam_bending2e.htm&quot; rel=&quot;nofollow&quot;&gt;a simply supported beam with a point load&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simple cases these are easy enough to do on paper, or a quick search for beam loading calculator will find an online too to do  the maths for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more complex cases or for more detailed analysis finite element analysis software makes life much easier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will also need to calculate the &lt;a href=&quot;https://en.wikipedia.org/wiki/Second_moment_of_area&quot; rel=&quot;nofollow&quot;&gt;Second Moment of Area&lt;/a&gt; (also called Moment of Inertia) for the beam you are interested in. For common industrial setions (tubes, I-beams etc) you can also just look it up for the particular section you are using. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final consideration is factor of safety. For obvious reasons you don't want a structure to be right on the edge of collapse as all materials have tolerances and all of these calculations are based on assumptions which may or may not be 100% valid for any given case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you want to calculate a safe working load as some percentage of the maximum load before failure. How high a percentage this is depends on how confident you are that you have accounted for all variables and how serious the consequences of failure are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there may be mandatory requirements in your jurisdiction for design of structures for particular applications especially buildings and other safety critical structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also important to add that the above assumes that the loads are static, moving loads can place much greater and more complex stresses on a structure and cyclic loads can also cause long term problems with &lt;a href=&quot;https://en.wikipedia.org/wiki/Fatigue_(material)&quot; rel=&quot;nofollow&quot;&gt;fatigue&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-03T18:34:56.987" LastActivityDate="2016-03-03T18:34:56.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7749" PostTypeId="1" CreationDate="2016-03-03T20:47:17.513" Score="5" ViewCount="166" Body="&lt;p&gt;One dimensional altitude control of a hovercraft. The model's dynamic equations that I came up with are, I treat the hover craft as a rigid particle and my goal is to have it hover at or reach a certain desired height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = ma$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$X(t) - mg = ma$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ma + mg = X(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking Laplace Transform yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m \cdot s^2 \cdot Y(s)  - mg \cdot U(s) = X(s)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y(s) = \frac{X(s)}{m \cdot s^2} + g \cdot U(s)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;strong&gt;X&lt;/strong&gt; is the input thrust and &lt;strong&gt;Y&lt;/strong&gt; is the output and &lt;strong&gt;U(s)&lt;/strong&gt;  is a constant unit step. However I can not seem to isolate for $$Y(s)/X(s)$$. I think it may have something to do with me including the gravity force in the dynamics equation when it may act as a disturbance but I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When creating a dynamic model of a system How do we handle the constant force of $$mg$$ do we treat it as a disturbance, and if so how do we go about doing that?&lt;/p&gt;&#xA;" OwnerUserId="4706" LastEditorUserId="4706" LastEditDate="2016-03-05T02:03:54.950" LastActivityDate="2017-02-14T22:33:00.550" Title="System Identification 1-d altitude control" Tags="|modeling|transfer-function|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7750" PostTypeId="1" CreationDate="2016-03-03T21:54:10.950" Score="2" ViewCount="1824" Body="&lt;p&gt;I'm designing a part, and I'm trying to figure out what distance a slot should be from the bottom of the part.  For simplicity in analysis, I'm approximating the region between the bottom of the slot and the bottom of the part as a beam that is fixed on both sides.  This beam will have a uniform load over the entire span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found this equation to find the maximum deflection of the beam $y_{max}$, given the load $w$, beam length $l$, elasticity modulus $E$, thickness $b$, and distance between the slot and the bottom of the part (or beam height) $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_{max}=-\frac{w l^4}{32bh^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out what $h$ ought to be, so I rewrote it in terms of $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h=-\frac{1}{2}\left(\frac{wl^4}{4Eby_{max}}\right)^{1/3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know what $l$, $E$, and $b$ are, and my plan is to write $h$ as a function of $w$, but I need to eliminate $y_{max}$.  I've tried finding it using the formula for bending stress ($\sigma=My/I$), but that resulted me in an identity for $y_{max}$.  I tried finding $y_{max}$ by solving the ODE $\ddot{y}=M/(EI)$, but that got me the first formula that's above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be awesome to have another formula for $y_{max}$, but a better/different method of determining what $h$ should be would be just as good.&lt;/p&gt;&#xA;" OwnerUserId="1660" LastActivityDate="2021-03-09T13:27:21.737" Title="Maximum deflection of a beam with both ends fixed and distributed load" Tags="|mechanical-engineering|statics|beam|deflection|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7751" PostTypeId="2" ParentId="7750" CreationDate="2016-03-03T22:05:41.660" Score="3" Body="&lt;p&gt;You need to decide whether the design limitation which determines h is max stress or max deflection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you decide deflection is more important insert the value for y which is the maximum allowable deflection for this part of your component. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If stress is more important (ie you don't care how much it moves but you don't want it to break) then you need to use the stress equation not the deflection one and solve for the yield stress (or acceptable working stress) your material.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If  you're not sure which is more important do it both ways and go with the most conservative result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note in this situation for a solid beam with fixed ends and a uniform load, the max tress is just the basic shear stress of [Force/cross sectional area] of the beam. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-03T22:23:27.823" LastActivityDate="2016-03-03T22:23:27.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7753" PostTypeId="2" ParentId="7746" CreationDate="2016-03-03T22:58:04.560" Score="5" Body="&lt;p&gt;The 1-1/4 drill is a reference to the &lt;em&gt;diameter&lt;/em&gt; of the drill. The hole will have a radius of 5/8.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-03-03T22:58:04.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7754" PostTypeId="1" AcceptedAnswerId="7757" CreationDate="2016-03-04T01:43:38.007" Score="0" ViewCount="83" Body="&lt;p&gt;Imagine a structure that is held above the water by pylons that are grounded on the ocean floor. In between these pylons is a pontoon that, when tides are rising holds and lifts a heavy weight. When the tide is falling the weight is supported by the structure, which can gradually let the weight down, and in doing so create electricity. Would this be a practical or even viable way to create electricity? Why or why not?&lt;/p&gt;&#xA;" OwnerDisplayName="user5319" LastEditorDisplayName="user5319" LastEditDate="2016-03-05T18:55:10.780" LastActivityDate="2016-03-05T18:55:10.780" Title="Would this be a practical way to convert tidal energy into electrical energy?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="7" ClosedDate="2016-03-04T12:58:10.053" ContentLicense="CC BY-SA 3.0" />
  <row Id="7757" PostTypeId="2" ParentId="7754" CreationDate="2016-03-04T10:44:52.803" Score="1" Body="&lt;p&gt;I agree with the comments about this question being a poor fit for the site, but one can nevertheless point out the common problem of tidal systems: small amount of vertical movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say that you wanted to build a relatively small, 1 megawatt power plant based on tidal power. You have tidal movements of, say, 2 meters every 6 hours. Thus the weight you would need is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{1 \mathrm{\,MW} \cdot 6 \mathrm{\,h}}{9.8 \mathrm{\frac{m}{s^2}} \cdot 2\mathrm{\,m}} = 10^9 \mathrm{\,kg}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you made it out of steel, it would cost approximately 500 million dollars. And at 10 cents per kWh, this gives a repayment time on investment of 5000 years.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-03-04T10:44:52.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7758" PostTypeId="1" CreationDate="2016-03-04T12:10:12.187" Score="1" ViewCount="43" Body="&lt;p&gt;I have a project that involes electrochromic materials and I am wondering if it's possible for an electrochromic material to turn from white color to black and vice versa. If the answer is yes, can you name that material?&lt;/p&gt;&#xA;" OwnerUserId="5324" LastEditorUserId="1832" LastEditDate="2016-03-04T12:26:54.613" LastActivityDate="2016-03-04T12:26:54.613" Title="Electrochromic materials" Tags="|materials|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7759" PostTypeId="2" ParentId="2920" CreationDate="2016-03-04T12:17:11.417" Score="1" Body="&lt;p&gt;Your math seems to be sound. It looks like you calculated everything correctly up to where you calculated $L$. At that point you should have all of the variables solved for and there should be no need to refine your numbers. &lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2016-03-04T12:17:11.417" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7760" PostTypeId="1" CreationDate="2016-03-04T13:56:14.773" Score="4" ViewCount="3581" Body="&lt;p&gt;How do you design a PID controller for MIMO systems in which the outputs outnumber the inputs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's typical to design a PID controller for MIMO industrial process with the same number of inputs and outputs, or with more inputs than outputs.  And very few references talk about systems with less inputs than outputs.  However, there are lots of chemical process where this is normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any analytical method except (square down ans nrga) to design a PID controller for system with more outputs than inputs?  I've considered a lot of references but haven't found anything!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l5lvR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l5lvR.jpg&quot; alt=&quot;Mimo Sys&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where Mimo sys matrixes are: (Matlab defiend programm)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = [-6.932e-2,17.41,-36.75,0,0,0,-6.0660,-31.54,0;&#xA;    -1.435e-4,2.719e-2,-1.411e-3,3.467e-1,0,-9.380e-1,7.139e-2,-1.691e-2,0;&#xA;    -4.537e-4,1.870e-3,-2.025e-1,0,1,0,-4.688e-2,7.563e-3,0;&#xA;    -1.304e-4,-7.179,-4.916e-1,-6.172e-1,-3.689e-2,7.631e-1,0,0,0;&#xA;    2.297e-5,0,-8.667e-1,4.393e-2,-1.947e-1,-2.026e-2,0,0,0;&#xA;    1.964e-5,4.263e-2,-1.329e-2,1.233e-3,1.579e-2,-1.600e-1,0,0,0;&#xA;    0,0,0,1,1.941e-1,2.771e-1,0,6.258e-2,0;&#xA;    0,0,0,0,8.192e-1,-5.736e-1,-5.612e-2,0,0;&#xA;    0,0,0,0,6.055e-1,8.648e-1,0,2.006e-2,0]&#xA;&#xA;&#xA;B = [0,0,-7.560,9.067e-4;&#xA;     -6.952e-3,1.293e-2,0,0;&#xA;     0,0,-3.425e-2,-9.577e-7;&#xA;     4.249,5.989e-1,0,0;&#xA;     0,0,-1.796,0;&#xA;     -7.287e-2,-2.877e-1,0,0;&#xA;     0,0,0,0;&#xA;     0,0,0,0;&#xA;     0,0,0,0]&#xA;&#xA;&#xA;C = [0,-5.758e-1,0,0,0,0,0,0,0;&#xA;     0,0,0,1,0,0,0,0,0;&#xA;     0,0,0,0,0,1,0,0,0;&#xA;     0,0,1,0,0,0,0,0,0;&#xA;     0,1,0,0,0,0,0,0,0;&#xA;     0,0,0,0,1,0,0,0,0;&#xA;     0,2.719e-2,-1.411e-3,3.467e-1,0,-9.380e-1,7.139e-2,0,0]&#xA;&#xA;&#xA;D = [-1.298e-1,-1.610e-1,0,0;&#xA;     0,0,0,0;&#xA;     0,0,0,0;&#xA;     0,0,0,0;&#xA;     0,0,0,0;&#xA;     0,0,0,0;&#xA;     -6.952e-3,1.293e-2,0,0]&#xA;&#xA; Gss = ss(A,B,C,D)&#xA; G = tf(Gss)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The forth input of Mimo sys is Constant.&lt;/p&gt;&#xA;" OwnerUserId="4412" LastEditorUserId="4412" LastEditDate="2016-03-09T16:18:26.170" LastActivityDate="2016-03-09T16:18:26.170" Title="How to design a PID controller for MIMO systems in which there are more outputs than inputs?" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7761" PostTypeId="2" ParentId="7579" CreationDate="2016-03-04T14:29:26.463" Score="1" Body="&lt;p&gt;1) One if you don't change temperature, infinite if you do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) It can't. Adding or removing heat can only be reversible when the temperature is not changing. This necessitates that both pressure and volume are changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Possible to say, however it would be incorrect. There are indeed an infinite number of reversible paths one could take between two states. (One can alternate between isothermal expansion/compression, and isentropic expansion/compression following any path that those allow). These paths can indeed produce different amounts of work. However, there is no guarantee that these paths will produce more work than an irreversible path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you're missing is that a reversible path can produce the most &lt;strong&gt;and least&lt;/strong&gt; work of all paths &lt;strong&gt;constrained by a highest and lowest temperature&lt;/strong&gt;. Usually what people care about is the work produced (or absorbed) by a &lt;strong&gt;cycle&lt;/strong&gt;. This will be the difference in work between two paths. As such reversible paths offer the highest difference, specifically the reversible path that goes to the highest temperature, then isothermally expands/compresses will have the greatest magnitude work &lt;strong&gt;of all paths that stay below that maximum temperature&lt;/strong&gt;, while the reversible path that cools to the lowest temperature and then isothermally expands/compresses will have the least magnitude of work &lt;strong&gt;of all paths that stay above that minimum temperature&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2016-03-04T14:29:26.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7763" PostTypeId="1" CreationDate="2016-03-04T14:35:30.660" Score="4" ViewCount="2293" Body="&lt;p&gt;From a typical ground investigation borehole log the tarmac thickness is recorded but no tests have been carried out on the road surfacing or formation level. I'm after an approximate bearing capacity of the road surface, including the contribution of the surface itself and the sub-formation under it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously heavy vehicles can be resisted, can I take a typical heavy vehicle and assume a minimum bearing capacity from the bearing stress under the vehicle or is there another/better more reliable method? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't carry out further insitu tests.&lt;/p&gt;&#xA;" OwnerUserId="5325" LastEditorUserId="282" LastEditDate="2016-03-22T00:22:30.717" LastActivityDate="2016-04-21T01:16:44.933" Title="How can the bearing capacity of an existing tarmac road be determined?" Tags="|civil-engineering|geotechnical-engineering|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7764" PostTypeId="2" ParentId="2768" CreationDate="2016-03-04T14:45:56.840" Score="1" Body="&lt;p&gt;The reason you're dividing the pressure's by the density is that in the naiver stokes equations that describe fluid flow have three types of terms: inertia terms that are multiplied by the fluid density, pressure gradient terms, and viscous terms that are multiplied by the fluid viscosity. If you divide the whole equation by the fluid density then you can combine your absolute viscosity and your fluid density to get the kinematic density. Then you only have one fluid property to worry about, which makes solving equations easier, but you have to divide your pressure by your fluid density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the density you want to divide by is the density of the fluid you're simulating flow for. If you're simulating multiple fluid densities in the same model you cannot use this simplification. However, it seems like you aren't simulating the air, so you can safely just use the fluid density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other thing to note is that it's only pressure gradients that matter in incompressible flow, so often times people will use gauge pressure (absolute pressure minus one atmosphere) as it doesn't make a difference in the flow results and remembering to add the one atmosphere everywhere is annoying. So if the pressure at your inlet is in gauge pressure than the pressure at the top of your tank should be 0 not one 1 atm.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2016-03-04T14:45:56.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7766" PostTypeId="1" CreationDate="2016-03-04T16:20:36.407" Score="8" ViewCount="511" Body="&lt;p&gt;Let's say that you are driving a Diesel Multiple Unit&lt;sup&gt;†&lt;/sup&gt; with either mechanical&lt;sup&gt;*&lt;/sup&gt; (i.e. an actual gearbox with multiple ratios) or hydraulic (i.e. two or more torque converters) transmission. When you move the “go faster” lever to a higher notch, what &lt;em&gt;exactly&lt;/em&gt; is being commanded? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibilities that I can think of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Higher mass flow of fuel into the engine cylinders, whatever happens after that, happens &lt;/li&gt;&#xA;&lt;li&gt;Engine governor set to a higher rpm, whatever happens after that, happens&lt;/li&gt;&#xA;&lt;li&gt;Something else&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This weirdly specific question does relate to a question presented by a client, but I'm not being paid to solve it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;†&lt;/sup&gt; My question is specifically about DMUs — and &lt;em&gt;not&lt;/em&gt; locos — examples of the kind of train I'm interested in include: &lt;a href=&quot;https://en.wikipedia.org/wiki/British_Rail_Class_150&quot; rel=&quot;noreferrer&quot;&gt;Class 150 Sprinter&lt;/a&gt; and  &lt;a href=&quot;https://en.wikipedia.org/wiki/British_Rail_Class_172&quot; rel=&quot;noreferrer&quot;&gt;Class 172 TurboStar&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*&lt;/sup&gt; Although unusual, and for a long time considered obsolete, there are still mechanical transmissions for this application in production. For example, the Voith DIWARail pack for small railcars, derived from a bus power pack, and I believe that ZF offer a similar solution.&lt;/p&gt;&#xA;" OwnerUserId="5323" LastEditorDisplayName="user16" LastEditDate="2016-03-04T17:15:57.197" LastActivityDate="2024-03-01T01:05:20.623" Title="On a Diesel/hydraulic DMU train what exactly does the “go faster” lever control?" Tags="|mechanical-engineering|control-engineering|hydraulics|rail|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7767" PostTypeId="2" ParentId="6537" CreationDate="2016-03-04T16:38:14.717" Score="1" Body="&lt;p&gt;As a cast part, there should be at least two pieces that form the mold. There should be parting line between these pieces. Dimensions on one side of the parting line would be considered outside dimensions while dimensions on the other side would e considered inside. If the parting line is not specified on the print, then it could indeed be ambiguous as the part may be possible to cast with the parting line on different sides of the feature. If this is the case then the drawing should be revised.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2016-03-04T16:38:14.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7769" PostTypeId="1" CreationDate="2016-03-04T18:00:17.510" Score="1" ViewCount="72" Body="&lt;p&gt;Repairs... My garage door has counterweights, and the weight of the door is connected to the counterweights with a steel cable which runs over two pulleys. The original (cast iron) pulleys had needle roller bearings, but lacked any lubrication, and probably incorrect termination/hardening of the outer ring caused excessive wear, and I had to replace them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found size-compatible replacements, but they have 'normal' ball bearings, which should be able to bear the load. I've been trying to find a way to tell if the axis will stand hold the narrower load though. The simplified configuration is now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oI65Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oI65Y.png&quot; alt=&quot;Pulley mounting&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White is the axle, blue is the fixed hardware, and green is the bearing of the new pulley. All measures are in mm. The extra space is because the original roller bearing was wider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the path the cable follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GpwaQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GpwaQ.png&quot; alt=&quot;Path of the cable&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked over many pages, and I'm not sure how to tackle this. The problem is not completely shearing stress. I've found precious little info on 'bending' limits of round bars (axle). The weight of the door is about 60 kg (same as weights, of course) and the load is distributed at each side of the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively I suspect this will hold, but then my mechanical intuition isn't as developed (I'm an electronics engineer). In short, will the (mild steel) 10 mm axis bear the weight? Or do I need some more hardened material?&lt;/p&gt;&#xA;" OwnerUserId="5331" LastEditorUserId="1832" LastEditDate="2016-03-04T18:55:43.077" LastActivityDate="2016-03-04T18:55:43.077" Title="Evaluating replacement pulley axle strength" Tags="|beam|bearings|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7770" PostTypeId="1" AcceptedAnswerId="7777" CreationDate="2016-03-03T05:15:29.027" Score="4" ViewCount="1478" Body="&lt;p&gt;In my mechanics of materials class we are required to be able to solve problems with bolts in shear. Unfortunately our textbook doesn't cover this subject we only had a little bit of information given on slides and I am having trouble finding other resources that expand on this concept. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example question for context:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/THEGu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/THEGu.png&quot; alt=&quot;Eccentric loaded bolts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far these problems are straight forward. The total shear on a bolt is equal to the direct shear plus the torsional shear. First you get the direct shear component by distributing the force evenly on each bolt so $\displaystyle\tau_D=\frac{P}{4A}$ where $A$ is the cross sectional area of the bolt parallel to the force. Then you calculate the torsional shear by finding the torque $T$ of the force $P$ about the centroid of the bolts and $\displaystyle\tau_T=\frac{Tc}{J}$ where $c$ is the distance of the bolt from the centroid and $J$ is the polar moment of inertia of the bolts. The direction of the torsional shear is found perpendicular to a line from the centroid to the bolt in the sense of the rotation of the torque.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have no trouble doing these types of problems but I'm curious what happens if the bolts are not all the same diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm guessing is that the torsional shear will change because the centroid and polar moment of inertia will change but the torque stays the same. Is that correct? Is the force in the direct shear component still evenly distributed on the bolts? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above example if one bolt has a larger diameter than the others would the direct shear on the bolts still be $\displaystyle\tau_D=\frac{P}{4A}$? (obviously $A$ is now different in the larger bolt) or would the force in the direct shear be proportional to the relative size of the bolts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain this to me to verify if my line of thinking is correct?&lt;/p&gt;&#xA;" OwnerUserId="5328" OwnerDisplayName="Devin Crossman" LastEditorUserId="33" LastEditDate="2016-03-04T18:59:55.863" LastActivityDate="2016-03-05T01:41:24.953" Title="Effect of varying diameter bolts on shear stress under eccentric loading" Tags="|mechanical-engineering|stresses|bolting|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7771" PostTypeId="1" CreationDate="2016-03-04T18:42:54.930" Score="0" ViewCount="2200" Body="&lt;p&gt;So I have found this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_\mathrm{suspension} = Kx + F_\mathrm{damper}V_\mathrm{contact}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$K$ = spring stiffness&lt;/li&gt;&#xA;&lt;li&gt;$x$ = the contact depth of spring(also known as spring compression)&lt;/li&gt;&#xA;&lt;li&gt;$F_\mathrm{damper}$ = damper strength&lt;/li&gt;&#xA;&lt;li&gt;$V_\mathrm{contact}$ = the speed at which the spring is compressing/decompressing&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this accurate equation for calculating car suspension force?&lt;/li&gt;&#xA;&lt;li&gt;What are the different factors in equations for different types of suspension setups found in multiple cars? This is meant that trucks and old cars use leaf springs while new cars mostly has coil springs, some cars has multiple springs or dampers etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5281" LastActivityDate="2016-03-04T22:39:48.927" Title="Car Suspension Force Equations" Tags="|automotive-engineering|wheels|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7772" PostTypeId="2" ParentId="7769" CreationDate="2016-03-04T18:46:09.713" Score="1" Body="&lt;p&gt;A good start would be to disregard the fact that the load is distributed over a small part of the shaft and look at the problem as a basic &quot;Three Point Bending&quot; of a simply supported beam with a center load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that load case, the Max Stress ($\sigma_{max}$) is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_{max} = \frac{P*L} {4Z} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $P$ is the load at the center of the shaft, $L$ is the distance between the supports and $Z$ is the section modulus. For a circular beam/shaft:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z = 0.78r^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mild steel, the yield strength is probably about 240MPa &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this analysis gives an acceptable result, you should be OK, since distributing the load over the bearing areas will result in lower maximum stresses.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2016-03-04T18:46:09.713" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7774" PostTypeId="2" ParentId="7770" CreationDate="2016-03-04T20:53:09.467" Score="0" Body="&lt;p&gt;This can get quite complex in practice as you also need to consider the fit of the bolts in the holes and the effect of tolerances on exact load distribution and stress concentrations in the plate and the assumptions inherent in the calculations you quote become increasingly doubtful. For example variations in hole size an placement mean that in your example it is entirely possible that one or more bolts could be completely unloaded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact in the sort of situation you are describing it is not really ideal to have the bolts loaded in shear at all. Instead the bolts should be preloaded in tension (eg by tightening to a known torque) and the shear loads are born by friction between the mating surfaces. This is essential in things like pipe flanges where any gapping of the joint would constitute a failure but also tends to give have advantages in other situations as you can a have a well understood and predictable load distribution without very tight tolerances on hole size and placement. This also tend to distribute the load much more evenly throughout the fixing rather than having stress concentrated on the point where the shank of the bolt touches the side of the hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are situations where it is appropriate to load bolts in shear but this tends to be when you are using them as a simple axle or pin etc (this is often seen in castors and shackles). In this case you almost always have only one bolt taking the load and often the bolt hole will be reinforced relative to the plate it is in with some sort of bushing or similar. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-04T20:58:54.133" LastActivityDate="2016-03-04T20:58:54.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7775" PostTypeId="2" ParentId="7771" CreationDate="2016-03-04T21:25:44.223" Score="1" Body="&lt;p&gt;This equation will give you the static force for any given displacement but on its own it may not tell you that much as suspension systems are by their nature dynamic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters this adds an additional inertia term to the RHS of the equation F = ma where m is the unsprung mass (wheel, hub etc) and a is the acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also any serious analysis of suspension behaviour needs to consider &lt;a href=&quot;https://en.wikipedia.org/wiki/Vibration&quot; rel=&quot;nofollow&quot;&gt;frequency response&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you an overview of the problem vehicle suspension has two distinct but interrelated functions the first is to isolate vibration in the wheels from bumps and irregularities in the road surface from the chassis of the vehicle for comfort and to stop it from shaking itself to bits. This aspect is broadly covered by the above equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second function is handling ie to control the weight distribution of the vehicle and the orientation of the tyre contact patch to the road as the vehicle brakes, accelerated and turns. Here the detailed geometry of the suspension and transmission of forces between all four wheels becomes very important. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, coil, leaf and torsion springs can be modelled by this equation. The choice of spring type for a particular vehicle depends on a wide variety of factors. In particular left springs may be non-linear and/or self damping. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-04T22:39:48.927" LastActivityDate="2016-03-04T22:39:48.927" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7777" PostTypeId="2" ParentId="7770" CreationDate="2016-03-05T01:41:24.953" Score="2" Body="&lt;p&gt;If the bolt diameters vary (rare in civil engineering constructions, we don't want the workers to get bolt sizes mixed up on the tenth floor from ground), then the centroid of the bolt areas will change, hence the value of J (all relative to the centroid).  Therefore the torque component of shear will change since the value of J and c (distance from centroid to bolt) will vary.&#xA;The direct shear component will change as well, since the total direct shear (P) must be shared prorata according to the areas.&#xA;Do not forget that the eccentricity e must be modified according to the new location of the centroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as a side note, this is the classical method based on superposition of the two components of an eccentric force into a force applied through the centroid, as well as a moment (torque).  It is considered conservative,... too conservative.&#xA;Kulak has done a good review of this method, as well as the ICR (instantaneous centre of rotation) method, which is considered realistic.  However, the ICR method requires iteration to find the centre of rotation, and hence is not suited for a first course.  The ICR method is realistic because it takes into account of a certain degree of slipping, and local failures in shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If and when you have time, here are some detailed readings:&#xA;&lt;a href=&quot;http://www.boltcouncil.org/files/2ndEditionGuide.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.boltcouncil.org/files/2ndEditionGuide.pdf&lt;/a&gt;&#xA;&lt;a href=&quot;http://personal.stevens.edu/~shassiot/SteelDesign/AISC-DesignExamples/pdf/RP/CrawfordKulak1968.pdf&quot; rel=&quot;nofollow&quot;&gt;http://personal.stevens.edu/~shassiot/SteelDesign/AISC-DesignExamples/pdf/RP/CrawfordKulak1968.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5339" LastActivityDate="2016-03-05T01:41:24.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7780" PostTypeId="1" CreationDate="2016-03-05T05:36:55.103" Score="2" ViewCount="387" Body="&lt;p&gt;There is a house for rent with a big barrel stove in the basement, the stove pipe is insulated where it comes through the floor however it is made of old sewer pipe, cast iron I imagine, it's about a 15-ft section, this doesn't seem like a good idea to me.  I can find lots on specs for stove pipe (gauge of steel etc) but nothing the actually says what &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; to use for stovepipe.  Would appreciated clarification.&lt;/p&gt;&#xA;" OwnerUserId="5340" LastEditorUserId="1832" LastEditDate="2016-03-05T10:28:53.897" LastActivityDate="2016-03-07T12:28:20.413" Title="Cast iron sewer pipe being used as stove pipe" Tags="|heat-transfer|" AnswerCount="2" CommentCount="3" ClosedDate="2016-03-09T00:56:58.133" ContentLicense="CC BY-SA 3.0" />
  <row Id="7781" PostTypeId="1" AcceptedAnswerId="7787" CreationDate="2016-03-05T09:19:33.870" Score="1" ViewCount="4356" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PzcDr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PzcDr.jpg&quot; alt=&quot;Dislocation motion&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this picture do we speak of (microscopic) plastic deformation if the edge dislocation has moved all the way to the right and emerges out of the surface? Or if you apply shear stress and stop when the edge dislocation has moved from A to B is it then plastically deformed aswell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nldvu.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nldvu.gif&quot; alt=&quot;Screw dislocation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about with this situation? Is it only plastically deformed when the dislocation line has moved all the way to the left and thus the screw dislocation emerging from the crystal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ow and one last small question:&#xA;'before and after the movement of a dislocation through some particular region of the crystal, the atomic arrangement is ordered and perfect; it is only during the passage of the extra half plane that the lattice structure is disrupted'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular region is that the region B to D, when the dislocation starts at A and moves untill emerges? And if the dislocation motion stops while the edge dislocation is at position B or C then this region isn't ordered and perfect? Because there is a linear defect in it?&lt;/p&gt;&#xA;" OwnerUserId="4117" LastActivityDate="2016-03-05T17:43:24.003" Title="Plastic deformation due to the motion of dislocations" Tags="|materials|metallurgy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7782" PostTypeId="1" CreationDate="2016-03-05T11:58:42.210" Score="3" ViewCount="7298" Body="&lt;p&gt;Apperently in the beginning of material science they theoretically calculated the strength of perfect crystals, which was way higher than the experimental strength. The difference was later ascribed to the presence of dislocations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible?&#xA;To strengthen a material you can strain-harden it by plastically deforming it and thus impose more dislocations into it. So the more dislocations the higher the strength will be; how then is a crystal with no dislocations stronger than one with dislocations?&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorDisplayName="user16" LastEditDate="2016-03-05T22:03:36.630" LastActivityDate="2016-03-05T22:03:36.630" Title="Theoretical strength of pure and perfect crystals are way higher than the strength of crystals with dislocation?" Tags="|materials|metallurgy|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7783" PostTypeId="2" ParentId="7780" CreationDate="2016-03-05T12:48:19.817" Score="1" Body="&lt;p&gt;Wow, that's unusual. But as long as it's a single segment, properly insulated from anything flammable and large enough diameter to provide proper draft for the stove, I guess there's nothing actually wrong with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I emphasize &quot;single segment&quot;, because cast iron sewer pipe segments are traditionally connected together with lead, and the heat of a fire could easily melt lead.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-03-05T12:48:19.817" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7784" PostTypeId="1" AcceptedAnswerId="7788" CreationDate="2016-03-05T14:00:00.670" Score="3" ViewCount="650" Body="&lt;p&gt;I think that some of you know what is rolling resistance. It is force that works in opposite direction to roll and is caused due to deformation of tire. It also affects car wheels.&#xA;There is one problem with it. I have found 2 equations around:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;$F_\mathrm{roll} = -V_\mathrm{car} * C_\mathrm{rr}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$F_\mathrm{roll} = W_\mathrm{load} * \sin\alpha * C_\mathrm{rr}$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, none of them seems to be right one. First one seems to be incomplete, inaccurate, but second one doesn't depend on speed making car shake when being in standstill.&lt;br&gt;&#xA;I need to know what is accurate(but still game-compatible, not too heavy) equation for rolling resistance.&lt;/p&gt;&#xA;" OwnerUserId="5281" LastActivityDate="2020-08-16T15:49:52.420" Title="Wheel rolling resistance" Tags="|automotive-engineering|deformation|wheels|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7785" PostTypeId="2" ParentId="4209" CreationDate="2016-03-05T14:35:49.707" Score="0" Body="&lt;p&gt;Some food for thought. Airtrax is an omni-directional drive system with special wheels to enable the kind of maneuverability. (I've only seen it used in low speed applications)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://m.youtube.com/watch?v=IlmKcohyXG0&quot; rel=&quot;nofollow&quot;&gt;https://m.youtube.com/watch?v=IlmKcohyXG0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-03-05T14:35:49.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7786" PostTypeId="2" ParentId="7782" CreationDate="2016-03-05T17:39:22.313" Score="5" Body="&lt;p&gt;Assume for the rest of my answer that, unless otherwise stated, the material in question is a macroscopic single crystal of some metallic element, free of volumetric defects (e.g. carbides, graphite, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary: single crystals with no dislocations are soft and yield readily because there are no dislocations to prevent the movement of newly introduced dislocations. Dislocations are easy to introduce to single crystals from their surfaces. For polycrystals, grain boundaries complicate introduction, but practically dislocations may be introduced through e.g. Frank-Read sources. Dislocations pile up on grain boundaries and other defects, causing strain hardening.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Theoretical Single Crystal Strength&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;There are two modes possible modes of plastic deformation in tension: cleavage and shear. Cleavage is movement of two portions of the crystal toward either the tension directions, away from the cleavage plane. Shear is movement of two portions of the crystal at an angle with respect to the tension directions, along a shear plane. Shear is more likely because atomic bonds do not need to be fully broken. Instead they can hop to their neighbors as shear progresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to Polyani and Orowan, the theoretical cleavage stress is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sigma_{\textrm{max}} \approx \frac{E}{\pi}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which for most metals centers around $50\ \textrm{GPa}$. The experimentally determined yield strengths of the strongest steels push just past $1\ \textrm{GPa}$ for comparison, and most metals are closer to $0.1\ \textrm{GPa}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to Frenkel, the theoretical shear stress is due to an entire plane of atoms hopping over a neighboring plane such that each atom moves by one atomic spacing to a new spot. The required stress is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau_{\textrm{max}} \approx \frac{G}{5.13}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and because $\nu\approx 1/3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;G = \frac{E}{2\left(1+\nu\right)}&#xA;\approx \frac{3E}{8}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau_{\textrm{max}} \approx 0.073E&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which for most metals centers around $5\ \textrm{GPa}$, which is closer to the experimental values and smaller than the cleavage strength by an order of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Practical Single Crystal Strength&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Dislocations allow atoms to move one &quot;row&quot; at a time, rather than requiring the entire plane to move in a coordinated fashion, in a fashion akin to an inchworm. The process of dislocation-based shear deformation is called slip, and the motion of an edge dislocation along an atomic plane is called glide. The required stress for glide is small compared with shearing an entire plane of atoms by several orders of magnitude. Edge dislocations are easy to introduce into a single crystal at the surface, and are the natural mode of shear deformation in metals. As deformation proceeds the dislocations move all the way from one surface to the other. It is impossible to &quot;coerce&quot; a single crystal into deforming without adding dislocations, because dislocations are so energetically easy to add. The question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;How then is a crystal with no dislocations stronger than one with dislocations?&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;has a simple answer: dislocation free crystals are not stronger! In fact they are softer and more easily yield to deformation, by virtue of their lack of dislocations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of a single crystal, assume the material is a practical polycrystal and thus has some obstacles to dislocation motion, such as grain boundaries and volumetric defects. Dislocations are added to internal grains through e.g. Frank-Read sources (&lt;a href=&quot;https://en.wikipedia.org/wiki/Frank-Read_Source&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;), as well as at surface grains by glide. The dislocations will tend to pile up on the obstacles instead of freely moving through the entire crystal. As new dislocations are added, the pile-up gets more severe, until new modes of dislocation motion, which require greater stress, activate, and dislocation motion proceeds again. The overall pile-up process is called strain hardening, and as you noted, is the reason materials with greater dislocation density have higher yield stress than those with low dislocation density.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Reference&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Most of this answer comes from a lecture module at University of Alabama in PDF format, located &lt;a href=&quot;http://bama.ua.edu/~mweaver/courses/MechBeh/N07.pdf&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. Primary sources for the equations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M. Polyani, Z. Phys. v. 7 (1921) p. 323&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E. Orowan, Rep. Prog. Phys., v. 12 (1949) p. 185&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J. Frenkel, Z. Phys. v. 37 (1926) p. 572&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but good luck locating copies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much more information on dislocations can be found at University of Cambridge DoITPoMS &lt;a href=&quot;http://www.doitpoms.ac.uk/tlplib/dislocations/index.php&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-03-05T20:46:37.040" LastActivityDate="2016-03-05T20:46:37.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7787" PostTypeId="2" ParentId="7781" CreationDate="2016-03-05T17:43:24.003" Score="3" Body="&lt;p&gt;Plastic deformation is defined as non-recoverable deformation. When a stress is applied to the material, it will first deform elastically. When the yield limit is reached, it will deform plastically which for metals means by dislocation motion. If during dislocation motion the stress is removed, the dislocation does not vanish, and the atoms stay in their new locations. Therefore, the answer to your questions is that any dislocation motion of any distance constitutes plastic deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your last question, when the dislocation has caused any strain, the lattice is imperfect. So the first two subfigures of the top image are imperfect, while the last is &quot;perfect&quot; in the sense of no lattice strain.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-03-05T17:43:24.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7788" PostTypeId="2" ParentId="7784" CreationDate="2016-03-05T23:31:40.463" Score="3" Body="&lt;p&gt;Before we dig into practical equations, I'll just say that tires are surprisingly complex in their behavior. There are a variety of equations that try to fit the experimental data reasonably well. Hard to say that any one is right or wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That having been said, the first equation is pretty inconsistent with the everything that I've ever seen. Weight on the wheel should almost certainly be included and rolling resistance rarely approaches 0 at 0 velocity in the literature.  Two forms cited by Gillespie [ 1] are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{F_{r}}{W} =  f_r = 0.1(1+V/100) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for low speeds and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{F_{r}}{W} =  f_r = f_o + 3.24 f_s(V/100)^{2.5} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;over a broader range of speeds. Were F_r, W, V and f_i are rolling resistance force, weight on the tire velocity and empirical constants. Units are lbf, lbf and mph (sorry!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term looks like an expression for increased rolling resistance when steering/slipping. $\alpha$ would be the slip angle of the tire. Something like that would be added to either of the above equation. To give you a sense of the what this looks like, check out this graph. Also for Gillespie:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fBqDd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fBqDd.png&quot; alt=&quot;Rolling resistance with slip angle and velocity&quot;&gt;&lt;/a&gt;&#xA;Lastly, with the above equations, you'll probably still have an issue with jittering around $V=0$ in a numerical simulation. Instead of using simply switching direction of force with the sign of velocity ($F_r*sign(V)$), try multiplying the force by $tahn(\beta V)$ instead, to smooth out the transition ($\beta$ is a parameter to control how sharp the transition is). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[ 1]Thomas, D. Gillespie. &quot;Fundamentals of vehicle dynamics.&quot; Society of Automotive Engineering Inc (1992): 168-193.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2016-03-06T16:26:09.727" LastActivityDate="2016-03-06T16:26:09.727" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7790" PostTypeId="1" CreationDate="2016-03-06T06:06:56.267" Score="3" ViewCount="820" Body="&lt;p&gt;The following question was given in a book as an autocad assignment.&#xA;The section depicted in the diagram is not a counter-bore or a spot face hole. But the professor is telling that the hole is a counter bore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oATUY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oATUY.jpg&quot; alt=&quot;The Question as give in the book&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the correction suggested by the professor in the blue pen. This is the section view of the front view along of A-A line.&#xA;Is the correction correct as I argued that this not true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oaWiK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oaWiK.jpg&quot; alt=&quot;Solution corrected&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3856" LastEditorUserId="33" LastEditDate="2016-03-06T14:58:52.260" LastActivityDate="2021-09-14T21:30:09.157" Title="Orthographic project (section view) dimensioning a counter bore?" Tags="|drafting|autocad|cad|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7791" PostTypeId="1" CreationDate="2016-03-06T06:34:41.390" Score="1" ViewCount="122" Body="&lt;p&gt;I'm trying to build a device to deliver a N-drop (N * .05ML) of liquid via a tube into a container.  The liquid might change from one application to the next, so it might be water in one, and oil in another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been using an electric solenoid valve that uses a gravity fed vessel into the solenoid input, and controlling the solenoid with an arduino to release a drop from the output end of the solenoid.  Depending on the viscosity of the liquid, I've been able to determine how long to open the valve for.  12.5 milliseconds for water gives me .05ML about 95% of the time, which is more than acceptable for my purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vessel which is to receive the liquid is below the solenoid,so I've been trying to use a gravity approach using plastic tubing.  The problem I'm running into is that the path is not a straight line down.  While there is always a downward slope, sometimes it is close to flat, so sometimes the drop just gets stuck in the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried tubing made of vinyl, silicone, of various diameters. When the N-Drop leaves the solenoid, it is released into a funnel connected to the tubing.  I use the funnel because this prevents a vacuum forming which will hold the drop with a vice like grip. With smaller diameter tubing the N-Drop forms a &quot;plug&quot; and makes its merry way down the tube.  When it gets stuck, another drop can push this one along.  I can prime the tube with drops, but this isn't ideal as evaporation may come into play.  With larger diameter tubing, instead of a plug, it's more like a blob inching it's way to the bottom.  Once again, on the less steep portions of the tube, the blob sometimes just gets stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's not much I can do with making the slope steeper.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you think of a way of ensuring that the drop makes its way through the tube 90% of the time?&lt;/p&gt;&#xA;" OwnerUserId="4334" LastEditorUserId="33" LastEditDate="2016-03-06T14:42:46.227" LastActivityDate="2016-03-07T08:17:20.220" Title="How can I deliver a measured amount of liquids through a tube?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7792" PostTypeId="2" ParentId="7790" CreationDate="2016-03-06T14:54:04.150" Score="1" Body="&lt;p&gt;The original drawing is ambiguous. There is no hidden line showing that the individual holes are counter bores. The portion of the drawing that is cut out also is in the area where the other portion of the counter bore would be shown. It could be clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of that being said, there is one way that you could assume that each of the holes has a counter bore: Look at where the dimensions are drawn from. The dimension shown in the cut out is drawn from the centerline of the hole. This would only make sense if it was dimensioning the radius of a counter bore. The dimension is connected to the centerline of the small hole even if the only dimension is to the outside of the part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end result is that the original drawing could be clearer, but without a counter bore, the dimension given doesn't make sense. &lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2016-03-06T14:54:04.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7793" PostTypeId="1" CreationDate="2016-03-06T16:21:46.400" Score="1" ViewCount="103" Body="&lt;p&gt;I am going to attempt an experiment to achieve 3d effect.&#xA;I read about autostereoscopy on wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to do this,&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bzF7l.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bzF7l.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already processed image in that format. see the image&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3D4Ex.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3D4Ex.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;every alternate pixel is of particular eye image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I create a black - transparent impression on a sheet then it would show a correct image of either left eye or right eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking of printing image(for example) shown below and cut the white areas on the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after that process I will align that sheet in front of my monitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bbS8H.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bbS8H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to ask is this a good idea or I am doing something wrong ?&lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2016-03-06T16:21:46.400" Title="experiment with interlacing to achieve 3d effect" Tags="|optics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7794" PostTypeId="1" CreationDate="2016-03-06T19:16:19.357" Score="-1" ViewCount="407" Body="&lt;p&gt;I have heard that car weight (for more friction = more stability and higher traction) is being increased when driving by a force called aerodynamic down-force which causes additional force on the car to push it (increasing the load on wheel), but at the same time doesn't increase car's weight (meaning that car can accelerate faster due to this force as it can exceed more force (according to equation $F_\mathrm{traction} = W\mu$)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to find out is how to calculate the &lt;strong&gt;increase of load&lt;/strong&gt; due to down-force. I don't need to calculate a force applied on car, but rather calculate, how much does load increase on each wheel due to down-force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I need to know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What constant values do I need to know.&lt;/li&gt;&#xA;&lt;li&gt;What are the equations.&lt;/li&gt;&#xA;&lt;li&gt;How to correctly split the down-force between wheels, or is it the same on all wheels?.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5281" LastEditorDisplayName="user16" LastEditDate="2016-03-07T15:16:45.313" LastActivityDate="2016-07-23T14:54:04.310" Title="Aerodynamical forces - downforce" Tags="|automotive-engineering|aerodynamics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7796" PostTypeId="2" ParentId="7794" CreationDate="2016-03-06T22:30:34.863" Score="1" Body="&lt;p&gt;Very broadly speaking aerodynamic downforce will be proportional to the square of speed relative to the surrounding air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although downforce effectively increases weight without increasing mass (in the sense of F=ma) it will increase drag, also broadly in proportion to the square of speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common principal aerodynamic elements of a car are a front wing, rear wing and diffuser (which creates downforce under the car by &lt;a href=&quot;https://en.wikipedia.org/wiki/Ground_effect_(cars)&quot; rel=&quot;nofollow&quot;&gt;ground effect&lt;/a&gt;). Ride height both overall and front to rear rake angle is very important in the function of the diffuser and undertray which has knock on effects for the suspension geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These forces will act at their respective centres of pressure and balancing them relative to the centre of mass and dynamic loads on the car is a crucial aspect of performance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as the forces directly produced by the wings themselves all of the bodywork will have an interrelated role in directing and controlling the airflow in open wheeled card the front wing is especially important in this respect. Similarly it is also essential to provide airflow for cooling or radiators, brakes ets as well as engine induction. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-07-23T14:54:04.310" LastActivityDate="2016-07-23T14:54:04.310" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7798" PostTypeId="1" CreationDate="2016-03-07T05:36:14.170" Score="2" ViewCount="2250" Body="&lt;p&gt;I want to add power of both motors in such a way that it increases the RPM on a wheel, to be used in a tricycle.&lt;/p&gt;&#xA;" OwnerUserId="5354" LastEditorUserId="33" LastEditDate="2016-03-07T14:44:03.257" LastActivityDate="2016-03-07T14:44:03.257" Title="How can I use 2 motors to increase rpm on one wheel?" Tags="|mechanical-engineering|motors|" AnswerCount="2" CommentCount="1" ClosedDate="2016-03-08T00:41:17.153" ContentLicense="CC BY-SA 3.0" />
  <row Id="7799" PostTypeId="2" ParentId="7798" CreationDate="2016-03-07T08:14:04.797" Score="0" Body="&lt;p&gt;basically to use energy of both motors at once you will need to store that energy somewhere. here comes flywheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drew rough diagram how the arrangement can look like in figure below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BLUE - motors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GREEN - worm and worm wheel\gear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;YELLOW - flywheel shaft&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LIGHT RED - flywheel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RED - output shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/coyaq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/coyaq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;though this mechanism can be a challenge to use in a tricycle , this principle of storing mechanical energy from multiple sources into flywheel and using all of it in single output shaft is used for ages in the IC engine.&lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2016-03-07T08:14:04.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7800" PostTypeId="2" ParentId="7791" CreationDate="2016-03-07T08:17:20.220" Score="1" Body="&lt;p&gt;Inspired by stuff seen in hospitals:&lt;br&gt;&#xA;Don't use a solenoid (won't work the same way when you have different fluids anyway), use a small dosing pump (stepper motor pushing on syringe). Have a tube with a smaller diameter (so time spent in tube is not so long) and connect tube directly to outlet of pumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you want different fluids, the simplest way would be one pump for each fluid feeding onto the same manifold.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-03-07T08:17:20.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7801" PostTypeId="1" AcceptedAnswerId="7871" CreationDate="2016-03-07T10:33:32.003" Score="3" ViewCount="1869" Body="&lt;p&gt;I have a question on my work which I cannot figure out. I have three chains which support a beam. On the beam acts a distributed load. I want to calculate how much load will go to each chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the sum of each force/moment is equal to zero. The magnitude of the distributed load ($w$ Newton/meter). The length of the beam $L$ (Meter) and the sub-distances ($x_a$, $x_b$, $x_c$, $x_d$ Meter). $F_{vt}$ (newton) represents the concentrated force of the distributed load ($F_v = w*l*0.5$) &#xA;and $F_{av}$ is the concentrated load in section $x_a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you go through all the point you get four formulas. The first one for point $a$ is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;\sum M_a = 0\\&#xA;F_b(x_b) * F_c(x_b+x_c) - (w *L)* \dfrac{L}{2} = 0&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in $x_a$ section forces are also action, which should a moment around point $a$. But how can I calculate the resultant force ($F_{av}$) for section $x_a$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried the following but I get a net force in this section of 0 N ($F_{av} = w\cdot\dfrac{0}{L} = 0$). So there is no force acting in section $x_a$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody help me out with this part? &#xA;&lt;a href=&quot;https://i.stack.imgur.com/VXRiK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VXRiK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4681" LastEditorUserId="4681" LastEditDate="2016-03-08T08:41:19.553" LastActivityDate="2016-03-15T14:23:19.560" Title="Distributed load with an offset?" Tags="|mechanical-engineering|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7802" PostTypeId="2" ParentId="7798" CreationDate="2016-03-07T12:11:35.660" Score="4" Body="&lt;p&gt;First, you should simply get a bigger motor.  Two motors adding to the same power will have more overhead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you must use two motors, then just connect them mechanically.  There is nothing special you need to do to have two identical DC electric motors add their torques to the same shaft.  The no-load maximum speed will still be the same, but the torque at all lower speeds will be doubled compared to a single motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes you aren't trying to run each motor independently but synchronously or phase controlled.  They can be brushless DC, but controlled at a higher level as if they were brushed DC motors.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-03-07T12:11:35.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7803" PostTypeId="2" ParentId="7780" CreationDate="2016-03-07T12:28:20.413" Score="2" Body="&lt;p&gt;I'm not a mechanical engineer, but my worry is that the high temperatures and possibly acidic condensates of combustion gasses could corrode the inside of the cast iron pipe.  You won't know it from the outside until one day enough flame or hot gasses get thru to catch something outside the pipe on fire.  Or just leaking combustion gasses would be bad for the inhabitants of the house.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-03-07T12:28:20.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7804" PostTypeId="2" ParentId="7066" CreationDate="2016-03-07T13:32:01.230" Score="3" Body="&lt;p&gt;First poisons ratio for isotropic materials applies equally between each pair of dimensions. That means that if the axial length is stretched $1\%$ both of the lateral dimensions would reduce by $\nu 1\%$. I don't know where you're getting the 0.25 number from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, this approximation only works when the strain is small. Lets say we have an orthogonal parallelepiped with dimensions $x$, $y$, and $z$. If we apply a stress $\sigma_x$ streching the x direction and let the y and z dimensions be stress free. Then we'll have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon_x=\frac{\sigma_x}{E}$$&#xA;$$\epsilon_y=-\nu\frac{\sigma_x}{E}$$&#xA;$$\epsilon_z=-\nu\frac{\sigma_x}{E}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the new dimensions would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x'=x(1+\epsilon_x)$$&#xA;$$y'=y(1-\nu\epsilon_x)$$&#xA;$$z'=z(1-\nu\epsilon_x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the new volume would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V'=x'\,y'\,z'= x\,y\,z\, (1+\epsilon_x)(1-\nu\epsilon_x)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the ratio of the new volume to the old volume would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{V'}{V}=(1+\epsilon_x)(1-\nu\epsilon_x)^2=1+(1-2\nu)\epsilon_x+(\nu^2-2\nu)\epsilon_x^2+\nu^2\epsilon_x^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the increase in volume, you can subtract the 1. Then there are three more terms, but since $\epsilon_x$ is assumed to be small then the square and cube terms should be dominated by the linear term. Thus those terms are usually neglected and you're left with $(1-2\nu)\epsilon_x$. If you have an incompressible solid then this value should always be zero, so then $\nu$ must equal $0.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us look at a more general example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V=V'$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is really saying:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{split}1=(1+\epsilon_x)(1+\epsilon_y)(1+\epsilon_z)=&amp;amp;1+ \\&#xA;&amp;amp; \epsilon_x+\epsilon_y+\epsilon_z + \\&#xA;&amp;amp; \epsilon_x \epsilon_y + \epsilon_y \epsilon_z + \epsilon_x \epsilon_z + \\ &amp;amp; \epsilon_x \epsilon_y \epsilon_z&#xA;\end{split}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but again with a assumption that the strains are small, the product of multiple strains should be really small, so this equation simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=\epsilon_x+\epsilon_y+\epsilon_z$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the general equations for stain in isotropic materials are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon_x=\frac1{E}(\sigma_x-\nu(\sigma_y+\sigma_z))$$&#xA;$$\epsilon_y=\frac1{E}(\sigma_y-\nu(\sigma_x+\sigma_z))$$&#xA;$$\epsilon_z=\frac1{E}(\sigma_z-\nu(\sigma_x+\sigma_y))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can sum these all together to get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=\epsilon_x+\epsilon_y+\epsilon_z=\frac1{E}(1-2\nu)(\sigma_x+\sigma_y+\sigma_z)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we see that one again to conserve volume $\nu=\frac12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now why didn't this work in your example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you applied the stretching both other directions should have contracted. In order for the second direction to not contract, that direction would also have have to be pulled. This pulling on the second direction would further contract the third direction, but it would also contract the first direction, so in order for that to stay at it's original displacement you'd have to pull even harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you specified that $\nu=-\frac{\epsilon_z}{\epsilon_x}$ but this is only true if there is only stress in the x direction which is not the case for your scenario (because there must have been a stress applied to keep the 4cm side at 4cm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally the strain you applied was too large. Even if you let the two lateral dimensions strain freely, you'd get a Poisson's ratio not equal to .5 just because the small strain approximation would no longer be valid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon_x=\frac{11}{7}-1\approx 0.57$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to conserve volume:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon_y=\epsilon_z=\sqrt{\frac{7}{11}}-1 \approx -0.20$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is quite a bit less than half.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2016-03-07T13:32:01.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7805" PostTypeId="2" ParentId="4209" CreationDate="2016-03-07T15:28:10.270" Score="1" Body="&lt;p&gt;Aside from the issue of tire wear on side-skidding wheels (which could conceptually be solved by using different materials for both wheels and road surfaces) the OP's proposed steering method has one huge disadvantage over &quot;conventional&quot; steering, and that is lack of precision. It necessarily involves a jump between static friction for straight-line driving and dynamic (sliding) friction when turning, and that &quot;jump&quot; will occur at an unpredictable point in time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not much of a problem for &lt;em&gt;military&lt;/em&gt; use of a tracked vehicle like a tank, but lack of precision steering when changing lanes when travelling at 70 mph on a busy highway is a very different matter.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-03-07T15:28:10.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7806" PostTypeId="1" CreationDate="2016-03-07T15:36:31.563" Score="4" ViewCount="331" Body="&lt;p&gt;I would like to identify the correct term for a long, lateral cut within a floor joist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based upon &lt;a href=&quot;https://engineering.purdue.edu/~jliu/courses/CE479/extras/Notching_&amp;amp;_Boring_Guide_A11.pdf&quot; rel=&quot;nofollow&quot;&gt;this guide&lt;/a&gt; and several other sources that I have come across, a &lt;code&gt;notch&lt;/code&gt; is a cut that is allowed to have a width up to $\frac{1}{3}$rd of the depth of the floor joist.  In other words, a 9&quot; joist can have a notch cut in it up to 3&quot; wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my particular case, I would like to make a cut several feet long in order to increase the available head room.  I would like to know the correct term so I can research joist reinforcement requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus points for providing additional references regarding US building codes related to this type of a modification.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2016-03-08T00:52:36.313" Title="What is the correct term for long cuts in a floor joist?" Tags="|structural-engineering|structures|terminology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7807" PostTypeId="1" CreationDate="2016-03-07T15:38:41.867" Score="2" ViewCount="3487" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/guor6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/guor6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the damping ratio associated with the pole at &lt;code&gt;A'? Using the values&#xA;of a and b from (a) above, determine the closed loop poles at&lt;/code&gt;A' and the&#xA;corresponding value of K for this pole. Do not read the value of `A' from the&#xA;gure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is the answer provided by the lecturer. I understand most of it except that I dont understand how did he get the expression highlighted in red. Could anyone explain to me how he did that?&lt;/p&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2017-03-03T20:48:25.430" Title="How do I find the closed loop pole at A?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7808" PostTypeId="1" CreationDate="2016-03-07T15:49:08.950" Score="1" ViewCount="4428" Body="&lt;p&gt;Why do we have to condense the gas after a turbine or other heat engine, to feed it back to the heater?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't it be more efficient to feed it to the heater as warm as it goes out of the thermal engine?&lt;/p&gt;&#xA;" OwnerUserId="4846" LastEditorUserId="2819" LastEditDate="2016-03-08T02:08:23.997" LastActivityDate="2016-03-08T07:57:30.733" Title="Why does a thermodynamic cycle need a condenser?" Tags="|thermodynamics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7809" PostTypeId="1" CreationDate="2016-03-07T16:11:40.147" Score="0" ViewCount="739" Body="&lt;p&gt;Trying to draw the free body diagram of this, any help would be appreciated.&#xA;The $y$ is the amplitude of vibration of ground. Up is the positive direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/T9E40.jpg&quot; alt=&quot;diagram of proposed design&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5360" LastEditorUserId="2246" LastEditDate="2016-03-08T19:44:17.527" LastActivityDate="2016-03-08T19:44:17.527" Title="Vibration absorber design: free body diagram of a mass-spring system" Tags="|mechanical-engineering|vibration|" AnswerCount="1" CommentCount="7" ClosedDate="2016-03-08T20:44:35.473" ContentLicense="CC BY-SA 3.0" />
  <row Id="7810" PostTypeId="1" AcceptedAnswerId="7972" CreationDate="2016-03-07T16:45:06.763" Score="1" ViewCount="507" Body="&lt;p&gt;Wikipedia says that &lt;a href=&quot;https://en.wikipedia.org/wiki/Anti-roll_bar&quot; rel=&quot;nofollow&quot;&gt;an anti-roll bar is a device that balances the weight of car between the wheels, preventing it from rolling&lt;/a&gt;. I want to simulate this in a game, but I don't know the equations behind it. I need to know how much weight the anti-roll bar transfers to each wheel. That means that I need to obtain a value that I can add to the load of wheel to manage it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W_{load} = W_{static} + W_{transfer} + W_{anti-roll}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The static load is dependant on weight distribution. I do it like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W_{static} = L_{0} / L * m * g / 2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $L_0$ is the longitudinal distance from the opposite (e.g., if checking the rear wheel, than this is the front wheel's distance) wheel joint axle to the centre of gravity, $L$ is wheelbase and $m$ is mass of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $W_{transfer}$ is acceleration-dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I obtain the $W_{anti-roll}$ value?&lt;/p&gt;&#xA;" OwnerUserId="5281" LastEditorUserId="368" LastEditDate="2016-03-08T20:42:45.833" LastActivityDate="2016-03-16T14:21:13.010" Title="Find the load applied to each wheel by the anti-roll bar" Tags="|automotive-engineering|applied-mechanics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7811" PostTypeId="2" ParentId="7806" CreationDate="2016-03-07T17:28:58.987" Score="3" Body="&lt;p&gt;In my experience, this is called a &lt;code&gt;rip&lt;/code&gt;. I've seen it done a lot (on the top of a beam section) to accommodate flooring thicknesses/slopes. Your reference actually mentions this, although briefly, on the first page under &lt;strong&gt;When a Notch Becomes a Rip&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've always interpreted the code (NDS, IBC) such that, and received plan check comments implying that, notching (and therefore ripping) in the center 1/3 of the member span is prohibited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way around these code requirements would be by re-qualifying the structural member using the (new) shorter depth or re-qualifying the floor system assuming a joist spacing of twice the actual spacing. The second method is extremely conservative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're doing a retrofit and neither of those approaches work, you need to start considering how you can reinforce the member after ripping it down to the size you need. There's nothing special about the process that sets it apart from the same basic approach you would take when reinforcing a member of any other material. I have accomplished this in the past by attaching steel plates or steel channel sections to the wood member. I've also accomplished this by simply sistering an additional wood member to both sides of the modified member. You're basically just trying to increase the member stiffness (i.e. It's moment of inertia) back up to around what it was before ripping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take care to deal with/check beam shear as well. Sometimes this can get tricky in retrofits because the beam support location (where beam shear is likely to be highest) is tougher to get to. It's not usually a concern if you only rip locally and don't add any additional load though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of a good reference, really. This is all mostly accomplished by basic analysis and design techniques covered in most introductory statics/mechanics books.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2016-03-08T00:52:36.313" LastActivityDate="2016-03-08T00:52:36.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7812" PostTypeId="2" ParentId="7809" CreationDate="2016-03-07T17:40:34.493" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;Isolate $m_\text{absorber}$. Following forces need to be introduced:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;gravity $m_\text{absorber}g$&lt;/li&gt;&#xA;&lt;li&gt;spring force $k_\text{absorber}(x_1-x_2)$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Isolate $m$. Following forces need to be introduced:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;gravity $mg$&lt;/li&gt;&#xA;&lt;li&gt;spring force $k_\text{absorber}(x_2-x_1)$&lt;/li&gt;&#xA;&lt;li&gt;spring force $k(x_2-y)$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Make sure to pay attention at the sign of the spring forces: depending on their orientation, you might need the opposite of the expressions mentioned above.&lt;/p&gt;&#xA;" OwnerUserId="3633" LastActivityDate="2016-03-07T17:40:34.493" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7813" PostTypeId="2" ParentId="7808" CreationDate="2016-03-07T18:45:10.957" Score="1" Body="&lt;p&gt;Somewhat counter intuitively, it is &lt;strong&gt;not&lt;/strong&gt; more efficient to use warmer air (or any working fluid) in the inlet to a heat engine.  Two similar questions were asked (and answered) regarding steam in a Rankine cycle (&lt;a href=&quot;https://engineering.stackexchange.com/questions/5307/why-use-steam-in-the-rankine-cycle?rq=1&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://engineering.stackexchange.com/questions/5308/why-do-we-need-to-condense-steam-in-the-rankine-cycle?rq=1&quot;&gt;here&lt;/a&gt;).  Hopefully these explain why this is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from reasons of efficiency, it would be difficult to recycle the exhaust from something like a jet engine or gas turbine because the working fluid is also the source of oxygen for the combustion reaction.  The oxygen content of the fluid at the inlet would decrease to the point where little or no heat could be generated.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-03-07T18:45:10.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7814" PostTypeId="2" ParentId="7801" CreationDate="2016-03-07T20:48:10.240" Score="0" Body="&lt;p&gt;A simple beam such as this one supported at three locations is statically indeterminate.  You have three unknown reactions, one for each support, but there are only two equations of static equilibrium available, hence you have one degree of indeterminacy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This type of beam is called a continuous beam, and there are different methods of calculating the reactions at each support.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-03-07T20:48:10.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7816" PostTypeId="1" CreationDate="2016-03-08T01:58:37.783" Score="1" ViewCount="312" Body="&lt;p&gt;What is the correct way to allow for lateral movement of a timing pulley in a transmission system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there are four pulleys, one of which has to move up and down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there standard practice for this sort of arrangement?  Or an alternative that would allow the the large central pulley to move up or down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O5mYR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O5mYR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="5310" LastEditorUserId="5310" LastEditDate="2016-03-10T05:32:15.327" LastActivityDate="2016-08-07T22:13:35.933" Title="Timing belt transmission with lateral pulley movement" Tags="|mechanical-engineering|engines|timing|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7817" PostTypeId="2" ParentId="7808" CreationDate="2016-03-08T02:04:00.500" Score="0" Body="&lt;p&gt;A condenser is an important part of a closed-loop phase change thermodynamic cycle. Thermodynamic cycles generate mechanical energy from the transfer of heat energy. Transfer is the key word here. The larger the differences in temperature in a closed loop system the more efficent it is. If there is no difference in temperature, there is no power to be extracted. I recommend reading the &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermodynamics wiki article&lt;/a&gt; for more information. Also have a look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Carnot_heat_engine#Carnot.27s_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carnot's theorem&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will explain a steam engine in layman's terms which may help. Have a look at the following image from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Steam_engine#Steam_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;steam engine wiki article&lt;/a&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KwstU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KwstU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At step 2 we can see heat is added to the fluid in the boiler making it a gas. Then in step 3 it expands through the turbine giving us power output. Then on step 4 we need the condenser to remove the heat and condense this expanded gas back into a fluid. That way the fluid can be easily pumped with a small amount of power input into the high pressure boiler. The pump and the turbine have roughly the same pressure drop across them, but since a much larger volume moves through the turbine, it produces more power than is consumed by the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely if we tried to not condense the gas and pump it back into the boiler, it would take more energy to pump it than we would get out of the turbine.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-08T02:04:00.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7818" PostTypeId="2" ParentId="7714" CreationDate="2016-03-08T04:01:12.070" Score="1" Body="&lt;p&gt;I'm sure the methods they are using and advertising in their video are extensively protected by patents that can be searched here: &lt;a href=&quot;https://www.google.com/?tbm=pts&amp;amp;gws_rd=ssl#tbm=pts&amp;amp;q=graphene%20production&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Graphene Production&quot; patent search&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that it is just a promotional video, and what they say is not necessarily true.&lt;br&gt;&#xA;At 1:11, &quot;...100 times better conductor than copper...&quot; Not likely:&#xA;&lt;br&gt;2500 W⋅m−1⋅K−1 Graphene/ 385 W⋅m−1⋅K−1 = only 6.5 times better it turns out.&lt;br&gt;&#xA;At 1:24 &quot;...and the middle is completely cool.&quot; I find it quite unlikely that even an exotic material like graphene can conduct heat without becoming that temperature ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://phys.org/news/2015-06-graphene-heat-transfer-riddle-unraveled.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This phys.org article&lt;/a&gt; explains that the crystals must be aligned. &quot;When two crystals are neatly lined up, heat transfer occurs just as predicted by theory. But if the two crystals have mis-aligned edges, the heat transfer is 10 times less.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past I have cut ice with low grade &lt;a href=&quot;https://en.wikipedia.org/wiki/Pyrolytic_carbon&quot; rel=&quot;nofollow noreferrer&quot;&gt;pyrolytic carbon&lt;/a&gt; that I originally procured for a diamagnetic demonstration. The very high conductivity is quickly evident and it works just like it does in this &lt;a href=&quot;https://www.youtube.com/watch?v=R5dwdZCKBZM&quot; rel=&quot;nofollow noreferrer&quot;&gt;youtube video&lt;/a&gt;.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/EAoLW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EAoLW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the &lt;a href=&quot;https://en.wikipedia.org/wiki/Graphene#Thermal_conductivity&quot; rel=&quot;nofollow noreferrer&quot;&gt;graphene wiki article&lt;/a&gt;:&#xA;&quot;Early measurements of the thermal conductivity of suspended graphene reported an exceptionally large thermal conductivity of approximately 5300 W⋅m−1⋅K−1,[129] compared with the thermal conductivity of pyrolytic graphite of approximately 2000 W⋅m−1⋅K−1 at room temperature.[130] However, later studies have questioned whether this ultrahigh value had been overestimated, and have instead measured a wide range of thermal conductivities between 1500 – 2500 W⋅m−1⋅K−1 for suspended single layer graphene.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as economic feasibility; there would have to be some major reductions in price and increases in conductivity to make it more economically advantageous than current geothermal fluid heat transfer systems; let alone other forms of power production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other references:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.extremetech.com/extreme/206573-mit-researchers-pioneer-technique-for-mass-producing-graphene&quot; rel=&quot;nofollow noreferrer&quot;&gt;MIT, Technique for mass-producing graphene&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://news.mit.edu/2015/manufacture-continuous-rolls-graphene-0521&quot; rel=&quot;nofollow noreferrer&quot;&gt;MIT, Continuous rolls of graphene&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.popularmechanics.com/science/a14651/this-scientist-invented-a-simply-way-to-mass-produce-graphene/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PopularMechanics, Mass producing graphene&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-08T04:01:12.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7821" PostTypeId="2" ParentId="7760" CreationDate="2016-03-08T06:33:43.087" Score="2" Body="&lt;p&gt;It really depends on the logic required for your system. I don't think there is a hard fast mathematical solution like there is for a standard one-input one-output PID. In many cases a PID may not even be the best logic for your system; especially if the outputs have a complex relationship with the input like a chemical reaction. PID's are attractive academically because you put a number in and get a number out regardless of the process. However the real world often benefits from controllers that consider the physics involved in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A control system with more outputs than inputs is going to be reliant on some internal equations and assumptions that tell it the physical rules it can anticipate. The controller could also be relying on open loop outputs in addition to the closed loop outputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you could give an example of one you have seen and we can assess it.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-08T06:33:43.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7822" PostTypeId="2" ParentId="7808" CreationDate="2016-03-08T07:57:30.733" Score="1" Body="&lt;p&gt;I'll try another approach that's maybe more intuitiv, and less 'thermodynamic'. That said, you should definitly learn some thermodynamics because it's a cool subject to know about!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a steam engine, you need to understand the concept of &lt;a href=&quot;https://en.wikipedia.org/wiki/Vapor_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;vapor pressure&lt;/a&gt;: Take a body of water and giv it enough time, there will be water vapor at a certain presure above the water surface. The higher the temperature, the higher the pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's look at the steam engine:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KwstU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KwstU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure difference between points 3 and 4 drive the turbine. For there to be a high pressure difference, you need a high temperature difference. So you need to cool after point 4.&lt;br&gt;&#xA;The steam also cools when expanding in the turbine, but that only goes so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simple explanation that hopefully helps, but you won't understand important things like steam engine efficiency with it, so again, look into thermodynamics by following the links in the other answers.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-03-08T07:57:30.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7825" PostTypeId="2" ParentId="7807" CreationDate="2016-03-08T12:06:37.487" Score="0" Body="&lt;p&gt;I don't have access to matlab right now, so plots that can illustrate my thinking will (if needed) follow later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what do we know? We know that the closed loop transfer function has 3 poles, 1 real pole and depending on K 2 complex conjugate poles. You now want to find the pole A &lt;sup&gt;†&lt;/sup&gt; for that $\zeta=1$. This means we need to find a pole with no imaginary parts and negative real parts (for a stable system). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it seems like it is sufficient to look at an open loop transfer function, I assume this is due to the fact that you could calculate K later to find the gain of your closed loop transfer function to yield you the corresponding poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s = -\zeta \omega_n \pm \omega_n \sqrt{1-\zeta^2}$ (I)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we yield $s = -\omega_n$ thus $\beta = \omega_n$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get possible complex conjugate poles we need to have the form of $(s+\beta)^2$ &#xA;There is an explanation which derives (I) from the fact that mostly these functions can be described with ODEs of order $n$ but I'm lacking the understanding to explain this to you so I'll just say this will have to do. I mainly apply these functions rather than deriving them. So bear with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(s+\alpha)(s+\beta)^2=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up we just let $(s+\alpha)$ be because we don't care for the poles of the left most part of the LHP because those poles always only have a real part as you can see from the root locus.&#xA;From the knowledge that $\zeta=1$ means no complex poles you can derive A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the sake of completion: $\beta &amp;gt; 0$ or as the lecturer wrote $\beta$ cannot $\in 0 &amp;lt; \beta &amp;lt; 1$ comes from the knowledge that the system shall not be unstable. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;†&lt;sub&gt; Why did you call A A'? Clearly A in the root locus is the point you are looking for.&lt;/sub&gt;&lt;/sup&gt;   &lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2016-03-08T12:06:37.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7826" PostTypeId="2" ParentId="7726" CreationDate="2016-03-08T12:07:26.553" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=nO5nOZVhx74&quot; rel=&quot;nofollow&quot;&gt;Another solution which is based on spur gears+springs rather than a clutch&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=-p8zrvU-cYE&quot; rel=&quot;nofollow&quot;&gt;Another solution which is based on spur gears+differentials rather than a clutch&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found using the keyword &quot;rotation rectifier&quot; (since it's the mechanical equivalent of an electrical rectifier in which alternating current is converted to direct current).&lt;/p&gt;&#xA;" OwnerUserId="1326" LastActivityDate="2016-03-08T12:07:26.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7827" PostTypeId="1" AcceptedAnswerId="7833" CreationDate="2016-03-08T15:12:43.860" Score="4" ViewCount="138" Body="&lt;p&gt;I recently went on a tour of a glass manufacturing facility who produce bottles for the prestige alcohol market. I was informed on the tour that red glass cannot be directly produced - any bottles that I have seen had made by applying a spray to a plain finished bottle as opposed to adding the colour into the molten glass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this?&lt;/p&gt;&#xA;" OwnerUserId="307" LastActivityDate="2016-03-08T21:07:12.697" Title="Why can't red flint glass be produced?" Tags="|materials|manufacturing-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7828" PostTypeId="1" CreationDate="2016-03-08T16:38:25.203" Score="1" ViewCount="282" Body="&lt;p&gt;I have translated the part illustrated below into Solidworks but I haven't been able to match the angle of projection used in the illustration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y9eP8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y9eP8.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I identify that angle used in the image, so that I can reproduce the figure as closely as possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawing is from an old book by Thomas French.&lt;/p&gt;&#xA;" OwnerUserId="5312" LastEditorUserId="368" LastEditDate="2016-03-08T16:47:40.533" LastActivityDate="2017-03-30T13:06:49.883" Title="How can I match the angle of projection from a book illustration?" Tags="|drafting|solidworks|cad|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7829" PostTypeId="2" ParentId="7810" CreationDate="2016-03-08T17:20:23.570" Score="0" Body="&lt;p&gt;The anti-roll bar &lt;em&gt;partially&lt;/em&gt; constrains the two wheels of a given axle to move up and down together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Partially&quot; is emphasized, because the coupling is deliberately less than 100% &amp;mdash; the bar is allowed to twist to some degree, acting as a torsion spring, in order to still allow the wheels to cope individually with bumps, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you make the antiroll bar stiffer, you'll get less rolling in turns, but the overall ride will feel &quot;harsh&quot; on bumpy surfaces. It will be very stiff in a race car on a track, and much less stiff in the family sedan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And BTW, the distance between axles has no relevance to how the antiroll bar works.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-03-08T17:20:23.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7830" PostTypeId="1" CreationDate="2016-03-08T17:24:33.407" Score="4" ViewCount="761" Body="&lt;p&gt;It's been a while since I've done any comprehensive work with structural dynamics, so let me know if I get anything wrong here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a horizontal beam that is clamped on one end and free on the other. It has the following parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;length $L$&lt;/li&gt;&#xA;&lt;li&gt;elastic modulus $E$&lt;/li&gt;&#xA;&lt;li&gt;moment of inertia $I$&lt;/li&gt;&#xA;&lt;li&gt;mass $M$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are no loads on this beam except for its own gravity, which can be modeled by a uniformly distributed load:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\psi = \frac{Mg}{L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using standard beam theory, we can calculate the deflection at points $x$ along the beam, as well as the maximum deflection $\delta_{max}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = \frac{\psi x^2}{24EI}(x^2 + 6L^2 - 4Lx)$$&#xA;$$\delta_{max} = \frac{\psi L^4}{8EI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a physical constraint that $\delta_{max}$ can be no more than $\epsilon$, so I propose the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spin the beam about the wall it's attached to at angular velocity $\omega$, which creates an axial centrifugal force. This should lower the value of $\delta_{max}$, and for all $\epsilon &amp;gt; 0$, I should be able to find $\omega$ meeting this constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm looking to derive is some function $\delta_{max}(\omega; \psi, L, E, I)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how can I incorporate the axial force? Each slice of beam $dm$ has centrifugal force equal to $\omega^2 x dm$ where $x$ is the distance from the clamp. I don't immediately see where to proceed from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is super appreciated! ^^&lt;/p&gt;&#xA;" OwnerUserId="5218" LastEditorUserId="33" LastEditDate="2016-03-08T20:58:08.873" LastActivityDate="2016-03-10T21:45:46.963" Title="Removing beam deflection through axial force" Tags="|mechanical-engineering|structures|dynamics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7832" PostTypeId="2" ParentId="7830" CreationDate="2016-03-08T19:49:47.040" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;My question is how can I incorporate the axial force?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Doing so would invalidate the assumptions made in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot;&gt;Bernoulli beam theorem&lt;/a&gt;, and therefore render your deflection equation invalid.  Per the linked Wikipedia article (emphasis mine),&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Euler–Bernoulli beam theory (also known as engineer's beam theory or classical beam theory) is a simplification of the linear theory of elasticity which provides a means of calculating the load-carrying and deflection characteristics of beams.  It covers the case for small deflections of a beam that is &lt;strong&gt;subjected to lateral loads only&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could model this in a finite element software if you wanted to really know the answer.  Unfortunately, the simplified approach you're outlining here won't work if you need to take axial load or torsion into account.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2016-03-08T19:49:47.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7833" PostTypeId="2" ParentId="7827" CreationDate="2016-03-08T20:59:32.113" Score="4" Body="&lt;p&gt;It seems likely that costs associated with the material prevents economical production of red glass. Rather than a matter of possibility, it may be a matter of viability. Without knowing more about how the glass is produced or the specific materials used for their bottles, it is difficult to say the exact reason. The best answer I can think of is that the only glass I am aware of which selectively transmits red light requires gold nanoparticles suspended in the silica glass network. Such glass is referred to as Cranberry Glass or Gold Ruby Glass (&lt;a href=&quot;https://en.wikipedia.org/wiki/Cranberry_glass&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't clear how much gold is required, though the abstract of &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1296207408001040&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt;, which uses an alternate production process, suggests about 0.2% by weight. The price of gold is about 40,000 USD/kg per the current spot price from &lt;a href=&quot;http://www.apmex.com/spotprices/gold-price&quot; rel=&quot;nofollow&quot;&gt;apmex.com&lt;/a&gt; as of 08 March 2016 at ~2:45 PM CST. Thus a glass bottle weighing 1 kg would have an added cost of about 80 USD from the gold alone. Expanding the process to mass-produce such bottles would require storing and shipping significant quantities of gold, which may add logistical and overhead costs for security. There would also be equipment requirements for producing colloidal aqua regia in the proper concentration and drying, handling, and mixing the nanoparticles into the liquid glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, flint glass (&lt;a href=&quot;https://en.wikipedia.org/wiki/Flint_glass&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;) contains zirconium and/or titanium. I can't think of any reason that either element would interfere with the cohesion of the gold particles, but I also can't strictly rule it out. If such a thing did happen, then Flint glass specifically might be impossible to make red by the use of gold nanoparticles.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-03-08T21:07:12.697" LastActivityDate="2016-03-08T21:07:12.697" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7834" PostTypeId="1" AcceptedAnswerId="7840" CreationDate="2016-03-08T21:57:18.477" Score="1" ViewCount="109" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3XnKn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3XnKn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So above I have a memory design including a 74LS138 3-to-8 multiplexer which enables 8 memory banks comprised of 2732s which are 4Kx8 in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why the memory addressing starts F8000 instead of F0000?&lt;/p&gt;&#xA;" OwnerUserId="5378" LastActivityDate="2016-03-10T02:54:39.383" Title="Why do these memory spaces start at F8000 instead of F0000?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7835" PostTypeId="1" AcceptedAnswerId="7836" CreationDate="2016-03-08T22:12:29.320" Score="3" ViewCount="90" Body="&lt;p&gt;I have a small assembly with several parts misaligned. By turning the grey wheel, the grey block is to rotate about its axis (Axis 1). The problem I have is that the grey wheel can't be turned further anti-clockwise given the current position. &lt;a href=&quot;https://youtu.be/ozMX3gkA3W8&quot; rel=&quot;nofollow noreferrer&quot;&gt;It seems to be jammed&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G9IDI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G9IDIt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Hbk1L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hbk1Lt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cEXqc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cEXqct.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red connecting rod can freely slide in the yellow bore, which itself can rotate with the rod in the grey bore. I'd expect the red rod to extend further and turn the yellow connector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know why the wheel might be jammed or how it could be modified to allow continuous rotation?&lt;/p&gt;&#xA;" OwnerUserId="4303" LastEditorUserId="4303" LastEditDate="2016-03-09T06:18:13.003" LastActivityDate="2016-03-09T06:18:13.003" Title="Unexpected jamming of a misaligned assembly" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7836" PostTypeId="2" ParentId="7835" CreationDate="2016-03-08T23:24:42.000" Score="2" Body="&lt;p&gt;It isn't clear what is constraining the motion of the gray ring (blue in the video), but it seems to be rotating on its own axis. If so, the mechanism jams because the pin at the end of the red rod hits the end of the slot in the orange L-bracket. Regardless of the position of the yellow piece WRT the gray bar, there is a fixed maximum distance between the yellow pin and the red pin.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-03-08T23:24:42.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7837" PostTypeId="1" AcceptedAnswerId="7848" CreationDate="2016-03-08T23:25:51.293" Score="3" ViewCount="4894" Body="&lt;p&gt;I want to perform a structural analysis on a portal frame using a finite element solver to get a displacements and reactions for a load. The software library only accepts loads applied to node, not to the member. The node loads have 6 parameters: 3 translation (x,y,z) and 3 rotation (xx,yy,zz). My load is a uniform load of 1 kN/m along one rafter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lhqwy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lhqwy.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to convert the uniform member load to point loads that I can apply at the nodes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Portal frame dimensions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;span: 10 m&lt;/li&gt;&#xA;&lt;li&gt;eave height: 5 m&lt;/li&gt;&#xA;&lt;li&gt;apex height: 6.34 m&lt;/li&gt;&#xA;&lt;li&gt;rafter length: 5.18 m&lt;/li&gt;&#xA;&lt;li&gt;roof pitch: 15°&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;EDIT: fixed apex height &lt;/p&gt;&#xA;" OwnerUserId="5380" LastEditorUserId="5380" LastEditDate="2016-03-10T03:35:07.010" LastActivityDate="2016-03-10T10:31:14.560" Title="Is it possible to convert uniform loads to point loads?" Tags="|beam|structural-analysis|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7838" PostTypeId="2" ParentId="7801" CreationDate="2016-03-08T23:38:43.467" Score="0" Body="&lt;p&gt;As has been said, this is a Statically Indeterminate Problem.  The force in each chain will depend on the elasticity of each chain, the bending elasticity of the beam, and how much each chain is tensioned.  One chain could be completely slack, and the other two could carry the whole load.   Or, theoretically, the central chain could carry the whole load.  In practice it would all depend on how much you choose to tension each chain.  A Finite Element Analysis would give you a solution for whatever beam section, and therefore elasticity, chain elasticity, and initial chain tensions you choose.&lt;/p&gt;&#xA;" OwnerUserId="5381" LastActivityDate="2016-03-08T23:38:43.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7839" PostTypeId="2" ParentId="7837" CreationDate="2016-03-09T00:27:26.350" Score="3" Body="&lt;p&gt;You can apply the total force of an evenly distributed load to the centre of the member. This will be a reasonable approximation in terms of the overall distribution of load in the structure as a whole but is less useful in determining the stresses and deflections in individual members and associated joints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a model based on point loads should, in general, be more conservative than  one with evenly distributed loads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, the case in your example looks like something where specific building codes would need to be considered rather than ad hoc analysis. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="368" LastEditDate="2016-03-09T00:41:41.180" LastActivityDate="2016-03-09T00:41:41.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7840" PostTypeId="2" ParentId="7834" CreationDate="2016-03-09T05:41:44.083" Score="3" Body="&lt;p&gt;Addresses are: A19 18 17 16 15 .... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This expands on Dave Tweed's correct but brief comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F8000 = 1111 &lt;strong&gt;1&lt;/strong&gt;000 0000 0000 0000    &amp;lt;- A15 = 1&lt;br&gt;&#xA;F0000 = 1111 &lt;strong&gt;0&lt;/strong&gt;000 0000 0000 0000    &amp;lt;- A15 = 0   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For F8000, A15 is high&lt;br&gt;&#xA;For F0000, A15 is low&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Dave Tweed notes, the decoder is only selected when all inputs to the '30 NAND gate are high.&lt;br&gt;&#xA;As A15 is one of the inputs it must be high for the decoder to be selected.&lt;br&gt;&#xA;As A19 18 17 16 15 must be high the applicable address is F8xxx as above .&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="86" LastEditDate="2016-03-10T02:54:39.383" LastActivityDate="2016-03-10T02:54:39.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7842" PostTypeId="1" CreationDate="2016-03-09T13:24:38.580" Score="4" ViewCount="2196" Body="&lt;p&gt;In a simple situation, one simply takes the average of outer diameter and inner diameter to gain the mean radius for calculating torque transmission through a contact face, such as a clutch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how would you work this out if there was no simple ID and OD to work with, say for example if the ID was broken by an array of holes as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/btOuV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/btOuV.png&quot; alt=&quot;Irregular Contact Face&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5388" LastEditorUserId="5388" LastEditDate="2016-03-09T17:04:06.700" LastActivityDate="2017-05-30T12:47:59.980" Title="How to Calculate Mean Radius of Irregular Contact Face (Torque Transmission)" Tags="|mechanical-engineering|mathematics|torque|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7843" PostTypeId="1" CreationDate="2016-03-09T14:55:11.330" Score="3" ViewCount="420" Body="&lt;p&gt;I have some time domain data which are responses of an aircraft to various inputs (example: time vs. bank angle for lateral stick pulse input). I need to compute the effective damping ratio and damped frequency of these &quot;signals&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I considered several options but none worked as well as I hoped.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Computing instantaneous frequency and damping ratio using Hilbert transform: This does not give single frequency and damping values. Also, I get negative values sometimes.&lt;/li&gt;&#xA;&lt;li&gt;Finding approximate second order system and using its parameters: Works for some test cases but I have test cases where the signal can not be represented well with second order system. &lt;/li&gt;&#xA;&lt;li&gt;Simply detecting the peaks (minima/maxima) in the signal and using them: In some test cases I don't have oscillations, i.e. no peaks. Besides I am not sure how to use the peaks to compute the damping ratio.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there any robust method to do this type of analysis?&lt;/p&gt;&#xA;" OwnerUserId="5391" LastEditorUserId="2246" LastEditDate="2016-03-09T15:10:16.557" LastActivityDate="2016-03-11T19:11:29.737" Title="Determining damped frequency and damping ratio from time domain data" Tags="|control-engineering|aerospace-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7844" PostTypeId="1" CreationDate="2016-03-09T15:01:03.503" Score="0" ViewCount="136" Body="&lt;p&gt;I have a tank of oil which I plan to heat 24/7 with constant 15 kW. I want to dissipate this energy using water as a coolant. I have the outside unit of an out of order split type air conditioner that could be used as a radiator to heat the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I dissipate this much energy on such a radiator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I exchange heat between the oil and the water? I thought of a heat exchanger used in floor heating. I saw some rated 70 kW and above. But I can't imagine how it works seeing it's small size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tank is about (1*1*1) m, filled with oil. I would use only as much oil as needed to immerse the electronics I want to keep cool at &amp;lt;50 °C. The energy to be dissipated is constant 15 kW.&#xA;The A/C radiator with a fan is installed ~5 m above ground, the tank being on the ground.&#xA;I want cooling to work also in the hottest summer, topping at about 40 °C on the hottest days of July.&#xA;The system must be closed, because I don't have infinite amount of water.&lt;/p&gt;&#xA;" OwnerUserId="5392" LastEditorUserId="5392" LastEditDate="2016-03-10T02:54:37.320" LastActivityDate="2016-03-11T20:39:06.557" Title="What is the most cost efficient way to cool a tank of oil with water?" Tags="|heat-transfer|cooling|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7846" PostTypeId="1" AcceptedAnswerId="7847" CreationDate="2016-03-09T20:09:46.613" Score="3" ViewCount="292" Body="&lt;p&gt;I'm planning to build a head-mounted camera stabilizer for pro GoPro camera. The idea is that when I tilt my head the camera would stay at a user-set angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using an accelerometer and a gyro to measure camera angle $\theta$ and angular velocity $\dot{\theta}$, and some motor to control the camera angle. I derived the following differential equation for the system (which is basically an inverted pendulum) where $l$ is length of the rod, $\theta$ is the angle of the camera due to gravity, $m$ is the mass, $g$ is the gravitational acceleration, $b$ is damping coefficient (due to friction) and $\tau$ is torque. $\tau_{motor}$ will be the input of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau_{total} = \tau_{motor} - \tau_{gravity} - \tau_{friction}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I\ddot{\theta} = \tau_{motor} - mgl\sin\theta - b\dot{\theta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ml^2\ddot{\theta} = \tau_{motor} - mgl\sin\theta - b\dot{\theta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ddot{\theta} = \frac{1}{ml^2}\tau_{motor} - \frac{g}{l}\sin\theta - \frac{b}{ml^2}\dot{\theta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here I'll continue designing the control algorithm using state-space methods such as state-feedback control (I choose camera angle $\theta$ and angular velocity $\dot{\theta}$ as states and $\tau_{motor}$ as input). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uKxCa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uKxCa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here's the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to test how my control system behaves when a person tilts his head (I'll be using Matlab and Simulink). How should I incorporate the tilting of the head (or rotating of the body of the motor since the motor will be attached to head) into my control design? I guess it can be thought as a disturbance to system (or another input)? How should it be incorporated to the system model (meaning the differential equation above)? Should there be maybe more differential equations? I thought it would be nice to give the system the the angle of the body of the motor (meaning the angle of the head) as function of time and observe how the control system works. In other words I would like to see how the controller works if I tilt my head for example back and forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, is this a good approach or should I just use another torque $\tau_{disturbance}$ to test the controller? I'm pretty sure that it isn't the same as tilting the motor body, but would perhaps lead to good enough model for the controller?&lt;/p&gt;&#xA;" OwnerUserId="5398" LastEditorUserId="1832" LastEditDate="2016-03-09T21:08:40.097" LastActivityDate="2017-09-25T17:09:46.047" Title="System model for motor controlled inverted pendulum with rotating motor body (used in camera stabilizer application)" Tags="|control-engineering|modeling|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7847" PostTypeId="2" ParentId="7846" CreationDate="2016-03-09T20:35:32.530" Score="3" Body="&lt;p&gt;There's a couple questions you can ask:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can you control the signal? If not, then it's either a noise or disturbance input. &lt;/li&gt;&#xA;&lt;li&gt;Is it affecting your system (plant), or is it affecting your measurements? If it's affecting the system, it's a disturbance input; if it's affecting the measurements, it's noise. Note that noise can be colored, biased, periodic, etc. That is, pretty much any signal can be noise or a disturbance. &lt;/li&gt;&#xA;&lt;li&gt;Does the disturbance have enough power that is effectively a position regulator?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Regarding point 3, if the power output of the human neck is comparable with the power output of your stabilization system, then its position will vary with reaction forces applied by your stabilization system. If the power output of the neck is significantly higher, then it becomes effectively rigid and is better modeled as a position regulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on these questions, I would guess that the power of your stabilizer is much less than the power of the neck, so I would treat the neck as a positioner. You can't control it, but it affects the plant, so it's a disturbance input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would include it with the definition of angular acceleration $\ddot{\theta}$, such that:&#xA;$$&#xA;\ddot{\theta} = \frac{1}{ml^2}\tau_{motor} - \frac{g}{l}\sin\theta - \frac{b}{ml^2}\dot{\theta} + \ddot{\theta}_{\mbox{head}} \\&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-03-09T20:35:32.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7848" PostTypeId="2" ParentId="7837" CreationDate="2016-03-09T22:01:39.423" Score="3" Body="&lt;p&gt;Not being able to apply loads along the member seems like a limitation of the software (without adding more nodes). If you can only apply loads at the nodes then the proper way to do this is to get the Fixed-End Reactions of the distributed load in terms of point loads AND MOMENTS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, imagine that your member that is subjected to the DL is fully fixed at each end and get the reactions. Then you switch the sign of the reactions and these are the loads you will need to apply at the nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So consider the member below. I will use &lt;a href=&quot;https://skyciv.com?cc=SE_SS&quot; rel=&quot;nofollow noreferrer&quot;&gt;SkyCiv Structural 3D&lt;/a&gt; software to show how this works. In red we have the DL load and in black are the reactions at the fixed ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LnZp6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LnZp6.png&quot; alt=&quot;Fixed End Reactions using SkyCiv Structural 3D&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we can apply those reactions (flipping the sign) to the actual structure now as you can see below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/INF1H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/INF1H.png&quot; alt=&quot;Solved structure with point loads and moments using SkyCiv Structural 3D&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we can actually use the same structural analysis software to verify that this is in fact the correct way to convert the DL to its equivalent nodal loads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uLfTx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uLfTx.png&quot; alt=&quot;Solved structure with distributed load using SkyCiv Structural 3D&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you'll notice that the reactions and displacements are identical between the models! Any minor differences are due to rounding of the reactions which were used as the loads. However, by modelling it with point loads and moments, the shear force and bending moment diagrams will only have the correct values at the nodes (not throughout the member obviously) due to the differences between the loads. If you're only looking for displacement and reactions then it is fine to model it like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; The dimensions of this frame did not match yours, so these values do not apply to you - it was just a simple example. The member I used was 3m across and 1m high. The software used was SkyCiv Structural 3D. You can signup for a &lt;a href=&quot;https://skyciv.com/free-signup?cc=SE_SS&quot; rel=&quot;nofollow noreferrer&quot;&gt;free account here&lt;/a&gt;. The free account can analyse structures with up to 5 members so your portal frame can easily be solved. Of course you won't need to worry about converting the DL to its equivalent nodal loads because the software can handle mid-member loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2016-03-10T10:31:14.560" LastActivityDate="2016-03-10T10:31:14.560" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7849" PostTypeId="2" ParentId="7837" CreationDate="2016-03-10T00:18:12.367" Score="2" Body="&lt;p&gt;If your FEM S/W can't handle distributed loads, throw it away and find something which does.  It's a pain having to convert DL to NL all the time.  Your software should make your life easier, not harder.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-03-10T00:18:12.367" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7850" PostTypeId="1" AcceptedAnswerId="7862" CreationDate="2016-03-10T02:33:49.343" Score="2" ViewCount="10740" Body="&lt;p&gt;So there's a permutation of rotation and translation each being either fixed or free for a column.  What do each mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the column is toe nailed to a plywood / beam underneath which is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's in a steel bracket nailed to a column below what is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it has a 2x6 plate of a wall on top nailed to it what is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does bracing in the middle of the column for either affect it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My case is a 8x8 wood post&lt;/p&gt;&#xA;" OwnerUserId="3037" LastEditorUserId="129" LastEditDate="2016-03-10T20:38:32.637" LastActivityDate="2016-03-10T21:15:58.097" Title="For a column, what does &quot;rotation fixed and translation fixed&quot; mean" Tags="|structural-engineering|structural-analysis|wood|columns|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7851" PostTypeId="1" CreationDate="2016-03-10T03:54:58.357" Score="4" ViewCount="645" Body="&lt;p&gt;Check the attached photo of a gasoline station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b8Jwp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b8Jwp.jpg&quot; alt=&quot;Gas Station&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two panel boxes at the center of the photo (in the corner of the station) are standard panels (IP-67), which enclose standard circuit breakers for the main gas pumps of the station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are less than 7 linear meters on separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to ask you, if you have experience on electrical EX-proof facilities: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there any mistake here? Or is everything right according to the EX-proof codes?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5400" LastEditorUserId="5400" LastEditDate="2016-03-28T05:15:56.547" LastActivityDate="2016-03-28T05:15:56.547" Title="Is an Explosion Proof Circuit Breaker Panel Required at a Gas Station?" Tags="|electrical-engineering|safety|gas|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7852" PostTypeId="1" CreationDate="2016-03-10T05:07:54.550" Score="4" ViewCount="326" Body="&lt;p&gt;I need a bearing that the inner bore will stay level and balanced when the out ring is rotated. My son with autism lacks the motor function to feed himself with a spoon. He can turn his wrist, when the spoon goes to his mouth its 90 degrees off. I need to be able to insert bearing into a handle and have a spoon in the bore that will stay level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CG0iT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CG0iTm.jpg&quot; alt=&quot;Sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5402" LastEditorUserId="1632" LastEditDate="2016-03-10T17:22:21.693" LastActivityDate="2016-08-08T11:20:24.137" Title="What type of bearing could be used to keep a utensil level while being moved from plate to mouth?" Tags="|mechanical-engineering|bearings|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7853" PostTypeId="1" AcceptedAnswerId="7861" CreationDate="2016-03-10T07:36:36.050" Score="4" ViewCount="3653" Body="&lt;p&gt;In structural dynamic mode analysis, one would have to carry out modal analysis in order to get different modes that contribute to the response of the seismic ( or wind) force. &lt;a href=&quot;http://www.vibrationdata.com/tutorials2/ModalMass.pdf&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a nice summary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a large structural model, there are so many modes, and we can't take them all. So we can only take the first few prominent modes ( with low frequencies), assuming that the modes will grow less and less important as the frequency becomes higher and higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Then the issue arises&lt;/strong&gt;: how to judge the relative importance of different modes ( whatever the term 'importance' may mean?) I find that the often-cited &lt;a href=&quot;https://en.wikipedia.org/wiki/Structural_dynamics#Modal_participation_factor&quot; rel=&quot;nofollow&quot;&gt;mode participation factor&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Gamma=\frac{\sum M \phi}{\sum\phi^T M \phi}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;doesn't work, because it depends on how we normalize the eigenvector, as one can readily see from the above formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other factor that allows us to judge whether a mode is more important or not?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastActivityDate="2016-03-10T18:38:08.277" Title="How to calculate the relative contribution of each mode in seismic dynamic mode analysis?" Tags="|structural-analysis|seismic|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7855" PostTypeId="1" AcceptedAnswerId="7856" CreationDate="2016-03-10T10:44:40.557" Score="6" ViewCount="9765" Body="&lt;p&gt;How can I find out the function equations in a diagram with 5 curves? Measuring by hand is very inaccurate and takes a lot of time. Any other possibilities?&lt;a href=&quot;https://i.stack.imgur.com/sZIQo.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sZIQo.png&quot; alt=&quot;Example: See uploaded Picture!&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5007" LastEditorUserId="33" LastEditDate="2016-03-10T13:50:42.683" LastActivityDate="2016-04-02T05:20:11.003" Title="How can I determine a function equation from a graph image?" Tags="|mathematics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7856" PostTypeId="2" ParentId="7855" CreationDate="2016-03-10T11:08:37.463" Score="12" Body="&lt;p&gt;I don't know how this function was obtained (or what it is supposed to represent). But you could try searching the literature to see if someone has published equations of the curves. However, looking at it I would suspect that there is not a straightforward expression for these curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best option I can think of is to use  &lt;a href=&quot;http://connectedresearchers.com/graph-digitizer-comparison-16-ways-to-digitize-your-data/&quot;&gt;graph digitization software&lt;/a&gt; to extract the data, then use a curve fit to obtain an equation for each line (lots of software can do this: Excel, MATLAB, etc). By this method you could get at least as much accuracy as reading values off the graph, and it could then be included in a computer program. &lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="2246" LastEditDate="2016-03-10T19:14:11.690" LastActivityDate="2016-03-10T19:14:11.690" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7857" PostTypeId="2" ParentId="7851" CreationDate="2016-03-10T13:29:13.673" Score="2" Body="&lt;p&gt;I see nothing in your picture to indicate that the circuit breakers need to be explosion proof.  They don't appear to be in a zone 0 location, nor would I expect that any explosion proof volume would extend thru the conduit to the circuit breakers.  That just doesn't make sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, cars aren't zone 0 rated in the first place, and neither are all kinds of things you are allowed to have in the car and on your person while filling up at a gas station.  The protection is between the gasoline and where you can get at it, not in the zone around the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your whole concept doesn't make sense.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-03-10T13:29:13.673" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7858" PostTypeId="2" ParentId="7855" CreationDate="2016-03-10T13:39:06.347" Score="4" Body="&lt;p&gt;You have a single-valued function dependent of two variables.  There are many ways to model that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know something about what this function represents, then going back to the physics might yield a useable equation with only a few coefficients.  Digitize a bunch of points, throw them at a least-squares error minimizer and see how close the result is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know nothing more about the physics behind the function, then the standard answer is a polynomial.  From the general look of the curves, I'd say you need at least a 3rd order polynomial (a cubic).  With two independent variables and the various crossover terms, that comes out to 10 coefficients to adjust to get these curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, I'd digitize 10 values for each curve, one where they hit each marked X division.  It really wouldn't take long to just sit down and do it.  That gives you 50 points, which should do a reasonable job of allowing a least-squares error solver to divine the coefficients for you.  For the polynomial case it would be trying to solve for 10 value.  In the physics modeling case, hopefully fewer.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-03-10T13:39:06.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7860" PostTypeId="2" ParentId="7852" CreationDate="2016-03-10T15:38:14.997" Score="1" Body="&lt;p&gt;I'm gonna assume you need the spoon's plate to work like a gyroscope, where whichever way your son moves &amp;amp; holds the spoon handle, the spoon's plate stays upright.&#xA;I think you need a gimbal bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard gimbal bearing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FJrAH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FJrAH.jpg&quot; alt=&quot;A standard gimbal bearing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ball bearing was used to stabilise a camera, and I think the lessons from the project could be applied in stabilizing your spoon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j9dNJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j9dNJ.jpg&quot; alt=&quot;gimbal camera stabilization&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5408" LastEditorUserId="5408" LastEditDate="2016-03-10T15:49:16.873" LastActivityDate="2016-03-10T15:49:16.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7861" PostTypeId="2" ParentId="7853" CreationDate="2016-03-10T16:11:17.317" Score="2" Body="&lt;p&gt;You should be normalizing your eigenvector so that the generalized mass matrix (defined by $ \hat{m} = \phi^T M \phi$) is the identity matrix, and therefore, the generalized mass of the &lt;em&gt;r&lt;/em&gt;th mode (defined by $ M_r = \phi^T_r M \phi_r$) has the value of 1. This should give you consistent modal participation factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modal mass participation ratio is widely used as the metric to determine the relative significance of modes in a modal response spectrum analysis. It has even been codified in many of the codes of record. For example, ASCE 7-05 (7-10 is at home, sorry), Section 12.9.1 states that,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The analysis shall include a sufficient number of nodes to obtain a combined modal mass of at least 90 percent of the actual mass in each of the orthogonal horizontal directions of the response considered in the model.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And to elaborate on what I think may be an underlying question in your query (and the code stipulation), you calculate total effective mass as, $$ M_{eff} = \Sigma\Bigg[\frac{\Gamma_i^2}{\hat{m}_{ii}} \Bigg] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $ M_{eff} $ is greater than 90% of your overall system mass, you've considered an appropriate number of modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the effective mass participation of each mode may be used as a guide to determine the relative &quot;importance&quot; of each node, $$ m_{eff,i} = \frac{\Gamma_i^2}{\hat{m}_{ii}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dynamics of Structures by Chopra is a really good reference if you are attempting to do this all by hand.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strike&gt;Edit: Check out this &lt;a href=&quot;http://www.vibrationdata.com/tutorials2/ModalMass.pdf&quot; rel=&quot;nofollow&quot;&gt;pretty good discussion and examples&lt;/a&gt; written by Tom Irvine.this&lt;/strike&gt; &lt;em&gt;Oops, this was your original link!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2016-03-10T18:38:08.277" LastActivityDate="2016-03-10T18:38:08.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7862" PostTypeId="2" ParentId="7850" CreationDate="2016-03-10T18:36:00.670" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So there's a permutation of rotation and translation each being either fixed or free for a column. What do each mean?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To say that a specific degree of freedom of a support is &quot;fixed&quot; is to say that the support cannot translate along, or rotate about, that axis. For example, if it is stated that the translation of a beam is fixed in the X axis at one end, that means that the beam cannot translate, or move, along the X axis (in the direction of +/- X) at that end. And similarly for rotation, if it is stated that the rotation of a beam is fixed about the X axis at one end, that means that the beam cannot rotate about the X axis at that end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we design things, we often assume an ideal, or perfect, boundary condition. For example, idealizing a beam as being &quot;simply supported,&quot; which means we assume it can rotate about either end and translate along its longitudinal axis at one end, usually enveloped the beam design forces of bending moment and shear force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality nothing we design is ever actually perfectly rigid (fixed) or perfectly free to translate or rotate, we make those assumptions to simplify the design process. In practice, we usually model support conditions that are a combination of fixed translations and free rotations, or vice versa. A very common boundary condition is called &quot;pinned.&quot; A pinned support is a support that is not allowed to translate (fixed for translation) but is allowed to rotate freely about any axis (free for rotation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer you specific questions, for an 8x8 post,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the column is toe nailed to a plywood / beam underneath which is it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pinned (free to rotate, not allowed to translate).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If it's in a steel bracket nailed to a column below what is it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Depends on the bracket and attachment to the bracket, but treating as pinned (free to rotate, not allowed to translate) will usually be conservative.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If it has a 2x6 plate of a wall on top nailed to it what is it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Pinned (free to rotate, not allowed to translate) .&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does bracing in the middle of the column for either affect it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2016-03-10T21:15:58.097" LastActivityDate="2016-03-10T21:15:58.097" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7863" PostTypeId="2" ParentId="7852" CreationDate="2016-03-10T19:17:16.287" Score="4" Body="&lt;p&gt;One difficulty here is that the masses involved are quite small so you need a very low friction bearing. Many industrial bearings are designed for high loads and speeds with quite close tolerances so in many cases they can be a little stiff under small forces even though this is small in the context of their intended application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case you might be better off with a plain bearing with quite loose tolerances, for example a polished stainless steel shaft in a Teflon bushing as low friction is a much higher priority in this case than eliminating radial play. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach would be to use a teflon rod with a hole down the middle as the handle and have the shaft of the spoon attached to a ground steel rod which passes through it. Using the full length of the handle as a long bearing will support it better and give you more room to work with rather than trying to find a small bearing which fits forward of the handle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will also help if you can 'crank' the shaft of the spoon so that the contents of the spoon are a little lower than the axis of the handle so you get a pendulum effect, similarly you would want to keep the centre of mass of the bowl of the spoon as low as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If might help if you made up a prototype in aluminium so that it is easy to bend by hand and quickly try different bend configurations to quickly work out what works best. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-12T20:04:47.047" LastActivityDate="2016-03-12T20:04:47.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7864" PostTypeId="1" AcceptedAnswerId="7869" CreationDate="2016-03-10T19:26:09.497" Score="3" ViewCount="22957" Body="&lt;p&gt;I am  bit confused. The slip direction is the direction of movement of a dislocation which is denoted by the burgers vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense in an edge dislocation, because the stress is applied perpendicular to the dislocation line and the movement of the line is in the direction of the applied stress. The burger vector is also perpendicular to the dislocation line for edge dislocation. So it makes sense that the direction of motion corresponds to the direction denoted by the burgers vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for screw dislocation, the motion of the dislocation is perpendicular to the applied stress and since the burger vector points in the direction of the dislocation line; the burger vector cannot point in same direction as the direction of movement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The professor even said that the direction of motion for a screw dislocation is perpendicular to the burgers vector. So how can a burger vector then correspond to the direction of slip when this direction of slip is the direction of movement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the points that confuse me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;slip direction: specific direction along which dislocation motion occurs&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;burger vector: direction corresponds to a dislocation's slip direction&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;screw dislocation: direction of motion is perpendicular to applied stress and the motion is perpendicular to the burger vector&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qHRls.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qHRls.gif&quot; alt=&quot;screw dislocation motion&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So these 3 things conflict with each other:&#xA;the motion of the dislocation is here perpendicular to the burger vector (and i have heard is always perpendicular for screw dislocations), so how can a burger vector denote the slip direction if the slip direction is the direction of the motion of the dislocation? For it to denote the motion of the dislocation shouldn't it align/be parrallel?&#xA;Or is that the mistake I am making, that a burger vector tells about  direction of the dislocation motion, but that it doesn't mean that the motion is in the same direction?&#xA;That there is just always a fixed relation between the two depending on the type of dislocation, but that the relation isn't always parrallel.&#xA;It is 90 degrees for screw and 0 degrees for edge related to the dislocation motion. So that indeed the burger vector says something about the direction of the motion but just that it doesn't say that they always in same direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because again clearly here the dislocation motion is not in the same direction as the direction of the burger vector;&#xA;The motion is from front to back, while the burger vector points in direction of the shear stress in this picture&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="4117" LastEditDate="2016-03-10T22:35:38.587" LastActivityDate="2019-01-30T21:17:51.067" Title="Relation between burgers vector and direction of motion for dislocations?" Tags="|materials|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7865" PostTypeId="2" ParentId="7816" CreationDate="2016-03-10T19:41:14.063" Score="1" Body="&lt;p&gt;It is common for timing belt systems to have a tensioner pulley, effectively an idler mounted on a sprung radial arm. In most situation this is just to take up slack from wear and stretch and generally ensure consistent tension throughout its life but if you allow enough travel it could certainly achieve what you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the amount of travel you require you might need a constant force spring to get best results and you would need to consider whether radial or liner travel for the idler/tensioner is best. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may even be that there is an off the shelf component that you can just bolt into the system. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-03-10T19:41:14.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7866" PostTypeId="1" AcceptedAnswerId="7887" CreationDate="2016-03-10T19:42:15.843" Score="1" ViewCount="2400" Body="&lt;p&gt;For my racing game I am using a simplified Pacejka Magical Formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mu_{lateral} = \mu_{0} * \mu_{1} * D * \sin(C * \arctan(B * x - E * (B * x - \arctan(B * x))))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\mu_{lateral}$ is sideways coefficient of friction&lt;/li&gt;&#xA;&lt;li&gt;$\mu_{0}$ is wheel friction coefficient&lt;/li&gt;&#xA;&lt;li&gt;$\mu_{1}$ is surface coefficient friction&lt;/li&gt;&#xA;&lt;li&gt;$B$ is wheel lateral stiffness&lt;/li&gt;&#xA;&lt;li&gt;$C$ is wheel lateral peak factor&lt;/li&gt;&#xA;&lt;li&gt;$D$ is wheel lateral shape factor&lt;/li&gt;&#xA;&lt;li&gt;$E$ is wheel lateral curvature&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, I don't know what are realistic values for $B, C, D, E$. I know well the first ones and I also assume that $C$ may be around &lt;code&gt;1.1&lt;/code&gt;. However, what's realistic stiffness, shape factor and curvature in sideways directions for different types of real-life tires?&lt;/p&gt;&#xA;" OwnerUserId="5281" LastActivityDate="2016-03-11T14:58:31.940" Title="Pacejka Formula - natural values" Tags="|automotive-engineering|friction|wheels|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7867" PostTypeId="2" ParentId="6048" CreationDate="2016-03-10T20:06:04.143" Score="1" Body="&lt;p&gt;The crucial piece of information is the torque curve of the engine ie. the function which relates torque to engine speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any gearbox, manual or automatic, has two primary concerns the first is to allow access to the desired level of torque at any given road speed, for example changing down to give extra torque for acceleration eg overtaking on a hill. The second consideration is to have the engine at the most fuel efficient RPM for a particular cruising speed. Many automatic gearboxes will automatically change down if the throttle is fully depressed as full throttle is interpreted as a demand for maximum torque.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also some circumstances where slightly different considerations apply. For example when traction is limited eg in the middle of a high speed corner or in wet or icy condition it may be desirable to select a higher gear than normal to limit torque and make the torque output less sensitive to throttle input to better control traction. &#xA;Similarly downshifting in advance of a hazard allows engine braking to contribute to overall braking effect but conversely too aggressive a downshift can lock the driving wheels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most basic system simply change up or down at set points to keep the RPM in a defined range. More sophisticated systems may also work in concert with other systems such as ABS, stability control or traction control. But in general the goal is to control the relationship between engine RPM and road speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also important to consider the gear ratios used as these many not necessarily be equal steps between all gears and gear ratios will be selected according to the characteristics of the engine and the expected use of the vehicle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most sophisticated systems will aim to emulate the sort of decisions that a driver makes when selecting a particular gear with a manual box. These can also be linked to engine mapping to create different modes related specific road conditions or to achieve a balance between performance and economy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example an economy mode might change up earlier to keep RPM in the most efficient rage while a 'sport' mode might attempt to keep RPM in the region of peak torque/power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In high performance cars these modes may also be related to differential and suspension settings and include modes for standing starts and overtaking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific characteristics of an engine's torque curve are often referred to 'drivability', it is generally desirable that an engine has a predictable response to throttle and a reasonably flat and linear torque curve or at least a broad RPM range where this is the case. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-11T20:48:35.690" LastActivityDate="2016-03-11T20:48:35.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7868" PostTypeId="2" ParentId="7844" CreationDate="2016-03-10T20:32:36.893" Score="1" Body="&lt;p&gt;The problem with using oil to cool the plant and the cooling the oil with a heat exchanger is that you are introducing a bottleneck into your system in that you are relying on convection to transfer heat from the hot surface to the heat exchanger through the oil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from anything else this greatly complicates the analysis and control of the problem. A big risk is that the average temperature of the oil could be within acceptable limits but you start to vapourise it at the hot surface which drastically drops the heat transfer rate and is difficult to detect or do anything about. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sort of context cost effectiveness can be as much about a simple and conservative solution which can be made to work with a fairly modest design input and this approach often ends up cheaper in the end than something which takes a lot of development and modelling to get working effectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using forced convection by circulating the oil itself greatly simplifies the problem and gives you a lot more direct control over the rate of heat transfer simply by increasing the flow rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you already have a cubic foot of oil in the tank the extra you would need to fill piping and a radiator is not huge, bearing in mind also that water used for cooling usually requires additives to prevent corrosion etc which in themselves may not be cheap. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-11T20:39:06.557" LastActivityDate="2016-03-11T20:39:06.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7869" PostTypeId="2" ParentId="7864" CreationDate="2016-03-10T20:59:54.020" Score="2" Body="&lt;p&gt;&lt;strong&gt;Summary: For an edge dislocation, the Burgers vector is parallel to dislocation motion. For a screw dislocation, the Burgers vector is parallel to the dislocation. The Burgers vector is always parallel to slip.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram below shows both edge and screw dislocations in an indealized cubic lattice. The edge dislocation is on the front face and the screw dislocation is on the right face. Burgers vectors are denoted with white-tipped arrows, and point in the same direction for the two dislocations shown. I have added red arrows that indicate applied shear. Slip occurs along the resolved shear direction, which in this case happens to also be the applied shear direction. Note that slip is the same for both dislocations, and the Burgers vector points in the same direction as slip in both cases. Note that the edge dislocation itself is perpendicular to slip while the screw dislocation is aligned with slip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If slip continued, the edge dislocation would proceed in the same direction as slip. The screw dislocation would recede away from the front plane, toward the back plane. It must move that direction in this case because otherwise the Burgers vector would have to get longer as strain proceeds and new edge dislocations are introduced, which is impossible. An analogy would be tearing a piece of paper. As you pull the ends of the tear further apart, the tip of the tear recedes along the length of the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3VRRZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3VRRZ.png&quot; alt=&quot;Dislocation diagram showing shear in red.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image is a modified version of an image found at &lt;a href=&quot;http://www.geology.um.maine.edu/geodynamics/AnalogWebsite/UndergradProjects2010/PatrickRyan/Pages/Background.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.geology.um.maine.edu&lt;/a&gt;. Original credit: Passchier and Trouw, pg 33 (2005).&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-03-10T23:00:27.807" LastActivityDate="2016-03-10T23:00:27.807" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7870" PostTypeId="2" ParentId="7830" CreationDate="2016-03-10T21:45:46.963" Score="3" Body="&lt;p&gt;This is well-known theory in finite element analysis. There are two different effects that contribute to the effective stiffness of the structure in addition to the elastic stiffness $K_e$ given by Euler-Bernoulli-Timoshenko beam theory, and you need to include both of them to get the right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Rotating the beam creates axial tension in the beam, which will be zero at the tip and increases to a maximum value at the rotation axis. This tension creates an additional stiffness that opposes lateral bending of the beam, in the same way that the tension in a guitar string affects its vibration frequency and hence the pitch of the note produced. (Of course in a guitar string, $K_e$ is negligible, unlike your beam). This stiffness term was traditionally called the &quot;geometric stiffness&quot; (for reasons which aren't worth explaining) but a more logical name is &quot;stress stiffness&quot; often denoted by $K_\sigma$, since it depends only on the internal stresses in the structure and not on the material properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, $K_\sigma$ for a &lt;em&gt;compressive&lt;/em&gt; stress is what causes Euler buckling in columns. At the &quot;critical load&quot; for buckling, $K_\sigma$ &quot;cancels out&quot; the elastic stiffness and when $K_e + K_\sigma = 0$, the structure can't resist the applied loads. For &lt;em&gt;tension&lt;/em&gt; stresses, $K_\sigma$ &lt;em&gt;increases&lt;/em&gt; the stiffness of the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) If the beam deflects from its original straight configuration, &lt;em&gt;the direction and/or the point of application of the centrifugal loads changes&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the beam rotates about a vertical axis and the beam deflects &lt;em&gt;vertically&lt;/em&gt;, the CF load on each particle of the beam is still in a &lt;em&gt;horizontal&lt;/em&gt; direction, and therefore the CF loads at the beam tip will create a bending moment that tends to &lt;em&gt;straighten out&lt;/em&gt; the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the beam deflects &lt;em&gt;tangentially&lt;/em&gt; (for example while you are ramping up the rotation speed from zero) the direction of the CF load is &lt;em&gt;radially away from the axis&lt;/em&gt;, and that generates a bending moment that tends to &lt;em&gt;increase&lt;/em&gt; the bending of the beam, not reduce it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual name for both these effects is &quot;follower forces&quot;, because the direction of the applied loads &quot;follows&quot; the deflection of the structure. The corresponding stiffness terms are usually called the &quot;load stiffness&quot; and denoted by $K_L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original equation for statics using only the elastic stiffness,&#xA;$$K_e u = F$$&#xA;becomes $$[K_e + K_\sigma(\omega) + K_L(\omega)]u = F.$$&#xA;In principle you could get expressions for $K_\sigma$ and $K_L$ for a simple situation (e.g. a rectangular straight beam) but in practice it's much easier to use a Finite Element package that includes these effects. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-03-10T21:45:46.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7871" PostTypeId="2" ParentId="7801" CreationDate="2016-03-10T22:31:26.170" Score="0" Body="&lt;p&gt;The simplest way to solve this problem would be to hinge the beam in the middle, so that you have two statically determinate beams.   Then there is no Bending Moment at B, so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fa = Fc = (w*2.5*2.5/2)/2 = 1.5625*w&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fb = w*5 - Fa – Fc = 1.875*w&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the central chain serve any purpose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a slight mistake in your formula – taking moments about A to the right-hand side, the moment due to the UDL is not w * L * L/2, but w*(Xb + Xc + Xd) * (Xb + Xc + Xd)/2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you keep it as a continuous beam, then I hope you do not design it “on the limit”, especially if it is a safety-critical situation (And 50kg dropping onto someone could harm them.)  I assume you will have a means of adjusting say the chain at B, before applying the UDL,  until the three chains all feel to be carrying equal, or near-equal loads.  Even 1mm may be enough to go from the chain at B being completely slack, to it carrying the total load on the beam&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a theoretical level, FEA is the way to go.  It is so easy to change the assumptions about chain elasticity, beam bending elasticity, pre-tensioning of each chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically, on the basis of the limited information you have given us, I don't think you will be far out if you assume the beam is hinged in the centre, even if it is continuous there, provided you adjust the chains before applying the UDL.&lt;/p&gt;&#xA;" OwnerUserId="5381" LastActivityDate="2016-03-10T22:31:26.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7872" PostTypeId="1" AcceptedAnswerId="7873" CreationDate="2016-03-11T01:51:25.283" Score="3" ViewCount="1327" Body="&lt;p&gt;Say I have a spring in the shape of the backbone of DNA. If I unwind the two parts which would have more potential energy when pushed all the way down, the double helical or both of the individual springs combined?&lt;/p&gt;&#xA;" OwnerUserId="5412" LastEditorUserId="2819" LastEditDate="2016-03-11T04:03:48.427" LastActivityDate="2016-03-11T04:03:48.427" Title="Would a double helix spring be more or less effective than two springs?" Tags="|design|springs|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7873" PostTypeId="2" ParentId="7872" CreationDate="2016-03-11T04:02:29.920" Score="4" Body="&lt;p&gt;I think two equal springs would test very close to a spring with double the spring constant. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Series_and_parallel_springs&quot; rel=&quot;nofollow noreferrer&quot;&gt;adding springs in parallel equation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two springs may be somewhat less material efficient than a single spring; because the cross section of a spring wire has an &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_area_moments_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;area moment of inertial&lt;/a&gt; (just like a beam) that provides the restoring force and the equation is not linear. Think of it like the strength of a cantilever wood 2x4 vs the lower cantilever strength of an equivalent mass of four wood 1x2 members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reason you may want to do it would be to achieve a desired spring constant with springs you already have or within what is available. However, most of the time this is done with nested springs because the ends are typically flattened and would not permit another spring without modification to the ends. Another issue would be spring travel and fill. Notice that the green and brown die springs below would not be able to even fit a duplicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xy81I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xy81I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;One spring design that does have space advantages over a standard coil is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Wave_spring&quot; rel=&quot;nofollow noreferrer&quot;&gt;wave spring&lt;/a&gt;. It has a shorter height than a coil spring of equal spring constant. Here are some &lt;a href=&quot;https://www.rotorclip.com/wave_spring_advantage3.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;wave spring applications&lt;/a&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Oa23L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oa23L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-11T04:02:29.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7875" PostTypeId="2" ParentId="7852" CreationDate="2016-03-11T06:58:47.800" Score="1" Body="&lt;p&gt;The first question any engineer, if faced with a new problem, is if it allready has been solved. I assume you talked to people with similar special needs (unable to bend wrist) or their caretakers and didn't come up with a good solution.&lt;/p&gt;&#xA;&lt;p&gt;I think the question for what kind of bearing you want is a sideshow, the first question is if your design is viable. I think not:&lt;/p&gt;&#xA;&lt;p&gt;I assume your son holds the spoon before him, lower arm more or less level with the table, shoves the spoon into the food. Then he tilts the lower arm upward or upward without compensating in the wrist, so the plate of the spoon ends up upright. I assume there is litte inward movement, because then you'd have spillage towards the handle. So to achieve your goal - have the meals less frustrating and messy - you need make sure the palte of the spoon stays level.&lt;/p&gt;&#xA;&lt;p&gt;Before you think about bearings, you need to consider which forces act on the spoon. In a spoon as in your drawing, the center of gravity of spoon + food will be close to the axis of rotation. Here's my proposal: the spoon attaches to a rod that runs the length of the handle. At the back side of the handle, the rod protrudesand makes an L-bend downward, where you attach a small weight. This will turn the spoon in the desired direction.&lt;/p&gt;&#xA;&lt;p&gt;For the actual build, I think &lt;a href=&quot;https://engineering.stackexchange.com/a/7863/61&quot;&gt;Chris Johns approach&lt;/a&gt; is best:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In your case you might be better off with a plain bearing with quite loose tolerances, for example a polished stainless steel shaft in a Teflon bushing as low friction is a much higher priority in this case than eliminating radial play.&lt;/p&gt;&#xA;&lt;p&gt;One approach would be to use a teflon rod with a hole down the middle as the handle and have the shaft of the spoon attached to a ground steel rod which passes through it. Using the full length of the handle as a long bearing will support it better and give you more room to work with rather than trying to find a small bearing which fits forward of the handle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="61" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-03-11T06:58:47.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7878" PostTypeId="1" CreationDate="2016-03-11T11:02:43.910" Score="5" ViewCount="463" Body="&lt;p&gt;In level unaccelerated flight we have relation &#xA;$$&#xA;W=L=\frac{1}{2}\rho\cdot V_{stall}^2\cdot S\cdot C_{l,max}&#xA;$$&#xA;taken from book &lt;a href=&quot;http://home.anadolu.edu.tr/~gdurmus/hyo413/book/Aircraft-Design-A-Conceptual_Approach.pdf&quot; rel=&quot;nofollow&quot;&gt;Daniel P Raymer &quot;Aircraft Design: A Conceptual Approach&quot;&lt;/a&gt; equation (5.5) at page 85.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: why does this formula use $C_{l,max}$ and not $C_{l,min}$? In fact in the case that the $C_l$ in use is not $C_{l,max}$, and the speed is little more than $V_{stall}$ (valued for $C_{l,max}$), it happens that the lift is not enough and stall happens on aircraft anyway. If using $C_{l,min}$ the calculation is more cautious.&lt;/p&gt;&#xA;" OwnerUserId="4916" LastEditorUserId="90" LastEditDate="2016-03-12T01:31:23.970" LastActivityDate="2016-04-12T19:02:32.827" Title="Questions about the level flight lift equation" Tags="|aerospace-engineering|safety|aerodynamics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7879" PostTypeId="1" AcceptedAnswerId="7959" CreationDate="2016-03-11T11:22:28.093" Score="7" ViewCount="2129" Body="&lt;p&gt;I'm working on a flood defence scheme in which a section of RC flood wall becomes sheet piles. The reasons for this aren't important, but the solution isn't to simply make all walls one or the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My issue is how to connect these two very different types of wall in a way that provides an expansion joint for the concrete but is also water-tight.&lt;/strong&gt; This isn't an issue for connecting individual concrete panels as a waterstop is simply cast into the expansion joint, providing a seal. However, sheet piles don't have anything to 'cast' onto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea so far is to design a sort of concrete (possibly mass, probably RC) 'pilaster' at the end of a section of sheet pile that will allow a water stop to be cast into. However, sheet piles deflect a fair amount which is making me wary of casting a rigid concrete section at the end of the sheet pile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm at a bit of a loss here and struggling to find a precedent of a connection like this. SURELY it's been done before? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've attached a quick hand sketch. I'm not too bothered about detailed calculations for this; I'm more interested in solving the buildability issues at the moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimensions are subject to minor alterations but roughly:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The RC wall penetrates 1.5 m below ground level&lt;/li&gt;&#xA;&lt;li&gt;Sheet pile penetrates 6 m below ground level&lt;/li&gt;&#xA;&lt;li&gt;RC wall stem is ~350 mm thick&lt;/li&gt;&#xA;&lt;li&gt;Sheet pile section is 450 mm thick (AU25 section)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IQBlT.png&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Flood defence design sketch&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IQBlT.png&quot; alt=&quot;&quot; title=&quot;Flood defence design sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3093" LastEditorUserId="368" LastEditDate="2016-03-15T20:09:53.253" LastActivityDate="2016-03-21T14:33:47.097" Title="Connections between RC retaining walls and sheet pile retaining walls" Tags="|civil-engineering|geotechnical-engineering|reinforced-concrete|retaining-wall|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7880" PostTypeId="1" CreationDate="2016-03-11T12:45:06.237" Score="8" ViewCount="8819" Body="&lt;p&gt;I am trying to work out how much energy (ideally electrical heating) I need to put into a heater to raise the temperature of water from say 10 degrees C to 40 degrees C, if the water is flowing at 10 litres per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the heater is 80% efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the specific heat of water comes into this at 4.18 J/g degree C but how do I work in the flow rate?&lt;/p&gt;&#xA;" OwnerUserId="5419" LastEditorUserId="368" LastEditDate="2016-03-11T17:58:36.533" LastActivityDate="2016-03-12T01:54:05.600" Title="Energy required to raise temperature of water flowing through a heater" Tags="|thermodynamics|heat-exchanger|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="7882" PostTypeId="2" ParentId="7880" CreationDate="2016-03-11T12:57:55.537" Score="9" Body="&lt;p&gt;You just need to convert the volumetric flow rate to a mass flow rate by multiplying by its density. This is easy for water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$10\ \mathrm{l/hr} \cdot 1\ \mathrm{kg/l} = 10\ \mathrm{kg/hr} = 10^4\ \mathrm{g/hr} = \frac{10^4}{3600}\ \mathrm{g/sec} = 2.78\ \mathrm{g/sec}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can multiply by the specific heat and the temperature rise to get the power required.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="368" LastEditDate="2016-03-11T17:11:18.950" LastActivityDate="2016-03-11T17:11:18.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7883" PostTypeId="2" ParentId="7880" CreationDate="2016-03-11T13:20:42.387" Score="4" Body="&lt;p&gt;$$P = \dot m \cdot C_p \cdot \Delta T $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $P$ is the power required in Watts ($Joules/sec$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot m$ is the mass flow rate ($kg/sec$) (&lt;em&gt;you'll have to convert the volumetric flow given to a mass flow rate&lt;/em&gt; by using $\dot m = Flow_{Vol} \cdot \rho$ where $\rho$ is the density of the fluid)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_p$ is the specific heat of the fluid ($ \frac{Joules}{kilogram \cdot K}$) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $ \Delta T$ is the temperature difference in $K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give you the power required to heat the water as it flows. The amount of energy will, of course depend on how long the water flows. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="86" LastEditDate="2016-03-11T15:35:24.380" LastActivityDate="2016-03-11T15:35:24.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7884" PostTypeId="1" CreationDate="2016-03-11T13:50:50.707" Score="7" ViewCount="212" Body="&lt;p&gt;My company was asked to design a custom product several years ago. The original requester never bought it, but we've sold several a year through distribution. It's been a small side-line for a while, but we'd like to understand the broader use and potential market for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This product sits between a motor controller and a permanent magnet motor. In the event a signal is received, perhaps indicating a light curtain is broken, this unit disconnects the motor from the controller and applies resistors across the motor leads. This stops the motor. This is clearly useful primarily for safety applications. Things I'd like to understand:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What alternate solutions exist?&lt;/li&gt;&#xA;&lt;li&gt;Who would be most likely so specify such solutions?&lt;/li&gt;&#xA;&lt;li&gt;How can I most effectively reach those people?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18" LastEditorDisplayName="user4139" LastEditDate="2018-05-30T00:16:29.707" LastActivityDate="2018-05-30T14:13:51.910" Title="Under what circumstances would one need or not need an electronic brake for a permanent magnet motor?" Tags="|electrical-engineering|motors|safety|magnets|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7885" PostTypeId="1" AcceptedAnswerId="8376" CreationDate="2016-03-11T13:53:04.200" Score="1" ViewCount="90" Body="&lt;p&gt;A cascaded PID controller is depicted in following image (courtesy of &lt;a href=&quot;http://www.controlglobal.com/assets/Media/0512/article_547_fig3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;controlglobal.com&lt;/a&gt;):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fCyjy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fCyjy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary PID tracks the setpoint $SP_1$ and delivers a setpoint $SP_2$ to the secondary PID.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wondered if it could be a problem if the secondary PID does not have a nice behaviour w.r.t. tracking its setpoint, while the total result of the cascaded PID controller is OK. This would mean that the overall tracking of $SP_1$ is OK, which is the goal of the total controller, while some internal controllers are not tuned too well.&lt;/p&gt;&#xA;" OwnerUserId="3633" LastActivityDate="2016-04-11T03:23:00.380" Title="Is it important to have a good behaviour inside internal loops of a cascaded PID controller?" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7886" PostTypeId="2" ParentId="7880" CreationDate="2016-03-11T14:52:33.233" Score="4" Body="&lt;ul&gt;&#xA;&lt;li&gt;10 litres per hour &lt;/li&gt;&#xA;&lt;li&gt;10 kg per hour&lt;/li&gt;&#xA;&lt;li&gt;10,000 g per hour&lt;/li&gt;&#xA;&lt;li&gt;166.66 g per minute&lt;/li&gt;&#xA;&lt;li&gt;2.77 g per second&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Every second, you need to raise the temperature of 2.77 g of water by 30 °C.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2.77 g of water&lt;/li&gt;&#xA;&lt;li&gt;4.18 J/g specific heat&lt;/li&gt;&#xA;&lt;li&gt;11.6 J every second for 1 degree&lt;/li&gt;&#xA;&lt;li&gt;348 J every second for 30 degrees&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming 80% efficient.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;348 J / 80% = 435 J&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You need 435 J/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Joules per second&lt;/em&gt; is a &lt;strong&gt;Watt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;435 W&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5422" LastEditorUserId="90" LastEditDate="2016-03-12T01:54:05.600" LastActivityDate="2016-03-12T01:54:05.600" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7887" PostTypeId="2" ParentId="7866" CreationDate="2016-03-11T14:58:31.940" Score="2" Body="&lt;p&gt;The actual parameters for real tyres tend to be heavily protected intellectual property, you won't find those anywhere. However, have a look at these various pages, they give indicative values of the various parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.edy.es/dev/docs/pacejka-94-parameters-explained-a-comprehensive-guide/&quot; rel=&quot;nofollow&quot;&gt;Pacejka ’94 parameters explained – a comprehensive guide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.racer.nl/reference/pacejka.htm&quot; rel=&quot;nofollow&quot;&gt;Pacejka's Magic Formula&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.biblioteca.uma.es/bbldoc/articulos/16512881.pdf&quot; rel=&quot;nofollow&quot;&gt;An Alternative Method to Determine the Magic Tyre Model Parameters Using Genetic Algorithms&lt;/a&gt; (uses a Michelin XZA 11R22.5)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="334" LastActivityDate="2016-03-11T14:58:31.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="7890" PostTypeId="2" ParentId="7878" CreationDate="2016-03-11T15:56:08.393" Score="0" Body="&lt;p&gt;My interpretation is simply that no idiot would try to decrease flight speed close to stall speed without first dropping all flaps, i.e. max lift coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general case, again if I understand that excerpt from the book, you could calculate the stall speed for any given wing configuration and use that lift coefficient.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-03-11T15:56:08.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7893" PostTypeId="1" CreationDate="2016-03-11T18:52:50.477" Score="0" ViewCount="540" Body="&lt;p&gt;I have tried recording RF data using several different software packages: SDR Console, SDR# and HDSDR. They all seem to produce WAV files. When I open the HDSDR file in Mathematica, it says the file is &quot;Integer16&quot; meaning 16-bit PCM data and that the sample rate is 2400000, which I guess means 2.4 million samples per second. The data has two channels. The first 10 elements of each of these channels is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;{0.00421143, -0.00283813, -0.00268555, -0.00271606, -0.00268555, &#xA;0.00488281, -0.00378418, -0.00241089, -0.00234985, 0.00363159}&#xA;&#xA;{0.00479126, -0.00332642, -0.00323486, -0.00338745, -0.00344849, &#xA;-0.00228882, 0.00415039, -0.00344849, -0.00344849, 0.0055542}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;From what I understand each of these values is a composite of two bytes, but why would it be a float if it Integer16?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how do I do determine the quadrature? How is the quadrature encoded?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2016-03-11T19:12:01.030" Title="How is RF data encoded in WAV files?" Tags="|rf-electronics|data|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7895" PostTypeId="2" ParentId="7893" CreationDate="2016-03-11T19:12:01.030" Score="1" Body="&lt;p&gt;SDR is normally done with I (in-phase) and Q (quadrature) channels. You can think of them as the two components of a complex number. These allow you to preserve the phase information through your signal processing chain, which is important for some kinds of modulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information about what this entails, you should browse our related &lt;a href=&quot;https://dsp.stackexchange.com/&quot;&gt;DSP.SE&lt;/a&gt; site.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="-1" LastEditDate="2017-04-13T12:47:29.797" LastActivityDate="2016-03-11T19:12:01.030" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7896" PostTypeId="1" AcceptedAnswerId="7897" CreationDate="2016-03-11T21:15:29.307" Score="3" ViewCount="241" Body="&lt;p&gt;I have been trying to model a custom gear to fit on to another physical gear. I having been using &lt;a href=&quot;http://www.rushgears.com/tech-tools/part-search/build-custom-gears&quot; rel=&quot;nofollow&quot;&gt;this 3d model gear generator&lt;/a&gt;. What I know about my &lt;a href=&quot;https://www.servocity.com/html/32_pitch_hitec_servo_gears.html#.VuMyM-IrLIU&quot; rel=&quot;nofollow&quot;&gt;physical gear&lt;/a&gt; is that it has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(what I believe to be a diametral) Pitch of 32&lt;/li&gt;&#xA;&lt;li&gt;Pitch Diameter of 1.5&quot; (38.1mm)&lt;/li&gt;&#xA;&lt;li&gt;Outer Diameter of 1.563&quot; (39.7002mm)&lt;/li&gt;&#xA;&lt;li&gt;20° Pressure Angle&lt;/li&gt;&#xA;&lt;li&gt;48 Teeth&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have access to a 3D printer and am trying to calculate a gear that will mesh with my physical gear given these specs. I printed a smaller first gear that happened to mesh perfectly with the physical gear. &lt;strong&gt;When I increased the number of teeth&lt;/strong&gt; (thus increasing the PD and OD of the gear in the generator) &lt;strong&gt;the teeth and distance between the teeth appeared to increase as well&lt;/strong&gt;, not meshing well with the physical gear at all.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;New gear specs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pitch Diameter of 1.2&quot; (30.48mm)&lt;/li&gt;&#xA;&lt;li&gt;Outer Diameter (Auto calculated, but I think it should be) 1.2&quot; + .063&quot; (30.48mm + 1.6002mm)&lt;/li&gt;&#xA;&lt;li&gt;Same Pressure Angle&lt;/li&gt;&#xA;&lt;li&gt;*No blacklash&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a way to calculate a gear that will mesh, or will I have to&#xA;  empirically derive the gear tooth and pitch by printing out gear after&#xA;  gear to see if it fits?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3305" LastActivityDate="2016-09-28T02:25:17.153" Title="3d model gear to fit another gear" Tags="|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7897" PostTypeId="2" ParentId="7896" CreationDate="2016-03-11T21:30:27.053" Score="4" Body="&lt;p&gt;It is important to remember that, when you design a set of meshing gears you need to keep the size and shape of the teeth the same, irrespective of the diameter. There is a temptation to design a gear template and simply scale the whole thing to get the diameter you want but this does not work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One implication of this is that for any particular design the range of available diameters available to you is limited by the need to have an integer number of teeth. For example if you start with a gear with diameter D=10 and N = 10 teeth then if you are locked into that ratio so you can't have a  diameter of 20.5 as that would imply 20.5 teeth and you can't have half a tooth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of &lt;a href=&quot;https://en.wikipedia.org/wiki/Gear#Standard_pitches_and_the_module_system&quot; rel=&quot;noreferrer&quot;&gt;module&lt;/a&gt; is widely used to simplify this design process&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-11T22:25:50.847" LastActivityDate="2016-03-11T22:25:50.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7898" PostTypeId="1" AcceptedAnswerId="7900" CreationDate="2016-03-11T21:35:03.963" Score="1" ViewCount="2378" Body="&lt;p&gt;I've heard that both of these coupling types introduce a degree of kinematic error in the driven shaft. What does it actually mean in practice? Is it that the driven shaft doesn't turn uniformly even when the input shaft does? But is this error arbitrary or the same amount of error is experienced for every revolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, in what sort of application should one be used but not the other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thought appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4302" LastActivityDate="2016-03-14T19:24:19.317" Title="Oldham coupling vs Universal joint" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7900" PostTypeId="2" ParentId="7898" CreationDate="2016-03-11T22:20:41.757" Score="1" Body="&lt;p&gt;Both of these types of joints do not provide a constant velocity to the output shaft because the resistance to rotation of the joint varies throughout one rotation. This is an inherent feature of the geometry rather than a materials or manufacturing issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a joint composed of two universal joints back to back will give a much better approximation of constant velocity, these are often found in steering columns. There are also designs which stack two universal joints concentrically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Universal joints do have the advantage that they are simple to manufacture and tend to be quite rugged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a  variety of designs of constant velocity (CV) joints used for different applications for example the &lt;a href=&quot;http://www.gknservice.com/gb/passenger_cars/constant_velocity_joints.html&quot; rel=&quot;nofollow&quot;&gt;CV joints&lt;/a&gt; used in front wheel drive vehicles often consist of a spherical inner and outer shell joined by ball bearings located in grooves in both shells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Key design considerations for selecting a particular joint type include : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The angle, or range of angles at which the joint needs to work &lt;/li&gt;&#xA;&lt;li&gt;The speeds and loads of the system&lt;/li&gt;&#xA;&lt;li&gt;Whether the joint will be subject to shock loading &lt;/li&gt;&#xA;&lt;li&gt;Acceptable levels of vibration &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-12T10:49:36.623" LastActivityDate="2016-03-12T10:49:36.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7901" PostTypeId="1" AcceptedAnswerId="7902" CreationDate="2016-03-11T23:18:31.990" Score="3" ViewCount="122" Body="&lt;p&gt;I love ejection seats and the flying contraptions they usually reside in.  In fact, I like them so much that I want to make my own, despite my profound lack of training and inexperience with mechanical engineering in general. After doing a little research into the materials for my ejection seat, I found a large collection of aluminum alloys to choose from.  Some are fairly simple to eliminate like the 1000, 2000, and 3000 series.  However, between the various grades in the 5000 and 6000 series, I have a lot more trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that 6061 is the most common alloy for reasons of weldability, corrosion resistance and a few properties.  Also, I intend to rivet the chair together because that's what I see in my detailed source pictures though I have a little experience with TIG welding so that may be an option too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the benign environment of my office, will it really matter which alloy I pick?  Can I just use common 6061 sheet or is there a better, cheaper alternative?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="5427" LastEditorUserId="122" LastEditDate="2016-03-12T02:20:03.823" LastActivityDate="2019-12-06T18:44:48.027" Title="Aluminum for an Ejection Seat Replica" Tags="|aluminum|welding|alloys|rivets|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7902" PostTypeId="2" ParentId="7901" CreationDate="2016-03-12T02:16:21.947" Score="3" Body="&lt;p&gt;In your situation (as long as you don't actually intend to use the thing!) 6061-T6 will be just fine. The main reason to pick it is that many more product forms are available in it than most other alloys, and it is relatively inexpensive. It is one of the strongest among the weldable aluminum alloys and very easy to machine (as most aluminum is.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one big exception is if any of your parts require you to bend the aluminum over a relatively sharp corner. 6061 will generally crack in this case, and 5052 is the most readily available alternative that bends a little better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do keep in mind that some or all of the parts in your source material are probably cast or forged aluminum, which you will not be able to visually replicate by machining from solid aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6061-T6 is heat treated, so if you do TIG weld it be aware that you decrease it's strength by more than 50%, even if you make a perfect weld which is not easy on aluminum. After welding it will be 6061-O, rather than 6061-T6 in the heat effected zone. Again, since you're just making a replica chair, this shouldn't be an issue.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-03-12T02:16:21.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7903" PostTypeId="2" ParentId="7878" CreationDate="2016-03-12T10:05:12.187" Score="1" Body="&lt;p&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt; I am no expert on aviation, I solely got this information via your paper and a little bit of research about the used variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all I can't follow your deduction. If you would have $V$ higher than $V_{stall}$ and $C_l$ lower than $C_{l,max}$ the equation should still hold. What you decrease with $C_l$ you increase squared via your velocity. So all in all the left side should not decrease and therefore $L$ should not decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore the paper states &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Equation (5.5) states that lift equals weight in level flight, and that at stall&#xA;  speed, the aircraft is at maximum lift coefficient. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So this is rather a deduction than an assumption to base design on. So for level flight for a given $V_{stall}$ you cannot decrease $V$ any further without increasing the angle of attack. However since you already are at $C_{l,max}$ you risk stall if you do decrease speed further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xQd3H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQd3Hm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is now in contradiction to Carls answer: The pilot would then have to extend flaps in order to fly slower than stall speed because the angle of attack cannot increase further. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Values range from about 1.2 to 1.5 for a plain wing with&#xA;  no flaps to as much as 5.0 for a wing with large flaps immersed in the&#xA;  propwash or jetwash. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If I understand the paper and my research correctly, the misconception here lies within the assumption that you would use the equation with given values and disregarding that the values are not independent from one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_l=\dfrac{L}{\frac{1}{2}\rho v^2S}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is just rearranged to solve for $L$. You do not choose $C_l$ but you determine it experimentally. See this &lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/lifteq.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One way to deal with complex dependencies is to characterize the dependence by a single variable. For lift, this variable is called the lift coefficient, designated &quot;Cl.&quot; This allows us to collect all the effects, simple and complex, into a single equation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I hope this shed some light on your question.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2016-03-12T10:05:12.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7904" PostTypeId="1" CreationDate="2016-03-12T15:18:00.803" Score="2" ViewCount="111" Body="&lt;p&gt;Pictured is an antique McCaskey Register Co Receipt Holder. I have tried to find a modern equivalent, but I have not had much luck. I am considering building my own, is the spring used a readily available component? What should I be looking for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G07P5.jpg&quot; alt=&quot;Front&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/7OswZ.jpg&quot; alt=&quot;Back&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5433" LastActivityDate="2016-03-12T15:18:00.803" Title="What type of spring is used in this antique receipt holder?" Tags="|springs|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7906" PostTypeId="1" CreationDate="2016-03-12T17:54:28.293" Score="3" ViewCount="498" Body="&lt;p&gt;As a follow up to &lt;a href=&quot;https://engineering.stackexchange.com/questions/7901/aluminum-for-an-ejection-seat-replica&quot;&gt;this question&lt;/a&gt;, I have a question about rivet hole management in Autodesk Fusion 360.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got two pieces, a flat sheet of aluminum and an inverted hat channel that I want to rivet together. The easy (and naive) way would be simply to specify a bunch of rivet holes in each piece while being careful to match the holes sizes and distances from each other.  And that works until I need/want to make changes to the rivet hole sizes or layout, then I have to go remember to change them on both pieces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am skeptical of any practice that relies on me remembering to do something in lots of places. I'd much rather create a kind of master rivet hole pattern that each piece inherits from.  That way, I make one change to the master and it's automatically propagated to all the places I need it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I link the rivet hole sizes and pattern together so if I change it one place, it changes it in the other?  I've seen &quot;Project geometry&quot; but can't seem to figure out how to make it do what I want.&lt;/p&gt;&#xA;" OwnerUserId="5427" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-09-10T16:08:52.583" Title="Rivet pattern management in Fusion 360" Tags="|aluminum|rivets|cad|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7910" PostTypeId="1" CreationDate="2016-03-13T06:16:55.767" Score="1" ViewCount="2059" Body="&lt;p&gt;I am building a machine that has to lift a load of approx 80 kg including the material and lifting mechanism's weight. The velocity requirement is quite low (around 4.5 cm/s or 0.045 m/s). The whole lifting assembly will run in a guiding channel and will have rolling contact with the guide. The beam shown in blue will be the one to be lifted. Also attached to it will be another assembly. The lifting approach i have chosen for this purpose is telescope lift (also called continuous lift), both assemblies will be lifted using a pulley system and steel wires being pulled by a motor sitting at the base frame. Also during the uplift,there will be a continuous decrease in pulling force due to outflow of material (about 25 kg). I don't want to use a speed control circuit. A friend of mine has suggested to use a very high torque output and over powered motor to cater for variation in load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that what type of motor is most suited for this job. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/TZXdC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TZXdC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5443" LastEditorUserId="2819" LastEditDate="2016-03-17T03:53:32.723" LastActivityDate="2016-04-18T08:28:00.037" Title="How to size a motor for a lifting mechanism?" Tags="|mechanical-engineering|motors|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7912" PostTypeId="1" CreationDate="2016-03-13T11:07:25.570" Score="3" ViewCount="27" Body="&lt;p&gt;I am trying to anderstand the following fact from this article (page 13):&#xA;&lt;a href=&quot;http://www.cell.com/neuron/pdf/S0896-6273(15)00596-6.pdf&quot; rel=&quot;nofollow&quot;&gt;How can single neurons predict behavior&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have a linear estimation of a stimulus:&#xA;$ \hat{s} = \mathbf{w}^T(\mathbf{r} - \mathbf{f}(s_0)) + s_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbf{w}$ is a vector of weights, $\mathbf{r}$ is a vector of responses responses of two neurons to a stimulus, $\mathbf{f}$ is the vector of average neural responses to the stimulus, and the stimuli (angels between $-\pi$ to $\pi$) are symetric (amount and location) around $s_0 = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone see why the following is true for the Pearson correlation (where $\Sigma$ is the covariance matrix of $\mathbf{r}$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Corr(\hat{s},r_k) = \frac{\langle \hat{s}r_k \rangle - \langle \hat{s} \rangle \langle r_k \rangle}{\sqrt{( \langle \hat{s}^2 \rangle - \langle \hat{s} \rangle^2) }\sqrt{( \langle r_k^2 \rangle - \langle r_k \rangle^2) }} = \frac{(\Sigma \mathbf{w})_k}{\sqrt{(\Sigma_{kk}\mathbf{w}^T \Sigma \mathbf{w})}} $$&lt;/p&gt;&#xA;" OwnerUserId="2040" LastEditorDisplayName="user16" LastEditDate="2016-03-13T15:32:22.043" LastActivityDate="2016-03-13T15:32:22.043" Title="Pearson correlation of neural responses with it's linear estimation" Tags="|statistics|biomedical-engineering|signal-processing|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7915" PostTypeId="1" AcceptedAnswerId="7917" CreationDate="2016-03-13T19:42:53.723" Score="3" ViewCount="176" Body="&lt;p&gt;The diagram is for three materials with different thermal conductivities. I think number 1 is the material with higher thermal conductivity because it transfers the heat faster so it gets to the steady temperature sooner than the other two. Am I right? If not or if my reasoning isn't correct, I appreciate your help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nh1zL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nh1zL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5452" LastEditorUserId="1832" LastEditDate="2016-03-13T22:46:46.860" LastActivityDate="2016-03-13T22:46:46.860" Title="Relationship between thermal conductivity and reaching the steady temperature?" Tags="|thermodynamics|heat-transfer|thermal-conduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7917" PostTypeId="2" ParentId="7915" CreationDate="2016-03-13T21:18:53.880" Score="2" Body="&lt;p&gt;$\dot{Q} = k A \frac{\Delta T}{l}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l$ thickness &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ surface area&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this follows that besides other factors the rate of heatflow is proportional to the thermal conductivity $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{Q} \propto k$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore a high thermal conductivity allows for a fast heat transfer, hence your interpretation that a fast change in temperature correlates with a high thermal conductivity is right.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2016-03-13T21:18:53.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7918" PostTypeId="1" CreationDate="2016-03-13T22:12:08.117" Score="1" ViewCount="201" Body="&lt;p&gt;I'm coming from a product design background and have &lt;em&gt;very&lt;/em&gt; modest knowledge about electrical engineering. Please bear with me...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently drawing a small wearable device and have been looking for a physical (un)latching mechanism. The mechanism does not need to be particularly fast (&amp;lt;1 s), strong (&amp;lt;1 N), nor have a large displacement (~1 mm). The mechanism will be unlocking an object within a bracelet, and will do so rarely - probably just once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the component needs to be quite small (ideal max height ~6 mm), off the shelf (manufacturing aim between 1000 and 5000 units), and be confined to the limited voltage and current from the batteries on wearable things (battery not yet defined).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked at miniature solenoids, geared motors, piezo stuff. But, I feel like there's some component or clever solution I'm missing..?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I acknowledge this as a kinda non-specific question, but it's sort of a blue pill | red pill decision at this point of development.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: adding simple sketch of desired functionality&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1qJKz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1qJKz.png&quot; alt=&quot;simple sketch of desired functionality&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user5458" LastEditorUserId="1586" LastEditDate="2016-03-22T18:01:52.503" LastActivityDate="2016-03-22T18:01:52.503" Title="(Un)latching component for wearable device" Tags="|mechanical-engineering|design|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7919" PostTypeId="1" CreationDate="2016-03-14T00:01:17.147" Score="0" ViewCount="3237" Body="&lt;p&gt;A neighbor recently asked me if his evaporative air conditioner was running correctly because he felt the blower drive motor and it was too hot to leave his hand on it for long.&lt;/p&gt;&#xA;&lt;p&gt;I was surprised to find nothing helpful to make a back of the envelope approximation of the operating temperature (the motor casing) of an AC induction motor commonly used as the centrifugal blower drive motor (via belt to pulley on the squirrel cage blower) inside the &amp;quot;swamp coolers&amp;quot; commonly used in the US Southwest states (with dry enough weather to make evaporative cooling effective).&lt;/p&gt;&#xA;&lt;p&gt;So, I read some thermodynamics and looked at some thermal modeling of AC induction motors and came up with the following rough approach (my question will be whether anyone has a better idea or other comment on this approach):&lt;/p&gt;&#xA;&lt;p&gt;In a simplified thermal circuit representation of a motor the various internal motor loss nodes are lumped circuit representations (with as much breakdown as you have data and computational capacity, e.g., Motor-Cad or the like). Because I have only typical values or nameplate ratings for the target blower motor I tried a rudimentary model where I ignore the internal motor thermal resistance and concentrate on the motor frame/housing/case thermal resistance. This seems reasonable since whatever the complexity internal to the motor, all of that heat is in series with the casing, so the casing can be considered as a single thermal resistance equal to the delta T in C across the motor casing divided by the motor heat loss (known from nameplate hp specified at a given rpm for a two speed motor and from general knowledge of typical efficiency for these typically split phase AC induction motors, the latter 60% at operating hp and proper load).&lt;/p&gt;&#xA;&lt;p&gt;The cooled air inside the AC enclosure was ambient for the problem and equal to 15 degrees C at the time of measurement. The motor casing was 37.78 degrees C on this occasion while producing 1/6 hp output at 60% efficiency, i.e., 82 watts heat loss. So &lt;span class=&quot;math-container&quot;&gt;$\dfrac{37.78 - 15}{82} = 0.2778 \text{ °C}/\text{W}$&lt;/span&gt; thermal resistance calculated for the motor housing. This was a 2-speed motor, i.e. two winding sets, one developing 1/6 hp at 1140 rpm the other 1/2 hp at 1725 rpm. We are operating at 1/6 hp and 1140 rpm in the instant problem. 115 VAC single phase.&lt;/p&gt;&#xA;&lt;p&gt;This thermal resistance calculated would be a combined convective and radiation resistance. It also includes the likely internal fins on the rotor (this being an ODP open drip proof housing and manufacturers typically include small fanning fins inside on the ends of the rotor shaft to produce an internal forced convection cool utilizing the air flow from ambient through the openings in the case). I looked at a typical rpm vs thermal resistance curve that researchers had developed empirically and derived a cubic polynomial regression fit that allowed me to estimate the 0 speed thermal resistance of the motor case (not true locked rotor, but just the static thermal resistance at a controlled current by the researchers). With that equation I estimated the thermal resistance of the case at 0 rpm without the internal fanning effect would be approx. 0.96626 deg. C/watt. For a motor casing estimated area of 0.1567626 square meters (cylinder) this gives a combined convective and radiation transfer coefficient of &lt;span class=&quot;math-container&quot;&gt;$\dfrac{1}{0.96626\cdot0.1567626} = 6.60182\text{ W}/\text{m}^2\text{°C} = h$&lt;/span&gt;. Typical combined heat transfer coefficients in this context for simple geometric shapes are in the range of &lt;span class=&quot;math-container&quot;&gt;$12 - 14\text{ W}/\text{m}^2\text{°C}$&lt;/span&gt;, so this is a bit low, but in the ballpark of expectations in rough initial estimate.&lt;/p&gt;&#xA;&lt;p&gt;I used the thermal resistance already calculated above and temperature at the occasion my neighbor believed his motor was too hot. That calculation gave me 44 °C for the motor plus ambient (it was 21 °C ambient within the AC enclosure on that occasion). A thermodynamic text opined that 44 °C was too hot to be tolerated on direct contact with skin to surface (depends on one's capacity for pain; theoretically it would require about 5 minutes at 50 °C to produce a major skin burn).&lt;/p&gt;&#xA;&lt;p&gt;The cubic polynomial modifying the measured casing thermal resistance dependent on rpm was of form &lt;span class=&quot;math-container&quot;&gt;$f(x) = ax^3 + bx^2 + cx + d$&lt;/span&gt; where the coefficients are &lt;span class=&quot;math-container&quot;&gt;$a = -0.00000124809$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b = 0.0067928076$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$c = -12.2709059880$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$d = 9884.78844117$&lt;/span&gt;&#xA;and within 1 - 5% fit within the range of 0 to 2500 rpm. The &lt;span class=&quot;math-container&quot;&gt;$f(x)$&lt;/span&gt; result should be divided by 10000 (1e4) to renormalize. Then you multiply that corrective factor times the 0 rpm thermal resistance given (0.96626 °C/W). &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is the rpm, e.g, 1140 in the present case with the motor running at &amp;quot;low speed&amp;quot; 1/6 hp. For example, &lt;span class=&quot;math-container&quot;&gt;$f(1140) = 2874.7921$&lt;/span&gt;. Divide that by 1e4 to get 0.2875. Multiply &lt;span class=&quot;math-container&quot;&gt;$0.2875$&lt;/span&gt; by the 0 rpm base thermal resistance of 0.96626 °C/watt to get 0.2779 °C/W case thermal resistance at 1140 rpm (as was experimentally measured).&lt;/p&gt;&#xA;&lt;p&gt;I should add that the motor is NEMA class F insulation so the motor is permitted to rise up to 105 °C casing temp in up to 40 °C ambient (combined ambient and motor 155 C permitted NEMA specifying the motor rise lower to allow for hot spots internal).&lt;/p&gt;&#xA;&lt;p&gt;I posted a more detailed answer (too long to post here) in pdf format (though a bit inelegant in form since I hadn't installed Tex on my systems back in 2016) at one of my websites:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.southwestpcsolutions.com/whyismyacmotorhot.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why Does My AC Motor Feel Hot?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5260" LastEditorUserId="5260" LastEditDate="2023-06-28T14:05:10.750" LastActivityDate="2023-06-28T14:05:10.750" Title="How to calculate temperature rise on AC induction motor in an evaporative cooler centrifugal blower?" Tags="|thermodynamics|motors|thermal-conduction|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="7921" PostTypeId="1" AcceptedAnswerId="7945" CreationDate="2016-03-14T04:46:00.937" Score="3" ViewCount="75" Body="&lt;p&gt;I'm running a simulation in a CAD on a rather simple assembly with several parts in which a motor effectively rotates a part. If the motor is set to constant 10RPM, the part rotates fine and the max. motor torque is around 500N-mm. However, if it's set to 5RPM, it gets stuck and the motor torque reaches 8000N-mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How come a higher RPM could avoid a lock-up? Isn't a higher torque less susceptible to rotational lock-ups, since it's the force to rotate an object?&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="2819" LastEditDate="2016-03-15T03:02:11.767" LastActivityDate="2018-02-21T15:09:21.337" Title="Why does a motor lockup at low RPM in my simulation?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7924" PostTypeId="1" AcceptedAnswerId="7946" CreationDate="2016-03-14T09:29:52.880" Score="3" ViewCount="889" Body="&lt;p&gt;I'm trying to model a part that rotates on a shaft. I measured empirically that it has a static torque of around 500 N·mm and a dynamic torque of around 400 N·mm. I want to simulate it in CAD but my software only takes static/dynamic friction as inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I convert the torque values to static/dynamic friction values for the part-shaft contact surface?&lt;/p&gt;&#xA;" OwnerUserId="4303" LastEditorUserId="368" LastEditDate="2016-03-14T16:53:32.550" LastActivityDate="2016-03-15T03:16:53.273" Title="Determine coefficients of friction based on the torque required to rotate a part on a shaft" Tags="|mechanical-engineering|torque|friction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7925" PostTypeId="1" CreationDate="2016-03-14T11:31:42.500" Score="2" ViewCount="2777" Body="&lt;p&gt;I have a system that consists of a centrifugal pump and a positive-displacement piston pump, and both pumps get water from one reservoir and then discharge this water back in the same reservoir.  The two pumps are connected in parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specs of each pump are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Centrifugal: $Q_{max} = 33 \frac{m^3}{h}$ , $H_{max} = 9.5m$ , Power= 1.5hp at 3000 rpm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Piston: $Q_{max} = 1.5 \frac{m^3}{h}$ , $H_{max} = 60m$ , Power= 1hp at 1450 rpm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I need to know what happens if I operated both pumps simultaneously, knowing that both pumps will face the same resistance in the system?&lt;/p&gt;&#xA;" OwnerUserId="3591" LastEditorUserId="1586" LastEditDate="2016-03-23T21:47:38.420" LastActivityDate="2017-05-30T16:04:35.563" Title="What happens if we operate a positive-displacement pump in parallel with a centrifugal pump?" Tags="|mechanical-engineering|fluid-mechanics|pumps|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7926" PostTypeId="2" ParentId="7925" CreationDate="2016-03-14T12:54:16.883" Score="3" Body="&lt;p&gt;Assuming the pumps are connected as in the following diagram, and that the inlet and outlet pipes are sized appropriately, the total flow rate will be the sum of each individual pump's flow rate.  Unless the pipes are severely undersized, you probably don't need to worry about backflow from one pump to the other.  &lt;a href=&quot;http://www.engineeringtoolbox.com/pumps-parallel-serial-d_636.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This link&lt;/a&gt; has some additional information that may be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LI0Au.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LI0Au.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-03-14T12:54:16.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7929" PostTypeId="2" ParentId="7906" CreationDate="2016-03-14T14:07:05.947" Score="2" Body="&lt;p&gt;I found this &lt;a href=&quot;http://fusion360.autodesk.com/learning/learning.html?guid=GUID-77275F36-81D1-49DD-B83A-1036C958B538&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; on how to use a single sketch to make multiple bodies from a single sketch.  It appears to do what I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The video transcript reads: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now that you've seen how 2D sketches drive 3D features, I'll switch over to another design to show you how a single sketch can be used to create multiple 3D features using different profiles. The sketch shown here has a few circular profiles and a tab extending at the top, along with a line on the left that will be used to create a Revolve feature. This will be the first 3D feature I'll create from this sketch, so I'll go ahead and enable the command from the &quot;Create&quot; drop-down menu and I can begin selecting closed profiles within the sketch. I can select as many as I need until I get the correct collective profile. As a side note, it's typically easiest to select all of your profiles from the top of the screen down to the bottom. I'll quickly finish this 3D feature by selecting the axis of revolution as well as change a few parameters for the feature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5427" LastActivityDate="2016-03-14T14:07:05.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7932" PostTypeId="2" ParentId="7919" CreationDate="2016-03-14T14:53:35.013" Score="1" Body="&lt;p&gt;If you want a back-of-the-envelope calculation, just use the heat convection equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{Q} = hA\Delta T&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{Q}$ is the rate of heat transfer, $h$ is the heat transfer coefficient, $A$ is the surface area, and $\Delta T$ is the temperature difference between the free stream air and the surface of the motor.  For your case, you already calculated that the rate of heat dissipation is $82 \,W$.  For convection, the heat transfer coefficient could be anywhere in a wide range, but $50 \frac{W}{m^2 K}$ is probably a good initial guess.  I'm assuming the motor housing is a cylinder $10\, cm$ long and $10\, cm$ diameter, so the surface area would be $0.047\, m^2$.  Solving for $\Delta T$ we get a $35^\circ C$ difference between air and motor, which would put the motor surface at $50^\circ C$ (assuming the air is at $15^\circ$).  This seems plausible based on your calculation and your neighbor's personal experience.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-03-14T14:53:35.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7934" PostTypeId="1" AcceptedAnswerId="7938" CreationDate="2016-03-14T15:39:05.923" Score="2" ViewCount="4564" Body="&lt;p&gt;I have a circular puck of stainless steel that is 9.5 mm in diameter.  In the center, there is already a 5 um pinhole.  My objective is to basically cut the steel along the dotted line, approximately 0.5 mm from the pinhole.  &lt;strong&gt;Can you advise on what techniques to can make this cut precisely?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CMzB6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CMzB6.png&quot; alt=&quot;Circular sheet of steel with 5 um pinhole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I want to do this is to be able to align two of these so that I can have 2 pinholes that are ~1 mm apart from one another, as in the following image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WVdRk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WVdRk.png&quot; alt=&quot;Two pinholes adjacent to one another&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5461" LastEditorUserId="5461" LastEditDate="2016-03-15T16:13:36.107" LastActivityDate="2024-01-29T12:48:34.783" Title="How can I cut a thin (1mm) sheet of stainless steel with high precision?" Tags="|steel|cutting|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7935" PostTypeId="2" ParentId="7898" CreationDate="2016-03-14T19:24:19.317" Score="0" Body="&lt;p&gt;The short answer is that universal joints are tolerant of angular misalignment while Oldham couplings are intended for use where the shafts are misaligned but parallel.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2016-03-14T19:24:19.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7938" PostTypeId="2" ParentId="7934" CreationDate="2016-03-14T21:50:49.977" Score="4" Body="&lt;p&gt;Calling a small disc that is 9.5 mm DIA and 1.0 mm thick &quot;a thin sheet of steel&quot; is a bit of a stretch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any machinist would simply clamp the disc in a suitable vise and mill away the unwanted part. Precision would be a couple orders of magnitude better than your requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have patience and only need a small number of them done, a pair of pliers and a grinding wheel could accomplish the same thing.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-03-14T21:50:49.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7939" PostTypeId="2" ParentId="7934" CreationDate="2016-03-14T22:21:11.177" Score="1" Body="&lt;p&gt;I would say that 0.25mm is not very difficult to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use Water jet cutter to be really accurate or if you don't have this possibility, you can simply use a mill or CNC with specific drill.&lt;/p&gt;&#xA;" OwnerUserId="1374" LastActivityDate="2016-03-14T22:21:11.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7940" PostTypeId="2" ParentId="7934" CreationDate="2016-03-14T22:48:16.573" Score="2" Body="&lt;p&gt;Probably the simplest way is to cut it a few mm oversized and grind or file off the excess. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-03-14T22:48:16.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7941" PostTypeId="1" CreationDate="2016-03-14T23:54:31.423" Score="1" ViewCount="271" Body="&lt;p&gt;Violin strings are either single core materials or of the wound variety with a single core of one material wound with a helix of another (usually heavier) material to give the string a lower resonant frequency. In analyzing the acoustic impedance of a wound string, which part will determine the acoustic impedance of the string , the core or the outside helix material? Or will the overall impedance be somewhere in between the two?&lt;/p&gt;&#xA;" OwnerUserId="628" LastEditorUserId="368" LastEditDate="2016-03-16T19:50:33.263" LastActivityDate="2016-07-14T22:18:16.973" Title="How can I determine the overall acoustic impedance of a string with different core and winding materials?" Tags="|mechanical-engineering|acoustics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7945" PostTypeId="2" ParentId="7921" CreationDate="2016-03-15T02:59:42.803" Score="3" Body="&lt;p&gt;CAD simulation is not real world. The model may break for a variety of reasons based on your geometry, mesh resolution, time step, software, and assumptions used in writing that software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you find something in a model that doesn't match reality, it is time to re-evaluate your model or find different software.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-15T02:59:42.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7946" PostTypeId="2" ParentId="7924" CreationDate="2016-03-15T03:16:53.273" Score="0" Body="&lt;p&gt;It is difficult to model with friction coefficients in this case, because the normal force is dependent on very small changes in dimensions and material strengths that your software may or may not support. Friction is by no means as easy as it is in the text books. The basic idea is find the force normal to the friction surface in question and calculate the coefficient with the &lt;a href=&quot;https://simple.wikipedia.org/wiki/Coefficient_of_friction&quot; rel=&quot;nofollow&quot;&gt;friction equation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be more accurate and easier to just iterate to a solution. First you guess at a friction coefficient, insert it into your simulation, run your simulation, see if your result is higher or lower than amount you measured, then guess a new value and repeat until you feel that you are are close enough to your measured value.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-15T03:16:53.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7947" PostTypeId="1" AcceptedAnswerId="7962" CreationDate="2016-03-15T04:32:55.640" Score="2" ViewCount="820" Body="&lt;p&gt;I've been looking at solutions to compress gases and getting stunned by the cost and maintenance requirements of existing systems to fill cylinders (whether for SCBA or chemical storage) to 2000-4000psi:  The cost of entry is over 4 figures, with commercial systems like those used in fire stations and SCUBA shops running into 5 figures.  And, as far as I can tell, they are all &lt;em&gt;direct mechanically driven&lt;/em&gt; multi-stage &lt;em&gt;pneumatic&lt;/em&gt; pistons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, sitting in my garage I have a manually-driven hydraulic bottle cylinder that can lift 20 tons, and typical hydraulic systems circulate hydraulic fluid at 5000-10000psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have in mind a very simple gas compressor: a single piston sealed on both ends.  On one side it is driven by a single hydraulic pump line.  On the other are two gas valves: An intake and an output.  Operation is simple:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Hydraulic pressure shut off.&lt;/li&gt;&#xA;&lt;li&gt;Gas intake valve open.&lt;/li&gt;&#xA;&lt;li&gt;Hydraulic flow reversed until piston has drawn in the maximum capacity of gas.&lt;/li&gt;&#xA;&lt;li&gt;Intake valve closed.&lt;/li&gt;&#xA;&lt;li&gt;Hydraulic pressure run until piston has reached maximum compression&lt;/li&gt;&#xA;&lt;li&gt;Output valve opened (if not one-way and hence automatic), then closed once pressure equalized with output cylinder.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the piston lacks the compression ratio to reach the desired psi, then the charged cylinder is switched to the input line and another cylinder to the output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I missing a reason this won't work?  Or does it already exist in some form close to this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, where and under what name might I find a long two-sided piston with pressure fittings on each end?&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2019-06-07T21:10:30.720" Title="Why no hydraulic piston gas compressors?" Tags="|hydraulics|compressors|compressed-gases|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7949" PostTypeId="2" ParentId="7910" CreationDate="2016-03-15T05:01:26.057" Score="1" Body="&lt;p&gt;An AC motor will turn at a constant speed with 1% to 5% slip provided the load is geared to be less than the listed maximum torque of the motor. An AC gear motor should bring it into the gear ratio range of your pulley system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approximate power required by your system can be calculated with the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html&quot; rel=&quot;nofollow&quot;&gt;gravitational potential energy equation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PE = mgh&lt;br&gt;&#xA;PE = 80kg * 9.81m/s^2 * 0.045 m&lt;br&gt;&#xA;PE = 35.3 joules&lt;br&gt;&#xA;Power = 35.3 joules/s = 35.3 Watts&lt;br&gt;&#xA;Power = 35.3/ (746W/hp) ~ 1/20hp&lt;br&gt;&#xA;Remember that cheap gear boxes are probably only 50% efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some motors that may fit your application:&lt;br&gt;&#xA;&lt;a href=&quot;https://www.zoro.com/dayton-ac-gearmotor-60-rpm-open-115v-1lpl6/i/G1850572/&quot; rel=&quot;nofollow&quot;&gt;Dayton 1/10hp 60rpm AC Gearmotor&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://www.zoro.com/dayton-ac-gearmotor-30-rpm-open-115v-1lpl8/i/G2136775/&quot; rel=&quot;nofollow&quot;&gt;Dayton 1/10hp 30rpm AC Gearmotor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how much travel you need, but it may be worth looking at &lt;a href=&quot;https://www.zoro.com/search?q=&amp;amp;categoryl2=Linear+Motion&amp;amp;categoryl3=Electro+Mechanical+Actuators&amp;amp;categoryl1=Bearings%2C+V-belts+%26+Power+Transmission+Equipment&quot; rel=&quot;nofollow&quot;&gt;prebuilt linear actuators&lt;/a&gt; instead of building your own pulley system.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-15T05:01:26.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7950" PostTypeId="2" ParentId="2599" CreationDate="2016-03-15T05:13:32.350" Score="1" Body="&lt;p&gt;Here is a &lt;a href=&quot;https://www.eia.gov/todayinenergy/detail.cfm?id=1490&quot; rel=&quot;nofollow&quot;&gt;good short article&lt;/a&gt; on the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says that reactors can be refueled in as little as 10 days, but outages typically last an average of two months because non-critical maintenance work is typically differed until a refueling outage. This makes sense, as most systems are safer and easier to work on when they are not running.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard non-critical maintenance on a steam power station would include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bearing replacement&lt;/li&gt;&#xA;&lt;li&gt;Replacement of corroded lines&lt;/li&gt;&#xA;&lt;li&gt;Cleaning scale from pipes, valves and heat excahngers&lt;/li&gt;&#xA;&lt;li&gt;Electrical breaker and relay replacement&lt;/li&gt;&#xA;&lt;li&gt;Control system upgrades&lt;/li&gt;&#xA;&lt;li&gt;VFD upgrades&lt;/li&gt;&#xA;&lt;li&gt;Sensors and pressure gauge, replacements&lt;/li&gt;&#xA;&lt;li&gt;Turbine inspection&lt;/li&gt;&#xA;&lt;li&gt;Boiler inspection&lt;/li&gt;&#xA;&lt;li&gt;Insurance inspections&lt;/li&gt;&#xA;&lt;li&gt;Lots of paperwork&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nrc.gov/reading-rm/doc-collections/cfr/part050/part050-0065.html&quot; rel=&quot;nofollow&quot;&gt;US NRC Regulations&lt;/a&gt; may be a good place to start looking for nuclear specific maintenance.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2016-03-15T23:36:31.717" LastActivityDate="2016-03-15T23:36:31.717" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="7953" PostTypeId="2" ParentId="7941" CreationDate="2016-03-15T11:36:57.513" Score="1" Body="&lt;p&gt;I'm not sure you understand what &quot;acoustic impedance&quot; is.  The string by itself has a resonant frequency defined by its elastic modulus, the applied longitudinal tension, and length.   If you really want to know how well  it couples to the air around it, I suspect you'll be diving into an ugly bit of math.  For one thing, unlike a speaker cone or a trumpet (for two examples), the string is coupled to the air along its full length, and primarily in two directions (the plane of vibration).  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-03-15T11:36:57.513" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7954" PostTypeId="2" ParentId="5905" CreationDate="2016-03-15T13:03:31.110" Score="1" Body="&lt;p&gt;I would like to add to Carltons answer and go a little more in depth on how to approach problems like that in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hydrostatics or Hydrodynamics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two possible solutions to the problem. You can either choose a reference frame that rotates with the fluids or that stays still and observes the system from the outside. This will determine if hydrostatics are allowed or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rotating reference frame&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The origin of the coordinate system (polar coordinates) is at the top of the cylinder, positive z axis direction is upward, against the gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is no movement in our system as we rotate the same way the fluid does, hydrostatics can be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\nabla p = \varrho \vec{f}$$&#xA;Due to the rotation of our reference frame we have to add the centrifugal force as an external force.&#xA;$$\vec{f} = \begin{pmatrix}&#xA;           \Omega^2r\\&#xA;           0\\&#xA;           -g&#xA;           \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&#xA;$$\begin{pmatrix}\frac{\partial p}{\partial r}\\&#xA;\frac{1}{r}\frac{\partial p}{\partial \varphi}\\&#xA;\frac{\partial p}{\partial z}&#xA;\end{pmatrix} = \varrho \begin{pmatrix}&#xA;           \Omega^2r\\&#xA;           0\\&#xA;           -g&#xA;           \end{pmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Successively Integration yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p(r,\varphi,z)=\varrho \Omega^2 \frac{r^2}{2} + C(\varphi,z) \tag{1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inserting in the second row&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{r}\frac{\partial \varrho \Omega^2 \frac{r^2}{2} + C(\varphi,z)}{\partial \varphi} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C'(\varphi,z) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $C(z)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inserting in the third row yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C'(z) = \frac{\partial C(z)}{\partial z} = - \varrho g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C(z) = - \varrho g z + D$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(r,z)=\varrho \Omega^2 \frac{r^2}{2} - \varrho g z + D$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the boundary condition that $p(r=0, z=0)=p_a$ it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(r,z)=p_a - \varrho g z +\varrho \frac{\Omega^2 r^2}{2} \tag{2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach is more flexible as you can account for any changes in the problemset without relying on a formula that is just 1 solution for 1 specific problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fixed reference frame&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we observe the system from the outside and we can no longer use hydrostatics for our problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here Navier-Stokes comes in handy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In polar coordinates for an inviscid, stationary flow the equation for the radial direction is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u\frac{\partial u}{\partial r} + \frac{v}{r}\frac{\partial u}{\partial \varphi} + w\frac{\partial u}{\partial z} - \frac{v^2}{r} = -\frac{1}{\varrho} \frac{\partial p}{\partial r} + f_r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\vec{v} = \begin{pmatrix}&#xA;           u\\&#xA;           v\\&#xA;           w&#xA;           \end{pmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As there are now no external forces acting on the system $f_r=0$ and we omit all zero terms that follow from $u=0$ and $z=0$ as we only have a velocity in $v$ direction, it follows that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$- \frac{v^2}{r} = -\frac{1}{\varrho} \frac{\partial p}{\partial r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\Omega = \frac{v}{r}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Omega^2 r = \frac{1}{\varrho} \frac{\partial p}{\partial r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which equals (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat with v and z direction and the same boundary condition you get to (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mOB6H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mOB6Hm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uwWU4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uwWU4m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prefer general solutions over specific solutions that why I felt the need to add to this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculating force&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can integrate over the walls of the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = \int_A p(r=R,z) R d \varphi dz$&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2016-03-15T14:10:01.257" LastActivityDate="2016-03-15T14:10:01.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7955" PostTypeId="1" AcceptedAnswerId="7958" CreationDate="2016-03-15T13:52:51.870" Score="6" ViewCount="2436" Body="&lt;p&gt;I am currently looking at a set of truss layouts produced by our truss manufacturing partner for a residential building project. I have reason to believe these layouts are generated using Alpine (Intelliview?) software. In any case, the design summary contains abbreviations which I cannot understand since I do not have a copy of the software. Please see below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wCPtr.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wCPtr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4HzmH.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4HzmH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://alpineitw.com/wp-content/uploads/2015/01/Read_A_Drawing.pdf&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; document R, U and W stand for reaction due to design load, uplift due to wind load and bearing width, respectively. TCLL, TCDL, BCLL and BCDL stand for top chord live load, Top chord dead load, Bottom chord live load and bottom chord dead load. However I cannot for the life of me determine what Rw, Rh, RL  and NCBCLL stand for. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="4808" LastEditorUserId="4808" LastEditDate="2016-03-15T14:25:21.067" LastActivityDate="2019-11-11T21:19:11.757" Title="Truss manufacturer drawings unclear" Tags="|structural-engineering|structures|structural-analysis|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="7956" PostTypeId="2" ParentId="7648" CreationDate="2016-03-15T14:07:24.480" Score="0" Body="&lt;p&gt;For this answer I'll assume &amp;quot;off-center&amp;quot; means the axis of rotation differs from the desired axis of rotation by angle &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The goal is to realign the axis of rotation, which can be accomplished by changing the rotational momentum vector.&lt;/p&gt;&#xA;&lt;p&gt;This goal can be achieved by applying forces to the axle.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nmKnm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nmKnm.png&quot; alt=&quot;Fly Wheel force diagram&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Applying symmetric forces as shown of magnitude &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; a distance of &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; from the center will add rotational inertia at a rate of &lt;span class=&quot;math-container&quot;&gt;$2FR$&lt;/span&gt; Which will rotate the axis of the Gyroscope at angular rotation rate of &lt;span class=&quot;math-container&quot;&gt;$\frac{2FR}{I\omega}$&lt;/span&gt; the axis of which will be perpendicular to the axis of the flywheel and parallel to the force being applied.&lt;/p&gt;&#xA;&lt;p&gt;This process will not change the magnitude of the angular velocity of the flywheel.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-03-15T14:07:24.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7957" PostTypeId="2" ParentId="7801" CreationDate="2016-03-15T14:23:19.560" Score="0" Body="&lt;p&gt;I have looked at the other solution that you mentioned.  That solution would be ok provided that the two assumptions hold. That is, the beam is not flexible, and the chains are adjusted before any load is applied, to be of equal length and carrying equal almost-zero loads.  Depending on the Load/Extension value of the chains, this may involve very small adjustments to their initial lengths, and resulting tensions.  Without the length and elastic modulus of the chains, and knowledge of the temperature stability of the support structure, it is all guesswork, but I expect that an adjustment in length of the centre chain of about 0.1 mm would be enough for it to go from carrying no load, to carrying the whole load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loads are so low, is an accurate answer really necessary?  Could you change it to only two chains?  Could you hinge the beam where the middle chain is attached?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at strand7.com, and their demo download.  This is a fully working Finite Element package, limited to 20 beams, so ok for your problem.  In my personal opinion it is the best FE package for user-friendliness. (Or - it is dead easy to understand and use.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry it took so long to come back.  I am currently struggling to understand how emails work, and how ubuntu works.&lt;/p&gt;&#xA;" OwnerUserId="5381" LastActivityDate="2016-03-15T14:23:19.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7958" PostTypeId="2" ParentId="7955" CreationDate="2016-03-15T14:41:12.790" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;However I cannot for the life of me determine what Rw, Rh, RL and NCBCLL stand for.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Given that the values Rw, Rh, and RL are under a Maximum Reactions header, I assume that they are the reaction components of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rw = reaction force due to wind (or possibly water?)&lt;/li&gt;&#xA;&lt;li&gt;Rh = reaction force due to horizontal load (or possibly hydrostatic?)&lt;/li&gt;&#xA;&lt;li&gt;RL = reaction force due to live load&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are given at support points $A$ and $I$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think NCBCLL is the construction live load, but I don't know what NCB stands for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another &lt;a href=&quot;http://www.dndb.info/PDF/truss-engineering-plans.pdf&quot;&gt;Alpine output sheet&lt;/a&gt; that might be of use as a reference.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are upper and lower dimensions for each portion of the span. I would like to understand why there are up to 4 distinct dimensions for each span.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's hard to tell exactly from the drawing, but these dimensions seem to be working from different sides of the truss members.  The dimensions on the top seem to be to the left side of the vertical members and the bottom dimensions seem to be on the right.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'd ask your truss supplier to clarify.  If they're using the software, they should know what the output means.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2016-03-15T14:41:12.790" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7959" PostTypeId="2" ParentId="7879" CreationDate="2016-03-15T18:35:44.477" Score="1" Body="&lt;h3&gt;Similar Situation&lt;/h3&gt;&#xA;&lt;p&gt;I would start by looking at a similar situation in other industries. For this I am thinking about waterproof expansion joints in bridge decks.&lt;/p&gt;&#xA;&lt;p&gt;The similarities are:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Relative movements of components&lt;/li&gt;&#xA;&lt;li&gt;Waterproof&lt;/li&gt;&#xA;&lt;li&gt;Steel and/or concrete&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;In general this seems to be a pretty good match.&lt;/p&gt;&#xA;&lt;h3&gt;Products&lt;/h3&gt;&#xA;&lt;p&gt;Some bridge decks have cast-in-place concrete which embeds the joint waterproofing similar to waterstops, but some methods apply a membrane to both sides.  This is usually done with enough slack in the membrane to allow for movement.&lt;/p&gt;&#xA;&lt;p&gt;I have used (but have no other affiliation with) two companies that specialize in this type of waterproofing:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://bridgepreservation.com/expansion-joints/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bridge Preservation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.stirlinglloyd.com/products/eliminator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stirling Lloyd&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Alternatives may be to use an epoxy adhesive to secure a strip of ~0.5 inch neoprene rubber to each side.&lt;/p&gt;&#xA;&lt;p&gt;The benefits are that these systems adhere to both concrete and steel. An example detail from Bridge Preservation is shown below:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HyFUZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HyFUZ.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h3&gt;Downsides&lt;/h3&gt;&#xA;&lt;p&gt;There are a couple of downsides to these methods.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They aren't designed to be structural, so some lagging may be required to support the retained soil/water.&lt;/li&gt;&#xA;&lt;li&gt;They are installed in the field, so construction will be delayed while the system is installed.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-03-16T02:45:11.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7960" PostTypeId="2" ParentId="7918" CreationDate="2016-03-15T19:04:20.027" Score="1" Body="&lt;p&gt;If I understand correctly; you would like component A to detach from component B. Should component A be held in place mechanically and trigger a signal or become a circuit-interrupt when detached? If so, you should research micro-witches. They can be configured for &quot;normally open&quot; or &quot;normally closed&quot; when purchased. Another idea would be to place conductive material on component A so that it acts as a &quot;fuse&quot; when detached.&lt;/p&gt;&#xA;" OwnerUserId="5478" LastActivityDate="2016-03-15T19:04:20.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7961" PostTypeId="1" CreationDate="2016-03-15T21:22:01.013" Score="0" ViewCount="324" Body="&lt;p&gt;I would like to be able to tilt a platform (the green cuboid below, pivoting around the red point) from a single solenoid (&lt;strong&gt;one&lt;/strong&gt; of the blue cylinders below). The solenoid cannot move (other than the shaft) and can only push (&lt;strong&gt;or&lt;/strong&gt; pull).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pc39d.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pc39d.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need something that will connect the solenoid to the platform so that one pulse of the solenoid will tilt the platform one way and an identical pulse would tilt the platform the other way. The platform can also be manually moved, so the solenoid must be able to cope with the platform being in a different position than expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas would be really well received! If I can clarify in any way, please let me know.&lt;/p&gt;&#xA;" OwnerUserId="5479" LastActivityDate="2016-03-16T16:11:39.430" Title="Actuate a tilting platform from a single point" Tags="|mechanical-engineering|actuator|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7962" PostTypeId="2" ParentId="7947" CreationDate="2016-03-15T21:22:43.753" Score="5" Body="&lt;p&gt;There are two aspects to this. First is that the main advantage of hydraulics is that is provides a  compact means of greatly multiplying &lt;em&gt;force&lt;/em&gt; but in compressing gasses you have to  a huge amount of work on the gas itself so the constraint is the power input so you are always limited by the capacity of the motor driving the compressor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that the gas in a cylinder with a volume of 1 cubic metre at 350 bar would occupy 350 cubic meters at atmospheric pressure. So if you have a cylinder with an intake volume of 1 litre that's 350 000 cycles to fill the cylinder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other very important factor is that high pressure compressed gasses whether for SCBA or industrial gasses is that they need to be clean and dry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conventional compressors and hydraulic cylinders are lubricated with oil and designed so that piston seals are slightly leaky to lubricate the seals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Providing oil free gas compression is a much greater technical challenge and requires high pressure dry sliding seals which is why it requires frequent rebuilds as the seals have a high wear rate. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-16T13:32:14.513" LastActivityDate="2016-03-16T13:32:14.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7964" PostTypeId="1" AcceptedAnswerId="8071" CreationDate="2016-03-16T01:23:30.243" Score="2" ViewCount="295" Body="&lt;p&gt;Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; is a hard, white material used for making firebricks and numerous molded ceramic parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I had a firebrick, took a core out of it using a thick-walled diamond hole saw and a drill press.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What lathe insert material would I then use to machine the outside of the Al&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; cylinder assuming I could take care of all the holes using diamond hole saws?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My budget for one cutting bit would be $500.  But that would take me many hours of work to raise, so I prefer something less expensive, since I'm probably going to break it, hence &lt;em&gt;cost effective&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3235" LastEditorUserId="368" LastEditDate="2016-03-16T16:42:33.053" LastActivityDate="2016-03-19T23:11:46.050" Title="What's the most cost-effective cutter bit material that can be used to machine aluminum oxide?" Tags="|materials|machining|ceramics|lathe|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7965" PostTypeId="1" CreationDate="2016-03-16T07:19:14.157" Score="54" ViewCount="21472" Body="&lt;p&gt;Despite having so much technological advancement, I wonder why cars still uses mirrors (right and left side of the driver) for the rear view. These mirrors can be easily replaced by cameras and monitors (display for the driver). I noticed such technology exists but it is restricted to only when you take the car in the reverse direction. If we use this technology while driving the car (i.e. in forward direction), I believe it will be helpful for the driver since multiple cameras installed on the rear side of the car will give consolidated view to the driver. Also it will save the hassle of monitoring both the mirrors while driving. I think this will have two benefits:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Since the mirrors will be replaced by cameras, it will reduce the wind resistance for car.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Weight of the car will be reduced negligibly assuming electronic circuits will be lighter than mirrors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So, my questions are&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why cameras (and obviously displays) are not used instead of mirrors? &lt;/li&gt;&#xA;&lt;li&gt;Does such technology exist? If it exists then why it is not widely used?&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5484" LastEditorUserId="282" LastEditDate="2016-03-22T00:22:37.483" LastActivityDate="2019-04-22T04:51:45.240" Title="Why do cars have mirrors for rear view rather than cameras?" Tags="|electrical-engineering|automotive-engineering|" AnswerCount="9" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7966" PostTypeId="2" ParentId="7965" CreationDate="2016-03-16T10:04:20.637" Score="32" Body="&lt;p&gt;The industry is already going down that path for the exact reasons you mentioned. There is an &lt;a href=&quot;http://www.wired.com/2014/04/tesla-auto-alliance-mirrors/&quot;&gt;article&lt;/a&gt; that provides information about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary the main reasons why we don't have it yet are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;State laws require mirrors and would have to change first&lt;/li&gt;&#xA;&lt;li&gt;Conventional mirrors are basically fail safe, imagine the camera system turning off on the motor way. The new systems have to match a mirror in terms of reliability and safety. Electronics can be manipulated, have a latency etc.&lt;/li&gt;&#xA;&lt;li&gt;Price point, conventional mirrors are pretty cheap and cars with cameras are not the basic configuration. Many people simply can't afford cars with expensive extras.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Mercedes Benz for example has some really nice safety features that use cameras already. 360 ° &lt;a href=&quot;http://www.mercedes-benz-intelligent-drive.com/com/en/3_parking/2_360-camera&quot;&gt;bird's eye parking&lt;/a&gt; assistance. Collision detection features that check your dead angle and if it is unsafe to change lanes a little red triangle lights up in your side mirror. Followed by acoustic warnings if you still try to change lanes up to point where the car brakes on side to put you back in your original lane. You even have active lane &lt;a href=&quot;http://www.mercedes-benz-intelligent-drive.com/com/en/1_driver-assistance-and-safety/7_active-lane-keeping-assist&quot;&gt;tracking&lt;/a&gt; already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technology is there already, we only adapt slowly to the change. It's the same with everything else, it takes years before new technology really reaches the rank and file.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2016-03-16T10:04:20.637" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7968" PostTypeId="2" ParentId="7965" CreationDate="2016-03-16T12:15:33.863" Score="115" Body="&lt;p&gt;This is another case of fancy new (SHINY!!!) technology being &lt;strong&gt;A Bad Idea&lt;/strong&gt;(TM) .  When you're driving, your eyes are focussed roughly at infinity, i.e. looking at objects more than 5 m away.  When you look at a rear view mirror, you are still looking at distant objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when you look at a camera display, you're focussing on the image screen, which is probably 30 cm away.  This means you need to constantly change focus to be able to look ahead and check the rear view monitor.  Not only is this physically exhausting for your eyes, it leads to a significant time period when you're not focussed on either area.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cameras for backing up, which (I hope) happens only at very slow speeds, may be helpful.   Blind spot coverage is generally an assistive system which does not provide direct view (just warning lights or sounds). &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-03-16T12:15:33.863" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7969" PostTypeId="2" ParentId="7961" CreationDate="2016-03-16T12:20:58.583" Score="3" Body="&lt;p&gt;Sounds like the mechansim in a &quot;clicky-top&quot; ballpoint pen.  You'll need a rotating cylindrical ratchet mechanism; the solenoid plunger moves inside this cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a drawing from &lt;a href=&quot;https://www.quora.com/How-does-a-click-pen-work&quot; rel=&quot;nofollow noreferrer&quot;&gt;quora.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0eqzH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0eqzH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-03-16T12:20:58.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7971" PostTypeId="1" CreationDate="2016-03-16T13:26:23.080" Score="1" ViewCount="111" Body="&lt;p&gt;Tritiated water should weigh more than regular water. Can they be centrifugally separated?&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2016-03-16T13:26:23.080" Title="Can tritiated water be separated from regular water with a centrifuge?" Tags="|waste-water-treatment|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7972" PostTypeId="2" ParentId="7810" CreationDate="2016-03-16T13:40:48.750" Score="0" Body="&lt;p&gt;I found my answer &lt;a href=&quot;http://www.edy.es/dev/2011/10/the-stabilizer-bars-creating-physically-realistic-stable-vehicles/#comment-30477&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. So I basically need to get the difference between compression in the two opposite wheels and multiply this difference by the bar's stiffness to calculate changes of suspension force on each side. Pretty simple. It is even simpler to do in my setup than what's described in the link, since I actually add this value to the suspension force.&lt;/p&gt;&#xA;" OwnerUserId="5281" LastEditorUserId="1832" LastEditDate="2016-03-16T14:21:13.010" LastActivityDate="2016-03-16T14:21:13.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7974" PostTypeId="2" ParentId="7965" CreationDate="2016-03-16T14:26:31.903" Score="22" Body="&lt;p&gt;Digital camera systems with LCD screens impose latency of &lt;em&gt;at least&lt;/em&gt; a frame on the images. Fine for slow moving parking cameras, maybe a problem at speed. An object moving at 60mph shown through a 30fps camera with a latency of 1 frame is 1m away from where it appears to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(CRT screens with analog TV cameras that have no digital in the signal path are pretty much the only way to avoid this, which would be an absurdly bulky system.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting the screen brightness right could be a problem when driving at night. The brightness and contrast needs to be roughly the same as ambient, otherwise having a bright rearview mirror would be a distraction. Whereas a mirror shows the same ambient illumination level as looking forwards through the windscreen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FoV and distortion could also be a problem. Objects would appear differently. This has a potential benefit - the elimination of the blind spot - but would require relearning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note on focusing: when you focus on an object in a mirror, you're focusing at the &lt;em&gt;apparent distance&lt;/em&gt;, not the distance of the mirror. So for a mirror 1m away from your eyes looking at a car 20m behind the mirror, you're focusing at 21m. This is not true for screens.&lt;/p&gt;&#xA;" OwnerUserId="1511" LastActivityDate="2016-03-16T14:26:31.903" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="7975" PostTypeId="1" AcceptedAnswerId="7978" CreationDate="2016-03-16T14:46:13.220" Score="0" ViewCount="53" Body="&lt;p&gt;Folks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so today in the Lab I came up with this problem (see picture). It shows a cross section of a cutting machine. There's a blade moving with a constant speed $\frac{dx}{dt}$ towards the center of the roll made up of wood. This roll is turning with a adjustable speed $w$. Let's assume we keep $w$ also constant. So the questions I ask myself are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How does the rotational speed w influence the thickness of the thickness of the fibers being cutted?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the mathematical function of the contact point (roll/blade)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do the fibers have a constant thickness?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/30UtU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/30UtU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through careful consideration of the problem, I found the answers to 1) and 3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes the thickness is influenced by $w$. For 3) it can be said that after one rotation of the roll, the fibers have a constant thickness. By observing the contact point on the wooden roll it can be seen that the point makes a spiral towards the center. Basically what is missing now, is the analytical/mechanical approach and mathematical formulation for this spiral.&lt;/p&gt;&#xA;" OwnerUserId="5499" LastEditorUserId="2246" LastEditDate="2016-03-16T18:48:46.047" LastActivityDate="2016-03-16T18:48:46.047" Title="What is the analytical approach for a fiber cutting machine?" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7977" PostTypeId="1" CreationDate="2016-03-16T15:14:48.340" Score="2" ViewCount="85" Body="&lt;p&gt;I am trying to design a project that Arduino would have connected water and salinity sensor, would be packed in a special box be dropped into ocean up to 500 meters. What sensors should I use and how should they be connected? I assume that I would need professional equipment, not Chinese stuff from eBay.&#xA;Hardware needs to be able to work under 50°C and up to 50 bar.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Budget is about up to 400 pounds for sensors.&lt;/li&gt;&#xA;&lt;li&gt;Resolution of pressure sensor 0 to 100 bar.&lt;/li&gt;&#xA;&lt;li&gt;Resolution of salinity sensor doesn't matter so much. Few percent's tolerance, just to measure it somehow.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="5503" LastEditorUserId="1832" LastEditDate="2016-03-16T15:43:28.547" LastActivityDate="2016-03-16T15:43:28.547" Title="Water pressure an salinity sensor connected to arduino" Tags="|electrical-engineering|design|" AnswerCount="0" CommentCount="1" ClosedDate="2016-03-16T17:18:04.910" ContentLicense="CC BY-SA 3.0" />
  <row Id="7978" PostTypeId="2" ParentId="7975" CreationDate="2016-03-16T15:33:07.557" Score="1" Body="&lt;p&gt;Your rotation speed $w$ is an &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_frequency&quot; rel=&quot;nofollow&quot;&gt;angular frequency&lt;/a&gt; (usually denoted with a small greek omega $\omega$, by the way). And its just what its name says: the change of angle of your roll in a certain amount of time. What is probably confusing is that the cutting velocity changes as the blade moves toward the rotation center of your roll. This is because the circumference depends on the radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotation period is $T=\frac{2 \pi}\omega$. The thickness of your fibers are defined by the distance the blade travels during one rotation period. With the velocity of your blade $\dot x$ this amounts to&#xA;$$&#xA;t= \dot x \cdot T = \frac{2 \pi \dot x}{\omega}&#xA;\quad\mbox{.}&#xA;$$&#xA;So as long as the ratio between angular velocity $\omega$ and blade velocity remains constant, the thickness is constant, too. This answers your questions 1 and 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trajectory of the contact point is best calculated in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Cylindrical_coordinate_system&quot; rel=&quot;nofollow&quot;&gt;cylindrical coordinate system&lt;/a&gt;:&#xA;$$&#xA;(r,\varphi) = (r_0-\dot x \cdot t,\omega \cdot t)&#xA;$$&#xA;with $r$ being the radial coordinate, $\varphi$ the angular coordinate and $r_0$ the outer radius of your roll. Here, $\varphi=0$ for $t=0$ &lt;a href=&quot;https://en.wikipedia.org/wiki/Without_loss_of_generality&quot; rel=&quot;nofollow&quot;&gt;w.l.o.g.&lt;/a&gt; If you need a different starting angle, just add e.g. a $\varphi_0$ to the $\varphi$ coordinate above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get cartesian coordinates, use the transformation&#xA;$$&#xA;(x,y) = (r \cdot \cos \varphi, r \cdot \sin \varphi)&#xA;\quad\mbox{.}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2016-03-16T15:33:07.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7982" PostTypeId="2" ParentId="7961" CreationDate="2016-03-16T16:11:39.430" Score="0" Body="&lt;p&gt;There is another mechanism I once found on a ballpoint pen which I made a simple sketch of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nsnGF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nsnGF.png&quot; alt=&quot;different ballpoint pen mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you press the bolt on the left side, the wedge is released from the top wall hook and pushed against the bottom wall. When you move the bolt far enough against the spring, the wedge will link with the hook on the bottom wall. Upon releasing the bolt, the spring on the right side will push the wedge forward, causing it to tilt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the ballpoint pen, the two hooks on upper and lower wall are not at the same location, but one is further left than the other. This causes the wedge to rest at two different positions, which is what you want in a ballpoint pen. But this is probably not important for your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably you can adapt this mechanism to your problem.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2016-03-16T16:11:39.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7983" PostTypeId="2" ParentId="7965" CreationDate="2016-03-16T16:48:49.083" Score="8" Body="&lt;p&gt;All the points mentioned by others are factors, but the true answer (i believe) is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why cameras (and obviously displays) are not used instead of&#xA;  mirrors?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The law (regulations on street automobiles) states that you need mirrors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The law will most likely not change to exclude physical mirrors because:&lt;br&gt;&#xA;like power brakes and power steering systems, it is a requirement that a physical connection is always available to actuate the brakes and steering in the case of any electronic or component failure for safety reasons. It must be a power &quot;assist&quot;, in that it can only &quot;assist&quot; the existing mechanical connection to make it easier on the driver to actuate, not replace it completely.&lt;br&gt;&#xA;In the event of any failure in the system controls or electronic components, the driver still has the ability to actuate the brakes and steering (albeit requiring much more effort than before).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically: the cameras and display are the electronic &quot;assist&quot; and the mirrors are the fail-safe physical devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to law, safety always comes first, then the optional nice-to-haves can have consideration like cameras and displays or power assisted systems.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does such technology exist? If it exists then why it is not widely&#xA;  used?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Technology definitely exists but durability and reliability (at consumer-level cost effectiveness) is where we are still struggling. Especially for automotive use, which means day-to-day relatively harsh conditions. It takes years to test new systems and approve them for use on automobiles.&lt;br&gt;&#xA;Usually when human lives are at risk, we have double or triple redundant systems depending on severity. The physical mirror serves as a (manual) redundant system.&lt;/p&gt;&#xA;" OwnerUserId="2430" LastEditorUserId="2430" LastEditDate="2016-03-16T18:58:43.677" LastActivityDate="2016-03-16T18:58:43.677" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="7985" PostTypeId="2" ParentId="7965" CreationDate="2016-03-16T17:56:52.550" Score="61" Body="&lt;p&gt;We still use mirrors because:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A mirror is cheap, simple, passive technology that works.  Pretty much the only failure mode is it breaking on impact.  (Though I have had a center rearview mirror fall once because the adhesive that held it to the windshield failed.)  Camera and screens are more expensive, more complex, active devices.  There are many more failure modes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A mirror allows you to observe with binocular vision, preserving 3D clues, like parallax.  A camera projects a single-viewpoint onto a screen giving a 2D projection.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A mirror allows you to see more by slightly shifting your viewing position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A mirror has much better contrast, color, and resolution than any camera and screen combo.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A mirror has effectively zero latency.  A camera and screen has a lag on the order of tens of milliseconds, which could conceivably make a difference in an emergency situation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Camera and screens (with good processing) could give many advantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;They could completely eliminate the glare of headlights or a low sun while still preserving contrast.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They could improve the aerodynamics of the car.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They could improve night vision by shifting infra-red spectrum into the visible range.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They could completely eliminate blind spots, even while allowing for radically different car shapes that might prove more aerodynamic or cheaper to manufacture.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;ETA:  &lt;a href=&quot;http://www.latimes.com/business/la-fi-hy-back-up-cameras-20160617-snap-story.html&quot; rel=&quot;noreferrer&quot;&gt;Backup cameras haven't stopped drivers from backing into stuff&lt;/a&gt;.  This is only tangentially relevant because backup cameras are for backing up but mirrors serve many other purposes.&lt;/p&gt;&#xA;" OwnerUserId="5513" LastEditorUserId="5513" LastEditDate="2017-11-30T23:26:17.543" LastActivityDate="2017-11-30T23:26:17.543" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="7987" PostTypeId="1" AcceptedAnswerId="7988" CreationDate="2016-03-16T18:15:13.300" Score="1" ViewCount="2087" Body="&lt;p&gt;I'm trying to do some sheet metal work in Autodesk Fusion 360 and it has proved frustrating in the extreme.  Maneuvering sketches and bodies to where I want them has proved extremely difficult.  I understand that Autodesk Inventor has far better support for sheet metal parts, which is great if I could afford Inventor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a workflow for making sheet metal parts in Fusion 360?&lt;/p&gt;&#xA;" OwnerUserId="5427" LastEditorUserId="1832" LastEditDate="2016-03-16T18:47:34.087" LastActivityDate="2016-05-15T20:19:21.087" Title="Making sheet metal parts in Autodesk Fusion 360" Tags="|cad|autodesk-inventor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7988" PostTypeId="2" ParentId="7987" CreationDate="2016-03-16T18:15:13.300" Score="2" Body="&lt;p&gt;This video about &lt;a href=&quot;https://knowledge.autodesk.com/community/screencast/af415e28-f8ee-4b0d-b83b-b6076c2a941d&quot; rel=&quot;nofollow&quot;&gt;Faux Sheet Metal&lt;/a&gt; parts was incredibly helpful.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Start with a set of sketches that describe your sheet metal project.&lt;/li&gt;&#xA;&lt;li&gt;Create patches based on those sketches.  Each patch can then be moved to it's proper location.&lt;/li&gt;&#xA;&lt;li&gt;After the patches are where they should be, then they can be extruded into bodies with the appropriate thickness. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;While this let's a user create parts that look like sheet metal parts, I believe there are significant differences between this process and a process that uses tools provided by Inventor or SolidWorks.&lt;/p&gt;&#xA;" OwnerUserId="5427" LastActivityDate="2016-03-16T18:15:13.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="7989" PostTypeId="1" AcceptedAnswerId="15605" CreationDate="2016-03-16T18:37:33.820" Score="3" ViewCount="2478" Body="&lt;p&gt;Shear areas have been obtained for a few different cross-sections. I-sections usually define their shear areas as equal to the product of the height of the section and the thickness of the web (assuming they fall under the thin-walled category).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what would the shear area for these sections be (assuming thin-walled theory applies)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4DFiq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4DFiq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should we adopt the smallest width which gives a continuous rectangle? But then the section to the right would have zero shear area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qHcep.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qHcep.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or should we adopt the width which gives the largest continuous rectangle by area?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wckOr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wckOr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there some other method?&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-03-16T19:46:28.757" LastActivityDate="2017-08-28T19:54:40.730" Title="Shear area of atypical section" Tags="|modeling|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="7991" PostTypeId="1" AcceptedAnswerId="7993" CreationDate="2016-03-16T22:57:03.320" Score="0" ViewCount="138" Body="&lt;p&gt;I'm building a catapult and I asked a question and got &lt;a href=&quot;https://engineering.stackexchange.com/a/7873/5412&quot;&gt;this&lt;/a&gt; response. Is there an equivalent for the wave spring but for angular motion like a torsion spring? Or any improved design on the torsion spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; An improved design by my opinion would be an increased spring constant in a decreased length&lt;/p&gt;&#xA;" OwnerUserId="5412" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-03-19T21:59:33.447" Title="Wave torsion spring?" Tags="|springs|projectiles|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7993" PostTypeId="2" ParentId="7991" CreationDate="2016-03-16T23:54:33.583" Score="2" Body="&lt;p&gt;The main advantage of wave springs compared to coil springs is that you can have the same stiffness and travel for less uncompressed length. This stems from the fact that a compression type coil spring needs to have an open helix to provide room for the spring to compress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In comparison the various types of torsion springs don't change their dimensions much as they are loaded and so the advantages of a wave spring don't really apply. In fact their relatively compact size is one of their main advantages in a lot of applications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would constitute an improve design depends very much on exactly what the constraints of the system as a whole are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few types which fall broadly under the category of torsion spring, including solid bars, helical springs (as in mouse traps) and bundles of fibres. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a long history of using bundles of fibres as torsion springs in catapults dating back at least to the roman &lt;a href=&quot;https://en.wikipedia.org/wiki/Ballista&quot; rel=&quot;nofollow&quot;&gt;balista&lt;/a&gt; and medieval &lt;a href=&quot;https://en.wikipedia.org/wiki/Mangonel&quot; rel=&quot;nofollow&quot;&gt;mangonel&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-19T21:59:33.447" LastActivityDate="2016-03-19T21:59:33.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7994" PostTypeId="1" CreationDate="2016-03-17T01:39:58.207" Score="1" ViewCount="2705" Body="&lt;p&gt;I'm studying Electronics Engineering so I have almost no background in mechanics, just a little of statics and dynamics. I have recently started an internship in an investigation center where they make microelectromechanical systems (MEMS). I was asked to know the deformation in $y$- and $x$-axes for a cantilever beam with a concentrated end load. I have more interest in the vectorial mechanics rather than the material mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done research in Minhang Bao's book &lt;em&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0444516166&quot; rel=&quot;nofollow&quot;&gt;Analysis and Design Principles of MEMS Devices&lt;/a&gt;&lt;/em&gt;, where I found that the maximum displacement at the free end is: &#xA;$$w(L) = \dfrac{4FL^3}{EI}$$&#xA;if the load is on the $y$-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I couldn't find any information related to the compression in the $x$-axis. I came up with the idea of using a similar triangles approach but think that won't be so accurate.&lt;/p&gt;&#xA;" OwnerUserId="5536" LastEditorUserId="2246" LastEditDate="2016-03-17T12:52:38.297" LastActivityDate="2016-03-17T12:52:38.297" Title="How determine the compression of a beam with a concentrated end load in the x axis?" Tags="|beam|deflection|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="7995" PostTypeId="2" ParentId="7994" CreationDate="2016-03-17T02:50:35.297" Score="1" Body="&lt;p&gt;It's not entirely clear what you mean, but I'm assuming you mean this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r9fvr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r9fvr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the axial deflection at the free end is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ w_x(L) = \dfrac{FL}{EA} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be trivially obtained by noting that the compressive stress is equal to&#xA;$$\sigma = \dfrac{F}{A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, by &lt;a href=&quot;https://en.wikipedia.org/wiki/Hooke%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hooke's Law&lt;/a&gt; we know that the relationship between stress and deformation (assuming linear elasticity) is equal to&#xA;$$\begin{align}&#xA;\sigma &amp;amp;= E\epsilon \\&#xA;\therefore \epsilon &amp;amp;= \dfrac{\sigma}{E}&#xA;\end{align}$$&#xA;where $E$ is the elastic modulus of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deflection at the free end is equal to&#xA;$$w(L) = \int_0^L\epsilon(x)\text{d}x$$&#xA;which, for a member under uniform stress such as this one, simplifies to&#xA;$$w(L) = \epsilon L$$&#xA;therefore, using the equations above, we get&#xA;$$ w_x(L) = \dfrac{FL}{EA} $$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-03-17T02:50:35.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="7998" PostTypeId="1" AcceptedAnswerId="8006" CreationDate="2016-03-17T04:39:32.493" Score="7" ViewCount="1772" Body="&lt;p&gt;I am trying to recreate the following model of a wooden roof truss provided to me by a truss manufacturer:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ObODF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ObODF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is twofold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the common/correct way to model the boundary conditions and internal fixity conditions (pin pin?) for the elements as well as the structure itself? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Given generic roof loading in pounds per square foot, how should I distribute the loads to the truss? It seems to me that applying distributed live and dead loads for the top and bottom chords on truss elements poses a stability problem, as it was under my impression &quot;idealized&quot; trusses can only resist axial loading and thus loads can only be applied at the nodes. However this is a real truss, so I expect things may be different.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4808" LastEditorUserId="33" LastEditDate="2016-03-18T01:06:53.707" LastActivityDate="2016-03-18T01:06:53.707" Title="How do you model a real-life truss in structural analysis software?" Tags="|civil-engineering|structural-engineering|modeling|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8000" PostTypeId="2" ParentId="7998" CreationDate="2016-03-17T06:35:51.213" Score="1" Body="&lt;p&gt;It's been a while since I thought about this, so I hope I'm close.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What? 1 fixed in each axis, I think?&lt;br&gt;&#xA;For calculating the truss, you have to treat the pins as.. pins... I believe.&lt;br&gt;&#xA;If you're concerned about any joints, then you can try to do a stress analysis on each joint, I guess. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Distribute the load to the nodes as you normally would for a beam and do the calculation for the truss.&#xA;Then do the beam analysis for the point load or whatever load you are distributing for analyzing the affected beam.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Sorry I can't comment yet.&lt;/p&gt;&#xA;" OwnerUserId="2602" LastEditorUserId="1832" LastEditDate="2016-03-17T10:28:38.827" LastActivityDate="2016-03-17T10:28:38.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8001" PostTypeId="1" AcceptedAnswerId="8039" CreationDate="2016-03-17T07:45:42.193" Score="4" ViewCount="311" Body="&lt;p&gt;I have thin weir, really just a sheet of metal with a smooth and straight upper edge. I want to know how much water will flow over this weir, when the water level on the higher side is a few cm above the edge.&lt;br&gt;&#xA;I'm interested in the special case when the water on the lower side i below the level of the weir:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/mceEg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mceEg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What formula describes this flow?&lt;/li&gt;&#xA;&lt;li&gt;What are important caveats?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-03-17T16:10:20.470" LastActivityDate="2018-08-02T01:51:30.060" Title="Formula to describe flow over thin weir" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8002" PostTypeId="2" ParentId="7934" CreationDate="2016-03-17T08:10:11.717" Score="3" Body="&lt;p&gt;The tool I would use is metal cutting scissors (or &lt;a href=&quot;https://en.wikipedia.org/wiki/Snips&quot; rel=&quot;nofollow&quot;&gt;Snips&lt;/a&gt;). It's remarkably easy to get a straight cut with these after a little practice. I would first make a cut far from the line (somewhere in the area you're going to discard), to get a feeling for a particular tool cutting steel of a particular thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With some experience it is possible to cut as close as 1mm to the line, which reduces the amount grinding work needed.&lt;/p&gt;&#xA;" OwnerUserId="5550" LastActivityDate="2016-03-17T08:10:11.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8003" PostTypeId="2" ParentId="7965" CreationDate="2016-03-17T08:32:13.060" Score="10" Body="&lt;p&gt;Cameras &lt;strong&gt;are&lt;/strong&gt; used as rear-view mirrors, simply not in mainstream passenger cars. For example, here's how rear-view is done in tramways:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iZNpE.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iZNpE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from improved wind resistance which you mentioned, cameras allow to cut maintenance costs. Vehicles without external mirrors can be easily washed by automated washing stations. On older tramways with mirrors, the cabin often has to be washed manually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important aspect is the IR sensitivity, which allows to use cameras in the dark, where mirrors would be inefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To sum it up, it is indeed possible to replace car mirrors with cameras, but this is not done for practical reasons.&lt;/p&gt;&#xA;" OwnerUserId="5550" LastActivityDate="2016-03-17T08:32:13.060" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8004" PostTypeId="1" CreationDate="2016-03-17T09:27:41.590" Score="15" ViewCount="266219" Body="&lt;p&gt;If a water pipe is 15 mm diameter and the water pressure is 3 bar, assuming the pipe is open ended, is it possible to calculate the flow rate or water velocity in the pipe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the calculations I have found appear to need 2 of these: diameter, flow rate, velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So more specifically can you calculate flow rate or velocity from water pressure and pipe diameter?&lt;/p&gt;&#xA;" OwnerUserId="5419" LastEditorUserId="2015" LastEditDate="2018-06-07T16:29:31.530" LastActivityDate="2021-08-04T07:53:33.217" Title="How to calculate flow rate of water through a pipe?" Tags="|mechanical-engineering|fluid-mechanics|pressure|hydraulics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8005" PostTypeId="2" ParentId="8004" CreationDate="2016-03-17T10:05:09.073" Score="3" Body="&lt;p&gt;&lt;strong&gt;General case&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic tools for this kind of questions would be Bernoulli's equation, in the case of water, for an incompressible fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{p}{\rho} + gz + \frac{c^2}{2} = const$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you stated correctly you would at least need to know the velocity for one point. You can extend Bernoulli with pressure drop terms or combine it with the continuity equation and / or make a momentum balance depending the complexity of the problem. To be clear: I mentioned this tools because they are used for this kind of problem, they will not help you solve yours without you knowing more parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possible prerequisites&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;you know that the flow is the result from the hydrostatic pressure out of a big enough tank&lt;/li&gt;&#xA;&lt;li&gt;you know $\eta$ and $N$ of the pump responsible for the fluid flow&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$\eta   \equiv \text{efficiency}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N \equiv \text{power}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically from what you currently stated, you cannot find out the velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Getting an estimate anyway&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could assume that the pressure at the entry is constant and no flow occurs there. Neglecting friction losses and height differences you would get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{p_{in}}{\rho} + gz + \frac{c_{in}^2}{2} = \frac{p_{out}}{\rho} + gz + \frac{c_{out}^2}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{p_{in}}{\rho} = \frac{p_{out}}{\rho} + \frac{c_{out}^2}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt{\frac{2(p_{in}-p_{out})}{\rho}} = c_{out} = 20 \frac{\mathrm{m}}{\mathrm{s}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{V} = cA = 10.60 \frac{\mathrm{L}}{\mathrm{min}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho \equiv 1000\frac{\mathrm{kg}}{\mathrm{m}^3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_{out} \equiv 1 \mathrm{bar}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \equiv \text{cross-sectional area of the pipe}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would do for a ballpark estimate. Alternatively you could get a bucket and measure how much water you can collect in a minute.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="90" LastEditDate="2016-03-17T23:09:13.953" LastActivityDate="2016-03-17T23:09:13.953" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8006" PostTypeId="2" ParentId="7998" CreationDate="2016-03-17T10:52:44.233" Score="6" Body="&lt;p&gt;The most common way to model this structure is as follows (ignore the fact that the proportions are a bit off):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CHPnw.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CHPnw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, all of the diagonals are pinned-pinned. You'll notice in the schematic, however, that the chords (including the diagonals from the supports to the top chord) are not segmented at every point of intersection with the diagonals. Indeed, the chords are probably as long as possible, presenting splices at points E, J and M simply due to constructability demands (usually defined by the size of a standard truck). These splices between beams composing the chords, however, are usually designed so as to behave as a fixed joint, which is why I didn't put pins anywhere along the chords themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the loads, you can just apply them as distributed loads. Since the individual member spans are quite small, almost no bending will occur and the vast majority of the internal forces will be axial regardless. Some analysis programs do allow you to apply distributed loads but set that they should be transformed into nodal forces, but there's really no need. And hell, the real truss &lt;strong&gt;&lt;em&gt;will&lt;/em&gt;&lt;/strong&gt; have that tiny bending, so you're more than welcome to include it in your analysis if you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exception of course are the chords. Since they are not pinned at every joint, they will suffer significant bending, given that the diagonals will &quot;deposit&quot; their axial loads as concentrated nodal loads along the chords' effective spans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally note that obviously the supports are different: one allows for horizontal displacement while the other doesn't.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-03-17T10:52:44.233" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8007" PostTypeId="1" AcceptedAnswerId="8015" CreationDate="2016-03-17T12:32:46.257" Score="9" ViewCount="699" Body="&lt;p&gt;Display screens connected to cameras are significantly different from windows or mirrors.  Looking through a window or mirror, we can focus on infinity, and we can move our vantage point to see a little different.  With a display screen showing live camera footage, we see the field of view from a single point (the camera location) and our focus is on the screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to develop a screen + camera that is much closer in behaviour to a window or a mirror?  I think the screen and camera would need to have the same surface area.  Both would be sensitive to direction, and when camera pixel $(C_i, C_j)$ on the camera receives a photon with frequency $\nu$ from angles $(C_\phi, C_\theta)$, the screen would send a corresponding photon at frequency $\nu$ from position $(S_i, S_j)$ to direction $(S_\phi, S_\theta)$, where $(S_\phi, S_\theta)$ are calculated from $(C_\phi, C_\theta)$ mimicking either window- or mirror-like behaviour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is such a device theoretically possible?  If yes, would such a device be technically feasible today?  If yes, has there been any serious work on such devices?  If it's theoretically possible but not feasible today, what would need to be developed before such devices may be on the horizon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should have a wide range of applications in &lt;a href=&quot;https://en.wikipedia.org/wiki/Telepresence&quot; rel=&quot;nofollow noreferrer&quot;&gt;telepresence&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Augmented_reality&quot; rel=&quot;nofollow noreferrer&quot;&gt;augmented reality&lt;/a&gt;, &lt;a href=&quot;https://engineering.stackexchange.com/q/7965/3599&quot;&gt;automotive engineering&lt;/a&gt;, and surely a lot of other fields.&lt;/p&gt;&#xA;" OwnerUserId="3599" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-03-17T18:06:52.870" Title="Can cameras+display screens with window/mirror-like optical behaviour exist?" Tags="|materials|optics|" AnswerCount="4" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8008" PostTypeId="2" ParentId="8007" CreationDate="2016-03-17T13:38:50.320" Score="0" Body="&lt;p&gt;&lt;strong&gt;3D Cameras&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3D cameras that consist of two cameras apart in order to allow depth perception has been around for a long time. The only floor is that displaying it to a pair of human eyes in a way the brain can understand can be difficult. Most efforts today focus on only showing one image to each eye, and allow the brain to focus on synchronizing the images to one coherent narrative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with that is either you need a display very close to the eyes, or a pair of polarized glasses.&lt;/p&gt;&#xA;" OwnerUserId="5569" LastActivityDate="2016-03-17T13:38:50.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8009" PostTypeId="2" ParentId="8004" CreationDate="2016-03-17T13:46:25.393" Score="13" Body="&lt;p&gt;&lt;strong&gt;Laminar Flow:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the flow in the pipe is laminar, you can use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Poiseuille Equation&lt;/a&gt; to calculate the flow rate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;Q=\frac{\pi D^4 \Delta P}{128 \mu \Delta x}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; is the flow rate, &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is the pipe diameter, &lt;span class=&quot;math-container&quot;&gt;$\Delta P$&lt;/span&gt; is the pressure difference between the two ends of the pipe, &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; is dynamic viscosity, and &lt;span class=&quot;math-container&quot;&gt;$\Delta x$&lt;/span&gt; is the length of the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your pipe is carrying water at room temperature, the viscosity will be &lt;span class=&quot;math-container&quot;&gt;$8.9\times 10^{-4} \, Pa\cdot s$&lt;/span&gt;.  Assuming the pipe is &lt;span class=&quot;math-container&quot;&gt;$5\, m$&lt;/span&gt; long and that the &lt;span class=&quot;math-container&quot;&gt;$3 \, bar$&lt;/span&gt; pressure is the gauge pressure, the flow rate is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;Q = \frac{\pi (0.015)^4(3\times 10^5\,Pa)}{128(8.9\times 10^{-4} \, Pa\cdot s)(5\,m)}=0.0084 \frac{m^3}{s} = 8.4 \frac{l}{s}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we calculate the Reynolds number for this flow rate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;V = \frac{Q}{A} = \frac{0.0084\frac{m^3}{s}}{\frac{\pi}{4}(0.015m)^2} = 48\frac{m}{s}&#xA;$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;Re = \frac{\rho D V}{\mu} = \frac{(1000\frac{kg}{m^3})(0.015m)(48\frac{m}{s})}{8.9\times 10^{-4}\, Pa\cdot s}= 8\times 10^{5}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...we see that this flow is well into the turbulent regime, so unless your pipe is very long, this method is not appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Turbulent flow:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For turbulent flow, we can use Bernoulli's Equation with a friction term.  Assuming the pipe is horizontal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{\Delta P}{\rho}+\frac{V^2}{2}=\mathcal{F}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\mathcal{F}$&lt;/span&gt; accounts for friction heating and is given in terms of an empirical friction factor, &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\mathcal{F} = 4f\frac{\Delta x}{D}\frac{V^2}{2}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friction factor, &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt;, is correlated to Reynolds number and pipe surface roughness.  If the pipe is smooth, like drawn copper, the friction factor will be about 0.003 in this case.  I got that value from &quot;Fluid Mechanics for Chemical Engineers&quot; by de Nevers, table 6.2 and figure 6.10.  I also assumed that the Reynolds number will be about &lt;span class=&quot;math-container&quot;&gt;$10^5$&lt;/span&gt;.  Substituting the equation for friction heating into Bernoulli's Equation and solving for velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;V=\sqrt{\frac{2 \Delta P}{\rho \left( 4f\frac{\Delta x}{D}+1 \right)}}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your pipe is some other material with a rougher surface, then this analysis will over-predict the flow rate.  I'd suggest looking for tables of friction factors for your particular material if you need higher accuracy.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="-1" LastEditDate="2019-05-29T21:03:15.413" LastActivityDate="2019-05-29T21:03:15.413" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8010" PostTypeId="2" ParentId="7965" CreationDate="2016-03-17T13:51:45.750" Score="7" Body="&lt;p&gt;Replacing all car mirrors with cameras is not a good idea. All it does is add unnecessary cost to a vehicle, and the cameras would actually not be quite as good as a standard mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existing automobile mirrors do their jobs quite well. They require zero electricity to operate, and they work when the vehicle is completely powered off. Mirrors are also very resistant to normal wear and tear. They don't break down, display erroneous images, and they even work somewhat when they are broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, cameras and their corresponding screens are far more inferior to a standard mirror. They are much more easily subjected to damage. In a vehicle, the sensitive electronics can fail over time due to stresses such as constant vibrations. Connections can come loose, etc. Using cameras can be somewhat disorienting. The screens would not typically be in a convenient, or logical place. It could be hazardous because the driver would be looking at a screen in the cockpit instead of looking around the car, and get a false sense of security. Anyone that has a backup camera knows that the camera becomes useless if it is at all obstructed. A small speck of dirt is all it takes to effectively block the view. The biggest drawback of cameras is that they do not operate at all when the vehicle is powered off. When a vehicle is parked in a city, typically the driver checks their mirror to make sure it is safe to exit the vehicle. If the vehicle is parked, and turned off, then they would lose the ability to check for traffic and could result in injury or the loss of life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cameras are great for certain purposes. The backup camera is very useful because it allows you to see objects which are not visible from standard mirrors. It lets you see how much room is behind you and is very useful when parking, and also attaching trailers. I could see other uses for cameras on vehicles such as on big trucks. Tractor trailers especially, have very large blind spots. The biggest blind spot is on the passenger side which is sometimes called the &quot;no zone&quot;. When vehicles pass trucks on the right, usually the driver cannot see where they are. The addition of cameras would help to greatly reduce the number of blind spots, and could result in the reduction of accidents.&lt;/p&gt;&#xA;" OwnerUserId="1681" LastActivityDate="2016-03-17T13:51:45.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8013" PostTypeId="2" ParentId="8007" CreationDate="2016-03-17T14:13:09.103" Score="1" Body="&lt;p&gt;Such a screen may be possible with similar technology to &lt;a href=&quot;https://en.wikipedia.org/wiki/Metamaterial&quot; rel=&quot;nofollow&quot;&gt;metamaterials&lt;/a&gt; which is known for the potential application as an 'invisibility cloak'. Some companies also claim to have achieved this for the military, but its effectiveness is questionable because all the PR around it uses still images and mock-ups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick would be to pick up the light from all directions and re-produce that same scatter on the other side (or from the screen). There are ways of &lt;a href=&quot;http://www.nanowerk.com/nanotechnology-news/newsid=39742.php&quot; rel=&quot;nofollow&quot;&gt;making things 'invisible' to some wavelengths using refraction&lt;/a&gt; to bend EM waves around a central object, but it's unlikely this would work for an arbitrarily placed 'screen', unless you could capture the input with a bundle of fibre optics and somehow reproduce it exactly at the other end (without 'twisting' of the outbound scatter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that seems quite hazy and far too under-developed for the practical application you are seeking here. Probably the best you could achieve would be a 3D lenticular screen with head/eye tracking so that it could manipulate the image according to the screen/head relative position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would only work for one person at a time with current technology, as far as I know. The input would then need to be processed into a 3D scene so that it could be re-displayed from other angles. This technology is reasonably mature and there are many technologies ranging from pure camera-based visible-light capturing with software processing, to active 3D scanning cameras that combine multiple active and passive inputs. Alternatively a tightly packed 2D array of cameras could be used and two suitable ones selected to match the relative head-screen orientation. Their field-of-view would still need to be manipulated according to the head-screen distance, this would most likely be easier to do digitally by cropping and scaling the image from a wide-angle lens.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2016-03-17T14:13:09.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8014" PostTypeId="1" AcceptedAnswerId="8023" CreationDate="2016-03-17T14:37:55.613" Score="1" ViewCount="188" Body="&lt;p&gt;This question originates from &lt;a href=&quot;https://engineering.stackexchange.com/questions/7955/truss-manufacturer-drawings-unclear&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://engineering.stackexchange.com/questions/7998/real-life-truss-modelling&quot;&gt;that&lt;/a&gt; question. I have realized the concerns boil down to the following figure.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GDSFZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GDSFZ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Simply put, is the truss above, with the displayed boundary/internal fixity conditions and distributed loading, structurally stable?&lt;/p&gt;&#xA;" OwnerUserId="4808" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-03-18T01:10:55.990" Title="How Do You Determine Truss Stability in a Modeling Program?" Tags="|structural-engineering|structures|structural-analysis|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8015" PostTypeId="2" ParentId="8007" CreationDate="2016-03-17T14:39:16.450" Score="5" Body="&lt;p&gt;The technology to do what you want has existed for decades, and it is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Holography&quot; rel=&quot;nofollow noreferrer&quot;&gt;holography&lt;/a&gt;.  The problem with common photographic sensors and displays is that they only record/reproduce amplitude information about the light.  In order to know e.g. what angle the ray came from, you need to also record the phase information of the light.  This is precisely what holography does.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image shown at the bottom you can see that two images of a single hologram taken from different angles show the mouse as if it is viewed from different angles.  There are parts of the scene visible from one angle that aren't even visible from the other angle such as the back of the mouse and the branch behind the mouse.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technologies needed to make real-time holograms (akin to a camera with a screen) are still in the R&amp;amp;D phases and are very rudimentary right now.   &lt;a href=&quot;https://en.wikipedia.org/wiki/Spatial_light_modulator&quot; rel=&quot;nofollow noreferrer&quot;&gt;Spatial light modulators&lt;/a&gt; provide a way of producing 2D holograms in real time.  &lt;a href=&quot;http://spie.org/newsroom/technical-articles/5425-a-real-time-3d-system-using-electronic-holography&quot; rel=&quot;nofollow noreferrer&quot;&gt;This group&lt;/a&gt; was able to record the hologram using a standard 4K camera with a lens array and used spatial light modulators to reproduce the hologram in real-time (not particularly well though).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Holo-Mouse.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vh2bMm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2016-03-17T18:06:52.870" LastActivityDate="2016-03-17T18:06:52.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8016" PostTypeId="1" AcceptedAnswerId="8020" CreationDate="2016-03-17T14:41:26.263" Score="4" ViewCount="514" Body="&lt;p&gt;What are the best materials and methods to keep the internal temperature of a device lower then the ambient temperature outside of the device? Based on the following information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Starting temp of device is 20°C&lt;/li&gt;&#xA;&lt;li&gt;Ambient temperature around the device is 200°C&lt;/li&gt;&#xA;&lt;li&gt;The time in this environment would be one hour (1hr), after which it would be removed&lt;/li&gt;&#xA;&lt;li&gt;The electronics in the device would be very low powered&lt;/li&gt;&#xA;&lt;li&gt;The size of the device would be 2&quot; square&lt;/li&gt;&#xA;&lt;li&gt;The device would be re-usable&lt;/li&gt;&#xA;&lt;li&gt;The device is in an oven, hot air surrounding it&lt;/li&gt;&#xA;&lt;li&gt;Inside the device is also air&lt;/li&gt;&#xA;&lt;li&gt;It is a simple sensor outputting readings to a bluetooth module&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The goal would be to keep the internal components below 125°C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions on how to achieve this would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="5572" LastEditorUserId="110" LastEditDate="2016-03-20T15:20:31.097" LastActivityDate="2016-03-20T15:20:31.097" Title="Suggestions on how to keep electronics cool in high ambient temperature location?" Tags="|electrical-engineering|thermal-insulation|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8017" PostTypeId="2" ParentId="2564" CreationDate="2016-03-17T15:01:12.027" Score="1" Body="&lt;p&gt;Using &lt;a href=&quot;http://home.hit.no/~hansha/documents/control/theory/kalmanfilter.pdf&quot; rel=&quot;nofollow&quot;&gt;this reference on linear discrete Kalman Filters&lt;/a&gt;, it looks like you can apply a standard observability model. Namely, for a linear Kalman Filter system defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align*}&#xA;x_{k+1} &amp;amp;= A x_k + B u_k \\&#xA;y_k &amp;amp;= C x_k + D u_k,&#xA;\end{align*}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the system is observable if $M_{obs}$ is full rank, where $M_{obs}$ is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xA;M_{obs} = \begin{bmatrix}&#xA;C \\&#xA;C A \\&#xA;\vdots \\&#xA;C A^{n-1}&#xA;\end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xA;\begin{bmatrix}&#xA;C \\&#xA;C A \\&#xA;\vdots \\&#xA;C A^{n-1}&#xA;\end{bmatrix} x_0 = &#xA;\begin{bmatrix}&#xA;y_0 \\&#xA;y_1 \\&#xA;\vdots \\&#xA;y_{n-1}&#xA;\end{bmatrix}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An EKF is just a linear Kalman Filter with Jacobians substituded in for $A$, $B$, $C$, $D$. By using an EKF, I assume that your state kinematics are adequately linearizeable, so observability for EKF should follow the same formulation as above.&lt;/p&gt;&#xA;" OwnerUserId="5573" LastEditorUserId="5573" LastEditDate="2016-03-17T16:02:28.430" LastActivityDate="2016-03-17T16:02:28.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8018" PostTypeId="2" ParentId="8007" CreationDate="2016-03-17T15:38:06.140" Score="1" Body="&lt;p&gt;Transmitting each individual &lt;em&gt;photon&lt;/em&gt; is infeasible, given the amount of computation that would be required, but technology for capturing some information about incoming light direction already exists and is used in the &lt;a href=&quot;https://illum.lytro.com/illum&quot; rel=&quot;nofollow&quot;&gt;Lytro&lt;/a&gt; &quot;light field&quot; camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding light field display does not exist, as far as I'm aware. The Lytro system uses a conventional display with postprocessing that lets you adjust focal point, depth of field, etc. after the photo has been taken.&lt;/p&gt;&#xA;" OwnerUserId="1511" LastActivityDate="2016-03-17T15:38:06.140" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8019" PostTypeId="2" ParentId="8014" CreationDate="2016-03-17T15:58:04.110" Score="0" Body="&lt;p&gt;@Wasabi is right. I fixed Z displacement and X and Y rotations at each and every node and it worked. The truss is structurally stable for given loading. However it is important to be careful with software assumptions and fix nodes where necessary to simulate a 2D model in a 3D program.&lt;/p&gt;&#xA;" OwnerUserId="4808" LastActivityDate="2016-03-17T15:58:04.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8020" PostTypeId="2" ParentId="8016" CreationDate="2016-03-17T16:06:53.680" Score="4" Body="&lt;p&gt;I can think of three ways you could solve this problem:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Insulate the device&lt;/li&gt;&#xA;&lt;li&gt;Add thermal mass to the device, e.g. make the case overly large and dense.&lt;/li&gt;&#xA;&lt;li&gt;Use some kind of cooling system, like a simple coil of tubing inside the device with coolant circulated through it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Insulation and thermal mass will slow the rate of heating.  If the device has a finite operating time in the hot environment, then there is certainly some amount of insulation/mass that can be added such that its internal temperature will not exceed 125 ℃ during that time.  The type and amount of material depends on a lot of factors; you'll need to provide more specifics about your setup for a more detailed answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cooling system approach may be overkill for what you need, but it has the advantage that it can extend your operating time indefinitely, and it is also easier to adjust on-the-fly.  I'm thinking cold water circulating through a tube, so you can just increase the water flow rate if you need more cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll of course want to minimize the power consumption of the electronics as much as possible, too, though I imagine you don't have much leeway there.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="90" LastEditDate="2016-03-17T22:39:48.117" LastActivityDate="2016-03-17T22:39:48.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8023" PostTypeId="2" ParentId="8014" CreationDate="2016-03-17T17:22:21.863" Score="3" Body="&lt;p&gt;As mentioned in the comments to the question, since you are using a 3D analysis tool, you need to fix rotations around the X and Y axes. If you had been using a 2D analysis tool, simple pinned and roller supports would be suficient since rotations are only allowed around the Z axis, but since this is a 3D model, rotations around the other axes must be restrained.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-03-17T17:22:21.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8024" PostTypeId="1" AcceptedAnswerId="8029" CreationDate="2016-03-17T18:03:53.943" Score="2" ViewCount="463" Body="&lt;p&gt;I want to know the heat conduction coefficient of a cup in order to investigate why tea cools at different rates in different containers. I know the cup is mostly plastic and paper. How do I determine the heat conduction? Do I use a known value?&lt;/p&gt;&#xA;" OwnerUserId="5410" LastEditorUserId="368" LastEditDate="2016-03-17T22:44:30.263" LastActivityDate="2016-04-22T10:40:10.963" Title="How can I determine the thermal conductivity of an open container?" Tags="|thermodynamics|heat-transfer|thermal-conduction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8026" PostTypeId="1" CreationDate="2016-03-17T18:38:28.370" Score="4" ViewCount="160" Body="&lt;p&gt;There are several tools available for monitoring electrical metrics (voltage, current, resistance, frequency). I have recently acquired all that I need for these studies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What tools are available for monitoring magnetic metrics (force, flux, reluctance, intensity, density)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am really looking for is a tool that can give me a 3 dimensional representation of field lines. I am looking for a tool that (perhaps) uses the Hall effect to generate a 3D digital plot of potentially complex field interactions involving multiple magnets (possibly in motion). Are there off-the-shelf tools available within the budget of an average hobbyist? If not, could I build a 3-dimensional array of Hall effect sensors that could be arranged around my experiment to capture fields and field changes over time? I am good with writing software but new to circuit creation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add-on question: Are there other tools you would recommend for magnetic studies by a hobbyist?&lt;/p&gt;&#xA;" OwnerUserId="5582" LastEditorUserId="5582" LastEditDate="2016-03-19T16:24:32.130" LastActivityDate="2016-03-19T16:24:32.130" Title="3D Representation of Magnetic Field Lines" Tags="|magnets|tools|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8027" PostTypeId="1" CreationDate="2016-03-17T19:05:03.627" Score="0" ViewCount="132" Body="&lt;p&gt;I saw &lt;a href=&quot;https://engineering.stackexchange.com/q/6294/2841&quot;&gt;this question&lt;/a&gt;, which was too broad, and I thought of a more specific question that can hopefully be answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &quot;normal&quot; (visible light spectrum) cameras, infrared cameras, and UV cameras, and there are also specialized &lt;a href=&quot;https://en.wikipedia.org/wiki/Full-spectrum_photography&quot; rel=&quot;nofollow noreferrer&quot;&gt;VNIR cameras&lt;/a&gt; that capture the visible spectrum plus some of the near-infrared range. As I understand it, the typical application of VNIR cameras use filters to &lt;em&gt;shift&lt;/em&gt; the NIR frequencies into the visible spectrum, to be captured by red, green, and blue sensors, producing false-color images that are essentially a composite of the NIR color information and the visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if, instead of concerning ourselves with the ability to compress the range of frequencies down into the visible spectrum for the sake of human eyesight, we designed the camera with the goal of recording the widest possible range of frequencies (for example, inventing a new image format with more than 3 channels per pixel - say, R, G, B, IR-A, IR-B, IR-C, UV-A, UV-B, UV-C, designing a sensor matrix of subpixels sensitive to those channels, etc.)? Accepting that we would have to substantially improve the manufacturing of the sensor array to achieve the same resolution as an ordinary RGB matrix, let's prioritize the minimizing of chromatic aberration over maximizing detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How far can we extend the frequency range of a single camera?&lt;/strong&gt; If all the light (or electromagnetic radiation) we capture still has to pass through the same aperture and lens(es) at the same time, could we get all of the infrared, visible, and UV spectra at once? Could we go further than that? What factors impose the hardest limitations on such an endeavor? If possible, please differentiate between something that's definitely, completely impossible, vs. just prohibitively expensive.&lt;/p&gt;&#xA;" OwnerUserId="2841" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-03-17T22:42:51.657" Title="What's the widest spectrum a single camera could process?" Tags="|optics|" AnswerCount="1" CommentCount="3" ClosedDate="2016-03-18T14:04:32.637" ContentLicense="CC BY-SA 3.0" />
  <row Id="8028" PostTypeId="2" ParentId="8026" CreationDate="2016-03-17T19:07:30.010" Score="3" Body="&lt;p&gt;There is indeed a 3 dimensional analogue of the iron filings experiment and it is often referred to as '&lt;a href=&quot;https://en.wikipedia.org/wiki/Ferrofluid&quot; rel=&quot;nofollow&quot;&gt;ferrofluid&lt;/a&gt;', a google search for the term will yield lots of interesting videos. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence it is small ferromagnetic particles suspended in some fluid with the right viscosity, again a search should yield plenty of homebrew recipes and it is certainly well within the scope of hobbyist research and even basic experiments should yield some visually interesting results, in fact ferrofluids are something of a nice medium for kinetic sculpture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limitation is that they are still subject to gravity and viscous forces so you won't necessarily be able to visualize complete loops of magnetic field patterns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible approach might be to use a suspension of very fine aluminium particles in air (in an enclosed space) to use induction to visualise field patterns via &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_levitation&quot; rel=&quot;nofollow&quot;&gt;magnetic levitation&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-03-17T19:07:30.010" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8029" PostTypeId="2" ParentId="8024" CreationDate="2016-03-17T19:30:28.260" Score="5" Body="&lt;p&gt;If you are interested in knowing why your tea is cooling in different times, you already answered the question yourself. It is due to the different thermal conductivities of the different items you use it to store.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{Q} = A U \Delta T \tag{1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the heat flux you want to know. This gives you a relationship between the heat that is transfered to the enviroment and the dependance on the type of your container. So you need to know the thermal transmittance $U$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in calculating specific values for different items the following approach should do:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Gather a list of all necessary heat transfer coefficients $\alpha_{ij}$ and thermal conductivity coefficients $\lambda_i$ In my opinion there is no need to calculate / measure these yourself, that's why books with tables of different coefficients exist. &lt;/li&gt;&#xA;&lt;li&gt;Create a simple but accurate enough model of your container. A cup can be regarded as a hollow cylinder for example. Make a sketch of it!&lt;/li&gt;&#xA;&lt;li&gt;Calculate the thermal transmittance $U$. Notice that different regions of your cup will have different heat transfer and thermal conductivity coefficients. For example the part of a coffee cup that is covered with a plastic lid and the lower part of the cup with no contact with the lid. Add them with the formulas for parallel and series (2). The attached image gives an example.&lt;/li&gt;&#xA;&lt;li&gt;Calculate the overall heat flux. A higher heat flux means a quicker cooling time.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$U=\frac{1}{R_{overall}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{c} \\&#xA;R_{overall} = \sum_{i=1}^n R_i &amp;amp;\text{in series} \tag{2}\\&#xA;\dfrac{1}{R_{overall}} = \sum_{i=1}^n \dfrac{1}{R_i} &amp;amp;\text{in parallel}&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_i$ are the resistances for heat transfer and conductivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{\lambda} = \frac{l}{\lambda A}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R_{\alpha} = \frac{1}{\alpha A}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l \equiv \text{length}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A \equiv \text{cross-sectional area}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RwrGI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RwrGIm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2016-04-22T10:40:10.963" LastActivityDate="2016-04-22T10:40:10.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8033" PostTypeId="2" ParentId="8027" CreationDate="2016-03-17T22:42:51.657" Score="2" Body="&lt;p&gt;A raw &lt;a href=&quot;https://en.wikipedia.org/wiki/Microbolometer&quot; rel=&quot;nofollow&quot;&gt;microbolometer array&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt; is completely insensitive to the wavelength (energy per photon) of the incoming radiation &amp;mdash; it simply measures the temperature rise that's due to the total incoming energy. We generally use them only for longwave IR, because they're the best technology for doing that, and we have better technologies for shorter wavelengths anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you could conceivably create a sensor that had multiple bolometers per pixel, each one looking through a suitable filter for a range of wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;Note that &lt;em&gt;complete cameras&lt;/em&gt; have intentionally limited response to short wavelengths because of the materials used in the lenses.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-03-17T22:42:51.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8034" PostTypeId="2" ParentId="8016" CreationDate="2016-03-17T23:50:15.643" Score="1" Body="&lt;p&gt;There are drawer sized data safes available that are fire rated for 30 - 60 minutes.  Perhaps you could inspect one of these for ideas on insulation /thermal mass.  I think that they utilise some sort of cheap ceramic like concrete.  Although 200 degrees isn't that much.  I would have thought that some form of fire rated foam might work.  It comes in spray cans like the common polyurethane foam filler stuff.  You'd just need a suitable (metal?) container to fill.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2016-03-17T23:50:15.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8035" PostTypeId="1" AcceptedAnswerId="9160" CreationDate="2016-03-18T06:54:59.630" Score="7" ViewCount="3257" Body="&lt;p&gt;I've seen two methods to describe porosity of porous media (specifically metal foams):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Porosity $\epsilon$ which is a ratio between volume of voids and total volume of the medium.&lt;/li&gt;&#xA;&lt;li&gt;PPI (Pores Per Inch) which is the number of pores in one linear inch.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Unfortunately while reading some papers about the relationship between volumetric heat transfer coefficient and porosity in metal foams, some authors describe the range of porosity they used with $\epsilon$ while the others use PPI density, so is there any relation to connect both terms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a similar question on &lt;a href=&quot;http://www.cfd-online.com/Forums/fluent/86659-porosity-ppi-how-convert.html&quot;&gt;cfd-online&lt;/a&gt; and someone answered with this formula but I am not sure where it came from or if it's even correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Pore density (ppcm)}=\frac{300}{D_{particle}}\sqrt{\frac{(1-\epsilon)^2}{\pi\epsilon}}$$&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2016-03-18T09:43:35.770" LastActivityDate="2017-11-05T11:37:43.000" Title="Is there a relation between porosity and PPI density in porous media?" Tags="|mechanical-engineering|heat-transfer|porous-medium|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8036" PostTypeId="1" CreationDate="2016-03-18T12:31:45.943" Score="5" ViewCount="8078" Body="&lt;p&gt;I had an argument with a friend about the safe location for openings in reinforced concrete beams. I've chosen position 1, a little above reinforcement steel, as at this location, the beam is subjected to tension in the lower fibers which is resisted by reinforcement steel and, assuming failure case, the concrete at lower fibers is cracked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other opinions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p2hmN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p2hmN.png&quot; alt=&quot;beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4382" LastEditorUserId="33" LastEditDate="2016-03-19T01:06:42.107" LastActivityDate="2016-03-19T14:55:21.640" Title="What are safe locations for openings in reinforced concrete beams?" Tags="|materials|structural-engineering|structures|reinforced-concrete|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8037" PostTypeId="2" ParentId="8036" CreationDate="2016-03-18T13:35:44.620" Score="3" Body="&lt;p&gt;This is the sort of thing that is best defined by each country's structural codes. I'm from Brazil, so I'm going to use the NBR 6118, our concrete code. That being said, the general gist behind all of the codes tends to be pretty similar.&lt;/p&gt;&#xA;&lt;p&gt;Doing a free translation, this is what is described in the Brazilian code:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;13.2.5.1 Openings that cross a beam along its width&lt;/p&gt;&#xA;&lt;p&gt;In any case, the minimal distance between an opening and the nearest beam-face must be greater than 5 cm (~2&amp;quot;) and twice the concrete cover adopted for the beam. The remaining section in this region, having discounted the area occupied by the opening, must be able to resist the design loads, as well as allow for a good concrete pour.&lt;/p&gt;&#xA;&lt;p&gt;So that no special check is required, the following conditions must all be satisfied:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;openings must be in the area under tension and at a distance of at least &lt;span class=&quot;math-container&quot;&gt;$2h$&lt;/span&gt; from the support, where &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is the height of the beam;&lt;/li&gt;&#xA;&lt;li&gt;the opening must be lesser than or equal to 12 cm and &lt;span class=&quot;math-container&quot;&gt;$\frac{h}{3}$&lt;/span&gt;;&lt;/li&gt;&#xA;&lt;li&gt;the minimal distance between aligned opening faces is &lt;span class=&quot;math-container&quot;&gt;$2h$&lt;/span&gt;;&lt;/li&gt;&#xA;&lt;li&gt;concrete covers must be satisfied and steel reinforcement is not disrupted.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So, of the positions you present, only position 1 is automatically valid (and maybe 3, depending on the beam's cross-section and where exactly it is, in relation to the support). Position 2 is in the compression area, and 4 and 5 are too close to the support.&lt;/p&gt;&#xA;&lt;p&gt;That being said, there's nothing to stop you from making openings at these other positions, so long as you certify that the structural integrity of the beam will not be affected.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-03-18T14:46:17.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8038" PostTypeId="2" ParentId="8036" CreationDate="2016-03-18T13:56:20.343" Score="2" Body="&lt;p&gt;Looking in ACI 318-14, &lt;em&gt;Building Code Requirements for Structural Concrete&lt;/em&gt;, I find the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WMW7b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WMW7b.png&quot; alt=&quot;22.5.1.7&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K9xnY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K9xnY.png&quot; alt=&quot;R22.5.1.7&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing a word search for &quot;opening&quot; only shows specific results for openings in one-way and two-way slabs and walls.  However, reading the above Commentary section R22.5.1.7, you can use the provisions of Chapter 23 to design using the strut-and-tie model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TMecm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TMecm.png&quot; alt=&quot;strut and tie&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you can use any rational analysis to determine how the hole in the concrete web affects the strength of the member.  For your cases (1) and (4), these appear to be in tensile regions of the beam, though (1) would be more ideal since there is less shear stress.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: concerning the strut-and-tie discussion in the comments and applicability to slender beams, the Code has this to say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JZMHC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JZMHC.png&quot; alt=&quot;23.1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, it appears that the strut-and-tie model can be used for slender beams so long as the member or local region are checked as an idealized truss following the provisions of that Chapter.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2016-03-18T20:47:01.657" LastActivityDate="2016-03-18T20:47:01.657" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8039" PostTypeId="2" ParentId="8001" CreationDate="2016-03-18T14:07:53.120" Score="4" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.engineeringtoolbox.com/weirs-flow-rate-d_592.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineering Toolbox&lt;/a&gt;, the flow rate over a weir is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Q=\frac{2}{3}C_d b \sqrt{2g}\, h^\frac{3}{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Q$ is the flow rate, $C_d$ is an empirical discharge coefficient, $b$ is the width of the weir, $g$ is gravitational acceleration, and $h$ is the head of fluid above the weir.  That same page has a few other formulas for various weir shapes.  The wild card here is certainly the discharge coefficient; it will depend on a lot of parameters.  &lt;a href=&quot;http://www.lmnoeng.com/Weirs/RectangularWeir.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; has a chart for determining the coefficient based on several dimensions of the weir and surrounding channel.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2018-08-01T15:48:55.457" LastActivityDate="2018-08-01T15:48:55.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8041" PostTypeId="1" CreationDate="2016-03-18T15:44:26.123" Score="1" ViewCount="94" Body="&lt;p&gt;Do the two conductors have to be separated (no current flowing) or two connected lines? Of course both connected to a generator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about the Maxwell equations I guess the two conductors are separated so we just have a static electric field between the two pieces of metal. But if that's the case why do many figures in the books also show a magnetic field? The magnetic field for what I know can only be generated if a variable current is flowing? I am a bit confused about this topic.&lt;/p&gt;&#xA;" OwnerUserId="5533" LastEditorUserId="1832" LastEditDate="2016-03-18T20:31:43.900" LastActivityDate="2016-03-18T20:31:43.900" Title="Why we need two conductors in a waveguide for TEM propagation?" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8042" PostTypeId="1" CreationDate="2016-03-18T16:14:34.230" Score="1" ViewCount="313" Body="&lt;p&gt;Raymer in his book Suggests a method to find Wing Loading W/S (for example we consider jet with 2 engines) (starting from specific thrust T/W, CLto, density/reference density etc.). In step 1: he suggests to use graphichs &quot;Fig. 5.4 Takeoff distance estimation&quot;: you enter takeoff distance (in ordinate), so .. in abscissa you can determine Takeoff Parameter (&quot;TOP&quot;) Step 2: From &quot;TOP&quot; you can find (W/S)=(TOP)sigmaCLto*(t/W) The questions are : in figure 5.4 you can find many diagrams: takeoff distance for a) &quot;Balanced field Length&quot; b) &quot;over 50 ft&quot; c) &quot;ground roll&quot;. Which one I have to use? What is the difference among them? Which curve you have to use, in order to calculate &quot;take off parameter&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Daniel P. Raymer, Aircraft design: a conceptual approach. Chapter 5 Thrust-to-weight ratio and wing loading, paragraph 5.3 Wing Loading, sub-paragraph Takeoff Distance, page 88-89 and &quot;Fig. 5.4 Takeoff distance estimation&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4916" LastEditorUserId="4916" LastEditDate="2016-05-06T09:35:47.473" LastActivityDate="2016-05-06T09:35:47.473" Title="In Raymer diagram “Take off parameter- Take off distance”, what is the difference among “balanced field length”, “over 50 ft”, “ground roll”?" Tags="|aerospace-engineering|aircraft-design|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8043" PostTypeId="1" AcceptedAnswerId="8067" CreationDate="2016-03-18T16:29:27.933" Score="9" ViewCount="4389" Body="&lt;p&gt;Consider a concrete column is under compression from top loading and also carries some shear stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If take a plane 2d element in the column with these stresses and rotate it to the point it gives out the it maximum normal stress then rotate it to give out maximum shear stress which both values should be higher than our original calculated stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't we compare those value to the concrete's compression and shear strength instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologize if my question is too simplistic, I'm still a civil engineering freshman.&lt;/p&gt;&#xA;" OwnerUserId="5624" LastActivityDate="2016-03-19T18:35:56.983" Title="Why don't we use principal maximum stresses when designing a structural element?" Tags="|mechanical-engineering|civil-engineering|structural-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8044" PostTypeId="2" ParentId="2363" CreationDate="2016-03-18T16:54:21.387" Score="2" Body="&lt;p&gt;Its most likely a combination of reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Steering behavior and traffic disruption:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having the axle at the very end of the bus :  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will drastically change the steering behavior or requirements by making the bus' front and rear axles drive along a smaller radius or tighter turn. This can cause more (quantity and/or severity) disruption to other traffic.&lt;br&gt;&#xA;The driver must drive way further forward before starting the turn, otherwise the rear wheels will quickly go beyond the lane the bus was in. This can mean drifting over the center of an intersection (possibly in danger of hitting traffic on the other side) if making a left, or hitting curbs (and possibly more damage to street structures, buildings, people, etc.) if making a right at an intersection. Naturally, having to drive further forward will block more lane(s)/traffic in that same direction until completion of the turn.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having the axle further up:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will allow the bus' front and rear axles to travel along a bigger radius. This allows the bus to make &quot;smoother&quot; turns, allows the bus to start turning sooner, and allows it to stay more within its own &quot;lane&quot; while turning.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;weight distribution and regulations:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used to work in the transportation industry.&lt;br&gt;&#xA;Axle/bogey location is move-able solely to compensate for weight distribution over the wheels. They must make sure that each wheel carries below a certain threshold level of weight in order to be allowed on the streets. If they can't make it work, then they add more axles to distribute it even more. This is enforced with weighing stations and the like.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;CalTrans&#xA;&lt;a href=&quot;http://www.dot.ca.gov/&quot; rel=&quot;nofollow&quot;&gt;http://www.dot.ca.gov/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;GENERAL RULE&#xA;&#xA;35550. (a) The gross weight on any one axle shall not exceed 20,000 pounds,&#xA;and the gross weight upon any one wheel, or wheels, supporting one end of an&#xA;axle, shall not exceed 10,500 pounds.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I believe the streets and freeways can only take so much localized pressure before they start to get damaged.&lt;/p&gt;&#xA;" OwnerUserId="2430" LastEditorUserId="2430" LastEditDate="2016-03-26T05:00:10.017" LastActivityDate="2016-03-26T05:00:10.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8045" PostTypeId="1" AcceptedAnswerId="8046" CreationDate="2016-03-18T17:35:16.133" Score="4" ViewCount="6818" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aj7XI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aj7XI.jpg&quot; alt=&quot;Increase yield strength&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that with strain-hardening you increase the yield and tensile strength. So if you unload at point D, then you also have cold worked it a bit right?&#xA;So the stress-strain curve will look different after point D and have a higher tensile point?&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="421" LastEditDate="2016-03-18T17:51:18.857" LastActivityDate="2016-03-18T21:04:13.217" Title="Does unloading beyond yield point also affect tensile strength?" Tags="|materials|metallurgy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8046" PostTypeId="2" ParentId="8045" CreationDate="2016-03-18T17:50:21.807" Score="5" Body="&lt;p&gt;&lt;strong&gt;Summary: If you are loading and unloading the same sample, the stress-strain curve will not change and the tensile strength will be the same, because stress and strain are calculated from the initial geometry. If you are comparing two otherwise-identical samples with different levels of cold work or strain hardening, their stress-strain curves will be different and they will have different tensile strengths.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you unload at D, then retest the sample (assuming necking has not yet occurred) it will effectively pick up where it left off. The stress-strain curve will follow the right-hand elastic curve (straight line) portion of the diagram back up to point D, and will then continue deforming plastically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, by unloading and then starting the test over you shift the origin of the plot to the right, to where the end of the unload line is. Note that from the new origin, D becomes the yield point, which is higher than before, and the elongation at failure is reduced because it is closer to the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you retest the same sample, the tensile strength does not increase. If you instead compare two samples of the same initial geometry and material, except one is as-annealed and the other is cold worked, the latter will have higher tensile strength. The reason for the discrepancy between the two scenarios (retesting vs two differently worked samples) is that plastic deformation has changed the geometry in the former scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To properly reflect the change in material properties when retesting would require measuring the narrower, longer sample and computing the new stress and strain curves from the new values after unloading. If that were done, the tensile strength recorded for the retested and remeasured sample would be higher than the initial test (stress increases as area decreases). Using the original geometric measurements would cause the sample to continue following the same stress strain curve. Another way of thinking about it is that retesting the same sample is like comparing two samples with different cold work &lt;em&gt;and&lt;/em&gt; different geometries, but basing the calculations on only one samples geometry, in such a way that the curves are identical.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-03-18T21:04:13.217" LastActivityDate="2016-03-18T21:04:13.217" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8047" PostTypeId="1" AcceptedAnswerId="8164" CreationDate="2016-03-18T20:33:18.630" Score="1" ViewCount="283" Body="&lt;p&gt;Using DePriester Chart and Given one of mole fractions ($z$), pressure and temperature we can acquire K-values for that properties and bubble and dew properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are sometimes that we get some K's and using the summation of $y/k$ we get some quantities that are not 1, like 0.9 or 1.2. My question is about those values, are they showing some physical property?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I wanted to calculate the dew point pressure at a given temperature and I take arbitrary pressures (including the dew point pressure) I get the following results.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[p] = psi    sum of y/k&#xA;- - - - - - - - - - - - &#xA;   100         0.828&#xA;   126         1.000&#xA;   150         1.174&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Do the values for the pressures above and below the dew point pressure have a physical meaning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, showing us that we are in two-phase, SH vapor or SC liquid region?&#xA;Or are they just showing there isn't VLE with that given property?&lt;/p&gt;&#xA;" OwnerUserId="5452" LastEditorUserId="1632" LastEditDate="2016-03-25T12:38:09.470" LastActivityDate="2016-03-25T12:45:33.087" Title="Physical meaning of some earned values using DePriester Chart" Tags="|thermodynamics|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8048" PostTypeId="1" AcceptedAnswerId="8050" CreationDate="2016-03-18T21:38:23.820" Score="2" ViewCount="195" Body="&lt;p&gt;I have a situation where a 5mm steel rod needs to be in contact and rolls over segments of a 80mm steel pipe's inner face. Would it be a good idea to loosely slip a Teflon bushing over the rod and use it as a roller? I haven't seen loose fitted bushings, so I'm not sure if it'd be right to let a bushing rotate like that. &lt;/p&gt;&#xA;" OwnerUserId="4301" LastActivityDate="2016-03-18T22:53:46.580" Title="Using a Teflon bushing as a roller" Tags="|mechanical-engineering|bearings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8049" PostTypeId="1" AcceptedAnswerId="15539" CreationDate="2016-03-18T22:52:11.110" Score="5" ViewCount="113" Body="&lt;p&gt;I'm curious about whether a typical liquid CPU cooler (like, e.g., this &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00A0HZMGA&quot; rel=&quot;nofollow&quot;&gt;Corsair H60&lt;/a&gt;) would be able to effectively cool a square sheet of aluminum with 15&quot; sides and a thickness of 1/8&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not even sure how to properly formulate the question so I'll describe the envisioned application. I'd attach the cooling plate to the bottom of the aluminum sheet and build a case for it with a cut-out for the fan. The result would (hopefully) be a refrigerated cooling pad for my constantly overheating laptop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is such an idea feasible? How powerful of a cooler would be needed to effectively chill such a sheet of aluminum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume an ambient room temperature of 70 degrees Fahrenheit and that the bottom of the laptop is roughly 160 degrees Fahrenheit. I'd like to reduce the temperature of the bottom of the laptop to 120 degrees Fahrenheit. The bottom of the laptop is aluminum and about 15&quot;x15&quot;. There is a gap between the aluminum sheet and the laptop bottom of about 1/8&quot;.&lt;/p&gt;&#xA;" OwnerUserId="5632" LastEditorUserId="368" LastEditDate="2016-03-21T15:21:47.227" LastActivityDate="2017-05-26T20:28:04.863" Title="How effectively would a typical liquid CPU cooler cool a 15&quot;x15&quot;x1/8&quot; sheet of aluminum?" Tags="|mechanical-engineering|cooling|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8050" PostTypeId="2" ParentId="8048" CreationDate="2016-03-18T22:53:46.580" Score="1" Body="&lt;p&gt;This is fine in principal, there are a few applications where a similar arrangement is used, in fact it is common for nylon wheeled castors to just have a plain bore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation the bushing will tend to travel along the rod as it rotates unless it is constrained so you will need something to locate it axially. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-03-18T22:53:46.580" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8052" PostTypeId="1" CreationDate="2016-03-19T07:49:19.047" Score="7" ViewCount="3567" Body="&lt;p&gt;I recently read in an &lt;a href=&quot;http://www.ancient-origins.net/news-history-archaeology/researchers-discover-secret-recipe-roman-concrete-020141&quot; rel=&quot;nofollow&quot;&gt;online article&lt;/a&gt;, that researchers have found the recipe of concrete used by imperial Roman architects. The cited &lt;a href=&quot;http://www.pnas.org/content/111/52/18484&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; however uses a recipe described by Vitruvius, which is 2,000 years old. So how exactly is this finding the recipe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, I actually want to ask if one can extract a detailed recipe from the article, understandable by a layman. Even though I read the article, I am not even close to an engineer and thus it is hard for me to judge if the article adds something new to the ancient recipe given by Vitruvius. In the end, I am looking for a step-by-step instruction approach, much like &lt;a href=&quot;http://www.ehow.com/how_5670473_make-ancient-roman-concrete.html#close&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;, maybe even more detailed (for example, in the article, the process to create the quicklime is not described with much detail, temperature missing, etc). I basically want to know, which is the most accurate formulation of the production process of ancient Roman concrete backed by science.&lt;/p&gt;&#xA;" OwnerUserId="5612" LastEditorUserId="2246" LastEditDate="2016-03-22T17:43:31.380" LastActivityDate="2018-03-21T20:33:15.917" Title="Recipe for Roman concrete" Tags="|civil-engineering|concrete|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8054" PostTypeId="1" AcceptedAnswerId="8056" CreationDate="2016-03-19T08:23:35.063" Score="13" ViewCount="5676" Body="&lt;p&gt;I found this strange looking bit in a bit set of uncommon bits, including Security Torx, hexalobular, tri-wing, spanner head, 12-point flange, Torq-set and others. It's the only one that I could not identify, what is it for?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6A1Be.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6A1Be.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SmZP6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SmZP6.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5641" LastEditorUserId="368" LastEditDate="2016-04-25T17:07:14.210" LastActivityDate="2021-01-01T12:23:26.383" Title="What is this Y-shaped screwdriver bit, and what is its purpose?" Tags="|mechanical-engineering|fasteners|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8056" PostTypeId="2" ParentId="8054" CreationDate="2016-03-19T12:13:31.727" Score="20" Body="&lt;p&gt;My first thought is that it might be intended to be a wing nut driver of some sort, but those are usually hollow cylinders with slots for the wings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah ... sure enough, it's described as such in &lt;a href=&quot;http://www.ebay.co.uk/itm/100-Piece-Security-Screwdriver-Drill-Driver-Bit-Set-With-Driver-Handle-Torx-PZ-/140873777112&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Ebay ad&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SfiAU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SfiAU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2016-03-19T16:53:21.270" LastActivityDate="2016-03-19T16:53:21.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8057" PostTypeId="1" CreationDate="2016-03-19T14:52:50.550" Score="6" ViewCount="208" Body="&lt;p&gt;I am simulating a turbine and have divided the mesh into domains: the piping before and after the blade and the turbine blades. The domain containing the blades rotates at 2000 rpm while the other domains are stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mesh is structure and the mesh (between the domains) is conformal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The turbine I am simulating is an axial-flow &lt;a href=&quot;https://en.wikipedia.org/wiki/Wells_turbine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wells turbine&lt;/a&gt;. The solution is steady state. This turbine has 8 blades (360 / 8 = 45° angle between blades). Due to the circumferential symmetry, only one-eighth of the annulus has been computed, imposing periodic boundary conditions in the tangential direction. In my model, the computational domain has been restricted in the axial direction to four chord lengths upstream and eight chord lengths downstream of the blade. Blades are symmetry (Naca00XX).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The computational domain is as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sspKU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sspKU.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According the following figures, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to obtain streamlines as in the following image from &lt;em&gt;Insight analysis of biplane Wells turbine performance&lt;/em&gt; (Fig. 7):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bvMDr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bvMDr.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dx.doi.org/10.1016/j.enconman.2012.03.023&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shaaban, S., &amp;amp; Hafiz, A. A. (2012). Effect of duct geometry on Wells turbine performance. &lt;em&gt;Energy Conversion and Management, 61&lt;/em&gt;, 51-58.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2016-03-23T17:37:28.577" LastActivityDate="2018-03-22T01:46:13.037" Title="Streamlines at the boundary of rotating and stationary domains are malformed" Tags="|mechanical-engineering|turbines|cfd|ansys|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8058" PostTypeId="2" ParentId="8036" CreationDate="2016-03-19T14:55:21.640" Score="0" Body="&lt;p&gt;First preference is #1 , as long as not too near to the edge (affecting rebar). No problem eliminate concrete at this location since the load is on rebar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second preference (if #1 means close to the edge / affecting rebar) is #3. #3 is less preferable since half of the hole is eliminating compressive concrete (upper half beam).&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2016-03-19T14:55:21.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8059" PostTypeId="1" CreationDate="2016-03-19T14:57:54.203" Score="3" ViewCount="174" Body="&lt;p&gt;I'm looking to implement a simple regenerative braking circuit in order to measure how much energy could be regenerated using a DC motor. My knowledge of electronics is very basic to say the least so I was looking for a bit of help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking to power the motor until it reaches it maximum velocity. I'll measure the input voltage and current using a daq device across the resistor R2 in the diagram In order to energy the input power and energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using a double pole double throw switch i'm going to connect the motor to these capacitors to store the generated emf from the motor whilst measuring the generated voltage and current across R2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this work? Apologies if this problem is of a trivial nature for this forum, like I said, my knowledge of electronics is pretty basic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WdhAq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WdhAq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3269" LastActivityDate="2017-11-23T05:41:08.953" Title="Regenerative Braking of DC motor" Tags="|mechanical-engineering|electrical-engineering|motors|regenerative-braking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8060" PostTypeId="2" ParentId="7934" CreationDate="2016-03-19T15:05:41.533" Score="0" Body="&lt;p&gt;CNC Milling should be precise enough (0.01 millimeter accuracy).&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2016-03-19T15:05:41.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8061" PostTypeId="2" ParentId="2363" CreationDate="2016-03-19T15:14:11.823" Score="1" Body="&lt;p&gt;Most busses in the U.S. use a rear engine/rear drive layout, with the engine located behind the rear axle for ease of maintenance.  There is no drive shaft per se, because the engine/transmission/rear axle are integrated.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-03-19T15:14:11.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8062" PostTypeId="2" ParentId="7222" CreationDate="2016-03-19T15:19:16.320" Score="2" Body="&lt;p&gt;I think that the MSJC (ACI530) actually contradicts itself, a little, with regard to biaxial bending of reinforced masonry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you pointed out, Section 2.2.3 (unreinforced masonry) points to the use of the Unity Equation. However, the commentary of Section 2.3.4.2.2 (reinforced masonry) explicitly says,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The interaction equation used in Section 2.2.3 is not applicable for reinforced masonry and is therefore not included in Section 2.3.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it seems that the omission of the unity equation from Section 2.3 is intentional. Further, it isn't clear what that statement means, exactly. Is it implying that the equation explicitly written in Section 2.2.3 ( $\frac{f_a}{F_a} + \frac{f_b}{F_b} \leq 1$ ) is not applicable to reinforced masonry, or that the Unity Equation in general is not applicable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I think that the language provided in Section 2.3.4.2.2 is fairly clear, or at least clear enough, to indicate how one might approach biaxial bending or (biaxial) bending and compression. It states,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The compressive stress in masonry due to flexure or due to flexure in combination with axial load, shall not exceed 0.45$f'_m$ provided that the axial compressive stress due to axial load component, $f_a$, does not exceed the allowable stress, $F_a$, in Section 2.2.3.1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to imply that, for axial force with bending, one would need to satisfy &lt;strong&gt;both&lt;/strong&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f_a}{F_a} \leq 1 \tag{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f_a + f_b}{0.45f'_m} \leq 1 \tag{2}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would then lead me to rationally assume that for pure biaxial bending we'd simply need to satisfy,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f_b^1 + f_b^2}{0.45f'_m} \leq 1 \tag{3}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is just a form of the Unity Equation. Hence, the code seems to contradict itself, kinda.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few final things worth noting:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I have a couple references, written to ACI380-11, that qualify masonry beams for biaxial bending using equation 3 (published by PPI),&lt;/li&gt;&#xA;&lt;li&gt;I see all sorts of stuff online, from reputable schools and organizations, using equation 3 to qualify reinforced masonry beams for biaxial bending,&lt;/li&gt;&#xA;&lt;li&gt;Don't forget that you need to check the combined stress in your steel as well. This conversation has largely focused on the stress in the masonry, but ensure that &#xA;$$\frac{f_s^1 + f_s^2}{F_s} \leq 1 \tag{4}$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2016-03-20T14:07:16.973" LastActivityDate="2016-03-20T14:07:16.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8063" PostTypeId="2" ParentId="8052" CreationDate="2016-03-19T15:48:13.743" Score="11" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;br&gt;&#xA;Before explaining why this type of Roman concrete is different, let me explain what typical concrete is. Concrete is composed of 4 primary materials:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Fine aggregate (sand)&lt;/li&gt;&#xA;&lt;li&gt;Coarse aggregate (gravel)&lt;/li&gt;&#xA;&lt;li&gt;Water&lt;/li&gt;&#xA;&lt;li&gt;Cement&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In typical concrete, the cement is &lt;a href=&quot;https://en.wikipedia.org/wiki/Portland_cement&quot; rel=&quot;nofollow noreferrer&quot;&gt;Portland cement&lt;/a&gt; therefore it is typically referred to as Portland Cement Concrete (PCC) to distinguish it from Asphalt Concrete, which also has fine and coarse aggregate, but uses Asphalt for the binder instead of Portland cement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;https://en.wikipedia.org/wiki/Portland_cement#Types&quot; rel=&quot;nofollow noreferrer&quot;&gt;several varieties&lt;/a&gt; of Portland cement that are given roman numbers I to V and are used for various purposes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type I - General purpose&lt;br&gt;&#xA;Type II - Moderate sulfate resistance&lt;br&gt;&#xA;Type III - High early strength&lt;br&gt;&#xA;Type IV - Low heat of hydration&lt;br&gt;&#xA;Type V - High sulfate resistant (to avoid a damaging chemical reaction &lt;a href=&quot;https://en.wikipedia.org/wiki/Alkali%E2%80%93silica_reaction&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASR&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cementicious materials&lt;/strong&gt;&lt;br&gt;&#xA;Portland cement is not the only binder that can be used in concrete, it is just very common, available (it is not a natural material rather it is manufactured), strong, and has reasonably good durability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other materials that react with water to form cementicious binders are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Pozzolan&quot; rel=&quot;nofollow noreferrer&quot;&gt;pozzolans&lt;/a&gt;. These can be various natural materials or often times are byproducts of other manufacturing processes. For example &lt;a href=&quot;https://en.wikipedia.org/wiki/Fly_ash&quot; rel=&quot;nofollow noreferrer&quot;&gt;fly ash&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Silica_fume&quot; rel=&quot;nofollow noreferrer&quot;&gt;silica fume&lt;/a&gt;, coal ash, slag, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These alternative pozzolans are often substituted in for Portland cement in various quantities. Typically it is not a full replacement, but rather a partial replacement. The exact ratio of Portland cement to pozzolan replacement vary by the mix design and pozzolan type. You can read more about them &lt;a href=&quot;http://www.concretecountertopinstitute.com/blog/2011/10/the-use-of-pozzolans-in-concrete/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Roman concrete&lt;/strong&gt;&lt;br&gt;&#xA;Cement and pozzolans react with water to make various chemical compounds that are primarily composed of calcium-aluminum-silicate-hydrate (CASH). The exact compounds that are formed depend on the type of cement used. This is the key part of why they believe the Roman concrete was so strong, and is the key focus of the article you linked. From the article&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The concrete foundations, walls, and vaulted ceilings are composed of decimeter-sized &lt;strong&gt;volcanic tuff and brick coarse aggregate (caementa) bound by volcanic ash–lime mortar&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So they just mentioned that they used &lt;a href=&quot;https://en.wikipedia.org/wiki/Tuff&quot; rel=&quot;nofollow noreferrer&quot;&gt;volcanic tuff&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Brick&quot; rel=&quot;nofollow noreferrer&quot;&gt;brick&lt;/a&gt; as the aggregate, and they used &lt;a href=&quot;https://en.wikipedia.org/wiki/Volcanic_ash&quot; rel=&quot;nofollow noreferrer&quot;&gt;volcanic ash&lt;/a&gt; mixed with &lt;a href=&quot;https://en.wikipedia.org/wiki/Lime_(material)&quot; rel=&quot;nofollow noreferrer&quot;&gt;lime&lt;/a&gt; as the pozzolan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that we want the cementitious material to form CASH compounds to be an effective binder. Volcanic ash is primarily $SiO_2$ or $AlO$ compounds, and lime is primarily $CaO$. When you mix those with water you have essentially&lt;br&gt;&#xA;$$CaO + AlO + SiO_2 + H_2O \to CASH$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the Romans did not have Portland cement at the time, so this was &lt;em&gt;complete replacement&lt;/em&gt; as far as the pozzolans are concerned. The authors found that a particular compound is formed during the reaction called strätlingite and note that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The strätlingite crystals in Imperial Roman mortar resemble microfibers that are added to the cement paste of present-day mortars and concretes to produce toughening—except that they crystallized in situ and reinforce interfacial zones, the most vulnerable component of the mortar fabric.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The authors emphasize the significance of these crystals&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Three principal differences exist between authigenic strätlingite crystals in Roman mortar and microfiber additives in present-day cement pastes.&lt;br&gt;&#xA;  First, radial spherulites of strätlingite grow preferentially in scoria interfacial zones, whereas microfiber additives remain in the cement paste and do not reinforce aggregate interfacial zones.&lt;br&gt;&#xA;  Second, strätlingite is resistant to corrosion, relative to glass and steel fibers in cement matrices.&lt;br&gt;&#xA;  Third, authigenic crystallization of dense strätlingite intergrowths occurs in the complex accretionary perimetral zones of scoriae, groundmass of scoriae, and cementitious matrix (Fig. 3F) long after portlandite was consumed at about 90 d of hydration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="34" LastEditorUserId="34" LastEditDate="2018-03-21T20:33:15.917" LastActivityDate="2018-03-21T20:33:15.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8064" PostTypeId="1" AcceptedAnswerId="8070" CreationDate="2016-03-19T15:52:00.510" Score="12" ViewCount="19844" Body="&lt;p&gt;This question is so fundamentally basic that I am almost embarrassed to ask but it came up at work the other day and and nearly no one in the office could give me a good answer. I was calculating the shear stress in a member using the equation, $\frac{Tr}{J_T}$ and noticed, that for a shaft with a circular cross section, $J_T = I_P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both $I_P$ and $J_T$ are used to describe an object's ability to resist torsion. $I_P$ is defined as, $ \int_{A} \rho^2 dA $ where $\rho$ = the radial distance to the axis about which $I_P$ is being calculated. But $J_T$ has no exact analytical equations and is calculated largely with approximate equations that no reference I looked at really elaborated on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, what is the difference between the Polar Moment of Inertia, $ I_P $, and the torsional constant, $ J_T $? Not only mathematically, but practically. What physical or geometric property is each a representation of? Why is $J_T$ so hard to calculate?&lt;/p&gt;&#xA;" OwnerUserId="129" LastActivityDate="2016-03-20T00:11:31.080" Title="What is the difference between the Polar Moment of Inertia, $ I_P $ and the torsional constant, $ J_T $ of a cross section?" Tags="|mechanical-engineering|structural-engineering|applied-mechanics|stresses|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8065" PostTypeId="2" ParentId="8054" CreationDate="2016-03-19T18:15:56.190" Score="4" Body="&lt;p&gt;It can also be used for fastening the 'eye' of hook and eye fasteners into wooden frames.&lt;/p&gt;&#xA;" OwnerUserId="5648" LastEditorUserId="272" LastEditDate="2016-03-21T12:04:41.850" LastActivityDate="2016-03-21T12:04:41.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8067" PostTypeId="2" ParentId="8043" CreationDate="2016-03-19T18:35:56.983" Score="8" Body="&lt;p&gt;The short answer is because it is too complicated/impossible to do so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram of the principal stress trajectories for an &lt;em&gt;uncracked&lt;/em&gt; concrete beam under both flexure and compression:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8g7N8.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8g7N8.gif&quot; alt=&quot;Stress Trajectories&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see the orientation and magnitude of the principal stresses will change depending on the point you are interested in and the applied loads. We know that concrete is weak in tension. So if we are looking at a location of principal tensile stress we can compare this to the tensile capacity of the concrete (which is often considered to be a function of $\sqrt{f_c'}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if the principal tensile stress exceeds the tensile capacity of the concrete? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well at that point the concrete may fail. But this doesn't mean the whole element will fail. It means that it will &lt;em&gt;crack&lt;/em&gt; at that location. But that is OK, that is what reinforcement is for!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we have a concrete element with a crack (or many cracks!), and reinforcement to hold the pieces together:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GFxBn.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GFxBn.gif&quot; alt=&quot;TensionCracks&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we now want to calculate our principal stresses, what is the state of stress at a particular point? We have some stress being carried by the reinforcement, some stress being carried by aggregate interlock along the cracks, some being carried by compression, and some voids where no stress can exist - how much goes into each mechanism? We can't simply use formulas like $\nu = \frac{VQ}{It}$ since this only applies to a uniform material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;We can't determine the state of stress with any reasonable certainty in a cracked concrete section$_1$.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what can we do now? Well, we do lots, and lots of &lt;a href=&quot;https://www.concrete.org/publications/internationalconcreteabstractsportal.aspx?m=details&amp;amp;ID=7920&quot; rel=&quot;noreferrer&quot;&gt;tests&lt;/a&gt; and then fit a design equation to the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned columns in your question. Columns are dominated by compressive stresses, so cracking is often not as much of an issue. However, there are still complicating factors which will make it difficult/impossible to determine the stress state. In fact, the commentary of &lt;a href=&quot;https://www.concrete.org/store/productdetail.aspx?ItemID=31814&quot; rel=&quot;noreferrer&quot;&gt;ACI 318&lt;/a&gt; says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The actual distribution of concrete compressive stress is complex and usually not known explicitly. ... The Code permits any particular stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So again, we are forced to take the easier route of assuming a simplified stress state and confirming that is safe according to tests. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty related to using these simplifications is incorporated into the safety factors used in building codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be much more satisfying to have a design methodology which is built on the principal stresses. This has apparently been tried in the past, but was always unsuccessful due to the difficultly in determining the stress state$_2$. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Kong, F. K., &amp;amp; Evans, R. H. (2013). Reinforced and prestressed concrete. Springer.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.concrete.org/publications/internationalconcreteabstractsportal.aspx?m=details&amp;amp;ID=7920&quot; rel=&quot;noreferrer&quot;&gt;ACI-ASCE Committee 326 (1962). Shear and Diagonal Tension&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2016-03-19T18:35:56.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8068" PostTypeId="2" ParentId="8064" CreationDate="2016-03-19T19:23:50.680" Score="0" Body="&lt;p&gt;Polar moment of inertia, Ip, is the resistance of a solid to be torsioned. However, rotational mass moment of inertia, J, is the inertia moment of a rotating solid. See &lt;a href=&quot;http://lpsa.swarthmore.edu/Systems/MechRotating/RotMechSysElem.html&quot; rel=&quot;nofollow&quot;&gt;this web&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand, J is the same as normal moment of inertia, but for rotating objects. &lt;/p&gt;&#xA;" OwnerUserId="5553" LastActivityDate="2016-03-19T19:23:50.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8070" PostTypeId="2" ParentId="8064" CreationDate="2016-03-19T22:10:54.570" Score="11" Body="&lt;p&gt;The torsion constant $J_T$ relates the angle of twist to applied torque via the equation: &#xA;$$ \phi = \frac{TL}{J_T G} $$&#xA;where $T$ is the applied torque, $L$ is the length of the member, $G$ is modulus of elasticity in shear, and $J_T$ is the torsional constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polar moment of inertia on the other hand, is a measure of the resistance of a cross section to torsion &lt;em&gt;with invariant cross section and no significant warping&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case of a circular rod under torsion is special because of circular symmetry, which means that it does not warp and it's cross section does not change under torsion. Therefore $J_T = I_P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a member does not have circular symmetry then we can expect that it will warp under torsion and therefore $J_T \neq I_P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which leaves the problem of how to calculate $J_T$. Unfortunately this is not straightforward, which is why the values (usually approximate) for common shapes are tabulated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of calculating the torsional constant is by using the &lt;a href=&quot;https://en.wikiversity.org/wiki/Introduction_to_Elasticity/Prandtl_stress_function&quot; rel=&quot;noreferrer&quot;&gt;Prandtl Stress Function&lt;/a&gt; (another is by using &lt;a href=&quot;https://en.wikiversity.org/wiki/Introduction_to_Elasticity/Warping_functions&quot; rel=&quot;noreferrer&quot;&gt;warping functions&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without going into too much detail one must choose a Prandtl stress function $\Phi$ which represents the stress distribution within the member and satisfies the boundary conditions (not easy in general!). It also must satisfy Poisson's equation of compatability: &#xA;$$ \nabla^2 \Phi = -2 G \theta $$ &#xA;Where $\theta$ is the angle of twist per unit length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have chosen the stress function so that $\Phi = 0$ on the boundary (traction free boundary condition) we can find the torsional constant by:&#xA;$$J_T = 2\int_A \frac{\Phi}{G\theta} dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; Rod of circular cross section&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the symmetry of a circular cross section we can take:&#xA;$$\Phi = \frac{G\theta}{2} (R^2-r^2) $$&#xA;where R is the outer radius. We then get:&#xA;$$J_T = 2\pi\int_0^R (R^2-r^2)rdr = \frac{\pi R^4}{2} = (I_P)_{circle}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt; Rod of elliptical cross section&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Phi = G\theta\frac{a^2 b^2}{a^2+b^2}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)$$&#xA;and&#xA;$$J_T = \int_A \frac{a^2 b^2}{a^2+b^2}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)dA = \frac{\pi a^3 b^3}{a^2+b^2} $$ &#xA;which is certainly not equal to the polar moment of inertia of an ellipse:&#xA;$$ (I_P)_{ellipse} = \frac{1}{4}\pi a b(a^2+b^2) \neq (J_T)_{ellipse}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since in general $J_T &amp;lt; I_P$, if you used the polar moment of inertia instead of the torsional constant you would calculate smaller angles of twist.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2016-03-19T23:23:06.797" LastActivityDate="2016-03-19T23:23:06.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8071" PostTypeId="2" ParentId="7964" CreationDate="2016-03-19T23:11:46.050" Score="2" Body="&lt;p&gt;In industrial applications probably the closest approximation is in dressing bonded abrasive wheels. These use abrasive particles including aluminum oxide but a different bonding medium which makes them a bit tougher than standard fire bricks. With this is mind a diamond tipped dressing tool would probably be a good thing to try. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I lathe turning stone scrapers are widely used. In this case a tough, abrasion resistant tool steel would probably be the best bet perhaps something like D2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for turning it is probably well worth sawing the outside to a hex shape to get the section closer to round before you start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach would be to use a belt linisher with a coarse ceramic belt and a cylindrical grinding jig. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if you just want a simple cylinder it would be logical to use core drills for both the inside and outside. In this case you might want to pre drill a pilot hole in the centre and fit a metal sleeve and use a plain pin instead of the pilot drill in the arbour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another big consideration is dust extraction to protect both the machine tool and your lungs a firebrick dust will wreak havoc on both. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-03-19T23:11:46.050" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8072" PostTypeId="2" ParentId="8064" CreationDate="2016-03-20T00:11:31.080" Score="3" Body="&lt;p&gt;This is almost a coincidence, and it is only true for solid or hollow circular cross sections. Of course shafts carrying torsion often &lt;em&gt;are&lt;/em&gt; circular, for reasons that are independent of the question! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torsion of a circular shaft is physically simple because of the symmetry of the circular shape. By symmetry, the stresses and strains at any point can only be a function of the radial distance from the centre line of the shaft. By Pythagoras' theorem, you can take an arbitrary pair of axes and express the radius as $r^2 = x^2 + y^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using that fact, you can convert the integral over the cross section into the sum of two integrals in the $x$ and $y$ directions, and again by symmetry those two integrals must be equal to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The form of the integrals happen to be exactly the same mathematical form as for the second moments of area of a circular beam, which leads to the result you asked about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't work for non-circular sections, because the stress distribution is not radially symmetrical. For example if you compare the torsion constant and polar moment of a solid square section, you will find the &quot;constants&quot; in the two formulas are different. The more the cross section deviates from a circle, the bigger the difference will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torsion constant for an complex shaped section (for example an I-beam) is hard to calculate because the stress distribution over the section is complicated, and there is no simple &quot;formula&quot; for it that you an integrate mathematically. Many of the formulas for torsion in engineering handbooks are based on simplified assumptions rather than &quot;exact&quot; mathematical solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in real life the &quot;errors&quot; are not too important, because when a torsional load is applied to a non-circular structure, the cross sections &quot;warp&quot;, i.e. &lt;em&gt;they no longer remain plane&lt;/em&gt;. In real life, the amount of warping is often unknown, because the restraints at the ends of the shaft affect it. If you really need an accurate estimate of the torsional stiffness of a non-circular component, you have to make a full 3-D model of the component itself and how it is fixed to the rest of the structure. If you make a model with that level of detail, there is not much point in reducing the answer to one number just so you can call it &quot;the torsional stiffness&quot;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-03-20T00:11:31.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8074" PostTypeId="1" AcceptedAnswerId="8076" CreationDate="2016-03-20T04:18:43.430" Score="4" ViewCount="661" Body="&lt;p&gt;I'm trying to use a 1500 RPM-rated electric motor to accelerate a (light, 25kg) shaft up to 50 000 RPM. At these speeds I figure using a belt system or roller chains is out of the question due to wear and friction; are these kinds of speeds possible to achieve reliably with a mechanical transmission or do I need to turn to hydraulic transmission? Or what is an efficient way of achieving this?&lt;/p&gt;&#xA;" OwnerUserId="5659" LastEditorUserId="90" LastEditDate="2016-03-20T06:02:30.880" LastActivityDate="2016-03-20T14:29:17.443" Title="Getting high RPM out of a low RPM motor" Tags="|gears|energy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8075" PostTypeId="1" CreationDate="2016-03-20T13:32:38.567" Score="3" ViewCount="40" Body="&lt;p&gt;What are good methods for black box model fitting? I have a set of experimental data, and a mathematical model (basically a set of differential equations) with adjustable parameters. The parameters should be adjusted so that the model reproduces the experimental data. I have some literature values to start, but simply with these values the curves produced by the model don't fit to the curves of the experiment (the literature values are to be regarded with a certain error anyway; the experimental data that I have to fit my model to is fortunately barely noisy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to get some results for two months, with results more than nothing but far from satisfactory. Lately what I've tried is an evolution algorithm, called &lt;a href=&quot;https://en.wikipedia.org/wiki/CMA-ES&quot; rel=&quot;nofollow&quot;&gt;CMA-ES&lt;/a&gt;, but it takes a lot of time, becasue it needs a few hundred iterations, and one iteration involves solving the set of DEs several times, which takes at least 15 minutes. (Now that the better server at the lab died, it takes 50 minutes per iteration...) It gave the best results so far, but these were again not satisfactory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you suggest some good methods to do the fitting?&lt;/p&gt;&#xA;" OwnerUserId="5666" LastActivityDate="2016-03-20T13:32:38.567" Title="Parameter fitting in transistor simulations" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8076" PostTypeId="2" ParentId="8074" CreationDate="2016-03-20T14:29:17.443" Score="2" Body="&lt;p&gt;It sounds like you need something like a &lt;a href=&quot;http://www.cutawaycreations.com/cutaways.php?project=pratt-whitney-pt6-turboprop-engine#/IMAGES/pratt-whitney-pt6-turboprop-engine/images/image4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;turboprop transmission&lt;/a&gt; operating in reverse. These are generally planetary gearboxes for good balance at high speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CWNvN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CWNvN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-03-20T14:29:17.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8077" PostTypeId="1" CreationDate="2016-03-20T14:39:53.847" Score="3" ViewCount="88" Body="&lt;p&gt;When you have a working prototype based on off the shelf parts (eg. Arduino), and you want to move to designing a finished product, what is the correct order of development?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Electrical Engineering? (PCB design, program, etc)&lt;/li&gt;&#xA;&lt;li&gt;Design?&lt;/li&gt;&#xA;&lt;li&gt;Mechanical Engineering?&lt;/li&gt;&#xA;&lt;li&gt;Others?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5572" LastActivityDate="2016-03-20T15:35:13.977" Title="What to do first when going from Prototype to Finished Product?" Tags="|product-engineering|" AnswerCount="1" CommentCount="4" ClosedDate="2016-03-20T17:33:41.683" ContentLicense="CC BY-SA 3.0" />
  <row Id="8078" PostTypeId="2" ParentId="8077" CreationDate="2016-03-20T15:13:24.237" Score="0" Body="&lt;p&gt;Using Arduino type DIY platforms is becoming very common in Rapid prototyping design, but they do have significant drawbacks. To over come them:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Develop a robust requirements document based on VOC (Voice of the Customer). This I believe is the most critical next steps. In corporate all the critical user testing feedback into the requirements document. (&lt;strong&gt;Scope&lt;/strong&gt;) &lt;/li&gt;&#xA;&lt;li&gt;Determine a realistic budget based on expected price points. This is key in determine parts selection, resource selection, etc. One could use an expensive FPGA or inexpensive micro controller to achieve the same end result. What is important is that not to overshoot the budget. (&lt;strong&gt;Cost&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Determine the time to market. Once again this will impact your component selection, resource selection etc. (&lt;strong&gt;Time&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After there is a good understanding of Cost, Time, and Scope (Also know is the triple constraints) then start working on product design. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Electrical design:&lt;/strong&gt; PCB design, electrical component selection (both electrical and mechanical), testing, manufacturing, compliance to regulations etc&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Software design:&lt;/strong&gt; Operating systems, software development platforms, software testing etc&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mechanical design:&lt;/strong&gt; Component selection (both electrical and mechanical), mechanical testing etc. Mechanical engineers work closely with electrical engineers in most cases. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Tool selection:&lt;/strong&gt; This includes any software for PCB design, software development, Mechanical CAD tools, any many more as it relates to finish product.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;User Interface Design:&lt;/strong&gt; If the end product has significant user interaction then it is important to consider Industrial Design into the design process.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Design for manufacturing and testing:&lt;/strong&gt; IMO this is very important step. Understanding the capabilities for the manufacture and design according the capabilities. If the manufacture is required to special equipment this will drive the cost of the product limiting the success. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The above should get you started.&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-03-20T15:35:13.977" LastActivityDate="2016-03-20T15:35:13.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8080" PostTypeId="1" AcceptedAnswerId="8081" CreationDate="2016-03-20T16:18:43.787" Score="1" ViewCount="35" Body="&lt;p&gt;I have an outer pipe and an inner pipe a lot smaller than the inside of the the outer pipe, and the head of the (mini) turbine will be connected to a metal rod. There will be wires connected to the head and it will be turning to the wind. How should I set up the wires?&lt;/p&gt;&#xA;" OwnerUserId="5669" LastEditorUserId="1832" LastEditDate="2016-03-20T16:27:16.847" LastActivityDate="2016-03-22T12:12:40.590" Title="How to make a wire turn with a turbine head" Tags="|design|" AnswerCount="1" CommentCount="2" ClosedDate="2016-03-20T23:15:50.903" ContentLicense="CC BY-SA 3.0" />
  <row Id="8081" PostTypeId="2" ParentId="8080" CreationDate="2016-03-20T16:29:28.590" Score="1" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Slip_ring&quot; rel=&quot;nofollow&quot;&gt;slip ring&lt;/a&gt; connector is the usual way to solve this problem. The electrical contact is made by fixed brushes which make sliding contact with a rotating ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point is to ensure that you maintain a good and consistent electrical contact. It is common to use brass slip rings and graphite contactors held in place with springs and designed to wear and be replaced periodically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphite has the advantage that it combines good electrical conductivity with self lubrication and is soft enough to wear to conform precisely to the shape of the surface of the slip ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also important that the enclosure is designed to prevent contamination of the contact surface by dust, oil, moisture etc. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-22T12:12:40.590" LastActivityDate="2016-03-22T12:12:40.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8084" PostTypeId="1" CreationDate="2016-03-20T22:03:33.817" Score="0" ViewCount="2019" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Sorry in advance if this question is excessively noobish or seems lazy. I have no engineering background and am unsure where to get started tackling this problem. Links to resources that could put me on the right path would be very much appreciated. I'm not looking for others to do my work for me; I just want to know what sorts of calculations I need to do.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm looking to build a cube-shaped structure, roughly 10 ft cubed, for suspending a series of seats (swings, sort of) for people to sit on. I'd planned on building the structure out of &lt;a href=&quot;http://www.tomcatglobal.com/Products/Box-truss/Core-truss-12-x-12-plated/&quot; rel=&quot;nofollow&quot;&gt;12&quot; aluminum trusses&lt;/a&gt; or buying one pre-made. &lt;a href=&quot;http://www.idjnow.com/global-truss-trussing-sq10x10.html?gclid=CjwKEAjw8bO3BRDp0bP_vL-7_lASJACL_d6w_2ublZnuCRMEnwEirCbZTkIN0Ln3Suvm5Ow69nU6kBoCo7zw_wcB&quot; rel=&quot;nofollow&quot;&gt;Here's an example&lt;/a&gt; of what I might buy or try to build, for visualization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd then loop rope around the horizontal trusses, using the rope to hang &quot;benches&quot;/&quot;swings&quot; from the structure. I'd attach the trusses to plates and use rebar to affix the plates (and the structure) to the underlying ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is -- How do I test whether this structure would be sturdy/safe?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If I'm planning on hanging, let's say, five seats from each horizontal segment, how can I calculate if the trusses would support that sort of vertical force?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How about the horizontal force caused by someone potentially swinging on one of the seats?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Horizontal forces caused by wind hitting any tarp (solid or perforated) I use to cover the structure?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If this calculation is too complex/multivariate for an amateur like myself to reliably tackle, is there somewhere I could turn for an engineer's feedback? &lt;/p&gt;&#xA;" OwnerUserId="5674" LastEditorUserId="1832" LastEditDate="2016-03-20T22:53:17.073" LastActivityDate="2016-03-21T02:16:14.160" Title="How to calculate a structure's ultimate load" Tags="|structural-engineering|structures|" AnswerCount="2" CommentCount="2" ClosedDate="2016-03-21T01:19:10.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="8085" PostTypeId="2" ParentId="8084" CreationDate="2016-03-20T23:29:38.827" Score="2" Body="&lt;p&gt;There is not a single safety calculation for a structure. Something built-up like this has multiple failure points, and each requires a separate calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, each fastener may shear, bend or get stripped. Each would be a different calculations (for each fastener). To totally inspect if the structure performs as expected you need a) a lot of calculations from an experienced engineer or b) a good FEA model from an professional engineer.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2016-03-20T23:29:38.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8086" PostTypeId="1" CreationDate="2016-03-20T23:32:25.690" Score="5" ViewCount="722" Body="&lt;p&gt;I know that forced convection requires some external bulk flow, but is there an equation to determine which form of Convection is dominant? I looked around and found the Richardson number $\mathrm{Ri}=\frac{\mathrm{Gr}}{\mathrm{Re}^2}$ but I don't know how to interpret the number or if it even applies to mass transfer.&lt;/p&gt;&#xA;" OwnerUserId="4785" LastEditorUserId="1632" LastEditDate="2016-03-21T00:13:25.640" LastActivityDate="2016-03-21T02:13:12.550" Title="How to determine if mass transfer convection is natural or forced?" Tags="|chemical-engineering|convection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8087" PostTypeId="2" ParentId="8084" CreationDate="2016-03-20T23:40:38.860" Score="1" Body="&lt;p&gt;In the first link you give there is a link to a PDF document which gives loading values for the trusses in various different load setups. For the sort of cubic structure you describe these should be perfectly valid as long as they are assembled as designed and this should give you a good starting point for an initial estimate for the load capacity of the structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will also need to incorporate some sort of safety factor: a number you multiply your expected load by to get the required capacity of the structure. For example, if you have a known load of 100 kg a safety factor of 3 would give you a required capacity of 300 kg. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This safety factor reflects your uncertainty about your analysis and gives some margin for unexpected circumstances, especially where the structure has safety implications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you are looking at prefabricated struts I would suggest that you find a reputable supplier and talk to them about your application. These things are designed to be versatile and general-purpose, so if you can describe your application in detail they should be able to give you specific advice.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="1832" LastEditDate="2016-03-21T02:16:14.160" LastActivityDate="2016-03-21T02:16:14.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8088" PostTypeId="2" ParentId="8086" CreationDate="2016-03-21T00:21:09.017" Score="5" Body="&lt;p&gt;The Richardson number $\mathrm{Ri} = \frac{\mathrm{Gr}}{\mathrm{Re}^2} = \frac{\text{buoyancy term}}{\text{flow gradient term}}$ &lt;a href=&quot;https://en.wikipedia.org/wiki/Richardson_number&quot; rel=&quot;noreferrer&quot;&gt;&lt;sup&gt;(1)&lt;/sup&gt;&lt;/a&gt; can be used to determine if forced or natural convenction dominate in thermal convection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{Re}$ = Reynolds number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathrm{Gr}$ = Grashof number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can neglect natural convection for $\mathrm{Ri} &amp;lt; 0.1$ and forced convection for $\mathrm{Ri} &amp;gt; 10$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't get into more detail about the number as you can learn more about it in the linked Wikipedia article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting question is: is this applicable to mass convection too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would lean towards saying yes, but I can't tell you for sure; it should be applicable in order to get at least a ballpark estimate.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Dimensional analysis of convective mass transfer yields a number of&#xA;  pertinent dimensionless groups that are, in general, analogous to&#xA;  dimensionless groups for convective heat transfer. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A close analogy exists between convective heat and convective mass&#xA;  transfer owing to the fact that conduction and diffusion in a fluid&#xA;  are governed by physical laws of identical form, that is, Fourier’s&#xA;  and Fick’s laws, respectively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Kreith, F.; Boehm, R.F.; et. al. “Heat and Mass Transfer”&#xA;&lt;em&gt;Mechanical Engineering Handbook&lt;/em&gt; p. 266 ff.&lt;a href=&quot;http://www.itiomar.it/pubblica/dispense/MECHANICAL%20ENGINEERING%20HANDBOOK/Ch04.pdf&quot; rel=&quot;noreferrer&quot;&gt;&lt;sup&gt;(2)&lt;/sup&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Give that chapter a read and decide for yourself what you can apply from thermal convection.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1832" LastEditDate="2016-03-21T02:13:12.550" LastActivityDate="2016-03-21T02:13:12.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8090" PostTypeId="1" AcceptedAnswerId="8105" CreationDate="2016-03-21T04:43:23.813" Score="6" ViewCount="5430" Body="&lt;p&gt;First a little background, I've been working in Michigan as an engineer since getting my BSME in 1998 (I graduated with my MSME in 2010). I've worked for a number of companies in a variety of industries as an engineer, but never under the direct supervision of a licensed PE since all of the work I've done has been considered under &quot;industrial exemption&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now however, I'm becoming less interested in simply being an employee and more interested in working for myself and being a PE would be advantageous. I'm reasonably sure I can pass the exam (with copious amounts of studying, naturally), but my concern is getting my experience verified. The PE's that I know, I've never worked with, either as a subordinate or as a peer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone had difficulty finding enough PE's to verify their experience when getting licensed? If so, were you successful in getting your experience verified? How do you recommend finding PE's to verify engineering work experience?  &lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="368" LastEditDate="2016-03-21T23:18:03.647" LastActivityDate="2016-03-22T00:07:32.740" Title="How can I verify my engineering work experience without having worked under a PE?" Tags="|licensure|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8091" PostTypeId="1" CreationDate="2016-03-21T11:37:17.140" Score="3" ViewCount="8179" Body="&lt;p&gt;I am trying to work out the acceptable pressure drop/increase on a leak test rig for a given leak rate. A simplified version of the test rig looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pfZ1j.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pfZ1j.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas (air) supply is at 1 bar above atmospheric pressure, and the inlet &amp;amp; outlet control valves + pressure sensor are part of the test rig and can be assumed to be leak-free. I am also working on the assumption that the process is isothermal.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Test 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The first test is to open the inlet control valve, with the outlet control valve closed and the valve of the thing under test also closed, and look for a pressure increase on the pressure sensor. At the start of the test, the pressure sensor is at atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my attempt, using $PV=nRT$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moles of gas gained downstream of the valve under test due to leakage past the valve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_2 = n_1 + \frac{LR * \Delta t * P_1}{RT}$ (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P_1$, $V_1$ and $n_1$ are the initial conditions ($P_1 = P_{atm}$), $LR$ is the leak rate past the valve under test and $\Delta t$ is the duration of the test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final pressure can then be worked out as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_2 V_2 = n_2 RT = (n_1 + \frac{LR * \Delta t * P_1}{RT}) RT$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $V_2 = V_1$ (the volume hasn't changed), we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_2 = \frac{P_1 V_1 + LR * \Delta t * P_1}{V_1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then obviously:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta P = P_2 - P_1 = \frac{LR * \Delta t * P_1}{V_1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that the right approach? Should I use $P_1 + 1 \text{bar}$ instead of $P_1$ in equation (1)?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Test 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now the whole thing is pressurised to 1 bar by opening the inlet control valve, the valve under test and closing the outlet control valve, before closing the inlet control valve, which is when the test starts. This is to look at leakage out of the &quot;thing&quot; through pneumatic fittings, etc... We are now looking at a pressure drop from 1 bar, instead of a pressure increase as in test 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using essentially the same approach, except that we're now looking at a number of moles of air lost in equation (1), which becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n_2 = n_1 - \frac{LR * \Delta t * P_1}{RT}$ (1a)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest follows as per test 1, so that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta P = P_2 - P_1 = -\frac{LR * \Delta t * P_1}{V_1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of course, this time $P_1 = P_{atm} + 1 \text{bar}$ and $LR$ is the leak rate out of the &quot;thing&quot; rather than past the valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, is that the right approach?&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-01-28T13:14:38.090" Title="Working out pressure drop from leak rate" Tags="|thermodynamics|pressure|gas|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8092" PostTypeId="2" ParentId="7879" CreationDate="2016-03-21T13:37:03.077" Score="0" Body="&lt;p&gt;We used a simple rubber between two concrete walls. I don't see any reason you can't use the same for a concrete steel wall connection. As these sorts of walls don't settle much you should be ok just coupling these two together without worrying about one settling and dragging the other. Just weld a flat sheet (possibly lined with rubber for water tightness) to the end of the steel one and lowered into the wet concrete. &lt;/p&gt;&#xA;" OwnerUserId="5092" LastEditorUserId="1832" LastEditDate="2016-03-21T14:33:47.097" LastActivityDate="2016-03-21T14:33:47.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8093" PostTypeId="2" ParentId="7763" CreationDate="2016-03-21T13:42:41.433" Score="1" Body="&lt;p&gt;You can't use the pressure from the heavy vehicle as your bearing capacity. The reason is that the pressure from small dimension (from tyres) is not equivalent to that of a large dimension (building footing) even though the net pressure immediately under them is the same. The pressure bulb penetration depth into soil is a lot bigger from the latter. Therefore, in road site investigations they only investigate the top metre of the soil and treat it if required. They hardly bother to see what's below that. While for a one storey to say 10 storey building you will drill down to 3 m to 10 m (depending on many things to work out this depth) respectively. Therefore, if you are using them only for the vehicle load, you should be fine. However, for bigger structures you need to investigate deeper.&lt;/p&gt;&#xA;" OwnerUserId="5092" LastEditorUserId="1832" LastEditDate="2016-03-21T14:36:42.453" LastActivityDate="2016-03-21T14:36:42.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8094" PostTypeId="2" ParentId="8091" CreationDate="2016-03-21T13:59:48.953" Score="2" Body="&lt;p&gt;The rate of gas leakage depends on the pressure difference, and in your setup it looks like the pressure difference is constantly decreasing as the test is running.  This means that the leak rate parameter ($LR$) in your derivation is the &lt;em&gt;average&lt;/em&gt; leak rate over the duration of the test.  In my experience, leak rate is usually specified as an instantaneous value at steady-state conditions (constant pressure difference).  I'm not sure how your leak rate spec is meant to be interpreted, but measuring it with an unsteady-state test rig is probably more difficult than with a steady-state setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, here's an equation you can use to correlate leak rate to pressure difference and hole area:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot m = \mu \cdot A \cdot \sqrt{\frac{\gamma}{\gamma -1} \left[ \left( \frac{p_1}{p_2} \right)^\frac{2}{\gamma} - \left( \frac{p_1}{p_2} \right)^\frac{\gamma + 1}{\gamma} \right]} \cdot \sqrt{2\cdot p_2 \cdot \rho_2}&#xA;$$&#xA;Source: &lt;a href=&quot;http://www.maceas.com/en/pdfs/leak-rate_calculation.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;MACEAS&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\dot m$ is mass flow rate&lt;/li&gt;&#xA;&lt;li&gt;$A$ is area of the leak&lt;/li&gt;&#xA;&lt;li&gt;$\gamma$ is the specific heat ratio of the gas (air = 1.4)&lt;/li&gt;&#xA;&lt;li&gt;$p_1$, $p_2$ are downstream and upstream pressures, respectively&lt;/li&gt;&#xA;&lt;li&gt;$\mu$ is a dimensionless friction coefficient that needs to be determined experimentally&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to use this in your current test setup, you'll need to integrate it over time and replace downstream pressure with an equation to account for the changing pressure.  Also take a look at &lt;a href=&quot;https://engineering.stackexchange.com/questions/7216/determining-mass-flow-rate-and-exit-velocity-of-compressible-flow/7219#7219&quot;&gt;this other, related question&lt;/a&gt; regarding the calculation of unsteady flow.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-03-21T13:59:48.953" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8096" PostTypeId="1" CreationDate="2016-03-21T17:15:44.290" Score="0" ViewCount="72" Body="&lt;p&gt;The task I got is to measure the girth of the abdomen continuously for a 24 hour period using a suitable method. I have thought to measure the impedance between two points in the abdomen and use the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roh=Resistance X (Area / Length)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this practically feasible?  Are there any better methods to achieve my task? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my &lt;a href=&quot;https://engineering.stackexchange.com/questions/7905/how-to-measure-abdominal-girth-measurement-of-the-distance-around-the-abdomen&quot;&gt;Previous question&lt;/a&gt;. I have gone down the impedance route for now. What I am envisioning is a belt worn around the waist to do the task.&lt;/p&gt;&#xA;" OwnerUserId="5434" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-03-22T15:21:41.120" Title="Measurement of girth of abdomen (automatic) using body impedance measurement" Tags="|electrical-engineering|measurements|biomedical-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2016-03-22T02:59:04.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="8097" PostTypeId="1" CreationDate="2016-03-21T17:33:57.827" Score="6" ViewCount="658" Body="&lt;p&gt;Given a cable (totally flexible) fixed at both ends, subjected to a vertical force $f(x)$ in his plane, with variable area $A(x)$, and variable elasticity $E(x)$ I want to find the differential equation of the vertical displacement $y(x)$ in his equilibrium position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the differential equation is:&#xA;$$\frac{d}{dx}\left( E(x)A(x)\frac{dy(x)}{dx} \right) = f(x)$$&#xA;But I can't get to that or something similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried the following:&#xA;The tension of the cable will be $\vec{T} = T\vec{u}$ where $\vec{u}$ is the vector tangent to the cable. Setting equilibrium in the direction of the cable we get: &#xA;$$ \vec T(s+\Delta s) - \vec T (s) + \vec f(x) = 0$$&#xA;so:&#xA;$$ \frac{d \vec T(s)}{ds} + \vec {f} (x) = 0 $$&#xA;For the basic equations of elasticity we know that &#xA;$$ T = A(x)E(x)\epsilon $$&#xA;But I don't know how to combine this information to get an equation with $x$ as an independent variable and $y$ as the dependent one.&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2017-07-25T13:33:56.063" Title="Differential equation of the vertical displacement of a cable" Tags="|structural-engineering|modeling|statics|mathematics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8098" PostTypeId="1" AcceptedAnswerId="8099" CreationDate="2016-03-21T17:47:11.380" Score="3" ViewCount="789" Body="&lt;p&gt;I have 8&quot; thick reinforced concrete walls and I would like to anchor my horizontal corner reinforcement with only 1 hook (I have only 1 layer of reinforcement in both walls). The problem is CSA A23.3 (same for ACI if I'm not mistaken) restricts minimum hook length to 6&quot; or $8d_b$ = 120mm = 5&quot; in my case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to put my reinforcement in the middle of my wall, there is no way for me to get that 6 inches. Can I simply assume that the bar is held by my vertical reinforcement (see picture)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o39uF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o39uF.png&quot; alt=&quot;CORNER&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5691" LastEditorUserId="2246" LastEditDate="2016-03-21T17:52:00.357" LastActivityDate="2016-03-21T17:55:58.260" Title="Basic Hook development length using CSA A23.3 provisions" Tags="|reinforced-concrete|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8099" PostTypeId="2" ParentId="8098" CreationDate="2016-03-21T17:55:58.260" Score="3" Body="&lt;p&gt;I'd agree, you cannot fit a &lt;em&gt;horizontally oriented&lt;/em&gt; 5&quot; 90° bar bend in an 8&quot; thick wall if you want your rebar curtain to be located at the center-line of the wall. However, there is no reason that the bent bars will not fit if you orient the bend vertically. The bars can be placed such that the 5&quot; 90° bar bend is oriented vertically.&lt;/p&gt;&#xA;" OwnerUserId="129" LastActivityDate="2016-03-21T17:55:58.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8100" PostTypeId="2" ParentId="8096" CreationDate="2016-03-21T19:02:24.453" Score="1" Body="&lt;p&gt;Measuring the resistance of a person's abdomen will not be a good method to determine their girth because the resistivity varies by more than two orders of magnitude.  According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Electric_shock#Body_resistance&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia article on electric shock&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The NIOSH states &quot;Under dry conditions, the resistance offered by the human body may be as high as 100,000 Ohms. Wet or broken skin may drop the body's resistance to 1,000 Ohms,&quot; adding that &quot;high-voltage electrical energy quickly breaks down human skin, reducing the human body's resistance to 500 Ohms.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, the resistance varies by a factor of ~100 while the measurement you are trying to make varies by a factor of ~2.  That gives a very terrible signal to noise ratio.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other ideas that might be more effective:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Optical: Use a camera against a scale or use lasers.&lt;/li&gt;&#xA;&lt;li&gt;Electrical: Determine the resonance frequency of their abdomen (I believe this will be less noisy, but I'm not 100% sure).&lt;/li&gt;&#xA;&lt;li&gt;Pressure waves: Bang their stomach with a quick tap and measure the time it takes the wave to propagate around to the back (this might not work well on muscular people).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="272" LastActivityDate="2016-03-21T19:02:24.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8101" PostTypeId="1" CreationDate="2016-03-21T19:04:39.523" Score="0" ViewCount="602" Body="&lt;p&gt;Raymer (see source below) affirms &quot;Thrust-to-weight ratio is closely related to maximum speed&quot;. Why? How can we demonstrate it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Daniel P. Raymer Daniel P. Raymer &quot;Aircraft Design: A conceptual Approach&quot; Chapter 5, Paragraph 5.2 &quot;THRUST-TO-WEIGHT RATIO&quot; Subparagraph &quot;Statistical Estimation of T/W&quot; page 80.&lt;/p&gt;&#xA;" OwnerUserId="4916" LastActivityDate="2016-09-30T17:43:08.737" Title="How can we prove that Thrust-to-weight ratio depends by max speed?" Tags="|aerospace-engineering|aircraft-design|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8102" PostTypeId="1" CreationDate="2016-03-21T21:43:41.333" Score="7" ViewCount="4064" Body="&lt;p&gt;I have a single angle connected to a double angle via a 3/16&quot; weld &lt;a href=&quot;https://i.stack.imgur.com/PRmzC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PRmzC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there is a factored, applied moment at that connection. I would like to ask how I can go about checking the capacity of this connection for moment? There is a lot of stuff in AISC for welds in Chapter J, partially and fully restrained moment connections but I am not sure where to start. What chapter/section of AISC codifies this kind of problem?&lt;/p&gt;&#xA;" OwnerUserId="4808" LastEditorUserId="33" LastEditDate="2016-03-22T13:49:22.273" LastActivityDate="2016-03-22T17:40:42.597" Title="How do you check a weld for applied moment?" Tags="|structural-engineering|structures|structural-analysis|welds|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8104" PostTypeId="2" ParentId="8102" CreationDate="2016-03-21T22:41:21.187" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What chapter/section of AISC codifies this kind of problem?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're looking for a specific code provision for this exact type of weld, you're out of luck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is possible to solve this problem using basic mechanics of materials principles.  For this case, I'm going to use elastic design.  You could also use the instantaneous center of rotation (ICR) method, but that's pretty complicated when you have loading in more than one direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to recognize is whether or not this joint is considered pinned or fixed.  If fixed, you have moments to contend with that introduce loading out of the plane of the weld.  First, let's introduce the equations that describe the weld section properties, as found in Blodgett's &lt;em&gt;Design of Welded Structures&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qJtm0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qJtm0.png&quot; alt=&quot;weld lines&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we call $x$ the axial direction of the member, $y$ the vertical, and $z$ transverse, you solve for the different weld force components like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_{xw} = \frac{F_x}{L_w} + \frac{M_y}{S_{yw}} + \frac{M_z}{S_{zw}}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_{yw} = \frac{F_y}{L_w} + \frac{M_x(0.5b)}{J_w}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_{zw} = \frac{F_z}{L_w} + \frac{M_x(0.5d)}{J_w}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $L_w$ is the total weld length and $ f_{xw}, f_{yx}, $ and $ f_{zw} $ represent the component forces of the weld about the noted axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the resultant weld force, you take&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_w = \sqrt{f_{xw}^2 + f_{yx}^2 + f_{zw}^2}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f_w$ is the resultant weld force.  This you compare to the allowable weld strength per the AISC code.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; this is a simplified approach that does not take flexibility of the joint into account.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2016-03-22T17:40:42.597" LastActivityDate="2016-03-22T17:40:42.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8105" PostTypeId="2" ParentId="8090" CreationDate="2016-03-22T00:07:32.740" Score="3" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;How can I verify my engineering work experience without having worked under a PE?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Many states offer the ability to substitute sufficient / substantial work experience for directly supervised work experience.  The key difference here is that the obligation is upon the applicant to demonstrate how the in lieu of work still meets or preferably exceeds the expectations of the professional board with regards to the quality of the work produced.  In other words, because you can't produce a PE to vouch for direct supervision then you have to demonstrate why the work ought to qualify.&lt;/p&gt;&#xA;&lt;p&gt;From what I can tell of &lt;a href=&quot;http://www.michigan.gov/lara/0,4601,7-154-72600_72602_72731_72865_73185-141914--,00.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Michigan's work experience requirements&lt;/a&gt;, you ought to be able to substitute sufficient work experience.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Work Experience Requirements&lt;/strong&gt;&lt;br /&gt;&#xA;All applicants must provide verification of at least 4 years of acceptable engineering work experience obtained after having received an acceptable bachelors degree. Work experience must be verified by five persons, three of whom must be licensed professional engineers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The devil is obviously in the details, and it would behoove you to contact that board for exact details.  But at first glance, it would appear that you'll be fine.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;To be a little more prescriptive in what you specifically need to do:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Contact the PE's that you've worked with before.  Find as many that you can that are willing to vouch for the quality of your work.  It's important to keep in mind that they are vouching more for your character and their perception of you following a particular process as opposed to direct supervision.  Some states call these &amp;quot;character reference&amp;quot; recommendations.&lt;/p&gt;&#xA;&lt;p&gt;Michigan expects at least 3 PE's to vouch for your work, you would be better off getting more than that if you can.  There's no harm in submitting more than 5 verifications as well, especially if they all hold PE's.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Start documenting the work experience that you've already completed.  Some states count a Masters of Engineering as an equivalent to one year of work experience.  In that documentation, call out the processes that you followed and make sure to note the review portion of those processes.  The state board is interested in work that demonstrates the full engineering process showing the beginning of the work through till the end and including the review to potentially improve the process itself.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Finally, really start studying the law and regulations surrounding the particular field of engineering that you wish to advertise your services within.  And let me be straight-up brutally honest for a moment:  Sealing a document merely means that you're willing to accept liability for a project should something that was reasonably foreseen go wrong.  That means you can personally be civilly or criminally held liable should there be a significant problem on a project.  Make sure you're comfortable with reasonable standards of care for the projects you want to engage so you can demonstrate due diligence in your offerings.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user16" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-03-22T00:07:32.740" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8107" PostTypeId="1" CreationDate="2016-03-22T01:38:26.853" Score="1" ViewCount="396" Body="&lt;p&gt;A hobby question. I have a load of 100 grams. The load incorporates all the components necessary to lift itself. I want to lift the load vertically (i.e. against gravity) at a speed of, say, 1 metre per second. I intend to use a small, hobby, toy-car type micro-motor/s, probably with some type of built in reduction gearbox, so as to increase torque.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Operating Range: 6 - 12VDC&lt;/li&gt;&#xA;&lt;li&gt;Torque @ Stall: 7 oz-in.@ 6VDC&lt;/li&gt;&#xA;&lt;li&gt;Torque @ Stall: 20 oz-in.@ 12VDC. Gear train damage can occur if stalled (locked)&lt;/li&gt;&#xA;&lt;li&gt;0.118” (3mm) Diameter “D” Shaft&lt;/li&gt;&#xA;&lt;li&gt;No load current: 40mA (6VDC)&lt;/li&gt;&#xA;&lt;li&gt;No load current: 100mA (12VDC)&lt;/li&gt;&#xA;&lt;li&gt;Stall current: 360mA (6VDC)&lt;/li&gt;&#xA;&lt;li&gt;Stall current: 1600mA (12VDC)&lt;/li&gt;&#xA;&lt;li&gt;No load speed: 240 RPM @ 6VDC&lt;/li&gt;&#xA;&lt;li&gt;No load speed: 460 RPM @ 12VDC&lt;/li&gt;&#xA;&lt;li&gt;Gear ratio: 50:1&lt;/li&gt;&#xA;&lt;li&gt;Motor size: See Schematics&lt;/li&gt;&#xA;&lt;li&gt;Shaft size: 0.118&quot; (3mm) Dia. x 0.393” (10mm) L&lt;/li&gt;&#xA;&lt;li&gt;Weight: 0.3 oz (8.5g)&lt;/li&gt;&#xA;&lt;li&gt;DC reversible motors&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I intend to use Lithium Ion batteries (TangsFire 18350 3.7V 1500mAh Eco - friendly Li - HP Rechargeable Batteries).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find a workable or perhaps optimal combination of motor(s), gear reduction, and battery supply that can do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total weight of 100 grams would include the weight of the components - batteries, motor(s) etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motors with a built in gearbox already attached weigh about 8.5 grams each.&#xA;The Lithium Ion batteries weigh about 15 grams each.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Motors&lt;/strong&gt;&lt;br&gt;&#xA;Specifications a likely motors (they have the reduction gearing already built in). I will likely use three of these micro-motors working together to provide more pulling power (i.e. torque?).&lt;br&gt;&#xA;&lt;a href=&quot;https://www.servocity.com/html/460_rpm_micro_gearmotorblocks.html&quot; rel=&quot;nofollow&quot;&gt;460 RPM Micro GearmotorBlocks&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Batteries&lt;/strong&gt;&lt;br&gt;&#xA;I plan to use 2 or 3 of these batteries: &lt;a href=&quot;http://www.gearbest.com/batteries/pp_91848.html&quot; rel=&quot;nofollow&quot;&gt;TangsFire 18350 3.7V 1500mAh Eco - friendly Li - HP Rechargeable Battery -  YELLOW 10&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Doable at all?&lt;/strong&gt; Will it pull a 100 gram load vertically upwards?&lt;br&gt;&#xA;The device will be attached to a string and it will hang down from a fixed point above the device. A bit like a mini-elevator.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;If it is doable&lt;/strong&gt;, what would be the &lt;strong&gt;maximum speed&lt;/strong&gt; that it could travel at in an upwards direction?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5697" LastEditorUserId="1586" LastEditDate="2016-03-23T21:47:51.920" LastActivityDate="2016-03-23T21:47:51.920" Title="Lifting a 100 gram load vertically - Best combination of motor/s, battery/ies and gearing?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8108" PostTypeId="2" ParentId="8101" CreationDate="2016-03-22T01:52:49.700" Score="2" Body="&lt;p&gt;First I'm going to use Wikipedia to disprove the statement:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Plane    |   T/W   |  Mach (max) &#xA;------------------------------ &#xA;Concorde |   0.373 |  2.04  &#xA;Typhoon  |   1.15  |  Mach 2 Class  &#xA;F-15C    |   1.07  |  2.5  &#xA;Harrier  |   1+    |  0.95&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There will be some irregularities because most of the planes in the list are military and exact numbers may not be known. Even taking that into account, there is no correlation between thrust-to-weight ratio and max speed. This is because at max speed, aerodynamic resistance (drag) is the biggest force. This is based on aircraft design, not engine size or plane weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what your quote was probably saying was something along the lines of &quot;... &lt;strong&gt;neglecting drag&lt;/strong&gt;, thrust to weight ratio is closely related to maximum speed&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is backed up by this &lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/fwrat.html&quot; rel=&quot;nofollow&quot;&gt;link to a NASA&lt;/a&gt; article on thrust to weight ratio where they make the exact same simplification of neglecting drag and talking only about horizontal acceleration.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="1293" LastEditDate="2016-03-22T07:03:27.957" LastActivityDate="2016-03-22T07:03:27.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8110" PostTypeId="1" CreationDate="2016-03-22T06:43:54.303" Score="7" ViewCount="1562" Body="&lt;p&gt;I'm trying to design a metal surface with micro-perforation for a product and light it from other side with LED, such that the perforated pattern appears only when lit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my previous prototype, which used laser drilling (20 micron holes) for perforation, there was considerable warping of the metal film (around 0.3 mm thick) which rendered the prototype unusable (aesthetically)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem I had with the prototype was that the light was not getting through perforation evenly i.e., I was not able to see the light from different angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On initial research about laser micro-drilling I found that the thermal conductivity of material plays important role in heat affected zone, which could result in warping. The material used for prototype was stainless steel, which has comparably lower thermal conductivity than aluminium. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What care needs to be taken to avoid or reduce the HAZ and hence the warping?&lt;/li&gt;&#xA;&lt;li&gt;For the visibility of light through perforation, will the hole size and spacing directly affect this?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am trying to achieve the appearance like shown below &#xA;&lt;a href=&quot;https://i.stack.imgur.com/71LyW.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/71LyW.jpg&quot; alt=&quot;Misfit Shine fitness band; the micro-perforation on dial is invisible when unlit.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Misfit Shine fitness band; the micro-perforation on dial is invisible when unlit.&lt;/p&gt;&#xA;" OwnerUserId="5700" LastEditorUserId="5700" LastEditDate="2016-03-22T10:38:22.733" LastActivityDate="2017-07-30T21:11:24.040" Title="Metal micro perforation" Tags="|metals|lasers|drilling|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8111" PostTypeId="1" AcceptedAnswerId="8117" CreationDate="2016-03-22T11:45:11.853" Score="4" ViewCount="209" Body="&lt;p&gt;How can the microstructure of steel after welding be determined in modern computer software?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that there are 'classic' methods of predicting microstructures like Schaeffler, DeLong or WRC constitutional diagrams. What I am wondering about is:  which algorithms are used in programs based on finite element method (FEM) computing (i.e. ESI Group software, particularly SYSWELD)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to find out in particular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. How does a FEM model made of austenite differ from the one that is made of ferrite?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Is a FEM model enough to determine the microstructure of a material, that is represented by that model? If not, what else do you need to examine transformations of microstructure during the welding process?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4084" LastEditorUserId="33" LastEditDate="2016-03-22T13:48:20.753" LastActivityDate="2016-03-22T16:07:37.537" Title="Determining the microstructure of steel after welding in modern computer software" Tags="|metallurgy|finite-element-method|structural-analysis|software|welds|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8113" PostTypeId="1" CreationDate="2016-03-22T12:40:19.693" Score="1" ViewCount="3004" Body="&lt;p&gt;What is &quot;redundancy,&quot; exactly? As far as I know a system/component is redundant if critical components are being duplicated to ensure a higher safety level while reducing availability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets take a look at a sensor. If it is a redundant sensor it has 2 channels, right? I have seen that some manufacturers say that their components have triple redundancy with 3 channels. How is this possible? As far as I get this triple redundancy means that there should be 4 channels working independently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example &lt;a href=&quot;http://www.mtssensor.de/fileadmin/medien/downloads/datasheets/Data_Sheet_G-Series_GT2_GT3_Analog_Redundant_551379_RevC_EN.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="5007" LastEditorUserId="2246" LastEditDate="2016-03-22T12:44:26.400" LastActivityDate="2024-03-29T04:00:59.090" Title="What does &quot;redundancy&quot; mean?" Tags="|electrical-engineering|sensors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8114" PostTypeId="2" ParentId="8113" CreationDate="2016-03-22T12:49:03.840" Score="5" Body="&lt;p&gt;Wikipedia says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In engineering, redundancy is the duplication of critical components&#xA;  or functions of a system with the intention of increasing reliability&#xA;  of the system, usually in the form of a backup or fail-safe.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In many safety-critical systems, such as fly-by-wire and hydraulic&#xA;  systems in aircraft, some parts of the control system may be&#xA;  triplicated, which is formally termed triple modular redundancy (TMR).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In reliability engineering, dual modular redundancy (DMR) is when components of a system are duplicated, providing redundancy in case one should fail.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think your confusion lies within the formal terms. If you think of a redudant system as a DMR system the term TMR makes more sense and takes away your confusion about the sensor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you add redundancy to the system (there is no such thing &quot;single&quot; redundancy) you can choose a DMR system or a TMR system. Usually people would think of a DMR system unless specified otherwise. So to keep consistent with your terms, what you would define as single redundant is the same as a DMR system. And then, if you have 3 modules in total, you speak of a TMR system. &#xA;It all refers to the total number of modules, not the number of added modules.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2016-03-22T15:48:09.593" LastActivityDate="2016-03-22T15:48:09.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8117" PostTypeId="2" ParentId="8111" CreationDate="2016-03-22T14:11:13.263" Score="6" Body="&lt;p&gt;&lt;strong&gt;Summary:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1) The answer to this question is difficult. You would need to know how austenite and ferrite behave in relation to what you are doing to them. You would also need to know their compositions, temperature field, etc. The results here could vary significantly depending on the specific parameters and how they change with time and with each other.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) Yes and no. You can get a statistical model of each FEM element at a macro scale to determine the statistical nature of the microstructure such as grain size, particle density, etc. Or you can model what happens at a microscale when welding to determine what the microstructure might look like under a microscope. To do either requires a lot of detailed thermodynamic data about all of the components present and their phases, as discussed below.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;More Specifically...&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;My approach to the problem would be to use statistical modeling to determine grain size, particle size and distribution, etc. The data would come from phase diagram data based on known compositions, together with assumed kinetic governing equations. The actual thermal and compositional kinetics are governed at a bulk scale by well known PDE models, but certain features such as size and distribution are determined by microscale kinetics. What phases appear and the order in which they appear are governed by the phase diagrams. We can generally assume that bulk kinetics provides an input into the microscale kinetic and thermodynamic relationships, but that the reverse is largely irrelevant. The microstructural data in turn provides an input into structure-property relationships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A complete model would tie all of these together more-or-less in that order, and would look something like:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use the bulk scale kinetics to get the temperature and composition profiles for each element.&lt;/li&gt;&#xA;&lt;li&gt;Use the temperature and composition profiles to generate statistical microstructural data for each element.&lt;/li&gt;&#xA;&lt;li&gt;Use the statistical microstructural data to determine properties.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The algorithms for common software are generally proprietary, so I can't say for sure, but I believe any package such as SYSWELD or MAGMA uses something along these lines. SYSWELD may even use the Schaeffler, DeLong, and WRC diagrams as part of its thermodynamic modeling. It depends to what degree they are making assumptions about the thermodynamic data and how much effort they've put into that part of their model.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Microstructural FEM&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;FEM may be used to model microstructural behavior at a microscale, such as mechanical and thermal behavior. Generally this is done by capturing a microstructural image representation, either by microscope (optical or SEM) for 2D or by computed x-ray tomography (CT) for 3D, and converting the representation into an FEM by identifying or segmenting different phases and their interfaces, and assigning appropriate (often anisotropic) material properties to each phase and each phase-pair interface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do all this, you need to be able to accurately segment phases, interfaces, and crystallographic orientation, which may take some expensive equipment and characterization work. Alternately, a phase field model may be used to attempt prediction of the microstructural morphology, and then relevant data captured from the phase field model. There are limitations in using a phase field model this way, which are discussed in the next section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a tool on &lt;a href=&quot;https://nanohub.org&quot; rel=&quot;noreferrer&quot;&gt;nanoHUB.org&lt;/a&gt; which does microstructural FEM for 2D images called &lt;a href=&quot;https://nanohub.org/tools/oof2&quot; rel=&quot;noreferrer&quot;&gt;OOF2&lt;/a&gt;. To use the tool you would need to create an account, and it is generally intended for educational purposes, but should give a general idea how a microstructural FEA might work in the 2D case. You might need to upload your own microstructure image, it's been awhile since I've used it and I've forgotten the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results of microstructural FEM are useful for determining how textured microstructures might behave. They are also useful for determining how microstructure can be linked to fatigue properties by identifying stress concentrations in the microstructure and how the microstructure might play a role in crack and void initiation.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Phase Field Models&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;To model phase transformation kinetics at a microscale, generally phase field models (&lt;a href=&quot;https://en.wikipedia.org/wiki/Phase_field_models&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;) are used. The models involve a number of moving parts, so to speak, but are sometimes faster and usually more robust for capturing moving interfaces than with traditional FEM models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary concept is that, for a field of elements containing two phases, the phase discrimintation of the entire field may be modeled with a scalar value varying from 0 to 1. If the value of an element is (very close to) 0 it is one phase, and (very close to) 1 the other phase. If it has an intermediate value it is part of an interface between the phases. Thus rather than a sharp interface as would be required with an FEM, the interface is modeled by assuming it is diffuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phase field models typically also track composition, temperature and free energy, and have a collection of governing equations which must be solved at each time step to determine the evolution of the next step. To use a phase field model thus requires knowledge of temperature dependent free energy curves and temperature dependent diffusion rates, both thermal and compositional, among other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to model diverse microstructural evolution phenomena with phase field models, including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;planar to dendritic solidification (&lt;a href=&quot;https://nanohub.org/tools/vkmlggs&quot; rel=&quot;noreferrer&quot;&gt;nanoHub.org tool&lt;/a&gt; and &lt;a href=&quot;http://solidification.mechanical.illinois.edu/Movies/Plapp/dendlowanis.webm&quot; rel=&quot;noreferrer&quot;&gt;solidification.org video&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;unstable cellular eutectic solidification (&lt;a href=&quot;http://solidification.mechanical.illinois.edu/Movies/Plapp/Eutectic_Colonies.webm&quot; rel=&quot;noreferrer&quot;&gt;solidification.org video&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;spinodal decomposition (&lt;a href=&quot;https://nanohub.org/tools/vkmlsd&quot; rel=&quot;noreferrer&quot;&gt;nanoHub.org tool&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;grain growth (&lt;a href=&quot;https://nanohub.org/tools/vkmlpsgg&quot; rel=&quot;noreferrer&quot;&gt;nanoHUB.org tool&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and certainly much more is possible, if challenging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unaware of any professional phase field modeling packages, as there are drawbacks limiting their usefulness outside academia and early-stage research. One limitation is that, depending on the model, the specific variables may not have a clear relationship with physical values that can be experimentally determined. Thus, sometimes the model parameters need to be adjusted systematically until the results &quot;look right.&quot; Additionally, getting useful information out of the model is another issue due to the same discrepancies between model parameter values and physical values. It is also possible to produce non-physical results quite easily without careful tailoring of governing equations to the specific model. Validation is another issue as performing the experiments is at best an expensive and time consuming process, and at worst virtually impossible depending on the specific parameters involved. Research is of course focused on reducing these issues, but because phase field models are relatively new (~10 years old at the earliest) there is much work yet to be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally phase field models are mostly useful for drawing a pretty picture of what a microstructure might look like without performing expensive experimentation and microscopic examination. They are also useful for creating animations of microstructural evolution. In the future their use may expand to predicting features such as statistical modeling and capturing data for FEM, but the limitations above restrict those uses.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Statistical Thermodynamic Models&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Generally, at a useful level, most engineers deal with bulk properties. After all, most engineers are designing products at a visible, macroscopic scale. As a result, the specifics of a tiny fraction of the microstructure isn't particularly useful directly to how the product might behave at a macroscopic level. Instead, we want to determine how the product behaves as a bulk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To model the bulk microstructural evolution and final properties of a material is usually done with a CALPHAD (Calculation of Phase Diagrams, &lt;a href=&quot;https://en.wikipedia.org/wiki/CALPHAD&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;) program. Microstructural CALPHAD models don't generate pretty pictures like in the previous two sections, but instead generate a statistical representation of certain classes of microstructure by generating a grain size or particle size and density distribution based on thermodynamic, kinetic, compositional, and temperature data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a model can be used in conjunction with a process FEM to determine local microstructural distributions for each finite element. &lt;a href=&quot;http://www.thermocalc.com/&quot; rel=&quot;noreferrer&quot;&gt;Thermocalc&lt;/a&gt; does thermodynamical statistical modeling and is a CALPHAD program. &lt;a href=&quot;http://www.magmasoft.com/en/&quot; rel=&quot;noreferrer&quot;&gt;MAGMA&lt;/a&gt; casting process simulation software combines statistical thermodynamic modeling with FEM in some of their alloy packages. A MAGMA user might then be able to predict some statistical data throughout the bulk of their product, and then generate scalar fields representing mechanical properties which vary over the product. It appears SYSWELD does the same thing for heat treatment and the welding process, probably by the general method described here.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;References&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://nanohub.org/&quot; rel=&quot;noreferrer&quot;&gt;nanoHub.org&lt;/a&gt; - A site with many computational tools and educational resources focused on nano-scale. Some information and tools are related to larger scale modeling, especially OOF2 (Object Oriented Finite Element Analysis, 2D) and VKML (Virtual Kinetics of Materials Laboratory) tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.solidification.org/&quot; rel=&quot;noreferrer&quot;&gt;solidification.org&lt;/a&gt; - A site with a number of neat movies of solidification processes, both experimental and phase field simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not endorse any linked sites or softwares, the links are only intended for referential and educational purposes.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-03-22T16:07:37.537" LastActivityDate="2016-03-22T16:07:37.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8118" PostTypeId="1" AcceptedAnswerId="8120" CreationDate="2016-03-22T15:36:35.213" Score="1" ViewCount="7852" Body="&lt;p&gt;I'm constructing 2.2 meters span, 1 m wide shelf. Want to make it cheap but thin.&#xA;I'm planning to use 18mm thick OSB board and reinforce it with a layer of stronger material stuck to it's bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first idea for reinforcement is 2 strips of 60*4 mm steel.&#xA;I'm not very familiar with the subject. I found OSB modulus of elasticity &lt;a href=&quot;http://article.sapub.org/10.5923.j.cmaterials.20140402.03.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; , transformed width of the OSB board to it's steel equivalent, that gave me 38mm. Then I ran it through L-beam moment of inertia online calculator, and beam calculator assured me that the sag would be much less than my acceptable 1cm. (The load is under 100kg)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second idea is to use glass fabric instead of steel. Glass itself is just 3 times less strong than steel, so that makes this option at least worth evaluating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence comes &lt;strong&gt;the question&lt;/strong&gt;:&#xA;Where do I find properties of the layer I'll get when I put fiber glass and pour epoxy on it? The only thing that's stated for the &quot;structural fiberglass fabric&quot; available on the market is it's ultimate tensile strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm talking about the glass fabric, not the plastic sheet reinforced with it:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1BCei.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1BCei.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5053" LastEditorUserId="33" LastEditDate="2016-03-22T16:18:35.610" LastActivityDate="2020-01-03T11:16:39.733" Title="How do I calculate the structural properties of a fiberglass laminated wood panel?" Tags="|materials|structural-analysis|composite|" AnswerCount="2" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8119" PostTypeId="1" AcceptedAnswerId="8122" CreationDate="2016-03-22T16:09:16.050" Score="1" ViewCount="438" Body="&lt;p&gt;I have modeled a beam as fixed-fixed. The reaction forces are -1494 lb.ft and 761 lb (vertical). There is no horizontal reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/drvLH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/drvLH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I know this beam is welded to another element. I am trying to design the weld for the necessary forces. How do I transfer the reaction forces at the fixed end to the weld? Should I invert the directions of the reactions or should I preserve the same directions? &lt;/p&gt;&#xA;" OwnerUserId="4808" LastActivityDate="2016-03-22T17:37:32.253" Title="How to transfer reaction forces of beam to a weld?" Tags="|structural-engineering|structures|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8120" PostTypeId="2" ParentId="8118" CreationDate="2016-03-22T16:47:40.990" Score="3" Body="&lt;p&gt;To determine the properties of a composite laminate, the properties of each component and their relative volume fractions must be known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The properties of glass fiber laminate may be determined using the rule of mixtures. Consider a cross-section of the fiberglass laminate as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lIdUO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lIdUO.jpg&quot; alt=&quot;Cross-section of fiberglass laminate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White regions are the fiberglass, black regions are the epoxy matrix. The white dots are transversely-sectioned fibers, and the white highly-eccentric ellipses are longitudinally-sectioned fibers. If we consider only the bottom portion of the image, so that all the fibers are aligned into and out of the image, then it appears that about 50% of the volume is occupied by fibers and 50% by matrix, which is a reasonable mix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the rule of mixtures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;E_c = E_f V_f + E_m V_m&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the subscript $c$ denotes composite bulk, $f$ denotes fiber, and $m$ denotes matrix, and $E$ is tensile modulus and $V$ is volume fraction. Then because the tensile modulus of the fibers are $E_f=70\ \textrm{GPa}$ and of the matrix is $E_m=2\ \textrm{GPa}$ (give or take a little), the composite modulus $E_c$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;E_c = (70\ \textrm{GPa})(0.5)+(2\ \textrm{GPa})(0.5) \\&#xA;E_c = 35\ \textrm{GPa} + 1\ \textrm{GPa} \\&#xA;E_c = 36\ \textrm{GPa}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For strength, the situation is a bit more challenging, and depends on elastic compatibility. Since we know the stiffness of the composite, $\sigma_{f,UTS}=2\ \textrm{GPa}$ for E-glass fibers gives the strain at failure by the relationship $\sigma_{f,UTS}/E_f=\varepsilon_f=0.029$. Similarly for the matrix assuming $\sigma_{m,UTS}=0.085\ \textrm{GPa}$ gives $\varepsilon_m=0.043$. Since the strain at failure is lower for the fibers, they fail first, by compatibility. The strength of the composite is then the product of strain at failure and composite modulus, or about $1\ \textrm{GPa}$. This result also assumes the interface doesn't fail before either of the components, hence the need for cleanliness and full wetting of the fibers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deflection may be calculated by assuming the shelf is a beam, and using the worst-case, i.e. 100kg centered load and simply supported at the ends, and not supported at the wall. The equation for the deflection is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\delta_{center} = \frac{FL^3}{48\sum EI} \\&#xA;= \frac{\left(980\ \textrm{N}\right)\left(8\ \textrm{m}^3\right)}{48\left[\left(36\times 10^9\ \textrm{N}\cdot\textrm{m}^{-2}\right)\left(8.4\times 10^{-9}\ \textrm{m}^4\right)+\left(2.5\times 10^9\ \textrm{N}\cdot\textrm{m}^{-2}\right)\left(5.8\times 10^{-6}\ \textrm{m}^4\right)\right]} \\&#xA;= \frac{7840\ \textrm{N}\cdot\textrm{m}^3}{7.1\times 10^{5}\ \textrm{N}\cdot\textrm{m}^{2}} \\&#xA;= 0.011\ \textrm{m} \\&#xA;\approx 1\ \textrm{cm}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is your desired value, but with no safety factor. I've assumed for simplicity that the neutral axis is at the midplane, which is inaccurate. The actual location is the elasticity-weighted centroid of the laminate area. The neutral axis should be shifted away from the laminate to compensate for its increased stiffness. The effect on the area moment of inertia of the laminate should be to increase it and with its greater stiffness should result in a smaller deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image comes from &lt;a href=&quot;http://aerospaceengineeringblog.com/composite-materials/&quot; rel=&quot;nofollow noreferrer&quot;&gt;aerospaceengineeringblog.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-03-22T17:18:14.270" LastActivityDate="2016-03-22T17:18:14.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8121" PostTypeId="1" CreationDate="2016-03-22T17:31:59.580" Score="3" ViewCount="97" Body="&lt;p&gt;Our organization has recognized the need to improve quality of both incoming material and outgoing products. Our products are primarily electrical/electronic but mechanical components are an integral part of the end product.  As result of sharing insight gained from &lt;a href=&quot;https://engineering.stackexchange.com/questions/3315/how-to-identify-low-quality-electronic-components-early&quot;&gt;this question&lt;/a&gt; and other public articles in process improvement, I was able to get management's attention concerning the possible need for additional training for team members in quality engineering.  So management asked me to provide a high level proposal for a quality engineering training program with a focus on six-sigma principles. This leads me to ask two questions which I believe are related:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are the most commonly used statistical tools in quality&#xA;engineering? &lt;/li&gt;&#xA;&lt;li&gt;What should be the focus when looking for a six sigma&#xA;consultant to train a manufacturing engineering organization?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-05-26T04:08:34.243" Title="Guidelines to develop quality engineering training content and select a trainer" Tags="|statistics|quality-engineering|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8122" PostTypeId="2" ParentId="8119" CreationDate="2016-03-22T17:37:32.253" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Should I invert the directions of the reactions or should I preserve the same directions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As long as you keep your directions consistent while solving for the weld forces, it won't matter whether you have them positive or negative.  Referring to my answer to your &lt;a href=&quot;https://engineering.stackexchange.com/a/8104/2246&quot;&gt;first question&lt;/a&gt; on weld strength, the $f_{xw}, f_{yw}, f_{zw}$ terms are all squared when solving for $f_w$, so directionality doesn't matter so long as you keep it consistent.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-03-22T17:37:32.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8123" PostTypeId="1" CreationDate="2016-03-22T18:52:53.873" Score="3" ViewCount="1003" Body="&lt;p&gt;It is my understanding that a mil is a one thousandth of an inch (1mil = 0.001”). Also, a mil is 25.4 microns, aka micrometers. Then, it is my understanding that Gauge is 100x mil (1mil-100Ga). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for some copper sheets for a project, and I am confused. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00LQ2ND32&quot; rel=&quot;nofollow noreferrer&quot;&gt;This Amazon&lt;/a&gt; listing: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/YlIdk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YlIdk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lists 0.0216&quot; (inch) as a 24Ga (Gauge). How does that work? Am I reading the units wrong? &#xA;0.0216&quot; = 21.6Mil = 2160Ga ? Is that right? &lt;/p&gt;&#xA;" OwnerUserId="1157" LastEditorUserId="2246" LastEditDate="2016-03-22T19:15:06.483" LastActivityDate="2016-03-22T23:01:23.693" Title="Are there different Gauge standards out there?" Tags="|materials|measurements|metrology|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8124" PostTypeId="2" ParentId="8121" CreationDate="2016-03-22T19:02:05.623" Score="2" Body="&lt;p&gt;Get an &lt;a href=&quot;http://asq.org/cert&quot; rel=&quot;nofollow&quot;&gt;ASQ&lt;/a&gt; certificate: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ASQ certification is a formal recognition by ASQ that an individual&#xA;  has demonstrated a proficiency within, and comprehension of, a&#xA;  specific body of knowledge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They have a bunch of &lt;a href=&quot;http://asq.org/cert/certifications&quot; rel=&quot;nofollow&quot;&gt;different&lt;/a&gt; certificates that are applicable to the different engineers/techs you guys have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Certified Quality Engineer (CQE) &lt;/li&gt;&#xA;&lt;li&gt;Certified Quality Inspector (CQI)&lt;/li&gt;&#xA;&lt;li&gt;Certified Quality Technician (CQT) &lt;/li&gt;&#xA;&lt;li&gt;Certified Reliability Engineer (CRE) &lt;/li&gt;&#xA;&lt;li&gt;Certified Quality Improvement Associate (CQIA) &lt;/li&gt;&#xA;&lt;li&gt;Certified Quality Process Analyst (CQPA) &lt;/li&gt;&#xA;&lt;li&gt;Certified Software Quality Engineer (CSQE)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Certified Six Sigma Green Belt (CSSGB)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Six Sigma Green Belt operates in support of or under the supervision of a Six Sigma Black Belt, analyzes and solves quality problems and is involved in quality improvement projects. A Green Belt is someone with at least three years of work experience who wants to demonstrate his or her knowledge of Six Sigma tools and processes.&lt;/p&gt;&#xA;" OwnerUserId="1157" LastActivityDate="2016-03-22T19:02:05.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8125" PostTypeId="2" ParentId="8123" CreationDate="2016-03-22T19:07:34.453" Score="2" Body="&lt;p&gt;Most gauge systems  like SWG aren't linear and are based on a volumetric or cross sectional area system of some sort. In the case of wire gauges it is usually derived from the number of drawing operations required to produce a particular size of wire, so the gauge number is not proportional to the diameter but rather inversely proportional to the cross sectional area and thus mass per unit length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically the gauge number, n comes from 1/n times some reference quantity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly the gauge of shot is the number of spheres which can be made from a pound of lead so it is a measure of the inverse of volume even though it is used to indicate a diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you want to relate gauge to diameter or thickness the easiest solution is to look it up in a table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that a large gauge number corresponds to a thinner wire. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-22T23:01:23.693" LastActivityDate="2016-03-22T23:01:23.693" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8126" PostTypeId="2" ParentId="8123" CreationDate="2016-03-22T19:23:16.510" Score="5" Body="&lt;p&gt;Gage (gauge) thickness are somewhat arbitrary.  Take for example the table below, excerpted from the AISC &lt;em&gt;Steel Construction Manual&lt;/em&gt;, 14th Ed.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LCQVs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LCQVs.png&quot; alt=&quot;sheet metal gages&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that gage thicknesses vary depending on material and coating but are approximately the same for a given gage number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.alascop.com/pdf/sheet_gauge_charts.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This PDF&lt;/a&gt; references the following for different material gages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aluminum (no official gauge standard) (&lt;a href=&quot;https://en.wikipedia.org/wiki/American_wire_gauge&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brown &amp;amp; Sharpe's Gauge Standard B&amp;amp;S or American Gauge A.W.G.&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Copper (&lt;a href=&quot;http://www.engineeringtoolbox.com/BWG-wire-gage-d_508.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Birmingham Gauge Standard B.W.G.&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Brass (Brown &amp;amp; Sharpe or American Gauge)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that the B&amp;amp;S and A.W.G. link shows a rather complicated formula for determining wire gauge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ d_n = 0.005\mbox{inch} * 92^\frac{36-n}{39} = 0.127\mbox{mm} * 92^\frac{36-n}{39} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the gauge number for gauges number 32 to 0.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2016-03-22T19:23:16.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8127" PostTypeId="1" AcceptedAnswerId="8128" CreationDate="2016-03-22T20:38:47.533" Score="2" ViewCount="249" Body="&lt;p&gt;I have a vehicle with 4 wheels. Say I replace the wheels with large helical gears (pinions). If I set the vehicle on linear helical racks, it should be able to travel in one dimension. Say I replace the racks with a similarly toothed ground surface, could the vehicle travel in 2 dimensions (without rotating the body of the vehical)? I think it would be possible with opposing front and rear wheel drives. It should operate like a &quot;screw&quot; in the second dimension. Would the ideal helix angle be 45 degrees (to support movement in both directions)? How will speed, Torque, and service life be affected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/17qHL.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/17qHL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3765" LastEditorUserId="33" LastEditDate="2016-03-22T21:43:31.757" LastActivityDate="2016-03-31T17:37:12.777" Title="Multi-Directional Rack and Pinion Drive - Vehicle Motion Using Helical Wheels" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8128" PostTypeId="2" ParentId="8127" CreationDate="2016-03-22T20:58:14.310" Score="1" Body="&lt;p&gt;No, it cannot work as you are imagining. Such a vehicle can roll to the left or right under its own power, and it could be pushed diagonally along the grooves by an external force, but it can't &quot;screw itself&quot; up and down. All four wheels must always turn in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should study &lt;a href=&quot;https://en.wikipedia.org/wiki/Omni_wheel&quot; rel=&quot;nofollow&quot;&gt;omnidirectional&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Holonomic_%28robotics%29&quot; rel=&quot;nofollow&quot;&gt;holonomic&lt;/a&gt; drive systems.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Actually, after giving it some thought, you could have &lt;em&gt;two&lt;/em&gt; sets of grooves in the surface at right angles to each other. If one pair of wheels meshed with one set of grooves, and the other pair mated with the other grooves, then you could get the kind of motion you're thinking of. But friction would be a huge issue for up-and-down motion.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2016-03-23T03:30:16.837" LastActivityDate="2016-03-23T03:30:16.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8131" PostTypeId="2" ParentId="8121" CreationDate="2016-03-22T21:20:39.407" Score="2" Body="&lt;p&gt;In the first order the most commonly used quality engineering tools are average, standard deviation, histograms, pareto charts, control charts.&#xA;In the second order tool such as Design for Experiments (DOE), ANOVA, student t-test, z-test, confidence testing, gauge R&amp;amp;R, Ishikawa diagram most likely are next set of tools that are most used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From software standpoint Microsoft excel can perform almost all statistically analysis, but &lt;a href=&quot;https://www.minitab.com/en-us/&quot; rel=&quot;nofollow&quot;&gt;Minitab&lt;/a&gt; is a better too. JMP from SAS is another good tool. A training program utilizing industry leading statistical software package should also be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a six sigma standpoint there are different levels, namely yellow belt, greenbelt, black belt, master black belt and so forth.  Primarily the two certification awarding bodies are &lt;a href=&quot;http://www.iassc.org/&quot; rel=&quot;nofollow&quot;&gt;IASSC&lt;/a&gt; and &lt;a href=&quot;http://asq.org/index.aspx&quot; rel=&quot;nofollow&quot;&gt;ASQ&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to selecting a consultant, I would reach out to referrals from peers in your industry. A search using your favorite search engine is a good start but I will request potential consultants to submit referrals. Also make sure you properly vet these consultants. A consultant with strong engineering background, might be something else to consider. &lt;/p&gt;&#xA;" OwnerUserId="282" LastActivityDate="2016-03-22T21:20:39.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8132" PostTypeId="1" AcceptedAnswerId="8140" CreationDate="2016-03-22T21:56:13.437" Score="3" ViewCount="882" Body="&lt;p&gt;I am trying to solve a fairly simple question but I'm kind of stuck on the technicalities:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Imagine that water is evaporating into initially dry air in the closed&#xA;  vessel shown schematically in Fig. 8.1-1(a). The vessel is isothermal&#xA;  at 25 °C, so the water’s vapor pressure is 3.2 kPa. This vessel has&#xA;  0.8 l of water with 150 cm&lt;sup&gt;2&lt;/sup&gt; of surface area in a total volume of 19.2 l. After 3 min, the air is five percent saturated. What is the mass&#xA;  transfer coefficient? How long will it take to reach ninety percent&#xA;  saturation?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GqX5G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GqX5G.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The answer starts with this: $$N_1=\frac{\text{Vapor concentration}\cdot\text{Air Volume}}{\text{Liquid Area}\cdot\text{Time}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far this makes sense. The solution then goes on to do this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N_1=\frac{0.05\cdot(\frac{3.2}{101})\cdot(\frac{1\ \mathrm{mol}}{22.4\ \mathrm{liters}})\cdot(\frac{273}{298})(18.4\ \mathrm{liters})}{(150\ \mathrm{cm^2})(180\ \mathrm{sec})}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the denominator but the numerator I'm not sure of. Overall there are 18.4/22.4 mol of gas and 5% of that will be water vapor, I get that. But what's with the temperature and pressure adjustment?&lt;/p&gt;&#xA;" OwnerUserId="4530" LastEditorUserId="368" LastEditDate="2016-03-22T23:10:00.807" LastActivityDate="2016-03-23T09:59:05.400" Title="How do temperature and pressure influence the rate of evaporation of water in a closed vessel?" Tags="|thermodynamics|pressure|chemical-engineering|process-engineering|vapor-pressure|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8133" PostTypeId="1" AcceptedAnswerId="8784" CreationDate="2016-03-22T22:08:47.683" Score="5" ViewCount="493" Body="&lt;p&gt;With linear bearings, could they be used to join one rod with another to form a cross-junction and rotate with around a 0.3N-mm torque? Suppose the rods are fairly thin, eg, 5mm in diameter. That is, could linear bearings bear a radial load? Would the bearings deform? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt; Here's a diagram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4F5I1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4F5I1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the blue rod is attached to a gear driven by a motor. The blue rod itself is also attached to the bearing's housing. The blue essentially rotates and drives the black, exerting force/stress on the bearing. The idea is to mount a slidable camera on the black which can be rotated by the green gear.&lt;/p&gt;&#xA;" OwnerUserId="4301" LastEditorUserId="4301" LastEditDate="2016-03-23T01:50:15.757" LastActivityDate="2016-11-04T02:48:59.067" Title="Could linear bearings bear axial load?" Tags="|mechanical-engineering|bearings|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8134" PostTypeId="1" AcceptedAnswerId="8135" CreationDate="2016-03-22T23:05:55.630" Score="10" ViewCount="2438" Body="&lt;p&gt;In the problem statement below, the question states to find the maximum load $\vec{P}$ that the truss can support. My method of approach was:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Draw a FBD for the entire structure.&lt;/li&gt;&#xA;&lt;li&gt;Generally I would identify all external forces, however for this problem I felt it was unnecessary because I felt they weren't needed if I knew already the forces in each member.&lt;/li&gt;&#xA;&lt;li&gt;So I then dove straight into joint $D$, and assumed each member connected to $D$ was in tension and experienced the maximum tensile force $\vec{T} = 1500$ lb.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, this led to the wrong answer when solving for $\vec{P}$. The solution manual instead found each external force in terms of $\vec{P}$ and started at joint $A$ to find $\overrightarrow{AD}$ and $\overrightarrow{AB}$ in terms of $\vec{P}$. In addition, in order to get numerical values, the solution manual assumed member $\overrightarrow{AB}$ was experiencing the maximum compression force of 660 lb. However, when I assume that member $\overrightarrow{AD}$ is experiencing the maximum tensile force, it doesn't work out the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is, conceptually, why must I find each each member's force in terms of $\vec{P}$, and why must one assume $\overrightarrow{AB}$ is experiencing maximum compression (but not $\overrightarrow{AD}$ in maximum tension)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I want to note that I do not need help with solving this problem, nor the math required. I simply am just looking for a conceptual answer as to why my approach did not work (i.e. only analyzing joint $D$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7jTDN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7jTDN.png&quot; alt=&quot;Problem Statement&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5715" LastEditorUserId="5715" LastEditDate="2016-03-22T23:26:10.640" LastActivityDate="2019-12-28T17:56:14.823" Title="Where did I go wrong conceptually when attempting to calculate the maximum force on a truss at a given joint?" Tags="|structures|statics|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8135" PostTypeId="2" ParentId="8134" CreationDate="2016-03-22T23:49:00.207" Score="9" Body="&lt;p&gt;The reason is that &lt;strong&gt;you assumed that the elements around node &lt;span class=&quot;math-container&quot;&gt;$\text{D}$&lt;/span&gt; will be the first to fail&lt;/strong&gt;. That is not the case. Indeed, it is the elements under compression (&lt;span class=&quot;math-container&quot;&gt;$\text{AB}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\text{BC}$&lt;/span&gt;) that will fail first. Also, &lt;strong&gt;you assumed that all the members around &lt;span class=&quot;math-container&quot;&gt;$\text{D}$&lt;/span&gt; will present the same axial force&lt;/strong&gt;, which is untrue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, here's your structure with a unitary load (the units are irrelevant):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oa3Bg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oa3Bg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here are the axial forces in each member under this unitary force (positive in tension):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NaJ1I.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NaJ1I.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these results are linearly proportional to the force applied, we can now find the concentrated load &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; at node &lt;span class=&quot;math-container&quot;&gt;$\text{D}$&lt;/span&gt; that each member can withstand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;P_{\text{AB}} = P_{\text{BC}} &amp;amp;= \dfrac{660}{0.943} = 700 \\&#xA;P_{\text{AD}} = P_{\text{CD}} &amp;amp;= \dfrac{1500}{0.687} = 2183 \\&#xA;P_{\text{BD}} &amp;amp;= \dfrac{1500}{1.333} = 1125 \\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the maximum load supported by the structure is the minimum of these values, which is 700 lb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show the validity of these results, let's now apply each of these forces and check the results in the relevant members:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/udRpv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/udRpv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1PvqI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1PvqI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xFQh5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xFQh5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;All results obtained with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis program.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T17:56:14.823" LastActivityDate="2019-12-28T17:56:14.823" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="8136" PostTypeId="5" CreationDate="2016-03-23T00:57:02.427" Score="0" Body="&lt;p&gt;Welding is the process of combining two pieces (typically metals) together. To successfully create welds, an experienced welder (person) with the proper tools can be required. Welding is a specialized profession much like carpentry or plumbing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not all welds require the same level of care and expertise.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2016-03-23T00:57:02.427" LastActivityDate="2016-03-23T00:57:02.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8137" PostTypeId="4" CreationDate="2016-03-23T00:57:02.427" Score="0" Body="Use this tag when the question deals with the design of welds or their uses. For questions about the physical process of making welds, use the [welding] tag." OwnerUserId="33" LastEditorUserId="33" LastEditDate="2016-03-23T00:57:02.427" LastActivityDate="2016-03-23T00:57:02.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8138" PostTypeId="1" CreationDate="2016-03-23T06:14:16.247" Score="5" ViewCount="124" Body="&lt;p&gt;I'm trying to figure out what angle to turn the front tires of my small robotic car (about a foot in length) so that it moves in a circle (on level ground) of radius $r$ (will be accommodating circles of different radii). In addition to being able to set my car's speed, it is equipped with a sensor which includes a gyroscope, accelerometer, and a magnetometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just looking for some kind of equation that relates the car's tire angle, the car's speed, the circle's radius, and maybe the angular velocity readings from the gyroscope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has a car-like configuration.&lt;/p&gt;&#xA;" OwnerUserId="5720" LastEditorUserId="1632" LastEditDate="2016-03-23T16:24:32.910" LastActivityDate="2016-03-23T16:24:32.910" Title="Circular motion and Car's Tire Angle" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8139" PostTypeId="1" AcceptedAnswerId="8330" CreationDate="2016-03-23T07:54:50.540" Score="5" ViewCount="133" Body="&lt;p&gt;I got a position error measured by a linear encoder like in the picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wcExZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wcExZ.jpg&quot; alt=&quot;Position error graph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain some possible reasons for this kind of shape to occur in control theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My system looks like table with four legs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four legs move up and down by using motors and have a linear encoder as control reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table should stay in its position after moving up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four servo motors actuating the four legs control the position by using linear encoder signals, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after moving the table up around 10 mm ~ 20 mm, a strange symptom has happened. One or two encoder signals look like square waves (sometimes one encoder signal, sometimes two, sometimes none).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple setup diagram is as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jVGFw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jVGFw.png&quot; alt=&quot;Diagram of table&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="5899" LastEditDate="2016-04-07T09:21:08.227" LastActivityDate="2016-04-07T10:11:00.463" Title="Square shaped position error" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8140" PostTypeId="2" ParentId="8132" CreationDate="2016-03-23T09:59:05.400" Score="5" Body="&lt;p&gt;$\frac{1}{22.4}\frac{\mathrm{mol}}{\mathrm{L}}$ is only valid for standard conditions. Use the ideal gas law to calculate the concentration for your conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s \equiv$ standard conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 \equiv$ your conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_s V_s = n_s R T_s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{p_s V_s}{n_s T_s} = R = const = \frac{p_2 V_2}{n_2 T_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{n_2}{V_2} = \frac{p_2 T_s n_s}{p_s T_2 V_s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$c = \frac{n_2}{V_2} = \frac{3.2 \mathrm{kPa}}{101 \mathrm{kPa}} \frac{273.15 \mathrm{K}}{298.15 \mathrm{K}} \frac{1 \mathrm{mol}}{22.4 \mathrm{L}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Insert in your equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_1 = \dfrac{0.05 \cdot c \cdot V_{Air}}{A \cdot t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_1 = \dfrac{0.05 \cdot \frac{3.2 \mathrm{kPa}}{101 \mathrm{kPa}} \frac{273.15 \mathrm{K}}{298.15 \mathrm{K}} \frac{1 \mathrm{mol}}{22.4 \mathrm{L}} \cdot 18.4 \mathrm{L}}{150 \mathrm{cm}^2 \cdot 180 \mathrm{s}}$&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2016-03-23T09:59:05.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8141" PostTypeId="2" ParentId="8138" CreationDate="2016-03-23T13:17:21.530" Score="3" Body="&lt;p&gt;As a beginning, it is highly encouraged to start by &lt;a href=&quot;https://sweetmfg.biz/uploads/files/tech-04understandingsteering-4.pdf&quot; rel=&quot;nofollow&quot;&gt;Ackerman's angle theory&lt;/a&gt; which models street car turning theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should define more your robot configuration, and decide which variables are dependent on each others and which not.&lt;/p&gt;&#xA;" OwnerUserId="5553" LastActivityDate="2016-03-23T13:17:21.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8142" PostTypeId="2" ParentId="2017" CreationDate="2016-03-23T16:52:07.227" Score="3" Body="&lt;p&gt;Q: ...how one can determine if a section of curve on a highway is classified as a &quot;spiral&quot; curve, and are there other advantages besides &quot;making it look good&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: The purpose of a spiral or transition curve has nothing to do with aesthetics.  They were originally introduced on railways for safety reasons in conjunction with superelevation as explained here: &lt;a href=&quot;http://www.abcls.ca/wp-content/uploads/pdfs/Notes_on_Transition_and_Terminal_Curves.pdf&quot; rel=&quot;nofollow&quot;&gt;Notes on Transitional and Terminal Curves&lt;/a&gt;.  On high speed highways, they permit a driver to enter/exit a circular curve with gradual increase/decrease in the deflection of the steering wheel.  The mathematical form of the spiral may vary from jurisdiction to jurisdiction, and may have changed in recent decades as hand-held calculators and other computing devices replaced the hard-copy field tables once used by surveyors.  One common form is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler_spiral&quot; rel=&quot;nofollow&quot;&gt;Euler Spiral or clothoid&lt;/a&gt;.  I have read that in India the usual transition curve is a third-order hyperbola, and I have been told that in Germany autobahns are designed as a continuous series of connected clothoids with no tangential sections or circular curves. &lt;/p&gt;&#xA;" OwnerUserId="5730" LastActivityDate="2016-03-23T16:52:07.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8143" PostTypeId="2" ParentId="2147" CreationDate="2016-03-23T18:16:44.033" Score="2" Body="&lt;p&gt;Compasses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The business part of a compass is out of sight.  This is a set of tiny magnets glued under the compass card and surrounding the jeweled pivot that supports the center of the card.&quot;  -The Annapolis Book of Seamanship, by John Rousmaniere&lt;/p&gt;&#xA;" OwnerUserId="5733" LastActivityDate="2016-03-23T18:16:44.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8144" PostTypeId="1" AcceptedAnswerId="8154" CreationDate="2016-03-23T18:48:08.183" Score="6" ViewCount="3343" Body="&lt;p&gt;Can a vertical round tube safely support a cantilevered arm mounted near the top of the tube?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ROUND VERTICAL TUBE:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;6061 Aluminum&lt;/li&gt;&#xA;&lt;li&gt;2 inch Outside diameter&lt;/li&gt;&#xA;&lt;li&gt;0.25 inch wall thickness &lt;/li&gt;&#xA;&lt;li&gt;3 feet height&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;ROUND VERTICAL TUBE / CANTILEVERED ARM INTERFACE:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://dectronusa.com/store/deluxe-2-pole-mount/&quot; rel=&quot;noreferrer&quot;&gt;this clamp&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(I don’t know the weight of the interface clamp but it is aluminum)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CANTILEVERED ARM:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;http://dectronusa.com/store/rigid-sections/&quot; rel=&quot;noreferrer&quot;&gt;Arm&lt;/a&gt; is 4 feet long with a working load of 100 lbs. at the end of the arm&lt;/li&gt;&#xA;&lt;li&gt;The arm itself weighs 11.4 lbs&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;ADDITIONAL NOTES:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The manufacturer of the articulated cantilevered arm (dectron USA) sells a Steel 2 inch OD, 0.125 inch wall thickness tube for this purpose. But their solution won't work for my application.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am not worried about the strength of the arm it is rated to support a 100 lb. working load.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Don’t think it matters but I have two 12” sections, one 6” and one 18” all rigid&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The only limit I want is 100 lbs. at the end of the arm other than that I want to use this thing with confidence for decades:)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Should I consider High-Strength 2024 Aluminum with the same dimensions instead?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;OR could I safely use THINNER walled tubing, 0.125 instead of 0.25 inch wall thickness, 6061 Aluminum, 2 inch OD?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for looking at this and your help&lt;/p&gt;&#xA;" OwnerUserId="5540" LastEditorUserId="5540" LastEditDate="2016-03-24T02:06:10.027" LastActivityDate="2016-03-24T22:50:47.467" Title="Strength of a Vertical Round Aluminum Tube with load cantilevered off it?" Tags="|structural-engineering|beam|aluminum|mechanical-failure|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8146" PostTypeId="1" AcceptedAnswerId="8147" CreationDate="2016-03-24T07:50:00.190" Score="5" ViewCount="45151" Body="&lt;p&gt;What are the criteria for choosing repeating variables in Buckingham's Pi theorem in dimensional analysis? In many problems, it's solved by taking D,V,H (Diameter, Velocity, Height) as repeating variables. Why do they take the above variables as repeating variables, when the problem also contains the following variables g,u (acceleration due to gravity, viscosity)?&lt;/p&gt;&#xA;" OwnerUserId="3966" LastEditorUserId="5899" LastEditDate="2016-04-07T08:38:03.950" LastActivityDate="2019-09-16T18:08:00.213" Title="Choosing of repeating variables in Buckingham's Pi theorem" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8147" PostTypeId="2" ParentId="8146" CreationDate="2016-03-24T12:46:42.887" Score="6" Body="&lt;p&gt;The repeating variables are any set of variables which, by themselves, cannot form a dimensionless group.  Diameter, velocity, and height cannot be arranged in any way such that their dimensions would cancel, so they form a set of repeating variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's assume that we suspect that a fluid we're studying behaves as a function of several variables including a characteristic length, velocity, viscosity, density, surface tension, etc., and we want to see what dimensionless numbers we can make out of them.  A possible choice of repeating variables would be length ($l$), velocity ($v$), and density ($\rho$) (in MKS units they would be $m$, $\frac{m}{s}$, $\frac{kg}{m^3}$), because they cannot be combined in any way to make a dimensionless group.  Now add surface tension ($\sigma$, units of $\frac{kg}{s^2}$) and combine all four variables to make a dimensionless group like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\rho l v^2}{\sigma}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the &lt;em&gt;Weber Number&lt;/em&gt;.  Another possibility would be to use viscosity instead of surface tension:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\rho l v}{\mu}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is of course the &lt;em&gt;Reynolds Number&lt;/em&gt;.  Ultimately, the choice of which combination to use out of all the possibilities comes down to whichever is more useful for the type of problem you're working on; Reynolds is good for turbulence and heat transfer, while Weber is more suitable for bubble and droplet formation.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-03-24T12:46:42.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8148" PostTypeId="1" AcceptedAnswerId="8151" CreationDate="2016-03-24T12:54:47.907" Score="0" ViewCount="242" Body="&lt;p&gt;I think there is an error in this questions solution and I want someone to check:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Example 8.1-4:Mass transfer from an oxygen bubble A bubble of oxygen originally 0.1 cm&#xA;  in diameter is injected into excess stirred water, as shown schematically in Fig. 8.1-1(d).&#xA;  After 7 min, the bubble is 0.054 cm in diameter. What is the mass transfer coefficient?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/blLBF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/blLBF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution goes on to state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d}{dt}(c_1\frac{4}{3}\pi r^3)=AN_1=-4\pi r^2k[c_1(\text{sat})-0]$$&#xA;$$\frac{dr}{dt}=-k\frac{c_1(\text{sat})}{c_1}=-0.034k$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't it be:&#xA;$$\frac{dr}{dt}=-3k\frac{c_1(\text{sat})}{c_1}=-0.102k$$&lt;/p&gt;&#xA;" OwnerUserId="4530" LastActivityDate="2016-03-24T17:07:05.910" Title="Mass Transfer from an Oxygen Bubble" Tags="|thermodynamics|chemical-engineering|process-engineering|vapor-pressure|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8149" PostTypeId="1" AcceptedAnswerId="8153" CreationDate="2016-03-24T13:53:53.253" Score="1" ViewCount="1523" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A steam power plant employs two adiabatic turbines in series. Steam&#xA;  enters the first turbine at 650 C and 7000 kPa and discharges from the&#xA;  second turbine at 20 kPa. The system is designed for equal power&#xA;  outputs from the two turbines, based on a turbine efficienct of 78%&#xA;  for each turbine. Determine the temperature and pressure of the system&#xA;  in its intermediate state between the two turbines. What is the&#xA;  overall efficiency of the two turbines together with respect to&#xA;  isentropic expansion of the steam from the initial to final state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How should I determine the pressure between two turbines?I have the &lt;a href=&quot;https://drive.google.com/file/d/0B2xNtfJLGsLwZ3ZnNWxQS283ZXM/edit&quot; rel=&quot;nofollow&quot;&gt;solution&lt;/a&gt; from a solution manual (prob. 8.6) but I just can't understand why it takes pressures around 700 KPa and does an interpolation.&#xA;what we know:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S3=S2=S1$ (if working isentropically) ($S$ is entropy)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x$ (Quality of exiting vapor)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:we could determine properties of the intermediate flow properties using one property($S$) just if it's two-phase, but it's assumed that the intermediate flow is superheated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure of problem's data and that's all we have.&lt;/p&gt;&#xA;" OwnerUserId="5452" LastEditorUserId="1293" LastEditDate="2016-03-24T16:29:19.680" LastActivityDate="2016-03-24T16:47:27.030" Title="Determining intermediate pressure between two turbines" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="1" CommentCount="14" ClosedDate="2016-03-25T08:22:46.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="8151" PostTypeId="2" ParentId="8148" CreationDate="2016-03-24T15:45:31.223" Score="3" Body="&lt;p&gt;Your approach is wrong because the left hand side represents the derivative of the composition of two time dependent functions. You completely ignored that fact and started cancelling terms. I try to give a simpler example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d 4x^3}{dx} = kx^2 &#xA;\Leftrightarrow 12x^2=kx^2&#xA; \Leftrightarrow 12=k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not the same as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d 4x}{dx} = k \Leftrightarrow 4=k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore use the chain rule&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{d(c \cdot V)}{dt} = V\frac{dc}{dt}+\frac{dV}{dt}c=-Ak\Delta c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would now assume a stationary diffusion, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V\frac{dc}{dt} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dV}{dt}c=-Ak\Delta c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A = \frac{dV}{dr}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dV}{dt}c=-\frac{dV}{dr}k\Delta c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this step I hope there are no mathematicians around, I cancel the differentials&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dr}{dt}c=-k\Delta c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dr}{dt}=-k\frac{\Delta c}{c}$&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2016-03-24T17:07:05.910" LastActivityDate="2016-03-24T17:07:05.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8153" PostTypeId="2" ParentId="8149" CreationDate="2016-03-24T16:47:27.030" Score="2" Body="&lt;p&gt;The answer in the solution manual simply assumes a set of four values of intermediate pressure (that somehow magically the actual intermediate pressure lies within!): &#xA;$$P_2 = \begin{pmatrix}725\\750\\775\\800\end{pmatrix} \text{kPa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for each pressure in the set, the one that results in an equal power output from the two turbines (or the closest value to $\triangle W = 0$) is the actual intermediate pressure (After some trials).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and we interpolate linearly to find the value of $P_2$ for which the work difference is zero:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2016-03-24T16:47:27.030" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="8154" PostTypeId="2" ParentId="8144" CreationDate="2016-03-24T17:48:02.750" Score="8" Body="&lt;p&gt;I'm going to run with this assuming the arm looks like the following diagram (I'm ignoring the 11.4 pounds of the bar for now to make the concepts easier to explain - that can be added later by assuming all 11.4 pounds run through the center of the bar.):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xLAkg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xLAkg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a pretty simple setup.  To figure out the loading on the shaft, we have to move the load from the tip of the arm to the center of the shaft. Fortunately, we can use the principle of &lt;a href=&quot;https://en.wikipedia.org/wiki/Couple_(mechanics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;force couples&lt;/a&gt;.  The basis is that a force off-center can be moved to the center, but you have to add a couple, or moment, equal to the force times the distance moved.  Here is what I came up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UjVz7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UjVz7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can apply some basic formulas.  We need to look at the cross section of the hollow tube and get some &lt;a href=&quot;http://www.amesweb.info/SectionalPropertiesTabs/SectionalPropertiesHollowCircle.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;basic properties&lt;/a&gt; of the cross section, as well as on the &lt;a href=&quot;http://asm.matweb.com/search/SpecificMaterial.asp?bassnum=MA6061t6&quot; rel=&quot;nofollow noreferrer&quot;&gt;material&lt;/a&gt; (we need yield strength $\sigma_y$, and $E$, the young's modulus).  The important sectional properties are the area, $A$, the area moment of inertia, $I_{xx}$, and the Section Modulus, $S_{xx}$.  Formulas for these can be described below (site also has a &lt;a href=&quot;http://www.amesweb.info/SectionalPropertiesTabs/SectionalPropertiesHollowCircle.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;calculator&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wP99P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wP99P.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can start looking at the four methods of failure:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Critical Vibration &lt;a href=&quot;https://en.wikipedia.org/wiki/Resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Natural Frequency)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Deflection&lt;/li&gt;&#xA;&lt;li&gt;Stress&lt;/li&gt;&#xA;&lt;li&gt;Buckling&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can eliminate &lt;strong&gt;vibration&lt;/strong&gt; immediately as a method of failure, but for reference, the critical frequency for a cantilevered beam with an end weight only is: (Ref &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071742476&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roark's Formulas for Stresses and Strains, Eighth Ed.&lt;/a&gt;, Table 16.8, Case 3a) $$f = \frac{1.732}{2\pi}\sqrt{\frac{EI_{xx}g}{WL^3}}$$&#xA;$$f = {0.2757}\sqrt{\frac{10000000 \frac{lbf}{in^2}*0.537in^4*386\frac{in}{s^2}}{100lbf*(36in)^3}} = 5.811 Hz$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Deflection&lt;/strong&gt; in z (down the pole) is minimal.  In this case then: $$\delta_z = \frac{W*L}{E*A}$$ $\delta_z = \frac{100lbf*36in}{10000000 \frac{lbf}{in^2}*1.374{in^2}} = 0.000262 in$ which is clearly negligible.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Deflection_(engineering)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deflection&lt;/a&gt; in x/y (across the body) is likely not.  In this case $$\delta_x = \frac{ML^2}{2EI_{xx}}$$ (Ref Roark again, Table 8.1 Case 3a).  $$\delta_x = \frac{5000 in*lbf * (36in)^2}{2*10000000\frac{lbf}{in^2}*0.537in^4} = 0.603 in$$  This is fairly significant and should be considered (the column will tip over just shy of  5/8&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stress&lt;/strong&gt; is also a concern here.  We need to look at the stress from the bending moment as well as from the downward force.  In this case, the formula is: $$\sigma = {F \over A} \pm \frac {M}{S_{xx}}$$ (Ref: &lt;a href=&quot;https://en.wikipedia.org/wiki/Bending#Large_bending_deformation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;).&#xA;$$\sigma = \frac{100 lbf}{1.374 in^2} \pm \frac{5000 lbf*in}{0.537in^3} = 72.78psi \pm 9310.98 psi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things to note - the bending part dominates over the downward portion, so this is the most important part to look at in a preliminary analysis.  The second is that this is pretty high, but not terrible. It has a 4.26 F.O.S, which is better than 1.8 typically used in most structures - but aluminum has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fatigue_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;fatigue life&lt;/a&gt; that needs to be investigated.  Repeated quick loading (pushing on it and pulling within seconds) could cause break, but it should have a decent life.  This, however, is a separate question entirely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Buckling&quot; rel=&quot;nofollow noreferrer&quot;&gt;Buckling&lt;/a&gt;&lt;/strong&gt; is where things get interesting - we need to account for both the moment buckling factor of safety and the downward force factor of safety separately, and combine. Read the Wikipedia article for more information about buckling, but suffice it to say that, with the exception of metal bars (like you have here), it's mostly empirical and relies on a lot of testing.  I'll put the formulas here to cover the details so you can get going on further testing if needed.  The formula for the downward force is: $$P' = 0.25*\frac{\pi^2*E*I}{L^2}$$ Source: Roark's again, Table 15.1, Case 1a.  This leads to $P' = 40895 lbf$, or a FOS of $\frac{40895lbf}{100lbf} = 409$.  Bending is $$M' = 0.72*\frac{Ert^2}{\sqrt{1-\nu}}$$ (Source is Roark's again, Table 15.2, Case 16) $\nu$ for aluminum is &lt;a href=&quot;http://www.engineeringtoolbox.com/poissons-ratio-d_1224.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;0.35&lt;/a&gt;.  This leads to $M' = 558312 in*lbf$, or FOS of $\frac{558312 in*lbf}{5000 in*lbf} = 111.6$.  Combining the two is $$\frac{1}{\frac{1}{111.6} + \frac{1}{409}} = 87.7$$  Therefore, buckling is not a method of failure. (Anything over 5 is usually acceptable without further analysis). &lt;strong&gt;Therefore, the 2&quot; aluminum tube is satisfactory for the given loads&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now onto the additional notes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I consider High-Strength 2024 Aluminum with the same dimensions instead?  &lt;strong&gt;You should be fine with this design&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I safely use THINNER walled tubing, 0.125 instead of 0.25 inch wall thickness, 6061 Aluminum, 2 inch OD?  &lt;strong&gt;This I leave as an exercise for you.  It should be pretty easy to investigate with the tools given - though note the deflection was already fairly high.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2016-03-24T22:50:47.467" LastActivityDate="2016-03-24T22:50:47.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8156" PostTypeId="1" CreationDate="2016-03-24T18:27:14.057" Score="4" ViewCount="1170" Body="&lt;p&gt;I was wondering if there is any equation that helps to calculate flue gas velocity through a pipe? Can I use the Poiseuille Equation to solve this problem by assuming the flue gas is acting as a newtonian fluid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pipe diameter is 3.66m, change in pressure is 3atm, length of the pipe is 10m, density of the flue gas is 0.811kg/m$^3$ at 150 ˚C, and the viscosity (C$_p$) is 0.0232.&lt;/p&gt;&#xA;" OwnerUserId="5742" LastEditorUserId="5742" LastEditDate="2016-03-25T16:59:18.257" LastActivityDate="2016-04-27T08:59:15.057" Title="Velocity of flue gases through a pipe" Tags="|mechanical-engineering|gas|compressed-gases|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8158" PostTypeId="1" CreationDate="2016-03-24T21:07:48.397" Score="-1" ViewCount="75" Body="&lt;p&gt;since the ground clearance of a sports car is low. Do we use small tires for them?&lt;/p&gt;&#xA;" OwnerUserId="5745" LastActivityDate="2016-03-24T22:56:53.850" Title="Are tires of a soprts car small in size for low ground clearance?" Tags="|mechanical-engineering|automotive-engineering|transportation|" AnswerCount="1" CommentCount="2" ClosedDate="2016-03-24T23:45:21.237" ContentLicense="CC BY-SA 3.0" />
  <row Id="8159" PostTypeId="2" ParentId="8158" CreationDate="2016-03-24T21:35:19.677" Score="1" Body="&lt;p&gt;Ground clearance has nothing to do with either wheel diameter or tyre profile. It is determined by suspension geometry and the clearance allowed by the body (particularly  wheel arches). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often low profile tyres are more about cosmetic appearance than actual performance. Note that F1 cars have tyres with  a larger aspect ratio than many standard road cars and still achieve small (and in some conditions zero) ground clearance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also an important difference between ride height and effective ground clearance. Ride height is simply the distance between the floor of the car and the road ground clearance determines the ability to traverse ramps and bumps without grounding and is affected by suspension travel and wheelbase. Short wheelbase vehicles will be less likely to ground on very rough terrain than longer wheelbase ones with equivalent ride height. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The placement and vulnerability of components such as prop shafts axles and sumps will also have an effect on effective ground clearance. In fact some extreme off road vehicles have steel or composite  plates protecting their undersides to allow them to slide over rocks without breaking anything important and similarly may have independent wishbone suspension at both front and rear compared to standard models with a solid rear axle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low profile tyres will tend to make the sidewall of the tyre stiffer which can be beneficial for traction and braking although this must be considered in the context of the overall suspension design as sidewall flex has an effect on both ride comfort and steering feel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low profile tyres can also allow for larger diameter brakes and more space for brake cooling ducts although this can also increase the unsprung mass of the wheels. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-03-24T22:56:53.850" LastActivityDate="2016-03-24T22:56:53.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8160" PostTypeId="1" CreationDate="2016-03-24T23:28:42.183" Score="10" ViewCount="4099" Body="&lt;p&gt;How can you calculate how much power can be generated by a motor when used as a generator (like a motor used in a wind turbine)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the power depend on how fast the magnet is rotated? If so, is there a equation that lets you know given the rpm, how much power should be generated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how can I account for limiting factors such as the strength of the magnet (moving a coil around a stronger magnet should give more power than moving coil around a weaker magnet)?&lt;/p&gt;&#xA;" OwnerUserId="5742" LastEditorUserId="1586" LastEditDate="2016-04-21T23:27:31.640" LastActivityDate="2016-05-12T15:54:42.420" Title="How to calculate Power of motor when used as a generator?" Tags="|electrical-engineering|motors|power|wind-power|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8163" PostTypeId="2" ParentId="7520" CreationDate="2016-03-25T09:34:29.563" Score="0" Body="&lt;p&gt;While not a singular material, and a very crude approximation of the concept, would an approach like this not allow more heat then sound to pass through?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z9O3X.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z9O3X.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5750" LastActivityDate="2016-03-25T09:34:29.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8164" PostTypeId="2" ParentId="8047" CreationDate="2016-03-25T12:45:33.087" Score="2" Body="&lt;p&gt;You can represent K as $K_i = \frac{p_i^{sat}}{p}$ for an ideal mixture using Raoult's law. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation directly yields the correspondance of $K$ and $p$. If the pressure is below the dew point pressure $K_i$ will be larger. Hence $\frac{z_i}{K_i}$ will be smaller and subsequently the sum over it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this tells you is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p$ yields $\sum \frac{z_i}{K_i} &amp;lt; 1$ pressure is below dew point pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p$ yields $\sum \frac{z_i}{K_i} &amp;gt; 1$ pressure is above dew point pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This is for a fixed temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you can make the same deductions with $K_i = \frac{y_i}{x_i}$ and the knowledge that at dew point the first drop of liquid forms.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2016-03-25T12:45:33.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8166" PostTypeId="1" CreationDate="2016-03-25T23:29:31.483" Score="-1" ViewCount="157" Body="&lt;p&gt;The chuck I have is:&#xA;&lt;a href=&quot;http://grizzly.com/products/0-1-2-x-JT33-Plain-Bearing-Drill-Chuck/G8628?utm_campaign=zPage&amp;amp;utm_medium=link&amp;amp;utm_source=direct&quot; rel=&quot;nofollow&quot;&gt;Jacob's 0-1/2&quot; JT #33&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lathe I have is:&#xA;&lt;a href=&quot;http://grizzly.com/products/8-x-16-Variable-Speed-Lathe/G0768?utm_campaign=zPage&amp;amp;utm_medium=link&amp;amp;utm_source=direct&quot; rel=&quot;nofollow&quot;&gt;G0768&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering what chuck arbor I need to buy / how to determine that.  Related: the chuck has threads in its hole, but I see no arbors with threads, so what's that all about?&lt;/p&gt;&#xA;" OwnerUserId="3235" LastActivityDate="2016-06-24T22:50:03.913" Title="What chuck arbor do I need for a Jacob's JT #33 chuck and G0768 grizzly lathe?" Tags="|machining|drilling|lathe|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8168" PostTypeId="1" CreationDate="2016-03-26T02:49:04.863" Score="1" ViewCount="54" Body="&lt;p&gt;I am trying to design mechanism that is sort of like a wind turbine (uses wind as &quot;fuel&quot;) but has a fan size that is in a pelton turbine. I was wondering if there is any math/equation that I can use to know how much power can be generated or how many rpms can the fan produce when attached to a generator. The equation of the wind turbine is not that applicable because the wind coming in will be going through a pipe (like in a peloton turbine, which has water), and the fan size is too small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an equation relating air velocity in a pipe and how much power that can generate?&lt;/p&gt;&#xA;" OwnerUserId="5742" LastActivityDate="2016-03-26T02:49:04.863" Title="How to design a fan that can be used to convert concentrated air to power" Tags="|mechanical-engineering|fluid-mechanics|pressure|turbines|" AnswerCount="0" CommentCount="1" ClosedDate="2016-03-27T19:24:17.323" ContentLicense="CC BY-SA 3.0" />
  <row Id="8169" PostTypeId="2" ParentId="7568" CreationDate="2016-03-26T08:06:18.140" Score="2" Body="&lt;p&gt;I am going to center in one of the most famous bridges of these conditions last years: San Francisco Bay Bridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This bridge is not planned for trains, and so they tested with huge hydraulic jacks (see &lt;a href=&quot;http://www.wired.com/2015/06/mystery-brand-new-bay-bridges-corroded-steel/&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;). This bridge is designed not to collapse during an earthquake, but just to suffer minor breakdowns that could be easily repaired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When earthquakes, one of the crucial points that was found is that the tower of the bridge must resist and not fall. And this is a key point in the current bridge, as it has individual and expensive basement and the main tower is divided in four pieces not to collapse completely (&lt;a href=&quot;https://www.newscientist.com/article/mg21228386-000-san-franciscos-huge-new-quake-proof-bridge/&quot; rel=&quot;nofollow&quot;&gt;see here&lt;/a&gt;). The bridge could be robust and indestructible, but aesthetically it would be much uglier and probably more expensive for the increase of concrete and other materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike more conventional suspension bridges, in which parallel cables are slung over towers and anchored at both ends in rock or concrete, San Francisco Oakland Bay bridge has only a single tower and a single cable that is anchored to the road deck itself, looping from the eastern end to the western end and back again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes no sense in defending Roman bridges among other. Romans just tested designs empirically till they realized that a particular model resisted, but the bridge engineering was not too big those times.&lt;/p&gt;&#xA;" OwnerUserId="5553" LastActivityDate="2016-03-26T08:06:18.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8170" PostTypeId="2" ParentId="1908" CreationDate="2016-03-26T12:44:16.187" Score="2" Body="&lt;p&gt;Submarine propulsor designers are least concerned about hub vortex cavitation as they are primarily concerned about cavitation inception, or the onset of cavitation, which first occurs in the tip vortex regions, where the velocities are highest.  Once cavitation occurs you are no longer operating in stealth mode. However, the hub vortex does waste a significant amount of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the fins looks to me to be a kind of hub stator used for diffusing the hub vortex and thereby increasing propulsor efficiency.  Take a look at this &lt;a href=&quot;http://articles.maritimepropulsion.com/article/IMPROVED-PROPULSIVE-EFFICIENCY-BY-WEAKING-THE-PROPELLER-HUB-VORTEX84713.aspx&quot; rel=&quot;nofollow&quot;&gt;article in maritime propulsion&lt;/a&gt;.  I would guess the covered picture could be some kind of &lt;a href=&quot;http://www.mol.co.jp/en/pr/2011/1145.html&quot; rel=&quot;nofollow&quot;&gt;propeller boss cap fin system&lt;/a&gt; which would even be more effective in improving propulsive efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spheres you refer to are a bit of a mystery to me.  They would create horseshoe vortices and convect momentum downward into the hub vortex region, which would slightly increase the pressure in the in hub vortex wake and help to alleviate flow separation when the hub sharply ends. However, it wouldn't seem to me this was their ultimate purpose, as generating secondary flows are not usually desirable unless there is another advantageous purpose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that there has been quite a bit of research the past 10-15 years in drag reduction by polymer or microbubble injection, I would expect a state-of-the-art submarine to be employ such a system. See for example this &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0894177708001386&quot; rel=&quot;nofollow&quot;&gt;paper by Nouri and Sarreshtehdari&lt;/a&gt; specifically about microbubble injection to reduce drag in the hub vortex. They are optimally placed for feeding into the hub vortex.&lt;/p&gt;&#xA;" OwnerUserId="511" LastActivityDate="2016-03-26T12:44:16.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8171" PostTypeId="2" ParentId="7568" CreationDate="2016-03-26T13:10:59.660" Score="12" Body="&lt;p&gt;You can never make something earthquake proof, but there are many things that can be done to resist earthquakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are long-span bridges built in earthquake zones. For example, the &lt;a href=&quot;http://www.pbs.org/wgbh/buildingbig/wonder/structure/akashi_kaikyo.html&quot; rel=&quot;noreferrer&quot;&gt;Akashi Kaikyo&lt;/a&gt; bridge in Japan is currently the longest span bridge in the world and is in a severe seismic zone. It is designed to withstand a magnitude 8.5 earthquake. In fact it was subjected to a magnitude 7.2 earthquake &lt;em&gt;during construction&lt;/em&gt;. galtor mentioned in another answer the San Francisco Bay bridge which has been retrofitted to improve it's seismic resistance. So designing bridges to resist strong earthquakes is certainly possible and has been done.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What can be done to improve the earthquake resistance of a bridge?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tuned_mass_damper&quot; rel=&quot;noreferrer&quot;&gt;Tuned mass dampers&lt;/a&gt; are used in tall buildings as well as in bridges to counteract the movement due to earthquakes as well as wind and other lateral loads. The Akashi Kaikyo bridge uses TMDs in the suspension towers for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.gerbusa.com/index.php?id=589&amp;amp;no_cache=1&amp;amp;tx_photogals_elementid=1601&amp;amp;tx_photogals_image=3&amp;amp;MP=503-593&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eOqC2m.jpg&quot; alt=&quot;Bridge TMD&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Base_isolation&quot; rel=&quot;noreferrer&quot;&gt;Base isolation&lt;/a&gt; is one of the most common techniques used to resist earthquake movement. These are devices which essentially separate the horizontal movement of the foundation from the rest of the structure by using some form of sliding bearings. If designed properly this can drastically reduce earthquake damage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dis-inc.com/media/retrofit-installation-photo.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BPwlnm.jpg&quot; alt=&quot;Base Isolation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://taylordevices.com/dampers-seismic-protection.html&quot; rel=&quot;noreferrer&quot;&gt;Seismic Dampers&lt;/a&gt; are also common. These are a range of devices which act to remove seismic energy from the structure similar to how shock absorbers on a car remove the vibration energy of the car going over a rough road. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.elemka.gr/en-us/bearings-seismic-isolation/elemka-bearings-seismic-isolation&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5c8Xwm.jpg&quot; alt=&quot;Seismic Damper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These technologies are well understood and are frequently used in bridges and buildings. There are more experimental techniques which are also possible such as: &lt;a href=&quot;http://openaccess.city.ac.uk/3635/&quot; rel=&quot;noreferrer&quot;&gt;rocking isolation&lt;/a&gt;, or active damping systems (computer controlled dampers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If desired these devices can also be used in combination in order to further enhance the earthquake response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In standard seismic design practice a structure is designed to accommodate some damage. This damage is, wherever possible, concentrated in elements which are more easily replaced (beams and braces), and which will not result in a disproportionate collapse if damaged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is certainly technically feasible to design large bridges to resist earthquake loading. Especially if there were no financial constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find this useful reading: &lt;a href=&quot;http://science.howstuffworks.com/engineering/structural/earthquake-resistant-buildings.htm&quot; rel=&quot;noreferrer&quot;&gt;How Earthquake-resistant Buildings Work&lt;/a&gt;. The techniques used on buildings can also be applied to bridges.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2016-03-27T20:49:15.960" LastActivityDate="2016-03-27T20:49:15.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8173" PostTypeId="1" AcceptedAnswerId="8175" CreationDate="2016-03-26T17:10:25.213" Score="5" ViewCount="221" Body="&lt;p&gt;Suppose the fit for a rod and a hole in a gear is supposed to be free running and the fit is 3mm H9/d9. For the specification of the hole for the manufacturing of the gear, is &quot;3mm H9&quot; already adequate since d9 only refers to the shaft tolerance which may not be pertinent to the manufacturing of the gear? Sometimes I see the fit specified on drawing sheets for some reasons, e.g., &quot;10mm H9/d9&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my understanding is correct, 3mm H9 essentially means 3mm - 3.025mm and 3mm d9 means min. 2.955mm - 2.98mm. So the clearance ranges from 0.02mm to 0.07mm - am I right?&lt;/p&gt;&#xA;" OwnerUserId="4303" LastEditorUserId="33" LastEditDate="2016-03-26T22:29:44.993" LastActivityDate="2016-03-26T22:29:44.993" Title="Specification of hole design tolerance" Tags="|design|tolerance|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8175" PostTypeId="2" ParentId="8173" CreationDate="2016-03-26T21:15:52.320" Score="3" Body="&lt;p&gt;Your clearance range is the minimum hole diameter minus max shaft diameter to max hole diameter minus minimum shaft diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer: Yes. Your math seems correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a useful tool: &lt;a href=&quot;http://www.amesweb.info/FitTolerance/FitTolerance.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.amesweb.info/FitTolerance/FitTolerance.aspx&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-03-26T21:15:52.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8176" PostTypeId="2" ParentId="8166" CreationDate="2016-03-26T22:13:32.933" Score="0" Body="&lt;p&gt;According to the spec for your lathe in the link you provided you need a MT2 arbour. What is a bit confusing is that that model of chuck is supposed to have a tapered fit to the arbour rather than a threaded one. JT#33 is the taper fit to the chuck and MT2 is the taper fit to the tailstock. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Threaded arbours certainly exist but you need to determine what thread your chuck has, that size is probably either 1/2 x 20 or 3/8 x 24. However taper mount chucks are far more common for lathes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a JT#33 chuck is what you want but it sounds like that's not actually what you have got. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-03-26T22:13:32.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8177" PostTypeId="1" CreationDate="2016-03-27T05:19:50.553" Score="2" ViewCount="20695" Body="&lt;p&gt;I was wondering how I can calculate the velocity of gas through a pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas, assuming ideal behavior, in a pipe (cylinder) of volume of 222.4 L has a pressure of 1.95 atmospheres and a density of 2.25 kg/m$^3$. Knowing that there will be turbulent flow, the equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V\ =\ \sqrt\frac{2\Delta P}{{\rho (4f\frac{\Delta x}{D}\ -\ 1)}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be used. However, because this is a pipe with no change in volume, the pressure should remain even, so can I plug in 2.25 kg/m$^3$ for pressure and find the velocity? The pipe length is 10 ft and diameter is 1 ft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to find density of the gas the ideal gas equation was used, relating pressure and temperature. For this gas, the temperature was 298 K.&lt;/p&gt;&#xA;" OwnerUserId="5742" LastEditorUserId="1293" LastEditDate="2016-03-27T06:06:14.660" LastActivityDate="2016-11-19T01:40:21.633" Title="How to calculate the velocity of gas in a pipe?" Tags="|fluid-mechanics|pressure|gas|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8179" PostTypeId="1" AcceptedAnswerId="8180" CreationDate="2016-03-27T17:15:19.463" Score="0" ViewCount="54" Body="&lt;p&gt;I am working on a project and I have some questions about setting up a passive heating system (thermosiphon) with a heat exchanger.&#xA;I will have a main boiler with a coil of copper tubing inside. This coil will be my heat exchanger. It will then continue to pass through an object which is the target of the passive heating cycle. My question is about the placement location of the coil's ends.&#xA;Please see the attached system drawings. Will either of these work? Also, to clear up the &quot;T&quot; in the lower-left: for the purposes of discussing the thermosiphon loop, you can pretty much ignore this. Check valves will be in place to ensure the return goes back into the exchanger coil.&#xA;Thanks in advance for the help&lt;img src=&quot;https://i.stack.imgur.com/dMULW.jpg&quot; alt=&quot;![enter image description here&quot;&gt;]&lt;a href=&quot;https://i.stack.imgur.com/dMULW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5755" LastActivityDate="2016-03-27T18:12:57.517" Title="Thermosiphon System" Tags="|heat-transfer|heat-exchanger|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8180" PostTypeId="2" ParentId="8179" CreationDate="2016-03-27T18:12:57.517" Score="1" Body="&lt;p&gt;In the broad sense your question is asked I'd say yes, this will work either way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things to consider that could be an indicator for you to go with #1 or #2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the actual design, the isolation of the copper coil in between the heater and the item to be heated you might try and keep the distance as short as possible to reduce heat loss. So this might favor #2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand #1 may be favorable if you need to increase heat transfer in the tank and want to increase the surface area for heat transfer.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2016-03-27T18:12:57.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8184" PostTypeId="1" CreationDate="2016-03-28T00:43:53.050" Score="7" ViewCount="483" Body="&lt;p&gt;Why do we need to shut down power generators if we are producing more power than the grid requires?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How it will damage the system if they are not shut down?&lt;/p&gt;&#xA;" OwnerUserId="5771" LastEditorUserId="2819" LastEditDate="2016-03-29T02:11:48.610" LastActivityDate="2016-03-29T02:11:48.610" Title="Why do we need to shut down power generators if we are producing more power than the grid requires?" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8188" PostTypeId="2" ParentId="2359" CreationDate="2016-03-28T04:24:47.380" Score="1" Body="&lt;p&gt;Well, this expression is the standard expression for the electric field energy $U$ on a dielectric with permitivity $\epsilon$ inside a volume $V$ but with some replacements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/electric/engfie.html&quot; rel=&quot;nofollow&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/electric/engfie.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Electric_potential_energy&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Electric_potential_energy&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;U_V=\frac{1}{2}\int_V \epsilon  E^2 dV&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last expression is known from Maxwell's. And proven in Wikipedia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally this expression comes in a volume integral on the dielectric $\epsilon_0 \epsilon_r$ -which you assumed to be a kind of cylinder of area $A$ and heigh $h$?- and for generic electric fields -which on this case you assumed to be constant $E_0$ on that cylinder?-. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence. you start with the expression for the energy in a volume, so the integral you must to solve is in the &lt;strong&gt;space, not in the time&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, you fill in the variables, on the above expression, make an average on a single frequency period $2\pi/\omega$ -again assuming your excitation is sinusoidaly varying with $\omega$ -, and you are trivially done.&lt;/p&gt;&#xA;" OwnerUserId="5400" LastEditorUserId="5400" LastEditDate="2016-03-28T04:42:17.700" LastActivityDate="2016-03-28T04:42:17.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8189" PostTypeId="2" ParentId="8184" CreationDate="2016-03-28T05:51:02.237" Score="5" Body="&lt;p&gt;The electricity grid has very very little capacitance. That means that in any and every second, the energy going into it, has to come out of it too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't come out in controlled, intended ways, it will come out in uncontrolled, unintended ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, by blowing up equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the routes for excess energy is to make synchronous spinning things on the grid, spin faster (i.e. the grid frequency increases) - some motors, for example. Some of these will quickly exceed their intended speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing that happens with rising supply, is that grid voltage increases. Some equipment will respond by using more energy, which in some cases can lead to faster burnout. Electrical equipment will tolerate some voltage variation, but after a point, increasing voltage will damage it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the excess electricity will get absorbed by some things getting hotter, and by some things going bang. Both on the demand-side, and on the supply-side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to give equipment manufacturers some certainty, limits are set on how far the grid frequency and voltage can vary away from their nominal set points. The system operator will adjust supply and demand to keep frequency and voltage within their limits. One way to stop frequency and voltage getting too high, is by reducing or stopping the output of some generators. Another way is to bring more demand online. All these rules are contained in the Grid Code: each system has its own.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2016-03-28T06:29:50.033" LastActivityDate="2016-03-28T06:29:50.033" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8190" PostTypeId="1" CreationDate="2016-03-28T07:12:31.950" Score="6" ViewCount="955" Body="&lt;p&gt;I have a double mass rotary model. The two masses are connected by an axis which together acts as a spring damper system of the fourth order. The system is actuated by a motor and the angle of rotation is measured by rotary encoders at each end of the mass. The input to the motor is milli-volts which then converts into corresponding torques. The output measured is angle of rotation of the mass. The actuation and the measurement is on the same side. &#xA;The system looks similiar to this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wPeU1.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wPeU1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transfer function from output to input of the system is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Plant = \large&#xA;{\frac{(4700 s^2 + 4393 s + 3.245 (10)^8)}{(s^4 + 7.574 s^3 + 1.202(10)^5 s^2)}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to determine the mass of the system from the transfer function or bodeplot? How can I relate the voltage, torque, mass and angle of rotation?&lt;/p&gt;&#xA;" OwnerUserId="5775" LastEditorUserId="1632" LastEditDate="2016-03-28T18:05:21.587" LastActivityDate="2016-04-28T17:32:49.657" Title="mass calculation from transfer function or bodeplot" Tags="|control-engineering|transfer-function|matlab|feedback-loop|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8191" PostTypeId="2" ParentId="8177" CreationDate="2016-03-28T08:16:31.570" Score="1" Body="&lt;p&gt;@idfka is correct there needs to be some pressure drop to overcome the friction and flow. In practice though for gas through 10 feet of pipe the pressure drop will not be in the measurable range unless the pipe is undersized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, you can't calculate the velocity with the inputs you have. Not sure what your objective is here but what you can do can assume some typical numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate pressure drop, you can assume a velocity of around 60-80 ft/sec for gas. This value will be different if you're expecting liquid, particles or if its corrosive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate velocity, you can assume a pressure drop of about 6 psi/100 ft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for using the ideal gas law, assuming you have the molecular weight, that should be fine as pressures are quite low. You can just use the inlet density in the equation you have as its not gonna change much along the pipe.&lt;/p&gt;&#xA;" OwnerUserId="1103" LastActivityDate="2016-03-28T08:16:31.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8192" PostTypeId="2" ParentId="8156" CreationDate="2016-03-28T08:47:48.723" Score="1" Body="&lt;p&gt;Are you sure your numbers are right? 3 atm of pressure drop is pretty large for 10 m length with that diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways what you need to do is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Guess a velocity, gas pipes are usually about 20 m/s but for flare headers it can go up about 90% of speed of sound. So the range is quite large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Calculate Reynolds number and find friction factor from a Moody diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Use the Darcy–Weisbach equation to calculate velocity. If this number is close enough to the guessed value in 1) then you're done. Otherwise use this new velocity in 1) and repeat until it converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps. you can use Hagen-Poiseuille if it is laminar flow (low Reynolds, forgot the range) but I doubt it's the case with that pressure drop. &lt;/p&gt;&#xA;" OwnerUserId="1103" LastActivityDate="2016-03-28T08:47:48.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8193" PostTypeId="1" AcceptedAnswerId="8226" CreationDate="2016-03-28T12:33:59.117" Score="7" ViewCount="503" Body="&lt;p&gt;Consider a fixed steam engine, such as in a Lancashire cotton mill or pumping in a Cornish mine.  Then consider a fixed diesel engine such as a large generator or ship's engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the steam engine have one or two large cylinders with long stokes, but the diesel has many cylinders with relatively short strokes?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2019-08-15T19:33:53.333" Title="Why do steam and diesel engines have differing cylinder configurations?" Tags="|steam|diesel|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8194" PostTypeId="2" ParentId="8184" CreationDate="2016-03-28T12:45:53.717" Score="2" Body="&lt;p&gt;There will also be environmental and economic reasons.  Since electricity can't be stored in this context, why make more when it's not needed?  You can decrease your emissions for the year, including carbon payments.  Emission legislation is so onerous these days (in the UK) that we're actually closing some power stations to save carbon.  You can also save fuel.  And if you never had excess capacity, you could never service or replace the generators.  Since it takes 12 hours to spin up a fossil fuel power station they can't be blipped on /off anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truth of my answer could be proved at finding out whether solar /wind /hydro power plants shut down at surplus times since they run for  (effectively) free. &lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2016-03-28T12:45:53.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8195" PostTypeId="1" AcceptedAnswerId="8200" CreationDate="2016-03-28T13:30:42.413" Score="11" ViewCount="17594" Body="&lt;p&gt;I'm using AutoCAD 2015, and I run into this all the time.  I'm given a group of DWG drawings, and I want to plot them into one monochrome PDF, but they do not have monochrome.ctb selected as the plot style.  Is there any way in Batch Plot to publish to PDF using monochrome without editing the individual DWG files?  If not, is there a way to edit the page setup of all the drawings together as a group?&lt;/p&gt;&#xA;" OwnerUserId="37" LastActivityDate="2016-03-28T19:03:04.040" Title="How do I batch plot drawings in monochrome without editing DWG?" Tags="|autocad|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8196" PostTypeId="1" AcceptedAnswerId="8197" CreationDate="2016-03-28T13:30:59.527" Score="7" ViewCount="1066" Body="&lt;p&gt;I am currently working on a project (or actually returning to a project) and need to have a heat exchanger boiler made. The hx boiler will be relatively small, a 10&quot; long cylinder around 4&quot; in diameter. One end can be capped, but the other should have a flange so I can work inside the unit. Inside, I will have a coil of copper tubing (heat exchange) and various other sensors/ports. I don't have access to a full machine shop to fabricate such a boiler myself so I am contacting a number of companies. My question really is in regards to the thickness of the steel required for the boiler. The boiler, at max pressure, will only be at 2.5 bar. Additionally, the material must be food-safe; the main boiler will contain water that will be used in food preparation. Given these working parameters, of what type and how thick should the stainless steel be? I will attach various rough sketches of the system for extra information.&#xA;Thanks in advance for the help!&#xA;&lt;a href=&quot;https://i.stack.imgur.com/g33zq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g33zq.jpg&quot; alt=&quot;Flange Cap&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/EnJ9c.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EnJ9c.jpg&quot; alt=&quot;Rough Cutaway&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5755" LastEditorUserId="5755" LastEditDate="2016-03-28T15:15:17.050" LastActivityDate="2016-03-30T22:41:24.980" Title="Questions about making stainless steel boiler" Tags="|materials|boilers|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8197" PostTypeId="2" ParentId="8196" CreationDate="2016-03-28T15:31:51.100" Score="7" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Based upon your local jurisdiction, they may require a permit, which requires a PE Stamped calculation for the work you are requesting, since the internal pressure exceeds 1 bar (the limit into which it goes into &lt;a href=&quot;https://www.asme.org/shop/standards/new-releases/boiler-pressure-vessel-code/pressure-vessels&quot; rel=&quot;nofollow&quot;&gt;ASME BPVC Section VIII&lt;/a&gt;).  See for example in &lt;a href=&quot;https://www.dir.ca.gov/dosh/pressure.html&quot; rel=&quot;nofollow&quot;&gt;California&lt;/a&gt; - anything which generates pressures over 15 psig requires a permit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As a PE who does design these vessels for a living, usually I use a pretty simple set of formulas to get a good estimate.  Treat the top, shell, and bottom as separate entities&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The shell can be treated using &lt;a href=&quot;https://en.wikipedia.org/wiki/Cylinder_stress&quot; rel=&quot;nofollow&quot;&gt;cylinder stress rules&lt;/a&gt;.  Typically use 20,000 psi as the maximum allowed stress (= 36,000 psi yield with 1.8 Factor of Safety)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The bottom (and for estimation purposes, the top) can be treated using a &lt;a href=&quot;http://www.efunda.com/formulae/solid_mechanics/plates/calculators/cpS_PUniform.cfm&quot; rel=&quot;nofollow&quot;&gt;simple supported plate&lt;/a&gt;.  Maximum deflection should be limited to 0.5% and the stress should be 20,000 psi.  Don't forget that those openings create &lt;a href=&quot;http://www.ijirset.com/upload/march/24_Stress%20Concentration.pdf&quot; rel=&quot;nofollow&quot;&gt;stress concentrators&lt;/a&gt; - which in itself is a big subject in the pressure vessel design world.  For estimation, you can take the principle of replacing the material removed - that is double the thickness around the hole for an area equal to one hole radius outwards.  From a cross section then, the material is the same.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You may find that other &lt;a href=&quot;https://en.wikipedia.org/wiki/Head_(vessel)&quot; rel=&quot;nofollow&quot;&gt;bottom head designs&lt;/a&gt; fit better, especially for reducing stress.  Ellipsoidal heads are almost always the same thickness as the shell.  Hemispherical heads follow &lt;a href=&quot;http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm&quot; rel=&quot;nofollow&quot;&gt;simple cylinder stress rules&lt;/a&gt; and are half that thickness.  Torispherical heads are usually double, and flat heads are typically four - eight times as thick.  That's why fire extinguishers are usually rounded bottoms on a &lt;a href=&quot;https://www.google.com/search?q=pressure%20vessel%20skirt&amp;amp;espv=2&amp;amp;biw=1600&amp;amp;bih=817&amp;amp;tbm=isch&amp;amp;imgil=9ZU_6RrB3q4PEM%253A%253BOPB16saQySFDqM%253Bhttp%25253A%25252F%25252Fampreva.technosoft.com%25252Fprojects%25252Findex%25252Ffeature%25252Fvessel7&amp;amp;source=iu&amp;amp;pf=m&amp;amp;fir=9ZU_6RrB3q4PEM%253A%252COPB16saQySFDqM%252C_&amp;amp;usg=__jxZLExBMFCfgBNlNxUgErShUQFs%3D&amp;amp;ved=0ahUKEwiNk7zr2OPLAhVDvoMKHcaLCS8QyjcISA&amp;amp;ei=aE35Vo2_O8P8jgTGl6b4Ag#imgrc=RA_Y0yYftWYOpM%3A&quot; rel=&quot;nofollow&quot;&gt;skirt.&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The joints between the bottom and shell, as well as the joint between the top and shell, need to be doubled due to the &lt;a href=&quot;https://books.google.com/books?id=5ODuCAAAQBAJ&amp;amp;pg=PA57&amp;amp;lpg=PA57&amp;amp;dq=pressure%20vessel%20design%20bending%20boundary%20layer&amp;amp;source=bl&amp;amp;ots=JYCoFZiQLJ&amp;amp;sig=BEtt_uVD5KPln2TPMm-ETdNLM-8&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwijueSR1uPLAhWruIMKHXW0CIQQ6AEIQTAG#v=onepage&amp;amp;q=pressure%20vessel%20design%20bending%20boundary%20layer&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;bending boundary layer&lt;/a&gt;.  This layer exists for $4\sqrt{R*t}$ from the actual joint.  The way to avoid this is to double the pressure vessel thickness from each of the above numbers for a uniform vessel thickness.  Again, you probably have to get a permit, and the PE who stamps the design will need to review these area especially&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For material selection I would recommend 304 Stainless steel - it is very common, easy to afford, and is considered biosafe per &lt;a href=&quot;http://www.astm.org/Standards/F899.htm&quot; rel=&quot;nofollow&quot;&gt;ASTM F899&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2016-03-28T18:21:20.007" LastActivityDate="2016-03-28T18:21:20.007" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8198" PostTypeId="1" CreationDate="2016-03-28T18:23:14.037" Score="4" ViewCount="48432" Body="&lt;p&gt;I am completing a question that is requiring calculation of the work done in a polytropic process. In the worked solutions the work equation seems to be reversed for the answer, as can be seen in the screenshot- the top row is p1v1-p2v2 insted of p2v2-p1v1. Any clue why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DeLRI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DeLRI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5780" LastEditorUserId="1632" LastEditDate="2016-03-28T19:05:37.583" LastActivityDate="2023-03-25T13:01:48.570" Title="Derivation of the equation of the work done in a polytropic process" Tags="|thermodynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8199" PostTypeId="1" CreationDate="2016-03-28T18:41:06.607" Score="5" ViewCount="130" Body="&lt;p&gt;To preface this question, almost all of my FEA experience has come from SolidWorks, so I haven't had to deal with these types of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm designing a truss structure in PATRAN composed of 1D line elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point, I have to do the following two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mesh each individual line to create nodes&lt;/li&gt;&#xA;&lt;li&gt;Apply properties (material, cross-section, stiffness) to curves&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sometimes I apply the properties first, but when I attempt to mesh PATRAN tells me &quot;Curve is in property &lt;em&gt;xyz&lt;/em&gt; which does not support Bar2.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I mesh first, but when I analyze the truss my results file is sometimes empty. I can't tell what distinguishes when it works and when it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which order should I do these?&lt;/p&gt;&#xA;" OwnerUserId="5781" LastActivityDate="2024-03-30T02:04:20.493" Title="PATRAN Order of Operations" Tags="|finite-element-method|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8200" PostTypeId="2" ParentId="8195" CreationDate="2016-03-28T19:03:04.040" Score="6" Body="&lt;p&gt;After playing around with it for a while, this seems to work:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In the Batch Plot dialog, add all the drawings/sheets you want to include:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y57FU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y57FU.png&quot; alt=&quot;Batch Plot dialog&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture above, you can see that different drawings have different default Page Setups.  None of these drawings have a page setup I can use.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Select the first drawing, and under Page Setup, select &quot;Import...&quot;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cbEuH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cbEuH.png&quot; alt=&quot;Select &amp;quot;Import...&amp;quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Choose a DWG file that does contain a page setup you like.  In my case, I'm looking for one that will plot the Extents of the model using the monochrome.ctb plot style.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your imported Page Setups are now available in the list.  Select the one you want.  In my case, I'm choosing the one called &quot;Imported: 11x17&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WYydj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WYydj.png&quot; alt=&quot;Select the Page Setup&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Select this Page Setup for each drawing.  You can select all the sheets at once (Ctrl-A, or Shift-Click) and then choose a page setup.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Save the Sheet List.  This will preserve your imported Page Setup for the entire list, so you won't need to do these steps in the future for these drawings.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Click the Publish button to plot the drawings.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This procedure doesn't make any changes to the DWG files.  In the future, you can load the Sheet List, and the imported Page Setups will still be selected.&lt;/p&gt;&#xA;" OwnerUserId="37" LastActivityDate="2016-03-28T19:03:04.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8202" PostTypeId="2" ParentId="8198" CreationDate="2016-03-28T19:05:02.660" Score="7" Body="&lt;p&gt;Stay away from running things by the rule book, you could have done the calculation yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W=\int_{1}^{2} pdV$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$pV^n=const=C$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W=\int_{1}^{2} \frac{C}{V^n}dV$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W= \left[\frac{C \cdot V^{1-n}}{1-n}\right]_1^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W= \frac{C \cdot V_2^{1-n}}{1-n}-\frac{C \cdot V_1^{1-n}}{1-n}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;multiply with $\frac{-1}{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W= \frac{- C \cdot V_2^{1-n}}{n-1}+\frac{C \cdot V_1^{1-n}}{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W= \frac{C \cdot V_1^{1-n}}{n-1}-\frac{C \cdot V_2^{1-n}}{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;replace with $CV_i^{-n}=p_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W= \frac{p_1 \cdot V_1- p_2 \cdot V_2}{n-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went the longer way instead of commenting on the switched denominator to add some value for other readers to this question.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2016-03-28T19:05:02.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8203" PostTypeId="1" CreationDate="2016-03-29T01:00:54.527" Score="4" ViewCount="3457" Body="&lt;p&gt;I want to calculate the force-displacement equation for a beam that is fixed at both ends. I think I should use the superposition principle but I am not very sure how to get started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that for, say, a beam with one fixed-end and applied load $P$, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;EI \frac{d^2v}{dx^2}&amp;amp;=\mu_0\\&#xA;EI \frac{dv}{dx}&amp;amp;=\mu_0x+C_1\\&#xA;EI v(x)&amp;amp;=\mu_0x^2/2+C_1x+C_2&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I solve this for a fixed beam on &lt;strong&gt;&lt;em&gt;both&lt;/em&gt;&lt;/strong&gt; ends?&lt;/p&gt;&#xA;" OwnerUserId="5785" LastEditorUserId="368" LastEditDate="2016-03-30T17:08:16.467" LastActivityDate="2016-03-30T17:08:16.467" Title="Deriving the deflection force equation for a beam that is fixed on both ends" Tags="|civil-engineering|beam|deflection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8204" PostTypeId="2" ParentId="8184" CreationDate="2016-03-29T02:03:56.667" Score="0" Body="&lt;p&gt;First, &lt;a href=&quot;https://en.wikipedia.org/wiki/Conservation_of_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;conservation of energy&lt;/a&gt; says that &lt;strong&gt;power input equals power output&lt;/strong&gt;. For all intensive purposes, &lt;strong&gt;the grid does not store any power&lt;/strong&gt;. Even if there were battery systems tied to the grid, they would be treated just like power generators and not really be part of the grid because of the DC to AC conversion process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Utilities in first world countries hold &lt;strong&gt;tight tolerances on AC grid power&lt;/strong&gt;. By design, the &lt;strong&gt;voltage and frequency change very little, regardless of the amount of power being transferred&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the United States a frequency 60Hz is held perfectly to the tolerance of the atomic clock in Colorado. Many low cost alarm clocks rely on this for time keeping. This very constant frequency allows AC motors to have a constant RPM, transformers to have predictable performance, and permits for much easier management of grid power over long distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To maintain this frequency, the utilities closely monitor and control the rpm of their turbines with with computer control algorithms. For example, if it cools off mid-day and lots of thermostatically controlled air conditioners turn off; the load on the system decreases. The computer algorithm will then quickly see that it needs to decrease steam input to the turbine to reduce power output in order to maintain 60Hz. It works just like &quot;cruise control&quot; in a car; it presses down on the accelerator as needed to maintain your set speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Voltage is also kept very constant; with consumer voltage ranging between 115VAC and 120VAC. In my experience the voltage will be different for different locations, but less than 0.1VAC variance at a location when not in a transient state. Regulation of the voltage is a little more complicated, but also computer controlled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from a power production stand point, no turbine ever needs to be turned &quot;off&quot;, but &lt;strong&gt;turbines on the grid do need to be throttled up and down as the demand load changes&lt;/strong&gt;. The logistics of this may result in some turbines being shut down because it is often more economic to operate a single turbine than it is to operate two turbines at lower output. For example one running at 60% capacity instead of two running at 30% capacity. Turbine operators and computer algorithms &lt;a href=&quot;https://en.wikipedia.org/wiki/Electric_power_transmission#Control&quot; rel=&quot;nofollow noreferrer&quot;&gt;smoothly handle these transitions&lt;/a&gt; so an end power user never notices a difference. There is always excess capacity ready to handle changes in load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the photo below (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Electrical_grid&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki&lt;/a&gt;) you can see many types of power production interconnected. Nuclear shields its fuel rods to regulate steam production, natural gas and coal reduce fuel input to regulate steam production, and hydro uses low loss valves to regulate water flow. Solar, with the proper electronics could regulate its output, but generally does not because it coincides with peak air conditioning loads and is always utilized 100%. &lt;a href=&quot;https://renewables.gepower.com/content/dam/gepower-renewables/global/en_US/documents/GEA30675A-IN%20Wind_1.7-103_Broch_R3_HR_NEW.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Modern wind turbines with adjustable pitch blades&lt;/a&gt; also throttle their production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0sFHO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0sFHO.png&quot; alt=&quot;Electricity Grid Schematic from wiki electrical grid article&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a controller went bad on a turbine attached to the main electrical grid, it would begin to lead (produce too much power) or lag(pull power from the grid). Once the amount of power it pulled or pushed was too much, it would trip its breakers because of over current and be safely disconnected from the grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible for an isolated single-turbine grid (like a back-country hydro turbine) with a control system failure to operate at too high of frequency or too high of voltage. This will destroy sensitive electrical equipment like DC power supplies and over speed AC motors. The main electrical grid however, has many systems in place to prevent that from happening.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-29T02:03:56.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8206" PostTypeId="1" CreationDate="2016-03-29T05:29:52.793" Score="0" ViewCount="57" Body="&lt;p&gt;&lt;strong&gt;How would one calculate the proper size of an electric motor to properly drive a soil sifter?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumptions:&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;Screen:&lt;/strong&gt;  flat, 18&quot;x24&quot;, 1/4&quot; to 3/4&quot; wire cloth (interchangeable screens), total weight (including frame) = 4 lbs&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Shake Info:&lt;/strong&gt; total screen movement/shake = 6&quot;, shake rate = 132 shakes per minute&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Load:&lt;/strong&gt;  15 lbs soil or gravel mix&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Other:&lt;/strong&gt; Belt driven&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt;  I'd hate to use a 3/4 hp motor when all I might need is a 1/50 hp motor.&lt;/p&gt;&#xA;" OwnerUserId="5778" LastEditorUserId="1586" LastEditDate="2016-04-04T16:58:14.893" LastActivityDate="2016-04-04T16:58:14.893" Title="How to calculate the right motor size &amp; type for a soil sifter?" Tags="|mechanical-engineering|motors|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8207" PostTypeId="1" AcceptedAnswerId="8225" CreationDate="2016-03-29T06:12:02.857" Score="4" ViewCount="498" Body="&lt;p&gt;I am designing a wind turbine to generate 1300 W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I modeled the output voltage of an AC generator using the following specifics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Number coil turns = 300.&lt;/li&gt;&#xA;&lt;li&gt;Copper wire gauge = 1.02 mm.&lt;/li&gt;&#xA;&lt;li&gt;Knowing the density of copper, I calculated the weight of the coil to be 4.39 kg.&lt;/li&gt;&#xA;&lt;li&gt;Magnetic field strength = 1.24 T. (&lt;a href=&quot;http://www.arnoldmagnetics.com/Portals/0/Files/Catalogs%20and%20Lit/Neo/151021/N38AH%20-%20151021.pdf?ver=2015-12-07-103811-373&quot; rel=&quot;nofollow&quot;&gt;here are the specs for the magnet&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Area = 1 m&lt;sup&gt;2&lt;/sup&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Rotational speed = 45 RPM.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I plugged these numbers into the output voltage equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon = NBA\omega\sin(\omega t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get a value of 1753 V. I am assuming that my generator will generate at least the 0.74 A to output a power of 1300 W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I modeled my turbine. I used the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_T = \frac{1}{2}\rho A v^3 C_p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to calculate rotor power. I assumed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;My turbine would be 30% efficient.&lt;/li&gt;&#xA;&lt;li&gt;The wind velocity would be 10 m/s.&lt;/li&gt;&#xA;&lt;li&gt;The area swept by blades would be 2.63 m&lt;sup&gt;2&lt;/sup&gt;.&lt;/li&gt;&#xA;&lt;li&gt;The density of air would be 1.25.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;After plugging these values in, I get a power yield of 493 W. However, I find that the rotor speed is well above 45 RPM. This equation gives the tip speed ratio of the turbine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda = \frac{wr}{v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that the optimum tip speed ratio of a three-blade turbine is 4.2, radius of the rotor blade is 0.915 m, and velocity of air is 10 m/s, that angular velocity would be 45.9 rad/s or 440 RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I get 1300 W from this turbine when the theoretical yield from power of the rotor is only 493 W, even though my rotor should turn at the correct RPM for the generator to produce 1300 W?&lt;/p&gt;&#xA;" OwnerUserId="5742" LastEditorUserId="368" LastEditDate="2016-03-30T17:20:41.730" LastActivityDate="2016-03-30T23:14:28.463" Title="Why does my model of a wind turbine show high RPM with low power output?" Tags="|mechanical-engineering|turbines|wind-power|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8208" PostTypeId="1" CreationDate="2016-03-29T07:11:29.587" Score="1" ViewCount="274" Body="&lt;p&gt;I'm working on a project to design a shell and tube heat exchanger. There's a mixture of 68.4% C,&lt;sub&gt;3&lt;/sub&gt;H&lt;sub&gt;6&lt;/sub&gt; and 31.4% C&lt;sub&gt;3&lt;/sub&gt;H&lt;sub&gt;8&lt;/sub&gt; being cooled from 135 $^\circ$F and 321 psia to 100 $^\circ$F and 285 psig. The volumetric flow rate of the hydrocarbon stream is 200 GPM. The cooling water on the outside is initially at 85 F and 60 psig and exit pressure of 50 psi. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if my process is correct, but I found the average of the inlet and outlet pressure and temperature in the hydrocarbon stream and then used those values in the ideal gas equation to find the densities of each hydrocarbon, I then took the average of the two densities and multiplied by the volumetric flow rate to find the mass flow rate of the hydrocarbon stream. I was then going to use the equation &#xA;$$Q =m^o C_{p1} (T_2-T_1)(0.684) + m^oC_{p2}(T_2-T_1)(0.314)$$ to find the heat transfer of the hydrocarbon stream and then balance it with the water stream. I'm not exactly sure how to find the specific heats of the hydrocarbons. I checked perry's handbook but couldn't find anything at the average temperature. I'm just looking for any advice on whether or not this would be a correct procedure and if so, where could I find the specific heats of the hydrocarbons.&lt;/p&gt;&#xA;" OwnerUserId="4785" LastEditorUserId="90" LastEditDate="2016-03-29T08:59:49.327" LastActivityDate="2016-03-29T08:59:49.327" Title="Heat Exchanger Design - finding necessary variables" Tags="|mechanical-engineering|heat-transfer|chemical-engineering|heat-exchanger|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8209" PostTypeId="2" ParentId="8208" CreationDate="2016-03-29T07:50:56.177" Score="2" Body="&lt;p&gt;Between 135 F and 100 F I don't think the heat capacity or density would vary much so using average properties should be fine. If you want to be more conservative you can use the higher Cp between inlet and outlet as this would yield the highest cooling water flow requirement. If you want to be more accurate for whatever reason then you could integrate the function Cp(T) from T1 to T2. The equation should be quite easy to integrate directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that mass in should be equal to mass out, ie volume flow*density at the inlet and outlet should be equal. Since you only have one number for volumetric flow rate, I would use the higher density of the two (inlet/outlet) to calculate the mass flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific heats of hydrocarbons as functions of temperature should be in Perry's somewhere at the back. Otherwise, I like using NIST Chemistry Webbook.&lt;/p&gt;&#xA;" OwnerUserId="1103" LastActivityDate="2016-03-29T07:50:56.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8210" PostTypeId="2" ParentId="8208" CreationDate="2016-03-29T08:10:58.587" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I found the average of the inlet and outlet pressure and temperature in the hydrocarbon stream and then used those values in the ideal gas equation to find the densities of each hydrocarbon&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why would you find the density of each hydrocarbon? you now have a &lt;strong&gt;mixture&lt;/strong&gt; flow and you need to find the thermal properties of this mixture for your heat transfer calculations (at the average temperatures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Density (to calculate mass flow rate):&#xA;$$ \frac{1}{\rho _{mix}} = \sum \frac{x_i}{\rho _i} $$&#xA;where $x$ is the mass fraction of component i in the mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: since the flow mass rations are close to 70% and 30%, taking arithmetic average of densities won't be quite accurate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Specific Heat:&#xA;$$Cp_{mix} = \sum x_i Cp_i$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally the equation becomes:&#xA;$$Q = m^o_{mix} Cp_{mix} (T_2 - T_1) $$&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2016-03-29T08:10:58.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8211" PostTypeId="2" ParentId="8203" CreationDate="2016-03-29T08:48:42.443" Score="2" Body="&lt;p&gt;Consider a horizontal beam of length $L$ along the x-axis with vertical displacement $v$. Also consider that is has an applied load $P$ at $\frac{L}{2}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{d^2v(x)}{dx^2} = \dfrac{M(x)}{EI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with finding the moment in terms of x. You should know that the moment reaction on the left end is $-PL/8$ and the force reaction is $P/2$ so for $ 0 \le x \le L/2$ the moment function is:&#xA;$$M(x) - \dfrac{PL}{8} + \dfrac{P}{2}x = 0$$&#xA;$$M(x) = \dfrac{PL}{8} - \dfrac{P}{2}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate once:&#xA;$$\dfrac{d^2v(x)}{dx^2} = \dfrac{PL}{8EI} - \dfrac{P}{2EI}x$$&#xA;$$\dfrac{dv(x)}{dx} = \theta(x) = \dfrac{PL}{8EI}x - \dfrac{P}{4EI}x^2 + C_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the slope, $\theta(x)$, at either end is zero because the ends are fixed which gives us the condition $\theta(0)=0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta(0) = 0 + 0 + C_1$$&#xA;$$\therefore C_1=0$$&#xA;$$\therefore \theta(x) = \dfrac{PL}{8EI}x - \dfrac{P}{4EI}x^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrate again:&#xA;$$v(x) = \dfrac{PL}{16EI}x^2 - \dfrac{P}{12EI}x^3  + C_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know the displacement at either end is zero so that gives the condition $v(0)=0$:&#xA;$$v(0) = 0 + 0 + C_2$$&#xA;$$\therefore C_2 = 0$$&#xA;$$\therefore v(x) = \dfrac{PL}{16EI}x^2 - \dfrac{P}{12EI}x^3$$&#xA;$$ = \dfrac{Px^2}{48EI}(3L-4x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation only applies between $ 0 \le x \le L/2$ but the example is symmetric.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2016-03-29T10:24:53.150" LastActivityDate="2016-03-29T10:24:53.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8216" PostTypeId="2" ParentId="8190" CreationDate="2016-03-29T14:25:02.110" Score="1" Body="&lt;p&gt;I'm going to derive the transfer function symbolically from the differential equations you provided in the hopes that you might be able to extract the rotational moment of inertia (not the mass directly, you'd have to know the radius of the rotating masses) from the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the equations of motion as I interpreted them from your comment. J1 is the mass that the torque is applied to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A75ea.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A75ea.gif&quot; alt=&quot;Equation of motion for mass 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S6c3W.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S6c3W.gif&quot; alt=&quot;Equation of motion for mass 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the Laplace transform and writing in matrix form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XUxXD.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XUxXD.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will use Cramers' method to determine the transfer function for mass 1, the mass to which the torque is applied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3RHWA.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3RHWA.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This transfer function Theta1/tT can be expressed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pn1Rr.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pn1Rr.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing this to your experimentally determined transfer function indicates that the rotational moment of inertia for mass 2 is 4700, and d is 4393. Using algebra you can solve for J1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/phUtZ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/phUtZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for J1 should give about 672.3891.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to check my math, here is a simple Matlab code I used:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;j1 = sym('J1');%moment of inertia mass 1&#xA;j2 = sym('J2');%moment of inertia mass 2&#xA;d = sym('d');%damping coefficient&#xA;k = sym('k');%spring constant&#xA;t = sym('T');%applied torque&#xA;s = sym('s');%laplace variable&#xA;&#xA;%system matrix&#xA;A = [j1*s^2+d*s+k -(d*s+k);-(d*s+k) j2*s^2+d*s+k]&#xA;&#xA;det(A)&#xA;&#xA;Acramer = [t -(d*s+k);0 j2*s^2+d*s+k]&#xA;&#xA;det(Acramer)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1174" LastEditorUserId="1174" LastEditDate="2016-03-29T17:03:43.130" LastActivityDate="2016-03-29T17:03:43.130" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8221" PostTypeId="1" AcceptedAnswerId="8325" CreationDate="2016-03-30T12:06:20.637" Score="10" ViewCount="375" Body="&lt;p&gt;I am making a testing facility for pumps. This facility has to be able to handle 500 m&lt;sup&gt;3&lt;/sup&gt;/h with at little turbulence as possible. The basin itself is 5 meters long, 2,5 meters wide and 2 meters tall. The water level in the basin is kept at 1,9 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basin is designed in two sections: the first is where the discharge flow comes into the basin (pipe ends about 1 meters in from the bottom). In that area the water flows freely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that it has to flow over the retaining wall (1.7m high), which should create a somewhat uniform flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5IV2J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5IV2J.png&quot; alt=&quot;overview&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there better solutions to do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other idea I have is a retaining wall somewhat like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0tz0T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0tz0T.png&quot; alt=&quot;wall idea&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes for three plates with cutout stripes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind this is to have a more uniform flow across the entire basin instead of a overflow, making it create a small bit of turbulence there.&lt;/p&gt;&#xA;" OwnerUserId="5799" LastEditorUserId="5799" LastEditDate="2016-03-31T11:14:40.150" LastActivityDate="2016-04-07T00:29:19.257" Title="Best way to get rid of turbulent flows?" Tags="|water-resources|turbulence|flow-control|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8222" PostTypeId="1" AcceptedAnswerId="8223" CreationDate="2016-03-30T13:44:17.563" Score="1" ViewCount="585" Body="&lt;p&gt;I have the situation from the following drawing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gHSf2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gHSf2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It represents a model for measuring the torque of a generator. &#xA;There is an arm connected to stator and, the other end of the arm to a bending type load cell. I want to model this system and have as an input the torque of the generator rotor and as output the torque that reaches the load cell. I want to model somehow how is the transfer of energy from the rotor to the load cell.&#xA;The load cell type is Hottinger Z6FC3. &#xA;The system is connected like in the figure below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q7XuX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q7XuX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new at this and don't know much mechanics. I don't know how to write the motion equations for the two bodies and connect them.&lt;/p&gt;&#xA;" OwnerUserId="5744" LastEditorUserId="5744" LastEditDate="2016-03-31T12:56:10.653" LastActivityDate="2016-03-31T15:02:43.283" Title="motion equation of rotor and stator" Tags="|mechanical-engineering|torque|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8223" PostTypeId="2" ParentId="8222" CreationDate="2016-03-30T15:28:36.170" Score="1" Body="&lt;p&gt;You need to apply &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow&quot;&gt;Newton's second law&lt;/a&gt; that states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Second law&lt;/strong&gt;:   The vector sum of the forces F on an object is equal to&#xA;  the mass m of that object multiplied by the acceleration vector a of&#xA;  the object: F = ma.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In your case, you need to replace force by torque, mass by inertia and acceleration by angular acceleration. Given that the torque provided by the rotational springs is the spring stiffness x the relative angle and that the torque provided by the rotational dampers is (in the first instance) the damping coefficient x the relative angular velocity, you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ T = (J_s + J_a) \ddot{\theta_r} + C_m (\dot{\theta_r} - \dot{\theta_s}) + K_m (\theta_r - \theta_s)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ J_{Gs} \ddot{\theta_s} + C_{Gs} \dot{\theta_s} + K_{Gs} \theta_s = C_m (\dot{\theta_r} - \dot{\theta_s}) + K_m (\theta_r - \theta_s)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\theta_r$ and $\theta_s$ are the rotational displacements of the rotor and the stator, respectively.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2016-03-31T08:11:19.860" LastActivityDate="2016-03-31T08:11:19.860" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8224" PostTypeId="1" AcceptedAnswerId="8233" CreationDate="2016-03-30T16:35:47.030" Score="12" ViewCount="379" Body="&lt;p&gt;We have obtained an old mercury-vapor &lt;a href=&quot;https://en.wikipedia.org/wiki/Diffusion_pump&quot;&gt;diffusion pump&lt;/a&gt; for high vacuum. However, we do not want to operate it with mercury in our lab due to health concerns (which is presumably the reason it was scrapped by the original owner). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, we plan to fill it with a silicone oil of the &lt;em&gt;DC-704&lt;/em&gt; type or similar. The pump has thermal regulation and a large baffle to stop oil vapor counter-propagation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anybody tried replacing mercury with oil? What are possible problems with this approach? &lt;/p&gt;&#xA;" OwnerUserId="5804" LastEditorUserId="368" LastEditDate="2016-03-30T17:03:04.037" LastActivityDate="2016-03-31T02:17:01.593" Title="Replace mercury with silicone oil in diffusion pump" Tags="|pumps|safety|vacuum|vacuum-pumps|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8225" PostTypeId="2" ParentId="8207" CreationDate="2016-03-30T20:03:50.443" Score="3" Body="&lt;p&gt;I don't think you really designed a 1300 W wind turbine. You &lt;em&gt;assumed&lt;/em&gt; your generator would be able to produce the necessary current but to be able to produce 1300 W you have to build a wind turbine big enough, given its efficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $C_p=0.3$ and $v=10\ \mathrm{m/s}$, you simply need an area of:&#xA;$$A=\frac{2P_T}{C_pv^3\rho}=7.07\ \mathrm{m^2}$$&#xA;$$\Rightarrow R=1.5\ \mathrm{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, you say $\lambda=4.2$ is optimal for a three-blade turbine, but it actually depends on the blade profile, which is optimized for a given &quot;design tip speed ratio&quot;. For more information about this, see chapter 3.7 of &quot;Burton, &lt;em&gt;Wind energy handbook&lt;/em&gt;, 2011&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &quot;Duquette, &lt;em&gt;Numerical Implications of Solidity and Blade Number on Rotor Performance of Horizontal-Axis Wind Turbines&lt;/em&gt;, 2003&quot; about influence of blade number on power coefficient.&lt;/p&gt;&#xA;" OwnerUserId="5806" LastEditorUserId="368" LastEditDate="2016-03-30T23:14:28.463" LastActivityDate="2016-03-30T23:14:28.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8226" PostTypeId="2" ParentId="8193" CreationDate="2016-03-30T20:31:07.110" Score="2" Body="&lt;p&gt;Steam engines can expand the steam only so much before the water vapor starts to condense.  In addition, the volume of the vapor increases significantly as it expands, making it difficult to obtain further work, as the cylinder size must increase to accommodate the volume of vapor passing through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most steam engines use three or four cylinders to expand the steam, and are referred to as &quot;triple expansion&quot; engines.  The steam enters the smallest, or high pressure cylinder, first out of the boiler.  Then, it leaves the HP cylinder and enters an intermediate pressure cylinder next, which is physically larger than the HP cylinder.  Finally, the steam enters the low pressure cylinder, which may be split into two cylinders because of the large volume of steam which must be handled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a diesel engine does not expand the combustion gases in multiple stages like a steam engine.  The fuel is injected and burned all in one cylinder.  Large diesel engines can therefore have their power output increased by adding additional cylinders.  These engines are usually designed in a modular fashion to make it relatively easy to increase the number of cylinders during construction.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-03-30T20:31:07.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8227" PostTypeId="1" CreationDate="2016-03-30T20:58:40.563" Score="0" ViewCount="602" Body="&lt;p&gt;I have an engineering background, I'm an EE, but I know almost nothing about mechanics and that's why I'm writing here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the challenge: how to control the rotation of multiple discs along the same axis? Assume you have a sliced cylinder. The ideal scenario would be to control every slice independently, but to make things easier I think it helps to consider that you can move them in a fixed number of combinations and on the same direction (e.g. a mechanism that enables to rotate the top and bottom cylinders at the same time and same direction - combination #1 - and the inner ones in the same way - combination #2 - but not as the same time as #1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea on how to accomplish this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I though I could use something like a &quot;key&quot; along the axis/shaft/rod. By moving it up/down it would lock different discs/slices. Not sure if this is a good idea. One issue I can see with this one, is that the slices that weren't locked by the &quot;key&quot; could be rotating freely and I want them to be steady.&lt;/p&gt;&#xA;" OwnerUserId="5807" LastActivityDate="2016-04-01T11:27:38.143" Title="Control the rotation of multiple discs along the same axis" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8228" PostTypeId="1" CreationDate="2016-03-30T21:47:58.830" Score="2" ViewCount="749" Body="&lt;p&gt;I am starting to learn how to use ANSYS APDL. So I don't have much knowledge in the software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have modelled a cantilever beam (with beam188 elements) and the only load that i have applied was the self weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I ask for the shear force diagram (element table) and plot the results, the SFD looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jc5Rn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jc5Rn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't it be linear? What happened?&lt;/p&gt;&#xA;" OwnerUserId="5808" LastEditorUserId="1832" LastEditDate="2016-03-31T02:07:48.453" LastActivityDate="2017-07-27T12:15:30.457" Title="Problems with shear force diagram in beam model" Tags="|beam|ansys|diagram|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8229" PostTypeId="2" ParentId="8196" CreationDate="2016-03-30T22:41:24.980" Score="0" Body="&lt;p&gt;In addition to the excellent answers on pressure vessels already posed a steam boiler has the additional complication that, as you are dealing with a change in phase under pressure there is a particular hazard with steam that you can get very rapid pressure spikes under certain conditions so the safety regulations for steam boilers are rather more stringent than for pressure vessels in general. For example interruption of the flow of water may cause a rapid pressure rise which may not be immediately apparent see &lt;a href=&quot;https://en.wikipedia.org/wiki/Steam_explosion&quot; rel=&quot;nofollow&quot;&gt;steam explosion&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will also find that your jurisdiction requires specific welding codes for welding of steam pressure vessels and/or individual testing and inspection. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-03-30T22:41:24.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8230" PostTypeId="2" ParentId="8193" CreationDate="2016-03-30T23:48:08.030" Score="4" Body="&lt;p&gt;The maximum efficiency of either the Diesel or a steam engine cycle  is (assuming you can treat steam as an ideal gas): $$\eta_{Diesel} = 1 - \frac{1}{(\frac{V_{Max}}{V_{Min}})^{k-1}} \left[\frac{r_c^k-1}{k(r_c-1)}\right] $$&#xA;Where $k$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_capacity_ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;heat capacity ratio&lt;/a&gt;, and $r_c$ is the cutoff volume ratio - i.e. the ratio of volumes after and before the combustion process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few things to note in this formula - Clearly it would seem that having a long cylinder stroke would be advantageous either way.  However, this would also require a very large engine, as the piston rod has to be longer.  Going to the second part, for a standard diesel combustion, the volume may expand to double at the most (i.e. $r_c = 2$,making the bracket = 1.17) between the onset of combustion and the end of combustion.  For steam, this isn't really the case.  As can be seen from this handy diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lQDWc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lQDWc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cutoff ratio approaches 20 before constant pressure injection is finished.  This makes the bracket about 2.0, really impeding efficiency.  However, if you've got the space for a boiler, condenser, reservoir and pump, you've got space for a really long piston rod to bump up the bottom term and bring the efficiency back to a reasonable number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what does this mean in terms of raw power?  $$\mathscr P = \eta*Q_{in} = \eta * \dot m C_v (T_{expand} - T_{comp})$$  Where $\dot m$ is the mass flow rate, $C_V$ is a constant based upon the gas used, and the $T$'s are the temperatures at the end of compression and the beginning of expansion, respectively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we have a similar efficiency, now the hard part is temperature.  A steam engine can't reach 1000K like a diesel can.  To makeup for the lack of power, it needs a lot of mass flow - big cylinders and double action really help. For the diesel, simply adding more of the already compact, high power, and fairly efficient engines will be easier than making a single big cylinder and dealing with the difficulties (such as the engine losing isoentropic compression / expansion), and removing that excessive heat from many distributed points (rather than a single massive point) makes a lot more sense due to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Square-cube_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;square cube law&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to summarize:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Higher cutoff ratio means steam engines need longer piston rods to maintain a higher efficency&lt;/li&gt;&#xA;&lt;li&gt;Since the temperature is higher, the diesel pistons need to be smaller diameter to allow for proper cooling to avoid damaging the materials.&lt;/li&gt;&#xA;&lt;li&gt;Since each &lt;em&gt;small&lt;/em&gt; diesel piston is able to perform efficiently and powerfully, it makes more sense to add a bunch of them then dealing with a single large chamber with inefficiencies.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2016-03-30T23:48:08.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8232" PostTypeId="2" ParentId="8227" CreationDate="2016-03-31T01:19:53.227" Score="1" Body="&lt;p&gt;I think you are looking for an &lt;a href=&quot;https://en.wikipedia.org/wiki/Electromagnetic_clutch&quot; rel=&quot;nofollow noreferrer&quot;&gt;electromagnetic clutch&lt;/a&gt;. There are obviously other types of clutches that would work, but electric ones are nicer when there are lots of units to control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z9WVx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z9WVx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;As far as keeping the other devices from freewheeling you could add some acceptable amount of friction, or you could add an &lt;a href=&quot;https://en.wikipedia.org/wiki/Electromagnetic_brake&quot; rel=&quot;nofollow noreferrer&quot;&gt;electromagnetic brake&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pVM46.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pVM46.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And like you mentioned, depending on your loads and system complexity it may be easier to directly drive each one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are probably other more creative ways to address your design, but you will have to share more of the details.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="86" LastEditDate="2016-03-31T01:26:58.170" LastActivityDate="2016-03-31T01:26:58.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8233" PostTypeId="2" ParentId="8224" CreationDate="2016-03-31T02:17:01.593" Score="4" Body="&lt;p&gt;I did some internet searching and did not find anyone that had done a direct swap out. Granted, probably the majority of these replacements occurred before the existence of the internet ;-). As I am sure you are already aware (but to be thorough), modern diffusion pump designs do use &lt;a href=&quot;http://www.vpcinc.ca/Products/Lubricants/Diffusion_Fluids.html&quot;&gt;synthetic working fluids&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will probably take some fine tuning of the temperatures of the heater and cooler, but I don't see any reason why it wouldn't work. If you have means of measuring the vacuum level, you could run through some different temperatures to empirically find its new ideal operating conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst case, the pumping efficiency will be less because the nozzle diameter was designed for higher density particles. I am not a vacuum physicist however, so who knows, it may work even better than before ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible issue is that it may have more &lt;a href=&quot;http://www.mtm-inc.com/av-20121127-diffusion-pumps.html&quot;&gt;backstreaming&lt;/a&gt; of the oil into the vacuum chamber resulting in unacceptable levels of oil contamination. Maybe the baffle you mentioned will take care of this; hard to say without some experimentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as long as no oxygen reaches the oil during operation, it wont fail catastrophically, so give it a try and let us know how it went!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference:&lt;br&gt;&#xA;&lt;a href=&quot;http://www.xtronix.ch/PDF/Diffusion%20Pump.pdf&quot;&gt;Good overview of diffusion pump working fluids&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://vacaero.com/information-resources/vacuum-pump-practice-with-howard-tring/1403-oil-diffusion-pump-controls.html&quot;&gt;Some good history or mercury as a diffusion pump working fluid&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-03-31T02:17:01.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8234" PostTypeId="1" AcceptedAnswerId="10507" CreationDate="2016-03-31T07:55:06.040" Score="4" ViewCount="974" Body="&lt;p&gt;&lt;strong&gt;Important note&lt;/strong&gt;: I am writing a structural FEM software from scratch, so I won't able to use existing commercial FEM package. I would need to know how all the important quantities are calculated from the first principles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would need to determine the &lt;strong&gt;accidental torsional effect&lt;/strong&gt; ( &lt;a href=&quot;https://drive.google.com/file/d/0B5oarfYUwEDrY1NBMU52M3FHVDA/view&quot; rel=&quot;nofollow noreferrer&quot;&gt;UBC97, 1631.5.6, pp-27&lt;/a&gt;, &lt;strong&gt;&lt;em&gt;this accidental torsion effect is needed for both response spectrum and linear time history analysis&lt;/em&gt;&lt;/strong&gt;) for seismic dynamic analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_t=F_xe_y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_x$ is the equivalent lateral static force in $x$ direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e_y$ is the accidental eccentricities in $y$ direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take note that all these involved quantities are evaluated &lt;strong&gt;&lt;em&gt;on floor basis&lt;/em&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My structures are quite irregular and general, such as this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/z6Mu3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a cantilever column or an one dimensional lumped mass model, the term &lt;strong&gt;on floor basis&lt;/strong&gt; is easy to interpret because there is only 1 node per floor, and the $F_x$ can only be applied on this node itself. Compare this lumped mass model with the above 2D structure FEM model and you can see the difference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SauGW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SauGW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when you have such a 2 dimensional irregular structure, analyzed using FEM, calculating $F_x$ and $M_t$ is no longer easy:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How should we obtain $F_x$, given that it is lateral &lt;em&gt;static&lt;/em&gt; force, but we are doing dynamic analysis?&lt;/li&gt;&#xA;&lt;li&gt;There are so many nodes on a floor, so on which node(s) should the $M_t$ be applied? &lt;/li&gt;&#xA;&lt;li&gt;In FEM, each beam, column, wall and slab are multiple-node elements. And each node has a internal force value  resulted from FEM analysis, what should I do with all these forces at different nodes? &lt;a href=&quot;https://engineering.stackexchange.com/questions/7662/how-to-derive-equivalent-static-load-for-irregular-structure-modeled-with-fem#comment14210_7673&quot;&gt;The comment here&lt;/a&gt; says that I should sum up all of the forces on each related node in order to obtain the correct $F_x$ ( I'm not sure whether I interpret it correctly or not). But I am highly skeptical of this approach, because won't that make $F_x$ dependent on the number of nodes, an artifact of how we mesh an element? How can a real physical quantity increases with the number of nodes we decide to mesh?&lt;/li&gt;&#xA;&lt;li&gt;If 3) is true, then what about the $F_x$ on the wall above a particular floor? Should we sum them in? What about the internal slab and beam? Do they have this force also?&lt;/li&gt;&#xA;&lt;li&gt;How should we intepret $M_t$? Is it internal member force to be resisted by beam/slab? If yes, if there are many slabs and beams into $x$ direction, then how should this $M_t$ be distributed? Or is it an external static force; so one would have to apply this external force on the structure in order to calculate the individual internal member force induced by it ( via the FEM equation $ku=F$)? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So the question is, how to calculate and interpret accidental torsion in a FEM model? I've tried to obtain a general way to reduce a 2 Dimensional general FEM structure into 1 Dimensional lumped mass model, so that I can apply the above formulation easily, but &lt;a href=&quot;https://engineering.stackexchange.com/a/7673/3353&quot;&gt;I was told that this wasn't possible&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-06-28T10:04:48.007" Title="Accidental Torsion and Equivalent Static Load for an irregular structure" Tags="|structural-engineering|structural-analysis|seismic|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8237" PostTypeId="1" CreationDate="2016-03-31T15:08:21.923" Score="2" ViewCount="64" Body="&lt;p&gt;Given a beam ($\rho, A, l$) rotating with angular velocity $\omega$, find the maximum stress in the beam and its location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mn9wd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mn9wd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the maximum stress occurs at the base, because the resultant force is the entire axial force. The total axial force is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = \int_0^l\frac{m}{l}\omega^2r\ dr = \frac{m\omega^2l}{2}$$&#xA;$$m = \rho Al$$&#xA;$$\sigma_{max} = \frac{F}{A} = \frac{\rho\omega^2 l^2}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I do this right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The follow-up question is, what happens when the beam is extremely thin: $t &amp;lt;&amp;lt; l$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to answer the follow-up.&lt;/p&gt;&#xA;" OwnerUserId="5818" LastActivityDate="2016-03-31T15:08:21.923" Title="Axially Distributed Load" Tags="|beam|" AnswerCount="0" CommentCount="2" ClosedDate="2016-04-01T13:42:55.233" ContentLicense="CC BY-SA 3.0" />
  <row Id="8238" PostTypeId="2" ParentId="8234" CreationDate="2016-03-31T15:14:10.133" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;How should we obtain $F_x$, given that it is lateral static force, but we are doing dynamic analysis?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$F_x$ is the sum of the lateral force for a story. If you've performed a dynamic analysis, this is computed (usually) by multiplying your nodal masses by their corresponding nodal accelerations. You should be able to easily output both these from your analysis results. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;There are so many nodes on a floor, so on which node(s) should the $F_x$ be applied?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question is ambiguous, but if you are asking which node $F_x$ would be applied to in order to capture accidental torsional effects (let's assume that you are using ELF since, as I mentioned above, you don't need to worry about accidental torsion if you are doing a dynamic analysis), the answer is, you don't apply $F_x$ to any node, really.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation, $$M_t = F_x e_y \tag{1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is only intended to be used to determine the magnitude of the torsional moment you need to account for in your design. It is not meant to be used to calculate a literal load you would directly apply to a model. In the past, we have accounted for torsional moments on structures (accidental and inherent) by calculating the torsional effect and then, via a rational process, applying loads to our model that would produce that torsional effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, lets say you have a very simple structure represented in your finite element model as a single 4 node, thin shell element. This structure sees a small torsional moment of 1 N/m. Your &quot;structure&quot; would look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;a href=&quot;https://i.stack.imgur.com/0wD7nm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0wD7nm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By decoupling the moment into a force applied at the 4 nodes, you might load your model like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;a href=&quot;https://i.stack.imgur.com/ic51R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ic51R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[Edit: Oops, drew arrows in wrong direction. Pretend the applied moment is rotating clockwise instead.]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no single correct way to account for any type of load in your models. An accidental torsional moment is no different. It is up to the engineer to use a rational process to determine how his/her mathematical models and methods will account for the loading applied to the structure being designed (in this case, accidental torsion).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;In FEM, each beam, column, wall and slab are multiple-node elements. And each node has a internal force value resulted from FEM analysis, what should I do with all these forces at different nodes? &lt;a href=&quot;https://engineering.stackexchange.com/questions/7662/how-to-derive-equivalent-static-load-for-irregular-structure-modeled-with-fem#comment14210_7673&quot;&gt;The comment here&lt;/a&gt; says that I should sum up all of the forces on each related node in order to obtain the correct $F_x$ ( I'm not sure whether I interpret it correctly or not). But I am highly skeptical of this approach, because won't that make $F_x$ dependent on the number of nodes, an artifact of how we mesh an element? How can a real physical quantity increases with the number of nodes we decide to mesh?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, proceed as I suggested the last time you asked this question. No, this doesn't make $F_x$ dependent on the number of nodes in your model. $F_x$, in this context, is dependent on the &lt;em&gt;tributary mass to each node.&lt;/em&gt; As you increase the number of nodes in your model, the amount of mass tributary to each node decreases (linearly), therefore, the force does not increase if you change the scale of your mesh discretization.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;4&quot;&gt;&#xA;  &lt;li&gt;If 3) is true, then what about the $F_x$ on the wall above a particular floor? Should we sum them in? What about the internal slab and beam? Do they have this force also?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This question is a little unclear and indicates a certain level of misunderstanding that I think is way beyond the scope of this answer. Refer to my answer to #2. It is up to the engineer to determine how to account for loading. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;5&quot;&gt;&#xA;  &lt;li&gt;How should we intepret $M_t$? Is it internal member force to be resisted by beam/slab? If yes, if there are many slabs and beams into xx direction, then how should this $M_t$ be distributed? Or is it an external static force; so one would have to apply this external force on the structure in order to calculate the individual internal member force induced by it ( via the FEM equation $ku=F$)?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please refer to my answer to #2. It is up to the engineer to determine how to account for loading. The torsional moment is resisted by the building structure as a whole. $M_t$ is not meant to simply be &quot;plugged in&quot; to your model/analysis.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I don't mean to sound mean spirited, but these are questions related to some very fundamental aspects of structural/stress analysis and code interpretation. My short answer here is in no way everything you need to know to understand the full implication and answers to your questions. I strongly urge you to consult with an experienced professional if these questions are the result of a real world analysis of something that will eventually be constructed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a quick glance at your included images suggests that you have a whole lot of inherent torsion in this structure as well. Make sure that your analysis accounts for that as well.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-04-05T00:19:14.723" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8239" PostTypeId="2" ParentId="2693" CreationDate="2016-03-31T16:41:06.290" Score="4" Body="&lt;p&gt;A subtle bit of wording - the exact length does not change, as pointed out by Air above in his answer.  However, the horizontal projection of the beam does change, as the shortest path between two points will always be a straight line.  Thus, by curving, the tip of the beam will have to move backwards a bit in the x direction, to account for the deflection in the y direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact change is minimal.  For a beam whose deflection $y(x)$ has been defined in terms of x, the change in &quot;length&quot; can be determined by the slope function $\theta(x) = y'(x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta L = -\frac{1}{2}\int^L_0(\theta(x))^2dx  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0071742476&quot; rel=&quot;nofollow&quot;&gt;(Ref Roark's formulas for stresses and strains, 8th edition)&lt;/a&gt;, Eq. 8.1-14.  Note it is always a shrinking of the beam.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2016-03-31T16:41:06.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8240" PostTypeId="1" CreationDate="2016-03-31T17:19:58.097" Score="4" ViewCount="198" Body="&lt;p&gt;On a different Stack, I encountered an answer that was making a comparison to nuclear powered naval ships. It was stated that the physical shielding being used was various kinds of metals. This seems... odd to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that water is probably being used to cool the reactors in ships. But, water is also excellent at shielding. Considering how abundant water is for the environments that naval ships operate in, do the ships use water for shielding? And if not, why?&lt;/p&gt;&#xA;" OwnerUserId="5822" LastActivityDate="2016-11-12T16:01:55.153" Title="Do ships powered by nuclear reactors use water for shielding?" Tags="|nuclear-engineering|naval-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8241" PostTypeId="2" ParentId="8127" CreationDate="2016-03-31T17:37:12.777" Score="0" Body="&lt;p&gt;Is this theoretical.  If so I will bow out, but if not, Airtrax Sidewinder Forklift already does this.  See the video.  &lt;a href=&quot;https://www.youtube.com/watch?v=vAiwLRGsNrE&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=vAiwLRGsNrE&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;techtooler&lt;/p&gt;&#xA;" OwnerUserId="5823" LastActivityDate="2016-03-31T17:37:12.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8242" PostTypeId="1" CreationDate="2016-03-31T18:13:00.430" Score="1" ViewCount="44" Body="&lt;p&gt;Has anyone used Freescale's (now NXP semiconductors) small engine ECU development kit? If so, please give me pointers on which model to choose to tune a 50cc Yamaha GY6 engine. Also as I'm still in college will the difficulty be too high ( i.e, would I require High level knowledge). &lt;/p&gt;&#xA;" OwnerUserId="5824" LastActivityDate="2016-03-31T18:13:00.430" Title="Small enigne ECU self development" Tags="|automotive-engineering|" AnswerCount="0" CommentCount="2" ClosedDate="2016-03-31T21:25:40.167" ContentLicense="CC BY-SA 3.0" />
  <row Id="8244" PostTypeId="2" ParentId="8240" CreationDate="2016-03-31T19:51:08.903" Score="4" Body="&lt;p&gt;The primary use/purpose of water in light water reactors is neutron flux moderation and cooling/steam generation. Water is only used for shielding during re-fueling. In land based PWR's (pressurized water reactors) the thickness of water to effectively protect workers from the radiation in the core when it is open for refueling is around 23 feet. Basically there's a 23 foot swimming pool (don't go swimming in one) between the fuel rods and the workers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shielding with various metals of different thicknesses is &lt;em&gt;much&lt;/em&gt; more economical for a ship and a power plant. While there is a lot of water available at sea, there's no need to drag around enough water inside the ship to shield people from the radiation. You may as well sink the boat. Metal shielding is the way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider that nuclear powered naval ships do not often refuel, maybe once every 20 years or never. Since the core will not be opened, no need to provide a fluid barrier.&lt;/p&gt;&#xA;" OwnerUserId="1174" LastActivityDate="2016-03-31T19:51:08.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8246" PostTypeId="1" AcceptedAnswerId="8259" CreationDate="2016-04-01T00:56:08.217" Score="4" ViewCount="7362" Body="&lt;p&gt;I have a 10 mm diameter rod of 99.5% pure tungsten that I want to drill a 0.1 mm hole in using my Grizzly G0768 lathe. The hole only needs to be between 0.1 mm and 1 mm deep and is for a 3D printer hot nozzle. I don't know if this will break the bit yet, but I will try it when the drill chuck comes in the mail. If the bit breaks, I'll move up to a 0.15 mm hole, and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't afford a high precision mill that costs more than $1,000. Assuming my lathe can't drill such a small hole, what other options do I have to accomplish this task on a budget? When I have gone shopping I have seen no indication of the stability / precision of the tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could send the pieces of tungsten off to get drilled, but I expect that might cost a fortune.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has experience making holes this tiny, I would like to know the approach you used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have nice $12, 0.1 mm PCB bits made in Germany from Tungsten Carbide that break if you so much as look at them. I also found some cheaper bits on eBay but they don't look like drill bits so I'm not sure if they'd even cut a hole.&lt;/p&gt;&#xA;" OwnerUserId="3235" LastEditorUserId="368" LastEditDate="2016-04-01T21:54:52.460" LastActivityDate="2018-08-14T05:23:23.260" Title="How can I cheaply drill a 0.1 mm hole in a small piece of tungsten?" Tags="|machining|drilling|lathe|" AnswerCount="7" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="8247" PostTypeId="1" CreationDate="2016-04-01T01:32:17.227" Score="4" ViewCount="481" Body="&lt;p&gt;I have a piston of diameter 0.7 mm exactly and a 0.7 mm drill bit.  Let's call the diameter of the piston $r_1$, the diameter of a hole at the end of a tube $r_0$, and the final-actual diameter of the tube hole $r$.  Then with fluid being pushed by the piston through the hole at the end ($r_0$) I calculate that I need:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\pi (r^2 - r_1^2) \lt \pi r_0^2 \\&#xA;\iff r^2 \lt r_1^2 + r_0^2 \\&#xA;\iff r \lt \sqrt{r_1^2 + r_0^2}&#xA;$$&#xA;in order to prevent back-flow in the negative direction of the piston movement.  I'm ignoring gravity.  With $r_0 = $0.05mm, $r_1 = $0.35mm, I come up with:&#xA;$r \lt \sqrt{(0.05)^2 + (0.35)^2} \approx 0.3536$.  So can I get a hole that's between 0.7 and 0.7072 mm (preferably closer to 0.7 mm) using a 0.7 mm bit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's okay if the tolerance is so tight that the piston gets stuck at room temp, because the piston is a rod of continually fed material that will be melted in the tube.&lt;/p&gt;&#xA;" OwnerUserId="3235" LastEditorUserId="368" LastEditDate="2016-04-05T22:11:25.830" LastActivityDate="2017-08-24T22:01:09.303" Title="What diameter hole will a twist drill bit of a given size actually make in a piece of metal?" Tags="|machining|drilling|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8248" PostTypeId="2" ParentId="8246" CreationDate="2016-04-01T03:50:00.287" Score="2" Body="&lt;p&gt;There is an instability (wobbliness) caused by uneven chuck jaws in my G0768 lathe.  I read online about it and was able to correct it by incrementally inserting standard paper pieces until the chuck was aligned enough.  (OTOH, the spindle is perfectly aligned it seems, by eye - can't see any movement in it).  So doing that I was able to drill a 0.25mm hole 2-3mm deep before the bit broke.  So I reckon with tissue paper I can get down to 0.1mm.  However, the lathe came with a 4-jaw independent chuck, so I just have to install that :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, a new Grizzly G0768 is capable with a little effort of drilling a 0.1mm hole 1-2mm deep into a piece of 99.5% Tungsten using carbide or tungsten carbide bits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.  Was able to drill that 0.25mm hole with the cheap PCB drillbit sets found on ebay for 7-10 USD.&lt;/p&gt;&#xA;" OwnerUserId="3235" LastActivityDate="2016-04-01T03:50:00.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8249" PostTypeId="2" ParentId="8246" CreationDate="2016-04-01T08:25:24.973" Score="5" Body="&lt;p&gt;I am afraid that drilling tungsten will be very tough with any available drill bit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also consider laser machining (in my country, there is at least one company offering even small batches with a very affordable price, comparable to the price of a single 0.1 mm drill bit - which you would almost certainly trash anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, you could build your own electroerosive milling machine and fit it with 100 μm tungsten wire from a light bulb; I guess that you could use it in many other cases later.&lt;/p&gt;&#xA;" OwnerUserId="5804" LastActivityDate="2016-04-01T08:25:24.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8250" PostTypeId="1" AcceptedAnswerId="8251" CreationDate="2016-04-01T09:14:06.017" Score="6" ViewCount="7577" Body="&lt;p&gt;I am working on a project to bring light into a bunker. The bunker is 3-4 meters below the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking into bringing natural light into the space, instead of using electricity. The space is difficult to access and light must therefore be transported by flexible cable for about 20 meters from the surface to the areas I want to light up. I am thinking of pipe around 1 cm diameter maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking into a cheap solution to create an optical pipe that is effective at transporting light.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I am wondering what techniques are typically used.&lt;/li&gt;&#xA;&lt;li&gt;What kind of pipes would you use ?&lt;/li&gt;&#xA;&lt;li&gt;Is it possible to fill the pipe with other materials to increase its effectiveness ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5834" LastEditorDisplayName="user16" LastEditDate="2016-04-01T13:59:32.823" LastActivityDate="2019-02-18T16:39:14.547" Title="Create a DIY light tube" Tags="|fiber-optics|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8251" PostTypeId="2" ParentId="8250" CreationDate="2016-04-01T10:24:16.437" Score="6" Body="&lt;p&gt;With those restrictions you will probably be stuck with using fibre optics, which are expensive. Something like the &lt;a href=&quot;http://www.parans.com/eng/sp3/&quot; rel=&quot;nofollow&quot;&gt;Parans SP3&lt;/a&gt; system for example. Making your own by purchasing fibers is a possibility but still not cheap and there are a variety of problems with actually implementing it (such as focusing enough light into the fiber and controlling heat issues). This is discussed in &lt;a href=&quot;https://www.youtube.com/watch?v=RiYOw5NsoKY&quot; rel=&quot;nofollow&quot;&gt;this Youtube video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a relatively inexpensive option called 'The Light Bandit' which had a &lt;a href=&quot;https://www.kickstarter.com/projects/1658782339/the-light-bandit-controlled-sunlight-for-your-home&quot; rel=&quot;nofollow&quot;&gt;Kickstarter campaign&lt;/a&gt; which unfortunately has failed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could somehow get a relatively straight run the much more economical sun tunnel type systems, such as &lt;a href=&quot;http://www.solatube.co.uk/&quot; rel=&quot;nofollow&quot;&gt;Solatube&lt;/a&gt; may be a possibility but they usually need a much larger diameter than 1cm to collect any reasonable amount of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some additional information which my be interesting/useful:&#xA;&lt;a href=&quot;http://www.aia.org/practicing/AIAB096065&quot; rel=&quot;nofollow&quot;&gt;Before Nightfall: Advanced Daylighting Technologies&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="2246" LastEditDate="2016-04-01T14:47:34.770" LastActivityDate="2016-04-01T14:47:34.770" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8253" PostTypeId="2" ParentId="8250" CreationDate="2016-04-01T11:17:06.283" Score="4" Body="&lt;p&gt;Considering your difficult geometry, you might look into solar panels above ground, with wires to the underground lights.  This has the added advantage that you can store some of the energy in batteries for use when there is no sunlight at the surface.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-04-01T11:17:06.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8254" PostTypeId="2" ParentId="8250" CreationDate="2016-04-01T11:17:08.737" Score="2" Body="&lt;p&gt;You could try adapting the &lt;a href=&quot;https://en.wikipedia.org/wiki/Liter_of_Light&quot; rel=&quot;nofollow&quot;&gt;litre of light&lt;/a&gt; principle, where a hole is made in the roof of a shed which is then plugged with a re-purposed 1.25 L soft drink bottle filled with water. It provides light in the interior of the shed while the sun is shining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a bottle was fitted to both ends of a internally reflective plastic tube and the whole system was filled with water, one bottle could act as the light collector, the tube would transport the light and the bottle at the other end of the tube would disperse the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottles don't have to be 1.25 L soft drinks bottle, but a proof of concept rig could be constructed cheaply this way. If it proves success then different &quot;bottles&quot; can be used.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-04-01T11:17:08.737" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8255" PostTypeId="2" ParentId="8227" CreationDate="2016-04-01T11:27:38.143" Score="0" Body="&lt;p&gt;A diagram would help.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sliced cylinder suggests parallel disks on a common shaft.&lt;br&gt;&#xA;&quot;Top and bottom&quot; suggests a vertical axle  - is it?&lt;br&gt;&#xA;Inner ones may also mean vertical.&lt;br&gt;&#xA;How many slices?&lt;br&gt;&#xA;Are they necessarily driven by the shaft?&lt;br&gt;&#xA;When top and bottom turn together is it always at the same speed as each other?   How many combinations are there?&lt;br&gt;&#xA;If disks are ABCD and you drive AD or BC is the roration speed the same in ll cases.&lt;br&gt;&#xA;Does it always go the same way ?&#xA;What separation between disks?&lt;br&gt;&#xA;How thick are the disks?  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A basic solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine discs are on a vertical axle. &#xA;A system that would work for some combinations from above is to have a shaft which can move vertically relative to the discs. The axle has dogs/splines/cams/protrusions/thingy's which can engage with a disk when they are in the plane of the disc but a disc can be disengaged by moving the axle vertically until the disc does not engage with a drive thingy. These are at simplest &quot;dog clutches&quot; or splined discs and shaft but many variants are possible. Rather than the shaft moving selectors could move engagement &quot;thingy&quot;s or the discs could if desired by slid on the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eg&#xA;dddd    dddd = disc&lt;br&gt;&#xA;          xx = bare shaft&lt;br&gt;&#xA;         tttt = thingy that couples disc to shaft   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thingys tttt are fastened to the shaft and move up and down with it.&lt;br&gt;&#xA;Discs dddd    dddd stay at same horizontal level.&lt;br&gt;&#xA;Disks are driven by shaft rotation when tttt is in line with disc.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assembly  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       xx &#xA;  dddd xx dddd  A&#xA;      tttt&#xA;      tttt&#xA;  dddd xx dddd  B&#xA;       xx&#xA;      tttt&#xA;  dddd xx dddd  C&#xA;       xx &#xA;       xx&#xA;  dddd xx dddd  D&#xA;      tttt&#xA;       xx&#xA;       xx&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As shown rotating the shaft drives no discs.&lt;br&gt;&#xA;Moving the shaft up one layer drives discs A &amp;amp; D&lt;br&gt;&#xA;Moving the shaft down  one layer drives discs B &amp;amp; C &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2016-04-01T11:27:38.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8258" PostTypeId="2" ParentId="8250" CreationDate="2016-04-01T14:05:53.680" Score="0" Body="&lt;p&gt;You can send your light via internal total reflection. You need to research a flexible material that's hollow inside (to fill with liquid) and the material's index of refraction match the formula for total internal reflection.&lt;/p&gt;&#xA;" OwnerUserId="5837" LastActivityDate="2016-04-01T14:05:53.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8259" PostTypeId="2" ParentId="8246" CreationDate="2016-04-01T17:59:50.400" Score="5" Body="&lt;p&gt;You really need to use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Indicator_(distance_amplifying_instrument)&quot; rel=&quot;nofollow&quot;&gt;dial indicator&lt;/a&gt; to set up a 4 jaw chuck to the sort of accuracy required for this job. This is fiddly compared to a self centering chuck but has the potential to give you better accuracy for equivalent quality parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly you need to make sure that the part you are drilling is faced off as square as possible in the chuck and don't remove it thereafter untill you are finished. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for small drill and high concentricity a collet holder rather than a jawed drill chuck will give you much better accuracy and for this sort of operation you need to make sure that your tailstock is properly concentric with the headstock spindle, which is a whole other question in itself but well covered in machining manuals and web forums. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of setup operation can be frustrating and time consuming but it does allow you to get much better accuracy out of a relatively inexpensive machine. Production machinery has a more precision 'built in' as a trade off between capital cost and productivity. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-01T18:49:43.680" LastActivityDate="2016-04-01T18:49:43.680" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8260" PostTypeId="2" ParentId="3061" CreationDate="2016-04-01T19:49:46.503" Score="0" Body="&lt;p&gt;North South lines were found using a Solar compass. Some are tripod mounted or mount on the top of the theodolite. Google solar compass for more info.&lt;/p&gt;&#xA;" OwnerUserId="5840" LastActivityDate="2016-04-01T19:49:46.503" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8261" PostTypeId="2" ParentId="8246" CreationDate="2016-04-01T20:22:55.587" Score="8" Body="&lt;p&gt;99.5% pure tungsten? A lot depends on what the last 0.5% is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ductile tungsten is made by adding a small amount of rhenium, for example. That makes it much more machinable, and improves your chances of drilling a hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then - a 0.1 mm hole is extremely small, and I would not try to make a hole like that mechanically. You can try electrochemical machining (ECM). Problem is - if you want to drill with ECM, you really need to be able to set up a flow of electrolyte, which would require a needle. The thinnest hypodermic needle I could find online is a 34 gage (0.19 mm OD) needle; if you coat the outside of it with a very thin lacquer you might have the potential to drill a 0.25 mm hole. Bigger than 0.1 mm, but maybe good enough for a first experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attach the needle to the negative side of a battery (power supply). Attach your tungsten to the positive side, with a resistor in series (needs some experimentation to get the right value; you don't want to drain the battery too quickly if you accidentally touch the two). Immerse the tungsten in a sodium chloride (kitchen salt) solution, and lower the needle to &lt;em&gt;just above&lt;/em&gt; the location of the hole. A current will start to flow that will etch away the tungsten. Lower the needle as the hole is formed. Try to maintain a flow of fresh electrolyte through the needle to keep the chemistry at the tip favorable for etching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to admit I have not tried this - but I believe it would work. See for example this diagram (from &lt;a href=&quot;http://nptel.ac.in/courses/112107077/module3/lecture15/lecture15.pdf&quot;&gt;this lecture&lt;/a&gt;) to get an idea of the setup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PWv68.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PWv68.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2016-04-01T20:22:55.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8262" PostTypeId="1" CreationDate="2016-04-01T21:49:55.497" Score="3" ViewCount="1537" Body="&lt;p&gt;The drag coefficient is given by the following expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_d = \frac{2F_d}{\rho v^2 A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a race car wing (which is simply an airplane wing upside down to generate downforce), what should we consider as the area? The cross-sectional area perpendicular to the flow or the wing surface area (parallel to the ground)?&lt;/p&gt;&#xA;" OwnerUserId="4659" LastEditorUserId="1586" LastEditDate="2016-04-04T16:57:49.597" LastActivityDate="2016-04-04T16:57:49.597" Title="What is the area when calculating the drag coefficient for a race car wing?" Tags="|aerodynamics|drag|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8263" PostTypeId="2" ParentId="8262" CreationDate="2016-04-01T23:07:04.640" Score="1" Body="&lt;p&gt;Most race car wings are a lot more complex than an aircraft wing as they are usually designed to manage airflow over the whole body of the car, particularly the wheels and also have to deal with turbulent flow both from preceding aerodynamic surfaces and cars which they are following as well as controlling vortex generation to control airflow over their own surfaces for more complex functions like separating downstream flows between aerodynamic surfaces and intakes for cooling and engine induction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in  an F1 car the complex elements of the &lt;a href=&quot;https://tianyizf1.wordpress.com/tag/front-wing/&quot; rel=&quot;nofollow&quot;&gt;front wing&lt;/a&gt; are not so much to generate downforce as to control the downstream flow, especially over and around the front wheels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that race car wings are primarily tuned by varying angle of attack which this formula does not consider at all. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-01T23:07:04.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8264" PostTypeId="2" ParentId="8262" CreationDate="2016-04-01T23:47:20.683" Score="2" Body="&lt;p&gt;When $C_d$ is an appropriate parameter for discussion, road vehicle aerodynamics uses the planform area of the wing. That's the area formed by the chord (see below) and total width of the wing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd use the same area in lift calculations for $C_L$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For complicated wings, or whole vehicles, where defining an area makes little sense, the common practice is to list the product $C_D* A$ (often written 'CDA') as a parameter, rather than attempting to decouple them. It works perfectly well in computing lift/drag forces and varies as $L^2$ for scale modeling purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wWAEZ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wWAEZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2016-04-01T23:47:20.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8265" PostTypeId="2" ParentId="8262" CreationDate="2016-04-02T05:03:09.177" Score="1" Body="&lt;p&gt;The area is named &quot;reference area&quot;. As long as it is defined by someone, you are free to choose any arbitrary value for A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make comparisons easier among different designs for a certain project, engineers define A and they do trade off studies with that reference. This way they can compare the performance of the wings for a certain car/plane model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand it's also possible to Not Fix the ref area value. This would allow benchmarking various designs for their aerodynamic efficiency. (A slightly larger wing could, in itself be more efficient). The problem with this approach is that the optimization can easily break the manufacturing interface rules, size rules, weight constraints, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, if the thing has some physical constraints (like car-wing attachment), it's almost certain that you would define the wing area first, as a number, and not change it throughout the design life for practical purposes. &lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2016-04-02T05:03:09.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8267" PostTypeId="1" AcceptedAnswerId="9161" CreationDate="2016-04-02T18:34:07.813" Score="5" ViewCount="190" Body="&lt;p&gt;Here is some background to the problem (in a stirred tank):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;With yield stress non-Newtonian (viscoplastic) fluids, it is possible to generate an agitated volume around the impeller, defined as a cavern, surrounded by a stagnant region where the shear stress is insufficient to overcome the apparent yield stress of the fluid.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes you can get a cylindrical cavern around the impeller, see the below image. &lt;a href=&quot;https://i.stack.imgur.com/YeL2m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YeL2m.png&quot; alt=&quot;Cylindrical Cavern around impeller&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;By performing a force balance between the applied torque, Γ and the shear stress acting on the surface of a cylinder, we can define the boundary by setting the shear stress equal to the yield stress τ = τY. The total torque is given by:&#xA;$$\Gamma = \frac{\pi}{2} \tau_{y}H_{C}D_{C}^2+\frac{\pi}{6}\tau_{y}D_{C}^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just can't get the second term. The first term I can get by doing:&#xA;$$\Gamma_{1}=\tau_y \cdot Area_{Curved} \cdot \frac{D}{2} = \pi \cdot \frac{D^2}{2} \cdot H_{c} \cdot \tau_{y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gets me the first term...but the second term I just can't get, this is what I'm doing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma_{2}=\tau_{y} \cdot Area_{Faces} \cdot \frac{D}{2} =\tau_{y} \cdot 2 \pi \cdot \frac{D^2}{4} \cdot \frac{D}{2} = \tau_{y} \cdot \pi \cdot \frac{D^3}{4} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Argh, so I'm getting D&lt;sup&gt;3&lt;/sup&gt;/4 instead of D&lt;sup&gt;3&lt;/sup&gt;/6 for the second term and I just can't work it out, if anyone can help I'd appreciate it.&lt;/p&gt;&#xA;" OwnerUserId="4530" LastEditorUserId="90" LastEditDate="2016-04-03T09:22:19.937" LastActivityDate="2016-07-03T15:08:35.013" Title="Working Out Shaft Torque in a Stirred Tank with a Cylindrical Cavern Formation (using a non Newtonian fluid)" Tags="|mechanical-engineering|fluid-mechanics|chemical-engineering|process-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8268" PostTypeId="2" ParentId="6739" CreationDate="2016-04-03T04:28:11.887" Score="3" Body="&lt;p&gt;I am assuming that your question is &quot;how to increase the flow rate&quot; and that your pump is a centrifugal pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For centrifugal pumps, the flow rate out of the pump is dependent on the pressure at the exit of the pump.  If the pressure is high, the flow rate will be lower.  If the pressure is low, the flow rate will be higher.  That dependency is typically illustrated for each pump in a graph of flow vs pressure which is called the pump's &quot;pump curve&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fluid systems with a centrifugal pump, the flow rate will be reduced by each element (tubing, fittings, valves, etc.) in the system.  Each element added to the system increases the pressure and reduces the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first step would be to find the pump curve for your pump and measure the pressure at the exit of the pump.  This will tell you two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;is the pump operating correctly?&lt;/li&gt;&#xA;&lt;li&gt;is the pump operating against a&#xA;relatively significant pressure?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the pump is not achieving the flow rate indicated on the pump curve from the pressure you are measuring, it may need to be rebuilt or replaced.  If the pump is generating its maximum flow rate already, it will need to be upgraded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pump is working correctly but not at maximum flow, you next step depends on the complexity of the system.  It may be easiest and cheapest to upgrade the pump to one which will generate a higher flow rate against the pressure you are seeing.  Or, it may be easiest and cheapest to investigate the other elements in the system and replace the problematic ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If upgrading the pump is not an option, you will need to determine where the restrictions are in your system.  The ideal way to do this is to place pressure gauges in between each element and look for the highest pressure difference.  I would start with one between pump and radiator, between radiator and intercooler, and between intercooler and pump.  Whichever element has the highest pressure drop will be the one which is causing the most restriction to the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other thoughts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I expect the interior loop is not reducing the flow rate substantially.&lt;/li&gt;&#xA;&lt;li&gt;If the tubing is a significant restriction, increasing the inner diameter of the tubing can increase the flow rate dramatically.  Reducing the length will increase the flow rate but may not have as dramatic an effect.&lt;/li&gt;&#xA;&lt;li&gt;Each valve and fitting reduces the flow rate.  If you suspect the fittings or valves, you might replace them with higher Cv (&lt;a href=&quot;https://en.wikipedia.org/wiki/Flow_coefficient&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Flow_coefficient&lt;/a&gt;) versions.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2408" LastEditorUserId="2408" LastEditDate="2016-04-04T04:15:51.643" LastActivityDate="2016-04-04T04:15:51.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8269" PostTypeId="1" CreationDate="2016-04-03T07:08:22.443" Score="2" ViewCount="108" Body="&lt;p&gt;Sorry for the weird question but I was wondering about the standard machining parameters for Aluminium (if there is even a standard). Particularly the values for a CNC process (preferably in SI) for a rough volume milling and one for a finishing cut. Google searches doesn't seem to be very helpful as I only need rough values for testing purposes. (CAD Package is CREO Parametrics 2.0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help!&lt;/p&gt;&#xA;" OwnerUserId="5850" LastActivityDate="2016-08-05T22:52:29.773" Title="Standard parameters for machining aluminium?" Tags="|materials|machining|aluminum|cnc|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8272" PostTypeId="1" AcceptedAnswerId="8274" CreationDate="2016-04-03T17:43:38.397" Score="4" ViewCount="14931" Body="&lt;p&gt;How does union joint with two male threaded ends work? Does one male end rotate 360 degrees around it axis? How is the inside of the union constructed? Can I connect two pipes with female threads using a union joint and assemble/disassemble it without too much pain? Specifically, I am looking at a stainless steel union joint.&lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="122" LastEditDate="2016-04-04T03:18:44.873" LastActivityDate="2016-04-04T06:30:31.727" Title="How does a union joint with two male threads work?" Tags="|piping|pipelines|joining|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8274" PostTypeId="2" ParentId="8272" CreationDate="2016-04-04T06:30:31.727" Score="3" Body="&lt;p&gt;Typically, most threads on pipes (in the USA at least) are male &lt;a href=&quot;https://en.wikipedia.org/wiki/National_pipe_thread&quot; rel=&quot;nofollow noreferrer&quot;&gt;NPT&lt;/a&gt; and all the couplings or fittings are female NPT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hZoVe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hZoVe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;You can connect two pipes with a coupling as long as the last pipe added is free to rotate about its axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M6Axg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M6Axg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;A union is used for easy removal of pumps and other serviceable equipment that would otherwise not be easily unthreaded. It is also used to attach pipes in which neither pipe can not be rotated about its axis (perhaps because it is assembled to other plumbing, etc). The following image is the best section view of a union I could find. Most of the ones I have used have a flat steel face that mates to a flat steel face with an oring (with no 45 degree bevel like shown in the image). The flat faces allow the equipment to easily be removed sideways out of the plumbing when the union is detached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JMM3K.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JMM3K.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.smithcooper.com/catalogue/group/2781/nameblock/Union&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stainless Steel Unions with datasheet links&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most hardware/plumbing stores sell pipe unions and will let you take them apart to see if they fit your needs. Seeing them first hand is the best way to get a feel for how they work.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-04-04T06:30:31.727" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8275" PostTypeId="1" CreationDate="2016-04-04T10:24:11.793" Score="4" ViewCount="909" Body="&lt;p&gt;We need to cut a very thin crack into a polymer plate for a research project. The closest thing to what we are looking for is &lt;a href=&quot;https://en.wikipedia.org/wiki/Electrical_discharge_machining&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wire EDM&lt;/a&gt; but since the polymer we are using is not conductive we cant use that method. Any suggestions? Is there an insulator version of Wire EDM?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;a is the crack length and we want the crack thickness to be less than    1 mm. It should be as thin as possible.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/POFtO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/POFtO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1110" LastEditorUserId="1110" LastEditDate="2016-04-04T18:53:19.087" LastActivityDate="2016-04-04T20:05:53.933" Title="How do I cut a crack into a polymer/plastic?" Tags="|materials|aerospace-engineering|manufacturing-engineering|plastic|composite|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8278" PostTypeId="1" CreationDate="2016-04-04T15:18:14.763" Score="1" ViewCount="276" Body="&lt;p&gt;I am simulating the acceleration of a car (FSAE) in a straight line. I have data of the Pacejka Curves ($F_x$ vs. $SR$) of the tire we will be using. The thing is that the curve doesn´t pass through the origin ($F_x=SR=0$). Instead the curve is shifted horizontally meaning that $F_x(SR=0)\neq0$. I have troubles finding the explanation. I am pretty sure it has to do with the rolling resistance of the tire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My model (simplified) is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The driving torque on the axle is resisted by $F_x(SR)$ which causes a countertorque. Resultant torque is what accelerates (rotationally) the axle. $F_x$ is what drives the car forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody explain me the meaning of a shifted Pacejka Curve?&lt;/p&gt;&#xA;" OwnerUserId="5860" LastEditorUserId="1832" LastEditDate="2016-04-04T15:52:31.510" LastActivityDate="2016-04-04T15:52:31.510" Title="How is rolling resistance accounted for in the Pacejka Curves?" Tags="|automotive-engineering|modeling|wheels|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8279" PostTypeId="1" CreationDate="2016-04-04T16:15:29.957" Score="8" ViewCount="104" Body="&lt;p&gt;I am working in a construction company which is building hydropower plants in Northern Norway. We are monitoring the Total Hydrocarbon Content (THC) of our process waste water. This particularly important during tunneling as the machines are likely to leak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment we are sending waste water samples to an analysis company and there are some disadvantages of that, including the price, the delay in information, and the risk of volatile hydrocarbons evaporating during transportation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to take weekly samples in the field and do the analysis myself. I am an environmental engineer with some background in the laboratory from my time as a student, but I have never tried to find THC before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been trying to find a simple and inexpensive way of measuring the THC in the water, but most of the instruments I have seen on various company websites are very sensitive (measuring in the ppm range) and meant to measure the THC continuously on some industrial flow (like &lt;a href=&quot;http://liquidgasanalyzers.com/product/hydrocarbon-voc-water-analyzer/&quot;&gt;this&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The measuring range should be something like 1-100 mg/L, and I simply cannot find any method or instrument that suits those needs. Do you know of any kind of instrument that can perform this analysis, or a manual method? I am looking for an inexpensive, portable, and simple solution.&lt;/p&gt;&#xA;" OwnerUserId="5862" LastEditorUserId="2246" LastEditDate="2016-04-04T16:32:34.813" LastActivityDate="2018-08-31T08:49:55.377" Title="Inexpensive and moderately accurate (field) method of measuring THC (Total Hydrocarbon Content) in water" Tags="|environmental-engineering|waste-water-treatment|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8281" PostTypeId="2" ParentId="8275" CreationDate="2016-04-04T16:59:04.807" Score="5" Body="&lt;p&gt;Since you used some specific words in your question, I will respond a certain way.  Cracks are generated by mechanical action, and cannot be machined in.  They may, in certain brittle materials, be a byproduct of machining.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a method like EDM will alter the properties of a polymer nearby, as it will heat the polymer.  If you are interested in introducing a crack for the purposes of fracture testing (see the reference to ASTM D5045 below), then it is imperative that you create a &quot;sharp&quot; crack that is representative of the material (and thus EDM is not a good choice).  The method for doing this involves machining a notch in the plastic at the test zone, and then using mechanical load to precrack the specimen.  The precrack needs to be measured, and then the fracture testing can commence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, when you said &quot;crack&quot; what you actually meant was that you need to cut a small opening in the polymeric material, I suggest you change your wording to &quot;machine a hole or slit&quot;.  In that case, I suggest a CNC end mill with a very small diameter bit, which will achieve tolerances approaching an EDM wire.  If kept cool enough, this will not alter the local properties of the polymer.  If this is not fine enough, then you are most certainly in need of a CNC laser cutter, which again will locally alter the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For method reference, see ASTM D5045-14: Standard Test Methods for Plane-Strain Fracture Toughness and Strain Energy Release Rate of Plastic Materials&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General information about material fracture: &lt;a href=&quot;https://en.wikipedia.org/wiki/Fracture&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Fracture&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Now that you have posted the drawing, I suggest an end mill. Bits are readily available that will machine that, particularly if it is not a hard material. You will want to work with a machine shop that does precision machining, and ask them specifically about tolerances. Also, I suggest you have them cut some samples prior to fabricating the final article so you can evaluate their tolerances. &lt;/p&gt;&#xA;" OwnerUserId="5865" LastEditorUserId="5865" LastEditDate="2016-04-04T20:05:53.933" LastActivityDate="2016-04-04T20:05:53.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8282" PostTypeId="1" CreationDate="2016-04-04T19:22:52.387" Score="4" ViewCount="75" Body="&lt;p&gt;I'm running an experiment and I need to have a container filled with water. The container will be placed inside a microwave oven, but I need to minimize any effect the microwaves will have on the water. How can I go about running my experiment? I'm thinking I need to insulate the container with some kind of material that will shield the water inside from the microwaves generated by the microwave oven.&lt;/p&gt;&#xA;" OwnerUserId="5866" LastEditorUserId="6790" LastEditDate="2020-05-01T17:39:19.060" LastActivityDate="2020-05-01T17:39:19.060" Title="A material to insulate a container so that water inside the container is not affected by microwaves?" Tags="|electrical-engineering|materials|rf-electronics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8283" PostTypeId="1" CreationDate="2016-04-04T20:24:53.547" Score="7" ViewCount="714" Body="&lt;p&gt;I received construction documents for an extensive residential remodel and looking at the engineering there's a glulam drop beam with a calculated member reaction that is 100% of allowed (3965 lbs @ 5 1/2&quot;). No other member comes close to 100%. &#xA;Does this have any practical implications? Is this common?&lt;/p&gt;&#xA;" OwnerUserId="5867" LastEditorUserId="368" LastEditDate="2016-04-06T15:52:52.597" LastActivityDate="2019-08-21T18:20:31.140" Title="What does it mean when actual member reaction is 100% of allowed?" Tags="|structural-engineering|beam|structural-analysis|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8284" PostTypeId="2" ParentId="8283" CreationDate="2016-04-04T20:46:40.070" Score="6" Body="&lt;p&gt;This means that the member is considered to be fully loaded for a failure mode which is being analyzed by that stress calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some things to (probably) keep in mind include:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Some conservatism likely exists in the analysis/design calculation. A member reported to be at 100% capacity is (probably) not actually at 100% capacity. This conservatism could include anything from assuming larger loads than actually exist, assuming smaller section sizes, assuming longer member spans, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The load case used to arrive at the conclusion that this member is at 100% capacity (probably) considers a transient load of some sort (like wind, seismic, dynamic water pressure, etc.) that isn't acting on the structure at all times.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Structural members are usually checked for several different &lt;a href=&quot;http://image.slidesharecdn.com/finalppt-150514043812-lva1-app6892/95/final-pp-comparative-investigation-on-shear-strength-prediction-models-for-sfrc-memberst-3-638.jpg?cb=1431578381&quot; rel=&quot;nofollow&quot;&gt;failure modes&lt;/a&gt;. An indication that a member is at 100% capacity for one failure mode (probably) doesn't mean that it is at 100% capacity for all failure modes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sometimes a member shown to be at 100% capacity based on the stress/design calculation may not be at 100% of it's &lt;em&gt;strength&lt;/em&gt; capacity. Many times structural members will be controlled by their &lt;a href=&quot;https://en.wikipedia.org/wiki/Serviceability_(structure)&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;serviceability&lt;/em&gt; requirements&lt;/a&gt; instead of their strength requirements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="129" LastEditorUserId="368" LastEditDate="2016-04-06T16:10:17.747" LastActivityDate="2016-04-06T16:10:17.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8285" PostTypeId="1" AcceptedAnswerId="8287" CreationDate="2016-04-04T22:05:14.957" Score="6" ViewCount="2926" Body="&lt;p&gt;I'm trying to understand the derivation of the turbulent kinetic energy equation, as described in this link: &lt;a href=&quot;http://www.diva-portal.org/smash/get/diva2:379743/FULLTEXT01.pdf&quot; rel=&quot;nofollow&quot;&gt;Evaluation of RANS turbulence models for flow problems with significant impact of boundary layers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm able to follow up to equation 2.26 on slide 11 (i.e. page 9), which states&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho \overline{\frac{\partial u_i'}{\partial t}u_i'} + \rho\left( \overline{u_j\frac{\partial u_i}{\partial x_j}u_i} -\bar{u}_j\frac{\partial u_i'}{\partial x_j}\bar{u}_i\right) = -\overline{\frac{\partial p'}{\partial x_j}u_i'} + \nu\overline{u_i'\nabla^2u_i'} + \rho \frac{\partial (\overline{u_i'u_j'})}{\partial x_j}\bar{u}_i,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the overbars denote the ensemble averages of the quantities under the bars.  Note that here, the average of a product is not necessarily equal to the product of the average of each term.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the link, using only averaging rules, the term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left( \overline{u_j\frac{\partial u_i}{\partial x_j}u_i} -\bar{u}_j\frac{\partial u_i'}{\partial x_j}\bar{u}_i\right)$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;can be simplified to become&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\overline{u_i'\frac{u_i'}{\partial x_j}}\bar{u}_j + \overline{\frac{\partial u_i'}{\partial x_j}u_i'u_j'} + \overline{\frac{\partial u_i'}{\partial x_j}u_j'}\bar{u}_i + \overline{u_j'u_i'}\frac{\partial u_i}{\partial x_j}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I substitute this directly into the previous equation, I get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho  \left( \overline{\frac{\partial u_i'}{\partial t}u_i'} + \overline{u_i'\frac{u_i'}{\partial x_j}}\bar{u}_j + \overline{\frac{\partial u_i'}{\partial x_j}u_i'u_j'} + \underbrace{\overline{\frac{\partial u_i'}{\partial x_j}u_j'}\bar{u}_i}_{\text{my 4th term}} + \overline{u_j'u_i'}\frac{\partial u_i}{\partial x_j}\right) = -\overline{\frac{\partial p'}{\partial x_j}u_i'} + \nu\overline{u_i'\nabla^2u_i'} + \rho \frac{\partial (\overline{u_i'u_j'})}{\partial x_j}\bar{u}_i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, their derivation yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho  \left( \overline{\frac{\partial u_i'}{\partial t}u_i'} + \overline{u_i'\frac{u_i'}{\partial x_j}}\bar{u}_j + \overline{\frac{\partial u_i'}{\partial x_j}u_i'u_j'} + \underbrace{\overline{\frac{\partial u_i'u_j'}{\partial x_j}}\bar{u}_i}_{\text{their 4th term}} + \overline{u_j'u_i'}\frac{\partial u_i}{\partial x_j}\right) = -\overline{\frac{\partial p'}{\partial x_j}u_i'} + \nu\overline{u_i'\nabla^2u_i'} + \rho \frac{\partial (\overline{u_i'u_j'})}{\partial x_j}\bar{u}_i.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference between my derivation and theirs is the 4th term on the left hand side of the equation.  I'm sure that their 4th term is correct, since it is supposed to cancel out with a term on the right hand side of the equation.  However, I can't seem to figure out how they obtained their 4th term on the LHS of the equation.  The link suggests that the chain rule and the incompressiblity assumption are involved, but I'm not sure how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, the overbar term $\overline{u_i'u_j'}$ is treated as a single term.  Thus, how can I apply the chain rule to  $\frac{\partial (\overline{u_i'u_j'})}{\partial x_j}$?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any hints to complete the missing steps would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="5899" LastEditDate="2016-04-07T15:36:45.420" LastActivityDate="2016-12-24T16:52:24.610" Title="Deriving the turbulent kinetic energy equation" Tags="|fluid-mechanics|turbulence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8287" PostTypeId="2" ParentId="8285" CreationDate="2016-04-04T23:00:55.917" Score="5" Body="&lt;p&gt;The two terms are equal due to the product rule and continuity equation. The use of the continuity equation might not be obvious, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In their notation, the Reynolds decomposition is $u_i = \overline{u}_i + u^\prime_i$. Taking the continuity equation and averaging leads to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial \overline{u}_j}{\partial x_j} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subtract the equation above from the non-averaged continuity equation to see that the fluctuations also are divergence free:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial u^\prime_j}{\partial x_j} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, apply the product rule to the term of interest (before averaging):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial u^\prime_i u^\prime_j}{\partial x_j} = u^\prime_j \frac{\partial u^\prime_i}{\partial x_j} + u^\prime_i \frac{\partial u^\prime_j}{\partial x_j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term is zero by continuity as seen above, and this completes the derivation. The averages commute, so the average of a derivative is the same as the derivative of the average.&lt;/p&gt;&#xA;" OwnerUserId="49" LastEditorUserId="49" LastEditDate="2016-04-05T01:37:59.697" LastActivityDate="2016-04-05T01:37:59.697" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8289" PostTypeId="1" CreationDate="2016-04-05T02:12:39.683" Score="13" ViewCount="3397" Body="&lt;p&gt;I was out walking and saw these strange looking spirals on some power lines.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/9o011.jpg&quot; alt=&quot;Power lines with spiral-shaped bird flight diverters&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I got home I did some research and discovered that they are to prevent birds form hitting the power lines, mainly used in conservation areas. I found &lt;a href=&quot;http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0032569&quot; rel=&quot;noreferrer&quot;&gt;this interesting study&lt;/a&gt; about which types of flight diverter are the most effective at preventing bird collisions. It mentions a few other types of diverter (plate, swivel, sphere), but the spiral seems to be the preferred shape to experiment with (big vs small) and the reason for this is not explained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand that a spiral wire has a good visibility to weight ratio, but is this the only reason for the shape? Would there be any current transmitted through (or induced in) the spiral, and would this shape help in any way to prevent any negative effects this might have? Or am I just thinking that there is an electrical reason because the spirals remind me of magnetic field lines, when really that's nothing to do with it?&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2020-07-25T13:24:17.633" Title="Why are bird flight diverters shaped like a spiral?" Tags="|electrical-engineering|power-electronics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8290" PostTypeId="1" CreationDate="2016-04-05T05:29:57.850" Score="4" ViewCount="431" Body="&lt;p&gt;A two-stroke engine equipped with a single cylinder having a bore of 12 cm and a stroke of 50 cm operates on an Otto cycle. At the beginning of the compression stroke air is at 100 kPa, 25 ℃. The maximum temperature in the cycle is 1100 ℃ . Use the PG model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the efficiency be if the clearance volume were 1200 cc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the Eqn: $$1-\frac{1}{r^{0.4}}$$  I keep getting 50.2%, but Mastering Engineering is telling me I'm wrong.  Does anyone see an error with this?  I'm lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have V&lt;sub&gt;1&lt;/sub&gt;=6854.84 which gives me a compression ratio of 5.71239 because V&lt;sub&gt;2&lt;/sub&gt; is 1200.  I must not have the right compression ratio because from there it should just be plug and chug but I can't see what I'm doing wrong.&lt;/p&gt;&#xA;" OwnerUserId="5874" LastEditorUserId="90" LastEditDate="2016-04-05T06:21:31.583" LastActivityDate="2017-10-03T09:05:33.417" Title="Otto Cycle Efficiency" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8291" PostTypeId="1" CreationDate="2016-04-05T08:16:35.647" Score="0" ViewCount="144" Body="&lt;p&gt;I am referring below diagram where pump and valves are used to allow waste, diluent water control in biomedical instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zF0Qk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zF0Qk.png&quot; alt=&quot;Diagram 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/glSnE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/glSnE.png&quot; alt=&quot;Diagram 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ex11n.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ex11n.png&quot; alt=&quot;Diagram 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am not able to get how single pump can have multiple port, I have seen pump with one suction inlet and one pressure outlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain me how pumps here having different ports (P1,P2,P3,P4,P5,P6)? I am not asking for flow, because it's dependent on application and no one can tell me by just looking at above image. But my question is about multiple ports for pumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="5877" LastEditorUserId="5877" LastEditDate="2016-04-07T11:05:35.137" LastActivityDate="2016-04-07T11:05:35.137" Title="Decoding pump- valve analogy for medical instrument" Tags="|mechanical-engineering|electrical-engineering|pumps|valves|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8292" PostTypeId="1" AcceptedAnswerId="8313" CreationDate="2016-04-05T08:42:50.100" Score="3" ViewCount="75" Body="&lt;p&gt;In Autocad 2015 for Mac, I have created a cylinder with a torus wrapped around it. The torus is 2 mm thick, with 1 mm embedded inside the cylinder and 1 mm outside. I want to subtract the torus from the cylinder, so I will be left with a semicircular groove around the cylinder. I have already done this once, but I had to remake the object so I deleted the original and started again - and now when I try to subtract the torus, it simply deletes the torus and leaves the cylinder untouched. What is going wrong, and how can I fix it?&lt;/p&gt;&#xA;" OwnerUserId="5878" LastEditorUserId="5878" LastEditDate="2016-04-20T06:33:06.327" LastActivityDate="2016-04-20T06:33:06.327" Title="How to subtract a region where two solids intersect?" Tags="|design|autocad|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8293" PostTypeId="1" AcceptedAnswerId="8295" CreationDate="2016-04-05T12:12:48.140" Score="3" ViewCount="474" Body="&lt;p&gt;Can anyone tell me how to go about this problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a square shaped laminate (polyethylene fiber and epoxy resin composite) that is constrained on one side. A gun is fired and the bullet hits the laminate. What kind of stresses result in the laminate? How should I go about their calculation? When I say stresses, I mean the stress tensor components $\sigma_{i,j}$. Also, can the strain tensor components $\epsilon_{i,j}$ be calculated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Textbooks and other sources don't show or tell how the stress components are measured. Also, in the above problem, the mass and the velocity of the bullet are known.&lt;/p&gt;&#xA;" OwnerUserId="4775" LastEditorUserId="1832" LastEditDate="2016-04-05T12:35:07.383" LastActivityDate="2016-04-05T17:04:51.467" Title="Impact load of a bullet on a laminate" Tags="|composite|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8294" PostTypeId="2" ParentId="8282" CreationDate="2016-04-05T12:16:29.770" Score="3" Body="&lt;p&gt;Unfortunately, you are in a bit of a bind.  You need a combination of things to successfully achieve your experiment, and I doubt you be able to achieve it using a commercial microwave oven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you need to build a faraday cage around the section you want to not be exposed to microwaves.  Faraday cages (&lt;a href=&quot;https://en.wikipedia.org/wiki/Faraday_cage&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Faraday_cage&lt;/a&gt;) provide a means to prevent RF energy from penetrating a fixed area by protecting that area with a conductive ground plane (to shunt the RF energy to common ground).  In fact, the microwave itself is a faraday cage, simply built to not allow the radiation to escape.  Unfortunately, commercial microwaves are not designed to allow you to put a faraday cage inside (they will arc), and I would be very cautious about trying to run a grounding line through the door (as this would prevent the oven itself from being an effective faraday cage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the faraday cage to work in your experiment (I will assume that your vessel has a cubic shape, just for the sake of discussion), one wall of the faraday cage also needs to be a semi-permeable membrane.  This will allow whatever mass transfer you are trying to evaluate, but still prevent radiation from entering the fluid inside the faraday cage.  While it is possible to make a faraday cage from a frequency selective surface such as the door of the microwave (&lt;a href=&quot;https://en.wikipedia.org/wiki/Frequency_selective_surface&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Frequency_selective_surface&lt;/a&gt;), it would require a custom design and fabrication to make it out of your semi-permeable membrane.  In addition, the metallic mesh used in the FSS would certainly have an impact on diffusion through the membrane, and that might not be acceptable to your study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from scratch, what you really want is a source of microwaves, a waveguide and a test area within the waveguide which will expose some of your materials to the RF energy, and keep some of it shielded.  This is doable with waveguides, which use good design to shape the flow of the RF energy, and would result in minimal exposure (but not zero) of your material.  There are many good references here (&lt;a href=&quot;http://www.microwaves101.com/encyclopedias/waveguide-primer&quot; rel=&quot;nofollow&quot;&gt;http://www.microwaves101.com/encyclopedias/waveguide-primer&lt;/a&gt;) but for your specific instance you need someone with detailed knowledge about waveguide design, and the ability to tailor it for your experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more fundamental question is whether you need to provide microwaves for your experiment because they are a source of heat, or because they are a source of RF radiation.  If you are simply using them for heating, I would suggest an alternative like heating elements/blankets or a specific heat source (burner, steam mantle).  If you are using the microwave for its RF energy, then you may be forced to pursue the complicated waveguide solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps, if you can post more detail I will provide edits.&lt;/p&gt;&#xA;" OwnerUserId="5865" LastActivityDate="2016-04-05T12:16:29.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8295" PostTypeId="2" ParentId="8293" CreationDate="2016-04-05T13:19:13.247" Score="3" Body="&lt;p&gt;Expanding on the comment by @grfrazee, it isn't as simple as with a static load. Several non-linear parameters change the response of the composite component.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Extremely high strain rate increases the observed brittleness and decreases the strain-at-failure and strain-at-rupture of the impacted materials.&lt;/li&gt;&#xA;&lt;li&gt;There is a non-obvious limitation on penetration depth of projectiles in materials, though for a thin plate this is may be irrelevant.&lt;/li&gt;&#xA;&lt;li&gt;Contact surfaces are constantly changing with time.&lt;/li&gt;&#xA;&lt;li&gt;Velocity is changing with time, though this may be a trivial decrease depending on relative energy and thickness.&lt;/li&gt;&#xA;&lt;li&gt;Force exerted is changing with time.&lt;/li&gt;&#xA;&lt;li&gt;Momentum and energy of the bullet and composite are changing with time.&lt;/li&gt;&#xA;&lt;li&gt;The composite may partially rupture, changing the local geometry, and its effects on stress and strain. The state of the geometry is also changing with time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming you've accounted for the non-linear change in material properties, the rest &lt;em&gt;could&lt;/em&gt; be modeled carefully using a non-linear dynamic structural finite element simulation. However it must be stressed that you will almost certainly need a dynamic mesh and careful application of boundary conditions. Once the simulation is complete, the simulation software should be able to output the stress and strain of the material in each direction, from which you could rebuild the stress and strain tensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative to this approach is to explore @grfrazee's answer and try to locate simpler models for deformation. It may be that a simple tensor model does not exist, in which case you would have to create your own from even simpler, 1D or 2D models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An experimental alternative, as pointed out by @grfrazee, is to shoot things and see what happens. Doing so is a bit more tricky than that statement might imply. Some possibilities for tracking live material deformation characteristics would include attaching strain gages to the component in a carefully determined configuration, or use a laser interferometer, or even perform the test inside a real-time x-ray machine. For the latter test, you would probably want to do something like dope the fibers with osmium tetroxide or a similar compound containing a similarly high-atomic-number element for contrast between the composite components.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-04-05T13:53:00.763" LastActivityDate="2016-04-05T13:53:00.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8297" PostTypeId="2" ParentId="8293" CreationDate="2016-04-05T13:47:09.647" Score="3" Body="&lt;p&gt;An additional complication is the behaviour of the bullet itself on impact, which is not only difficult to model but can vary a lot according to the construction of the bullet, it's geometry and mass, and the angle and energy of impact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a low-velocity, unjacketed lead bullet might squash against the impact surface with little or no penetration, which is a reasonable approximation of an inelastic collision and so the best modelling approach might be to consider its momentum and here it is certainly possible, although not trivial, to construct fairly credible finite element model based on dynamic stress and strain. Indeed you may even be able to get meaningful order of magnitude numbers with manual calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand a high-velocity, armour-piercing bullet might well pass through cleanly and transfer relatively little energy to the target. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third case is that fragmentation or tumbling  of the bullet or target.  In high energy impacts you may even observe fluid-type behaviour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the actual contact surface between the bullet and the target there will also be shock waves that propagate through the target, which can cause high stresses and failures (eg. spalling) and can eject material from the back side of the target even if the bullet itself does not penetrate it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the big difficulty with modeling this sort of situation is that you are dealing with strain rates far outside the assumptions underlying the material properties and equations used for static and most dynamic modelling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way firing a bullet at a target is a fundamentally different condition from pushing it through a target slowly regardless of how much force is applied. Indeed in practical impact testing it is more usual to think in terms of energy than stress and strain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact when it gets right down to real world applications standards for body armour will usually specify resistance to specific rounds (in terms of caliber, propellant and bullet type) at a specific range. Simply because there isn't a measurable physical property which is generally descriptive of this sort of performance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to think carefully about what you are trying to determine. In the context of armour something which generates a wide cone of debris and fragmentation on impact may actually be far worse than just allowing the bullet to pass through intact. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-05T17:04:51.467" LastActivityDate="2016-04-05T17:04:51.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8298" PostTypeId="2" ParentId="8292" CreationDate="2016-04-05T15:07:39.570" Score="2" Body="&lt;p&gt;Often this sort of problem is caused by some small and non-obvious error in your drawing, for example if a point hasn't snapped to exactly where you expected to one or more lines aren't exactly in the right place. Particularly if you end up with open loops or small gaps between lines which should intersect. You can also have problems if you have several lines or points on top of each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most CAD software has a tool to diagnose these sorts of problems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another common problem is picking/entity selection errors, these can sometimes crop up from minor conflicts with graphics processing eg some software prefers Nvidia to Radeon graphics hardware which can sometimes cause odd problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the first thing is to look at your basic geometry and make sure that it is what it is supposed to be. The second is to look back through your workflow and make sure that you have done everything properly. It is very easy with CAD to slip into bad habits where you do things which are a bit iffy and work in some circumstances but not others. Similarly is is very easy when learning to get yourself into a wrong mode or layer or assign an unwanted property to an object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Autocad is also well supported by online forums so it may also be worth asking your question there as well. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-06T11:01:56.360" LastActivityDate="2016-04-06T11:01:56.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8300" PostTypeId="2" ParentId="8221" CreationDate="2016-04-05T15:31:46.783" Score="2" Body="&lt;p&gt;In wind tunnels usually meshes and/or honeycombs are used to reduce turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to really achieve low turbulence it's probably not enough to add a single component to your facility. I rather suspect you need to evaluate the overall design. Some kind of diffusors at the pipe end(s) might be helpful. If you look at the flow behind bridge pillars, you can see that turbulence can be preserved over long time scales also in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is non-trivial.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2016-04-05T15:31:46.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8301" PostTypeId="1" CreationDate="2016-04-05T16:08:44.887" Score="6" ViewCount="1494" Body="&lt;p&gt;I am making a die for cutting arbitrary shapes from thin (0.1–0.5 mm) copper foil/sheet using a manual arbour press. The product will be an oak leaf shape, approximately life-size, for use as part of a sculpture project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty experienced at general fabrication and metalwork and have access to a well equipped fabrication workshop including TIG welding, lathe, manual die grinding, manual forging and heat treatment, etc. (but not CNC milling). However, I don't know what would be an effective and reasonably low cost way to make a die for this sort of process. Is there a reasonable way to make this in-house or is it better to just get a die made by a specialist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accuracy of sizing isn't that important, the main priority is aesthetically smooth and well-finished edges. Visual appearance and minimal finishing are the main goals. I will be cutting in batches of around 700.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outline will be a stylised oak leaf, so a continuous loop with various lobes, where my first priority is smooth transition between radii rather than exact tolerances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The options I've considered so far are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A knife-type tool (perhaps pressing onto a rubber block) &lt;/li&gt;&#xA;&lt;li&gt;A shearing tool with a positive and negative die (like a hole punch)—in this case I am particularly interested in how to work out the optimum tolerances between the top and bottom tools for a given sheet thickness (probably 0.1 mm annealed copper)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="1586" LastEditDate="2016-04-20T00:24:49.173" LastActivityDate="2016-05-03T00:36:43.767" Title="How to manufacture cutting dies for metal foil?" Tags="|mechanical-engineering|manufacturing-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8302" PostTypeId="1" AcceptedAnswerId="8318" CreationDate="2016-04-05T18:25:28.110" Score="4" ViewCount="49" Body="&lt;p&gt;How do people design height gaining based on nested poles ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this picture (I made them) :&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JQUC2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JQUC2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to achieve height based on these 3 poles, how they should behave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this how the behave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J9JdM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J9JdM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had been searching over the internet I haven't found the related material, and it's kinda hard what to type. I'm doing some tasks on designing a simple machine (no foundation knowledge but with will) and my English is sort of limited, especially to technical stuff but I hope you understand it .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Imagine like there's somehow an auto increasing portable stair to get over the tree, the poles(both sides) are consisted of  poles fragment on each of them, how the fragments behave in this case. &lt;/p&gt;&#xA;" OwnerUserId="5883" LastEditorUserId="5883" LastEditDate="2016-04-05T19:23:46.317" LastActivityDate="2016-04-05T19:23:46.317" Title="Principal on gaining more height based on nested poles?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ClosedDate="2016-04-07T15:09:56.787" ContentLicense="CC BY-SA 3.0" />
  <row Id="8305" PostTypeId="1" CreationDate="2016-04-05T20:15:18.450" Score="1" ViewCount="304" Body="&lt;p&gt;Which ASME standard defines Class 150 Nipples - material stainless steel? What is the pressure-temperature rating for such a fitting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2WUMZ.jpg&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Stainless steel nipple&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2WUMZ.jpg&quot; alt=&quot;&quot; title=&quot;Stainless steel nipple&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have ASME standards available but I don't believe ASME B16.11 has any mention of double-threaded hexagonal nipples. I am interested especially in pressure/temperature dependence.&lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="368" LastEditDate="2016-04-07T00:33:19.863" LastActivityDate="2016-04-07T00:33:19.863" Title="What is the pressure-temperature rating for a stainless steel Class 150 nipple?" Tags="|piping|pipelines|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8306" PostTypeId="2" ParentId="8301" CreationDate="2016-04-05T20:21:43.977" Score="3" Body="&lt;p&gt;I'd recommend approaching this as a two-step process of cutting the shape and then preparing the edges. In my experience the cut edges of 5 mil annealed copper (0.005&quot;, about the size you're looking at) are always sharper than I want to be casually handling, even without burrs. It's also a bit of a delicate edge to ease because of how malleable thin, annealed copper is.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Cutting&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I suspect die-cutting is the way to go here. Most projects I've seen or done with this material have been smaller-scale, where it's practical to cut one to a few pieces using a variety of hand snips. These snips give you basically a zero-clearance shear and make the cleanest cut (rotary cut-off wheels, especially hand-held, are by far the worst) but the edge still demands attention. I expect this would be comparable to the positive/negative die cutting method you've considered. With a thinner foil, you might be able to get results with a knife edge but 5+ mil is thick enough, and copper ductile enough, that I'd expect more of an embossing effect if you tried it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The challenge is that you're going to want very little clearance between the positive and negative dies. For example, using &lt;a href=&quot;http://www.flat-stock.com/~store/Merchant2/merchant.mv?Screen=PROD&amp;amp;Store_Code=T&amp;amp;Product_Code=DICLI-DIE_CLEARANCE&amp;amp;Category_Code=DICLI&quot; rel=&quot;nofollow&quot;&gt;this die clearance calculator&lt;/a&gt; suggests 5% of thickness as the &quot;regular&quot; clearance for any reasonable set of parameters (C110 H01 being the closest material option). That would be 5–25 µm clearance on your die. I'm hardly an expert on die-cutting but this doesn't sound like something you're going to achieve in-house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't mean you need to resort to specialty fabrication, though, especially if perfection is not the goal. My advice, since your stock is inexpensive, is to make a prototype in-house ASAP with as reasonably tight a fit as you can achieve between your positive and negative dies, using your available tools and whatever method you're most comfortable with. It's very possible that will serve your purpose. (You'll need to keep your negative die aligned on the press, of course.)&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Edge Preparation&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Depending on the results of your cutting process, your edges might get rolled a bit. If this happens, I'd try placing the cutout concave-down on a sheet of medium-fine grit sandpaper and giving it a few strokes with light pressure. This might be sufficient, since a workable die shouldn't produce any burrs, and it's &quot;quick and dirty&quot; enough to not be infeasible for a batch of several hundred cutouts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't get any dishing, that seems like a bit of an issue, as your edge isn't exposed all along one plane. But believe it or not, you can do quite a bit with something as simple as a scouring pad, when working with a material as soft as annealed copper. If you need a bit more abrasive power that will still fit nicely into the nooks of your shape without creating any sharp nicks or dents, try a sanding sponge. If you need to get into smaller nooks or you're worried about bending/crimping your cutouts using a pad or sponge, a small round file can also do the job, though likely more painstaking labor with more aggressive action (you could accidentally cut a notch in the edge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possibility that a bit of die-roll would actually help in the edge preparation step might seem to imply that more clearance is actually helpful. To some extent that may even be true; however, the more stretching-deforming action you have along your edge, the less likely you are to get a clean cut without burrs. As the edge rolls, it will become work-hardened. The irregular shape of your oak leaf cutout won't help here, as stresses may be less consistent along the path of the cut, and you may start to move more in the direction of tearing/ductile fracture. So I would still start by trying for the smallest-clearance dies you can quickly prototype.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again, it's possible that you'll decide the rough edge that results from your cutting method isn't too sharp for your purposes after all. It doesn't sound like you're producing a piece that will be handled very frequently or by very many people after it's finished and you're going to have to handle the assembly with care simply by virtue of working with such thin copper. Yet another reason to start by prototyping and then address problems as they appear, rather than ahead of time.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-04-05T22:29:13.977" LastActivityDate="2016-04-05T22:29:13.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8309" PostTypeId="1" CreationDate="2016-04-06T01:10:40.813" Score="3" ViewCount="10321" Body="&lt;p&gt;I am working on a design for a small lathe that could double as a horizontal drilling machine like a Westhoff.&lt;/p&gt;&#xA;&lt;p&gt;My plan is to have specific geared speeds, rather than continuously variable speed.&lt;/p&gt;&#xA;&lt;p&gt;I am wondering what the best type of motor would be. Cost is no object, but it needs to be compact, no more than 6&amp;quot; in diameter and hopefully less.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Startup torque is not required, because in this type of application you can always start the motor unloaded; in fact, even if the motor had zero startup torque I could potentially add a small starter motor to start the main motor, if that would be better.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It would run on normal AC household current&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My thought is to find a motor with a high peak torque rating and then just run the motor at that ideal speed, using gears to adjust speed as necessary&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Obviously, the more horsepower the better&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Some additional points to consider:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For the class of lathe/mill intended, the target HP would be between 1 HP and 2 HP&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A lighter or more compact motor is desirable, so that the overall footprint of the lathe is minimized. Therefore, a motor that has relatively high HP for its weight/size is desirable.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;High-end lathes use a slipping clutch to protect the motor and gearing from excessive loads. The clutch is set to a torque such that the maximum possible load would not break any of the teeth in any of the gears.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A single speed motor is fine because a high-end lathe/mill will have a gearing system that determines the RPM of the machine and allows it to be varied.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is important that the motor have strong inertial force because when metal is being cut sometimes a hard inclusion will be present so that the load on the motor can unexpectedly spike up for a fraction of a second when the cutter hits the inclusion.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is important for the lathe's speed to be as consistent as possible because if the lathe has variability in speed as the load varies, then what will happen is the quality of the finish of the cut will be degraded. What is wanted is for that that motor to keep turning at exactly the same speed regardless of the variability of the load.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cost is not a factor, because other parts of the lathe will be much more expensive to construct than the motor, so the motor is a relatively minor part of the cost to construct the lathe.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;What type of motor best fits these characteristics?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2021-08-10T03:27:23.120" LastActivityDate="2021-08-11T10:26:33.030" Title="Type of motor for a small lathe" Tags="|motors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8310" PostTypeId="1" CreationDate="2016-04-06T03:56:22.517" Score="1" ViewCount="231" Body="&lt;p&gt;I am trying to connect to a FTP server on a raspberry pi and transfer a file onto the server. I am using the SIM5320A chip on a breakout module and I am having trouble getting the module to connect to the server. I have forwarded the proper FTP port (21) on the router that the pi is connected to and I also have the vsftpd set up on the pi (which is a FTP linux program). I have tested the ftp server by transferring something from an ftp client on my computer to the ftp server on the pi and it works. I will post the commands I used as well as some of my thoughts. Please help me out and give me feedback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AT+CGSOCKCONT=1,&quot;IP&quot;,&quot;fast.t-mobile.com&quot; (something I am not sure about is the apn, does anyone know if this one for tmobile is correct? I can't find that much information online)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AT+CFTPSERVER=&quot;xxx.xxx.xxx.xxx&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AT+CFTPPORT=21&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AT+CFTPUN=&quot;*******&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AT+CFTPPW=&quot;*******&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AT+CFTPPUT=&quot;/files/fromcell.txt&quot; - At this point i should be prompted to enter the text that is going to be in this txt file after I save, but instead of prompting me, the module waits while the red light on it flashes 200 ms on and 200 ms off which is supposed to mean that the data connection I requested is active. It then gives me an error that says &quot;+CFTPPUT: 215&quot; and this error code means &quot;failed to connect socket&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any insight on what the APN is and how it should be set? I can't seem to find much else information on how to set it online. &lt;/p&gt;&#xA;" OwnerUserId="5885" LastActivityDate="2016-04-06T03:56:22.517" Title="AT Commands for FTP Server Connection on SIM Modem" Tags="|communication|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8311" PostTypeId="1" AcceptedAnswerId="8335" CreationDate="2016-04-06T05:07:05.793" Score="3" ViewCount="7310" Body="&lt;p&gt;Is there a big difference between salt water and normal tap water in terms of thermal properties (heat capacity, for example)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, can anyone give me a table of thermal and thermodynamic properties of salt water? &lt;/p&gt;&#xA;" OwnerUserId="5886" LastEditorUserId="5899" LastEditDate="2016-04-07T08:38:13.013" LastActivityDate="2016-04-07T13:28:20.833" Title="Thermal properties of salt water" Tags="|fluid-mechanics|thermodynamics|heat-transfer|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8313" PostTypeId="2" ParentId="8292" CreationDate="2016-04-06T06:12:49.343" Score="1" Body="&lt;p&gt;To be honest this came entirely from chance and I still have NO idea what caused the problem. The answer by @ChrisJohns will probably be a more likely cause of the issue, as for me I'm simply going to put it down to a bug, because this is how I fixed it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turn it off and on again. Make your torus/etc and put it in place, but if subtract doesn't work, undo the subtraction, save, quit, reopen and try it again, one subtraction at a time. Like I said, no idea what the cause is, so if this still doesn't work take a look at @ChrisJohns answer&lt;/p&gt;&#xA;" OwnerUserId="5878" LastActivityDate="2016-04-06T06:12:49.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8314" PostTypeId="2" ParentId="8311" CreationDate="2016-04-06T08:23:57.040" Score="1" Body="&lt;p&gt;The thermal conductivity of seawater is 0.6 W/mK at 25 °C and a salinity of 35 g/kg. The thermal conductivity decreases with increasing salinity and increases with increasing temperature.&lt;/p&gt;&#xA;" OwnerUserId="5744" LastActivityDate="2016-04-06T08:23:57.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8315" PostTypeId="1" AcceptedAnswerId="16672" CreationDate="2016-04-06T10:41:47.753" Score="2" ViewCount="776" Body="&lt;p&gt;Suppose there is a cubic material with an internal heat source ($\Delta q / \Delta t =$ Constant), and is immersed in a sufficiently large amount of water. Now I would like to use finite difference method to simulate the steady state temperature distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a natural convection boundary condition within the range of laminar flow can be assumed, and the maximum temperature should not exceed 100 $^\circ$C. As the surface temperature distribution is not uniform, each surface node should have its unique convective heat transfer coefficient ($h$), as this parameter depends on the temperature difference between the solid wall and the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But by looking up some heat transfer textbooks, I have never seen people address to the problem in such a way. I only found some treatments and correlation equations on the scenario of uniform temperature distribution of different geometries, while in my simulation, the temperature of different surface nodes are different. Giving it a second thought, I realized that convection heat transfer coefficient is dependent upon fluid properties, which will make it extremely complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is it possible to perform such calculations with finite difference method? Is there a method to at least estimate the convective heat transfer coefficient?&lt;/p&gt;&#xA;" OwnerUserId="5891" LastActivityDate="2017-08-08T19:20:16.087" Title="Convective heat transfer coefficient in finite difference method boundary conditions" Tags="|thermodynamics|heat-transfer|chemical-engineering|thermal-conduction|convection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8316" PostTypeId="1" AcceptedAnswerId="8320" CreationDate="2016-04-06T13:07:24.357" Score="9" ViewCount="4706" Body="&lt;p&gt;I have about 12 bottles of water. If I want to minimize the energy consumption of my refrigerator, should I put these bottles in the fridge or leave them out?&lt;/p&gt;&#xA;" OwnerUserId="424" LastEditorUserId="368" LastEditDate="2016-04-07T00:04:29.120" LastActivityDate="2023-06-12T10:42:22.420" Title="Difference in energy consumption between empty fridge and fridge filled with bottles of water" Tags="|thermodynamics|energy-efficiency|refrigeration|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8317" PostTypeId="1" CreationDate="2016-04-05T18:01:14.163" Score="0" ViewCount="16" Body="&lt;p&gt;How do people design &lt;em&gt;height gaining&lt;/em&gt; based on nested poles ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this picture (I made them) :&#xA;&lt;a href=&quot;https://i.stack.imgur.com/FC5OU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FC5OU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to achieve height based on these 3 poles, how they should behave? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this how the behave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bkAuA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bkAuA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had been searching over the internet I haven't found the related material, and it's kinda hard what to type. I'm doing some tasks on designing a simple robot (no foundation knowledge but with will) and my English is sort of limited, especially to technical stuff but I hope you understand it .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Imagine like there's somehow an auto increasing portable stair to get over the tree, the poles(both sides) are consisted of  poles fragment on each of them, how the fragments behave in this case. &lt;/p&gt;&#xA;" OwnerUserId="5883" OwnerDisplayName="Bravo" LastActivityDate="2016-04-06T13:29:32.873" Title="Principal on gaining more height based on nested poles?" Tags="|design|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2016-04-06T13:42:38.033" ContentLicense="CC BY-SA 3.0" />
  <row Id="8318" PostTypeId="2" ParentId="8302" CreationDate="2016-04-05T22:31:16.990" Score="1" Body="&lt;p&gt;&lt;em&gt;Telescoping&lt;/em&gt; is the English term for this and a search for &lt;a href=&quot;https://www.google.com/search?q=telescoping%20mechanisms&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;gws_rd=cr#q=telescoping%20mechanisms&amp;amp;tbm=isch&amp;amp;pws=0&amp;amp;gws_rd=cr&quot; rel=&quot;nofollow&quot;&gt;telescoping mechanisms&lt;/a&gt; will give lots of info on the various approaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A motor + transmission mechanism is called an actuator. So searching for &lt;a href=&quot;https://www.google.com/search?q=telescoping%20actuator&amp;amp;biw=1280&amp;amp;bih=610&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&quot; rel=&quot;nofollow&quot;&gt;telescoping actuator&lt;/a&gt; will show you devices that automatically extend with some sort of external or internal control.&lt;/p&gt;&#xA;" OwnerUserId="7744" OwnerDisplayName="hauptmech" LastActivityDate="2016-04-06T06:26:31.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8319" PostTypeId="2" ParentId="8316" CreationDate="2016-04-06T13:49:25.360" Score="6" Body="&lt;p&gt;There are two aspects to this.&lt;/p&gt;&#xA;&lt;p&gt;First if your fridge is filled with anything other than air then you will save the energy associated with cold air escaping whenever you open the door.&lt;/p&gt;&#xA;&lt;p&gt;However if you put bottles of water in your fridge it will take a lot more work by the fridge to cool them down than the equivalent volume of air as a litre of water at room temperature contains a lot more heat than a litre of air at the same temperature so more work is required from the fridge to cool a litre of water than a litre of air.&lt;/p&gt;&#xA;&lt;p&gt;If you intend to put the bottles in the fridge at some point it is certainly better to fill the fridge now rather than put them in individually. But if you don't actually need the water to be cool ever it would be better to fill the spare space with empty bottle or blocks of styrofoam.&lt;/p&gt;&#xA;&lt;p&gt;This is because of the relative &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_capacity&quot; rel=&quot;nofollow noreferrer&quot;&gt;specific heat capacities&lt;/a&gt; of air and water. The specific heat capacities of air and water by mass are 1 and 4 KJ/Kg·K, respectively, but water is around 1,000 times denser than air so the heat capacities by volume are different by a factor of 4,000.&lt;/p&gt;&#xA;&lt;p&gt;So it is obvious that the energy wasted by loss of air when you open the fridge is fairly trivial compared to the work needed to cool the same volume of water.&lt;/p&gt;&#xA;&lt;p&gt;Typically a fridge will use around 1 unit of mechanical power for every 3 units of cooling power.&lt;/p&gt;&#xA;&lt;p&gt;So overall there is a small advantage in minimising the volume of free air in a fridge but it is only worth doing this by filling it with things that you want to refrigerate anyway or solids (or containers or foams) which have a low volumetric heat capacity.&lt;/p&gt;&#xA;&lt;p&gt;This only applies to losses from opening the door though, once the contents are at a stable temperature it makes no difference to energy consumption what is actually in it, assuming that is is reasonably well-sealed.&lt;/p&gt;&#xA;&lt;p&gt; &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="42024" LastEditDate="2023-06-12T10:42:22.420" LastActivityDate="2023-06-12T10:42:22.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8320" PostTypeId="2" ParentId="8316" CreationDate="2016-04-06T14:21:19.033" Score="10" Body="&lt;p&gt;The fridge works by pumping heat from its interior to its exterior. The amount of work it needs to do is proportional to the amount of heat entering the interior (and also the temperature difference between interior and exterior).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ways that heat gets into the interior:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Putting warm objects into the fridge.&lt;/li&gt;&#xA;&lt;li&gt;Heat leaking through the walls of the fridge.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you put a warm bottle of water into the fridge, there is a &lt;em&gt;transient&lt;/em&gt; load associated with cooling that water down. However, once the water is cool, it has &lt;em&gt;no further effect&lt;/em&gt; on the work that the fridge must do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat leaks into the fridge at a rate that is proportional to the temperature difference between outside and inside. This leakage is &lt;em&gt;constant&lt;/em&gt;, and is not affected in any way by any objects that may be inside the fridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if the fridge is empty, you should just turn it off. If you want the water cool, you should put it inside the fridge and leave it on. However, you should not turn the fridge off and on frequently &amp;mdash; there's a huge transient load associated with cooling down the interior walls, etc. of the fridge itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, every time you open the door, some of the cool air inside is replaced with warm air from outside. This also represents a small transient load, but if you're opening the door a lot, it can add up.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-04-06T14:21:19.033" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8321" PostTypeId="2" ParentId="8316" CreationDate="2016-04-06T15:19:46.797" Score="-2" Body="&lt;p&gt;The added weight of the water bottles will cause the contact patch between the feet of the refrigerator and the floor to deform ever so slightly, thus making better thermal contact between the feet and the (presumably warmer) floor.  Hence, the addition of the water bottles will increase the heat loss through the feet by an amount so small as to be undetectable by even the most sensitive heat flux sensors.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-04-06T15:19:46.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8322" PostTypeId="2" ParentId="8283" CreationDate="2016-04-06T15:24:14.653" Score="2" Body="&lt;p&gt;For a light weight design (e.g. an aircraft) you would try to achieve 100% of the allowable loads for &lt;em&gt;every&lt;/em&gt; part of the structure, under some operating condition (not necessarily the &lt;em&gt;same&lt;/em&gt; condition for every part, or course). But in building design, other considerations may be more important - for example using consistent sized components could minimize cost (by ordering in bulk) and/or simplify the construction. If a set of beams are all specified as the same size, you can't install the &quot;wrong size&quot; in some particular location. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common components like beams are manufactured in a standard range of sizes. If the full range of different sizes in the relevant building codes and standards are not available locally &quot;off the shelf&quot;, it may be cheaper and/or quicker to use over-sized beams rather than find a supplier for the minimum size that is actually required. Of course that will reduce the load factors in the oversized parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the &lt;em&gt;stiffness&lt;/em&gt; of some parts of the may be more important than their &lt;em&gt;strength&lt;/em&gt;. For example it would be a bad idea to design a floor that was perfectly safe to carry the expected loads, but which flexed up and down when you walked across the room.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-04-06T15:24:14.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8323" PostTypeId="2" ParentId="8316" CreationDate="2016-04-06T20:34:50.743" Score="6" Body="&lt;p&gt;If you have unused space in the fridge, then it's better to fill it with something than to leave it air.  This is because every time you open the door, less air will be exchanged with the room, so less to cool again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're not going to open the fridge, then it doesn't matter.  As others have already pointed out, putting the water bottles in the fridge represents a one time energy cost.  Once they are cold, they cost no further energy, and reduce the volume of cold air lost when the door is opened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you are really after less energy usage and you have excess space in the fridge, the best long term answer is to get a smaller fridge.  Get one that is just large enough to hold what you need to keep cold.  In smaller fridges, the cold box has a smaller surface area, which is where heat comes in when everything is at steady temperature and the door isn't opened.  All else being equal, the fridge with the smallest cold box surface area will require the least power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course all else isn't equal.  There are some cost/performance/longevity tradeoffs in the cooling system, but mostly it's about the insulation.  That is also a cost tradeoff, so different models have different power requirements for the same cold box.  At least here in the US, sellers are required to list power consumption figures for refrigerators.  Look for the one that requires the least power of those that have the minimum size cold box you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other things you can do to minimize power consumption:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Make sure the radiator has good ventillation.  The cooling system is a heat pump that moves heat from the cold box to the outside of the fridge.  The radiator is where it gets rid of that heat.  If airflow to the radiator is blocked, then the heat pump has to work harder (make the radiator hotter) for the same cooling.  This costs power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radiator is usually a metal &quot;mesh&quot; sort of thing on the back of the fridge.  Modern fridges are designed so that there is some minimum space for air flow with the fridge backed up all the way against at wall.  Moving it out from the wall a little and making sure that air can flow in at the bottom and come out easily at top can help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Keep the radiator clean.  Dust is a good insulator, which will make the cooling system work harder to push the exhaust heat thru the dust layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Put the fridge in a colder place if you can.  The colder the radiator, the less hard the cooling system has to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Minimize the time the door is open.  If you have several things to move in and out, open the door once, move everything in quick succession, then close the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-04-06T20:34:50.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8324" PostTypeId="1" AcceptedAnswerId="37328" CreationDate="2016-04-06T23:33:31.797" Score="15" ViewCount="160060" Body="&lt;p&gt;I'm trying to calculate the clamping force resulting from torquing a nut and bolt to a particular level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found this formula in various forms in a lot of places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T = KDP$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$T$ = Torque (in-lb) &lt;/li&gt;&#xA;&lt;li&gt;$K$ = Constant to account for friction (0.15 - 0.2 for these units)&lt;/li&gt;&#xA;&lt;li&gt;$D$ = Bolt diameter (inches)&lt;/li&gt;&#xA;&lt;li&gt;$P$ = Clamping Force (lb)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I applied this to my problem&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$T = 0.6\text{ N-m} = 5.3\text{ in-lb}$&lt;/li&gt;&#xA;&lt;li&gt;$D = 3\text{ mm} = 0.12\text{ in}$&lt;/li&gt;&#xA;&lt;li&gt;$K = 0.2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This gives $P = \dfrac{T}{KD} = 220\text{ lb} = 100\text{ kg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I have two questions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The result seems way too high. I'm using a tiny M3 bolt, and not much torque. I can't see how this would result in 100 kg force. Can anyone see the error?&lt;/li&gt;&#xA;&lt;li&gt;The formula doesn't take account of thread pitch. I would expect a fine thread to give more clamping force for the same torque. Is there a formula that accounts for thread pitch?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2135" LastEditorUserId="1832" LastEditDate="2016-04-07T10:22:21.993" LastActivityDate="2023-06-05T22:37:48.563" Title="Calculation of Clamping force from bolt torque" Tags="|mechanical-engineering|applied-mechanics|torque|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8325" PostTypeId="2" ParentId="8221" CreationDate="2016-04-07T00:29:19.257" Score="2" Body="&lt;p&gt;Assuming you are wanting to minimize turbulence in the region of the pump, you would probably need to use flow straighteners somewhere upstream of the pump.  You would have to study the optimal diameter and length of the straighteners.  MIT simply used &lt;a href=&quot;https://en.wikipedia.org/wiki/Flow_straightener#A_low-cost_handmade_flow_straightener&quot; rel=&quot;nofollow&quot;&gt;drinking straws&lt;/a&gt; to do this in a wind tunnel.  Here are a couple of the more well known papers on the subject:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://fluidsengineering.asmedigitalcollection.asme.org/article.aspx?articleid=1433483&quot; rel=&quot;nofollow&quot;&gt;Reducing Water Tunnel Turbulence by Means of a Honeycomb&lt;/a&gt; by Lumley and McMahon&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://fluidsengineering.asmedigitalcollection.asme.org/article.aspx?articleid=1423349&quot; rel=&quot;nofollow&quot;&gt;Control of Free-Stream Turbulence by Means of Honeycombs: A Balance Between Suppression and Generation&lt;/a&gt; by Loehrke and Nagib&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="511" LastActivityDate="2016-04-07T00:29:19.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8326" PostTypeId="1" CreationDate="2016-04-07T05:06:02.333" Score="8" ViewCount="207" Body="&lt;p&gt;I am about to do a project about &lt;a href=&quot;https://en.wikipedia.org/wiki/MIT_Engineering_Systems_Division#Overview&quot; rel=&quot;nofollow noreferrer&quot;&gt;engineering systems&lt;/a&gt; in a Systems Engineering course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to sketch an engineering system in which clinical decision support systems are being used to diagnose patients by a medical doctor/physician. It doesn't have to be fully correct. I just need to show that I understand how different actors/subsystems interact with each other, and who have economic/political/social interest in this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to sketch something, but I'm not sure if it makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xTZSZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xTZSZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that a patients contacts his/her doctor. They cooperate to figure out what is wrong with the patient. The doctor enters some information into the (intelligent) decision support system (=IDSS). The IDSS has a user interface, an inference engine and a knowledge base. The knowledge base gets its information both from doctors entering stuff but also from the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for this to work, there must be some software developers who build, maintain and control the system, but also some insurance companies who will cover any costs if something goes wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outside of the system, there are students or researchers who potentially can benefit from using the system, both by learning but also by searching for patterns in the huge amount of data. There are also politicians who are interested in this system since the system potentially is more efficient and reduces human mistakes. But there are also hackers or IT criminals who want to gain access to this data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is kind of a fluffy assignment, but I just need some good advice, so I really hope that you will help me out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I forgot something in this system? Is there something that doesn't make sense at all?&lt;/p&gt;&#xA;" OwnerUserId="5858" LastEditorUserId="282" LastEditDate="2016-06-09T01:21:23.207" LastActivityDate="2016-06-09T01:21:23.207" Title="Sketching an engineering system in healthcare" Tags="|systems-design|systems-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8327" PostTypeId="1" CreationDate="2016-04-07T08:31:32.847" Score="2" ViewCount="26" Body="&lt;p&gt;We plan a conveyor that will dump something into containers. A level gauge on the tip of the container will measure the heap below. The conveyor is placed on a turntable and will e rotated horizontally a piece when the heap below is at a certain height, to make use of more of the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system will look something like this but only drop of to one side:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/80Ttq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/80Ttq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;em&gt;Image source: &lt;a href=&quot;http://www.dbtechnologiesbv.nl/products/turntable-conveyor&quot; rel=&quot;nofollow noreferrer&quot;&gt;DB Technologies BV&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to decide between either a continuos signal for the position of the turntable, or discrete signals at every point the conveyor will be. we are talking about 6-7 position signals in that case (3 per container plus maybe one so the conveyor is out of the way when both containers are moved). From the mechanical or process standpoint, I don't see a need for a continuos signal, 3 positions per container are fine enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that this amount of signalers is even easier to implement than a continous sensor. Is this assumption correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an argument from the control and measurement system side of things for a continous signal?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-04-07T08:36:41.827" LastActivityDate="2016-04-07T08:36:41.827" Title="Decission between several position switches and one continuos position signal for controling rotating conveyor" Tags="|mechanical-engineering|control-engineering|measurements|positions-measurement|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8328" PostTypeId="1" CreationDate="2016-04-07T09:31:01.547" Score="5" ViewCount="3375" Body="&lt;p&gt;I would like to ask what is the difference (if any) between normalization and non-dimensionalization. I will try to explain what I have done until now with an equation that I would like to work with and you can make your comments on this specific example. The mass balance is given by:&#xA;$$\frac{\partial C_{i}}{\partial t}=-u_s\frac{\partial C_{i}}{\partial z}+\rho_b \sum_{k=1}^{N_{reac}} \nu_{j,k} R_{i,k}$$&#xA;If we choose as characteristic values the:&#xA;$$L_o, F_o, T_o, P_o, m_o$$&#xA;$$u_o = \frac{F_o Rg T_o}{P_o L_o^2}$$&#xA;$$C_o = \frac{P_o}{Rg T_o}$$&#xA;length, molar flowrate, temperature, pressure, mass accordingly, and by substituting then we get:&#xA;$$\frac{u_o C_o}{L_o}\frac{\partial C^*_{i}}{\partial t^*}=-\frac{u_o C_o}{L_o}u_s^*\frac{\partial C^*_{i}}{\partial z^*}+\frac{m_o}{L_o^3}\rho_b^* \frac{F_o}{m_o}\sum_{k=1}^{N_{reac}} \nu_{j,k} R^*_{i,k}$$&#xA;where $^*$ denotes the dimensionless (?normalized?) variables. After some more calculations we get:&#xA;$$\frac{\partial C^*_{i}}{\partial t^*}=-u_s^*\frac{\partial C^*_{i}}{\partial z^*}+\frac{F_o}{L_o^2 u_o C_o}\rho_b^* \sum_{k=1}^{N_{reac}} \nu_{j,k} R^*_{i,k}$$&#xA;BUT&#xA;$$\frac{F_o}{L_o^2 u_o C_o}=\frac{F_o}{L_o^2 \frac{F_o Rg T_o}{P_o L_o^2} \frac{P_o}{Rg T_o}}=1$$&#xA;So is this equation normalized? or non-dimensinalized? Shouldnt there be some Peclet / Reynolds etc number to work with? If for example I try to change the inlet conditions then this number ($\frac{F_o}{L_o^2 u_o C_o}$) will still be equal to 1. Can anyone point out a book or a site where I could read and understand more about this subject?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="4655" LastActivityDate="2022-05-02T13:09:03.150" Title="Normalization and Non-Dimensionalization of a mass balance equation" Tags="|fluid-mechanics|chemical-engineering|modeling|numerical-methods|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8329" PostTypeId="2" ParentId="8324" CreationDate="2016-04-07T09:47:09.477" Score="10" Body="&lt;p&gt;That figure is about right for a low tensile bolt. See also this &lt;a href=&quot;http://www.engineersedge.com/calculators/torque_calc.htm&quot; rel=&quot;noreferrer&quot;&gt;calculator&lt;/a&gt; and &lt;a href=&quot;https://www.fastenal.com/content/feds/pdf/Torque-Tension%20Chart%20for%20Metric%20Fasteners.pdf&quot; rel=&quot;noreferrer&quot;&gt;this table&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a reality check if we approximate to a cross sectional area of 7 mm&lt;sup&gt;2&lt;/sup&gt; and a load of 1000 N that gives a tensile stress of 140 MPa which is below yield even for low tensile steels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular context, where torque is known, the thread pitch doesn't come into it as you are calculating based on the relationship between torque, friction and tension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fine thread will (all else being equal) be stronger than a coarse one. Some methods involve calculating clamping force by tightening the bolt by a predetermined angle and here pitch does matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A screw thread is essentially a variation of a wedge or inclined plane and can provide very high mechanical advantage before you even consider the leverage of the wrench/driver used. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-11T21:31:31.347" LastActivityDate="2016-04-11T21:31:31.347" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8330" PostTypeId="2" ParentId="8139" CreationDate="2016-04-07T10:11:00.463" Score="2" Body="&lt;p&gt;Your system is not reaching the target value. Possibly this is due to static friction in the actuator mechanism, which causes it not to move until the force exceeds some threshold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you are using a PID controller, the residual error will cause the I error term to accumulate slowly. This will cause the actuator force to increase until it is enough to cause actual movement. It seems that this movement will cause a small jerk, which will move the table further than intended and cause opposite error value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way you can test this is to change the target height in small steps (e.g. 1 µm) and see what is the smallest distance you can reliably move. You can then either choose to improve the actuator, or to set this as the deadband value in your controller.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-04-07T10:11:00.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8331" PostTypeId="2" ParentId="8291" CreationDate="2016-04-07T10:25:43.917" Score="3" Body="&lt;p&gt;My first guess for any medical pump is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Peristaltic_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;peristaltic pump&lt;/a&gt;, as they can pump one fluid then another without contamination (by swapping lines).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Couple this with the fact that they can also be multi-channel pumps, and I'd almost guarantee that this is what it is. I've seen maybe one instance of non-peristaltic multiple pumps on one shaft (not counting turbo chargers and jet engines).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ohe0K.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ohe0K.gif&quot; alt=&quot;24 channel peristaltic pump&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-04-07T10:25:43.917" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8333" PostTypeId="1" AcceptedAnswerId="8351" CreationDate="2016-04-07T12:20:47.343" Score="4" ViewCount="2461" Body="&lt;p&gt;I am measuring the rotational speed of a shaft using a disk with 60 cuts and an optical fork. The signal from the optical fork is sent to a frequency to voltage converter and after that to a data acquisition system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the working principle of this frequency to voltage converter? Is it counting pulses?&lt;/p&gt;&#xA;" OwnerUserId="5744" LastEditorUserId="5899" LastEditDate="2016-04-07T15:50:50.977" LastActivityDate="2016-04-18T14:16:32.190" Title="How does a frequency to voltage converter work?" Tags="|measurements|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8335" PostTypeId="2" ParentId="8311" CreationDate="2016-04-07T12:38:26.333" Score="3" Body="&lt;p&gt;MIT provide a table of water properties at a range of salinity and temperature values: &lt;a href=&quot;http://web.mit.edu/seawater/&quot; rel=&quot;nofollow&quot;&gt;http://web.mit.edu/seawater/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also excel/matlab correlations as well.&lt;/p&gt;&#xA;" OwnerUserId="2616" LastEditorUserId="2616" LastEditDate="2016-04-07T13:28:20.833" LastActivityDate="2016-04-07T13:28:20.833" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8336" PostTypeId="2" ParentId="8328" CreationDate="2016-04-07T12:49:19.280" Score="1" Body="&lt;p&gt;Sorry with not being familiar with all of your variables. If it is a mass balance as you say, then you can't get a Reynolds or Péclet number. A Reynolds number appears in the dimensional analysis of the impulse balance (the Reynolds number contains viscosity which originates from the stress tensor for Newtonian fluids). I'm also not very familiar with Péclet, but as it contains thermal diffusivity I suppose it appears in the dimensional analysis of the energy equation.&lt;/p&gt;&#xA;&lt;p&gt;From my understanding, non-dimensionalizing is a special case of normalizing. In dimensional analysis you normalize a variable using a characteristic value of this variable (to receive equations with unitless variables). The idea behind dimensional analysis is that relationships in physics should not depend on your measurement units. This is used by measuring quantities in multiples of the respective characteristic value picked.&lt;/p&gt;&#xA;&lt;p&gt;And sorry, I could only point you to German books on dimensional analysis. Good books on fluid mechanics usually also have an introduction to dimensional analysis. (I prefer Schade, Strömungslehre, 2013, e-ISBN 978-3-11-029223-7.) an American author book well know as &lt;a href=&quot;https://www.academia.edu/40559428/Fluid_Mechanics_by_Yunus_Cengel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cengle, Fluid Dynamics&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="90" LastEditDate="2022-05-02T13:09:03.150" LastActivityDate="2022-05-02T13:09:03.150" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="8337" PostTypeId="2" ParentId="8333" CreationDate="2016-04-07T15:08:05.943" Score="2" Body="&lt;h3&gt;TL;DR&lt;/h3&gt;&#xA;&lt;p&gt;The history of FVCs, basically and chronologically, goes as follows:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a charge-dispenser made of a voltage limiter, a capacitor, and diodes;&lt;/li&gt;&#xA;&lt;li&gt;Two Op-Amps and capacitor&lt;/li&gt;&#xA;&lt;li&gt;Two Op-Amps and capacitor plus bleed resistor&lt;/li&gt;&#xA;&lt;li&gt;LM131&lt;/li&gt;&#xA;&lt;li&gt;Cascading two or more fast Sallen-Key filters&lt;/li&gt;&#xA;&lt;li&gt;Phase-Locked loop&lt;/li&gt;&#xA;&lt;li&gt;Digital&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;You don't say which frequency to voltage converter (FVC) you are using, or what it is based upon (LM331, LM131, AD650, ADVFC32, LM2907/LM2917 etc. There are many ways to accomplish frequency to voltage conversion - there are both analogue and digital methods, and within those camps there are again various ways of achieving the same thing.&lt;/p&gt;&#xA;&lt;p&gt;If you want to see how the FVC came about then here is an interesting &lt;em&gt;historical&lt;/em&gt; explanation, from basics: &lt;a href=&quot;http://electronicdesign.com/digital-ics/whats-all-frequency-voltage-converter-stuff-anyhow&quot; rel=&quot;nofollow noreferrer&quot;&gt;What's All This Frequency-To-Voltage Converter Stuff, Anyhow?&lt;/a&gt; It is written by Robert A. Pease, the guy who designed the LM131 for National, so this information is straight from the horse's mouth, so to speak. A &lt;a href=&quot;http://electronicdesign.com/datasheet/whats-all-frequency-voltage-converter-stuff-anyhow-pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PDF&lt;/a&gt; is also available from there.&lt;/p&gt;&#xA;&lt;p&gt;I will attempt to summarize the article below:&lt;/p&gt;&#xA;&lt;h2&gt;Analogue&lt;/h2&gt;&#xA;&lt;h3&gt;First Version&lt;/h3&gt;&#xA;&lt;p&gt;Thirty years ago, a guy asked him if he could show him how to make a Frequency-to-Voltage converter (FVC), while he was working at at George A. Philbrick Researches. He designed a charge-dispenser made of a voltage limiter, a capacitor, and diodes. Evidently, it worked pretty well.&lt;/p&gt;&#xA;&lt;h3&gt;Second Version&lt;/h3&gt;&#xA;&lt;p&gt;In 1964 he put a new version into the old Philbrick Applications Manual.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jaPbS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jaPbS.jpg&quot; alt=&quot;FVC 1965&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The first amplifier has a limited output voltage. The p-p voltage&#xA;across the capacitor is pretty well established:&lt;/p&gt;&#xA;&lt;p&gt;V p-p = 2Vz + 2Vd - 2Vd&lt;/p&gt;&#xA;&lt;p&gt;So, the charge (Q = C × V p-p) flows through the feedback resistor of&#xA;the second amplifier. The output voltage will be, on the average:&lt;/p&gt;&#xA;&lt;p&gt;Vout = Rf × C × V p-p × f&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Third version&lt;/h3&gt;&#xA;&lt;p&gt;A few years later, he got into the Voltage-to-Frequency Converter (VFC) business and at the same time, he came up with an improved circuit for an FVC (see figure 2).&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The input comparator is set up to accommodate TTL&#xA;signals, but if you put a resistor from the + input to -15 V, you can&#xA;accommodate symmetrical signals; a resistor from the + input to ground&#xA;will cut down the hysteresis and let you handle small signals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OvVkq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OvVkq.jpg&quot; alt=&quot;FVC 1972&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The real improvement in this FVC was the bleeder resistor, the 3.3 MΩ added to the right end of the capacitor.&lt;/p&gt;&#xA;&lt;h3&gt;LM131&lt;/h3&gt;&#xA;&lt;p&gt;After he left Philbrick, he joined National and designed the LM131 voltage-to-frequency converter&lt;sup&gt;3&lt;/sup&gt;, using completely different ideas than any of the Philbrick circuits.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It used Q = I × T, rather than the Q =&#xA;C × V employed by all of the Philbrick ones. It didn't need ±15 V; it&#xA;could run on +15 or +30 or +12 or +5 V—much easier to apply. BUT, it&#xA;still had the same constraint when you used it as an F-to-V converter:&#xA;If you want low ripple, it's hard to get fast response.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Cascading two or more fast Sallen-Key filters&lt;/h3&gt;&#xA;&lt;p&gt;In 1978, he wrote an application note on how to improve the response time of an FVC—in the Linear Apps Handbook.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I showed how to cascade&#xA;two or more fast Sallen-Key filters to give reasonably quick response,&#xA;yet filter out the ripple at 24 dB per octave.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h3&gt;Phase-Locked loop&lt;/h3&gt;&#xA;&lt;p&gt;In 1979, he wrote another App Note showing how to use a phase-locked loop to make a quicker F-to-V converter, about 2 ms.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;That's about 10&#xA;cycles of the new frequency—a further 20:1 improvement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Digital&lt;/h2&gt;&#xA;&lt;h3&gt;Fast clock and digital counter&lt;/h3&gt;&#xA;&lt;p&gt;Recently, a guy asked him how to make a 60-Hz FVC with quick response&#xA;and negligible lag or delay.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I told him that the standard procedure is&#xA;to use a fast clock and a digital counter. But the number of counts&#xA;collected during one period is linearly proportional to the period of&#xA;the signal, and you might have to do some digital computations to&#xA;convert that to a signal representing the frequency. Then I realized&#xA;that a &amp;quot;multiplying&amp;quot; DAC can be used to divide in a reciprocal mode.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;He built it up and it worked. This Frequency-to-Voltage converter settles in one cycle of the frequency and uses only a small number of parts.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TcBpw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TcBpw.jpg&quot; alt=&quot;FVC - CMOS logic&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The digital logic generates a couple of pulses at the time of each&#xA;rising edge of the incoming frequency (you could use some kind of dual&#xA;one-shot multivibrator, but I didn't have any of those around). The&#xA;first pulse loads the data from the CD4040 into the DAC (the pulse&#xA;also disables the path from the clock to the counter to avoid any&#xA;confusion from rippling in the counter). Then the second pulse resets&#xA;the counter.&lt;/p&gt;&#xA;&lt;p&gt;The MDAC has storage registers built in, so the data from the counter&#xA;is fed right in to the DAC when the WRITE-2-bar pulse is applied. The&#xA;MDAC isn't connected in the normal way, with the variable resistance&#xA;in the input path. The fixed resistor is in the input, and the&#xA;impedance controlled by the Digital code is connected as the feedback&#xA;resistor. This permits the multiplying DAC to act as a divider, so the&#xA;reciprocal function is done neatly—not in the digital realm, and not&#xA;in the analog world, but on the cusp between them. (More on this in a&#xA;few months). The LM607BN was chosen for the op amp because you need&#xA;low offset. It's cheap, Vos is only 25 µV typical (60 µV max.), and&#xA;you don't need a trimmer pot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I seriously recommend that you read the &lt;a href=&quot;http://electronicdesign.com/digital-ics/whats-all-frequency-voltage-converter-stuff-anyhow&quot; rel=&quot;nofollow noreferrer&quot;&gt;original article&lt;/a&gt;, as my abridged version has, unfortunately, had to leave out most of the salient technical facts.&lt;/p&gt;&#xA;" OwnerUserId="5899" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-04-18T14:16:32.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8338" PostTypeId="1" CreationDate="2016-04-07T15:29:22.833" Score="2" ViewCount="407" Body="&lt;p&gt;I have a question regarding valves: Let's assume a district heating system where you want to control the mass flow in a sub-station with a valve.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is (in normal cases) the diameter of the valve (fully open) smaller than the diameter of the pipe that is going to the house (substation)?&lt;/li&gt;&#xA;&lt;li&gt;Is this the reason, why a valve has a nominal pressure drop at full opening? Because when the diameter of the pipe and the valve would be equal, then the valve would have no pressure-drop when it is fully open right?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5906" LastEditorUserId="5899" LastEditDate="2016-04-07T17:50:12.547" LastActivityDate="2016-04-07T19:41:37.697" Title="Fluid dynamics: valve and pressure drop" Tags="|pressure|valves|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8339" PostTypeId="1" CreationDate="2016-04-07T17:43:37.100" Score="3" ViewCount="945" Body="&lt;p&gt;I am designing a system wherein an assembly has to be moved up and down( 4 inches) at regular intervals ( every 5 mins) using a Linear Actuator. I wanted constrain the DOF and ensure that the entire assembly moves straight up and down( 1-2 degree misalignment is acceptable). The entire assembly is a cantilever and Hence, I wanted to have some bearings to restrict the moment and ensure straight vertical motion.The weight of the entire assembly is about 50lb and the distance from the support frame (for cantilever) is about 7 inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, is it okie to have sleeve bearings(Red color in picture below) (Ultra-Low-Friction Dry-Running Flanged Sleeve Bearings from McMaster:  &lt;a href=&quot;http://www.mcmaster.com/#standard-sleeve-bearings/=11vp8eo&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mcmaster.com/#standard-sleeve-bearings/=11vp8eo&lt;/a&gt; ) which serves as a guide for the motion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zaOSx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zaOSx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above Image, the fixed frame is rigid and the moving assembly( in light pink) slides on the sleeve bearing carriages attached to the fixed frame( in dark pink color)using the guide block attached to the moving assembly( in green). However, since it is a cantilever I want to provide additional support at the other end .Cost &amp;amp; compactness is a factor for me in this design , so are normal sleeve bearings( &lt;strong&gt;in RED color- ultralow friction&lt;/strong&gt; ) good for support the assembly at the other end . &lt;/p&gt;&#xA;" OwnerUserId="3179" LastActivityDate="2018-02-27T12:44:19.373" Title="Use of sleeve bearings for linear motion" Tags="|mechanical-engineering|civil-engineering|bearings|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8340" PostTypeId="2" ParentId="8339" CreationDate="2016-04-07T18:29:33.690" Score="1" Body="&lt;p&gt;I'm not sure that the sleeve bearings are adding very much to that design. In a cantilever the most important load is usually the bending moment at the supported end and they don't do really anything to support that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;50lb cantilevered to 7 inches isn't by any means unmanageable  and If I were you I would just design the carriage to take all of the load. Here one possibility is to use similar sleeve bearings in a pair of vertical shafts (2 per shaft) which should effectively constrain it in all directions apart from vertical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative way to do this would be to have nylon wheel running in channel shaped rails &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-07T18:29:33.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8341" PostTypeId="2" ParentId="8338" CreationDate="2016-04-07T19:41:37.697" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;While there is some variation on valve type (gate, ball, needle etc.), the answer is yes, the valve body and internals constrict the aperture for flow even when the valve is in the full open position.  This is visually evident when looking at a cross section of a typical valve body (a great example of which is shown here: &lt;a href=&quot;http://news.cision.com/the-nanosteel-company/i/valve-body-cross-section-is-one-example-of-many-large-scale-industrial-part-designs-which-can-be-for,c1227018&quot; rel=&quot;nofollow&quot;&gt;http://news.cision.com/the-nanosteel-company/i/valve-body-cross-section-is-one-example-of-many-large-scale-industrial-part-designs-which-can-be-for,c1227018&lt;/a&gt;).  In addition, most valves (all but ball valves really) force the fluid to significantly shift its path as it travels through the valve, which causes significant pressure drop as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The answer to this question is &quot;mostly&quot;.  As a matter of clarification, recall that there is a pressure drop across a length of pipe.  In the case of an open valve, there is some pressure drop associated with the frictional losses of the fluid across the surface.  However, the vast majority of the pressure losses are due to the constriction of the flow (and sometimes the introduction of drag centers into the fluid) as it flows through the valve body.  This can best be seen by looking at the values presented for valve recovery coefficients (higher is more &quot;pipelike&quot;), a small example of which are at the bottom of this page: &lt;a href=&quot;http://www.mycheme.com/calculating-liquid-pressure-drops-across-control-valves/&quot; rel=&quot;nofollow&quot;&gt;http://www.mycheme.com/calculating-liquid-pressure-drops-across-control-valves/&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5865" LastActivityDate="2016-04-07T19:41:37.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8344" PostTypeId="1" CreationDate="2016-04-08T03:53:45.160" Score="0" ViewCount="909" Body="&lt;p&gt;I'm working on a heat exchanger design project and I'm stuck at one of the first steps - energy and rate balance. In the problem, we have a hydrocarbon stream getting cooled by a water stream. We are given the following values for the Hydrocarbon stream: Vdot=200GPM, T1=135 F, T2=100F, P1=321 psia, with minimum pressure drop of 285 psig, X1=0.684 C3H6, X2=0.316 C3H8. And for the water stream we are given: T1=85 F, P1= 60psig, with an allowable pressure drop of 50psi. I found the values for mCp(T2-T1) for the hydrocarbon stream, but I'm only given values for the inlet temperature and pressure of the cooling water stream. I found the specific heat of the water stream, but I'm still left without the mass flow rate and exit temperature. Can anyone provide some advice on how to obtain one of these values so I can balance both of the streams? Any help would be greatly appreciated. Thank you! &lt;/p&gt;&#xA;" OwnerUserId="4785" LastEditorUserId="4785" LastEditDate="2016-04-08T05:35:07.217" LastActivityDate="2016-04-08T08:51:22.623" Title="Finding mass flow rate and exit temperature in heat exchanger stream" Tags="|mechanical-engineering|heat-transfer|chemical-engineering|heat-exchanger|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8346" PostTypeId="1" CreationDate="2016-04-08T07:54:46.560" Score="2" ViewCount="1165" Body="&lt;p&gt;First of all my english writing is not so good, but I can read it without problems. &#xA;I've got this mechanical system and I need to get some transfer functions. We made all the force diagrams and got the respective equations, they're eight. Now  we need to get the transfer function of $\dot{x}_1\over v_1$, $x_2\over v_3$, $\ddot{x}_3\over v_4$ and $x_4\over v_7$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assuming that the equations are correct, what we were trying to do was for the first T.F clear all $X_2(s)$ from the second equation and replace it on the first equation, then we clear all $sB_1$, $B_2X_1(s)$ from the new equation that we just got from replacing $X_2(s)$, then we're trying to make $sB_1$, $B_2X_1(s)\over V_1(s)$ but there's no way to clear $V_1(s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very difficult to understand, and I don't know if I explained myself, but if someone can tell us what we're doing wrong I'll be really grateful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I0G9a.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I0G9a.jpg&quot; alt=&quot;System&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9Knl2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Knl2.jpg&quot; alt=&quot;Equations&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5916" LastEditorUserId="1832" LastEditDate="2016-05-04T10:26:13.347" LastActivityDate="2022-06-23T06:21:45.553" Title="Mechanical System Transfer Function - Systems Modelling" Tags="|mechanical-engineering|control-engineering|systems-engineering|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8347" PostTypeId="2" ParentId="8344" CreationDate="2016-04-08T08:51:22.623" Score="1" Body="&lt;p&gt;In heat exchanger design, the outlet temperature of the water is usually one of the factors that has to be decided upon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has an upper bound limit of the lowest temperature of your HC stream so 100°F (as you want it to cool the HC stream), but practically you want it lower than this so there is a temperature gradient between the two fluids at all points. In this case you might want to look at a water outlet temperature of 90°F or 95°F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which of these is better will depend on the economics of your system as the first requires more water, but a smaller HEX (as you have a higher driving force) whereas the second needs less water but a larger HEX.&lt;/p&gt;&#xA;" OwnerUserId="2616" LastActivityDate="2016-04-08T08:51:22.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8348" PostTypeId="2" ParentId="8333" CreationDate="2016-04-08T15:55:08.267" Score="2" Body="&lt;p&gt;As a general guide:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method is to provide a pulse of fixed width every time a slot passage occurs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pulse widths are set such that at  maximum pulse rate the output is either JUST 100% or somewhat less than 100%. For example.&lt;br&gt;&#xA;If the max shaft speed = 600 RPM&lt;br&gt;&#xA;then max disk RPS = 10 rps&lt;br&gt;&#xA;so there are 60 slots x 10 rps = 600 pulses per second max. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each pulse is set to 1 mS wide&lt;br&gt;&#xA;then at full speed the output is high 0.600 of the time and low 0.400 of the time.&lt;br&gt;&#xA;If the pulses are set to say 5V high&lt;br&gt;&#xA;then Vmax = 0.6 x 5V = 3V.&lt;br&gt;&#xA;So, output is 3V at 600 RPM&lt;br&gt;&#xA;or 3/600 V/RPM&lt;br&gt;&#xA;or 5 mV per RPM. &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2016-04-08T15:55:08.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8349" PostTypeId="1" AcceptedAnswerId="8504" CreationDate="2016-04-08T17:48:51.833" Score="2" ViewCount="100" Body="&lt;p&gt;I'm finding myself using control charts and I'm unclear on a very basic point. I've read page after page of texts, but they all focus on how to make the initial chart rather than walking you through how to update it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make things easier, I'll set the situation:&#xA;I'm making an Individual Control Chart. Every thirty points, the chart is updated. In January, I put points 1-30 on the chart. February, I put points 31-45 on, and March points 45-60. So, the chart needs to be updated now. Where do I go from here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming none of the data points were out of control, here's what I think happens:&#xA;I think points 1-30 are removed from the chart. I think the chart has the average and control limits calculated using data points 31-60.  I think that future points (61-90) are plotted on the newly updated chart until data point 90 is reached, at which point the chart is updated, points 31-60 dropped, and the average and limits recalculated using points 61-90.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that the limits for updated versions of the chart should be calculated using xbarr +/- 3s/c4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, using the power of randbetween, I've made some sample charts and sample data. The first is for the month of January, where data was taken every day, giving 31 data points, and at the end of the month, the chart that follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Date    Value   Average        SD         UCL           LCL&#xA;1/1/2016    9   6.161290323 2.887868582 14.82489607 -2.502315423&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a bunch of values for the dates stretching out over the next few months (I can post all 60 rows if really desired).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/axyOd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/axyOd.jpg&quot; alt=&quot;sample_chart_01&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, so I've plotted the first 30 datapoints. So far so good. In February, only 15 samples are sampled for data. The datapoints are added to the chart for a total of 45 datapoints on the chart:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/jpN71.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jpN71.jpg&quot; alt=&quot;sample_chart_02&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In March, another 15 samples are sampled and this brings the total number of datapoints up to 60:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YEZQb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YEZQb.jpg&quot; alt=&quot;sample_chart_03&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now at this point, I know that the chart must be updated. But there are a few ways I can do it. One of which is to remove the January data (the first 30 datapoints) both from the chart and from the calculations and only plot datapoints 31-60, with datapoints 61-90 being plotted on this as well (at which time, we would go through the process again):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XaqvF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XaqvF.jpg&quot; alt=&quot;sample_chart_04&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or I could calculate the average, SD, and the upper and lower limits based on datapoints 31-60, but plot datapoints 1-60:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Ev6s8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ev6s8.jpg&quot; alt=&quot;sample_chart_05&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or I could re-calculate the average, SD, and upper and lower limits on datapoints 1-60 AND plot datapoints 1-60:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sIHcM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sIHcM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these give different limits, different charts, and all of them seem equally valid because nowhere does it seem to discuss what you do to actually &lt;em&gt;update&lt;/em&gt; the chart once it has been made.&lt;/p&gt;&#xA;" OwnerUserId="5919" LastEditorUserId="5919" LastEditDate="2016-04-12T12:40:38.560" LastActivityDate="2017-06-19T01:47:07.233" Title="How do I update an individual control chart?" Tags="|statistics|quality-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8350" PostTypeId="2" ParentId="8309" CreationDate="2016-04-08T18:45:51.513" Score="4" Body="&lt;p&gt;If money is no object then there are a lot of advantages to using a DC motor for  a lathe. The ability to have continuously variable speed control is a huge advantage for a lathe. Compared to other machine tools a lathe has the specific problem that the circumference (and thus proper cutting speed inevitably changes as you work. Consider turning a 30mm bar down to 10mm, the circumference changes by a factor of 3 so any fixed cutting speed is going to be a compromise, the alternative being changing gears mid job. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also very useful to be able to fine tune cutting speeds on the fly when cutting threads or turning difficult materials. Being able to precisely adjust cutting speeds also lets you get more effective capacity out of a given motor power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative is an AC induction motor, these are cheaper and don't require electronic control but by the same token don't allow for speed control in single phase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't much to choose between the two in terms of startup torque. Larger AC motors are generally split phase which gives high starting torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You limitation on power will be down to what your supply can deliver. Single phase motors generally go up to about 4hp but this requires a 240V 20A supply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give you some sense of power requirements  a 500W lathe with a DC brushless motor will cope fairly well with drilling holes up to about 16mm diameter and just about manage 25mm holes with care in metals with good machinability (ie leaded or sulphurated mild steels or brass) and cuts up to about 1mm at a single pass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a small lathe (say 120mm chuck) a 1200W motor would probably be ideal and that's about the limit for a domestic power supply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general DC motors tend to be a bit more compact for this sort of power requirement than AC induction motors also the ability to add speed control greatly simplifies the transmission design, although having a few gears is still an advantage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My experience is that for a small lathe running off a single phase supply a DC brushless motor is a lot more usable and efficient in terms of overall production rates than an equivalent AC one. A 3 phase machine gives the best of both worlds but that is not an option here. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-08T18:45:51.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8351" PostTypeId="2" ParentId="8333" CreationDate="2016-04-08T18:47:15.140" Score="4" Body="&lt;p&gt;There are a number of ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Russell described, one way is to make a fixed size pulse each period, then low pass filter the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is to make a frequency-dependent filter.  Feed in a signal with fixed amplitude and varying frequency.  The filter varies the amplitude as a function of frequency.  You then make a AC to DC amplitude converter, often called a &quot;detector&quot;.  Such a detector is at the heart of a AM radio, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have a bunch of narrowly tuned circuits in parallel, then see which one has the highest response to the input signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can sample the input signal, convert it to digital values, and run a FFT (fast fourier transform) on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are expecting only a single input frequency, you can tweak a oscillator to attempt to mimic that frequency.  If the oscillator is voltage-controlled and you do this in a feedback loop, then the control signal into the oscillator becomes the voltage signal indicating the input frequency.  This may sound far fetched, but this method is common enough to have a name.  It's called a &lt;i&gt;phase locked loop&lt;/i&gt; (PLL), and you can get PLL chips that do most of this for you.  Most newer (last 40 years or so) FM radios work on this principle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Old FM radios used a different technique, called a &lt;i&gt;ratio detector&lt;/i&gt;.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-04-08T18:47:15.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8352" PostTypeId="2" ParentId="8346" CreationDate="2016-04-08T19:27:06.283" Score="2" Body="&lt;p&gt;First off, the equations you have are not correct. For example, if equations 1 and 5 are added it results in $V_1+V_5=0$. The l.h.s of the first four equations should be $V_1-V_5$, $V_2-V_6$, $V_3-V_7$, and $V_4-V_8$. Then you are have four equations in the four unknowns $X_1$, $X_2$, $X_3$, and $X_4$ after taking Laplace transforms with zero initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general method is to solve for these four unknowns and express them in terms of the $V_i$ and then you get the transfer functions for each input-output pair. The derivatives can then be obtained by multiplying with powers of the Laplace variable $s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you need just 4 transfer functions, I think it can be done a bit more easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first transfer function, you are interested only in $V_1$. Set all the other $V_i$ to zero and solve for $\frac{X_1}{V_1}$. Since you want the derivative just multiply the result by $s$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second transfer function, set all the $V_i$ except $V_3$ to zero and solve for $\frac{X_2}{V_3}$. Multiply the result by $s^2$ to obtain the desired transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you get the idea.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2016-04-11T13:51:25.943" LastActivityDate="2016-04-11T13:51:25.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8353" PostTypeId="1" CreationDate="2016-04-08T20:03:45.767" Score="0" ViewCount="579" Body="&lt;p&gt;For a while I have thought it would be interesting if someone could develop a 'reciprocating electric' engine.  It would use the same principle as maglev trains to move a magnet up and down in a reciprocating motion, driving a crankshaft via connecting rods. I don't know if it would have any advantage over a pure rotary electric motor, perhaps the ability to control torque better. Also the power stroke can be in both the up and down directions. The biggest advantage is that it would give electric motors 'character' and 'coolness' in the same way car enthusiasts compare the various engine configurations of internal combustion engines (V8, V6, Flat 6, etc).  Does anyone know if this has been developed?  If not this could be a good college engineering project or masters thesis.&lt;/p&gt;&#xA;" OwnerUserId="5922" LastEditorUserId="368" LastEditDate="2016-05-17T14:44:59.177" LastActivityDate="2016-05-17T14:44:59.177" Title="V8 electric engine" Tags="|mechanical-engineering|electrical-engineering|automotive-engineering|electric-vehicles|" AnswerCount="1" CommentCount="2" ClosedDate="2016-04-08T21:37:28.143" ContentLicense="CC BY-SA 3.0" />
  <row Id="8354" PostTypeId="2" ParentId="8353" CreationDate="2016-04-08T20:19:00.803" Score="3" Body="&lt;p&gt;It is certainly possible as each cylinder could effectively be a solenoid (i'm actually pretty sure I have seen exactly this but can't remember where atm) ....however the reciprocating nature of IC engines is one of their big disadvantages as it introduces inherent  vibration and puts big transient loads on certain parts (especially connecting rods). In comparison the rotor of an electric motor is a much more desirable state of affairs from a mechanical perspective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, from a performance perspective, if you put aside to problems of power density for mobile electric power, you would go for electric motors over reciprocating engines every time as they are quite, light, simple and have flat torque curves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want V8 sound it is better to just have a V8 soundtrack and speakers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you say this would be an excellent Mech or Electrical Engineering undergraduate project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit yes, this is a thing &lt;a href=&quot;https://www.youtube.com/watch?v=6HND4x22jXM&quot; rel=&quot;nofollow&quot;&gt;V8&lt;/a&gt; (link to youtube video) and &lt;a href=&quot;https://www.youtube.com/watch?v=uhYEdD94vH0&quot; rel=&quot;nofollow&quot;&gt;V12&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-08T23:19:24.350" LastActivityDate="2016-04-08T23:19:24.350" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8355" PostTypeId="1" AcceptedAnswerId="8361" CreationDate="2016-04-08T21:15:48.993" Score="5" ViewCount="5562" Body="&lt;p&gt;Does anyone know how to determine the moment of inertia of the filled circular sector shown below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sqZ9I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sqZ9I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently I have to use $dA=r\ d\theta\ dr$, $y= r\sin(\theta)$ and end up getting $I_x=\frac{r^4}{8}(\alpha-\sin(\alpha))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have no idea how they got that, I have never had this.&#xA;Could someone step by step explain how to get to $I_x$ with this question and how to integrate something that has two different differentials.&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="272" LastEditDate="2016-04-09T19:50:09.430" LastActivityDate="2016-04-11T15:38:18.780" Title="Determine the moment of inertia of a filled circular sector" Tags="|statics|mathematics|moments|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8358" PostTypeId="1" CreationDate="2016-04-09T03:58:30.270" Score="2" ViewCount="133" Body="&lt;p&gt;Not sure how to ask a question usefully, as I'm not an engineer, but here goes. I have a harpsichord, essentially a sort of triangular slab (I'm assuming a general answer, though I could provide specifics) for which I've turned 3 tapered legs, each set in some distance from the edges, and attached securely with 4 screws. But the instrument wobbles extremely easily: if you push it on a side with one finger it's easy to move it half an inch or more. So I'd like it to be &quot;stable&quot; instead of wobbly. Each leg is 2.75&quot; square at the top and at the bottom is a circle with diameter .675&quot;. Here's an early version of the leg (its final dimensions are slightly different), but very close to the present version: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WaTX4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WaTX4.jpg&quot; alt=&quot;leg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the legs in context:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J5Vjb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J5Vjb.jpg&quot; alt=&quot;harpsichord&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: By what factor do I need to scale the thickness of the present legs (height is constant) to make the instrument &quot;stable?&quot; I'm not sure how to define stable, but the point is, it should only wobble with &quot;significant&quot; effort, not minimal, one-fingered effort. I'm pretty sure that if the thickness was increased by a factor of 10 it wouldn't wobble easily, as it would be resting on tree trunks, practically. But what's a reasonable number? 1.5, 1.7, 2.3? I have &lt;em&gt;no&lt;/em&gt; idea, but I'd be grateful for your suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit in response to the answer: I'd like to avoid redesigning with bracing. And I've already made a large plate, which is the bottom of the instrument. Assuming a rigid connection with the legs, wouldn't increasing the thickness improve stability?&lt;/p&gt;&#xA;" OwnerUserId="5925" LastEditorUserId="5925" LastEditDate="2016-04-09T20:23:29.093" LastActivityDate="2019-12-28T18:22:25.773" Title="By what factor should I scale some tapered legs to make my harpsichord stable?" Tags="|mechanical-engineering|structural-engineering|design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8359" PostTypeId="2" ParentId="8358" CreationDate="2016-04-09T04:46:46.610" Score="2" Body="&lt;p&gt;The thickness of the legs is not the important thing. There are plenty of historical instruments with legs no thicker than yours, for example see &lt;a href=&quot;http://www.baroque-music.com/harpsichord.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is how the legs are attached to the instrument. The easiest way to make a stable design is to put some bracing between the legs, so the legs effectively form a stable &quot;trestle table&quot; and the instrument just sits on top of it. &lt;a href=&quot;http://www.earlymusicshop.com/product.aspx/en-GB/1000452-ems-english-harpsichord&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a simple example - admittedly, with legs somewhat less elegant than yours. I would expect there would be bracing pieces at the top of the legs, as well as the ones you can see at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course people sometimes took this idea to extremes, as in &lt;a href=&quot;http://groups.csail.mit.edu/mac/users/bob/harpsichord.php&quot; rel=&quot;nofollow&quot;&gt;this example&lt;/a&gt; (copied from a 17th century design).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could try making a large &quot;plate&quot; (say 6 or 7 inches square) securely attached to the top of each leg (e.g. with several recessed screws plus glue joining the plate to the square top the existing leg) and then fix the instrument to the plate with a bolt at each corner. &lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="1832" LastEditDate="2016-04-09T15:09:53.343" LastActivityDate="2016-04-09T15:09:53.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8360" PostTypeId="1" CreationDate="2016-04-09T05:25:48.853" Score="6" ViewCount="734" Body="&lt;p&gt;Pressure gauges are protected from steam condensation in a regular boiler by using a siphon where steam condenses and prevents steam from going further up to the gauge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in a vacuum steam boiler when the boiler is re pressurized after use the pressure would push the water in the siphon up to the vacuum gauge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I currently dont use any siphon and my vacuum gauges are far off (from condensation I guess) after a few boils, how can I prevent that ? &lt;/p&gt;&#xA;" OwnerUserId="5926" LastEditorUserId="2015" LastEditDate="2017-10-20T19:26:17.647" LastActivityDate="2017-10-20T19:26:17.647" Title="How to protect a pressure gauge from condensation in a vacuum steam boiler?" Tags="|mechanical-engineering|vacuum|steam|instrumentation|siphon|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8361" PostTypeId="2" ParentId="8355" CreationDate="2016-04-09T12:39:09.307" Score="5" Body="&lt;p&gt;&lt;strong&gt;Since you actually asked for the moment about the $x$ axis.&lt;/strong&gt; Calculating the moment of inertia about the $x$ axis is a fair deal more complicated than calculating it about the $z$ axis as in my other answer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start with, we will recognize that the symmetry about the $x$ axis lets us only work on the top half and then multiply by a factor of 2 in the end.  Now, only looking at the top half we can break the piece up into two sections: $I_1$ is on the left and is a triangle and $I_2$ is on the right and is a right triangle with a circular hypotenuse.  Since this is clearly a homework problem, I'm going to skip the algebra steps and just show you the core parts of the problem (i.e. I'm going to use Mathematica to do the brute force algebra and integration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/btcBe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/btcBe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;$I_1$ Calculation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;The moment of inertia is given by &#xA;$$&#xA;I_1=\rho\iint y^2\ dydx,&#xA;$$&#xA;where $\rho$ is the mass density per unit area, which looks simple enough.  The difficulty is just in getting the correct limits of the double integral.  For a a given position along the $x$ axis, the limits of $y$ range from $0$ to $x\tan(\alpha/2)$.  And we will integrate $x$ from 0 to $r_0\cos(\alpha/2)$.  This gives&#xA;$$&#xA;\begin{align}&#xA;I_1&amp;amp;=\rho\int_0^{r_0\cos(\alpha/2)}dx\int_0^{x\tan(\alpha/2)}dy\ y^2\\&#xA;&amp;amp;=\rho\int_0^{r_0\cos(\alpha/2)}dx\ x^3\tan^3(\alpha/2)\\&#xA;&amp;amp;=\rho\frac{r_0^4}{12}\sin^3(\alpha/2)\cos(\alpha/2)&#xA;\end{align}&#xA;$$&#xA;This simplifies to $\frac{\rho}{12} bh^3$ which is the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_area_moments_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;well known value for the moment of inertia of a triangle&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;$I_2$ Calculation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This one goes the same way as the last one, but the limits and the integration are more difficult.  This time $x$ will vary from $r_0\cos(\alpha/2)$ to $r_0$.  Over that range $y$ will vary from 0 to $\sqrt{r_0^2-x^2}$.  Putting this into the integral gives&#xA;$$&#xA;\begin{align}&#xA;I_2&amp;amp;=\rho\int_{r_0\cos(\alpha/2)}^{r_0}dx\int_0^{\sqrt{r_0^2-x^2}}dy\ y^2\\&#xA;&amp;amp;=\frac{\rho}{3}\int_{r_0\cos(\alpha/2)}^{r_0}dx\ (r_0^2-x^2)^{3/2}\\&#xA;&amp;amp;=\frac{\rho}{96}r_0^4\left(6\alpha-8\sin(\alpha)+\sin(2\alpha)\right)&#xA;\end{align}&#xA;$$&#xA;That last integral was quite tricky, and I ended up just plugging it into Mathematica.  I'm sure it is possible to find it in the standard integral tables though.  &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;All together now&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Finally, putting everything together and working through some trig identities simplifies the whole thing to&#xA;$$&#xA;2(I_1+I_2)=\frac{\rho}{8}r_0^4(\alpha-\sin\alpha)&#xA;$$&#xA;Which is again &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_area_moments_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;what we expect from the standard table for a circular segment&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2016-04-09T20:00:11.673" LastActivityDate="2016-04-09T20:00:11.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8362" PostTypeId="1" CreationDate="2016-04-09T16:46:13.033" Score="0" ViewCount="10294" Body="&lt;p&gt;I would like to know what is a velocity response spectrum, PseduoVelocity Response spectrum, and how they are different from one another? Could you also do the same for acceleration and Pseduoacceleration? Also the explanation of this graph:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MOSga.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MOSga.png&quot; alt=&quot;Graph for PseduoVelocity response spectrum &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5931" LastEditorUserId="110" LastEditDate="2016-04-09T19:26:53.833" LastActivityDate="2024-03-26T01:30:55.763" Title="What is a Velocity, Pseudovelocity, Acceleration, Pseudoacceleration response spectrum?" Tags="|structural-engineering|structural-analysis|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8363" PostTypeId="1" AcceptedAnswerId="8392" CreationDate="2016-04-09T18:30:49.700" Score="5" ViewCount="2133" Body="&lt;p&gt;What is the physical meaning of using negative frequencies while making Nyquist plots?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that we do a mapping from $s-$ plane to $f(s)$ plane, and we move along a contour. What I don't get is that Is the Nyquist plot purely a mathematical construct? Is there no relation between the frequency on plot and actual frequency? If there is, then what is the point of plotting negative frequencies,Is it just so that we could count the number of encirclements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to grasp the idea behind it.&lt;/p&gt;&#xA;" OwnerUserId="1155" LastEditorUserId="1155" LastEditDate="2016-04-10T05:23:02.927" LastActivityDate="2016-04-12T14:27:47.697" Title="Why are negative frequencies included in the nyquist plot?" Tags="|control-theory|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8364" PostTypeId="2" ParentId="8358" CreationDate="2016-04-09T22:18:39.727" Score="2" Body="&lt;p&gt;If the legs are simply screwed at the joint, then they should be considered pinned to the base of the instrument. This means that the angle between a leg and the base (initially equal to 90°, with the leg vertical and the base horizontal) can change: under a horizontal force such as your finger, the leg can move away from the vertical while the base remains horizontal. This is because a simple screw setup as you've described isn't capable of forcing the leg and the base to move in unison; small relative motions are allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the joint were a perfect pin, then the structure would be completely unstable: any horizontal force whatsoever would cause the structure to collapse. The reason this isn't the case is because the screws do generate some level of &quot;fixity&quot; between the leg and the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the most precise way of describing the joint is neither as fixed (no relative rotation, therefore there is a full transfer of moment from one to the other) nor as pinned (full relative rotation, so no moment transfer), but as partially fixed: relative rotations are allowed between the leg and the base, but there will be some moment transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, looking at two legs, the structure can be modelled thusly (&lt;em&gt;see note regarding this modelling at bottom of post&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uOVKZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uOVKZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black bars represent the original shape, while the blue lines represent the deformed shape due to the horizontal load. The value pointed out is the horizontal deflection. It is unitless because this model is utterly symbolic: I don't know the real distances, actual materials used, the thickness of the floor of the instrument, how the screws were placed, etc. Just use these values to compare (in broad terms) how the solutions described below will improve your situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as @alephzero mentioned in their answer, there are a few solutions to this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One they describe is by strengthening the joint between the legs and the base. The extra screws and glue would &lt;strong&gt;&lt;em&gt;try&lt;/em&gt;&lt;/strong&gt; to turn this partially-fixed joint into a fully-fixed (or at least more-fixed) one, which could dramatically reduce how flexible the structure is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zWgPa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zWgPa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option would be to create bracing in the structure. @alephzero suggests bracing at the feet of the legs. Here's the result considering partial-fixity between the legs and the base and the legs and the bracing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/puzJ0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/puzJ0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a very similar result can be obtained with bracing that is placed just below the joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kcbct.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kcbct.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It becomes clear that the stiffest structure is when you make the joint between the leg and the base of the harpsichord perfectly fixed. This is however quite difficult and I don't believe @alephzero's suggestion would be sufficient. You'd probably need to use steel L-shapes or wooden wedges (&lt;a href=&quot;https://www.google.com/search?q=wood+corner+brace&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;see here&lt;/a&gt; and &lt;a href=&quot;http://www.hobbithouseinc.com/personal/woodpics/images/cornerbrace.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for examples) on all four sides of each of the feet in order to achieve anything close to full fixity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I believe the solution of creating bracing just below the base of the instrument is also a good alternative. This is the concept behind tables where there's bracing just below the tabletop (usually touching the tabletop, even). &lt;a href=&quot;http://needforseat.website/wp-content/uploads/2015/09/wooden-kitchen-table-solid-wood-tables-at-the-galleria-picture.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;This one&lt;/a&gt;, for example (and some others can be seen &lt;a href=&quot;https://www.google.com.br/search?q=wooden+table&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). Would they do better with bracing at the feet? Yes, but this already helps tremendously.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note on partial-fixity modelling&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I created these models in &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool. It's my go-to for simple models such as these, but it doesn't allow for partial fixity, so I had to improvise. You'll notice that the joints between the leg and the base have a second node right next to them. The bar segment between these nodes was given a moment of inertia orders of magnitude smaller than that of the horizontal bar representing the base. This allows these short segments to deform with great ease, simulating partial fixity to some extent. Obviously, the results shown here depend greatly on the ratio of the segment's and the horizontal bar's moment of inertia. If the segment's inertia were equal to the bar's, the joint would behave as if fully fixed. If it were equal to zero, (the model wouldn't run, but even if it did) the joint would be perfectly pinned and the structure would be unstable. I don't know how one would convert this improvised method to an actual partial-release hinge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here are the results for different ratios. In the models above I considered a very low ratio (0.15%) since such connections are usually modeled as pinned. You'll notice, however, how for ratios as of probably around 5% the braced solutions perform better than a perfectly fixed joint with no bracing. Obviously, a fixed joint &lt;strong&gt;with&lt;/strong&gt; bracing would perform even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{array}{r|r|r|r}&#xA;R &amp;amp; \text{Current} &amp;amp; \text{Brace Foot} &amp;amp; \text{Brace Base} \\&#xA;0.01\% &amp;amp; 2071 &amp;amp; 1035 &amp;amp; 1035 \\&#xA;0.1\%  &amp;amp; 210  &amp;amp; 104  &amp;amp; 104 \\&#xA;1\%    &amp;amp; 23.5 &amp;amp; 11.2 &amp;amp; 11.3 \\&#xA;2\%    &amp;amp; 13.1 &amp;amp; 6.1  &amp;amp; 6.1 \\&#xA;4\%    &amp;amp; 8.0  &amp;amp; 3.4  &amp;amp; 3.5 \\&#xA;10\%   &amp;amp; 4.9  &amp;amp; 1.9  &amp;amp; 2.0 \\&#xA;20\%   &amp;amp; 3.8  &amp;amp; 1.3  &amp;amp; 1.4 \\&#xA;50\%   &amp;amp; 3.2  &amp;amp; 0.9  &amp;amp; 1.0 \\&#xA;75\%   &amp;amp; 3.1  &amp;amp; 0.9  &amp;amp; 0.9 \\&#xA;90\%   &amp;amp; 3.1  &amp;amp; 0.8  &amp;amp; 0.9 \\&#xA;\end{array}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:22:25.773" LastActivityDate="2019-12-28T18:22:25.773" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8366" PostTypeId="1" CreationDate="2016-04-10T06:40:07.427" Score="2" ViewCount="717" Body="&lt;p&gt;I'm designing a shell and tube heat exchanger with a hydrocarbon stream (31.6% propane and 60.4% propylene) entering at 135F and 321psi, and leaving at 100F being cooled by water at 85F and 74.7 psi. As of now I have the hydrocarbon stream on tube side and water stream on shell side, mainly due to the high pressure of the hydrocarbons and it being at a higher temperature. Is this a decent enough basis for my decision? Any help would be greatly appreciated, thank you! &lt;/p&gt;&#xA;" OwnerUserId="4785" LastActivityDate="2016-04-11T21:13:05.030" Title="Fluid allocation for shell and tube heat exchanger" Tags="|mechanical-engineering|heat-transfer|chemical-engineering|heat-exchanger|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8367" PostTypeId="1" AcceptedAnswerId="8368" CreationDate="2016-04-10T07:54:08.047" Score="-1" ViewCount="130" Body="&lt;p&gt;I'm an engineering enthusiast but I don't yet have a formal education in engineering and I'm hoping you guys can tell me about the implications of using a metal road design in terms of durability and cost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My (vague) idea is (for 2 lanes of road) a set of 3 rails laid down as a vertical layout for a length of road, similar to railroad tracks, which supply power for a self-deploying set if metal road panels. The horizontal panels are designed to drain water and grooved for tire traction. The way the panels self deploy isn't really important, it's just a mechanism that rides on the three rails and lays down a load of the metal panels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The self deployment isn't really important it's just to lower the construction cost and explain the 3 rail design I had in mind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm purposely being vague when I say metal, expecting answers to provide the kind of metal that would be optimal because I don't know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just that we spend billions across the globe replacing roads, and while metal would obviously cost a lot more than cement, I'm really curious about whether the world would save money by building very durable roads which need to be replaced less often and which for the most part might be replaced systematically the same way they're deployed, requiring only a small fraction of the maintenance of a traditional road replacement.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the real maintenance cost of this imaginary design is unknown, my question is purely about the difference in cost for materials and the durability so I can have some basis for speculation about the viability of said design. &lt;/p&gt;&#xA;" OwnerUserId="3746" LastEditorUserId="6790" LastEditDate="2018-05-07T11:42:30.670" LastActivityDate="2018-05-07T11:42:30.670" Title="What would be the implications of a metal highway design?" Tags="|civil-engineering|highway-engineering|roadway|" AnswerCount="1" CommentCount="4" ClosedDate="2016-04-11T22:21:50.310" ContentLicense="CC BY-SA 3.0" />
  <row Id="8368" PostTypeId="2" ParentId="8367" CreationDate="2016-04-10T09:37:50.087" Score="7" Body="&lt;p&gt;Asphalt, which is essentially crushed aggregate bound with bitumen has a lot of advantages for road building. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is porous and so can drain by allowing water to pass through it as well as running off the surface. &lt;/li&gt;&#xA;&lt;li&gt;Because it is made up of particles of crushed masonry it has good friction characteristics in all directions (ie you get the same grip&#xA;turning as you do braking) which isn't greatly affected by wear. ie the 'grip' is largely inherent in the material rather than depending on a particular surface coating or treatment. &lt;/li&gt;&#xA;&lt;li&gt;Asphalt is pretty easy to lay as it can be continuously poured onto a prepared subsurface either by specialist plant or just by shoveling out out of the back of a flatbed truck. &lt;/li&gt;&#xA;&lt;li&gt;In road building it is the foundation layers of the road which are the expensive bit and even then the cost is largely associated with the planning, design, surveying and infrastructure (utilities, signage, bridges, culverts, drainage etc) rather than the materials. &lt;/li&gt;&#xA;&lt;li&gt;It can be repaired reasonably easily by planing off the top layer and relaying or even adding a dressing layer over the old surface.&lt;/li&gt;&#xA;&lt;li&gt;properly made repairs should be more or less seamless and permanent.&lt;/li&gt;&#xA;&lt;li&gt;it is possible to lay lower grade temporary surfaces during major works quickly and cheaply. &lt;/li&gt;&#xA;&lt;li&gt;serviceable temporary repairs for potholes etc can be made very cheaply&lt;/li&gt;&#xA;&lt;li&gt;Trenches for utilities etc are easy to dig and fill in. &lt;/li&gt;&#xA;&lt;li&gt;Because the binder is relatively soft small defects can self-repair, especially in hot weather. &lt;/li&gt;&#xA;&lt;li&gt;Expansion joints are not required. &lt;/li&gt;&#xA;&lt;li&gt;Asphalt can be laid over an uneven surface and doesn't require precise leveling as it doesn't have discrete joints. &lt;/li&gt;&#xA;&lt;li&gt;Deterioration of asphalt roads over time is inconvenient but it is unlikely to lead to catastrophic failures in contrast to eg steel&#xA;railway lines. &lt;/li&gt;&#xA;&lt;li&gt;Asphalt isn't subject to much in the way of corrosion or environmental degradation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are some very obvious potential problems with a road with a steel deck&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The surface would need to be textured to provide grip and water drainage. This would be ground away fairly rapidly by dirt and grit carried onto the road and the action of tyres and so would need renewal. &lt;/li&gt;&#xA;&lt;li&gt;Joints between sections would need to be well aligned to provide an adequate surface. &lt;/li&gt;&#xA;&lt;li&gt;How would they be joined ?  Steel structures are generally either bolted or welded but both are expensive on this sort of scale compared to just raking out hot asphalt. &lt;/li&gt;&#xA;&lt;li&gt;You would still need to prepare a stable substructure. &lt;/li&gt;&#xA;&lt;li&gt;The construction would need to accommodate thermal expansion and expansion joints in bridges etc add a lot of noise and vibration when driven over. &lt;/li&gt;&#xA;&lt;li&gt;even mild steel is moderately  expensive and requires a lot of energy to produce. &lt;/li&gt;&#xA;&lt;li&gt;Most structural steels need protection from corrosion by painting, coatings etc which require regular maintenance when exposed to the elements. This will be seriously exacerbated in regions where roads need to be salted in the winter. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a real world comparison we can look at the &lt;a href=&quot;http://www.transport-watch.co.uk/facts-sheet-7-rail-versus-road-capital-costs-track&quot; rel=&quot;nofollow&quot;&gt;relative cost per km&lt;/a&gt; of railway track vs multi-lane roads. This puts rail at around 6 times more expensive per km than roads. Note also that a large proportion of the cost of these projects is associated with structures such as bridges and signs/signalling as well as the cost of the land itself, rather than the actual road surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://www.dot.ca.gov/hq/esc/oe/cost_index/historical_reports/PI_2012_qtr4.pdf&quot; rel=&quot;nofollow&quot;&gt;this document&lt;/a&gt; for estimates of civil engineering materials by weight. Note that it puts &lt;strong&gt;asphalt as  94 dollars per ton&lt;/strong&gt; and &lt;strong&gt;structural steel at  2586 dollars per ton.&lt;/strong&gt; This doesn't necessarily tell you a lot about the total costs of what would necessarily by very different approaches but illustrates that asphalt is a very inexpensive bulk material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth considering that even in the case of steel bridges the road surface is almost invariably asphalt or concrete except where the structure is temporary or take very low traffic. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-11T22:55:15.873" LastActivityDate="2016-04-11T22:55:15.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8369" PostTypeId="1" CreationDate="2016-04-10T17:38:28.133" Score="1" ViewCount="3250" Body="&lt;p&gt;How to add force in certain point. I want to see influence of point force on my structure. (See image) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That force is for example weight of some tank or something similar inside my structure. I don't have suports defined yet. How to define force on x,y,z point in ANSYS Workbrench? Is that possible? Model (geometry) will be imported as STP file  to ANSYS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does someone have idea how this can be accomplished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oDTl7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oDTl7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5855" LastActivityDate="2016-04-11T02:24:55.723" Title="Ansys Workbench - adding force in point" Tags="|finite-element-method|structural-analysis|ansys|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8370" PostTypeId="1" CreationDate="2016-04-10T19:48:45.907" Score="4" ViewCount="817" Body="&lt;p&gt;I am building a small ship (or very large sailboat) that will have a concrete ballast at the base of the keel. I am looking to increase the weight of the concrete (lb/ft&lt;sup&gt;3&lt;/sup&gt;), but I also want to maximize the impact resistance of the concrete to resist cracking/shattering should the boat hit a rock hard vertically due to waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplified dimensions of the concrete portion of the keel will be 11.5&quot; W x 18&quot; D x 40' L. This will essentially be bolted on to the bottom of a laminated wood (fir) keel which is 11.5&quot; W x 12&quot; D x 40' L (simplified dimensions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine the best concrete recipe and reinforcement scheme for this particular usage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to determine:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;optimal size of rebar to use;&lt;/li&gt;&#xA;&lt;li&gt;optimal size(s) of aggregate;&lt;/li&gt;&#xA;&lt;li&gt;how many parts of cement, sand, and aggregate to mix;&lt;/li&gt;&#xA;&lt;li&gt;and the optimal amount of rebar to include.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general heavier is better; I am not looking to lighten the concrete. I know concrete recipes are a topic all on their own; what I need is a recipe to start with that will produce a strong and heavy concrete that resists cracking from a heavy impact. The boat will displace 100 gross tones of water and may be subjected to waves with crest-to-trough wave height of about 6' in areas where there is an unlikely yet possible danger of striking underwater obstructions. It doesn't need to support being continuously jackhammered, but support get hit perhaps say 5 times over it's expected lifetime of say 20 years (just to give you an idea of infrequency). I am not planning to strike bottom, but it can and does happen despite all cautions taken.&lt;/p&gt;&#xA;" OwnerUserId="5941" LastEditorUserId="6790" LastEditDate="2020-05-02T19:36:03.237" LastActivityDate="2020-05-02T19:36:03.237" Title="How to design a reinforced concrete ship keel?" Tags="|reinforced-concrete|marine-engineering|ships|naval-engineering|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8374" PostTypeId="2" ParentId="7325" CreationDate="2016-04-10T22:45:28.883" Score="1" Body="&lt;p&gt;There is no legal requirement to follow any specified design method or standard. From &#xA;&lt;a href=&quot;https://www.gov.uk/guidance/ce-marking&quot; rel=&quot;nofollow&quot;&gt;https://www.gov.uk/guidance/ce-marking&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is up to you to ensure that your product complies with the&#xA;  essential requirements of the relevant EU legislation. The use of&#xA;  harmonised standards remains voluntary. You may decide to choose other&#xA;  ways to fulfil these essential requirements. If you don’t follow the&#xA;  safety requirements of a standard as it is written you will need to&#xA;  show that your product is as safe, by presenting the relevant&#xA;  documentation.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-04-10T22:45:28.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8375" PostTypeId="2" ParentId="8369" CreationDate="2016-04-11T02:24:55.723" Score="2" Body="&lt;p&gt;So if the point isn't a point on your geometry then you can add a &lt;strong&gt;Remote Force&lt;/strong&gt;. It is in the loads menu. You will then need to choose a Scoping Method and choose the faces/features of your geometry that the load will be transferred to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally I think you may need to create that point where the load is applied in your geometry so you can select it as the location of the Remote Force.  &lt;/p&gt;&#xA;" OwnerUserId="1848" LastActivityDate="2016-04-11T02:24:55.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8376" PostTypeId="2" ParentId="7885" CreationDate="2016-04-11T03:23:00.380" Score="2" Body="&lt;p&gt;In general, poor performance of PID2 will cause poor performance in PID1, in both disturbance rejection and setpoint tracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some special instances where poor performance of PID2 in the form of high frequency oscillation is attenuated by G1, so the performance of PID1 is not significantly affected. &lt;/p&gt;&#xA;" OwnerUserId="5945" LastActivityDate="2016-04-11T03:23:00.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8378" PostTypeId="1" CreationDate="2016-04-11T04:41:31.130" Score="5" ViewCount="1653" Body="&lt;p&gt;I am trying to build a protective enclosure for Microsoft's Kinect v2, which emits (and perceives) IR in the 827–850 nm range. In order to fully enclose the device without blocking those signals, I need some transparent and hard material that will let infrared pass through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried acrylic, glass and crystal from the local glass cutter, but I can't get good results from any of those materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best result I got was when I stuck the pane flush to the Kinect's front, otherwise the IR camera just can't see anything. However, technically I can't do that because I need to allow the Kinect to tilt, and I can't have the pane always flush to the Kinect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The enclosure will be completely opaque apart from the front pane (which is what I'm looking the material for).&lt;/p&gt;&#xA;" OwnerUserId="5947" LastEditorUserId="368" LastEditDate="2016-04-11T16:42:10.843" LastActivityDate="2016-04-11T17:14:45.793" Title="Transparent, hard material that lets infrared through" Tags="|materials|optics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8379" PostTypeId="1" CreationDate="2016-04-11T07:46:12.783" Score="9" ViewCount="367" Body="&lt;p&gt;I have a design with two parts (at the moment):&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a fixed 10cm ring with vertical tracks&lt;/li&gt;&#xA;&lt;li&gt;a freely rotating block with a cut that may freely slide along that tracks:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SVoCv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SVoCvm.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I'm trying to find a low profile way to ensure that whenever the cut in the moving block faces upward vertically, the block is pushed up (almost) touching the inner face of the ring. That is, it returns to its &amp;quot;home position&amp;quot; whenever the cut faces up:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZFQ7v.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZFQ7vm.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The block is just driven by a crank. Indeed, the crank can freely shift vertically. However, i'm hoping the block (along with the crank) to return to the home position whenever the cut points upwards.&lt;/p&gt;&#xA;&lt;p&gt;The only way I could think of is to have a part like a leg sticking out from the bottom of the block so that it pushes against the ring or an external track (near the bottom of the ring) whenever the cut nears facing upward. The issue is that this leg will require quite a bit of extra space, e.g., when the cut faces down but the block is at the top, the leg will point upward. The method should not constrain the block's movement along the tracks or rotation.&lt;/p&gt;&#xA;&lt;p&gt;Does anyone know of a better way?&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="41730" LastEditDate="2023-05-16T01:11:52.013" LastActivityDate="2024-02-10T03:11:36.417" Title="Forcing a part into &quot;home position&quot;" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="8381" PostTypeId="2" ParentId="8378" CreationDate="2016-04-11T09:28:21.380" Score="5" Body="&lt;p&gt;Acrylic and glass do let 800-900 nm IR radiation through quite well (see, e.g., &lt;a href=&quot;http://www.plasticgenius.com/2011/05/infrared-and-ultraviolet-transmission.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, your problem is probably that part of the light gets reflected at each of the surfaces. This is unavoidable, though it can be reduced by some anti-reflective coatings, but those are not available off-the-shelf as they need to be designed according to material and wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further problems will be caused by the refraction, which will distort the path of the light. This may interfere with the calibration and focusing of the Kinect's optical parts.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastEditorUserId="368" LastEditDate="2016-04-11T16:44:11.637" LastActivityDate="2016-04-11T16:44:11.637" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8383" PostTypeId="2" ParentId="8355" CreationDate="2016-04-11T13:38:30.193" Score="10" Body="&lt;p&gt;Consider an infinitesimal element of area $r d\theta  dr$ which is at a distance $r \sin (\theta)$ from the $x$ axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d9NGX.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d9NGX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its moment of inertia is $r d\theta  dr (r \sin (\theta ))^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment of inertia about the $x$ axis of the complete sector:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I_x = \int _0^{r_0}\int _{-\frac{\alpha }{2}}^{\frac{\alpha&#xA;   }{2}}r^3\sin ^2(\theta) d \theta dr = \frac{1}{8} r_0^4 (\alpha -\sin (\alpha ))$$&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2016-04-11T15:38:18.780" LastActivityDate="2016-04-11T15:38:18.780" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8385" PostTypeId="2" ParentId="8378" CreationDate="2016-04-11T17:14:45.793" Score="1" Body="&lt;p&gt;As jpa points out, the issue is not with the transparency of the materials that you are considering.  I suspect that the issue is most likely caused by light being reflected directly from the IR emitter into the camera.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intensity of light specularly reflected from a surface very close to the camera will be orders of magnitude brighter than the faint amount of light diffusely reflected from clothing and other objects which are much farther away.  This will saturate the image sensor in the IR camera and make it impossible to see the IR pattern from other objects.  Imagine trying to see the light of a small LED through a 100 W light bulb.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am correct, then the solution is fairly easy.  You simply need to tilt the transparent material at a fairly steep angle so that the reflected light is sent to a part of the enclosure where you can put a material which strongly absorbs IR.  Exactly how steep the angle needs to be depends on the details of the emitted IR pattern and the position of the screen, but I suspect that it will need to be at least 25&amp;deg;.  If you tilt the material all the way to ~55&amp;deg; then you may be able to take advantage of &lt;a href=&quot;https://en.wikipedia.org/wiki/Brewster%27s_angle&quot; rel=&quot;nofollow&quot;&gt;Brewster's angle&lt;/a&gt; to significantly reduce the amount of reflected light, but the effectiveness depends on the polarization state of the emitted light and the direction of the tilt.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2016-04-11T17:14:45.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8386" PostTypeId="2" ParentId="8366" CreationDate="2016-04-11T21:13:05.030" Score="2" Body="&lt;p&gt;This is a good start for your decision and you can probably get away with it however you should factor in everything that influences that decision.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cleaning&lt;/li&gt;&#xA;&lt;li&gt;Corrosion&lt;/li&gt;&#xA;&lt;li&gt;Fouling&lt;/li&gt;&#xA;&lt;li&gt;Temperature&lt;/li&gt;&#xA;&lt;li&gt;Viscosity&lt;/li&gt;&#xA;&lt;li&gt;Pressure&lt;/li&gt;&#xA;&lt;li&gt;Pressure Drop&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Higher Temperatures and a higher pressure favor allocating the fluid to the tube site.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Put a high-pressure fluid on the tube side. This usually minimizes exchanger cost. The smaller tube diameter has a higher pressure rating for the same metal thickness compared to the larger diameter shell.&lt;a href=&quot;http://www.chemicalprocessing.com/articles/2013/shell-and-tube-heat-exchanger-best-practices/&quot; rel=&quot;nofollow&quot;&gt;&lt;sup&gt;(1)&lt;/sup&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the temperatures are high enough to require the use of special alloys placing the higher temperature fluid in the tubes will reduce the overall cost.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chemical Engineering Design, 12 Heattransfer Equipment,  R K Sinnott &lt;a href=&quot;https://books.google.de/books?id=DJaxUL3numgC&amp;amp;pg=PA660&amp;amp;lpg=PA660&amp;amp;dq=hotter%20fluid%20tube%20shell&amp;amp;source=bl&amp;amp;ots=2d07kx7Y7v&amp;amp;sig=zrMEPX3pglMNVpICrtpMUkoOLi0&amp;amp;hl=de&amp;amp;sa=X&amp;amp;ved=0ahUKEwjjvZbr8YXMAhXH7BQKHWjADKUQ6AEIJzAB#v=onepage&amp;amp;q=hotter%20fluid%20tube%20shell&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;&lt;sup&gt;(2)&lt;/sup&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I wouldn't consider 135 F &quot;hot&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is another article &lt;a href=&quot;https://www.quora.com/Chemical-Engineering-Why-we-pass-cold-fluid-to-shell-side-and-hot-fluid-to-tube-side&quot; rel=&quot;nofollow&quot;&gt;&lt;sup&gt;(3)&lt;/sup&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all I would advise you to work through the sources I provided and this should help you with your choice. Without knowing more I would probably choose the same allocation.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2016-04-11T21:13:05.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8390" PostTypeId="2" ParentId="8316" CreationDate="2016-04-12T00:46:04.800" Score="0" Body="&lt;p&gt;Many factors play in this scenario:&lt;br&gt;&#xA;1- the shape of the bottles and their material thermal insulation index.&lt;br&gt;&#xA;2-how you place the bottles on the shelves: do they promote air current or impede it.&lt;br&gt;&#xA;3- the frequency of opening the fridge door.&lt;br&gt;&#xA;As many have mentioned there is an initial load when room temperature bottles cool of.   but after that point everything else being equal it boils down two cases; penetration of heat into fridge due to design and manufacturing exigencies and material thermal properties of fridge without bottles versus contribution of bottles acting like the vanes of a cold radiator letting the stream of transient air wash around it.   So if the shape and placement of bottles doesn't encourage entrant air circulate easily it would help saving energy a bit.&lt;br&gt;&#xA;ultimately if you fill up the entire available space with water past the initial pick of energy the chronic use of energy will be smaller.&lt;br&gt;&#xA;we could look at this fridge as revers of a cooling tower. if we don't want to interact thermally with ambient air we make it hard by placing the bottles of water in a way that shields the other parts of the fridge from convection or ventilation. this concept has been used for centuries in middle eastern architecture, eg, passive air-conditioning.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2016-04-12T02:10:08.830" LastActivityDate="2016-04-12T02:10:08.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8391" PostTypeId="1" AcceptedAnswerId="8396" CreationDate="2016-04-12T01:49:34.647" Score="4" ViewCount="1430" Body="&lt;p&gt;Why does the AISC handbook only list one figure for Section Modulus for shapes such as Angle(L) steel?  Shouldn't there be two? The steel is asymmetric.  Is the figure given just the Delta to the bottom or something; figuring a design only needs to be concerned with Tension?&lt;/p&gt;&#xA;" OwnerUserId="5967" LastActivityDate="2016-04-13T18:07:50.973" Title="AISC Section Modulus numbers" Tags="|steel|geometry|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8392" PostTypeId="2" ParentId="8363" CreationDate="2016-04-12T02:42:01.687" Score="3" Body="&lt;p&gt;A complete Nyquist diagram can tell if the closed loop system is stable. Namely when going from negative infinite imaginary frequency to positive infinite imaginary frequency and back again to negative infinite imaginary frequency over the D contour you encircle the entire right half plane on the input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When drawing the Nyquist diagram of a system, then each unstable zero and pole will cause a clockwise- and anticlockwise-encirclement of the origin respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you write the open loop as a quotient of two polynomials&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;L(s) = C(s) H(s) = \frac{N(s)}{D(s)},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the closed loop becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{L(s)}{1 + L(s)} = \frac{N(s)}{D(s) + N(s)}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw the Nyquist diagram of $L(s)$ and look at the encirclements of the minus one point you are basically looking at the unstable zeros and poles of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1 + L(s) = \frac{D(s) + N(s)}{D(s)}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of clockwise encirclements of the minus one point will be equal to the number of unstable roots of $D(s)+N(s)$ minus the number of unstable roots of $D(s)$. The first are also the number of unstable poles of the closed loop system and the second are the number of unstable poles of the open loop system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every physical system the part of Nyquist diagram of the negative frequencies is a mirror image of the positive frequencies. So it is not required to draw them both, but it can help identify the total number of encirclements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Also be aware of encirclements you get infinity. For example $1/s^3$ does have one clockwise encirclement of the minus one point.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2016-04-12T02:42:01.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8393" PostTypeId="2" ParentId="8391" CreationDate="2016-04-12T04:15:51.723" Score="0" Body="&lt;p&gt;When using this &lt;a href=&quot;http://www.engineersedge.com/materials/aisc_structural_shapes/aisc_structural_shapes_viewer.htm&quot; rel=&quot;nofollow&quot;&gt;this tool&lt;/a&gt; which is based on the AISC shapes database there are two values provided. If it is an equal angle section then $S_x = S_y$ and $Z_x = Z_y$, so perhaps that's why you're only seeing one value.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastActivityDate="2016-04-12T04:15:51.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8394" PostTypeId="2" ParentId="8370" CreationDate="2016-04-12T08:51:42.150" Score="2" Body="&lt;p&gt;You intend to submerge a rectangular concrete section in a fairly corrosive environment (ocean water). The section is 11.5&quot; wide and 18&quot; deep. During the 20 year life span of the boat, you anticipate 6 heavy impact events. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concrete keel is intended to act as a beam which is to spread the impact load evenly onto the laminated wood hull to such an extent that the hull does not sustain any damage. This means that under full impact load the concrete beam (keel) can deflect only a limited amount, before it causes damage to either the points where it is fixed to the boat, or the boat itself. This deflection is a function of load as well as geometry and design of the beam. The beam will require structural steel (the amount can be calculated, based on maximum impact loads).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steel mentioned above needs to be protected to avoid corrosion. This means it needs to be encased in min. 2&quot; of concrete. Should an impact occur, this concrete cover most likely will be compromised due to spalling of the the concrete due to the impact. No matter what you do, you will have a 100 ton load dropping 6 feet onto a very small area during impact. This is in the realm of large breakers which are used to demolish concrete buildings. Once the steel is exposed, rust will form and cause further spalling of concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only sustainable solution is to take a 16-17&quot; steel pipe with a fairly generous wall thickness and fill it with mass concrete to get the ballast. This way your keel is impact resistant, and you have a long beam which spreads the load evenly to your wooden hull.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="2637" LastEditDate="2016-04-12T10:44:10.127" LastActivityDate="2016-04-12T10:44:10.127" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8395" PostTypeId="2" ParentId="6832" CreationDate="2016-04-12T12:20:11.130" Score="4" Body="&lt;p&gt;In this answer, I'm going to use the sign convention that torques and angular velocities are positive if their direction points to the right (left would be an equally valid alternative for this answer too) using the Right Hand Rule, i.e. seen as clockwise if looking from the left hand side. Any gears spinning in the opposite direction will have a negative angular velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EQUILIBRIUM APPROACH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach to determining the torques on each gear would be to draw the free body diagram of an individual gear with all the forces (meshing gears, shaft reaction) and external torques (shaft torque) exerted onto that gear. Then, by neglecting inertial forces, use equilibrium to write an expression relating the unknown teeth forces and external torques. An example of this is done for the gear marked with a 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jAAgs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the directions of the forces are completely arbitrary, while the input torque respects the sign convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying equilibrium, and eliminating the shaft force (only interested in gear forces and external torques), we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{II} = T_{in}/r_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r_1$ is the gear pitch radius (approximately) or the distance from the gear centre to the pressure line (accurate, but not needed if ultimately interested in external torques). Repeating this for all gears, it should then be possible to relate the external torques to each other, and determine the ratios of each relative to the input torque, the sign of which ratio indicating the direction of the torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some gear bodies, there will only be forces acting on it. The following example of a gear train should highlight this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQAyl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free Body Diagrams:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EFIPi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how the middle gear has no external torque acting on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VIRTUAL POWER APPROACH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second approach can be simpler to use, especially if the angular velocities are already known. First of all, a single algebraic equation relating the angular velocities of gears with external torques should be obtained. For simple gear trains, this equation has only two angular velocities, and takes the form:&#xA;$$\omega_{out} = G \omega_{in}$$&#xA;where $G$ is the gear ratio to be determined. For most planetary systems, there can be three angular velocities. An example I will use is the speed rule for a simple &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Epicyclic_gearing&quot; rel=&quot;nofollow noreferrer&quot;&gt;planetary gearbox&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/m2A2y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_s = (1+R)\omega_c - R\omega_a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where subscripts 's','c' and 'a' refer to the sun, carrier and annulus gears. $R$ is the ratio of teeth of the annulus gear to that of the sun gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we need a power balanced equation: power in = power out. This is where the sign convention comes in handy: since the conventional directions of torque and angular velocity are the same, multiplying the external torque acting on a gear with its angular velocity will give you the value of power in. Therefore, the power balance equation will be the sum of all the power-in products, equal to zero. i.e. In general:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i T_i \omega_i = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my example, this is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_c \omega_c + T_s \omega_s + T_a \omega_a=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simple gear train, it would now be very simple to substitute the angular velocity equation into the power balance equation, but not immediately so for angular velocity equations with more than two velocities. To solve such equations, you need to set one of the angular velocities to zero, as if holding that gear still, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set $\omega_a=0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ang. Vel. Equation: $\omega_s = (1+R)\omega_c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power Balance: $T_c \omega_c + T_s \omega_s=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute Ang. Vel. Equation into power balance:&#xA;$T_c/T_s = -(1+R)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat for setting another velocity to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems like cheating by forcibly setting angular velocities to zero, but it is still valid as the equilibrium solution (i.e. The external torque values) never depended on the angular velocities to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, whichever method was used, a quick check can be made to see if all the torques satisfy global equilibrium. Thanks to the sign convention, global equilibrium is achieved if:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_i T_i = 0$$&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2016-04-12T12:20:11.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8396" PostTypeId="2" ParentId="8391" CreationDate="2016-04-12T13:20:35.333" Score="4" Body="&lt;p&gt;To quote &lt;a href=&quot;https://en.wikipedia.org/wiki/Section_modulus&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The elastic section modulus is defined as $S = \dfrac{I}{y}$, where $I$ is the second moment of area (or moment of inertia) and $y$ is the distance from the neutral axis to any given fibre. &lt;strong&gt;It is often reported using $y = c$, where $c$ is the distance from the neutral axis to the most extreme fibre&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The tables only show the section modulus for the fiber farthest from the centroid. &lt;a href=&quot;http://www.engineersedge.com/materials/aisc_structural_shapes/aisc_structural_shapes_viewer.htm&quot; rel=&quot;nofollow&quot;&gt;This properties viewer&lt;/a&gt; shows L8x8x1-1/8 with $d = b = 8 \text{ in}$, $\overline y = 2.4 \text{ in}$ (from the horizontal leg), $I = 98.1 \text{ in}^4$, and $S = 17.50 \text{ in}^3$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;S_{top} &amp;amp;= \dfrac{I}{d-\overline y} \approx 17.50 \\&#xA;S_{bot} &amp;amp;= \dfrac{I}{\overline y} = 40.875 \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can easily get from one section modulus to another, though (but it might just be easier to calculate it with the traditional equations):&#xA;$$S_{bot} = S_{top}\dfrac{d-\overline y}{\overline y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth stating however that the critical fiber may not be the one farthest from the centroid. While the farthest face will always have the largest (in magnitude) stress, the near face may be the controlling factor in design. After all, it may be that the lower compressive stress on the near face may lead to buckling before the higher tensile stress on the far face reaches the material's ultimate strength.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-04-13T18:07:50.973" LastActivityDate="2016-04-13T18:07:50.973" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8397" PostTypeId="2" ParentId="8363" CreationDate="2016-04-12T14:27:47.697" Score="1" Body="&lt;p&gt;The negative frequencies are not physical, neither do they convey any new information. The value at the negative frequency is the complex conjugate, and hence the plot is symmetric about the real axis. If you know one you can get the other as a conjugate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value in plotting the negative values (or going around the entire contour) is in doing the stability analysis. &lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-04-12T14:27:47.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8398" PostTypeId="1" AcceptedAnswerId="8401" CreationDate="2016-04-12T15:34:23.993" Score="3" ViewCount="8528" Body="&lt;p&gt;The two terms seem interchangeable and I can't find a strict definition for both, so what's the difference between them if there is any?&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="1293" LastEditDate="2016-04-12T15:42:14.333" LastActivityDate="2021-11-17T01:31:55.533" Title="What's the difference between impeller and rotor in centrifugal pump?" Tags="|mechanical-engineering|pumps|turbomachinery|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8399" PostTypeId="1" CreationDate="2016-04-12T15:47:04.183" Score="2" ViewCount="79" Body="&lt;p&gt;The specs say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Admittance shall be by means of a&#xA;  locked door and/or key operated switch,&#xA;  which in order to provide compatibility&#xA;  within Local Authorities, shall be a &lt;strong&gt;Yale&#xA;  Key, 10½ Section, change 900&lt;/strong&gt;. Access to&#xA;  level 1 facilities shall be gained without&#xA;  opening the main controller door(s).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What exactly is the meaning of the bolded section? I guess it defines &lt;em&gt;some&lt;/em&gt; kind of key (and associated lock), but how do I read it? How can I even begin searching for a lock fulfilling these specs?&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2016-04-12T15:47:04.183" Title="What is &quot;Yale Key, 10½ Section, change 900&quot;" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8400" PostTypeId="2" ParentId="7878" CreationDate="2016-04-12T19:02:32.827" Score="0" Body="&lt;p&gt;Reading this section (5.3) of the book carefully, you will note that equation 5.5 is a stepping stone.  The author is pointing out that, at stall speed, there is a relationship between Cl, the weight of the aircraft and velocity.  He goes on to conceptualize the ability to use lift devices to modify Cl in equation 5.7.  Given that this is aircraft conceptual design, he is providing insight into the relationships that must be developed by the engineer to move from a vehicle gross weight, a takeoff/landing speed and a wing lift coefficient.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wing lift coefficient will later be turned into a relationship between angle of attack and Cl (a polar), and from there into a tool by which airfoils may be selected to provide optimal performance of lift and drag over various regions of flight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand your confusion, based on where he goes with the rest of the section I think equation 5.5 would have been better stated if he had rearranged it to solve for Cl, rather than stating the equation in terms of lift force required.  However, your statement about Cl,min ignores the definition of stall.  At the point of stall (angle and velocity combination), there is a decrease in lift coefficient either way (increased or decreased angle).  As such, the minimum condition for flight (and hence the minimum takeoff velocity) is given by the relationship in equation 5.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stall_(fluid_mechanics)#.E2.80.9CStall_speed.E2.80.9D&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Stall_(fluid_mechanics)#.E2.80.9CStall_speed.E2.80.9D&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5865" LastActivityDate="2016-04-12T19:02:32.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8401" PostTypeId="2" ParentId="8398" CreationDate="2016-04-12T19:23:59.393" Score="4" Body="&lt;p&gt;An impeller is the type of rotor found in a &lt;a href=&quot;https://mechanicalboost.com/centrifugal-pump-an-overview/&quot; rel=&quot;nofollow noreferrer&quot;&gt;centrifugal pump&lt;/a&gt;, in fact another name for &amp;quot;centrifugal pumps&amp;quot; is &amp;quot;impeller pumps&amp;quot;. Many different types of pump have rotors.&lt;/p&gt;&#xA;&lt;p&gt;The pump &amp;quot;rotor&amp;quot; is simply a term the rotating components of the pump. Gear pumps, sliding vane pumps, swash-plate pumps, and circumferential piston pumps all have rotors, but none of them have impellers&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="-1" LastEditDate="2021-07-22T15:58:32.410" LastActivityDate="2021-07-22T15:58:32.410" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="8403" PostTypeId="1" CreationDate="2016-04-13T00:15:56.013" Score="2" ViewCount="239" Body="&lt;p&gt;I'm studying the Hilbert transform and its properties. I found out that Hilbert transform can be considered as a linear time-invariant system. How can I show this?  I guess this should be trivial as my book does not show it.&lt;/p&gt;&#xA;" OwnerUserId="5983" LastEditorUserId="1586" LastEditDate="2016-05-09T23:06:09.883" LastActivityDate="2016-05-09T23:06:09.883" Title="Hilbert transform LTI" Tags="|electrical-engineering|control-engineering|telecommunication|signal|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8404" PostTypeId="2" ParentId="8403" CreationDate="2016-04-13T02:03:37.003" Score="1" Body="&lt;p&gt;A linear operator $\Theta$ is an operator acting on functions $F$ and $G$, with coefficients $a$ and $b$, such that the following equality holds:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Theta(aF+bG) = a \Theta(F) + b \Theta(G)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put into words, order of operations does not matter with respect to the linear operator, multiplication of functions by coefficients, and addition of functions with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show that the Hilbert Transform is a linear operator, apply the principle given above with the Hilbert Transform in place of $\Theta$.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-04-13T02:03:37.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8405" PostTypeId="1" CreationDate="2016-04-13T09:38:29.400" Score="4" ViewCount="1316" Body="&lt;p&gt;Acetone must be recovered from a waste process stream, which also contains chloroform , so that it can be reused in the chemical process.&#xA;    The chemical process requires however that before the recovered acetone can be reused in the process that it meets certain criteria. The distillation column to be used for this separation is already in place and has the equivalent to 12 theoretical trays (not including the reboiler)&#xA;    The waste stream is a mixture containing 60 mole % acetone and 40 mole % chloroform. The chemical process  requires the top product to contain not less than 90 mole % acetone in order for it to meet specification required for reuse into the chemical process. It is also required that the bottom product contains  50 mole % acetone. As part of the setup of the existing distillation column there is a heat exchanger on the feed line into the column that allows the feed to enter at its bubble point and at a rate of 100 mol/hr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aM4ZV.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have tried to calculate the equilibrium curve which looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/olw8O.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure though if this curve can be used for the calculation&lt;/p&gt;&#xA;" OwnerUserId="5987" LastEditorUserId="1832" LastEditDate="2016-04-13T10:36:15.893" LastActivityDate="2017-06-30T06:39:33.977" Title="Calculating the minimum reflux ratio for separation of Acetone and Chloroform" Tags="|chemical-engineering|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8406" PostTypeId="1" CreationDate="2016-04-13T09:42:18.050" Score="0" ViewCount="55" Body="&lt;p&gt;I am doing this daylighting project for which we must bring the light far down into a bunker. I have found &lt;a href=&quot;https://i.stack.imgur.com/eUtt4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eUtt4.jpg&quot; alt=&quot;Optical fiber&quot;&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard and read that the quantity of light that you bring in a cable decreases, 60% of light for 10m cable, 40% for 20m cable etc.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Will a thicker cable bring more light in ?&lt;/li&gt;&#xA;&lt;li&gt;If yes, does it make sense to start, where we capture the light, with a thick cable, as the distance increases, use thinner cables. The light from the thicker cable is concentrated into the thinner cables with convex lenses out of the thicker cable and then into the thinner cable using concave lenses. The final are where we want to put the light might need thin cables.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1wHhe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1wHhe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5834" LastActivityDate="2016-04-15T15:36:18.947" Title="Daylighting, how to improve coverage" Tags="|energy-efficiency|solar-energy|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8407" PostTypeId="1" AcceptedAnswerId="8414" CreationDate="2016-04-13T11:58:42.207" Score="4" ViewCount="7560" Body="&lt;p&gt;I am trying to solve the problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The critical resolved shear stress for iron is 27 MPa. Determine the&#xA;  maximum possible yield strength for a single crystal of Fe pulled in&#xA;  tension.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think that this is related to the equation $\sigma_{yield} = \dfrac{\tau_{crit}}{\cos{\Phi}\cos{\gamma}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The minimum stress necessary to introduce yielding occurs when a&#xA;  single crystal is oriented such that $\Phi=\gamma=45°$; under these&#xA;  conditions, $\sigma_{yield}= 2\tau_{crit}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I tried check my answer, the solution manual said I should use $\sigma_{yield}= 2\tau_{crit}$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In order to determine the maximum possible yield strength for a single&#xA;  crystal of Fe pulled in tension, we simply employ Equation 7.5 as&#xA;  $\sigma_y=2\tau_{crss}=(2)(27\ \mathrm{MPa})=54\ \mathrm{MPa}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't get this. Doesn't this calculate the minimum yield strength, not the maximum? The solution manual is considering the possibility of the iron crystal being pulled in tension in a direction that is most favourable and makes one of the slip systems reach the critical value the quickest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to know the maximum possible yield strength shouldn't you calculate the situation in which tension is applied in a direction that makes $\cos\Phi\cos\gamma$ as small as possible? So, basically, in a direction in which the resolved shear stress is just a fraction of the applied tensile stress?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution manual gets the yield strength when you pull the single iron crystal in such a direction that the resolved shear stress is the largest fraction possible of the applied stress. But if you reorient the single crystal, couldn't it withstand a way larger applied tensile stress and thus have a higher yield strength?&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="368" LastEditDate="2016-04-13T18:02:04.377" LastActivityDate="2016-04-13T20:20:09.843" Title="Determine the maximum possible yield strength for a single crystal of Fe pulled in tension" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8408" PostTypeId="1" AcceptedAnswerId="15767" CreationDate="2016-04-13T12:32:50.223" Score="4" ViewCount="4303" Body="&lt;p&gt;I know that an hydraulic jump can occur only when the Froude number is higher than 1 (supercritical flow), and I am familiar with the Bélanger equation. However, I feel like I have a fundamental misunderstanding about hydraulic jumps: if a jump can occur when Froude number is larger then 1, what prevents the flow in rivers from &quot;jumping&quot; immediately whenever its Froude number becomes slightly more than 1? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean how can a fluid develop a very high Froude number without making a hydraulic jump? According to what I read, in industrial applications such jumps are specially induced on the fluid to dissipate its energy. So I want to understand how it's done.  &lt;/p&gt;&#xA;" OwnerUserId="3284" LastEditorUserId="368" LastEditDate="2016-04-13T18:04:51.593" LastActivityDate="2019-04-30T12:01:47.793" Title="Where exactly will a hydraulic jump occur?" Tags="|fluid-mechanics|hydraulics|open-channel-flow|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8409" PostTypeId="1" AcceptedAnswerId="8422" CreationDate="2016-04-13T12:34:26.537" Score="1" ViewCount="1261" Body="&lt;p&gt;I have trasfer functions of a plant and a controller in laplace domain. I checked for the closed loop response by applying a step response. The system is found to be stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked the response of the system for the same step reference in discrete domain. I can see that the closed loop response of the discrete system is unstable. The transfer function is converted to discrete form by c2d option in matlab with a sampling time of 1ms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldnt the response of the system be same in both continuous and discrete domain (atleast for high sampling frequency)?&lt;/p&gt;&#xA;" OwnerUserId="5775" LastActivityDate="2016-04-14T13:30:10.160" Title="Closed loop response of a discrete system" Tags="|control-engineering|transfer-function|matlab|feedback-loop|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8410" PostTypeId="2" ParentId="7925" CreationDate="2016-04-13T13:33:49.773" Score="0" Body="&lt;p&gt;You need to provide pump curves, especially for the centrifugal pump, and a system curve for the pipework. Or better: don't provide them, do the work yourself. Here's how:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't want to push water into the pressure side of your centrifugal pump. This happens when the head (static + dynamic) at the outlet is higher then the max. head the pump can deliver. Find the static head and the dynamic head for different flowrates (=system curve). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume pipes from both pumps will join shortly after the pumps and the hydraulic losses are neglible. If you think the piping between pumps and T-piece is significant, please provide a P&amp;amp;ID sketch of your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway you don't even need to find your exact working point, just look a your system curve at the combined flow rate of both pumps and see where the static + dynamic head are. If lower than the max head for your centrifugal pump, you are clear. The actual working point of the system will be a bit lower than this, but as I read your question you don't need to know this exactly.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-04-13T13:33:49.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8411" PostTypeId="1" CreationDate="2016-04-13T13:36:29.900" Score="3" ViewCount="43" Body="&lt;p&gt;We need to position skips to within a few (~10?) cm on level ground. I've seen positioning systems for roll on/off containers, these were horizontal guide rails on the ground and presumably the container is pushed between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skips are lowered onto the ground from above:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/V7qT7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V7qT7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;em&gt;Source: &lt;a href=&quot;http://images.google.de/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;source=imgres&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0ahUKEwin7ZO6yIvMAhUFzRQKHfCADsMQ5RMIBDAA&amp;amp;url=http%3A%2F%2Fwww.loacker.cc%2Fhu%2Fstoffstroeme%2Fdienstleistungen%2Fcontainersysteme%2Fabsetzcontainer.html&amp;amp;usg=AFQjCNHCG5Bi92__niOkHFGNHsHdiBPM4w&quot; rel=&quot;nofollow noreferrer&quot;&gt;Loacker&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance, a few bars jutting from the ground at an angle (to form a funnel) could do the trick. From looking at videos the skips swing when beeing set down. The bars would need a coating with PE or a similar soft(er than metal) and tough material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to invent something as I'm sure this is a solved problem. So far my googling hasn't brought me to a solution. What to do when your skip needs exact placements? &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="2819" LastEditDate="2016-04-14T00:18:33.680" LastActivityDate="2016-04-14T00:18:33.680" Title="Positioning system for skips" Tags="|mechanical-engineering|civil-engineering|transportation|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8414" PostTypeId="2" ParentId="8407" CreationDate="2016-04-13T17:18:11.337" Score="3" Body="&lt;p&gt;&lt;strong&gt;The book should probably have said &quot;minimum possible yield strength&quot; or &quot;maximum possible resolved shear stress.&quot; Note that the maximum yield strength of any real single crystal does not depend on one slip system and thus can not be infinite despite what &lt;a href=&quot;https://en.wikipedia.org/wiki/Schmid&amp;#39;s_Law&quot; rel=&quot;nofollow&quot;&gt;Schmid's Law&lt;/a&gt; might imply. The maximum yield strength occurs at a set of angles &quot;furthest&quot; from all of the slip systems and is more difficult to calculate.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There appears to be confusion about what the Schmid's Law actually means. Schmid's Law relates applied tensile stress to resolved shear stress. In other words, a way of transforming an applied tensile stress to a shear stress in a given plane and along a given direction within that plane, based on the angles of the plane and direction with tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yield strength of a single crystal is related, by Schmid's Law, to the critical resolved shear stress necessary to cause slip. The shear stress necessary to cause slip for a given slip system is a material property. The applied tensile stress on a single crystal is related, also by Schmid's Law, to the resolved shear stress on a given slip system, via the angles between slip plane and tension direction, and slip direction and tension direction. Slip plane and slip direction together make a slip system. Note that yield strength (a material property) is related to critical stress (a material property), and applied stress is related to resolved stress. When the applied stress is such that the resolved shear stress equals the critical resolved shear stress, then the applied stress is equal to the yield strength, and the crystal slips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the yield strength of a single crystal in a given direction, first pick a slip system (plane and direction), then pick the direction of tension. Finally determine what tensile stress is required to cause the actual resolved shear stress to equal the critical resolved shear stress. Schmid's Law allows you to convert tensile stress to actual resolved shear stress. Then replace actual resolved shear stress with critical resolved shear stress, and work backward to determine the yield strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your other question, yes it is possible for the yield strength to vary depending on the choice of slip system and tension direction. Single crystals have anisotropic mechanical properties. Consider a (not-real) system with only one slip plane. If that one slip plane is oriented perpendicular or parallel to the direction of tension, slip is impossible because the actual resolved shear stress is 0 in both cases no matter what amount of tension is applied. In a sense, the theoretical tensile yield strength &lt;em&gt;by slip&lt;/em&gt; of such a non-physical system is infinite in those directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All real crystal systems have slip systems such that slip can occur in every tensile direction, though the tensile strength may vary based on orientation. Which slip system becomes active depends on which system's resolved shear stress exceeds its critical resolved shear stress first, which in turn depends on the relative angles the systems make with the choice of tension direction. It is also important to note many crystalline materials behave isotropically in bulk due to individual grains being oriented randomly, unlike in a single crystal which has a dominant orientation.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="2246" LastEditDate="2016-04-13T20:20:09.843" LastActivityDate="2016-04-13T20:20:09.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8416" PostTypeId="2" ParentId="8411" CreationDate="2016-04-14T00:17:00.930" Score="2" Body="&lt;p&gt;I could not find anything in an internet search either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A steel wedges like you show should work fine. The taller the wedges the easier it will be for the driver to position. The swinging will stop when the dumpster strikes the wedges; perhaps they will need additional anchoring or tied together to repeatedly handle this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polyethylene like you mentioned, would help protect the paint and consequential corrosion. Polyethylene tubing (pex) sliced and slid over steel bar that forms the wedge shape would be an economical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure treated wood, crude as it may seem, may also be a viable solution because it will be easier on the dumpster and absorbs impact energy. Anchored sand bags could also be used. They would have to occasionally be replaced, but would offer very good dampening of the swinging motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have control over the positioning equipment you could install dampening on the chains so that swinging is not an issue. Something like this &lt;a href=&quot;https://www.zoro.com/norton-door-closers-hydraulic-door-closer-8301/i/G0035603/&quot; rel=&quot;nofollow&quot;&gt;door dampener&lt;/a&gt; attached to one of the chains on either side (for symmetry) would quickly slow any swinging.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-04-14T00:17:00.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8417" PostTypeId="2" ParentId="8406" CreationDate="2016-04-14T01:01:21.747" Score="1" Body="&lt;p&gt;I am not an optics expert, but since light inside an optical fiber is contained by reflection, the only limit is the heat dissipation of the fiber and surrounding materials. Perhaps someone smarter than I can comment on how diffraction comes into play at the higher intensities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good baseline would be to look at &lt;a href=&quot;http://www.bodor.com/parts/metal-laser-cutter-parts/1112.html&quot; rel=&quot;nofollow&quot;&gt;metal cutting cnc lasers&lt;/a&gt; and what diameter of fiber optic line they are using. The one in the link is 2,000 W at (just guessing) about 6 mm diameter fiber. This is for 1064 nm, but most of the solar spectrum that is transmissible through glass is at a shorter wavelength so you would be able do do 2,000 W or better. Sunlight is roughly 1,000 W per square meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The economic constraint to this however is your ability to focus that much sunlight into the small diameter cable and at an angle reasonably perpendicular to the polished face of the optical fiber. This &lt;a href=&quot;http://www.himawari-net.co.jp/e_page-index01.html&quot; rel=&quot;nofollow&quot;&gt;optical fiber sun tracking system&lt;/a&gt; looks like a good solution. So ultimately the diameter of fiber you choose will be determined based on your fiber cost per foot, focusing system cost, and depth of bunker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often well drilling systems have a minimum diameter on the order of 6&quot;. If you are already drilling/digging a vertical hole to your bunker, putting in a 6&quot; diameter &lt;a href=&quot;http://www.homedepot.com/p/VELUX-10-in-Sun-Tunnel-Tubular-Skylight-with-Rigid-Tunnel-and-Low-Profile-Plastic-and-Metal-Flashing-TSR-010-0000/205048142&quot; rel=&quot;nofollow&quot;&gt;mirrored light tube&lt;/a&gt; would be a much easier and lower cost solution. If you still need to keep the nukes, zombies, and radiation out just use some extra thick glass at both ends ;-)&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="368" LastEditDate="2016-04-15T15:36:18.947" LastActivityDate="2016-04-15T15:36:18.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8418" PostTypeId="2" ParentId="2674" CreationDate="2016-04-14T01:38:38.843" Score="1" Body="&lt;p&gt;Had a similar design need. Found a free software tool called Beamboy2.2  With the engineering data you have this simple piece of software can be a great testbed for your solutions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The software accommodates cantilever beam setup so you can test specifically for the deflection of the steel based on your specific loading conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link I downloaded from is &lt;a href=&quot;http://www.geocities.ws/richgetze/beamb22.zip&quot; rel=&quot;nofollow&quot;&gt;http://www.geocities.ws/richgetze/beamb22.zip&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I set up a beam model that goes from hitch to end of frame. Representing one support rail so halve your loads. I design at 2.5 times actual loads to accommodate the effects of bumps and dips. 2.5 is what the better commercial shops use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use $\ell/300$ as my deflection limit measured from the support at the axle to the end of the frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose for liability I should wave and scream that after you use this to see what works and what you want to build you &lt;strong&gt;&lt;em&gt;must&lt;/em&gt;&lt;/strong&gt; take it to a real live engineer. I am not one. Nor is this software. But the math seems accurate and I'm sure you'll find it useful.&lt;/p&gt;&#xA;" OwnerUserId="5997" LastEditorUserId="1832" LastEditDate="2016-04-14T02:23:29.240" LastActivityDate="2016-04-14T02:23:29.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8420" PostTypeId="2" ParentId="8362" CreationDate="2016-04-14T12:13:23.080" Score="2" Body="&lt;p&gt;First off, a spectrum is a grouping of the maximum response of many analysis. So if you have a spectrum of displacement for a particular model/loading, this means that for a model that has a period of 0.1 seconds, you compute the full displacement response of your model/loading and extract only the maximum displacement value. Then repeat for a model of 0.2 seconds, 0.3 seconds, 0.4 seconds, etc. until you have the value for every period. You can then plot the displacement spectrum for you specific loading.&#xA;Now for the difference between spectrum and pseudo spectrum, if you use Duhamel integral to calculate the displacement spectrum, you get the following equation:&#xA;$$\text{max}[D(t)]=\text{max}\mid -\frac{1}{\omega_D} \int_0^t \ddot u_g(\tau)e^{-\xi\omega(t-\tau)}sin(\omega_d)(t-\tau) d\tau \mid $$ &#xA;In civil engineering, since damping is usually under 20%, we can reasonable assume that $$ \omega_d \approx \omega \text{ and } \xi^2 \approx 0$$&#xA;To get the pseudo velocity spectrum, you can simply integrate the displacement spectrum. With the preceding simplifications, this leads to:&#xA;$$\text{max}[V(t)]=\text{max}\mid - \int_0^t \ddot u_g(\tau)e^{-\xi\omega(t-\tau)}cos(\omega)(t-\tau) d\tau \mid $$&#xA;As we know that sine and cosine have the same maximum value, you can say that the two equations are a fraction of $\omega$ appart. You can thus say that the pseudo spectrum of velocity, that is the spectrum that was not calculated by computing the maximum value of each velocity response curve, can be found by multipling the displacement spectrum by $\omega$:&#xA;$$S_v = \omega S_d$$&#xA;We can do the same procedure for pseudo acceleration:&#xA;$$S_a = \omega^2 S_d$$&#xA;For civil engineering practice, it is reasonable to assume that the pseudo spectrums = the actual spectrums, but don't take my word on it, get yourself a structural dynamics book and read away on this very interesting topic.&lt;/p&gt;&#xA;" OwnerUserId="5691" LastActivityDate="2016-04-14T12:13:23.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8422" PostTypeId="2" ParentId="8409" CreationDate="2016-04-14T13:30:10.160" Score="2" Body="&lt;p&gt;The plant and controller:&#xA;$$\text{sys}=\frac{4700 s^2+4393 s+3.245\times 10^8}{s^4+7.574 s^3+120200. s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$pid=0.287\, +0.008 s+\frac{0.5}{s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed-loop system obtained as $\frac{pid*sys}{1+pid*sys}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$csys=\frac{37.6 s^4+1384.04 s^3+2.59961\times 10^6 s^2+9.31337\times 10^7 s+1.6225\times&#xA;   10^8}{1. s^5+45.174 s^4+121584. s^3+2.59961\times 10^6 s^2+9.31337\times 10^7&#xA;   s+1.6225\times 10^8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poles are all in the left-hand plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \{-11.8643\pm \, 346.642 i,-9.80897\pm \, 25.3345 i,-1.82737\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus as expected, the response to a unit-step is stable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oB32V.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oB32V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zero-order hold approximation for a sampling period of 1 ms:&#xA;$$ \frac{-0.0371234 z^4+0.144584 z^3-0.213662 z^2+0.141976 z-0.0357747}{-1. z^5+4.83687&#xA;   z^4-9.46905 z^3+9.38322 z^2-4.70688 z+0.955831}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poles are all within the unit-circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \{0.929426\pm \, 0.335734 i,0.989921\pm \, 0.0250846 i,0.998174\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, as expected, the response to a unit-step is stable (and the plot is essentially the same):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L8bev.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L8bev.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-04-14T13:30:10.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8423" PostTypeId="1" AcceptedAnswerId="8425" CreationDate="2016-04-14T15:17:09.750" Score="4" ViewCount="2021" Body="&lt;p&gt;Given a nonlinear system, such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;x_1' &amp;amp;= x_2 \\&#xA;x_2' &amp;amp;= −x_1^3 + u \\&#xA;y    &amp;amp;= x_2&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I check the &lt;strong&gt;zero-state observability&lt;/strong&gt; of the system?&lt;/p&gt;&#xA;" OwnerUserId="6006" LastEditorUserId="368" LastEditDate="2016-04-14T18:04:53.207" LastActivityDate="2016-04-14T18:07:39.113" Title="How can I check whether a nonlinear system is zero-state observable?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8425" PostTypeId="2" ParentId="8423" CreationDate="2016-04-14T17:36:47.273" Score="2" Body="&lt;p&gt;I've found the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check if a system is &lt;strong&gt;zero state observable&lt;/strong&gt;, put $u=0$ and check whether $x=0$ when $y=0$. If yes, it is &lt;strong&gt;zero-state observable&lt;/strong&gt;. Otherwise not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the given system, by putting $u=0$ and $y=0$, we see that $x_2=0$, therefore $x'_2=0$ and thus $-x_1^3=0$ or $x_1=0 \implies x=0$. Thus it is &lt;strong&gt;zero-state observable&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6006" LastEditorUserId="368" LastEditDate="2016-04-14T18:07:39.113" LastActivityDate="2016-04-14T18:07:39.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8426" PostTypeId="1" CreationDate="2016-04-14T21:08:29.913" Score="2" ViewCount="3940" Body="&lt;p&gt;I have heard it said that the Laplace transform has &quot;many applications&quot;. But the only &quot;applications&quot; that I ever came across are those that require one to solve either some kind of differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you provide an application of the transform, where the transform of the function allows one to solve a problem, but which is not differential equation related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Fourier transform has applications to signal processing, where one sends/receives signals by transforming and inverting them afterwards. This application with the Fourier transform has no interest in any differential equation solution. Is there a similar application for Laplace?&lt;/p&gt;&#xA;" OwnerUserId="443" LastActivityDate="2016-04-15T12:05:10.343" Title="Applications of the Laplace Transform" Tags="|mathematics|" AnswerCount="1" CommentCount="2" ClosedDate="2016-04-17T16:26:51.823" ContentLicense="CC BY-SA 3.0" />
  <row Id="8428" PostTypeId="1" CreationDate="2016-04-15T00:23:22.680" Score="4" ViewCount="1730" Body="&lt;p&gt;I've been working on this problem for years (literally) and haven't yet found a workable, scalable, pragmatic, and cost-effective solution that would allow me to selectively rotate one or more of 80 discs closely aligned on a single, rotating shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I have 80 identical &quot;discs&quot;, each of which is approximately 2 mm wide, spaced out with approximately 2 mm between each disc, concentrically mounted along a shaft (which can be anywhere from 5-20 mm in diameter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately my background is in electronics and not in mechanics or mechanical engineering, and I'm unaware as to whether or not this can be considered a &quot;solved problem&quot; at this scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried a number of different approaches (both conceptually and in prototype) but have constantly been stymied by the tight specifications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conceptually, something as simple as a gear translating rotation in one axis to another (perpendicular) axis, attached to each &quot;disc&quot; and actuated by a linearly-moving motor would do the trick:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sQ3HM.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sQ3HM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One attempt I've used is to have a rotating &quot;indexer&quot; fixed to a spindle in the middle of the shaft (with the rest of the shaft being free to spin around the spindle), then moving that linearly while subjecting it to rotational movement when aligned with the disc in question (with a matching cutout in each disc mated to the shape of the shaft/indexer), but the need for extreme precision in order for it to move freely along the axis made it unworkable (along with the fact that the shaft/spindle now needed to be twice the width of the working surface. In the example below, the indexer would be 9b, used to rotate 11H without rotating 11G or 11I:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wEnPh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wEnPh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know one solution would be a clutch of sorts - if one exists that would fit on/in each of these discs. Then the clutches could be mounted directly to the shaft, and the clutch for each disc (somehow, as of yet that's a big question mark) engaged or disengaged as needed when the shaft is turning to translate that motion to each disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had an idea to use a &quot;magnetic clutch&quot; (if that is a thing) where rectangular strips of magnets would be mounted into channels in the shaft, then tiny magnets would be embedded around the periphery of the cutout in each disc. The idea was that when not &quot;locked&quot; in place by some external mechanism, each disc would (by magnetic force) rotate with a spinning shaft. When externally locked, the disc would then hold in place with the &quot;clutch&quot; disengaged as the shaft spun freely under it. Unfortunately, it seemed that even with neodymium magnets, the friction from the entire system was too much for the magnets and the discs were permanently in an &quot;unlocked&quot; position and would not spin with the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the obvious solution of using pure friction. I have leeway in what material the discs and shaft are composed of, but I'm not sure if that's something that can be considered a reliable solution as it should last for hundreds of thousands of &quot;disengaged rotations&quot; without wearing out or losing friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I might be overcomplicating the problem. Bottom line: tiny discs, one rotating shaft. Selectively (preferably in parallel, but one-at-a-time is fine) rotating each of the discs, without needing tolerances only achievable in a factory and at a affordable price (let's say under $100).&lt;/p&gt;&#xA;" OwnerUserId="3837" LastEditorUserId="3837" LastEditDate="2016-04-15T00:52:01.467" LastActivityDate="2016-11-15T12:56:52.093" Title="Selectively rotating (engaging/disengaging) one of 80 couplings on a small shaft?" Tags="|mechanical-engineering|design|architecture|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8429" PostTypeId="2" ParentId="8405" CreationDate="2016-04-15T02:40:04.440" Score="0" Body="&lt;p&gt;Seems like you have an azeotropic mixture over there means you can't separate it to get a bottom product &amp;lt; 40 % acetone with a straightforward distillation. However, since you don't, ie the feed, top and bottoms compositions as well as feed conditions (saturated liquid) all put you to the right of this point (of x=y on the equilibrium line) you should be able to use McCabe Thiele method for this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do &#xA;a) find Rmin that would yield you your top product composition. Then with this Rmin you can calculate what the bottoms composition would be with 12 stages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) If you can't meet the bottoms composition with Rmin from a) (and I think you won't) then start increasing R till you get the bottoms composition you want with 12 stages.&lt;/p&gt;&#xA;" OwnerUserId="1103" LastActivityDate="2016-04-15T02:40:04.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8431" PostTypeId="2" ParentId="659" CreationDate="2016-04-15T06:19:27.493" Score="-2" Body="&lt;p&gt;Al tape is also good option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some epoxy mixed with metal powder (Iron powder) is another method.&lt;/p&gt;&#xA;" OwnerUserId="6012" LastActivityDate="2016-04-15T06:19:27.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8432" PostTypeId="1" AcceptedAnswerId="8444" CreationDate="2016-04-15T06:20:07.400" Score="2" ViewCount="4179" Body="&lt;p&gt;I am trying to perform simple 2D CFD analysis on Fluent having 3 adjacent zones(inlet, porous and outlet).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YawGb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YawGb.png&quot; alt=&quot;Zones figure&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The problem I have is whenever I select non-equilibrium thermal model in porous zone in &lt;em&gt;Cell Zone Conditions&lt;/em&gt; (as follows)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dBto0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dBto0.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I get the following error:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A duplicate of this zone cannot be generated automatically, as it&#xA;contains a non-conformal mesh interface.&lt;/p&gt;&#xA;&lt;p&gt;You must manually make a copy&#xA;the porous fluid zone and define it as a solid zone before you enable&#xA;the non-equilibrium thermal model.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;AFAIK, a non-conformal mesh interface is an interface which does not have equal number of nodes on both sides, but it seems to me (visually) that I have an equal number of nodes on both sides (between zones):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qPY6B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qPY6B.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;So how can I make sure that I have conformal mesh interface between the fluid zones and the porous zone?&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-08-30T09:51:06.940" Title="How to force creating a conformal mesh interface between zones in Fluent?" Tags="|cfd|porous-medium|fluent|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8434" PostTypeId="1" CreationDate="2016-04-15T08:28:37.670" Score="4" ViewCount="841" Body="&lt;p&gt;I made a block diagram using simulink. When I solve it with ode45 and ode23 it gives me different graphs. Is it normal? Should I use the proper one?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m1=1.3608;&#xA;m2=0.096;&#xA;r=0.0592;&#xA;I2=0.0002175;&#xA;k1=186.3;&#xA;g=9.81;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is what the model looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fYiMQ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fYiMQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the ode45 solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ToVAJ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ToVAJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't post the ode23 solution but it is very different.&lt;/p&gt;&#xA;" OwnerUserId="6014" LastEditorUserId="334" LastEditDate="2016-04-15T11:17:46.737" LastActivityDate="2018-04-29T22:01:56.117" Title="Is it normal to get different solutions with different ODE methods in MATLAB simulink?" Tags="|matlab|simulink|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8436" PostTypeId="2" ParentId="8426" CreationDate="2016-04-15T11:50:32.777" Score="4" Body="&lt;p&gt;I think you are neglecting the broad range of applications included in solving 'some kind of differential equation.'  At the very basic level of physics, everything we know about the world is expressed in terms of differential equations.  Some examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Maxwell%27s_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electromagnetism&lt;/a&gt; (gives rise to most topics in electrical engineering)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Classical_mechanics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mechanics&lt;/a&gt; (gives rise to most topics in mechanical engineering)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Einstein_field_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gravitation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quantum Mechanics&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At the much more complex level of engineering, many of the problems are still expressed in terms of differential equations.  To give you a feel, here are some questions from this site:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/5665/differential-equations-of-a-simplified-loading-bridge&quot;&gt;Differential equations of a (simplified) loading bridge&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/2935/developing-a-stochastic-differential-equation-model-for-concrete-fibers&quot;&gt;Developing a stochastic differential equation model for concrete fibers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/8097/differential-equation-of-the-vertical-displacement-of-a-cable&quot;&gt;Differential equation of the vertical displacement of a cable&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Laplace transform, in particular, is used widely to 'solve some kind of differential equation' in these applications:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikibooks.org/wiki/Control_Systems/Transforms&quot; rel=&quot;nofollow noreferrer&quot;&gt;Control systems analysis&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikibooks.org/wiki/Electronics_Communication/Signal_Processing/Signal_Transformation/Laplace_Transform&quot; rel=&quot;nofollow noreferrer&quot;&gt;Analog and digital communication&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikibooks.org/wiki/Circuit_Theory/Laplace_Transform&quot; rel=&quot;nofollow noreferrer&quot;&gt;Analyzing electrical circuits&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Laplace_transform#Example_1:_Solving_a_differential_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radioactive decay&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="272" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-04-15T12:05:10.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8437" PostTypeId="2" ParentId="8428" CreationDate="2016-04-15T12:01:41.137" Score="1" Body="&lt;p&gt;One possible idea:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Lock the discs on the shaft using either wax or &lt;a href=&quot;https://en.wikipedia.org/wiki/Hot-melt_adhesive&quot; rel=&quot;nofollow&quot;&gt;hotglue&lt;/a&gt;. Making the shaft out of plastic may make the glue bond better.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wrap a few turns of &lt;a href=&quot;https://en.wikipedia.org/wiki/Resistance_wire&quot; rel=&quot;nofollow&quot;&gt;resistance wire&lt;/a&gt; around the shaft next to each disc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Run normal cable from the resistance wire back to control electronics through the center of the shaft.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Apply power to melt the glue/wax on each disc you want to be free of the shaft rotation. It will take some time to melt it, and you will want to adjust the power so that it will not overheat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-04-15T12:01:41.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8438" PostTypeId="2" ParentId="8428" CreationDate="2016-04-15T12:11:17.547" Score="3" Body="&lt;p&gt;A mechanical solution using movable rods:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Have 7 rods with pieces of rubber around them. The rods should be positioned so that you can move them leftwards or rightwards. When the rubber piece touches the disc, it should stop the rotation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Place the rubber pieces on either the left side or the right side of each disc. This way when you move a rod to the left, it will lock some discs and release others.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Arrange so that rod 1 when actuated will release the discs 0, 2, 4, 6, ... and lock discs 1, 3, 5, 7, .... Rod 2 should release 0, 1, 4, 5, ... and lock 2, 3, 6, 7, ....&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To rotate a single disc, actuate the rods according to the binary number of the disc. For example, to leave disc 7 free to rotate, actuate rods 1, 2 and 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-04-15T12:11:17.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8439" PostTypeId="1" AcceptedAnswerId="8440" CreationDate="2016-04-15T16:09:16.783" Score="1" ViewCount="725" Body="&lt;p&gt;As from the derivation we are able to see that the implicit density based solver (such as BTCS) is unconditionally stable for any courant number. So what is it, which limits the upper boundary of CFL while using it in practical applications?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need this for simulations in ANSYS Fluent.&#xA;I have been told that non-linear effects start coming in, but I don't know what exactly are the non-linearities?&lt;/p&gt;&#xA;" OwnerUserId="1155" LastActivityDate="2016-04-15T20:56:20.963" Title="What limits my Courant number in Implicit Density based Solver?" Tags="|cfd|ansys|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8440" PostTypeId="2" ParentId="8439" CreationDate="2016-04-15T19:16:28.930" Score="1" Body="&lt;p&gt;Analytical stability results for backward-time, centered space (BTCS) discretizations only apply to the first order (linear) hyperbolic partial differential equation.  The only reason why they are applicable to the navier stokes equation is because navier stokes tends to exhibit behaviors similar to this first order linear wave equation in some cases (e.g. slow moving laminar flow).  The stability analysis is much easier to perform on linear equations such as the 1st order wave equation and provides somewhat of an estimate of timestepping stability in the non-linear hyperbolic PDE case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The navier stokes equations have nonlinearities in the momentum equation (i.e. the nonlinear advection term $\nabla\cdot(\vec{u}\otimes\vec{u})$).  As this non-linear term becomes stronger, the stability estimate based on BTCS discretization (for the 1st order linear wave equation) becomes less and less precise.  Unfortunately, non-linear equations are extremely difficult (impossible?) to analytically derive stability conditions for different timestepping schemes.  In practice, you will observe some deviations from the stability conditions for BTCS for the first order wave equation.  Explicit timestepping schemes often need to be much smaller than 1 in order to observe convergence, while backward timestepping will have some limit to how large the time stepsize can be.  Usually, this limit is larger than that which an explicit timestepping scheme would allow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heuristics on stability criterion for the navier stokes equations are generally obtained empirically via trial-and-error and are highly problem dependent.&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2016-04-15T20:56:20.963" LastActivityDate="2016-04-15T20:56:20.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8442" PostTypeId="1" CreationDate="2016-04-15T21:38:09.203" Score="4" ViewCount="474" Body="&lt;p&gt;I'm working on simulating compressible fluid flow over a flat plate with turbulence.  The energy equation for compressible flow with turbulence requires modeling a turbulent heat flux of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q_T=\frac{\mu_T c_p}{Pr_T}\frac{\partial T}{\partial x_j}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mu_T, c_p,Pr_T$ are the turbulent viscosity, specific heat (constant pressure) and turbulent Prandtl number.  Clearly, all three of these paramters vary throughout the domain and require a model to determine their values as a function of the flow variables (density $\rho$, velocity $u$, pressure $p$, and temperature $T$.  The turbulent viscosity is estimated using the variables of the specifically chosen turbulence model.  The specific heat $c_p$ is a function of temperature, and can be estimated by interpolating over the most recent temperature values.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have no idea how to estimate the turbulent Prandtl number as a function of the temperature, velocity, density, or pressure.  The only relevant discussion I've found on this issue is in Wilcox (2006)'s &lt;em&gt;Turbulence Modeling for CFD&lt;/em&gt; (chapter 5, section 5.4.2, pg 250), where it states &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A constant value for $Pr_T$ is often used and this is usually satisfactory for shock-free flows up to low supersonic speeds, provided the heat transfer rate is not too high.  The most common values assumed for $Pr_T$ are 0.89 or 0.90, in the case of a boundary layer.  Heat transfer predictions can usually be improved somewhat by letting $Pr_T$ vary through the boundary layer.  Near the edge of a boundary layer and throughout a free shear layer, a value of the order of 0.5 is more appropriate for $Pr_T$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;While modeling $Pr_T$ as a constant is OK for some cases, I want to know if there is a standard model to prescribe its value as a function of the flow variables $T,u,\rho,$ and $p$.  Preferably, this model should ascribe values near 0.9 within the boundary layer and 0.5 throughout a free shear layer, as described by Wilcox.  How should I prescribe the turbulent Prandtl number throughout the domain as a function of the flow variables?&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2016-04-15T21:38:09.203" Title="How should I set the turbulent prandtl number in a turbulent compressible flow simulation?" Tags="|cfd|turbulence|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8443" PostTypeId="2" ParentId="8428" CreationDate="2016-04-16T11:10:28.190" Score="2" Body="&lt;p&gt;Depending on the rotational speed and the torque required, you may be able to use something that resembles a &lt;strong&gt;pin tumbler lock&lt;/strong&gt; (I can't put more than 2 links in a post so no link to Wikipedia..)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The inner, rotating shaft will have holes in it, into which small cylindrical pins will be inserted:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/u41rN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u41rN.png&quot; alt=&quot;rotating shaft&quot;&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(pins are red, shaft is black)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;An inner rod with some sort of a bump on it moves inside the shaft (either the whole rod moves or only the bump.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When the bump pushes the pins outside they fit into holes inside a disc and force it to rotate with the shaft:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QP8Uz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QP8Uz.png&quot; alt=&quot;engaging and disengaging&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(inner rod is gray, disc is blue, disc pins are orange and springs are purple)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;When the bump moves, the rotating shaft's pins are brought back to their disengaged position by springs inside the disc's holes (the pins inside the disc are not really necessary, they are there to show the resemblance to a lock).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That's just general idea and maybe it's a little gimmicky, but it is another approach.&lt;/p&gt;&#xA;" OwnerUserId="6022" LastActivityDate="2016-04-16T11:10:28.190" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8444" PostTypeId="2" ParentId="8432" CreationDate="2016-04-16T12:18:52.753" Score="1" Body="&lt;p&gt;I found the issue addressed in Fluent user guide:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Parts are groups or collections of bodies. Parts can include multiple&#xA;bodies and are then referred to as multibody parts. &lt;em&gt;&lt;strong&gt;If your&#xA;geometry contains multiple parts then each part will be meshed with&#xA;separate meshes with no connection between them, even if they&#xA;apparently share faces.&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;You can convert a geometry which has multiple parts into one with a&#xA;single part by using the Form New Part functionality in the ANSYS&#xA;DesignModeler application. Simply select all of the bodies and then&#xA;select Tools &amp;gt; Form New Part. If you have an external geometry file&#xA;that has multiple parts that you wish to mesh with one mesh, then you&#xA;will have to import it into the DesignModeler application first and&#xA;perform this operation, rather than importing it directly into the&#xA;ANSYS Meshing application.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;From DesignModeler:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Xkg9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Xkg9.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The whole geometry is now a one part (still with separate zones) with a conformal interface between the zones.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-04-16T12:18:52.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8445" PostTypeId="2" ParentId="2419" CreationDate="2016-04-16T12:33:07.207" Score="1" Body="&lt;p&gt;It's all about the fact that spherical shell would not experience bending if the load is evenly distributed over the surface (which is the case when pressure is the only load). The very idea of using shells is to create them of such shape that stresses due to bending would be low, and the load bearing capacity of the shell would be provided by the tensile forces (or by compression) arising in the shell (tensile/compression forces lead to much lower stresses than bending moments do). An interesting example devoted to load bearing capacity of shells is presented on the following webpage (it is devoted to so-called &quot;shallow&quot; shells, which have low curvature, i.e. they closely resemble plates, but can withstand much larger loads, exactly because compression in the material plays much bigger role than bending in this case):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://members.ozemail.com.au/~comecau/quad_shell_shallow_shell.htm&quot; rel=&quot;nofollow&quot;&gt;http://members.ozemail.com.au/~comecau/quad_shell_shallow_shell.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5992" LastEditorUserId="5992" LastEditDate="2016-04-16T13:50:08.100" LastActivityDate="2016-04-16T13:50:08.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8448" PostTypeId="1" AcceptedAnswerId="8461" CreationDate="2016-04-16T17:24:27.867" Score="2" ViewCount="782" Body="&lt;p&gt;I am curious to know the difference between a lead compensator (lead filter) and a PD controller with a low pass filter. From the theory I can see that the transfer function of both these controllers are same. By choosing appropriate gains and placing zeros and poles at right frequencies, we can obtain same bodeplot (magnitude and phase plot). So, what is the real difference between the two? Why use different names for the same controllers? &lt;/p&gt;&#xA;" OwnerUserId="5775" LastEditorUserId="110" LastEditDate="2016-04-16T19:53:56.383" LastActivityDate="2016-04-18T02:06:44.497" Title="Is lead filter same as PD combined with a low pass filter?" Tags="|control-engineering|transfer-function|feedback-loop|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8450" PostTypeId="1" AcceptedAnswerId="8451" CreationDate="2016-04-16T20:09:47.670" Score="3" ViewCount="539" Body="&lt;p&gt;I was very surprised when i found out that Newton's solution to the problem of solid of minimal resistance was a body with somewhat strange nose cone - the end of the nose cone is  a plane perpendicular to the body's axis. So my question is, was Newton's calculation confirmed and used somehow by space programs?&lt;/p&gt;&#xA;" OwnerUserId="3284" LastActivityDate="2016-04-20T01:24:31.857" Title="Was Isaac Newton's solution to the problem of solid of least resistance comfirmed in space programs?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8451" PostTypeId="2" ParentId="8450" CreationDate="2016-04-16T20:56:52.903" Score="5" Body="&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton got some things spectacularly right, and other things spectacularly wrong. In this case, he didn't remotely understand the &lt;em&gt;real&lt;/em&gt; physics of the problem - and neither did anybody else until a few hundred years after his death. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You shouldn't believe everything in Newton's writings on alchemy either - nor his predictions of the date of the end of the world based on his analysis of the Bible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Longer answer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The great importance of Newton's work attacking this type of problem in &lt;em&gt;Principia&lt;/em&gt; was not that he got almost all the details wrong, but that &lt;strong&gt;he attempted to solve the problems at all&lt;/strong&gt; using mathematical modelling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Newton's time, there was no clear understanding of the difference between energy and momentum. The greatest achievement of &lt;em&gt;Principia&lt;/em&gt; was in pinning down the concept of &quot;momentum&quot;, and the relationship between &quot;force&quot; and &quot;rate of change of momentum&quot;, as given by Newton's second law. And the mathematical edifice that Newton built on that foundation was of course spectacularly right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with no real concept of energy, Newton's ventures into fluid mechanics were much less successful. I very much doubt whether anyone working in the space program would be naïve enough to consider use Newton's models - and most likely they would know nothing about them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show the sort of errors in Newton's fluid dynamics models, it might be instructive to start with a simpler problem which he also considered in &lt;em&gt;Principia&lt;/em&gt;, namely to calculate the speed of sound in air. Newton attempts to solve this by comparing the vibrations of the air with the vibrations of a pendulum (which he could both analyze and measure experimentally), which was a good way to start. But he had no notion of the concept of the internal energy stored in the fluid via the motion of individual molecules, which is related to the temperature of the fluid. In fact, in Newtons' time temperature measurement had barely progressed beyond the subjective and qualitative human notions of &quot;hot&quot; and &quot;cold.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using modern terminology, he incorrectly assumed that all the internal energy in the fluid was accounted for by changes in its pressure and volume. This led him to the equation of state $PV = \text{constant}$, instead of the correct adiabatic relationship $PV^{\gamma} = \text{constant}$ where $\gamma \approx 1.4$ for air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After deriving the properties of air from experiment, this error produced a calculated value for the speed of sound that was about 10% slower than what had been measured. (His measured speed of 339 m/s was probably accurate, but there is no record of the temperature, humidity, wind speed, etc, when the measurement was made).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To account for this discrepancy, Newton makes the wild assumption (&lt;em&gt;Principia&lt;/em&gt; Vol 2 Section 8 proposition 50) that if the particles making up the air are rigid spheres of finite size, and separated from each other by 10 times their diameter, then (by analogy with a &quot;Newton's cradle&quot; device with gaps in between the balls) the speed of sound would be about 10% higher than his calculated result. (It is perhaps unkind to wonder if he had forgotten his grand declaration &quot;Hypotheses non fingo&quot; - &quot;I don't make speculative assumptions!&quot;) He doesn't attempt any mathematical justification of that idea, nor does he explain how the air particles align themselves with the direction in which the sound is travelling!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then goes on to consider the effect of water vapor on the speed of sound. He seems to consider the particles of water vapor to be simply inert bodies that occupy some proportion of the volume of the air, and therefore that water vapor increases the speed of sound further by adding more &quot;balls&quot; to his supposed &quot;cradle&quot;. Of course this logic is actually backwards, but Newton had no idea that the molecular weight of water is about half that of air, and the presence of water thus reduces the density of the air while having little effect on its compressibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His analysis of the air resistance of a projectile (&lt;em&gt;Principia&lt;/em&gt; Book 2, Section 7, Proposition 34, theorem 28) is equally far from the truth. He attempts to do this by considering the change of momentum of the projectile as it collides with stationary particles of air and knocks them aside. He assumes the collisions are perfectly elastic, and therefore (in modern terminology) no mechanical energy would be converted into heat. That means he is assuming the fluid is inviscid. But the correct mathematical analysis of inviscid irrotational flow gives the result that the drag force is &lt;strong&gt;zero&lt;/strong&gt; for &lt;strong&gt;any&lt;/strong&gt; shape of body! Newton made a second mistake, in ignoring what happens to the air &lt;em&gt;behind&lt;/em&gt; the projectile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, Newton's model is not too unrealistic for a completely different scenario, namely piercing a hole through a (fairly loose) granular material like earth. Indeed, the shape of the nose of a typical &quot;mole&quot; tool used for trenchless tunneling is the simple pointed cone shape that Newton claims would give the least drag, as shown here: &lt;a href=&quot;http://en.terra-eu.eu/underground-piercing-tools/2-16-produkte&quot; rel=&quot;nofollow&quot;&gt;http://en.terra-eu.eu/underground-piercing-tools/2-16-produkte&lt;/a&gt;. But any similarity between that scenario and the analysis of a hypersonic space capsule is purely conincidental.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-04-20T01:24:31.857" LastActivityDate="2016-04-20T01:24:31.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8453" PostTypeId="1" CreationDate="2016-04-17T11:01:20.453" Score="1" ViewCount="316" Body="&lt;p&gt;If I have a closed loop second order transfer function such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T(s) = \frac{10-s}{0.3s^2+3.1s+(1+24K_{c})}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I still use this formula for overshoot (when a step input is applied) ?:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{A}{B}=e^{\frac{-\pi \zeta}{\sqrt{1-\zeta^2}}}$$&lt;/span&gt;&#xA;Where &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; is the step input size&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you can but I'm not sure, can someone confirm?&lt;/p&gt;&#xA;" OwnerUserId="4530" LastEditorUserId="18238" LastEditDate="2020-05-10T02:32:48.043" LastActivityDate="2022-06-01T09:22:19.180" Title="Second Order Transfer Function Question regarding Overshoot Formula" Tags="|electrical-engineering|control-engineering|chemical-engineering|process-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="8454" PostTypeId="1" CreationDate="2016-04-17T12:26:55.570" Score="0" ViewCount="599" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5kdVD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5kdVD.png&quot; alt=&quot;velocity diagram request&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone confirm the velocity of B, given point C has velocity of 3 units/s down and point A has a velocity of 5 units/s upwards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach has been to resolve for the tangent velocity component of B with respect to A, and tangent component of velocity B with respect to C, then found the resultant B using the two aforementioned tangents?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="6037" LastActivityDate="2016-04-18T15:04:08.360" Title="Mechanical Linkage Question - Velocity Diagram" Tags="|mechanical-engineering|applied-mechanics|dynamics|mathematics|turbomachinery|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8455" PostTypeId="1" CreationDate="2016-04-17T13:31:39.263" Score="5" ViewCount="121" Body="&lt;p&gt;I have a boat that uses a propeler above the water. The propeler blows the wind to the back of the boat and it makes the boat go straight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one problem with this approach: I am using a big propeller and I need to move the motor high so the propeller does not touch the water. As high as I move the motor there is an undesirable increase in the torque generated by the motor related to it's base. I mean, if I move the motor higher, it makes the boat sink the front part of it, if I move the motor down it eases the sink of the front of the boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I said it all to ask this question: can you see the red lines and the right angle (in red) at my image? If I make that angle (instead of 90) to become 130 will it reduce the force that makes the front of the boat sink? Or making the angle 40, will ease this force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to remember one thing: when we increase/decrease the angle (from 90) the height of the force becomes lower and lower, so it should reduce the torque but it also creates another force (vertical component) that can make the bottom of the boat &quot;fly&quot; or sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought a lot about this and I have no idea if I should increase or decrease the angle (red) in order to prevent the front of my boat to sink cause when I increase a lot the power of my motor my boat gets very fast but the front of it eventually gets inside the water and the bloat flips, which is terrible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made some experiments today and I tried an angle of 70 degree, in that case, when speed is low the back of the boat sinks but when I go fast the tip (front) sinks. When I tried and angle of about 110 degree and the speed is low the tip (front) of the boat almost sinks with low speed but for some reason when it goes fast the front does not sink much more. What's going on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9pzR8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9pzR8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6038" LastEditorUserId="1586" LastEditDate="2016-04-19T22:54:28.750" LastActivityDate="2017-01-02T15:20:56.863" Title="What should be the angle of the motor to the base?" Tags="|torque|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8456" PostTypeId="1" CreationDate="2016-04-17T14:23:18.373" Score="1" ViewCount="210" Body="&lt;p&gt;For a electrical switching device which opens and closes frequently and forcefully. Would a combination of a soft polymer matrix and a conductive metal fibre be a good choice for this? While polymer in itself is not known for having good wear resistance, since it is opened and closed forcefully, using an elastomer could take the energy from the forceful closing and just deform very shortly. The metal matrix inside is protected and could be the charge carrier. Are my assumptions correct? What other ways are there to achieve what I mentioned? &lt;/p&gt;&#xA;" OwnerUserId="6039" LastActivityDate="2016-05-18T17:27:29.660" Title="Design a composite material with good wear resistance and electrical conductivity" Tags="|materials|metals|composite|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8457" PostTypeId="2" ParentId="8456" CreationDate="2016-04-17T14:49:53.947" Score="2" Body="&lt;p&gt;One problem is that you may not have good contact between adjacent metal fibres. In general resin/fibre composites aim to completely coat the fibres so it is entirely possible that a metal/plastic composite could have very low conductivity, depending on how the fibres are orientated. Equally fine metal fibres will have a layer of oxidation on the surface which will further decrease conductivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't really see any reason why a metal composite would be better than solid metal for this sort of application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the spring/damping  characteristics of an elastomer it would probably be better and simpler to mount a brass contactor on a polymer leaf spring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanical (as opposed to semiconductor) electrical contacts are most commonly made from brass as it has a good balance of conductivity, mechanical properties (reasonable hardness, toughness, ductility and strength) and corrosion resistance. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-18T17:27:29.660" LastActivityDate="2016-05-18T17:27:29.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8460" PostTypeId="2" ParentId="8453" CreationDate="2016-04-17T23:46:41.303" Score="0" Body="&lt;p&gt;I think you should still be able to find the inverse laplace transforms of the closed loop TF, at set values of Kc. Such that the overshoot can be determined in the time domain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the step input in the s-domain,  $\frac{B}{s}$? &lt;/p&gt;&#xA;" OwnerUserId="6049" LastActivityDate="2016-04-17T23:46:41.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8461" PostTypeId="2" ParentId="8448" CreationDate="2016-04-18T02:06:44.497" Score="3" Body="&lt;p&gt;A lead filter implies that the zero has a lower frequency than the pole. While a PD controller with a low-pass-filter does not necessarily imply that order. Also a lead filter (usually) does not have the zero and pole to far apart from each other, meaning that the bode diagram does not get very close to the asymptotes of +1 slope for magnitude and 90° for phase between the zero and pole. While the PD and lowpass in general can be spaced very far apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you are right that there is theoretically not really a difference, besides that the PD and lowpass has an additional degree of freedom, due to the P action. Namely for both you have to place a zero and a pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However when implementing both you might get different results due to the way you might implement each filter when discretizing them. But usually there is only a significant difference when you get close to your Nyquist frequency.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2016-04-18T02:06:44.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8462" PostTypeId="1" AcceptedAnswerId="8471" CreationDate="2016-04-18T03:04:12.720" Score="2" ViewCount="1787" Body="&lt;p&gt;I very much ruined 3 $10 diamond hole saws making this ceramic insulator out of a firebrick and Ryobi drill press.  Drilling out the large core took 1-2 hours and a lot of arm strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I will turn to custom ceramic shops or my own castable refractory molds. I am concerned that the custom ceramic shops will be too expensive though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is a refractory cement available to me that can withstand the melting point of titanium?  I.e. there is a Tungsten hotnozzle about 6mm x 10mm that goes at the bottom of this ceramic tube.  It's for a 3D printer that prints metal (hopefully!).  So, I don't need the entire piece to withstand that temp, but just at the end point and radiating around that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I search for castable refractory cement, it seems like only one product type comes up.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4UurB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4UurBt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ULAgo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ULAgot.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3235" LastEditorDisplayName="user16" LastEditDate="2016-04-18T15:46:56.520" LastActivityDate="2020-04-13T07:25:04.250" Title="What is the highest-temp castable refractory cement available?" Tags="|metallurgy|casting|ceramics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8463" PostTypeId="2" ParentId="7910" CreationDate="2016-04-18T08:28:00.037" Score="-1" Body="&lt;p&gt;To calculate the needed torque, you could use the principle of virtual labor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would write this as a comment, if I had enough reputation.&lt;/p&gt;&#xA;" OwnerUserId="6052" LastActivityDate="2016-04-18T08:28:00.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8464" PostTypeId="1" CreationDate="2016-04-18T09:29:40.817" Score="0" ViewCount="87" Body="&lt;p&gt;For my birthday in February, my Partner got me an iMac.  (I specced it out to include what I thought was enough computing versus cost.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a Physical Therapist who comes to our home and treats us.  She is also an Ergonomics expert.  I have the Mac on a desk, which has a keyboard tray / drawer, so that is good.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the keyboard sits about 5/8 inches too low on the desk, and I need something to raise it up.  (Currently, I am using a book that we found that just happened to meet the height requirement; Unfortunately, it doesn't meet the dimension requirements for this keyboard, which is the Apple Mac US English Keyboard with the Numeric Keypad.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need is something that has just the slightest amount of 'give' to it, like rubber or something similar.  I would like to be able to contact whomever sells the material and order a piece cut to the dimensions I specify.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For your reference, the dimensions are as follows: &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Height:  5/8 inches&lt;/li&gt;&#xA;&lt;li&gt;Depth:  4.5 inches &lt;/li&gt;&#xA;&lt;li&gt;Width:  16 15/16 inches&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would appreciate very much if you could point me in the direction of a material that I might consider.&lt;/p&gt;&#xA;" OwnerUserId="6053" LastEditorUserId="1832" LastEditDate="2016-04-18T11:18:17.090" LastActivityDate="2016-04-18T14:09:27.197" Title="What material can I use for a 5/8 inch keyboard lift that can be cut to fit under my keyboard?" Tags="|materials|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8465" PostTypeId="2" ParentId="8464" CreationDate="2016-04-18T11:09:19.457" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Nitrile_rubber&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nitrile rubber&lt;/a&gt; is probably what you want (typically 40 to 70 Shore A), but &lt;a href=&quot;https://en.wikipedia.org/wiki/Neoprene&quot; rel=&quot;nofollow noreferrer&quot;&gt;Neoprene rubber&lt;/a&gt; might have more &quot;give&quot;. &lt;a href=&quot;https://en.wikipedia.org/wiki/Silicone_rubber&quot; rel=&quot;nofollow noreferrer&quot;&gt;Silicone rubber&lt;/a&gt; is also an option, although it's likely to be more expensive. Probably worth trying to get a couple of free samples to see which one suits you best.  Here is an example of different materials and shore hardnesses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JZCSr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JZCSr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a vendor, I would suggest Googling as it's probably best selecting somebody close to where you live, or at least in the same country. There are likely to be loads, on top of the generic suppliers like &lt;a href=&quot;http://uk.rs-online.com/web/&quot; rel=&quot;nofollow noreferrer&quot;&gt;RS&lt;/a&gt;. If you want it cut to the right dimensions (and I think that the thickness here is going to be the most difficult to find), you probably will need to go to a specialist.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2016-04-18T11:09:19.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8466" PostTypeId="1" CreationDate="2016-04-18T13:02:25.190" Score="4" ViewCount="768" Body="&lt;p&gt;I have an interference fit problem of two perfect cylinders modelled in 2D using symmetry. The overall interface pressure is around 20MPa lower than the hand calculations and I get non-uniform displacement / pressure near the edges of the model. Any suggestions on how to remove these raisers would be greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've attached the pressure results.&#xA;[&lt;img src=&quot;https://i.stack.imgur.com/9rBjI.png&quot; alt=&quot;Pressure distribution&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6056" LastActivityDate="2024-01-08T22:06:15.720" Title="Interference fit problem in ANSYS Workbench 16.1 - how do I get rid of pressure/displacement raisers?" Tags="|structural-engineering|finite-element-method|ansys|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8467" PostTypeId="2" ParentId="8454" CreationDate="2016-04-18T13:52:23.380" Score="1" Body="&lt;p&gt;This is how to approach this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UaYZm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UaYZm.png&quot; alt=&quot;sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rule is that the relative rotation center between two bodies (point &lt;em&gt;B&lt;/em&gt;) lies on the line connecting the instant centers of rotation of each body (points &lt;em&gt;D&lt;/em&gt; and &lt;em&gt;E&lt;/em&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point &lt;em&gt;D&lt;/em&gt; is located to the left of &lt;em&gt;A&lt;/em&gt; because a positive rotation makes &lt;em&gt;A&lt;/em&gt; go upwards with $$\omega_{AB} = \frac{v_A}{a}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly point &lt;em&gt;E&lt;/em&gt; is located to the right of &lt;em&gt;C&lt;/em&gt; causing a positive rotation for the point to move down and $$\omega_{CB} = \frac{v_C}{c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to color code the motion in order to make it more clear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I48XG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I48XG.png&quot; alt=&quot;sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;em&gt;B&lt;/em&gt; belongs to both &lt;em&gt;AB&lt;/em&gt; and &lt;em&gt;CB&lt;/em&gt; it means that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_B = \omega_{AB} (g+h) = \omega_{CB} h $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are four equations to be used with the geometry of &lt;em&gt;a&lt;/em&gt;, &lt;em&gt;c&lt;/em&gt;, &lt;em&gt;g&lt;/em&gt; and &lt;em&gt;h&lt;/em&gt; to be used for the solution. I had to consider the following angles to solve this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tAC1o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tAC1o.png&quot; alt=&quot;sketch3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got $\theta_C = 0.403696$ and $\theta_A = 0.11990$. By projecting the various lengths along the &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; axes I get the following relationships&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned} &#xA;  g \cos(\theta_B) + (200-a) &amp;amp; = c \\&#xA;  g \sin(\theta_B) &amp;amp; = 150 \\&#xA;  (g+h) \cos(\theta_B) + (200-a) &amp;amp;= 250 \cos(\theta_C) \\&#xA;  (g+h) \sin(\theta_B) &amp;amp; = 250 \cos(\theta_A)&#xA;\end{aligned} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the kinematics (first set of equations) we substitute $a=\frac{5}{\omega_{AB}}$, $c=\frac{3}{\omega_{CB}}$, $h=\frac{v_B}{\omega_{CB}}$ and $g=v_B \left( \frac{1}{\omega_{AB}} - \frac{1}{\omega_{CB}} \right)$ above to get four equations for four unknowns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution is&lt;/p&gt;&#xA;&#xA;&lt;blockquote class=&quot;spoiler&quot;&gt;&#xA;  &lt;p&gt; $$\begin{aligned} \theta_B &amp;amp;= 0.585446 \\ \omega_{AB} &amp;amp; = 0.014515 \\ \omega_{CB} &amp;amp; = 0.036685 \\ v_B &amp;amp;= 6.51982 \end{aligned} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2016-04-18T13:52:23.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8468" PostTypeId="2" ParentId="8466" CreationDate="2016-04-18T14:00:06.687" Score="0" Body="&lt;p&gt;What are your constraints like? from the proximity it seems that you might be over constrained or at least to have uniform radial stiffness near the edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option for might be to turn on the &lt;code&gt;Normal-Projected Normal&lt;/code&gt; option&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6AzqD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6AzqD.png&quot; alt=&quot;pic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get smooth results you need to try out different contact formulations also, but in my experience you will never get 100% smooth. The best I have ever achieved is pressure results within 10% of the average.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2016-04-18T14:00:06.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8469" PostTypeId="2" ParentId="8464" CreationDate="2016-04-18T14:09:27.197" Score="0" Body="&lt;p&gt;It might also be worth looking at EVA foam. This is widely used by theatrical prop and costume makers as it very easy to work with and is easily available in the form of exercise mats etc. This gives you the advantage that you can handle it in a shop to see if it feels right. You can cut it with a craft knife and it accepts most general purpose adhesives, in case you need to stack several layers to get the right thickness. You may also find a better range of colours than other materials, if that is important to you. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-18T14:09:27.197" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8470" PostTypeId="2" ParentId="8454" CreationDate="2016-04-18T15:04:08.360" Score="0" Body="&lt;p&gt;Assume that the link AB makes an angle $a[t]$ with the horizontal and the link CB makes an angle $c[t]$ with the horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equating the position of B (x and y coordinates) computed w.r.t both A and C, we get two equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 250 \cos (a(t))+200=250 \cos (c(t)) $$&#xA;$$ 250 \sin (c(t))+150=250 \sin (a(t)) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations can be solved for $a[t]$ and $c[t]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a(t)=\tan ^{-1}\left(\frac{3+4 \sqrt{3}}{3 \sqrt{3}-4}\right),\ c(t)=\tan&#xA;   ^{-1}\left(\frac{4 \sqrt{3}-3}{4+3 \sqrt{3}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity of B (x and y coordinates) can also be computed in two ways. So we get two more equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -250 c'(t) \sin (c(t))=-250 a'(t) \sin (a(t))$$&#xA;$$ 250 c'(t) \cos (c(t))+\text{vc}=250 a'(t) \cos (a(t))+\text{va}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations can be solved for $a'[t]$ and $c'[t]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a'(t)=-\frac{\left(\sqrt{3}-4\right)&#xA;   (\text{va}-\text{vc})}{1250},\ c'(t)=\frac{\left(4+\sqrt{3}\right)&#xA;   (\text{va}-\text{vc})}{1250}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity of B can now be computed as&#xA;$$ \left\{-250 c'(t) \sin (c(t)),250 c'(t) \cos (c(t))+\text{vc}\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left\{-250 a'(t) \sin (a(t)),250 a'(t) \cos (a(t))+\text{va}\right\} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the values of $a[t]$, $c[t]$, $a'[t]$, $c'[t]$ and $va=5$, $vc=-3$ gives the same result in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left\{-3.60, 5.43\right\}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point B is moving upwards and to the left.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-04-18T15:04:08.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8471" PostTypeId="2" ParentId="8462" CreationDate="2016-04-18T15:28:59.677" Score="3" Body="&lt;p&gt;There are a number of approaches to this, depending on exactly what you want to achieve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;http://www.castreekilns.co.uk/dense-castable-189-c.asp&quot; rel=&quot;nofollow&quot;&gt;dense castable refractories&lt;/a&gt; which will operate up to 1800C. For small quantities the best source is likely to be a ceramic kiln supplier. Industrial foundry supplier may have a wider range of products but tend to deal in larger quantities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible alternative is graphite, this needs to be shielded from oxygen to prevent oxidation, but then again so does molten titanium so you would need to do that anyway as titanium is very prone to porosity and embrittlement due to absorbing atmospheric gasses, even at temperatures well below its melting point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zirconia based refractories can also work well at very high temperatures. One option is to use a sofer insulating refractory, or indeed rigidised ceramic wool which is coated with a layer of zirconium slurry, this is the principal behind ceramic shell investment casting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible that any shielding gas used will also provide a significant cooling effect and as such you may find that the case temperature is significantly less than the molten metal temperature, especially as pressurised argon will give you a significant cooling effect for free. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside I'm not sure how much work you have done on the whole concept of extruding molten titanium through a nozzle but it seems a bit iffy to me as I am not aware of any current process which does anything remotely like this and for absolute certain in TIG welding any contact between the tungsten electrode and the base metal is normally considered a defect as the base/filler will instantly alloy with the tungsten and contaminate it. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-18T22:13:17.867" LastActivityDate="2016-04-18T22:13:17.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8473" PostTypeId="1" AcceptedAnswerId="13857" CreationDate="2016-04-18T17:02:57.040" Score="3" ViewCount="179" Body="&lt;p&gt;I have some questions in the basics of machinery terminology &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can a spring ever be a link? Most books (eg. Ham et al. 58, Shigley pg. 5) consider links to be rigid. However, I have come across articles using springs as links.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are joints always the same thing as kinematic pairs, or do they ever differ in any particular circumstances?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is mechanism a subset of kinematic chain or is chain a subset of mechanism? Some books mention if there is no fixed link, then the combination cannot be called a mechanism but can be a kinematic chain.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&#xA;" OwnerUserId="6061" LastEditorUserId="272" LastEditDate="2016-04-19T11:53:24.287" LastActivityDate="2017-02-22T12:58:18.640" Title="Questions on machinery terminology" Tags="|mechanical-engineering|terminology|kinematics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8474" PostTypeId="2" ParentId="659" CreationDate="2016-04-18T19:21:07.303" Score="-1" Body="&lt;p&gt;Try using a t a u (thermocouple attachment unit) they normally work &lt;/p&gt;&#xA;" OwnerUserId="6064" LastActivityDate="2016-04-18T19:21:07.303" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8475" PostTypeId="1" CreationDate="2016-04-18T19:23:28.183" Score="2" ViewCount="90" Body="&lt;p&gt;I have downloaded the following part file of a servo motor as IGES from the vendor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eWYXx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eWYXx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have then built an assembly that uses several of these motors, as well as a number of idlers/hubs, brackets, machine screws, and whatnot. I've used the standard &quot;constrain mate to face&quot; and &quot;constrain mate axis&quot; constraints, as well as inserting bolts from content center, and bolted connections for bolt+nut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as it turns out, the &quot;horn&quot; part of this servo (the aluminum wheel on the front with the eight threaded holes) is modeled as a static part of the mesh in the IGES file. But, in real live, this moves/articulates. I'd like my model to move/articulate the same as the real thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I get there with the minimum amount of re-work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest I can think of right now is to break off the horn from the motor housing, and create two part files, then create a sub-assembly out of those part files, and then delete each of the motors in the bigger assembly and re-add them. Given the amount of constraints and bolt connections on those things, that would be about as bad as starting over from scratch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easier/better way to achieve what I want?&lt;/p&gt;&#xA;" OwnerUserId="1889" LastActivityDate="2016-04-18T19:23:28.183" Title="Inventor: How to replace static part with articulation?" Tags="|mechanical-engineering|modeling|simulation|cad|autodesk-inventor|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8476" PostTypeId="2" ParentId="35" CreationDate="2016-04-18T19:49:00.730" Score="1" Body="&lt;p&gt;There are three key temperature which may affect the properties of steel. As mentioned in the accepted answer the recrystallization temperature is the most significant as it can potentially affect any steel, especially ones which have been cold worked to improve their properties or have a high alloy content eg stainless, chrome-moly steels and some castings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second consideration is the tempering temperature. This only applies to steels which have been heat treated, generally cutting tools, dies,  springs and certain other very high strength/hardness parts. The tempering range can vary between 180 and 300 C or up to 600C for high speed steels. Heating above the tempering temperature will remember  the steel and consequently soften it. this is usually only a concern for finished components although some types of stock are supplied hardened and tempered, typically high alloy tool steels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final concern is that very high temperatures, approaching the melting point of the steel may cause the growth of very large crystals of even deep oxidation of the surface. This is a concern for all grades but particularly those containing chromium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, while many hot rolled steel grades can be hot worked without any loss of mechanical properties some have a narrow working window (red short) and the manufacturers data should be consulted in all cases. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-18T19:49:00.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8477" PostTypeId="1" AcceptedAnswerId="8479" CreationDate="2016-04-19T01:02:51.577" Score="3" ViewCount="225" Body="&lt;p&gt;What kinds of sensors can be used to monitor the condition of a hard disk?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to be able to detect when a hard disk is not operating correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unexpected vibrations may be caused by spindle problems (probably low frequency), or by head crashing (high frequency).&lt;/p&gt;&#xA;" OwnerUserId="2169" LastEditorUserId="1586" LastEditDate="2016-04-19T22:24:43.450" LastActivityDate="2016-04-19T22:55:46.530" Title="What sensors should be used to analyze vibration of a hard disk?" Tags="|mechanical-engineering|electrical-engineering|motors|sensors|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8478" PostTypeId="1" CreationDate="2016-04-19T02:43:39.147" Score="3" ViewCount="2916" Body="&lt;p&gt;When calculating the factor of safety for a fillet weld that is loaded in shear what should be the maximum allowable stress? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some books suggest using the distortion energy (Ssy = 0.577Sy for the maximum allowable stress) while others use maximum shear stress theory (Ssy = 0.3Sut for the maximum). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are both of these ways correct? Which one is used more in the field?&lt;/p&gt;&#xA;" OwnerUserId="6047" LastActivityDate="2017-11-02T06:10:12.847" Title="Strength of Welded Joints" Tags="|mechanical-engineering|design|mechanical-failure|welding|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8479" PostTypeId="2" ParentId="8477" CreationDate="2016-04-19T02:47:24.767" Score="3" Body="&lt;p&gt;There are two types of sensors commonly used in hard disks to detect vibration and head crashing. They are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Accelerometer&lt;/strong&gt;: Accelerometers are used to detect sudden vibrations. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Sudden Motion Sensors:&lt;/strong&gt; These are mostly used in Apple Notebooks to detect a sudden drop of a laptop and to disengage the hard drive. SMS system parks the hard drive and restart the hard drive when stability is detected. SMS include a triaxial accelerometers. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Active_hard-drive_protection&quot; rel=&quot;nofollow&quot;&gt;Active hard-drive protection&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sudden_Motion_Sensor&quot; rel=&quot;nofollow&quot;&gt;Sudden Motion Sensor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://Mac%20notebooks:%20About%20the%20Sudden%20Motion%20Sensor&quot; rel=&quot;nofollow&quot;&gt;Mac notebooks: About the Sudden Motion Sensor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.analog.com/library/analogDialogue/archives/39-11/hdd.html&quot; rel=&quot;nofollow&quot;&gt;Using Dual-Axis Accelerometers to Protect Hard Disk Drives&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eetimes.com/document.asp?doc_id=1271337&quot; rel=&quot;nofollow&quot;&gt;Accelerometers protect hard drives&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="1586" LastEditDate="2016-04-19T22:55:46.530" LastActivityDate="2016-04-19T22:55:46.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8480" PostTypeId="1" CreationDate="2016-04-19T08:16:50.427" Score="-2" ViewCount="163" Body="&lt;p&gt;Here's the situation: a thin concrete wall needs to go up, but it's a very large diameter, say 1 km. The wall thickness is on the order of 20 cm. The wall only needs to be maybe 3 meters high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming a giant single mold shaped like a ring is not practical. Instead, there are multiple molds, shaped as pieces of arc, that the wet cement is poured into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is, how do you make the wall uniform? I'm thinking the concrete will shrink as it cures, so there will be a gap between each poured piece of wall arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some way around this? How do you incrementally extend the wall without gaps? I'm trying to imagine some separater, like a plank of wood, blocking one end of the mold. Once the next mold is set up and poured, you can pull out the plank. But is that practical? How do you seal the plank against the mold so there are no leaks? Besides that, how would you press the sides of the mold together hard enough so there aren't any leaks there either?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edits:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I forgot to mention this is &lt;strong&gt;non-reinforced concrete.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The environment is kinda cold. I'd say yearly ranges would see &lt;strong&gt;-25 to +25 Celsius.&lt;/strong&gt; However, I would very much love to hear about thermal expansion and how to handle that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IDK anything about ground conditions. I'm just imagining normal dirt. Someone mentioned a possible need of footers but I can't see that. Remember this wall is very thin. 20 cm thick and 3 meters high. &lt;strong&gt;That sounds pretty light to me&lt;/strong&gt;, but if necessary, I would consider a flaring base for a snowshoe effect. Drilling to the bedrock for footers seems pretty expensive, so some kind of built-in foundation via a flaring base would be my first thought.&lt;/p&gt;&#xA;" OwnerUserId="3510" LastEditorUserId="3510" LastEditDate="2016-04-19T12:07:23.700" LastActivityDate="2016-04-19T12:07:23.700" Title="How to construct a very large circular wall made of concrete?" Tags="|structural-engineering|concrete|construction-management|" AnswerCount="1" CommentCount="11" FavoriteCount="0" ClosedDate="2016-04-19T14:24:00.097" ContentLicense="CC BY-SA 3.0" />
  <row Id="8481" PostTypeId="2" ParentId="8480" CreationDate="2016-04-19T10:57:07.573" Score="4" Body="&lt;p&gt;You can build this wall as you would any other. Given its size, the fact that it's a ring becomes irrelevant, which means that for all intents and purposes, it can be designed and constructed with the same concepts as that of a linear wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be demonstrated thusly: the wall is 1000 meters in diameter, which gives it a perimeter of approximately 3141 meters ($p = d\pi$). Now, a common maximum for small angle approximations (where angles can be ignored) is 10 or 14 degrees (&lt;a href=&quot;https://en.wikipedia.org/wiki/Small-angle_approximation#Error_of_the_approximations&quot; rel=&quot;nofollow&quot;&gt;source&lt;/a&gt;). So, conservatively, the maximum length of an arc of wall which could be designed as a straight wall is 87 meters ($\ell = \dfrac{p}{360}*\theta$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, walls can't be built with such dimensions. Not only would they be impractical to build, they would also be very prone to severe cracking due to drying shrinkage and thermal expansion during their service lifetimes. Walls must be designed with control joints which effectively turn a wall into a series of wall segments. Each of these segments will be much shorter than 87 meters and can therefore be designed as if they were straight instead of curved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These control joints will need to properly treated for waterproofing of course. &lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-04-19T11:36:28.283" LastActivityDate="2016-04-19T11:36:28.283" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8482" PostTypeId="1" CreationDate="2016-04-19T13:16:37.407" Score="2" ViewCount="89" Body="&lt;p&gt;What's the proper first letter designator for a torque guard in a DIN or ISO compliant P&amp;amp;ID? I find fist-letter designators for a few physical values (L for level, S for speed etc.) but none for force or torque. As of now I'd use WS+ (W for weight, mass) as the closest but I'm not entirely happy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is either non defined in the codes or nobody bothered to copy the designator to where I can find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I'm fine with an answer of the type: &quot;I did X, a dozen people looked at my drawings, no one was confused&quot;&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-04-20T05:55:02.530" LastActivityDate="2016-04-20T05:55:02.530" Title="Proper designator for a torque guard in a P&amp;ID" Tags="|cad|diagram|pi-diagram|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8483" PostTypeId="5" CreationDate="2016-04-19T13:21:08.717" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-04-19T13:21:08.717" LastActivityDate="2016-04-19T13:21:08.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8484" PostTypeId="4" CreationDate="2016-04-19T13:21:08.717" Score="0" Body="For questions relating to piping &amp; instrumentation diagrams" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-04-19T14:03:40.837" LastActivityDate="2016-04-19T14:03:40.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8485" PostTypeId="1" CreationDate="2016-04-19T13:21:14.067" Score="0" ViewCount="2118" Body="&lt;p&gt;In, &lt;em&gt;&lt;a href=&quot;http://www.cientificosaficionados.com/libros/cohetes.pdf&quot; rel=&quot;nofollow&quot;&gt;How to Design, Build, and Test Small Liquid-Fuel Rocket Engines&lt;/a&gt;&lt;/em&gt;, by Krzycki, page 18 has the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t_w = \dfrac{PD}{16000}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the following:&#xA;Pressure ($P$) in combustion chamber is 500 psi.&#xA;Mean Diameter of Combustion Chamber ($D$) is 2 inches. &#xA;Working Stress (as given in the book) 8,000 psi for the Copper.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chamber Wall Temperature of 100°C (I know that's low, help me out here, I'm new). &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Solve for Combustion Chamber wall thickness.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Working Stress is listed as about 8,000 psi in the book, is working stress the same as proportional limit?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;While I am a mechanical engineering student, this question isn't homework. I'm a hobbyist with a passion for rocketry. I know this is my first post here, I don't want to get off on the wrong foot. Please substitute different values if that helps! I was hoping someone could walk me through this. Specifically what I don't understand, if I multiply a pressure variable (500 psi) by a diameter (2 inches), what unit of measure is the answer? Obviously $$500*2=1000$$ But 1000 what? $$1000/16000=0.0625$$ .... Is 0.0625 inches? That's 1.5mm and that doesn't seem thick enough at all.... &lt;/p&gt;&#xA;" OwnerUserId="6076" LastEditorUserId="2246" LastEditDate="2016-04-19T14:18:08.330" LastActivityDate="2016-04-19T14:32:13.180" Title="Chamber Wall Thickness for Liquid Fuel Rocket" Tags="|mechanical-engineering|materials|pressure|aerospace-engineering|metals|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8486" PostTypeId="2" ParentId="8485" CreationDate="2016-04-19T14:10:40.103" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Solve for Combustion Chamber wall thickness.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Eqn. (22) of the text you reference shows the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = \frac{PD}{2t_w}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which mechanical engineers will recognize as the equation for stress in a pipe wall given an internal pressure.  This equation may be rewritten to solve for wall thickness $t_w$, as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  t_w = \frac{PD}{2S}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading the paragraph above (23), the allowable stress $S$ is limited to 8,000 psi.  The author rewrites (23) to solve for the wall thickness $t_w$, as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  t_w = \frac{PD}{16000\mathrm{psi}}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but he drops the units out of the equation.  This is generally a thing to avoid since it leads to confusions like what you're experiencing, especially for someone unfamiliar with whatever unit system the author is working in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if you have an internal pressure of 500 psi and a diameter of 2 inches,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  t_w = \frac{(500\mathrm{psi})(2\mathrm{in})}{16000\mathrm{psi}} = 0.0625\mathrm{in} = \frac{1}{16} \mathrm{in}. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that an internal pressure stresses the vessel in membrane tension only, this is reasonable with the pressure and diameter provided.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Working Stress is listed as about 8,000 psi in the book, is working stress the same as proportional limit?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Working stress is given as a fraction of the proportional limit (a.k.a. &lt;a href=&quot;https://en.wikipedia.org/wiki/Yield_%28engineering%29&quot; rel=&quot;nofollow&quot;&gt;yield stress&lt;/a&gt;).  The working stress includes a factor of safety against yield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I do a search for copper pipe alloy, it looks like typical copper tube one might buy at a home store is alloy C10200, C10300, C10800, C12000, or C12200 (&lt;a href=&quot;http://www.copper.org/applications/plumbing/techref/tpf_stds/tube_pipe_stds.html&quot; rel=&quot;nofollow&quot;&gt;source here&lt;/a&gt;).  Referring to the ASME Boiler &amp;amp; Pressure Vessel Code, 2013, Section II, Part D, Table Y-1, these material alloys have a yield strength of 9,000 psi and a &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultimate_tensile_strength&quot; rel=&quot;nofollow&quot;&gt;tensile strength&lt;/a&gt; of 30,000 psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it seems that your reference material is using 89% of yield or 27% of tensile strength as the working stress.  Whether or not this is acceptable is up to you.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2016-04-19T14:32:13.180" LastActivityDate="2016-04-19T14:32:13.180" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8487" PostTypeId="2" ParentId="4040" CreationDate="2016-04-19T17:54:43.123" Score="2" Body="&lt;p&gt;CircuitWorks doesn't simulate the electronics; it just integrates between some common PCB design software formats and SolidWorks for mechanical information. For example, maximum component height, location of heat sinks and mounting holes, keep-out areas, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like what you want is a plug-in that drives the mechanical constraints in SolidWorks (motor torque, etc) based on electronics simulation output. I haven't seen this particular thing done, but what generally happens in these kinds of integrations is that some product is found, or some in-house tool is written, that takes the output of one of the simulations, and converts it to the right kind of input for the other simulation. Often, the tools aren't set up to do real-time feed-back, and thus you have to go to batch-based interchange, where you provide some specific stimuli at specific times, and &quot;render it out&quot; in a batch, and then go back over it and verify that the right thing happened at the right time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, perhaps more common in more mundane projects, the mechanical guys and electronics guys agree on some particulars (shape, size, etc) and each build their own thing, meeting every once in a while to verify assumptions, and the first integration test is run in the first actually-built prototype, rather than inside computers. Experience will tell you what things actually matter in the interface specified between the two systems, and may provide a checklist to go through to make sure you don't forget anything. (&quot;Your batteries develop HOW much heat when charging !?&quot;)&lt;/p&gt;&#xA;" OwnerUserId="1889" LastActivityDate="2016-04-19T17:54:43.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8488" PostTypeId="1" AcceptedAnswerId="8490" CreationDate="2016-04-19T18:56:22.023" Score="4" ViewCount="5988" Body="&lt;p&gt;In a tensile test the peak load was 69kN, but the point of failure load was 60kN. Why is it lower?&lt;/p&gt;&#xA;" OwnerUserId="5780" LastActivityDate="2016-04-19T19:44:46.680" Title="Why is the point of failure load lower than the peak load?" Tags="|mechanical-engineering|structural-engineering|civil-engineering|stresses|" AnswerCount="1" CommentCount="2" ClosedDate="2016-04-20T06:47:46.237" ContentLicense="CC BY-SA 3.0" />
  <row Id="8489" PostTypeId="1" AcceptedAnswerId="8495" CreationDate="2016-04-19T19:44:19.940" Score="5" ViewCount="1908" Body="&lt;p&gt;I have a tube, at the bottom of the tube is burning wood and an air inlet. How can i increase the speed of the exhaust gas coming out of the tube? I thought about &quot;bottle-necking&quot; the exhaust pipe to make it go faster (Bernoulli's principle) but is there any other way?&lt;/p&gt;&#xA;" OwnerUserId="3538" LastEditorUserId="1586" LastEditDate="2016-04-19T22:25:01.170" LastActivityDate="2016-04-22T10:50:39.043" Title="How to increase the speed of exhaust gas?" Tags="|fluid-mechanics|thermodynamics|combustion|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8490" PostTypeId="2" ParentId="8488" CreationDate="2016-04-19T19:44:46.680" Score="12" Body="&lt;p&gt;That is due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Necking_(engineering)&quot;&gt;necking&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a bar is under tension, two opposing mechanisms take place:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The bar loses cross-sectional area since it attempts to retain its volume while being stretched (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Poisson%27s_ratio&quot;&gt;Poisson's ratio&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Once the bar begins to suffer plastic deformations, it undergoes &lt;a href=&quot;https://en.wikipedia.org/wiki/Work_hardening&quot;&gt;stress hardening&lt;/a&gt; and it's elastic modulus rises.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So long as the increase in elastic modulus is greater than the loss of area, the stress (or force) measured will increase. At a certain point, however, the material starts to lose area faster than it hardens, which means that the force required to continue deforming the element starts to diminish, until the element reaches its ultimate deformation capacity and snaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you'd been measuring the force and dividing by the element's original cross-sectional area to get stress results, then these would have the same profile as the force results: increasing up to the peak load and then dropping to point of failure. This is known as &lt;strong&gt;&lt;em&gt;engineering stress&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, you use special equipment to measure the actual cross-sectional area as it changes during the test, then you'll see that the stress actually increased during the entire test due to stress hardening. This is known as &lt;strong&gt;&lt;em&gt;true stress&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a graph showing the difference between engineering and true stress. The red line represents engineering stress and the blue, true stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PPrjr.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PPrjr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Hooke%27s_law#/media/File:Stress_v_strain_A36_2.svg&quot;&gt;Wikipedia&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the name implies, engineers tend to use engineering stress for the simple reason that it is infinitely easier to measure and design with. And as the diagram above demonstrates, the difference between engineering and true stress in the elastic range where we spend 95% of our time is very small.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-04-19T19:44:46.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8491" PostTypeId="2" ParentId="8489" CreationDate="2016-04-19T20:17:17.830" Score="3" Body="&lt;p&gt;If the tube is acting as a chimney, then i would first try a longer tube, and add lots of insulation to the outside of it too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hotter gases inside it would be less dense and want to rise faster, the longer tube gives the gases more time to accelerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically stuff to increase the draft of the chimney&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2016-04-19T20:17:17.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8492" PostTypeId="1" CreationDate="2016-04-19T21:08:48.960" Score="0" ViewCount="477" Body="&lt;p&gt;I placed a normal cooling fan 45cm away from a small turbine generator. i had 3 turbines, one with 4 blades, 8 blades, and 12 blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The one with 12 blades always rotated the fastest, even after changing any possible variables such as height or distance from the fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some research and apparently turbines with higher number of blades rotate more slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why did my experiment show that more blades on a turbine gives more RPMs? Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="3538" LastActivityDate="2024-04-02T14:01:55.240" Title="Why did my result show &quot;more blades on turbine = more RPM&quot;?" Tags="|fluid-mechanics|airflow|wind-power|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8493" PostTypeId="1" CreationDate="2016-04-19T20:11:05.607" Score="0" ViewCount="583" Body="&lt;p&gt;There is a small &lt;strong&gt;13' long&lt;/strong&gt; x 6' wide, &lt;strong&gt;1,000 lbs&lt;/strong&gt; (&lt;em&gt;total weight of everything&lt;/em&gt;) all-fiberglass white pontoon boat with dual &lt;strong&gt;flat-bottom fiberglass pontoons&lt;/strong&gt;. (Each flat pontoon bottom is 2' in width, with an empty 2' space between the two, running the length of the pontoon boat). The boat is in a lake and sits with &lt;strong&gt;2.5&quot; of it being below the waterline&lt;/strong&gt;.  There is a &lt;strong&gt;2 HP electric motor&lt;/strong&gt; mounted on the back of the boat. It is tiltable and extends beyond the 13' length of the boat itself. Its propeller is on shaft that places it nearly two feet below the water line when submerged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boat needs to be driven on to a foam-filled drive-on/drive-off floating dock. The floating dock is secured to a bulkhead at one end and tapers/slopes downward for about a foot at its loading end, where the two pontoon noses of the boat would commence movement on to the dock. The dock has a flat top surface area of 13' x 8' and has &lt;strong&gt;3&quot; of freeboard&lt;/strong&gt; (i.e., height of dock that is above the water line) without the boat on it. The rigid top surface area of the floating dock has a permanent friction-reducing industrial coating or surface. (The &lt;strong&gt;co-efficient of friction&lt;/strong&gt; would approximate that of outdoor carpeting.)    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;QUESTION: Will the 2 HP electric motor be sufficient to power/drive the 1,000 lbs, 13' boat on to the floating dock?&lt;/strong&gt; If so, how much of the computed capacity margin to do so will be needed to do the work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ASSUMPTIONS:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) The 2 HP motor must drive at least 12' of the 13' long boat on to the floating dock. Power is controlled with a throttle handle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Although the boat could approach the dock with some kinetic energy, moving at 2 or 3 MPH, the worst case scenario is that the boat is static in the water, with its nose only several feet from the sloped end of the dock, before the 2 HP motor starts its attempt to drive the boat on to the dock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) The expanded polystyrene (EPS) foam filled dock is rated to keep up to a 1,500 lbs load entirely above the water line. With a 1,000 lbs boat on it, the dock would still have freeboard of 1&quot;.  &lt;/p&gt;&#xA;" OwnerUserId="6121" OwnerDisplayName="Bill D." LastActivityDate="2016-04-20T18:24:19.117" Title="Can a 2 HP motor move a 1,000 lbs flat-bottom boat across a flat material surface?" Tags="|friction|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8495" PostTypeId="2" ParentId="8489" CreationDate="2016-04-19T22:36:01.440" Score="5" Body="&lt;p&gt;The speed of exhaust gases from a stack is governed (approximately) by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V=C\sqrt{2gH\frac{T_i-T_o}{T_i}}$, where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C$ is the discharge coefficient (typically between 0.65-0.7, but can vary based on the inlet and outlet geometry)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$g$ is the acceleration due to gravity, approximately $9.81 m/s^2$ or $32.2 ft/s^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H$ is the height of the chimney&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_i$ is the average temperature of the gas in the stack (in $K$ or $^{\circ}R$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_o$ is the temperature of the air outside&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have several ways to increase the velocity of the gases, the easiest of which should be &lt;strong&gt;increasing the length of your chimney&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Flue-gas_stack#Flue-gas_flow-rate_induced_by_the_draft&quot; rel=&quot;nofollow&quot;&gt;Flue-gas stack&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stack_effect&quot; rel=&quot;nofollow&quot;&gt;Stack effect&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Chimney#Chimney_draught_or_draft&quot; rel=&quot;nofollow&quot;&gt;Chimney (draft)&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3015" LastEditorUserId="3015" LastEditDate="2016-04-21T04:48:59.193" LastActivityDate="2016-04-21T04:48:59.193" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8496" PostTypeId="1" CreationDate="2016-04-19T23:53:29.510" Score="4" ViewCount="5216" Body="&lt;p&gt;What are the strongest brazing materials for titanium?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main materials I would like to use would be copper or silver, in which case I would like to ask which would form a stronger bond with titanium.  Otherwise what are the strongest bond forming materials for brazing nickel titanium wire to titanium?&lt;/p&gt;&#xA;" OwnerUserId="6084" LastEditorUserId="6790" LastEditDate="2020-05-02T20:17:19.547" LastActivityDate="2020-05-02T20:17:19.547" Title="Strongest brazing materials for titanium?" Tags="|metals|welding|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8503" PostTypeId="2" ParentId="186" CreationDate="2016-04-20T12:05:51.250" Score="0" Body="&lt;p&gt;See comments. Based on start point of:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Super 13cr is defined as a low-carbon stainless steel. The chemical composition specified from suppliers such as Sumitomo specifies Fe min 0%- Max 0%, C is to be below 0,03.&#xA;    Commonly used in oil and gass applications to resist sour environments and some H2S. But it's expensive as... 4 chickens, in solid gold.&#xA;&#xA;http://www.howcogroup.com/materials/mechanical-tubing-octg/grade-super-13-cr-13-5-2-tube.html&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="6095" LastEditorUserId="6095" LastEditDate="2016-04-21T10:38:33.133" LastActivityDate="2016-04-21T10:38:33.133" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8504" PostTypeId="2" ParentId="8349" CreationDate="2016-04-20T14:16:48.793" Score="1" Body="&lt;p&gt;Out of desperation, I got in contact with some very nice statisticians at NIST. They walked me through the answer to my question, which boils down to: control charts are an empirically derived chart with strong statistical backing. However, because of this by its very nature, there is no one right answer. It depends on how historically stable your process is, whether or not new changes need to be accounted for, etc. You have to have an understanding of your data, an understanding of what is relevant and how that will affect your data in order to correctly decide. It's a judgement call based on someone who understands the process enough to be able to call when the limits and such are clearly incorrect and must be done a different way. Hope this helps anybody in the future needed help on control charts.&lt;/p&gt;&#xA;" OwnerUserId="5919" LastActivityDate="2016-04-20T14:16:48.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8505" PostTypeId="1" AcceptedAnswerId="17627" CreationDate="2016-04-20T15:02:14.247" Score="4" ViewCount="1874" Body="&lt;p&gt;I am analyzing an industrial ventilation system that at some point operates two industrial size blowers in parallel. They often show oscillating behaviour: one blower briefly draws more power / generates more flow, and the other performs less. Then the roles reverse. The only practical, known way to reduce the oscillation is to throttle down the vane valve in front of the fan and then open the valve again. The blowers are almost (but not completely) identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seems that both blowers affect each other in some unstable way. Is this a known phenomena? I'm not finding anything when searching for fan/blower/pump oscillation/resonance/hysteresis.&lt;/p&gt;&#xA;" OwnerUserId="6099" LastEditorUserId="2015" LastEditDate="2017-10-20T19:20:33.537" LastActivityDate="2017-10-20T19:20:33.537" Title="Why do parallel fans/blowers show an oscillatory behavior?" Tags="|mechanical-engineering|fluid-mechanics|pumps|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8506" PostTypeId="2" ParentId="8496" CreationDate="2016-04-20T15:28:50.737" Score="5" Body="&lt;p&gt;To give at least some context to this. Titanium is generally considered 'exotic' for welding purposes as it is quite sensitive to both temperature change and contamination by atmospheric gasses, even at temperatures significantly below its melting point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like aluminium, titanium is quite reactive but protects itself from corrosion by formation of a stable oxide surface and any welding, brazing or soldering process will inevitably involve removing this layer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titanium can be TIG welded to itself without &lt;em&gt;too much&lt;/em&gt; difficulty but this definitely comes under the category of 'advanced' welding in terms of both operator skill and equipment needed. In particular it needs good shielding on both sides of the weld. Indeed in some cases it may be preferable to weld it under an entirely inert atmosphere.   The weldability also depends on the alloy with high alloy multi-phase alloys being more difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Titanium can be brazed with the right filler and flux. Because of its reactivity vacuum brazing is an attractive option. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might get away with using a low temperature silver solder with a fairly active flux but in general high strength brazed joints will need to be done under vacuum or an inert atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind though that titanium will burn in air if it gets hot enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your application you might want to to consider mechanical methods such as rivets or screws or look at adhesive joints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : brazing wires. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one hand brazing can be a good way to attach wires to much larger metal parts because there is much less chance of melting the wire and it can form a neat and convenient joint. However it is less suitable for highly stressed applications for a couple of reasons. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The brazing filler usually has a significantly lower strength than the base metal. This may not matter in certain types of joints (eg bicycle frames) where the joint is designed so that the contact area is much larger than the section of the members being joined but joining wire to a flat surface is not so good. &lt;/li&gt;&#xA;&lt;li&gt;Brazed joints are strongest when there is a relatively small and consistent gap between the surfaces being joined. This is hard to achieve with wire. &lt;/li&gt;&#xA;&lt;li&gt;Wire derives much of its strength from the drawing process (ie work hardening) so heating to brazing temperatures may well significantly reduce its strength. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Certainly brazing load bearing wires is not something that is seen very often. It is much more common to see some sort of mechanical (eg pressed, crimped or spliced) terminator....think bike brake cables and guitar strings. I'm not saying that it won't work but it doesn't sound like the best way to proceed to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possibility would be to put an eye or ferrule on the end of the wire and TIG weld an attachment point  to the joint, this is actually not that far off how biological tendon-bone attachments work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of steel vs titanium, you will certainly find steel a lot easier to work with, especially in the proof of concept stage of development where it is likely that you will be making a lot of changes to the prototypes. There is also a certain amount of middle ground between mild steel and titanium in the shape of high strength steels like chrome-moly alloys. It is certainly safe to say that there are a lot of difficulties associated in manufacturing in titanium. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-27T19:16:55.840" LastActivityDate="2016-05-27T19:16:55.840" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8507" PostTypeId="2" ParentId="8301" CreationDate="2016-04-20T15:40:43.347" Score="2" Body="&lt;p&gt;In the interest of anybody else who has a similar problem I have been experimenting with rule die cutting which seems to be quite a promising approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using 12mm plywood for the die form, cut with a fret saw and various thicknesses of cold rolled steel shim stock for the cutting edge. In initial tests this works OK on 0.1mm copper foil and produces acceptable edge finish. In fact using soft wood or rubber to cut onto it rolls the edges over somewhat which is an advantage in this application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also gives clean cuts on paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially used 0.6mm shim stock ground with a single beveled cutting edge. Although this is nowhere near as hard as heat treated tool steel it is at least easy to hone to a polished cutting edge. I used diamond and then ceramic stones to achieve this. The 0.6mm shim is perhaps a bit on the thick side for this and is difficult to bend to radii less than about 40mm. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-20T15:40:43.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8508" PostTypeId="1" AcceptedAnswerId="8513" CreationDate="2016-04-20T15:59:05.797" Score="4" ViewCount="746" Body="&lt;p&gt;We want to monitor hard disks for present problems or pending failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the Fast Fourier Transform (FFT) of a given hard disk in question, then, be compared with a reference FFT for a perfectly working one, and then conclude hard disk may be faulty or about to fail if its FFT differs significantly from the reference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FFT would be used to detect unexpected frequencies, or harmonics, that do not normally occur in a 'healthy' hard disk.  FFT would input vibrations from hard disk, and output the collection of vibration's constituent frequencies.  This would be compared to the frequency domain function of 'healthy' hard disk.&lt;/p&gt;&#xA;" OwnerUserId="2169" LastEditorUserId="2246" LastEditDate="2016-04-20T19:47:18.103" LastActivityDate="2016-04-23T10:56:45.493" Title="Can Fast Fourier Transform be used to monitor for fault vibrations in a hard disk?" Tags="|mechanical-engineering|electrical-engineering|motors|sensors|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8509" PostTypeId="2" ParentId="8493" CreationDate="2016-04-20T17:59:32.043" Score="0" Body="&lt;p&gt;What you need to know:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the coefficient of friction between the boat and the dock? (Both static and dynamic)&lt;/li&gt;&#xA;&lt;li&gt;What is the total surface area of the contact patch between boat and dock? (So you can calculate the drag force)&lt;/li&gt;&#xA;&lt;li&gt;What is the efficiency of the propeller into the water? (So you can calculate the push force)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And, the coefficient of friction may change based on the exact pressure the boat (weight over suface area) puts on the dock. Also, depending on lubrication -- is there barnacles or slick seaweed in the way? If it's frozen, is there slick ice? Is it always homogenous water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You haven't provided any of this information, and thus we can't answer the question from a mathematical point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a practical point of view, why does it matter? Could you have a steel cable and a winch on the dock, that allows you to hand crank it up when the motor can't do it? Could you electrically power that winch if needed? (With enough gearing and low-loss gearbox, a small motor might be sufficient.)&lt;/p&gt;&#xA;" OwnerUserId="1889" LastActivityDate="2016-04-20T17:59:32.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8510" PostTypeId="2" ParentId="8493" CreationDate="2016-04-20T18:24:19.117" Score="1" Body="&lt;p&gt;For a winch a 2HP would probably be adequate for dragging a 1000lb boat across a smooth surface and this is certainly in the normal range of winch motors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the difference between a winch and a propellor is that a winch can be geared to provide the torque required whereas propellers generally operate at much higher RPM and in a fairly narrow range at that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you already have an electrical power supply on the boat it may be easier just to have a separate electric winch which won't be especially heavy or expensive. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-20T18:24:19.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8511" PostTypeId="1" CreationDate="2016-04-20T21:59:58.347" Score="0" ViewCount="1811" Body="&lt;p&gt;I am working on designing a pressure cooker. I've have tried to find, using two different models, the wall thickness based on the following assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Cylindrically shaped container&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Made of stainless steel&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Internal pressure of 20 psi&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The  internal pressure is what sets a lower limit on the thickness&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From the two aforementioned models I have found a wall thickness of 0.1mm and 0.05mm respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wall thickness I get seems to be very thin. My guess is that the internal pressure is not what makes ordinary pots thick but actually it's the resistance to impacts occuring during regular usage that sets the limit on how thin the walls can be. Is this correct? Is there a rigorous framework for estimating the required thickness from this criterion? I haven't been able to find litterature on this.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The models are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple pipe: $\sigma=\frac{pR}{t}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Annaratone: $s=\frac{p D_e}{2f+p}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p$ - pressure&#xA;$R$ - radius&#xA;$t$ - thickness&#xA;$D_e$ - Outer Diameter&#xA;$f$ - basic allowable stress&lt;/p&gt;&#xA;" OwnerUserId="5980" LastEditorUserId="5980" LastEditDate="2016-04-22T10:20:23.900" LastActivityDate="2016-04-22T18:56:59.773" Title="Can anyone verify this wall thickness of a pressure cooker?" Tags="|mechanical-engineering|mechanical-failure|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8512" PostTypeId="2" ParentId="8511" CreationDate="2016-04-21T00:22:22.233" Score="0" Body="&lt;p&gt;Assuming it is a typical &quot;kitchen-stove-sized&quot; pressure cooker, your wall thicknesses seem to be the right order of magnitude, for the &lt;strong&gt;cylindrical&lt;/strong&gt; part of the cooker, if your criterion is &quot;what is the minimum thickness to withstand the pressure&quot;. The flat base will most likely need to be thicker, otherwise it would bulge outwards under the pressure and become unstable. You also need to consider the stress concentration where the flat base joins to the cylindrical sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But those are not the only criteria. As well as impact resistance, you should think about other issues like &quot;what would happen if the all the water is boiled off&quot;, &quot;what would happen if the safety valve sticks and doesn't open&quot;, &quot;what thickness do you need to ensure a good seal between the pan and the lid&quot;, &quot;what happens if the seal fails when the cooker is fully pressurised&quot;, &quot;what happens if somebody tries to pressurize the cooker using oil instead of water&quot;, etc, etc, (Note: those are results of a couple of minutes brainstorming, not an exhaustive list!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling &quot;pressure cooker safety standard&quot; threw up this standard, which doesn't actually cover your specification, but might point you in the right direction to search further: &lt;a href=&quot;http://ulstandards.ul.com/standard/?id=136&quot; rel=&quot;nofollow&quot;&gt;http://ulstandards.ul.com/standard/?id=136&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-04-21T00:22:22.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8513" PostTypeId="2" ParentId="8508" CreationDate="2016-04-21T01:16:26.180" Score="2" Body="&lt;p&gt;&lt;strong&gt;Disclaimer:&lt;/strong&gt; Although I have studied FFT, I have not applied FFT to a practical application. Therefore I would differ a direct response to someone else. Regardless I would really like to see the end result too.&lt;/p&gt;&#xA;&lt;p&gt;But I can offer you following two application examples and source code to access data from three axis ADXL345 accelerometer to help you with your endeavors. Examples were developed on ARM Cortex A8 and ARM Cortex M4 processors. The first one uses a Beaglebone Black / Embedded Linux platform and the second is a bare metal design. All X, Y, and Z data points are captured. These are used to calculate Roll and Pitch.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Three axis ADXL345 accelerometer implementation using a Embedded Linux platform&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=bd7ULY1frA4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Accelerometer Application on Embedded Linux&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Three axis ADXL345 accelerometer implementation using a Embedded Linux platform&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/6ZFy4zMGLgg&quot; rel=&quot;nofollow noreferrer&quot;&gt;TIVA - ADXL345 Accelerometer Demo Using I2C&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Both the designs use I2C bus to capture the data. So digitally sampled data is available. Therefore code has to modified to apply FFT. With little effort you should be able determine frequencies, harmonic's etc.&lt;/p&gt;&#xA;&lt;p&gt;Your are welcome to fork the source code and use them as you wish.&lt;/p&gt;&#xA;&lt;p&gt;Also below are few interesting graph on that might be of help.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Rotational Vibration Safeguard (RVS)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ip5XL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ip5XL.jpg&quot; alt=&quot;Rotational Vibration Safeguards&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Simple Free fall profile for a HDD&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1VFsH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1VFsH.jpg&quot; alt=&quot;Simple Free fall profile&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;A Traditional Protection Algorithm&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Traditionally, the HDD protection algorithm has been based on&#xA;free-fall modeling, as explained below, in which the outputs of the&#xA;sensors contained in the accelerometers can be easily captured by a&#xA;digital oscilloscope or other data sampling system.&lt;/p&gt;&#xA;&lt;p&gt;A “test sled” can be assembled using two ADXL320 dual-axis&#xA;accelerometers. The axes of the accelerometers are aligned with the X,&#xA;Y, and Z axes, as depicted in figure below (Figure 4), thereby providing&#xA;values of acceleration along the X, Y, and Z coordinates. (The Y1 output is&#xA;redundant and is not used.) The outputs of the coordinate axes are&#xA;sampled by a 12-bit ADC contained in an ADuC832 precision analog&#xA;microcontroller, which integrates the sampled data and feeds it to an&#xA;internal 8052-compatible core processor. The sampled data is then&#xA;transferred to the computer—via an RS-232 interface—for analysis.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VN2yV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VN2yV.jpg&quot; alt=&quot;Test Sled&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The figure below (Figure 5) shows the sequence of responses sensed by the two &amp;gt; sensors. Values X and Y are supplied by one accelerometer, the values Z and Y1&#xA;are supplied by the other accelerometer. Note also that the plot is&#xA;divided into four consecutive intervals labeled: “static,” “rollover,”&#xA;“free-fall drop,” and “impact.” The sampling interval, shown along the&#xA;X axis, is determined by the ADC, which is clocked at 200 Hz for each&#xA;variable, or one sample of each variable every 5 milliseconds. The&#xA;Y-axis scale represents the values delivered by the 12-bit ADC in the&#xA;ADuC832 smart-transducer front end, plotted for all four axes.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RTjPK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTjPK.jpg&quot; alt=&quot;Protection Algorithm&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://github.com/mahengunawardena/Tiva_I2C_Nokia_ADXL345&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tiva_I2C_Nokia_ADXL345 source code&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://github.com/mahengunawardena/BeagleboneBlack_ADXL345_GUI&quot; rel=&quot;nofollow noreferrer&quot;&gt;BeagleboneBlack_ADXL345_GUI source code&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/3348/calculating-pitch-yaw-and-roll-from-mag-acc-and-gyro-data/3350#3350&quot;&gt;Calculating pitch, yaw, and roll from mag, acc, and gyro data&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.hgst.com/sites/default/files/resources/WP_RVS_25March.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rotational Vibration Safeguard (RVS)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.dell.com/downloads/global/vectors/2008_freefallprotection.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dell Raises the bar in Shock-Resistant Hard Drives&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.analog.com/library/analogDialogue/archives/39-11/hdd.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Using Dual-Axis Accelerometers to Protect Hard Disk Drives&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-04-23T10:56:45.493" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8514" PostTypeId="1" CreationDate="2016-04-21T04:06:31.687" Score="2" ViewCount="45" Body="&lt;p&gt;I am interested in creating a screen to display moving images (similar to moving desktop background), however I am looking for the thinnest possible material available i can use as the screen to display the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally working in the same way as a laptop screen however much thinner (E.g only using the components necessary to display the image, no housing).  The material can be any at this stage, such as plasma, LCD, LED, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intention is to create something that works in a similar way to wallpaper, where it can be applied directly to a wall, and for it to be connected wirelessly to a transmitter in order for it to display a moving image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any answers would be much appreciated. Thank you&lt;/p&gt;&#xA;" OwnerUserId="6104" LastEditorUserId="6104" LastEditDate="2016-04-21T16:00:57.313" LastActivityDate="2016-04-21T17:53:58.610" Title="What is the thinnest possible depth of any given material required to display an image?" Tags="|electrical-engineering|telecommunication|" AnswerCount="1" CommentCount="4" ClosedDate="2016-04-22T18:08:14.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="8515" PostTypeId="2" ParentId="6013" CreationDate="2016-04-21T04:29:11.887" Score="3" Body="&lt;p&gt;I just thought that I would clarify something that Fred said. In an emergency shutting down the nuclear chain reaction is the easy part. The main issue is the decay heat. The decay heat comes from the residual radioactive components in the core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This decay heat starts off at approximately 7% of the total reactor power. However, it drops off exponentially with time. In about an hour it is down to about 1%. This may not seem like a lot of heat to remove however when considering a 3GWth (~1GWe) that is 30MW that has to be removed for an extended period. Without electricity providing circulation of cooling fluid etc. This is very hard!! &lt;/p&gt;&#xA;" OwnerUserId="4065" LastActivityDate="2016-04-21T04:29:11.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8516" PostTypeId="1" CreationDate="2016-04-21T06:33:49.657" Score="2" ViewCount="169" Body="&lt;p&gt;I’m in a position where I need to order high volumes of nuts and bolts for a project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read through a South African National Standard, SANS 2001-CS1, in which it states that when I use hot dip galvanized fasteners the nut should be of a higher class than the associated bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please specify whether this is accurate and what the reason will be for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your assistance will be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6107" LastEditorUserId="2246" LastEditDate="2016-04-21T13:03:02.557" LastActivityDate="2016-04-21T13:50:22.630" Title="Hot Dip Galvanized Nuts" Tags="|fasteners|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8519" PostTypeId="2" ParentId="8511" CreationDate="2016-04-21T10:32:57.587" Score="5" Body="&lt;p&gt;Those figures are about right for the bare minimum wall thickness for the walls of a cylinder however, as you say, this would be inadequate in practice for a number of reasons. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Safety concerns dictate  that steam pressure vessels are designed with a large factor of safety. You certainly don't want any sort of pressure vessel to be operating anywhere near the yield stress of the material. &lt;/li&gt;&#xA;&lt;li&gt;The hoop stress in an ideal cylinder is one thing but the design of the ends (ie lid and base) also have a significant effect on the stresses and pressure on the ends will introduce tensile as well as hoop stresses to the cylinder walls. &lt;/li&gt;&#xA;&lt;li&gt;Stress raisers from fittings, catches etc and manufacturing artefacts may cause local stress concentrations well in excess of the calculated figures. &lt;/li&gt;&#xA;&lt;li&gt;The pot needs to be robust enough to stand up to everyday use and handling. &lt;/li&gt;&#xA;&lt;li&gt;Even when the stresses are at an acceptable level deflections may not be, especially in terms of sealing the lid. To put it another way being strong enough does not guarantee that something is rigid enough. &lt;/li&gt;&#xA;&lt;li&gt;Depending on the material, fatigue, work hardening,  corrosion embrittlement, thermal stresses,  etc etc may be factors. &lt;/li&gt;&#xA;&lt;li&gt;From a business perspective it may be preferable  to adopt  a more conservative design in order to simplify quality control, especially where material costs are a small proportion of overall costs and the risks associated with potential liability or loss of reputation are high. &lt;/li&gt;&#xA;&lt;li&gt;A product made from heavier gauge material may be seen as better quality by consumers. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;EDIT &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The criterial I have listed are, by their nature, a bit fuzzy. The general approach to this sort of problem is to first do your best estimate based on calculations of the conditions you expect in normal use and then apply a &lt;a href=&quot;http://www.engineeringtoolbox.com/factors-safety-fos-d_1624.html&quot; rel=&quot;nofollow&quot;&gt;factor of safety&lt;/a&gt;. This makes allowances for uncertainty in your calculations, variations in actual material properties and wear and tear over the life of the product as well as unforeseen conditions (to an extent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general a larger factor of safety is used where you are less certain about your model either because comprehensive calculations are impractical or conditions of use are difficult to predict. Larger factors of safety are needed when a failure would have safety implications (eg for pressure vessels, lifting equipment etc) but can be mitigated by more rigorous quality control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that factors of safety for aerospace are often significantly smaller than for general structures despite the fact that aerospace requires very high levels of reliability and redundancy. The reason for this is that weight/mass savings are so economically important that it is worth spending a lot of resources in the design stage to achieve reliability by testing and very detailed analysis rather than just being generous with loading calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast if you are designing a bridge it usually works out cheaper just to err on the side of caution, especially as you are often choosing between standardised parts, or at least parts fabricated from standard sections rather than designing every single component from scratch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is good practice for pressure vessels to be practically proof tested at a pressure in excess of their designed operating pressure. In many cases this is a regulatory requirement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finite element analysis software can be a useful tool for optimising designs but it needs to be treated with care and an understanding of its limitations. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-22T18:56:59.773" LastActivityDate="2016-04-22T18:56:59.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8521" PostTypeId="2" ParentId="186" CreationDate="2016-04-21T12:12:55.117" Score="7" Body="&lt;p&gt;Iron and carbon have an interaction which make them different from most engineering alloys. This is to do with both the relative size of C and Fe atoms and their chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon atoms are just the right size to insert themselves into the crystal lattice of iron, this strains the lattice enough that it is somewhat harder and stronger than pure iron. However the really important part is that the presence of carbon allows steel to be heat treated. Here it is heated above a critical temperature at which the crystal structure changes and if it is cooled rapidly the carbon content prevents it from returning to it's 'normal' structure at room temperature and instead forms a multi-phase structure which is highly stressed but chemically stable and as such is very hard with a high tensile strength. This can be further modified by controlled reheating to partially reverse this transformation and produce a material with controllable strength hardness and toughness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above is a quick overview and there are whole books on the detailed behaviour of steels as the iron-carbon system can exist is several different states with different crystal structures and various micro-structural combinations of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This type of heat treatment is pretty much unique to steel and certainly very different from the way that most alloys behave and is a result of the specific interaction between iron and carbon and depends on the fact that iron can exist as both body centred and face centred cubic crystals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also achieved by very low concentrations of carbon, usually less than 1.2%  or so. In fact only about 0.7% carbon by mass is soluble in iron and any surplus will tend to form carbides or precipitate out as graphite (as in cast iron). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various metal carbides in use (such as tungsten carbide) but these are really ceramics rather than solid solution alloys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also at least one type of stainless steel (H1) which is precipitation hardened and contains nitrogen instead of carbon. This is a different hardening mechanism to that of carbon steel. The purpose of eliminating carbon is to improve corrosion resistance, especially in salt water. I've only ever encountered this is a blade steel in knives. There are also low carbon stainless steels but these are not hardenable by heat treatment and are designed for improved weldability. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-21T15:41:46.090" LastActivityDate="2016-04-21T15:41:46.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8522" PostTypeId="2" ParentId="8516" CreationDate="2016-04-21T13:32:45.417" Score="1" Body="&lt;p&gt;&lt;strong&gt;I'd like to preface this with the statement that I am entirely unfamiliar with &lt;em&gt;any&lt;/em&gt; South African standards, so take my educated guess for what it is.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I was able to find what I think is the correct version of &lt;a href=&quot;http://documents.tips/documents/sans-2001-cs1-2005-construction-works-part-cs1-structual-steelwork-.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;SANS 2001-CS1:2005&lt;/a&gt; online.  I'm assuming your confusion stems from the seemingly contradictory nature of Section 4.5.1.1 with Section 4.5.1.6, which are copied below:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;4.5.1.1 Bolt/nut Combinations&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The combination of bolts and nuts shall not be less than the following:&lt;/p&gt;&#xA;&lt;p&gt;a) class 4.8 bolts or screws with class 4 nuts in accordance with SANS 1700-5-2; or&lt;/p&gt;&#xA;&lt;p&gt;b) class 8.8 bolts with class 8 nuts in accordance with SANS 1700-5-2.&lt;/p&gt;&#xA;&lt;p&gt;Any bolt assembly which seizes when being tightened shall be replaced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;4.5.1.6 Galvanized nuts&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Nuts that are to be hot-dip galvanized shall be of a higher class than the associated bolt or screw.  Nuts shall be checked after being galvanized for free running on the bolt and shall be re-tapped to the permissible deviations if necessary to ensure a satisfactory tightening performance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My guess is one of two things:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;either the hardness of the nut is expected to decrease after the hot-dip galvanization process (which dips the nut in molten zinc, thereby tempering it a little), thus &amp;quot;downgrading&amp;quot; it, or&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;galvanized nuts are sold to a different class system, so you have to make sure your nut class exceeds the bolt class (I don't know how things work in South Africa, so this is a wild guess).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Section 4.5.1.1 shows that a &amp;quot;matching&amp;quot; nut is used with  the corresponding bolt.  You want a nut that's stronger than the bolt, such that if the bolt fails (fractures), it's very apparent that the failure has occurred.  If a thread fails, it's much less noticeable, and hence more dangerous since that bolt assembly cannot be counted on.  See &lt;a href=&quot;http://www.theboulderco.com/products-specifications/product-specifications-02.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this PDF&lt;/a&gt; for more information on the topic.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-04-21T13:32:45.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8523" PostTypeId="2" ParentId="8514" CreationDate="2016-04-21T17:53:58.610" Score="2" Body="&lt;p&gt;Let's assume that you are using OLEDs (organic light emitting diodes) which can emit a photon when charged with an electron. You could build a single molecule pixel (just nip down to the corner store and pick up an electron microscope with atomic manipulator). However the Real challenge is the wiring and insulation. the real question is how are you going to separate your pixels electrically while powering and addressing them. by the time you figured out how to do the wiring, the lights are only a heat dissipation issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current thinking on OLED containers is dual transparent conductive plates sandwiching your organics. since both glass and plastic can be produced a thickness under 1mm there is no reason other than heat dissipation to expect thickness to be much greater than 3mm. On the other hand heat dissipation is a big issue and proportional to the size of the display. &lt;/p&gt;&#xA;" OwnerUserId="803" LastActivityDate="2016-04-21T17:53:58.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8524" PostTypeId="1" CreationDate="2016-04-21T18:47:45.950" Score="2" ViewCount="322" Body="&lt;p&gt;I have a question regarding a pipe-network:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pxFV7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pxFV7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Pressure A &amp;gt; B&#xA;diameters of the outside pipes are bigger than the pipe with the massflow m2,m3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The brown rectangle is a valve (the diameter of the valve is smaller than the pipe-diameter). I know from the continuity equation, that the volume flow rate in a pipe is constant. But how is that relation within a network? Is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;massflow: m2 &amp;lt; m3(through the valve)?&#xA;massflow m1 &amp;lt; m2 (smaller diameter)?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5906" LastEditorUserId="7083" LastEditDate="2016-08-06T20:57:10.290" LastActivityDate="2016-08-06T20:57:10.290" Title="Pipe-network pressure - mass flow relation" Tags="|fluid-mechanics|pressure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8525" PostTypeId="1" CreationDate="2016-04-21T19:29:20.990" Score="8" ViewCount="1168" Body="&lt;p&gt;I need to have some beams fabricated with copes on the ends.   They will be set on top of existing steel trusses, and used to replace undersized purlins to shore up existing rafters (hence the need for the copes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My structural engineer specified 4&quot; long copes.  There is a cope at each end.  The distance between the edge of the flanges on the steel trusses is 186&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D6KQj.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D6KQj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is a matter of practicality.  If I have the fabricator make the coped length of the beam exactly 186&quot;, I feel like I may have a difficult time getting the beams in place.   So I'd like to have a little tolerance to give me room to slide the beam above one truss and then pivot it into place above the other.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much tolerance should I allow for to make this easy on myself, the installer of the beam, without diminishing the integrity of the design?  In the CAD drawing below, I have just arbitrarily chose to show that the coped end-to-coped end distance should give me 1/4&quot; of play on each end of the beam.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MNzyN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MNzyN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to overthink this, but these suckers are heavy and installing them is going to be fun enough without having to fight tolerance problems.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a tolerance rule of thumb for this situation?   What tolerance would you use?&lt;/p&gt;&#xA;" OwnerUserId="3830" LastEditorUserId="2246" LastEditDate="2016-04-21T19:52:43.353" LastActivityDate="2020-03-14T01:07:51.523" Title="Tolerance for Beam With Cope End" Tags="|structural-engineering|beam|detailing|" AnswerCount="3" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="8526" PostTypeId="5" CreationDate="2016-04-21T19:53:33.113" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-04-21T19:53:33.113" LastActivityDate="2016-04-21T19:53:33.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8527" PostTypeId="4" CreationDate="2016-04-21T19:53:33.113" Score="0" Body="Questions dealing with detailing of members for fabrication or construction drawings." OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2016-04-21T22:00:11.470" LastActivityDate="2016-04-21T22:00:11.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8528" PostTypeId="1" CreationDate="2016-04-21T21:39:29.233" Score="2" ViewCount="214" Body="&lt;p&gt;I faced a logical dilemma, while designing and simulating Active suspension system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I have planned:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;My system will be controlling actuators of a vehicle, using PI.&lt;/li&gt;&#xA;&lt;li&gt;It is 2nd order mechanical system with mass-spring-damper-actuator.&lt;/li&gt;&#xA;&lt;li&gt;In Matlab and Simulink, I am using Transfer Functions for analysis (to manage overshoot, settling time, etc.) and State Space representation (to design observer/estimator).&lt;/li&gt;&#xA;&lt;li&gt;Sensors used are ONLY accelerometers, that will produce acceleration of the suspension displacement. However, I require a displacement, which will be Observed/Estimated by designed estimator and fed back to compare with the reference point.&lt;/li&gt;&#xA;&lt;li&gt;I cannot use double integration as drift occurs.&lt;/li&gt;&#xA;&lt;li&gt;I have 2 states from 2nd order system. (x1 = displacement and x2 = velocity)&lt;/li&gt;&#xA;&lt;li&gt;We are dealing with LTI SISO system.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My mechanical system matrices are as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = [0 1; (-ks/ms) (-bs/ms)]; &#xA;B = [0; (1/ms)];&#xA;C = [1 0];&#xA;D = [0];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I use an Observer/Estimator &lt;strong&gt;&lt;em&gt;states&lt;/em&gt;&lt;/strong&gt; to estimate the suspension displacement (in Simulink) from the acceleration that is provided from the accelerometer, if in my opinion, acceleration state does not exist? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know, If you would like to see some of my code or derivations, but I think this is one of the &lt;em&gt;Logical Understanding&lt;/em&gt; problems, so any clarification will help to get me back on track.&lt;/p&gt;&#xA;" OwnerUserId="6114" LastEditorDisplayName="user16" LastEditDate="2016-06-29T11:53:57.913" LastActivityDate="2016-07-29T12:06:58.267" Title="Active suspension with accelerometers and displacement observer/estimator problem" Tags="|mechanical-engineering|control-theory|matlab|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8529" PostTypeId="2" ParentId="186" CreationDate="2016-04-21T22:25:14.800" Score="3" Body="&lt;p&gt;&lt;strong&gt;Summary: The Fe-C system, and thus steel, is unique due to a eutectoid transformation from a high-solubility phase to a low solubility phase that allows for a wide variety of microstructures and properties which are highly and relatively easily tunable. Other first-row transition metals have different, and less exploitable, behavior when alloyed with carbon.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fe-C is the only first-row transition metal-carbon system that has a eutectoid transformation in its phase diagram.The eutectoid transformation changes austenite to ferrite and cementite on cooling. Austenite has high carbon solubility, and ferrite has low carbon solubility. I am picking on first-row transition metals as they tend to have chemical behavior &quot;close&quot; to that of steel, with similar cost, density, and other &quot;obvious&quot; properties (with the exception of scandium, which is extremely rare and expensive), and examining all 70+ metals is a fair amount of work for this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nature of the eutectoid transformation allows for many microstructures and thus a high degree of tunable properties. Consider a eutectoid steel austenitized and cooled at varying rates:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If cooled slowly, a moderately ductile, moderately strong pearlite microstructure forms. Pearlite results from a cooperative nulceation and growth process as carbon leaves austenite during its transformation to ferrite, forming alternating lamellae of ferrite and cementite.&lt;/li&gt;&#xA;&lt;li&gt;If cooled moderately rapidly and then held isothermally for a period of time, a much harder bainite microstructure forms. The kinetics of bainite formation are not well understood, but the microstructure is a less-organized arrangement of cementite and ferrite, again resulting from carbon coming out of solution as austenite transforms into ferrite.&lt;/li&gt;&#xA;&lt;li&gt;If cooled extremely rapidly, an extremely strong and hard martensite microstructure forms. Martensite formation is a diffusionless process in which carbon is trapped in austenite while it transforms to a BCC structure, distorting the lattice into a strained BCT structure which is difficult to strain further, hence its high strength. By altering the quantity of carbon and being creative with heat treatment schedules, a wide array of microstructural combinations are available.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With appropriate alloying and heat treatment, it is possible to have a steel with retained austenite, ferrite, pearlite, bainite and martensite all in the same material. Such complex microstructures are impossible in other first-row transition metal-carbon systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the wide heat-treatability and wide array of microstructures and properties are entirely due to the presence of a eutectoid transformation which takes a high-solubility phase to a low-solubility phase. The eutectoid transformation itself is due to a phase change from austenite (FCC) to ferrite (BCC) and the resulting significant loss of carbon solubility. &lt;strong&gt;The answer to your question is effectively no&lt;/strong&gt;, there are no other alloys (of which I am aware) that behave like steel during processing. The answer to your alternate question is that carbon has less useful and less exploitable effects on other first-row transition metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are the Fe-C, Ni-C and Mn-C phase diagrams for comparison. Note that the Fe-C phase diagram stops at 0.2 a/a C while the others go to 1.0 a/a C. Ni-C has no eutectoid, only a eutectic transformation, and thus can only be precipitation hardened. Any other microstructure formation would have to occur during solidification. Mn-C phase diagram has a eutectoid, but it goes from a high-solubility phase to another high-solubility phase, which means that extremely large amounts of carbon would be present in the lower temperature phase (nearly 10% a/a C compared with less than 1% a/a C in steel), which would result in extreme brittleness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bXw6p.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bXw6p.png&quot; alt=&quot;Fe-C Phase Diagram&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZuNb8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZuNb8.png&quot; alt=&quot;Ni-C Phase Diagram&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LpYYv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LpYYv.png&quot; alt=&quot;Mn-C Phase Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-04-22T04:16:54.383" LastActivityDate="2016-04-22T04:16:54.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8530" PostTypeId="2" ParentId="8508" CreationDate="2016-04-21T23:48:27.060" Score="1" Body="&lt;p&gt;In principle this might work. For example similar methods are used in real time to detect developing faults in jet engines, by processing the signals from vibration sensors on the engine. (The details of the methods used are confidential, however).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem you will face is that you are not measuring the accelerations of an isolated disk drive, but a disk that is one component of a bigger mechanical system (a complete desktop or laptop computer, a server rack, etc). You will need some way to separate the (probably small) vibrations of the disk itself from other vibration sources, for example cooling fans operating intermittently and/or at variable speeds. Even with an isolated disk drive, you would probably do better to use &lt;em&gt;all&lt;/em&gt; the data that you can easily capture - for example the disk-head-positioning data coming from the disk controller would give you a measure of the forces that are &lt;em&gt;exciting&lt;/em&gt; the vibrations, to correlate with measuring the disk's &lt;em&gt;response&lt;/em&gt; to those forces.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-04-21T23:48:27.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8531" PostTypeId="2" ParentId="8508" CreationDate="2016-04-22T06:27:54.850" Score="0" Body="&lt;p&gt;Im pretty confident in that you can see some changes in the fourier analysis if you have a sensitive enough sensor in the right place and the failure is mechanical. Similar methods are used to measure the health of bearings with success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you do not know what you are looking for. Unless your a very skilled simulation guy or can find some analysis which tells you what your looking for then you need some empirical data. Go forth and measure a bunch of drives and try to see how they differ. Get hold of ones that have failed and see what the difference is. Then try to predict which drive is going to brake and monitor that.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2016-04-22T06:27:54.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8532" PostTypeId="2" ParentId="8524" CreationDate="2016-04-22T08:33:19.693" Score="5" Body="&lt;p&gt;Actually the continuity equation means that at steady state, the mass flowrate in a pipe must be constant (as mass cannot be created or destroyed not volume) but the volume can vary.&lt;br&gt;&#xA;so:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m2=m3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If this fluid is incompressible such (as water) the volume flowrate would be the same, but for a compressible fluid such as a gas the volumetric flowrate at point 3 would be higher than point 2.  (the pressure drop would lead to an expansion of the gas)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure drop in both branches must be the same.  Pressure drop is proportional to velocity so you need more volume in a larger cross-sectional area pipe to produce the same velocity (velocity = volumetric flowrate / cross sectional area).  The valve will act as an obstruction resulting in even more pressure drop for a given velocity.  Therefore you need much more fluid to travel through the large diameter, unobstructed branch so&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m1&amp;gt;m2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2616" LastActivityDate="2016-04-22T08:33:19.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8533" PostTypeId="1" CreationDate="2016-04-22T09:39:04.877" Score="20" ViewCount="550" Body="&lt;p&gt;&lt;em&gt;Note: this question was totally rephrased and expanded on, addressing the comments and questions posed to the first version.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The de-facto standard source for runoff calculations in South Africa is the &quot;SANRAL Runoff Manual&quot; (&lt;a href=&quot;http://www.nra.co.za/content/Drain5.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.nra.co.za/content/Drain5.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to design a weir for a small earth dam in an &lt;strong&gt;arid area&lt;/strong&gt; with a small (smaller than 200 ha), fairly flat catchment area in &lt;strong&gt;southern Africa&lt;/strong&gt;. The idea behind the dam is to collect not more than 10 000 cubic metres of water for small stock watering purposes for the immediate time after the rainy season, somewhat reducing the dependency on ground water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To design the dam wall and the spillway, one needs flow data through the catchment area in question. The rational method provides this formula: &#xA; $Q=\frac{CIA}{3,6}$ &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; ...(3.8)&lt;br&gt;&#xA;where: $Q \hskip{1.5em}= $ peak flow (m&lt;sup&gt;3&lt;/sup&gt;/s)&lt;br&gt;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;$C \hskip{1.7em}= $ run-off coeficient (dimensionless)&lt;br&gt;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;$I\hskip{2em} = $ average rainfall intensity over catchment (mm/hour)&lt;br&gt;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;$A \hskip{1.7em}= $ effective area of catchment (km&lt;sup&gt;2&lt;/sup&gt;)&lt;br&gt;&#xA;&amp;emsp;&amp;emsp;&amp;emsp;$3,6\quad = $ conversion factor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In arid areas rainfall events often are thunderstorms of fairly short duration, often less than an hour. These events are separated by long dry periods. Rainfall is generally measured on a daily basis. These recorded values are daily rainfall values, but they say nothing about the duration of the rainfall events. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rainfall intensity is determined by dividing rainfall by the duration of the rainfall event. Thus, the shorter the rainfall event, the higher the rainfall intensity for the same amount of total rainfall. But since rainfall is not really measured by the second, minute or hour it is impossible to accurately determine rainfall intensity in areas where the rainfall events are short. In fact the margin for error increases tremendously as the duration of rainfall events decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using rainfall this data (with a large margin of error) for large catchment &#xA;areas is not recommended, since the rainfall does not have a &quot;temporal distribution for at least a duration equal to the time of Concentration&quot; (page 3.17). Q would be over-estimated and thus dams and spillways would be over-designed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite is true for small catchment areas, where the time of concentration is less than the duration of a rainfall event. The margin of error in the rainfall intensity calculation will become unacceptably critical. Especially for very small dams they will either be under designed, or they will be financially unfeasible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an alternative method to determine runoff in arid areas, where rainfall events are of short duration and accurate rainfall data (rainfall and duration of rainfall events) is not available?&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="5053" LastEditDate="2016-05-11T19:44:43.480" LastActivityDate="2024-03-25T14:09:15.043" Title="Is there a method to determine the runoff for arid areas in developing countries?" Tags="|civil-engineering|water-resources|hydrology|" AnswerCount="4" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="8534" PostTypeId="1" AcceptedAnswerId="8538" CreationDate="2016-04-22T10:23:38.033" Score="0" ViewCount="375" Body="&lt;p&gt;I'm trying to figure out what kind of pump I should use to boil water at low temperature. Here is my idea of the setup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cQb6C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cQb6C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dark blue water comes in, the pump sucks down the light-blue water. This will reduce the pressure in the top, forcing the dark blue water start boiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water then condenses and gets pumped out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important condition is of course the temperature of the incoming water. This can be calculated using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Antoine_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Antoine equation&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p =10^{\left(8.07131 – \frac{1730.63}{233.426+T}\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p$ is pressure in mmHG, and $T$ is temperature in degrees Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out what how I can translate the boiling of water to psi. I get confused because as some water gets pumped away, that will be replaced by newly condensed water, thus increasing the pressure again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming I should use a positive-displacement pump. Please let me know if this is a bad idea for some reason.&lt;/p&gt;&#xA;" OwnerUserId="6054" LastEditorUserId="2246" LastEditDate="2016-04-22T13:53:19.737" LastActivityDate="2016-04-22T20:45:09.447" Title="Can I use a water-pump to lower pressure and boil water?" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8535" PostTypeId="2" ParentId="2464" CreationDate="2016-04-22T10:46:29.933" Score="1" Body="&lt;p&gt;You can use discretization of the problem into $N$ points, such that you only have to determine a finite number of parameters (assuming $f$ and $g$ are somewhat continuous functions). For the derivative and integration you can use Euler method, higher order methods can be used, but make the problem harder to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reformulation gives:&#xA;$$&#xA;h = \frac{t_1}{N-1}, \quad \vec{x} = [x_1, x_2, \dots, x_N], \quad \vec{y} = [y_1, y_2, \dots, y_N],&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\max_{\vec{x}, \vec{y}} &amp;amp; &amp;amp; \sum_{n=1}^{N-1} f(h (n-1), x_n, y_n) h\\&#xA;s.t. &amp;amp; &amp;amp; x_{n+1} = x_n + g(h (n-1), x_n, y_n) h, &amp;amp; &amp;amp; n = 1, 2, \dots, N-1&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have to add the boundary constraints to the equality constraints of the optimization problem. You can use multiple different methods to solve this problem, for example if you have access to Matlab you could use &lt;a href=&quot;http://nl.mathworks.com/help/optim/ug/fmincon.html?refresh=true&quot; rel=&quot;nofollow&quot;&gt;fmincon&lt;/a&gt;, which minimizes the cost function which can be fixed by adding an minus sign in front of the sum. Often you also have to supply an initial guess, which might also affect the solution, since different guesses might converge to different local maxima. By increasing $N$ you should get a more and more accurate solution, but it will probably take longer to solve. It might converge faster if you use the solution of a problem with less points and interpolate them and then use that as an initial guess for the problem of the larger number of points. &lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2016-04-22T10:46:29.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8536" PostTypeId="2" ParentId="8489" CreationDate="2016-04-22T10:50:39.043" Score="1" Body="&lt;p&gt;i am not sure about bottelnecking what does it means but the only natural way to increase the exhaust velocities is to place a &lt;strong&gt;NOZZLE&lt;/strong&gt; at the outlet as it converts the pressure energy into kinetic energy(&lt;strong&gt;From Steady Flow Energy Equation)&lt;/strong&gt; and to have more kinetic energy you should have high rate of combustion that will produce high pressure resulting into high velocities through a nozzle.&#xA;for the non natural processes you can use a centrifugal/reciprocating pump to achieve higher rate of discharge&lt;/p&gt;&#xA;" OwnerUserId="4889" LastActivityDate="2016-04-22T10:50:39.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8537" PostTypeId="2" ParentId="7622" CreationDate="2016-04-22T11:03:20.897" Score="1" Body="&lt;p&gt;I assume that you also want to minimize some function of $t$, $T$, $\vec{x}(t)$ and $w(t)$. In this case it depends on how you define a solution to this problem. If you define a solution as minimization of that function, then you will have multiple solutions if there are multiple local minima, but the global minima would be the best solution. If there are multiple solutions with the same functional value as the global minima, then that would just mean that there are multiple &quot;trajectories&quot; from $\vec{x}_0$ to $\vec{x}_1$ which are equally good. You would have to choose one of them or adjust your cost function.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2016-04-22T11:03:20.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8538" PostTypeId="2" ParentId="8534" CreationDate="2016-04-22T11:17:18.753" Score="5" Body="&lt;p&gt;You can use a water pump to evaporate water. But you shouldn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The evaporation chamber needs to be closed, except for the suction pipe. Otherwise when you lower pressure water (or air) will flow in, raising pressure again. So you can create small batches of vapor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water will evaporate where pressure is lowest, this is likely to be your pump inlet. &lt;a href=&quot;https://en.wikipedia.org/wiki/Cavitation#Pumps_and_propellers&quot; rel=&quot;nofollow&quot;&gt;Not what you want&lt;/a&gt;. You need to make sure that the ambient pressure in the evaporation chamber is lower than vapor pressure at whatever temeprature your water has. At the same time, ambient pressure + hydraulic head between level in chamber and pump inlet needs to be higher &lt;a href=&quot;https://en.wikipedia.org/wiki/NPSH&quot; rel=&quot;nofollow&quot;&gt;by a margin&lt;/a&gt;, else you have evaporation in your pump. Not good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you have low pressure, low temperature vapor. What do you want to do with it and how do you transport it? If you use a blower, the same blower could also create the vacuum for your evaporation and save you all the pump headaches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could create vapor with a water pump in a batch process, but I fail to see an aplication where this makes sense.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-04-22T20:45:09.447" LastActivityDate="2016-04-22T20:45:09.447" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8540" PostTypeId="1" CreationDate="2016-04-22T13:05:33.537" Score="0" ViewCount="2007" Body="&lt;p&gt;I'm trying to find the right joint for a small but sensitive application. The joint should be around 10mm-20mm long with a diameter/height of &amp;lt;10mm. Because of the high sensitivity (thus a need for a low friction joint), do ball joints or universal joints typically have lower friction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see that camera stabilisers tend to use universal joints - do universal joints tend to have lower friction then?&lt;/p&gt;&#xA;" OwnerUserId="4302" LastActivityDate="2016-04-24T20:52:58.487" Title="Low friction application: Ball joints or Universal joints?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8541" PostTypeId="1" AcceptedAnswerId="8546" CreationDate="2016-04-22T14:25:04.783" Score="6" ViewCount="211" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PTClt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PTClt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hello, I've been asked to show that:&#xA;$$d\Gamma=\tau 2\pi r^2 dr$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\Gamma$ is the torque applied to the shaft and $\tau$ is the shear stress of the sample fluid at radius $dr$. So, I know that $d\Gamma= Force\cdot dr$ and that $Force = \tau \cdot Area$ and this is where I start encountering issues. If we're looking at some radius $dr$, isn't the area $2\pi dr $ or just the circumference of the local circle? So I end up with $d\Gamma = \tau 2 \pi dr\cdot dr$ which isn't right, can someone help?&lt;/p&gt;&#xA;" OwnerUserId="4530" LastActivityDate="2016-05-31T01:29:46.593" Title="Force Balance Question" Tags="|mechanical-engineering|fluid-mechanics|applied-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8542" PostTypeId="1" CreationDate="2016-04-22T14:40:32.433" Score="2" ViewCount="76" Body="&lt;p&gt;In steam, when we say it's 0.9 quality, we mean that 90% is gas and 10% is liquid. But, as for specific volumes, will the specific volume of the gas be the inverse of the mass of the gas over the volume of the whole mixture or of the part of the mixture that the gas occupies? What is that part? In other words, is it correct to say,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\large{\frac{V_{total}}{m_{(total)}}\ =\ \frac{V_{total}}{m_{(gas)}}+\frac{V_{total}}{m_{(liquid)}}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or what? What is the equation?&lt;/p&gt;&#xA;" OwnerUserId="4846" LastEditorUserId="5899" LastEditDate="2016-04-23T17:22:37.997" LastActivityDate="2016-09-20T20:07:43.640" Title="Definition of non saturated thermodynamic mixture?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8544" PostTypeId="1" CreationDate="2016-04-22T16:25:59.647" Score="7" ViewCount="135" Body="&lt;h2&gt;I have a dry film moly-graphite lubricant (CoF=.075) and a lubricious CVD coating (CoF=.05).&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The CoF of each of these lubricious materials was determined using the same ASME standardized test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The standardized test consists of:&lt;/em&gt;  a plain steel object is coated in the lubricant to be tested; it is placed on a plain steel plate;   the plate is slowly tilted until the object starts to slide; the angle at which sliding occurs determines the CoF (a lower angle correpsonds to a lower CoF). &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;If I used &lt;strong&gt;&lt;em&gt;both&lt;/em&gt;&lt;/strong&gt; the coating and the lubricant together, would that lower the CoF further?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I.E. If I took the steel test object, and I coated it with the CVD coating, then on top of that I applied the dry-film moly-graphite lubricant, would that combination have a lower CoF than either material alone?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;If yes, How much? If no, Why not?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This seems like a relatively straight forward question, but I've yet to get a clear answer from anyone.  Most people seem to drag other variables into the problem statement, but I'd like to know, all other variables remaining equal, can we answer this question - or is it truly impossible?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The following assumptions apply:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;em&gt;only&lt;/em&gt; variable that changes between tests is the lubricant used.  All other factors remain exactly the same in each test. E.g. surface finish, temperature, interface load, materials being tested, etc. all remain exactly the same between tests. &lt;/li&gt;&#xA;&lt;li&gt;The CVD coating is permanent and non-removable.  The moly-lubricant can be applied over the CVD coating and/or to the other part; either way is perfectly acceptable, so whichever scenario allows you to answer the question is fine by me. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;PS:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This question was triggered by a problem I ran into at work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not a homework assignment - I've been out of school for almost a decade now :)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It started off as a simple question that no one in the office could answer, so I turned to the internet, because I'm a curious george. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, my attempts for help on &lt;em&gt;other&lt;/em&gt; internet sites have been met mostly with hostility and wild assumptions about the problem that were not stated in the get go.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got one answer here so far, and I realized some assumptions regarding outside factors were being made that I didn't anticipate, so I revised the question to clarify what assumptions should apply in any attempt to answer the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to know, if all other factors remain exactly the same, what effect to expect from adding a lubricant to an already extremely lubricious material. &lt;/p&gt;&#xA;" OwnerUserId="2598" LastEditorUserId="2598" LastEditDate="2016-04-26T13:33:18.517" LastActivityDate="2018-08-15T20:59:22.547" Title="Adding a lubricant to an already lubricious material, how is CoF affected?" Tags="|mechanical-engineering|applied-mechanics|modeling|friction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8546" PostTypeId="2" ParentId="8541" CreationDate="2016-04-22T18:37:21.133" Score="7" Body="&lt;p&gt;The problem here is that $dr$ is not the radius, $r$ is! Your expression for the infinitesimal area should be $2\pi r dr$: this is for a thin loop of radius $r$. (Excuse the crude diagram):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X3XXO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $r$ is a coordinate which is the radius of the loop, and $dr$ is a infinitesimal increment of $r$, i.e. the thickness of the loop. Since the loop is infinitely thin , the loop is like a rectangle curved around, so it's area will be the circumference of the circle, $2\pi r$, times it's thickness, $dr$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uqyUb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps :)&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2016-04-22T18:43:32.943" LastActivityDate="2016-04-22T18:43:32.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8547" PostTypeId="1" AcceptedAnswerId="8664" CreationDate="2016-04-22T18:47:45.563" Score="4" ViewCount="708" Body="&lt;p&gt;An uncold-worked brass specimen of average grain size 0.009 mm is heated to 600 degrees Celsius for 1000s, what is the average grain size based on this graph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pRS7L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pRS7L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to tackle this problem? I don't know if I am using the graph correctly.&#xA;My approach was:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1000s is 16.67 min, $\log(16.67) = 1.22$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\frac{22}{100}$ between 10 and 10&lt;sup&gt;2&lt;/sup&gt;. Which is about 0.67 cm to the right of 10 on x-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets estimate this corresponds to 10&lt;sup&gt;-1.333&lt;/sup&gt; on the y-axis of the 600 degrees celcius graph line. So 0.046 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this does not seem correct, because I did not incorporate the 0.009 mm starting position. Which corresponds to heat treatment time of about 1 min at 600 degrees so should I add 1 min to the 1000s? And thus look look at the point on the graph at 17,76 min?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer in book seems way off, they talk about average grain size 0.2 mm. To me that seems like they looked at the graph of 700 degrees Celsius.&lt;/p&gt;&#xA;" OwnerUserId="4117" LastEditorUserId="2246" LastEditDate="2016-04-22T20:55:25.913" LastActivityDate="2016-05-03T12:45:39.837" Title="Estimate the grain size after grain growth" Tags="|materials|metallurgy|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8548" PostTypeId="2" ParentId="8540" CreationDate="2016-04-22T21:51:44.363" Score="4" Body="&lt;p&gt;It depends exactly what you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ball joints and universal joints aren't directly comparable as universal joints can transmit torque (as in drive shafts) whereas ball joints just provide location. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There isn't an inherent difference in friction between the two as this will depend on manufacturing tolerances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big advantage of a ball and socket joint is that is can provide precise location of a pivot point but allow free movement in rotational axes. On the other hand a universal joint moves more freely in some planes than others as it is the combination of two axes of rotation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It really depends whether you just want rotation about a fixed point (as in a tilt/pan mount for a camera)  or transmission of axial torque (as in a drive shaft). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often &lt;em&gt;ball joints&lt;/em&gt; are designed with a moderate amount of friction just by being a tight fit, however there are are also &lt;a href=&quot;https://en.wikipedia.org/wiki/Spherical_bearing&quot; rel=&quot;nofollow&quot;&gt;spherical bearings&lt;/a&gt; which give rotation about a point in 3 axes and may give you more options. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-24T20:52:58.487" LastActivityDate="2016-04-24T20:52:58.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8549" PostTypeId="1" CreationDate="2016-04-23T00:09:27.530" Score="8" ViewCount="232" Body="&lt;p&gt;Rio de Janeiro, Brazil is currently stunned by the fact that yesterday a span of a brand new (innaugurated in January) bicycle bridge along the coast collapsed when struck by a wave, killing some people that were on it at the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few videos have surfaced:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.instagram.com/p/BEes18YhvHy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This one&lt;/a&gt; is of the moment it collapsed, but the man behind the camera didn't notice what happened until a few moments later, so you don't can't clearly see the wave impacting the structure, but you can see how the waves are redirected and slam against the structure almost vertically.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=lMnJoyEXojY&quot; rel=&quot;nofollow noreferrer&quot;&gt;This one&lt;/a&gt; is from after the collapse, but clearly shows the aftermath. At 2:15 you can see an example of another wave running up the rocks, spraying up higher than the road. At 2:30 you can see the columns (including their topping, where the beam was supported are in perfect condition and don't seem to present any damage.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://g1.globo.com/rio-de-janeiro/noticia/2016/04/parte-da-ciclovia-desaba-em-sao-conrado-zona-sul-do-rio.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; shows a video representation of what is believed to have happened: the waves lifted the beam off the pillars and caused a rigid-body rotation around its own axis. This, along with the mint condition of the tops of the columns implies that the project probably did not adopt a beam-column connection which could resist tension, which would therefore have impeded the beam's &quot;liftoff&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, my actual question is: how does one calculate a structure to resist such a load? I've done some searching and have found some articles (&lt;a href=&quot;http://www.citg.tudelft.nl/fileadmin/Faculteit/CiTG/Over_de_faculteit/Afdelingen/Afdeling_Waterbouwkunde/sectie_waterbouwkunde/people/personal/gelder/publications/papers/doc/paper34b.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;[A]&lt;/a&gt; &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.248.6639&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;[B]&lt;/a&gt; &lt;a href=&quot;https://www.flow3d.com/home/resources/case-studies/coastal-and-maritime-case-studies/computation-wave-forces-coastal-bridges&quot; rel=&quot;nofollow noreferrer&quot;&gt;[C]&lt;/a&gt; &lt;a href=&quot;https://icce-ojs-tamu.tdl.org/icce/index.php/icce/article/view/6998/pdf_1&quot; rel=&quot;nofollow noreferrer&quot;&gt;[D]&lt;/a&gt;) regarding wave action on bridges, but they all consider the more common case of a wave moving in a horizontal direction striking the side of a bridge. Now, how should one translate this to this case, with the wave being thrust (and possibly sped up) vertically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any codes which consider such cases? Also, more generally, are there any codes which define even standard wave action? International codes are fine. (I'm leaning a bit on &lt;a href=&quot;https://engineering.meta.stackexchange.com/a/111/1832&quot;&gt;Rick Teachey's position on the &quot;recommendations/finding stuff&quot; meta post&lt;/a&gt; for this part).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-03-16T16:42:56.367" LastActivityDate="2016-05-12T19:14:18.607" Title="Vertical wave action on bridge" Tags="|structural-engineering|civil-engineering|structures|structural-analysis|waves|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8550" PostTypeId="1" AcceptedAnswerId="8551" CreationDate="2016-04-23T13:21:24.327" Score="3" ViewCount="553" Body="&lt;p&gt;With gravity camera stabilisers like below, how come counterweights are needed to stabilise the load (camera)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CGmS7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CGmS7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it to ensure that the handle joins at the centre of gravity? If so, why does it have to be at the centre of gravity?&lt;/p&gt;&#xA;" OwnerUserId="4302" LastActivityDate="2016-04-25T11:20:05.453" Title="How come counterweight is needed to stabilise camera?" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8551" PostTypeId="2" ParentId="8550" CreationDate="2016-04-23T13:34:30.553" Score="7" Body="&lt;p&gt;The mass at the bottom is actually to make sure that the center of gravity of the whole assembly is &lt;em&gt;directly below&lt;/em&gt; the pivot (ball joint) at the top of the handle. Otherwise, it will tilt or try to flip over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass at the bend (at the same level as the handle) increases the moment of inertia in the horizontal plane to add stability to the pointing direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two masses together also increase the total &quot;sprung mass&quot; to reduce the resonant frequency of the entire assembly and reduce vibration. (Your arm provides the &quot;spring&quot; and &quot;damper&quot; for the system.)&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-04-23T13:34:30.553" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8552" PostTypeId="1" AcceptedAnswerId="8563" CreationDate="2016-04-23T16:34:08.953" Score="1" ViewCount="3781" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dzdEC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dzdEC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flux in the gas can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N_{1}=k_{p}(p_{1}-p_{1i})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the flux in the liquid can be written as:&#xA;$$N_{1}=k_{x}(x_{1i}-x_{1})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we know that $$x^*_{1}=\frac{p_{1}}{H}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I use this to find the overall mass transfer coefficient?&#xA;I know that the driving force will just be $(p_{1}-Hx_{1})$ but what's $K_{g}$ and more importantly, how is it derived?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'll give the best answer to anyone who can even give a hint&lt;/p&gt;&#xA;" OwnerUserId="4530" LastEditorUserId="4530" LastEditDate="2016-04-24T11:10:12.790" LastActivityDate="2016-05-05T00:21:25.923" Title="Derivation of Overall Mass Transfer Coefficient" Tags="|heat-transfer|chemical-engineering|process-engineering|diffusion|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8553" PostTypeId="2" ParentId="8544" CreationDate="2016-04-23T16:46:28.283" Score="5" Body="&lt;p&gt;You can't calculate friction coefficients this way. At the atomic-scale level, the friction coefficient depends on the interactions of the two materials across the interface. If you measure the behaviour of the interface between materials A and B, and A and C, you can't say anything about the interface between B and C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, friction coefficients are dependent on the surface finish, temperature, relative velocity, interface pressure, etc., etc. The only way to measure an &quot;accurate&quot; value is to do an experiment that matches the &quot;real life&quot; conditions as closely as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: The claim that friction depends on the interface pressure refers to the (very simple) Coulomb &lt;strong&gt;model&lt;/strong&gt; of friction - which has the advantage that it is simple enough to use in high-school-level mechanics questions, but is only &quot;accurate&quot; for a very limited range of materials and operating conditions.)&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="368" LastEditDate="2016-04-27T20:02:03.263" LastActivityDate="2016-04-27T20:02:03.263" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8554" PostTypeId="2" ParentId="8550" CreationDate="2016-04-23T18:27:10.280" Score="1" Body="&lt;p&gt;When the pivot point is right at the center of gravity, the camera will not tilt if you move the support left or right - there will be no net torque on the camera. If you have the COG just below the pivot (ball joint), you make the system stable (that is, it will naturally want to be upright) but lateral movement will result in a tilt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closer to the pivot your center of gravity is, the less the camera will tilt - but also, the longer it will take to come back to vertical if you tilt it.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2016-04-23T18:27:10.280" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8555" PostTypeId="2" ParentId="8542" CreationDate="2016-04-23T18:33:26.610" Score="1" Body="&lt;p&gt;No, the specific volume of the steam as for any other property is calculated as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v = v_f + xv_{fg}$$&#xA;where $v_f$ is the specific volume of saturated liquid, $v_{fg}$ is the  difference between the specific volume of saturated vapor and specific volume of saturated liquid ($v_g - v_f$) (You can obtain them from your steam tables) and $x$ is the dryness factor(steam quality).&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2016-04-23T18:33:26.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8556" PostTypeId="2" ParentId="8550" CreationDate="2016-04-23T19:17:42.943" Score="2" Body="&lt;p&gt;A camera system is more sensitive to rotational shake than any translation. That is because translation causes a shift proportional to the noise but rotation causes an error proportional to the distance to the target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GqjQ2.png&quot; alt=&quot;Camnoise&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: Example of noise in a plane, note that in 3 dimensions there are more rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of momentum, means that more mass is slower to move around. When it comes to rotation the distribution of this mass affects the way something resists rotation. We call this moment of inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inertia brings &quot;slowness&quot; to the system. Net effect is that the harder it is to rotate, the more force you need for the rotation to happen. Thus the system rotates less from the noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By putting the weight far away from the joint you make the moment of inertia bigger. Meaning its harder to rotate (around the axis that has furthest distance to weight) thus damping the system more. At the same time the overall weight of the system does not need to grow as much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For similar reasons, you want the center of gravity be below the ball joint so that the camera stays upright. Or more exactly, the system will orient so that the center of gravity is under the ball joint, so you want to design it so that this benefits you. Also you want to avoid having a lot of weight around the ball joint on a horizontal plane so that turning the camera horizontally is not too heavy. This way, the system prefers motions that are more natural for your camera work.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="86" LastEditDate="2016-04-25T11:20:05.453" LastActivityDate="2016-04-25T11:20:05.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8557" PostTypeId="1" CreationDate="2016-04-24T01:43:11.437" Score="0" ViewCount="1841" Body="&lt;p&gt;In my class, we are currently trying to understand the relationship between gear ratio and motor power consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given two gears, one attached to a motor, and the other attached to a string carrying a mass, I am trying to identify the gear ratio that minimizes the power consumption of the motor given a vertical distance travelled by the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We analytically determined the motor constant, the motor resistance, the radius of the gear about the motor, and mass to be lifted, and the voltage applied to the motor, and the goal is to use this information to determine the gear ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the two relevant motor equations as well as the various relationships provided by the gear ratio, but I am unsure where to begin with this problem.  If someone could provide a push in the right direction, I have a bunch of ideas in my head and could hopefully take it from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two motor equations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;V &amp;amp;= iR + K\omega \\&#xA;T &amp;amp;= \left(i-i_0\right)K&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="5355" LastEditorUserId="1832" LastEditDate="2016-04-24T03:15:52.683" LastActivityDate="2017-09-22T03:53:46.783" Title="Relationship Between Gear Ratio and DC Motor Power Consumption" Tags="|motors|gears|power|energy-efficiency|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8560" PostTypeId="1" AcceptedAnswerId="8562" CreationDate="2016-04-24T15:09:23.040" Score="4" ViewCount="7632" Body="&lt;p&gt;I read in a documentation to some adhezive: &lt;em&gt;the tensile strength is 13 N/mm&lt;sup&gt;2&lt;/sup&gt;&lt;/em&gt;. How many kilograms can the 10x10mm bonding hold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to calculate it by the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;F &amp;amp;= m/g \\&#xA;\therefore m &amp;amp;= Fg&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the strength is in N/mm&lt;sup&gt;2&lt;/sup&gt; e.g. that is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;p = F/A \\&#xA;\therefore F = pA&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;substituting the $F$ in $m = Fg$, I got&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = pAg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using real numbers,&#xA;- tensile strength = 13 N/mm&lt;sup&gt;2&lt;/sup&gt; = 13*10e6 N/m&lt;sup&gt;2&lt;/sup&gt;&#xA;- A = 10x10mm (100mm&lt;sup&gt;2&lt;/sup&gt; = 0.0001m&lt;sup&gt;2&lt;/sup&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = 13*10e6 * 0.0001 * 9.81 = 12753 \text{ kg} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result doesn't look right - the 10x10mm bonding imho can't hold 13.000kg weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did I make a mistake? So the question is: how to convert the 13 N/mm&lt;sup&gt;2&lt;/sup&gt; tensile strength to kilograms (e.g. how many kilos I can hang to the bonded 10x10mm plates?)&lt;/p&gt;&#xA;" OwnerUserId="2617" LastEditorUserId="1832" LastEditDate="2016-04-24T20:19:45.833" LastActivityDate="2016-04-24T21:05:33.117" Title="Tensile strength to weight conversion" Tags="|structural-engineering|stresses|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8561" PostTypeId="2" ParentId="8560" CreationDate="2016-04-24T15:47:00.280" Score="1" Body="&lt;p&gt;Performing the same calculations with the same values I obtain approximately 1275 kg. Hence there must be at least an arithmetic error in your calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Dave Tweed pointed out in &lt;a href=&quot;https://engineering.stackexchange.com/a/8562/421&quot;&gt;his answer&lt;/a&gt;, this is not the correct calculation to obtain the tensile force required to cause failure. The correct calculation results in about 130 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That isn't unreasonable if the adhesive is loaded purely in tension. However, if the load is slightly off center, or there is a torque, then the adhesive will be loaded unevenly and begin to fail at one end first. Once the adhesive has failed at a corner or edge, the remaining adhesive will have higher stress and failure will occur more rapidly. Adhesives tend to fail by peeling more readily than in pure tension. Peel strength is generally a more useful measure of adhesive strength, as adhesives tend to fail by peel long before they fail in tension.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-04-24T19:21:35.483" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8562" PostTypeId="2" ParentId="8560" CreationDate="2016-04-24T16:01:47.937" Score="4" Body="&lt;p&gt;Pay attention to your unit analysis. You multiplied by the acceleration due to gravity, rather than dividing by it. Unit analysis would have shown this to be nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$13 \frac{N}{mm^2} \cdot 100 mm^2 = 1300 N = 1300 \frac{kg \cdot m}{s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, to get kg, you need to divide by an acceleration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1300 \frac{kg \cdot m}{s^2}}{9.81 \frac{m}{s^2}} = 132.5 kg$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: In terms of the symbolic analysis, your very first statement was wrong: Force equals mass &lt;em&gt;times&lt;/em&gt; acceleration, so you should have written it as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = m\cdot g \text{  and  } m = \frac{F}{g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, when you combine that with the equation for pressure, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m = \frac{pA}{g}$$&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2016-04-24T21:05:33.117" LastActivityDate="2016-04-24T21:05:33.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8563" PostTypeId="2" ParentId="8552" CreationDate="2016-04-24T17:36:42.213" Score="2" Body="&lt;p&gt;Regarding $k_p$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you've got $N_1$ a flux in $mol/ s \cdot m^2 $ and a partial pressures $p_i$ in $Pa$. $k_p$ has got to connect those, and make the units work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly what $k_p$ is depends on the flow of the gas. If the gas is stationary, mass transfer will by diffusion alone. Unfortunately, that process doesn't have a steady-state solution in a semi-infinite domain. If you had a thickness for the gas layer ($L_x$), $k_p$ would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{D_{AB}}{L_x R T} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $D_{AB}$, $R$, and $T$, are the diffusion coefficient, universal gas constant and temperature. $D_{AB}$ is a property of the species that you're tracking ($A$) and the remaining species that make up the gas ($B$). You can look those up. In a pinch, I assume $2 \cdot 10^{-5}$ for air-like gases.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, go with the idea that you've got some kind of flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With flow over a surface you'll end up with boundary layer, a thin region where velocity and species concentrations vary between their values at the surface and in the free stream. That flow is generally going to enhance the mass transfer, by actually moving things around a lot faster than diffusion (diffusion is wicked slow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ct3VN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ct3VN.png&quot; alt=&quot;velocity boundary layer over a flat plate, from wikipedia article on boundary layers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to cook up some solutions to the boundary layer equations and actually derive $k_p$. But that's tedious, only works in a few situations and the results are still approximate. What you actually want to do is look up a correlation. Like these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z9RwH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z9RwH.png&quot; alt=&quot;table taken from https://www.cpp.edu/~tknguyen/che313/pdf/chap3-1.pdf&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those'll give you the Sherwood number based on the Reynolds number and Schmidt number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Sh = \frac{k_g L}{D_{AB}},  Re = \frac{u L}{\nu}, Sc = \frac{\nu}{D_{AB}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you aren't familiar, the only thing that might be tricky here is choosing $L$. It's the characteristic length of the flow configurations. Usually, it's just the length of the object in the direction of flow. Except inside pipes, use the diameter.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2016-04-24T17:36:42.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8564" PostTypeId="1" AcceptedAnswerId="8567" CreationDate="2016-04-24T19:21:57.257" Score="2" ViewCount="1722" Body="&lt;p&gt;The precast concrete stairs at the home I recently purchased do not have a footing.  Lack of a footing coupled with drainage issues has caused them to sink over time.   I'm going to raise them and get them properly supported underneath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do that I want to clamp them near their center of gravity (along the length), so that I can jack them up back to where they belong and minimize the risk of breaking the concrete (I'll continuously add temporary support underneath them as I slowly jack them up).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have dug out the bottom of the steps and determined that they are monolithically poured and solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the measurement along the length of the steps where the center of gravity will be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zcsx6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zcsx6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  This is not homework.  Here are the stairs in question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AsSPU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AsSPU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see how they are pulling away from the foundation and have rotated forward and dropped.  The sidewalk slab in front of it has also sunk and needs to be raised, so it is worse than it seems in this photo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attempted to solve this by hand because it seems trivial.   I'm glad I asked, because the solution I came up with was wrong.  I assumed I could reduce it to working with the ratio of cross sections which are simply three rectangles (since they all have the same width).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I take the area of each step individually I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;17.5 \times 18 &amp;amp;= 315 \\&#xA;14 \times 11   &amp;amp;= 154 \\&#xA;47.5 \times 3  &amp;amp;= 142.5&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that if I add the bottom two steps together, they are a smaller combined area (and thus volume) than the large step [pretty sure this is where I went wrong ... when you multiply by width this is no longer the case].  So it appeared to me at the time that the COG would be somewhere in the large step.  I'll denote the COG as $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I then used this to give me $c$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$18 \times c = 18 \times (17.5 - c) + 154 + 142.5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That worked out to 16.98 (17 inches for all practical purposes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, after seeing @Chris Johns solution, I realize that what I had in my brain is not correct, and anybody with sinking stairs should ignore everything above.&lt;/p&gt;&#xA;" OwnerUserId="3830" LastEditorUserId="1832" LastEditDate="2016-04-24T21:00:32.103" LastActivityDate="2016-04-24T21:43:42.640" Title="Center of Gravity of Precast Concrete Steps" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8565" PostTypeId="2" ParentId="8564" CreationDate="2016-04-24T19:42:36.033" Score="2" Body="&lt;p&gt;From the left hand side in your drawing I make it x = 23.5&quot;  y = 6.32&quot; ie more or less in the centre of the second step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that as you have done a CAD drawing of it already there is probably a tool in the software  to calculate the CoG for you (that's how I did it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's going to break anywhere the inside corner between the first and second steps is probably the most likely place. So try to avoid putting to much bending load on the bottom step. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For academic interest the way to do this by hand is to break the areas down into convenient blocks as you have done and calculate moments of each individual centre of mass about the CoG. By definition the sum of the moments about the CoG will be zero. Here even though the bottom step is thin, it has significant effect because it extends a long way from the CoG. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also be interested in &lt;a href=&quot;https://en.wikipedia.org/wiki/Concrete_leveling&quot; rel=&quot;nofollow&quot;&gt;foam injection leveling&lt;/a&gt; this is a method of leveling, jacking and stabilising concrete by injecting expanding PU foam underneath it. Here the gas pressure produced by the foaming reaction lifts the slab and the foam itself creates a stable base. I can't vouch for the usefulness of this method as I've only come across it in passing but it seems like it is worth considering in this situation. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-24T21:43:42.640" LastActivityDate="2016-04-24T21:43:42.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8566" PostTypeId="1" CreationDate="2016-04-24T20:37:59.007" Score="8" ViewCount="662" Body="&lt;p&gt;*I have updated this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using two codes for modeling incompressible, two-dimensional flow out of a tank.  My fluid dynamics experience is pretty basic; I'm using the codes to understand a chemical process.  The codes haven't been used much, so I'm running a benchmark/test (running both codes on same flow problem) to make sure I understand them.  Unfortunately, I can't post the codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem:&#xA;In the test, I want a pressure gradient to drive flow (a constant pressure at the bottom of the tank and 0 at the tank outlet).  So far, velocities calculated from one code are much higher than velocities from the other.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One code uses the Navier-Stokes equations in dimensionless form solved by Galerkin finite elements with a perturbation method (penalty formulation).  This works by eliminating pressure from the equations.  I managed to track down the developer and he ran an example for me.  Our results are identical, so I'm not making a mistake with the inputs.  The other code uses a volume of fluid method and pressure is not eliminated.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm now wondering if the dimensionless code (where pressure is eliminated) can handle pressure boundary conditions correctly.  If anyone has any ideas about this, I'd love to hear them.  Thanks a lot.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit for Dan&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the tank geometry I'm dealing with, plus the dimensionless groups that (according to the documentation) are used in the code that is based on dimensionless Navier-Stokes.  The dimensionless parameters are the ones with the bars over the letters. $\rho$ is fluid density; $\eta$ is fluid viscosity. I know it is possible to nondimensionalize pressure by using velocity in the nondimensionalization, but I don't know the velocity at the inlet so don't want to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing in the documentation about how to deal with a pressure boundary condition.  I have used the pressure dimensionless term, but with $A$, not $a$, since I want to define an inlet boundary condition.  This gives me a pressure that is too high because it leads to velocities that are greater than in the other code.  I hope this is clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pgu6q.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pgu6q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1416" LastEditorUserId="1416" LastEditDate="2016-05-27T16:39:23.817" LastActivityDate="2016-05-27T16:39:23.817" Title="Incompressible 2D pressure-driven flow: for Navier-Stokes equations in nondimensional form, how should I express pressure boundary conditions?" Tags="|fluid-mechanics|statics|finite-element-method|mathematics|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8567" PostTypeId="2" ParentId="8564" CreationDate="2016-04-24T21:35:02.720" Score="4" Body="&lt;p&gt;The equation to find the center of gravity of an object which can be subdivided into smaller parts is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{CG} = \dfrac{\sum \overline{y}_iA_i\rho_i}{\sum A_i\rho_i}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\overline{y}_i$, $A_i$ and $\rho_i$ are the &lt;a href=&quot;https://en.wikipedia.org/wiki/Centroid&quot;&gt;centroid&lt;/a&gt;, area and specific weight of each part, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the specific weight of all parts is equal, and so the equation can be simplified to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{CG} = \overline{y} = \dfrac{\sum \overline{y}_iA_i}{A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, in such a case, the center of gravity of the object is equal to its centroid ($\overline{y}$). Also, to be clear, the $A$ in the denominator is the object's total area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in this example, the solution is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\overline{x}_1A_1 &amp;amp;= \left(\dfrac{17.5}{2}\right) \times \left(18 \times 17.5\right) \\&#xA;\overline{x}_2A_2 &amp;amp;= \left(17.5 + \dfrac{14}{2}\right) \times \left(10.5 \times 14\right) \\&#xA;\overline{x}_3A_3 &amp;amp;= \left(17.5 + 14 + \dfrac{47.5}{2}\right) \times \left(3.0 \times 47.5\right) \\&#xA;A &amp;amp;= \left(18 \times 17.5\right) + \left(10.5 \times 14\right) + \left(3.0 \times 47.5\right) \\&#xA;\color{red}{\overline{x}} &amp;amp;= \color{red}{\dfrac{\sum \overline{x}_iA_i}{A} = 23.5} \\&#xA;\overline{y}_1A_1 &amp;amp;= \left(\dfrac{18}{2}\right) \times \left(18 \times 17.5\right) \\&#xA;\overline{y}_2A_2 &amp;amp;= \left(\dfrac{10.5}{2}\right) \times \left(10.5 \times 14\right) \\&#xA;\overline{y}_3A_3 &amp;amp;= \left(\dfrac{3.0}{2}\right) \times \left(3.0 \times 47.5\right) \\&#xA;\color{red}{\overline{y}} &amp;amp;= \color{red}{\dfrac{\sum \overline{y}_iA_i}{A} = 6.3}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your center of gravity is 23.5&quot; from the left-hand side of your image (from the exterior face of the upper step) and 6.3&quot; from the bottom of your image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, don't think that just because you raise the steps from their center of gravity that they won't get damaged. If you just do that, then the steps may become like a dual-cantilever, in which case your thin &quot;step&quot; (if you can call it that) will risk breaking apart due to its low height and therefore low moment of inertia. Now, if what you actually want is help on raising your steps without damaging them, I suggest you ask another question referencing this one.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-04-24T21:35:02.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8568" PostTypeId="2" ParentId="2534" CreationDate="2016-04-24T23:28:12.867" Score="1" Body="&lt;p&gt;Your absolute minimum responsibility is to inform your customer that you consider the intended usage to be unsafe and that you do not recommend or endorse using it in this way. This should be done in writing and be acknowledged by the customer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next level depends on context. It is not impossible that your customer has good reason to do this and has their own procedures to ensure that it is used in a safe way. However if you believe that they are just cutting corners then we all know that it would not be ethical to supply them with the means to do it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to do this properly then the only way is to have an in-depth discussion with your customer to ensure that they will use your product responsibly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the correct ethical approach  goes somewhat beyond discharging your nominal legal responsibilities and in any case you may still end up being legally liable even if you tick the boxes on something you know to be dangerous. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-24T23:28:12.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8574" PostTypeId="1" CreationDate="2016-04-25T12:25:00.210" Score="3" ViewCount="1038" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pot-in-pot_refrigerator&quot; rel=&quot;nofollow&quot;&gt;The Wikipedia article &lt;em&gt;Pot-in-pot refrigerator&lt;/em&gt;&lt;/a&gt; says that sand in the outer pot somehow affects the cooling efficiency of refrigerator. I don't understand what the sand does in this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, compare filling the outer pot with water vs. filling the outer pot with both water and sand. Would these have some significant difference in cooling efficiency, meaning how fast heat is transferred between the inner pot, the middle layer and the outside air?&lt;/p&gt;&#xA;" OwnerUserId="424" LastEditorUserId="368" LastEditDate="2016-04-25T18:49:01.130" LastActivityDate="2016-05-26T13:55:36.850" Title="What does the sand do in a pot-in-pot refrigerator?" Tags="|thermodynamics|heat-transfer|refrigeration|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8577" PostTypeId="1" CreationDate="2016-04-25T18:58:22.307" Score="1" ViewCount="198" Body="&lt;p&gt;I studied about the effect of turbine blade thickness on flow separation in following. The blade section is Naca 00 XX in this turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper states that if blade thickness increases, separation on the blade delays. Why does this happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper also states that if the blade thickness increases, there is higher pressure on the suction surface and hence this gives a delayed separation. Why does this happen? Why is there high pressure on the suction surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paper also states: Blade (A) has lower torque than blade (E). Why?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2016-08-08T04:02:58.190" LastActivityDate="2018-01-26T07:15:04.383" Title="Why do thicker turbine blades delay flow separation?" Tags="|mechanical-engineering|fluid-mechanics|turbines|" AnswerCount="1" CommentCount="2" ClosedDate="2018-01-27T16:10:40.730" ContentLicense="CC BY-SA 3.0" />
  <row Id="8579" PostTypeId="1" CreationDate="2016-04-25T19:15:43.933" Score="5" ViewCount="46" Body="&lt;p&gt;All the methods I know of for calculating bedload discharge are for canals of rectangular cross-section (constant depth of flow). The bedload is first computed for a unit of width (using the constant water depth) and then multiplied by the width. However, I have an almost triangular cross-section, where the depth is not constant. How can I calculate bedload discharge in this case?&lt;/p&gt;&#xA;" OwnerUserId="6185" LastEditorUserId="368" LastEditDate="2016-04-25T23:12:50.030" LastActivityDate="2016-04-25T23:12:50.030" Title="Bedload discharge for canals of triangular cross-section" Tags="|civil-engineering|hydraulics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8581" PostTypeId="2" ParentId="8549" CreationDate="2016-04-25T21:37:29.537" Score="-1" Body="&lt;p&gt;I think in this particular case the tensions force on the bridge should be calculated by regarding the bottom surface of the bridge as a water pump's impeller or the drag induced on the tip of a submarine or torpedo. The Entire bottom of the bridge geometry should offer the least resistance to flow of wave break. Dynamic drag forces are very similar to already exiting methods to design submarines! if the geometry of the abutment is potentially capable of creating explosive impact that has to be accounted for too.     Surely there will be new research and code added for this particular case in the future.   It is only that this is a new phenomenon.   I had designed a concrete deck including a suspended swimming pool supported by 6 concrete cassions 45&quot; deep into bedrock. And after some years visiting the job I noticed some of the caissons had cracks on top where they met the slab which was due to seasonal uplift and settlement force of clay soils. One has to think about all eventualities when designing theses kind of structures!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-04-25T21:37:29.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8582" PostTypeId="1" CreationDate="2016-04-25T22:50:47.613" Score="2" ViewCount="108" Body="&lt;p&gt;As you may know, as eccentricity gets smaller, the location of apoapsis/periapsis becomes less well-defined.  Rather equivalently, I've noticed that if eccentricity is &amp;lt; 4e-7 when converting from Keplerian elements to Cartesian and back again, significant errors start to creep in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know how small of an eccentricity my code should realistically be expected to handle.  I know these issues are handled well by equinoctial elements, but would rather avoid that complication if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Orbital_eccentricity#Examples&quot; rel=&quot;nofollow&quot;&gt;Wikipedia says&lt;/a&gt;, &quot;Neptune's largest moon Triton has an eccentricity of 1.6 × 10&lt;sup&gt;−5&lt;/sup&gt;, the smallest eccentricity of any known body in the Solar System.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's very fine and interesting, but what about artificial satellites?  What is the smallest eccentricity achieved by one of those?&lt;/p&gt;&#xA;" OwnerUserId="6188" LastEditorUserId="6188" LastEditDate="2016-04-26T01:29:14.557" LastActivityDate="2016-07-25T02:46:33.613" Title="What is the smallest eccentricity achieved by an artificial satellite?" Tags="|aerospace-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8583" PostTypeId="2" ParentId="8582" CreationDate="2016-04-26T00:18:47.513" Score="1" Body="&lt;p&gt;How about zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are artificial satellites which are placed in circular orbits, particularly geostationary orbits where the satellite stays over the same spot on the earth's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Geosynchronous_orbit&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Geosynchronous_orbit&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-04-26T00:18:47.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8585" PostTypeId="1" CreationDate="2016-04-26T04:39:14.753" Score="2" ViewCount="310" Body="&lt;p&gt;I was wondering if I was to have two objects both made of isotropic materials such as steel and aluminum and the two objects were the same geometry and I was to apply the same load and boundary conditions to each, would one of the materials have a higher stress?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that because stress = load / area then the stress for steel would be exactly the same as for aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then there is the relation between Young's modulus $E = \frac{\mbox{stress}}{\mbox{strain}}$. So it is difficult for me to justify my first thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be interesting to hear some thoughts on this.&lt;/p&gt;&#xA;" OwnerUserId="2940" LastEditorUserId="2246" LastEditDate="2016-04-26T13:59:13.853" LastActivityDate="2016-04-26T13:59:13.853" Title="Isotropic materials and levels of stress" Tags="|materials|applied-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8588" PostTypeId="1" CreationDate="2016-04-26T07:35:40.253" Score="5" ViewCount="460" Body="&lt;p&gt;Conductive paint by conventional technique using gum and graphite powder does not stick to the PDMS surface and comes off once the paint dries. How else can I make a conductive connection over the PDMS surface so that it can stick to it nicely? &lt;/p&gt;&#xA;" OwnerUserId="6194" LastEditorUserId="1586" LastEditDate="2016-05-09T23:06:32.057" LastActivityDate="2021-08-26T19:11:30.580" Title="How to make a conductive paint over PDMS?" Tags="|electrical-engineering|lithography|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8590" PostTypeId="2" ParentId="8585" CreationDate="2016-04-26T07:54:23.423" Score="4" Body="&lt;p&gt;In general the answer is yes, however for particular cases the stresses could be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your stress &quot;definition&quot; (stress = load/area) is valid for uni-axial stress state, for example a rod subjected to an axial force. In this case, within the elastic range stresses in isotropic materials would be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this valid only if we accept some common engineering assumptions such as using &lt;a href=&quot;http://www.eqsgroup.com/all-about-steel/difference-between-true-stress-and-engineering-stress.asp&quot; rel=&quot;nofollow&quot;&gt;engineering stresses&lt;/a&gt;. This means that we do not take into account how the cross-section changes under the loading due to Poisson effect. However, if we take this into account and dealing with &lt;a href=&quot;http://www.eqsgroup.com/all-about-steel/difference-between-true-stress-and-engineering-stress.asp&quot; rel=&quot;nofollow&quot;&gt;true stresses&lt;/a&gt; there will be a difference due to the difference in Poisson's ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when engineering stresses are used there are cases where the stresses will be different due to the different Poisson's ratio. For example internal forces and stresses in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Kirchhoff%E2%80%93Love_plate_theory&quot; rel=&quot;nofollow&quot;&gt;plate&lt;/a&gt; are influenced by the Poisson's ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is geometrical non-linearity where the change of global geometry under loading influences the distribution of internal forces, stresses (the change in cross-section is typically not considered). Since the deformations are dependent on the Young's modulus you will get different stresses for different materials. A sagging cable is simple example for this.&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2016-04-26T07:54:23.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8591" PostTypeId="2" ParentId="8574" CreationDate="2016-04-26T11:34:49.100" Score="0" Body="&lt;p&gt;Looking up some data on the Intartubezz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All units are    $\frac{W}{m-K}$&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Water at 20° C      0.6         &#xA;&#xA;Sand, dry          0.15 - 0.25      &#xA;Sand, moist        0.25 - 2      &#xA;Sand, saturated    2 - 4&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So you can see that the combination of sand and water is better than either alone.&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="615" LastEditDate="2016-04-26T13:15:03.120" LastActivityDate="2016-04-26T13:15:03.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8593" PostTypeId="2" ParentId="8574" CreationDate="2016-04-26T13:09:53.027" Score="1" Body="&lt;p&gt;This type of cooler works by evaporation ie the latent heat of vapourisation of the water draws energy from its immediate surroundings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can feel this effect for yourself if you ever get a low boiling point liquid like acetone on your skin (not recommended on bare skin but you can still feel the effect through thin gloves). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below its boiling point water can only evaporate from it's surface (in fact this is the definition of boiling) so placing it in a porous medium, in this case an unglazed pot greatly increases the surface area available for evaporation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again you can see this in the fact that a glass of water will take a long time to evaporate but of you pour it out onto a flat surface it will disappear quite quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly this works bet if the porous medium is damp rather than completely soaked as the idea is to have a thin film of water which is constantly replenished as it evaporates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wet sand provides additional surface area and also is an effective way of keeping the outer pot damp  but not saturated while allowing conduction of heat from the inside of the vessel to the outside surface. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-26T13:09:53.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8594" PostTypeId="1" CreationDate="2016-04-26T13:46:55.677" Score="4" ViewCount="726" Body="&lt;p&gt;Does precipitation hardening really only make the material &lt;strong&gt;harder&lt;/strong&gt; while Solid Solution strengthening makes it &lt;strong&gt;stronger&lt;/strong&gt; altogether (thus increasing the area under the stress-strain-curve &lt;strong&gt;edit:&lt;/strong&gt; excuse the mistake here: the area under the stress-strain curve denotes the &lt;strong&gt;toughness&lt;/strong&gt;, not strength)? If so, then what is the reason behind this? Don't both mechanisms work by impeding dislocation gliding (one through precipitates and the other ones through introduced strains)?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In this &lt;a href=&quot;http://www.azom.com/article.aspx?ArticleID=2819&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;, it is mentioned that &quot;The high tensile strengths of precipitation hardening stainless steels come after a heat treatment process [...]&quot;, which suggests that hardening indeed also increases strength. Is there a specific reason for why the names were chosen in this manner?&lt;/p&gt;&#xA;" OwnerUserId="6039" LastEditorUserId="272" LastEditDate="2016-04-26T16:39:47.263" LastActivityDate="2016-05-01T14:01:03.063" Title="Why are the processes called &quot;Precipitation Hardening&quot; and &quot;Solid Solution Strenghtening&quot;?" Tags="|materials|design|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8595" PostTypeId="2" ParentId="8594" CreationDate="2016-04-26T14:16:08.353" Score="4" Body="&lt;p&gt;For most metal alloys tensile strength and hardness are effectively different expressions of the same property. Although this does depend on exactly how hardness is measured and some related properties such as abrasion resistance can diverge from this especially in things like stainless steel where you have hard carbide grains embedded in a relatively soft matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also cases where components can have a hard outer layer and a more ductile core (eg white/grey iron castings) or case hardened steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly where an alloy can undergo changes in hardness/strength during the manufacturing process it is conventional to talk about hardening. Often this sort of terminology dates back to a period when the empirical properties of materials were well understood but not necessarily in a rigorous scientific way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the specific terms in the question, Solid Solution Strengthening is the general term for the phenomenon the at solid solution of two elements will be stronger/harder than the pure base metal because the alloying element disrupts slipping of dislocations. Precipitation hardening is a specific process used to achieve this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the choice of strengthening vs hardening is just down to slightly different context as in manufacturing 'strengthening' can be ambiguous (as it could relate to structural design as well) whereas 'hardening' clearly relates to a material property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in a materials science context 'strengthening' is a bit more precise, as tensile strength is a better defined property than hardness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also reasonable to suppose that precipitation hardening is named by analogy to older processes like quench hardening of steel and, work hardening where the increase in hardness is the most immediately obvious outcome of the process. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-26T14:33:37.287" LastActivityDate="2016-04-26T14:33:37.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8596" PostTypeId="1" CreationDate="2016-04-26T15:19:13.473" Score="3" ViewCount="999" Body="&lt;p&gt;I want to use a woven carbon fiber tube in cloth form to act as an acoustic absorber in a 40 mm internal diameter pipe. The bad news is that it has to carry a variety of solvent/air mixtures at temperatures up to 80 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it is cut to length the ends fray horrendously. Now, an obvious way to stop this would be to put some tape around the end. Unfortunately I cannot find a tape that can withstand alcohols, ketones, hydrocarbons and chlorinated hydrocarbons. The solvents are why I am using carbon fiber in the first place. I could move to glass fiber but I expect I would encounter a similar problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I stop the cut ends of the mat from fraying?&lt;/p&gt;&#xA;" OwnerUserId="3417" LastEditorUserId="368" LastEditDate="2016-04-26T15:44:24.490" LastActivityDate="2023-10-01T16:40:10.090" Title="How can I stop carbon fibre mat from fraying?" Tags="|materials|carbon-fiber|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8597" PostTypeId="1" CreationDate="2016-04-26T15:36:45.973" Score="5" ViewCount="175" Body="&lt;p&gt;I am trying to understand network interdiction. I have read these papers, but it is still hard to get a clear understanding of what it exactly means:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.buehrer.ece.vt.edu/papers/BlindNetworkInterdiction.pdf&quot; rel=&quot;noreferrer&quot;&gt;Blind Network Interdiction Strategies - A Learning Approach&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rutcor.rutgers.edu/pub/rrr/reports2012/04_2012.pdf&quot; rel=&quot;noreferrer&quot;&gt;Network interdiction – models, applications, unexplored directions&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Could someone explain it to me in layman's terms, using minimal technical language?&lt;/p&gt;&#xA;" OwnerUserId="6201" LastEditorUserId="368" LastEditDate="2016-04-26T16:16:38.733" LastActivityDate="2016-04-26T17:12:39.557" Title="Explanation of network interdiction in layman terms" Tags="|computer-engineering|networking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8598" PostTypeId="2" ParentId="8596" CreationDate="2016-04-26T15:53:29.610" Score="3" Body="&lt;p&gt;One solution would simply to hem it as you would any other textile if it's just for the ends of a tube &lt;a href=&quot;http://www.holiday-crafts-and-creations.com/how-to-do-blanket-stitch.html&quot; rel=&quot;nofollow&quot;&gt;blanket stitch&lt;/a&gt; would be easy enough to do by hand and should do the job. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-26T15:53:29.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8599" PostTypeId="2" ParentId="8597" CreationDate="2016-04-26T17:12:39.557" Score="3" Body="&lt;p&gt;The first paragraph from &lt;a href=&quot;http://rutcor.rutgers.edu/pub/rrr/reports2012/04_2012.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;your second reference&lt;/a&gt; seems pretty clear.  The first sentence states that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Network interdiction is the monitoring or halting of an adversary’s activity on a network.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your confusion probably arrises from thinking of the word 'network' in too specific a way.  To understand this more clearly, it is probably best to consider a military example, which seems to be &lt;a href=&quot;https://en.wikipedia.org/wiki/Interdiction&quot; rel=&quot;nofollow noreferrer&quot;&gt;the most common usage&lt;/a&gt; for the term.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The map below shows the presence of ISIS in Iraq and Syria during 2014.  The territory under their control can be thought of as a network.  The major cities and towns can be thought of as nodes (vertices), and the connections between them (mostly roads in this case) can be though of as links (edges).  The US military probably has the goal of &lt;strong&gt;network flow interdiction&lt;/strong&gt; which seeks to disrupt the flow of goods through the network in the most effective way possible.  This would prevent the various territories under ISIS control from being able to transport arms, food, and other goods to each other.  Trying to identify the most effective way to do so is one of the problems that network interdiction tries to solve.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hardnoxandfriends.com/2014/10/02/isis-is-just-one-mile-away-from-baghdad/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xnYWX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2016-04-26T17:12:39.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8600" PostTypeId="1" CreationDate="2016-04-26T17:30:24.890" Score="5" ViewCount="5691" Body="&lt;p&gt;The textbook &quot;Materials Science and Engineering An Introduction&quot; mentions that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Precipitation hardening and the treating of steel to form tempered martensite are totally different phenomena, even though the heat treatment procedures are similar; therefore, the processes should not be confused. The principal difference lies in the mechanisms by which hardening and strengthening are achieved.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I read through both chapters but I cannot see the difference. Both processes work by a solubility difference of B in A at different temperatures, then quench it to prevent the solute from diffusing out and then speed up the aging process through heat, which allows precipitates of the solute (any solute or Cementite in the case of Martensite) to form, which impede the movement of dislocations and thus make it very hard, yet through being in a rather ductile matrix improving the overall toughness. Where is the difference between the two?&lt;/p&gt;&#xA;" OwnerUserId="6039" LastEditorUserId="33" LastEditDate="2016-04-26T18:05:26.613" LastActivityDate="2016-04-26T18:20:29.677" Title="Difference between Precipitation Hardening and Tempered Martensite" Tags="|materials|stresses|metals|mechanical-failure|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8601" PostTypeId="2" ParentId="8600" CreationDate="2016-04-26T17:59:36.387" Score="1" Body="&lt;p&gt;The fundamental difference is that precipitation hardening involves the distribution of alloy atoms in the crystal structure of the base metal. This occurs in most alloys including mild steel. Here the difference in size between the atoms in the alloy (or particles in a composite) strains the crystal structure and makes it more difficult for dislocations to move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hardening of steel on the other hand involves the creation of  completely new structure. This is possible because iron can exists both body centred (ferrite)  and face centred cubic crystals (austenite). Austenite is only stable at high temperatures and has a higher carbon solubility than ferrite. On slow cooling austenite will transform back to ferrite and any excess carbon will form cementite (iron carbide)...this is essentially precipitation hardening (forming pearlite) and is the default state of high carbon steels. However if austenite is cooled rapidly by quenching there is not enough time for cementite to form and the dissolved carbon is trapped in a highly stresses structure (martensite). This is what allows high carbon steels to achieve very high hardness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most cases the resulting material will be too hard and brittle to be useful and tempering is used to transform some of the martensite to pearlite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And additional complication is that these processes in steel also produce specific microstructures (eg martensite may show needle-shaped structures under a microscope). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that wasn't enough then there is an additional microstructure in the form of bainite, formed by quenching to an intermediate temperature and holding for a period. Bainite is essentially a more strained and structured form of pearlite (ie ferrite and cementite) with a distinct laminar structure. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-26T17:59:36.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8602" PostTypeId="2" ParentId="8600" CreationDate="2016-04-26T17:59:39.903" Score="4" Body="&lt;p&gt;&lt;strong&gt;Summary: Both processes depend on relative solubility of solute between high and low temperature solvent phases. How each process proceeds at sub-microscopic scale is very different. Precipitation hardening is diffusion-limited, while martensite formation is diffusionless.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precipitation hardening is a &lt;strong&gt;diffusion-limited&lt;/strong&gt; process in which (generally substitutional) solute atoms are first made to dissolve in a solvent above a eutectic or eutectoid temperature, then forced to precipitate from solution below the eutectic or eutectoid temperature. Precipitation is the result of a nucleation and growth process of a secondary phase, which relies on diffusion of solute atoms to nucleation sites. The relative rates of nucleation and growth are controlled by temperature. The size and dispersion of the secondary phase is controlled by time, with coarsening occurring as time proceeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Martensite is the result of a &lt;strong&gt;diffusionless&lt;/strong&gt; process in which (generally interstitial) solute atoms are trapped in a transforming lattice, causing a lattice distortion. In steel, the lattice changes from FCC to BCC when transforming from austenite to ferrite. If the transformation is rapid enough, then there is not enough time for diffusion of carbon and consequent nucleation and growth of the secondary cementite phase. The carbon becomes trapped in the lattice as it changes from FCC to BCC. If carbon is beyond its solubility limit in ferrite, the lattice will eventually distort into a body-centered tetragonal (BCT) configuration. The degree of martensite formation is not controlled by time, but by temperature. Lower temperatures yield greater fraction of martensite formed. Extent of martensite formation is not controlled by diffusion because the lattice transformation to BCT occurs at the speed of sound in the material; much more rapidly than any diffusion process in solids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time-temperature-transformation (TTT) or isothermal transformation (IT) diagram below will help illustrate both concepts. The diagram is for an AISI 1050 steel, which is effectively just 0.5% by weight of carbon, a bit of manganese for preventing sulfur embrittlement, and the remainder is iron. At any temperature above the $A_3$ temperature, all of the carbon is fully dissolved in iron, which is fully austenitic as austenite is the stable solvent phase. Below the $A_3$ temperature austenite begins transforming into ferrite, which is now the stable solvent phase. As the iron transforms, its carbon solubility limit drops from about 0.75% by weight to about 0.02% by weight (see the Fe-C phase diagram below). Carbon therefore begins diffusing out of iron. If the process is allowed considerable time to occur while held at say, 600 C, then pearlite will form. Pearlite is a lamellar microstructure consisting of ferrite, the solvent phase, and cementite, a stoichiometric compound which is about 6.7% by weight of carbon. This is, essentially, precipitation hardening. For many materials, the secondary phase precipitates at the grain boundaries, which is undersirable. That is why a solutionization step is generally required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead the steel is cooled rapidly enough to bypass the &quot;nose&quot; of the curved portion of the diagram, then when it reaches the $M_s$ temperature, martensite will begin to form, as sufficient carbon will still be in the lattice to cause distortion to BCT as it transforms. If cooling continues to $M_f$, then virtually the entire material will have transformed to martensite. Note that room temperature is a bit above $M_f$, so not all of the material will transform to martensite if it is only cooled to room temperature. No matter how long it is held at room temperature, the amount of martensite will not change. In contrast, if we had held the material at 600 C for a long time, the cementite in our pearliet would tend to coarsen and eventually form spherical particles as diffusion proceeds, in a process unsurprisingly called spheroidization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FYwCA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FYwCA.jpg&quot; alt=&quot;AISI 1050 TTT Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image taken from &lt;a href=&quot;http://practicalmaintenance.net/?p=1345&quot; rel=&quot;nofollow noreferrer&quot;&gt;practicalmaintenance.net&lt;/a&gt;, which has a nice discussion of development of TTT diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1nrCu.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1nrCu.gif&quot; alt=&quot;Fe-C Phase Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image taken from &lt;a href=&quot;http://ocw.mit.edu/courses/materials-science-and-engineering/3-11-mechanics-of-materials-fall-1999/related-resources/&quot; rel=&quot;nofollow noreferrer&quot;&gt;mit.edu&lt;/a&gt; open courseware.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-04-26T18:20:29.677" LastActivityDate="2016-04-26T18:20:29.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8604" PostTypeId="2" ParentId="8594" CreationDate="2016-04-26T18:49:27.293" Score="2" Body="&lt;p&gt;To answer your other question about choice of names, it appears that solid solution strengthening may have originally been referred to as solid solution hardening in the literature. See for example &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/000161606390213X&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt; by RL Fleishcher dating to 1963 in Acta Metallurgica, 11.3, pp. 203-209, entitled &quot;Substitutional solution hardening.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precipitation hardening has a fairly storied history which is detailed in &lt;a href=&quot;http://core.materials.ac.uk/search/detail.php?id=2108&quot; rel=&quot;nofollow&quot;&gt;this article&lt;/a&gt;, a lecture on precipitation hardening located at &lt;a href=&quot;http://core.materials.ac.uk&quot; rel=&quot;nofollow&quot;&gt;core.materials.ac.uk&lt;/a&gt;. From the article, Wilm discovered in 1906 that an Al-Cu-Mg alloy was found to have hardened when left to sit at room temperature before testing. The as-yet-unknown process was called age hardening, due to the material increasing in hardness with age. Later work in the late 1930s determined the hardening was due to the formation of what are now known as Guinier-Preston (GP) zones, which are nano-scale arrangements (wires or plates) of solute atoms which, as you noted, impede dislocation motion. It wasn't until the advent of transmission electron microscopy in the 1950s that GP zones were actually directly observed, and the full precipitation hardening process discovered and understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite the fact that the material is strengthened in a materials engineering sense, it appears that the terms age hardening and precipitation hardening have stuck. Oddly, solid solution strengthening appears to have changed from solid solution hardening.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-04-26T18:49:27.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8606" PostTypeId="2" ParentId="8596" CreationDate="2016-04-27T00:49:20.043" Score="3" Body="&lt;p&gt;Stitching it like Chris Johns mentioned would be a good solution. Carbon fiber is typically sold in a &lt;a href=&quot;https://www.acpsales.com/Carbon-Fiber-Tow-and-Cords.html&quot; rel=&quot;nofollow&quot;&gt;tow&lt;/a&gt; (untwisted bundle of fibers), but with a large gauge needle you might be able to get it to work on a standard sewing machine. There will probably be some fuzz to manage as fibers are broken in the process, but would be much easier than by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be a bit more expensive, but a &lt;a href=&quot;http://zoltek.com/products/panex-30/yarns/&quot; rel=&quot;nofollow&quot;&gt;carbon fiber yarn&lt;/a&gt; would likely behave much better in the sewing machine. Also &lt;a href=&quot;http://www.goodfellow.com/catalogue/GFCat4J.php?ewd_token=mizEwEZ308h0A5MIznOqnkOLS2jT11&amp;amp;n=m6PFwHm8y4EKWwVwsXVXFWIWkb17IM&quot; rel=&quot;nofollow&quot;&gt;Goodfellow carbon fiber yarn&lt;/a&gt;. &lt;br&gt;&#xA;Reference book: &lt;a href=&quot;https://books.google.com/books?id=uI-pAgAAQBAJ&amp;amp;pg=PA421&amp;amp;lpg=PA421&amp;amp;dq=carbon%20fiber%20textiles%20sewing%20aerospace&amp;amp;source=bl&amp;amp;ots=SDcsTsxOfC&amp;amp;sig=aAQWTFl0Z0ZjI52CqZhDm3BuBLw&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwia0ISgx63MAhVExWMKHYX3BeoQ6AEIMzAD#v=onepage&amp;amp;q=carbon%20fiber%20textiles%20sewing%20aerospace&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Design and Manufacture of Textile Composites&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are likely a number of contractors that will preform this stitching. Their prices are likely very high, but they may be a good resource if it proves to be more difficult.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.kuka-robotics.com/usa/en/solutions/solutions_search/L_R199_Sewing_of_carbon_fiber_mats.htm&quot; rel=&quot;nofollow&quot;&gt;Kuka Robotics&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://laystitch.com/fibers.html&quot; rel=&quot;nofollow&quot;&gt;LayStitch Technologies&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option would be to pinch the ends of the carbon fiber fabric between graphite, metal, or other material to mechanically hold it (rivets or other). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, less advisable but possible quick fix solution would be &lt;a href=&quot;http://www.oreillyauto.com/site/c/detail/WAL0/35958/N0073.oap&quot; rel=&quot;nofollow&quot;&gt;Muffler Putty (Exhaust Cement)&lt;/a&gt; sold at local auto parts stores.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-04-27T00:49:20.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8608" PostTypeId="1" CreationDate="2016-04-27T01:53:02.220" Score="4" ViewCount="1369" Body="&lt;p&gt;I was considering some theoretical limits on transferring heat from one system at to another to another. It is well known that if you have two systems at temperatures $T_1$ and $T_2$, respectively, with $T_2&amp;gt; T_1$ then you need to provide work to transfer heat to the system at temperature $T_2$. Now, that work is then coming from some different thermodynamic process that we choose not to describe. But leaving that out is actually compromising the ideal solution as extracting work first and then having to dump that at the system at $T_2$ is not necessarily optimal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This led me to consider three systems at temperatures $T_1$, $T_2$, and $T_3$, respectively, with $T_1&amp;lt;T_2&amp;lt;T_3$ where we want to pump heat to the system at temperature $T_3$ but now there is no external source of work. In the limit of infinite heat capacities, we then find that ratio between the heat supplied to the system at $T_3$ and the heat extracted from the system at $T_2$ in the optimal reversible case will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta = \frac{T_3}{T_2}\frac{T_2 - T_1}{T_3 - T_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See below for the derivation. So, in principle at least, one can construct a heat pump that heats a house using outside environments at two different temperatures without the need for work to be provided. E.g. in winter the outside temperature near the ground could be -10 °C while at some depth below the ground it could be 10 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is then how to realize this in practice. Presumably one would need some sort of a Stirling machine. What can one expect about the amount of heat that can be extracted from the environment per unit time in some realistic setting?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To derive the efficiency formula, one can consider a Carnot process that extracts work from the systems at temperature $T_1$ and $T_2$ and another Carnot process that acts as a heat pump between the systems at temperatures $T_2$ and $T_3$ using the work provided for by the first Carnot process. However, a simpler derivation is to consider any arbitrary process that adds an amount of heat of $q_i$ to system $i$, the First law then implies that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^3 q_i = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reversibility implies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{i=1}^3 \frac{q_i}{T_i} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for the $\eta \equiv -\frac{q_3}{q_2}$ then leads to the formula given above.&lt;/p&gt;&#xA;" OwnerUserId="6206" LastEditorUserId="368" LastEditDate="2016-04-27T23:14:59.607" LastActivityDate="2017-05-26T16:02:30.890" Title="Heat pump driven by a Stirling engine that requires no external work" Tags="|thermodynamics|heating-systems|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8609" PostTypeId="1" CreationDate="2016-04-27T05:22:45.843" Score="1" ViewCount="2547" Body="&lt;p&gt;In which case speed of car will be faster: if wheels are connected to dc motor directly or with gears?&#xA;We are making battery powered car in which case our car will be faster?&lt;/p&gt;&#xA;" OwnerUserId="6209" LastEditorUserId="6209" LastEditDate="2016-04-27T17:46:46.103" LastActivityDate="2016-07-26T19:46:08.107" Title="connecting wheels to dc motor directly or with gears" Tags="|mechanical-engineering|electrical-engineering|automotive-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8610" PostTypeId="1" AcceptedAnswerId="10353" CreationDate="2016-04-27T06:04:40.573" Score="3" ViewCount="264" Body="&lt;p&gt;How is the position of a train modeled in a computer control system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of two basic approaches but I am not sure which (if any) are used in the real world.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Lat, Long&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Ex: (-42, 23)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the train is modeled by its real world  position. The only issue I could see with this is extra work would have to be done at some stage to determine what track the train is on.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Distance, Track&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Ex: (20m, Track 34-B)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the train would be modeled by its distance from the start of the track, and the track it was on. This isn't exactly as intuitive right off the bat but I think it makes more sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only ask because earlier I was reading some about PTC and as I was reading this popped into my head.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Edit:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As brought up by ratchet freak it most likely depends on the sensors on a track.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is anyone out there who is or currently was part of the development of such a PTC system who could speak to which method is used.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if there is a link to a detailed specification of a PTC system (May exist because the conversion of railways is partially managed by the government).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is probably something similar to (Distance, Track) for data gathered by on track sensors and then GPS is used as a fallback which can be converted into (Distance, Track).&lt;/p&gt;&#xA;" OwnerUserId="6211" LastEditorUserId="1586" LastEditDate="2016-05-09T23:47:18.327" LastActivityDate="2016-06-30T23:57:29.247" Title="How to notate the position of a train?" Tags="|civil-engineering|software|rail|transportation|" AnswerCount="2" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8611" PostTypeId="2" ParentId="1794" CreationDate="2016-04-27T08:14:27.253" Score="4" Body="&lt;p&gt;The answer is &lt;strong&gt;no&lt;/strong&gt;, and the phrasing is key.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;assuming large quantities of reasonably priced materials&quot; (are available)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then yes, it would certainly be possible to manufacture a knife made from a single section of diamond. As @nick-t commented, &lt;a href=&quot;https://en.wikipedia.org/wiki/Diamond_knife&quot; rel=&quot;nofollow&quot;&gt;Diamond knives exist&lt;/a&gt;&#xA;and are used in ultramicrotomy, the sectioning of a substance into extremely thin slices for use in transmission electron microscopy. The shaft is not diamond in this case, and the diamond blade is surrounded by metal as a mounting bracket, but these are solvable differences which mostly exist for cost reasons. Assuming large quantities of gem-quality diamond are reasonably priced and available, no problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, isn't solvable:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...a &lt;em&gt;functional&lt;/em&gt; knife...    &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(emphasis added)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamond doesn't create a functional knife. Ceramic knives made of aluminum oxide or other very hard materials exist, you have likely used them. They are not good (compared to steel) as they exhibit microchipping almost invariably. The same problem vexes the users of glass knives and obsidian ones; diamond knives are intended to delay that problem, and do, but this is in the hands of professionals doing surgery, or in a machine. In a normal, out-in-the-world usage scenario, a diamond knife would fail near-instantly. The crystal lattice of the diamond, which gives the diamond its hardness, also causes this fragility; so the problem is inherent to diamonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.catra.org/pages/products/kniveslevel1/slt.htm&quot; rel=&quot;nofollow&quot;&gt;CATRA testers&lt;/a&gt; are widespread in the knife industry, at least among larger manufacturers. They are also widely known to be inaccurate in testing some steels, and especially ceramics; the problem arises because a human cannot cut as precisely as a machine can, and cannot entirely prevent lateral forces. A diamond knife would no doubt score a wonderful result in CATRA testing, as do ultrahard steels; but in reality, they would not be functional.&lt;/p&gt;&#xA;" OwnerUserId="6212" LastActivityDate="2016-04-27T08:14:27.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8612" PostTypeId="2" ParentId="8609" CreationDate="2016-04-27T11:52:46.493" Score="1" Body="&lt;p&gt;This depends on the total gear ratio between motor and wheels. You can obtain faster speeds at the wheels but then the torque will be lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two gears ($i=1, 2$) with $N_i$ teeth, following holds for the speeds $\omega_i$ and torques $T_i$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;${\omega_2\over\omega_1}={N_1\over N_2}={T_1\over T_2}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3633" LastEditorUserId="3633" LastEditDate="2016-04-27T11:57:55.393" LastActivityDate="2016-04-27T11:57:55.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8613" PostTypeId="1" AcceptedAnswerId="8718" CreationDate="2016-04-27T13:56:56.740" Score="6" ViewCount="1773" Body="&lt;p&gt;I have acquired an old loading dock built in 1930 that has toe-bearing rafters in addition to undersized purlins.  As you can imagine, many of the rafters have sagged.  Also, because of the toe-bearing design of the rafter, the building has no eaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many articles about avoiding toe-bearing design.   But there aren't many articles about remedying the situation in retrospect.   Here is a drawing of the existing situation.   The small white block in the middle of the rafter is a purlin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XR0n7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XR0n7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here are some questions and thoughts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Inexpensive Fix&lt;/strong&gt; - The articles state that you could use some type of joist hanger to possibly fix the problem, but it's not clear to me at all what type of joist hanger would help me.  This would be a low cost way to fix the issue. I could:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Jack up the sagging rafters&lt;/li&gt;&#xA;&lt;li&gt;Sister up a new rafter, and &lt;/li&gt;&#xA;&lt;li&gt;Use a joist hanger to help prevent cracks in the re-entrant corner.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is there a Simpson Strong-tie (or some other brand) hanger that would help solve this issue?&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Medium cost Fix&lt;/strong&gt; - I would really like eaves, so I was thinking another much more involved approach might be to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Support the existing rafter with a false wall,&lt;/li&gt;&#xA;&lt;li&gt;Lower the wall by half a block, and then&lt;/li&gt;&#xA;&lt;li&gt;Sister up a new rafter that has a proper birdsmouth cut that will allow the rafter to sit on the heel.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The sistered rafter could now extend beyond the wall, giving me an eave.  Here's a picture of what I am thinking about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NZzJa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NZzJa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:  Ugh, it's not clear from the angle I posted, but the sistered rafter is sitting on it's heel on the lowered block/sill plate.   It's not hanging in the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A properly engineered sistered rafter can take the load from the purlin on down, can't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;  Based on a comment, here are more details:  The existing rafters are 2x6, 24&quot; centers.   The pitch is 4/12.   Just from visual inspection, the re-entrant corner on the birdsmouth appears to be right around half the thickness of the 2x6, so it seems like it's effectively a 2x3.  I believe the purlin is about 9' from the birdsmouth, but I would need to remeasure that, if it's critical.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Most expensive fix&lt;/strong&gt; - Replace the entire roof with prefabbed trusses.  Seems like overkill, but that sort of depends on the answers to the questions above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Better/cheaper/easier ideas?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3830" LastEditorUserId="3830" LastEditDate="2016-04-27T21:18:10.693" LastActivityDate="2020-03-27T06:40:51.933" Title="Fixing a Roof with Toe-bearing Rafters" Tags="|structural-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8614" PostTypeId="2" ParentId="8610" CreationDate="2016-04-27T15:37:31.793" Score="3" Body="&lt;p&gt;In most cases a position on the track will be most useful as GPS probably won't have high enough resolution to indicate which line it is on if there are several tracks in parallel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also many signalling systems work by sectors where only one train is allowed to enter a given stretch of track at any given time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly the precise location doesn't really matter that much as you know that it is on a track so its is much more useful to know which track, which direction and how far along. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly it is a lot easier to convert from relative track position to absolute position than the other way around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something else to consider is that, in the UK at least the motorways have location marking system based on marker posts every 100 metres (or yards ?) which gives the number and distance to the next junction. This means it is very easy for anybody to give the emergency services or breakdown assistance a very accurate location for any incident and similarly emergency vehicles can tell exactly where they are relative to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be astonished if most railway networks don't use a similar system. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-27T15:37:31.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8615" PostTypeId="2" ParentId="2908" CreationDate="2016-04-27T16:07:47.447" Score="1" Body="&lt;p&gt;I debated between Answer or Comment on this, but ultimately I think it is more of an answer - albeit an imperfect answer. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It sounds to me like the primary problem you are trying to solve is preventing material transfer from the balls/cups etc. during the ball milling process. I don't think you necessarily need to change materials, and perhaps could even get away with just adding &lt;strong&gt;a PVD or CVD coating&lt;/strong&gt; to the existing components. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;DLC (diamond-like-carbon) coating pops to mind first,&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;however I think there are many coatings that could potentially serve you well.  DLC coatings are very hard and wear resistant (as the &quot;diamond-like&quot; name would suggest).  They do not flake or chip off, and while I don't know the end use or the nature of the &quot;no contamination&quot; requirement, these coatings are completely inert to almost all organic and synthetic chemicals, and completely inert to the human body as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this application, I would think a ta-C or possibly a ta-C:H coating could work well.  Another DLC that is supposed to be extremely hard and wear resistant, but that I have no personal experience with, is UNCD (ultrananocrystalline diamond). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also find more traditional tool coatings, such as TiN or TiAlN could work well for you - I couldn't make a recommendation as to the best coating for your application.  I'm not a coatings engineer, just an engineer that has had good experience using these types of coatings for my own applications in the past :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The main benefit of 440C is it's high hardenability&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;With a maximum hardness of around 60 rockwell C, 440C is a stainless that can rival many tool steels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like you are using this material in the annealed condition, which is questionable to me. This material is typically chosen for it's high hardenability - it's not (to the best of my knowledge) commonly used in the annealed state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like you chose this material because it's commonly used in other similar designs; I wonder what condition is this material used in those designs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can get your hands on some 440C components from one of those other designs I would do a rockwell test to see whether or not it's heat treated.  I would put money on it being heat treated. Annealed material will probably be in the 20's-30's rockwell C, while heat treated material is going to be in the 50's.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do intent do to a rockwell test, try doing in on a component with a flat as spheres can be difficult to test and give inaccurate readings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a place to start with heat treat, I find &lt;a href=&quot;http://cartech.ides.com/datasheet.aspx?i=103&amp;amp;e=72&quot; rel=&quot;nofollow&quot;&gt;Carpenter's data sheets&lt;/a&gt; are usually pretty reliable.  Their recommendations for 440C are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;HARDEN:&lt;/strong&gt;  Heat to 1850/1950ºF (1010/1066ºC); soak; quench in warm oil or cool in air. Hardness will be ≈60HRC. Do not overheat or you will not be able to attain maximum hardness.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;TEMPER:&lt;/strong&gt; To remove peak stresses and yet retain maximum hardness, temper at least one hour at 300/350ºF (149/177ºC).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you are looking to heat treat, I would probably start there. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;I do not think 316 would be a good choice&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As it's a much softer (gummier) material than 440C, and I think it would exasperate the material transfer problem. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Now I said this was an imperfect answer,&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;because I don't know that it directly answers your question.  It offers an avenue to a possible solution, however it is incomplete as you'll need to discuss with a coating expert the exact needs of your application and see what they say. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, &lt;strong&gt;&lt;em&gt;I have no knowledge of how these coatings perform at cryogenic temperatures&lt;/em&gt;&lt;/strong&gt;, or what coating would work best with the abrasives used in your ball milling process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I have used DLC coatings to solve some unique problems where I work, and they have allowed me to do things I don't think I could have found an alternative for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also wanted to say that finding a reputable and reliable coating vendor was the hardest part of adapting this technology for me.  As that can be quite a daunting challenge, I offer a suggestion only as a place to start, and with no further endorsement other than I've personally had good experience with Oerlikon Balzers.  I do not know the policy on recommending vendors, and I have no affiliation with any vendors - so feel free to edit this answer to remove the vendor name if it violates any policies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of who you choose to do business with, &lt;strong&gt;&lt;em&gt;I would highly recommend speaking with one of their applications engineers&lt;/em&gt;&lt;/strong&gt; about the specific needs for your application, and seeing what coating they recommend. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Closing thoughts&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;PVD/CVD coatings are tribological in nature - they affect the interface with other  materials, but do not change the properties of the base substrate.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we talk about bearing design, there is a property commonly referred to as embedability.  This basically refers to a bearing materials ability to absorb (or embed within itself) foreign materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing more about the nature of your design, my gut tells me that you would not want this to happen, as it would mean your balls/cups etc. would be &lt;em&gt;embedding&lt;/em&gt; some amount of the alloys that you were trying to grind.  It seems unlikely to me that this would be desirable.  This is another reason I think 316 would be a poor choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reduce embedability, you want a harder substrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendation to you would be to heat treat the 440C components first, then apply a PVD coating afterwards, to obtain the best combination of properties for your application.  I would definitely talk to a coatings engineer about what coatings will work for your application (e.g. temperatures, material compatibility, etc.)&lt;/p&gt;&#xA;" OwnerUserId="2598" LastEditorUserId="2598" LastEditDate="2016-04-27T19:43:06.363" LastActivityDate="2016-04-27T19:43:06.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8616" PostTypeId="2" ParentId="8609" CreationDate="2016-04-27T16:09:47.960" Score="0" Body="&lt;p&gt;Your first consideration is whether the motor provides enough torque to both allow the vehicle to start from a standstill ( a practical vehicle will also need to be able to do this on a moderately steep slope) and to provide acceptable acceleration across a range of speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often top speed is the icing on the cake in terms of performance as torque at moderate speeds is generally much more important for overall performance. Similarly there is no point in having a high top speed if you can only achieve it on a 10 mile long drag strip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand there is no point in having a lot more torque than traction (although a bit more is usually better than a bit less). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric motors have fairly flat torque curves so for a light vehicle direct drive is not grossly unreasonable and the final drive ratio can be fine tuned a bit by the tyre diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also depends a bit on what you mean by 'fast' as the performance requirements for something like land speed record attempt are very different from a karting circuit or rallycross. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-27T16:09:47.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8617" PostTypeId="1" CreationDate="2016-04-27T16:42:27.613" Score="2" ViewCount="1062" Body="&lt;p&gt;So we are developing a plastic case, where we want light to be diffused through the case. The light is kept 2-3 mm away from the inside wall. &#xA;I wanted to know what thicknesses allow passage of light, &#xA;and also what colors allow better diffusion. &#xA;I have seen a couple of products which are in White. &#xA;But otherwise there is usually a light pipe for the diffusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Material most likely will be ABS+PC or just ABS.&#xA;presently design is 1mm. we are thinking that particular area reduce thickness a bit and keep a provision for a cylinder to block light from diffusing to other areas. &#xA;Any idea on what mix would be the best? &#xA;or if color will also affect the diffusion.  I cant recollect any black abs part which diffuses light. Only white I have seen.&lt;/p&gt;&#xA;" OwnerUserId="6215" LastEditorUserId="6215" LastEditDate="2016-04-28T17:23:51.757" LastActivityDate="2021-12-05T16:05:54.793" Title="thickness and other specifications PC + ABS for allowing light through?" Tags="|mechanical-engineering|design|plastic|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8618" PostTypeId="2" ParentId="7524" CreationDate="2016-04-27T17:53:44.553" Score="2" Body="&lt;p&gt;You didn't specify cost of glue. If the glue is of no cost (other than end to end joints as you mentioned) then my inclination is to think outside the box here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could shred the toothpicks into fibers (pull them apart, grind them up, whatever) and then use the glue to create a composite matrix using the fibers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could get a lot of stringy fibers from very few toothpicks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know all the rules, but If you can bring stuff in other than just a hairdryer, then I would bring in a food saver type vacuum bagger and some 10 or 20 lbs foam, and some Vaseline. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carve a negative form for the bridge into the foam and coat it with vaseline (as a mold release agent). Mix up the wood fibers and glue and spread this mixture out into the form.  Slip the whole thing into a food saver bag and turn on the vacuum pump to apply pressure while it cures.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the hair dryer to apply heat over the whole thing to help cure while it's in the bag. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, once your WFRG (wood-fiber-reinforced-glue) bridge is cured you would remove the whole thing from the bag, and pop the bridge out of the mold.  Done.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I didn't see time limit listed in your question, but I would imagine all of this would be &lt;em&gt;way&lt;/em&gt; faster than building a more traditional bridge gluing individual toothpicks together, so I assume the time allotment would be sufficient.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, depending on the rules about bringing stuff in, you might be able to make the mold ahead of time and bring it with you, so all you have to do during the construction phase is shred wood, mix with glue, apply, and cure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something I learned at an early age is it's better to ask forgiveness than ask permission.  If the rules don't specifically ban you from bringing in this outside stuff then I would say it's fair game.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I believe this fiber-reinforced-matrix construction could work well.  However I've never done it myself, so I'm not positive.  It's similar in concept to fiberglass or carbon fiber, GRFC, or FRP, and conceptually would work the same way and offer the same benefits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally, fiber-reinforced products work because the fibers are suitably strong (overall, not individually) to resist great tensile loads, while the matrix binder (in this case the glue) provides adequate support to hold the fibers in place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it's your homework assignment, I would do your research on FRP's (fiber reinforced plastics) and other composite materials to determine whether or not this will indeed work for you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An H-Beam or I-Beam form I think would prove suitable.&lt;br&gt;&#xA;You could also look at adding gussets and ribbing within your foam mold to increase strength and rigidity at areas of stress concentration. &#xA;Looking at the rules you've written, I don't see how this would violate any of them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have access to 3d modeling software and FEA software at your school, you could do an optimization study to give you a good idea of how to build the bridge and where to add gussets to assure structural integrity.  Many FEA suites offer the ability to apply composite materials.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can look at the way many plastic playsets and slides are constructed.  These are typically roto-molded plastic, not composites, but the beam construction and ribbing could give you some hints as to the approach to take. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Finally, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're mileage may vary, but I'll tell you this - if I was a teacher, who did this project every year and watched countless students create countless iterations of the same basic bridge design using girder construction, I would be pretty impressed if someone brought in a completely fresh approach to the problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This attitude of thinking differently has served me well in my career, it may or may not work for you.  Just a thought. &lt;/p&gt;&#xA;" OwnerUserId="2598" LastEditorUserId="2598" LastEditDate="2016-04-27T20:21:48.573" LastActivityDate="2016-04-27T20:21:48.573" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8619" PostTypeId="1" AcceptedAnswerId="8623" CreationDate="2016-04-27T18:13:24.710" Score="22" ViewCount="50161" Body="&lt;p&gt;I have assembled a small DIY drip-irrigation system for my terrace garden. Please have a look at the attached image. I switch on a small pump to start the drip-irrigation system and then switch it off. But even after that, the water keeps flowing through the system and stops only when I physically lift the pump out of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I stop this water flow without requiring any physical action?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that I plan to automate the switch-on/off of the pump using a timer so that it functions without requiring my physical presence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dcnMT.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dcnMT.jpg&quot; alt=&quot;Drip Irrigation System Setup&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;========================================================================&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried the suggestion of &quot;Making a pinhole in the highest point of the pipe.&quot; It worked like a charm and solved the issue I was facing. Completely loved it, more so, because it does not involve procuring new stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also like the option of using a solenoid-valve, but have not tried it yet. Will use it when I install such a system again where the tank is at a considerably higher place than the plants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you everyone who took the effort to write an answer, and that too with detailed explanations.&lt;/p&gt;&#xA;" OwnerUserId="6217" LastEditorUserId="6217" LastEditDate="2016-05-04T14:52:17.570" LastActivityDate="2024-02-24T01:49:18.940" Title="How to stop water flow in a siphon?" Tags="|fluid-mechanics|siphon|" AnswerCount="12" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8620" PostTypeId="2" ParentId="8619" CreationDate="2016-04-27T18:20:13.157" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How can I stop this water flow without requiring any physical action?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could add in a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B007N0J98E&quot; rel=&quot;nofollow noreferrer&quot;&gt;solenoid valve&lt;/a&gt; downstream of your pump that switches off when the pump shuts off.  These can be found for a reasonable price online.  You should be able to wire it to the same switch as the motor and once de-energized, it will close itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bDsaYm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bDsaYm.jpg&quot; alt=&quot;solenoid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2016-04-27T18:20:13.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8621" PostTypeId="2" ParentId="8619" CreationDate="2016-04-27T18:20:22.303" Score="8" Body="&lt;p&gt;There are two ways to stop a siphon:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Put a valve inline with the siphon that simply blocks the flow, &quot;freezing&quot; the state of the system. Flow will resume as soon as the valve is opened again, without needing the pump to re-prime the siphon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Put a valve in the top of the siphon that allows air into the tubing, breaking the suction that keeps the siphon flowing. Close the valve and operate the pump to restart the flow.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the second option, you can get an automatic valve, known as a &quot;siphon breaker valve&quot; or &quot;antisiphon valve&quot; that admits air into the system as soon as the pressure in the tubing drops below atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a third option is to rearrange things so that you don't create a siphon in the first place.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2016-04-27T20:34:23.673" LastActivityDate="2016-04-27T20:34:23.673" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8622" PostTypeId="2" ParentId="8619" CreationDate="2016-04-27T18:25:07.597" Score="7" Body="&lt;p&gt;An alternative answer which requires no physical action &lt;em&gt;in the system&lt;/em&gt; is to move the tank or the garden so that the full water line level is below the level of all of the outlets. You may have to scale up your pump for the additional head required to lift the water. I am also assuming the outlets aren't submerged within the garden after irrigation, since then you'd be flooding your garden. If this is the case, then you'll have to move the tank lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, this requires an initial physical action to set up the system this way.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-04-27T18:25:07.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8623" PostTypeId="2" ParentId="8619" CreationDate="2016-04-27T19:33:49.177" Score="25" Body="&lt;p&gt;You want to let air into the pipe when you switch the pump off, without letting water out. Here's a few ways to do this.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Make a pinhole in the highest point of the pipe. You will lose a bit of water this way, but if it is above the tank, the water will drip back in (provided it doesn't spray too far.) You could even put it just under the tank lid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Install a tee and riser at the highest point of the pipe. this will need to be high enough to avoid the pump pressure pumping water out of the riser.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Install a vacuum breaker valve. This is basically a non-return valve, letting air into the pipe but not letting water out.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;direct the outlet of the pump to a high point via a short tube. Collect the water in a funnel and feed the water from the funnel to your plants. Funnels can be made from the top of plastic bottles, cut off and inverted (you may have problems with thread compatability though.) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Knowing the backpressure in the line will help make the best selection, especially for options 2 and 4. I assume that for good distribution to your various plants, you have restrictions at the various outlets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For option 1 (the cheapest option here) your pinhole should be similar in size to one of the plant outlets. This should avoid wasting too much water, while allowing the incoming air to break the syphon (if it is too small and/or the flow of water is fast, the current may be strong enough to carry all the air bubbles away. A wider pipe section at the high point will help avoid this.)&lt;/p&gt;&#xA;" OwnerUserId="1575" LastEditorUserId="1575" LastEditDate="2016-04-27T21:03:24.213" LastActivityDate="2016-04-27T21:03:24.213" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8624" PostTypeId="1" AcceptedAnswerId="9104" CreationDate="2016-04-27T20:22:54.610" Score="5" ViewCount="703" Body="&lt;p&gt;I have been working with an old code for modeling incompressible, 2D viscous flow out of a tank to understand a chemical process.  There isn't much documentation and I'm not a fluid dynamicist, so I have been trying to check that things are working as they should by comparing results with those from another code.  The comparison is great when I use a velocity inlet condition.  However, there is a major discrepancy when I use a pressure inlet condition.  In both cases, the outlet pressure is zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have spent several weeks trying to figure out what the problem is with the pressure inlet condition (and have an active question about it) because I have pressure inlet values that I want to use in my problem.  I don't know velocity inlet values.  But I have to acknowledge I've hit a wall with figuring out my pressure inlet boundary condition problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be a dumb question, but I would really appreciate it if someone would help me confirm if I can transform my pressure values into an inlet velocity condition.  If I can, maybe I can simply use an inlet velocity and my problems will be over!  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that:&#xA;$$P = \dfrac{1}{2}\rho U^2 + \rho gz$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ is fluid density, $U$ is velocity, $g$ is the acceleration of gravity, and $z$ is depth of the tank inlet below the tank outlet at 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So if I simply plug in my inlet pressure value and rearrange, will I get a valid inlet velocity that I can use for a velocity boundary condition?&lt;/strong&gt;  My concern is that I'm dealing with viscous flow and I think this expression is related to the Bernoulli equation, which does not account for viscous flow.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm right and I can't use this expression to calculate an inlet velocity, does anyone know if there's an alternative?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I say I probably can't is that I've tried it and the results from the two codes don't match.  I'm just trying to figure out where my problem is--if it's likely to be a bug in one of the codes, which code is the problem, or if I'm making some mistake when inputting my boundary condition values.&lt;/p&gt;&#xA;" OwnerUserId="1416" LastEditorUserId="1416" LastEditDate="2016-05-30T08:25:37.233" LastActivityDate="2016-06-29T20:27:06.983" Title="How can I translate a pressure boundary condition to a velocity boundary condition for incompressible, viscous flow?" Tags="|fluid-mechanics|statics|mathematics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8626" PostTypeId="1" CreationDate="2016-04-28T03:39:57.837" Score="5" ViewCount="190" Body="&lt;p&gt;What de facto standards or otherwise exist that prescribe a style for freehand technical lettering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the ubiquity of CAD there is probably a diminished need for standardized handwriting.  Technical lettering was not part of my undergraduate engineering training yet was frequently mentioned and occasionally required by professors.  When &quot;required&quot;- the criteria presented were few; simply capital case, sufficiently 'black', and ruled (drawn on/between lines).  I am aware that true or traditional technical lettering has much greater complexity such as prescribing the order of strokes for drawing characters and their proportion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally have sought to apply block lettering for readability and general aesthetic in handwritten drafts, but for lack of formal training apply personal habits of using dashed '7's and Z's and serif'ed 'I's- all of which I suspect are faux pas.  Nonetheless I'm curious as to where to look for historical consensus to develop my drafting.  &lt;/p&gt;&#xA;" OwnerUserId="5783" LastActivityDate="2016-04-29T01:25:40.787" Title="(de facto) Standards for freehand technical lettering?" Tags="|drafting|standards|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8627" PostTypeId="1" AcceptedAnswerId="13373" CreationDate="2016-04-28T04:07:31.670" Score="1" ViewCount="334" Body="&lt;p&gt;I am learning by myself structural dynamics but confused about the concept of scaled ground motion data. I can't get the reason why the acceleration data should be scaled. I mean, isn't it better to use the original data &quot;as it is&quot; to compute the response of structures so that the results can reflect actual response of a specific structure? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT :&lt;/strong&gt; Is it the case that when we design a structure, scaled data should be used, while unscaled data can be utilized when we conduct post-seismic-analysis of a structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please summarize the situations when we should use scaled or unscaled data?&lt;/p&gt;&#xA;" OwnerUserId="6225" LastEditorUserId="6225" LastEditDate="2016-04-28T14:20:08.957" LastActivityDate="2017-01-22T03:47:00.487" Title="Why is it useful to scale seismic ground motion data" Tags="|structural-engineering|dynamics|seismic|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8628" PostTypeId="2" ParentId="8619" CreationDate="2016-04-28T08:01:54.297" Score="2" Body="&lt;p&gt;If your pump lifts the water into a hopper, from which it flows under gravity to the plants, this will provide the air gap mentioned in some of the other answers.  The hopper could basically be a funnel pushed into the end of the hosepipe.  It should be sized and positioned (height) so that it doesn't overflow too much, and over the top of the water container so that any overflow falls back in.  This could be quite an easy retrofit depending on how your system is set up already.&lt;/p&gt;&#xA;" OwnerUserId="433" LastActivityDate="2016-04-28T08:01:54.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8630" PostTypeId="2" ParentId="8617" CreationDate="2016-04-28T12:31:16.363" Score="1" Body="&lt;p&gt;The answer isn't as simple as thickness alone. The material must be designed in such a way as to allow light but not images to pass through whatever designed thickness is acceptable. First figure out the mechanical requirements, then select a material which has a crystallization fraction that is both manufacturable and allows the required amount of light through at that thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unaware of a general formula for thickness versus transmission of light through polymers since processing history plays a large role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some tradeoffs to consider...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Increasing thickness means slower cooling rates and thus larger crystal fraction, more crystallinity, and less transmission of more scattering)&lt;/li&gt;&#xA;&lt;li&gt;Less crystallinity means more brittle behavior, lower material stiffness, and higher thermal distortion and creep rates&lt;/li&gt;&#xA;&lt;li&gt;Thicker walls mean more global strength and stiffness (as opposed to material strength and stiffness)&lt;/li&gt;&#xA;&lt;li&gt;Choice of material affects the ability to form crystallites, and all thermal and mechanical properties&lt;/li&gt;&#xA;&lt;li&gt;Opaque filler can be used to reduce transmission but can increase brittleness and strength, may require testing to optimize.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A common choice for electronics housings is ABS which always scatters light due to in-situ rubber microparticles and high impact toughness and dimensional stability at room temperature. Again, due to processing history effects, there is no general formula for thickness versus transmission, so you will probably need to test.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-04-28T12:31:16.363" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8631" PostTypeId="1" CreationDate="2016-04-28T12:36:28.033" Score="1" ViewCount="1829" Body="&lt;p&gt;What are the dimensions for the cutouts in an electrical outlet faceplate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to add several outlets in a sheet metal covering but I cannot find the dimensions or a dxf to use for the shape.  The closest I've found is the following image from &lt;a href=&quot;http://www.archtoolbox.com/materials-systems/electrical/electricalfaceplates.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.archtoolbox.com/materials-systems/electrical/electricalfaceplates.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem is that it gives rectangular dimensions without providing the corner radius and spacing between the two outlets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BlUIT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BlUIT.png&quot; alt=&quot;faceplate dimensions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6227" LastActivityDate="2016-04-28T12:36:28.033" Title="dimensions for electrical outlet faceplate cutout" Tags="|design|" AnswerCount="0" CommentCount="3" ClosedDate="2016-04-29T03:21:43.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="8632" PostTypeId="2" ParentId="8626" CreationDate="2016-04-28T15:44:27.647" Score="2" Body="&lt;p&gt;I don't know if there is a formal &quot;standard&quot; for lettering on hand drawings.&#xA;The best I could tell you to do is consult an older drafting text or the book &quot;Architectural Graphic Standards&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0471700916&quot; rel=&quot;nofollow&quot;&gt;http://www.amazon.com/Architectural-Standards-American-Institute-Architects/dp/0471700916&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many draftsmen had their own personal styles for lettering, and as long as it was legible, there were no problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some drafting shops would use lettering templates, like Leroy lettering drafting tools, to maintain a standard style of lettering for their drawings.  These templates could be used for pencil or ink drawings.  Some of these tools can be found second-hand on sites like e-bay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Technical_lettering&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Technical_lettering&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-04-28T15:44:27.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8633" PostTypeId="2" ParentId="8619" CreationDate="2016-04-28T17:22:50.410" Score="1" Body="&lt;p&gt;How about using two water tanks.  Pump sits in tank 1 and pumps twice a day for 5 minutes into tank 2.  Tank 2 has the siphon connected and waters the plants via drip irrigation as before.  The drip irrigation will only drip as much water as the pump pushes from tank 1 to tank 2 in the five minute interval. The siphon part can remain 'charged' as the water level in tank 2 reaches the water level at plant height. &lt;/p&gt;&#xA;" OwnerUserId="6231" LastActivityDate="2016-04-28T17:22:50.410" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8634" PostTypeId="2" ParentId="306" CreationDate="2016-04-28T18:27:32.060" Score="1" Body="&lt;p&gt;To add to other answers, pneumatic tire absorbs most of the minor impacts smoothing the load variations on studs, control arm and bearings. Once fitted with hard tire, you will have to redesign the suspension to cope with much more instantaneous stress from small bumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when traveling over gravel, pneumatic tires have the most contact patch due to absorption of high spots(gravel). &lt;/p&gt;&#xA;" OwnerUserId="6205" LastActivityDate="2016-04-28T18:27:32.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8635" PostTypeId="2" ParentId="2980" CreationDate="2016-04-28T18:38:33.403" Score="5" Body="&lt;p&gt;To add to @ratchetfreak's answer, there 2 more reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Manufacturing process&lt;/strong&gt;&#xA;It's much easier and more reliable to make a continuous surface that us bevelled, that an angled surface from rubber material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Stress distribution&lt;/strong&gt;&#xA;For the same reason why airplane windows are rounded, car tires are beveled. Stress is much better distributed when surface s continuous. By introducing angles, you would concentrate stress in those areas leading to premature failure. &lt;/p&gt;&#xA;" OwnerUserId="6205" LastActivityDate="2016-04-28T18:38:33.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8636" PostTypeId="1" CreationDate="2016-04-28T18:44:02.633" Score="2" ViewCount="522" Body="&lt;p&gt;Currently in the process of developing a 3d printed prosthetic hand and I am in need of advice on the type of rubber (or other material) we should use at the finger tips to give it more grip. The hand is capable of closing with a good amount of force, but it has a hard time holding heavier solid objects (like a glass cup, for instance) due to the lack of friction. I figure we only need small patches of some material at the contact points on each finger tip and potentially some in the palm, but I'd like to get insight from people who know more than I do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are considering using the inner tube of a bicycle and cutting it into small patches, but we are not sure if we would be better off with something thicker to allow the clutched objects to sort of sink in to the rubber more.&lt;/p&gt;&#xA;" OwnerUserId="6234" LastEditorUserId="6234" LastEditDate="2016-04-29T01:43:53.483" LastActivityDate="2016-05-02T08:09:23.383" Title="Easily accessible material for maximum grip on a prosthetic hand?" Tags="|materials|3d-printing|biomedical-engineering|biomechanics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8637" PostTypeId="1" AcceptedAnswerId="9053" CreationDate="2016-04-28T18:53:31.517" Score="4" ViewCount="4595" Body="&lt;p&gt;Some time ago, I was going through some exercises and I came across a odd question, for which my atomic species degrees of freedom did not match my degrees of freedom using molecular species. I showed my lecturer and he couldn't see why this was the case. The (summarized) question is as follows:&#xA;2 reactions occur in a reactor: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\text{C}_2\text{H}_6\text{O} + \text{H}_2\text{O} &amp;amp;\rightarrow 2\text{CH}_3\text{OH} \\&#xA;2\text{CH}_3\text{OH} &amp;amp;\rightarrow \text{C}_2\text{H}_4 + 2\text{H}_2\text{O}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input feed of water and $\text{C}_2\text{H}_6\text{O}$ is specified and the selectivity of the first equation. Reactions do not go to completion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there are 4 unknowns ($\text{H}_2\text{O}$, $\text{C}_2\text{H}_4$, $2\text{CH}_3\text{OH}$, and $\text{C}_2\text{H}_6\text{O}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing the atomic analysis:&#xA;$$\text{DoF} = 4 - 3(\text{C},\text{H},\text{O}) - 1 (\text{Selectivity}) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing the molecular species:&#xA;$$\text{DoF} = 4 + 2 - 4(\text{H}_2\text{O}, \text{C}_2\text{H}_4, 2\text{CH}_3\text{OH}, \text{C}_2\text{H}_6\text{O}) - 1 (\text{Selectivity}) = 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you see where the problem is?&lt;/p&gt;&#xA;" OwnerUserId="6233" LastEditorUserId="1832" LastEditDate="2016-04-28T19:02:16.983" LastActivityDate="2016-05-28T20:09:44.523" Title="How does one calculate degrees of freedom in a reactor with parallel reactions?" Tags="|chemical-engineering|mathematics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8638" PostTypeId="1" CreationDate="2016-04-28T19:02:44.090" Score="4" ViewCount="787" Body="&lt;p&gt;I'm not an engineer, but I build custom audio equipment and systems as a hobby. My question to you is: from a computer power supply (PSU) which converts AC power into DC of various voltages (12&amp;nbsp;V-5&amp;nbsp;V) would it be possible to power a Bluetooth receiver which uses 5&amp;nbsp;V 1&amp;nbsp;A power from a wall plug USB socket, directly from the PSU? &#xA;I've tried using the red (5&amp;nbsp;V) wire with the ground to a female USB cable which I plug into the Bluetooth receiver, but to no avail. The part I don't know about is the 1&amp;nbsp;A. What is the relevance of this?&lt;/p&gt;&#xA;" OwnerUserId="6236" LastEditorUserId="1832" LastEditDate="2016-04-28T19:20:39.257" LastActivityDate="2016-04-29T01:41:15.783" Title="Computer psu to directly power usb port" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="5" ClosedDate="2016-04-30T12:31:50.130" ContentLicense="CC BY-SA 3.0" />
  <row Id="8640" PostTypeId="2" ParentId="7328" CreationDate="2016-04-28T20:00:53.170" Score="1" Body="&lt;p&gt;&lt;em&gt;Completely changed answer, taking the first comments into account&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the dictionary definitions for gravel, stone and slag gives a few clues: (from www.dictionary.com)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;gravel:&lt;/strong&gt; &quot;small stones and pebbles, or a mixture of these with sand.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;stone:&lt;/strong&gt; &quot;a rock or particular piece or kind of rock, as a boulder or piece of agate.&quot; &lt;em&gt;or&lt;/em&gt; &quot;a piece of rock quarried and worked into a specific size and shape for a particular purpose&quot; &lt;em&gt;or&lt;/em&gt; &quot;a small piece of rock, as a pebble.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;slag:&lt;/strong&gt; Also called cinder. the more or less completely fused and vitrified matter separated during the reduction of a metal from its ore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Stone&quot;&lt;/strong&gt; and &lt;strong&gt;&quot;gravel&quot;&lt;/strong&gt; are descriptive of the size / particle distribution or even the source of a material, while &lt;strong&gt;&quot;slag&quot;&lt;/strong&gt; is a type of material, irrespective of particle size or even shape. I think he intended to specify where the material would be sourced i.e. &quot;rock&quot; (from a quarry), &quot;gravel&quot; (from a gravel pit) and &quot;slag&quot; (waste product from heavy industry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author then chose to specify &quot;crushed&quot; materials because he probably needs the properties of a crushed material (ragged edges). &lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="2637" LastEditDate="2016-04-29T16:08:47.043" LastActivityDate="2016-04-29T16:08:47.043" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8641" PostTypeId="2" ParentId="8619" CreationDate="2016-04-28T20:27:42.437" Score="5" Body="&lt;p&gt;1) If you have a siphon system, why use a pump? Use a valve to turn the siphon flow on/off. Much cheaper than running a 40W pump. A simple gate valve at the crest will work perfectly in conjunction with a foot check valve at the inlet and another check valve at the outlet to prevent reverse movement of air into the siphon when it is shut off leaving the siphon primed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If your going to use a pump, make the pump do the work. Rearrange your system to get rid of the pipe sloping down, so all the piping slopes up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) A 1 psi check valve at the outlet of the pump should break open when the pump starts, and close even under the siphon suction.&lt;/p&gt;&#xA;" OwnerUserId="6238" LastEditorUserId="-1" LastEditDate="2020-01-24T17:44:44.550" LastActivityDate="2020-01-24T17:44:44.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8642" PostTypeId="2" ParentId="8636" CreationDate="2016-04-28T23:52:22.897" Score="1" Body="&lt;p&gt;One very easy solution is to spray the required surfaces with latex based contact adhesive. This is fairly tacky when cured and the spray will also give a bit of texture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively liquid latex is cheap and easily available and can be applied by dipping the fingertips and/or palm into it. Without fillers or vulcanisation it is not as durable as some other rubbers but it is convenient and easy to use for prototyping. Or you could use something like &lt;a href=&quot;http://www.plastidip.com/home_solutions/Plasti_Dip&quot; rel=&quot;nofollow&quot;&gt;Plasti Dip&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is self amalgamating tape used for electrical insulation etc which is butyl rubber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the hand is human hand shaped you could also consider using gloves.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-29T00:00:58.617" LastActivityDate="2016-04-29T00:00:58.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8644" PostTypeId="2" ParentId="8638" CreationDate="2016-04-29T01:09:06.883" Score="1" Body="&lt;p&gt;I did a quick google search on current rating on a 5V supply on a ATX power supply here is data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kd4jW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kd4jW.jpg&quot; alt=&quot;Power Rating&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base on this data your 550W power supply should be able to source sufficient current to power the Bluetooth receiver. Your problem might not be power rating but how you are using the PSU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand ATX power supplies have a &lt;code&gt;PS_ON&lt;/code&gt; which needs to be pulled down to switch on the PSU. When you pull down the &lt;code&gt;PS_On&lt;/code&gt; the &lt;code&gt;PWR_OK&lt;/code&gt; signal line goes high. After this point you can source current from the other supply lines. The a power supply is connected to the mains the 5VSB is hot. You might be able to power your device using the 5VSB source too. This is the purple wire, you might want check with a multi meter to verify.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the power sequencing diagram for a PSU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RYeSK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RYeSK.jpg&quot; alt=&quot;ATX Power supply timing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some notes about &lt;code&gt;PS_ON&lt;/code&gt; and &lt;code&gt;PWR_OK&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U0WIQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U0WIQ.jpg&quot; alt=&quot;PS_ON description&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iphC7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iphC7.jpg&quot; alt=&quot;PWR_OK description&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the connector pin out that might be of helpful to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VZS6f.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VZS6f.jpg&quot; alt=&quot;ATX PSU connector layout&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are few references with more insight that might be of assistance to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HTH&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/a/176061/50587&quot;&gt;Troubleshooting PCB of an ATX power supply?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/a/186858/50587&quot;&gt;Can a power supply still provide a voltage if it is broken?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://web2.murraystate.edu/andy.batts/ps/PowerRatings.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Representative ATX 12 V. Power Supply Ratings (amps)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2016-04-29T01:41:15.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8645" PostTypeId="2" ParentId="8626" CreationDate="2016-04-29T01:25:40.787" Score="2" Body="&lt;p&gt;There is an ISO standard for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latest version is (probably) &#xA;&quot;ISO 3098-1:2015 Technical product documentation -- Lettering -- Part 1: General requirements&quot;. &lt;a href=&quot;http://www.iso.org/iso/home/store/catalogue_tc/catalogue_detail.htm?csnumber=65679&quot; rel=&quot;nofollow&quot;&gt;http://www.iso.org/iso/home/store/catalogue_tc/catalogue_detail.htm?csnumber=65679&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some equivalent local standards, for example DIN 6776, and I believe there is an ANSI standard that dates back to the 1930s. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-04-29T01:25:40.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8646" PostTypeId="2" ParentId="2980" CreationDate="2016-04-29T03:46:17.790" Score="4" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you press a cylinder on a flat surface the contact pressure concentrates on the edges and becomes very high. We call that edge stress condition. If the cylinder has a profile (curved instead of straight) it causes the pressure distribution to be more even and hence lower in value.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to account for the slight angles at which tires approach the road and not cause the tire to roll on one edge only you need a profile.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some roads are not convex, but rather concave, and a flat cylinder would not roll well with the edges in contact only.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In fact, the restoring moment (response torque to slip) is proportional to the width of the contact patch, and when edge loading occurs it throws off the handling of the car. Ever try to steer a car with a deflated tire? It is very hard to do.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1051" LastEditorUserId="368" LastEditDate="2016-04-29T23:40:13.920" LastActivityDate="2016-04-29T23:40:13.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8647" PostTypeId="1" AcceptedAnswerId="8649" CreationDate="2016-04-29T04:48:01.353" Score="1" ViewCount="1091" Body="&lt;p&gt;I am IT engineer not a refrigeration expert, pardon my basic question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a NH3 or other absorption cycle work with a generator temperature of say 30C, and condenser temperature of 10C by maintaining correct pressure level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the technical papers I read on the internet, talk about requirement of 80C+ temperature for generator and 25-30C for condenser. This would make sense if one has access to hot water source like geo thermal, or any other free heat source. But if I have access to low temperature cool water of 10C, and relatively stable heat source of 25C+ (say atmospheric temperature), can a absorption refrigeration system be designed to harvest that energy. One of the main objective is to have very low electrical energy input, so compression system may not work out. The second objective is to have a -3C or lower evaporator temperature to make salty ice and store it, and use it when the cold water source is not available. If at all its possible, what major issues to keep in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess even the standard off the shelf absorption systems can be made to use 30C heat, by first passing it through a heat pump to increase that temperature to 80C+. But that additional stage also need lot of electricity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestion to effectively harvest this low temperature source would be of great help.&lt;/p&gt;&#xA;" OwnerUserId="6026" LastActivityDate="2016-04-29T07:31:41.190" Title="absorption refrigeration with low generator temperature?" Tags="|hvac|refrigeration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8648" PostTypeId="1" AcceptedAnswerId="8651" CreationDate="2016-04-29T06:19:07.040" Score="3" ViewCount="1132" Body="&lt;p&gt;From what I know, for a nice rectangle slab with 4 edges bounded by beams, one can distribute the slab loading to adjacent beam via the &lt;a href=&quot;http://www.engr.uky.edu/~gebland/CE%20382/CE%20382%20PDF%20Lecture%20Slides/CE%20382%20L3%20-%20System%20Laoding.pdf&quot; rel=&quot;nofollow&quot;&gt;Tributary Area method&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if the slab is irregular, or there are arbitrary line loads on the slab, or the slab is not of uniform method, or not all sides of the slabs are covered by beams? How should the slab's dead load and live load, and any load on slab be distributed to beams? Is there a general way?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="3353" LastEditDate="2016-04-29T09:43:27.427" LastActivityDate="2016-04-29T10:27:31.200" Title="What is the general way for the slab loading be distributed to adjacent beam?" Tags="|structural-analysis|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8649" PostTypeId="2" ParentId="8647" CreationDate="2016-04-29T07:31:41.190" Score="0" Body="&lt;p&gt;Basically to vaporise the NH3 in generator and pressurise it, you need to add heat that will be greater than the latent heat of ammonia due to fact that water is present it will take some part of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To vaporise and seperate water and NH3 at lower temperature you will need to add more amount of heat (red line) with respect to heat you will add to higher temperature (blue line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;further more the temperature range you have provided is not practical. I mean the heat added to the aqua ammonia by just atmosphere would be insufficient to vaporise the refrigerant. let alone it to pressurise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to add heat to it you need to give it some external heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:- the diagram below is for ammonia refrigerant I have drawn basic diagram on the left. I didn't find the diagram for aqua ammonia strong solution. Actual curve of the solution may vary&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jAQqM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jAQqM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2016-04-29T07:31:41.190" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8650" PostTypeId="1" AcceptedAnswerId="8666" CreationDate="2016-04-29T09:50:51.820" Score="8" ViewCount="2479" Body="&lt;p&gt;From &lt;a href=&quot;http://www.bgstructuralengineering.com/BGASCE7/BGASCE7005/BGASCE70504.htm&quot;&gt;ASCE 7-05 code&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ASCE 7-05 Section 4.6 states &quot;The full intensity of the appropriately&#xA;  reduced live load applied only to a portion of a structure or member&#xA;  shall be accounted for if it produces a more unfavorable effect than&#xA;  the same intensity applied over the full structure or member.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The article then goes on to demonstrate how can we calculate the pattern of live loading for a few simple, text-book cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem now is, what if the configuration is not as simple? In real life the beam configuration, the support condition can be very different than from textbook examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to obtain the most critical pattern of live loading for the most general situation? Is there an algorithm for this?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="368" LastEditDate="2016-04-29T23:47:19.647" LastActivityDate="2019-12-28T17:55:44.043" Title="General solution for the most critical pattern of live loading" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8651" PostTypeId="2" ParentId="8648" CreationDate="2016-04-29T10:27:31.200" Score="2" Body="&lt;p&gt;&lt;em&gt;Extension of my comment to an answer, feel free to edit and expand it!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most general method would be to use a numerical model consisting the slab as well, thus the load transferring mechanism is captured by the model. Within the finite element (FEM) paradigm plate elements would suffice. With today's computers and FEM applications it can be easily solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know about your background and experience but using plate elements with line elements to model slabs with beams is quite widespread. That is why I just mentioned FEM without any further details. If you could provide more details on your experience or formulate more specific questions, e.g. how to model the eccentricity of beams, beam slab connection, then we might be able to give equally specific answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two practitioners' guides on &lt;strong&gt;modelling of reinforced concrete slabs&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://publications.lib.chalmers.se/records/fulltext/176734/local_176734.pdf&quot; rel=&quot;nofollow&quot;&gt;Recommendations for finite element analysis for the&#xA;design of reinforced concrete slabs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.scribd.com/doc/19184786/How-to-Design-r-c-Flat-Slabs-Using-Finite-Element-Analysis&quot; rel=&quot;nofollow&quot;&gt;How to Design r.c. Flat Slabs Using Finite Element Analysis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding &lt;strong&gt;how the tributary approach compares with FEM plate model&lt;/strong&gt; for a simple case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://s-frame.com/index_files/Files/Approaches%20to%20Area%20Load%20Distribution%20in%20S-FRAME.pdf&quot; rel=&quot;nofollow&quot;&gt;A Case Study Comparing Two Approaches for Applying Area Loads: Tributary Area Loads vs Shell Pressure Loads&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on their simple example for a particular slab-to-beam stiffness ratio:&#xA;The tributary area method is on the safe side and yields to about 20% overestimation of maximal bending moment and shear force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are based on a single, simple example thus the results are not conclusive, at best only indicative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the structure type was not mentioned and maybe you have other than a building structure in mind, here are two relevant books on &lt;strong&gt;bridge decks&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.scribd.com/doc/72761049/Bridge-Desk-Analysis&quot; rel=&quot;nofollow&quot;&gt;O'Brien and Keogh: Bridge Deck Analysis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.scribd.com/doc/76442313/Bridge-Deck-Behaviour-E-C-Hambly&quot; rel=&quot;nofollow&quot;&gt;Hambly: Bridge Deck Behaviour&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1352" LastActivityDate="2016-04-29T10:27:31.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8652" PostTypeId="2" ParentId="8650" CreationDate="2016-04-29T13:22:31.543" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How to obtain the most critical pattern of live loading for the most general situation? Is there an algorithm for this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as I've been able to determine, finding the worst-case combinations of live loads is part based on previous experience, part engineering judgment, and part iteration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, you can make an educated guess as to which pattern of live loading will produce the worst case beam moments and reactions (not that one pattern will not necessarily produce the maximum moment &lt;em&gt;and&lt;/em&gt; reaction simultaneously).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the structure gets more complex, it becomes harder to determine the &quot;correct&quot; pattern of live loading to maximize the structure response.  This is where iteration and experience come into play.  The link you include also discusses using influence lines to help determine your locations for live loading, which is a good thing to study up on.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2016-04-29T13:22:31.543" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8653" PostTypeId="2" ParentId="8637" CreationDate="2016-04-29T13:27:22.057" Score="2" Body="&lt;p&gt;The difference is due to you missing an unknown for the molecular degree of freedom analysis.  You need two selectivity unknowns as you have two equations, which will progress at different rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. The selectivity for the first equation will tell you how much methanol is present, but you need another selectivity variable to tell you how much of that methanol has been converted to ethene and water.&lt;/p&gt;&#xA;" OwnerUserId="2616" LastActivityDate="2016-04-29T13:27:22.057" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8655" PostTypeId="1" CreationDate="2016-04-29T14:00:09.510" Score="5" ViewCount="330" Body="&lt;p&gt;A synthesis gas mixture ($\text{H}_2:\text{CO}$ with molar proportion 2:1) is needed for methanol synthesis. The final mixture must be available in a gas cylinder of 50 L, say cylinder $M$, which contains, initially, pure $\text{H}_2$ at 5 bar. The carbon monoxide which will compose the final mixture is available in a smaller cylinder (20 L), say cylinder $m$ in a volumetric proportion of 9:1, at 120 bar. Both cylinders are, initially, at 303 K. A 1/8” steel tube must be used with a globe valve placed 1 m from the small cylinder $m$ and 4 m from the cylinder $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one calculate the pressure and temperature of both cylinders in the process where the gas transfer is done opening totally and suddenly the globe valve, after 15 s? The dynamics of the problem seem to be difficult to implement.&lt;/p&gt;&#xA;" OwnerUserId="6250" LastEditorUserId="1832" LastEditDate="2016-04-29T14:04:15.827" LastActivityDate="2018-07-19T07:02:58.297" Title="Thermodynamics of gas exchange between gas cylinders" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8658" PostTypeId="1" CreationDate="2016-04-29T20:50:40.460" Score="0" ViewCount="165" Body="&lt;p&gt;Below is the circuit for a shadow alarm. Please tell me the role of potentiometer(VR1) used in the circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VF5hq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VF5hq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6257" LastEditorUserId="6257" LastEditDate="2016-04-30T05:24:52.687" LastActivityDate="2016-04-30T22:17:15.797" Title="what is the role of potentiometer in this circuit?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="2" ClosedDate="2016-04-30T04:04:43.027" ContentLicense="CC BY-SA 3.0" />
  <row Id="8659" PostTypeId="2" ParentId="8658" CreationDate="2016-04-29T21:36:22.937" Score="1" Body="&lt;p&gt;It looks like it is there to provide adjustment for the sensitivity of the LDR in switching the adjacent transistor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not at all unusual for analogue sensors which switch a  transistor especially when they are  bit noisy. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-30T22:17:15.797" LastActivityDate="2016-04-30T22:17:15.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8660" PostTypeId="2" ParentId="8658" CreationDate="2016-04-29T21:46:24.907" Score="0" Body="&lt;p&gt;It appears that this is some sort of light-activated switch and/or alarm.  The reason I say &quot;appears&quot;, is that the output of the light sensor is going into a processor, and that processor controls a relay and a buzzer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without documentation or the source code, we can only assume that the relay and buzzer are activated as a result of the light sensor input, although there is no way to tell from the schematic what the policy is, whether the relay is activated in low light or bright light, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potentiometer adjusts the light sensor output voltage level for specific light levels.  It appears that the signal on the emitter of T1 is being fed into a digital input.  At some voltage level, that input will switch from being seen as low by the code to being seen as high.  The potentiometer adjusts at what light level that happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All around, this is a crappy circuit for various reasons, but that's beyond the scope of this question.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-04-29T21:46:24.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8661" PostTypeId="1" AcceptedAnswerId="8677" CreationDate="2016-04-29T21:46:45.403" Score="3" ViewCount="10029" Body="&lt;p&gt;In a bike frame build, are there any differences between hexagonal cross-section tube instead of the typical round tubes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that both are made from the same aluminium alloy and have the same wall thickness, are there differences in weight, strength, or durability? Does the hexagonal shape have any particular advantages over round tube?&lt;/p&gt;&#xA;" OwnerUserId="6258" LastEditorUserId="421" LastEditDate="2016-04-29T22:23:23.037" LastActivityDate="2016-04-30T20:06:37.160" Title="Properties of an hexagonal shape vs round tube shape" Tags="|structural-engineering|design|stresses|structural-analysis|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8662" PostTypeId="2" ParentId="8594" CreationDate="2016-04-30T00:22:11.330" Score="2" Body="&lt;p&gt;This started as a comment but got too long. &#xA;I think you have some good answers already regarding the naming of precipitation hardening, so my answer attempts to address some of the other semantic issues it appears you are struggling with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Area under the curve is term which refers to a stress-strain curve, which (as it sounds) is a graphical representation of a materials behavior under stress.  This is a measure of the materials &lt;strong&gt;strength&lt;/strong&gt;, and does not necessarily indicate toughness, or hardness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strength is a materials ability to resist a load (known as stress).  When a load is imparted to a material, it will undergo a deformation. The measured amount of deformation is called strain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A stress-strain curve therefore is simply a plot of the amount of strain (deformation of the material) corresponding to various levels of stress (load).  Typically this is plotted with strain along the horizontal axis and stress along the vertical.  There are different ways these curves may be expressed e.g. engineering strain vs true strain, but I won't go into that here. I mention it as something to be aware of if you're going to be reading and comparing curves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I mentioned this is a measure of material's strength, and I'll attempt to give a brief explanation of this here.  For metallic materials, a stress-strain curve is typically generated from a uniaxial tensile test, which measures the materials ability to withstand an ever increasing load in one direction under tension.  The standard test consists of rigging a (typically cylindrical) sample of material between two clamps, one clamp is stationary, and the other is attached to a platform which moves away from the stationary jaw.  This movement apart applies tension to the sample.  This tensile load is increased until the material breaks, and the highest load achieved before rupture is known as the ultimate tensile strength of the material.  (So UTS refers to the maximum amount of tensile stress a material can withstand before failure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from ultimate tensile strength (UTS) the other common measure of strength you'll see is yield strength (YS). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With metallic materials there are typically two distinct regions in the stress-strain curve; elastic and plastic.  As load is increased from 0 up until the yield point the strain will increase proportionally at a linear rate relative to stress.  At loads (stress) of this intensity the material behaves elastically, which is to say that when the load is removed, the material will return back to its original shape (stress=0, strain=0). At some point, the stress will become great enough to overcome the material's elastic limit, meaning permanent deformation occurs (stress=0, strain&gt;0). This permanent deformation is known as plastic deformation, and so the area under the curve in this region is known as the plastic region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point at which this transition occurs is known as the material's yield point. Typically, it's not distinct enough to measure the exact point at which yield occurs, so an approximation known as offset yield is typically used.  You'll commonly see material specifications refer to 0.2% offset yield (or some other percent offset) which refers to the approximated yield point. I won't get into the methods for offsetting to determine approximate yield, but it's easy enough to google if you're interested. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing regarding material strength; so far we've only discussed tensile strength, which is the most common measurement of metallic materials, however other measurements of strength exist as well.  For example, ceramics and cements are not commonly measured under tension but rather under compression, which is known as compressive strength.  This again will be graphed as a stress-strain curve, but is a different test involving compression of the material until failure.  I just thought it worth mentioning as it relates to the understanding of the word &quot;strength&quot; in materials science.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to &lt;strong&gt;hardness.&lt;/strong&gt; &#xA;Hardness is typically measured as the materials ability to resist indentation. Various standards exist for testing hardness, such as Rockwell, Vickers, brinnell (all common with metallics) and shore (common with rubbers and plastics). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While hardness is measured as resistance to indentation, it's commonly used to judge a materials resistance to abrasion and wear, and can can be manipulated to increase a materials performance in certain applications or to improve machinability/workability of the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hardness and strength commonly have a positive correlation as the mechanisms which increase a material's hardness typically also increase the material's strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Toughness&lt;/strong&gt; typically has a negative correlation to hardness.  Toughness is a measure of a material's ability to resist fracture under load.  Brittleness is the opposite of toughness, and typically as a material becomes more hard, it also becomes more brittle.  This is not the rule, just typical for many metallics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Toughness is a little more difficult to quantify than strength, as there are many ways for a material to to be loaded to failure and for fracture to occur.   Some common indicators of toughness are impact resistance (charpy is a common standard test),  fracture resistance/resistance to crack growth (plain-strain fracture toughness is a common test), and ductility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ductility is similar to toughness, but is not necessarily related to a materials ability to resist fracture, but is rather a measurement of its ability to flow and deform under load.  Ductility is commonly expressed as reduction of area or percent elongation; both are measures of plastic deformation before failure, typically established from the tensile test. &lt;/p&gt;&#xA;" OwnerUserId="2598" LastEditorUserId="2598" LastEditDate="2016-05-01T14:01:03.063" LastActivityDate="2016-05-01T14:01:03.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8664" PostTypeId="2" ParentId="8547" CreationDate="2016-04-30T03:56:51.300" Score="1" Body="&lt;p&gt;The recrystallization temperature for brass is &amp;lt;400 Celsius (Harding et al., 1980), so the original microstructure won't affect the results here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your process seems good, and my guess would be the answer in the book was either from the 700 Celsius line or 1000 min on the 600 Celsius line.&lt;/p&gt;&#xA;" OwnerUserId="6259" LastActivityDate="2016-04-30T03:56:51.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8665" PostTypeId="2" ParentId="8528" CreationDate="2016-04-30T04:20:20.017" Score="1" Body="&lt;p&gt;If you can apply a sinusoidal displacement to the suspension and measure the signal from the accelerometer, you can avoid some of the problems with drift in the accelerometer. You can filter the accelerometer signal to remove low frequencies, because the signal you want it the sinusoidal one. Then integrate twice to get displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=C\sin{\omega t}$$&#xA;$$v=-C \omega \cos{\omega t}$$&#xA;$$d=-C\omega^2 \sin{\omega t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have the magnitude of your displacement with the drift removed by filtering it out.&lt;/p&gt;&#xA;" OwnerUserId="6259" LastActivityDate="2016-04-30T04:20:20.017" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8666" PostTypeId="2" ParentId="8650" CreationDate="2016-04-30T04:22:34.813" Score="9" Body="&lt;p&gt;As mentioned in the linked text and in &lt;a href=&quot;https://engineering.stackexchange.com/a/8652/3353&quot;&gt;@grfrazee's answer&lt;/a&gt;, the secret is influence lines. Or, more generically, influence surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters, let's stick to influence lines, since they are far easier to describe. An influence line is a diagram for a given point on an object composed of unidimensional beam elements. It describes the internal force that will occur on that point due to a unit load applied at different points along the entire structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, a simply supported beam has the following bending-moment influence line for the point at a quarter-span (I'm mostly going to talk about bending moment influence lines here, but the general gist of things applies to other forces as well):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hj6GL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hj6GL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if a concentrated unitary vertical load (say, 1&amp;nbsp;kN) were applied at that point, it would cause a bending moment at that point equal to 0.75&amp;nbsp;kNm (or 7.5&amp;nbsp;kNm if the load were 10&amp;nbsp;kN). If, on the other hand, the unit load were applied at midspan, the moment felt at the quarter-span would be equal to 0.50&amp;nbsp;kNm. And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also tells you that the worst-case scenario for this point is to have the entire structure loaded. This can be seen by the simple fact that all values on the influence line are positive, therefore a load applied at any point on this beam will increase the internal forces occurring at the quarter-span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, is an isostatic structure which can be trivially solved. Once you move into hyperstatic (statically indeterminate) structures, things get messy. For instance, take a look at this relatively simple hyperstatic beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dqnry.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dqnry.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a relatively simple structure, but it's already impossible to find a closed-form solution for the precise best location of loads. For non-trivial structures, influence lines are a pain.&lt;sup&gt;1&lt;/sup&gt; You can however notice one important thing: at the supports the value is null and shifts from positive on one side to negative on the other. This happens in every structure. If, instead of supports you had columns, then the value at the columns won't actually be equal to zero due to the deformability of the column. That being said, the result is usually very close to zero, so you can usually consider columns as perfectly rigid (i.e. as normal supports) with barely any loss of accuracy (assuming reasonable layouts). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you're only dealing with distributed loads (such as in a building), this is the only rule you need to find your solution: if you're looking for the maximum positive (tension on the bottom fiber) bending moment, apply loads on the span in question, don't apply loads on the neighboring spans, apply on the ones neighboring those, etc. In this case, the actual values of the influence line are irrelevant, all that matters is the sign (positive or negative) at each span. Basically, here's the rule in graphic form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/waPiy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/waPiy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what if you're building a bridge and you need to take into consideration the position of the load train, which is composed of concentrated loads? Complicating matters, the position of the load train usually has a lower (if any) distributed live load, which means there's an interaction between these two parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, looking at the second figure, where would you put the load train? It is pretty intuitive that you want to place it near the maximum value (in this case 0.3704). But what if you have an even number of wheels or if your load train is asymmetric? Do you want to put the truck's load center at the maximum? Do you want to certify that the heaviest wheel is at the maximum? Is your uniform load so freakishly high that, actually, you're better off putting the truck far away where it won't reduce the result due to your uniform load?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even worse, what if you're actually looking for your negative bending moment envelope? Then you know you want your truck at the neighboring span, where the influence line sign is negative, but once again, where do you put it? You'll need to derive the equation of that curve in order to find the point of the maximum value (its not at the middle of that span), and then you'll still have the same issues described above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all possibilities which can't be reduced to a closed form solution for a generic structure. So you need to rely on software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What most programs actually do is &lt;strong&gt;cheat&lt;/strong&gt;. They approximate the solution by doing a moving-load analysis. First they use influence lines as described above to find out where to put the uniform loads. Then, for the load train itself, they simply place it at one spot, calculate the results, move it a certain distance (usually user-defined), calculate the new results, and repeat. It then gets the worst case and adopts that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method is obviously flawed because, if you use a step size equal to, say, one meter, you don't know if that maximum value found is the &lt;em&gt;true&lt;/em&gt; maximum or if there was a certain point between the tested steps which would have given a higher result (there almost certainly is). So it's up to the user to define a step size such that the difference between the actual result and the one obtained is negligible (I usually use a step size &lt;strong&gt;&lt;em&gt;at most&lt;/em&gt;&lt;/strong&gt; equal to one tenth of the smallest span, preferably significantly smaller than that).&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This entire answer, however, has relied on influence lines. These are useful for linear structures such as simple beam systems and even some bridges. But if you have a truly three-dimensional structure, influence lines don't cut it and must be generalized to influence surfaces. These are no more than the three-dimensional version of influence lines. However, like all such things, influence surfaces are orders of magnitude harder to obtain. Every program I know which can calculate them brute-forces it: they apply a concentrated force at each node, one at a time, and see what happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, as far as distributed loads go, the same approach suggested above (apply on one span, skip its neighbors, apply on the next ones, etc) can also be applied successfully for influence surfaces as well. In this case it becomes something of an approximation since the boundaries between slabs are usually just beams which are quite flexible for vertical displacements (relative to columns or actual supports). This means that, unlike the case for influence lines, where the influence line value at the supports is equal (or almost) to zero, the value at the slab supports (the beams) is not necessarily so. That being said, the error is usually reasonable (especially considering the low influence values for the slabs other than the one being studied).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, it is quite common to simply assume for buildings (&lt;strong&gt;&lt;em&gt;not bridges&lt;/em&gt;&lt;/strong&gt;) that the worst case is with the entire structure under load, without considering influence lines. This is assumed knowing it is false and goes against safety (not loading neighboring slabs would result in a larger positive bending moment that that obtained by loading the entire structure), but it is equivalent to assuming that the value of the influence line on neighboring slabs is so small that it can be considered equal to zero. The validity of such an assumption is dependent on each structure's configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by &lt;a href=&quot;https://engineering.stackexchange.com/questions/8650/general-solution-for-the-most-critical-pattern-of-live-loading#comment16068_8666&quot;&gt;@Arpi in the comments to this answer&lt;/a&gt;, it's also worth mentioning that this all assumes linear behavior. If your analysis is non-linear, then everything falls apart. Non-linearity breaks everything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;All figures here were created with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; It's actually quite easy to determine influence lines yourself if you have an analysis software, even if it doesn't calculate them itself. For bending moments, place a hinge on the desired point and apply equal and opposite bending moments to each side of the hinge such that they create a unit rotation in the deformed configuration. That deformed configuration is your influence line. This same idea (&lt;a href=&quot;https://en.wikipedia.org/wiki/M%C3%BCller-Breslau%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Müller Breslau Principle&lt;/a&gt;, which is based on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Betti%27s_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maxwell-Betti reciprocal work theorem&lt;/a&gt;) can be applied to find the other forces' influence lines as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; The Ftool software used to draw these figures actually uses a genetic algorithm to find the optimal load train position. It isn't analytical and is actually itself something of an approximation, but it is for all intents and purposes exactly correct. The article which developed this method can be found &lt;a href=&quot;https://en.wikipedia.org/wiki/Betti%27s_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; if anyone's interested.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T17:55:44.043" LastActivityDate="2019-12-28T17:55:44.043" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="8667" PostTypeId="1" CreationDate="2016-04-30T04:39:19.717" Score="1" ViewCount="315" Body="&lt;p&gt;So here's my situation: I'm a mechanical engineering student working on a product innovation for one of my courses. The product itself is fairly heavy into embedded programming, which I'm comfortable with. What I'm less comfortable with is the small electrical engineering bit I need to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, I need to be able to control (turn on or off) a high-speed (USB) data line via transistor (or SSR, if you want to be pedantic). When a high signal is sent to the (base or gate), the line is enabled, and data can flow. When there is a low present, the communication is disabled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I'm dealing with very low current flow, my inclination is to use an NPN transistor, mostly because I have a few handy. Is there a better transistor or special rating I should be looking for for this application?&lt;/p&gt;&#xA;" OwnerUserId="3992" LastActivityDate="2016-08-29T16:25:07.913" Title="Transistor choices for switching high speed data communication lines" Tags="|electrical-engineering|embedded-systems|equipment-selection|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8668" PostTypeId="1" AcceptedAnswerId="8701" CreationDate="2016-04-30T05:13:05.843" Score="3" ViewCount="18941" Body="&lt;p&gt;Which position of the water pump in the below diagram provide maximum water output to the overhead tank?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5dd2s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5dd2s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Each floor is 12 feet in height.&lt;/li&gt;&#xA;&lt;li&gt;I already have the pump (Centrifugal Regenerative Self Priming 1HP Peripheral Type, Self Priming - 180sec At 3m STATIC SUCTION LIFT) and would like to know which is the best location to install it for maximum water intake to the overhead tank. The water Supply is only available for a limited period everyday and the goal is to achieve maximum water to the overhead tank.&lt;/li&gt;&#xA;&lt;li&gt;The Water supply line provides water from a bigger (8inch) main line but water pressure from it is not adequate to pump up water on top level. (Water pressure only enough to supply water to first floor and that too with reduced pressure.)&lt;/li&gt;&#xA;&lt;li&gt;The water pipe from main supply to pump and from pump to overhead tank is 1/2 inches wide.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6262" LastEditorUserId="6262" LastEditDate="2016-04-30T12:18:45.257" LastActivityDate="2018-09-18T09:21:09.553" Title="Water pump at top floor or bottom?" Tags="|pressure|pumps|piping|pipelines|building-design|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8672" PostTypeId="1" AcceptedAnswerId="8674" CreationDate="2016-04-30T08:44:34.647" Score="4" ViewCount="3434" Body="&lt;p&gt;I have found that drill bits are always slightly undersized. For example, 10mm bits tend to be more like 9.8mm diameter. And in aluminium, the hole ends up being about 9.9mm. The same seems to be the case for conical stepped drill bits&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found it useful for inserting ball bearings, since they can be hammered in place without being damaged and will stay firmly in place. Also, bolts tend to be slightly undersized so will have a more snug (but still loose) fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the actual reason for drill bits being undersized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oddly, hole saws tend to be slightly oversized. A 15mm hole saw's teeth outer diameter measures to be 15.05mm and makes a 15.3mm hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a term for drill bits which are slightly oversized instead? Or hole saws which produce slightly undersized holes? (assuming they exist)&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-04-30T19:48:36.230" Title="Why are drill bits always slightly undersized?" Tags="|drilling|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8674" PostTypeId="2" ParentId="8672" CreationDate="2016-04-30T10:03:38.917" Score="3" Body="&lt;p&gt;They short answer is that they aren't. Exact sizes will, of course depend on manufacturing tolerances but 0.2mm is a lot undersized by normal standards, although for jobber drills this won't actually matter much.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example metric &lt;a href=&quot;http://www.kasthurimmc.com/tap-drill-chart.html&quot; rel=&quot;nofollow&quot;&gt;tap drill sizes&lt;/a&gt; are generally specified to at least 0.1mm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible explanation for measuring drills as undersized is that the shank often gets worn, especially if it has slipped in a chuck and it is difficult to accurately measure the diameter across the flutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual size of the hole produced by a given drill is a different matter and will depend on the drilling setup and the material being drilled. Twin flute twist drills tend to produce slightly triangular or hexagonal holes, especially when the material being drilled is significantly thinner than the diameter of the drill, but this is more to do with flexing then the actual diameter of the drill bit.  This tends to make them functionally undersized in terms of fitting bolts etc and for general fabrication it is normal practice to go 0.5-1.0mm above nominal size for a clearance hole for a metric bolt (notwithstanding any additional issues with fit and clearance between sets of holes etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With twin flute twist drills in general it tends to be the roundness, and in some cases straightness, of holes which is the limiting factor on accuracy. Better tolerances are usually achieved by using a reamer, a multi-flute drill/mill or by boring with a lathe, mill or specialist drilling machine. There are also specialist drills eg &lt;a href=&quot;https://en.wikipedia.org/wiki/Gun_drill&quot; rel=&quot;nofollow&quot;&gt;gun drills&lt;/a&gt; for use where the hole depth is very much larger than the hole diameter. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-04-30T19:48:36.230" LastActivityDate="2016-04-30T19:48:36.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8675" PostTypeId="2" ParentId="8668" CreationDate="2016-04-30T13:41:28.173" Score="6" Body="&lt;p&gt;Think about the pressures.  With the pump above the supply line, the pressure at the input of the pump will be lower than the supply.  At roughly 2 PSI per foot, you're down about 18 PSI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since air pressure is about 15 PSI, the system will at least work as long as the supply pressure is over 3 PSI, but that's still not a good situation.  If the &quot;supply&quot; is a open pool, then it won't work at all since even a full vacuum can't get the water that high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With low input pressure, the pump will be more prone to cavitation.  That's bad for the mechanics and a waste of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, put the pump at the bottom, not the top.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2016-05-01T12:53:12.863" LastActivityDate="2016-05-01T12:53:12.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8676" PostTypeId="2" ParentId="8661" CreationDate="2016-04-30T19:07:00.243" Score="2" Body="&lt;p&gt;I can't think of much advantage to using a hexagonal tube, except perhaps cosmetic - but this is subjective. Hexagonal tube is likely to be much more expensive as it is a non-standard shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is assumed that both the round and hexagonal tubes are extruded (which is likely for a bike frame) we can ignore any cold-forming or welding issues. For them to be the same strength (and weight) they would have to have the same cross-sectional area. The axial stiffness is $EA$, where $E$ is Young's modulus which depends on the material (assumed to be the same in this case) and $A$ which is the cross sectional area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hexagon with an equal cross-sectional area to a circle of radius $r$ and (thin) wall thickness $t$ must have side lengths of: $a=\frac{1}{3} \left(\pi  r+\sqrt{3} t\right)$ if the wall thickness is assumed to be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While bending is not a major factor in bike frames we can compare the bending stiffness of the two shapes anyway. Bending stiffness is $EI$ where $I$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Second_moment_of_area&quot; rel=&quot;nofollow&quot;&gt;second moment of area&lt;/a&gt; of the cross-section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I$ of a (thin) circle is: $I_c = \pi  r^3 t$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I$ of a hexagon (equal area and wall thickness as circle) is: $I_h =\frac{5}{18} t \left(9 a^3-9 \sqrt{3} a^2 t+12 a t^2-2 \sqrt{3} t^3\right) = \frac{5}{54} t \left(\pi ^3 r^3+3 \pi  r t^2\right)$ after substituting $a$ from above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can immediately see that the hexagon has a lower second moment of area in comparison to the circle. In fact the second moment of area of the hexagon is about 91% of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also do a simple buckling comparison by considering the Euler buckling load: $ \frac{\pi^2 EI}{L^2} $, where $L$ is the length of the tube. Again, the circle is more resistant to buckling since its $I$ is larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this basic analysis shows that a hexagon of equal area and wall thickness would be weaker under bending and have a lower buckling load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some factors which I have not considered: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;localised buckling of tubes under bending (not a straightforward analysis - especially for the hexagon). For example the &lt;a href=&quot;https://en.wikipedia.org/wiki/Brazier_effect&quot; rel=&quot;nofollow&quot;&gt;Brazier effect&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Impact resistance. The hexagon may be more resistant to localised impact, particularly if that impact occurs at a corner. This is probably offset by the fact that a circle would deflect projectiles better. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general a large diameter circular tube is the most efficient shape under axial loading.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2016-04-30T19:32:57.833" LastActivityDate="2016-04-30T19:32:57.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8677" PostTypeId="2" ParentId="8661" CreationDate="2016-04-30T20:06:37.160" Score="1" Body="&lt;p&gt;An additional factor is that any section which has defined corners will tend to concentrate stress at the corners rather than it being evenly distributed throughout the section. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With tubing this can be a double effect that you potentially  have work hardening from the manufacturing process concentrated at the corners as well. If the section has a uniform wall thickness then it certainly means that some of the material is effectively wasted as the corners will encounter yield before the flats and at worst it can lead to crack propagation points and fatigue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fairly good rule of thumb in structures that any sort of discontinuity represents at best an inefficiency and at worst a potential failure point. An ideal structure will have a smoothly varying section with section size proportional to the stresses on it, in fact bones are an excellent example of this. Although this is, in most cases, impractical for fabricated frame structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are other pragmatic considerations to be taken into account for example square or rectangular tube is much easier to join than round as square tube can just be mitred to the required angle with a saw whereas round tube needs to be notched with a specialist machine or painstakingly hand fitted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems likely to me that hex tube would be a fairly poor compromise between  square and round/elliptical section tube. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-04-30T20:06:37.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8678" PostTypeId="2" ParentId="8636" CreationDate="2016-05-01T04:40:07.310" Score="0" Body="&lt;p&gt;A simple way to test out the effect would be to wrap or stretch some household rubber bands around the fingers/grippers to increase friction.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-05-01T04:40:07.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8679" PostTypeId="1" AcceptedAnswerId="8680" CreationDate="2016-05-01T08:15:12.563" Score="3" ViewCount="555" Body="&lt;p&gt;On a recent project for a outdoor shed the roof support structure did not look like one of the conventional roof truss designs I have encountered. See snapshot below. (The span is 10 m with the support members 250x125 mm beams)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BHh47.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BHh47.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this still a truss? Or what is this sort of support framework called? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like a textbook King Post Truss (to me) but without the vertical member. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Emd0H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Emd0H.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6277" LastActivityDate="2016-05-01T10:11:49.457" Title="Steel Roof Truss Shape Question" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8680" PostTypeId="2" ParentId="8679" CreationDate="2016-05-01T09:43:58.460" Score="5" Body="&lt;p&gt;This looks like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Collar_beam&quot;&gt;Collar Beam Roof&lt;/a&gt; (or sometimes a Collar Tie Roof). It is often used in lightly-loaded residential construction. One popular use is to form Cathedral Ceilings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking this is not a truss since the structure relies &lt;em&gt;primarily&lt;/em&gt; on bending of the rafters to support the roof loads. By definition &lt;a href=&quot;https://en.wikipedia.org/wiki/Truss&quot;&gt;trusses&lt;/a&gt; are assumed to consist entirely of tension/compression elements and bending of these elements is minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the higher up the collar beam is the more bending would be present in the rafters which can result in some outwards movement of the walls. This is not likely to be an issue in a shed but could lead to some drywall cracking in a finished interior.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2016-05-01T10:11:49.457" LastActivityDate="2016-05-01T10:11:49.457" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8682" PostTypeId="2" ParentId="8667" CreationDate="2016-05-01T13:09:16.140" Score="1" Body="&lt;p&gt;You don't want to physically interrupt USB data lines.  These have controlled impedances, are bi-directional, and differential.  These issues makes &quot;switching&quot; the lines difficult.  In addition, DC resistance is used as signalling to the host when a device is attached, and what kind of device it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is unclear what &quot;turning on and off&quot; the data lines of USB means anyway.  Are you trying to stop acting like a device at all, or just stop transferring data?  If the former, then the host will have to go thru enumeration when the device is &quot;turned on&quot; again.  Note that this can be a slow process and is highly dependent on the operating system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a USB device that Windows 7 recognizes quickly, indicated by making a short noise.  I don't know what the OS does, but it can take a few seconds before the device is actually available to a program.  This is much faster on other machines, so it's not the device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should really consider doing this at the logical level, not electrically.  Let whatever microcontroller that is connected to the USB shut down its USB peripheral if you want to make the device appear to completely go away.  Or, depending on your system and what &quot;on/off&quot; really means, this micro can just ignore any new received data and not send any new data to the host.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall you need to think about what you are really trying to accomplish instead of asking about this &quot;solution&quot; someone dreamed up.  Basically, that's not how USB works.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2016-05-01T13:14:30.353" LastActivityDate="2016-05-01T13:14:30.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8683" PostTypeId="1" AcceptedAnswerId="8684" CreationDate="2016-05-01T13:58:36.513" Score="4" ViewCount="668" Body="&lt;p&gt;I am trying to estimate the volume of excavation needed for the foundations of a structure from the drawings provided. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawings clearly give the dimensions of the RC / cement footing but I suppose the actual excavation will need to be larger than this? How much larger should I plan for in the costing calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any rules of thumb? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue that comes to mind is that although the RC / cement footing structures are rectangular or various dimensions, any practical excavation would have a slight slope in the walls that increases the excavation volume? Or is that not so significant?&lt;/p&gt;&#xA;" OwnerUserId="6277" LastEditorUserId="1832" LastEditDate="2016-05-01T14:12:43.720" LastActivityDate="2016-05-01T14:27:33.910" Title="Foundation Excavation: How much larger than the RCC footing?" Tags="|structural-engineering|foundations|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8684" PostTypeId="2" ParentId="8683" CreationDate="2016-05-01T14:27:33.910" Score="3" Body="&lt;p&gt;I don't know of any code which defines this, but a one meter &quot;working perimeter&quot; around the footing is probably more than enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the slope of the excavation walls, that's trickier to define with a blanket statement. The problem is that the slope is necessary to give the soil stability. Unfortunately, the soil's stability is highly dependent on its properties (&lt;a href=&quot;https://en.wikipedia.org/wiki/Cohesion_(geology)&quot; rel=&quot;nofollow&quot;&gt;cohesion&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Angle_of_repose&quot; rel=&quot;nofollow&quot;&gt;angle of repose&lt;/a&gt;, etc), so the requisite angle to stabilize the excavation is also a function of these properties. Therefore, a blanket statement of &quot;a slope of X is adequate&quot; can't be given without knowledge of the type of soil being discussed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, for non-problematic soils (i.e. non-expansive and non-collapsible soils, with a reasonable SPT, etc), a long-term excavation can usually be given a 2:3 slope (two vertical for three horizontal), so adopting such a slope for a temporary excavation would be very conservative. Depending on the soil, I believe even a 1:1 slope could be possible for a temporary excavation.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-05-01T14:27:33.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8687" PostTypeId="2" ParentId="8636" CreationDate="2016-05-02T08:09:23.383" Score="0" Body="&lt;p&gt;Some leather variants give good grip, and are used for similar purpose on e.g. gloves. The 'grippiness' of the leather depends a lot on how exactly it has been finished and which side of the skin is outwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A local sewing supply store might be able to help you pick a suitable piece of leather.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-05-02T08:09:23.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8688" PostTypeId="1" AcceptedAnswerId="8700" CreationDate="2016-05-02T09:08:33.140" Score="0" ViewCount="2438" Body="&lt;p&gt;I need to modify a chuck's jaws so they can handle high temperatures, but will not scratch the workpiece. (Glass). The load will be very small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will probably use aluminium jaws and either coat them with, or attach to them, this material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example items being held:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Borosilicate glass tube, 30mm long, 6mm diameter, 5g weight&lt;/li&gt;&#xA;&lt;li&gt;Borosilicate glass tube, 50mm long, 30mm diameter, 50g weight&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The tube will be held horizontally by the chuck, and should be supported such that the other end will not droop by more than 2mm. That end is held in a propane bunsen flame, on the edge of the hottest part of the flame: a cone 15-20mm height, with a base ~12mm, 1200-1400°c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not have figures on what the chuck jaws will have to withstand, but the air passing over the tip of the jaws probably wouldn't be more than about 250°c, and the temperature of the glass at that point will be about the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of issue might be the heat radiated from the flame - so a heat reflecting material will help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since the rest of the chuck will be barely above room temperature, it may help to have thermally conductive material to dissipate heat from the tips of the jaws to the chuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plastics seemail like an obvious choice for softness, but high temperature plastics seem hard to find. Silicone (260°c) and PTFE (204°c) are barely temperature resistant enough, but so far seem like the best choice, since higher temperature polymers like Vespel (300-400°c) are very hard to come by. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What alternatives are there? Metals? Composites?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only other approach I might use is a heat shield directly in front of the chuck, but it would be awkward, as it would have to fit quite snugly around the workpiece, different shields would be needed for different size workpieces, and would be time consuming if batches of workpieces need to be processed. I can't imagine how this could be automated for large batches. Even then, scratching the workpiece could be a problem.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-05-02T12:13:17.580" LastActivityDate="2016-05-03T15:00:52.603" Title="Materials which withstand high temperatures, and are relatively soft, for chuck jaws" Tags="|materials|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8691" PostTypeId="1" AcceptedAnswerId="8694" CreationDate="2016-05-02T12:10:35.960" Score="-1" ViewCount="189" Body="&lt;p&gt;I am an electrical engineer, making a hobby project on an automated unwanted-material-removal system in grains and pulses (rice, wheat, etc). I am well versed with programming and electrical stuff, but I am stuck at one point with the mechanical design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My system uses a camera located on top of a conveyor belt. The camera scans the belt and identifies various impurities on the belt (using an intelligent machine learning algorithm). A brush is positioned further down the belt that pushes the impurities off the belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea for the system is to simply make the device sit in any container, with the belt part projecting out of the container. A tumbler below the belt will collected the purified grains. Sort of the things that you see in rock crushing plants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for this, I need to design a mechanical system that lifts the rice and puts / pours onto the conveyor belt. I am unable to come up with any solid design, given that I am from an electrical engineering background. And the system should be low powered, dust resistant and should work with one slowly rotating motor (or maybe a fast motor with a set of step-down gear system)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest me a good starting point for such systems or what design I can use to achieve the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The quantity of rice or grains moved in a unit time is very small, say a handful.&lt;/p&gt;&#xA;" OwnerUserId="5372" LastEditorUserId="5372" LastEditDate="2016-05-02T16:53:31.160" LastActivityDate="2016-05-02T16:53:31.160" Title="Mechanism to continously lift grains and put them on a conveyor belt" Tags="|mechanical-engineering|design|structures|" AnswerCount="1" CommentCount="1" ClosedDate="2016-05-02T16:51:30.853" ContentLicense="CC BY-SA 3.0" />
  <row Id="8692" PostTypeId="1" AcceptedAnswerId="8695" CreationDate="2016-05-02T12:13:16.833" Score="9" ViewCount="9427" Body="&lt;p&gt;One of my garage door torsion springs broke again.  It makes me wonder why some sort of counterweight system isn't used instead.  Torsion springs seem to have a 90 day warranty generally and are &quot;rated&quot; for up to 15,000 cycles.  It seems like there is a lot less stress on a system that uses counterweights, &lt;a href=&quot;http://hermco.ca/products/2000-series/&quot; rel=&quot;noreferrer&quot;&gt;this company&lt;/a&gt; has a counterweight system for high use places (condo buildings or businesses with large garages or car washes) with warranties for 100,000-500,000 cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it initial cost?  Space required?  Just to keep repairmen in business?  &lt;/p&gt;&#xA;" OwnerUserId="6291" LastActivityDate="2022-06-06T12:43:02.580" Title="Why aren't counterweights used more for garage doors?" Tags="|springs|building-design|" AnswerCount="9" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8693" PostTypeId="2" ParentId="8688" CreationDate="2016-05-02T13:07:58.050" Score="1" Body="&lt;p&gt;If you are unsure exactly what heat the material will be exposed to I would begin with Silicone rubber.  It is relatively cheap and should withstand temperatures up to 260 degrees Celsius with few complaints.  Plus you can usually pick exactly what stiffness you desire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The selection on McMaster Carr is pretty good and for just a few bucks you will know whether or not you need to go searching for a more exotic material.&lt;/p&gt;&#xA;" OwnerUserId="4697" LastActivityDate="2016-05-02T13:07:58.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8694" PostTypeId="2" ParentId="8691" CreationDate="2016-05-02T13:43:36.343" Score="2" Body="&lt;p&gt;There are several ways to lift grains: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;auger/screw conveyor&lt;/li&gt;&#xA;&lt;li&gt;cup elevator - like a vertical conveyor belt with cups mounted to it, cups tilt near top&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There are also systems with little plates on a string, beeing pulled through a pipe (i forgot the name)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of the above are used in large scale grain handling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, given that the quantitiy is 'a handful', I would consider another approach:  Consider a hopper above a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotary_feeder&quot; rel=&quot;nofollow&quot;&gt;rotary feeder&lt;/a&gt; - a bunch of chambers on a wheel. Bulk goods fall into the chamber when it's on top, the wheel turns slowly and the bulk goods fall from the chamber when it's in the bottom position. This allows the dosing of bulk goods. You can use a rotary feeder with a large hopper and forgoe the  conveyor.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-05-02T13:43:36.343" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8695" PostTypeId="2" ParentId="8692" CreationDate="2016-05-02T13:54:06.987" Score="8" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It makes me wonder why some sort of counterweight system isn't used instead.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd guess it's because residential garages are normally a bit limited on space and the counterweights need somewhere to go.  Torsion springs are nice because they take up relatively little room on the ceiling of the garage, as opposed to counterweights, which would need space to travel vertically &lt;em&gt;and&lt;/em&gt; a footprint in which to travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a builder was thinking ahead, he might be able to find a way to get counterweights inside of a wall, similar to what is done with counterweighted windows.  This could be an issue later on if a different garage door is used that requires different counterweights since you'd have to open a wall to replace them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A duty cycle of 15,000 equates to about 10 years if a door is opened four times a day, every day of the year.  Most automatic garage door openers &lt;a href=&quot;http://www.phoenixazgaragedoorrepair.com/garage-door-repair/1536/what-is-the-life-expectancy-of-a-garage-door/garage-door-blog/&quot; rel=&quot;noreferrer&quot;&gt;can be expected to last 10-15 years&lt;/a&gt;.  You didn't mention if you have an automatic opener, but it would make sense to tie the replacement of both of these items to the same timeframe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Safety is another concern.  With torsion springs, if one breaks, it stays within the shaft/tube that it's mounted to.  If an extension spring breaks, it can go flying anywhere in a garage (unless restrained by safety cables).  If a counterweight rope breaks, the weight can fall on you, your child, your pet, or your expensive car.  In all cases, if the supporting mechanism fails, the door can fall onto something, assuming it's not also restrained by an automatic door opener.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Just to keep repairmen in business?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm sure planned obsolescence is part of the issue here, but not necessarily the main culprit.  All mechanical devices have a service life, and it's usually a balancing act between service life and cost when it comes to specifying a system.  I can't see why a counterweight system would cost more than a spring system, but I'm not a garage door installer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could always ask your local garage door installer if they offer a counterweight system for a residential garage.  I can't seem to find anyone online offering these for residential garage doors, though.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2016-05-02T16:39:34.777" LastActivityDate="2016-05-02T16:39:34.777" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8697" PostTypeId="1" AcceptedAnswerId="8698" CreationDate="2016-05-02T18:17:17.700" Score="2" ViewCount="174" Body="&lt;p&gt;When building a 12 m tall industrial shed with a peaked roof it is desirable that the sides of the shed should stay open as much as possible for ventilation / safety / access concerns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, an entirely open side from top to bottom might cause a lot of rain exposure when the rain comes in angled?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a trade-off possible? Perhaps to enclose sides till a certain distance below the roof edge? Any heuristics about how much? (I've shown 4000 mm in the sketch below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fnqTL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fnqTL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contents inside the shed are relatively robust / waterproof &amp;amp; hence absolute rain protection is not needed. &lt;/p&gt;&#xA;" OwnerUserId="6277" LastEditorUserId="6277" LastEditDate="2016-05-02T18:28:21.793" LastActivityDate="2016-05-02T18:54:01.203" Title="Siding for Industrial Shed: Rain protection vs Ventilation" Tags="|structures|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8698" PostTypeId="2" ParentId="8697" CreationDate="2016-05-02T18:54:01.203" Score="3" Body="&lt;p&gt;This will very much depend on the exposure of the building to wind as this is what will drive the rain in at an angle. So you really need to consider the prevailing wind direction and how sheltered the building is by other nearby structures etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If may be that if the prevailing wind direction is fairly consistent you can have one side more open than another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to have panels which can be folded  down when required eg if you have 4000mm of fixed cladding, you could have the same in fold-down form (this arrangement is fairly common in barns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it depends what access you need obviously forklifts will need more height clearance than pedestrians. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also consider using strip curtains for at least part of the height. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-02T18:54:01.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8699" PostTypeId="1" CreationDate="2016-05-02T20:32:00.077" Score="1" ViewCount="84" Body="&lt;p&gt;at the begining,in this models inner diameter 2in,outer diameter 4 inc.I want to find out changes of the shape after compression as inc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yXa7F.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yXa7F.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;R9.jpg&lt;/p&gt;&#xA;" OwnerUserId="6297" LastEditorUserId="6297" LastEditDate="2016-05-02T21:22:23.493" LastActivityDate="2016-05-02T21:22:23.493" Title="how to find minor and major dimension after compression on ANSYS" Tags="|mechanical-engineering|structural-engineering|civil-engineering|ansys|" AnswerCount="0" CommentCount="0" ClosedDate="2016-05-03T12:31:26.743" ContentLicense="CC BY-SA 3.0" />
  <row Id="8700" PostTypeId="2" ParentId="8688" CreationDate="2016-05-02T20:32:20.853" Score="4" Body="&lt;p&gt;Carbon fibre or kevlar textile might work as a soft covering on an aluminium backing plate &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of metals high purity annealed copper &lt;em&gt;might&lt;/em&gt; be soft enough on it's own. You can also get self adhesive copper foil tape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also be worth considering using semi-consumable sleeves or collets. For example a hardwood dowel, bored to the appropriate diameter with a couple of slits along its length might do the trick. Even if it chars a bit at the end they might last long enough to be worthwhile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disposable sleeves also have the advantage that there is less risk of picking up swarf or grit that could scratch the  glass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly it might be worth looking at O rings as a cheaper way to protect the rod than manufacturing jaws from stock, or maybe even use PTFE tape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might also be worth trying some dipping the glass in a natural or synthetic rubber to protect it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of heat shielding maybe look at small, disposable aluminium pie tins you could just punch a hole in the middle and suspend it by the glass rod itself. this might give just enough protection to make an otherwise marginal material viable. Similarly blowing air across the chuck with a portable fan or shop vac on blow mode might do just enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible option is &lt;a href=&quot;http://www.tufnol.com/materials-full/glass-fabric-grades/tufnol-grade-10g-50.aspx&quot; rel=&quot;nofollow&quot;&gt;Tufnol&lt;/a&gt;, this is a phenolic-resin-based composite, laminated with glass, paper, or other textiles and has somewhat better heat resistance than a lot of other composites and is reasonably cheap and available as sheet stock. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="2246" LastEditDate="2016-05-03T15:00:52.603" LastActivityDate="2016-05-03T15:00:52.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8701" PostTypeId="2" ParentId="8668" CreationDate="2016-05-02T22:33:31.360" Score="0" Body="&lt;p&gt;Your main aim is to draw more water in short time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;flow rate&lt;/em&gt; through a system such as yours is a balance between the pumps abilities (head curve) and system friction loss (which increases with flow rate). To increase the flow rate - reduce system losses - replace pipelines with those of large diameter (say 1 inch).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;location&lt;/em&gt; of the pump does not influence the flow rate - as long as adequate suction pressure is available. To ensure the later, keeping the pump at the bottom is better (as discussed in other answers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want to raise the water &lt;em&gt;higher&lt;/em&gt; with a single pump, then consider keeping it on the top (provided the pressure in main line is sufficient to provide the list 14.7 psi = 10.33 m (34 ft) of lift)&lt;/p&gt;&#xA;" OwnerUserId="6299" LastActivityDate="2016-05-02T22:33:31.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8702" PostTypeId="2" ParentId="5372" CreationDate="2016-05-03T00:11:07.917" Score="2" Body="&lt;p&gt;The reason sensor error is distributed among all those variables is because different applications have drastically different requirements. Let;s take out the errors you can't do anything about: the non-repeatable errors and hysteresis. They aren't linear or stationary, and as such, it's hard to compensate for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many sources of non-linear noise, one of which is temperature effects. Those, if you so choose, you could map out the non-linearity in detail and calibrate the sensor to remove much of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ultimately, it comes down you what it is you want to detect. Sensor datasheets will provide a noise density. The larger the measurement bandwidth, the more noise the system will let in. So match the sensing bandwidth to the signal you want to detect. Understand the temperature effect and calibrate/compensate if necessary. That's as general an answer as your original question.&lt;/p&gt;&#xA;" OwnerUserId="6301" LastActivityDate="2016-05-03T00:11:07.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8703" PostTypeId="2" ParentId="8301" CreationDate="2016-05-03T00:36:43.767" Score="2" Body="&lt;p&gt;I'm new here, however at this sort of business for a long time.  Yes die cutting is great and has low $ cost for short and med. runs. I have even made a blanking die (steel rule or clicker) for 5 parts and smacked them out with a hammer and block of UMHW. There are jewelry grade prototype systems (kits) see RioGrande for blanking small parts from thin material.  For limited runs I have sent my drawings out for acid etch blanking.  Just remember to let them know what piece you want back, so you don't get a really a really nice sheet full of fancy holes.  The acid thing is also great when it comes to the tough materials.&lt;/p&gt;&#xA;" OwnerUserId="6302" LastActivityDate="2016-05-03T00:36:43.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8704" PostTypeId="2" ParentId="8692" CreationDate="2016-05-03T01:40:46.303" Score="0" Body="&lt;p&gt;I think the many reasons include space, complexity in linkage and the hardware store can cram it in a short fiber mass-produced one size fits all box.  I have never looked at spring loaded doors as the paragon of safety having installed and demolished a freighter load of them with many scary near enough misses regardless of proper form of action.  Honestly I think counter weights are great.  I have owned houses with window sash weights beautifully hidden, screen doors closed by weighty junk hung on fishing line, however they are light compared to garage doors. Most municipalities feel a permit and fee is in order regardless of the quality of junk (oops) product sold to and installed by the licensed &quot;Professional&quot; My guess?&lt;/p&gt;&#xA;" OwnerUserId="6302" LastActivityDate="2016-05-03T01:40:46.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8708" PostTypeId="1" CreationDate="2016-05-03T07:01:08.160" Score="1" ViewCount="193" Body="&lt;p&gt;Are there any heuristics / thumb-rules to evaluate whether a shed-structure has been efficiently designed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be specific, I have an industrial shed, 12 m tall, 270 m2 floor area (longest span 10 m)  &amp;amp; the structural design submitted needs approx. 22 tons of structural steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any thumb rules to judge if this is reasonably efficient? Or has the designer lazily just submitted an overly conservative design?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a non-snow location with no roof access &amp;amp; no winches / gantry cranes the roof loads ought to be really light. Intuitively 22 tons seemed a lot of steel for this to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If adding more details or even the entire design would help judging I can do that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps others can give me numbers from other projects? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rWs9m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rWs9m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SNZ1r.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SNZ1r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6277" LastEditorUserId="64" LastEditDate="2016-05-04T07:05:01.830" LastActivityDate="2016-05-04T07:05:01.830" Title="Heuristics for Efficient Structure (Shed) Design" Tags="|structural-engineering|structures|engineering-economics|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8709" PostTypeId="1" AcceptedAnswerId="8741" CreationDate="2016-05-03T08:27:38.007" Score="4" ViewCount="834" Body="&lt;p&gt;I've started playing around with brazing pieces of extruded aluminium sections to join them together for strong joins as an alternative to bolting them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pieces are made of 6082-T6 aluminium alloy. I've found that the heating needed to braze softens the aluminium enormously, to the point where 3mm thick sections can be effortlessly deformed with pliers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done a lot of searching and I need to either naturally or artificially age-harden it (precipitation hardening). But how long it takes to naturally harden, or the precise process needed to artificially harden it (temperature &amp;amp; time) escapes me. Also, quenching might help, but I haven't found any definitive answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I re-harden the alloy?&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-05-04T05:15:59.343" LastActivityDate="2016-05-05T20:29:16.353" Title="6082 Aluminium softened by brazing. How to harden?" Tags="|materials|brazing|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8712" PostTypeId="2" ParentId="8692" CreationDate="2016-05-03T11:18:59.930" Score="1" Body="&lt;p&gt;I'll just list some reasons counterweights are bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You need vertical space reserved for them, as opposed to the unused ceiling space for springs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If a cable fails, both the door and the counterweight will fall down. Fast. And hurt something.  (extension springs are required to have a retention cable run thru them, and torsion springs have nowhere to go)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Counterweights have to dangle freely (unless you're planning on installing a very expensive raceway); which implies high probability of interference or jamming.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-05-03T11:18:59.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8713" PostTypeId="2" ParentId="8708" CreationDate="2016-05-03T11:26:03.797" Score="4" Body="&lt;p&gt;Reading your question, my understanding is that you trust your engineer to have done at least a reasonable job designing the structure. You aren't worried the structure might collapse, that the engineer might have forgotten a load case or that they might have chosen inefficient cross-sections types (i.e, they didn't use rectangular columns). I understand you are only concerned with knowing whether the structure, as is, is inefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are therefore two ways the structure might be inefficient: the structural plan is poorly designed and/or the cross-sections adopted for each element are excessive and therefore inefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 10-meter span sounds pretty reasonable at a glance, as does the elevation presented, so I don't think we need to concern ourselves with the &quot;poorly designed plan&quot; scenario. This leaves us only to check the cross-sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel elements are often designed according to a &quot;utilization rate&quot;. This is a value which goes from 0.0 to 1.0, where 0.0 means the element is not under any loading under any circumstance and 1.0 means that under some load combination the element is having to resist forces exactly equal to its design strength. Obviously, a utilization rate above 1.0 means the element risks failure (though I believe some codes allow rates slightly above 1.0 in some situations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy way to check your engineer's work is to look at their design calculations for each element and check the utilization rates (they may also be expressed in terms of &quot;this element suffers $X$ and can resist $Y$, $X &amp;lt; Y \therefore \text{OK!}$&quot;, in which case the rate is equal to $\frac{X}{Y}$). If they are all close to 1.0, then the structure isn't overly conservative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly counts as &quot;close to 1.0&quot; is tricky to define. Steel elements have discrete cross-sections; you can choose between section A or B. In some cases, it may be that section A is insufficient (utilization rate above 1.0), while section B is quite a bit bigger, in which case the engineer will have no choice but to select B, even though its rate isn't very efficient. Therefore, unless you're an engineer yourself and can actually redo your engineer's math testing other possible cross-sections, I suggest you be generous in what you consider &quot;close to 1.0&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if all of the elements are requiring your generosity, then that probably means that a slight modification to the plan would improve significantly your global design. For instance, a 10-meter span between the supporting frames shown in the elevation may mean that the columns fell under the &quot;section A or B&quot; scenario described above. In such a case, it may be that reducing that span just a bit allows you to improve significantly the efficiency of your columns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, if you have a continuous element, you might have different rates for different parts of the element. For example, a continuous beam with multiple spans of different lengths might have a rate for each span. The engineer must then decide between making a uniform beam which is equal for all spans (and which will therefore be a bit inefficient in the smaller spans) or use different cross-sections for each span. The latter choice may be more material-efficient (since you optimize the cross-section for each span), but may very well actually be more expensive due to the increased complexity of the joints. The decision between these two choices is not trivial, so I honestly suggest accepting your engineer's better judgement (if you have any confidence in it). The same applies to columns: those at the corners will usually have to resist lower forces than the central ones, so the engineer must decide between adopting one cross section for all columns or different sections for each one. In this case, it is very common to adopt one section for all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are truly concerned about the design, another option would be to hire a second engineer to do a quick &quot;quality control&quot; check of your project.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-05-03T11:26:03.797" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="8714" PostTypeId="2" ParentId="8547" CreationDate="2016-05-03T12:45:39.837" Score="2" Body="&lt;p&gt;Grain growth is the movement of grain boundaries by diffusion to reduce the grain boundary area. This graph tells that at 600 degrees, a treatment of one minute shall give the y coordinate of the 600 degrees graph. So treating a sample for 1000s shall mean heating the sample at 600 from time t=0.0. The starting point is inherent in the process itself. &lt;/p&gt;&#xA;" OwnerUserId="6194" LastActivityDate="2016-05-03T12:45:39.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8715" PostTypeId="1" CreationDate="2016-05-03T12:50:53.860" Score="1" ViewCount="921" Body="&lt;p&gt;I was trying to use the new interface of ANSYS Workbench to couple ANSYS with Excel without learning Python. I'm having problems because I can only set one parameter in Excel, like one cell. Once I managed to put an array, but I can't do this right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to do is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem has: input x, input y, output z, output w. What I am doing is using the interface in ANSYS, exporting the data, copying to another Excel spreadsheet and after that passing to Origin for better graphs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to do is have four columns in Excel, put something like 20 values in input x, input y, then update the Workbench project with all design points, and my other two columns in Excel get filled automatically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I can't put an array as a parameter in ANSYS, I am missing the automation of what I want.&lt;/p&gt;&#xA;" OwnerUserId="5097" LastEditorUserId="2246" LastEditDate="2016-05-03T17:28:56.577" LastActivityDate="2016-05-03T17:28:56.577" Title="Linking ANSYS workbench with excel" Tags="|finite-element-method|ansys|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8718" PostTypeId="2" ParentId="8613" CreationDate="2016-05-03T19:01:43.220" Score="1" Body="&lt;p&gt;OK, so after having a couple more roofing gurus come out, I just had one with an &quot;ahha moment&quot; that seems so obvious now looking back on it.  So far this seems to be the best solution to my particular problem, and may help somebody else in the future with a toe-bearing rafter situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is simply build a new roof on top of the old one.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HaBR7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HaBR7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By running new 2x6 rafters from the ridge down over the existing sheathing, I get the overhang I want, get my R-19 insulation, don't have to try and take the sag out of the old rafters, don't have to worry about an expensive demolition.  I will plan on shimming the new rafters over the purlins where the old rafters have sagged, to make sure the new rafters are supported by the purlins from Day 1 and don't start sagging too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some questions that come to mind with this solution are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;will the existing steel ridge beam handle the weight of the new roof and any snow loads?&lt;/li&gt;&#xA;&lt;li&gt;how best to fasten the new rafters at the tail end?&lt;/li&gt;&#xA;&lt;li&gt;can the new rafters remove some stress off the old rafters?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Anyhow, thought I'd share this contractor's simple, yet clever solution.&lt;/p&gt;&#xA;" OwnerUserId="3830" LastEditorUserId="1832" LastEditDate="2016-05-03T20:34:27.027" LastActivityDate="2016-05-03T20:34:27.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8719" PostTypeId="1" CreationDate="2016-05-03T19:12:21.807" Score="3" ViewCount="4177" Body="&lt;p&gt;Is there a material that has higher kinetic friction than static friction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume it's sliding on any surface you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better if you come up with one that has this property on a standard flooring material: steel, wood, smooth concrete, tile, asphalt, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also better if the material has this property at a macro scale.&lt;/p&gt;&#xA;" OwnerUserId="5924" LastEditorUserId="6790" LastEditDate="2018-05-07T12:11:21.653" LastActivityDate="2022-09-09T00:45:13.917" Title="Is there any material that has a higher kinetic friction than static friction?" Tags="|materials|friction|tribology|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8720" PostTypeId="1" CreationDate="2016-05-03T20:31:13.103" Score="1" ViewCount="315" Body="&lt;p&gt;For vehicles that cannot steer (built to travel in a straight line) does having the wheels toe inwards help with straight line stability? That is, will it make the vehicle go straighter with slight construction asymmetries and frame warping?&lt;/p&gt;&#xA;" OwnerUserId="5924" LastEditorUserId="90" LastEditDate="2017-11-03T01:23:53.070" LastActivityDate="2017-11-03T01:23:53.070" Title="Does toe-in in the wheels of a car help with straight-line stability without steering?" Tags="|automotive-engineering|wheels|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8721" PostTypeId="2" ParentId="8719" CreationDate="2016-05-03T20:33:04.237" Score="0" Body="&lt;p&gt;I dont think so. as kinetic force is always lesser than static.&#xA;Here is a similar discussion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/541/can-the-coefficient-of-static-friction-be-less-than-that-of-kinetic-friction&quot;&gt;Similar question&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link is about the theory behind the static and kinetic friction. the coefficient of static friction can never be less than the coefficient of kinetic friction. Having greater kinetic than static friction just doesn't make any sense.&lt;/p&gt;&#xA;" OwnerUserId="6215" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2016-05-04T14:21:48.840" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8722" PostTypeId="2" ParentId="8719" CreationDate="2016-05-03T20:40:19.253" Score="5" Body="&lt;p&gt;Friction is not a property of a material, it is a property of the interface between two objects, which may be the same or different materials. One interface I can think of that has that property is a thin layer of grease -- the force required to move increases with velocity.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-05-03T20:40:19.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8723" PostTypeId="2" ParentId="8719" CreationDate="2016-05-03T21:18:08.417" Score="3" Body="&lt;p&gt;Sliding friction must, logically always be less than static friction. This becomes obvious when you look at it in terms of forces rather than coefficients. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a block resting on a flat surface. You apply a small force to one side of the block to try to slide it along. Initially it doesn't move as the reaction force provided by friction increases in line with the force you exert. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say that applying 10N of force is just enough to overcome static friction and get the block moving. Now if the sliding friction becomes greater the 10N force is no longer enough to overcome it and so it stops and your back to a static condition.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at this is that friction is a reaction force so it can never be greater than the force applied otherwise you would be effectively getting work from nowhere.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it is certainly possible to have materials whose friction properties change as &lt;em&gt;a result of&lt;/em&gt; friction, for example carbon-carbon brakes are most effective at quite high temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also viscous drag which is proportional some function of speed although it is not really very meaningful to talk about static friction in this context. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-03T21:18:08.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8724" PostTypeId="1" CreationDate="2016-05-04T01:37:55.930" Score="1" ViewCount="622" Body="&lt;p&gt;I had a look at this &lt;a href=&quot;http://theconstructor.org/practical-guide/hydraulic-coefficients-of-orifice/2029/&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is the orifice is located on the bottom of the tub, instead of the side. The water jet will project downwards, instead of with an X and Y component (as mentioned in the article).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other methods of calculating the coefficient of contraction of a orifice where the orifice is located on the bottom of the container or do I just have to estimate ~0.62 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://drive.google.com/file/d/0B55SLbVVgR4XaGV4U2txWHRMcUk/view&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; to see my experiment.&lt;/p&gt;&#xA;" OwnerUserId="6316" LastEditorUserId="1832" LastEditDate="2016-05-04T01:57:28.767" LastActivityDate="2016-05-04T16:52:11.707" Title="Practical method of calculating coefficient of contraction?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8725" PostTypeId="1" CreationDate="2016-05-04T02:32:27.230" Score="2" ViewCount="1632" Body="&lt;p&gt;The title pretty much sums up the question, but basically I need to create 3 sizes of gears. I have tried 2 shapes, the first being basic rectangles. However, according to my Engineering teacher, neither of those designs will work. So now I am going to try a spur gear, but I have no idea how to get the right angle and everything else, especially seen as the shape of each spur will change with the gear diameter. So how do I go about designing a spur gear in autocad? Any (free) auto cad plugins or calculators are appreciated as well.&lt;/p&gt;&#xA;" OwnerUserId="5878" LastActivityDate="2016-08-20T22:18:17.177" Title="Working out functional gear in AutoCad" Tags="|design|gears|drafting|autocad|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8726" PostTypeId="1" AcceptedAnswerId="8729" CreationDate="2016-05-04T04:04:37.290" Score="7" ViewCount="720" Body="&lt;p&gt;I understand that the concept of line balancing is such that each station takes exactly the same amount of time to process one unit. Clearly, this is an idealization and real manufacturing plants will have bottleneck stations and underutilized stations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if someone has an idea on what numbers are typical in various industries? For instance, if there is a 10% take difference between the slowest and fastest stations, is the line considered balanced enough? &lt;/p&gt;&#xA;" OwnerUserId="6319" LastEditorUserId="1586" LastEditDate="2016-05-12T19:33:18.600" LastActivityDate="2016-05-12T19:33:18.600" Title="Line balancing - How far do companies go?" Tags="|manufacturing-engineering|engineering-economics|" AnswerCount="2" CommentCount="1" ClosedDate="2016-05-04T13:57:55.910" ContentLicense="CC BY-SA 3.0" />
  <row Id="8727" PostTypeId="2" ParentId="8726" CreationDate="2016-05-04T05:19:00.797" Score="2" Body="&lt;p&gt;It will depend on the industry. Competition drives new processes, techniques and technology that improve efficiency. Also, the larger the market, and market share of a company, the greater the economies of scale. Also, the certainty of a particular station being able to process parts at a precise speed will depend on factors such as its complexity and the nature of the parts or materials used to make it, changing perhaps from one hour to the next, or one day to the next, for example, between 80% and theoretically at least, 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a lot of laboratory glassware is surprisingly still largely hand-made. Tools and techniques have improved, and no doubt will continue doing so. But automation always becoming cheaper. But the bespoke nature means it is unlikely to ever warrant factory lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But on the other side of the scale, although I don't have figures, common hardware such as nuts and bolts are produced on such large scales, and the materials so consistent, and the process so completely automated, that the utilisation of posts is with little doubt above 90%. But this is then open to debate exactly what the definition here of efficiency is. For example, if it is the capacity of an average person with a given skill set, or a given machine at full speed, the economy of a machine at a certain speed: full speed may incur higher energy consumption per part, dramatically increase wear on its parts, or reduce quality of parts produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how far do companies go? It's down to their bottom line; whatever makes more money.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-05-04T05:57:58.640" LastActivityDate="2016-05-04T05:57:58.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8728" PostTypeId="1" CreationDate="2016-05-04T07:18:48.193" Score="3" ViewCount="246" Body="&lt;p&gt;We are looking for a solution to hold two Power cables on a 3D printed part. This part is installed in a vibrating environment. So we need the cables to be pressed not to move and not to transfer vibrations to electrical components. We have thought about a C shape holder, but it would surely not maintain cables enough. We also looked for components sold on the market but nothing meets our requirements. Here is a picture to help you to understand. Thank you in advance !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aSNkS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aSNkS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6308" LastActivityDate="2016-05-11T10:20:06.410" Title="How to hold electrical wires on a part?" Tags="|electrical-engineering|3d-printing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8729" PostTypeId="2" ParentId="8726" CreationDate="2016-05-04T07:25:27.883" Score="9" Body="&lt;p&gt;I can answer for the Chemicals Sector I'm familiar with: Overall line balancing is quite poor between processing operations. 10% difference would be surprisingly good. 50%  difference isn't uncommon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several reasons for this I can think of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lot of equipment comes in discrete sizes. e.g. Compressors, Centrifuges etc So often the next size higher means you have a lot of slack built in that equipment from Day-0.&lt;/li&gt;&#xA;&lt;li&gt;Often the exact max capacity of a complex piece of equipment like a distillation column is not known with 100% accuracy. So designers incorporate a factor of safety in mentioning the rated capacity (that will become a contractual obligation). Once plant is operational it can be pushed to max capacity which is often higher than the on-paper capacity the equipment was designed for. Ergo now other pieces of equipment become bottlenecks. And you have an unbalanced line. &lt;/li&gt;&#xA;&lt;li&gt;Some equipment like Heat Exchangers or Plug Flow Reactors will foul over their working life and hence become less effective. There can be a wide margin between their Day-0 throughput and 6 months down the line. As a result, other equipment not suffering from such in-service derating will end up showing excess capacity. &lt;/li&gt;&#xA;&lt;li&gt;Often Capex pricing of specific equipment is highly non-linear with respect to capacity. e.g. You could be adding 50% extra capacity at only 10% extra price. So designers end up having unbalanced lines simply by hedging their risks by over-specifing where the cost penalty isn't high.&lt;/li&gt;&#xA;&lt;li&gt;Over the long life of a plant (e.g. 20 yrs) some innovations come up where with modest investment a equipment can be significantly boosted in throughput. e.g. Better generations of catalysts. The parts of the train that &lt;em&gt;cannot&lt;/em&gt; be innovated upon then become bottlenecks. &lt;/li&gt;&#xA;&lt;li&gt;The lead times for new equipment commissioning can be long. e.g. 3 years. So sometimes management will knowingly add larger equipment, consciously realizing that the new equipment will stay underutilized for a while till the other parts of the line can be upgraded.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Employees &amp;amp; site-engineers come up with operating innovations that debottleneck a specific equipment. Suddenly the rest of the line is underutilized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;8.Chemical processing lines can make multiple products. e.g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A--&gt;B--&gt;C--&gt;D--&gt;E&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;where both C &amp;amp; E are salable products. The plant was designed line-balanced assuming you'd sell 500 tons of E &amp;amp; 300 tons of C every month. Unfortunately 5 years down the line the market for C is bad. So parts of the line pre-C become underutilized although the rest of the line is still balanced. &lt;/p&gt;&#xA;" OwnerUserId="6277" LastActivityDate="2016-05-04T07:25:27.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8730" PostTypeId="1" CreationDate="2016-05-04T08:00:15.360" Score="2" ViewCount="1833" Body="&lt;p&gt;Usually a hinge looks like this. And pretty straightforward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[![Example of a hinge design][1]][1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are trying to develop something where I add steps, and I control the angle adjustments in those steps. Lets say 15deg, and a total of 90 degrees usage. &#xA;Let's say a monitor, you can tilt it around 15-30 degrees in one dimension, similarly, the application is to basically limit the angles to 6 in the step of 15 deg. So the user decides 0,15,30,45,60,75,90 deg (max)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By adding notches like you had mentioned, would it not restrict the motion backwards? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l4rKo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l4rKo.png&quot; alt=&quot;example of how the design looks like&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of mechanism will be used. My guess is a rachet has to be there. &#xA;But I can't seem to get the design in my head, nor any other product which has this.&#xA;Also for a tight clearance, I think 0.05mm would do, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Ethan I have a name for it. Detent hinges are a viable option. &#xA;But I am afraid if it will add a lot of time to design a new one. Or if it should be picked up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked a few of the designs. I did not really understand the working of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7UceB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7UceB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6215" LastEditorDisplayName="user748" LastEditDate="2016-10-01T23:11:04.600" LastActivityDate="2023-07-25T14:00:01.247" Title="Repositionable ratchet hinge design" Tags="|mechanical-engineering|design|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8731" PostTypeId="2" ParentId="8728" CreationDate="2016-05-04T08:07:02.553" Score="7" Body="&lt;p&gt;A clamp and strain relief are used all over the place. You'll see this everywhere from domestic electrical plugs to the cable exit of power tools. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of cable and sheath/insulation is a factor too. For example, vehicle power cables are designed to handle vibrations (but also high temperature and to be light weight, so maybe not the perfect compromise for your use). Thinner cables might help you, so optimise that, perhaps. Similarly, the extra flexibility of finer stranded conductors would help, as might using multiple wires for each power rail, and using individual wires instead of multi-core cables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing the strain by way of a &lt;a href=&quot;https://www.google.co.uk/search?q=cable%20machine%20holder&amp;amp;safe=off&amp;amp;client=tablet-android-samsung&amp;amp;prmd=sivn&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwjsz8Tb_L_MAhVFaRQKHc-ABfEQ_AUICCgC&amp;amp;biw=1024&amp;amp;bih=768#safe=off&amp;amp;tbm=isch&amp;amp;q=cable%20drag%20chain&quot; rel=&quot;nofollow&quot;&gt;cable drag chain&lt;/a&gt; etc will mean the clamp can be less substantial, strain reliefs can be more flexible, and wear on the cable sheath reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with automotive and aviation electrics, use rubber grommets where cables pass through enclosures. Use cable sleeving on long runs subject to abrasion. Soldering wires can weaken strands when flexed, so crimp or fasten them instead. Do &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; tin (solder) wires that enter screw terminals, as it will loosen with time due to the softness of the solder; the same goes for crimp terminals. Instead, use bootlace ferrules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your position, it sounds like you'll need to experiment. Try a design, and when there's a failure, improve the design to address that specific weakness. You might need to use a tougher PCB, a tougher cable clamp, tougher connectors, more flexible cables, tougher sheaths. Look at other design solutions: connections between loud speakers' voice coils and terminals, other 3D printers... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When all else fails, glue-gun the heck out of it. It has been done! Electronics can be encapsulated in specialist epoxies to prevent vibrations weakening components' leads (among other reasons). Just remember to allow for heat dissipation - most circuit designs will have some components that are at their limit already.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-05-11T10:20:06.410" LastActivityDate="2016-05-11T10:20:06.410" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8732" PostTypeId="2" ParentId="8730" CreationDate="2016-05-04T09:13:50.430" Score="5" Body="&lt;p&gt;I agree: since a monitor is quite heavy in relation to the strength of the person adjusting its position, a sprung notch system where torque can move the arm from one position to another wouldn't work. However, the notches could hold a key in place until a lever is used to release the key. A ratchet system would work if the weight of the monitor exerts a torque only in one direction. As it happens, I have a monitor stand which does exactly that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/edEiI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/edEiI.jpg&quot; alt=&quot;monitor stand picture&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the monitor is pushed up, the ratchet system clicks to hold it in place. The lever is pressed to lower it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearances in that design are incredibly tolerant. Exactly which part were you referring to? As long as the latch won't let the monitor drop, you could have clearances of 10mm if you wanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The post important concern should be how safe it is to use. If heavy monitors are used, the screen would be dangerous if the ratchet fails. Also, consider every safety aspect you can. Fingers getting caught in mechanisms,  wires getting steered in mechanisms, children playing with it, you name it. You will need to CE mark it, which means you can be held personally responsible if someone gets injured while using it, or indirectly.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-05-05T12:09:21.090" LastActivityDate="2016-05-05T12:09:21.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8736" PostTypeId="2" ParentId="8725" CreationDate="2016-05-04T11:08:09.430" Score="4" Body="&lt;p&gt;Gear design from scratch is a fairly major process in itself and so the vast majority of gears are made to standardised tooth profiles. Metric gears are usually designed to the &lt;a href=&quot;http://www.metrication.com/engineering/gears.html&quot; rel=&quot;nofollow&quot;&gt;module system&lt;/a&gt;. This specifics gears according to pitch circle diameter and number of teeth (the module number is the ratio between the two). So a larger module means finer teeth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the fiddly calculations of tooth profile, clearances etc are covered by the standard so you don't have to reinvent the wheel every time you design a part with gears in it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves the designer to specify thing like the thickness of the gear (which determines its torque capacity) as well as the arrangements for fixing it to a shaft. Although in practice this often means picking an off the-shelf gear set rather than getting them manufactured as stock gears will cover almost all requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the module system you need to work out your gear rations by number of teeth, you can't just have arbitrary diameters as for obvious reasons you can't have fractions of teeth on meshing spur gears. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For assembly drawings it is common to draw meshing gears as circles of the appropriate PCD and use a note to specify the module. AutoCad probably has a feature to draw the teeth for you based on this information but I don't know for sure. In any case, in modern context,  it is not usually necessary to draw the detail of a gear in an engineering drawing as it will be specified elsewhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also be aware that for meshing spur gears the dimensions and profiles of the teeth must be the same regardless of the diameter of the gear ie you cant just draw a template gear of one size and scale it to give different rations. Similarly for a given tooth profile there will be a lower limit to the diameter which will physically accommodate the number of teeth required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an academic project, properly specifying stock gears should be perfectly adequate  as long as you set out the criteria used (unless of course the assignment is to design the gear profile form scratch, in which case you would hope that the basics would have been covered by tuition...) &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-08-20T22:18:17.177" LastActivityDate="2016-08-20T22:18:17.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8737" PostTypeId="2" ParentId="8720" CreationDate="2016-05-04T11:39:57.030" Score="1" Body="&lt;p&gt;I rather doubt it.  Unless the resistance to drift for each tire is significantly asymmetric, i.e. each tire resists yaw to its side more than the other tire applies yaw to its opposite side, the car will still wander.&lt;/p&gt;&#xA;&lt;p&gt;By way of comparison, take a look at train wheels.  The train rides on two rails (duh), and the wheels' contact region is actually bevelled inwards.  As a result, if the train drifts to the left, the left wheels rise slightly on the left rail, while the right wheels drop proportionately.  .  I don't see a way to do that with car wheels, since they are soft &lt;em&gt;and&lt;/em&gt; have no rails to run up/down on.&lt;/p&gt;&#xA;&lt;h2&gt;edit&lt;/h2&gt;&#xA;&lt;p&gt;Per the &amp;quot;edit2&amp;quot; below, I've removed the erroneous claim to gravity.  Both referenced pages below clearly state that it's a matter of linear distance differences due to wheel radius at the contact point.&lt;/p&gt;&#xA;&lt;p&gt;Good explanation and images at Wiki's  &lt;a href=&quot;https://en.wikipedia.org/wiki/Hunting_oscillation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hunting Oscillation&lt;/a&gt; page.&lt;/p&gt;&#xA;&lt;p&gt;EDIT: I've really addressed camber more than toe-in, but my impression is that overall the comments concerning straight-line tracking are still appropriate. If there's a greasemonkey who can correct me, please do.&lt;/p&gt;&#xA;&lt;p&gt;EDIT2 It's not gravity! If the train drifts to the left, the left wheels contact the rail at an increased diameter so try to rotate more slowly, but their speed of rotation is locked to the right wheels (which are trying to speed up) by the axle, so the wheel set steers to the right. &lt;a href=&quot;https://en.wikipedia.org/wiki/Wheelset_(rail_transport)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Wheelset_(rail_transport)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-11-02T15:04:02.410" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8738" PostTypeId="2" ParentId="7208" CreationDate="2016-05-04T14:44:34.320" Score="0" Body="&lt;p&gt;When material A of coefficient of friction $\mu_A$ and material B of $\mu_B$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If material A is being rubbed with material B.&#xA;The requirement for abrasion and heat generated would affect both parts.&#xA;Hence if the application was for grinding, the tool itself would get affected. &#xA;So it would make sense to use a harder material to smooth the softer material. &#xA;For a steel part, you could use diamond, or even steel itself, but here the point of contact is different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also a lot of other factors come into play, &#xA;if you have rollers, and a part going in might be hot rolled, hence there is temperature difference or material property difference which helps roller material stay in its own geometry. &#xA;Also rollers have to be designed so as to take care of camber.&lt;/p&gt;&#xA;" OwnerUserId="6215" LastEditorUserId="1832" LastEditDate="2016-05-04T17:18:15.413" LastActivityDate="2016-05-04T17:18:15.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8740" PostTypeId="2" ParentId="8725" CreationDate="2016-05-04T16:06:26.753" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Involute_gear&quot; rel=&quot;nofollow noreferrer&quot;&gt;Involute spur gears&lt;/a&gt; have &lt;a href=&quot;https://en.wikipedia.org/wiki/Involute&quot; rel=&quot;nofollow noreferrer&quot;&gt;involute shape&lt;/a&gt;. This means that the point of contact between meshing teeth will move in a straight line (called the line of action). Visualized it will look like a string is unwound from one wheel and wound on another wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H9lEv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H9lEv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;image captured from &lt;a href=&quot;http://woodgears.ca/gear_cutting/template.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;woodgears.ca free gear template page&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the number of teeth and the angle of that line that will decide the shape of the spurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a cheap solution an option? If so there is a gear generator on &lt;a href=&quot;http://woodgears.ca/gear/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;woodgears.ca&lt;/a&gt; that can export to DXF. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. No affiliation to woodgears.ca. It's just a site I know that had a decent gear generator.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2016-05-04T16:06:26.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8741" PostTypeId="2" ParentId="8709" CreationDate="2016-05-04T16:41:08.890" Score="4" Body="&lt;p&gt;I found an article that discusses the heat treatment of 6082:&#xA;&lt;a href=&quot;http://www.journalamme.org/papers_vol32_2/3224.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.journalamme.org/papers_vol32_2/3224.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the interest of having a standalone answer, here's a summary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll be able to recover quite a lot of strength by aging between 130 C and 190 C. The authors recommend treating at 190 C for 6 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some overaging will occur at 190 if you let it sit too long, but the effect is minimal, even at 20 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Higher temperatures can be used (e.g. 1 hour at 220 C) but the risk of overaging increases a lot. Personally, I'd avoid that unless I was pretty confident that my oven would maintain a constant temperature and not swing +/- 10 degrees around the set point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &#xA;Starrise points out solutionizing is a potential step. You'd want to get the part above 575 C for a while and then quench before aging. That would get the entire part to a nice, uniform state. The only downside is a potential for warping when you quench.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If solutionizing isn't an option, postweld aging will still buy back a bit of strength, but it won't be nearly as good as getting true T6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're only able to age, &lt;a href=&quot;http://maxal.com/files/QuickSiteImages/Maxal_Guide_for_Aluminum_Wldg_9-12_doc.pdf&quot; rel=&quot;nofollow&quot;&gt;Maxal&lt;/a&gt; actually recommends starting with T4 and then aging to T6 after welding. That'll get the smallest possible reduction in strength.&lt;/p&gt;&#xA;" OwnerUserId="68" LastEditorUserId="68" LastEditDate="2016-05-04T17:12:01.153" LastActivityDate="2016-05-04T17:12:01.153" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8742" PostTypeId="2" ParentId="8724" CreationDate="2016-05-04T16:52:11.707" Score="1" Body="&lt;p&gt;Unfortunately, these coefficients are not really something that you can derive. They're the combined effect of viscosity, surface tension etc. interacting with the geometry of the orifice. They're usually found experimentally and given either as a single value or best fit function of relevant parameters such Reynolds number, hole diameter/ thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've got enough information about setup, you could go looking for an empirical correlation that matches what you've got. If not, go with the single value that you have already.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2016-05-04T16:52:11.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8743" PostTypeId="1" CreationDate="2016-05-04T19:28:38.113" Score="0" ViewCount="87" Body="&lt;p&gt;The situation is this: At a current project, there's a pair of (parallel, redundant) radial blowers to raise the gas pressure for the downstream consumers (CHP, gas burner). All vibrating machinery should be isolated from the pipeline by compensators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the installed compensators leak. New, better (stainless steel vs. thin rubber) ones have been reordered but will take a few weeks. We need to commission other machinery that depends on the blowers. So the contractor suggested the following interim workaround: Weld pieces of pipe so that the blowers are connected without compensators, install the compensators in a few weeks when they arrive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, everybody knows that machinery needs compensators, but nobody I've asked so far has seen a blower operate without. So no one knows for sure what the problems are, and how grave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my thoughts on possible risks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The vibrations could damage welds &lt;/li&gt;&#xA;&lt;li&gt;The vibrations could damage gaskets&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think both are unlikely. Nevertheless, here's the mitigation so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The blowers are equipped with variable frequency drives, ramp up will be done slowly to avoid startup-shakiness&lt;/li&gt;&#xA;&lt;li&gt;If there's massive vibration with the blowers running, we'll reconsider&lt;/li&gt;&#xA;&lt;li&gt;After the provisional pipeworks are done, and again after the compensators are installed, there will be a pressure test&lt;/li&gt;&#xA;&lt;li&gt;Further down the road, 10% of the welds will be X-rayed. I will make sure that many of those checks will cluster near the blowers &lt;/li&gt;&#xA;&lt;li&gt;I'm thinking about having someone check the gaskets of the flanged connections near the blowers with leak detection spray after every or every few operation days&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;Did I miss a risk that's not covered by the mitigation mentioned above?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some more facts and considerations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;pipeline upstream and downstream is stainless steel with welded and flanged connections&lt;/li&gt;&#xA;&lt;li&gt;upstream and downstream, there's an arch after the compensator&lt;/li&gt;&#xA;&lt;li&gt;downstream, the pipe can move a few cm (not mounted to the wall or anything for 2m)&lt;/li&gt;&#xA;&lt;li&gt;upstream, the pipe is bolted to the floor shortly before the blowers&lt;/li&gt;&#xA;&lt;li&gt;We are talking relatively small radial blowers - 120Nm&lt;sup&gt;3&lt;/sup&gt;/h @ 20mbar $\Delta P$ maximum&lt;/li&gt;&#xA;&lt;li&gt;Max. pressure in the pipeline will be 40-50mbar over pressure&lt;/li&gt;&#xA;&lt;li&gt;The room this is happening in is Ex-Zone 1 or 2, all equipment (including the blowers) is Atex rated and there's a gas warning installed&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastEditorUserId="2246" LastEditDate="2016-05-04T20:01:53.457" LastActivityDate="2016-05-04T20:01:53.457" Title="Operation of radial blower without compensators in gas pipeline - What problems to expect and how to mitigate?" Tags="|mechanical-engineering|pipelines|gas|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8745" PostTypeId="2" ParentId="8709" CreationDate="2016-05-04T21:12:12.280" Score="1" Body="&lt;p&gt;While brazing can be useful for ad-hoc repairs of aluminium AC TIG or MIG welding is vastly preferable for structural applications and in the long run is likely to be less hassle than re heat treating it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of TIG welding in this context is twofold: a) you can use filler metal which is as strong or stronger than the base metal and with an alloy composition optimized for welding; b) you can keep the heat input into the metal to a minimum as the arc is a much higher temperature than a brazing flame and much more concentrated so the heat goes into he weld pool only and not the bulk  material. Also TIG welds are cleaned by the arc itself rather than requiring fluxing, which itself requires more heat put into the base metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From personal experience you can TIG weld aluminum extrusions with little noticeable loss of tensile strength, this is in the application of structural racking for commercial vehicles. I also have expedience of annealing and cold working extruded aluminium for similar applications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously your mileage may  vary depending on application but I would certainly recommend looking seriously at TIG welding for this sort of application &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="368" LastEditDate="2016-05-05T20:29:16.353" LastActivityDate="2016-05-05T20:29:16.353" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8746" PostTypeId="2" ParentId="8552" CreationDate="2016-05-05T00:21:25.923" Score="1" Body="&lt;p&gt;Convert both the pressure and concentration terms as mole fractions.&#xA;Now you have y for gas phase mole fraction and x for liquid phase mole fraction.&#xA;Henry's law is taken as am equilibrium relation in your question, so the equilibrium diagram(y vs x) is just a straight line through the origin with slope H'.&#xA;Taking the gas side mass transfer coefficient on mole fraction basis as ky.&#xA;So, on the equilibrium diagram, the bulk point is given as M(x0,y0) and interfacial concentration as N(xi,yi).&lt;strong&gt;[Note: N must be on the equilibrium curve as the interface concentrations are always in equilibrium.]&lt;/strong&gt;&#xA;Now, we take a quantity (yb*) that is defined as the mole fraction of solute  in gas phase that can remain in equilibrium with (xb), the mole fraction in bulk liquid phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = ky(yb-yi) = kx(xi-xb) = Kg(yb-yb*)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(yb-yb*) = (yb-yi) + ((yi-yb*)/(xi-xb))*(xi-xb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N/Kg = N/ky = H'*N/kx &lt;em&gt;(since yb&lt;/em&gt; is in equilibrium with xb, by definition)*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This give the overall gas phase coefficient Kg  = (1/ky) + (H'/kx).&#xA;Overall liquid phase coefficient can be determined similarly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE : The equilibrium diagram described above is used when the equilibrium relationship isn't linear.&lt;/p&gt;&#xA;" OwnerUserId="6331" LastActivityDate="2016-05-05T00:21:25.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8747" PostTypeId="2" ParentId="8453" CreationDate="2016-05-05T00:28:08.363" Score="0" Body="&lt;p&gt;Yes you can, but only for underdamped response.&#xA;The formula is derived from a second order system response, the one in your question looks to be a first order system but under a PI controller.&#xA;That usually isn't relevant for the formula, given that the parameters are so arranged that the response is underdamped.&lt;/p&gt;&#xA;" OwnerUserId="6331" LastEditorUserId="6331" LastEditDate="2016-05-05T16:59:15.620" LastActivityDate="2016-05-05T16:59:15.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8751" PostTypeId="2" ParentId="8246" CreationDate="2016-05-05T15:42:52.373" Score="3" Body="&lt;p&gt;Actually they have a machine call fine hole machine or call Superdrill machine in the market, and use a type of copper tubing with diameter 0.08mm. Or, can use copper tube 0.08mmm as electrode, machining using electrode discharge machine (EDM).&lt;/p&gt;&#xA;" OwnerUserId="6341" LastActivityDate="2016-05-05T15:42:52.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8752" PostTypeId="1" CreationDate="2016-05-05T17:59:39.293" Score="2" ViewCount="257" Body="&lt;p&gt;I was wondering what is the smallest possible rocket to get to LEO. I know that rocket fuel accounts for 90% the total mass of the rocket. Lets say I have 500g of payload and my rocket is made from an aluminium shell. How much fuel would I need to get to LEO and how tall and wide would the rocket need to be?&lt;/p&gt;&#xA;" OwnerUserId="6342" LastActivityDate="2016-05-05T20:21:57.693" Title="Smallest possible rocket to get to LEO?" Tags="|aerospace-engineering|" AnswerCount="1" CommentCount="5" ClosedDate="2016-05-05T20:26:40.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="8753" PostTypeId="2" ParentId="8752" CreationDate="2016-05-05T20:21:57.693" Score="2" Body="&lt;p&gt;Keep in mind that most of the fuel is used to lift fuel, not payload. The amount of fuel required to lift a given payload is a complicated relationship involving many factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One aspect that limits how small a rocket can be is the cube-square relationship between fuel mass and drag: As the linear dimensions of the rocket scale down, fuel quantity goes down with the cube of the scale factor, but drag only goes down with the square of it. Eventually you get to a point where you can't carry enough fuel to even get out of the atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-05-05T20:21:57.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8754" PostTypeId="2" ParentId="8405" CreationDate="2016-05-05T20:56:03.400" Score="0" Body="&lt;p&gt;You cannot use Mc cabe Thiele because at Rmin it requires infinite number of steps.&#xA;Constantly changing R to see what sticks sounds like a huge amount of work.&#xA;Maybe go with Lewis Sorel method, it's tedious but gives a stage wise calculation and since you know the number of trays already, it will give you more definite results. &lt;/p&gt;&#xA;" OwnerUserId="6331" LastActivityDate="2016-05-05T20:56:03.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8755" PostTypeId="2" ParentId="8692" CreationDate="2016-05-06T03:44:09.710" Score="5" Body="&lt;p&gt;Seems like to me the reason is a counterweight provides a constant offset to the door weight and that weight changes as the door is brought back to horizontal. I suppose there would be a way to design a variable counterweight mechanism but that would end up more complicated(costly) than the springs.   &lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2016-05-06T03:44:09.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8756" PostTypeId="1" AcceptedAnswerId="8758" CreationDate="2016-05-06T08:11:28.473" Score="9" ViewCount="2671" Body="&lt;p&gt;For example, metal tape measures can be extended and stay straight when in a &quot;u&quot; orientation, but collapse the other way round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I presume the same phenomenon is why metal shelving has sheet metal on top, and flanges on the bottom rather than the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm guessing&lt;/strong&gt;: it's because of the direction of compression/tension, and buckling. Since a long piece of material being compressed is more likely to buckle than a short piece. In one orientation, the buckling can only happen when the walls buckle across their height but in the other orientation, the compression acts across the entire length of the channel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both a common-sense approach, and a mathematical approach are welcome; I assume formulas are well established, using &quot;moment of inertia&quot; values for beam profiles or similar.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-05-06T15:14:17.363" Title="&quot;u&quot; vs &quot;n&quot; orientation of channel profiles: why difference in strength?" Tags="|applied-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8757" PostTypeId="1" AcceptedAnswerId="8762" CreationDate="2016-05-06T08:49:27.767" Score="2" ViewCount="73" Body="&lt;p&gt;I want to compare strength of a metal before and after heat treatment, to assess to effectiveness of that treatment. (See my previous question, &lt;a href=&quot;https://engineering.stackexchange.com/questions/8709/6082-aluminium-softened-by-brazing-how-to-harden&quot;&gt;Aluminium softened by brazing. How to harden?&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I want to find the yield point (the strain a material can take without being deformed, ie the range where it remains elastic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two ideas that use a similar setup. Shove a length of material (constant width, height, length, shape) in a vice. Apply a force to the end. Fasten a $1 digital luggage scale to the other end. Hang a container of water or sand, adding a little each time. Deflection could be measured as the distance from that end to some reference point with a digital caliper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, either:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Apply an increasing force, until the force is no longer proportional to the deflection (since up till the yield point, materials bend according too hooke's law). Or,&lt;/li&gt;&#xA;&lt;li&gt;Apply a force and release. Repeat with increasing forces until the material no longer returns to its original shape.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The problem is, to achieve any real accuracy, both 1 and 2 above would involve 10's of manual iterations per assessment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides buying or hiring a rheometer, are there any practical alternatives?&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-05-06T23:02:21.570" Title="Low-tech ways of assessing relative strength of a material (yield point)" Tags="|measurements|yield-point|strength|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8758" PostTypeId="2" ParentId="8756" CreationDate="2016-05-06T11:31:26.920" Score="9" Body="&lt;p&gt;You are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A channel section in a &quot;u&quot; or &quot;n&quot; orientation is obviously asymmetric around its transversal x-axis. This means that its centroid is not located at the midpoint of its height. Instead, the centroid will be closer to its web than to the opening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the stress at a given fiber at a distance $y$ from the centroid is equal to&#xA;$$\sigma = \dfrac{My}{I}$$&#xA;in bending, this means that the stresses in the web will be lower than the stresses at the ends of the flanges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using your metal tape measure example, an open tape measure without other supports will behave like a cantilever, with its rotation fixed at the &quot;mouth&quot; of the tape measure. It will therefore be under negative bending moment (compression on bottom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the tape measure is very slender, it is more sensitive to buckling due to compression than to simple collapse due to tension. Therefore, in order to avoid buckling, you want to have your tape measure in a &quot;u&quot; orientation, which will reduce the compressive stresses, while increasing the tensile stresses. In an &quot;n&quot; orientation, the stresses are inverted, with low tension and high compression (and therefore lower resistance to buckling).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-05-06T15:14:17.363" LastActivityDate="2016-05-06T15:14:17.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8760" PostTypeId="1" AcceptedAnswerId="8793" CreationDate="2016-05-06T14:50:46.373" Score="5" ViewCount="1845" Body="&lt;p&gt;I am performing a calculation in which I need to plot a generated response spectrum to a seismic tripartite graph.  However, I am having a heck of a time finding a blank tripartite graph online.  A low-resolution example is shown below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PhEeg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PhEeg.png&quot; alt=&quot;tripartite graph&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;How might I go about generating a blank seismic tripartite graph?  As far as I know, Excel does not have this functionality, since you need to be able to plot four separate axes in 2D with two of the axes at an angle.  All axes are in logarithmic scale.&lt;/p&gt;&#xA;&lt;p&gt;The unit system should be as such:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Spectral displacement: inches&lt;/li&gt;&#xA;&lt;li&gt;Spectral velocity: in/sec&lt;/li&gt;&#xA;&lt;li&gt;Spectral acceleration: % of &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Frequency: Hz&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Otherwise, a link to a high-resolution graph would be acceptable.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Edit January 29, 2021&lt;/p&gt;&#xA;&lt;p&gt;I have figured out how to make one of these graphs using LaTeX.  Please see &lt;a href=&quot;https://www.glennfrazee.com/eng-tools/tripartite-paper&quot; rel=&quot;nofollow noreferrer&quot;&gt;this post on my website&lt;/a&gt; for details.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2021-01-29T17:52:59.080" LastActivityDate="2021-01-29T17:52:59.080" Title="Generating a Blank Seismic Tripartite Graph" Tags="|seismic|frequency-response|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8761" PostTypeId="1" CreationDate="2016-05-06T16:09:24.680" Score="2" ViewCount="1295" Body="&lt;p&gt;I'm building a shelf out of an Oriented Strand Board (OSB-3), reinforced with steel strip attached to it's bottom side. Please check my calculations for deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OSB thickness is $h_{OSB}=18 \text{ mm}$, it's width $b_{OSB}=500 \text{ mm}$, I googled OSB's modulus of elasticity to be $E_{OSB}=3.5 \text{ GPa}$.&#xA;For steel strip it's $h_{st}=4 \text{ mm}$, $b_{st}=60 \text{ mm}$, $E_{st}=200 \text{ GPa}$. The shelf span $L=2000 \text{ mm}$. Load is uniformly distributed $q=500 \text{ N/m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I approach the task with utmost utilitarianism, using online tools and off-the shelf formulas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I transform area of OSB into steel:&#xA;$$b_{eq}=\dfrac{E_{OSB}}{E_{st}} \cdot b_{OSB}=\dfrac{3.5}{200}\cdot 500=8.75 \text{ mm}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I find moment of inertia of uniform inverted T-beam from step 1, using an &lt;a href=&quot;https://skyciv.com/free-moment-of-inertia-calculator/#&quot; rel=&quot;nofollow&quot;&gt;on-line calculator&lt;/a&gt;:&#xA;$$I=16079 \text{ mm}^4$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I obtain max deflection at center&#xA;$$d=\dfrac{5qL^4}{384E_{st}I}=\dfrac{5\cdot500\text{ N/m}\cdot(2000\text{ mm})^4}{384\cdot200\text{ GPa}\cdot16079\text{ mm}^4}=32 \text{ mm}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does OSB modulus of elasticity look legit?&lt;/li&gt;&#xA;&lt;li&gt;Did I miss something, are the calculators and formulas appropriate here?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5053" LastEditorUserId="5053" LastEditDate="2016-05-30T11:47:01.040" LastActivityDate="2016-05-30T11:47:01.040" Title="Composite beam calculation" Tags="|mechanical-engineering|materials|beam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8762" PostTypeId="2" ParentId="8757" CreationDate="2016-05-06T16:10:29.800" Score="1" Body="&lt;p&gt;Determining the yield point for aluminium alloys can be difficult as it is not as well defined as steel and the stress/strain curve tends to gradually peel away from the straight elastic line rather than there being a distinct cut-off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a low tech experiment ultimate tensile strength is easier to measure as you are more likely to observe an obvious failure rather than interpolating from a  graph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly using a cantilever loading setup (i.e. bending it with a mass hung off one end) is a bit easier to arrange and should make the failure point a bit more obvious. This makes it more difficult to derive actual numbers for stress but it will certainly make direct comparison easier, especially if you have a known 'good' sample to compare it to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility for this sort of application is to use a hardness tester, very accurate one's aren't cheap but it's possible that a cheap one may do the trick, depending on the range you are looking at. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if  you are looking at a practical performance test for fabricated assemblies then  basic load test which is reasonably close to real world conditions is as good a bet as anything. &#xA;Bearing in mind also that stress raisers from any joining method may have an effect on practical strength so you may as well check if brazing is any better than bolts for your application. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-06T23:02:21.570" LastActivityDate="2016-05-06T23:02:21.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8763" PostTypeId="1" AcceptedAnswerId="8765" CreationDate="2016-05-06T17:18:34.523" Score="0" ViewCount="3518" Body="&lt;p&gt;Hello I was watching a video recently about aircraft carriers and how they use compressed air to propel aircraft off the ship. It go me thinking if I had, maybe a 50cm x 50 cm x 50 cm spacecraft that was put into orbit. Would I be able to use compressed air as thrust to orientate the space craft? or even better would I be able to use it to give the space craft a greater distance from the Earth?&lt;img src=&quot;https://i.stack.imgur.com/fr0dE.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;img src=&quot;https://i.stack.imgur.com/v93bG.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="6351" LastEditorUserId="-1" LastEditDate="2021-04-28T04:20:50.037" LastActivityDate="2021-04-28T04:20:50.037" Title="Can Compressed air be used as a thrust component in a spacecraft to propel the craft through space or to orient the craft?" Tags="|aerospace-engineering|compressed-air|" AnswerCount="2" CommentCount="2" ClosedDate="2016-05-07T13:48:06.183" ContentLicense="CC BY-SA 4.0" />
  <row Id="8764" PostTypeId="2" ParentId="8761" CreationDate="2016-05-06T17:42:32.590" Score="1" Body="&lt;p&gt;The modulus of elasticity depends on the grade of OSB you're adopting. The code BS EN 300:2006 has a table of OSB properties, including the modulus of elasticity. &lt;a href=&quot;http://www.osb-info.org/technical.html&quot; rel=&quot;nofollow&quot;&gt;This site&lt;/a&gt; describes OSB and presents all of these properties for each grade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OSB has four grades, each with a different modulus of elasticity:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;OSB/1 - General purpose boards and boards for interior fitments (including furniture) for use in dry conditions ($E = 2.5\text{ GPa}$)&lt;/li&gt;&#xA;&lt;li&gt;OSB/2 - Load-bearing boards for use in dry conditions ($E = 3.5\text{ GPa}$)&lt;/li&gt;&#xA;&lt;li&gt;OSB/3 - Load-bearing boards for use in humid conditions ($E = 3.5\text{ GPa}$)&lt;/li&gt;&#xA;&lt;li&gt;OSB/4 - Heavy-duty load-bearing boards for use in humid conditions ($E = 4.8\text{ GPa}$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So your adopted modulus is adequate if you're using OSB/2 or 3, but if you're making furniture, you're probably using OSB/1, in which case you should adopt $E = 2.5\text{ GPa}$. This would increase the deflection to 41.25&amp;nbsp;mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible issue is certifying that there is in fact deformation compatibility between the OSB and the steel strip. Simple glue may not be sufficent to guarantee this. If the steel and the OSB are allowed to deform independently, even just a little, that will dramatically reduce the stiffness and increase the deflection of the section.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-05-06T17:42:32.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8765" PostTypeId="2" ParentId="8763" CreationDate="2016-05-06T17:45:21.970" Score="3" Body="&lt;p&gt;In the case of aircraft carriers the compressed air is generally a working fluid for a mechanical catapult rather than providing direct thrust like a rocket (steam is also used for the same job on carriers using steam plant). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that compressed gas can be useful in providing small amounts of thrust, although air may not be the best choice because of its oxygen content and an inert gas may be preferred. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of overall power to weight ratio you generally get more out of a solid/liquid fuel system as you are going from a high density phase to a gas so you don't have the additional weight of a pressure vessel. Note that for standard industrial pressurized gas containers the container may have as much mass as the gas it contains. So even if you don't have actual combustion some sort of chemical gas generation may be preferable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall compressed gas is convenient, portable and simple but has fairly mediocre power density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth considering that simple 'bottle rockets' get much more thrust when they use water to provide reaction mass while the compressed air gives the energy storage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with pure gas thrust is that the rate of conversion of potential to kinetic energy is limited by the local speed of sound and tends to be quite lossy. For jet engines this is mitigated by the fact that there is an unlimited supply of air to push through the engine from the immediate environment, rockets using fuel/oxidiser use sheer brute force of energy density of the propellant at the expense of overall efficiency and the fact that a rocket is essentially very simple so the dead weight of the engine itself is very low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly where hydrogen of oxygen are used as fuel or oxidiser they are generally liquefied rather than pressurised. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So overall there isn't a general answer, compressed gas is viable for thrust on a small scale, as for small orbital adjustments but it is less attractive for larger thrust requirements. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-06T22:16:33.347" LastActivityDate="2016-05-06T22:16:33.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8767" PostTypeId="2" ParentId="8763" CreationDate="2016-05-06T19:49:51.580" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Simplified_Aid_For_EVA_Rescue&quot; rel=&quot;nofollow&quot;&gt;SAFER&lt;/a&gt;, used by NASA for EVAs (&quot;Space Walks&quot;), uses compressed &lt;a href=&quot;https://www.nasa.gov/missions/shuttle/f_saferspacewalk.html&quot; rel=&quot;nofollow&quot;&gt;nitrogen&lt;/a&gt; for thrust.  Air is mostly nitrogen, and the properties are fairly similar.  So, essentially it not only could be done, but even has been done.&lt;/p&gt;&#xA;" OwnerUserId="2066" LastActivityDate="2016-05-06T19:49:51.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8769" PostTypeId="1" AcceptedAnswerId="11888" CreationDate="2016-05-07T05:16:00.917" Score="1" ViewCount="1840" Body="&lt;p&gt;Is there a way to derive a formula of modulus of rupture (flexural strength) vs compressive strength?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know is there any mathematical reason for the $\sqrt{f_{c} ^{'}}$ placed somewhere in the formula?&lt;/p&gt;&#xA;" OwnerUserId="252" LastEditorUserId="1832" LastEditDate="2016-05-07T11:46:32.390" LastActivityDate="2016-10-16T23:07:05.530" Title="Is there a way to derive a formula of modulus of rupture vs compressive strength?" Tags="|structural-engineering|civil-engineering|concrete|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8771" PostTypeId="1" AcceptedAnswerId="8772" CreationDate="2016-05-07T10:25:50.643" Score="6" ViewCount="297" Body="&lt;p&gt;I know comparatively little about general engineering methodologies, but it strikes me that the (now outdated) &quot;waterfall&quot; method is possibly very apt for mechanical engineering. Perhaps advances in rapidly changing software engineering could also be applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are software methodologies influencing other engineering disciplines? Or vice versa? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia's page on &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Engineering&quot; rel=&quot;noreferrer&quot;&gt;engineering&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Koen argues that the definition of what makes one an engineer should not be based on what he produces, but rather how he goes about it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which vaguely implies that development processes are to some degree ad-hoc, given that neither the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Engineering#Methodology&quot; rel=&quot;noreferrer&quot;&gt;methodology&lt;/a&gt; section of that page, nor the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Engineering#Problem_solving&quot; rel=&quot;noreferrer&quot;&gt;problem solving&lt;/a&gt; subsection of the engineering page make any references to development stages in the precise ways that is widely encouraged in software engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why the apparent lack of processes?&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="1586" LastEditDate="2016-05-10T17:57:31.810" LastActivityDate="2016-05-24T16:29:31.303" Title="Are latest software engineering methodologies, such as Agile, being adopted by other engineering disciplines?" Tags="|project-management|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8772" PostTypeId="2" ParentId="8771" CreationDate="2016-05-07T10:59:10.703" Score="4" Body="&lt;p&gt;It is my observation and practical experience that most engineering product development processors are industry and institution based. Waterfall though old are still used in some industries. But thanks to the software industry and globalization agile methodology is fast gaining acceptance is engineering discipline. The automotive industry is one that has had to change.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The US automotive industry in the past used to take for ever produce new products. But automotive manufactures from east as force the US manufactures to fast adopt, or be left behind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are legitimate barriers, primarily cost. Unlike software some tools are very expensive. Interrupting engineering manufacturing process to make modifications has significant impact in productivity or in other words long downtime. On such situations the good old waferfall methodology is the only viable option. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to advances in technology such as 3D printing, quick turn PCB, advanced CAD tools agile methodology is fast been adopted in engineering product development processes. The follow graphic from &lt;a href=&quot;http://www.eetimes.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;eetimes&lt;/a&gt; better explains the shift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Mc7e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Mc7e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another example from stage-gate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xZugA.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xZugA.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some references where you might finds specific public examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.stage-gate.com/publications_winning_4ed.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Winning at New Products (This is a great read)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eetimes.com/document.asp?doc_id=1279137&quot; rel=&quot;nofollow noreferrer&quot;&gt;Agile hardware development – nonsense or necessity?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.stage-gate.com/resources_stage-gate_greeninnovation.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Going Green in Product Innovation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-05-23T10:07:45.387" LastActivityDate="2016-05-23T10:07:45.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8773" PostTypeId="1" AcceptedAnswerId="8775" CreationDate="2016-05-07T12:35:26.483" Score="3" ViewCount="235" Body="&lt;p&gt;What are the differences between the processes of 3D printing with plastics versus metals? Why is metal such a problem? There are plenty of metals with low melting points. And AFAIK metal can be sprayed and directed in a number of ways such as with electric arcs and similar.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="1586" LastEditDate="2016-05-09T23:06:02.787" LastActivityDate="2016-05-09T23:06:02.787" Title="Why is 3D printing with metals so uncommon?" Tags="|mechanical-engineering|3d-printing|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8774" PostTypeId="1" CreationDate="2016-05-07T12:48:31.627" Score="3" ViewCount="1566" Body="&lt;p&gt;I am using &lt;a href=&quot;http://openbuildspartstore.com/eccentric-spacer/&quot; rel=&quot;nofollow&quot;&gt;eccentric spacer nuts&lt;/a&gt; to create a tight fit between wheels and a rail of a linear sliding mechanism. (Expand the product description in the link to see a picture of it in use).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They aren't actually nuts: they have no thread. But the hexagonal part of its exterior allows the spacer's off-centre hole to be adjusted with respect to the external cylindrical section with a spanner, thereby positioning a shaft going through it relative to the structure holding them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there alternative mechanisms for adjusting the distance between opposite wheels? They are very effective, but are £2 per piece from the UK, and it really adds up. They only seem to be sold by suppliers of the open-rail and v-slot systems, so I'm hoping, either:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;These accent rich spacer nuts are known as something else in other systems, at a lower price, or...&lt;/li&gt;&#xA;&lt;li&gt;There are other mechanisms that achieve the same.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-05-07T16:38:35.817" LastActivityDate="2018-09-18T10:40:15.110" Title="Alternative to eccentric spacer nuts" Tags="|mechanisms|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8775" PostTypeId="2" ParentId="8773" CreationDate="2016-05-07T13:09:57.627" Score="4" Body="&lt;p&gt;Metals are used. Albeit, in machines that are much more expensive and complex than your run of the mill FDM style machines that you're likely referring to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason you don't see many or any metal printers(&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Selective_laser_sintering&quot; rel=&quot;nofollow&quot;&gt;SLS&lt;/a&gt;) and &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Direct_metal_laser_sintering&quot; rel=&quot;nofollow&quot;&gt;DMLS&lt;/a&gt; are likely due to patents and costs. They use lasers and special powdered materials with a binding material them. See Wikipedia links. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, when the patent for FDM type plastic printers expired, the market flooded with lower cost machines and parts. When the patents for SLS expire, expect to see more of those type of machines hit the market.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That all said, there is a small team in Europe called Sinterit working on an affordable SLS machine.  &lt;a href=&quot;http://www.sinterit.com/&quot; rel=&quot;nofollow&quot;&gt;http://www.sinterit.com/&lt;/a&gt; I'm not sure it can print metal. Again, maybe due to power requirements/patents/etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be others working on them too. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-05-07T13:09:57.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8776" PostTypeId="2" ParentId="8773" CreationDate="2016-05-07T14:41:46.223" Score="0" Body="&lt;p&gt;I am not an expert on this topic but Metal 3D printing is available. Here are some images of 3D printed faucets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AFaiI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AFaiI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PjAri.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PjAri.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-05-08T11:24:02.277" LastActivityDate="2016-05-08T11:24:02.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8777" PostTypeId="2" ParentId="8771" CreationDate="2016-05-07T17:49:46.337" Score="5" Body="&lt;h3&gt;Waterfall&lt;/h3&gt;&#xA;&lt;p&gt;Waterfall is the basic method that a lot of engineering is done with already. I would say that &amp;quot;waterfall&amp;quot; was the original method for software engineering because software engineering came about much later than the other areas of engineering, e.g. software already borrowed from other areas of engineering.&lt;/p&gt;&#xA;&lt;h3&gt;Agile (and others)&lt;/h3&gt;&#xA;&lt;p&gt;Agile is used in software engineering and not in other engineering areas because there are big differences in risk. Agile sounds terrible (impossible) in construction. It also will not work in areas where there are actual consequences for problems (and a change is harder than an update away).&lt;/p&gt;&#xA;&lt;h3&gt;Middle ground&lt;/h3&gt;&#xA;&lt;p&gt;There are lots of middle areas between the two extremes, so not everything will fall into these areas, e.g. rapid prototyping could be considered a form of agile.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-05-11T01:05:15.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8779" PostTypeId="1" CreationDate="2016-05-07T19:44:15.370" Score="3" ViewCount="86" Body="&lt;p&gt;In a standard 4 stroke vertical internal combustion engine, Does the downward piston movement compress gas below it? If so, does it affect efficiency?&lt;/p&gt;&#xA;" OwnerUserId="908" LastEditorUserId="1293" LastEditDate="2016-05-08T08:19:21.813" LastActivityDate="2016-05-08T08:19:21.813" Title="What happens at the other side of the piston?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8780" PostTypeId="2" ParentId="8779" CreationDate="2016-05-07T19:52:23.360" Score="6" Body="&lt;p&gt;In a single-cylinder engine, yes, although the crankcase is usually vented. The total change in volume and/or pressure isn't all that great anyway. And any energy absorbed by compressing the air is returned on the next up-stroke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In multi-cylinder engines, there are pistons moving up when other pistons are moving down, so the change in total crankcase volume is negligible.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-05-07T19:52:23.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8781" PostTypeId="2" ParentId="8779" CreationDate="2016-05-07T19:53:51.793" Score="2" Body="&lt;p&gt;Not as such, for a start the space in the crank case is quite a large volume compared to the cylinder bore and will in any case usually have a breather system of one sort or another. The most important reason for this is to prevent oil from the crankcase and sump from being forced back past the piston rings into the combustion chamber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is still a certain amount of resistance just from displacing this air, although this may be quite small compared to the resistance from induction fuel/air into the cylinder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Induction is a small but not insignificant part of overall efficiency. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-07T19:53:51.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8782" PostTypeId="1" CreationDate="2016-05-07T20:09:31.817" Score="0" ViewCount="389" Body="&lt;p&gt;I have instructions for using aluminium brazing rod &quot;HTS-2000&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Clean the area to be worked with a wire brush. Heat the work surface until it is hot enough to melt the rod. Open the pores of the work surface so the rod can penetrate the surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have never heard of metal having &quot;pores&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is this opening of pores? And is there some minimum/maximum temperature/duration for this heating? (Obviously lower temperature than melting point of the base metal, and higher than the filler rod)&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-05-07T20:36:32.583" LastActivityDate="2016-05-08T02:40:52.370" Title="Aluminium brazing instructions: heat thoroughly to &quot;open the pores&quot; before tinning with rod" Tags="|aluminum|brazing|heat-treatment|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8783" PostTypeId="2" ParentId="4172" CreationDate="2016-05-07T22:09:46.197" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.materialsviews.com/no-escape-for-rust-non-destructive-detecting-of-corrosion-under-coatings/&quot; rel=&quot;nofollow&quot;&gt;This article&lt;/a&gt; written on  May 02 2012 states that no non-destructive methods exist currently, other than a scanning Kelvin probe  (which requires long measurement times).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it states that &lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/maco.201005918/abstract;jsessionid=EC9DDC474D028522B0836F76B2023710.f01t04&quot; rel=&quot;nofollow&quot;&gt;research is in progress into using Pulsed Phase Thermography&lt;/a&gt;. According to the article it's got teething problems but has proved that it's possible.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-05-07T22:09:46.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8784" PostTypeId="2" ParentId="8133" CreationDate="2016-05-07T22:36:08.083" Score="1" Body="&lt;p&gt;The bearing should have figures on the moment loads it can take. Failing that, use two bearings and allow them enough movement so effectively each is just supporting a purely radial load.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-05-07T22:36:08.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8785" PostTypeId="1" CreationDate="2016-05-08T01:57:06.110" Score="1" ViewCount="1342" Body="&lt;p&gt;What are the common filler materials used in PP, PE, LDPE, HDPE, and PVC-moulded materials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a background - I own a limestone quarry and selling calcium carbonate in powder form. I want to make value added products from CaCO3, which I came across the filler master batches. But I am not sure what type of plastics use that. Answer to this will help me identify a particular sector and do targeted market analysis as I understand different end users require different grades of CaCO3 fillers.&lt;/p&gt;&#xA;" OwnerUserId="6360" LastEditorUserId="1832" LastEditDate="2016-05-09T21:02:55.123" LastActivityDate="2016-08-11T20:03:23.763" Title="What is the common filler used in plastic injection moulding?" Tags="|materials|chemical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8786" PostTypeId="2" ParentId="8782" CreationDate="2016-05-08T02:40:52.370" Score="1" Body="&lt;p&gt;Aluminium reacts strongly with the oxygen in the air to form a thin stable layer of aluminium oxide which covers the surface and stops any further reaction. You have to remove this oxide layer, by wire brushing, before you can braze the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminium is a soft metal, and the wire brushing will inevitably scratch and damage the metal surface. This is a good thing, because it increases the surface area for the brazing material to bond with the aluminum. You need to apply enough braze to fill up the &quot;pores&quot; that you create when cleaning the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the aluminium part was originally made by a &quot;low tech&quot; process such as sand casting, the surface will indeed be &quot;porous&quot; in the sense that the surface roughness from the original sand mould will still be there, and there may be some sand still embedded in the metal surface until you remove it by brushing. Of course the roughness tends to get filled up with dirt, as well as by the aluminium oxide layer, and becomes less obvious until you clean up the part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brazing rod melts at a lower temperature than the aluminium. The metal is hot enough when you can remove the heat source, and the brazing rod will melt when it touches the metal. Heating for longer doesn't achieve anything, and may melt the aluminium  - which is &lt;em&gt;not&lt;/em&gt; what you are trying to do!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you haven't done this before, get some scrap aluminium and practice before you do the &quot;real&quot; job. The heating time depends very much on your heat source and the size of the aluminum part. It may only take (literally) a few seconds to get a thin aluminium plate hot enough to braze. Heating a solid block of metal with a small-sized gas torch will take much longer. The basic technique is to apply heat, remove the gas torch, apply the braze, and if it doesn't melt immediately apply the torch again and repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have any scrap aluminium available, you can get some cheaply from places like waste disposal sites, scrap car breaker's yards, etc. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-05-08T02:40:52.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8789" PostTypeId="2" ParentId="481" CreationDate="2016-05-08T07:47:17.993" Score="0" Body="&lt;p&gt;to Calculate the Horse Power for the Hydraulic system&#xA;we Need following data&#xA;1) G P M of the Pump ( Vane / gear / piston pump )&#xA;2) Working pressure of the hydraulic oil in P S I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore ; H.p = 0.0005833 x P.S.I x G.P.M&lt;/p&gt;&#xA;" OwnerUserId="6363" LastActivityDate="2016-05-08T07:47:17.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8792" PostTypeId="1" AcceptedAnswerId="16766" CreationDate="2016-05-08T10:23:14.167" Score="4" ViewCount="124" Body="&lt;p&gt;I'm dealing with hot combustion products flowing through heat exchangers and I'd like to perform some calculations regarding the flow properties (velocity, temperature, pressure and density) as it passes trough the heat exchanger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've a 1D model already built and I'd like to test it and validate it, but I'm lacking some critical information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any place where I can find the absolute viscosity [Pa*s] as a function of temperature (for pressures of 1atm, or close to it) for some select gases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know the viscosity between 300K and 2500K for gaseous CO2, H2O, N2 and O2. In polynomial form (something similar to the NASA/JANAF polynomials) would be nice, but in table would also be good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already tried NIST, but it doesn't have that information for the temperature range I'm interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to run a macro for FLUENT, Star ccm+ or OpenFOAM and get those values from those programs? I have legal access to all of them.&lt;/p&gt;&#xA;" OwnerUserId="6364" LastActivityDate="2017-08-15T09:41:17.007" Title="Viscosity of select species at atmospheric pressure for very large temperatures?" Tags="|fluid-mechanics|modeling|combustion|heat-exchanger|data|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8793" PostTypeId="2" ParentId="8760" CreationDate="2016-05-08T10:32:00.890" Score="2" Body="&lt;p&gt;You can try running the following R script which produces the figure below.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9kJAX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9kJAX.png&quot; alt=&quot;tripartite graph paper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;require(ggplot2)&#xA;require(data.table)&#xA;&#xA;# The constant value grid lines&#xA;freqs = unique(c(seq(0.1, 0.2, by = 0.05), seq(0.2, 1.0, by = 0.1),&#xA;      seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0),&#xA;      seq(10.0, 20.0, by = 5.0), seq(20.0, 100.0, by = 10.0),&#xA;      seq(100.0, 200.0, by = 50.0), seq(200.0, 1000.0, by = 100.0)))&#xA;vels = unique(c(seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0),&#xA;      seq(10.0, 20.0, by = 5.0), seq(20.0, 100.0, by = 10.0),&#xA;      seq(100.0, 200.0, by = 50.0), seq(200.0, 1000.0, by = 100.0)))&#xA;accs = unique(c(seq(0.001, 0.002, by = 0.0005), seq(0.002, 0.01, by = 0.001),&#xA;      seq(0.01, 0.02, by = 0.005), seq(0.02, 0.1, by = 0.01),&#xA;      seq(0.1, 0.2, by = 0.05), seq(0.2, 1.0, by = 0.1),&#xA;      seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0),&#xA;      seq(10.0, 20.0, by = 5.0), seq(20.0, 100.0, by = 10.0),&#xA;      seq(100.0, 200.0, by = 50.0), seq(200.0, 1000.0, by = 100.0),&#xA;      seq(1000.0, 2000.0, by = 500.0), seq(2000.0, 10000.0, by = 1000.0)))&#xA;disps = unique(c(seq(0.0001, 0.0002, by = 0.00005), seq(0.0002, 0.001, by = 0.0001),&#xA;          seq(0.001, 0.002, by = 0.0005), seq(0.002, 0.01, by = 0.001),&#xA;          seq(0.01, 0.02, by = 0.005), seq(0.02, 0.1, by = 0.01),&#xA;          seq(0.1, 0.2, by = 0.05), seq(0.2, 1.0, by = 0.1),&#xA;          seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0),&#xA;          seq(10.0, 20.0, by = 5.0), seq(20.0, 100.0, by = 10.0),&#xA;          seq(100.0, 200.0, by = 50.0), seq(200.0, 1000.0, by = 100.0),&#xA;          seq(1000.0, 2000.0, by = 500.0), seq(2000.0, 10000.0, by = 1000.0)))&#xA;&#xA;# Horizontal grid lines (const vel = in/s)&#xA;vel_grid = data.table()&#xA;freqmin = min(freqs)&#xA;freqmax = max(freqs)&#xA;for (vel in vels) {&#xA;  xx = c(freqmin, freqmax)&#xA;  yy = c(vel, vel)&#xA;  label = c(paste0(&quot;vel&quot;,vel), paste0(&quot;vel&quot;,vel)) &#xA;  color = &quot;vel&quot;&#xA;  local_dt = data.table(x = xx, y = yy, label = label, color = color)&#xA;  vel_grid = rbind(vel_grid, local_dt)&#xA;}&#xA;&#xA;# Vertical grid lines (freq = cycles/sec)&#xA;freq_grid = data.table()&#xA;velmin = min(vels)&#xA;velmax = max(vels)&#xA;for (freq in freqs) {&#xA;  xx = c(freq, freq)&#xA;  yy = c(velmin, velmax)&#xA;  label = c(paste0(&quot;freq&quot;,freq), paste0(&quot;freq&quot;,freq)) &#xA;  color = &quot;freq&quot;&#xA;  local_dt = data.table(x = xx, y = yy, label = label, color = color)&#xA;  freq_grid = rbind(freq_grid, local_dt)&#xA;}&#xA;&#xA;# -45 grid lines (constant acc = omega * vel = (2*pi*freq) radians/sec * vel in/sec  &#xA;#                              = (2*pi*freq)*vel in/sec^2 = (2*pi*freq)*vel/g  Gs)&#xA;gravity = 32.2*12.0  # in/s^2&#xA;acc_grid = data.table()&#xA;for (acc in accs) {&#xA;  for (freq in freqs) {&#xA;    f = freq&#xA;    vel = acc*gravity/(freq*2*pi)&#xA;    if (vel &amp;lt; velmin) {&#xA;      vel = velmin&#xA;      f = acc*gravity/(vel*2*pi)&#xA;    }&#xA;    if (vel &amp;gt; velmax) {&#xA;      vel = velmax&#xA;      f = acc*gravity/(vel*2*pi)&#xA;    }&#xA;    xx = c(f)&#xA;    yy = c(vel)&#xA;    label = c(paste0(&quot;acc&quot;,acc)) &#xA;    color = &quot;acc&quot;&#xA;    local_dt = data.table(x = xx, y = yy, label = label, color = color)&#xA;    acc_grid = rbind(acc_grid, local_dt)&#xA;  }&#xA;}&#xA;&#xA;## The const acc=1 line is the disp axis&#xA;acc_1 = acc_grid[acc_grid$label == &quot;acc1&quot;]&#xA;disp_axis = data.table()&#xA;for (disp in disps) {&#xA;  acc = 1.2&#xA;  omega = sqrt(acc*gravity/disp)&#xA;  freq = omega/(2*pi)&#xA;  vel = omega*disp&#xA;  xx = c(freq)&#xA;  yy = c(vel)&#xA;  label = c(paste0(&quot;disp_axis&quot;,acc)) &#xA;  color = &quot;disp_axis&quot;&#xA;  local_dt = data.table(x = xx, y = yy, z = disp, label = label, color = color)&#xA;  disp_axis = rbind(disp_axis, local_dt)&#xA;}&#xA;&#xA;# +45 grid lines (const disp = vel/omega = vel/(2*pi*freq) in)&#xA;disp_grid = data.table()&#xA;for (disp in disps) {&#xA;  for (freq in freqs) {&#xA;    f = freq&#xA;    vel = disp*freq*(2*pi)&#xA;    if (vel &amp;lt; velmin) {&#xA;      vel = velmin&#xA;      f = vel/(disp*2*pi)&#xA;    }&#xA;    if (vel &amp;gt; velmax) {&#xA;      vel = velmax&#xA;      f = vel/(disp*2*pi)&#xA;    }&#xA;    xx = c(f)&#xA;    yy = c(vel)&#xA;    label = c(paste0(&quot;disp&quot;,disp)) &#xA;    color = &quot;disp&quot;&#xA;    local_dt = data.table(x = xx, y = yy, label = label, color = color)&#xA;    disp_grid = rbind(disp_grid, local_dt)&#xA;  }&#xA;}&#xA;&#xA;## The const disp=0.1 line is the acc axis&#xA;disp_1 = disp_grid[disp_grid$label == &quot;disp0.1&quot;]&#xA;acc_axis = data.table()&#xA;for (acc in accs) {&#xA;  disp = 0.12&#xA;  omega = sqrt(acc*gravity/disp)&#xA;  freq = omega/(2*pi)&#xA;  vel = omega*disp&#xA;  xx = c(freq)&#xA;  yy = c(vel)&#xA;  label = c(paste0(&quot;acc_axis&quot;,acc)) &#xA;  color = &quot;acc_axis&quot;&#xA;  local_dt = data.table(x = xx, y = yy, z = acc, label = label, color = color)&#xA;  acc_axis = rbind(acc_axis, local_dt)&#xA;}&#xA;&#xA;# Join vertical and horizontal&#xA;grid = rbind(vel_grid, freq_grid, acc_grid, disp_grid)&#xA;&#xA;### Axis label points&#xA;disp = 0.08&#xA;acc = 1000&#xA;omega = sqrt(acc*gravity/disp)&#xA;freq = omega/(2*pi)&#xA;vel = omega*disp&#xA;acc_axis_label = data.table(x = c(freq), y = c(vel), label=&quot;Spectral acceleration (g)&quot;)&#xA;&#xA;disp = 200&#xA;acc = 0.8&#xA;omega = sqrt(acc*gravity/disp)&#xA;freq = omega/(2*pi)&#xA;vel = omega*disp&#xA;disp_axis_label = data.table(x = c(freq), y = c(vel), label=&quot;Spectral displacement (in)&quot;)&#xA;&#xA;# Plot horizontal grid&#xA;xticks = freqs[c(TRUE, FALSE)]&#xA;yticks = vels[c(TRUE, FALSE)]&#xA;&#xA;plt = ggplot() +&#xA;      geom_line(data = grid, aes(x = x, y = y, group = label, color = color),&#xA;                size=0.25) +&#xA;      geom_line(data = acc_1, aes(x = x, y = y), color = 1, size=0.5) + &#xA;      geom_line(data = disp_1, aes(x = x, y = y), color = 1, size=0.5) + &#xA;      xlab(&quot;Frequency (Hz)&quot;) + &#xA;      ylab(&quot;Spectral velocity (in/sec)&quot;) + &#xA;      scale_color_discrete(guide = FALSE) +&#xA;      scale_x_log10(limits = c(freqmin, freqmax),&#xA;                    expand = c(0,0),&#xA;                    breaks=xticks,&#xA;                    labels= c(&quot;0.1&quot;,&quot;0.2&quot;,&quot;0.4&quot;,&quot;0.6&quot;,&quot;0.8&quot;, xticks[xticks &amp;gt;= 1])) +&#xA;      scale_y_log10(limits = c(velmin, velmax),&#xA;                    expand = c(0,0),&#xA;                    breaks=yticks) +&#xA;      coord_fixed() +&#xA;      theme_bw() +&#xA;      theme(panel.grid.major = element_blank(),&#xA;            panel.grid.minor = element_blank()) +&#xA;      annotate(&quot;text&quot;, x = acc_axis$x, y = acc_axis$y, label = as.character(acc_axis$z), &#xA;               size = 3, angle = -45) + &#xA;      annotate(&quot;text&quot;, x = disp_axis$x, y = disp_axis$y, label = as.character(disp_axis$z), &#xA;               size = 3, angle = 45)  +&#xA;      annotate(&quot;text&quot;, x = acc_axis_label$x, y = acc_axis_label$y, label = acc_axis_label$label,&#xA;               size = 4, angle = 45)  +&#xA;      annotate(&quot;text&quot;, x = disp_axis_label$x, y = disp_axis_label$y, label = disp_axis_label$label,&#xA;               size = 4, angle = -45) &#xA;print(plt)&#xA;ggsave(&quot;seismic_tripartite_paper1.pdf&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2016-05-08T10:32:00.890" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8794" PostTypeId="1" AcceptedAnswerId="10231" CreationDate="2016-05-08T11:19:50.903" Score="4" ViewCount="1230" Body="&lt;p&gt;I have an thin uniform rectangular steel plate supported on 3 sides and free on one, and  supporting a uniform load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to determine what thickness of plate I need to keep deflection below a given value, but the deflection may not be &quot;small&quot; compared to the plate thickness, so standard formulae for maximum deflection may not be accurate, and I'm stuck on how to check my answer, or how to calculate it more accurately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plate is a flat rectangular plate (stainless 316 or mild, undecided) anything from 1.5 -15mm thick with an unsupported area at its edge of 1250x500mm total 0.625m^2. It is simply supported along the 500-1250-500 edges and free on the other 1250 edge. The unsupported area carries a static load of 4800 N/m^2 (approx 360kg spread uniformly over the unsupported area) plus its self weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3 supported edges are unrestrained simple supports that can slide or rotate; they don't resist any movement except in the Z-direction (a bit like it's resting across the  3 edges of a &quot;u&quot; shaped pit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is the thickness of steel plate I need, to ensure maximum deflection (at the middle of the free edge?) stays under possible values of 3mm / 5mm / 10mm (the most likely permitted deflections or at least a good selection to choose between)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I am guessing solutions could be thicknesses around 1.5 - 5mm, which means that the deflection might not be &quot;small&quot; compared to its thickness and the usual simple calculation may not be very accurate or trustworthy. But I'm not sure....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks - and any hints how I can work this out myself appreciated but not essential :)&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2016-05-08T11:30:08.053" LastActivityDate="2017-11-21T19:37:13.977" Title="Deflection of thin plate with 1 free edge and deflection &gt; thickness" Tags="|deflection|numerical-methods|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8796" PostTypeId="2" ParentId="8794" CreationDate="2016-05-08T15:39:52.180" Score="1" Body="&lt;p&gt;The small deflection approximation is not based on deflection vs thickness, it is based on deflection over span. So for your maximum case of 10mm deflection on a 1250mm span, you are only looking at a 0.8% deflection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at is from a statics point of view. Where the Deflection is one side of a triangle and half the span in the hypotenuse. This would give you an angle of 0.5 degrees. Then you can look at how much force is being transferred into the part horizontally instead of vertically with $F_{_{x}}=F*sin\theta$. Therefore, with your maximum deflection case you are only going to get about $0.008*F$ transferring into the plate in the non-normal direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the normal equations are fine for you what you are trying to do.&lt;/p&gt;&#xA;" OwnerUserId="6369" LastActivityDate="2016-05-08T15:39:52.180" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8797" PostTypeId="1" AcceptedAnswerId="8799" CreationDate="2016-05-08T15:55:22.193" Score="-1" ViewCount="1163" Body="&lt;p&gt;Do you know if there is some sort of equipment that can measure the distance to a sound? What practical application can such an equipment have? The reason I'm asking is because I just got an idea of how to make such a machine. But it's really not hard to figure out so I guess there already exists such things. I didn't find anything on google though. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6367" LastActivityDate="2016-05-08T22:18:57.947" Title="Measuring the distance to a sound" Tags="|acoustics|" AnswerCount="3" CommentCount="3" ClosedDate="2016-05-09T01:23:30.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="8798" PostTypeId="2" ParentId="8797" CreationDate="2016-05-08T16:26:38.147" Score="1" Body="&lt;p&gt;You may be thinking of a device with multiple microphones some distance apart (enough that you can measure the delay between each one), which you could then use to triangulate the direction and distance of the sound source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a system was used to track submarines during the second world war. &lt;a href=&quot;https://en.wikipedia.org/wiki/Hawker_Siddeley_Nimrod&quot; rel=&quot;nofollow&quot;&gt;RAF Nimrod&lt;/a&gt; planes would drop buoys into the ocean which had hydrophones, and they would transmit the sound they heard back to the plane. The staff aboard could then decide where to drop another buoy in order to get the best angle for triangulating the submarine position as it moved around trying to avoid detection. You can read about this more on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Anti-submarine_warfare&quot; rel=&quot;nofollow&quot;&gt;anti-submarine warfare&lt;/a&gt; Wikipedia page. Since then, the US Navy deployed a more permanent system called &lt;a href=&quot;https://en.wikipedia.org/wiki/SOSUS&quot; rel=&quot;nofollow&quot;&gt;SOSUS&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes - there does already exist a device, or rather a network of devices, for finding the direction and distance of a sound source. I don't know if there is a smaller singular device for doing this same thing on land, but it's certainly viable.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2016-05-08T16:26:38.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8799" PostTypeId="2" ParentId="8797" CreationDate="2016-05-08T16:33:15.460" Score="1" Body="&lt;p&gt;Yes you can do this in fact, there are acoustic devices used by the military to &lt;a href=&quot;https://en.wikipedia.org/wiki/Boomerang_(countermeasure)&quot; rel=&quot;nofollow&quot;&gt;detect the location of sniper&lt;/a&gt;s. However, this is something of a special case and it relies on detecting not just one sound but a well understood acoustic signature. In the case of a bullet you have the sound of it being fired and the sound of the sonic boom as it passes. In fact crude rules of thumb have been used to get rough distances for as long as supersonic ammunition has been used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly,  because this is a particular application you can also infer a lot of extra information about the likely speed and trajectory of a bullet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky part is not so much the hardware as the processing required to process the information into something useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously with familiar sounds human hearing is pretty good at determining distance and direction, although this can be hampered by echos and the general acoustic environment.direction can, in general be measure by the small delay in receiving the sound between two or more receivers and the distance by how loud the sound is (as long as you know how loud the sources is). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the principal is fairly simple and essentially similar to active radar and sonar systems. The difficulty comes in the fine detail of specific applications particularity in accounting for complexities  like changes in speed of sound due to variations in air density and interaction of sound with obstacles. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-08T16:46:25.360" LastActivityDate="2016-05-08T16:46:25.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8800" PostTypeId="2" ParentId="8773" CreationDate="2016-05-08T17:31:24.147" Score="4" Body="&lt;p&gt;There are a number of practical processes for 3D printing in metals using various powder sintering techniques as described in other answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it is difficult to use the sort of direct extrusion processes that you see with plastics because most metal alloys don't have the same sort of viscous, self adhesive phase that thermoplastics do. Most molten metals are fairly fluid, cast iron, steel, bronze and aluminium have a viscosity not too far off water when melted if there is an intermediate phase it is usually a slush mixture of solid and liquid phases rather than something which can be extruded and adheres to itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that consumer 3D printing is designed for relatively inexpensive low volume production, so trying to do the same thing with metal is likely to be self defeating. In most cases it is likely to be easier to print a pattern in wax and plastic and investment cast it if you want it in metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly in the case of low melting point alloys it is just as cheap and almost  easy to make a silicone mould from a master and cast it plus the fact that modern resins may actually be superior to low melt metals in most respects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example of this is is the models used for table top games. A decade ago these were injection moulded plastic for simple models and white metal for more detailed ones. Now they are almost exclusively cast resin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact that metals are inherently a bit more versatile than plastics in terms of rapid prototyping and fabrication. So there are more competing options for low volume manufacturing of metal parts than plastic ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of direct equivalents to plastic filament extrusion in metal MIG welding is not that far off inn terms of process and a robotic version of the process is certainly reasonably plausible, however as already discussed the flow characteristics of metal mean that MIG certainly doesn't have anything like the resolution of plastic extrusion as there is a minimum voltage/current which is required to strike and arc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact that a decent MIG welder which produces anything like a consistent weld bead would start at around £500 with the 3 axis control on top of that, plus the fact that MIG generates a lot of fumes, spatter and UV radiation which needs to be managed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TIG welding offers a bit more precision but is equivalently more expensive for the basic machinery and an automated setup would require additional wire feed equipment. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-09T11:41:49.923" LastActivityDate="2016-05-09T11:41:49.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8805" PostTypeId="2" ParentId="8797" CreationDate="2016-05-08T22:18:57.947" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Do you know if there is some sort of equipment that can measure the distance to a sound [source]?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A familiar example of this is estimating the distance to a lightning strike. The light arrives nearly instantaneously. The sound travels at 340 m/s at sea level (at standard temperature and pressure). By measuring the time between the light and sound arriving we can estimate the distance to the source. It's roughly 3 s/km or 5 s/mile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without the light source the problem becomes more difficult. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One microphone doesn't give you any location information. &lt;/li&gt;&#xA;&lt;li&gt;Two microphones may tell you the 'line' that the source is on but not the range. e.g. For two microphones 100 m apart a difference in time of 0.1 s (34 m) gives a location of the sound somewhere on an infinite curve that passes between them.&lt;/li&gt;&#xA;&lt;li&gt;A third microphone is required to triangulate the source.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The system will have to deal with background noise, echos, changes is sound &lt;em&gt;ground&lt;/em&gt; velocity due to wind, etc.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2016-05-08T22:18:57.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8806" PostTypeId="1" AcceptedAnswerId="8812" CreationDate="2016-05-09T00:39:42.670" Score="1" ViewCount="146" Body="&lt;p&gt;I was intrigued to read about so-called &lt;a href=&quot;https://www.sciencedaily.com/releases/2015/07/150713095732.htm&quot; rel=&quot;nofollow&quot;&gt;&quot;heat storage ceramic&quot;&lt;/a&gt; and was wondering how the energy stored in such way compares to energy stored in a NMC Li-Ion battery, if the latter is used to generate heat (the article lists it a 230kJ L&lt;sup&gt;-1&lt;/sup&gt;). I know it is not an apples-to-apples comparison, but I am curious about any ballpark assessment, especially in terms of energy stored per kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason for my interest is potential use in electric vehicles in cold climates. I drive an EV in Quebec, and I know that when the temperature is significantly below freezing, I can expect to spend 5-7KWh per 100km of driving simply to heat the interior. If energy density of &quot;heat-storage ceramic&quot; is significantly higher per weight or volume, it could make sense to install heat storage &quot;batteries&quot; alongside electrical batteries.&lt;/p&gt;&#xA;" OwnerUserId="6373" LastActivityDate="2016-05-09T19:26:26.400" Title="Energy density of &quot;heat storage ceramic&quot; vs. NMC li-ion batteries" Tags="|energy-storage|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8807" PostTypeId="1" CreationDate="2016-05-09T02:07:27.307" Score="4" ViewCount="715" Body="&lt;p&gt;In &lt;a href=&quot;https://youtu.be/KuG_CeEZV6w?t=132&quot; rel=&quot;nofollow&quot;&gt;this popular YouTube video&lt;/a&gt;, a man attempts to fold a piece of paper more than seven times using a hydraulic press.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the seventh fold was applied pressure from the press, the man shows the audience that the paper became brittle, and breaks apart almost how styrofoam does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper isn't usually brittle, so what exactly took place such that the material properties of the piece of paper wildly changed?&lt;/p&gt;&#xA;" OwnerUserId="6374" LastActivityDate="2019-11-18T11:07:24.990" Title="Why does paper under pressure become brittle?" Tags="|materials|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8808" PostTypeId="2" ParentId="8807" CreationDate="2016-05-09T08:05:19.370" Score="4" Body="&lt;p&gt;In short, Paper is made from wood and it has properties of wood. While in thin sheet of paper we cannot feel significant difference. but when many papers are combined and pressed(see the book video on same channel), it behaves more like wood. but due to cutting into thin sheets it will not retain same strength of wood.&lt;/p&gt;&#xA;" OwnerUserId="2116" LastEditorUserId="2116" LastEditDate="2016-05-09T17:49:38.373" LastActivityDate="2016-05-09T17:49:38.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8812" PostTypeId="2" ParentId="8806" CreationDate="2016-05-09T19:26:26.400" Score="1" Body="&lt;p&gt;The energy density of lithium Ion batteries is between 900 kJ/L and 2430 kJ/L according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Lithium-ion_battery&quot; rel=&quot;nofollow&quot;&gt;Wikipedia&lt;/a&gt;.  This is at least a factor of 4 higher than the energy density you give for the heat storage ceramic so no gains in terms of space.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some &lt;a href=&quot;http://global.kyocera.com/prdct/fc/list/tokusei/hiju/index.html&quot; rel=&quot;nofollow&quot;&gt;example ceramics&lt;/a&gt; have a specific gravity ranging from 2.5 kg/L to 6.0 kg/L.  At the lower end, the heat storage ceramics will have a specific energy of 92 kJ/kg while the lithium ion batteries range between 100 and 265.  So, if the heat storage ceramics have a very low specific gravity, then you might save a little bit of weight over low density lithium ion batteries, but the gains won't be substantial.&lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2016-05-09T19:26:26.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8814" PostTypeId="1" CreationDate="2016-05-10T10:26:44.763" Score="4" ViewCount="957" Body="&lt;p&gt;Here's a column top which holds a driveway overbridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nHekP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nHekP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The column is a reinforced concrete cylinder with an inverse truncated cone on top - so it expands sideways at its top. The column top holds some narrow thunk and that thunk holds the driveway floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the thunk cross section square is much, much smaller than the column top cross section square, so it looks like the column top is overbuilt and the top is too wide for no reason - the thunk could rest on the column top even if the column had no expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would this design be used instead of a cylinder?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="374" LastEditDate="2016-05-10T10:51:04.357" LastActivityDate="2016-05-11T13:36:27.560" Title="Why would a column expand before accepting load through a narrow thunk?" Tags="|structural-engineering|structures|bridges|reinforced-concrete|columns|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8815" PostTypeId="1" CreationDate="2016-05-10T13:38:06.237" Score="2" ViewCount="491" Body="&lt;p&gt;I am trying to figure out the best way to mould/ create a closed shape (It is a 500mm long quadcopter arm) that has a foam core, and a rigid good quality carbon fibre outer.&#xA;Oh and to add to the fun, I need a hole through the center of the foam core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Said shape must be structural and resistant to twist and bend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial thought is to create 2 halves of the foam core on a 3 axis router, with the hole present, stick them together, then somehow wrap them with carbon... epoxy, jam into a mould? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously there will be an amount of finishing involved, but thats OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process muse be repeatable and yield a near-identical item each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ETA:&#xA;Further reading suggests that Vacuum assisted resin transfer moulding may be the answer. I would need to create a (very stiff) mould slightly larger than the foam core, pre-attach/ wrap the core with carbon/kevlar fabric, and then mould around. &#xA;I will need to somehow stop the central hole from filling with resin of course!&lt;/p&gt;&#xA;" OwnerUserId="6393" LastEditorUserId="6393" LastEditDate="2016-05-10T14:06:44.647" LastActivityDate="2016-06-11T13:47:27.650" Title="How can I mould a foam-core'ed carbon fibre shape?" Tags="|aerospace-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8816" PostTypeId="2" ParentId="8814" CreationDate="2016-05-10T13:38:41.527" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The column top holds some narrow thunk and that thunk holds the driveway floor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think we have some nomenclature that needs to be cleared up here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I believe you're calling a &quot;thunk&quot; is actually a &lt;em&gt;bearing plate&lt;/em&gt;.  This is a little steel platform that transfers the load from the bridge to the column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sTlwS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sTlwS.jpg&quot; alt=&quot;bearing plate&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;(&lt;a href=&quot;http://civildigital.com/types-functions-bearings-bridges/&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expanded portion at the top of the column is called a &lt;em&gt;capital&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lTo8Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lTo8Z.jpg&quot; alt=&quot;capital&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;(&lt;a href=&quot;http://www.aboutcivil.org/reinforced-concrete-building-elements.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note that the thunk cross section square is much, much smaller than the column top cross section square, so it looks like the column top is overbuilt and the top is too wide for no reason - the thunk could rest on the column top even if the column had no expansion.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have two different materials here that have different strengths.  Steel has usually 4-10 times the compressive strength that concrete does.  Thus, a much smaller cross-section (in terms of area) of steel can carry a load than a concrete cross-section can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why the column capital is much wider than the bearing plate, this is likely because the bearing area is being distributed into the column cross-section, similar to the image below (excerpt from the &lt;em&gt;PCI Design Handbook&lt;/em&gt;, 7th Edition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2E1Qr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2E1Qr.png&quot; alt=&quot;bearing dist&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why would this design be used instead of a cylinder?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The inverted tapered cone for the capital could either be a design choice (to increase the bearing strength) or an aesthetic choice.  We have no way of really knowing without looking at the calculations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be a wild shot, but they could have flared the top of the column to make placing concrete easier.  Kind of like pouring something into a funnel.  However, I kind of doubt it.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastEditorUserId="2246" LastEditDate="2016-05-11T13:36:27.560" LastActivityDate="2016-05-11T13:36:27.560" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8817" PostTypeId="1" CreationDate="2016-05-10T16:40:02.797" Score="6" ViewCount="3862" Body="&lt;p&gt;In the movie Apollo 13 there's a scene when the LEM is losing power; when they switch the module on, one of the astronauts reads the levels and says that one battery has no Volts but 2 Amps. What did he mean?&lt;/p&gt;&#xA;" OwnerUserId="6396" LastEditorUserId="368" LastEditDate="2016-05-11T18:10:55.427" LastActivityDate="2016-05-11T18:10:55.427" Title="In the Apollo 13 mission, how could the batteries have &quot;no Volts&quot; with two Amps of current?" Tags="|electrical-engineering|battery|current|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8818" PostTypeId="2" ParentId="8814" CreationDate="2016-05-10T17:01:09.587" Score="1" Body="&lt;p&gt;Columns like this are often connected to the horizontal elements of the structure they support by bearings which are designed to support the vertical load from the weight of the structure but not transfer horizontal loads to the columns. This ensures that the columns are loaded only in compression and allows movement of the structure they support (in this case the steelwork of the bridge deck) without putting unwanted loads on the columns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design of the bearing can vary and depends on the design requirements and may be eg  rocker, pinned, mechanical  thrust bearings or elastomer based. Clearly for a long span bridge the range of movement can be quite significant, especially is allowances are made for resistance to impacts or earthquakes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few reasons why the top of the column might be flared. If the bearing is bolted on then bolted connections in concrete require a minimum distance for the edge to spread their load effectively and to mitigate corrosion. As mentioned in other answers the exact solution will come down to detailed design choices and analysis but the short answer is that it is to accommodate the bearing &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-10T17:01:09.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8819" PostTypeId="1" CreationDate="2016-05-10T17:34:12.667" Score="4" ViewCount="1796" Body="&lt;p&gt;I am new to using ANSYS. I use the student version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to find what is the difference between two .db or .dbb  files? I have two files say &lt;code&gt;a.db&lt;/code&gt; and &lt;code&gt;b.db&lt;/code&gt;, which are supposed to be the same exact model for a simple beam structure with one point load on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One gives as solution for a deflection, which is a little bit different from the other file.  When I open the .db file inside ANSYS APDL GUI and look at all the listings in the GUI, I can't see any difference any where. I looked at material properties, nodes, elements, solver type, etc... and they are the same. But when I solve the model I get a very small difference in result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly there should be a better way to compare two &lt;code&gt;.db&lt;/code&gt; files than having to open each and look at each listings one by one and do manual comparison like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a command to list to a text file, everything inside the db file, all the tables, etc.. so that one can use some &lt;code&gt;diff&lt;/code&gt; utility for example to see what is the difference? or is there is some other ANSYS specific tool to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the student version of ANSYS on windows, it is ANSYS APDL mechanical version 16.02.&lt;/p&gt;&#xA;" OwnerUserId="934" LastEditorUserId="282" LastEditDate="2016-12-01T11:28:55.083" LastActivityDate="2021-04-10T06:06:05.523" Title="How to compare two ANSYS mechanical APDL databases? (.db or .dbb files)" Tags="|mechanical-engineering|ansys|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8820" PostTypeId="2" ParentId="8785" CreationDate="2016-05-10T20:02:47.980" Score="1" Body="&lt;p&gt;There are a number of reasons for adding filler material to polymers for casting or moulding, for  example :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Improving tensile strength, impact toughness and crack resistance : here short strand fibres such as glass or carbon may give the best performance but any powder filler will help to some extent. &lt;/li&gt;&#xA;&lt;li&gt;To increase bulk : especially with more expensive reins for obvious reasons cheap and easily handled materials which don't much compromise mould-ability and mechanical properties are preferred eg fumed silca, calcium carbonate powder, talc etc &lt;/li&gt;&#xA;&lt;li&gt;To simulate other materials : especially metals and some types of stone by adding fine powder of the appropriate material. &lt;/li&gt;&#xA;&lt;li&gt;To reduce density : especially hollow glass beads &lt;/li&gt;&#xA;&lt;li&gt;To control opacity/UV resistance &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Adding any inert filler will also tend to increase cure/demould  times and reduce the heat generated and associated problems in exothermic cure multi-part resin systems although this is obviously less of an issue for thermoplastics. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-13T18:39:51.690" LastActivityDate="2016-05-13T18:39:51.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8821" PostTypeId="2" ParentId="8814" CreationDate="2016-05-10T20:13:10.483" Score="2" Body="&lt;p&gt;I find such a column hard to justify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by @grfrazee and @ChrisJohns, what you are calling a &quot;thunk&quot; is usually called a bearing plate, and these are responsible for transmitting the force from the superstructure (beams) to the mesostructure (columns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by @ChrisJohns, these bearing plates may well be bolted, in which case a minimal distance to the edge of the column is necessary. That being said, these distances are usually quite small, and at a glance it seems to me that the smaller column diameter would be sufficient to satisfy that distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@grfrazee defines the flaring as a capital, but I'm not sure that's what we're seeing here. Capitals are mostly used when a column supports a slab directly (as opposed to supporting a beam which supports a slab). Their purpose is to increase the contact area and therefore decrease the punching force on the slab. Given that these columns are not supporting slabs (and even if they were, the bearing plate forces a concentration of punching force regardless), I don't think the flaring of the column is meant to be a capital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@grfrazee's image (repeated below) also raises another problem with this flared column head. By increasing the size of the column, they have increased the size of area $A_2$ in the figure below. This increases the bearing strength at the head, but if this compression were the ruling factor, you'd have the same problem away from the &quot;capital&quot;, once the section is reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2E1Qr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2E1Qr.png&quot; alt=&quot;a&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio $\dfrac{A_1}{A_2}$ (as opposed to $\dfrac{A_2}{A_1}$ when calculating bearing strength) is used when calculating the splitting forces at the head of the column, and the smaller this ratio (or, the larger $A_2$ is in proportion to $A_1$), the greater the splitting forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So honestly, I can't really see a valid motivation behind using such a column.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="2246" LastEditDate="2016-05-11T13:32:12.790" LastActivityDate="2016-05-11T13:32:12.790" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8822" PostTypeId="2" ParentId="8728" CreationDate="2016-05-10T20:13:31.507" Score="2" Body="&lt;p&gt;I would suggest modifying your 3D printed part so that you can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Cable_lacing&quot; rel=&quot;nofollow&quot;&gt;cable lacing&lt;/a&gt; to secure the wires. It's how NASA builds wires into cables and harnesses for space vehicles. If it can withstand vibrations during launch, I'm pretty sure it would work for your application. I was lucky and learned from a retired NASA technician, but you should be able to figure it out from the NASA standard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference chapter 9 of NASA-STD-8739.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2016-05-10T20:13:31.507" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8823" PostTypeId="1" CreationDate="2016-05-10T21:46:41.963" Score="3" ViewCount="229" Body="&lt;p&gt;After designing an aircraft idea I had in &lt;a href=&quot;http://www.xflr5.com/xflr5.htm&quot; rel=&quot;nofollow&quot;&gt;XFLR5&lt;/a&gt;, I noticed that it had a really high coefficient of moment (C&lt;sub&gt;m&lt;/sub&gt;) compared to other wings of similar size. What does this mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I played around with my center of gravity and the coefficient went down, but never to the level of the other aircraft. What relation does this have with the aerodynamic center?&lt;/p&gt;&#xA;" OwnerUserId="6401" LastEditorUserId="368" LastEditDate="2016-05-11T18:06:24.543" LastActivityDate="2017-06-25T15:21:37.803" Title="What does high coefficient of moment mean for an aircraft?" Tags="|aerospace-engineering|airfoils|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8824" PostTypeId="1" AcceptedAnswerId="22147" CreationDate="2016-05-11T01:27:22.703" Score="5" ViewCount="10007" Body="&lt;p&gt;What are the relative pros and cons between aluminum and steel fasteners when riveting together sheet aluminum (50XX or 60XX) when high amounts of vibration are anticipated? Assuming that the usual guidelines for sheet thickness, hole size and separation are followed, which metal would be the better choice? I don't anticipate atmospheric corrosion to be an issue. &lt;/p&gt;&#xA;" OwnerUserId="5427" LastActivityDate="2018-06-08T22:53:20.410" Title="Aluminum vs Steel rivets/fasteners in a high vibration environment" Tags="|steel|aluminum|vibration|rivets|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8825" PostTypeId="1" AcceptedAnswerId="8852" CreationDate="2016-05-11T06:54:03.970" Score="2" ViewCount="1313" Body="&lt;p&gt;I am setting up an experimental station for which I am using a small gear pump to pump water. I have noticed that on the discharge port of the pump, a short row of bubbles comes out at high flows and low pressures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, at 650 ml/min and 2 bar of system pressure I might observe a relatively long row of bubbles coming out from the port. By increasing the system pressure up to 7 bar, the row shortens and eventually disappears. The same happens if I decrease the flow. I have checked the feed, and it does not contain any bubbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has led me to believe I am having some degree of cavitation in my pump, is this correct?&lt;/p&gt;&#xA;" OwnerUserId="6405" LastEditorUserId="368" LastEditDate="2016-05-11T18:14:12.940" LastActivityDate="2019-02-15T15:38:00.460" Title="Small bubbles on a pump discharge" Tags="|mechanical-engineering|pressure|pumps|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8826" PostTypeId="1" AcceptedAnswerId="8827" CreationDate="2016-05-11T12:29:14.987" Score="3" ViewCount="840" Body="&lt;p&gt;I am reading this story on artist Ugo Rondinone's sculpture in the Nevada desert (&lt;a href=&quot;http://www.nytimes.com/2016/05/15/arts/design/building-an-artists-magic-mountains-to-draw-visitors-to-the-desert.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Building an Artist’s ‘Magic Mountains’ to Draw Visitors to the Desert&lt;/a&gt;) and simply wondering (as an artist myself): how did they make those columns stay up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't see any information on dimensions but from the photo with the artist below, one might guess that these columns are abut 25 feet high?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming there is a solid concrete footing for the columns to stand on and some preparation of the &quot;joints&quot; between the stones:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is the weight of the stones enough to keep them stacked and resistant to wind and weather (as such weather as there might be in the desert)?&lt;/li&gt;&#xA;&lt;li&gt;Is it likely that the stones are drilled to incorporate some steel armature?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I understand that the stones could just be balanced, but given that this is a funded and likely insured project, I find that hard to believe likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U9XgI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U9XgI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DMD1I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DMD1I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="896" LastEditorUserId="272" LastEditDate="2016-05-11T14:50:54.540" LastActivityDate="2016-05-13T21:36:31.687" Title="How do the columns in the 'magic mountains' sculpture stay up?" Tags="|structural-engineering|structures|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8827" PostTypeId="2" ParentId="8826" CreationDate="2016-05-11T13:47:39.193" Score="11" Body="&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=Cjuf14_2cCw&quot;&gt;This marketing video&lt;/a&gt; shows moments of their construction.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At 0:40 you can see them all upright. They aren't painted, but you can tell it's the same installation.&lt;/li&gt;&#xA;&lt;li&gt;At 1:37 and 1:46 you can see guys drilling into the boulders&lt;/li&gt;&#xA;&lt;li&gt;From 2:16 to 2:24 you can see the base is also being drilled and, most clearly, steel rods are being slid into the rocks.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So it is clear that the rocks are indeed incorporating steel reinforcement at the joints to avoid toppling over due to wind. Whether this reinforcement runs through the entire column or if the steel exists only around the joints is hard to tell. The former may be surmised by the length of the rods being inserted in the video, but the latter would be far easier to execute.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-05-11T13:47:39.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8828" PostTypeId="1" AcceptedAnswerId="8837" CreationDate="2016-05-11T14:28:22.510" Score="5" ViewCount="95" Body="&lt;p&gt;I am working on a project for college, and I am stuck on one idea that I don't really know how to represent it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically what I want to do is something like a pipe, that if a put a kind of cylinder on inside from the top side, the cylinder will reach the end by force of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KPQfy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KPQfy.png&quot; alt=&quot;cylinder going down&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if I turn the pipe, so I put the bottom on top and the top on the bottom, that cylinder must be stopped/blocked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9FlQq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9FlQq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my idea was to put some-kind of corrugated pipe as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LA64U.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LA64U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way the pipe will be one way only. As creating a pipe like this for a school project will be too hard, my questions is if is there already a pipe like this, if not if you have any other idea to achieve this.&lt;/p&gt;&#xA;" OwnerUserId="6410" LastEditorUserId="282" LastEditDate="2016-05-12T01:09:27.360" LastActivityDate="2016-05-12T12:22:45.047" Title="One way plastic rubber pipe" Tags="|mechanical-engineering|pipelines|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8829" PostTypeId="2" ParentId="8817" CreationDate="2016-05-11T14:31:43.470" Score="9" Body="&lt;p&gt;The part of the movie transcript you're talking about (&lt;a href=&quot;https://sfy.ru/?script=apollo13&quot;&gt;taken from here&lt;/a&gt;) is where Fred Halse says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Batt B, no volts, amps are okay. Batt C, s***. No volts,&#xA;  only two amps. They may die before the main chutes open.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To make sense of it, I think we need to ground ourselves with the real Apollo 13 and basic electronics. A battery can have zero volts across it and current flowing if it is being short circuited. Compare this to the &lt;a href=&quot;http://apollo13.spacelog.org/search/?q=Volts&quot;&gt;actual Apollo transcripts which mention &quot;volts&quot;&lt;/a&gt;: there are moments shortly following the catastrophy where zero volts is mentioned:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;d02 h07, m55, s20: &lt;a href=&quot;http://apollo13.spacelog.org/02:07:55:20/#log-line-201320&quot;&gt;I believe we've had a problem here.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;d02 h07 m58 s07: &lt;a href=&quot;http://apollo13.spacelog.org/02:07:58:07/#log-line-201487&quot;&gt;Okay, AC 2 is showing zip. I'm going to try to reconfigure on that, Jack.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;d02 h07 m58 s33 &lt;a href=&quot;http://apollo13.spacelog.org/02:07:58:33/#log-line-201513&quot;&gt;It's reading about 25-1/2, MAIN B is reading zip right now.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From what I gather, those are due to the destruction of the fuel cells (batteries). So no current would be flowing either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they quickly swap around connections between fuel cells, main buses and AC inverters, and I cannot find any further mention of zero volts. The voltages are usually around 30V, which makes sense since they were designed to be nominally at 28V DC.  (Aside: What caused the accident in the first place: at some point parts of the design were changed from 65V DC. &lt;a href=&quot;http://history.nasa.gov/ap13fj/08day3-problem.htm&quot;&gt;Explanation here&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the movie transcript, your quote is said straight after:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;KEN MATTINGLY&#xA;  - Re-entry interface in one minute and thirty seconds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the zero volt 2 amp movie quote vaguely corresponds in content and timing with the real transcript where battery and parachutes are mentioned shortly preceeding reentry:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;d05 m22 h22 s56: &lt;a href=&quot;http://apollo13.spacelog.org/05:22:22:56/#log-line-512576&quot;&gt;Okay, Odyssey; Houston. Just for your information, it looks as though battery C will deplete around main chute time; that's expected; you've got plenty of amp-hours in the other batteries.&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So I think that is your answer. Hollywood was trying to give the impression of the batteries having just about enough juice left. Technically it had little to do with volts and amps, but rather amp-hours; the energy left in the batteries.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-05-11T16:39:12.527" LastActivityDate="2016-05-11T16:39:12.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8830" PostTypeId="2" ParentId="8815" CreationDate="2016-05-11T14:51:52.283" Score="0" Body="&lt;p&gt;If you can go with a circular cross-section, you could use a pre-manufactured CFRP tube, e.g. &lt;a href=&quot;https://www.acpsales.com/Tubes-Carbon-Fiber.html&quot; rel=&quot;nofollow&quot;&gt;one of these&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cross-section shape is important, there is &lt;a href=&quot;https://www.acpsales.com/Carbon-Fiber-Tapes-and-Sleeves.html&quot; rel=&quot;nofollow&quot;&gt;CFRP sleeves&lt;/a&gt; you could probably wrap around your custom core, working in &quot;Positivbauweise&quot; (positive design?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly, the hole should be in longitudinal direction. For such a hole, I would apply the CFRP to a solid foam core and drill a hole in a second step. Use styrofoam or similar as core material that you can easily remove later. Also with some plastic cores, you can use acetone to remove the core entirely in case you don't need it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If surface finish is really important, you will have to go with VARTM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Sorry for the links being from the same supplier, but those were the first ones I found. I'm sure you can get similar products from lots of other suppliers.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2016-05-11T14:51:52.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8832" PostTypeId="1" AcceptedAnswerId="8836" CreationDate="2016-05-11T15:11:35.843" Score="-1" ViewCount="98" Body="&lt;p&gt;I am currently MS student in artificial intelligence (AI) and working on my thesis on reinforcement learning, which robot should learn by itself how to control a quadcopter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short, I have come up with a method for such purposes, but before I jump into quadcopter and deal with its non-linear controlling problems, which could take a month to program, I need to test the method to see if it's really a solution for this kind of problem. I've defined a simplified version of my problem, such as following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Simplified scenario:&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I have a ball &lt;strong&gt;dropped&lt;/strong&gt; at an altitude and the program must learn the &lt;strong&gt;force&lt;/strong&gt; it needs to apply on the ball to hold it still in the air at a pre-defined altitude.&lt;br&gt;&#xA;  (i.e if target altitude is higher than the altitude that ball's dropped, it should up-force the ball until it reaches the altitude, otherwise, it should moderately stabilize the ball in the target altitude).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In this scenario it doesn't matter how the force will apply to the ball, i.e. there is no rope attached or etc. it just gets applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last time I read something related to physics was 6-7 years ago, so the motion equation of this problem is beyond my specialty, but it seems to me that the defined scenario is a classic physic problem.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Questions:&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What are the equations to formulate the scenario (falling ball with an external force besides gravity) to be able to write its simulator?&lt;/li&gt;&#xA;&lt;li&gt;What about adding some noise to the problem, like air resistance factor or wind, how would the equations look then?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate if someone help a C.S. fellow here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;PS: Please let the equations be simple as they could be, I am not quite good at reading complicated physic equations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations-and-Free-Fall&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; site and details @Jodes kindly explained, I have come up with following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = {1 \over 2}at^2 + v_0t + x_0&#xA;$$&#xA;where $a = g + \sum {F_i \over m_{\text{ball}}}$ and $F_i$ are the external forces applied to the ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm right, the noises could also be combined to the equation as external forces&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does this formulation valid for the defined scenario?&lt;/li&gt;&#xA;&lt;li&gt;If so, how to make it work in 2D and 3D environment(does simply vectorizing the $\vec{x}, \vec{a}, \vec{v}$ will do the job?)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6412" LastEditorUserId="6412" LastEditDate="2016-05-11T17:23:46.403" LastActivityDate="2016-05-11T19:43:21.027" Title="Need motion equations for falling ball for simulating" Tags="|mechanical-engineering|simulation|experimental-physics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8833" PostTypeId="1" CreationDate="2016-05-11T15:24:58.857" Score="1" ViewCount="762" Body="&lt;p&gt;I am looking to build a detachable hammock stand structure for camping. I was originally thinking about making one out of PVC but I am in the UK and it isn’t cost effective as it looks like it needs to be imported. I am therefore looking for alternative materials that are durable enough to bear a heavy load (up to two adults) but are light enough to carry when detached. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions? I have been looking at polycarbonate tubes but I am unsure if it would be rigid enough with weight pulling on it and what diameter the tube would have to be. Any ideas are greatly appreciated!       &lt;/p&gt;&#xA;" OwnerDisplayName="user6413" LastActivityDate="2016-05-11T15:24:58.857" Title="Suitable tube materials for making a light-weight, heavy-load bearing structure" Tags="|materials|structural-engineering|structures|" AnswerCount="0" CommentCount="4" ClosedDate="2016-05-11T18:19:53.437" ContentLicense="CC BY-SA 3.0" />
  <row Id="8834" PostTypeId="2" ParentId="8825" CreationDate="2016-05-11T15:33:19.457" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have noticed that on the discharge port of the pump, a short row of bubbles comes out at high flows and low pressures. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This does indeed sound like &lt;a href=&quot;https://en.wikipedia.org/wiki/Cavitation&quot;&gt;cavitation&lt;/a&gt; to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cavitation occurs usually in areas with a rapid change in pressure (from high to low, usually).  As you say, when you increase the system pressure, the bubbles disappear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it this way - you either have too fast or too low pressure of a flow.  The fluid can't go through the outlet side fast enough, so bubbles form as the fluid tries to keep up.  By decreasing the flow or increasing the pressure, the fluid can keep up with itself on its way out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should try to balance your system such that cavitation is minimized.  It's not good for the pump parts.&lt;/p&gt;&#xA;" OwnerUserId="2246" LastActivityDate="2016-05-11T15:33:19.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8836" PostTypeId="2" ParentId="8832" CreationDate="2016-05-11T18:50:07.560" Score="1" Body="&lt;p&gt;The force equation of an object in a uniform gravitational field with a mysterious compensating force is &#xA;$$&#xA;ma=F_c-mg&#xA;$$&#xA;Where $m$ is the mass of the object, $g$ is the acceleration due to gravity, $F_c$ is the force applied by your controller, and $a$ is the net acceleration.  For simulation purposes, you probably want to express this as a differential equation&#xA;$$&#xA;\frac{d^2}{dt^2}y=\frac{F_c}{m} - g.&#xA;$$&#xA;If you want to add a random fluctuating force due to noise, $F_n$ then you can simply add it in on the RHS of the equation&#xA;$$&#xA;\frac{d^2}{dt^2}y=\frac{F_c}{m} - g + \frac{F_n}{m}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2016-05-11T19:43:21.027" LastActivityDate="2016-05-11T19:43:21.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8837" PostTypeId="2" ParentId="8828" CreationDate="2016-05-11T18:59:25.930" Score="5" Body="&lt;p&gt;Your idea would not be too hard to construct. The one direction tines can be installed through the side of the pipe. Any springy rod material could be used; plastic rod, spring steel, etc. Rubber would probably not be a good choice since it has higher friction and less restoring force than similarly sized plastic or spring steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drill holes in the side of the pipe and insert the tines. Depending on the scale and wall thickness of the pipe you may need additional support to keep your tines at the angle without failure. This may require a gusset on each tine or a larger diameter pipe to anchor to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 45 degree angle and a passage slightly smaller than your object like you show would be a good starting point. You may have to change the geometry if you require changes in the passing or blocking forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like mentioned in the comments, it is helpful if you can provide a bigger picture understanding of the design problem. There is often an easier solution that may not have been considered.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-05-11T18:59:25.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8841" PostTypeId="1" CreationDate="2016-05-12T11:13:00.547" Score="2" ViewCount="5436" Body="&lt;p&gt;There are several standards which all are valid when it comes to mechanical shock, bump and vibration, both for products in use and packaged and being transported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a few:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ISO 13355:2001 - Packaging -- Complete, filled transport packages and unit loads -- Vertical random vibration test&lt;/li&gt;&#xA;&lt;li&gt;ISO 2244:2000 - Packaging -- Complete, filled transport packages and unit loads -- Horizontal impact tests&lt;/li&gt;&#xA;&lt;li&gt;ASTM D6344 – 04 (2009), Standard Test Method for Concentrated Impacts to Transport Packages&lt;/li&gt;&#xA;&lt;li&gt;ISTA Air ride&lt;/li&gt;&#xA;&lt;li&gt;IEC 60068-2-27:2008 - Environmental testing - Part 2-27: Tests - Test Ea and guidance: Shock&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;End customer might have specific requirements, but what would a good baseline be? Is there a de facto/popular standard that everyone tests according to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Product is more expensive than your average consumer product, but far from a space shuttle.&lt;/p&gt;&#xA;" OwnerUserId="6424" LastEditorUserId="368" LastEditDate="2017-12-14T16:45:56.437" LastActivityDate="2018-04-09T10:36:55.327" Title="Which vibration standard is the most common for transported packages?" Tags="|standards|vibration|transportation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8842" PostTypeId="2" ParentId="8828" CreationDate="2016-05-12T12:08:47.947" Score="1" Body="&lt;p&gt;This is known as a &quot;valve&quot; (duh :-) ).  If you have enough room in your pipe, cut a circular piece of similar pipe whose diameter is a bit less than the pipe's diameter.  then mount this curved plate with a hinge on one side and mount a small stop on the other side. You want it set up so that when the pipe is in the &quot;go&quot; orientation, the plate falls down due to gravity and since it's curved, it goes flat against the pipe's interior.  In the &quot;stop&quot; config, the plate falls against the stop and blocks the pipe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies for poor 3D rendering of the &quot;potato chip&quot; - shaped valve plate.  Left picture is in blocking mode; right picture shows the tube with the other end up, in 'passthru' mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eW7Ud.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eW7Ud.jpg&quot; alt=&quot;Tube up/down&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="615" LastEditDate="2016-05-12T12:22:45.047" LastActivityDate="2016-05-12T12:22:45.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8843" PostTypeId="2" ParentId="8841" CreationDate="2016-05-12T12:44:13.433" Score="1" Body="&lt;p&gt;In general, an appropriate transit test should reflect the particular shipping environment which the product will be subjected to, usually determined by the size, weight and shipping mode(s). See my answers here: &lt;a href=&quot;https://engineering.stackexchange.com/q/2068/2061&quot;&gt;Are there standard vibration and shock specs for shipment of a product?&lt;/a&gt; and here: &lt;a href=&quot;https://engineering.stackexchange.com/q/5923/2061&quot;&gt;Vibration testing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience, the best test spec is not simply selecting one of the published standards and going with it as is, though that will be much easier and may be a better place to start. The optimum test will be instead based on different components of established standards and should contain references those standards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ISTA standard contains a variety of protocols from which you can pick and choose based on your needs. If you are looking for a standard that's easy to understand and use, ISTA is it. You should review their guide here:&#xA;&lt;a href=&quot;http://www.ista.org/forms/ISTAGuidelines.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISTA Guidelines&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another consideration is access to appropriate test equipment. How are you going to perform these tests? In general, if you have access to the equipment necessary to perform the test, you should already have the knowledge to select appropriate tests. If you don't, then you'll need to have testing done by someone who does and they should be able to advise you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, forewarned is forearmed. Knowing that you're shipping via a UPS type of carrier and not knowing the package dimensions and weight of your product, I'd suggest a test consisting of the following as a starting point: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;24 hr Thermal conditioning (X units at -34C &amp;amp; X units at +66C)&lt;/li&gt;&#xA;&lt;li&gt;10 pt drop test (height determined by pkg weight) Onto the bottom vertex of the carton which includes the carton seam. Each of the three edges radiating from that vertex. Flat on each of the 6 faces. (UPS)&lt;/li&gt;&#xA;&lt;li&gt;Random vibration (typically 0.52 gRMS) for 1 hour on each of the three axes with a vibration profile according to ASTM D4728, ISTA or ideally a profile developed from data collected by monitoring your particular distribution stream.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The above assumes that the items will be shipped individually, shipping a palletized load is a different animal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you suggest, the end customer might have their own standards or requirements to comply with a different published standard. In that case, I suggest that you have a dialogue with them about testing for shipping and handling and make sure that whatever tests you perform meet or exceed their standard and that they agree to any deviation from their standard.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: I spent 7 years testing and developing test standards for a wide variety of consumer product/packaging for shipping/handling via all manner of transport modes.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-05-12T20:16:28.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8845" PostTypeId="1" CreationDate="2016-05-12T13:03:51.360" Score="3" ViewCount="1234" Body="&lt;p&gt;What’s the composition of the (slighlty sticky) substance we find on foam earplugs that allow them to be easily moldable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This substance slows down the expansion of the foam, so it makes the foam a little bit more rigid and help the foam to stick into the ear canal. (It might also have some insulation property in combination with the foam.) This substance get out when you wash the earplugs with soap, it's probably sprayed on the foam, or impregnated on the foam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea ?&lt;/p&gt;&#xA;" OwnerUserId="6427" LastEditorUserId="6427" LastEditDate="2016-05-13T03:35:13.297" LastActivityDate="2023-10-05T14:14:33.257" Title="foam earplugs: what’s the composition of the (non washable) substance that slow down their expansion?" Tags="|chemical-engineering|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8846" PostTypeId="1" CreationDate="2016-05-12T13:38:44.493" Score="5" ViewCount="412" Body="&lt;p&gt;I have to design a PI-controller with zero order hold for the plant &#xA;$G_s(s) = \frac{0.24}{(s+4)(s+12)}$ to meet the specifications of $\omega_n=5\,s^{-1}$ and $\zeta = 0.6$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I choose a $T=0.12\,s$ which leads to a approx. 13 samples per oscillation  (and $T \leq 0.125 T_d$)and applied the Z.O.H. to&#xA;$H_o(s)G_s(s)=G(s)\approx \frac{1}{1+0.5 T s}\frac{0.24}{(s+4)(s+12)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the transfer function of the PI-controller:&#xA;$G_D(s) = K_c\frac{a + s}{s}$ I get the characteristic equation with &#xA;$H_c(s) = \frac{G(s)}{1+G_D(s)G(s)}$ to $\frac{4s}{s^3 + 32.667 s^2 (314.672 + 4 K_c)s + 800.016 + K_c a 4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the desired system of:&#xA;$\frac{1}{(s^2 + 2 \omega_n \zeta s + \omega_n^2)(s + \alpha)}= \frac{1}{s^3 + 32.6s^2 + 184.6 s + 665}$ where I choose  $\alpha=26.6$ with $\alpha \approx 4 \ldots 5 \omega_n$ (hence, the $s^2$-term is equal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I did the comparison of coefficents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$314.672 + 4 K_c = 184.6 \rightarrow K_c=-32.518$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$800.916 + K_c a 4 = 665 \rightarrow a = 1.038 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically this should be now a proper $K_c$ and $a$. But simulation shows undesired behavior (step response goes to infinity). Should I introduce another pole? Is the general approach right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Remark: I edited the problem considerable with help from &lt;a href=&quot;https://electronics.stackexchange.com/questions/25635/design-of-a-pi-controller&quot;&gt;Design of a PI controller&lt;/a&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="6429" LastEditorUserId="6429" LastEditDate="2016-05-12T17:55:26.200" LastActivityDate="2024-02-21T02:03:25.407" Title="PI-Controler: Comparison of characteristic equations" Tags="|control-theory|pid-control|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="8847" PostTypeId="2" ParentId="8815" CreationDate="2016-05-12T13:47:24.020" Score="1" Body="&lt;p&gt;If your part is above a certain diameter (the hole in the middle) you can fit a vacuum bag within the hole between the two parts of the foam core do your lay up on the core put it into the negative mold and close the vacuum bag (which has now a toroidal shape). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each case the dimensions/tolerances of the foam core need to be adapted to the layup thickness / volume fraction of matrix/fibers.&lt;/p&gt;&#xA;" OwnerUserId="6429" LastActivityDate="2016-05-12T13:47:24.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8848" PostTypeId="2" ParentId="8549" CreationDate="2016-05-12T14:38:26.077" Score="3" Body="&lt;p&gt;I understand this question is about an event in Brazil and not the US. However, section 3.7.4 (Wave Loads) of the 2012 AASHTO LRFD Bridge Design Specifications simply says: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wave action on bridge structures shall be considered for exposed structures where the development of significant wave forces may occur. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(This isn't a summary; that's literally the entire section.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The commentary recommends referring to the &lt;em&gt;Shore Protection Manual&lt;/em&gt; from the Coastal Engineering Research Center. Copies of the manual (at least older editions) seem to be available via a quick Google search; it is very lengthy and I am not familiar with it myself, so it may or may not have specific recommendations for calculation of load cases for vertical wave action. Perhaps someone more familiar with the manual can edit my answer with more information. &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2016-05-12T19:14:18.607" LastActivityDate="2016-05-12T19:14:18.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8849" PostTypeId="1" CreationDate="2016-05-12T15:34:57.603" Score="4" ViewCount="164" Body="&lt;p&gt;I have this &lt;a href=&quot;http://www.inventorwizard.nl/gallery/air%20radial%20engine/Air%20Radial%20Engine.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;6 cylinder radial engine&lt;/a&gt; designed in Solidworks.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ex4YT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ex4YT.jpg&quot; alt=&quot;Rendering&quot;&gt;&lt;/a&gt;&#xA;Rendering&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LCoYf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LCoYf.png&quot; alt=&quot;Blueprint&quot;&gt;&lt;/a&gt;&#xA;Blueprint (Source:Inventor Wizards)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I want to perform some motion analysis on this engine to make some prediction about it's feasibility.&#xA;So by now I have successfully completed the motion study simulation and generated plots of piston acceleration against time.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UauQU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UauQU.png&quot; alt=&quot;Acc-Time Plot&quot;&gt;&lt;/a&gt;&#xA;Plotted for a crankshaft speed of 200RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can i use some method or this data to predict acceleration of large scale model of this engine?&lt;/p&gt;&#xA;" OwnerUserId="6430" LastEditorUserId="6430" LastEditDate="2016-05-13T16:02:34.793" LastActivityDate="2016-05-13T16:02:34.793" Title="How to compare small and large scale model of an IC engine to predict force, torque, rpm etc generated by actual engine?" Tags="|design|simulation|mechanical-failure|solidworks|cad|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8850" PostTypeId="2" ParentId="8160" CreationDate="2016-05-12T15:54:42.420" Score="2" Body="&lt;p&gt;When used as a generator, a motor can generate approximately the same power that it requires for operation as a motor at rated speed and torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the voltage produced by a generator is directly proportional to speed. Power is proportional to voltage multiplied by current. The maximum current that can be drawn from a generator is generally a fixed value that is not determined by speed. Therefore the power that can be produced by a generator is directly proportional to speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effect of changing the strength of the magnetic field (field excitation) is generally to increase or decrease the voltage generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of factors that determine a generator's limits for safe operation. Those factors are generally similar between motor and generator operation. They need to be considered for each different type of machine, AC and DC, wound-field and permanent-magnet, induction and synchronous, plus all of the sub-categories and design variations.&lt;/p&gt;&#xA;" OwnerDisplayName="user6431" LastActivityDate="2016-05-12T15:54:42.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8851" PostTypeId="1" CreationDate="2016-05-12T20:46:25.113" Score="1" ViewCount="37" Body="&lt;p&gt;I have an application in which I need to adjust the position of 2 perforation wheels along the y axis to regulate their distance from a film web edge.  Though the speed of this positioning will be quite slow (low FT), I am concerned about damaging the perforation wheels as they are pressed against a steel roller with significant force to ensure cutting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figure there a couple ways to prevent damage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Reduce force (Fp) on perforation wheel during movement.&lt;br&gt;&#xA;-Interrupts continuous operation of the machine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Spin the perforation wheel about the z axis proportionally to the speed of movement along the y axis, sort of like steering.  I would need to sync the two wheels together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Do nothing: The wheel self corrects by deflecting in the y direction by some small angle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Context: The total travel of the perf wheels is unlikely to exceed an inch in total.  I do not currently know the pressure on the wheels, I don't know of an easy way to measure it either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the machine in its current form is in continuous use at the moment I cannot really trial many things for real.  Thanks for the advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q0GAV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q0GAV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4697" LastActivityDate="2018-06-10T05:17:33.890" Title="Help avoiding damage to perforation wheels" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8852" PostTypeId="2" ParentId="8825" CreationDate="2016-05-12T21:21:39.637" Score="5" Body="&lt;p&gt;As well as 'proper' cavitation which is a fairly high energy phenomena and certainly a concern in terms of wear on the working surfaces of the pump it is possible that pressure changes are causing dissolved gasses in the water (most likely air) to come out of solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressures and flow-rates here aren't huge but having said that gear pumps can produce very high local pressures. It would be interesting to know if you get the same bubbles when the pump is connected to a short tube to an open outlet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unknown is if this is happening on the surfaces of the pump with enough energy to be a problem or if it is happening downstream of the pump itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could also be indicative of a small leak somewhere, note that a leak which admits small bubbles of air may not be very noticeable in terms of the amount of fluid which is leaking. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-12T21:21:39.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8855" PostTypeId="1" AcceptedAnswerId="8856" CreationDate="2016-05-13T02:49:53.967" Score="0" ViewCount="513" Body="&lt;p&gt;I would like to know how to calculate unit mass of beam having uniform cross section, made by shell, by using Patran or Nastran, not by manual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there be any functions to calculate mass or unit mass?&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="4624" LastEditDate="2016-05-13T03:59:32.083" LastActivityDate="2016-05-13T03:59:32.083" Title="Calculation of unit mass of beam shell by using patran or nastran" Tags="|modeling|simulation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8856" PostTypeId="2" ParentId="8855" CreationDate="2016-05-13T03:11:03.830" Score="1" Body="&lt;p&gt;You can do it in Nastran using the &lt;strong&gt;Grid point weight generator&lt;/strong&gt; option. Look up PARAM,GRDPNT in the documentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Patran probably has something equivalent, but the Nastran output is &quot;guaranteed&quot; to be consistent with the model that you actually analysed.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-05-13T03:11:03.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8858" PostTypeId="1" AcceptedAnswerId="8859" CreationDate="2016-05-13T10:41:53.537" Score="4" ViewCount="347" Body="&lt;p&gt;In an answer to &lt;a href=&quot;https://engineering.stackexchange.com/questions/8160/how-to-calculate-power-of-motor-when-used-as-a-generator&quot;&gt;this question&lt;/a&gt;, for a motor used as a generator, the following was said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The maximum current that can be drawn from a generator is generally a fixed value that is not determined by speed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What equation(s) show this? &lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-05-13T14:32:02.127" Title="Why is the current that can be drawn from an electric generator constant?" Tags="|electromagnetism|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8859" PostTypeId="2" ParentId="8858" CreationDate="2016-05-13T11:31:21.653" Score="2" Body="&lt;p&gt;The current that a generator can produce is limited by the wire used in its construction. At some current level, the I&lt;sup&gt;2&lt;/sup&gt;R losses in the wire will cause excessive heating inside the generator, as well as a voltage drop at the output.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-05-13T11:31:21.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8860" PostTypeId="2" ParentId="8858" CreationDate="2016-05-13T11:40:19.717" Score="2" Body="&lt;p&gt;That statement about the current being independent of speed is just plain wrong.  This should be obvious by thinking about the limiting case.  When the generator isn't spinning at all, it won't produce any current, but it will produce current at some finite speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily prove to yourself that the statement is wrong.  Take a simple brushed DC motor.  These also work backwards as generators.  Connect it to a ammeter.  That's essentially shorting its output while measuring the current.  Now spin the motor (generator).  The current will be proportional to the spinning speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't follow the link to where this quote came from.  Perhaps there are various qualifying statements around it.  However, taken by itself without some specific context, that statement is wrong.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-05-13T11:40:19.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8861" PostTypeId="1" AcceptedAnswerId="8864" CreationDate="2016-05-13T12:23:02.843" Score="0" ViewCount="82" Body="&lt;p&gt;I'm currently filling out an application for the Massachusetts FE &amp;amp; EIT certification. Section 22 requests information regarding memberships to professional organizations. Unfortunately, I'm not a member of any and I'm afraid of leaving this section blank. Is anyone familiar with how leaving this section blank would impact my chances of getting certified?&lt;/p&gt;&#xA;" OwnerUserId="6441" LastActivityDate="2016-05-13T15:16:28.500" Title="How difficult is attaining EIT certification?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2016-05-13T17:07:31.180" ContentLicense="CC BY-SA 3.0" />
  <row Id="8864" PostTypeId="2" ParentId="8861" CreationDate="2016-05-13T15:16:28.500" Score="1" Body="&lt;p&gt;I am not from Mass nor do I plan to go through the EIT process again.  Unless they state a requirement for you to complete X task, they should not be able to deny you access to the test based on not having completed the task.  Personally, I believe they are more compiling statistics with that information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joining professional organizations are a great thing and will help your business communications and to develop contacts.&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2016-05-13T15:16:28.500" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8866" PostTypeId="1" AcceptedAnswerId="8868" CreationDate="2016-05-13T18:23:49.623" Score="1" ViewCount="1303" Body="&lt;p&gt;In order to give you a good idea of what I want to accomplish suppose you have a telescope mounted on a single rod extruding from the ground. The telescope is free to move such that you can tilt it to point at a particular altitude but can never point towards the ground. In order to move the telescope you have a wheel which can only be rotated in a single direction (say clockwise). Is there a gear configuration that would make a point alternate rotations every 180 degrees? and also have it configured such that the power loss is consistent throughout the entire cycle so that no half a cycle takes more power than the other? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an illustration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mh2NG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mh2NG.png&quot; alt=&quot;Img&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1084" LastActivityDate="2016-05-13T21:24:48.367" Title="Is there a way rotate a wheel in a single direction but have opposing motions every 180 degrees?" Tags="|mechanical-engineering|gears|applied-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8867" PostTypeId="2" ParentId="8845" CreationDate="2016-05-13T18:30:24.297" Score="1" Body="&lt;p&gt;Mouldable earplugs are often made from PU foam which can be self skinning ie the foam structure is created in a mould (as opposed to being cut from bulk material). I n this case the smooth skin is just a change in the density of the material, created as it is forced against the surface of the mould rather than a separate coating. again in this case the 'slow recovery' is a material property of the foam rather than any additional coating. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-13T18:30:24.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8868" PostTypeId="2" ParentId="8866" CreationDate="2016-05-13T18:55:05.107" Score="3" Body="&lt;p&gt;A crank (or 4 bar) type linkage can be forced to do this. In a crank mechanism there is a a point at bottom dead centre and top dead centre where the rotation could go in either direction regardless of the direction of the driving wheel/arm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could have a stop which prevents the driven arm from rotating past 180 degrees, at which point it will naturally reverse direction. Note that steam trains have opposite wheels driven by cranks 90 degrees out of phase precisely to prevent this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also use half gears (ie gears with no teeth on half of their circumference) along  with an idler gear on one to achieve a similar effect.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also lots of different cam and multi-bar linkages which will produce reciprocating motion depending on your exact requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between the myriad of possible solutions will depend to a large extent on the function which relates input to output, particularly how linear it is throughout the range of movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this sort of application it is also well worth looking through old patents as there are thousands and thousand of mechanical linkage patents  out there for almost every imaginable mechanical application, many of which have now expired. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-13T21:24:48.367" LastActivityDate="2016-05-13T21:24:48.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8871" PostTypeId="2" ParentId="8845" CreationDate="2016-05-13T20:52:06.010" Score="5" Body="&lt;p&gt;That sticky oily substance found in the ear plug foam is the plasticizer leaching out. The plasticizer is unlikely to be added on after the foam is made, I think it would be mixed in with the source polymer before it gets foamed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually the more plasticizer there is, the softer the foam gets, and the slower the foam recovers from being compressed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is likely happening when you wash them, is that the soap is removing the oil based plasticizer from the foam.&#xA;A short wash should only remove plasticizer from the surface of the material, but since it is a foam, internally it has a high surface area, making more plasticizer available to be washed away.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2016-05-13T20:52:06.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8872" PostTypeId="1" CreationDate="2016-05-13T19:47:14.960" Score="1" ViewCount="62" Body="&lt;p&gt;I've used a few types of mechanically actuated variable pitch props on my boat, so I know their performance is good.  However, their strictly mechanical and also self-contained nature makes their operation unreliable.  I'm looking into the feasibility of an electrically actuated one for my own use, and I have a few questions.  Since a prop must bolt on to an existing shaft drive, it has to be fully self-contained- no levers or wires connected externally.  Lots of challenges here, but the biggest that I'm having trouble finding info on is the g-force question.  At 4k rpms, say 2&quot; off center, that's suddenly 900 gs of acceleration.  Will any servo or controller be able to handle that?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to measure the RPM internally as well, though potentially I could mount a magnet on the drive to count passes/rpms that way.  But an accelerometer on board would be neater.  Again, could it handle the gs?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll certainly entertain discussion of moving everything to the center (not ideal), but even then any part of a component that extends to 1&quot; from center would see 500 gs.&lt;/p&gt;&#xA;" OwnerUserId="6477" OwnerDisplayName="Eric" LastEditorUserId="2659" LastEditDate="2016-05-13T23:20:48.420" LastActivityDate="2016-05-13T23:20:48.420" Title="Servo, control, rpm sensor in high g environment" Tags="|mechanical-engineering|electrical-engineering|embedded-systems|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8873" PostTypeId="2" ParentId="8826" CreationDate="2016-05-13T21:36:31.687" Score="1" Body="&lt;p&gt;I do this for a living and I would certainly say that having a single steel rod running the full height of the sculpture would be the idea way to do it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method means that all of the major loads on the structure apart form its own weight in compression are supported by steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have individual pins for each joint means that any side loads are transmitted into the stone as shear of bending loads which is far form idea for natural stone (or even concrete for that matter). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another very important consideration is the effect of water ingress into the structure and how this could cause problems with freeze/thaw cycles and corrosion. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-13T21:36:31.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8874" PostTypeId="1" CreationDate="2016-05-13T22:10:45.067" Score="5" ViewCount="5217" Body="&lt;p&gt;Ziegler–Nichols tuning creates a &quot;quarter wave decay&quot;. What does that mean? I cannot find anything online!&lt;/p&gt;&#xA;" OwnerUserId="4846" LastEditorUserId="1586" LastEditDate="2016-05-25T18:08:58.343" LastActivityDate="2022-10-24T01:35:55.753" Title="what is a quarter wave decay response after Ziegler–Nichols tuning?" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8875" PostTypeId="1" CreationDate="2016-05-14T02:12:28.440" Score="1" ViewCount="127" Body="&lt;p&gt;I was wondering how a minute repeater works - it is a complication in a watch that chimes the hours, quarter hours, and minutes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if you engaged the repeater, and the time was 2:47, the watch would chime &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dong, dong, ding-dong, ding-dong, ding-dong, ding, ding&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What are the mechanisms needed in order to make this happen? There are probably many ways to build one. How would one describe the workings of the simplest minute repeater possible?&lt;/p&gt;&#xA;" OwnerUserId="1879" LastEditorUserId="1879" LastEditDate="2016-05-15T14:03:07.377" LastActivityDate="2016-05-16T22:11:20.430" Title="How does a minute repeater work?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8876" PostTypeId="2" ParentId="417" CreationDate="2016-05-14T06:54:17.650" Score="3" Body="&lt;p&gt;I made my own wind tunnel,but it was NASA's visualization one. Incense did not work as a smoke source for me, but you might have better luck. What I ended up using for smoke was dry ice and hot water. One drawback to this is that the fluid density is higher than air, so the condensed air drops in a horizontal wind tunnel. The solution to this for me was to have the wind tunnel vertical.&lt;/p&gt;&#xA;" OwnerUserId="6285" LastActivityDate="2016-05-14T06:54:17.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8879" PostTypeId="1" AcceptedAnswerId="8883" CreationDate="2016-05-14T13:21:42.823" Score="1" ViewCount="2069" Body="&lt;p&gt;From an Inspection and Test Checklist for flange heaters (the number is the same on both lines):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Kxp9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Kxp9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wonder what the difference between &quot;Design Drawing&quot; (or is it &quot;Drawings&quot;?) and &quot;Manufacturing Drawing(s)&quot; is. Is it that Mfg. Dwg. merely represents some part of Design Dwg. (sheet no.3) that is copied for use in the production shop? Or are there some additions\changes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is a very basic question. I tried to translate this document and dropped the translation because I found it quite hard to understand many of the terms used, especially in the condensed context of a checklist. But I remembered some terms, and it made me curious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I was told on a translators' site that it could possibly mean &quot;Manufacturer's Drawing Number&quot; in this context. I wonder if that's true. &lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="349" LastEditDate="2016-05-15T01:54:43.137" LastActivityDate="2016-05-15T01:54:43.137" Title="Difference between &quot;Design Dwg.&quot; and &quot;Mfg.Dwg.&quot;" Tags="|drafting|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8881" PostTypeId="1" CreationDate="2016-05-14T15:31:52.353" Score="2" ViewCount="1083" Body="&lt;p&gt;In a description for ballscrews on a CNC, I saw two types of error listed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Incremental error: &amp;lt; 0.0006&quot;/ft&lt;/li&gt;&#xA;&lt;li&gt;Cumulative error: &amp;lt; 0.0008&quot; at 35&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are these types of error measuring? What is the difference between them?&lt;/p&gt;&#xA;" OwnerUserId="6453" LastEditorUserId="1832" LastEditDate="2016-05-14T16:16:06.630" LastActivityDate="2016-05-14T22:50:36.423" Title="What is the difference between incremental and cumulative error?" Tags="|mechanical-engineering|tolerance|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8883" PostTypeId="2" ParentId="8879" CreationDate="2016-05-14T18:50:38.413" Score="5" Body="&lt;p&gt;This describes the difference. &lt;a href=&quot;http://www.vista-industrial.com/blog/engineering-drawings-vs-manufacturing-drawings/&quot;&gt;http://www.vista-industrial.com/blog/engineering-drawings-vs-manufacturing-drawings/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;design drawing&lt;/strong&gt; or &lt;strong&gt;engineering drawing&lt;/strong&gt; is a &lt;em&gt;complete specification&lt;/em&gt; of the finished product - in other words, it shows the important properties of each part (dimensions, materials, specifications for electrical components, etc), and how the parts work together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;manufacturing drawing&lt;/strong&gt; has a more limited function. It tells you how to &lt;em&gt;make&lt;/em&gt; each a particular part, but nothing else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example of a heater, the &lt;em&gt;design&lt;/em&gt; drawing would specify the sizes of the screws or bolts that fix different parts of the heater together, but the heater company won't make any &lt;em&gt;manufacturing&lt;/em&gt; drawings for the screws or bolts - they will buy those parts ready made. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the company that &lt;em&gt;makes&lt;/em&gt; the bolts and screws &lt;em&gt;will&lt;/em&gt; have detailed manufacturing drawings showing exactly &lt;em&gt;how&lt;/em&gt; to make a bolt from a plain metal rod, or whatever type of raw material they use, but they won't have any engineering drawings that show how the bolts are used after they have sold them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often one &quot;drawing&quot; consists of several pages, or &quot;sheets&quot; (before computers, these were literally several physical sheets of paper). In your example, perhaps the design drawing was the first sheet, and the manufacturing drawing for each component of the heater was on a separate sheet - for example the manufacturing drawing for the &quot;shell&quot; was sheet 3 of the complete set. A complete set of drawings for a complex device may contain hundreds or even thousands of sheets.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-05-14T18:50:38.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8884" PostTypeId="1" AcceptedAnswerId="8896" CreationDate="2016-05-14T19:24:55.537" Score="2" ViewCount="681" Body="&lt;p&gt;I'm a 3rd-year bachelor student and I may be mistaken but from what I have understood:&#xA;In control theory we assume that the studied systems have a State Variable Description. Then from there we can derive that for such systems the transfer function must be proper.&#xA;But, to have a State Variable Description means that the system can be described by a simple differential equation or equivalently by a system of differential equations.&#xA;I am not really convinced that all physical systems can be described by such equations. Although I don't know what kind of other description there could be, I would like to find some strong argument. I guess it must have to do with the equations of fluid or Newtonian dynamics but if someone could be more specific I would appreciate it.&lt;/p&gt;&#xA;" OwnerUserId="6458" LastEditorUserId="1832" LastEditDate="2016-05-14T21:04:11.147" LastActivityDate="2016-05-16T02:40:45.580" Title="Control theory - Differential equations and state variable description" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8885" PostTypeId="1" AcceptedAnswerId="8969" CreationDate="2016-05-14T21:58:10.683" Score="1" ViewCount="156" Body="&lt;p&gt;I am a complete solidworks novice trying to create the part pictured in the orthographic schematic below but i am really struggling. I hoped someone might be able to guide me in creating this part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/70wxT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/70wxT.jpg&quot; alt=&quot;Sideplate schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I began by creating a sketch as follows but i get stuck after this. I'm not sure how to align the top two circles in the center of the plate or how to create the rounded edge on the bottom of the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zv1hq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zv1hq.jpg&quot; alt=&quot;progress&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated! &lt;/p&gt;&#xA;" OwnerUserId="6459" LastEditorUserId="6459" LastEditDate="2016-05-14T22:09:13.457" LastActivityDate="2016-05-21T14:01:04.997" Title="Creating a side plate in Solid Works" Tags="|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8886" PostTypeId="2" ParentId="8881" CreationDate="2016-05-14T22:50:36.423" Score="2" Body="&lt;p&gt;The &quot;incremental error&quot; is the relative error in moving between two positions. (But only the USA would think of specifying a dimensionless quantity in &quot;inches per foot&quot; just to confuse people.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the screw is commanded to move a distance $x$ (in any length unit!) the distance actually moved will be between 0.99995$x$ and 1.00005$x$ (because 0.0006/12 = 0.00005)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the cumulative error after several movements (or one long movement) is less than this: the above formula would give between 34.99825&quot; to 35.00175&quot; for a commanded total movement of 35&quot;, but the second specification says the actual distance moved will be between 34.99920&quot; and 35.00080&quot;, which is about half the worst incremental error.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-05-14T22:50:36.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8887" PostTypeId="1" CreationDate="2016-05-15T02:39:56.353" Score="3" ViewCount="113" Body="&lt;p&gt;Consider an arm of length $D$ supporting $W$ weight. The arm is welded to the plate with bolts $d$ inches apart. How do I determine the thickness of the steel plate 6 in wide by a length supported every $D_1$ inches that the arm is bolted to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H7pbO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H7pbO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W$ is nominally 300 lbs.&lt;/p&gt;&#xA;" OwnerUserId="6456" LastEditorUserId="1832" LastEditDate="2016-05-15T06:02:50.297" LastActivityDate="2016-05-15T15:38:53.590" Title="Steel thickness required to mount cantilevered arm" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8888" PostTypeId="2" ParentId="8887" CreationDate="2016-05-15T06:24:49.087" Score="2" Body="&lt;p&gt;This will depend greatly on the stiffness of the remaining structure (the supports indicated around the $D_1$ span, whatever those rest on, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the remaining structure can be considered relatively stiff, but that the plate's connection to its supports (around $D_1$) is hinged, then the structure is isostatic and can be trivially solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the bending moment, for example (this obviously depends on the actual values of $D$ and $D_1$) (ignore the units, they're irrelevant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ObVMn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ObVMn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the supports can be considered fixed (not allowing rotations, then the bending moment diagram is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gWQ7d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gWQ7d.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it happens, regardless of the chosen condition, the maximum bending moment on the plate in this example is 150 kNm (or whatever unit you desire). The maximum stress applied on the plate will be equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma = \dfrac{My}{I}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M$ is the moment, $y$ is the distance to the fiber farthest from the centroid (in this case, $y = \frac{t}{2}$, where $t$ is the thickness of the plate) and $I$ is the section's inertia (in this case, $I = \dfrac{bt^3}{12}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding appropriate safety factors to both the applied load and the maximum allowable stress in your steel, you can therefore reverse-engineer the necessary thickness of your plate. This ignores the bending moment due to the structure's self-weight, which is obviously a function of the plate's thickness, but it works as a very decent first approximation (and, actually, in this layout, the plate's thickness shouldn't change the bending moment anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other verifications such as for shear strength, but these follow a similar concept as shown for bending moment. You must also check the connection itself, but that requires more details regarding the connection. &lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-05-15T06:24:49.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8889" PostTypeId="1" CreationDate="2016-05-15T09:50:37.270" Score="5" ViewCount="1473" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LJxwQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LJxwQ.png&quot; alt=&quot;Stress-strain curve i am working with&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stress-strain curve presents a linear elastic range, followed by yielding and accompanied by a decrease in toughness the material goes into the plastic region. As it strains more and more the material then suddenly snaps! Without necking... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fracture occurring without necking isn't something I would expect from a ductile material, in fact it's a trait of a brittle material. But, the material can't be brittle with all that strain in the plastic region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which is it?&#xA;Question: Would this material fall into a &quot;mostly ductile&quot; category instead of ductile?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have four stress-strain curves i am working with at the moment. Three of them are similar to this one however, they show a decline in stress when necking before fracture. This diagram does not decline in stress before fracture. They were all tested on the same tensile stress machine (a Shimadzu machine i believe). Part of the assessment is to determine the possible types of materials each curve could be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Because this material doesn't neck so i'm wondering if there are any materials which have a stress-strain curve such as this. I was thinking some metals may strain and then just snap or perhaps timber. I'm really not sure.&lt;/p&gt;&#xA;" OwnerUserId="6462" LastEditorUserId="6463" LastEditDate="2016-05-16T09:49:33.420" LastActivityDate="2017-08-17T19:32:19.153" Title="Do all ductile materials experience necking before fracture?" Tags="|mechanical-engineering|materials|structural-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8890" PostTypeId="1" AcceptedAnswerId="8893" CreationDate="2016-05-15T11:53:24.247" Score="0" ViewCount="490" Body="&lt;p&gt;it is a simple question but I am new to this software. Both in &lt;em&gt;Postprocessor&lt;/em&gt; and within &lt;em&gt;Measure Characteristic&lt;/em&gt; options there are only angular velocities and accelerations - and no angle (orientation) itself. I saw &lt;em&gt;Angle&lt;/em&gt; and &lt;em&gt;Orientation Measure&lt;/em&gt; options but this must be something more advanced, since all I need is generalised coordinate of a part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it possible to obtain set of generalised coords against time in table form instead of plot? If so, please explain how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maverick&lt;/p&gt;&#xA;" OwnerUserId="6464" LastActivityDate="2016-05-15T14:30:06.123" Title="ADAMS/View - how to plot part orientation?" Tags="|mechanical-engineering|software|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8893" PostTypeId="2" ParentId="8890" CreationDate="2016-05-15T14:30:06.123" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You can export a Adams/Solver dataset (.adm) that will contain all the variables that were solved in a ascii format. (file export)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;you can export a results file.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can also export measures in the post processor as a ascii tabular data. In the file export dialog in case you want to specify what format the measure is to be had in.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2016-05-15T14:30:06.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8894" PostTypeId="2" ParentId="8889" CreationDate="2016-05-15T15:13:18.957" Score="2" Body="&lt;p&gt;The diagram is representing &lt;em&gt;true stress&lt;/em&gt;, as opposed to the &lt;em&gt;engineering stress&lt;/em&gt;. So as not to copy-paste, see &lt;a href=&quot;https://engineering.stackexchange.com/a/8490/1832&quot;&gt;my answer to a related question&lt;/a&gt; for a description of these different stress measurements. Basically, &lt;em&gt;true stress&lt;/em&gt; takes into account the reduced cross-sectional area due to necking, therefore the stress measurement never drops. Meanwhile, &lt;em&gt;engineering stress&lt;/em&gt; considers the original area throughout the test, even though necking has reduced the actual area, leading to a reduction in the apparent requisite stress to further deform the member.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-05-16T11:05:11.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8895" PostTypeId="2" ParentId="8887" CreationDate="2016-05-15T15:38:53.590" Score="0" Body="&lt;p&gt;To come up with the internal forces/moments you can use Free Body Diagrams, or you can use &lt;a href=&quot;https://forceeffect.autodesk.com/frontend/fe.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;AutoDesk ForceEffect&lt;/a&gt; which is free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PsEBM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PsEBM.png&quot; alt=&quot;fbd&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NUzJb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NUzJb.png&quot; alt=&quot;res&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with full results for each element in the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WbRaN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WbRaN.png&quot; alt=&quot;table&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the bolt connection to the beam is the weak spot so I would check the shear forces on the plate near the connections and I would make sure there is sufficient area to spread the support loads.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2016-05-15T15:38:53.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8896" PostTypeId="2" ParentId="8884" CreationDate="2016-05-16T02:40:45.580" Score="4" Body="&lt;p&gt;A professor of mine, long ago, once quipped:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Give me a word... any word at all... and I'll show you how the root&#xA;  of that word is a PDE!&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;A touch of Philosophy&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;All physical systems are modeled by some set of mathematical equations, be they differential, algebraic, integral, stochastic, etc.  The simplest models of physical systems only consider variations in the time variable (i.e. ordinary differential equations).  While these models tend to be very simplified compared to the &quot;true physics&quot;, they are often just &quot;good enough&quot; for specific engineering applications because we, as engineers, often only care about the input-output behavior of the system over time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Advantages of ODE Models:&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;Ignoring all variations with respect to other variables (especially space) simply makes things easier to both compute and analyze (in comparison to other types of differential equations). You can take advantage of a great deal of analysis tools, including impulse &amp;amp; frequency response, bode plots, etc...  If your system of ODE's are linear or otherwise linearizable around a setpoint, you can often use standard control techniques like PID, LQR, and LQG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Disadvantages of non-ODE models&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;I emphasize that ODE's are simpler to analyze insofar as they require the minimal mathematical complexity required of an engineer.  There are, of course, more complicated models of physical systems (e.g. partial differential equations) that are likely more reflective of the true physics in comparison to an ODE model.  However, the analysis tools that you learned for ODE's (a.k.a. classical control theory) no longer apply.  Instead, you will likely have to use a set of tools that fall under the category of optimal control theory.  For most problems of interest to engineers, optimal control requires fairly sophisticated numerical methods (usually adjoint-based optimization).  The situation gets even more complicated if you include stochasticity in any of the variables or parameters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, control of anything other than an ODE is both very complicated to do in practice and is often unnecessary to begin with.  Non-ODE control is a very highly specialized field and is generally not taught as a part of core engineering classes.  If you ever get the chance to control anything other than an ODE in practice, it will likely come as a part of state-of-the-art scientific research rather than run-of-the-mill engineering applications.&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2016-05-16T02:40:45.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8898" PostTypeId="1" AcceptedAnswerId="8904" CreationDate="2016-05-16T04:26:40.373" Score="1" ViewCount="1172" Body="&lt;p&gt;I am checking the leak rate of a carbon seal installed on an aircraft engine bearing. The test cart has a reading of pounds per hour but my manual says the leakage should not be more than 0.08 cubic metres (2.5 cubic feet) per minute of free air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since pounds are a weight measure and cubic metres/feet are volume measures, I can't compare them directly. How do I convert the volumetric flow rate to a mass flow rate?&lt;/p&gt;&#xA;" OwnerUserId="6470" LastEditorUserId="368" LastEditDate="2016-05-16T22:17:26.133" LastActivityDate="2016-05-16T22:21:54.577" Title="How do I convert cubic metres/feet per minute to a pounds per hour flow rate?" Tags="|airflow|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8903" PostTypeId="1" CreationDate="2016-05-16T11:09:13.863" Score="7" ViewCount="415" Body="&lt;p&gt;I'm trying to solve this problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given the transfer function:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$ G(z) = \frac{z^{2}+2z+1}{z^{3}-z^{2}-5z+1} $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Design a deadbeat observer for this transfer function and compute the&#xA;  error dynamics $e(0)=e$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that to design a deadbeat observer, you should first convert the transfer function to an observable canonical form, then design your observer to place the poles of the transfer function at 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $e(k+1)=(A-LC)e(k)$, which I'm guessing I have to use. I can find the matrix $(A-LC)$ but not any of the other matrices. This is where I'm stuck; how do I compute the error dynamics?&lt;/p&gt;&#xA;" OwnerUserId="3637" LastEditorUserId="368" LastEditDate="2016-05-16T22:06:11.760" LastActivityDate="2017-01-09T04:58:14.720" Title="Computing error dynamics from state space deadbeat observer" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8904" PostTypeId="2" ParentId="8898" CreationDate="2016-05-16T12:20:36.797" Score="3" Body="&lt;p&gt;There are several pieces of additional information you need to go from units of volume/time to units of force/time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For gasses, temperature and pressure matter.  Air is mostly nitrogen and oxygen.  These have atomic masses of 14 and 16 respectively.  Both form diatomic molecules, so air is a mixture of gasses with molecular masses of 28 and 32.  Since there is more nitrogen than oxygen, I'm going to say &quot;air&quot; has a molecular mass of 29.  Close enough given your precision of &quot;0.08&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At Standard Temperature and Pressure (STP; 1 atm, 0 &amp;deg;C), a mole of ideal gas occupies 22.4 liters.  Let's say you're really asking about 21 &amp;deg;C.  That's 294 &amp;deg;K as apposed to 273 &amp;deg;K at STP.  The volume will therefore be (22.4 l)(294 &amp;deg;K)/(273 &amp;deg;K) = 24.1 l.  Since that's one mole, it has a mass of 29 g.  Put another way, we've decided the density of your &quot;air&quot; is 29 g/24.1 l.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 1000 liters in a cubic meter, so your 0.08 cubic meters is 80 liters, which has a mass of 96.2 g.  There are 60 minutes in one hour, so we can compute the mass flow rate in your final time units, which is 5.77 kg/h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining problem is to convert kg to pounds.  Kg is a unit of mass, and pounds is a unit of weight.  The conversion depends on gravity.  You didn't say if whether this is on the surface of the earth, on the internal space station, or somewhere else.  I'll assume the surface of the earth where 1 kg of mass weighs about 2.2 pounds.  (5.77 kg)(2.2 lb/kg) = 13 lb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final answer is therefore 0.08 cubic meter per minute of air at 1 atmosphere, 21 &amp;deg;C, on the surface of the earth is about 13 pounds/hour.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2016-05-16T22:21:54.577" LastActivityDate="2016-05-16T22:21:54.577" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8906" PostTypeId="1" CreationDate="2016-05-17T07:29:32.787" Score="5" ViewCount="91" Body="&lt;p&gt;Consider an ordinary staircase with a banister rail supported by a wall at one end and a steel post of constant profile at the other end. The base of the post is embedded rigidly in the fabric of whatever structure the stairs are part of.  The post must not bend beyond a set (small) amount, permanently deform, or fail to support the guard rail, in the event that someone falls downstairs or against it, and the maximum deflection of the post in the event of a fall is in any case to be small compared to its length (rough indication: 0.5–4.0% or say ~5–40 mm vs ~1000 mm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rail and post, and any fixings, are each thick enough and well enough secured, that they will not shear or become loosened by any plausible impact of a person (or persons, including anything they carry) falling down the stairs and against them at any point.  Within these criteria, it's desired to make the post with a narrow profile, and therefore minimally thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a pure engineering perspective (ignoring any building codes for now) is it sufficient to reason that any collision/impact will be elastic and therefore the minimum profile for the post can be derived by:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Estimating a maximum KE and PE for the falling persons,&lt;/li&gt;&#xA;&lt;li&gt;Setting a maximum desired deflection of the post.&lt;/li&gt;&#xA;&lt;li&gt;Reasoning on a pure energy basis that the post can be treated as a simple/ideal deflected cantilever beam, and its minimum profile is that which satisfies strain energy of the steel at maximum desired deflection &gt;= maximum KE+PE of likely falling objects?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It seems to make sense and yet I am not too familiar with impulsive impact (more familiar with statics), and I'd like to check if there's a reason this would be an underestimate or must be treated with caution. (For example maybe a falling object would have CoG slightly higher than the actual rail height? Or other reasons might exist that this would lead to an insufficient result?)&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="368" LastEditDate="2016-05-17T17:55:27.217" LastActivityDate="2016-08-15T19:25:01.230" Title="Can the thickness of a guard post be derived purely from energy principles?" Tags="|structural-engineering|beam|deflection|columns|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8908" PostTypeId="1" AcceptedAnswerId="8915" CreationDate="2016-05-17T07:47:08.880" Score="7" ViewCount="9565" Body="&lt;p&gt;I have a structure, which contains only frame elements ( beam and column).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8zIej.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8zIej.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I know, if I apply point load at the columns on both left and right hand side,and if the beam-column joint is rigid, then the column at the center will experience extra moment, because the beam moment will be transmitted to the center column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the center column will take moment because of the beam's existence, am I right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to eliminate this extra moment taken by the center column, can I then set the beam-column join connection as pin? How would the structural behavior changes?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastActivityDate="2019-12-28T18:02:49.657" Title="Can I reduce the column moment due to beam, by setting the beam end to pin?" Tags="|structural-engineering|structural-analysis|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8910" PostTypeId="1" CreationDate="2016-05-17T12:17:39.203" Score="-1" ViewCount="382" Body="&lt;p&gt;I have a study question that I am not able to solve:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A cylindrical electrical resistance is used to heat water.&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Radius is 2 mm and the length is 80 cm.&lt;/li&gt;&#xA;  &lt;li&gt;Electrical conductivity of the resistance is 3x10&lt;sup&gt;3&lt;/sup&gt; mOhm&lt;sup&gt;-1&lt;/sup&gt; while the thermal conductivity is 4 W/mK.&lt;/li&gt;&#xA;  &lt;li&gt;Water temperature is 30 &amp;deg;C and the convection coefficient is 100 W/m&lt;sup&gt;2&lt;/sup&gt;K.&lt;/li&gt;&#xA;  &lt;li&gt;Maximum allowable surface temperature of the resistance is 150 &amp;deg;C.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;  &#xA;  &lt;p&gt;a.    Derive the radial temperature distribution inside the resistance&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;b.    Calculate the maximum electrical current in A &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;c.    Calculate the resistance and heat generation rate &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;d.    Calculate the central temperature&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h3&gt;What I could establish so far&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Now, I got stuck, because I can not find the heat generation due to electricity, though everthing else comes of by finding the T(x). I do not want an explicit answer though, just a way of how I can solve it.&lt;/p&gt;&#xA;" OwnerUserId="6486" LastEditorUserId="368" LastEditDate="2016-05-19T23:07:27.633" LastActivityDate="2016-05-19T23:07:27.633" Title="Heat Transfer, with energy generation" Tags="|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="8911" PostTypeId="1" CreationDate="2016-05-17T13:58:40.707" Score="2" ViewCount="61" Body="&lt;p&gt;I am a student and I work in the friction stir welding of a hardened aluminium alloy 7075-T6, and I want to confirm about this information before I put it in the article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found in the literature that there are two ways of calculating the ratio of speed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ratio=advancing speed/(rotational speed*pin diameter)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the other is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ratio=Rotational speed/advancing speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what should I choose?&lt;/p&gt;&#xA;" OwnerUserId="6487" LastEditorUserId="90" LastEditDate="2016-05-17T14:47:40.330" LastActivityDate="2016-05-17T14:47:40.330" Title="how to calculate the speed ratio (Rotational speed to advancing speed) in friction stir welding" Tags="|friction|welding|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8912" PostTypeId="2" ParentId="8906" CreationDate="2016-05-17T14:25:40.730" Score="1" Body="&lt;p&gt;Using a purely energy-based approach, while theoretically possible, is not recommended. Not because it wouldn't work, but simply because most codes don't use such methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that railings, as well as any structural elements, are ruled by codes. In Brazil, the relevant code is the NBR 14718, which defines the static and dynamic horizontal and vertical forces that a railing must resist. Could you transform these forces into energy? Sure, probably. But you can't simply &quot;estimate a maximum kinetic and potential energy for a falling person&quot; and use that, because you won't know if your values are sufficient to satisfy the codes. Well, unless you do your estimates, convert them into forces and check that they are sufficient... at which point, why not just use the forces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The codes also define factors of safety, which a purely theoretical method such as the one you're suggesting can't generate. The Brazilian code at least also defines the allowable deflections at $\dfrac{L}{250}$, where $L$ is the railing's span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then there's the issue of dynamic impact factors. From a purely energy perspective I don't believe this can be easily solved. After all, a load applied instantaneously will generate a temporary deflection which is twice as large as the permanent (elastic) deflection.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-05-17T14:25:40.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8913" PostTypeId="5" CreationDate="2016-05-17T14:51:24.240" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-05-17T14:51:24.240" LastActivityDate="2016-05-17T14:51:24.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8914" PostTypeId="4" CreationDate="2016-05-17T14:51:24.240" Score="0" Body="Questions about the design and operation of electric drive vehicles, especially on-road vehicles." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-05-17T14:51:24.240" LastActivityDate="2016-05-17T14:51:24.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8915" PostTypeId="2" ParentId="8908" CreationDate="2016-05-17T15:17:16.840" Score="5" Body="&lt;p&gt;Yes this is correct - if you want to eliminate the moments in the columns then a pinned connection between the beam and columns will achieve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To demonstrate this I have recreated your structure with some imaginary point loads and dimensions in a structural analysis software. The columns are fully fixed at the base. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagrams on the left the images below have &lt;strong&gt;fully-fixed connections&lt;/strong&gt;, while on the right the columns are &lt;strong&gt;pinned connections&lt;/strong&gt; to the beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lz5TT.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lz5TT.png&quot; alt=&quot;SkyCiv 3D Structural Analysis Software Input&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/np5DW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/np5DW.png&quot; alt=&quot;SkyCiv 3D Structural Analysis Software Bending Moment Diagram&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yBlPd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yBlPd.png&quot; alt=&quot;SkyCiv 3D Structural Analysis Software Axial Force Diagram&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TFSzW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TFSzW.png&quot; alt=&quot;SkyCiv 3D Structural Analysis Software Deflection Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most importantly you will notice zero moment in the columns in the right-hand case (pinned connections). Generally this increases axial force and we can see this in the center column. Notice also that the max total displacement (vector sum of X and Y displacement) is larger in the pinned model. When there is a moment acting on the columns, this causes the columns to deflect sideways, whereas in the pinned model there is no sideways deflection (of course no buckling is considered since this is just a linear static analysis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps! I did this really quickly in &lt;a href=&quot;https://skyciv.com&quot; rel=&quot;noreferrer&quot;&gt;SkyCiv Structural 3D analysis software&lt;/a&gt; if anyone is wondering. If you sign up for a free account a small structure like this can be solved in &lt;a href=&quot;https://skyciv.com/3d-structural-analysis-software-features&quot; rel=&quot;noreferrer&quot;&gt;SkyCiv Structural 3D&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2016-05-18T01:27:28.387" LastActivityDate="2016-05-18T01:27:28.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8916" PostTypeId="5" CreationDate="2016-05-17T17:57:21.770" Score="0" Body="&lt;p&gt;Questions on the subject of home improvement and the role of engineers in residential design and construction work. Technical questions about specific home improvement projects should not use this tag; instead, use tags that categorize the project itself.&lt;/p&gt;&#xA;&lt;p&gt;For advice on home improvement / DIY, see our sister site &lt;a href=&quot;https://diy.stackexchange.com/&quot;&gt;Home Improvement Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="6767" LastEditDate="2020-11-14T09:42:14.663" LastActivityDate="2020-11-14T09:42:14.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8917" PostTypeId="4" CreationDate="2016-05-17T17:57:21.770" Score="0" Body="Questions on the subject of home improvement and the role of engineers in residential design and construction work. Technical questions about specific home improvement projects should not use this tag; instead, use tags that categorize the project itself. For advice on home improvement / DIY, see our sister site Home Improvement Stack Exchange." OwnerUserId="368" LastEditorUserId="6767" LastEditDate="2020-11-14T09:42:20.083" LastActivityDate="2020-11-14T09:42:20.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="8918" PostTypeId="2" ParentId="8908" CreationDate="2016-05-17T18:34:26.057" Score="6" Body="&lt;p&gt;As @pauloz1890 has already stated in &lt;a href=&quot;https://engineering.stackexchange.com/a/8915/1832&quot;&gt;his answer&lt;/a&gt;, if you add a hinge to a beam-column connection, you will eliminate the bending moment on your column (assuming no horizontal loads are applied).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, your rendering shows a frame composed of concrete beams and columns. The joints between the beams and columns are clearly monolithic. If this is the accurate representation of your structure and it cannot be modified, then your model must remain as it is, &lt;strong&gt;without a hinge&lt;/strong&gt; (and with bending moment on the columns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you can only add this hinge to your model if you can also add the hinge to the actual structure. There are many ways to do this. If your beams are cast-in-place, then you'll most likely need to adopt a Freysinnet or Menasger hinge (there are a few others, but these are the most common). If, on the other hand, your beams are pre-cast, then there are many different types of bearings which can be used (elastomeric or mechanical).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FVvAD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FVvAD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;sub&gt;&lt;a href=&quot;https://www.researchgate.net/publication/276089627_Concrete_hinges_in_bridge_engineering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; (Click &quot;Download full text&quot;, free))&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you choose to modify your structure to allow for hinges, you need to decide how the beams will behave around the central column. Will you have two independent spans (a full hinge), or will you have a continuous beam over both spans (only the column is hinged, this is the example given by @pauloz1890)? This will affect the internal forces on your beams and the axial force distribution between the columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MXqfI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MXqfI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that these models all assume fixed supports at the base of the columns. If a structure is fully hinged at the base and at the joints between all the columns and beams, then it will become hypostatic (unstable). Therefore, in order to define hinged connections between your columns and beams, you must guarantee that the base will be fixed against rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Diagrams obtained with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:02:49.657" LastActivityDate="2019-12-28T18:02:49.657" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="8919" PostTypeId="1" CreationDate="2016-05-18T06:06:02.397" Score="-1" ViewCount="438" Body="&lt;p&gt;I have to design a science display board consisting of 20 panels 2m x 2m high and wide, which can fold up as a giant pamphlet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty is how to load and unload them from a lorry in less than an hour. The display boards fold out of the back of a lorry and down a ramp where they have to go oblique to one another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know a strong, simple, stable hinge which is fast to unfold from a lorry, which doesn't make the panels fall over?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zgE0K.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zgE0K.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2016-05-19T13:25:59.737" Title="Hinge joint for flat panels that can also flex horizontally?" Tags="|mechanical-engineering|structures|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8920" PostTypeId="1" AcceptedAnswerId="8932" CreationDate="2016-05-18T07:05:56.193" Score="3" ViewCount="4059" Body="&lt;p&gt;My question might seem a simple one, but it has been troubling me for a while.&#xA;How many hours in a day do conventional electrical generators work (the ones in large power plants)?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do they by any method conserve or store the power before it reaches the consumer?&lt;/p&gt;&#xA;" OwnerUserId="6493" LastEditorUserId="163" LastEditDate="2016-05-18T19:50:44.983" LastActivityDate="2016-06-14T15:44:47.340" Title="Working hours of power plant generators" Tags="|electrical-engineering|energy|power-engineering|energy-storage|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8921" PostTypeId="1" CreationDate="2016-05-18T07:11:25.840" Score="5" ViewCount="680" Body="&lt;p&gt;Walking into a building, how does one find out where the pipes are, how strong the walls are (currently) and other key structural integrity measures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;without breaking walls or utilising harmful-to-human emmissions (e.g.: strong x-rays)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some ideas I have (mostly from geophysics and medicine):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Get floor-plan and other information from maintenance folk&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetometer&quot;&gt;Magnetometer&lt;/a&gt; to detect metal and current decay&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sonar&quot;&gt;Sonar&lt;/a&gt;—possibly &lt;a href=&quot;https://en.wikipedia.org/wiki/Infrasound&quot;&gt;infrasonic&lt;/a&gt;; but more likely &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultrasound&quot;&gt;ultrasonic&lt;/a&gt;; e.g.: &lt;a href=&quot;https://en.wikipedia.org/wiki/Synthetic_aperture_sonar&quot;&gt;synthetic aperture sonar&lt;/a&gt;—to figure out location and density of objects through the walls. &lt;em&gt;(they may of changed since plan was written&lt;/em&gt;)&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Palpation&quot;&gt;Palpation&lt;/a&gt; for more obvious (but less objective) measures.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ocular_tonometry&quot;&gt;Tonometry&lt;/a&gt; could objectively be applanated  across a larger area.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Electrical_impedance_tomography&quot;&gt;Electrical impedance tomography&lt;/a&gt; could be used for this also.&lt;/li&gt;&#xA;&lt;li&gt;Alternatively some other kind of &lt;a href=&quot;https://en.wikipedia.org/wiki/Tomography&quot;&gt;tomography&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6492" LastActivityDate="2019-06-26T17:05:35.167" Title="Seeing through walls and noninvasive structural integrity metrics" Tags="|structural-engineering|civil-engineering|structures|sensors|measurements|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8922" PostTypeId="1" AcceptedAnswerId="8940" CreationDate="2016-05-18T07:19:07.603" Score="0" ViewCount="25331" Body="&lt;p&gt;Assume that I am writing a new FEM software for structural analysis, &lt;em&gt;it shouldn't really matter whether this is a frame or frame+shell element analysis tool.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that &lt;a href=&quot;https://www.quora.com/What-is-the-difference-between-primary-secondary-and-tie-beam-and-where-are-located-in-building-and-how-can-we-identify-them-and-what-are-their-purpose/answer/Asafa-Khan-%D8%A7%D8%B3%D8%A7%D9%81%D8%A7-%D8%AE%D8%A7%D9%86?srid=36H4&quot; rel=&quot;nofollow noreferrer&quot;&gt;primary and secondary beams have their own function&lt;/a&gt;, and are designed for different use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GYtIU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GYtIU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Primary Beam&lt;/strong&gt;: A horizontal beam connecting columns (simply supported&#xA;  or shear connected.) Function: It will transfer the load from&#xA;  secondary beam(if present) to the columns.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Secondary Beam&lt;/strong&gt;: A horizontal beam connecting primary beams (simply&#xA;  supported or shear connected.) Function: It will transfer the load to&#xA;  the primary beam and not directly connected to the columns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In FEM, everything is just frame element, and all the beams and columns are equal, so whether one beam should resist more loads ( ie primary beam) than the other is completely up to the outcome of FEM procedure. But as engineers, sometimes we &lt;em&gt;just&lt;/em&gt; want &lt;em&gt;this&lt;/em&gt; one beam to take more loads than the neighboring ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I modify my FEM engine to do what I want?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="3353" LastEditDate="2016-05-28T03:15:14.960" LastActivityDate="2016-05-28T03:15:14.960" Title="How to distinguish primary and secondary beam in FEM formulation?" Tags="|finite-element-method|structural-analysis|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8923" PostTypeId="1" AcceptedAnswerId="8964" CreationDate="2016-05-18T07:27:10.757" Score="6" ViewCount="5414" Body="&lt;p&gt;Assuming that I have a beam that spans across a few supports. I can design the beam as &lt;a href=&quot;http://machinedesign.com/whats-difference-between/what-s-difference-between-beam-diagrams&quot;&gt;a continuous beam (a beam that is loaded and has more than two supports)&lt;/a&gt;, or individual beams that each encompass between two supports. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I know, the design of continuous beam is always more economical than beam with supports. Why this is the case? I understand that this is because the forces and moment in continuous beam are less than in beams with multiple supports. But I am looking for a deeper, more physical explanation on &lt;em&gt;why&lt;/em&gt; the forces are smaller in continuous beam. &lt;/p&gt;&#xA;" OwnerUserId="3353" LastActivityDate="2021-06-26T13:57:40.393" Title="Why the design of continuous beam is always more economical than beam with supports?" Tags="|structural-analysis|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8924" PostTypeId="2" ParentId="8921" CreationDate="2016-05-18T09:15:39.803" Score="7" Body="&lt;p&gt;Depending on the region and structure of building the techniques might vary. Below are few suggestions &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Termite inspection is another popular method to determine the structure integrity of building.&lt;/li&gt;&#xA;&lt;li&gt;A handheld radar scanner from DeWalt might also do the job as describe &lt;a href=&quot;https://youtu.be/hmXZZy4L9Lg?t=24s&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;li&gt;Depending on the type of pipes a metal detector might be another good option.&lt;/li&gt;&#xA;&lt;li&gt;Non-Destructive Testing is another method used for evaluating structural integrity of bridges. I don't see this technique can be used in building too. The &lt;a href=&quot;http://www.mistrasgroup.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mistrasgroup&lt;/a&gt; specializes in this area and they have white papers on this topic. &lt;/li&gt;&#xA;&lt;li&gt;In order to trace a hot water lines, you could run the hot water of a while and used use an infra-red camera. Chance are the cold water lines might be around the same area. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sifau.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sifau.jpg&quot; alt=&quot;Infra-red Camera&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Handheld Radar Scanner&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ypCxJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ypCxJ.jpg&quot; alt=&quot;Handheld Radar Scanner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure there other professional methods and I leave it to expert to comment on those techniques. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.flir.com/instruments/building/display/?id=49420&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thermal Imaging for HVAC &amp;amp; Plumbing&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.lowes.com/pd_488153-70-DCT418S1_0__?productId=4777992&quot; rel=&quot;nofollow noreferrer&quot;&gt;DEWALT 12-Volt Max Hand Held Radar Scanner&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/hmXZZy4L9Lg&quot; rel=&quot;nofollow noreferrer&quot;&gt;DEWALT 12v Hand Held Radar Scanner DCT418&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://diy.stackexchange.com/questions/14714/is-there-a-way-to-trace-water-pipes-through-walls&quot;&gt;Is there a way to trace water pipes through walls?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.orkin.com/termites/termite-structural-damage/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Termite Structural Damage&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.diyfixit.co.uk/magazine-diy-tools-a-products/pipe-and-cable-detectors.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pipe and cable detectors&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/1906/can-the-structural-integrity-of-bridges-be-measured-in-real-time-for-safety&quot;&gt;Can the structural integrity of bridges be measured in real-time for safety?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://lifehacker.com/5947450/whats-the-least-intrusive-way-to-find-out-whats-in-a-wall&quot; rel=&quot;nofollow noreferrer&quot;&gt;What's the Least Intrusive Way to Find Out What's In a Wall?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-05-27T21:38:10.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8925" PostTypeId="2" ParentId="4113" CreationDate="2016-05-18T10:21:08.417" Score="1" Body="&lt;p&gt;When a cubic EOS is solved at the critical point, it MUST have 3 equal roots, because the parameter equations were derived in order to satisfy the critical constraints (inflection point with zero slope at the critical point).&#xA;E.g. for the R-K Eos Zc=1/3&#xA;(Giorgio Soave)&lt;/p&gt;&#xA;" OwnerUserId="6495" LastActivityDate="2016-05-18T10:21:08.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8926" PostTypeId="2" ParentId="8920" CreationDate="2016-05-18T10:51:13.197" Score="3" Body="&lt;p&gt;There isn't a single answer that covers large power plants around the world. There's way too much variation in generation plant, and way too much variation in the way that plant is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some plants will operate for just a few minutes occasionally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some will run 24/7 except when closed for maintenance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, absolutely, the energy is stored before it reaches the consumer - but that storage happens &lt;em&gt;before&lt;/em&gt; generation of electricity. Either in the form of fuel stores, or in a hydro reservoir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in general, such generators are part of a grid that might span a region of a country, or even several countries - and the grid pools both generation and consumption. So just because you've got electricity, does not mean that the generator closest to you is working: your electricity might have travelled a thousand km to get to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, absolutely, the energy is stored before it reaches the consumer - but almost all that storage happens &lt;em&gt;before&lt;/em&gt; generation of electricity. Either in the form of fuel stores, or in a hydro reservoir. There is some on-grid electricity storage too, but this is absolutely tiny in comparison (several orders of magnitude). It's almost always in the form of pumped hydro storage - so surplus electricity is used to lift a large volume of water a few hundred metres; the water sits in a reservoir; and is then released through turbines later to generate electricity again. This is about 80% efficient. And it's very rarely sited with a generator - it's sited where there are some suitable mountains (or, in at least one case, coastal cliffs).&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2016-05-18T17:29:41.007" LastActivityDate="2016-05-18T17:29:41.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8927" PostTypeId="2" ParentId="8920" CreationDate="2016-05-18T12:13:30.850" Score="0" Body="&lt;p&gt;This answer is with reference to power generating plants which feed the grid, not &quot;small&quot; generators which just power a local building.  The latter, of course, can be turned off when there's no demand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, power plants which use hydropower can be 'turned up' or 'turned down' relatively easily by shunting the water flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power plants which depend on local heating, especially by burning fuel, however have extremely long ramp-up and ramp-down times (many hours), so by and large they end up dumping excess electrical power at night or other times of low demand.  There are a few plants which try to recover some of this by doing things like using the excess electrical power to pump water into a reservoir, and using that water to generate power during peak demand.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-05-18T12:13:30.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8928" PostTypeId="2" ParentId="8920" CreationDate="2016-05-18T12:20:51.273" Score="5" Body="&lt;p&gt;Nuclear power plants are always working, except circa one month/year for maintenance. It's mainly because stopping the nuclear reaction and cooling the reactor down is a very long and difficult process (a few weeks). So &quot;turning on&quot; and &quot;turning off&quot; a Nuclear Power Plant (NPP) is not as simple as pushing a button.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary circuit (containing the water heated by the nuclear reactions) is always connected to the secondary circuit (which is a classical Rankine cycle), and the steam of the secondary circuit, which is used to make the turbines turn, and therefore the alternator, thus producing electricity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the processes for stocking the electricity produced by the NPPs are very limited. I know that there are some kind of big batteries, but their capacity is limited. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a big problem in my home country (France), where 75% of the electricity is produced by nuclear energy (we are the most &quot;nuclearized&quot; country in the World), and the electrical consumption is - obviously - much smaller in August than in December. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consequence is that in August, we produce about 70 GW of electrical power and use only 55 GW of it, and therefore we are forced to sell electricity to other countries because we can't stock it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the renewable sources: I know that for photovoltaics (PV) systems, there are small systems of batteries. This is much simpler because the power is way smaller than nuclear power plants (about 1-10 kW by PV system). Moreover, the intermittency of photovoltaics (you don't produce electricity at night, while night is the moment when you consume the most) make it really necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For windpower, you can read a bit about CAES (compressed air energy storage), which is a very promising way to stock energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for thermal power plants (fuel, charcoal) and dams, as someone said previously, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the energy is stored before it reaches the consumer - but that storage happens before generation of electricity. Either in the form of fuel stores, or in a hydro reservoir.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6497" LastEditorUserId="1832" LastEditDate="2016-05-18T13:44:29.133" LastActivityDate="2016-05-18T13:44:29.133" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8929" PostTypeId="2" ParentId="8923" CreationDate="2016-05-18T15:14:17.457" Score="4" Body="&lt;p&gt;The reality is a bit more complicated, and depends greatly on the materials being used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take the following structure as our example, where the middle support may or may not be hinged (creating a beam which is either continuous or not):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bacZ7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bacZ7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is symmetric, I can actually simplify it to these two cases. The one on the left assumes the structure was continuous, the one on the right, that it was hinged:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oBnuX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oBnuX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by observing that the continuous beam is strictly worse than the simply-supported one with regards to shear force. It increases the maximum shear force for no benefit. So, now that we've gotten that out of the way, let's move on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we notice that the maximum bending moment (ignoring sign) is the same in both diagrams. If your section is symmetric around its transversal x-axis, then its resistance to positive and negative bending moments is the same, so this is irrelevant. If your section isn't symmetric, then this question can't be reasonably answered because the &quot;right&quot; answer depends greatly on the actual section being used. Therefore I will now assume that the section is symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're dealing with a steel beam, the continuous beam is best. It'll have to resist the same forces in either case, but this diagram is actually friendlier to a steel beam because it implies in a larger $C_b$, a coefficient which controls lateral-torsional buckling (LTB) given by the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_b = \dfrac{12.5M_{max}}{2.5M_{max} + 3M_a + 4M_b + 3M_c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where all moments are in absolute value and $M_a$, $M_b$ and $M_c$ are the bending moments at one-quarter, one-half, and three-quarters of the span, respectively. In the continuous beam all three of these are smaller than in the simply-supported one, implying in a larger $C_b$ and therefore a higher strength against LTB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're dealing with a reinforced concrete structure, however, I'd argue that the simply-supported one may be more efficient. The reason is that LTB usually isn't relevant (reinforced concrete beams tend to be far less slender than equivalent steel beams), but you can control where you put your reinforcement. The maximum necessary reinforcement will be equal in both situations since the maximum bending moment is the same. However, the continuous beam will require you to properly reinforce both faces of the beam, while the simply-supported one will allow you to properly reinforce only the bottom face and just add the minimal reinforcement on the top face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the best answer for concrete structures is dependent on the actual case. After all, the simply-supported beam will have to reinforce against the maximum bending moment twice: once for each span, while the continuous beam only suffers that once. However, the continuous beam needs to truly reinforce both faces of the beam, while the simply-supported one can get by with minimal reinforcement on the top face. Depending on the actual situation, the better solution may change. If the positive bending moment on the continuous beam actually ends up being equal (or close to) the minimal reinforcement, then the continuous beam will be better. On the other hand, if this reinforcement is significant, the simply-supported beam may come out on top.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The text above describes cost-optimization on the beams themselves, without taking into consideration the costs of implementation. This is where continuous beams shine. They are trivial to construct. Simply-supported beams, on the other hand, raise many problems: you'll need expansion joints to close the gap between the beams and you'll need twice the joints (bearing pads or what have you), among others issues.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-05-18T15:14:17.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8930" PostTypeId="2" ParentId="8920" CreationDate="2016-05-18T15:37:02.567" Score="2" Body="&lt;p&gt;A typical conventional power plant may consist of multiple generating units rather than one big huge unit.  By having multiple units, the plant is better able to respond to changes in demand during the day or to shut down generating units for maintenance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By having power distributed by a sophisticated electrical grid, if a plant in one region of the country must go off line, power can be purchased from other operators and sent to the affected area over the power grid until the local plant can come back on line.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-05-18T15:37:02.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8931" PostTypeId="1" AcceptedAnswerId="8958" CreationDate="2016-05-18T16:06:30.137" Score="4" ViewCount="509" Body="&lt;p&gt;I'm working on a 2D-model (in the &lt;em&gt;xy&lt;/em&gt;-plane) in NASTRAN and am getting convergence problems. The main cause of these convergence problems is a &quot;high or negative matrix factor diagonal ratio.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've attached a picture of a setup similar to the one I'm working on. The elements in &lt;em&gt;white&lt;/em&gt; are rod elements; the elements in &lt;em&gt;pink&lt;/em&gt; are bar elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qt2QL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qt2QL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loads and BCs are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Angular velocity of 60 rad/sec about the &lt;em&gt;y&lt;/em&gt;-axis&lt;/li&gt;&#xA;&lt;li&gt;Gravitational acceleration of 9.8 m/s$^2$ in the &lt;em&gt;-y&lt;/em&gt;-direction&lt;/li&gt;&#xA;&lt;li&gt;$123456$ constraints on the top and bottom nodes to prevent them from moving or rotating&lt;/li&gt;&#xA;&lt;li&gt;$123$ constraints on the inner-most node to prevent it from moving&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is some text from the f06 file:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yfzmp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yfzmp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nXkkI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nXkkI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nodes 13 and 27 are where the rods connect to the bars, and Nodes 20 and 24 are just two random nodes on the right-hand side of the pink section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the bar properties, there is an option called &quot;Pinned DOF @ Node 1,&quot; which may or may not be helpful here. The options are UX, UY, UZ, RX, RY, RZ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried experimenting with various boundary conditions, but I can't get convergence in either the linear static case or the nonlinear static case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally high matrix factor diagonal ratios imply that I have a very stiff element connected to a not-very-stiff element, but Nodes 20 and 24 &lt;strong&gt;aren't&lt;/strong&gt; where the rod/bar intersection happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how I might go about diagnosing the problem? I think I'm missing boundary conditions somewhere, but I'm not entirely sure where that is.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;UPDATE&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Whoops, I forgot to explain what behavior I &lt;em&gt;expect&lt;/em&gt; from this contraption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I have a really high angular velocity (and perfectly normal gravity), I would expect the pink nodes to move rightward and downward due to the loads. The only thing stopping them from moving indefinitely are the white rods. Since these aren't beams and don't transmit moment stiffness, all these rods can really do is extend in length and change angle, since they're required to be a straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will result in a shape that looks similar to the pink curve, but probably shifted rightward a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ran the case with PARAM,BAILOUT,-1 (linear static), and I got the following clearly incorrect deflection ($10^7$ m deflection): &lt;strong&gt;image removed since it didn't provide any useful information and it took up too much space&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;UPDATE #$2$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've pinpointed the problem, I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/goCN7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/goCN7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/52vj0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/52vj0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blue elements are rods, pink elements are bars. The first case converged (nonlinear static) and gave results pretty close to what I expected (deflections were slightly high, but my angular velocity is slightly high, too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second case failed to converge, and it had a problem with my new node - it told me that it's having diagonal ratio problems in the T2 direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's something NASTRAN is doing wrong with how these two bar elements are connected at this node. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="5218" LastEditorUserId="5218" LastEditDate="2016-05-19T18:05:36.753" LastActivityDate="2016-05-20T14:23:24.650" Title="FEA: What boundary conditions am I missing?" Tags="|structural-engineering|finite-element-method|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8932" PostTypeId="2" ParentId="8920" CreationDate="2016-05-18T17:23:04.207" Score="2" Body="&lt;p&gt;There isn't currently any practical means of storing electrical energy on a large scale. There are technologies which can store energy for specific applications but certainly nothing which would have much of an impact on national level distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For obvious reasons, in a distributed power grid the power demand will vary considerably throughout any given day and over the course of  year and a lot of analytical effort goes into predicting this demand in great detail, even to the extent of keeping track of popular television programmes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different types of generating plant has different abilities to match demand depending on how quickly it can be turned on and off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general terms steam turbine systems (eg nuclear, coal, oil fired etc) are not easy to shut down and tend to stay on more of less permanently, only being shut down for maintenance, these provide the base load above which demand fluctuates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas turbines are a bit more agile and can be switched on at relatively short notice to meet peak demands, the downside being that this can be a more expensive way to generate electricity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is for neighbouring countries to buy power off each other to help to balance supply and demand across a wider geographical area. There may also be schemes for industrial users with high energy demands to stop or reduce production at peak times in return for a discount. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that effective prediction and management of energy supply and demand is a big factor in the overall efficiency of energy usage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly it is usual for countries (or equivalent geographical areas) to have integrated electrical distribution networks such that all of the major power plants are contributing to a grid system. This helps to ensure redundancy and flexibility in meeting demand so that if one generator goes offline for scheduled maintenance or unexpected failure the slack can be picked up by reserve capacity elsewhere. Obviously in very large areas this may be broken down into regions but conversely there is a lot of buying and selling of electricity throughout Europe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously there are a lot of variables here and changes in government policy in terms of taxation, subsidy and general regulation will have a significant effect as well as local conditions of geography and energy usage. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-14T15:44:47.340" LastActivityDate="2016-06-14T15:44:47.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8934" PostTypeId="2" ParentId="8919" CreationDate="2016-05-18T18:17:23.533" Score="1" Body="&lt;p&gt;There are lots of ways to approach this depending on what your exact requirements are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first question is whether you need an actual 'hinge' or if you just want to be able to fix the boards at a specific angle. In the latter case screw or bolt in brackets at the appropriate angle may be the simplest solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want permanently fixed hinges to allow the whole thing to fold up then &lt;a href=&quot;http://www.wickes.co.uk/Wickes-Piano-Hinge-Nickel-600mm/p/159784&quot; rel=&quot;nofollow&quot;&gt;piano hinges&lt;/a&gt; are a good solution as they are neat and low profile and can be cut to the length you require from long strips. The downside in this situation is that they will only allow you to fold in one direction (ie not past 180 degrees). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another common solution is 'lift off' &lt;a href=&quot;http://www.daro-ind.com/cms/assets/_core/image/liftoffnew.gif&quot; rel=&quot;nofollow&quot;&gt;half pin hinges&lt;/a&gt; which can be easily assembled and disassembled as they just slot together without tools or additional fixings. This also gives you a lot of versatility as each board can be transported as a separate unit and you can assemble them in any combination or configuration you want. They will let you fold through close to 360 degrees, depending on exactly how they are positioned, the downside being that they will stick out on the sides of the boards and leave a gap between them when assembled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be able to fold the panels completely flat and  in both directions then you may need a &lt;a href=&quot;http://www.hafele.co.uk/shop/Images/400_300/92790049P1.JPEG&quot; rel=&quot;nofollow&quot;&gt;double action hinge&lt;/a&gt;. If these were designed as a pair of half pin hinges joined by a short bar/link then this would also make assembly and disassembly easier as you can just drop the link in without having to lift the whole panel. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-19T13:25:59.737" LastActivityDate="2016-05-19T13:25:59.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8935" PostTypeId="1" AcceptedAnswerId="8938" CreationDate="2016-05-18T20:34:22.630" Score="4" ViewCount="2125" Body="&lt;p&gt;I have a &lt;em&gt;seemingly&lt;/em&gt; simple structure and I need to perform a hand calculation to check that the stresses in the members are below the allowable values.  In its simplest form, the structure is a rectangular frame made from rectangular tube welded at the corners and suspended from four equal length cables attached at the corners.  The cables meet above the center of the rectangular frame and attach to a hook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For loading purposes, each member of the frame can be considered to have the same uniformly distributed load along its entire length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it looks something like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TSZp1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TSZp1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I have been poring over static, mechanics, structural analysis and other books and resources trying to find the methodology for analyzing this but all of these resources seem to stop at planar structures where the loading is in the plane of the structure.  Either this is a difficult problem, I am overthinking things, or I have been out of practice with respect to statics for too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have so far is a free body diagram where I have replaced each cable with a spherical moveable support (the 3D equivalent of a roller support).  Each corner is free to rotate in all three directions and to translate in the plane of the frame but not out of it.  So there is a single vertical reaction force at each corner (which is actually the vertical component of the cable tension).  This is an assumption.  Please let me know if I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The degree of static indeterminancy is &#xA;$(6m + R) - 6j$, where $m$ is the member count, $j$ is the joint count and $R$ is the number of unknown reaction forces.  Plugging in the values leads to an answer of 4.  However, I think I can actually reduce that to 1 since (due to symmetry) the reaction force at each corner is identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is where I am stuck.  What is the next step? Or is this the point at which people say &quot;let the computer solve it&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3440" LastEditorUserId="3440" LastEditDate="2016-05-18T20:49:12.677" LastActivityDate="2019-04-14T11:56:11.547" Title="Structural Analysis of a Space Frame" Tags="|structural-analysis|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8936" PostTypeId="2" ParentId="8935" CreationDate="2016-05-18T20:57:56.037" Score="0" Body="&lt;p&gt;IMO there is much missing information, so it's hard to know what the next step should be. Without knowing, for example, how the pieces are joined, or how exactly the cables are attached, I wouldn't know what I'd need to calculate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But without more information, I would treat the pieces as simple beams. Then taking each in isolation: apply distributed load, and the compression from the cables, solve to find deflection and stress. So not a strictly static / rigid body analysis, but not terribly hard I imagine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMO the real question is how far you need or want to go? &lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-05-18T21:05:17.890" LastActivityDate="2016-05-18T21:05:17.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8937" PostTypeId="2" ParentId="8935" CreationDate="2016-05-18T23:38:30.730" Score="0" Body="&lt;p&gt;It's not clear what the difficulty is with analyzing this frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the frame is supported by cables, each cable ideally will support only a tension load. Since the frame is loaded evenly, one can make use of symmetry in decomposing it to find the loads in the cable(s).  By the equations of statics, the vertical component of the tension in each cable must equal the weight of the portion of the frame attached to that cable.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-05-18T23:38:30.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8938" PostTypeId="2" ParentId="8935" CreationDate="2016-05-19T03:11:38.960" Score="3" Body="&lt;p&gt;You state in &lt;a href=&quot;https://engineering.stackexchange.com/questions/8935/structural-analysis-of-a-space-frame/8937#comment16492_8936&quot;&gt;a comment&lt;/a&gt; beneath &lt;a href=&quot;https://engineering.stackexchange.com/a/8936/1832&quot;&gt;@Jodes' answer&lt;/a&gt; that since the structure is welded, that the joint can be considered fixed. The veracity of that claim is somewhat arguable (especially if done &quot;artisanally&quot;), but let's go with it. After all, if the joint were considered hinged, then the solution for each beam would be trivial and equal to a simply-supported beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, thankfully, the actual solution even if we consider the beams to be fully fixed to each other is still basically the same as if we'd considered them simply supported. The reason for this is the bi-symmetry of the structure (including its loading).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call the long beams $L_1$ and $L_2$, and the short ones $S_1$ and $S_2$. Let's pretend the structure is hinged. In this case, due to the structure's symmetry, we know that both $L$ bars will deform equally, as will the $S$ bars. Therefore, the $L$ bars will both present the same rotations around their transversal x-axes at the extremities. Unfortunately, the structure isn't hinged, so these rotations will be transferred to the $S$ beams as well. However, since both extremities will transfer the same rotations, the $S$ beams will behave as if this is a rigid-body motion, and will therefore not suffer any torsion. Obviously, the same applies the other way around: the $S$ bars will be deformed by the loads applied and will impose a rotation on the $L$ bars. These, however, will be rotated equally on both ends and will therefore simply do a rigid-body rotation, without suffering torsion. Since the $L$ and $S$ beams are perpendicular to one another, they also don't transfer bending moment between themselves, meaning the bending moments at the extremities will also be null. The end result is therefore identical to that of a series of simply-supported beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that you don't have to just trust me, here's a simple model. The purple diagram is the bending moment, which you can see is identical to that of four simply-supported beams. The orange diagram is the torsion suffered by each beam. It may seem large, but if you look at the scale on the bottom right you'll notice it's insignificant (maximum and minimum values equal to zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZkM8G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZkM8G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Diagram created with &lt;a href=&quot;http://www.autodesk.com/products/robot-structural-analysis/overview&quot; rel=&quot;nofollow noreferrer&quot;&gt;Autodesk Robot&lt;/a&gt;, a 3D structural analysis tool.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-05-19T03:11:38.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8939" PostTypeId="1" AcceptedAnswerId="8942" CreationDate="2016-05-19T10:09:23.347" Score="2" ViewCount="43" Body="&lt;p&gt;&lt;a href=&quot;http://www.reactivetools.com/datasheets/reflex_lite.pdf&quot; rel=&quot;nofollow&quot;&gt;This description&lt;/a&gt; of the Reflex Lite packer by Reactive states the following among its advantages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;No mechanical delay required&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could that mean? The packer has no built-in system of delayed activation or delayed swelling control? Or &quot;there is no need for complex procedures when installing the packer&quot;?&lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="349" LastEditDate="2016-05-19T10:34:27.307" LastActivityDate="2016-05-19T13:02:51.043" Title="Meaning of &quot;no mechanical delay required&quot; in packer description (oil drilling)" Tags="|drilling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8940" PostTypeId="2" ParentId="8922" CreationDate="2016-05-19T10:14:08.553" Score="1" Body="&lt;p&gt;You have some options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Remove/Ignore the secondary beams from your analysis.&lt;/strong&gt; If they aren't there to take bear major loads, then remove them completely and perform your analysis on the primary beams in isolation. This could potentially be performed by reducing the Young's modulus of the secondary beams, to simulate artificially suppress their effect on displacements of the primary beams as a result of loading.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Perform nonlinear contact analysis&lt;/strong&gt; Set gaps between the primary and secondary beams. Load the primary beams and set contact conditions between the two. The secondary beams will only start bearing load after the primary beams have deformed and make contact with the secondary beams.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;2. Explanation&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using FE methods, very few models can be analysed using linear static modelling, they are simply too complex. A linear static models can be computed very fast and usually need only be calculated once. Why only once? - consider that a linear static models deformation will be linearly proportional to the force applied. If for example an applied load x leads to a deformation y, it's safe to assume that a load of 2x will deform the part by 2y etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very few models can be accurately modelled in this way because in real life, there are many causes of &lt;strong&gt;nonlinearity.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically 3 causes of nonlinearity are considered: Material nonlinearity (look at the stress strain for rubber or steel - both nonlinear); geometric nonlinearity (deformations can alter the location and orientation of loads and internal strains) and &lt;strong&gt;contact nonlinearity&lt;/strong&gt;. Contact nonlinearity is best explained with a simple diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NbVvv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NbVvv.png&quot; alt=&quot;Contact Nonlinearity&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus by providing a physical gap between the primary and secondary beams and setting up contact conditions, your FEM software will necessarily consider a contact nonlinearity condition. This might not accurately describe your model, but applying this method could force the simulation to only load the secondary beams after the primary beams have deformed by a certain margin.&lt;/p&gt;&#xA;" OwnerUserId="2688" LastEditorUserId="2688" LastEditDate="2016-05-19T10:50:13.640" LastActivityDate="2016-05-19T10:50:13.640" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="8942" PostTypeId="2" ParentId="8939" CreationDate="2016-05-19T13:02:51.043" Score="1" Body="&lt;p&gt;I've just got a reply from the company:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some of our competitors use an applied outer layer of material to delay their swell curves (mechanical delay). At Reactive, we select the correct polymer for the well, so no fabricated delay system is required.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Voila. &lt;/p&gt;&#xA;" OwnerUserId="349" LastActivityDate="2016-05-19T13:02:51.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8943" PostTypeId="2" ParentId="8922" CreationDate="2016-05-19T13:56:52.033" Score="1" Body="&lt;p&gt;Your description of what primary and secondary beams do is dramatically different to the definition given in the link you present (which is itself correct).&lt;/p&gt;&#xA;&lt;p&gt;To quote the answer in your link:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Primary Beam:&lt;/strong&gt;&lt;br /&gt;&#xA;A horizontal beam connecting columns (simply supported or shear connected.)&lt;br /&gt;&#xA;&lt;em&gt;Function&lt;/em&gt;: It will transfer the load from secondary beam(if present) to the columns.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Secondary Beam:&lt;/strong&gt;&lt;br /&gt;&#xA;A horizontal beam connecting primary beams (simply supported or shear connected.)&lt;br /&gt;&#xA;&lt;em&gt;Function&lt;/em&gt;: It will transfer the load to the primary beam and not directly connected to the columns.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That is the only difference. Secondary beams take in the loads from the slab and transfer them to the primary beams, which then transfer them to the columns.&lt;/p&gt;&#xA;&lt;p&gt;So all you need to do in your FEA software is allow the user to define how these loads are applied. &lt;em&gt;&lt;strong&gt;You don't need to create different categories of beam elements (primary vs. secondary).&lt;/strong&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;You are calling your software a &lt;em&gt;&lt;strong&gt;frame&lt;/strong&gt;&lt;/em&gt; analysis tool. This implies that 2D elements like plates and shells aren't available, only 1D beam elements. This therefore means that the user will have to take some decisions as to how area loads which are applied to the slab (which you can't model, since you don't have 2D elements) are transferred.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If your program is simple, the user will have to apply the linear distributed loads on each beam by hand. In this case, there's no problem: the user applies the relevant loads on the secondary beams which then transfer them to the primary beams.&lt;/li&gt;&#xA;&lt;li&gt;Your program may also include &amp;quot;area loads&amp;quot;, where the user defines an area and a load and the program automatically decides how to distribute this area load to the beams, usually through tributary areas. So long as the user can define the beams that are to be considered for these area loads, that's also not a problem. You just define the area load and then tell the program to only consider the secondary beams (though, depending on the layout, letting some of the load go directly to the primary beams would actually be more correct).&lt;/li&gt;&#xA;&lt;li&gt;If your program is actually a full structural analysis tool with 2D elements (and is therefore more than just a &lt;em&gt;&lt;strong&gt;frame&lt;/strong&gt;&lt;/em&gt; analysis tool) as well to model the slab, that's still fine. If the user can tell the slab not to take the specific beams into consideration while meshing (such as your intended primary beams), they won't share any nodes and therefore won't have any direct load transfer. The slab will mesh joining itself only to the remaining beams, transferring loads to them, who will then transfer the loads to the primary beams.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Also, the image in your OP is a poor example of primary and secondary beams because, structurally, they are identical. The secondary ones are a bit shorter, but they share the same top-level with the primaries. This means that when the slab transfers loads, it makes no distinction between them. Look at the tributary areas for each slab, it's clear the vast majority of the loads will naturally go to the secondaries, with only a small triangle of load going to each of the primaries. Therefore, when you model them there should be no distinction between them: the loads should go where they naturally wish to go.&lt;/p&gt;&#xA;&lt;p&gt;However, if you're modelling something like the roof below, then you'll need to control how the loads are distributed. After all, all the loads coming from the roof need to go through the purlins. None of the loads may go directly to the rafters.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tWqF9m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tWqF9m.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt; &lt;sup&gt;&lt;a href=&quot;http://www.civilengineeringx.com/building/bce/Steel-roof-truss.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-05-20T11:33:27.873" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8944" PostTypeId="1" CreationDate="2016-05-19T15:35:59.340" Score="4" ViewCount="232" Body="&lt;p&gt;I am currently interested into the diverse Downlink LTE channels broadcasted over air after some (QPSK, 16QAM...) modulation. More precisely their waveform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if, with a good filtering of the noise, we could characterize the LTE signal in time domain, so we could correlate a waveform with a specific downlink channel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another words: is the LTE signal waveform dependant from the data it is carrying or the antenna it comes from? And if yes, which part of the signal in time domain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will distinguish two cases here, first the channels carrying information (sequence of bits), and the Reference Signal (RS) which exists only at PHY layer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We know that transport channel like PBCH are periodic in time (i.e occupying 1st to 4th symbol of the #1 slot in each RF, repeating the same MIB during 4 RF -&gt; %40ms). This is the same thing for the PDCCH controlling channel, with the specificity that the number of allocated symbols can vary from 1 up to 4 symbols (which i guess makes the potential correlation harder).&lt;br&gt;&#xA;I find these channels potentially interesting because of the redundant data they will carry over time (at least i think it's redundant for PBCH - information about the antenna configuration for PBCH, less sure about the PDCCH). Unlikely to the PDSCH channel which carries variable data over time in its payload, and therefore that I'm judging more difficult to characterize (maybe i'm wrong?).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;RS is also redundant in time. It's used to help UE figuring out the DL power. Thought, I don't know what we can expect of its waveform.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I don't forget SSCH and PSCH, broadcasted every 5 Sub-Frames. Maybe they are also characterizable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hAgfc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hAgfc.png&quot; alt=&quot;LTE Radio Frame of 1.4MHZ, cyan stands for PBCH, red for RS, orange for PDCCH&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should it be relevant, I don't have the MathWorks® LTE System Toolbox, even though I'm pretty sure it could be helpful.&lt;/p&gt;&#xA;" OwnerUserId="6507" LastEditorUserId="7083" LastEditDate="2016-07-21T00:41:29.833" LastActivityDate="2016-07-21T00:41:29.833" Title="LTE channel characterization from Time Domain waveform" Tags="|electrical-engineering|telecommunication|lte|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8945" PostTypeId="2" ParentId="8910" CreationDate="2016-05-19T17:56:49.657" Score="1" Body="&lt;p&gt;Using &lt;a href=&quot;http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html&quot; rel=&quot;nofollow&quot;&gt;This tutorial&lt;/a&gt;, I calculate the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume surface temperature of 150°C. Use water temperature of 30°C, and convection coefficient of 100W/m^2K. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Surface area =  2 x PI x 2x10^-3 x 0.8 = 1.0053 cm^2&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Power = 100 x 0.010053 x (150 - 30) = 120.636W&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The resistance:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Resistance = (3 x 10^3 x 0.8) / 0.010053 = 239 kΩ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The current flowing through it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I = sqrt (P / R) = sqrt ( 121 / 239k ) = 22.5mA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-05-19T17:56:49.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8946" PostTypeId="1" AcceptedAnswerId="8962" CreationDate="2016-05-19T21:15:09.907" Score="3" ViewCount="2042" Body="&lt;p&gt;I am trying to find the internal forces in the member at point G in the following illustration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ReGKC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ReGKC.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is different from the solution in the textbook and I think it has to do with how I'm looking at the 1500 lb external force acting on point &lt;strong&gt;A&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've included the 1500 lb force in my free body diagrams for both &lt;strong&gt;AB&lt;/strong&gt; and &lt;strong&gt;AE&lt;/strong&gt;. I'm getting the feeling that it should only be acting on one of these members; however, I am unsure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the load at the joint act on only one member, or on all members in the joint?&lt;/p&gt;&#xA;" OwnerUserId="6512" LastEditorUserId="368" LastEditDate="2016-05-20T19:56:53.560" LastActivityDate="2016-05-24T13:22:59.070" Title="Does an external force at a joint between two members act on one member or on both?" Tags="|structural-engineering|civil-engineering|statics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8948" PostTypeId="1" AcceptedAnswerId="8954" CreationDate="2016-05-20T00:08:06.637" Score="-5" ViewCount="125" Body="&lt;p&gt;In the electric power equation $P=RI^2$, $P$ represents the product of $RII$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand the fundamental anatomy of the $RII$ term. More precisely, why does the current, $I$ appear twice as a factor? Does this represent two different dimensions of $I$, in the sense of a measurable, albeit temporal or spatial quantity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, is one factor &quot;speed&quot; and the other one &quot;direction?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:. This &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_velocity#Particle_in_three_dimensions&quot; rel=&quot;nofollow&quot;&gt;link&lt;/a&gt; about angular velocity seems to support my question regarding direction, speed, and dimensions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers  &lt;/p&gt;&#xA;" OwnerUserId="6515" LastEditorUserId="6515" LastEditDate="2016-05-27T13:49:47.387" LastActivityDate="2016-05-27T13:49:47.387" Title="In the equation for power as a function of current and resistance, what do the two factors for current represent?" Tags="|electrical-engineering|circuits|current|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8950" PostTypeId="1" CreationDate="2016-05-20T05:55:01.250" Score="1" ViewCount="468" Body="&lt;p&gt;I have an application in which I need to detect a button press on top of a platform which has a load (0.2kg ~ 0.5kg). The press is executed on the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally I think what want is a tactile button with a holding force of greater than the load + margin. However I cant find one which such holding force and low profile (which is important for the application).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vTUDw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vTUDw.jpg&quot; alt=&quot;Tactile switch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm considering building a small mechanism on top of the detection circuitry to provide the holding force (this click mechanism also allows for other detection methods).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The behavior I'm looking for is as follows (ranges and forces are rough estimates): &lt;a href=&quot;https://i.stack.imgur.com/CjGMk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CjGMk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm avoiding load cells and other analog reading methods due to energy requirements (device is battery powered with a small battery).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically what I want/need is either how to design such a mechanism or if it is possible to source some sort of &quot;non-linear low profile spring&quot; in specific force thresholds.&lt;/p&gt;&#xA;" OwnerUserId="6517" LastEditorUserId="6517" LastEditDate="2016-05-20T11:48:22.647" LastActivityDate="2016-05-20T11:48:22.647" Title="&quot;Click&quot; mechanism" Tags="|springs|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="8951" PostTypeId="1" CreationDate="2016-05-20T06:42:04.953" Score="1" ViewCount="71" Body="&lt;p&gt;Considering the two options&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;build internal wall on top of 10&amp;nbsp;cm deep concrete slab reinforced with 7.7&amp;nbsp;mm smooth steel wire mesh&lt;/li&gt;&#xA;&lt;li&gt;build internal wall on top of 30&amp;nbsp;cm deep concrete floor joist reinforced with a 24x24&amp;nbsp;cm steel cage made of 12&amp;nbsp;mm deformed steel rebar&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What effect do&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the deeper concrete&lt;/li&gt;&#xA;&lt;li&gt;the greater amount of steel&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;in the floor joist have in supporting the weight of the wall (which I assume is treated as a compressive load)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases assume that the concrete used is the same grade. The wall is not load-bearing, as the overall structure is a reinforced concrete frame.&lt;/p&gt;&#xA;" OwnerUserId="6518" LastEditorUserId="1832" LastEditDate="2016-05-20T13:26:52.187" LastActivityDate="2016-06-19T13:45:50.500" Title="What effect does the weight of a wall have on the steel in the reinforced concrete below?" Tags="|steel|concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8952" PostTypeId="1" CreationDate="2016-05-20T09:14:02.037" Score="0" ViewCount="31" Body="&lt;p&gt;I have to draw a part sectional view of a valve for a project. A 3D view of the valve is supplied and a line AB denotes where the part section is to be. It says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The part section is to be to the left of AB&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean to section &lt;em&gt;part&lt;/em&gt; of the area left of AB and not section the right of AB. Or does this mean to section &lt;em&gt;all&lt;/em&gt; of the area left of AB and not section the right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="6520" LastActivityDate="2016-05-21T18:43:03.320" Title="Interpretting a Part Section" Tags="|drafting|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8953" PostTypeId="1" AcceptedAnswerId="8955" CreationDate="2016-05-20T11:01:47.737" Score="1" ViewCount="713" Body="&lt;p&gt;I want to calculate how heavy a mobile hydraulic post table base must be to support an adjustable monitor arm and monitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monitor (iMac) Specs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Height: 20.3 inches (51.6 cm)&lt;/li&gt;&#xA;&lt;li&gt;Width: 25.6 inches (65.0 cm)&lt;/li&gt;&#xA;&lt;li&gt;Weight: 30 pounds (9.54 kg)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.northerntool.com/shop/tools/product_200652091_200652091&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mobile Hydraulic Post Table Specs&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ship Weight: 120.0 lbs&lt;/li&gt;&#xA;&lt;li&gt;Load Capacity: 300 lbs&lt;/li&gt;&#xA;&lt;li&gt;Platform Size: 16 x 16 in&lt;/li&gt;&#xA;&lt;li&gt;Raised Height: 49 in&lt;/li&gt;&#xA;&lt;li&gt;Lowered Height: 31 in&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ergotron.com/ProductsDetails/tabid/65/PRDID/787/language/en-US/Default.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Monitor Arm Specs&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Product Weight: 17.5 lbs (8 kg)&lt;/li&gt;&#xA;&lt;li&gt;20 inches (51 cm) of vertical adjustment&lt;/li&gt;&#xA;&lt;li&gt;62 inches (157 cm) of side-to-side motion&lt;/li&gt;&#xA;&lt;li&gt;Screen ≤ 46&quot;&lt;/li&gt;&#xA;&lt;li&gt;Capacity 14–30 lbs&lt;/li&gt;&#xA;&lt;li&gt;Lift 20&quot;&lt;/li&gt;&#xA;&lt;li&gt;Tilt 80°, Pan 360°, Rotation 90˚&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The one capability that I think could cause the post table to collapse is the 62 inches (157 cm) of side-to-side motion. I've included a drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0bTZT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0bTZT.jpg&quot; alt=&quot;Workstation sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6510" LastEditorUserId="1832" LastEditDate="2016-05-20T12:34:43.207" LastActivityDate="2016-05-20T16:29:55.547" Title="Calculate mobile base weight to support adjustable arm" Tags="|structures|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8954" PostTypeId="2" ParentId="8948" CreationDate="2016-05-20T12:52:43.060" Score="5" Body="&lt;p&gt;Firstly, by temporal I think you mean spatial. Secondly if you consider acceleration, the units are m/s^2. Seconds-squared are not spatial, but they are temporal. But they needn't be either. Consider E=mc^2. c is the speed of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your formula comes from these two formulas:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P = I V&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, the power consumed by a device is equal to the voltage across it multiplied by the current flowing through it.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;V = I R&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which means the voltage across a resistive element is equal to the current flowing through it multiplied by its resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you substitute the second into the first, you get:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;P = I V = I ( I R )  = I^2 R&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you look at the equivalent SI units for power, volts, current and resistance, you will find the above to be consistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this doesn't answer your question, you really need to be clearer!&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-05-20T12:59:26.610" LastActivityDate="2016-05-20T12:59:26.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8955" PostTypeId="2" ParentId="8953" CreationDate="2016-05-20T13:17:38.163" Score="0" Body="&lt;p&gt;To calculate this, we need to know what tipping moment your table will suffer. This is pretty simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the arm totally extended, it is $31''$ long (half of $62''$). You seem to be clamping the arm to the table, so I'm going to assume that the distance from the screen at the tip of the arm to the center of the table is equal to $31 + \frac{16}{2} = 39''$. The pivot point when tipping, however, will be one of the wheels, so we can subtract the distance from the post to one of the wheels. This isn't given, so I'm going to assume that distance is around $5''$, which gives us a total arm of $34''$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now we have to calculate the tipping moment, considering the weight of the screen and of the arm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M = 30\cdot34 + 17.5\cdot\left(3 + \dfrac{31}{2}\right) = 1344\text{ lb-in}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is already resisted by the weight of the post table (120&amp;nbsp;lb&lt;sup&gt;1&lt;/sup&gt;), which can be considered concentrated at the post, generating therefore a moment equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M = 120 \cdot 5 = 600\text{ lb-in}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You therefore have here an excess of $1344-600 = 744\text{ lb-in}$ which you'll need to counter with additional weights. We can find the necessary weight by distributing it uniformly on the base, which gives us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W = \dfrac{744}{5} = 148.8\text{ lb}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is however only the nominal weight that's necessary. I would put in quite a bit extra weight to be sure the structure will hold in unexpected situations (someone bumping against the screen or some such).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table states it can support 300 lb, so this is still acceptable:&#xA;$$148.8 + 17.5 + 30 = 196.3 &amp;lt; 300\text{ lb}$$&#xA;Hell, I  personally don't see why not simply put 250&amp;nbsp;lb of weight, just as a safety precaution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to check the clamping of the arm to the table. For this, however, we need more information regarding the table (thickness, material) and of the clamp.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; This is actually the ship weight, which means the table will be lighter. Get the correct weight and repeat these calculations. Make sure to also check the actual distance from the post to the wheels, which I assumed here at $5''$.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-05-20T16:29:55.547" LastActivityDate="2016-05-20T16:29:55.547" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8956" PostTypeId="2" ParentId="8951" CreationDate="2016-05-20T13:28:35.483" Score="1" Body="&lt;p&gt;The first thing to note is that the wall will not act as a compressive load, but as a load which generates shear and bending on the supporting structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, the joist will undoubtedly be the stronger option. Both shear and bending strength are highly proportional to height, so the taller and better-reinforced joist will certainly have an easier time withstanding the wall's weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, if there are architectural reasons such that the wall would be better-placed directly on the slab, without a supporting joist directly below, that's still fine. Just do the math and make sure the slab can take it (well, you should do the math for the joist too, of course).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-05-20T13:28:35.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8957" PostTypeId="1" CreationDate="2016-05-20T13:35:00.447" Score="1" ViewCount="214" Body="&lt;p&gt;I'm using AutoCAD 2015.  While I'm drawing, I spend most of my time on the Home ribbon tab.  I like the Zoom ribbon button, but it is on the View tab.  Is there any way to move the Zoom button to the Home tab, so I don't have to switch ribbon tabs to hit it?  Alternatively, is there any way to add the Zoom button (or at least just a Zoom Extents or a Zoom All button) to the bottom toolbar, or somewhere where it will be visible all the time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that I can do a Zoom Extents just by typing &quot;Z-Enter-E-Enter&quot;, but I like my toolbar button.  :)&lt;/p&gt;&#xA;" OwnerUserId="37" LastActivityDate="2016-05-26T17:43:14.560" Title="How can I get the Zoom button to be visible while I'm on the Home tab?" Tags="|autocad|cad|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8958" PostTypeId="2" ParentId="8931" CreationDate="2016-05-20T14:17:56.780" Score="3" Body="&lt;p&gt;After a long discussion with the OP in comments and chat, it emerged that the basic cause of the problem is the geometry of the structure. The curved &quot;beam&quot; has a length/diameter ratio of about 50,000:1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this answer &lt;a href=&quot;https://engineering.stackexchange.com/questions/5311/stiffness-method-fem-slender-members/5321&quot;&gt;Stiffness Method FEM - Slender Members&lt;/a&gt; points out, the relative size of the axial and bending stiffness of a beam is of the order of $L^2/d^2$, or around $10^9$ for the OP's structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For practical purposes the &lt;em&gt;elastic&lt;/em&gt; bending stiffness of the OP's beam is zero, and it will resist bending only because of the &lt;em&gt;stress&lt;/em&gt; stiffness created by the axial tension in it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach to modeling this would be to start with a &lt;em&gt;straight&lt;/em&gt; configuration of the &quot;beam&quot;, modelled by rod elements not beams, to eliminate the vanishingly small elastic bending stiffness from the model. A nonlinear stress analysis could then start by applying a tension force to pre-stress the structure, and then change the loading in increments to the actual loading conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea would be to perform a nonlinear &lt;em&gt;dynamic&lt;/em&gt; analysis, and let the inertia of the structure constrain its behaviour under the initial loads. You would have to include some fictitious damping in the model to produce a final steady-state solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A naïve approach to overcome the $10^9$ ratio of axial to bending stiffness would be to use short beam elements so that the stiffness ratios within each element were of similar magnitude. But to achieve that, the model would have to consist of say 10,000 or 100,000 elements, which is not likely to be a sensible way to attack the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my own experience I would not choose Nastran to analyse such extremely nonlinear behaviour. A FEA package that was designed &quot;from the ground up&quot; to deal with nonlinear analysis, such as Abaqus, would be a safer bet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be worth searching the literature for similar types of problems - for example the mechanics of cables for mooring floating structures in the offshore oil and gas industry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP might get some qualitative insight into the structural behaviour by increasing the diameter of the beam (probably by a few orders of magnitude) to get the model to run.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-05-20T14:23:24.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8959" PostTypeId="1" CreationDate="2016-05-20T15:05:17.970" Score="5" ViewCount="432" Body="&lt;p&gt;I am designing an enclosure, and top and bottom should be linked thanks to snap fit. I have to use 3D-printing to make both parts. The material is ABS. I made some trials, but they keep breaking. I am using the design in the following picture. &lt;a href=&quot;https://i.stack.imgur.com/SdacA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SdacA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; Do you have any idea about how I can strengthen them or something else to prevent them from breaking ? Thank you for your help !&lt;/p&gt;&#xA;" OwnerUserId="6308" LastActivityDate="2016-05-21T16:26:07.757" Title="Snap fit on an enclosure" Tags="|design|3d-printing|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8960" PostTypeId="1" AcceptedAnswerId="8963" CreationDate="2016-05-20T16:06:21.337" Score="4" ViewCount="181" Body="&lt;p&gt;I am having trouble finding a 3.4 µH through hole inductor that is in stock right now. I was wondering if I can just place 2 inductors that add up to 3.4 µH in series to solve this problem. I am not sure if this would introduce any problems, if so what kind of problems and how can I mitigate them.&lt;/p&gt;&#xA;" OwnerUserId="6522" LastEditorUserId="1832" LastEditDate="2016-05-20T16:40:29.037" LastActivityDate="2016-05-21T13:52:10.260" Title="Circuit Design Inductor Problem" Tags="|electrical-engineering|power-electronics|circuits|circuit-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8961" PostTypeId="2" ParentId="8959" CreationDate="2016-05-20T16:12:55.230" Score="4" Body="&lt;p&gt;3D prints are weak in the direction of each layers plane. You could try to make the part in a different orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second thing to try is to use a coarser print. My friend tested mechanical properties of different 3D prints and concluded that the printers were worse the smaller the layer size was. Even materials that technically better didnt do well in a real test since the failure mode is not in the material itself but rather the bond between layers. Finer layers having more seeds for the error to start from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also strengthen the print with some other material by sewing or taping something on the part.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2016-05-20T16:12:55.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8962" PostTypeId="2" ParentId="8946" CreationDate="2016-05-20T19:57:08.080" Score="2" Body="&lt;p&gt;&lt;strong&gt;An externally applied load is only ever applied once;&lt;/strong&gt; it is then transferred through the connection (via internal forces) to all other members of the joint. If there are multiple members connected at the point of application, you can make an arbitrary choice of where to apply that load, as long as you stick with that choice for the entire analysis. (This is correct for static analysis of trusses and frames, at least.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find it sometimes helps to &quot;zoom in&quot; conceptually on that linkage and go through the transfer of force step-by-step, with as many free-body diagrams as necessary, even including the pin itself if you're at all uncertain about how force transfers from member to member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll start with the wood member, with the pin hole exaggerated for convenience and cutting at G to save me some time sketching:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/loIHH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/loIHH.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've applied the external force here. My force balance for this FBD will include the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Sigma F_y = F_{A,y} + F_{G,y} - 1,500\ \mathrm{lb_f} = 0 \tag 1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point on, &lt;strong&gt;the value of the externally applied force is reflected in the values of&lt;/strong&gt; $F_{A,y}$ &lt;strong&gt;and&lt;/strong&gt; $F_{G,y}$. Now I'll draw the FBD of the pin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dmr0p.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dmr0p.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A note on FBD conventions: I like to draw every force, the first time I define it on a diagram, in the positive direction. It doesn't &lt;em&gt;really&lt;/em&gt; matter which way you draw it the first time, as long as you always respect the initial definition in any later step, but in this case it works out nicely for analysis by inspection. Here's the equation in the y-direction anyway:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Sigma F_y = F_{rod,y} - F_{A,y} = 0 \tag 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can look at the other member in the joint, which is the metal rod. You can consider this to show a crimped end rather than the forked &lt;a href=&quot;https://en.wikipedia.org/wiki/Clevis_fastener&quot; rel=&quot;nofollow noreferrer&quot;&gt;clevis&lt;/a&gt; end shown in the original illustration, if you like; it makes no difference to the analysis but saves time sketching the FBD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sq3hX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sq3hX.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why didn't I draw the external 1,500 lbf load on this third diagram? The answer is simple:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We have defined $F_{rod,y}$ originally in terms of all forces present in the previous FBD of the connection;&lt;/li&gt;&#xA;&lt;li&gt;We have defined those forces present in the FBD of the connection all in terms of the forces present in the previous FBD of the first member;&lt;/li&gt;&#xA;&lt;li&gt;The external force was applied in the FBD of the first member.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Our equations are all linked by shared variables, so the effect of the external force when applied just once propagates through to all of the other members and connectors at that joint—and potentially to the entire frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearrange equation $(1)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{A,y} = 1,500\ \mathrm{lb_f} - F_{G,y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute into equation $(2)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{rod,y} - (1,500\ \mathrm{lb_f} - F_{G,y}) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $F_{rod,y}$ is applied directly to the metal rod in the third FBD and its definition takes into account the external load we already applied in the first FBD, you should now be able to see that there's no need to apply the external load to more than one member in a joint. If you did apply it again, you'd be double-counting and your analysis would be incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, as long as we're not concerned the pin will fail, we can skip drawing its FBD and just apply the third law action/reaction pairs directly between the members of the joint.&lt;/p&gt;&#xA;" OwnerUserId="368" LastActivityDate="2016-05-20T19:57:08.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8963" PostTypeId="2" ParentId="8960" CreationDate="2016-05-20T21:50:24.460" Score="1" Body="&lt;p&gt;Correct - putting inductors in series will result in an effective inductance equal to their sum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is the extra wiring, and/or change in the wires' paths. Moving two wires closer together makes them act like a small capacitor. Long wires act like a resistance, and (obviously) loops act like inductors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.4uH is relatively large, so you might not have much of a problem since inductors tend to have a large tolerance, +/- 10% is typical. Just keep the inductors close together, and modify the surroundings as little as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other issues to consider though, such as internal resistance. (An ideal inductor has an internal resistance of zero ohms). Try to match the series resistance in the same way as inductance. Aim for the sum to equal that of the one you're replacing. This is going beyond my expertise, but I suspect it won't matter much; it depends on the circuit.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-05-20T21:57:12.460" LastActivityDate="2016-05-20T21:57:12.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8964" PostTypeId="2" ParentId="8923" CreationDate="2016-05-20T22:41:38.290" Score="6" Body="&lt;p&gt;I see another answer has given input on the strength side of things, where the economics isn't as clear, so I will add that one of the crucial aspects that usually make continuous beams economical is deflection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given load the deflection of a continuous beam is smaller (usually &lt;em&gt;much&lt;/em&gt; smaller) than what it would be for a simply supported beam of the same section and span due to the addition of end restraint(s) and their influence on the curvature distribution. For a problem governed by deflection limits in the service state (and many if not most are) this means that a continuous beam can use a smaller section for the same deflection compared with a simply supported beam. Smaller section means less material and less expense.&lt;/p&gt;&#xA;" OwnerUserId="6523" LastActivityDate="2016-05-20T22:41:38.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8965" PostTypeId="2" ParentId="8959" CreationDate="2016-05-20T23:05:59.003" Score="3" Body="&lt;p&gt;In addition to joojaa's excellent advice, you might try reducing the height of the interference and/or increasing the length of the flexure to which it is mounted &amp;mdash; such changes would reduce the internal stress in the plastic to manageable levels.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-05-20T23:05:59.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8966" PostTypeId="2" ParentId="5461" CreationDate="2016-05-21T05:31:39.067" Score="0" Body="&lt;p&gt;Best used for metal o ring : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas filled metal O Ring for high temperature and moderate pressure applications. OGI for internal pressure and OGE for external pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In oil and gas applications metal seals are highly preferred. In applications like deep sea where the pressures are extreme and the environment is corrosive special seals are made. At times they are gold plated for excellent performance. In power generation these seals are used to seal various parts of turbine section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type of seals used:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Metal O-ring - Shaped as well as circular seals.&lt;/li&gt;&#xA;&lt;li&gt;C-ring - Shaped as well as circular seals.&lt;/li&gt;&#xA;&lt;li&gt;Spring energized C-rings - Shaped as well as circular seals.&lt;/li&gt;&#xA;&lt;li&gt;Radial / CA seals – circular in shape.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;click here more information about metal o rings &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://htmsfutom.com/metal-o-ring-manufacturer.php&quot; rel=&quot;nofollow&quot;&gt;http://htmsfutom.com/metal-o-ring-manufacturer.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.facebook.com/futomengineerspvtltd/&quot; rel=&quot;nofollow&quot;&gt;https://www.facebook.com/futomengineerspvtltd/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6526" LastEditorUserId="1832" LastEditDate="2016-05-21T09:26:57.813" LastActivityDate="2016-05-21T09:26:57.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8967" PostTypeId="2" ParentId="7568" CreationDate="2016-05-21T13:34:02.087" Score="5" Body="&lt;p&gt;Actually very long bridges (and super-tall buildings) often have less issues with earthquakes than their smaller brethren. This is due to them generally being much more flexible and therefore having lower fundamental periods which makes them less susceptible to resonance in their fundamental modes. The fundamental modes are the swaying patterns that include the majority of the structural mass. An extreme simplification would be that the main structure sways so slowly that it barely notices the quick movements of an earthquake. A bit like a large ship in small waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general &quot;mid-size&quot;-structures, with fundamental frequencies between say 1Hz and 10Hz, are usually much more harshly affected as there is a much larger risk of fundamental resonance leading to very large load effects. For very large and slender structures the wind engineering is generally a bigger challenge than the earthquake engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the piers and abutments and their connections to the main bridge deck are critical as they usually are much stiffer than the bridge as a whole. And given the amount invested and the potentially horrific consequences of a failure of a large structure a lot of effort will of course be put into performing and checking (and triple checking) the earthquake engineering of every part of the structure. I'm just pointing out that the problems aren't simply proportional to scale, larger structures aren't necessarily harder to &quot;earthquake-proof&quot; than smaller ones.&lt;/p&gt;&#xA;" OwnerUserId="6523" LastActivityDate="2016-05-21T13:34:02.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8968" PostTypeId="2" ParentId="8960" CreationDate="2016-05-21T13:52:10.260" Score="1" Body="&lt;p&gt;3.4 &amp;micro;H is not a standard value, but 3.3 &amp;micro;H is.  There are plenty of those out there, even in thru-hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inductors aren't generally accurate enough to justify making a 3.4 &amp;micro;H version when a 3.3 &amp;micro;H version is already in production.  The extra 3% more isn't worth another product to manufacture, and for customers isn't worth another part to stock.  You better have a really good reason you are committing your company to stock a new inductor with a non-standard value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to step back and examine why you need such a unusual inductance value so accurately that 3% less actually matters.  Usually you determine the rough inductance you need, find a suitable part, then design the details of the circuit around that.  You can get resistors and timing components at much higher accuracies than inductors, for example, so you pick exact values for those last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also examine why you want thru-hole.  Unless this is a high-current high-power part that is therefore physically heavy, thru-hole makes little sense.  Thru-hole will limit you to parts designed 10 years ago or longer.  New development is obviously in surface mount parts, except for physically large inductors.  You haven't said anything about the saturation current requirement or maximum resistance you can tolerate.  &quot;3.4 &amp;micro;H&quot; is only one part of the spec for a inductor.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-05-21T13:52:10.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8969" PostTypeId="2" ParentId="8885" CreationDate="2016-05-21T14:01:04.997" Score="1" Body="&lt;p&gt;I would highly recommend the first couple of solidworks tutorials. Creating something when you don't understand what you're doing generally leads to underdefined,  incorrect parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But search for centreline (a construction line) or right click on a normal line to give you some geometry to align the top two circles with. A filet is what you want for the rounded edges. Also the green boxes are constraints,  and I can see you have some you don't want, like the left and right uprights have been forced to be vertical. To add constraints, right click on the parts you want to constrain and select what you need. To delete them, click on the green squares and press delete &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also your sketch is currently floating in space, not fixed to the geometry like it should be. To fix this, select the origin and a circle (hold shift) and select make concentric. Basically you need to do a tutorial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a line is fully defined, it will turn black. When the whole sketch is fully defined, you can move on, go to the features toolbar, and select extruded boss/Base. If you move on without fully defining the sketch, you can accidently drag lines out of place without knowing. Which is the worst to try to hunt down and fix. &lt;/p&gt;&#xA;" OwnerUserId="1342" LastActivityDate="2016-05-21T14:01:04.997" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8971" PostTypeId="2" ParentId="8959" CreationDate="2016-05-21T16:26:07.757" Score="1" Body="&lt;p&gt;Also in addition to joojaa's advice, you want to print in a direction such that the cross section of the hook is parallel to your bed, ie, it prints up lots of L shapes, and there's no weak point (in the two other directions either the connection between the case and the L will break, or the curve of the L). Having three snaps makes it so that one of the snap's cross section will always have to be perpendicular to the bed base, and so this one will always be weaker. I would fix in one direction by a slide / overlap / pin, and clip in the other. Undertolerance and test to find your correct fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your design also uses features based on injection moulding techniques ie: single sided mould release by having the material to hold your screws and (pins?) going all the way down to the Base of your container. In 3d printing, you're not constrained by this, and you can print on top of nothing using cheaper support material. But chamfer the edges underneath or that will be weak. &lt;/p&gt;&#xA;" OwnerUserId="1342" LastActivityDate="2016-05-21T16:26:07.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8972" PostTypeId="2" ParentId="8952" CreationDate="2016-05-21T16:39:25.647" Score="1" Body="&lt;p&gt;A picture of this question would be quite useful, or a scribble something like it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm assuming the line is drawn on a 2D drawing of the part. Imagine that this part is lying on the table in front of you so that you see the exact same view of it as drawn. If you take a super sharp knife and cut along the dotted line AB, all the way down the piece like cutting a potato in half, that's your section. Then you take the LEFT piece and place it so that you are looking at the cross section you just made. Then you draw this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cut the part all the way along the dotted line, so if this extends all the way through the part, you draw the part as above. However, if it extends part way though, it's like you made the cut (all the way down to the table, but not all the way through the potato) and then tore off the right chunk - you'll have a section view for part of it, and a complete view for the rest of it, with an interface in between where you've torn it. &lt;/p&gt;&#xA;" OwnerUserId="1342" LastActivityDate="2016-05-21T16:39:25.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8973" PostTypeId="2" ParentId="8952" CreationDate="2016-05-21T18:43:03.320" Score="0" Body="&lt;p&gt;Although a line is the conventional way to indicate a section it may be easier to visualise it if you  think of it as a plane passing through the line. If you are asked for a left or right section this indicates that the plane is vertical, top or bottom indicates that the plane is horizontal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plane can be thought of as cutting the part with one side or the other effectively ceasing to exist as far as the drawing is concerned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of this is to show internal detail of an assembly or part and is common for complex core castings, engines, gearboxes etc where internal parts may be contained within a casing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases actual parts may be physically cut as illustrative tools for museums and teaching aids eg &lt;a href=&quot;http://previews.123rf.com/images/gortan123/gortan1231212/gortan123121200016/16859227-Automatic-gearbox-cross-section-on-black-Stock-Photo-transmission-automatic-car.jpg&quot; rel=&quot;nofollow&quot;&gt;http://previews.123rf.com/images/gortan123/gortan1231212/gortan123121200016/16859227-Automatic-gearbox-cross-section-on-black-Stock-Photo-transmission-automatic-car.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the red pained parts are material which has been cut by the sectioning plane. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-21T18:43:03.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8974" PostTypeId="2" ParentId="8921" CreationDate="2016-05-21T19:05:49.653" Score="1" Body="&lt;p&gt;As already mentioned there are off the shelf tools for detecting various items which may be buried in interior walls. Timber studding, metal pipes and current carrying wires can all be detected and differentiated reasonably easily by fairly cheap detectors in close proximity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Handheld detectors are available in DIY or more professional specs and are generally used when you need to drill into a wall to ensure that you either have a secure fixing point (eg studding rather than plasterboard) or to avoid hitting pipes or electrical wires. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assessing structural integrity and strength is a bit of a different problem. Apart from anything else you need some idea of what you are looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in a timber structure rot, fungal attack or termites may be the main concern, in steel structures you may be looking for corrosion, fatigue cracking, degradation of protective coatings or creep; in concrete cracks and corrosion of steel reinforcement are potential concerns as well as environmental and chemical degradation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An inspection of a structure may also look at whether is has been constructed as designed eg have corners been cut by using inferior materials or omitting fasteners and whether the dimensions specified in the design have been followed correctly. There are also potential situations where apparently minor deviations form the design can have serious consequences for example the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hyatt_Regency_walkway_collapse&quot; rel=&quot;nofollow&quot;&gt;Hyatt Regency walkway collapse&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example illustrates that a simple evaluation of material properties may not &#xA;guarantee that a structure is sound or safe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of structures as a whole measuring resonant frequency may be a useful metric of overall structural integrity (especially for tall buildings and bridges which sway measurably with wind loads) here lasers or ultrasound may be able to provide very precise measurements although this does of course depend on having a baseline measurement to compare against. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also be useful to measure changes in dimensions over time particularity where is is suspected that buildings may b e sinking into their foundations here again very precise positional measurements are necessary. A very crude version of this is  to simply glue a glass microscope slide over a crack in a structure if it breaks over time that indicates that the crack is growing or shrinking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition drones, robots or endoscopes may be used to make visual inspections of otherwise inaccessible parts of a structure. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-21T19:05:49.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8975" PostTypeId="2" ParentId="6450" CreationDate="2016-05-21T20:38:51.327" Score="1" Body="&lt;p&gt;There are several theories of failure in mechanics of materials. Among them we use shear theory and Distortion energy theory which arises from plasticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes failure is the difference between maximum and minimum stresses. If all the three are equal, materials will never fail. If we take the minimum radial stress as zero, it is conservative.Taking it into account the design is not much influenced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Circumferential stress is like $ pr/t,$ radial stress is like $p,$ When stress ratio is $r/t\approx 15  $ ..negligible.&lt;/p&gt;&#xA;" OwnerUserId="1270" LastEditorUserId="1270" LastEditDate="2016-05-21T22:56:50.073" LastActivityDate="2016-05-21T22:56:50.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8976" PostTypeId="1" CreationDate="2016-05-22T02:45:48.530" Score="2" ViewCount="1183" Body="&lt;p&gt;I've been studying gimbal lock and am puzzled by something in regards to the Apollo 11 mission mentioned on &lt;a href=&quot;http://en.wikipedia.org/wiki/Gimbal_lock&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;They preferred an alternate solution using an indicator that would be triggered when near to 85 degrees pitch.&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;Near that point, in a closed stabilization loop, the torque motors could theoretically be commanded to flip the gimbal 180 degrees instantaneously. Instead, in the LM, the computer flashed a 'gimbal lock' warning at 70 degrees and froze the IMU at 85 degrees&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;— Paul Fjeld, Apollo Lunar Surface Journal&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What I don't understand is this instantaneous flip of a gimbal by 180 degrees to prevent a gimbal lock. Which gimbal? And how would that change anything? I'm struggling to conceptualize how it would prevent the lock.&lt;/p&gt;&#xA;" OwnerUserId="6535" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-11-03T06:36:23.057" Title="Apollo 11 gimbal lock flip" Tags="|mechanical-engineering|measurements|aerospace-engineering|instrumentation|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8977" PostTypeId="1" AcceptedAnswerId="8981" CreationDate="2016-05-22T07:10:37.957" Score="4" ViewCount="11276" Body="&lt;p&gt;When a material does not show a distinct yield point then 0.2% of strain is considered and a line is drawn parallel to the elastic line and the corresponding stress is called proof stress. My question is why are they taking 0.2% of strain? Is that an assumption and does it differ with materials&lt;/p&gt;&#xA;" OwnerUserId="6538" LastActivityDate="2020-09-18T04:05:50.620" Title="Why 0.2% of strain is considered while taking proof stress?" Tags="|materials|civil-engineering|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8979" PostTypeId="1" CreationDate="2016-05-22T10:05:54.760" Score="1" ViewCount="150" Body="&lt;p&gt;Good day, I am a &lt;strong&gt;mechanical engineering student&lt;/strong&gt;, we conducted &lt;strong&gt;three-point bending test for a metal specimen&lt;/strong&gt; and I am wondering why &lt;strong&gt;the temperature of the specimen did not increase&lt;/strong&gt; as the UTM certainly &lt;strong&gt;did work on the specimen&lt;/strong&gt;, therefore some &lt;strong&gt;thermal energy increase should be present&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in comparison when we conducted tensile test to another specimen in which we realized that the metal became appreciably hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it was because &lt;strong&gt;bending involved tension and compression&lt;/strong&gt;, reasoning that because they are opposite of the other, while tension releases thermal energy because of the pulling away (rupture) of interatomic bonds, compression accepts thermal energy because it needs for the compression of the bonds? Therefore no (or minimal) thermal energy increase could be observed after bending.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't research anything related to this phenomena on the internet. (Also we had &lt;strong&gt;never conducted yet a compressive test&lt;/strong&gt;, so I don't know if compressing the metal reduces its temperature). I also know the physics of thermal expansion (metal expands with increase in temperature), I am not just sure if it applies the other way around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just want to know if you think my hypothesis is correct, or if not could you provide a link or explanation? Thank You&lt;/p&gt;&#xA;" OwnerUserId="6540" LastActivityDate="2016-06-05T02:13:59.977" Title="No increase in temperature during bending?" Tags="|mechanical-engineering|materials|civil-engineering|metals|" AnswerCount="1" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8980" PostTypeId="2" ParentId="7736" CreationDate="2016-05-22T11:25:30.107" Score="0" Body="&lt;p&gt;There is, it looks like this here:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3ELWP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3ELWP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know it this is standardized, but it is used quite often like this. In university the questions in exam looked sometimes like this.&lt;/p&gt;&#xA;" OwnerUserId="6052" LastEditorUserId="1324" LastEditDate="2016-05-22T11:48:35.707" LastActivityDate="2016-05-22T11:48:35.707" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="8981" PostTypeId="2" ParentId="8977" CreationDate="2016-05-22T11:58:03.210" Score="6" Body="&lt;p&gt;It is not even an assumption but somewhat arbitrarily chosen and not universally adopted either. In some codes and for some materials 0.1% is chosen and in others 0.05% and sometimes x% total strain is used rather than the x% proof stress. Also proof stress is not x% of strain, it is 0.2% remaining strain after unloading. So the 0.2% proof stress is the stress where after unloading you end up with a permanent elongation of 0.2% of your specimen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of 0.2% is a compromise between being easily measurable with simple equipment while being exact enough for most engineering purposes.&lt;/p&gt;&#xA;" OwnerUserId="6523" LastActivityDate="2016-05-22T11:58:03.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8984" PostTypeId="1" CreationDate="2016-05-22T18:28:54.837" Score="-1" ViewCount="137" Body="&lt;p&gt;The tension spring broke on our garage door causing it to fall. Afterward, the garage looked as it does in the pictures below. The most straightforward explanation to me is that the impact of the fall caused the door to bend between the bottom two panels causing the damage seen. Assuming a person falling into the door could have caused the damage, it's also a slight possibility that a guest could have fallen back into it while playing ping pong and we didn't notice the bending until after the door fell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Given the pictures, how likely is it that that type of bowing could be caused by someone/something impacting the garage door?&lt;/strong&gt; The damage seems like exactly what I'd expect from the several hundred pound door falling on itself and if that weren't the case, it really doesn't seem likely that, if it could withstand that impact without damage, then someone falling into it would be able to contort it to that extent. Also given that it's bowing across the entire width and not bent around a localized area make it seem even less plausible that it could have some lateral impact on it. Then again, I don't know anything about the structural integrity of garage doors and if they're much better at withstanding a drop than something hitting it from the side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rYtcm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rYtcm.jpg&quot; alt=&quot;garage door bowing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More images can be found &lt;a href=&quot;https://i.stack.imgur.com/aovHR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6547" LastEditorUserId="110" LastEditDate="2016-05-23T01:10:59.800" LastActivityDate="2016-05-23T12:42:42.373" Title="Could this garage door's bowing be caused by someone falling into it?" Tags="|structural-engineering|structures|structural-analysis|" AnswerCount="2" CommentCount="4" ClosedDate="2016-05-23T15:21:16.587" ContentLicense="CC BY-SA 3.0" />
  <row Id="8987" PostTypeId="2" ParentId="8976" CreationDate="2016-05-22T21:40:40.550" Score="0" Body="&lt;p&gt;What they are talking about is the difference of mathematics and reality. The mathematics can somewhat easily solve each position, but cannot interpolate between these positions very well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case there is a limit to what positions the mechanism can reach turning one way. The mathematics has no problem in solving what the position should be. It's just that for the physical structure cannot move to that position with a small move. So the speed of the mechanism would need to be very very fast which it obviously isn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in the gimbal lock your movement directions are severely limited so you might need to unwind the entire mechanism from one end to the other which is not what you want to do. It's all fine if you have infinitely fast gimbals like your mathematical solution needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a bit like a early GPS system would tell you to turn left onto the road that is on the bridge above you. Sure the roads cross here but your car can not fly/teleport to the bridge - you should have taken a exit hundreds or thousands of meters ago that allowed you to eventually reach that road. So the system simply does not know the move is illegal.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorDisplayName="user748" LastEditDate="2016-10-03T03:03:53.093" LastActivityDate="2016-10-03T03:03:53.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8988" PostTypeId="1" CreationDate="2016-05-23T00:45:29.570" Score="1" ViewCount="55" Body="&lt;p&gt;I was wondering what would be more efficient means of producing clean water. The first method could be though a distillation process where by you boil contaminated water and collect the water vapor (similar to &quot;sling shot&quot; water purifier), the next method would be to cool plates and collect the condensation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first method would require you to head up a lot of water just to cool it back down again, but the second method is possible highly inefficient as you are cooling a lot of air just o get a small amount of water vapour to condensate. Any one know of any calculations or studies that have addressed this question?&lt;/p&gt;&#xA;" OwnerUserId="3270" LastActivityDate="2016-05-23T00:45:29.570" Title="Efficiency of water distillation vs active Condensation Trap" Tags="|water-resources|refrigeration|" AnswerCount="0" CommentCount="1" ClosedDate="2016-05-23T13:38:07.397" ContentLicense="CC BY-SA 3.0" />
  <row Id="8991" PostTypeId="1" CreationDate="2016-05-23T04:10:23.157" Score="1" ViewCount="70" Body="&lt;p&gt;I want to model the following case(Attached picture)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/syn0w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/syn0w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This case have reciprocating and rotation motion. Is it possible that this case models in Fluent or CFX  without writing Code?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How i set this reciprocation motion in CFX? Is it possible that give me related paper and useful links and tutorials about this case? I found documentations about this case but don't found tutorial about this case. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="282" LastEditDate="2016-05-31T23:16:19.523" LastActivityDate="2016-05-31T23:16:19.523" Title="How do I set Reciprocating and Rotation motion in Fluent or CFX?" Tags="|mechanical-engineering|cfd|waves|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8992" PostTypeId="1" CreationDate="2016-05-22T11:42:52.103" Score="5" ViewCount="144" Body="&lt;p&gt;I recently bought two 12 V DC motors from Amazon. After I received them today, I tried testing them. One of the two motors run fine but the other motor shows erratic behaviour. It gets very hot within 10-15 seconds of running, and then stops. It also doesn't run as fast as the other motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During testing I directly connected a 8.6 V LiPo battery to the motor and it had sparks coming out of it. I then soldered wires to the motor's contacts and added another cell to the pack making it 11.98 V. The sparking stopped, and the motor started working, but I noticed this behaviour. There's also very high resistance when I try to rotate the motor shaft by hand, while on the other motor, it turns very easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if I damaged my motor, or if it was already damaged when I received it. How can I determine the cause of the problem?&lt;/p&gt;&#xA;" OwnerDisplayName="YaddyVirus" LastEditorUserId="368" LastEditDate="2016-06-22T16:14:42.753" LastActivityDate="2016-10-20T19:01:47.640" Title="How can I tell if a motor is defective?" Tags="|mechanical-engineering|electrical-engineering|motors|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8993" PostTypeId="2" ParentId="8992" CreationDate="2016-05-22T11:50:09.867" Score="1" Body="&lt;p&gt;You cannot damage a DC motor just by applying a lower then expected voltage to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the motor shaft is harder to rotate then on similar motor, it probably has a mechanical damage. However, a short circuit in the winding could also be the cause for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to measure the electrical resistance between the two connectors of the motor and compare the value with the healthy one. If it is the same value then you have a mechanical issue, if not then you most probably have a short circuit somewhere in the windings. &lt;/p&gt;&#xA;" OwnerUserId="3700" OwnerDisplayName="50k4" LastActivityDate="2016-05-22T11:50:09.867" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="8994" PostTypeId="2" ParentId="8992" CreationDate="2016-05-23T11:26:07.723" Score="1" Body="&lt;p&gt;It sounds like your bad motor has a mechanical problem.  For some reason, the rotor is not spinning as freely as it should.  This may be due to a bad bearing, something misaligned, etc.  Return it for a new one or get your money back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you never applied more than the rated value of 12 V to your motor, you did nothing that damaged it.  There was something originally wrong with this motor.  At worst, there is some cascading damage, but that is all a result of the original problem that was present when you received the motor.  Again, return it.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-05-23T11:26:07.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8995" PostTypeId="2" ParentId="8984" CreationDate="2016-05-23T11:30:27.630" Score="0" Body="&lt;p&gt;I don't think your picture is definitive one way or the other.  If there are some scrape marks, transferred paint, etc at the center of the bow, then it would make it more likely something bumped into the door.  Since there don't appear to be any, either scenario is plausible.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-05-23T11:30:27.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="8996" PostTypeId="2" ParentId="8984" CreationDate="2016-05-23T12:42:42.373" Score="2" Body="&lt;h2&gt;Too low to be a human, maybe a car did it&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;After looking at the other pictures, the bulge seems to be too low to have been done by a human.  If someone had fallen into the door from a fall (pingpong), I'd expect the bulge to be higher on the door since if the person was most of the way fallen over, the force of their fall would be going down, not out as we see in the picture.  Also, the bowing would be more pointed because humans aren't that wide, so the impact would be more focused in a small area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bump look like it's at the right height and the right width to be a car impact. From the picture where the SUV is visible, it's easy to draw a horizontal line from the bumper to the bulge. You can see that there's no significant bowing on any of the other panels so the impact was delivered only at that height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that this bulge was only discovered after the spring broke, it may have been there for years.  A human is unlikely to have done that damage.&lt;/p&gt;&#xA;" OwnerUserId="5427" LastActivityDate="2016-05-23T12:42:42.373" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="8997" PostTypeId="1" AcceptedAnswerId="8998" CreationDate="2016-05-23T13:29:31.567" Score="1" ViewCount="344" Body="&lt;p&gt;I want to know how much I can over-drive a small electric motor. Its speed at the rated voltage is too slow, so I want to find a higher voltage that won't blow it. It will only be used intermittently, but I figure if I run it for 100x the max expected duration with the expected load then it will probably survive repeated intermittent use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, safety concerns aside, how can I apply a relatively constant load? I can't think of a reliable way of using friction, such as attaching a disk and hold something on it, vaguely resembling a vehicle brake disk. I could attach a weight via a pulley but it would quickly run out of length. I could use a second motor and load that electrically to increase the torque, which is IMO the best idea, but I'm wondering if there are any other mechanical solutions? I think I'm missing something really obvious.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-05-24T18:21:23.950" Title="Apply constant load to an electric motor - low tech way" Tags="|friction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="8998" PostTypeId="2" ParentId="8997" CreationDate="2016-05-23T13:49:57.660" Score="2" Body="&lt;p&gt;Probably the easiest way to add a reasonably constant mechanical load to a motor is a fan. Because aerodynamic forces are exponential in respect to speed this should be consistent enough for evaluation purposes and has the advantage that you don't have the problems of wear and adjustment associated with solid friction brakes. Here the obvious solution for a small motor is to scavenge the fan  from a cheap PC cooler. If the fan  has an enclosure you can even vary the load by the simple expedient of using duct tape to restrict the flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a small motor a fan in air should be fine, for higher power applications a paddle wheel immersed in a fluid is more compact,albeit more complex to set up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be able to measure of vary the load then a &lt;a href=&quot;https://en.wikipedia.org/wiki/Band_brake&quot; rel=&quot;nofollow&quot;&gt;friction band brake&lt;/a&gt; is a good solution as you can measure the tension in the band to give a crude measure of torque and even a simple setup should at least give you a consistent range even if you can't derive precise figures for power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also get cheap handled non-contact tachometers to measure rpm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more detailed measurements of load then you could always connect the motor to a generator and measure the power generated. This can give fairly reliable direct measures of power output and a simple optical or magnetic tachometer will give you rpm and thus torque. With the right instruments this setup also allows for sampling of variations over time. With some basic programming and electronics  knowledge building a data logger to a PC is realistic (especially with a legacy parallel port).   &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-24T18:21:23.950" LastActivityDate="2016-05-24T18:21:23.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9000" PostTypeId="1" AcceptedAnswerId="9019" CreationDate="2016-05-23T15:10:42.097" Score="2" ViewCount="1205" Body="&lt;p&gt;I do not have any engineering background but I am studying some mathematics to try and understand and solve this problem. I have a mesh (M1) in cylindrical coordinates and this deforms into a mesh (M2) (again in cylindrical coordinates). I need to compute strain in radial, longitudinal and circumferential directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I have done the following computations on each point of the mesh. The deformation gradient F is computed:&lt;br&gt;&#xA;$$F = x·X^T·(X·X^T)^{-1}$$&#xA;where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x$: the deformed cylindrical coordinates (3x1 matrix)  &lt;/li&gt;&#xA;&lt;li&gt;$X$: the undeformed cylindrical reference coordinates (3x1 matrix)  &lt;/li&gt;&#xA;&lt;li&gt;$X^T$: transpose of $X$  &lt;/li&gt;&#xA;&lt;li&gt;$X^{-T}$ : inverse of transpose of $X$  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Langrangian finite strain tensor $E$ is then computed:&lt;br&gt;&#xA;$$E = \dfrac{1}{2}(F^T·F-I)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$F^T$ : transpose of $F$  &lt;/li&gt;&#xA;&lt;li&gt;$I$ : identity matrix (3x3)  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I then take the diagonal of $E$ and this gives me the principal strains in the orthogonal cylindrical coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&#xA;Am I doing it right?&#xA;Do I need to do some further computations on the diagonal of $E$ to get the principal strains at each point of this mesh that is deforming?&lt;/p&gt;&#xA;" OwnerUserId="6557" LastEditorUserId="6557" LastEditDate="2016-05-24T12:06:27.690" LastActivityDate="2019-05-25T07:38:04.900" Title="Deformation gradient, strain tensor from cylindrical coordinates" Tags="|structural-engineering|structural-analysis|finite-element-method|kinematics|deformation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9001" PostTypeId="1" CreationDate="2016-05-23T15:38:22.747" Score="6" ViewCount="166" Body="&lt;p&gt;I want to break this PCB with some know force of X newtons in the bottom left tip of the PCB. The problem is that in Static Dynamics I just get infinite deformation and in explicit dynamics I really don't know what to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone knows how to do this? I want to see the part crack away from the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://s33.postimg.org/l823k1m8f/figure_2.png&quot; alt=&quot;my problem&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="5097" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.303" LastActivityDate="2016-06-03T00:19:53.330" Title="How to simulate a crack in a PCB?" Tags="|electrical-engineering|structural-engineering|structural-analysis|finite-element-method|ansys|" AnswerCount="0" CommentCount="3" FavoriteCount="0" ClosedDate="2018-06-05T17:18:31.227" ContentLicense="CC BY-SA 3.0" />
  <row Id="9005" PostTypeId="1" CreationDate="2016-05-24T12:42:19.373" Score="2" ViewCount="831" Body="&lt;p&gt;Two gears are spinning is opposite directions, I would like a third gear to spin in one direction (the direction is not important), and this third gear must be the sum of the motions of the two gears. The speeds of the two gears can vary. If they were spinning at constant speed, this would be easy for me, however they are always changing. &lt;/p&gt;&#xA;" OwnerUserId="6569" LastActivityDate="2016-05-24T14:53:32.167" Title="Is there a way to add together the rotations of two spinning gears, which are spinning in opposite directions, into one rotation?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9006" PostTypeId="2" ParentId="8946" CreationDate="2016-05-24T13:22:59.070" Score="-1" Body="&lt;p&gt;Using the method of virtual work, assume that $AE$ is given a small deflection $d\theta$ downwards and that the unknown force is $Y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 1500 (6+2+2) d\theta + Y (6+2) d\theta + \left(\int_0^6 \frac{300 l}{6}l \, dl\right)d\theta =0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Rightarrow 15000  + 8 Y+ 3600 =0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Rightarrow  Y = -2325 $$&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-05-24T13:22:59.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9009" PostTypeId="2" ParentId="9005" CreationDate="2016-05-24T14:53:32.167" Score="4" Body="&lt;p&gt;The device you are looking for is called a differential like @Ethan48 commented. There are two ways of using the differential the typical use  is to split one powered shafts power into two separate shafts. This is used in cars and that is the typical place people have heard of the concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the nature of mechanical couplings it can also be used in reverse to sum the movement of two axes into one. So you can use it to connect two driving shafts into one output.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2016-05-24T14:53:32.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9010" PostTypeId="2" ParentId="8771" CreationDate="2016-05-24T16:29:31.303" Score="2" Body="&lt;p&gt;Software engineering is a relatively newer form of engineering so borrows more than leads - the agile methodologies borrowed heavily from &lt;a href=&quot;https://en.wikipedia.org/wiki/Lean_manufacturing&quot; rel=&quot;nofollow&quot;&gt;lean manufacturing&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Toyota_Production_System&quot; rel=&quot;nofollow&quot;&gt;Toyota&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Kanban&quot; rel=&quot;nofollow&quot;&gt;Kanban&lt;/a&gt;. The Wiki summary of Toyota principles is very close to the &lt;a href=&quot;http://www.agilemanifesto.org/&quot; rel=&quot;nofollow&quot;&gt;agile manifesto&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1069" LastActivityDate="2016-05-24T16:29:31.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9012" PostTypeId="1" CreationDate="2016-05-24T17:57:11.357" Score="1" ViewCount="274" Body="&lt;p&gt;In &lt;a href=&quot;https://engineering.stackexchange.com/questions/8931/fea-what-boundary-conditions-am-i-missing&quot;&gt;this previous quesiton I posted&lt;/a&gt;, alephzero kindly noted that many of the convergence problems I was having were due to extremely high slenderness ratios that NASTRAN did not feel comfortable executing. This is all well and good, and while the accepted answer to the question indicates I may need a change in either my model or my software, I seem to have found an interesting quirk about my structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following two images. The following is true about both:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The left single &lt;strong&gt;beam&lt;/strong&gt; has an extremely high stiffness&lt;/li&gt;&#xA;&lt;li&gt;The right single &lt;strong&gt;rod&lt;/strong&gt; has an area of $10^{-7}$ m$^2$&lt;/li&gt;&#xA;&lt;li&gt;The bottom &lt;strong&gt;beams&lt;/strong&gt; are composed of $1000$ elements&lt;/li&gt;&#xA;&lt;li&gt;The left-most two nodes are fixed in &lt;strong&gt;123456&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;An angular velocity of $18 \frac{\text{rad}}{\text{sec}}$ was applied&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Just because I have a thousand elements does mean I necessarily fixed the slenderness problem, but I might have made a dent in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/93UIA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/93UIA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gyfoY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gyfoY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first image, the $1000$ elements are lined horizontally and the maximum deflection is on the order of a millimeter, with a curved shape that agrees with my expectation (right-most node tries to move to the right, the rod it's connected to is too stiff to extend, moves in an arc upward keeping the rod the same length, rest of the elements conform to this new curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, with almost none of the parameters changed in the second image, I get deflections of about one kilometer. All that changed was the initial geometric state, where I start with a curve instead of a horizontal line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this seem like normal behavior? Naively, I would expect that changing the initial geometry slightly (the right-most node moved up $30$ cm or something from the first to the second image) shouldn't cause massive fluctuations in the deflection, so is there some modeling error I'm not considering here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The .f06 file does complain about a few of the rightward nodes having problems in the second translational direction, and this is clear looking at the deflection, but I don't know &lt;em&gt;why&lt;/em&gt; these nodes are having problems.&lt;/p&gt;&#xA;" OwnerUserId="5218" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-05-25T20:29:13.713" Title="FEA: Changes in initial geometry cause divergence of solution" Tags="|structural-engineering|finite-element-method|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9014" PostTypeId="1" CreationDate="2016-05-24T20:08:30.240" Score="1" ViewCount="84" Body="&lt;p&gt;I'm a ropeway engineer with a lifetime of experience. I would like to build a zipline in a public place (officials agreed to support me). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I get safety standards for constructing a zipline? I live in Georgia (country) and it's a new thing here, so I was thinking about adopting European standards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just don't know where to find it or which company to contact.&lt;/p&gt;&#xA;" OwnerUserId="6578" LastEditorDisplayName="user16" LastEditDate="2016-05-25T12:33:32.840" LastActivityDate="2016-05-25T12:33:32.840" Title="Safety standards involved in building a zipline" Tags="|structural-engineering|standards|" AnswerCount="0" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9015" PostTypeId="2" ParentId="9012" CreationDate="2016-05-25T01:00:17.393" Score="2" Body="&lt;p&gt;Any set of nonlinear equations may have a several mathematical solutions (and often an unknown number of solutions), some of which may not correspond to anything physical. Your first solution does appear to correspond to something physically realistic. The second one clearly isn't physical, unless the material is linear elastic up to strains of 1000 or more, but the mathematics doesn't know anything about your material behaviour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving nonlinear equations is usually done by an iterative procedure. If the unconverged solution becomes physically unrealistic during an iteration, it often stays that way for the rest of the solution process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As was discussed in the other thread, the first iteration, with the internal stresses all zero, corresponds to a linear solution, and for the curved structure the linear displacements will be huge. After that, what happens might be correct mathematically, but it isn't &quot;engineering&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the straight structure, any divergence from &quot;straight&quot; will probably be suppressed automatically on the first iteration. You then have a situation where the structure has the right order of magnitude of tension along the beam, which will give the right order of magnitude of bending stiffness, and the subsequent iterations will converge to a physically sensible answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be instructive to plot the deformed shape after each iteration for the two solutions, to see how the convergence progressed in each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to investigate what the convergence criteria mean physically for your problem. IIRC Nastran has a choice of checking the size of the displacement increments, and/or the change in internal strain energy of each element, and/or the residual loads at each node point. It's possible that your second solution has &quot;converged&quot; according to the criterion you selected (i.e. the default one, if you didn't specify it) but not according to the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the first solution would converge to the same result with far fewer elements - e.g. 100 elements, or maybe even only 10, instead of 1000. That would be another an indication that it corresponds to something physically real.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Yet another idea for a solution strategy&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the OP has found a way to get a converged solution starting from a straight configuration of the beam, that suggests a possible way to analyse the curved beam: First do a load step that applies loads straighten out the beam, then do a second load step which simultaneously removes those loads and applies the &quot;real&quot; loads to the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the loads required to straighten the beam, you could apply non-zero prescribed displacements to the model (using &lt;code&gt;SPC&lt;/code&gt; input, not &lt;code&gt;SPC1&lt;/code&gt;) and request output of the reaction forces with &lt;code&gt;SPCF=ALL&lt;/code&gt; in the case control. The required non-zero displacements in the input are simply the distances between the corresponding grid points in the curved and straight models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that Nastran fill fail if you constrain &lt;em&gt;every&lt;/em&gt; degree of freedom in the model - you have to let it calculate something for itself! Probably, fixing all the translations but not the rotations would be enough. If that doesn't work, you can create a dummy element that is not connected to anything else in the model, fix one end of it, and apply no loads to it. Nastran will then be happy to calculate that the other end of it doesn't move, and give you the reaction forces to all the other constraints.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-05-25T20:29:13.713" LastActivityDate="2016-05-25T20:29:13.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9016" PostTypeId="1" AcceptedAnswerId="9017" CreationDate="2016-05-25T04:42:05.683" Score="0" ViewCount="2713" Body="&lt;p&gt;My question has two parts.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is BSc engineering and Btech engineering ?&lt;/li&gt;&#xA;&lt;li&gt;Which has the highest validity in electronic and computer fields?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt; PS : Specially in North American region.&lt;/p&gt;&#xA;" OwnerUserId="6579" LastEditorUserId="1586" LastEditDate="2016-05-25T18:08:53.750" LastActivityDate="2016-05-27T21:30:48.917" Title="BTech or BSc engineering" Tags="|electrical-engineering|computer-engineering|" AnswerCount="1" CommentCount="0" ClosedDate="2016-05-25T12:29:39.370" ContentLicense="CC BY-SA 3.0" />
  <row Id="9017" PostTypeId="2" ParentId="9016" CreationDate="2016-05-25T11:13:07.850" Score="2" Body="&lt;p&gt;I think Bachelor of Technology (B-Tech) is mostly a designation used by Indian Institute of Technology (IIT) and National Institute of Technology (NIT). &lt;a href=&quot;https://www.quora.com/What-is-the-difference-between-a-Bachelor-of-Engineering-and-a-Bachelor-of-Technology&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; write up describes &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bachelor of Engineering as knowledge oriented&lt;/li&gt;&#xA;&lt;li&gt;Bachelor of Technology as skill oriented&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In USA some engineering schools offer two engineering tracks per discipline. The traditional engineering track is more theory base content and is referred to as BSc Engineering. On the contrary the technology track offers more hand on and has more laboratory work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is best you look at syllabus for both traditional and technology track programs from a few recognized programs. Here are a few for you to get started.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.purdue.edu/ECE/Academics/Undergraduates/UGO/pdf/BSEE%20Sample%20Plan.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purdue BS EE Plan of Study&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://tech.purdue.edu/sites/default/files/ECET-fall-2011.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purdue BS EET Plan of Study&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.purdue.edu/ME/Academics/Undergraduate/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purdue BS ME Plan of Study&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://tech.purdue.edu/new-albany/degrees/mechanical-engineering-technology/plan-of-study&quot; rel=&quot;nofollow noreferrer&quot;&gt;Purdue BS MET Plan of Study&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://engineering.siu.edu/elec/undergraduate/undergraduate-courses.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Southern Illinois University at Carbondale EE courses&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://engineering.siu.edu/tech/undergraduate/engineering-technology/courses.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Southern Illinois University at Carbondale EET courses&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you follow an ABET accredited program then you can take both Fundamentals in Engineering exam and Professional Engineer exam. If you pass both the exams per the requirements you can obtain Professional Engineer designation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both traditional and technology track are equally recognized. In my experience most students with technology based concentration find work in the area of Product Design, Product Development, Product Testing and Product Service in the respective discipline. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some references that have a broader and in-depth explanation. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Engineer_in_Training&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineer in Training&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.quora.com/What-is-the-difference-between-a-BTech-and-a-BSc&quot; rel=&quot;nofollow noreferrer&quot;&gt;What is the difference between a BTech and a BSc?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.quora.com/What-is-the-difference-between-a-Bachelor-of-Engineering-and-a-Bachelor-of-Technology&quot; rel=&quot;nofollow noreferrer&quot;&gt;What is the difference between a Bachelor of Engineering and a Bachelor of Technology?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/405/what-are-the-major-differences-between-engineering-degrees-and-engineering-techn?lq=1&quot;&gt;What are the major differences between Engineering degrees and Engineering Technology degrees in terms of employability?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-05-27T21:30:48.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9018" PostTypeId="1" AcceptedAnswerId="9025" CreationDate="2016-05-25T12:49:20.793" Score="3" ViewCount="1540" Body="&lt;p&gt;I am designing a product that requires internal mixing of a sealed container. If you need to visualize the container, imagine some quasi-oil drum that holds about 20&amp;nbsp;L of fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking to use as little electrical input as necessary so I will be using a manual hand turner with some sort of paddles attached to mix the liquid appropriately. However, the seal on the joint or input valve connecting the manual crank to the tank &lt;strong&gt;must&lt;/strong&gt; be as airtight as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I make the hinge connecting the crank to the tank as airtight as possible?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6582" LastEditorUserId="368" LastEditDate="2016-05-25T16:32:29.137" LastActivityDate="2016-05-25T20:18:40.273" Title="How can I make a rotating hinge airtight?" Tags="|mechanical-engineering|seals|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9019" PostTypeId="2" ParentId="9000" CreationDate="2016-05-25T13:00:12.080" Score="1" Body="&lt;p&gt;I'm not sure what the interpretation of your deformation gradient is. To my knowledge the deformation gradient gives the local deformation, i.e. the deformation around a certain point. To obtain this deformation, local information (i.e. information in one point only) is not enough, but you also require some information on the surroundings. The formula I know is&#xA;$$&#xA;F_{ij}=\frac{\partial x_i}{\partial X_j}&#xA;$$&#xA;with $F_{ij}$ being the deformation gradient, and $x_i$ and $X_j$ are local coordinates in the deformed and undeformed configuration, respectively. Thus to calculate the continuous $F_{ij}$, you need the differentials of the local coordinates $\partial x_i$ and $\partial X_j$. I'm no expert on discrete meshes, but one possibility is to calculate $F_{ij}$ for each element in your mesh by using finite differences $\Delta x_i$ and $\Delta X_j$. If unsure you should probably ask again more precisely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your finite strain tensor calculation seems correct to me. Note that when the deformation gradient is calculated for an element, the finite strain tensor is also for that element. Again I'm no expert on the interpretation of the results. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Finite_strain_theory#Physical_interpretation_of_the_finite_strain_tensor&quot; rel=&quot;nofollow&quot;&gt;wikipedia article on finite strain theory&lt;/a&gt; seems to have some answers for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When calculating in curvilinear coordinate systems, things usually become a bit more complicated than in cartesian coordinates. However, since cylindrical coordinates are locally cartesian, your calculation is fine. For more complex curvilinear coordinate systems you would need to evaluate your equations using co- and contravariant bases.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2016-05-25T13:00:12.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9021" PostTypeId="2" ParentId="8328" CreationDate="2016-05-25T15:03:15.250" Score="0" Body="&lt;p&gt;When you want to work with non-dimensional variables you should also look at the boundary and initial conditions of the problem. That way you can check whether non-dimensional groups appear in the boundary conditions as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That say, to make an equation non-dimensional means to replace dimensional variables like $C$ with non-dimensional ones like $\Theta=\frac{C}{C_0}$. Now, when making the equation non-dimensional you may choose your quantity $C_0$ to be anything at all as long as it has the same units as $C$. There are many ways to make an equation non-dimensional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you normalize or scale an equation (I am assuming scaling in my vocabulary is what you mean), you pick $C_0$ so that the new non-dimensional variable has a domain from 0 to 1. This is often accomplished in mass transport problems by also subtracting by a reference value: $\Theta = \frac{C-C_R}{C_0-C_R}$. You decide on the variables $C_0$ and $C_R$ so that your new variable $\Theta$ stays in a range $0 \leq \Theta \leq 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scaling (or normalizing) has the advantage that it is more likely to reduce the number of parameters in your problem. In addition, it has the advantage of making your derivatives also scaled (hopefully, and with some care). That means that your derivatives will also have order of magnitude one. If this is accomplished (sometimes not trivial), any non-dimensional groups can be probed to see if they are much larger or much smaller than one to verify their importance to the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good source for information on non-dimensional and scale techniques to solve transport problems is Analysis of Transport Phenomena by Deen.&lt;/p&gt;&#xA;" OwnerUserId="1031" LastActivityDate="2016-05-25T15:03:15.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9022" PostTypeId="2" ParentId="9018" CreationDate="2016-05-25T15:29:12.037" Score="4" Body="&lt;p&gt;If the torque requirement isn't too high you can use a magnetic coupling between the outside crank and the inside paddles, like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_stirrer&quot; rel=&quot;nofollow&quot;&gt;magnetic stirrer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2016-05-25T15:29:12.037" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9023" PostTypeId="2" ParentId="5653" CreationDate="2016-05-25T15:42:25.543" Score="1" Body="&lt;p&gt;The way you have solved your problem you have treated the concentration at the surface of the sphere as known ($y_{B,\text{surf}}$). Notice that in your final answer, if you plug in $r=r_\text{sphere}$ all you will get is $y_{B,\text{surf}}$. Instead, your boundary condition at the surface should be something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;N_{B,r}=-K_1P_B^{0.5}=-K_1y_B^{0.5}P^{0.5}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you are equating the flux at the surface of the catalyst particle (where the reaction is happening) to the reaction rate. Rearranging you can write that at $r=r_\text{sphere}$, $y_{B,\text{surf}}$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left(\frac{N_{B,r}}{-K_1P^{0.5}}\right)^{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you could solve the problem to find the value of $N_{B,r}$ which is constant at steady state according to your equations. You may get a transcendental equation of $N_{B,r}$ that requires numerical or graphical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One caveat, this is all based on a film-model of mass transport and heterogeneous reaction. It means that you will need some experimental data to correlate the rate of reaction to the thickness of the film model, $\delta$.&lt;/p&gt;&#xA;" OwnerUserId="1031" LastActivityDate="2016-05-25T15:42:25.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9024" PostTypeId="1" CreationDate="2016-05-25T19:05:51.527" Score="2" ViewCount="1180" Body="&lt;p&gt;I am trying to solve for the deflection $w$ of a rectangular membrane with tension $T$ acting on two opposite edges. These two edges are fixed i.e. $w = 0$. The other two opposite edges are free. The size of the membrane is $L$x$L$. There is a pressure $P$ acting on an area $\frac{L}{3}$x$\frac{L}{3}$ at the centre of the membrane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A schematic is presented below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NZAJt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NZAJt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tension was acting on four edges and they were fixed. The equations and boundary conditions could be written as: &#xA;$$T\frac{\partial^2w}{\partial x^2}+T\frac{\partial^2w}{\partial y^2} = p$$&#xA;with boundary conditions:&#xA;$$w(x=0) = w(x=L) = w(y=0) = w(y=L) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know how to modify these equations and boundary conditions for my problem.&lt;/p&gt;&#xA;" OwnerUserId="6585" LastEditorUserId="1832" LastEditDate="2016-05-25T19:25:16.413" LastActivityDate="2016-05-31T17:40:18.853" Title="Solid mechanics: Deflection of a rectangular elastic membrane" Tags="|mechanical-engineering|applied-mechanics|beam|deformation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9025" PostTypeId="2" ParentId="9018" CreationDate="2016-05-25T20:18:40.273" Score="4" Body="&lt;p&gt;What you're looking for is a mechanical face seal, which are used to allow rotating shafts to penetrate into a sealed volume such as a pump, mixer, compressor etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important thing to remember about these seals is that all such seals leak to some degree. The exact seal you need will be driven by your specific requirements such as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How much leakage of the process fluids can you tolerate?&lt;/li&gt;&#xA;&lt;li&gt;What is the process fluid?&lt;/li&gt;&#xA;&lt;li&gt;What are the pressures and temperatures?&lt;/li&gt;&#xA;&lt;li&gt;Operating speed&lt;/li&gt;&#xA;&lt;li&gt;Budget&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the higher end are engineered and stock mixer seals and cartridge seals like these from &lt;a href=&quot;https://www.flowserve.com/Products/Seals/Mixer&quot; rel=&quot;nofollow&quot;&gt;Flowserve&lt;/a&gt;. Those can easily run into the tens of thousands of dollars depending on your specific requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also less expensive OEM seals designed for things like swimming pool pumps. Such as &lt;a href=&quot;https://www.flowserve.com/Products/Seals/OEM-%26-Special-Duty/ci.PacSeal%2Cen_US.standard--------&quot; rel=&quot;nofollow&quot;&gt;these, also from Flowserve&lt;/a&gt;. Again, depending on the specifics of your application you might get away with a sub $100 solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the low end of the scale is simply using one or more radial lip seals that ride on the shaft, usually on either side of the bearing for the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also eliminate the need for a seal completely by using a magnetic drive mixer like these from &lt;a href=&quot;http://www.magnasafe.com/&quot; rel=&quot;nofollow&quot;&gt;Magnasafe&lt;/a&gt;. I have never seen a hand operated model though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can give you further guidance, but not without more detail from your end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your project involves any significant gas pressure, I strongly suggest that you following the ASME code for pressure vessels or contract with someone who is qualified and incorporating safety features to limit the possibility that the vessel could become over pressured and turn into a bomb. People continually underestimate the energy contained in relatively small volumes, like 20l, of pressurized gas and the consequences can be lethal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclosure: I used to work as an product development engineer in the Seals division of Flowserve. I cited their literature because I'm more familiar with their products. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2016-05-25T20:18:40.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9026" PostTypeId="1" AcceptedAnswerId="25335" CreationDate="2016-05-25T21:43:58.773" Score="5" ViewCount="193" Body="&lt;p&gt;How come axial compressors always narrow and turbines always widen towards the rear? From what I know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The speed of sound of air increases with temperature, therefore increases (decreases) as it goes through the compressor (turbine).&lt;/li&gt;&#xA;&lt;li&gt;The compressor and turbine blades at each stage should be meeting the air at just below the local speed of sound (more precisely critical Mach number).&lt;/li&gt;&#xA;&lt;li&gt;The rotational speed of all stages of a given spool is constant.&lt;/li&gt;&#xA;&lt;li&gt;The cross-section through which the air is flowing is &lt;span class=&quot;math-container&quot;&gt;$\pi(r_e^2-r_i^2)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$r_e$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$r_i$&lt;/span&gt; are the exterior and interior radii, respectively.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So shouldn't compressors widen and turbines narrow towards the rear, with the interior varying more than the exterior? And shouldn't turbines be much larger (~2x) than the compressors they drive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can obviously see civil engines don't look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RR Trent 1000 (Courtesy of aviationgazette)&#xA;&lt;img src=&quot;https://www.aviationgazette.com/wp-content/uploads/2016/03/4-trent-1000-design.jpg&quot; alt=&quot;RR Trent 1000&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GE GEnX (Courtesy of www.mikejamesmedia.com)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/886Hb.jpg&quot; alt=&quot;GE GEnX&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything I'm not taking, or taking too much, into account in my analysis?&lt;/p&gt;&#xA;" OwnerUserId="6589" LastEditorUserId="6767" LastEditDate="2019-01-01T15:56:45.960" LastActivityDate="2019-01-02T00:17:44.037" Title="Turbocompressor spool geometry" Tags="|geometry|turbomachinery|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9028" PostTypeId="2" ParentId="9026" CreationDate="2016-05-26T00:14:04.463" Score="3" Body="&lt;p&gt;The compressor is compressing the gas, so the same amount of air (on a mass or molar basis) occupies less volume.  If the goal is to keep the velocity approximately constant, the cross sectional area has to get smaller.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2016-05-26T00:14:04.463" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9029" PostTypeId="1" CreationDate="2016-05-24T16:36:00.257" Score="1" ViewCount="32" Body="&lt;p&gt;The following turboprop arrangement is given (see picture below). It has a single shaft, and all stages have been numbered (from &amp;quot;a&amp;quot; to &amp;quot;8&amp;quot;).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Uehnp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uehnp.jpg&quot; alt=&quot;Turboprop-Single-Shaft&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;PROBLEM:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I am asked to express the specific thrust in terms of &lt;span class=&quot;math-container&quot;&gt;$\theta\left(\equiv\dfrac{T_{03}}{T_a}\right)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pi_c\left(\equiv\dfrac{p_{03}}{p_{02}}\right)$&lt;/span&gt;, and the ambient conditions (pressure &lt;span class=&quot;math-container&quot;&gt;$p_a$&lt;/span&gt;, temperature &lt;span class=&quot;math-container&quot;&gt;$T_a$&lt;/span&gt;, flight speed &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;INFORMATION/DATA AVAILABLE FOR THIS PROBLEM&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Nozzle is choked (&lt;span class=&quot;math-container&quot;&gt;$p_8=p_a$&lt;/span&gt;) and it develops NO thrust (&lt;span class=&quot;math-container&quot;&gt;$u_8=U$&lt;/span&gt;, that is, the gas velocity at the exit matches the flight speed, &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Flight speed, &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Turboprop components are &lt;em&gt;NOT IDEAL&lt;/em&gt;, therefore all efficiencies are given: shaft, &lt;strong&gt;t&lt;/strong&gt;urbine, &lt;strong&gt;c&lt;/strong&gt;ompressor, &lt;strong&gt;g&lt;/strong&gt;earbox, &lt;strong&gt;p&lt;/strong&gt;ropeller, &lt;strong&gt;b&lt;/strong&gt;urner and &lt;strong&gt;d&lt;/strong&gt;iffuser (&lt;span class=&quot;math-container&quot;&gt;$\eta_{m}$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$\eta_{t}$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$\eta_{c}$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$\eta_{g}$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$\eta_{p}$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$\eta_{b}$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$\eta_{d}$&lt;/span&gt;, respectively).&lt;/li&gt;&#xA;&lt;li&gt;Ideal gas (&lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$c_p$&lt;/span&gt; are constant throughout the main engine core).&lt;/li&gt;&#xA;&lt;li&gt;Fuel heating value is known (&lt;span class=&quot;math-container&quot;&gt;$Q_r$&lt;/span&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Propeller work coefficient is known (&lt;span class=&quot;math-container&quot;&gt;$C_{wP}$&lt;/span&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Ambients conditions are known (&lt;span class=&quot;math-container&quot;&gt;$T_a$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$p_a$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;MY ATTEMPT&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$p_8=p_a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u_8=U$&lt;/span&gt;, the total thrust developed is: &lt;span class=&quot;math-container&quot;&gt;$$F=\dot{m_f}U+F_{p}$$&lt;/span&gt; where:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\dot{m_f}$&lt;/span&gt; is the fuel mass flow rate.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$F_{p}$&lt;/span&gt; is the thrust developed by the propeller.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I am being asked for the specific thrust, that is: &lt;span class=&quot;math-container&quot;&gt;$$F_s\equiv\dfrac{F}{\dot{m}}$$&lt;/span&gt;&lt;span class=&quot;math-container&quot;&gt;$$\boxed{F_s=f\cdot U+\dfrac{F_{p}}{\dot{m}}}$$&lt;/span&gt; where we have introduced a parameter: &lt;span class=&quot;math-container&quot;&gt;$f\equiv \dot{m_f}/\dot{m}$&lt;/span&gt;, known as &lt;em&gt;fuel ratio&lt;/em&gt;. Note that the fuel ratio here is the only unknown quantity, since &lt;span class=&quot;math-container&quot;&gt;$$\dfrac{F_{p}}{\dot{m}}=\dfrac{\frac1U \dot{m}c_pT_a\eta_pC_{wP}}{\dot{m}}=\dfrac{c_pT_a\eta_pC_{wP}}{U}$$&lt;/span&gt; is all known from the data available.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, we need to come up with an expression that relates the fuel ratio, &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt;, with &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pi_c$&lt;/span&gt; (and with the ambien conditions, of course). Remember again that &lt;span class=&quot;math-container&quot;&gt;$\theta\equiv\dfrac{T_{03}}{T_a}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\pi_c\equiv\dfrac{p_{03}}{p_{02}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;To achieve this, I decide to use the energy balance on the burner, to find that:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\boxed{f=\dfrac{h_{04}-h_{03}}{\eta_bQ_r-h_{04}}=\dfrac{c_p\left(T_{04}-\theta T_a\right)}{\eta_bQ_r-c_pT_{04}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In this last expression, all of the parameters are known &lt;strong&gt;except&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$T_{04}$&lt;/span&gt;. Note that this temperature, &lt;em&gt;known as the stagnation temperature at the turbine inlet&lt;/em&gt;, is usually a design parameter (and is &lt;em&gt;usually&lt;/em&gt; allowed to be used as a parameter in the resulting expression of &lt;span class=&quot;math-container&quot;&gt;$F_s$&lt;/span&gt;). However this is not the case for the present problem, in which we will need to write &lt;span class=&quot;math-container&quot;&gt;$T_{04}$&lt;/span&gt; in terms of (again) &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pi_c$&lt;/span&gt; and the ambient conditions.&lt;/p&gt;&#xA;&lt;p&gt;So I proceed to apply the energy balance on the shaft, to relate powers: &lt;span class=&quot;math-container&quot;&gt;$$\dot{W_T}=-\dfrac{1}{\eta_m}\dot{W_C}-\dfrac{1}{\eta_m \eta_g}\dot{W_P}$$&lt;/span&gt; From the definition of the propeller efficiency, we know that: &lt;span class=&quot;math-container&quot;&gt;$\dot{W_P}=\eta_p\dot{W_P}^{\text{ideal}}=\eta_p\dot{m}c_pT_aC_{wP}$&lt;/span&gt;, and therefore this term is known from the given data. So we can say that: &lt;span class=&quot;math-container&quot;&gt;$$h_{04}-h_{05}=\dfrac{1}{\eta_m}\left(h_{03}-h_{02}\right)+\dfrac{\eta_p}{\eta_m \eta_g}c_pC_{wP}\dot{m}T_a$$&lt;/span&gt; Since the diffuser is considered to be adiabatic, we can rest assured that: &lt;span class=&quot;math-container&quot;&gt;$T_{02}/T_a=1+U^2/\left(2c_pT_a\right)$&lt;/span&gt;, a known quantity. This allows us to algebraically manipulate the former equation to obtain: &lt;span class=&quot;math-container&quot;&gt;$$\boxed{T_{04}=c_pT_a\left[\dfrac{T_{05}}{T_a}+\dfrac{1}{\eta_m}\left(\theta-\dfrac{U^2}{2c_pT_a}-1\right)+\dfrac{\eta_p}{\eta_m \eta_g}c_pC_{wP}\dot{m} \right]}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Again, a new parameter arises: &lt;span class=&quot;math-container&quot;&gt;$T_{05}$&lt;/span&gt;. Remember that this is the total temperature at the nozzle inlet (or the turbine exit). But since the nozzle is adiabatic, we can state that: &lt;span class=&quot;math-container&quot;&gt;$$T_{05}=T_{08}\Longrightarrow T_{05}=T_8\left(1+\dfrac{\gamma-1}{2}M_8^2\right)=\ldots=T_8\left(1+\dfrac{U^2}{2c_pT_8}\right)$$&lt;/span&gt; Therefore: &lt;span class=&quot;math-container&quot;&gt;$$\boxed{T_{05}=T_8+\dfrac{U^2}{2c_p}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here is where I stop. Note that &lt;span class=&quot;math-container&quot;&gt;$T_8$&lt;/span&gt; is unknown, and needs to be related with &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\pi_c$&lt;/span&gt; and the triad &lt;span class=&quot;math-container&quot;&gt;$\{T_a, p_a, U\}$&lt;/span&gt;. It is worth noting that the parameter &lt;span class=&quot;math-container&quot;&gt;$\pi_c$&lt;/span&gt; hasn't been used yet, which suggests me to follow a reverse flow-path approach.&lt;/p&gt;&#xA;&lt;p&gt;So: how do I find &lt;span class=&quot;math-container&quot;&gt;$T_8$&lt;/span&gt; here (temperature at the nozzle exit), and how do I account for the remaining design parameter, &lt;span class=&quot;math-container&quot;&gt;$\pi_c$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerDisplayName="Jose Lopez" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-01-28T22:04:38.443" Title="Turboprop: Specific thrust as a function of two independent parameters" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9030" PostTypeId="2" ParentId="8121" CreationDate="2016-05-26T04:08:34.243" Score="1" Body="&lt;p&gt;Apart from current answers, I think a quality engineer must also  be aware of the FMEA processes (DFMEA as well as PFMEA in general)... In fact , for a Mfcg company, (if it is not designing the product), PFMEA (along with the relevant other processes like control plans) are essential too.... SQC is also  important.... If inspection is being done on sampling basis, the logic of deciding sampling plans must also  be understood...&#xA;(source; personal experience in a automobile design and mfcg company)&lt;/p&gt;&#xA;" OwnerUserId="6592" LastActivityDate="2016-05-26T04:08:34.243" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9032" PostTypeId="1" AcceptedAnswerId="9040" CreationDate="2016-05-26T10:32:13.610" Score="1" ViewCount="320" Body="&lt;p&gt;How is the armature in an electric motor energized (i.e. main conditions on signal as function of position)? Let's consider the following parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$n_a$,$n_f$ the number of armature and field pole pairs, respectively&lt;/li&gt;&#xA;&lt;li&gt;$\gamma_{a,k}={\pi k\over n_a}$, $\gamma_{f,k}={\pi k\over n_f}$ the position of armature and field pole $k$ ($k\in \{1,\ldots,2n\}$)&lt;/li&gt;&#xA;&lt;li&gt;$\gamma$, the angular position of the shaft&lt;/li&gt;&#xA;&lt;li&gt;$I(\gamma)$, the current in armature as a function of position ($\forall\:\theta\in\mathbf{R}\::\:I(\theta+\pi)=-I(\theta)$)&lt;/li&gt;&#xA;&lt;li&gt;$I_k(\gamma)=I(n_f×(\gamma-\gamma_k))$, the current in armature pole $k$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The most important question is: should $I_k(\gamma_k)=0$ (i.e. should the current commutate whenever a field pole passes an armature pole)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I created the gif below to illustrate. The vectors represent the magnetic moments: blue is armature, yellow is field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aEoHU.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aEoHU.gif&quot; alt=&quot;Abstract motor model&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6589" LastEditorUserId="6589" LastEditDate="2016-05-26T11:15:22.127" LastActivityDate="2016-05-27T03:24:00.297" Title="Energizing of armature as function of position in electric motor" Tags="|electrical-engineering|motors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9033" PostTypeId="2" ParentId="9032" CreationDate="2016-05-26T12:52:15.253" Score="0" Body="&lt;p&gt;Most commonly (in induction motors) it's the other way around, the armature is energized in a arbitrary frequency coming straight from the 3-phase mains power and the rotor can do nothing but follow it along. This is done with 3 sets of coils in the stator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This fixes the rotation speed to a multiple of the mains frequency. If you need a different rotation speed you can add a gearing reduction as needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in a universal engine the rotor is energized using commutator. The field created by the stator follows the input current (stays constant if DC) and the rotor will make its own field offset to that field to create torque.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2016-05-26T12:52:15.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9034" PostTypeId="2" ParentId="8957" CreationDate="2016-05-26T13:37:57.543" Score="1" Body="&lt;p&gt;AutoCAD has a very nice ability to customize the user interface. Essentially you build a custom tab with the commands that you would like in your own configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per comments: &lt;a href=&quot;https://knowledge.autodesk.com/support/autocad/learn-explore/caas/CloudHelp/cloudhelp/2016/ENU/AutoCAD-Core/files/GUID-27573791-DCEE-4751-A18A-7E54E773BEF6-htm.html&quot; rel=&quot;nofollow&quot;&gt;https://knowledge.autodesk.com/support/autocad/learn-explore/caas/CloudHelp/cloudhelp/2016/ENU/AutoCAD-Core/files/GUID-27573791-DCEE-4751-A18A-7E54E773BEF6-htm.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="205" LastEditorUserId="205" LastEditDate="2016-05-26T17:43:14.560" LastActivityDate="2016-05-26T17:43:14.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9035" PostTypeId="2" ParentId="9024" CreationDate="2016-05-26T14:25:01.280" Score="0" Body="&lt;p&gt;For the sake of a solution:&lt;br&gt;&#xA;$w$ can be considered a function only of $x$ since two sides are bound. Therefore ${\partial w\over\partial y}=0$ therefore&#xA; ${\partial^2w\over\partial y^2}=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have three regions ($[0;{L\over3}],[{L\over3};{2L\over3}],[{2L\over3};L]$, called $R_1$,$R_2$ and $R_3$):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x\in R_1\implies P=0 \\&#xA;\implies {\partial^2w_1\over\partial x^2}(x)=0\implies{\partial w_1\over\partial x}(x)=A_1\\ \implies w_1(x)=A_1x+B_1$&lt;/li&gt;&#xA;&lt;li&gt;$x\in R_2\implies P\ne 0 \\&#xA;\implies {\partial^2w_2\over\partial x^2}(x)={P\over T}\implies{\partial w_2\over\partial x}(x)={P\over T}x+B_2\\ \implies w_2(x)={P\over2T}x^2+B_2x+C$&lt;/li&gt;&#xA;&lt;li&gt;$x\in R_3\implies P=0 \\&#xA;\implies {\partial^2w_3\over\partial x^2}(x)=0\implies{\partial w_3\over\partial x}(x)=A_3\\ \implies w_3(x)=A_3x+B_3$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, you need to join the functions together (continuity, derivatives and function itself), which will take an enormous amount of time in this case (you're basically dealing with a spline here).&lt;/p&gt;&#xA;" OwnerUserId="6589" LastEditorUserId="6589" LastEditDate="2016-05-31T17:40:18.853" LastActivityDate="2016-05-31T17:40:18.853" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9037" PostTypeId="1" AcceptedAnswerId="9043" CreationDate="2016-05-26T18:31:47.350" Score="-1" ViewCount="573" Body="&lt;p&gt;I am asked to design a timing/timer pulley based on a timer belt which comes in standard size. How do I do the calculation to find the number of teeth, depth and width of groove etc?&lt;/p&gt;&#xA;" OwnerUserId="6278" LastEditorUserId="1832" LastEditDate="2016-05-26T22:13:23.613" LastActivityDate="2016-05-26T23:05:17.890" Title="How to design a timer/timing pulley?" Tags="|mechanical-engineering|design|gears|pulleys|" AnswerCount="1" CommentCount="1" ClosedDate="2016-05-27T14:59:28.310" ContentLicense="CC BY-SA 3.0" />
  <row Id="9038" PostTypeId="1" AcceptedAnswerId="9056" CreationDate="2016-05-26T19:20:03.217" Score="3" ViewCount="854" Body="&lt;p&gt;I have a frame, similar to a container frame, with two vessels inside. Each vessel is supported at four points on two beams, as shown in this mockup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LMlp4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LMlp4m.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using ANSYS to simulate the loaded frame. How should I represent the impact of the weight of the vessels on the frame so that my structural analysis is correct? Should I divide the total weight of the vessels by four and place that force on each supporting plate or use remote force to represent the weight of each vessel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know only the mass and center of gravity of vessels. I don't have full models of the vessels, just a model of the frame.&lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="282" LastEditDate="2016-05-27T19:26:38.273" LastActivityDate="2016-05-27T19:26:38.273" Title="How to represent a load supported at multiple points on multiple beams?" Tags="|mechanical-engineering|structural-engineering|structural-analysis|ansys|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9039" PostTypeId="1" AcceptedAnswerId="9041" CreationDate="2016-05-26T19:50:57.563" Score="2" ViewCount="936" Body="&lt;p&gt;I have heard of lasers being used to cut metal, however all laser cutting machines I've found can't cut metal, I'm guessing because it reflects the light?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is different in the designs of such machines between those that can and can't cut metal? Is it just accepted that energy is lost in reflection, and has a safe way of absorbing reflections?&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="282" LastEditDate="2016-05-27T19:28:24.860" LastActivityDate="2018-07-06T18:39:35.837" Title="How to handle reflections when cutting metal with a laser?" Tags="|lasers|cutting|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9040" PostTypeId="2" ParentId="9032" CreationDate="2016-05-26T19:54:49.427" Score="1" Body="&lt;p&gt;Synchronous motors can be operated with a VFD in the same way that induction motors are usually operated. To start the motor, the stator is energized at a very low frequency, 1 Hz or so. The motor is then accelerated by increasing the frequency at a rate that is limited by the torque with which it is desired to accelerate the load or the torque that the system has a the capacity to deliver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque can be more tightly controlled by measuring the torque angle and controlling the armature current to provide the desired torque. The torque angle is the angle between the rotor field and the stator field. To control torque, the phase of the armature current would be adjusted so that the moving position of the magnetic poles created by the armature current would lead (for positive torque) the moving position of the field poles as indicated by position encoder feedback. This is called a &quot;flux vector&quot; torque control strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an electric vehicle, for example, the &quot;gas pedal&quot; could provide a torque command. Pressing the pedal down would call for more torque and accelerate the vehicle. Lifting your foot from the pedal would call for less torque (or perhaps braking torque) and decelerate the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between the physical stator poles and the rotor poles is not relevant. The relevant thing is the angle between the rotating stator field and the rotor field.&lt;/p&gt;&#xA;" OwnerDisplayName="user6431" LastEditorDisplayName="user6431" LastEditDate="2016-05-27T03:24:00.297" LastActivityDate="2016-05-27T03:24:00.297" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9041" PostTypeId="2" ParentId="9039" CreationDate="2016-05-26T21:40:19.173" Score="7" Body="&lt;p&gt;A typical &quot;hobbyist laser cutter&quot; with about a 50W power rating won't get you very far cutting metal except for metal foil, but there is no big deal about cutting up to 1 inch thick mild steel plates or tubes with a few kilowatts of laser power. There are plenty of videos on YouTube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you need enough power to cut quickly before the (high) thermal conductivity of the metal heats up with work piece and causes distortion, cracking, locked-in stress concentrations, etc. Aluminium needs higher power to cut than steel, because of its higher thermal conductivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical thermal conductivity of hobbyist materials like wood, plastics etc is less than 1 W/m.K. Stainless steel about 15, high and low carbon steels about 50, aluminium about 200, and copper about 400.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The price tag for say a 5kW laser cutter for &quot;industrial-scale&quot; metal cutting is in the £100,000 to £500,000 range, so you won't find many for sale on Ebay.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-05-26T21:40:19.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9042" PostTypeId="1" CreationDate="2016-05-26T22:47:50.810" Score="0" ViewCount="10749" Body="&lt;p&gt;Here is the model (yellow line is 1ft) of the gantry crane I used to build the real-life model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Kye5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Kye5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a little wobbly now that I've built it. I haven't put the wheels on the ends yet. Can you think of any way to make it more stable and strong? I'm afraid of traversing a drop off with a load like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uxSOj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretend there are wheels and a chain hoist. For the moment, I've put some screws through with the intention of later drilling holes for some long bolts at yet-to-be-determined locations. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="6605" LastEditorUserId="1832" LastEditDate="2016-05-27T18:53:32.427" LastActivityDate="2021-08-27T19:59:13.593" Title="1 Ton Wooden Gantry Crane - Structural Advice" Tags="|structural-engineering|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9043" PostTypeId="2" ParentId="9037" CreationDate="2016-05-26T23:05:17.890" Score="0" Body="&lt;p&gt;There are two aspects to this problem: mechanical (number of teeth) and geometric (width and depth of groove).&lt;br&gt;&#xA;Solving the mechanical part  requires:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p$ the pitch (distance from the face of one tooth to the next) (probably mm)&lt;/li&gt;&#xA;&lt;li&gt;$v$ the speed of the belt (m/s)&lt;/li&gt;&#xA;&lt;li&gt;$f_i$ the required rotational speed of an element $i$ driven by the belt (Hz)&lt;/li&gt;&#xA;&lt;li&gt;$n_i$ the number of teeth on element $i$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This gives us the useful rule that, for any element $i$&#xA;$$2\pi r_i f_i = v$$&#xA;from which we deduce that&#xA;$$r_i = {v\over2\pi f_i}$$&#xA;and since $r_i = p n_i$, we can deduce that&#xA;$$n_i = {r_i\over p}={v\over2\pi p f_i}$$&#xA;so we already have the radius and number of teeth for each element $i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the geometric part requires knowing the shape of the tooth (round, square, etc.), which depends entirely on the belt.&lt;/p&gt;&#xA;" OwnerUserId="6589" LastActivityDate="2016-05-26T23:05:17.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9044" PostTypeId="2" ParentId="8874" CreationDate="2016-05-27T00:18:55.417" Score="1" Body="&lt;p&gt;The response of a system can be seen as a sum of it's stationary response and it's transient response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transient response disappears after some time. It has an oscilatory behavior, even if it is just an exponential (see complex Fourier series: every periodic function is a sum of exponentials).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;quarter wave decay&quot; we mean that, after doing Ziegler-Nichols' tuning, the transient response will (usually) decay four times faster. Thus, it's period is a quarter of the untuned response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: you increase the frequency of the transient response, also increasing the rate at which it's energy is dissipated. This makes the period of existence of the transient response a quarter of the original one. The downside is that this creates a high overshoot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ziegler-Nichols aims to reduce the time that it takes to reach the permanent (set point) response, while reducing the overshoot to 1/4 of the untunned response. This is done by adjustments in the proportional gains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this &lt;a href=&quot;https://en.wikipedia.org/wiki/File:PID_Compensation_Animated.gif&quot; rel=&quot;nofollow&quot;&gt;GIF in wikipedia&lt;/a&gt; that illustrates well that concept.&lt;/p&gt;&#xA;" OwnerUserId="6606" LastEditorUserId="6606" LastEditDate="2016-07-26T13:39:42.437" LastActivityDate="2016-07-26T13:39:42.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9045" PostTypeId="1" AcceptedAnswerId="9086" CreationDate="2016-05-27T03:28:25.240" Score="1" ViewCount="169" Body="&lt;p&gt;About this time of year in my region people often look at the air conditioning compressor pumping heat to the outside air on one side of their house while their swimming pool heater tries to pump heat into the water on the other side, and wonder why they're paying twice to move heat.  Inevitably someone thinks of submerging their compressor coils, but then they run the numbers and realize that it won't be too many days before their swimming pool is uncomfortably hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this leads to an interesting engineering question: What is the practical breakeven point between air-source and water-source heat pumps (for cooling indoor air)?  I.e.: Is it possible, due to its higher thermal conductivity, for a water sink &lt;em&gt;at a higher temperature&lt;/em&gt; to be more efficient for cooling than an air sink &lt;em&gt;at a lower temperature&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically: Suppose we have an outdoor compressor with surface area &lt;em&gt;x&lt;/em&gt; pumping &lt;em&gt;y&lt;/em&gt; BTUs of heat.  I am assuming that because water has something like 20 times the thermal conductivity of air that it would be a more efficient heat sink even when it's somewhat hotter than the air sink.  (Is that correct?)  But &lt;em&gt;how much more efficient&lt;/em&gt;?  I.e., at what temperature differential is air as efficient as water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Or is this not a practical question?  For example, we run fans to force more air through air-cooled compressors.  That takes energy, but maybe it raises the effective thermal conductivity to the point that it is breakeven with a convective water sink ... even including the energy to drive the fan?)&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="77" LastEditDate="2016-05-27T16:12:09.990" LastActivityDate="2016-05-29T20:00:55.557" Title="Can it be more efficient to pump heat into warmer water than cooler air?" Tags="|thermodynamics|heat-transfer|hvac|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9046" PostTypeId="1" CreationDate="2016-05-27T08:03:15.197" Score="-3" ViewCount="164" Body="&lt;p&gt;I want to build a drone that can carry around 7lbs. Is it possible to make a drone that can carry that much weight using electric motors without it being over /$500? &lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2016-05-29T02:48:12.573" LastActivityDate="2016-05-29T02:48:12.573" Title="diy drone w/ carrying capacity of 7lbs?" Tags="|mechanical-engineering|design|applied-mechanics|" AnswerCount="1" CommentCount="5" ClosedDate="2016-05-27T13:09:46.743" ContentLicense="CC BY-SA 3.0" />
  <row Id="9047" PostTypeId="1" CreationDate="2016-05-27T10:26:05.927" Score="1" ViewCount="896" Body="&lt;p&gt;I am thinking of a straight thermally isolated pipe, and incompressible flow through it. &#xA;Due to friction, pressure energy drop of fluid, (increase of internal energy), can be calculated via the Darcy–Weisbach equation. &#xA;If I look at the internal energy of fluid as the average velocity of molecular  movement, and pressure energy as impact of molecules on the pipe:&#xA;how come that one of these energies can raise and another drop, in the incompressible flow?&#xA;Because in an other situation where we start to apply external heat to a pipe, we expect both pressure and internal energy of the fluid to rise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pressure energy I am referring to $\frac{p}{\rho}$, and as thermal (internal) energy to $c_v T$. All for a mass of 1 kg.&#xA;By using the Kinetic Molecular Theory principles, I thought that I can visualise the process of flow as I mentioned: temperature increase - molecule speed increase, which leads to a higher pressure exerted on walls of pipe. &#xA;But in reality, as the pressure drops, the temperature increases which messes my approach.&#xA;Can I take in consideration that the density of any fluid (for example water), isn't constant (with drop of pressure and temperature rise, density lowers at some decimal), and therefore the pressure drops, even with temperature rise, which fits my visualising process, or is this all too much of trying to simplify the situation?&#xA;Any answer appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6611" LastEditorUserId="3633" LastEditDate="2016-06-30T14:02:44.640" LastActivityDate="2016-06-30T14:02:44.640" Title="How does pressure energy convert to thermal in incompressible flow?" Tags="|fluid-mechanics|thermodynamics|heat-transfer|energy|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9048" PostTypeId="1" CreationDate="2016-05-27T11:09:02.793" Score="0" ViewCount="32" Body="&lt;p&gt;I want to mount some equipment into a &lt;a href=&quot;https://en.wikipedia.org/wiki/Polyethylene&quot; rel=&quot;nofollow&quot;&gt;polyethylene&lt;/a&gt; (PE) shaft. I can't bolt or drill directly into the walls, but the supplier of the shaft can weld little PE blocks onto the wall: 50x50x50 mm, with bolts pressed into the blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have about 30kg worth of piping and valves etc, most of the weight will be borne by the ground actually but I need an additional attachment point higher up. My idea was to have two of these blockse mounted on the walls, with a steel rail between them paralell to a section of pipe to attach the pipe via one or two hangers. The shaft is 1 m across, the horizontal rail will be about 0,8m. I will have shear on the blocks but no moment. The wall thickness of the shaft is at least 10mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I don't need exact structural calculations, I need to know if this is feasible so i can leave it to the contructor to figure out, or if I need to spend more brainpower on the issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand there are plate welds and butt welds in the PE world? In either case, how much load can expect one my little blocks to bear?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-05-27T11:20:31.513" LastActivityDate="2016-05-27T11:20:31.513" Title="Help me guesstimate the loads I can apply to a little PE block welded to a PE wall" Tags="|mechanical-engineering|welds|polyethylene|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9049" PostTypeId="5" CreationDate="2016-05-27T11:09:41.150" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-05-27T11:09:41.150" LastActivityDate="2016-05-27T11:09:41.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9050" PostTypeId="4" CreationDate="2016-05-27T11:09:41.150" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-05-27T11:09:41.150" LastActivityDate="2016-05-27T11:09:41.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9051" PostTypeId="2" ParentId="9046" CreationDate="2016-05-27T11:30:58.863" Score="2" Body="&lt;p&gt;Yes it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wanna carry roughly 3.5 kg. Let's make the drone with payload 10kg heavy. For a quad rotor, with TWR of 2, I can get the 4 motors for 1200\$ (public price from T-motor U8 Pro), I don't know the price of the propellers, let's say 400\$. The whole makes 1kg. Adding 3 kg for the frame (I roughly scale down from a 25kg drone) and 2500\$ (here, I have no idea how much it would cost), we've got 2.5kg for Lithium Cells, it's roughly a 10Ah 12S battery, for 500\$. We may add 200\$ for electronics and sensors (and roughly no weight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've got your drone for less than 5000\$, flying 20 minutes. It excludes the cost of the time you'll spend in development, and also the cost of the prototypes you'll need to correct the approximations I made, unless you accept performances far from this first approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you plan to design your own propeller, you'll probably need 100000$ for CFD tools, prototypes, and measurement tools. If you want a cheap prototype, use off-the-shelf components. If you plan to build thousands of drones, then the prices above are irrelevant, and you should consider some serious R&amp;amp;D.&lt;/p&gt;&#xA;" OwnerUserId="2664" LastActivityDate="2016-05-27T11:30:58.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9052" PostTypeId="2" ParentId="8177" CreationDate="2016-05-27T11:42:30.707" Score="0" Body="&lt;p&gt;You can't just arbitrarily plug in the density of the air into the formula and say it's a pressure.  What did you learn about units?  Are the units of density and pressure the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of the pipe is irrelevant to calculating the flow velocity over a given length.  You do need to know the pressure at the inlet and the pressure at the outlet of the pipe.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-05-27T11:42:30.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9053" PostTypeId="2" ParentId="8637" CreationDate="2016-05-27T14:28:23.340" Score="1" Body="&lt;p&gt;The answer to your questions is that you calculate the DOF for a reactor with parallel reactions the same way you would any other process. Your problem is unrelated to the fact that you have parallel reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you have a different number of DOF for molecular species vs atomic species is because you have one less equation than you think for the atomic species balance. Remember that the equations resulting from your three atomic balances should be independent. If they are not, you have to count one less equation. The way to check for this (short of trying to solve the problem and failing) is to write the three atomic balance equations and check whether you can get one of them by algebraic manipulations of the other two. Let us label $n_1$ and $n_2$ as the input molar flow rate of diethyl ether (DE) and water (W), respectively; and $n_3$, $n_4$, $n_5$, and $n_6$ as the output molar flow rate of DE, W, ethylene (E), and methanol (M), respectively. Then the atomic balances read:&#xA;$$&#xA;\begin{align}&#xA;\text{(1) Oxygen: } &amp;amp; n_1+n_2=n_3+n_4+n_6 \\&#xA;\text{(2) Hydrogen: } &amp;amp; 6n_1+2n_2=6n_3+2n_2+4n_5+4n_6 \\&#xA;\text{(3) Carbon: } &amp;amp; 2n_1=2n_3+2n_5+n_6&#xA;\end{align}&#xA;$$&#xA;If you take two times equation (1) and add two times equation (3) you get equation (2) (verify this). This means that the set of equations is not independent and you can only count two atomic species in your balance instead of three. Both approaches then give you 1 DOF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, the selectivity should be defined as:&#xA;$$&#xA;\frac{\text{moles of desired product}}{\text{moles of undesired product}}&#xA;$$&#xA;which mean that it should not be defined with respect to the first reaction but with respect to the entire reactor.&lt;/p&gt;&#xA;" OwnerUserId="1031" LastEditorUserId="1031" LastEditDate="2016-05-28T20:09:44.523" LastActivityDate="2016-05-28T20:09:44.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9054" PostTypeId="1" CreationDate="2016-05-27T16:05:09.583" Score="2" ViewCount="194" Body="&lt;p&gt;Noise canceling headphones use a microphone to capture noise in order to cancel the same. The microphone will capture whatever acoustic signal reaches it. Since it is in the vicinity of the speaker, it must capture the sound coming from the speaker apart from the noise, particularly low frequency components. This should result in not only removal of noise but also of some portions of the signal. How it is ensured that only noise is removed?&lt;/p&gt;&#xA;" OwnerUserId="6614" LastEditorUserId="6614" LastEditDate="2016-06-27T01:54:21.273" LastActivityDate="2016-06-28T16:48:28.960" Title="Why noise canceling headphones do not attenuate the signal itself?" Tags="|sound-isolation|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9055" PostTypeId="1" CreationDate="2016-05-27T16:49:37.560" Score="3" ViewCount="38" Body="&lt;p&gt;I want to discuss the &lt;a href=&quot;https://www.researchgate.net/publication/260994815_Parametric_effects_on_kinetics_of_esterification_for_biodiesel_production_A_Taguchi_approach&quot; rel=&quot;nofollow&quot;&gt;effect of parameters on the kinetics of esterification for bio-diesel production&lt;/a&gt;. In esterification process, temperature effect and molar ration are two important parameters. During conversion of oliec acid with methanol to bio-diesel, temperature contributes more than molar ratio ( happen before 70 to 80 % conversion to bio-diesel). But after 70 to 80% conversion molar ratio contribute more than temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why this happens. Why the contribution of temperature and molar ratio reverse after 70 to 80% conversion to bio-diesel?&lt;/p&gt;&#xA;" OwnerUserId="6564" LastEditorUserId="1586" LastEditDate="2016-05-28T02:17:56.680" LastActivityDate="2016-05-28T02:17:56.680" Title="Parameter effects on kinetics of esterification for bio-diesel production" Tags="|thermodynamics|chemical-engineering|kinematics|renewable-energy|petroleum-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9056" PostTypeId="2" ParentId="9038" CreationDate="2016-05-27T19:23:29.883" Score="1" Body="&lt;p&gt;If the vessels (and the load distribution of whatever they're containing) are truly symmetric (or close enough), then yes, you can simply replace them with four equal concentrated loads at the points where they are supported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the vessels aren't symmetric but can be reasonably considered to have a pretty uniform stiffness (they don't have two really strong legs and two weak ones, for example), then you can use the vessels' centers of gravity to calculate the load distribution via this equation:&#xA;$$R = P\cdot\left(1-\dfrac{c_x}{L_x}\right)\cdot\left(1-\dfrac{c_y}{L_y}\right)$$&#xA;where $c_x$ and $c_y$ are the distance in the $x$ and $y$ axes, respectively, from the desired support to the center of gravity, and $L_x$ and $L_y$ are the spans between supports in the respective axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the center of gravity is perfectly centered between all supports, then $\dfrac{c_x}{L_x}=\dfrac{c_y}{L_y}=\dfrac{1}{2}$ and $R = \dfrac{P}{4}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, the center of gravity is at coordinate $\left(\frac{L}{3}, \frac{L}{3}\right)$ (so, close to the bottom-left support), then the load will be distributed $\frac{4}{9}$ at the bottom-left support, $\frac{2}{9}$ at the top-left and bottom-right supports, and only $\frac{1}{9}$ at the top-right support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the vessels are totally asymmetric, then you need to know the actual vessel geometry to calculate the load distributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simple assumption which is perfectly correct for static analyses. If you need to deal with vibrations or other dynamic behaviors, then there may be fluctuations on how much load goes to each support.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-05-27T19:23:29.883" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9057" PostTypeId="2" ParentId="9042" CreationDate="2016-05-27T19:30:45.297" Score="4" Body="&lt;p&gt;I have some experience in designing gantry cranes in steel (although not  timber) and I would say that for the sort of loads you are talking about this does seem a bit marginal, especially given that the material properties of a going length of timber are inherently less certain  than those of structural steel sections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot will depend on how the joints are made, again wood is that bit less predictable than welded or bolted joints in steel so you need extra margin of safety to accommodate this. But suffice to say that if it feels 'wobbly' this is not a good sign. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the situation in the second picture of your question is just bad. This imposes a huge bending load on the upright legs and makes the whole system inherently unstable. If you want to do this you need adjustable legs to keep the legs vertical and the beam horizontal. This sort of lean would not be acceptable for any sort of gantry crane, not matter how well constructed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sort of design the likely failure points are the joints between the beam and the legs and buckling of the legs themselves. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-27T19:30:45.297" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9058" PostTypeId="1" AcceptedAnswerId="9059" CreationDate="2016-05-27T19:41:43.003" Score="2" ViewCount="2244" Body="&lt;p&gt;I have a very large antique mirror with some dense heavy wood framing it.  I've stripped out some home electrical wire to use the copper ground wire as the cable that ties to the back of the mirror frame and holds it up on the wall.  My question is how can I tell if the copper wire is strong enough to rely on and is it possible it will snap given time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I search the internet for load bearing on copper wire it refers to the wrong kind of load i.e. electrical which is not what I want to know.&lt;/p&gt;&#xA;" OwnerUserId="6617" LastActivityDate="2016-05-27T22:17:01.737" Title="Is it safe to hold up a large heavy mirror with copper wire?" Tags="|structural-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9059" PostTypeId="2" ParentId="9058" CreationDate="2016-05-27T19:47:30.990" Score="8" Body="&lt;p&gt;Copper wire is not ideal as it is soft, has low tensile strength and  can &lt;a href=&quot;https://en.wikipedia.org/wiki/Creep_(deformation)&quot;&gt;creep&lt;/a&gt; under load over time and if you are reusing old electrical wire there is no way to know whether it has small cracks in it which could exacerbate this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best option is to get some picture hanging wire which is designed for the purpose and is cheaply and easily available from any hardware supplier. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-27T20:14:55.203" LastActivityDate="2016-05-27T20:14:55.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9060" PostTypeId="2" ParentId="9054" CreationDate="2016-05-27T20:23:32.573" Score="3" Body="&lt;p&gt;It's not clear what you are asking.  Noise cancelling headphones measure the noise to be cancelled, then actively produce the negative of that noise at just the right volume so that the two cancel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no issue of &quot;recording&quot; anything here.  The microphone data is used real time, with perhaps a tiny delay to account for the propagation of the waves it measured across the distance to the emitter.  The speed of sound is roughly 3 &amp;micro;s/mm, so even if the mic and speaker were 10 mm apart (a lot), then that would only be 30 &amp;micro;s of &quot;recording&quot; before the recorded information would be discarded.  In practise, this would only require a single sample anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &quot;attenuate the same&quot;, it is unclear what you are talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better answer requires a better question.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-05-27T20:23:32.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9061" PostTypeId="2" ParentId="9054" CreationDate="2016-05-27T20:49:34.733" Score="0" Body="&lt;p&gt;Consider that the noise is being 'cancelled' by your perception because of the speaker in your ear. Most headphones of this type also have some sort of material to absorb vibrations so that the microphone couldn't pick up any sound coming from the speaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the microphone could somehow collect sound from the speaker, the antiphase sound through the speaker would be out of phase with the original sound that was picked up by the microphone, and the microphone would then not pick up any background sound. The computer in the headphones would not produce any antiphase sounds, then the microphone would pick up the ambient noise again in the next iteration. This would simply result in very poor quality headphones that would probably hardly cut any ambient noise at all by human perception.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the speaker in any headphone that does not have some sort of noise absorption material (passive noise cancelling) is going to be far enough away from the microphone that the microphone couldn't pick up the sound. You shouldn't be listening to anything greater than 80dB for a prolonged period of time, and the inverse square law&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{I}{{I}'}= \left ( \frac{{d}'}{d}  \right )^{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shows pretty clearly that simply the distance between the speaker and the microphone isn't enough to reduce the dB at the microphone substantially.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Simply&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It would be terrible headphone design, so they don't make them that way. They use noise absorbing material such as padding on over-ear headphones and rubber on earbuds.&lt;/p&gt;&#xA;" OwnerUserId="6619" LastActivityDate="2016-05-27T20:49:34.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9062" PostTypeId="2" ParentId="9058" CreationDate="2016-05-27T22:17:01.737" Score="1" Body="&lt;p&gt;You can find a table of mechanical properties of various gages of copper and aluminum wiring here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.okonite.com/engineering/conductor-properties.html&quot; rel=&quot;nofollow&quot;&gt;Wire Mechanical Properties&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just have to Google the right terms.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-05-27T22:17:01.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9066" PostTypeId="1" AcceptedAnswerId="9074" CreationDate="2016-05-28T03:10:06.480" Score="4" ViewCount="6310" Body="&lt;p&gt;Inspired by &lt;a href=&quot;https://engineering.stackexchange.com/q/8923/3353&quot;&gt;this question&lt;/a&gt;, and also &lt;a href=&quot;https://engineering.stackexchange.com/a/8918/3353&quot;&gt;this answer&lt;/a&gt; to a separate question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two beam models, one with two independent spans (a full hinge), another is a continuous beam over both spans (only column is hinged), the below is the diagram of both beams, and their force/bending moment distribution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MXqfI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MXqfI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I derive from first principles of static analysis or from rules of thumb that continuous beams have a smaller deflection?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2019-12-28T17:56:42.333" Title="Why does a continuous beam have less deflection than a pair of simply supported beams?" Tags="|structural-engineering|beam|structural-analysis|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9069" PostTypeId="1" AcceptedAnswerId="9070" CreationDate="2016-05-28T12:19:31.523" Score="0" ViewCount="173" Body="&lt;p&gt;I have recently studied that the mmf produced by the stator in an electrical machine varies as a rectangular wave over the air gap periphery of the machine;&#xA;And to make the wave similar to a sinusoidal wave we use techniques such as distributing the winding,short pitching the coils or using a current sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is,Why should I go through this process?Why not leave it as a rectangular wave,Since these processes reduce the mmf peak?&lt;/p&gt;&#xA;" OwnerUserId="6493" LastActivityDate="2016-05-28T13:24:10.430" Title="MMF in AC windings" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9070" PostTypeId="2" ParentId="9069" CreationDate="2016-05-28T13:24:10.430" Score="1" Body="&lt;p&gt;You could actually leave the MMF rectangular (in fact &lt;a href=&quot;https://en.wikipedia.org/wiki/Brushless_DC_electric_motor&quot; rel=&quot;nofollow&quot;&gt;brushless DC motors&lt;/a&gt; are synchronous motors with a rectangular MMF) but there is a drawback: the torque induced by a rectangular MMF under steady conditions is much less constant. There is a torque ripple in the form of harmonics (the sinusoidal harmonics that are in the rectangular MMF).&lt;/p&gt;&#xA;" OwnerUserId="5806" LastActivityDate="2016-05-28T13:24:10.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9071" PostTypeId="1" CreationDate="2016-05-28T13:46:31.260" Score="1" ViewCount="603" Body="&lt;p&gt;I use the Maxon planetary gear #203115 for a project. Here are its characteristics (from the datasheet):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$n=12:1$&lt;/li&gt;&#xA;&lt;li&gt;3 stages&lt;/li&gt;&#xA;&lt;li&gt;$C_\max=7.5 \mathrm{Nm}$ &lt;/li&gt;&#xA;&lt;li&gt;$\eta_\max=\eta(C_\max)=0.81$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Maxon precises that the efficiency depends on the torque applied but is independent but is unaffected by the motor speed. They give a schematic curve:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7XUWW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7XUWW.png&quot; alt=&quot;efficiency_curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I assume that the losses come only from dry friction, I find:&#xA;$$\eta(C)=\frac{C-C_{frict}}{C}$$&#xA;$$\eta_\max=\frac{C_\max-C_{frict}}{C_\max} \Rightarrow C_{frict}=C_\max(1-\eta_\max)=1,425 \mathrm{Nm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately these equations give an unrealistically low efficiency for low applied torque:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Ktl6w.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ktl6w.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I model the efficiency more precisely ? &lt;/p&gt;&#xA;" OwnerUserId="5806" LastActivityDate="2016-05-28T13:46:31.260" Title="Planetary gear efficiency for a variable load" Tags="|gears|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9072" PostTypeId="1" CreationDate="2016-05-28T14:45:19.517" Score="1" ViewCount="635" Body="&lt;p&gt;I know piezo film has one of the lowest efficiency to date when it comes to converting mechanical energy (or stress in the context of piezoelectricity) and motor (magnet and coiled copper wire) seems to be quite efficient. But is there something more efficient than electromagnetic induction? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6627" LastEditorUserId="6628" LastEditDate="2016-05-30T13:24:52.373" LastActivityDate="2016-05-30T13:24:52.373" Title="Is there a more efficient transducer to convert mechanical energy into electrical energy than electromagnetic induction?" Tags="|mechanical-engineering|electrical-engineering|motors|power-electronics|" AnswerCount="1" CommentCount="2" ClosedDate="2016-05-29T21:19:38.123" ContentLicense="CC BY-SA 3.0" />
  <row Id="9073" PostTypeId="1" AcceptedAnswerId="9075" CreationDate="2016-05-28T16:24:08.417" Score="2" ViewCount="333" Body="&lt;p&gt;Why does positive phase and gain margins cause any system to be stable whereas negative phase and gain margins cause the system to be unstable ?&lt;/p&gt;&#xA;" OwnerUserId="6571" LastActivityDate="2016-06-01T14:32:17.737" Title="Condition for system stability" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9074" PostTypeId="2" ParentId="9066" CreationDate="2016-05-28T17:36:09.643" Score="10" Body="&lt;p&gt;For a simple visual demonstration, take one of the spans in your example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is fully hinged, then each span can be represented as a simply-supported beam. If the beam is continuous, then each span can be represented as fixed-and-pinned (or at least pinned and partially fixed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VLcdi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VLcdi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference between these two models is the moment reaction at the fixed support. Therefore, we can replace the fixed support with a pinned one and a concentrated bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pgXSI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pgXSI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This bending moment reaction is what causes the reduction of deflection in a continuous beam. It causes an uplift of the beam which counters some of the deflection due to the ordinary load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DDXKm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DDXKm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you want a more numeric explanation, look no further than the fact that a beam's bending moment is equal to the second derivative of its deflection. The negative bending moment caused by the beam's continuity therefore implies in a reduction of the beam's curvature and therefore, a reduction of its maximum deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;All diagrams obtained via &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T17:56:42.333" LastActivityDate="2019-12-28T17:56:42.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="9075" PostTypeId="2" ParentId="9073" CreationDate="2016-05-28T22:54:51.043" Score="3" Body="&lt;p&gt;This is linked to the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Nyquist_stability_criterion&quot; rel=&quot;nofollow&quot;&gt;Nyquist stability criterion&lt;/a&gt;, which comes down to the number of encirclements of the minus one point of the open-loop in the Nyquist diagram. Namely the number of unstable closed-loop poles will be equal to the number open-loop unstable poles minus the number of counterclockwise encirclements of the minus one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A negative phase margin means that you enter the unit disk, with the minus one point to your right (which is a part of a clockwise encirclement of the minus one point). You can only prevent an encirclement by decreasing in phase and go to the positive real axis outside the unit disk, to close the loop from the Nyquist contour, if you add unstable poles to the open-loop system. Adding this also still makes the closed-loop unstable. However the opposite does not have to be true, that positive phase margin means that the closed-loop will be stable. Namely when the open-loop system is unstable it is possible to have positive phase margin, while the closed-loop system is unstable. So positive phase margin is necessary but not sufficient for closed-loop stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if a negative gain margin has any practical meaning, because multiplying by a negative number also induces 180° of phase shift. So in order to identity such a margin you need to find a crossing of a multiple of 360°. You can get that the closed-loop will become unstable when you multiply the open-loop by a negative number, however the existence of a negative gain margin does not mean that the closed-loop will be unstable.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastEditorUserId="1765" LastEditDate="2016-06-01T14:32:17.737" LastActivityDate="2016-06-01T14:32:17.737" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="9076" PostTypeId="1" AcceptedAnswerId="9078" CreationDate="2016-05-29T12:50:13.717" Score="5" ViewCount="242" Body="&lt;p&gt;I'm building a APRS tracker for a high altitude balloon and am tracing the voltage as it's not powering up. I'm confused why it seems that this capacitor's ground appears to not be connected to anything. &lt;a href=&quot;https://i.stack.imgur.com/2Qdzt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;topside of board&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tTwld.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tTwld.jpg&quot; alt=&quot;underside of board&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6631" LastActivityDate="2016-05-29T14:13:40.527" Title="Why is capacitor 2 not grounded on this board?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9077" PostTypeId="2" ParentId="251" CreationDate="2016-05-29T12:55:11.687" Score="1" Body="&lt;p&gt;This question has no clear answer in its current form. The strip method is a lower bound approach (meaning that you approach the optimum solution from the safe side) based on beam analogy. So in theory (given endless ductility, etc.) you could arrange your strips in any way you want and you would always end up with a safe (but possibly horrible in all other aspects) solution. The recommendations given for the application of the strip method are to keep redistribution needs and service state-behavior in check as well as finding economical solutions. There's a lot of different recommendations in the strip method that vary between cases, corner angles for load dividers, moment ratios, widths of column strips vs mid-strips, widths of edge strips, reinforcement distributions within the strips, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to learn more about the strip method, and if you're going to apply it you really should, Hillerborg's book &quot;Strip Method Design Handbook&quot; is a good place to start. I personally find both the strip method and the strut and tie method to be very simple and elegant manifestations of applied engineering judgment.&lt;/p&gt;&#xA;" OwnerUserId="6523" LastActivityDate="2016-05-29T12:55:11.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9078" PostTypeId="2" ParentId="9076" CreationDate="2016-05-29T14:13:40.527" Score="5" Body="&lt;p&gt;Actually, your picture clearly shows the pad to be connected to the top plane.  Most likely that's ground.  With the top plane being ground, how else would you expect pads to be connected to ground?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can clearly see the connections between the pad and the plane at the 12, 3, 6, and 9 hour clock positions.  The reason the pad isn't connected to the plane all around is because that would make hand-soldering difficult.  With copper connected everywhere, the plane would act like a large heat sink, preventing the pad from getting hot enough for solder to flow properly.  These arrangements of isolated connections and gaps are often referred to as &lt;i&gt;thermals&lt;/i&gt; in PC board lingo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see such thermals on the pad immediately right of the one you marked, on the right-most oval pad at lower right, and on the right pad in the row of 7 at bottom.  All these pads are electrically connected to the top layer, which is largely a plane, most likely ground.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-05-29T14:13:40.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9079" PostTypeId="1" AcceptedAnswerId="9081" CreationDate="2016-05-29T15:59:54.010" Score="2" ViewCount="41" Body="&lt;p&gt;I once saw on TV a factory dipping a product into a tank of paint and it magically emerged with a crisp pattern. (It was an Archery bow, and the pattern was army-style camouflage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was I deceived or is it actually possible?&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="7083" LastEditDate="2018-01-28T22:18:16.703" LastActivityDate="2018-01-28T22:18:16.703" Title="Multicolor dip-painting techniques" Tags="|manufacturing-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9081" PostTypeId="2" ParentId="9079" CreationDate="2016-05-29T16:22:23.210" Score="2" Body="&lt;p&gt;This sounds like &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrographics_(printing)&quot; rel=&quot;nofollow&quot;&gt;hydrographic printing&lt;/a&gt;. In short a design is printed onto a water soluble film and floated in a container of water to apply it to a surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is generally used to apply complex and detailed patterns to curved and irregularly shaped surfaces as it conforms well to contours, especially as a cheaper and more repeatable alternative to hand painting. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-29T16:22:23.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9083" PostTypeId="1" AcceptedAnswerId="9085" CreationDate="2016-05-29T17:45:56.153" Score="2" ViewCount="1889" Body="&lt;p&gt;I would like to find out if it is possible to reduce the divergence of a red laser pointer beam from 40  degrees to 1 degree.       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Ruslan wrote on October 3 2013 that because the active zone of a red diode laser has diameter of order of several micrometers the divergence of a red laser pointer beam is currently 40 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to Heisenberg's uncertainty principle $\Delta\times\Delta p \gtrsim \hslash/2$, one can't really make a quantum have zero momentum in any direction. So you can't say that photons go in the same direction - this is just a simplified description of laser operation. In reality, the thinner the beam, the higher the divergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare e.g. a DPSS laser (e.g. green laser pointer) with a diode laser (e.g. a red laser pointer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a DPSS laser the active material will have diameter of order of hundreds of micrometer, and the exiting beam will start from even smaller diameter for various reasons. The divergence is quite small: if you remove the collimating lens, your light image from a green laser pointer will be several centimeters after the light goes several meters. Divergence angle would be $\lambda/d = 532\ \mathrm{nm}/100\ \mu\textrm{m} \approx 0.3°$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try doing the same with a red laser pointer, you'll see that its light diverges quite a lot: after going several centimeters in direction of propagation, it'll already give image of several centimeters. The reason for this is that active zone of diode laser has diameter of order of several micrometers. This makes output beam quite thin, making Δx small and thus Δp high, and this is what leads to high divergence. Divergence angle would be $\lambda/d = 640\ \mathrm{nm}/1\ \mu\textrm{m} \approx 40°$. Actual angle would depend on which transverse direction you select, because active zone is ∼10× longer in one direction than in another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the thicker your starting laser beam, the more collimated it is, so if you manage to make a (visible wavelength) laser with beam starting at 1cm thickness, you'll have almost perfectly collimated laser beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if there is a method to enlarge the active zone of a red diode laser from a diameter of several micrometers to 30 micrometers. Also, could a lens with short focal length be used to reduce the divergence of a red laser pointer beam from 40 degrees to 1 degree?&lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="6350" LastEditDate="2016-05-30T09:23:20.093" LastActivityDate="2016-05-30T16:15:56.820" Title="Is it possible to reduce the divergence of a red laser pointer beam from 40 degrees to 1 degree?" Tags="|optics|lasers|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9084" PostTypeId="2" ParentId="9072" CreationDate="2016-05-29T19:28:27.847" Score="1" Body="&lt;p&gt;In engineering, efficiency is defined as output power divided by input power. Weight, the number of components and other factors may be related to efficiency in some way, but efficiency as an end result is measured only by input and output power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most efficient machines for converting mechanical power to electrical power are a wound-field synchronous generators. They are used for automotive alternators, engine-generator sets and power station generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first automotive alternators that replaced DC generators in cars in about 1960 were rated at about 400 watts. Today's automotive alternators can produce 1000 watts or more. The maximum efficiency of automotive alternators is currently about 85%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large power station generators produce hundreds of millions of watts. Recently manufacturers began offering power station generators rated 2000 million watts. The largest synchronous generators have an efficiency of 98 or 99 percent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, efficiency increases as machine power rating increases because the physical size makes it easier and because the economic reward is larger for higher power machines. In general, synchronous generators will be 85 to 99 percent efficient with the efficiency being fairly consistently proportional to the power rating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Permanent-magnet synchronous generators using the latest magnetic materials can be more efficient than wound-field synchronous generators. They generally require a voltage regulating system that cuts into the efficiency gain, but they could still be more efficient at lower power ratings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other types of generators are generally less efficient, but they can have other advantages that cause them to be used in various situations.&lt;/p&gt;&#xA;" OwnerDisplayName="user6431" LastEditorDisplayName="user6431" LastEditDate="2016-05-29T19:54:49.803" LastActivityDate="2016-05-29T19:54:49.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9085" PostTypeId="2" ParentId="9083" CreationDate="2016-05-29T19:37:21.323" Score="3" Body="&lt;p&gt;Of course, a lens with a short focal length can be used. The easiest way is to take the standard lens that comes with most laser pointers and move it a bit closer to the laser crystal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the beam waist of the output beam to be exactly at the lens aperture, then you'll have to do a bit more work and find a suitable microlens. For a beam with 1 deg ($\theta=0.017$ rad) divergence, you'd need the beam to have a diameter $d\approx \lambda/\theta=37~\mathrm{\mu m}$ at the lens. The lens would be at a distance of 52 microns from the laser crystal. So that would be rather difficult to achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: clarification on beam waist and divergence. Here is what a Gaussian laser beam looks like this (image by DrBob on Wikipedia): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/53/GaussianBeamWaist.svg/500px-GaussianBeamWaist.svg.png&quot; alt=&quot;Gaussian beam&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a typical laser, the output coupler is at the narrowest part, the beam waist. The divergence angle $\Theta$ is a fixed parameter of the beam, no matter how far you propagate it. If you put a lens in the beam waist or within the Rayleigh length $z_R$, you can't decrease the far-field divergence angle $\Theta$, although you can focus it with a short focal length and achieve an even higher divergence after the focus. If you're not interested in the beam diameter near the beam waist, only in the spot size at 100 m away, then I wouldn't bother with microlenses and just use a regular lens at a few millimeters from the laser aperture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't comment on modifying the diameter of the laser crystal itself; if it is possible at all, it will require advanced semiconductor processing techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way: using Heisenberg to describe beam divergence as a function of beam diameter and wavelength doesn't make much sense, because Planck's constant cancels out once you write out $\Delta p$. The rule of thumb $\theta=\lambda/d$ works just as well for acoustic waves, which are not at all quantum mechanical.&lt;/p&gt;&#xA;" OwnerUserId="6350" LastEditorUserId="6350" LastEditDate="2016-05-30T06:53:59.163" LastActivityDate="2016-05-30T06:53:59.163" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9086" PostTypeId="2" ParentId="9045" CreationDate="2016-05-29T20:00:55.557" Score="2" Body="&lt;p&gt;The hot-side air coil of an AC installation (the part that heats the outdoor air) is a compromise between effectivity of heat transfer from the refrigerant to the air, the size, and the cost. Heat transfer to air typically comes with a significant thermal resistance, for two reasons. Firstly, air is a bad heat conductor. Secondly, air has a low heat capacity, so the air will heat up as it passes through the air coil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick Google didn't give me typical values of the effective thermal resistance of the air coil. If you have an IR thermometer, you could measure the temperature of the coil and compare it to the air temperature at the outdoor inlet. It wouldn't surprise me if the coil is 20 K higher in temperature than the outdoor air, assuming that the AC is running at full power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put a coil designed for air into water (assuming that corrosion etc. is not a problem), with some means of circulating the water, I think you can safely assume that the coil temperature will be very close to the water temperature. So, the break-even point would be when the water temperature is, say, 19 K higher than the air temperature.&lt;/p&gt;&#xA;" OwnerUserId="6350" LastActivityDate="2016-05-29T20:00:55.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9088" PostTypeId="2" ParentId="5967" CreationDate="2016-05-29T22:13:28.857" Score="1" Body="&lt;p&gt;Comments by @Mr.P nudged me to realize that there is a redistribution of equivalent loads due to losses which cannot be trivially encompassed by Lin's method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To demonstrate this, take the following simply-supported beam (ignore all concepts of units or scale here, this is a thought exercise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tE6mi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tE6mi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bending moment diagram solely due to prestress for this beam will be the following polygonal diagram (before any losses, assuming &lt;span class=&quot;math-container&quot;&gt;$P\cos\theta \approx P$&lt;/span&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0E1RB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0E1RB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to the diagram obtained by a concentrated vertical load at midspan equal to &lt;span class=&quot;math-container&quot;&gt;$F = 2P\sin\theta \approx 40$&lt;/span&gt;, which is balanced out by the two vertical loads at the supports, each equal to &lt;span class=&quot;math-container&quot;&gt;$P\sin\theta \approx 20$&lt;/span&gt; (in the opposite direction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, let us now consider friction losses. Let's assume they cause a 10% reduction to the stress in the tendon at midspan. This implies that the bending moment at midspan will also be reduced by 10%, and will therefore equal 90. However, the diagram's profile is no longer polygonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That can be easily observed by looking at the bending moment at any other point. In an isostatic structure, the bending moment is simply equal to &lt;span class=&quot;math-container&quot;&gt;$P \cdot e$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; is the distance between the cable and the centroid. Looking at quarter-span, the bending moment before losses was equal to &lt;span class=&quot;math-container&quot;&gt;$100\cdot0.5=50$&lt;/span&gt;, or exactly half of the mid-span moment. To calculate after losses, however, we need to calculate the force at this point. Simplifying things considerably, let's assume the loss here is half of that at midspan, so only 5%.&lt;sup&gt;1&lt;/sup&gt; In that case, the bending moment at quarter-span will be equal to &lt;span class=&quot;math-container&quot;&gt;$95\cdot0.5=47.5$&lt;/span&gt;, which is &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; equal to half of the mid-span moment of 90.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the bending moment diagram from support to midspan becomes of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M = Pe\left(\dfrac{2x}{L}-0.1\left(\dfrac{2x}{L}\right)^2\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; is the distance of the cable to the centroid at midspan, &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the span, and &lt;span class=&quot;math-container&quot;&gt;$0.1$&lt;/span&gt; represents the 10% loss of prestress at midspan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yZgr4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yZgr4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting the first and second derivatives of this equation gives us the equivalent concentrated load at midspan and the uniform load distributed along the entire span, respectively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;M' = Q &amp;amp;= Pe\left(\dfrac{2}{L} - 0.2\dfrac{4x}{L^2}\right) \\&#xA;Q\left(\dfrac{L}{2}\right) &amp;amp;= \dfrac{1.6Pe}{L} \therefore F = 2Q\left(\dfrac{L}{2}\right) = 32 \\&#xA;M'' = q &amp;amp;= \dfrac{0.8Pe}{L^2} = 0.8&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the equivalent loading which generates the correct (simplified) bending moment diagram after friction losses is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fhpli.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fhpli.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies in a new uniform load which did not appear before losses, a representation of the load redistribution described by @Mr.P. The equivalent concentrated load at midspan is also no longer equal to &lt;span class=&quot;math-container&quot;&gt;$F = 2P\sin\theta$&lt;/span&gt;, which would have resulted in &lt;span class=&quot;math-container&quot;&gt;$F \approx 36$&lt;/span&gt;. The prestress, however, is balanced, since &lt;span class=&quot;math-container&quot;&gt;$32 + 0.8\cdot10 = 40$&lt;/span&gt;, as expected.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The same reasoning can be applied to parabolic cables. For the following beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cwQKa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cwQKa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the uniform equivalent load is &lt;span class=&quot;math-container&quot;&gt;$q = \dfrac{8Pe}{L^2} = 2$&lt;/span&gt;, which generates a total upwards force of &lt;span class=&quot;math-container&quot;&gt;$2\cdot32=64$&lt;/span&gt;, to be cancelled out by the concentrated downwards forces of 32 at each support. The bending moment diagram before any losses is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ql0sn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ql0sn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, assuming once again a 10% loss due to friction at midspan, and that the friction-loss profile is linear, the bending moment diagram becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RMmyU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RMmyU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has the following cubic equation (with &lt;span class=&quot;math-container&quot;&gt;$x=0$&lt;/span&gt; at the midspan):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M = 2.00\cdot128\left(1-\left(\dfrac{2x}{L}\right)^2\right)\left(0.9+0.1\dfrac{2x}{L}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;getting the first and second derivatives, I can find the equivalent loading, which in this case is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vx4em.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vx4em.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given what we saw with the first example, it comes as no surprise that there's an increase in the distributed load near the supports. It also makes intuitive sense that the distributed load is reduced near the midspan. How that concentrated load at midspan comes to be, however, I have no idea. However, once again the prestress is self-balanced: &lt;span class=&quot;math-container&quot;&gt;$\dfrac{1.8+2.4}{2}\cdot32-3.2=64$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer therefore answers the question posed of how prestressing with losses is self-balanced: the equivalent load is redistributed, but the total value is not modified. That being said, I cannot explain how to calculate this redistribution in a general case because, well, I don't know how that's done.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;Though friction losses are usually quite linear (or polygonal), this is a poor assumption. After all, the 10% loss at midspan must include the effect of the concentrated angle change at that point. At quarter-span, the losses are only due to linear friction loss, and will therefore probably be substantially lower than 5%. That being said, we can just state that 10% is at the point immediately before the angle change, where the bending moment approaches 100 but where only linear friction losses have occurred.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;All diagrams obtained with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:22:08.407" LastActivityDate="2019-12-28T18:22:08.407" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="9089" PostTypeId="1" CreationDate="2016-05-30T04:58:12.643" Score="0" ViewCount="102" Body="&lt;p&gt;What is the meaning of sentence &quot;Measured Data has 30 PPM accuracy&quot;?&#xA;I want to know what is the meaning of PPM accuracy.&lt;/p&gt;&#xA;" OwnerUserId="6638" LastActivityDate="2016-05-31T13:25:06.140" Title="Data accuracy measurment" Tags="|electrical-engineering|instrumentation|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9090" PostTypeId="2" ParentId="9089" CreationDate="2016-05-30T07:02:31.620" Score="4" Body="&lt;p&gt;ppm = parts per million, analogous to percent = parts per 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 1 ppm error is equivalent to 0.0001 % error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of accuracy, it's, like percents, a dangerous unit to use, since it is not always clear whether it refers to a fraction of the maximum signal, to a fraction of the actual measured value, or to an absolute error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: you measure the concentration of something with a device that can measure concentrations between 0 and 10000 ppm. A data point shows a concentration of 1000 ppm. The accuracy is stated to be 10 ppm. Now the reader has to figure out which of the following is meant:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;10 ppm of full scale means that the value is 1000 +/- 0.1 ppm.&lt;/li&gt;&#xA;&lt;li&gt;10 ppm of the actual value means that the value is 1000 +/- 0.01 ppm.&lt;/li&gt;&#xA;&lt;li&gt;10 ppm absolute error means that the value is 1000 +/- 10 ppm.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6350" LastEditorUserId="272" LastEditDate="2016-05-31T13:25:06.140" LastActivityDate="2016-05-31T13:25:06.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9092" PostTypeId="1" AcceptedAnswerId="9093" CreationDate="2016-05-30T09:49:19.243" Score="4" ViewCount="3889" Body="&lt;p&gt;Boiling temperature of Acetylene is -84°C and that of Petrol is 95°C (at Atmospheric Pressure). So the spark is efficient to ignite the compressed Acetylene and run the Engine. But why wont that work ? Is any other factor that I miss or misunderstood ?&lt;/p&gt;&#xA;" OwnerUserId="6640" LastActivityDate="2018-03-12T01:47:46.983" Title="Why wont Petrol engine run on Acetylene?" Tags="|automotive-engineering|fuel-economy|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9093" PostTypeId="2" ParentId="9092" CreationDate="2016-05-30T10:25:12.847" Score="12" Body="&lt;p&gt;There are certainly a few reasons why acetylene is not a very practical fuel for an IC engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the most important is that it is inherently unstable and tends to explode under pressure, this makes it much more difficulty to store as it must be dissolved in acetone in an inert matrix, unlike say propane which can just be liquefied. This means that refueling would probably involve physically swapping containers and is exacerbated by the fact that it is quite aggressive on  things like hoses and seals. These added complications with safe storage also take a big bite out of it's practical energy density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With any Otto cycle fuel the requirement is not just that it can be ignited but that it can be ignited at the right  time and a lot of chemistry goes into formulating gasoline which ignites reliably when a spark it applied but not before. Premature ignition (ie. during the compression phase)can wreak havoc on a piston engine, even worse is premature ignition in fuel pumps or injectors. Acetylene also burns significantly hotter than gasoline which presents a problem for combustion chamber design (hydrogen presents similar problems). Similarly piston engines require controlled combustion rather than violent detonation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that carrying pressurised acetylene around in a vehicle would represent a serious hazard, far more so than a tank full of gasoline. Gasoline is very unlikely to actually explode in a fire whereas acetylene tanks have been known to level entire buildings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact that acetylene is expensive and energy inefficient to produce as it is obtained by decomposing calcium carbide, which itself must be manufactured, as opposed to fuels obtained by fractioning petroleum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while it might not be impossible to develop and engine which ran on acetylene there isn't really any reason why you would want to. Bearing in mind that the main reason acetylene is used is for its very high flame temperature with oxygen and it doesn't have significantly greater energy density than other liquid or gaseous fuels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside to this one niche use for acetylene is in lights for things like caving where it is generated on demand by dripping water into a supply of calcium carbide (not carbonate as I originally wrote)  and burned to produce bright white flame, as this is a reasonably compact way to produce a long duration, bright light. Although the advent of LED lights and lithium batteries have rendered this, if not entirely obsolete then at least less attractive in many circumstances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main application for acetylene is in oxy/fuel welding. Here alternative fuel gasses such as propane can't provide the correct flame chemistry for the process to be self shielding to exclude oxygen from the base metal ie it is relatively easy to set up an oxy/acetylene flame to consume all of the available oxygen while maintaining a flame temperature hot enough for welding. However for similar processes such as brazing and flame cutting propane is often preferred as it has adequate flame temperature but is significantly cheaper as well as being safer and easier to handle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that hydrogen presents some of the same problems as acetylene in terms of storage, handling and combustion.  But is of more interest as hydrogen &lt;em&gt;potentially&lt;/em&gt; represents a non-petrochemical alternative to fuels like gasoline and natural gas which can be produced either from surplus electrical power or bio-reactors. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-05-30T14:40:22.243" LastActivityDate="2016-05-30T14:40:22.243" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9096" PostTypeId="2" ParentId="9089" CreationDate="2016-05-30T12:58:18.723" Score="0" Body="&lt;p&gt;PPM or &lt;em&gt;parts per million&lt;/em&gt; is usually used to measure the concentration of a substance in a mixture, generally when the substance being measure is a small fraction of the total. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common usage is for defining acceptable or safe levels of pollutants, toxins etc in water or air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most logical interpretation of the 30 PPM accuracy claim is that the device has an accuracy of +/- 30 PPM ie if it reads 100 PPM the true value lies in the range of 70 to 130 PPM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be interpreted as a &lt;em&gt;range&lt;/em&gt; of 30 PPM in which case the true figure would be 15 PPM plus or minus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible that it could mean that the device can detect concentrations &lt;em&gt;above&lt;/em&gt; 30 PPM which might be more likely in the case of something which gives an alarm when a certain threshold is reached. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-05-30T12:58:18.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9097" PostTypeId="1" CreationDate="2016-05-30T14:41:36.780" Score="0" ViewCount="560" Body="&lt;p&gt;I'm going to make &lt;a href=&quot;http://www.abeautifulmess.com/2013/02/josh-modern-stool.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;such a stool&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5YxyJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5YxyJ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Originally it's made of thick plywood, and I want to switch to a thinnest possible OSB to make it as cheap and light as possible.&lt;/p&gt;&#xA;&lt;p&gt;It's a bar stool of 70 cm height, 33 cm width. Front plate is 16° inclined from vertical.&lt;/p&gt;&#xA;&lt;p&gt;My gut tells me that the weakest elements are regions on the sides of the opening(The left one is encircled on the image). I have to keep the opening, it's bottom edge acts as a footrest.&lt;/p&gt;&#xA;&lt;p&gt;At the narrowest point the width of this area is &lt;span class=&quot;math-container&quot;&gt;$b=5$&lt;/span&gt; cm, opening height is &lt;span class=&quot;math-container&quot;&gt;$L=28$&lt;/span&gt; cm.&lt;/p&gt;&#xA;&lt;p&gt;I calculate that &lt;span class=&quot;math-container&quot;&gt;$h=12~\text{mm}$&lt;/span&gt; OSB would suit me.&lt;/p&gt;&#xA;&lt;p&gt;My first question: Can I consider it as centrally loaded column?&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I calculate slenderness ratio:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\lambda = \frac{L_{eff}}{r}=\frac{L_{eff}}{\sqrt{I/A}}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$I=\dfrac{bh^3}{12}=7200~\text{mm}^4$&lt;/span&gt;. So &lt;span class=&quot;math-container&quot;&gt;$\lambda={28~\text{cm}}/{\sqrt{7200~\text{mm}^4/(50~\text{mm}\cdot12~\mathrm{mm})}}=81$&lt;/span&gt;. It tells me that it's a long column. So the load would be limited by buckling.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Critical load &lt;span class=&quot;math-container&quot;&gt;$F=\dfrac{E_{OSB}I\pi^2}{L^2}=\dfrac{3~\mathrm{GPa}\cdot7200~\mathrm{mm}^4\cdot\pi^2}{(28~\mathrm{cm})^2}=270~\mathrm{kgf}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Second question: Am I right that the stool made of 12 mm OSB would be able to hold over a ton of static load?&lt;/p&gt;&#xA;&lt;p&gt;Third question: You can notice screwheads on the picture. How can I estimate the load they need to withstand?&lt;/p&gt;&#xA;" OwnerUserId="5053" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-06-05T16:53:51.180" Title="Do I understand column buckling?" Tags="|structural-engineering|stresses|strength|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="9098" PostTypeId="1" CreationDate="2016-05-30T15:20:07.767" Score="5" ViewCount="598" Body="&lt;p&gt;Why Silicon can be used in many engineering materials like oil (Silicone oil), grease, caulk, rubber(PDMS), resin? What property of Silicon allows so?&lt;/p&gt;&#xA;" OwnerUserId="6194" LastActivityDate="2016-06-03T00:57:24.560" Title="Silicon and silicones" Tags="|materials|plastic|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9099" PostTypeId="2" ParentId="9098" CreationDate="2016-05-30T16:06:08.310" Score="12" Body="&lt;p&gt;Silicon forms bonds in a similar way to carbon, and has a similar outer-shell electron orbital configuration. Carbon, as you may well know, has the ability to covalently bond with up to four lone pairs of electrons on neighboring atoms in a generally tetrahedral arrangement, giving carbon a large versatility in molecular configuration and conformation. The versatility of carbon allows for a wide range of biomolecules, polymers, and other complex organic molecules. Silicon has a similar bonding arrangement, i.e. tetrahedral, except that silicon-silicon bonds are less strong and less flexible than carbon-carbon bonds, and silicon naturally tends to form the highly stable diamond silicon lattice. The diamond silicon lattice is brittle and semi-conducting, but not terribly useful for polymeric materials. Indeed, I have not heard of successful attempts to form economical, purely silicon-based polymers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as you noted, most silicon-based polymer materials are silicone, or perhaps more accurately polysiloxane. Polysiloxanes are materials which contain a repeating, alternating configuration of silicon and oxygen atoms along the primary backbone. Polysiloxanes are generally mechanically flexible molecules which retain the configurational and conformational advantages of the tetrahedral bonding of silicon.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-05-30T16:06:08.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9101" PostTypeId="1" AcceptedAnswerId="10531" CreationDate="2016-05-30T17:50:17.300" Score="1" ViewCount="586" Body="&lt;p&gt;I'm trying to understand the implications for buckling and column sizing in the drawings below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3000mm 120x60 RHS steel column supports a 350mm horizontal steel beam. The column base and right side of the beam are both fixed but can pivot. They are joined by a continual fillet weld &quot;T&quot; join, so the junction can rotate but will retain its angle. The only load is a vertical static force on the left-most 100mm of the horizontal beam.  The column can be placed at any position under the beam, as shown in the 3 drawings (nearer to the right would be better). The thickness of the beam is thick enough that it will not deflect noticeably or materially.  The thickness of steel in the RHS column is to be based on calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The column's required cross-section clearly depends on the axial compressive force, which varies according to the column's position.  But the eccentric load is a concern and I can't be quite sure if I am right to ignore it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assuming&lt;/strong&gt; that the beam is sufficiently rigid and won't noticeably deflect, &lt;strong&gt;then&lt;/strong&gt; when the column position changes to the right (as shown), does the column experience only an increased axial compressive force, or should I also consider an additional bending moment imposed on the column as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reason for thinking I can ignore it and only consider the axial compressive load on the column, is that wherever the force applies on the beam, the column only experiences it as an axial load. But I'm not 100% sure since if the beam deflects even slightly (as it must), this must cause bending of the column (as they are welded) and therefore a bending moment in the column (however slight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this effect significant compared to the increased axial compressive force, or is it trivial and dominated by the increase in axial compression? Do I need to allow for an additional bending moment in my buckling calculations, in this situation, to ensure the column profile is adequate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xl49Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xl49Z.png&quot; alt=&quot;Drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(NOTE - I am based in Europe so any specific calculations would be Eurocode not North America code based. That said, the question isn't really about specific code calculations and it shouldn't matter)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="6366" LastActivityDate="2016-06-30T03:21:03.700" Title="Buckling of a cantilever with midway support" Tags="|beam|columns|buckling|moments|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9102" PostTypeId="1" AcceptedAnswerId="9110" CreationDate="2016-05-30T18:42:35.683" Score="0" ViewCount="100" Body="&lt;p&gt;&lt;strong&gt;How do you design against an earthquake?&lt;/strong&gt; &#xA;Do one initially consider earthquake in early design stages or is earthquake mitigation considered later in the design stage after conceptual stage? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the advantages and disadvantages of the above-mentioned approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to find enough material online on this topic can someone guide me on this so that I can further my research.&lt;/p&gt;&#xA;" OwnerUserId="6632" LastActivityDate="2016-05-31T07:06:07.003" Title="How do you design against an earthquake?" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ClosedDate="2016-05-31T14:23:00.070" ContentLicense="CC BY-SA 3.0" />
  <row Id="9103" PostTypeId="2" ParentId="9098" CreationDate="2016-05-30T18:59:05.720" Score="8" Body="&lt;p&gt;Starrise gave an answer describing silicone from an engineering/chemical perspective, but I disagree that this is the reason for silicone to be so widely used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real reasons why it is used so widely: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It is &lt;strong&gt;versatile&lt;/strong&gt;. It is used as a grease (like other greases, e.g. petroleum based ones) because it provides good lubrication. It is used as a caulk (like other caulks, e.g. cement or acrylic based caulks) because it makes a good sealant and can be squeezed between two tiles and can cure. It is used as in rubbers, because it has the necessary qualities like being flexible and insulating. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is &lt;strong&gt;inert&lt;/strong&gt;. This is extremely important! Organic materials are rarely durable, but even heavily processed organic-based synthetics or &#xA;inorganic ones tend to change a lot over time. Fats go rancid, plastics become brittle, petroleum based greases can get quite thick over time. Silicone doesn't change much with heat, UV light, time, or contact with oxygen, and is also quite resistant to many chemicals which would stain, corrode or otherwise damage a material. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It has an extremely &lt;strong&gt;low toxicity&lt;/strong&gt; (at least the finished product, I don't know about the production process). You can bake in it, smear it on your hair, drink it, or implant it in your body, and everything goes well. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The raw materials are &lt;strong&gt;plentiful&lt;/strong&gt;, literally &quot;like sand at the sea&quot; as the German proverb goes. No need to worry about sourcing rare elements, or the declining oil reserves. No bad association with &quot;you are killing our nature by mining/deforesting/whatever&quot;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It makes &lt;strong&gt;user friendly products&lt;/strong&gt; which are both useful and attractive. If you make a caulk out of it, it cures after air exposure, without stinky fumes. If you sell it as a baking form, it releases the cake without fuss. If you make a lubricant out of it, it doesn't have to smell or look a dirty brown. It takes colors well, is easily shaped, has a clean pleasant feel when touched, does not smell, easy to clean, somewhat water repellent, does not tear too easily, does not wear out too much. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course, the atomic structure described by the other answer is, in a sense, the reason why silicone happens to have the qualities described here. But while it is further back in the chain of causality, it is not really as important for the question you asked. Because, if another atomic structure had produced a material of equal utility and marketability, we would have used it just as happily (and do so, for other applications where slightly different qualities are needed). &lt;/p&gt;&#xA;" OwnerUserId="4865" LastActivityDate="2016-05-30T18:59:05.720" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9104" PostTypeId="2" ParentId="8624" CreationDate="2016-05-30T19:56:24.633" Score="1" Body="&lt;p&gt;It's difficult to tell how your problem should be approached with the information that you provide, for example in what sense you find a discrepancy. I assume your system is as you described in the question &quot;&lt;a href=&quot;https://engineering.stackexchange.com/questions/8566/incompressible-2d-pressure-driven-flow-for-navier-stokes-equations-in-nondimens&quot;&gt;Incompressible 2D pressure-driven flow: for Navier-Stokes equations in nondimensional form, how should I express pressure boundary conditions?&lt;/a&gt;&quot;, with the picture that I copy at the bottom of this post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flow is upwards. Normally, the way to replace a velocity boundary condition by a pressure boundary condition is by trial and error until you have the pressure that will generate the same average velocity over your inlet plane. If that is what you did, but you found a different pressure than with the other calculation method, then something is wrong with at least one of the calculation methods and you have to a systematic comparison between the two models, checking all parameters that you can plot and figure out which fundamental parameter is wrong. For example, you could have forgotten to switch on the effect of gravity in one of the codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your question is: how do I calculate the pressure to set at the inlet from just the viscosity and a given inlet velocity, then you are using the wrong approach. The pressure at the inlet will be the result of two contributions: gravity ($P_g=\rho g h$) and flow resistance at the outlet pipe. For that, you need to take the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydraulic_diameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydraulic diameter&lt;/a&gt; $D_H=2a$ of the outlet pipe, get the average outlet velocity $U_o$ and calculate the Reynolds number, $\mathit{Re}=\rho U_o D_H/\eta$. The pressure drop at the outlet will generally have several contributions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Bernoulli: $\Delta p=b\rho U_o^2/2$, where $b$ is a factor to account for the velocity profile of the fluid at the outlet; $b=1$ for a uniform velocity and $b=2$ (I think) for a parabolic velocity profile. This is sometimes called: &quot;exit loss&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Friction along the pipe length, also called &quot;major loss&quot;: &#xA;$$ \Delta p_{\mathrm{major}}={fL\over D_H} {\rho U_o^2\over 2}, $$&#xA;where $f$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Darcy-Weisbach friction factor&lt;/a&gt;, which depends on the Reynolds number.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Inlet loss, also called &quot;minor loss&quot;,&#xA;$$ \Delta p_{\mathrm{minor}} = \xi{\rho U_o^2\over 2}, $$&#xA;where $\xi$ is the minor loss coefficient.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Tabulated values for all these losses in various pipe geometries are in &lt;a href=&quot;http://pbadupws.nrc.gov/docs/ML1220/ML12209A041.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Idelchik, Handbook of Hydraulic Resistance&lt;/a&gt; (warning: 21 MB download).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what happens inside your vessel, but if you're only interested in the relation between flow rate and pressure, I wouldn't even bother with trying to run a CFD code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pgu6q.jpg&quot; alt=&quot;flow schematic&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6350" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-05-30T19:56:24.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9105" PostTypeId="2" ParentId="9101" CreationDate="2016-05-30T23:06:31.963" Score="0" Body="&lt;p&gt;The general rule for engineering is that if you can't dismiss it based upon experience or a trivial side calculation, calculate it. It may very well be trivial, it may not be. But until you're willing to risk your license on it, calculate it.  Ultimately, every eccentric load, at some point, induces enough moment that the moment + compression load will cause buckling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably don't have to include it in the formal write up, you can just state your recommendation in the notes without justification. Most likely, nobody will comment on it. But if it's concerning you, calculate it, make sure that you're satisfied it'll pass all of the methods of failure, and state that it is your recommendation. &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2016-05-30T23:06:31.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9106" PostTypeId="1" AcceptedAnswerId="9182" CreationDate="2016-05-30T23:07:27.710" Score="2" ViewCount="318" Body="&lt;p&gt;Double diffraction is where a light beam passes through an grating with distance D1 between ridges and then passes through an grating with distance D2 The image after passing through diffraction gratings is a function of the product of two Fourier tranforms in the frequency domain. In other words, the product of two Fourier transforms in the frequency domain is equivalent to the convolution of the two Fourier integrals in the time domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we match spatial coherence length for a given wavelength with the spacing between the diffraction grating ridges, what is the intensity of the image after passing through two cascaded diffraction gratings each with distinct spacing between grating ridges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask this question is to examine the best linear combination of 2 cascaded diffraction grating where the 2 unique diffraction gratings are centered around 2 of the primary colors, red green and blue. For example, suppose we want to filter an orange laser pointer in the far field(i.e 100 meters) to protect pilot's vision from the orange laser pointer or dual wavelength laser pointer upon takeoff or landing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the diffraction gratings would be centered at a fixed visible light wavelength. Would it be necesssary for  one of the cascaded diffraction grating to be centered around a variable visible light wavelength harnessing the Pockels effect where a varying electric field modulates the electro-optic behavior of the diffraction grating  aperture radius?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse my novice use of terminology and suggest how I should modify my presentation if you have time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Chih-Hsien Jason Lin, Ph.D and Dr. Yu-wha Lo , Ph.D of the Cornell Research Foundation filed U.S. Patent 5760960 A, &quot;Cascaded self-induced holography&quot;,  in June 1998 which harnesses Young's Interference experiment with diffraction gratings. This excellent research may be used to deflect and filter laser pointer beams in the visible light region. Cambridge University Press published a book , Optical coherence and quantum optics, by Professor Leonard Mandel and Professor  Emil Wolf of the University of Rochester,around the same time analyzing the complex spatial coherence function from Young's two slit interference experiment using Fourier integrals. Pages 171 to Page 176 of this seminal book directly address the analysis of spatial coherence and optical filters for Young's two slit interference experiment &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;FIG. 2 illustrates that the size of the central opaque region 40 defines a target exposure area 76 on the substrate 60 by shading the target area 40 from direct illumination by the light source 72. This allows only the interference patterns from the cascaded grating pairs 56 and 58 to reach the target area 76. Preferably, the size of the target area 76 can be from a few millimeters to a centimeter square, and is comparable to the exposure size for an optical stepper.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The linearly polarized monochromatic light from the light source 72 indicated at 80 is first diffracted by the first and third mask gratings 44, 50 on the top surface of the mask 10. Most of the photon energy is contained in the first order diffraction beams labeled as +1 on the right and -1 on the left. Other diffractive beams and the zero order transmitted beams are not shown in FIG. 2 because they do not play a role in the final grating formation. When the +1 and -1 diffraction beams reach the second and fourth mask gratings 48 and 54 on the bottom of the mask 10, two new diffracted beams labeled as +2 and -2 are generated. As shown in FIG. 2, the target area 76 is the area where only the desired +2 and -2 beams can interfere, thereby exposing a grating 90 which has a period of approximately 1/4 of the periods in the mask gratings 44, 48, 50 and 54.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A5qcI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A5qcI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/i1aAN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i1aAN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LYnkg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LYnkg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grover A. Swartzlander wrote in 2000 in the attached article , &quot;Peering into Darkness with a vortex spatial filter&quot; , at the bottom of this question that &quot;It is sometimes desireable to attenuate the intense glare of a bright coherent beam of light to enhance the detection of an incoherent background signal or a weak nearly collinear source&quot;. I would like to extend this research to protecting airline pilots from the effects of laser pointers shone from the ground at planes on takeoff and landing. The reason I wish to enhance the detection of a weaker incoherent  background signal is to image this signal on a Planar transparent display.&lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="2651" LastEditDate="2016-06-03T07:09:52.453" LastActivityDate="2016-06-08T07:21:35.970" Title="What happens when very bright coherent light and weaker incoherent background light passes through doubly cascaded diffraction gratings?" Tags="|optics|signal-processing|" AnswerCount="1" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="9107" PostTypeId="2" ParentId="8541" CreationDate="2016-05-31T01:29:46.593" Score="3" Body="&lt;p&gt;Your Average MechEng's answer is fantastic and it is often how I explained the area of a thin circular shell to my students. But the mathematical derivation of such formula might be of interest to some. Using the above answer's diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X3XXO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X3XXO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we are interested in computing the area between the two dashed lines. The inner circle has area $\pi r^2$ while the outer circle has area $\pi (r+\Delta r)^2$. The area between the two is obtained by subtraction:&#xA;$$&#xA;\begin{align}&#xA;\text{Area} &amp;amp;= \pi (r+\Delta r)^2 - \pi r^2 \\&#xA;&amp;amp;= \pi (r^2 + 2r\Delta r + \Delta r^2 - r^2) \\&#xA;&amp;amp;= \pi (2r\Delta r + \Delta r^2)&#xA;\end{align}&#xA;$$&#xA;Taking the limit as $\Delta r$ goes to zero yields the desired answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, it does not seem like any of these two answers resolve your issue, although they get you closer.&lt;/p&gt;&#xA;" OwnerUserId="1031" LastActivityDate="2016-05-31T01:29:46.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9108" PostTypeId="1" CreationDate="2016-05-31T02:40:29.730" Score="0" ViewCount="1396" Body="&lt;p&gt;Refer to the chapter 14 of the Wilson book &quot;Dynamic Analysis of structures&quot;: &lt;a href=&quot;http://www.edwilson.org/book/14-EIGEN.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;calculation of stiffness and mass orthogonal vectors&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table 14.6 shows the &lt;strong&gt;lateral mass participation&lt;/strong&gt; and &lt;strong&gt;vertical mass participation&lt;/strong&gt; for a frame structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/50dfA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/50dfA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand how can one tell whether an eigen mode contributes to lateral mass or vertical mass. From my understanding, mass participation &lt;a href=&quot;http://www.vibrationdata.com/tutorials2/ModalMass.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;for each mode is determined by the general mass and the eigenvector of the mode&lt;/a&gt;, and eigenvector of the mode usually consists of contribution from all the $x$, $y$ and $z$ direction, so how can one tell whether the mode is contributed to lateral or vertical direction?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastActivityDate="2017-02-27T02:12:41.217" Title="The distinction between lateral mass participation and vertical mass participation" Tags="|structural-analysis|modal-analysis|eigenvalue-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9109" PostTypeId="2" ParentId="9024" CreationDate="2016-05-31T04:33:13.127" Score="3" Body="&lt;p&gt;This answer expands upon the approach suggested by @setun-90 and shows that the algebra gets quite complicated.  I'd suggest a Green's function approach instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumptions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;isotropic&lt;/li&gt;&#xA;&lt;li&gt;deflection of membrane is small&lt;/li&gt;&#xA;&lt;li&gt;applied tension large so that its variation due to deflection is small&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Governing equation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the tension in the $x, y$-directions be $T_x, T_y$ per unit length, respectively.&#xA;Let $w$ be the lateral deflection.&#xA;The resultant in the $w$-direction due to the tension $T_x$ on the $dy$-edge of a &#xA;small element of area $(dx \times dy)$ is&#xA;$$&#xA;  R_x := (T_x dy) \left(\theta_x + \frac{\partial \theta_x}{\partial x}\, dx\right) - &#xA;  (T_x dy)\, \theta_x = T_x\, \frac{\partial \theta_x}{\partial x}\, dx dy&#xA;$$&#xA;where we have used $\sin\theta_x \approx \theta_x$ for small $\theta_x$ and&#xA;$&#xA;  \theta_x := \frac{\partial w}{\partial x} \,.&#xA;$&#xA;Similarly, the resultant in the $w$-direction due to $T_y$ is&#xA;$$&#xA;  R_y := (T_y dx) \left(\theta_y + \frac{\partial \theta_y}{\partial y}\, dy\right) - &#xA;  (T_y dx)\, \theta_y = T_y\, \frac{\partial \theta_y}{\partial y}\, dx dy&#xA;$$&#xA;where &#xA;$&#xA;  \theta_y := \frac{\partial w}{\partial y} \,.&#xA;$&#xA;Therefore, the total lateral force due to $T_x$ and $T_y$ is&#xA;$$&#xA;  R = R_x + R_y = \left(T_x\, \frac{\partial \theta_x}{\partial x} +&#xA;                        T_y\, \frac{\partial \theta_y}{\partial y}\right) dx dy \,.&#xA;$$&#xA;This force is balanced by the applied lateral pressure, $p(x,y)$, and we have&#xA;$$&#xA;   p(x, y)\, dx dy + \left(T_x\, \frac{\partial \theta_x}{\partial x} +&#xA;                        T_y\, \frac{\partial \theta_y}{\partial y}\right) dx dy = 0 \,.&#xA;$$&#xA;Plugging in the definitions of $\theta_x$ and $\theta_y$, and eliminating $dx dy$, we get the governing equation&#xA;get&#xA;$$ \boxed{&#xA;   p(x, y) + T_x\, \frac{\partial^2 w}{\partial x^2} +&#xA;                        T_y\, \frac{\partial^2 w}{\partial y^2} = 0 \,.}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case where $T_x = T$ and $T_y = 0$&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the governing equation reduces to&#xA;$$&#xA;  \frac{\partial^2 w}{\partial x^2}  = -\frac{p(x,y)}{T} \,.&#xA;$$&#xA;Integrating once, we have&#xA;$$&#xA;  \frac{\partial w}{\partial x}  = -\int\frac{p(x,y)}{T}\,dx + F(y) + F_c&#xA;$$&#xA;where $F(y)$ is an integration function and $F_c$ is an integration constant.&lt;br&gt;&#xA;Integrating again&#xA;$$\boxed{&#xA;   w(x, y) = \int\left[-\int\frac{p(x,y)}{T}\,dx + F(y) + F_c\right] dx + G(y) + G_c }&#xA;$$&#xA;where $G(y)$ is another integration function and $G_c$ is an integration constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case where $p(x,y)$ is a discontinuous function&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the case where&#xA;$$&#xA;  p(x,y) = \begin{cases}&#xA;              0 &amp;amp; \text{for} \quad 0 \le x &amp;lt; \tfrac{L}{3}, 0 \le y \le L \\&#xA;              0 &amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3},&#xA;                    0 \le y &amp;lt; \tfrac{L}{3} \\&#xA;              P &amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3} ,&#xA;                    \tfrac{L}{3} \le y \le \tfrac{2L}{3}\\&#xA;              0 &amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3},&#xA;                    \tfrac{2L}{3} &amp;lt; y \le L \\&#xA;              0 &amp;amp; \text{for} \quad \tfrac{2L}{3} &amp;lt; x \le L, 0 \le y \le L &#xA;           \end{cases}&#xA;$$&#xA;Then, &#xA;$$&#xA;   \int \frac{p(x,y)}{T}\,dx = \begin{cases}&#xA;              0 &amp;amp; \text{for} \quad 0 \le x &amp;lt; \tfrac{L}{3}, 0 \le y \le L \\&#xA;              0 &amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3},&#xA;                    0 \le y &amp;lt; \tfrac{L}{3} \\&#xA;              \tfrac{Px}{T} &amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3},&#xA;                    \tfrac{L}{3} \le y \le \tfrac{2L}{3}\\&#xA;              0 &amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3},&#xA;                    \tfrac{2L}{3} &amp;lt; y \le L \\&#xA;              0 &amp;amp; \text{for} \quad \tfrac{2L}{3} &amp;lt; x \le L , 0 \le y \le L &#xA;           \end{cases}&#xA;$$&#xA;This leads to&#xA;$$&#xA;  \frac{\partial w}{\partial x}  = \begin{cases}&#xA;              F_1(y) + F_{c1} &amp;amp; \text{for} \quad 0 \le x &amp;lt; \tfrac{L}{3}, 0 \le y \le L \\&#xA;              F_2(y) + F_{c2} &amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3},&#xA;                    0 \le y &amp;lt; \tfrac{L}{3} \\&#xA;              -\tfrac{Px}{T} + F_3(y) + F_{c3} &amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3},&#xA;                    \tfrac{L}{3} \le y \le \tfrac{2L}{3}\\&#xA;              F_4(y) + F_{c4} &amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3},&#xA;                    \tfrac{2L}{3} &amp;lt; y \le L \\&#xA;              F_5(y) + F_{c5} &amp;amp; \text{for} \quad \tfrac{2L}{3} &amp;lt; x \le L , 0 \le y \le L &#xA;           \end{cases}&#xA;$$&#xA;From the second integration:&#xA;$$\boxed{&#xA;  w(x,y)  = \begin{cases}&#xA;              [F_1(y) + F_{c1}]\,x + G_1(y) + G_{1c} &amp;amp; \text{for} \quad 0 \le x &amp;lt; \tfrac{L}{3}, 0 \le y \le L \\&#xA;              [F_2(y) + F_{c2}]\,x + G_2(y) + G_{2c}&amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3},&#xA;                    0 \le y &amp;lt; \tfrac{L}{3} \\&#xA;              -\frac{Px^2}{2T} + [F_3(y) + F_{c3}]\,x + G_3(y) + G_{3c} &amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3},&#xA;                    \tfrac{L}{3} \le y \le \tfrac{2L}{3}\\&#xA;              [F_4(y) + F_{c4}]\,x + G_4(y) + G_{4c}&amp;amp; \text{for} \quad \tfrac{L}{3} \le x \le \tfrac{2L}{3},&#xA;                    \tfrac{2L}{3} &amp;lt; y \le L \\&#xA;              [F_5(y) + F_{c5}]\,x + G_5(y) + G_{5c}&amp;amp; \text{for} \quad \tfrac{2L}{3} &amp;lt; x \le L , 0 \le y \le L &#xA;           \end{cases}}&#xA;$$&#xA;The 20 unknown quantities in this solution need 20 boundary conditions, making the algebra quite complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Green's function approach&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative is to use the Green's function approach in which we start with an assumed sinusoidal lateral load&#xA;$$&#xA;  p(x,y) = \sum_{m=1}^\infty \sum_{n=1}^\infty p_0 \sin \frac{m\pi x}{a} \sin \frac{n\pi y}{b}&#xA;$$&#xA;where $a$ and $b$ are the dimensions of the membrane.  We can then specialize the solution to the case where the applied load is a point load at the location $(\xi,\eta)$.  Let $K(x,y,\xi,\eta)$ be this solution for a point load.  Then the total deflection can be computed using&#xA;$$&#xA;  w(x,y) = \iint_A p(\chi,\eta) K(x, y, \chi, \eta) \,d\chi d\eta \,.&#xA;$$&#xA;Details of this approach can be found in Section 29 of &quot;Theory of Plates and Shells&quot; by Timoshenko and Woinowsky-Krieger.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="3488" LastEditDate="2016-05-31T07:50:31.350" LastActivityDate="2016-05-31T07:50:31.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9110" PostTypeId="2" ParentId="9102" CreationDate="2016-05-31T07:06:07.003" Score="0" Body="&lt;p&gt;The general shape of the building will affect earthquake resistance. I think L-shaped buildings are quite bad. You also need to know whether the soil is capable of supporting a heavy, tall building during an earthquake. If you want to keep the cost under control, better make a decision early in the design process on where to put supporting walls and structures so that you get sufficient resistance for a minimum amount of material.&lt;/p&gt;&#xA;" OwnerUserId="6350" LastActivityDate="2016-05-31T07:06:07.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9111" PostTypeId="1" AcceptedAnswerId="9112" CreationDate="2016-05-31T09:33:22.507" Score="1" ViewCount="864" Body="&lt;p&gt;I want to know what torque a stepper motor can hold when stationary. They seem to only have one figure in specifications for holding torque which, I presume, is while coils are energised. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this normally handled in usage of stepper motors? Is there a way of calculating holding torque while coils are not energised? I know braking can be done by shorting the coils, but it's obviously not as effective as just holding one coil energised to maintain position. Is one coil usually just kept energised to hold position with maximum torque? Is there a limit to how long one coil can remain energised?&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="1586" LastEditDate="2016-05-31T23:16:09.977" LastActivityDate="2016-06-01T10:54:12.950" Title="Stepper motor holding torque without coils energized?" Tags="|electrical-engineering|motors|torque|stepper-motor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9112" PostTypeId="2" ParentId="9111" CreationDate="2016-05-31T11:18:56.863" Score="2" Body="&lt;p&gt;The value Detent Torque or &lt;a href=&quot;https://en.wikipedia.org/wiki/Cogging_torque&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cogging Torque&lt;/a&gt; is what you should look for in the specification sheet. The detent torque is related to stepper motor power loss. The following is a reference from Minebea' web site.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Detent Torque&lt;/strong&gt;: amount of torque that the motor produces when it is not energized. No current is flowing through the windings.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is an excerpt from &lt;a href=&quot;https://www.openimpulse.com/blog/wp-content/uploads/wpsc/downloadables/57BYGH420-Stepper-Motor-Datasheet.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;57BYGH420 Stepper Motor Datasheet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K3WMU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K3WMU.jpg&quot; alt=&quot;Detent Torque Specification&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Referances:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cogging_torque&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cogging torque&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nmbtc.com/step-motors/engineering/torque-and-speed-relationship/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stepper Motor Speed and Torque Relationship&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ewh.ieee.org/soc/es/Nov1998/12/BEGIN.HTM&quot; rel=&quot;nofollow noreferrer&quot;&gt;In-depth Learning of Cogging/Detenting Torque through Experiments and Simulations&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.geckodrive.com/support/step-motor-basics.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Step Motor Basics Guide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-06-01T10:54:12.950" LastActivityDate="2016-06-01T10:54:12.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9113" PostTypeId="2" ParentId="9047" CreationDate="2016-05-31T12:25:19.610" Score="1" Body="&lt;p&gt;You need to differentiate between average velocity and average speed of molecules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the liquid is not moving, then average velocity will be zero, but average speed will not be. &lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-05-31T12:25:19.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9115" PostTypeId="1" AcceptedAnswerId="9138" CreationDate="2016-05-31T13:53:02.977" Score="0" ViewCount="1490" Body="&lt;p&gt;Is there an energy related reason why, in the production of ammonia, air is not introduced in the primary reformer? In this case we don't need the second reformer. I could think of a few reasons like the fact that in the secondary reformer hydrogen gas is needed to react with the oxygen to water, which can react with $\text{CH}_4$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is not really energy related?&lt;/p&gt;&#xA;" OwnerUserId="6654" LastEditorUserId="368" LastEditDate="2016-06-02T19:21:02.990" LastActivityDate="2016-06-02T19:21:02.990" Title="Why isn't air added to the primary reformer in ammonia production?" Tags="|chemical-engineering|industrial-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9116" PostTypeId="1" AcceptedAnswerId="9166" CreationDate="2016-05-31T14:02:49.563" Score="5" ViewCount="524" Body="&lt;p&gt;This question might sound stupid as I'm more of a software guy and never really had to do control engineering (besides the 101 classes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have a second order system I could approximate, with the following denominator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s^2 + a_1s + a_0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Both $a_1$ and $a_0$ are positive, which means, for a 2&lt;sup&gt;nd&lt;/sup&gt; order system, that it is stable.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this system's input is a motor's command (not a speed but a PWM cycle value), and the output is the robots angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, my angle will go to $+\infty$ with a constant input, which means my system will never stabilize. &lt;strong&gt;So the system should not be stable.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, plotting my transfer function's output to a constant input using scipy gives me the expected result (an angle which ends up growing linearly as the speed stabilizes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what am I missing here? Is my model completely wrong? Or did I miss something on the definition of &quot;stability&quot; itself? It would make sense if I worked on the speed but the variable I work on is the angle.&lt;/p&gt;&#xA;" OwnerUserId="6655" LastEditorUserId="1832" LastEditDate="2016-05-31T23:09:55.813" LastActivityDate="2016-06-03T18:52:51.583" Title="A stable transfer function which diverges?" Tags="|control-engineering|control-theory|stability|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9118" PostTypeId="2" ParentId="9066" CreationDate="2016-05-31T18:43:28.833" Score="1" Body="&lt;p&gt;A beam carries load by bending under it, deflection. We say load creates shear and shear over span creates moment whis is resisted by the beam by reacting in a way that produces opposite moment by producing a coupled force of tension and compression on the far fibers of the cross section of beam.&lt;br&gt;&#xA; A simplified way of looking at it is to imagine the beam is a solid, unyielding material not capable of accepting any strain under any loading.&lt;br&gt;&#xA; Now we cut this beam into many small sections and connect them back together by stringing two elastic ropes one passing over the top edge and the other the bottom edge of the beam, leaving small gaps in between the sections!&lt;br&gt;&#xA;Now if you load the beam with continuous loading the elastic rope on the bottom will expand under the  tension load and the top rope would be compressed to let the top of the gaps close. when we look at the profile the beam now is deflected down.&lt;br&gt;&#xA;In case of a beam spanning two bays continuously the rope on top over the column will now be under tension, because the beam would want to bend in concave curve, effectively lifting part of the adjoining ends of beam, reducing the virtual span of the beam on either side.&lt;br&gt;&#xA;If we add one more elastic rope to the top over the column area it will pull the top parts of beam more and again will lift the mid span deflection a bit more.&lt;br&gt;&#xA;in extreme case if we add enough ropes to the top we can reduce the deflection on mid-span to an arbitrary small value.  This is basically the idea for finite element analysis!     &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-05-31T18:43:28.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9119" PostTypeId="1" AcceptedAnswerId="9120" CreationDate="2016-05-31T19:09:18.750" Score="4" ViewCount="527" Body="&lt;p&gt;Sorry if this is in the wrong place; I haven´t posted in SE Engineering before, and the site said that SEE is also for students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I´m a High School senior and I´m debating between Computer Engineering and Computer Science. Can someone give me practical ¨real world¨ differences between the two? I &lt;em&gt;think&lt;/em&gt; I´m more interested in embedded systems, VR, AI, and electronics than just software (though who really knows until I actually study it!) Especially virtual reality and AI because they´re such new, promising fields. If I want to work in those fields would either degree be more applicable? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your input! I´d really appreciate some replies from outside sources or personal experience.&lt;/p&gt;&#xA;" OwnerUserId="6659" LastActivityDate="2016-05-31T22:07:05.180" Title="Computer engineering vs Computer Science?" Tags="|computer|" AnswerCount="1" CommentCount="7" ClosedDate="2016-05-31T23:32:13.793" ContentLicense="CC BY-SA 3.0" />
  <row Id="9120" PostTypeId="2" ParentId="9119" CreationDate="2016-05-31T21:55:20.320" Score="0" Body="&lt;p&gt;Computer Engineering tend to go more deeper into Electrical engineering topics such as electronics, transistors, micro processor design, embedded systems as well as some computer programming topic. Labs tend to be more hands on working with electronic components, as well as using electronic measurement instruments such as oscilloscopes and logic analyzers   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computer Science programs tend go deeper into topic such as programming languages, database design, object oriented programming and so forth. The exposure to electronics are much lighter in a computer science programs. Labs tend to mostly focused on writing software programs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest that you look at computer engineering and computer science curriculum from different universities. This is should help you get better understanding of the differences between programs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the undergraduate curriculum requirement for &lt;a href=&quot;http://www.ece.illinois.edu/academics/ugrad/curriculum/ce-curriculum-07.asp&quot; rel=&quot;nofollow&quot;&gt;Computer Engineering&lt;/a&gt; and &lt;a href=&quot;https://cs.illinois.edu/current-students/undergraduates/undergraduate-curriculum-requirements#BSENG&quot; rel=&quot;nofollow&quot;&gt;Computer science&lt;/a&gt; at &lt;a href=&quot;http://illinois.edu/&quot; rel=&quot;nofollow&quot;&gt;University of Illinois&lt;/a&gt;. I suggest that you do side by side comparison between a both the programs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are more interested in Embedded Systems, I suggest that you follow a Computer Engineering program. I also suggest following some of the free online course. These will give you better first hand experiences on what is best for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck&lt;/p&gt;&#xA;" OwnerUserId="282" LastEditorUserId="282" LastEditDate="2016-05-31T22:07:05.180" LastActivityDate="2016-05-31T22:07:05.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9121" PostTypeId="1" CreationDate="2016-05-31T22:41:15.397" Score="3" ViewCount="533" Body="&lt;p&gt;This is a table with two wings that fold down, and two legs &lt;em&gt;per side&lt;/em&gt; that swing out for support. I'm pretty confident about this design being stable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EpfwN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EpfwN.png&quot; alt=&quot;first table design&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The green section at the top is the only surface that remains in place. The vertical green section at the back is another wing on the opposite side, shown folded down. For the yellow wing, the legs are shown fully rotated outward to their stop position.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The total measurement of the surface is 22&quot; (wing) + 6&quot; (stationary section) + 22&quot; (other wing) = 48&quot;. The total height is 24&quot;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an alternate design. I would like to find a way to reduce the overall weight and only include one swing-out leg (other wing removed for illustrative purposes):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5Y1dP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Y1dP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The swing-out leg is shown here fully extended to its resting position. It would have some kind of ball/socket mechanism to lock in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the single leg in the alternate design could swing out &lt;em&gt;from&lt;/em&gt; the center to form a T-shaped frame underneath, which is probably more expected:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t8FEv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t8FEv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how can I calculate this table's stability? I worry that in the second design, there would be stability issues in the corners. I'm an amateur builder and not sure how to go about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oak would be the surface material, with the legs being 1&quot; square steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a similar table from Crate and Barrel that I can't link to due to being a new user (&lt;a href=&quot;https://i.stack.imgur.com/rSDdw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;[A]&lt;/a&gt;, &lt;a href=&quot;https://i.stack.imgur.com/6SrFp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;[B]&lt;/a&gt;):&lt;/p&gt;&#xA;" OwnerUserId="6660" LastEditorUserId="-1" LastEditDate="2016-06-01T22:03:20.117" LastActivityDate="2018-02-23T05:24:01.933" Title="Stability issues in gateleg table design" Tags="|mechanical-engineering|structural-engineering|product-engineering|industrial-engineering|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9122" PostTypeId="2" ParentId="9121" CreationDate="2016-06-01T02:58:27.577" Score="1" Body="&lt;p&gt;This question is in regards to stability, so that's what I'm going to focus on (as opposed to strength, which would require far more information regarding materials, dimensions and connections).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your table has three configurations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both wings down&lt;/li&gt;&#xA;&lt;li&gt;One wing (left or right) up, the other down&lt;/li&gt;&#xA;&lt;li&gt;Both wings up&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Stability can be measured in three ways: how much horizontal or vertical (upwards or downwards) force is required to tip a structure over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fOhkx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fOhkx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without a doubt the most unstable of these configurations is with both wings down. In this case, the table is supported basically only by the four central feet. Obviously, very little horizontal force is required to tip it over on its long side. In fact, the force required is equal to&#xA;$$F_h = \dfrac{Ww}{2h}$$&#xA;where $W$ is the total weight of the table, $w$ is the width of the supports (in this case, 6&quot;), and $h$ is the height of the table. Therefore, we can see that the stability is proportional to the table's weight and to how wide its supporting base is, and is inversely proportional to the table's height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For downwards vertical loads, the equation is&#xA;$$F_{v-} = \dfrac{Ww}{2e^-}$$&#xA;where $e^-$ is the distance between the farthest point outside of the supports and the nearest support. Given that the feet of the table are basically right on the outer perimeter of the table, the value of $e$ is close to zero, therefore $F_{v-}$ is necessarily quite high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For upwards vertical loads, the equation is almost the same&#xA;$$F_{v+} = \dfrac{Ww}{2e^+}$$&#xA;but $e^+$ is the distance between the farthest point outside of the supports and the opposing support. $F_{v+}$ is always lower than $F_{v-}$, but arguably also less likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations can be applied for the other configurations as well. The new legs dramatically increase the value of $w$ in the direction perpendicular to the 39&quot; side, increasing $F_h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For downwards vertical loads, the design of those rotating legs matters. If they open as in the first drawing, then that's fine: once again the legs cover basically the entire perimeter of the table, leaving little space to flip the table and therefore requiring a massive load to do so. If it's a single leg swinging to the middle of the wing, then that's more problematic. In one direction (perpendicular to the 39&quot; side), you'll have $e^- = 11''$ (half of 22&quot;). In the other, you'll strictly speaking still be fine due to the main legs, but that will put a lot of strain on the hinges connecting the wings to the main segment (this is the instability you mention &quot;in the corners&quot;). Also, this second design requires at least two hinges per wing, preferably as far as possible from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For upwards vertical loads, the same issues apply to the second design, but to a lesser extent. $e^+$ will equal $6+22=28''$ or $2\times 22+6=50''$ if one or both wings are open, respectively.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-06-01T02:58:27.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9123" PostTypeId="1" CreationDate="2016-06-01T04:35:26.707" Score="1" ViewCount="695" Body="&lt;p&gt;I am trying to determine the maximum torque that can be applied to a 6061 Aluminium tube with O.D. = 55&amp;nbsp;mm and I.D. = 53&amp;nbsp;mm. The yield strength of 6061 Aluminium I have taken as 55&amp;nbsp;MPa and using the formula&#xA;$$T = \frac{tJ}{r}$$ and the polar moment of inertia of a hollow cylinder&#xA;$$J = \pi\frac{D^4 - d^4}{32}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comes out to 247&amp;nbsp;Nm. I'm highly skeptical of this number as it seems quite absurd that an aluminium tube with a 1&amp;nbsp;mm thick wall can handle that much torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been quite a few years since I've done mechanics so I'm sure I'm missing something (or several things).&lt;/p&gt;&#xA;" OwnerUserId="3605" LastEditorUserId="1832" LastEditDate="2016-06-01T10:50:52.077" LastActivityDate="2016-06-01T10:50:52.077" Title="Maximum torque on thin walled aluminium shaft" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9124" PostTypeId="2" ParentId="9123" CreationDate="2016-06-01T09:48:57.487" Score="1" Body="&lt;p&gt;I suspect a thin walled shaft like that will fail through buckling long before it reaches the shear stress limit.  Especially if it's long and doesn't have thicker butting at the ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be some guidelines/rules of thumb you can look up to get an idea what performance is reasonable in this case, but certainly the shear limit is only useful as the theoretical maximum.&lt;/p&gt;&#xA;" OwnerUserId="6665" LastActivityDate="2016-06-01T09:48:57.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9125" PostTypeId="1" AcceptedAnswerId="9127" CreationDate="2016-06-01T10:29:36.307" Score="3" ViewCount="837" Body="&lt;p&gt;I'm building something to dry glass tubes. It's raised a question which I'd like some background knowledge on before I draw up requirements for the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I want to dry something with heat alone, then obviously in a room environment (50% humidity, relatively free airflow, atmospheric pressure) heating something above 100°c would quickly evaporate fairly small amounts of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, a powerful enough jet of air would blast off most water, quickly evaporating any remains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But say I use a heated airflow. Say the heater power is fixed, what would the relationship be between speed of drying and airflow? I think this is a hard question to answer without specifics, but perhaps one question in isolation could be answered:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, for small amounts of water, and a fixed heater power: if airflow is doubled, then the difference in temperature between that airflow and room temperature would be halved. Similarly, if the airflow is halved, the difference in temperature would be doubled. What would the effect on rate of evaporation be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this is impossible to answer without more information, in which case, how might this problem be solved? A compromise would have to be established on criteria like cost of higher pressure air flow generation,  noise levels, power consumption, energy efficiency, rate of drying, surface area of objects to be dried, etc etc etc. Might this be better off just developed with experimentation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any level of guidance would be appreciated&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-06-04T13:02:40.323" Title="Optimum airflow for drying with heated air" Tags="|drying|evaporation|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9126" PostTypeId="2" ParentId="9066" CreationDate="2016-06-01T11:29:05.870" Score="0" Body="&lt;p&gt;To add to the answers already given, you are only scratching the surface with your 2 span beams. One you get to 3 or more spans, you will see that your bending moment is also reduced. &#xA;Problems that may arise from having multiple continuous spans include having to restraint your beam's bottom chord, since it will be in compression at the supports. In steel construction, this is usually done by tying the open web steel joists to the column top / beam bottom chord.&lt;/p&gt;&#xA;" OwnerUserId="5691" LastActivityDate="2016-06-01T11:29:05.870" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9127" PostTypeId="2" ParentId="9125" CreationDate="2016-06-01T17:33:51.740" Score="1" Body="&lt;p&gt;If you are only interested in temperature, flow rate, and drying rate, then you could consider the following quantities for a given air temperature $T$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How much heat you need for evaporation (based on how much water is on the glassware)?&lt;/li&gt;&#xA;&lt;li&gt;How much heat you need to bring the entire mass of glassware to that temperature?&lt;/li&gt;&#xA;&lt;li&gt;What volume of air do you need to carry the moisture of one batch, assuming saturated vapor and how much heat do you need to heat all that air? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This will give you the the minimum amount of energy needed to dry the glassware. To increase the rate of evaporation, I would put priority on increasing the water-carrying capacity of the resulting air, since evaporation rate is primarily driven by the difference between the actual water content of the air and the maximum water content. Specifically, the rate of evaporation is&#xA;$$Q = K (c_{\mathrm{max}} - c), $$&#xA;where $K$ is a proportionality constant, $c_{\mathrm{max}}$ is the temperature-dependent maximum water content of the air (e.g., in kg/m3 units) and $c$ is the actual water content. For your system, you could start by taking $c$ as the final water content of the air after all the glassware has dried.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, consider increasing the flow velocity of the air without increasing the volume of air. This will increase the proportionality constant $K$. Turbulent flow, as opposed to laminar flow, is a good thing here.&lt;/p&gt;&#xA;" OwnerUserId="6350" LastEditorUserId="6350" LastEditDate="2016-06-04T13:02:40.323" LastActivityDate="2016-06-04T13:02:40.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9128" PostTypeId="2" ParentId="9026" CreationDate="2016-06-01T17:48:53.773" Score="0" Body="&lt;p&gt;The most probable explanation why gas generators aren't designed the way I proposed is because they are designed around a constant &lt;strong&gt;air speed&lt;/strong&gt;, not &lt;strong&gt;Mach number&lt;/strong&gt;.&lt;br&gt;&#xA;Maybe, in current compressor (turbine) stages, the rotors (stators) accelerate the air too much for the stators (rotors) to slow down, thus the external radius decreases (increases) to compensate and keep the relative air speed at the next stage constant.&lt;/p&gt;&#xA;" OwnerUserId="6589" LastActivityDate="2016-06-01T17:48:53.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9130" PostTypeId="2" ParentId="9108" CreationDate="2016-06-01T21:33:42.123" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;From my understanding, mass participation for each mode is determined&#xA;  by the general mass and the eigenvector of the mode, and eigenvector&#xA;  of the mode usually consists of contribution from all the $x$, $y$ and&#xA;  $z$ direction, so how can one tell whether the mode is contributed&#xA;  to lateral or vertical direction?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is correct in general, but for a simple and very symmetrical frame model like your drawing it is quite likely that the &lt;em&gt;lowest&lt;/em&gt; frequency modes of the response will either be a mainly lateral &quot;swaying&quot; or &quot;twisting&quot; motion, or a mainly vertical &quot;pogo stick&quot; motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that situation, you can decide whether the modal mass participation is &quot;mainly vertical&quot; or &quot;mainly lateral&quot; simply by looking at plots of the mode shapes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;higher&lt;/em&gt; frequency mode shapes will tend to be more localized, and have more cross-coupling between &quot;lateral&quot; and &quot;vertical&quot; motion, but in a simple model they may be too inaccurate to be of much practical value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating more &quot;accurate&quot; values instead of the zeros shown in the table may not add much value to understanding how the structure behaves, especially in a text-book example, because the differences between the real structure and the simple model may be greater than the errors in taking the &quot;small&quot; mass participation factors as zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After splitting the modes into two groups in this way, you can (approximately) consider the effect of vertical and lateral base loading independently.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-06-01T21:33:42.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9131" PostTypeId="1" AcceptedAnswerId="9132" CreationDate="2016-06-01T21:33:53.423" Score="4" ViewCount="112" Body="&lt;p&gt;I asked &lt;a href=&quot;https://engineering.stackexchange.com/questions/9125/optimum-airflow-for-drying-with-heated-air&quot;&gt;a question&lt;/a&gt; where a comment pointed out:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Requirements are things that you should be drafting to figure out what you want, before you know exactly how you are going to solve it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was doing it backwards. I was going to define requirements after I had investigated design ideas to weigh up compromises that might be made. What I called requirements were flexible, depending on the cost vs effectiveness of design ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I do not know what I want before I consider design ideas, if the possible criteria are not requirements, then what are they? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is the product fits into a larger system, ultimately the goal is profit from a business perspective, but considering the whole system's economics while designing the one product would be too complicated, so the requirement for &quot;maximum profit&quot; doesn't help!&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-06-04T22:18:52.157" Title="What if project requirements depend on discoveries while designing?" Tags="|project-management|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9132" PostTypeId="2" ParentId="9131" CreationDate="2016-06-01T22:16:36.163" Score="2" Body="&lt;p&gt;History tends to teach us that attempting to design something without having a clear idea of what you ultimately want to achieve rarely works out well. Equally the definition of the problem to be solved should address the underlying problem, not just be an outline of a proposed solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is, of course, a perfectly valid place for open-ended or speculative research and development, but this isn't quite the same thing as actually designing something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, if the requirements of a design seem very flexible it's because they haven't been defined well enough. The requirements should define what you want to achieve not a wishlist of 'features' which may or may not assist in solving the core problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course context matters a lot. For example if you are a member of a design team it may well be that another designer will give you a very specific set of specifications. Say you are designing an alternator for a specific engine the lead designer may well tell you specifics like maximum current capacity, duty cycle, average current capacity, voltage tolerance, maximum mass, environmental durability metrics and life expectancy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if you are developing an overall design concept for a car you might be negotiating rather more diffuse goals with strategic marketing people who might very well have rather more abstract objectives in mind and it is when it comes to 'designing' a marketable product rather than a strictly engineering a solution to a well defined problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ultimately comes down to the fact that the goal of '&lt;em&gt;what can I design that people will buy at a profitable price for me&lt;/em&gt;' is not a strictly engineering problem. In the same way that '&lt;em&gt;how can I improve the reliability of this pump without increasing its cost too much&lt;/em&gt;', very much is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly anyone who has worked in any kind of freelance capacity will be well aware that what a client says they &lt;em&gt;want&lt;/em&gt; is often a statement of how they imagine the problem should be solved rather than a definition of what they actually want the design to achieve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that there is a good argument that one of the defining skills of engineering is the ability properly identify what the real problem/need is without presupposing a particular solution and similarly to be able to differentiate between problematic symptoms and their underlying cause. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-02T12:51:29.367" LastActivityDate="2016-06-02T12:51:29.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9133" PostTypeId="2" ParentId="9131" CreationDate="2016-06-01T23:40:07.667" Score="1" Body="&lt;p&gt;In context of the question, the goal is to engineer a product to maximize the return on investments (ROI). In order to achieve this one has to engineer the product meet Voice of the Consumer/Customer (VOC) which essentially is the requirements. The hardest part is to understand VOC, or &lt;strong&gt;&quot;What is the problem consumers are seeking a solution?&quot;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no one size fits all answer to effective requirements gathering. This depend on many things such as the industry, business objectives, size of the market, and many more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Successful companies such as Apple, Samsung are masters in understanding VOC. In most of these companies the requirements are generated by the Marketing team based on consumer insights. These consumer insights help create requirements documents which in turn are handed over to research and development to generate engineering specifications. In turn research and development develop functional prototype use for consumer testing. Consumer testing helps further refines the requirement before handing fairly well defined set of requirement including cost, and go to market timing to product engineering to fully engineer the product for commercialization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary &lt;strong&gt;critical design requirements&lt;/strong&gt; should be defined upfront, but it is acceptable to refine some during the design and development process. There are many good write-ups and organizations focusing on this topic and some of them are listed below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/0465025781&quot; rel=&quot;nofollow&quot;&gt;Winning at New Products: Creating Value Through Innovation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.pdma.org/&quot; rel=&quot;nofollow&quot;&gt;The Product Development and Management Association (pdma)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-06-02T12:08:39.673" LastActivityDate="2016-06-02T12:08:39.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9134" PostTypeId="1" CreationDate="2016-06-02T01:03:22.170" Score="-1" ViewCount="114" Body="&lt;p&gt;I am trying to figure out the best way to connect an &lt;a href=&quot;http://www.meas-spec.com/downloads/DP86.pdf&quot; rel=&quot;nofollow&quot;&gt;O-ring mounted differential pressure sensor&lt;/a&gt; into 1/8&quot; female NPT. Ideally would like to do it with something like aluminum tubing, but I'm not sure what the best adapters are. I don't have the ability to custom fabricate a solution so I need an off the shelf solution.&lt;/p&gt;&#xA;" OwnerUserId="6672" LastEditorUserId="368" LastEditDate="2016-06-02T19:00:09.860" LastActivityDate="2016-08-02T14:59:56.263" Title="How to mount an O-ring differential pressure sensor in a pipe" Tags="|sensors|hydraulics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9135" PostTypeId="2" ParentId="9131" CreationDate="2016-06-02T05:55:25.920" Score="2" Body="&lt;p&gt;Requirements come in 2 flavours, client requirements and technical requirements. Client requirements are converted to technical requirements because they are usually too abstract to be used as absolute metrics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the client might say &quot;light,&quot; but your technical goal is something more tangible like &quot;0.5 kg or less.&quot; It is important to understand that something might have been lost in translation here and that this is a slight corruption of the message so it's easier for you to work with the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now while exploring the design space you might need to specify secondary considerations. These then may become specifications to secondary engineering tasks, requirements for subcomponents if you wish. The thing to notice is that the client has changed, it's no longer your original client it is your design (or worse, yourself) that has transformed into a client.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mixing up client wishes is fine as long as you are still fulfilling the original client goals. However, dragons lie this way. You may ultimately lose sight of what the clients want. Carcasses of many businesses lie on this road. So, best not go there and keep your design realities separated from your requirements. Do not mix requirements you made up with real requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many companies will do this on multiple levels. Managers try to redefine the clients so that work is easier on them. Engineers, designers and sales specify too tight or to lose requirements. Often they bring in themselves into the process thinking their needs are client needs. Sometimes this is good sometimes bad, but ultimately you do not know if they are or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to finally answer the question: These are, most likely, not requirements but design realities, limitations and so on. The client needs something, the technical solutions something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not the requirements you are looking for. Move along.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="368" LastEditDate="2016-06-02T19:31:10.333" LastActivityDate="2016-06-02T19:31:10.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9136" PostTypeId="1" AcceptedAnswerId="9137" CreationDate="2016-06-02T07:20:25.903" Score="-1" ViewCount="583" Body="&lt;p&gt;I'm looking for a rubber thread that can be stretched elastically to 10 or 20 times of its original length. Does such material exist? Which is the best one in this respect?&lt;/p&gt;&#xA;" OwnerUserId="6675" LastActivityDate="2016-06-02T14:17:56.763" Title="What is the most stretchable rubber thread?" Tags="|materials|" AnswerCount="2" CommentCount="5" ClosedDate="2016-06-02T19:15:16.117" ContentLicense="CC BY-SA 3.0" />
  <row Id="9137" PostTypeId="2" ParentId="9136" CreationDate="2016-06-02T12:46:17.380" Score="1" Body="&lt;p&gt;A quick lmgtfy reveals a &lt;a href=&quot;http://www.livescience.com/22945-super-stretchy-material-also-super-strong.html&quot; rel=&quot;nofollow&quot;&gt;hydroelastomer&lt;/a&gt; made of alginate polyacrylamide and water with a stretch ratio over 20:1.  I believe that article refers to &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3642868/&quot; rel=&quot;nofollow&quot;&gt;this paper&lt;/a&gt; . &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-06-02T12:46:17.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9138" PostTypeId="2" ParentId="9115" CreationDate="2016-06-02T14:01:14.270" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.owlnet.rice.edu/~ceng403/nh3ref97.html&quot; rel=&quot;nofollow&quot;&gt;This report from Rice University covers design and economics of synthesis gas reformers.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction 1: $CH_4+H_2O&amp;lt;-&amp;gt;CO+3H_2$ (Steam reforming)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction 2: $CO+H_2O&amp;lt;-&amp;gt;CO_2+H_2$ (Water gas shift)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaction 3: $CH_4+3/2O_2&amp;lt;-&amp;gt;2H_2O+CO$ (Combustion)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reactions 1 &amp;amp; 2 are endothermic and occur in the primary reformer. Reaction 3, the combustion reaction, is exothermic and occurs along with reactions 1 and 2 in the secondary reformer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Among many other reasons, air is not introduced in the primary reformer so that we get the most methane conversion to hydrogen with the smallest reformer possible. The link above explains sizing considerations between primary and secondary reformers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The secondary reformer is a much simpler vessel and has the primary purpose to burn up extra methane that has slipped out of the primary reformer. The heat from the exothermic combustion reaction is recovered and used for the primary reforming. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the air question from the comments, there are two ways to get nitrogen into the process. One method is using an air separation unit. The other is to put air in the secondary reformer, combust the methane and oxygen, and keep the left over nitrogen in the syngas for ammonia conversion. &lt;/p&gt;&#xA;" OwnerUserId="3569" LastActivityDate="2016-06-02T14:01:14.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9139" PostTypeId="2" ParentId="9136" CreationDate="2016-06-02T14:17:56.763" Score="1" Body="&lt;p&gt;This is quite a broad question, but in an engineering context the main properties you are looking for are &lt;a href=&quot;https://en.wikipedia.org/wiki/Elastic_modulus&quot; rel=&quot;nofollow&quot;&gt;elastic modulus&lt;/a&gt; which is stress/strain is how much a material stretched under load and &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultimate_tensile_strength&quot; rel=&quot;nofollow&quot;&gt;ultimate tensile strength&lt;/a&gt; OR &lt;a href=&quot;https://en.wikipedia.org/wiki/Yield_(engineering)&quot; rel=&quot;nofollow&quot;&gt;yield strength&lt;/a&gt; which are both measures of how much load a material will sustain before it fails. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Materials with a high modulus of elasticity are broadly classified as 'rubbers', this may include 'natural' rubbers (derived from latex) or various silicone, vinyls,  polyurethanes etc etc. This category of materials is usually referred to in materials science as &lt;em&gt;elastomer&lt;/em&gt;s, although the term &lt;em&gt;rubber&lt;/em&gt; is also widely used &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-06-02T14:17:56.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9140" PostTypeId="1" CreationDate="2016-06-02T14:23:54.443" Score="6" ViewCount="636" Body="&lt;p&gt;I'm interested into the power radiated out of an eNodeB (LTE/4G) radio base station: which parameters does this power depend on, if they are constant or relying on the environment...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this in order to be able to predict the rated power transmitted out of the BS antenna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been seeking in the 3GPP technical requirements and sheets, but nothing conclusive about it until now, specificaly for LTE signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It shall differ from the others radio standards such as GSM or UMTS since LTE uses OFDM which implies multiple carriers instead of one. My guess is that those multiple carriers should in theory reduce the rated power per carrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still have some certitudes about it, those being that the output power depends on the cell size, also that traffic-load, somehow, is taken into account (potential small cells may be deployed for capacity boost).&lt;/p&gt;&#xA;" OwnerUserId="6507" LastEditorUserId="7083" LastEditDate="2016-07-21T00:30:50.287" LastActivityDate="2016-07-21T00:30:50.287" Title="How does LTE eNodeB set the Base Station output power?" Tags="|electrical-engineering|telecommunication|wireless-communication|lte|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9141" PostTypeId="1" CreationDate="2016-06-02T15:14:21.627" Score="2" ViewCount="26251" Body="&lt;p&gt;As I know so far, mortar and grout are mixtures of cement paste and fine aggregates, but both of them play different roles in the construction site. Is it true that grout has more strength compared to mortar? What are the differences between them?&lt;/p&gt;&#xA;" OwnerUserId="6422" LastEditorUserId="368" LastEditDate="2016-06-02T19:03:15.747" LastActivityDate="2016-08-24T10:23:15.340" Title="What is the difference between grout and mortar?" Tags="|materials|civil-engineering|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9144" PostTypeId="2" ParentId="9141" CreationDate="2016-06-02T15:39:18.640" Score="1" Body="&lt;p&gt;In general grout is used for filling gaps between courses of bricks, blocks or tiles usually once they are in place for a combination of  environmental resistance and cosmetic appearance. The main purpose or mortar is the structural bonding of laid blocks as they are laid,  usually  a combination of adhesion, load distribution and levelling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice this covers a huge range of materials, of which cement based ones are a fraction. For example there are a whole category of materials based on lime which can be used as mortars, grouts, renders and restoration/repair putty or indeed as castable media. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reference to the comment from Ethan48 you could extend this definition by saying that mortar tends to be a stiffer and more plastic formulation used as a sort of 'glue' whereas grout is finer and (relatively) more fluid for use in filling gaps in assembled structures. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-04T17:27:32.910" LastActivityDate="2016-06-04T17:27:32.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9145" PostTypeId="1" CreationDate="2016-06-02T17:08:59.297" Score="2" ViewCount="231" Body="&lt;p&gt;While solving the Navier-Stokes equation using &lt;strong&gt;SIMPLE&lt;/strong&gt; algorithm, we solve the transport equation for additional scalar at the last step according to the SIMPLE's procedure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to add a new scalar, reaction progress variable $c \ \epsilon \ [0,1] $, to model the extent of burning in case of diffusion flames, but I am not able to understand how is this variable coupled with the temperature equation?How do we recalculate the temperature once we have solved for $c$ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we proceed according to cycle below, can someone please expand upon the transport equation part (STEP 4)--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CGCL1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CGCL1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we solve for &lt;strong&gt;Energy (Enthalpy)&lt;/strong&gt; equation first and then find out &lt;strong&gt;$h$&lt;/strong&gt; and then, &lt;strong&gt;$c$&lt;/strong&gt; or is it something else? Because as far as I understand, I know that &lt;strong&gt;$c$&lt;/strong&gt; is used to calculate Temperature of the cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I am wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: $c$ is the reaction progress variable denoting the extent or fraction of reaction, $h$ is enthalpy, $u$ is speed, and the $2nd$ and $3rd$ term on RHS below denote source terms independent of $c$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial (\rho c)}{\partial t}  + \frac{\partial(\rho u c)}{\partial x} = \frac{\partial(\rho D_{eff} \frac{\partial(c)}{\partial x})}{\partial x} + \omega_{deflagration} + \omega_{detonation} $$&lt;/p&gt;&#xA;" OwnerUserId="1155" LastEditorUserId="1155" LastEditDate="2016-06-13T02:18:32.960" LastActivityDate="2016-06-13T02:18:32.960" Title="How is the reaction progress variable $c$ coupled with Temperature equation while solving the SIMPLE algorithm in CFD?" Tags="|fluid-mechanics|cfd|combustion|diffusion|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9146" PostTypeId="2" ParentId="9141" CreationDate="2016-06-02T18:39:17.360" Score="4" Body="&lt;p&gt;Mortar is used when you want materials to adhere to one another, as in brick laying or tile placement. Grout however is a material that is meant to flow, sometimes under pressure. The main purpose of using grout is to serve in situations where you need to fill gaps or holes that exist due to construction techniques (such as concrete cooling in huge masses of concrete or structural support on climbing formwork). The purpose of using both materials are different, although they are in essence made of the same components. As for resistance I would say it depends on the situation, I wouldn't say that grout has more strength compared to mortar without having a bit of context. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you might define a bit better what you mean by &quot;strength&quot;, one material can be more resistant to compression while the other more resistant to traction, so there isn't a direct comparison between the materials without having a bit of context on what applications you are using it for.&lt;/p&gt;&#xA;" OwnerUserId="6685" LastActivityDate="2016-06-02T18:39:17.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9147" PostTypeId="1" CreationDate="2016-06-02T20:04:10.420" Score="0" ViewCount="250" Body="&lt;p&gt;I have a model of a cantilever (fixed on one end, and free on all other boundaries) that bends in response to an internal stress field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the amount of force this cantilever can exert on an external object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experimental analogue, as far as I know, is the blocking force--if you start with the cantilever in a flat state and physically obstruct the bending of the cantilever, how much force must you exert in order to keep the cantilever flattened out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional Info&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model I'm using is a PDE that describes a non-linear, hyperelastic solid. For a given state of deformation, I can calculate the stress tensor and potential energy density throughout the cantilever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I have been using the finite element method to find the state of deformation in static equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="6686" LastActivityDate="2016-06-03T00:16:40.170" Title="Calculate force that can be exerted by a bending actuator" Tags="|mechanical-engineering|applied-mechanics|modeling|finite-element-method|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9149" PostTypeId="1" AcceptedAnswerId="9159" CreationDate="2016-06-02T22:54:38.580" Score="4" ViewCount="188" Body="&lt;p&gt;I need to control the temperature with high precision, and for this task would be good a temperature controller with pt-100 sensor, something like this one &lt;a href=&quot;http://www.ink-bird.com/product/detail/p/PID_Temperature_Controller_ITC-100/id/7&quot; rel=&quot;nofollow&quot;&gt;PID Temperature Controller ITC-100&lt;/a&gt;. As I see that, it can be very simple: there is an input (pt100 sensor), an output - to the heater element (my heating element needs 12V supply). However, I still have some things which are unclear for me:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is it generally supposed that there is a power source for the heater&#xA;element which I also put in the circuit or usually the temperature&#xA;controller gives the voltage at the output ? Or is that voltages&#xA;supposed for the external SSR, not my heater?&lt;/li&gt;&#xA;&lt;li&gt;In many sources I have seen circuits with external SSR. However,&#xA;such temperature controllers already have internal relay. Should it&#xA;be possible without any problem just wire output to my heater&#xA;(because in all schemes output goes to the external relays)?&lt;/li&gt;&#xA;&lt;li&gt;Wires soldered to the pt sensor shouldn't influence much to the&#xA;measurement, right?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6688" LastEditorUserId="6688" LastEditDate="2016-06-03T05:28:24.930" LastActivityDate="2016-06-04T03:03:10.873" Title="Controlling the temperature with high precision" Tags="|temperature|heating-systems|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9150" PostTypeId="2" ParentId="9097" CreationDate="2016-06-03T00:06:22.643" Score="1" Body="&lt;p&gt;&lt;strong&gt;1st Question:&lt;/strong&gt; To answer if you can consider it a centrally loaded column or not think of the effects an eccentric load have on your section. An eccentric load will have the same effect as a normal load applied together with a bending load. The normal load is of the same module of your eccentric load and the bending is proportional to the distance to the center. Since the load is not too far away from the center of the section I would say you can approximate it by considering centered load, but that would need some structural analysis. You could model as central load and see what safety factor you have by using the material you want, if you have some space than i think you should be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to make sure though you can calculate two different scenarios, one where you have a central load applied and other where you have a bending load applied with value $B = P\times l$ where $l$ is the distance from your point load to the center of the cross-section. That would cause and increase of tension with a magnitude of $$\sigma = \frac{B\times\frac{h}{2}}{I}$$ it being traction on the outside and compression on the inside. Putting together both tensions (normal and bending) you would have a total tension of $$\sigma = \frac{P}{A} \pm \frac{P\times l\times\frac{h}{2}}{I}$$ you should evaluate in which extreme of your cross-section which signal of the second factor is appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this expression, two things point to the fact that the second term will be irrelevant: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$I$ is much greater than $A$&lt;/li&gt;&#xA;&lt;li&gt;$l$ and $\frac{h}{2}$ are small compared to the other values&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is enough to convince me that modelling as a centered load is appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you must still be aware of the critical load to avoid buckling, but looking at your calculations it shouldn't be a problem (if one section needs $270 kgf$ to buckle I would expect it take more than a ton to cause buckling in one section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2nd Question:&lt;/strong&gt; You should know (maybe get this info with the manufacturer) what is the admissible traction and compression tensions on your OSB board and calculate how much load that is. Always think of tension rather than load because if you have a smaller cross-sectional area the same load will cause greater tensions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it you based yourself on your previous calculations to make this affirmation, you are not correct. You would be correct saying that the stool made of 12 mm OSB is able to withstand more than 1 ton without suffering buckling, but it would fail by mechanisms other than buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3rd Question:&lt;/strong&gt; The function of the nails in this piece is simply to keep the piece together i would say, nails and bolts are solicited when the piece needs to resist shearing stress. In your case, the top nails will not be solicited by the person sitting, because the load is transmitted by contact to the lower parts of your structure. The lateral nails will only need to resist a component of the person's weight that tries to &quot;open&quot; the legs of your stool, if you want to calculate that you should calculate how much of the person's weight becomes horizontal force that the nails must resist. If you have a value of that force, divide it by the number of nails and you should know to how much effort each nail will be subject. You would then need to know that this effort is normal to the nail as in trying to rip it of the board. Instinctively i wouldn't expect you to have problems with this (since the angle of the legs is close to vertical) and I would disregard this analysis. Not blindly though, but considering that the effort in the nails would be small.&lt;/p&gt;&#xA;" OwnerUserId="6685" LastActivityDate="2016-06-03T00:06:22.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9151" PostTypeId="2" ParentId="9147" CreationDate="2016-06-03T00:16:40.170" Score="1" Body="&lt;p&gt;Maybe you've tried this, but if you don't require an exact solution, you can simulate the experiment on finite element analysis, by restricting the other end of the beam and solving for the reaction force there. This reaction force is exactly what you want.&lt;/p&gt;&#xA;" OwnerUserId="6685" LastActivityDate="2016-06-03T00:16:40.170" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="9153" PostTypeId="2" ParentId="9098" CreationDate="2016-06-03T00:57:24.560" Score="1" Body="&lt;p&gt;The key characteristic of silicone polymers that make it so versatile is that they are not a single compound. Just like polyurethanes are extremely versatile organic polymers, polysiloxanes are extremely versatile organic/inorganic polymers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The backbone of the polysiloxane alternates silicon and oxygen atoms and as such is inorganic. But much of what gives this family of molecules its versatility is the variety of organic compounds on the Si atoms in the backbone:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CADBi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CADBi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Siloxane.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Siloxane&lt;/a&gt; by Sei.  &lt;a href=&quot;https://creativecommons.org/licenses/by-sa/3.0/deed.en&quot; rel=&quot;nofollow noreferrer&quot;&gt;CC BY-SA 3.0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the R groups can be a wide variety of organic group. If both R groups are methyl you get the vary famous polydimethylsiloxane. Cross-linkers can also be included. These can be sensitive to humidity so that they cure at room temperature (caulk) or fancier systems that cure in the lab to produce silicone rubbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question is that silicones are not a single substance that is super versatile, but a family of polymers that can vary in chemistry and structure, providing a wide range of controllable properties. Their main advantage over purely organic polymers are well listed in @rumtscho's answer.&lt;/p&gt;&#xA;" OwnerUserId="1031" LastActivityDate="2016-06-03T00:57:24.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9156" PostTypeId="2" ParentId="9149" CreationDate="2016-06-03T08:58:53.767" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;Wires soldered to the PT sensor shouldn't influence much to the measurement, right?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These PT sensors are based on electrical resistance, so extending the leads will cause a &lt;em&gt;slight&lt;/em&gt; change in the readings. You can minimise this with good connections and thick wire. (&lt;strong&gt;Thermocouples are different&lt;/strong&gt;, you would need to use the right sort of wire for those.)&lt;/p&gt;&#xA;" OwnerUserId="6665" LastActivityDate="2016-06-03T08:58:53.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9157" PostTypeId="2" ParentId="9116" CreationDate="2016-06-03T09:05:57.763" Score="2" Body="&lt;p&gt;I'm also a software guy and not completely master of the subject, but I tried to model your system. Each step is shown, so you can catch mistakes I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the schematic diagram of DC motor:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KWRR4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KWRR4.png&quot; alt=&quot;DC Motor Schematic Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kirchoff's voltage law for electrical circuit:&#xA;\begin{equation}\label{eq:kirchoff} \tag{1}&#xA;V_s = V_l + V_r + V_e \\&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's $2^{nd}$ law of motion:&#xA;\begin{equation}\label{eq:newton} \tag{2}&#xA;\sum{F} = ma&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation for Electro-Motive-Force (EMF):&#xA;\begin{equation}\label{eq:emf} \tag{3}&#xA;V_e = K_e \dot{\theta}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation for electro-mechanical convertion of tourque:&#xA;\begin{equation}\label{eq:torque} \tag{4}&#xA;\tau = K_ti&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using equations \ref{eq:kirchoff} &amp;amp; \ref{eq:emf}:&#xA;$$&#xA;u(t) = L \frac{di}{dt} + R i + K_e \dot\theta&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using equations \ref{eq:newton} &amp;amp; \ref{eq:torque}:&#xA;$$&#xA;K_t i = J \ddot\theta&#xA;$$&#xA;Laplace transfer of equations:&#xA;$$&#xA;U(s) = s L I(s) + R I(s) + s K_e \Theta(s) \\&#xA;K_t I(s) = s^2 J \Theta(s)&#xA;$$&#xA;$I(s)$ is common:&#xA;$$&#xA;I(s) = \frac{U(s) - s K_e \Theta(s)}{s L + R } \\&#xA;I(s) = \frac{s^2 J \Theta(s)}{K_t}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{U(s) - s K_e \Theta(s)}{s L + R } = \frac{s^2 J \Theta(s)}{K_t} \\&#xA;K_t U(s) - s K_t K_e \Theta(s) = s^3 J L \Theta(s) + s^2 J R \Theta(s) \\&#xA;K_t U(s) = s^3 J L \Theta(s) + s^2 J R \Theta(s) - s K_t K_e \Theta(s) \\&#xA;K_t U(s) = \Theta(s) ( s^3 J L + s^2 J R - s K_t K_e )&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transfer function of system with voltage input and position output:&#xA;$$&#xA;\frac{\Theta(s)}{U(s)} = \frac{K_t}{s^3 J L + s^2 J R - s K_t K_e}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to investigate voltage-speed relationship:&#xA;$$&#xA;s(\frac{\Theta(s)}{U(s)}) = s(\frac{K_t}{s^3 J L + s^2 J R - s K_t K_e}) \\&#xA;\frac{s\Theta(s)}{U(s)} = \frac{K_t}{s^2 J L + s J R - K_t K_e})&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying step input, which is PWM with 100% duty cycle:&#xA;$$&#xA;\Theta(s) = \frac{\alpha}{s} \frac{K_t}{(s^3 J L + s^2 J R - s K_t K_e)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be good to investigate stability of a system with impulse input instead of step input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With mobility analogy, Voltage &amp;amp; Velocity are cross variables, Current &amp;amp; Force are through variables, you can also investigate the system as RLC series circuit as shown below.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/F6eui.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F6eui.png&quot; alt=&quot;RLC Series Circuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $u(t)$ input $V_c$ output, it is voltage-velocity relation and $u(t)$ input $\int{v_c(t)dt}$ for voltage-position relation. So it adds 1 more degree to your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PjGm1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PjGm1.png&quot; alt=&quot;Voltage-Velocity Relation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1599" LastEditorUserId="1599" LastEditDate="2016-06-03T09:25:18.077" LastActivityDate="2016-06-03T09:25:18.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9159" PostTypeId="2" ParentId="9149" CreationDate="2016-06-03T12:40:31.700" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;1.Is it generally supposed that there is a power source for the heater element which I also put in the circuit or usually the temperature controller gives the voltage at the output ? Or is that voltages supposed for the external SSR, not my heater?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;2.In many sources I have seen circuits with external SSR. However, such temperature controllers already have internal relay. Should it be possible without any problem just wire output to my heater (because in all schemes output goes to the external relays)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Typically, the decision to use the controller's internal power source, internal relay or an external relay would be based on the requirements for the heater. If switching the heater's load is within the capacity of the internal power or relay, it's simpler to just use that. (Remember, simpler is almost always better since there are fewer things to fail.) On the other hand, if you have a 480V heater, you're probably going to need an external relay, mechanical or SSR. In this case, the internal relay can handle 250V/3A or 120V/6A. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wires soldered to the pt sensor shouldn't influence much to the measurement, right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can extend the leads on these sensors. There will be some effect on the measurement due to the added resistance of the extensions. How much this affects your measurement is going to depend on how much resistance you add with your extension wires. I would expect this to be minimal unless you're running a really long extension. It's likely that any such error can be compensated for when calibrating the system when it's setup in its final configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot of information on these controllers here: &lt;a href=&quot;http://ink-bird.com/asset/file/ITC-100_%20Manual_%20V1.0_6.30.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ink-bird.com/asset/file/ITC-100_%20Manual_%20V1.0_6.30.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at Section 7 of the manual I linked above, it shows sample system diagrams of the various modes in which the controller can be set up. It also indicates which model (There are 3) of the controller they are using for that setup.  &lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2016-06-03T14:54:17.883" LastActivityDate="2016-06-03T14:54:17.883" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="9160" PostTypeId="2" ParentId="8035" CreationDate="2016-06-03T13:07:24.007" Score="2" Body="&lt;p&gt;The answer is system specific. That is, unless you know something else about the structure of your porous material, there is no clear relation between pore density and pore size. A simplified 2D example may look like this (where the circles are pores):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IGpUB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IGpUB.png&quot; alt=&quot;same ppi different porosity&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both of these samples the pore density in ppi is the same, but the porosity is different. They are approximately 4 ppi samples but in the one on the left the porosity is about 0.5 and the one on the right the porosity is much less than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have confirmed this through a quick search of literature on foams like the one you describe. &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0017931001002204&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; one (behind a paywall) for example has a table where they list their samples and some of them have the same ppi but different porosities. In their works they describe their porous material as being made of pores with diameters $d_p$ and fibers of diameter $d_f$. Given that additional information, I guess someone could develop a relationship between pore density and pore size.&lt;/p&gt;&#xA;" OwnerUserId="1031" LastActivityDate="2016-06-03T13:07:24.007" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9161" PostTypeId="2" ParentId="8267" CreationDate="2016-06-03T13:19:19.883" Score="0" Body="&lt;p&gt;For your curved surfaces, the lever arm for the calculation of torque is a constant $D/2$. That is not true in the case of the cavern's faces. In that case the lever arm is a variable ($r$). To calculate the torque you would need to integrate the stress over the area times the lever arm. It should look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\int\int\tau_y\cdot r\, dA&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the double integral is over the area where the force acts. Be careful to pick the appropriate differential area for a constant z surface in cylindrical coordinates. I did the calculation and the answer matches the one you state.&lt;/p&gt;&#xA;" OwnerUserId="1031" LastActivityDate="2016-06-03T13:19:19.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9163" PostTypeId="2" ParentId="9134" CreationDate="2016-06-03T13:52:21.510" Score="0" Body="&lt;p&gt;That sensor is available in a package with NPT fittings from the manufacturer. Why not simply use one of those? You can find links to them here: &lt;a href=&quot;http://www.meas-spec.com/product/pressure/DP86.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.meas-spec.com/product/pressure/DP86.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not one of those, there are countless other pressure transducers that do have NPT fittings. I'm sure you can find one using the search engine of your choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you have a very good reason, there's no need to re-invent the wheel.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2016-06-03T13:52:21.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9164" PostTypeId="1" CreationDate="2016-06-03T14:20:14.717" Score="0" ViewCount="105" Body="&lt;p&gt;I was just wondering why when a system is stable there is a frequency response $\left(H(p) \rightarrow H(j\omega)\right)$ and when the system is unstable the frequency response doesn't exist?&lt;/p&gt;&#xA;" OwnerUserId="6458" LastEditorUserId="1832" LastEditDate="2016-06-03T19:19:26.087" LastActivityDate="2016-06-03T19:19:26.087" Title="Stability implies frequency response existance" Tags="|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9165" PostTypeId="2" ParentId="9164" CreationDate="2016-06-03T15:33:45.270" Score="1" Body="&lt;p&gt;Both stable and unstable systems have a frequency response. So your statement is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When creating a bode plot, the difference is in the phase at the crossover frequency, that is to say the frequency at which the magnitude is equal to 0 decibel or gain of 1. A stable system will have a phase larger then -180 degrees, where a system which is unstable has a phase smaller then -180 degrees. This has to do with the stability criterium of Nyquist. &lt;/p&gt;&#xA;" OwnerUserId="3357" LastActivityDate="2016-06-03T15:33:45.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9166" PostTypeId="2" ParentId="9116" CreationDate="2016-06-03T17:07:19.447" Score="5" Body="&lt;p&gt;Without any more info, I think your problem arises from the values of &lt;span class=&quot;math-container&quot;&gt;$a_0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$a_1$&lt;/span&gt;. The answer is a little involved, so a bit of systems background is necessary.&lt;/p&gt;&#xA;&lt;h1&gt;The short answer&lt;/h1&gt;&#xA;&lt;p&gt;Your system should be stable, but I don't think you are simulating your system long enough to see it stabilize. Calculate your settling time and simulate it for at least that long to see the complete response transient.&lt;/p&gt;&#xA;&lt;h1&gt;The long answer&lt;/h1&gt;&#xA;&lt;p&gt;The characteristic equation (i.e. denominator of the transfer function) of a general 2nd order system has the following convenient form:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$s^2 + 2 \zeta \omega_n s + w_n^2 $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In this form, &lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt; is known as the damping ratio and &lt;span class=&quot;math-container&quot;&gt;$\omega_n$&lt;/span&gt; is the natural frequency. As you noted this system is guaranteed to be stable if &lt;span class=&quot;math-container&quot;&gt;$w_n^2 &amp;gt; 0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$2 \zeta \omega_n &amp;gt; 0$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In systems engineering we have several rules of thumb for determining the transient response of such systems. One of these is called the settling time &lt;span class=&quot;math-container&quot;&gt;$t_s$&lt;/span&gt;. It is a measure of how long it takes for the amplitude of the system's oscillations to decrease below a certain threshold, usually 2% or 5% of the steady state value (i.e. the value at which the system stabilizes). You can calculate your settling time using the following approximation (this one is for the 2% criterion):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$t_s = \frac{4}{\zeta \omega_n}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Therefore, if you do not simulate your system until the settling time, you may see a system that does not appear to stabilize. Note that if &lt;span class=&quot;math-container&quot;&gt;$\omega_n$&lt;/span&gt; is very small (therefore your &lt;span class=&quot;math-container&quot;&gt;$a_0$&lt;/span&gt; parameter is very small) the settling time could be very long.&lt;/p&gt;&#xA;&lt;p&gt;For your reference, you can calculate &lt;span class=&quot;math-container&quot;&gt;$\omega_n$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt; from your parameters via the following:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega_n = \sqrt{a_0} \qquad \zeta = \frac{a_1}{2 \omega_n}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h1&gt;A demonstration&lt;/h1&gt;&#xA;&lt;p&gt;Check out what happens when I simulate a second order system with &lt;span class=&quot;math-container&quot;&gt;$\omega_n = 0.01$&lt;/span&gt; rad/s and &lt;span class=&quot;math-container&quot;&gt;$\zeta = 0.9$&lt;/span&gt; (i.e. &lt;span class=&quot;math-container&quot;&gt;$a_0 = 0.0001$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$a_1 = 0.0180$&lt;/span&gt;). Its settling time is approximately 444 seconds. The initial conditions are &lt;span class=&quot;math-container&quot;&gt;$\phi = 0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\frac{d\phi}{dt} = 0$&lt;/span&gt;. I applied a unit step input (i.e. a constant motor input equal to 1). The following figures show the same system with the exact same conditions, but simulated for different lengths of time. This was all done in MATLAB Simulink, my preferred math simulator.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sD0Lh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sD0Lh.png&quot; alt=&quot;Simulating for too little time&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KpTGU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KpTGU.png&quot; alt=&quot;Simulating for an appropriate amount of time&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As you can see in the first image it appears that my system is unstable, but in the second image it becomes clear that it is simply taking a very long time to stabilize.&lt;/p&gt;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-06-03T18:52:51.583" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="9167" PostTypeId="1" CreationDate="2016-06-03T11:09:46.767" Score="2" ViewCount="2280" Body="&lt;p&gt;Okay, so I've noticed that a lot of people grind coffee, and a lot of companies grind it on an industrial scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having produced some of the drink we called coffee, a huge proportion of the mass of the beans is simply thrown away, discarded. It's biodegradable, at least, but it's still waste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could coffee grounds, dried out and compressed, be used as a building material?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it can, it would have these benefits:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A building material from a wasted resource&lt;/li&gt;&#xA;&lt;li&gt;The source material is plentiful, it's everywhere.&lt;/li&gt;&#xA;&lt;li&gt;No additional mining or agriculture would be required to produce it.&lt;/li&gt;&#xA;&lt;li&gt;Like wooden housing, it would bio-degrade if left abandoned. So it's a low environmental impact.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Questions on this topic: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can it be done? &lt;/li&gt;&#xA;&lt;li&gt;What challenges would it present? &lt;/li&gt;&#xA;&lt;li&gt;What practical issues are keeping this from happening?&lt;/li&gt;&#xA;&lt;li&gt;What would it be like living in a coffee house?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="AJFaraday" LastActivityDate="2016-06-03T17:15:34.143" Title="Why don't we make bricks out of coffee grounds?" Tags="|building-physics|" AnswerCount="1" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9168" PostTypeId="2" ParentId="9167" CreationDate="2016-06-03T15:01:26.377" Score="4" Body="&lt;p&gt;Bricks are primarily made of sand and clay. The sand acts as a filler, and clay is a substance that has some rather remarkable qualities. Coffee grounds do not exhibit the qualities of sand or clay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, people have suggested using coffee grounds to improve various qualities of bricks. Coffee grounds exhibit one quality that neither sand nor clay possess: Coffee grounds burn. The coffee grounds in a brick made with a mix of sand, clay, coffee grounds, and lime will burn upon being fired, thereby increasing the porosity of the fired brick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since coffee grounds burn (they burn quite nicely, better than wood), another use for spent coffee grounds is artificial logs for fireplaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://link.springer.com/article/10.1617/s11527-015-0525-6&quot; rel=&quot;nofollow&quot;&gt;P. Muñoz Velasco, et al., &quot;Eco-fired clay bricks made by adding spent coffee grounds: a sustainable way to improve buildings insulation,&quot; &lt;em&gt;Materials and Structures&lt;/em&gt; 49.1-2 (2016): 641-650&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.com/patents/US5910454&quot; rel=&quot;nofollow&quot;&gt;Rodney K. Sprules, &quot;Coffee-based solid fuel composition&quot;, US patent 5910454 A&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="219" OwnerDisplayName="David Hammen" LastActivityDate="2016-06-03T15:01:26.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9170" PostTypeId="1" AcceptedAnswerId="9185" CreationDate="2016-06-03T18:08:33.200" Score="1" ViewCount="203" Body="&lt;p&gt;What is the name of the angle if I have larger wheels on the rear axles compared to the front, this would make the cars body and chassis not parralel to the road, unlike if the wheels were the same size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words the body and chassis would be at an angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this angle have a certain technical name?&lt;/p&gt;&#xA;" OwnerUserId="6697" LastActivityDate="2016-06-08T17:31:09.870" Title="Entire Car Chassis &amp; Body Angle" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9171" PostTypeId="2" ParentId="5323" CreationDate="2016-06-03T21:15:07.557" Score="5" Body="&lt;p&gt;The solution is fairly straightforward.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Short answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The system is controllable without any modifications. You made a small mistake calculating the new controllability matrix: you are missing the first column of $\bar{M}$, which should be $\begin{bmatrix}0 &amp;amp; I\end{bmatrix}^T$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Full proof&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let's consider the second system, and define some augmented matrix placeholders:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix}x' \\ y'\end{bmatrix} = \begin{bmatrix}A &amp;amp; B\\ 0 &amp;amp; 0\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix}&#xA;+ \begin{bmatrix}0 \\ I\end{bmatrix} u =&#xA;\tilde{A}\tilde{x} + \tilde{B}u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to find out if this system is controllable, we have to prove that its controllability matrix $\tilde{M}$ has full rank (i.e. all its rows or columns are linearly independent). The expression for the controllability matrix is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tilde{M} = \begin{bmatrix}\tilde{B} &amp;amp; \tilde{A}\tilde{B} &amp;amp; \tilde{A}^2\tilde{B}&amp;amp; \dots &amp;amp; \tilde{A}^{n+m-1}\tilde{B}\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start to solve for each term in terms of $A$ and $B$, a pattern begins to emerge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tilde{A}\tilde{B} = \begin{bmatrix} B \\ 0 \end{bmatrix} \qquad&#xA;\tilde{A}^2\tilde{B} = \begin{bmatrix} AB \\ 0 \end{bmatrix} \qquad&#xA;\tilde{A}^3\tilde{B} = \begin{bmatrix} A^2 B \\ 0 \end{bmatrix}  \dots \qquad&#xA;\tilde{A}^p\tilde{B} = \begin{bmatrix} A^{p-1} B \\ 0 \end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we can rewrite $\tilde{M}$ as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tilde{M} = \begin{bmatrix}0 &amp;amp; B &amp;amp; AB &amp;amp; A^2B &amp;amp; \dots &amp;amp; A^{n-1}B &amp;amp; A^n B &amp;amp; \dots &amp;amp; A^{n+m-2}B\\ I &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \dots &amp;amp; 0 &amp;amp; 0 &amp;amp; \dots &amp;amp; 0\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the middle of the first row is actually $M$, which we can then substitute into $\tilde{M}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tilde{M} = \begin{bmatrix}0 &amp;amp; M &amp;amp; A^n B &amp;amp; \dots &amp;amp; A^{n+m-2}B\\ I &amp;amp; 0 &amp;amp; 0 &amp;amp; \dots &amp;amp; 0\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have this matrix in a workable form, let's look at the elements to see if we can make a statement about its rank. Since $M$ occupies the first $n$ rows of $\tilde{M}$, and we know that $M$ has full rank, we can conclude that those rows are linearly independent with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to ask, are the final $m$ rows linearly independent with respect to each other and the first $n$ rows? Focusing on the first column only, we see the null matrix $0$, followed by the identity matrix $I$. The identity matrix has full rank, and all its rows are also linearly independent of the null matrix. Therefore, without looking at the other columns we can conclude that the last $m$ rows are linearly independent with respect to each other and the rest of the rows in $\tilde{M}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, all rows of $\tilde{M}$ are linearly independent, therefore it is full rank, therefore the system is controllable.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="368" LastEditDate="2016-06-21T16:24:54.777" LastActivityDate="2016-06-21T16:24:54.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9172" PostTypeId="2" ParentId="9170" CreationDate="2016-06-04T01:48:57.097" Score="2" Body="&lt;p&gt;In many cases, automotive engineers borrow from the field of aviation to define vehicle rotation about the three major axes.  Using their terminology, the angle you refer to would be called &quot;pitch&quot;.  In addition to larger-diameter rear tires, one could say a vehicle &quot;pitches forward&quot; under heavy braking.  The other axes are yaw (as if the car were on a turntable) and roll (the tilting experienced during fast cornering).&lt;/p&gt;&#xA;" OwnerUserId="6699" LastActivityDate="2016-06-04T01:48:57.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9174" PostTypeId="1" CreationDate="2016-06-04T02:00:51.163" Score="0" ViewCount="3794" Body="&lt;p&gt;I am interested in becoming an industrial engineer. I know the basics of Java and Python but I cannot program well. I am wondering whether it is still possible for me to become an industrial engineer. &lt;/p&gt;&#xA;" OwnerUserId="6701" LastActivityDate="2016-06-04T12:36:10.500" Title="Amount of programming needed to become a decent industrial engineer" Tags="|industrial-engineering|" AnswerCount="3" CommentCount="2" ClosedDate="2016-06-04T14:42:58.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="9175" PostTypeId="2" ParentId="9149" CreationDate="2016-06-04T03:03:10.873" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;Wires soldered to the pt sensor shouldn't influence much to the measurement, right?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A &quot;pt100&quot; sensor has a resistance of about 100 ohms (hence its name!) and changes resistance by about 0.385 ohms per degree C. Your question doesn't say what you mean by &quot;high precision&quot; but those two numbers should give you an idea of what lead resistance is acceptable for your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use long leads and still get &quot;high precision&quot;, it would be better to use a 3-wire RTD probe, as shown in the ITC-100 documentaton. That configuration uses a Wheatstone bridge type of circuit, and only the &lt;em&gt;difference&lt;/em&gt; in resistance between the three leads contributes to the error. Three-wire probes can use leads up to about half a kilometer long. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-06-04T03:03:10.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9176" PostTypeId="2" ParentId="9174" CreationDate="2016-06-04T03:03:50.557" Score="1" Body="&lt;p&gt;Industrial engineering is not one of those professions where the engineers need to be particularly good programmers, unless you plan on developing industrial engineering software, in which case, it could be a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because IE involves optimization of industrial processes and procedures, you probably will find yourself learning how to use various computer tools to assist in these tasks, much like other engineers would learn how to use spreadsheets or CAD programs, but you would not be expected to develop these tools yourself from scratch.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-06-04T03:03:50.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9177" PostTypeId="2" ParentId="9174" CreationDate="2016-06-04T03:56:15.567" Score="-1" Body="&lt;p&gt;All higher level functions in engineering start to resemble programming jobs more and more as time goes by. Many students want to avoid hearing this, after all they did not choose computer science (CS). Do you need it?  No not really, but it can be extremely useful for many careers and valuable in general. Same could be said for most of your subjects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets take the CAD as an example. CAD is asymptotically closing in on programming. Sure you can use CAD in a stupid manner but parametrical CAD design is allready a weird graphical programing language in its own right. When you start taking more advantage of it to make Design Automation and increasing your own productivity your smack middle in programing land. Not computer science mind you but programming nonetheless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case point analytical software like FEM and other simulation packages usually require you to be atleast somewhat programmer savy to get into those optimisation problems that currently arent being solved. You need to be able to run many simulations and touchup larger and lager sets of data. Sure using simulation data management helps but this is moving towards scripting related things all the time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same applies to calculations they are all moving to computers and you would certainly call the things programs. I know people tend to use spreadsheets for many of these tasks, but there is a limit to how practical a spreadsheet is in many contexts. The advice i would give you is to learn to graph in python or matlab it sidesteps many problems and mistakes you will make in excel. This also helps you understand physics better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other areas such as documentation, requirements are strongly moving into this direction as well. Automation is also some sort of programing branch and you should be atleast somewhat familiar with those things. Already most but the softest of soft Ph.D level works involve programming, most master levels follow soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do not nessecerily have to be a stellar programer, thats what the cs majors are for. You just need to be the guy who can whip up a program to solve a problem in the domain of your speciality, because CS graduates lack this. This is also something unversities in general approach badly so taking more courses in computer sciece wont help you much. But do take a algorithms 101 course, that is useful if you ever need to build a machine to separate or sort things and teaches basics of what Compuer Science as opposed to programming is really about.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2016-06-04T09:04:29.683" LastActivityDate="2016-06-04T09:04:29.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9178" PostTypeId="2" ParentId="9174" CreationDate="2016-06-04T12:36:10.500" Score="0" Body="&lt;p&gt;This is actually a complex question. Learning the syntax of Java or Python specifically has a decent risk of not being especially helpful at least in the sense that your next workplace may not actually use those languages. What does stand a chance of being helpful is the process of abstracting your eningeering problem into something that can be turned into code - this will challenge your understanding of the underlying eningeering principles and make you a better engineer. But to do this you need to have reached a level in programming where converting an engineering task into code is a feasible objective ( to a degree it is not necessary to succeed at this task to obtain this benefit but it is necessary to be able to attempt it seriously).&#xA;What is also useful is to learn enough of a language that you can consider the semantics ahead of the syntax - then you can effectively generalise your knowledge of programming into situations which are unrelated to the language(s) you learned which may well not be the ones you are required to code in. This is related to the other answers references to programming skills increasing CAD or spreadsheet skills.&lt;/p&gt;&#xA;" OwnerUserId="6704" LastActivityDate="2016-06-04T12:36:10.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9180" PostTypeId="1" CreationDate="2016-06-04T13:50:08.483" Score="1" ViewCount="46" Body="&lt;p&gt;While discharging water from a short tube, I noticed that a vortex began to form on the top of the discharge tube causing the jet issuing from a short tube to contract (less than the cross sectional area of the short tube). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why does the jet contract below the short tube?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the vortex forms the velocity is at a maximum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_1v_1=A_2v_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the area must be a minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also observed that the contracted jet maintains this state until all the water is discharged from the container. Could this be explained in terms of head loss caused by the vortex?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ovw1B.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ovw1B.jpg&quot; alt=&quot;Jet before&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xPKaS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xPKaS.jpg&quot; alt=&quot;Jet after vorcity formation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6316" LastEditorUserId="1031" LastEditDate="2016-06-04T23:42:50.497" LastActivityDate="2016-08-07T23:35:23.957" Title="Why does a vortex at a tube's entry causes a contraction below a discharging container's short tube?" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9181" PostTypeId="1" CreationDate="2016-06-04T15:59:53.920" Score="3" ViewCount="147" Body="&lt;p&gt;I am designing a DC-DC boost converter using an SMD LM2623 and an inductor which is also SMD of 4.7&amp;nbsp;μH. I want to first build a prototype but I can't use an ordinary breadboard. Could someone please advise me whether there is a way I can mount SMD components for testing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, I have failed to get an electronic simulator having LM2623, does someone know any simulator which can help me do my circuit analysis?&lt;/p&gt;&#xA;" OwnerUserId="6093" LastEditorUserId="1832" LastEditDate="2016-06-04T16:49:42.317" LastActivityDate="2016-07-06T00:51:22.757" Title="SMD Bread board and electrical simulation software" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9182" PostTypeId="2" ParentId="9106" CreationDate="2016-06-04T16:20:13.757" Score="2" Body="&lt;p&gt;The original question was along the lines of: &quot;what happens if light passes through cascaded gratings&quot;. Before that question can be answered, you have to understand what a single grating does. See the image below: a single beam of light will be split up into diffraction orders, numbered $n=0, \pm 1, \pm 2, \ldots$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vsHwo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vsHwo.png&quot; alt=&quot;Generic grating with laser beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the incident light beam is a mixture of wavelengths, then all diffraction orders other than 0 will have diffraction angles $\theta$ depending on the wavelength and the order, i.e., $\sin\theta=n\lambda/d$, where $d$ is the grating line pitch and $\lambda$ the wavelength. This is for a transmissive grating with normal incidence. If the incidence is close to normal, the formula is approximately correct; for large angles, the formula gets a bit more complicated (&lt;a href=&quot;https://en.wikipedia.org/wiki/Diffraction_grating#Theory_of_operation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;). For a reflective grating, the angles are roughly half of this. If there is no solution for $\theta$ for a given value of the $n$, it means that that diffraction order does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no such thing as &quot;centered on a particular wavelength&quot; for a grating. At most, you can have &quot;optimized for a particular wavelength&quot; in the sense that the ratio between intensities of the diffraction orders has a particular value for a particular wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A grating can be designed such that the zeroth order is suppressed, like in the picture below. Generally, a high suppression is only possible for a single wavelength-angle combination. In this picture, the incident angle is 0 deg (normal incidence).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UDU23.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UDU23.png&quot; alt=&quot;grating without 0th order&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may think that suppressing the zeroth order is a good thing if you want to protect aircraft pilots against laser beams. However, if you consider the scenario where the laser beam can have any angle and may hit the grating at any position, then you'll see that this approach will not really help the pilot. In the following picture, laser beam B1 would have hit the pilot's eye if there hadn't been a grating and laser beam B2 would not have hit the pilot's eye. With the grating in place, the pilot's eye gets hit by the $n=-1$ diffraction from beam B2. For this grating, which has only  the orders $n=\pm 1$, you roughly double the chance of a random beam hitting the pilot and any hit will be at half of the original intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fHKQv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fHKQv.png&quot; alt=&quot;Two beams on a grating&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we understand what a single grating does, we can look at the case of cascaded gratings, which should speak for itself. Now, each incident beam will result in four new beams -- or even more if the gratings allow other orders than $n=\pm1$. If the gratings have different line pitches $d_1$ and $d_2$, then there will be different diffraction angles $\theta_1$ and $\theta_2$ for the first and the second grating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vUsDX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vUsDX.png&quot; alt=&quot;One beam, two gratings&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can arrange the gratings in such a way that you put some object (maybe a third grating) at the location where beams B1b and B2a cross. Depending on the exact wavelength and path lengths of the two paths, the beams may interfere destructively or constructively. The patent and the paper that you refer to are along those lines. You should bear in mind that in such applications, the incident angle of the beam incident on the first grating and the distances between the gratings and the position of the detector should be exactly right within a fraction of the wavelength; otherwise it won't work. For the application that you have in mind (aircraft cockpits), this is not the way to go. Keep in mind that &quot;background light&quot; and &quot;foreground light&quot; is a subjective distinction. All that matters for the grating is the angle, the wavelength, and the position on the grating where a light ray hits the grating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, if you replace the cockpit windows by diffraction gratings, the light from the horizon and the landing strip will be diffracted as well and will make it nearly impossible to see objects on the outside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to protect a pilot against the most common high-power laser pointers (532 nm wavelength), you should be using &lt;a href=&quot;https://en.wikipedia.org/wiki/Dielectric_mirror&quot; rel=&quot;nofollow noreferrer&quot;&gt;dielectric mirrors&lt;/a&gt; as filters. Such mirrors have a coating that is a stack of thin layers, tuned to reflect light of a certain wavelength. Schematically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PDRFq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PDRFq.png&quot; alt=&quot;Dielectric mirror&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can be designed to reflect only over a very narrow wavelength range, in which case they are called &quot;notch filter&quot;. You can buy these off the shelf, for example &lt;a href=&quot;http://www.edmundoptics.eu/optics/optical-filters/notch-filters/od-4-notch-filters/67110/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Edmunds OD4 notch filter&lt;/a&gt;. A disadvantage for pilot protection is that the reflecting wavelength will depend on the angle of incidence $\alpha$. The only way to make them reflect 532 nm wavelength over a wide range of angles is by making them reflect over a large range of wavelengths. If you do that for green light, you will probably end up with something resembling very dark sunglasses, which only transmit deep red and deep blue light and block orange-yellow-green-cyan.&lt;/p&gt;&#xA;" OwnerUserId="6350" LastEditorUserId="6350" LastEditDate="2016-06-08T07:21:35.970" LastActivityDate="2016-06-08T07:21:35.970" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="9183" PostTypeId="1" AcceptedAnswerId="9184" CreationDate="2016-06-04T16:22:06.553" Score="5" ViewCount="1239" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Choked_flow#Mass_flow_rate_of_a_gas_at_choked_conditions&quot; rel=&quot;noreferrer&quot;&gt;According to compressible flow theory&lt;/a&gt;, a downstream pressure of 0.528 times upstream pressure is sufficient to choke a flow device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do industrial grade pressure regulators throttle a wide range of pressures (say 150&amp;nbsp;bar to 20&amp;nbsp;bar)?&lt;/p&gt;&#xA;" OwnerUserId="6705" LastEditorUserId="1832" LastEditDate="2016-06-04T16:47:31.253" LastActivityDate="2016-06-05T07:10:16.720" Title="How do gas flow regulators avoid choking?" Tags="|fluid-mechanics|pressure|airflow|gas|compressible-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9184" PostTypeId="2" ParentId="9183" CreationDate="2016-06-04T16:39:01.293" Score="6" Body="&lt;p&gt;The short answer is that they don't. Choking is not a bad thing. For a gas flow regulator, it's actually a good thing, because under choking conditions, the mass-flow rate is no longer dependent on the downstream pressure. The mass-flow rate is still dependent on the upstream pressure and the size of the aperture connecting the high and low-pressure sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a reduction valve that is designed to keep the downstream pressure constant, rather than the mass flow rate, a mechanism is needed to open and close the aperture depending on whether the downstream pressure is below or above the pressure set point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; if you're worried about energy losses in the gas stream, due to viscous dissipation: don't. If a &lt;a href=&quot;https://en.wikipedia.org/wiki/Perfect_gas&quot; rel=&quot;nofollow&quot;&gt;perfect gas&lt;/a&gt; (that's an ideal gas with an additional condition for the specific heat) enters a thermally insulated throttle device at low speed and exits it at low speed (through a pipe with a large diameter), then its temperature will be the same as when it entered. One way is to look at the enthalpy of the gas, $H=m C_V T+ pV$, which must be conserved if no energy is added or removed along the way. Because $pV=nRT$, this condition can only be met if the gas has the same temperature before and after. Note that I mentioned &quot;low speed&quot;, which is what you typically want in a pressure regulator. In the nozzle, where the speeds are sonic, the energy balance will also include kinetic energy of the gas jet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its may take a few moments for you to wrap your head around this (at least, for me it did, years ago).&lt;/p&gt;&#xA;" OwnerUserId="6350" LastEditorUserId="6350" LastEditDate="2016-06-05T07:10:16.720" LastActivityDate="2016-06-05T07:10:16.720" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="9185" PostTypeId="2" ParentId="9170" CreationDate="2016-06-04T17:43:36.887" Score="1" Body="&lt;p&gt;The fixed (unloaded), or typical angle of the floor of the car is usually referred to as &lt;em&gt;rake angle&lt;/em&gt; and can be a combination of suspension geometry (including relative wheel/tyre diameters) and the shape of the floor itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is particularly significant in cars which use ground effect to generate down-force where the floor, in conjunction with the front wing and a rear diffuser are designed to accelerate air under the floor of the car to create an area of low pressure by increasing the space under the floor and hence the cross sectional area of the flow from front to rear. Here the space enclosed between the floor and the road surface effectively forms an expanding nozzle although one of the technical challenges which limits this is to seal the gap between the floor and the road at the sides. There are examples where this has been achieved with flexible skirts (especially in early ground effect racing cars) although now it is more often achieved by precise control of airflow, especially vortices, off the front wing and other bodywork. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as the tyre diameter this can be controlled by the suspension geometry and spring/damper characteristics and it is perfectly possible to have a level, or even actively adjustable rake angle regardless of relative tyre diameters.  Indeed, cars with active suspension may adjust this angle in order to control the centre of pressure of downforce generated front to rear and/or to adjust the balance between downforce and drag. Some degree of control can also be achieved by flexible or movable sections of aerodynamic bodywork. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may also be expressed in terms of front to rear &lt;em&gt;ride height&lt;/em&gt; or &lt;em&gt;ground clearance&lt;/em&gt; (especially in an off-road context).&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-08T17:31:09.870" LastActivityDate="2016-06-08T17:31:09.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9186" PostTypeId="2" ParentId="9131" CreationDate="2016-06-04T22:18:52.157" Score="0" Body="&lt;p&gt;I think I wrote that comment; I think I also added quite a few examples of what requirements could look like. In that question, you asked something along the lines of &quot;what is the best way to ...&quot;. But &quot;best&quot; does not exist if you don't give any guidance on how to evaluate the 'goodness' of a particular solution. What's &quot;best&quot; for me can be something else than &quot;best&quot; for you. That's why I wrote: &quot;to figure out what you want&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In such an early stage of the development process, requirements will generally not be final and may not be very detailed. And it's fine if further investigations lead to the conclusion that the requirements were not realistic and need to be modified. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Requirements and design can be cascaded:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Requirement: make a profit with a factory producing N widgets per year at production cost P per widget, for a capital investment C.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Design: the factory should have places to a) shape the widgets out of raw materials, b) clean them, c) dry them, d) package them. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Requirement: the drying place may cost C/4 and should handle N widgets per year. The widgets come in clean and wet, go out clean and dry. Operation costs of the drying place is max P/10 per widget.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Design: we could do it with a 100 kW heater, blowing 5000 m3/h of hot air over the widgets.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Ideally, you break down the engineering tasks such that at the bottom of the chain, different engineers can be working on the different subsystems without having to be aware of all the problems that the other engineers are dealing with. In practice, it will often turn out that some of the choices at steps 2 and 3 were not right and the design (#2) and requirements (#3) have to be tweaked. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in particular if the work is split up over multiple people, having good requirements upfront, even if not final, reduces the probability that someone comes with a design that provokes a reaction along the lines of: &quot;But that's a ridiculous solution; technically it meets the requirements, but that's not what I meant!&quot;&lt;/p&gt;&#xA;" OwnerUserId="6350" LastActivityDate="2016-06-04T22:18:52.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9187" PostTypeId="2" ParentId="9181" CreationDate="2016-06-05T00:24:10.487" Score="2" Body="&lt;p&gt;Most simplest and cost effective method is the purchase a LM2623 Evaluation Board board. The Evaluation board looks something like the following. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GK95O.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GK95O.jpg&quot; alt=&quot;LM2623 Evaluation Board image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you contact your Texas Instrument Sales representative you can almost get the board free or for a very nominal price. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternate option is to look for similar DC-DC converter break out boards. They might not have a exact one-one mapping but might do a job for simple prototyping experiment. Below is an alternate from Pololu &lt;a href=&quot;https://www.pololu.com/product/2570&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adjustable 4-12V Step-Up Voltage Regulator U3V50ALV&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oAdNi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oAdNi.jpg&quot; alt=&quot;Adjustable 4-12V Step-Up Voltage Regulator&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attach a few references that you might want research further. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to use surface mount to thru hole converters. Below is an example. Below is an example of one such from Jameco. Issue is at the end your cost might be the same as purchasing a EVM of Texas Instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6qXEC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6qXEC.jpg&quot; alt=&quot;Protoboard Soic To Dip 6 Board 2-SOT 23 2-SOT 323 353 2-SC70&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many options. All you have to do is to make the necessary compromise as pertaining to your project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.pololu.com/product/2571&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pololu Adjustable 9-30V Step-Up Voltage Regulator U3V50AHV&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.pololu.com/category/132/step-up-voltage-regulators&quot; rel=&quot;nofollow noreferrer&quot;&gt;Regulators and Power Supplies - Step-Up Voltage Regulators&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/tool/lm2623ev&quot; rel=&quot;nofollow noreferrer&quot;&gt;General Purpose, Gated Oscillator Based, DC DC Boost Converter - LM2623EV&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.jameco.com/webapp/wcs/stores/servlet/StoreCatalogDrillDownView?langId=-1&amp;amp;storeId=10001&amp;amp;catalogId=10001&amp;amp;categoryName=cat_6520&amp;amp;subCategoryName=Electronic%20Design%20%2F%20Prototyping%20Systems%20%2F%20Prototype%20Adapters&amp;amp;category=652030&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electronic Prototyping Adapters&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.adafruit.com/categories/570&quot; rel=&quot;nofollow noreferrer&quot;&gt;Power Converter Boosters&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-07-06T00:51:22.757" LastActivityDate="2016-07-06T00:51:22.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="9188" PostTypeId="1" CreationDate="2016-06-05T00:24:32.660" Score="2" ViewCount="40" Body="&lt;p&gt;I made a DC-DC converter its  ideal transfer function is:&#xA;$$  V_0(D) = DV_i  $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took measurements of its response and then made a linear regression, I got this result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V_0(D) = 0.98DV_i -0.42 $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite similar to the expected value. Now, can I say that the system has a linear response?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;strictly speaking, the system response should be zero when the input is zero, but in this case, if D = 0:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V_0 = -0.42 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the correct way to refer at the response of this system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a plot of the theoric response, red, vs experimental values green.&#xA;The 0.42 basically represents an offset because the losses of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LdA5J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LdA5J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6590" LastEditorUserId="6590" LastEditDate="2016-06-05T01:43:48.277" LastActivityDate="2016-06-05T01:43:48.277" Title="Can a system with a response Y(x) = ax + c be called linear?" Tags="|modeling|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="9189" PostTypeId="2" ParentId="9188" CreationDate="2016-06-05T01:27:52.570" Score="1" Body="&lt;p&gt;Well, I wouldn't say the result is quite similar, I don't have any information on scale or units of measurement but the $-0.42$ value might be significant. You should analyze other data, such as the correlation factor, and decide if the linear model you use to regress the data is appropriate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question: no. I think that, based only on what you have presented, you cannot conclude your system has a linear response. You should try to fit other models to prove they don't fit as well as a linear one and also present some more data. That doesn't mean it's not linear though only that you need to do some more analysis before concluding that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; By the way, if a system does have a $Y(x) = ax + c$ response it is by definition linear, I think your question is whether your system does have such a response.&lt;/p&gt;&#xA;" OwnerUserId="6685" LastActivityDate="2016-06-05T01:27:52.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="9190" PostTypeId="2" ParentId="9180" CreationDate="2016-06-05T01:40:32.897" Score="1" Body="&lt;p&gt;On the second picture, it does seem like there is air inside the pipe. Is that correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is so, the reason there is less area outside the pipe is that the water inside the tube was only on the walls of the tube, with an empty space (air) in its core. Now when the water leaves the tube the water is not attracted to the tube's walls anymore and by surface tension cluster together on the center. Although it seems like the area is smaller that is not the case, there was air on the center of the tube and after the water leaves there's no air anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="6685" LastActivityDate="2016-06-05T01:40:32.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10187" PostTypeId="2" ParentId="8979" CreationDate="2016-06-05T02:13:59.977" Score="2" Body="&lt;p&gt;First of all, I don't think likely that your hypothesis is true. I don't think the heat generated in tension would be &quot;absorbed&quot; in compression. That seems unlikely and I've never heard of such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I don't think termal expansion is what is going on when you run a tensile test and the metal gets hot. The heat that is liberated if you 'stretch' a specimen is not working through the same mechanism that a metal raises volume when heated. Those are two different effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I think is going on is something that someone already suggested in the comments: when you ran the tensile test, you had a big part of your specimen getting stretched thus generating significant heat. When you did the bending test only some localized areas of your specimen suffered significant traction, and it probably was not as much traction as it was subject on the pure tensile test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the small areas of your specimen that did suffer traction did also generate heat, but it was less heat than you were expecting. If you should ran the same test and measure the heat on the outside of the area that is being directly being bent (suffering traction) I think you would detect a temperature change.&lt;/p&gt;&#xA;" OwnerUserId="6685" LastActivityDate="2016-06-05T02:13:59.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10188" PostTypeId="1" AcceptedAnswerId="10194" CreationDate="2016-06-05T05:39:29.983" Score="2" ViewCount="1246" Body="&lt;p&gt;I know this might sound ridiculous, but I am attempting to make a fan designed to induce lift of a 200 lbs object. By 'fan', I mean like the fans that lift a drone off the ground. I can only have a maximum of 3 fans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to lift the object at least 5 feet of the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know what type of fans I would need, and/or the RPM of the fans? I am little bit confused on what to ask since there aren't any other questions like this. Also I am a beginner to engineering.&lt;/p&gt;&#xA;" OwnerUserId="6709" LastActivityDate="2016-06-05T15:54:48.893" Title="What type of fan would I need to induce the lift of a 200 lbs object?" Tags="|materials|design|" AnswerCount="2" CommentCount="3" ClosedDate="2016-06-05T23:11:48.507" ContentLicense="CC BY-SA 3.0" />
  <row Id="10189" PostTypeId="2" ParentId="7033" CreationDate="2016-06-05T09:38:53.497" Score="0" Body="&lt;p&gt;Endurance limit decreases with frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YIOkN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YIOkN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6710" LastActivityDate="2016-06-05T09:38:53.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10191" PostTypeId="1" AcceptedAnswerId="10192" CreationDate="2016-06-05T14:10:25.717" Score="0" ViewCount="89" Body="&lt;p&gt;It seems like pitch angle could be computed from &lt;a href=&quot;http://43oh.com/2010/11/msp430-based-quadcopter-ez430-rf2500/&quot; rel=&quot;nofollow&quot;&gt;just gyro data&lt;/a&gt;, which uses 2 gyros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, isn't pitch computed from acceleration rather than gyro data?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;pitch = 180 * arctan (accelerationX/sqrt(accelerationY*accelerationY + accelerationZ*accelerationZ))/PI&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where no gyro data is needed. So how could a copter be balanced with just gyro data?&lt;/p&gt;&#xA;" OwnerUserId="4302" LastActivityDate="2016-06-05T14:52:12.423" Title="Balancing a copter with just gyro data" Tags="|mechanical-engineering|sensors|" AnswerCount="1" CommentCount="1" ClosedDate="2016-06-05T23:11:28.443" ContentLicense="CC BY-SA 3.0" />
  <row Id="10192" PostTypeId="2" ParentId="10191" CreationDate="2016-06-05T14:52:12.423" Score="0" Body="&lt;p&gt;It depends on the type of gyro.  The cheap and available electronic gyros produce a signal proportional to the rotation speed.  You don't get absolute angles.  Even if you started stationary and level, then integrated from there, the answer would be useless in a just a few seconds due to offset drift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some gyros are more accurate, but cost &lt;i&gt;much&lt;/i&gt; more.  Other gyros use a spinning flywheel as a reference and do produce angle signals, but those are also much more expensive, not to mention heavier and larger.  NASA can afford to put them on spacecraft and commercial jetliners used to use them for long over-water hauls where there weren't navigation beacons, but you're not going to put one on your copter.  Even then, the mechanical gyros drift over time too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, &lt;i&gt;inertial navigation&lt;/i&gt; fundamentally relies on integration to find position, so has a ever-increasing error envelope with time.  This is one reason the astronauts on the way to the moon and back had to do regular position fixes from star sightings and the like.  The inertial navigation provided the immediate position and orientation, but the accumulated drift had to be occasionally reset from absolute measurements using other methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of cheap MEMs gyros (the only kind you can reasonably put on your copter) they are really only good to tell you the angular velocity.  The absolute vertical direction is determined by measuring the gravity vector, and you can measure the absolute angle about the vertical axis with a Hall effect magnetometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, of course you use GPS for absolute location.  It doesn't tell you orientation, but once you're moving you can find heading from multiple GPS readings.  GPS is used for the long term data, and inertial navigation only to fill in the short term information that is too detailed for GPS.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-06-05T14:52:12.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10193" PostTypeId="2" ParentId="10188" CreationDate="2016-06-05T15:24:17.143" Score="0" Body="&lt;p&gt;One way or another, you need to find a way to develop about 1000 Newtons of upward force by accelerating a mass of air downward. You can make tradeoffs between the amount (mass) of air and the acceleration you give it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Human-powered_helicopter&quot; rel=&quot;nofollow&quot;&gt;Here is one solution&lt;/a&gt; that uses a very large area and low velocity/acceleration, in order to match the capabilities of a human being as a power source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use other power sources, the system can be afford to be less efficient, leading to other solutions that use higher velocities/accelerations and smaller areas.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-06-05T15:24:17.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10194" PostTypeId="2" ParentId="10188" CreationDate="2016-06-05T15:26:19.233" Score="0" Body="&lt;p&gt;Start out by doing a momentum ballance.  You want to lift 200 pounds, so that means you need to push down on air with 200 pounds (890 N).  That means you need to move (890 * Kg * m/s) of air every second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this gives you a tradeoff.  You could, for example, push 10 kg of air per second down at 89 m/s, 100 kg per second of air down at 8.9 m/s, etc.  In theory, you get the same lift force either way.  Basically you can push a little air a lot, or a lot of air a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, consider the power requirement.  The kinetic energy of a moving mass is (&amp;frac12; * m * V&lt;sup&gt;2&lt;/sup&gt;).  Note that annoying V&lt;sup&gt;2&lt;/sup&gt; term.  It says that pushing a little air a lot requires more power than pushing a lot of air a little.  This is why sail planes have such wide wings.  Power usage is a primary design consideration, so the large area being swept across by the wings each time unit allows for pushing a lot of air, which requires imparting less velocity to the air, which saves power since that's proportional to the square of the velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a start on your numbers, let's pick a air velocity and see what we get.  I'm going to pick about half the speed of sound as example, so 165 m/s.  This means you have to move 5.4 kg of air per second.  At 20 &amp;deg;C, that's about 4500 liters of air per second.  The power you'd be imparting to the moving air is 73 kW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of that jet (for now assuming constant velocity crossection just to get a rough feel where we're at) is (4.5 m&lt;sup&gt;3&lt;/sup&gt;/s)/(165 m/s) = 0.027 m&lt;sup&gt;2&lt;/sup&gt;.  That's a circular jet  186 mm in diameter, or about 7.3 inches.  Of course the flow crossection won't be constant, so the diameter will need to be bigger, and the power higher because some air will be faster than the average and some slower, but the faster will require disproportionately more power.  I don't want to get into those fluid dynamics details here.  I'll arbitrarily say the area needs to be about 50% larger, so 9 inches in diameter.  I'll also round up the 73 kW to 100 kW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have as a very rough starting point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;165 m/s average jet speed&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;9 inch jet diameter&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;100 kW power required&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there you can scale things accordingly.  The area is inversely proportional to the jet speed, and the diameter goes with the square root of the area.  The power goes with the jet speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's see what we get with 1/10 the jet speed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;16.5 m/s average jet speed&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;28 inch jet diameter&lt;br&gt;&#xA;&amp;nbsp;&amp;nbsp;10 kW power required&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, these are very rough numbers with a lot of simplifying assumptions.  The point is to give you a ballpark idea, and a basis for making relative tradeoffs from.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2016-06-05T15:54:48.893" LastActivityDate="2016-06-05T15:54:48.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10195" PostTypeId="1" CreationDate="2016-06-05T15:40:43.770" Score="2" ViewCount="552" Body="&lt;p&gt;I have a passing interest in water desalination projects. One of the challenges of desalination is simply moving the water from the ocean to the desalination plant. Typically the plant is built above the height of sea level, possibly by tens to hundreds of meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a highly efficient mechanical method already devised, to use the weight of the waste water returning to the ocean, to lift more supply water to the desalination plant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since purified water is removed by the desalination plant, the brackish return water will be slightly lower in volume than the supply, and there will be losses from viscous drag in the pipes, so there's not a huge amount of energy recovery available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it seems like some useful percentage of energy saving / energy recovery should be possible, using the stored potential energy of the lifted return water, to offset the energy requirements of lifting the supply water.&lt;/p&gt;&#xA;" OwnerUserId="3126" LastActivityDate="2016-08-15T06:21:40.290" Title="Transporting water uphill at the highest efficiency possible?" Tags="|energy-efficiency|hydrology|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10196" PostTypeId="2" ParentId="9097" CreationDate="2016-06-05T16:53:51.180" Score="1" Body="&lt;p&gt;While the calculations you've done are correct for an ideal structure, we are not dealing with an ideal structure here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one, you don't take into consideration the fact that the column is inclined. This will generate a static bending moment along the entire column, meaning you aren't dealing with buckling under compression, but under flexo-compression, which reduces the buckling load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, your $L_{eff}$ assumes that the column is pinned on both ends. This is a conservative estimate, since there will be some rotational stiffness on both ends, but it's fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the biggest flaw in your calculation is that you assume that the buckling length of your &quot;column&quot; is equal to the height of the opening, but that is incorrect. The correct buckling length is probably more like the distance from the midpoint of the lid (where you sit) to the midpoint of the bracing beam near the ground. You then need to perform a buckling analysis for a column with a changing cross-section (from a solid section to a section with an opening to a solid section, with a changing width throughout). It this can be done analytically (no such guarantee), it will be a huge pain. This would be best done by letting a FEM program deal with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When modelling the column this way, the boundary conditions will probably be fixed at the base (due to the high stiffness caused by the tall bracing) and pinned at the top (the simple screw connection will offer little resistance to small rotations), which reduces your $L_{eff}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as @Mr.P already mentioned in a comment beneath the OP, this calculation you've done is for Euler buckling, which is the theoretical limit for a perfect specimen. We are not dealing with perfect specimens in the real world, so designing anything by the Euler load is non-conservative (your actual structure will resist less than this load). In structures there are a multitude of codes which describe how to take imperfections into consideration, but I'm not sure if they are valid or practical for a small stool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best solution is probably as suggested by @alephzero in a comment beneath the OP: woodworking rules of thumb. I'm not a DIYer, so I don't know what these are, but I'm sure they exist and will give you a good notion of what's reasonable and what isn't. Also, there's no solution that's better than a prototype. Go ahead and build the stool. Sit (and stand) on it every which way, put some ridiculous weight on it (stacks and stacks of books and then sit on top of them), see how sturdy it is. Also make sure to test how &quot;flexible&quot; it is. A wobbly stool might not actually collapse, but the person sitting on it won't feel very safe on it.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-06-05T16:53:51.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10197" PostTypeId="2" ParentId="10195" CreationDate="2016-06-05T16:55:29.540" Score="0" Body="&lt;p&gt;You can construct two elevator shafts with the cables connected via a pulley. If you fill one elevator with with processed water to go down and fill the other one with fresh sea water, then you just need a small electric engine to offset the weight difference. This will be as efficient as you can get it in terms of wasted energy. But the capital costs and the low throughput make it rather unpractical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find this a contrived example, but the losses of viscous drag that you mention yourself can be reduced to an arbitrarily low value by choosing a sufficiently large pipe diameter. It will cost a lot, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you need to specify what the capital cost is allowed to be for a given throughput.&lt;/p&gt;&#xA;" OwnerUserId="6350" LastActivityDate="2016-06-05T16:55:29.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10198" PostTypeId="1" AcceptedAnswerId="10209" CreationDate="2016-06-05T19:37:09.907" Score="2" ViewCount="774" Body="&lt;p&gt;How can we calculate the minimum bend radius(the radius which it starts to buckle) of a annular cylindrical tube? Intuitively I guess that for two tubes having same wall thickness but different outer diameters, the larger diameter one will have larger bend radius. Is there an analytical way to confirm this?&lt;/p&gt;&#xA;" OwnerUserId="6715" LastActivityDate="2019-11-02T16:32:12.577" Title="minimum bend radius" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10199" PostTypeId="1" CreationDate="2016-06-05T23:09:51.907" Score="4" ViewCount="6989" Body="&lt;p&gt;I understand how water content helps compaction, by being kind of a lubricant between the soil particles, that helps them slide into position. But I don´t really understand how it starts affecting negatively after surpassing the optimum water content, but not surpassing the zero void line.&lt;/p&gt;&#xA;" OwnerUserId="6716" LastEditorUserId="90" LastEditDate="2022-08-03T00:50:41.490" LastActivityDate="2022-08-03T00:50:41.490" Title="How does excess water makes soil compaction more difficult?" Tags="|civil-engineering|geotechnical-engineering|soil|soil-mechanics|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10203" PostTypeId="1" AcceptedAnswerId="10208" CreationDate="2016-06-06T05:24:31.350" Score="0" ViewCount="326" Body="&lt;p&gt;If the angle corresponding to the open loop transfer function of a system at infinite frequency is smaller than that at zero frequency, the polar plot curve in the clockwise direction with increase in frequency. Why is it so ?&lt;/p&gt;&#xA;" OwnerUserId="6571" LastActivityDate="2018-02-27T03:41:39.273" Title="Polar plot direction with increase in frequency" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10208" PostTypeId="2" ParentId="10203" CreationDate="2016-06-06T16:55:26.753" Score="1" Body="&lt;p&gt;This is just how the phase component of of a complex value, returned by the transfer function, is defined. Phase is the angle between the positive real axis and the complex value in the counterclockwise direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you start at some phase $\phi_1$ at a frequency of zero and go to phase $\phi_2$ at an infinite frequency, where $\phi_1&amp;gt;\phi_2$, then on average the polar plot should rotate in the clockwise direction (the direction in which the phase decreases).&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2016-06-06T16:55:26.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10209" PostTypeId="2" ParentId="10198" CreationDate="2016-06-06T17:53:53.677" Score="2" Body="&lt;p&gt;There is an analytical solution to the buckling of a tube under flexure which was first proposed by &lt;a href=&quot;http://rspa.royalsocietypublishing.org/content/116/773/104&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brazier in 1927&lt;/a&gt;. There is also a more detailed and helpful discussion of Brazier's solution in Calladine's book &lt;a href=&quot;http://dx.doi.org/10.1017/CBO9780511624278.017&quot; rel=&quot;nofollow noreferrer&quot;&gt;'Theory shell structures'&lt;/a&gt;. (Apologies for the pay-walled links ...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't repeat the analysis here but there are two main points which are are considered in the analysis: Firstly, consideration of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Brazier_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brazier effect&lt;/a&gt; which is the flattening (ovalisation) of a tube when it is put under bending. You can see this effect by experimenting with a straw.  Secondly, the local buckling condition of a cylindrical shell segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results of this analysis are summarized by Calladine&lt;a href=&quot;http://dx.doi.org/10.1017/CBO9780511624278.017&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt; in the following plot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/udoX3m.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/udoX3m.gif&quot; alt=&quot;Brazier Plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a dimensionless plot with the following dimensionless parameters:&#xA;$s_{cr}$ is the critical stress for local buckling, $s$ is the extreme fibre stress, $m$ is the bending moment, $\zeta$ is a flattening parameter, and $\frac{1}{c}$ is the radius the tube is bent to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius at which buckling occurs is the the radius corresponding to the intersection of the $s_{cr}$ and $s$ curves as shown on the plot. You can calculate the actual (dimensioned) radius of curvature by reading $c$ from the plot and unraveling the dimensionless parameters. In which case you get the minimum bend radius R:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R = \frac{3^\frac{1}{2}r^2}{2 c t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t$ is the tube thickness, and $r$ is the radius of the tube. Therefore the minimum bend radius increases with the square of the tube radius. There will be lots of other practical aspects which will influence the bending of a tube in reality which are not considered in this analysis.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2016-06-06T17:53:53.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10210" PostTypeId="2" ParentId="10198" CreationDate="2016-06-06T18:33:58.150" Score="1" Body="&lt;p&gt;There is also a practical element in that there are quite a few different methods for bending tube ranging from a completely unsupported bend through simple jigs, &lt;a href=&quot;http://www.baileigh.com/pipe-bender-rdb-125&quot; rel=&quot;nofollow&quot;&gt;draw benders&lt;/a&gt; (which provide external support) to  serious industrial NC &lt;a href=&quot;http://www.baileigh.com/mandrel-benders-mb-nce-1-series&quot; rel=&quot;nofollow&quot;&gt;mandrel benders&lt;/a&gt; which support the tube from the inside during bending and can give sophisticated control of strain rate and other parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a difference between creating a bend at a point and rolling a continuous radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice the behaviour can vary significantly even from one batch of stock to the next so any calculations need to be treated with some degree of caution.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-06-06T18:33:58.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10211" PostTypeId="1" CreationDate="2016-06-06T21:49:51.100" Score="1" ViewCount="8936" Body="&lt;p&gt;I am experiencing serious reduction in torque when using PWM to reduce speed - I'm temporarily resorting to 10Hz, and even then, 50% mark-space gives me (at most) half the torque of fully on. I'm assessing the torque by seeing what weight can be lifted via a worm gear. I've tried up to 1kHz and it gets worse with higher frequencies. Yet supposedly PWM should give the full torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What &lt;em&gt;in theory&lt;/em&gt; could be causing this? I bumped into something mentioning inductance of windings, but inductance, when driving a 6000 RPM dc motor with 10Hz PWM shouldn't have any effect on torque? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is the inductance, then I suppose I'll need to drive the motor with a constant current source, which I suppose will mean driving the motor above its rated voltage for low speeds, assuming it won't die.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-06-06T22:09:14.483" LastActivityDate="2019-04-17T01:01:55.373" Title="Dc motors: relationship between PWM frequency and Torque" Tags="|motors|pwm|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10212" PostTypeId="2" ParentId="7586" CreationDate="2016-06-06T21:57:43.123" Score="0" Body="&lt;p&gt;Concrete and Steel behave in drastically different ways: concrete is assumed not to be able to bear any tensile stress as soon as first cracks appear while steel is ideal material for tensile stress. Optimally it would be ideal to have thin tiny strands of steel as opposed to few large ones, kind of similar to how fiberglass works homogenously.&lt;br&gt;&#xA;Specially since we know our design assumptions are highly simplified to make the job of engineering practical.  Things like climate and deferential sun exposure, humidity, permeating of corrosive agents through cracks in a member, all impact the behavior of a member. Also as mentioned above the area and contact surface of steel bar are related in a disadvantage ratio of power of two for area of the bar that transfers tension through skin friction to concrete.&lt;br&gt;&#xA;On the other hand practical requirements during the course of construction such as bars not easily be deformed under the foot traffic of workers or accommodation space to allow passage of transient pipes or conduits should be permitted. that's were engineering codes come into play to make a uniform and predictable game plan. so the mechanical engineer or hardware designer know what they specify or expect will work.   &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-06-06T21:57:43.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10213" PostTypeId="2" ParentId="3087" CreationDate="2016-06-07T00:12:34.110" Score="2" Body="&lt;p&gt;I recommend you to learn control theory while implementing what you learn directly. &#xA;a useful example to do so ,this course for mobile robots using control theory &#xA;&lt;a href=&quot;https://www.coursera.org/learn/mobile-robot/home/info&quot; rel=&quot;nofollow&quot;&gt;https://www.coursera.org/learn/mobile-robot/home/info&lt;/a&gt; &#xA;and in each lecture you can implement what you learn on a simulator on MATLAB and you can even implement it on a hardware; small mobile robot. &#xA;-You can also read or use as a reference &quot;Modern Control theory, for William Brogan&quot; i believe it is the best control theory reference i ever seen. &#xA;and finally GOOD LUCK :)&lt;/p&gt;&#xA;" OwnerUserId="6734" LastActivityDate="2016-06-07T00:12:34.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10217" PostTypeId="1" CreationDate="2016-06-07T08:09:19.593" Score="6" ViewCount="48085" Body="&lt;p&gt;I made this kind of knurl on shaft :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9PkJV.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9PkJV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I am not sure how to be define on the drawing. I found some examples but not sure if they are correct. On the examples there type (GV) and depth of knurl and no more information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be very thankful if somebody show me drawings or books with correct info.&lt;/p&gt;&#xA;" OwnerUserId="6739" LastEditorUserId="1832" LastEditDate="2016-06-07T10:41:39.817" LastActivityDate="2016-06-16T06:56:06.147" Title="How to define Knurl on drawing?" Tags="|mechanical-engineering|modeling|machining|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10219" PostTypeId="1" CreationDate="2016-06-07T09:34:07.637" Score="0" ViewCount="801" Body="&lt;p&gt;A bar of length L and of uniform cross-section area A and second moment of area 1 is subjected to pull P.If  the young's modulus of elasticity of the bar material is E,the expression for &lt;strong&gt;strain energy&lt;/strong&gt; stored in the bar will be&#xA;1)P2L2/2AE  2)PL2/2EI&#xA;3)PL2/AE    4)P2L/AE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(Note:The  P2L2 and 2AE in the option is P squared,L squared and 2 times AE respectively)&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="6538" LastActivityDate="2016-07-07T12:02:50.617" Title="Strain Energy Problem" Tags="|materials|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10220" PostTypeId="2" ParentId="10219" CreationDate="2016-06-07T11:17:22.897" Score="1" Body="&lt;p&gt;5) None of the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strain energy stored in the bar will be $$U = \frac{1}{2}P\Delta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $$\Delta = \frac{PL}{AE}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&#xA;$$U=\frac{P^2L}{2AE}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_I/BookSM_Part_I/08_Energy/08_Energy_02_Elastic_Strain_Energy.pdf&quot; rel=&quot;nofollow&quot;&gt;Strain Energy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-06-07T11:17:22.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10221" PostTypeId="2" ParentId="10217" CreationDate="2016-06-07T11:30:17.933" Score="-1" Body="&lt;p&gt;Knurls are usually made by rolling or cutting. A tool with a free-spinning rotating head (or opposed pair of heads) is pressed into the part on a lathe and the pattern is transferred (in negative) from the head. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the pattern itself is fixed by the design of the tool head so you need to specify the tool type which gives you the pattern you want. Specific tool manufactures will have their own conventions for naming/specifying  patterns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically knurled patterns will be either straight (ridges parallel to the long axis of the part) or a diamond pattern (ridges crossing at 45 degrees). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut knurls tend to be a bit crisper than rolled ones but have some limitations on how they can be applied ie the tool needs to be advanced axially from one end of the knurl as opposed to rolling where the tool is advanced radially. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-06-07T11:30:17.933" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10222" PostTypeId="1" AcceptedAnswerId="10240" CreationDate="2016-06-07T13:40:14.700" Score="1" ViewCount="975" Body="&lt;p&gt;I'm dealing with an application in which I use small DC brush geared motors to move a robot. The motors are driven by a standard FET driver. I do have the option to coast or brake the motor after every operation. Brake would be preferable since the robot would stop directly after a &quot;forward&quot; or &quot;backward&quot; operation. What I'm wondering is: would the brake operation affect negatively the lifespan of the motor or gearhead? Or is this effect negligible?&lt;/p&gt;&#xA;" OwnerUserId="3685" LastActivityDate="2016-06-08T11:56:05.797" Title="DC brush geared motor: coast vs brake effects on motor/gearhead lifespan" Tags="|motors|robotics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10223" PostTypeId="1" AcceptedAnswerId="10301" CreationDate="2016-06-07T13:40:54.763" Score="4" ViewCount="153" Body="&lt;p&gt;Would adding a shade (i.e. a piece of plastic) at the back of an A/C increase the efficiency of a window unit A/C? I'm not talking about covering the fins but rather placing a sheet of plastic above to create shade without obstructing the air flow. Or does the forced air easily suck off the heat regardless of how much the sun is shining on the unit?&lt;/p&gt;&#xA;" OwnerUserId="6742" LastEditorUserId="1832" LastEditDate="2016-06-07T13:56:29.777" LastActivityDate="2019-04-12T10:21:14.110" Title="Will a shade hood increase a/c efficiency?" Tags="|heat-transfer|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10224" PostTypeId="1" AcceptedAnswerId="10225" CreationDate="2016-06-07T13:42:18.907" Score="11" ViewCount="4466" Body="&lt;p&gt;Here's a photo of a bearing plate where a bridge reinforced concrete beam meets the earthfill&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ccKdX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ccKdX.jpg&quot; alt=&quot;bearing plate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bridge span is about 20 meters long and consists of two reinforced concrete beams each resting on two bearing plates like the one shown - one plate for each end of each beam, total four plates. The bridge holds a railway track designed for 25 tons per axle cars. The bearing plate is made of cast iron (or maybe steel) and consists of two large parts joined through a hinge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;25 tons per axle cars means the bridge bears something like several hundred tons when a train is passing which we can assume causes at least one hundred tons per bearing plate shown. Yes, I just ignored the bridge weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only the plate upper and lower surfaces are rather small but the plate further concentrates the accepted load and transfers it onto the hinge through even smaller surface. Basically this rather small hinge alone accepts more than one hundred tons. And this is designed on purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the load deliberately concentrated instead of being distributed or at least transferred through some part with uniform section?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="374" LastEditDate="2016-06-08T11:56:06.423" LastActivityDate="2022-06-16T05:20:23.530" Title="Why would a bearing plate that further concentrates load onto small surface be used?" Tags="|structural-engineering|civil-engineering|steel|bridges|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10225" PostTypeId="2" ParentId="10224" CreationDate="2016-06-07T13:57:53.380" Score="22" Body="&lt;p&gt;Because bridges and other structures are not static objects. They must be allowed to flex under varying loads and also accommodate changes in length from thermal expansion. The hinge pin allows changes in angle. and the sliding joint between the upper hinge plate and the flat plate on the bottom of the beam allows changes in length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the connections were rigid, these forces could destroy the structure over time.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-06-07T13:57:53.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10226" PostTypeId="2" ParentId="10223" CreationDate="2016-06-07T14:03:36.663" Score="2" Body="&lt;p&gt;The case temperature has negligible effect on the operation of an air conditioner. (It's already a light color, right?) There's so much air running through the unit that the heat transferred through the case is a tiny fraction of the heat transferred directly to the air. Indeed, any heat flowing in through the case is mostly transferred directly to air flowing right back out again.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-06-07T14:03:36.663" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10227" PostTypeId="2" ParentId="10224" CreationDate="2016-06-07T14:22:40.593" Score="10" Body="&lt;p&gt;The reason is pretty simple. Steel is significantly stronger than concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays we have high-performance concretes with $f_c &amp;gt; 100~\text{MPa}$ (and ultra-high-performance, which is substantially higher), but most ordinary structures don't use such high strength concrete. This bridge seems relatively weathered, so the concrete is probably at most $f_c &amp;gt; 40~\text{MPa}$ (probably even less).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel, on the other hand, nowadays has at least $f_y &amp;gt; 250~\text{MPa}$, often even more. I don't believe steel strength has evolved as quickly as concrete (correct me if I'm wrong), so the steel used on that bridge is probably at least equal to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steel on that bridge is therefore some 7-8 times stronger than the concrete. So, whatever area the concrete requires to safely transfer the load to the steel (via the plates), the steel actually needs far less, so it can safely reduce its own dimensions. Buckling is controlled by the bracing all around the hinge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why a hinge is used at all, that has to do with how the bridge was designed, as described in @DaveTweed's answer.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-06-07T14:22:40.593" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10229" PostTypeId="1" CreationDate="2016-06-07T15:37:34.653" Score="3" ViewCount="351" Body="&lt;p&gt;Suppose we have a horizontal rigid bar hinged at one end to the wall. An inclined straight link hinged at one end to the same wall, is pinned to the rigid bar at its center. A load P is acting at the free end of the horizontal link. For static analysis we draw the free body diagrams for both the bodies and equate the net force and moment to zero. How will the static force analysis change if we replace the inclined straight link with a curved link? Will it depend on the nature of the curve?&lt;/p&gt;&#xA;" OwnerUserId="6061" LastEditorUserId="282" LastEditDate="2016-06-09T00:43:12.083" LastActivityDate="2019-12-28T18:02:28.203" Title="Replacing a straight rigid link by a curved rigid link" Tags="|mechanical-engineering|statics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10230" PostTypeId="2" ParentId="4209" CreationDate="2016-06-07T17:54:10.607" Score="1" Body="&lt;p&gt;Meet the skid-steer loader:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/m6M4o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m6M4o.png&quot; alt=&quot;Skid-steer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a versatile beast of a machine, packing lots of power, multiple attachments and (most relevantly to the topic at hand) great maneuverability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does exactly what you suggested. It has stationary tires, which turn at different angular velocities in order to turn the vehicle, like a tank. It's easy to turn in a full circle without changing the machine's position. This gives it a number of advantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It can readjust its position when loading and unloading things, if (for example) the forks need to move a few inches to the left or right.&lt;/li&gt;&#xA;&lt;li&gt;It can move a bucket sideways to smooth out dirt or gravel.&lt;/li&gt;&#xA;&lt;li&gt;Its turn radius is 0, meaning it can get into tight spaces not designed for vehicles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Compared to a compact track loader (a similar machine that uses tracks instead of tires), the skid steer has &lt;a href=&quot;http://foresternetwork.com/daily/construction/construction-equipment/wheels-vs-tracks-the-debate-over-compact-loaders-part-i/&quot; rel=&quot;nofollow noreferrer&quot;&gt;these advantages and disadvantages&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lower maintenance and repair cost&lt;/li&gt;&#xA;&lt;li&gt;Lighter and faster&lt;/li&gt;&#xA;&lt;li&gt;Operates poorly if at all in muddy or snowy conditions&lt;/li&gt;&#xA;&lt;li&gt;Affects the ground more&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A car has little to no need for any of these advantages, as they're not used to load, unload or landscape (though attaching a trailer would be much easier), and generally travel on pathways designed for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There actually exist accessories for wheeled vehicles that provide mini tracks for each wheel, to be used in mud or snow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zC45x.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zC45x.jpg&quot; alt=&quot;Truck tracks&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even so, the basic steering functionality is still by directed turning rather than skid steering. In fact, there are even machines that are designed for tracks that steer in the same manner:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QOqAX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QOqAX.jpg&quot; alt=&quot;Curbing machine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If even some tracked machines steer in the same manner that cars do, instead of by skidding, there must be some advantage, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One advantage (the above curbing machine steers conventionally for this reason) is that it's much easier to form a smooth curve. While computer systems could be used to determine each tire's rotation speed to produce a given turn radius, it's much simpler to simply turn the wheels. Also, as pointed out by @alphazero, one is that skidding reduces traction significantly, which can lead to uncontrollable situations at high speed. There's also the matter of wear and tear resulting from the skidding (and protip, try to turn the steering wheel in your car only when moving, if practical).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cars generally steer by turning their wheels instead of skidding because few of the advantages of skid steering have much application to a car but the disadvantages range from expensive to complex to disastrous.&lt;/p&gt;&#xA;" OwnerUserId="5493" LastActivityDate="2016-06-07T17:54:10.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10231" PostTypeId="2" ParentId="8794" CreationDate="2016-06-07T18:02:11.207" Score="3" Body="&lt;p&gt;You are correct that this problem falls within the category of 'large-deflection' problems since the deflection is larger then about $\frac{t}{2}$. To correctly analyse a plate under large-deflections would require solving the &lt;a href=&quot;https://en.wikipedia.org/wiki/F%C3%B6ppl%E2%80%93von_K%C3%A1rm%C3%A1n_equations&quot; rel=&quot;nofollow&quot;&gt;Föppl–von Kármán equations&lt;/a&gt;, which is generally not possible except for some very specific cases. Numerical solutions to some cases have been published (for example in &lt;a href=&quot;https://www.amazon.co.uk/Design-Transverse-Loading-Monographs-Constrado/dp/0258969911&quot; rel=&quot;nofollow&quot;&gt;this book&lt;/a&gt;), unfortunately your case does not appear to be one of them. Using non-linear finite element analysis may be the only option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, &lt;em&gt;if the edges were restrained from sliding&lt;/em&gt; membrane action will occur (some of the load will be carried by direct tension in the plate and not just bending) and the stiffness of the plate would be higher than that predicted by small deflection theory [Roark (2002) pg. 448]. Therefore, if a small displacement analysis predicts the displacement will be less than your requirements, a large-displacement analysis would result in a smaller deflection and therefore also meet your requirements. Note in this case stresses for a given load will also be less than predicted by small-displacement theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roark, R. J., Young, W. C., &amp;amp; Budynas, R. G. (2002). Roark's formulas for stress and strain. New York: McGraw-Hill.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2016-06-07T18:02:11.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10232" PostTypeId="1" AcceptedAnswerId="10234" CreationDate="2016-06-07T19:23:17.447" Score="2" ViewCount="1451" Body="&lt;p&gt;I am trying to measure the height [level] of the water in a box that goes from -3°C (saltwater) with water and 150°C dry. Objects will be placed inside the box but will not take up the whole area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a sensor that is robust enough to go through those temperatures and measure accurately. I was thinking of using a float switch on a worm gear, but the motor couldn't stand the heat... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other methods/sensors are there? I also have a clear tube that runs outside of the tank meant for visually inspecting the height of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/2695/can-i-use-a-ultrasound-sensor-to-measure-water-level&quot;&gt;Can I use a ultrasound sensor to measure water level?&lt;/a&gt; is a closely related question but focuses upon ultrasound.  That won't work for my particular case because the contents of the box are effectively unknown.  And capacitive techniques won't work either as the box will be grounded.&lt;/p&gt;&#xA;" OwnerUserId="6652" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2018-04-14T04:43:11.530" Title="How to measure height of water" Tags="|sensors|distance-measurement|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10233" PostTypeId="2" ParentId="10229" CreationDate="2016-06-07T20:25:04.120" Score="5" Body="&lt;p&gt;I understand your structure is as follows (where the circular version is just an example of a &quot;curved&quot; link):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fhPCC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fhPCC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that is clear is that this structure is isostatic, therefore the external reactions will be equal in both cases. However, the internal forces within the bars will obviously change due to the change in geometry. In this example, for instance, here is a comparison of the forces in each case (obviously dependent on the actual forces and dimensions):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/29WSk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/29WSk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first case functions like a truss-member, with uniform axial load and no shear force or bending moment. The curved case, however, generates shear force and therefore bending moment. This also reduces the axial load over the entire member, but it is worth noting that the maximum value is equal to the value seen in the first case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, however, the horizontal bar presents the same internal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Diagrams created with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:02:28.203" LastActivityDate="2019-12-28T18:02:28.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10234" PostTypeId="2" ParentId="10232" CreationDate="2016-06-07T21:52:24.737" Score="0" Body="&lt;p&gt;I didn't understand quite clearly what you want, but it sounds definitely like ultrasound. Although you mention that it probably won't work, I would still try it first. Another possibility would be to use radioactive material (gamma-source). But it might be difficult to get this stuff. The radioactive type is used for measurement of fill height in bottles. Though I don't know how they do it exactly.&lt;br&gt;&#xA;I recommend to contact companies, which build bottle filling machines. They probably have ready to use solutions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning the laser: Any laser should be able to do this. The cheapest should be the best suited. You don't need a high quality beam.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this doesn't help you, please make a drawing, to make your question more understandable.&lt;/p&gt;&#xA;" OwnerUserId="6052" LastEditorDisplayName="user748" LastEditDate="2016-10-04T04:38:51.207" LastActivityDate="2016-10-04T04:38:51.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10235" PostTypeId="2" ParentId="10232" CreationDate="2016-06-08T00:14:00.863" Score="2" Body="&lt;p&gt;How about using a &lt;a href=&quot;http://www.molex.com/molex/products/family?key=capacitive_fluid_level_sensors&quot; rel=&quot;nofollow noreferrer&quot;&gt;Capacitive Fluid Level Sensor&lt;/a&gt; from Molex. The sensing element can be affixed outside of the tank. Here is an image of the sensing element affixed outside of the a container. . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vVgz2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vVgz2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an expert from the product website. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Capacitive Fluid Level Sensors easily mount to the outside of a container and measure through almost any non-metallic material. They are custom designed and optimized for a range of applications such as medical devices, pump controls, beverage machines etc. A design can use a traditional printed circuit board for a flat surface, or a thin, flexible circuit to accommodate curved surfaces or space-constrained applications. The sensor includes customized embedded software which can be configured for auto-calibration for easy installation, or manual calibration to maximize the accuracy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With regard to robustness, you will have to work with the vendor. This offering is fairly new and I don't have any details beyond what is published on the company websites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/HY_ahbqIdEM?t=11s&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molex - Capacitive Fluid Level Sensors (Video)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.literature.molex.com/SQLImages/kelmscott/Molex/PDF_Images/987651-4852.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sensing the future of appliance design&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.literature.molex.com/SQLImages/kelmscott/Molex/PDF_Images/987651-2043.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;Capacitive Fluid Level Sensors&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/0du-QU1Q0T4&quot; rel=&quot;nofollow noreferrer&quot;&gt;How Capacitive Liquid Level Sensors Work&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/190000/water-level-sensor-circuit/190035#190035&quot;&gt;Water Level Sensor Circuit&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2016-07-07T01:48:23.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10236" PostTypeId="1" AcceptedAnswerId="10242" CreationDate="2016-06-08T06:40:02.610" Score="2" ViewCount="599" Body="&lt;p&gt;Let's say you build a square reinforced concrete deck at height of, say, 3 meters, with the deck supported by its own steel reinforcement and by four reinforced concrete piers (call them, A, B,C, and D) below each corner of the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you then build four infill walls, A-B, B-C, C-D, D-A, are those walls under compressive load? From one side, the weight of the concrete slab would say yes, but on the other side the fact that deck was supported prior to the walls, maybe not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a difference, in the context of the loads transmitted to a wall below, between a vertical stack as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wall - reinforced concrete ring beam - slab&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wall - reinforced concrete deck with reinforcement designed to support the deck by itself and the piers alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example, without the ring of beams:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/lijJP.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="6518" LastEditorUserId="1832" LastEditDate="2016-06-08T13:48:20.970" LastActivityDate="2016-06-13T15:04:20.663" Title="Does a concrete slab (deck) cause compressive load in a wall beneath it?" Tags="|beam|concrete|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10238" PostTypeId="2" ParentId="10217" CreationDate="2016-06-08T08:59:16.357" Score="2" Body="&lt;p&gt;Because I am not sure what are the standard dimensions of the knurl I made this kind of defining and sent to supplier, hope they will advice if something is not clear&#xA;&lt;a href=&quot;https://i.stack.imgur.com/jP6RC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jP6RC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6739" LastActivityDate="2016-06-08T08:59:16.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10239" PostTypeId="2" ParentId="10223" CreationDate="2016-06-08T09:37:12.890" Score="1" Body="&lt;p&gt;Yes.&#xA;If we calculate by refrigerant chart, efficiency is better if temperature difference is smaller (because the vapor-liquid dome is smaller to top, comparing low pressure work with high pressure work, with the same low pressure line length, the higher high pressure the smaller high pressure line length - thus efficiency smaller). So if condenser is colder, efficiency will be better.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastEditorUserId="4630" LastEditDate="2019-04-12T10:21:14.110" LastActivityDate="2019-04-12T10:21:14.110" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10240" PostTypeId="2" ParentId="10222" CreationDate="2016-06-08T11:56:05.797" Score="2" Body="&lt;p&gt;The forces of breaking a motor by shorting its electrical connections are the same as starting the motor with full voltage applied.  Actually, they'd only be the same in a ideal motor.  In any real motor the breaking forces would be a little less due to the less than 100% efficiency of the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor torque is proportional to the current.  During electronic breaking, that current is the generator voltage applied to the coils resistance.  During startup, that current is the driving voltage applied to the coils resistance.  Unless the mechanical load is providing significant power on its own, the breaking current, and hence the breaking torque, won't exceed the initial stall current when the full voltage is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all a long way of saying that the maximum mechanical stress is more likely at startup with full voltage, than when electrically breaking by shorting the motor connections.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-06-08T11:56:05.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10241" PostTypeId="2" ParentId="10236" CreationDate="2016-06-08T13:46:03.927" Score="0" Body="&lt;p&gt;Since you are building infill walls you don't expect the wall to resist compression. You must calculate the structure without considering the walls as resisting elements, only as weigh to the beam below it. Usually in this cases the mortar between the bricks is expected to be flexible enough that if the structure deforms the wall is able to deform with it and not resist any significant tensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for this purpose when you build an infill wall with ceramic bricks you always leave an empty space between the wall and the structure above it (~2cm) that is only filled when the whole structure is built (upper levels) and the initial deformations of the reinforced concrete have occurred. Even so, when you do fill this space, you will want to fill it with some sort of flexible filling (sometimes mortar is enough) so that the future deformations of the structure are absorbed by the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So usually you don't expect the wall to be able to support any tensions, but there will be some due to the deformation of the reinforced concrete structure. For that reason brick walls are considere flexible and must be able to accommodate this displacements without resisting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I've said is considering you don't want the wall to be structure components (you did say they were infill walls).&lt;/p&gt;&#xA;" OwnerUserId="6685" LastActivityDate="2016-06-08T13:46:03.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10242" PostTypeId="2" ParentId="10236" CreationDate="2016-06-08T13:46:09.113" Score="2" Body="&lt;p&gt;It depends on the order of operations for the construction of the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start assuming the structure is as simple as the image you've given us: a single story with a slab supported by columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the columns and slab are built, their formwork is removed and only then the wall is built, the wall shouldn't suffer any stresses. This is because the removal of the slab's formwork allowed it to deform itself into a stable configuration. When your wall comes up to the slab, it will simply close the gap, but won't have to support the slab since, well, the slab is already supporting itself. This will only be true in the short term, however. Over time, creep will try to further deflect the slab, and the wall will end up resisting that movement, generating a compressive force. This can be reduced by increasing the time between removal of the formwork and the raising of the wall, thereby increasing the amount of creep which occurs without harming the wall (this effect would be minimal, though).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the columns and slabs are built and the wall is raised before their formwork is removed, then the wall will suffer compression. In this case, when the formwork is removed, the wall will (try to) impede the deflection of the slab and will therefore be compressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the slab is actually supported by a ring of beams, the same effects apply as described above. That being said, the beams will dramatically increase the stiffness of the slabs, and will therefore reduce the deflection of the structure above the wall. This will reduce the compression suffered by the wall (either immediately or due to creep), but not eliminate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the structure actually has multiple floors, then the entire structure should be built before any walls are fully raised. The walls should also be raised starting from the top floor, and then proceeding downwards. This way, the load due to the wall of an upper floor will already have deformed the structure before a wall is raised. This way we can guarantee that a wall will not participate in supporting the walls above it. If the walls are made of brick, these can be raised along with the structure, so long as they leave a gap between their top course (row of bricks) and the bottom of the slab. Once the entire structure is raised, that gap can be filled (starting from the top floor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @AndyT well mentioned in a &lt;a href=&quot;https://engineering.stackexchange.com/questions/10236/does-a-concrete-slab-deck-cause-compressive-load-in-a-wall-beneath-it#comment18108_10242&quot;&gt;comment&lt;/a&gt; below, regardless of when the wall is raised, it will participate to some extent in resisting loads due to live/accidental loads.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-06-13T15:04:20.663" LastActivityDate="2016-06-13T15:04:20.663" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10243" PostTypeId="2" ParentId="10229" CreationDate="2016-06-08T17:50:17.670" Score="0" Body="&lt;p&gt;This depends whether you are just talking about a rigid body diagram/model where the structure is idealised. Here it is normal to assume that any links are perfectly rigid and massless and the model is only interested in the distribution of forces (as opposed to stresses or deflections) in the structure or mechanism and the relative position of point nodes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case it doesn't really matter what shape the links are as the forces will be transmitted between nodes just the same regardless of the path the links follow as long as the points they join are the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously once you start to consider the mass, stiffness and strength of links their geometry very much does matter but having said that if you design straight and curved links to have equivalent stiffness and strength they will behave in the same way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where you may well start to see significant differences is in situations where you have high strains and strain rates and where dynamic forces and vibrations are significant, fro example in something like a cricket bat or golf club as opposed to something like a linkage in a hydraulic digger. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-06-08T17:50:17.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10244" PostTypeId="1" AcceptedAnswerId="10245" CreationDate="2016-06-08T18:14:35.397" Score="2" ViewCount="1111" Body="&lt;p&gt;I've been translating a truck scale description in a business offer letter, and the specification table there, listing the five parts comprising the prospective order, says on line five:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cable for connecting the load cell to the weighing terminal, &lt;strong&gt;bare end&lt;/strong&gt;. Length: 30m.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the Russian original, the phrase &quot;открытый конец&quot; (literally &quot;open end&quot;) most likely means that the cable has no connector attached to it. It probably comes out of the load cell, and then is cut at this length, 30m. The Buyer can attach the necessary connector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be a naturally-looking phrase for describing the same &quot;bare end&quot; in a similar English document? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://ell.stackexchange.com/questions/93187/phrase-for-describing-a-cable-without-a-connector-at-the-end&quot;&gt;Also asked on ELL SE&lt;/a&gt;) &lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="-1" LastEditDate="2017-04-13T12:55:21.033" LastActivityDate="2016-06-09T00:43:19.927" Title="Phrase for describing a cable without a connector at the end" Tags="|terminology|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10245" PostTypeId="2" ParentId="10244" CreationDate="2016-06-08T18:22:49.310" Score="5" Body="&lt;p&gt;I think that 'bare end' gets the meaning across pretty well. Although in the context of a specification I would be tempted to spell out precisely that the cable is not supplied with a connector or terminator as a footnote eg : &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cable for connecting the load cell to the weighing terminal, bare end.&#xA;  Length: 30m.* &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;*Cables are supplied as cut and will require a suitable connector to be fitted to the free end before use.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would also add that, from the perspective of a customer,  I would look for an indication that a braided cable has been at least prepared in some way that it won't be frayed or unravelling when I receive it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you don't specify whether this is an electrical/data cable or a load bearing one (from the fact that it comes off a load cell I've assumed that it is something like a winch cable) Obviously if it's something exotic or specialized like oilfield wire-line or HT power transmission that's a whole different matter. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-08T19:54:29.620" LastActivityDate="2016-06-08T19:54:29.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10248" PostTypeId="2" ParentId="10244" CreationDate="2016-06-08T19:44:11.367" Score="1" Body="&lt;p&gt;If there is more than one conductor in the cable, the term 'flying leads' is often used.&lt;/p&gt;&#xA;" OwnerUserId="3440" LastActivityDate="2016-06-08T19:44:11.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10251" PostTypeId="2" ParentId="10244" CreationDate="2016-06-09T00:33:45.123" Score="3" Body="&lt;p&gt;I'm guessing this is an electrical cable, not a mechanical cable. The &quot;load cell&quot; is the part that actually measures the load, and presumably the &quot;weighing terminal&quot; is the electronics that displays the measurements to the operator -  possibly inside a building, to protect the electronics and the operator from the weather etc, while the actual truck weighing mechanism may be some distance away outside. Connecting the load cell to something with 30m of wire rope doesn't seem like a reasonable design for a &quot;truck scale.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Bare end&quot; or &quot;bare ended&quot; is understandable, but a more formal general term would be &lt;strong&gt;unterminated.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more specific terms for particular forms of &quot;unterminated&quot; wire and cable - see the &quot;wire termination options&quot; and &quot;cable termination options&quot; here: &lt;a href=&quot;http://www.dsmt.com/resources/termination-options/&quot; rel=&quot;nofollow&quot;&gt;http://www.dsmt.com/resources/termination-options/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of those options only apply to a cable that is already cut to the correct length, and has been prepared ready to connect to something - but that doesn't seem to be the situation that you are describing.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-06-09T00:43:19.927" LastActivityDate="2016-06-09T00:43:19.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10254" PostTypeId="1" CreationDate="2016-06-09T17:08:03.427" Score="2" ViewCount="225" Body="&lt;p&gt;This will probably sound like a silly question, but here goes....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have an actuator, with a torque rating of 1&amp;nbsp;Nm. I then connect it to a wheel, which starts rotating at a constant speed. From Newton's second law, I know that Net Force = Mass * Acceleration, or in this case, Net Torque = Mass * Angular Acceleration. Given the input torque of 1&amp;nbsp;Nm, this would mean that the wheel would have a net torque of 1&amp;nbsp;Nm applied to it. Therefore, the wheel should experience an acceleration of 1&amp;nbsp;Nm/mass. However, the wheel doesn't keep on accelerating -- it just rotates at a constant speed. So why doesn't the wheel accelerate, when the actuator is applying a fixed torque?&lt;/p&gt;&#xA;" OwnerUserId="2727" LastEditorUserId="1832" LastEditDate="2016-06-09T17:52:37.403" LastActivityDate="2016-06-11T13:19:17.340" Title="Understanding net torque from an actuator" Tags="|mechanical-engineering|control-engineering|dynamics|friction|actuator|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10255" PostTypeId="1" AcceptedAnswerId="10256" CreationDate="2016-06-09T20:14:45.080" Score="2" ViewCount="293" Body="&lt;p&gt;I'd like to drill several very small (~0.015&quot;) but relatively long (~1&quot;) holes into a bar of stainless steel, as shown in the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uhBjJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uhBjJ.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a quick check on McMaster and all the drilling lengths are much shorter (&amp;lt;&amp;lt;0.5&quot;) than I need. I'm not sure if it's feasible after all. How can I source this part, or machine it myself?&lt;/p&gt;&#xA;" OwnerUserId="6770" LastEditorUserId="368" LastEditDate="2016-06-09T21:25:38.973" LastActivityDate="2016-06-09T22:02:58.600" Title="How can I drill a long, narrow hole into the end of a steel circular bar?" Tags="|mechanical-engineering|steel|machining|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10256" PostTypeId="2" ParentId="10255" CreationDate="2016-06-09T21:57:49.603" Score="1" Body="&lt;p&gt;It should be technically feasible using EDM (electrical discharge machining). There are purpose-built EDM &quot;hole popper&quot; drilling machines made for this type of work - the actual drilling time would be about 5 seconds or less per hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your OP says drilling &quot;into&quot; a bar to a depth of 1in, so it's not clear the holes are completely through 1in thick metal, or 1in deep blind holes in a thicker part. Laser drilling would be possible for through holes, but there might be problems getting an accurately controlled depth for blind holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither of these options is a &quot;do-it-yourself&quot; machining operation, of course.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-06-09T22:02:58.600" LastActivityDate="2016-06-09T22:02:58.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10257" PostTypeId="1" CreationDate="2016-06-09T23:33:50.507" Score="5" ViewCount="849" Body="&lt;p&gt;In general, will screws tightened at room temperature become tighter or looser at very low temperature (5 K)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, the screw should loosen because it will contract radially, while the surrounding material will contract slightly less (or even expand, in the limit of a very large screw radius and a very thin ring of surrounding material).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the screw should tighten because the screw should contract linearly along its axis, making the threads tighter, while the surrounding material should contract slightly less because it has other boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a general answer, or does it depend on material, screw size, and/or the exact geometry?&lt;/p&gt;&#xA;" OwnerUserId="6771" LastActivityDate="2016-06-10T00:34:56.953" Title="Do bolts become tighter or looser at cryogenic temperature?" Tags="|mechanical-engineering|bolting|thermal-expansion|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10258" PostTypeId="2" ParentId="10257" CreationDate="2016-06-10T00:34:56.953" Score="4" Body="&lt;p&gt;The bolt tension could increase, decrease, or stay the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on the thermal stress distribution in the &lt;em&gt;whole&lt;/em&gt; structure, which is likely to be eventually supported by something that remains at (approximately) room temperature. This is essentially the same situation as any bolted joint that will carry a load. The &quot;pre-stress&quot; put into the bolts by tightening them when the structure is assembled must be sufficient to accommodate the &quot;working stress&quot; in the joint, whatever causes it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are ways to join structures that allow for thermal expansion without creating thermal stresses. They would involve sliding joints, and/or connecting links that can pivot to accommodate the expansion/contraction, etc. A simple example is the tail-pipe of a car, which hangs &quot;freely&quot; from suspension points on links that can pivot as it expands, rather than being rigidly bolted to the car at both ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the bolts are a different material from the parts they are connecting, that is an additional complication. What happens depends on the relative sizes of the coefficients of thermal expansion for the materials. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-06-10T00:34:56.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10260" PostTypeId="2" ParentId="10254" CreationDate="2016-06-10T06:50:00.873" Score="2" Body="&lt;p&gt;The torque rating for the actuator depends on the speed it's running at - as it approaches its &quot;top speed&quot; it can't deliver the full rated torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasons depend a lot on the actuator type. For example with a DC electric motor the coil induces a back-EMF at higher speeds causing a tail off on the torque curve. For other actuator types there will be other losses which affect their performance. (All in addition to friction in bearings, etc.)&lt;/p&gt;&#xA;" OwnerUserId="6665" LastActivityDate="2016-06-10T06:50:00.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10261" PostTypeId="1" AcceptedAnswerId="10264" CreationDate="2016-06-10T06:53:08.840" Score="2" ViewCount="773" Body="&lt;p&gt;When there is one rectangular beam like below image,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3kce0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3kce0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how the beam is deflected according to thicknesses of top plate and bottom plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concerning cross section patterns are like below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L7HG3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L7HG3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions are as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thicknesses of both side plates are same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 1 : thickness of top &amp;lt; thickness of bottom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 2 : thickness of top = thickness of bottom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type 3 : thickness of top &gt; thickness of bottom&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which type is less deformed? and Why?&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="4624" LastEditDate="2016-06-10T08:54:52.453" LastActivityDate="2016-06-14T13:42:11.753" Title="Deflection of beam, according to thickness of top plate and bottom plate" Tags="|mechanical-engineering|structural-engineering|civil-engineering|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10262" PostTypeId="1" CreationDate="2016-06-10T07:48:35.453" Score="0" ViewCount="69" Body="&lt;p&gt;I have observed that plastic toys that are stored in the plastic blisters, which is transparent, tends to stay the same over a long time. There are no chemical reaction taking place even after decades of storage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically asking about the plastic toys like Hot Wheels and Matchbox. I would like to know what is the chemical compound used for such stable use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for reading my question.&lt;/p&gt;&#xA;" OwnerUserId="6775" LastActivityDate="2016-06-10T12:30:03.977" Title="What is the plastic blister chemical compound that stores plastic toys?" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10263" PostTypeId="2" ParentId="10223" CreationDate="2016-06-10T10:19:36.857" Score="2" Body="&lt;p&gt;I wrote to the manufacturer of the particular unit that I recently purchased and posed the question and this was their response:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Thank you for contacting **********, ***********.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I will be happy to assist with answering your question about the&#xA;  efficiency of the air conditioner.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If you install the air conditioner where direct sunlight will be, it&#xA;  will possibly causing the appliance to run a little longer.  The&#xA;  appliance will not be affected by performance if you decide to not&#xA;  install it in a shaded area.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I hope this information is helpful to you and if you need additional&#xA;  assistance please reply to this email. Thank you for being a valued&#xA;  Frigidaire customer.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sincerely,&lt;/p&gt;&#xA;  &#xA;  &lt;hr&gt;&#xA;  &#xA;  &lt;p&gt;Correspondence Specialist&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And I also received this follow-up:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Per your manual it states “Install the room air conditioner on the&#xA;  shady side of your home. A window that faces north is best because it&#xA;  is shaded most of the day.” So this will be energy efficient to store&#xA;  it in a shaded area. Due to the fact that air conditioners also work&#xA;  based on outside temperatures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6742" LastEditorUserId="6742" LastEditDate="2016-06-10T12:58:20.340" LastActivityDate="2016-06-10T12:58:20.340" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10264" PostTypeId="2" ParentId="10261" CreationDate="2016-06-10T10:55:20.237" Score="3" Body="&lt;p&gt;The deflection of a simply-supported beam under a concentrated load at midspan, regardless of the cross-section, is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = \dfrac{PL^3}{48EI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only thing that changes in your cases is the inertia, $I$. If Type 3 is the same as Type 1, only inverted, then their deflection should be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sum of the top and bottom thicknesses is equal in all the cases (for instance, Type 1 is $1 + 3$, Type 2 is $2 + 2$, and Type 3 is $3 + 1$), then the inertia of Type 2 will be greater than that of Type 1 and 3. To prove this, we need to dive into the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the following dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1TcvC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1TcvC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can calculate all the properties. Type 2 is merely the case where $t_1=t_2$. The centers of gravity of these cross-sections are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\overline{y} &amp;amp;= \frac{h_1b_1\frac{h_1}{2} - h_2b_2\left(t_1 + \frac{h_2}{2}\right)}{h_1b_1-h_2b_2} \\&#xA;&amp;amp;= \frac{h_1b_1\frac{h_1}{2} - h_2b_2\left(t_1 + \frac{1}{2}\left(h_1-t_1-t_2\right)\right)}{h_1b_1-h_2b_2} \\&#xA;&amp;amp;= \frac{h_1b_1\frac{h_1}{2} - h_2b_2\frac{h_1}{2} - h_2b_2\left(\frac{t_1-t_2}{2}\right)}{h_1b_1-h_2b_2} \\&#xA;&amp;amp;= \frac{\frac{h_1}{2}\left(h_1b_1 - h_2b_2\right)}{h_1b_1-h_2b_2} - \frac{h_2b_2\left(\frac{t_1-t_2}{2}\right)}{h_1b_1-h_2b_2} \\&#xA;&amp;amp;= \frac{h_1}{2} - \frac{h_2b_2\left(t_1-t_2\right)}{2\left(h_1b_1-h_2b_2\right)}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Type 2, this simplifies to $\frac{h_1}{2}$, as expected, since $t_1=t_2$, zeroing the second term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the inertia of these sections is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \frac{b_1h_1^3}{12} + b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - \frac{b_2h_2^3}{12} - b_2h_2\left(t_1 + \frac{h_2}{2}-\overline{y}\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Type 2, this simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_2 = \frac{b_1h_1^3}{12} - \frac{b_2h_2^3}{12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to see that Type 2 will always have a larger inertia, we need to find the difference between $I_{1,3}$ and $I_2$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\Delta I &amp;amp;= I_{1,3} - I_2 \\&#xA;&amp;amp;= b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - b_2h_2\left(t_1 + \frac{h_2}{2}-\overline{y}\right)^2 \\&#xA;&amp;amp;= b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - b_2h_2\left(t_1 + \frac{1}{2}\left(h_1-t_1-t_2\right)-\overline{y}\right)^2 \\&#xA;&amp;amp;= b_1h_1\left(\frac{h_1}{2}-\overline{y}\right)^2 - b_2h_2\left(\frac{1}{2}\left(h_1+t_1-t_2\right)-\overline{y}\right)^2 \\&#xA;&amp;amp;= b_1h_1\left(\frac{h_2b_2\left(t_1-t_2\right)}{2\left(h_1b_1-h_2b_2\right)}\right)^2 - b_2h_2\left(\frac{t_1-t_2}{2} + \frac{h_2b_2\left(t_1-t_2\right)}{2\left(h_1b_1-h_2b_2\right)}\right)^2 \\&#xA;&amp;amp;= \left(\frac{t_1-t_2}{2}\right)^2\left(b_1h_1\left(\frac{h_2b_2}{h_1b_1-h_2b_2}\right)^2 - b_2h_2\left(1 + \frac{h_2b_2}{h_1b_1-h_2b_2}\right)^2\right) \\&#xA;&amp;amp;= \left(\frac{t_1-t_2}{2}\right)^2\left(b_1h_1\left(\frac{h_2b_2}{h_1b_1-h_2b_2}\right)^2 - b_2h_2\left(\frac{h_1b_1}{h_1b_1-h_2b_2}\right)^2\right) \\&#xA;&amp;amp;= \left(\frac{t_1-t_2}{2\left(h_1b_1-h_2b_2\right)}\right)^2\left(b_1h_1\left(h_2b_2\right)^2 - b_2h_2\left(h_1b_1\right)^2\right) \\&#xA;\Delta I &amp;amp;= \underbrace{\left(\frac{t_1-t_2}{2\left(h_1b_1-h_2b_2\right)}\right)^2}_{\geq0}\underbrace{b_1h_1b_2h_2}_{&amp;gt;0}\underbrace{\left(h_2b_2- h_1b_1\right)}_{&amp;lt;0} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\Delta I$ is therefore the product of three values, two of which are positive (or zero) and one of which is negative (since $h_1$ and $b_1$ are greater than $h_2$ and $b_2$, respectively), $\Delta I \leq 0$. This therefore means that $I_{1,3}$ (the inertia of Types 1 and 3) will always be lesser than $I_2$ (the inertia of Type 2). Therefore, the deflections under Type 2 will be smaller than those of Types 1 and 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this above assumes that the material is isotropic and has the same behavior under both tension and compression. Something like reinforced concrete, where the concrete under tension is useless since it'll just crack, Type 3 is probably the best, since it puts the most concrete in the compression zone and leaves the steel to resist the tension.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;This answer was initially incorrect, stating that Types 1 and 3 would have a larger inertia. I assumed the parallel axis component would not be so significant, and we all know what happens when one assumes. After seeing &lt;a href=&quot;https://engineering.stackexchange.com/a/10292/1832&quot;&gt;@kamran's answer&lt;/a&gt;, I double-checked my work and realized I was sorely mistaken. I apologize for the inconvenience.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-06-14T13:42:11.753" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10266" PostTypeId="2" ParentId="10254" CreationDate="2016-06-10T11:59:33.610" Score="5" Body="&lt;p&gt;There are two things going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, even if this &quot;actuator&quot; can produce constant torque, the torque required to keep the load spinning will be at least in part a function of the spinning speed.  There will be some friction and other forces that increase with increased speed.  &lt;i&gt;Viscous friction&lt;/i&gt; increases linearly with speed, and other effects, like air resistance, increase even more strongly with speed.  Therefore, even with constant torque, eventually the friction forces rise to exactly oppose the torque and the object no longer speeds up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, your &quot;actuator&quot; probably can't supply constant torque, even over its rated speed range.  For a electric motor, for example, torque decreases with speed at a fixed applied voltage.  The motor has a finite unloaded speed that is mostly proportional to applied voltage.  That is the speed at which it's externally available mechanical torque drops to 0.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-06-10T11:59:33.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10267" PostTypeId="2" ParentId="10262" CreationDate="2016-06-10T12:30:03.977" Score="3" Body="&lt;p&gt;To answer a question like this I would look up blister packaging and see what materials are commonly used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Blister_pack#Materials&quot; rel=&quot;nofollow&quot;&gt;Blister packaging article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the article I have linked above suggests some possibilities: PVC (polyvinyl chloride), PVDC (Polyvinylidene_chloride) and PCTFE (Polychlorotrifluoroethylene).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is the most likely material is PVC by the way (it seems to be very common in packs I have looked at).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also look for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Resin_identification_code&quot; rel=&quot;nofollow&quot;&gt;recycling code&lt;/a&gt; embossed on the material, for example PVC is a triangle with a &quot;3&quot; inside it. (The code is not always put on packs though.)&lt;/p&gt;&#xA;" OwnerUserId="6665" LastActivityDate="2016-06-10T12:30:03.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10269" PostTypeId="1" CreationDate="2016-06-10T14:12:35.873" Score="5" ViewCount="168" Body="&lt;p&gt;I was just reading about the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fuzzy_control_system#Antilock_brakes&quot;&gt;anti-lock brake system example&lt;/a&gt; on the wiki page for fuzzy control system. The principle seems to be simple: instead of just assigning a SINGLE category (cold, cool, nominal, warm or hot) to the temperature of the brake, it translates the temperature into a continuous scale (fuzzy set), like 70% cold and 30% cool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see the advantage of this over the 0-1 logic because now the input is smoother than the 0-1 logic and so the output will also be smoother. But if you have the temperature available, which is already in a continuous scale, why not just control directly based on the temperature? Why do you need the extra step to convert the temperature into fuzzy set and then do the control? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed fuzzy control seems to be a hot topic a couple of decades ago. So maybe it's just because the controllers (in applications like cars and washing machines) were only capable of controlling based on rules rather than doing PID control?&lt;/p&gt;&#xA;" OwnerUserId="6780" LastActivityDate="2016-08-21T22:03:07.897" Title="Fuzzy control of anti-lock brake system" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10273" PostTypeId="2" ParentId="10254" CreationDate="2016-06-11T13:19:17.340" Score="0" Body="&lt;p&gt;The torque rating of the motor is basically the maximum torque that the motor can generate, generally at standstill(ie. 0rpm).the performance characteristics of a dc motor would look like this &lt;a href=&quot;https://i.stack.imgur.com/i3uoj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i3uoj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6791" LastActivityDate="2016-06-11T13:19:17.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10277" PostTypeId="1" AcceptedAnswerId="10287" CreationDate="2016-06-12T13:19:03.080" Score="2" ViewCount="450" Body="&lt;p&gt;How does the Michelson interferometer measure the self-coherence function of coherent light in an incoherent background? &lt;a href=&quot;http://discovery.ucl.ac.uk/2670/&quot; rel=&quot;nofollow&quot;&gt;&lt;em&gt;Detection of coherent light in an incoherent background&lt;/em&gt; (Coutinho et al., 1999)&lt;/a&gt; is still the best analysis and implementation of this concept despite considerable progress in semiconductor fabrication and photonics since its publication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask this question is because airplane pilots need to see the landing strip, ground landmarks and the horizon while at the same time harmful visible light laser rays are filtered.&lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="368" LastEditDate="2016-06-14T15:18:10.073" LastActivityDate="2016-06-18T15:23:39.723" Title="How does the Michelson interferometer measure the self-coherence function of coherent light in an incoherent background?" Tags="|electrical-engineering|optics|lasers|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10278" PostTypeId="1" AcceptedAnswerId="10279" CreationDate="2016-06-12T15:56:42.520" Score="3" ViewCount="3707" Body="&lt;p&gt;I have a hole that contains metal shavings after I have tapped it. Is there a common method to clean out metal shavings from between the threads of the tapped hole?&lt;/p&gt;&#xA;" OwnerUserId="2602" LastEditorUserId="33" LastEditDate="2016-06-14T13:54:30.867" LastActivityDate="2019-03-06T05:09:16.940" Title="Cleaning Shavings from Tapped Hole" Tags="|fasteners|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10279" PostTypeId="2" ParentId="10278" CreationDate="2016-06-12T16:09:02.783" Score="5" Body="&lt;p&gt;There are a few ways to do this obviously it's easier to clean a through hole than a blind one. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Run a clean second or plug tap through the hole. If you keep a tap just for this purpose it also has the benefit of keeping one 'fresh' tap which clean clean up any burrs of defects from worn taps when initially cutting the thread. &lt;/li&gt;&#xA;&lt;li&gt;Flush the hole with cutting fluid, suds  or degreaser. &lt;/li&gt;&#xA;&lt;li&gt;Scrub the thread with a cylindrical nylon or brass brush of the appropriate diameter, a brush mounted in a cordless drill can be an effective way to clean multiple threads. &lt;/li&gt;&#xA;&lt;li&gt;Cut a slot down the length of a bolt or stud of the appropriate size and carefully run it into the thread &lt;/li&gt;&#xA;&lt;li&gt;Hold the part inverted and tap with a nylon or hide hammer. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Edit : good point about compressed air. A probe type air duster (either from a compressor or a hand held aerosol)  or even a manual blower can be effective, especially where threads are hard to get at by other means. This tends to work best when the hole is reasonably free of oil and grease so flushing with something like a solvent or alkaline degreaser first may help. Obviously care needs to be taken not to blow solvents or debris from the hole into your own (or anybody else's) face and eyes. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-14T15:49:42.357" LastActivityDate="2016-06-14T15:49:42.357" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10280" PostTypeId="1" CreationDate="2016-06-12T18:16:42.813" Score="1" ViewCount="146" Body="&lt;p&gt;Why is electricity used to power most things? Is it because it is the most widely available? Why can't some other form of energy be used?&lt;/p&gt;&#xA;" OwnerUserId="6806" LastActivityDate="2016-06-13T12:55:58.073" Title="Why is electricity used for powering our device?" Tags="|power|" AnswerCount="1" CommentCount="3" ClosedDate="2016-06-12T22:28:05.307" ContentLicense="CC BY-SA 3.0" />
  <row Id="10281" PostTypeId="2" ParentId="10280" CreationDate="2016-06-12T18:58:34.310" Score="4" Body="&lt;p&gt;Electricity is a convenient way to generate power centrally and distribute it to where it is needed according to demand. Central power generation by technologies like steam and gas turbines means that you can build large power plants with the associated gains in efficiency and economies of scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrical power is also very versatile:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Electric motors have good performance characteristics (good power to weight ratio, flat torque curves etc), are generally reliable, having few moving parts and are easy to control. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat and especially light can be generated directly from electrical current with very simple technology. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electronics and micro-electronics are the most effective way by far that we have of miniaturizing logic and information processing. These inherently require electrical power to work. You could have mechanical computers but they are &lt;em&gt;much&lt;/em&gt; harder to miniaturise. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electrical power is clean and mostly quiet at the point of use.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is easy and reasonably efficient to convert voltage for different applications. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Electrical power does have some disadvantages though. It is generally an expensive way to produce heat locally so oil, gas or other fuels may may more economic sense in many situations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ability to store electrical power is limited so while batteries are excellent for small, low powered self contained devices, electricity is less good for things which need to be mobile and use a lot of power. This also means that the infrastructure needs to be able to quickly adjust output according to overall demand. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-13T12:55:58.073" LastActivityDate="2016-06-13T12:55:58.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10282" PostTypeId="1" CreationDate="2016-06-12T22:47:45.177" Score="0" ViewCount="122" Body="&lt;p&gt;I am using 123D Design to build a model for 3D printing.  What I want to build is a large 2D rectangle with several smaller rectangles inside. Then, I want to extrude the larger rectangle, but leave the smaller ones as holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that I could subtract the smaller rectangles from the larger one and then extrude the remainder.  However, when I use the subtract tool, I cannot select sketches - just solids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on how to do this?  I would very much like to know how to do it 2D first, and then extrude the remainder.  However, if I must transform this to 3D first, I'll be willing to learn that instead.  Much appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YMU8X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YMU8X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4403" LastActivityDate="2016-10-11T16:14:33.323" Title="How do I &quot;subtract&quot; sketches in CAD?" Tags="|mechanical-engineering|modeling|3d-printing|cad|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10283" PostTypeId="2" ParentId="10282" CreationDate="2016-06-12T23:40:41.010" Score="1" Body="&lt;p&gt;Sketches aren't usually part of the actual model until you use them to define a solid feature. In general when you extrude a sketch with closed loops inside it the inner loops will be excluded from the extrusion or, depending on the software, you will have the option which closed loops are selected as part of the extrusion dialogue. So it is quite likely that what you want will happen by default, depending on how your particular software package deals with selections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some software packages will allow you to define 2-dimensional surfaces but many don't as they can make models difficult to manage and you can only work with points, lines and volumes (solids). In many cases there is a distinct separation in the workflow between sketches used to define solid features and the features themselves. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-11T16:14:33.323" LastActivityDate="2016-10-11T16:14:33.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10284" PostTypeId="1" AcceptedAnswerId="10286" CreationDate="2016-06-13T03:20:48.750" Score="2" ViewCount="162" Body="&lt;p&gt;In Russia, there's a special term for a pipe that is not circular in its section: &quot;a profiled pipe&quot; (профильная труба). There's even a Wikipedia page for it: &lt;a href=&quot;https://ru.wikipedia.org/wiki/%D0%9F%D1%80%D0%BE%D1%84%D0%B8%D0%BB%D1%8C%D0%BD%D0%B0%D1%8F_%D1%82%D1%80%D1%83%D0%B1%D0%B0&quot; rel=&quot;nofollow noreferrer&quot;&gt;профильная труба&lt;/a&gt;. Such pipes come in a wide variety of sizes and usually have an oval, square or rectangular section (&quot;profile&quot; in Russian), but other kinds of sections are possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder what is the common cover-all technical term describing &quot;any kind of non-circular pipe&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uWE6V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uWE6V.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example of usage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pervouralsk New Pipe Plant produces a range of non-circular pipes, including pipes of variable cross-section.   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://ell.stackexchange.com/questions/93589/looking-for-a-common-term-for-non-circular-pipe&quot;&gt;Also asked on ELL SE&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="-1" LastEditDate="2017-04-13T12:55:21.033" LastActivityDate="2016-06-13T10:28:42.117" Title="Looking for a common term for &quot;non-circular pipe&quot;" Tags="|terminology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10286" PostTypeId="2" ParentId="10284" CreationDate="2016-06-13T10:28:42.117" Score="4" Body="&lt;p&gt;For structural purposes these are usually called &lt;a href=&quot;https://en.wikipedia.org/wiki/Hollow_structural_section&quot; rel=&quot;nofollow&quot;&gt;Hollow Structural Sections&lt;/a&gt; (HSS). For non-structural purposes &lt;em&gt;Mechanical Tubing&lt;/em&gt; is often used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Europe the main body that deals with structural design of these is &lt;a href=&quot;http://www.cidect.com/en/&quot; rel=&quot;nofollow&quot;&gt;CIDECT&lt;/a&gt;, who have a series of useful design guides. The equivalent would probably be the &lt;a href=&quot;https://steeltubeinstitute.org&quot; rel=&quot;nofollow&quot;&gt;Steel Tube Institute&lt;/a&gt; in the USA who also deal with non-structural components.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2016-06-13T10:28:42.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10287" PostTypeId="2" ParentId="10277" CreationDate="2016-06-13T12:11:56.300" Score="2" Body="&lt;p&gt;As you change the relative arm lengths of a Michelson interferometer, the transmission (or reflection) coefficient of the interferometer ranges from $T=0$ to $T=1$ for coherent light, but, if designed properly, will always have a transmission of $T=0.5$ for incoherent light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we define the length of the two arms of the Michelson to be $L_1$ and $L_2+\delta\ell$, where $L_1$ and $L_2$ are macroscopic distances on the order of meters and $\delta\ell$ is a microscopic distance on the order of micrometers, then the interference properties of the Michelson only depend on $\delta\ell$ for highly coherent light like a laser.  However, the interference properties of the Michelson for incoherent light depend on the relative macroscopic distances.  If $|L_1-L_2|\gg L_c$, where $L_c$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Coherence_length&quot; rel=&quot;nofollow&quot;&gt;coherence length&lt;/a&gt; of the incoherent light, then the Michelson will not display any interference regardless of $\delta\ell$.  &lt;/p&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2016-06-13T18:33:17.073" LastActivityDate="2016-06-13T18:33:17.073" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="10288" PostTypeId="1" AcceptedAnswerId="10307" CreationDate="2016-06-13T15:07:50.037" Score="0" ViewCount="103" Body="&lt;p&gt;I would like to get my hands on a mechanically-set timer (wind up is fine, so that it's self-powered) which sends a small electrical pulse to a mag-lock once the timer goes off, thus causing the mag-lock to open. Is there anything out there that does this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what would be the cheapest / easiest way to replicate such function, i.e. some mechanical way to produce a small electrical impulse once a mechanical timer finishes. I don't mind electronic methods, but since this is for heavy outdoor use I'd prefer a purely mechanical setup.&lt;/p&gt;&#xA;" OwnerUserId="6815" LastEditorUserId="6815" LastEditDate="2016-06-13T15:54:21.257" LastActivityDate="2016-06-14T17:30:31.600" Title="Is there a device like this?" Tags="|electrical-engineering|magnets|timing|" AnswerCount="2" CommentCount="3" ClosedDate="2016-06-16T11:10:49.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="10289" PostTypeId="1" AcceptedAnswerId="10290" CreationDate="2016-06-13T16:26:19.057" Score="5" ViewCount="8067" Body="&lt;p&gt;For reinforcing in concrete, anchorage / lap lengths are determined by the multiplying the diameter of the rebar with a factor. Generally a factor of 40-45 is accepted as the de-facto standard. According to EC2, the factor falls within the range from 34 and 66, depending on concrete quality, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my understanding that the factor is in linked to the amount of shear resistance required to act against a force pulling out a rebar at its ultimate tensile stress. The factor is thus the height of the shear cone created. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When lapping a bar of larger diameter with a smaller one, my explanation above would be used to argue the case that the smaller diameter is taken to determine lap length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found various sources where people have said the smaller bar should be used. I can find no code references defining it though. I would assume that one goes for the larger diameter to be safe, but the extra steel obviously increases cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="2637" LastEditDate="2018-02-23T10:11:35.023" LastActivityDate="2020-01-04T00:55:24.420" Title="What is the splice length for two reinforcement bars in tension of different diameter?" Tags="|structural-engineering|reinforced-concrete|reinforcement|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10290" PostTypeId="2" ParentId="10289" CreationDate="2016-06-13T18:10:14.713" Score="4" Body="&lt;p&gt;The use of the smaller diameter is &lt;em&gt;theoretically&lt;/em&gt; correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a splice? It is when tension force in a rebar is transferred to the concrete via adhesion and then to another rebar. Since concrete is very weak against tension, large lap lengths are required such that the stress at any point in the concrete is not excessive. However, once the splice is over, all the tension from the first rebar has been transferred to the other one, which must also be capable of withstanding that same tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if you are splicing two rebars of different diameters, the maximum tensile force that can be transferred via the splice is equal to the yield strength of the smaller rebar. If the tensile force is larger (say, equal to the yield strength of the larger rebar), then the smaller rebar will rupture immediately after the splice. So there's no reason to design your splice to withstand a force larger than the yield strength of the smaller bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you are splicing multiple rebars in the same position, going from a few large diameter bars to many smaller ones, an incremement to the splice length may be required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure below, for instance, assume this is a splice section, with the tension from the large red bars being transferred to the small blue ones. The arrows roughly represent the furthest distance that tension needs to travel from each bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aEeyh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aEeyh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The codes usually define a maximum distance between lapping rebars. The Brazilian code NBR-6118 and EC2 both put that limit at $4\phi$ or 50&amp;nbsp;mm. In this case, there's the question of which $\phi$ to use, and I honestly don't know. However, both codes do allow distances larger than this limit so long as the splice length is increased accordingly. So in cases such as the one pictured above, an increase to the splice length may be necessary to satisfy this requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reasons I can't quite fathom, though, the Brazilian code explicitly states that splices between different diameters &lt;strong&gt;must adopt the larger diameter&lt;/strong&gt;, so that's what I'm stuck with. &lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-06-13T18:10:14.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10291" PostTypeId="2" ParentId="10223" CreationDate="2016-06-13T18:57:40.147" Score="1" Body="&lt;p&gt;The housing under direct sun will be much warmer but its contribution is minimized by some factors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-1 the surface area of the housong compared to densely placed fins of the radiator/condenser.&lt;br&gt;&#xA;  -2 the fan positioned in a way to create the greatest current of air blowing through the fins, getting chilled, while stagnating around the walls of the box.&lt;br&gt;&#xA;  -3 in some models the the flow of hot air near the walls of unit has been impeded by partitioning the space in a way to insulate the current of chilled air.&lt;br&gt;&#xA;  -4 choosing the window were the inflow of air is from a shaded area has more impact. Consider trees and their moving shadow in the hottest time of the day.  In Los Angeles the difference between shade and sun radiation is 10 to 15 degrees. So if we use tributary source from a shaded area it makes a big difference.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2016-06-13T19:04:52.930" LastActivityDate="2016-06-13T19:04:52.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10292" PostTypeId="2" ParentId="10261" CreationDate="2016-06-13T22:32:15.560" Score="3" Body="&lt;p&gt;The deflection in type 2 is the least. Type one and type 3 will deform more because of the fact that neutral axis if the beam will move nearer to top or bottom.&lt;br&gt;&#xA;When this happens the new strain on the narrower flange if the beam will be bigger resulting in more elongation at that flange. Meaning more deflection of the beam.&lt;br&gt;&#xA;One crude way of looking at it is the new neutral axis will consume some of the web of the beam from the thick side. Hence the beam bends more to stress the left over cross section more to make up for its reduced area. It can easily be seen in drawing of beams cross section stress/ strain digram which i tried to draw on my phone by finger!  &lt;a href=&quot;https://i.stack.imgur.com/EqHCf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EqHCf.jpg&quot; alt=&quot;cross section if the beam.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-06-13T22:32:15.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10293" PostTypeId="2" ParentId="10288" CreationDate="2016-06-13T22:56:51.947" Score="0" Body="&lt;p&gt;A mag lock typically uses 12 volts at low current to power the electromagnet to remain locked. The unlocking mechanism trips the power to the electromagnet. Mechanical delay relays do exist, but you would need one that has a normally closed relay that would be opened by the mechanical delay.&lt;/p&gt;&#xA;" OwnerUserId="3569" LastActivityDate="2016-06-13T22:56:51.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10294" PostTypeId="2" ParentId="10199" CreationDate="2016-06-14T01:36:17.610" Score="6" Body="&lt;p&gt;Beyond a certain level of saturation, the bulk modulus of water starts playing an important role.  Since water is nearly incompressible at moderate pressures, the pore pressure increases rapidly and equilibrates with the externally applied force when we try to compress the water.  This makes it difficult to compact fully saturated soils and partially saturated soils with a large water content under confined conditions.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2016-06-14T01:36:17.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10298" PostTypeId="1" CreationDate="2016-06-14T11:41:28.280" Score="1" ViewCount="761" Body="&lt;p&gt;A question I ask myself now and again is if it is technically feasible to build battery powered construction equipment, like large excavators - say 250kW motive power at the shaft of the central hydraulic power pack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an electrical engineer so I'm not sure how the motor, battery and everything in between would look like. It doesn't seem trivial:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;250 kW motive power is roughly 280kW electrical power. Most batteries deliver single digit voltage and are connected in series to deliver 12V or whatever is needed. DC motors at this power &lt;a href=&quot;http://w3app.siemens.com/mcms/infocenter/dokumentencenter/ld/Documentsu20Catalogs/dc-motor/da12-2008-en.pdf&quot; rel=&quot;nofollow&quot;&gt;seem to need between 420 and 800V&lt;/a&gt;, depending on speed and torque. So I'm either talking about stacking quite a lot of battery units (with their internal resistance!) or using a DC-DC converter (do these exist for 280kW? How (in)efficient are they?) or looking at a different motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided a 250kW 12V motor exists, I'd be talking at about 23kA current. This seems quite collosal to me (but maybe I'm wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way could be to mount several motors on one shaft, but then the inactive motors would likely act as brakes and add their own headaches (or not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, for me as a non electrical engineer it's not trivial to envision such a power train. How could it look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An answer should:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;provide &lt;em&gt;a&lt;/em&gt; workable solution, not neccessrily &lt;em&gt;the&lt;/em&gt; best - proof what is optimal is surely bexond the scope of an answer&lt;/li&gt;&#xA;&lt;li&gt;list all the major components - motor, battery, converters, other major stuff I forgot - with effiencies in this particular use case&lt;/li&gt;&#xA;&lt;li&gt;provide a decent estimate of overall efficiency (input: charging battery, output: motive energy provided)&lt;/li&gt;&#xA;&lt;li&gt;Energy density of battery and smallest viable battery to deliver the power&lt;/li&gt;&#xA;&lt;li&gt;ballpark estimates of weight and cost of major components are appreciated but not neccessary&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastEditorUserId="1586" LastEditDate="2016-06-28T00:09:39.883" LastActivityDate="2016-06-28T00:09:39.883" Title="How to power a large electric motor from a battery?" Tags="|electrical-engineering|power-electronics|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10299" PostTypeId="1" AcceptedAnswerId="10304" CreationDate="2016-06-14T12:35:35.440" Score="0" ViewCount="146" Body="&lt;p&gt;I have seen this video pretending to decrease by 5° the temperature inside home using an air-conditioner based on the &lt;a href=&quot;https://physics.stackexchange.com/questions/7868/why-does-the-air-we-blow-exhale-out-from-our-mouths-change-from-hot-to-cold-depe&quot;&gt;air entrainment&lt;/a&gt; principle (see the &lt;a href=&quot;https://youtu.be/jPuh8IFbIzQ?t=1m6s&quot; rel=&quot;nofollow noreferrer&quot;&gt;video here at 1m6s&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT : The video is now private but you can see a picture of the air-conditioner &lt;a href=&quot;http://www.iflscience.com/environment/these-diy-ecofriendly-air-conditioners-are-cooling-down-bangladesh/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such an air-conditioner really work?&lt;/p&gt;&#xA;" OwnerUserId="6827" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2016-06-15T09:39:21.007" Title="Is it possible to use air entrainment for air conditioning" Tags="|heat-transfer|airflow|temperature|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10300" PostTypeId="1" AcceptedAnswerId="10332" CreationDate="2016-06-14T14:35:00.417" Score="5" ViewCount="3124" Body="&lt;p&gt;I only have basic metal fabrication tools and do not own a mill or a lathe, but would like one. I took apart my cheap drill press to study whether or not I can convert it into a milling machine. The spindle features a 4 tooth 9mm x 9mm tooth male splined section that mates with a female splined bushing press fit into the pulley's bore (see parts numbers 62 and 71 in diagram). The entire spindle fits inside the quill tube (see part 68) creating the quill assembly (see photo below).  The spline allows the spindle to extend downwards and upwards in the head casting while also allowing the spindle to rotate with maximum torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I am considering replacing the entire spindle with an 16mm O.D.  ER20 tool holder. This will give me a more ridgit hold on the cutting tool as Jacobs Taper cannot handle axial forces required for milling. I am also planning on replacing the existing bearings with either angular contact bearings or tapered bearings.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to maintain the quill's retract and extend mechanism. To do this, I will have to somehow jointhe end of the ER toolholder to a splined shaft that will fit in the female end. Currently, I can't seem to find any off the shelf spline to 16mm coupling. I am not even sure what they are called.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any solutions for coupling or attaching a spline shaft to a plain old shaft? What is it called and do they exist?  Finally, I am not completely married to using a spline drive mechanism in order to extend and retract the quill, so I welcome any suitable alternative mechanisms. I saw a ball spline, but wasn't sure if it works the same way as the existing mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a photo of a similar quill:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F5fCS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F5fCS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N5KkC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N5KkC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vu1Ti.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vu1Ti.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2248" LastEditorUserId="2248" LastEditDate="2016-06-17T00:09:52.520" LastActivityDate="2016-06-23T01:12:07.697" Title="Are there solutions for coupling a spline shaft to a round diameter shaft" Tags="|mechanical-engineering|gears|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10301" PostTypeId="2" ParentId="10223" CreationDate="2016-06-14T15:30:55.370" Score="1" Body="&lt;p&gt;There will obviously be some performance boost if you shade the fins/housing, but it will be negligible, since the ambient temperature &lt;em&gt;around&lt;/em&gt; the a/c unit will not be affected by shading the unit itself.  the ambient temperature on the north side of a structure can be considerably lower than the south or west side - my house has a carport on the north side of the house and I frequently see temperatures that are 10 degrees cooler than the south side of my house.  &lt;/p&gt;&#xA;" OwnerUserId="6830" LastActivityDate="2016-06-14T15:30:55.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10302" PostTypeId="2" ParentId="8920" CreationDate="2016-06-14T15:37:16.457" Score="0" Body="&lt;p&gt;&lt;em&gt;How many hours in a day do conventional electrical generators work (the ones in large power plants)?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer; it depends on the constraints of the generators and the priorities of the system operator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying area is known as &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Unit_Commitment_Problem_in_Electrical_Power_Production&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unit Commitment&lt;/a&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A family of mathematical optimization problems where the production of&#xA;  a set of electrical generators is coordinated in order to achieve some&#xA;  common target, usually either match the energy demand at minimum cost&#xA;  or maximize revenues from energy production.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming that you are referring to thermal generators, the constraints &lt;a href=&quot;https://en.wikipedia.org/wiki/Unit_Commitment_Problem_in_Electrical_Power_Production#Types_of_production_units&quot; rel=&quot;nofollow noreferrer&quot;&gt;include&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;minimum up/down time  &lt;/li&gt;&#xA;&lt;li&gt;ramp up/down rate  &lt;/li&gt;&#xA;&lt;li&gt;modulation/stability &lt;/li&gt;&#xA;&lt;li&gt;start-up/shut-down ramp rate&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Do they by any method conserve or store the power before it reaches the consumer?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No. At least not over the time frames that unit commitment is considered (of the order of hours/half hours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over much shorter timeframes (seconds) energy is stored in the rotating mass of the generators. This is described in answers to &lt;a href=&quot;https://engineering.stackexchange.com/questions/2245/quantifying-inertia-on-the-electricity-grid&quot;&gt;this question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4476" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-06-14T15:37:16.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10303" PostTypeId="1" CreationDate="2016-06-14T15:53:10.557" Score="1" ViewCount="81" Body="&lt;p&gt;I want to assess how much time it takes for a single step of a stepper motor to settle. &#xA;Starting from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau=J\frac{d²\theta}{dt²}$$&#xA;Taking&#xA;$$\tau=K_m cos(\theta)\approx K_m \theta$$&#xA;Yields&#xA;$$K_m \theta -J\frac{d²\theta}{dt²}=0$$&#xA;Or in Laplace domain&#xA;$$K_m\theta(s)-J(s²\theta(s)-s\theta(t=0)-\dot{\theta}(t=0))=0$$&#xA;However the initial conditions are:&#xA;$$\theta(t=0)=0$$&#xA;$$\dot{\theta}(t=0)=0$$&#xA;Therefore&#xA;$$\theta(s)(K_m-Js²)=0$$&#xA;Now what :( ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do have the control systems toolbox of matlab, for the following steps, but honestly I don't know what they are. &lt;/p&gt;&#xA;" OwnerUserId="2982" LastEditorUserId="2982" LastEditDate="2016-06-14T16:27:29.703" LastActivityDate="2016-06-14T16:27:29.703" Title="Can't solve the 2nd order differential equation of a stepper motor single step position" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10304" PostTypeId="2" ParentId="10299" CreationDate="2016-06-14T15:54:37.173" Score="1" Body="&lt;p&gt;This device is not really working using the principles referenced in the other SE article.  In this case the air inside the hut is being heated by the sun and this is really about improving ventilation and letting that heated air out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, however, it would perform worse than a fully open window from a heat transfer point of view, however, it would also act as a screen for privacy/exclusion reasons.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have doubts it works much better with the bottles than without them (ie. just a series of holes).  The article mentions the bottles are used to funnel cold air in but holes would allow that too. Also, the bottles would cause more interference with the drawing out of hot air if the wind is blowing parallel to the wall than simple holes would.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL:DR It would certainly help cool the air inside, but I think simpler methods would work better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: Apparently this question has appeared on multiple SE see: &lt;a href=&quot;https://physics.stackexchange.com/questions/261284/how-does-the-eco-cooler-air-conditioner-really-work&quot;&gt;Physics.SE&lt;/a&gt; and &lt;a href=&quot;https://skeptics.stackexchange.com/questions/34302/does-the-eco-cooler-device-cool-the-air-in-a-home-without-electricity&quot;&gt;Skeptics.SE&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2616" LastEditorUserId="-1" LastEditDate="2017-04-13T12:46:46.810" LastActivityDate="2016-06-15T09:39:21.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10305" PostTypeId="1" CreationDate="2016-06-14T16:25:09.363" Score="2" ViewCount="116" Body="&lt;p&gt;I am designing a sculpture based on a Gothic window which will consist of a steel frame with 3 inset panels in cast resin. The panels will be the 3 larger interior ares in the drawing with detail on both faces. Each face will be moulded separately and then joined in a 'clamshell' type arrangement with the gap filled with rigid foam (possible with additional internal reinforcement). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The panels will have a lot of surface detail (based on a pattern made by primary school children) and the exterior surface will be metal filled resin (probably aluminium) and approximately 20mm thick, backed internally with several laminations  of glass mat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These panels will be bolted to the steel structure via weld brackets. They are not inherently structural but will be subject to light loads form flexing of the main structure plus wind loads etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the past I have used polyester resin for this application but had problems with casting defects largely caused by the high viscosity of the resin plus metal powder. I have had much better results with PU casting resin for smaller, well supported castings but I'm not sure how it would bear up under long term outdoor exposure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am looking for nay thoughts on the relative merits of PU vs polyester for this application. The mouldings should have a life of at least 10 years outdoors in a temperate climate (UK). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kRkFO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kRkFO.jpg&quot; alt=&quot;drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="7083" LastEditDate="2018-01-28T22:45:48.153" LastActivityDate="2018-06-13T16:23:44.630" Title="PU vs Polyester resin for exterior moulding" Tags="|materials|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10307" PostTypeId="2" ParentId="10288" CreationDate="2016-06-14T17:25:45.413" Score="1" Body="&lt;p&gt;In principal it should not be very difficult to adapt an off the shelf mechanism like a mechanical alarm clock or kitchen timer. They will already have some sort of latching mechanism to activate the bell after the elapsed time so it should be a simple matter to adapt this to close an electrical contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that I'm not entirely convinced that a mechanical system is necessarily going to be any more rugged or reliable than an electronic one. As already mentioned in other answers a mag-lock will need a reliable power supply and a timer circuit is simple to achieve with either digital or analogue electronics and can be potted of placed in an IP rated enclosure to make it pretty much indestructible compared to any kind of clockwork system. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-14T17:30:31.600" LastActivityDate="2016-06-14T17:30:31.600" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10308" PostTypeId="1" CreationDate="2016-06-14T20:34:45.867" Score="1" ViewCount="38" Body="&lt;p&gt;I am building a simple &lt;a href=&quot;http://www.instructables.com/id/DIWire-Bender/&quot; rel=&quot;nofollow&quot;&gt;DIWire Bender&lt;/a&gt;.  Included in their Google Code Site materials list is the &lt;a href=&quot;http://catalog.orientalmotor.com/item/stepping-motors/pk-series-stepping-motors/pk264a2a-sg10?&amp;amp;amp%3Bseo=110&quot; rel=&quot;nofollow&quot;&gt;OrientalMotor 2 Phase Stepper Motor&lt;/a&gt;, which will be responsible for feeding and bending the wire.  Each motor is priced around $200, and in total they exceed my budget.  I am searching for similar but less expensive solutions below 100 bucks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Which variables in this spec list would be critical in choosing a similar motor?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Motor Type :  2-Phase&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Holding Torque : 368 oz-in&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Shaft/Gear Type   : Spur Gear&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Gear Ratio (X:1) : 10 :1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Shaft : Single&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Type : Geared&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Output Step Angle : 0.18 °&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Connection Type   : Bipolar (Series) / Unipolar&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Current per Phase (A/phase)   : 1.4 [Bipolar (Series)] / 2 [Unipolar]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Voltage (VDC) : 3.9 [Bipolar (Series)] /2.8 [Unipolar]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Resistance (Ω/phase) : 2.8 [Bipolar (Series)] / 1.4 [Unipolar]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Shaft Runout      0.05 mm (0.002 in.) T.I.R.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Concentricity     0.075 mm (0.003 in.) T.I.R.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Perpendicularity      0.075 mm (0.003 in.) T.I.R.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Radial Play       0.025 mm (0.001 in.) maximum of 5 N (1.12 lb.)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Axial Play        0.075 mm (0.003 in.) maximum of 10 N (2.2 lb.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6832" LastActivityDate="2016-06-14T20:34:45.867" Title="What Specs are Important in Choosing a Similar Stepper Motor?" Tags="|stepper-motor|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10309" PostTypeId="1" AcceptedAnswerId="10310" CreationDate="2016-06-14T23:16:16.193" Score="5" ViewCount="390" Body="&lt;p&gt;I am learning about Variable Reluctance ( VR ) Stepper Motors.  As stated in my reference, they &quot;have no detent torque.&quot;  What does this mean?&lt;/p&gt;&#xA;" OwnerUserId="6832" LastActivityDate="2016-06-14T23:41:15.280" Title="What is 'No Detent Torque'?" Tags="|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10310" PostTypeId="2" ParentId="10309" CreationDate="2016-06-14T23:41:15.280" Score="6" Body="&lt;p&gt;Detent torque is the torque while coils are not energised. Permanent magnets and the iron stator hold the motor in a given position without any power needing to be applied, although this force is usually much lower than the full torque - in my experience about 10%.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-06-14T23:41:15.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10312" PostTypeId="1" CreationDate="2016-06-15T12:44:50.297" Score="1" ViewCount="61" Body="&lt;p&gt;Power braking (increasing power while brake applied) at start help to start car at higher speed, but on other hand we lose traction due to tire slippage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this increase acceleration (as shown in movies)? If yes, then why don't F1 racers do it?&lt;/p&gt;&#xA;" OwnerUserId="908" LastEditorUserId="1832" LastEditDate="2016-06-15T13:31:38.257" LastActivityDate="2016-06-15T13:34:30.653" Title="Does power braking increases acceleration?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10313" PostTypeId="2" ParentId="10312" CreationDate="2016-06-15T13:34:30.653" Score="1" Body="&lt;p&gt;There are a number of potential reasons for braking under power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Braking generally gives finer, more consistent and more immediate control of torque than throttle position and so it may be useful for making fine adjustments to the torque applied to the wheels while keeping engine revs in a desirable range and avoiding the instability potentially associated with lifting off the throttle while turning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Braking also affects weight distribution so can be used, in conjunction with brake  bias adjustment, to mitigate under-steer and improve turn-in, especially in rallying, by transferring load from the rear to the front wheels. In fact the &lt;a href=&quot;https://en.wikipedia.org/wiki/McLaren_MP4/12&quot; rel=&quot;nofollow&quot;&gt;Mclaren MP4/12&lt;/a&gt; had a system which used an additional brake pedal applied to one wheel only (depending on turn direction) to assist steering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another specific application in F1 was during the era of exhaust blown diffuses. In this case the high velocity airflow generated by the exhaust was used to generate down-force for  there was a clear advantage in keeping revs high during cornering, holding off a significant amount of power with the brake if necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of traction from a standing start there is a complex interplay between clutch, throttle and brake as well as engine and throttle mapping and gear ratios. Modern F1 cars also have dual hand clutch controls (although regulations have changed this year) so this obviously makes balancing brake and throttle somewhat easier than with a foot operated clutch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general a good racing start involves balancing two competing considerations firstly maximising available torque by keeping revs in an optimum range and secondly avoiding wheel-spin (although a small amount of tyre slip may be beneficial). &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-06-15T13:34:30.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10314" PostTypeId="1" AcceptedAnswerId="10323" CreationDate="2016-06-15T14:03:34.193" Score="2" ViewCount="85" Body="&lt;p&gt;How do I recognize a good Hydraulic Servovalve? It is probably the frequency response but what do I have to look for in these diagrams?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct to assume that the higher the frequency at a given amplitude ratio (i.e. -6dB) the better the dynamic behaviour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the phase shift?&lt;/p&gt;&#xA;" OwnerUserId="5007" LastActivityDate="2016-06-16T08:24:22.533" Title="How do I recognize a good Hydraulic Servovalve?" Tags="|hydraulics|" AnswerCount="1" CommentCount="3" ClosedDate="2016-06-18T04:17:48.163" ContentLicense="CC BY-SA 3.0" />
  <row Id="10315" PostTypeId="1" AcceptedAnswerId="10436" CreationDate="2016-06-15T14:47:56.670" Score="2" ViewCount="662" Body="&lt;p&gt;Gypsum is used to prevent immediate stiffen and reduce heat loss of cement. If gypsum has already been added into cement,do we still need to undergo curing process after that?&lt;/p&gt;&#xA;" OwnerUserId="6422" LastActivityDate="2016-07-25T14:21:34.567" Title="If gypsum has been added into clinker during the production of cement,is curing process still needed to be carried out?" Tags="|civil-engineering|construction-management|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10316" PostTypeId="2" ParentId="10217" CreationDate="2016-06-15T19:14:17.333" Score="4" Body="&lt;p&gt;It looks to me like your guess was pretty good. The required information is only the type of knurl, the extents of the knurl, and the pitch. It is conventional to show the representative pattern, but not necessarily in true scale or projection. The following images are from a technical drawing textbook (Giesecke et al.):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3V4UC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3V4UC.png&quot; alt=&quot;Technical drawings of three different knurl patterns labeled a, b, and c&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It goes on to specify that this is enough information for hand gripping purposes, but if the knurl is intended for a press-fit, a minimum diameter after knurling should be listed. This insures that there is enough upset formed by the knurling process to assure your press-fit is secure. In this situation it is obviously also important to include the toleranced diameter of the part before the knurl is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not recommend calling out the depth of the knurl. The machinist will likely have to adjust this as it is the only option they have to make sure the pattern aligns with itself on successive revolutions. At the most, I would specify a minimum depth, but not a target depth or a maximum depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a matter of style, I would recommend calling out the knurl on the face view, not the end view so it is more clear what it references. Additionally it may be hard for a machinist to knurl right up to that shoulder on the right of the knurled area. If you could hold the knurling back even 1mm that would probably help. You can probably safely assume that the machinist will know to only knurl the cylindrical faces, not the flats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the US, ANSI B94.6 has a tolerance class system to ensure that the pre- and post- knurl OD tolerances are compatible, but I haven't seen any reference to how it handles metric units like you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a little more information on knurls and how they are calculated and formed in the Machinery's Handbook machining section, but it is very inch-centric.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-06-15T19:14:17.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10318" PostTypeId="1" AcceptedAnswerId="10337" CreationDate="2016-06-15T21:52:51.107" Score="1" ViewCount="147" Body="&lt;p&gt;My translation:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Internal pipe diameter inspection&lt;/p&gt;&#xA;&lt;p&gt;This kind of inspection is carried out on a by-order basis using a certified gauge (for sizes 40 mm and above) of the go/no go type, with both sides of the pipe subjected to measurement along the length specified in the regulatory documentation. Pipes intended for use in shock-absorbers are inspected using the special instrument called a bore gauge. Instrumental inspection of internal pipe diameter is carried out for &lt;strong&gt;critical pipes&lt;/strong&gt; and relies on the use of ultrasonic equipment.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The original Russian phrase is &amp;quot;трубы ответственного назначения&amp;quot; and could be translated as &amp;quot;pipes used in critically important roles (applications)&amp;quot;: when a malfunction may lead to injuries, deaths or major industrial accidents, or to the shutdown of the whole plant, with the resulting financial losses.&lt;/p&gt;&#xA;&lt;p&gt;Is the phrase &amp;quot;critical pipes&amp;quot; a proper choice, or are there some other phrases to signal the same meaning? Would &amp;quot;critical duty pipes&amp;quot; fit?&lt;/p&gt;&#xA;&lt;p&gt;The word-for-word of the original is &amp;quot;truby (pipes) otvetstvennogo (of high responsibility) naznacheniya (purpose)&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-06-17T10:17:28.103" Title="Is &quot;critical pipe&quot; a proper term for a pipe used in a critically important role?" Tags="|terminology|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10319" PostTypeId="1" CreationDate="2016-06-15T21:58:52.010" Score="-4" ViewCount="201" Body="&lt;p&gt;What are the requirements to take the fundamentals of engineering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance could a physics major, computer science major, IT major or something thing else take the exam?  How does accreditation factor into it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also bear in mind I'm not asking about EIT or EI.  Just a seat for the exam.  I understand it will vary from state to state so whether or not the NCEES puts any restrictions on the exam would be nice to know.&lt;/p&gt;&#xA;" OwnerUserId="6849" LastActivityDate="2016-06-17T00:53:10.397" Title="Requirements for seat a on FE" Tags="|licensure|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10321" PostTypeId="1" CreationDate="2016-06-16T06:37:03.213" Score="2" ViewCount="138" Body="&lt;p&gt;Generally, how would I go about finding the maximum distributed pressure which a flat rectangular panel of material, say glass or acrylic, can withstand given thickness, length, and width?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a glass panel 10cm by 10cm by 1cm thick. How much pressure in Pa could the panel withstand before failing? Something like a door on a vacuum chamber.&lt;/p&gt;&#xA;" OwnerUserId="6854" LastActivityDate="2016-06-16T14:32:24.013" Title="failure point of flat panel under pressure" Tags="|pressure|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10322" PostTypeId="2" ParentId="10217" CreationDate="2016-06-16T06:56:06.147" Score="4" Body="&lt;p&gt;The (European) standard for knurls is DIN 82. The standard includes a proposed drawing representation as well. Its intention is to describe hand gripping aid knurls. As Ethan48 says, for fitting you need a more precise definition, at least for the outer surface (diameter, cylindricity, runout, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One remark to your design: look for knurling tools on the net, you might need some distance between the knurling and the shoulder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sZEBU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sZEBU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6855" LastActivityDate="2016-06-16T06:56:06.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10323" PostTypeId="2" ParentId="10314" CreationDate="2016-06-16T08:16:12.033" Score="2" Body="&lt;p&gt;You can consider the frequency response as response of a low pass filter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one rough example, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When cut-off freq(freq at -6dB) is 50Hz, If you want to control the valve as 100Hz speed(roughly open and close in 0.01 second), the valve is not suitable for your application, because the valve will filter your input from near 50Hz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to control the valve on 5Hz(open and close in 0.2 second), the valve will be sufficient, in terms of valve response, because 5Hz is in the bandwidth of the filter.&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="4624" LastEditDate="2016-06-16T08:24:22.533" LastActivityDate="2016-06-16T08:24:22.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10324" PostTypeId="2" ParentId="10321" CreationDate="2016-06-16T09:30:49.757" Score="2" Body="&lt;p&gt;If you wanted to solve this from first principles (difficult - especially for rectangular plates) you would usually use &lt;a href=&quot;https://en.wikipedia.org/wiki/Kirchhoff%E2%80%93Love_plate_theory&quot; rel=&quot;nofollow&quot;&gt;Kirchhoff plate theory&lt;/a&gt;, which assumes small deflections ($\delta &amp;lt; t/2$) as well as some other simplifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Results have been published for common cases. &lt;a href=&quot;https://books.google.co.uk/books?id=BnxUAAAAMAAJ&amp;amp;redir_esc=y&amp;amp;hl=en&quot; rel=&quot;nofollow&quot;&gt;Roark's formulas for stress and strain&lt;/a&gt; has an extensive compilation of plate formulas for different loading and boundary conditions. You can find these published in lots of other places too. For example on &lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html&quot; rel=&quot;nofollow&quot;&gt;Roymech&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your case of a square plate with a uniform load over the entire plate, and assuming simply supported sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum stress in the plate is:&#xA;$$ \sigma_{max} = \frac{0.2874 q}{t^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $t$ is the thickness of the plate, $q$ is the load per unit area, and $ \sigma_{max}$ is the maximum stress in the plate. If we take a very basic failure criteria of $\sigma_{max} = \sigma_{fail}$ then the applied load at failure is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ q_{max} = \frac{\sigma_{fail} t^2}{0.2874 } $$&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2016-06-16T14:32:24.013" LastActivityDate="2016-06-16T14:32:24.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10325" PostTypeId="2" ParentId="10319" CreationDate="2016-06-16T11:16:40.577" Score="0" Body="&lt;p&gt;The following section of the flowchart might answer some of the questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YIH2C.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YIH2C.jpg&quot; alt=&quot;FE Waiver Chart&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are references that you might find useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.bpelsg.ca.gov/applicants/flowchart_for_eit_waiver.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Do I qualify for the FE waiver?&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/5985/110&quot;&gt;What is the sequence of exams to become a Professional Engineer in California?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/7239/which-us-states-require-a-bachelors-in-engineering-to-become-licensed?rq=1&quot;&gt;Which US states require a bachelors in Engineering to become licensed?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.nspe.org/resources/blogs/pe-licensing-blog/alternate-pathways-pe-how-states-vary-their-engineering-education&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alternate Pathways to the PE: How States Vary in Their Engineering Education Licensure Requirements&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-06-16T11:26:41.587" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10326" PostTypeId="1" CreationDate="2016-06-16T14:33:22.923" Score="4" ViewCount="258" Body="&lt;p&gt;On of the more simple throttles is a channel, which is described by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation&quot; rel=&quot;nofollow&quot;&gt;Hagen-Poiseuille&lt;/a&gt; equation:&#xA;$$\Delta P = \frac{8\mu L Q}{\pi r^4}$$&#xA;This equation, when rearranged, tells us that the Flowrate $Q$ is linearly dependent on $\Delta P$. Do passive non-linear pneumatic devices exist, which are not linearly dependent on $\Delta P$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, is there a pneumatic equivalent to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermistor&quot; rel=&quot;nofollow&quot;&gt;Thermistor&lt;/a&gt; - which has a non-constant $R(U)$-dependency, and as such a nonlinear Flow vs Voltage curve?&lt;/p&gt;&#xA;" OwnerUserId="46" LastEditorUserId="46" LastEditDate="2016-06-16T15:12:24.897" LastActivityDate="2024-01-14T21:24:07.503" Title="Do non-linear passive pneumatic throttles exist?" Tags="|pneumatic|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10327" PostTypeId="2" ParentId="10300" CreationDate="2016-06-16T18:24:30.470" Score="0" Body="&lt;p&gt;This is general advice rather than a direct answer but given the circumstances you have described my feeling is that this sort of modification may be more trouble than it is worth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of this is that a drill press doesn't necessarily need to be all that rigid in order to work well. With sensible work holding you can get perfectly acceptable results for general drilling with even quite a basic machine and you may very well find that any gains in rigidising the spindle are lost by the overall tolerances of the machine as a whole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even quite an expensive pillar drill is not by any means the same thing as a vertical mill. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart form anytime else if you are prepared to do this level of modification you might well be better off spending your time refurbishing a small used lathe as that will probably give you more benefit in the long run for similar effort. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-06-16T18:24:30.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10328" PostTypeId="2" ParentId="1829" CreationDate="2016-06-16T18:36:24.120" Score="0" Body="&lt;p&gt;As mentioned these are 'teeth per inch' threads. they find their way onto metric thread gauges because BSP threads are still the standard in many countries for tings like gas fittings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is partly because of legacy issues where it would simply be too much hassle to change over millions of fittings on things like industrial gas bottles and the associated regulators and fittings as opposed to nuts and bolts which have a much higher turnover and in many cases will allow a straight swap for the nearest equivalent metric size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage is that discourages bodges using standard nuts and bolts for gas fittings which have specific safety requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also BSP type fittings are generally so specialist that there is little advantage in standardising to metric as they will only ever be used to mate to each other. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-06-16T18:36:24.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10330" PostTypeId="1" CreationDate="2016-06-16T21:46:51.873" Score="1" ViewCount="2744" Body="&lt;p&gt;1) Under &lt;em&gt;Visualization&lt;/em&gt; in Abaqus, contours for displacements-U1 and U2 are available. How could I generate contours for displacements in x-y direction, that is Ux and Uy? Similarly how can I get contours for stresses Sigma_x, Sigma_y and Sigma_xy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) To find homogenized elastic constants of a representative volume element of a fiber reinforced composite (that means to get homogenized E1,E2,E3,G12,G23,G13,v12,v23,v13), I need to find elastic constants of each element. I also need to get volume of each element. How can I get this two information (elastic constants and volume) for each element? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) In Abaqus, in the representative volume element of fiber reinforced composite, fiber is aligned parallel to z axis. How could I rotate the axis so that the fiber would be aligned parallel to x axis? Noteworthy that I am modeling 2D planar problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate your kind help. Thanks in advance!  &lt;/p&gt;&#xA;" OwnerUserId="6861" LastEditorUserId="7461" LastEditDate="2017-05-13T13:42:41.707" LastActivityDate="2017-05-13T13:42:41.707" Title="How to Plot Stress Contours in Abaqus?" Tags="|mechanical-engineering|abaqus|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10331" PostTypeId="2" ParentId="10319" CreationDate="2016-06-17T00:53:10.397" Score="0" Body="&lt;p&gt;The only requirement imposed by NCEES to sit the FE is that you pay the exam fee. NCEES addresses this generally on &lt;a href=&quot;http://ncees.org/exams/fe-exam/&quot; rel=&quot;nofollow&quot;&gt;their page about the FE&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h3&gt;Exam fees and requirements&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;An exam fee of $225 is payable directly to NCEES during the&#xA;  registration process. Some licensing boards may require examinees to&#xA;  file a separate application and pay an application fee as part of&#xA;  their approval process in order to be qualified for seating for an&#xA;  NCEES exam. Your licensing board may have additional requirements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And again, more strongly-worded, on the third page of &lt;a href=&quot;http://ncees.org/exams/cbt/examinee-guide/&quot; rel=&quot;nofollow&quot;&gt;the NCEES Examinee Guide&lt;/a&gt; (emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Eligibility to sit for an NCEES exam varies by state/territory and is&#xA;  determined by each licensing board. &lt;strong&gt;It is your responsibility to&#xA;  review your licensing board requirements at &lt;a href=&quot;http://ncees.org/exams/ncees-member-boards/&quot; rel=&quot;nofollow&quot;&gt;ncees.org/boards&lt;/a&gt;&#xA;  before registering for an exam.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, how do you figure out the exact requirements in your state? Let's look at an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Massachussetts&lt;/strong&gt; requires that an &lt;a href=&quot;https://www.pcshq.com/?page=MA%20Application%20Form%20FE%203.pdf&quot; rel=&quot;nofollow&quot;&gt;additional application form&lt;/a&gt; be submitted to &lt;em&gt;and processed by&lt;/em&gt; &lt;a href=&quot;http://www.mass.gov/ocabr/licensee/dpl-boards/en/&quot; rel=&quot;nofollow&quot;&gt;their state board&lt;/a&gt;. This form is titled &lt;strong&gt;BOARD APPLICATION FOR THE FE AND/OR EIT CERTIFICATION&lt;/strong&gt; and seems as though it can actually be used in the way you describe, by selecting the third option in response to question 21:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;21&quot;&gt;&#xA;  &lt;li&gt;Upon successful completion of the fundamentals exam, select the classification below that best describes your eligibility for&#xA;  Engineer-in-Training certification:&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(c) Eligibility for EIT certification insufficient due to formal&#xA;  education requirements or experience requirements; no certification&#xA;  will be issued upon successful completion of the fundamentals exam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This doesn't seem to strictly agree with the state regulations, though—specifically &lt;a href=&quot;http://www.mass.gov/courts/docs/lawlib/250-299cmr/250cmr3.pdf&quot; rel=&quot;nofollow&quot;&gt;250 CMR 3.01(7)&lt;/a&gt;, which reads:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Once the Board has reviewed the applicant's education, Work&#xA;  Experience, character and experience references, college transcripts&#xA;  and documentation, the Board may schedule the engineering applicant&#xA;  for an interview; and the Board shall schedule all land surveying&#xA;  applicants for the mandatory oral exam. Once the Board has determined&#xA;  that all applicable statutory requirements have been met, the&#xA;  applicant will be scheduled to sit for the applicable written&#xA;  examinations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This language &lt;em&gt;assumes&lt;/em&gt; that the applicant is pursuing certification; it's not written to accommodate what you might call an &quot;exam tourist&quot; (if there is such a thing). But there's also 3.10(3):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Individuals who have passed the respective FE or FS Exam prior to&#xA;  meeting the other certification requirements noted above must contact&#xA;  the Board's administrator for the proper filing requirements for&#xA;  certification.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nearly all of the explicit requirements set out in that chapter are concerning the PE/PS exams rather than the Fundamentals exams; the only explicit requirement I notice concerning the Fundamentals is that you submit the state application. Practically speaking, it's possible the body that reviews that application has internal policies and procedures that would result in your application being denied, in which case you might have to appeal directly to the reviewing body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short, just read the applicable state regulations. If something is ambiguous or hard to understand, contact the state licensing board directly. Depending on your motivation for sitting the exam without an engineering background, however, they may not be inclined to help you and you might have to just try submitting the application and see if they let you in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other end of the spectrum, &lt;strong&gt;California&lt;/strong&gt; does not require any separate application, in which case you would just register with NCEES and pay the fee. If you live in such a state, you can create an account on the NCEES website and go through all the necessary steps to register for the exam during the open registration period. You will need to agree to certain terms, that you are who you say you are, etc., and pay the fee to NCEES. You will be prompted to provide information about your education and it's important that you answer truthfully. On the day of the exam, bring a printout of your registration and a photo ID.&lt;/p&gt;&#xA;" OwnerUserId="368" LastActivityDate="2016-06-17T00:53:10.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10332" PostTypeId="2" ParentId="10300" CreationDate="2016-06-17T01:50:20.720" Score="1" Body="&lt;p&gt;In general, I would warn you that you aren't likely to get very good rigidity out of this system because by maintaining the quill mechanism and adding the length of the tool holder plus an extra joint (of any type) you are adding more cantilever to what sounds like already fairly sloppy tool. In addition, be warned that the manufacturer of the tool certainly won't endorse any project like this and any joint you implement yourself may fail, so you're going to be spinning a pretty heavy piece of metal pretty fast at your own risk. Personally, I wouldn't recommend it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, you asked about ways to couple the spline shaft with the tool holder's cylindrical end, here are some options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly you could buy an adapter. Spline shaft adapters &lt;a href=&quot;http://www.hubcityinc.com/documents/7R-HubCityFarmProductsAndAccessories.pdf&quot; rel=&quot;nofollow&quot;&gt;are made&lt;/a&gt;, but usually only for much larger shafts. In addition, I'm not sure if that spline profile will be standard or specific to your drill press manufacturer. &lt;a href=&quot;http://www.hayescouplings.com/index.php/products/misc_products&quot; rel=&quot;nofollow&quot;&gt;Other people&lt;/a&gt; will make you a custom spline adapter, which may be expensive but would get the job done. Most reasonably equipped local machine shops could also make an adapter. The problem with connecting spline-to-spline is that with the cutting forces, you'll have significant axial and radial forces to resist. Splines are good at transmitting torque, and reasonable at resisting radial force, but don't resist axial forces at all. For that reason, pick a spline adapter design where the female spline rigidly cplamps to the spline with a thru-bolt, not one of the designs that only has a set screw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another strategy would be to try to develop as much clamping force on the outside of the spline shaft as you can, completely forgoing any contact with the radial faces of the spline. To do this you could use a regular shaft coupling like &lt;a href=&quot;http://www.climaxmetal.com/Catalog_Downloads/climax/ClimaxCatalog2013-MISCC.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://shaft-collars-couplings.staffordmfg.com/viewitems/metric-rigid-shaft-couplings/rigid-shaft-couplings-one-piece-split-clamp-type-w&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. Standard rigid shaft couplings with different sized bores at each end are rare though, so you may still need to get one made custom. It is common to find couplings where one end is a finished bore and the other end is unfinished to allow you to bore it out yourself, but I haven't found any small enough for your application. If you do decide to do this, the bore needs to be cut by a machinist on a lathe, don't attempt to use the drill press for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last option, though I'm still not confident you could find parts small enough is a taper lock shaft coupling like the Fenner RM12 shown &lt;a href=&quot;http://shaft-collars-couplings.staffordmfg.com/viewitems/metric-rigid-shaft-couplings/rigid-shaft-couplings-one-piece-split-clamp-type-w&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; or the Climax C600M-9 shown &lt;a href=&quot;http://catalog.climaxmetal.com/item/keyless-rigid-couplings/metric-keyless-rigid-coupling-series-c600/c600m-9&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;. &lt;a href=&quot;http://www.climaxmetal.com/KLD-600.html&quot; rel=&quot;nofollow&quot;&gt;They&lt;/a&gt; have compatible sizes for 9 mm and 11 mm shafts, but don't seem to make it up to 16mm in one go, so you may still need one special bushing. These type of couplers work on a taper principle, much like a collet and produce probably the strongest and most rigid clamping force of any of the options here. Because these systems are modular (the inside tapered bushing works with multiple flanges) you are most likely to be able to find the sizes you need as stock parts. I have not searched exhaustively for suppliers that may stock your specific sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, I think that this is a very risky idea and may well result in failure or even injury. I suspect that to join such mismatched shafts you will need a part custom-machined, which as you know is not a cheap proposition. If you do need something custom machined, you might as well get a whole new shaft machined that has the 9mm spline profile and the tool holder itself, but by the time you pay for that, you probably could have bought a small used mill with much less hassle and risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course other options like drilling a through hole radial to the shaft and inserting a bolt or pin, or welding the two shafts together, but neither will give you nearly the rigidity or concentricity you need, and they are likely even more dangerous that the other options.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-06-17T01:50:20.720" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10333" PostTypeId="1" CreationDate="2016-06-17T03:33:26.870" Score="1" ViewCount="632" Body="&lt;p&gt;My textbook says for mode combination, the peak value $r$ of each response quantity is determined by combining the peak values $r_n$ of the modal contributions to the &lt;strong&gt;&lt;em&gt;same response quantity&lt;/em&gt;&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7QBN4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7QBN4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, this means that we have to calculate the response (be it displacement, the difference in displacement, the forces or moment) for each mode first, before we apply the modal combination method such as SQSS, CQC and ABS in order to obtain the total response. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also verified by the numerical example in the textbook. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the textbook example, all the members are 1-element members, easy to interpret the above clause in the textbook. What about members (beams, columns) that are of more than one element? In FEM we often create a single member out of multiple elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we combine the element response into member response on per mode basis first, before we combine all the member modal responses into total member response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or do we combine the total modal response for each element, and then only combining the element response into member response?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, is the modal combination done at element level or at structural member level?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="1832" LastEditDate="2016-06-17T12:15:29.813" LastActivityDate="2016-10-20T18:14:35.690" Title="How to handle modal combination in response spectrum analysis for member forces for member with more than 1 element?" Tags="|structural-analysis|seismic|modal-analysis|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10335" PostTypeId="1" AcceptedAnswerId="10345" CreationDate="2016-06-17T09:33:47.453" Score="21" ViewCount="8861" Body="&lt;p&gt;In a recent incident in London&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Network Rail said an empty train had travelled past a red signal, which resulted in an automatic derailment. No one was injured. &lt;a href=&quot;https://www.theguardian.com/uk-news/2016/jun/16/derailed-train-causes-severe-disruption-at-paddington-station&quot;&gt;[link]&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The derailment has caused quite a bit of damage, and a lot of travel disruption along this track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reading of the National Rail statement is that the derailment was a feature of the system, a response to the signal being passed at danger. While I'm sure it caused less damage than a train collision, it still seems dangerous and expensive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things like &lt;a href=&quot;https://en.wikipedia.org/wiki/Train_stop&quot;&gt;train stops&lt;/a&gt;, to trigger the brakes exist, or one could imagine diverting the train into a sand trap. Why weren't options like these used instead of derailing?&lt;/p&gt;&#xA;" OwnerUserId="3174" LastEditorUserId="3174" LastEditDate="2016-06-17T12:36:43.147" LastActivityDate="2016-10-11T14:58:49.907" Title="Why would a train automatically derail if a signal is passed at danger?" Tags="|safety|rail|" AnswerCount="7" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10336" PostTypeId="1" AcceptedAnswerId="10352" CreationDate="2016-06-17T10:09:59.127" Score="3" ViewCount="2362" Body="&lt;p&gt;My translation of &lt;a href=&quot;http://www.percey.ru/upload/pntz_tech_catalog.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a document published by the Pervouralsk New Pipe Plant&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Inspection of pipe-end root face (“face ring”)&lt;/p&gt;&#xA;&lt;p&gt;Inspection of this type is carried out as required by the regulatory documentation using a ruler or a gauge.&lt;/p&gt;&#xA;&lt;p&gt;Bevel angle inspection&lt;/p&gt;&#xA;&lt;p&gt;Inspection of this type is carried out as required by the regulatory documentation using a [&lt;strong&gt;protractor&lt;/strong&gt;? &lt;strong&gt;goniometer&lt;/strong&gt;?] or a gauge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;What tool do we usually use to inspect bevel angles at the end of a pipe?&lt;/p&gt;&#xA;&lt;p&gt;The Russian word used is &amp;quot;угломер&amp;quot;, literally &amp;quot;angle-measurer&amp;quot;. According to Multitran, it could be translated in dozens of ways.&lt;/p&gt;&#xA;&lt;p&gt;(I'm translating the word &amp;quot;truba&amp;quot; as pipe despite some saying that &amp;quot;tube/tubing&amp;quot; is a more proper word: the reason is, the plant uses &amp;quot;pipe&amp;quot; throughout its English-language website)&lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2023-08-31T18:47:19.443" Title="What tool does one use to inspect pipe bevel angles?" Tags="|terminology|piping|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10337" PostTypeId="2" ParentId="10318" CreationDate="2016-06-17T10:17:28.103" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.dictionary.com/browse/critical?s=t&quot; rel=&quot;nofollow&quot;&gt;Critical (meaning 8)&lt;/a&gt;: &lt;em&gt;of essential importance&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Critical pipes&quot; sounds fine to me, meaning &quot;the ones which are absolutely essential&quot;. It doesn't clash with any engineering meaning (which is, I assume, your reason for asking here).&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2016-06-17T10:17:28.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10338" PostTypeId="2" ParentId="10335" CreationDate="2016-06-17T12:21:44.643" Score="21" Body="&lt;blockquote&gt;While I'm sure it caused less damage than a train collision&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Exactly.  This is a last-resort response intended to avoid a full-on collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A derailed train causes some damage and is a mess to clean up.  A collision between two trains will likely cause much more damage, be even harder to clean up, and quite likely get people killed.  It can also impact the surrounding area if hazardous chemicals were being transported.  Derailing a train in a (somewhat) controlled way makes it much less likely that a tanker car, for example, breaks and spills its contents, compared to a much less predictable and more violent collision between two trains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triggering the brakes is a totally different mechanism that isn't as fail safe.  For one thing, it assumes the braking system of the train is functioning.  For another, it requires logic on the train to detect the problem.  That's not so desirable if this logic on the train failed and is what caused the problem in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derailing a train is a last resort, and as such needs to be fail-safe.  It also needs to not be triggered except when all else has already failed.  This condition needs to be independently detected, and then independently acted upon from the normal train systems.  Automatic braking might be a reasonable safety system, but derailment by external means would still be a last resort when that didn't work for whatever reason.  Basically, automatic braking and automatic derailment are two independent and different system, with the latter something you probably want regardless of what system is in place for the former.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="368" LastEditDate="2016-06-17T20:21:28.783" LastActivityDate="2016-06-17T20:21:28.783" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10339" PostTypeId="2" ParentId="10335" CreationDate="2016-06-17T14:11:03.140" Score="11" Body="&lt;h3&gt;Why derail?&lt;/h3&gt;&#xA;&lt;p&gt;If the train went past a red signal, then all of the other safety systems already failed! The method of last resort is to make absolutely sure that the train will not travel down the track.  That is done by derailing it.&lt;/p&gt;&#xA;&lt;h3&gt;How did it get to this point?&lt;/h3&gt;&#xA;&lt;p&gt;The real question is how did it get to the point of passing a red signal? At that point, something bad is already happening, so without knowing which bad thing is happening, there needs to be &lt;strong&gt;one&lt;/strong&gt; solution that can work for every situation.  That is what was used here.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-06-17T14:11:03.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10340" PostTypeId="2" ParentId="7291" CreationDate="2016-06-17T14:28:28.370" Score="1" Body="&lt;p&gt;These opposed cylinder engines have the same working of a 2-stroke engine .&#xA;4 Stroke engine is more efficient primarily because of the presence of valves which precisely control the flow of charge into the chamber and exit the exhaust gases with proper timing which is hard to achieve by ports in a 2 stroke engine. The mileage  of 4 stroke is greater and moreover running a turbocharger with extra crankshaft and gearing mechanisms will require   a high power diesel engine that reduces the effectiveness of turbocharger due to low speeds hence it further decrease efficiency . Due to two piston compression there is high compression ratio and produces lot of heat and there is very less space for implementing an effective cooling system this will reduce the thermal efficiency. Hence even though it can produce more power than a 4 stroke it has a lot of efficiency losses.  &lt;/p&gt;&#xA;" OwnerUserId="6867" LastActivityDate="2016-06-17T14:28:28.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10341" PostTypeId="1" CreationDate="2016-06-17T14:33:14.500" Score="2" ViewCount="43099" Body="&lt;p&gt;I have a (maybe stupid) question which is making me crazy: what is the difference between a saturated steam and saturated liquid state? I mean the physical difference not the difference on a Thermodynamic diagram (such as the greater entropy of the sat steam w/r/t the sat liquid).&lt;/p&gt;&#xA;" OwnerDisplayName="user6267" LastEditorUserId="1832" LastEditDate="2016-06-17T15:30:45.747" LastActivityDate="2020-07-19T04:02:39.217" Title="Difference between Saturated Liquid and Saturated Steam" Tags="|thermodynamics|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10342" PostTypeId="2" ParentId="1929" CreationDate="2016-06-17T14:50:50.273" Score="1" Body="&lt;p&gt;Glow plugs are used on small diesels, like those in cars.  For larger diesel engines, like those in locomotives, the engines use either compressed air or hydraulic starters to spin the engine over until it starts running.  You can see some of these engines being started on You Tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the engine does not start the first time, usually there is a means for someone to use a hand pump to recharge either the air bottle or the hydraulic fluid reservoir to get another start.  These are for emergencies, typically.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-06-17T14:50:50.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10344" PostTypeId="2" ParentId="10335" CreationDate="2016-06-17T15:23:09.420" Score="4" Body="&lt;p&gt;It should probably be noted that this incident occurred on a section of multiple track railway just outside a very large passenger terminus. Based on the photograph linked to above, I'd estimate the train ended up close to Westbourne Bridge, at 51.518867, -0.183969. The signalling along this track is famous for its complexity, and had lead to at least one notable rail accident with multiple fatalities, &lt;a href=&quot;https://en.wikipedia.org/wiki/Ladbroke_Grove_rail_crash&quot; rel=&quot;nofollow&quot;&gt;the Ladbroke Grove rail crash&lt;/a&gt; in 1999.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such there has obviously been a very high incentive on ensuring that the track arrangement in the area causes signals passed at danger to terminate safely and without any possibility of trains fouling other lines (the concept of flank protection), especially the main up and down lines carrying high speed trains. Evidently the track engineers decided that for this signal and for a train travelling at the speed that one was, that would result in a derailment of the train.&lt;/p&gt;&#xA;" OwnerUserId="6873" LastEditorUserId="6873" LastEditDate="2016-06-17T15:28:12.270" LastActivityDate="2016-06-17T15:28:12.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10345" PostTypeId="2" ParentId="10335" CreationDate="2016-06-17T15:34:41.847" Score="29" Body="&lt;p&gt;Firstly, the incident happened as the train was leaving a siding passing a shunt signal. These provide less authorisation than a normal signal does, even when they are not at danger (the train can permit &lt;em&gt;as far as the line is clear or the next signal&lt;/em&gt;, there is no guarantee the track ahead is clear).&lt;/p&gt;&#xA;&lt;p&gt;Now, in the UK, there are four train protection systems: AWS, TPWS, ATP, and ETCS. However, these are all primarily designed for trains on normal running lines, rather than those on sidings. I'll examine each of these in turn insofar as they protect signals:&lt;/p&gt;&#xA;&lt;h3&gt;AWS&lt;/h3&gt;&#xA;&lt;p&gt;The Automatic Warning System (AWS) is a simplistic system using magnets/electromagnets to signal an alarm to the train driver which must be acknowledged within 3–4 seconds or an emergency brake application is initiated, which per the rulebook they may not override. (It &lt;em&gt;can&lt;/em&gt; be overridden through use of an isolating switch/cock, but without someone else in the cab one would be breaking the rules by leaving their seat to override it—regulations forbid the switch from being anywhere they can reach.)&lt;/p&gt;&#xA;&lt;p&gt;Magnets are positioned 150–250m away from the signal, to give the driver a chance to view the signal before acknowledging it. In the case of a siding, it is assumed that the train will be parked there for some duration of time, and given sidings are normally roughly the same length as the trains that typically park in them, the train would not pass over a magnet if one were installed when leaving the siding. As a result, magnets are not installed for sidings.&lt;/p&gt;&#xA;&lt;h3&gt;TPWS&lt;/h3&gt;&#xA;&lt;p&gt;The Train Protection &amp;amp; Warning System (TPWS) is a comparatively modern system, dating from the 90s, which causes an emergency brake application when a train passes a signal at danger or when approaching a signal above a set speed (designed to bring any train to a stop within the &amp;quot;safe overrun distance&amp;quot;, i.e., prior to any junction on the track). It was designed to be a cheaper to rollout solution than ATP (below), while stopping the majority of accidents ATP would stop.&lt;/p&gt;&#xA;&lt;p&gt;In principle, it could be used to protect sidings, as given the typically low speed limit in sidings, one would typically be able to have only the loop at the signal, as the stopping distance would be short. It's largely not been used to protect sidings, likely because such incidents are rare, leading the cost/benefit to conclude it's not worthwhile.&lt;/p&gt;&#xA;&lt;h3&gt;ATP&lt;/h3&gt;&#xA;&lt;p&gt;Automatic Train Protection (ATP) is really a group of systems, two of which were installed in the UK as part of trials before a proposed national rollout, which ultimately never happened due to an estimated £1 billion cost. These systems are designed to prevent a train from ever passing a signal at danger.&lt;/p&gt;&#xA;&lt;p&gt;Essentially, two systems developed elsewhere were installed: the Belgian TBL1 on the Great Western Main Line, including Paddington station; and SELCAB, which was a development of the German LZB, used on the Chiltern Line. Neither TBL1 nor SELCAB (nor LZB) has ever been used to protect sidings. (LZB especially is very expensive to install, as it requires a continuous wire along the track.)&lt;/p&gt;&#xA;&lt;p&gt;However, trains operating on these lines are not required to have the systems installed (at least trains fitted with the equipment that run on the Great Western Main Line are required to be taken out of service if the system fails), and the train that derailed was not equipped with it (though obviously given the sidings not being equipped this isn't a cause).&lt;/p&gt;&#xA;&lt;h3&gt;ETCS&lt;/h3&gt;&#xA;&lt;p&gt;The European Train Control System is a system that is starting to be rolled out in the UK, currently only active on the Cambrian Line which was used as an experimental implementation. There's plenty written about this, but because it's not (yet) installed on the GWML I'm not going to discuss it here.&lt;/p&gt;&#xA;&lt;h3&gt;So… derailment?&lt;/h3&gt;&#xA;&lt;p&gt;A number of other mechanisms exist to protect the line. Traditional ones are catch points (where you have points to direct a train away from other lines that may be occupied, typically with some short section of track beyond) and derailers (designed to immediately derail a train, primarily used around places like depots where movement speeds are low).&lt;/p&gt;&#xA;&lt;p&gt;In this case, the train passed over catch points which were set to protect the mainline. While derailing a train causes disruption, had it run into a busy commuter train the outcome could've been a disaster.&lt;/p&gt;&#xA;&lt;p&gt;The other options you mentioned are train stops and diverting the train into a sand trap. Train stops are pretty much not used on mainline rail systems because the mechanical parts limit the possible speed of they can be used at limiting their use to low-speed track like sidings and depots, where incidents are relatively rare. A sand trap is essentially something you can put following catch points (and is something often used, or a sand bank) but this requires space for it, which in a congested area near to a station is unlikely to exist.&lt;/p&gt;&#xA;&lt;p&gt;Ultimately, a lot of this comes down to cost/benefit of various solutions, and the fact that catch points have frequently been installed in such locations for over a century, and do avoid crashes in the rare cases when the signal is passed at danger. If somewhere frequently had signals passed at danger (&amp;quot;frequently&amp;quot; by railway standards, that is!), I would expect another approach to be used.&lt;/p&gt;&#xA;" OwnerUserId="6870" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-06-17T15:34:41.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10346" PostTypeId="2" ParentId="10335" CreationDate="2016-06-17T15:47:48.383" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Why would a train automatically derail if a signal is passed at danger?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When the cost of derailment is lower than the cost of a train collision, and the possibility of train collision exists, then a &lt;a href=&quot;https://en.wikipedia.org/wiki/Derail&quot;&gt;train derailment device&lt;/a&gt; may be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A red signal means danger, stop.  If a train continues past that point it is likely to encounter another train or similar danger, and result in a collision or uncontrolled derailment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Controlled derailment costs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One damaged train and all its train stock&lt;/li&gt;&#xA;&lt;li&gt;Injury/fatality to those on train and those in derailment path&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Train collision costs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Two damaged trains and all their train stock&lt;/li&gt;&#xA;&lt;li&gt;Train line/tracks unavailable for use until trains cleared&lt;/li&gt;&#xA;&lt;li&gt;Track damage&lt;/li&gt;&#xA;&lt;li&gt;Injury/fatality of those one two trains and those in path of uncontrolled derailments in both directions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Things like train stops, to trigger the brakes exist, or one could imagine diverting the train into a sand trap. Why weren't options like these used instead of derailing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First, a train stop would not necessarily save the train.  Once beyond a red light, the train is in danger of collision even if it does stop, as the oncoming train may not have enough warning to also stop, particularly in adverse weather conditions and around track bends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Situations like this are very rare, and so the cost of upgrading all derailment devices to sand traps or spurs is even more than the cost of derailment.  These devices are placed as a last resort, and are only used after several other systems have failed and a collision or significant damage/death is imminent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the money to upgrade such devices to non-derailing safety catches is better spent on preventing them before they trigger than on saving the trains after they trigger.&lt;/p&gt;&#xA;" OwnerUserId="784" LastActivityDate="2016-06-17T15:47:48.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10347" PostTypeId="2" ParentId="10335" CreationDate="2016-06-17T15:50:49.960" Score="5" Body="&lt;p&gt;A derailer is used rather than a device to trip the brakes because there is no guarantee there are any brakes to trip. These things are not intended to stop trains moving under power with their air compressors pounding and their brake reservoirs at pressure. They are intended to stop loose strings of cars which have started rolling and are headed for the main line. &lt;/p&gt;&#xA;" OwnerUserId="6874" LastEditorUserId="6874" LastEditDate="2016-06-28T20:01:16.263" LastActivityDate="2016-06-28T20:01:16.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10348" PostTypeId="2" ParentId="10335" CreationDate="2016-06-17T17:28:30.563" Score="4" Body="&lt;p&gt;The short answer is because it's cheaper and there's less loss of life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at a simple example. Now, I only really know about US trains, so forgive me if some of this doesn't translate exactly correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First a train can only go where the rails are. It's not an automobile where you can pull off to one side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually there are &quot;side lines&quot; that allow trains to pass, or allow cars to sit without blocking traffic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Signaling is two fold (mostly). There are the lights next to the track that signal, essentially, that the next segment of track is clear and to proceed. There are &quot;in cab&quot; signals in the engines that signal that the next segment of the path is clear and to proceed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different areas have different coverage with signals. Some well traveled lines have the nice in-cab automated signals. Some areas have the lights by the side of track. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Signals are places far enough apart that (with speed limits) a train should be able to stop before sitting the next segment&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;====|==&amp;lt;T=|=====|=&amp;lt;T==&#xA;    G     R     G&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In that figure the R (red) signal should stop a train from entering the section of track that is occupied by a train. The G (green) signals that the next segment is clear. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;====|&amp;lt;T==&amp;lt;T|=====|====&#xA;    G      R     G&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This figure is an accident waiting to happen.  The second train WILL run into the first (as far as the system is concerned).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, there are many safety system in place to make sure that the rails stay clear and there are no accidents. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; =================================\=/==========================&#xA; ====|====T&amp;gt;=====|===========|=====X=====|=========|==&amp;lt;T=======&#xA;     R           R           R           R         R &#xA;     5           4           3           2         1    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Let's say this is a normal track segment and were looking at the signals from the train headed left. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;First, were all ready in an &quot;high alert situation&quot; because (for our example) &lt;code&gt;&amp;lt;T&lt;/code&gt; should not be on that track. It's true it could divert at the junction but (for this example) that's not normal, the trains are too close and something is already wrong.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Signals are turned all Red. Meaning STOP NOW. Auto brake systems kick in. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;=================================\=/==========================&#xA;====|====T&amp;gt;=====|===========|=====X=====|===&amp;lt;T====|===========&#xA;    R           R           R           R         R&#xA;    5           4           3           2         1&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;On crap, &lt;code&gt;T&amp;gt;&lt;/code&gt; stopped, but &lt;code&gt;&amp;lt;T&lt;/code&gt; isn't stopping. &lt;/li&gt;&#xA;&lt;li&gt;Divert &lt;code&gt;&amp;lt;T&lt;/code&gt;!!!! (note sometimes the diversion isn't possible)&lt;/li&gt;&#xA;&lt;li&gt;Evacuate &lt;code&gt;T&amp;gt;&lt;/code&gt; (note, not possible unless it's a passenger train)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;=================================\=/==========================&#xA;====|====T&amp;gt;=====|===========|===&amp;lt;TX=====|=========|===========&#xA;    R           R           R           R         R &#xA;    5           4           3           2         1&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;OMG, It didn't divert. &lt;/li&gt;&#xA;&lt;li&gt;Call emergency crews, get them to &lt;code&gt;T&amp;gt;&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;A crash WILL happen. Do everything you can to minimize damage.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;&#xA;=================================\=/==========================&#xA;====|====T&amp;gt;=====|===||==&amp;lt;T==|=====X=====|=========|===========&#xA;    R           R           R           R         R &#xA;    5           4           3           2         1&#xA;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;9&quot;&gt;&#xA;&lt;li&gt;Auto derailment at ||&lt;/li&gt;&#xA;&lt;li&gt;No lives lost, Damage in money, crappy press, but no loss of life. Clean up crew is a repair crew not a hearse.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Usually, an engine under power would not have this problem. They are basically set to stop at any serious failure. &lt;/li&gt;&#xA;&lt;li&gt;Wagons may have this problem. A train can be some loose wagons that got away from a rail yard. The brakes on wagons are like on semi-trucks it takes air pressure to disengage them, with no pressure they should engage.  So if there moving without an engine there are no brakes.&lt;/li&gt;&#xA;&lt;li&gt;Trains usually take a very long time to get up to speed, and a very long time to stop. Miles, not feet.&lt;/li&gt;&#xA;&lt;li&gt;When trains are running in the same direction they can run closer together. The trailing train just has to be going slower than or at the same speed as the leading train.&lt;/li&gt;&#xA;&lt;li&gt;There is always enough distance between trains to stop, but in a crammed line, stopping may take several segments, and can be done at the same time. Train 1 slows and Train 2 slows. This allows more trains per line.&lt;/li&gt;&#xA;&lt;li&gt;With two trains going towards each other on the same line, you are supposed to keep enough distance between the trains that both can stop if the junction is missed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6876" LastEditorUserId="6874" LastEditDate="2016-10-11T14:58:49.907" LastActivityDate="2016-10-11T14:58:49.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10349" PostTypeId="1" AcceptedAnswerId="10364" CreationDate="2016-06-17T18:53:02.733" Score="7" ViewCount="8461" Body="&lt;p&gt;Why are nuts 6-sided? Why not 4 or 8?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YA5ZH.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YA5ZH.jpg&quot; alt=&quot;standard 6-sided nut&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6880" LastActivityDate="2016-06-20T20:08:14.820" Title="Why are nuts 6-sided?" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10350" PostTypeId="2" ParentId="10349" CreationDate="2016-06-17T19:44:45.180" Score="4" Body="&lt;p&gt;The main reason to prefer hexagons over square or octagon section in general is that hex bar can be held in a  3-jaw self centering chuck which makes it convenient for machining operations. Hex bar is certainly the most common stock section after round and square/rectangular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that a hexagon has a natural draft angle so it will tend to be easier to drop forge in closed dies than square or octagonal sections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do occasionally see square nuts in some applications for example in captive nuts which sit in a cage or channel, nuts designed to be welded to plates are also sometimes square. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also older fasteners predating mass production are more likely to be square. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-20T20:08:14.820" LastActivityDate="2016-06-20T20:08:14.820" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10351" PostTypeId="2" ParentId="10336" CreationDate="2016-06-17T22:00:50.653" Score="1" Body="&lt;p&gt;Based on your translation, I think the appropriate term here is &lt;strong&gt;protractor&lt;/strong&gt;. It's directly analogous to the mention of &quot;ruler&quot; in the preceding section, being an instrument that measures continuously in standard units rather than a tool that confirms some aspect of geometry without making an explicit measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The context makes clear that the specific type of protractor would be a bevel protractor—in just the same way that &quot;gauge&quot; in the context of &quot;Bevel angle inspection&quot; refers to a bevel gauge and not, say, a gauge for gapping spark plugs. Or, since there's not really any practical difference between a bevel gauge and a bevel protractor that I'm aware, some more specialized sort of gauge designed just for pipe bevels. If I needed to routinely check pipe bevels to make sure they conformed to some set of standard offerings, I could do much worse than to use a custom set of gauges that fit on the bevel of the pipe - I'm thinking of some kind of inverted, tapered plug gauge, but I don't know that it has any more specific name. These could easily be manufactured in-house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing is that gauges are often specialized tools that &lt;em&gt;classify&lt;/em&gt; but do not directly &lt;em&gt;measure&lt;/em&gt;. &lt;a href=&quot;https://en.wikipedia.org/wiki/Wire_gauge&quot; rel=&quot;nofollow&quot;&gt;Wire gauges&lt;/a&gt; are a good example—you stick the wire into holes on the instrument until you find the smallest one that it'll fit through, and then you record the &lt;em&gt;gauge number&lt;/em&gt; marked by that hole. In the corresponding classification system, that gauge number is associated with a measurement—but the gauge doesn't actually measure continuously, it only demonstrates that the quantity you're measuring lies in between one gauge and the next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you stick a piece of wire into your &lt;a href=&quot;https://en.wikipedia.org/wiki/American_wire_gauge&quot; rel=&quot;nofollow&quot;&gt;AWG&lt;/a&gt; tool and find that the smallest hole it fits into is the #12 hole, you can call that a 12-gauge wire—but that's not the same as measuring the diameter of the wire to be 2.053 mm (the size of a #12 gauge hole). You've only really established the diameter of the gauge hole as an upper bound, plus the diameter of the next smaller hole as a lower bound, for the actual diameter of the wire.&lt;/p&gt;&#xA;" OwnerUserId="368" LastActivityDate="2016-06-17T22:00:50.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10352" PostTypeId="2" ParentId="10336" CreationDate="2016-06-17T22:28:16.157" Score="2" Body="&lt;p&gt;Based on your translation of 'root face' I'm assuming that the bevel they are inspecting is a bevel that has been prepared for a groove weld between two pipes. If that's the case, the typical measurement tool, at least in America, would be a weld inspector's gauge like &lt;a href=&quot;http://www.galgage.com/BridgeCam.pdf&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt;. There are various designs of gauge that do the job, but a Bridge Cam gauge is the most likely. It allows measurement of the bevel angle, the root face, and the alignment of the joint using different parts of the gauge. Under most codes, all of these are subject to inspection at the fit-up stage. It also allows measurement of assorted other things after the weld is complete, including undercut and convexity/overfill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part you use to measure the bevel can more generally be described as a protractor, but you wouldn't really want to tell an inspector or a welder to use a protractor. Generally a protractor refers to a larger device that is very difficult to use on thinner substrates or in tight spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the question of pipe vs tube, I think you are correct in using 'pipe' for your application. In general american english, pipe refers to a round hollow section designed for fluid (water, air, oil, etc) to flow through. Tube refers to a hollow section that could be round, rectangular or another shape, but generally isn't designed for things to flow through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do keep in mind that if the whole pipe is cut at an angle, that would be called a miter, not a bevel. A bevel specifically refers to an angled cut on the thickness of the material, in  this case the thickness of the pipe wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically in the structural fabrication world, pipe refers to a section that is manufactured in sizes corresponding to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nominal_Pipe_Size&quot; rel=&quot;nofollow&quot;&gt;Nominal Pipe Size&lt;/a&gt; standard. In this system, the pipe called 1&quot; pipe actually has a significantly larger outside and inside diameter for historic reasons. Tube would refer to a steel section made to any other round size (usually a round number of fractional inches.)&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-06-17T22:28:16.157" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10353" PostTypeId="2" ParentId="8610" CreationDate="2016-06-17T22:45:27.183" Score="7" Body="&lt;p&gt;Speaking personally (as someone who deals with this sort of thing for a US Class I on a regular basis) -- operational and maintenance personnel find railroad locations the same way you and I find highway locations -- milepost + track name, and named locations called &quot;stations&quot; in US railroad parlance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The former are not far off in usage from their highway counterparts -- albeit not always one mile apart due to changes in the track alignment.  In areas where more than one track is present, the various tracks are named as well -- this may be something like &quot;Track Two&quot; for a mainline track in a multiple main territory or something more obscure like &quot;50&quot; or &quot;Piglet Lead&quot; for tracks in yards and industrial areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latter may be akin to a named/signed highway junction, or simply a named point along the tracks marked by a sign.  These are more commonly used by higher-level people, such as folks in Customer Service, as they don't care as much about what precisely the train is doing at a given station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Communication Based Train Control (CBTC) systems may either use GPS lat/long and then convert it to a track name and milepost based on an internal database of the railroad's layout (this is what Positive Train Control or PTC does), or use beacons set in the track that act as &quot;electronic mileposts&quot; (the European Rail Traffic Management System/ERTMS approach, at least at Level 1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, along with this all, you need to know what route (subdivision) you are on to begin with -- that's akin to the highway number on a road.  It doesn't do you any good at all to know you're at milepost 100.000000 on the main track if you don't have the foggiest clue what subdivision you're in!&lt;/p&gt;&#xA;" OwnerUserId="964" LastEditorUserId="964" LastEditDate="2016-06-30T23:57:29.247" LastActivityDate="2016-06-30T23:57:29.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10354" PostTypeId="2" ParentId="10349" CreationDate="2016-06-18T02:17:43.877" Score="7" Body="&lt;p&gt;Six sides is a good compromise between being able to handle high torque and being able to get tools onto and off of it easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4-sided nuts can handle more torque, but are a pain to work with, especially in confined spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuts with more than 6 sides are too easy to &quot;round off&quot; when a tool slips (perhaps because of poor manufacturing tolerances), rendering them useless.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-06-18T02:17:43.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10355" PostTypeId="1" CreationDate="2016-06-18T03:44:03.527" Score="2" ViewCount="6423" Body="&lt;p&gt;I'm looking to build myself a cheap water fountain that filters out gunk for my cat. Here is how it will work (I think)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MtPJi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MtPJi.png&quot; alt=&quot;Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway just wondering if the pump &lt;em&gt;has to be submerged&lt;/em&gt; in water in order for it work properly. I was hoping not because the bowl is fairly small and doesn't need to be huge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So do submersible water pumps need to be submersed in water to function properly?&lt;/p&gt;&#xA;" OwnerUserId="6887" LastActivityDate="2017-01-21T20:36:08.667" Title="Do submersible water pumps need to be submersed in water to function properly?" Tags="|pumps|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10356" PostTypeId="2" ParentId="10341" CreationDate="2016-06-18T05:59:42.893" Score="2" Body="&lt;p&gt;The difference is just that:  one is in the liquid phase and the other is in the vapor phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A material like water has a certain temperature at which it boils.  This temperature changes with ambient pressure, such that a low pressure corresponds to a low boiling point temperature and a high pressure increases the boiling point temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a saturation condition is reached, the liquid phase and the vapor phase are in equilibrium with one another, i.e., both phases can exist simultaneously.  If a small amount of energy is added to saturated liquid, it turns into vapor at constant temperature.  Likewise, if a small amount of energy is removed from saturated vapor, it will condense to liquid at const. temp.&lt;/p&gt;&#xA;" OwnerUserId="5216" LastActivityDate="2016-06-18T05:59:42.893" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10358" PostTypeId="1" CreationDate="2016-06-18T07:19:26.243" Score="1" ViewCount="147" Body="&lt;p&gt;Does electromagnetic shielding using conductor metal sheet enclosure attenuates magnetic field of the electromagnetic wave as well? If yes, what is the relation? I know it does attenuate electric field of the electromagnetic wave.&lt;/p&gt;&#xA;" OwnerUserId="6614" LastEditorUserId="6614" LastEditDate="2016-06-22T02:16:04.907" LastActivityDate="2016-06-22T02:16:04.907" Title="Does electromagnetic shielding using conductor metal sheet enclosure attenuates magnetic field as well?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10359" PostTypeId="1" CreationDate="2016-06-18T08:04:44.890" Score="2" ViewCount="284" Body="&lt;p&gt;Historically, the switches on regular railways were switched manually using a lever by dedicated staff to route trains correctly before getting replaced by an automatic electronic system. A similar system is in place for trams in modern times, but how was it done before this was possible, with horse-drawn trams and at the beginning of electrification? It would obviously not be possible to have separate staff be responsible for switching the switches like on railways in cities as they are often located in the middle of roads, in addition to being very numerous in more complex systems. &lt;strong&gt;How did tram switches operate before being equipped with an automatic electronic system?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6892" LastActivityDate="2016-06-20T11:49:28.527" Title="Historically, how did tram switches work?" Tags="|mechanical-engineering|engineering-history|rail|transportation|traffic-intersections|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10361" PostTypeId="2" ParentId="10359" CreationDate="2016-06-18T14:33:42.757" Score="1" Body="&lt;p&gt;I don't see why the simplest explanation isn't the best, i.e. &lt;strong&gt;manual switches&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Freight trains used to have people on board whose entire job was to manually walk and throw switches. This would work just the same for trams. Also, trams don't usually travel at high speeds, to it wouldn't slow things down much at all to have someone walk to manually throw the switch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't state why you think that manual labor was not possible. Especially in older times, labor was the cheapest part of any process.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2016-06-18T14:33:42.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10362" PostTypeId="2" ParentId="10277" CreationDate="2016-06-18T15:23:39.723" Score="0" Body="&lt;p&gt;The order of the dark fringe of the coherent source to lock to is the first order because the self coherence function's first minimum corresponding to the sinc function's  envelope is located at an interval of lambda\4 multiplied by the distance d where lambda equals wavelength and d is equal to the distance between a fixed mirror's image and an moving mirror's image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information of the greatest interest is the spectrum in one branch of the interferometer called the interferogram's inverse Fourier transform. As a result,&#xA;the interferometer's signal as a function of the optical path difference and the spectrum as a function of wavenumber, also known as 1/wavelength,form a Fourier transform pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to explore this topic, please read chapter 6 , Fourier Trahsform Spectroscopy , in the book written by the Finnish professors, J. Kauppinen and J. Partanen , titled Fourier Transforms in Spectroscopy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I enjoyed chatting with the New York genius expert, @Chris Mueller , about this topic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="2651" LastActivityDate="2016-06-18T15:23:39.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10363" PostTypeId="1" AcceptedAnswerId="10369" CreationDate="2016-06-18T16:20:02.683" Score="-2" ViewCount="127" Body="&lt;p&gt;Let's say that the engine is spinning at 7000 rpm and the transmission is spinning at 2000 rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you calculate the rpm at which the two will meet when the clutch pedal is released and how much rpm will be transferred each second?&lt;/p&gt;&#xA;" OwnerUserId="6897" LastActivityDate="2016-06-18T21:19:56.930" Title="Car clutch rpm transfer" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2016-06-19T04:56:10.127" ContentLicense="CC BY-SA 3.0" />
  <row Id="10364" PostTypeId="2" ParentId="10349" CreationDate="2016-06-18T16:43:32.510" Score="14" Body="&lt;p&gt;There are 3, 4, 8, and 12-sided nuts, and most exotically, 5-sided nuts all in use for specific applications, but the 6-sided hex nut is most popular because it offers a good trade off between a bunch of factors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ease of tightening/loosening, especially in tight spaces:&lt;/strong&gt;&#xA;There are six distinct orientations you can grip a hex nut in, each 60 degrees apart. This means that if you can swing your wrench through a 60 degree arc, you can tighten the nut enough to remove the wrench and place it on the next pair of flats. In fact most open end wrenches have the jaws offset 15 degrees form the handle so that  by flipping the wrench you can tighten the nut with as little as 30 degrees of access. (Box wrenches often have 12 points, allowing the same principle without the flipping.) A four sided nut would have a significant drawback that even with an offset wrench head, you would need 45 degrees of access to tighten the nut. Additionally, it's much easier to put an open end wrench on a hex nut as the two adjacent faces on the approach side guide the wrench towards the flats. By contrast, if you approach a square nut with the wrench flats not quite parallel to the nut flats, the wrench can very easily jam. Of course when there is space for access with a ratchet these considerations don't make much difference, but as people try to make machines smaller and lighter, designers often leave only the minimum access required to reach fasteners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Torque transmitted&lt;/strong&gt;&#xA;The more points the nut has, the more likely it is that the bearing surface of the point will fail in compression and 'round over.' This problem is exacerbated by the fact that some gap has to be left between the wrench and the nut to allow for manufacturing and alignment tolerances, which significantly decreases the contract area between the wrench and the nut. In practice, it's pretty hard to round over a hex nut unless you are using an adjustable wrench (where slop in the mechanism can lead to a bigger gap between the nut and the wrench and less parallel wrench faces.) A square  nut is even harder to round over, and in this aspect is more desirable than a hex nut. 12-point nuts do exist, but if the faces were flat they would round over very easily, so instead they are manufactured with points that increase the size of the torque-transmitting face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Joint Properties&lt;/strong&gt;&#xA;Standard square nuts have a larger bearing area than their standard hex counterparts. This sometimes makes them preferable for a connection to a soft substrate like wood as they are less likely to pull through the material. Because square nuts can accommodate a greater gap between the wrench and the nut, they can be used in captive situations like weld-on cage nuts, or self-aligning server rack nuts. One common use for square nuts is on blind, low strength connections to soft wood. The bolt can be tightened from one side and the nut will embed itself in the wood, with the large faces preventing it from cutting a counter-bore into the wood. By contrast a hex nut would typically cut a circle in the wood and spin. Oversized hex nuts, thick washers, and tension control bolts all allow alternate solutions to these problems using hex nuts, but generally at a higher manufacturing cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Material Efficiency&lt;/strong&gt;&#xA;In order to maintain the minimum amount of material between the threads and the edge of the nut, square nuts have a fair amount of 'wasted' volume in the corners that increases the amount of metal per nut for a given screw size. As you increase the number of flats, the shape of the nut becomes closer to a circle and more material efficient. (By the same logic, when laying out a connection hex nuts take up less space whereas square nuts have to be placed further apart.) For a few nuts, this difference is negligible, but for a company that buys or produces nuts by the truckload, it would add up very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ubiquity&lt;/strong&gt;&#xA;Even if these trade-offs changed, the simple fact that most people who work on machinery and structures have tools designed around 6-sided nuts would provide a big dis-incentive to change. Auto mechanics, for example have to buy a whole special set of tools when a car manufacturer decides to use an unusual fastener. In the case of the 5-sided nut, the very reason it is used is to be tamper resistant because very few people own wrenches that fit it, and adjustable tools designed for even-numbered polygons won't work. It is mainly used for fire hydrant fittings and valves.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-06-18T16:43:32.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10366" PostTypeId="1" CreationDate="2016-06-18T18:34:24.157" Score="3" ViewCount="85" Body="&lt;p&gt;While writing a report on data broadcasting I came across &lt;a href=&quot;http://roscansat.com/files/data/RXQ2-433.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; paper where I found description on broadcast operation and one-to-many as shown below&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;broadcast operation; where a single master address many RXQ2 modules&#xA;  concurrently. (using many RXQ2 modules set to the same address)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;one-to-many; A network consisting a master and many slaves (the&#xA;  receivers all have the same address)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is what is the difference between these two configurations of data transition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;em&gt;broadcast operation&lt;/em&gt; seem to be transmitting data synchronously and as for the &lt;em&gt;one-to-many&lt;/em&gt; there was nothing to it. I could not find any information on this online as well.&lt;/p&gt;&#xA;" OwnerUserId="6899" LastActivityDate="2016-07-23T12:16:12.517" Title="Difference between broadcasting operation and one-to-many in RF transceiver" Tags="|data|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10367" PostTypeId="2" ParentId="10355" CreationDate="2016-06-18T20:19:56.803" Score="0" Body="&lt;p&gt;That's not the problem here.  You would need to put the filter ahead of the pump, as is done in any aquarium setup, or the pump itself will clog.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said,  you absolutely should be changing the cat's water at least once a day, so worrying about minor dust or gunk is irrelevant.  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-06-18T20:19:56.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10368" PostTypeId="1" AcceptedAnswerId="10376" CreationDate="2016-06-18T20:58:48.550" Score="0" ViewCount="431" Body="&lt;p&gt;I'm working on building a 3D Zoetrope like &lt;a href=&quot;https://www.youtube.com/watch?v=RjSxrVXsfVM&quot; rel=&quot;nofollow&quot;&gt;the Disney Zoetrope in this video&lt;/a&gt;. The part that has me stumped is the bearing. I bought a lazy susan bearing but it's a bit loud, doesn't have any mount options, and doesn't spin as freely as I'd like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also see turntable bearings that seem like essentially the same thing but with better mounting options. My worry with those is their size. I have a 4 foot platter and most bearings are quite small, which is why I went with the lazy susan bearing at 24&quot; diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thrust bearings have also come up but the ones I've seen don't have any mount options. They seem constructed exactly as I need them with the bearings being sandwiched vertically (top to bottom), instead of horizontally between an outer and inner ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also been reading that certain ball bearings have a a radial and/or angular thrust capacity. And that this may provide the rotation I need, though again the mounting options seem really limited for a turning table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been googling for days and seem to have a mental block. I don't know how I would stabilize a 4 foot platter if the bearing is small and centered, like the turntable bearings I found. So, what kind of bearing would be best for this? How would I mount it (if its not super obvious)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of using a pulley to drive the zoetrope, though I'd prefer a direct connection with the motor drive shaft in the middle of the platter.&lt;/p&gt;&#xA;" OwnerUserId="6902" LastEditorUserId="368" LastEditDate="2016-06-20T21:52:08.580" LastActivityDate="2016-10-15T21:03:50.803" Title="Quiet Turntable Bearing for 3D Zoetrope" Tags="|mechanical-engineering|bearings|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10369" PostTypeId="2" ParentId="10363" CreationDate="2016-06-18T21:19:56.930" Score="1" Body="&lt;p&gt;You need a speed-vs-torque curve for the engine, which will generally have a negative slope. Then you need a speed-vs-torque curve for the transmission and whatever load is connected to it, which will generally have a positive slope. Plot both curves on the same graph, and the point where they cross is the RPM you'll have when the clutch is engaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If either of the curves is not fixed &amp;mdash; e.g., throttle changes on the engine, or tire slippage on the transmission, the problem becomes much more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to determine how the RPMs of the two parts change &lt;em&gt;as the clutch engages&lt;/em&gt;, you'll need information about the moments of inertia of the various parts, along with data about how the clutch pressure is changing with time.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-06-18T21:19:56.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10370" PostTypeId="1" CreationDate="2016-06-18T23:23:29.783" Score="1" ViewCount="140" Body="&lt;p&gt;What are the various lumped element dynamic system models called?  By lumped element dynamic system models, I mean the various nodal models comprised of things that look like capacitors, inductors, resistors, etc.  In other words, the models where these &lt;a href=&quot;https://en.wikipedia.org/wiki/Analogical_models#Dynamical_analogies&quot; rel=&quot;nofollow&quot;&gt;analogies occur&lt;/a&gt;. For example, for electrical systems, we just call them circuits.  In heat mass transfer, I sometimes see them called thermal nodal models or thermal circuits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I'm trying to study &lt;a href=&quot;https://en.wikipedia.org/wiki/Bond_graph&quot; rel=&quot;nofollow&quot;&gt;bond graphs&lt;/a&gt;, which talks about these analogies, but has its own notation.  Generally, in the bond graph literature, I see various diagrams and how they relate to bond graphs, but I'm not familiar with the various diagrams outside of electrical circuits.  For example, I see similar diagrams for mechanical systems with dampers.  I see diagrams for hydraulic components.  I'm pretty sure that there's even some kind of diagram that uses chemical potential for for current and molar flow for voltage.  The problem is that I'm not familiar with these diagrams and notation.  I'd look them up, but I'm not sure what they're called and I'm having a difficult time determining the common use names for the lumped element models in various domains.&lt;/p&gt;&#xA;" OwnerUserId="6904" LastActivityDate="2016-06-18T23:23:29.783" Title="What are the various lumped element dynamic system models called?" Tags="|mechanical-engineering|electrical-engineering|modeling|dynamics|circuits|" AnswerCount="0" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10371" PostTypeId="1" AcceptedAnswerId="10372" CreationDate="2016-06-18T23:49:12.970" Score="0" ViewCount="1178" Body="&lt;p&gt;What are some basic, open source aerodynamics or CFD simulation software that can import Blender or Sketchup models? I am looking for software that is easy-to-use, free, and suitable for educational purposes.&lt;/p&gt;&#xA;" OwnerUserId="6085" LastActivityDate="2016-06-19T02:38:46.480" Title="Simple, open source aerodynamics simulation software that can import Blender or Sketchup models?" Tags="|fluid-mechanics|aerospace-engineering|simulation|computer-aided-design|aerodynamics|" AnswerCount="1" CommentCount="3" ClosedDate="2016-06-19T04:53:18.987" ContentLicense="CC BY-SA 3.0" />
  <row Id="10372" PostTypeId="2" ParentId="10371" CreationDate="2016-06-19T02:38:46.480" Score="1" Body="&lt;p&gt;CFD requires a mesh of the model.  To my knowledge, blender and sketchup do not do this.  You may need an intermediary mesh generator to realize any cfd simulations, freeware or otherwise.  Beware that mesh generators produce a plethora of different file formats which may or may not be readable in your cfd simulator of choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for a good open source cfd simulator, i'd recommend OpenFOAM.  Beware: it is neither simple to understand or use and i don't recommend it unless you really have a good grasp of fluid dynamics snd numerical discretization and solver methods&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2016-06-19T02:38:46.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10373" PostTypeId="2" ParentId="7428" CreationDate="2016-06-19T03:10:16.970" Score="1" Body="&lt;p&gt;It is possible to export the Pro|E/Creo model to a external FEA application and do it there. At the same time it makes me wonder a bit, if your load position stays the same and your assuming the material is deforming linearly then doubling load is just going to double the deflections. Seems a bit pointless to recalculate given that this linearity and small deflections are assumed by your FE solver. See the results you get just assume the shape is a linear spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scripting with VBA is certainly possible if you have the VB API installed. The VB API is disabled by default so you need to go and activate it first. This is explained in length in the VB API folder under your install (it is in my experience not installed by default nearly anywhere). It is good to know that VB API is actually a COM bridge so it works fine in any COM capable language like C# ane python.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However to my knowledge there are no API calls for the wizard so you would need to go trough the screenscraping route.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2016-06-19T07:28:49.560" LastActivityDate="2016-06-19T07:28:49.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10376" PostTypeId="2" ParentId="10368" CreationDate="2016-06-19T10:01:16.330" Score="2" Body="&lt;p&gt;I would suggest getting an unbraked hub assembly intended for a lightweight trailer. You can get the stub axle, bearings and hub as a set and it gives you a convenient flange mounting for your main platter. As long as the platter itself is reasonably stiff and well balanced it shouldn't need any additional support. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recently used a similar arrangement for a rotating plinth for a trio of bronze and glass statues with a total rotating mass of around 150 kg with a diameter of 1.0 metres. &lt;a href=&quot;https://i.stack.imgur.com/GQXYX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GQXYX.jpg&quot; alt=&quot;turntable &quot;&gt;&lt;/a&gt; In this case I just cut the stub axle to the required length and welded it to a 6mm thick steel plate which was bolted into a corresponding plate in the framework of the base, this allowed it to be squared with shims. Note that the castors visible are for moving the whole thing and not related to the rotating mechanism. I was able to stand on the end  of one of the rotating arms without any problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the drive you could have a pneumatic or rubber wheel directly driving the underside of the platter. This also gives you to opportunity to fine tune the gear ratio according to where on the radius you put the drive wheel. This is probably the easiest drive solution to implement. You would probably want the motor and drive wheel on a sprung or counter-weighted pivot a bit like the belt drive on a lathe. Here a motor with an integrated gearbox would be make life easier as you won't need to provide a large reduction ratio. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you could fix a pulley to the hub itself (note that the axle is fixed and it is the hub which rotates). &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-15T21:03:50.803" LastActivityDate="2016-10-15T21:03:50.803" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10377" PostTypeId="1" AcceptedAnswerId="10385" CreationDate="2016-06-19T11:38:42.577" Score="0" ViewCount="162" Body="&lt;p&gt;On the pictures below is a tool called &quot;bridge cam gauge&quot;, mentioned &lt;a href=&quot;https://engineering.stackexchange.com/a/10352/349&quot;&gt;in this answer&lt;/a&gt;. I wonder why the words &quot;bridge&quot; and &quot;cam&quot; are used in its name. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;a href=&quot;https://en.wiktionary.org/wiki/cam#Noun&quot; rel=&quot;nofollow noreferrer&quot;&gt;found a definition&lt;/a&gt; of &quot;cam&quot; that might be relevant:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A curved wedge, movable about an axis, used for forcing or clamping two pieces together.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As I understand, the 'cam' here is the flat roundish beaky plate that you can rotate about its hinge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about &quot;bridge&quot;? &lt;a href=&quot;https://en.wiktionary.org/wiki/bridge#Noun&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiktionary offers a variety of senses&lt;/a&gt;, which makes it hard for a non-native speaker of English. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UMgy9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UMgy9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PjDAG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PjDAG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://english.stackexchange.com/q/333286/48335&quot;&gt;Also asked on ELU SE&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. It is also called &quot;cam type weld gauge&quot;&lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-06-19T19:13:35.687" Title="What is the etymology of &quot;bridge cam gauge&quot;?" Tags="|terminology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10379" PostTypeId="1" AcceptedAnswerId="10382" CreationDate="2016-06-19T13:42:21.757" Score="3" ViewCount="30045" Body="&lt;p&gt;Why do we design foundations at SLS and not ULS?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use ULS wouldn't we be taking a higher factor of safety?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;SLS = Serviceability limit state. It considers how the structure is actually expected to behave, generally by using partial factors of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ULS = Ultimate limit state. It considers complete structural failure, utilising partial factors to increase loads and decrease strength.&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="255" LastEditDate="2016-06-20T08:54:20.403" LastActivityDate="2019-04-18T14:20:09.940" Title="why are foundations designed at SLS" Tags="|civil-engineering|foundations|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10380" PostTypeId="2" ParentId="10358" CreationDate="2016-06-19T16:32:34.320" Score="3" Body="&lt;p&gt;A (electric) conductive sheet does not shield the magnetic field at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use an iron/steel sheet, which has a high magnetic permeability then it would reduce the magnetic field (which prefers to stay inside the iron/steel rather than go back outside again ;-) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get the situation where the magnetic field is created by the electric current, and by careful arrangement of the conductors you keep it's field totally inside the outer one. Coaxial cable being one case in point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are getting interference from a magnetic field it is normal to use a high permeativity shield (iron , steel, Mu metal) or to add a ferrite core around the electric cable (the go/return pair) so that there is minimal induction into the cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In what context do you need to know (e.g. 'homework', electronic noise rejection, large magnetic field rejection, etc.)?&lt;/p&gt;&#xA;" OwnerUserId="6868" LastActivityDate="2016-06-19T16:32:34.320" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10381" PostTypeId="2" ParentId="6096" CreationDate="2016-06-19T16:44:41.367" Score="0" Body="&lt;p&gt;At that frequency you are likely to also be suffering from a lot of parasitic components, such as capacitance between different conductor, inductance along wires, mutual inductance between wires and the resistance of the wires, chassi and shield them selves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will depend on your application what is the best technique (including a good understanding of what's going on!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How have you decided you have a magnetic field that is both at 125 kHz, and that it needs shielding (alone)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming it's part of a larger circuit, have a look at the work of Henry Ott, and of Bill Whitlock (Jensen Transformers) - they have a lott of good stuff about shielding (and theory applies at all frequencies ;-).&lt;/p&gt;&#xA;" OwnerUserId="6868" LastActivityDate="2016-06-19T16:44:41.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10382" PostTypeId="2" ParentId="10379" CreationDate="2016-06-19T16:57:10.860" Score="6" Body="&lt;p&gt;No. In fact, SLS frequently leads to a higher factor of safety than ULS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because ULS deals with whether or not the structure will collapse. It does not care whether the structure will stand by generating absurd deformations, massive cracking (in the case of reinforced concrete structures), or if it will vibrate like a guitar string whenever you take a step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SLS, on the other hand, does care about such things. It makes it so that your structure must not only stand upright, it must also not have excessive deformations, cracks or vibrations. This frequently means that a larger cross-section or more steel reinforcement is necessary to satisfy SLS than would be necessary according to ULS. This therefore leads to a larger safety factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, this isn't always the case. After all, ULS uses a reduced material strength and increased loads, while SLS uses the characteristic values for everything. Though SLS's underlying assumptions are quite conservative, in many cases ULS is still the controlling factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Foundations, however, deal with soil. The main concern here tends to be how the soil will behave under load. Once fully loaded, the structure will sink into the soil due to the applied compression. That is an unavoidable fact. The question is whether your structure will sink by one meter or one millimeter. That is the sort of question that is best answered by SLS. After all, so long as the structure is still standing, sinking by one meter is fine according to ULS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it's not as if foundations have no factor of safety. In fact, theirs is usually the highest safety factor in the entire structure. One usually defines that the soil's safety factor (ultimate strength divided by applied stress) must be greater than 2, at the very least. Concrete structures (other than their foundations) usually have a total safety factor of around 2, and steel structures of around 1.7.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="255" LastEditDate="2016-06-20T08:55:56.223" LastActivityDate="2016-06-20T08:55:56.223" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10383" PostTypeId="1" AcceptedAnswerId="10387" CreationDate="2016-06-19T17:28:00.140" Score="0" ViewCount="1738" Body="&lt;p&gt;I have a heating system with which I heat up 10 liters of water by 24.2°C in 610&amp;nbsp;seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My system consumed 0.5&amp;nbsp;KW of electric energy in this time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I used this formula:&#xA;Specific heat of water * volume * temperature difference / time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I get 1,66 KW (created)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;efficiency 332%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? Or must I use some other formula to calculate efficiency?&lt;/p&gt;&#xA;" OwnerUserId="6914" LastEditorUserId="1832" LastEditDate="2016-06-19T17:51:35.583" LastActivityDate="2016-06-20T00:57:48.650" Title="Calculate efficiency of water heating" Tags="|heat-transfer|energy|energy-efficiency|heating-systems|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10385" PostTypeId="2" ParentId="10377" CreationDate="2016-06-19T18:37:27.650" Score="4" Body="&lt;p&gt;See here &lt;a href=&quot;http://www.newmantools.com/gauge/wghowto.htm#wg4&quot; rel=&quot;nofollow&quot;&gt;http://www.newmantools.com/gauge/wghowto.htm#wg4&lt;/a&gt; (gauge type WG-4) for how it is used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works the same way as a cam. The rotating part (marked &quot;undercut or reinforcement&quot; in your second picture) has a pointed end (marked with the arrow) that &quot;follows&quot; the profile of the parts to be welded like the follower on a conventional cam, as in the graphic in your Wikipedia link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the &quot;bridge&quot; part just means that the gauge has two &quot;feet&quot; that are in contact with the part being measured, and &quot;bridges the gap&quot; between them. See &lt;a href=&quot;http://marinenotes.blogspot.co.uk/2012/08/sketch-and-describe-bridge-gauge-how-is.html&quot; rel=&quot;nofollow&quot;&gt;http://marinenotes.blogspot.co.uk/2012/08/sketch-and-describe-bridge-gauge-how-is.html&lt;/a&gt; for a different type of &quot;bridge gauge&quot;, which doesn't have a cam.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-06-19T18:37:27.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10386" PostTypeId="2" ParentId="10383" CreationDate="2016-06-19T19:39:43.090" Score="0" Body="&lt;p&gt;The formula's you use seem correct. The value for efficiency however is impossible as you can never reach an efficiency of 100% let alone over 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Efficiency = ( Usefull energy / Total energy ) * 100% or ( Usefull power / total power) * 100%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy = Cp (assume constant) * M * (Tf - Ti) in which M is equal to V * rho&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power = Energy / Time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, either the time it took to heat up is invalid or the system you use has a lot more power than it says it does.&lt;/p&gt;&#xA;" OwnerUserId="6918" LastEditorUserId="6918" LastEditDate="2016-06-19T19:52:19.580" LastActivityDate="2016-06-19T19:52:19.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10387" PostTypeId="2" ParentId="10383" CreationDate="2016-06-20T00:57:48.650" Score="2" Body="&lt;p&gt;You are confusing units, which is making your question difficult to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrical energy is measured in kW-&lt;em&gt;hours&lt;/em&gt; (or Joules), not kW. If your heater consumed 500 W of &lt;em&gt;power&lt;/em&gt; for 610 s, then you used 305 kJ of &lt;em&gt;energy&lt;/em&gt; to heat your water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, if 10 l (10 kg) of water was raised by 24.2 &amp;deg;C, then it absorbed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4.186 \frac{J}{g ^\circ C} \cdot 10 kg \cdot 24.2 ^\circ C = 1013 kJ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;of &lt;em&gt;energy&lt;/em&gt;. Clearly, something is wrong with your description of what actually happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the heater actually used 0.5 kWh of &lt;em&gt;energy&lt;/em&gt; (i.e., 2.95 kW for 610 seconds), this corresponds to 1800 kJ, which would indicate that the process is only about 1013/1800 = 56.3% efficient. You lost quite a lot of the incoming energy in the form of heat that ended up somewhere other than in the water.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-06-20T00:57:48.650" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10394" PostTypeId="2" ParentId="10359" CreationDate="2016-06-20T11:40:21.107" Score="1" Body="&lt;p&gt;I think you don't understand how the switch can work if there is no place to put the lever in the middle of a street. The answer is simple: you only put the socket and every driver brings his own lever. This system is still used today when the electric drive malfunctions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=3LzGIyB50TU&amp;amp;t=3m40s&quot; rel=&quot;nofollow&quot;&gt;Here's a video &lt;/a&gt;where it can be seen how it's done.&#xA;We can see the driver putting the lever in dedicated socket. &lt;/p&gt;&#xA;" OwnerUserId="1931" LastEditorUserId="1931" LastEditDate="2016-06-20T11:49:28.527" LastActivityDate="2016-06-20T11:49:28.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10395" PostTypeId="2" ParentId="10326" CreationDate="2016-06-20T15:59:25.327" Score="0" Body="&lt;p&gt;So you want a valve that, once set for a particular flow rate, gives the same flow rate for a variety of differential pressures? If so, I think the only thing you're going to get to do that is a &lt;a href=&quot;http://asahi-america.com/valves-and-actuation/manual-valves/constant-flow-valves&quot; rel=&quot;nofollow&quot;&gt;constant flow valve&lt;/a&gt;. They're analog, but not exactly passive. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-06-20T15:59:25.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10396" PostTypeId="1" CreationDate="2016-06-20T18:34:34.840" Score="0" ViewCount="638" Body="&lt;p&gt;I'm want to make a 2 meters wide sphere that should be circular enough as a giant hamster ball! I could make it of some sort of strong plastic or steel. It should have a hatch on top that can be opened or closed so that anyone can enter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: what I'm asking for is the topology of the sphere and how can I build it in a way that makes it completely rounded when assembled. I'm asking about the parts and assembling it&lt;/p&gt;&#xA;" OwnerUserId="6889" LastEditorUserId="6889" LastEditDate="2016-06-20T23:04:21.790" LastActivityDate="2016-06-20T23:04:21.790" Title="How do I fabricate a sphere?" Tags="|steel|structures|" AnswerCount="2" CommentCount="1" ClosedDate="2016-06-20T21:30:36.987" ContentLicense="CC BY-SA 3.0" />
  <row Id="10397" PostTypeId="2" ParentId="10396" CreationDate="2016-06-20T19:40:17.623" Score="1" Body="&lt;p&gt;If you want a hamster ball I would say that a good approach would be to construct it from a series of rings in steel tube, rather like the lines of latitude and longitude on a globe with the gaps filled in with mesh. IIRC correctly a prop like this was used on  the &lt;em&gt;Gladiator&lt;/em&gt; gameshow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fairly tricky fabrication to pull off as there are a lot of tube joints to be made and you will almost certainly need to construct a reasonably sophisticated jig to keep everything concentric and you need to consider carefully how you will handle the 'poles' where you will have a lot of tubes meeting at one point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large diameter circles of steel tube can be made using a &lt;a href=&quot;http://www.baileighindustrial.co.uk/metalworking/roll-benders&quot; rel=&quot;nofollow&quot;&gt;roll bender&lt;/a&gt;  without too much difficulty although making a dozen of so with adequately consistent radius for this sort of application is not trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously any deviations from spherical will make it more difficult to handle as it will tend to pull in one direction or just move erratically. It also goes without saying that there are some fairly significant safety considerations which would need to be addressed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other alternative would be to mould it in some sort of composite material like fibreglass. Here the challenge is to produce a 2 m diameter spherical mould. The most sensible approach would be to make two hemispheres and bond them together. You may be able to find something off the shelf which would serve as a mould but I can't imagine what off hand. Failing that you would probably need to construct a plywood hemispherical framework,  skin it with plaster and then &lt;a href=&quot;http://3.bp.blogspot.com/-gDYFqSQIUAM/T1IbFFU1ESI/AAAAAAAAAWo/9LhJKNl6LQc/s1600/13.jpg&quot; rel=&quot;nofollow&quot;&gt;turn&lt;/a&gt; it to an accurate hemisphere using a former mounted on a spindle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fibreglass structure of this size would also need to be reinforced with internal ribs or foam between two layers of laminate to be adequately stiff without being too heavy. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-06-20T19:40:17.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10399" PostTypeId="2" ParentId="10396" CreationDate="2016-06-20T20:50:46.603" Score="1" Body="&lt;p&gt;Look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Platonic_solid&quot; rel=&quot;nofollow&quot;&gt;regular polyhedra&lt;/a&gt; for some ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it might be reasonable to construct a vacuum-forming mold for plexiglass sheets that creates a 5-sided section of a spherical surface. 12 of these could be assembled into a sphere, with one of them removable to create the hatch.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-06-20T20:50:46.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10401" PostTypeId="1" CreationDate="2016-06-20T21:47:12.470" Score="13" ViewCount="10739" Body="&lt;p&gt;I'm taking a thermodynamics class and had the following question. I was going to ask my professor directly, but it seems like a stupid question with a simple answer, so I thought I'd try my luck here in fear of embarrassment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When air flows through a pipe, its stagnation enthalpy doesn't change. For a calorically perfect gas, we have that the enthalpy varies linearly with the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_0 = \text{constant}= h + \frac12u^2$$&#xA;$$h = c_pT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's look at air. Air has a specific heat at constant pressure of $1000\ \mathrm{\frac{J}{kg\cdot K}}$. Room temperature air is about $300 \text{ K}$. Wind speeds are usually below $15\ \mathrm{\frac{m}{s}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewriting our equations, we can say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$c_p T_0 = c_pT + \frac12u^2$$&#xA;$$1000\ \mathrm{\frac{J}{kg \cdot K}} \cdot 300 \text{ K} = 1000\ \mathrm{\frac{J}{kg \cdot K}} \cdot T + \frac12(15\ \mathrm{\frac{m}{s}})^2$$&#xA;$$300000 \frac{J}{kg} = 1000T\ \mathrm{\frac{J}{kg}} + 112.5\ \mathrm{\frac{J}{kg}}$$&#xA;$$T = 299.9 \text{ K}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, moderate wind speeds don't change the temperature at all. Why, then, does moving air outside feel significantly colder than stagnant air?&lt;/p&gt;&#xA;" OwnerUserId="6940" LastEditorUserId="368" LastEditDate="2016-06-21T15:49:54.490" LastActivityDate="2022-07-26T15:24:18.627" Title="Why does moving air feel cold?" Tags="|mechanical-engineering|thermodynamics|hvac|temperature|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10402" PostTypeId="2" ParentId="10401" CreationDate="2016-06-20T22:03:34.230" Score="2" Body="&lt;p&gt;In still air, the main heat transfer mechanisms are diffusion and natural convection.  If an object is at a higher temperature than the still air, heat will flow from the object to the air.  In essence, the air immediately surrounding the object would be at a higher temperature than just the air itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In moving air, you also have diffusion and natural and forced convection as heat transfer mechanisms.  Forced convection works to remove the heated air in the immediate vicinity around that object (due to diffusion) and supply more air at the original temperature of the air more rapidly than the other mechanisms.&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="368" LastEditDate="2016-06-21T15:28:30.363" LastActivityDate="2016-06-21T15:28:30.363" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10403" PostTypeId="2" ParentId="10401" CreationDate="2016-06-20T22:14:55.487" Score="5" Body="&lt;p&gt;The air feeling cold is really your skin being cooled by forced convection and evaporation of sweat.&lt;br&gt;&#xA;With no air movement a boundary layer of hotter air forms over the skin and so because of the smaller temperature difference the rate of heat loss decreases.&lt;br&gt;&#xA;Moving air over the skin disrupts this boundary later allowing unheated air to be in contact with the skin thus increasing the rate of heat loss and so the skin cools more.&#xA;Evaporation is a change of state which requires latent heat which is provided by skin which thus cools more.&#xA;See &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Wind_chill&quot; rel=&quot;noreferrer&quot;&gt;wind chill&lt;/a&gt; and for the &quot;opposite&quot; effect &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Heat_index&quot; rel=&quot;noreferrer&quot;&gt;heat index&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6941" LastActivityDate="2016-06-20T22:14:55.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10404" PostTypeId="2" ParentId="10401" CreationDate="2016-06-20T22:20:39.483" Score="27" Body="&lt;p&gt;There are a couple reasons. First of all, it's important to note that the sensation of warmth or coolness is only indirectly related to temperature. &lt;strong&gt;The receptors in your skin that deal with temperature are mainly sensitive to heat transfer and changes in temperature&lt;/strong&gt;, not so much absolute temperature values. For example, here's an interesting excerpt from &lt;a href=&quot;http://www.britannica.com/science/thermoreception&quot; rel=&quot;nofollow noreferrer&quot;&gt;the EB article on thermoreception&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Cold receptors respond to sudden cooling with a transient increase in&#xA;  discharge frequency (called the dynamic response) that is directly&#xA;  related to the prior temperature and the magnitude and rate of the&#xA;  temperature decrease. If the cooler temperature is maintained, the&#xA;  discharge frequency adapts to a frequency of static discharge that is&#xA;  directly related to the cooler temperature.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the sensation of coolness has to do with how quickly heat is being transferred away from the skin. Heat transfer occurs in three modes: radiation, conduction and convection. It's that last one that's important, because convection relies on motion; with no motion, there's only radiation and conduction. Air is a pretty good insulator, making conduction less effective; and it's transparent across a wide spectrum, meaning there's no significant radiative heat exchange. And your skin has many tiny hairs (and perhaps larger hairs, depending on the person) that work &lt;em&gt;against&lt;/em&gt; any minor convective flows, as from a draft or small disturbance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, &lt;strong&gt;in the absence of convection (moving air), your skin will locally warm the air right around itself, and that air won't be very quickly replaced by cooler air&lt;/strong&gt;. As it warms, it conducts even less heat away from your skin (because the smaller temperature differential is a weaker driver).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the much more significant factor in most cases is probably &lt;strong&gt;increased &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/sweat.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;evaporative cooling&lt;/a&gt;&lt;/strong&gt;. Just as that layer of air around your skin conducts heat and is warmed by your skin, it also evaporates moisture and becomes more humid. (Your skin can always lose some amount of moisture to dry air, even if you don't feel sweaty.) Just as heat transfer will be reduced as the air warms up and approaches your body temperature, so too will evaporation be reduced as the air immediately surrounding your body gets slightly more humid. But when the air is moving, it gets much more effective at evaporating moisture from your skin. You can read about &lt;a href=&quot;https://en.wikipedia.org/wiki/Perspiration#Mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;the mechanism of perspiration&lt;/a&gt; for some more background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases, the moving air acts relatively more like a constant-energy &lt;em&gt;sink&lt;/em&gt; because as your body contributes heat energy and/or moisture, those higher-energy molecules move away from the interface with your skin and are replaced by more cool, dry air. From an analytical perspective, if the air is moving quickly enough you don't have to account for it getting warmer or more humid over time as it exchanges heat and moisture with your skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;a href=&quot;https://engineering.stackexchange.com/questions/10401/why-does-moving-air-feel-cold/10404#comment18345_10401&quot;&gt;this comment points out&lt;/a&gt;, it's important to recognize that while perspiration is specifically a cooling mechanism, convection works both ways; if the surrounding air is warmer than your skin, a breeze will make it feel even hotter. If you're really interested in this topic, UC Berkeley's Center for the Built Environment has a neat &lt;a href=&quot;http://comfort.cbe.berkeley.edu/&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermal comfort tool&lt;/a&gt; that you can play with, which goes into much more detail as far as individual and environmental variables.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-06-21T15:43:26.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10406" PostTypeId="2" ParentId="10300" CreationDate="2016-06-21T04:14:23.540" Score="1" Body="&lt;p&gt;The really quick and dirty solution: weld the current chuck to the current taper. That will ensure it won't come flying off. Then you can do light milling and start making parts for a stronger/better machine without having to buy any new bits for your drill press! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't bother throwing a lot of money at this as drill presses are generally not rigid enough to make 'good' milling machines. Any welder should be able to put some large tac-welds (I'd do four 90 degrees apart) on that in about 5 minutes which would prevent it ever coming off (be careful about doing too much welding as you don't want to introduce too much heat into the part). &lt;/p&gt;&#xA;" OwnerUserId="2768" LastEditorUserId="2768" LastEditDate="2016-06-21T22:02:34.347" LastActivityDate="2016-06-21T22:02:34.347" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10407" PostTypeId="1" CreationDate="2016-06-21T07:45:25.643" Score="1" ViewCount="1008" Body="&lt;p&gt;We know that for constant heat flux $\dot{q}$ Nusselt number can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Nu = \frac{hL}{k}$ with $\dot{q}=h(T_f-T_s)$, therefore $Nu =\frac{\dot{q}L}{k(T_f-T_s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm a little bit stuck on writing a definition for a constant wall temperature, lets say in a duct with $T_s=const.=500K$ where $\dot{q}$ is not a known constant from boundary conditions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one derive $Nu$ in this case?&lt;/p&gt;&#xA;" OwnerUserId="6945" LastActivityDate="2016-07-04T12:32:22.577" Title="Nusselt number definition for a constant wall temperature" Tags="|fluid-mechanics|heat-transfer|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10408" PostTypeId="2" ParentId="10407" CreationDate="2016-06-21T08:37:25.393" Score="1" Body="&lt;p&gt;The Nusselt number is the ratio between convective and conductive heat transfer, which cannot refer to the duct wall (as it is solid and no convection).  So the thermal conductivity $k$, should be for the fluid in the duct (air?) not the duct wall.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat transfer coefficient $h$, is across the boundary layer.  This can be calculated if you know the heat transfer rate, but otherwise there are correlations to estimate it.  Frequently these correlations actually calculate the Nusselt number itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correlations are usually specific to a particular geometry and flow type, for example for circular ducts and turbulent flow, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nusselt_number#Dittus-Boelter_equation&quot; rel=&quot;nofollow&quot;&gt;Dittus-Boelter&lt;/a&gt; is often used.&lt;/p&gt;&#xA;" OwnerUserId="2616" LastActivityDate="2016-06-21T08:37:25.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10409" PostTypeId="1" CreationDate="2016-06-21T09:45:33.820" Score="1" ViewCount="961" Body="&lt;p&gt;I need some approximate figures for how many litres of oxygen per minute a typical car engine requires. How would I estimate this?&lt;/p&gt;&#xA;" OwnerUserId="4846" LastEditorUserId="368" LastEditDate="2016-06-21T16:16:44.567" LastActivityDate="2016-06-21T20:00:18.617" Title="How can I estimate the rate of consumption of oxygen in a car engine?" Tags="|automotive-engineering|combustion|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10411" PostTypeId="2" ParentId="10409" CreationDate="2016-06-21T11:09:11.217" Score="4" Body="&lt;p&gt;Do the math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a reference car and get its fuel milage at some speed.  From that you can compute the amount of gasoline used in one minute.  Now look at the chemistry of gasoline and figure out how much oxygen your one minute worth of gasoline requires to completely burn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more details going on in a internal combustion engine, like nitrogen reacting with oxygen, not all carbon fully reacting with oxygen, etc.  However, the above should be a reasonable starting point.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;My car gets 50 miles/gallon at 55 miles/hour (it's a Honda Civic Hybrid, actually gets a bit more than that on average).  That's 1.1 gallons/hour = 0.0183 gal/minute = 69.4 ml/min.  The density of gasoline is about 750 g/l, so we have about 52 g of gasoline used every minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll approximate gasoline as being pure octane, which is C&lt;sub&gt;8&lt;/sub&gt;H&lt;sub&gt;18&lt;/sub&gt;.  Since carbon has a atomic weight of 12 and hydrogen of 1, that of octane is 114.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ideal combustion, each carbon atom combines with two oxygen, and every two hydrogen with one oxygen.  One octane molecule will therefore combine with 25 oxygen atoms.  The atomic weight of oxygen is 16, so the total atomic weight of the oxygen required for one octane molecule is 400.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that 400 units of oxygen will be used for each 114 units of octane by mass.  That means our 52 g of gasoline require 182 g of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question now comes down to what the volume of 182 g of oxygen is.  If you just wanted to know the quantity of oxygen, then mass is more useful anyway.  I'll therefore determine the volume of typical &quot;air&quot; that is required to get that much oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxygen is diatomic, so has a molecular weight of 32 in air.  That means our 182 g of oxygen is 5.7 moles.  Let's say air is 21% oxygen, so that means we need 27.2 moles of air.  At 20 &amp;deg;C and 1 atm, a mole of ideal gas has a volume of 24 liters.  Therefore, the final answer within all the specific conditions picked above is that the engine will suck in about 650 liters of air per minute.  Since some of these conditions can vary quite widely, and the fuel efficiency used is well off to the efficient end for current cars, I'll round up and say it takes 1 cubic meter of air per minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there wasn't any engineering here.  This was just plugging in numbers that are all easily looked up.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2016-06-21T20:00:18.617" LastActivityDate="2016-06-21T20:00:18.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10413" PostTypeId="2" ParentId="10409" CreationDate="2016-06-21T12:54:15.993" Score="2" Body="&lt;p&gt;You need to do a &quot;mass balance&quot;. Here's roughly how:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many litres of fuel per hour does the car use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work out the mass of this volume of fuel (I assume we can look up the density somewhere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, and assuming the fuel is gasoline C&lt;sub&gt;8&lt;/sub&gt;H&lt;sub&gt;18&lt;/sub&gt;. Convert the known mass to &lt;a href=&quot;https://en.wikipedia.org/wiki/Mole_(unit)&quot; rel=&quot;nofollow&quot;&gt;moles&lt;/a&gt;. (This is the tricky step that a chemist would know.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the equation of combustion of gasoline, work out how many moles of O&lt;sub&gt;2&lt;/sub&gt; you need. Then convert that to a mass of oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air happens to be 79% nitrogen and 21% oxygen &lt;strong&gt;by volume&lt;/strong&gt; and weighs a little over a kilogram per cubic metre (we'd have to look it up). So work out how many kilos of air you need and then its volume in cubic metres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I haven't done this in over twenty years so I'll leave you to work out how to find the mass and volume of air from the oxygen mass. It'll require juggling a few numbers but it shouldn't be too hard.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above we're assuming the engine mix uses the ideal (&quot;stoichiometric&quot;) quantity of air and there's full combustion with no side reactions like carbon monoxide, oxides of nitrogen etc.  So it's an approximate figure.&lt;/p&gt;&#xA;" OwnerUserId="6665" LastActivityDate="2016-06-21T12:54:15.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10414" PostTypeId="1" AcceptedAnswerId="10416" CreationDate="2016-06-21T14:17:18.867" Score="0" ViewCount="67" Body="&lt;p&gt;I have an idea for a sports equipment innovation &quot;a new product&quot; that a track and field athlete can use to practice and train his sport. The technology involves a laser beam. I'm wondering if my idea is feasible and/or if it has already been done because we think the idea is good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is that when training pole vault or height jump, the athlete uses a laser beam instead of a bar, and the laser beam measures if the athlete succeeds. Has it already been done or is it not feasible?&lt;/p&gt;&#xA;" OwnerUserId="6955" LastActivityDate="2016-06-21T14:44:54.597" Title="Feasibility of product idea" Tags="|mechanical-engineering|design|product-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2016-06-21T19:00:52.143" ContentLicense="CC BY-SA 3.0" />
  <row Id="10415" PostTypeId="1" AcceptedAnswerId="10419" CreationDate="2016-06-21T14:37:41.353" Score="2" ViewCount="2676" Body="&lt;p&gt;I am working with a piece of equipment that has a vacuum sealed chamber.&#xA;The outer wall (ring) of the chamber is shown in the image below.&#xA;This ring is bolted to a bottom plate and a top plate is raised up and down using pneumatics.&#xA;When the chamber is sealed the differential pressure between the inside and outside is nominally maintained at 10 psid.&#xA;The system also has a laminar flow meter and once vacuum is pulled and the chamber is for the most part isolated, we can read about 1.5cc/min flow (leakage) in the chamber.&#xA;Maximum flow readings for all other system components other than the chamber seal is below 0.0.03 cc/min, so I am convinced the issue is with the sealing of the chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image can be seen a gland for an o-ring.  This is an existing design, not of our own, so only minimal alterations to the design may be allowed.&#xA;We will be looking at the flatness and finish on the top plate, but there has always been some variability on leakage from cycle to cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this is my question...&#xA;&lt;strong&gt;What is the likelyhood that adding a second o-ring in series with the first, with a minimal gap between them, may reduce the total leakage rate of the chamber?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U4Lhu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U4Lhu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6956" LastActivityDate="2017-09-27T18:15:42.563" Title="Series O-ring in Vacuum Application" Tags="|mechanical-engineering|vacuum|seals|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10416" PostTypeId="2" ParentId="10414" CreationDate="2016-06-21T14:44:54.597" Score="2" Body="&lt;p&gt;It's certainly feasible.  I don't know if it has been done.  I don't think it's a great idea, for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What would the athlete be aiming for?  It won't be as visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In the actual sports, you're allowed to touch the bar, as long as it doesn't fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)  You may be affected (put-off) by the presence of a physical bar in a competition, when you trained without one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would potentially shorten the time between attempts, but it would be less representative of the real thing.&lt;/p&gt;&#xA;" OwnerUserId="6872" LastActivityDate="2016-06-21T14:44:54.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10417" PostTypeId="2" ParentId="10355" CreationDate="2016-06-21T14:52:43.930" Score="1" Body="&lt;p&gt;A submersible water pump will operate when not submerged, but running dry will usually reduce the life of the pump because the water is typically the cooling source for the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my answer to your question would be Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Might I suggest looking at Pioneer Pet #3026A 12V Pump &amp;amp; Transformer, which appears to be an inline pump.&#xA;But I think you may have more challanges.&#xA;Pet fountain systems are not usually filtered.&#xA;Having a filter usually requires a stronger more costly pump...so most inexpensive pet fountain system I have seen just circulate the water.&lt;/p&gt;&#xA;" OwnerUserId="6956" LastEditorUserId="6956" LastEditDate="2016-06-22T20:01:04.213" LastActivityDate="2016-06-22T20:01:04.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10419" PostTypeId="2" ParentId="10415" CreationDate="2016-06-21T15:48:58.933" Score="2" Body="&lt;p&gt;Simply adding another o-ring in series will not give much improvement in the leak rate.  This is because the leak rate of the inner o-ring will prevent the internal space (between the two o-rings) from being brought to any reasonable vacuum level so the pressure differential will remain the same on the inner o-ring.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you can use a separate pump to pump down the cavity between the two o-rings, then you will greatly reduce the effective leak rate into the main chamber.  The separate pump must be able to hold a good vacuum in the space between the o-rings for this to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, perhaps quicker, solution might be to use vacuum grease to create a better seal with the single o-ring.  To quote from a &lt;a href=&quot;https://www.lesker.com/newweb/flanges/flanges_technicalnotes_gaskets_1.cfm?pgid=0&quot; rel=&quot;nofollow&quot;&gt;Kurt J. Lesker technical note&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The debate about greasing o-rings is ongoing. Typically, the same vacuum level can be achieved with or without grease. However, when not using grease, the o-ring must be dust- and fiber-free, and the metal surfaces touching the o-ring must have at least a 32 rms finish. Grease permits sealing in less pristine conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="272" LastEditorUserId="272" LastEditDate="2016-06-22T11:53:40.493" LastActivityDate="2016-06-22T11:53:40.493" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10421" PostTypeId="1" AcceptedAnswerId="10479" CreationDate="2016-06-21T23:27:49.040" Score="4" ViewCount="589" Body="&lt;p&gt;I'm writing code for a microcontroller that will ramp up the speed of a stepper motor as quickly as possible, for a jig that I'm building, that moves a workpiece from one position to another along a linear rail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is not about modifying the mechanical or electrical system around the stepper motor, but simply the rate at which the stepper motor is stepped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a desired speed, it will ramp up the speed of a stepper motor (and so the speed of the workpiece) from stationary. (The stepper motor will not have any kind of feedback to the controller other than a switch at each end so it can periodically establish its position, and stop if it's lost/gained too many steps in one direction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'll need to implement/write a function which gives the optimal speed $s$ (revolutions per second) at time $t$ for maximum acceleration. (Where $s(t=0)=0$) which the microcontroller will use to determine how fast to accelerate the workpiece / stepper motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function might look like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s = kt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $k$ is an optimized constant, $t$ is in seconds, $s$ is revolutions per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum maximum acceleration will be specified by choosing a value of $k$. It will be determined by experiment rather than calculation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;What I want to know is:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;One (or more) formulas that I might use. I would then attempt various contants, varying them so I find values that result in good acceleration.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;A strategy for determining those constants.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;E.g. using the above formula: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;start with some value of $k=k_0$&lt;/li&gt;&#xA;&lt;li&gt;Attempt #1: Run the motor with that value of $k$, starting from $t=0$. Record the speed at which the motor failed to accelerate.&lt;/li&gt;&#xA;&lt;li&gt;Attempt #2: Double $k_0$, call it $k_1$. Repeat step 2.&lt;/li&gt;&#xA;&lt;li&gt;Attempt #3: Use half of $k_0$ as $k_{-1}$. Repeat step 2.&lt;/li&gt;&#xA;&lt;li&gt;If $k_1$ gives the highest speed, double it as $k_2$.&lt;/li&gt;&#xA;&lt;li&gt;If $k_{-1}$ gives the highest speed, half it as $k_{-2}$.&lt;/li&gt;&#xA;&lt;li&gt;If $k_0$ gives the highest speed, followed by $k_1$, take the average as $k_{0.5}$&lt;/li&gt;&#xA;&lt;li&gt;If $k_0$ gives the highest speed, followed by $k_{-1}$, take the average as $k_{-0.5}$&lt;/li&gt;&#xA;&lt;li&gt;Repeat step 2.&lt;/li&gt;&#xA;&lt;li&gt;And so on&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Further thoughts&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I think the above formula might work well assuming there is the equivalent of a flywheel attached to the motor, and:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there is zero friction&lt;/li&gt;&#xA;&lt;li&gt;the motor exerts constant torque at any speed&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly assuming:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the motor and system has zero mass / inertia&lt;/li&gt;&#xA;&lt;li&gt;the motor exerts constant torque at any speed&lt;/li&gt;&#xA;&lt;li&gt;the only resistance to movement is friction&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then the motor can still accelerate to full speed immediately. (So $k$ could be infinite). And if there is inertia, assuming friction forces are constant, (ie independent of speed), assuming I'm correct so far, then the formula is still good; $k$ will just be a lower value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that leaves questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is a typical relationship between stepper motor speed and torque? Clearly it tails off as speed increases, otherwise there would be no limit to its speed. So it clearly won't be taken into account by the above formula. I am investigating &lt;a href=&quot;http://www.orientalmotor.com/technology/articles/article-speed-torque-curves-for-step-motors.html&quot; rel=&quot;nofollow&quot;&gt;this link&lt;/a&gt; which goes into much detail, but perhaps there are factors which are more significant than others, or there are more basic models that give good approximations.&lt;/li&gt;&#xA;&lt;li&gt;What other factors could be significant that would can not be modeled within the above formula?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-06-24T07:39:46.370" LastActivityDate="2016-06-26T15:18:47.197" Title="Optimising driving speed of stepper motor for maximum acceleration by trial and error" Tags="|mechanical-engineering|applied-mechanics|stepper-motor|optimal-control|acceleration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10423" PostTypeId="2" ParentId="10415" CreationDate="2016-06-22T05:16:02.710" Score="4" Body="&lt;p&gt;I'm assuming that the piece with the groove and the mating pieces are metal or a dense molded plastic. If this was a 3d printed piece, I'd be worried about porosity of the parts too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second O-ring won't do much for you honestly. If your leak rate is that high, I'd question the design of the O-ring groove and O-ring selection. Does the O-ring sit proud of the groove when installed? If so, by how much? How much stretch is in the O-ring when installed? What's the ratio of the groove cross section to the nominal cross section of the O-ring when installed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common mistake is to design such a groove based on the nominal cross section of the O-ring and not accounting for the fact that this shrinks as the O-ring stretches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mating surfaces must be relatively flat, in order to control the gap. if there's an excessive gap and the O-ring material is too soft, the pressure can push the O-ring material into the gap and cause leaks. The best resource I've found for designing this type of static seal interface is the &lt;a href=&quot;http://www.parker.com/portal/site/PARKER/menuitem.6c94058cc6466c6315731910237ad1ca/?vgnextoid=a141a35c7575e210VgnVCM10000048021dacRCRD&amp;amp;vgnextfmt=default&quot; rel=&quot;nofollow&quot;&gt;Parker O-Ring Handbook&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to ask; Why use an O-ring? Why not a flat gasket? Or even a liquid gasket replacement like Loctite 510 or one of the hundreds of other sealants?&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2016-06-22T05:16:02.710" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10424" PostTypeId="1" CreationDate="2016-06-22T05:32:17.407" Score="-1" ViewCount="32" Body="&lt;p&gt;Our lab has created several multivariate models to predict sample properties from infrared spectra. However, different spectrometers often give different responses, and often the model created on one instrument cannot be used with another without first using some method to transfer the calibration to the second instrument. I need a good method for transferring a multivariate model to a second spectrometer so that it still gives accurate predictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already come across a few methods:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;create entirely new model for second instrument&lt;/li&gt;&#xA;&lt;li&gt;piecewise direct standardization&lt;/li&gt;&#xA;&lt;li&gt;adding samples to the model that have been measured with the second instrument&lt;/li&gt;&#xA;&lt;li&gt;linear fit to the model output for the second instrument to correct the output (slope-bias)&lt;/li&gt;&#xA;&lt;li&gt;transfer using &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0169743911002048&quot; rel=&quot;nofollow&quot;&gt;virtual standards&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I have tried these, but the only ones that work for our models require several standard samples (samples with known property values or spectra) to be measured on the second instrument, which ends up taking almost as much time as simply creating a new calibration model. I would prefer a method that does not require this hassle. and I am hoping that someone can offer a better solution.&lt;/p&gt;&#xA;" OwnerUserId="6968" LastEditorUserId="368" LastEditDate="2016-06-22T16:10:56.033" LastActivityDate="2016-06-22T16:10:56.033" Title="How can I transfer a multivariate model between spectrometers without the use of standard samples?" Tags="|chemical-engineering|spectrometry|calibration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10425" PostTypeId="1" CreationDate="2016-06-22T06:46:18.980" Score="4" ViewCount="262" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5DkQL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5DkQL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the photo above I can name the first 3 intersection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Regular intersection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) T intersection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Skewed intersection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having a hard time on identifying the fourth intersection, what do you call that one?&lt;/p&gt;&#xA;" OwnerUserId="3703" LastEditorUserId="272" LastEditDate="2016-06-22T11:50:50.670" LastActivityDate="2016-06-25T02:19:54.683" Title="How do you call the intersection with barrier and you can always right turn?" Tags="|terminology|highway-engineering|traffic-intersections|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10426" PostTypeId="1" CreationDate="2016-06-22T07:39:53.123" Score="3" ViewCount="1672" Body="&lt;p&gt;What would be the best solution for attaching polyurethane foam to a lycra sheet of material. I was thinking a chain or pad stitch or possibly glue, but not sure if sewing would be  the best alternative or another approach might work better to secure firmly these two types of materials together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or alternative, how to bond polyurethane foam with a leather sheet?&lt;/p&gt;&#xA;" OwnerUserId="6969" LastEditorUserId="282" LastEditDate="2016-06-25T05:44:27.643" LastActivityDate="2018-10-16T06:00:17.990" Title="How to attach lycra / leather to Polyurethane foam?" Tags="|materials|manufacturing-engineering|plastic|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10427" PostTypeId="1" CreationDate="2016-06-22T08:16:26.807" Score="0" ViewCount="158" Body="&lt;p&gt;I would like to be able to save, for example, a SOLIDWORKS(or any 3D-CAD program) assembly in some format that includes all the joint local/global locations and the parts global location and orientation. Is there standard file format for that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question is there a CAD file format that includes mass and inertia data with graphics?&lt;/p&gt;&#xA;" OwnerUserId="6971" LastEditorUserId="6971" LastEditDate="2016-06-23T12:04:27.703" LastActivityDate="2016-06-23T12:04:27.703" Title="How can I save mass, inertia, joint location and part orientation data from any 3D-CAD program to a CAD file?" Tags="|simulation|cad|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10428" PostTypeId="2" ParentId="10427" CreationDate="2016-06-22T09:07:39.403" Score="2" Body="&lt;p&gt;This sounds like the exact synopsis of my masters thesis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically a format that can contain this info exits. STEP could include this information, but since the exporter has some serious leeway into implementation so no export actually contains all you ask for and even if it did theres no guarantee your importer would understand it even if included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, exporting this information is pretty easily done yourself. The only thing that is hard to implement is the geometry export, but for that you can use any of the existing formats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you do is you write a script that querries each parts transformation matrix, mass info and joint positions then dump it into a  (json, xml etc file) file next to your normal export. Then write a importer for that data in the opposite end. This should not need to be terribly hard, ive done this in Creo and Catia. Creo having by far the least accessible API took about 2 weeks to implement once i figured out that this was indeed what i wanted to do. In contrast for Catia less than a week. Reading it in to say Adams or Simulink was pretty trivial by comparasion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way the creo simulink exporter works exactly like this.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2016-06-22T18:14:10.830" LastActivityDate="2016-06-22T18:14:10.830" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10430" PostTypeId="2" ParentId="10424" CreationDate="2016-06-22T12:50:36.917" Score="0" Body="&lt;p&gt;If you know the sets of coefficients $R_k$ which adjust the output for each wavelength band  for both spectrometers, just take the corrected set from one spectrometer, divide by its correction coeffs, and multiply by the correction coefs from the other one.  (or apply the full inverse of whatever correction algorithm was used).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two spectrometers have different bands, then do a spline interpolation fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what the problem is here.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-06-22T12:50:36.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10431" PostTypeId="2" ParentId="10425" CreationDate="2016-06-22T14:58:48.247" Score="3" Body="&lt;p&gt;In UK terminology, your separate turning lane would be called a &quot;filter lane&quot; ( see e.g. &lt;a href=&quot;http://www.collinsdictionary.com/dictionary/english/filter-lane&quot; rel=&quot;nofollow&quot;&gt;Collins dictionary definition&lt;/a&gt;). This definition doesn't mean that you can &lt;em&gt;always&lt;/em&gt; turn in that lane, it means that this lane is exclusively for turning; traffic in it may or may not have to wait for a green light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might therefore be tempted to call it a &quot;crossroads with filter lanes&quot;. (I would call your situation &lt;strong&gt;a&lt;/strong&gt; a &quot;crossroads&quot; rather than a &quot;regular intersection&quot;. But then I'd also use the word &quot;junction&quot; rather than &quot;intersection&quot;...&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2016-06-22T14:58:48.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10433" PostTypeId="1" CreationDate="2016-06-22T16:09:04.713" Score="3" ViewCount="111" Body="&lt;p&gt;I'm encountering some troubles while trying to solve this exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider a radio link between a geostationary satellite and a ground station. We send a binary signal (represented by two voltage levels), which got filtered by a rised cosine filter, frequency modulated to $f_p = 4~\text{GHz}$ with modulation index $I_f = 9$. We have to guarantee a signal-to-noise ratio after demodulation (let's call it SNRAD) of 15&amp;nbsp;dB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exercise asks to determine the bandwidth $B$ of the modulating signal known that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At the receiver input the noise power, in the same bandwidth of the received signal, is $W_N = -100~\text{dBm}$;&lt;/li&gt;&#xA;&lt;li&gt;The system temperature is $T_s = 290~\text{K}$;&lt;/li&gt;&#xA;&lt;li&gt;The text of the exercise recalls that the bandwidth of the whole frequency modulated signal is 4 times the Carson's Bandwidth: $B_{tot} = 4B_c = 4B\left(I_f + 1\right)$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Do someone have any suggestions about how to solve this exercise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT. I've tried to use the SNRAD information to obtain the SNR of the received signal, but I don't know if it is a correct way to use this information. By the way, as you will notice, the number that I obtain in this way seems incorrect since it is too low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$SNRAD=3 I_f ^2 SNR = 3 ~ \cdot ~ 81 ~ \cdot ~ SNR = 243 ~ \cdot ~ SNR$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which in dB becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ SNRAD_{dB} = 15 \text{dB} = 23.86 \text{dB} + SNR_{dB}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which I obtain something strange:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ SNR_{dB} = 15 \text{dB} - 23.86 \text{dB} = -8,86 \text{dB}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second approach I thought to calculate the bandwidth is to use the information given about the noise power level in the same bandwidth of the received signal. Again I don't know formulas which link the two things (noise power level and bandwidth) and the SNR obtained before seems too low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT2. I forget to try the classical formula which link the noise power level with the bandwidth considered. In this case we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W_N = F k_B T B$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F = \frac{T_s + T_0}{T_s}$ . Now we have another problem, we have two unknown variables: the received signal power level (S) and the bandwidth of that signal (B) which need to find to solve the exercise. In fact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$SNR_{dB} = -8.86 \text{dB} = S_{dB} - W_{N,dB}$ .&lt;/p&gt;&#xA;" OwnerUserId="6978" LastEditorUserId="6978" LastEditDate="2016-07-08T15:17:27.430" LastActivityDate="2016-07-08T15:17:27.430" Title="FM transmission exercise" Tags="|electrical-engineering|telecommunication|wireless-communication|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10435" PostTypeId="2" ParentId="10269" CreationDate="2016-06-22T20:14:09.263" Score="2" Body="&lt;p&gt;In your &lt;a href=&quot;https://en.wikipedia.org/wiki/Fuzzy_control_system#Antilock_brakes&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt; it mentions that the micro controller for the ABS is making decisions based on the &quot;states&quot; of the brakes, one of which is temperature. The micro controller may be be applying different control gains depending on the states, which is a form of adaptive control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple PID controller may not be able to be tuned to handle (perform well enough) a wide range of states, so there is some advantage to using FL in combination with traditional methods. FL is by no means the only way to do adaptive control, but it is fairly easy to implement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to be more direct in answering your question, it sounds like whoever designed the ABS in that example chose not to implement control on a continuous range of input states (i.e. temperature) because it was a lot easier than coming up with an alternative (adaptive controller, gain scheduler, or PID).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be a lot of complex issues when designing a controller that has to deal with a system that can alternate between linear and non-linear response regions or uncertainty in observations. There is some simple beauty to a FLC in that it can be easily setup to deal with such situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an interesting read, try this &lt;a href=&quot;http://personales.upv.es/asala/publics/papers/C24ALCA98.pdf&quot; rel=&quot;nofollow&quot;&gt;paper&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1174" LastActivityDate="2016-06-22T20:14:09.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10436" PostTypeId="2" ParentId="10315" CreationDate="2016-06-22T20:31:00.240" Score="1" Body="&lt;p&gt;When gypsum is added to the concrete mix it changes some of its &#xA;characteristics.Curing is the process in which the concrete is protected from loss of moisture and kept within a reasonable temperature range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you were using gypsum alone, you would still need to add water: much less than concrete, but still...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the curing process, it depends on several factors(temperature, humidity, materials)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer your question, yes you should still carry out curing process. This is something that you just can't escape. &lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2016-06-22T20:31:00.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10437" PostTypeId="2" ParentId="10300" CreationDate="2016-06-23T00:11:53.310" Score="0" Body="&lt;p&gt;I would tend to agree with Chris Johns on this one - it's probably not a great use of your time and may not provide satisfactory results - but that's not what you asked... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To directly answer your questions: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any solutions for coupling or attaching a spline shaft to a plain old shaft? What is it called and do they exist? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Yes, there are. A press-fit, or interference fit, may be a suitable means of attaching the two components.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you're not familiar with the concept, a press fit (aka &lt;a href=&quot;https://en.wikipedia.org/wiki/Interference_fit&quot; rel=&quot;nofollow&quot;&gt;interference fit&lt;/a&gt;) means you press a pin into a hole that is slightly smaller than the pin (creating interference between the components which holds them together).  In the case of metal-to-metal were talking a size difference of few tenths' of a thousandth of an inch [.0001] to a few thousandths [.001] of an inch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you provide more information regarding the materials of construction (presumably some sort of hardened tool steel for the shaft, not sure about your collar) and actual dimensions (actual meaning you measured the parts with a micrometer, or other suitable measuring device, and not the nominal dimensions provided by the manufacturer or parts description) then you could use an online fit calculator such as &lt;a href=&quot;http://www.engineersedge.com/calculators/mechanical-tolerances/force-fit-tolerances.htm&quot; rel=&quot;nofollow&quot;&gt;this one&lt;/a&gt; provided by Engineers Edge to get an idea of the interference needed - maybe look to the FN5 class for example.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm suspecting you're going to need a significant interference to withstand the loads expected, so you'll probably need a hydraulic press or a good source of heat, or maybe a combination of the too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, if you know the material of construction, you can use the coefficient of thermal expansion to determine the temperature change necessary to cause the female member to expand large enough to slip the male member in, and then when it cools and contracts it will create the interference necessary to hold the parts in place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option you might consider, is that you can press fit the pieces together (I recommend the shrink fit method above to maintain concentricity) and since this is a splined shaft fitting into a round hole, you may be able to braze the shaft to the coupling (filling in the gaps in the teeth with brazing material) which would be extremely strong.  Again without knowing more about the base materials I don't know if this is a suitable suggestion, but it's worth considering.  Many cutting tools used for milling operations which have carbide inserts are brazed, and it's proven to be a reliable and strong connection for the job. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard disclosure:  you should consult an engineer experienced with this type of application (someone specializing in machine design, or machine maintenance might be a good place to start) before taking numbers off some website or the words of some guy who doesn't have all the information about the application.  For one - the numbers suggested on the site I linked to are based off an ANSI standard that provides recommendations for common applications of holding a round pin in a round hole - none of which describes your application.  And for two - you're playing with some pretty large forces and heavy loads in a milling operation, and a catastrophic failure could send an extremely sharp large chunk of carbide, cobalt, or high-speed-steel straight at your face, eye, throat, or child-rearing organs which would not be very fun at all. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Finally, I am not completely married to using a spline drive mechanism in order to extend and retract the quill, so I welcome any suitable alternative mechanisms. I saw a ball spline, but wasn't sure if it works the same way as the existing mechanism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Bridgeport mills typically use an R8 quill.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I don't know if it's feasible, but if you're going down this road you might look at adapting one of these to your drill press. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly - my personal opinion only - is that the amount of time and money you'll spend on this project, which may or may not have a good outcome, is not going to be worth it.  Have you considered just purchasing a used bridgeport (or cheaper chinese knock-off) mill?   I'm not sure where you're located, but I just did a quick search on craigslist and in my area there are at least 5 ad's for 2500 USD or less, and a couple for 1200 USD or less.  I would seriously consider what you want to do with your mill, and decided when it's worth spending the money on a proper machine, and until then, you can always outsource your milling work to a local shop or even an online company like rapid-machining. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best of luck to you!&#xA;-Steven&lt;/p&gt;&#xA;" OwnerUserId="2598" LastEditorUserId="2598" LastEditDate="2016-06-23T01:12:07.697" LastActivityDate="2016-06-23T01:12:07.697" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10438" PostTypeId="2" ParentId="10366" CreationDate="2016-06-23T11:56:36.067" Score="1" Body="&lt;p&gt;A broadcast operation involves flashing the data just like a notice board, and whoever requires the data, can read it from the node or terminal broadcasting it. Hence, any node which comes in the radio range of the node broadcasting the data, can see the information or data. In case of RF transceiver, if the master is broadcasting, all slaves in its radio range will be able to read the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas, in one- to- many operation, also known as multicasting, there is a group formed called multicast group. This group includes all those nodes to which the particular data is intended and hence, this data cannot be read by any other node which is not included in the group even though, it is in the radio range. In multicasting, each data packet from the master consists of an address header which consists of multicast address. This address is common for all members included in the multicast group. So, the packet can only be received or inferred by those members whose address match the packet address header, i.e., this operation provides a sort of authorization to some nodes over the others.&lt;/p&gt;&#xA;" OwnerUserId="6645" LastActivityDate="2016-06-23T11:56:36.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10439" PostTypeId="1" CreationDate="2016-06-18T09:43:50.083" Score="6" ViewCount="573" Body="&lt;p&gt;Suppose we have a curved beam of a rectangular cross-section $A$, Young's modulus $E$, and a constant radius of curvature $R$. We now expose the beam to a vertical point load of magnitude $P$ at its free end:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5QOBC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5QOBC.png&quot; alt=&quot;Curved beam loaded at free end&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I now want to know is where the beam will snap (I am assuming perfectly brittle material behaviour, and that the beam material is far stronger in compression than in tension). For this simple case, this is relatively straight forward (the beam will snap very close to the point of force application), but I want to know what happens when $P$ is shifted along the negative x-axis closer to the clamped end of the beam. For clarity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mXhld.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mXhld.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Etc. pp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, as $P$ is shifted along the x-axis, the dominant type of stress will shift from bending to compression; if $P$ is applied very close to the clamped point, it's essentially a buckling problem, while it's similar to the bending of a straight cantilever beam when $P$ is applied at the free end. Now what happens in between, and how does this transition take place? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut feeling is that as $P$ is shifted (for clarity, I don't mean to imply any dynamic loading here), the point at which the beam will snap will increasingly move away from the point of force application, so if we were to start loading the beam at its far end until it snaps, then move the point of force application to the new far end and so on, we would end with snapped fragments of the beam which &lt;em&gt;increase in length&lt;/em&gt;. However, I have thus far failed to come up with a sensible model that would allow me to predict whether this is what is actually happening, and would be grateful for any hint.&lt;/p&gt;&#xA;" OwnerUserId="6987" OwnerDisplayName="David" LastEditorUserId="6987" LastEditDate="2016-06-25T07:22:15.890" LastActivityDate="2016-06-25T07:22:15.890" Title="Fragmentation of a curved cantilever beam under compression" Tags="|structural-engineering|structural-analysis|statics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10440" PostTypeId="1" CreationDate="2016-06-23T15:14:21.477" Score="-1" ViewCount="226" Body="&lt;p&gt;Read this &lt;a href=&quot;http://www.codecogs.com/library/engineering/fluid_mechanics/pipes/mouthpiece/external-mouthpiece.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary: The article claims the &lt;strong&gt;coefficient of discharge&lt;/strong&gt; decreases with increase in &lt;strong&gt;length&lt;/strong&gt;, &quot;due to greater frictional resistance between the walls of the mouthpiece and the flowing water.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Verified this experimentally (albeit with vertical set-up)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/R79tx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R79tx.png&quot; alt=&quot;Experimental set-up&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The losses due to friction increase with pipe length. Reducing the coefficient of discharge very slightly.&lt;/p&gt;&#xA;" OwnerUserId="6316" LastEditorUserId="26654" LastEditDate="2022-08-31T02:08:47.940" LastActivityDate="2022-08-31T02:08:47.940" Title="Fluids: What is the relationship between discharge coefficient and external tube length?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10441" PostTypeId="1" AcceptedAnswerId="10444" CreationDate="2016-06-23T16:20:16.697" Score="1" ViewCount="93" Body="&lt;p&gt;I'm designing an environmentally sealed chamber which will need to have its oxygen level controlled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chamber has a volume of roughly 1.5 liters. I don't expect pressure to vary too much ... maybe .8 to 1.2 bar. I'm not sure about flow rates, but we would want fairly fine control over the O2 concentration, so maybe a maximum of around .1 L/s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To that end, our intent is to have a supply of nitrogen (or other inert gas) and a supply of oxygen. I've never worked with gas canisters of any sort before, but I need to be able to feed controlled amounts of nitrogen and oxygen into the chamber as well as control a relief valve of some sort, all electronically. What sort of valves/controls would I need to allow a controlled amount of gasses in/out of the chamber?&lt;/p&gt;&#xA;" OwnerUserId="6989" LastEditorUserId="6989" LastEditDate="2016-06-23T17:01:18.060" LastActivityDate="2016-06-25T18:00:06.877" Title="Device (electronic valve?) for controlling O2/Nitrogen levels in sealed chamber" Tags="|design|compressed-gases|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10442" PostTypeId="2" ParentId="10439" CreationDate="2016-06-23T17:22:06.273" Score="4" Body="&lt;p&gt;Your assumptions are incorrect. The breaking point will almost always be at the fixed point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the internal forces in your structure with loading at two different positions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oSKkE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oSKkE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial force is always greatest at the support, and its value is the same regardless of where the load is applied. Bending moment is also greatest at the support, but the shorter lever arm in the second case leads to lower bending moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the most critical point will be at the support, regardless of where the load is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possible wrinkle is if you're dealing with reinforced concrete (RC) or some other material where the resistance to compression is far greater than to tension. An RC structure might actually fail at some other point, since the compression may (up to a point) actually help since it will reduce the tension suffered due to bending. Therefore, it's possible that the critical point in such a structure is somewhere with sufficient bending moment, and not enough compression to mitigate it.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-06-23T17:22:06.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10443" PostTypeId="1" CreationDate="2016-06-23T18:10:27.220" Score="1" ViewCount="1551" Body="&lt;p&gt;I want to stamp my own fishing spinner blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://lurepartsonline.com/Online-Store/Colorado-Blades/Premium-Finish-Colorado-Blades.html&quot; rel=&quot;nofollow&quot;&gt;http://lurepartsonline.com/Online-Store/Colorado-Blades/Premium-Finish-Colorado-Blades.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specs are usually 1mm thick, 3x2&quot; oval curved oval. Ideally, I would want to be able to adjust the size and shape, or apply grooves/patterns to the blade. I assume I can buy bulk sheet metal from a hardware store.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking I could possibly get 3d-printed casts for different designs, or possibly prebuilt shapes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm mostly a programmer, not a mechanic, so I'm just looking for a general idea of what is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this feasible, and how much should I expect to pay in machinery?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen some hydraulic presses go for under $1000, but it's a private hobby of mine and I could probably pay well over that if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to stress that this is an at home hobby and I don't want to rent machinery or order custom designs from a factory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize cutting the metal won't necessarily give a perfect shine, so any addition points on finishing/painting would be appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a video of how these things are generally produced commercially:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=zjovDNw1GcY&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=zjovDNw1GcY&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6991" LastActivityDate="2016-06-23T19:13:48.447" Title="Hand-pressed metal stamping - fishing spinner blades" Tags="|machining|metals|manufacturing-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2016-06-24T01:44:44.143" ContentLicense="CC BY-SA 3.0" />
  <row Id="10444" PostTypeId="2" ParentId="10441" CreationDate="2016-06-23T18:29:20.053" Score="2" Body="&lt;p&gt;It sounds like you might be looking at a similar setup  to purge gas for arc welding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you would typically a &lt;a href=&quot;http://www.r-techwelding.co.uk/argon-regulator/&quot; rel=&quot;nofollow&quot;&gt;regulator&lt;/a&gt; which screws into the valve fitting of the gas bottle which provides coarse control for flow-rate and also has a pressure gauge so you can see how much gas is left in the bottle. For fine control applications like TIG welding it is also common to add a &lt;a href=&quot;http://www.r-techwelding.co.uk/argon-flow-meter/&quot; rel=&quot;nofollow&quot;&gt;combined flow meter and valve&lt;/a&gt; downstream of the regulator (usually screwed into the regulator outlet) these are typically of  a 'peashooter'type   in the range of 0-25 litres per minute. In my experience this will allow for straightforward adjustments in the range of +/- 0.5 litre per minute with good consistency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this is manually set  but with welding the gas flow is turned on and off by a solenoid valve so you could simply add an off the shelf  welder solenoid into the loop and hook it up to your control system. &#xA;&lt;em&gt;Note that these solenoid valves can be a bit leaky over time so you may also want a more positive manual isolation valve in the system, although turning off the bottle vale will achieve the same thing&lt;/em&gt; . So you preset the flowrate  and control the duration of pulses of gas on demand rather than attempt to precisely adjust a constant flow-rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has the advantage that it is all off the shelf and easily available components and the requisite fittings and hoses are readily available through retail outlets. Without knowing your exact requirements it is hard to know if this sort of setup would be ideal but it does have the advantage of being a cheap and convenient way to set up a proof of concept prototype at the least. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;One thing to be wary of&lt;/strong&gt; is that the reactivity of oxygen usually demands that oxygen specific parts (especially hoses, seals and regulators) are used. But if you are using inert gas to fine tune the mixture this shouldn't be a major problem. Oxygen is routinely used in industry and it's not absurdly dangerous but you should certainly familiarise yourself with the risks and proper precautions associated with oxygen specifically and bottled gas in general as some of the hazards are not intuitively obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also applies to the chamber itself, any seals, adhesives, hoses etc should be rated for exposure to pure oxygen and you may need to employ a check valve to keep oxygen out of your inert gas lines. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-25T18:00:06.877" LastActivityDate="2016-06-25T18:00:06.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10445" PostTypeId="2" ParentId="7690" CreationDate="2016-06-23T19:05:49.510" Score="-1" Body="&lt;p&gt;I perform this kind of analysis all the time when running pressure vessels and other scrubbers, using LISA:&lt;a href=&quot;https://i.stack.imgur.com/rHoSu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rHoSu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, I have the option to add lump mass onto individual nodes (for my &quot;live loads&quot;, or other non structural dead weight), and my material has a density option available:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kwSMb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kwSMb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, I set a 1g gravity load downwards, and then plan my seismic loading based upon code.  ASCE for scrubbers and vessels has a few variations, depending on the designers discretion.  I use the most conservative one with this method however, and go with a &lt;em&gt;horizontal&lt;/em&gt; gravity in one of the main directions, equal to SDS * I / R.   Since LISA is not a non-linear FEA, this does not get P-Delta effects, but still gets displacements and stresses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8Ymfo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Ymfo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the scrubber has shifted slightly to the left under the seismic loading (the internal vacuum is doing most of the displacement though).  The software has automatically lumped the mass of the individual structural elements (as given in the &quot;density&quot; dialog for material) to the middle of the individual elements.  I place the live load mass directly on top of the applicable elements.  This way, the gravity load will affect them completely.  Then, it spits out the displacements and stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I don't have P-delta effects from my linear FEA, I will typically go back to the drawing board, and make a lumped mass model for buckling analysis.  While the FEA is done in fine strokes and small elements, this is done by hand instead.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2016-06-23T19:05:49.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10446" PostTypeId="2" ParentId="10443" CreationDate="2016-06-23T19:08:01.030" Score="1" Body="&lt;p&gt;I have been working on something similar myself recently, albeit in somewhat thinner material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 1 mm thick metal stock it is well worth considering getting your flat  blanks laser or water jet cut. In most countries there are plenty of local suppliers who will be happy to do this for you based on either a hand drawing or a vector graphics file and it isn't usually very expensive especially as they may be able to get volume discounts on materials. This is quite likely to be more economical than setting up to stamp them yourself from sheet material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of forming you may well be able to get away with a fairly cheap manual press, something like a fly or arbour press as it looks like the dishing is fairly shallow. as you say you will need dies to do this and here you have  a few options. 3-D printing may not be an ideal option as most consumer processes use plastics which aren't remotely up to the sort of forces involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sort of fairly flat shapes you have linked to it may be practical to grind your own dies from tool steel blanks with  either a hand held die grinder, pendant drill or a milling machine again you also have the option to get them CNC machined by a third party as long as you can produce a 3-D model of the form you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course doing it yourself gives you more flexibility to fine tune the forms as you go along. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also consider casting the dies, setting up a home bronze foundry is certainly within the realms of an ambitious hobbyist if you are prepared to invest a moderate amount of time and money and forging them in steel is also realistic option. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For thin metal is is also perfectly viable to cold work them to shape on open dies or using a pitch bowl. This is more labour and skill intensive per unit but don't require a large investment in equipment  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For finishing a polishing lathe can be used for manual finishing and de-burring, I would think you would be looking at textile mops, fleece (scotchbrite) wheels and possibly unitised wheels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A barrel polisher is also worth looking at for more efficient bulk finishing and polishing of small items. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to work with aluminium a home anodising setup is also worth considering. I don;t know if photo etching would be useful here but that is also a reasonable hobby technology for fine metalworking. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-23T19:13:48.447" LastActivityDate="2016-06-23T19:13:48.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10448" PostTypeId="2" ParentId="10224" CreationDate="2016-06-24T03:22:05.400" Score="0" Body="&lt;p&gt;If you go back to your basic engineering courses and look at you bending moment diagrams for beams, quite often they will be illustrated with pin roller supports.  Pinned at 1 end only allowing rotation, and roller connection at the other allowing for rotation and horizontal translation.  This makes the beam statically determinant.&lt;/p&gt;&#xA;&lt;p&gt;When this bridge was originally built, elastomeric/rubber bearing pads and a few others did not exist as an option.  This design emulates our formulas that we use for design, or rather the formulas work with this arrangement.  So this type of configuration is good.  It allows us to use our formulas as intended, keeps the design simple, and used the technology of the time.  In addition, as mentioned in other posts, it allows for rotation at the support as a result of live load, variations in dead, or sag after removal of shoring (assumed shored construction.  If the girder were lifted into place after casting, the large steel plate embedded into the concrete allows for some inaccuracies in measurement of the span and even placement.  It also allows for support should the beam shift slightly due to vibration or earthquake.&lt;/p&gt;&#xA;&lt;p&gt;Note, you will also see a similar configuration with steel girders on various types of bearings.  I believe the term shoe or shoe plate will get used, though that may be more for buildings than bridges.&lt;/p&gt;&#xA;&lt;p&gt;On an aside&lt;/p&gt;&#xA;&lt;p&gt;When it comes to &amp;quot;RAIL&amp;quot; bridges, the vast majority in North America designed to AREMA will consist of simple spans whether it be single span or multi-span bridges.  I found this statement funny while on my AREMA course as I just finished inspecting about a dozen rail bridges in my city with the vast majority not following this rule.  In highway bridges you will tend to see continuous for live load and these bridges as a result are not statically determinant.&lt;/p&gt;&#xA;" OwnerUserId="6993" LastEditorUserId="6993" LastEditDate="2022-06-16T05:20:23.530" LastActivityDate="2022-06-16T05:20:23.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10449" PostTypeId="1" CreationDate="2016-06-24T03:51:48.807" Score="0" ViewCount="183" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aV0y9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aV0y9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For A. I know that the formula for belt is simply&#xA;$L=\frac{Pi(D_a+D_b)}{2}+2C+\frac{(D_b-D_a)^2}{4C}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives me $L= 116.99&quot;$ since C is equal to $50$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For B. However I'm stuck and can't get the speed of the rack, I think it moves to right based on my inspection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I got stuck after solving rpm of Pulley B which is $40$rpm from from velocity ratio of $3:1$ from $(N_aD_a)=(N_bD_b)$ now i know that shaft C is turned 3 times for every 2 times of B. My problem is I need to get the diameter of gear B to relate it to Gear C that drives the same as gear 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please include your formulas and show your solution.&lt;/p&gt;&#xA;" OwnerUserId="6994" LastActivityDate="2017-03-24T14:16:38.030" Title="Kinematics Mechanics Find the length of the belt and the speed of the rack" Tags="|mechanical-engineering|design|kinematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10450" PostTypeId="1" AcceptedAnswerId="10456" CreationDate="2016-06-24T04:42:01.133" Score="2" ViewCount="1807" Body="&lt;p&gt;In engineering classes we made the assumption that structural supports (fixed, roller, etc.) are rigid. Thus, the displacements of a beam ,for example, at a fixed support are set to zero in the horizontal and vertical directions as boundary conditions. However in real situations shouldn't there be some displacement due to deformation in the support? Wouldn't that affect stresses/strains in both the beam and the support considerably? &lt;/p&gt;&#xA;" OwnerUserId="6995" LastActivityDate="2016-06-24T20:48:47.560" Title="How can we assume that structural supports are rigid?" Tags="|structural-engineering|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10454" PostTypeId="2" ParentId="10450" CreationDate="2016-06-24T11:34:52.927" Score="0" Body="&lt;p&gt;Depending on the level of engineering course you are taking, as previously mentioned this can be a simplified case, and most engineers love to keep things simple.  In addition, if you start tossing in various movements, your solution for the beam can go from being statically determinant to indeterminant.  As a result you may not have the math courses or engineering formulas developed yet to handle those situations.  You will eventually come across differential settlement where one or more supports will drop by different amounts.  In addition you will be taking on cases where both end of the beam are restrained and the beam contracts.  These are real issues that as a structural engineer you will deal with.  And at that time you will have your complete degree to help you, along with analysis programs, and a myriad of resources available on the net and elsewhere to assist you.&lt;/p&gt;&#xA;" OwnerUserId="6993" LastActivityDate="2016-06-24T11:34:52.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10456" PostTypeId="2" ParentId="10450" CreationDate="2016-06-24T17:55:21.120" Score="3" Body="&lt;p&gt;Like many things in engineering this one often comes down to a judgment on if this effect is &lt;em&gt;signficant&lt;/em&gt;. By significant we would mean does it change the results enough for us to worry about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few examples may be worth considering.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A simple beam under uniform load. Take:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bTNlUm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bTNlUm.png&quot; alt=&quot;Simply supported&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ w = 10\ \mathrm{kN/m}  \\&#xA;L = 2\ \mathrm{m} \\&#xA;E = 30\ \mathrm{GPa} \\&#xA;I = 6.75 \times 10^{-4}\ \mathrm{m^4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which the standard deflection formula gives:&#xA;$$  \frac{5wL^4}{384 E I} = 0.103\ \mathrm{mm} $$&#xA;Which assumes small deflections and therefore no horizontal movement of the supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately there is no simple analytical way to consider the horizontal displacement. Using non-linear finite element analysis the maximum vertical deflection is $0.108\ \mathrm{mm}$ and the right support moves by $1.4 \times 10^{-5}\ \mathrm{mm}$. Clearly the horizontal movement is very, very small and considering it doesn't make enough of a difference in the deflection or stresses for us to worry about. For beams with realistic sizes, materials, and loads this will usually be the case.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A support settlement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A support settlement is one situation where even small displacements of the supports need to be considered. For example a fixed-fixed beam where one end is displaced by $\Delta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wsUYIm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wsUYIm.png&quot; alt=&quot;support settlement&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the resulting bending moment in the beam due to the support displacement is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M= \frac{6EI\Delta}{L^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the same properties as above and a displacement of $\Delta=1\ \mathrm{mm}$ results in $M=30\ \mathrm{kNm}$. Which, if our beam was made of concrete, could be enough to cause it to crack. In this case we definitely would want to consider the displacement of the support.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some other cases which displacements of the supports would be considered:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In tall buildings the effect of axial displacement of the columns is considered (it is not &lt;em&gt;significant&lt;/em&gt; in low rise buildings)&lt;/li&gt;&#xA;&lt;li&gt;When temperature fluctuations are considered, the fact that an element may not actually be free to move is considered, as this will result in (sometimes very significant) stresses. This is a common source of cracking in concrete.&lt;/li&gt;&#xA;&lt;li&gt;When second-order effects are considered, the so called &lt;a href=&quot;https://wiki.csiamerica.com/display/kb/P-Delta+effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;$P-\delta$ effect&lt;/a&gt;, where the structure is in a displaced configuration (due to wind, earthquake, etc), and the loads are then applied.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="368" LastEditDate="2016-06-24T20:48:47.560" LastActivityDate="2016-06-24T20:48:47.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10457" PostTypeId="2" ParentId="10450" CreationDate="2016-06-24T18:26:40.050" Score="1" Body="&lt;p&gt;I recently saw a picture on Facebook of a t-shirt quite similar to this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9FjjF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9FjjF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is as good a definition of engineering as any. And that's because you are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To say that any support is perfectly rigid is a &lt;strong&gt;lie&lt;/strong&gt;, since there will always be some deformation of the support. But, then again, stating that a material is perfectly elasto-plastic (&lt;a href=&quot;https://en.wikipedia.org/wiki/Hooke%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hooke's Law&lt;/a&gt;) is also a &lt;strong&gt;lie&lt;/strong&gt;, since micro-imperfections will always make a member deform in a slightly chaotic (as in, not perfectly predictable) manner. &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler-Bernoulli beam theory&lt;/a&gt; is a &lt;strong&gt;lie&lt;/strong&gt; since it ignores shear deformations. &lt;a href=&quot;https://en.wikipedia.org/wiki/Timoshenko_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Timoshenko beam theory&lt;/a&gt; (which does consider shear) is also a &lt;strong&gt;lie&lt;/strong&gt;, since it considers perfectly elastic material behavior, which we saw above is a &lt;strong&gt;lie&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is not a single &quot;theory&quot; (in engineering) which is not a &lt;strong&gt;lie&lt;/strong&gt;, and that is simply because engineering doesn't care about providing a &lt;em&gt;perfect&lt;/em&gt; description of reality. Engineers just want a theory that's &lt;em&gt;good enough&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/6020/why-do-we-even-use-engineering-stress&quot;&gt;That's why we use engineering stress&lt;/a&gt; (as opposed to true stress). That's why we use Euler-Bernoulli or Timoshenko beam theory. And here I'm talking about the fundamental laws governing structural engineering. If we can accept simplifying assumptions here, in the bedrock of our field, we can accept them on a case-by-case basis as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's why we usually consider supports to be rigid. Because it's a simplification which makes our lives infinitely easier and which is &lt;em&gt;good enough&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you create a complex model, where you consider the stiffness of the columns? Sure. But the columns are resting on the foundations, so you'll need to consider the stiffness of those as well, and they are resting on the soil, so you'll need to consider the effect of that as well. There comes a time when you need to decide whether it's worth it. Is the deformation of the supports really going to make a difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have reason to believe that your supports may be significantly flexible (for example, your foundations are resting on a complex soil), then yes, you absolutely &lt;strong&gt;must&lt;/strong&gt; consider the effect that will have on your structure. If, however, there is no reason to believe the support's deformations will be all that relevant, then, well, it's not relevant and there's no reason to consider it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, as a detail, in the simple case of isostatic structures (such as simply-supported beams), small support deformations don't cause any stress increase at all. Well, actually, that's a &lt;strong&gt;lie&lt;/strong&gt; since it arises from Euler-Bernoulli theory. But if you pretend that Euler-Bernoulli is true (you really should), then yeah, no internal forces emerge.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-06-24T18:26:40.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10459" PostTypeId="2" ParentId="10425" CreationDate="2016-06-25T02:19:54.683" Score="2" Body="&lt;p&gt;This is probably an example of a &lt;em&gt;channelized intersection&lt;/em&gt;, where you have &lt;a href=&quot;http://safety.fhwa.dot.gov/intersection/conventional/signalized/fhwasa13027/ch11.cfm#e1114&quot; rel=&quot;nofollow&quot;&gt;right turn bays&lt;/a&gt; that split from the main road just before the intersection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://safety.fhwa.dot.gov/intersection/conventional/signalized/fhwasa13027/ch11.cfm&quot; rel=&quot;nofollow&quot;&gt;Individual Movement Treatments - FHWA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="848" LastActivityDate="2016-06-25T02:19:54.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10460" PostTypeId="1" AcceptedAnswerId="10491" CreationDate="2016-06-25T03:05:52.790" Score="3" ViewCount="1538" Body="&lt;p&gt;For those of us in the US, we're used to seeing roadways with yellow stripes separating traffic traveling against our direction of travel, and white stripes separating traffic traveling in the same direction. Of course, &lt;a href=&quot;http://mutcd.fhwa.dot.gov/htm/2009r1r2/part3/part3b.htm&quot; rel=&quot;nofollow&quot;&gt;Chapter 3B&lt;/a&gt; of the Manual on Uniform Traffic Control Devices requires this. However, I noticed that countries such as England, Japan, Russia, Australia, etc., don't distinguish these features by color. What particular rationales are there for using the color yellow to separate opposing directions of travel on roadway facilities?&lt;/p&gt;&#xA;" OwnerUserId="848" LastActivityDate="2016-06-27T14:35:53.223" Title="Why do North American Roadways Use Yellow Instead of White to Separate Opposing Traffic?" Tags="|civil-engineering|highway-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10461" PostTypeId="2" ParentId="10460" CreationDate="2016-06-25T03:50:56.360" Score="1" Body="&lt;p&gt;Some regions of north america such as the northern US and Canada receive snow.  Depending on the snow fall, level of maintenance, traffic, and a few other factors, lines can become obscured to a degree with snow.  White lines are much harder to pick out than yellow lines.  I personally feel its far more important to know when you are about to accidentally cross over a median line (centre yellow line) and into potentially on coming traffic than it is to accidentally cross over a white line and into an adjacent lane flowing in the same direction.&lt;/p&gt;&#xA;" OwnerUserId="6993" LastActivityDate="2016-06-25T03:50:56.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10462" PostTypeId="2" ParentId="4238" CreationDate="2016-06-25T04:11:27.043" Score="0" Body="&lt;p&gt;Looking at the the street view provided by the OP I would say that vertical profile constraints are at play here.  The ramp needs a certain vertical clearance near its start and near its end. Rather then have the road return to ground level, the designers may have opted to use one vertical curve over the entire ramp that suited their design needs.  The consistent vertical curve should ensure a comfortable ride for the user as well as provide plenty of sight distance to ensure they have enough time to brake if need be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to what has already been said about required clearance for over sized load, there may be a major industry on this route that routinely ships large/tall objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar to the need for formwork/falsework clearances, there may be an allowance for future maintenance or rehabilitation work to be performed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having done more than a few bridge inspection, I can dream and say they left plenty of clearance for an inspector to work under the bridge in a bucket  truck over live traffic without the need for a lane closure!  Pipe dream that one!&lt;/p&gt;&#xA;" OwnerUserId="6993" LastActivityDate="2016-06-25T04:11:27.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10463" PostTypeId="1" AcceptedAnswerId="10493" CreationDate="2016-06-25T08:14:36.540" Score="5" ViewCount="364" Body="&lt;p&gt;We cast a bridge's concrete deck and kept samples for concrete strength tests at 7 and 28 days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We took a test at 7 days, and the test result were over the design strength (designed for 30&amp;nbsp;MPa, result was 33&amp;nbsp;MPa). The control engineer insisted that we have to also do the 28-day test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems pointless to me, as I believe concrete gains strength during curing, and doesn't lose it. Can the results of the 28-day test be worse than those of the 7-day test?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I am asking this question because I want to get paid for the job. The control engineer (who signs the documents) is the one who is insisting that we have to wait until the 28-day test.&lt;/p&gt;&#xA;" OwnerUserId="7011" LastEditorUserId="255" LastEditDate="2016-06-27T15:56:11.667" LastActivityDate="2016-06-27T16:00:25.063" Title="Can concrete lose strength during curing?" Tags="|concrete|strength|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10466" PostTypeId="1" CreationDate="2016-06-25T13:36:19.393" Score="1" ViewCount="124" Body="&lt;p&gt;I'd like to calculate the max contact pressure in the below assembly, just before contact loss between the beam and the wall.&#xA;The beam is clamped to the wall with a pretensioned screw, and sustaining a force at his gravity center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming all bodies dimensionnaly stable (no distorsion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder about few questions...&#xA;how to calculate the maximum force at center of gravity before contact loss ?&#xA;how to work out the composed pressure profile...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some sketches that may help understanding the problem :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5K3J9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5K3J9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VZEoh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VZEoh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot for any kind of help (calculation steps...).&#xA;Have a good day !&lt;/p&gt;&#xA;" OwnerUserId="7013" LastEditorUserId="86" LastEditDate="2016-06-25T14:28:47.697" LastActivityDate="2016-06-25T14:28:47.697" Title="Contact pressure - screw pretension" Tags="|mechanical-engineering|pressure|systems-design|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10468" PostTypeId="2" ParentId="10426" CreationDate="2016-06-25T17:01:22.663" Score="1" Body="&lt;p&gt;Sewing may work, depending on the context, the problem here is that the stitches are quite likely to tear through the foam and may well work loose over time through friction causing the thread to effectively saw through it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adhesive may be the most effective solution, this will be easier with something like leather as there is less likelihood of it soaking through to the surface of the fabric. Here something like a latex based contact adhesive should work reasonably well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more porous textiles you could look at bonding a lining of something like canvas or calico to the foam first and then stitching into that or using double sided tape or Velcro. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all cases I would advise doing some tests with small samples to get a feel for how different methods will perform. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-06-25T17:01:22.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10469" PostTypeId="1" AcceptedAnswerId="10476" CreationDate="2016-06-25T21:46:41.830" Score="2" ViewCount="656" Body="&lt;p&gt;I am building a battery discharge system for LiPo batteries which accepts a charged battery and discharges it at a CONSTANT CURRENT regardless of the battery voltage, until a threshold is reached. The threshold is the LiPo storage voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of the system is to measure actual delivered power under variable battery environmental conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I have is regarding the load itself and the feedback system required  to vary the load to meet the constant current requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the recommended load?   e.g. motor or heater&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could the load resistance be varied in direct proportion to the battery voltage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any circuit references would be valuable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(sorry I'm an ME not an EE)&lt;/p&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="1832" LastEditDate="2016-06-26T00:39:22.103" LastActivityDate="2016-06-26T12:37:47.910" Title="Constant current load under varying input voltage" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10470" PostTypeId="1" AcceptedAnswerId="10471" CreationDate="2016-06-26T00:26:13.563" Score="3" ViewCount="583" Body="&lt;p&gt;I'm asking this here because I really cannot find a better place to ask this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Near my house (in the US), there is a pair of parallel train tracks operated by Union Pacific, primarily for cargo-carrying trains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along the track, there are various types of signals (for the trains): some are circular with three lights, some are linear with three lights (like a traffic light), and I think I've seen some with two lights. There's one pole on each side of the track, each with its own lights. Sometimes, on a pole, there are two sets of the lights - one above the other (I.e. Two traffic lights on the same vertical pole, and another identical one for the other set of tracks).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do these signals mean? Why do they vary? How do I read them?&lt;/p&gt;&#xA;" OwnerUserId="1177" LastEditorUserId="33" LastEditDate="2016-06-26T00:31:57.030" LastActivityDate="2017-01-10T21:58:00.270" Title="Train Signalling - What Do They Mean" Tags="|rail|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10471" PostTypeId="2" ParentId="10470" CreationDate="2016-06-26T02:10:04.340" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/North_American_railroad_signals&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/North_American_railroad_signals&lt;/a&gt; is a general introduction to the physical types of signals and the different signalling systems in use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://signals.jovet.net/rules/index.html&quot; rel=&quot;nofollow&quot;&gt;http://signals.jovet.net/rules/index.html&lt;/a&gt; has a &quot;common reference guide&quot; which describes the physical forms of different types of signals, plus guides to the signal codes used by different rail companies, including Union Pacific.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-06-26T02:10:04.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10474" PostTypeId="1" CreationDate="2016-06-26T11:58:18.463" Score="0" ViewCount="111" Body="&lt;p&gt;My mechanical &amp;amp; civil engineering usage is gone as the last time I did that was in Freshman year &amp;amp; my major went in another direction, but I've tried to be as thorough as possible using a prior question.&lt;br&gt;&#xA;&lt;em&gt;PS: If this has to go to Physics SE, please let me know or move it.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this question I posted on DIY:&lt;br&gt;&#xA;&lt;a href=&quot;https://diy.stackexchange.com/questions/87834/what-kind-of-extruded-aluminum-sections-to-replace-main-wooden-frame-on-this-diy&quot;&gt;What kind of Extruded Aluminum sections to replace main wooden frame on this DIY project?&lt;/a&gt;,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I received a &lt;a href=&quot;https://diy.stackexchange.com/a/87840/15762&quot;&gt;very detailed &amp;amp; interesting answer&lt;/a&gt; here from &lt;a href=&quot;https://diy.stackexchange.com/users/51443/ben-welborn&quot;&gt;Ben Welborn&lt;/a&gt; pointing out various factors around angles, lengths loads, torque, Center of Gravity/ Mass etc.&lt;br&gt;&#xA;I still have to completely digest the math/ calculations, which I'll have to do with my uncle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That question was centered around the vertical sections, whereas he did help with determining some of the detailed loading parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, I am curious to find &quot;&lt;strong&gt;In what position would a human be exerting Max Load on the horizontal rods, and what would that load be?&lt;/strong&gt;&quot;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Constraints:&lt;/strong&gt;&lt;br&gt;&#xA;Human body weight: 200 lbs for simplicity. No added plates, dumbbells, barbells etc.&lt;br&gt;&#xA;Rods are 36-40&quot; wide and 1-2&quot; in Diameter if that helps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Position types:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No aerials &amp;amp; similar momentum; i.e. Handclap pull-ups. &lt;/li&gt;&#xA;&lt;li&gt;Static positions as shown or what you guys can think of, &lt;/li&gt;&#xA;&lt;li&gt;Dynamic positions/ movements; Hanging V-up, i.e. Straight vertical Hang and full leg sit up from the waist, but in steady control (I guess the 90 degree L-position would be max leverage), NO Crossfit style momentum. &lt;/li&gt;&#xA;&lt;li&gt;No purposefully destructive positions/ movements as suggested in one comment; &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;The worst case scenario I believe would be to have Žydrūnas Savickas at the top of your structure (or at the midspan between two supports), resting his feet on the wall, grabbing one of the rungs and then trying to do a 524kg deadlift.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please do not constrain yourself to only the given examples of what people are able to do, but I'm sharing examples to kick start things. Positions vary in being 1 or 2 handed/ legged, both on same or different horizontal rods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Which of these or other positions/ leverage you might think of would be position of Max Load?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jlxqk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jlxqk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="6649" LastEditorUserId="-1" LastEditDate="2017-04-13T12:22:17.377" LastActivityDate="2016-07-23T21:12:08.430" Title="Positions of Max Load a human would exert on horizontal rods of such a system by positioning leverage &amp; bodyweight?" Tags="|mechanical-engineering|structural-engineering|structures|stresses|wood|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10475" PostTypeId="2" ParentId="10460" CreationDate="2016-06-26T12:35:03.153" Score="3" Body="&lt;p&gt;As mentioned there are relatively few situations in the UK where this would apply as multi-lane roads tend to have central barriers or other distinctive features. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a system of coloured &lt;a href=&quot;http://www.drivingtesttips.biz/motorway-studs.html&quot; rel=&quot;nofollow&quot;&gt;reflective studs&lt;/a&gt; used on major roads which is designed to indicate different types of lane divisions at night. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;White : ordinary lane divisions &lt;/li&gt;&#xA;&lt;li&gt;Amber : the edge of the central reservation (the area containing the barrier separating opposing lanes of traffic. &lt;/li&gt;&#xA;&lt;li&gt;Green : indicates where slip roads entering of leaving merge with the main traffic lanes. &lt;/li&gt;&#xA;&lt;li&gt;Red : indicate the left hand (inside) boundary of the road, either the outside of the hard shoulder (emergency lane) or  the edge of the road itself where no hard shoulder is present. Red studs may also be used on some more minor roads. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Virtually all single carriageway  roads in the UK have while reflective studs marking the centre line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On roads without a central barrier double white lines may be used on certain stretches to indicate that traffic may not cross or straddle them (effectively no passing). These are generally found on fast stretches of road with limited visibility where overtaking would be dangerous. In some cases an additional overtaking lane will alternate between directions every few miles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yellow lines are generally used in urban area to indicate traffic restrictions especially for parking, waiting and stopping as in double yellow lines and box junctions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the road system in north america is significantly different to the UK as distances are much longer.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-26T12:48:29.440" LastActivityDate="2016-06-26T12:48:29.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10476" PostTypeId="2" ParentId="10469" CreationDate="2016-06-26T12:37:47.910" Score="1" Body="&lt;p&gt;If you want a constant current load, design a constant current load.  As you say, this is a circuit that draws the same current over some range of applied voltage.  That's your spec right there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various ways to achieve this.  The conceptually simple is to get a measure of the current, then use feedback that adjusts a pass element to keep the current constant.  Some current sinks do work this way, especially if a separate supply is available to power the control electronics.  For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fXEfO.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fXEfO.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All current entering at SINK goes thru R1.  The voltage across R1 is therefore proportional to the current being sunk.  Due to how the feedback is arranged, the opamp adjusts the gate of Q1 to maintain the same voltage across R1 as Vref.  Vref is therefore the control input that sets how much current to sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are details of stability, bandwidth, and accuracy beyond the scope of a simple answer here.  However, such circuits do work and are used in reality.  In this case, you'd want a separate power supply for the opamp in most cases.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choosing R1 is a tradeoff.  You want it large for good signal to noise ratio, considering the opamp offset voltage as one source of noise.  On the other hand, you want R1 low for lower minimum voltage the current sink can operate at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another less accurate but much simpler current sink is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UwumW.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UwumW.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This exploits the property of a bipolar junction transistor where the collector current is largely independent of the collector voltage over a decent range of collector voltages.  As before, all of the sink current flows thru R1.  In this case, the base current also flows thru R1, but that's small compared to the collector current due to the gain of the transistor.  For example, if the transistor has a gain of 50, then about 98% of the current thru R1 is the sink current and the other 2% is the base current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The voltage on R1 will be Vref minus the B-E drop of the transistor.  At first approximation, that can be considered constant, around 700 mV.  This kind of current sink is also used in real life when a few percent accuracy and some load on Vref are OK.  This is the case often enough to make this a useful circuit.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-06-26T12:37:47.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10477" PostTypeId="1" CreationDate="2016-06-26T12:55:55.493" Score="5" ViewCount="10532" Body="&lt;p&gt;&lt;a href=&quot;http://www.ft.com/cms/s/0/2cfd8d2c-2816-11e6-8ba3-cdd781d02d89.html&quot; rel=&quot;noreferrer&quot;&gt;Why bamboo is the ‘green steel’ of 21st-century Asian architecture&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.dezeen.com/2015/11/04/bamboo-fibre-stronger-than-steel-dirk-hebel-world-architecture-festival-2015/&quot; rel=&quot;noreferrer&quot;&gt;Bamboo fibre is stronger and cheaper than steel, says Dirk Hebel&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.bentrideronline.com/messageboard/showthread.php?t=110976&quot; rel=&quot;noreferrer&quot;&gt;Carbon Fiber Vs Steel Vs Aluminum Vs Bamboo Pros and Cons&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://answers.yahoo.com/question/index?qid=20140109082613AAhwUs6&quot; rel=&quot;noreferrer&quot;&gt;Is bamboo stronger than steel?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are their comparative strengths in various directions and orientations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been reading a lot online but can't seem to find definitive comparison that talks numbers and specific comparison of numbers in various directions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there specific experiment(s) or research whose findings can be outlined here? &lt;/p&gt;&#xA;" OwnerUserId="6649" LastEditorUserId="6649" LastEditDate="2017-11-16T03:24:13.393" LastActivityDate="2019-09-16T23:14:54.710" Title="Steel vs Bamboo strength types/ weight and in various directions &amp; orientations? Specific, Tensile, Compressive, Shear, Lateral etc?" Tags="|materials|steel|wood|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10478" PostTypeId="2" ParentId="10477" CreationDate="2016-06-26T15:08:30.540" Score="9" Body="&lt;p&gt;Stating categorically that bamboo is stronger than steel is a bit like stating that automobiles are faster than planes. On its face, it is a bit shocking, and seems wrong. But a rocket-powered automobile is certainly going to be faster than a one-seat propeller plane over a short distance on a controlled track. Then again, the same rocket car is going to lose in a ground-speed race to an SR-71 blackbird by a factor of about 2.5. Note that each vehicle has its advantages and disadvantages aside from just measurement of speed. So it goes with broad material comparisons. Bamboo may be strong, but there are steels that will almost certainly be stronger. Bamboo fibers may be stronger than the bamboo itself, or even stronger than some steels, but will probably still not approach the strength of the strongest steels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest strength steels have isotropic strength near $2\ \textrm{GPa}$, and have ductilities of $3\%$ to $10\%$. Kevlar fibers are near $3\ \textrm{GPa}$ but only along their length. Their transverse strength is closer to $50\ \textrm{MPa}$, and they have limited extensibility, closer to $1\%$. All of these values are available with some internet sleuthing and in many textbooks. Bamboo fiber data is harder to find, but the sources &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0263822305001923&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1359835X10000126&quot;&gt;here&lt;/a&gt; (both links end at ScienceDirect, and are peer-reviewed journal articles) have interesting results. The former link has either $350\ \textrm{MPa}$ or $550\ \textrm{MPa}$ depending on preparation method, with no reported variance (though there is probably considerable variance, it was not reported). The latter link gives $650\pm175\ \textrm{MPa}$ and $800\pm100\ \textrm{MPa}$ for different preparation methods. Note that these values are all along the fiber lengths. No values were found for transverse strength, though it is probably an order of magnitude lower than longitudinal strength if the mechanisms are similar to those of non-biological polymers. And, at any rate, the fiber strengths are misleading since the fibers must be mixed into a polymer-matrix composite to be of any use (except perhaps as ropes), which will result in strength values lower than those of the fibers alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring the strength of bamboo plants is challenging due to the variation in size and shape, and the difficulties of finding proper measuring tools, or even of what geometric values to measure. I was unable to find any information on the subject of bamboo plants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with all material selection problems, there are other tradeoffs worth considering depending on application and requirements. Some things to think about include ecological factors (bamboo is renewable), environmental factors (bamboo is flammable), loading factors (steel can be made ductile and its high properties are isotropic), and weight considerations (bamboo is much less dense). Another important factor is processing: steel is always initially made by casting to a generic shape then by working, or directly to shape by casting. The process is intensive but relatively straightforward and well understood. To form a bamboo composite one must separate the fibers from the remaining plant material, chop them to size, coat them with sizing to ensure compatibility with the matrix material, mix with matrix resin, and then finally form into shape. Some of these steps and how they affect properties are not well understood yet.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-06-26T15:08:30.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10479" PostTypeId="2" ParentId="10421" CreationDate="2016-06-26T15:18:47.197" Score="2" Body="&lt;p&gt;One way might be to try to model and understand the physical problem. You are however clear in your question that you would prefer an empirical approach where constants are tuned to experience data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that the easiest and most general way is to use a piece-wise approximation of the velocity curve. Generically this may be made by using &lt;em&gt;basis&lt;/em&gt; functions $b_i(t)$ such that the velocity $s$ at any time is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ s(t) = \sum_{i=0}^{N}b_i(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple, and in this case likely sufficient, approach may be to use piese-wise linear velocity (corresponding to constant acceleration over a time interval). That is, divide the time into a finite number of time steps $t_0, t_1, ..., t_N$ and at each such time define a &lt;em&gt;parameter&lt;/em&gt; $v_i$ which is the velocity that should be used at that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your algorithm for the engine could then, for example, at any given time $t$ do a &lt;em&gt;bisection search&lt;/em&gt; in your $t_i$ array and do &lt;em&gt;linear interpolation&lt;/em&gt; between $(t_j, v_j)$ and $(t_{j+1},v_{j+i})$ to calculate the output $s(t)$ at any time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are likely to find that you do not need to divide the timeline into more than a few (order of 10 steps and thus order of 10 parameters) pieces in order to achieve a solution that is acceptably close to the optimal. You may also experiment with having non-constant time steps such that for example the time step increase when the time increases and there is less change in the acceleration the system can achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also see &lt;a href=&quot;https://calcopedia.com/s/demo/accelerationofamass&quot; rel=&quot;nofollow&quot;&gt;this sample example&lt;/a&gt; for an illustration of how close to an optimal solution it may be possible to get only with a few (10) time steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additional advantage of such an approach is that you can easily tune your velocity in one part of the process without having to adjust the parameters to maintain the response at other times. This leads us to a suggested approach for tuning the parameters:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use a constant acceleration of the entire range ($v_i$ increase linearly) and tune it such that you can reach close to the terminal velocity (see the sample).&lt;/li&gt;&#xA;&lt;li&gt;Tune each $v_i$ parameter individually, that is, first adjust $v_1$ such that you &lt;em&gt;just about&lt;/em&gt; can sustain that acceleration in the time interval $[t_0, v_1)$.&lt;/li&gt;&#xA;&lt;li&gt;Next you can do the same process for $v_2$ etc. etc. (and notice that once you have configured $v_i$ you can go on to tune $v_{i+1}$ without ever having to go back and adjust $v_i$&lt;/li&gt;&#xA;&lt;li&gt;You have now constructed a linear approximation of the optimal velocity.&lt;/li&gt;&#xA;&lt;li&gt;If you have areas where there are large changes, or the given velocity is ok in the start of a block but fails in the end you can always refine the &quot;mesh size&quot; in that area by introducing one additional $(t,v)$ pair.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The tuning of each parameter based on experiment may be based on a procedure as you suggest, i.e. doubling the $v_i$ if you encounter no failure, else bisecting between the highest &lt;em&gt;know ok&lt;/em&gt; and the lowest &lt;em&gt;known not ok&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7027" LastActivityDate="2016-06-26T15:18:47.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10482" PostTypeId="2" ParentId="9054" CreationDate="2016-06-26T22:26:18.977" Score="0" Body="&lt;p&gt;Noise cancelling headphones work by measuring external nose, inverting the signal produced and adding it to the speaker output. This either produced a net effect of 'quiet' or eliminates noise from the audio feed. Separating and audio input from external noise is an inherent part of this and can be achieved through microphone placement, acoustic insulation, electronic/digital filtering or any combination of the above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact if the system 'knows' what the desired audio output is it can be added after any nose cancellation takes  place and it is also possible to predict what frequencies are likely to be problematical in terms of noise in advance, depending on context. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the simple answer is that the whole system is designed so that the external noise and desired audio output are separated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noise cancellation tends to work less well for low and mid frequencies, apart form anything else because there is often a residual need or verbal communication and allowing audible warnings to work, especially in an industrial context. But having said that it is often high frequency noise which is most annoying and hazardous in industrial applications to there is a lot of scope for specific tuning with filters and parametric and graphic EQ especially where software allows this to adapt to ambient conditions. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-06-28T16:48:28.960" LastActivityDate="2016-06-28T16:48:28.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10484" PostTypeId="1" CreationDate="2016-06-27T07:10:58.950" Score="1" ViewCount="101" Body="&lt;p&gt;Plants running the Haber Bosch process provide the vast majority of nitrogen used in fertilizers. Collectivly, they consume about &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Haber_process#Economic_and_environmental_aspects&quot; rel=&quot;nofollow&quot;&gt;3–5% of the world's natural gas production [...] (around 1–2% of the world's annual energy supply)&lt;/a&gt;&lt;/em&gt;. Other sources stress that natural gas use accounts for roughly half the energy budget of these plants, the other half beeing electricity to run the turbocompressors. With casual googling, I only found numbers at the global level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to have more exact numbers for the natural gas and energy consumption of a typical Haber Bosch plant, relative to the fertilizer output.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-06-27T07:10:58.950" Title="What is the actual natural gas and electricity consumption of a Haber Bosch plant?" Tags="|chemical-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10485" PostTypeId="1" CreationDate="2016-06-27T08:43:20.523" Score="1" ViewCount="466" Body="&lt;p&gt;I am using a long ranged ultrasonic sensor to measure milk height in milk silos. Milk is stirring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Milk Silos Design: Shows two type of silos both are stirred in different direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vPFXF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vPFXF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7034" LastEditorUserId="1586" LastEditDate="2016-06-28T00:59:22.963" LastActivityDate="2016-06-28T00:59:22.963" Title="How can I increase my ultrasonic sensor accuracy?" Tags="|electrical-engineering|wireless-communication|data|distance-measurement|iot|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10486" PostTypeId="1" CreationDate="2016-06-27T09:29:15.553" Score="2" ViewCount="102" Body="&lt;p&gt;I have a bucket full of water with a hole in bottom, lying on a spring like in the picture I have attached. Is this system considered to be linear or not? &lt;a href=&quot;https://i.stack.imgur.com/dux9om.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dux9om.png&quot; alt=&quot;my system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7035" LastEditorUserId="9" LastEditDate="2016-06-28T09:07:21.607" LastActivityDate="2018-03-28T22:15:12.460" Title="Harmonic systems, system linearity" Tags="|control-theory|systems-engineering|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10487" PostTypeId="2" ParentId="10449" CreationDate="2016-06-27T09:44:55.230" Score="1" Body="&lt;p&gt;You don't need the diameters. The text says &quot;C turns 3 times for every 2 turns of B.&quot; Thus knowing the rotation speed of B you can calculate the rotation speed of C. The rotation speed $f$ is the amount of revolutions $N$ per time interval $\Delta t$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f=\frac{N}{\Delta t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio between two rotation speeds of two connected gears 1 and 2 is the same as the ratio between the amount of revolutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{f_1}{f_2} = \frac{N_1}{\Delta t} \cdot \frac{\Delta t}{N_2} = \frac{N_1}{N_2}$$&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2016-06-27T09:44:55.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10488" PostTypeId="1" AcceptedAnswerId="10492" CreationDate="2016-06-27T11:58:14.507" Score="1" ViewCount="59" Body="&lt;p&gt;What kind of resource provides summary data on how given kind of aluminum yields to machining? Are there norms, or such?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explaining:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different kinds of metals yield to machining differently; e.g. MO-59 brass has a small addition of lead, which makes it leave a clean, smooth cut; in contrast pure copper will pull, crease, drag into threads; where the milled surface of brass left a mirror shine, copper left a fleece-like mess of metal turned into myriads of tiny wires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data for a few materials I worked with, I got through word-on-mouth, or own experimentation. I &lt;em&gt;know&lt;/em&gt; resources exist, from which professionals learn this in an organized manner, but I don't even know what keywords to look for.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2016-10-08T18:11:20.947" Title="Where can I find how good given aluminum is for machining?" Tags="|metallurgy|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10491" PostTypeId="2" ParentId="10460" CreationDate="2016-06-27T14:35:53.223" Score="2" Body="&lt;p&gt;The most likely reason yellow is used as a color was probably the result of standardization efforts between the 1960s and 1970s (NCHRP Report 484, p8). Since the United States is a very influential country in North America, most other countries bordering them emulated its system to a degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, the United States has considered moving to an all-white pavement marking system (&lt;a href=&quot;http://onlinepubs.trb.org/onlinepubs/nchrp/nchrp_rpt_484.pdf&quot; rel=&quot;nofollow&quot;&gt;NHCRP Report 484&lt;/a&gt;). The NCHRP (National Cooperative Highway Research Program) conducted a feasibility study for all-white pavement markings, documented as Report 484. At the time the report was conducted, another study was underway (Project 5-18) to determine the effectiveness of yellow pavement markings. The &lt;a href=&quot;http://onlinepubs.trb.org/onlinepubs/nchrp/nchrp_rrd_328.pdf&quot; rel=&quot;nofollow&quot;&gt;summary&lt;/a&gt; as well as the &lt;a href=&quot;http://onlinepubs.trb.org/onlinepubs/nchrp/nchrp_w125.pdf&quot; rel=&quot;nofollow&quot;&gt;full report&lt;/a&gt; can be found on the Transportation Research Board.&lt;/p&gt;&#xA;" OwnerUserId="848" LastActivityDate="2016-06-27T14:35:53.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10492" PostTypeId="2" ParentId="10488" CreationDate="2016-06-27T15:20:10.933" Score="1" Body="&lt;p&gt;Basic internet search on machinability of metals turns up several hits.&#xA;The one below has a comprehensive rating system and includes comparison for several aluminum alloys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.quakerchem.com/wp-content/uploads/pdf/skill_builders/no10_machinability_ratings.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.quakerchem.com/wp-content/uploads/pdf/skill_builders/no10_machinability_ratings.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6956" LastActivityDate="2016-06-27T15:20:10.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10493" PostTypeId="2" ParentId="10463" CreationDate="2016-06-27T16:00:25.063" Score="3" Body="&lt;p&gt;With the exception of anything extremely unusual: the concrete will continue to gain strength until 28 days. In fact, it will continue to gain strength for something like 50 years, but the main strength gain happens early on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motivation for waiting for the 28-day test is therefore nothing to do with strength. Far more likely: the later you are paid, the better it is for the main contractor. This helps the contractor manage cash flow, and/or allows them to make interest by having money in their bank account for longer. I have no idea what your contract says in terms of when you get paid, but it is completely normal for the control engineer to pay you as late as he is able.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2016-06-27T16:00:25.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10495" PostTypeId="1" CreationDate="2016-06-27T19:04:03.947" Score="5" ViewCount="51730" Body="&lt;p&gt;I am trying to figure out the minimum gate width required for parking a car inside the house which has a 15 feet road in front of it. Please refer to the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hB1jS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hB1jS.png&quot; alt=&quot;Plot Layout&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plot is 20 feet wide. I have assumed 6 inch wall and 2 feet X 2 feet columns. These are just rough figures. I am yet to do proper calculation to come up with technically correct figures. Vehicle under consideration is 17 feet X 6 feet as shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have referred to AASHTO - Geometric design of Highways and Streets document to come up some calculation. Here is the turning path for a passenger class design vehicle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/86Agy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/86Agy.png&quot; alt=&quot;P class vehicle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Road is in pink for good visibility of dimensions. Plot is on the right side of the road where the curved path leads to. I mentioned the vehicle to be of dimensions 17 feet X 6 feet but according to the code, P class design vehicle has a size of 19 feet X 7 feet and I am using it directly. I have kept the gate columns right next to the road so as to get maximum area inside house boundary. Right now, they are 19.5 feet apart but ultimately they will be kept at 20 feet. To keep things on a safer side, I assumed the arcs to be circular and took the inner radius as 14 feet (instead of 14.4 feet) and outer as 26 feet (instead of 25.5 feet) and drawn this figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kJ8sz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kJ8sz.png&quot; alt=&quot;Design&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am doing this for the first time and hence want it to be verified by you guys. By this design, I am getting a gate width of almost 16 feet. Are these calculations correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way I can reduce the gate width? Assuming that driver doesn't need to go into the house in one swing. He can make a partial turn, then reverse a bit and then align the vehicle properly to get it inside the gate. One thing that comes to mind is that vehicle under consideration is smaller than P class design vehicle (17'x6' as compared to 19'x7') but I don't know how to carry out the complete calculations for a given vehicle dimension. Also I am not sure whether reducing the dimensions of design vehicle will be a good idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) This house is located in a very less crowded street (very few vehicles passing by) and one can always use full road width when trying to park the vehicle inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Moving the gate away from property line can be one way of reducing the gate width but property owner will be losing a good chunk of land which doesn't seem like a great idea. Also, the gate columns will have to be moved 11 feet inside to get a gate width of 13 feet - not a good idea at all.&lt;/p&gt;&#xA;" OwnerUserId="4756" LastEditorUserId="4756" LastEditDate="2016-06-28T04:42:10.677" LastActivityDate="2020-06-18T04:37:46.023" Title="Gate width calculation for Passenger car design vehicle" Tags="|civil-engineering|transportation|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10496" PostTypeId="1" AcceptedAnswerId="10499" CreationDate="2016-06-27T20:36:51.760" Score="3" ViewCount="10688" Body="&lt;p&gt;What is the most common material gas pipes made of? &#xA;I read in a unreliable website it's steel, is it true? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also how good is steel at stopping or slowing down the leaking of gas? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas I am using or planning on using is Argon which is denser than air.&lt;/p&gt;&#xA;" OwnerUserId="7041" LastEditorDisplayName="user16" LastEditDate="2016-06-27T23:09:44.830" LastActivityDate="2016-06-28T00:07:59.033" Title="What material is used to make gas pipes?" Tags="|gas|pipelines|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10497" PostTypeId="2" ParentId="10496" CreationDate="2016-06-27T22:32:10.850" Score="1" Body="&lt;p&gt;What is the context of your question? You are asking about natural gas but then mention argon. Natural gas and argon are very different substances with different uses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer depends a lot on your environment, the composition of the natural gas, and the operating pressure. Most old natural gas pipelines were made with steel. However, steel pipelines are prone to corrosion and must have some kind of corrosion mitigation program in place (eg. pigging, chemical injection, etc.) This doesn't really apply if the pipe has an internal lining. Polyethylene is popular for lower pressure natural gas because it doesn't experience the same kind of corrosion problems steel does. However, it doesn't have the same kind of mechanical strength as steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your pipeline is leaking I don't think it really matters what material you've chosen; it is leaking and the material isn't going to make it better or mitigate it. There is no such pipeline that will seal itself and stop the leak. Pipelines are also not porous enough that your process fluid would slowly seep out of it. If a pipe has been compromised enough to be leaking, the only way to stop the leak is to repair or replace the pipe. You should think more about what material and operating practices would prevent a leak from occurring in the first place; not about what material might slow or stop the leak once it has occurred. &lt;/p&gt;&#xA;" OwnerUserId="7044" LastActivityDate="2016-06-27T22:32:10.850" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10498" PostTypeId="1" AcceptedAnswerId="10510" CreationDate="2016-06-27T23:09:25.320" Score="5" ViewCount="205" Body="&lt;p&gt;I need to build an air tight system that can hold gas (Argon Gas). I see that most pipes use steel. Is steel tight enough to hold that gas? Argon is very dense, so I think it should have no problem. &lt;/p&gt;&#xA;" OwnerUserId="7041" LastEditorUserId="33" LastEditDate="2016-07-06T20:51:16.330" LastActivityDate="2016-07-09T23:37:57.823" Title="Is steel air (argon) tight?" Tags="|gas|pipelines|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10499" PostTypeId="2" ParentId="10496" CreationDate="2016-06-28T00:07:59.033" Score="5" Body="&lt;p&gt;In my admittedly limited experience working with plumbers, natural gas pipes are typically made from galvanized steel for interior work, or from black steel pipe for underground. Last-leg fittings and tubes which connect the stub-outs to various appliances are sometimes made from copper and brass. Other materials are possible, as noted &lt;a href=&quot;http://www.engineeringtoolbox.com/approved-gas-pipes-d_1112.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. Some things to consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When joining two pipes made from dissimilar metals (including steel and cast iron), always use a dielectric union to decrease the risk of galvanic corrosion between the two metals.&lt;/li&gt;&#xA;&lt;li&gt;In US jurisdictions, exterior pipes must be painted to reduce the risk and rate of weathering.&lt;/li&gt;&#xA;&lt;li&gt;In the US, natural gas typically contains a small amount of methyl or ethyl mercaptan. The additive has a foul smell which makes leaks readily detectible by most people. These compounds contain sulfur and will react with copper to form copper sulfide, eventually corroding through the pipe. Some copper pipes of types K and L are approved for use with gas. Only properly approved copper pipes should be used with natural gas, with appropriate fittings.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For argon, the first point is still relevant, though the last is likely irrelevant if the argon is pure. The second point may be relevant if you need any pipe runs outdoors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for leaking, the pipes themselves should not leak, typically it is joints that leak. If you need high pressure, ensure that the schedule of your pipe is sufficiently high (i.e. wall thickness) and that it is rated for the required pressure. As with all pressure vessels, ensure you have appropriate mechanical safety valves and a pressure regulator to decrease the risk of explosion. If you are worried about diffusion through the pipe, unless you plan to heat the argon to many hundreds of degrees C (which you will need specialized pipe for), the diffusion rates will probably be trivial. Focus on the pipe joints first for leak reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only need to transport small quantities of argon over short distances, as in a lab setting, copper and brass flare fittings are most likely sufficient if the pressures aren't too high. If you need to transport large quantities, you should consult an professional engineer.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-06-28T00:07:59.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10500" PostTypeId="1" CreationDate="2016-06-28T00:10:50.080" Score="1" ViewCount="72" Body="&lt;p&gt;Please excuse my ignorance in advance, (I'm in cs not EE), I am trying to figure out how can I get the distance from a car to a wall, what form of technology could I use IR sensors, ultrasound, etc.(that is the most cost effective and easy to use), when its all said and done I just want it to send a numerical value of the distance from the car to the wall.&lt;/p&gt;&#xA;" OwnerUserId="7045" LastEditorUserId="282" LastEditDate="2016-12-01T07:52:34.093" LastActivityDate="2016-12-01T07:52:34.093" Title="What can I use to calculate distance?" Tags="|electrical-engineering|automotive-engineering|computer-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2016-06-29T05:51:27.353" ContentLicense="CC BY-SA 3.0" />
  <row Id="10501" PostTypeId="2" ParentId="10500" CreationDate="2016-06-28T05:08:42.103" Score="2" Body="&lt;p&gt;There are a couple of solutions available. You will have to decide a budget and the required accuracy to settle for one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Laser distance sensor - These will give you pretty accurate measurements over a wide range of distances. You can take a look at the product catalog here: &lt;a href=&quot;https://www.sick.com/us/en/product-portfolio/c/PRODUCT_ROOT&quot; rel=&quot;nofollow&quot;&gt;https://www.sick.com/us/en/product-portfolio/c/PRODUCT_ROOT&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Ultrasonic distance sensor - These are way cheaper than laser ones and will give you decent accuracy over a shorter distance. This one from sparkfun will give you a range of about 6 meters: &lt;a href=&quot;https://www.sparkfun.com/products/639&quot; rel=&quot;nofollow&quot;&gt;Ultrasonic Range Finder - LV-MaxSonar-EZ1&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Infrared distance sensor - These are even cheaper but you won't get great accuracy. They work in even lower distance ranges as compared to previous two. Here is a link for your reference: &lt;a href=&quot;https://www.sparkfun.com/products/8958&quot; rel=&quot;nofollow&quot;&gt;Infrared Proximity Sensor Long Range - Sharp GP2Y0A02YK0F&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything boils down to what exactly you want - how much accurate, what distance range and how cheap.&lt;/p&gt;&#xA;" OwnerUserId="4756" LastEditorUserId="282" LastEditDate="2016-06-29T02:39:12.793" LastActivityDate="2016-06-29T02:39:12.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10502" PostTypeId="1" CreationDate="2016-06-28T06:28:48.560" Score="2" ViewCount="151" Body="&lt;p&gt;I'm quite new to using 3D Drawing programs, except a bit of work in Google's Sketchup.&#xA;As I'm trying to draw a model that should be 3D printed, and heared that Sketchup is kinda bad practice for this, I started using a different program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've started drawing in 123D Design, this was quite working, only now I ran into a problem. I've made a mistake, as the right border is to small, and I need to stretch this. This should be quite easy if the corners weren't rounded. But I cannot find any way to stretch this border with 3mm while the corner is rounded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same has to be done for the rear plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I couldn't find out how to do this in 123D Design, I've also tried doing it with Autocad. In here, I found the Stretch command, but this doesn't seem to do what I wish to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The selected part in the image is the part i want to stretch by 3mm to the right.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2GwAz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2GwAz.jpg&quot; alt=&quot;Stretched selection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the probably little simple question, but I couldn't find how to do this by myself.&lt;/p&gt;&#xA;" OwnerUserId="7047" LastEditorUserId="7047" LastEditDate="2016-06-28T07:58:53.307" LastActivityDate="2016-06-28T07:58:53.307" Title="Stretch part of object in Autocad or 123D Design" Tags="|3d-printing|autocad|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10504" PostTypeId="2" ParentId="1" CreationDate="2016-06-28T08:11:41.310" Score="1" Body="&lt;p&gt;We have dry cement bulkers truck running on local roads, feeding dry cement to mixer factory where it is turned into wet mixed cement. Pump air into the truck to 'blow' cement out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;http://youtube.com/watch?v=p74BAe-OYq0&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt; shows cement and sand are 'blow' by air. Powder (cement/sand) are placed in closed container. Air blow in and powder come out on, says, 2 to 3 inch diameter, tubing. Powder can be transferred in construction site, at horizontal and vertical distance of tens of meters. You can see the blue air-compressor in the video. Here is a similar &lt;a href=&quot;http://youtube.com/watch?v=p74BAe-OYq0&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7048" LastEditorUserId="9" LastEditDate="2016-06-28T09:12:52.693" LastActivityDate="2016-06-28T09:12:52.693" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10505" PostTypeId="1" CreationDate="2016-06-28T08:28:20.057" Score="2" ViewCount="219" Body="&lt;p&gt;I have an equation of a PI controller. What means here the velocity constant &quot;R&quot; exactly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m(t)=K(e(t)+R{\int}e(t)dt)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R=1sec^{-1}$$&lt;/p&gt;&#xA;" OwnerUserId="7035" LastEditorUserId="7035" LastEditDate="2016-07-03T06:00:09.460" LastActivityDate="2016-07-04T19:38:34.207" Title="What means the velocity constant of the integration in PI controller formulas?" Tags="|pid-control|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10506" PostTypeId="2" ParentId="357" CreationDate="2016-06-28T08:47:00.820" Score="1" Body="&lt;p&gt;Lithium battery is more &quot;dangerous&quot;. It is restricted (Wh limited) as passenger carry-on and banned as cargo, by many airlines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you design a new system, do not 'choose to use' Lithium if your product does not require small size/weight per watt hour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cellular phone designers and consumers, see size/weight/operating hours are key-to-sale and accept the 'risk'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vehicles, can afford bigger battery size/weight, and non-lithium battery are widely used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dream-liner switch to Lithium and result was less than smooth. Cellphone and external battery pack catching fire are regular news items, for many years already. Actual failure rate is not 'high', consider the number of battery in use and the risk is considered 'acceptable', by many. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use Lithium, if you really need. Do not use it when you can afford a bit bigger size/weight in your design.&lt;/p&gt;&#xA;" OwnerUserId="7048" LastEditorUserId="7048" LastEditDate="2016-06-28T10:31:04.750" LastActivityDate="2016-06-28T10:31:04.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10507" PostTypeId="2" ParentId="8234" CreationDate="2016-06-28T10:04:48.007" Score="0" Body="&lt;p&gt;There is a paper on how to account for &lt;a href=&quot;https://wiki.csiamerica.com/download/attachments/9536133/Fahjan%2C%20Accidental%20eccentricity.pdf?version=1&amp;amp;modificationDate=1338950028768&amp;amp;api=v2&quot; rel=&quot;nofollow&quot;&gt;accidental eccentricity in dynamic modal analysis of buildings&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The methodology is based on dynamic modal superposition technique by&#xA;  applying of accidental torsion to global force vectors related to each&#xA;  modal shape. For this purpose, the global displacement vectors are&#xA;  computed for each modal shape at the first step. Then, corresponding&#xA;  global force vectors are computed by using the global mass matrix, the&#xA;  eigenvalues of the system and the global displacement vectors. For&#xA;  each mode shape, the accidental torsional moments of the global force&#xA;  vector are updated by the required amount of eccentricity in either&#xA;  direction. Static analysis is carried out to find the modified global&#xA;  modal displacement vectors and the internal forces for each member for&#xA;  each modal shape. The nodal displacements and the internal force&#xA;  resultants can be combined by using standard modal combination&#xA;  techniques.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3353" LastActivityDate="2016-06-28T10:04:48.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10510" PostTypeId="2" ParentId="10498" CreationDate="2016-06-28T16:54:36.477" Score="5" Body="&lt;p&gt;Steel is routinely used for high pressure storage of industrial argon. For practical purposes it is certainly 'gas tight' although there may be a certain amount of atomic level diffusion into the material in the long term but this is more about contamination than actual leakage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in comments the quality of mechanical and welded joints is also critical for pressurised gas seals. There are numerous standards which deal with this in a variety for circumstances. In the UK BSP threaded fittings are commonly used for most gasses, including argon as well as a variety of compression and quick release fittings for low pressure applications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For prototype applications coded TIG welding will generally be preferred for fabrication of the vessel itself and attachment points for fixtures and fittings in terms of both gas tightness and structural integrity. Mass produced gas containers usually use coded automated processes for pressure vessel fabrication in conjunction with in-line quality control testing. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-07-09T23:37:57.823" LastActivityDate="2016-07-09T23:37:57.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10511" PostTypeId="2" ParentId="10330" CreationDate="2016-06-28T17:05:45.660" Score="1" Body="&lt;p&gt;(1) In Abaqus/CAE the stress at each integration point is given by the field variable $S$. Note that in Abaqus, like many or even most FE solvers, the stress computed is the Cauchy (true) stress. So, for a material with local material directions given by the Global basis: $\sigma_{x}=S_{11}$, $\sigma_{y}=S_{22}$, etc. You will need to make sure that stresses are being written to the ODB during the analysis to visualize them in CAE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) This question is a bit more complicated. Generally, when talking about constitutive modeling the material parameters are independent of length scale. This is kinda the point of continuum scale modeling. If the material properties are varying spatially in your model you will need to either implement field variables or a user defined material (either a UMAT for Abaqus/Standard or a VUMAT for Abaqus/Explicit). I don't have any experience specifically with fiber reinforced composites in Abaqus, but the set of parameter you are describing in your question are for a general linear elastic, anisotropic material, which may or may not be able to describe the homogenized response of the material you are interested in. If that is the case, I know that it is possible to construct layered composites within Abaqus, but I can only point you to the Abaqus User Manual for more information. As for specific material constants, the best place to look is in the literature or, better yet, go run some experiments. This might not be the answer you are looking for, but each material is different. Given the anisotropy of the material you are going to want to look for data from multiple loading configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also not quite sure why you would need the volume for each element, but if your structure can be meshed uniformly them it might be possible to calculate element volumes prior to the analysis. If your mesh is irregular the answer is a bit more complicated. One possible solution I can think of is extracting integration point coordinates within the UMAT framework for each element, then based on all the integration points for the element determine the element nodal coordinates based on the associated quadrature rules. Another possibility is to look at the input file (.inp) and determine the nodal coordinates for each element, using them to calculate the element volumes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) You can assign local material orientations in the &lt;em&gt;Property&lt;/em&gt; menu of CAE. The symbol looks like a yellow &lt;em&gt;L&lt;/em&gt; with red axes. Within the Material Orientation menu you can create local directions based on any coordinate system. In the Definition drop down select &lt;em&gt;Coordinate system&lt;/em&gt;, then in the &lt;em&gt;CSYS&lt;/em&gt; option create a new coordinate system. This can be down by clicking the coordinate system icon.&lt;/p&gt;&#xA;" OwnerUserId="7056" LastActivityDate="2016-06-28T17:05:45.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10512" PostTypeId="1" AcceptedAnswerId="10515" CreationDate="2016-06-28T18:28:29.157" Score="-1" ViewCount="50" Body="&lt;p&gt;Fire broke out from wing of an airliner after emergency landing. All were safely evacuated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this video, &#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=cKLbjPmOIq4&quot; rel=&quot;nofollow&quot;&gt;video&lt;/a&gt;, fire engines, standing by on run way as pilot have maydayed, rushed pass the airliner, positioned themselves in a 'formation' at up wind position, then, advanced toward the downwind airliners and started spraying foam.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a documentary program, during a drill, the first fire engine started spraying foam as soon as it got within 'shooting range'. That is, saving  extra time, 48 seconds in this real case, for all engines moving into upwind formation position.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the pro and con of these two approaches?&lt;/p&gt;&#xA;" OwnerUserId="7048" LastActivityDate="2016-06-28T23:08:40.850" Title="Fighting airplane fire on runway" Tags="|fire-sprinkler|" AnswerCount="1" CommentCount="0" ClosedDate="2016-06-29T05:52:22.493" ContentLicense="CC BY-SA 3.0" />
  <row Id="10513" PostTypeId="1" CreationDate="2016-06-28T19:29:43.597" Score="1" ViewCount="23" Body="&lt;p&gt;When storing a large amount of polymer material, afterwards we are experiencing bad flow properties during processing. We expect it is due to storing the material for long periods of time at high temperature and humidity in our warehouse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to do some small scale test to reproduce the conditions and test a number of additives to enhance the flow properties of the material. We want to have a relatively small amount ($3\,\mathrm{kg}$) of material in a box and add weights to simulate the weight of the $500\,\mathrm{kg}$ box.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are there any standardized test for these types of compaction test?&lt;/li&gt;&#xA;&lt;li&gt;How do I determine the weights needed to simulate the weight of a $500\,\mathrm{kg}$?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;An attempt at answering question 2 was made by assuming we want the same pressure on the large box as on the test scale. In the large box, assuming the material a hydrostatic fluid, it experiences a pressure at the bottom of the box:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P = \rho_b g h \approx 10\,\mathrm{kPa} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I have assumed a bulk material density $\rho_b=500\,\mathrm{kg/m^3}$, gravitational acceleration $g\approx10\,\mathrm{m/s^2}$ and a material height $h=2\,\mathrm{m}$. At the test scale we then need to provide the same pressure by applying a force:&#xA;$$ F=m_tg=PA_t \rightarrow m_t={{PA_t}\over{g}}=\rho_bA_th=50\,\mathrm{kg}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for test box with area $A_t=0.1\,\mathrm{m^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polymer is a TPE (SEBS) with about 10% oil by weight, Shore 60 hardness but flexible when compounded.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastEditorUserId="368" LastEditDate="2016-06-29T17:15:51.457" LastActivityDate="2016-06-29T17:15:51.457" Title="Designing a good downscaled compaction test" Tags="|materials|product-testing|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10514" PostTypeId="1" CreationDate="2016-06-28T23:00:50.330" Score="1" ViewCount="79" Body="&lt;p&gt;My team owns a Point Grey camera that we need to attach a wide angle lens to. The wide angle lens we have has a cable allowing for autoiris and p-iris control, but the camera doesn't have the autoiris/p-iris port. Is there another way to connect the iris controller? Can we get away with not using it at all?&lt;/p&gt;&#xA;" OwnerUserId="7059" LastEditorUserId="368" LastEditDate="2016-06-29T17:19:34.703" LastActivityDate="2016-06-29T17:19:34.703" Title="How can I use a p-iris/autoiris lens on a camera without a connection port?" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="3" ClosedDate="2016-06-29T07:55:17.227" ContentLicense="CC BY-SA 3.0" />
  <row Id="10515" PostTypeId="2" ParentId="10512" CreationDate="2016-06-28T23:08:40.850" Score="1" Body="&lt;p&gt;The first priority for any fire crew is to be part of the solution, and not become part of the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Staying downwind of a kerosene fire would put them in the likely path of a lot of hot gas and smoke, even if the flames don't reach them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, there's an infamous story about a similar incident that happened in the company I work for, many years ago. They were carrying out a test that involved a decommissioned aircraft fuselage. They had a fire crew present as a safety precaution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something went wrong with the test, resulting in a fuel leak on to the ground followed by a fire. Unfortunately, the ground was sloping, and the burning fuel flowed gently downhill, heading straight towards the fire truck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a video of the fire truck reversing away from the incident as fast as possible, which gets shown regularly to trainees as a classic example of &quot;how &lt;strong&gt;&lt;em&gt;not&lt;/em&gt;&lt;/strong&gt; to do a risk assessment and make a safety plan&quot;...&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-06-28T23:08:40.850" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10519" PostTypeId="1" AcceptedAnswerId="10526" CreationDate="2016-06-29T13:02:14.323" Score="1" ViewCount="58" Body="&lt;p&gt;I would like to know with which existing types of power plants(thermal,wind,etc.) can compressed air energy storage(CAES) be used efficiently? And how can it aid power plants like thermal(a brief explanation is sufficient)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know it requires multiple compressors and hence can be used with a gas turbine power plant. But why would a gas turbine need CAES since it can readily obtain air and compress it at any time?&lt;/p&gt;&#xA;" OwnerUserId="6493" LastEditorUserId="6493" LastEditDate="2016-06-30T16:54:17.670" LastActivityDate="2016-06-30T16:54:17.670" Title="Compatibility Of CAES" Tags="|energy-storage|" AnswerCount="2" CommentCount="2" ClosedDate="2016-07-01T11:18:08.463" ContentLicense="CC BY-SA 3.0" />
  <row Id="10523" PostTypeId="1" AcceptedAnswerId="10525" CreationDate="2016-06-29T14:03:33.373" Score="6" ViewCount="1546" Body="&lt;p&gt;Why do CAES systems have intercoolers after every compression stage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't it a waste of work, since the air will cool anyway in the storage cavern?&lt;/p&gt;&#xA;" OwnerUserId="6493" LastEditorUserId="368" LastEditDate="2016-06-29T17:07:03.727" LastActivityDate="2016-06-29T19:14:25.733" Title="Why does compressed air energy storage use intercoolers?" Tags="|energy-storage|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10524" PostTypeId="2" ParentId="10519" CreationDate="2016-06-29T14:12:27.863" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But why would a gas turbine need CAES since it can readily obtain air and compress it at any time?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Power plants that are operated by solar power can't &quot;obtain&quot; compressed air at any time as you say. For example solar tower power plants can't operate at all time of day without energy storage system, so at times of solar peak when the plant can deliver more power than its actual load, it stores this excess power for the times when the solar power input is minimum (or vanishes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you find this picture of an hourly performance of a hybrid gas turbine useful (check the power solar to turbine curve along the hours of the day).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Source: Algeria: a case study for solarised gas turbine systems - Gertig et al.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BQJIt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BQJIt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2016-06-29T14:12:27.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10525" PostTypeId="2" ParentId="10523" CreationDate="2016-06-29T14:31:33.097" Score="17" Body="&lt;p&gt;Not just CAES system, intercooling is desired for multi stage compressors in general, if you checked this diagrams of isentropic, polytropic and isothermal compression processes between the same pressure limits you can see that the curve that requires minimum power (area to the left of the process curve) for the same inlet and outlet conditions is the curve on the left (isothermal compression)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7tQLc.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7tQLc.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, by removing sufficient heat from the compression process say by constant cooling of the compressor casing you can get close the isothermal path. But since cooling the casing won't be much effective in removing heat another technique is used which is stage intercooling, and here in this figure you can see the saved power by splitting the compression process into two stages with intercooling in between:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ja04f.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ja04f.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Isn't it a waste of work, since the air will anyway cool in the storage cavern.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I hope it's clear now that it's not a waste of work (at the contrary). And it won't be much of a storage system without a proper thermal insulation to minimize the heat loss during the hours of storage.&lt;/p&gt;&#xA;&lt;p&gt;(Figures are from Cengel's book of Thermodynamics - An Engineering Approach)&lt;/p&gt;&#xA;" OwnerUserId="1293" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-06-29T14:37:20.970" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10526" PostTypeId="2" ParentId="10519" CreationDate="2016-06-29T14:43:40.443" Score="1" Body="&lt;p&gt;Classical CAES plants (all two of them currently in operation) need a heat source to generate power: The gas cools with expansion, thus loosing pressure, they need to heat it to realise all the potential energy. One way around his is to combine the CAES with a gas turbine, the air is mixed with gas and undergoes combustion in the turbine. So in effect the CAES provides the work otherwise neccessary for compression of the combustion air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably could use this with other heat engines. But CAES are built for grid balancing where you want to adjust the output of your plant quickly, and quickest engines in this regard are gas turbines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the system in place in the plants Huntdorf (Germany) and McIntosh (USA).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrast with adiabatic CAES, where the heat from compression is stored and reused when generating power. The furthest developed project along these lines, ADELE in Staßfurt (Germany), was canceled recently though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://de.wikipedia.org/wiki/Druckluftspeicherkraftwerk#Druckluft-Gas-Kombikraftwerk&quot; rel=&quot;nofollow&quot;&gt;wikipedia Druckluftspeicherkraftwerk&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-06-30T05:41:01.670" LastActivityDate="2016-06-30T05:41:01.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10527" PostTypeId="2" ParentId="10523" CreationDate="2016-06-29T19:14:25.733" Score="1" Body="&lt;p&gt;It would be a waste of work, if the air did &lt;em&gt;not&lt;/em&gt; cool in the storage cavern! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as you say, it does cool, hence you'll have less pressure available when expanding the air than the pressure you needed to work against when compressing it. The excess pressure could be avoided by compressing in isothermal mode, and &lt;a href=&quot;https://engineering.stackexchange.com/a/10525/5531&quot;&gt;as Algo said&lt;/a&gt; you approximate isothermal by putting an intercooler between two isentropic compressors.&lt;/p&gt;&#xA;" OwnerUserId="5531" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-06-29T19:14:25.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10528" PostTypeId="1" CreationDate="2016-06-29T20:04:48.070" Score="0" ViewCount="187" Body="&lt;p&gt;I am newbie in the area of clutches. I am making a motorized cable reel (DC motor) which I needed to make it easy for anyone to pull the cable reel out manually without much effort. Only pulling the cable out is manual. The retraction of the cord is done automatically via the DC motor. So, based on my research online, one way bearings mounted on my &lt;a href=&quot;http://www.skf.com/group/industry-solutions/two-three-wheelers/applications/engine/start-system/skf-one-way-clutch/index.html&quot; rel=&quot;nofollow&quot;&gt;DC motor sprocket&lt;/a&gt; seem to be the way to easily/freely pull the cord reel out manually without too much effort. But based on my understanding of this one way bearing, I would need disengage this one way bearing clutch while I need to retract the cord back in the opposite direction. If this statement is correct, I was wondering how I should disengage this clutch for retracting the cable back in. Do I have to use a powered electromagnetic clutch (something like &lt;a href=&quot;http://ogura-clutch.com/electromagnetic_clutch.php&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;) while will get disengaged while powered, and thus allow the cord to get retracted back in easily? Is this the only way of doing this?&lt;/p&gt;&#xA;" OwnerUserId="6120" LastEditorUserId="9" LastEditDate="2016-07-01T04:35:53.603" LastActivityDate="2016-07-01T04:35:53.603" Title="Does a one way clutch for motorized cable reel exist?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10529" PostTypeId="1" CreationDate="2016-06-29T20:30:25.727" Score="0" ViewCount="59" Body="&lt;p&gt;I am curious how car bodies were made back in the 1950s. Specifically the micro-car bodies like the Isetta. What techniques and tools were used back then? They seem fairly simple. Could a body for an Isetta be built in a small shop? &lt;/p&gt;&#xA;" OwnerUserId="4214" LastActivityDate="2016-07-01T23:52:12.870" Title="What type of shop could make an isetta body?" Tags="|manufacturing-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10530" PostTypeId="2" ParentId="10529" CreationDate="2016-06-29T21:47:13.520" Score="1" Body="&lt;p&gt;I don't think there was anything special about the manufacturing. Pressed steel sheets for the panels which were welding together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest cost specific to the Isetta would have been the dies used to press the panels. The presses could have been used for any similar-sized panels. The original Italian manufacturers also produced motor scooters and refrigerators, not &quot;full size&quot; cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some pictures of the UK Isetta assembly line here: &lt;a href=&quot;http://www.isetta.org.uk/page16.html&quot; rel=&quot;nofollow&quot;&gt;http://www.isetta.org.uk/page16.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The UK assembly line was constructed from scratch in just 3 weeks, in a factory space which had previously been used to repair railway engines. That suggests there was nothing particularly novel about the processes used. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-06-29T21:47:13.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10531" PostTypeId="2" ParentId="9101" CreationDate="2016-06-30T03:21:03.700" Score="0" Body="&lt;p&gt;A general rule of thumb is that if the center of the vertical load is within the middle 1/3 of the column (kern), the eccentricity of the load can be ignored.  You will have to calculate this for each possible position of your load.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2016-06-30T03:21:03.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10532" PostTypeId="2" ParentId="10528" CreationDate="2016-06-30T05:39:52.570" Score="1" Body="&lt;p&gt;Why do you need a clutch at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just let the motor free-wheel when it is not retracting the cable. The cable cannot be pushed back anyway, as it will flex instead of coiling.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-06-30T05:39:52.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10533" PostTypeId="2" ParentId="357" CreationDate="2016-06-30T06:06:47.570" Score="3" Body="&lt;p&gt;Here is a good resource that has very detailed information about Li-Ion hazards: &#xA;&lt;a href=&quot;http://www.prba.org/wp-content/uploads/Exponent_Report_for_NFPA_-_20111.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lithium-Ion Batteries Hazard and Use Assessment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(&lt;em&gt;Lithium-Ion Batteries Hazard and Use Assessment&lt;/em&gt;. Celina Mikolajczak, et al. July 2011. Fire Protection Research Foundation)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will summarize it below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;What is the chemical composition of Lithium Ion and Lithium Polymer battery?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cathode material varies, most commonly either LiCoO2 or LiFePO4 are used. Anode is usually graphite (carbon) based. The largest fire hazard is the electrolyte: this consists of lithium salts dissolved in an organic solvent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The organic solvent is comparable to gasoline in flammability. There is some ongoing research to non-flammable electrolytes, but they are not yet in common use, and probably do not work as well as current electrolytes. A non-flammable electrolyte would greatly reduce the fire risk, but the high temperature could still ignite e.g. plastic parts nearby.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Why do rechargeable Lithium Ion &amp;amp; Lithium Polymer batteries explode or catch fire?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The batteries store a large amount of energy. If this energy is accidentally released by e.g. short-circuit or physical damage, it will heat up the battery. In addition, when the temperature exceeds about 150°C, the lithium in the battery will start to react chemically and release even more heat, which is called a thermal runaway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the electrolyte liquid heats, it will expand, and at some point the pressure exceeds the strength of the battery casing. The batteries are designed with an over-pressure vent, which should allow the electrolyte to leak out relatively slowly. If the vent malfunctions, the only way out is for the battery casing to explode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter how the electrolyte got out, it is flammable and the battery casing is probably very hot at this point. This will often cause the electrolyte to ignite, releasing even more energy and possibly causing other battery cells nearby to overheat also.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Why cannot the rechargeable battery explosions and catching fire be prevented?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They are prevented, to a very large extent. Compared to the amount of li-ion batteries currently in use in the world, very few fires occur. Most of the fires that do occur are from low-quality products that do not have the safety systems such as protection circuit and often also overload the batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ultimately it is similar to &quot;why cannot house fires be prevented&quot;. When you have flammable materials and energy in the same place, eventually they'll combine to cause a fire.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastEditorUserId="8055" LastEditDate="2018-07-17T14:56:39.670" LastActivityDate="2018-07-17T14:56:39.670" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10534" PostTypeId="2" ParentId="3010" CreationDate="2016-06-30T08:56:12.527" Score="-1" Body="&lt;p&gt;PVC pipes for outdoor sewerage systems are intended primarily for buried gravity sewers and industrial wastewater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need cast iron and grates that covers and protects the equipment. &#xA;Also for the efficiency of drainage, channel drains are very effective and safe, especially made from stainless steel.&lt;/p&gt;&#xA;" OwnerUserId="7075" LastEditorUserId="368" LastEditDate="2016-06-30T15:48:32.470" LastActivityDate="2016-06-30T15:48:32.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10536" PostTypeId="1" CreationDate="2016-06-30T21:16:05.197" Score="7" ViewCount="130" Body="&lt;p&gt;So I'm taking a compressible flow class soon, and I've been thinking beforehand about the topic. A substance like water can easily be pressurized, but its density doesn't change easily, meaning there has to be some microscopic force between molecules  in equilibrium with the pressure force. What force would this be though? Would it be Van der Waals force perhaps?&lt;/p&gt;&#xA;" OwnerUserId="7081" LastActivityDate="2016-07-01T13:28:40.343" Title="What makes something like water so hard to change its density?" Tags="|fluid-mechanics|compressible-flow|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10537" PostTypeId="1" AcceptedAnswerId="12465" CreationDate="2016-06-30T23:05:17.100" Score="2" ViewCount="276" Body="&lt;p&gt;Please note that the question is &quot;Why&quot;, not &quot;How&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sure I'm getting this wrong, but I'm stuck and I don't know why! If a reverse thrust is applied during an aircraft landing wouldn't the total amount of work necessary to pull the air inside the turbine (hence pushing the aircraft forward) be canceled by the amount of work used to push the air forward (i.e. pushing the aircraft backwards)? I mean, why does it work? Why does the net work push the aircraft backwards and help to stop the plane?&lt;/p&gt;&#xA;" OwnerUserId="7086" LastEditorUserId="86" LastEditDate="2016-07-01T14:20:07.770" LastActivityDate="2016-11-24T00:19:04.933" Title="Why does aircraft reverse thrust actually work?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10538" PostTypeId="2" ParentId="10536" CreationDate="2016-07-01T02:16:10.367" Score="6" Body="&lt;p&gt;Two factors contribute to the incompressibility of liquids. The first is the electrostatic repulsion between the electron fields of neighboring molecules keeping the molecules from getting closer to each other without increasing force. The second is that intermolecular spacing is similar to that of solids made from the same material, due to intermolecular attraction and in turn due to some form of secondary bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that no conventional material is totally incompressible, and greater hydrostatic force results in higher density. This is reflected in a material's bulk modulus, and the relationship between hydrostatic stress, bulk modulus, and volumetric strain. For example, water has a bulk modulus of approximately 3.12 GPa, meaning that an application of 3.12 MPa hydrostatic stress would result in a volume decrease of about 0.1%, and thus a density increase of close to the same amount. Other liquids have similar bulk moduli. For many applications this is a negligible change and can be safely ignored, and hence liquids are generally treated as incompressible.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2016-07-01T02:16:10.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10539" PostTypeId="2" ParentId="10537" CreationDate="2016-07-01T04:44:14.410" Score="5" Body="&lt;p&gt;It works because burning the fuel inside the engine adds a lot of energy to the exhaust air that more than compensates for the work done sucking in and compressing the intake air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you shut off the fuel supply, it won't work &amp;mdash; just as you suspect.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-07-01T04:44:14.410" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10541" PostTypeId="1" CreationDate="2016-07-01T13:18:12.030" Score="2" ViewCount="88" Body="&lt;p&gt;I need to be able to pull a blow molding clamp out of the machine onto a temporary frame for maintenance. The clamp weighs 20,000 lb but the starting force to get it moving is only 365 lb, so can I use a 1500 lb winch to pull the clamp out onto the roll-out frame? The clamp is sitting on 8 V-Grooved wheels and running on rails.&lt;/p&gt;&#xA;" OwnerUserId="7094" LastEditorUserId="368" LastEditDate="2016-07-01T20:49:32.100" LastActivityDate="2016-07-04T20:01:21.993" Title="Can I use a 1500 lb winch to pull a clamp on wheels that weighs 20,000 lb?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10542" PostTypeId="2" ParentId="10536" CreationDate="2016-07-01T13:28:40.343" Score="0" Body="&lt;p&gt;For comparison, I found &lt;a href=&quot;https://www.quora.com/Could-you-give-me-a-list-of-the-most-compressible-liquids&quot; rel=&quot;nofollow noreferrer&quot;&gt;a table&lt;/a&gt; of some compressible liquids.  I don't know the original source.  Just &lt;strong&gt;look&lt;/strong&gt; at Hg!&lt;br&gt;&#xA; BTW, I've seen notes indicating that liquid H and He are pretty compressible, but a quick search didn't turn up easily translatable data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7PjoN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7PjoN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-07-01T13:28:40.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10543" PostTypeId="2" ParentId="6108" CreationDate="2016-07-01T13:31:06.890" Score="0" Body="&lt;p&gt;You are comparing an internal combustions engine with an external combustion engine. Both have benefits and tradeoffs. Practical efficiencies are limited by the basic engine design and the materials of construction. You are describing a gas turbine exhaust driven turbine which has a high power to weight ratio which is good for airplanes, but is maintenance intensive. External combustion in a boiler to feed a steam plant is much more reliable but requires heavy machines which is fine for a base loaded electrical power generation plant - in that case you want reliability and the easy ability to scale up power production by burning more fuel as the base load changes.&lt;/p&gt;&#xA;" OwnerUserId="7095" LastActivityDate="2016-07-01T13:31:06.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10545" PostTypeId="1" AcceptedAnswerId="10546" CreationDate="2016-07-01T16:51:05.703" Score="2" ViewCount="130" Body="&lt;p&gt;I was wondering, basicly every radio-transsmission works with electromagnetic waves which travels at the same speed whether I modulate the information with a low frequency or high frequency such as done with LI-FI.&#xA;What makes Li-fi faster? what makes any wireless transmission faster/slower if eventually I send my information with electromagnetic wave which travels at the same speed?&#xA;What makes lifi special?&lt;/p&gt;&#xA;" OwnerUserId="7096" LastActivityDate="2016-07-01T18:40:16.370" Title="Why is Li-Fi faster than regular wifi?" Tags="|telecommunication|wireless-communication|wifi|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10546" PostTypeId="2" ParentId="10545" CreationDate="2016-07-01T18:24:58.783" Score="2" Body="&lt;p&gt;The wave propagation speed is only relevant when you start approaching the theoretical limit of that type of wave's ability to transmit information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an analogy, consider people having a conversation in a room full of air. They're using sound waves to transmit information. The speed at which those waves propagate is the same for every speaker, in that particular medium. Does that imply that everyone transmits information at the same speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is no, of course not. The information is overlaid on the sound wave in a particular way, and we call that speech. You can transmit the same information at the same speed across a wide range of frequencies. You can speak faster or slower without varying the pitch of your voice. (In fact, &lt;a href=&quot;https://www.youtube.com/watch?v=4X4Fy8YqysY&amp;amp;t=130&quot; rel=&quot;nofollow noreferrer&quot;&gt;some people can transmit speech faster than you can receive it.&lt;/a&gt;) You can use different languages that might allow you to communicate particular ideas at different rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theoretical maximum speed at which information can be transmitted via audible speech is only loosely related to the speed of propagation of an audio wave. &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/tables/soundv.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Put the same conversation underwater and the waves propagate four or five times faster&lt;/a&gt; but you don't even notice because the time required for the wave to propagate is so many orders of magnitude less than the time required to encode and decode a single datum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Digital signals using EM waves are a much more specialized technology, compared to the human systems that enable verbal communication. But we're still far from any theoretical limit on the rate at which EM waves could possibly transmit information. Each of the &lt;a href=&quot;https://en.wikipedia.org/wiki/IEEE_802.11#Protocol&quot; rel=&quot;nofollow noreferrer&quot;&gt;802.11 protocols&lt;/a&gt; used by your Wi-Fi device has a theoretical maximum bitrate, based on multiple elements of its design. At one point those were measured in Mb/s, now they're measured in Gb/s and there's plenty of room to keep going. &lt;a href=&quot;https://physics.stackexchange.com/q/56240/82691&quot;&gt;A question over on Physics SE asks about the upper limit for fiber optics&lt;/a&gt; (which also use EM waves, of course) and you can see from the answers that, at Gb/s for Wi-Fi or Li-Fi, we're still many orders of magnitude away from hitting a ceiling on transfer speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why the wave propagation speed isn't as meaningful as you expect. As far as what makes Li-Fi faster than Wi-Fi, I'm not even sure that's the case in isolation; practically speaking, what seems to be driving the development of the former is &lt;a href=&quot;https://en.wikipedia.org/wiki/Bandwidth_allocation#Bandwidth_limitations&quot; rel=&quot;nofollow noreferrer&quot;&gt;over-saturation of the frequencies used by the latter&lt;/a&gt;. That's more of an issue of congestion than anything else, which means that if Li-Fi had been developed first, the situation might be reversed, and popular science mags might be writing the same BS headlines about Wi-Fi instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.informationweek.com/mobile/lifi-wireless-at-light-speed-comes-with-drawbacks/a/d-id/1323344&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Wireless at light speed.&quot;&lt;/a&gt; Give me a break.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2016-07-01T18:40:16.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10547" PostTypeId="2" ParentId="10541" CreationDate="2016-07-01T20:46:53.117" Score="3" Body="&lt;p&gt;If your 365 lb force figure is right, then yes this will work. Make sure you are including force to accelerate as well as force to overcome rolling resistance and gravity if the rails aren't perfectly level. Also make sure you have a mechanism in pace to slow and stop the clamp - it's no good to get it moving if it keeps on moving and hurts someone.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="368" LastEditDate="2016-07-01T20:50:01.620" LastActivityDate="2016-07-01T20:50:01.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10548" PostTypeId="2" ParentId="10529" CreationDate="2016-07-01T23:46:08.580" Score="1" Body="&lt;p&gt;For the Isetta you can see that the overall body shape is quite flat and boxy, without the more pronounced compound curves of other cars of the period. This makes it relatively easy to produce the panels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as die formed pressings there are various manual methods for forming sheet metal body work using tools like power hammers, swage tools  and &lt;a href=&quot;https://en.wikipedia.org/wiki/English_wheel&quot; rel=&quot;nofollow&quot;&gt;english wheels&lt;/a&gt;. These are still used today for small volume and custom car bodies although they are highly skilled processes they are certainly within the scope of a small shop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also the start of the period where you begin to see to see proper sports cars with specifically designed light weight space frame chassis and aluminium and, later, fibreglass body-shells, often using fairly standard engines and transmissions as opposed to the pre-war tradition of very basic ladder chassis with bigger and bigger engines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally there was a distinct separation between the engineering aspects of the, engine, chassis and drive train of a car and the coachwork and interior of a car but the pre and post war period started to see this becoming more integrated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were a lot of small volume cars made in the 1950's as the post war economy in Europe meant that there was quite a lot of surplus material and industrial capacity but not much cash for investment in new tooling so it is common to see cars with relatively advanced design but with quite simple manufacturing processes, often making use of post-war surplus materials and parts. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-07-01T23:52:12.870" LastActivityDate="2016-07-01T23:52:12.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10552" PostTypeId="1" CreationDate="2016-07-02T14:17:11.403" Score="0" ViewCount="240" Body="&lt;p&gt;In the picture below, the stepper motors are fixed to stay in place. They have small pulleys that rotate the bigger pulleys. If building such a system, how should the bigger pulley be attached? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the image it seems the smaller U-shaped piece where the camera is mounted is fixed to the bigger pulley. This is simple with some screws but how is the pulley connected to the frame? Does the same method apply to the other side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i1zIL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i1zIL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7102" LastActivityDate="2016-09-01T18:31:17.477" Title="Attaching timing belt pulley" Tags="|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10553" PostTypeId="2" ParentId="3633" CreationDate="2016-07-02T17:05:29.173" Score="0" Body="&lt;p&gt;A spring loaded piston actuating a rack, and turning a pinion gear.  Smaller the pinion gear, relative to the length of the rack... the more rotations achieved in a given stroke.  And also the greater the piston pressure required...&lt;/p&gt;&#xA;" OwnerUserId="7103" LastActivityDate="2016-07-02T17:05:29.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10554" PostTypeId="2" ParentId="443" CreationDate="2016-07-03T00:03:42.383" Score="-4" Body="&lt;p&gt;The draw back with the Studor Single Pipe System is very apparent to me. PAPAs aside (they can be used on a dual line drainage systems) there is no separation of Grey water from Black. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the Black (waste) line gets clogged up you will be walking in that waste in your shower. The use of a single line drain is very un-sanitary period! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's amazing that the Mexicans use a single line without venting even though the public drainage system hasn't been installed. All of their Hospitals are built this way! Very un-sanitary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story of &quot;SHIT&quot; is that the second &quot;SHIT&quot; will eat the first &quot;SHIT&quot;. The volume of &quot;SHIT&quot; will never gain until you add water to it period! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimalizing the water in the waste/cess-pit involves venting it for evaporation purposes &amp;amp; reducing the volume of water that allows the &quot;SHIT&quot; to gain in volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of water needed in the &quot;Crapper&quot; (the tank behind the toilet) becomes the issue to produce the &quot;scouring&quot; affect that you prescribed for cleaning the line. Evaporating the access water at the waste/cess/systern/pit should be the key issue. Not dumping it into a public drainage system that inevitably deposits it into the ocean or water table. Waste itself can be great fertilizer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Skyscraper construction there should be 4 lines. 1 for grey water drainage &amp;amp; venting, 2 for waste drainage &amp;amp; venting, 3 for venting the waste systern (very large cess-pit) &amp;amp; 4 for venting the grey water systern before it overflows into the public water system or onsite RO unit (reverse osmosis plant). They should all go to above roof level, though some have the systerns in a wing far less high than the original skyscraper those reductions in height of the venting adds up to back pressure problems. They need to be equal in height &amp;amp; diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cross venting from the Grey water line to the Black water line has been the norm &amp;amp; has been sustained with a street-t &amp;amp; costly reducers for many years. Clean outs on every floor to facilitate plumber access with snakes at waste height in the wall with street-t's also prove that water &amp;amp; waste do not reach terminal velocity even when the water falls from a hundred feet up, the diameter being the key. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see the two lines being separate completley &amp;amp; marked so. Pinching pennies on the plumbing system is guaranteed to prevent property appreciation &amp;amp; very un-sanitary environmental conditions. Maybe also enhancing the likelihood of Humankind's predictable demise, drowning in their own &quot;SHIT&quot;.   &lt;/p&gt;&#xA;" OwnerUserId="7107" LastActivityDate="2016-07-03T00:03:42.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10557" PostTypeId="2" ParentId="10552" CreationDate="2016-07-03T17:23:18.107" Score="1" Body="&lt;p&gt;In some ways this arrangement isn't completely ideal because you are potentially putting a significant bending moment on the bearings which support the larger pulleys especially in the lower, horizontal pulley. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally you would have the driven pulleys mounted in a Y shaped yoke with a bearing on each arm and the shaft fixed to the pulley and the frame element it is driving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that the short cylindrical elements (most visible in the top right) are bearing/housing units but they could just be spacers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that this isn't necessarily a real problem in this sort of application as long as the frame is stiff enough and the loads on the bearings are within acceptable limits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this design it looks like the moving parts of the frame are fixed directly to the bosses on the pulleys with set screws with bearings on the opposite end of the shaft only. It's not clear if these are moving element bearings or just plain bushings. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-07-03T17:23:18.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10559" PostTypeId="1" CreationDate="2016-07-04T03:19:01.943" Score="1" ViewCount="308" Body="&lt;p&gt;I'm currently studying control systems specifically the mass-spring-damper system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I7N51.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I7N51.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is described by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{y}=(b/m)\dot{y}+(k/m){y}=(b/m)\dot{u}+(k/m){u}$$&#xA;Where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b:$ coeffiecient of friction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k:$ coefficient of stiffnes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m:$ mass of spring&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From class we were told that &#xA;$$error = e(t) =y_{ref}-y(t)$$&#xA;Taking the first and second derivative&lt;br&gt;&#xA;$$\dot{e}(t)=-\dot{y(t)} $$&#xA;$$\ddot{e}(t)=-\ddot{y(t)}$$&#xA;After this we substituted with $e(t)$ &#xA;$$\ddot{e}(t)=-(b/m)\dot{e}(t)-(k/m){e}(t)+r(t)+d/m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then $r(t)$ is defined as $r(t)= ky_{ref}+mk_pe(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is where does the term $r(t)$ and $d$ come from. The example is taken from Ogata's book, but the dynamics of the error isn't covered in the example of the boook. Where can I read more about this so I can understand this equation.&lt;/p&gt;&#xA;" OwnerUserId="7106" LastActivityDate="2016-07-04T03:19:01.943" Title="Theory of steady state error and dynamics of error for automatic control systems" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10560" PostTypeId="1" CreationDate="2016-07-04T06:28:58.647" Score="2" ViewCount="349" Body="&lt;p&gt;I'm using an F2802x microcontroller and I saw in Embedded Coder in Simulink that I have Digital output block. When I use this block, how does it work?   &lt;/p&gt;&#xA;" OwnerUserId="7116" LastEditorUserId="7083" LastEditDate="2016-08-13T23:06:37.027" LastActivityDate="2016-12-22T12:45:18.357" Title="Why we use GPIO for embedded systems?" Tags="|electrical-engineering|embedded-systems|simulink|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ClosedDate="2016-12-30T00:08:38.210" ContentLicense="CC BY-SA 3.0" />
  <row Id="10561" PostTypeId="1" AcceptedAnswerId="10562" CreationDate="2016-07-04T08:30:20.437" Score="1" ViewCount="56" Body="&lt;p&gt;Are there any recommended clamps for holding samples of a viscoelastic material in place? My goal is to have the clamps hold the rectangular samples along lines drawn parallel to edges of the samples for elastic testing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hand screw clamps appear to be completely inappropriate as the material just squishes out of position. But, I think most clamps would suffer this weakness. &lt;/p&gt;&#xA;" OwnerUserId="7119" LastActivityDate="2016-07-05T05:23:33.780" Title="clamps for viscoelastic materials" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10562" PostTypeId="2" ParentId="10561" CreationDate="2016-07-04T11:31:18.300" Score="1" Body="&lt;p&gt;Indeed, clamps are not very appropriate for materials with a high Poisson's ratio for exactly that reason. A few better alternatives may be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Drill a hole through the sample near each extremity. Pass a steel bar through the hole and then clamp onto the bar. This has the problem of generating significant shear stresses around the holes during the experiment, such that the rupture may be due to shear instead of tension as expected. The mode of rupture should make it easy to identify how the element ruptured.&lt;/li&gt;&#xA;&lt;li&gt;If the material is castable (liquid which hardens), a modified version of the above can be used, where multiple small steel threads in both directions perpendicular to the force are embedded into the material during casting. The clamps can then be tied to the threads. This helps distribute the transfer force, reducing the shear stress at any given point.&lt;/li&gt;&#xA;&lt;li&gt;Depending on the material's properties, something like concrete's &lt;a href=&quot;http://www.buildingresearch.com.np/services/mt/mt2.php&quot; rel=&quot;nofollow&quot;&gt;split cylinder test&lt;/a&gt; may be possible.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.buildingresearch.com.np/services/mt/mt2.php&quot; rel=&quot;nofollow&quot;&gt;Flexure tests&lt;/a&gt; may also work, though their results may not be exactly equal to those of true tension tests.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-07-04T11:31:18.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10563" PostTypeId="1" CreationDate="2016-07-04T11:54:13.367" Score="1" ViewCount="477" Body="&lt;p&gt;A closed steam boiler in a coffee machine idles at around 124 °C, the steam and water inside at 1.2 barg. When a steam valve is opened, steam rushes out through a steam wand inserted in milk, to aerate and heat it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to determine is the flow rate of that steam, but all thermodynamic info on steam boilers I'm finding is for industrial situations and too complex for my very basic understanding of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm disregarding any further heating from the boiler when the valve is open and am just assuming an infinitely big boiler with unlimited pre-generated steam &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking the steam tube cross sectional area is important, as would be the length and type of tube for losses. Can someone put me on the right track as far as formulae go.&lt;/p&gt;&#xA;" OwnerUserId="7121" LastActivityDate="2016-07-06T07:54:26.973" Title="Rate of flow of steam going from one pressure to another" Tags="|thermodynamics|steam|boilers|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10564" PostTypeId="2" ParentId="10407" CreationDate="2016-07-04T12:32:22.577" Score="0" Body="&lt;p&gt;If heat transport is mainly by diffusion (conduction), a general definition of heat transport is:&#xA;$$\left.q\right|_w=-k \left. {{dT}\over{dx}}\right|_w = h\left(\left.T\right|_w-T_f\right)$$&#xA;The subscript $w$ implies that the derivative and the temperature is defined at the wall. Furthermore, by convention, it is assumed the temperature difference is positive, i.e. the wall temperature is higher than the fluid temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the heat flux is unknown but you are able to measure the temperature field then you may calculate the Nusselt number through:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Nu}=\frac{hL}{k}=-\frac{L}{\left.T\right|_w-T_f}\left.{{dT}\over{dx}}\right|_w$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to use you original equation and measure the heat input into the system by a heat balance. This is done by integrating the total energy at the entrance (usually known from boundary conditions, i.e. imposed enthalpy, temperature, etc.), the total energy at the outlet and finding the difference which must be the energy input into the system by the walls. This, divided by the heat transport surface area, gives you the mean heat flux and from it can be determined a mean Nusselt number. &lt;/p&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2016-07-04T12:32:22.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10565" PostTypeId="2" ParentId="10563" CreationDate="2016-07-04T12:46:52.150" Score="3" Body="&lt;p&gt;You stated that most things you found are too complex for your basic understanding, is it safe to assume that the answer you are seeking may be an educated guess? If so I would go with Bernoulli&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g z_1 + \frac{c_1^2}{2} = g z_2 + \frac{c_2^2}{2} + \int_1^2 \frac{dp}{\varrho}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume isothermal state change. Also assume $c_1 = 0$ if you say you regard the boiler as infinitely big. Properties of the steam could be acquired via books or online tables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly you need the mentioned geometry of the tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details use the instationary form and add friction losses if you like. I personally would neglect this.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2016-07-06T07:54:26.973" LastActivityDate="2016-07-06T07:54:26.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10566" PostTypeId="2" ParentId="10563" CreationDate="2016-07-04T13:01:39.227" Score="3" Body="&lt;p&gt;You can use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation&quot; rel=&quot;nofollow&quot;&gt;Hagen-Poiseuille&lt;/a&gt; equation to determine the flow rate as a result of a pressure gradient. A caveat in using this equation is that the flow has to be laminar, i.e. inertial effects can be neglected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is best used in the form:&#xA;$$Q=\frac{1}{128}\frac{\pi d^4}{\mu}\frac{\Delta P}{L}$$&#xA;from which you can see that the volumetric flow rate $Q$ is dependent (indeed) on the cross sectional area throught the tube diameter $d$, the length $L$ of the tube and the pressure drop $\Delta P$ applied along the tube. Furthermore, the type of fluid is important through the viscosity $\mu$; a more viscous fluid would be more difficult to force through the tube at the same pressure gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation already accounts for frictional losses at the tube walls (hence the required pressure drop) but it doesn't take into account any bends in the tubing or the type of tubing. However, I expect their influence to be negligible as long as most of the tubing is straight and rough enough to allow the use of the no-slip condition but not so much to induce flow instabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned earlier, after determining the flow rate from the information you have, you will need to check that the flow is indeed laminar. If not the equation cannot be used. You can do this by calculating the Reynolds number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Re}=\frac{\rho vd}{\mu}\approx\frac{\rho Q}{\mu d}&amp;lt;1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the above condition holds you are fine. If not then the approach suggested by &lt;strong&gt;@idkfa&lt;/strong&gt; is best used.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2016-07-04T13:01:39.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10569" PostTypeId="2" ParentId="10563" CreationDate="2016-07-04T15:38:34.797" Score="2" Body="&lt;p&gt;I don't recommend using the Hagen-Poiseuille equation as the flow is most probably turbulent judging from the given pressure difference. and since the flow rate will be mainly governed by the pressure difference between the boiler and the wand outlet (say ambient) I think @idkfa 's answer is enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if it you have a situation where the length of the pipe/friction losses are of significance, you can use the Unwin-Babcock equation which is an extension of the Darcy-Weisbach equation to be applicable for steam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \Delta P = \frac{0.0001306\,QL}{d^3\rho _ {steam}} (1 + \frac{3.6}{d})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta P$ is the pressure drop in psi, $Q$ is the volumetric flow rate, $L$ is the pipe length, $d$ is the pipe diameter and $\rho _{steam}$ is the density of the steam. However, the equation is not always guaranteed to give accurate result for all cases of steam flow so you should use the Moody chart alongside (giving it the upper hand as a judge of the pressure drop result).&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2016-07-04T15:38:34.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10571" PostTypeId="2" ParentId="10505" CreationDate="2016-07-04T19:38:34.207" Score="2" Body="&lt;h1&gt;Quick Answer&lt;/h1&gt;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; in the PI controller equation would best be described as a &amp;quot;weighting factor&amp;quot; that determines the influence of the integral term relative to the proportional term. The larger &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; gets, the more the integrated error affects the output of the controller compared to the instantaneous error.&lt;/p&gt;&#xA;&lt;h1&gt;More explanation&lt;/h1&gt;&#xA;&lt;p&gt;Often PI controllers are expressed in the following form:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$u(t) = K e(t) + K_I \int e(t)dt$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; is called the proportional gain, and &lt;span class=&quot;math-container&quot;&gt;$K_I$&lt;/span&gt; is called the integral gain. You tune the controller by changing &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K_I$&lt;/span&gt; until the system behaves the way you want.&lt;/p&gt;&#xA;&lt;p&gt;The equation given in the question expresses &lt;span class=&quot;math-container&quot;&gt;$K_I$&lt;/span&gt; as a multiple of &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;, i.e. &lt;span class=&quot;math-container&quot;&gt;$K_I = RK$&lt;/span&gt;. Therefore if you substitute it back into the controller equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$u(t) = K e(t) + RK \int e(t)dt = K (e(t) + R \int e(t)dt)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, instead of tuning &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K_I$&lt;/span&gt; separately, you can look at the process as tuning the proportional gain &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;, then deciding how much influence the integral term has by adjusting &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;. This way of expressing the gains might be useful depending on how you approach the problem of tuning the controller. Really it's just mathematical semantics, you can use whichever form makes more sense to you.&lt;/p&gt;&#xA;&lt;h1&gt;Aside&lt;/h1&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; has units of &lt;span class=&quot;math-container&quot;&gt;$\text{s}^{-1}$&lt;/span&gt; (inverse seconds) in order to make the equation have consistent units. For example if &lt;span class=&quot;math-container&quot;&gt;$e(t)$&lt;/span&gt; is a distance, then &lt;span class=&quot;math-container&quot;&gt;$\int e(t) dt$&lt;/span&gt; has units of distance * time. Therefore, &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; must have units of 1/time in order to cancel the time in the integral term.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-07-04T19:38:34.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10572" PostTypeId="2" ParentId="10541" CreationDate="2016-07-04T20:01:21.993" Score="0" Body="&lt;p&gt;F= m.a, it all depends on what acceleration you pull it with. So 20k/1500= 13.33 ft/s^2 roughly 40% of gravity acceleration. Use 60% of that as a conservative number. You could pull hypothetically unlimited load with any winch no matter how small the strength as long as the load is not locked in place, as dislodgement force could be very high, and your winch has spring loaded bracket properly adjusted to break the start jerking. Pulling is different than lifting! You deal more with friction than weight.&lt;br&gt;&#xA;As an example NASA's 450 ton space station can be dragged to different orbit( called a boost) or to correct position for docking with the Shuttle by tiny rockets with a thrust of something in the order of hundreds of lb. force. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-07-04T20:01:21.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10575" PostTypeId="2" ParentId="10561" CreationDate="2016-07-05T05:23:33.780" Score="0" Body="&lt;p&gt;If practical rolling the strap sample a few times around a double cylinder mechanism which are designed to lock into each other with tension similar to seat belt or the old military uniform service belts would be the best setting.&lt;br&gt;&#xA;This clamp on elastic material would not perform evenly, it will allow more slip on the ends. So it has to be installed far enough to allow for redistribution of strain and has to be calibrated for establishment of the origin datum after loading.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-07-05T05:23:33.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10577" PostTypeId="1" CreationDate="2016-07-05T10:06:41.793" Score="1" ViewCount="185" Body="&lt;p&gt;I have slightly inclined (50 cm over 6 meters) metal roof over my studio. The roof has bad thermal insluation -- 15 cm of mineral wool between sheet metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The location is Kiev, Ukraine. Ambient temperatures extremely rarely exceed 32 °C, and most of the activities in the studio occur after 6 p.m., when it's never over 30 °C. I lived through a year without AC, with only fans for air intake and circulation.&lt;br&gt;&#xA;With ventilation turned off the hottest time inside is 7 to 8 p.m.. With ventilation on, and 20 people inside the temperature inside is 1-2 °C more then ambient air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to cool the inside for a few degrees by pouring a water on the roof. &#xA;My questions are:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can I hope for outside roof surface to get as low as a wet-bulb&#xA;temperature? &lt;a href=&quot;http://erml.org/oct/images/b/stud.id-dia.gif&quot; rel=&quot;nofollow&quot;&gt;The graph&lt;/a&gt; gives me 25 °C at warmest hour of&#xA;ambient 32 °C 55 % humidity.   &lt;/li&gt;&#xA;&lt;li&gt;Which method of water distribution is more effective:&lt;br&gt;&#xA;  a) Drip tube. Would be easiest to install. The drawback being the risk of water joining into a few streams instead of uniformly wetting the surface.&lt;br&gt;&#xA;  b) Misters.&lt;br&gt;&#xA;  c) Sprinklers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5053" LastEditorUserId="5053" LastEditDate="2016-07-05T11:37:06.947" LastActivityDate="2016-08-05T05:23:00.397" Title="Evaporative roof cooling" Tags="|thermodynamics|hvac|evaporation|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10578" PostTypeId="2" ParentId="10577" CreationDate="2016-07-05T11:52:04.773" Score="1" Body="&lt;p&gt;What is the general range of relative humidity?   You won't get anywhere unless the air is reasonably dry.&lt;br&gt;&#xA;Next, you want to maximize the evaporative area or most of the water will just roll off the roof. Essentially, you are looking at turning your roof into a radiator aka  heat exchanger.  This will involve bonding a lot of finstock or evaporative towers to the existing structure.  And even if you do that, the existing insulation will hamper the flow of heat inside the room to the cooled roof layer.&lt;br&gt;&#xA;All in all, you're probably going to be much better off with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Evaporative_cooler&quot; rel=&quot;nofollow&quot;&gt;swamp cooler&lt;/a&gt; set up with a fan just outside a window. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-07-05T11:52:04.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10579" PostTypeId="2" ParentId="10560" CreationDate="2016-07-05T12:05:33.473" Score="2" Body="&lt;p&gt;&lt;strong&gt;GPIO&lt;/strong&gt; stands for &lt;strong&gt;G&lt;/strong&gt;eneral &lt;strong&gt;P&lt;/strong&gt;urpose &lt;strong&gt;I&lt;/strong&gt;nput &lt;strong&gt;O&lt;/strong&gt;utput. Lets assume you want to generate a pulse at pin 39 as in the below image. So when you wire the pulse to pin 39 the underline software will generate the necessary code to configure pin 39 to be an output pin enabling the pin to output a pulse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aU0rS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aU0rS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the code stack between Simulink and Microcontroller. So when you create a Digital Output block the software will essentially convert and configure your Digital Output block into a Digital output pin. The code will convert into binary code to be able to loaded into the micro controller. Below is a high level diagram that explains the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lJOWo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lJOWo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://focus.ti.com/download/tidc/pdfs/asia/2008/tw_T2_2_Joey_Chen_resized.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;C2000 Digital Signal Controllers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.mathworks.com/matlabcentral/fileexchange/54929-embedded-coder-target-for-energia/content/html/TIVAgettingstarted.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Embedded Coder Target for Energia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-07-05T13:07:48.853" LastActivityDate="2016-07-05T13:07:48.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10581" PostTypeId="2" ParentId="10440" CreationDate="2016-07-05T17:56:36.933" Score="0" Body="&lt;p&gt;It seems to me the function of the mouthpiece is to allow the flow to fully develop again. Assuming the length is exactly enough for the flow to be fully developed, then increasing the length will have no impact on the flow and thus the discharge coefficient. Shortening the mouth piece will result in a underdeveloped flow and the discharge coefficient will change, with the limiting value being the case where no mouthpiece is used.  &lt;/p&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2016-07-05T17:56:36.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10582" PostTypeId="1" AcceptedAnswerId="18268" CreationDate="2016-07-05T21:10:10.850" Score="10" ViewCount="8532" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&lt;p&gt;The Bailey Bridge panel bridge system has been around since WWII.  It has been credited as being one of the top 3 technological advancements that allowed the allies to win the war.  After the war, the surplus of Bailey Bridge parts were sold off and distributed around the world.  While these bridges were intended to be for &lt;em&gt;temporary&lt;/em&gt; use (military temporary use and civilian temporary use are two different things), some of these old bridges can still be found in use today.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yIZO3.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yIZO3.jpg&quot; alt=&quot;Bailey schematic&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BflbI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BflbI.png&quot; alt=&quot;Bailey Schematic with tapered transom&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;A quick search on google will return all sorts of beautiful modern bridges that are not Bailey Bridges.  To me, a Bailey Bridge is a very specific type of panel bridge from WWII or post war sold by Bailey-Uniflote.  However listening to colleagues, and even looking at the google search results, it appears that Bailey Bridge is just another word for Panel Bridge.  Similar to Xerox and Photocopy, or Kleenex and tissues.  So this has muddy some of my search efforts thus far.&lt;/p&gt;&#xA;&lt;h2&gt;The Problem&lt;/h2&gt;&#xA;&lt;p&gt;The problem I face is trying to determine which of the 4 (maybe more) types of panel bridge systems I may be looking at as they can be very similar in component layout at a quick glance and even some basic measurements.&lt;/p&gt;&#xA;&lt;p&gt;So far my research has told me that some of the distinguishing factors for these panel type bridges are:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Panel size&lt;/li&gt;&#xA;&lt;li&gt;Panel diagonal members (I, C, HSS/TUBE)&lt;/li&gt;&#xA;&lt;li&gt;Deck width&lt;/li&gt;&#xA;&lt;li&gt;Transom location&lt;/li&gt;&#xA;&lt;li&gt;Transom size&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Panel Size&lt;/h2&gt;&#xA;&lt;p&gt;The original Bailey Bridge had a panel length of 10'0&amp;quot; (3.048 mm) from centres of connecting pin holes and a height of 4'9&amp;quot; (1.448 mm) from centres of connecting pin holes.  The overall height was 5'1&amp;quot;.  Anything outside of these dimension would indicate a newer panel system such as an Acrow 700xs or Mabey Compact 200 or the like.&lt;/p&gt;&#xA;&lt;h2&gt;Diagonal Members&lt;/h2&gt;&#xA;&lt;p&gt;In the Bailey Bridge panel there are two diamond shapes.  The diagonals of these shapes were originally I sections.  At some later time these sections changed to C-Channels and rumour has it HSS or tube sections were used by certain manufactures.  Anyone have any confirmation on how to identify or rule out potential bridge type based diagonal section type?  Maybe even some date ranges for the various section type?&lt;/p&gt;&#xA;&lt;h2&gt;Deck width&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The original Bailey Bridge has a roadway width of _____ and a clear width between trusses of ____.&lt;/li&gt;&#xA;&lt;li&gt;The standard widened Bailey Bridge has a roadway width of 10'9&amp;quot; (3.28 m) and a clear width between trusses of 14'3&amp;quot; (4.34 m).&lt;/li&gt;&#xA;&lt;li&gt;The Extra Wide Bailey Bridge has a roadway width of 13'9&amp;quot; (4.19 m) and a clear width between trusses of 15'8&amp;quot; (4.77m).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I am not sure what the standard deck width is on an Acrow 300 or a Mabey 100.&lt;/p&gt;&#xA;&lt;h2&gt;Transom Location&lt;/h2&gt;&#xA;&lt;p&gt;On a Bailey Bridge, the transoms are located adjacent to the panel's end vertical and middle vertical.  This can been seen in the following diagram and photo:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hfwaJ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfwaJ.png&quot; alt=&quot;Bailey Transom Location Diagram&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gW1AC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gW1AC.png&quot; alt=&quot;Bailey Single Transom Location Photo&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;If the transoms need to be double up to carry a heavier load then there are 4 transoms per bay.  Two either side of the middle vertical and 1 inside of each end vertical as shown in the following photo:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JvA5E.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JvA5E.png&quot; alt=&quot;Bailey Double Transom Location Photo&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;For an Acrow 300 style panel bridge, the transom is located at the base of panel's diamond as seen in the following photo:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wstre.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wstre.png&quot; alt=&quot;Acrow 300 Bridge Transom Location&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Mabey 100 transoms are located ?????&lt;/p&gt;&#xA;&lt;h2&gt;Transom Size&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A standard Bailey Bridge has a transom length of 18'0&amp;quot; (5.49 m) with a constant 10&amp;quot; depth.  There will be 3 holes near the bottom flange to allow for 3 panel lines.&lt;/li&gt;&#xA;&lt;li&gt;A Standard Widened Bailey Bridge has a transom length of 19'11&amp;quot; (6.1 m) with a 12 in depth between under the deck and a 10&amp;quot; depth at the panel lines.  There will be 4 holes near the bottom flange to allow for 4 panel lines.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hv3b6.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hv3b6.jpg&quot; alt=&quot;Wide DD Bailey Transom&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An Extra Wide Bailey Bridge has a transom length of 19'11&amp;quot; (6.1 m) with a 12 in depth between under the deck and a 10&amp;quot; depth at the panel lines.  There will be 3 holes near the bottom flange to allow for 3 panel lines.&lt;/li&gt;&#xA;&lt;li&gt;It also appears that transom with a tapered flange also exist while maintaining a uniform section depth&lt;/li&gt;&#xA;&lt;li&gt;There is also the M2 and M3 Transoms identified in the US Bailey Bridge Field Manual&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Acrow 300 and Mabey 100 transoms are?????&lt;/p&gt;&#xA;&lt;h2&gt;SUMMARY&lt;/h2&gt;&#xA;&lt;p&gt;What are the key factors that distinguish Bailey Bridges?&lt;/p&gt;&#xA;&lt;p&gt;So far I can only tell Acrow 300 by its transom position.&lt;/p&gt;&#xA;&lt;p&gt;What about:&lt;/p&gt;&#xA;&lt;p&gt;UK Bailey&lt;br /&gt;&#xA;US Bailey&lt;br /&gt;&#xA;Mabey 100&lt;/p&gt;&#xA;&lt;p&gt;Just found this really &lt;a href=&quot;http://www.thinkdefence.co.uk/2012/01/uk-military-bridging-equipment-the-bailey-bridge/&quot; rel=&quot;noreferrer&quot;&gt;nice write up on the Bailey Bridge&lt;/a&gt; but it still does not completely answer my questions.&lt;/p&gt;&#xA;" OwnerUserId="6993" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2021-01-28T20:45:42.493" Title="Determining the difference between US Bailey Bridge, UK Bailey Bridge, Mabey Compact 100 and Acrow 300" Tags="|bridges|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10583" PostTypeId="1" CreationDate="2016-07-06T03:22:41.653" Score="1" ViewCount="178" Body="&lt;p&gt;I am holder of Engineering card, now I moving to USA and would like to avoid at least hustle about translating all diplomas and deal with transcripts. Engineering card issued by Feani have Bachelor/Master studies listed on it, is it enough for USA, Texas to be more precise?  As i understand I can't avoid EIT and PE test even if I am registered engineer in my country but avoiding transcripts would be nice start. &lt;/p&gt;&#xA;" OwnerUserId="7142" LastEditorUserId="61" LastEditDate="2016-07-06T05:48:39.857" LastActivityDate="2016-07-06T05:48:39.857" Title="Is Engineering card issued by Feani recognized in USA?" Tags="|civil-engineering|licensure|" AnswerCount="0" CommentCount="5" ClosedDate="2016-07-07T00:15:51.237" ContentLicense="CC BY-SA 3.0" />
  <row Id="10584" PostTypeId="1" AcceptedAnswerId="10616" CreationDate="2016-07-06T06:44:36.443" Score="2" ViewCount="10885" Body="&lt;p&gt;I have a pressure regulator bringing an inlet pressure of 150 bar down to 10 bar which is then lead into a plenum followed by a nozzle of area 340 mm^2, exposed to atmospheric conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weirdly enough at both points the necessary condition of pressure ratio for choking is satisfied!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://wikimedia.org/api/rest_v1/media/math/render/svg/7d4f2f098327a3e772615e3e2ba44bd4697d1ebc&quot; alt=&quot;pressure ratio&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical exit pressure below which choking occurs (p*), is 5.8 bar at the nozzle exit and 79.7 bar at the regulator exit. How do I explain the fact that the flow chokes only at the nozzle? &lt;/p&gt;&#xA;" OwnerUserId="6705" LastEditorUserId="6705" LastEditDate="2016-07-06T10:03:46.853" LastActivityDate="2016-07-13T14:08:34.067" Title="How to determine the choke point in a compressed gas system?" Tags="|fluid-mechanics|thermodynamics|aerospace-engineering|compressed-gases|compressible-flow|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10586" PostTypeId="2" ParentId="10498" CreationDate="2016-07-06T10:12:40.373" Score="2" Body="&lt;p&gt;Just adding on to Mr.Johns's answer, do try to make your threaded fittings according to NPT (National Pipe Thread) or BSPT (British Standard Pipe Taper). Both have a tapered profile which will significantly reduce leakages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Learned it the hard way here. All the best!&lt;/p&gt;&#xA;" OwnerUserId="6705" LastActivityDate="2016-07-06T10:12:40.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10587" PostTypeId="1" CreationDate="2016-07-06T12:38:32.073" Score="1" ViewCount="130" Body="&lt;p&gt;A laptop's antenna receives signals from a Wi-Fi router in two ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;signal falling directly on the antenna from the router, say 1&amp;nbsp;m away&lt;/li&gt;&#xA;&lt;li&gt;signal coming from same router but bouncing off the walls and traveling some distance, say 10&amp;nbsp;m.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore, at any point of time, the antenna receives multiple signals at the same frequency. How is the resulting interference taken care of?&lt;/p&gt;&#xA;" OwnerUserId="6614" LastEditorUserId="6614" LastEditDate="2016-07-06T19:58:19.470" LastActivityDate="2016-07-08T03:52:34.330" Title="Interference between direct signal and reflected signal" Tags="|electrical-engineering|wireless-communication|wifi|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10588" PostTypeId="1" CreationDate="2016-07-06T13:37:38.037" Score="3" ViewCount="614" Body="&lt;p&gt;I have a 3 m x 1 m x 1 m cuboid weighing approximately 10,000 kg. I am trying to resolve the vertical forces that will support the cuboid if the cuboid must rest at a 45 degree angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which of the two solutions below (if either) is best for calculating the resolving forces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases I assume a 2-D cross section of the cuboid where the resolving forces act at the bottom two corners. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The first method is to treat the cuboid as a beam and the weight as an continuously distributed load.&lt;/strong&gt; This results in an even distribution of the load between the two resolving forces being even irrespective of the angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The second method is to assume the weight to be a point load at the center of mass.&lt;/strong&gt; Then use the sine rule to determine the position the force acts along the base line of the beam. This gives an uneven distribution of the loads with weighting slightly towards the lower corner but seems to be dependent on thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reasonably confident it is the second but I want to know the mathematical explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qdpeA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qdpeA.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7149" LastEditorUserId="368" LastEditDate="2016-07-06T23:54:59.823" LastActivityDate="2016-07-11T20:02:43.707" Title="Does thickness impact resolving loads for an inclined beam?" Tags="|mechanical-engineering|statics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10589" PostTypeId="1" AcceptedAnswerId="10594" CreationDate="2016-07-06T14:02:42.530" Score="5" ViewCount="189" Body="&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Designing the behavior of controllers is an important task for many applications. Every application needs its own requirements fulfilled. For example a CNC machine needs a controller with a strict asymptotic convergence to the set value. An overshooting behavior can’t be tolerated due to the fact that once removed material can’t be replaced by the same tool. Other examples for devastating results with an overshooting controller behavior are autopilots landing an aircraft or shuttles docking the ISS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any real applications where a controller with an overshooting behavior can not only be tolerated but leads to an overall suitable performance for the system? I am looking for a specific example not just field of tasks where it might be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thoughts of my own&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are systems where an overshooting control behavior can be tolerated. Like if a robot needs to stay on a specific path on the way to its destination with a given tolerance. A controller that controls the distance to this given path can let the robot wiggle a little around the given trajectory as long as it keeps it inside the tolerance. But if you can design a controller with an asymptotic converge to its trajectory you would prefer it to get rid of this wiggle. I can’t think of any real application where you would prefer an overshooting controller over an asymptotic one if you had the choice.&lt;/p&gt;&#xA;" OwnerUserId="7055" LastEditorUserId="368" LastEditDate="2016-07-06T20:24:05.043" LastActivityDate="2016-07-07T13:35:01.813" Title="When are overshooting controllers preferred over asymptotic ones?" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10590" PostTypeId="1" AcceptedAnswerId="10591" CreationDate="2016-07-06T14:16:18.777" Score="0" ViewCount="731" Body="&lt;p&gt;What is the difference in temperature between a  surface in the sun as supposed to one in the shade and how is it calculated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance:&#xA;What will the  temperature be if a metal sheet lies in the sun and then what will its temperature be if the same sheet lies in a shaded area?&lt;/p&gt;&#xA;" OwnerUserId="7092" LastActivityDate="2016-07-06T14:29:08.807" Title="Surface in the sun vs. surface in the shade" Tags="|temperature|solar-energy|radiation|" AnswerCount="1" CommentCount="2" ClosedDate="2016-07-07T13:18:59.420" ContentLicense="CC BY-SA 3.0" />
  <row Id="10591" PostTypeId="2" ParentId="10590" CreationDate="2016-07-06T14:29:08.807" Score="0" Body="&lt;p&gt;As Dr. Bad Science Ben Goldacre says, &quot;I think you'll find it's a bit more complicated than that.&quot;  It's not hard to dig up solar spectral power density data - which depends heavily on latitude and season, of course -- but how a &quot;metal&quot; sheet behaves depends on tons of parameters.&lt;br&gt;&#xA;Different metals (elemental or mixtures) have different spectral absorptivities as well as specific heat (how much energy per unit volume per unit temperature change). Further, for example, a roughened surface may have a higher net absorption.&lt;br&gt;&#xA;In summary, you'll need a lot more specific info, including sheet thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the shade, any material will reach thermal equilibrium with the local atmosphere.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-07-06T14:29:08.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10592" PostTypeId="2" ParentId="10587" CreationDate="2016-07-06T14:36:14.060" Score="3" Body="&lt;p&gt;Wifi uses &lt;a href=&quot;https://en.wikipedia.org/wiki/Orthogonal_frequency-division_multiplexing&quot; rel=&quot;nofollow&quot;&gt;Orthogonal frequency-division multiplexing (OFDM)&lt;/a&gt;. OFDM is resilient to delay spreading which essentially is multipath RF signals or signal bouncing off walls per your question. Here is better explanation from &lt;a href=&quot;http://www.revolutionwifi.net/revolutionwifi/2015/3/how-ofdm-subcarriers-work&quot; rel=&quot;nofollow&quot;&gt;PHY Basics: How OFDM Subcarriers Work&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An advantage of OFDM is that by using a reduced symbol rate of 250,000 symbols per second the negative effects of multipath distortion are reduced. Since each symbol occupies more time, there is more resilience to delay spread which is caused by multipath when signal reflections cause multiple copies of the same transmitted symbol to arrive at the receiver at slightly different times. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suggest that you scan and read the references below for multipath and you will get will better reads to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://bwrcs.eecs.berkeley.edu/Classes/EE225C/Lectures/Lec16_ofdm.pdf&quot; rel=&quot;nofollow&quot;&gt;Introduction to ODFM&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://complextoreal.com/wp-content/uploads/2013/01/ofdm2.pdf&quot; rel=&quot;nofollow&quot;&gt;Orthogonal Frequency Division Multiplex&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eetimes.com/document.asp?doc_id=1225472&quot; rel=&quot;nofollow&quot;&gt;OFDM kills multipath distortion&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.radio-electronics.com/info/rf-technology-design/ofdm/ofdm-basics-tutorial.php&quot; rel=&quot;nofollow&quot;&gt;OFDM Orthogonal Frequency Division Multiplexing Tutorial&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-07-08T03:52:34.330" LastActivityDate="2016-07-08T03:52:34.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10594" PostTypeId="2" ParentId="10589" CreationDate="2016-07-06T21:04:37.043" Score="6" Body="&lt;p&gt;For some systems, the salient criterion is settling time to within some error band.  Sometimes you can get faster settling by allowing earlier overshoot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a system to get to within some minimum error of before starting a process, you probably don't care what the system is doing before the process is started, only that it gets there as quickly as possible since you're waiting until then.  It's not hard to come up with scenarios where settling time is important, and overshoot before settling largely irrelevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's say some process has to occur at a nominal 200 &amp;deg;C, &amp;plusmn;10 &amp;deg;C.  When you're powering on the oven, you care that it settles to the 190-210 &amp;deg;C range quickly.  It's going to get there faster than a asymptotic controller if allowed to overshoot to 210 &amp;deg;C, then oscillate with ever-decreasing amplitude around 200 &amp;deg;C.  The same holds true if you change the set point, and you can't do anything useful until it gets to within some tolerance of the new setpoint.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2016-07-07T13:35:01.813" LastActivityDate="2016-07-07T13:35:01.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10595" PostTypeId="1" CreationDate="2016-07-06T21:52:43.853" Score="5" ViewCount="1474" Body="&lt;p&gt;We have an extended Chevy express with a 24&quot; high-top, on which we intend to install roof-mounted PV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really concerned with how to properly install the panels. We've selected six 100 W Renogy panels and a 400 Ah LiFePO4 battery pack, but it might be too much for our roof. I'm worried that if we are going 80 mph up I-5 and hit a 40 mph cross-breeze we could tear the roof off the van, rip the panels off the roof, destabilizing and roll the vehicle, etc. How can I avoid or minimize these risks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was initially thinking of laying three panels long-ways on each side of the van, with a ~20&quot; section down the middle I can walk down and hang out or service the panels, and where the roof vent will go. Each section then would only be 22&quot; wide (140&quot; long) with space on both sides for air to dissipate to either side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My plan is to build my own rack with bent heavy conduit and bolts. The rack should be pretty strong. But could unwieldy pressures blow out the panels? I don't know what thresholds they should withstand. The product shows 2400 Pa high winds and 5400 snow, but what does that correlate to in driving down the road?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that with the current layout, the panels must go over each side by 4.5 inches. To me, this means they can be more affected by wind. If the roof was just flat instead of the bubble top, mounting might be much easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZR8eg.jpg&quot; rel=&quot;noreferrer&quot; title=&quot;Solar diagram V1&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZR8eg.jpg&quot; alt=&quot;&quot; title=&quot;Solar diagram V1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7152" LastEditorUserId="368" LastEditDate="2016-07-07T19:11:14.033" LastActivityDate="2021-06-25T04:26:25.913" Title="How can I add roof-mounted solar panels to a van without making it unstable at highway speeds?" Tags="|automotive-engineering|airflow|photovoltaics|" AnswerCount="4" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10596" PostTypeId="2" ParentId="10588" CreationDate="2016-07-06T23:02:43.940" Score="3" Body="&lt;p&gt;The more exact solution is the second one. Usually beams aren't very wide or high compared to their length, so if they are inclined, taking their self weight as a linear load is acceptable. As you mentioned, the distribution of the load depends on the height of your cross-section and on the angle of your &quot;cuboid.&quot; You can see this by making an extreme case where the right reaction needs to be reversed to hold up your cuboid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UVccQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UVccQ.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case it is clear that the linear load would not apply. If you want to resolve the reactions, simply do a summation of your vertical forces (only dead weight at the center of mass) and a summation of moments.&lt;/p&gt;&#xA;" OwnerUserId="5691" LastEditorUserId="368" LastEditDate="2016-07-06T23:49:48.847" LastActivityDate="2016-07-06T23:49:48.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10598" PostTypeId="1" CreationDate="2016-07-07T01:32:52.520" Score="1" ViewCount="76" Body="&lt;p&gt;I am currently looking into attaching a wire onto Silicon Nitride igniter, which is being used as a resistive heater. The problem I am experiencing is that the Silicon nitride is extremely hard to bond things on as is, and in addition, the bonding agent has to be able to tolerate temperatures up to 1K Celsius. I have already experimented with ceramic epoxy and found difficulty.&#xA;Alternatively, we have already considered using a silicon Carbide igniter, however, all commercial igniters have bases that are too big for our uses.&#xA;It is possible for us to use filaments and create a base, however, we would like to avoid the hassle if possible.&#xA;What would be a good way to do this?&lt;/p&gt;&#xA;" OwnerUserId="7156" LastActivityDate="2016-07-07T01:32:52.520" Title="What is a good way to bond stuff to silicon nitride?" Tags="|epoxy|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10600" PostTypeId="1" AcceptedAnswerId="10604" CreationDate="2016-07-07T05:31:12.577" Score="1" ViewCount="9804" Body="&lt;p&gt;I'm building a 3D scanning (photogrammetry) device with a rotating platform, a bunch of wheels and a stepper motor. What I want to know is how much weight that motor could rotate on the platform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently the platform weights just over 2kg and is just under 40cm wide (see the image below). I estimate that the objects I will be scanning will weigh no more than 5kg (just to be safe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the link to that servo motor says it has a 'stall torque' of 15kg - cm at 6V. I have no idea what that means. Is that enough? Will it be able to rotate the my platform?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R1qu2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R1qu2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7158" LastEditorUserId="110" LastEditDate="2016-07-07T11:09:13.257" LastActivityDate="2016-07-07T11:25:24.313" Title="How much weight can my stepper motor rotate?" Tags="|stepper-motor|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10601" PostTypeId="1" CreationDate="2016-07-07T08:40:47.183" Score="1" ViewCount="251" Body="&lt;p&gt;Found a model for a simulated centrifugal pump in the paper of Lewin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ac.els-cdn.com/096706619500015M/1-s2.0-096706619500015M-main.pdf?_tid=bb9b04d0-441b-11e6-89e2-00000aab0f6b&amp;amp;acdnat=1467879838_27ac1eebc37201611d63caf374d4736e&quot; rel=&quot;nofollow&quot;&gt;Lewin,1993&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question arises in the simulation study the author is doing on pages 419/420.&#xA;He assumes that the pressure of the pump is given by&#xA;$$P=30+k2*F^2,$$&#xA;where $F$ is the water flow and $k2$ a time-varying resistance paramter.&#xA;Is in such a model assumed that the flow $F$ is constant?&#xA;I was wondering because considering the plot of $P$ and $F$ in &lt;strong&gt;figure 4&lt;/strong&gt; it looks like the flow $F$ is varying with time (should be the line in the middle of the plot).&lt;/p&gt;&#xA;" OwnerUserId="7160" LastActivityDate="2016-07-07T08:40:47.183" Title="Is the flow in a centrifugal pump always constant?" Tags="|mechanical-engineering|pumps|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10602" PostTypeId="1" CreationDate="2016-07-07T10:00:00.107" Score="3" ViewCount="179" Body="&lt;p&gt;Today, we came up with the idea of creating a 3 dimensional matrix where I  simultaneously stagger 3 times 5 times 5 Fabry Perot band-reject interferometeric filters with ten degree wide bands of angle of incidence insensitivity , red, green or blue wavelength selectivity as well as optical path difference equal  0 degrees, 10 degrees, 20 degrees, and wavelength in arithmetic progression to cover the incidence angle range of 0 to 50 degrees, the spectrum from 450 nm to 750 nm and the optical path difference range of 0 to 40 microns.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;In the diagram below, &#xA;&#xA;   deltapsi_0 = 0 degrees of phase delay &#xA;   deltapsi_1 = 10 degrees of phase delay&#xA;   deltapsi_2 = 20 degrees of phase delay&#xA;   deltapsi_3 = 30 degrees of phase delay&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As described in Coutinho et al's paper &quot;Detection of Coherent Light in &#xA;an Incoherent Background&quot; with URL discovery.ucl.ac.uk/2670/1/2670.pdf, the purpose of the Fabry Perot interferometer is to measure the self-coherence function of the incident radiation to give an interferogram, and uses a narrow band filter to produce the  sinc envelope necessary to measure the change in position of the self-coherence function minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Coutinho et al's paper &quot;Detection of Coherent Light in &#xA;an Incoherent Background&quot;, the authors use an electro-mechanical scanning Michelson interferometer to measure the self-coherence function of the incident radiation. However, the use of an electro-mechanical scanning Michelson interferometer limits the tuning speed required to protect pilots from incoming laser pointer rays within an 100 microsecond interval. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in 1999,  Brazilian Navy Officer Coutinho and IEEE Fellow Griffith used a Fourier Transform to measure the position of the first minima of the self coherence function. The disadvantages of the Fourier transform are that it requires the processing of a large amount of data and a reference laser is required to calibrate the path difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subsequently, in 2007, Brazilian Navy Officer  Coutinho and IEEE Fellow Griffith got rid of the Fourier transform by using Interferogram Phase Step Shift(IPSS) where an interferometer measures self-coherence in the vicinity of the first mimimum with a much shorter scan interval and a signal processing algorithm to extract the position of the phase step function discontinuity in the fringe carrier, proportional to self-coherence. The most problematic aspect of this is that it requires preknowledge of the coherent laser's spectral characteristics within a narrowband width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 3 dimensional matrix block diagram shown below, we simultanously measure the self coherence function amplitude at 5 different optical pathlength  differences and curvefit a quintic polynomial function representing the self coherence function's sinc envelope for the purpose of measuring the change in position of the self-coherence function minimum. By doing the operation simulatenously and in parallel we hope to speed up the tuning speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@hsinghal pointed out to me that as far as Fabry-Perot filters are concerned they are highly sensitive on the incident angle and wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below, I show the formulas for the dependence of the Fabry-Perot filter's reflectivity and transmitivitty coefficients on the angle of incidence courtesy of the European Journal of Physics authors Juan J Monzón and Luis L Sánchez-Soto&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uImSk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uImSk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am hoping to apply Dong Guoyan,et al.'s ideas which the authors wrote in this 2013 paper &quot;Zero phase delay with relax incident condition in photonic crystals(PhC)&quot;. with the URL &lt;a href=&quot;https://www.osapublishing.org/oe/viewmedia.cfm?uri=oe-21-24-29860&amp;amp;seq=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.osapublishing.org/oe/viewmedia.cfm?uri=oe-21-24-29860&amp;amp;seq=0&lt;/a&gt; that &quot;Since the PhC structure can be engineered easily and have more design flexibility, it is convenient to manipulate the frequency and direction of EMW propagating in PhC with zero phase delay by changing the parameters of PhC configuration or the incident angle. Due to the prosperous nanofabrication technology, this method  for  engineered  control  of  the  phase  delay  in  PhCs  may  have  significant  potential  applications  in  chip–scale  transmission  lines,  information  processing  and  the  design  of  new  type optical devices&quot; to building 10 degree phase delay devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to use the transparent Indium Tin Oxide thin film multilayer structure from Christopher J.Stolz's paper Improved ITO Laser Resistance by imbedding within a Fabry Perot filter with this URL [&lt;a href=&quot;https://www.dropbox.com/s/9h99qzwpm9z6sxd/Scans%207.4.16.pdf?dl=0]&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.dropbox.com/s/9h99qzwpm9z6sxd/Scans%207.4.16.pdf?dl=0]&lt;/a&gt;. Please remember that you have to create or use a free Dropbox account to view this paper. The reason being is that we wish to deploy such a filter above the cockpit window is to increase resistance to laser damage and minimize the angle of incidence effect where the band rejection filter spectra is shifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing the laser resistance will reduce the life-cycle costs of installing a filter to block red, green or blue laser pointers multiple times over a 6 month period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read this article, [&lt;a href=&quot;https://news.slashdot.org/story/16/07/05/2230241/japan-says-yes-to-mirrorless-cars]&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://news.slashdot.org/story/16/07/05/2230241/japan-says-yes-to-mirrorless-cars]&lt;/a&gt;, Japan Says Yes to Mirrorless Cars, you realize that in a few years Japan will also say yes to planes with Planar(@TM) video displays. As a result, this topology can be installed  outside the airplane cockpit window instead of inside the interior cavity of a Boeing 757 cockpit window with 6 inch thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is whether I can use this geometrical arrangement of optical devices to band-reject coherent laser light in an incoherent background with 30 db attenuation and 10 nanometer effective notch bandwidth while &#xA;  1) preserving the out-of-band field of view for pilots within 100 meters of the airport &#xA;  2) provide wide angle of incidence coverage from 0 degrees to 50 degrees.&#xA;  3) generating an output image in less than 100 microsecond time &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How difficult would it be to fabricate such an 3 dimensional matrix&#xA;to meet the global demand from the aviation industry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might we improve upon the weaknesses in this proposed design?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because we are sampling the interferogram  at 5 unique values of optical path difference , we do not need to find a Fabry Perot  scanning interferometers with 100 microsecond tuning speed. Please note that our 3 dimensional array of Fabry Perots and optical phase delay elements can simultaneously handle red , green and blue wavelengths stretching from 450 to 750 nm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l7tTy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l7tTy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/UFAJX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UFAJX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="2651" LastEditDate="2016-07-18T10:08:14.267" LastActivityDate="2016-07-18T10:08:14.267" Title="3D matrix for staggering 5 ten degree wide bands of angle of incidence shifted by target wavelength and optical path difference in arithmetic series" Tags="|optics|aircraft-design|electronic-filters|" AnswerCount="0" CommentCount="18" ContentLicense="CC BY-SA 3.0" />
  <row Id="10604" PostTypeId="2" ParentId="10600" CreationDate="2016-07-07T11:25:24.313" Score="3" Body="&lt;p&gt;The limit on the stepper motor is the torque it can produce.  This has nothing to do with the weight it is rotating, except that the weight may cause more friction, which the motor has to overcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass does matter somewhat, but only in a rotational sense, and only to determine how fast the motor's torque will speed up the rotation.  The proper term for this &quot;rotational mass&quot; is &lt;i&gt;moment of inertia&lt;/i&gt;.  The torque divided by the moment of inertia about the rotating axis gives you the angular acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another limit is what the motor bearings can handle.  The motor datasheet should give you the maximum axial and radial forces the motor is rated for.  In your case, the axial force is the weight of the platform and everything on it, minus the weight taken up by the wheels.  That may be hard to determine, since it's not clear how much of the weight is supported by the wheels and how much by the motor shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end in this case, the limiting factors will be the friction and therefore the torque required just to get the platform rotating at all, and how fast you want to get it up to speed.  The former depends on too many things that are hard to know up front.  Build it and see, then get a motor with enough torque to reliably overcome the friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also use gears or a belt drive to effectively turn a fast but low-torque motor into a slow but high-torque motor.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-07-07T11:25:24.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10605" PostTypeId="1" CreationDate="2016-07-07T12:08:11.423" Score="0" ViewCount="113" Body="&lt;p&gt;I'm trying to model subcritical flow in an open channel with a side weir of certain small length at the middle of the main channel, so a certain amount of fluid discharges through the weir. I need to match the two parameters - Froude number and flow depth just &lt;em&gt;at the beginning of the side weir&lt;/em&gt; with experimentally measured values.&#xA;I tried modelling with upstream depth of water and velocity corresponding to required Froude number upstream of weir, at the inlet, which was kept at about 10m from the weir. However the flow depth reduces significantly by the time it reaches the weir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I accurately model to obtain the required values of flow depth and Froude number upstream near the side weir?&lt;/p&gt;&#xA;" OwnerUserId="7163" LastActivityDate="2016-12-06T03:43:16.907" Title="How to model subcritical open channel flow?" Tags="|simulation|cfd|open-channel-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10606" PostTypeId="2" ParentId="1987" CreationDate="2016-07-07T13:09:08.297" Score="2" Body="&lt;p&gt;The critical parameters depends on the application. Beside Low Power, other critical parameters need to be considered. These may include code size, serial communication ports etc. Below is analysis that was preformed to choose MSP430 microcontroller. For this particular application active mode and standby mode power consumption was critical, together with the number of serial ports, and storage size.  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;╔══════════════╦═══════╦═══════╦═════╦═════════╦══════════════════════╦══════╦══════╦══════╦══════╦════════╦══════╦════════╦══════╗&#xA;║ Part Number  ║ Cost  ║ Flash ║ Ram ║ USB Ram ║ Active Mode (uA/MHz) ║ LPM0 ║ LPM1 ║ LPM2 ║ LPM3 ║ LPM3.5 ║ LPM4 ║ LPM4.5 ║ UART ║&#xA;╠══════════════╬═══════╬═══════╩═════╩═════════╬══════╦════════╦══════╬══════╩══════╩══════╩══════╩════════╩══════╩════════╩══════╣&#xA;║              ║       ║          (KB)         ║      ║ Flash  ║ Ram  ║                             uA                            ║&#xA;╠══════════════╬═══════╬═══════╦═════╦═════════╬══════╬════════╬══════╬══════╦══════╦══════╦══════╦════════╦══════╦════════╦══════╣&#xA;║ MSP430F5528  ║ $3.55 ║  128  ║  8  ║    2    ║  290 ║   460  ║  260 ║  83  ║      ║   7  ║  1.6 ║        ║  1.3 ║  0.18  ║   2  ║&#xA;╠══════════════╬═══════╬═══════╬═════╬═════════╬══════╬════════╬══════╬══════╬══════╬══════╬══════╬════════╬══════╬════════╬══════╣&#xA;║ MSP430FR5737 ║ $1.55 ║  15.5 ║  1  ║         ║  81  ║   270  ║  200 ║  175 ║      ║  61  ║  6.3 ║        ║  5.9 ║  0.32  ║   2  ║&#xA;╠══════════════╬═══════╬═══════╬═════╬═════════╬══════╬════════╬══════╬══════╬══════╬══════╬══════╬════════╬══════╬════════╬══════╣&#xA;║ MSP430FR5969 ║ $2.35 ║   63  ║  2  ║         ║  103 ║   270  ║  130 ║  80  ║  35  ║  0.9 ║  0.6 ║   0.4  ║  0.5 ║   0.2  ║   2  ║&#xA;╚══════════════╩═══════╩═══════╩═════╩═════════╩══════╩════════╩══════╩══════╩══════╩══════╩══════╩════════╩══════╩════════╩══════╝&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;MSP430FR5969 has low LPM3(Standby mode) current and low Active mode current, but has less storage space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y3G1k.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y3G1k.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/product/MSP430F5529&quot; rel=&quot;nofollow noreferrer&quot;&gt;MSP430F5529 16-Bit Ultra-Low-Power Microcontroller&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/product/MSP430FR5737&quot; rel=&quot;nofollow noreferrer&quot;&gt;MSP430FR5737 24 MHz ULP microcontroller&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/product/MSP430FR5969&quot; rel=&quot;nofollow noreferrer&quot;&gt;MSP430FR5969 16 MHz Ultra-Low-Power Microcontroller&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://software-dl.ti.com/trainingTTO/trainingTTO_public_sw/MSP430_LaunchPad_Workshop/v4/Chapters/MSP430m07_LPM.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;MSP430 Low Power Optimization&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.rs-online.com/designspark/electronics/eng/blog/who-has-the-lowest-power-mcu&quot; rel=&quot;nofollow noreferrer&quot;&gt;Who has the lowest power MCU?&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eembc.org/ulpbench/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ULPBench™ Scores&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/32653/how-to-choose-a-low-power-microcontroller&quot;&gt;How to choose a low power Microcontroller?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.silabs.com/Support%20Documents/TechnicalDocs/picking-the-right-microcontroller-based-on-low-power-specs.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to Pick the Right Microcontroller Based on Low-Power&#xA;Specifications&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.embedded.com/design/mcus-processors-and-socs/4413368/Choosing-low-power-microcontrollers-by-the-numbers&quot; rel=&quot;nofollow noreferrer&quot;&gt;Choosing low-power microcontrollers by the numbers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.tablesgenerator.com/text_tables&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tables Generator&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2016-12-04T02:25:25.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10607" PostTypeId="2" ParentId="1987" CreationDate="2016-07-07T13:52:03.220" Score="1" Body="&lt;p&gt;When low power consumption is important, you have to look at the whole picture.  A microcontroller with low Joules/cycle is only one thing to look at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good start would be to pick a battery and micro combination where the micro can run directly from the battery over the useful voltage range of the battery.  Many PIC microcontroller, for example, can run from 1.8 or 2.0 V up to 5.5 V.  That nicely covers a single lithium ion cell, two or three primary &quot;1.5 V&quot; cells in series, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important strategy is to let the micro sleep as much as possible.  If you only need to check something once a second, then maybe you can have the micro wake up for 1 ms every second.  That means you need to look at sleep current carefully.  It also means you want one with a built-in RC oscillator to minimize startup time, assuming a few percent clock rate accuracy is good enough.  There are lots of tradeoffs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You didn't give enough information to have some idea of the computing power required, so we can't say what would be adequate.  But in general, take a look at any Microchip PIC with the &quot;XLP&quot; (extra low power) feature.  In particular the 16F1xxx series has some impressively low power devices.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-07-07T13:52:03.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10609" PostTypeId="1" AcceptedAnswerId="10611" CreationDate="2016-07-07T17:12:18.673" Score="1" ViewCount="238" Body="&lt;p&gt;In civil engineering calculation I have seen weight of IPE 300 profile expressed in kN/m&lt;sup&gt;2&lt;/sup&gt; unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IPE  300 0.42 kN/m&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know IPE 300 is approximately 42.4 kg/m. How did they calculated that value expressed in kN/m&lt;sup&gt;2&lt;/sup&gt;?&lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="1832" LastEditDate="2016-07-07T18:21:05.397" LastActivityDate="2016-07-07T18:21:05.397" Title="IPE Beam weight" Tags="|mechanical-engineering|civil-engineering|beam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10610" PostTypeId="2" ParentId="10595" CreationDate="2016-07-07T17:59:25.797" Score="0" Body="&lt;p&gt;I would suggest building a support structure on the interior under the roof to help root the panels as well as distribute any force from driving more evenly. to help prevent air pockets and high upward pressure i would angle the panels with the front part being on the downward slope. this will increase drag but you should not have to worry about the panels or roof ripping off. also consider cutting off the bubble top and welding a flat piece of metal.&lt;/p&gt;&#xA;" OwnerUserId="6158" LastActivityDate="2016-07-07T17:59:25.797" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10611" PostTypeId="2" ParentId="10609" CreationDate="2016-07-07T18:17:47.880" Score="3" Body="&lt;p&gt;The density of structural steel is usually adopted as 7850&amp;nbsp;kg/m&lt;sup&gt;3&lt;/sup&gt; (or thereabouts). According to &lt;a href=&quot;http://www.b2bmetal.eu/i-sections-ipe-specification&quot; rel=&quot;nofollow&quot;&gt;this source&lt;/a&gt;, the cross-sectional area of an IPE 300 section is 53.8&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;. The linear mass of such a section is therefore 42.2&amp;nbsp;kg/m, which converts to a linear weight of 0.422&amp;nbsp;kN/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;m &amp;amp;= 42.2\ \text{kg/m} \\&#xA;w &amp;amp;= mg \\&#xA;&amp;amp;= 42.2 * 10 = 422\ \text{N/m} = 0.422\ \text{kN/m}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've seen anyone use 0.42&amp;nbsp;kN/m&lt;sup&gt;2&lt;/sup&gt;, then you've seen someone doing it wrong and not paying attention to dimensional analysis.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-07-07T18:17:47.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10612" PostTypeId="1" AcceptedAnswerId="10629" CreationDate="2016-07-07T18:19:33.907" Score="0" ViewCount="204" Body="&lt;p&gt;I got the Ikea SKARSTA desk &lt;a href=&quot;http://www.ikea.com/us/en/catalog/products/S49084965/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ikea.com/us/en/catalog/products/S49084965/&lt;/a&gt;, it's a sit stand desk that uses a manual shaft to rise and lower the desk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BuzIv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BuzIv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I plan to make the desk motorized using a satellite dish motor and a satellite positionner to control the motor (I removed the linear actuator part and kept only the motor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LUelV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LUelV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QfWsf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QfWsf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I'm facing is that the space between shaft and tabletop is about 1cm, so I'm not sure how can I transmit the rotation from the motor to the shaft (which is simply a long Allen key).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bdE2j.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bdE2j.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked into pulley systems and cogs but couldn't find suitable pieces, please advise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update Jul 9:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k2mDW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k2mDW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is how to transmit the rotation from the motor (green) to shaft (orange), motor can be placed at any position relative to table top but shaft isn't.&lt;/p&gt;&#xA;" OwnerUserId="7167" LastEditorUserId="7167" LastEditDate="2016-07-09T00:55:34.007" LastActivityDate="2016-07-09T00:55:34.007" Title="Transmit rotation in a tight space" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10613" PostTypeId="1" AcceptedAnswerId="10622" CreationDate="2016-07-07T18:47:41.170" Score="0" ViewCount="61" Body="&lt;p&gt;I have a 2&amp;nbsp;mm piece of UV transparent acrylic sheet. The datasheet defines it as 80% transparent at 300&amp;nbsp;nm. However, when I use it with different temperatures (4-50°C), the transmittance characteristics are totally different. It is much more transparent at higher temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know if this is true - that the light transmittance of acrylic varies with temperature? And what might be the reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific, I am interested in the 250-600&amp;nbsp;nm wavelength range.&lt;/p&gt;&#xA;" OwnerUserId="7168" LastEditorUserId="1832" LastEditDate="2016-07-07T20:31:19.223" LastActivityDate="2016-08-07T14:00:32.913" Title="Light transmittance of acrylic with temperature" Tags="|electrical-engineering|materials|optics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10614" PostTypeId="2" ParentId="10595" CreationDate="2016-07-07T19:09:46.850" Score="1" Body="&lt;p&gt;There are a few different issues here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly is the integrity of the frame itself. As well as the loads from the weight of the panels and aerodynamic drag it will be subject to fairly significant vibration and weather exposure over it's life so corrosion and fatigue are potential concerns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing is establishing decent mounting points to the vehicle. Ideally  you want to be bolting into structural members in the body of the vehicle rather than just single skin panels, even if this involves fabricating brackets to link the rack mounting points to something solid. Also the top section of extra high roof vans are often fibreglass or moulded plastic so you may well need to add an internal frame to extend the load down to structurally rigid parts of the vehicle. If you are lucky you may find unused seat-belt mounting points (if the same body is used for a people-carrier) which can be used. Otherwise you may have to drill into a pillar and back it with a plate and captive nut. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final consideration is aerodynamic forces. Here the main concern is to avoid generating lift by getting airflow underneath the panels. A simple spoiler to direct airflow over the top of the rack will help a lot here but ideally you would want to have the whole thing sealed all the way around and perhaps have it slightly higher at the back. If you can create a smooth blend between the panels and the roof of the van you should avoid any serious issues. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-07-09T21:58:35.783" LastActivityDate="2016-07-09T21:58:35.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10615" PostTypeId="1" CreationDate="2016-07-07T19:37:54.410" Score="1" ViewCount="21" Body="&lt;p&gt;I am trying to do a differential surveying project with some outdated equipment, specifically:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An old AllegroCX personal handheld computer (Win 4.1) with SurvCE loaded on it to collect data and establish communication &lt;/li&gt;&#xA;&lt;li&gt;Two Novatel ProPak V3 Recievers (one for base, one for rover)&lt;/li&gt;&#xA;&lt;li&gt;Two external radio communicators&lt;/li&gt;&#xA;&lt;li&gt;Two pinwheel style antennas&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am able to establish a connection with the rover's Novatel reviever and AllegroCX using a serial cable, but I cannot get it to find the base receiver using radio communications. Is this possible wirelessly, or do I need to set up the base station via serial cable as well? Do I need to have a PC connected to the base receiver?&lt;/p&gt;&#xA;" OwnerUserId="7169" LastEditorUserId="1832" LastEditDate="2016-07-07T20:28:57.727" LastActivityDate="2016-07-07T20:28:57.727" Title="Establishing DGPS Communications With SurvCE and Novatel Recievers" Tags="|surveying|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10616" PostTypeId="2" ParentId="10584" CreationDate="2016-07-08T02:54:00.620" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How do I explain the fact that the flow chokes only at the nozzle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To explain why the flow only chokes at a single point, it is important to remember that for a system at steady state, there can only be a single flow.  All of the fluid entering the system must leave.  Since there can only be a single flow rate through the system, you must then consider where that maximum flow can occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The following equations describe the flow through a frictionless nozzle where the expansion occurs adiabatically and isenthropically.  They are from Perry's page 6-23.  The actual flow through an orifice is usually handled by a flow coefficient since the flow through an orifice will be less than a frictionless nozzle.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\frac{p^*}{p_0} &amp;amp;= \left(\frac{2}{k+1}\right)^{k/(k-1)} \\&#xA;\frac{T^*}{T_0} &amp;amp;= \frac{2}{k+1} \\&#xA;\frac{\rho^*}{\rho_0} &amp;amp;= \left(\frac{2}{k+1}\right)^{1/(k-1)} \\&#xA;G^* &amp;amp;= p_0 \sqrt{\left(\frac{2}{k+1}\right)^{(k+1)/(k-1)}\left( \frac{kM_w}{RT_0} \right)} \\&#xA;w^* &amp;amp;= G^*A \\&#xA;V = V^* = c^* &amp;amp;= \sqrt{\frac{kRT^*}{M_w}}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the formulas above, the $^*$ represents the condition at choked flow and the $_0$ condition is inlet. The other variables are defined as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$p$ = pressure&lt;/li&gt;&#xA;&lt;li&gt;$T$ = temperature&lt;/li&gt;&#xA;&lt;li&gt;$\rho$ = density&lt;/li&gt;&#xA;&lt;li&gt;$G$ = mass velocity (mass flow per unit area)&lt;/li&gt;&#xA;&lt;li&gt;$w$ = mass flow&lt;/li&gt;&#xA;&lt;li&gt;$A$ = nozzle exit area&lt;/li&gt;&#xA;&lt;li&gt;$V, c$ = exit velocity&lt;/li&gt;&#xA;&lt;li&gt;$R$ = gas constant&lt;/li&gt;&#xA;&lt;li&gt;$M_w$ = molecular weight&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Choked flow occurs when the downstream pressure is less than the critical pressure or the pressure ratio is less than the critical ratio. This is shown in equation 1 and repeats your initial question.  Once you know the flow will be choked, you can then use the remaining equations.  Looking at the equation for the mass velocity, $G^*$, you can see that choked flow is a function of gas composition $(k,M_w)$ and inlet conditions $(T_0,p_0)$ and that changing downstream conditions has no effect on the mass velocity.  To get to the mass flow rate $w^*$ you must also consider the orifice area $A$.  With those variables known, you can determine which orifice will create the limiting flow rate.  This can become an iterative process as changing upstream conditions may then limit downstream components.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have a pressure regulator bringing an inlet pressure of 150 bar down to 10 bar&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Since you have a pressure regulator, this tells you something about the dynamics of the system.  The pressure regulator presents a variable sized orifice to the process until it is fully opened.  Once it is fully open, it behaves like a fixed orifice size.  Since the pressure regulator is able to adjust to maintain a downstream condition, it will not be the limiting component until it is wide open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the regulator is partially open, the system has established a steady state condition wherein:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the flow out of the system (to atmosphere) is the maximum flow through the system&lt;/li&gt;&#xA;&lt;li&gt;the position of the regulator (e.g. what % open) is such that its exposed flow area provides exactly the same flow as the outlet to atmosphere; this flow is a function of the temperature, upstream pressure, and composition of the gas at that choke point&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6681" LastEditorUserId="6681" LastEditDate="2016-07-13T14:08:34.067" LastActivityDate="2016-07-13T14:08:34.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10618" PostTypeId="1" CreationDate="2016-07-08T08:29:27.280" Score="4" ViewCount="695" Body="&lt;p&gt;I have a circuit that consumes enough power to keep running from a powerbank. However, to save energy I'm putting it in low energy mode at night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem with powerbanks, they turn off when not enough power is consumed. My proposed solution was creating a separate low-power circuit that times how long the powerbank is off and at the right moment provides a short burst of power. The devices in my circuit then turn on, the powerbank detects this and also turns on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the devices in my circuit are Arduinos or Arduino compatible running on 5V. The circuit pulls a maximum of 200mA. I have no physical input on the system as it is placed in a different location. Everything needs to be automated. It preferably has to use the powerbank due to portability and the desire of having no external wires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a safe method for an external circuit, to turn on another circuit in an automated way?&lt;/p&gt;&#xA;" OwnerUserId="7175" LastActivityDate="2018-02-28T22:52:38.343" Title="External circuit turning on powerbank" Tags="|electrical-engineering|power-electronics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10620" PostTypeId="2" ParentId="10618" CreationDate="2016-07-08T12:53:16.440" Score="1" Body="&lt;p&gt;The problem is the Powerbank detects when nothing is drawing power, assumes nothing is connected to it and shuts down. Your circuit doesn't draw enough current to be detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution - program your controller to wake up from its low power state once a minute, flash an LED for a few seconds to draw some current, then go to sleep again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You will need to do some experiments to determine if once a minute is enough, and also if a single LED is enough load.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard quite a few people on the internet discuss this: for example &lt;a href=&quot;http://www.dorkbotpdx.org/blog/paul/battery_pack_load&quot; rel=&quot;nofollow&quot;&gt;here is a solution someone invented&lt;/a&gt; that draws current off one of the Powerbank's USB outputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Personally I avoid this by just making little packs of, say, ten NiMh AA cells but they're more bulky of course.)&lt;/p&gt;&#xA;" OwnerUserId="6665" LastActivityDate="2016-07-08T12:53:16.440" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10622" PostTypeId="2" ParentId="10613" CreationDate="2016-07-08T13:10:22.523" Score="0" Body="&lt;p&gt;I'm tempted to say that you've measured this change, so what's the question here, but there are a couple things you can do to validate your tests.  Most important: run your source &amp;amp; detector in exactly the same setup but without the acrylic sheet in place, and see if the response varies with temperature.  You want to be sure that only the sheet is causing the observed change.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, there are less critical things to check: is your source polarized? Does the sheet move at all as the temperature changes? Acrylics often have stress lines which affect the polarization of light passing thru them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, do the deepest Google search you can for info or tech papers on thermal sensitivity of acrylics.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-07-08T13:10:22.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10623" PostTypeId="1" CreationDate="2016-07-08T13:30:41.737" Score="3" ViewCount="119" Body="&lt;p&gt;According to the following figure, boundary layer becomes progressively thicker with downstream distance x . Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OtsPg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OtsPg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an other question as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When turbine blade thickness increases, what dose happen about blade boundary layer thickness? Dose BL thickness increase about blade?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am grateful that guide me about this question and if it is possible, please give me useful links and papers about these questions.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="282" LastEditDate="2016-07-09T18:44:14.503" LastActivityDate="2017-11-23T06:47:23.320" Title="Question about bounday layer thickness?" Tags="|mechanical-engineering|fluid-mechanics|turbines|cfd|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10624" PostTypeId="1" AcceptedAnswerId="10625" CreationDate="2016-07-08T14:42:04.260" Score="1" ViewCount="480" Body="&lt;p&gt;I want to prototype a wah-wah guitar pedal using an arduino, a potentiometer, and some meccano I have lying around.&#xA;I thought of two ways to do this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the first one is to put a pinion on the pot, and connect a rack to the pedal itself; however I can already see that it can't be as simple as that, because the rack to pinion angle will change as the pedal gets pushed more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the second one is to use several gears, the first one on the axle of the pedal, the last one on the potentiometer, so that the full motion of the pedal results in a full potentiometer rotation; I don't like that one much, as I don't think I have the requisite gear ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having absolutely no background in mechanics, that's it for me; I don't even know the name of the concepts I might be needing. Can Stackexchange Engineering provide some pointers for making this at home, with no complex tools beyond a sander, a dremel, and such?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your attention.&lt;/p&gt;&#xA;" OwnerUserId="7179" LastActivityDate="2018-03-30T17:03:01.550" Title="Building a wah-wah pedal" Tags="|gears|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10625" PostTypeId="2" ParentId="10624" CreationDate="2016-07-08T15:33:10.380" Score="3" Body="&lt;p&gt;Using a potentiometer is a bad idea in the long term, because typical electronic pots are not designed to handle high mechanical loading or very frequent movement. Compare how may times in a day (or even in a minute) you are likely to operate the pedal, compared with how many times you adjust the volume pot on a radio or a similar device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better idea would be to make a &quot;contact-free&quot; mechanism that outputs the position of the pedal. One effective home-built solution is to use an analog optical system. All you need is a light source (e.g. an LED) and a photodiode, separated by a moving vane with a variable-width slot in it to vary the amount of light that reaches the photodiode as you move the pedal. Write some software for the Arduino to convert the photodiode output into whatever you need to control the sound generation part of the project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't require a complicated mechanical design. The moving vane can be fixed directly onto the pedal itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This type of system was used for the various control pedals on pipe organs long before digital electronics was available - often with a small light bulb powered at well below its rated voltage to increase its life span, and a &quot;photocell&quot; similar to those used in pre-electronic photographic light meters as the sensor. They worked reliably for decades, with no maintenance whatever.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-07-08T15:40:05.770" LastActivityDate="2016-07-08T15:40:05.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10626" PostTypeId="1" AcceptedAnswerId="10627" CreationDate="2016-07-08T16:04:15.720" Score="1" ViewCount="152" Body="&lt;p&gt;I'd like to prototype a kind of device to move a rope like a sinusoidal, or similar waveform, it doesn't have to be precise about the waveform, the goal is hide this mechanism in order to show a floating waveform in space for an exhibition. I think to use an electric motor to vibrate one side of the robe and to fix the other side, but I'm not an expert in this field. How to convert the rotatory movement to vibrate the rope? Any hints on how to do this? Thank you&lt;/p&gt;&#xA;" OwnerUserId="1957" LastActivityDate="2016-07-10T02:41:33.787" Title="motor to move rope sinusoidally" Tags="|motors|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10627" PostTypeId="2" ParentId="10626" CreationDate="2016-07-08T16:49:53.373" Score="3" Body="&lt;p&gt;If you are looking for a way to convert constant-speed rotary motion into a sinusoidal transverse motion, the simplest option is probably a &lt;a href=&quot;https://en.wikipedia.org/wiki/Scotch_yoke&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scotch Yoke&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scotch yoke is mechanically quite simple, so it shouldn't be too hard to build for your application. You can make the piston long so that the scotch yoke and motor are hidden in a box or something. Tie one end of the rope to the piston, fix the other end to a post and you have your waveform. See my simple sketch for an idea of how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HmTEx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HmTEx.jpg&quot; alt=&quot;A simple sketch of a scotch yoke&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A limitation of this choice is that your oscillation amplitude is determined by the length of the arm/slot, therefore you can't change your amplitude once you have built your device. However if all you want is to move your rope in a general sinusoid pattern it should be ok if the amplitude is fixed.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="6695" LastEditDate="2016-07-08T17:26:03.300" LastActivityDate="2016-07-08T17:26:03.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10629" PostTypeId="2" ParentId="10612" CreationDate="2016-07-08T19:40:22.613" Score="1" Body="&lt;p&gt;You can couple to the hex shaft using a suitable socket wrench socket. (If you don't want to destroy the original shaft, get the correct size Allen wrench and cut off the short arm.) Then you can use standard socket wrench accessories to change the direction of the axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick search turned up options such as the MFE-6 flexible shaft or the M-140A universal joint in the &lt;a href=&quot;http://www.snaponindustrialbrands.com/DSN/wwwsnaponindustrialbrandscom/Content/PDF/Snap-on%20Industrial%20Brands%20-%20CAT3%20227.pdf&quot; rel=&quot;nofollow&quot;&gt;Snap-On Tools catalog&lt;/a&gt;. Other vendors have similar offerings.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-07-08T19:40:22.613" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10631" PostTypeId="1" CreationDate="2016-07-09T14:01:54.650" Score="2" ViewCount="51" Body="&lt;p&gt;I have a 0.5&amp;nbsp;kW single-phase AC motor. What happens if the supply is 220&amp;nbsp;V but can't supply the full current needed? Will the motor torque be lower? Or will it be slower?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it be any different if this motor is a water pump motor?&lt;/p&gt;&#xA;" OwnerUserId="7188" LastEditorUserId="1832" LastEditDate="2016-07-09T14:10:55.837" LastActivityDate="2016-07-09T21:52:14.910" Title="What happens if current provided to mono phase motor is less than rated power" Tags="|electrical-engineering|motors|power|solar-energy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10632" PostTypeId="2" ParentId="8478" CreationDate="2016-07-09T18:01:23.367" Score="0" Body="&lt;p&gt;For most cases, yield strength (and also thickness) of weld metal is equal or higher than the yield stress (and also thickness) of base metal. In practical we may think they're equal. Because it's pointless to make calculations for weld metal which is 300 MPa when base metal yields in 235 MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a document called PQR (Procedure Qualification Record). In PQR, company makes both non-destructive and destructive tests on a welded specimen and use this document as proof for it's qualified welds. You can read yield and tensile stress values on PQR. You may find values of weld metal may be lower than values of base metal. But that's an exceptional. Let's think weld metal is at least as strong as the adjoining base metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the point is &quot;welding&quot; is a service provision which is defined in ISO 9001 7.5.2. That means you can never be sure welding process is grateful even if you have some controls and tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 0.3 Sut or 0.577 Sy they're all attitudes for precaution and they're all true. Sorry, I don't know which one is used more in field.&lt;/p&gt;&#xA;" OwnerUserId="7191" LastActivityDate="2016-07-09T18:01:23.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10634" PostTypeId="1" CreationDate="2016-07-09T21:06:27.237" Score="3" ViewCount="333" Body="&lt;p&gt;I'm having a hard time understanding the benefits of VCD (like Slingshot) over a conventional boiler in terms of volume of condensate produced per unit of energy input.  I've found quite a few resources explaining (in too much detail for me to fully grasp) the operation of VCDs (&lt;a href=&quot;http://www.aquaback.com/technology/how-it-works&quot; rel=&quot;nofollow&quot;&gt;[A]&lt;/a&gt;, &lt;a href=&quot;http://www.ide-tech.com/wp-content/uploads/2013/09/The-Mechanical-Vapor-Compression-38-Years-of-Experience.pdf&quot; rel=&quot;nofollow&quot;&gt;[B]&lt;/a&gt;), but I haven't come across (or understood maybe) any explaining why compressing the vapor is either necessary or more than marginally advantageous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I think I understand: inlet water comes into the system, flows through a heat exchanger to be preheated by outflowing condensate, bringing the inlet water as close to the boiling temp as practical. It's then boiled by heat exchanged from the compressed vapor (above atmospheric pressure and normal boiling/condensing temp), producing low pressure vapor that feeds into the compressor. Let me know if I've got this wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see how this would be a closed system and could be very efficient, but it seems to me (erroneously, I assume) that a non-compression boiler could be (very nearly) as efficient. Is the higher temperature difference between the compressed vapor and inlet water the main advantage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen several sources focus on how the compression stage raises the condensation point. Beyond increasing the temp difference across the heat exchangers, it seems like this would also push the point (or zone) where the vapor condenses further downstream in the heat exchanger and a larger share of its length would be filled with vapor than without compression. Is there an advantage to this? Seems like the lower mass (vapor vs. liquid) would transfer less total heat, even with the larger temp difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure I'm looking at this wrong, but I can't see where. Any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="7193" LastEditorUserId="1832" LastEditDate="2016-07-09T21:42:35.790" LastActivityDate="2017-03-15T20:55:26.867" Title="Is Vapor Compression Distillation inherently more efficient than a thermal-only boiler-condensor?" Tags="|heat-transfer|steam|boilers|evaporation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10635" PostTypeId="2" ParentId="8478" CreationDate="2016-07-09T21:34:16.397" Score="2" Body="&lt;p&gt;A lot depends on the loading situation and the quality standard of the weld. As mentioned in another answer the yield stress of the filler metal is often higher than that of the base metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this needs to be qualified by the fact that the welding process can, in some circumstances, modify the material properties of the base metal in the &lt;em&gt;heat affected zone&lt;/em&gt; (especially for cold worked or high alloy steels). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also  the question of weld defects such as incomplete penetration, cold starts, inclusions, porosity etc etc as well as the fact that some alloys require specific pre and/or post weld treatment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;very&lt;/strong&gt; general terms weld defects are often related to crack propagation problems and so are more of a concern with structures where fatigue and other cyclic or impact related modes of failure are a concern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Critical welding applications are well covered by application/industry specific standards which cove the whole process as well as testing and inspection procedures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For good quality welds in low alloy steels with good penetration and no major defects it is generally reasonable to consider a welded joint as if it was a continuous piece of metal unless you're working to a code which states otherwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that I recently quoted for a maintenance job for aluminium alloy generator cages where welds are a known failure point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this is all highly contextual which is why we have specific welding codes which define procedures and testing protocols and specific situational rather than having a single fudge factor for 'welds'. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-09-07T23:21:46.330" LastActivityDate="2016-09-07T23:21:46.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10636" PostTypeId="2" ParentId="10624" CreationDate="2016-07-09T21:51:14.203" Score="0" Body="&lt;p&gt;One way to do this is to have a toothed pulley connected to the pot and have a toothed belt wrapped around it at about 90 degrees held in tension by a spring. So that it comes down vertically from the footplate of the pedal and then wraps around the pulley to the horizontal plane to be joined to the back of the base of the pedal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in another answer more modern pedals which directly control some parameter eg wah/volume/pitch often use optical sensors as they are more durable and less likely to introduce noise. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-07-09T21:51:14.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10637" PostTypeId="2" ParentId="10631" CreationDate="2016-07-09T21:52:14.910" Score="3" Body="&lt;p&gt;Torque is proportional to current.  If you can't supply the full current, then the torque will be lower than it would be otherwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest current draw is at startup when the motor is not turning.  This means the motor will produce lower than intended torque when it tries to get going.  If this torque isn't enough to get the motor moving, then it will just sit there getting hot, and the current requirement won't decrease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point is that you supply can't be 220 V when it's unable to supply the current the load is trying to draw.  Something has to give.  This could be the supply putting out lower than 220 V, or it could be popping a fuse or breaker, or possibly melting or otherwise destroying something.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-07-09T21:52:14.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10639" PostTypeId="2" ParentId="10626" CreationDate="2016-07-10T02:41:33.787" Score="1" Body="&lt;p&gt;Following on what Forward Ed said, a crank and connecting rod seems like a pretty straight-forward design. I think the motion of the rope will approach a sine wave as the connecting rod is made longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the free end of the rope might need some attention, too. If you want a travelling sine wave, I think it'll have to have a specific mass added to it (the free end) so that the end doesn't whip around (too little mass), or hang there near motionless (too much mass). Kind of like the kinetic equivalent of a terminating resistor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a standing sine wave, I think the free end could be held stationary (or maybe nearly-anchored, with just a little bit of stretch), but the relationship between the motor speed and length of rope becomes important. Only certain lengths of rope would produce a standing wave for a given motor speed.&lt;/p&gt;&#xA;" OwnerUserId="7193" LastActivityDate="2016-07-10T02:41:33.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10643" PostTypeId="2" ParentId="10623" CreationDate="2016-07-10T18:31:37.220" Score="4" Body="&lt;p&gt;The greater the length of the flat plate, the more contact area there is with the fluid.  Consequently, the fluid near the start of the plate has had less viscous friction forces acting on it than the end of the plate.  So, the end of the plate has more fluid slowed down than the front.  The width of the boundary layer must necessarily increase with the amount of fluid that had been slowed down by viscous friction forces with the wall.  The overall motion of the fluid just pushes more of the slower fluid downstream.&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2016-07-10T18:31:37.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10644" PostTypeId="1" CreationDate="2016-07-11T03:38:50.163" Score="1" ViewCount="53" Body="&lt;p&gt;I have an old casio casitone CT-310&#xA;Rating: DC 9V 8.5W&#xA;Battery 1.5V x 6&#xA;or AC adaptor AD-5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking around the house for a power supply to use.&#xA;I found the following:&#xA;Output: 9V DC 200mA&#xA;from a hand held mini vacuum cleaner charging base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the keyboard must demand about 945mA (8.5W / 9V) but the supply delivers only 200mA correct? What will happen if I try to use it? will it be very quiet? or do nothing? or get damaged?&lt;/p&gt;&#xA;" OwnerUserId="7205" LastActivityDate="2016-07-11T05:26:41.770" Title="substituting power supplies safe?" Tags="|electrical-engineering|power|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10646" PostTypeId="2" ParentId="10644" CreationDate="2016-07-11T05:26:41.770" Score="4" Body="&lt;p&gt;If you get lucky, the power supply will either shut down because it is overloaded, or deliver a very reduced voltage. Either way, the keyboard won't work properly, or won't work at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you get unlucky, the power supply will overheat, catch fire, and burn your house down. (I'm not joking - at least about the chance of it catching fire).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;wall wart&quot; rated at 9V and at least 1A (and preferably a bit more current, so you are not running it at 100% of its rated output) should work fine.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-07-11T05:26:41.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10649" PostTypeId="1" CreationDate="2016-07-11T14:48:33.127" Score="0" ViewCount="109" Body="&lt;p&gt;I keep reading about how great the energy density is for hypothetical fuel cells, but as far as I can tell these objects do not exist in any practical form outside of laboratory prototypes and &quot;educational&quot; kits. I keep reading statements like &quot;our company is developing... blah blah&quot;. When I do a search on Amazon for fuel cells the only thing that comes up is &quot;kits&quot;, books on &quot;how to build your own cells&quot; (LOL), and racing fuel cells (which are just special gas tanks, not hydrogen fuel cells).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, practical fuel cells actually exist or it is just a bunch of hype and wishful thinking? By &quot;practical&quot; I mean a fuel cell of 500-2000 watts for a home solar system and includes cost as a factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After further searching around I did find some practical fuel cells and found out why they are not readily available: because a 500 W cell that would be enough to power, say, a computer, costs \$10,000. For \$10,000 I can buy enough electricity to power the computer for hundreds of years. So, the practicality of these devices appears to be limited to the International Space Station and things like that.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2016-07-11T15:28:06.493" LastActivityDate="2016-07-11T15:47:26.950" Title="Do commercial fuel cells actually exist or it is just hype?" Tags="|energy-storage|battery|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10650" PostTypeId="1" AcceptedAnswerId="10652" CreationDate="2016-07-11T14:57:05.657" Score="3" ViewCount="1363" Body="&lt;p&gt;I have general problem of understanding how forces are transfered from 3d members problem to plane problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R5EB9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R5EB9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example what would be 2D representation of forces F1 and F2 acting on horizontal A beam, and what will be effect of beam A,B,C,D and column 1 and 2 on horizontal A beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I represent horizontal beam A in 2D so I can do calculations of bending moment, shear forces etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All joints can be considered as welded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;  (After Mr. Wasabis answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can replace influence of force F1 with reactions of beam B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0nD2X.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0nD2X.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so reactions are influencing horizontal A (beam C and F2 are igonred for simplicity). Is this OK - With moment Ms1 in plane perpendicular to horizontal A causing torsion of horizonta A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HoHRl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HoHRl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="5855" LastEditDate="2016-07-11T17:30:53.593" LastActivityDate="2019-12-28T18:21:40.247" Title="3D problem to 2D plane problem" Tags="|mechanical-engineering|structural-engineering|civil-engineering|structures|structural-analysis|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10651" PostTypeId="2" ParentId="10649" CreationDate="2016-07-11T15:42:21.993" Score="4" Body="&lt;p&gt;Fuel cells have hit the start of commercial success in Japan, where &lt;a href=&quot;https://fuelcellsworks.com/archives/2015/09/23/ene-farm-installed-120000-residential-fuel-cell-units/&quot; rel=&quot;nofollow&quot;&gt;sales so far have been at least 120,000&lt;/a&gt;. Given they only became viable commercial products very recently, that's pretty good uptake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're generally being used there as domestic CHP systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are still open questions about their long-term reliability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, they've grown beyond being a product that only existed as prototypes and unique one-off applications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to whether they are economically viable - well, that depends on a lot of local factors. Available grants, feed-in tariffs, your retail price of electricity and of any heat supply, the ratio of your electricity demand to your heat demand, the price of carbon, the grid's carbon intensity, the carbon intensity of the fuel cell's fuel supply, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not yet clear where their main markets will be: whether it will be mobile or stationary applications; and for domestic, commercial or industrial usage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend this peer-reviewed paper &lt;a href=&quot;http://www.brunogpollet.com/wp-content/uploads/2016/04/Current-status-of-fuel-cell-based-combined-heat-and-power-systems-for-residential-sector.pdf&quot; rel=&quot;nofollow&quot;&gt;&quot;Current status of fuel cell based combined heat and power systems for residential sector&quot;&lt;/a&gt; by Ellamla et al from 2015 for a thorough overview of the particular market segment named in the title.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fuel-cell vehicle sales have only been in the hundreds to date. There may be a market for hydrogen-powered fuel-cell freight vehicles; though they would appear to have comprehensively lost the car / automobile market to battery electric vehicles.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2016-07-11T15:47:26.950" LastActivityDate="2016-07-11T15:47:26.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10652" PostTypeId="2" ParentId="10650" CreationDate="2016-07-11T16:38:40.850" Score="2" Body="&lt;p&gt;This is quite a broad question, but I'll attempt an answer (be sure to read the last section if the rest seems a bit daunting).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters, it depends greatly on how true to reality this model you have shown us is. Is it really just a box skeleton with some loads applied or is there a slab or some other plate structure which might help to redistribute the loads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, where are the structure's supports (foundations)? I'm here going to assume they're only beneath the columns.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's start by assuming there is no slab and the given model is an adequate representation of the structure, meaning it is a hollow, open box composed purely by beam elements. In this case, &lt;span class=&quot;math-container&quot;&gt;$F1$&lt;/span&gt; will obviously be supported purely by beam B and &lt;span class=&quot;math-container&quot;&gt;$F2$&lt;/span&gt;, by beam C. These beams will themselves be supported by the &quot;horizontal A&quot; and (unnamed) &quot;horizontal B&quot; beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, these &quot;horizontal&quot; beams are flexible, since they deform under load. That means that beams B and C must be modelled as if resting on springs instead of rigid supports. Since these &quot;horizontal&quot; beams are of equal length and (I assume) cross-section, however, this ends up not making any difference, so feel free to model the supports as rigid if you wish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all the members are assumed fixed (due to being welded), the torsional stiffness of the &quot;horizontal&quot; bars will create a small rotational stiffness on B and C's supports (which is often ignored, since it's so small).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then calculate the support reactions for beams B and C and then create models for the &quot;horizontal&quot; beams with those support reactions as applied concentrated forces (including concentrated torsion moments representing the bending moment reactions found in the model for B and C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again the doubt regarding the fixity of the model appears: are the &quot;horizontal&quot; beams pinned or fixed to the foundations beneath the columns? If they're fixed, then we need to know how the foundation behaves:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does the foundation behave as a stiff support for rotations, meaning the &quot;horizontal&quot; beams must be modelled as totally fixed? In this case, the support will forbid all of the deformations from the &quot;horizontal&quot; beams, meaning nothing will be transferred to beam A. In this case, beam A is not affected by &lt;span class=&quot;math-container&quot;&gt;$F1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F2$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Is the foundation flexible for rotations, meaning the &quot;horizontal&quot; beams must also be modelled as having rotational springs at its supports?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the latter case or if the &quot;horizontal&quot; beams can be said to be pinned to the supports, then beam A will suffer some internal forces. That's because the &quot;horizontal&quot; beams will have suffer, at their extremities (where A is situated), rotations due to supporting beams B and C, as can be seen here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dGGaG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dGGaG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the loading isn't symmetric, the angle found for each &quot;horizontal&quot; beam at A's supports will be different. This will impose a concentrated torsion on A's extremities. If the loading were perfectly symmetric, with the loads applied at B and C's midspans, then the imposed rotations on A would be the same at both supports, generating no torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies to the torsional rotation suffered by the &quot;horizontal&quot; beams due to being fixed to beams B and C, as described previously. In this case, this will generate bending moment on A (which won't cancel out, even if the loading were symmetric).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating these concentrated torsions and bending moments on A, you can solve for the internal forces.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now let's look at the case where there is in fact a slab which may help to redistribute the loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loads are applied directly over beams B and C, which will therefore certainly absorb the vast majority of the loads. However, since the beams are elastic (can deform under load), some of the force can be transferred to the remaining beams as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking a longitudinal cut (parallel to the &quot;horizontal&quot; bars), we can (over)simplify the slab to the following beam (looking at just one of the forces for now):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z9CxS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z9CxS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the beams' and slab's stiffness, the fraction of redistribution will change. For instance, here are the deformations and reactions for different beam stiffnesses (the latter case is probably more realistic):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G9cma.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G9cma.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a drastic oversimplification, ignoring the influence of the &quot;horizontal&quot; beams, which should also get some load redistributed to them. But it shows how some load can get transferred directly to A. That being said, the load would be transformed into a distributed load along A (or a segment of A), but I admit to not being entirely sure how to adequately make that transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than this, the possible load-transfer described in the previous section is still possible: apply the fraction of the load that is in fact supported by beams B and C, get the support reactions, send them to the &quot;horizontal&quot; beams, etc.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Having said all this, it is most likely entirely acceptable to assume that beam A will not be in any way affected by forces &lt;span class=&quot;math-container&quot;&gt;$F1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F2$&lt;/span&gt;. Whatever transfer there may be, via slab or imposed deformations, will probably be minimal. The examples given above for redistribution via the slab aren't very good. In fact, the proportion is probably closer to a few percentage points being transferred to A &lt;em&gt;at most&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost without a doubt, the main thing you'll need to deal with in beam A is the beam's own self-weight (and that of the slab, if it exists).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;All diagrams made with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:21:40.247" LastActivityDate="2019-12-28T18:21:40.247" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="10654" PostTypeId="1" CreationDate="2016-07-11T19:12:54.263" Score="1" ViewCount="24" Body="&lt;p&gt;I am trying to solve the Saint-Venant torsion problem with FEA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read some classic text and stuck in finding the constant value of stress function on multiple boundaries (i.e. closed section with holes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many classic texts, it is shown that the value of constant can be determined by evaluate the following equation at each boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dmYVu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dmYVu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gq41t.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gq41t.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone show me how to compute Phi at each boundary based on above equations? &#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7214" LastActivityDate="2016-07-11T19:12:54.263" Title="Constant on multiply-connected boundaries for Saint-Venant torsion" Tags="|structural-analysis|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10655" PostTypeId="2" ParentId="10588" CreationDate="2016-07-11T19:54:36.477" Score="0" Body="&lt;p&gt;We usually disregard the small change in CG of the sloped beams but in this case the change is significant and only your 2nd option is correct.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CoBA8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CoBA8.jpg&quot; alt=&quot;the inclined beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2016-07-11T20:02:43.707" LastActivityDate="2016-07-11T20:02:43.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10656" PostTypeId="1" CreationDate="2016-07-11T20:15:05.267" Score="4" ViewCount="781" Body="&lt;p&gt;I'm using commercial software to simulate unsteady incompressible flow of air through a 1m long rectangular pipe with cross section 0.2mx1m.  I have a velocity inlet condition (u=100m/s, $\nabla p=0$) and a pressure outlet condition (p=0, $\nabla u=0$).  For simplicity, I'm running the simulation in 2D on the x-y plane, with the pipe oriented horizontally and with wall boundary conditions on the bottom of the pipe (along y=0) and a symmetry condition at the top of the pipe (at y=0.1).  I used a symmetry condition so as to reduce the number of cells in my mesh and thus speed up computation time.  I also made sure to bias my mesh toward the wall boundary to ensure proper capturing of the boundary layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I run the simulation to a steady state, I observe that the maximum velocity within the pipe is slightly larger than the inlet velocity ($|u_{max}|\approx 101$).  This is counter-intuitive to me.  I expect the velocity to be bounded by the inlet velocity.  It doesn't make sense to me that the fluid velocity should accelerate at all.  Therefore, I ask:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the same thing happen in real life?  If fluid enters a rectangular pipe of similar dimensions and length at a particular inlet velocity, should I expect a velocity profile within some cross-section of the pipe to exceed the inlet velocity?  If so, what causes the fluid velocity to increase within the pipe (from a physics/mechanical engineering perspective)?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is there a problem with the way I setup the problem (i.e. a numerical issue that I should be aware of for this problem)?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2016-07-11T20:34:07.577" LastActivityDate="2016-08-07T23:15:45.933" Title="Should I expect fluid velocity within a pipe to increase beyond the inlet velocity?" Tags="|fluid-mechanics|cfd|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10657" PostTypeId="2" ParentId="10656" CreationDate="2016-07-11T21:26:36.543" Score="3" Body="&lt;p&gt;The &lt;em&gt;mean&lt;/em&gt; velocity of the fluid must be constant, if the fluid is incompressible and the area is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since you are modelling a viscous fluid, there will be a velocity profile over a cross section of the pipe, with zero velocity at the pipe walls. So the velocity at the center of the pipe will be greater than the mean velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably you specified that the inlet velocity was constant over the whole area of the pipe, so the mean velocity was 100 m/s. Strictly speaking, that velocity profile is inconsistent with the physics of the problem, because the velocity at the pipe wall is zero not 100 m/s, but for many purposes the inconsistency isn't very important. Of course if you were trying to model exactly what happens at the pipe inlet it &lt;em&gt;would&lt;/em&gt; be important, and you would need to find a different way to specify the boundary conditions.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-07-11T21:26:36.543" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10658" PostTypeId="1" CreationDate="2016-07-12T02:05:41.000" Score="0" ViewCount="2912" Body="&lt;p&gt;I need a vacuum to provide 21&quot; Hg but looking online many list an ultimate vacuum of 10Pa but 1&quot;Hg = 3300Pa? I know I am missing something and just want to make sure I get it right. Thanks for the help! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ebay.com/itm/Single-Stage-4CFM-1-3HP-Rotary-Vane-Deep-Vacuum-Pump-HVAC-AC-Air-tool-R134-R410a-/371397348640?hash=item567901cd20:g:BHMAAOSwf-VWXedU&quot; rel=&quot;nofollow&quot;&gt;http://www.ebay.com/itm/Single-Stage-4CFM-1-3HP-Rotary-Vane-Deep-Vacuum-Pump-HVAC-AC-Air-tool-R134-R410a-/371397348640?hash=item567901cd20:g:BHMAAOSwf-VWXedU&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7216" LastActivityDate="2016-07-12T16:03:41.613" Title="Vacuum Pump hg&quot; vs Pa units?" Tags="|mechanical-engineering|pumps|vacuum|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10659" PostTypeId="2" ParentId="10650" CreationDate="2016-07-12T06:09:06.960" Score="0" Body="&lt;p&gt;As a first estimate and assuming all the loads and stresses are below working stress limits and assuming the supports are simple pin connection:&#xA;Beam A support reactions are&#xA;$$R1=\dfrac{P_b(l-a) + P_c(b)}{l}$$&#xA; where $P_b$ is the reaction of beam B on beam A and $P_c$ is reaction of beam C on beam A and $a$ and $b$ are the distance of the beams $b$ and $c$ from the respective ends of beam $a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the end reactions of beam $b$ or $c$ on beam $a$ are simply Pb = (F1 x distant of F1 to far end of b)/its length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are available in any engineering hand book or quick charts. However if we were to consider the welds to be fixed joints then it turns into an indeterminate structure! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;there are approximate methods or computer programs available but the significant difference is that you have to have the member properties and loads as a starting point of analysis because then the entire structure comes into action and plays some role in supporting the applied load. i.e. If the beam across from beam a is too stiff and connected to two stiff columns with correct welded connections hypothetically in extreme case it can support the majority of loads and let beam a relieved of most of the load. Beam a can even be removed entirely and beams b and c act cantilever from the other beam! &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2016-07-12T11:16:19.773" LastActivityDate="2016-07-12T11:16:19.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10660" PostTypeId="1" CreationDate="2016-07-12T11:27:03.723" Score="2" ViewCount="281" Body="&lt;p&gt;Land survey precision tape calibration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proper use of measuring tapes (we call them chains) for precision work requires temperature correction and specific tension. Basic steel tapes have a coefficient of thermal expansion (CTE) of approx. 12 ppm per deg C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calibration tapes are made of Invar, and a similar alloy commercially sold as Lovar. These alloys have a CTE of about 1 ppm (?) per deg C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: what is the true (accepted/published) CTE of 36Ni/64Fe steel (Invar)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2: what is the CTE and composition of Lovar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a K&amp;amp;E Lovar 100 foot 'standard' tape. (It has only 3 marks (0, 50, and 100).)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have found re: CTE of Lovar is anywhere from 0.5 to 5 ppm per deg. Some cite better than Invar, some say an economical alternative to Invar, or not as good as Invar, or somewhere in between Invar and steel. &lt;/p&gt;&#xA;" OwnerUserId="7220" LastActivityDate="2019-12-02T22:14:24.220" Title="CTE of Invar steel" Tags="|mechanical-engineering|civil-engineering|thermodynamics|heat-transfer|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10661" PostTypeId="1" AcceptedAnswerId="10677" CreationDate="2016-07-12T15:02:01.857" Score="0" ViewCount="154" Body="&lt;p&gt;I have two parts that should be bolted together with four M6 bolts. The tolerances of the bottom part are quite big and the holes for this part are fixed to Ø6. The tolerances of both parts are fixed due to manufacturing and lowering of costs.&#xA;I need to determine the size of the holes of the top part in order for them to fit together. I have attached a photo showing the tolerances of both parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am in doubt how to calculate the size of the holes. For instance for the left side, is this the correct way?&#xA;$$\begin{align}&#xA;\text{Diameter} &amp;amp;= \text{Delta}_{dim} + \text{Tol}_{dist} + D_{shaft} \\&#xA;&amp;amp;= (211.5-210)+3.5 + 6 = 1.5 + 3.5 + 6 \\&#xA;&amp;amp;= Ø11&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RKbIO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RKbIO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7223" LastEditorUserId="1832" LastEditDate="2016-07-12T16:45:36.240" LastActivityDate="2016-08-13T13:59:25.620" Title="Dimensioning of hole from tolerances" Tags="|mechanical-engineering|design|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10662" PostTypeId="2" ParentId="10660" CreationDate="2016-07-12T15:37:31.707" Score="2" Body="&lt;p&gt;Presumably the measured properties here are credible, since NASA was comparing  Lovar and Invar &quot;for optical, instrument and electrical&#xA;applications in particular where dimensional stability is critical&quot; - for example space telescopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20150018888.pdf&quot; rel=&quot;nofollow&quot;&gt;http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20150018888.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the caveat mentioned in the report: the measured properties depend on how the material was manufactured and heat treated, so the NASA data might not be applicable to your tapes.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-07-12T15:43:45.173" LastActivityDate="2016-07-12T15:43:45.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10663" PostTypeId="1" CreationDate="2016-07-11T18:35:54.860" Score="4" ViewCount="832" Body="&lt;p&gt;I'm trying to build a laser distance sensor using triangulation (With this as my main inspiration, &lt;a href=&quot;https://sites.google.com/site/todddanko/home/webcam_laser_ranger&quot; rel=&quot;nofollow noreferrer&quot;&gt;Webcam Based DIY Laser Rangefinder&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yzyVZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yzyVZ.jpg&quot; alt=&quot;Laser distance measurement using triangulation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My plan is to use a linear CCD array as the imager, such as the Toshiba TCD132D, for detecting the absolute distance to an object. I'll basically try to make a higher resolution version of the Parallax TSL1401-DB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to my limited knowledge of optics and lenses, i have a hard time figuring out what the optical setup should be. Note that the distance sensor should be able to measure distances from 20&amp;nbsp;cm to 5&amp;nbsp;m, with a precision of +/- 0.5&amp;nbsp;cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that the object would cause the laser beam to have a diffuse reflection, and that some kind of convex lens should be used to focus the diffuse reflections on a few pixels in the linear array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would prefer an off the shelf board mounted lens, but since the linear array sensor is 1.4&amp;nbsp;cm wide, none of the lenses i have found would be a good fit. Does any of you have any suggesting regarding which optics to use, and preferably any literature or tutorials that would make me understand this topic a bit clearer?&lt;/p&gt;&#xA;" OwnerDisplayName="Asmus Bjerregaard Hansen" LastEditorUserId="1832" LastEditDate="2016-07-12T16:47:59.507" LastActivityDate="2016-07-12T16:47:59.507" Title="Optics for a triangulation laser distance sensor" Tags="|optics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10664" PostTypeId="2" ParentId="10658" CreationDate="2016-07-12T16:03:41.613" Score="2" Body="&lt;p&gt;I think your confusion is in absolute vs gauge pressure.  Atmospheric pressure is about 101 kPa, or 101,000 Pa, or about 30 inches of Hg.  Vacuum is commonly measured relative to atmospheric pressure, and represents a negative gauge pressure but still a positive absolute pressure.  You want 21 inches Hg vacuum - i.e. -21 inches gauge, or about 9 inches absolute.  The 10 Pa &quot;ultimate&quot; vacuum is in absolute units, not gauge, and is about 0.003 inches Hg.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2016-07-12T16:03:41.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10665" PostTypeId="1" CreationDate="2016-07-12T16:34:32.043" Score="4" ViewCount="88" Body="&lt;p&gt;At a very high level what are required technical analysis to &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;determine the need to for a new 4G LTE cellular transmitter in US?&lt;/li&gt;&#xA;&lt;li&gt;find find the optimal location for the transmitter in US?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back Story:&lt;/strong&gt; The leading US based cellular service provider wants to install a cellular  transmitter in our neighborhood on public property. The residence are oppose to the idea. In all accounts 4G LTE cellular coverage is very good. Below is coverage map from the vendor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KhciP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KhciP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vendor during public hearing claimed that that cellular signals are very poor and shared a some measurement data including Reference Symbol Received Power (RSRP) data for 4G LTE. It appears that the vendor is not upfront of all a facts, only shares selected information. &lt;/p&gt;&#xA;" OwnerUserId="7083" LastEditorUserId="282" LastEditDate="2016-10-13T10:16:17.527" LastActivityDate="2016-10-13T10:16:17.527" Title="Technical Analysis to determine optimal Cellular to transmitter location" Tags="|electrical-engineering|wireless-communication|lte|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10667" PostTypeId="1" AcceptedAnswerId="10668" CreationDate="2016-07-12T18:05:06.050" Score="2" ViewCount="116" Body="&lt;p&gt;I have to build a device to apply a pure normal force over a force plate. The requirement is that it &lt;em&gt;doesn't&lt;/em&gt; apply any shearing (by means of friction) on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is: for any given device structure, there will always appear some deformation, specially under load, and so I don't want to depend on geometric alignment, I would like to have my indenter to have an &lt;em&gt;appropriate joint&lt;/em&gt; with the proper types of degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I though about a ball bearing roller, but that would only decouple the force along one direction, since it could still transmit force by friction along the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/neZAk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/neZAk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the recommended way to eliminate coplanar reaction forces, so that only normal force is transmitted and coplanar forces are neutralized?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think lubricant-based solutions don't fit well, since they are always quantitative in nature. I think a qualitative solution involving mechanical decoupling by a proper joint is better, similar to the ball-bearing one, but with an additional degree of freedom.&lt;/p&gt;&#xA;" OwnerUserId="2120" LastActivityDate="2016-07-12T19:20:14.703" Title="Device to apply pure normal force over planar surface, without friction-based coplanar forces" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10668" PostTypeId="2" ParentId="10667" CreationDate="2016-07-12T19:20:14.703" Score="2" Body="&lt;p&gt;If you apply a force which can be decomposed to a non-zero horizontal component, that component has to be transferred onwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you only want the vertical component to be transferred to your plate, you'll need to create a secondary structure to absorb the possible horizontal component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible inspiration are &lt;a href=&quot;https://en.wikipedia.org/wiki/Slide_plate&quot; rel=&quot;nofollow noreferrer&quot;&gt;sliding plate bearings&lt;/a&gt;. These allow the free translation of an object along the horizontal plane, meaning no horizontal forces are transferred to the underlying structure (other than inevitable friction, which is low due to the adopted materials). Obviously, you'll want to limit these translations at some point, so you'll need to design a container which restrains the plate's movements. When the plate touches the container, a horizontal force will appear between them, so the container will have to rest on something other than the structure you're trying to protect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'll forgive my horrible sketch, this is what I mean:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Un76.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Un76.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The low-friction pad allows the upper plate to move freely in both directions. If an inclined force is applied to the upper plate, it will move until it touches the bounding structure, at which point the horizontal component of the force will be absorbed by the wall (or whatever supports the bounding structure, which can't be the same as whatever it is you're trying to protect), while the vertical component is transferred directly to the underlying structure.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-07-12T19:20:14.703" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10669" PostTypeId="1" CreationDate="2016-07-12T20:52:38.837" Score="2" ViewCount="807" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V9Run.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V9Run.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I have invented something and I need to have some external broaches made but I'm being asked for some dimensions I simply do not understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need 4 external broaches. They all have the same major diameter 7.6&amp;nbsp;mm. They all have an angle of 90° between each spline. I do not need any radius in the root between the splines and the peaks can be sharp too. I need spline counts of 18, 20 24, and 28.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am told by the toolmaker they cannot programme their machines until I give them numbers relating to pin diameter and a distance between pins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've Googled till my fingers bleed but nowhere can I find a definitive answer to the dimensions I need, how to work it out or even to describe what these things are and their relevance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope the attached helps&lt;img src=&quot;https://i.stack.imgur.com/iLKiN.jpg&quot; alt=&quot;![enter image description here&quot;&gt;]&lt;a href=&quot;https://i.stack.imgur.com/iLKiN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7231" LastEditorUserId="7231" LastEditDate="2016-07-13T14:37:24.127" LastActivityDate="2016-07-13T14:37:24.127" Title="Straight sided serration dimensioning" Tags="|mechanical-engineering|gears|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10671" PostTypeId="1" CreationDate="2016-07-13T05:00:43.053" Score="2" ViewCount="89" Body="&lt;p&gt;I am interested in conducting a research project involving the design of a tool to aid astronauts to perform experiments on lunar surfaces, but I am having difficulties coming up with appropriate key words or questions to ask to begin my research. The specific scenario of focus is to design a tool that can act as an anchor in a sand-like medium on the surface of the moon; the goal is to design a tool with a method to dig/lodge/bury itself underneath the soil surface in a way such that removing the tool requires a maximal load. The research I will conduct is for the purpose of selecting an optimal geometry for the tool (i.e. my initial concept is a helical drill). As this phase of my project, I will assume the design will be rely on geometry alone, however, I may also incorporate applied forces within the tool in some way (i.e. pneumatics) to aid anchoring. I am interested in this topic in order to submit a proposal to the NASA Micro-g NeXT challenge (you can see their webpage for more details). Is this design problem related to fluid mechanics, and if so, what sub-discipline of fluid mechanics is concerned with modelling the behavior of dry, granular media such as sand?&lt;/p&gt;&#xA;" OwnerUserId="7233" LastEditorUserId="7233" LastEditDate="2016-07-14T15:08:29.667" LastActivityDate="2016-07-14T15:08:29.667" Title="What engineering research field is concerned with the mechanics of adhesion; more specifically, how objects behave in soil media such as sand?" Tags="|mechanical-engineering|fluid-mechanics|aerospace-engineering|research|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10672" PostTypeId="2" ParentId="10671" CreationDate="2016-07-13T05:33:16.053" Score="3" Body="&lt;p&gt;Your problem doesn't fit neatly into one field of engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adhesion and setting of a chemical grout will come under fluid mechanics and possibly mechanical engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mechanical anchoring device fits into mechanical engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How the soil and/or rock behaves when the anchor is embedded into it will be covered by the field of geomechanical/geotechnical engineering.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-07-13T05:33:16.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10673" PostTypeId="2" ParentId="10671" CreationDate="2016-07-13T05:58:22.383" Score="1" Body="&lt;p&gt;I would say mechanical engineering with specialization in nano-technologies and tribology. On Uni I've had a few subjects covering adhesion like Tribology, Nano-technologies etc. However also physicists and chemical engineers cover this phenomena as well, but only mechanical engineers and partially physicists cover the mechanical aspect of it. &lt;/p&gt;&#xA;" OwnerUserId="7234" LastActivityDate="2016-07-13T05:58:22.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10674" PostTypeId="1" CreationDate="2016-07-13T06:12:58.273" Score="1" ViewCount="2442" Body="&lt;p&gt;I'm using Ansys workbench 15 and new to it. I want to perform ball bearing stress analysis but I'm unable to apply bearing load to it. The problem is when I select faces and click on apply, it again shows no selection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any body tell me what is the problem and how to perform ball bearing analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boundary conditions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;No separation type contact between grooves and bearings ball (grooves of both inner and outer races.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fixed support to outer race.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Fig. is attached showing contacts.&lt;a href=&quot;https://i.stack.imgur.com/Ej49l.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ej49l.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/ruDKR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ruDKR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part 1 indicates groove of inner race and part 10 indicate grooves of outer race.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;contact body - grooves&lt;/p&gt;&#xA;&#xA;&lt;p&gt;target body - balls&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to apply radial load on bearing using bearing load at the center of inner race. My load is nearly 3000 N with rotational speed of 30 rad/sec.&#xA;I used Solidworks software to create my 3D model and then save it as .igs file and import it into Ansys.&lt;/p&gt;&#xA;" OwnerUserId="7235" LastEditorUserId="7235" LastEditDate="2016-07-13T09:29:27.477" LastActivityDate="2020-10-31T06:21:40.553" Title="In ansys workbench, unable to apply bearing load" Tags="|ansys|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10675" PostTypeId="1" AcceptedAnswerId="14362" CreationDate="2016-07-13T07:36:24.107" Score="3" ViewCount="148" Body="&lt;p&gt;I am making an aircraft in a simulator (using JSBSIM). Note that I am not a professional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manual states that it is limited below certain calibrated airspeed to a certain positive angle of attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried different schemes to make this work, but I have not been successful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the simplest was to just freeze the elevator when the limit was reached. But in for example a loop, the alpha will increase even with frozen elevators as the plane goes through the loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another was to use a feedback of the exceeded alpha with a gain into reducing the elevator actuation. But that makes the plane pitch jerk. It should be smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope someone can help me out. I prefer a solution that does not use PID controller as they are a pain to calibrate. Block diagrams or text help is very appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7236" LastEditorUserId="7236" LastEditDate="2017-03-22T10:53:41.787" LastActivityDate="2017-03-22T10:59:04.253" Title="How can I limit a simulated aircraft's angle of attack?" Tags="|control-engineering|aerospace-engineering|simulation|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10676" PostTypeId="1" CreationDate="2016-07-13T08:51:01.177" Score="3" ViewCount="58" Body="&lt;p&gt;I want to know how is exactly modulated a LTE subcarrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I'm working on Synchronization Signals (P-SS, S-SS) the subcarriers I'm talking about are centered around DC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have coded the algorithms that output diverse information (Zadoff-Chu sequences, m-sequences...) that will be carried by the specific 62 subcarriers (1.4MHz BW) corresponding to both PSCH and SSCH. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm now willing to modulate the information on the subcarriers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that this formula is some kind of sinusoidal signal, i.e. for PSS:&#xA;s(t) = sin(2π&lt;em&gt;k&lt;/em&gt;Fe/f + φ) factor the data I want to map on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found what seems to be the general equation of the subcarrier, s(t)=A(t)exp(|wt + phi(t)|) giving a classic s(t)=A(t)cos(wt + Phi(t))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I haven't found anything precise until now on OFDM documentation about the complete modulation.&lt;/p&gt;&#xA;" OwnerUserId="6507" LastEditorUserId="7083" LastEditDate="2016-07-21T00:31:25.083" LastActivityDate="2016-07-21T00:31:25.083" Title="What is PSS waveform formula?" Tags="|telecommunication|lte|modulation|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10677" PostTypeId="2" ParentId="10661" CreationDate="2016-07-13T09:15:32.040" Score="1" Body="&lt;p&gt;Tolerancing is about stacking the min and max tolerances of each part, if you take the left side as you said, you have 211.5+-3.5 and 210+-1:&#xA;Bottom = min 208 max 215&#xA;Top =    max 211 min 209&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the worst case scenarios, which is the min - max you get 6mm at one end of the tolerances and and 3mm at the other between the 2 holecentres, so the holes can be as much as 6mm offset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with 6mm variation on a 6mm hole, i think your top plate will need at least a 12mm hole to fit every time...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you could make that smaller but you'd have to change the dims on on of the plates, if you can.&lt;/p&gt;&#xA;" OwnerUserId="7237" LastActivityDate="2016-07-13T09:15:32.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10679" PostTypeId="1" CreationDate="2016-07-13T12:33:29.720" Score="1" ViewCount="34" Body="&lt;p&gt;I study a model for plate bending due to tangential live load applied to its upper surface $x_3=h$. After deformation the load remains tangential to the deformed upper surface. What would then be the contribution of the load to the plate energy? My guess is&#xA;$$&#xA;\int_{\Gamma_h}\boldsymbol\sigma_h:\nabla{\bf r}_hd\boldsymbol x,&#xA;$$&#xA;but not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any guess, reference or hint?&lt;/p&gt;&#xA;" OwnerUserId="7241" LastActivityDate="2016-07-13T12:33:29.720" Title="Live load contribution in total energy of elastic plate" Tags="|mechanical-engineering|mathematics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10680" PostTypeId="2" ParentId="10595" CreationDate="2016-07-13T14:26:07.420" Score="2" Body="&lt;p&gt;I work for a company that, builds custom truck bodies and upfits vans, though we do not work with GM/Chevy vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best source of information will be in the &lt;a href=&quot;https://www.gmupfitter.com/pages/body-builder-manuals&quot; rel=&quot;nofollow&quot;&gt;Body Builder Guide&lt;/a&gt; that GM provides to upfitters and others that use their vehicles as a base for their own custom vehicles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll also want to review GM's Best Practices (found at the same link). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you'll want to tie in whatever structure you create to the structure of the van itself, following the Body Builder's Guide and the Best Practices. Deviate from those at your peril/liability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not use metal conduit as a structural material. It's not intended for that kind of use. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2016-07-13T14:26:07.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10681" PostTypeId="1" CreationDate="2016-07-13T15:24:22.513" Score="9" ViewCount="2755" Body="&lt;p&gt;Everywhere I look, be it a PID, lead, lag control or anything else, there are Simulink schematics with transfer functions. This is all nice for system response simulation, however currently I have to implement a PID control with a filtered derivative term into a strictly certified medical device, which is controlled by a microchip with a C++ code controlling it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we take the PID controller in its frequency domain as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C(s) = k_\mathrm{P} + \frac{k_\mathrm{I}}{s} + k_\mathrm{D}s = \frac{y(s)}{e(s)},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can implement that as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y(t) = k_\mathrm{P}e(t) + k_\mathrm{I}\int_{0}^{t}e(\tau)d\tau + k_\mathrm{D}\dot{e}(t), \quad e(0) = 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which we can write into pseudo code as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;integral += error*dt&#xA;derivative = (error - prevError) / dt&#xA;y = kp*error + ki*integral + kd*derivative&#xA;prevError = error&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, now we take the filtered PID control as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C(s) = k_\mathrm{P} + \frac{k_\mathrm{I}}{s} + k_\mathrm{D}\frac{sN}{s + N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best I can think of is to create inverse Laplace as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y(t) = k_\mathrm{P}e(t) + k_\mathrm{I}\int_{0}^{t}e(\tau)d\tau + k_\mathrm{D}\left( N\delta(t) - N^2e^{-Nt} \right)e(t), \quad e(0) = 0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but what does that really represent? Integrating via $dt$ is one thing, but all I can see in the $e^{-Nt}$ is the fact that after few seconds, no matter what I do, I am going to have just another proportional constant. Should I reset the time at some point? And we haven't even written that into C++ yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the correct approach?&lt;/p&gt;&#xA;" OwnerUserId="7242" LastActivityDate="2016-07-14T13:55:04.297" Title="Writing a filtered derivative term of the PID controller into the C++ code" Tags="|mechanical-engineering|control-engineering|applied-mechanics|control-theory|pid-control|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10683" PostTypeId="1" CreationDate="2016-07-13T21:36:09.483" Score="5" ViewCount="125" Body="&lt;p&gt;The central heat and air in the house I'm renting is unable to keep the house below 80 °F during the hot summer here (100 °F days).  We've had 3 HVAC techs out to look at it and they all say there's &quot;nothing wrong with the system&quot; and that it's &quot;sized appropriately.&quot;  The latest said perhaps it's just a combination of small factors (old house, leaky windows, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The property owner is going to provide me with two window units to use - one in the bedroom and one in the living room.  Would it be more efficient (in electricity cost) to turn off the central AC entirely and rely on the window units, or better to run all three systems?&lt;/p&gt;&#xA;" OwnerUserId="7246" LastEditorUserId="368" LastEditDate="2016-07-13T22:12:59.860" LastActivityDate="2016-07-14T05:13:35.770" Title="What's more energy efficient: fewer air conditioners working at max capacity or more working at part capacity?" Tags="|hvac|energy-efficiency|cooling|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10686" PostTypeId="2" ParentId="10683" CreationDate="2016-07-14T03:05:15.213" Score="2" Body="&lt;p&gt;Test it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any answer without a test is really just a fancy guess. And a test will answer your question definitively and quantitatively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your test needs to be as controlled as possible. Which means you will need to turn off all the other major power draws (or keep them on and constant — &lt;em&gt;like the refrigerator&lt;/em&gt; — during the test.) You also need to make sure you have the same starting temperature AND ambient temperature and humidity when you run your test trials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After you account for all the control factors, just record your starting meter reading. Cool to your desired temperature. Then record your finishing meter reading. Then repeat for the other configurations you wish to test.&lt;/p&gt;&#xA;" OwnerUserId="3820" LastActivityDate="2016-07-14T03:05:15.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10688" PostTypeId="2" ParentId="10683" CreationDate="2016-07-14T05:13:35.770" Score="1" Body="&lt;p&gt;Many new air-conditioners have a SEER efficiency rating. &lt;a href=&quot;https://en.wikipedia.org/wiki/Seasonal_energy_efficiency_ratio&quot; rel=&quot;nofollow&quot;&gt;SEER&lt;/a&gt;, Some cities have rewards for higher efficiency units. Generally every thing else being equal having more units means better control and lower electricity bill.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In average households the ducting system is not efficient and balanced, Partially due to the architectural layout of the rooms, partitions and spaces and the fact that the farther downstream of ducting the more loss of pressure and warmer outflow.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other advantage of multi-unit system is you don't waste energy keeping spaces not in use cool.&#xA;In modern homes they split the central air-conditioner into several units depending on the layout of the floors and utility of spaces!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-07-14T05:13:35.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10689" PostTypeId="2" ParentId="10671" CreationDate="2016-07-14T06:02:38.770" Score="1" Body="&lt;p&gt;You seem to be describing a tent peg. This should be in civil engineering or geotechnics. My search phrases would be ground achors for large tents and the like. You could also draw a picture of how your anchor will look like and what forces will act on it and ask for similar use cases, this would give you more ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CE/geotechnics deals with friction on parts embedded in ground and the like, so it should be your friend.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-07-14T06:02:38.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10690" PostTypeId="1" AcceptedAnswerId="10692" CreationDate="2016-07-14T07:14:12.777" Score="0" ViewCount="155" Body="&lt;p&gt;I want to power an Arduino project for several weeks on a remote location. At the location, data is gathered and sent over the internet. The project runs on 5V and draws 200mAh max. I need a power source that runs on its own, without cables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do so, I was looking into big batteries. My idea was to use a battery like &lt;a href=&quot;http://www.produktinfo.conrad.com/datenblaetter/1300000-1399999/001372559-da-01-en-LATERNENBATTERIE_ALKALI_4LR25SZ__6V_50AH.pdf&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. It has a capacity of 50Ah and an output of 6V which I can downstep with a voltage regulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have several questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How safe is this battery? Can directly touching the contacts cause harm?&lt;/li&gt;&#xA;&lt;li&gt;Can my project draw 200mAh or will the battery push out more due to its high capacity?&lt;/li&gt;&#xA;&lt;li&gt;And if it's dangerous, how can I protect myself and my system?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hope my information in this post is sufficient, thanks for any responses.&lt;/p&gt;&#xA;" OwnerUserId="7175" LastActivityDate="2016-10-17T09:18:52.337" Title="Low voltage, high ampere battery safety" Tags="|power-electronics|battery|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10691" PostTypeId="1" CreationDate="2016-07-14T07:17:19.263" Score="0" ViewCount="200" Body="&lt;p&gt;I want to use a computer to initiate a game on a physical pinball game. (This is to see if I can get the computer to learn to play pinball using machine vision and deep Q networks.) I want to pull the shooter rod out that is used to send the ball into play. You pull back on the shooter rod and release it to send the ball up to the top of the play field. I need to create a hardware solution that can grip and release this so that it strikes the ball under computer control. You can see what it looks like in the image. I can use the computer to run a servo or electromagnet to initiate it somehow but I'm not sure what configuration would be easiest/best to implement. Although it would be nice, its not really important to control how far the shooter rod is pulled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me I need a thing that can pull it back and then maybe a solenoid to release the grip suddenly. What kind of hardware solution do you think would work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wOduS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wOduS.jpg&quot; alt=&quot;pinball image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7249" LastEditorUserId="7249" LastEditDate="2016-07-14T23:44:03.650" LastActivityDate="2016-07-14T23:44:03.650" Title="How to pull the shooter rod on a pinball game using a servo of some kind" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10692" PostTypeId="2" ParentId="10690" CreationDate="2016-07-14T08:10:25.957" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;How safe the battery is depends on the build quality. Make sure you get a good one (maybe more expensive...). Also the conditions in which the battery (and the Arduino) will operate matter: humidity temperature, ...&lt;/li&gt;&#xA;&lt;li&gt;Your project draws the current of 200mA. A battery never 'pushes'. (for that matter, NO source pushes power... Tension [Volt] STANDS, Current [Ampere] RUNS.) Consider this: the power grid is capable of delivering... thousands(???) of amperes. good thing it doesn't do that to every appliance in your house :p&lt;/li&gt;&#xA;&lt;li&gt;it's not dangerous, however...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;which I can downstep with a voltage regulator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;be careful. 1V doesn't see much, but voltage regulator has to dissipate the excess power. this means: 1V x 200mA (the 200mA runs of course also through the regulator) means that 0,2W has to be dissipated by the regulator. Make sure it can get rid o this power. Also, this is power that is lost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;maybe it is better to get a 12V battery (which are also more compact) and get a descent DC/DC converter (+90% efficiency is easily available) which gives you a nice 5V output. &lt;/p&gt;&#xA;" OwnerUserId="7247" LastEditorUserId="1586" LastEditDate="2016-09-12T18:31:34.443" LastActivityDate="2016-09-12T18:31:34.443" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10694" PostTypeId="2" ParentId="10691" CreationDate="2016-07-14T09:42:00.387" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LxHqM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LxHqM.jpg&quot; alt=&quot;Pinball shooter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not the most elegant solution, or probably the best, but simple(ish):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd attach a lever to the puller to extend it out beyond the side (or just create space between the handle and the machine, then use a motor to spin a second lever with a bearing on it that pulls the lever back. as it spins through 90 degrees it'll release so no need for a solenoid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as there's space, you should be able to adjust the length of the motor lever to lengthen or shorten the travel, and again if you leave space the motor can just spin right around to catch the lever next time (more space than there is in my sketch!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any of that make sense?&lt;/p&gt;&#xA;" OwnerUserId="7237" LastActivityDate="2016-07-14T09:42:00.387" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10696" PostTypeId="2" ParentId="10681" CreationDate="2016-07-14T10:49:11.893" Score="1" Body="&lt;p&gt;Well, I have no idea whether this is correct, but the only thing that came to my mind after the night's sleep and a morning shower is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My inverse Laplace is wrong, because I did not account for the error input $e(t)$ that is time variant. Therefore, if we look at the derivative part of the transfer function of the controller, we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_\mathrm{D}(s) = \frac{sN}{s + N}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we take whatever the error input there is and recreate the output as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_\mathrm{D}(s) = \frac{sN}{s + N}e(s) \implies sy_\mathrm{D}(s) + Ny_\mathrm{D}(s) = Nse(s),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be transformed into time domain as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{y}_\mathrm{D}(t) + Ny_\mathrm{D}(t) = N\dot{e}(t).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution for this differential equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_\mathrm{D}(t) = \mathrm{constant} \times e^{-Nt} + e^{-Nt}\int_0^tN\dot{e}(\tau)e^{N\tau}d\tau, \quad y_\mathrm{D}(0) = 0, \quad e(0) = e_0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be implemented into the code, but I don't fancy the absolute time $t$ being there, because we are talking embedded system with finite memory and about 120 MHz clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other option is to discretize the differential equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{y_\mathrm{D}^n - y_\mathrm{D}^{n - 1}}{\Delta t} + Ny_\mathrm{D}^{n} = N\frac{e^n - e^{n - 1}}{\Delta t}, \quad n \in \mathbb{N},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y_\mathrm{D}^n(1 + N\Delta t) = N\left( e^n - e^{n - 1} \right) + y_\mathrm{D}^{n - 1}, \quad y_\mathrm{D}^0 = 0, \quad e^0 = e_0,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the time-step, &lt;em&gt;not&lt;/em&gt; power to the $n$. From this, we are able to compute the $y_\mathrm{D}^n$, sum it to the other two constants of the controller and feed that to the input of the actuators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the problem is that I am currently facing the fact that an embedded system will have to solve a differential equation, real-time. I don't know whether this is the correct approach in the industry, so I would be glad if anyone could point out a different solution. The problems I see with this are many - discretized equations are basically forward Euler, may this cause some problems? Shouldn't Runge-Kutta be used?&lt;/p&gt;&#xA;" OwnerUserId="7242" LastEditorUserId="7242" LastEditDate="2016-07-14T11:00:12.537" LastActivityDate="2016-07-14T11:00:12.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10697" PostTypeId="2" ParentId="10691" CreationDate="2016-07-14T12:50:57.433" Score="1" Body="&lt;p&gt;Well, since IAAP (physicist) &lt;strong&gt;and&lt;/strong&gt; IAAPF (pinball fanatic), I guess I'll chime in.  There's really no substitute for a locking pawl here.  You probably should replace the human-rated handle on the shooter shaft with something more like a right circular cylinder.  The pawl &amp;amp; solenoid combo are spring-loaded, so driving the entire mechanism forward allows the pawl to engage the afore-mentioned cylinder. Use a motor or whatever to pull the shaft back as far as desired, then fire the solenoid to release the pawl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lame attempt at diagram follows....&lt;a href=&quot;https://i.stack.imgur.com/CWDYH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CWDYH.png&quot; alt=&quot;plunger operator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-07-14T12:50:57.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10699" PostTypeId="2" ParentId="10681" CreationDate="2016-07-14T13:55:04.297" Score="1" Body="&lt;p&gt;Please check the following reference &lt;a href=&quot;http://www.ifac-control.org/publications/list-of-professional-briefs/pb_wittenmark_etal_final.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.ifac-control.org/publications/list-of-professional-briefs/pb_wittenmark_etal_final.pdf&lt;/a&gt;, especially page 50. I think this should help you completely.&lt;/p&gt;&#xA;" OwnerUserId="3357" LastActivityDate="2016-07-14T13:55:04.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10701" PostTypeId="1" CreationDate="2016-07-14T18:21:09.833" Score="1" ViewCount="51" Body="&lt;p&gt;One of the criteria for getting an engineering license is working under a licensed engineer for a period of time. This seems like a chicken-and-egg problem. How were the first engineering licenses issued?&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2016-07-14T18:21:09.833" Title="How were the first engineering licenses issued?" Tags="|engineering-history|licensure|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10702" PostTypeId="2" ParentId="4146" CreationDate="2016-07-14T18:57:27.273" Score="2" Body="&lt;p&gt;Based on the angled surfaces on the left, I am assuming that the option of a pass-through core on that side of the tool is not viable.  In that case. a collapsible core may be able to create the feature you are describing.  Additionally, if this is on a diameter, as I assume, if you &quot;break&quot; the latches on the diameter enough, you can do this with a partial unscrewing core (think two latches at 180 degrees apart and two gaps between them.  The feature that creates this rotates so that the steel is out from underneath the latch and then is retracted back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a couple of ideas.  Feel free to contact me if you have any questions.  This is what the company I work for does.&lt;/p&gt;&#xA;" OwnerUserId="6733" LastActivityDate="2016-07-14T18:57:27.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10703" PostTypeId="2" ParentId="2353" CreationDate="2016-07-14T19:01:00.277" Score="0" Body="&lt;p&gt;There are many stock options available on the open market to do this.  One of the things my company has done is to create it's own line of enclosures (we have a 5x7&quot;, a 4x5&quot;, and a 3x4&quot; version).  These are then used to &quot;prototype&quot; your final assembly to get it in the hands of your market target.  Once you have that feedback and demand is being generated, you know better how to design what the market needs with regard to your product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to contact me if you have any other questions.  I would be happy to help point you in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="6733" LastActivityDate="2016-07-14T19:01:00.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10705" PostTypeId="1" CreationDate="2016-07-15T02:11:07.163" Score="2" ViewCount="13250" Body="&lt;p&gt;I need to make a planetary gear system such as the following:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Izeyc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Izeyc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sun and planet gears will be stationary (so that they can only rotate around their centers, not around the sun gear). My goal is to make it so that whenever the sun gear rotates $90^{\circ}$, each planet gear will rotate $-180^{\circ}$ and the ring gear will rotate $-90^{\circ}$. I am fairly certain from my own simulations that a $2:1$ ratio for the sun-to-planet gear system will produce the $90^{\circ}:-180^{\circ}$ rotation I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Will a $2:1$ ratio make the sun and planet gears rotate in a $90^{\circ}:-180^{\circ}$ ratio?&lt;/li&gt;&#xA;&lt;li&gt;How many teeth are needed on the ring gear (or what is the gear ratio) to make the gears rotate in a $90^{\circ}:-180^{\circ}:-90^{\circ}$ fashion?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7261" LastActivityDate="2016-07-16T16:13:39.953" Title="Calculating the amount of teeth on ring gear in planetary gear system" Tags="|gears|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10708" PostTypeId="2" ParentId="7072" CreationDate="2016-07-15T07:38:47.300" Score="1" Body="&lt;p&gt;For a sleeve bushing, you want a soft slippery material with sufficient yield strength to withstand any deformation of the hole surface during operation. For the shaft, you want a hard smooth material with sufficent yield strength to withstand being deformed/bent in order to ensure that the surfaces of the shaft spin true to the axis of rotation at all times during operation. You want as close a possible fit as you can get, you want it to feel slightly snug with a small amount of friction. Spin the parts while not under any load and the rotating surface of the shaft will lap the bushing until there is zero friction and it helps to move the parts in and out along the axis slightly to eliminate as much material as you can from the bushing surface until it rotates freely. If you do all these steps and tolerance your parts correctly and use the right materials, you will have a perfect rotating bushing joint. Make sure to factor in the loads it will be operating under, and make sure to follow the guidelines above, and these kinds of joints can withstand a few hundreds of pounds with very low friction comparable to a ball bearing and gets smoother and looser as time goes on and the bushing continues to wear. You just want to prevent deformation and reduce friction.&lt;/p&gt;&#xA;" OwnerUserId="7263" LastEditorUserId="7263" LastEditDate="2016-07-15T07:43:54.870" LastActivityDate="2016-07-15T07:43:54.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10710" PostTypeId="1" AcceptedAnswerId="10714" CreationDate="2016-07-15T08:17:18.590" Score="4" ViewCount="2530" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cbxRR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cbxRR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I am sorry if this question has been asked before or if the answer is obvious, but I cannot seem to find the answer. A lot of the structural engineering problems we do involve calculating the elongation of certain types of beams/cables etc while under tensile stress. In particular there was a relatively straightforward question that asked for the total elongation of a structure comprised of two connected steel cables which have different cross sectional areas. Each of the two cables had consistent cross sectional area, and the structure was placed under an axial load of 800N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for the total elongation was simple with delta = (FL)/(EA) for each cable and adding the two deltas. My confusion however is with why this is considered the total elongation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the structure has undergone this elongation, it will become a 'new' structure, with a different length and different cross=sectional area for each cable. My question is, if the load continues to be applied after this 'initial' elongation (delta1+ delta1), then wouldn't the new structure still be under direct stress, and thus based on the new E of each of the now deformed cables, would each deformed cable undergo deformation again? And would this process not repeat until the object yields?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that in tensile testing a specimen's stress-strain graph will be  obtained by deforming the specimen until it has yielded. This is why I am confused because under load, would these cables not just deform until they yielded? And if so, wouldn't total elongation then be based on the final length just before the yield point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The two steel cables in the questions are connected together, with the cross-sectional areas of the two cables being distinct, i.e. a smaller circle stuck onto a larger one&lt;/p&gt;&#xA;" OwnerUserId="5469" LastEditorUserId="5469" LastEditDate="2016-07-15T09:54:35.627" LastActivityDate="2022-12-22T04:31:49.093" Title="How does a constant load affect the elongation of a steel cable?" Tags="|steel|stresses|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10711" PostTypeId="2" ParentId="10710" CreationDate="2016-07-15T08:50:29.830" Score="0" Body="&lt;p&gt;&lt;strong&gt;For your specific case:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total elongation = elongation of cable 1 plus elongation of cable 2. The word &quot;total&quot; is used to make sure it's obvious that it doesn't just want elongation of cable 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Short term effects:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The L in the formula is &quot;undeformed length without F&quot;. Therefore you don't need to recalculate L and apply the formula again.&#xA;The cross-sectional area A is similarly the &quot;area without F&quot;.&#xA;E does not change.&#xA;Therefore delta = (FL) / (EA) only needs to be done once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long term effects:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the long term, steel can relax. (This is similar in concept to creep in concrete, if you have heard of that.) The micro-structure rearranges, making the steel element longer. This additional deformation is plastic (non-recoverable when the load is removed) and is much smaller than the initial elastic deformation.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2016-07-15T11:26:06.640" LastActivityDate="2016-07-15T11:26:06.640" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10713" PostTypeId="1" CreationDate="2016-07-15T09:37:50.307" Score="0" ViewCount="98" Body="&lt;p&gt;I want to do something like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/t3IOt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t3IOt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I have 3 questions that need to be answered:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is it possible to make this kind of lens?&lt;/li&gt;&#xA;&lt;li&gt;If it is possible, what are the equations/measurements needed to make it?&lt;/li&gt;&#xA;&lt;li&gt;Are there another alternative to make cylindrical image flat?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Additional note: Please no &quot;image processing&quot; or programming to flatten the image because I need to take a very high and precise image of the surface and I want to prevent any kind of blur from the camera input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="7264" LastActivityDate="2016-07-15T18:09:38.387" Title="Flatten cylindrical image using lens" Tags="|optics|" AnswerCount="1" CommentCount="3" ClosedDate="2016-07-16T13:13:40.417" ContentLicense="CC BY-SA 3.0" />
  <row Id="10714" PostTypeId="2" ParentId="10710" CreationDate="2016-07-15T13:51:02.053" Score="3" Body="&lt;p&gt;This answer will assume the applied load is not enough to damage your structure (the stresses generated are lower than the elastic yield stress). I am also going to ignore the distinction between &lt;a href=&quot;https://engineering.stackexchange.com/a/8490/1832&quot;&gt;engineering and true stress&lt;/a&gt;. Both are common basic assumptions in any structural analysis class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with your reasoning is that you're pretending that a bar under load is the same as when it is unloaded. It isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the bar deforms so that the system can reach a state of equilibrium. The instant you apply the load to the bar, the system is unbalanced. There is a force applied by the rope on the bar, but the bar is unable to react according to Newton's Third Law (for every action, there is an equal and opposite reaction). The system is therefore momentarily dynamic, and the bar is stretched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the bar is stretched, this generates an internal stress state, which is felt by the rope as a reaction to the force it is applying on the bar. At some point in time, this internal stress state (and therefore the reaction to the rope's applied force) will be equal to the force applied by the rope and the system will reach equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the fundamental laws of structural analysis is &lt;a href=&quot;https://en.wikipedia.org/wiki/Hooke%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hooke's Law&lt;/a&gt;, which states that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma = E\epsilon$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be reworked into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;F_I &amp;amp;= EA\epsilon \\&#xA;F_I &amp;amp;= \frac{EA}{L}\delta&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F_I$ is the internal force in the beam (equal to the product of the stress and cross-sectional area) and $\delta$ is the beam's &lt;strong&gt;&lt;em&gt;total&lt;/em&gt;&lt;/strong&gt; displacement, in relation to it's original length $L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to then get the new length $\overline{L} = L+\delta$ and apply this equation again, you'd get a value of $F_1$ which is greater than the applied external force, which would mean the system is unbalanced, at which point the bar would shorten so as to reach a state of equilibrium once again.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-07-15T13:51:02.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10716" PostTypeId="2" ParentId="10713" CreationDate="2016-07-15T17:48:55.060" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, but it is going to be very expensive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A mirror lense could also work the same way as a lense. Ive also done this digitally with one camera and 3 flat mirrors to read cylindrical labels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably also cylindrical line cameras exist, and you can also sweep a straight one (think table scanner in arc. This is 2-3 magnitudes cheaper than a lense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A digital solution with 3 or 4 cameras is prolbably  2-3 magnitudes cheaper than a custom lense and as good quality.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Your going to get problems with lighting your target if you use a very wide angle reader.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2016-07-15T18:09:38.387" LastActivityDate="2016-07-15T18:09:38.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10717" PostTypeId="2" ParentId="10705" CreationDate="2016-07-15T18:07:04.243" Score="2" Body="&lt;p&gt;For the first question: yes you are correct, provided indeed the centres of the planet gears stay fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the other question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the needs you are specifying for planetary gearing is in conflict with each other, i.e. the problem is over-constrained. You can define a speed ratio between two types of gears (e.g. sun and planet) but not for three gear types; the third gear will already be determined in order to satisfy the gear ratio between two gear types while also satisfying geometric constraints: that is, the annulus (ring) gear must have a number of teeth equal to that of the sun gear plus twice that of a planet gear; gear pitch radius is proportional to number of teeth as teeth size on meshing gears must be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are initially six unknowns here: angular velocities of the sun, planet and annulus gears; and the number of teeth of the sun, planet and annulus gears:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_s, \omega_p, \omega_a, N_s, N_p, N_a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Six unknowns require six independent equations for a unique solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinematics, or the relations of speeds of gears, must be valid, and so this enforces 1 of the 6 equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_s N_s + \omega_p N_p = \frac{1}{2} \left( \omega_s N_s + \omega_a N_a\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometry must be satisfied; i.e. the sun and planet gears must be able to fit inside the annulus gear: equation 2 of 6:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N_a = N_s + 2 N_p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you wish to constrain the motion so that the centres of the sun and planet gear do not move; equation 3 of 6, noting pitch radius is proportional to tooth number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_s N_s = -\omega_p N_p$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equations 4 and 5 come from specifying speed ratios:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_p = -2 \omega_s$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_a = -\omega_s$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, it's tempting to think: oh the problem's under-constrained, we still need one more equation. However, whenever you substitute equations into each other, two variables will cancel out in the process: one of the angular velocities and one of the tooth numbers. This is okay: it is the ratios of speeds between gears that matter, not the absolute speeds; same for teeth numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With two unknowns dropping out, this leaves us with 5 equations and 4 unknowns: the problem is over-constrained. This means that, apart from pure coincidence, contradictions will arise whenever you try to solve the equations. An example would be like trying to solve the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x + y = 3$$&#xA;$$x + 2y = 5$$&#xA;$$2x + y = 5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two of the equations give a solution, the third creates a contradiction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you resolve this over-constraining? Remove one of the five equations! You can't remove the kinematic or geometric equations since they are fundamental laws for planetary gears. Instead, remove the constraint where the centres of the planet gears must be fixed, or enforce a speed ratio between only two gear types. Then you get four equations, four unknowns.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2016-07-15T18:13:15.573" LastActivityDate="2016-07-15T18:13:15.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10718" PostTypeId="1" CreationDate="2016-07-15T18:29:20.273" Score="2" ViewCount="90" Body="&lt;p&gt;Is there any way to simulate sound transmission properties of the human body with solid materials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While there are several sound sources inside the human body (heartbeat, air flow, bubbles moving inside the intestines) only a very small amount is transferred to the sourrounding air. It can however be tapped on by a stethoscope in a quite efficient fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this process comparable to sound in and over water, where sounds sources underwater can be heard well underwater but almost nothing leaves the water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to create a solid body that would transmit almost no sound to the air, but can be tapped on with a stethoscope? &lt;/p&gt;&#xA;" OwnerUserId="3603" LastEditorUserId="9" LastEditDate="2016-07-15T19:44:02.743" LastActivityDate="2016-08-17T14:11:28.090" Title="How to simulate sound emission properties of the human body?" Tags="|sound-isolation|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10719" PostTypeId="2" ParentId="10705" CreationDate="2016-07-15T18:29:41.240" Score="0" Body="&lt;p&gt;Have a look at the excellent &lt;a href=&quot;https://woodgears.ca/gear/planetary.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Woodgears&lt;/a&gt; site. Here he explains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Working out the tooth counts for planetary gears is actually not that complicated, so I initially neglected to mention how to do it. But having received the question a number of times, I'll elaborate.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For convenience, let's denote R, S, and P as the number of teeth on the gears.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/44JSa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/44JSa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;R Number of teeth in ring gear&lt;/li&gt;&#xA;&lt;li&gt;S Number of teeth in sun (middle) gear&lt;/li&gt;&#xA;&lt;li&gt;P Number of teeth in planet gears&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The first constraint for a planetary gear to work out is that all teeth have the same pitch, or tooth spacing. This ensures that the teeth mesh.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The second constraint is:&#xA;   $$   R = 2 × P + S $$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;That is to say, the number of teeth in the ring gear is equal to the number of teeth in the middle sun gear plus twice the number of teeth in the planet gears.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the gear at left, this would be 30 = 2 × 9 + 12&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your drawing you have constrained R and S so you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P = \frac {R - S}{2} = \frac {24 - 12}{2} = 6 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The article also covers &quot;Working out planetary gear turns ratios&quot; and from this we learn that the ratio of angle turned by the planets relative to the ring is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ T_P = \frac {R T_R}{R+S} = \frac {24T_R}{24+12} = \frac {2}{3}T_R $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where Tp is the angle of rotation of the planets and Tr is the angle of rotation of the ring. So the planets will turn 2/3 that of the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't get the ratios you wish for.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another way of looking at this is that if the planets rotate by a number of teeth on the sun gear they will &quot;throw&quot; the ring by double that number of teeth. (It may be easier to imagine this on a linear rack and pinion with the bottom rack (sun) fixed and the upper rack (ring) moveable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we rotate the planets one complete turn around your sun they will have traveled 12 teeth. The ring will have been thrown by 24 teeth but its diameter is 4/3 times that of the planet's centre so its rotation will be 24 * 3 / 4 = 18 teeth or 1.5 times that of the planets.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2016-07-16T16:13:39.953" LastActivityDate="2016-07-16T16:13:39.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10720" PostTypeId="1" CreationDate="2016-07-15T20:45:50.607" Score="2" ViewCount="4045" Body="&lt;p&gt;I need a vacuum pump to provide at least 21&quot; Hg of vacuum. Looking at the cheap rotary pumps on ebay, &lt;a href=&quot;http://www.ebay.com/itm/1-3HP-3CFM-R410a-R134-HVAC-AC-Air-Tool-Freon-Charge-Rotary-Vane-Deep-Vacuum-Pump-/322102194296?hash=item4afec97078:g:Qr4AAOSwcL5XL9gr&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; one seems to be way more than fit for the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way though I can regulate it though to provide lets say around 15-20&quot; Hg? This isn't a vacuum chamber or something and I don't want to have a crazy amount of vacuum.&lt;/p&gt;&#xA;" OwnerUserId="7269" LastEditorUserId="9" LastEditDate="2016-07-15T22:17:04.123" LastActivityDate="2016-07-18T16:44:52.273" Title="How to regulate a vacuum pump?" Tags="|measurements|vacuum|tools|vacuum-pumps|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10722" PostTypeId="1" AcceptedAnswerId="10732" CreationDate="2016-07-16T00:58:48.913" Score="0" ViewCount="286" Body="&lt;p&gt;To experienced supercar owners/engineers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best fuel for a turbocharged 6 liter V8, 7.6 liter V10, or 9.2 liter V12 (Audi R8 or Lamborghini Aventador type cars) in terms of speed/horsepower efficiency and minimization of wear on the engine? Mileage efficiency can be discarded (it can do 4 gallons per mile... not a problem :) ).&lt;/p&gt;&#xA;" OwnerUserId="7272" LastActivityDate="2016-07-16T20:21:53.227" Title="Best fuel for supercar engines" Tags="|energy-efficiency|fuel-economy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10725" PostTypeId="2" ParentId="10195" CreationDate="2016-07-16T05:20:55.270" Score="1" Body="&lt;p&gt;Hydroelectric dams approximate your description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the below schematic diagram of a hydroelectric power plant. As you can see, the kinetic energy of the water flow spins a turbine which powers a generator to produce electric power. That electric power can be used to (partially) power a pump to pump the water to whatever elevation is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sLgVo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sLgVo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2016-07-16T05:45:54.543" LastActivityDate="2016-07-16T05:45:54.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10726" PostTypeId="2" ParentId="10718" CreationDate="2016-07-16T05:28:00.443" Score="2" Body="&lt;p&gt;Use a &lt;em&gt;ballistics dummy&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ballistics dummies are designed and constructed to approximate the mechanical properties of the human body. They consist primarily of a molded ballistics gel designed for the same purpose. They are used most commonly for ballistics testing done by forensics and crime labs.&lt;/p&gt;&#xA;&#xA;Ballistics Dummy&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eKzAV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eKzAV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2016-08-17T14:11:28.090" LastActivityDate="2016-08-17T14:11:28.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10727" PostTypeId="1" CreationDate="2016-07-16T09:16:00.810" Score="-1" ViewCount="45" Body="&lt;p&gt;I am doing a project on rain detection and closing of windows. It is school based project. I am using an arduino, a rain sensor and a motor. Kindly advice on the right motor to use. And also, do I need a motor controller?&lt;/p&gt;&#xA;" OwnerUserId="7277" LastEditorUserId="110" LastEditDate="2016-07-16T12:28:39.943" LastActivityDate="2018-02-20T17:02:52.140" Title="Project on rain detection and home automation system" Tags="|power-electronics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10730" PostTypeId="2" ParentId="10710" CreationDate="2016-07-16T16:30:36.587" Score="0" Body="&lt;p&gt;This is a good question, and shows that the OP is &lt;em&gt;thinking&lt;/em&gt; about what he/she is being taught!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is perfectly possible to model the situation the way the OP proposes. The definition of &quot;engineering strain&quot; that is taught in most first courses on &quot;strength of materials&quot; is not the only way (and often is not the most useful way) to measure strain, but if students are only told about one way to measure strain, they may end up in a situation where &quot;if the only tool you have is a hammer, you have to solve every problem as if it was a nail&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP's idea leads to the formulating mechanics using &lt;em&gt;logarithmic strain&lt;/em&gt; and &lt;em&gt;true stress&lt;/em&gt;. Some industry-standard computer software (for example Abaqus) uses this formulation for &lt;em&gt;every&lt;/em&gt; problem, because &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it is no more complicated than using engineering stresses and strains&lt;/li&gt;&#xA;&lt;li&gt;it is applicable to a wider class of problems where the displacements and strains are &lt;em&gt;not&lt;/em&gt; small, and&lt;/li&gt;&#xA;&lt;li&gt;in the limit where the displacements and strains &lt;em&gt;are&lt;/em&gt; small, the results are identical for all practical purposes. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The third bullet point explains one reason why this formulation is &lt;em&gt;not&lt;/em&gt; taught in a first course - the other reason being that it requires more mathematics (including calculus) to understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Deformation_(mechanics)&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Deformation_(mechanics)&lt;/a&gt; gives a summary of several different ways to measure strains (some of which are even more general than logarithmic strain). Googling for &quot;logarithmic strain&quot; and &quot;true stress&quot; will find plenty of other relevant material.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-07-16T16:30:36.587" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10731" PostTypeId="2" ParentId="10727" CreationDate="2016-07-16T17:30:51.703" Score="1" Body="&lt;p&gt;You are getting ahead of youself.  First you have to gather the specs and requirements.  Only then can you choose parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, motors come with different power and torque ratings.  Without knowing how much force it takes to open and close a window, how far the travel is, and how fast you want the window opened/closed, you don't know what the motor needs to do.  You therefore can't pick one.  In addition to knowing what it takes to actuate the window, you also have to decide on the mechanical coupling of the motor to the window.  That may cause a different torque/power tradeoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd probably start by thinking of the mechanical coupling first.  How are you going to take the rotary motion of a motor to ultimately move the window.  Then you can take a stab at finding a motor.  This could be a iterative process.  For example, let's say you use a rack and pinion.  You lay out the basic mechanical system, but then pick a pinion gear size to match a available motor.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-07-16T17:30:51.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10732" PostTypeId="2" ParentId="10722" CreationDate="2016-07-16T20:21:53.227" Score="1" Body="&lt;p&gt;Use the fuel specified by the manufacturer in the car's manual and/or near the gas cap. Likely a High Octane Fuel based on your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're referring to Octane levels... The higher the Octane, the harder it is to ignite the fuel. In the case of High Compression Engines(typically in sports cars like you mention in your question) lower octane fuel can detonate on the compression stroke before the piston actually reaches the point where the spark plug fires. This is not good. Higher octane fuels will not pre-detonate under higher compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Newer&lt;/em&gt; engines can handle lower octane, but performance will suffer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more here: &lt;a href=&quot;http://www.contactmagazine.com/Issue54/EngineBasics.html&quot; rel=&quot;nofollow&quot;&gt;A bit about early detonation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-07-16T20:21:53.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10733" PostTypeId="1" CreationDate="2016-07-16T20:30:43.633" Score="4" ViewCount="1620" Body="&lt;p&gt;I am designing an machine that is going to use a pneumatic cylinder to accelerate a projectile. I have a tentative cylinder specified, and am working on the control and supply side. I have verified that I can generate enough force, have enough air pressure and volume, etc. The next thing I need to figure out is roughly how fast I can expect the cylinder to be moving by the end of its stroke. Is there a way to estimate this that takes into account fluid dynamics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I know:&#xA;the cylinder has a 4&quot; bore and a 6&quot; stroke.&#xA;The air pressure available is approximately 100psi (I expect I'll probably need to regulate it down to moderate the speed.)&#xA;The inlet port on the cylinder is 3/8 NPT, fed from a large tank via a short piece of 3/8&quot; schedule 40 piping for a true ID of approx .493&quot;.&#xA;The outlet port is also 3/8&quot; NPT venting directly to the atmosphere. (I will probably add a muffler, but we can ignore it for the purposes of this estimate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplistically, I know that once the full inlet air pressure is acting on the piston, the force on the cylinder is the effective area (12.6 square inches) times the air pressure (100 psi) for a force of 1,260 pounds-force. The mass of my projectile + the piston is around .3 slugs, so my acceleration should be roughly 4,200 feet per second squared. From that, I get a final velocity of roughly .875 feet per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this an appropriate way to make the estimate, or am I right in assuming that the cylinder will take an appreciable amount of time to reach full force and my actual final velocity will be lower than this? Is there a conventional way to estimate this behavior of the cylinder? If not, is there a way I can calculate what the pressure acting on the cylinder will be as a function of time to plug in to my model above?&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2016-07-16T20:38:44.023" LastActivityDate="2016-07-16T20:38:44.023" Title="Estimating Maximum Speed for a Pneumatic Cylinder" Tags="|fluid-mechanics|pressure|compressed-air|pneumatic|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10734" PostTypeId="2" ParentId="6006" CreationDate="2016-07-17T00:49:54.690" Score="1" Body="&lt;p&gt;&lt;strong&gt;Non-Linear Control Approach&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your approach is correct! I just want to point it out some tips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1^{st}$: Try to get the canonical form for the generalized coordinates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(q) \ddot{q}$ + $C(q,\dot{q}) \dot{q}$ + $g(q)$ = $u$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were $M(q)$ is the inertia matrix, $C(q,\dot{q})$ is the Coriolis matrix, $g(q)$ is the gravitational vector and, $u$, is your input for the controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2^{nd}$: The goal is to make the resulting dynamics of the system an double integrator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{q} = \ddot{q}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the resultant dynamic is decoupled!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3^{rd}$: Using the set of equations: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(q) \ddot{q}$ + $C(q,\dot{q}) \dot{q}$ + $g(q)$ = $u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u$ = $M(q) a_{q}$ + $C(q,\dot{q}) \dot{q}$ + $g(q)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{q} = \ddot{q}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just gonna need to choose the variable $a_{q}$ for the controller. (Substitute those $u$'s to verify the double integrator)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4^{rd}$: Defining the error as $e = q - q_{d}$, were $q_{d}$ is your desired joint position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\ddot{e} + K_{1}\dot{e} + K_{0}e$ = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explicit form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a_{q} = \ddot{q}_{d} - K_{1} \big( \dot{q}- \dot{q}_{d} \big) + K_{0} \big( {q}-{q}_{d} \big) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your matrices $K_{0}$ and $K_{1}$ are symmetric and positive-defined, the error dynamic is asymptotically stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/onEz1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/onEz1.jpg&quot; alt=&quot;System + COntroller&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see more, I suggest the reading of this &lt;a href=&quot;https://books.google.com.br/books/about/Robot_Modeling_and_Control.html?id=wGapQAAACAAJ&amp;amp;source=kp_cover&amp;amp;redir_esc=y&quot; rel=&quot;nofollow noreferrer&quot;&gt;book&lt;/a&gt;. ( The figure is from there ).&lt;/p&gt;&#xA;" OwnerUserId="4499" LastEditorUserId="4499" LastEditDate="2016-07-22T01:35:48.043" LastActivityDate="2016-07-22T01:35:48.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10736" PostTypeId="1" CreationDate="2016-07-17T02:20:12.407" Score="-1" ViewCount="109" Body="&lt;p&gt;I wonder whether the P34 &lt;a href=&quot;http://www.wired.com/2014/12/well-didnt-work-crazy-plan-bring-6-wheeled-cars-f1/&quot; rel=&quot;nofollow&quot;&gt;died too soon&lt;/a&gt;. Basically, I'm wondering whether the additional pair of wheels could have conferred the car any advantages that, if solved, would have given it an advantage over 4-wheeled F1 cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6-wheeled F1 cars have had a very brief existence but what if R&amp;amp;D continued? Would it have overtaken the 4-wheeled F1 cars?&lt;/p&gt;&#xA;" OwnerUserId="7119" LastEditorUserId="7119" LastEditDate="2016-07-19T11:11:38.770" LastActivityDate="2016-07-19T11:25:29.093" Title="Can a six-wheeled F1 car have a better acceleration profile than a typical F1 car?" Tags="|mechanical-engineering|automotive-engineering|wheels|acceleration|" AnswerCount="2" CommentCount="1" ClosedDate="2016-07-20T01:00:14.273" ContentLicense="CC BY-SA 3.0" />
  <row Id="10737" PostTypeId="1" CreationDate="2016-07-17T02:51:36.590" Score="0" ViewCount="760" Body="&lt;p&gt;I am designing a mixer for some 3D printer resin vats and am stuck on the best way to approach the problem. At the beginning I thought I should use three mixers in a line, all spinning in opposite directions so as to compliment each other (like an egg beater but with an extra mixer). But the more I research the less sure I am that this would be better than one, large mixer. Without doing a complex CFD I have no way of knowing which would be better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am hoping that someone with a bit more knowledge on fluid mechanics can inform me on the pro's and con's of each, and which approach would be better to mix fluid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I appreciate all the answers but they seem to avoid the question. I would just like to know the advantages of using a single mixer vs multiple mixers with a slightly viscous fluid. &lt;/p&gt;&#xA;" OwnerUserId="4189" LastEditorUserId="4189" LastEditDate="2016-07-27T10:15:56.260" LastActivityDate="2016-07-30T06:30:57.963" Title="How to best mix a viscous fluid, multiple mixers or a single mixer?" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10739" PostTypeId="1" AcceptedAnswerId="10748" CreationDate="2016-07-17T04:11:51.407" Score="1" ViewCount="1856" Body="&lt;p&gt;I used a very inexpensive circuit based on the TP4056 chip and built a Lithium Ion battery charger for 18650 batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, the batteries I use have so much capacity that the circuit, which is limited to only charging at 1 amp, takes many hours to charge my batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to do, is build a circuit that I can add to the one I purchased. This new circuit would (ideally) sample the current flow from the charger circuit to the battery, and multiply it by two.  My batteries can handle a charge current of 3 amps safely. All I want to do is charge them at 2 amps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attached a diagram that I made hopefully clearly illustrating what I am looking for. Obviously any and all of it is subject to modification based on new information, knowledge etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rkshO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rkshO.jpg&quot; alt=&quot;Existing and desired circuit diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7283" LastEditorUserId="7083" LastEditDate="2016-07-21T00:30:23.817" LastActivityDate="2016-07-21T11:00:50.387" Title="Looking for a current multiplier circuit - if possible" Tags="|electrical-engineering|consumer-electronics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10740" PostTypeId="2" ParentId="7622" CreationDate="2016-07-17T04:16:33.373" Score="0" Body="&lt;p&gt;&lt;strong&gt;Optimal Control Theory&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;I will talk about the &lt;em&gt;&lt;strong&gt;Hamilton-Jacobi-Bellman&lt;/strong&gt;&lt;/em&gt; equation. But, for more information, you can read this &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/0486434842&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;book&lt;/a&gt;. There are information about &lt;em&gt;&lt;strong&gt;Dynamic Programming&lt;/strong&gt;&lt;/em&gt; and &lt;em&gt;&lt;strong&gt;Calculus of Variation&lt;/strong&gt;&lt;/em&gt; for optimal control problems.&lt;/p&gt;&#xA;&lt;p&gt;Suppose you have a system with the dynamic described by the state equation:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\dot{\vec{x}} = f(\vec{x}, \vec{u})$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;were &lt;span class=&quot;math-container&quot;&gt;$\vec{x}$&lt;/span&gt; is the state variables and &lt;span class=&quot;math-container&quot;&gt;$\vec{u}$&lt;/span&gt; the control inputs.&lt;/p&gt;&#xA;&lt;p&gt;Usually, to control a system, we wish to minimize a performance measure:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$J = h(\vec{x}(T), \vec{u}(T)) + \int_{0}^{T} g(\vec{x}(t), \vec{u}(t), t) dt $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;were h(&lt;span class=&quot;math-container&quot;&gt;$\cdot$&lt;/span&gt;) and g(&lt;span class=&quot;math-container&quot;&gt;$\cdot$&lt;/span&gt;) are known as the terminal cost and functional, respectively.&lt;/p&gt;&#xA;&lt;p&gt;For your first question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What is known about how many solutions for &lt;span class=&quot;math-container&quot;&gt;$\vec{u}$&lt;/span&gt;  exist for such a problem?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The number of feasible control inputs, &lt;span class=&quot;math-container&quot;&gt;$\vec{u}$&lt;/span&gt;, are not relevant for the problem. The question we are looking for the answer is: there exist an optimal control law, &lt;span class=&quot;math-container&quot;&gt;$\vec{u}^{*}$&lt;/span&gt;, such that the performance measure is satisfied?&lt;/p&gt;&#xA;&lt;p&gt;For your second question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;How do engineers address/interpret the existence of multiple solutions?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;To deal with those feasible  solutions, we try to solve the minimization problem.&lt;/p&gt;&#xA;&lt;p&gt;Defining the &lt;em&gt;&lt;strong&gt;Hamiltonian&lt;/strong&gt;&lt;/em&gt; as:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mathcal{H} = g(\vec{x}(t), \vec{u}(t), t) + \frac{\partial J}{\partial x}\big(  f(\vec{x}, \vec{u}) \big)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Is necessary and, for no boundaries conditions, sufficient to show that:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{\partial\mathcal{H}}{\partial \vec{u}} = 0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial^{2}\mathcal{H}}{\partial \vec{u}^{2}} &amp;gt; 0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The solution of this partial derivative of the &lt;em&gt;&lt;strong&gt;Hamiltonian&lt;/strong&gt;&lt;/em&gt; will give you the optimal control law, &lt;span class=&quot;math-container&quot;&gt;$\vec{u}^{*}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;I hope this is useful. Sorry if it is a little to much abstract.&lt;/p&gt;&#xA;&lt;p&gt;Best regards.&lt;/p&gt;&#xA;" OwnerUserId="4499" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-07-17T04:16:33.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10741" PostTypeId="1" CreationDate="2016-07-17T06:07:01.230" Score="-1" ViewCount="76" Body="&lt;p&gt;How we can generate free electricity constantly?&#xA;with the help of magnetic field we can generate free electricity but its amount is very less, also with the help of eccentric load we can generate free electricity but its speed is very less....&#xA;but i think with the help of hybrid system like combination of gravity effect, ‘the rule of shoulder of Archimedes’, solar system and combination of capacitor bank we can constantly generate free electricity for one home???&lt;/p&gt;&#xA;" OwnerUserId="7285" LastActivityDate="2016-07-17T14:42:16.643" Title="Can we use Gravitation Force for Generation of Free electricity?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ClosedDate="2016-07-17T23:55:41.350" ContentLicense="CC BY-SA 3.0" />
  <row Id="10744" PostTypeId="1" AcceptedAnswerId="10745" CreationDate="2016-07-17T11:03:32.877" Score="2" ViewCount="1276" Body="&lt;p&gt;I want to build a small remote-controlled submarine, and have been thinking about how to handle the ballast tanks. The simplest option seems to be a piston shaft, which fundamentally needs some kind of (sealed) linear actuator with sufficient force to overcome water/air pressure at reasonable depths (say 0m to 50m).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is I don't have an infinite supply of power available on-board and all the linear actuators I found use ludicrous amounts of it, such as over 1 amp at 12 volts... but I only need to move the piston every now and then, most of the time it just needs to sit wherever I tell it to and not move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, once  a linear actuator has been &quot;actuated&quot; into place, can I de-power it and expect it to hold up to its maximum rated load? Does it depend on the design of the actuator? If not, what are some engineering approaches I can use to mechanically lock the piston into place until I need to unlock it, so that I can trade off precious electrical power for mechanical strength?&lt;/p&gt;&#xA;" OwnerUserId="5659" LastActivityDate="2016-07-17T11:36:39.193" Title="Do linear actuators need to stay powered once in place? If not, how can I lock them?" Tags="|mechanical-engineering|motors|actuator|linear-motors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10745" PostTypeId="2" ParentId="10744" CreationDate="2016-07-17T11:36:39.193" Score="3" Body="&lt;p&gt;Some linear actuators are self-locking &amp;mdash; they can't be back-driven by force on the output connection. A leadscrew with a fine enough thread pitch is one example, and since you don't need a lot of speed for your application, wold probably be a good choice.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-07-17T11:36:39.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10746" PostTypeId="2" ParentId="10741" CreationDate="2016-07-17T11:44:53.863" Score="1" Body="&lt;p&gt;No, the physics doesn't work that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both gravity and magnetism are examples of &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Conservative_vector_field&quot; rel=&quot;nofollow&quot;&gt;conservative vector fields&lt;/a&gt;, which means (among other things) that the net change in energy for any closed path in the field is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you can't extract power from them via any continuous (repetitive) process. You can only use them to assist with the conversion of power from one form to another, such as with a motor or generator.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-07-17T11:44:53.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10747" PostTypeId="2" ParentId="10736" CreationDate="2016-07-17T12:03:21.740" Score="0" Body="&lt;p&gt;The premise that increasing the area of the contact between the tires and the road somehow increases traction is fundamentally flawed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As any first-year physics student can tell you, the available frictional force is proportional only to the coefficient of friction between the two materials and the total force holding them together &amp;mdash; the area of the contact drops out of the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only advantages to adding more wheels would have to be due to secondary effects, such as being better able to &lt;em&gt;maintain&lt;/em&gt; contact in dynamic situations such as driving on a rough surface. And any such advantages would have to be balanced against the disadvantages associated with the extra complexity, power losses and weight.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-07-17T12:03:21.740" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10748" PostTypeId="2" ParentId="10739" CreationDate="2016-07-17T13:02:49.223" Score="2" Body="&lt;p&gt;Theoretically TP4056 is capable of charging at 1.4A. Practically your design is limited to the capability of the USB source par the picture or your external source. You need to change the resistor on Program pin on the TP4056 to the desired charge current &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5mwiF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5mwiF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per the circuit above the resistor is R4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also from the LTC4056 data sheet: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;PROG (Pin 5):&lt;/strong&gt; Charge Current Programming Pin. Provides&#xA;  a virtual reference voltage of 1V for an external&#xA;  resistor (RPROG) connected between this pin and ground&#xA;  to program the battery charge current. In constant current&#xA;  mode the typical charge current is 915 times the current&#xA;  through this resistor (ICHG = 915V/RPROG). Current is&#xA;  limited to approximately 1.4mA (ICHG of approximately&#xA;  1.4A).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Based on the maximum available Li-Ion battery capacity you should be able fully charge the battery under 4hrs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Word of caution:&lt;/strong&gt; Although you might want to charge at 2A the battery chemistry might not recommend such a high charge rate. Read the battery chemistry specification very carefully. Improper use can be dangerous. For example &lt;a href=&quot;https://batterybro.com/products/lg-mh1-inr18650mh1-3200mah-10a&quot; rel=&quot;nofollow noreferrer&quot;&gt;LG Chem 18650 MH1&lt;/a&gt; has a standard charge current of 1.5A and Max charge current of 3.1A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a chart on charging Li-Ion batteries. In most cases it is recommend that the battery be charged 0.8C. Also as the battery reaches full capacity the charge rate also drops as described by the chart below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n5GrF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5GrF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The advised charge rate of an Energy Cell is between 0.5C and 1C; the complete charge time is about 2–3 hours. Manufacturers of these cells recommend charging at 0.8C or less to prolong battery life. Most Power Cells can take a higher charger. Charge efficiency is about 99 percent and the cell remains cool during charge.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reference: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://cds.linear.com/docs/en/datasheet/405642f.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;LTC4056-4.2 - Linear Li-Ion Charger with Termination in ThinSOT&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.electroschematics.com/10551/tp4056-lipo-battery-charger-rc-toys/&quot; rel=&quot;nofollow noreferrer&quot;&gt;TP4056 LiPo Battery Charger For RC Toys&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://batteryuniversity.com/learn/article/charging_lithium_ion_batteries&quot; rel=&quot;nofollow noreferrer&quot;&gt;Charging Lithium-ion&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-07-17T16:50:44.677" LastActivityDate="2016-07-17T16:50:44.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10750" PostTypeId="2" ParentId="10741" CreationDate="2016-07-17T14:42:16.643" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can we use Gravitation Force for Generation of Free electricity?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. See hydro-electricity. This initially uses solar power to evaporate water which later falls as rain. If the water is collected at a height suitably high enough to drive a turbine we use gravity to drive the turbine.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How we can generate free electricity constantly? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Free in what way? Solar and hydro are free in that we don't pay for the fuel. We do pay for the hardware.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... with the help of magnetic field we can generate free electricity but its amount is very less, ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is a poorly worded sentence and makes no sense.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... also with the help of eccentric load we can generate free electricity but its speed is very less.... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So is this.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... but i think with the help of hybrid system like combination of gravity effect, ‘the rule of shoulder of Archimedes’, solar system and combination of capacitor bank we can constantly generate free electricity for one home???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And this.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2016-07-17T14:42:16.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10752" PostTypeId="2" ParentId="10736" CreationDate="2016-07-17T17:38:52.107" Score="3" Body="&lt;p&gt;The complication here is that the simple relationship between wight (or downforce assisted equivalent) coefficient of friction and traction is only true up to a point and in performance tyre  that point is where the thermal and mechanical loads on a tyre start to change its mechanical properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Performance tyres are generally designed to work best with a small amount of slip angle ie just at the point where static friction transitions to sliding friction so in an F1 context the contact patch of the tyre very much does make a difference. In this context the 'grip' is effectively the maximum load a tyre can tolerate before it transitions from static (rolling) friction) to sliding friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not so much about the 'classical' theory of static friction as that ability of the tyre compound to deal with the amount of energy being put into it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 6 wheeled car was not so much about the fact that 4 drive wheels are better than two but a clever exploitation of the regulations which limited rear tyre size but neglected to specify that you could only have two axles, something witch was very quickly changed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the 2017 F1 regulations increase the width of the tyres for the explicit reason of increasing performance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary while a larger tyre contact patch does not increase grip/traction in general it very much does increase the &lt;strong&gt;upper limit&lt;/strong&gt; of grip where the limiting factor is the breakdown of the tyre compound.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-07-19T11:25:29.093" LastActivityDate="2016-07-19T11:25:29.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10754" PostTypeId="1" AcceptedAnswerId="10758" CreationDate="2016-07-17T22:49:27.983" Score="5" ViewCount="4185" Body="&lt;p&gt;The vast majority of wind turbines I have ever seen feature 3 blades, as shown here:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/InIXy.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/InIXy.jpg&quot; alt=&quot;3&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Recently while driving through western new York state, however, I passed by several turbines featuring only two blades, as shown here:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G5FMx.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G5FMx.jpg&quot; alt=&quot;2&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;What advantages and disadvantages would a 2-bladed wind turbine have over 3-bladed versions?&lt;/p&gt;&#xA;" OwnerUserId="5953" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-04-25T21:18:38.833" Title="Advantages/Disadvantages of two-bladed windmill?" Tags="|power-engineering|wind-power|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10755" PostTypeId="2" ParentId="2733" CreationDate="2016-07-17T23:18:19.463" Score="1" Body="&lt;p&gt;I've got a few of the circular motors in my junk box and took one apart. With what I've found on the net, and what I see in the unit I have, here is what I found.  The circular units are for very small wattage requirements and usually are applied in equipment, like the fish tank pump and microwave turntable discussed here. The unit I have came out of a commercial printer as a feed roller driver. The round units are low cost and do not need a winding armature; a simple round winding into which is set a ceramic rotor on a bearing. Without a magnetic conductor, or armature, there is no room to put a shaded pole copper ring. That means the motor needs some method to insure it rotates in the proper direction. And that is where a simple mechanical cog comes into play. The rotor has a feature that looks like a bow-tie and above that a small gear, which drives #1 reduction gear. This #1 gear has a friction arm below it. As the #1 gear turns, the arm turns with the gear until it hits a stop, and the gear keeps turning.  If the motor starts the &quot;right&quot; way, the arm moves away from the bow-tie feature. If the motor started the &quot;wrong&quot; way, the arm moves toward the bow-tie and the design is such that at the prescribed angle the rotor/bow-tie slams into the arm. The angle seems away from a magnetic cogging pole, about in the middle between 2 magnetic &quot;bumps.&quot; When the impact occurs, you can see the rotor bounce off the arm and instantly starts in the &quot;right&quot; direction, which moves the arm out of the way. &lt;/p&gt;&#xA;" OwnerUserId="7292" LastActivityDate="2016-07-17T23:18:19.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10756" PostTypeId="2" ParentId="10720" CreationDate="2016-07-18T01:05:42.240" Score="0" Body="&lt;p&gt;So you just need to add a sensor that will turn off the pump when the pressure is in your acceptable range and turns it on again when the pressure rises too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more volume is in the chamber the more time it will take for the pump or a leak to change the pressure. You can use that to increase the time between duty cycles.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2016-07-18T01:05:42.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10757" PostTypeId="2" ParentId="10754" CreationDate="2016-07-18T01:17:22.267" Score="3" Body="&lt;p&gt;One of the advantages of two blades over three blades is less materials are required to make the blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disadvantage of a two bladed rotor compared to a three bladed rotor is due to the interruption to airflow and additional turbulence that the pole, that the wind turbine sits on, produces. This becomes apparent when one of the blades is directly in line with the pole. The pole interrupts the turbulence dissipation of the blade that is in line with it. This can increase vibration within the whole system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three bladed rotors can produce smoother rotation because they have two blades at the top of the pole when the third blade is in line with the pole.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-07-18T01:17:22.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10758" PostTypeId="2" ParentId="10754" CreationDate="2016-07-18T02:52:30.400" Score="5" Body="&lt;p&gt;A 2-bladed wind turbine is less stable mechanically than 3 (or more) blades. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the two blades are in line, it is much easier to twist the hub of the turbine in the direction along the line of the blades than to twist it at right angles to the blades. If the turbine is being rotated because the wind direction changes, this will cause an unbalanced twisting force on the hub (and the pole) at twice the rotation frequency of the blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This unwanted twisting of may feed back into the blades causing them to vibrate. If the speed of the twisting matches the natural vibration frequency of the blades, this could lead to catastrophic mechanical failure. As a consequence of this, 2-blade turbines usually have to be rotated slower than those with more blades, if the wind direction changes, but the wind blowing in an off-axis direction also produces the same unwanted twisting effects. Catch-22!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hub mechanism and/or the pole itself may need to be designed stronger or stiffer to resist these unwanted twisting forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 3 or more blades, the mechanical behaviour of the turbine is actually the same in &lt;em&gt;all&lt;/em&gt; possible orientations of the blades. Even if there are even number of blades, for example 4, there is nothing &quot;special&quot; mechanically about the 90-degree angle between two adjacent blades but economically 4 blades are more expensive to make than three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanical issues with 2-blade turbines can be &quot;designed out,&quot; but the optimum number of blades to generate the maximum &lt;em&gt;power&lt;/em&gt; for a given size of turbine is typically 4, ignoring considerations like the extra weight of more blades. As such, a 3-blade design is often the best &quot;minimum cost&quot; compromise between a simple and reliable design, high power output, and low weight.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="-1" LastEditDate="2019-04-25T21:18:38.833" LastActivityDate="2019-04-25T21:18:38.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10759" PostTypeId="1" CreationDate="2016-07-18T06:30:05.680" Score="2" ViewCount="304" Body="&lt;p&gt;I am working on an idea for my capstone project. I am trying to get an idea of a device, sensor or reader that can detect the front traffic light situation. so that it can tell when the light is green or red. I am still not sure what is the best way to detect the light? Is there any possibilities that the sensor can tell the light color without using any corresponding sensors or devices in each single traffic light? I was trying to find how driverless cars can see the traffic lights but still no ideas yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate your participation. &lt;/p&gt;&#xA;" OwnerUserId="7295" LastEditorUserId="9" LastEditDate="2016-07-23T01:08:56.500" LastActivityDate="2016-07-23T01:08:56.500" Title="How can a sensor detect the traffic lights?" Tags="|traffic-light|" AnswerCount="2" CommentCount="1" ClosedDate="2016-07-18T20:56:14.630" ContentLicense="CC BY-SA 3.0" />
  <row Id="10760" PostTypeId="1" CreationDate="2016-07-18T09:49:03.550" Score="1" ViewCount="160" Body="&lt;p&gt;I am trying to find again a small plastic damper that I was looking at last year, but forgot the exact name for that type and now I am unable to find it again. Quite frustrating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is pretty much a plastic piston with a small hole at the end, which acts as a linear damper and that is pushed back by a spring. Pretty much like those RC shock absorbers, but far simpler and cheaper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to use it for slowing down a small lid and would be the best solution, but can't find the exact name. &lt;a href=&quot;https://i.stack.imgur.com/0TfnQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0TfnQ.jpg&quot; alt=&quot;something similar to this, but far simpler and cheaper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7297" LastActivityDate="2016-09-08T16:43:46.683" Title="Axial damper type?" Tags="|mechanical-engineering|pneumatic|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10762" PostTypeId="1" CreationDate="2016-07-18T11:14:51.950" Score="1" ViewCount="112" Body="&lt;p&gt;I know a &lt;a href=&quot;https://en.wikipedia.org/wiki/PID_controller&quot; rel=&quot;nofollow&quot;&gt;PID controller&lt;/a&gt;. Recently, I heard someone mention a PIAD controller. What is this? Is it a kind of extension of a regular PID controller?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the control engineering litterature, I have only found a &lt;a href=&quot;https://www.researchgate.net/profile/Dingyu_Xue/publication/224055997_A_comparative_introduction_of_four_fractional_order_controllers/links/0fcfd50a9b59b40dd8000000.pdf&quot; rel=&quot;nofollow&quot;&gt;$\text{P}\text{I}^λ\text{D}^µ$&lt;/a&gt; controller yet.&lt;/p&gt;&#xA;" OwnerUserId="3633" LastActivityDate="2016-07-19T17:35:19.830" Title="What is a PIAD controller?" Tags="|control-engineering|pid-control|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10763" PostTypeId="2" ParentId="10759" CreationDate="2016-07-18T12:05:27.527" Score="3" Body="&lt;p&gt;This will be done with a video camera. Writing software to process images is easier than it used to be. People are writing software on Raspberry pi (and I think, impressively, even Arduino)  that do this sort of thing. And given that your application is a driverless car, it would need similar capabilities anyway. MATLAB software is actually capable of virtually writing the code for you. The difficult bits anyway. You'll need to buy various add-ons. While this is the easiest solution, bare in mind that &lt;strong&gt;this is still a hard problem.&lt;/strong&gt; The tools are not for the faint hearted or the impatient. See here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://uk.mathworks.com/products/computer-vision/&quot; rel=&quot;nofollow&quot;&gt;Computer Vision System Toolbox&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can't imagine it having been done any way besides processing images from a camera. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that answers your question?&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2016-07-18T12:17:17.497" LastActivityDate="2016-07-18T12:17:17.497" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10764" PostTypeId="1" CreationDate="2016-07-18T12:07:25.413" Score="1" ViewCount="195" Body="&lt;p&gt;I'm working with a custom made mini computer, &lt;a href=&quot;http://www.ti.com/lsds/ti/processors/sitara/arm_cortex-a8/am335x/overview.page&quot; rel=&quot;nofollow&quot;&gt;am335x&lt;/a&gt; based. &#xA;It comes with 2 GPIO ports that we use to provide services to our customers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to change one of those specific ports, that is already being attached to a specific device, to use an internal pull down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The port is always with value 1 when no voltage is provided, and I need to change that because when that device provide no voltage, it has the same internal state has when it provides 0V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the information of the specific pin:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    root@be905g:~# cat /sys/kernel/debug/pinctrl/44e10800.pinmux/* | grep &quot;pin 20&quot;&#xA;    pin 20 (44e10850.0)&#xA;    pin 20 (44e10850.0): 48302180.eqep (GPIO UNCLAIMED) function pinmux_eqep1 group pinmux_eqep1&#xA;    pin 20 (44e10850.0) 0000002e pinctrl-single &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I saw that the &quot;0x2e&quot; value that is associated to the pin should be &quot;0x27&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've modified the DTS file and recompiled, and even with 0x27 in memory the port behaves as having a pull up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used the command:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    mem_util 0x44e10850&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and it returned 0x27, but still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how the pin description in the &quot;.dts&quot; file I decompiled from the shipped &quot;.dtb &quot;looks:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    pinmux_eqep1 {&#xA;                pinctrl-single,pins = &amp;lt;0x50 0x2e 0x54 0x36&amp;gt;;&#xA;                linux,phandle = &amp;lt;0x40&amp;gt;;&#xA;                phandle = &amp;lt;0x40&amp;gt;;&#xA;            };&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7298" LastEditorUserId="7083" LastEditDate="2016-07-21T00:15:39.653" LastActivityDate="2016-07-21T00:15:39.653" Title="am335x use gpio with pull down" Tags="|electrical-engineering|embedded-systems|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10765" PostTypeId="2" ParentId="10759" CreationDate="2016-07-18T12:10:24.220" Score="1" Body="&lt;p&gt;The state of any traffic lights is just one of the many elements that are extracted from the scene in front of the vehicle using video cameras and &lt;a href=&quot;http://opencv.org/&quot; rel=&quot;nofollow&quot;&gt;machine vision&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Software identifies objects such as other vehicles, pedestrians and other potential obstacles, plus information such as road edges, lane markings, signs and traffic signals.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2016-07-18T12:34:01.810" LastActivityDate="2016-07-18T12:34:01.810" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10766" PostTypeId="2" ParentId="10720" CreationDate="2016-07-18T13:46:40.090" Score="3" Body="&lt;p&gt;The device you're looking for is called a vacuum regulator and sits between the vacuum source and the process. &lt;a href=&quot;https://www.equilibar.com/vacuum-regulators/&quot; rel=&quot;nofollow&quot;&gt;This page&lt;/a&gt; from Equilibar shows some examples and provides links that explain how they work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use a vacuum breaker, which operates more or less like a pressure relief valve, but a vacuum regulator will offer more precise control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a variety of low cost vacuum regulator and vacuum breaker options on eBay. I have never used any of those, but I have used a few high end, electrically controlled versions on automated systems. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2016-07-18T16:44:52.273" LastActivityDate="2016-07-18T16:44:52.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10767" PostTypeId="1" AcceptedAnswerId="47053" CreationDate="2016-07-18T14:27:39.313" Score="5" ViewCount="46236" Body="&lt;p&gt;I have come across various values for maximum flow velocity of water in pipes.&#xA;At university we were told that 2 m/s is the upper limit where flow becomes turbulent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In various planning reports I have read recently, 1 m/s is stated as the upper limit. I could calculate the exact velocity at which water in a specific size and type of pipeline will no longer flow in a laminar way, but for infrastructure planning purposes, this exercise is not required. What would a reasonable value for potable water be? Alternatively, what would the value for raw water (river) be?&lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2024-03-26T22:19:10.010" Title="Rule of thumb maximum flow velocity of water in pipes" Tags="|civil-engineering|water-resources|" AnswerCount="8" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10769" PostTypeId="1" AcceptedAnswerId="10775" CreationDate="2016-07-18T18:23:18.170" Score="1" ViewCount="1703" Body="&lt;p&gt;I have an application that is drawing fluid from a reservoir via a standard dip tube. This application has a special requirement that once the fluid is depleted from the reservoir, the pump pulling liquid from the reservoir needs to shut off before drawing air into the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function of this type of valve would be more or less the opposite of a floating shutoff valve used in toilet tanks where the floating mechanism closes the valve once the tank is filled. My application would do the opposite, where the float closes a valve on the dip tube or shuts of the pump once the tank is empty (or close to it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know if this type of valve exists on the market, or is it something I need to design on my own? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7301" LastActivityDate="2023-12-28T14:34:38.490" Title="Reverse float shutoff valve" Tags="|pumps|hydraulics|valves|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10770" PostTypeId="1" CreationDate="2016-07-18T18:29:00.017" Score="4" ViewCount="3230" Body="&lt;p&gt;I know some biologist that are &lt;a href=&quot;http://redpath-staff.mcgill.ca/hendry/Foster2007JEB21,263.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;taking curvature measurements&lt;/a&gt; on photos with softwares to understand the variation on beak morphology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if there is a way to build a mechanical tool (or if one exist already) to measure the curvature of a &lt;a href=&quot;http://laser.physics.sunysb.edu/~brandon/journal.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;circle's (or ellipse) arc&lt;/a&gt;. The arc is not always of the same length, so it would need to be adjustable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a constrain (see the grey line): &#xA;&lt;a href=&quot;https://i.stack.imgur.com/lNyt1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lNyt1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't fit a &lt;a href=&quot;http://www.sharpfine.com/Calipers/digital-radius-caliper.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;big calliper&lt;/a&gt; on the bird's head. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the radius of the circle that correspond to the beak curvature&lt;/li&gt;&#xA;&lt;li&gt;how &quot;steep&quot; is that curvature (probably a lack of vocabulary on my side but see the image down to see what I mean by &quot;steep&quot;)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5Rxkv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Rxkv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this image as the different birds in the image above. Bird #4 would have a flatter beak than bird #1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to be able to calculate the &quot;steepness&quot; even if the x, y or z are not of the same length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea of bringing a mechanical tool is to calculate rapidly and conveniently a number and add it directly to a dataset. It's much faster than to look at photos and analyse with a software. So if there is a mechanical tool designed for this, it would be super useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the image is just a visualization of the problem. I would like to measure the beak of &lt;a href=&quot;http://www.hdouglaspratt.com/Art_Gallery/HawaiianHoneycreepers/hawaiian_honeycreepers_evolution_radiation.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;different birds on the field&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7284" LastEditorUserId="7284" LastEditDate="2016-07-20T14:39:19.510" LastActivityDate="2020-01-15T11:17:51.750" Title="Can we build a tool that measure curvature on an arc?" Tags="|measurements|tools|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10771" PostTypeId="1" CreationDate="2016-07-18T19:24:15.657" Score="1" ViewCount="32" Body="&lt;p&gt;Generally, there is a talk about noise margin and noise immunity in digital circuits. Both terms seem interrelated but some texts differ from this. My query is whether these two are entirely different terms? If so, then is there a way to quantify noise immunity?&lt;/p&gt;&#xA;" OwnerUserId="7302" LastEditorUserId="1832" LastEditDate="2016-07-18T19:58:25.743" LastActivityDate="2016-07-19T11:11:22.537" Title="Is there a way to quantify or graphically represent noise immunity? Is noise immunity different than noise margin? If yes, how?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10772" PostTypeId="2" ParentId="10770" CreationDate="2016-07-18T20:50:37.133" Score="6" Body="&lt;p&gt;For a circle, yes. Such a tool exists:&#xA;&lt;a href=&quot;http://www.sharpfine.com/Calipers/digital-radius-caliper.html&quot; rel=&quot;nofollow&quot;&gt;Digital Radius Caliper&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also &lt;a href=&quot;http://www.cutwel.co.uk/measuring-tools/small-tool-instruments/miscellaneous-gauges/insize-digital-radius-gauge-2183-series?utm_source=google&amp;amp;utm_medium=shopping&amp;amp;utm_campaign=purenet_feed&amp;amp;VariantId=2183&amp;amp;gclid=CjwKEAjwn7e8BRCUqZiP_vnrtBkSJAC_lp4HZdoiTLI8lmAI6p1BbZQq1Dd8J7dQVCB9ml7Yx7SMlhoCQeLw_wcB&quot; rel=&quot;nofollow&quot;&gt;Digital Radius Gauges&lt;/a&gt;, which may be useful for smaller curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a general curve, you can't measure the curvature exactly since it varies continuously along the curve. You could perhaps assume a region is approximately circular and measure that. But depending on the size of the tool and the rate of curvature change this could be wildly inaccurate.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2016-07-19T10:16:10.147" LastActivityDate="2016-07-19T10:16:10.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10773" PostTypeId="2" ParentId="10770" CreationDate="2016-07-18T21:45:39.797" Score="4" Body="&lt;p&gt;How about a &lt;strong&gt;Contour Gauge&lt;/strong&gt;:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IaAic.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IaAic.jpg&quot; alt=&quot;Contour Gauge&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here:&#xA;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Profile_gauge&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia Article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Make your measurement&lt;/li&gt;&#xA;&lt;li&gt;Trace it&lt;/li&gt;&#xA;&lt;li&gt;Scan/digititize it&lt;/li&gt;&#xA;&lt;li&gt;Analyze it (spline/curve fit) with some type of CAD software. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2016-07-19T11:25:32.913" LastActivityDate="2016-07-19T11:25:32.913" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10774" PostTypeId="2" ParentId="10710" CreationDate="2016-07-18T22:14:29.800" Score="1" Body="&lt;p&gt;The entire structure acts like a spring; it stretches/expands until the point when it is strained enough to carry the load. It will not stretch any more if you keep applying the force because it is at equilibrium.  Like the scale at home we weigh ourselves with. By stepping on the scale you apply force to the spring, it will contract or expand till its loaded sufficiently to support your weight. It won't keep expanding if you don't get off it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can test the equilibrium by jumping on the scale or adding a short impact load to your example by adding a 20 kg load and removing it fast. the system start vibrating reacting to this perturbation and depending on the mechanical damping this vibration will die out over a few periods.&#xA;The beauty of using elastic materials for structural members is just that: the resist large loads by expanding a bit and go back to their original position after removal of the force/load. No member will start to carry any load before it has been deformed in one way or the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar configuration of what you show here is used as suspension spring for cars: a soft but long spring in series with a short but hard spring. the soft spring takes most of road's bumps with soft reaction and comfort for passengers but the hard spring comes into action when the is big pot holes on the road!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-07-18T22:14:29.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10775" PostTypeId="2" ParentId="10769" CreationDate="2016-07-19T05:38:36.457" Score="0" Body="&lt;p&gt;As I understand your problem, you have a pump mounted outside the tank that draws fluid via a dip tube, and you want to protect this pump from running dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to turn off the pump anyway once the tank is empty, so I would use a level switch (float, conductive, swing ... any will do) to shut off the pump and have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Check_valve&quot; rel=&quot;nofollow&quot;&gt;check valve&lt;/a&gt; on the line somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many switches can be mounted above the tank so that you can service them without getting your hands wet. The valve should also be mounted dry. Make sure the pipe always ends under the water level, I would always leave 1,5 time the pipe diameter water in the tank and have 1x pipe diameter space between pipe end and floor.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-07-19T05:38:36.457" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10776" PostTypeId="2" ParentId="10771" CreationDate="2016-07-19T11:11:22.537" Score="1" Body="&lt;p&gt;In my experience, the two refer to the same thing.  However, &quot;noise immunity&quot; is a concept, and &quot;noise margin&quot; is a quantity.  A salesman might say how the product has exceptional noise immunity, but the spec sheet would list the noise margin in volts, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of digital circuits, the noise margin is the difference between the nominal voltage level of a signal, and the nearest voltage at which it could be interpreted to be in a different state.  For example, let's say a digital input has a logic low threshold of 1.2 V and a logic high threshold of 2.7 V.  With 3.3 V in, the noise margin is 600 mV.  If the 3.3 V signal dips by 600 mV, it is no longer guaranteed to be interpreted as a logic high.  A 0 V signal has 1.2 V of noise margin in this example.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-07-19T11:11:22.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10779" PostTypeId="2" ParentId="10762" CreationDate="2016-07-19T17:35:19.830" Score="2" Body="&lt;p&gt;The A could be acceleration. The acceleration part is then used as feedforward or feedback in addition to the ordinary P, I and D parts. Take for instance a height controller for a quadcopter. For an ordinary PID controller the equation would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = K_p \cdot e + K_i \cdot \int e\ dt + K_d \cdot \dot{e}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $e = h_{d} - h$ (error between desired height and actual height above ground) and $F$ is the wanted up-down (heave) force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens if the desired height is 2 m and you let go of the quadcopter at the same height? It probably hits the ground, as the integral term of the controller never had time to wind up, unless you have very high gains. If you modify the PID controller to include a very basic feedforward of the acceleration we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = K_p \cdot e + K_i \cdot \int e\ dt + K_d \cdot \dot{e} + m a$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the quadcopter mass and $a$ is the acceleration. When you let go of the quadcopter at 2 m the equation reduces to $F = mg$, and the wanted heave force immediately cancels the gravity. The integral part doesn't have to be used. The acceleration has be either measured or calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I see an A together with PID in my field (motion control) it usually means acceleration, but it could of course be something else in another field.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2016-07-19T17:35:19.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10781" PostTypeId="1" AcceptedAnswerId="10786" CreationDate="2016-07-20T03:40:14.553" Score="0" ViewCount="5507" Body="&lt;p&gt;In our dynamics class, it is highly recommend to us to use D'Alembert's approach rather than the traditional approach to solving dynamic problems. For example, consider the problem of a block resting on a surface (we know the cof): Rather than calculating the frictional force of the block on the slope and comparing that to the component of weight acting down the slope, we draw the free body diagram including the inertial force and directly solve for a, which will tell is if the block is moving or not. My question is why this method is recommended over the traditional approach? It seems the answer is usually with regards to making it easier to solve more complex problems, but I don't quite understand why this trivial manipulation of Newton's second law makes it much easier to solve problems? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry if this has been asked before, but I couldn't quite find a satisfactory answer online!&lt;/p&gt;&#xA;" OwnerUserId="5469" LastActivityDate="2016-07-20T13:52:26.780" Title="What is the advantage of using D'Alembert's principle when dealing with dynamic systems?" Tags="|dynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10782" PostTypeId="1" AcceptedAnswerId="10822" CreationDate="2016-07-20T05:28:43.407" Score="2" ViewCount="211" Body="&lt;p&gt;I have a 1.5 mm thick plastic sheet part which hast to be bent at a joint 0.3 mm thick by ~ 10 Deg. I know (found from design drawing) compressed and stretched lengths of the joint. How can I find the length of the joint in non-bent (flat) position so that when it is bent the compressed and stretched joint lengths have the initial known lengths ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rH10e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rH10e.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7314" LastActivityDate="2016-07-23T21:38:30.843" Title="Finding initial length of a bended plastic sheet (Bend Allowance)" Tags="|mechanical-engineering|structural-engineering|materials|plastic|deformation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10783" PostTypeId="1" AcceptedAnswerId="10785" CreationDate="2016-07-20T06:27:43.213" Score="3" ViewCount="1287" Body="&lt;p&gt;I have a plate of aluminum 5 mm thick to which I would like to attach a hook. The plate runs roughly vertically, but only spans around 18 cm from anchor point to anchor point. The hook should be able to withstand upwards of 1000 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've drilled a 12 mm hole through the plate and my strategy is to insert the hook bolt in it and tighten it using a nut on each side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in order to keep the aluminum from deforming, I'm going to need one or two steel washers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, whether it is necessary to use a washer on both sides or whether one side is sufficient?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one is enough, does it make any difference, which side it rests on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning goes like this: The job of the washer is to keep the hook from bending the aluminum by spreading the force over a large area. As long as a single washer is taut against the aluminum – held there by the nut – the hook won't be able to bend either way.&lt;/p&gt;&#xA;" OwnerUserId="7317" LastActivityDate="2016-07-20T20:40:27.213" Title="Washers on both inside and outside?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10784" PostTypeId="2" ParentId="4090" CreationDate="2016-07-20T06:30:22.747" Score="-1" Body="&lt;p&gt;You were using the wrong unit for density. It must be in slugs/ft&lt;sup&gt;3&lt;/sup&gt;, not in lbm/ft&lt;sup&gt;3&lt;/sup&gt;. By the way, 32.2 lbm = 1 slug.&lt;/p&gt;&#xA;" OwnerUserId="7316" LastEditorUserId="86" LastEditDate="2016-07-20T12:39:06.597" LastActivityDate="2016-07-20T12:39:06.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10785" PostTypeId="2" ParentId="10783" CreationDate="2016-07-20T13:39:10.283" Score="6" Body="&lt;p&gt;While your setup will probably work either way, it's advisable to use a washer on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A washer does help prevent local bending as you note, but that isn't really its main purpose. The biggest reason to use a washer is to prevent the nut from cutting into or shearing through the base material. In your case where the hook bolt and nut are steel, and the plate is softer aluminum, this is especially a concern. As you tighten the connection, without a washer the nuts are likely to dig into the aluminum and start to act like a drill. It also wouldn't be too hard for the nut without a washer to be pulled through the plate (causing a shear failure in the shape of the nut) if there isn't a washer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in a similar case, using a shouldered eye bolt where the bolt, the substrate, and the nut are the same material, it would be customary to only use the washer on the nut side, because the bolt is not designed to ever be loaded in compression, only in tension against the nut. Because of this the bolts are designed with a shoulder that isn't as big as a washer, but meets the needs of the bolt for the angles it is rated at. Washers may be added when required for thread length or orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that based on your description it sounds like you may be side-loading the hooks which is generally not how they are designed to be used. If you want to make a separate question with the plate, the hook selection, the geometry, and the load, you could make sure you are using the hardware correctly. 1000 kg is a very serious load and if it's going to be anywhere near people, the setup should be evaluated according to standard rigging practices in your country.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="122" LastEditDate="2016-07-20T20:40:27.213" LastActivityDate="2016-07-20T20:40:27.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10786" PostTypeId="2" ParentId="10781" CreationDate="2016-07-20T13:43:01.747" Score="3" Body="&lt;p&gt;I don't really understand how D'Alembert's principle &quot;helps&quot; in solving the problem in your OP, but I assume your teacher wants you to practise using it in simple situations before you try more complicated ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real value of working in a &lt;em&gt;non-inertial coordinate system&lt;/em&gt; is when such a system makes it easier to describe the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, suppose you want to model the dynamics of an aircraft. Planes can move in any direction in 3-dimensional space, and the nose of the plane can also be &lt;em&gt;pointing&lt;/em&gt; in any direction - not necessarily in the same direction that the plane is moving (see any video of an aerobatics display for examples!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to described the forces created by the plane's engines and control surfaces (rudder, tailplane, ailerons) is to use a coordinate system &lt;em&gt;fixed to the plane&lt;/em&gt;. Since this coordinate system may be accelerating and/or rotating relative to the ground, it is not an inertial coordinate system and you have to include the d'Alembert forces depending on how it is moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the dynamics of rotating machinery. Suppose you want to model the vibrations of the blades of a wind turbine, while it is rotating at constant angular velocity. The easiest way to set up the model is in a coordinate system that rotates with the blades, including the d'Alembert forces. In general the blade vibration frequencies will &lt;em&gt;not&lt;/em&gt; have any simple relation to the speed of rotation of the whole windmill, and they will depend on the rotation speed. In fact, because of the Coriolis forces on the system, even in the rotating system the blades don't vibrate back and forth in simple harmonic motion in a straight line, like the systems you study in a first dynamics course. Each point on the blade actually moves in around an elliptical path whose center is at the &quot;non-vibrating&quot; position. Trying to describe that motion, and set up the equations of motion, in an inertial coordinate system fixed relative to the ground would be much more complicated than using a coordinate system relative to the blade itself. And if you use a computer model to solve the dynamics of the system, the easiest way to visualise the results is to make an animation in the same rotating coordinate system - i.e. to display what you would see if you were &quot;sitting on the blade&quot; as it rotates.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-07-20T13:48:21.777" LastActivityDate="2016-07-20T13:48:21.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10788" PostTypeId="1" CreationDate="2016-07-21T01:06:53.360" Score="1" ViewCount="169" Body="&lt;p&gt;PCB module is about the size of a nickel and has an ARM processor and a sensor chip or two, plus a connector with 5 or 6 wires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Module must be firmly attached to device because device will be vibrating.  Environment of device may vary: hot/cold, wet/dry, wind/dust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cannot modify the target device in any way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I can think of is epoxy the bottom (non-circuit side) of PCB to device, which will be metal or plastic, possibly painted.  Specifically, coat entire bottom of PCB with epoxy and stick to device.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface of device is flat.  Device surface may be larger or smaller than module diameter, but always big enough for epoxy to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If surface of device is conductive, then standoffs would be epoxied between PCB and device surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot think of another method to attach module.&lt;/p&gt;&#xA;" OwnerUserId="2169" LastEditorUserId="7083" LastEditDate="2016-10-25T08:34:22.480" LastActivityDate="2016-10-25T08:34:22.480" Title="Is epoxy the best way to permanently attach a small PCB to a mechanism that will be vibrating?" Tags="|mechanical-engineering|electrical-engineering|materials|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10789" PostTypeId="1" AcceptedAnswerId="10793" CreationDate="2016-07-21T04:29:35.613" Score="4" ViewCount="1970" Body="&lt;p&gt;I am referring to the answer &lt;a href=&quot;https://engineering.stackexchange.com/a/8915/3353&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that on a simple structure &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lz5TT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lz5TT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(The diagrams on the left the images below have fully-fixed connections, while on the right the columns are pinned connections to the beams.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A: The deflection on beam is bigger in pinned model, but the deflection on beam/column joint is bigger in fixed model&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TFSzW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TFSzW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B: The moment on column in pinned model is 0&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/np5DW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/np5DW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C: The axial load on column in pinned model is bigger, because the moment is being transformed into axial load (?? is this reasoning true??)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yBlPd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yBlPd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;These conclusions are true for this particular simple model, but are these three conclusions always true on any general model?&lt;/li&gt;&#xA;&lt;li&gt;And why? Can it be explained in terms of loading flow and statics? Or this is just FEM behavior that we can't explain more intuitively?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-07-21T12:40:42.563" Title="Is pin column-beam joint always have less moment, more deflection compared to fixed joint?" Tags="|structural-engineering|structural-analysis|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10790" PostTypeId="2" ParentId="10739" CreationDate="2016-07-21T11:00:50.387" Score="3" Body="&lt;p&gt;What you ask for won't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if you were to create a current amplifier as you describe, it would require a separate power source.  You can't cheat physics.  Twice the current out at the same voltage means twice the power.  The additional power has to come from somewhere other than the input current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have additional power available, it would be better to use it directly to create a higher current charger than trying to amplify the output of a low current charger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the charger is managing the battery by creating a particular current and voltage profile over time.  The charger will get the wrong idea of what the battery is doing with a current amplifier between it and the battery.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-07-21T11:00:50.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10791" PostTypeId="1" AcceptedAnswerId="10794" CreationDate="2016-07-21T12:24:01.647" Score="2" ViewCount="113" Body="&lt;p&gt;I'm spec'ing out a conveyor belt and want a control mechanism to make sure that it runs straight. I know there exists a product for this since I've seen the function in an older tender, but don't have a clear idea how the device looks like and what it does - googling *&quot;Geradelaufkontrolle Förderband&quot;' didn't yield anything helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a device that gives me an alarm when the belt runs off course, what is it called, and how does it work?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorDisplayName="user16" LastEditDate="2016-07-22T00:36:57.963" LastActivityDate="2016-07-22T00:36:57.963" Title="How to determine / control if a conveyor belt runs straight?" Tags="|mechanical-engineering|sensors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10793" PostTypeId="2" ParentId="10789" CreationDate="2016-07-21T12:40:42.563" Score="2" Body="&lt;p&gt;Let me start by answering your second question: models such as this one, which involve only one-dimensional beam elements, are 100% analytical and can therefore always in theory be understood intuitively. There is no &quot;FEM behavior&quot; for such models. Sometimes the models may get complex with lots of bars and whatnot, which may make &quot;intuitive explanations&quot; more difficult, but the result will always be analytical.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's start by looking at statement &lt;strong&gt;B&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's take a look at beam 4 in your model (the left-most beam). More specifically, it's bending moment diagram. As you've noticed, the pinned model displays zero moment at the left-most column. This is the very definition of a hinge and is expected behavior. The moment on the beam at the central column is non-zero because the beam itself is not hinged, but the central column is hinged and therefore displays zero moment at the node.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, on to statement &lt;strong&gt;A&lt;/strong&gt;, starting by looking at the beam's deflection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's keep looking at the bending moment diagram. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory#Static_beam_equation&quot; rel=&quot;nofollow&quot;&gt;beam equation&lt;/a&gt; tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\partial^2 }{\partial x^2}\left(EI\frac{\partial^2 w}{\partial x^2}\right) = q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which also tells us that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EI\frac{\partial^2 w}{\partial x^2} = M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that is: bending moment (divided by stiffness $EI$) is the second derivative of deflection. From calculus, we know that the second derivative of any function described the function's curvature. So bending moment describes deflection's curvature, which describes the &quot;acceleration&quot; with which the beam's tangent (the first derivative of deflection, and therefore bending moment's integral) changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the more balanced a bending moment diagram is between positive and negative bending moment, the more the total &quot;acceleration&quot; cancels itself out, implying in smaller tangent changes, and therefore smaller deflections. So yes, a fixed node will always lead to smaller deflections&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To answer the matter of the node's displacement, we first need to explain point statement &lt;strong&gt;C&lt;/strong&gt;. For that we need to look at beam 4 in isolation. To do so, we need to replace the surrounding beams with elastic supports which describe their stiffness.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The vertical supports' stiffness will be equal to the columns' axial stiffness (the node with the central column will also have a tiny addition due to the other beam's stiffness against imposed transversal displacements)&lt;/li&gt;&#xA;&lt;li&gt;The horizontal supports' stiffness will be equal to the columns' stiffness against imposed transversal displacements&lt;/li&gt;&#xA;&lt;li&gt;The rotational supports' stiffness will depend on the boundary conditions. If hinged, then the outer node will have zero stiffness and the central node will have a stiffness equal to the other beam's stiffness against imposed rotations. If fixed, then both nodes will have the columns' stiffness against imposed rotations, adding the other beam's stiffness as well for the central node.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, basically, the only difference between the hinged and fixed cases is in the rotational stiffness (as would intuitively be expected). This increased stiffness, however, causes the node to pull in a greater proportion of all forces, thereby increasing the axial forces in your outer column and reducing them in the central column in the fixed model.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Returning to the issue of the node's deflections, they are now easy to explain. After all, in the fixed model the column suffers more axial forces, naturally increasing the vertical deflections. But it also suffers bending moment, which generates horizontal deflections as well as a tiny bit of additional vertical deflection.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-07-21T12:40:42.563" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10794" PostTypeId="2" ParentId="10791" CreationDate="2016-07-21T14:28:59.000" Score="2" Body="&lt;p&gt;&lt;strong&gt;Alignment Monitoring&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many switch systems that can be mounted to the frame that will detect when a conveyor is beyond the acceptable limits of alignment.  A google of &quot;Conveyor Belt Alignment Sensor&quot; brings this within the first page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.go4b.com/usa/technical-support/product-manuals/belt-misalignment/touchswitch-belt-alignment-sensor-manual.pdf&quot; rel=&quot;nofollow&quot;&gt;http://www.go4b.com/usa/technical-support/product-manuals/belt-misalignment/touchswitch-belt-alignment-sensor-manual.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sensors will assist with major changes in alignment and sudden failures but they do not compare with the ability of a trained belt manager who walks the belts at least every shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Equipment Install&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the travel rollers on a conveyor system can be set to correct for a minorly inconsistent alignment of the conveyor structure it adds additional wear to the bearings, rollers and belting.  The additional wear will also reduce the efficiency of your conveyor if that is a concern to your operation.  In all situations I recommend laser levels for the install of conveyor structure (even in semi-mobile installs) If this is a cross country type conveyor system that will deal with variable terrain crossing, vertical and horizontal curves or surface to elevated structure transitions using a knowledgeable surveyor is critical to alignment and the foundations should not be underestimated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Operating Alignment&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When major components on the conveyor are replaced or at the completion of the install the start up and training of the conveyor is important to insure that there isn't a failure due to premature wear.  This site has a good set of basic knowledge about training and individual adjustment for travel rollers. &#xA;&lt;a href=&quot;http://www.shippbelting.com/BeltTraining.aspx&quot; rel=&quot;nofollow&quot;&gt;http://www.shippbelting.com/BeltTraining.aspx&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="205" LastActivityDate="2016-07-21T14:28:59.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10795" PostTypeId="1" CreationDate="2016-07-21T16:52:20.457" Score="0" ViewCount="227" Body="&lt;p&gt;I want to encase a circuit in a ~2-3&quot; cube of some kind of visually and Bluetooth transparent material.  What material will do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like:&#xA;1. half fill a cube shaped mold with liquid or jelly material&#xA;2. add circuit&#xA;3. fill up mold so that circuit is 'floating' in the middle and sealed.&#xA;4. wait for material to dry&#xA;5. circuit is now sealed within sold, transparent cube, and can be controlled via Bluetooth from iPhone app.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to demonstrate that the circuit with battery and Bluetooth communication can be controlled wirelessly from iPhone management app.&lt;/p&gt;&#xA;" OwnerUserId="2169" LastEditorUserId="9" LastEditDate="2016-07-23T01:07:02.203" LastActivityDate="2016-07-23T01:07:02.203" Title="How to hermetically seal a circuit in a visual &amp; bluetooth transparent solid cube ~2-3&quot;?" Tags="|mechanical-engineering|materials|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10797" PostTypeId="2" ParentId="10795" CreationDate="2016-07-21T17:25:59.483" Score="4" Body="&lt;p&gt;The process of encasing electronics in a material is called &lt;em&gt;potting&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various resin materials can be used, such as polyester resin or epoxy.  Discussion of the types of resin can be found in this article on &lt;a href=&quot;http://www.pottingsolutions.com/my%20site/Technology/resin%20type.htm&quot;&gt;Potting Compound Types&lt;/a&gt;.  Several sources have recommended epoxy as the easiest material to begin with, but it is &lt;em&gt;much&lt;/em&gt; more expensive than polyester resin, so epoxy may not be practical if you want a deep casting.  Examples of some specific products include this &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0018N9E3M&quot;&gt;clear polyester casting resin&lt;/a&gt; and &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B000X5G6O6&quot;&gt;3M epoxy potting compound&lt;/a&gt;.  Note the small 1.5 fl oz size of the epoxy compound versus the 32 fl oz of the polyester compound.  Some casting resins may not be designed specifically for electronics, and are often marketed for various kinds of arts and crafts, but might work fine for your purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potting electronics can be a complex process, since it can trap the heat produced by electronics and can cause physical stress due to thermal expansion/contraction.  A useful introduction to some of these issues can be found in the article &lt;a href=&quot;http://www.pottingsolutions.com/my%20site/Technology/potting_hints.htm&quot;&gt;Potting Hints&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7338" LastActivityDate="2016-07-21T17:25:59.483" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10799" PostTypeId="1" AcceptedAnswerId="10810" CreationDate="2016-07-21T22:56:32.703" Score="1" ViewCount="69" Body="&lt;p&gt;Using WiFi or bluetooth frequencies 5.8GHz and 2.4-2.485GHz respectively, I want to place a hermetically sealed circuit at the bottom of a gold fish bowl, and want to know if the water will significantly attenuate WiFi or Bluetooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gold fish bowl would be probably 6&quot; to 1', or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I want to know rate of signal absorption per distance through intervening medium of water.&lt;/p&gt;&#xA;" OwnerUserId="2169" LastEditorUserId="2169" LastEditDate="2016-07-22T00:07:53.167" LastActivityDate="2016-07-22T17:25:58.253" Title="Will the water in a gold fish bowl signficantly abosorb Bluetooth or WiFi if transciever is at bottom of bowl?" Tags="|electrical-engineering|materials|telecommunication|rf-electronics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10800" PostTypeId="1" AcceptedAnswerId="10817" CreationDate="2016-07-22T04:03:06.923" Score="6" ViewCount="2828" Body="&lt;p&gt;Following on from &lt;a href=&quot;https://engineering.stackexchange.com/a/10793/255&quot;&gt;this answer&lt;/a&gt;, I am reading about the difference between hinge and pin connections &lt;a href=&quot;http://www.tboake.com/SSEF1/pin.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, and I don't quite understand about it. I think I would need to see the mathematical representation ( the boundary condition) involving pin and hinge connection, in order to actually understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I represent, in mathematical terms, the pin and hinge connection? And how can it be represented in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Direct_stiffness_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stiffness matrix method&lt;/a&gt;, as boundary conditions (maybe)?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-07-25T10:36:59.510" Title="Mathematical representation of hinge and pin connection" Tags="|structural-engineering|structural-analysis|statics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10801" PostTypeId="2" ParentId="616" CreationDate="2016-07-22T05:36:58.683" Score="-1" Body="&lt;p&gt;You are KIND OF RIGHT !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IN THIN PLASTIC DEFORMATION ONLY&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OD of steel washer = OD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thickness of Material (Pastic) = Tp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear Area = (OD+2.5*Tp)*pi()*Tp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear force is Shear Area * MPa = KN force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plastic deformation forms a &quot;bulb&quot; effect same goes for concrete piles&#xA;in the ground ! the diameter x 2.5 is the effective area for the end of the pile.&#xA;This gets adjusted should the depth be less than 2.5 x the diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Garret Krampe&lt;/p&gt;&#xA;" OwnerUserId="7341" LastActivityDate="2016-07-22T05:36:58.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10802" PostTypeId="2" ParentId="10800" CreationDate="2016-07-22T07:39:36.620" Score="0" Body="&lt;p&gt;If you compare their construction, it is very similar. They both resist shear and axial loading but have no bending momentum. I guess the real difference is in the type of constructions where they're used and in the way they're bolted. My thinking is that pin connections are used for &quot;heavy-duty&quot; steel construcions, while hinges are used for light structures. Also hinges are connected with multiple bolts...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="7234" LastActivityDate="2016-07-22T07:39:36.620" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10803" PostTypeId="1" CreationDate="2016-07-22T07:52:00.657" Score="0" ViewCount="466" Body="&lt;p&gt;Not sure if I'm applying the formula correctly. I'm trying to find the deflection in a cantilever beam that is loaded by a uniformly-distributed load. The formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=- \frac{wx^2}{24EI}(x^2+6L^2-4Lx) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \text{Deflection} \\&#xA;x= \text{Distance from wall} \\&#xA;w=3\ \text{kN/m} \\&#xA;E=200\ \text{GPa} \\&#xA;I=986e6\ \text{mm}^4\\&#xA;L=10.5\ \text{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for $x=10.5\ \text{m}$ I'm getting:&#xA;$$v=- \frac{3000(10.5)^2}{24*200e9*986e6}(10.5^2+6*10.5^2-4*10.5*10.5) \\&#xA;v=-2.31e-14$$&lt;/p&gt;&#xA;" OwnerUserId="7342" LastEditorUserId="1832" LastEditDate="2016-07-22T10:55:52.830" LastActivityDate="2016-07-22T10:55:52.830" Title="Cantilever beam load formula" Tags="|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10804" PostTypeId="2" ParentId="10803" CreationDate="2016-07-22T09:28:37.987" Score="3" Body="&lt;p&gt;You've converted w into N/m, E into N/m2, but you've left I in mm4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, dividing I by 1e12, your answer multiplies by 1e12, giving v = 0.0231m, or 23.1mm. Which sounds like a sensible answer.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2016-07-22T09:28:37.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10805" PostTypeId="1" CreationDate="2016-07-22T11:39:35.873" Score="1" ViewCount="54" Body="&lt;p&gt;I need to measure luminance vs angle characteristics of an led lamp. I have access to a lux meter and a luminance meter. The problem is, that the luminance meter can only measure luminance within 1 degree, and the lamp is about 100cm x 2cm with LEDs about every 4cm, so the diffusor is not evenly lit. How should I go about measuring the luminance of the whole surface? Should I do many measurements along the lamp and do some kind of average? I even have a robot that could move the luminance meter by increments of 1mm or so. Or maybe should I measure the light with the lux meter and somehow calculate the luminance from that and the surface area?&lt;/p&gt;&#xA;" OwnerUserId="7346" LastEditorUserId="9" LastEditDate="2016-07-23T01:05:57.187" LastActivityDate="2016-07-23T01:05:57.187" Title="How to measure the luminance an unevenly lit object?" Tags="|optics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10806" PostTypeId="2" ParentId="10800" CreationDate="2016-07-22T14:28:12.343" Score="4" Body="&lt;p&gt;I'm not sure if the use of the term 'hinge' in this context is common in engineering design (at least I haven't heard of it being used in this way).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the website you link to they seem to be referring to the difference between steel connections that freely allow rotations ('pins') and those where connection rotation is restrained in practice ('hinges') - they claim in this case because more than one bolt is used. I feel that this is a poor explanation of what is happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common connection type for steel structures are &lt;a href=&quot;http://www.steelconstruction.info/Simple_connections&quot; rel=&quot;nofollow&quot;&gt;Simple Connections&lt;/a&gt;. These connections are assumed to not transfer any bending. Engineers typically call such connections 'pins' regardless if they are built with a physical pin or not. Mathematically these are modelled as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^2 w}{dx^2} = M = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where w is the displacement as a function of the distance along the beam (x), and M is the bending moment. Because of the way these connections are constructed a small rotation at the connection will be allowed due to: tolerance in the bolt holes, shear deformation of the bolts, etc. Since small deflections are assumed (a common assumption in engineering) this means there will only be a small error by assuming that the connection is free to rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if there is a requirement to accommodate large rotations. For example, in cable structures, or perhaps where large thermal deformations are expected. The connection can be constructed with a physical pin to accommodate this. However, mathematically the boundary condition is the same (M = 0). Note that pins are also used in practice for ease of assembly and aesthetic purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to know the difference in terms of mathematical boundary conditions between a simple bolted connection and a true pin you could perhaps state it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;True Pin: $$\frac{d^2 w}{dx^2} = M = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple bolted connection: $$\frac{d^2 w}{dx^2} = M  \approx 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a reasonable amount of engineering judgement which goes into deciding if a connection can be designed assuming zero bending moment at the connections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative to these connections is &lt;a href=&quot;http://www.steelconstruction.info/Moment_resisting_connections&quot; rel=&quot;nofollow&quot;&gt;'fixed' or 'moment resisting'&lt;/a&gt; connections. These are designed (and constructed) such that bending is transferred through the connections ($ M \neq 0$).&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2016-07-22T14:36:14.917" LastActivityDate="2016-07-22T14:36:14.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10807" PostTypeId="1" AcceptedAnswerId="10863" CreationDate="2016-07-22T16:14:55.467" Score="2" ViewCount="709" Body="&lt;p&gt;I have an application that requires torque limiting and I need to determine if VFD torque limiting will provide sufficient protection. I have used VFDs for torque limiting through gearboxes before with success, but I am hoping to find a way to calculate the result as opposed to just gut feeling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10HP motor&lt;br&gt;&#xA;1775 rpm &lt;br&gt;&#xA;69.5:1 gearbox reduction (im assuming 50% efficient)&lt;br&gt;&#xA;25.5 rpm at driving sprocket&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at some frequency drives to get an indication of their torque/amp sensitivity. Ive seen 0.1A, 0.01A, 0.1% of torque and 0.01% of torque as control resolutions, but have not found any literature specifically addressing if this resolution can be reasonably expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation I am gearing down, so inertia, backlash and controller reaction time will not be an issue.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2016-08-01T21:58:28.993" LastActivityDate="2016-08-02T01:20:54.927" Title="Torque sensitivity of a VFD induction motor through a gearbox?" Tags="|control-engineering|motors|torque|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10808" PostTypeId="2" ParentId="10788" CreationDate="2016-07-22T16:15:13.337" Score="1" Body="&lt;p&gt;If you can't modify the surface you are attaching to then it sounds like an adhesive is your only practical option. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best adhesive for the task will depend a lot on the target surface. Epoxy is among the strongest classes of engineering  adhesives but it can be sensitive to application conditions and surface preparation. It also tends not to be great at filling large gaps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One attractive alternative is construction type grab/gap filling adhesives. These tend to be reasonably tolerant of adverse conditions and adhere well to a wide variety of surfaces. Their high viscosity also means that they give immediate adhesion and combined with the fact that they are often air curing means that they are quite convenient to use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously if you glue something to a painted surface the bond strength will be limited by the adhesion of the paint to the underlying surface. Similarly, some types of plastic are very resistant to adhesion (especially things like polypropylene) and may require specialist adhesives and/or surface preparation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also double sided and foam backed tapes which can be good for this sort of application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with most adhesive applications you really want to select the best product for the specific circumstances of the job so you would probably do well to talk directly to a specialist adhesive manufacturer or dealer as there are thousands of specialist products available. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="7338" LastEditDate="2016-07-22T17:41:14.067" LastActivityDate="2016-07-22T17:41:14.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10809" PostTypeId="1" CreationDate="2016-07-22T17:17:16.363" Score="0" ViewCount="2330" Body="&lt;p&gt;Is it possible for essential oils to leak from sealed bottles that are being transported by airplane in the cargo hold?&lt;/p&gt;&#xA;" OwnerUserId="7356" LastEditorUserId="9" LastEditDate="2016-07-23T01:05:08.673" LastActivityDate="2020-02-11T09:46:28.660" Title="Do bottles leaking due to air pressure in an airplane cargo hold?" Tags="|pressure|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10810" PostTypeId="2" ParentId="10799" CreationDate="2016-07-22T17:25:58.253" Score="4" Body="&lt;p&gt;There is information out there, but probably the best way to find out is to actually try it.  Grab a small Bluetooth or Wifi device and seal it in a waterproof enclosure or bag, and submerge it to the bottom of the fish bowl under water.  See what happens.  This should provide more valuable insights than the simple theoretical RF attenuation of water per meter, since in the fish bowl you have an irregular geometry and also the device will have a short path for RF out the bottom of the bowl.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some references that might be useful as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A forum thread reported that &lt;a href=&quot;http://www.societyofrobots.com/robotforum/index.php?topic=8408.0&quot; rel=&quot;nofollow&quot;&gt;Bluetooth under water doesn't work&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A white paper from Laird Technologies says, &quot;The 5 GHz wave form is attenuated by common building materials to a greater degree than the 2.4 GHz wave form. &lt;strong&gt;On the other hand, the 2.4 GHz wave&#xA;form is optimally absorbed by water.&lt;/strong&gt;&quot; (&lt;a href=&quot;http://www.digikey.com/Web%20Export/Supplier%20Content/Laird_776/PDF/laird-wireless-optimizing-operation-5ghz.pdf&quot; rel=&quot;nofollow&quot;&gt;Optimizing Operation at 5 GHz&lt;/a&gt;, p. 4)&lt;/p&gt;&#xA;" OwnerUserId="7338" LastActivityDate="2016-07-22T17:25:58.253" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10811" PostTypeId="1" CreationDate="2016-07-22T18:03:13.617" Score="4" ViewCount="455" Body="&lt;p&gt;I have a aluminium plate of radius 50mm and thickness 4mm. How much torque is needed to rotate it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case additional information is required :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a RS 380 motor, but I guess torque required is irrespective of motor. &#xA;How do you calculate torque of other round plate sizes? Like 60mm or 55mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is approximatively the exact torque for 50mm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qF84Q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qF84Q.jpg&quot; alt=&quot;aluminium plate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[![motor][2]][2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kccmN.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kccmN.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7358" LastEditorUserId="7358" LastEditDate="2016-07-23T19:19:56.427" LastActivityDate="2016-07-27T23:54:14.730" Title="What is the required torque to rotate a 50mm aluminimum plate of weight 100g?" Tags="|mechanical-engineering|motors|torque|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10812" PostTypeId="2" ParentId="8976" CreationDate="2016-07-22T18:27:17.337" Score="0" Body="&lt;p&gt;An older style guidance systems had a gyroscope that would point in one location in space (for example, it might point at the North Star).  Then, the rest of the vehicle would rotate around.  This gyro was connected to the vehicle using gimbals that allowed rotations around each axis.  See some of the figures on the wiki page  &lt;a href=&quot;https://en.wikipedia.org/wiki/Gimbal&quot; rel=&quot;nofollow&quot;&gt;for examples&lt;/a&gt;.  The computer could then &quot;figure out&quot; which way the spacecraft is pointing by using angle sensor on each of the gimbals.  Gimbal lock can occur when two of these gimbals are both aligned along the same axis.  I believe that the Apollo IMU had a mechanical gyro with two gimbals on it (they can be though of as pitch and roll in a typical attitude).  But, if the axis of each gimbal is lined up with the axis of the gyro, then there are multiple solutions to a rotation from that point.  Mechanically, certain gimbals can end up stuck at that point, and the math can end up with essentially a solution that has a divide by zero.&lt;br&gt;&#xA;More modern systems use strap down gyros (bolt the gyro to the vehicle and measure the applied torques) or &quot;laser gyros&quot; (no physical gyro just changes in light going in a circle) but those were not feasible in the mid-1960s.  &lt;/p&gt;&#xA;" OwnerUserId="2066" LastActivityDate="2016-07-22T18:27:17.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10813" PostTypeId="1" CreationDate="2016-07-22T22:07:24.670" Score="0" ViewCount="53" Body="&lt;p&gt;My contractor did not the work properly on the structure of my house.  He did not support the beam correctly.  I hired an engineer to check it and he sent his recommendation.  I asked my contractor to correct his work as per the drawings and he said he will send his expert to check it.  I was wondering what kind of expert can overule the engineer recommendation? If this expert says that everything is fine, who is responsible? &lt;/p&gt;&#xA;" OwnerUserId="7360" LastActivityDate="2016-07-30T15:50:32.187" Title="Structural Correction requires" Tags="|structural-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10814" PostTypeId="2" ParentId="10809" CreationDate="2016-07-22T23:56:04.110" Score="-1" Body="&lt;p&gt;Definitely possible, air pressure varies with altitude. Air pressure at sea level (approximate pressure at which the bottles were sealed) is higher than the cabin pressure of a plane (correct me if I'm wrong but commercial cabin pressure is normally somewhere in the region of air pressure at 6,000ft, when flying at 36,000ft) so there will be a pressure differential that could cause the seal to fail, and the contents to leak out in order to equalise the pressure difference.&#xA;If the bottles are upright the air should escape before the oil though!&lt;/p&gt;&#xA;" OwnerUserId="7237" LastActivityDate="2016-07-22T23:56:04.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10817" PostTypeId="2" ParentId="10800" CreationDate="2016-07-23T11:55:59.383" Score="7" Body="&lt;p&gt;I believe you have chosen a poor reference. Indeed, that link has absolutely nothing to do with structural analysis, but rather a specific implementation in steel structures.                    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pinned support is a boundary condition which restricts all displacements but allows the structure to rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hinge (more formally, an internal hinge), on the other hand, is a description of the structure's behavior. Specifically, it removes the rotation compatibility between bars around a node (between all bars or a subset of them), thereby increasing the structure's degrees of freedom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more visual demonstration of the difference, here's a structure with four pinned supports, along with its deformed configuration under the effect of a uniform load. You'll notice that the supports allow the structure to rotate around them, however one beam's inclination at a given node must be equal to the inclination of the neighboring beam on the same node, meaning there may be no discontinuity in the derivative of the deflection at the support (compatibility of rotations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sfa9R.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sfa9R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now two variations on the same structure with the same loading, but with an internal hinge at different locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fQBHW.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fQBHW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, an internal hinge is placed on the same position as one of the supports. This removes the compatibility of rotations around that node, leading to a clear discontinuity in the derivative of the deflection around that support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yPKKz.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yPKKz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the hinge is placed in the middle of the central span, leading to a clear discontinuity in the derivative of the deflection at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hinges can also be placed to just one side of a node. For instance, look at the deflection of this frame under a uniform vertical load and a concentrated horizontal load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NvOGv.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NvOGv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The columns and beams all maintain their original perpendicularity. Specifically at the central column, notice that the beams maintain their angular compatibility (no discontinuity in the derivative of the deflection).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, we add hinges to the tops of the columns:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FtWPe.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FtWPe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beams and columns no longer remain perpendicular in their deflected state. The beams over the central column, however, maintain their angular compatibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, the nodes were fully hinged, then each span would behave independently like a simply-supported beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/77nrI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/77nrI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In the direct stiffness method, the fundamental equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{q\}=K\{d\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pinned support defines the contents of $\{d\}$, by setting the degrees of freedom of the given node to $(d_x, d_y, \theta) = (0, 0, \theta)$ (in the case of a 2D structure). An internal hinge, on the other hand, increases the dimensions of the calculation, since it increases the number of degrees of freedom. So, instead of a node having merely $(d_x, d_y, \theta)$, it will have $(d_x, d_y, \theta_1, \theta_2, \theta_3, ...)$, where $\theta_i$ represents the rotation on different sides of the hinge.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="255" LastEditDate="2016-07-25T10:36:59.510" LastActivityDate="2016-07-25T10:36:59.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10819" PostTypeId="1" AcceptedAnswerId="10823" CreationDate="2016-07-23T19:13:19.997" Score="1" ViewCount="50" Body="&lt;h2&gt;Ive been wondering about how small of a cooler you can build.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Although this probably is just really low-level engineering, i have a rather particular question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my idea, the cooler has to be as small as possible, &lt;em&gt;rather efficient&lt;/em&gt; and &lt;em&gt;not use a chemical compound or reaction that is not locally reversible&lt;/em&gt;. It has to be an fully working circle that, theoretically, keeps working, solely relying on electricity for power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I understand how certain cooling systems work, I could not find a satisfying example, since mobile phones, tablets and other handheld devices rely on, &lt;em&gt;if any&lt;/em&gt;, air cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what I'm searching for woul'd be a small cooler that can reach slightly lower temperatures than the surrounding area, lets say about 5 °C, maybe less or even more, if it just gets &lt;strong&gt;below average surrounding temperatures&lt;/strong&gt; (room-temperatures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be achievable in a &lt;strong&gt;watch-like size&lt;/strong&gt;?&#xA;And &lt;em&gt;how efficient&lt;/em&gt; woul'd that be, &lt;strong&gt;running on electrical reloading&lt;/strong&gt; only?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;:&#xA;The heat would, as far as i understand thermodynamics, not be a big problem.&#xA;A short-range &lt;em&gt;&quot;relocation&quot;&lt;/em&gt; (idk the term) &lt;em&gt;about a few centimeters&lt;/em&gt; to a &lt;em&gt;small vent&lt;/em&gt; would be enough.&lt;/p&gt;&#xA;" OwnerUserId="7373" LastEditorUserId="7373" LastEditDate="2016-07-26T16:27:43.733" LastActivityDate="2016-07-26T16:27:43.733" Title="How small can coolers get?" Tags="|electrical-engineering|heat-transfer|cooling|" AnswerCount="1" CommentCount="1" ClosedDate="2016-07-24T15:38:44.920" ContentLicense="CC BY-SA 3.0" />
  <row Id="10820" PostTypeId="2" ParentId="10474" CreationDate="2016-07-23T21:12:08.430" Score="1" Body="&lt;p&gt;The load that exert an human supported on two rods depends chiefly on:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The weight of the human.&lt;/li&gt;&#xA;&lt;li&gt;How apart from the rods he is (measured in the horitzontal direction).&lt;/li&gt;&#xA;&lt;li&gt;How close are the two rods that support him.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From the photographs you posted, those exerting a smaller load are those in which the person just hangs vertically from one or several bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest load that I can imagine would be a person in an horitzontal position holding to two contiguous rods with his feet (see picture), although I doubt if there exist a gymnast strong enough to suport himself in such a demanding position.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pyB5r.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pyB5r.jpg&quot; alt=&quot;If the man raised his hands to the left of the image, the load would be even higher&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If somebody could stand this way, the horizontal load exerted to the rods would be about five times his weight (assuming that the gymnast has his centre of masses about his belly button - around 1 m above ground - and that rods are spaced about 20 cm).&lt;/p&gt;&#xA;" OwnerUserId="7371" LastActivityDate="2016-07-23T21:12:08.430" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10821" PostTypeId="1" CreationDate="2016-07-23T21:16:13.737" Score="5" ViewCount="558" Body="&lt;p&gt;When running CFD simulations with turbulence on, I often need to specify inlet parameters such as the turbulent intensity, turbulent viscosity ratio and/or the turbulent length scale.  If these parameters are already known for a given experiment, I have no problems validating experimental data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm curious to know how these parameters are measured experimentally.  How can I directly measure the turbulent kinetic energy and turbulent viscosity from a single experiment?  Due to the inherent randomness of turbulence,  would I need a lot of repeated experiments to measure it or can it be done in a single experiment?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this typically done in practice? Knowing that there is a trade-off between accuracy and cost, what would be a low cost measurement technique that is minimally intrusive to the overall flow while providing accuracy to within an order of magnitude with the fewest number of repeated experiments?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="1293" LastEditDate="2016-07-24T05:43:35.917" LastActivityDate="2016-08-24T21:20:59.313" Title="How to measure turbulence parameters experimentally?" Tags="|fluid-mechanics|measurements|cfd|turbulence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10822" PostTypeId="2" ParentId="10782" CreationDate="2016-07-23T21:38:30.843" Score="1" Body="&lt;p&gt;If we can assume that the material behaves in an elastic linear way and that the joint is only loaded by bending, then as you commented the neutral line lies in the middle and that deformation on the two extreme fibres is equal but in opposite directions. Then, original length is just the average of both deformed lengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, to get this result we must assume too that the conditions for the Euler–Bernoulli beam theory holds, and according to your diagram we might approaching its limits (short span compared to width, not very small deformation...) but I think it's still a good approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the material is not linear elastic, location of neutral fibre depends on the constitutive equation of the material (the stress-strain diagram), and the original length may not be the average of both deformed lengths, specially if the constitutive equation is not symmetrical.&lt;/p&gt;&#xA;" OwnerUserId="7371" LastActivityDate="2016-07-23T21:38:30.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10823" PostTypeId="2" ParentId="10819" CreationDate="2016-07-23T23:02:27.583" Score="4" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermoelectric_cooling&quot; rel=&quot;nofollow&quot;&gt;Peltier cooler&lt;/a&gt; would fit the requirements you set out. They can be compact enough to fit the 'wristwatch sized' requirement and use electrical power to achieve the cooling effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do still need some way of shedding heat from the hot side of the cooler and this may require either a passive heat exchange, heat pipe etc depending on the application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The performance will depend not just on the temperature difference you want to achieve but also on the heat flux required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common comparison would be with the coolers used in laptops which tend to be heat pipe based. Typically a modern CPU would generate up to about 100W of heat, depending on load and performance and look to limit internal temperatures on the hot side to about  40-50 deg C. This isn't exactly the same as the refrigeration you require but should give you some sense of what is possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a bit more space to play with water cooling is now pretty mainstream for high performance PCs although most most units are substantially bigger than wristwatch size they will give substantially better performance than a laptop heat spreader. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-07-25T21:21:29.700" LastActivityDate="2016-07-25T21:21:29.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10825" PostTypeId="1" AcceptedAnswerId="10826" CreationDate="2016-07-24T16:31:49.647" Score="0" ViewCount="923" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hshwjm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hshwjm.jpg&quot; alt=&quot;this&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the picture above, in order to calculate the moment of load acting on the beam we take moment $M=\frac{W}{L}×L×\frac{L}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't we calculate the moment as $M=\frac{W}{L}×L×L$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we need to take the distance from center of gravity to a point as the distance?&lt;/p&gt;&#xA;" OwnerUserId="6422" LastEditorUserId="3010" LastEditDate="2016-07-24T19:20:52.767" LastActivityDate="2016-07-24T20:23:21.733" Title="Why do we need to take distance of center gravity of a distributed load to a point in order to calculate moment?" Tags="|civil-engineering|beam|construction-management|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10826" PostTypeId="2" ParentId="10825" CreationDate="2016-07-24T20:23:21.733" Score="0" Body="&lt;p&gt;A point force produces a moment with respect to a point equal to the product of the force by the distance. To compute the moment produced by a system of forces, we can compute the moment for each force and sum the individual moments, or we can just sum the forces and multiply them by the average distance, therefore getting the same moment. Finding the average distance is the same as finding the centre of gravity of the system of forces, since the distance to the center of gravity is the same as the average distance to the forces in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm understanding you, in your formula you are using the distance to the most distant point with distributed load, but most of the load is acting at a shorter distance and therefore producing less moment that the value you are accounting for. If we use the distance to the centre of gravity, some loaded points would be closes to reference points (producing less moment) and some will be further (producing more moment) but the global result will be right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, beware that the moment you gave in your question is valid for a cantilever beam with uniform load, and that's not what is in your drawing shows. Your drawing shows a simple supported beam, that is, supported at both ends. The maximum moment in a simple supported beam is $M=\frac{W}{L}\times\frac{L^2}{8}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just a last note: beware that the definition of moment I've used is valid when the direction of the force is perpendicular to the distance to the point, as usually happens when computing moments in beams. If they are not perpendicular, definition of moment should be adjusted.&lt;/p&gt;&#xA;" OwnerUserId="7371" LastActivityDate="2016-07-24T20:23:21.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10827" PostTypeId="2" ParentId="10821" CreationDate="2016-07-24T20:49:06.063" Score="2" Body="&lt;p&gt;One method is to use a hot wire anemometer. &lt;a href=&quot;http://www.dantecdynamics.com/docs/support-and-download/research-and-education/practicalguide.pdf&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; (about 50 pages, ignoring the product-specific stuff at the end) is a good read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, a hot wire anemometer is simply a metal wire (often platinum-coated tungsten wire with a diameter of a few micrometers) with an electrical current passed through it. The rate of heat loss from the wire depends on the air flow velocity over it, and the resistance of the wire changes with temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple way to use the probe is to supply either a constant voltage or constant current, and measure the change in electrical resistance. A more common method in practice is to use an electrical circuit with a feedback loop that attempts to keep the resistance of the wire (and hence its temperature) constant. The time-dependent electrical power supplied to the wire is then equal to the amount of cooling from the air flow, which can be converted into the time-dependent flow velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot wire probes are physically small (the length of the &quot;hot wire&quot; is of the order of 1 or 2 mm), and because of their low thermal inertia they can measure at frequencies up to 10kHz or higher to resolve short timescale features of turbulent flow.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-07-25T21:00:15.843" LastActivityDate="2016-07-25T21:00:15.843" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10828" PostTypeId="1" CreationDate="2016-07-25T04:15:39.827" Score="-1" ViewCount="3180" Body="&lt;p&gt;I have a drawing that has true position called out on many many holes.  However, there are several holes that have true position (boxed dimensions)called out but no datum reference or control box.  They are called out true position relative to another hole that has true position datum control boxes.  How do I put a tolerance on these?  Do I use the datum from the referenced hole?&lt;/p&gt;&#xA;" OwnerUserId="7380" LastActivityDate="2016-08-05T23:23:34.530" Title="What is the tolerance of True Position without Datum Callout" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10830" PostTypeId="1" CreationDate="2016-07-25T10:40:08.763" Score="0" ViewCount="942" Body="&lt;p&gt;Adjustable selfie stick are two separate tubes that combined together... but how can they be able to lengthen and shorten... I want to know how it is made in the inside?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already googled &quot;how adjustable stick work&quot; &amp;amp; &quot;how adjustable selfie stick work&quot; but couldn't find any explanation...&lt;/p&gt;&#xA;" OwnerUserId="3252" LastActivityDate="2016-07-25T11:08:38.223" Title="How does adjustable selfie stick work?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10831" PostTypeId="2" ParentId="10830" CreationDate="2016-07-25T11:08:38.223" Score="2" Body="&lt;p&gt;Most adjustable selfie sticks probably use the same mechanism often used in telescope tripod legs (aka telescopic pole). Basically it is two hollow tubes one smaller than other, that fits snuggly in the other tube. You then have usually have a ring at the join that allows you to tighten a element that closes the gap to stop the items sliding, though other locking types also exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not being one standardized part many different form factors may exist. Some which undobtedly depart from the design presented above. One could easily use a similar mechanism to that of a collapsible umbrella to similar effect by having multiple notches. Combinations are endless.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2016-07-25T11:08:38.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10832" PostTypeId="2" ParentId="10788" CreationDate="2016-07-25T11:50:22.333" Score="1" Body="&lt;p&gt;Depending on temp, and inertial effects, you may consider solder. Fatigue limit of epoxy is not always reliable. Wire connection to the board should not be solder, embrittlement of the conductor should be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, have you considered, with the above mentioned, using Bluetooth instead of wired connectors.&lt;/p&gt;&#xA;" OwnerUserId="7386" LastActivityDate="2016-07-25T11:50:22.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10834" PostTypeId="2" ParentId="10315" CreationDate="2016-07-25T13:47:52.007" Score="2" Body="&lt;p&gt;This a bit late but I will add to the answers already given, with some additional technical and historical perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Portland cement contains four Bogues compounds - C3A, C2S, C3S and C4AF.  Of these, one is not hydraulic - C3A.  C3A (tricalcium aluminate) was introduced to Portland cement during the 1920's as it was found to reduce the heat required for production, thereby reducing the cost of manufacture.  It was &lt;em&gt;not&lt;/em&gt; introduced for engineering purposes, as some may suggest.  Unfortunately, as it is not hydraulic, C3A will dissolve with time if immersed in water, and it is susceptible to sulphate attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During hydration, however, C3A produces about 3 or 4 times more heat than other Bogues compounds - leading potentially to flash setting and cracking problems in the hardened concrete.  Accordingly, gypsum, or CaSO4 (calcium sulphate), is inter-ground with the clinker to control this flash setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The addition of CaSO4 (particularly the sulphate portion), however, is again detrimental.  Accordingly, Portland cement standards limit the maximum SO4 content.  Thus we see that the addition of C3A necessitates the addition of gypsum and both are detrimental to long term performance, in my view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some will argue that the addition of C3A will have benefit with respect to binding chlorides and resisting steel corrosion.  I have not found this to be the case in my projects and research.  But this view is subject to some controversy.  I prefer to eliminate C3A and gypsum when possible but this will increase the cost of Portland cement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we return to your core question:  adding C3A, which will also require the addition of gypsum, will &lt;em&gt;make the concrete more sensitive during construction and will require greater attention to curing, not less&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastEditorUserId="3266" LastEditDate="2016-07-25T14:21:34.567" LastActivityDate="2016-07-25T14:21:34.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10835" PostTypeId="2" ParentId="10199" CreationDate="2016-07-25T15:39:27.267" Score="2" Body="&lt;p&gt;For a given compaction energy, this is a volumetric problem.  Keep in mind that we are working with a given volume (say 1 cubic metre) of dry particles (solids), water and air, thus:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Compacted soil = dry particles + water + air&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Dry particles, for example sand (cohesionless soil), as oven dry, can be compressed to form a granular skeleton that contains air within the voids between particles - but air has no mass (weight). The density that we can get will depend upon the grading and shape of the particles, and how tightly they will align under compaction force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we add water to the dry particle mixture and compact it.  Water has mass (as compared to air).  So we can gradually fill this dry air void structure with water - displacing the air - and the overall density of the mixture will increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we increase the water content too much, it will start to displace the soil particles within the given total volume.  So the lesser density component, water, will now be a greater proportion of the mixture. This will cause the mixture density to decrease overall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why there is an optimum moisture content.  For a given compaction effort, there will be an optimum moisture content, beyond which the density starts to drop, because soil particles are displaced by the lower density component, water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clays, these same general concepts will apply but there are other factors at play, such as the type of clay and it's affinity for moisture, the alignment of the platelet structure and other features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, there will be air within the compacted soil structure, even at the optimum moisture content.  The higher compaction energy will remove more of the air, for example modified proctor tests yield a higher density than standard proctor tests, but there will still be a small percentage of air in any case.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastActivityDate="2016-07-25T15:39:27.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10836" PostTypeId="2" ParentId="10737" CreationDate="2016-07-25T19:05:51.027" Score="0" Body="&lt;p&gt;In my experience the three main issues with mixing viscous resins are :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) 'dead' spaces where you don't get adequate mixing, especially at the sides and corners of the mixing vessel. Here having some sort of scraper which closely conforms to the shape of the container with blades which redirect the fluid into the main flow can help a lot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Introducing air bubbles into the mixture which can be transferred into the casting or extrusion. Vacuum treatment can be effective in getting rid of these but this is not always convenient and it is better to avoid them if possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Ease of cleaning. This is a major issue for time sensitive mixes, less so for photo-setting resins but even so you may well want to clean or just dispose of parts in contact with the resin between batches or in the case of contamination. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One thing I would suggest is to look closely at the dispensers used for &lt;a href=&quot;http://www.fischer.co.uk/PortalData/10/Resources/support/sales-documents/documents/1.Chemical_(V3)web.pdf&quot; rel=&quot;nofollow&quot;&gt;chemical masonry fixings&lt;/a&gt;. These are often two part, very viscous resin systems (eg filled polyester) which are dispensed by extruding the two parts through a disposable nozzles with an internal spiral channel which is very effective at mixing the two parts for direct injection into a cavity. They are usually dispensed from a cylindrical cartridge containing both parts in separate compartments but the principal could be adapted to continuous  or batch extrusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may even be that the existing nozzles (which you can buy separately) will work as they are for your application and you just need to design the extruder. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-07-25T19:05:51.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10837" PostTypeId="2" ParentId="10737" CreationDate="2016-07-25T21:00:36.243" Score="1" Body="&lt;p&gt;Consider ultrasonic (or acoustic) mixing for this. It may be the simple solution you're after. This can also be used to de-gas the liquid. If you do some googling, you will find plenty of videos, photos, equipment and scholarly papers about it. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-07-25T21:00:36.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10838" PostTypeId="1" AcceptedAnswerId="10842" CreationDate="2016-07-26T00:44:35.567" Score="0" ViewCount="50" Body="&lt;p&gt;I have an old vertical and horizontal cutting machine that is working manually it is kind of useless because it is hard to make accurate measurements. I want to use two digital sensor (with output for any controller) to find the position with less than 1mm error on two dimensions, left-right (with rack and pinion mechanism) 0 to 2,5 m and up-down (with roller screw mechanism) 0 to 1m. What kind of sensor do I have to use under 300 euros?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4gLrh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4gLrh.jpg&quot; alt=&quot;rack and pinion mechanism&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/EyRiX.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EyRiX.gif&quot; alt=&quot;roller screw mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7392" LastEditorUserId="7392" LastEditDate="2016-07-26T00:58:21.930" LastActivityDate="2016-07-26T03:00:35.737" Title="How to find the positioning of cutter machine with high accuracy with digital sensor?" Tags="|control-engineering|sensors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10839" PostTypeId="2" ParentId="2688" CreationDate="2016-07-26T00:48:56.807" Score="0" Body="&lt;p&gt;I can see the partial derivative, $\frac{\partial M(x_1)}{\partial F_{RB}}=0$&#xA;  on your X1 section is actually not correct, it should be &quot; = x&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy hand calculating method of these simple problems is using moment distribution method. Using symmetry we start from left hand span and assume fix ends initially then release them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;RA = RB ql/2 and MA = MB = ql^2/12.&lt;br&gt;&#xA;now we release the joint A half of the moment is carried to joint be so now the MB = ql^2/12 + ql^/24 = 3ql^2/24 = ql^2/8.&lt;br&gt;&#xA;Now we calculate final RA after moment distribution: RA = ql/2 - ql/8 = 3ql/8.&lt;br&gt;&#xA;We set the RC = 3ql/8 too by symmetry. Now RB = 2ql - 6ql/8 = 10ql/8. 10/8 ql = 5/4 ql.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-07-26T00:48:56.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10840" PostTypeId="2" ParentId="8908" CreationDate="2016-07-26T01:26:27.413" Score="0" Body="&lt;p&gt;If you add hinge to center post you are still going to have loads carried by the center post. The only way to isolate the center post from any load on the lateral posts is to have the 2 side post-beam joints pin connection as well as the center post. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-07-26T01:26:27.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10841" PostTypeId="1" CreationDate="2016-07-26T02:41:38.887" Score="2" ViewCount="79" Body="&lt;p&gt;Is there any recommended/suggested/required penetration rate for a miniature cone penetrometer? A 20mm/s is for a standard size of cone penetrometer, what about for miniature one?&lt;/p&gt;&#xA;" OwnerUserId="7395" LastEditorUserId="90" LastEditDate="2022-08-03T00:20:58.060" LastActivityDate="2023-12-26T04:04:15.487" Title="Penetration rate for miniature cone" Tags="|geotechnical-engineering|soil|soil-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10842" PostTypeId="2" ParentId="10838" CreationDate="2016-07-26T03:00:35.737" Score="0" Body="&lt;p&gt;Buy yourself a DRO (Digital Read Out) kit from eBay/aliexpress or somewhere similar. They are within or close to your budget and should give you significantly more precision and accuracy than 1mm.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UFakE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UFakE.jpg&quot; alt=&quot;DRO kit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-07-26T03:00:35.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10844" PostTypeId="1" CreationDate="2016-07-26T12:12:33.937" Score="3" ViewCount="399" Body="&lt;p&gt;when doing the normal RC design of an RC beam, it is determined that compression steel reinforcement is required, does a prestressed tendon in the compression zone contribute to the area of compression steel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly it could contribute, but with a reduced contribution because it already has an axial load within it.&lt;/p&gt;&#xA;" OwnerUserId="5325" LastActivityDate="2016-08-16T21:09:20.250" Title="do prestressing tendons located in the compression zone contribute to compression reinforcement?" Tags="|structural-engineering|reinforced-concrete|prestressed-concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10845" PostTypeId="2" ParentId="10844" CreationDate="2016-07-26T14:32:40.327" Score="4" Body="&lt;p&gt;The short answer would be &quot;yes&quot;, but there is a lot of things to be aware:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First of all, you should check that the building code you are following allows you to do so.&lt;/li&gt;&#xA;&lt;li&gt;You should check adherence between steel and concrete. Since the ratio between strength and lateral surface is a lot higher for tendons than for passive reinforcing steel, adherence conditions are a lot worse. Furthermore, beware that some kinds of tendons (unbonded?) don't have any adherence at all and can't be counted as reinforcing.&lt;/li&gt;&#xA;&lt;li&gt;Most building codes require that longitudinal bars of reinforcing in compression zones to be inside the cage - with some layout of other bars - to avoid buckling. Placing of tendons not always complies with such a requirement.&lt;/li&gt;&#xA;&lt;li&gt;To evaluate the contribution of tendons, please remember that deformation of steel and concrete must be compatible, and that ultimate deformation of high tensile steel is way larger than that of concrete in compression. Then, in your calculations you can't just replace tendons by the rebar of the same ultimate resistance.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And about your last sentence, contribution of steel is not reduced because of axial load, but it's just the other way: in the compression zone, the load introduced to steel by bending is in opposite direction of the load introduced by prestressing. However, prestressing do reduce the contribution of concrete, because both prestressing and bending cause compression to concrete in the compression zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just as an end note: prestressing in the compression zone should be avoided if possible, but you probably already know that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I'm not used to write about reinforced concrete in English. If my vocabulary is too weird to be properly understood, please comment and I'll try to fix it.&lt;/p&gt;&#xA;" OwnerUserId="7371" LastEditorUserId="7371" LastEditDate="2016-08-16T21:09:20.250" LastActivityDate="2016-08-16T21:09:20.250" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10846" PostTypeId="2" ParentId="8609" CreationDate="2016-07-26T19:46:08.107" Score="0" Body="&lt;p&gt;The question can't be answered with the information given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gears convert one combination of (speed x torque) to a different combination of (speed x torque).  For ideal gears, the (speed x torque) product is the same from input to output.  Real gears have some loss due to friction, so the (speed x torque) product of the output will be a little lower than that of the input.  Speed times torque is power, which must be conserved.  The difference between the output power and the input power is the power lost in the gears, which ends up as heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this it should be clear that just throwing &quot;gears&quot; in a linkage doesn't mean the result will be higher speed.  First, the gears can be arranged to provide higher or lower speed depending on the gear ratio.  However, even if the gears increase speed from their input to output, the resulting decrease in torque may ultimately result in lower speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the highest speed, the motor must be optimally matched to the load.  For any particular (speed x torque) product, optimal matching is the point where changing the speed/torque balance results in less speed.  If gears provided more torque at lower speed, then the result would be slower because the lower torque isn't needed and the limit becomes the speed capability.  If gears provided more speed at lower torque, then the result would be slower because the smaller torque can't move the load as fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gears between the motor and a drive shaft can result in higher top speed &lt;i&gt;if that matches the load better&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally bare electric motors have high speed and low torque compared to what the drive shaft of your car probably wants.  The right gearing that reduces speed but increases torque will probably help.  To know what the gear ratio should be, you have to know some parameters, or experiment.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-07-26T19:46:08.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10848" PostTypeId="1" AcceptedAnswerId="10849" CreationDate="2016-07-26T20:48:06.300" Score="2" ViewCount="4350" Body="&lt;p&gt;I understand that poles on the left hand complex plane (LHP) make a continuous linear dynamic system stable. What's so significant about imaginary poles on the LHP that make a system stable? What does it mean to have poles and zeros on the LHP verses the right hand plane (RHP)?&lt;/p&gt;&#xA;" OwnerDisplayName="user7211" LastEditorUserId="191" LastEditDate="2016-07-27T03:28:45.870" LastActivityDate="2016-07-27T15:39:46.297" Title="In Control Systems Engineering, why do imaginary poles and zeros on the LHP indicate stability?" Tags="|mechanical-engineering|electrical-engineering|control-engineering|control-theory|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10849" PostTypeId="2" ParentId="10848" CreationDate="2016-07-26T21:14:29.290" Score="6" Body="&lt;p&gt;Only the poles in the LHP are necessary for stability. This is because the transient response of a LTI system will consists of a linear combinations of $e^{p_i t}$. If a pole is complex, $p_i=\rho_i+i \sigma_i$, you can use Euler's formula, such that the contribution to the transient response can be written as,&#xA;$$&#xA;e^{\rho_i t} \left(\cos(\sigma_i t)+i \sin(\sigma_i t)\right).&#xA;$$&#xA;A complex pole will always occur in pairs (complex conjugate) such that their combination will be real. These transient responses will only die out in time if $\rho_i&amp;lt;0$. This is the same as all poles in the LHP.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2016-07-26T21:14:29.290" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10850" PostTypeId="2" ParentId="8845" CreationDate="2016-07-27T04:27:37.003" Score="0" Body="&lt;p&gt;It might not be an answer, but it's an interesting explanation &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Washing of foam earplugs does not have a major effect on strength or&#xA;  attenuation, but... foam earplugs do retain moisture. You can never&#xA;  &quot;squeeze&quot; it all out. And that has two negative after-effects:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;First, the moisture causes the foam to swell in size a bit (affecting&#xA;  fit and expansion time).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A way to test that hypothesis would be to wash a new earplug in water, without soap to see if it affects its structure. If it does that would confirm the moisture hypothesis which could increase the size of bubbles inside which could affects its compression and soundproofing properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the following question would be: &lt;strong&gt;how can we efficiently dry the foam?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: &lt;a href=&quot;http://www.hearforever.org/blogs/01/24/2011/qa-washing-foam-earplugs-good-or-bad-idea&quot; rel=&quot;nofollow&quot;&gt;http://www.hearforever.org/blogs/01/24/2011/qa-washing-foam-earplugs-good-or-bad-idea&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="6427" LastActivityDate="2016-07-27T04:27:37.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10851" PostTypeId="1" AcceptedAnswerId="10861" CreationDate="2016-07-27T09:40:15.913" Score="2" ViewCount="175" Body="&lt;p&gt;I'm looking for a device that breaks a circuit once the voltage of my battery gets too low. This to protect the battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: I have a 12V battery, once it reaches ~11V the huge voltage drop begins, but the device that is connected to the battery still runs and draws current. I need another device that can detect the voltage goes below 11V and then automatically breaks the circuit, preventing the circuit to draw more current from the battery. Basically, it shuts down the whole circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since English is not my native language and I'm not aware of any kinds of systems in my own language, I'm having a hard time finding something like this. Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: or can I make my own?&lt;/p&gt;&#xA;" OwnerUserId="7175" LastEditorUserId="7175" LastEditDate="2016-07-27T09:58:34.217" LastActivityDate="2016-07-29T17:02:06.233" Title="Looking for a device that breaks the circuit if the voltage gets too low" Tags="|circuits|battery|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10853" PostTypeId="1" CreationDate="2016-07-27T11:44:14.133" Score="3" ViewCount="140" Body="&lt;p&gt;PSS is based on 62 values from a Zadoff-Chu sequence. Three different sequences can be used by an LTE Cell, to represent the Physical Cell ID within the Cell ID group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is the following, $u$ being an index $[0;2]$ corresponding to a Physical Cell ID.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{cases}&#xA;\partial u(n)=e(−\dfrac{j\pi}{63} un(n+1))&amp;amp;\text{ for }n=0,1,…,30 \\&#xA;\partial u(n)=e(−j\pi u(n+1)(n+2)63)&amp;amp;\text{ for }n=31,32,…,61&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now my question:&lt;/strong&gt; is there any kind of QAM or QPSK applied to each of the 62 values we get from the sequence, before they are mapped to the time/frequency grid (Radio frame)?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I know that Secondary Synchronization Signal (composed of 31-length&#xA;M sequence) is modulated by BPSK before being mapped.&lt;/li&gt;&#xA;&lt;li&gt;I've found that and OFDM subcarrier has for formula $s(t) = A\cos(wt     + \Phi(t))$. I still haven't found yet where this intervenes.&lt;/li&gt;&#xA;&lt;li&gt;For a 1.4&amp;nbsp;MHz bandwidth, a 126 points IFFT is performed, if you&#xA;know what that really means (physically speaking) beyond switching&#xA;from frequency to time-domain.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the end I would like to get the proper antenna output signal carrying a PSS, to study it. &lt;/p&gt;&#xA;" OwnerUserId="6507" LastEditorUserId="1832" LastEditDate="2016-08-02T00:44:15.640" LastActivityDate="2016-08-02T00:44:15.640" Title="How is modulated the Primary Synchronization Signal in LTE?" Tags="|electrical-engineering|telecommunication|wireless-communication|lte|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10856" PostTypeId="1" AcceptedAnswerId="10860" CreationDate="2016-07-27T13:26:01.567" Score="5" ViewCount="2131" Body="&lt;p&gt;What kind of simplified structural system would a Leonardo bridge use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried approximating the system to a simple statically determined beam (see below) but our calculations did not give any realistic results. I believe it is due to a negligence of the potential strain in each beam. Maybe using a Spring Load Constant combined with nodes would help?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/3zrdT.jpg&quot; alt=&quot;&quot;&gt;&#xA;A &lt;a href=&quot;http://www.leonardodavincisinventions.com/leonardo-da-vinci-models/leonardo-da-vincis-self-supporting-bridge/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Leonardo bridge&lt;/a&gt; is a bridge constructed without any kind of binding, but just using uniform length pieces of Wood. It can be constructed in this manner:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0V7RI.gif&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own approximation is rough but each of the connections cannot be considered nodes because none of the pieces are loose or movable.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/oprqM.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. It would greatly simplify my calculations if any answers would consider a statically determined system even if it is a simplification. &lt;/p&gt;&#xA;" OwnerUserId="7412" LastEditorUserId="6767" LastEditDate="2019-01-02T18:47:38.103" LastActivityDate="2019-01-02T18:47:38.103" Title="What kind of simplified structural system would a Leonardo bridge use?" Tags="|structural-engineering|civil-engineering|structures|bridges|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="10859" PostTypeId="2" ParentId="10848" CreationDate="2016-07-27T15:39:46.297" Score="0" Body="&lt;p&gt;Briefly, poles in the left half plane (R &amp;lt; 0) symbolize ringing that dies down over time, at R = 0 ringing that stays the same, and in the right half plane (R &gt; 0) ringing that increases in amplitude over time.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-07-27T15:39:46.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10860" PostTypeId="2" ParentId="10856" CreationDate="2016-07-27T19:48:15.120" Score="2" Body="&lt;p&gt;We will solve for n members with these assumptions.&lt;br&gt;&#xA;members are L ft long by 2H, (H=L/20) thick and assuming the notch depth at H on the center to accommodate next member linkage.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loads are applied uniformly spread vertically over the exposed top of the bridge at m. ton/ meter.&lt;br&gt;&#xA;The structure as we add to members looks more and more like an arch.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of which is deduced from similar triangles H/L = L/R&lt;br&gt;&#xA;$R=L^2/h$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here we apply as an approximation the arch formulas with one caveat the moment will be converted to compression and tension on the joints by this formula&lt;br&gt;&#xA;$T=C=M/H$ and also shear converts to vertical joint loads.&lt;br&gt;&#xA;There are many ways to solve a pin end support arch but an starter is to assume one support on rollers and solve then superimpose the lateral load needed to bring the roller end back in. I have some hand books if you need, but here is a link.&lt;br&gt;&#xA;   &lt;a href=&quot;http://www.civilengineeringx.com/bdac/stresses-in-arches/&quot; rel=&quot;nofollow&quot;&gt;http://www.civilengineeringx.com/bdac/stresses-in-arches/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-07-27T19:48:15.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10861" PostTypeId="2" ParentId="10851" CreationDate="2016-07-27T21:32:11.133" Score="4" Body="&lt;p&gt;The simple answer is that you can get off the shelf modules which does exactly what you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are often called 'split charge modules' and are designed for installations like boats and motor-homes which have batteries charged off an engine alternator but prevent the starter battery form being fully discharged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one example &lt;a href=&quot;http://www.12voltplanet.co.uk/voltage-sensitive-relay-12v-140a.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.12voltplanet.co.uk/voltage-sensitive-relay-12v-140a.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are not expensive and are easy to fit with basic knowledge of electronic systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also more sophisticated systems which allow for conditioning and management of auxiliary batteries &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used this brand before : &lt;a href=&quot;https://i.stack.imgur.com/EHlE2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EHlE2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't remember the exact specification but they would certainly be in the 1A sort of range for a 12V system. I suggest this one in particular as it is easy to disassemble so you could discard the case and screw terminals to make it a lot more compact to package with the rest of your circuit. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-07-29T17:02:06.233" LastActivityDate="2016-07-29T17:02:06.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10862" PostTypeId="2" ParentId="10811" CreationDate="2016-07-27T23:54:14.730" Score="1" Body="&lt;p&gt;Force is required to accelerate mass and overcome friction force. So the torque required in steady rpm only as big as the friction.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2016-07-27T23:54:14.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10863" PostTypeId="2" ParentId="10807" CreationDate="2016-07-28T01:21:12.773" Score="3" Body="&lt;p&gt;Induction motors don't have a consistent current vs. torque relationship. In order accurately control or limit torque, VFDs must determine an equivalent circuit for the motor then monitor and control the electrical parameters that determine the torque. I believe that VFDs that have a good torque limiting capability will have a specification stating the accuracy with which they can do that. They will only be able to state what can be done at the motor shaft. The variations in torque lost in the gearbox will significantly degrade the torque limiting or controlling accuracy. I suspect that finding out what to expect from the gearbox will be the most important and most difficult task in this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you state what VFD models you have experience with and/or are considering for use in this situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re link in comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you click the &quot;Specifications&quot; tab, you will see &quot;PowerFlex 700 AC Drive Technical Data,&quot; a document that you should download. On page 4 of that document, you will find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jtj6b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jtj6b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might find another manufacturer that can do better, but I doubt any will do a lot better. When a speed-controlled drive goes into torque limit, the motor will slow down until it reaches an operating point where the limiting value of torque is sufficient to drive the load. If no such point is reached, the drive will be at a standstill. At that point, the torque should not exceed the setpoint, but it may be less than the setpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Re Answer Posted by Asker&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went through your answer, made a diagram and added some notes as shown below. The way I interpreted your numbers, my calculations gave slightly different results, but I put yours on the diagram. If you you used &quot;best estimate&quot; losses, I would suggest that you also do the calculations using the highest and lowest losses that you think might be possible. I believe that the lowest losses will show the smallest margin between normal operating torque and the failure level torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that you might get better performance by sizing the drive for the maximum desired operating torque. You calculations seem to indicate it is sized at the failure level torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f9Fjh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f9Fjh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user6431" LastEditorDisplayName="user6431" LastEditDate="2016-08-02T01:20:54.927" LastActivityDate="2016-08-02T01:20:54.927" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="10864" PostTypeId="1" CreationDate="2016-07-28T03:27:04.010" Score="3" ViewCount="21259" Body="&lt;p&gt;I am reading a document about beam end release &lt;a href=&quot;http://download.autodesk.com/us/algor/userguides/mergedProjects/setting_up_the_analysis/Nonlinear/Loads_and_Constraints/end_release.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An end release will allow either or both ends of a beam element to&#xA;  rotate about or translate along one or more of the local axes of the&#xA;  beam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And the article continues with the following diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a8xkW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a8xkW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(a) Fixed-fixed beam with a hinge point at 1 and 2.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/USt8r.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/USt8r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(b) The theoretical rotation or slope of the beams. Note how the result is discontinuous at the hinge points.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The understanding I get from the above is that if we release at a beam end, then it will become hinge ( and hence the rotation at the end is not continuous)-- and that's it. Is it true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is true, then I don't understand &lt;a href=&quot;https://risa.com/news/what-is-the-difference-between-a-boundary-condition-and-an-end-release/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the moment released term&lt;/a&gt; in RISA software:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hqcwb.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hqcwb.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not even sure whether the beam end release and the moment end release are connected, and if yes, how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are the beam end release and moment end release modeled mathematically?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="3353" LastEditDate="2016-07-28T08:14:33.447" LastActivityDate="2016-08-02T23:51:38.630" Title="What does &quot;beam end release&quot; actually mean, and how it is modeled in mathematical terms?" Tags="|structural-engineering|structural-analysis|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10865" PostTypeId="1" CreationDate="2016-07-28T06:29:06.503" Score="0" ViewCount="355" Body="&lt;p&gt;I know that the frequency and rotational speed of a spinning shaft or wheel is measured using a device that senses a piece of reflective tape on one of the rotating components. I am thinking a laser device might be best suited but haven't been able to find the correct device. What is this device called?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a fairly high rotational speed (3600 RPM). &lt;/p&gt;&#xA;" OwnerUserId="4984" LastEditorUserId="2061" LastEditDate="2016-08-01T14:43:13.373" LastActivityDate="2016-08-01T14:43:13.373" Title="How do I sense reflective tape such as on a spinning wheel?" Tags="|sensors|measurements|lasers|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10866" PostTypeId="1" AcceptedAnswerId="10950" CreationDate="2016-07-28T07:40:16.567" Score="1" ViewCount="1602" Body="&lt;p&gt;In &lt;a href=&quot;https://en.wikipedia.org/wiki/Direct_stiffness_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;direct stiffness method&lt;/a&gt;, the beam and column connection are  assumed to be done through center-line beam and column:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x9bAY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x9bAY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And hence we can represent them easily using frame elements ( 1D element) and do not have to model the surface contacts or using area elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in reality, beam and column may not always be connected via centerline: eg: such configuration is not uncommon&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VnElg.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VnElg.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is easy to see that the offset of beam shall induce additional moment on the supporting column. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I modify my stiffness matrix in order to accommodate the offset of beam from the column center location? The frame representation of beam and column is just too convenient to give up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purpose of this question, assume that I am writing a frame analysis software from first principle ( yes, from the equations on the wiki page that I link to), and hence can't make use of any existing software package &lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="3353" LastEditDate="2016-07-28T08:41:32.083" LastActivityDate="2016-08-04T13:28:58.927" Title="How can I model beam column connection that doesn't connect at centerline?" Tags="|structural-engineering|structural-analysis|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10867" PostTypeId="2" ParentId="10866" CreationDate="2016-07-28T08:36:09.107" Score="2" Body="&lt;p&gt;There are two main options for modelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 1&lt;/strong&gt; - Instead of just modelling the beam and the column, add a third member, the little beam that causes the offset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Option 2&lt;/strong&gt; - Use offset geometry (if available within the software you are using). I'm not 100% sure how this works (and it may differ from software package to software package): possibly it does modify the stiffness matrix, possibly the software introduces a dummy element to connect the beam to the column, much like Option 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture you have provided, the beam to cause the offset looks very small compared to the main beam and the column. If you went down the Option 2 route, you would not get forces and moments in that member in order to design it, and I suspect it might get overloaded. So, for the picture provided, I would recommend Option 1.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="3353" LastEditDate="2016-08-04T13:28:58.927" LastActivityDate="2016-08-04T13:28:58.927" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="10868" PostTypeId="2" ParentId="10864" CreationDate="2016-07-28T08:40:45.190" Score="3" Body="&lt;p&gt;Mathematically, a release is achieved by giving a stiffness of zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A release in rotation is the same as a release in moment: the only way to ensure rotation continuity from one member to another is to transfer moment between them. Hence a moment end release is a stiffness of zero against rotation about the relevant axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, you can release a beam end in any degree of freedom (hence the final option from RISA). In practice, the vast majority of beam end releases used are moment releases, hence your first source has conflated the two.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2016-07-28T08:50:04.910" LastActivityDate="2016-07-28T08:50:04.910" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="10869" PostTypeId="1" AcceptedAnswerId="10874" CreationDate="2016-07-28T10:12:42.750" Score="0" ViewCount="45" Body="&lt;p&gt;From what I know, the key to reducing losses due to eddy currents in a core is to reduce the path that they take, as is shown in this &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/d/d1/Laminering_av_k%C3%A4rna.svg&quot; rel=&quot;nofollow&quot;&gt;diagram illustrating solid vs. laminated cores&lt;/a&gt; (courtesy of Wikipedia).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would taking this one step further and making the core a bundle of wires/rods instead of a stack of laminations reduce the losses further? If so, then why aren't there any commercial designs using it?&lt;/p&gt;&#xA;" OwnerUserId="6589" LastEditorUserId="6589" LastEditDate="2016-07-28T10:18:48.687" LastActivityDate="2016-07-28T13:42:20.970" Title="Inductor/Transformer core design - reducing losses from eddy currents" Tags="|electrical-engineering|electromagnetism|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10872" PostTypeId="2" ParentId="10865" CreationDate="2016-07-28T12:56:07.393" Score="0" Body="&lt;p&gt;Hall effect sensor - put a small magnet on the wheel and the sensor mounts alongside - your counting circuit will receive  a &quot;pulse&quot; each time the magnet goes past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optical sensor - you arrange an infra red light emitting diode (or similar light source) beside the wheel, and at just one point on the wheel a hole is drilled through to let the light pass through the wheel itself. Then a photodiode or some other light sensor is on the other side.  Again, it gives a signal each time the hole pops round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There are other types of sensor I guess, but these are the most common ones you'll find schematics for.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you need some sort of electronic counter to count pulses, or maybe measure the time interval between each pulse, and convert that to a &quot;speed&quot; measurement. You don't say anything about a counter - but I'd guess one of the cheaper Arduino devices will do it. (They have  a lot of circuits for this sort of thing if you dig through their website.) &lt;a href=&quot;http://playground.arduino.cc/Learning/Tachometer&quot; rel=&quot;nofollow&quot;&gt;Here's an example&lt;/a&gt; to show the principle.&lt;/p&gt;&#xA;" OwnerUserId="6665" LastActivityDate="2016-07-28T12:56:07.393" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10874" PostTypeId="2" ParentId="10869" CreationDate="2016-07-28T13:42:20.970" Score="0" Body="&lt;p&gt;What stands between each lamination ? Isolator. If you switch from lamination to wires, you'll get much more isolator, and therefore less metal in the same volume of core. Doing so, you'll get a less efficient core, you'll have to increase its size, increasing cost, wire length on the coils, losses in the wires and that's often a bigger drawback than the benefit in metal losses reduction. I'm also doubtful about the mechanical resistance of such core.&lt;/p&gt;&#xA;" OwnerUserId="2664" LastActivityDate="2016-07-28T13:42:20.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10880" PostTypeId="2" ParentId="10864" CreationDate="2016-07-29T08:10:51.920" Score="2" Body="&lt;p&gt;Beam End Release and Moment End Release are the same thing: it can be thought as converting a fixed beam end to a pivoted one (see later), relaxing the constraint on rotation while enforcing a new constraint on bending moment. The idea of applying constraints to beam ends is useful in understanding the mathematical implication in changing a fixed beam end to a pivoted one, and is discussed below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any continuous element of a beam, for small deflections, is governed by the following differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d^2}{dx^2}\left( E(x) I(x) \frac{d^2 u}{dx^2}\right) + q(x) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $E(x)$ is the Young's Modulus of the beam, $I(x)$ is the Second Moment of Area about the axis of bending for the beam's cross section, $u(x)$ is the upward displacement of the beam, and $q(x)$ is the downward force per unit length acting on the beam. $x$ is a coordinate such that, for a beam of length $L$, $x=0$ at one end, and $x=L$ at the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is a forth order differential equation, and so it requires four boundary equations. This is done by applying two constraints at each end. There are three different types of beam end, each with different constraints to be applied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SmQmu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SmQmu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FIXED END&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where the end of the beam is rigidly clamped to e.g. a wall. This end allows both shear forces and bending moments to be transmitted from the beam to the wall. The end of the beam here cannot rotate nor vertically displace. This is like the beam ends shown in the first diagram of your question before releasing either end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the end at $x=0$ was a fixed end, the following two constraints are applied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No vertical displacement&#xA;$u(0) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No rotation&#xA;$\theta(0) = \frac{du(0)}{dx} = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PIVOTED END&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is your moment-released end: the beam still may not vertical displace at the end, but it can rotate. However, since it may rotate, the beam can no longer transmit bending moments to the support, so a bending moment of zero must be set at the end. Hence moment-released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the end at $x=0$ was pivoted, the following constraints are applied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No vertical displacement&#xA;$u(0) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No bending moment&#xA;$M(0) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $M(x)=-E(x)I(x) \frac{d^2 u}{dx^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;FREE END&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This type of beam end is connected to nothing: it is free to displace and rotate at the ends. However, no bending moments or shear forces can be transmitted from the end since there is nothing attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the end at $x=0$ is free, the following constraints are applied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No bending moment&#xA;$M(0) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No shear force&#xA;$S(0) =0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $S(x) = -\frac{d}{dx} \left(E(x)I(x) \frac{d^2 u}{dx^2} \right)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a cantilever is a beam with one fixed end and one free end, and a simply supported beam has both ends pivoted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon setting two constraints to each end, the general solution to the differential equation can be obtained, allowing the vertical displacement, rotation, curvature, bending moments and shear forces to be determined.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2016-07-29T12:51:47.693" LastActivityDate="2016-07-29T12:51:47.693" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10881" PostTypeId="1" AcceptedAnswerId="10892" CreationDate="2016-07-29T14:05:38.183" Score="3" ViewCount="1161" Body="&lt;p&gt;I have tested a gear box for many hours. Where the gears are made of a hardened steel alloy and the housing is 7000 series aluminum. Now I drained the gear box of its gear oil and I need to weight the metal shavings in that oil. So I can put in fresh oil and test the gear box for a couple more hours. And then do the test all over again. Can anyone give me a good idea as to how I can accurately separate the oil with out losing the tiny pieces of metal? I thought I could filter the oil through coffee filters then give it a thorough cleaning with hot water and soap, then drying it in the oven. Can anyone tell me if this is a good idea or can you come up with a better one?&lt;/p&gt;&#xA;" OwnerUserId="7443" LastEditorUserId="77" LastEditDate="2016-07-30T20:43:33.050" LastActivityDate="2018-06-13T02:27:30.893" Title="How can I separate gear oil from metal shavings to weigh worn metal?" Tags="|gears|fluid-filtration|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10883" PostTypeId="2" ParentId="10737" CreationDate="2016-07-30T06:30:57.963" Score="0" Body="&lt;p&gt;Generally a one rotor mixer can be designed to mix as well as a multiple rotor mixer. You also generally employ some form of stator to keep the volume from rotating as a single mass without turbulence. This can be done by simply making the volume not round or not putting the mixing rotor in the center of the volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If particulate is settling out like mud, that will not reintroduces into the stream by the motion of fluid alone, a larger rotor for scraping may be required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Chris Johns mentioned, bubbles creation can be an issue. To avoid this you need to lower the mixer rotor rpm and increase the distance below the surface level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mixer design I recommend considering is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_stirrer&quot; rel=&quot;nofollow&quot;&gt;magnetic stirring plate&lt;/a&gt;. It can be placed under any thin container and has no seals that can fail. They are typically high rpm, but a lower rpm one could be construed. The magnetic coupling design also allows for experimenting with different locations in the mixing chamber.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-07-30T06:30:57.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10885" PostTypeId="1" CreationDate="2016-07-30T09:58:04.577" Score="2" ViewCount="177" Body="&lt;p&gt;Hairdryers are capable of electrocuting people and are used in generally wetter conditions and run on mains power. Trimmers are used on dry bodies and use safe DC power from an adapter. Why are hairdryers not using adapters, especially when they are used in more dangerous conditions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can guess these reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Trimmer cables are detachable from the body and harder to insulate&lt;/li&gt;&#xA;&lt;li&gt;Trimmers have chargeable variants, so they are designed to run on DC&lt;/li&gt;&#xA;&lt;li&gt;Trimmers will be unpleasantly heavy with integrated chargers&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="444" LastActivityDate="2023-11-16T15:27:56.493" Title="Why do hairdryers use line power and beard trimmers use adapters?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10886" PostTypeId="1" AcceptedAnswerId="10888" CreationDate="2016-07-30T10:06:42.380" Score="0" ViewCount="74" Body="&lt;p&gt;Recently, I read an October 2006 IEEE JOURNAL OF LIGHTWAVE TECHNOLOGY, article titled, High-Sensitivity Detection of Narrowband Light in a More Intense Broadband Background Using Coherence Interferogram Phase written by the esteemed Ricardo C. Coutinho, Member, IEEE, David R. Selviah, Member, IEEE, and Hugh D. Griffiths, Fellow, IEEE from highly respected University College of London.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this excellent article which is available publicly, it describes&#xA;  that the frame rate of  the scan is 150 microns per seconds in steps&#xA;  of 0.1 micron which is equivalent to 1500 seconds time duration for&#xA;  detecting the presence of a coherent laser light in an incoherent&#xA;  light background with averaging &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Page 3656 &quot;A 10-μm scan length, corresponding to a 20-μm path&#xA;  difference range, was chosen, and this was scanned at a speed of 150&#xA;  μm/s in steps of 0.1 μm.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I am wrong about this calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, tonight I read in the July 25 2016 Optics Express journal, an article titled &quot;Optical digital Coherent detection technology enabled flexible and ultrafast quantitative phase imaging&quot; by Yuan-Hua Feng et al. from Peking University Electrical Engineering Department. This article is currently open and availabe at the Optical Society Of America web portal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Feng and his colleagues report that &quot;we present a digital coherent time-stretch microscopy system that is capable of delivering ultrafast QPI of fixed cells in a scanning rate up to 100 MHz without averaging&quot; . In addition, these researchers state that &quot;bebefiting from a digital coherent receiver, we can retrieve intensity and phase information with high quality simultaneously. Because of the improved Signal to Noise (SNR) of the  proposd imaging method, the amplitude and phase images were processed without averaging. So we can achieve an effective line-scan rate as high as 100 MHz.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these papers are very important to use because we wish to protect pilot's vision from increasingly sophisticated laser pointers on the ground 150 meters away aimed at the Boeing 747-787 cockpit windows. If we can detect the presencen of coherent laser light in an incoherent background within 10 milliseconds of arrival , we can then activate an electro-optic shuttter to block the laser ray and then funnel the recovered incoherent light signal to an Planar(@TM) transparent display.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is which of the two papers is most effective for accomplishing the aforementioned purpose. In addition, how is the digital coherent receiver able to achieve line-scan rates as high as 100 MHz? Please use equations to explain this if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is greatly appreciated.  &lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="2651" LastEditDate="2016-07-30T10:18:10.080" LastActivityDate="2016-07-30T11:43:24.233" Title="How is an digital coherent receiver able to achieve 100 Mhz line-scan rate for detecting coherent light in the presence of incoherent light?" Tags="|optics|lasers|digital-communication|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10887" PostTypeId="1" AcceptedAnswerId="22175" CreationDate="2016-07-30T10:14:37.237" Score="4" ViewCount="187" Body="&lt;p&gt;I recently flew on a A380 and I saw this interesting socket. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BznWK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BznWK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain to me what is the purpose of these 4 golden connections?&lt;/p&gt;&#xA;" OwnerUserId="2070" LastEditorDisplayName="user4139" LastEditDate="2018-06-10T09:25:34.087" LastActivityDate="2018-06-11T03:46:11.887" Title="What is that airplane headphone socket?" Tags="|electrical-engineering|aircraft-design|audio-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10888" PostTypeId="2" ParentId="10886" CreationDate="2016-07-30T11:43:24.233" Score="1" Body="&lt;p&gt;You are wrong about that calculation. Pay attention to the unit analysis!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have 150 &amp;micro;m/s and 0.1 &amp;micro;m/step, then you divide the two to get the step rate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{150 \frac{\mu m}{s} }{ 0.1 \frac{\mu m}{step}} = 1500 \frac{step}{s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the time taken per step is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{ 0.1 \frac{\mu m}{step}}{150 \frac{\mu m}{s} } = 0.667 \frac{ms}{step}$$&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-07-30T11:43:24.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10889" PostTypeId="1" CreationDate="2016-07-30T11:47:19.447" Score="1" ViewCount="345" Body="&lt;p&gt;I want to make an inflatable seat that is electronically controlled. I am specifically looking to be able to estimate the level of inflation of an airbag using small electronic sensors. The airbag is to be filled using a DC pump, and a solenoid will be used to control exhaust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used the term &quot;level of inflation&quot; because I am using and airbag that somebody is sitting on, so the air will be getting compressed by the persons weight and movement. This will obviously cause variations in pressure and volume. In addition, I would like to use this measurement/information to inflate the airbags to the same levels at a later date. However, I won't necessarily be starting with a completely deflated bag every time I inflate the bag to this measured level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes some sense - apologies if it's too convoluted. I'm happy to answer further questions.&lt;/p&gt;&#xA;" OwnerUserId="7451" LastActivityDate="2016-12-31T14:09:52.390" Title="Measuring the level of inflation of an airbag" Tags="|pressure|sensors|pumps|airflow|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10890" PostTypeId="1" AcceptedAnswerId="10899" CreationDate="2016-07-30T13:44:05.873" Score="3" ViewCount="564" Body="&lt;p&gt;I have trial version of STAAD.Pro software and I am trying to figure out why my calculation are not same as manual one. I have modeled simple structure. But I don't want to have beams 1,27,26,25,24. It is intended that all that beams are one beam and beams 6,11,16,21 are welded to that beam and central beam (there are also beams 3,5,10,15,20 and they are in reality one single beam). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that influencing results (Continues load is added to 8 middle beams). I have merged beams 1,27,26,25,24 into single beam , and beams 3,5,10,15,20 into single beam-  and beams 2,7,12,17,22 into single beam also - but than it says that I have multiple structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Top view of structure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KVVsR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KVVsR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain how to model intended? &lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="7083" LastEditDate="2016-08-06T20:57:03.180" LastActivityDate="2016-08-06T20:57:03.180" Title="Staad - structure design" Tags="|mechanical-engineering|structural-engineering|civil-engineering|software|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10891" PostTypeId="2" ParentId="10885" CreationDate="2016-07-30T15:30:41.803" Score="4" Body="&lt;p&gt;This has nothing to do with safety.  (At least in the U.S., &lt;a href=&quot;http://www.cpsc.gov/vi-VN/Business--Manufacturing/Business-Education/Business-Guidance/Household-Electrical-Products/Hand-Supported-Hair-Dryers/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the CPSC requires all hand-held hair dryers to have an integral GFI&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;p&gt;Many trimmers on the market still run directly off line power.&lt;/p&gt;&#xA;&lt;p&gt;Battery-powered/cordless trimmers are the only reason you will find any that use a power converter: The motor is built to run off of DC, provided by the on-board battery.  The battery has to be charged from a DC supply.&lt;/p&gt;&#xA;&lt;p&gt;You &lt;em&gt;might&lt;/em&gt; find trimmers that do not feature batteries but do require a DC adapter.  This is typically done when a manufacturer wants to produce a wired model, and realizes that it is cheaper to simply replace the battery and charging circuit with a transformer than to manufacture a separate AC motor.&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="77" LastEditDate="2023-11-16T15:27:56.493" LastActivityDate="2023-11-16T15:27:56.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="10892" PostTypeId="2" ParentId="10881" CreationDate="2016-07-30T15:40:25.493" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://www.dartmouth.edu/~chemlab/techniques/vfiltration.html&quot; rel=&quot;nofollow&quot;&gt;Standard lab approach is vacuum filtration&lt;/a&gt;: Get a Buchner funnel and filter paper sized to capture the smallest particles you care to measure.  Now:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Weigh filter.&lt;/li&gt;&#xA;&lt;li&gt;Place filter in funnel and wet with solvent.&lt;/li&gt;&#xA;&lt;li&gt;Connect funnel to vacuum filter flask.&lt;/li&gt;&#xA;&lt;li&gt;Turn on vacuum.&lt;/li&gt;&#xA;&lt;li&gt;Pour sample into funnel.&lt;/li&gt;&#xA;&lt;li&gt;Continue to rinse sample with a volatile solvent until it is clean.  For rinsing off oil you could use acetone.&lt;/li&gt;&#xA;&lt;li&gt;Wait a few moments for solvent to evaporate.&lt;/li&gt;&#xA;&lt;li&gt;Weigh filter with sample.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="77" LastEditorUserId="77" LastEditDate="2016-07-30T18:35:52.023" LastActivityDate="2016-07-30T18:35:52.023" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="10893" PostTypeId="2" ParentId="10813" CreationDate="2016-07-30T15:50:32.187" Score="1" Body="&lt;p&gt;If a &lt;em&gt;licensed&lt;/em&gt; civil engineer signs or stamps his calculations, then &lt;em&gt;he&lt;/em&gt; is liable if his calculations prove to be incorrect.  Licensed engineers are required to carry professional liability insurance specifically to cover that liability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, if someone else pays for a design, and you are dubious it is adequate, just make sure that the engineer who signed off on it has an active license and insurance policy.  If it fails then you claim against the insurance.&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2016-07-30T15:50:32.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10894" PostTypeId="1" AcceptedAnswerId="56510" CreationDate="2016-07-30T17:37:20.110" Score="4" ViewCount="150" Body="&lt;p&gt;Why don't cellphones use unidirectional antennas instead of omnidirectional (single dipole) antennas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be accomplished by using a unidirectional array of dipole antennas (specifically, a broadside array) placed at different locations in the phone, where each antenna's phase could be changed in such a way to minimize radiation directed into the user's head and maximize radiation sent in the direction of the cell tower, producing a radiation pattern as is shown in the picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, with digital electronics, arbitrary phase-delays could be introduced to change the orientation of the major lobe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cellphone batteries would last longer and there would be less adverse health effects due to EMF exposure.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QJDZR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QJDZR.png&quot; alt=&quot;Unidirectional Array&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6085" LastEditorUserId="6085" LastEditDate="2016-07-30T21:32:13.217" LastActivityDate="2023-10-01T22:19:21.407" Title="Why don't cellphones use unidirectional antennas?" Tags="|electrical-engineering|wireless-communication|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10896" PostTypeId="2" ParentId="10890" CreationDate="2016-07-31T12:31:23.040" Score="2" Body="&lt;p&gt;In short, by having them each as single members or beams should not be influencing the result provided that you have connected them together as fully fixed/welded (this is the member end condition - there should be no releases at the ends of the members so that the software will treat the individual members as one continuous member).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining them into one single member is not the correct way to model it. They need to be individual members with the correct member end conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why your manual calculations are not matching the software is anyone's guess - either your calculations are wrong or you're not modelling it correctly in the software. Perhaps you should provide your calculations or more info about the structure and how it is both loaded and supported, then we can probably help you further.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastActivityDate="2016-07-31T12:31:23.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10897" PostTypeId="2" ParentId="10885" CreationDate="2016-07-31T14:37:46.890" Score="3" Body="&lt;p&gt;First, it's incorrect that beard trimmers always use some adapter between them and line power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the main reason for hair dryers not using adapters is the large power they require.  At 1 to 2 kW, a separate DC power supply would be large and expensive.  At such power, you want to use the line voltage directly to minimize the current.  Lower voltage would require higher current at the same power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is cheaper and better for the user to make the hairdryer safe to use line power, than to push the safety issues onto a separate power supply.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-07-31T14:37:46.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10898" PostTypeId="2" ParentId="10330" CreationDate="2016-07-31T14:43:53.153" Score="1" Body="&lt;p&gt;In a 2d abaqus analysis you can &lt;em&gt;only&lt;/em&gt; rotate the material orientation in-plane, in other words you &lt;em&gt;can not&lt;/em&gt;  &quot;transform&quot; the material 3- direction into the plane of the analysis. (Which seems to be what you are asking in #3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution to this problem is to &quot;manually&quot; transform the material definition, i.e. reordering the values on the &lt;code&gt;*Elastic,ENGINEERING CONSTANTS&lt;/code&gt; line.  (and i'm sure you know the poisson ratios are not symmetric, so its not quite a trivial reordering)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to #2 You can request element volume as an output field variable, EVOL&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2016-08-01T15:58:46.047" LastActivityDate="2016-08-01T15:58:46.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10899" PostTypeId="2" ParentId="10890" CreationDate="2016-07-31T17:48:42.030" Score="3" Body="&lt;p&gt;In structural analysis, one represents a real &lt;strong&gt;beam&lt;/strong&gt; (or column, or any other &quot;unidimensional&quot; element) is represented by one or more &lt;strong&gt;bars&lt;/strong&gt;. A bar is defined as a connection between two nodes. If you join all the horizontal bars into one, then you'll be stating that the left-most and right-most nodes (belonging to bars 1 and 24, respectively) are connected, but you'll lose the information regarding how the real beam is influenced by the vertical bars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some programs that allow you to define the real beam as a single bar and then internally break it down into multiple bars when it detects the existence of other elements (such as the vertical bars) which should connect to it (I know &lt;a href=&quot;http://www.autodesk.com/products/robot-structural-analysis/overview&quot; rel=&quot;nofollow&quot;&gt;Autodesk Robot&lt;/a&gt; does this, and there are probably others), but clearly STAAD is not one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, your current bar layout is correct, since it adequately demonstrates that the beams are connected. You must then define the appropriate releases to describe how the beams are connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, however, no releases are required anywhere. Since the horizontal bars represent a single beam, no releases are necessary. This means the bars must deform as one. And the vertical bars are all welded to the horizontals, so they should also deform and rotate in unison. Therefore, no releases are necessary there either.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-08-02T13:07:21.787" LastActivityDate="2016-08-02T13:07:21.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10901" PostTypeId="1" AcceptedAnswerId="10921" CreationDate="2016-07-31T21:49:08.563" Score="3" ViewCount="4155" Body="&lt;p&gt;In digital or discrete control, instead of continuous analog-type variables, a controller samples a process variable signal at regular discrete time intervals. So, effectively, the process variable does not change as far as the system is concerned between those samples, or if it does, the change won’t be detected until the next sample. And it will be abrupt if it changes at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would think that the faster the sampling rate (the more samples in the same time interval, or shorter times between samples), the more closely the discrete control would emulate ideal continuous control. Apparently that is not the case. In fact, as I understand it, increasing the sampling rate past some critical point can actually make a discrete time or digital control system unstable, as it can push the transfer function poles out close to and even outside the unit circle of stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be counter intuitive, at least to me. Is there a simple explanation for this that I’m missing here?&lt;/p&gt;&#xA;" OwnerUserId="5393" LastEditorUserId="1832" LastEditDate="2016-08-01T00:15:47.557" LastActivityDate="2016-08-02T17:56:06.930" Title="Discrete Control System Process Sampling Rate" Tags="|control-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10902" PostTypeId="1" CreationDate="2016-07-31T22:47:13.843" Score="0" ViewCount="1144" Body="&lt;p&gt;I need to provide an estimation of the BHP of a fan used in a HVAC system (water-source heat-pump). I have only limited information, the fan nameplate motor is 1 HP, it will run at 750 cfm is rated at 0.5 in w.c. of ESP. I don't have any information about its efficiency (motor or total). What could be a good estimation for its BHP?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since most motor reach their peak efficiency around 75% load I was thinking about doing 1 HP * 0.75 / eff and using 85% as its efficiency. Would this be a bad estimation?&lt;/p&gt;&#xA;" OwnerUserId="7466" LastEditorUserId="-1" LastEditDate="2016-08-01T09:38:59.837" LastActivityDate="2017-05-12T18:33:05.733" Title="Fan BHP estimation" Tags="|mechanical-engineering|hvac|power|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10903" PostTypeId="1" CreationDate="2016-07-31T23:05:24.813" Score="1" ViewCount="147" Body="&lt;p&gt;I was hoping there exists a material like hard rubber used in bushings, but has the property to stiffen when exposed to vibration with high frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my application is something like a small dampening suspension on a system with stiff wheels so in high speeds the dampening will get stiffer due to the higher frequency vibrations that occurs from the small wheel rolling on a surface like asphalt. So the system is soft in low speeds and stiffer in higher speeds. Do you guys know of any material or system with these properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;br&gt;&#xA;Honestly, I was looking for something small and hopefully something not reliant on a power supply to change the ride characteristics of a skateboard/longboard truck depending on its speed. My theory is that if the bushing on the truck gets stiffer in higher speeds the ride is more stable and less prone to speed wobbles. ( Stiff bushings/dampener requires more weight to turn the board = less wobble in higher speeds)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this is not that difficult to make using something mechanical with an electric speed sensor, but a better solution would be if such a material exists or if its possible to make a bushing molded in some material that stiffens when a small electrical current is applied.&lt;/p&gt;&#xA;" OwnerUserId="7467" LastEditorUserId="7467" LastEditDate="2016-08-02T12:48:58.153" LastActivityDate="2016-12-08T17:04:54.670" Title="Material for suspension that stiffens in higher speeds?" Tags="|springs|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10904" PostTypeId="2" ParentId="10865" CreationDate="2016-08-01T14:13:09.273" Score="2" Body="&lt;p&gt;The &quot;laser device&quot; you're looking for is called a Laser Tachometer, though other illumination sources will work. I have had good success with devices from Monarch Instruments. Specifically the ROS-P line of optical sensors (I see they have a &lt;a href=&quot;http://www.monarchinstrument.com/product.php?ID=46&quot; rel=&quot;nofollow&quot;&gt;laser version&lt;/a&gt; ) and the &lt;a href=&quot;http://www.monarchinstrument.com/product.php?ID=24&quot; rel=&quot;nofollow&quot;&gt;PLT200 Pocket Laser Portable Tachometer&lt;/a&gt;. The laser devices do work very well in more challenging environments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my typical application using the PLT200, there's a small piece of reflective tape attached to the rotating shaft and the device is aimed at the target. The shaft speed is displayed on the screen, which gives you a number to check against what's expected. For longer term testing, I've also set these up on a fixed mount and used their data port which will output a TTL pulse train with one pulse/tape mark/rev to the data acquisition system of my choosing. While the device has its own frequency to speed conversion for the display, the output is just the pulsetrain so you'll need to do that on your own. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am in no way associated with Monarch and I'm sure there are other manufacturers, but I have always had good success with these.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2016-08-01T14:21:18.973" LastActivityDate="2016-08-01T14:21:18.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10905" PostTypeId="2" ParentId="10903" CreationDate="2016-08-01T14:39:16.263" Score="0" Body="&lt;p&gt;This sort of thing could be accomplished using &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetorheological_damper&quot; rel=&quot;nofollow&quot;&gt;Magnetorheological (magnetic) dampers&lt;/a&gt; which are already used in a variety of vehicles (such as the &lt;a href=&quot;http://newatlas.com/go/5752/&quot; rel=&quot;nofollow&quot;&gt;Audi TT&lt;/a&gt;). These use shock absorbers filled with a fluid containing iron particles. When a magnetic field is applied these particles align making the response stiffer. The stiffness can also be varied by adjusting the strength of the applied magnetic field. You could use whatever input you want for the controller: speed, vibration frequency, etc. The downside is this type of system is relatively expensive/complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more detailed description of how these work is available &lt;a href=&quot;http://www.autoevolution.com/news/how-magnetorheological-suspension-works-8947.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2016-08-01T14:39:16.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10906" PostTypeId="2" ParentId="10889" CreationDate="2016-08-01T16:30:22.627" Score="1" Body="&lt;p&gt;Some sort of strain gauge on the surface of the airbag would probably be your best bet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you can make the following relations to get to the 'inflation level of the airbag':&lt;/p&gt;&#xA;&#xA;&lt;p&gt;strain gauge reading -&gt; airbag's surface curvature -&gt; airbag inflation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are a few steps in between those I mentioned above, and some fairly difficult math will be required to relate the external curvature to the inflation level of the airbag. But that is the general idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at this image to get an idea of how a strain gauge works and see if it looks like something you could use.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/EWsne.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EWsne.jpg&quot; alt=&quot;strain gauge description&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture from strain gauge wikipedia page.&lt;/p&gt;&#xA;" OwnerUserId="7474" LastActivityDate="2016-08-01T16:30:22.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10909" PostTypeId="1" AcceptedAnswerId="10911" CreationDate="2016-08-01T19:54:58.853" Score="0" ViewCount="693" Body="&lt;p&gt;I'm interested in designing a transmission to achieve a constant-speed, variable-torque output from variable-speed, variable-torque input -- something like a CVT, but applied backward. Is such a thing possible? Do any examples of a transmission like this exist commonly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm happy to answer any clarifying questions, as I'm imagining this doesn't include all the needed information. Thank you in advance for your patience!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I should mention that the specific effect that I'm looking for is that when the input speed increases, the output speed remains constant, but the output &lt;em&gt;torque&lt;/em&gt; increases proportionally.&lt;/p&gt;&#xA;" OwnerUserId="7478" LastEditorUserId="7478" LastEditDate="2016-08-02T11:30:03.400" LastActivityDate="2016-08-02T13:38:03.537" Title="Transmission for constant-speed, variable-torque output from variable-speed, variable-torque input?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10911" PostTypeId="2" ParentId="10909" CreationDate="2016-08-01T21:24:34.360" Score="1" Body="&lt;p&gt;In an early solution to air conditioning of railroad cars the variable-speed motion of the axles was transmitted through such a coupling to provide constant speed drive for the A/C compressor. There are several mechanical and electromechanical CVT type mechanisms that could be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find information about the various CVT technologies, you should also search &quot;mechanical adjustable-speed drive&quot; and &quot;mechanical variable-speed drive.&quot; The term CVT is primarily used for vehicle components. The ASD and VSD terms are used with industrial equipment. In industry, electrical equipment is usually used, but there are mechanical alternatives. Note that descriptions of the equipment that you find will be be the direction opposite to what you want, but the reverse alternative can usually be achieved with the same or similar equipment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;EDIT: I should mention that the specific effect that I'm looking for is that when the input speed increases, the output speed remains constant, but the output torque increases proportionally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can regulate speed or you can regulate torque, but you can not regulate both simultaneously. The characteristics of the load must dictate one or the other. Perhaps you should explain the system more completely and add a diagram.&lt;/p&gt;&#xA;" OwnerDisplayName="user6431" LastEditorDisplayName="user6431" LastEditDate="2016-08-02T13:38:03.537" LastActivityDate="2016-08-02T13:38:03.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10912" PostTypeId="2" ParentId="10807" CreationDate="2016-08-01T21:57:44.853" Score="0" Body="&lt;p&gt;Lots of assumptions. Some gearbox data and empirical motor current data would go a long way to predicting the performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;emsp;  &amp;emsp;  &amp;emsp;  Sensorless vector mode  &lt;br&gt;&#xA;&amp;emsp;  &amp;emsp;  &amp;emsp;  Startup torque will be ignored by the drive &lt;br&gt;&#xA;10  &amp;emsp;  &amp;emsp;  hp Powerflex 700 VFD    &lt;br&gt;&#xA;10  &amp;emsp;  &amp;emsp;  hp motor    &lt;br&gt;&#xA;1775    &amp;emsp;  &amp;emsp;  rpm &lt;br&gt;&#xA;29.6    &amp;emsp;  &amp;emsp;  ft lbs  &lt;br&gt;&#xA;69.5    &amp;emsp;  &amp;emsp;  gear box ratio  &lt;br&gt;&#xA;200 &amp;emsp;  &amp;emsp;  ft lbs assumed constant gearbox torque consumption, referenced on output    &lt;br&gt;&#xA;50% &amp;emsp;  &amp;emsp;  worst case planetary gearbox efficency (in additon to constant) &lt;br&gt;&#xA;26  &amp;emsp;  &amp;emsp;  rpm @ driving sprocket  &lt;br&gt;&#xA;928 &amp;emsp;  &amp;emsp;  ft lbs @ driving sprocket   &lt;br&gt;&#xA;95% &amp;emsp;  &amp;emsp;  estimated chain efficency   &lt;br&gt;&#xA;19  &amp;emsp;  &amp;emsp;  rpm @ driven sproket    &lt;br&gt;&#xA;1176    &amp;emsp;  &amp;emsp;  ft lbs @ driven sprocket    &lt;br&gt;&#xA;&amp;emsp;  &amp;emsp;  &amp;emsp;      &lt;br&gt;&#xA;&amp;emsp;  &amp;emsp;  &amp;emsp;  Full 10 hp has previously damaged mixing rotor  &lt;br&gt;&#xA;1176    &amp;emsp;  &amp;emsp;  ft lbs Estimated emperical damaging torque  &lt;br&gt;&#xA;700 &amp;emsp;  &amp;emsp;  ft lbs, desired torque limit    &lt;br&gt;&#xA;553 &amp;emsp;  &amp;emsp;  ft lbs, pre gearbox, desired torque limit   &lt;br&gt;&#xA;18.8    &amp;emsp;  &amp;emsp;  ft lbs seen by motor, desired torque limit  &lt;br&gt;&#xA;500 &amp;emsp;  &amp;emsp;  ft lbs Estimated Normal opperational torque, (motor needs amp'd for emperical data here)    &lt;br&gt;&#xA;395 &amp;emsp;  &amp;emsp;  ft lbs, operational torque  &lt;br&gt;&#xA;14.2    &amp;emsp;  &amp;emsp;  ft lbs seen by motor, operational torque    &lt;br&gt;&#xA;&amp;emsp;  &amp;emsp;  &amp;emsp;      &lt;br&gt;&#xA;15.4%   &amp;emsp;  &amp;emsp;  Torque as range of total    &lt;br&gt;&#xA;&amp;emsp;  &amp;emsp;  &amp;emsp;      &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0.1% percent of motor rated toque, torque resolution. It is not clear if this requires torque feedback or not.&#xA; &lt;a href=&quot;http://literature.rockwellautomation.com/idc/groups/literature/documents/um/20b-um002_-en-p.pdf&quot; rel=&quot;nofollow&quot;&gt;Page 22 of powerflex700 programming manual&lt;/a&gt;    &lt;br&gt;&#xA;No closed loop torque sensor will be used as they are prohibitively expensive   &lt;br&gt;&#xA;    &lt;br&gt;&#xA;Worst published torque regulation on the drive is +/-5% in the &lt;a href=&quot;http://literature.rockwellautomation.com/idc/groups/literature/documents/td/20b-td001_-en-p.pdf&quot; rel=&quot;nofollow&quot;&gt;powerflex700 technical data&lt;/a&gt; &lt;br&gt;&#xA;    &lt;br&gt;&#xA;15.4% &gt; 5%, VFD torque limiting will be satisfactory in this application    &lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-08-01T21:57:44.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10913" PostTypeId="1" CreationDate="2016-08-01T23:32:53.843" Score="3" ViewCount="798" Body="&lt;p&gt;Can someone tell me what type of equipment is used to measure RF power levels transmitted by a cellular LTE base station? I would like to gain some insight into how the follow data was obtained.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nhD6R.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nhD6R.jpg&quot; alt=&quot;Cellular RF power level&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back Story:&lt;/strong&gt; The local company wants to install a cellular base station in our neighborhood. They have shared the above data set indicating that the cellular coverage is poor in a certain area. But cellular service provider's web site indicate excellent coverage. Information doesn't tally up.&lt;/p&gt;&#xA;" OwnerUserId="7083" LastEditorUserId="282" LastEditDate="2016-08-16T03:51:25.770" LastActivityDate="2018-05-24T14:29:08.617" Title="How to measure transmit power level from LTE base station?" Tags="|electrical-engineering|wireless-communication|lte|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10914" PostTypeId="2" ParentId="10913" CreationDate="2016-08-02T03:39:36.593" Score="1" Body="&lt;p&gt;On iPhones the signal strength indicator can be changed to a numerical value as indicated by the image below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QaBeM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QaBeM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above can be done by dialing *3001#12345#*. The phone will enter into field test mode. The value indicated is in dBm. Lower dbm value indicates a poor signal. A dBm value closer to 0 is a strong signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly on a Android phone the same infomation can be found at About device -&gt; Status (Kit Kat) depending the Android version. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nu6m9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nu6m9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cnet.com/how-to/force-iphone-to-show-numerical-value-for-signal-strength/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Force iPhone to show numerical value for signal strength&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ubersignal.com/field-test-mode&quot; rel=&quot;nofollow noreferrer&quot;&gt;Field Test Mode: What it is and How To Enable it on Your Phone&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-08-04T02:48:27.587" LastActivityDate="2016-08-04T02:48:27.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10915" PostTypeId="1" AcceptedAnswerId="10917" CreationDate="2016-08-02T07:21:55.107" Score="4" ViewCount="107" Body="&lt;p&gt;I am looking to build a specific kind of lift for a project I'm working on but I don't know the name.  I am willing to hire a mechanical engineer to help me design and build it (or buy one if it already exists) but I'd like to do as much research as possible before deciding on the next steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have in mind is like a ferris wheel, save that it is shaped more like a stretched rubber band than a circle (i.e. it has vertical sides and a short circular path at the top and bottom).  Instead of carts that people sit in I'd just like a flat steel floor, say 4'x8', for loading cargo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a switch I'd like the flat &quot;carts&quot; to move into a loading space, load from the front, then be sent up to make room for the next cart.  Maximum load should be about 1,000lbs per cart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this type of device called?  I know similar things have been built for parking cars, but I don't want something quite so large and I'm interested in storing smaller things than cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry if this isn't more descriptive, please ask if I can clarify anything and I'll be happy to.&lt;/p&gt;&#xA;" OwnerUserId="7481" LastEditorUserId="1586" LastEditDate="2016-08-02T23:07:34.523" LastActivityDate="2016-08-02T23:07:34.523" Title="Lift-type identification - What class of lifts are ferris wheels in?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10916" PostTypeId="2" ParentId="10915" CreationDate="2016-08-02T09:46:03.117" Score="5" Body="&lt;p&gt;You probably mean a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bucket_elevator&quot; rel=&quot;nofollow&quot;&gt;Bucket Elevator&lt;/a&gt;. They come in several types some of which fulfill your need. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image in wikipedia is one shaped more like a escalator but there are more vertical bucket elevators. Bucket elevators have many uses from water filtration to grain transport.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2016-08-02T09:51:23.637" LastActivityDate="2016-08-02T09:51:23.637" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10917" PostTypeId="2" ParentId="10915" CreationDate="2016-08-02T13:59:03.660" Score="4" Body="&lt;p&gt;At the scale you describe (8'x4') this would be similar to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Paternoster&quot; rel=&quot;nofollow&quot;&gt;Paternoster&lt;/a&gt;, except that it would need to stop to load and unload the carts. Paternosters (and bucket elevators) move continuously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Health and Safety regulations have closed down most of them (especially those that can be used by the general public) because of the risk of accidents when entering or leaving. YouTube has some videos, including what happens when you ride around the top or the bottom - which is designed to be safe.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-08-02T13:59:03.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10921" PostTypeId="2" ParentId="10901" CreationDate="2016-08-02T17:56:06.930" Score="2" Body="&lt;p&gt;Based on the reference cited in the question comments, this particular situation occurs when you implement a controller designed in the continuous time domain on a discrete time system with measurement or model uncertainties.&lt;/p&gt;&#xA;&lt;p&gt;Designing the controller in continuous time means that the sample rate is not taken into account/is assumed to be infinitely fast. Therefore the controller can behave very differently for different sample rates, even if the control gains remain unchanged. In particular, increasing the sample rate can destabilize the system by making it more sensitive to noise/uncertainty in the measurements (especially with PD and PID controllers).&lt;/p&gt;&#xA;&lt;h1&gt;A simplified example&lt;/h1&gt;&#xA;&lt;p&gt;You have a proportional-derivative controller, with a derivative gain &lt;span class=&quot;math-container&quot;&gt;$K_d = 1$&lt;/span&gt; (the proportional part is not really important for this example). Your control system has a sensor which gives you a measurement of your error with a random uncertainty up to &lt;span class=&quot;math-container&quot;&gt;$w = 0.1$&lt;/span&gt; units. At a certain moment in time the true error signal is changing at a constant rate of &lt;span class=&quot;math-container&quot;&gt;$\frac{de}{dt} = 10$&lt;/span&gt; units/s.&lt;/p&gt;&#xA;&lt;p&gt;First you implement your controller with a sample time of &lt;span class=&quot;math-container&quot;&gt;$T=0.1$&lt;/span&gt; s. You estimate the current rate of change using a first order approximation, and it just so happens that you experience the maximum error in this timestep:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{de}{dt}_{measured} = \frac{e(k)-e(k-1)}{T} = \frac{\frac{de}{dt}T + w}{T} = \frac{de}{dt} + \frac{w}{T} = 10 + \frac{0.1}{0.1} = 11$&lt;/span&gt; units/s&lt;/p&gt;&#xA;&lt;p&gt;Therefore the output of the derivative part of your controller is &lt;span class=&quot;math-container&quot;&gt;$11K_d=11$&lt;/span&gt;, i.e. 10% higher than the ideal output of the continuous controller (&lt;span class=&quot;math-container&quot;&gt;$10K_d=10$&lt;/span&gt;). Notice that the contribution of the error term to your controller output is inversely proportional to sample time, therefore smaller sample times (faster sample rates) increase its effect.&lt;/p&gt;&#xA;&lt;p&gt;Then you decide, faster must be better, so you decrease the sample time to &lt;span class=&quot;math-container&quot;&gt;$T=0.01$&lt;/span&gt; s. Now, in the same situation, the estimated rate of change of error is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{de}{dt}_{measured} = 10 + \frac{0.1}{0.01} = 20$&lt;/span&gt; units/s&lt;/p&gt;&#xA;&lt;p&gt;The output of the derivative part of the controller is now &lt;span class=&quot;math-container&quot;&gt;$20K_d = 20$&lt;/span&gt; i.e. almost double what it was before, and 100% higher than the ideal output. Your controller is now behaving more aggressively even though the true rate of change of error has not increased.&lt;/p&gt;&#xA;&lt;p&gt;The problem is exaggerated in this example by using a first-order approximation, but more complex approximations of the derivative will only mitigate, not eliminate, the problem. The fact is that changing the sample rate of the system can have a negative impact in the presence of any kind of measurement uncertainty. Now, whether your controller performs better or worse when you increase your sample rate depends on the relative magnitude of the uncertainty versus the rate of change of the error signal, the relative magnitudes of your derivative and proportional gains, and the type of approximation you use for your derivatives.&lt;/p&gt;&#xA;&lt;h1&gt;Conclusion&lt;/h1&gt;&#xA;&lt;p&gt;If you are designing a system where the controller cannot be approximated as continuous and there are measurement or other uncertainties present, increasing the sample rate will not necessarily increase performance because the influence of the uncertainty on the control input will increase.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-08-02T17:56:06.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10922" PostTypeId="1" CreationDate="2016-08-02T20:56:16.767" Score="3" ViewCount="454" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8Pf6c.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Pf6c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using only the diagram above (Not using the questions in the diagram), i am asked to find the components of F2 along F1 and F3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble with this question. I can easily resolve any of the vectors into their x and y components but I am not 100% certain how to resolve it along the other vectors. Am I correct in saying that to resolve F2 along F1 we draw a line down directly from the head of F2 to F1 and then take 750*Cos(45) to get the component parallel to F1. What should I do to resolve F2 along F3? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is given as 1240N and 884N. Any hints on how to get to those answers would be much appreciated. Perhaps I am not understanding what is meant by resolving a vector along another vector?&lt;/p&gt;&#xA;" OwnerUserId="7492" LastEditorUserId="-1" LastEditDate="2016-08-03T11:04:33.073" LastActivityDate="2016-08-03T15:25:08.160" Title="Finding the components of a vector along two other vectors" Tags="|statics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10923" PostTypeId="1" CreationDate="2016-08-02T22:21:32.620" Score="1" ViewCount="68" Body="&lt;p&gt;Over years of flying small airplanes I have seen fatigue cracks and frayed brackets and connections.&lt;br&gt;&#xA;I wonder if there is an easy way to search the propagation of stress wave in machinery and connection brackets and parts. I am not concerned here with vibrations so much.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we have a bracket holding the block of an engine to another part and at short intervals of time it undergoes huge stresses of impact of rotation of the engine cylinders, shafts, etc.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stress wave propagates inside the bracket and reflects and interfere with its refractions and echo until it dies out by dispensing its energy into friction or transferring to other parts!&lt;br&gt;&#xA;I know there is a large volume of empirical tests involved but I am concerned with a mathematical modeling scheme. Thank you! &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-08-05T07:28:31.977" Title="stress wave propogation in metals" Tags="|stresses|shock|fatigue|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10924" PostTypeId="2" ParentId="10864" CreationDate="2016-08-02T23:51:38.630" Score="-2" Body="&lt;p&gt;A beam could move in 3 axis x,y,z and bend in 3 axis and rotate along the x axis (torsion) . In all and every one of these cases the support can act in a variety of ways, allow free movement, allow no movement, allow resisted movement, impart pre configured hard constraint such as is the case for pre/post tensions re-bars or none flexible pre designed displacement to load the member in a desired setting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Releasing one end means to release one or several or all of these constrains. I have worked with Risa it is a good FEM software but 25 years ago the version i had did not have a good menu of release. Such as spring loaded constrain, or pre set hard displacement.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-08-02T23:51:38.630" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10925" PostTypeId="1" AcceptedAnswerId="10932" CreationDate="2016-08-03T00:08:51.947" Score="7" ViewCount="1489" Body="&lt;p&gt;In historical or modern buildings, flat ceilings are typically made of concrete, wood, metal, solid bedrock (for carved tunnels), or generally a material that can withstand pushing forces (pressure) and pulling forces (tension).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ceiling made from bricks / masonry would typically be vault-shaped, because otherwise it would crash. This is what we find in a lot of historic buildings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can so far only remember flat brick ceilings from imaginary buildings e.g. in a roleplaying context or in video games. But I wonder if there are cases where such a thing can be realistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, I can imagine two ways to achieve a flat ceiling with masonry:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;By using very good cement, so the entire thing would behave like one solid piece of rock or concrete.&lt;/li&gt;&#xA;&lt;li&gt;By imitating a vault, in a way similar to how a Fresnel lens imitates a regular concave lens:&lt;br&gt;&#xA;Use slightly V-shaped bricks, and stack them like this, more or less: `\\\|||///``.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can this ever work?&lt;/li&gt;&#xA;&lt;li&gt;Does this technique have a name?&lt;/li&gt;&#xA;&lt;li&gt;How far can you go with this, what are the limitations?&lt;/li&gt;&#xA;&lt;li&gt;Are there historical examples?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2640" LastEditorUserId="2640" LastEditDate="2016-08-03T14:55:16.147" LastActivityDate="2016-08-03T16:07:11.283" Title="Can masoned ceilings be flat?" Tags="|civil-engineering|masonry|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10926" PostTypeId="2" ParentId="2307" CreationDate="2016-08-03T00:33:33.127" Score="0" Body="&lt;p&gt;Make a reverse Farnsworth Fusor! A little helium gas in a mylar balloon with special considerations made to polarize the metal interior as a capacitor with the ground being your gas feed tube! The plasma will accelerate from the ground as positive helium nuclei to neutralize on the negative balloon walls. Because electrons streaming to ground act with little force and the ground wire can be insulated until a plasma stream develops you don't have to worry about the electrostatic attraction to ground overcoming the strong negative plates self repulsion and the strong helium plasma pressure! &lt;/p&gt;&#xA;" OwnerUserId="7493" LastActivityDate="2016-08-03T00:33:33.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10927" PostTypeId="1" CreationDate="2016-08-03T01:26:54.650" Score="2" ViewCount="141" Body="&lt;p&gt;I am building a device to remind myself of keeping my spine straight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a way to implement the &quot;reminder&quot; feature and the control package, but  I have not found a way to measure changes in curvature in a small form factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about using a simple light encoder with a marked rod that slides through it, but that would make it much too large to stay hidden under clothes. Rotary encoders suffer from the same problem. Plus, to get a reliable value, I would need multiple individual sensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way I can think of is using something similar to &lt;strong&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=EkPeF-sYQ-0&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;&lt;/strong&gt; and monitoring changes in pressure in the different tubes, but this seems unviable to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A perfect solution would be something directly implemented in the flexible surface... Some sort of flexible PCB based solution perhaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I detect changes in curvature?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7494" LastActivityDate="2016-08-03T08:17:53.927" Title="How to measure changes in curvature?" Tags="|mechanical-engineering|sensors|measurements|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10928" PostTypeId="2" ParentId="10925" CreationDate="2016-08-03T02:22:23.427" Score="5" Body="&lt;p&gt;I had a project around 1976 in the middle east with similar situation. Because of a sudden shortage of cement and re-bars we had to modify the plans for a few thousand homes in several villages that were totally destroyed by an earthquake.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We talked with local masons and people in the industry and modified the plans to take advantage of locally available building materials. Which were burned clay brick and their traditional mortar which was a mixture of gypsum and lime and clay mud mixed with straw.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We adopted the ceilings to two typical modules, rectangular of 10x 16 feet and squares of 20x20.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases we used arches of very shallow rise to width like 3% for the rectangular and 5% for squares, both of them supported buy concrete beams on brick walls. Because of the concern about the future quakes the buildings were re-enforced with concrete vertical tie columns connecting roofs and ceilings to the foundation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The square ceiling were covered by a pattern of four triangle gently raising up and meeting at a key brick at the center of the room like a big X.&#xA;The ceilings were surfaced to a level finish by light weight gypsum plaster.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;we had to ask the old mason masters to train the young ones the skill and art of laying the bricks without shoring by slapping the new row into a slanted bed of mud covering the last row of brick while making sure there was solid brick to brick connection on each row! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many quakes, some in the order of 7.8 Richter, have shaken that same area. As much as I know no significant damage has happened.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below you can see two sketches. One is looking up perspective view of the shallow vault ceiling, the other one front view of shallow barrel ceiling. The trick of laying the bricks without forms is too mix the mortar with just the right proportion so it dries with enough strength to hold the dead weight and at the same time give the mason time to work on the bricks. The vault proved to be exceptionally strong and easy to build. The bricks were approx. 2x4x8 inches burned into 400 psi compression strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nPj5k.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nPj5k.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gUlJ3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gUlJ3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2016-08-03T11:19:45.043" LastActivityDate="2016-08-03T11:19:45.043" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10929" PostTypeId="2" ParentId="10927" CreationDate="2016-08-03T02:23:03.383" Score="2" Body="&lt;p&gt;Some ideas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You try to could measure length on the surface or possibly between 2 points instead as a proxy. The arched back is longer than the unarched one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact when measuring bending of metal you use &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Strain_gauge&quot; rel=&quot;nofollow&quot;&gt;strain gauges&lt;/a&gt; to do just this. So in fact this would probably do it you just need a suitable gauge, container and attachment.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You may want to measure relative of two points position instead of arc &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a good posture has different center of balance so you might measure that instead, this would mean you could instrument a chair for example.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2016-08-03T02:23:03.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10930" PostTypeId="2" ParentId="10927" CreationDate="2016-08-03T08:17:53.927" Score="0" Body="&lt;p&gt;My solution wouldn't be some thing that clings to the spine - it sounds far too much of an inconvenience to put the thing on!  I think monitoring general upright posture is more useful anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest a mini harness or shoulder holster that goes on over the arms and holds a slim package against the &lt;strong&gt;upper back&lt;/strong&gt;, between the shoulder blades. This would hold a typical hobby accelerometer/gyro as well as the inevitable electronics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose would be to monitor the angle of the shoulders, effectively logging the wearer's stance or deportment, whether standing or sitting.  It's to make sure the user is not slumping forward (and perhaps not twisting too).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A period of logging and analysis would be needed first to decide upon the preferred angles. Then once this is set it can send a reminder (using a vibrating motor) when the wearer is slouching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional benefit - with recording, this might be useful for joggers and activity fans to monitor their daily footstep or activity count, in a device that is less affected by arm movements. The vibrating motor might be useful for a silent alert of various goals too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I accept this is not a specific measure of spine curvature. I suggest it as a more general measure of good posture, in a fairly compact package.)&lt;/p&gt;&#xA;" OwnerUserId="6665" LastActivityDate="2016-08-03T08:17:53.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10931" PostTypeId="2" ParentId="10925" CreationDate="2016-08-03T10:45:04.687" Score="1" Body="&lt;p&gt;Not sure if this counts, but the proprietary Halfen concealed lintel system* achieves flat brick ceilings. It may not count because the masonry is only decorative, not structural - it actually uses steel for the structural support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See e.g. &lt;a href=&quot;https://www.google.co.uk/search?site=&amp;amp;tbm=isch&amp;amp;source=hp&amp;amp;biw=1684&amp;amp;bih=847&amp;amp;q=halfen%20concealed%20lintel&amp;amp;oq=halfen%20conceal&amp;amp;gs_l=img.3.0.0i30j0i24.505.4985.0.6168.22.10.4.7.11.0.125.920.7j3.10.0....0...1ac.1.64.img..2.13.803...0j0i8i30.gcDkGA8rYHQ#imgrc=BR4O7NKROwPjYM%3A&quot; rel=&quot;nofollow&quot;&gt;google image of flat brick arch with halfen concealed lintel&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Other proprietary systems are probably available from other manufacturers. I'm not sure there's a recognised name for this technique, possibly &quot;concealed lintel&quot; might suffice.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The problem with creating true flat-masonry ceilings is that masonry is very poor in tension - the brick-mortar joints take compressive load but have little to no tension capacity. With a flat ceiling the bottom face of the ceiling will be in tension due to bending stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically some sort of mortar/epoxy/glue which has tension capacity, and also achieves a tension bond to bricks could work.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2016-08-03T14:59:10.670" LastActivityDate="2016-08-03T14:59:10.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10932" PostTypeId="2" ParentId="10925" CreationDate="2016-08-03T11:46:33.590" Score="5" Body="&lt;p&gt;What you're looking for is called a &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Jack_arch&quot; rel=&quot;nofollow&quot;&gt;jack arch&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top and bottom are flat, but the joints are angled in such a way that the the parabolic &quot;path of compression&quot; is entirely contained within the layer of masonry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loading (e.g., the outward force at the ends) is higher than with a thinner parabolic arch or vault, simply because the extra material is additional dead load.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2016-08-03T16:07:11.283" LastActivityDate="2016-08-03T16:07:11.283" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10933" PostTypeId="1" CreationDate="2016-08-03T11:55:18.943" Score="1" ViewCount="393" Body="&lt;p&gt;I measured the vibration (jerk) on the left and right side of different motorcycles. I tested different engine types: Boxer engine, 3 cylinder, and V4. I measured the vibration on the same route, while driving for approx 30 minutes. I take the average jerk for every measurement. For every measurement the jerk on the right side of the motorcycle is higher than on the left side. I switched my sensors, since I thought this would be due to my sensors, but the results are the same. The difference is about 5-10%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd thought that the vibration would be more or less symmetrical, but it seems to be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you think of a reason why the vibration differs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to comments and answers here is some additional information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I ran the experiment with different participants on different motorcycles&lt;/li&gt;&#xA;&lt;li&gt;It seems that the difference of jerk occurred while standing, as well as driving (I cannot measure new data right now - but evaluated my existing datasets)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://goo.gl/PiyucD&quot; rel=&quot;nofollow&quot;&gt;This is the route I used&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;The motorcycles used in this experiment are regular ones you can buy at any retailer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7500" LastEditorUserId="7500" LastEditDate="2016-08-05T08:17:33.870" LastActivityDate="2016-08-05T08:17:33.870" Title="Motorcycle Engine Vibration - Difference between left and right side" Tags="|mechanical-engineering|automotive-engineering|motors|measurements|vibration|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="10934" PostTypeId="2" ParentId="10922" CreationDate="2016-08-03T15:25:08.160" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IEWUF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IEWUF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the geometry $a_1=45 {}^{\circ}$ and $a_2=\tan ^{-1}\left(\frac{3}{4}\right)$ are known and $c_1$ and $c_2$ need to be determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can be obtained by solving the following two equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 750 \cos \left(a_1\right)-c_2 \cos \left(a_2\right)=c_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ c_2 \sin \left(a_2\right)+750 \sin \left(a_1\right)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give $$ c_1=875 \sqrt{2}=1237.44$$ and $$ c_2=-625 \sqrt{2}=-883.883$$&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-08-03T15:25:08.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10935" PostTypeId="1" CreationDate="2016-08-03T15:41:11.210" Score="2" ViewCount="357" Body="&lt;p&gt;&lt;strong&gt;the problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm prototyping for a musical instrument based on sliding sticks (see &lt;a href=&quot;https://vimeo.com/170450482&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://vimeo.com/170450482&lt;/a&gt;) but I'm struggling to make a robust, yet smooth clamping mechanism for the sticks. The challenge is to hold the (hard wood) sticks in such a way that they won't fall to the ground, yet are as easy and smooth as possible to move by hand. (And of course it needs to be cheap and easy to produce.) I have thought of and tried a number of solutions, but up until now none seem sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;insufficient solution one&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution used in &lt;a href=&quot;https://youtu.be/vHZtDkUsfNo?t=25s&quot; rel=&quot;nofollow noreferrer&quot;&gt;the first functional prototype&lt;/a&gt; uses flat springs on each side of the board. While it keeps the stick in place, the springs are not easily adjustable, making the stick move not that smooth. Also they were  relatively difficult to manufacture consistently.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/1IJce.jpg&quot; alt=&quot;solution 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;insufficient solution two&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to find a more adjustable solution, we make clamps with springs for&#xA;the second functional prototype. While this allowed for good adjustablilty, this solution had too much wiggle, still didn't &lt;em&gt;feel&lt;/em&gt; smooth and was even more intensive to manufacture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8Kb8.jpg&quot; alt=&quot;solution 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;insufficient solution three&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing that things needed to get more smooth, we tried to use felt to slide the stick against. This worked yet it proved difficult to make the hole exactly the right size for the sticks plus felt. Above that there were concerns that the felt would get more and more compact over time, as such losing their clamping force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fOhpV.jpg&quot; alt=&quot;solution 3&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;possible solutions 4?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal solution seems to be in the image below; to use felt, but being able to adjust it's position. Yet I have not been able to find an adjustable mechanism which stays in place and keeps consistent over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6kBwo.jpg&quot; alt=&quot;solution 4&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there things I am missing? I am sure there are similar mechanisms out there in the world, but I haven't been able to find anything .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So my question is:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What would be a better mechanism for this problem?&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="7504" LastActivityDate="2016-08-03T15:41:11.210" Title="Designing a clamping mechanism for linear sliding stick" Tags="|springs|bearings|mechanisms|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10936" PostTypeId="2" ParentId="10933" CreationDate="2016-08-03T15:55:17.680" Score="1" Body="&lt;p&gt;There are quite a few variables that you seemingly haven't fully considered when running this experiment. It was a good idea to change the sensors to test any discrepancies there, but you also should consider your route and your riding/bio-mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if you ride the same route every time, but are constantly turning left, one side is going to jerk more than the other due to the weight distribution. Or if you naturally lean more to one side than the other while riding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would keep working to disprove this hypothesis, instead of thinking of reasons why it exists, and when you get more supporting evidence, go back to the drawing board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a good test would be 5 minutes (not including acceleration) on level, paved road. Bring some friends/whoever along and have everyone do this ride a couple times. (A convenience sample isn't always a good representation of the actual 'population' but at least it'll remove some of the questions regarding your riding style.)&lt;/p&gt;&#xA;" OwnerUserId="7473" LastActivityDate="2016-08-03T15:55:17.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10938" PostTypeId="1" AcceptedAnswerId="10940" CreationDate="2016-08-03T16:42:47.380" Score="2" ViewCount="23948" Body="&lt;p&gt;I understand that for a rectangular c-s the shear stress distribution is parabolic and the max shear stress occurs at the neutral axis and has a value of 1.5V/A. Where V is the 'applied shear force' and A is the cross-sectional area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this in turn then means that the shear force at this point is equal to 1.5V ( 1.5 times larger than the applied shear force) - which seems a tad strange physically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this due to the fact the the average shear force (average shear stress x cs area) is equal to the applied shear force? This is the only way it makes sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fpDrt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fpDrt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7476" LastActivityDate="2016-08-03T18:10:44.263" Title="Maximum Shear stress in Beams" Tags="|mechanical-engineering|structural-engineering|statics|stresses|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10939" PostTypeId="2" ParentId="10938" CreationDate="2016-08-03T17:12:17.867" Score="1" Body="&lt;p&gt;V/A represents the average shear stress, that is the total shear force resisted by the whole section, which has an area of A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we understand, if the stress distribution is uniform, the maximum shear stress will be equal to the average stress.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we have a parabolic distribution of shear stress, some areas will be stressed less and others more than the average stress V/A, as shown in the figure posted.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the expression tau(max)=(3/2)(V/A) shows is that the worst case of the stress (not force) is 50% higher than the average.&lt;/p&gt;&#xA;" OwnerUserId="5339" LastActivityDate="2016-08-03T17:12:17.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10940" PostTypeId="2" ParentId="10938" CreationDate="2016-08-03T17:13:23.637" Score="6" Body="&lt;p&gt;You've got your terms confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum shear &lt;strong&gt;stress&lt;/strong&gt; at the midpoint is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau_{max} = 1.5\frac{V}{A} = 1.5\overline\tau$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\dfrac{V}{A}=\overline\tau$, which is the average shear &lt;strong&gt;stress&lt;/strong&gt; along the entire section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the only viable comparison to be made, stress to stress. And having a maximum stress greater than the average stress is totally reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your doubt, however, is that &quot;the shear &lt;strong&gt;force&lt;/strong&gt; at this point is equal to $1.5V$&quot;. That is not the case. There is no shear force at any point in the section. There is only a shear stress. The entirety of the shear stress must then be integrated over the area to obtain the shear force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be thinking &quot;stress is just equal to force divided by area, so can't I just do&#xA;$$\begin{align}&#xA;\tau &amp;amp;= \frac{V}{A} \\&#xA;\therefore \tau_{max} &amp;amp;= \frac{V_{max}}{A} \\&#xA;\tau_{max} &amp;amp;= 1.5\frac{V}{A} \\&#xA;\frac{V_{max}}{A} &amp;amp;= 1.5\frac{V}{A} \\&#xA;V_{max} &amp;amp;= 1.5V&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and prove that the shear force at the midpoint is greater than the applied shear force?&quot; But I already beat you to it. After all, as I mentioned at the start, $\dfrac{V}{A}$ gives you the &lt;strong&gt;average&lt;/strong&gt; stress along the section. So $\dfrac{V_{max}}{A}$ is equivalent to the following stress profile, which clearly isn't the one you're expecting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Iu6b3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iu6b3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-08-03T17:13:23.637" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10941" PostTypeId="1" AcceptedAnswerId="10947" CreationDate="2016-08-01T17:03:23.443" Score="3" ViewCount="2003" Body="&lt;p&gt;first post so bear with me please!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Studying Mech Eng and in all notes provided and I have come across online the stress/strain distribution is shown as I have sketched belo (see pic). Surely this is only a representation of the right hand side of the beam (for hogging as shown) as the left would be as I have sketched below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in assuming this? Why do they always just show the first distribution? Why only show the 2 triangles and not the full square? What is the distribution in the middle of the beam (assuming that is where a single point load is applied)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this is clear enough to understand!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for you help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/40n9V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/40n9V.jpg&quot; alt=&quot;bending stress/strain&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Additionally, I feel this is where my confusion orginated from:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/NQ6dY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NQ6dY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now understand the stress distribution is constant throughout the length of the beam (although unsure as to why it's shown in that particular direction). But can someone please explain why the above image then makes sense. Is this only valid on the right hand side?&lt;/p&gt;&#xA;" OwnerUserId="7476" OwnerDisplayName="massey95" LastEditorUserId="1832" LastEditDate="2016-08-03T20:26:10.823" LastActivityDate="2016-09-09T17:46:43.620" Title="Beams in Bending: Bending stress / strain distrution" Tags="|structural-engineering|structural-analysis|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10942" PostTypeId="2" ParentId="10941" CreationDate="2016-08-01T19:15:50.917" Score="3" Body="&lt;p&gt;You seem to be confusing the &lt;em&gt;stress&lt;/em&gt; distribution in the beam with the &lt;em&gt;forces&lt;/em&gt; on either side of a cut in the beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axial stress component is positive (tension) on the outside of the curve and negative (compression) on the inside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you cut the beam, the stress creates equal and opposite forces on the two cut faces. This is the same basic idea as tension in a string, which is equivalent to equal and opposite forces acting on the ends of the string. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing two separate diagrams for the forces doesn't add any more information to the one diagram for the stress. For example you could cut the beam at &lt;em&gt;any&lt;/em&gt; angle, and the same find the normal and shear forces on the cut surface from the stress distribution (using Mohr's circle, for example). The single diagram showing the stress distribution contains &lt;em&gt;all&lt;/em&gt; the information about the state of the structure.&lt;/p&gt;&#xA;" OwnerUserId="683" OwnerDisplayName="alephzero" LastActivityDate="2016-08-01T19:15:50.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10943" PostTypeId="1" CreationDate="2016-08-03T17:42:35.820" Score="-1" ViewCount="173" Body="&lt;p&gt;I just read the 2015 excellent paper , &quot;&lt;a href=&quot;https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-23-19-24308&amp;amp;id=326598&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tunable microwave photonic notch filter based on sliced broadband optical source&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to downconvert a microwave photonic notch filter to a visible light photonic notch filter? What is the tuning speed of this microwave photonic filter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT: August 4 2016 9:26 P.M Frank . The tuning speed of this analog microwave photonic filter is limited primarily by the response time of the balanced photodiode. For example, the  Newport Nanosecond Photodetector, Silicon, 350-1000 nm, 0.8 mm Diameter, 8-32 / M4 Model: 1621 with URL, [&lt;a href=&quot;https://www.newport.com/p/1621]&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.newport.com/p/1621]&lt;/a&gt;, has a rise time of 1 nanosecond.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT: August 6 2016 6:00 P.M. Frank . Today's laser pointers are almost exclusively continuous wave)(CW) laser pointers . Continuous wave laser pointers do not need chirp compensation. Currently, pulsed lasers are used for high speed telecommunications because narrow pulse widths are needed. In the near future, it is estimated that CW laser pointers will be replaced by pulsed laser pointers which require chirp compensation similar to fiber Bragg grating (FBG) with variable delay line(VDL). Chirping is defined as a waveform with a continuously increasing or decreasing frequency spectra where frequency equals speed of light divided by wavelength.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT August 7 2016 7:00 P.M. Frank There are two types of balanced photodetectors. Free space and Fiber, which is faster and smaller. The ThorLabs PBD430A Fiber model takes the difference between two signals , subtracting them across the spectrum for visible or infrared light . The PDB430A has two different frequency ranges 0 Hz to 400 MHz and 30Khz to 350 Mhz. ThorLabs told me last Friday that their PDB430A has a switching speed fast enought to respond to laser pointer strikes within 50 millisecond , thereby protecting the pilots from permanent damage.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT August 7 2016 7:00 P.M switching frequency = speed of light (3.0 x 10^8 meters/second divided by wavelength 532 nanometers = 500 Terahertz switching time which is so fast one cannot see fluctuations of the electric field]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT August 8 2016 2:30 A.M.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper is very important because we wish to protect pilot's vision from increasingly sophisticated laser pointers on the ground 150 meters away aimed at the Boeing 747-787 cockpit windows. If we can detect the presence and wavelength  of a coherent laser light in an incoherent background within 10 milliseconds of arrival , could I then activate a tunable band-reject  filter based on this paper  centered at the visible red, green or blue wavelengths?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, I read with great interest in this paper that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The frequency omega2 can be tuned by the MZI without changing the filter shape according to Eq. (9). Another slicing way is to directly use the FPOP to shape the spectrum of the BOS. In this way, the interferometer is not needed in the approach, but a FPOP with high resolution is required.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In December 2009, Yuuki Watanabe and Toshiki Itagaki wrote in SPIE Journal of Biomedical Optics Volume 14 Issue6 JBO Letters 2009, &quot;Real-time display on Fourier domain optical coherence tomography(OCT) system using a graphical processing unit&quot; that a display rate of 27.9 frames per second for processed images(2048 FFT size X 1000 lateral A-scans) is achieved in our OCT system using a line scan CCD camera operated at 27.9 kHz using the C and C++ programming language&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given today's embedded microprocessors, could we boost the FDOP display rate of 27.9 frames per second to 500 frames per second to meet Boeing and Airbus passenger plane pilot's expectations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore , I read with great interest in the paper that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In this paper, two filters are necessary. One filter is used to filter the BOS; the other one is used to split the single laser from the BOS. In order to quickly verify the principle, we use two waveshapers in the experiment. It results in a high system cost. Moreover, a way to reduce the cost is to use fiber Bragg gratings (FBG) with optical circulators to work as the filters. Take the second filter as an example, we can send the light into the FBG via an optical circulator. When the Bragg wavelength is just equal to the laser wavelength, the reflected light only includes the laser light and the transmitted light only contains the BOS part.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I estimate the performance loss associated with replacing the waveshaper with fiber Bragg gratings and optical circulator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This paper is a very nice example of out-of-the box engineering because it has no moving elements or electro-optic Pockels or Kerr effect. Rather, the authors subtract a single frequency source from an allpass optical filter producing a band-reject filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6LxOT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6LxOT.jpg&quot; alt=&quot;Fig. 1 The schematic diagram. SSB: Single-side-band modulator.&quot;&gt;&lt;/a&gt;&#xA;  The schematic diagram of the proposed microwave photonic filter (MPF) is illustrated in the above figure. The single-side-band modulator is a fancy name for a microwave frequency mixer. The optical power and optical angle frequency of the laser source is     and    , respectively. The spectrum-sliced broadband optical source (BOS) is first confined within the optical angle frequency range between  1   and  2   ( 1 &amp;lt; 2  ) via the Fourier domain optical processor (FDOP), and then sliced via a Mache-Zehnder interferometer (MZI) which is constructed by two 50/50 optical couplers and a variable delay line (VDL). The FDOP works as a programmable optical filter. The single-frequency light and the broadband light are coupled together and then single-side-band modulated to carry the radio frequency (RF) signal. The modulated light is fed into a dispersion fiber, and then launched into a wavelength division multiplexer (WDM). The WDM splits the light into two parts: Port B only includes the laser part; Port C only includes the BOS part. A balanced photodetector (BPD) is employed to detect the two parts, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me if I have stated anything incorrect. The use of equations is appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vC8fl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vC8fl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="2651" LastEditDate="2016-08-09T06:04:26.047" LastActivityDate="2016-08-09T06:04:26.047" Title="Is it possible to adapt the design and principle of operation for a tunable microwave photonic notch filter to a visible light photonic notch filter?" Tags="|materials|optics|lasers|photonics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10944" PostTypeId="2" ParentId="10890" CreationDate="2016-08-03T17:47:01.937" Score="0" Body="&lt;p&gt;I think it is imperative to first determine if the cross-beam to main-beam connection is a moment-(resisting-)connection.  I am not sure if the cross beams are (or need to be) moment-connected to the main beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the given case, beams 6,11,16,... 8,13,18... appear to be cross beams that carry local floor loads.  It is not generally necessary to make them moment connected.  Usually they are welded to a pair of angles, which will fit to some pre-drilled holes on the web of the main beams for ease of erection.  Individual beams can be transported in different trucks.  If this is the case, the connection is usually considered a pinned connection.&#xA;The following is what they look like: &lt;a href=&quot;http://www.engr.mun.ca/~adluri/courses/steel/ppt%20files1/Topic%20-Connections%20-typical%20joints.pdf&quot; rel=&quot;nofollow&quot;&gt;Typical Steel Connections&lt;/a&gt;&#xA;(scroll to about 1/3 down the page and look for beam-to-beam connections)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are &quot;welded&quot; to the main beams and are expected to transmit moments, they would likely look like these, and obviously cost more in fabrication: &lt;a href=&quot;http://www.steelconstruction.info/Moment_resisting_connections#Welded_beam-to-column_connections&quot; rel=&quot;nofollow&quot;&gt;Moment resisting connections&lt;/a&gt;&#xA;(scroll to the last fifth of the page to see beam-to-beam connections).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a complication with perpendicular beam-to-beam moment connections in the sense that the beam-depths are usually different. In this case, the main beams are likely to be deeper than the cross beams.  So again, moment connections are not usually called for.  In any event, if beam 6 is moment connected to beams 1 and 27, the latter will provide very little torsional resistance, so I doubt if it is worthwhile to have them moment-connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more reason that the cross beams are not moment-connected to the main ones would be if all these beams are welded in the shop, it would become a single piece that may not be transportable by trucks to the building site using highways.  On the other hand, welding in the field would create extra quality assurance expenses, x-ray inspection, etc.&lt;/p&gt;&#xA;" OwnerUserId="5339" LastEditorUserId="7083" LastEditDate="2016-08-06T20:56:58.790" LastActivityDate="2016-08-06T20:56:58.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10945" PostTypeId="2" ParentId="10938" CreationDate="2016-08-03T18:10:44.263" Score="1" Body="&lt;p&gt;If we need to calculate how much shear a rectangular beam can take this is the formula.&lt;br&gt;&#xA;V= 2/3 [A x tau(allowable)].  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We find the allowable tau in charts readily available but for lumber under normal humidity it is around 80-90 psi.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-08-03T18:10:44.263" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10946" PostTypeId="1" CreationDate="2016-08-03T18:27:38.293" Score="2" ViewCount="80" Body="&lt;p&gt;Is solving stochastic HJB a practical method for stochastic optimal control in engineering field or it is just a method used in finance?&lt;/p&gt;&#xA;" OwnerUserId="7506" LastEditorUserId="77" LastEditDate="2016-08-03T20:26:34.150" LastActivityDate="2016-08-04T10:58:32.263" Title="Is Hamilton Jacobi Bellman optimization used in engineering control?" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="0" CommentCount="4" ClosedDate="2016-12-10T01:42:45.983" ContentLicense="CC BY-SA 3.0" />
  <row Id="10947" PostTypeId="2" ParentId="10941" CreationDate="2016-08-03T20:24:55.747" Score="3" Body="&lt;p&gt;Before discussing any such matters, we must define our internal force sign conventions. It is common to define them thusly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TfxCv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TfxCv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, tension is positive, compression is negative. Shear forces are positive upwards when on the left side of the cut, but downwards on the right side of the cut (a good way to remember is that positive shear rotates the beam clockwise). Bending moment is positive if it generates tension on the bottom of the section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When an external bending moment is applied to a beam, it resists deformations by generating an internal stress state. This stress state is such that it can be described by an internal bending moment which is equal and opposite to the external bending moment, cancelling it out and allowing the beam to remain at rest in this new deformed configuration. It also generates zero total axial force (under pure bending).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be trivially seen from the deflection, your beam is under negative bending moment. This means the top fibers of the beam are elongated and the bottom fibers are shortened, meaning the internal stress state goes from tension at the top to compression at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we cut the beam at midspan, we can choose to look at either the left or right side of the cut, which will give us:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BUYDP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BUYDP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that both sides must be equivalent, since they represent the exact same things in the exact same spot, only looking at different sides of it. And indeed, they are exactly the same, once we remember the sign convention.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both sides show tension (positive) in the top, compression (negative) in the bottom. Good.&lt;/li&gt;&#xA;&lt;li&gt;On the left side of the cut, the tension on top and compression in the bottom would generate a clockwise rotation, which is negative. On the right side of the cut, they would generate a counter-clockwise rotation, which is also negative. Good.&lt;/li&gt;&#xA;&lt;li&gt;I didn't represent shear in the image because it isn't relevant to the current question, but we'd expect the forces to either side to be equal and opposite.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, there is no benefit to displaying the stress distribution on both sides. If you have one, it is all you need and the other is entirely redundant. Now, this also means that one is not more correct than the other. So, if you're going to write a diagram, should you do it with positive axial forces going to right (as in the right side of the cut and in the example you gave in your edit) or to the left? It's entirely up to you. It is customary to have positive axial forces going to the right because, well, usually positive values go to the right.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-08-03T20:24:55.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10949" PostTypeId="1" CreationDate="2016-08-04T02:50:28.130" Score="0" ViewCount="3839" Body="&lt;p&gt;Gearboxes usually have a reducing ratio like 200:1, which means that they take a high-speed input and convert it into a high-torque output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would they work if I connected them in reverse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an &lt;em&gt;approximate&lt;/em&gt; ratio at which a human could no longer power one in this configuration (without the use of a large lever)?&lt;/p&gt;&#xA;" OwnerUserId="2801" LastActivityDate="2016-08-04T21:55:19.630" Title="Using a gearbox in reverse" Tags="|gears|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10950" PostTypeId="2" ParentId="10866" CreationDate="2016-08-04T04:09:00.873" Score="4" Body="&lt;p&gt;Use a stiff element, &lt;a href=&quot;https://engineering.stackexchange.com/a/10867/3353&quot;&gt;as explained by AndyT&lt;/a&gt;, to model the offset might create numerical issues for the finite element model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some alternatives:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Alternative 1:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Build &lt;a href=&quot;https://engineering.stackexchange.com/questions/10866/how-can-i-model-beam-column-connection-that-doesnt-connect-at-centerline/10867?noredirect=1#comment19260_10867&quot;&gt;on the comment by alephzero&lt;/a&gt;, one can model the offset as Multi-Point Constraint (MPC). An explanation is provided &lt;a href=&quot;http://what-when-how.com/the-finite-element-method/modelling-techniques-finite-element-method-part-3/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The basic idea of using MPC is to create a set of MPC equations that&#xA;  gives the relation between the DOFs of the two separated nodes. It&#xA;  assumes that the two corner nodes are connected by a rigid body. MPC&#xA;  equations are then derived using the simple kinematic relations of the&#xA;  DOFs on the rigid body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rJmIs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rJmIs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cjhBJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cjhBJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Q2Aay.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q2Aay.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/U88rT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U88rT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Alternative 2:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Another similar approach is to use the &lt;a href=&quot;http://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.Ch20.d/IFEM.Ch20.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Direct Fabrication of offset beam stiffness ( section 20.4.3)&lt;/a&gt;, where the stiffness matrix is explicitly modified to incorporate the offset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GMDTd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GMDTd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lOkib.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lOkib.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-08-04T04:09:00.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10951" PostTypeId="2" ParentId="7336" CreationDate="2016-08-04T04:59:53.067" Score="0" Body="&lt;p&gt;&quot;can I operate the pump without water in its casing?&quot;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;YES, you can use a self-oiling sliding vane pump. They &lt;em&gt;will&lt;/em&gt; pull a vacuum, allowing water to fill the centrifugal pump case. We use it and can prime from water source up to 12' below the pump.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All the usual variables apply regarding pump sizes vs. net suction head, etc.&lt;/p&gt;&#xA;" OwnerUserId="7513" LastActivityDate="2016-08-04T04:59:53.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10952" PostTypeId="1" CreationDate="2016-08-04T07:48:53.967" Score="2" ViewCount="3004" Body="&lt;p&gt;How can I calculate the torque that is needed to keep the steering wheel at the same steering angle during turning at given speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A car with &lt;a href=&quot;https://en.wikipedia.org/wiki/Ackermann_steering_geometry&quot; rel=&quot;nofollow&quot;&gt;Ackermann steering geometry&lt;/a&gt; weights 1000 kg. It is going though a curve - the steering angle is constant 5 degrees. Car's velocity is 10 m/s. Steering ratio is 15:1 (15 degrees on steering wheel means 1 degree of actual steering angle). What torque do I need to apply to the steering wheel to hold those constant 5 degrees on steering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know the torque because I want to robotize steering of a passenger vehicle and since torque will be applied by human driver's hands I need to supply such torque by an electric motor (probably a stepper with a required &quot;holding torque&quot; specification).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the torque could probably somehow measured on the steering wheel (with a torque wrench) with power steering off but it would be nice to be able to at least roughly calculate it. I actually need to know all torques for all car's normal operation scenarios (turning when standing still, when turning in speed, when going through a curve etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may have forgotten few parameters for the example. Say car's width is 1.5m (tire to tire) and distance between front and read axle is 2m.&lt;/p&gt;&#xA;" OwnerUserId="414" LastEditorUserId="414" LastEditDate="2016-08-04T12:36:34.790" LastActivityDate="2018-06-10T11:51:54.690" Title="Steering wheel torque when car is turning" Tags="|torque|robotics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10953" PostTypeId="2" ParentId="10949" CreationDate="2016-08-04T09:14:30.650" Score="1" Body="&lt;p&gt;Generally yes, if you use the original output shaft as the new input shaft then you would be achieving the opposite (i.e. converting high-torque input to high-speed output).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In answer to your second question, it depends on the value of torque. Theoretically if a lever is long enough you can &quot;turn&quot; anything. Say you limited the lever length to 1m and the fact that a human can probably utilise their whole mass with gravity plus extra effort (like jumping) to produce around 1000N of force, then they could produce 1 kN-m of torque. I'm just speculating without extra detail.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2016-08-04T21:55:19.630" LastActivityDate="2016-08-04T21:55:19.630" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10955" PostTypeId="2" ParentId="10949" CreationDate="2016-08-04T11:59:39.090" Score="3" Body="&lt;p&gt;It depends on the type of gearing. Gearboxes for winches and the like typically are designed (for example worm gears in the drive chain) so that they can not possibly be turned in reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand a manual car/truck gearbox gets pushed in reverse every time you downshift&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2016-08-04T11:59:39.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10956" PostTypeId="1" CreationDate="2016-08-04T14:35:29.840" Score="2" ViewCount="73" Body="&lt;p&gt;I would like to model a plane trajectory, here are my assumptions and conclusions so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;we denote the plane's speed with $v(t)$, its trajectory $x(t)$, the total trajectory time will be $\tau$&lt;/li&gt;&#xA;&lt;li&gt;we have start and end times of flights as well as distance, so we have $\tau$ and $\int_0^\tau v(t)dt$&lt;/li&gt;&#xA;&lt;li&gt;the plane has an acceleration phase, let's call it $p_1$&lt;/li&gt;&#xA;&lt;li&gt;the plane has a plateau phase, $p_2$, where it reaches its cruise speed, $v_c$&lt;/li&gt;&#xA;&lt;li&gt;the plane has a deceleration phase, $p_3$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The plane speed over time is&#xA;$v(t) = \begin{cases}&#xA;v_1(t) &amp;amp; t \in p_1 \\&#xA;v_c  &amp;amp; t \in p_2\\&#xA;v_2(t) &amp;amp; t \in p_3 \\&#xA;0 &amp;amp; \text{else}&#xA;\end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $p_1 = [0, t_1], p_2 = [t_1, t_2]$ and $p_3 = [t_2, \tau]$, we assume&#xA;$t_1 = \tau - t_2$ (essentially, same amount of time accelerate as to decelerate) and we rewrite $p_1 = [0, t_1], p_2 = [t_1, \tau- t_1]$ and $p_3 = [\tau-t_1, \tau]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We 'll now further assume $v_2(t) = v_1(t - \tau)$ (the deceleration shape is the inverted acceleration shape).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at this point we are left with determining $v_1(t)$. The model has $m$ parameters for $v_1$ and $t_1$, $v_c$ (note that $t_1$ and $v_c$ can be included as a parameter in $v_1$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are assumptions realistic, make sense, anything missing?&lt;/li&gt;&#xA;&lt;li&gt;How would you go about modelling $v_1(t)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Possible answer 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make the assumption that $v_1(t)$ is linear:&#xA;$v_1(t) = \frac{v_c}{t_1}t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall model has two parameters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible answer 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go with a sigmoid-like function for $v_1$ but with a bounded input, I thought $\sin$ with the right phase would be good for this since it also has derivatives of zero at $\frac{dv_1(0)}{dt} = \frac{dv_1(t_1)}{dt} = 0$, this leads to:&#xA;$v_1(t) = v_c\frac{1-\cos{(\omega t)}}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega = 2\pi/t_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again two parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I don't want to have $v_c$ as a parameter/constant, I thought of introducing $\alpha = \frac{\omega}{v_c}$ thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_1(t) = \frac{\omega}{2\alpha}(1-\cos{(\omega t)})$&lt;/p&gt;&#xA;" OwnerUserId="7517" LastEditorUserId="7083" LastEditDate="2016-08-05T21:52:03.610" LastActivityDate="2016-08-05T21:52:03.610" Title="How can I model a plane's trajectory in a simple way?" Tags="|modeling|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10957" PostTypeId="1" AcceptedAnswerId="10958" CreationDate="2016-08-04T16:01:08.197" Score="4" ViewCount="25744" Body="&lt;p&gt;I'm building a set of monkey bars for my kids and want to use aluminum tubing (6061 T6) for the rungs. The tubes will be 3 feet long and 1 inch outside diameter. Supported on both ends with a circular metal pipe bracket. I have a choice of various wall thicknesses, all the way up to a full solid round bar. How do I calculate the minimum wall thickness I need to safely support (without any permanent deformation) something like 200 pounds at the middle point of the rung?&lt;/p&gt;&#xA;" OwnerUserId="7519" LastActivityDate="2016-08-04T19:12:18.360" Title="How much weight can an aluminum tube beam support?" Tags="|beam|statics|aluminum|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10958" PostTypeId="2" ParentId="10957" CreationDate="2016-08-04T19:12:18.360" Score="6" Body="&lt;p&gt;We use this equation for a simply supported beam loaded at its center with 200lbs.  M= P x L/4 =200 x 3 /4 = 150 lbs.ft&lt;br&gt;&#xA;for calculating the tube stress we convert this to 150 x 12 = 1800 lbs.inch  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a hollow round tube we calculate section module which is  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Z = 0.78 (R_o^4 -R_i^4)/R_o$ assuming you have 16 gauge tube with 0.065 inch walls, Z=0.1838   .    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have $sigma = M/Z so 42000 (allowable aluminum stress) = M/Z $  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = 42000 x 0,1838/4= 1929 lbs$&lt;br&gt;&#xA;Usually in applications like this which is dynamic loading they use a safety factor of 3 so you can expect this tubing to be able to support approximately&lt;br&gt;&#xA; $ 600 lbs$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then again there are many standards of aluminum. For this illustration I assume 2024-T3 which has hi strength, you need to find out what you have.&lt;br&gt;&#xA;This answer is just an illustration and you should check with the manufacturer about their tests on strength and ductility. Also connections and wall construction has to be checked.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-08-04T19:12:18.360" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10959" PostTypeId="1" CreationDate="2016-08-04T21:17:49.707" Score="3" ViewCount="151" Body="&lt;p&gt;&lt;strong&gt;The Basics:&lt;/strong&gt;&#xA;I am trying to come up with an equation describing the motion of a particle resting on a non-stationary surface, given the motion of the surface. Imagine setting a coin on a board and then sliding the board back and forth in an oscillating motion. At a certain rate the coin will start to slip, but will not be completely stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Known Variables:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Static Friction &lt;/li&gt;&#xA;&lt;li&gt;Dynamic Friction&lt;/li&gt;&#xA;&lt;li&gt;Motion of the Surface (i.e. the board, x(t), v(t), a(t))&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other Details:&lt;/strong&gt;&#xA;The motion of the board is non-linear and not a 'nice' function (it is a set of data points). I plan to determine the motion of the particle discretely using Scilab (poor man's MatLab). I am only considering 1D motion, forward and backward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background Info:&lt;/strong&gt; I'm a mechanical engineer, but it has been a while since I've used any significant amount of dynamics, calculus, or differential equations. I've tried to research this on my own for a long time, but it seems like I need to re-learn those subjects before even attempting this problem. I'm willing to work for it, but not quit my job and go back to school...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intent is to use this to model the performance of a machine that I'm working on so that I can optimize the design. I've tried writing an algorithm that looks at the acceleration between each data point and determine whether or not it is slipping. If it is not, I would add the motion traveled by the surface to the position of the particle. That works OK as a very rough estimate, but does not take into account the inertia of the particle or the existing velocity of the particle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a relatively simple way to do this?&lt;/p&gt;&#xA;" OwnerUserId="7520" LastActivityDate="2016-08-05T14:36:06.180" Title="Calculating the Motion of a Particle on a Moving Surface Due to the Friction Between Them" Tags="|mechanical-engineering|applied-mechanics|dynamics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="10960" PostTypeId="2" ParentId="10577" CreationDate="2016-08-05T05:23:00.397" Score="0" Body="&lt;p&gt;One way to accomplish evaporative roof cooling is to install a green roof (also called bio roof). Basically moss, grass or other plants growing on the roof. The plants and soil will retain the water until it evaporates, and also increase the surface area available for evaporation. Usually this is done on flat roofs, but it might work for such slightly slanted roof also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cooling effect of green roofs is widely recognized, though I couldn't find any actual measurements on how close to wet bulb temperature you would get:&#xA;&lt;a href=&quot;https://www.epa.gov/heat-islands/using-green-roofs-reduce-heat-islands&quot; rel=&quot;nofollow&quot;&gt;https://www.epa.gov/heat-islands/using-green-roofs-reduce-heat-islands&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-08-05T05:23:00.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10961" PostTypeId="2" ParentId="10943" CreationDate="2016-08-05T06:00:54.237" Score="-1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I discovered this morning that one can adapt the Tsinghua University&#xA;  professor's novel design and analysis microwave photonic notch filter&#xA;  to a visible light photonic notch filter by applying the excellent&#xA;  Princeton University paper titled &quot;High frequency modulation&#xA;  capabilities and quasi single-sideband emission from a quantum cascade&#xA;  laser&quot; with URL &#xA;  [&lt;a href=&quot;https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-22-19-23439&amp;amp;id=301440]&quot; rel=&quot;nofollow&quot;&gt;https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-22-19-23439&amp;amp;id=301440]&lt;/a&gt;.&#xA;  Here is an excerpt from that 2014 Princeton University Department of&#xA;  Electrical Engineering paper;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Abstract&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both intensity- (IM) and frequency-modulation (FM) behavior of a directly modulated quantum cascade laser (QCL) are measured from 300 Hz to 1.7 GHz. Quantitative measurements of tuning coefficients has been performed and the transition from thermal- to electronic-tuning is clearly observed. A very specific FM behavior of QCLs has been identified which allows for optical quasi single sideband (SSB) modulation through current injection and has not been observed in directly modulated semiconductor lasers before. This predestines QCLs in applications where SSB is required, such as telecommunication or high speed spectroscopy. The experimental procedure and theoretical modeling for data extraction is discussed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;© 2014 Optical Society of America&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Introduction&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since the first demonstration of the single-mode, room temperature and continuous wave operating quantum cascade laser (QCL) less than a decade ago [1], a number of applications were triggered that use their advantages as monolithic mid-infrared (mid-IR) semiconductor lasers. One of the most significant applications is the spectroscopic molecular sensing, which can be several orders of magnitude more sensitive than in the near infrared. These monolithic, tunable mid-IR lasers allow development of both compact and highly sensitive trace-gas detection systems. Tunable laser spectroscopy with semiconductor laser sources inherently relies on the wavelength tuning capability through variation of laser injection current. However, since this process is not arbitrarily fast this imposes some limitations on the applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were only a few reports in the literature on the experimental results of the (small-signal) modulation response of directly modulated QCLs. Examples include [2] (intensity modulation, IM) [3], (frequency modulation (FM) up to 100 kHz), and [4] (only THz QCLs). Significantly more extensive are reports on the FM response in diode lasers [5,6].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main differences between QCLs and conventional interband DFB diode lasers (e.g., those operating at telecom wavelengths) are expected from the thermal implications of significantly thicker active regions and the fundamentally different carrier dynamics in QCLs. The latter is due to the ultrashort intersubband carrier lifetimes, according to which QCLs can have relaxation frequencies in the 100 GHz range. This has been shown theoretically by Shore et al. [7,8]. Hence, it is very interesting to have experimental studies in the high frequency regime. Furthermore, the electronic tuning (at f &gt;&gt;10MHz) could be quantified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this paper the FM to IM ratio for a CW-DFB-QCL emitting at ~9 µm is presented for a wide range of modulation frequencies up to 1.7 GHz. The variation of the FM-IM ratio with bias current and its influence on the optical spectrum of the laser radiation is analyzed.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Optical frequency tuning mechanisms and emitted spectrum&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The tuning of optical frequency of a semiconductor laser through injection current is mainly caused by two physical effects [9,6]: firstly the Joule heating of the active region results in thermally induced changes of the refractive index, which causes frequency tuning of the emitted laser radiation, and secondly the refractive index variation from the electro-optic light-carrier interaction also affects the laser output frequency. Since heat conduction is an inert process, the low frequency behavior is mostly dominated by thermal tuning, whereas the electronic tuning is governed by the carrier dynamics and for diode lasers dominates in the MHz and GHz range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Frank's  EDIT August 9 2016 2:01A.M The tuning speed of this analog microwave photonic filter is limited primarily by the response time of the balanced photodiode. For example, the Newport Nanosecond Photodetector, Silicon, 350-1000 nm, 0.8 mm Diameter, 8-32 / M4 Model: 1621 with URL, [&lt;a href=&quot;https://www.newport.com/p/1621]&quot; rel=&quot;nofollow&quot;&gt;https://www.newport.com/p/1621]&lt;/a&gt;, has a rise time of 1 nanosecond.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if I have made any mistakes.&lt;/p&gt;&#xA;" OwnerUserId="2651" LastEditorUserId="2651" LastEditDate="2016-08-09T06:01:44.847" LastActivityDate="2016-08-09T06:01:44.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10962" PostTypeId="2" ParentId="10923" CreationDate="2016-08-05T07:28:31.977" Score="2" Body="&lt;p&gt;I think this is more a comment than an answer, but I have too small reputation to comment (poor me!). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summarizing, you are asking: how dynamic loads propagate inside a structure?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would underline that, in my perspective, telling that you are not concerned about vibrations, is a wrong starting point (fatigue crack derives exactly from them); for a vibration, you have a variable strain and consequently a variable stress and consequently you face fatigue problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For static cases, a very interesting technique is the so called &quot;photoelasticity&quot;, that correlate the stress to the &quot;pictorial&quot; view by means of the so called &quot;birefringence&quot; (I am not really an expert, do not ask me more, but it is really impressing seeing it!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For dynamic cases, situation becomes more and more complex and further  important with the development of composite materials, more prone to fatigue problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several attempts are numerical, and they fundamentally obtain operating loads from modal analysis, operational modal analysis and general in-situ tests. Then, this loads are traduced into a refined model for stress analyses. &#xA;Several techniques, depending on materials, boundary conditions and loads type, can be used for transforming them in stresses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, fatigue tests, in particular for standard &quot;metal&quot; airplanes start very soon with the project, starting from parts to the entire mockup model of the aircraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, there are several projects for deducing the state of stress during operating condition. Consider that, i.e., if you mount an extensometer in one part of you airplane, you measure the strain state and consequently the stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several attempt are made with FBG and OF in general for deriving stress state during operating conditions. &#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=_CdULw4-j_o&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=_CdULw4-j_o&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there are also technique for damage detection based on something similar to the one you suggest, but they are still at they early stages: they seem to be promising, but till now they worked with very simple geometries and boundary conditions (nothing similar to your aircraft brackets...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I gave you some interesting references...&lt;/p&gt;&#xA;" OwnerUserId="7505" LastActivityDate="2016-08-05T07:28:31.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10963" PostTypeId="1" AcceptedAnswerId="10964" CreationDate="2016-08-05T08:09:19.887" Score="1" ViewCount="3413" Body="&lt;p&gt;I have a continuous beam across a center column (not sure whether I've draw it correctly or not)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yvCEb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yvCEb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/q/9066/3353&quot;&gt;The Displacement and force diagram is as shown:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3ic7e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3ic7e.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first diagram is the bending moment, second the axial force, the third is the displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the boundary condition at $R_A$, $R_B$ and $R_C$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I can infer, it seems to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w(0)=w(L)=w(2L)=0$ (corresponds to the deflection at the three support)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M(0)=M(2L)=0$, or $\frac{d^2w(0)}{dx^2}=\frac{d^2w(2L)}{dx^2}=0$ (corresponds to the moment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I suspect that I am still missing some boundary conditions in order to derive the complete displacement/force diagram for the continuous beam. Are there any boundary conditions that I've missed?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-08-05T12:11:55.573" Title="The boundary condition(s) for a continuous beam" Tags="|structural-engineering|structural-analysis|beam|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10964" PostTypeId="2" ParentId="10963" CreationDate="2016-08-05T09:00:56.407" Score="3" Body="&lt;p&gt;Well you are missing the compatibility of slopes at the middle support:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dw(L^-)}{dx} = \frac{dw(L^+)}{dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of symmetric geometry &amp;amp; loading, the slope of the beam at the middle support will be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the bending moment has no derivative at x=L, you will need to derive the deflections of the two halves separately and 'join' them with compatibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; derivation of beam deflection formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow&quot;&gt;Euler-Bernoulli beam equation&lt;/a&gt; (assuming constant EI), and taking x from the outer supports towards the center:&#xA;$$ q = EI \frac{dw^4}{dx^4} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating four times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ EI \frac{d^3 w}{dx^3} = q x + A $$&#xA;$$ EI \frac{d^2 w}{dx^2} = q \frac{x^2}{2} + Ax +B $$&#xA;$$ EI \frac{d w}{dx} = q \frac{x^3}{6} + A\frac{x^2}{2} + Bx + C $$&#xA;$$ EI w = q \frac{x^4}{24} + A\frac{x^3}{6} + B\frac{x^2}{2} +Cx + D$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noticing the problem is symmetric the boundary conditions are:&#xA;$$ w(0)=w(L) = 0$$&#xA;$$ \frac{d w(L)}{dx} = 0 $$&#xA;$$ \frac{d^2 w(0)}{dx^2} = 0 $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore we can immediately see that:&#xA;$ B = D = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now have two equations with two unknowns (A,C). Solving we find:&#xA;$$ A = -\frac{3 L q}{8}$$&#xA;$$ C = \frac{L^3 q}{48}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now substitue all the constants back into the equation for w. Simplifying results in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ w = \frac{q x}{48EI}  (L-x)^2 (L+2 x) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is identical to the result referenced &lt;a href=&quot;http://www.engineersedge.com/beam_bending/beam_bending22.htm&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; (note their coordinate system has x=0 at the centre). Also notice how this is exactly the same result as a &lt;a href=&quot;http://www.engineersedge.com/beam_bending/beam_bending11.htm&quot; rel=&quot;nofollow&quot;&gt;propped cantilever&lt;/a&gt;. This is due to symmetry, meaning that the beam slope at the center is zero (which is the same boundary condition as a cantilever support).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also substitute into the bending moment equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M = EI \frac{d^2 w}{dx^2} = \frac{1}{8} q x (4 x-3 L) $$&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2016-08-05T11:16:36.230" LastActivityDate="2016-08-05T11:16:36.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10965" PostTypeId="1" CreationDate="2016-08-05T10:13:35.430" Score="3" ViewCount="30" Body="&lt;p&gt;I have a zinc lock box that looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UCurl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UCurl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where there's a top and a bottom part. Notice that there're two interesting features (circled in red):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ptfyE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ptfyE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After cutting it up slightly:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rAMqz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rAMqz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/k0llA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k0llA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The section view (of the right feature) looks like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C9adq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C9adq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the part circled in blue previously looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qvKfv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qvKfv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are those features results of some staking operation? If so, is it common for staking to produce such clean, squarish dents (red arrow) even on the lower part? My guess is that these might be designed for holding the two parts together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've now cut another piece off to hopefully aid better understanding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pa2Kn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pa2Kn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l7liO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l7liO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The features highlighted by red arrows don't seem to be regular shapes. The corners of the deformations in the upper component seem to have very small radii, but the features in the lower component that &quot;bite&quot; into the upper component look quite rough. Could this mean that the dents in the upper part were part of the cast and not deformations created by staking into the lower part?&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="4302" LastEditDate="2016-08-06T05:06:49.363" LastActivityDate="2016-08-06T05:06:49.363" Title="Joining zinc components" Tags="|casting|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10966" PostTypeId="1" CreationDate="2016-08-05T10:44:23.700" Score="2" ViewCount="223" Body="&lt;p&gt;It is quite frustrating to have multiple adapter for connecting to desktop. For example, if I want to connect my phone to desktop I need Bluetooth/wifi; why can't I just just use the Mobile's main telecommunication. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was the case with mobile charger wherein we had to find the correct charger to charge the mobile, now no more it is the case; we can do it with MicroSD charger which is now generally available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can't we connect the mobile to desktop (or other way round) each other without the need to have Bluetooth or Wifi (We do have Wireless Modem from telecom companies which work on Mobile's main telecommunication. &lt;/p&gt;&#xA;" OwnerUserId="7527" LastEditorUserId="1586" LastEditDate="2016-08-10T03:19:47.863" LastActivityDate="2016-10-03T03:03:44.897" Title="Why not only GSM (3G/4G/..)? Why do we need Bluetooth, Wifi,etc...?" Tags="|electrical-engineering|control-engineering|embedded-systems|telecommunication|wireless-communication|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10967" PostTypeId="2" ParentId="7535" CreationDate="2016-08-05T11:32:44.037" Score="1" Body="&lt;p&gt;TL;DR: Go to &lt;code&gt;Dimstyle&lt;/code&gt; and change your text alignment setting to &lt;code&gt;ISO standard&lt;/code&gt; and/or play around with font settings.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The dimension style of the second picture looks like the default for a new AutoCAD file - the style of the first picture has been altered, so you'll have to change the setting of the second file to match.&lt;/p&gt;&#xA;&lt;p&gt;To alter the style of your dimensions, you need to use the aptly named Dimension Style settings. I'm using AutoCAD 2014, but I imagine it's a similar process for other versions.&lt;/p&gt;&#xA;&lt;p&gt;You can try and find the toolbar button if you want, but it's also easily reachable by beginning to type &lt;code&gt;'dimstyle'&lt;/code&gt; while you have nothing else selected and selecting the appropriate option, which should open a box that looks like this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sTeGX.png&quot; alt=&quot;Start&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;As you can see from the picture, the circle dimension created by the radius tool is in the style of Picture 2 from the question.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Click &lt;code&gt;New...&lt;/code&gt; and then &lt;code&gt;Continue&lt;/code&gt; to create your new style.&lt;br /&gt;&#xA;A new window will open at this point where you can customise the various aspects of your dimensions. Having access to the original file allows you to compare the two styles and make the appropriate changes - other users looking at this question won't, so I'll make a note of what to do next.&lt;/p&gt;&#xA;&lt;p&gt;There are two main differences between the two - the text being horizontal, and the font.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The font&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Open the &lt;code&gt;Text&lt;/code&gt; tab on the the New Dimension Style window. There will be a heading labeled &lt;code&gt;Text style&lt;/code&gt;. Click the small box with 3 dots in.&lt;/li&gt;&#xA;&lt;li&gt;On the right side of the screen, click the button &lt;code&gt;New...&lt;/code&gt;. Name it something appropriate, then double click it in the Styles menu on the left to make sure you're editing the correct one.&lt;/li&gt;&#xA;&lt;li&gt;Click the drop-down box under &lt;code&gt;Font Name&lt;/code&gt;. Select any font to suit, &lt;code&gt;Arial Unicode MS&lt;/code&gt; matches Picture 1 fairly well.&lt;/li&gt;&#xA;&lt;li&gt;Click apply to return to the previous screen. Make sure your text style is the one selected in the drop-down box.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The text alignment&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Go to the text tab once again.&lt;/li&gt;&#xA;&lt;li&gt;In the box labeled &lt;code&gt;Text Alignment&lt;/code&gt;, make sure the button for &lt;code&gt;ISO standard&lt;/code&gt; is pressed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your screen should look something like this. Note the differences on the preview screen.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Paxcl.png&quot; alt=&quot;Done Box&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;After hitting ok, you'll have to redraw your dimension in the new style - they don't automatically update.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fvg9bm.png&quot; alt=&quot;Final&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7526" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-08-05T11:32:44.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10968" PostTypeId="2" ParentId="7210" CreationDate="2016-08-05T12:19:43.640" Score="1" Body="&lt;p&gt;I probably do not get exactly the point, but the Timoshenko model pratically has been developed for extending the theory with the contribution of shear effects (that is usually negligible for slender beams) only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torsion is orthogonal to bending and consequently you are not going to consider it inside your calculation of bending deformation energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as for reference, check it out the vlasov beam theory for considering the effect of torsion-axial displacement coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other more smart models can take into account also coupling of torsion and bending and so on, but they are at least an extension of the Timoshenko model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this can be of help.&lt;/p&gt;&#xA;" OwnerUserId="7505" LastActivityDate="2016-08-05T12:19:43.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10969" PostTypeId="2" ParentId="8455" CreationDate="2016-08-05T12:36:59.737" Score="1" Body="&lt;p&gt;In order to decrease the moment that makes your boat front sink, you can decrease the arm of the propeller force (M = F x b). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compute the torque applied by the propeller around the CG and check if it is a good idea. A brief sketch suggests that if you increase the angle you increase the torque (same force!) and you decrease the contact force with the water and if you decrease the angle you decrease the torque but you increase the contact force. So also the contact point becomes important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing I can propose is to change the shape of the boat keel in order to make it glide at high velocity (consider the boat as an airfoil and you get the point!) keeping the same configuration for the propeller.&lt;/p&gt;&#xA;" OwnerUserId="7505" LastActivityDate="2016-08-05T12:36:59.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10971" PostTypeId="2" ParentId="10959" CreationDate="2016-08-05T14:36:06.180" Score="1" Body="&lt;p&gt;Modelling static friction can be a real mess. There isn't an analytical solution to this problem since static friction works on an if/else kind of logic. However, you can model it numerically. I would suggest the following approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you have precomputed discretely sampled surface trajectories. Let $a(k)$ and $v(k)$ be the known acceleration and velocity of the surface at timestep $k$. Let $m$ be the mass of the particle and let $a_p(k)$ and $v_p(k)$  acceleration and velocity of the particle at timestep $k$. Let $F_{s,max}$ and $F_k$ be the known maximum static friction and kinetic friction. You have two equations for acceleration of the particle that you can switch between based on the acceleration and velocity of the plate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If $v_p(k) - v(k) = 0$ and $ma(k)&amp;lt;F_{s,max}$ then&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$a_p(k) = a(k)$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Else/otherwise&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$a_p(k) = F_k/m$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now that you have $a_p(k)$, numerically integrate to get $v_p(k+1)$ and $x_p(k+1)$. Loop over the entire set of data points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the logic: if the particle is not moving relative to the surface then it is experiencing the force of static friction, which will keep it moving at the same acceleration as the surface (up to a limit, dictated by the particle's inertia). When the inertia of the particle exceeds the maximum static force the particle will start to slip, therefore it will now only experience the kinetic friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: although in theory you can use the condition $v_p(k) - v(k) = 0$, numerically speaking it's unlikely that you will ever actually get exactly $0$, so I would suggest using a condition like $|v_p(k) - v(k)| &amp;lt; v_{min}$ instead. $v_{min}$ is an arbitrarily small velocity that means the particle has &quot;practically stopped&quot;.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2016-08-05T14:36:06.180" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10972" PostTypeId="1" AcceptedAnswerId="11029" CreationDate="2016-08-05T17:11:14.880" Score="4" ViewCount="700" Body="&lt;p&gt;I am trying to determine if it is possible to &quot;physically&quot; determine the volume of liquid (such as water) in a non-solid container like a fluid bladder like &lt;a href=&quot;http://www.interstateproducts.com/pillow_tanks.htm&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; or &lt;a href=&quot;http://atlinc.com/pillow.html&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that it is not possible to keep accurate track of the difference between how much fluid is placed in the bladder and how much is taken out, and that a &quot;physical&quot; measurement must be made (where &quot;physical&quot; can include using a sensor, or using some sort of actual measurement of the dimensions of the bag to compute the volume).&lt;/p&gt;&#xA;" OwnerUserId="7531" LastActivityDate="2016-10-05T16:38:05.107" Title="Determine volume of liquid in a fluid bladder" Tags="|fluid-mechanics|measurements|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="10974" PostTypeId="1" CreationDate="2016-08-05T20:13:08.667" Score="2" ViewCount="161" Body="&lt;p&gt;In order to have continuous service a mobile devices switches to base station with higher signal strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Is it possible using the mobile device (Android or iPhone) to determine to LTE base station to which the device has established a connection? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I am interested only for LTE service/devices&lt;/p&gt;&#xA;" OwnerUserId="7083" LastEditorUserId="282" LastEditDate="2016-08-16T03:51:00.003" LastActivityDate="2016-08-16T03:51:00.003" Title="How can one determine the LTE base station providing service to a mobile device?" Tags="|electrical-engineering|wireless-communication|lte|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10975" PostTypeId="2" ParentId="9" CreationDate="2016-08-05T22:48:57.023" Score="1" Body="&lt;p&gt;Ive been cutting, plating ,sawing, drilling aluminium and castings for over 30 years Alumininium alloys cool rapidly after the material is poured and rolled to form a standard plate thickness and depending upon the cooling rate a &quot;grain&quot; structural size variation is formed within the material . larger grains form at the outer layer and reduce in size accordingly to the cooling of the material and the compression of the rollers  , the material grain size delta has inherent  boundary stresses and removing material from each of the faces unevenly can cause warp-age due to  stress  imbalances , now this would tend to be on smaller proportionately thinner plates under a inch thick say, back in the days of manual machining plate Aluminium  we used to skim one face with a fly cutter turn over the  plate, break the skin on the 2nd face and then do a &quot;spring cut of thous(0.005)or 0.12 mm  by fully opening vice and then applying a light clamping force this then sets an unstressed  Flat surface ,also if the machine you are using is not rigid enough !! . or your feeds and speeds could be wrong and the swarf is not taking the heat away from the work surface , there is a multitude of possible causes without seeing a dimensioned atricle drawing and your production methods its impossible to point the finger in one direction&lt;/p&gt;&#xA;" OwnerUserId="5718" LastActivityDate="2016-08-05T22:48:57.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10977" PostTypeId="2" ParentId="10828" CreationDate="2016-08-05T23:23:34.530" Score="-1" Body="&lt;p&gt;box dimensions we usually work to are a 2 thou tolerance +/_ 0.001 off target dimension, this might help&#xA;&lt;a href=&quot;http://help.solidworks.com/2012/English/SolidWorks/sldworks/c_Dimension_Tolerance_and_Precision.htm&quot; rel=&quot;nofollow&quot;&gt;http://help.solidworks.com/2012/English/SolidWorks/sldworks/c_Dimension_Tolerance_and_Precision.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5718" LastActivityDate="2016-08-05T23:23:34.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10978" PostTypeId="1" AcceptedAnswerId="11014" CreationDate="2016-08-05T23:58:18.947" Score="9" ViewCount="1629" Body="&lt;p&gt;I am trying to simulate a winch as a speed-regulated motor that works through a gearbox to lift a mass. The output of the gearbox is a drum, which rotates to accumulate cable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel comfortable &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_moments_of_inertia#Moments_of_inertia&quot; rel=&quot;noreferrer&quot;&gt;converting the mass to a moment of inertia&lt;/a&gt; and I also feel comfortable with converting &lt;em&gt;that&lt;/em&gt; moment of inertia (output-side) to the moment of inertia &quot;seen&quot; by the motor (input-side) with &lt;a href=&quot;https://www.physicsforums.com/threads/inertia-change-through-a-gearbox-question.387779/&quot; rel=&quot;noreferrer&quot;&gt;the gearbox ratio&lt;/a&gt;. With a simple simulation, I have no problem writing the equations of motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My complication comes when I want to model &quot;stretch&quot; in the cable. I thought I could do this by simply putting a spring of arbitrary stiffness between the winch drum and the mass, as pictured below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zza1q.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zza1q.png&quot; alt=&quot;Spring Inertia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this model, for the sake of simulation, I'm assuming I know the &quot;drum height&quot;, which would be how far the drum has turned multiplied by the drum radius, and the height of the load. The spring force would be $k(\phi r - y)$, but &lt;em&gt;how do I apply this to the motor&lt;/em&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a motor model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\Theta}{V} = \frac{K_T}{R_a Js+K_T K_b}&#xA;$$&#xA;and a PI controller model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{V}{\Theta_\mbox{error}} = \frac{k_p \left(s + \frac{k_i}{k_p} \right)}{s}\\&#xA;$$&#xA;where $\Theta$ is the motor speed, $V$ is terminal voltage, $J$ is the inertia of the load and machinery, and $R_a$, $K_T$, and $K_b$ are the motor armature resistance, torque constant, and back EMF constant, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interaction I'm interested in studying occurs when the PI controller is tuned to the &lt;em&gt;anticipated&lt;/em&gt; load inertia $J$, which would be found with the motor, gearbox, drum, and load mass, but the system actually &quot;sees&quot; the springy-mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplification is done by setting the $k_i/k_p$ ratio equal to $K_TK_b/R_aJ$, giving:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\Theta}{\Theta_{\mbox{error}}} = \frac{V}{\Theta_{\mbox{error}}}\frac{\Theta}{V} = &#xA;\left( &#xA;\frac{k_p \left( s + \frac{K_TK_b}{R_aJ} \right) }{s}&#xA;\right) \left( \frac{\frac{K_T}{R_aJ}}{s+\frac{K_T K_b}{R_aJ}} \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note I can leave $k_p$ as variable because the ratio $k_i/k_p$ can be set to whatever I want via $k_i$ as long as $k_p$ isn't zero.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in an &lt;em&gt;ideal&lt;/em&gt; world, where the value of &quot;total&quot; inertia $J$ is known in advance, the pole cancels, and the entire system reduces to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\Theta}{\Theta_{\mbox{error}}} =&#xA;\left( \frac{k_p}{s} \right) \left( \frac{\frac{K_T}{R_aJ}}{1} \right) \\&#xA;$$&#xA;$$&#xA;\frac{\Theta}{\Theta_{\mbox{error}}} =&#xA;\frac{1}{\frac{R_aJ}{k_pK_T}s}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, $\Theta_{\mbox{error}} = \Theta_{\mbox{ref}} - \Theta_{\mbox{out}}$, so, with algebra:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\boxed{\frac{\Theta_{\mbox{out}}}{\Theta_{\mbox{ref}}} = \frac{1}{\frac{R_aJ}{k_pK_T}s + 1}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, kind of sorry to shotgun so much detail, but I wanted to impress on anyone reading that I feel confident with all of my steps so far and that I have spent &lt;em&gt;considerable&lt;/em&gt; effort working on this problem. Now, again to my question - I want to simulate &lt;em&gt;stretch&lt;/em&gt; in the cable between the drum and the load, but I'm not sure how to use the spring force to modulate load inertia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thought I had was to try to fake an &quot;equivalent mass&quot;, by assuming:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F = m_{\mbox{equivalent}}a \\&#xA;m_{\mbox{equivalent}} = \frac{F_{\mbox{spring}}}{a} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this doesn't feel right, and I'm not sure what I would use for acceleration $a$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm frustrated to be this far along on the problem and getting stumped by what seems like should be an easy issue, but I really can't think of a way to approach this problem. I think if I could frame it correctly I could work out the mechanics, but it's the force-to-inertia conversion I feel like needs to be made that has me stumped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, for the record, I have also tried back-tracking my motor model to include the load torque. This gives seemingly-reasonable results, but in the end I subtract the load torque from the motor torque to get net torque, then &lt;em&gt;apply that net torque to the total inertia&lt;/em&gt; to get motor acceleration. That feeds on down the line and, again, I'm not sure that I'm treating total inertia correctly. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2018-08-14T14:30:18.010" Title="How does load force impact load inertia?" Tags="|motors|dynamics|springs|transfer-function|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="10979" PostTypeId="2" ParentId="10978" CreationDate="2016-08-06T01:11:25.467" Score="4" Body="&lt;p&gt;Stretch in the spring delta $ Y = A.sin(\omega.t) = A.sin\sqrt(k/m) . t $ &#xA;So the delta Y is not constant but if you are interested in delt Y_max  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;delta $Y max = m/k$, by Hooks law.&lt;br&gt;&#xA;Because your system doesn't accelerate except at the beginning and end assuming the pulley starts and stops suddenly that's you maximum. Any gradual start/stop acceleration will have to be subtracted from the acceleration of spring which is&lt;br&gt;&#xA;$ - \omega^2 . t$&lt;br&gt;&#xA;$ \omega = \sqrt(k/m)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;looking at free body diagram of mass&lt;br&gt;&#xA;As you noted force is $ K(\phi.r - y)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m.dx^2/dt^2 = -K(\phi.r-r)$&lt;br&gt;&#xA;divide both sides by K we get:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m/K.dx^2/dt^2 +\phi.r=y$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega^2.dx^2/dt^2 +\phi.r =y$       &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this will help.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2016-08-06T16:51:39.957" LastActivityDate="2016-08-06T16:51:39.957" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="10980" PostTypeId="1" AcceptedAnswerId="10995" CreationDate="2016-08-06T02:18:54.640" Score="2" ViewCount="4295" Body="&lt;p&gt;Is there a formula or method to calculate the pressure of failure of a cylindrical container of known material, thickness, and size?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="1832" LastEditDate="2016-08-06T05:17:06.943" LastActivityDate="2022-04-06T13:28:31.670" Title="Method of Calculating pressure of failure in a cylindrical container?" Tags="|pressure|statics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10981" PostTypeId="2" ParentId="10980" CreationDate="2016-08-06T04:16:48.053" Score="2" Body="&lt;p&gt;There is a simple formula for calculating the stress in thin-walled cylindrical pressure vessels. In the circumferential direction, the formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sigma_c=\frac{pr}{t}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in the longitudinal direction (along the length of the cylinder):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sigma_x=\frac{pr}{2t}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$p =$ the pressure inside the cylinder,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$r =$ the inner radius of the cylinder, and&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$t =$ the thickness of the wall $(r/t \geq 10)$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With this formula, it is a simple matter to calculate at what pressure the vessel will fail, based on its material properties and whatever factor of safety you like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\sigma_c$ and $\sigma_x$ are assumed to be constant throughout the wall of the cylinder, and that the walls of the cylinder are subjected to tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;em&gt;Mechanics of Materials Ninth Edition&lt;/em&gt; by R.C. Hibbeler.&lt;/p&gt;&#xA;" OwnerUserId="7465" LastEditorUserId="7465" LastEditDate="2016-08-06T06:14:25.980" LastActivityDate="2016-08-06T06:14:25.980" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="10983" PostTypeId="2" ParentId="10966" CreationDate="2016-08-06T13:35:28.867" Score="1" Body="&lt;p&gt;Several reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly: They're designed for different purposes. It's basically the same reason individual power tools are usually better than all-in-ones. Sacrafices in one aspect are used for gains in another. For example, bluetooth is optimised for flexibility at the expense a bit of bandwidth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also different frequencies have different properties. Higher frequencies tend not to go through obstacles so well, but can provide higher bandwidth. Mixing protocols on one frequency won't work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Protocols such as GSM are designed to cope with more simultaneous users than WiFi, which is required because if the range it needs to work over to be practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about the requirements if Bluetooth, WiFi and GSM, and tried to merge them into one, you'd find the electronics would be more expensive, things would interfere with each other more, and there would be no more space left in the airwaves&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-08-06T13:35:28.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10984" PostTypeId="1" AcceptedAnswerId="10985" CreationDate="2016-08-06T14:04:41.607" Score="-2" ViewCount="39" Body="&lt;p&gt;This question is a chemical engineering question related to ecology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the list of possible threats of human feces to the environment? What about toilet paper? How do engineers deal with this problem? I mean, sure, there are sewers, but sewers consume energy, and process more than sit and toilet paper, also soap and other detergents. I want to know, these specific cases:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;how would you treat dung&lt;/li&gt;&#xA;&lt;li&gt;how would you treat dung and paper (toilet paper and paper towels)&lt;/li&gt;&#xA;&lt;li&gt;how would you treat dung and paper (toilet paper and paper towels) and soap&lt;/li&gt;&#xA;&lt;li&gt;how would you treat dung and paper (toilet paper and paper towels) and soap and detergents/washing products&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this logical order, I would assume that treating these wastes increases in complexity as we go down the list. Treating something further down is sufficient for treating something further up, but treating something further up alone (perhaps in some situations the extras provided by entries further down the list are not needed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is the question of whether what you eat affects the ability of human feces to act as fertilizer, and why would house our core dung be better than human dung? How would it be chemically different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your insight.&lt;/p&gt;&#xA;" OwnerUserId="7542" LastActivityDate="2016-08-06T15:53:26.067" Title="Disposal of dung, related products, and cleaning products, prioritized and case by case" Tags="|chemical-engineering|waste-water-treatment|waste-disposal|" AnswerCount="1" CommentCount="1" ClosedDate="2016-08-06T20:52:23.660" ContentLicense="CC BY-SA 3.0" />
  <row Id="10985" PostTypeId="2" ParentId="10984" CreationDate="2016-08-06T15:53:26.067" Score="2" Body="&lt;p&gt;The main issue is the spread of pathogens, human faeces can harbour some nasty diseases, notably e-coli, hepatitis and cholera. Accumulation can also provide an environment where pathogens can survive for  a long time and there is a significant risk of them leeching into water supplies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a problem with any animal waste that if it accumulates in water supplies it creates excessive nutrient levels which promote the growth of algae and bacteria which, as well as being a health hazard can over-grow and crowd out other organisms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has also been suggested that metabolic products from drugs and things like hormone based contraceptives can cause environmental problems. Although it is obviously more complex to establish causality with this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the above reasons much more care needs to be exercise when using human waste as fertiliser than with manure from other animals. However if properly handled, composting can be a simple and effective way to make it safe. This both allows the more accessible nutrients which tend to promote unwanted microorganisms to be broken down and, if done properly can reach temperatures where harmful organisms are killed off, at least to safe levels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more technical solution is something like an anaerobic digester, these are becoming increasingly popular, especially in industrial agriculture to create revenue from crop and animal waste and have the advantage that they make use of the energy in the waste as well as the minerals and nitrates. Here controlled conditions and selected microorganisms are used to generate methane from the waste slurry. It is even potentially possible that they could be engineered to produce plastics or other chemical feed-stocks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously when you are talking about human waste you also need to take social and cultural behaviour into account. Most people in the developed world are now deeply ingrained with the idea of getting rid of their bodily waste instantly and, ideally without even seeing it and any significant change in this behaviour may be a hard sell, at least at first, unless you can offer some very significant benefits. This is, of course a very new thing and it is not that long ago that there was  thriving industry in human waste for all sorts of industrial purposes. and there are some parts of the world where it is actually a challenge to persuade people to use toilets, even when they are available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course another major source of waste is that we tend to spend huge resources on making water safe to drink and then use most of it just for flushing toilets, irrigation etc and only drink a tiny percentage of it. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-08-06T15:53:26.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10986" PostTypeId="1" CreationDate="2016-08-06T17:04:14.697" Score="1" ViewCount="746" Body="&lt;p&gt;Please I need some help. If I have a driving shaft with 10&amp;nbsp;Nm torque and 1000&amp;nbsp;RPM, and I add two 3-inch pulleys on the shaft to transmit power to another shaft with a pair of 6-inch pulleys mounted on it and connected to the initial 3&quot; via a belt, will I have transmitted a total torque of 40&amp;nbsp;Nm to this new bar at 500 RPM?&lt;/p&gt;&#xA;" OwnerUserId="7545" LastEditorUserId="1832" LastEditDate="2016-08-07T15:24:29.493" LastActivityDate="2016-10-27T02:15:34.190" Title="Torque and RPM power transmission" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10987" PostTypeId="1" AcceptedAnswerId="11049" CreationDate="2016-08-06T19:23:54.370" Score="2" ViewCount="1453" Body="&lt;p&gt;All mobile phones have bars or some other type of indicator displaying signal strength. As mobile devices move away from the base station the signal strength becomes weak and at some point becomes out of range or establishes connections with a different tower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; For a LTE base station is it possible to determine the relationship between the signal strength and distance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumption:&lt;/strong&gt; LTE base station is located in a plain field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back story: A cellular service provider is planning on installing a LTE base station using equipment from &lt;a href=&quot;http://www.jmawireless.com/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;JMA wireless&lt;/a&gt;. The model number of the equipment is &lt;a href=&quot;http://www.jmawireless.com/images/uploads/pdfs/X7CQAP-FRO-645-V_DataSheet.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;X7CQAP_FRO_645&lt;/a&gt;. Below is a H-plane pattern plot for 750MHz frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1jskY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1jskY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone provide insight as how to interpret the above plot in relation to signal strength verses distance? Do I need more information?&lt;/p&gt;&#xA;" OwnerUserId="7083" LastEditorUserId="282" LastEditDate="2016-08-16T03:51:08.957" LastActivityDate="2016-08-16T03:51:08.957" Title="How to determine signal strength verses distance for LTE base stations?" Tags="|electrical-engineering|wireless-communication|lte|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10989" PostTypeId="2" ParentId="10966" CreationDate="2016-08-06T20:55:53.317" Score="1" Body="&lt;p&gt;Several reasons but mostly bandwidth and speed.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It makes sense to keep all communication channels as short and local as possible. This reduces likelihood of interference, channel contention and cost.&lt;/li&gt;&#xA;&lt;li&gt;It offers improved bandwidth and data transfer speed.&lt;/li&gt;&#xA;&lt;li&gt;The local solution requires no maintenance. Once you've paid for your WiFi or Bluetooth device that's it. Once you go to someone else's tower you are going to have to pay.&lt;/li&gt;&#xA;&lt;li&gt;Communications can be private.&lt;/li&gt;&#xA;&lt;li&gt;Communications can be low power.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2016-08-06T20:55:53.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10990" PostTypeId="2" ParentId="10972" CreationDate="2016-08-06T21:44:39.463" Score="1" Body="&lt;p&gt;You may be familiar with &lt;strong&gt;LIDAR&lt;/strong&gt;. This is how land topography is put into digital format to be used in various 3D models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some kind of &quot;point cloud&quot; device that could penetrate the bladder material but reflect off of the fluid contained within? If you could find such a device, you could create a point cloud of the fluid mass and calculate the fluid volume by importing the point cloud into volume/topography calculation software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be like an x-ray machine that penetrates the bladder, but bounces off the water. The whole bladder would have to be scanned to create the point cloud of the fluid surface. You could then do a volume difference calculation between the 3D fluid surface and the bottom, flat, supporting surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.3dreshaper.com/en/software-en/support-software-en/practical-exercises&quot; rel=&quot;nofollow&quot;&gt;Check out the ore and liquid cubature examples&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7549" LastEditorUserId="7549" LastEditDate="2016-08-06T21:56:32.343" LastActivityDate="2016-08-06T21:56:32.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="10991" PostTypeId="2" ParentId="10980" CreationDate="2016-08-07T00:41:50.267" Score="1" Body="&lt;p&gt;If you draw the free body diagram of half of cylinder like an arch and draw pressure force vectors, it has pressure force radiating around the entire surface out with equal magnitude but fanning out 180 degrees. The upward projection of these vectors will sum up to: $p\cdot 2r = \text{lateral force}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This force has to be carried by 2x thickness of the wall x unit length. Hence we get&#xA;$$pr = \sigma t$$&#xA;But for the ends we have&#xA;$$\frac{p\pi r^2}{2rt}= \frac{p\pi r}{2}$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However we have to calculate for buckling due to higher exterior pressure as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very rough estimate for buckling is to assume the entire external pressure is supported at the edges of the container acting in fixed end moment joints as the abutments of an arch if the container is long ($l&amp;gt;4r$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are attachments or stiffeners buckling has to be modeled considering the total geometry. We have seen one to many twisted frozen can of coke left in freezer. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2016-08-07T05:13:17.630" LastActivityDate="2016-08-07T05:13:17.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10992" PostTypeId="2" ParentId="10760" CreationDate="2016-08-07T03:32:21.763" Score="1" Body="&lt;p&gt;Are you thinking of &lt;a href=&quot;https://en.wikipedia.org/wiki/Coilover&quot; rel=&quot;nofollow&quot;&gt;coilovers&lt;/a&gt;? They look very similar to the image you posted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coilovers are shock absorbers/dampers with a spring wrapped around the outside. They appear in many vehicle suspensions. &lt;/p&gt;&#xA;" OwnerUserId="7465" LastEditorUserId="6695" LastEditDate="2016-09-08T16:43:46.683" LastActivityDate="2016-09-08T16:43:46.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10995" PostTypeId="2" ParentId="10980" CreationDate="2016-08-07T12:31:41.883" Score="3" Body="&lt;p&gt;In this answer, I aim to extend the discussion to thicker walled cylinders too. If you want to skip the maths, the pressures are summarised at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;DEFINED TERMS:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;$\sigma_{rr}$ - radial component of stress&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_{\theta\theta}$ - circumferential component of stress&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_{zz}$ - axial component of stress&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this answer, I will assume that failure occurs in the curved walls of the cylinder. Different failure modes, could be considered, but that is beyond the scope of my answer. It is in my intuition that the curved walls are the weak points of failure, but I will leave the discussion of that to more qualified answers.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;FIRST YIELD&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When it comes to determining the failure of thick walled cylinders (defining failure to occur the instance yield occurs anywhere in the cylinder), Lamé's Equations are useful. These equations give the stresses that lie in the plane of the circular cross section of the cylinder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EYsTB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EYsTB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the rotational symmetry axis is moving out of the page, normal to the cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lamé's Equations are derived from the governing equilibrium equation for an axisymmetric solid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial}{\partial r} \left( r \sigma_{rr}\right) = \sigma_{\theta\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $r$ is the distance of the point where stresses are being considered from the axis of rotational symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lamé's equations themselves are given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_{rr} = A - \frac{B}{r^2}$$&#xA;$$\sigma_{\theta\theta} = A + \frac{B}{r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $A$ and $B$ are arbitrary constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a cylinder with internal pressure $p$, no external pressure, an internal radius of $a$ and thickness $t$, these equations become:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_{rr} = \frac{p}{m^2-1}\left(1-\frac{m^2a^2}{r^2}\right)$$&#xA;$$\sigma_{\theta\theta} = \frac{p}{m^2-1}\left(1+\frac{m^2a^2}{r^2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $m=1 + \frac{t}{a}$ (i.e. ratio of outer to inner radii). Note that as $t$ gets smaller, $m$ approaches 1 and the stresses approach that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the final stress component, it is reasonable to assume that, away from the edges of the cylinder, the axial stress is constant in the walls. This stress will oppose the pressure acting on the cylinder ends:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_{zz} = \frac{p}{m^2-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a rough plot of the three stresses as function of $r$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MSSWI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MSSWI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the three components of stress known, a suitable &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Yield_surface&quot; rel=&quot;nofollow noreferrer&quot;&gt;yield criterion&lt;/a&gt; should be used. Usually Tresca or von Mises are suitable for metals like steel. Here, I will assume Tresca, which states that yield first occurs at a point on the cylinder whenever the following is satisfied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\max\left(|\sigma_{rr}-\sigma_{\theta\theta}|,|\sigma_{rr}-\sigma_{zz}|, |\sigma_{\theta\theta}-\sigma_{zz}|\right)=\sigma_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\sigma_y$ is the yield stress of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By looking at the plot, it can be seen that yield will occur on the inside of the cylinder, so that at yield:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$|\sigma_{rr} - \sigma_{\theta\theta}| = \frac{2m^2}{m^2-1} p=\sigma_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, failure by first yield occurs when:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=\frac{m^2-1}{2m^2}\sigma_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum pressure before first yield increases with thickness, up to a maximum of $\frac{\sigma_y}{2}$ at infinite thickness.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;COLLAPSE&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Even when yield first occurs, the walls of the cylinder won't burst open, i.e. collapse has yet to occur. The point at which collapse occurs should occur after first yield: would you ever expect an infinitely thick cylinder to burst? The collapse pressure is useful to determine since it is at this point that the cylinder fails to retain its pressurised contents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When first yield occurs, there is still regions in the cylinder that have yet to yield, and so the internals are retained. As you increase the pressure beyond that of first yield, you get a plastic region of yielded material that forms at the inner face of the cylinder and creeps along the cylinder thickness towards the outer face:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KTliT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KTliT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collapse occurs as soon as the plastic zone reaches the outer face of the cylinder. The issue in trying to calculate this load is that elasticity no longer applies: Lamé's equations, which are only valid for linear elastic cylinders, will not be valid. However, the governing equilibrium equation still applies: equilibrium must be upheld regardless whether the material law is linear or not. Since the plastic zone has extended through the entire cross section, less us assume that the stresses abide by Tresca's criterion throughout the cylinder (this assumption is valid according to the Lower Bound Theorem, which returns an estimate of the collapse pressure which must be less than the actual collapse pressure, giving a conservative estimate):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_{\theta\theta} - \sigma_{rr} = \sigma_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now substitute this into the equilibrium equation to give:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r \frac{\partial \sigma_{rr}}{\partial r} = \sigma_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this and applying boundary conditions then gives a collapse pressure of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p = \ln{(m)} \sigma_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly, for all positive $m$, $\ln (m)$ is greater than or equal to $\frac{m^2-1}{2m^2}$, so first yield must always occur before collapse. Only when $m$ approaches 1 (thin-wall) do the pressures for first yield and collapse approach being equal.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;IN SUMMARY&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For failure by first yield, the pressure must be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=\frac{m^2-1}{2m^2}\sigma_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For failure by collapse, the pressure must be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p=\ln{(m)} \sigma_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $m$ is the ratio of external to internal radii.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a rough plot for both:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DEJUu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DEJUu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $m$ increases, the first-yield limit increases to a maximum of $\frac{\sigma_y}{2}$, while the collapse limit increases to infinity. For the collapse limit to be double that of the first-yield limit, $m$ needs to be around 2.2.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2016-08-07T21:38:55.183" LastActivityDate="2016-08-07T21:38:55.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10996" PostTypeId="1" CreationDate="2016-08-07T13:56:16.840" Score="0" ViewCount="49" Body="&lt;p&gt;If we need to measure the Electrostatic field of a celestial body due to its interaction with space plasmas, What kind of instruments can be used to achieve this?&lt;/p&gt;&#xA;" OwnerUserId="7555" LastActivityDate="2017-02-03T20:35:50.490" Title="How would you measure the electrostatic field of a celestial body from its surface?" Tags="|electrical-engineering|instrumentation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10997" PostTypeId="2" ParentId="10996" CreationDate="2016-08-07T15:18:41.140" Score="1" Body="&lt;p&gt;One way to measure electric fields directly is called a &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Field_mill&quot; rel=&quot;nofollow&quot;&gt;field mill&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses moving vanes to impose an AC modulation on the field and a differential sensing setup that allows most sources of error to be zeroed out, something like how a chopper-stabilized opamp works.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-08-07T15:18:41.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="10998" PostTypeId="1" CreationDate="2016-08-07T16:10:33.070" Score="1" ViewCount="82" Body="&lt;p&gt;I'm having a difficult time identifying a resistor's value as I can't find the golden / silver tolerance line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I can't see a silver or gold line here! They're both brown on both sides&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/NIZCp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NIZCpt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e3aBt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e3aBtt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7557" LastEditorDisplayName="user16" LastEditDate="2016-08-07T16:58:52.030" LastActivityDate="2016-08-07T23:48:19.597" Title="Help with reading resistor color codes" Tags="|electrical-engineering|circuits|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="10999" PostTypeId="2" ParentId="10998" CreationDate="2016-08-07T16:55:35.730" Score="1" Body="&lt;p&gt;Based upon the pictures, it appears that the bands are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brown, red, black, black, brown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using one of many readily available &lt;a href=&quot;http://www.ohmslawcalculator.com/resistor-colour-chart&quot; rel=&quot;nofollow&quot;&gt;color code charts&lt;/a&gt; or &lt;a href=&quot;http://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-resistor-color-code-5-band&quot; rel=&quot;nofollow&quot;&gt;calculators&lt;/a&gt; that are online, it appears that you have a 120 ohm resistor with 1 % tolerance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st band, 1st digit - brown - 1&lt;br&gt;&#xA;2nd band, 2nd digit - red - 2&lt;br&gt;&#xA;3rd band, 3rd digit - black - 0&lt;br&gt;&#xA;4th band, multiplier - black - 1&lt;br&gt;&#xA;5th band, tolerance - brown - 1%.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice the additional gap between the last black band and the second brown band.  That helps identify which band is the tolerance band.  You'll also notice that the red band is sitting on the shoulder whereas the last black band is spaced a little bit away from the shoulder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if my interpretation of that photo is wrong, and the order actually is brown, black, black, red, brown then you'll have a completely different value (10k ohm in this case) for the resistor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when there's any doubt about the resistor, you ought to use an &lt;a href=&quot;https://en.wikipedia.org/wiki/Ohmmeter&quot; rel=&quot;nofollow&quot;&gt;ohmmeter&lt;/a&gt; to validate what you think the value ought to be.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2016-08-07T23:48:19.597" LastActivityDate="2016-08-07T23:48:19.597" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11000" PostTypeId="1" CreationDate="2016-08-07T17:10:36.247" Score="1" ViewCount="1059" Body="&lt;p&gt;Once the turbo is on, more air can participate to the combustion, which also requires more fuel. Then, it's clear that the turbo can give more power (at the price of more consumption). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1: Can this be used as well to improves economy ? (i.e. less fuel for same  RPM than without turbo)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2:  How does the gain behave with RPM, passed the initial threshold ? Does it make the total efficiency curve of the engine more flat, or less flat ?&lt;/p&gt;&#xA;" OwnerUserId="7547" LastEditorUserId="7547" LastEditDate="2016-08-09T19:53:37.560" LastActivityDate="2016-08-09T21:24:09.557" Title="Does the turbo-compressor increase the efficiency of car engine?" Tags="|automotive-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11001" PostTypeId="1" CreationDate="2016-08-07T17:22:43.107" Score="3" ViewCount="862" Body="&lt;p&gt;I want to use a NPN (2n4401) and PNP (2n3906) with both base connected to a single microcontroller output in order to switch voltage to two inputs off and on.  I'm not sure how to create the circuit I need.  Tried a basic switching circuit without getting expected results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are additional details:&#xA;I am using an attiny85 to control the circuit, 0-3.3v at input to base pins of the transistors. I'm switching two lines of an led display (ltd-2601). I want to switch the two lines every `10ms in order to drive the display. So the inputs to the 2601 would be 0v and 3.3v for 10ms and then 3.3v and 0v for 10ms.  That's what I hope to accomplish using the npn and pnp transistors together (shared base pin) driven by the single line of the microcontroller. Basically, when the micro-controller line is low the npn output would be 3.3v and the pnp 0v and vice versa when line is high.  The goal here is to use only 1 line of the attiny to switch the display rather than two lines.&lt;/p&gt;&#xA;" OwnerUserId="7560" LastEditorUserId="7083" LastEditDate="2016-08-14T06:27:46.153" LastActivityDate="2016-08-14T06:27:46.153" Title="Use npn (2n4401) and pnp (2n3906) connected to single microcontroller to switch" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11003" PostTypeId="2" ParentId="11001" CreationDate="2016-08-07T19:11:06.127" Score="1" Body="&lt;p&gt;Here is an example circuit configuration where a &lt;a href=&quot;http://www.ti.com/lit/ug/spmu296/spmu296.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;TM4C123G (Tiva)&lt;/a&gt; is driving a NPN transistor to switch a voltage On and Off. I hope this help you to get started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pepzS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pepzS.jpg&quot; alt=&quot;Tiva Micro driving NPN transistor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are two example circuits for high active and low active. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/halmw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/halmw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.edx.org/course/embedded-systems-shape-world-utaustinx-ut-6-03x&quot; rel=&quot;nofollow noreferrer&quot;&gt;Embedded Systems - Shape The World (Chapter 8)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ermicro.com/blog/?p=423&quot; rel=&quot;nofollow noreferrer&quot;&gt;Using Transistor as a Switch&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-08-10T03:38:52.287" LastActivityDate="2016-08-10T03:38:52.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11005" PostTypeId="1" CreationDate="2016-08-07T21:30:00.507" Score="2" ViewCount="260" Body="&lt;p&gt;I wanted to evaluate the safety of a walking aid device (walker). For example, if the way it is constructed and whenever I handle it, there are no risk of rotation, around its center of mass, turning over, i.e imbalance/instability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a walker with 4 wheels. A person performs around 20% of its body weight for holding the handles of the walker. When the person pushes the walker, will it be a possibility for the walker to turn over, is it unstable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there is a theoretically way to calculate that? And virtually? Can I use ANSYS to check that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7548" LastActivityDate="2017-05-03T16:26:42.773" Title="Equilibrium of rigid bodies" Tags="|mechanical-engineering|design|statics|safety|stability|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11006" PostTypeId="1" CreationDate="2016-08-07T22:36:28.950" Score="1" ViewCount="63" Body="&lt;p&gt;I simulated an axial turbine(Wells turbine). Now I have faced with strange problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Performance of any turbine can be improved if the losses are identified and minimized. Several losses can be defined by flow separation, small or large vortices creation or flow circulation, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, when the turbine have lower efficiency than others turbines, it means that this turbine have more losses than others. In other words, the mentioned turbine have more flow separation, vortices or flow circulation than others. This again means that mentioned turbine have more entropy generation than others because entropy generation include velocity gradient terms and when separation and circulation(losses) increase, entropy generation also increases. Is my statement right?am i right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, i designed an improved blade and a thicker blade(constant cord) than to base blade. . My results show that the Improved blade have better efficiency and lower entropy generation than to base blade. These results are logical. the thicker blade have lower efficiency and more losses than base blade(thinner blade) . And so, this cause higher entropy generation. But according to my results,  the thicker blade have lower efficiency(it is logical) and lower entropy generation(I think that it is not logical) than to base blade. &lt;strong&gt;Is it possible that entropy generation decreases  with decreasing efficiency?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am grateful that guide me about this case. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2016-08-07T22:36:28.950" Title="Is it possible that entropy generation decreases with decreasing turbine efficiency(First Low efficiency )?" Tags="|mechanical-engineering|fluid-mechanics|turbines|cfd|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11007" PostTypeId="2" ParentId="10656" CreationDate="2016-08-07T23:15:45.933" Score="0" Body="&lt;p&gt;Paul,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;At first glance&lt;/strong&gt;, it seems to me that the &lt;strong&gt;average velocity&lt;/strong&gt; a good distance down along the pipe should be less than the &lt;strong&gt;average velocity&lt;/strong&gt; at the inlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I believe this is because inlets usually act as constrictions. The cross sectional area of flow would be decreased at the inlet. Within the pipe, the cross sectional area of flow would be larger, so the &lt;strong&gt;average velocity&lt;/strong&gt; would be slower.&lt;/p&gt;&#xA;" OwnerUserId="7549" LastActivityDate="2016-08-07T23:15:45.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11008" PostTypeId="2" ParentId="10998" CreationDate="2016-08-07T23:27:59.870" Score="-2" Body="&lt;p&gt;I happen to have 3 similar resistors (from the Arduino kit):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h4okm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h4okm.jpg&quot; alt=&quot;3 different resistors&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't resist to measure them, so from left to right they 220 ohm, 1k ohm and (as I assumed) 10 k ohm.&lt;/p&gt;&#xA;" OwnerUserId="7194" LastActivityDate="2016-08-07T23:27:59.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11009" PostTypeId="2" ParentId="9180" CreationDate="2016-08-07T23:35:23.957" Score="0" Body="&lt;p&gt;I think flow rate and surface tension are the two, main ingredients here. You can see the same effect at your kitchen tap when you vary from high flow to low flow. The contraction is readily visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suspect that low velocities allow surface tension more time to act and contract the water boundary. Higher flows may tend to dominate the surface tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a good analogy would be how the velocity profile within a pipe during laminar flow is decidedly different from the velocity profile during turbulent flow. In turbulent flow, it's decidedly more &quot;square&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7549" LastActivityDate="2016-08-07T23:35:23.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11012" PostTypeId="2" ParentId="7852" CreationDate="2016-08-08T10:28:38.420" Score="0" Body="&lt;p&gt;I've been miming eating with spoon without wrist movement and I'm not quite sure I understand the problem correctly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my understanding using a gimbal bearing would more of a psychological torture device, and the solution is simple! The spoons handle only needs a dog leg in its handle,say 45 degrees, and/or the users nose length from rear of spoon to the handles upper bend of dog leg for obvious reasons, from there you could have a standard bearing fixed to either end of the remaining handle and tubing cut to desired length with inside diameter appropriate to the outside diameter of the bearings then slide/press bearings into tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HOWEVER!!! It will be imperative that consideration is made as to the action of loading the spoon with food... If the user targets the food with a forward trajectory it would work fine, BUT!, should the need arise to drag the spoon from the rear of plate/bowl/dish to the front or vice versa as you would for the proper etiquette of soup consumption, this design is flawed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to circumventing this flaw is to limit rotation by way of two keys counter sunk into each end of tube handle immediately behind the bearings ends that face the inside of the tube and two more keys fixed to the inner shaft/axle between the tubing keys to limit rotation. I would also suggest making the rotational travel limiting keys customisable by identical multiple fixing points for two opposing outer keys.&lt;/p&gt;&#xA;" OwnerUserId="7565" LastEditorDisplayName="user16" LastEditDate="2016-08-08T11:20:24.137" LastActivityDate="2016-08-08T11:20:24.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11013" PostTypeId="1" AcceptedAnswerId="13034" CreationDate="2016-08-08T14:41:56.830" Score="5" ViewCount="3624" Body="&lt;p&gt;I want to make a small air filter using standard axial pc fans in order to suck smoke like that have been produced in soldering process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I assume filtering will make it difficult. Am i wrong? The air have to be forced through the granulated carbon filter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I researched about radial fans which produced spesifically for this job(suction), but they have disadvantages like price, vibration and noise. They are at least 60dba. May not suitable to be used in room. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main purpose is increasing suction of air, not the air flow. I can go with minor flow. (if my expression is not true please correct)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, can a group of fan in same direction, increase the suction of air? Can it force the air to pass through carbon filter about 1-2cm thickness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j7at7.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j7at7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7567" LastActivityDate="2019-04-13T04:08:49.467" Title="Can air suction be increased by consecutive pc fans in small air tunnel?" Tags="|airflow|aerodynamics|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11014" PostTypeId="2" ParentId="10978" CreationDate="2016-08-08T15:29:40.580" Score="6" Body="&lt;p&gt;Let's first compute the model. The control design is a separate effort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque applied to the drum is $n T_M $, where n is the gear ratio and $T_M$ is the output produced by the motor. $T_M= K_T i(t)$, where $K_T$ is a proportionality constant and $i(t)$ is the motor current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can write the equations for the mechanical system:&#xA;$$ m y''(t)+m g-k (y(t)-r \theta (t))=0$$&#xA;$$ J \theta ''(t)+k r (y(t)-r \theta (t))=n K_T i(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here m is the mass and k is the spring constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To write the motor equation, we need to determine the back emf. The back emf is proportional to the motor speed and to write it in terms of the drum speed we multiply it also with the gear ratio n.&#xA;$$L i'(t)+R i(t)+n K_b \theta '(t)=V(t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $V(t)$ is the applied voltage, $L$ is the inductance, $R$ is the resistance, and $K_b$ is the proportionality constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These three equations have $V(t)$ as the input and $i(t)$,$\theta (t)$, and $y(t)$ as states/outputs. This can be used to obtain the state-space model or transfer-function model. (The following were obtained using Mathematica) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zy42R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zy42R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the control design can begin...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there has been some confusion about the inertia to be used, let me clarify the answer. I am going to assume one set of gears in the gearbox - a gear with inertia $J_1$ on the drum side and a gear with inertia $J_2$ on the motor side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the answer above I neglected the inertia of the gears. The only change that needs to be done now is modify the second equation as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(J+J_1\right) \theta ''(t)+k r (y(t)-r \theta (t))=n i(t) K_T$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the equation to describe the transient dynamics of the motor shaft is also desired then it is an additional equation involving $\theta_M$(rotation of the motor shaft), the inertia $J_2$, etc. However, this is not necessary if the objective is to control the drum position.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2016-08-10T19:45:51.197" LastActivityDate="2016-08-10T19:45:51.197" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11015" PostTypeId="1" CreationDate="2016-08-08T18:32:17.003" Score="3" ViewCount="309" Body="&lt;p&gt;I'm working with some flexures (circularly symmetric tube with pins at the ends).  Their bending stiffness matches my FEA.  However, I'm concerned that they are buckling, reducing their overall axial stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a way to measure the axial stiffness in situ; the flexures are part of a larger system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a lot of mechanical measurement equipment either, but I do have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Accelerometer&lt;/li&gt;&#xA;&lt;li&gt;Laser distance measurement device&lt;/li&gt;&#xA;&lt;li&gt;Oscilloscope&lt;/li&gt;&#xA;&lt;li&gt;Masses, scales, ...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I can model the buckling, but the rest of the team relies on measurement and doesn't believe models.  So, I need to come up with a method of measuring buckling of the flexure in compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zEFb5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zEFb5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7569" LastEditorUserId="7569" LastEditDate="2016-08-08T19:18:31.550" LastActivityDate="2017-01-07T13:33:27.153" Title="Methods to Measure Buckling without Expensive Equipment" Tags="|mechanical-engineering|measurements|buckling|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11016" PostTypeId="2" ParentId="11001" CreationDate="2016-08-09T01:15:26.377" Score="1" Body="&lt;p&gt;Either transistor inverts when used in the common emitter configuration.  This is how they are used if you feed the base from a digital output (via a resistor), and the emitter is grounded in the NPN case, or tied to the supply in the PNP case.  The difference is which direction they will actively drive.  They each go to high impedance in the other direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want two digital signals, each the inversion of the other, controlled from a single digital output, then use a inverter chip.  Single inverters come in the same SOT-23 package small transistors do.  Using a inverter is both simpler and will yield better results.  All you need is the inverter chip and it's bypass cap.  The transistor solution requires at least a transistor and a resistor, so at best takes the same space.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-08-09T01:15:26.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11017" PostTypeId="2" ParentId="10760" CreationDate="2016-08-09T01:57:05.307" Score="1" Body="&lt;p&gt;You're looking for a &quot;shock absorber&quot; or &quot;damper&quot;. You could prefix your search with &quot;miniature&quot;. Try McMaster for a start &lt;a href=&quot;http://www.mcmaster.com/#hydraulic-dampers/=13n5nu9&quot; rel=&quot;nofollow&quot;&gt;http://www.mcmaster.com/#hydraulic-dampers/=13n5nu9&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-08-09T01:57:05.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11018" PostTypeId="1" CreationDate="2016-08-09T10:48:07.987" Score="3" ViewCount="62" Body="&lt;p&gt;How do shoreline nuclear facilities lower the temperature of wastewater generated by their cooling systems before it enters the ocean?&lt;/p&gt;&#xA;" OwnerUserId="7579" LastEditorUserId="368" LastEditDate="2018-05-31T15:26:48.160" LastActivityDate="2018-07-01T13:10:01.993" Title="Nuclear wastegate thermal dissipation systems?" Tags="|environmental-engineering|nuclear-technology|nuclear-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="11019" PostTypeId="2" ParentId="5407" CreationDate="2016-08-09T11:42:23.870" Score="1" Body="&lt;p&gt;The practical effects of polarity of quite complex and depend on the process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense TIG is where the 'pure' effects of polarity are clearest as the arc creation is entirely separate from the filler feed. Here DC- (electrode -ve) is used for most metals except for aluminium and magnesium alloys. This polarity puts the majority of the energy into the weld puddle which is both more efficient and helps prevent the tungsten electrode from overheating. For aluminium the problem is that the thick, stable oxide layer needs to be cleaned off as welding progresses lest it contaminate the weld. Here reverse polarity is used as the electron flow tends to rip the oxide off the surface. You can actually see this happening in a TIG weld and a fresh weld will have a 'frosted' area around the weld bead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice most modern machines now use a modified AC current for aluminium as this allows more adjustment of the arc properties. Specifically the balance between +ve and -ve current flow balances cleaning effect with electrode temperature and adjusting the AC frequency controls the width of the arc (which in turn affects penetration). The most sophisticated machines also a range of different wave-forms (square, sine, trapezium, triangular etc) which also affect penetration and heat distribution. Note also that in AC TIG the end of the electrode will melt and form  a ball slightly larger than the electrode diameter the fast cycling AC stops it from falling off into the weld. In DC welding the tungsten should not melt at all.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In MIG welding the fact that the filler wire is also the electrode means that heat is transferred to the puddle in a different way and the shape and shape and size of the metal droplets formed is very important. This favours putting more heat into the electrode as it is constantly melting and physically transferring that energy to the weld puddle in any case. So solid wire MIG welders conventionally run with reverse polarity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In MMA things are more complex again as the composition of the electrode coating interacts with the arc polarity and the results are really a function of the particular electrode type and most electrodes will have a recommended polarity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This subject is also further confused by the fact that conventional current is actually backwards in its sign convention compared to the rest of physics (ie the actual flow of electrons is from -ve to +ve) which doesn't make this any easier to understand. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-08-09T12:12:16.597" LastActivityDate="2016-08-09T12:12:16.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11021" PostTypeId="2" ParentId="11000" CreationDate="2016-08-09T15:00:27.513" Score="2" Body="&lt;p&gt;The turbo increases the amount of air that is introduced to the engine. Mass airflow sensors see this increase in air and signal the fuel injectors to add proportionally more fuel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the net result of having &lt;em&gt;effectively&lt;/em&gt; larger cylinders - the engine with the turbo can &quot;process&quot; the same amount of fuel per cycle as a larger engine with no turbo. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight of the turbo is very small relative to the weight of the engine, so this means there's an increase in the power-to-weight ratio of the engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be the answer you're after, but it's not clear because &quot;efficiency&quot; isn't well-defined when referring to automotive applications. There's &quot;fuel economy&quot; - how far you can go per unit of fuel, and there's &quot;fuel efficiency&quot; - how much mechanical power you get per unit of fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most fuel efficient point of an engine is generally wide-open throttle under maximum or near-maximum load (think accelerating from a stop). Fuel economy is dictated more by driving style, which includes &lt;em&gt;minimizing&lt;/em&gt; acceleration rates, putting fuel &lt;em&gt;economy&lt;/em&gt; at odds with fuel &lt;em&gt;efficiency&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from that standpoint, if you were to drive a turbo-charged vehicle the same way you would drive the non-turbo counterpart, you would get the same accelerations at a lower throttle percentage. This would seem to imply better fuel economy at the cost of lower fuel efficiency.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-08-09T15:00:27.513" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11022" PostTypeId="2" ParentId="11000" CreationDate="2016-08-09T21:24:09.557" Score="3" Body="&lt;p&gt;The key thing here is that a  turbo extracts energy from the exhaust which would otherwise be lost to the environment, this means that it has the &lt;strong&gt;potential&lt;/strong&gt; to improve the thermal efficiency of an engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At its simplest a turbocharger uses a paired turbine and compressor to pressurise the air inducted into the cylinders. Normally aspirated engines just use atmospheric pressure to 'fill' the cylinder which costs a certain amount of engine power as the descending piston needs to 'suck' air through the inlet port via  ducting and filters, not a lot of power but a bit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition having a higher pressure air charge in the cylinder means a greater mass of oxygen is available so a greater mass of fuel can be burned or the same mass of fuel can be burned more completely so you can get more energy per cycle or extract more energy from the fuel. This is especially the case at higher RPM where the ability to get enough air into the combustion chamber in the time available becomes a limiting factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also means that you can get more power out of a smaller capacity engine block for relatively  small overheads in terms of extra ancillary equipment especially for moderate boost pressures.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that turbos can have a significant role to play in hybrid power trains, particularly as they can substantially increase the power to weight ratio of a given capacity engine at relatively constant RPM. It is also possible to use a turbo for electrical power input/output when connected to  motor/generator here the turbo can output surplus power to the electrical storage system and similarity electrical power can be imputed to keep the turbine spinning in its optimum range to reduce or eliminate turbo lag. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a real world example of this is current generation F1 engines which are now getting close to 1000bhp peak output with a thermal efficiency of around 50% (better than any IC road car by some margin) with a capacity of 1.6 litres (from which typical consumer engines generate about 100bhp) albeit at the expense of a service life measured in tens of hours. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-08-09T21:24:09.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11023" PostTypeId="2" ParentId="11001" CreationDate="2016-08-09T21:48:45.650" Score="0" Body="&lt;p&gt;I am not too sure of there reason behind using the transistors. I am sure there is a good reason, below is a another suggestion. Here below is a quick sketch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OsdwH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OsdwH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will need an additional inverter logic gate IC. I random picked &lt;a href=&quot;http://www.nxp.com/documents/data_sheet/74HC05.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;74HC05&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2016-08-09T21:48:45.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11024" PostTypeId="1" AcceptedAnswerId="11026" CreationDate="2016-08-10T00:05:22.390" Score="2" ViewCount="1753" Body="&lt;p&gt;What material is the viewing screen of a microwave made of?  I have been researching and all I find is that the holes on the screen are too small to let the bulky microwave radiation to pass through.&lt;/p&gt;&#xA;" OwnerUserId="7201" LastActivityDate="2016-08-10T06:19:38.973" Title="What is the screen of a microwave made off?" Tags="|materials|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11026" PostTypeId="2" ParentId="11024" CreationDate="2016-08-10T06:19:38.973" Score="2" Body="&lt;p&gt;The shielding in a microwave must be an electrically conductive material (usually thin steel), because it reflects microwaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The viewing window is also made of an electrically conductive material (thin steel), but has holes cut in it that are large enough to permit shorter wavelengths like light to pass through (so we can view it), but reflects long wavelengths like microwaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Microwave_oven#Design&quot; rel=&quot;nofollow&quot;&gt;Microwave Oven Design on Wiki&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-08-10T06:19:38.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11027" PostTypeId="2" ParentId="11013" CreationDate="2016-08-10T06:36:20.393" Score="2" Body="&lt;p&gt;Fans in series will increase the maximum effective pressure. If the filter is of high resistance this will result in an increase in flow for the system. If there is negligible resistance, like if the filter was taken out, one fan will have the same flow rate as 3 fans in series. There is a lot going on here that will be too much to explain in one question. I recommend reading on fan curves, maximum pressure and maximum flow rate if you would like to broaden your understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.zoro.com/dayton-axial-fan-115vac-4-1116in-h-4-1116in-w-4wt49/i/G2433523/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zoro.com&lt;/a&gt; does a good job of posting engineering specifications for the fans and blowers they sell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your product design; solder fans already exist. If you would rather build your own I recommend gleaning some information from their designs like the light weight filter they use or fan dimensions and flow rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B011J6JCN4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Soldering fan on amazon, 1&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0072HZ1IA&quot; rel=&quot;nofollow noreferrer&quot;&gt;Soldering fan on amazon, 2&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JmGUE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JmGUE.jpg&quot; alt=&quot;solder fan&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-08-10T06:36:20.393" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11028" PostTypeId="2" ParentId="10966" CreationDate="2016-08-10T06:40:45.823" Score="0" Body="&lt;p&gt;Different frequency ranges have different characteristics.  Using Phone frequencies is like shouting really really loud. Other phones are shouting too, so you  need to take turns shouting. But this does not really answer why you could not do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;WiFi is a different frequency band that has a much shorter range than GSM, it also penetrates objects less. This is a good thing because it means that the number of networks that overlap is smaller. Of you were to use Mobile ranges then the systems would have a pretty long reach which would make you see quite  many more personal stations. This would be a mess, since you would lose out on connection speed as you would need to share your space with more users.  You are  already seeing problems with WiFi in densely populated buildings with lots of home networks. Imagine multiplying that problem by including every station in a multi kilometer range. This can give you some idea why you would want to avoid using Mobile ranges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really convinced Bluetooth is needed. However it is designed to be very very cheap so it's easy to add to quite many places. It is also pretty low powered so that is a plus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second reason is the phone frequencies are owned by the phone company, in some cases they paid thousands of millions of dollars for the ownership. In essence using this range is like trespassing in a factory. The owners will not look at this very favorably. Simply it is illegal, you can end up having a lengthy and costly talk with government officials if you mess up on this space too much. Since you are using a wavelength that has quite considerable range your unlicensed use will be heard for several kilometers, the phone company will know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a third reason. Innovation and developmental reasons. If we all converged to using one standard then there would be a much slower development cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the end what you ask would be possible. Just not very practical, and more expensive for you as end user. It is also a question of how we have chosen to allocate resources, it may be that selling the ranges to private operators ultimately will be a bad thing.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorDisplayName="user748" LastEditDate="2016-10-03T03:03:44.897" LastActivityDate="2016-10-03T03:03:44.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11029" PostTypeId="2" ParentId="10972" CreationDate="2016-08-10T07:00:56.947" Score="3" Body="&lt;p&gt;Here are some options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Weight&lt;/strong&gt;- Reliable if the density is known. Bladder can be hung from a load cell, or placed on a surface supported by load cells, and volume can be derived continuously.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Bag in tank&lt;/strong&gt; - One easy way to calculate volume is to place an awkward shape in a tank of water and measure the increase in volume. Simply inflate the bladder submersed in water in a straight wall tank and an ultrasonic (or other) level transmitter will be able to accurately give you an continuous volume measurement. &lt;a href=&quot;https://en.wikipedia.org/wiki/Archimedes%27_principle&quot; rel=&quot;nofollow&quot;&gt;Archimedes' principle&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Open Loop Calibration Curve&lt;/strong&gt; - Reliable if the bladder is designed such that it always grows in a predictable fashion (not one side then the other). A proximity sensor or distance sensor (possibly ultrasonic or laser) would be used to target a location on the bladder that changes the most during fill and empty (center of a whoopie cushion for example). You would then take an empty measurement, record the distance, fill with 10ml,  record the distance, fill with 10 more ml, record the distance, and so on...  The cure you create can then be used in a computer or PLC to continuously derive your volume from the distance sensor input.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;3D Scanning&lt;/strong&gt; - I would only recommend this option if the others were not feasible, the geometry was unpredictable, and a continuous measurement was not needed. Multi-photograph reverse rendering, laser scanning, and LIDAR(as Inquisitive mentioned) are all possible options. The computer scans the object over a period of seconds or minutes then analytically computes the volume.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-08-10T07:00:56.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11030" PostTypeId="1" AcceptedAnswerId="11052" CreationDate="2016-08-10T08:40:45.043" Score="-1" ViewCount="736" Body="&lt;p&gt;What is the highest pressure possible today?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a sketchy idea of using a weather balloon with the lightest possible pressure vessel that can pack as much helium as possible and would like some pointers on where to lookup to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to have a weather balloon liftoff with a composite pressure vessel dynamically controlling/adjusting the gas ?&#xA;What would be the ideal pressure vessel in terms of highest pressure possible?&#xA;Weight of the vessel plus pumping the gas back in the cylinder in flight? Is it doable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;some numbers please&lt;/p&gt;&#xA;" OwnerUserId="7591" LastActivityDate="2016-08-11T19:08:54.447" Title="Highest pressure helium(lifting gas)can be stored at?" Tags="|fluid-mechanics|gas|pressure-vessel|" AnswerCount="1" CommentCount="1" ClosedDate="2016-08-12T23:07:21.367" ContentLicense="CC BY-SA 3.0" />
  <row Id="11031" PostTypeId="2" ParentId="11015" CreationDate="2016-08-10T10:33:47.300" Score="2" Body="&lt;p&gt;In principle the axial load in an element is related to its natural frequency. Think of it like a guitar string - increasing the tension in the string raises the pitch (frequency). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be possible to attach an accelerometer and then strike the element with a hammer to get it vibrating and measure the natural frequency (or even measure with the laser - if you can get displacement-time with reasonable resolution). If there is a sudden change in axial load (due to buckling) it may be possible to detect this through a change in natural frequency. It would be useful to do some tests before installing the element, or possibly FEA, to see if this is workable and what to expect to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will no doubt be many issues with practically doing this. Especially since in your case the length is not constant. I'm also not completely certain it will work. However, this sort of thing has been successfully tried before which you can read in this &lt;a href=&quot;http://eprints.unife.it/389/&quot; rel=&quot;nofollow&quot;&gt;thesis&lt;/a&gt; or this &lt;a href=&quot;https://www.newscientist.com/article/dn3679-bad-vibes-reveal-risk-of-rail-track-buckling/&quot; rel=&quot;nofollow&quot;&gt;article&lt;/a&gt; about evaluating the stress in rail tracks.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2016-08-10T10:44:13.880" LastActivityDate="2016-08-10T10:44:13.880" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11032" PostTypeId="1" CreationDate="2016-08-10T11:42:54.820" Score="1" ViewCount="92" Body="&lt;p&gt;hey guys I'm having trouble with a convolution problem here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A continuous-time linear system has the input/output relationship&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{-\infty}^t (t-\lambda-2)  x(\lambda) d{\lambda}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $1 ≤ t ≤ 2$, determine the output y(t) arising from the input $x(t) = u(t)−2u(t−1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the graph is going to be something like this right&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EJxZP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EJxZP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(probably ignore the axis labels it's wrong I know; I'm just providing a sketch for the $x(\lambda)$ function)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really confused about how to proceed&lt;/p&gt;&#xA;" OwnerUserId="7593" LastActivityDate="2016-08-10T11:42:54.820" Title="Signals and systems convolution problem?" Tags="|electrical-engineering|mathematics|signal|" AnswerCount="0" CommentCount="1" FavoriteCount="0" ClosedDate="2016-08-11T21:17:53.200" ContentLicense="CC BY-SA 3.0" />
  <row Id="11033" PostTypeId="1" CreationDate="2016-08-10T12:25:35.177" Score="-1" ViewCount="45" Body="&lt;p&gt;Recently, we experienced complete outages for around 8 hours, due to construction on a nearby road. We have decided to install a backup battery and solar array to prevent this from happening in the future.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question One&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Is this feasible? Would a gas generator be better? Would a battery be able to power a house with average power consumption in the US for one day, which would be about 33 kWh?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question Two&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Where would I be able to purchase such a battery, and a solar array that would fit the specifications above?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The battery must be at least 33 kWh in capacity&lt;/li&gt;&#xA;&lt;li&gt;This array, with zero outside power supply, should power an average US home for one day (24 hr) at average usage.&lt;/li&gt;&#xA;&lt;li&gt;This battery should be able to charge by solar panels, perhaps installed on the roof or on another place (location is not relevant).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7596" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-08-10T13:25:39.940" Title="What tools to use for a solar powered backup?" Tags="|electrical-engineering|battery|energy-storage|solar-energy|" AnswerCount="1" CommentCount="2" ClosedDate="2016-08-11T10:52:38.143" ContentLicense="CC BY-SA 3.0" />
  <row Id="11034" PostTypeId="2" ParentId="11033" CreationDate="2016-08-10T13:08:34.533" Score="2" Body="&lt;p&gt;It sounds like what you are after is increased energy security. You want a backup which would have a very low load factor - it wouldn't get much use each year. You want fast response, times of high power, and a decent amount of storage. But it sounds like you don't need absolute continuity of power - it's ok if power goes out for a few seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Almost all entities in your position that I'm acquainted with, use a diesel generator.&lt;/strong&gt; It would meet your needs, be very cheap to buy and own, and be fairly easy to maintain (you might need to run it for a few hours each month to keep it in decent shape). It will be smelly, noisy and environmentally damaging &lt;em&gt;in operation&lt;/em&gt;; but if you barely use it, that's much less of an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just need to spec it to be able to comfortably produce the maximum power draw you need - that might for example involve an electric shower, a kettle and a bunch of appliances on all at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted &lt;em&gt;absolute&lt;/em&gt; continuity of power, then a battery-powered UPS (uninterruptible power supply; it's charged from the mains power) for the critical kit would be an appropriate &lt;em&gt;addition&lt;/em&gt; to the diesel genset. It would not be cost-effective to have it for &lt;em&gt;all&lt;/em&gt; your kit: only for the kit that you absolutely need continuity of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PV is a completely different issue, tangential to your requirements. You might want it, for all sorts of reasons; but &lt;em&gt;not&lt;/em&gt; for the use-case in your question. &lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2016-08-10T13:25:39.940" LastActivityDate="2016-08-10T13:25:39.940" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11035" PostTypeId="2" ParentId="10978" CreationDate="2016-08-10T16:13:12.913" Score="1" Body="&lt;p&gt;I think the approach of Suba Thomas gives a good model: start with the sum of forces at the load and the sum of moments at the drum. Then determine the needed motor model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial motor model of chuck needs a stiff system where a single value for the moment of inertia can be calculated, while the goal of the model is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The interaction I'm interested in studying occurs when the PI controller is tuned to the anticipated load inertia $J$, which would be found with the motor, gearbox, drum, and load mass, but the system actually &quot;sees&quot; the springy-mass. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One note about the inertia in the drum moment equation of Suba Thomas: &#xA;Do not forget the inertia of the motor increased to the drum. Depending on the chosen motor, its influence can be significant. &#xA;So I would choose $J = J_{motor} * i^2 + J_{drum}$&lt;/p&gt;&#xA;" OwnerUserId="7590" LastEditorUserId="7590" LastEditDate="2016-08-10T16:18:55.947" LastActivityDate="2016-08-10T16:18:55.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11036" PostTypeId="1" AcceptedAnswerId="11039" CreationDate="2016-08-10T18:06:16.120" Score="3" ViewCount="39" Body="&lt;p&gt;Why do some variable speed motor switches have the &lt;em&gt;highest&lt;/em&gt; speed setting next to the &lt;em&gt;off&lt;/em&gt; setting instead of the &lt;em&gt;lowest&lt;/em&gt; speed setting next to the off setting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, most 3-speed household air-circulating / cooling fans will have 3 speeds with 3 being the highest. And the sequence will be: Off &gt; 3 &gt; 2 &gt; 1. Why isn't it Off &gt; 1 &gt; 2 &gt; 3? What design considerations are taken into account?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, most rheostat settings (e.g., light dimmer knobs/switches) have the dimmest setting closest to the off position.&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2016-08-11T12:35:53.033" LastActivityDate="2016-08-11T12:35:53.033" Title="Design decisions for variable speed motor switch settings (vs., say, rheostats)" Tags="|design|motors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11037" PostTypeId="1" AcceptedAnswerId="11045" CreationDate="2016-08-10T21:30:05.160" Score="4" ViewCount="180" Body="&lt;p&gt;The other day I was in the waiting room of a doctor's office in a new building as part of UCLA medical complex. A new construction as part of multi million dollar extension of existing medical complex in west Los Angeles.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I felt a small tremor and because of California's proximity to San Andreas fault immediately suspected an earth-quake! When the tremor did not go on as I expected and died out I realized it must have been caused by the impact of steps of another patient. I paid more attention and noticed the vibrations are very noticeable specially when the stride of the person resonates with natural frequency of the floor.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my question re. undesired vibration of beams and floors under foot traffic.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider we have a beam of steel WF 12x120 with 20 feet  span with I(xx) = 1070 inch^4.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the deflection under the foot of a man weighing 180 lbs.&lt;br&gt;&#xA;Lets assume he is wearing work-boots with negligible flexibility and when he lands on his foot he exert equivalent momentum as if he fell down 2 inches?&lt;br&gt;&#xA;And what is the vibration of this beam if we assume 5% damping?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a comparison static deflection of this beam under same load is. &#xA;0.001614 inch = $Wl^3/48EI$&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2016-08-11T17:14:46.467" LastActivityDate="2016-08-12T10:51:37.613" Title="Undesired vibration of a beam under foot traffic" Tags="|beam|vibration|deflection|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11038" PostTypeId="1" AcceptedAnswerId="11040" CreationDate="2016-08-11T01:14:26.430" Score="0" ViewCount="141" Body="&lt;p&gt;I have encountered a type of unit that I cannot understand. The unit describes the rate of permeation through a substance:(cm^3 . mm)/m^2 . d . atm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please explain how to interpret these units? Please attempt to leave the answer in simple - to - understand terms as chemistry, not engineering, is my primary field.&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2016-08-11T12:11:41.337" Title="Understanding Permeation Units" Tags="|pressure|measurements|gas|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11039" PostTypeId="2" ParentId="11036" CreationDate="2016-08-11T03:34:18.070" Score="3" Body="&lt;p&gt;An electric motor needs more current on startup to overcome inertia. While the low setting will likely start a brand new fan (example: unplugging it and plugging it back in on the low setting), it may not start an old fan that has more friction from worn bearings, etc. Even on a new fan, internal heat generated from this slow startup on the low setting may reduce the life of the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the friction was high and the low setting did not start the fan, the motor would burn up rather quickly because the current would stay high and there would be no cooling from rotation. If that same worn out fan was started on high it may have been fine because the current to maintain rpm on the low setting would be much less than startup and the cooling would be present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So essentially the Off &gt; 3 &gt; 2 &gt; 1 sequence is a very low cost way to ensure that 95% of the time the user toggles through the full power setting 3 to get to the lower power settings. This brief full power on startup helps get the motor to full speed quickly and increase the overall reliability of the product.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-08-11T03:34:18.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11040" PostTypeId="2" ParentId="11038" CreationDate="2016-08-11T03:46:08.900" Score="2" Body="&lt;p&gt;$$&#xA;\frac{\left(\mbox{amount of gas}\right)\left(\mbox{thickness of membrane}\right)}&#xA;{\left(\mbox{membrane area}\right)\left(\mbox{time}\right)\left(\mbox{differential pressure of gas}\right)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://onlinelibrary.wiley.com/doi/10.1002/app.1975.070190915/abstract&quot; rel=&quot;nofollow&quot;&gt;Units of gas permeability constants&lt;/a&gt; (source of the formula)&lt;br&gt;&#xA;&lt;a href=&quot;http://www.soarnol.com/eng/solution/solution040507.html&quot; rel=&quot;nofollow&quot;&gt;Gas Barrier Introduction&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://en.labthink.com/en-us/literatures/detailed-introduction-to-material-permeability-indexes.html&quot; rel=&quot;nofollow&quot;&gt;Introduction to Material Permeability Indexes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="1633" LastEditDate="2016-08-11T12:11:41.337" LastActivityDate="2016-08-11T12:11:41.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11042" PostTypeId="2" ParentId="11037" CreationDate="2016-08-11T04:14:47.787" Score="0" Body="&lt;p&gt;I am not an expert in these type calculations, but I do know it very much depends on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Natural_frequency&quot; rel=&quot;nofollow&quot;&gt;natural frequency&lt;/a&gt; of the floor as a system; not just assumptions on one of the supports. Typically knowing and avoiding the natural frequency at any energy input level is more important than knowing the energy inputs at various frequencies. It may be that the natural frequency of the floor system, at a particular location/area on the floor, coincided with the frequency of a person walking (~1 to 2Hz) in which case there would be very little dampening (as you may have witnessed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be addressed by adding additional mass to the floor in the problem location in order to increase inertia and lower the natural frequency. To determine how much mass to achieve a natural frequency of 0.5Hz for example will require experiment and/or some serious calculations and/or FEA analysis. Maybe that is why there are so many of those oddly placed giant potted plants in modern architecture ;-)&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-08-11T04:14:47.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11043" PostTypeId="1" CreationDate="2016-08-11T06:46:44.943" Score="0" ViewCount="55" Body="&lt;p&gt;I'm building a small microscope through which I need to pump fluid samples to image.  Currently I'm using a &lt;a href=&quot;http://ibidi.com/xtproducts/en/ibidi-Labware/Channel-Slides/m-Slide-I-Luer-Family&quot; rel=&quot;nofollow&quot;&gt;flow chamber from Ibidi&lt;/a&gt; that is perfect (length of a standard microscope slide x 5mm x 0.1mm), except that it's really not designed to be reused- one side of the channel is very thin material and cracks easily under pressure or vacuum, making it difficult to clean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also not looking to break the bank. The Ibidi channels are around $20 apiece, which would again be perfect if they were more durable for reuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="7602" LastEditorUserId="1832" LastEditDate="2016-08-11T10:44:09.803" LastActivityDate="2017-01-03T20:41:06.477" Title="Reusable 100um x 5mm flow chamber" Tags="|optics|microfluidics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11044" PostTypeId="1" AcceptedAnswerId="11066" CreationDate="2016-08-11T07:27:17.320" Score="1" ViewCount="143" Body="&lt;p&gt;Can someone please explain if the hypocycloid described below would degenerate to a line. If possible, can someone provide an actual plot of the entire question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scenario is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A circle starts rolling on a flat surface. After it completes one revolution, the path starts curving up in the arc of a circle till it becomes vertical and then the path becomes vertical. During the curved part of the path, the circle performs exactly half a revolution and then completes the other half of the revolution on the vertical path. Plot the locus of the point which was initially in contact the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;The problem is 2-D. Everything happens in a plane (I've copied the question exactly, except the original asks you to plot on paper). By vertical, it actually means upwards, parallel to the vertical edge of the paper and similarly for horizontal.&lt;/p&gt;&#xA;" OwnerUserId="7604" LastEditorUserId="7604" LastEditDate="2016-08-11T16:52:13.267" LastActivityDate="2016-08-13T10:42:00.013" Title="Hypocycloid degenerates to straight line" Tags="|drafting|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11045" PostTypeId="2" ParentId="11037" CreationDate="2016-08-11T11:26:05.143" Score="2" Body="&lt;p&gt;An exact theoretical analysis of this is rather involved so I will take an approximate approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can idealise the beam as a single degree of freedom (SDOF) dynamic system. The theory behind this can be found in a variety of dynamics textbooks. For example: Biggs - Introduction to Structural Dynamics (1964).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the beam has uniform mass and stiffness we can obtain a Load-mass factor $K_{LM} = 0.49$ for a point load in the center of a simply supported beam. This converts the mass and load of a distributed beam into an equivalent SDOF spring-mass system by assuming a mode shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the central point load it is straightforward to show that the stiffness $k = \frac{48EI}{L^3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then calculate the circular natural frequency of the equivalent SDOF system: &#xA;$$\omega_n = \sqrt{\frac{k}{K_{LM}M}}$$&#xA;where $M$ is the total mass of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the damped natural frequency:&#xA;$$\omega_d = \omega_n\sqrt{1-\zeta^2}$$&#xA;where, $\zeta$ is the damping ratio (in this case we take this as 0.05 since you specify 5% damping).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the situation you describe, we can consider the system as having an initial velocity, which we will determine by conservation of momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Velocity of the man when he impacts the beam:&#xA;$$v_f = \sqrt{2 a d} = \sqrt{2(32.2 \frac{ft}{s^2})(\frac{2}{12}ft)}=3.3\frac{ft}{s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If he 'sticks' to the beam the initial velocity of the beam-man system is found by conservation of momentum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_i (0.49 M+m)= v_f m \rightarrow v_i = v_f \frac{m}{m+0.49 M}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The displacement of a damped single degree of freedom system subject to an initial velocity is (see a dynamics text for derivation):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(t)=\frac{v_i e^{-\zeta \omega_n t}}{\omega_d} \sin{\left(\omega_d t\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so with&#xA;$$v_i = 3.3 \frac{ft}{s}\frac{180lb}{180lb+0.49 (120\frac{lb}{ft})(20ft)}=1.59 \frac{in}{s}$$&#xA;$$k = 48 \frac{EI}{L^3} = \frac{48(29000 ksi)(1070in^4)}{\left[(20ft)(12\frac{in}{ft})\right]^3}=108 \frac{kips}{in}$$&#xA;$$\omega_n = \sqrt{\frac{k}{K_{LM}M+m}}=\sqrt{\frac{108\frac{kips}{in}(386\frac{in}{s^2})}{0.49 (2400lb)+180lb}} = 175 \frac{1}{s}$$&#xA;$$\omega_d = \omega_n\sqrt{1-\zeta^2} =  175\sqrt{1-0.05^2}=174.8\frac{1}{s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and we can now calculate the beam mid-span displacement x(t):&#xA;$$ x(t)=\frac{v_i e^{-\zeta \omega_n t}}{\omega_d} \sin{\left(\omega_d t\right)} = \frac{1.59 \frac{in}{s} e^{-0.05 (175\frac{1}{s}) t}}{174.8\frac{1}{s}} \sin{\left(174.8\frac{1}{s} t\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The response is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6PKPa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6PKPa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequency of the loading, for example due to walking, will have a large effect on the response. Especially if the frequency of loading is close to the natural frequency of the element (&lt;a href=&quot;https://en.wikipedia.org/wiki/Resonance&quot; rel=&quot;nofollow noreferrer&quot;&gt;resonance&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For floor vibrations you may find the &lt;a href=&quot;http://www.steelconstruction.info/Floor_vibrations&quot; rel=&quot;nofollow noreferrer&quot;&gt;steelconstruction.info page on this topic&lt;/a&gt; interesting.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2016-08-12T10:51:37.613" LastActivityDate="2016-08-12T10:51:37.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11047" PostTypeId="1" AcceptedAnswerId="11050" CreationDate="2016-08-11T16:07:40.253" Score="0" ViewCount="740" Body="&lt;p&gt;Is it possible for a weather balloons filled with helium gas to lift the weight of itself and a helium tank(highest psi possible) which can fit the volume of the lifting gas in it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a typical weather ballon material is used as the ascending vehicle would the balloon with the helium pumped back in the tank fall down to the earths surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. In water the weight as well as the surface area for that weight determine its buoyancy - ships etc. A 5tons of dense block steel would sink in water but spread around over wide enough area it should float instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't the same thing apply to the difference between a helium gas filling the  larger volume of the balloon vs the high pressure compact volume storage vessel?&lt;/p&gt;&#xA;" OwnerUserId="7591" LastEditorUserId="7591" LastEditDate="2016-08-11T17:10:23.823" LastActivityDate="2016-08-11T17:19:35.677" Title="Will helium gas lift more weight in air when it is spread over large volume compared to equal weight heavily compressed small volume?" Tags="|aircraft-design|compressed-gases|pressure-vessel|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11048" PostTypeId="2" ParentId="10974" CreationDate="2016-08-11T16:47:45.623" Score="3" Body="&lt;p&gt;You can see this information if you enter &quot;field test mode&quot;. Google that for device specific instructions. It should show you the name or serial number of the base station you are connected to, but I'm not sure if you can use that to find it's physical location, although I haven't tried.&lt;/p&gt;&#xA;" OwnerUserId="7608" LastActivityDate="2016-08-11T16:47:45.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11049" PostTypeId="2" ParentId="10987" CreationDate="2016-08-11T17:00:03.427" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For a LTE base station is it possible to determine the relationship&#xA;  between the signal strength and distance?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ideally yes. Practically, probably not very reliably. Given the following pieces of information and assumptions, you can find the distance between any transmitter and receiver:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The transmitting antenna's radiation pattern and orientation (gain)&lt;/li&gt;&#xA;&lt;li&gt;The receiving antenna's radiation pattern and orientation (gain)&lt;/li&gt;&#xA;&lt;li&gt;The transmitter's power&lt;/li&gt;&#xA;&lt;li&gt;An accurate measure of received power&lt;/li&gt;&#xA;&lt;li&gt;Assume no obstructions (as you have)&lt;/li&gt;&#xA;&lt;li&gt;Frequency of signal&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are the terms needed in &lt;a href=&quot;https://en.wikipedia.org/wiki/Friis_transmission_equation&quot; rel=&quot;noreferrer&quot;&gt;the Friis Transmission equation&lt;/a&gt; to solve for Distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot you show is likely a gain plot along the horizontal plane in dB relative to an &lt;a href=&quot;https://en.wikipedia.org/wiki/Isotropic_radiator&quot; rel=&quot;noreferrer&quot;&gt;isotropic antenna&lt;/a&gt; (equal power in every direction). Every 3 dB represents a doubling of gain (and hence energy) transmitted in that direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, you can find a 'rough' measure of your phones received power by entering field test mode (google for instructions). The rest of the information is going to be tough to come by.&lt;/p&gt;&#xA;" OwnerUserId="7608" LastActivityDate="2016-08-11T17:00:03.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11050" PostTypeId="2" ParentId="11047" CreationDate="2016-08-11T17:19:35.677" Score="2" Body="&lt;p&gt;As a reality check lets consider industrial gas tanks as the specifications are easily available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A BOC size D hydrogen tank has a gross weight of 57kg and contains 23kg of hydrogen at 230bar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At atmospheric pressure hydrogen has a density of 0.09 kg per cubic metre so you get about 255 cubic metres of gas at atmospheric pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give lift of about 250 kg, so substantially more than the gross weight of the tank. You could also improve on this a bit by improving the performance of the cylinder eg a composite tank could significantly reduce the 'dead' weight of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is in compressing the gas back into the tank which would require a high pressure compressor plus a power supply to run it. This is probably not entirely beyond the realms of possibility but is going to eat substantially into the payload and probably isn't really worth it as the only benefit of this system compared to just venting the gas is saving the cost of the gas itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the short answer is that yes it is probably possible (albeit likely very expensive) but there aren't any real advantages in doing it this way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking at something a bit larger and more sophisticated like a passenger airship then, maybe as you already have an on-board power supply and the overheads of the compressors and tanks may be worthwhile, but it would take a fairly serious feasibility study of an actual design concept to say for sure. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-08-11T17:19:35.677" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11051" PostTypeId="1" AcceptedAnswerId="11055" CreationDate="2016-08-11T18:03:27.203" Score="0" ViewCount="357" Body="&lt;p&gt;So the below question might sound like a homework question but in reality it is an oversimplified explanation of what I'm trying to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we have a perfect machine that pushes out 600cfm into a box. On the other end of the box is another machine that only consumes 300cfm from the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the pressure in the box and how do I calculate that? I've looked into Boyle's Law and something just doesn't seem right about the results I'm getting so I'm trying to make sure that this is the right law I should be looking into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's relevant, assume a volume of 360in&lt;sup&gt;3&lt;/sup&gt; and temperature of 310&amp;nbsp;K.&lt;/p&gt;&#xA;" OwnerUserId="4325" LastEditorUserId="1832" LastEditDate="2016-08-11T18:39:18.620" LastActivityDate="2016-08-14T00:16:58.433" Title="Relationship between Pressure and volume with Air" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11052" PostTypeId="2" ParentId="11030" CreationDate="2016-08-11T19:01:24.020" Score="0" Body="&lt;p&gt;It sounds like what you really want to know is what is the best ratio of mass of gas to mass of container. Pressure is part of this but by no means the whole story. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Industrial gasses are typically stored in portable tanks at around 250 bar, although higher pressures of 300 bar or more are becoming more common. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These cylinders are typically mass produced in steel and are designed for long lives in under fairy adverse conditions and as such have quite generous factors of safety. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind you could certainly get better performance by using higher performance materials and trimming safety factors in favour of more stringent quality control. Also, while cylindrical tanks are the most convenient shape for industrial use they aren't the most efficient in terms of rated pressure to weight ratio. For example toroidal composite tanks are a lot lighter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no overriding reason to go for &lt;em&gt;very&lt;/em&gt; high pressures in this context as the mass of the gas is the same however much you compress it. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-08-11T19:08:54.447" LastActivityDate="2016-08-11T19:08:54.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11054" PostTypeId="2" ParentId="11051" CreationDate="2016-08-12T00:53:09.180" Score="1" Body="&lt;p&gt;600 - 300 = 300 cfm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are pushing 300 cubic feet per minute into a 0.20833 cubic foot (or 5.89925 liter) box. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the resulting pressure of 300 cubic feet per minute into a 0.20833 cubic foot box? Assuming that the pressure is an unstoppable force, going into an indestructible box, the pressure will increase variably and periodically until you have a solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, at first, generally, gasses will compress into liquids, once the liquid state is reached it becomes &quot;incompressible&quot;... so the pressure will then escalate  quickly. For simplicity, lets just use nitrogen as a guinnea pig. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposing that you can perfectly remove heat from the system to maintain 310 kelvin (98.33 °F?), as nitrogen becomes liquid the pressure will be maintained about 45,000 psi (give or take a few hundred psi, until the box is full). For understanding, consider how the boiling point of water is maintained until it is all gone- the same thing is true of the boiling point of nitrogen- the pressure is also maintained (until the box is full, since you are reversing the boiling process).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mole of gas is 22.4L, so you are compressing about 13.4 moles per minute into 5.9 liters. The density of liquid nitrogen is 0.807 kg/L... and one mole (of N2) weighs about 28 grams. So, 5.9 L x 0.807 kg tells us the net weight of the full box of liquid nitrogen... about 4.76 kgs. 4.76kg / 28 g tells us how many moles of liquid N2 will fit in the box. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 170 moles of liquid N2 will fit in the box... how long do we have before it's full? At 13.4 moles per minute, you have about 12 minutes and 41.2 seconds before the pressure really escalates. Then, liquid nitrogen will compress into solid nitrogen. Making all the same assumptions, at 98 °F, the pressure will rise sharply to about 75,000 psi, and there it will stay pretty steady until the box is full of solid N2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid nitrogen weighs 1.027 kg per liter. Simply stated, the proportional difference in density will tell us the proportional volume left in the vessel. So 100 x (1.027 - 0.807) / 1.027 = the percentage of room left in the vessel (before it becomes full of solid N2). 21.42% of 5.9 L is 1.264 liters. 1.264 L x 1.027 kg per liter tells us how many kilograms of solid nitrogen we can add before the box is full of solid nitrogen. 1.3 kg of solid nitrogen / 28 g per mol is 46.43 moles (neglecting rounding errors). At 13.4 moles per minute being added to the box, you will have about 3 minutes and 28 seconds before essentially the box is full of solid N2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point it is even more incompressible and the pressure will be applied toward the nuclear fusion of nitrogen. Now I don't know the pressure at which nitrogen will fuse, but for what it's worth you are talking about binding energies... and there's talk of it on the physics SE. In a practical way, the binding energy is measurably the difference in mass between the atomic number and the representative isotope. Remember, mass is energy. For example, the energy (by weight) of N14 is actually 14.0030740052 g/mol. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the binding energy is something like 7.7 MeV or about 1.23368e-12 joules per nitrogen atom and you have about 206.5 moles of solid N2; that's about 2.5x10^26 atoms. And that's about 3x10^14 joules - holding a box of nitrogen together. For silicon, the binding energy is 8.45 MeV (1.3534777464e-12 joules/atom). Now, since 2 nitrogen atoms will fuse (in my mind) to become 1 silicon atom, that's half the atoms. So, 14.4 MeV - 8.45 MeV is how much energy is absorbed turning one turning 2 atoms of N into 1 atom of Si; and you now have 1.25x10^26 atoms- that's 9.53295e-13 joules x 1.25x10^26. So, you need 1.2x10^14 joules to turn 5.9L of solid N2 into silicon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for convenience let's totally bend physics to our whim and say one joule equals 0.003245 psi. So, 1.2x10^14 / 0.003245 equals the total psi before nitrogen &quot;alchemizes&quot; into silicon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 3.67x10^16 psi, you will have silicon. How long will this take? Well I'm not sure, but we can be sure that the pressure going from a liquid to solid went up by about 30,000 psi in probably well under a minute. So, assuming that it's pretty close to the &quot;incompression rate/limit&quot; let's just pick a nice easy figure like 100,000 psi per minute. Now, 3.67x10^16 / 1x10^5 = 3.67x10^11 minutes... or about 700,000 years (give or take maybe 300,000 years). But this is really messy because some of it will become silicon but some of it won't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming nitrogen is &lt;em&gt;only&lt;/em&gt; becoming silicon, the density of silicon is 2.3296 kg/L. Now, you have to fill about half of the box with silicon and don't forget that after another say 700,000 years nitrogen is once again being reduced to about half it's volume. Eventually all of the silicon will fuse to become nickel. I think it's safe to say the amount of energy or psi for this has never actually been measured... so I will just leave the rest of these calculations alone. The density of nickel is  8.912 kg/L. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I just want to make mention of this because it seems interesting to note that nickel should fuse to become barium, but as the density of barium is only 3.51 kg/L (which is only slightly greater than than 1/2 of 8.912)- at this point you will convert nickel to barium without much leftover volume in the box! You may proceed relatively quickly (remember the void turns to silicon then nickel and then barium) at this step and move right on to Copernicium (aka ununbium). The density of Copernicum is unknown, but it should be about 23.7 kg/L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that, I just don't know. Maybe it becomes a black hole or something. But to really answer the question, the limit of the pressure might be 100,000 psi per minute (when compressing a solid). Otherwise it will be the vapor pressure of the liquid or the melting pressure of the solid (perhaps for a few minutes, I guess &lt;em&gt;that&lt;/em&gt; depends on the substance).&lt;/p&gt;&#xA;" OwnerUserId="7610" LastEditorUserId="7610" LastEditDate="2016-08-12T01:12:05.057" LastActivityDate="2016-08-12T01:12:05.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11055" PostTypeId="2" ParentId="11051" CreationDate="2016-08-12T02:09:58.383" Score="0" Body="&lt;p&gt;You can use the ideal gas law at first for a rough estimate.&#xA;Assuming we start at atmospheric pressure, and the flow rate is in standard cubic feet per minute:&#xA;$$P = 14.7 psia$$&#xA;$$V = 360 in^3 = 0.2083 ft^3$$&#xA;$$R = 10.731 \frac{ft^3*psi}{R*lbmol}$$&#xA;$$T = 310K = 558^\circ R$$&#xA;Starting point in the box at time = 0:&#xA;$$n=\frac{PV}{RT}=0.0005115 mol$$&#xA;Flow rate for 300 cfm converted to moles per second:&#xA;$$\frac{14.7 psi * (\frac{300 cfm}{60s/min})}{10.731}*558^\circ R = 0.01227 mol/s$$&#xA;After one second of pushing 0.01227 moles of air into a box that already contains 0.0005 moles of air:&#xA;$$n = 0.0128 mol$$&#xA;$$P = \frac{0.0128 mol * 10.731 * 558R}{0.208 ft^3} = 367.5 psi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation, the box fills up so fast that the outlet flow would increase drastically to a certain point until you either deadhead the blower or get a box failure. Or as @Ben Welborn suggested, you may create nuclear fusion resulting in a wormhole between the engineering SE and physics SE&lt;/p&gt;&#xA;" OwnerUserId="3569" LastEditorUserId="3569" LastEditDate="2016-08-14T00:16:58.433" LastActivityDate="2016-08-14T00:16:58.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11057" PostTypeId="1" CreationDate="2016-08-12T03:30:24.047" Score="2" ViewCount="87" Body="&lt;p&gt;If experimental data is obtained from a proportional solenoid valve, specifically flow vs current (assuming inlet pressure is constant), the fit is nearly quadratic in nature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Andersen's &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/1575241641&quot; rel=&quot;nofollow&quot;&gt;Analysis and Design of Pneumatic Systems&lt;/a&gt; predict that the flow area of an orifice is linearly proportional to the flow rate for constant inlet pressure. But for a valve, the magnetic characteristics must also contribute a relationship between applied current and displacement of the valve poppet which changes flow area of the orifice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the electromagnetic element the main factor leading to the quadratic flow vs current relation, or are other factors involved?&lt;/p&gt;&#xA;" OwnerUserId="820" LastEditorUserId="820" LastEditDate="2016-08-12T14:23:00.627" LastActivityDate="2016-08-12T14:23:00.627" Title="What specific characteristic of proportional solenoid valves makes the flow to current behavior quadratic?" Tags="|modeling|valves|flow-control|electromagnetism|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11059" PostTypeId="1" AcceptedAnswerId="11061" CreationDate="2016-08-12T12:02:18.813" Score="1" ViewCount="165" Body="&lt;p&gt;Someone told me to avoid over-sizing when using DC Motors. Why? Is this because of sparking and higher wear (brushes) while the motor is running at a lower power rating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain?&lt;/p&gt;&#xA;" OwnerUserId="5007" LastEditorUserId="7083" LastEditDate="2016-08-20T17:52:52.233" LastActivityDate="2016-08-20T17:52:52.233" Title="Why to avoid over sizing a DC Motor?" Tags="|electrical-engineering|motors|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11060" PostTypeId="2" ParentId="11059" CreationDate="2016-08-12T14:34:01.513" Score="2" Body="&lt;p&gt;Overdimensioning is never good because there is a problem called, &quot;tolerance stack up&quot;. &lt;a href=&quot;https://en.wikipedia.org/wiki/Tolerance_analysis&quot; rel=&quot;nofollow&quot;&gt;Tolerance stack up&lt;/a&gt; is essentially predicament where you consider way too much tolerances and thus throwing everything out of proportion. In other words, too much modification can create a large amount of ambiguity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if I wanted to create a particular amount of magnetism in the motor, I have change &quot;x&quot;. Okay, now that I have changed &quot;x&quot;, I now need to change &quot;y&quot; because &quot;x&quot; changed. Well now I have to change &quot;z&quot; because of &quot;y&quot;, et cetera, et cetera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why we learn about optimization in Calculus and Matrix Algebra in order to combat that problem :)&lt;/p&gt;&#xA;" OwnerDisplayName="user7211" LastActivityDate="2016-08-12T14:34:01.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11061" PostTypeId="2" ParentId="11059" CreationDate="2016-08-12T15:49:09.110" Score="4" Body="&lt;p&gt;This answer assumes that you meant to ask about &lt;em&gt;over-sizing&lt;/em&gt; a DC motor rather than &lt;em&gt;over-dimensioning&lt;/em&gt; it. over-dimensioning is when you provide too many dimensions such that they might be in conflict with each other based on drafting errors, rounding, or tolerances. For example if you draw a 3 story building and indicate that each floor is to be 10 feet tall, but you also dimension that the overall building must be 31 feet tall, nobody would know where to put the extra foot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over-sizing is when you pick an item that is bigger than you theoretically need. For example if structural analysis indicates that you need a 12&quot; tall beam, but you decide to specify a 14&quot; beam to be safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over-sizing motors is common - it is typical in the mechanical world to specify a motor 25 - 50% larger than your calculations show you need as a rule of thumb. Over-sizing motors is useful because there could be some variation in the load the motor supports. A pump may have to run a little faster than it was designed to, or a bearing could be improperly lubricated, increasing friction, for example. In this case, it is desirable for the motor to be able to function even though conditions are not exactly as designed. It is also usually costly to replace a motor with a larger motor, so the relatively small cost of buying a larger motor in the first place is usually worth it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also some down sides to over-sizing motors though, which is why it's unusual to see a motor 100% or 200% bigger than it theoretically needs to be. One obvious issue is cost. Bigger motors are more expensive and buying a bigger motor than you ever need means spending money that didn't have to be spent. There's also an issue with efficiency. Most decently sized motors are more efficient in the top half of their speed range. If they are used at very low speeds, they will waste more energy as heat. This is compounded by the fact that there is more rotating mass in the motor, so it takes more energy to start and stop the machine. In addition, some larger fans have cooling fans attached to the motor shaft to blow cool air across the motor, preventing overheating. At very low speeds, this fan does not move quickly enough to provide cooling air, so the motor is more likely to overheat which will damage the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other wrinkle is that variable speed controllers have limited resolution, and if your useful range for the motor is all in the lowest 10% of its speed range, your controller might not be able to regulate speed as carefully as you need it to. This is a constraint that can be designed around, but adds complexity and limitations to your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So motor size is always a trade-off between the risks of having a motor that is too small to move the load and a motor that is too big and wastes power or overheats. Most industries have settled in the 25 - 50% range but there are certainly valid exceptions where efficiency or reliability are strongly controlling factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.farnell.com/datasheets/62656.pdf&quot; rel=&quot;nofollow&quot;&gt;This Farnell datasheet&lt;/a&gt; has a good overview of these principles, and most motor vendors also have tools or guides that will help you with motor sizing. Some even publish efficiency curves for their motors.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-08-12T15:49:09.110" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11066" PostTypeId="2" ParentId="11044" CreationDate="2016-08-13T10:27:32.207" Score="0" Body="&lt;p&gt;I discovered that this indeed hoes happen. Since radius of the path is twice of the disc, the locus is a straight line there. Used GeoGebra for the purpose!&#xA;&lt;a href=&quot;https://s10.postimg.org/f340jevft/S2_ED_6.gif&quot; rel=&quot;nofollow&quot;&gt;Animation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7604" LastEditorUserId="7604" LastEditDate="2016-08-13T10:42:00.013" LastActivityDate="2016-08-13T10:42:00.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11068" PostTypeId="2" ParentId="10661" CreationDate="2016-08-13T13:59:25.620" Score="2" Body="&lt;p&gt;The most straightforward solution to this problem is to drill the holes in the second plate at the time of assembly, using the first plate as a template. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that it really does sound like there is a manufacturing issue which needs to be addressed if you have such large tolerance on a part this small. Even with the most basic manufacturing process I can't imagine that this can't be improved on without undue cost. Even just making a template for marking the hole centres with a marker pen should improve matters substantially, bearing in mind that having to drill 12 mm clearance holes for an M6 fastener and having to obtain and use oversized washers is going to add a significant amount of faff to the whole process. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-08-13T13:59:25.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11077" PostTypeId="1" CreationDate="2016-08-14T01:55:24.540" Score="1" ViewCount="56" Body="&lt;p&gt;How do I convert $\dfrac{\text{cm}^3\cdot\text{mm}}{\text{m}^2 \cdot \text{day} \cdot \text{atm}}$ for permeation of gas through EVOH to molecules of gas permeating per second as a function of thickness of EVOH?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, I am trying to use the permeation units to determine how much a container will be emptied of gas via permeation after a given amount of time.&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="9" LastEditDate="2016-08-14T15:42:15.573" LastActivityDate="2016-08-14T15:42:15.573" Title="How to convert permeation units to molecules/second?" Tags="|pressure|measurements|gas|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11079" PostTypeId="1" CreationDate="2016-08-14T04:12:44.213" Score="2" ViewCount="61" Body="&lt;p&gt;I stop receiving TV signals whenever their is cloud cover outside. On the other hand I have never had any such problems with my internet connection or GPS system. I cannot understand why and how often (or easily) these different technologies get affected by bad weather conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question(s) would be:    &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How often and why does satellite broadcasting get affected due to weather conditions?&lt;/li&gt;&#xA;&lt;li&gt;How often and why does point-to-point (using satellites) get affected?&lt;/li&gt;&#xA;&lt;li&gt;How often and why does point-to-point (using line of sight) get affected?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By how often I mean how much worse weather conditions have to be for communication to be affected.&lt;/p&gt;&#xA;" OwnerUserId="7633" LastEditorUserId="1832" LastEditDate="2016-08-14T04:18:46.233" LastActivityDate="2016-08-14T04:18:46.233" Title="Effect of bad weather on P2P Communication or Broadcasting?" Tags="|telecommunication|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11080" PostTypeId="1" CreationDate="2016-08-14T12:44:33.533" Score="4" ViewCount="917" Body="&lt;p&gt;I have two pipes, blue &amp;amp; yellow wherein blue is wider than yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need something (lock) that will prevent the yellow pipe from falling if the lock is tight, and at the same time is able to loosen so that I can adjust the yellow pipe up or down whenever I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I will hang under the yellow pipe is &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/5/57/Logitech-usb-speakers.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;computer speaker&lt;/a&gt;, one in my left and right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nM7Qu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nM7Qu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speaker Weight: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Left: 239g (8.4oz), Right: 343g (12.1oz)&lt;/p&gt;&#xA;" OwnerUserId="3252" LastEditorUserId="3252" LastEditDate="2016-08-15T06:59:22.970" LastActivityDate="2016-08-17T12:54:36.153" Title="Where can I find a lock that can do this?" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11081" PostTypeId="2" ParentId="11080" CreationDate="2016-08-14T15:23:19.910" Score="1" Body="&lt;p&gt;If you don't mind only being able to adjust the pipe in steps, the locks that can often be seen on steel tube scaffolding or medical crutches might be a perfect solution to this problem! Here's a picture of what I'm talking about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xFIiI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xFIiI.jpg&quot; alt=&quot;https://neeraliparbhu.files.wordpress.com/2012/09/img_8789.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, if the pipe isn't too thin, you might try cutting a thread into the pipes. Unfortunately, these may loosen over time if the fit is too loose or the pitch is too large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also just drill a hole into the side of the outer pipe and put a wingnut through it, this will allow you to adjust on the fly. I don't like these very much because they tend to slip unless you really tighten them up, but then you might crush the pipe on the inside. If the speaker is fairly small, however, it might work.&lt;/p&gt;&#xA;" OwnerUserId="7465" LastEditorUserId="7468" LastEditDate="2016-08-17T09:23:42.063" LastActivityDate="2016-08-17T09:23:42.063" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11082" PostTypeId="1" CreationDate="2016-08-14T17:28:01.087" Score="2" ViewCount="130" Body="&lt;p&gt;I was going through some literature in power electronics relating to sliding mode control. I noticed that they have separately dealt with the reaching condition and then the existence condition for the sliding mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: Doesn't it imply that if the reaching condition is met, the existence condition is also automatically met?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $s$ is the sliding surface, Reaching Condition: $s\dot{s} \leq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Existence Condition: $\lim\limits_{s\to0} s\dot{s} \leq 0$&lt;/p&gt;&#xA;" OwnerUserId="7638" LastEditorUserId="3633" LastEditDate="2016-08-17T09:40:51.013" LastActivityDate="2016-08-17T09:40:51.013" Title="Sliding Mode Control" Tags="|control-engineering|control-theory|power-electronics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11083" PostTypeId="1" AcceptedAnswerId="11086" CreationDate="2016-08-14T21:12:11.847" Score="2" ViewCount="7129" Body="&lt;p&gt;Suppose I have three 2D drawings prepared in AutoCAD LT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That 2D drawings are supposed to represent &lt;em&gt;top&lt;/em&gt;, &lt;em&gt;left&lt;/em&gt; and &lt;em&gt;front&lt;/em&gt; projections of the 3D solid body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I view that 3D solid body?&lt;/p&gt;&#xA;" OwnerUserId="4670" LastEditorUserId="1293" LastEditDate="2016-08-15T07:28:13.397" LastActivityDate="2016-08-15T10:08:55.123" Title="Convert 2D projections to 3D View in AutoCAD" Tags="|drafting|autocad|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11085" PostTypeId="1" CreationDate="2016-08-15T02:11:28.943" Score="3" ViewCount="126" Body="&lt;p&gt;I am doing a robotics project. The goal is to make an autonomous car start from one point and stop at another point around 10 +-1 meters away. I need to make a steering mechanism. I want to be as close as 3 cm away from the point, so GPS probably won't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My ideas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I cannot have a physical line taped to the floor, so I can't use line following robotics code. But, could I take some sort of electronic device and hover it above the point then remove the electronic device and have the car go straight to where the electronic device was? How would I do this and what would I use?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have already thought of rotary encoders, but I still need a way to keep the car dead straight for 10-ish meters. My current RC car veers about 20 cm either way. Could I used a gyroscope/accelerometer/imu to help keep the car on a straight line? How would I do this/are the gyro/accel sensors sensitive to notice super fine changes in angle? How will I line the car up perfectly straight?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please let me know your thoughts! &lt;/p&gt;&#xA;" OwnerUserId="7642" LastEditorUserId="7083" LastEditDate="2016-08-20T17:53:09.727" LastActivityDate="2016-08-20T17:53:09.727" Title="Autonomous RC car trying to stop at a point" Tags="|mechanical-engineering|electrical-engineering|control-engineering|automotive-engineering|sensors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11086" PostTypeId="2" ParentId="11083" CreationDate="2016-08-15T06:55:46.183" Score="3" Body="&lt;p&gt;I suggest you to watch these videos first:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=ZFSV47I7SBg&quot; rel=&quot;nofollow&quot;&gt;AutoCAD 2D to 3D Conversion Trick&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://knowledge.autodesk.com/support/autocad/learn-explore/caas/CloudHelp/cloudhelp/2016/ENU/AutoCAD-Core/files/GUID-ECFB3220-6484-4D0B-BB7E-B06AD9F4E856-htm.html&quot; rel=&quot;nofollow&quot;&gt;Learn how to take 2D drawing designs and ideas and turn them into 3D objects.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;http://acronymonline.org/2d-flat-geometry-3d-autocad/&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Also:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to Convert a 2D Model to 3D Using AutoCAD&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;by Darrin Koltow, studioD, &lt;code&gt;http://smallbusiness.chron.com/convert-2d-model-3d-using-autocad-31833.html&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Use AutoCAD's &quot;Extrude&quot; and &quot;Revolve&quot; commands to turn 2-D designs into 3-D models. &quot;Extrude&quot; is how 3-D modelers to refer to the technique of stretching a 2-D shape into 3-D space. AutoCAD performs this expansion by first extending a new axis at right angles to the 2-D axes on which your 2-D design sits. It then makes a copy of the 2-D shapes you are expanding at a higher location on the axis, while keeping the original shapes at the base of the axis. After converting your 2-D design to 3-D, display it with realistic lighting and shadows using the &quot;Render&quot; command on the Render tab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click the &quot;File&quot; menu and select the &quot;Open&quot; command from the context menu. Navigate to an AutoCAD file with a 2-D model that you want to convert to 3-D and double-click on it. AutoCAD will load the file for you to convert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type &quot;Perspective 1&quot; to indicate that you want to view your design with perspective, which means that parallel lines will appear to converge as they do in the physical world. This option is more realistic than parallel projection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click the “cube” icon at the top right of the canvas and then drag the mouse until the top, right and front sides of the cube are visible. This changes the viewpoint from 2-D to 3-D, allowing you to see the three-dimensionality of the 3-D form that you'll create from the 2-D model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click the drop-down list at the top of the application window, then click the &quot;3-D modeling&quot; item to change the workspace to one that displays tools for creating and editing 3-D objects. Click the &quot;Home&quot; tab to access the &quot;Modeling&quot; panel, which displays the &quot;Extrude&quot; tool for converting rectangular 2-D shapes into block-like 3-D forms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click the &quot;Extrude&quot; button; then click a shape of your 2-D model that you want to turn into a block form, as opposed to a cylindrical or spherical form. Press &quot;Enter.&quot; The shape you selected will expand into 3-D space and the top of the box formed from the shape will stick to your cursor. Drag the cursor until the box reaches the height you want, and then click the mouse to end the extrusion. Use the &quot;Extrude&quot; command on the remaining parts of the 2-D model that you want to expand into blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 6&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click the &quot;Surface&quot; tab and go to the &quot;Create&quot; panel to find the &quot;Revolve&quot; command, which makes cylindrical, spherical and other round forms from 2-D shapes. Click this command, and then click a shape in your 2-D model that you want to convert to a cylinder or other round form. Type the axis around which you want to revolve the shape, which can be &quot;x,&quot; &quot;y,&quot; or &quot;z.&quot; Press &quot;Enter&quot; to perform the revolution, then revolve the other parts of your design that you want to have round 3-D forms to complete the conversion of your 2-D design to a 3-D model.&lt;/p&gt;&#xA;" OwnerUserId="7468" LastEditorUserId="7468" LastEditDate="2016-08-15T10:08:55.123" LastActivityDate="2016-08-15T10:08:55.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11087" PostTypeId="2" ParentId="11085" CreationDate="2016-08-15T11:58:57.527" Score="1" Body="&lt;p&gt;Another possible solution is to use a Magnetometer. This will help you keep a Autonomous steer the vehicle with reference to magnetic north.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can use IR sensor to look obstacles in the front or rear to navigate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also look at &lt;a href=&quot;http://users.ece.utexas.edu/~valvano/Volume1/E-Book/C14_ADCdataAcquisition.htm#14_5&quot; rel=&quot;nofollow&quot;&gt;Robot Car Controller&lt;/a&gt;. You will find some good ides to get started&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you have imagined you can use a gyroscope/accelerometer/imu. The youtube video &lt;a href=&quot;https://www.youtube.com/watch?v=6ZFy4zMGLgg&quot; rel=&quot;nofollow&quot;&gt;TIVA - ADXL345 Accelerometer Demo Using I2C&lt;/a&gt; shows demo accelerometer use case. You can find link to the complete source code. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sparkfun.com/products/10530&quot; rel=&quot;nofollow&quot;&gt;SparkFun Triple Axis Magnetometer Breakout - HMC5883L&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://github.com/mahengunawardena/Tiva_I2C_Nokia_ADXL345/blob/master/Tiva_i2c.c&quot; rel=&quot;nofollow&quot;&gt;Tiva_I2C_Nokia_ADXL345&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-08-20T11:23:18.887" LastActivityDate="2016-08-20T11:23:18.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11088" PostTypeId="2" ParentId="11080" CreationDate="2016-08-15T13:29:22.177" Score="4" Body="&lt;p&gt;For this problem, I think that the way a bike saddle works would be ideal. You have a slit in the outer pipe, allowing it to bend inwards slightly when tightened to grip the inner pipe. Then you have an adjustable tension ring around the outer pipe, with a quick-release lever to grip with an appropriate pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IHhuU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IHhuU.png&quot; alt=&quot;Diagram of saddle with adjustment mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can buy these clamps separately, so it should just be a case of finding the right size of pipes and clamps. A Google search for &lt;a href=&quot;https://www.google.co.uk/search?q=seat+post+quick+release+clamp&quot; rel=&quot;nofollow noreferrer&quot;&gt;seat post quick release clamp&lt;/a&gt; is a good place to start.&lt;/p&gt;&#xA;" OwnerUserId="95" LastActivityDate="2016-08-15T13:29:22.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11089" PostTypeId="1" AcceptedAnswerId="11090" CreationDate="2016-08-15T13:47:29.620" Score="6" ViewCount="237" Body="&lt;p&gt;I realise that &quot;bomb&quot; can refer to all sorts explosives, but I am mainly interested in air-dropped bombs. Ever since reading about the &lt;a href=&quot;https://en.wikipedia.org/wiki/1967_USS_Forrestal_fire&quot;&gt;USS Forrestal fire&lt;/a&gt;, I wondered about the safety of bombs when stored or transported in general- but in plain English: suppose I were able to hold a bomb in my hands- say, the B61 Nuclear bomb or the GBU-10 Paveway II- would it be okay to drop it on the ground? How often could I drop it before it damages the bomb? Is there a minimum speed/impact required that could be reached without using an aircraft?&lt;/p&gt;&#xA;" OwnerDisplayName="user7646" LastActivityDate="2016-08-17T17:16:05.297" Title="Is it safe to drop a bomb?" Tags="|mechanical-engineering|safety|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11090" PostTypeId="2" ParentId="11089" CreationDate="2016-08-15T15:11:47.157" Score="10" Body="&lt;p&gt;In general a modern commercial/military explosive device will not explode if dropped. The most common commercial/military explosives such as &lt;a href=&quot;https://en.wikipedia.org/wiki/C-4_(explosive)&quot;&gt;C4&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Trinitrotoluene&quot;&gt;TNT&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/ANFO&quot;&gt;ANFO&lt;/a&gt; are catagorised either as &lt;a href=&quot;https://en.wikipedia.org/wiki/Explosive_material#Secondary_explosive&quot;&gt;Secondary&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Explosive_material#Tertiary_explosive&quot;&gt;Tertiary&lt;/a&gt; explosives, which means they are fairly insensitive to shock/fire and require a severe shock (usually from a &lt;a href=&quot;https://en.wikipedia.org/wiki/Detonator&quot;&gt;detonator&lt;/a&gt;) in order to explode. Tertiary explosives being the most stable. However, &lt;em&gt;if the detonator is attached&lt;/em&gt; then they can be very sensitive. Which is why detonators are never attached until just before detonation and are always stored separately from the bulk of the explosive material. The shock required to set off an explosive is different for each explosive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, I was involved in a commercial explosive test involving 3 simultaneous detonations of C4. One of the detonators failed while the other two charges successfully detonated. This sent the third charge of C4 flying through the air due to the shock waves of the other two explosions (and could be seen on high-speed video) and it did not explode. Obviously a dangerous situation to locate and diffuse the unexploded charge, but everything was fine in the end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principal there is no limit to how often you can drop/reshape the explosive. However, common sense would obviously require explosives to be handled very carefully.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2016-08-15T15:11:47.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11091" PostTypeId="1" CreationDate="2016-08-15T15:20:13.810" Score="2" ViewCount="1102" Body="&lt;p&gt;I am struggling calculating the correct torque needed to rotate a windmill object at the required speed. Suppose I have a windmill:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        ^&#xA;        |&#xA;     &amp;lt;- o -&amp;gt;&#xA;        |&#xA;        v&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with a motor at the centre, and the four blades are 2.4m long, each weighing about 10kg. I'd like to calculate the torque needed to rotate the windmill at around 5 rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be fine calculating the torque needed if there were just one blade, but my intuition tells me that since everything is balanced and symmetric, ANY non-zero torque would be sufficient. However, this answer seems &quot;too easy&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an engineer, just a meagre pure mathematician, so any pointers or tips (or even answers!) would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="7648" LastActivityDate="2016-08-15T18:03:21.597" Title="Calculating correct motor torque for windmill" Tags="|mechanical-engineering|motors|torque|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11092" PostTypeId="2" ParentId="11091" CreationDate="2016-08-15T18:03:21.597" Score="1" Body="&lt;p&gt;Torque, $\tau$, is required for rotational acceleration, $\tau = I\alpha$, and for viscous losses, $\tau = b\omega$. Here I'm using $\phi$ for angular position, $\omega$ for angular velocity, and $\alpha$ for angular acceleration. $I$ is the moment of inertia &lt;em&gt;as &quot;seen&quot; by whatever you care about&lt;/em&gt;, and $b$ is damping constant, again &lt;em&gt;as seen by whatever you care about&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the windmill blades, they will have some &lt;a href=&quot;https://en.wikipedia.org/wiki/Moment_of_inertia&quot; rel=&quot;nofollow&quot;&gt;moment of inertia&lt;/a&gt; about their centers of mass. You can use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallel_axis_theorem&quot; rel=&quot;nofollow&quot;&gt;parallel axis theorem&lt;/a&gt; to &quot;convert&quot; a moment of inertia from about the center of mass to about some other useful point in space - this would be the center of the hub. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to note here that the hub is likely to have its own drum &lt;em&gt;to which&lt;/em&gt; the blades attach - the blades don't all meet in the exact middle like a bunch of orange slices, which means that the center of the hub (about which the blades rotate) is probably a phantom or &lt;em&gt;projected&lt;/em&gt; point in space. Use &lt;em&gt;that&lt;/em&gt; distance for the parallel axis theorem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, once you have the moment of inertia for everything, then you need the viscous damping coefficient. This can be very difficult to measure, depending on the size of your windmill, but you can try to determine it by starting the windmill at a known speed and then removing the actuating device and timing how long it takes for the speed to decay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Net torque is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau_{\mbox{net}} = \tau_{\mbox{inputs}} - \tau_{\mbox{outputs}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inputs are the things that act on the windmill, the outputs are the things that are acted on by the windmill. Generally, the only thing acting on the windmill would be the torque applied by the wind, but this is a complex relationship that has a lot to do with wind speed, air quality, and blade shape. I would imagine you already have some modeled relationship, so we can just leave that as $\tau_{\mbox{wind}}$. Everything else described mathematically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau_{\mbox{net}} = \tau_{\mbox{wind}} - \tau_{\mbox{generator}} - \tau_{\mbox{acceleration}} - \tau_{\mbox{viscous}} \\&#xA;$$&#xA;$$&#xA;\tau_{\mbox{net}} = f(\mbox{wind}) - \tau_{\mbox{generator}} - I\alpha - b\omega \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generator torque can be approximated by examining the power output of the generator (I'm assuming this is, in fact, a power generating windmill and not a grain windmill or something else). If the output power of the generator is $P = IV$, where $P$ is power in Watts, $V$ is voltage (in volts) and $I$ is current in amps, then you could approximate the generator torque via:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_{\mbox{out}} = \eta P_{\mbox{in}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the output power is the electrical side, the input power is the mechanical side, and eta ($\eta$) is a constant denoting the efficiency of the power conversion process. I would probably put this around 0.8 (80%) if it were my simulation, but you could do anything from 0.7 to 0.95 and probably make a justifiable case. In the end, though, you want your simulations to be as accurate as possible, so if you're using an off-the-shelf (OTS) generator, ask for the efficiency of the device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of stating the above conservation of power formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_{\mbox{out}} = \eta P_{\mbox{in}} \\&#xA;IV = \eta \tau_{\mbox{generator}}\omega \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the &lt;em&gt;mechanical&lt;/em&gt; power formula, $P=\tau\omega$. You can reformulate this as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau_{\mbox{generator}} = \frac{IV}{\eta\omega} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, finally,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\tau_{\mbox{net}} = f(\mbox{wind}) - \frac{IV}{\eta\omega} - I\alpha - b\omega \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This net torque gets applied to the windmill shaft, which causes everything attached to accelerate or decelerate (again, from $\tau = I\alpha$). Integrate the acceleration for velocity and again for position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're trying to control the speed, the only two things you really have control over are your wind inputs (by varying blade pitch) and your generator output (by varying output voltage). If you're connected to a &quot;sufficiently large&quot; electrical &quot;body&quot; (like the national power grid), then a very slight change in output voltage should make dramatic differences on how &quot;loaded&quot; or &quot;unloaded&quot; the generator is. For a small setup, though, you might want to consider adding bank of power resistors to ensure there is always sufficient &quot;demand&quot; for your electrical power, otherwise the automated algorithm may over-voltage everything attached in an attempt to increase generator output power. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-08-15T18:03:21.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11094" PostTypeId="1" AcceptedAnswerId="11095" CreationDate="2016-08-15T18:57:04.163" Score="-2" ViewCount="395" Body="&lt;p&gt;I know the stack exchange does not help on hw problems, but please understand that I am just self learning these topics, I'm not in a class, and I'm using my dads old engineering mechanics book... Now my book says I'm completely wrong on a VERY simple vector problem (&lt;strong&gt;Question 2.6&lt;/strong&gt;). This bothers me and want to see if you guys come up with their answer of: 18.95°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Answer: 52.02°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/83O0Q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/83O0Q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7652" LastEditorUserId="1832" LastEditDate="2016-08-15T19:17:35.673" LastActivityDate="2016-08-15T19:17:35.673" Title="Simple statics vector analysis problem" Tags="|statics|" AnswerCount="1" CommentCount="3" ClosedDate="2016-08-16T01:22:17.773" ContentLicense="CC BY-SA 3.0" />
  <row Id="11095" PostTypeId="2" ParentId="11094" CreationDate="2016-08-15T19:13:20.627" Score="3" Body="&lt;p&gt;What steps did you take to try to solve the problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b-b is the only &quot;output&quot; for vertical components in force F. I'll call the force in b-b $F_{b-b}$ and the force in F $F_F$. This means that the vertical component of force in $F_{b-b}$ must equal the vertical component of force in $F_F$. That is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F_F \sin{\alpha} = F_{b-b} \sin{60^{\circ}} \\&#xA;$$&#xA;Then, just solve for $\alpha$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sin{\alpha} = \frac{F_{b-b}}{F_F} \sin{60^{\circ}} \\&#xA;\alpha = \mbox{asin}\left({\frac{F_{b-b}}{F_F} \sin{60^{\circ}}}\right) \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, plug in the numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\alpha = \mbox{asin}\left(\frac{150}{400}\sin{60^{\circ}}\right) \\&#xA;\alpha = \mbox{asin}\left(0.375*0.866\right) \\&#xA;\alpha = \mbox{asin}(0.3248) \\&#xA;\boxed{\alpha = 0.3308 \mbox{ rad} \\&#xA;\alpha = 18.95^{\circ}}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-08-15T19:13:20.627" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11097" PostTypeId="1" CreationDate="2016-08-16T09:12:27.337" Score="2" ViewCount="1891" Body="&lt;p&gt;I'm trying to calculate the safety factor against separation of members in a design that looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QZCB1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QZCB1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to find the bolt and cylinder stiffness, but I don't know how to find the stiffness of the effective loaded area of the plates. I'm used to calculate the stiffness of hollow cones as represented here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0QEd5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0QEd5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this is not the case. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="7344" LastActivityDate="2016-08-16T09:12:27.337" Title="Stiffness of members in a bolted joint" Tags="|mechanical-engineering|bolting|stiffness|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11098" PostTypeId="1" CreationDate="2016-08-16T09:32:58.790" Score="3" ViewCount="255" Body="&lt;p&gt;I've been wondering if roads and stairways laced with piezoelectric generators be a viable method in clean energy generation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently unfamiliar with all engineering aspects of piezoelectricity except for the fact that it generates electricity given differences in pressure. We generate pressure all the time using our bodies, dropping our feet downstairs, and our automobiles generate pressure on our roads all the time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1. So I'm wondering if current piezoelectrical generators could be able to generate electricity through the transmitted pressure of our paved roads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2. If that is possible, would the generators be able to stand such pressures casually? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q3. How much electrical generation could I expect from a 50 step stair ladened with such a device? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q4. If it's plausible in terms of engineering, is it economically viable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would love to hear an answer, thanks!  &lt;/p&gt;&#xA;" OwnerUserId="7658" LastActivityDate="2016-08-16T11:58:40.147" Title="Electricity generation via piezoelectric roads, stairways, is it plausible?" Tags="|electrical-engineering|infrastructure|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11099" PostTypeId="2" ParentId="11098" CreationDate="2016-08-16T10:52:20.230" Score="5" Body="&lt;p&gt;Q1: Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2: Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q3: It depends but generally not much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q4: No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without getting into the technical details of how you recover the energy, consider it from a conservation of energy perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, you can generate energy from roadways or stairs in a number of ways, but that energy has to come from somewhere.&#xA;If you make the road generate energy from the cars traveling over it, then you will reduce the fuel economy of the cars driving on it. If you take energy from stairs, you will make them harder to walk up. By the time you've collected enough energy to be useful, you've normally had a significant impact on the practicality of the road/stairs you are using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roads do heat up a small amount when driven on; this is a mixture of heat generated in the road surface and heat transferred from the tires, which heat up when driven.  This heat is effectively wasted and and so could in theory be used to generate usable energy with no penalty. That gives you a theoretical upper limit on the &quot;free&quot; energy that you could collect. However the total heating effect is only a few degrees and spread over a huge area, the inefficiencies involved gathering this energy (either before or after it becomes heat) and making use of it mean it's simply not a practical idea from an economic point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything you do to gain more energy than that will make it harder for the cars to roll on the road and so reduce their efficiency. At that point, you are burning fuel to generate electricity, you may as well cut out the middle man and use a generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one time when cars want to lose energy, when they are slowing down. Conventional cars throw the extra energy away as heat in the brakes. If you could come up with a road surface that only tried to remove energy from the car while it is slowing, then that would be a potentially practical energy generation system. So far the most practical system for doing this has proven to be to build the equipment to do this into the car rather than the road and call it a hybrid car.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastEditorUserId="86" LastEditDate="2016-08-16T11:58:40.147" LastActivityDate="2016-08-16T11:58:40.147" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11100" PostTypeId="1" CreationDate="2016-08-16T12:34:47.020" Score="3" ViewCount="117" Body="&lt;p&gt;We have a plate capacitor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/oIri9.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture shows this plate capacitor where the space between the electrodes is filled with a dielectric plate with &lt;em&gt;dielectric constant&lt;/em&gt; $E_r = 8$:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Determine the length of $h$ to which should extract dielectric plate,&#xA;in order to reduce the capacity of capacitor by 4 times&lt;/li&gt;&#xA;&lt;li&gt;What will be the percentage decrease in capacitance if we pull the plate to a length of $a/5$?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7660" LastEditorUserId="282" LastEditDate="2016-10-13T10:16:28.307" LastActivityDate="2016-10-13T10:16:28.307" Title="Effect of Dimensions on Plate Capacitor" Tags="|electrical-engineering|materials|" AnswerCount="2" CommentCount="1" ClosedDate="2016-08-17T13:29:54.757" ContentLicense="CC BY-SA 3.0" />
  <row Id="11101" PostTypeId="1" AcceptedAnswerId="11103" CreationDate="2016-08-16T12:48:42.397" Score="1" ViewCount="214" Body="&lt;p&gt;I'm looking to rivet two sheets of aluminium together with the pop type.  I've researched that 3.2mm rivets should go into 3.3mm holes.  (Who has a 3.3mm drill bit?)  That's not a lot of clearance, which I understand is part of the concept to allow adequate development of shear capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, how do I practically do this for a row of 20 rivets?  I don't think that I can pre drill 20 holes to a tolerance of +/- 0.05mm.  Certainly not on my cheap 250W pillar drill.  The &quot;How to Rivet&quot; demonstrations on line only show the basic use of a rivet gun on one rivet.  How would I join two sheets with a row of pop rivets?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2016-08-16T13:27:08.300" Title="How to install multiple pop rivets" Tags="|mechanical-engineering|rivets|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11102" PostTypeId="2" ParentId="11100" CreationDate="2016-08-16T13:10:28.377" Score="2" Body="&lt;p&gt;Since we know from basic:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C = ε.A/H&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which is in our case:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C = ε.(b.a)/d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In order to reduce the capacity of capacitor for 4 times&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;b = b/4 (Divide length by 4)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;a = a/4 (Divide width by 4)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;d = d.4 (Multiply height by 4)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;or play with constant ε (find some material which has ε' as ε/4)&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For what percentage will decrease the capacitance, if we divide the plate's width by 5 ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that you can use the same equation: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C' = ε.(b.(a/5))/d&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then you can get: &lt;code&gt;(C'/C)*100 = %20.&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that it gets smaller as the %20 of the original value.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C&lt;/strong&gt; (Capacity) = &lt;strong&gt;ε&lt;/strong&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Vacuum_permittivity&quot; rel=&quot;nofollow&quot;&gt;Vacuum permittivity&lt;/a&gt;) . &lt;strong&gt;A&lt;/strong&gt; (Area) / &lt;strong&gt;H&lt;/strong&gt; (Height)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, please take a look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Capacitor&quot; rel=&quot;nofollow&quot;&gt;Capacitor Wiki.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7468" LastEditorUserId="7468" LastEditDate="2016-08-16T13:20:57.910" LastActivityDate="2016-08-16T13:20:57.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11103" PostTypeId="2" ParentId="11101" CreationDate="2016-08-16T13:27:08.300" Score="5" Body="&lt;p&gt;For large production runs, the base metal may be punched, stamped, laser cut, water jet, etc but for a DIY project, drilling is going to be your best bet. 'Match-drilling' is best, where you drill both layers at the same time ensuring that the two holes line up. Depending on the material and thickness, you could clamp, tack weld, or tape the two layers together before drilling the holes. If the geometry of your joint is conducive, you may be able to do a couple of rivets and then drill the rest of the holes on the already-assembled sandwich, assuring the layers line up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pop rivets will work in slightly larger holes too, if necessary, but with reduced holding power. If you have to oversize a hole, it's better to do so on the front side then on the back (blind) side. If you have access to the back side, rivet washers are also available for this situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you match drill and you're still having trouble with tolerances, it may be that your drill press setup is not rigid enough. If that's the case you may want to check out &lt;a href=&quot;https://engineering.stackexchange.com/questions/160/how-do-i-keep-my-bit-from-wandering-on-a-drill-press/161#161&quot;&gt;this question&lt;/a&gt; for some strategies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drill bits are generally available in every .1mm increment up to 10mm without being too exotic.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-08-16T13:27:08.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11104" PostTypeId="1" CreationDate="2016-08-16T15:57:46.390" Score="5" ViewCount="4885" Body="&lt;p&gt;I'm trying to design a stylus for use with multiple touchscreen technologies, but I'm not sure what materials I should start with (specifically for the tip). Right now I'm investigating a few materials, but I'd like more background on how other styli on the market are made today (particularly for capacitive-touch).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For any off-the-shelf stylus, what is the material used for the tip? (typically it is a soft black rubber).&lt;/li&gt;&#xA;&lt;li&gt;There is probably a coating on the tip material to make it more lubricious (lower coefficient of friction) on [typically] glass screens. What would that coating be?&lt;/li&gt;&#xA;&lt;li&gt;How are they made conductive to facilitate capacitive functionality? Is the coating material conductive, or is the rubber material doped with something to make it conductive? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7663" LastActivityDate="2017-03-06T15:53:06.037" Title="Stylus Materials" Tags="|materials|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11105" PostTypeId="1" AcceptedAnswerId="11106" CreationDate="2016-08-16T16:27:58.430" Score="4" ViewCount="124" Body="&lt;p&gt;I'm hoping this is a relatively simple question to answer, but I just couldn't find it on Google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to air conditioning (or any cooling system that relies on compressing/evaporating), how do you actually control the 'cooling' temperature? After compressing/pressurizing the refrigerant, I understand that it becomes cool due to releasing heat, however, how do you control the level of coolness? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is temperature directly related the amount of pressure being applied to the refrigerant? I.e, more pressure leads to lower temperature, less pressure leads to higher temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="7664" LastActivityDate="2016-08-16T22:08:55.193" Title="How do you actually control the cooling level of AC?" Tags="|cooling|compressed-air|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11106" PostTypeId="2" ParentId="11105" CreationDate="2016-08-16T22:08:55.193" Score="4" Body="&lt;p&gt;The temperature of evaporated refrigerant gas is constant as a property of refrigerant material. And usually it is considerably below the thermostat setting in the controlled space. The design of the system relies on the circulation of cooled air and its gradual warming up to deliver the cool air at desired temperature so that it is still cool enough to be effective and not too cool to be uncomfortable. Thermostats cut off the power to the compressor pump which is the starting point of refrigeration four phase cycle.&lt;br&gt;&#xA;Thermostats have built in delays as to switch the system back on or off with enough laps to allow the system get back to equilibrium and prevent freezing of valves or undue heat at the start up of the pump.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along this path of delivery of the cool air there are incidental materials such as partitions, ducting, furniture, any material with high thermal index, that act as a moderator of temperature by absorbing heat from ambient air and delivering it back when the air coming from the unit it too cold. So the system's sudden switch on and off cycles are tempered into gradients of temperature that are hard to notice to occupants! &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-08-16T22:08:55.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11107" PostTypeId="2" ParentId="2589" CreationDate="2016-08-17T00:35:34.057" Score="1" Body="&lt;p&gt;Yes use Aluminum 5052 H32 for sheet metal parts with bends. r =or&gt; T.&#xA;I do it like this, get the lengths of the straight lines and set aside.&#xA;The t/T is a little more than .50, say .53 until you get the real number.&#xA;K= t/T, radius of the neutral line is inside r + t = r + .53T&#xA;For a 90 degree bend, length of the bend is 2*pi*(r+t)/4 = pi*(r+t)/2 = pi*(r+.53T)/2, for any angle of bend, length of the bend is 2*pi*(r+t)*angle/360&#xA;length = staight lengths + length of bend, continue to add more bends if needed&#xA;eg: angle 1-1/4in x 2-1/4in outside dimensions, 90deg bend, 1/8in thk aluminum 1/8in inside radius, remove t and r to get straights 1in and 2in add 2*pi*(.125+.53(.125))in/4 = 3in + .3in = 3.3in  &lt;/p&gt;&#xA;" OwnerUserId="7666" LastActivityDate="2016-08-17T00:35:34.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11108" PostTypeId="2" ParentId="11100" CreationDate="2016-08-17T07:04:38.137" Score="0" Body="&lt;p&gt;The plate system above can effectively thought as two parallel capacitors.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both will have b length and d thickness. Then leftside one will have h width and  Er=1 as electric constant, while the rightside one will span by a-h and score Er=4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, simply enough, in the usual approximation for flat plate capacitors (i.e. give up edges and consider inner field constant and outer one naught )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C=C_\text{L}+C_\text{R}=\epsilon_0\frac{h\,b}{d}+\epsilon_0 \epsilon_\text{r}\frac{(a-h)b}{d}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which by some algebra can be rearranged as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C=\epsilon_0\frac{a\,b}{d}\left(\epsilon_\text{r}-(\epsilon_\text{r}-1)\frac{h}{a}\right )&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be read as &lt;strong&gt;one a by b capacitor with average relative electric constant linearly changing from Er to 1 as h/a ratio spans 0 to 1&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7668" LastActivityDate="2016-08-17T07:04:38.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11110" PostTypeId="2" ParentId="11080" CreationDate="2016-08-17T12:54:36.153" Score="0" Body="&lt;p&gt;I guess you need something like a blockade lock. You might get some idea through &lt;a href=&quot;https://www.alibaba.com/product-detail/High-Security-container-push-steel-bolt_60437017946.html?spm=a2700.7724857.0.0.K9tfl3&amp;amp;s=p&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of assumptions I am making here. One of them is that is a one-time lock that you need, like the ones they use to seal containers. I don't know how frequently you'll need to separate the two parts. If it is very rarely, like once in an year or so, such locks might do the trick.&lt;br&gt;&#xA;My computer speakers are way too heavy (since they are old) for this kind of a lock, but the ones you have linked in the questions, don't really look that heavy. This kind of lock should do the trick for you.&lt;/p&gt;&#xA;" OwnerUserId="2189" LastActivityDate="2016-08-17T12:54:36.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11111" PostTypeId="1" AcceptedAnswerId="11146" CreationDate="2016-08-17T13:02:39.060" Score="2" ViewCount="628" Body="&lt;p&gt;I am relatively new (not in terms of registration but in terms of activity) on this site. I don't know if this question is suited here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen two rail locomotives connected one after another and those two (assuming) hauling the rakes. While this is true, I didn't see a loco-pilot (In India, the engineer who drives a train locomotive is called a Loco-Pilot) in the second locomotive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that second locomotive is on neutral gear and not throttling at all, then it makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If its a case of two locomotive hauling the rakes together, is it a valid assumption that the two have synced to the same speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The real question is in regards with, how the amount of power required to haul the rakes is divided among the number of locomotives connected?&lt;/p&gt;&#xA;" OwnerUserId="2189" LastEditorUserId="2189" LastEditDate="2016-08-18T03:55:32.150" LastActivityDate="2016-08-19T01:48:36.123" Title="Two railway locomotives connected one after another" Tags="|rail|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11112" PostTypeId="2" ParentId="11111" CreationDate="2016-08-17T13:12:43.123" Score="2" Body="&lt;p&gt;&lt;strong&gt;Yes!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your assumption is actually true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these two locomotives connected with solid and undeformable link, we can say they are sharing the same speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OCp0f.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OCp0f.jpg&quot; alt=&quot;Link paper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other condition the &lt;strong&gt;LOCO CONNECTION&lt;/strong&gt; must deform.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here another strong locomotive link can be seen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iD2BC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iD2BC.png&quot; alt=&quot;LOCO CONNECTION&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7468" LastEditorUserId="7468" LastEditDate="2016-08-17T13:18:35.793" LastActivityDate="2016-08-17T13:18:35.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11113" PostTypeId="1" AcceptedAnswerId="11136" CreationDate="2016-08-17T15:19:28.057" Score="2" ViewCount="124" Body="&lt;p&gt;I have been doing research into blood pressure monitors, and I was wondering when the cuff tightens, how do the transducers actually measure the systolic/diastolic pressure in the arteries.&lt;/p&gt;&#xA;" OwnerUserId="5065" LastEditorUserId="63" LastEditDate="2016-08-18T17:02:43.000" LastActivityDate="2016-08-20T17:34:48.353" Title="Transducers in cuff-style blood pressure monitors" Tags="|electrical-engineering|sensors|biomedical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11114" PostTypeId="1" AcceptedAnswerId="11135" CreationDate="2016-08-17T16:16:11.830" Score="4" ViewCount="4405" Body="&lt;p&gt;One night, raining, crossing a street, I noticed strange thing about one of the stopped cars: In front of that car's headlamps, it rained &lt;em&gt;dashed lines&lt;/em&gt;. I assumed that the car used LED for lights, running on AC, all with the same phase. (I couldn't recognize the model of the car.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what benefit do the AC headlights give? Most car lights run on battery so you can use it even when the engine is off. What powers AC car lamps when the engine is off?&lt;/p&gt;&#xA;" OwnerUserId="7672" LastActivityDate="2016-08-19T20:37:32.063" Title="Why do these car headlights run on AC?" Tags="|electrical-engineering|automotive-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11115" PostTypeId="2" ParentId="11114" CreationDate="2016-08-17T16:49:15.680" Score="1" Body="&lt;p&gt;I would say the the lights were not LED's, but HID lights. The flickering you observed is not due to the difference between using AC vs DC for the headlamps, but due to voltage fluctuations in the output of the ballast. Some of the cheap ballasts don't provide a stable output voltage and there's noticeable flicker. Think of a florescent light in a building where the ballast is going bad. This is pretty common with cheap or aging HID ballasts.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2016-08-19T20:37:32.063" LastActivityDate="2016-08-19T20:37:32.063" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11116" PostTypeId="2" ParentId="11114" CreationDate="2016-08-17T16:51:05.530" Score="6" Body="&lt;p&gt;LEDs, especially high output ones are often run off pulsed DC, this is a convenient and efficient way to control current and thus brightness, especially in a vehicle electrical system where the voltage may vary somewhat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't identical to AC which implies a switching the polarity of the current but rather just a series of on/off pulses. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-08-17T16:51:05.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11117" PostTypeId="2" ParentId="11089" CreationDate="2016-08-17T17:16:05.297" Score="2" Body="&lt;p&gt;As mentioned already the bulk of the actual explosives used in munitions ect is pretty stable in itself. Clearly bombs need to be stored, handles and transported, often in difficult conditions so large quantities of sensitive explosives would be as much of a hazard to the user as the enemy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precise method for detonating a bomb will vary according to its intended use and may include impact fuses (sometimes with a time delay), timed fuses, altitude fuses, proximity detection (especially for air to air missiles)  etc etc. For example bombs intended to create large craters may have an impact fuse with a short delay to enable them to penetrate the ground for some distance before detonating. Conversely anti-personnel weapons may be most effective detonated some distance above the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically the detonator and associated mechanism won't be fitted untill it is required for use and there will be safety procedures to ensure that  this is only done at the correct time, including visual indicators that a detonator is fitted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition there may be other safeguards, for example a safety pin which prevents the detonator from actuating unless it is removed. For impact fuses this could be as simple as a cap over the fuse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may also be additional systems to ensure that a bomb cannot detonate untill it has been properly released from the aircraft. A simple way to do this is to have a small turbine at the nose of the bomb which is operated by the airflow over is as it falls and must turn a certain number of times before the bomb is armed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is somewhat more complex for artillery shells which need to be sensitive enough to detonate when required but not be set off by the forces involved in firing them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A bomb will usually require a detonator to be fitted before it is capable of exploding at all. &lt;/li&gt;&#xA;&lt;li&gt;There will be a variety of additional safety features intended to ensure that the bomb isn't detonated untill the desired time, varying depending on the intended use and sophistication of the design. &lt;/li&gt;&#xA;&lt;li&gt;The type of fuse used will depend on where the bomb is intended to be detonated relative to its target. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-08-17T17:16:05.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11118" PostTypeId="1" CreationDate="2016-08-17T19:19:15.433" Score="6" ViewCount="9481" Body="&lt;p&gt;I was given formulas for general evaporation, for a spill calculator I am building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are formulas, that are supposed to approximate evaporation, are for a number of different chemicals. Molecular weight is pulled from the chemical itself, same with vapor pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a number of different spills at this location. Some spills will go directly onto asphalt/gravel or soil. We do have a number of different areas that have secondary containment, which is referred to as a &quot;Berm&quot;, so that is the reason we have two formulas below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some of them:&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;General Evaporation Rate Formula From Asphalt&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$E =\frac {0.284 v^{0.78} M^{0.667}AP}{82.05T}$$&#xA;Where:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$E$ is Evaporation in pounds per minute  &lt;/li&gt;&#xA;&lt;li&gt;$v$ is the average windspeed&lt;/li&gt;&#xA;&lt;li&gt;$M$ is the molecular weight in g/mol&lt;/li&gt;&#xA;&lt;li&gt;$A$ is the surface area in feet squared&lt;/li&gt;&#xA;&lt;li&gt;$P$ is the vapor pressure in mmHg&lt;/li&gt;&#xA;&lt;li&gt;$T$ is the temperature in Kelvin&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Evaporation Rate from Berm&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;$$E =\frac {0.284 \cdot \text{MW}^{0.667}(\text{Average windspeed})^{0.78}(\text{Surface Area})(\text{vapor pressure})}{82.05 \cdot \text{temperature in K}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface Area Equations:&lt;br&gt;&#xA;&lt;strong&gt;Scenario 1&lt;/strong&gt;:&lt;br&gt;&#xA;$l$ = Length of the liquid spilled&lt;br&gt;&#xA;$w$ = Width of liquid spilled&lt;br&gt;&#xA;$l \cdot w = A$&lt;br&gt;&#xA;$A - (\pi \cdot (\frac {\text{Diameter of tank in berm}}{2})^2 \cdot (\text{How many tanks in berm}))$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario 2&lt;/strong&gt;: &lt;em&gt;If we have 3 tanks in the berm&lt;/em&gt;&lt;br&gt;&#xA;$l$ = Length of the liquid spilled&lt;br&gt;&#xA;$w$ = Width of liquid spilled&lt;br&gt;&#xA;$l \cdot w = A$&#xA;$A - (\pi \cdot ((\frac {\text{Diameter of tank in berm}}{2})^2)) \cdot (\text{Number of tanks of that diameter in berm}))-\pi (\frac{(\text{Diameter of tank in berm})}{2})^2 \cdot (\text{Number of tanks of that diameter in berm}))- \pi(\frac{(\text{Diameter of tank in berm})}{2})^2 \cdot (\text{Number of tanks of that diameter in berm}))$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't trust the formulas given, because they were given by another student assigned to this task. These formulas are not found in any workbook, website, etc. I am always skeptical about using other people's work without documentation making sure they have proof to backup their claims that this stuff will work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking precisely this: are these formulas suitable to use as an approximation for evaporation? If not, what would be a better formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an engineer, I am a computer science/mathematics major. This project has nothing to do with school though. It is a work project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Research&lt;/strong&gt;&lt;br&gt;&#xA;The closest thing I can find to an equation I can use is from &lt;a href=&quot;http://www.engineeringtoolbox.com/evaporation-water-surface-d_690.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but I am unable to decipher the characteristics it is asking since it is extremely technical in that field. I'm willing to do some research but I still have yet to find an acceptable formula/equation I can implement in this program. Another example is &lt;a href=&quot;http://www.mmshah.org/publications/ASHRAE%202014%20Evaporation%20paper.pdf&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; where equation #7 is the formula it says to use for spills but I have no idea how it works.&lt;/p&gt;&#xA;" OwnerUserId="7674" LastEditorUserId="7674" LastEditDate="2016-10-24T19:11:52.043" LastActivityDate="2016-10-24T19:11:52.043" Title="How can I calculate the evaporation rate for a hazmat spill?" Tags="|thermodynamics|measurements|environmental-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11119" PostTypeId="2" ParentId="11111" CreationDate="2016-08-17T21:17:42.843" Score="4" Body="&lt;p&gt;Well, yes, of course they have to be going the same speed; otherwise one would pull away from the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect your real question has to do with how the throttling is done. Locomotives that are designed to be ganged in this fashion have a mechanism that allows their throttles to be &quot;slaved&quot; to the master throttle located where the driver is. That way, throttling up or throttling down affects &lt;em&gt;all&lt;/em&gt; of the available power from all of the locomotives.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-08-17T21:17:42.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11120" PostTypeId="1" AcceptedAnswerId="11123" CreationDate="2016-08-18T00:26:27.083" Score="3" ViewCount="1186" Body="&lt;p&gt;Say I have a simple reinforced concrete frame (3as shown below) with all beam ends pinned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d5TLN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d5TLN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;( the keyplan)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the detailing output:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C3iMx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C3iMx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;The beam elevation detail ( left), beam cross section detail ( right)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9agh9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9agh9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;Column cross section&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that due to the pins, all beam end moments will be 0. However, at the column joint, will there be any moment? I think no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one of my engineering friend think that because this is a &lt;em&gt;reinforced concrete&lt;/em&gt; beam, and we design our beam reinforcement bars for the designed moment, this designed moment should somehow be transferred to column (even though the frame analysis shows that there is no moment at this beam joint), and thus the column joint will have moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In other words, the presence of reinforcement bars in the beam means that we need to design for moment in the column at the joint, as the reinforcement will create a moment in the column, even though in the analysis this &quot;reinforcement induced&quot; moment is simply not there ( and the analysis shows us that column moment at joint is a 0). Is this a common practice ( as stipulated by convention or demanded by code of practice), or is this something completely unheard of?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could he possibly be right?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="3353" LastEditDate="2016-08-19T08:55:00.987" LastActivityDate="2016-08-19T08:55:00.987" Title="Reinforced concrete design: will there be extra nominal moment on column in a frame structure?" Tags="|structural-engineering|reinforced-concrete|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11121" PostTypeId="1" AcceptedAnswerId="11127" CreationDate="2016-08-18T00:37:09.913" Score="1" ViewCount="431" Body="&lt;p&gt;&lt;em&gt;I am asking this because some of my engineer friends said that they won't want to analyze beam as continuous beam for RC building structures, because it is &quot;unsafe&quot; to do so, whether it has anything to do with less deflection or no, I don't know.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His logic is that &lt;a href=&quot;https://engineering.stackexchange.com/questions/9066/why-continuous-beam-has-less-deflection-than-beam-with-support&quot;&gt;continuous beam predicts less deflection&lt;/a&gt; ( and hence gives more economical design), is his claim valid? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder how does practicing engineers view the modelling and analyzing/designing beam spans as continuous beams ( instead of beams with many supports) when it comes to &lt;strong&gt;reinforced concrete building structures&lt;/strong&gt;? Do they generally feel that it is OK to design beam spans as continuous beam ( and hence more economical because the reinforcement bars used will be less), or do they share my friend's opinion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What does the building code of practice ( or common design practice) across the world say about continuous beam design, do they discourage or encourage it? Any reason not to design beam spans as continuous beam?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2018-04-01T19:38:19.140" Title="Is continuous beam design deemed &quot;unsafe&quot; by any code of practice/design practice for reinforced concrete building structures?" Tags="|structural-engineering|reinforced-concrete|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11123" PostTypeId="2" ParentId="11120" CreationDate="2016-08-18T02:15:47.207" Score="2" Body="&lt;p&gt;Just as there is conservation of energy and mass in the universe, so must you also think of &quot;conservation of internal forces&quot;. I totally made this term up, it doesn't exist. What I mean by that is that forces and moments never simply appear or disappear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you've stated, your beam will present zero bending moment at nodes 4 and 6, where it connects with the left and right columns, respectively. Now, for those columns to present any bending moment at those nodes, the bending moment would have to come from somewhere, in such a way as to explain the discontinuity which would be present at the node (column at node 4 with bending moment, beam at the same node without bending moment). In this case, there is nowhere for such a bending moment to appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, no, the left and right columns will also present zero bending moment at the nodes. The central column depends on how it's linked to the beam. If the column is hinged to the continuous beam, then it'll present zero bending moment. Otherwise, if the column is fixed to the beam, then it will absorb some of the beam's negative bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that this has nothing to do with the structure's material or with structural design. The solution for internal forces is defined entirely by structural analysis, which is mostly agnostic to such considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how to calculate the reinforcement for a column with zero bending moment, one adopts minimal reinforcements defined in codes or, if your country's codes have no such definitions, common practice.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After your edit showing your actual situation, however, let me quote a comment I previously wrote beneath this answer: &quot;A fixed node [...] wouldn't have the narrow throat, but would be a monolithic connection with almost all the column and beam rebar passing through.&quot; This connection has no throat and is entirely monolithic (where does the beam end and the column begin? You can't tell because the connection is, well, both) and is therefore clearly, beyond a shadow of a doubt, a fixed joint. Your model shouldn't have any hinges on those connections and, therefore, the beam and the column should both present a (possibly small) bending moment at that point (negative bending moment on the beam, which generates tension on the top fiber. In the column, this bending moment will generate a tension on the external fiber). The bending moment in both will be equal.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-08-19T01:32:57.327" LastActivityDate="2016-08-19T01:32:57.327" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11124" PostTypeId="1" CreationDate="2016-08-18T05:17:50.100" Score="1" ViewCount="717" Body="&lt;p&gt;I'm struggling with the practical application of a PID control algorithm when the set point does not require any external force to stay in position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have a geared electric motor that has a mass attached perpendicular to the motor's shaft. The mass is fixed to the shaft so it will rotate with the shaft without slipping. There is a sensor on the mass that knows its angle relative to horizontal (an accelerometer in my case), which is used for error feedback. The control system attempts to drive the motor such that the mass is always balanced vertically above the shaft, as pictured below. The motor itself is attached to a robotic arm that may rotate around arbitrarily, so the motor will work to keep the mass vertical at all times.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8JbGI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8JbGI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key thing to note here is that once the motor has achieved the set point (the mass is vertical above the shaft), it does not require any more work from the motor. No additional force is required to hold it there. In fact, there's enough back-drive friction in the gearing that even if the mass is, say, 10 degrees from vertical, it'll still stay put.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is in regards to correctly implementing a PID control loop for this kind of system. Here's the situation that has me confused:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the mass has come to rest deflected 5 degrees from vertical. In the PID loop, the P component is very small because the error is very small. So the integrator winds up to do most of the work of removing this steady-state error. It does its job and adds just enough power into the motor to drive the mass to the vertical position. Once the vertical position is achieved, no additional force is required to hold it there. But that integrator component still has a non-zero value and will keep driving the motor, making it push the mass passed the set point and to the other side. An inevitable oscillation about the set point will then occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there special considerations in a PID algorithm for this case? Or am I just fundamentally misunderstanding how a standard PID algorithm will react to the system I described above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;br&gt;&#xA;My original question was misusing the word &quot;stable&quot;. Thank you to Carl Witthoft for pointing that out. The focus of my question, however, is how to properly implement a PID loop when the system requires that the control forces drop to zero as the position error approaches zero.&lt;/p&gt;&#xA;" OwnerUserId="6996" LastEditorUserId="6996" LastEditDate="2016-08-18T17:13:06.420" LastActivityDate="2016-08-18T20:07:42.800" Title="PID position control when setpoint does not require continuous force" Tags="|control-theory|pid-control|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11126" PostTypeId="2" ParentId="11120" CreationDate="2016-08-18T08:24:47.567" Score="2" Body="&lt;p&gt;You seem confused between how a structure behaves in real life and how it behaves in a structural analysis model. They key point you seem to be missing is that &lt;em&gt;one must reflect the other&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, in real life, there will be reinforcement continuity between the beam and the column, &lt;em&gt;then the analysis model should not use pin-ended beams&lt;/em&gt;; it should use fully-connected beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, in the design, the analysis model is based on pin-ended beams, &lt;em&gt;then the detailing should be done to avoid moment-fixity&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2016-08-18T08:24:47.567" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11127" PostTypeId="2" ParentId="11121" CreationDate="2016-08-18T08:30:02.597" Score="4" Body="&lt;p&gt;You question starts with (emphasis mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Given that continuous beam &lt;strong&gt;predicts&lt;/strong&gt; less deflection&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not correct. A continuous beam does not &lt;em&gt;predict&lt;/em&gt; less deflection, a continuous beam &lt;em&gt;results in&lt;/em&gt; less deflection (than a pair of simply supported beams).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly to &lt;a href=&quot;https://engineering.stackexchange.com/questions/11120/reinforced-concrete-design-will-there-be-extra-nominal-moment-on-column-in-a-fr&quot;&gt;this question&lt;/a&gt;, you seem confused between how a structure behaves in real life and how it behaves in a structural analysis model. With reference to &lt;a href=&quot;https://engineering.stackexchange.com/a/11126/255&quot;&gt;my answer there&lt;/a&gt;: if you design it as continuous in your analysis model, then you need to detail the reinforcement to achieve this. Hence analysing a beam as continuous is safe &lt;em&gt;as long as you detail it to be continuous&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For additional safety in design, you &lt;em&gt;could&lt;/em&gt; design a beam as both pin-ended &lt;em&gt;and&lt;/em&gt; continuous. i.e. you could carry out a pin-ended analysis, then carry out a continuous analysis, and then envelope the results. This makes &lt;em&gt;some&lt;/em&gt; sense: if a continuous beam fails in hogging, it will then behave as a pin-ended beam for further load (provided it has rotational capacity at the joint). However, this would generally be thought of as overdesign: you're designing in extra safety but at extra cost, so unless you absolutely need a paranoid level of safety then you're wasting money.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-08-18T11:49:27.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11128" PostTypeId="1" CreationDate="2016-08-18T10:07:57.483" Score="1" ViewCount="245" Body="&lt;p&gt;If I need to get a pressure of &lt;strong&gt;6 bar&lt;/strong&gt; at the exit of a nozzle for cleaning purpose, how can I determine the following (assuming nozzle outlet dia to be 5 mm):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;strong&gt;pump capacity&lt;/strong&gt; to provide such a pressure.&lt;/li&gt;&#xA;&lt;li&gt;The &lt;strong&gt;diameter of pipe&lt;/strong&gt; connecting pump to nozzle assuming total &lt;strong&gt;5 m&lt;/strong&gt; length.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for mathematically precise answers, but just need to get overall idea of how to calculate the pump and connection specifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some additional information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The working fluid is water.&lt;/li&gt;&#xA;&lt;li&gt;The nozzle is used to clean a plate which would be kept at distance of 200mm from the nozzle exit.&lt;/li&gt;&#xA;&lt;li&gt;The working pressure is specified as 6 bar, which I believe is the stagnation pressure at the nozzle exit.&lt;/li&gt;&#xA;&lt;li&gt;The nozzle dia is 5 mm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7686" LastEditorUserId="7686" LastEditDate="2016-08-19T07:18:30.340" LastActivityDate="2016-08-20T01:41:11.997" Title="Pressure-Flowrate relation at Nozzle" Tags="|pressure|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11130" PostTypeId="1" AcceptedAnswerId="11163" CreationDate="2016-08-18T12:35:51.037" Score="4" ViewCount="8790" Body="&lt;p&gt;I'm my engineering drawing course, do I need to mention and show every dimension in the diagram if I've drawn everything to scale and also written the scale below?&lt;/p&gt;&#xA;" OwnerUserId="7604" LastActivityDate="2016-10-03T03:03:36.927" Title="Do I need to mention every dimension on a to-scale drawing?" Tags="|drafting|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11131" PostTypeId="2" ParentId="11130" CreationDate="2016-08-18T12:58:04.483" Score="7" Body="&lt;p&gt;I am not a professional engineer yet, but in my engineering design courses we were taught to show certain dimensions. For example, you choose a datum line and show certain dimensions. You don't need to label every single dimension, but you should show your dimensions in a way such that the dimensions which are NOT shown can be calculated arithmetically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you have a single line, with 4 points starting at A and ending at D with B and C somewhere along the line, and A is your datum, then you could just show the dimensions for A-D, A-B and A-C.Simple arithmetic will then allow you to calculate the distance from B-C, for example (by taking A-C minus A-B). I believe this is done so that the person looking at your drawings doesn't* need to get a ruler out to check the dimension!&lt;/p&gt;&#xA;" OwnerUserId="5469" LastEditorUserId="5469" LastEditDate="2016-08-18T17:18:06.723" LastActivityDate="2016-08-18T17:18:06.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11132" PostTypeId="2" ParentId="11130" CreationDate="2016-08-18T13:00:38.107" Score="4" Body="&lt;p&gt;Yes, with the caveat as supplied by masiewpao's answer that dims which can be calculated directly from supplied dims can be left off.&lt;br&gt;&#xA;The reason you must write in the dims is that &quot;to scale&quot; is purely qualitative. Nobody ever puts a scale on drawing to guesstimate the dimensions.  The reasons why not should be obvious :-)&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-08-18T13:00:38.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11133" PostTypeId="2" ParentId="11124" CreationDate="2016-08-18T13:06:54.060" Score="1" Body="&lt;p&gt;You are confusing the definition of &quot;stable,&quot; I suspect.  Yes, the weight will remain vertical in the absence of applied forces, but no, that is not the stability you're looking for (with apologies to Obi-Wan).  This is known as an &quot;unstable equilibrium&quot; because any force, or if you want to get fancy, quantum fluctuations as described by the Heisenberg Uncertainty Principle(1) , will cause the system to leave equilibrium.&lt;br&gt;&#xA;By comparison, if the weight is hanging vertically below the rod, that's a &quot;stable equilibrium&quot; because any minor displacement will be counteracted by gravitational force, returning the system to the equilibrium point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, yes, you'll need some sort of controller to keep the weight pointing up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Yes, Heisenberg is silly in this context but there are plenty of physics students who are given macro problems like this just to demonstrate the magnitude of the effect.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-08-18T13:06:54.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11134" PostTypeId="2" ParentId="11130" CreationDate="2016-08-18T14:32:14.347" Score="3" Body="&lt;p&gt;It very much depends on the context. Drawings are there for a purpose ie to communicate information and you need to keep that specific purpose in mind when producing the drawing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if you are producing a drawing for the manufacture of a part then you need to think about what information is needed for the manufacturing process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from experience that it is possible to have a drawing which isn't actually missing any information as such but the dimensions have been assigned in such a way as to be very unhelpful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is difficult to generalise but you need to think carefully about the specific information that the person using the drawing needs as far as is practical this should be presented clearly on the drawing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in another answer it is often best to present dimensions relative to a few well defined datum lines or points rather than stacking them on top of each other but again the context and especially the manufacturing process involved makes a difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having to scale by measuring from a drawing is usually bad practice as it tends to introduce extra errors, especially if a drawing is reproduced or handled a lot. any dimensions not shown should be easy to deduce by simple addition or subtraction. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-08-18T14:32:14.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11135" PostTypeId="2" ParentId="11114" CreationDate="2016-08-18T14:48:46.210" Score="3" Body="&lt;p&gt;The headlight LEDs are using DC, but the voltage regulation on all modern power supplies is PWM or &lt;a href=&quot;https://en.wikipedia.org/wiki/Pulse-width_modulation&quot; rel=&quot;nofollow&quot;&gt;Pulse Width Modulation&lt;/a&gt;. The electronics switch to full on quickly and full off quickly so there is very little loss in the switching components. This is used in lots of electronics, but in the application you saw it is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/DC-to-DC_converter&quot; rel=&quot;nofollow&quot;&gt;DC DC converter&lt;/a&gt;, and permits the car to go from a variable battery voltage to a constant led voltage (or &lt;a href=&quot;https://en.wikipedia.org/wiki/Constant_current#Constant_current_power_supplies&quot; rel=&quot;nofollow&quot;&gt;constant current&lt;/a&gt; for higher quality LED power supplies).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually a high enough frequency is chosen that it is not visible or displeasing to the eye. If you can estimate the velocity of the rain drop and length of the illuminated path you can calculate the frequency they were using. Velocity in m/s divided by the length in meters for example.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-08-18T14:48:46.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11136" PostTypeId="2" ParentId="11113" CreationDate="2016-08-18T15:29:33.327" Score="2" Body="&lt;p&gt;This is actually a really neat question, and it involves a few engineering and anatomical concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what happens when an automatic blood pressure cuff measures your blood pressure:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The cuff inflates until it reaches a pressure just above systolic pressure (the maximum pressure your blood reaches during a heartbeat). When it reaches this pressure, the main artery in your arm gets closed off because its pressure is not enough to resist the compression of the cuff.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The cuff then starts to deflate slowly. There is a pressure sensor that measures pressure fluctuations in the cuff. Once the cuff reaches your systolic pressure, the artery opens up just enough to let some blood through. Since the artery is mostly closed, this causes tiny vibrations as the blood squeezes through (this is because the blood flow is &quot;turbulent&quot;). These vibrations are transmitted through your skin and into the air in the cuff. The pressure sensor is sensitive enough detect these tiny vibrations in the air of the cuff once they start, and it records the current pressure of the cuff as the &quot;systolic pressure&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The pressure in the cuff keeps decreasing, until the tiny vibrations stop (the blood flow becomes &quot;laminar&quot; instead of turbulent). This is the &quot;diastolic pressure&quot;, i.e. the minimum pressure during your heartbeat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It turns out that doctors do the exact same thing with a manual cuff and a stethoscope: they pump up the cuff until they can't hear any gurgling noises in the stethoscope, then they slowly release until they hear the first gurgling noise (systolic pressure reached). They keep releasing pressure until they can't hear the noise any more, at which point they've found the diastolic pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've actually performed these measurements myself in a Biomechanical Engineering Measurements course. Cool stuff!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1121444/&quot; rel=&quot;nofollow&quot;&gt;Oscillatory Blood Pressure Monitoring Devices&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="1832" LastEditDate="2016-08-20T17:34:48.353" LastActivityDate="2016-08-20T17:34:48.353" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11140" PostTypeId="2" ParentId="11130" CreationDate="2016-08-18T19:09:50.857" Score="7" Body="&lt;p&gt;My general philosophy is that you don't need to dimension &lt;em&gt;everything&lt;/em&gt;, you just need to dimension every &lt;em&gt;unique feature&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, say you have a bolt pattern, with one hole in each corner. You give the dimension for every offset for the holes, but this is more information than required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I do it is to show bounding dimensions for the plate, dotted lines to indicate vertical and horizontal centerlines (no dimensions there as it's understood they're half the bounding dimensions), then just show the distance from each centerline to one hole. For the hole, have a callout that gives the hole diameter and tolerance and/or threading requirements. Again, if all four holes are the same, then just use, &quot;4x 6mm Thru All&quot; or similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Machinists are some of the smartest people on the planet. My preference would be to give a clear, concise set of drawings to them and then, if I've forgotten a dimension, they can write back and ask. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to recap, I gave 5 dimension - bounding X, bounding Y, centerlineY-Y, centerlineX-X, and hole diameter. That gives everything to produce the plate and four holes. If you have fillets on the corners (try to avoid ever making something with sharp edges unless you need to!) then that's one more dimension - 6 total - four eight features. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get away with this because you dimension the &lt;em&gt;unique&lt;/em&gt; features. One hole and one fillet on a square plate - the rest are clearly symmetric. The only time I would start adding more dimensions than that is if there are so many features that it could be confusing as to which features are members of a pattern and which aren't, or if there are offsets or jigs in the base material that would make patterning more difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is the case, then again my personal preference is to have one sheet per pattern or set of unique features. If you've got a part that has through holes, threaded holes, and some slots or something, then one sheet gives bounding dimensions for the part plus dimensions for the through holes, the next sheet gives dimensions for the threaded holes, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 minutes of a machinist's time is a couple reams of paper, so keep the drawings clean by using as many sheets as you need to in order to dimension all of the &lt;em&gt;unique features&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of parting notes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use the machinists. Explicitly ask for a critique on every part and drawing you produce. You should, if at all possible, talk to them in advance about what you want to do and how you're planning on doing it. See the earlier comment: machinists are brilliant people. &lt;/li&gt;&#xA;&lt;li&gt;This goes along with the one pattern-per-sheet rule of thumb - don't ever, if at all possible, make a drawing for an 8.5x11 sheet of paper. Use the 11x17 or larger format. Again, the more legible you can make your work, the better for everyone involved. I think, if you ever see a set of your own drawings on 11x17, you'll never go back.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-08-18T19:09:50.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11141" PostTypeId="2" ParentId="11124" CreationDate="2016-08-18T19:24:38.090" Score="2" Body="&lt;p&gt;You are correct about the effect of the Proportional and Integral terms in your controller, but don't forget the Derivative part! While the Integral term tends to add energy to a system, the Derivative term subtracts energy by trying to slow down the rate of change of error. It acts like friction or damping in the system. Derivative terms are usually necessary for systems which don't have much physical damping, otherwise they would oscillate a lot before coming to a stop. This is why generally speaking you will see PD and PID controllers, but rarely just P and never (to my knowledge) I control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you fix your problem? Well you have two options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first option is to increase the derivative gain. This will reduce oscillations but it will also slow the response of your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second option is to limit the integral windup. This will prevent the integral term from getting too large, meaning your torque will stop increasing after a certain point. This way you can use a smaller derivative gain since the integral term will not get so large. I would start with a saturation function that causes $K_I \int e(t)dt \leq m g l$ (where $K_I$ is the integral gain, $e$ is the error, $m$ is the mass, $g$ is acceleration due to gravity, and $l$ is the length of the arm on which the mass is attached). This is a worst-case estimate of how large your integral term would have to be. Start from there and try decreasing it until you get the performance you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will probably need to use a combination of option one and two to achieve the best performance.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2016-08-18T19:24:38.090" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11142" PostTypeId="2" ParentId="11124" CreationDate="2016-08-18T20:07:42.800" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Let's say the mass has come to rest deflected 5 degrees from vertical&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How did the mass come to rest with the PID controller enabled? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The P component is very small because the error is very small.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The proportional &lt;em&gt;error&lt;/em&gt; is very small, but the proportional &lt;em&gt;gain&lt;/em&gt; make the  &quot;P component&quot; be however large you want it to be. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Once the vertical position is achieved, no additional force is required to hold it there.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The mass must have &lt;em&gt;moved&lt;/em&gt; into position. Having mass and a speed (with which it attained the correct position) means that it has &lt;em&gt;inertia&lt;/em&gt;. The inertia then requires that either a force is applied to remove the speed or that the mass must pass vertical. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;An inevitable oscillation about the set point will then occur.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To be more blunt than CarlWitthoft, could you put the mass in position with such precision that it would stay there forever? The answer is no, for the reason he mentioned: such a position is inherently unstable. Given enough time, it will fall over. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Am I just fundamentally misunderstanding how a standard PID algorithm will react to the system I described above?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't think you're misunderstanding the PID controller, I think you're beginning to see some of the drawbacks of PID control. The same integral error that prevents &quot;droop&quot; if you were trying to control the reference position to be 90 degrees (parallel to ground) causes you to overshoot when you try to control the load to be vertical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible to overcome the shortcoming in this (relatively simple) case by adding a &quot;feed-forward&quot; term. That is, you could calculate the &lt;em&gt;anticipated&lt;/em&gt; load torque, based on load angle, mass, and the length of the arm to which the mass is attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load torque you are calculating is the &lt;em&gt;static&lt;/em&gt; torque required to hold the load in a position. It does not provide any torque to &lt;em&gt;move&lt;/em&gt; the load to a new position, or any control to &lt;em&gt;maintain&lt;/em&gt; the position, if you had any modelling errors (error measuring mass, effective moment arm, etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;seeding&quot; the motor with the rough current (torque) it needs to maintain the load at &lt;em&gt;any&lt;/em&gt; angle, the PID controller could then be tuned to a set of &quot;universal&quot; gains that should then behave the same at any angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to summarize:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PID control is most effective when the underlying system isn't changing. When the system changes (the required torques change, for example), then either the control gains need to change or the performance degrades. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use a compensatory controller (feed-forward control, for example) to &quot;normalize&quot; the underlying system, promoting a more versatile &quot;robust&quot; PID controller. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record, I hate the word robust.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-08-18T20:07:42.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11143" PostTypeId="1" CreationDate="2016-08-18T23:25:13.207" Score="5" ViewCount="112" Body="&lt;p&gt;In a &lt;a href=&quot;http://www.businessinsider.com/heres-why-the-us-military-is-replacing-the-m16-2015-10&quot;&gt;news article today&lt;/a&gt; the reporter claims that modern improvements in &quot;ammunition design&quot; have made long rife barrels unnecessary for assault weapons. Here is the paragraph:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;The 20-inch barrel isn’t needed anymore.&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;One of the biggest differences between the M4 and the M16A4 is the&#xA;  barrel length: 14.5 inches versus 20 inches, respectively. The M16A4’s&#xA;  longer barrel allowed for a higher muzzle velocity and a longer&#xA;  effective range; conversely the shorter M4 barrel limited its&#xA;  performance in both categories. But improvements in ammunition design&#xA;  have enabled the M4 to close the gap with its longer forefather.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The Marine Corps adopted a new 62-grain, 5.56×45 mm Special Operations&#xA;  Science and Technology round. The SOST round is designed to perform&#xA;  out of barrels as short as 10.5 inches, so the M4 has no difficulty&#xA;  shooting out to the extreme end of effective range, negating the&#xA;  advantage of the long M16A4 barrel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What improvements in ammunition design are they referring to? Why would a bullet be equally accurate coming out of short barrel?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2016-08-20T01:55:57.670" Title="Improvements in ammunition design?" Tags="|firearms|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11144" PostTypeId="1" CreationDate="2016-08-19T01:09:36.257" Score="1" ViewCount="106" Body="&lt;p&gt;In these units: cm³.20μ/m².24hr.atm - (definition) cubic centimeters for a 20 micron film thickness per square meter per 24 hours under a pressure difference of 1 atm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what does the &quot;Pressure difference of 1 atm&quot; mean?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="1832" LastEditDate="2016-08-19T01:41:19.387" LastActivityDate="2016-08-19T04:14:23.797" Title="Pressure Differential Analysis" Tags="|pressure|gas|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11146" PostTypeId="2" ParentId="11111" CreationDate="2016-08-19T01:48:36.123" Score="4" Body="&lt;h2&gt;Multiple Unit control (or at least how it's typically done)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you are wondering &lt;em&gt;how&lt;/em&gt; you can get two locomotives to do the same thing when there's nobody in the second locomotive, this is done using a technique called Multiple Unit (MU) control in the US and Multiple Working in British practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While older locomotives and self-propelled carsets (especially outside the US, as the 27-pin AAR-standard MU connector was developed from the electrical MU system originally implemented by EMD in their wildly successful E and F series units) used a variety of incompatible systems for this, most modern locomotive designs worldwide have settled on the AAR system.  This is in essence a discrete-electrical control system for all basic electrical functions on the locomotive -- the locomotive air braking system is handled through auxiliary air hoses that are connected along with the 27-pin MU jumper cables, while the MU connectors on newer locomotives also carry an AAR-standardized communications bus for advanced, computerized functions (such as detailed alarms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result of this jumpering, each locomotive is effectively synchronized to the lead locomotive's commands -- the crew of the lead locomotive sets up the controls the way they want, and the MU cables transmit that setup to the rest of the consist.  The &quot;reverse&quot; and &quot;forward&quot; lines are also transposed at certain points (in each jumper, and also in the connectors on the &lt;em&gt;front&lt;/em&gt; of a locomotive) to allow locomotives to run correctly when MUed in while facing &quot;backwards&quot; -- in fact, it is rare (at least in North American freight practice) to see locomotive consists where there isn't at least one backwards facing locomotive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the operator's perspective -- they are sitting in the cab of one jumbo locomotive with the combined capabilities of all the locomotives MUed together behind them.  Furthermore, the MU wires can be trainlined down passenger coaches to a cab car for push-pull passenger operations, or connected to a receiver box for remote control of the locomotive via radio -- this last one is used for Distributed Power for heavy trains or mountain grades as well as for improvising a Remote Controlled Locomotive out of a regular locomotive for yard use.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Sometimes, this isn't always the case, though&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;However, not every consist (rake over where the OP is) of locomotives is MUed throughout.  Some older or smaller locomotives, as well as many engines dedicated to switching duty, may have incompatible (the Brits are plagued by this) or no MU controls, and will wind up &quot;dead in train&quot; if they need to be hauled a long distance.  A locomotive can also end up &quot;dead in train&quot; if it needs to be moved somewhere else, but its motive power simply isn't needed to get the train from point A to point B -- too much motive power on the head-end of a train will cause it to derail or break-in-two from excessive drawbar forces.  Finally, a locomotive may have broken down somewhere on the line; this means that the locomotive will be hauled &quot;dead&quot; to a shop so that it can be repaired.&lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2016-08-19T01:48:36.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11147" PostTypeId="1" CreationDate="2016-08-19T02:26:01.910" Score="0" ViewCount="191" Body="&lt;p&gt;By introducing many new gears and changing the tire size I am altering the torque delivered to a rear bicycle wheel from an electric motor. This then changes the &quot;&lt;strong&gt;power where rubber meets pavement&lt;/strong&gt;&quot;. &quot;&lt;strong&gt;Forward thrust?&lt;/strong&gt;&quot; I don't know what the proper name is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have googled around, but find myself just plain confused and unable to sort through how to evaluate this by using the information I have - which consists of the ratios and wheel sizes. &lt;a href=&quot;http://www.maelabs.ucsd.edu/mae_guides/machine_design/machine_design_basics/Mech_Ad/mech_ad.htm&quot; rel=&quot;nofollow&quot;&gt;This web page&lt;/a&gt; seems to explain it all, but for the life of me I can't follow it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yeah, I guess I'm a dummy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not have an actual measure of the motor's starting torque, yet. Have yet to figure out how to affix my torque wrench to the motor without hurting the motor or myself. Motor torque will be a variable named in the equation, so a value won't be needed at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to determine how much MORE torque I am applying to the wheel with the new gear setup versus the original gear setup. This is so I can decide if I have increased the delivered torque enough to accomplish the goal of moving &lt;strong&gt;&lt;em&gt;~180lbs&lt;/em&gt;&lt;/strong&gt; with relative ease compared to the original gearing. The original gearing and wheel size was insufficient for the task and by looking at the percentage or ratio of the increase I should be able to ballpark whether I have the design right to move the weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current ratios were set up so that the max speed of the wheel in high gear at the motor's preferred speed of &lt;strong&gt;&lt;em&gt;2700RPM&lt;/em&gt;&lt;/strong&gt; will be at the legal limit of &lt;strong&gt;&lt;em&gt;22KPH/15MPH&lt;/em&gt;&lt;/strong&gt;. I thought that that &lt;strong&gt;&lt;em&gt;should&lt;/em&gt;&lt;/strong&gt; put my starting power in the lowest gear somewhere good enough to get the bike and the weight rolling along - but I cannot figure out how to verify that.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Original Gearing/Wheel setup:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor &gt; 5.81 reduction direct mount on a 11.5&quot; diameter wheel (that's&#xA;actual rolling diameter, it is nominally a 12&quot;) .&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;New Gearing/Wheel Setup:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor &gt; 5.81 reduction &gt; 3-Speed transmission reductions &gt;&#xA;final-drive 1.5 reduction fixed to 20.5&quot; diameter wheel (that's&#xA;approximate rolling diameter - it is nominally a 21&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3-Speed gearing ratios are 1.33, 1, and .75 - these are not&#xA;changeable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New gearing combined ratios:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st gear - 11.05, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd gear - 6.68, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3rd gear - 8.86 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;all fixed to the same 20&quot; wheel.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If I should be calling torque '&lt;strong&gt;power&lt;/strong&gt;' when I talk about it at the wheel instead of the motor, please let me know what the distinction is and if I need to make them distinct in the equation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Again:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an equation to use that allows me to vary the 5.81 reduction and the 1.5 final reduction and see the changes in torque/power at the wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;At its core the question is:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much more starting grunt do I get with &lt;code&gt;XXX ratio&lt;/code&gt; and &lt;code&gt;Xwheel&lt;/code&gt; than I did with &lt;code&gt;AAA ratio&lt;/code&gt; and &lt;code&gt;Awheel&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="7381" LastEditorUserId="7381" LastEditDate="2016-08-21T22:55:52.200" LastActivityDate="2016-08-25T04:16:58.230" Title="Equation for comparing force delivered to road at wheel based on varying gear ratios?" Tags="|gears|torque|power-engineering|wheels|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11148" PostTypeId="2" ParentId="11147" CreationDate="2016-08-19T04:13:46.030" Score="1" Body="&lt;p&gt;The power concept is that power in equals power out (ignoring friction losses).  Power = torque * rpm.  Take the original setup first.  If the gear reduction is 5.81, then the speed of the $rpm_{motor}=5.81*rpm_{gear}$.  Since the power has to be the same, $rpm_{motor}*torque_{motor}=rpm_{gear}*torque_{gear}=5.81*rpm_{gear}*torque_{motor}$.  If you cancel out $rpm_{gear}$, then $torque_{gear}=5.81*torque_{motor}$.  This is conservation of energy at work and the basis of mechanical advantage: you go 5.81 times slower but you get 5.81 times more torque.  To figure out the torque gain for your more complex power train, multiply through all the gear reductions.  For example, $5.81*1.33*1.5=11.6$, so the torque increase from the motor to the wheel is 11.6.  Since your original torque increase was 5.81, this is double.&lt;/p&gt;&#xA;" OwnerUserId="7692" LastActivityDate="2016-08-19T04:13:46.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11149" PostTypeId="2" ParentId="11144" CreationDate="2016-08-19T04:14:23.797" Score="1" Body="&lt;p&gt;One atmosphere pressure difference means the pressure one one face of this membrane is higher than the other face by one atmosphere.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the driving force for tests to measure the permeability of membranes for certain gases under constant pressure gradient and constant temperature. &lt;a href=&quot;http://www.separationprocesses.com/Membrane/MT_Chp03b2.htm&quot; rel=&quot;nofollow&quot; title=&quot;link&quot;&gt;check here&lt;/a&gt;. Pressure difference is to test static mechanical permeability. Many other test are needed for each chemical or substance.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-08-19T04:14:23.797" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11150" PostTypeId="1" CreationDate="2016-08-19T04:55:16.463" Score="6" ViewCount="323" Body="&lt;p&gt;I have a DC motor system with three states.&#xA;The states are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$x_1$ = position (degrees), &lt;/li&gt;&#xA;&lt;li&gt;$x_2$ = velocity (degrees per second)&lt;/li&gt;&#xA;&lt;li&gt;$x_3$ = current (Amps).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My plant input $u(t)$ is measured in volts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dynamics are such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;\frac{dX}{dt} = AX + Bu(t) \\&#xA;Y = CX + Du(t)&#xA;\end{gather}$$&#xA;And:&#xA;$$\begin{align}&#xA;A &amp;amp;= \left[\begin{matrix}&#xA;0 &amp;amp; 1                 &amp;amp; 0     \\&#xA;0 &amp;amp; -\dfrac{b_m}{J_m} &amp;amp; \dfrac{K_m}{J_m} \\&#xA;0 &amp;amp; -\dfrac{K_m}{L}   &amp;amp; -\dfrac{R}{L}&#xA;\end{matrix}\right] \\&#xA;B &amp;amp;= \left[\begin{matrix}&#xA;0 \\ &#xA;0 \\&#xA;\dfrac{1}{L}&#xA;\end{matrix}\right] \\&#xA;C &amp;amp;= \left[\begin{matrix}&#xA;1 &amp;amp; 0 &amp;amp; 0&#xA;\end{matrix}\right] \\&#xA;D &amp;amp;= [0]&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $J_m$ is the rotor inertia, $b_m$ is the viscous damping, $R$ is the armature resistance, $k_m$ is the motor constant, and $L$ is the inductance.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My objective is to control the position of my DC motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;rank&lt;/strong&gt; of my observability matrix is equal to the number of states in my system, so the system is fully observable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q = \text{rank}&#xA;\left(\left[\begin{matrix}&#xA;C \\&#xA;CA \\&#xA;CA^2&#xA;\end{matrix}\right]\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am measuring the position using an encoder and hope to employ a reduced order observer to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partitioning the A matrix into quadrants yields the measured and observed dynamics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A = \left[\begin{matrix}&#xA;A_{11} &amp;amp; A_{12} \\&#xA;A_{21} &amp;amp; A_{22}&#xA;\end{matrix}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case:&#xA;$$\begin{align}&#xA;A11 &amp;amp;= [0] \\&#xA;A12 &amp;amp;= \left[\begin{matrix}&#xA;1 &amp;amp; 0&#xA;\end{matrix}\right] \\&#xA;A21 &amp;amp;= \left[\begin{matrix}&#xA;0 \\&#xA;0&#xA;\end{matrix}\right] \\&#xA;A22 &amp;amp;= \left[\begin{matrix}&#xA;-\dfrac{b_m}{J_m} &amp;amp; \dfrac{K_m}{J_m} \\&#xA;-\dfrac{K_m}{L}   &amp;amp; -\dfrac{R}{L}&#xA;\end{matrix}\right]&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the &lt;strong&gt;Observability&lt;/strong&gt; of my reduced state observer partition is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_{Obs} = \text{rank}&#xA;\left(\left[\begin{matrix}&#xA;A_{12} \\&#xA;A_{12}A_{22}&#xA;\end{matrix}\right]\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rank of this is 2 (full rank).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen in other projects that the following condition must be satisfied:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;Q_{Obs} &amp;amp;\approx Q-(\text{number of states}-\text{number of measured states}) \\&#xA;2 &amp;amp;\approx 3-(3-1) \\&#xA;2 &amp;amp;\approx 1&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How many states need to be measured in a reduced order observer?&lt;/li&gt;&#xA;&lt;li&gt;Why is it that my system is fully observable, but I cannot measure all the states of my system?&lt;/li&gt;&#xA;&lt;li&gt;If I measure both position and velocity it seems that $Q_{Obs}=1$. Making the equation:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$Q_{Obs} \approx Q-(\text{number of states}-\text{number of measured states})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;simplify to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;1 &amp;amp;\approx 3-(3-2) \\&#xA;1 &amp;amp;\approx 2&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean the system is not observable without measuring all three states?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4.&lt;/strong&gt; If I assume the mechanical time constant is much slower than the electrical time constant (poles are closer to 0 in the real axis):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Can I measure only position and estimate velocity while neglecting the electrical dynamics?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will this provide a robust estimator when implemented?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7693" LastEditorUserId="7693" LastEditDate="2016-08-19T20:45:11.920" LastActivityDate="2016-09-18T21:15:15.943" Title="How many states need to be measured for a reduced order observer?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11152" PostTypeId="1" CreationDate="2016-08-19T09:56:33.407" Score="2" ViewCount="287" Body="&lt;p&gt;I have an 4x4 which I have heavily modified, the only thing missing is mechanical differential locks. I think it wouldn't be too hard to fit a four channel abs pump from another vehicle and produce my own controller. Which would give me more or less the same ability as diff locks plus a few more features without the hefty price tag.&#xA;Features I intend to have are.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Individually locking a wheel on demand (just like fiddle brakes at the touch of a button for all 4 wheels)&lt;/li&gt;&#xA;&lt;li&gt;Two potentiometers for automatic control of the amount of slip to allow between left and right wheels.&lt;/li&gt;&#xA;&lt;li&gt;Automatic air stop. For detecting front wheel air (on jumps etc)to slow them to a min speed when we detect a run away difference between front and rear.(this will stop breakage of front drive components when they hit the ground).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I plan to monitor wheel speed with 4 hall effect sensors and abs rings fitted to each wheel. The control unit will be an arduino and will count pulses on interrupt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electronics and programming side of things I am quite proficient at.&#xA;The hydraulic side of things no so much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that these abs pumps have 4 channels and two feeds usually labelled mc1 mc2, I assume one for front and one for rear from the master cylinder.&#xA;each channel is controlled by a three way valve. This is where my knowledge runs out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question then is how are these valves usually controlled in terms of positions and electrical connection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whats the maximum / typical frequency these valves actuate at or is it variable depending on rotational speed?&lt;/p&gt;&#xA;" OwnerUserId="7698" LastActivityDate="2016-08-19T09:56:33.407" Title="ABS Traction control / fiddle brakes" Tags="|electrical-engineering|hydraulics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11153" PostTypeId="2" ParentId="11143" CreationDate="2016-08-19T11:30:01.793" Score="1" Body="&lt;p&gt;Looks like this report, and the links therein, are seriously lacking in technical info. I did a quick lmgtfy and found some info on the munition in question at &lt;a href=&quot;http://www.thefirearmblog.com/blog/2010/02/17/usmc-adopt-new-5-56mm-mk318-mod-0-ammunition/&quot; rel=&quot;nofollow&quot;&gt;the firearm blog&lt;/a&gt;. Quoting: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;SOCom developed the new rounds for use with the Special Operations&#xA;  Force Combat Assault Rifle, or SCAR, which needed a more accurate&#xA;  bullet because its short barrel, at 13.8 inches, is less than an inch&#xA;  shorter than the M4 carbine’s. Using an open-tip match round design&#xA;  common with some sniper ammunition, SOST rounds are designed to be&#xA;  “barrier blind,” meaning they stay on target better than existing M855&#xA;  rounds after penetrating windshields, car doors and other objects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd guess this round would still have greater range &amp;amp; accuracy when launched from a long-barrel rifle, but it exceeds the Marine Corps' specs for range and accuracy.  There appear to be a large number of non-round reasons why the M16 is being replaced, including maneuverability, burst mode operation, and so on.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-08-19T11:30:01.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11155" PostTypeId="1" AcceptedAnswerId="11192" CreationDate="2016-08-19T17:23:33.173" Score="2" ViewCount="50" Body="&lt;p&gt;I'd like to mock up a quick prototype of a chassis assembly for a type of cart, which, in production, would be composed of CNC bent aluminum extrusions (hollow, approximately 5/8&quot; circular profile), and connected using plastic joints. I'm wondering if anyone could recommend methods for prototyping the tubular components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering using PVC, but I would like to use something I could bend and shape angles into relatively easily.&lt;/p&gt;&#xA;" OwnerUserId="7703" LastActivityDate="2016-08-25T18:33:32.773" Title="Prototyping tubular components" Tags="|prototyping|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11157" PostTypeId="2" ParentId="11150" CreationDate="2016-08-19T19:03:32.260" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;It is not always states that are measured. The measurements $y$ are combinations of states in general or a subset. In the reduced order observer, in this case, you have one measurement. &lt;/li&gt;&#xA;&lt;li&gt;The system being fully observable means that using the available measurement, all the states can be observed or indirectly measured.&lt;/li&gt;&#xA;&lt;li&gt;The equation does not make sense. But as I mention above you don't have to directly measure all the states.&lt;/li&gt;&#xA;&lt;li&gt;You can measure just the position and velocity and not neglect the electrical dynamics. Use the complete dynamics to estimate the current. This is the purpose of the observer.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the observability rank $Q_{Obs}$ is tripping you. Observability is a property of the system not the observer. 'Observability of my reduced state observer' is nonsense speak. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With one measurement the system is observable. That means using the knowledge of the system behavior ($A$, $B$, $C$, and $D$) together with the input signal $u$ and the measurement $y$ all the states can be estimated. The reduced state observer goes one step ahead: because one of the states is already measured that need not be estimated, only the remaining two need to estimated. You wouldn't use a reduced state observer if the measurement was very noisy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the system is observable with one measurement then it is definitely so with two. Again, if you have enough confidence in those measurements, they can be used directly and a reduced state observer can be used to estimate just the remaining state.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2016-08-19T21:05:51.903" LastActivityDate="2016-08-19T21:05:51.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11159" PostTypeId="2" ParentId="7019" CreationDate="2016-08-19T21:42:18.570" Score="2" Body="&lt;p&gt;Yes, it can be removed but it is a delicate task requiring some experience...I first did this over 50 years ago.  You should be drawing on some heavy velum that can withstand the erasing.  You will need an erasing shield and electric eraser with the coarse dark grey eraser to gently sand a layer of paper off with the ink.  It's easy to erase a hole.  If you're going to do a lot of ink drafting you may want to look into mylar!   Goid luck drafting ....computers suck!&lt;/p&gt;&#xA;" OwnerUserId="7708" LastActivityDate="2016-08-19T21:42:18.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11160" PostTypeId="1" CreationDate="2016-08-20T01:21:22.417" Score="0" ViewCount="882" Body="&lt;p&gt;I have a question about gas valve selection and working out the CV. The equation I have needs the mass flow in kg/hr, but I have been given the volumetric flow rate in litres/minute. How do I go about converting this? Thanks&lt;/p&gt;&#xA;" OwnerUserId="7710" LastActivityDate="2016-08-20T15:51:46.063" Title="Natural gas mass flow conversion" Tags="|gas|compressible-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11161" PostTypeId="2" ParentId="11128" CreationDate="2016-08-20T01:41:11.997" Score="1" Body="&lt;p&gt;If you know the specifications (pipe smoothness) of the pipe, then you can use the Darcy-Weisbach equation to calculate the pressure drop of the water across your 5m pipe. It is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta P = f \frac{L}{D} \frac{\rho V^{2}}{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where P is your pressure, f is the friction factor, L is the length of your pipe, D is the diameter, rho is the density of water and V is the flow velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can use the pump duty equations in order to work out what power your pump needs to be in order to deliver 6 bar at the nozzle at your required velocity. &lt;/p&gt;&#xA;" OwnerUserId="6521" LastActivityDate="2016-08-20T01:41:11.997" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11162" PostTypeId="2" ParentId="11143" CreationDate="2016-08-20T01:55:57.670" Score="2" Body="&lt;p&gt;The main reason that the SOST round performs as well as the legacy M855 in terms of range and accuracy (all while being shot from a shorter barrel) is because it was &lt;em&gt;designed&lt;/em&gt; to be fired from the shorter barrel from the get-go.  This mostly comes down to adjusting the propellant geometry and burn rate so that it all burns up before the bullet exits the gun.  The M855 propellant was designed to burn up within the 20 inches of bullet travel it got in the M16.  Firing it from a 14 or 10 inch barrel puts it way outside its original design envelope, so it's no wonder that it's lacking in performance when fired from the M14 or the SCAR.  There's also a trade-off between barrel wear and muzzle velocity; you can use a hotter propellant that will give you more velocity, at the expense of shorter barrel lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does raise the question as to why anyone used a 20-inch barrel in the first place, if the same performance can be had from a 10.5-inch barrel.  The reason is that the legacy M855 round has much better armor-piercing capabilities than the SOST.  That article leaves this part out, and in doing so doesn't make a fair comparison between the two rounds.  The M855 was designed during the cold war, where it would likely be used against enemies wearing some body armor or against light-armored vehicles.  For defeating those kinds of targets, you need velocity.  The M855 fired from an M16 has a few hundred feet per second over the SOST fired from the SCAR as far as I can tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the performance improvements of ammo is of course due to advancements in technology, but very often it's just the result of re-optimizing a few system components to adjust to changing requirements.  I think that's the main reason for the perceived improvement of the SOST over the M855.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-08-20T01:55:57.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11163" PostTypeId="2" ParentId="11130" CreationDate="2016-08-20T02:12:11.733" Score="5" Body="&lt;p&gt;As with many questions, the real answer is &amp;quot;it depends.&amp;quot; The biggest variables are what kind of engineer you are, who your audience is, and what industry you're working in. In a class setting, it is probably a good idea to consult your textbook or your teacher for their opinions in the context of the class, but here are some general factors in making the decision when you don't have someone to ask. In industry, if I know the department or company that is going to do the actual work, I do usually ask them what information they need and try to tailor the drawing to them. More formally many companies will have internal policies or follow consensus standards like &lt;a href=&quot;https://www.asme.org/products/codes-standards/y145-2009-dimensioning-and-tolerancing&quot; rel=&quot;noreferrer&quot;&gt;ASME Y145&lt;/a&gt; or the &lt;a href=&quot;http://www.iso.org/iso/iso_catalogue/catalogue_ics/catalogue_ics_browse.htm?ICS1=1&amp;amp;ICS2=100&quot; rel=&quot;noreferrer&quot;&gt;ISO ICS 01.100&lt;/a&gt; series. But often, you won't have that luxury. Maybe you are drawing something to go out to competitive bid, or for a project that has no manufacturing pipeline set up yet. In this case you have to make decisions based on the context. A number of other answers here provide good guidelines for relatively conventional machined parts, but don't necessarily address other types of drawings you may need to produce.&lt;/p&gt;&#xA;&lt;p&gt;The general hierarchy of dimensioning goes like this:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Dimension critical features:&lt;/strong&gt; These are the items that must be right for your object to serve its purpose. For example if you are drawing a sprocket the center bore must be correct, or it will not fit on the shaft. For mechanical drawings, you should pay particular attention to the tolerances on these dimensions and make sure they are very clear. Make sure you are dimensioning what really matters to you, even if this isn't the most convenient way to dimension it for your audience. In architecture and some other disciplines, you can also add &amp;quot;HOLD TO&amp;quot; to the dimension, to indicate that other dimensions may have to be adjusted slightly to make this dimension correct.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Dimension all functional requirements:&lt;/strong&gt; These are all things that matter to your design, but are not as essential as your critical features. You should make sure that they are dimensioned and clear, but you don't have to draw extra attention to them like you do with a critical dimension. For example on the sprocket, the thickness of the sprocket needs to be machined correctly for it to fit, but it won't be a very tight tolerance.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Dimension things your audience needs to know:&lt;/strong&gt; In many designs there are some aspects that are arbitrary. Maybe in your situation the thickness of the sprocket hub doesn't matter at all - it's under very little load and going in the middle of a big empty shaft for example. All the same, you should dimension it so that the person making it doesn't have to make the choice for you. If you want to allow them to make some economical decisions (eg start with a material size they have in stock) you could chose to put a very wide tolerance on the dimension. If it impacts what the next person has to do, you should provide some information.&lt;/p&gt;&#xA;&lt;p&gt;Note the this section can be a little different for different industries. Machinists for example are used to having a part fully defined on a drawing so there's no guesswork for them to do. On the other hand if you are drafting a stud wall for a carpenter to frame out in a building, you would typically show them the minimum stud spacing, any special conditions (doorways, windows, etc.) but not every other detail. For example, it typically wouldn't matter to the engineer which side they started the studs from and where they ended as long as there is no gap greater than you specified. It might be easier for them to start from the north side because that's where the lumber is, or easier to start from the south side because the adjacent wall has already been framed. They might get clever and start the pattern centered on the wall to save one stud's worth of material. As long as the drawing contains all of the dimensions that matter to you, it's OK to leave it under-defined in this situation. For this step, it is really helpful to understand the process that someone else will be using to make your part whether it's machined, welded, carved, molded, etc.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Dimension things that other audiences would like to know (optional:)&lt;/strong&gt; Often there is useful information you could provide, even though it doesn't matter to the person you're sending the drawing to. In the sprocket example, the machinist won't care what the pitch diameter is if you've already given them the profile to cut. All the same, you might decide to add it. This information will be useful if another mechanical designer wants to use your sprocket and comes across your drawing first. Generally since these dimensions are just for information, not an instruction to the next person, they will be indicated as reference dimensions. That should be done by placing them in parenthesis or adding the abbreviation 'REF' after the word.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Once all of these dimensions are listed, and the appropriate tolerances, materials, processes, etc are specified, your drawing should be complete. Now it's possible that if your object is complicated, the drawing will be cluttered with lots of dimensions that may seem redundant. There are a few options to reduce the total number of dimensions but retain the important information. This list is certainly not exhaustive, but it should get you started in most applications. Note that some of these make more work for your audience, so they may not always be received warmly.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Patterns:&lt;/strong&gt; If your features follow a grid pattern, it is sufficient to dimension one object in each row/column rather than each item. Similarly if your objects are all distributed rotationally around a point, you could dimension the circle they fall on and the spacing between them. Then as long as one starting position is given, the next person can construct the rest of the pattern.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DFLBU.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DFLBU.png&quot; alt=&quot;Object with grid dimensions.&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XP5TB.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XP5TB.png&quot; alt=&quot;Object with grid dimensions and alignment lines.&quot; /&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;When the pattern is not obvious, it's a good idea to provide dotted lines that clarify which features are in the pattern and which aren't.&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Typical:&lt;/strong&gt; If your drawing includes many features that match (entirely or in some aspect) you may dimension one and add &amp;quot;TYPICAL&amp;quot; or just &amp;quot;TYP&amp;quot; after to indicate that everything that looks like that, is the same as that.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Similarly if a feature appears more than one place, you can describe the number - give the details on one location and then specify &amp;quot;8 PLACES&amp;quot; for example.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XqwAf.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XqwAf.png&quot; alt=&quot;Object with typical dimensions.&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/laoMy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/laoMy.png&quot; alt=&quot;Object with places dimensions.&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Symmetry:&lt;/strong&gt; If your object is symmetrical, you need only dimension one half of it, and then show a center line.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/reYom.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/reYom.png&quot; alt=&quot;Object with a centerline.&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Profile-controlled Geometry:&lt;/strong&gt; When a feature or part is truly too complex to dimension clearly, even after you have considered detail and auxiliary views, your best bet may be a profile tolerance. With this option, you will provide a (digital) model of the curve or surface to your audience, and your drawing will show the shape without any actual dimensions. A profile tolerance block should be used, which indicates that the dimensions are controlled by the profile and gives your audience information on what the tolerance is. Using a profile dimension on an object that you could dimension reasonably easily would come across as lazy and unprofessional, so save this for truly complex shapes! More information on profile tolerances is available &lt;a href=&quot;http://www.tec-ease.com/gdt-terms.php&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.gdandtbasics.com/profile-of-a-surface/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. Note that even when you aren't using  profile controlled geometry, you may still give a copy of your digital file to the shop you are working with as a courtesy to speed up their process. In general if you provide a properly dimensioned drawing, it will be considered to supersede the computer geometry unless you specifically agree otherwise.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Lastly, there are a number of good textbooks on drafting that will go into much more detail with better examples and style guidance. One that I would recommend is Technical Drawing with Engineering Graphics (Giesecke et al.)&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-08-20T02:12:11.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11164" PostTypeId="2" ParentId="11130" CreationDate="2016-08-20T05:49:54.717" Score="2" Body="&lt;p&gt;One should never ever measure things from a drawing. There are several reasons for this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The printer was misconfigured, and the output is no longer in scale. This happens a lot. Even if not the printer has a limited resolution so very small details may be off anyway.&lt;/li&gt;&#xA;&lt;li&gt;The measurement may be inaccurate maybe the dimension in question is 10.1mm but the machinist measured from the drawing and interpreted it as 10mm. Bad! Also for longer spans the measurement may be a bit skewed etc.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The drawing is an acceptance document, what you're documenting is the acceptance criteria. You should thus have all the dimensions that matter to you marked out. Otherwise you may need to pay for all manufacturing defects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should not mark out all dimensions just so much that the dimension  loop is complete. So to not introduce conflicting information in the drawing. If you just add those dimensions put them in parentheses. This may be useful for the machinist if the acceptance criteria is not really useful for machining for some reason (this can be bad for very accurate parts though).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you totally leave out a dimension loop then you are leaving things to the machinists judgement. While this may not be bad as such, you're letting somebody else do the thinking in a way that produces something spectacularly different from your needs. Most shops would not accept such jobs but your local machinist might.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But no you should try to add as few dimensions as possible that captures what your needs are. And suplement with as much data as the manufacturer needs to get things done.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorDisplayName="user748" LastEditDate="2016-10-03T03:03:36.927" LastActivityDate="2016-10-03T03:03:36.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11165" PostTypeId="1" CreationDate="2016-08-20T13:00:55.643" Score="2" ViewCount="427" Body="&lt;p&gt;I want to basically make a pin matrix controller either by using spring, electromagnets or small motors. Spring being the most viable option. Something like what’s shown in the image. I'm pretty new to arduino and micro-controllers in general so any input would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pplxi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pplxi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7712" LastEditorUserId="7083" LastEditDate="2016-08-20T17:52:59.257" LastActivityDate="2016-08-20T17:52:59.257" Title="Make a prototype of a braille tablet using arduino" Tags="|electrical-engineering|embedded-systems|" AnswerCount="0" CommentCount="2" ClosedDate="2016-08-21T16:15:22.997" ContentLicense="CC BY-SA 3.0" />
  <row Id="11167" PostTypeId="2" ParentId="11160" CreationDate="2016-08-20T15:51:46.063" Score="2" Body="&lt;p&gt;You can probably assume the natural gas is an ideal gas, then apply the corresponding equation of state on any arbitrary control volume of the flow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;pV=mRT&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p$ is pressure, $V$ is volume, $m$ is mass, $R$ is the &lt;em&gt;specific&lt;/em&gt; gas constant, and $T$ is temperature.  $R$ for methane is $518.3 \frac{J}{kg \cdot K}$, which might be close enough.  You will also need to know pressure and temperature, then just solve the equation of state for mass.  Don't forget to convert your units of time as well.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2016-08-20T15:51:46.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11168" PostTypeId="1" CreationDate="2016-08-20T16:21:15.503" Score="0" ViewCount="108" Body="&lt;p&gt;I have an application in which speed of the shaft is 0.54&amp;nbsp;rpm. The input speed of the motor (induction 4 pole AC motor) is 1400&amp;nbsp;rpm. So the gear reduction is enormous. I had two helical redactors coupled together and I read that in this situation  when you have such big reduction (2 stage reduction) and super low speed you have to limit the input torque. I don’t understand this – why is this necessary? The start up torque of AC motor is 2-4 times higher than the nominal torque, how to start this application? Can someone tell me why in big reduction systems it is necessary to limit input torque? &lt;/p&gt;&#xA;" OwnerUserId="7714" LastEditorUserId="1832" LastEditDate="2016-08-20T17:36:33.513" LastActivityDate="2016-08-22T14:38:47.383" Title="Torque limiting application" Tags="|mechanical-engineering|gears|manufacturing-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11174" PostTypeId="1" AcceptedAnswerId="11193" CreationDate="2016-08-21T14:18:38.430" Score="2" ViewCount="378" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OSkwJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OSkwJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I convert this into a mechanical network? like a circuit network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my attempted work is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EfGBF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EfGBF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is this schematic correct?&lt;/p&gt;&#xA;" OwnerUserId="1363" LastEditorUserId="1363" LastEditDate="2016-08-21T15:11:21.587" LastActivityDate="2016-08-22T18:44:43.950" Title="How do I convert this mechanical system into a mechanical network?" Tags="|control-engineering|control-theory|systems-engineering|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11176" PostTypeId="1" CreationDate="2016-08-21T16:03:53.637" Score="1" ViewCount="374" Body="&lt;p&gt;When a thermally isolated vessel of liquid water is depressurized to vacuum, part of water will evaporate, remainder will freeze - total heat being split between latent heat of ice and enthalpy of vaporization. I wanted to calculate the proportions - how much will evaporate, and how much will freeze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is the values of latent heat and enthalpy of vaporization wander quite a bit with pressure, and they are quite hard to find for anything else than 1 bar. I managed to obtain the value for steam (2501KJ/kg), but I'm still not successful with ice - there is a multitude of tables and documents that list a variety of materials at 1 bar, so if there is one that follows the pressure, I was unable to find it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - what would be the latent heat of fusion of water at about 612 Pa?&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2016-08-21T16:03:53.637" Title="Latent heat of fusion of water near triple point" Tags="|heat-transfer|pressure|data|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11177" PostTypeId="1" CreationDate="2016-08-21T17:17:22.483" Score="2" ViewCount="709" Body="&lt;p&gt;So, first of all, I am a High School student and I wanted you to explain me something (in a more &quot;concrete&quot; way, if you know what I mean): I had an idea the other day to make a car with a wind turbine attached to it, so that the more the car runs, the more wind gets into the turbine and more energy is generated, wich powers the car. I know this is some kind of &quot;moto perpetuo&quot; so it wouldn't work anyway, so I asked my physics teacher about it and he said It cannot work because of the first law of thermodynamics, and he asked me a way to calculate the energy generated by the turbine. I just wanted a concrete way of showing it doesn't work, like, using data from wind turbines that exist, cars that exist and so on. But how do I calculate the energy generated?&lt;/p&gt;&#xA;" OwnerUserId="7726" LastActivityDate="2024-03-31T19:25:25.963" Title="Why isn't it possible to build a car moved by wind power?" Tags="|thermodynamics|automotive-engineering|motors|wind-power|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11178" PostTypeId="1" CreationDate="2016-08-21T17:27:55.820" Score="2" ViewCount="722" Body="&lt;p&gt;IANAE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to make an 1/8&quot; steel rotary switch shaft into 1/4&quot; so that it will accept the much more common 1/4&quot; shaft knobs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use McMaster-Carr 92510A422:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XJLRb.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XJLRb.gif&quot; alt=&quot;mcmaster-carr 92510A422&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is an aluminum unthreaded spacer that has an ID of 0.115&quot;, will heating it expand it enough to fit the 0.125&quot; shaft?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 13&quot; drill press but otherwise, I do not have any ability to mill things so I need to start with off-the-shelf parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just realized, the nut that goes over the shaft has an ID of only 0.2175&quot;. So it would not fit over the new shaft.&lt;/p&gt;&#xA;" OwnerUserId="7727" LastEditorUserId="7727" LastEditDate="2016-08-21T18:34:53.900" LastActivityDate="2016-12-19T21:46:01.467" Title="Heat Fit Aluminum Sleve onto Shaft" Tags="|aluminum|heat-treatment|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11179" PostTypeId="2" ParentId="11177" CreationDate="2016-08-21T17:53:38.397" Score="6" Body="&lt;p&gt;The fundamental problem in this case is that the turbine in not powered by 'wind' as such but by the relative motion of the car through the air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not 'free' energy, because the turbine must do work to generate energy it must also exert a net force opposing the motion of the car ie drag. So any energy you generate with the turbine must ultimately be provided by the car's engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also no matter how well designed the turbine is some energy is wasted (second law of thermodynamics) so you will always be worse off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that there are some similar situations where you can improve the efficiency of a car with additional equipment. The difference it that, to work they much somehow harvest energy which would otherwise be wasted by the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One well established method is a turbocharger. This is a turbine powered not by air flowing over the body but by exhaust gasses. Conventionally turbos are used to pressurise air inducted into the engine whcih can certainly increase power and potentially improve thermal efficiency. They can also be  connected to motor-generators which (in hybrid vehicles at least) can harvest electrical energy from the exhaust (see current F1 engines for more details). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is regenerative braking. Here instead of using friction between brake pads and disks to slow the car down (which converts kinetic energy to heat which is lost to the surroundings)  braking is achieved by using the drive train to drive a generator connected to a load (usually a battery) capable of storing energy. Again current F1  technology has used this as part of a strategy to achieve fairly spectacular improvements in thermal efficiency . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the turbine method you suggest &lt;em&gt;could&lt;/em&gt; be used as a form of regenerative braking as in this case you actually want the drag from the turbine to slow the car down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases the key point is that you are &lt;strong&gt;harvesting energy which would otherwise be wasted&lt;/strong&gt; and thus improving the overall efficiency of the system. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-08-21T17:53:38.397" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11180" PostTypeId="2" ParentId="11178" CreationDate="2016-08-21T18:12:04.953" Score="2" Body="&lt;p&gt;Shrink fitting depend on the fact that you have a shaft and collar which are sized to an interference fit when cold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem with aluminium is that you have a fairly narrow window for heating it before it first anneals and then melts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rough calculation  suggests that that change in diameter with heating to 200C is in the order of thousandths of an inch which is somewhat less that the tolerance of your part (although this is back of the envelope stuff so I may be wrong but is suspect that its not too far out). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case I think that for a knob you will be much better off using an adhesive,  epoxy or a &lt;a href=&quot;http://www.loctite.co.uk/retaining-9171.htm&quot; rel=&quot;nofollow&quot;&gt;bearing retaining adhesive&lt;/a&gt; should be plenty strong enough and a lot more convenient.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-08-21T18:12:04.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11182" PostTypeId="1" AcceptedAnswerId="11185" CreationDate="2016-08-22T06:07:00.957" Score="-1" ViewCount="1770" Body="&lt;p&gt;I think this is very fundamental question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering why 0.00259 is used as weight-mass conversion ratio, when inch/pound unit is used in Nastran.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain the reasons?&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="4624" LastEditDate="2016-08-22T06:15:37.317" LastActivityDate="2016-08-22T09:24:46.120" Title="Where does weight mass conversion ratio '0.00259' come from in Nastran?" Tags="|simulation|nastran|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11183" PostTypeId="2" ParentId="11168" CreationDate="2016-08-22T08:54:20.003" Score="2" Body="&lt;p&gt;Because the torque is multiplied by the same factor that you reduce the speed with. For large reductions this can create a lot of torque. If the output is seized up then this results in large forces which will start damaging things like bending or shearing shafts, stripping gears or cracking the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common way to limit torque is to include a sacrificial plastic gear that gets stripped when torque becomes too large. This is then the designated weak point just like the fuse in an electrical system.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2016-08-22T14:38:47.383" LastActivityDate="2016-08-22T14:38:47.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11184" PostTypeId="1" CreationDate="2016-08-22T09:12:14.907" Score="3" ViewCount="57644" Body="&lt;p&gt;Could anyone please explain what is the difference between active and passive earth pressure when we talk about weight retaining walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both are lateral pressure exerted by the soil on a retaining wall, but why do we call one active and the other passive ?&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditDate="2016-08-24T08:06:57.593" LastActivityDate="2016-10-29T20:56:59.570" Title="Active vs passive earth pressure" Tags="|civil-engineering|geotechnical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11185" PostTypeId="2" ParentId="11182" CreationDate="2016-08-22T09:24:46.120" Score="5" Body="&lt;p&gt;It's the inverse of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravitational_acceleration&quot;&gt;acceleration of gravity&lt;/a&gt; g in imperial units. &#xA;$$&#xA;1 g = 386.1 in/s^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;1/g = 0.00259 s^2/in&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need that since:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;weight = mass * g&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1125" LastActivityDate="2016-08-22T09:24:46.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11186" PostTypeId="2" ParentId="11184" CreationDate="2016-08-22T10:41:02.473" Score="0" Body="&lt;p&gt;Active earth pressure is the one that is exerted by the soil that tends to overturn or slide the retaining wall. It is unfavorable to stability of the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Passive earth pressure is the one exerted from the other side and that tends to stabilize it. Generally, it is applied near the &quot;toe&quot; of the wall.&lt;/p&gt;&#xA;" OwnerUserId="413" LastActivityDate="2016-08-22T10:41:02.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11188" PostTypeId="2" ParentId="11177" CreationDate="2016-08-22T12:50:32.017" Score="1" Body="&lt;p&gt;You are correct that you can't create energy by moving faster, since the only thing &quot;wind&quot; will do is oppose your motion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's certainly possible to build a vehicle which rolls on wheels and is powered either by a sail or a turbine-shaped mechanism.  In fact, there's the famous (and argument-generating) design which allows a vehicle to go dead down-wind faster than the wind.  See, for example, &lt;a href=&quot;https://vimeo.com/10477373&quot; rel=&quot;nofollow&quot;&gt;DDWFTW&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, these designs are completely impracticle for on-road navigation.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-08-22T12:50:32.017" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11189" PostTypeId="2" ParentId="11177" CreationDate="2016-08-22T13:19:53.373" Score="0" Body="&lt;p&gt;It does work - with some limitations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a sand yacht, in something like a 15mph breeze on the beam (normal to the direction of travel). As the sand yacht starts moving, the &lt;a href=&quot;http://www.noughtinfinity.net/WBH/images/pdf/sb_technical_paper.pdf&quot; rel=&quot;nofollow&quot;&gt;apparent wind&lt;/a&gt; increases and shifts forward. Sheet in, to sail close hauled. and (anecdotally) the thing accelerates like a rocket and you can be doing 50mph+ (generally with no brakes) with an apparent wind that feels almost dead ahead (you can calculate its speed and direction from simple geometry)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no magic involved, you are initially extracting power from the 15mph wind, and visiting a larger volume of it (tapping more available energy) when at speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Designing a wind turbine - which is really just a set of sails proceeding normal to the wind direction, you find the apparent wind increases as the blade starts turning - the &quot;tip speed ratio&quot; determines teh most efficient angle to set the blade and is often 6:1, or 90mph for the same breeze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Limitations come with other wind directions. If the wind is dead ahead, a sail cannot extract power from it, so you would have to proceed at an angle to the actual wind. Tacking across all lanes of the highway is considered anti-social and quite possibly dangerous, unless everybody else is doing the same. This is where a wind turbine geared to drive the wheels could succeed, and again, the apparent wind sped will increase as you move to windward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most difficult direction, ironically, is the one you might think was easiest - with the wind dead aft. You are simply blown along, until the apparent windspeed falls towards zero. No further acceleration through this windless point is possible, though I have met someone who claimed that you could extract energy to sail downwind faster than true windspeed, provided you could sail &quot;round&quot; this point rather than through it. Never seen it done though...Here's &lt;a href=&quot;http://www.google.co.uk/patents/US2804038&quot; rel=&quot;nofollow&quot;&gt;one of the patents&lt;/a&gt; if you're interested...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/05/bauer_tacking.pdf&quot; rel=&quot;nofollow&quot;&gt;Related work...&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2016-08-22T13:19:53.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11192" PostTypeId="2" ParentId="11155" CreationDate="2016-08-22T18:12:22.377" Score="2" Body="&lt;p&gt;&lt;em&gt;EDIT - I've solved this issue after a bit of searching for similar application requirements, so I'll post my solution:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying a bit of heat using a heat gun to sand-filled PVC pipe allowed me to slowly bend the pipe into my desired shape. The sand prevents the PVC from collapsing inward as it is bent.&lt;/p&gt;&#xA;" OwnerUserId="7703" LastEditorUserId="7703" LastEditDate="2016-08-25T18:33:32.773" LastActivityDate="2016-08-25T18:33:32.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11193" PostTypeId="2" ParentId="11174" CreationDate="2016-08-22T18:44:43.950" Score="2" Body="&lt;p&gt;I think what you have is correct, provided the friction and spring constants are inverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am more comfortable with the force-voltage analogy and would create the equivalent one as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6npCG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6npCG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-08-22T18:44:43.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11202" PostTypeId="1" CreationDate="2016-08-23T20:22:07.280" Score="0" ViewCount="8566" Body="&lt;p&gt;If I have a pipe with external radius R and internal radius r. Maximum compressive force will be given by the same formula Fmax=σS as for cylinder or different? where $S = πR^{2}-πr^{2}$, and σ is compressive strength of material. In other words, if I have some shape with the same mass per meter length as cylinder, maximum compressive force will be the same or different? For example, if I have solid cylinder from steel with 10 cm radius, area will be 314 square cm, and if I have pipe from the same steel with external radius 20 cm and internal radius 17.32 cm, area and mass per meter length will be the same as cylinder 10 cm radius. What can handle more compressive force pipe or cylinder or it will be the same? If it is different, maybe someone can give me useful links and formulas?&lt;/p&gt;&#xA;" OwnerUserId="7687" LastActivityDate="2016-10-24T11:07:20.963" Title="Maximum compressive force for pipe vs solid cylinder" Tags="|mechanical-engineering|materials|strength|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11203" PostTypeId="1" AcceptedAnswerId="11209" CreationDate="2016-08-23T21:14:00.753" Score="0" ViewCount="3119" Body="&lt;p&gt;Is the mass flow rate constant in a thermodynamic cycle? More specifically a Rankine or Carnot cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if the mass flow rate of the turbine is 100 kg/s, is the mass flow rate for the boiler / condenser / compressor or pump also 100 kg/s?&lt;/p&gt;&#xA;" OwnerUserId="4442" LastActivityDate="2016-08-25T05:35:09.407" Title="Is the mass flow rate constant in a thermodynamic cycle?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11204" PostTypeId="1" AcceptedAnswerId="11210" CreationDate="2016-08-23T22:33:55.477" Score="6" ViewCount="5546" Body="&lt;p&gt;I'm trying to power and agitator (alternating circularly back and fourth) with a crank. I've seen a bunch of complicated mechanisms that do this, but I'm looking for the mechanically simplest method.&lt;/p&gt;&#xA;" OwnerUserId="539" LastActivityDate="2023-10-31T22:58:01.083" Title="Is there a mechanism that can turn circular motion into alternating circular motion?" Tags="|gears|pistons|mechanisms|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11205" PostTypeId="1" AcceptedAnswerId="11207" CreationDate="2016-08-24T01:01:17.020" Score="0" ViewCount="103" Body="&lt;p&gt;Approximately how many types of stress are known/defined? Ex. Piola stress,  Cauchy stress ...&lt;/p&gt;&#xA;" OwnerUserId="252" LastActivityDate="2016-08-24T01:41:20.457" Title="Approximately how many types of stress are there?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11207" PostTypeId="2" ParentId="11205" CreationDate="2016-08-24T01:41:20.457" Score="4" Body="&lt;p&gt;Your examples aren't really &quot;types&quot; of stress, but different ways to &lt;em&gt;measure&lt;/em&gt; stress. The real world structure just responds the way it responds - it doesn't worry about what sort of math you used when you tried to model its behaviour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stress_measures&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Stress_measures&lt;/a&gt; lists a handful different stress measures that have acquired &quot;names&quot;. Cauchy stress (&quot;true stress&quot;) and Second Piola-Kirchhoff stress are probably the most commonly used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't really make sense to talk about a measure of stress in isolation,  without also considering how to measure strain, and how to model the constitutive equations that give the relationship between stress and strain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example 2-PK stress works nicely with Green strain, which is a convenient way to define the behavior of a body which has &quot;small&quot; elastic deformations superimposed on arbitrary large rigid body motions - in particular, large rigid body rotations, where approximations like $\sin \theta \approx \theta$ are not appropriate. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-08-24T01:41:20.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11209" PostTypeId="2" ParentId="11203" CreationDate="2016-08-24T06:48:15.370" Score="3" Body="&lt;p&gt;&lt;strong&gt;Is the mass flow rate constant in a thermodynamic cycle?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;strong&gt;general&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Depends.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here is how;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one of your components effect the pumping/boiling etc. of course your mass flow rate will change, it may change periodicly or it may not. Also there can be leakage or something that interacts with your fluid, of course that will cause to decrease in your mass flow rate with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But mostly, in general thermodynamic cycles, you don't need to worry. Because they build with simple components (boiler/pump/turbine/condenser) and they are not a factor that mass flow rate depends. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If we be &lt;strong&gt;specific&lt;/strong&gt; about Rankine and Carnot cycles; your assumptions is actually &lt;strong&gt;TRUE!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the &lt;strong&gt;Rankine Cycle&lt;/strong&gt;;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g2zfs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g2zfs.png&quot; alt=&quot;Rankine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The working fluid in a Rankine cycle follows a closed loop and is&#xA;  reused constantly. (&lt;em&gt;Wikipedia&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That &quot;&lt;em&gt;reused constantly&lt;/em&gt;&quot; means the total mass of the fluid is constant in cycle, therefore you have to use the same fluid in the same system. Unless there is a malfunction at one of your parts, you can be sure that you mass flow rate is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the &lt;strong&gt;Carnot Cycle&lt;/strong&gt;;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KsTKT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KsTKT.png&quot; alt=&quot;Carnot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Carnot cycle is a &lt;strong&gt;theoretical thermodynamic cycle&lt;/strong&gt; proposed by Nicolas Léonard Sadi Carnot in 1824 and expanded upon by others in the 1830s and 1840s. It provides an &lt;strong&gt;upper limit on the efficiency&lt;/strong&gt; that any classical thermodynamic engine can achieve during the conversion of heat into work, or conversely, the efficiency of a refrigeration system in creating a temperature difference (e.g. refrigeration) by the application of work to the system. It is not an actual thermodynamic cycle but is a theoretical construct. (&lt;em&gt;Wikipedia&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So we know it is &lt;em&gt;theoretical&lt;/em&gt;. But again, if we speaking about &quot;&lt;strong&gt;a simple closed system&lt;/strong&gt; &lt;em&gt;(control mass analysis)&lt;/em&gt;&quot;, yes, the Carnot Cycle will be running with constant mass flow rate because of the same reason of Rankine Cycle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Don't forget to read &lt;em&gt;Mark&lt;/em&gt;'s comment about the real life application issues that happens while trying to hold mass flow rate constant.&lt;/p&gt;&#xA;" OwnerUserId="7468" LastEditorUserId="7468" LastEditDate="2016-08-25T05:35:09.407" LastActivityDate="2016-08-25T05:35:09.407" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11210" PostTypeId="2" ParentId="11204" CreationDate="2016-08-24T07:50:06.767" Score="11" Body="&lt;p&gt;What about a crank-rocker mechanism?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/djkxX.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/djkxX.gif&quot; alt=&quot;crank-rocker mechanism&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://www2.udec.cl/%7Emrazeto/sismecanicos/erdman/CONTENT/CHAPS/CROCK.GIF&quot; rel=&quot;nofollow noreferrer&quot;&gt;udec.cl&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;When I am looking for inspiration for mechanisms, I have a look in the book Mechanisms and mechanical devices, by Neil Sclater.&lt;/p&gt;&#xA;" OwnerUserId="7590" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-02-21T08:25:54.043" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="11211" PostTypeId="1" AcceptedAnswerId="11214" CreationDate="2016-08-24T08:11:11.353" Score="0" ViewCount="1852" Body="&lt;p&gt;A 2&amp;nbsp;m long beam fixed in one side with a 2000&amp;nbsp;kg concentrated load at free end. Profile of the beam is HEB100 material S235.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if the beam will break under this load. If yes, what is the maximum load it can handle and how do you calculate it?&lt;/p&gt;&#xA;" OwnerUserId="7757" LastEditorUserId="1832" LastEditDate="2016-08-24T10:23:09.667" LastActivityDate="2016-08-24T10:29:08.603" Title="Maximum deflection of a beam, fixed in one end and concetrated load at free end" Tags="|structural-engineering|civil-engineering|steel|structures|beam|" AnswerCount="2" CommentCount="2" ClosedDate="2016-08-24T15:34:31.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="11212" PostTypeId="2" ParentId="4035" CreationDate="2016-08-24T09:29:39.203" Score="0" Body="&lt;p&gt;You can use ANSYS Electromagnetics Suite to simulate your project. There are some videos on Youtube showing how to do it using ANSYS Maxwell and Simplorer.  In &lt;a href=&quot;https://www.youtube.com/watch?v=LW9TcpkZaZY&quot; rel=&quot;nofollow&quot;&gt;this example&lt;/a&gt;, a magnet is moving with a constant speed inside a coil.&lt;/p&gt;&#xA;" OwnerUserId="3715" LastActivityDate="2016-08-24T09:29:39.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11213" PostTypeId="2" ParentId="11211" CreationDate="2016-08-24T09:37:04.617" Score="1" Body="&lt;p&gt;To know whether the beam will break or not, you need to calculate the maximum stress in it and compare it to the maximum stress that your material can bear. For S235, it will be most likely 235MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the maximum stress is located in the section where the beam is fixed. If it breaks, it will break there. To calculate it, start by calculating the bending momentum $M$ in this point then calculate the stress using &#xA;$\sigma = \frac{M*y}{I}$ where $I$ is the section modulus. It should be given for your HEB100. And $y$ is the half the  height of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the maximum stress in your beam, compare it to 235MPa.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditDate="2016-08-24T10:00:07.503" LastActivityDate="2016-08-24T10:00:07.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11214" PostTypeId="2" ParentId="11211" CreationDate="2016-08-24T09:56:45.893" Score="1" Body="&lt;p&gt;Bending moment at the fixed end for a cantilever with point load at free end is given by: $PL$ where $P$ is the concentrated load and $L$ is the beam's length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M = 2000\ \text{kg} \cdot \dfrac{0.01\ \text{kN}}{1\ \text{kg}} \cdot 2\ \text{m} = 40\ \text{kNm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum stress is $\sigma = \dfrac{M}{Z}$, where $Z$ is the section modulus ($Z = \dfrac{I}{y}$).&#xA;$$\sigma = \frac{40\ \text{kNm}\times10^6}{89.9\ \text{cm}^3\times10^3} = 444\ \text{N/mm}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stress is well above the the yield of the beam and it will deform plastically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, it is above the plastic yield too, so it will break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't account for lateral torsional buckling because the further calculation is only required if the beam passes this simple test.&lt;/p&gt;&#xA;" OwnerUserId="7758" LastEditorUserId="1832" LastEditDate="2016-08-24T10:29:08.603" LastActivityDate="2016-08-24T10:29:08.603" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11215" PostTypeId="2" ParentId="9141" CreationDate="2016-08-24T10:23:15.340" Score="0" Body="&lt;p&gt;Assuming cementitious binder, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Grout = cement:water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mortar = cement:sand:water(:lime)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concrete = cement:aggregate:sand:water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My dictionary states that people mistakenly refer to grout as having sand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mortars are usually weaker than the product it binds so that they get damaged first because mortar is usually cheaper to repair.&#xA;Grout is usually as strong as the material it binds.&lt;/p&gt;&#xA;" OwnerUserId="7758" LastActivityDate="2016-08-24T10:23:15.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11217" PostTypeId="1" CreationDate="2016-08-24T14:37:15.433" Score="2" ViewCount="2949" Body="&lt;p&gt;I'm trying to re-remember how to resolve forces in pin joint frames. I seem to have gotten stuck rather quickly and the examples I've been through don't replicate the same issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my truss I'm working on (terrible image, sorry)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/eExJw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eExJw.png&quot; alt=&quot;Pin Joint Frame&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When trying to calculate the reaction force at B1, B2 and B3 I get different answers depending on where I take moments about from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment about B1,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(B2) = 2 x F(B3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 x F(B3) + 3 x F(B2) = -10 x 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(B3) = +2.5kN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(B2) = +5kN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take Moment about B2, &#xA;F(B1) = F(B3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 x F(B3) + 3 x F(B1) = -10 x 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(B1) = +5kN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(B3) = +5kN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment about B3, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(B2) = 2 x F(B1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 x F(B1) + 3 x F(B2) = -10 x 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(B1) = +2.5kN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F(B2) = +5kN&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different answers for the each restraints.&#xA;Can someone please advise where I am going wrong?&#xA;Thanks&#xA;Rich&lt;/p&gt;&#xA;" OwnerUserId="7761" LastEditorUserId="7761" LastEditDate="2016-08-24T14:53:11.830" LastActivityDate="2019-12-28T18:10:49.850" Title="Resolving forces in a simple truss" Tags="|civil-engineering|structures|structural-analysis|" AnswerCount="2" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11219" PostTypeId="1" CreationDate="2016-08-24T17:53:31.087" Score="1" ViewCount="1384" Body="&lt;p&gt;Perhaps similar to reasons for having a better braking system at the front of a vehicle as when braking, the weight shifts to the front, creating a greater downward force on the front wheels that further increases the traction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When accelerating, the converse seems to be true (?), the weight must shift to the rear, pushing down on the rear wheels. Surely this&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;reduces the traction in general for the front wheels when accelerating&lt;/li&gt;&#xA;&lt;li&gt;Would get only worse the greater the acceleration. I imagine with a rear wheel drive car, the traction would increase with greater acceleration.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I personally haven't driven a rear wheel car yet. However, it feels very uneasy when moving off in a front wheel drive where the weight seems to offload from the front and seem to make the wheels more likely to slip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from historical reasons, what are some good reasons that FWD cars are widely used today?&lt;/p&gt;&#xA;" OwnerUserId="7676" LastEditorUserId="7676" LastEditDate="2016-08-24T18:09:36.867" LastActivityDate="2022-08-30T03:18:14.920" Title="Are front wheel drive vehicles inherently flawed?" Tags="|automotive-engineering|" AnswerCount="7" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11223" PostTypeId="2" ParentId="11219" CreationDate="2016-08-24T18:50:52.120" Score="8" Body="&lt;p&gt;Do the math.  Wheel slippage isn't a issue unless you are trying to accelerate exceptionally hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, front wheel drive cars usually have better accelerating traction because the load on the front wheels is usually higher than on the back wheels.  This is because the engine is heavy compared to other components, and when it's in front, there is more load on the front wheels than on the back.  Accelerating does change this balance slightly, but in most cases not enough to make the load on the front wheels less than on the back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some cars have rear engines, which also feature rear drive for this reason.  The old VW Beetle is a classic example.  They were known to have good traction in snow, compared to other rear drive cars of the same era.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-08-24T18:50:52.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11224" PostTypeId="2" ParentId="11219" CreationDate="2016-08-24T18:54:29.073" Score="4" Body="&lt;p&gt;There is no real general answer to this as the drive and handling characteristics will vary from vehicle to vehicle, depending particularly on weight distribution and suspension design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the effect of weight transfer tends to be more significant under braking than acceleration as most cars are capable of generating higher braking loads than acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also some very important advantages to FWD cars &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Because there is no prop shaft to take drive to the rear you save space in the cabin. This goes hand in hand with a transverse engine layout (a logical consequence of FWD) and is a big factor in making cars more compact overall and with more usable internal space. &lt;/li&gt;&#xA;&lt;li&gt;FWD lends itself well to designing vehicles around a general purpose modular floor-pan. This allows manufacturers to make a lot of different models around one set of running gear and floor (a bit like a truck ladder chassis). While preserving the advantages of monocoque construction. &lt;/li&gt;&#xA;&lt;li&gt;For general everyday driving the handling of a FWD is generally safer as loss of grip through under-steer tends to me more progressive than overseer through loss of traction at the rear wheels and easier to correct as well as less likely to provoke a dangerously inappropriate reaction in an inexperienced driver. &lt;/li&gt;&#xA;&lt;li&gt;Driven steering wheels can improve handling in certain types of corner as the drive is pulling the vehicle in the desired direction as opposed to RWD which is acting at some tangent to the steering direction. &lt;/li&gt;&#xA;&lt;li&gt;Having the engine weight over the drive wheels improves traction in general driving, especially in slippery conditions. FWD cars generally handle better in snow, ice and mud. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;RWD is generally only an advantage in cars with a very high power to weight ratio, especially mid engined cars where weight distribution is not biased towards the front. The main reason for this is that when cars are expected to handle steering and traction/braking loads simultaneously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that this only really matters in driving conditions where tyres are approaching the limit of grip eg in racing which is not what the majority of production cars are designed for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also bear in mind that up untill the point where the drive wheels actually lose traction/grip it make no difference and this is generally a lot more manageable (especially for non-expert drivers) in a FWD car where you just don't accelerate quite as quickly as opposed to a RWD car where it can end up swapping ends. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary for the vast majority of consumer cars FWD is just plain better and there are plenty of fairly large cars with pretty decent performance which are FWD. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get into the realms of out and out sports cars things are a bit different but then again most proper sports cars are mid-engined in any case (note also that most 'front engined' performance cars have  the mass of the engine well behind the front axle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anything the trend has been for larger and more powerful card to be FWD as time has gone on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that the Mini, one of the first mass produced FWD cars was extremely successful in rallying in the 1960s. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-12-29T13:19:21.813" LastActivityDate="2016-12-29T13:19:21.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11225" PostTypeId="2" ParentId="11177" CreationDate="2016-08-24T19:41:17.027" Score="-1" Body="&lt;p&gt;Actually, it is theoretically possible to build a wind-powered vehicle.  You would use a vertical windmill that can accept wind from any direction and it would charge a battery and power the vehicle's electric motor.  Now of course this requires that there is wind, and that the wind creates sufficient energy to charge the battery and power the vehicle.  The battery is needed for when there is not enough wind to power the vehicle.  How much wind is needed and how fast one might go will require some engineering.  It may turn out that there is no place with sufficient wind to overcome the losses inherent in the system.  I wouldn't plan on  taking the vehicle on the freeway.       &lt;/p&gt;&#xA;" OwnerUserId="7768" LastActivityDate="2016-08-24T19:41:17.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11226" PostTypeId="2" ParentId="11147" CreationDate="2016-08-25T04:16:58.230" Score="0" Body="&lt;p&gt;If we define the variables as&#xA;$$T_m = Torque.at.motor.shaft.in.FtLbs$$&#xA;$$R_1 = original.decimal.gear.ratio$$&#xA;$$R_2 = new.decimal.gear.ratio$$&#xA;$$W_1 = radius.of.original.wheel.in.feet,$$&#xA;$$W_2 = radius.of.new.wheel.in.feet$$&#xA;$$I = increase.as.percentage$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then this equation should show the percent increase from the original ratio and wheel to the new ratio and wheel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = 100 \frac{(T_mR_1/W_1)-(T_mR_2/W_2)}{T_mR_1/W_1}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I answered this myself since the question got some activity then got ignored. Thanks to Janus and Chuck for comments that led to me building the equation. It would be greatly appreciated if anyone would confirm that this is the right way to calculate this based on info in the original question.&lt;/p&gt;&#xA;" OwnerUserId="7381" LastActivityDate="2016-08-25T04:16:58.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11227" PostTypeId="2" ParentId="11219" CreationDate="2016-08-25T06:59:52.167" Score="5" Body="&lt;p&gt;Most cars have the engine in the front. This is for two main reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;So that weight is over the front steering wheels and traction is maintained.&lt;/li&gt;&#xA;&lt;li&gt;So that in a front-end collision the kinetic energy of the engine and heavy drive train parts smash into the brick wall (dissipating that energy) as opposed to crushing the passengers between the heavy engine and the brick wall.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A front engine car has superior traction with front wheel drive because there is more weight over the front tires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many classic sports cars are still rear wheel drive for the higher off-the-line traction you mentioned and lower weight(front wheel drive is more complex because the same wheels must both power and steer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, your front wheels still must steer (You never want to steer with the rear wheels because you will lose traction during braking; a much more severe problem). So if you have enough acceleration for the front wheels to lose traction you will lose steering; again probably more severe than a lower high-end acceleration.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2016-08-25T07:06:50.820" LastActivityDate="2016-08-25T07:06:50.820" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11229" PostTypeId="2" ParentId="11202" CreationDate="2016-08-25T09:36:42.483" Score="2" Body="&lt;p&gt;In my opinion the comment of Ethan48 should give you enough information to answer your question. Hereby a bit more elaboration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a column under axial compression the stress is uniformly distributed*. Thus the stress is indeed given by $\sigma = F/S $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When designing a column under axial compression, you do not only check whether the stress is not too high, but also check if the column will not buckle. &#xA;The formula for euler buckling is $F_{buckling} = {(\pi^2* E*I)} / {(K*L)^2}$&#xA;Have a look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Buckling&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia page&lt;/a&gt; for details on this formula and how to use it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force at which a column buckles depends on the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_second_moments_of_area&quot; rel=&quot;nofollow&quot;&gt;area moment of inertia $I$&lt;/a&gt;. For a circular hollow section with the same area as a circular solid section, $I$ is larger. Thus, with all else being equal, the solid column will buckle under a smaller load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow&quot;&gt;the Wikipedia page on beam theory&lt;/a&gt; for more background and details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Depending on how the column is loaded, it takes some distance for the force to distribute itself uniformly. So the maximum stress could well be higher than the standard formula gives. Never forget that theory is a simplification of reality. &lt;/p&gt;&#xA;" OwnerUserId="7590" LastActivityDate="2016-08-25T09:36:42.483" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11231" PostTypeId="2" ParentId="11217" CreationDate="2016-08-25T15:30:50.303" Score="3" Body="&lt;p&gt;As @mg4w already mentioned in a comment, this is a statically indeterminate structure. This means that the structure can't be trivially solved as you have tried (unless you use a simplifying assumption, as I describe towards the end). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To show the problem, these are global equilibrium equations we have at our disposal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= H_{top} + H_{mid} + H_{bot} = 0 \\&#xA;\sum F_y &amp;amp;= V_{top} + V_{mid} + V_{bot} - 10 = 0 \\&#xA;\sum M_{mid} &amp;amp;= -3.00H_{top} + 3.00H_{bot} - 3.00\times10 = 0 \\&#xA;&amp;amp;= -H_{top} + H_{bot} + 10 = 0 \\&#xA;\therefore H_{top} &amp;amp;= H_{bot} + 10&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but this is as far as we can go. I could substitute &lt;span class=&quot;math-container&quot;&gt;$H_{bot}$&lt;/span&gt; in the &lt;span class=&quot;math-container&quot;&gt;$\sum F_x = 0$&lt;/span&gt;, but that wouldn't get me anywhere. We're stuck. That's the trouble with hyperstatic (statically indeterminate) structures. Global equilibrium won't get you anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it can be solved by hand, but it takes a bit of work. Here's your structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rinYx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rinYx.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by simplifying your structure as far as we can to keep things clean once we reach the point where trivial methods will get stuck. For a start, observe that bars 4 and 7 will never take on any load. These are truss bars (hinged on both ends) and therefore only take axial load. For them to take an axial load, they would have to deform along their axes, but since both extremities are constrained, they can't deform and therefore will never take a load. So we can remove them from our model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the region where the load is applied can be trivially solved. The load is vertical and only bar 2 has a vertical component, so it will have to absorb the entirety of the external force. So the vertical component of the axial force in bar 2 is equal to -10&amp;nbsp;kN. The horizontal component will have to be such that the resultant force is parallel to the bar, so we can find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}&#xA;\frac{-10}{h} = \tan\theta_2 = \frac{3}{-2} \\&#xA;\therefore h = 6.67\ \text{kN} \\&#xA;N_2 = \sqrt{10^2 + 6.67^2} = 12.02\ \text{kN (compression)}&#xA;\end{gather}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if bar 2 is generating a horizontal force of 6.67&amp;nbsp;kN on that node, bar 1 must balance that out, generating a horizontal force of -6.67&amp;nbsp;kN, which puts the bar in tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can now remove bars 1 and 2 from the model as well, replacing them with their internal forces. We therefore end up with the following model (the supports are hinged, so no need for the hinge &quot;ball&quot;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xjYw5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xjYw5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the horizontal load applied by bar 1 is simply absorbed by the support and we can mostly forget about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since bars 3 and 6 have the same tangent (in modulus), the vertical force of 10&amp;nbsp;kN carried by bar 2 will be equally divided between them. Just as with bar 2, this will generate horizontal forces within them. Using the same method as above, we find that &lt;span class=&quot;math-container&quot;&gt;$h = \pm1.67\ \text{kN}$&lt;/span&gt; (positive for bar 6, negative for bar 3). The resultant axial force in the bars due to this load is therefore equal to 5.27&amp;nbsp;kN (tension for bar 3, compression for bar 6). Since the horizontal components generated by bars 3 and 6 cancel themselves out, bar 5 is unaffected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now we get to the point where we get stuck: the 6.67&amp;nbsp;kN horizontal force applied by bar 2. Bars 3, 5 and 6 all have horizontal components and can therefore participate in absorbing this force. We therefore have to figure out how the force is parcelled out between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bars can be replaced with springs which resist the displacement of the nodes. The stiffness of the springs for truss bars is equal to &lt;span class=&quot;math-container&quot;&gt;$K = \dfrac{EA}{L}$&lt;/span&gt;. The springs work according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Hooke%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hooke's Law&lt;/a&gt;, which basically states that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F = K\delta$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; is the deformation of the bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the node will clearly only deform in the horizontal direction. But bars 3 and 6 are inclined, so their stiffness is only partial, so we need to get the horizontal component of that stiffness, which is equal to &lt;span class=&quot;math-container&quot;&gt;$\overline{K}_3 = \overline{K}_6 = \dfrac{EA}{L}\cos\theta$&lt;/span&gt; (where &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the angle of bars 3 and 6 with the horizontal axis. The sign doesn't matter for cosine). The deformation felt by bars 3 and 6 is also inclined, so we also have to get just the horizontal component of that deformation, which gives us &lt;span class=&quot;math-container&quot;&gt;$\overline{\delta}_3 = \overline{\delta}_6 = \delta\cos\theta$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, Hooke's Law for bars 3 and 6 for horizontal deformations actually looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F = K\delta\cos^2\theta = \frac{EA}{L}\cos^2\theta\delta$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total stiffness felt by the node is therefore equal to (assuming equal &lt;span class=&quot;math-container&quot;&gt;$EA$&lt;/span&gt; for all bars):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;K &amp;amp;= K_3 + K_5 + K_6 \\&#xA;K_3 = K_6 &amp;amp;= \frac{EA}{L_3}\cos^2\theta \\&#xA;K_5 &amp;amp;= \frac{EA}{L_5}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what fraction of the horizontal force will go to bar 5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;f_5 &amp;amp;= \dfrac{\dfrac{EA}{L_5}}{\dfrac{EA}{L_5} + 2\dfrac{EA}{L_3}\cos^2\theta} \\&#xA;f_5 &amp;amp;= \dfrac{\dfrac{1}{L_5}}{\dfrac{1}{L_5} + 2\dfrac{1}{L_3}\cos^2\theta} \\&#xA;f_5 &amp;amp;= \dfrac{1}{1 + 2\dfrac{1}{\sqrt{10}}\left(\dfrac{1}{\sqrt{10}}\right)^2} = 0.9405 \\&#xA;\therefore f_3 = f_6 &amp;amp;= \frac{1 - 0.9405}{2} = 0.0297&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, bar 5 gets 94.05% of the horizontal load and bars 3 and 6 get 2.97% each. Bar 5 therefore has an axial compression of &lt;span class=&quot;math-container&quot;&gt;$0.9405\times6.67 = 6.27\ \text{kN}$&lt;/span&gt;, while the horizontal component of the axial force in bars 3 and 6 is equal to &lt;span class=&quot;math-container&quot;&gt;$0.0297\times6.67 = 0.20\ \text{kN}$&lt;/span&gt;. Using the same method as used previously for bars 2, 3 and 6, we can also find the vertical components for bars 3 and 6 as equal to &lt;span class=&quot;math-container&quot;&gt;$v = 0.20\times\dfrac{\pm3}{1} = \pm0.60\ \text{kN}$&lt;/span&gt; (positive for bar 3, negative for bar 6). This generates a resultant axial force on bars 3 and 6 of 0.63&amp;nbsp;kN (compression in both).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding the axial forces in bars 3 and 6 due to both vertical and horizontal components of the forces applied by bar 2, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{alignat}{2}&#xA;N_3 &amp;amp;= 5.27 - 0.63 = 4.64\ \text{kN}&amp;amp;&amp;amp;\text{ (tension)} \\&#xA;N_6 &amp;amp;= -5.27 - 0.63 = -5.90\ \text{kN}&amp;amp;&amp;amp;\text{ (compression)} \\&#xA;\end{alignat}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, the forces in each of the bars is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{alignat}{2}&#xA;N_1 &amp;amp;= 6.67\  \text{kN}&amp;amp;&amp;amp;\text{ (tension)} \\&#xA;N_2 &amp;amp;= 12.02\ \text{kN}&amp;amp;&amp;amp;\text{ (compression)} \\&#xA;N_3 &amp;amp;= 4.64\  \text{kN}&amp;amp;&amp;amp;\text{ (tension)} \\&#xA;N_4 &amp;amp;= 0.00\  \text{kN}&amp;amp;&amp;amp; \\&#xA;N_5 &amp;amp;= 6.27\  \text{kN}&amp;amp;&amp;amp;\text{ (compression)} \\&#xA;N_6 &amp;amp;= 5.90\  \text{kN}&amp;amp;&amp;amp;\text{ (compression)} \\&#xA;N_7 &amp;amp;= 0.00\  \text{kN}&amp;amp;&amp;amp; \\&#xA;\end{alignat}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the reactions, we can just use these internal components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the top support, we have bar 1 and bar 3.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{alignat}{2}&#xA;V_{top} &amp;amp;= 5.00 - 0.60 &amp;amp;&amp;amp;= 4.40\ \text{kN} \\&#xA;H_{top} &amp;amp;= 6.67+1.67-0.20 &amp;amp;&amp;amp;= 8.14\ \text{kN} \\&#xA;\end{alignat}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the middle support, we only have bar 5, so it only has a horizontal component of -6.27&amp;nbsp;kN.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;V_{mid} &amp;amp;= 0.00\ \text{kN} \\&#xA;H_{mid} &amp;amp;= -6.27\ \text{kN} \\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the bottom support, we only have bar 6.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{alignat}{2}&#xA;V_{bot} &amp;amp;= 5.00 + 0.60 &amp;amp;&amp;amp;= 5.60\ \text{kN} \\&#xA;H_{bot} &amp;amp;= -1.67-0.20 &amp;amp;&amp;amp;= -1.87\ \text{kN} \\&#xA;\end{alignat}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice that these reactions satisfy the global equilibrium equations. And here's the computer model to check our work (allowing for errors of rounding):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/61USX.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/61USX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, there is a way to trivially solve this problem by adopting a simplifying assumption. All you have to do is assume that bar 5 is axially rigid, meaning it will not deform at all. The validity of such an assumption is questionable if the bar all have the same &lt;span class=&quot;math-container&quot;&gt;$EA$&lt;/span&gt;. However it is less than a third of the length of bars 3 and 6, meaning it is more than three times as stiff. If that's enough for you to consider it rigid (I personally believe that's only acceptable for things at least one order of magnitude stiffer than its neighbors, but I don't know of any code &quot;suggestions&quot; on the matter), the solution becomes trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since bar 5 is rigid, the central node won't suffer any horizontal displacements from the horizontal force applied by bar 2. Therefore, bars 3 and 6 won't deform and won't be affected by that component, meaning bar 5 will absorb the entirety of that horizontal component. And that's that. Obviously, this will change the results for bars 3 and 6, as well as the reactions. For comparison, here's the result in this case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xfkUi.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xfkUi.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;All figures obtained via &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:10:49.850" LastActivityDate="2019-12-28T18:10:49.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11232" PostTypeId="2" ParentId="11219" CreationDate="2016-08-25T18:10:48.177" Score="1" Body="&lt;p&gt;I think a point that has been passed over and really should be considered isn't related to the technical merits of FWD verses RWD but is economic. The emergence of the modern FWD automobile is directly related to the auto industry reducing the auto unit cost. It's just less expensive in materials, time and labor for the auto manufactures to produce FWD cars. The FWD configuration consists of engine, transmission, differential and drive shafts assembled into a single unit. It's typically installed in the vehicle with four bolts by one individual. It's analogous to the emergence of the uni-body construction eliminating the frame as a separate assembly. There are technical reasons for doing this but the economic aspects can't be overlooked. &lt;/p&gt;&#xA;" OwnerUserId="7778" LastActivityDate="2016-08-25T18:10:48.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11234" PostTypeId="1" CreationDate="2016-08-25T20:43:35.417" Score="2" ViewCount="209" Body="&lt;p&gt;I want to modify the coil springs in my car to lower it 1&quot; without raising the spring rate, so simply trimming the overall length is not an option.  My idea is to remove material from along the top edge the active coils.  Obviously, making the rod diameter smaller will make it weaker so it will compress more under a given load.  This is where I will get my lowering.  I want a progressive rate, so I'm thinking I should remove material only from the first one or two coils.  In my head, this will yield a softer ride for normal driving on average highway irregularities when the spring is being compressed those first 2 inches or so.  Yet, it will be more resistant to the compression when under more load, such as on larger dips/humps and/or when the vehicle is loaded more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I were to remove material from the spring:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Will the spring compress more and lower the car without increasing firmness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Will I actually get a progressive rate if I remove the material as described?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) What would be the difference between removing the material from, say, one coil versus spreading it along two or even three coils?  (The spring in question has 5 active coils)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer:  I would remove material with a flap disc for better control and to minimize heat and to create a smooth surface to avoid stress risers.  I would also gradually taper in and out of the maximum amount removed from the coils instead of immediately digging in at the start and finish of the modified areas.&lt;/p&gt;&#xA;" OwnerUserId="7781" LastActivityDate="2016-08-25T20:43:35.417" Title="Modifying coil spring to alter rate progressively" Tags="|springs|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11235" PostTypeId="1" AcceptedAnswerId="11297" CreationDate="2016-08-26T00:18:41.877" Score="0" ViewCount="508" Body="&lt;p&gt;I am curious about calculating stress at various points along a barrel as a bullet is being fired through it. The interference fit (press/friction fit) equations immediately come to mind, but I am wondering if these equations are appropriate to use because of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(i) The non-linear pressure curve of the gas behind a fired bullet&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ii) The bullet is moving through the barrel (dynamic system)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to some interfernce fit equations I'm refering to,&#xA;&lt;a href=&quot;http://www.engineersedge.com/calculators/machine-design/press-fit/press-fit-equations.htm&quot; rel=&quot;nofollow&quot;&gt;http://www.engineersedge.com/calculators/machine-design/press-fit/press-fit-equations.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pointers?&lt;/p&gt;&#xA;" OwnerUserId="7783" LastEditorUserId="7783" LastEditDate="2016-08-26T00:24:39.690" LastActivityDate="2016-08-30T05:53:31.593" Title="Are &quot;interference fit&quot; equations appropriate for calculating barrel stress?" Tags="|stresses|dynamics|friction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11236" PostTypeId="1" CreationDate="2016-08-26T00:36:53.580" Score="4" ViewCount="13118" Body="&lt;p&gt;I need to calculate $K$ or equivalent stiffness of a beam that has a varying $I$ across its length. $I$ has been given to me as a function of $x$, or length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been unable to find a proper equation or set of equations describing how to obtain $K$ for this situation. I've found equations for constant $I$ values across the length of beams. Is there an equation I am simply unaware of, or some general procedure for deriving $K$ for variable $I$ values?&lt;/p&gt;&#xA;" OwnerUserId="7767" LastEditorUserId="1832" LastEditDate="2016-08-26T18:03:46.437" LastActivityDate="2020-02-26T14:09:13.037" Title="Determining stiffness of a beam w/varying moment of inertia" Tags="|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11237" PostTypeId="1" CreationDate="2016-08-26T01:38:57.803" Score="0" ViewCount="144" Body="&lt;p&gt;I've been studying mechanical simulation with nastran/patran, nowadays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I knew weight density is used in imperial unit system, instead of mass density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel using weight density is tedious, because I have to put a conversion factor to change weight density into mass density, when I perform dynamic simulation, always.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just had used SI unit in my whole life, so that I don't know which advantage there is in using weight density, compared to using mass density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain why weight density is used in imperial unit, instead of mass density? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then, what is the advantage of weight density, compared to mass density?&lt;/p&gt;&#xA;" OwnerUserId="4624" LastActivityDate="2016-08-26T05:51:29.823" Title="What is pros of weight density?" Tags="|mechanical-engineering|unit|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11238" PostTypeId="2" ParentId="11237" CreationDate="2016-08-26T05:44:23.797" Score="1" Body="&lt;p&gt;There isn't a profound reason for this. It is just that in US customary units (don't call them &quot;Imperial&quot; any more - first, Britain no longer has an empire, and second, the UK no longer uses them in any serious engineering work) force, pressure, and stress are measured in weight units, but mass and density are measured in mass units. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a &quot;weight unit of mass&quot;, the &lt;em&gt;slug&lt;/em&gt;, but it is based on lengths measured in feet, not in inches, so it still needs another conversion factor of 12 applied to it for most engineering work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is generally less error-prone to include the conversion factor in the input to software as &quot;the acceleration due to gravity&quot; (or in Nastran the reciprocal of that number, for some perverse reason) rather than do all the conversions by hand. Unless you are working in the space industry (e.g. your are modelling a the deployment of a lunar lander) the value of the &quot;magic conversion factor&quot; is always the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average engineer in the USA knows the density of mild steel is about 0.28lb/in^3, but probably wouldn't recognize 7.25e-4 as the density of anything!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there is an analogous situation in SI units. If you want to include the weight of an object as a force in your model, the weight of a mass $m$ kilograms is $mg$ Newtons. But there are many engineering situations where the weight of an object is negligible compared with the other forces acting on it, and can be ignored.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-08-26T05:51:29.823" LastActivityDate="2016-08-26T05:51:29.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11239" PostTypeId="1" AcceptedAnswerId="11256" CreationDate="2016-08-26T06:37:02.433" Score="0" ViewCount="298" Body="&lt;p&gt;I am working on chess board using &lt;code&gt;SolidWorks 2016&lt;/code&gt; and I've created two rectangles, outer one and inner one with &lt;code&gt;8x8&lt;/code&gt; grid using &lt;code&gt;Linear Sketch Pattern&lt;/code&gt;:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7WF3O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7WF3O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;Why is my sketch &lt;strong&gt;under defined&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDENDUM&lt;/strong&gt;:&lt;br&gt;&#xA;It seems I've managed to manage relations of inner rectangle - &lt;strong&gt;I've added dimensions between lines and added &lt;code&gt;Perpendicular&lt;/code&gt; relation to base two lines in centre of coordinate system&lt;/strong&gt; - so I think now it is fully defined, however, whole sketch is still &lt;strong&gt;under defined&lt;/strong&gt;:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2ulLR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2ulLR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Why?&lt;/p&gt;&#xA;" OwnerUserId="6290" LastEditorUserId="6290" LastEditDate="2016-08-26T07:53:33.167" LastActivityDate="2016-08-28T07:25:10.377" Title="Under defined sketch - inner rectangle for chessboard" Tags="|solidworks|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11240" PostTypeId="2" ParentId="11236" CreationDate="2016-08-26T09:44:37.680" Score="6" Body="&lt;p&gt;You could insert the variable $I(x)$ into the integral equation for the rotation and the deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First determine your model. Then determine the equation of the moment $M(x)$. Then enter this in the equation of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rotation: $\theta = \int \frac{M(x)}{E*I(x)}dx $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solve this equation (or let &lt;a href=&quot;http://www.wolframalpha.com/&quot; rel=&quot;noreferrer&quot;&gt;wolfram alpha&lt;/a&gt; do it for you) , add the relevant boundary conditions (such as $\theta(0) = 0$ for a clamped beam) and then solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;deflection: $\nu = \int \theta dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and add the relevant boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="7590" LastEditorUserId="7590" LastEditDate="2016-08-26T10:04:52.353" LastActivityDate="2016-08-26T10:04:52.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11242" PostTypeId="1" AcceptedAnswerId="11243" CreationDate="2016-08-26T11:33:52.650" Score="1" ViewCount="847" Body="&lt;p&gt;Why we study continuous time systems if in the real life we control them using micro-controller which is discrete time .. ?&lt;/p&gt;&#xA;" OwnerUserId="4780" LastActivityDate="2016-08-26T13:47:29.603" Title="Continuous time and Discrete time systems" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11243" PostTypeId="2" ParentId="11242" CreationDate="2016-08-26T13:47:29.603" Score="5" Body="&lt;p&gt;A controller is built around a &lt;em&gt;physical&lt;/em&gt; system. What is the open-loop behavior of the system? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to sample the physical (continuous-time!) system in order to provide feedback to your discrete controller. What sampling rate should you be using for that microcontroller? &lt;a href=&quot;http://www.wescottdesign.com/articles/Sampling/sampling.pdf&quot;&gt;Are you sure&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the physical system modeled and the sampling mechanism designed, how do you develop the controller? Well, with Laplace transforms you could do PID control or state feedback control, but all of those use continuous time differentiators ($s$) or integrators ($1/s$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do you perform differentiation or integration in a discrete control system? Easy - the Z transform! But, &lt;a href=&quot;https://en.wikipedia.org/wiki/Z-transform&quot;&gt;which Z-transform&lt;/a&gt;? Do you use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bilinear_transform&quot;&gt;Tustin Bilinear Z-transform&lt;/a&gt;? Well, that's great but it has some &lt;a href=&quot;https://en.wikipedia.org/wiki/Bilinear_transform#Frequency_warping&quot;&gt;frequency warping&lt;/a&gt;. You could maybe avoid that by using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Matched_Z-transform_method&quot;&gt;Z-transform that maps your poles correctly&lt;/a&gt;, or you could use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Impulse_invariance&quot;&gt;Z-transform that maps the impulse response&lt;/a&gt;, or you could do something &lt;a href=&quot;https://en.wikipedia.org/wiki/Advanced_Z-transform&quot;&gt;more advanced&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point being that, like all of engineering, there are trade-offs. Every method of converting a continuous time controller to a discrete time &lt;strong&gt;representation&lt;/strong&gt; of that controller introduces unwanted aspects. &lt;em&gt;It's up to&lt;/em&gt; you &lt;em&gt;to understand the various drawbacks and apply the correct approach.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, how can you understand a drawback unless you have a &quot;perfect&quot; system by which to compare all the others? This is a similar argument for studying ideal/frictionless systems in physics - you need to understand what the &lt;em&gt;best possible outcome&lt;/em&gt; is in order to understand your limitations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevermind the fact that you could also implement an analog controller. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-08-26T13:47:29.603" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11244" PostTypeId="1" CreationDate="2016-08-26T16:33:32.183" Score="3" ViewCount="4730" Body="&lt;p&gt;In gearing, according to &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Pressure_angle&quot; rel=&quot;nofollow&quot; title=&quot;Wikipedia&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;http://www.hessmer.org/gears/InvoluteSpurGearBuilder.html&quot; rel=&quot;nofollow&quot;&gt;Dr. Rainer Hessmer's involute gear generator&lt;/a&gt;, the most popular pressure angles in the industry are 14.5°, 20° and 25°, with 20° being the most abundant. I can appreciate why 20° and 25° might be common values since they are nice round numbers. This then brings up the question; why 14.5°?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the lower pressure angle of 14.5° will come with some advantages over 20°, such as keeping the lines of action of the forces on the teeth further from the gear axis, reducing the reaction force of the shaft for the same torque transmitted. What I don't get is why a rather peculiar, un-round number such as 14.5° is a common choice? Was there a practical reason behind it, such as the minimum pressure angle before undercutting occurs for gears of a typical number of teeth, or was the value applied when a large quantity of these 14.5° gears existed prior to a standard pressure angle having existed? Perhaps it is a relic of an older standard of gears?&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2016-08-26T16:33:32.183" Title="Why is 14.5° a commonly used pressure angle in gears?" Tags="|gears|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11245" PostTypeId="2" ParentId="11236" CreationDate="2016-08-26T18:01:31.517" Score="3" Body="&lt;p&gt;The short answer is you can't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slightly longer answer is you can, but the solution is specific to the adopted cross-section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The much longer answer will now demonstrate why this is the case by means of an example. In this example, we're just going to take a simple cantilever beam (fixed at &lt;span class=&quot;math-container&quot;&gt;$x=0$&lt;/span&gt;) of span &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;. The beam will have a variable stiffness &lt;span class=&quot;math-container&quot;&gt;$E \cdot I(x)$&lt;/span&gt;, since it is a simple rectangular cross-section, with a height that varies (linearly) from &lt;span class=&quot;math-container&quot;&gt;$h_0$&lt;/span&gt; at the fixed end to &lt;span class=&quot;math-container&quot;&gt;$h_1$&lt;/span&gt; at the free end. The applied load will be a simple concentrated load &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; at the free end. What is the deflection at the free end?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @jos already mentioned in their answer, we use the beam equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an isostatic beam, so the bending moment equation can be trivially obtained as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M = P(L-x)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this must then be divided by the beam's stiffness and the result must be integrated to obtain the beam's tangent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\theta = \int\limits_0^L\frac{P(L-x)}{E \cdot I(x)}\text{d}x$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we can already see the problem. In ordinary beams we can just remove &lt;span class=&quot;math-container&quot;&gt;$EI$&lt;/span&gt; from the integral and go on our merry way. Here, however, &lt;span class=&quot;math-container&quot;&gt;$I(x)$&lt;/span&gt; will also need to be integrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what's the equation for &lt;span class=&quot;math-container&quot;&gt;$I(x)$&lt;/span&gt;? In this case, it will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I(x) = \frac{b \cdot h(x)^3}{12} = \frac{b \cdot \left(h_0 + \left(h_1-h_0\right)\dfrac{x}{L}\right)^3}{12} = \frac{b \cdot \left(h_0 + \Delta h\dfrac{x}{L}\right)^3}{12}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we &quot;just&quot; have to find &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}&#xA;\theta = \frac{1}{E}\int\frac{12P(L-x)}{b \cdot \left(h_0 + \Delta h\dfrac{x}{L}\right)^3}\text{d}x \\&#xA;\theta = \frac{6PL^3}{Eb\Delta h^2}\left(\frac{-\Delta hL+2\Delta hx+h_0L}{(\Delta hx+h_0L)^2} + C_1\right)&#xA;\end{gather}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Already getting ugly, right? Now, we just have to integrate this once more to get the deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}&#xA;\delta = \frac{6PL^3}{Eb\Delta h^2}\int\limits_0^L\left(\frac{-\Delta hL+2\Delta hx+h_0L}{(\Delta hx+h_0L)^2} + C_1\right)\text{d}x \\&#xA;\delta = \frac{6P}{Eb\Delta h^2}\left(\dfrac{L(\Delta h+h_0)}{\Delta h(\Delta hx+h_0L)}+\dfrac{2\ln(\Delta hx+h_0L)}{\Delta h}+C_1x+C_2\right)&#xA;\end{gather}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to find &lt;span class=&quot;math-container&quot;&gt;$C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_2$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}&#xA;\theta(0) = 0 = \frac{-\Delta hL+h_0L}{(h_0L)^2} + C_1 \\&#xA;\therefore C_1 = \frac{\Delta hL-h_0L}{(h_0L)^2} \\&#xA;\delta(0) = 0 = \dfrac{L(\Delta h+h_0)}{\Delta hh_0L}+\dfrac{2\ln(h_0L)}{\Delta h}+C_2 \\&#xA;\therefore C_2 = -\dfrac{L(\Delta h+h_0)}{\Delta hh_0L}-\dfrac{2\ln(h_0L)}{\Delta h} \\&#xA;\end{gather}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\therefore \delta = \frac{6P}{Eb\Delta h^2}&amp;amp;\left(\dfrac{L(\Delta h+h_0)}{\Delta h(\Delta hx+h_0L)}+\dfrac{2\ln(\Delta hx+h_0L)}{\Delta h}+ \\&#xA;\frac{\Delta hL-h_0L}{(h_0L)^2}x-\dfrac{L(\Delta h+h_0)}{\Delta hh_0L}-\dfrac{2\ln(h_0L)}{\Delta h}\right)&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there you have it. To check my work, I created a beam with &lt;span class=&quot;math-container&quot;&gt;$h_0 = 200\ \text{mm}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$h_1 = 100\ \text{mm}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b = 100\ \text{mm}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$L = 10\ \text{m}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$E = 10000\ \text{MPa}$&lt;/span&gt;, and applied a load &lt;span class=&quot;math-container&quot;&gt;$P = 1\ \text{kN}$&lt;/span&gt; at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I simulated the beam by cutting it into discrete segments, each with a different uniform cross-section. To test the sensitivity, I made two models: one with ten segments and one with twenty. Here are the results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VswCL.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VswCL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the equation obtained above, I got 81.78&amp;nbsp;mm, for an error of 1% compared to the models. Given those are approximations, that looks pretty good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we can then obviously obtain the stiffness of the beam: &lt;span class=&quot;math-container&quot;&gt;$1\ \text{kN}/81.78\ \text{mm} = 12.2\ \text{kN/m}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now that we've gone through all this, what can we learn? Basically, that this is really hard. And the solutions aren't elegant. This was just about the most trivial cross-section ever and things already got messy. I mean, those are logarithms, for crying out loud. What if the section varied both in height &lt;strong&gt;and&lt;/strong&gt; width? What if the section didn't vary linearly, but polynomially (or something else)? What if we were dealing with an asymmetrical I-section, with six varying variables (&lt;span class=&quot;math-container&quot;&gt;$t_{f,top}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b_{f,top}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$t_{f,bot}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b_{f,bot}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$t_{w}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$h_{w}$&lt;/span&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the most common solution when dealing with variable sections is to do as I did: split your beam into discrete segments, each with a different interpolated cross-section. Some professional FEA programs make this easy and some even have variable sections built in, but I don't know how they're implemented.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2020-02-26T14:09:13.037" LastActivityDate="2020-02-26T14:09:13.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11246" PostTypeId="2" ParentId="11219" CreationDate="2016-08-26T18:18:47.677" Score="0" Body="&lt;p&gt;FWD cars were created to save space, and create dealer income. In years past most people could and did service most of their vehicle. The industry didn't want people to service their own vehicles, so they decided to make them non-serviceable by the general public. Without specialized equipment much of the FWD car is non-serviceable by consumers. They are throw away vehicles. That's why you do not see many of them after they get much age on them. Their drive train is made of thin aluminum castings and are NOT nearly as durable as cast iron that is still used in most rear wheel drive cars and trucks today. Real Sports cars use rear wheel drive because FWD cars do not handle well if one wheel breaks traction. That includes driving on ice and snow. Handling and performance  is what makes a sports car. If you know how to drive, a RWD vehicle is very unlikely to swap ends as was stated. It's to bad that so many people would rather open their wallet, rather than their toolbox. Or listen to people who have never worked on anything. &lt;/p&gt;&#xA;" OwnerUserId="7788" LastActivityDate="2016-08-26T18:18:47.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11248" PostTypeId="2" ParentId="11204" CreationDate="2016-08-26T23:23:21.427" Score="5" Body="&lt;p&gt;One solution is to use a system with a ring gear that only has teeth on half of its inner circumference, like conceptually shown in the following image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f4KOq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f4KOq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three gears:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ring gear (red)&lt;/li&gt;&#xA;&lt;li&gt;Alternating gear (blue)&lt;/li&gt;&#xA;&lt;li&gt;Output gear (orange)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The alternating gear and output gears can be identical, and the sum of the numbers of teeth of the alternating and output gears should equal that to the ring gear if it had all its teeth instead of half. Otherwise the two inner gears will not properly fit inside the ring gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mechanism works by rotating the ring gear, and the desired alternating motion will be obtained from the output gear. At any one time, the ring gear meshes will one of the inner gears. The inner gear that meshed will the ring gear will rotate in the same direction as the ring gear, causing the other inner gear to rotate in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's say we rotate the ring gear CW. If the ring gear meshes with the output gear, the output gear will also turn CW. CW in, CW out. Now, if the ring gear meshes with the alternating gear, the alternating gear turns CW, forcing the output to be ACW. CW in, ACW out: the output shaft has alternated in direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The meshing of the ring gear switches between one inner gear to the other every half turn of the ring gear. It is worth noting that the output gear's speed does not change except when the meshing alternates, at which point the speed reverses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One issue with this concept as-is is that one revolution of the ring gear brings about one full turn of the output gear one way, followed by one full turn the other way: this rotation may be too much for your agitation mechanism. To resolve this, a gear reduction stage should be added to reduce the output rotation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lMTtQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lMTtQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gear has been fitted onto the side of the output gear (also orange to indicate its attachment), and this gear meshes with a larger gear (green). The green gear will turn slower and less far than the old orange output gear, and therefore the green gear becomes the new output. Therefore, the amount of rotation of the final output (green) that occurs before the direction alternates, in degrees, $\theta$, can be calculated using the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta=360° \cdot N_1/N_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $N_1$ is the number of teeth on the gear attached to the side of the old orange output gear, and $N_2$ is the number of teeth on the new green output gear. In the image just above, $N_1=12$ and $N_2=36$, so the rotation angle before direction alternation is 120°. Note that if $N_1 \gt N_2$, you can get more than one output revolution before alternation (although this might not be suitable if the output is subject to high forces resisting motion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, it should be said that it is possible that, as the meshing transitions from one inner gear to the other, there may be a point where both inner gears are in mesh with the ring gear, jamming the gearset and making mesh alternation impossible. To resolve this, finer gear teeth should be used, as well as removing/filing down one or two of the end teeth on the ring gear to eliminate this dual meshing region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ALTERNATIVE SETUP (Credit to joojaa's comment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(@joojaa Please let me know if I've misinterpreted your comment)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative configuration that does away with the ring gear would be like as shown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dRKwq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dRKwq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where we have the follow gears:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Output gear (orange)&lt;/li&gt;&#xA;&lt;li&gt;2x Large gears (dark red and dark blue)&lt;/li&gt;&#xA;&lt;li&gt;2x Small gears (red and blue)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The small gears should have equal radii, as should the large gears. Either one of the red or blue gear sets should be driven by the crank. The large gears are in constant mesh, so this ensures that the red and blue gear sets rotate in opposite directions. The small gears have half the teeth removed and aligned such that only one of the small gears are in mesh with the output. Like before, as the mesh alternates, so does the direction of rotation of the output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the rotation of the output gear before alternation is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta=180° \cdot N_s/N_o$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $N_s$ and $N_o$ are the numbers of teeth on one of the small gears and the output gear respectively. Also, having selected the numbers of teeth for the small and output gears, the big gears must have the following number of teeth (so that the gears fit together):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N_b = N_s + N_o$$&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2016-08-28T16:55:12.423" LastActivityDate="2016-08-28T16:55:12.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11249" PostTypeId="2" ParentId="14" CreationDate="2016-08-27T09:59:53.630" Score="1" Body="&lt;p&gt;In order to build a passive house you have to limit the final energy use for cooling to 15 kWh/m² per year and your primary energy use for HVAC and lighting has to be less than 90 kWh/m² per year minus the deduction for compactness, so let's say around 70 kWh/m².&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess getting to 15 kWh/m² final energy use for cooling (i.e. the heat extracted from spaces including all the insulation losses) in a climate like Saudi Arabia is almost impossible. In a classic building technique it would mean very thick insulation and small windows in order to limit solar heat loads through fenestrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other criterion however, the primary energy use, would not be that difficult to meet because it represents the balance between the primary energy used and generated on-site by renewables (e.g. PV panels). So if your vapour compression cooling system uses 100 kWh/m² per year and your PV panels produce 40 kWh/m² per year (per m² of building net surface, not per surface of panels) you get to 60 kWh/m² in total which is quite good. If your PV panels create 100 kWh/m² then your balance is 0 and you have a net zero energy building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in conclusion, it does not make really sense to apply the 15 kWh/m² rule in Saudi Arabia. This rule (like the whole passive house concept) was made up in Germany and is a completely political choice which does not have any meaning (it could also be 10 or 25 kWh/m²). What is more important is the net primary energy use and this one you can keep in the passive limits. In fact, to build a building that has a low net primary energy balance Saudi Arabia is a good place because you have plenty of sun and you can generate a lot of green electricity on site with PV panels, something that is not possible in cold climates where the solar energy is not so abundant.&lt;/p&gt;&#xA;" OwnerUserId="7792" LastActivityDate="2016-08-27T09:59:53.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11250" PostTypeId="1" CreationDate="2016-08-27T10:53:06.920" Score="1" ViewCount="90" Body="&lt;p&gt;I.e. when you wait at a traffic light, without stopping your combustion engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;( My long term curiosity is about some empirical raw approximation of the consumption in liters/minute. Easy to measure at road speed, but not at rest - or very slow speed. Ultimately I'ld like to get orders of magnitudes, for main factors isolated, but here for a start a global number at rest is all I seek for. ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To settle numbers if needed, I can give the setup for my (French) car: gasoil, electronic injection, 130 ch din, 2.0 liters. But I'm curious for other setups as well. &lt;/p&gt;&#xA;" OwnerUserId="7547" LastEditorUserId="7547" LastEditDate="2016-08-30T07:32:26.780" LastActivityDate="2017-01-30T14:47:07.750" Title="What is the consumption of a still car?" Tags="|automotive-engineering|" AnswerCount="2" CommentCount="2" ClosedDate="2016-08-29T14:29:47.880" ContentLicense="CC BY-SA 3.0" />
  <row Id="11251" PostTypeId="1" CreationDate="2016-08-27T12:25:53.830" Score="1" ViewCount="57" Body="&lt;p&gt;I try to understand the interaction of (real) engine + gearbox on the consumption. So I purposely don't want to account for friction on air and floor for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the engine alone, the optimal efficiency is obtained for some given rotation speed ( about 2000 RPM, I guess ), then the consumption per turn increases above as well as below this optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we couple a gear box, as the wheels velocity slowly increase (slowly for quasi-stationarity) we have to change gear. During this slow process the engine thus cycles between under-rev, optimal, over-rev several times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;→ at the end, might it be that the consumption per km/h varies non monotonically with the wheels velocity ? (again, not accounting for air and floor friction. Could be tested on rolls.)&lt;/p&gt;&#xA;" OwnerUserId="7547" LastActivityDate="2017-01-25T07:21:50.290" Title="Is consumption of car (engine+gearbox) monotonic with (steady) speed?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11254" PostTypeId="2" ParentId="11250" CreationDate="2016-08-28T04:15:19.670" Score="1" Body="&lt;p&gt;I recommend against using MPG. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Consider the engine at idle and the fact that it must stay at operating temperature. Consider convection and radiation emitted from the engine, and then the 20% efficiency. Using that, you should be able to deduce a possible fuel consumption. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to try the real deal, remember that engines pull intake plenum vacuum at idle which makes the air less dense and makes the engine appear smaller in order to consume a balanced air-fuel ratio based on mass. That, you can definitely solve, depending on the displacement. Air is .00238 slugs per cubic inch. 14.7:1 by mass for fuel consumption. Some engines pull up to 22 in-hg of vacuum at idle. Good luck. &lt;/p&gt;&#xA;" OwnerUserId="7801" LastActivityDate="2016-08-28T04:15:19.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11255" PostTypeId="2" ParentId="11251" CreationDate="2016-08-28T04:45:07.270" Score="-1" Body="&lt;p&gt;A guy on youtube named &quot;guns cars and digits&quot; explained this indirectly with his camshaft design video and his electric vehicle range video. You need his drag force and power equations and you can find them on wiki.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can close the intake valve after bottom dead center. When you do that, there are certain circumstances where the air goes back into the intake plenum. Every cylinder head and camshaft combo has its own &quot;Air slew&quot; function to define this. &#xA;Airflowresearch or Brodix describe their cylinder head flow, and the conditions for which it's possible. This will tell you when the cylinder fills up completely and makes the proportionate amount of torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take those results and plug them into the horsepower equation. If your engine is making more power than drag and uphill force require, you accelerate. If you don't, you slow down. You burn the fuel no matter what, depending on throttle position and load.&lt;/p&gt;&#xA;" OwnerUserId="7801" LastActivityDate="2016-08-28T04:45:07.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11256" PostTypeId="2" ParentId="11239" CreationDate="2016-08-28T07:25:10.377" Score="1" Body="&lt;p&gt;I've managed to solve the problem. I've deleted all relations and then selected &lt;code&gt;Display/Delete Relations-&amp;gt;Fully define sketch&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6290" LastActivityDate="2016-08-28T07:25:10.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11257" PostTypeId="2" ParentId="11250" CreationDate="2016-08-28T14:06:14.287" Score="1" Body="&lt;p&gt;As already mentioned mpg is pretty much meaningless when stationary as it is zero regardless of how much fuel is actually used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the key thing is that there is a certain base load on a running engine which generally doesn't change that much with speed. This is comprised of a few different things: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Induction load : each cylinder much suck air into the combustion chamber through valves, ducts and filters this costs a certain amount of energy. This tends to be a bit larger at part throttle (for gasoline engines, not diesel)  as the throttle valve restricts airflow. &lt;/li&gt;&#xA;&lt;li&gt;Internal friction : IC engines have a lot of moving parts and bearing surfaces on crank bearings, con-rods, valves, piston rings etc etc all contribute to load on the engine. &lt;/li&gt;&#xA;&lt;li&gt;Exhaust : this is assisted by residual pressure in the combustion chamber but even so back pressure form the exhaust system including mufflers, catalytic converters etc will generate some resistance on the exhaust stroke. Turbos can potentially reclaim some energy form the exhaust but don't really do much at idle. &lt;/li&gt;&#xA;&lt;li&gt;Combustion : the combustion chamber has a fixed volume so there is a minimum amount of fuel required for proper ignition. So at low revs combustion efficiency may be reduced. This is helped a lot by active engine management which can adjust air/fuel ration and ignition timing in respect to rpm. &lt;/li&gt;&#xA;&lt;li&gt;Auxiliary Systems : fuel pumps, oil pumps water pumps, air conditioning and alternator all put a load on the engine. Indeed there may be more load on the cooling system when stationary as the only airflow through the radiator is from the fan. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that these loads, although relatively small in terms of overall engine power are more or less constant regardless of the external load on the engine and so take a much bigger bite out of the thermal efficiency at low speeds as they are the main load. This does not necessarily mean that the engine itself  is working 'less efficiently' as such at lower speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some care is needed here with analysis as an idling engine, by conventional measures has infinitely bad fuel economy and thermal  efficiency as both are measures of useful work vs energy consumed and when stationary you are doing no 'useful work' but still consuming energy. So it is much more useful to look at fuel consumer per second in  these conditions. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-01-30T14:47:07.750" LastActivityDate="2017-01-30T14:47:07.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11258" PostTypeId="1" CreationDate="2016-08-28T14:41:28.460" Score="2" ViewCount="204" Body="&lt;p&gt;I need advice on how to optimize a finite element model of a transversely isotropic material. I have modeled the material with 5 nominal moduli in Abaqus and performed an eigenvalue extraction. Additionally I have done a modal analysis of the modeled part, have a full set of eigenvalues and have separated bending, torsional and in-plane modes. So far so good.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial moduli are just a starting point as the measurement is done on coupons and the material itself is sensitive to the measurement method. What I would like to do is use my initial eigenvalue extraction in conjunction with my FRF results to calibrate the material characteristics in the model input.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done some searching and it appears as though a DOE approach and creating a response surface may be an option. I am concerned about the time required(FEA jobs) for a full factorial DOE with 5 factors however. Additionally, I have investigated the Matlab optimization toolbox and read about using sequential quadratic programming for this correlation. I am comfortable with Matlab but have never used that toolbox and do not know how to create a transfer function between the inputs and outputs. Finally, I know about iSight but must find an alternative.&lt;/p&gt;&#xA;" OwnerUserId="7803" LastEditorUserId="7461" LastEditDate="2017-05-14T15:10:37.140" LastActivityDate="2017-05-14T15:10:37.140" Title="FRF based material constant optimization" Tags="|mechanical-engineering|materials|finite-element-method|modal-analysis|abaqus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11259" PostTypeId="2" ParentId="11258" CreationDate="2016-08-28T16:20:42.527" Score="1" Body="&lt;p&gt;I have only done this sort of optimization using non-commercial software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than trying to fit &quot;the FRF&quot; (i.e. the complete shape of all your measured response functions) I would start by estimating the frequencies and mode shapes from the measured FRFs and fitting just those parameters. In fact a good first attempt would be just to fit the frequencies, and use the mode shapes extracted from the FRFs only to identify (manually) which modes in the measured data correspond to which modes in the FE model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The efficient way to do this is to use an optimization algorithm that uses gradient information. If the model is &quot;close enough&quot; to the real structure, that should converge rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to get the gradients, or sensitivity coefficients - i.e. $$\frac{\partial\, \text{frequency}_i}{\partial\, \text{property}_j}, \quad \text{for } i = 1,n, \ j = 1,m$$&#xA;is to use a FE package that will calculate them directly as part of the analysis, based on the sensitivity of the element stiffness and mass data to the material properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't use Abaqus so I don't know if it (or iSight) has that capability. Trying to estimate the gradients by doing several runs of the FE model with different material properties is likely to be expensive (and possibly inaccurate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then use a gradient-based optimization algorithm to find the &quot;best&quot; values of the properties to fit the measured data. Re-run the FE model with those values, rinse and repeat.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-08-28T16:20:42.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11262" PostTypeId="1" AcceptedAnswerId="11268" CreationDate="2016-08-28T22:18:27.297" Score="0" ViewCount="37" Body="&lt;p&gt;In short I'm designing a 'Humane fly swatter' for my University application into bachelor of Industrial Design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to utilise the energy of swatting a fly to capture air through an intake, compress it and ejecting it at a higher velocity in turn drawing more air through the intake creating a slight vacuum, enough to draw the fly into a holding container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reiterate I'm not looking to create a big suction from the intake itself. Think of the swatter more as a butterfly net. You are capturing the fly into the funnel and from within the funnel the air velocity would have increased enough to suck the fly the remainder of the way into the holding container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple visualisation of air being compressed and ejected through a formed funnel. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/liLqT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/liLqT.jpg&quot; alt=&quot;Essentially this effect&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm wondering if anyone could at the bare minimum provide me with some keywords and concepts that I can use to help me research ways of achieving this and to use this information to inform my concept designs. My biggest issues isn't understanding the concepts but not having to vocabulary to be able to search for what I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="7805" LastActivityDate="2016-08-29T02:17:24.267" Title="What are some concepts I can look into to aide in creating a 'compression funnel'(?) without the aide of a motor?" Tags="|vacuum|aerodynamics|compressible-flow|drag|" AnswerCount="1" CommentCount="0" ClosedDate="2016-08-30T10:41:26.583" ContentLicense="CC BY-SA 3.0" />
  <row Id="11263" PostTypeId="1" CreationDate="2016-08-29T00:22:02.170" Score="7" ViewCount="20196" Body="&lt;p&gt;Copper is around 14 times more conductive than lead according to &lt;a href=&quot;https://www.bluesea.com/resources/108/Electrical_Conductivity_of_Materials&quot; rel=&quot;noreferrer&quot;&gt;this source&lt;/a&gt;. If this is the case why are car battery terminals and connectors built using Lead? I saw both types of replacement connectors at the automotive store and am interested in the advantages of each? Thanks for any information. &lt;/p&gt;&#xA;" OwnerUserId="7785" LastEditorUserId="7430" LastEditDate="2017-02-23T22:32:42.053" LastActivityDate="2023-04-24T13:41:02.497" Title="Copper is more conductive, so why use lead battery terminals?" Tags="|automotive-engineering|metallurgy|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11268" PostTypeId="2" ParentId="11262" CreationDate="2016-08-29T02:17:24.267" Score="0" Body="&lt;p&gt;Why not just make a butterfly net? Or do you mean that the collection vessel is solid? If so, hate to burst your bubble but &lt;a href=&quot;https://www.youtube.com/watch?v=u4CQCNfe-KY&amp;amp;t=35s&quot; rel=&quot;nofollow&quot;&gt;it's not going to work&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name of the effect you're looking for is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Venturi_effect&quot; rel=&quot;nofollow&quot;&gt;Venturi effect&lt;/a&gt;, but it doesn't work quite how you seem to want it to from your description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of air that flows through the smaller diameter of your funnel has to go somewhere - where exactly is that? If it's a bottle, see the video I linked above - it's not going to work. If it's a mesh, then again, you've just made a more complicated butterfly net. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hate to shoot an idea down without providing alternatives, but I don't see a way to make workable what you have described. I was motivated to get college degrees in engineering in part so I could critique my own designs, so hopefully you'll let this motivate you (to try something different and/or take some engineering classes) instead of getting discouraged. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-08-29T02:17:24.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11272" PostTypeId="2" ParentId="6539" CreationDate="2016-08-29T12:50:35.717" Score="0" Body="&lt;p&gt;For a very long time, the rate-limiting factor for HDDs is the speed at which they spin. With modern drives with cacheing, etc, this is less true, but essentially manufacturers push every drive to what they deem safe maximum operating speeds. Any higher and the bearings would wear out much quicker, seals would get too hot and rupture, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the speed of the drive you have is likely the speed you should be spinning it at even if you dont care about reading/writing data.&lt;/p&gt;&#xA;" OwnerUserId="7817" LastActivityDate="2016-08-29T12:50:35.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11275" PostTypeId="1" CreationDate="2016-08-29T17:02:55.720" Score="0" ViewCount="1737" Body="&lt;p&gt;Say you have two identically-weighted boxes of materials, same size, both solid, both the same temperature. The only difference is that one is made out of metal and one is made out of plastic. Why does the metal feel hotter on your hand?&lt;/p&gt;&#xA;" OwnerUserId="7821" LastActivityDate="2016-08-29T20:46:15.913" Title="Metal vs Non-metal heat conduction" Tags="|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11276" PostTypeId="2" ParentId="6539" CreationDate="2016-08-29T17:14:04.223" Score="1" Body="&lt;p&gt;The other answers give some information as far as how fast hard drives can run during normal use, but your application doesn't involve normal use, so you might be able to crank some higher speeds out of them. If you want to take a rough theoretical approach to this problem, you can look at the stresses in a spinning disk and compare them to the strength of the hard drive platter material (aluminum in your case, but some platters are made from glass and other materials).&lt;/p&gt;&#xA;&lt;p&gt;Assuming the aluminum part of the platter is bearing most of the load, and it can be approximated as a solid disk (any hole in the middle is very small), and ignoring stresses generated by the fluid flow in your Tesla Turbine (possibly not a safe assumption, but very complicated otherwise) then the stress in the disk can be calculated as follows (from &lt;a href=&quot;http://www.engineeringtoolbox.com/stress-rotation-disc-ring-body-d_1752.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineering Toolbox&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ &#xA;\sigma = \frac{\omega^2 r^2 \rho}{3}&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; is the stress (N/m^2), &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; is the angular velocity (rad/s), &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the radius of the disk (m), and &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the density of the disk material (kg/m^3).&lt;/p&gt;&#xA;&lt;p&gt;Rearrange for &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; and you have:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\omega_{max} = \sqrt{\frac{3\sigma_{yield}}{r^2 \rho}}&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\omega_{max}$&lt;/span&gt; is the maximum theoretical velocity at which the disk would start to break apart and &lt;span class=&quot;math-container&quot;&gt;$\sigma_{yield}$&lt;/span&gt; is the yield stress of the disk material.&lt;/p&gt;&#xA;&lt;h1&gt;Example&lt;/h1&gt;&#xA;&lt;p&gt;If you have a 3.5'' (0.0889 m) disk, made of some aluminum alloy with yield stress &lt;span class=&quot;math-container&quot;&gt;$\sigma_{yield} = 24.1 \times 10^6$&lt;/span&gt; Pa (I wasn't sure which alloy is used in HD construction, so I picked the lowest yield strength I could find from &lt;a href=&quot;http://www.engineeringtoolbox.com/properties-aluminum-pipe-d_1340.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aluminum Alloys&lt;/a&gt;), and density &lt;span class=&quot;math-container&quot;&gt;$\rho = 2700$&lt;/span&gt; kg/m^3, then you would have a maximum rotation speed of:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\omega_{max} = \sqrt{\frac{3(24.1 \times 10^6)}{(0.0889/2)^2 (2700)}} \approx 3680 \text{ rad/s} \approx 35200 \text{ rpm}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Obviously this is very fast, you wouldn't actually want to run your drives that fast due to many factors. Consider that even below the yield stress materials can stretch due to creep and, as other answers have mentioned, higher speeds will cause increased wear in the bearings and other parts. However, this would be a good way to estimate a hard upper limit on your drive speed.&lt;/p&gt;&#xA;&lt;p&gt;Given the number of assumptions and approximations I made in this analysis, I would feel comfortable with a safety factor of 4 (i.e. &lt;span class=&quot;math-container&quot;&gt;$\sigma_{max} = \sigma_{yield}/4$&lt;/span&gt;) which would suggest that my example drive could run with low risk of the platters shattering at a speed of approximately 17,600 rpm, albeit with increased wear.&lt;/p&gt;&#xA;&lt;p&gt;I suggest you look further into your application to figure out what parameters are relevant for you.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-04-01T14:57:54.013" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11281" PostTypeId="1" CreationDate="2016-08-29T20:28:20.927" Score="1" ViewCount="151" Body="&lt;p&gt;When learning about Shape Memory Alloys (SMAs) I was told that the mechanism that allows for strain recovery in SMAs is the martensite to austenite transformation that occurs at low temperature around 90C. This transformation is named for the transformation that occurs in heat treatment in steels since the crystal structures of both are identical, although the transformation in steel occurs around 727C. My question is why is the transition temperature in NITINOL SMA so much lower than the transformation temperature of steels?&lt;/p&gt;&#xA;" OwnerUserId="7827" LastActivityDate="2016-08-29T20:28:20.927" Title="Why is the martensite to austenite transition temperature in Shape Memory Alloys so much lower than in steel?" Tags="|materials|metallurgy|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11282" PostTypeId="2" ParentId="11263" CreationDate="2016-08-29T20:36:58.353" Score="17" Body="&lt;p&gt;Yes, copper is more conductive than lead, but that is not necessarily the primary criterion for selecting the connector material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For car batteries, making sure there's a good connection between the two pieces of metal (the stud on the battery and the connector on the wire) is more important, and lead wins out here because it is so much more malleable (soft) and adapts itself more readily to any mismatch between the shapes of the individual pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second reason the battery stud is made of lead is that lead is the primary component of &lt;em&gt;all&lt;/em&gt; the metal parts of the battery, and there's no compelling reason to use anything else.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-08-29T20:36:58.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11283" PostTypeId="2" ParentId="11275" CreationDate="2016-08-29T20:46:15.913" Score="2" Body="&lt;p&gt;Because it &lt;em&gt;IS&lt;/em&gt; hotter &amp;mdash; where you're touching it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you touch a hot surface, heat flows from the surface to your hand. This both heats your hand and cools the surface. When the surface is made of a material with poor thermal conductivity, the spot where you touched it remains cooler than the rest of the material because the heat cannot flow in fast enough to replenish what was lost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the metal (material with high thermal conductivity), the heat flows faster and the spot where you touch it stays hotter (higher temperature).&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-08-29T20:46:15.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11284" PostTypeId="1" CreationDate="2016-08-29T22:35:22.167" Score="2" ViewCount="318" Body="&lt;p&gt;As far as I've researched, turbulence has a chaotic nature that can't truly be predicted or analyzed. If we took into account all viscous and inertial forces, air resistance, fluid density, etc., why couldn't we predict it?&lt;/p&gt;&#xA;" OwnerUserId="7828" LastEditorUserId="191" LastEditDate="2016-08-30T02:24:51.300" LastActivityDate="2016-08-30T11:37:57.607" Title="Why can't we predict turbulence deterministically?" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11285" PostTypeId="1" AcceptedAnswerId="11304" CreationDate="2016-08-30T00:09:45.777" Score="5" ViewCount="3859" Body="&lt;p&gt;I think I understand how heat pipes work, and thus the following one has me flummoxed:-&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kpb5X.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kpb5X.png&quot; alt=&quot;Chinese heat pipe&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;It's from a typical bulk order Chinese web site, but I've seem many similarly contorted arrangements.  They are very common in the over clocking /modding community.  There is never discussion as to which way the heat sink must be oriented for efficient operation.&lt;/p&gt;&#xA;&lt;p&gt;My example seems ridiculous.  With an U bend in the middle, I don't see how heat sinks like this can evaporate at the hot end and condense at the cold end.  Surely the condensate will just pool in the U bend?  Even with condensate wicking, it must be easier to wick downwards with gravity than upwards fighting gravity.&lt;/p&gt;&#xA;&lt;p&gt;Are these types of heat sinks just a con?  Overclocking /modding heat sinks are never spec'd with a deg. C /W rating or recommended orientation.  This would not do in the engineering world.  Can it work equally effectively in any orientation?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2021-01-29T09:11:36.223" Title="Do heat pipes work in any orientation?" Tags="|heat-transfer|cooling|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11287" PostTypeId="1" CreationDate="2016-08-30T01:42:03.910" Score="1" ViewCount="100" Body="&lt;p&gt;&lt;strong&gt;QUESTION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a business project, I'm looking into the viability of a greenhouse situated in Canada's far northern town of Iqaluit. The problem is how to mathematically simulate the internal air temperature of the greenhouse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;WHAT I ALREADY HAVE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;detailed 3D renderings of the greenhouse&lt;/li&gt;&#xA;&lt;li&gt;all materials with their corresponding thermodynamic values such as K-values&lt;/li&gt;&#xA;&lt;li&gt;climate data that shows the approximate intensity of solar radiation in Iqaluit, angles of sunlight throughout the day and hourly temperature data.&lt;/li&gt;&#xA;&lt;li&gt;data for the required environment for certain crops to thrive, such as humidity, required sunlight, temperature fluctuations, pH, etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CONTEXT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What make this greenhouse investigation unique is that the greenhouse is a much smaller affordable design that is meant to grow a small supplementary source of fresh produce for low-income families which often suffer from food insecurity in northern towns such as Iqaluit due to prohibitively high costs for fresh produce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would also like to strive for simplistic, mechanical systems that are easier to maintain and less prone to failure. A major concern of the group as of now is by how much could the internal air temperature plummet by at night?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;GOAL&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal in mind is to use the information we have so far to determine:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it possible to grow vegetables such as leafy greens within the current design WITHOUT an external power source other than sunlight? (A greenhouse that does not require electric heating is a major goal) How does the internal air temperature vary with the current design?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In the case that the current design is not sufficient for healthy crop growth, what modifications are necessary? (Many potential solutions have been considered):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Have a solar powered heater that charges through the day and intermittently provides heat through the night.&lt;/li&gt;&#xA;&lt;li&gt;Greatly increase thermal insulation surrounding the greenhouse through the use of a thick air-inflated wrap that surrounds the entire structure in addition to thicker polycarbonate panels for construction.&lt;/li&gt;&#xA;&lt;li&gt;Use a setup of concave reflective panels surrounding the greenhouse to increase the amount of heat collected during the day. I have already done some calculations but my numbers ended up far off. Equations I have worked with so far are $Q = mc\Delta T$ and $\dfrac{Q}{t} = \dfrac{A\Delta T}{\text{Thermal Heat Transfer Coefficient}}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;CONCLUSION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the problem demands looking at every single material and also examining the characteristics of the air inside as an ideal gas, as well as the poly-carbonate and it's emissivity/albedo (black body) characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any steps or hints in the right direction with equations or methods of calculation would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="3608" LastEditorUserId="1832" LastEditDate="2016-08-30T02:20:05.627" LastActivityDate="2016-08-30T12:20:36.783" Title="Thermodynamics of an Arctic Greenhouse" Tags="|thermodynamics|environmental-engineering|solar-energy|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11290" PostTypeId="2" ParentId="11284" CreationDate="2016-08-30T02:37:03.737" Score="-5" Body="&lt;p&gt;I think that you've answered your own question. Turbulence is a truly chaotic system at the atomic level, and chaotic systems by definition cannot be predicted.  Minute variations in initial conditions develop into widely different outcomes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This boils down to the disparity between the Heisenberg and Einstein interpretations of quantum physics.  Whilst there are those that still believe in predestination, the current scientific consensus is that there are natural underlying forces that act totally randomly and can only be described loosely by stochastic means.  This is made even more problematic due to the Observer Effect /Uncertainty Principle at the quantum level where you can't accurately measure some particles without affecting them.  So the more you measure, the more uncertainty is introduced.  If you can't understand how this could be, that's okay because no one else can exactly.  This means that a total quantum model of the convection currents in your coffee cup cannot be constructed.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2016-08-30T02:37:03.737" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11297" PostTypeId="2" ParentId="11235" CreationDate="2016-08-30T05:53:31.593" Score="1" Body="&lt;p&gt;Indeed, you can use the same equations for calculating the stresses, as the barrel essentially is a thick walled pipe with internal pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the equations you linked:&#xA;(7) &quot;Radial Stress Casued by axial force&quot; (8) &quot;Circumferential Stress Caused by Axial force&quot; are the right ones. (there seems to be a mistake though: the correct term for (8) would be &quot;Circumferential Stress Caused by &lt;strong&gt;internal pressure&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nonlinearity of the pressure: the pressure will be the highest at the chamber. Assuming an even wall thickness, check the stresses here and you are ok. In case of changing barrel outer diameter, you shuld look for benchmark pressure curves, and check the stresses at multiple locations along the barrel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dynamics: in my opinion this concerns the fatigue life of the barrel. To calculate the lifetime in terms of number of shots fired and survival probability, you will need the Haigh-diagram of the chosen material. But I believe the wear will be the limiting factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: this paper seems to be dealing with the same issue:&#xA;&lt;a href=&quot;http://www.slideshare.net/JoshuaRicci/design-of-a-rifle-barrel&quot; rel=&quot;nofollow&quot;&gt;http://www.slideshare.net/JoshuaRicci/design-of-a-rifle-barrel&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7843" LastActivityDate="2016-08-30T05:53:31.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11298" PostTypeId="1" AcceptedAnswerId="11342" CreationDate="2016-08-30T06:32:56.180" Score="1" ViewCount="1050" Body="&lt;p&gt;This morning, as with every morning, I had my coffee. However, today it was burnt, because I slightly overfilled the water. I use an Italian Percolator on a gas top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CHetv.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CHetv.gif&quot; alt=&quot;Diagram of the Italian Percolator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZELnx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZELnx.jpg&quot; alt=&quot;Picture of the Italian Percolator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, you put it on to boil, and once you hear the water bubbling, the coffee is ready, and chamber C will be full of coffee.&#xA;As you can see, you fill chamber A with water, loosely fill B with coffee, and the water then rises through B, up through the red pipe, and flows over into C. However, as the picture shows, there is a small valve on A. If you fill the water above this valve, you get horrible coffee. When you hear it bubbling, very little coffee has risen into chamber C, and you have to wait a long time for it to do this, while water also escapes from between chambers B and C (which screw together). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does this happen? I cannot think of why there should be a valve to release what is presumably steam. Surely the machine should work regardless? Why does the steam have to be released for the mechanism to work properly - is it linked to the pressure of the system?&lt;/p&gt;&#xA;" OwnerUserId="7845" LastEditorUserId="1832" LastEditDate="2016-09-02T21:04:20.703" LastActivityDate="2016-09-06T03:13:44.437" Title="How does the Italian Percolator work?" Tags="|heat-transfer|pressure|heating-systems|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11299" PostTypeId="1" CreationDate="2016-08-30T07:37:33.427" Score="1" ViewCount="1665" Body="&lt;p&gt;In the control there are two kinds of margin   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Phase Margin&lt;/li&gt;&#xA;&lt;li&gt;Gain Margin&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are plenty of examples available to find about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is my doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is stability margin? I was studying &lt;a href=&quot;https://www.google.co.in/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=4&amp;amp;ved=0ahUKEwj6i9XZv-jOAhXJqo8KHcQ2AqEQFggxMAM&amp;amp;url=https%3A%2F%2Fwww.cds.caltech.edu%2F~murray%2Fcourses%2Fcds101%2Ffa04%2Fcaltech%2Fam04_ch8-3nov04.pdf&amp;amp;usg=AFQjCNEcY90ZcB0Wf-cNC1qmyCru7BPiiw&amp;amp;sig2=ALCFsHQJEv451XtOnLneJw&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt; pdf. On &lt;strong&gt;page-13&lt;/strong&gt;, just before the &lt;strong&gt;Ziegler-Nichols' Tuning Methods&lt;/strong&gt; topic, there is a paragraph which contains the stability margin values. But it does not explain how to find these stability margin values. Can anyone explain the steps to find &lt;strong&gt;the stability margin&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance :-) &lt;/p&gt;&#xA;" OwnerUserId="7847" LastActivityDate="2018-01-27T19:55:29.607" Title="How to find the stability margin of a transfer function?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11300" PostTypeId="2" ParentId="11299" CreationDate="2016-08-30T10:06:22.590" Score="1" Body="&lt;p&gt;The easiest way to visualize both margins is to plot the openloop transfer function, also called a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Nyquist_plot&quot; rel=&quot;nofollow&quot;&gt;Nyquist plot&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the gain margin is the gain by which you can multiply the transfer function such that it will cross the minus one point. This can be found by finding where the transfer function crosses the negative x-axis. Since applying a gain to the transfer function scales the entire plot uniformly, therefore the gain margin is one over the distance between the negative x-axis crossing and the origin. On a side note: it is possible that the gain margin is infinite if the transfer function does not cross the negative x-axis (excluding the origin itself). This concept can also be translated to a bode plot of the open loop transfer function, but now you have to find the point(s) where the phase is -180° (plus or minus a hole number of times 360°), since that translates to the negative x-axis. The gain margin itself can be found by taking one over the modulus (transforming the dB to normal gain) at those frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phase margin is the angle by which the Nyquist plot can be rotated clockwise around the origin before the transfer function would cross the minus one point, or in other words the angle between a point which crosses the unit circle (relative to the origin), the origin and the minus one point. In a bode plot of the transfer function the unit circle translates to the 0 dB line of the magnitude plot, so the phase margin can also be found by looking at the phase at the frequencies of the 0 dB crossings relative to the -180°  (plus or minus a hole number of times 360°), since that translates to the phase of the minus one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can actually define a third margin, namely the modulus margin, which is the smallest distance between the transfer function and the minus one point in the Nyquist plot. This also translates to one over the highest value of the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Sensitivity_(control_systems)&quot; rel=&quot;nofollow&quot;&gt;sensitivity function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2016-08-30T10:06:22.590" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11301" PostTypeId="2" ParentId="11285" CreationDate="2016-08-30T11:20:53.060" Score="2" Body="&lt;p&gt;Orientation is in fact irrelevant.  Heat pipes are used &lt;a href=&quot;http://www.nasa.gov/mission_pages/station/research/news/heat_pipes.html&quot; rel=&quot;nofollow&quot;&gt;in space&lt;/a&gt;, among other places. Quoting from guess-what-reference-page, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At the hot interface of a heat pipe a liquid in contact with a&#xA;  thermally conductive solid surface turns into a vapor by absorbing&#xA;  heat from that surface. The vapor then travels along the heat pipe to&#xA;  the cold interface and condenses back into a liquid – releasing the&#xA;  latent heat. The liquid then returns to the hot interface through&#xA;  either capillary action, centrifugal force, or gravity, and the cycle&#xA;  repeats.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-08-30T11:20:53.060" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11302" PostTypeId="1" CreationDate="2016-08-30T11:29:43.917" Score="0" ViewCount="25" Body="&lt;p&gt;Still in the long term purpose of understanding the various factors of consumption of my car, I recently discovered that I have about 15-20% difference between the liter/km (and indeed, liters) given by the car computer vs the value I obtain by filling the tank quite to the same max level each time divided by km done since the last filling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&gt; Where is the problem ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a few measures on more than 40l,  I see about 7l of difference between liters used according to the computer vs the gas pump. So I don't think imprecision of my &quot;filling at same level&quot; could explain. Idem for evaporation (plus it's in the wrong order, since the computer see more liters consumed than what I do have to refill).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&gt; On which kind of sensors is based the consumption evaluation? tank level, or some accumulated flux at injection? Or more indirect? Are some values empirically estimated rather than measured, or accumulated with bad precision? Are my sensors faulty, or is it quite a common imprecision?&lt;/p&gt;&#xA;" OwnerUserId="7547" LastEditorUserId="86" LastEditDate="2016-08-30T16:24:11.770" LastActivityDate="2016-09-05T18:57:22.957" Title="reliability of consumption indicator" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11303" PostTypeId="2" ParentId="11284" CreationDate="2016-08-30T11:32:19.763" Score="4" Body="&lt;p&gt;We &lt;strong&gt;can&lt;/strong&gt; predict and analyse turbulence, and frequently do. That's one of the things that Computational Fluid Dynamics models do, and what weather forecasting models do, and what climate models do, ... and so on. All of these models have some skill at predicting and analysing turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we can't do is predict the exact, specific patterns of turbulence in all circumstances. That's because we currently don't have a bullet-proof method to always get nicely-behaved solutions to the Navier-Stokes equations. Whether such a method can exist or not, is literally &lt;a href=&quot;http://www.claymath.org/millennium-problems/navier%E2%80%93stokes-equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;a million-dollar question&lt;/a&gt;: that is to say, whoever proves it one way or another stands to win one million US dollars. It's possible that it's a question that will never be solved. You can find &lt;a href=&quot;https://physics.stackexchange.com/q/56496/4066&quot;&gt;relevant matter on our sister physics site&lt;/a&gt;. Whether the underlying system is mathematically chaotic - and thus deterministic - or instead is a stochastic (random) process - is an open question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that doesn't prevent us from making useful predictions about turbulence; nor does it prevent us from analysing turbulence. The more information we get about starting conditions, the more skill the models have in prediction.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2016-08-30T11:37:57.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11304" PostTypeId="2" ParentId="11285" CreationDate="2016-08-30T11:56:48.047" Score="6" Body="&lt;p&gt;Orientation does often matter.  As Carl's answer mentions, the liquid can get from the condenser to the hot interface via capillary action, but most common heat pipes are designed assuming gravity will do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Capillary action is much more effective in space where there is no gravity, but produces very little flow when it has to work against gravity.  Therefore, even heat pipes designed to transport the liquid via capillary action in space need to be oriented correctly on earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure the radiator is above the hot interface, &lt;i&gt;especially&lt;/i&gt; where there is no datasheet available.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-08-30T11:56:48.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11305" PostTypeId="2" ParentId="11287" CreationDate="2016-08-30T12:20:36.783" Score="3" Body="&lt;p&gt;There are way too many details to get into here.  However, do the basic first order analysis.  Determine whether there is enough solar power over a day to keep the greenhouse at some temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you know all the dimensions and insulation values.  Pick a plausible temperature differential between inside and outside, and compute the power required to maintain that temperature differential.  This is all straight forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now compare that to the area required to intercept that amount of solar power.  At your latitude, use maybe 800 W per square meter from unobstructed sun perpendicular to a flat black panel.  Then consider expected number of sunlight hours per day, and cosine of angle to the sun over the day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the average sun power received doesn't meet or exceed the average power required, the system simply won't work and there is no point going further.  Redesign or go home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the average received power is sufficient, then you look into whether sufficient heat storage is reasonable for what is required.  You first have to decide how long of a no-sun interval you want to design for.  Obviously there won't be sun every night, but you also have to consider cloudy days.  What length of bad weather occurs infrequently enough for it to be acceptable for the plants to die or supplemental heat to be required?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that overall, any useful solution won't be cheap or small due to the large thermal storage required.  You can store more heat in the same volume if it can be stored as a phase change in some material.  This would need to be a liquid-solid transition to be practical.  Unfortunately, the water liquid-solid transition temperature is not convenient.  I've heard of paraffin used for this, but a few cubic meters of paraffin is probably too expensive.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-08-30T12:20:36.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11306" PostTypeId="1" AcceptedAnswerId="11308" CreationDate="2016-08-30T13:32:33.817" Score="3" ViewCount="909" Body="&lt;p&gt;To maximize the piezoelectric transducer reception/transmission power we must choose the correct backing/matching layer materials, I'm working on a project to measure a steel tank liquid level (non invasive from the bottom, sensor must be glued) and we have the electronic board finished and the &lt;a href=&quot;https://www.steminc.com/PZT/en/piezo-electric-disc-50x21mm-r-1-mhz&quot; rel=&quot;nofollow noreferrer&quot;&gt;ceramic piezoelectric disk&lt;/a&gt;, but we are lacking of knowledge of choosing this material that I mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tank that we'll measure is a 3 meters tall, 8mm steel bottom wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I read the material selection must take in consideration the acoustic impedance, am I right? Also someone recommend using closed cell foam as backing material and stainless steel as matching layer, but we're not sure about the thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do anyone have recommendations regarding the correct materials/layer thickness for this application?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yFq68.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yFq68.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7850" LastEditorUserId="1586" LastEditDate="2017-11-01T22:15:10.453" LastActivityDate="2017-12-31T22:27:26.977" Title="Piezoelectric matching/backing layer materials" Tags="|mechanical-engineering|electrical-engineering|materials|acoustics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11307" PostTypeId="1" CreationDate="2016-08-30T17:33:16.490" Score="4" ViewCount="529" Body="&lt;p&gt;I did a couple of quick calcs to better understand the phenomenon of biaxial bending. One thing that is currently tripping me up is the sign convention. For example, if we have a rectangular beam as follows &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iEO70.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iEO70.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $b = 15''$ and $h = 30''$ and a coordinate system established as show with origin at exactly the middle. $I_x = 33750\text{ in}^2$ and $I_y = 8437.5\text{ in}^2$. Furthermore we have a moment $M_x = 100\text{ lb.in}$ occuring about the +x axis and a moment $M_y = 100\text{ lb.in}$ occuring about the +y axis. Intuitively, we can tell $M_x$ is causing compression at the bottom and tension at the top of this section. Likewise we can also say $M_y$ is causing compression at the right and tension at the left of the section. However, when we run the numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\sigma_{top} &amp;amp;= \frac{M_xy}{I_x} = \frac{(100\text{ lb.in})*(15\text{ in})}{33750\text{ in}^4} &amp;amp;&amp;amp;= 0.044\text{ psi} \\ &#xA;\sigma_{bot} &amp;amp;= \frac{M_xy}{I_x} = \frac{(100\text{ lb.in})*(-15\text{ in})}{33750\text{ in}^4} &amp;amp;&amp;amp;= -0.044\text{ psi} \\&#xA;\sigma_{right} &amp;amp;= \frac{M_yx}{I_y} = \frac{(100\text{ lb.in})*(7.5\text{ in})}{8437.5\text{ in}^4} &amp;amp;&amp;amp;= 0.088\text{ psi} \\ &#xA;\sigma_{left} &amp;amp;= \frac{M_yx}{I_y} = \frac{(100\text{ lb.in})*(-7.5\text{ in})}{8437.5\text{ in}^4} &amp;amp;&amp;amp;= -0.088\text{ psi} \\&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is highly confusing to me that $\sigma_{bot}$ and $\sigma_{right}$, which are both in compression, have opposing signs. Equally confusing is the fact that $\sigma_{top}$ and $\sigma_{left}$, which are both in tension, have opposing signs. Can anyone bring light to this basic mechanics of materials issue?&lt;/p&gt;&#xA;" OwnerUserId="4808" LastEditorUserId="1832" LastEditDate="2016-08-30T18:30:12.913" LastActivityDate="2016-08-31T11:25:29.073" Title="Biaxial Bending Sign convention" Tags="|structural-engineering|structural-analysis|applied-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11308" PostTypeId="2" ParentId="11306" CreationDate="2016-08-30T22:18:39.910" Score="2" Body="&lt;p&gt;I am assuming that you are using the &lt;a href=&quot;http://www.ti.com/product/TDC1000-Q1&quot; rel=&quot;nofollow noreferrer&quot;&gt;TDC1000 Analog Front End (AFE)&lt;/a&gt;. This is because you are using an image snippet from the the image below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hUq5N.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hUq5N.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so I suggest that you use &lt;a href=&quot;https://en.wikipedia.org/wiki/Cyanoacrylate&quot; rel=&quot;nofollow noreferrer&quot;&gt;cyanoacrylate&lt;/a&gt; (CYA) glue also known as crazy glue. It is important that the transducer has good mechanical coupling to the stainless steel material. I have tried a material similar to closed cell foam and did not have much success. I believe this is because the foam material attenuates the acoustic signal. You also might want to consider roughen the attachment area with a sandpaper. Below are some images to help you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a2EVa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a2EVa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cwGpt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cwGpt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider adding some hot glue around the transducer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Otafs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Otafs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/thqDI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/thqDI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Completed assembly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you might need a High Voltage boost circuit to measure 3 meters of liquid level. Note density might also be a factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other factor to consider in improving piezoelectric transducer reception/transmission power is to create a cavity as shown in the below images&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n9Fhm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n9Fhm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7eEHq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7eEHq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lit/an/snaa266a/snaa266a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to Select and Mount Transducers in Ultrasonic Sensing for Level Sensing and Fluid ID?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/2697/110&quot;&gt;Can I use a ultrasound sensor to measure water level?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/q/1883/110&quot;&gt;What is Radial and Thickness Mode Vibration in a Piezo Electric Ceramic Disc Transducer?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-09-03T01:03:00.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11311" PostTypeId="1" CreationDate="2016-08-31T07:51:44.183" Score="0" ViewCount="162" Body="&lt;p&gt;I know that the current title can be hard to understand. Let me try to explain our problem: We do manufacture and service industrial machines and we are currently updating our spare parts manuals for our distributors. In those manuals we do name some of the parts based on their placement for different planes on the machine like &quot;top-bottom&quot; part, &quot;front-back&quot; part, and of course the most problematic one &quot;left-right&quot; part. (It's problematic because people tend to name parts on the sagittal plane from their standpoint which reverses the placement of the parts. &quot;Left&quot; and &quot;Right&quot; do switch in that case.) When we name the parts on the sagittal plane (I'm not sure of the term for the plane. I've just looked it up) we use the driver's seat orientation which means we imagine ourselves standing in the machine facing its front side and deciding on the left-right nomenclature based on this position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to educate our worldwide distributors and service technicians about this nomenclature and every time we explain it with sentences like the one above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find out if this orientation has a standard naming which is accepted worldwide. Terms like &quot;Driver's seat placement&quot; or &quot;Driver's seat orientation&quot; doesn't do the job most of the time and needs further explanation.&lt;/p&gt;&#xA;" OwnerUserId="4839" LastEditorUserId="4839" LastEditDate="2016-08-31T08:49:58.707" LastActivityDate="2016-08-31T18:27:27.317" Title="What's the term used for machine parts orientation naming on the sagittal plane and from inside the machine?" Tags="|terminology|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11313" PostTypeId="2" ParentId="11302" CreationDate="2016-08-31T11:11:29.190" Score="2" Body="&lt;p&gt;The measure at the pump should certainly be pretty accurate, at least in terms of volume as they typically use positive displacement type pumps (ie each cycle of the pump delivers a known volume) and are calibrated for the purposes of trading standards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However one complication is that although fuel is sold by volume it is its mass which is the true indicator of how much energy is consumed. Tanks at filling stations are usually underground and so should be at a fairly constant temperature but the fuel tank of a vehicle can heat up and cool down a lot so the fuel expands and contracts, changing its volume. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the computer is measuring fuel consumption via the flow through the injectors it is entirely plausible that the absolute value of volume flow-rate isn't very accurate, again because what the ECU really needs to know is the air/fuel ratio by mass. Here the volume flow-rate is a convenient measure but will actually be adjusted by feedback with other sensors so it's more interested in whether more or less fuel is required by the engine than the absolute quantity. Again it is likely that the fuel will be substantially warmer by the time it gets to the injectors than it was when dispensed at the filling station. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is measuring the level in the tank then it may just not be a very accurate sensor as it's main purpose is to give the driver a rough idea of how much fuel is left and changes in level and fuel sloshing around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two measures you have which should be pretty reliable are the volume dispensed at the pump and the car's odometer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally you should keep a running total of both ie when you fill up note the mileage and the volume of fuel added. That way your calculations are based on the most reliable data easily available and you should smooth out some of the rounding errors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately volume is a convenient way of selling fuel but for very precise measurements of efficiency and fuel economy you really need to be dealing in mass. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-09-05T18:57:22.957" LastActivityDate="2016-09-05T18:57:22.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11314" PostTypeId="2" ParentId="11307" CreationDate="2016-08-31T11:25:29.073" Score="3" Body="&lt;p&gt;This confusion seems to come from the difference between the sign convention for moments and the arbitrarily chosen coordinate system of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_and_moment_diagram#Normal_convention&quot; rel=&quot;nofollow noreferrer&quot;&gt;most common sign convention&lt;/a&gt; for moments, a positive bending moment causes the beam to bend in a U shape (compression at the top, tension at the bottom):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YedGEm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YedGEm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the way you have drawn your coordinate system, the bending moment about the x-axis causes tension on the top and compression on the bottom (from the positive bending moment convention this would be a &lt;em&gt;negative&lt;/em&gt; moment). The bending moment about the y-axis causes tension on the left, and compression on the right. As you have noticed, this causes tension to be positive for bending about the x axis and negative for bending about the y-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have mixed two different bending moment sign conventions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the coordinate system of the beam can be selected arbitrarily, one way to avoid this confusion is to use a coordinate system which matches with the positive bending moment sign convention:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MFhE9m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MFhE9m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could take the absolute value of $\frac{My}{I}$ then add the sign afterwards in order to ensure that tensile stresses are positive and compressive stresses are negative (or vice versa as long as you are consistent).&lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2016-08-31T11:25:29.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11315" PostTypeId="2" ParentId="11311" CreationDate="2016-08-31T12:56:41.677" Score="1" Body="&lt;p&gt;There are globally accepted names for these positions, and they're even color-coded!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a really big issue, consider using &lt;a href=&quot;https://en.wikipedia.org/wiki/Port_and_starboard&quot; rel=&quot;nofollow&quot;&gt;port and starboard&lt;/a&gt;. This is used in nautical and aviation industries world-wide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I kept myself straight on the terms was by letter count:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{lr}&#xA;\mbox{Fewer Letters} &amp;amp; \mbox{More Letters} \\&#xA;\mbox{Port} &amp;amp; \mbox{Starboard} \\&#xA;\mbox{Left} &amp;amp; \mbox{Right} \\&#xA;\mbox{Red} &amp;amp; \mbox{Green}&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for left and right to have meaning, as you have discovered, you have to define &quot;forward,&quot; because port means left and starboard means right &lt;em&gt;as you face forward&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are naming the parts based on how they will be accessed for maintenance and usage (from the operator's point of view), then it would probably be wise to put port on the operator's left and starboard on the operator's right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, nit-picking here, these are left and right &lt;em&gt;as you face forward&lt;/em&gt;, meaning that, if you do go that route, the technically correct thing to do would be to have what is further away (radially) from the operator be forward and what is closer to the operator be aft or rear. As &lt;em&gt;that&lt;/em&gt; may be confusing, you may want to flip the port/starboard directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could also be reasonably explained in that you (the operator) are facing a machine (vessel) that is also facing you. In that case, port is right, starboard is left, forward is close to the operator, aft is away from the operator. That's probably the way I would do it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And again, they're color-coded; port/red, starboard/green. So you can color code your machine and/or maintenance parts (or part packaging) that shows where the part goes. You don't need to color-code the entire machine - ships and aircraft just have running lights on the respective sides that are colored. You could get by just by posting color swatches somewhere visible. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-08-31T12:56:41.677" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11316" PostTypeId="1" CreationDate="2016-08-31T14:59:05.737" Score="5" ViewCount="839" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZtARu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZtARu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Might there a difference in design approach for punching shear for the two scenarios? Or something to pay particular attention to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the slope of the shear plane (normally 1:1) the same for both scenarios? I would assume that it is, since the two design codes I usually use (Eurocode and SABS) do not differentiate between the two scenarios.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2019-02-06T02:02:45.557" Title="Is there a difference between punching shear for columns on slabs vs slabs on columns?" Tags="|structural-engineering|shear|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11317" PostTypeId="1" AcceptedAnswerId="11325" CreationDate="2016-08-31T17:30:34.840" Score="2" ViewCount="94" Body="&lt;p&gt;I was watching a video on the ti products out currently for wireless microcontrollers, and they mentioned a &quot;multi-standard&quot; CC2650 mcu here &lt;a href=&quot;http://bcove.me/uekqgi7q&quot; rel=&quot;nofollow&quot;&gt;http://bcove.me/uekqgi7q&lt;/a&gt;. What makes a microcontroller &quot;multi-standard&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7862" LastEditorUserId="7862" LastEditDate="2016-09-02T20:45:20.403" LastActivityDate="2016-09-02T20:45:20.403" Title="What is a multi-standard mcu?" Tags="|electrical-engineering|embedded-systems|wireless-communication|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11318" PostTypeId="2" ParentId="7100" CreationDate="2016-08-31T18:19:47.650" Score="0" Body="&lt;p&gt;I think u are wrong assuming that velocity is variable along the reactor.&#xA;The velocity of gas is only the effect of pressure difference according to Darcy-Weissbach putted into Stoke's momentum balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mass balance to be correct, I'd assume that density is changed over the reactor with constant velocity. &#xA;Namely it depends on the pressure loss, but it is mainly dependent on the ratio of L/d (reactor length to r. diameter), so if it's long enough, the velocity can be changed somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't considered deeply your issue. Could u attend? &lt;/p&gt;&#xA;" OwnerUserId="7858" LastActivityDate="2016-08-31T18:19:47.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11319" PostTypeId="2" ParentId="11311" CreationDate="2016-08-31T18:27:27.317" Score="3" Body="&lt;p&gt;Whether or not there is &quot;standard naming convention&quot; is irrelevant, if your customers don't know about that standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And unless you are supplying to people whose work involves anatomy, I doubt they will ever use technical terms like &quot;saggital plane!&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest answer is: draw a nice big diagram on Page 1 of your catalog (or even on the covers) with some clear labels - and then &lt;em&gt;always&lt;/em&gt; use the same terminology as your diagram.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-08-31T18:27:27.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11320" PostTypeId="1" AcceptedAnswerId="11324" CreationDate="2016-08-31T19:34:28.647" Score="4" ViewCount="8598" Body="&lt;p&gt;If I have a small gear that I am turning a large gear with, from what I know that gives me the ability to move heavier things.  Like a winch of sorts.  Why does turning that little gear allow that mechanical advantage.  I saw a video that said gears were really lots of levers. I can’t visualize the lever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a car, for example, why is first gear more powerful that fifth for initial movement?  I don't know how to visualize it.  In order for me to lift a car with a lever, I need a fulcrum close to the car and an effort arm fairly long (I guess).  So where's the fulcrum and effort arm in the gear moving the car here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my question worded correctly for what I ask in the body?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:  Am I correct that gears in a car transmission are mostly for ratio aspect of things?  if I look at two gears together, I can see the ratio and one turning twice as much as another.  But what is missing is the longer arm that eased the effort.  Is that &quot;effort arm&quot; replaced with a strong source of power (an engine) to make up for the missing long effort arm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:  This is the video I saw that made me think about it, &lt;a href=&quot;https://youtu.be/odpsm3ybPsA?t=3m5s&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/odpsm3ybPsA?t=3m5s&lt;/a&gt;  I started it at the point I'm asking about.  It's pretty quick for the part I'm talking about, maybe 2 minutes.  You see that guy turning the crank to move that little gear?  See the long crank, the &quot;effort arm&quot;?  That looks like a lever to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I saw images somewhere, that had similar ideas and it was showing the crank, like on a well crank, being the &quot;circle&quot; or wheel moving the axle which turns the small gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a car though you don't have that big wheel to turn first.  You have the small drive shaft.  Is there a bigger wheel or gear I am missing?  One that is rotating before the first small gear that turns the big one?  edit:  Is this the flywheel?&lt;/p&gt;&#xA;" OwnerUserId="7865" LastEditorUserId="7865" LastEditDate="2016-08-31T22:07:34.400" LastActivityDate="2016-09-01T13:17:31.103" Title="Why do small gears used to turn big gears allow more force with less effort?" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11321" PostTypeId="1" AcceptedAnswerId="11338" CreationDate="2016-08-31T19:38:51.583" Score="1" ViewCount="1799" Body="&lt;p&gt;What shape of object will roll down the fastest through the decline piece of wood if travelling  with a straight line assuming there are frictionless ramp to stop them from going to the side given the same amount of material with their smoothness being the same? Could it be sphere or a partial sphere with cross section trimmed off? Please also including the distribution of the mass. Is the mass focused in the center given much faster velocity to roll down? What about the 3D shape generate from other curve for example, elliptical or hyperbola, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please show your math formulation...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, I know that a hollow cylinder will roll down slower a solid-centered cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;interesting video: &lt;a href=&quot;https://www.youtube.com/watch?v=cB8GNQuyMPc&quot; rel=&quot;nofollow&quot;&gt;https://www.youtube.com/watch?v=cB8GNQuyMPc&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting thing to learn from this video is that the density doesn't matter, bigger cylinder VS smaller cylinder are the same, even the weight doesn't matter in his experiments... &lt;/p&gt;&#xA;" OwnerUserId="252" LastEditorUserId="252" LastEditDate="2016-08-31T21:51:39.937" LastActivityDate="2016-09-01T16:52:17.553" Title="What shape of object will roll down the fastest through the decline piece of wood on Earth?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11322" PostTypeId="2" ParentId="11320" CreationDate="2016-08-31T21:06:58.593" Score="2" Body="&lt;p&gt;Consider a lever, one side of the rod is longer and one shorter. When you rotate the thing around he fulcrum then your lever ends move a different amount. We can thus say that you concentrate the effort of a big movement to a smaller one this is mechanical advantage. We could formulate this as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum M = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or since $M = F l$ we can say that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_1 l_1 = F_2 l_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_2 = F_1 \frac{l_1}{l_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also formulate this as the length of paths taken but since length of a circular arc is $2\pi r$ they cancel out since the same factor is in both radius terms. Now you can essentially think of 2 gears as circles that can not slip in relation to each other therefore the smaller rotates faster and produces more force than the larger one the formula is the same as above. So if you put a force in on a side that has double the radius you get that the force the other end is doubled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F_{out} = F_{in} \frac{r_{in}}{r_{out}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since gears must mesh relative to the radius, circumference and diameter we can just count the number of teeth per revolution instead of radiuses as it produces same thing. So replace $r$ with $z$ if you wish.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2016-08-31T21:19:17.903" LastActivityDate="2016-08-31T21:19:17.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11324" PostTypeId="2" ParentId="11320" CreationDate="2016-08-31T21:12:54.340" Score="4" Body="&lt;p&gt;The work-energy principle says that the change in energy in a system is equal to the work done to the system. Conservation of energy means that the sum of inputs must equal the sum of outputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work, in turn, is a force times a distance. That is, $W = Fd$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the sum of inputs and sum of outputs have to be equal and there's no friction, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(Fd)_{\mbox{in}} = (Fd)_{\mbox{out}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's talk about the lever first. Assuming both ends of the lever are rigid, motion of certain distance on the input &lt;em&gt;requires&lt;/em&gt; motion of a certain distance on the output. How far the output moves for a particular input motion is determined by the location of the fulcrum. FYI, the &lt;em&gt;study&lt;/em&gt; of how physical constraints determine input and output motions is called &quot;kinematics.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the input and output distances are fixed by physical constraints, and the input force is a given (you supply a force of $X$), then the only thing in the equation that can change to &quot;balance&quot; the input work and the output work is the output force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the output force varies as required to keep $(Fd)_{\mbox{out}}$ equal to $(Fd)_{\mbox{in}}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this all makes sense so far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lever doesn't actually move strictly up-and-down, though. It &lt;em&gt;rotates&lt;/em&gt; about the fulcrum. The actual distance the input traverses is $L_1\theta$, and the output moves $L_2\theta$, where $L_1$ is the length of the lever from the input side to the fulcrum, $L_2$ is the length of the lever from the output side to the fulcrum, and $\theta$ is the angle of how much the lever rotated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Define the &lt;em&gt;arc length&lt;/em&gt;, or distance actually traveled by the input or output end of the lever to be $s$. The input moves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;s_1 = L_1\theta \\&#xA;$$&#xA;The output moves:&#xA;$$&#xA;s_2 = L_2\theta \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you divide the output by the input, you can see that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{s_2}{s_1} = \frac{L_2\theta}{L_1\theta} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thetas cancel, and you're left with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{s_2}{s_1} = \frac{L_2}{L_1} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be restated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\boxed{s_2 = \left(\frac{L_2}{L_1}\right)s_1} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The output distance traveled is equal to the input distance times the &lt;em&gt;ratio&lt;/em&gt; of lever arm lengths. You can plug this back into the work equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F_1 s_1 = F_2 s_2 \\&#xA;F_1 s_1 = F_2 \left(\frac{L_2}{L_1}\right)s_1 \\&#xA;$$&#xA;Cancel the $s_1$:&#xA;$$&#xA;F_1 = \left(\frac{L_2}{L_1}\right)F_2  \\&#xA;\boxed{F_2 = \left(\frac{L_1}{L_2}\right)F_1} \\ &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, these two boxed equations show that the output &lt;em&gt;distance&lt;/em&gt; changes by $L_2/L_1$, but the output &lt;em&gt;force&lt;/em&gt; changes by $L_1/L_2$. The output force can &lt;em&gt;only&lt;/em&gt; go up if the output distance goes down, and vice-versa. This ability to &quot;exchange&quot; force for distance is referred to as &quot;mechanical advantage.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, considering this, where a lever can't flip &quot;around-the-world&quot; because it would hit the ground or fall off the fulcrum, a pulley or gear &lt;em&gt;can&lt;/em&gt; rotate continuously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where before, for the lever, the amount of output motion was dependent on the lengths of the lever arms, here the &quot;levers&quot; are actually gears, and their &quot;lengths&quot; are their radii. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, just like before:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;s_2 = \left(\frac{r_2}{r_1}\right) s_1 \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference here is the change from lengths to radii. The total angular distance traversed, $s$, still keeps the same form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the output gear's radius is very large and the input gear's radius is very small, you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;s_2 = \left(\frac{\mbox{big}}{\mbox{small}}\right)s_1 \\&#xA;s_2 = \left(\mbox{really big}\right) s_1 \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, revisiting the work equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;W = Fd \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but, as discussed, the distance traveled isn't quite linear, it's the &lt;em&gt;arc&lt;/em&gt; the lever takes about the fulcrum, because the lever &lt;em&gt;rotates&lt;/em&gt; about the fulcrum. So you could say instead, that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;W = Fs \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, from the definition of arc length:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;s = L\theta \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, you could substitute:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;W = FL\theta \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can view or group this two ways - the first is as I did the substitutions here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;W = F(L\theta) \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but, you could also group that to read:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;W = (FL)\theta \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a force times a lever arm? A torque. So you can rewrite the equation as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;W = \tau \theta \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the analogy between linear systems and rotational systems - a force is akin to a torque, and a distance is akin to an angular span. Important to note here that $\theta$ isn't the particular angle the system is currently at, it's the angle &lt;em&gt;through which&lt;/em&gt; the system traversed. That is, $\theta = \theta_2 - \theta_1$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, hopefully the explanation of how the linear (lever) and rotational (pulley or gear) frames are related makes sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, more to your question, the &quot;force multiplier&quot; effect that you get with a lever arm, or gear, etc. is a &lt;em&gt;tradeoff&lt;/em&gt; between applied force and applied &lt;em&gt;distance&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do the same amount of work to lift a 1000 pound rock up 1 foot as you do to lift a 1 pound rock up 1000 feet. If you don't have the strength to lift the 1000 pound rock directly, you can use &lt;em&gt;mechanical advantage&lt;/em&gt; to &lt;em&gt;trade&lt;/em&gt; the 1000 pounds for the 1000 feet.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;:EDIT:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I drew a picture that hopefully illustrates the relationship between gears and levers. A lever has a fulcrum, which provides reaction forces, and a length on either side of the fulcrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gear (pulley, etc.) is like a lever with equal lengths on either side of the fulcrum that is lashed/tied/bound to another lever. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hoRs3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hoRs3.png&quot; alt=&quot;Gear Transmission&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lashing that &quot;ties&quot; the two levers together is referred to as the gear mesh. Two teeth come into contact with one another, and that physical contact causes one &quot;lever&quot; (gear) to &lt;em&gt;push&lt;/em&gt; the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll add a little more information, in the hopes that more detail will help cement the analogy rather than confuse you. Just like the example I've mentioned - two levers lashed together, if you lash them together too tightly then they're not actually able to move at all. The same can happen with a gear mesh - if the gears are too close together, the mesh is too tight and the assembly won't spin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, if the lashing that binds the levers is too loose, then when you change direction there will be some dead band where the lashing is sagging. The input is able to rotate freely before the lashing snaps taught again, at which point the output starts to move. Again, the same thing happens in real gear systems - if the gears are too far apart, or the teeth are too narrow, then there is a void between one pair of teeth and the next. This is referred to as &lt;a href=&quot;https://en.wikipedia.org/wiki/Backlash_(engineering)&quot; rel=&quot;nofollow noreferrer&quot;&gt;backlash&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this might be too much information, but my hope is that you can understand that two gears are like two levers that have been tied together. If the binding (backlash) is too tight then the gears can't move, but if it's too lose then the output tends to get jerked around a lot as the &quot;lashing&quot; goes through the slack-taught-jerk-slack cycle, like a series of flicks instead of a continuous push.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2016-09-01T13:17:31.103" LastActivityDate="2016-09-01T13:17:31.103" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11325" PostTypeId="2" ParentId="11317" CreationDate="2016-08-31T22:43:31.320" Score="0" Body="&lt;p&gt;I think &quot;multi-standard&quot; for CC2650 that the microcontroller support, BLE, Zigbee, 6lowpan, and RF4CE. It is the same SOC, and you load different software stacks support different wireless protocols. &lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2016-08-31T22:43:31.320" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11326" PostTypeId="1" CreationDate="2016-09-01T02:26:52.787" Score="0" ViewCount="64" Body="&lt;p&gt;Assume the motor on the vertical reel is fixed, and the entire base of the reel rotates on the bottom plate. If both of these hose reels weigh the same amount, will the vertical reel require the same amount of force to rotate as the horizontal reel? Or could the vertical reel be rotated using a less powerful motor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/99YvU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/99YvU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7869" LastActivityDate="2016-09-01T11:06:58.907" Title="Is the same amount of force required to rotate these reels?" Tags="|mechanical-engineering|motors|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11327" PostTypeId="1" CreationDate="2016-09-01T09:09:43.833" Score="2" ViewCount="261" Body="&lt;p&gt;I need to use a mathematical model of the crank in a two-stroke (diesel) engine. Since a real two-stroke engine endures anharmonic motion, I need to implement higher order harmonics - preferably into a second order ODE, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ddot x +2\zeta\omega_n\dot x+\omega_n^2x=\mu r\omega^2\cos (\omega t+\phi)+\Gamma\omega_n^2\cos(\omega t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do one model the anharmonics that a two-stroke diesel engine suffers from by using higher order harmonics? Is it adequate to implement e.g.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sum _{i=1}^m\kappa_i\omega_{n,m}^2\cos(i\omega t+\phi_i)$$&lt;/p&gt;&#xA;" OwnerUserId="7873" LastEditorUserId="9" LastEditDate="2016-09-04T17:35:00.903" LastActivityDate="2016-09-05T14:28:28.140" Title="Mathematical modelling of the crank in a two-stroke (diesel) engine" Tags="|mechanical-engineering|modeling|dynamics|mathematics|vibration|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11329" PostTypeId="2" ParentId="8819" CreationDate="2016-09-01T10:37:45.697" Score="1" Body="&lt;p&gt;I don't know if there is a direct method for this. (I suppose you already check command reference PDF). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But one possible solution could be run both models and compare both the output txt file. You can use a website to highlight the differences between both txt files. &#xA;I'm afraid there is no other elegant solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards,&lt;/p&gt;&#xA;" OwnerUserId="7874" LastActivityDate="2016-09-01T10:37:45.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11330" PostTypeId="2" ParentId="6646" CreationDate="2016-09-01T10:54:19.383" Score="2" Body="&lt;p&gt;You can easily get the structural rotation around X axis for a given node (N) using the next APDL command: &lt;code&gt;*get,Parameter_output,NODE,N,ROT,X&lt;/code&gt;. You can substitute &lt;code&gt;X&lt;/code&gt; by &lt;code&gt;Y&lt;/code&gt; or &lt;code&gt;Z&lt;/code&gt; depending the axis you want reference to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find more information on the *GET command in the Command Reference Manual about this case on page 742.&lt;/p&gt;&#xA;" OwnerUserId="7874" LastEditorUserId="1832" LastEditDate="2016-09-01T11:16:52.030" LastActivityDate="2016-09-01T11:16:52.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11332" PostTypeId="2" ParentId="11326" CreationDate="2016-09-01T11:06:58.907" Score="2" Body="&lt;p&gt;Unless you're doing something very wrong, the majority of the torque generated by the motor will be used to overcome the force required to pull the hose onto the reel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, assuming that the application for both of these reels is the same &amp;mdash; i.e., they're reeling up the same kind of hose under the same circumstances &amp;mdash; the torque required will be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any secondary effects related to the reel bearings, etc. should be miniscule relative to that.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-09-01T11:06:58.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11333" PostTypeId="2" ParentId="11327" CreationDate="2016-09-01T12:30:25.667" Score="0" Body="&lt;p&gt;Try searching for the term &quot;reciprocating mass balancing&quot;, google returns several useful-looking results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case they aren't detailed enough, I would suggest using the oldschool paper-based literature. My favourite in this topic is &quot;Kraefte, Momente und deren Ausgleich in der Verbrennungskraftmaschine&quot;, Springer. Looking at your name the language may be ok for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole topic is way too broad for a post, but if you need specific equations I can look them up for you. &lt;/p&gt;&#xA;" OwnerUserId="7843" LastActivityDate="2016-09-01T12:30:25.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11334" PostTypeId="1" CreationDate="2016-09-01T14:07:33.517" Score="1" ViewCount="7672" Body="&lt;p&gt;I purchased an evaporator holding the following specifications:&#xA;Cooling Capacity = 2 kW&#xA;Size = 66 X 53 X 43&#xA;What I need to know is how to choose a compressor just to gain this full cooling capacity?&lt;/p&gt;&#xA;" OwnerUserId="3591" LastEditorUserId="2819" LastEditDate="2016-09-02T02:17:51.423" LastActivityDate="2020-03-04T17:59:52.870" Title="How to size a refrigeration compressor based on an evaporator size?" Tags="|thermodynamics|refrigeration|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11335" PostTypeId="2" ParentId="11321" CreationDate="2016-09-01T14:12:27.027" Score="2" Body="&lt;p&gt;Let's assume the ramp &lt;em&gt;does&lt;/em&gt; have friction so things will roll, but not so much that it's &quot;sticky.&quot;  Then you simply need to calculate the torque being applied, which is a function of the radial distribution of mass for each object.&lt;br&gt;&#xA;This is a standard set of problems in any first-year mechanics text book, so I won't dive into the math of calculating the angular moments for different shapes.&lt;br&gt;&#xA;Just for a quick example:  given constant total mass,  a (infinitely thin) cylinder has all its mass one radius away from the axis.  If you build an object of the same total mass, but with a massless cyclinder of same radius and all mass concentrated along the cylinder's axis, far less torque is required to accelerate it and it'll get going much faster than the first cylinder.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, please take a read thru some introductory physics texts, or peek at  good online sites such as &lt;a href=&quot;http://physics-help.info/physicsguide/mechanics/rotational_dynamics.shtml&quot; rel=&quot;nofollow&quot;&gt;physicsguide&lt;/a&gt;  and &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html#rotcon&quot; rel=&quot;nofollow&quot;&gt;hyperphysics&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, you might want to re-post to physics.SE&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-09-01T14:12:27.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11337" PostTypeId="2" ParentId="10903" CreationDate="2016-09-01T15:34:45.217" Score="1" Body="&lt;p&gt;Dampers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resistive forces are inertias (m), dampers (b), and springs (k), and the equation is generally written:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F_{\mbox{applied}} = m\ddot{x} + b\dot{x} + kx \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That derivative term - $\dot{x}$ - means that the damper produces a force proportional to the &lt;em&gt;speed&lt;/em&gt; of the object. Wind resistance doesn't do anything if you're not moving, but gets larger &lt;em&gt;as you move faster&lt;/em&gt;. Dampers are the same way - they provide more force the faster something moves. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-09-01T15:34:45.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11338" PostTypeId="2" ParentId="11321" CreationDate="2016-09-01T16:52:17.553" Score="2" Body="&lt;p&gt;A solid cylinder with a protrusion like a cam if positioned such that the rolling down starts from the inclined tip of the cam will go down faster. The cam should be formed sharp so as not to change the rotational inertia, J, of the mass too much. The cylinder should be placed in an initial tilted imbalance, ready to fall down.&lt;br&gt;&#xA;The length of the ramp should be long enough to let the cylinder roll several times. This will average out the wobbling up and down of CG of the roller.&#xA;This geometry will promote rapid start of rotation and faster roll down. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-09-01T16:52:17.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11340" PostTypeId="2" ParentId="11327" CreationDate="2016-09-02T01:57:17.767" Score="2" Body="&lt;p&gt;You don't need a differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write the kinematics for the piston as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}&#xA;  x &amp;amp; = r \sin(\varphi) - \ell \sin(\beta) =0 \\&#xA;  y &amp;amp; = r \cos(\varphi) + \ell \cos(\beta)&#xA;\end{aligned} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Where $r$ is the crank radius, $\ell$ the conrod length, $\varphi$ the crank angle and $\beta$ the conrod angle. Piston is constrained at $x=0$ and the vertical position is $y$&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is solved for the conrod angle $\beta=\arcsin\left(\frac{r}{\ell} \sin \varphi \right)$ and the piston position $y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now start differentiating the above with the chain rule $\frac{{\rm d}\square{}}{{\rm d}t} = \frac{\partial \square{}}{\partial \varphi} \dot{\varphi} + \frac{\partial \square{}}{\partial \beta} \dot{\beta}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{aligned} 0 &amp;amp; = r \dot{\varphi} \cos\varphi - \ell \dot{\beta} \cos \beta \\ \dot{y} &amp;amp;= -r \dot{\varphi} \sin\varphi  -\ell \dot{\beta} \sin\beta \end{aligned}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is solved for the conrot rot. speed&#xA; $$\boxed{\dot{\beta} =\dot{\varphi} \frac{r \cos\varphi}{\ell \cos\beta}}$$ &#xA;and the piston speed &#xA;$$\boxed{\dot{y}=-r (\dot\varphi+\dot\beta)\sin\varphi}$$. Rinse and repeat&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{aligned}&#xA;0 &amp;amp; = -\ell \ddot{\beta} \cos \beta + \ell \dot{\beta}^2 \sin\beta + r \ddot{\varphi} \cos\varphi - r \dot{\varphi}^2 \\&#xA;\ddot{y} &amp;amp; = -\ell \dot{\beta}^2 \cos\beta -\ell \ddot{\beta} \sin\beta -r \dot{\varphi}^2 \cos\varphi -r \ddot{\varphi} \sin \varphi&#xA;\end{aligned}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to get the conrod rot. acceleration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\boxed{\ddot{\beta} = -\frac{\dot{\beta} (\dot{\varphi}^2-\dot{\beta}^2)\tan \varphi}{\dot\varphi}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and piston acceleration (&amp;lt;= what you are looking for)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\boxed{\ddot{y} = -r ( \ddot\varphi + \ddot\beta) \sin\varphi - r \dot\varphi (\dot\varphi+\dot\beta)\cos\varphi }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;&#xA;If the crank speed was constant at $\dot\varphi = \Omega$ the result can be expanded out in terms of $\varphi$ only as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\ddot{y}}{\Omega^2 r} = -\cos \varphi - r \frac{r^2 \cos^4\varphi + 2 (\ell^2-r^2)\cos^2\varphi-\ell^2+r^2}{(\ell^2-r^2 \sin^2\varphi)^\tfrac{3}{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now that you know the piston acceleration, you can do sum of forces on the piston (including gas pressure) to find the wristpin reaction load and the piston side load. Typically the conrod is split into rotating mass and reciprocating mass and added to the piston mass in this step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be more elaborate you will need a system of equations considering the mass moment of inertia of the conrod. The unknowns are the 6 pin reaction force components and 1 piston side load and 1 crank torque. The equations are 2 for the piston, 3 for the connecting rod and 3 for the crank. A total of 8 equations and 8 unknowns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is completely solvable fast as I have done so in a simulation at work.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastEditorUserId="1051" LastEditDate="2016-09-05T14:28:28.140" LastActivityDate="2016-09-05T14:28:28.140" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11341" PostTypeId="2" ParentId="11334" CreationDate="2016-09-02T02:15:20.543" Score="5" Body="&lt;p&gt;When designing a refrigeration system, the first thing you need to know is the temperature you wish to heat/cool something to and the temperature of the source/sink you are pumping to or from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Carnot%27s_theorem_(thermodynamics)&quot; rel=&quot;noreferrer&quot;&gt;Carnot Equation&lt;/a&gt; to solve for your theoretical best efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know that, you deturmine what capacity you need. Like 2kW for example if I know I need to cool a 2kW laser, or 100W if I know I need to cool a 100W processor. You then include your inefficiency and design for 20% or more above what you need as a safety factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From that point, you can go on to sizing your compressor, working fluid, operating pressure, condenser, evaporator, air flow rates, etc. Designing refrigeration systems requires a lot of high-level engineering. I recommend investigating your original objective, researching refrigeration systems, and posting back here with a new question if you need help at that point.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-02T02:15:20.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11342" PostTypeId="2" ParentId="11298" CreationDate="2016-09-02T02:53:11.167" Score="1" Body="&lt;p&gt;I have never used one, but I read about it on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Moka_pot&quot; rel=&quot;nofollow&quot;&gt;Italian Percolator (Moka Pot) Wikipedia Page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The valve is a pressure relief just like on a pressure cooker; it is for safety and probably not involved with the burnt result. You may want to clean it with some vinegar to be sure it is in good operating condition, but it is probably fine since other pots have been turning out well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further down the wiki page it says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the lower chamber is almost empty, bubbles of steam mix with the upstreaming water, producing a characteristic gurgling noise. This &quot;strombolian phase&quot; allows a mixture of superheated steam and water to pass through the coffee, which leads to undesirable results, and therefore brewing should be stopped as soon as this stage is reached.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Perhaps filling the water higher changes your timing, and higher temperature steam is reaching your coffee grounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had to guess, without your anecdote, I would say that for the same amount of time, an overfilled pot would result in a more dilute, lower temperature brew, and an underfilled pot would result in dark burnt brew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have to time yourself, sacrifice a few cups, and/or get the &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B018QHQSB8&quot; rel=&quot;nofollow&quot;&gt;thermocouple meter&lt;/a&gt; out ;-)  Good Luck!&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2016-09-06T03:13:44.437" LastActivityDate="2016-09-06T03:13:44.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11343" PostTypeId="1" CreationDate="2016-09-02T03:20:04.623" Score="1" ViewCount="54" Body="&lt;p&gt;A Hazardous Chemicals Waste Recycling company &amp;amp; KOH production, ABANDONED the site, they cut (looks like with a diamond saw) the concrete floor (25cm height) at about the middle of the warehouse, in the size/shape of a bed, for what reasons?  and, some Circular Cuttings (diameter of a round plate)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the PURPOSE of ALL these CUTTINGS?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also did cuttings on the asphalt floor outside the building, rectangular shape (approx 400m2 &amp;amp; 300m2 area), and replaced it with NEW ASPHALT. Few months after they left the site, the NEW ASPHALT is covered with rusty powder. How is the RUSTY powder forming? chemical reaction??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also BLACK OIL forming above the OLD asphalt floor? How come it's rising from the floor? chemical reaction??&lt;/p&gt;&#xA;" OwnerUserId="7883" LastEditorUserId="7083" LastEditDate="2016-09-03T01:02:42.880" LastActivityDate="2016-09-03T01:02:42.880" Title="Hazardous Chemicals Waste Recycling company &amp; KOH production, ABANDONED the site" Tags="|environmental-engineering|" AnswerCount="0" CommentCount="6" ClosedDate="2016-09-03T18:34:17.673" ContentLicense="CC BY-SA 3.0" />
  <row Id="11344" PostTypeId="1" CreationDate="2016-09-02T08:59:12.997" Score="1" ViewCount="859" Body="&lt;p&gt;I am asked to design a pile cap which also serves as a collision barrier. What special design considerations need to be addressed in order to raise pile caps above ground level?&lt;/p&gt;&#xA;" OwnerUserId="7887" LastEditorUserId="1832" LastEditDate="2016-09-02T11:17:42.720" LastActivityDate="2016-09-02T11:17:42.720" Title="Does pile cap need to be buried?" Tags="|civil-engineering|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11347" PostTypeId="2" ParentId="11344" CreationDate="2016-09-02T10:41:06.380" Score="3" Body="&lt;p&gt;A pilecap is something to distribute loads into piles. There are no particular &quot;rules&quot; for pilecaps that mean it needs to be treated differently above or below ground. It just needs to be designed like any other structural element: for all loads that could be applied on it, with all resistances that may occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you'll need to design it for collision loads, any vertical loads coming onto it, with resistance from the piles and from soil that is against any below-ground parts.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2016-09-02T10:41:06.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11348" PostTypeId="1" AcceptedAnswerId="11349" CreationDate="2016-09-02T14:56:54.387" Score="2" ViewCount="210" Body="&lt;p&gt;According to this figure whats the difference between leading edge flap and slat and why the slat come with plain wing and TE flap come alone?&#xA;thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Webiz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Webiz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7892" LastActivityDate="2016-09-02T22:09:58.857" Title="Whats the difference between leading edge and slat?" Tags="|aerospace-engineering|aerodynamics|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11349" PostTypeId="2" ParentId="11348" CreationDate="2016-09-02T15:56:27.980" Score="1" Body="&lt;p&gt;This comes down to the fact that flaps are fundamentally changing the geometry of the wing so with flaps deployed the wing is a different shape with correspondingly different characteristics. So when a flap is deployed it is essentially a modification of the main wing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand the primary role of slats (distinct form LE flaps) is to control the airflow over the wing, specifically to delay flow separation, and as such don't directly contribute lift themselves but allow the main wing to operate at a higher angle of attack before it stalls.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some aerofoils also have slots which allow airflow between low and high pressure sides of the wing with the intention of either reducing drag or keeping the airflow attached or both. It's not uncommon to see flaps with slot gaps between the training edged of the main wing and the leading edge of the flap itself. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-09-02T22:09:58.857" LastActivityDate="2016-09-02T22:09:58.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11350" PostTypeId="1" CreationDate="2016-09-03T02:01:24.380" Score="1" ViewCount="256" Body="&lt;p&gt;How to derive Darcy's law only in the case of saturated soil in mathematical/theoretical way without having to derive all the cases? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only find the generalized theoretical proof for all cased but not its more simple version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is its simple version I ask you to derive(This is coming from Introduction to geotechnical engineering by Das):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=ki$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $v$, $k$ and $i$ are the discharge velocity, hydraulic conductivity and hydraulic gradient, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Darcy's law is a constitutive equation that describes the flow of a fluid through a porous medium. The law was formulated by Henry Darcy based on the results of experiments on the flow of water through beds of sand, forming the basis of hydrogeology, a branch of earth sciences. But then I was only able to find the theoretical derivation of the generalized Darcy's law&lt;/p&gt;&#xA;" OwnerUserId="252" LastEditorUserId="252" LastEditDate="2016-09-05T04:59:43.093" LastActivityDate="2016-09-05T04:59:43.093" Title="How to derive Darcy's law only in the case of saturated soil without derived all cases?" Tags="|civil-engineering|mathematics|soil|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11352" PostTypeId="1" CreationDate="2016-09-03T06:40:01.400" Score="1" ViewCount="110" Body="&lt;p&gt;I want to fit a Hub motor ($15kW$ BLDC motor) to my car to convert it into a hybrid vehicle.&#xA;Would it be possible for me to use the same motor (as a generator) to charge my battery pack, while the vehicle runs on it's normal Internal Combustion engine?&lt;/p&gt;&#xA;" OwnerUserId="7898" LastEditorUserId="13311" LastEditDate="2018-02-24T17:16:10.883" LastActivityDate="2018-09-23T04:03:14.667" Title="Charging a battery pack using a motor" Tags="|mechanical-engineering|electrical-engineering|automotive-engineering|motors|battery|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11353" PostTypeId="1" AcceptedAnswerId="11354" CreationDate="2016-09-04T00:30:48.187" Score="0" ViewCount="462" Body="&lt;p&gt;I am working on a project where 1L of cold water (20C) has to be heated (80C) very quickly (30 seconds). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were a few designs I was considering&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Heater cartridge (similar to what you would find in a 3D printer) - I can get up to 800W&lt;/li&gt;&#xA;&lt;li&gt;Peltier - 10 &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00IKDL22O&quot; rel=&quot;nofollow&quot;&gt;peltiers&lt;/a&gt; would add up to about 700W, but would require a power supply that could handle 60A, which would be difficult to find for a reasonable price&lt;/li&gt;&#xA;&lt;li&gt;Nichrome wire - The challenge with nichrome is how to transfer the heat to water. I was thinking of getting insulated nichrome and rapping that around a metal container that would hold the water. &lt;/li&gt;&#xA;&lt;li&gt;Blow torch or other flame - If this is the best method, I would need help calculating the 'wattage'. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I don't think any of these methods would work because the wattage is too low in any manner. I know water heating is typically measured in kW, but on 120V AC, the circuit would break at 3.6kW (30A circuit breaker). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is what is the best method of heating the water with the given parameters (not limited to the ideas I had)?&lt;/p&gt;&#xA;" OwnerUserId="2164" LastEditorDisplayName="user16" LastEditDate="2016-09-04T01:32:49.463" LastActivityDate="2016-09-04T01:32:49.463" Title="How do you quickly heat cold water?" Tags="|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11354" PostTypeId="2" ParentId="11353" CreationDate="2016-09-04T00:53:26.280" Score="4" Body="&lt;p&gt;The specific heat of water is 4.186 J/g&amp;deg;C, so heating 1000 grams of water by 60 &amp;deg;C requires about 250 kJ of energy. Applied over a 30-s period, this represents almost 8.4 kW of power &amp;mdash; assuming that none of the power is lost elsewhere along the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach would be to pre-store the heat that you need for the water in some sort of thermal reservoir. For example, you could have a large block of metal (aluminum is cheap and has a high specific heat of 0.900 J/g&amp;deg;C) that you keep at 80 &amp;deg;C (or somewhat higher) using a heater and a thermostat. The total thermal capacity of the reservoir needs to be significantly higher than that of the water by about 10:1, so you'd need about 50 kg of aluminum. Pass the water through channels in or over the surface of the aluminum, and it will heat up in short order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also use a reservoir of hot water (10 l or so) and pass the water you want to heat through a heat exchanger (e.g., coiled tubing) that's submerged in the reservoir.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-09-04T00:53:26.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11357" PostTypeId="1" CreationDate="2016-09-04T16:58:06.350" Score="4" ViewCount="111803" Body="&lt;p&gt;I have a question that is annoying me for a long time. I know that I can calculate the moment of inertia of a rectangular cross section around a given axis located on its centroid by the following formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GCIeM.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GCIeM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also know that more generically, the moment of inertia is given by the integer of an area times the square of the distance from its centroid to the axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets say I have a rectangular section with a height of 200&amp;nbsp;mm and a width of 20&amp;nbsp;mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use the formulas of the first method, in relation to an x axis parallel to the width:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_x=\frac{bh^3}{12}=\frac{20\cdot200^3}{12}=1333.33\text{ cm}^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the second method, why do I get different results when calculating twice the area of half a section, multiplied by the square of the distance from its centroid to the x axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_x= 2A_{half\ section}d^2 = 2\cdot(200/2\cdot20)*(200/4)^2= 1000\text{ cm}^4$$&lt;/p&gt;&#xA;" OwnerUserId="7907" LastEditorUserId="1832" LastEditDate="2016-09-04T19:19:49.167" LastActivityDate="2017-01-18T21:12:37.257" Title="Moment of Inertia of a Rectangular Cross Section" Tags="|mechanical-engineering|structural-engineering|geometry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11358" PostTypeId="1" AcceptedAnswerId="11395" CreationDate="2016-09-04T18:40:39.393" Score="3" ViewCount="164" Body="&lt;p&gt;I am creating a 24 hour blood pressure monitor that is designed to be sleek and comfortable, I have decided on The Honey Well Series NBP AN transducer (see more info here: &lt;a href=&quot;http://www.mouser.com/ds/2/187/honeywell-sensing-basic-board-mount-pressure-senso-740338.pdf&quot; rel=&quot;nofollow&quot;&gt;Basic Board Mount Pressure Sensors&lt;/a&gt;, p13) I want to connect that to a board so that the data can be transmitted wirelessly, or by cable to a phone or computer, how would I go about doing that? &lt;/p&gt;&#xA;" OwnerUserId="5065" LastEditorUserId="7083" LastEditDate="2016-09-13T05:49:42.203" LastActivityDate="2016-09-17T13:35:34.307" Title="Trasmitting data from pressure transducer to phone/web server" Tags="|electrical-engineering|pressure|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11359" PostTypeId="2" ParentId="11358" CreationDate="2016-09-04T21:10:22.807" Score="3" Body="&lt;p&gt;One such way transfer data is to use a Internet of Things (IOT) platform. You will need at minimum a microcontroller and wifi module. Currently some semiconductor vendors produce a both the microcontroller in a system on a chip (SOC) form factor. Either way you can access pressure sensor data via the microcontoller use the wifi connection to transmit the data to the web. Take a look at following two option from Texas Instruments. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/product/CC3100&quot; rel=&quot;nofollow&quot;&gt;Internet-of-Things Solution for MCU Applications CC3100&lt;/a&gt; + &lt;a href=&quot;http://www.ti.com/ww/en/launchpad/launchpads-connected-ek-tm4c123gxl.html&quot; rel=&quot;nofollow&quot;&gt;TIVA TM4C123&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/tool/cc3200-launchxl&quot; rel=&quot;nofollow&quot;&gt;SimpleLink Wi-Fi CC3200 LaunchPad (Both microcontroller and Wifi in one package)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are also other vendor that offer similar solutions. I will listed few below under references. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other options, microcontroller + cellular, Microcontroller + BLE, BLE to smartphone + internet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest that you do a little research to narrow down the options. At which point you can post a question specific to your option so detail solution can be provided.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://markelthinkslearnscreates.wordpress.com/tag/wifi/&quot; rel=&quot;nofollow&quot;&gt;CONNECT TIVA CONNECTED LAUNCHPAD TO EXOSITE CLOUD USING TI SIMPLELINK WIFI CC3100&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://community.thingspeak.com/tutorials/electric-imp/data-storage-and-retrieval-with-thingspeak/&quot; rel=&quot;nofollow&quot;&gt;Data Storage and Retrieval with ThingSpeak for Electric Imp Connectivity&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.adafruit.com/products/2680&quot; rel=&quot;nofollow&quot;&gt;Adafruit.io Internet of Things Feather ESP8266 - WiFi Starter Kit&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.seeedstudio.com/EMW3165-CortexM4-based-WiFi-SoC-Module-p-2488.html&quot; rel=&quot;nofollow&quot;&gt;EMW3165 - Cortex-M4 based WiFi SoC Module&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sparkfun.com/products/13865&quot; rel=&quot;nofollow&quot;&gt;SparkFun IoT Wifi Starter Kit&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://starterkit.att.com/&quot; rel=&quot;nofollow&quot;&gt;IoT Starter Kit from AT&amp;amp;T&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-09-07T09:33:44.197" LastActivityDate="2016-09-07T09:33:44.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11361" PostTypeId="2" ParentId="11357" CreationDate="2016-09-04T22:06:31.663" Score="7" Body="&lt;p&gt;You have misunderstood the parallel axis theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment of inertia of an object around an axis is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \iint\limits_R\rho^2\text{d}A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ is the distance from any given point to the axis. In the case of a rectangular section around its horizontal axis, this can be transformed into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;I_x &amp;amp;= \int\limits_{-b/2}^{b/2}\int\limits_{-h/2}^{h/2}y^2\text{d}y\text{d}x \\&#xA;I_x &amp;amp;= \int\limits_{-b/2}^{b/2}\left.\dfrac{1}{3}y^3\right\rvert_{-h/2}^{h/2}\text{d}y\text{d}x \\&#xA;I_x &amp;amp;= \int\limits_{-b/2}^{b/2}\dfrac{1}{3}\dfrac{h^3}{4}\text{d}x \\&#xA;I_x &amp;amp;= \left.\dfrac{1}{3}\dfrac{h^3}{4}x\right\rvert_{-b/2}^{b/2} \\&#xA;I_x &amp;amp;= \dfrac{bh^3}{12}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what if we wanted to get the inertia around some other axis at a distance $r$ from our centroid? In this case, all we have to do is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \iint\limits_R(\rho+r)^2\text{d}A$$&#xA;$$I = \iint\limits_R\left(\rho^2 + 2\rho r + r^2\right)\text{d}A$$&#xA;$$I = \iint\limits_R\rho^2\text{d}A + 2r\iint\limits_R\rho\text{d}A + r^2\iint\limits_R\text{d}A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first component $\iint\limits_R\rho^2\text{d}A$ is simply equal to the original moment of inertia. The second component $2r\iint\limits_R\rho\text{d}A$ is equal to zero since we're integrating around the centroid (it'll become a function of $y^2$, which when integrated from $-h/2$ to $h/2$ gives zero). The third component is equal to $Ar^2$. So, in the end, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I' = I + Ar^2$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, if you want to calculate the moment of inertia of a rectangular section by considering each of its halves (half above the centroid, half below), you need to do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;I_{half} &amp;amp;= \dfrac{b\left(\dfrac{h}{2}\right)^3}{12} \\&#xA;I'_{half} &amp;amp;= I_{half} + b\left(\dfrac{h}{2}\right)\left(\dfrac{h}{4}\right)^2 \\&#xA;&amp;amp;= \dfrac{bh^3}{96} + \dfrac{bh^3}{32} = \dfrac{bh^3}{24} \\&#xA;I_{full} &amp;amp;= 2I'_{half} = \dfrac{bh^3}{12}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the original value for the full section. QED.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-09-04T22:06:31.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11362" PostTypeId="1" CreationDate="2016-09-05T01:50:44.353" Score="0" ViewCount="1051" Body="&lt;p&gt;Is there a limit to the size of ships that can have hydrofoils (would an aircraft carrier work)? Would there be cavitation if the load was high enough, and to what extent would using larger foils help, and would they stop providing any advantage over non-hydrofoils at some point?&lt;/p&gt;&#xA;" OwnerUserId="863" LastActivityDate="2016-09-06T03:11:53.573" Title="How Big can Hydrofoils get?" Tags="|marine-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11366" PostTypeId="1" CreationDate="2016-09-05T16:04:09.333" Score="1" ViewCount="161" Body="&lt;p&gt;I am new to ansys and I am running ansys 17.1 on a macbook (on a parallel desktop) and I have a problem I can't solve. When I import or create a geometry, I can only see the silhouette of it in the &quot;Study&quot; tab (see picture 1). The contour results are not plotted either. When I try to edit the geometry instead, the part looks just fine (see picture 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone ever had this problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jMkFL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jMkFL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture 2:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tKkMX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tKkMX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2437" LastActivityDate="2016-09-05T16:04:09.333" Title="Ansys: issue with geometry visualisation" Tags="|ansys|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11367" PostTypeId="1" AcceptedAnswerId="11368" CreationDate="2016-09-05T16:24:06.867" Score="4" ViewCount="1831" Body="&lt;p&gt;A translator &lt;a href=&quot;http://www.multitran.ru/c/m.exe?a=4&amp;amp;MessNum=336813&amp;amp;l1=1&amp;amp;l2=2&quot; rel=&quot;nofollow noreferrer&quot;&gt;posted&lt;/a&gt; this diagram of a weld joint on a translator's forum (the figure below). What does the abbreviation &quot;B.C.&quot; mean on the diagram?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GQS5n.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GQS5n.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ImM77.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ImM77.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mMdu0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mMdu0.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="368" LastEditDate="2016-09-07T14:54:05.350" LastActivityDate="2016-09-07T14:54:05.350" Title="What does the abbreviation &quot;B.C.&quot; mean on a weld joint diagram?" Tags="|terminology|welds|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11368" PostTypeId="2" ParentId="11367" CreationDate="2016-09-05T18:05:46.603" Score="6" Body="&lt;p&gt;I'm not aware of BC as standard English nomenclature for welding, but I have a guess for what they mean. The fact that there is a period after the B but not the C makes me wonder if something was cut off or whited out in the original.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think they mean backgouge (B&lt;strong&gt;G&lt;/strong&gt;) which would be a common process in a double sided vee groove weld it it needed to be a complete joint penetration (CJP) weld. This is sometimes indicated with a dotted line like that, although strictly of they backgouge as shown they should have drawn a u groove weld instead of a vee groove on the top side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There welder will bevel the bottom side using a grinder or plate beveler. Then they will weld the bottom side, but the root of the bottom weld will be contaminated by air from the other side of the weld since there is no backer or backing gas and thus no shielding on the far side. On the top side, using a grinder or a carbon electrode they will remove all of the contaminated root metal and then begin the seconds side weld. This weld if protected on the back side by the metal of the first weld, yielding a sound weld through the whole depth of the joint.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-09-05T18:05:46.603" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11369" PostTypeId="1" CreationDate="2016-09-05T19:19:04.813" Score="0" ViewCount="48" Body="&lt;p&gt;The Peclet number express the ratio between the heat transferred by advection and by conduction (diffusion). If it is sufficiently high the heat transferred by conduction can be neglected with respect to the other term. In a book I've found that the first principle of thermodynamic applied to a tube, under the assumption of high Pe, is the well known formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=mC_p(T_{in}-T_{out})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should this be modified if the previous assumption is not fulfilled?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dE}{dt}=\sum m(h+z+v^2/2)+Q-L$$&lt;/p&gt;&#xA;" OwnerDisplayName="user6267" LastEditorUserId="2559" LastEditDate="2017-02-19T16:33:28.067" LastActivityDate="2017-02-19T16:33:28.067" Title="Peclet Number and first principle for open system" Tags="|thermodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11371" PostTypeId="1" CreationDate="2016-09-06T00:21:58.987" Score="1" ViewCount="148" Body="&lt;p&gt;I have noticed that nichrome is a common resistance wire used for various heating purposes. I have seen it used inside what seem like glass tubes (like on an electric stove) and just floating in coils (like in hair dryers), but how would it be made in a sheet form?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that it is not insulated, which means that if it touches another piece of metal, the current would flow through that metal rather than the nichrome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would want the material it is wrapped in to be thermally conductive, but not electrically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen designs like this which put an insulator around the nichrome, but I don't see how that is effective at distributing the heat produced by the nichrome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to achieve an even as possible distribution of heat at about 150C. My power supply is going to be a 12V supply that can do up to 15A (so current is not much of an issue here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yc6x6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yc6x6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2164" LastEditorUserId="2164" LastEditDate="2016-09-06T20:11:49.547" LastActivityDate="2016-09-06T20:11:49.547" Title="How are most sheet style nichrome-based heating elements constructed?" Tags="|thermodynamics|heat-transfer|design|thermal-conduction|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11372" PostTypeId="1" CreationDate="2016-09-06T01:19:28.007" Score="1" ViewCount="38" Body="&lt;p&gt;If I have a measurement $A$ with uncertainty $\Delta A$, then what is the error if I do a calculation like $\frac{A}{\sqrt{3}}$?&lt;/p&gt;&#xA;" OwnerUserId="7922" LastEditorUserId="163" LastEditDate="2016-09-06T07:36:54.693" LastActivityDate="2016-10-06T09:00:21.247" Title="How do measures of error or uncertainty change, when the underlying variable is transformed?" Tags="|experimental-physics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11373" PostTypeId="2" ParentId="11362" CreationDate="2016-09-06T03:11:53.573" Score="2" Body="&lt;p&gt;There is no hard limit on the size of an airfoil/hydrofoil. A ships hull is a hydrofoil. This is how a sailboat &lt;a href=&quot;https://en.wikipedia.org/wiki/Sailing_into_the_wind&quot; rel=&quot;nofollow&quot;&gt;sails into the wind&lt;/a&gt; for example. A &lt;a href=&quot;https://en.wikipedia.org/wiki/Keel&quot; rel=&quot;nofollow&quot;&gt;keel&lt;/a&gt; is a large hydrofoil used in addition to the boats hull for directional stability or to counteract the force of the sails on a sailboat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrofoil&quot; rel=&quot;nofollow&quot;&gt;Hydrodynamic lifting&lt;/a&gt; of vessels for higher speed and higher efficiency is possible too, and have some very promising applications. However, like everything, there are economic/practicality constraints, especially when going big.&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;Adding hydrofoils to an aircraft carrier, for example, would require:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lots of engineering, prototyping, money, more money, and risk.&#xA;&lt;li&gt;Multiple long cord (maybe 10ft) hydrofoils added 20ft below the lowest part of the hull. They would be designed to lift half of the hull or maybe even all of the hull out of the water at cruising speed.&#xA;&lt;li&gt;More exposed hull at cruise speed; more vulnerable.&#xA;&lt;li&gt;Can the structure withstand failure of one or multiple hydrofoils at cruise speed.&#xA;&lt;li&gt;Deep hydrofoils limit traversable depth at lower speeds.&#xA;&lt;li&gt;Hydrofoils would be more easily damaged in a combat situation and&#xA;    would need to be discardable, or other emergency provisions.&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;Benefits of this would be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;This would reduce power consumption at top speed&#xA;&lt;li&gt;Increase top speed with current powerplant.&#xA;&lt;li&gt;Better rough water stability at cruise speed if hydrofoils have a computer controlled leveling system.&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;While such an aircraft carrier might be advantageous to current designs, it comes with significant opportunity cost. Would that money be better spent on higher performance jets, better nuclear power systems for ships, or even peace-making efforts ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These engineering efforts may make more sense for more cost conscious industries such as bulk transportation. Again, lots of details to contend with. Some &lt;a href=&quot;http://newatlas.com/mitsubishi-air-lubrication-system/21196/&quot; rel=&quot;nofollow&quot;&gt;new ships&lt;/a&gt; use bubbles to reduce drag; so there are opportunities in this area. &lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-06T03:11:53.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11374" PostTypeId="1" AcceptedAnswerId="11380" CreationDate="2016-09-06T07:15:13.193" Score="1" ViewCount="154" Body="&lt;p&gt;I'm doing plant design and one of the reactors is a furnace that uses methane to burn. I want to calculate the amount of methane that is required. I have the energy needed and the calorific value of methane but now I need a average thermal efficiency. Just to be clear I want the ratio of energy produced over how much was put in. Thanks&lt;/p&gt;&#xA;" OwnerUserId="7927" LastActivityDate="2016-09-06T12:47:39.657" Title="Thermal efficiency of a furnace reactor?" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11375" PostTypeId="2" ParentId="11372" CreationDate="2016-09-06T07:21:29.487" Score="0" Body="&lt;p&gt;I'f I understand correctly the error should be (delta A)/sqrt(3). Then the percentage error would be that over the actual answer.&lt;/p&gt;&#xA;" OwnerUserId="7927" LastActivityDate="2016-09-06T07:21:29.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11376" PostTypeId="2" ParentId="6517" CreationDate="2016-09-06T07:31:29.870" Score="1" Body="&lt;p&gt;When your doing a problem like this you start by looking at the reaction that you want to happen. I'm sure you will find lots of information on the oxidation of nickle.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find a temperature at which the reaction occurs (kinetics).&lt;/li&gt;&#xA;&lt;li&gt;Find out how much catalyst or nickle is in the reactor.&lt;/li&gt;&#xA;&lt;li&gt;Calculate the oxygen needed trough basic stoichiometry.&lt;/li&gt;&#xA;&lt;li&gt;The percentage of oxygen in the feed is up to you. The amount will affect the efficiency. There should be literature on it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hope this helps or points you in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="7927" LastEditorUserId="7927" LastEditDate="2016-09-07T18:23:11.417" LastActivityDate="2016-09-07T18:23:11.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11377" PostTypeId="2" ParentId="11372" CreationDate="2016-09-06T07:34:08.617" Score="2" Body="&lt;p&gt;In general, if the calculation is multiplication by a constant, then you multiply the uncertainty by the same constant. That's assuming that &quot;uncertainty&quot; refers to something like standard deviation, rather than to variance: if it's variance, then you multiply by the &lt;em&gt;square&lt;/em&gt; of the constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's addition of a constant, it doesn't change the uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's doing anything fancy like taking a log or raising to a power, or including some other random variable or more than one observation of your existing variable, then it gets complicated: and as a general approach, one can run a Monte Carlo simulation to get a numeric answer.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2016-09-06T08:03:07.820" LastActivityDate="2016-09-06T08:03:07.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11378" PostTypeId="1" CreationDate="2016-09-06T08:18:45.093" Score="1" ViewCount="309" Body="&lt;p&gt;On the electromagnet setup pictured below, only the core part is magnetic (when powered, it attracts a piece of steel) but not the 2 &quot;wings&quot; outside that are supposed to channel the magnetic flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ulASj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ulASj.jpg&quot; alt=&quot;setup&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5926" LastActivityDate="2016-09-06T14:20:58.917" Title="why is the outside of an electromagnet not magnetic?" Tags="|magnets|electromagnetism|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11380" PostTypeId="2" ParentId="11374" CreationDate="2016-09-06T12:47:39.657" Score="1" Body="&lt;p&gt;The efficiency depends on how well-insulated the reactor is, and how complete the burn is. But broadly speaking, something between 80% and 100%. It also depends on whether you're recovering heat by condensing the water vapour - if not, you want to use the LHV (rather than the HHV) of methane (or equivalently, de-rate the HHV by 10%)&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2016-09-06T12:47:39.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11381" PostTypeId="1" CreationDate="2016-09-06T12:57:56.687" Score="1" ViewCount="102" Body="&lt;p&gt;I was surprised that flight recorders only have enough battery to send out signals for 30 days. I'm guessing that it blindly sends out signals that help it to be located. But instead, why aren't they made to only send out a signal when they receive a ping? Surely, then, they could last for years?&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="7083" LastEditDate="2018-01-28T22:45:51.740" LastActivityDate="2018-01-28T22:45:51.740" Title="Why don't flight recorders transmit only when they receive a ping?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11382" PostTypeId="2" ParentId="11378" CreationDate="2016-09-06T13:02:25.450" Score="2" Body="&lt;p&gt;The strength of an electromagnet's field will typically be strongest on the inside of the coil (talking about a normal electromagnets). This is true for the geometry of the magnet you've shown in the question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you run a current through the coil, magnetic fields (perhaps not the best way to phrase this) are produced along the path of the coil that are always perpendicular to the coil. You can see an illustration of this below in an image on Wikipedia's page on electromagnets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gfvXz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gfvXz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken from: &lt;a href=&quot;https://en.wikipedia.org/wiki/Electromagnet#/media/File:Magnetic_field_of_wire_loop.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Electromagnet#/media/File:Magnetic_field_of_wire_loop.svg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can observe is that in the center of the coil the magnetic fields produced along the path of the coil tend to converge (geometry dependent). On the outside of the coil these fields diverge from each other. It should be easy to see that the field strength inside the coil will be stronger than the outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The configuration of your electromagnet is a little confusing. The coil is oriented perpendicular to the long axis of the core, which means the &quot;wings&quot; as you call them don't really get a strong field as they are outside the coil. If the coil was rotated 90 degrees so that the core was aligned with the long axis of the coil then I would agree that the field should be strong throughout the whole core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm no physicist, this is the easiest way I can explain it. If someone else can produce a more technical answer you'd benefit more from that.&lt;/p&gt;&#xA;" OwnerUserId="1174" LastEditorUserId="1174" LastEditDate="2016-09-06T14:20:58.917" LastActivityDate="2016-09-06T14:20:58.917" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11383" PostTypeId="2" ParentId="11381" CreationDate="2016-09-06T14:06:51.650" Score="2" Body="&lt;p&gt;Think about it a bit: you'd need a monstrous piece of gear to be able to &lt;strong&gt;receive&lt;/strong&gt; a ping from a distant source on a search platform. Transmitters are not the same as receivers.&lt;br&gt;&#xA;  The primary design goal of the flight recorder assembly is crash survival.  Further, after 30 days, there's pretty much no chance of finding survivors (with the notable exception of the   &lt;a href=&quot;https://en.wikipedia.org/wiki/1972_Andes_flight_disaster&quot; rel=&quot;nofollow&quot;&gt;Andes 1972 crash &lt;/a&gt;.  The world's a big place.  Aircraft occasionally disappear and are found years later pretty much by blind luck.  For that matter,  cargo ships sink all the time and even given their bulk and the theoretical ability to put a very large, long-lived beacon on them, they're not found either. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-09-06T14:06:51.650" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11384" PostTypeId="2" ParentId="5566" CreationDate="2016-09-06T16:45:25.387" Score="0" Body="&lt;p&gt;In simple terms earth pressure is both very similar and very different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertical earth pressure is given by:&#xA;Density x height x gravity.&#xA;Here the density depends on the material, which varies with type of soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal earth pressure is where it diverges from the simple water model. The percentage of the vertical force applied horizontally depends on the ability of the soil to support and transfer load.&#xA;Usually this is a simple coefficient for granular  material (around 0.5)  and for cohesive takes account of the shear strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are theories, such as silo theory, that reduce the volume of soil acting on an point base on failure planes.&lt;/p&gt;&#xA;" OwnerUserId="7758" LastActivityDate="2016-09-06T16:45:25.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11385" PostTypeId="2" ParentId="11369" CreationDate="2016-09-06T20:00:21.747" Score="1" Body="&lt;p&gt;If the convection cannot be neglected, then you need to add Newton's law term to the heat balance. The second term is the law of conservation of energy and Q includes all the heat applied to the body, so no modification should be made.&lt;/p&gt;&#xA;" OwnerUserId="7933" LastActivityDate="2016-09-06T20:00:21.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11386" PostTypeId="1" CreationDate="2016-09-06T20:09:55.583" Score="2" ViewCount="964" Body="&lt;p&gt;Ultrasonic transducers emit sound at a frequency greater that 20 KHz. What are the ways the frequency of sound emitted by such a transducer can be verified?&lt;/p&gt;&#xA;" OwnerUserId="6614" LastEditorUserId="282" LastEditDate="2016-10-13T08:54:54.597" LastActivityDate="2016-10-13T08:54:54.597" Title="Determining frequency of an ultrasonic transducer" Tags="|electrical-engineering|sensors|acoustics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11389" PostTypeId="2" ParentId="11386" CreationDate="2016-09-06T23:43:29.157" Score="3" Body="&lt;p&gt;If you're talking about a &quot;bare&quot; transducer (i.e., not a module with its own driver), you can just hook it up to an oscilloscope, hit it with a narrow pulse of voltage, and look for the decaying waveform of its self-resonance. Measure the frequency of that waveform.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-09-06T23:43:29.157" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11390" PostTypeId="1" AcceptedAnswerId="11398" CreationDate="2016-09-07T02:37:13.727" Score="-3" ViewCount="300" Body="&lt;p&gt;Much power is used to get rocket off the platform on the initial launch? Could electric engines used like booster rockets work similar in the way they launch planes of an aircraft carrier? The electric turbines would disconnect at the top of the launch platform that could also be made taller. Could it save fuel by creating lift for the first 1000 ft or so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UYZFc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UYZFc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; supplement &lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-07-18T15:57:30.910" LastActivityDate="2018-07-18T15:57:30.910" Title="What prevents the use of electric engines to assist with spacecraft launches? (Space Shuttle Booster Supplement)" Tags="|electrical-engineering|aerospace-engineering|turbines|experimental-physics|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="11391" PostTypeId="2" ParentId="11381" CreationDate="2016-09-07T07:45:27.357" Score="2" Body="&lt;p&gt;An &lt;a href=&quot;https://en.wikipedia.org/wiki/Emergency_position-indicating_radiobeacon_station&quot; rel=&quot;nofollow&quot;&gt;ELT or Emergency Locator Transmitter&lt;/a&gt; is built to be a reliable emergency device to locate a wrecked aircraft relatively quickly by satellite and ground triangulation. The goal of locating the aircraft is to save human lives; so 30 days is usually sufficient. Finding remains and satisfying people's curiosity is not part of the current design criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transmitting on an interval is the most reliable method, as it starts transmitting on the emergency frequency as soon as a plane goes down; not when someone decides to start looking for it. Satellites and ground towers are always listening on this frequency and will triangulate the signal as quickly as possible. Think of it like your house is on fire and you are trapped inside. You would yell &quot;help, help, help&quot; loudly and continuously instead of listening at the door for bystanders before you yell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To extend the amount of time an ELT signal lasts could be done by:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt; Adding a larger battery (obviously)&#xA;&lt;li&gt; After the 30 days go into a &quot;Marco Polo&quot; mode, waiting to receive a signal before transmitting like you mentioned. Remember it still takes some power to listen.&#xA;&lt;li&gt; After the 30 days decrease how often the signal is transmitted logarithmically. If the ELT synchronized with the aircraft clock it could emit signals at predictable times so that search personnel would be ready to listen. This would allow for very very low power use during the waiting period.&#xA;&lt;li&gt; Adding water pressure sensing capability to the unit so that it only sends sonar pulses after being submerged to a certain depth.&#xA;&lt;li&gt; Adding multiple directional antennas instead of an omnidirectional antenna. Sensors in the unit could fire the antennas that are not pointed at the ground increasing signal strength and reducing power consumption.&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-07T07:45:27.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11394" PostTypeId="1" AcceptedAnswerId="11402" CreationDate="2016-09-07T17:50:28.137" Score="7" ViewCount="12383" Body="&lt;p&gt;Defined the &lt;strong&gt;Gear Ratio&lt;/strong&gt; ($i$) as the ratio between the number of teeth on the pinion (driving gear) ($N_{1}$) and the number of teeth on the gear (driven gear) ($N_{2}$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this better, if better at all, having non-integer values of $i$? Could anyone please illustrate using arguments exploring the fabrication and performance points of view?&lt;/p&gt;&#xA;" OwnerUserId="7942" LastEditorUserId="7942" LastEditDate="2016-09-10T15:20:29.160" LastActivityDate="2022-12-03T16:57:38.983" Title="Why are spur gear ratios not always whole numbers in practice?" Tags="|mechanical-engineering|design|applied-mechanics|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11395" PostTypeId="2" ParentId="11358" CreationDate="2016-09-07T18:25:26.820" Score="1" Body="&lt;p&gt;Since your sensor has an analog output. If your comfortable with prototype level work, get an Arduino that has analog inputs then connect a bluetooth module such as the HC-05. Then, you can transmit that data using some code that I wrote to send multiple sensors in columns. You can modify this code to your needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jmaturner.com/sensors.html&quot; rel=&quot;nofollow&quot;&gt;sending sensor data - serial over bluetooth&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your next task would be to write software, but you can view this data on a PC or phone using a terminal program like Teraterm.&lt;/p&gt;&#xA;" OwnerUserId="5890" LastEditorUserId="5890" LastEditDate="2016-09-17T13:35:34.307" LastActivityDate="2016-09-17T13:35:34.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11396" PostTypeId="2" ParentId="11394" CreationDate="2016-09-07T19:05:27.603" Score="3" Body="&lt;p&gt;It should be fairly obvious that any gear can only have an integer number of teeth. Also for gears to mesh properly the tooth profile on all meshing gears needs to be similar if not identical. So, in practice the limitation is that individual gears need integer teeth and there is some pretty much fixed ratio between the number of teeth and radius which rules out some ratios for a given tooth profile and size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it is entirely possible to have non integer gear &lt;strong&gt;ratios&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spur gears are usually designed on the module system, this is based on the ratio of N teeth to the diameter of the gear and thus eliminates the irrational pi from the specification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only applies to conventional spur gears though, with pulleys you can have any arbitrary ratio you want and it is certainly not impossible to fudge tooth sizes even in spur gear or chain drives to get arbitrary ratios at the expense of increased backlash and/or wear. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-09-07T19:05:27.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11398" PostTypeId="2" ParentId="11390" CreationDate="2016-09-07T21:24:57.217" Score="5" Body="&lt;h1&gt;Electric turbine engines&lt;/h1&gt;&#xA;&lt;p&gt;First off, &amp;quot;electric turbine engines&amp;quot; don't really exist. What you're probably thinking of is an electric version of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Jet_engine#Turbine_powered&quot; rel=&quot;noreferrer&quot;&gt;jet turbine engine&lt;/a&gt;, which is a type of combustion engine (it burns fuel). Air is sucked in the front, compressed by a compressor, then mixed with fuel and burned in a combustion section. A turbine extracts some of the energy from the air and uses it to power the compressor, while the rest of the air shoots out the back of the engine to provide thrust. The important thing to understand is that the turbine is extracting energy from the air, not generating thrust. The burning fuel adds energy to the air which is converted into thrust by the structure of the jet engine. The turbine is just a way of making the jet engine more compact, so it doesn't need a separate power source for the compressor.&lt;/p&gt;&#xA;&lt;p&gt;The closest electric engine equivalent that I could find would be something like the &lt;a href=&quot;https://en.wikipedia.org/wiki/Airbus_E-Fan&quot; rel=&quot;noreferrer&quot;&gt;Airbus E-Fan&lt;/a&gt; (see also the &lt;a href=&quot;http://www.airbusgroup.com/int/en/corporate-social-responsibility/airbus-e-fan-the-future-of-electric-aircraft/technology-tutorial/E-Thrust.html&quot; rel=&quot;noreferrer&quot;&gt;Airbus website&lt;/a&gt;). It uses an electric motor to spin a fan, providing thrust. It does not use a turbine, it's more like a fancy propeller.&lt;/p&gt;&#xA;&lt;p&gt;Apparently &lt;a href=&quot;http://www.scientificamerican.com/article/impossible-electric-airplane-takes-flight/&quot; rel=&quot;noreferrer&quot;&gt;Airbus is hoping to use the E-Fan on commercial passenger jets in the future&lt;/a&gt;, albeit using a gas-powered generator to provide electricity, so it might be able to provide enough thrust to launch a small rocket.&lt;/p&gt;&#xA;&lt;h1&gt;Will this work for your application? (Edit: Space Shuttle booster replacement)&lt;/h1&gt;&#xA;&lt;p&gt;There are two problems with replacing the solid fuel booster rockets of the space shuttle with electric fans:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The booster rockets generate &lt;a href=&quot;https://en.wikipedia.org/wiki/Space_Shuttle_Solid_Rocket_Booster&quot; rel=&quot;noreferrer&quot;&gt;13,800 kN of thrust &lt;em&gt;each&lt;/em&gt; at liftoff&lt;/a&gt;, for a total of 27,600 kN. By comparison, the E-Fan only produces 750 N, therefore you would need 37,000 electric engines to get the same thrust.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The booster rockets burn for 2 minutes and are ejected when the shuttle reaches an &lt;a href=&quot;https://en.wikipedia.org/wiki/Space_Shuttle#Solid_rocket_boosters&quot; rel=&quot;noreferrer&quot;&gt;altitude of 46 km&lt;/a&gt;. At that height the atmosphere is &lt;a href=&quot;https://en.wikipedia.org/wiki/Atmosphere_of_Earth#/media/File:Comparison_US_standard_atmosphere_1962.svg&quot; rel=&quot;noreferrer&quot;&gt;so thin&lt;/a&gt; that an E-Fan wouldn't be able to produce any thrust. In fact, since the E-Fan works by &amp;quot;pushing&amp;quot; air through its duct, its performance would decrease as the atmosphere thinned, whereas the solid boosters do not need air to operate.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;h1&gt;Conclusion&lt;/h1&gt;&#xA;&lt;p&gt;You can't replace the space shuttle solid rocket boosters with electric fans/engines, even if you could solve the battery/fuel problem. The electric fans simply do not produce enough thrust. Even if they did they still require an atmosphere to work properly, but the space shuttle needs its boosters to take it most of the way out of the atmosphere. Neat idea, but I'm afraid it just won't work.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-08-09T14:29:09.297" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11400" PostTypeId="2" ParentId="11394" CreationDate="2016-09-07T23:46:06.480" Score="2" Body="&lt;p&gt;The primary reason is it promotes longer gear life. If the gear ratio were an integer number and a tooth was damaged, that tooth would wear on the same few opposing teeth until failure. A non integer ratio causes a damaged tooth to wear across all opposing gear teeth. This spreads the effective wear out, vice focusing it like an integer ratio would.&lt;/p&gt;&#xA;" OwnerUserId="1052" LastActivityDate="2016-09-07T23:46:06.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11401" PostTypeId="1" AcceptedAnswerId="11403" CreationDate="2016-09-07T23:53:09.057" Score="2" ViewCount="625" Body="&lt;p&gt;I am not sure if this is the right place to ask it, but this is a question that I thought today, and it gave me some curiosity to understand. Imagine that a car will curve, we can say the turn is a bit tight , what are the factors that can help it to flip? I was wondering about some aspects:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the car has mass, it has inertia, so while it is curving it tends to keep the motion in the same direction that it was instants before the turn. right? So, if the car has more mass, it has more inertia, and since there is friction, one heavier car would flip easier then one lighter, considering that all other possible variables were equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Center of gravity, a car with an higher center of gravity would flip easier. The whole inertia of the car distributed to higher heights would be further of the tires(where friction acts), creating angular momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The car being thin because it has less surface contact with the ground;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The car being lighter. This opposes what I've said in &quot;a)&quot;, but a heavier car is more difficult to get off the road. A lighter car has more instability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I wrong? In what I'm wrong? What do you think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for helping. :)&lt;/p&gt;&#xA;" OwnerUserId="7947" LastActivityDate="2016-09-08T06:18:50.210" Title="What aspects can help a car flip?" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11402" PostTypeId="2" ParentId="11394" CreationDate="2016-09-08T00:58:07.087" Score="9" Body="&lt;p&gt;Gear ratios can be referred to as &lt;em&gt;hunting&lt;/em&gt; and &lt;em&gt;non-hunting&lt;/em&gt; ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a non-hunting ratio, any one gear tooth will contact its corresponding gears' teeth in the same place every time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a hunting ratio, the number of teeth in the gear set have no common factors between them. This way as the pitch diameters 'roll' over one another it takes a very long time for the same teeth to once again make contact. As previously mentioned, this is much better for wear, vibration and noise.&lt;/p&gt;&#xA;" OwnerUserId="2732" LastEditorUserId="77" LastEditDate="2016-09-10T15:49:48.327" LastActivityDate="2016-09-10T15:49:48.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11403" PostTypeId="2" ParentId="11401" CreationDate="2016-09-08T01:56:36.993" Score="1" Body="&lt;p&gt;A car will flip as soon as the acceleration vector from its center of mass points outside the &quot;footprint&quot; defined by connecting the tire contact patches with lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not this occurs has a lot to do with how much lateral force the tires can generate through their friction with the ground &amp;mdash; or if they catch on something like a curb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, the mass has little to do with it, because it affects the &quot;grip&quot; of the tires in the same proportion that it contributes to the inertia of the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometry of the vehicle &amp;mdash; the length and width of the wheelbase vs. the height of the center of gravity &amp;mdash; is the primary factor determining stability. A car with a wide/long stance and a low CG will generally slide rather than flip, unless the wheels catch on something.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-09-08T01:56:36.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11404" PostTypeId="2" ParentId="11401" CreationDate="2016-09-08T02:05:57.917" Score="0" Body="&lt;p&gt;A heavier car will not flip easier than a lighter one, while going through a curve (unless you are considering strong side winds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand why this is the case we have to look at what makes a car flip. A car flips because the sum of moments is not zero. Normally when you push against the side of a car, its weight will get distributed differently between the tires, which creates a moment which opposes the moment created by you pushing against the car. When the entire weight/normal force of the car has to be placed at the tires on one side of the car, then the car will be at the tipping point before it starts to flip. When a car goes through a curve at high speed, then the &quot;push&quot; from the side can be replaced by a centrifugal force (which is a pseudo force, only present in the reference frame of the car). However both the centrifugal force and the maximum opposing moment, due to gravity, are both proportional to the mass of the car. So changing the mass of the car, assuming identical mass distributions, does not affect how easier it is for the car to flip in a curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The properties that do affect the ease with which a car can flip going through a curve are the height of the center of mass of the car and the width between the tires of the car. Namely the height of the center of mass of the car acts as the lever arm of the centrifugal force and (half) the width between the tires acts as the lever arm of gravity, the bigger the arm the bigger the moment the force generates.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2016-09-08T02:05:57.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11405" PostTypeId="1" CreationDate="2016-09-08T02:45:55.260" Score="0" ViewCount="7525" Body="&lt;p&gt;I want to know the function of counterweights on crankshaft. I heard it is to able the piston to go back up, to repeat the explosion stroke. But doesnt the piston on one cylinder goes back up because the piston in other cylinder is pushed down? This makes the first cylinder go down and up.&lt;/p&gt;&#xA;" OwnerUserId="7947" LastActivityDate="2016-09-08T04:58:48.070" Title="Counterweight of Crankshaft" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11406" PostTypeId="2" ParentId="11405" CreationDate="2016-09-08T04:58:48.070" Score="3" Body="&lt;p&gt;The counterweighs have two main functions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Reduce the bending load on the crankhsaft. Imagine a 4 cylinder engine with a flat plane crankshaft. The first cylinder is at the top dead center (TDC), the second is at the bottom dead center (BDC). The conrod and the piston of the first cylinder &quot;pulls&quot; the crankshaft up, the second down, while it is fixed at the main bearings. The crankshaft will be deformed at the main bearings, which means a lot of stress at the fillets next to the bearings, and has a negative effect on the lifetime of the bearing shells.&#xA;If you add counterweights to the crankshaft at each cylinder, and adjust the size of these carefully, they will pull the crankshaft with the same force as the conrod and piston, but in the opposite direction, and the crankshaft will have no bending loads&lt;/li&gt;&#xA;&lt;li&gt;In most engines the majority of the forces is countered internally (eg. a 4 cylinder engine has two pistons moving up, 2 moving down, so the engine itself will remain in place). But in some cases, like in a single cylinder engine, or in some 3 cylinder engines the internal forces are not in balance. For these engines you need additional counterweights, so the engine will not shake when the cranktrain is rotating.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In reality it is slightly more complicated, as you will have forces with rotating and alternating directions. You adjust the counterweight size to cancel only a certain amount of the mass forces of the conrod (the &quot;rotating mass&quot;), and not the piston. By how much is a compromise between costs, quality, weight, speed and size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the forces in the crank mechanism are not exactly sinusoid (inherent from the finite length of the conrod). With counterweights on the crankshaft you can influence the balance only in the first order. For higher frequencies you need additional measures, like extra balancer shafts in the crankcase or in the oilpan. They rotate with twice the speed as the crankshaft, cancelling the second order forces.&lt;/p&gt;&#xA;" OwnerUserId="7843" LastActivityDate="2016-09-08T04:58:48.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11407" PostTypeId="2" ParentId="11401" CreationDate="2016-09-08T06:18:50.210" Score="0" Body="&lt;p&gt;When a car makes a turn with a radius of R it is subject to an acceleration force, vector, called centripetal force $$ F=m V^2/R $$&lt;br&gt;&#xA;this force is applied to the center of gravity of the car and is directed horizontally opposite of center of turn.&lt;br&gt;&#xA;This Vector adds to the vector of weight of the car which is directing vertically down.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of these two vectors is a vector drifting out and down diagonally, call it: R fore roll .&lt;br&gt;&#xA;If R falls outside of the rectangular area between four wheels the car will start to roll. But it takes a few seconds for it to start from just lifting the inside tires off the road to pass the critical angle of no return. Experienced drivers could maneuver the car back to balance during those early moments.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A car with lower CG and wider wheel-base needs much larger centripetal force to go off balance.&#xA;A car with stronger roll control rods will be more stable but will have a hard ride. Fancy cars have computer controlled roll prevention!&lt;br&gt;&#xA;having worn out tires on the rear promotes skidding and roll. Road condition on a tight turn such as bumps or puddles promotes roll. Finally sudden jerky braking to avoid a pot-hole is dangerous too. &#xA;Some race cars have suspension designed to lower the CG and open the wheel-base in a tight roll!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-09-08T06:18:50.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11408" PostTypeId="2" ParentId="494" CreationDate="2016-09-08T07:38:26.507" Score="2" Body="&lt;p&gt;A prime example of welders that would perform such welds would be a Lincoln PowerMig 210, Hobart Handler 210 MVP. This is just to name a few. Miller and Everlast also have the &quot;MVP=Multi Variable Process&quot; units. Keep in mind most Multiprocessing welders have a lower production time or more clearly defined as &quot;Duty Cycle time&quot; due to the Multi-variable process of using 120v or 240v. All of these units have a max thickness close to production grade but are more suitable for the DIY person. I personally use a Lincoln PowerMig 210. Not that it's any better but the all around characteristics of welding GMAW=MIG welding MIG also referred to as Metal Inert Gas, SMAW=Shielded Metal Arc Welding also referred to as Arc Welding, FCAW=Flux Core Arc Weld, FCAW-S=Shielded Gas Flux Core, GTAW=Gas Tungsten Arc Welding also referred to as TIG in both AC and DC electrode negative or positive, And Aluminum Spool gun Mig welding. These are very versatile machines capable of handling most DIY jobs with a high quality of weld deposition. The welding process of FCAW-S has the highest deposition rate and is used more commonly in high production areas. Downside of these units would be on average 3/8&quot; max carbon steel or 1/4&quot; Aluminum thickness.&lt;/p&gt;&#xA;" OwnerUserId="7948" LastActivityDate="2016-09-08T07:38:26.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11409" PostTypeId="1" AcceptedAnswerId="11412" CreationDate="2016-09-08T09:43:13.013" Score="9" ViewCount="876" Body="&lt;p&gt;I've recently been on &lt;a href=&quot;https://en.wikipedia.org/wiki/R%C3%BCgen&quot; rel=&quot;noreferrer&quot;&gt;Rügen&lt;/a&gt; and I have seen this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DkHq6.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DkHq6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have an idea what this is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(To give Google a chance: A ship with four &quot;towers&quot; made out of steel on it. They looks a bit like high-voltage electrical towers.)&lt;/p&gt;&#xA;" OwnerUserId="3061" LastActivityDate="2016-09-08T15:07:44.237" Title="What kind of ship is this?" Tags="|ships|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11412" PostTypeId="2" ParentId="11409" CreationDate="2016-09-08T14:34:19.593" Score="12" Body="&lt;p&gt;It's a &lt;strong&gt;heavy-lift vessel&lt;/strong&gt;, carrying four identical structures as cargo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very likely they are &lt;a href=&quot;http://www.4coffshore.com/windfarms/jacket-or-lattice-structures-aid5.html&quot; rel=&quot;noreferrer&quot;&gt;jacket foundations for offshore wind turbines&lt;/a&gt;. The excellent &lt;a href=&quot;http://www.4coffshore.com/windfarms/vessels.aspx?catId=3&quot; rel=&quot;noreferrer&quot;&gt;4C Offshore has a database of such vessels used on offshore wind farms&lt;/a&gt;. It also has a list of all offshore windfarms under construction by country, and tells you which vessels are working on it, so given that you know where the photo was taken, you may able to work out which farm it's supplying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think they're going to &lt;strong&gt;&lt;a href=&quot;http://www.4coffshore.com/windfarms/wikinger-germany-de47.html&quot; rel=&quot;noreferrer&quot;&gt;Wikinger&lt;/a&gt; offshore wind farm&lt;/strong&gt;. That's using jacket foundations, is being built as I write this post, and is &lt;a href=&quot;http://www.4coffshore.com/offshorewind/index.html?lat=54.834&amp;amp;lon=14.068&amp;amp;wfid=DE47&quot; rel=&quot;noreferrer&quot;&gt;about 20 km north-east of Rügen&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a jacket foundation &lt;em&gt;in situ&lt;/em&gt; at &lt;a href=&quot;http://www.4coffshore.com/windfarms/alpha-ventus-germany-de01.html&quot; rel=&quot;noreferrer&quot;&gt;alpha ventus windfarm&lt;/a&gt;, courtesy of &lt;a href=&quot;http://eandt.theiet.org/magazine/2014/05/windfarm-foundations.cfm&quot; rel=&quot;noreferrer&quot;&gt;the IET&lt;/a&gt;:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/t8aoz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t8aoz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2016-09-08T15:07:44.237" LastActivityDate="2016-09-08T15:07:44.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11413" PostTypeId="1" CreationDate="2016-09-08T16:26:02.397" Score="2" ViewCount="11360" Body="&lt;p&gt;Does anyone know how to derive the deflection formula $\delta = \dfrac{-0.0068wL^4}{EI}$ for a continuous beam spans over 4 supports?&lt;/p&gt;&#xA;" OwnerUserId="7953" LastEditorUserId="1832" LastEditDate="2016-09-08T16:53:42.840" LastActivityDate="2023-08-11T04:24:23.987" Title="Deflection Formula for Continous Beam" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11415" PostTypeId="1" CreationDate="2016-09-08T16:51:35.847" Score="1" ViewCount="66" Body="&lt;p&gt;Recently some friends and I conceived of a method for visualizing in real-time the pitch/yaw/roll of an aircraft where a model aircraft is mounted inside a gimbal; then, the controller for the gimbal would receive telemetry from a flight simulator and update the position of the aircraft to show you how steep the aircraft is climbing or what compass direction it is heading. (see &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/5/5c/Gimbal_lock_airplane.gif&quot; rel=&quot;nofollow&quot;&gt;this diagram for an idea of what we are talking about.&lt;/a&gt;) Not free-floating, but with a motor at each axle to keep the position correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We began our search for this by scouring the internet but haven't found any examples of a motorized gimbal like this. There are several difficulties we have encountered in designing our own, also. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a safe and effective way to translate control signals and power through an axle so that the inner gimbal ring motors will receive enough power to turn a balance 3D model on all axes? We considered wireless communication for the motors and using batteries but the batteries would add a lot of weight to each ring which would need to be balanced out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, has this idea already been proposed and dismissed for being impractical? I don't want to reinvent the wheel (no pun intended) but also don't know the correct search terms to find a product like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any help you can give me,&lt;/p&gt;&#xA;" OwnerUserId="7954" LastActivityDate="2016-09-08T16:51:35.847" Title="Feasibility of a motorized 3-axis gimbal for visualization" Tags="|design|motors|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11417" PostTypeId="1" CreationDate="2016-09-08T17:47:16.283" Score="4" ViewCount="120" Body="&lt;p&gt;UPC barcodes are comprised of a series of black and white bars. Bars can range in width from 1 to 4 (1 being the thinnest and 4 being the thickest). Each digit, 0 through 9 is encoded using exactly 4 bars:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Digit | Bar Thickness Sequence&#xA;  0   |        3-2-1-1&#xA;  1   |        2-2-2-1&#xA;  2   |        2-1-2-2&#xA;  3   |        1-4-1-1&#xA;  4   |        1-1-3-2&#xA;  5   |        1-2-3-1&#xA;  6   |        1-1-1-4&#xA;  7   |        1-3-1-2&#xA;  8   |        1-2-1-3&#xA;  9   |        3-1-1-2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;How were the bar widths assigned to the numbers 0 through 9?  I notice that each series of 4 bars sums to 7, but other than that I am not seeing an obvious pattern?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: The numbers a UPC barcode encodes can have special meanings (product classifcation, vendor, checksum, etc.)  This question is only about how the bar widths (the barcode encoding) was derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one of the references I found: &lt;a href=&quot;http://www.adams1.com/upccode.html&quot; rel=&quot;nofollow&quot;&gt;All About UPC Barcode &amp;amp; EAN Barcode&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="7957" LastEditorUserId="7083" LastEditDate="2016-09-13T05:50:15.093" LastActivityDate="2016-09-18T19:53:43.937" Title="How were the sequence of bar widths assigned to digits 0-9 for UPC barcodes?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11418" PostTypeId="2" ParentId="11413" CreationDate="2016-09-08T20:07:44.517" Score="8" Body="&lt;p&gt;I assume you mean a beam with three spans of equal length $L$ under the effect of a uniform distributed load, like the one displayed below. Also, I assume you're asking about the maximum displacement on the lateral (non-central) spans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/49wg6.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/49wg6.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a statically indeterminate structure, so this'll take some work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters, we can notice this structure is symmetric, and can therefore be simplified to the following (the rectangle represents a rotational restraint, to represent that there's no rotation at the central midspan):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LzALs.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LzALs.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, is &lt;strong&gt;still&lt;/strong&gt; statically indeterminate. So now let's pretend that rotational restraint doesn't exist, which turns this into an isostatic simply-supported beam with cantilever. Actually, let's get rid of the cantilever as well, and replace it with a concentrated force and moment at the support, equal to $M=-\dfrac{q\left(\frac{L}{2}\right)^2}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pARUG.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pARUG.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now let's calculate the displacement at the middle of the simply-supported span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we need to calculate the reactions (still considering the cantilever):&#xA;$$\begin{align}&#xA;\sum V &amp;amp;= V_1 + V_2 + \frac{3}{2}qL = 0 \\&#xA;\sum M_1 &amp;amp;= LV_2 + \frac{9}{8}qL^2 = 0 \\&#xA;\therefore V_2 &amp;amp;= -\frac{9}{8}qL \\&#xA;V_1 &amp;amp;= -\frac{3}{8}qL&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have that, we can start integrating our way up to the deflection.&#xA;$$\begin{align}&#xA;Q &amp;amp;= \int\limits_0^L q\text{d}x \\&#xA;&amp;amp;= qx + C \\&#xA;Q(0) &amp;amp;= C = V_1 = -\frac{3}{8}qL \\&#xA;M &amp;amp;= -\int\limits_0^L Q\text{d}x \\&#xA;&amp;amp;= -(\frac{q}{2}x^2 - \frac{3}{8}qLx + C) \\&#xA;M(0) &amp;amp;= C = 0 \\&#xA;EI\theta &amp;amp;= \int\limits_0^L M\text{d}x \\&#xA;&amp;amp;= -\frac{q}{6}x^3 + \frac{3}{16}qLx^2 + C_1 \\&#xA;EI\delta &amp;amp;= \int\limits_0^L \theta\text{d}x \\&#xA;&amp;amp;= -\frac{q}{24}x^4 + \frac{1}{16}qLx^3 + C_1x + C_2 \\&#xA;\delta(0) &amp;amp;= C_2 = 0 \\&#xA;\delta(L) &amp;amp;= -\frac{q}{24}L^4 + \frac{1}{16}qL^4 + C_1L = 0 \\&#xA;\therefore C_1 &amp;amp;= -\frac{q}{48}L^3 \\&#xA;\therefore \delta &amp;amp;= \frac{q}{EI}\left(-\frac{x^4}{24} + \frac{Lx^3}{16} -\frac{L^3x}{48}\right)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This, however, is not our deflection equation, since we still need to deal with that rotation restriction at the end of our cantilever, which we'd ignored so far. To do that, we need to get the rotation at the free end of the cantilever. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;M_c &amp;amp;= q\left(\dfrac{x^2}{2} - \dfrac{Lx}{2} + \dfrac{L^2}{8}\right) \\&#xA;EI\theta_c &amp;amp;= \int\limits_0^L M_c\text{d}x \\&#xA;&amp;amp;= q\left(\dfrac{x^3}{6} - \dfrac{Lx^2}{4} + \dfrac{L^2x}{8} + C\right) \\&#xA;EI\theta_c(0) &amp;amp;= C = EI\theta(L) \\&#xA;&amp;amp;= qL^3\left(-\frac{1}{6} + \frac{3}{16} - \frac{1}{48}\right) = 0 \\&#xA;EI\theta_c\left(\frac{L}{2}\right) &amp;amp;= q\left(\dfrac{\left(\frac{L}{2}\right)^3}{6} - \dfrac{L\left(\frac{L}{2}\right)^2}{4} + \dfrac{L^2\left(\frac{L}{2}\right)}{8}\right) \\&#xA;&amp;amp;= \dfrac{qL^3}{48}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if there were no rotational restriction, the cantilever'd have a rotation of $\dfrac{qL^3}{48}$ at its free end. The rotational restriction must therefore apply a concentrated bending moment sufficient to cancel out that rotation. Now, if a concentrated bending moment $M'$ is applied at the end of the cantilever, it will generate a uniform diagram along the cantilever and a linear diagram along the simply-supported span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we need to redo everything all over again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum V &amp;amp;= V_1 + V_2 = 0 \\&#xA;\sum M_1 &amp;amp;= LV_2 + M' = 0 \\&#xA;\therefore V_2 &amp;amp;= -\frac{M'}{L} \\&#xA;V_1 &amp;amp;= \frac{M'}{L} \\&#xA;Q &amp;amp;= \int\limits_0^L 0\text{d}x \\&#xA;&amp;amp;= C \\&#xA;Q(0) &amp;amp;= C = V_1 = \frac{M'}{L} \\&#xA;M &amp;amp;= -\int\limits_0^L Q\text{d}x \\&#xA;&amp;amp;= \frac{M'}{L}x + C \\&#xA;M(0) &amp;amp;= C = 0 \\&#xA;EI\theta &amp;amp;= \int\limits_0^L M\text{d}x \\&#xA;&amp;amp;= \frac{M'}{2L}x^2 + C_1 \\&#xA;EI\delta &amp;amp;= \int\limits_0^L \theta\text{d}x \\&#xA;&amp;amp;= \frac{M'}{6L}x^3 + C_1x + C_2 \\&#xA;\delta(0) &amp;amp;= C_2 = 0 \\&#xA;\delta(L) &amp;amp;= \frac{M'}{6L}L^3 + C_1L = 0 \\&#xA;\therefore C_1 &amp;amp;= -\frac{M'L}{6} \\&#xA;\therefore \delta &amp;amp;= \frac{M'}{EI}\left(\frac{x^3}{6L} -\frac{Lx}{6}\right)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now for the cantilever:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;EI\theta_c &amp;amp;= \int\limits_0^L M_c\text{d}x \\&#xA;&amp;amp;= M'x + C \\&#xA;EI\theta_c(0) &amp;amp;= C = EI\theta(L) \\&#xA;&amp;amp;= \frac{M'L}{3} \\&#xA;EI\theta_c\left(\frac{L}{2}\right) &amp;amp;= \frac{5M'L}{6}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can find the requisite applied bending moment at the end of the cantilever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\frac{5M'L}{6} &amp;amp;= \dfrac{qL^3}{48} \\&#xA;\therefore M' &amp;amp;= \dfrac{qL^2}{40}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Almost there!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all that's left to do is sum up the deflection equations for the simply-supported span. We end up with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\delta &amp;amp;= \frac{q}{EI}\left(-\frac{x^4}{24} + \frac{Lx^3}{16} -\frac{L^3x}{48}\right) + \frac{M'}{EI}\left(\frac{x^3}{6L} -\frac{Lx}{6}\right) \\&#xA;&amp;amp;= \frac{q}{EI}\left(-\frac{x^4}{24} + \frac{Lx^3}{16} -\frac{L^3x}{48}\right) + \frac{q}{EI}\left(\frac{Lx^3}{240} -\frac{L^3x}{240}\right) \\&#xA;&amp;amp;= \frac{q}{EI}\left(-\frac{x^4}{24} + \frac{Lx^3}{15} -\frac{L^3x}{40}\right) \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum value, however, is not at the midpoint. So now we have to find the minimum via the derivative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta' = \frac{q}{EI}\left(-\frac{x^3}{6} + \frac{Lx^2}{5} -\frac{L^3}{40}\right) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then use &lt;a href=&quot;http://www.wolframalpha.com/input/?i=diff(-x%5E4%2F24+%2B+L*x%5E3%2F15+-L%5E3*x%2F40,x)&quot; rel=&quot;noreferrer&quot;&gt;Wolfram Alpha&lt;/a&gt; to find that the minimum occurs at $x\approx0.44604 L$. Throw that into the full equation for the deflection and you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = -\dfrac{0.00688421327975355744qL^4}{EI}$$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-09-08T20:07:44.517" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11419" PostTypeId="1" CreationDate="2016-09-08T20:10:10.660" Score="4" ViewCount="1304" Body="&lt;p&gt;Is there an alternative for silicone rubber sponge? The reason I was thinking of this material was because I needed something that was&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Flexable&lt;/li&gt;&#xA;&lt;li&gt;Thermally Conductive&lt;/li&gt;&#xA;&lt;li&gt;Electrically Insulated&lt;/li&gt;&#xA;&lt;li&gt;Waterproof&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Silicone rubber seems very expensive, so I was looking for a cheaper option that would still meet the above requirements. &lt;/p&gt;&#xA;" OwnerUserId="2164" LastActivityDate="2023-07-13T21:02:21.577" Title="Cheaper alternative to silicone rubber sponge?" Tags="|materials|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11420" PostTypeId="1" AcceptedAnswerId="11427" CreationDate="2016-09-08T20:30:42.273" Score="0" ViewCount="67" Body="&lt;p&gt;I am a fairly recent graduate (I graduated December of 2015 with a Bachelor's of Science in Computational Science) and since graduating I have had an increasing desire to get involved in engineering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What areas of engineering employ aspects of computational science?&lt;/p&gt;&#xA;" OwnerUserId="7960" LastEditorUserId="2819" LastEditDate="2016-09-09T02:25:30.353" LastActivityDate="2016-09-09T02:42:30.383" Title="What are some overlaps between Computational Scientist and Mechanical Engineering?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ClosedDate="2016-09-09T10:25:48.633" ContentLicense="CC BY-SA 3.0" />
  <row Id="11421" PostTypeId="1" AcceptedAnswerId="11429" CreationDate="2016-09-08T21:01:05.640" Score="3" ViewCount="45" Body="&lt;p&gt;I am unsure of how to interpret the uncertainty given for a Multimeter I have been using. Everything is listed as follows ±(0.8%+10). I have never seen uncertainty listed this way, with a +&quot;x&quot; after the percent uncertainty. What range does this correspond to exactly, -10.8% to 10.8% or is it -0.18% to 0.18%? Here is a link to the datasheet: &lt;a href=&quot;http://www.smart-prototyping.com/image/data/3_equipment/1_electronic/multimeters/100409%20-%20Victor%20VC830L%20Manual.pdf&quot; rel=&quot;nofollow&quot;&gt;Digital Multimeter Datasheet&lt;/a&gt;. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="6582" LastActivityDate="2016-09-09T11:18:36.610" Title="How to interpret this Uncertainty?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11422" PostTypeId="1" CreationDate="2016-09-08T21:06:49.780" Score="2" ViewCount="14124" Body="&lt;p&gt;Is miter joint in welding - butt joint or corner joint ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My example includes 45 degrees miter (90 degrees) angle on two SHS (Square Hollow Sections) - 50x50x3 [mm] Working with EN/ISO norms regarding welding and joint preparation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My opinion is that horizontal welds are butt welds and vertical corner welds (fillet) - Just my opinion. I am not sure about it, that is why I seek guidance. See image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K6BuR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K6BuR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will these welds be designated and shown on a technical drawing ?&lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="2559" LastEditDate="2016-09-09T16:37:41.807" LastActivityDate="2024-01-27T17:19:54.990" Title="What type is miter joint - Welding" Tags="|mechanical-engineering|welding|welds|joining|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11423" PostTypeId="2" ParentId="11422" CreationDate="2016-09-08T21:18:40.460" Score="2" Body="&lt;p&gt;Assuming you're talking about plates: it would depend on you code, but usually it's the angle between the two joints that matters. If you mean a 90 degree joint with two 45 degree miter angles, I would still treat it as a corner joint. If there is a 10 degree miter angle on each piece so the dihedral angle in the joint is 160 degrees, it would be considered a butt joint.&lt;/p&gt;&#xA;&lt;p&gt;In AWS D1.1 for example, the notes of table 3.3 specify that and dihedral angle from 135-180 is a butt joint and any dihedral angle from 45 to 135 can be considered a corner joint. In the 2008 version of the code it's note j - I don't have another year handy to cross reference. Note that in some scenarios you may need to account for a Z loss factor on the acute side.&lt;/p&gt;&#xA;&lt;p&gt;If you are talking about tubes, this is significantly more complicated and we'll need more details, especially what type of tube and what code you are working to.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;In reference to your edit that this is a tubular mitered L connection : I believe the proper thing would be to specify a PJP weld around the two flat sides and the outside corner and a fillet on the inside corner. For the purposes of weld qualification, the two flat PJPs would be considered butt welds and the outside corner would be considered a corner weld. This might mean that you want to provide different bevel types at each side. It is also acceptable to specify a PJP weld on the three sides and allow the contractor to chose the bevel type that is best for them as long as they obtain the effective throat you list (or the code minimum)&lt;/p&gt;&#xA;&lt;p&gt;The following is from the &lt;a href=&quot;https://steeltubeinstitute.org/hss/faq-category/welding/&quot; rel=&quot;nofollow noreferrer&quot;&gt;FAQ&lt;/a&gt; at the steel tube institute.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;QUESTION:&lt;/p&gt;&#xA;&lt;p&gt;Are different AWS D1.1 prequalified welds required around the perimeter of a 45o miter butt joint used to connect two pieces of HSS at 90 degrees? Or, can the same weld be used around the entire perimeter? In either case, what is the prequalified weld typically used? A partial-joint-penetration groove weld?&lt;/p&gt;&#xA;&lt;p&gt;ANSWER:&lt;/p&gt;&#xA;&lt;p&gt;In direct answer to your question, edge preparation (beveling) of the HSS would be required along 3 of the 4 edges of the miter joint, in order to successfully accomplish PJP welds along those 3 edges.&#xA;The capacity of such welds is limited, and the welds themselves (to matched box sections) need to be made very carefully. For such miter joints, especially if there is reasonable applied loading on the HSS members, it is recommended that the two HSS each be separately welded (usually by fillet welds) to a 45o stiffening plate. This type of “knee connection” is shown and discussed on pages 67 to 69 of CIDECT Design Guide No. 3 – “Design Guide for Rectangular Hollow Section (RHS) Joints under Predominantly Static Loading”, 2nd. edition, 2009.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I'm not familiar with euronorms, but I imagine they would be similar since the underlying principles are the same.&lt;/p&gt;&#xA;" OwnerUserId="122" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-09-09T17:54:44.147" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11427" PostTypeId="2" ParentId="11420" CreationDate="2016-09-09T02:42:30.383" Score="2" Body="&lt;p&gt;There are several places Computational Science and Mechanical Engineering overlap. Computation is a tool that many engineers use on a daily basis to make design decisions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Computational_fluid_dynamics&quot; rel=&quot;nofollow&quot;&gt;Computational Fluid Dynamics (CFD)&lt;/a&gt;- Learn &lt;a href=&quot;http://www.openfoam.com/&quot; rel=&quot;nofollow&quot;&gt;OpenFOAM&lt;/a&gt; and do some contracting.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Finite_element_method&quot; rel=&quot;nofollow&quot;&gt;Finite Element Analysis (FEA)&lt;/a&gt;- Learn OpenFoam or &lt;a href=&quot;http://gmsh.info/&quot; rel=&quot;nofollow&quot;&gt;GMSH&lt;/a&gt; and do some contracting.&lt;/li&gt;&#xA;&lt;li&gt;Computation is the only way to solve many ugly differential equations, and a much easier route to solving many of the solvable ones. Engineers know math, but we don't necessarily like math and are not awesome at it. Large engineering firms benefit from in-house math, computational, and data handling expertise.&lt;/li&gt;&#xA;&lt;li&gt;Find a research firm or supercomputing contractor that hires lots of engineers and computational scientists.&lt;/li&gt;&#xA;&lt;li&gt;Any other &lt;a href=&quot;https://en.wikipedia.org/wiki/Computational_science#Related_fields&quot; rel=&quot;nofollow&quot;&gt;computational science related field&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-09T02:42:30.383" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11428" PostTypeId="2" ParentId="7736" CreationDate="2016-09-09T09:42:44.497" Score="1" Body="&lt;p&gt;Even though I can't find any standardized literature, the theory behind of what you seem to be looking for is the so called &lt;strong&gt;systems modelling&lt;/strong&gt;. Peter E. Wellstead published (can be accessed here: www.control-systems-principles.co.uk) his efforts in &quot;formulating a simple unified approach to system modelling&quot; that could set a framework for handling mechanical systems using the exchange of energy of it's elements as the unifying key point. The result of this approach is a simple and manageable method to model lumped systems.This method uses differential equations and a set of graphical representation alternatives, namely network methods (topological  network used traditionally in electrical circuits) or bond graph methods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of the topological representation as requested in your question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UuSJi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UuSJi.png&quot; alt=&quot;Example of a Mechanical System represented by a Network &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his book you'll find other examples of more complex mechanical systems: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ball and the beam problem&lt;/li&gt;&#xA;&lt;li&gt;An automotive test bed&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I don't yet have permission to publish links but the mentioned book can be found at www&quot;dot&quot;control-systems-principles&quot;dot&quot;co&quot;dot&quot;uk .&lt;/p&gt;&#xA;" OwnerUserId="7942" LastEditorUserId="7942" LastEditDate="2016-09-09T09:51:57.457" LastActivityDate="2016-09-09T09:51:57.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11429" PostTypeId="2" ParentId="11421" CreationDate="2016-09-09T11:18:36.610" Score="1" Body="&lt;p&gt;It means that there are two parts to the error.  One part is proportional to the reading (0.8%), and the other part is a absolute error (10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if a reading is 174, then there is 174 x 0.8% = 1.4 proportional error, plus the 10 fixed error for a total error of 11.4.  A reading of 174 therefore means the actual value is in the range of 162.6 to 185.4.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-09-09T11:18:36.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11431" PostTypeId="2" ParentId="11394" CreationDate="2016-09-09T11:41:50.203" Score="2" Body="&lt;p&gt;reading these answers i think there is some misunderstanding re: what is meant by integer ratios. For most gear sets (whether spur, helical, etc) the ratio is the ratio of the number of teeth which is obviously an integer ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example a typical example differential might have a 41 tooth ring and 11 tooth pinion for a ratio of &lt;code&gt;41:11&lt;/code&gt;.  That is typically expressed in decimal  which is &lt;code&gt;3.72727272...:1&lt;/code&gt; which is typically rounded off to &lt;code&gt;3.73:1&lt;/code&gt;.  The actual ratio remains a ratio of integers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you have a transmission with a sequence of gears you would multiply all the integer ratios and still have an overall integer ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit aside to the question, in this example common teeth repeat contact only every 11 revs of the big gear, and every tooth combination meshes an equal number of times. The reason cited by @ahipsterpeterpan may be why Chevy chose prime numbers for the tooth count.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2016-09-09T11:41:50.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11433" PostTypeId="2" ParentId="11417" CreationDate="2016-09-09T14:03:32.990" Score="3" Body="&lt;p&gt;@hazzey is on the right track.  This is kind of the reverse of &lt;a href=&quot;https://en.wikipedia.org/wiki/Gray_code&quot; rel=&quot;nofollow&quot;&gt;Gray Code&lt;/a&gt;, where the intent was to avoid bad data during meta-states of the encoder.  Since a bar code is static, we want to maximize the signal difference between &lt;strong&gt;all&lt;/strong&gt; values, not just neighbors, &lt;strong&gt;and&lt;/strong&gt; we want to minimize the total physical space -- hence the paucity of &quot;4&quot; widths.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll point out that there are &lt;a href=&quot;https://en.wikipedia.org/wiki/Barcode&quot; rel=&quot;nofollow&quot;&gt;a number of different implementations&lt;/a&gt; in use today. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-09-09T14:03:32.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11434" PostTypeId="2" ParentId="10941" CreationDate="2016-09-09T16:42:56.003" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;understand that the force diagram is just a representation and depends on many presuppositions, on the other hand if you consider it in an actual practice the results are different than that of standard diagrams, but we can say that the fluctuations are very minimal. Still, it has a different output every time you run it.&lt;/li&gt;&#xA;&lt;li&gt;showing one diagram is enough to explain what is happening throughout the beam, more or less with a factor of safety.&lt;/li&gt;&#xA;&lt;li&gt;now if you want to remember the simple concept of sagging and hogging and to clear the positive and negative confusions, you go through &lt;a href=&quot;https://youtu.be/9CJDi0HTpL4&quot; rel=&quot;nofollow&quot;&gt;this video&lt;/a&gt; this will sure take exactly 2 min 5sec of your life, but trust me you will remember this sign conventions for the rest of your life.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7970" LastEditorUserId="1832" LastEditDate="2016-09-09T17:46:43.620" LastActivityDate="2016-09-09T17:46:43.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11435" PostTypeId="2" ParentId="11422" CreationDate="2016-09-09T16:52:06.117" Score="1" Body="&lt;p&gt;I would agree with your assumption on the weld types. That being said, you could probably just specify a V-groove weld all the way around the joint like this(where xx is the size of the weld) If necessary, you can provide as many details as you need into the callout:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3tri8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3tri8.png&quot; alt=&quot;Groove Weld all around&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could something like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dvBSM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dvBSM.png&quot; alt=&quot;Fillet+groove&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, you could just use 4 separate weld callouts for each of the 4 sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm following this chart:&#xA;&lt;a href=&quot;https://app.aws.org/mwf/attachments/64/225364/AWSWeldSymbolchart.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://app.aws.org/mwf/attachments/64/225364/AWSWeldSymbolchart.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-09-09T16:52:06.117" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11436" PostTypeId="2" ParentId="11422" CreationDate="2016-09-09T17:59:48.263" Score="0" Body="&lt;p&gt;For all practical purposes you have two butt welds, a fillet weld and an outside corner weld. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often for general fabrication these will be treated as separate welds. Indeed if the work is assembled in a jig they may not all be accessible at the same time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sort of situation weld order is very important as this type of mitre joint is very prone to distortion and SHS isn't the most dimensionally precise of materials. In particular they may be supplied in lengths which are not as straight as you would ideally like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in terms of design you properly don't need to worry too much about specifying the welds beyond the general standard/code you want to apply...more important to the fabricator is the geometric tolerances for the joint ie what are the important dimensions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that for this size of square tube getting accurate mitre cuts is not trivial so specification of acceptable root gaps in respect to overall dimensions is very relevant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly in most applications some dimensions matter more than others so it's important to be clear what your priorities are in your tolerances.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally if you are not conversant with welding codes just TELL your supplier what you want to achieve in plain language.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-09-09T18:09:31.663" LastActivityDate="2016-09-09T18:09:31.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11437" PostTypeId="1" CreationDate="2016-09-09T18:53:05.310" Score="3" ViewCount="559" Body="&lt;p&gt;I'm trying to understand how an oscillating sprinkler like this, is able to change directions. Does it still use a four bar linkage mechanism like a traditional oscillating sprinkler?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H5V1o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H5V1o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7869" LastActivityDate="2016-09-15T05:24:48.497" Title="How does an oscillating sprinkler like this change directions?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11438" PostTypeId="1" CreationDate="2016-09-09T19:29:26.220" Score="2" ViewCount="1472" Body="&lt;p&gt;I need to count digital pulses with a width of 30&amp;nbsp;ns and display the updated count to the user every second. The pulses have already been through a comparator and a pulse shaper. Resolution isn't a concern, my only concern is the counting the number of 30&amp;nbsp;ns pulses while picking up minimal noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easy way to use a PIC microcontroller or a special circuit to execute the count? Could I use a PIC or circuit to execute the count and relay the count to a raspberry pi for further processing/display?&lt;/p&gt;&#xA;" OwnerUserId="7972" LastEditorUserId="1832" LastEditDate="2016-09-09T21:09:11.407" LastActivityDate="2016-09-10T10:15:15.213" Title="What's the easiest way to count high speed pulses?" Tags="|electrical-engineering|equipment-selection|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11442" PostTypeId="1" AcceptedAnswerId="11458" CreationDate="2016-09-09T23:09:23.537" Score="1" ViewCount="1472" Body="&lt;p&gt;So I'm trying to create a system that automatically runs evaporative coolers to keep a warehouse at a desired temperature, given the internal and external temperature and humidity. I can get the coolers to run automatically and read the data I want quite easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what I need to know: is there a way to calculate how much a cooler cools down a room? I can find out the volume of the room and the airflow of the coolers, so an equation that requires those will be fine. Is there an equation that describes how much a cooler cools a room down?&lt;/p&gt;&#xA;" OwnerUserId="7975" LastActivityDate="2016-09-13T14:04:30.083" Title="How much does an evaporative cooler cool down a room" Tags="|airflow|cooling|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11443" PostTypeId="2" ParentId="11438" CreationDate="2016-09-09T23:19:07.157" Score="0" Body="&lt;p&gt;One way to count the pules is to use an interrupt trigger on positive or negative edge. You can use a the PIC or even you might be better of using the raspberry pi it self. With regard to minimizing noise I would suggesting placing an appropriate filter circuit to filter out the noise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another one to consider is Capture and Compare mode. I have not used this but I read about this briefly. I think PIC microcontroller support Capture and Compare. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly with regard to noise, it is best that you understand the source of the noise and characteristic of the noise. Then filter out the any noise that affects your design. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ww1.microchip.com/downloads/en/DeviceDoc/ccpwm.pdf&quot; rel=&quot;nofollow&quot;&gt;CCP Module Implementation and Functionality&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ww1.microchip.com/downloads/en/DeviceDoc/31014a.pdf&quot; rel=&quot;nofollow&quot;&gt;Compare/Capture/PWM (CCP)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-09-10T00:22:48.817" LastActivityDate="2016-09-10T00:22:48.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11445" PostTypeId="2" ParentId="11438" CreationDate="2016-09-10T02:45:22.037" Score="0" Body="&lt;p&gt;I don't know PICs, but they, like most microcontrollers have timer/counter registers which can be set up as input from a GPIO to up-count those digital pulses on a rising edge, falling edge, or both edges. You need to make sure your processor frequency is higher than your input frequency otherwise you can miss some counts. Also, depending on the size of the counter register, 8, 16, 32 bit, etc you could run into an overflow condition. In that case. You need to trigger an interrupt(ISR) when the counter overflows to maintain the overflow count. For example, In an 8 bit counter register, it will overflow and reset to 0 after 255 pulses have been counted. The counter can trigger an interrupt on an overflow condition and that interrupt routine can simply do something like overflowCount+=1 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you read your total pulse count to the user, you would need to do:totalPulses= counter register value+(255*overflowCount) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know for how long this will count, but you could overflow your counter variable if you don't use a large enough variable type. Maybe use a 64bit Long type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read through your PIC's data sheet and you should find an example of a counter. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2016-09-10T02:51:31.937" LastActivityDate="2016-09-10T02:51:31.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11446" PostTypeId="1" AcceptedAnswerId="11448" CreationDate="2016-09-10T04:25:08.780" Score="2" ViewCount="1095" Body="&lt;p&gt;The core question is &quot;What Ingress Protection (IP) rating is required for an item to be dishwasher safe?&quot;  Is the answer different for residential versus commercial dishwashers?  Is the practical residential answer different than the regulated answer (e.g., for food prep or medical settings).  Does the use of dishwashing detergent invalidate the rating in practice?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, IP ratings are not cumulative from IPX6 (jets) or less to IPX7 (immersion) or more, so an item might technically need a dual rating, but I've rarely seen that in my very limited research. Are both needed?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty of performing some tests appears to vary (e.g., 1 meter immersion is pretty simple compared to water jet set at a specific flow rate, angle, and turning rate).  Might this lead to some certifications not being sought?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My most immediate interest is shopping for a cooking thermometer, but it seems applicable to a variety of other equipment like small speakers (picnic disaster) and phones (baby vomit) and keyboards.  I know at least some hospital/medical keyboards are specified as IP66 or IP67 and also washable and dishwasher safe (and presumably &quot;coffee-safe&quot; and &quot;Coke-safe&quot;).  Some flexible membrane keyboards are described similarly but are not rated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple queries such as &quot;What IP rating is dishwasher safe&quot; or &quot;Is IP66 dishwasher safe&quot; were not directly fruitful.  Several manufactures describe some of their products as dishwasher safe that are rated variously at IP66 or IP67 or IP57 (but not other comparably rated products even from the same manufacturer).&lt;/p&gt;&#xA;" OwnerUserId="7978" LastActivityDate="2016-09-10T05:31:06.890" Title="IP Rating Required to be Dishwasher Safe (Commercial and Residential)" Tags="|materials|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11447" PostTypeId="1" CreationDate="2016-09-10T05:28:17.597" Score="1" ViewCount="55" Body="&lt;p&gt;I have a &lt;a href=&quot;http://www.despatch.com/lbb.aspx&quot; rel=&quot;nofollow&quot;&gt;Despatch LBB 1-69&lt;/a&gt; lab oven in my shop. It is a single-phase, 110-volt, 30-amp unit. It is wired to a 220 volt, 2-phase, 40-amp panel (on a single 110 volt phase), which is fed by the 100-amp main service. All wiring is appropriate or larger for the task at hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During operation, the element is switched on until the unit approaches the set point, then switches to a 1 or 2 Hz (approximate, not measured) pulse and maintains this mode until the heater is switched off or the set point is changed. The &quot;on&quot; period is substantially shorter than the &quot;off&quot;, perhaps 10% or less of the period duration. It makes for a very stable temperature but I'm experiencing a nasty side effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the sun has gone down, there's a noticeable flicker in the garage's florescent lights, that coincides perfectly with the oven's operation, pulsing away. More problematically, every light on the same phase throughout the entire place displays the same behaviour. This is unacceptable, especially heading into the darker winter period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were a fluid circuit, I'd add an accumulator to reduce some or all of the pressure fluctuations but clearly it's not. Would an appropriately sized capacitor (or bank) be a solution? Could they act as a sufficient reservoir to smooth out the fluctuations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if this plays in to anything, but the voltage in the facility is quite high - on the order of 123 to 126 VAC at any given time.&lt;/p&gt;&#xA;" OwnerUserId="7979" LastActivityDate="2016-09-10T05:28:17.597" Title="How to solve power surge caused by curing oven PWM cycle?" Tags="|electrical-engineering|power-engineering|heating-systems|electrical-grid|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11448" PostTypeId="2" ParentId="11446" CreationDate="2016-09-10T05:31:06.890" Score="2" Body="&lt;p&gt;What you've found is, pretty much, a good guide to the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IP ratings are not designed to measure whether something is dishwasher safe or not. That's not their purpose. And so they just don't map onto whether or not something is dishwasher-safe or not. That is to say, something might be IP66K and not be dishwasher-safe; something else that is IP55 might be dishwasher-safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can see how one might expect &quot;what IP-rating is dishwasher-safe?&quot; to be meaningful question - but it's not. It's a different kind of rating.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2016-09-10T05:31:06.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11452" PostTypeId="2" ParentId="11438" CreationDate="2016-09-10T10:15:15.213" Score="1" Body="&lt;p&gt;30ns is too fast for PIC microcontrollers that I know of. For example, the PIC16F77 requires a period of one instruction cycle plus 40ns, so at maximum clock frequency the period must be at least 200+40ns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would use a 4-bit counter IC (eg 74169, synchronous 4-bit binary up/down counter) which would divide by 16. It's output would be fed into the PIC's timer input pin (T0CKI for the 16F77). But this will only count multiples of 16. If you want a precise count while pulses are being received it will get tricky because not only would those 4 bits need to be latched, but they'd have to be synchronised with the PIC's timer/counter.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-09-10T10:15:15.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11454" PostTypeId="1" AcceptedAnswerId="11465" CreationDate="2016-09-10T14:37:04.407" Score="0" ViewCount="4878" Body="&lt;p&gt;Consider I have a pole stuck in the ground as illustrated below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kTPSN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kTPSN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a force F1 is exerted on the pole, its bottom end will push against the ground it's stuck in; the ground then exerts a resistance force. Intuitively, if F1 is very strong then the pole's upper end will move in F1's direction and if F1 is very weak, the ground resistance Fr keeps the pole in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given some measures of the pole, specifically, how deep it is (L2) and what radius r it has (it is a cylinder), I would like to be able to determine the magnitude required by F1 to move the pole's end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have asked a similar question, with more physical details on another stack exchange site, too. &lt;a href=&quot;https://physics.stackexchange.com/questions/279345/compute-forces-exerted-on-a-pole-in-ground&quot;&gt;Here&lt;/a&gt; is the link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the ground is made of only one material, with uniform density, disregarding any &quot;dirty&quot; effects, such as moisture, or temperature. The length of the pole, its radius (it is a cylinder), and its depth L2 are all constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's view the pole as a lever with its pivot where air, ground, and pole meet. Then we have F1 * L1 = F2 * L2 when the lever is in balance. Now, I would like to find out at what force F1 the pole starts moving; the pole is moving when F1 * L1 &gt; Fr* L2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem would be solved if we find a means to compute Fr. My idea was that as the pole (in the ground)pushes towards the right at every height 0 &gt; L &gt;= L2 (L = 0 on dotted area where air and ground meet), there is a resistance force Fr(L) at each height (view illustration &lt;a href=&quot;https://physics.stackexchange.com/questions/279345/compute-forces-exerted-on-a-pole-in-ground&quot;&gt;here&lt;/a&gt;). For big L this force is small while for small L the force grows. I assume Fr(L) must grow in a linear fashion as it does with levers (F1 * L1 = F2 * L2). Assuming that, Fr could be computed by F2 = 0∫L2 Fr(L) dL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution need not be that exact, the force need only be computed approximately: the force will be used by game simulation the physics need only appear real. All that is required is a means to compute what resistance force Fr the pulling force F1 must overcome to move the pole.&lt;/p&gt;&#xA;" OwnerUserId="7985" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2016-09-12T21:53:06.110" Title="How much force can a pole in ground take before falling?" Tags="|mechanical-engineering|simulation|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11455" PostTypeId="1" CreationDate="2016-09-10T22:12:15.140" Score="1" ViewCount="150" Body="&lt;p&gt;When selecting bearings, engineers should be concerned also about specifying the lubricants that will provide an optimal performance. In order to do that, the viscosity ratio is the parameter to defined in order to select the lubricant. Viscosity ratio is defined as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \kappa = \frac{\nu}{\nu_{1}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$ \nu $ is the operating viscosity of the lubricant [$mm^2/s$]&lt;/li&gt;&#xA;&lt;li&gt;$\nu_{1}$ is the rated viscosity depending on the bearing mean diameter and rotational speed [$mm^2/s$]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From literature, it is said that a viscosity ratio greater than one can extend the bearing's life. If $ 0.4 \lt \kappa \lt 1$ it is recommended to use additives. While if $ \kappa \lt 0.4$ it is compulsory to use them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the reasons behind this?&lt;/p&gt;&#xA;" OwnerUserId="7942" LastEditorUserId="1832" LastEditDate="2016-09-10T23:00:19.027" LastActivityDate="2016-09-10T23:00:19.027" Title="Why viscosity ratio should be greater then 1 for longer bearing life?" Tags="|mechanical-engineering|automotive-engineering|design|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11456" PostTypeId="1" AcceptedAnswerId="11459" CreationDate="2016-09-11T05:17:37.653" Score="3" ViewCount="838" Body="&lt;p&gt;I am supposed to be re-drawing this image to scale&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vhBjZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vhBjZ.png&quot; alt=&quot;Original Drawing&quot;&gt;&lt;/a&gt; However when I Try to connect the arc R11 with arc R26 using the radius R22 I get a negative # in my calculations. i.e. R26-R22=(-4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JQARV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JQARV.png&quot; alt=&quot;Using this formula&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a different way to draw this? I have tried nearly everything and have completed the rest of the drawing except this one point, Am I missing something here?&lt;/p&gt;&#xA;" OwnerUserId="7993" LastActivityDate="2016-09-11T13:58:39.030" Title="Drawing the Connection of 2 Arcs with Radius" Tags="|mechanical-engineering|design|drafting|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11457" PostTypeId="1" CreationDate="2016-09-11T10:24:33.833" Score="1" ViewCount="64" Body="&lt;p&gt;While solving some exercises for an exam, I faced a problem with a two-stage adiabatic reactor, producing methanol (exothermal reaction), with intermediate cooling. Both stages achieve equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two of the questions were:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Inlet temperature at the second stage&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Load at the intermediate cooling exchanger&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have the inlet temperature in the first reactor, so I can couple a heat balance and a mass balance to obtain the temperature at the outlet of the first stage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I calculate the inlet temperature at the second stage?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first idea has been to suppose it equal to the first inlet temperature, but I cannot find any evidence to back it up. On the contrary, multiple charts have shown the inlet temperatures increasing after each stage. How else can I obtain it?&lt;/p&gt;&#xA;" OwnerUserId="7994" LastActivityDate="2016-12-10T20:28:34.983" Title="Define inlet temperature in multi-stage adiabatic reactor" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11458" PostTypeId="2" ParentId="11442" CreationDate="2016-09-11T13:55:33.367" Score="5" Body="&lt;p&gt;The capacity of an evaporative cooler depends on the relative humidity. The lowest theoretical temperature that the cooler can reach is the wet bulb temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Evaporative_cooler&quot; rel=&quot;nofollow&quot;&gt;wikipedia article&lt;/a&gt; has some discussion on the design and shows how it works with psychometric charts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the design you'd look at a 'typical' hot day, compare wet bulb to dry bulb for that day and assume a certain efficiency (e.g. it reaches 80% from dry bulb to wet bulb). The energy in the supply air stream would be:&#xA;$$&#xA;Q = \dot{m}\Delta_Tc_p\eta&#xA;$$&#xA;i.e.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy = massflow x temperature drop x specific_heat x efficiency&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta_T$ being the temperature difference before and after the evap cooler using your assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q$ would be your cooling capacity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HTH&lt;/p&gt;&#xA;" OwnerUserId="75" LastEditorUserId="1633" LastEditDate="2016-09-13T14:04:30.083" LastActivityDate="2016-09-13T14:04:30.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11459" PostTypeId="2" ParentId="11456" CreationDate="2016-09-11T13:58:39.030" Score="5" Body="&lt;p&gt;Arc 22 does not connect with arc 26 there is a straight horizontal line in between. The hint to this fact is in the width dimension of 44 that would not exist otherwise. There are further hints in the image itself, the small construction line segment shows that its a round of a corner. With a trained eye you can also see in the image the the curvature ends before connect. I have drawn a clarifying image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ySbBz.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ySbBz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: Showing construction piece by piece and type (color).&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2016-09-11T13:58:39.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11460" PostTypeId="1" AcceptedAnswerId="11462" CreationDate="2016-09-11T15:25:49.287" Score="1" ViewCount="131" Body="&lt;p&gt;Why is it not recommend to have pump shut off head lower than its operating head?&#xA;Is the answer related to dropping curve by any chance?&#xA;Also why is the discharge valve closed when starting the pump operation?&#xA;Any help on this appreciated&lt;/p&gt;&#xA;" OwnerUserId="7372" LastEditorUserId="7372" LastEditDate="2016-09-11T15:31:45.477" LastActivityDate="2024-03-08T14:37:59.947" Title="Centrifugal pumps working" Tags="|mechanical-engineering|fluid-mechanics|chemical-engineering|pumps|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11461" PostTypeId="2" ParentId="11457" CreationDate="2016-09-11T18:24:32.623" Score="1" Body="&lt;p&gt;This sounds like a problem out of Fogler. I would assume it's the same as the first stage. There is no way to calculate it seeing as you must find the load of the heat exchanger. One of those two variables must be specified to get the other one, so if they don't give the inlet temperature you have to assume it. Not sure what charts your talking about. Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="7927" LastActivityDate="2016-09-11T18:24:32.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11462" PostTypeId="2" ParentId="11460" CreationDate="2016-09-11T18:25:54.417" Score="3" Body="&lt;p&gt;To address the first point without further clarification from your side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you mean unstable pump curves with that. Otherwise I wouldn't know how you could achieve a higher operating head than shut off head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yA7iR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yA7iRm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result would be an oscillating behavior which is not desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part of your question the main reasons I know are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;lower load on the driver during start up&lt;/li&gt;&#xA;&lt;li&gt;avoiding recirculation if the operating pressure in the pipe is high&lt;/li&gt;&#xA;&lt;li&gt;maintaining a constant pressure in the pipe, i.e. the valve is beeing opened when the discharge pressure is equal or exceeds the pipe pressure&lt;/li&gt;&#xA;&lt;li&gt;avoiding a dry pump, especially if the flow on the suction can be interrupted&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In general there are a lot of information on the two questions and a good book or two could prove helpful if you would like to achieve in depth knowledge. Although there should be a lot of solid information in the web too.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2016-09-11T18:25:54.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11463" PostTypeId="2" ParentId="11454" CreationDate="2016-09-11T19:03:10.793" Score="-1" Body="&lt;p&gt;The first scenario is that the pole is very firmly anchored into the ground in a hard and reliant surface to a good depth for example if it is embedded in a large mass of concrete. Here the pole will bend over its whole length until the elastic limit in reached an then fail at some point colse to the ground surface and form a '[plastic hinge&lt;a href=&quot;https://en.wikipedia.org/wiki/Plastic_hinge&quot; rel=&quot;nofollow&quot;&gt;1&lt;/a&gt;' and then continue to bend at this point untill it fails completely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second scenario it that the pole itself can tolerate substantially more force than the ground in which it is embedded. This is a lot more complex to model as 'dirt' can have very variable physical properties depending on its composition and condition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a dry clay soil with lots of rubble in it can behave in a similar way to concrete and it will resist any force well up to a certain point at which it will suddenly crack and fail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of game simulation if you want a simple method it would probably make sense to consider that the pole pivots somewhere near ground level and just pick a maximum value for Fr and compare moments F1*L1 and Fr*L2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more detail you could say that you set a value A for F1 which is able to start shifting the pole and a higher value B which knocks it over completely at which it is no longer fixed to the ground (ie becomes a detached object lying on the ground) and any vale in between causes a certain angle of lean. ie if A is 100 force units and B is 200 force units it leans over at 45 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach allows you an easy way to adjust values without having to run a complex soil mechanics simulator in your game. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another simple and physically reasonable approach would be to assume that the pole has a permanently attached plug of material around it below ground level and just treat that as a dead weight, as if it was placed in a bucket of concrete siting on the ground. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-09-11T19:03:10.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11464" PostTypeId="2" ParentId="11454" CreationDate="2016-09-11T21:30:13.440" Score="4" Body="&lt;p&gt;this is a typical engineering problem and has an equation approved by IBC numbered 18-1.&lt;br&gt;&#xA;$$d = 0.5 A{1 + [1 + (4.36h/A)]1/2}$$&#xA;The calculation process is I first calculate A =2.3* post load/(allowed passive soil bearing value*diameter of caisson)&lt;br&gt;&#xA;Then you plug A into EQ 18-1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The post resists overturning momentum mainly by rotating around a pivot &lt;strong&gt;under the soil surface&lt;/strong&gt;, usually near 1/3 total embedment depth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have a lateral force acting at height H trying to turn the pole clockwise the top part of the pole embedded in dirt will compress the soil to right with a triangular stress area and the rest of embedded pole will press the dirt to left.&lt;br&gt;&#xA;This way of analysis has been tested in many different types of soil and shown to be a good assumption.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there are limitations and conditions that needs to be observed. I recommend you google the IBC code 18-1. Just bear in mind this code is not concerned with vertical loads or torsion. &lt;a href=&quot;https://i.stack.imgur.com/D1Loz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D1Loz.jpg&quot; alt=&quot;diagram of flag pole and soil stresses.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2016-09-11T23:22:00.447" LastActivityDate="2016-09-11T23:22:00.447" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11465" PostTypeId="2" ParentId="11454" CreationDate="2016-09-11T21:31:41.663" Score="2" Body="&lt;p&gt;Depending on the kind of engineer you ask you will get many different options to approach this - your question is pretty open-ended. As a Geotechnical Engineer I'd say your pole can be looked at as a retaining wall. Now, if the ground is modelled as real &quot;dirt&quot; it will have some give. What would happen is that the pole will rotate about some point in the ground (again - you can go into this in more depth) and also bend (or not if rigid). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this rotation causes a the rod to displace the soil, which exerts &quot;passive&quot; or &quot;active&quot; lateral pressure, depending on whether it is stressed further of allowed to relax (i.e. rod moving into soil region or moving away from it). These pressures are given by the coefficients of passive and active earth pressure ($K_{p}$ and $K_{a}$) and the vertical effective stress ($\sigma_{v}' = hg\rho$). The coefficients are found from the angle of friction of the soil ($\phi'$) as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K_{a} = \dfrac{1-\sin \phi'}{1+\sin \phi'}&#xA;$$&#xA;and&#xA;$$&#xA;K_{p} = \dfrac{1+\sin \phi'}{1-\sin \phi'}&#xA;$$&#xA;So passive pressure would be $P_{p} = K_{p}\sigma'$ and so on. A rigid rod will then rotate around the point where all the forces on it are balanced.&#xA;You can assume some value of $\phi'$ and be on your way. These pressure will grow linearly with depth, so the resulting forces from the different active/passive wedges will act at 1/3 from the bottom of each wedge. Everything can now be expressed in terms of the distance between the pivot and the ground level. Equilibrium around the pivot will give you its location and then the sizes of the forces can be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OG7ln.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OG7ln.png&quot; alt=&quot;Moments with passive pressure ONLY&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture above is how I would approach this (neglecting the active soil wedges, if included they mirror the passive ones with a limiting pressure as given above). From equilibrium about the orange point you can find its depth and then the values of the pressures (with appropriate Fr and soil parameter $\phi'$ assumed). Or you can assume a value of x and find the Fr that causes movement. Many options possible.&lt;/p&gt;&#xA;" OwnerUserId="7997" LastEditorUserId="7997" LastEditDate="2016-09-12T21:53:06.110" LastActivityDate="2016-09-12T21:53:06.110" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11466" PostTypeId="1" CreationDate="2016-09-11T22:09:44.193" Score="1" ViewCount="91" Body="&lt;p&gt;Computers get hot.  So usually we try to make them less hot.  But what if instead of trying to cool down a computer we used that heat for something else?  Not necessarily any thing big like a whole room; maybe just something like a pet's cage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is funneling the heat generated by a computer (any kind; desktop, server rack, etc.) a practical thing to do?  Are there any reliable stories of it being done successfully?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7998" LastEditorUserId="1586" LastEditDate="2016-09-12T18:31:54.513" LastActivityDate="2016-09-12T18:31:54.513" Title="Have there been successful attempts at using a computer's heat to warm something up?" Tags="|electrical-engineering|heat-transfer|heating-systems|computer|computer-hardware|" AnswerCount="1" CommentCount="4" ClosedDate="2016-09-13T00:10:38.227" ContentLicense="CC BY-SA 3.0" />
  <row Id="11468" PostTypeId="1" CreationDate="2016-09-11T23:44:16.573" Score="0" ViewCount="634" Body="&lt;p&gt;In this video, the counter shaft is not rotating while a gear slides into place. But if the counter shaft was rotating, wouldn't the gear sliding over have a difficult time engaging with a gear on the counter shaft? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=BMNjxqRYHks&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=BMNjxqRYHks&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HXi8X.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HXi8X.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7869" LastEditorUserId="7869" LastEditDate="2016-09-12T23:40:10.307" LastActivityDate="2016-09-12T23:40:10.307" Title="How do gears engage and mesh in a sliding mesh gearbox?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11469" PostTypeId="2" ParentId="11468" CreationDate="2016-09-12T00:14:08.340" Score="2" Body="&lt;p&gt;Yes if the shafts were rotating then it would be difficult - this is why the gears are engaged when the shafts are not transmitting torque or rotating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to change gears during operation then a constant mesh gearbox is clearly a better option. Usually the gears are not locked to the shaft and they have &lt;strong&gt;dogs&lt;/strong&gt; on their sides. These gears transmit torque by engaging with a hub (that also has dogs) that slides along the shaft but is splined with the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the video below starting from 3:00. The purple components are the sliding hubs I'm talking about. They have dog teeth that interact with the dog teeth on the sides of the gears. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/QPaUJfA1KsY?t=3m2s&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/QPaUJfA1KsY?t=3m2s&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2016-09-12T00:21:01.117" LastActivityDate="2016-09-12T00:21:01.117" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11470" PostTypeId="1" CreationDate="2016-09-12T00:15:44.240" Score="0" ViewCount="137" Body="&lt;p&gt;I have a C02 tank that is filled at 1000 psi. it's got a duel pressure regulator that allow you to dial it down to 45 PSI. Is there a simple way to regulate pressure from 1000 psi to 45 ( non adjustable ). I understand the theory of how a pressure regulator works, just wondering if there is an inline way to do this.&lt;/p&gt;&#xA;" OwnerUserId="539" LastActivityDate="2016-09-15T05:07:17.900" Title="Is there and easy way to go from 1000 psi to 45 psi without a complicated pressure regulator" Tags="|pressure|valves|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11474" PostTypeId="2" ParentId="11466" CreationDate="2016-09-12T12:11:00.583" Score="1" Body="&lt;p&gt;This makes little sense to do on the scale of a single computer that may draw 100 W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this has been done on the scale of whole buildings.  Buildings that are designed expecting high power dissipation in some rooms but not others often have air conditioners or heat pumps for cooling/heating individual rooms or small zones.  These tie to building-wide water circulation systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat pumps that heat/cool individual rooms or zones take/dump the pumped heat from/to water circulating to a central pool.  A building-wide system keeps this pool in a controlled temperature range.  When the building needs overall heat, then the heat added to the pool from those zones that are being cooled reduces the deliberate heat that is added to the pool by the central system.  So yes, in that case the heat from computers, and anything else that dissipates power, is re-used to warm other parts of the building.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-09-12T12:11:00.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11476" PostTypeId="1" CreationDate="2016-09-12T14:53:48.833" Score="2" ViewCount="802" Body="&lt;p&gt;I have a microcontroller similar to the one in the accompanying picture. I do not want to solder wires to the microcontroller contacts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it some kind of printed circuit board on which a microcontroller can be placed without having to solder wires? I cannot find anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/waeqe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/waeqe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8004" LastEditorUserId="90" LastEditDate="2016-09-13T12:16:53.727" LastActivityDate="2016-09-13T12:16:53.727" Title="Bread board for STM32" Tags="|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11477" PostTypeId="2" ParentId="11476" CreationDate="2016-09-12T16:43:58.990" Score="1" Body="&lt;p&gt;Based on the image I believe that the GPIO pins are already mapped to 0.1&quot; (2.54mm) male connector pins. If my assumption is correct you have few options&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use 0.100&quot; (2.54 mm) Female Header such as one one below. Then you can solder the male pin on the female header.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://a.pololu-files.com/picture/0J887.150.jpg?bd32478f1db0e7691dbe8214a498ee0c&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://a.pololu-files.com/picture/0J887.150.jpg?bd32478f1db0e7691dbe8214a498ee0c&quot; alt=&quot;0.100&amp;quot; (2.54 mm) Female Header&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Use Male/Female Jumper Wires like the ones below &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AlFzT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AlFzT.jpg&quot; alt=&quot;Male/Female Jumper Wires&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above you should be able route the signal to an external PCB. &lt;/p&gt;&#xA;" OwnerUserId="1586" LastEditorUserId="1586" LastEditDate="2016-09-12T16:52:28.997" LastActivityDate="2016-09-12T16:52:28.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11480" PostTypeId="2" ParentId="11476" CreationDate="2016-09-13T00:04:09.597" Score="1" Body="&lt;p&gt;If you don't want to solder wires to the microcontroller contacts looking into STM32 expansion boards. They look something like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wXFir.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wXFir.png&quot; alt=&quot;STM32F4 Discovery Shield&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above one is called STM32 discovery shield. You can attach other peripherals can create a development board like the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CX7ns.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CX7ns.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some links to find other alternative solutions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.mikroe.com/stm32/stm32f4-discovery-shield/&quot; rel=&quot;nofollow noreferrer&quot;&gt;STM32F4 Discovery Shield&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://hackaday.com/2013/02/19/impressive-dev-boards-for-your-stm32-dev-boards/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Impressive Dev Boards for your STM32 Dev board&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.st.com/content/st_com/en/products/evaluation-tools/product-evaluation-tools/mcu-eval-tools/stm32-mcu-eval-tools/stm32-3rd-party-evaluation-tools/stm32f4dis-ext.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;STM32F4DISCOVERY extension board&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2016-09-13T00:04:09.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11483" PostTypeId="2" ParentId="3538" CreationDate="2016-09-13T11:41:34.460" Score="0" Body="&lt;p&gt;Round profiled vessels fail at vacuum collapse pressures far below those designed for internal pressures. If more vacuum resistance is required, corrugated designs are preferred. Thin bellows survive upto 30% of external pressure as vacuum.&lt;/p&gt;&#xA;" OwnerUserId="1270" LastActivityDate="2016-09-13T11:41:34.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11487" PostTypeId="2" ParentId="11417" CreationDate="2016-09-13T14:42:57.573" Score="0" Body="&lt;p&gt;Just an observation, but first note there are only 20 combinations of 1-4 that add to 7, i.e. half are used and half are invalid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any one-bit error (A bar read as too wide or narrow) would be an obvious error, with the sequence no longer summing to 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A one bit &lt;em&gt;shift&lt;/em&gt;, that is a bar read too wide and one of the adjacent spaces correspondingly too narrow, would add to 7 and appear valid, however if you look at the sequences selected for use you will find that every possible shift like that results in an unused sequence and would so be identified as an error.   ( Example  if &lt;code&gt;2-1-2-2&lt;/code&gt; was read as &lt;code&gt;1-2-2-2&lt;/code&gt; or &lt;code&gt;2-2-1-2&lt;/code&gt; or &lt;code&gt;2-1-3-1&lt;/code&gt; that would be an error ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course that doesn't tell us anything about how the specific sequence to number mapping was made. &lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2016-09-13T14:54:27.903" LastActivityDate="2016-09-13T14:54:27.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11489" PostTypeId="1" CreationDate="2016-09-13T14:55:45.390" Score="1" ViewCount="1838" Body="&lt;p&gt;I've already got a gear that has its teets as in the picture:&#xA;I want to create the mechanism from the picture, but it's harder than I expected.&#xA;I think my gear is not the problem. But I'm not able to create the gray part, the gear anchor.&#xA;Is there any way to calculate the shape and/or sizes that matter for the anchor?&#xA;I tried a lot of things but I can't find the ideal shape for the anchor points that hook between the teeths of the gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h78sh.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h78sh.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8014" LastEditorUserId="615" LastEditDate="2016-09-14T14:20:57.253" LastActivityDate="2016-09-14T14:20:57.253" Title="How to design a gear escapement like in an old pendulum clock?" Tags="|mechanical-engineering|design|gears|applied-mechanics|timing|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11490" PostTypeId="1" AcceptedAnswerId="11492" CreationDate="2016-09-13T14:59:59.553" Score="7" ViewCount="7211" Body="&lt;p&gt;I am working on project involving designing a recoil spring coil of diameter 10.2 mm and wire diameter 1.25 mm. I tried using a solid core wire but it failed due inadequate tensile strength. Dimensions can only be altered by 5 mm, which is not sufficient to make the solid core wire work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor suggested looking into braided wire rather than a solid core wire. I considered 6x7 wire rope but they are too big. After some googling, I found out there are recoil springs with braided wires with 3 strands. I know it is going to be somewhat less than 3x the strength individual wires. I searched handbooks, reference books, etc. but I could not find any established relations/formulae between strands and tensile strength. How can I determine the tensile strength of a braided steel wire based on the number of strands?&lt;/p&gt;&#xA;" OwnerUserId="8015" LastEditorUserId="368" LastEditDate="2016-09-13T16:15:50.337" LastActivityDate="2017-11-12T16:42:43.330" Title="How to calculate tensile strength of braided steel wire?" Tags="|mechanical-engineering|materials|steel|springs|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11491" PostTypeId="1" CreationDate="2016-09-13T15:36:36.047" Score="1" ViewCount="3475" Body="&lt;p&gt;I am trying to solve the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A pump feeds two hoses, each of which is 45 m long and is fitted with a nozzle. Each nozzle has a coefficient of velocity of 0.97 and discharges a 37.5 mm diameter jet of water at 24 m/s when the nozzle is at the same level as the pump. If the power lost in overcoming friction in the hoses is not to exceed 20 per cent of the hydraulic power available at the inlet end of the hoses,&lt;br&gt;&#xA;  calculate (a) the diameter of the hoses, taking f = 0.007, and (b) the power required to drive the pump if its efficiency is 70 percent and it draws its water supply from a level 3 m below the nozzle.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here are my workings for part (a) of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Head lost at nozzle $H_p = H - 4fLv^2/ D2g$&lt;br&gt;&#xA;or Hp = Head at inlet to hoses - frictional head loss in each hose.&lt;br&gt;&#xA;$C_v = 0.97$ and Jet Velocity $V = C_v \sqrt {2g * Hp}$, &#xA;so then I worked out $H_p$ from this to be 31.208 m.&lt;br&gt;&#xA;For continuity of flow, flow from hoses = flow from nozzle, so $v = V (d^2/ D^2)$ and $v = 0.03375/ D^2$.&lt;br&gt;&#xA;I have then put this in the frictional head loss formula $4fLV^2/ D2g$ and this gives $1.43521875 * 10^-3/ D^5 * 19.62$.&lt;br&gt;&#xA;Note $2g = 19.62$ and $H = 5 * \text{frictional head loss}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lastly I put this in the formula $H_p = H - 4fLv^2/ D2g$. I solved for $D$ and I got 98.7 mm for the hose diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this on the correct path for Part (a) of the question?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I work out the power for Part (b)?&lt;/p&gt;&#xA;" OwnerUserId="8013" LastEditorUserId="368" LastEditDate="2016-09-13T16:00:28.170" LastActivityDate="2017-03-12T17:47:18.493" Title="How can I calculate the power to drive a water pump based on discharge from two hoses?" Tags="|fluid-mechanics|power|pipelines|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11492" PostTypeId="2" ParentId="11490" CreationDate="2016-09-13T16:12:01.670" Score="5" Body="&lt;p&gt;First off, the tensile strength is a property of the &lt;strong&gt;material&lt;/strong&gt;, not the spring itself. Strength, often denoted as $\sigma = F/A$, can be considered how much force a material can withstand for a given cross-section. If you want something to be able to withstand more force the simplest approach is to increase the cross-section or get a stronger material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, I think you are mistaking the purpose of braided wire. Generally braiding wires doesn't increase their ability to withstand forces. Assuming the net diameter of the braided wire is equal to the diameter of a solid one, the cross-sectional area of the solid cable will be larger, suggesting that it can withstand more force. What braiding does do is make the wire &lt;strong&gt;more resistant to wear&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it this way: once a crack forms in a solid wire, it can spread across the whole thing, resulting in failure. If a crack forms in a single strand of wire in a braided wire, then only the individual strand will fail, leaving the other strands to take up the load. Keep in mind this doesn't prevent the braided wire from failing, it just gives you a bit more use (and visible signs of damage) before the wire fails completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out &lt;a href=&quot;http://www.ctspring.com/press/2012/03/stranded-and-shaped-wire-springs&quot; rel=&quot;noreferrer&quot;&gt;this page&lt;/a&gt; about stranded recoil springs. The author says the stranded springs have an improved lifetime in applications with high impact stresses and velocities. The author also claims there are currently no accurate programs for calculating the stresses on braided wire.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2016-09-13T16:12:01.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11493" PostTypeId="1" AcceptedAnswerId="11494" CreationDate="2016-09-13T16:27:57.263" Score="7" ViewCount="12880" Body="&lt;p&gt;So let's say I have some heat generating machine be it a combustion engine or a fridge cooled by a fluid. I don't have a single radiator that is big enough for the amount of heat being generated but I do have a couple of smaller ones so I guess I could connect them together.&#xA;Should I put the radiators in series (one connected to the next one and so on) or in parallel (having the coolant intake pipe split into all of the radiators) and why? What would be the most efficient set up?&lt;/p&gt;&#xA;" OwnerUserId="2231" LastActivityDate="2019-03-08T21:26:10.110" Title="Radiators in series or parallel?" Tags="|thermodynamics|heat-transfer|cooling|heat-exchanger|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11494" PostTypeId="2" ParentId="11493" CreationDate="2016-09-13T17:35:49.070" Score="7" Body="&lt;p&gt;The efficiency of any radiator (heat exchanger) is a function of the temperature difference between the two fluids in question. All else being equal, a heat exchanger with a greater temperature differential will transfer more heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each radiator will have a temperature gradient across it. (Here I'm talking about how much the temperature of each fluid changes as it passes through the heat exchanger.) If you hook them in parallel, each one will be getting 1/N of the flow, but they will all have the same temperature gradient from input to output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you hook them in series, all of the flow will be going through all of them, but each one will have only roughly 1/N of the overall temperature difference across it &amp;mdash; with the hottest one also having the highest differential, because it's transferring more heat to the other fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you can make this &quot;series or parallel&quot; decision independently for each of the two fluids. There is a total of four different ways you can configure them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, I don't think it really makes any practical difference in terms of the thermodynamics. Personally, I would be inclined to hook them in parallel+parallel (i.e., parallel paths for both fluids) &amp;mdash; partly because I like that kind of symmetry, and partly because of secondary considerations such as maintenance. When you have a parallel connection with individual shutoff valves, you can repair or replace one radiator without shutting down the system altogether. You can either run at reduced capacity, or design the radiator system to have N+1 redundancy in the first place.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-09-13T17:35:49.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11495" PostTypeId="2" ParentId="11489" CreationDate="2016-09-13T17:49:13.823" Score="4" Body="&lt;p&gt;The mechanism is called an &quot;escapement&quot;, and part of its function is that the gear imparts energy to the swinging teeth in order to keep the pendulum (or whatever) moving in spite of any frictional losses it might have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in your diagram, the gear teeth impinge on a curved surface on each of the swinging teeth. Note also that as a result, the swinging teeth are not symmetrical. Instead, each one has its curved surface in a particular angular relationship to the radius of the gear at that point. It is this curve that allows the rotating gear to give the swinging mechanism a &quot;push&quot; on each stroke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this enough of a hint to get you going in the right direction? (pun intended! :-)&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2016-09-13T20:54:42.370" LastActivityDate="2016-09-13T20:54:42.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11496" PostTypeId="2" ParentId="11493" CreationDate="2016-09-14T00:52:06.780" Score="-1" Body="&lt;p&gt;If we consider two equal size radiators connecting them parallel is more effective. ∆Q/∆t = -K×A×∆T/x, where ∆Q/∆t is the rate of heat flow; -K is the thermal conductivity factor; A is the surface area; ∆T is the change in temperature and x is the thickness of the material (∆T/x is called the temperature gradient and is always negative because of the heat of flow always goes from more thermal energy to less). Wikipedia.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way we maintain a higher slope of heat exchange rate due to maintaining original delta T across the two radiators.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-09-14T00:52:06.780" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11497" PostTypeId="2" ParentId="11490" CreationDate="2016-09-14T01:16:26.810" Score="2" Body="&lt;p&gt;This is because the strands are continually curved and thus the stress distribution inside of them is higher near the centerline (higher curvature) than on the outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the contact forces add stresses in the off-axis directions which raise the internal stress level beyond $\sigma = \frac{T}{A}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The math is quite complex, but not unwieldy. Stranded cables offer superior strength for their weight compared to a solid wire of the same overall diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. The curvature of a wire whose centerline prescribes a circle of $r$ and has a helix with pitch (distance/angle) of $p$ is $$\rho = \frac{r^2+p^2}{r}$$&lt;/p&gt;&#xA;" OwnerUserId="1051" LastEditorUserId="1051" LastEditDate="2016-09-14T01:23:38.083" LastActivityDate="2016-09-14T01:23:38.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11501" PostTypeId="1" CreationDate="2016-09-14T13:58:19.493" Score="3" ViewCount="2736" Body="&lt;p&gt;I looking the specific meaning of fatigue damage in structures. I have one structure and which is subjected to repeated loading cycles. I want to see is this design is safe from fatigue point of view.&lt;/p&gt;&#xA;" OwnerUserId="8029" LastEditorUserId="33" LastEditDate="2018-08-24T14:20:58.133" LastActivityDate="2018-08-24T14:20:58.133" Title="What is the unit of measurement of fatigue damage in a structure?" Tags="|mechanical-engineering|structural-engineering|civil-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="11502" PostTypeId="2" ParentId="11501" CreationDate="2016-09-14T14:37:57.970" Score="1" Body="&lt;p&gt;Typically you might use &lt;a href=&quot;http://aerospacengineering.net/?p=166&quot; rel=&quot;nofollow&quot;&gt;Miner's Rule and compute the Damage Fraction&lt;/a&gt; which is the percentage of predicted life consumed. It's a simplified model though and there are better ways using software like &lt;a href=&quot;http://alta.reliasoft.com/?_ga=1.78432724.1766762803.1473862375&quot; rel=&quot;nofollow&quot;&gt;ALTA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2016-09-14T14:37:57.970" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11504" PostTypeId="1" AcceptedAnswerId="11506" CreationDate="2016-09-14T16:05:53.570" Score="4" ViewCount="827" Body="&lt;p&gt;I've seen some of the robotics competitions use pneumatic actuators for brute force actions ( like a hammer or lift ), but nothing precise. Is it possible to use pneumatics in a more precise scenario?, I'm assuming this is how pneumatic robotic arms operate, but I'm a little unclear on how they operate so precisely. &lt;/p&gt;&#xA;" OwnerUserId="539" LastEditorUserId="2819" LastEditDate="2016-09-15T02:49:58.230" LastActivityDate="2016-09-15T02:54:21.207" Title="Are precision pneumatics possible?" Tags="|control-engineering|robotics|pneumatic|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11505" PostTypeId="2" ParentId="11504" CreationDate="2016-09-14T16:34:02.407" Score="3" Body="&lt;p&gt;Yes, it's possible. &#xA;See here:&#xA;&lt;a href=&quot;http://www.eng-tips.com/viewthread.cfm?qid=213974&quot; rel=&quot;nofollow&quot; title=&quot;Eng Tips Thread&quot;&gt;Position Control of a Pneumatic Actuator&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this youtube video here: &lt;a href=&quot;https://www.youtube.com/watch?v=CyFDYbK1ebc%20%22Youtube%20Video%20of%20Actuator%20with%20Position%20Control&quot; rel=&quot;nofollow&quot;&gt;Pneumatic Actuator With Position Control on YouTube&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many solutions use a servo controlled pneumatic valve in conjunction with a position sensor on the actuator to control pressures on both sides of the piston to maintain a specific position or motion profile.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-09-14T16:34:02.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11506" PostTypeId="2" ParentId="11504" CreationDate="2016-09-15T02:54:21.207" Score="5" Body="&lt;p&gt;While pneumatic systems can be controlled in a non-discrete manner as GisMofx mentioned, that is generally not the norm because the control system to achieve that movement is more expensive that comparative electric actuators. The result is still less precise and less responsive. This is the reason that all CNC machines are electric. Pneumatic systems are also much less energy efficient than electrical or even hydraulic power systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pneumatic actuators are very hard to precisely position because air is compressible. The flow has to change with inertia and a varying load adding dependencies to the control logic. Pushing a constant load cart back and forth on a rail is hard enough; dynamically loaded systems would be very difficult and the reaction speed would be much lower than hydraulic or electric actuators. So dollar for dollar, an electric actuator will always be more precise than a pneumatic actuator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The low-cost pneumatic cylinder is offset by the higher cost associated with more complex programming and the expensive bi-directional proportional valve to control the airflow. Hydraulics also use an expensive bi-directional proportional valve, but are much easier to motion control because the load does not greatly affect the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general rule:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Electronic for highest precision (discrete or motion control)&lt;/li&gt;&#xA;&lt;li&gt;Hydraulic for high force (discrete or motion control)&lt;/li&gt;&#xA;&lt;li&gt;Pneumatic for cheap (discrete)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-15T02:54:21.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11507" PostTypeId="2" ParentId="11493" CreationDate="2016-09-15T04:29:48.123" Score="2" Body="&lt;p&gt;&lt;strong&gt;Assumptions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;Radiator&quot; means a forced-air air-to-fluid heat exchanger.&lt;/li&gt;&#xA;&lt;li&gt;Radiators in either setup will pull from their own fresh air source (not from the exhaust of another radiator).&lt;/li&gt;&#xA;&lt;li&gt;Ignoring radiator design, natural convection, and internal fluid turbulence effects.&lt;/li&gt;&#xA;&lt;li&gt;Flow through parallel radiators is perfectly equal.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;The efficiency would be the same either way. I invented some fluid temp numbers to help visualize. Flow follows the arrows. Air temp would be 20C for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Series&lt;/strong&gt;&#xA;&lt;br&gt;40C -&gt; Radiator1 -&gt; 34C -&gt; Radiator2 -&gt; 30C&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Radiator1 is exhausting more heat than Radiator2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Parallel&lt;/strong&gt; (new names for radiators for comparison)&#xA;&lt;br&gt;40C -&gt; RadiatorA -&gt; 30C&#xA;&lt;br&gt;40C -&gt; RadiatorB -&gt; 30C&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Both A and B are exhausting the same heat. Both will exhaust less heat&#xA;than Radiator1 but more than Radiator 2. The net of both systems will be the same.&lt;/li&gt;&#xA;&lt;li&gt;RadiatorA gives the same temperature drop as Radiator1&amp;amp;2 combined because it has half the flow and half the cooling surface area.&lt;/li&gt;&#xA;&lt;li&gt;The temperature gradient efficiency difference that is made obvious by the less efficient Radiator2 and more efficient Radiator1 is present in both A&amp;amp;B radiators. If we were able to sample the center of A or B we would get the same temperature as between 1 and 2.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other Design Considerations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Series benefits&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The primary benefit of series radiators is that you can guarantee that the flow through each radiator is equal. This is necessary for optimum efficiency. In a parallel system, you can make all hose lengths identical and have all the same fittings (minor losses) for each path, but it is not a guarantee.&#xA;&lt;li&gt; The second benefit of series, is that increased flow velocity increases turbulence inside the radiator. It may cause a measurable increase in total heat transfer if the fluid is not as good of thermal conductor like oil.&#xA;&lt;li&gt; It takes less fittings to plumb radiators in series. This means less install labor and less potential leak locations.&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;strong&gt;Parallel benefits&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Increased flow velocity in the series configuration also increases pressure drop, pumping energy requirements, and heat added to the fluid from this pumping energy input (it all has to go somewhere).&#xA;&lt;li&gt;Parallel has the ability to isolate a radiator for service during operation as Dave Tweed mentioned. But this is a minor benefit, because with a few more plumbing additions, series can be isolated and serviced during operation as well.&#xA;&lt;li&gt;It is easier to compare radiator efficiencies when running in parallel. When one radiator has fouled from internal or external contamination it is easy to see that it has less of a differential than the other without doing any math.&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-15T04:29:48.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11508" PostTypeId="2" ParentId="11470" CreationDate="2016-09-15T05:07:17.900" Score="2" Body="&lt;p&gt;As far as engineering systems go, it doesn't get much simpler that a pressure regulator; single or double. The mechanical feedback is solid solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pressure_regulator#Operation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiki link for pressure regulator operation&lt;/a&gt; (for future readers reference)&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kGmH2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kGmH2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways to regulate pressure with various tradeoffs.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For a fixed flowrate you could size an orifice to reduce the pressure to 45psi. This is not technically regulation however because any change in temperature or flowrate will greatly change the pressure. That would not be a safe or reliable solution.&lt;/li&gt;&#xA;&lt;li&gt;An orifice and a pressure relief similar to a &lt;a href=&quot;http://vlab.amrita.edu/?sub=1&amp;amp;brch=282&amp;amp;sim=1207&amp;amp;cnt=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;zener power regulator&lt;/a&gt; where the orifice would restrict the flow and the pressure relief would bleed any unused gas to atmosphere. That would be a very wasteful solution.&lt;/li&gt;&#xA;&lt;li&gt;Use a high-speed electric on-off valve in combination with an electronic controller to turn the flow on and off at different duty cycles to hit a target pressure measured by a pressure transducer on the output stream. This is similar to &lt;a href=&quot;https://en.wikipedia.org/wiki/Pulse-width_modulation&quot; rel=&quot;nofollow noreferrer&quot;&gt;PWM in electronics&lt;/a&gt;. It has been explored for &lt;a href=&quot;http://hydraulicspneumatics.com/other-technologies/pwm-unconventional-way-regulate-pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;use in hydraulics&lt;/a&gt;. The main problem with this method is that it is expensive and it is hard to dampen all the pressure oscillations for a steady output.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-15T05:07:17.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11509" PostTypeId="2" ParentId="11437" CreationDate="2016-09-15T05:24:48.497" Score="3" Body="&lt;p&gt;Most mass-produced devices have utility patents on them. I was able to locate the one (or a very similar one) for your device:&#xA;&lt;a href=&quot;https://patents.google.com/patent/EP0826427A2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;EP19970306109&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/B5yQo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B5yQo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;By pushing the arm 19 in one direction the rocker member 17 is moved to the position illustrated in Figure 8 in which the inlet 15 is open&#xA;  to enable water to enter the inlet 15 whilst the inlet 14 is blocked&#xA;  by the member 17...&quot;&#xA;  &lt;br&gt;&lt;br&gt;&quot;When water reaches the impeller 8 through the inlet 14, the&#xA;  water flow drives the impeller in one rotational direction and when &#xA;  water reaches the impeller 8 through the inlet 15 the water flow drives the impeller 8 in the other rotational direction.&quot;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/BMFqs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BMFqs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-15T05:24:48.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11510" PostTypeId="2" ParentId="11501" CreationDate="2016-09-15T07:32:59.117" Score="5" Body="&lt;p&gt;Damage calculations use S-N curves, which define the &lt;strong&gt;number of cycles to failure&lt;/strong&gt;, N(S), &lt;strong&gt;for stress range&lt;/strong&gt; S. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The damage value itself can be thought of as the proportion of the fatigue life that is used up by 1 cycle of stress range S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D(S) = \begin{cases}&#xA;\dfrac{1}{N(S)} &amp;amp;\text{if }S &amp;gt; FL \\&#xA;0               &amp;amp;\text{if }S \leq FL&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$FL$ is the defined endurance limit below which no damage occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 3 models of mean stress effects Goodman, Soderberg and Gerber. Each one has its own formulas. This is so because mean stress effects are handled by modifying each stress range according to a formula dependent on the mean stress level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T-N curves are handled in a similar way. A T-N curve defines the &lt;strong&gt;number of cycles to failure&lt;/strong&gt;, N(T), &lt;strong&gt;for effective tension range&lt;/strong&gt; T. There is no analogue of endurance limit for T-N curves. Likewise there are no analogues of stress concentration factor and thickness correction factor.&#xA;As for S-N curves, damage is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D(T) = \dfrac{1}{N(T)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The summation of damage is then performed in an identical manner to that performed for S-N curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So that would be your unit: number of cycles before failure.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8037" LastEditorUserId="1832" LastEditDate="2016-09-15T11:13:27.053" LastActivityDate="2016-09-15T11:13:27.053" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11511" PostTypeId="2" ParentId="11493" CreationDate="2016-09-15T07:39:57.857" Score="0" Body="&lt;p&gt;The answers above look a little over-complex.&#xA;The problem is pretty simple: how much overall heat you can transfer from your source to the radiators. Since no detailed information is available, I can generally say that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The best setup is the setup that offers &lt;strong&gt;the most radiating surface in contact with the heat-generating surface&lt;/strong&gt; (in your case, if the coolant intake pipe can touch all of the radiators, it's the best case). &lt;/li&gt;&#xA;&lt;li&gt;If you can only have one of the radiators in actual contact with the heat generating surface, then the same rule applies between the 1st and 2nd radiator: set the second one up so it has maximum shared surface with the 1st one. That most likely happens when they are in parallel.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8037" LastActivityDate="2016-09-15T07:39:57.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11514" PostTypeId="2" ParentId="3317" CreationDate="2016-09-15T11:17:28.090" Score="0" Body="&lt;p&gt;You might have got the answer if not,&#xA;come to your direct question for plotting actual bending moment. Just choose the element option key for &quot;quadratic&quot; instead of linear and re run the problem and plot the BMD.&lt;/p&gt;&#xA;" OwnerUserId="8043" LastActivityDate="2016-09-15T11:17:28.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11515" PostTypeId="1" CreationDate="2016-09-15T12:42:01.567" Score="3" ViewCount="148" Body="&lt;p&gt;I work as a research assistant in the field of water resources management.&#xA;We build small scale physical models of rivers and make experiments with sediment. The model rivers are about 20m long and 3m wide, the water depth is from 0 to half a meter. The bottom is either concrete or gravel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One important task is to quantitatively measure the bedform that results after a model run. (i.e. where the gravel got washed away and where it accumulated)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering whether there is a chance to use a medical ultrasound device&#xA;to record the topography of the (river) bottom.&#xA;I'm hoping that these devices deliver a 2D profile of the riverbed and if I track to position + orientation of the ultrasound probe I can put together a 3d topography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone ever done some experiments like that? I mean alienated a ultrasound device? or can anyone tell me why my idea is theoretically bad / good? Where could be problems? &#xA;Is it possible to get spatial information out of an ultrasound image?&#xA;I mean in the image the ground should become apparent very clear as there is only water between the sensor and the ground.&#xA;I'm aiming at an accuracy of the resulting topography of about 5mm.&lt;/p&gt;&#xA;" OwnerUserId="8044" LastEditorDisplayName="user748" LastEditDate="2016-10-04T00:23:27.117" LastActivityDate="2017-01-03T07:37:59.607" Title="Could a medical ultrasound device be used for recording the topography of a small river?" Tags="|distance-measurement|medical-devices|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11516" PostTypeId="1" CreationDate="2016-09-15T20:58:16.917" Score="1" ViewCount="2362" Body="&lt;p&gt;Retention/Detention Ponds are often added to a site-design to mitigate storm-water runoff. &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0008884606002122&quot; rel=&quot;nofollow&quot;&gt;Enhanced Porosity Concrete (EPC), aka Pervious Concrete&lt;/a&gt;, can be used as a parking-lot pavement that allows storm-water to flow through it. If a detention-pond was covered with a filter-fabric, and clean-gravel was used to fill the pond on top of the filter-fabric, then EPC could be used to cover the gravel such that a parking-lot would be created on top of the pond, and using the same foot-print as the pond. The gravel would support the pavement loads and the water-storage capacity of the gravel-filled pond would be a function of the amount of void-space in the gravel.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because parking-lots and detention-ponds both take up a large part of land area of a building-site it would seem that combining the two in the same foot-print would save money and land area for the building-site developer. The pavement would also keep debris out of the pond and create a safer environment.&lt;/p&gt;&#xA;" OwnerUserId="7950" LastEditorUserId="368" LastEditDate="2016-09-16T15:31:07.717" LastActivityDate="2016-09-18T02:59:34.720" Title="Can a retention/detention pond and a parking-lot be constructed in the same site foot-print?" Tags="|civil-engineering|concrete|environmental-engineering|water-resources|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11517" PostTypeId="2" ParentId="10902" CreationDate="2016-09-15T21:33:31.307" Score="1" Body="&lt;p&gt;One way to do it would be to first estimate the internal static pressure ($ISP$) of your system (due to coils etc...) and add it to the maximum external static pressure ($ESP$) to get the total static pressure ($TSP$, in in. w.c). Then, since you know the airflow ($Q$, in cfm) I would calculate the ideal fan power ($P_{fan}$, in W) required as follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_{fan}= Q * TSP * \frac{249}{2118}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can calculate the brake horsepower ($BHP$, in HP) by using the fan efficiency ($μ$, in %), if you don't have it, use 60% (to stay conservative).&lt;/p&gt;&#xA;" OwnerUserId="8047" LastEditorUserId="8047" LastEditDate="2017-05-12T18:33:05.733" LastActivityDate="2017-05-12T18:33:05.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11518" PostTypeId="1" CreationDate="2016-09-15T21:45:36.860" Score="3" ViewCount="1006" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YPwBT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YPwBT.png&quot; alt=&quot;Hydraulic Line&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi all!,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently working on a hydraulic project. One of the procedure requires me to setup a line as shown in the picture. I am fairly new to this field, I am not sure what does the symbol highlighted in red represents. I have used google image try to find it, no results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Sean &lt;/p&gt;&#xA;" OwnerUserId="8048" LastEditorUserId="8048" LastEditDate="2016-09-15T23:00:23.010" LastActivityDate="2016-09-16T02:02:25.600" Title="Hydraulic Symbols" Tags="|fluid-mechanics|hydraulics|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11519" PostTypeId="1" CreationDate="2016-09-15T23:26:30.187" Score="1" ViewCount="1027" Body="&lt;p&gt;I have an electrical connector made of zinc-aluminum alloy (exact mixture unknown) that I would like to braze to a steel tube. Can this be done, what kind of material(s) would work for this application?&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2018-04-11T02:07:01.937" Title="Brazing Zinc-Aluminum Alloy to Steel" Tags="|welding|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11520" PostTypeId="2" ParentId="11518" CreationDate="2016-09-16T02:02:25.600" Score="4" Body="&lt;p&gt;I worked in a hydraulics design field for two years and found that unfortunately there isn't a standard way to draw many components. Each manufacturer seems to choose their own route. Designers are left to redraw or interpret components and sometimes hybrid symbols are created making the problem worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Information about what your project does and what you think the component does would be helpful. Also knowing the vendor or a recommended manufacturer would be helpful in hunting it down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess is that it is a poorly drawn check valve. The little spring drawn on it, most likely means that it is spring actuated (spring closed) and has some backpressure. Here are some check valve symbols it is similar to. I suggest you evaluate your hydraulic circuit and see if a check valve makes sense in that location before fabrication.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://www.iso.org/obp/ui#iso:grs:14617-8:8.4.5.2.5:X2115&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISO Spring-loaded ball-type non-return valve&lt;/a&gt;&lt;br&gt;&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/XBwA9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XBwA9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Check_valve&quot; rel=&quot;nofollow noreferrer&quot;&gt;Check Valve Symbol on Wiki&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kAH7B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kAH7B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://gallery.proficad.com/symbols/others/p-id/valves/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ProfiCAD Symbol Library&lt;/a&gt;&lt;br&gt; &#xA;  &lt;a href=&quot;https://i.stack.imgur.com/wcVcm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wcVcm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.conceptdraw.com/diagram/valve-symbols&quot; rel=&quot;nofollow noreferrer&quot;&gt;ConceptDraw symbol library&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/mMiig.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mMiig.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-16T02:02:25.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11521" PostTypeId="2" ParentId="11516" CreationDate="2016-09-16T02:07:36.817" Score="4" Body="&lt;p&gt;Yes, &lt;a href=&quot;https://en.wikipedia.org/wiki/Stormwater_detention_vault&quot; rel=&quot;nofollow&quot;&gt;Underground Detention&lt;/a&gt; is a thing. That Wikipedia article even mentions placing them under parking lots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact method that you propose is a little different than preconstructed vaults, but it is the same in the end. Pervious concrete on top of gravel with voids serves the same purpose but with much lower capacity. It is not really speeding up the infiltration of water into the ground.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2016-09-16T02:07:36.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11522" PostTypeId="2" ParentId="11519" CreationDate="2016-09-16T16:58:58.387" Score="1" Body="&lt;p&gt;Zinc/aluminium alloys are not uncommon in die cast parts and it should be possible to braze/solder. You would need a filler rod designed for 'white metal' or 'multi-metal' filler and the appropriate flux. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that brazing/soldering unusual materials is often a bit tricky as you need to judge the temperature of both parts within a fairly narrow window with little in the way of visual clues and the low melting point of zinc means they you have to be very gentle with the heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally I would look at a mechanical connector as the first choice for something like this, even if it involves brazing or welding an intermediate bracket, stud etc to the steel. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-09-16T16:58:58.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11524" PostTypeId="1" AcceptedAnswerId="11536" CreationDate="2016-09-13T12:57:52.313" Score="3" ViewCount="495" Body="&lt;p&gt;I have engineering data emerging from more or less complex constructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a regular base I need outlines from construction data to perform geometric calculations on them. The construction data comes from different CAD packages (e.g. Inventor or SolidWorks). I basically need certain closed polygons stored in a way I can read it with primitive methods like sloppy designed PERL programs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The polygons are planar. I.e. the data I want to process is 2D. Basically it often is an outline of a 3D object, e.g. a housing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my favourite input format for such extracted polygon data is a CSV-file or plain text with whatever separators are available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As most CAD-SW can import 3D-Models saved in STEP, I practically can choose which software I use to save the polygon data. The outcome is, I will probably end up having DXF or DWG files, which are the most primitive CAD-files I can get from that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves me with getting my polygons out of that goddamn DXF into a CSV file. Does anyone have an (preferably) easy way to do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum: For increased clarificaiton, I describe my task. I have to design PCBs which fit nicely into oddly shaped housings. While my EDA-sw exhibits a fairly usable routine for importing DXF-data, my self programmed autoplacer does not. So in my EDA file, I may have a nice outline, but this does not help me with placing the components inside that polygon. Hence I need the polygon data separately in my autoplacer.&lt;/p&gt;&#xA;" OwnerUserId="3786" OwnerDisplayName="Ariser" LastEditorUserId="3786" LastEditDate="2016-09-17T22:05:32.660" LastActivityDate="2016-10-05T12:49:36.153" Title="Method to extract closed curve from 3D CAD into CSV or TXT" Tags="|cad|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11525" PostTypeId="1" AcceptedAnswerId="11529" CreationDate="2016-09-17T09:20:06.580" Score="3" ViewCount="329" Body="&lt;p&gt;I've recently been reading through introductory texts on the mechanical properties of materials such as &lt;a href=&quot;http://link.springer.com/book/10.1007%2F978-94-007-4342-7&quot; rel=&quot;nofollow&quot;&gt;this&lt;/a&gt;. Most start with a description of various material constants such as Young's modulus, Poisson's ratio and a zoo of other properties, and try to determine these properties through a study of dislocations and interatomic interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I read about the very specific test conditions for these kinds of properties, it's not immediately obvious how well these kinds of coefficients and measurements translate to the real world. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I can find coefficients describing hardness, toughness, etc. from laboratory experiments under specific experimental conditions, but it's not obvious that these coefficients can give a more than qualitative description description of how well my nail can hold together a desk, how well my steel truss can hold up a roof, etc., which are ultimately the questions I might be interested in. These coefficients seem to describe very specific geometries, and it seems that I can only get answers about more general geometries using FEA, leaving these kinds of measurements as useful for 1) qualitative descriptions and 2) feeding parameters into FEA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think some of the confusion stems from the seemingly arbitrary nature of all these parameters. It appears that someone designed a test and came up with a new parameter every time they were interested in a behavior under different conditions, and there's no minimal degrees-of-freedom type argument that prevents me from coming up with my own test and modulus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, what is the argument that the mechanical properties covered by the usual texts (Young's, Poisson's, hardness, toughness, etc.) can give me a complete mechanical description of a system under arbitrary geometries and loads? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking about the parameter space more abstractly, each test can be represented as a material-dependent function that accepts a given material geometry and load geometry and maps it onto strain (throwing out hysteresis and time dependent phenomena for simplification). If we represent our inputs in R^n and our outputs in, say, R^m, what is the minimum number of moduli and constants needed to define this function? What's the argument I won't open up a textbook and find that it defines a new and non-redundant mechanical parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if such a well defined relationship exists, where might I find out more about how to go from the parameters derived from well defined experiments (stretching a cylinder, squeezing a flat plate, etc.) and more general material and stress geometries?&lt;/p&gt;&#xA;" OwnerUserId="8066" LastEditorUserId="8066" LastEditDate="2016-09-17T17:12:47.133" LastActivityDate="2019-03-15T10:00:53.267" Title="Value of Mechanical Constants in Real-World Applications" Tags="|materials|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11526" PostTypeId="2" ParentId="11525" CreationDate="2016-09-17T15:01:02.757" Score="0" Body="&lt;p&gt;It is because materials science is fundamentally a empiric science. The models we have are adequate at best and can not be built exactly from first principles. Too many things going on at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example of this I know of a big engine manufacturer that has engines made out of material X on the market, has had for decades and they work well. But they can not model this, if they model it with best of knowledge they get that the engine should last for a fraction (Weeks or days as opposed to multiple decades) of what they actually do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by all means if you can make accurate prediction with your own model go for it. Just do not expect people to use your stuff unless you can actually show a consistent history of being right.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2016-09-17T15:01:02.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11527" PostTypeId="1" CreationDate="2016-09-17T13:25:14.520" Score="0" ViewCount="95" Body="&lt;p&gt;I have a water channel with a flow of 1.5 m3 per second. What would the potential energy be if I install a waterwheel with fins almost touching the sides and bottom?&lt;/p&gt;&#xA;" OwnerDisplayName="Greengrower" LastActivityDate="2016-11-17T12:14:56.520" Title="Water wheel potential" Tags="|power|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11528" PostTypeId="2" ParentId="11527" CreationDate="2016-09-17T14:08:23.060" Score="0" Body="&lt;p&gt;Mechanical and electrical energy is the product pairs of variables, like torque.angle, force.distance, volts.charge, pressure.volume, and so on. If you only have one of those, it's like the sound of one hand clapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your flow rate is volume per second, so now all you have to do is define the pressure difference available to get the power (energy per second). Keep all your units in SI, and it will just come out correctly (Pascals, m3, watts, Joules). &lt;/p&gt;&#xA;" OwnerUserId="8092" OwnerDisplayName="Neil_UK" LastActivityDate="2016-09-17T14:08:23.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11529" PostTypeId="2" ParentId="11525" CreationDate="2016-09-17T17:41:25.130" Score="3" Body="&lt;p&gt;Generally, the first-order model parameters you discuss (specifically elastic constants) relate the complete stress state of an infinitesimal cube of a homogeneous material to its complete strain state. There are 21 such elastic constants forming a symmetric rank four tensor. Left-multiplying the elastic tensor against the symmetric rank two strain tensor (or its inverse against the stress tensor) gives the stress tensor (or strain tensor). As with all physical models in every field the model of elastic material behavior is wrong (e.g. during plastic deformation), but useful under certain assumptions. Notably, those of linear elasticity, continuum homogeneity, and uniform standard environmental conditions. The many tests you write of are useful for determining those 21 elastic constants by experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model is still resonably useful when those assumptions are close to true but not quite perfect. The model can also be extended to include the effects of variable temperature, etc. Given resonably accurate values, a computer can be fed those values for use in deformation calculations in FEA simulation software which may be used to estimate deformations in arbitrary geometric domains subject to nearly arbitrary boundary conditions. The results of the simulations can be used to guide design decisions in real world products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, other mathematical models may require other physical values such as elastic limits, deviation from linearity, strain rate effects, temperature effects, plastic deformation models, etc. Models have also been developed which correlate physical properties with microstructure, and microstructure with processing history and with material composition. This allows for variation of the physical properties over the geometric domain based on differential processing parameters, e.g. corners solidifying faster than hotspots in a casting, resulting in more accurate simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without physical parameters and fundamental models none of this would be possible.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2016-09-18T15:13:43.233" LastActivityDate="2016-09-18T15:13:43.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11531" PostTypeId="1" CreationDate="2016-09-17T18:57:21.180" Score="0" ViewCount="8535" Body="&lt;p&gt;*Originally posted in Thermodynamics.. reposted here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Helloooo everyone! Self proclaimed jet engine designer here~ Don't worry I won't be designing airliner engines anytime soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know when air passes through a diffuser or a duct of progressively larger cross section it gets slowed down, compressed with an increase in temperature. And when the passage is of progressively smaller cross section the opposite would happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now.. a great many online publications for laymen explain the above fact clearly.. sometimes with nice illustrations but they don't go as far to detail the calculations. And I find specialized literature to be rather difficult to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example what would the outcome be if air is discharged out of a compressor (at 300K 300m/s 1.5 bar) and passes through a diffuser that has an exit area of twice that of inlet area? I don't think it will be so simple because all three properties I've mentioned must change simultaneously..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could an expert thermodynamitician explain this please? Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="8071" LastActivityDate="2019-06-02T22:39:35.230" Title="How to calculate inlet and exit area of a gas turbine engine compressor diffuser?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11532" PostTypeId="2" ParentId="2783" CreationDate="2016-09-17T20:43:52.743" Score="1" Body="&lt;p&gt;The answer should depend on what strains are you interested in. For longitudinal strains, 0.001 (that is, 1 mm/m) is a usual figure to predimension joints in bridges, and that is quite independent from loading, since most of strain comes from shrinkage and thermal deformation - as the Wikipedia article you quoted says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in strains caused by actions, you should look at the most bended sections and how are they dimensioned. Bended sections are dimensioned with two limit states in mind - sometimes explicit, sometimes implicit, depending on the method used:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Maximum strain in concrete about 0.0035 in compression.&lt;/li&gt;&#xA;&lt;li&gt;Maximum strain in rebar steel near the elastic limit - that is, about 0.002 depending on steel grade.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the bridge is properly dimensioned, those strains shouldn't be exceeded even under dynamic loads if the bridge is expected to remain undamaged.&lt;/p&gt;&#xA;" OwnerUserId="7371" LastActivityDate="2016-09-17T20:43:52.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11533" PostTypeId="1" CreationDate="2016-09-17T20:53:07.360" Score="1" ViewCount="199" Body="&lt;p&gt;I'm in the design process for a cooling system. There is the need for some valves to control the flow of cooling fluid. In most machine states there's a constant fluid flow through either of two branches of the circuitry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially I was working with 2/2 directional control valves with direct acting solenoid. Having low system pressure I'm forced to use direct acting valves. A power consumption of 4 W seems to be the lower boundary for a valve with ND 2mm. This is a waste of energy, since most of the time the valves stay in a constant state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then changed to 3/2 valves, so that only one state will consume energy. But 3/2 direct acting valves are exceptionally expensive and are available with small diameters only. I thought about ball valves, but those in turn are manufactured with diameters starting from 1/2&quot; and have big gearboxes making them expensive as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So finally: Is there a possiblilty to switch a liquid between two brances of a circuit which I didn't think/know of?&#xA;Tubing has mostly 8mm inner diameter, fluid is water/glycol mixture with temperatures between 10 and 90 degrees. Voltage available for controlling valves is 12V.&#xA;Perhaps I'm just missing the affordable ball valve manufacturer or some tricky workaround? &lt;/p&gt;&#xA;" OwnerUserId="3786" LastActivityDate="2018-01-13T03:12:21.067" Title="Reducing energy consumption of valves" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11534" PostTypeId="2" ParentId="11533" CreationDate="2016-09-17T23:39:04.910" Score="0" Body="&lt;p&gt;Have you thought about an electromagnetic valve? Needs power only when moving. What I am envisioning would be a piston type 90° to the flow and watertight when closed. The other possibility would be a thermostat.&lt;/p&gt;&#xA;" OwnerUserId="8073" LastActivityDate="2016-09-17T23:39:04.910" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="11535" PostTypeId="2" ParentId="11516" CreationDate="2016-09-18T02:59:34.720" Score="1" Body="&lt;p&gt;Underground storage can be used, but it is much more expensive. There are a variety of options such as concrete tanks, R-tanks, or piped storage. Whatever you go with, you need to ensure that you are able to collect stormwater from your site and convey it to your tanks. Using pervious pavement is usually difficult since it's hard to drain a large area to your pervious surface. Using inlets is more common. Lastly, when draining underground storage, deeper is more expensive so you should keep it shallow. This makes it tougher to route storms with appropriate release rates, not the same as designing a basin. So due to cost, normal basins are frequently used unless you have significant site restraints.&lt;/p&gt;&#xA;" OwnerUserId="40" LastActivityDate="2016-09-18T02:59:34.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11536" PostTypeId="2" ParentId="11524" CreationDate="2016-09-18T08:32:38.557" Score="2" Body="&lt;p&gt;It seems you are interested in a flat slice of your CAD model. While you could use a 3D file and slice it yourself that seems like a bit overkill as the CAD application is perfectly capable of doing the slices for you.&lt;/p&gt;&#xA;&lt;h1&gt;quick and dirty&lt;/h1&gt;&#xA;&lt;p&gt;Ok, so each CAD has a 2D drawing mode, you can save that drawing out as &lt;code&gt;dxf&lt;/code&gt; or &lt;code&gt;pdf&lt;/code&gt; both are easy to parse. If you don't happen to find a good tool for this turn the &lt;code&gt;pdf&lt;/code&gt; into &lt;code&gt;svg&lt;/code&gt; that can be easier to parse. This approach can also be done quick and dirty by leveraging such tools as Inkscape or Illustrator. Lets do an example because its easy to do:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uVHLf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uVHLf.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: Quick and dirty export drawing as &lt;code&gt;pdf&lt;/code&gt;/&lt;code&gt;svg&lt;/code&gt; then isolate and read points from that file I &lt;a href=&quot;https://graphicdesign.stackexchange.com/questions/73629/create-2-morphable-svgs/73636#73636&quot;&gt;used modified version this tool&lt;/a&gt; to dump coordinates from &lt;code&gt;pdf&lt;/code&gt;. You should be able to do that in Perl easily. &lt;a href=&quot;http://pastebin.com/wm8fTdCX&quot; rel=&quot;nofollow noreferrer&quot;&gt;data available here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Proper Method&lt;/h1&gt;&#xA;&lt;p&gt;It is possible to access both SolidWorks and Inventor trough a COM bridge so you can access the CAD applications data model directly from your Perl code.This has several benefits but mostly not needing to parse intermediate files. You could select the relevant edges and just traverse them directly from the CAD. Now I only have access to SolidWorks at work but  similar approach works in inventor as i have done it.&lt;/p&gt;&#xA;&lt;p&gt;I had some extra time at work to do some quick &lt;code&gt;VBA&lt;/code&gt; code for SolidWorks. The code takes all the lines of a closed sketch, sorts them into polygon (with a naive N^2 algorithm) order and prints them in the &lt;code&gt;VBA&lt;/code&gt; debug console.&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;Option Explicit&#xA;&#xA;Sub main()&#xA;    Dim swApp                   As SldWorks.SldWorks&#xA;    Dim swModel                 As SldWorks.ModelDoc2&#xA;    Dim swPart                  As SldWorks.PartDoc&#xA;    Dim swSelMgr                As SldWorks.SelectionMgr&#xA;    Dim swFeat                  As SldWorks.Feature&#xA;    Dim swSketch                As SldWorks.Sketch&#xA;    Dim numLines                As Long&#xA;    Dim vLines                  As Variant&#xA;    Dim dict                    As New Collection&#xA;    Dim i                       As Variant&#xA;&#xA;    Set swApp = CreateObject(&amp;quot;SldWorks.Application&amp;quot;)&#xA;    Set swModel = swApp.ActiveDoc&#xA;    Set swPart = swModel&#xA;    Set swSelMgr = swModel.SelectionManager&#xA;    Set swFeat = swSelMgr.GetSelectedObject5(1)&#xA;    Set swSketch = swFeat.GetSpecificFeature2&#xA;&#xA;    &#xA;    numLines = swSketch.GetLineCount2(1) 'Exclude crosshatch lines&#xA;    vLines = swSketch.GetLines2(1) 'Exclude crosshatch lines&#xA;&#xA;    Dim startP, endP, line As Variant&#xA;    &#xA;    For i = 1 To numLines - 1&#xA;        line = Array(Array(vLines(12 * i + 6) * 1000, _&#xA;                           vLines(12 * i + 7) * 1000), _&#xA;                     Array(vLines(12 * i + 9) * 1000, _&#xA;                           vLines(12 * i + 10) * 1000))&#xA;        dict.Add (line)&#xA;    Next i&#xA;    &#xA;    startP = Array(vLines(6) * 1000, _&#xA;                   vLines(7) * 1000)&#xA;    endP = Array(vLines(9) * 1000, _&#xA;                 vLines(10) * 1000)&#xA;&#xA;    pp startP&#xA;    pp endP&#xA;    &#xA;    For i = 1 To dict.Count - 1&#xA;        endP = NextPoint(dict, endP)&#xA;        pp endP&#xA;    Next i&#xA;End Sub&#xA;&#xA;Sub pp(point As Variant)&#xA;    Debug.Print &amp;quot; &amp;quot; &amp;amp; Str(point(0)) &amp;amp; &amp;quot;, &amp;quot; &amp;amp; Str(point(1))&#xA;End Sub&#xA;&#xA;Function NextPoint(dict As Collection, point As Variant) As Variant&#xA;    Dim i As Variant&#xA;    For i = 1 To dict.Count&#xA;&#xA;        Dim data, startRP, endRP As Variant&#xA;        data = dict.Item(i)&#xA;        startRP = data(0)&#xA;        endRP = data(1)&#xA;    &#xA;        If endRP(0) = point(0) And endRP(1) = point(1) Then&#xA;            dict.Remove (i)&#xA;            NextPoint = startRP&#xA;            Exit Function&#xA;        End If&#xA;        If startRP(0) = point(0) And startRP(1) = point(1) Then&#xA;            dict.Remove (i)&#xA;            NextPoint = endRP&#xA;            Exit Function&#xA;        End If&#xA;        &#xA;    Next i&#xA;End Function&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;Since vba is calling COM you ca code you can use nearly any language for example perl implements Win32::OLE that can do the job.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/riuLR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/riuLR.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Image 2&lt;/strong&gt;: Example part with simple one loop sketch results in &lt;a href=&quot;http://pastebin.com/raw/uEETZNjd&quot; rel=&quot;nofollow noreferrer&quot;&gt;this output&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Epilogue&lt;/h1&gt;&#xA;&lt;p&gt;If you really want to export 3D polygon data and do the slicing manually then i would export either OBJ or STL. But this would be way down on my list of approaches mainly because all other approaches are simpler.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-10-05T12:49:36.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11537" PostTypeId="1" CreationDate="2016-09-18T11:39:51.010" Score="1" ViewCount="1028" Body="&lt;p&gt;BACKGROUND: I'm burying a 3 foot x 4 foot drainage box (2 foot tall) in my driveway to hold/distribute excess gutter flow (1&quot; of rain produces ~21cuft). I've constructed the walls with brick, laid on their edge so that the water can drain out through the 10 holes in each.  I need to construct a cover that, when itself covered with ~3&quot; of crush&amp;amp;run will support a potentially heavy vehicle driving over it.  To further support the cover, I'm laying 1.25&quot; iron pipe across the top (counter sunk into the brick), at 1 foot in from each side, so that the top is supported on one foot centers (horizontally: earth, then 3&quot; gap to brick wall, then 12&quot; gap to pipe, then 12&quot; gap to pipe, then 12&quot; gap to brick wall, then 3&quot; gap to earth).  I could pour a 4&quot; slab over the hole, but then it would be nearly impossible to lift, should the need arise.&#xA;QUESTION:  If I fill 5' long 1.25&quot; pvc pipe with quikrete and a rebar, and then covered the top with them (placed tightly), then covered the top with the 3&quot; of crush&amp;amp;run, would that be strong enough to support a heavy truck (UPS, Dump truck, etc.)?  (This way, if I needed access, I could just dig up a few of the pipe.)&lt;/p&gt;&#xA;" OwnerUserId="8068" LastActivityDate="2016-09-18T20:46:07.880" Title="Shear strength of reinforced concrete filled PVC pipe" Tags="|structural-engineering|reinforced-concrete|concrete|shear|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11538" PostTypeId="1" CreationDate="2016-09-18T11:54:09.447" Score="4" ViewCount="276" Body="&lt;p&gt;I am working on the vibration of the continuous system, I have seen lots of books on vibrations which talks about the natural frequency and mode shapes of the continuous system. I am interested in finding when particular time varying load is applied on continuous system how do we know which modes of the systems are excited and why?  &lt;/p&gt;&#xA;" OwnerUserId="8029" LastActivityDate="2016-09-20T22:52:55.013" Title="Which mode does get excited under action of dynamic loading?" Tags="|dynamics|vibration|modal-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11539" PostTypeId="1" AcceptedAnswerId="12428" CreationDate="2016-09-18T11:57:12.440" Score="1" ViewCount="2090" Body="&lt;p&gt;I have read a few ISO standards for performing the so-called 'burst pressure' test in which a filter or an element of one is tested to failure. I am curious whether such a failure can occur under normal fuel line pressures (e.g. after filter clogging). To break down:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is a typical fuel line pressure?&lt;/li&gt;&#xA;&lt;li&gt;Can a blocked/clogged filter 'burst' under such pressure?&lt;/li&gt;&#xA;&lt;li&gt;If not, what is the worst that could happen?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8079" LastEditorUserId="7083" LastEditDate="2016-09-22T00:55:59.897" LastActivityDate="2023-11-16T14:44:55.787" Title="Is a filter burst possible under fuel line pressure?" Tags="|pressure|diesel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11543" PostTypeId="2" ParentId="11537" CreationDate="2016-09-18T20:46:07.880" Score="2" Body="&lt;p&gt;A sketch would be helpful. But if my understanding of what you want to do is correct, you want to cover a 3 by 4 feet sump pit by light material which you could remove in case it is needed.&lt;br&gt;&#xA;The pvc pipe filled with quickrete or even high strength concrete or any other mix of epoxy and cement is going to fail because it basically hardly can sopprt its own weight let alone dead and live load surcharge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;your brick walls are going to fails as well.&lt;br&gt;&#xA;you need to contact the building department of your city and ask what are the codes and if they have a basic hand out which details what you need to do.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually they expect you to make a foundation on competent soil at the depth they specify, or or concrete pad and build the wall with adequate reinforcement and specified to take the traffic load which runs up to 1000 or more lbs/square foot.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cover can be made of steel plate with stiffener beams under it.&lt;br&gt;&#xA;The entire cover and its attachment to the walls and the pit have to be engineered.&lt;br&gt;&#xA;Otherwise you are setting a trap for somebody to fall into and get seriously hurt!&lt;br&gt;&#xA;Alternatively if there is enough space to build the pit outside of traffic pattern you can use lighter live loads, 100 lbs/sqr.ft.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-09-18T20:46:07.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11544" PostTypeId="2" ParentId="11538" CreationDate="2016-09-18T21:49:16.267" Score="5" Body="&lt;p&gt;Take the Fourier Transform of the time varying driving force, this will give the frequency content of the driving force. Multiple modes of vibration can be driven at once, and will superpose with each other, but time varying driving forces with a frequency content that are high at frequencies near a particular resonant frequency will mainly drive the resonant frequency's corresponding vibration mode. Note that for a perfectly sinusoidal (in time) driving force will have an infinite (Dirac delta) frequency content at the frequency of the sinusoid, and zero for all other frequencies. A perfect impulse (i.e. infinite force applied for infinitesimal time) will have a frequency content equal over all frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this can be done is because superposition applies to vibrating systems, assuming vibrations are small enough. You can split any time varying driving force into a the sum of multiple sinusoids: consider what modes each sinusoid drives, and superpose all the effects together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MODAL PARTICIPATION&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this section, I will only consider 1D vibrating systems governed by the wave equation. There exist cases, such as a transversely vibrating cantilever beam, that are governed by &lt;em&gt;wave-like&lt;/em&gt; equations which I won't cover here unless it's of interest. Since the governing equation is the wave equation, the mode shapes will be sinusoidal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to determine what amounts of each mode is being driven it is useful to write the deflected shape, $w(x,t)$, as the sum of each of the modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w(x,t) = \sum_{i} \alpha_i (t) u_i(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\alpha_i (t)$ is the amplitude of the $i^{th}$ mode, which can vary in time, and $u_i(x)$ is the $i^{th}$ mode shape of unit amplitude. This expression is then substituted into the governing wave equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the governing wave equation for a vibrating tense string is given as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu \frac{\partial^2 w}{{\partial t}^2} - T \frac{\partial^2 w}{{\partial x}^2} = f(x,t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $w(x,t)$ is the transverse displacement $\mu$ is mass-per-unit-length of the string, $T$ is the tension in the string, and $f(x,t)$ is the distributed transverse force-per-unit-length acting on the string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This equation can be adapted to other wave-governed systems by replacing the variables appropriately).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LzZI0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LzZI0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the sum of modes expression into the wave equation gets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mu \sum_i{(\ddot \alpha_i + \omega_i^2 \alpha_i) u_i(x)} = f(x,t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\omega_i$ is the resonant frequency corresponding to the $i^{th}$ mode. Then, by multiplying by $u_j(x)$ and integrating along the whole domain with respect to $x$, noting that some terms cancel out (the integral of $u_i(x) u_j(x)$ is zero for $i\ne j$ due to &lt;em&gt;orthogonality&lt;/em&gt; between different mode shapes), we get the following differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot \alpha_j + \omega_j^2 \alpha_j =  \frac{2}{\mu L} \int_0^L f(x,t) u_j(x) dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $L$ is the length of the 1D domain (in this case, the length of the string).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this all means is that, given the distributed force-per-unit length on the system, solving the above differential equation will give the amplitude of the $j^{th}$ mode as a function of time, and hence quantify how much that mode is present at any point in time. If the distributed force is sinusoidal, after a while (i.e. at steady state) the modal amplitude will also vary sinusoidally at the same frequency. For example if:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f(x,t) = f_0(x) \sin(\omega t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha_j(t) = \alpha_{j,0} \sin(\omega t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\alpha_{j,0} = \frac{\frac{2}{\mu L} \int_0^L f_0(x) u_j(x) dx}{\omega_j^2 - \omega^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how the modal amplitude seems to rocket to infinity if the system is sinusoidally driven at the resonant frequency. This is to be expected for undamped resonance: in reality, damping will prevent such unphysical responses, but this has be omitted from the scope of this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to modify the modal amplitude differential equation so that the force applied is a point force instead of a distributed force. Therefore, if a point force $F(t)$ is applied at $x=x_0$, then (substituting $f(x,t) = F(t) \delta(x -x_0)$, where $\delta(x)$ is the Dirac delta function):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot \alpha_j + \omega_j^2 \alpha_j =  \frac{2}{\mu L} F(t) u_j(x_0) $$&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2016-09-20T22:52:55.013" LastActivityDate="2016-09-20T22:52:55.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11545" PostTypeId="1" CreationDate="2016-09-18T23:17:44.647" Score="5" ViewCount="667" Body="&lt;p&gt;I have a system like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H7bbm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H7bbm.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I have to estimate the height of the lower tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, I'm considering that I have a non-viscous, incompressible fluid in steady flow.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Base apertures&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The outlet velocity of liquid when draining a tank or a container can be expressed as:&#xA;\begin{equation}&#xA;v = C_v \sqrt{2gH}&#xA;\label{eq:baseAperture1}&#xA;\end{equation}&#xA;where $v$ is the outlet velocity $[m/s]$, $C_v$ is the velocity coefficient (water: 0.97), $g$ is the acceleration of gravity ($g=9.81 m/s^2$) and $H$ is the height $[m]$.\&#xA;The volume flow can be expressed as:&#xA;\begin{equation}&#xA;V = C_d A \sqrt{2gH}&#xA;\label{eq:baseAperture2}&#xA;\end{equation}&#xA;where $V$ is the volume flow $[m^3/s]$, $C_d$ is the discharge coefficient ($C_d = C_c C_v$ where $C_c$ is the contraction coefficient (sharp edge aperture: 0.62, well rounded aperture: 0.97)) and $A$ is the area aperture $[m^2]$\\&#xA;I know the volume flow and the height, so, using the first equation I get the outlet velocity $v$. Using the second equation I can get the area A of the outlet hole (I have two possible results because I don't know if the hole has a sharp edge aperture or a well rounded aperture).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Bernoulli&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{d}{ds} \left(\frac{v^2}{2} + \frac{p}{\rho} + g h\right) = 0&#xA;\end{equation}&#xA;where $v$ is the flow speed, $p$ is the pressure, $\rho$ is the density, $g$ is the gravity and $h$ is the height.\&#xA;From the previuos equation, I get:&#xA;\begin{equation}&#xA;\frac{v^2}{2g} + \frac{p}{\gamma} + h = constant&#xA;\end{equation}&#xA;where $\gamma = \rho g$\&#xA;For a tank with a small orifice in a lateral, close to the bottom, I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wRNGH.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wRNGH.png&quot; alt=&quot;Bernoulli equation flow tank&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Bernoulli part 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Making a lot of maths, I get these two equations:&#xA;\begin{equation}&#xA;\dot{h}_1 = -\frac{A_{d1} \sqrt{2g}}{A_{D1}} \sqrt{h_1} + \frac{K_P}{A_{D1}}V_P&#xA;\end{equation}&#xA;\begin{equation}&#xA;\dot{h}_2 = \frac{A_{d1}\sqrt{2g}}{A_{D2}} \sqrt{h_1} - \frac{A_{d2} \sqrt{2g}}{A_{D2}} \sqrt{h_2}&#xA;\end{equation}&#xA;where $h_1$ is the height of the upper tank, $h_2$ is the height of the lower tank, $A_{d1}$ is the area of the orifice of tank 1 and $A_{d2}$ is the area of the orifice of tank 2, $g$ is the gravity, $A_{D1}$ is the area of the upper tank and $A_{D2}$ is the area of the lower tank, and $V_P$ is the voltage of a pump. In my exercise, $Q_P = V_P K_P$ where $Q$ is the discharge. &lt;strong&gt;Is $Q_P$ the volume flow?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to linearize that equations, using the Jacobian, and solving that equations but I don't have the areas $A_{D1}$ and $A_{D2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I get $A_{D1}$ and $A_{D2}$?&lt;/p&gt;&#xA;" OwnerUserId="8089" LastActivityDate="2022-04-22T11:01:20.160" Title="Fluid dynamics of non interacting tanks in series" Tags="|electrical-engineering|control-engineering|chemical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11547" PostTypeId="1" CreationDate="2016-09-19T06:19:12.747" Score="1" ViewCount="20" Body="&lt;p&gt;How the patterns or widths will be given to numbers while encoding data to barcodes like for 1 it will be 2221 and for 2 it is 2122. I didn't understand how these widths are given&lt;/p&gt;&#xA;" OwnerUserId="8091" LastActivityDate="2016-09-19T06:19:12.747" Title="Encoding data to barcodes" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="1" ClosedDate="2016-09-19T11:54:05.237" ContentLicense="CC BY-SA 3.0" />
  <row Id="11548" PostTypeId="1" CreationDate="2016-09-19T12:18:34.533" Score="3" ViewCount="154" Body="&lt;p&gt;I am tasked to design a bridge pile cap. Shear links are required. However due to heavy reinforcement used, contractor requested to hook the links between inner rebars to ease construction. Is this allowed in British Standard? How should this be taken into consideration if it is allowed?&lt;/p&gt;&#xA;" OwnerUserId="7887" LastActivityDate="2016-09-19T12:18:34.533" Title="Can shear links be designed to hook between inner layers of top and bottom reinforcement?" Tags="|structural-engineering|foundations|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11550" PostTypeId="1" AcceptedAnswerId="11551" CreationDate="2016-09-19T13:33:29.537" Score="1" ViewCount="65" Body="&lt;p&gt;I have the following model:&#xA;$$ x[k+1] = Ax[k] + Bu[k]$$&#xA;$$ y[k]   = x[k] $$&#xA;where $x \in \mathbb{R} $ is the state, $A \in \mathbb{R}$, $u \in \mathbb{R}^4$ are the inputs and $B \in \mathbb{R}^4$ and $y[k]$ is the output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From theory this can be solved using:&#xA;$$ y[k] = A^k x[0] + \sum_{i=0}^{k-1} A^{k-1-i}Bu[i]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find the time instance $k$ when $y[k]$ has a specific value?&lt;/p&gt;&#xA;" OwnerUserId="5948" LastEditorDisplayName="user748" LastEditDate="2016-10-01T23:10:44.300" LastActivityDate="2016-10-01T23:10:44.300" Title="Find the instance when discrete time model has a specific value" Tags="|electrical-engineering|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11551" PostTypeId="2" ParentId="11550" CreationDate="2016-09-19T15:00:39.520" Score="3" Body="&lt;p&gt;The answer depends on your initial value and your input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For specific initial values and inputs there are easy solutions, but in general I think the fastest way is to let the computer solve this for you.&#xA;So crank up your matlab / python / mathematica / excel / other tool of choice&#xA;and let the brute force of computation serve you.&lt;/p&gt;&#xA;" OwnerUserId="7590" LastActivityDate="2016-09-19T15:00:39.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11552" PostTypeId="1" CreationDate="2016-09-19T19:24:37.603" Score="4" ViewCount="4138" Body="&lt;p&gt;I'm attempting to calculate the clamping force from a u-bolt. This arises from an existing support configuration that is being re-qualified due to the installation of new equipment. I am stuck with this configuration unless it turns out to be truly impossible to re-qualify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{4cm}$&lt;a href=&quot;https://i.stack.imgur.com/WD0pl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WD0pl.png&quot; alt=&quot;Figure 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\hspace{7cm}$&lt;strong&gt;Figure 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The u-bolts, as shown in Figure 1, are supporting a vertical load. This vertical load is being resisted only by the clamping force which exists from the 2 u-bolts around the vertically oriented 2&quot; dia. Schedule 40 pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial thoughts are to approach the qualification as I would for a bolt and nut in a slotted hole. That is to say, use the installation torque to calculate the bolt preload, and then use the preload along with an appropriate coefficient of friction to calculate the clamping load. This approach seems inadequate to me, but I don't really have a good reason why I feel that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone ever dealt with a similar issue? What method was used to calculate the clamping force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="129" LastEditorUserId="129" LastEditDate="2016-09-19T19:42:24.110" LastActivityDate="2022-11-15T12:07:34.913" Title="Clamping force from U-bolt around pipe" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11553" PostTypeId="2" ParentId="8193" CreationDate="2016-09-19T19:53:42.440" Score="1" Body="&lt;p&gt;Early steam engines weren't actually steam engines at all but were rather atmospheric engines and the steam was a convenient way of generating a partial vacuum in the cylinder by condensation. This is typically the case for the really huge Cornish beam engines. As they were driven by less than one atmosphere of pressure differential they needed very large cylinders to do a useful amount of work. Also they did work on what would conventionally be called the compression stroke which is why they use an overhead rocker beam rather than a connecting rod/crank arrangement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get into the realms of high pressure steam engines the difference in scale compared to IC engines is less pronounced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another aspect is that steam engines are typically driven by a fairly large reservoir of pressurised steam (ie a boiler). By contrast the hot working fluid for an IC engine is, by definition, generated by discrete combustion within the cylinder itself so there is a limited amount of energy available to do work per cycle. By contrast with steam (or compressed air) you can continue to do work on a piston for an indeterminate length of cylinder, limited only by the volume of your pressurised gas reservoir. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice there are limits as boring very long cylinders starts to get technically difficult and expensive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that there is a crossover in scale between steam and IC engines indeed it is possible to convert a IC engine to run on steam. Air cooled early VW engines have been converted in this way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/LJq2Hc_mXFI&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/LJq2Hc_mXFI&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-09-19T19:53:42.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11554" PostTypeId="1" AcceptedAnswerId="11556" CreationDate="2016-09-20T03:23:10.430" Score="2" ViewCount="3097" Body="&lt;p&gt;From what I've read it seems that hydro turbines can theoretically achieve 100% energy conversion from kinetic to electrical energy. However, I just wanted to double check if this was true.&lt;/p&gt;&#xA;" OwnerUserId="8097" LastEditorUserId="8097" LastEditDate="2016-09-20T15:26:04.050" LastActivityDate="2023-12-24T01:13:30.570" Title="What is the theoretical maximum energy efficiency of hydro turbines?" Tags="|turbines|energy-efficiency|renewable-energy|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11556" PostTypeId="2" ParentId="11554" CreationDate="2016-09-20T08:08:05.003" Score="5" Body="&lt;p&gt;Something can only be 100% efficient when a lot of generous assumptions are taken into account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;No friction (turbine surface, turbulence, nozzle, or bearings)&#xA;&lt;li&gt;Perfect cup design (100% 180degree jet reversal)&#xA;&lt;li&gt;A dry wheel (no water is spun off)&#xA;&lt;li&gt;Not counting the gravitational potential to drop the water away from the wheel in the efficiency calculation.&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pelton_wheel&quot; rel=&quot;noreferrer&quot;&gt;Pelton wheel&lt;/a&gt; is 100% efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I worked in alternative energy for several years and, 90% is the best is have seen (Pelton, oversized penstock, ~400ft head). 50% would probably be a good across-the-board average for currently installed microturbines in the western United States. The lower the head(upstream pressure), the larger and more expensive a turbine and supporting components have to be to achieve the same efficiency. Also, the energy to remove the water from the turbine becomes appreciable. So generally, the lower the head the lower the efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a graph of different turbine efficiencies. Pelton is a common choice for variable flow streams because a single turbine has a relatively high efficiency across a wide range of flowrates. Efficiency is very design and resource dependent so use this graph for a general understanding only (&lt;a href=&quot;http://slideplayer.com/slide/330622/&quot; rel=&quot;noreferrer&quot;&gt;image source&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Pelton wheel is a good turbine to start learning the mechanics before you look at more complex turbines. &lt;a href=&quot;http://www.learnengineering.org/2013/08/pelton-turbine-wheel-hydraulic-turbine.html&quot; rel=&quot;noreferrer&quot;&gt;Pelton Turbine - Working &amp;amp; Design Aspects&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CtWvC.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CtWvC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-20T08:08:05.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11557" PostTypeId="1" CreationDate="2016-09-20T09:41:56.003" Score="2" ViewCount="2066" Body="&lt;p&gt;I am currently designing a small pneumatic system that pumps air into an airbag, holds the air in, then releases it to the atmosphere. Each pump in the system actually feeds two sets of airbags and solenoids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have decided to use normally closed (NC) solenoids as the air has to be held in the airbag for a long time (and I don't want to deplete my battery by energising a normally open solenoid for all of that time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set up is as follows: I have a dc pump, connected to a 2/2 NC solenoid valve, which feeds into the air bag. The second solenoid is connected to this line (using some kind of T connector) in order to release the pressure from the airbag when necessary. As mentioned before, the dc pump actually feeds two sets of airbags and solenoids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: The first solenoid in the set up only needs to let air pass in one direction, however when it is closed it will likely experience a pressure differential in different directions at different times. Can a NC, single direction solenoid valve stay closed when the pressure differential is in either direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.zonhen.com/spec/ZHV/ZHV-0419.jpg&quot; rel=&quot;nofollow&quot;&gt;Here&lt;/a&gt; is a link to one of the solenoids that I am considering (warning - its a chinese site, so may run a little slowly)&lt;/p&gt;&#xA;" OwnerUserId="7451" LastActivityDate="2016-10-20T14:01:18.880" Title="How strict is the direction of air flow in a single direction pneumatic solenoid valve?" Tags="|pressure|pumps|valves|pneumatic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11558" PostTypeId="2" ParentId="11557" CreationDate="2016-09-20T13:18:59.617" Score="1" Body="&lt;p&gt;That datasheet is a little light on details, so I can't answer for sure how that valve will do, but in general, when a solenoid valve is closed it will not pass fluid in either direction. This should be true up to a reasonable pressure and usually there is a rated leakage rate. So if I understand your situation right that you only need it to pass in the 'correct' direction, but you need it to not pass in both directions, you should be OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that many solenoid valves will not pass in the 'wrong' direction when they are open. In some cases they won't pass at all because they rely on air pressure in addition tot he force from the solenoid to actuate. In other cases they will pass, but much less efficiently because the geometry of the valve is designed to let the medium flow smoothly the other way, and it is not optimized for reverse flow. Usually there is not an additional check valve in line unless it is explicitly listed in the diagram, so it would be a bad idea to count on the valve not passing, even if that's the usual behavior. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, where the air pressure is used to actuate the valve, you'll also need a minimum pressure differential for the valve to operate even in the manufacturer's intended direction. So while it will hold the back-pressure just fine, you won't be able to switch it's state until you establish a few PSI of forward pressure. These kinds of valves are sometimes advertised as 'energy efficient' because they use less electrical power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, this is probably a case of you get what you pay for. If those are the only valves you can afford or fit in your project or whatever, buy them and test them (or try to contact the manufacturer if you dare.) But if you have options, it's well worth a few more dollars to buy a valve that is from a 'main line' vendor that has more detailed documentation and support. Like most industries there is a lot of formal knowledge in the fluid power industry, and also a lot of people hacking things together and getting unpredictable results. The more you know about your components, the more you know about your system and the more you know about your system the better your chance of getting your product to work reliably.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-09-20T13:18:59.617" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11559" PostTypeId="1" CreationDate="2016-09-20T13:58:58.527" Score="1" ViewCount="51" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Adsorption/Bio-oxidation_process&quot; rel=&quot;nofollow&quot;&gt;AB process&lt;/a&gt; is wastewater treatment process that works like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'A Stage': Wastewater is (after mechanical treatment) subjected to a high load activated sludge process where also a lot of contaminats are adsorbed on the sludge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;'B Stage': Low load activated sludge process with further BOD destruction and nitrification and denitrification in anoxic zones.  A part of the effluient of the B vessels (before settlement tanks) is recirculated to the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each stage has it's own set of settlement tanks and return sludge, so the biology is adapted to the specific conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question: I've seen the claim (and actuall don't doubt it) that the B-stage is more effective (Per volume available) when the effluent of the A-Stage is added in a cascading way: Say (in a rectangular, long tank) 50% of effluent at the inflow end, 50% in the middle. Return sludge is all added at the beginning of the B stage. A carbon source can be added if required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What is the process reason why cascading is more effective? In a cascading low load reactor, is there still recirculation?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-09-20T13:58:58.527" Title="Why have cascading vessels in the 'B' part of the 'AB' wastewater treatment process?" Tags="|process-engineering|environmental-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11560" PostTypeId="2" ParentId="11554" CreationDate="2016-09-20T15:24:04.990" Score="1" Body="&lt;p&gt;Yes, it's true in the sense that there's no other theoretical upper limit on conversion that's below 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are such theoretical limits that are strictly less than 100% for PV; one for single-junction (the Shockley Queisser limit), a higher one for two junction, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there is a theoretical limit on heat engines - the Carnot efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydro has no such theoretical limit other than the standard hard 100% limit: all hydro is strictly less than 100%.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2016-09-20T15:24:04.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11561" PostTypeId="1" AcceptedAnswerId="11569" CreationDate="2016-09-20T16:40:57.600" Score="5" ViewCount="686" Body="&lt;p&gt;I'm looking at a motor datasheet where the moment of inertia is specified in units of &quot;oz in sec^2&quot;. I would think that inertia should have dimensions of mass * length^2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some further Googling and found that some automatic unit conversion calculators (&lt;a href=&quot;http://www.numberfactory.com/nf%20inertia.htm&quot; rel=&quot;nofollow&quot;&gt;example&lt;/a&gt;) let you convert between seemingly contradictory units - the site I've linked lists both kg-m^2 to kg-m-s^2 as acceptable units for moments of inertia. What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="3232" LastEditorUserId="3232" LastEditDate="2016-09-20T20:04:39.903" LastActivityDate="2016-09-21T13:54:55.610" Title="Contradictory units for moments of inertia" Tags="|applied-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11562" PostTypeId="1" AcceptedAnswerId="18066" CreationDate="2016-09-20T20:56:06.317" Score="2" ViewCount="769" Body="&lt;p&gt;First time poster here. First of all please excuse my english mistakes but most of all my terminology and ignorance on the subject. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to the question! A friend of mine asked me if I could help him convert his ceramic 3d printer extruder from compressed air to a piston system (driven by a leadscrew linear actuator). As of now, he told me, he uses 4 bar of pressure to make the clay extrude at the desired rate and that's all I know plus the nozzle and cartridge internal diameter (respectively 1 and 65 mm). What I'm looking for is a stepper motor/lead screw combination that can exert enough linear force to move the clay through the nozzle (the rate of extrusion would be given by the stepping rate of the motor). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I've done as &quot;homework&quot; but I'm a firmware developer so I'm hoping to get some guidance in solving this problem. Since the nature of this problem is not only practical but also not needing particular precision, what I've done are essentially ballpark, or back of the envelope, calculations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;converted bar to N/mm&lt;sup&gt;2&lt;/sup&gt; (4&amp;nbsp;bar -&gt; 0.4&amp;nbsp;N/mm&lt;sup&gt;2&lt;/sup&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;multiplied N/mm&lt;sup&gt;2&lt;/sup&gt; by the cartridge cross sectional area ($0.4\text{ N/mm}^2 \cdot 32.5^2\pi\text{ mm}^2$) to get the total force exerted on the clay of ~1320&amp;nbsp;N (I'll round this to 1400&amp;nbsp;N to have some margin)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Knowing the force needed I picked a lead screw from a catalog to get some specs for the next calculations, what I chose is a screw of 10&amp;nbsp;mm diameter and 2&amp;nbsp;mm/rev lead&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;using &lt;a href=&quot;https://en.wikipedia.org/wiki/Leadscrew&quot; rel=&quot;nofollow&quot;&gt;this Wikipedia article&lt;/a&gt;, I calculated the needed torque to raise a load of 1400&amp;nbsp;N, assuming that's the clay's resistance according to step #2. (I used the raise formula with the stainless steel coefficient of friction to purposely oversize the system a bit to have some overhead). So the terms were:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;F &amp;amp;= \text{load on the screw} &amp;amp;&amp;amp;= 1400\text{ N} \\&#xA;d_m &amp;amp;= \text{mean diameter} &amp;amp;&amp;amp;= 0.01\text{ m} \\&#xA;\mu  &amp;amp;= \text{coefficient of friction} &amp;amp;&amp;amp;= 0.2 \\&#xA;l &amp;amp;= \text{lead} &amp;amp;&amp;amp;= 0.002\text{ m}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{Torque} = \frac{Fd_{m}}{2} \left( \frac{l+ \pi \mu d_{m}}{ \left( \pi d_{m}- \mu l \right)} \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note I also used the formula with the lead angle and angle of friction to be sure of my results for which I used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;\phi &amp;amp;= \text{angle of friction} &amp;amp;&amp;amp;= 11.3° \\&#xA;\lambda &amp;amp;= \text{lead angle} &amp;amp;&amp;amp;= 3.6433°&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Torque} = \frac{Fd_{m}}{2} tan(\Phi + \lambda)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a total result of 1.86&amp;nbsp;Nm of torque (from both formulae!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well that's all nice and gives an answer that I like (a torque of 1.9&amp;nbsp;Nm is quite common for steppers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the fact that I like it doesn't mean that it's correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes me suspicious is that 1400&amp;nbsp;N of force is more or less what's needed to hold up a 140&amp;nbsp;kg man (please bear with me on that... I'm trying to use some &quot;everyday experience&quot; to get to a correct answer) and I can extrude the clay from the cartridge with my bare hand... I'm sure I'm not able to put 140&amp;nbsp;kg on that! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I'm asking is essentially if I followed the right path in tackling this problem. &lt;/p&gt;&#xA;" OwnerUserId="8102" LastEditorUserId="8102" LastEditDate="2016-09-22T15:13:39.697" LastActivityDate="2017-11-17T13:25:43.023" Title="Calculate proper stepper torque to convert clay extruder from compressed air to leadscrew actuator. Is this the correct way?" Tags="|mechanical-engineering|pressure|motors|torque|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11563" PostTypeId="2" ParentId="11552" CreationDate="2016-09-20T21:13:10.843" Score="1" Body="&lt;p&gt;I had to design these once to hold up a conveyor in a farm building. I didn't have that much load on there, so I specified to tighten the 4 bolts to 70% of their ultimate load (as required in CSA S16 for slip-critical bolts) in an X pattern. I then used 75% of the value of the ultimate load in tension for my bolt to find my normal load on the post, just to take into account any creep or fatigue that could arise. With the normal load, the area of my plate and the coefficient of friction, I found how much I could withstand.&lt;/p&gt;&#xA;&lt;p&gt;I guess you should also check the tube for a thin point load if you have a lot of load; this wasn't my case!&lt;/p&gt;&#xA;" OwnerUserId="5691" LastEditorUserId="9448" LastEditDate="2022-11-15T12:07:34.913" LastActivityDate="2022-11-15T12:07:34.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11564" PostTypeId="2" ParentId="11531" CreationDate="2016-09-20T22:22:51.173" Score="1" Body="&lt;h1&gt;I notice a couple of issues with your problem:&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Underspecified:&lt;/strong&gt; The ratio of the exit to the entrance areas is not enough information to provide an answer. You need to provide another property, such as the pressure, temperature, or isentropic efficiency of the diffuser.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;More of a fluids problem than thermodynamics:&lt;/strong&gt; Thermodynamics is more about energy transfer from heat and work, while your problem seems to be more about fluid speed. That's okay though, since fluids and thermo are tightly connected.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Symbolic solution&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Despite the problem being underspecified for a numeric answer, you can still set up the problem, but first you need to state some simplifying assumptions. My solution assumes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Steady state:&lt;/strong&gt; the flow does not vary with time.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Inviscid fluid:&lt;/strong&gt; the air has negligible viscosity, losing negligible energy to friction.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Ideal gas:&lt;/strong&gt; the air behaves as an ideal gas.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Constant potential energy:&lt;/strong&gt; you didn't mention a height change, so the inlet and exit are presumably at about the same height, and thus have the same potential energy.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Density&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Density can be found from the ideal gas law, which you might remember from chemistry as:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P \cdot V = n \cdot \bar R \cdot T$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is pressure, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is temperature, &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; is volume, &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is the number of moles of gas, and &lt;span class=&quot;math-container&quot;&gt;$\bar R$&lt;/span&gt; is the ideal gas constant in units of Joules per mole per Kelvin. In thermo we can replace the &lt;span class=&quot;math-container&quot;&gt;$\bar R$&lt;/span&gt; with a gas constant measured in kilograms that is unique to each gas. For air, &lt;span class=&quot;math-container&quot;&gt;$R_{air}=286.9 ~ J \cdot kg^{-1} \cdot K^{-1}$&lt;/span&gt;. The gas law then becomes:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P \cdot V = m \cdot R_{air} \cdot T$$&lt;/span&gt;&#xA;Recalling the definition of density &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; as the ratio of mass to volume, you can divide both sides by &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; and then rearrange, producing an equation for density:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\rho = \frac{P}{R_{air} \cdot T}$$&lt;/span&gt;&#xA;For the values you gave, &lt;span class=&quot;math-container&quot;&gt;$\rho_i = 1.74 ~ kg \cdot m^{-3}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conservation of mass&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Because the flow is steady, mass flow into the diffuser must equal mass flow exiting:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\dot m_i=\dot m_e$$&lt;/span&gt;&#xA;From fluids, mass flow is the product of density, area, and velocity, so the mass flow equation becomes:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\rho_i \cdot A_i \cdot v_i = \rho_e \cdot A_e \cdot v_e$$&lt;/span&gt;&#xA;You stated that the exit area is twice that of the inlet area, so this equation becomes:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\rho_i \cdot v_i}{2} = \rho_e \cdot v_e$$&lt;/span&gt;&#xA;For the values you gave, &lt;span class=&quot;math-container&quot;&gt;$\rho_e \cdot v_e = 261 ~ kg \cdot m^{-2} \cdot s^{-1}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Compressible Bernoulli Equation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We have changing pressures and velocities along the streamline of an inviscid fluid, so Bernoulli's equation is the next tool we pull out. We need the compressible form since the air density is likely to change quite a bit between inlet and exit. The compressible Bernoulli equation is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{v^2}{2} + \frac{\gamma}{\gamma -1} \cdot \frac{P}{\rho} + g \cdot z = constant$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; is fluid velocity. &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; is the gas's ratio of specific heats, which for air is about 1.4. I'll refer to the &lt;span class=&quot;math-container&quot;&gt;$\gamma/(\gamma - 1)$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;, because it's annoying to type otherwise. The constant I will refer to as &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; (there's no significance to me choosing &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; by the way, I just needed letters and who cares). &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is gravitational acceleration and &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is height. We can neglect &lt;span class=&quot;math-container&quot;&gt;$g \cdot z$&lt;/span&gt; because we are assuming that potential energy is constant. Using these simplifications, we can rewrite the equation into a more manageable form:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{v^2}{2} + \alpha \cdot \frac{P}{\rho} = \beta$$&lt;/span&gt;&#xA;We know the properties at the inlet, so we can calculate beta, which comes out to be &lt;span class=&quot;math-container&quot;&gt;$\beta = 347,000 ~ m^2 \cdot s^{-2}$&lt;/span&gt; for the values you gave.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why it's underspecified&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;At present, we have 2 equations to use:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\rho_e \cdot v_e = \frac{\rho_i \cdot v_i}{2} = 261 ~ kg \cdot m^{-2} \cdot s^{-1}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$and$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{v_e^2}{2} + \alpha \cdot \frac{P_e}{\rho_e} = \beta = 347,000 ~ m^2 \cdot s^{-2}$$&lt;/span&gt;&#xA;Unfortunately, we have 3 unknowns: &lt;span class=&quot;math-container&quot;&gt;$v_e, P_e,$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\rho_e$&lt;/span&gt;. Algebra teaches us that this is a craptacular situation, as the number of unknowns must equal the number of equations in order to solve the system. You might think that we could use our gas equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\rho_e = \frac{P_e}{R_{air} \cdot T_e}$$&lt;/span&gt;&#xA;but unless we define &lt;span class=&quot;math-container&quot;&gt;$T_e$&lt;/span&gt; we've just changed our situation to 3 equations with 4 unknowns.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Wrapping up&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I hope I've been able to help with understanding some of the math. If you're wondering why I said that this is more a fluid dynamics question than a thermo one, notice that I didn't mention entropy, enthalpy, work, that sort of thing. Perhaps I could have used them, but based on the info you provided I didn't see it (maybe with exit temperature and/or pressure, enthalpy could be used).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please correct me anywhere you find a mistake!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;BS in Mechanical Engineering from Virginia Tech (Go Hokies!)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bernoulli&amp;#39;s_principle#Compressible_flow_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia - Bernoulli equation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/individual-universal-gas-constant-d_588.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineering Toolbox - individual gas constant&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8104" LastEditorUserId="6767" LastEditDate="2019-06-02T22:39:35.230" LastActivityDate="2019-06-02T22:39:35.230" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="11565" PostTypeId="2" ParentId="11539" CreationDate="2016-09-21T01:04:56.023" Score="0" Body="&lt;p&gt;Without knowing what fuel system this question originates on, your first point cant be addressed because there are a variety of different types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For points two and three:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A blocked/clogged filter will essentially create a pressure drop in the down stream pipe. The pressure in the pipe under blockage conditions will be based on how the fuel is being delivered. If its under a gravity flow scenario then you could perform some simple calculations to determine the pressure head of the pipe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If its being pumped the pump curve will give you a likely maximum pressure that could be present in the pipe. If this pressure is not higher than the ISO test parameters no damage will occur. &lt;/p&gt;&#xA;" OwnerUserId="8105" LastActivityDate="2016-09-21T01:04:56.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11567" PostTypeId="1" CreationDate="2016-09-21T02:45:21.060" Score="1" ViewCount="85" Body="&lt;p&gt;I am using a noise sensor (ADMP401) along with Arduino Mega. I use FFT to convert the noise sensor analog output to frequency domain, then remove the first 30 bins (to remove some board noise), then calculate the rms value of the remaining bins. So, during a period of 10 seconds (which is my polling period), I have a set of 12 rms values from which I take the median value to get the ambient noise equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its working fine and I can see the output value responding to noises. I usually get an output of around 7 to 10 for silent conditions. The problem is, for some units, this base value of 7 to 10 shifts to around 2 to 3 after running for some time and then will remain at that lower level. Still the sensor will respond to noises. But since this shifting is happening, I am unable to correlate the output to dB levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using openmusiclabs FFT algorithm for arduino. Has anyone experienced this kind of behavior while working with noise sensors and FFT?&lt;/p&gt;&#xA;" OwnerUserId="7929" LastEditorUserId="7083" LastEditDate="2016-09-22T00:56:14.323" LastActivityDate="2016-09-23T13:42:24.360" Title="FFT output for noise sensor shifts after some time" Tags="|electrical-engineering|sensors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11568" PostTypeId="1" AcceptedAnswerId="11572" CreationDate="2016-09-21T04:02:29.673" Score="4" ViewCount="1412" Body="&lt;h1&gt;Introduction&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I have a file with .rpt extension from ABAQUS which is output of a model. This .rpt contains stresses of elements and displacements of nodes in plane text format. There is an item with this file which i was not able to find in ABAQUS manual fluently. I wanted to find out stresses in this . this is original model and it is completely with STRI3 element.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kNTNy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kNTNy.jpg&quot; alt=&quot;Original Abaqus Model&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a section of output file:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xkgpp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xkgpp.png&quot; alt=&quot;.rpt file content&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does give S11, S22 (normals) and S12 (shear) stresses in defined location.&#xA;I believe each element does have 3 points which for integration are used. These points are also known as &quot;Integration Points&quot;. In the text file above it seems each element does have 3 integration points. &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Actual Question&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;What are exact location of integration points of an STRI3 element, if this STRI3 element does have 3 nodes: N1, N2, N3 with known coordinates (X1,Y1,Z1), (X2,Y2,Z2), (X3,Y3,Z3).&lt;/p&gt;&#xA;" OwnerUserId="8108" LastEditorUserId="7461" LastEditDate="2017-05-15T16:07:44.223" LastActivityDate="2017-05-15T16:07:44.223" Title="What are exact location of Integration Points in STRI3 element in ABAQUS?" Tags="|finite-element-method|abaqus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11569" PostTypeId="2" ParentId="11561" CreationDate="2016-09-21T05:37:24.083" Score="8" Body="&lt;p&gt;I spent some time searching on this one, and surprise surprise, Imperial units strike again!&lt;/p&gt;&#xA;&lt;p&gt;The good old &lt;a href=&quot;https://en.wikipedia.org/wiki/Pound_(force)&quot; rel=&quot;noreferrer&quot;&gt;&amp;quot;pounds-force pounds-mass issue&amp;quot;&lt;/a&gt; masked with ounces. This is why any calculation more involved than sizing a pulley should be done in metric, lol.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.orientalmotor.com/technology/articles/motor-sizing-calculations.html&quot; rel=&quot;noreferrer&quot;&gt;Oriental Motor- Basics of motor control&lt;/a&gt;, explains the following:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Units of Measure for Moment Inertia&lt;/p&gt;&#xA;&lt;p&gt;The units of inertia are commonly used in two ways, oz-in&lt;sup&gt;2&lt;/sup&gt; and oz-in-sec&lt;sup&gt;2&lt;/sup&gt;. The former includes gravity, the latter only mass.&lt;/p&gt;&#xA;&lt;p&gt;Theoretically, inertia is factor of mass so it should not include gravity, however, practically we can not easily measure mass on the earth.&lt;/p&gt;&#xA;&lt;p&gt;Oriental Motor commonly provides inertia in oz-in&lt;sup&gt;2&lt;/sup&gt;. Then, when we calculate the Acceleration Torque in Torque Calculation we divide the total the total inertia by the gravity from.&lt;/p&gt;&#xA;&lt;p&gt;Gravity = 386 in/sec&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;oz-in&lt;sup&gt;2&lt;/sup&gt; = Inertia based on weight&lt;/li&gt;&#xA;&lt;li&gt;oz-in-sec&lt;sup&gt;2&lt;/sup&gt; = inertia based on mass&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Calculation for oz-in&lt;sup&gt;2&lt;/sup&gt; to oz-in-sec&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ir2Tn.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ir2Tn.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So an oz-in-sec&lt;sup&gt;2&lt;/sup&gt; &lt;strong&gt;does&lt;/strong&gt; exist because it's an ounce-force not an ounce-mass.&lt;/p&gt;&#xA;&lt;p&gt;A kg-m-s&lt;sup&gt;2&lt;/sup&gt; &lt;strong&gt;does not&lt;/strong&gt; exist, because kg is mass not force. You could convert oz-in-sec&lt;sup&gt;2&lt;/sup&gt; with Newtons, but I wouldn't even go there, lol.&lt;/p&gt;&#xA;&lt;p&gt;The best course of action is to follow Oriental Motor's guide to get to oz-in&lt;sup&gt;2&lt;/sup&gt;. Then convert to metric kg-m&lt;sup&gt;2&lt;/sup&gt; and don't look back ;-)&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-09-21T13:54:55.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11571" PostTypeId="1" AcceptedAnswerId="11574" CreationDate="2016-09-21T08:48:32.060" Score="3" ViewCount="309" Body="&lt;p&gt;Concerning this subject, I consulted the following Wikipedia page: &lt;a href=&quot;https://en.wikipedia.org/wiki/Universal_joint#Equation_of_motion&quot; rel=&quot;nofollow&quot;&gt;Universal joint&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the elementary kinematics equation of the U-joint is stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, a little bit further in the article &lt;a href=&quot;https://en.wikipedia.org/wiki/Universal_joint#Double_Cardan_Shaft&quot; rel=&quot;nofollow&quot;&gt;Double Cardan Shaft&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the configuration described consists of 2 U-joints, the previously mentioned equation is reversed, i.e:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tan \gamma _{1}=\cos \beta \tan \gamma _{2}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tan \gamma _{2}=\cos \beta \,\tan \gamma _{1}\qquad \tan \gamma _{4}=\cos \beta \,\tan \gamma _{3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not intend to doubt the validity of the info on the article, but this hit me as a pretty weird thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these relationships correct?&lt;/p&gt;&#xA;" OwnerUserId="8113" LastEditorUserId="7083" LastEditDate="2016-09-22T00:56:07.013" LastActivityDate="2016-09-22T00:56:07.013" Title="Universal joint vs Double Universal Joint (CV configuration) kinematics" Tags="|automotive-engineering|kinematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11572" PostTypeId="2" ParentId="11568" CreationDate="2016-09-21T12:31:34.197" Score="2" Body="&lt;p&gt;The STRI3 element is a shell element has 3 integration points according to the ABAQUS manual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g654T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g654T.png&quot; alt=&quot;STRI3 Element in ABAQUS&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forget 3D coordinates (global X,Y,Z) for the moment. Just think of the shell's local 2D coordinate system in x,y. Take the isoparametric representation of any triangle as having coordinates (0,0), (1,0) and (0,1) as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iMIim.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iMIim.png&quot; alt=&quot;IsoP Triangle Element&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integration points in ($\xi$, $\eta$) coordinates for this element are:&#xA;$$A = (\frac{1}{6}, \frac{1}{6})$$&#xA;$$B = (\frac{2}{3}, \frac{1}{6})$$&#xA;$$C = (\frac{1}{6}, \frac{2}{3})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your I-beam example you have provided it is probably fairly simple as they all look like right-triangles and match the same shape as isoparametric representation of the triangle. The integration points are 1/6 or 2/3 along the base/height depending on which integration point you're looking at. So you probably wouldn't need to worry about shape functions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If your triangles were not simple right-triangles and lied anywhere in space then you would need to do the following...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can relate the ($\xi$, $\eta$) coordinate system to the ($x$, $y$) coordinate system with these shape functions:&#xA;$$N_1 = 1 - \xi -\eta$$&#xA;$$N_2 = \xi$$&#xA;$$N_3 = \eta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At integration point A we have $\xi = 1/6$ and $\eta = 1/6$. Therefore:&#xA;$$N_1 = 2/3$$&#xA;$$N_2 = 1/6$$&#xA;$$N_3 = 1/6$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if our triangle had ($x$, $y$) vertices of $(0,0)$, $(3,2)$ and $(1,3)$ then the coordinates of A are:&#xA;$$ x_A = \sum N_ix_i = N_1x_1 + N_2x_2 + N_3x_3 = \frac{2}{3}(0) + \frac{1}{6}(3) + \frac{1}{6}(1) = \frac{2}{3}$$&#xA;$$ y_A = \sum N_iy_i = N_1y_1 + N_2y_2 + N_3y_3 = \frac{2}{3}(0) + \frac{1}{6}(2) + \frac{1}{6}(3) = \frac{5}{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Divj5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Divj5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have very cruedly drawn this example above and you can see that the point coordinates $(2/3,5/6)$ match the position for A in the diagram quite well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can repeat this for B and C. Then you would have the coordinates of A, B and C, which are the integration points in the shell's local coordinate system. Then you would have to use a transformation matrix based on the shell's vertices to convert those local x,y coordinates to the global 3D coordinate system in X,Y,Z.&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2016-09-21T12:39:19.047" LastActivityDate="2016-09-21T12:39:19.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11573" PostTypeId="2" ParentId="11571" CreationDate="2016-09-21T13:05:37.570" Score="0" Body="&lt;p&gt;I think the confusion is that, for the double joint, $tan\gamma_2$ is the output angle, not the angle of the intermediate shaft relative to the source $tan\gamma_1$ .  If you hack through the coordinate systems on the way from the first joint to the second joint, it should come out correctly. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-09-21T13:05:37.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11574" PostTypeId="2" ParentId="11571" CreationDate="2016-09-21T15:06:59.997" Score="1" Body="&lt;p&gt;The first equation seems correct, and when it is used again it seems to be incorrectly reversed as you point out. But, it did not matter in the end because the mistake was made twice. With the correct equation we still get the same final result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations&#xA;$$ \tan \left(\gamma _1\right)=\cos (\beta ) \tan \left(\gamma _2\right) $$&#xA;$$ \tan \left(\gamma _3\right)=\cos (\beta ) \tan \left(\gamma _4\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;together with $\gamma _3=\gamma _2+\frac{\pi }{2}$ and $\tan \left(\gamma +\frac{\pi }{2}\right)=\frac{1}{\tan (\gamma )}$, gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan \left(\gamma _4\right) = \frac{1}{\cos (\beta&#xA;   )} \tan \left(\gamma _3\right) =&#xA;\frac{\tan \left(\gamma _2\right)}{\tan \left(\gamma _1\right)}\frac{1}{\tan \left(\gamma _2\right)}=&#xA;\frac{1}{\tan \left(\gamma _1\right)}=&#xA;\tan \left(\gamma _1+\frac{\pi }{2}\right) $$&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-09-21T15:06:59.997" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11577" PostTypeId="1" AcceptedAnswerId="11580" CreationDate="2016-09-21T19:01:15.747" Score="-1" ViewCount="80" Body="&lt;p&gt;I am trying to get a better understanding of measurements and instrumentation before starting an online class next week. I have been reading a text book and came across a concept question I am stuck on a little bit. It was this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Design a measurement system that could read temperature, output it to a  and regulate the temperature of a room. You have a sensor, space heater, and a computer.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another similar one was: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How would you get the temperature of an oven to a digital output.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know some of the parts. The sensor would be a sort of thermocouple. You would filter the input and sample it. Do some sort of analog to digital conversion... There are obviously details and parts I'm missing. Putting all of this together is kind of killing my brain though as I just don't get it yet. I would love if someone had the know-how to break it down a little and help me grasp this a little better. &lt;/p&gt;&#xA;" OwnerUserId="8120" LastEditorDisplayName="user16" LastEditDate="2016-09-22T00:31:41.033" LastActivityDate="2019-05-13T00:18:37.553" Title="Designing a simple measurement system" Tags="|mechanical-engineering|measurements|signal-processing|" AnswerCount="1" CommentCount="0" ClosedDate="2016-09-26T11:37:12.940" ContentLicense="CC BY-SA 3.0" />
  <row Id="11578" PostTypeId="1" CreationDate="2016-09-21T19:53:40.317" Score="3" ViewCount="158" Body="&lt;p&gt;We are currently using a liquid vacuum filler machine that vacuum fills nail polish. There is a Gast 1/4 horsepower motor and a Gast rotary vane pump that according to the gast website pulls 20&quot; of vacuum at a flow volume of 4CFM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to know if it is possible to switch to a Venturi vacuum pump to accomplish the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are four filling heads on the machine.&lt;/p&gt;&#xA;" OwnerUserId="8121" LastEditorUserId="1832" LastEditDate="2016-09-21T20:04:08.350" LastActivityDate="2020-08-02T17:08:10.787" Title="Liquid Vacuum Filler machine using Venturi Pump" Tags="|mechanical-engineering|fluid-mechanics|vacuum|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11579" PostTypeId="2" ParentId="11578" CreationDate="2016-09-21T21:27:31.477" Score="1" Body="&lt;p&gt;Yes, you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out this manufacturers table for reference: &lt;a href=&quot;http://www.2linc.com/venturi_vacuum.htm&quot; rel=&quot;nofollow&quot;&gt;Air Venturi Vacuum Generator Pump for Workholding, Fixturing and Clamping&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the 1/4 npt sized part you can achieve 22-26&quot;Hg at 4.5 scfm vacuum flow which requires 70-80psi compressed air input at 5scfm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, your liquid would need to be delivered before flowing into the venturi. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trade off is compressed air supply for an electric motor. That's probably more of a financial/business/equipment decision that's out of this question's current scope. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="7083" LastEditDate="2016-09-22T00:55:47.327" LastActivityDate="2016-09-22T00:55:47.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11580" PostTypeId="2" ParentId="11577" CreationDate="2016-09-21T22:30:34.767" Score="0" Body="&lt;p&gt;Those are extremely vague questions, so you could go to varying levels of detail. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;This is a feedback control question. The short answer: connect temperature sensor and space heater to computer in a room Develop a control method/program on computer to read temperature and switch on/off heater as control method determines necessary. There are many control methods that can be used. Check out this basic Wikipedia article: &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Temperature_control&quot; rel=&quot;nofollow noreferrer&quot;&gt;Temperature control&lt;/a&gt; I suggests you read about P, PI, and PID control. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here's another interesting and light article: &lt;a href=&quot;https://www.instrumart.com/pages/283/temperature-controller-basics-handbook&quot; rel=&quot;nofollow noreferrer&quot;&gt;Temperature Controller Basics Handbook&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Find a suitable thermocouple that can withstand the temperatures of the oven. I like to over-rate. Probably a K type with 800F minimum rating. Either get an analog or digital thermocouple amplifier chip. For example, I suggest a max31855k digital output. Connect that and some type of display to a microcontroller an read the sensor once every second and write the value to the display. The Internet has plenty of examples and sample code..especially for Arduino. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I suggest you get an Arduino and some thermocouple kits and lcd kits... which you can probably get it all for around $30(or less!) and you can make a temperature controlled thing. See the code and how it works. Experiment. It will give you a much better understanding of how these things interact. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="6767" LastEditDate="2019-05-13T00:18:37.553" LastActivityDate="2019-05-13T00:18:37.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11581" PostTypeId="1" CreationDate="2016-09-22T01:21:15.187" Score="4" ViewCount="1168" Body="&lt;p&gt;In qualifying the results of a Finite Element (FE) beam  model, I'm realizing that I dont understand FE beam elements as well as I should.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there a procedure to determine what order beam element is required?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Specifically, for the 2D tip-loaded cantilever beam shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3U1fy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3U1fy.jpg&quot; alt=&quot;Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analytical solution given by &lt;strong&gt;Euler-Bernoulli Beam Theory&lt;/strong&gt; is used to qualify my FE results, where:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;$y(x) = -\frac{F}{6EI_{zz}}(3Lx^{2}-x^{3}) \; \rightarrow \; \therefore y(x=L) = -\frac{FL^{3}}{3EI_{zz}}$&lt;/li&gt;&#xA;  &lt;li&gt;$\sigma_{xx} = \pm \frac{M(x)c}{I_{zz}}$&lt;/li&gt;&#xA;  &lt;li&gt;$\tau_{xy} = \frac{F}{2I_{zz}}(\frac{h^{2}}{2I_{zz}}-y^{2}) \; \rightarrow \; \therefore \tau_{xy}(y) = \frac{3}{2}\frac{V}{A}$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am testing &lt;a href=&quot;http://www.ansys.stuba.sk/html/elem_55/chapter4/ES4-188.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Beam 188&lt;/a&gt; elements in ANSYS using Linear (2 Node) and Quadratic (3 Node) beam elements.  For both linear and quadratic elements, I increment the number of elements required until the FE model convergences with the analytical (Euler-Bernoulli) solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For the single element, linear Beam 188 solution:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/diEFS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/diEFS.png&quot; alt=&quot;b188L-1E&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;The maximum displacement is ~7% less than the analytic solution!&lt;/li&gt;&#xA;  &lt;li&gt;The maximum bending stress is 50% less than the analytic solution!&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I thought that the stiffness matrix for a linear beam element was formulated from the &lt;a href=&quot;http://www.efunda.com/formulae/solid_mechanics/beams/theory.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;differential beam equations&lt;/a&gt;, making it an exact solution for a beam subjected to point loads.  As I understand it, higher order elements are required for distributed loading,curved beams, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the stress gradient cannot be captured without additional elements, but is it possible to calculate stress along the beam from the beam equations and interpolated (shape function) values? &lt;strong&gt;Still, are the nodal displacements accurate for a single element?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are linear beam element solutions ever exact and when are they used in practice?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="2538" LastEditorUserId="2538" LastEditDate="2016-09-22T04:48:03.717" LastActivityDate="2020-09-02T07:57:11.690" Title="Are linear beam element solutions ever exact and when are they used in practice?" Tags="|beam|finite-element-method|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11583" PostTypeId="1" CreationDate="2016-09-22T12:09:12.810" Score="0" ViewCount="1367" Body="&lt;p&gt;I want to know whether the torque required by motor depends on load. If it does, then how can I calculate it?&lt;/p&gt;&#xA;" OwnerUserId="8128" LastEditorUserId="7083" LastEditDate="2016-09-26T23:57:25.107" LastActivityDate="2017-02-24T18:54:43.080" Title="How to calculate torque required by a motor depending on load?" Tags="|motors|torque|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11584" PostTypeId="2" ParentId="11581" CreationDate="2016-09-22T14:20:48.230" Score="0" Body="&lt;p&gt;You have to understand that discretization with both linear and quadratic elements converge in the limit.  The accuracy you acheive depends on both the order and size of the element.  More elements should result in greater accuracy.  You will never acheive &quot;exact&quot;, analytical results.  The only way to tell if your solution makes sense is to keep refining your mesh and verifying that the error keeps getting smaller (watch out, and make sure you don't refine the mesh too much though... this can lead to loss of precision due to numerical subtraction of nearly equal values from assembling an fem system from mesh points that are too close to each other).&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2016-09-22T14:20:48.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11585" PostTypeId="1" AcceptedAnswerId="11591" CreationDate="2016-09-22T16:11:26.437" Score="1" ViewCount="38" Body="&lt;p&gt;I'm interested in bringing in a GHT thread compatible product, but my sample's diameter was slightly smaller. GHT has a large (pitch to pitch) diameter of 1.0625 inch. The sample I received has a diameter of 1.0415 inch on the male side, and it doesn't leak and screws in perfectly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this diameter difference meaningful for the long term reliability of my product?&lt;/p&gt;&#xA;" OwnerUserId="8130" LastActivityDate="2016-09-22T19:08:36.840" Title="Plumbing / Thread Compatibility" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11586" PostTypeId="1" CreationDate="2016-09-22T16:18:11.087" Score="0" ViewCount="178" Body="&lt;p&gt;In a power electronics converter is there any difference between reactive power input to the converter and reactive power consume by converter? (In case source current waveform is square in nature).&#xA;If yes then how we can calculate the difference between the two reactive powers?&lt;/p&gt;&#xA;" OwnerUserId="8131" LastActivityDate="2017-05-22T08:28:38.203" Title="Difference between Reactive power input and Reactive power consume by Load?" Tags="|power-electronics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11587" PostTypeId="1" AcceptedAnswerId="11593" CreationDate="2016-09-22T16:46:15.580" Score="1" ViewCount="1103" Body="&lt;p&gt;I am currently designing a motorized zipline trolley that will carry a person on a horizontal cable, so I am trying to calculate the friction on the wheels to determine what kind of motors I need. I initially approached the problem using a static friction model using the coefficient of static friction between two steels. However, my professor told me to replace my model with a rolling friction model, so I took the coefficient of rolling friction between two steels. This greatly reduced the calculated value of friction; however, I feel like there should be more friction because the weight of the person will cause deflection in the cable, so a portion of the wheels' circumference will be in contact with the cable. I want to assume no slipping for the wheels, so I don't know how to account for the static friction while keeping the rolling friction model. How can I calculate the additional frictional force that I am not modeling?&lt;/p&gt;&#xA;" OwnerUserId="2804" LastActivityDate="2016-09-23T02:16:46.873" Title="How do I model the friction on a rolling pulley wheel, which holds weight, on a horizontal cable?" Tags="|friction|wheels|pulleys|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11588" PostTypeId="1" AcceptedAnswerId="11599" CreationDate="2016-09-22T17:08:29.327" Score="3" ViewCount="7408" Body="&lt;p&gt;If you have a saturated steam (quality=1.0) volume  that is compressed adiabtically will the steam condense to follow the saturation curve or superheat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found in an old textbook: &lt;a href=&quot;https://books.google.com/books?id=o5pBAAAAIAAJ&amp;amp;pg=PA444&amp;amp;lpg=PA444&amp;amp;dq=compressing+saturated+steam&amp;amp;source=bl&amp;amp;ots=62g5CGqi45&amp;amp;sig=Ox5v-8gszNyoHnRrnU73gjVJhk4&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwiC3qzmvqPPAhUUImMKHVEHD00Q6AEIUzAH#v=onepage&amp;amp;q=compressing%20saturated%20steam&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;The Principles of Thermodynamics (1899)&lt;/a&gt; on Google the following passage:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For compression, adiabatic:&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;If we start with pure saturated steam, without admixture of water, it will be superheated by compression.&lt;/li&gt;&#xA;  &lt;li&gt;If the initial steam weight is greater than that of the water, steam is generated by the compression.&lt;/li&gt;&#xA;  &lt;li&gt;If there is more water than steam, steam is condensed during compression.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am still at a bit of a loss for this makes physical sense and how this is properly captured in an energy balance. (i.e., $\frac{\text{d}U}{\text{d}t} = -\frac{p\text{d}V}{\text{d}t}$)&lt;/p&gt;&#xA;" OwnerUserId="5432" LastEditorUserId="1832" LastEditDate="2016-09-22T17:50:25.697" LastActivityDate="2016-09-23T13:49:56.430" Title="Compressing saturated steam" Tags="|thermodynamics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11591" PostTypeId="2" ParentId="11585" CreationDate="2016-09-22T19:08:36.840" Score="0" Body="&lt;p&gt;GHT relies on a gasket to seal, not a tight fit between the threads. This makes it more resistant to dirt and also cheaper to manufacture. I would guess that your sizes will work fine and you shouldn't be concerned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the sizes you measured, the radial clearance is only .0105 (ten thou.) That might be a little loose for precision threads, but anything smaller than about 2 thou clearance becomes very difficult to assemble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All that said, the Machinery's Handbook has information on hose threads and gives some tolerances. It's on page 1873 (at least in the 28th edition.) Depending on the material and processes, they list a minimum major diameter for the male thread as low as 1.035. &lt;a href=&quot;http://machiningproducts.com/html/NPSH-NH-NHR-Thread-Dimensions.html&quot; rel=&quot;nofollow&quot;&gt;This&lt;/a&gt; website also lists the appropriate dimensional information in case you don't have the handbook. The official standard is &lt;a href=&quot;https://www.asme.org/products/codes-standards/b1207-1991-hose-coupling-screw-threads-inch&quot; rel=&quot;nofollow&quot;&gt;ANSI/ASME B1.20.7&lt;/a&gt;. If you are specifying compatible threads for a production run, you should probably consider buying it or at least referencing it on your drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On another type of thread, like NPTF where the threads are the sealing component, tolerances will be tighter. Fittings that have a tapered or flared metal-to-metal surface to seal usually have thread tolerances somewhere in the middle of these two.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-09-22T19:08:36.840" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11592" PostTypeId="2" ParentId="11581" CreationDate="2016-09-22T20:15:57.540" Score="1" Body="&lt;p&gt;I'm not familiar with ANSYS, but the basic reason would seem to be that the ANSYS beam is &lt;em&gt;not&lt;/em&gt; exactly the same formulation as the OP's version of Euler beam theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many FE systems have beam elements formulated using Euler-Timoshenko beam theory which includes some additional terms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP's link to the documentation implies this, in the quote&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These elements support an elastic relationship between transverse&#xA;  shear forces and transverse shear strains. Users may specify&#xA;  transverse shear stiffnesses using real constants.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If your input generator allows you to define the section as a geometric shape (i.e. a rectangle) the relevant factors will probably be calulcated automatically. There will probably be some way to switch the extra terms off, if you dig deeper into the documentation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, the extra terms  should make the beam more flexible, i.e. the tip deflection of the FE model should have been bigger than the Euler beam theory value, not smaller as the OP reported. Maybe there are other differences in the formulation, or the OP did a non-linear analysis instead of a linear one, or whatever....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the &quot;50% discrepancy between the maximum bending stress&quot;, the OP's plot seems to show constant stress along the length of the beam which is obviously wrong. This may be an artefact of the way ANSYS presents the output, and the plotting program displays it. The stress at the &lt;em&gt;mid-point&lt;/em&gt; along the length of the beam should of course be 50% of the stress at the fixed end. You always need to take care to understand &lt;em&gt;exactly&lt;/em&gt; what data your &quot;pretty colored and/or animated output pictures&quot; are based on, before you actually believe them! &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-09-22T20:15:57.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11593" PostTypeId="2" ParentId="11587" CreationDate="2016-09-23T02:16:46.873" Score="1" Body="&lt;p&gt;There are a two items I think you are wondering about, the effect of static friction and the extended length of the contact area between the rope and the pulley. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start up torque on a motor can be greater than the design torque of the motor, so your professor may be suggesting to ignore the rolling friction because of that. I would say it's a valid assumption to assume the motor can overcome the static friction if it's overcoming the dynamic friction. Check out motor-torque curves on engineering toolbox for example. I can't remember the complete explanation and terms for this, but when you start a motor there is not a strong magnetic field produced by the rotor, because it is not moving yet, and so the magnetic field that is being produced by the AC current in the stator gets stronger and produces more torque than when the rotor is spinning at its nominal rating. That's why you have high inrush current when starting a motor, no matter what (assuming you have no starter controls on it). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the contact area of the rope on the pulley is irrelevant. You have assumed no slip for starters, so the total frictional force between the rope and the pulley that would be altered by contact length is not going to change. Draw a free body diagram of the pulley. You will have to have tension in the rope between the two pulleys, and you can theoretically make that tension to whatever you want. The tension in the rope (due to the tension you set as well as the mass of the trolley load) will act as the torque on the pulley of the motor. That is the torque the motor has to overcome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be some deflection in the rope due to the load, which you can solve for. Then it's just the added tension in the rope from the load that the motor has to account for since the rope tension is always perpendicular to the pulley. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break that out into components to and balance then solve for the new tension. I'm on my phone right now and can't enter the equation. But I will try once I have access. &lt;/p&gt;&#xA;" OwnerUserId="1143" LastActivityDate="2016-09-23T02:16:46.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11594" PostTypeId="2" ParentId="11583" CreationDate="2016-09-23T02:21:00.603" Score="1" Body="&lt;p&gt;Yes you can find required torque of motor if you know the load. Torque created by the load acts on the motor shaft and the motor shaft must overcome that torque to operate, otherwise it would cease to rotate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For no load conditions there is a no load amp draw. This represents the internal losses of the motor (bearing friction, major and minor losses) that, at a minimum, the motor must overcome to operate at its design speed. &lt;/p&gt;&#xA;" OwnerUserId="1143" LastActivityDate="2016-09-23T02:21:00.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11595" PostTypeId="2" ParentId="11588" CreationDate="2016-09-23T08:39:55.510" Score="4" Body="&lt;p&gt;Take a look on this T-V water diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oe9TA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oe9TA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;If we start with pure saturated steam, without admixture of water, it&#xA;  will be superheated by compression.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Suppose we have a steam located at point C or G or K where dryness factor = 1 (no water), further compressing/pressurizing of the steam will make the substance end up in the superheated vapor region (outside the saturation dome).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;If the initial steam weight is greater than that of the water, steam is generated by the compression.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we have a point between point B and C at P = 0.1 MPa with a dryness factor of 0.8 (near point C) for example meaning that we have greater amount of vapor than water, further compressing (say to FG line, 1 Mpa) will lead us to a point that could be very close to saturated vapor line (higher dryness factor than 0.8) or in the superheated region, which means steam (more precisely vapor) is generated in both cases.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;If there is more water than steam, steam is condensed during&#xA;  compression.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is the opposite situation of case 2, if we have a point near B, say having dryness factor of 0.2, further compressing (say to FG line, 1 Mpa) will lead us to a point on the FG line with dryness factor that is lower than 0.2, meaning less vapor content (condensation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope all is clear now.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2016-09-23T08:39:55.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11597" PostTypeId="1" CreationDate="2016-09-23T12:33:27.777" Score="0" ViewCount="194" Body="&lt;p&gt;In Automating Manufacturing systems by Hugh Jack, Chapter 11: Flowchart based design, 11.6: practice problems, question number 3. I've attached a photo of the book's solution (link 1) and an other one of my own solution (link 2). My problem with the book's solution is that after the start block, assuming the door is already closed, then the user would have to press the button twice according to the flowchart for the door to open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Design a garage door controller using a flowchart. The behavior of the garage door controller is as follows, - there is a single button in the garage, and a single button remote control. - when the button is pushed the door will move up or down. - if the button is pushed once while moving, the door will stop, a second push will start motion again in the opposite direction. - there are top/bottom limit switches to stop the motion of the door. - there is a light beam across the bottom of the door. If the beam is cut while the door is closing the door will stop and reverse. - there is a garage light that will be on for 5 minutes after the door opens or closes.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link 1:&lt;a href=&quot;http://www.4shared.com/photo/YPA5d2kYce/image.html&quot; rel=&quot;nofollow&quot;&gt;http://www.4shared.com/photo/YPA5d2kYce/image.html&lt;/a&gt;&#xA;Link 2:&lt;a href=&quot;http://www.4shared.com/photo/9yyZRsj8ba/image.html&quot; rel=&quot;nofollow&quot;&gt;http://www.4shared.com/photo/9yyZRsj8ba/image.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right? Or what am I missing? Do I have to assume a certain position of the object whenever faced with such type of questions?&lt;/p&gt;&#xA;" OwnerUserId="8146" LastActivityDate="2016-10-23T22:53:07.467" Title="Flowcharts in PLCs, garage door design problem" Tags="|electrical-engineering|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11598" PostTypeId="2" ParentId="11567" CreationDate="2016-09-23T13:42:24.360" Score="1" Body="&lt;p&gt;I figured out something strange while testing this. We tested the units in a reverberation chamber which is almost completely sound insulated for almost close to 24 hours and the base range never shifted to lower level. Also, the values were very stable. Took it back again to our deployment place and the shift happened after around 2 hours. The values are unstable too; it's moving up and down in small levels. (All the test done with no noise, may be the air conditioner and a PC humming noise was there in the background). Does that mean that the noise sensor is behaving differently in different environments; may be picking up some frequencies ? Clueless. &lt;/p&gt;&#xA;" OwnerUserId="7929" LastActivityDate="2016-09-23T13:42:24.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11599" PostTypeId="2" ParentId="11588" CreationDate="2016-09-23T13:49:56.430" Score="3" Body="&lt;p&gt;An ideal adiabatic compression of steam is a reversible process. This means that on the state of the fluid will follow lines of constant entropy. An alternative view is this is the opposite of an ideal turbine where the initial assumption is ds = 0 (adiabatic expansion in reverse).&lt;/p&gt;&#xA;&lt;p&gt;For the T-s diagram constant entropy will put you at steam or liquid depending on the condition of the steam-water.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If x &amp;gt; 0.5 then compression will lead to steam&lt;/li&gt;&#xA;&lt;li&gt;If x &amp;lt; 0.5 then compression will lead to liquid&lt;/li&gt;&#xA;&lt;li&gt;If x ~ 0.5 then compression will stay at roughly 50/50 mixture until supercritical water conditions are met (the maximum of the saturation curve).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Or for the Mollier P-h diagram. If you start at &amp;quot;B&amp;quot;, compressing the fluid will follow the red line and go to superheated steam.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qvzu6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qvzu6.jpg&quot; alt=&quot;Temperature-Entropy Diagram for Water&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LZjil.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LZjil.jpg&quot; alt=&quot;Mollier Pressure-Enthalpy Diagram of Water&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5432" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-09-23T13:49:56.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11602" PostTypeId="1" AcceptedAnswerId="11604" CreationDate="2016-09-24T00:59:01.567" Score="2" ViewCount="2764" Body="&lt;p&gt;The theoretical formula for my beam deflection is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$  v(x) =&#xA;\begin{cases}&#xA;-\frac{Px}{48EI}(3L^2-4x^2),  &amp;amp; 0\le x \le {L\over2}\\&#xA;-\frac{P(x-L)}{48EI}(L^2-8Lx+4x^2), &amp;amp; {L\over2}\lt x \le L&#xA;\end{cases}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to derive the formulae for the slope $v'$, curvature $v''$ and $v'''$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this just a simply case of taking the first, second, and third derivative of the original equation? I know absolutely nothing about structural beams...&lt;/p&gt;&#xA;" OwnerUserId="7342" LastActivityDate="2016-09-24T02:31:48.050" Title="Help an electrical engineering student with a beam deflection question" Tags="|mechanical-engineering|structural-engineering|mathematics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11603" PostTypeId="2" ParentId="11586" CreationDate="2016-09-24T02:02:09.887" Score="1" Body="&lt;p&gt;Not sure if this is more of a comment or an answer but I hope it helps. Reactive power is never really consumed, so it's a misleading term. In inductive loads the reactive power goes into the magnetic field that is constantly switching directions. (like in electric motors for example). In capacitive loads the reactive power goes into the magnetic field between the two plates (like in capacitor banks). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I deal more with buildings and large equipment, rather than the electronic components, but an electrical utility company must size the feed into the building to accommodate the real power and the reactive power, which is by definition the apparent power (vector sum of real and reactive power). But the reactive power is constantly being fed back and forth because the magnetic fields are changing directions according to the frequency of the electricity provided. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I believe they are the same thing. The reactive power input is the same as consumed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With real power it's different because it's tangible. A resistive electric heater turns the electricity into heat so it's easier to conceptualize that it's 'consumed' but the power input is equal to the power consumed one way or another. But with real power you can't send it back. It's already been spent as heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrical utility companies will charge you based on your electricity consumption (units of kWh typically, which is a unit of energy) AND they will charge you based on your peak power draw because the more reactive load you have, the larger they have to size their equipment to accommodate BUT they don't charge you for consumption of reactive power because that does not occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it's a different application, but the general concept holds true and it really helped me to understand he difference. &lt;/p&gt;&#xA;" OwnerUserId="1143" LastActivityDate="2016-09-24T02:02:09.887" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11604" PostTypeId="2" ParentId="11602" CreationDate="2016-09-24T02:31:48.050" Score="4" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first derivative of the deflection is equal to the tangent of the deflection, which for small deflections can be approximated as equal to the angle of rotation of the beam at each point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second derivative (times $EI$) is the bending moment along the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third derivative (times $EI$) is the shear force along the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth derivative (times $EI$) is the distributed load along the beam.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-09-24T02:31:48.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11605" PostTypeId="2" ParentId="11491" CreationDate="2016-09-24T02:42:13.323" Score="1" Body="&lt;p&gt;This is the methodology for calculating the power. I do not agree with Carl Witthofts comment because it does not take into account the major and minor losses or the potential energy (from height difference), and I think he is acknowledging that by saying oops.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find your flow rate (velocity times area x 2 for each hose!)&lt;/li&gt;&#xA;&lt;li&gt;Find your major loss head(from the length of both hoses).&lt;/li&gt;&#xA;&lt;li&gt;Find your minor loss head(from the nozzle and whatever other components there are, if any).&lt;/li&gt;&#xA;&lt;li&gt;Find your head from the height difference of 3 m.&lt;/li&gt;&#xA;&lt;li&gt;Add heads from 2, 3, and 4 together to get your total pressure.&lt;/li&gt;&#xA;&lt;li&gt;Fluid Power is equal to your total flow times your total pressure drop from (5)&lt;/li&gt;&#xA;&lt;li&gt;Your motor input power is equal to Fluid Power/0.7 (efficiency)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To account for the 20% maximum hose power loss portion,&#xA; 8. Isolate your major losses (hose losses) only and calculate the power to overcome those losses only.&#xA; 9. Take that power and compare it to the power calculated in 7.&#xA; 10. If it's less than 20% of the power, you're good! If not, then increase the power of the motor until your power to overcome the hose losses is equal to 20% of the motor power.&lt;/p&gt;&#xA;" OwnerUserId="1143" LastActivityDate="2016-09-24T02:42:13.323" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11606" PostTypeId="1" CreationDate="2016-09-24T07:06:25.433" Score="0" ViewCount="591" Body="&lt;p&gt;Is it possible to boil the water simply using vacuum pump?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there any chances of ice formation?&lt;br&gt;&#xA;If yes, then how we can calculate the heat given to the water to avoid ice formation?&lt;/p&gt;&#xA;" OwnerUserId="8131" LastEditorUserId="2819" LastEditDate="2016-09-25T18:46:46.373" LastActivityDate="2016-09-26T08:21:38.077" Title="Evaporation of water using a vacuum pump?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11607" PostTypeId="1" CreationDate="2016-09-24T08:41:27.923" Score="3" ViewCount="6609" Body="&lt;p&gt;Please explain with theoretical concepts and a brief example.&lt;/p&gt;&#xA;" OwnerUserId="8151" LastEditorUserId="1832" LastEditDate="2016-09-24T10:18:39.213" LastActivityDate="2023-01-22T05:20:05.270" Title="What is the difference between tolerance and uncertainty in measurement?" Tags="|mechanical-engineering|applied-mechanics|measurements|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11608" PostTypeId="2" ParentId="11607" CreationDate="2016-09-24T09:57:58.070" Score="5" Body="&lt;p&gt;Tolerance is a property of a design. Tolerance is a acceptable deviation from plan that is designed as part of your acceptance criteria. You document this in your drawing/contract either as a general tolerance or specific demand for one feature in your part. If a part meets your tolerance then you have agreed to accept it as is if not then the part is rejected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rejection usually leads to making a new part, fixing it or in some situations you have quality classes and use lower quality products for less demanding jobs. Later is typical in semiconductor industry and lcd panel manufacturing. Specifying tighter tolerances means that it takes more effort to produce the components leading to higher cost. On the otherhand you pay more for parts but you may gain this back in turn by having less problems in assembly and use. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Measurement uncertanty is a property of physics, and measurement. Things change all the time and on the lowest level this is bounded by Heisenberg uncertainty principle, but you need not go this deep to notice it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a simple experiment of measuring the length of a pen. No matter how many times you try you always get slightly different results because of how your experiment is set up. There are ways to make your experiment better, but as you increase accuracy your setup becomes more combersome to conduct. Imagine how much a 1 degree temperature difference affects the length of your pen, or how much the movement of your house due to traffic affects it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measurement uncertainty is just a intervall that tells how reliable your measurement result is. Its not part of your design but part of your measurement apparatus. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Measurement uncertainty is not linked to tolerance although the need for tolerance is caused by much the same reasons as measurement uncertainty. Tolerance is usually quite much higher than measurement  uncertainty, even if some tolerances are very hard to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note if you put very stringent tolerances on your design then you also need to verify those. So you need better equipment and better planned experiments to verify this. And if you do not then what is the purpose of having such a high tolerance?&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2016-09-24T14:24:25.617" LastActivityDate="2016-09-24T14:24:25.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11609" PostTypeId="2" ParentId="10788" CreationDate="2016-09-24T13:49:18.000" Score="1" Body="&lt;p&gt;One technique that might work is to use velcro and 3M Very High Bond Tape (sticky on both sides) to adhere the velcro strips to the circuit board and the vibrating member. This will give you the advantage of having the board easily removable. Try an experiment with this and see if you can shake the board loose, VHB tape is Very sticky, so much so that 3M classifies it as a mechanical fastener , not just plain old tape ;-)&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2016-09-24T13:49:18.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11611" PostTypeId="1" AcceptedAnswerId="11612" CreationDate="2016-09-24T21:41:43.437" Score="2" ViewCount="684" Body="&lt;p&gt;I recently &lt;a href=&quot;http://m.gloucestershirelive.co.uk/elmbridge-court-burger-plans-on-track/story-29745204-detail/story.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;read&lt;/a&gt; about a newly proposed 'burger' lane across a roundabout.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r2ybd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r2ybd.png&quot; alt=&quot;burger lane&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;From what I can see,  it bypasses the main point of a roundabout, a continuous circle of traffic with well defined rights of way. The only reason I can imagine one is to stop a tributary to the roundabout clogging the other entrances,but I don't understand how that would be achieved.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;What's the purpose of a 'burger' lane in a roundabout?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1500" LastEditorUserId="1500" LastEditDate="2022-02-25T16:58:29.983" LastActivityDate="2022-03-23T12:05:20.233" Title="What's the purpose of a 'burger' lane in a roundabout?" Tags="|civil-engineering|transportation|traffic-intersections|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="11612" PostTypeId="2" ParentId="11611" CreationDate="2016-09-25T00:44:54.060" Score="2" Body="&lt;p&gt;There are several of these near where I live (in the UK) but I've never heard the term &quot;burger lanes&quot; before. We don't seem to have a name for them - they just are!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They serve the same basic function as a roundabout with a flyover bridge, except that all the roads are on the same level. The intersections of the burger lane and the roundabout are controlled by traffic lights.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have an advantage where the main traffic flow is different at different times of day (e.g. during the commuting rush hours) - for example traffic can move faster across the &quot;burger lane&quot; than negotiating the sharper curves entering and leaving the roundabout itself, if the roundabout is relatively small and there is no space available to enlarge it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example here's one on Google Maps where there is a &quot;burger lane&quot; on the north-east bound carriage way of the dual-carriageway A52 (to avoid driving round most of the roundabout) but not on the south-west bound carriageway, which only passes one intersection on the roundabout - and that road, &quot;Town St,&quot; has a negligible amount of traffic compared with the A6007 &quot;Ilkeston road&quot; on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.uk/maps/@52.9359805,-1.2496093,332m/data=!3m1!1e3&quot; rel=&quot;nofollow&quot;&gt;https://www.google.co.uk/maps/@52.9359805,-1.2496093,332m/data=!3m1!1e3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-09-25T00:56:10.167" LastActivityDate="2016-09-25T00:56:10.167" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11613" PostTypeId="1" CreationDate="2016-09-25T00:49:11.057" Score="2" ViewCount="727" Body="&lt;p&gt;From what I understand a smoke alarm works by emitting some sort of radioactive particle and then counting how many of these particles hit a receiver. When there is smoke in between, it blocks some of the particles and so the alarm goes off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it gets older the radioactive particles per minute that are being released will reduce, because some of it has already been emitted and so there's less of it to emit. (half life and such).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that when it get's older and emits less particles that it becomes more sensitive?&lt;/p&gt;&#xA;" OwnerUserId="8155" LastActivityDate="2016-09-25T11:41:24.683" Title="Do smoke alarms become more sensitive as they get older?" Tags="|electrical-engineering|consumer-electronics|radiation|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11614" PostTypeId="1" AcceptedAnswerId="11620" CreationDate="2016-09-25T01:02:12.057" Score="3" ViewCount="1389" Body="&lt;p&gt;It's common in submarine movies that when suffering some kind of shock (like a near miss by a torpedo or falling deep into a trench), water will start bursting into the cabin and the crew will run around turning little wheels or cranks to stop the leaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this realistic? What are those wheels &lt;em&gt;doing&lt;/em&gt;? Is it even plausible for a submarine to leak without a catastrophic failure?&lt;/p&gt;&#xA;" OwnerUserId="8156" LastEditorUserId="6790" LastEditDate="2020-05-04T08:02:42.200" LastActivityDate="2022-03-28T07:36:08.853" Title="Do submarines really have cranks for closing leaks?" Tags="|mechanical-engineering|naval-engineering|submarines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11615" PostTypeId="2" ParentId="11613" CreationDate="2016-09-25T01:10:27.067" Score="3" Body="&lt;p&gt;The isotope used in smoke detectors, americium-241, has a half-life of 432 years. It would take a very long time for simple radioactive decay to affect the operation of a smoke detector.&lt;/p&gt;&#xA;" OwnerUserId="8156" LastActivityDate="2016-09-25T01:10:27.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11619" PostTypeId="2" ParentId="11613" CreationDate="2016-09-25T11:41:24.683" Score="2" Body="&lt;p&gt;That isn't how they work at all. The smoke does not &quot;block&quot; the radiation in that way. And even if it did, a decay in the rate of emission would make the detector &lt;em&gt;more&lt;/em&gt; sensitive, not less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory is that uncontaminated air is relatively hard to ionize, while air with smoke in it is easily ionized, for example, by an alpha particle emitted by a small sample of americium-241. Therefore, the heart of the detector is an ionization cell that has two electrodes that have a voltage bias across them, and the americium is located inside this cell. If there is smoke in the air, the radiation will ionize it, and there will be a measurable current between the two electrodes in the form of pulses created by the ionization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average current is proportional to the rate at which the americium emits particles, so it will go down as the americium decays, but as Brian Gordon says, the half-life is very long (432 years), so it would have no real effect over the useful lifetime of the detector as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the current peaks caused by individual ionization events will always have the same amplitude, so the actual &quot;sensitivity&quot; of the detector to a particular concentration of smoke will be affected to an even lesser degree, depending on exactly how the detector electronics &quot;integrates&quot; the events to decide whether to trigger the alarm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it could be programmed to look for a surge that is, say, 3&amp;times; the &quot;background&quot; rate, in which case, the sensitivity would not be affected at all, even as the background rate decays with time.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-09-25T11:41:24.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11620" PostTypeId="2" ParentId="11614" CreationDate="2016-09-25T12:09:29.007" Score="5" Body="&lt;p&gt;I think that the &quot;cranks&quot; you're talking about are the vent valves that allow air circulation between compartments inside the submarine, which are normally left open in order to allow the air purification equipment to provide clean air to each compartment and remove the stale air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a leak from the outside occurs in one compartment, these valves would allow it to flood the entire submarine. Therefore, it is best if the valves for that compartment can be closed, which would allow the rest of the compartments to continue to operate normally. Second best is to close the valves of all of the other compartments, which will allow the air system to flood in addition to the affected compartment and prevent all compartments from getting new air. In this case, it is necessary to surface ASAP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in deep-water events, all of this is for naught, because the bulkheads between compartments are not nearly as strong as the pressure hull, and they will fail regardless of the state of the valves. The only way a crew can survive for any significant length of time is if the pressure hull is still intact. Anything to the contrary you see in movies is mostly &quot;dramatic license&quot;.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-09-25T12:09:29.007" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11623" PostTypeId="2" ParentId="11606" CreationDate="2016-09-25T19:03:10.977" Score="1" Body="&lt;p&gt;Yes, it is possible to boil water at room temperature with a vacuum pump. It is &lt;a href=&quot;https://www.google.com/search?q=boiling%20water%20in%20a%20vacuum&quot; rel=&quot;nofollow&quot;&gt;widely published&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boiling removes heat energy from the water which can lead to freezing if the temperature gets low enough. Again, &lt;a href=&quot;https://www.google.com/search?q=boiling%20water%20in%20a%20vacuum%20ice&quot; rel=&quot;nofollow&quot;&gt;published many places&lt;/a&gt;. This &lt;a href=&quot;https://youtu.be/2IOyJa8NSk0?t=175&quot; rel=&quot;nofollow&quot;&gt;youtube video&lt;/a&gt; shows it well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avoiding ice formation during this process is a thermodynamic balance between heat input to the water from its surroundings and heat loss via low-pressure evaporation. I recommend asking this in a new question with some specific details, like pressure, temperature, container type, temperature, and your purpose and we might be able to help.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-25T19:03:10.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11624" PostTypeId="1" AcceptedAnswerId="11625" CreationDate="2016-09-25T20:31:42.800" Score="1" ViewCount="98" Body="&lt;p&gt;I have designed a product and I found that there is a patent on that specific product. There also seems to be many patents and for the smallest or most obvious use of items. I am wondering if anyone has had experience creating a product and finding there are patents on it and then how do you counter with that? How can you be sure you 100% violate it?&lt;/p&gt;&#xA;" OwnerUserId="8164" LastEditorUserId="8164" LastEditDate="2016-10-27T20:53:12.890" LastActivityDate="2016-10-27T20:53:12.890" Title="Patent for product I designed" Tags="|product-engineering|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11625" PostTypeId="2" ParentId="11624" CreationDate="2016-09-25T21:02:25.247" Score="1" Body="&lt;p&gt;The patent of interest is currently in application stage. Review the marking in the below image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WSpuO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WSpuO.jpg&quot; alt=&quot;Patent Application - Example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general from an engineering product design stand point the first to investigate is if the patent is active. Two scenario to look for is as follows. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the patent currently maintained by the author, if the author is not maintaining the patent, then you will not be infringing on patent, and you should not have to worry about.&lt;/li&gt;&#xA;&lt;li&gt;Is the patent active. All utility patents are valid for only 17 years from the date the patent was granted. If the patent has expired you should not have to worry about infringing on the patent.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If the patents are active then you need to carefully investigate are the claims. Is your engineering design infringing on any of the claims? Looks like their is 20 claims listed in the referenced patent. Below is claim 1 (primary claim) for the patent of interested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rZVEb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rZVEb.jpg&quot; alt=&quot;Patent Claim&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you believe your product is infringing on the patent application, you will need to hire a patent lawyer to investigate work around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://patents.stackexchange.com/questions/13539/how-to-apply-for-a-patent-in-the-us-eu/13540#13540&quot;&gt;How to apply for a patent in the US/EU?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.beemlaw.com/video-gallery/the-three-types-of-patents#.V-sjN_krLIW&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Three Types of Patents&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:54:31.463" LastActivityDate="2016-09-28T01:57:34.010" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11628" PostTypeId="2" ParentId="2510" CreationDate="2016-09-26T02:14:09.227" Score="3" Body="&lt;p&gt;In theory (remember this is theory, and I think that's what you want) you would find the part of the screw that would fail first due to the applied force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The screw could fail in three ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The screw head is pulled off and then there is nothing holding the plate to be screw. (It could theoretically break other ways, but the largest force will probably be perpendicular to the head face compared to some force applied to a cross section). &lt;/li&gt;&#xA;&lt;li&gt;The threads shear and the screw becomes a pin basically. &lt;/li&gt;&#xA;&lt;li&gt;The screw shaft snaps. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So you use the screw material to get your elastic modulus for tension and for shear. Then you calculate the total area the force would be applied against (the screw shaft area for tension or the combine area of thread contact for shear). You need to know the applied force and then you calculate be resulting pressure experienced by the screw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pressure is greater than the modulus, it will fail in that method. That's basically a high level summary. There could be torque applied to the head as well if the plate isn't pulling straight against the screw head bottom face. &lt;/p&gt;&#xA;" OwnerUserId="1143" LastActivityDate="2016-09-26T02:14:09.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11629" PostTypeId="1" AcceptedAnswerId="11633" CreationDate="2016-09-26T08:21:10.333" Score="-2" ViewCount="136" Body="&lt;p&gt;I am looking about things will make phone emit more radiation and I found some case&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bad mobile network signal&lt;/li&gt;&#xA;&lt;li&gt;Turn on Bluetooth and Wifi&lt;/li&gt;&#xA;&lt;li&gt;Moving fast&lt;/li&gt;&#xA;&lt;li&gt;When start dial&lt;/li&gt;&#xA;&lt;li&gt;When start send message&lt;/li&gt;&#xA;&lt;li&gt;Battery (not affect)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I do not know enough or not? Anything else can affect phone emit radiation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be great appreciated.&lt;/p&gt;&#xA;" OwnerUserId="8171" LastActivityDate="2016-09-26T12:07:24.790" Title="What thing will make phone emit more radiation?" Tags="|electrical-engineering|radiation|" AnswerCount="1" CommentCount="9" ClosedDate="2016-09-26T15:02:41.760" ContentLicense="CC BY-SA 3.0" />
  <row Id="11632" PostTypeId="1" AcceptedAnswerId="11635" CreationDate="2016-09-26T08:28:20.920" Score="5" ViewCount="713" Body="&lt;p&gt;We have standby diesel generators that are only run a few times a year. How long can we reasonably keep the diesel stored on site before using or disposing of it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is accepted that there is a finite shelf life particularly after the UK has started using B5 (5% biodiesel) but I have been unable to find any guidance or papers indicating how long it is recommended to store it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it depends on how well it is stored (contaminants etc.) but any best practice guidance / documents / journal articles that anyone could point me to would be much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2019-11-07T17:58:15.973" Title="How long can UK diesel be stored?" Tags="|diesel|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11633" PostTypeId="2" ParentId="11629" CreationDate="2016-09-26T08:54:15.370" Score="6" Body="&lt;p&gt;Generally the term emissions or RF energy is used rather than radiation. While radiation or even non-ionizing radiation can be used it causes excessive worry and panicking in people who don't understand the differences between types of radiation. No one claims that a lightbulb is flooding the room with radiation but it is, we normally call this radiation light. And the light from a bulb is both more energy that a phone emits and a lot closer in frequency to the dangerous types of radiation than anything you'll get from a phone. In other words if a phone is bad for your health then you need to spend your life in a dark room because the sun or any artificial light will clearly kill you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things can be split into two groups - Intentional radiation and unintentional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intentional radiation is when the phone is using a radio transmitter. So any extra radio activity will generate more emissions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When in standby the phone will exchange data with the network at a fairly low duty cycle, just an occasional I'm still here message. Anything that requires more communication will increase the radio activity. So any phone calls, texts, data etc... will increase the amount of information the phone needs to send and so increase the emissions.&#xA;There is also a hand off every time you move between cells which generates a small amount of data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Independent of how busy the phones transmitter is the the phone adjusts its transmit power based on the signal strength so low signal strength will cause it to transmit at a far higher signal power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lots of data in an area with weak signal will put out a lot of energy (and drain your battery)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other radios such as WiFi and bluetooth will also transmit however they are lower power. WiFi will reduce the amount of the phone needs to use the cell network for data and uses a lower power than the phone data link. All things being equal wifi can result in a net decrease in transmitted RF energy however people normally send more data when on wifi. Bluetooth is very low power, unless you are using something that requires constant data, e.g. using a bluetooth headset, the power it radiates is negligible in comparison to the other radios.&#xA;As with the cell phone radio, total emissions are very dependent on how busy the radio links are. These are battery powered systems, when not in use they are designed to be as low power as possible. Radio transmitters use a lot of power and so are kept powered down as much of the time as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally there are unintended emissions, these are far far lower power than the intentional transmissions but can be measured. All electronic systems emit small amounts of RF energy, the busier they are the more they emit which means anything that uses the phones CPU is causing it to emit a small amount of RF radiation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit - &#xA;One minor addition: Connected cables can have an impact (headphones, chargers etc...). While they don't cause any extra emissions directly they can end up acting as antennas and so increasing the efficiency with which the already existing signals are transmitted. This is primarily going to impact the unintentional transmissions but it could also have minor impacts on the intentional transmissions.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastEditorUserId="7659" LastEditDate="2016-09-26T12:07:24.790" LastActivityDate="2016-09-26T12:07:24.790" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11634" PostTypeId="1" AcceptedAnswerId="11893" CreationDate="2016-09-26T09:44:37.810" Score="1" ViewCount="79" Body="&lt;p&gt;I'm trying to model a non-linear, static SISO system with&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;input variable $u$&lt;/li&gt;&#xA;&lt;li&gt;disturbance variable $z$ (can be controlled for identification purposes but not during actual operation)&lt;/li&gt;&#xA;&lt;li&gt;output variable $x$&lt;/li&gt;&#xA;&lt;li&gt;system behaviour $N\{x,u,z\}$ is hysteresis afflicted for both $u$ and $z$&lt;/li&gt;&#xA;&lt;li&gt;$u$ and $z$ influence the hysteresis characteristics of each other &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm searching for any literature that deals with this kind of problem. &#xA;It should contain approaches on how to model such a system and if possible a way to invert it, because ultimately I'm trying to linearize the overall behaviour $N^{-1}\{N\} = 1$. &lt;/p&gt;&#xA;" OwnerUserId="8170" LastActivityDate="2016-10-17T12:01:27.357" Title="Modelling a system with multiple hysteresis behaviours and mutual interference" Tags="|control-engineering|modeling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11635" PostTypeId="2" ParentId="11632" CreationDate="2016-09-26T15:02:10.077" Score="6" Body="&lt;p&gt;The internet has a range of values from 3 months to a  year, depending largely on the political/economic interest of the site in question.  Here's a couple quotes that you might want to follow up on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From  &lt;a href=&quot;http://www.springboardbiodiesel.com/storing-biodiesel-fuel&quot;&gt;springboardbiodiesel&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The truth is that all fuels will degrade over time. In fact, the EPA&#xA;  reports that ULSD diesel has a shelf life of between 3-6 months.&#xA;  Biodiesel, too, has a shelf life that can vary significantly, but with&#xA;  the proper fuel management, biodiesel's shelf life can be extended&#xA;  dramatically.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The biggest factors that affect biodiesel storage life include:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Microbial contamination Chemical contamination Exposure to light&#xA;  Temperature Exposure to air The type of feedstock And the additives&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From  &lt;a href=&quot;https://www.bellperformance.com/bell-performs-blog/biodiesel-storage-life&quot;&gt;BellPerformance&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All fuels degrade over time.  The old school diesel fuels from the&#xA;  fifties could be kept relatively fresh for 1-2 years. The advent of&#xA;  ultra low sulfur diesel cut this storage to 3-6 months. When you blend&#xA;  biodiesel into the mix, the figure changes again. So there's no one&#xA;  stock answer to the question. A fuel's storage life is going to depend&#xA;  on a host of factors related to storage conditions.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The influencers of storage life for both conventional diesel and&#xA;  biodiesel would be contamination with microbes and/or chemicals,&#xA;  light, storage temperature, oxygen exposure, and the type of biodiesel&#xA;  feedstock that the fuel was made from.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From  &lt;a href=&quot;http://www.associatedpetroleum.com/pdf/AE-BioDiesel-MythsFacts.pdf&quot;&gt;AssociatedPetroleum&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most fuel today is used up long before six months, and many petroleum&#xA;  companies do not recommend storing petroleum diesel for more than six&#xA;  months. The current industry recommendation is that biodiesel be used&#xA;  within six months, or reanalyzed after six months to ensure the fuel&#xA;  meets ASTM specifications (D-6751). A longer shelf life is possible&#xA;  depending on the fuel composition and the use of storage enhancing&#xA;  additives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;At the risk of getting booted off this forum,  &lt;strong&gt;YMMV&lt;/strong&gt;  :-)&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-09-26T15:02:10.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11637" PostTypeId="1" CreationDate="2016-09-26T18:09:40.820" Score="3" ViewCount="412" Body="&lt;p&gt;I have a beam experiencing a pure bending moment. The setup is similar to this image from &lt;a href=&quot;http://slideplayer.com/slide/5275312/&quot; rel=&quot;nofollow noreferrer&quot;&gt;these lecture notes&lt;/a&gt; (with different values).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jwirp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jwirp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the flexure formula, we can determine that the stress (and thus the strain) at any point along the top of the beam between C and D should be the same:&#xA;$$&#xA;\sigma_{max} = \frac{Mc}{I}&#xA;$$&#xA;where $M$ is the moment, $c$ is the distance from the neutral axis, and $I$ is the area moment of inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've a couple of strain gauges along the top of the beam measuring strain in the axial direction. These gauges are reading different values at different locations between C and D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this expected? Is it a limitation of the model or are the gauges malfunctioning (or laid up improperly)?&lt;/p&gt;&#xA;" OwnerUserId="8178" LastEditorUserId="1832" LastEditDate="2016-09-26T18:39:54.607" LastActivityDate="2023-01-14T12:09:19.323" Title="Different measured strain in beam experiencing pure bending" Tags="|mechanical-engineering|civil-engineering|stresses|moments|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11638" PostTypeId="1" CreationDate="2016-09-26T20:09:59.690" Score="0" ViewCount="281" Body="&lt;p&gt;All the documentations I could find goes upto only 1000 kcmil. Please direct to a reference for larger cables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a method to extrapolate from smaller cable sizes? &lt;/p&gt;&#xA;" OwnerUserId="8180" LastActivityDate="2024-03-21T10:05:45.653" Title="What is the rated tensile strength of 2000 kcmil stranded copper conductor?" Tags="|metallurgy|metals|power-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11639" PostTypeId="2" ParentId="10986" CreationDate="2016-09-27T01:19:42.200" Score="1" Body="&lt;p&gt;The pulley shaft on the driving shaft is only required to determine the rotational speed of the second pulley on the free shaft. Since the motor can only produce 10Nm of torque, you know the torque and the pulley size is irrelevant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What agentp was referring to is conservation of energy, meaning in this case the power of your driving shaft is transmitted to the free shaft, assuming no losses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power equation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power = Torque x Rotational Velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can solve for the power of the drive shaft and the rotational velocity. The rules for pulley or gear diameter to speed ratio is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D1/D2 = N2/N1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which can be determined by inspection in this case as the pulley with diameter twice as big will logically rotate half as fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then substitute those values into the equation and solve....which results in 20Nm&lt;/p&gt;&#xA;" OwnerUserId="1143" LastActivityDate="2016-09-27T01:19:42.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11640" PostTypeId="1" AcceptedAnswerId="11671" CreationDate="2016-09-27T03:46:39.173" Score="0" ViewCount="275" Body="&lt;p&gt;I need to get a motor from an electric motorbike to drive a single-speed BMX Bike chain, and I haven't worked with bikes/sprockets before so I don't know where to start (I was only supposed to be doing the throttle and electronics, not sure how I ended up with the chain drive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I need a sprocket with as few teeth as possible (~10-14 or less) and can be mounted onto a 9.5mm shaft double flattened to 8.4mm, and 4.5mm long, and have the teeth fit a standard single-speed BMX bike chain.(measurements are from a set of calipers, motor doesn't have any specs :P) I have a sprocket from the motor, but it isn't the correct tooth size/spacing for a bike chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this something that exists or can be made on a part-by-part basis, or do I need to weld something up?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://pbs.twimg.com/media/CtVKT0PVYAEhSyA.jpg&quot; alt=&quot;Picture of the motor shaft measurements, I guess twitter isn&amp;#39;t a good way to host images&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8182" LastEditorUserId="3305" LastEditDate="2016-10-03T13:27:01.603" LastActivityDate="2016-10-03T13:27:01.603" Title="Designing and fabricating custom gears and sprockets" Tags="|mechanical-engineering|motors|bicycles|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11643" PostTypeId="1" AcceptedAnswerId="11645" CreationDate="2016-09-27T06:16:59.847" Score="0" ViewCount="47" Body="&lt;p&gt;I'm searching for the calculation of the quality of the estimated model output compared to the measured data. I'm working with the System Identification Toolbox. You can see an example here &lt;a href=&quot;http://de.mathworks.com/help/ident/ref/compare.html&quot; rel=&quot;nofollow&quot;&gt;http://de.mathworks.com/help/ident/ref/compare.html&lt;/a&gt; at the first graph. Is the percentage calculated with an integral or any other method?&lt;/p&gt;&#xA;" OwnerUserId="8185" LastEditorUserId="1832" LastEditDate="2016-09-27T10:27:24.327" LastActivityDate="2016-09-27T12:09:18.140" Title="System Identification Toolbox - How is the quality of the model output calculated?" Tags="|electrical-engineering|control-engineering|matlab|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11645" PostTypeId="2" ParentId="11643" CreationDate="2016-09-27T08:16:23.267" Score="-1" Body="&lt;p&gt;I found the answer. It is at the same page at Output Arguments --&gt; Fit.&#xA;The fit is calculated (in percentage) using:&#xA;$$&#xA;fit=100*(1-\frac{||y-\hat{y}||}{||y-mean(y)||})&#xA;$$&#xA;where $\hat{y}$ is the estimated model and $y$ the measured data&lt;/p&gt;&#xA;" OwnerUserId="8185" LastEditorUserId="8185" LastEditDate="2016-09-27T12:09:18.140" LastActivityDate="2016-09-27T12:09:18.140" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11647" PostTypeId="2" ParentId="11640" CreationDate="2016-09-27T11:07:45.320" Score="0" Body="&lt;p&gt;That sort of shaft arrangement is often used for circular saws. Here the blade (or in your case the sprocket is clamped between two washers by a nut on the threaded part of the shaft. The inner washer has a slot matching the flats on the shaft instead of a round hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also need a spacer, slightly thinner than the sprocket to keep it concentric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you set the torque on the nut  right this means that the sprocket should slip rather than rounding off the flats on the shaft if it is overloaded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MeYMf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MeYMf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-09-27T11:07:45.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11648" PostTypeId="2" ParentId="11638" CreationDate="2016-09-27T12:11:37.423" Score="0" Body="&lt;p&gt;Is this the table you're looking for?  &lt;a href=&quot;http://ecat.burndy.com/Comergent/burndy/documentation/section%20o-reference.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Burndy&lt;/a&gt;, see page O-30 . If I read it correctly, you should be able to plot strength vs cmil for the values in that table and generate a smooth curve which should predict the strength at 2E6 cmil.&lt;/p&gt;&#xA;&lt;h2&gt;ETA&lt;/h2&gt;&#xA;&lt;p&gt;I took the Burndy data, ran a &lt;code&gt;lowess&lt;/code&gt; smoother followed by a linear regression and got the coefficients&#xA;&lt;code&gt;tensile = 5.72476410e-02*(cmils) +   4.75077150e+03&lt;/code&gt; , which produces a value for &lt;code&gt;2E6 cmil&lt;/code&gt; of  &lt;code&gt;1.1925e+05&lt;/code&gt;  (that's breaking strength in pounds)&lt;/p&gt;&#xA;&lt;p&gt;Use with caution :-), as that's a serious amount of extrapolation.&lt;/p&gt;&#xA;&lt;p&gt;I found a value at &lt;a href=&quot;http://www.prioritywire.com/brochures/Utility%20Catalog.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PriorityWire&lt;/a&gt; (PDF) which says 34 200  pounds for Aluminum cable.  I suspect that it's rare anyone would go to the incredible expense of using Cu for a strength member, which is why nobody measures the value.  Certainly Cu can support its own weight for any run length you can imagine.&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-09-27T12:42:02.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11649" PostTypeId="2" ParentId="11637" CreationDate="2016-09-27T13:22:40.760" Score="2" Body="&lt;p&gt;What variation in value are you experiencing?&#xA;Theoretically, the curvature or strain of the beam should be constant in between load points. Minor variation could be expected, but major probably indicates a problem.&#xA;Assuming all strain gauges are calibrated thermo corrected etc then it could be a faulty strain gauge or poor adhesive/preparation.&lt;/p&gt;&#xA;" OwnerUserId="7758" LastActivityDate="2016-09-27T13:22:40.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11651" PostTypeId="2" ParentId="11637" CreationDate="2016-09-27T16:51:43.583" Score="2" Body="&lt;p&gt;If the loads you apply are too big, you can get local crushing of the beam, or plastic deformation of the beam at the loading points instead of uniform elastic bending.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But without more details of your experimental setup, it's a complete guess whether this is actually the cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic check is whether the beam returns to its original state when the loads are removed - i.e. it is still straight, and the gauges all read zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course if this is a lab experiment and some previous student had permanently bent the beam before you used it, you won't get the expected results!)&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-09-27T16:51:43.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11652" PostTypeId="2" ParentId="11637" CreationDate="2016-09-28T00:34:14.967" Score="2" Body="&lt;p&gt;I would add to the above answers:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beam could have sustained strain hardening from the manufacturing, cutting, preparation or any strain history that has left local changes in its elasticity modulus even if it has been straightened and looks within acceptable tolerances.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-09-28T00:34:14.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11653" PostTypeId="2" ParentId="2510" CreationDate="2016-09-28T01:20:29.440" Score="0" Body="&lt;p&gt;Many cities have either their own code on allowable load for nails and bolts or have adopted a uniform building code.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Los Angeles for example the building department has published these codes on their site, or give them to you as a hand out for free.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;American Wood Council has many pages of guidelines and information on their site.&lt;a href=&quot;http://www.awc.org/codes-standards/calculators-software/connectioncalc&quot; rel=&quot;nofollow&quot;&gt;their calculator&lt;/a&gt; The allowable load for wood connections depends on many factors:&lt;br&gt;&#xA; wood structural rating, its humidity content, species, storage history as well as mechanical strength and type of fastener (screw, bolt, lag bolt, prefabricated fasteners and kits) and the mode of their failure( shear, bending moment, pull out, etc..) and alos the function of the connection, there are load factors for seismic loads or shear loads, etc.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For nails there are concerns as to the ductility of nail, its size and length, the distance between nails, the grain of wood and its angle and many more factors. All of these cases have been extensively tested and tabulated and are available from your City or the hardware store or the manufacturers site.&lt;br&gt;&#xA;A deputy structural engineer inspecting the construction site is responsible for checking the nailing and proper application of their use and will ask the builder to correct or remove not complying nails, fasteners or members and re-installing them correctly.&lt;br&gt;&#xA;You have to discuss your case with an engineer in your building department. In most cases it is free consultation.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2016-09-28T01:25:41.290" LastActivityDate="2016-09-28T01:25:41.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11654" PostTypeId="2" ParentId="7896" CreationDate="2016-09-28T02:25:17.153" Score="1" Body="&lt;p&gt;Gears will only mesh together when they both share the same Diametral Pitch. There are many online descriptions for this term, that is what you want to learn about. Here is an example of one: &lt;a href=&quot;https://www.bostongear.com/pdf/gearology/chapter02.pdf&quot; rel=&quot;nofollow&quot;&gt;https://www.bostongear.com/pdf/gearology/chapter02.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8189" LastActivityDate="2016-09-28T02:25:17.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11656" PostTypeId="1" AcceptedAnswerId="11663" CreationDate="2016-09-28T03:18:27.297" Score="1" ViewCount="1129" Body="&lt;p&gt;It seems to me that &lt;em&gt;abrasive saw&lt;/em&gt; is a term mainly used in the US (the &lt;a href=&quot;https://en.wikipedia.org/wiki/Abrasive_saw&quot; rel=&quot;nofollow&quot;&gt;Wikipedia article on it&lt;/a&gt; has no links to pages in any other languages), whereas &lt;a href=&quot;https://en.wikipedia.org/wiki/Angle_grinder&quot; rel=&quot;nofollow&quot;&gt;angle grinder&lt;/a&gt; has some 20+ translations. So, can someone clarify whether discs for abrasive saws are interchangeable with (thin/cutting) discs for angle grinders and what other differences there might be between these two tools (even if just in the US)?&lt;/p&gt;&#xA;" OwnerUserId="480" LastActivityDate="2016-09-28T13:38:30.663" Title="Difference between an angle grinder and an abrasive saw" Tags="|cutting|tools|grinding|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11659" PostTypeId="2" ParentId="11656" CreationDate="2016-09-28T06:38:47.753" Score="2" Body="&lt;p&gt;In general they are not interchangable. But if they use the same speed and clamping method then they might be. Best to check with the manufacturer.&#xA;Some discs are meant to be used only on the outside edge for cutting, whereas others might be used side on (say, 35 degrees) to grind.&#xA;We have problems in the UK of untrained people using cutting blades for grinding and having the blade shatter.&lt;/p&gt;&#xA;" OwnerUserId="7758" LastEditorUserId="7758" LastEditDate="2016-09-28T13:38:30.663" LastActivityDate="2016-09-28T13:38:30.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11663" PostTypeId="2" ParentId="11656" CreationDate="2016-09-28T13:32:51.207" Score="2" Body="&lt;p&gt;Generally an abrasive saw is a bench-top tool. It looks like a miter saw but has a grinding disc instead of a toothed blade. The dist will be on the order of 12-16 inches in diameter for a typical saw. Angle grinders are handheld tools with much smaller discs - usually between 4 and 8 inches in diameter. While there are come very large angle grinders and some very small abrasive saws, usually the wheels will be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For each, a range of wheels are available. All abrasive saw wheels are designed for cutting on only the outside edge, and will be thick enough to support them selves over their diameter. They will be made of slightly different abrasives, optimized for the metal that you are cutting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some wheels for angle grinders are designed to cut along the outside edge as well. These are usually called 'cutting wheels' or 'cutoff wheels.' Because of the smaller diameter, they can bu much thinner than abrasive saw wheels without shattering too easily. Being thin allows them to cut faster, reduces waste from kerfs, and makes them lighter. For angle grinders, other kinds of blades are also available for surface grinding. These wheels are meant to cut on one flat face for blending welds, cleaning metal, contouring, and other surface work. These wheels will be thicker so that they can handle the loads applied. There are a bunch of other types of wheels from flap discs and soft pad grinders to wire wheels for other specific purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in general, event though they might sometimes be the same diameter in a few cases, discs for abrasive saws and cutoff grinders will not be interchangeable. In any case, be sure to check that the rated rotational speed (RPM) for the abrasive blade matches or exceeds the spindle speed of your tool.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-09-28T13:32:51.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11664" PostTypeId="1" CreationDate="2016-09-28T16:25:43.143" Score="7" ViewCount="7223" Body="&lt;p&gt;I am looking for calculator or chart for slip fit and press fit for non standard size parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have oversized dowel pin 10mm +0.03/+0.01 diameter. I need to create a slip fit hole for that oversize dowel pin. What the size and the tolerance of the hole should be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is I have an undersized plastic bushing which diameter is 20mm -0.05/-0.02 diameter. I need to create press fit hole for that bushing. What the size and the tolerance of the hole should be? &lt;/p&gt;&#xA;" OwnerUserId="8198" LastActivityDate="2024-03-01T08:02:15.267" Title="How to find press fit/ slip fit for non standard size parts?" Tags="|mechanical-engineering|tolerance|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11665" PostTypeId="1" AcceptedAnswerId="11667" CreationDate="2016-09-28T18:42:01.667" Score="-2" ViewCount="59" Body="&lt;p&gt;Yesterday I asked this questions on physics part of stackexchange, but I was redirected to engineering side of the force ;-). We all know about the rapidly developing technology of 3D printing. However, this technology is still very very VERY SLOW. For printing objects larger than a matchbox you have to wait days. I wonder if technique and physics will ever allow us to create objects directly from the atoms, by manipulating the setting, structure and connectivity of atoms of a given element. If possible, how can we control the atoms so as to create a fast 3D printer?&lt;/p&gt;&#xA;" OwnerUserId="8200" LastEditorUserId="1832" LastEditDate="2016-09-28T18:55:07.973" LastActivityDate="2016-09-28T18:59:43.757" Title="Is the construction of a high-speed 3D printer (replicator from Star Trek) possible?" Tags="|mechanical-engineering|structural-engineering|3d-printing|" AnswerCount="1" CommentCount="4" ClosedDate="2016-09-29T11:43:41.627" ContentLicense="CC BY-SA 3.0" />
  <row Id="11667" PostTypeId="2" ParentId="11665" CreationDate="2016-09-28T18:59:43.757" Score="4" Body="&lt;p&gt;A replicator is a very long way from 3D printing. Although 3D printing is often endowed in the media with quasi magical properties all it really is is multi axis CNC extrusion and not fundamentally different from any other CNC manufacturing process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key thing about the replicators in star Trek is that they can produce any material on demand from energy/elementary particles including very complex organic compounds (eg Capt Picard is often seen ordering 'Earl grey Tea : hot'). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sort of context fabricating the shape of the object is fairly trivial compared to synthesizing the matter it is composed of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have been experiments in fabricating atomic scale objects using electron microscopes but this is still a long way away form making up complex molecules in bulk on demand. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-09-28T18:59:43.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11668" PostTypeId="2" ParentId="11118" CreationDate="2016-09-28T19:15:39.177" Score="3" Body="&lt;p&gt;The evaporation formula you have is a modification of the EPA method for calculating the evaporation rate. See here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.epa.gov/epcra/technical-guidance-hazardous-analysis-emergency-planning-extremely-hazardous-substances&quot; rel=&quot;nofollow&quot;&gt;Technical Guidance for Hazards Analysis&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;look at Appendix G. The modified form I found in a paper by John Barry in CEP magazine. Somebody took the time to scan it and has it posted on the internet:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aiche.org/resources/publications/cep/2005/january/estimating-rates-spreading-and-evaporation-volatile-liquids&quot; rel=&quot;nofollow&quot;&gt;Estimating Rates of Spreading and Evaporation of Volatile Liquids&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Search for the title for a freely available version. Note that your berm and non-berm equations are essentially the same, just with careful calculations of the surface area. Unfortunately, I don't know enough about the equations to gauge their accuracy or expected error. They seem to be based on basic theories defining the evaporation rate as:&#xA;$$&#xA;ER=kAP^*&#xA;$$&#xA;where $k$ is the mass transfer coefficient, $A$ is the surface area of the spill, and $P^*$ is the vapor pressure. Calculation of the mass transfer coefficient typically requires an empirical correlation and is where the molecular weight, wind speed, and temperature dependence comes in. Note also that the vapor pressure is temperature dependent and you should treat it appropriately.&lt;/p&gt;&#xA;" OwnerUserId="1031" LastEditorUserId="1031" LastEditDate="2016-09-29T15:10:23.630" LastActivityDate="2016-09-29T15:10:23.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11669" PostTypeId="2" ParentId="11664" CreationDate="2016-09-28T19:32:21.297" Score="0" Body="&lt;p&gt;You will need to decide what standard you want to base your estimates off of and the level of tolerance. There is also the expense and manufacturability to consider. Here is a calculator I found. The site has others as well. I am sure an internet search will lead to other calculators and guidelines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.engineersedge.com/calculators/mechanical-tolerances/ansi-fit-tolerances.htm&quot; rel=&quot;nofollow&quot;&gt;https://www.engineersedge.com/calculators/mechanical-tolerances/ansi-fit-tolerances.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7989" LastActivityDate="2016-09-28T19:32:21.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11671" PostTypeId="2" ParentId="11640" CreationDate="2016-09-29T04:19:16.630" Score="0" Body="&lt;p&gt;There are quite a few online websites you can find that specialize in the design and fabrication of gears/sprockets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a really nice website for generating custom sprockets and gears is &lt;a href=&quot;http://www.rushgears.com/tech-tools/part-search/build-custom-gears&quot; rel=&quot;nofollow&quot;&gt;Rush gears&lt;/a&gt; because they have a nice sprocket/gear online model generator. You can download the models of gears in CAD format and even order the gear to be made by them in a bunch of materials. Just click the Click to build CAD models and you can adjust a bunch of different specifications.&lt;/p&gt;&#xA;" OwnerUserId="3305" LastEditorUserId="3305" LastEditDate="2016-10-02T22:39:25.860" LastActivityDate="2016-10-02T22:39:25.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11682" PostTypeId="2" ParentId="7591" CreationDate="2016-09-29T19:07:46.640" Score="2" Body="&lt;p&gt;I cannot give you the solution by using transfer functions. However I can give you a general form by using the state space representation. I will do it for a &lt;strong&gt;square&lt;/strong&gt; system, i.e. the number of inputs and outputs is equal. For a system with $n$ inputs and $m$ outputs it is getting more messy and a lot harder to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system&#xA;$$\dot{x}=f(x)+ g_1(x)u_1 + \ldots + g_m(x)u_m$$&#xA;with outputs&#xA;$$y_1 = h_1(x), \ldots, y_m = h_m(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First introducing the Lie Derivative. The Lie Derivative of $h$ with respect to $f$ or along $f$ is&#xA;$$L_{f}h(x) = \frac{\partial h}{\partial x}f(x)$$&#xA;For example, the following notation is used:&#xA;$$\begin{split} &#xA;L_g L_f &amp;amp;= \frac{\partial (L_f h)}{\partial x}g(x)\\&#xA;L_f^2 h(x) &amp;amp;= L_f L_f h(x) &amp;amp;= \frac{\partial (L_f h)}{\partial x}f(x)\\&#xA;L_f^kh(x) &amp;amp;= L_f L_f^{k-1}h(x) &amp;amp;= \frac{\partial (L_f^{k-1})}{\partial x}f(x)&#xA; \end{split}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Introducing the notion of relative degree with respect to each output.&#xA;Consider the $i$-th output and differentiate it with respect to time:&#xA;$$\dot{y}_i = L_fh_i(x) +L_{g_1}h_i(x)u_1 +\ldots L_{g_m}h_i(x)u_m$$&#xA;This expression depends explicitly on at least one input if (for all $x$):&#xA;$$\left(L_{g_1}h_i(x),\ldots,L_{g_m}h_i(x) \right)\neq (0,\ldots,0)$$&#xA;If so, the $i$-th output has relative degree $k_i = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the relative degree $k$ per output $i$ if&#xA;$$\left(L_g,L_f^{k_i-1}h_i(x),\ldots,L_{g_m}L_f^{k_i-1}h_i(x)\right) \neq (0,\ldots,0)$$ &#xA;for all $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is now Input-Output Linearised (hence decoupled) when applying the following feedback&#xA;$$u(x) = -A^{-1}(x)N(x)+A^{-1}(x)v$$&#xA;with the decoupling matrix $A(x)$, vector $N(x)$ and new input vector $v$.&#xA;Where&#xA;$$A(x) = &#xA;\begin{pmatrix}&#xA;L_{g_1}L_f^{k_1-1}h_1(x) &amp;amp; \ldots &amp;amp; L_{g_m}L_f^{k_1-1}h_1\\&#xA;\vdots &amp;amp; &amp;amp; \vdots\\&#xA;L_{g_1}L_f^{k_m-1}h_m(x) &amp;amp; \ldots &amp;amp; L_{g_m}L_F^{k_m-1}h_m\\&#xA;\end{pmatrix},\quad&#xA;N(x) = &#xA;\begin{pmatrix}&#xA;L_f^{k_1}h_1(x)\\&#xA;\vdots\\&#xA;L_f^{k_m}h_m(x)&#xA;\end{pmatrix}&#xA;$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence $A(x)$ needs to be invertible for all $x$. If you want the transfer functions, simply apply Laplace. &lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2016-09-29T19:07:46.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11684" PostTypeId="1" AcceptedAnswerId="11689" CreationDate="2016-09-29T19:26:51.813" Score="4" ViewCount="310" Body="&lt;p&gt;sorry for the odd title. We're working on a project to find a cheaper and safer way to install high voltage power lines in the Scottish highlands. One of my group mates had a really cool idea to instead of actually digging underground, you could just lay the cables and cover them with dirt/some other material which we haven't thought about at the moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking because I assume the industry would currently be using the best available technology to them, and they don't use this method; they lay them underground. I'm trying to figure out the potential drawbacks to this proposed method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Initially i thought it could be regulations, i.e. minimum depths etc, however, as far as I can tell from official documentation, there are only guidelines. In general, cables are laid between 0.45m -1m below ground, but not always. For example from a cable map I found, there was an 11kV line just 35mm undeground in Edinburgh. Official documentation seems to imply depths are situational, and as long as a risk assessment is performed, shallower or deeper depths can be allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) In terms of cost, digging through the rocky terrain of the Scottish highlands would pose significant challenge and cost. Getting heavy vehicles out onto that sort of terrain isn't easy. Using some sort of packing material, such as dirt, would surely be a cheaper option? There would of course be costs incurred due to rough terrain for transport trucks etc, but I don't see how it could cost more than digging through the terrain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potential drawbacks I can think of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Using a mound structure would pose issues such as erosion. However, maintenance (which must be carried out on cables anyway) shouldn't be too difficult, especially since faults in the structure would be easy to spot since it is overground. It may require more regular maintenance due to weathering, but that would be offset by the fact that each maintenance trip wouldn't involve unearthing the ground?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Potential damage to cables. I don't even know if this is a real drawback (in comparison to the existing system since cable damage isn't very uncommon). I read some case studies from one of the regulatory bodies, and it seems damage came mostly when projects were undertaken without consulting the relevant authorities for cable maps. I assume the mound layout would be very similar, and probably wouldn't encounter a greater frequency of damage to the cables, but this is pure conjecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) It would be an easier target for malicious activity since it would be easier to get to without the need for heavy machinery. However, cable maps aren't very difficult to get, and someone could just as easily target shallow cables using the current system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, does anyone have any criticisms of the potential solution? Or does anyone know of this idea having been implemented but not worked, or was determined to not be feasible? Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="5469" LastEditorUserId="191" LastEditDate="2016-09-30T19:41:29.753" LastActivityDate="2016-10-06T17:54:57.763" Title="Why shouldn't we put high voltage cables physically underground?" Tags="|power-engineering|electrical-grid|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11685" PostTypeId="1" CreationDate="2016-09-30T03:09:56.033" Score="3" ViewCount="418" Body="&lt;p&gt;What are the prospects of seeing rocket engines become the standard for commercial aircraft?  It might be a dumb question; I really don't know all that much about the topic.  The reason I ask is that rocket powered aircraft are much faster—and who doesn't like speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concerns are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I don't know whether there any designs that would be economical.  I understand rockets burn a tremendous amount of fuel and that is obviously a limiting factor.&lt;/li&gt;&#xA;&lt;li&gt;There could be safety concerns.  Just because we can do something doesn't necessarily make it a good idea.  We have seen the struggles Musk has had with his rockets.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We keep hearing about all these super trains that can get us from Chicago to NY in an hour, even though the whole idea of traveling on a track at those speeds seems quite dangerous.  However, up in the air there is nothing to hit, which is a different story.  &lt;/p&gt;&#xA;" OwnerUserId="8226" LastEditorUserId="368" LastEditDate="2016-09-30T15:51:39.757" LastActivityDate="2017-04-28T05:43:33.133" Title="Are rocket engines practical for commercial aircraft?" Tags="|aircraft-design|transportation|rocketry|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11686" PostTypeId="1" CreationDate="2016-09-30T03:21:29.980" Score="0" ViewCount="40920" Body="&lt;p&gt;I am looking at an ammonia refrigeration unit and trying to understand how loading and unloading solenoid valves work? and also how to modulate the compressor motor over a period of time.&lt;/p&gt;&#xA;" OwnerUserId="8227" LastActivityDate="2017-03-26T06:55:54.540" Title="How does loading and unloading work in an air compressor?" Tags="|mechanical-engineering|pressure|compressed-air|compressors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11687" PostTypeId="2" ParentId="11685" CreationDate="2016-09-30T03:27:27.913" Score="1" Body="&lt;p&gt;To answer first let's look at the different types of &quot;rocket&quot; engines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - Solid Fuel: The issue here is that one a solid fuel rocket is ignited it can't be shut off or throttled very effectively. Imagine the pilot ignites the engine and the aircraft WILL (and I mean WILL) start moving, continue moving and get into the air regardless if he wants to stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 - Liquid Fuel: In general, for best efficiency you'd want cryogenic fuels like Liquid Hydrogen but Kerosene also works which add a huge issue in terms of weight (for the tanks) as well as volatility of the fuels and the cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 - Hybrid solid/liquid: Again, solid fuel is difficult/impossible to stop the chemical reaction and difficult to throttle.&lt;/p&gt;&#xA;" OwnerUserId="8225" LastActivityDate="2016-09-30T03:27:27.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11689" PostTypeId="2" ParentId="11684" CreationDate="2016-09-30T05:30:19.780" Score="4" Body="&lt;p&gt;Your four drawbacks are very good reasons for not laying cables on the ground and just covering them with soil or &quot;stuff&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Erosion will be a major issue. Exposed cables would be a safety hazard to humans and wildlife. It would also increase the risk of cable degradation and failures. Countering the affects of erosion, but covering the cable, will also add to the operating cost of the cable because the uncovered parts of the cable would need to be covered over again, repeatedly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another impact will be that the covered cable on the land surface will present an obstacle to other forms of land use. Crossing points will need to be constructed over the cable, to allow people to access the land on the other side of the cable. This would add additional cost to laying the cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Trans-Alaska_Pipeline_System&quot; rel=&quot;nofollow&quot;&gt;Trans-Alaska Pipeline&lt;/a&gt;, which has been placed of the surface, has an impact on the migration of wildlife&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By burying a cable, it prevents people and animal from fiddling with the cable and thus prevents damaging infrastructure &amp;amp; personal injury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a buried cable is out if sight and out of mind to most people and thus has a better chance of not being interfered with and being able to function as it should for a long a possible.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-09-30T05:30:19.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11690" PostTypeId="2" ParentId="11685" CreationDate="2016-09-30T08:39:26.590" Score="4" Body="&lt;p&gt;Economically a rocket engine will always lose out to a jet engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'll ignore solid fuel rockets, they are impractical for commercial air travel due to their fixed thrust and inability to turn on and off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A both liquid rockets and and jets need fuel but a rocket engine also needs oxygen, the jet engine pulls this from the air. So for the same amount of fuel burnt your cost of fueling a rocket is far higher, you need to supply two fuel tanks instead of one. You also need to lift twice as much mass into the air with you which means you need to use more fuel for the same trip since your aircraft is now heavier at takeoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People are working on the idea of a rocket engine that can use oxygen from the air until it gets above the atmosphere (e.g. &lt;a href=&quot;http://www.reactionengines.co.uk/sabre.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.reactionengines.co.uk/sabre.html&lt;/a&gt;) in order to minimize the amount of oxygen you need to carry. But that is intended more for space planes than conventional aircraft. If you were to stay in the atmosphere it would make the fuel costs a lot closer but it adds a lot of extra complexity which will have an impact on operating costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason you are asking is for faster travel times. You can go a lot faster with a jet engine than any airliner in operation today. Concorde could give most fighters from it's time a run for their money when it came to cruising speed. There is no reason why you couldn't build a jet airliner that went even faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end it doesn't matter how you power your super fast airliner, jet, rocket, big elastic bands, you always hit the same problem. Pushing something through the air at those speeds creates a LOT of drag. You end up burning a lot more fuel to got the same distance if you go fast which means the operating costs are a lot higher. While some people are willing to pay the extra price it's not a big enough market to make it pay. &lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-09-30T08:39:26.590" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11691" PostTypeId="2" ParentId="11685" CreationDate="2016-09-30T11:04:01.093" Score="6" Body="&lt;p&gt;At a fundamental level, it should be clear that throwing stuff out the back is going to be less economical than sucking in stuff in the front and throwing it out the back faster.  For one thing, the former has to carry around a lot of stuff at takeoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possible way to save you from the fundamental issue above is to throw the stuff out the back so fast that you don't need much of it.  In fact, rocket engines strive to do this.  If you were able to throw stuff out the back at near the speed of light, then a rocket engine would look pretty good.  With our current technology, especially what we can reasonably carry on a plane, we can't get the speed anywhere near fast enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another fundamental issue favoring jets over rockets is that jets only have to carry half their chemical reactants.  The other half is available in the atmosphere they fly thru.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For flying around earth's atmosphere, rockets are unlikely to be competitive with jets and propellers any time soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, rockets are sometimes used in special cases where the extra cost is worth it.  This is done routinely by the military to allow large aircraft to take off from small runways, for example.  In this case, the rockets are only to assist takeoff, and only last seconds.  That's enough to get the plane going fast enough so that the wings make more lift than the plane's weight.  The regular plane engines are also used at full power.  The rockets add just enough thrust to allow the necessary short takeoff distance.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-09-30T11:04:01.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11693" PostTypeId="2" ParentId="8101" CreationDate="2016-09-30T17:43:08.737" Score="0" Body="&lt;p&gt;I think the issue here is we are trying to figure out how T/W (Thrust to Weight ratio) is a function of maximum speed. However, it is the maximum speed that is a function of T/W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T/W will determine the efficiency of the thrust of the aircraft which (in conjunction with aerodynamic drag which is a function of velocity) will determine the maximum speed of the aircraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That deals with the Thrust term. Now let's talk about the weight. The aircraft must also generate enough Lift to balance the Weight. The faster the velocity the higher the lift force (to a point) but that will also increase the drag which will lower the maximum speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when we talk about T/W we need to remember that at steady-level flight T/W = L/D which will be indicative of the maximum speed.&lt;/p&gt;&#xA;" OwnerUserId="8225" LastActivityDate="2016-09-30T17:43:08.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11694" PostTypeId="2" ParentId="11684" CreationDate="2016-09-30T18:58:18.730" Score="2" Body="&lt;p&gt;I do not know the laws in your country, but the key is &quot;risk assessment&quot;. And the risk we are talking about is the risk of killing someone, not the risk of the cable being damaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irrespective of laws, it is likely the project will be signed off by a professional engineer and protected by your companies malpractice insurance. If someone dies as a result of gross negligence, the professional engineer(s) go to prison and the company may lose its malpractice insurance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only place I could see it being ok to put high voltage lines on the surface is in a military controlled area where it is continuously monitored and kept free of people and equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have heard of lines being buried at shallower depths, but concrete is poured above them to ensure that even heavy equipment will not dig them up. Your risk assessment would have to include what the area is used for, number of people that will be in proximity to it on a daily basis, development that will occur in the area over the line's lifespan, soil work-ability, failure detection, and monitoring methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a simultaneous construction of a service road may be needed to move the equipment. You may be able to crush some excavated rock on site and use that to cover the cable in low areas; reducing the total amount of trenching required. Coarsely crushed rock would not have as many erosion issues and would be more difficult to excavate accidentally; reducing risk. The road would need blocked from normal traffic and periodically driven by service trucks to check for erosion and unapproved human activity.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-09-30T18:58:18.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11695" PostTypeId="2" ParentId="11686" CreationDate="2016-09-30T22:06:08.137" Score="1" Body="&lt;p&gt;Many compressors are driven by induction motors for efficiency, which means that they do not have the starting torque required to get going against the full back pressure of the system. This is true of both air compressors and refrigeration compressors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air compressors usually have a check valve between the compressor itself and the reservoir tank and an automatic valve that vents the line between the compressor and the check valve to atmosphere whenever the compressor is stopped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In small refrigerators, the system is simple enough that the pressure equalizes through the expansion valve within a few seconds or minutes of the compressor shutting off without any other intervention. You just have to make sure that you don't try to restart the compressor too soon after its last shutdown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In larger refrigeration systems, explicit solenoid valves are used to isolate the compressor from the rest of the system and to equalize the pressure between input and output so that it can start up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to &quot;modulation&quot;, nearly all refrigeration systems use &quot;bang-bang&quot; controllers &amp;mdash; either full on or full off, with the duty cycle modified as needed in order to maintain a particular temperature. It's rare to need anything more sophisticated than that.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-09-30T22:06:08.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11699" PostTypeId="1" CreationDate="2016-10-01T16:45:02.170" Score="3" ViewCount="537" Body="&lt;p&gt;I'm trying to develop a finite conical shell element with six dof at each node. I want to express the rotations $\phi$, $\theta$ and $\psi$ in terms of the displacements $u$, $v$ and $w$. The displacements are functions of $x$ and $y$.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ETBDs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ETBDs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Using the geometric relations and small angle approximation, I have found $\phi\approx \tan(\phi)=\frac{dw}{dv}$, $\theta\approx \tan(\theta)=\frac{dw}{du}$ and $\psi\approx \tan(\psi)=\frac{dv}{du}$. But I'm not yet convinced that this is the way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know any other general approach to express the rotations in terms of the displacements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I have attached a picture of the geometry of the cone and its coordinate system. $x$ is the meriodional coordinate along the generator of the cone which is presumed to be zero at the apex of the cone. $y$ is the circumferential coordinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qTf2w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qTf2w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8241" LastEditorUserId="8241" LastEditDate="2016-10-02T09:00:31.630" LastActivityDate="2017-04-18T11:38:12.293" Title="How to calculate rotations from displacements?" Tags="|mechanical-engineering|structures|finite-element-method|mathematics|robotics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11700" PostTypeId="1" CreationDate="2016-10-01T18:14:11.707" Score="3" ViewCount="1286" Body="&lt;p&gt;Pardon me for kind of a stupid question, but I keep wondering. Traditionally, tower cranes' cabins are located at the top of the tower. I can understand why it have been necessary up to ~1990s, when it was the only way for a human to get reliable real-time visuals of what's going on. But today, with the advent of camera and monitor technologies (when cameras and LCD monitors are dirt-cheap), what is the exact problem with moving the crane operator's cabin to the ground level and make all visuals using a system of cameras?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching the Internet for the answer, I've encountered quite a few of companies that produce camera systems for cranes, but it looks like that majority of them propose to use cameras as &lt;em&gt;additional&lt;/em&gt; viewing aids (i.e. &quot;view from the hook&quot;, &quot;view from the pulley&quot;, etc), but not replacing the sky cabin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From kind of delitant point of view, having a cabin up high actually poses a bunch of problems, i.e.:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;safety: fallen crane = very real chance of death or major injuries&lt;/li&gt;&#xA;&lt;li&gt;comfort: it's very hard to maintain proper air conditioning and temperatures up high&lt;/li&gt;&#xA;&lt;li&gt;time consumption: a way up to the work place takes 20 minutes for an operator, and takes &lt;em&gt;lots&lt;/em&gt; of physical endurance to do so&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is it really feasible in modern world to continue using sky cabins on the tower cranes? Am I missing something obvious, or is it just very persistent tradition that nobody dares to break?&lt;/p&gt;&#xA;" OwnerUserId="8244" LastActivityDate="2016-10-02T18:17:44.270" Title="Tower crane with a cabin on ground level" Tags="|pulleys|building-physics|construction-management|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11701" PostTypeId="2" ParentId="11700" CreationDate="2016-10-01T21:05:03.130" Score="3" Body="&lt;p&gt;It is not trivial to replace human vision with a camera in this sort of situation. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it is difficult to reproduce good depth perception with cameras &lt;/li&gt;&#xA;&lt;li&gt;the human eye can change focus quickly and intuitively, in order to scan between near and distant objects you would need to be constantly refocusing the camera and potentially missing information out of the current depth of field this is one reason why cameras aren't a good replacement for rear view mirrors). &lt;/li&gt;&#xA;&lt;li&gt;humans have an innate sense of orientation and where they are looking in space. Trying to reproduce a wide field of view with multiple remote cameras can be very disorientating. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This isn't so much ab out the load itself and its path from its start point to end point but overall situational awareness and being able to pick up emerging hazards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These sorts of systems are just about possible, the F35 warplane uses a panoramic camera system with a VR helmet display but is is extremely expensive to implement and has had a lot of problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address your particular points. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if a crane falls down this is very bad,  whether or not someone is in the cab, as a result it should be very rare so having an operator in the cab is not a major safety concern. &lt;/li&gt;&#xA;&lt;li&gt;cranes need power anyway so laying on heating and air conditioning is not, in itself a big problem. Besides the people working on the ground on the rest of the site will be out in the weather anyway, that's just a fact of the construction industry. &lt;/li&gt;&#xA;&lt;li&gt;many construction jobs are physically demanding, but for a crane operator climbing up to the cab may be the only exertion they do all day. If its a real problem you can just fit an elevator to the crane. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the short answer is that it would be technically very difficult to adequately replace a crane operator with a remote system and there is no compelling reason to do so. The main reason for putting the operator up there in the first place is that they have a very good field of view, which is difficult to reproduce with cameras.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-02T18:17:44.270" LastActivityDate="2016-10-02T18:17:44.270" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11702" PostTypeId="1" AcceptedAnswerId="11714" CreationDate="2016-10-01T21:44:39.463" Score="3" ViewCount="311" Body="&lt;p&gt;I have a Nema 17 stepper motor that does 200 steps per revolution, or 400 800 and 1600 micro-steps if in micro-stepping mode for convenience sake I would like to somehow translate the 200 steps into 360 steps. What gear ratio / micro stepper configuration do I need to convert 200 steps to 360 steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone wants an easy way to create an involute spur gear, I found this page, converted the to DXF in illustrator then extruded in Rhino!&#xA;&lt;a href=&quot;http://geargenerator.com/&quot; rel=&quot;nofollow&quot;&gt;http://geargenerator.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="539" LastEditorUserId="539" LastEditDate="2016-10-04T15:49:25.677" LastActivityDate="2016-10-04T15:49:25.677" Title="How do I mechanically convert 200 steps into 360 discreet degrees?" Tags="|mechanical-engineering|gears|stepper-motor|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11703" PostTypeId="2" ParentId="3473" CreationDate="2016-10-01T22:30:31.923" Score="0" Body="&lt;p&gt;the difference in densities is essential.  A bubble of air will have much more of a buoyant force than a bubble of oil in the water.&lt;/p&gt;&#xA;" OwnerUserId="8246" LastActivityDate="2016-10-01T22:30:31.923" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11704" PostTypeId="2" ParentId="11700" CreationDate="2016-10-01T23:55:39.950" Score="2" Body="&lt;p&gt;From 1924 to 1932 &lt;a href=&quot;http://www.railwaywondersoftheworld.com/sydney-harbour.html&quot; rel=&quot;nofollow&quot;&gt;Sydney Harbour Bridge&lt;/a&gt; was constructed. It used two creeper cranes on both sides of the arch during construction to hoist people &amp;amp; materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crane operators couldn't see what they were hoisting, &quot;&lt;strong&gt;they were given orders by telephone from officials working some 400 ft below&lt;/strong&gt;&quot;. One of the pictures in the above link shows inside the control cabin of one of the cranes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Had current camera technology existed when the bridge was being constructed, they most likely would have used it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enabling crane drivers to directly see what is being hoisted is ideal, but not necessary, provided correct safety measures are used. It all comes down to safety and risk mitigation. &lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-10-01T23:55:39.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11706" PostTypeId="1" AcceptedAnswerId="11723" CreationDate="2016-10-02T02:38:10.793" Score="1" ViewCount="916" Body="&lt;p&gt;A 200×250 mm panel of mass 20 kg is supported by hinges along edge AB. Cable CDE is attached to the panel at C, passes over a small pulley at D, and supports a cylinder of mass m. Neglect the effect of friction.&lt;/p&gt;&#xA;&lt;p&gt;For those who cannot see the picture. The axis are oriented as follows, x+ is pointing out and to the right, y+ is pointing up and z+ is out and to the left.&#xA;&lt;span class=&quot;math-container&quot;&gt;$A=[0,0,.25]$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$ B=[0,0,0]$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$ C=[.2sin(\Theta),-.2cos(\theta),.125]$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$ D=[.2,.1,0]$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I have &lt;span class=&quot;math-container&quot;&gt;$\sum F_x=0=A_x+B_x+\frac {mg}{\lVert CD\rVert}(.2-.2sin(\theta))$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sum F_y =0=A_y+B_y-192.2+\frac {mg}{\lVert CD\rVert}(.1+.2cos(\theta))$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sum F_z=0=A_z+B_z+\frac {mg}{\lVert CD\rVert}(-.125)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I am having trouble deciding where I want to make my moment in order to create a system I can solve for m in terms of &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vH61D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vH61D.png&quot; alt=&quot;Static System in quesiton&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8249" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-10-03T17:21:58.290" Title="How/where to take moments to solve a static equilibrium equation" Tags="|statics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11707" PostTypeId="2" ParentId="11700" CreationDate="2016-10-02T03:35:19.807" Score="5" Body="&lt;p&gt;The crane operator at the top of the tower doesn't just have a view of the load being lifted. He/she can see &lt;em&gt;the entire surrounding area&lt;/em&gt; and take account of any potential hazardous situations that might develop, for example other workers moving about who are not aware of what the crane is intending to do, vehicles being driven into potentially dangerous positions, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the crane operator being able to see the actual load isn't necessary at all. Cranes are often driven by someone who can't see the load, by watching hand signals given by someone who can see it - and sometimes there may be a chain of two or three people signalling to each other to get a line of sight from the load to the crane driver.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend of mine was on the &quot;receiving end&quot; of a fairly extreme one-off example of that sort of operation. He wanted an expensive ($100,000) full size grand piano moving into a room on the second floor of his house, but there was no way to get it up the stairs. The solution was to use a mobile crane parked on the road, lifting the piano across the garden and &lt;em&gt;over the roof of the house,&lt;/em&gt; and lowering it onto a second-floor balcony on the &quot;blind side&quot;. The crane operators didn't think there was anything difficult or unusual about that operation - especially when they discovered that the legs could be removed from the piano (moving pianos was not their usual line of business!) making it much easier to handle than they thought it would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you could argue that a global view could be provided by an array of cameras and a bank of monitors - but two eyes and one brain are a cheaper and more reliable solution.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-10-02T03:35:19.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11710" PostTypeId="1" CreationDate="2016-10-02T09:07:32.880" Score="2" ViewCount="171" Body="&lt;p&gt;We're working on a project that require some mechanical engineering and electronics engineering. In this project we've to model this overshot &lt;a href=&quot;https://en.wikipedia.org/wiki/Water_wheel&quot; rel=&quot;nofollow noreferrer&quot;&gt;water wheel&lt;/a&gt; (so the water comes from the left top side and flows over the wheel to a lower place, so that the wheel will spin at a certain velocity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we've to get a startup to model this mathematically, we're looking for a so complete model (we're looking for as much as important inputs and as outputs we're looking for the torque and the power of the wheel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wzaLc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wzaLc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8251" LastEditorUserId="1293" LastEditDate="2016-10-02T09:35:06.933" LastActivityDate="2016-10-02T09:35:06.933" Title="Mathematic modeling of an overshot water wheel" Tags="|applied-mechanics|modeling|mathematics|building-physics|" AnswerCount="0" CommentCount="1" ClosedDate="2016-10-03T03:00:56.450" ContentLicense="CC BY-SA 3.0" />
  <row Id="11711" PostTypeId="2" ParentId="11686" CreationDate="2016-10-02T22:13:41.837" Score="1" Body="&lt;p&gt;Load/Unload and Modulation are different control schemes for rotary-screw compressors (either for air, ammonia or other gases). Both methods use a slider valve to reduce the capacity of the compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modulation controls this slider valve to maintain a specific system pressure, it is however very inefficient when compared with modern VFD controls. Load/Unload uses the same slider valve, but operates with the valve all the way closed or all the way open. In this fashion, the compressor is operating at its maximum effiency when loaded, and wasting less power than when it is idle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am actually having a contractor convert one of our 350HP Quincy air compressors from modulation to load/unload on Monday. The air system already has VFD compressors for keeping the pressure stable. Switching the compressor over to load/unload will increase its overall efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Load/unload is much better than start/stop. While it does use more power, it greatly reduces wear and tear on the compressor and supporting electrical components.&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image shows the slider vavle (&lt;a href=&quot;http://www.ref-wiki.com/technical-information/145-compressors/31164-screw-compressors.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image Source&lt;/a&gt;):&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0SnvH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0SnvH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotary-screw_compressor#Load.2Funload&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotary-screw wiki article&lt;/a&gt; explains load/unload and modulation operation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Load/unload&lt;/strong&gt;&lt;br&gt;&#xA;  In a load/unload control scheme, the compressor remains continuously powered. However, when the demand for compressed air is satisfied or reduced, instead of disconnecting power to the compressor, a device known as a slide valve is activated. This device uncovers part of the rotor and proportionately reduces capacity of the machine down to typically 25% of the compressors capability, thereby unloading the compressor. This reduces the number of start/stop cycles for electric motors over a start/stop control scheme in electrically-driven compressors, improving equipment service life with a minimal change in operating cost. This scheme is utilised by nearly all industrial air-compressor manufacturers. When a load/unload control scheme is combined with a timer to stop the compressor after a predetermined period of continuously unloaded operation, it is known as a dual-control or auto-dual scheme.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Modulation&lt;/strong&gt;&lt;br&gt;&#xA;  Instead of starting and stopping the compressor, a slide valve as described above modulates capacity to the demand. While this yields a consistent discharge pressure over a wide range of demand, overall power consumption may be higher than with a load/unload scheme, resulting in approximately 70% of full-load power consumption when the compressor is at a zero-load condition.&#xA;  &lt;br&gt;&#xA;  Due to the limited adjustment in compressor power consumption relative to compressed-air output capacity, modulation is a generally inefficient method of control when compared to variable-speed drives. However, for applications where it is not readily possible to frequently cease and resume operation of the compressor (such as when a compressor is driven by an internal-combustion engine and operated without the presence of a compressed-air receiver), modulation is suitable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;This &lt;a href=&quot;https://www.youtube.com/watch?v=XAzST5n1FMo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Youtube video&lt;/a&gt; shows the slider valve in operation.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-10-02T22:13:41.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11713" PostTypeId="1" CreationDate="2016-10-03T06:53:20.687" Score="1" ViewCount="374" Body="&lt;p&gt;I have a continuous beam supported on columns. I need the brickwork above to act as load bearing wall for the next floor above. How would I model this in a FEM suite like SUMO (Prokon) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In SUMO I can insert concrete shear walls, but no brick walls. Since brick walls offer no tensile restraint, I can't use concrete shear walls.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2016-10-03T06:53:20.687" Title="How would I model a load bearing brick wall supported on a beam in FEM" Tags="|structural-engineering|structural-analysis|reinforced-concrete|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11714" PostTypeId="2" ParentId="11702" CreationDate="2016-10-03T07:08:50.240" Score="2" Body="&lt;p&gt;You want a transmission with a ratio R such that $ R \cdot 200 / 360 $ gives an integer number of steps per degree. Then in controlling software you can program it to take e.g. 5 steps to move one degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Brian Drummond mentioned in a comment, 36:20 is one possible ratio, giving 1 step per degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other options:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;9:5, equal to 36:20 except smaller gears.&lt;/li&gt;&#xA;&lt;li&gt;9:1, gives 5 steps per degree, so greater torque but slower speed&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As to how to implement the transmission, you can do it with either gears or with belts. Some of the ratios above are very basic, I'm sure you could even find a premade 9:1 part.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-10-03T07:08:50.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11717" PostTypeId="1" CreationDate="2016-10-03T10:15:25.687" Score="1" ViewCount="1263" Body="&lt;p&gt;I think that putting the OLs of all the motors in series { figure (1) } provides better protection such that if any one of the motors is overloaded the circuits trips preventing the rest of the motors from working during the fault condition. But then, connecting them in parallel { figure (2) } provides better selectivity such that when a fault occurs, it's much quicker to determine the point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is better? Which is conventionally used?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y34tr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y34tr.jpg&quot; alt=&quot;control circuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8146" LastEditorUserId="110" LastEditDate="2016-10-03T11:33:30.447" LastActivityDate="2016-12-04T19:40:52.817" Title="Series or parallel overloads in control circuits controlling several motors?" Tags="|electrical-engineering|control-engineering|power-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11719" PostTypeId="1" AcceptedAnswerId="11737" CreationDate="2016-10-03T14:24:49.003" Score="0" ViewCount="1008" Body="&lt;p&gt;I want to know if it is possible to control the gearbox of a car with a manual transmission, specifically one with the shift stick in the steering wheel column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bZH21.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bZH21.jpg&quot; alt=&quot;steer column&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is using linear actuators or hydraulic cylinders to control the gas / clutch pedals and &quot;some kind of mechanism&quot; to control the shifting stick directly or removing the gear stick and installing the mechanism on the gearbox directly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is this idea feasible or has it been done before? Or do all autonomous cars use automatic or &quot;tiptronik&quot; gearbox?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;NOTE: I have taken a look at this post before &lt;a href=&quot;https://engineering.stackexchange.com/questions/5838/remote-start-on-a-car-with-manual-transmission&quot;&gt;Remote start on a car with manual transmission&lt;/a&gt;, but did not find a satisfying answer&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="8262" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-10-04T11:24:49.333" Title="Remote control of a manual transmission" Tags="|control-engineering|automotive-engineering|gears|" AnswerCount="1" CommentCount="0" ClosedDate="2016-10-04T11:31:16.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="11721" PostTypeId="1" CreationDate="2016-10-03T15:28:50.207" Score="2" ViewCount="263" Body="&lt;p&gt;I used a commercial CFD software to simulate incompressible turbulent flow of air over a flat plate.  I setup a 2D domain 1.5m long in the x direction and .2m in the y direction.  The boundary conditions are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Left side (x=0): is a velocity inlet&lt;/strong&gt;&lt;br&gt;&#xA;u = 100m/s&lt;br&gt;&#xA;$\nabla p\cdot n = 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Right side (x=1.5) is a pressure outlet&lt;/strong&gt;&lt;br&gt;&#xA;$p = 0$ Pa&lt;br&gt;&#xA;$\nabla u\cdot n = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bottom Side (y=0) is a wall boundary&lt;/strong&gt;&lt;br&gt;&#xA;u=0m/s&lt;br&gt;&#xA;$\nabla p\cdot n = 0$   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Top Side (y=0.2) is a symmetry boundary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For turbulence, I enabled a $k-\omega$SST model and prescribed turbulent intensity of 5% and a viscosity ratio $\frac{\mu_t}{\mu}=10$ at the inlet.  My program automatically sets conditions at all other boundaries automatically.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My mesh consists of 300 quadrilateral elements in the x-direction and 100 quadrilateral elements in the y-direction.  To ensure that the mesh is sufficiently refined to capture the boundary layer, I biased the mesh so that the elements connected to the wall boundary are 4000 times smaller than the elements connected to the symmetry boundary (i.e. the elements connected to the wall boundary are approximately $4*10^{-6}$m )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a time step of 0.01s, I ran my simulation up to 30s to ensure a steady state was achieved.  In the end, I was surprised to find that my boundary layer thicknesses $\delta$ were smaller than the well-known 1/7 power law, given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\delta}{x} = \frac{0.16}{(Re_x)^{\frac{1}{7}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\frac{u x}{\nu}$, where $x$ is the horizontal distance from the beginning of the plate, and $\nu$ is the kinematic viscosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that I should be able to achieve a close match to this 1/7 power law with my simulation independently of how I choose the turbulent conditions at the inlet.  But my boundary layer ended up being almost 30% smaller than the 1/7th power law all across the plate and I'm not entirely sure why.  Is there a minimum turbulent viscosity ratio and turbulent intensity required to achieve 1/7th power law?&lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2020-04-19T23:35:22.987" Title="Obtaining the correct flat plate turbulent boundary layer thickness in a CFD simulation" Tags="|fluid-mechanics|cfd|turbulence|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11723" PostTypeId="2" ParentId="11706" CreationDate="2016-10-03T17:21:58.290" Score="0" Body="&lt;p&gt;Since you are only interested in the mass $m$, all you have to do is to set the moment about the hinge AB to zero for equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you have to consider only the two forces $M g$ (panel weight) and $m g$, since only they contribute to the relevant moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you wish, I can work out the complete answer, but probably this is all you need to arrive at the answer yourself.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, my calculations give me the result for $m$ as $$ \frac{5 M \sin (\theta )}{\sin (\theta )+2 \cos (\theta )}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot shows that $m$ increases with $M$ and $\theta$ as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vrzui.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vrzui.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting confirmation is that the current configuration will not work for $\theta \gtrsim 2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0o1rz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0o1rz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-10-03T17:21:58.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11724" PostTypeId="1" CreationDate="2016-10-03T17:47:07.670" Score="2" ViewCount="275" Body="&lt;p&gt;I was working on designing some robotic grippers and I find I have to make a pretty large device to have anywhere near the same power as the human hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this other engineer's experience? Is the human hand, cubic centimeter for cubic centimeter just really strong compared to a machine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess this translates down to how ants can pick up &quot;boulders&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2016-10-04T11:09:57.873" Title="Strength of human hand compared to robot" Tags="|mechanical-engineering|robotics|" AnswerCount="2" CommentCount="1" ClosedDate="2016-10-05T22:04:23.910" ContentLicense="CC BY-SA 3.0" />
  <row Id="11725" PostTypeId="1" CreationDate="2016-10-03T18:42:44.873" Score="2" ViewCount="1670" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X1KJE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X1KJE.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;How is such a one piece car chassis from carbon fiber made at one piece?&lt;/p&gt;&#xA;&lt;p&gt;Do they make the parts individually and bond it some way that it is extremely durable bond?&lt;/p&gt;&#xA;&lt;p&gt;What bonds are really durable for carbon fiber that they can allow for carbon fiber to behave in stress tests as one piece?&lt;/p&gt;&#xA;&lt;p&gt;Since carbon fiber needs to be vacuumed to be cured, doing this one piece has some challenges how can this even be possibly made one piece?&lt;/p&gt;&#xA;&lt;p&gt;NOTE: Any book or resources on this topic would be greatly appreciated it&lt;/p&gt;&#xA;" OwnerUserId="8164" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2022-05-22T04:39:42.140" Title="How is a carbon fiber chassis made one piece?" Tags="|mechanical-engineering|materials|carbon-fiber|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11726" PostTypeId="1" AcceptedAnswerId="11761" CreationDate="2016-10-03T19:03:37.673" Score="0" ViewCount="1432" Body="&lt;p&gt;For example, if your transfer function is defined as :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{H(s)}{Q(s)} = \frac{1}{5s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the gain and time constant be in this case (since it's usually in the form of $\dfrac{k}{Ts+1}$ where $k$ and $T$ are the time constant and gain, respectively)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is the transfer function for a tank being filled, but the exit stream is a pump so it is constant. Where $H$ and $Q$ are deviation variables for output and input, respectively)&lt;/p&gt;&#xA;" OwnerUserId="8264" LastEditorUserId="1832" LastEditDate="2016-10-03T20:18:30.517" LastActivityDate="2016-10-07T11:43:00.600" Title="What is the gain and time constant of a system with a constant output?" Tags="|chemical-engineering|process-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11727" PostTypeId="1" CreationDate="2016-10-03T19:19:46.113" Score="2" ViewCount="127" Body="&lt;p&gt;I am trying to select new hose cam locks to connect to cam locks on an existing pipe system. There seem to be a lot of different types of cam locks as I've searched for them, so I am unsure how I distinguish between them. Are there different swages, groove radii, or other variables that I need to measure so I can match up new cam locks, or are cam locks all the same and only vary by diameter? If there are measurable variables, is there a chart available that lists these variables (e.g. like the charts for measuring steel C-channel weights)?&lt;/p&gt;&#xA;" OwnerUserId="2733" LastEditorUserId="2733" LastEditDate="2016-10-03T19:26:58.703" LastActivityDate="2016-11-02T20:23:06.790" Title="Matching up cam lock or cam and groove coupling" Tags="|mechanical-engineering|piping|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11728" PostTypeId="2" ParentId="11727" CreationDate="2016-10-03T19:25:49.203" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Camlock connectors are capable of carrying liquids, powders and granules and are interchangeable with all other cam and groove couplings made to the same standard (A-A-59326A – formally MIL-C-27487),[1] ensuring compatibility with couplings made to the same specification.&#xA;  &lt;a href=&quot;https://en.wikipedia.org/wiki/Camlock_(fluid_fitting)&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Camlock_(fluid_fitting)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2733" LastActivityDate="2016-10-03T19:25:49.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11729" PostTypeId="1" CreationDate="2016-10-03T19:47:31.747" Score="6" ViewCount="8933" Body="&lt;p&gt;Screens on TVs and phones et cetera are always rectangular. Why? Why are they not round? Eyes are round. The Sun, the source of almost all of our light, is round. Telescopes have round lenses and mirrors. If a photon has a shape it would be round since it is described by sinus waves and interferometry. Sure, we have two eyes but that gives a 3D illusion of an oval rather than of a rectangular field of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the corners of our unnatural rectangular computer screens' user interfaces, there are hidden stuff like the time of day and some abort/undo button. Stuff you don't want to see unless you actively search for it. Why have corners on a screen? Who came up with that idea? Cathode ray oscilloscopes were invented with circular displays, how did they acquire horns and corners?&lt;/p&gt;&#xA;" OwnerUserId="4958" LastActivityDate="2020-11-12T01:07:46.730" Title="Why aren't screens round instead of rectangular?" Tags="|design|optics|engineering-history|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11730" PostTypeId="1" CreationDate="2016-10-03T21:10:34.677" Score="2" ViewCount="249" Body="&lt;p&gt;I am installing an acrylic shelf, 75-1/4&quot; L x 10&quot; W x 1-1/2&quot; deep.&#xA;It is fixed to the wall at the short ends. Should I be worried about deflection in the center of the shelf when weight is added? Is there anything I can do to calculate this?&lt;/p&gt;&#xA;" OwnerUserId="8266" LastActivityDate="2016-10-03T21:10:34.677" Title="Deflection of Acrylic Shelf" Tags="|structural-engineering|construction-management|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11731" PostTypeId="2" ParentId="11729" CreationDate="2016-10-03T23:03:31.520" Score="4" Body="&lt;p&gt;X-Y addressing and raster scanning are both much simpler when addressing a rectangular area. So either you need to lose (waste) the corners or adopt a fundamentally different addressing scheme.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There HAVE been polar displays on round screens so it's not impossible - just very inconvenient... look at 1940--1950s radar screens for some examples. Maybe some vector graphics displays in the 1970s, GEC 4000? but my memory is hazy on those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have failed to find any 1970s vector graphic displays with non-rectangular screens, but this &lt;a href=&quot;http://www.theatlantic.com/technology/archive/2013/01/the-never-before-told-story-of-the-worlds-first-computer-art-its-a-sexy-dame/267439/&quot; rel=&quot;nofollow&quot;&gt;slightly NSFW image&lt;/a&gt; from 1956 shows that round CRT displays were not exclusively the domain of polar plots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably radar practice influenced the choice of display on this project, which may have been a polar plot at heart, augmented with information drawn in vector graphics form, and therefore not bound by the regularity of raster scanning. The linked video &lt;a href=&quot;https://www.youtube.com/watch?v=iCCL4INQcFo&quot; rel=&quot;nofollow&quot;&gt;(IBM Sage Computer Ad, 1960)&lt;/a&gt; doesn't actually show any polar plots, just vector graphics and a precursor to touch screen interaction.&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="435" LastEditDate="2016-10-04T12:27:46.713" LastActivityDate="2016-10-04T12:27:46.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11732" PostTypeId="1" CreationDate="2016-10-04T00:00:02.647" Score="1" ViewCount="49" Body="&lt;p&gt;As the highway engineers here may already know, U.S. highway 71 through most of Missouri was recently redesignated as a section of Interstate 49 after the completion of grade separation work at several locations.  However, having been on the road recently, I noticed one thing that is missing from it that most rural Interstates have: rest areas.  While private facilities are available at most exits from that stretch of road, the lack of rest areas on an over 100 mile stretch of road when most Interstates have them at roughly 50 mile intervals strikes me as an oversight by MoDOT, and left me wondering if it was a substandard highway for that reason alone.  Furthermore, I reviewed the 2017-2021 Transportation Improvement Plan for the MoDOT Southwest District, and there were no projects in there to add rest areas to I-49.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, are there any standards or guidance materials for the placement/location of rest areas (with or without services) on newly constructed or designated rural Interstates?  Or can a highway be built or upgraded to &quot;Interstate standard&quot; without such provisions without drawing a peep from standards or guidance material?&lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2016-10-04T00:00:02.647" Title="Are there standards or guidance for the placement and location of rest areas on newly constructed or designated U.S. Interstates?" Tags="|design|highway-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11735" PostTypeId="2" ParentId="11724" CreationDate="2016-10-04T07:59:37.157" Score="0" Body="&lt;p&gt;Human hand is certainly a very efficient and compact machine. It is so versatile. It can pick up a miniature gem of a diamond barely visible and place it in an expensive watch; same hand when fisted can break a slab of concrete or kill a cow.&lt;br&gt;&#xA;Hand doctors have tables of range of motion and power from each joint and finger or combination of them. At our gym many regular looking guys can exert 120 lbs. to the grip machine.&lt;br&gt;&#xA;The miracle of power and flexibility of hand is its complex set of tendons and ligaments which allow for great range of tasks, from fine work of a brain surgeon to blacksmith workers or marshal art athletes.&lt;br&gt;&#xA; There are definitely specialized tools, levers and machines that are designed to extend the strength of and spectrum of its capability hand and they are getting more efficient and take advantage of modern light weight material. But non are as compact and multifaceted as our hand. Hand has this extreme range of dexterity because its muscles can turn from holding to stabilizing to active to passive, or even sometimes just add bulk under the active muscle so its arc of flexion is optimized, for wider lever arm advantage.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And all this time we have not even scratched the surface of subject of our nerve network in hands and how they read the input as far as touch, pressure, heat, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;our hand is a respectable example of nature's perfection.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-10-04T07:59:37.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11736" PostTypeId="2" ParentId="11724" CreationDate="2016-10-04T11:09:57.873" Score="2" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;I find I have to make a pretty large device to have anywhere near the same power as the human hand.&lt;/p&gt;&#xA;&lt;p&gt;Is the human hand, cubic centimeter for cubic centimeter just really strong compared to a machine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Keep in mind that power and strength (force) are very much separate things.&lt;/p&gt;&#xA;&lt;p&gt;It is easy to make a very strong and small gripper, by using worm gearing to magnify the force of a small motor. The limit is the strength of the materials used. Human bones and tendons are pretty strong, but there are engineered materials that are even stronger for the same size.&lt;/p&gt;&#xA;&lt;p&gt;As for power (work done per unit of time), one cannot look at the hand in isolation. Like Andrew mentioned in a comment, the muscles are actually in your arm. And the cooling and energy supply are all over your body. To compare, a hydraulic gripper can deliver very high power in a small space because it offloads part of the work to the hydraulic pump supplying it.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-10-04T11:09:57.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11737" PostTypeId="2" ParentId="11719" CreationDate="2016-10-04T11:24:49.333" Score="2" Body="&lt;p&gt;What you've described is not that far from a computer-controlled automatic transmission. In fact, so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Semi-automatic_transmission&quot; rel=&quot;nofollow&quot;&gt;semi-automatic transmission&lt;/a&gt; seems like a quite good match.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biggest difference is usually the use of a torque converter instead of a clutch. A friction clutch requires a certain amount of &quot;feel&quot; for the car, to know how fast to release it for a smooth start or gear change. A torque converter is more forgiving of timing, which makes it easier to control automatically.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And as for retrofitting such a mechanism to an existing car: it surely has been done, for example in remote control cars for movies. But in general there would be little point to do it that way, when automatic and computer controlled gearboxes are available off-the-shelf.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-10-04T11:24:49.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11738" PostTypeId="2" ParentId="10232" CreationDate="2016-10-04T11:25:29.127" Score="1" Body="&lt;p&gt;I do work for a company that makes sensors that might be applicable.  These sensors are used, among other places, to measure levels in ballast tanks of ocean-going ships.  The possible gotcha for your case is the size of the tank.  These sensors are meant for tanks depths of a few meters at least, up to 30 meters or so.  When the tank is too small, the errors in the system usually become too significant.  A few mm difference of liquid height in a ballast or fuel tank on a ship is irrelevant, especially considering the ship isn't sitting perfectly still and the liquid always sloshes around a bit anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best description of the technology I found on their web site is at &lt;a href=&quot;http://www.jowa-usa.com/jowausaproductslanding.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jowa-USA &#xA;Marine Products&lt;/a&gt;.  Imagine a resistance wire wrapped around a conducting rod, but with insulators in the right places so that at rest the wire doesn't touch the rod.  When external pressure is applied, like from being submersed in a liquid, the wire is squished against the rod and makes electrical contact.  The resistance between the rod and the wire at the top of the sensor then tells you the distance from the top to the first contact, from which you can derive the level of the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course this is all packaged and encased to be impervious to various fluids.  Different materials are used depending on the fluids.  The two most common applications are for ballast tanks (seawater) and fuel tanks (diesel fuel).  These things last many years on average.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="8055" LastEditDate="2017-03-06T17:30:38.183" LastActivityDate="2017-03-06T17:30:38.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11741" PostTypeId="2" ParentId="11729" CreationDate="2016-10-04T11:59:10.103" Score="13" Body="&lt;p&gt;The earliest CRT displays were in fact round.  However, that was more due to manufacturing tradeoffs of making large glass envelopes with vacuum inside than a desire for the display area to be round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rectangular makes more sense than round for a variety of reasons:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;It's what people want to look at.  We use rectangular pieces of paper.  Photographs are generally rectangular.  Contrary to what you say, we don't perceive our world in a visual circle.  Our visual attention area is wider than tall.  Consider that rectangular paintings have been the norm from the beginning.  Sure, circular paintings exist, but far far more of them are rectangular.  Especially when each canvas was made by hand, the artist could have chosen pretty much any shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Rectangular shapes are easier to handle.  Imagine if all the papers on your desk were circular.  They would be difficult to stack, file, and above all, line up a stack together.  When manufacturing a frame for something like a picture a rectangular shape is easier.  Rectangles can be tiled with no gaps, whereas this is impossible with circles.  Building circular windows in houses would be more difficult and labor-intensive than the rectangular ones we use now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rectangular frames of various sized can be built from the same linear stock, just cut to different lengths.  With different circular sizes, lots of different curvatures would need to be stocked.  Linear shapes are much easier to saw, extrude, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this says why display screens should be rectangular, but when everything else has a good reason to be rectangular, then having display screens circular would be annoying.  You want to be able to go back and forth between display screens and the rest of the world that works on rectangular visual snippets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Rectangular devices have orientation.  Something like a circular smart phone would be a pain since you'd always pick it up at some arbitrary orientation, then would have to turn it to get the top up.  Yucc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Having a narrower and wider dimension can be useful.  Think of a cell phone in your pocket.  Pocket openings need to be kept narrow, else things would fall out too easily and there'd be other undesirable tradeoffs in the garment.  Different limitations apply to pocket depth.  The result is that pockets are usually deeper than wide.  A circular cell phone, for example, would be limited in size by the pocket width.  A rectangular cell phone can have more area by allowing the long dimension to be wider than the pocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Pixels are arranged in a rectangular pattern.  This simplifies manufacturing arrays of pixels, like in LCD monitors, and results in the simplest math we have come up with for addressing pixels in software and physically addressing them electrically.  When images are rectangular-aligned arrays of pixels, it naturally follows that images are rectangles too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you propose arranging pixels in a circular display?  If you leverage the existing manufacturing technology, then they will be in a rectangular pattern, but with awkward short rows and columns at the edges.  If you lay out pixels in a circular pattern, you will have to develop all new manufacturing technology, and the software and hardware to address them will be a nightmare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rectangular displays make sense.  They were not chosen by laziness or arbitrarily.  They are what you inherently want.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-10-04T11:59:10.103" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11742" PostTypeId="2" ParentId="11725" CreationDate="2016-10-04T13:12:54.897" Score="3" Body="&lt;p&gt;First off, think about good old-fashioned injection molds or even lost-wax molds.  It's relatively easy to cast complicated single-piece objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of carbon-fiber fab, it depends a bit on what you call &quot;one piece.&quot; Usually the structure is built up out of intricately cut 'fabric' which is coated with bonding fluids to produce an integral single structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's some pretty decent explanations of one process &lt;a href=&quot;http://www.theturboforums.com/content/155-Home-Fabricated-Carbon-Fiber-Panel&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; .  It appears that if you have a good mold, you can place the fabric however you like &amp;amp; cure it in place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may enjoy this &lt;a href=&quot;https://www.youtube.com/watch?v=kaoq8Mc4xxw&quot; rel=&quot;nofollow&quot;&gt;BMW video&lt;/a&gt; on their processes. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-10-04T13:12:54.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11745" PostTypeId="2" ParentId="7807" CreationDate="2016-10-04T18:03:02.937" Score="1" Body="&lt;p&gt;As the gain $K$ is varied and the root loci reach A, two of the branches (green and blue) meet at A and there are two roots there, which your lecturer calls $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third root is somewhere on the red locus, which is also real and the root is denoted by $\alpha$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the characteristic equation of the closed loop system is $(s+\alpha) (s+\beta)^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term is then expressed as the standard second order system, which is not used in the rest of the answer.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-10-04T18:03:02.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11746" PostTypeId="2" ParentId="11725" CreationDate="2016-10-04T20:17:06.603" Score="1" Body="&lt;p&gt;One approach is to manufacture a core from materials such as rigid foam, expanded metal (aluminium honeycomb etc) other similar low density but rigid materials along with any inserts and attachment points and then cover the whole thing with the appropriate pre-preg cloth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can then be wrapped in plastic film to seal it and then vacuumed and cured in an autoclave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laying up over a core may not give quite as smooth a surface finish as laying up in a machined mould but allows for much more complex shapes, and it is entirely possible to use polished moulds on some surfaces but not all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example image in the question not that while the exterior surfaces such as the roof are smooth a lot of the interior surfaces have noticeable texture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can see this in the following video : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/u4BrWI5Xi0g&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/u4BrWI5Xi0g&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-06T14:42:00.867" LastActivityDate="2016-10-06T14:42:00.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11749" PostTypeId="1" AcceptedAnswerId="11752" CreationDate="2016-10-05T01:54:33.990" Score="1" ViewCount="711" Body="&lt;p&gt;I'm trying to design a jewellery  box, with a mirror in the lid. When open, the lid should rest at an angle of about 135 degrees from original closed position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been learning about 4 bar mechanisms, they sound appropriate (frame = box, coupler = lid) BUT in my case, the crank won't go through a full 360 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been using a &lt;a href=&quot;http://www.mekanizmalar.com/fourbar.html&quot; rel=&quot;nofollow&quot;&gt;simulator&lt;/a&gt; with crank=150, coupler=160, rocker=frame=225 but I need a way to model the start / finish positions to ensure everything lines up right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an app, or even a page full of math, covering the use of 4 bar mechanisms as a hinge? (I'm new to the mechanical engineering field, any help appreciated)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; The techniques of designing joint positions are apparently what I'm looking for. &lt;strong&gt;Mechanism synthesis&lt;/strong&gt; (2- or 3-position synthesis) is the term to look for. It would seem I'm also going to need a full-blown CAD program, as compared to simple web-based apps I've been using&lt;/p&gt;&#xA;" OwnerUserId="8277" LastEditorUserId="8277" LastEditDate="2016-10-06T04:18:28.780" LastActivityDate="2016-10-06T04:18:28.780" Title="4 bar mechanism simulator" Tags="|mechanisms|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11750" PostTypeId="1" AcceptedAnswerId="11754" CreationDate="2016-10-05T04:49:31.110" Score="1" ViewCount="196" Body="&lt;p&gt;Is there a specific way to figure out the gear ratio I would need for a wind up mechanism that could last around one year? Would I start out with a really high tooth and step down too a small enough ratio for a complete cycle to last a year? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to devise a polynomial function, then use calculus to solve for the gear ratios, number of teeth, and time?&lt;/p&gt;&#xA;" OwnerUserId="746" LastEditorUserId="746" LastEditDate="2016-10-05T04:56:56.860" LastActivityDate="2016-10-05T11:08:05.080" Title="Time duration for a mechanical wind up mechanism?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11751" PostTypeId="2" ParentId="11515" CreationDate="2016-10-05T06:45:38.200" Score="2" Body="&lt;p&gt;I've had the possibility to test several different diagnostic ultrasound devices in a big bucket of water with gravel at the bottom! The result is interesting: At first, for all devices you have to use a &quot;curved sound head&quot;, the linear one doesn't work at distances further than a few centimeters. With cheap devices I could hardly see the bottom or the gravel. With more expensive 3D / 4D devices the bottom and gravel became very clear and with the software we could measure distances with mm accuracy. The problem which however remains, is that the data cannot easily be exported for further processing. And there are to few parameters so adjust (not generally!, there are a lot... but for this special purpose). For example a tested &quot;4D device&quot; records a 3D model of the object when you scan it from different angles (e. g. for getting images of babys in 3D), but the time over which the images are taken is fixed and so the area of the riverbed which could be recorded is limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically the principle works - you can record the riverbed with mm accuracy but the devices are to much like &quot;black boxes&quot;.&lt;/p&gt;&#xA;" OwnerUserId="8044" LastActivityDate="2016-10-05T06:45:38.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11752" PostTypeId="2" ParentId="11749" CreationDate="2016-10-05T08:39:57.733" Score="5" Body="&lt;p&gt;Reading your question leads me to belive that you are not really after a simulator &lt;em&gt;at all!&lt;/em&gt; Instead you are looking for a &lt;strong&gt;mechanism synthesis&lt;/strong&gt; method, which allows you to design the mechanism. A simulator only shows how the thing would work given measurements but does not help you into placing the joints that you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case one of the simpler synthesis methods is probably going to suffice, although there are many I would suggest that you use either &lt;strong&gt;2 position&lt;/strong&gt; or &lt;strong&gt;3 position synthesis&lt;/strong&gt; which are easy to make on your own all you need is a ruler and a compass, or a suitable drawing application/CAD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3 position synthesis is a special case of the 2 position synthesis. So by teaching the 3 position method i will in fact teach you both. I do suggest you draw this is some CAD application as it makes it easy for you to parametrise the values as you may want to fiddle with the joint positions a bit to get a better result (any MCAD can do this). Let get started*:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Draw a linear span of what connect the joints on the lid. (You may want to draw the lid also so you can see how the entire thing behaves). This will be your first position, i suggest you start with lid closed. Name the joints so you know whet connects where (I will name the joints B1 and C1 for first position. I start with B so they the designed joints will get the name be A and D)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uC3an.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uC3an.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1:&lt;/strong&gt; Draw the first position (lid closed). Name the joints and position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Draw the second position, say where the lid rests in open position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IItLv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IItLv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 2:&lt;/strong&gt; Draw the second position (lid open)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Connect the joints with lines. B1 to B2, C1 to C2. Find the midpoint on the connecting line and draw a perpendicular line at the midpoint. The joint A now need to be on any point at the perpendicular line that hits midpoint B1 and B2 while the joint D needs to be on the other line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eoeMT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eoeMT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 3:&lt;/strong&gt; Joints should be anywhere on line A or line D, you pick.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Due to the set being over constrained and the fact that i can not control the descent direction. I can take a 3rd position and repeat the above method. This will give me an answer that is fully defined. Since i want the lid to approach quite vertically it would be best to position it close to B1. Due to practical considerations on where the joint can be I'm going to place this into a virtual position that happens before open. &lt;strong&gt;Repeat steps (2-3)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B0nGd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B0nGd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 4&lt;/strong&gt;: The second constraint forces the joints A and B to be in the position where the 2 constraints cross.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You still need to check that you havent got a degenerate solution. That's it. After doing this a few times i'm sure you have the ultimate positions. Go search your own solution. There are more sophisticated methods, but ultimately there is no perfect way to choose. It is always a bit open to human intuition, there are more solutions possible (but that's good this makes them patenttable). But 3 points synthesis should suffice for you. Like i said this is the place where you crack open your CAD as it makes it easy for you to explore the space &lt;/p&gt;&#xA;&#xA;&lt;p&gt;* I am drawing in illustrator so i can not easily remake the positions but if you use a CAD then it would be easy to just drag things until your position is good enough.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2016-10-05T14:31:38.877" LastActivityDate="2016-10-05T14:31:38.877" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11754" PostTypeId="2" ParentId="11750" CreationDate="2016-10-05T11:08:05.080" Score="4" Body="&lt;p&gt;Starting with the gear ratio is the wrong way to look at this problem.  Start by determining the total energy you need for one year.  No amount of gearing can fix it if the fully wound coil doesn't hold the energy you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know the energy you need, you look for a spring coil that can store at least a little more than that.  Only then are you ready to consider gear ratio.  The spring spec will give you the number of turns over the coil's useful discharge life.  Only you can tell how many turns of some shaft you need to run your device for one year.  The overall gear ratio is then the latter divided by the former.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-10-05T11:08:05.080" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11755" PostTypeId="1" AcceptedAnswerId="11764" CreationDate="2016-10-05T13:36:14.820" Score="0" ViewCount="5917" Body="&lt;p&gt;I have the formula for the translational stiffness which is : &#xA;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?k=%5Cfrac%7BA%5Ccdot&amp;space;E%7D%7Bl%7D&quot; title=&quot;k=\frac{A\cdot E}{l}&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what I would like to find is the rotational stiffness. Is there any formulas to convert from one to an other knowing the pitch of the screw and the diameter ?&lt;/p&gt;&#xA;" OwnerUserId="1374" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.330" LastActivityDate="2016-10-05T23:31:31.967" Title="Convert translational stiffness to rotational stiffness of a screw" Tags="|mechanical-engineering|bolting|stiffness|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11757" PostTypeId="1" AcceptedAnswerId="11765" CreationDate="2016-10-05T16:06:04.300" Score="1" ViewCount="66" Body="&lt;p&gt;I understand that the inside of a carbonated aluminium beverage can is pressurised. When the tab is pulled up,a rivet is lifted. Does this cause the can to depressurise? So then the can can be pierced with less force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="8286" LastActivityDate="2016-10-05T22:08:20.363" Title="When beverage can stay-on tabs are lifted do they depressurise the can by lifting the rivet?" Tags="|mechanical-engineering|pressure|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11759" PostTypeId="2" ParentId="11684" CreationDate="2016-10-05T17:16:22.857" Score="1" Body="&lt;p&gt;When we had to run HV lines to a building, we put them in steel pipes 6 ft deep, covered the pipes with 2 feet of sand, then laid a foot of concrete on that and painted it red, and filled the trench with earth.  People dig, maps get lost.&lt;/p&gt;&#xA;" OwnerUserId="7768" LastActivityDate="2016-10-05T17:16:22.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11760" PostTypeId="2" ParentId="11717" CreationDate="2016-10-05T18:19:15.777" Score="1" Body="&lt;p&gt;We typically use overload relays in parallel. A typical single-line diagram for a motor control centre (switchboard which supplies motors) is shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W1rhE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W1rhE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle is: &lt;strong&gt;A fault on one motor should not affect the rest of plant, unless it's critical to the operation of the overall process.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never seen motor overload relays used in series.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If a fault on a motor &lt;em&gt;does&lt;/em&gt; affect the overall process, we'd use an auxiliary contact of the overload relay to create an interlock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a large motor might need a grease pump to maintain lubrication to its bearings. If the grease pump stops working, we need to stop the motor. We would do this by taking an auxiliary contact off the grease pump's overload relay, and wiring it into the starter circuit for the large motor. This way, if the grease pump suffers a fault, the auxiliary contact will drop out, and the large motor will stop.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastEditorUserId="1001" LastEditDate="2016-10-05T18:25:32.840" LastActivityDate="2016-10-05T18:25:32.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11761" PostTypeId="2" ParentId="11726" CreationDate="2016-10-05T19:42:23.560" Score="1" Body="&lt;p&gt;$G(s) = \frac{1}{s}$ is called an integrator. It continuously integrates your signal and therefore it has no time constant, hence $\tau = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your system $\frac{H(s)} {Q(s)} = \frac{1}{5s}$ can be written as an integrator multiplied with a gain of $\frac{1}{5}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the pole-zero gain is $k_{pz} = \frac{1}{5}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The static gain of your function $k_s = \lim_{s\to 0} \frac{1}{5s} = \infty$ or you might say there is no static gain.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastEditorUserId="8216" LastEditDate="2016-10-07T11:43:00.600" LastActivityDate="2016-10-07T11:43:00.600" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11764" PostTypeId="2" ParentId="11755" CreationDate="2016-10-05T20:06:53.863" Score="3" Body="&lt;p&gt;The translational stiffness is written as&#xA;$$ k_l = \frac{A\,E}{\ell},$$ &#xA;where the stiffness $k_l$ is in $\left[\frac{N}{m}\right]$, the area $A$ is in $\left[m^2\right]$, the young's modulus is in $\left[\frac{N}{m^2}\right]$ and the length $\ell$ is in $\left[m\right]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotational stiffness is written as&#xA;$$ k_r = \frac{G\,J}{\ell},$$&#xA;where the stiffness $k_r$ is in $\left[\frac{Nm}{rad}\right]$, the second moment of area (or torsion constant) $J$ is in $\left[m^4\right]$, the rigidity modulus is in $\left[\frac{N}{m^2}\right]$ and the length $\ell$ is in $\left[m\right]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming linear elasticity the following relation between the rigidity modulus and the young modulus holds&#xA;$$ G = \frac{E}{2(1+v)},$$&#xA;where $v$ is the poison ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second moment of area (or torsion constant) for a circle is&#xA;$$ \iint\limits_{R} r^2\,dA =&#xA;   \int_0^{2\pi}\int_0^r r^2\left(r\,dr\,d\theta\right) = &#xA;   \int_0^{2\pi}\int_0^r r^3\,dr\,d\theta =&#xA;   \int_0^{2\pi} \frac{r^4}{4}\,d\theta = \frac{\pi}{2}r^4 =&#xA;   \frac{A\,r^2}{2}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\,$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the rotational stiffness of a cylinder can be rewritten as function of $k_l$&#xA;$$ k_r = \frac{\frac{E}{2(1+v)} \frac{A\,r^2}{2}}{\ell} =&#xA;   \frac{A\,E}{\ell} \frac{r^2}{4(1+v)} =&#xA;   k_l \, \frac{r^2}{4(1+v)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation does not hold for any other shape! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking at a screw, there is a thread at the outside which has a significant effect on the rotational stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, leaving half the diameter open in the cylinder, the stiffness will only decrease $0.0625\%$ which is insignificant. Increasing the outer radius will increase the stiffness significant.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastEditorUserId="8216" LastEditDate="2016-10-05T23:31:31.967" LastActivityDate="2016-10-05T23:31:31.967" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="11765" PostTypeId="2" ParentId="11757" CreationDate="2016-10-05T22:08:20.363" Score="0" Body="&lt;p&gt;Yes. You are correct. The can is first depressurized as the rivet is lifted allowing the lever action of the tab to pop open the scored can opening&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch this short video showing this is detail: &lt;a href=&quot;https://youtu.be/ekv0kprA3AY&quot; rel=&quot;nofollow&quot;&gt;https://youtu.be/ekv0kprA3AY&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-10-05T22:08:20.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11766" PostTypeId="1" CreationDate="2016-10-05T22:59:52.570" Score="-2" ViewCount="185" Body="&lt;p&gt;Aluminum 6061-T6 has a density of 2.7 g/cm^3, a Young's modulus&#xA;of 69 GPa, and an ultimate tensile strength of 310 MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a &quot;spring&quot; were to be made by pulling a 10cm long cylinder of&#xA;material in tension with a desired stiffness of 5000kN/m, what&#xA;should the diameter (in millimeters) be for an Aluminum&#xA;&quot;spring&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dont want the answer for this question but the FORMULA for working it out as i am totally unsure how to derive the diameter from the info given..&lt;/p&gt;&#xA;" OwnerUserId="8293" LastEditorUserId="8292" LastEditDate="2016-10-06T08:33:35.310" LastActivityDate="2016-10-09T00:03:08.527" Title="Spring Diameter Question" Tags="|springs|elastic-modulus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11767" PostTypeId="1" AcceptedAnswerId="11771" CreationDate="2016-10-06T00:31:28.227" Score="0" ViewCount="845" Body="&lt;p&gt;I know the ratio at which the cross sectional area changes to length is Poisson's ratio but is this why necking is more apparent in some materials than others or is the necking effect only dependent on the material's ductility? &lt;/p&gt;&#xA;" OwnerUserId="8294" LastActivityDate="2016-10-06T14:21:25.330" Title="Does Poisson's effect explain why the necking effect is more apparent in some materials during a tensile test?" Tags="|mechanical-engineering|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11769" PostTypeId="1" AcceptedAnswerId="11785" CreationDate="2016-10-06T05:06:49.520" Score="0" ViewCount="144" Body="&lt;p&gt;I am new to thermal management.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to lift about 250&amp;nbsp;W of heat from hot end of thermoelectric cooler/peltier cooler (TEC). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of hot side of TEC is 30°C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use some kind of heat exchanger/chiller to accomplish this task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P = h S (T_s-T_f) $$&#xA;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P$ = heat to be removed = 250&amp;nbsp;W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h$ = heat transfer coefficient &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S$ = area of contact between hot side of TEC &amp;amp; fluid (air/water) used&#xA;for convective heat transfer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_s$ = temperature of hot end&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_f$ = temperature of fluid (air/water)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know how to calculate $h$. If I calculate that, it will give me temperature of fluid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I compute $h$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And is my approach correct?&lt;/p&gt;&#xA;" OwnerUserId="7162" LastEditorUserId="1832" LastEditDate="2016-10-06T13:38:53.647" LastActivityDate="2017-09-05T22:11:16.900" Title="Lifting 250 watts of heat using some heat exchanger/heat sink" Tags="|mechanical-engineering|thermodynamics|heat-transfer|heating-systems|heat-exchanger|" AnswerCount="1" CommentCount="3" ClosedDate="2016-10-08T20:06:30.597" ContentLicense="CC BY-SA 3.0" />
  <row Id="11771" PostTypeId="2" ParentId="11767" CreationDate="2016-10-06T14:21:25.330" Score="0" Body="&lt;p&gt;It is important to think about the underlining physics of a tensile test. What material are you testing? What is the geometry you are testing? How much global strain is being applied? With these questions in mind, I think it is necessary to think about this problem both in terms of elastic and plastic deformations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the stress-strain behavior for two materials, one with no plasticity (traditionally rubbers) and one that exhibits plasticity (traditionally metals). First let's focus on the elastic material. It is possible to imagine a situation were differences in Poisson's ratio manifests as differences in necking behavior. One might consider there to be necking in a simple rectangular tensile specimen. In this situation the boundary conditions are such that displacements at the ends of the specimen are zero. Therefore, during deformation a curved edge profile might develop. For perfectly elastic deformations this is where it is clear that differences in Poisson's ratio might affect the relative necking between two materials. If the Poisson's ratio is zero, then we would not expect there to be any geometric changes in the sample--the rectangular specimen just gets longer. However, on the other end of the spectrum, if the material is incompressible, $\nu = 0.5$, the initial and final configurations must have the same volume. Geometrically this means, given the fixed zero displacement boundary conditions of our test, there needs to be some distribution of deformation, necking, within the specimen.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of plasticity the picture gets a bit more complicated. The short answer is that the final geometry of a tensile specimen will be dependent on its material properties. This includes both the plastic and elastic behavior (including the Poisson's ratio) of the material. Materials that are more ductile have a higher strain to failure, and this feature, when comparing elastically similar yet plastically dissimilar materials, predominately drives differences in necking behavior. If you have two materials, one which can withstand 2% strain and one 20% (assuming $\epsilon_{yield} = 1\%$), each increment of strain the second material can withstand relative to the first will contribute to differences in the final geometries of the specimens. In this example the failure of the material is driving the final configuration of the neck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, it is important to realize that one necking starts to occur things are no longer as simple as they appear. Once the specimen starts to neck the stress state with in the gage is, if it ever really was, no longer uniform. This means that at the smallest section of the neck stresses increase (dropping crosssectional area) and new components of stress start popping up. &lt;/p&gt;&#xA;" OwnerUserId="7056" LastActivityDate="2016-10-06T14:21:25.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11772" PostTypeId="1" CreationDate="2016-10-06T16:31:05.340" Score="1" ViewCount="224" Body="&lt;p&gt;I'm working on developing a speed controller for a motor that is geared to rotate a mass. Currently, the motor in question has a step voltage applied to it and moves the mass at a relatively slow speed. As it has been explained to me, applying a step voltage to a motor would create a large inductive load due to a high value of dv/dt, but as a novice in motors, I have been unable to find any constitutive relationship defining this rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, if I am attempting to ramp from a slow wind-up speed, to a high operating speed, and finally to a slow wind-down speed - what are some things I should take into consideration when designing the controller for this device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am definitely a novice in the field of controller design and am just beginning to learn all of this. I will be using labview.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="8302" LastActivityDate="2016-11-28T12:55:49.647" Title="Motor Controller Design Considerations" Tags="|mechanical-engineering|control-engineering|motors|pid-control|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11773" PostTypeId="2" ParentId="11684" CreationDate="2016-10-06T17:54:57.763" Score="0" Body="&lt;p&gt;Aside from safety and maintainability concerns, there are also costs with insulation and capacitive losses to consider. See for example: &lt;a href=&quot;https://www.quora.com/Why-arent-underground-cables-used-for-power-transmission-in-India&quot; rel=&quot;nofollow&quot;&gt;https://www.quora.com/Why-arent-underground-cables-used-for-power-transmission-in-India&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7622" LastActivityDate="2016-10-06T17:54:57.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11775" PostTypeId="1" CreationDate="2016-10-07T03:58:39.780" Score="2" ViewCount="8111" Body="&lt;p&gt;When we do detailing for RC structures, the rebars in a beam have to anchor into a column. For an external column (not at the roof), the rebars are usually bent in 90 deg to provide sufficient anchorage length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I read the design manuals, books, etc, the rebars are always bent to the far face of the beam. E.g. The top bars in the beam bend to the bottom and the bottom bars bend to the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May I know why is the bending direction always like that?&lt;/p&gt;&#xA;" OwnerUserId="8306" LastActivityDate="2016-10-15T13:42:00.760" Title="Reasons for bending direction of rebars in beams" Tags="|structural-engineering|civil-engineering|reinforced-concrete|detailing|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11776" PostTypeId="1" CreationDate="2016-10-07T05:43:53.230" Score="2" ViewCount="660" Body="&lt;p&gt;Is the accuracy interval the tolerance for the instrument?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the precision interval indicate a single standard deviation? &lt;strong&gt;And more importantly, does the precision assume a perfect operator (so the variance is due simply to randomness) or does it take into account an operators reading/technique?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask is because I did a bunch of measurement using a 1 mL micropipette with a &quot;precision&quot; of 0.002 mL and got a standard deviation of 0.0026 and am curious if that additional 0.0006 error is because of my poor technique compounding upon the instruments &quot;inherent&quot; uncertainty.&lt;/p&gt;&#xA;" OwnerUserId="442" LastActivityDate="2016-10-08T05:21:12.323" Title="What does the accuracy and precision values for a pipette mean?" Tags="|measurements|statics|tolerance|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11778" PostTypeId="1" CreationDate="2016-10-07T12:05:12.887" Score="7" ViewCount="3568" Body="&lt;p&gt;I am designing an aquatic center on the top floor of a multi-story building. The critical beams and columns to be designed are supporting a 25&amp;nbsp;m lap pool and I am wondering whether to apply a live-load or a dead-load safety factor to the water load acting on the structural elements.&lt;/p&gt;&#xA;" OwnerUserId="6624" LastEditorUserId="368" LastEditDate="2016-10-10T16:03:56.303" LastActivityDate="2016-10-10T23:08:20.057" Title="Should I model the load from a pool at the top floor of a multi-story building as live or dead?" Tags="|structural-engineering|civil-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11779" PostTypeId="2" ParentId="11778" CreationDate="2016-10-07T12:20:12.360" Score="6" Body="&lt;p&gt;Given that the water level will probably remain almost entirely stable, dead loading is the way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the only time I imagine the water level would change is if the pool is emptied for cleaning, but that is quite a rare event and, in fact, would &lt;em&gt;reduce&lt;/em&gt; the applied load, so no worry. Sure, when the pool is then being refilled there would be some dynamic action, but filling up a pool takes a lot of time, so the &quot;dynamic factor&quot; should be negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the reason safety factors are greater for live loads is due to their inherent variability. You may design a bridge for a 45-ton truck, but who knows how heavy trucks will be in 40 years' time? Meanwhile, dead loads are far more consistent: in 40 years' time, the bridge will weigh the same as it does now. And unless your material is defective or your construction is stupendously shoddy, the self-weight you use in your analysis will probably be within a few percentage points of the true weight. You can therefore place a lower safety factor for dead loads, while adopting large ones to cover the uncertainty in live loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @AndyT mentioned in a comment, there may be elements that are relieved by the pool. An empty pool in such cases would not have this relieving effect. Relieving dead loads are usually fully considered (maybe slightly reduced). Relieving live loads are ignored, causing no relief. For these elements, one may decide to consider the empty-pool case as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Ethan48 mentioned in a comment, it may be relevant to calculate the structure with the pool both empty and full. The empty pool could be relevant for the columns, when checking the &quot;maximum bending moment (due to wind, for example) and minimal axial compression&quot; case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, perhaps the most reasonable approach is to consider the load as a hybrid: with a dead load factor when unfavorable but a live load factor (0.0) when favorable.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This mostly applies to somewhat &quot;simple&quot; structures. If your building requires more complex calculations, then things might get messy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I'd guess this may have an important effect in seismic calculations, since the water (or lack thereof) would represent a large mass (or lack thereof) at the very top of the structure. As well, as mentioned by @CarlWitthoft in a comment below, if the building is slender and therefore vulnerable to wind-induced dynamics, the dynamic motion of the water could come into play and therefore act more like a live load.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-10-10T14:07:44.363" LastActivityDate="2016-10-10T14:07:44.363" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11780" PostTypeId="1" CreationDate="2016-10-07T13:18:14.347" Score="3" ViewCount="263" Body="&lt;p&gt;If you carefully look at LED in flashlights, they are blinking at fast rate despite the batteries are DC. It's even more obvious on flashlights with separate R,G,B LED (to allow the control of color) when you move it fast.&#xA;Why is it so ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;( It's also the case for &quot;power on&quot; indicators in many devices, but here it could just be a low quality DC low voltage from AC 220V. )&lt;/p&gt;&#xA;" OwnerUserId="7547" LastEditorUserId="7083" LastEditDate="2016-10-25T09:18:40.373" LastActivityDate="2016-10-25T09:18:40.373" Title="Why are LED blinking?" Tags="|electrical-engineering|consumer-electronics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11782" PostTypeId="2" ParentId="11775" CreationDate="2016-10-07T14:54:28.127" Score="0" Body="&lt;p&gt;The reason is simple geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top bars have the most clear space to bend down. They can't bend up (out of the beam) and bending to one side or the other means that they are likely to interfere with other bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't mean that the bar have to bend that way.  If there is enough room to maintain clear spacing, the bars could bend any way.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2016-10-07T14:54:28.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11783" PostTypeId="1" CreationDate="2016-10-07T17:29:01.437" Score="1" ViewCount="1943" Body="&lt;p&gt;In an effort to learn Embedded System Programming I recently installed Keil uVision 5.0. I am using a STM32F development board. At this point all I am interested in is getting &lt;a href=&quot;http://mightydevices.com/?p=121&quot; rel=&quot;nofollow&quot;&gt;STM32 L1 Tutorial #1: Hello, World! (LED Blink)&lt;/a&gt; code complied and loaded into the microcontroller. While attempting to build the code I got the following errors. What is the cause of this error?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*** Using Compiler 'V5.06 update 3 (build 300)', folder: &#xA;'C:\Doc\Programs\Keil52\ARM\ARMCC\Bin'&#xA;Build target 'Target'&#xA;Error instantiating RTE components&#xA;Error #540: 'Keil::Device:Startup:1.0.0' component is not available for target 'Target'&#xA;Target not created.&#xA;Build Time Elapsed:  00:00:00&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="8055" LastEditorUserId="7083" LastEditDate="2016-10-25T12:47:29.663" LastActivityDate="2016-10-25T12:47:29.663" Title="How to resolve Keil uVision 5.0 compilation errors?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11784" PostTypeId="2" ParentId="11775" CreationDate="2016-10-07T17:30:24.147" Score="0" Body="&lt;p&gt;The reason for bending the top bars down and the bottom ones up are to create a box which would act as a added ductile component to hold as a cage the end part of the beam even in cracked condition.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a major earthquake or in case of a catastrophic failure concrete can break into small chunks losing the stiffness. Bending of rebars in a this way will hold the cracked joint together stopping the building from collapse, providing precious time for evacuation. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-10-07T17:30:24.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11785" PostTypeId="2" ParentId="11769" CreationDate="2016-10-07T17:31:57.223" Score="1" Body="&lt;p&gt;This is a convective heat transfer problem you're asking about. Calculating the coefficient isn't normally possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can do is read up on stuff like fin geometry on heat sinks, and CFD.  The long-term solution here is to derive an equation for your system, something like P = f(X) where X is (for example) the length of a fin. Then, perform a CFD simulation for X = 1 to get P for that particular geometry. Then you can scale your fin until you get the right value for P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's get a bit more complicated because you probably shouldn't use dimensioned parameters for something like this.  Instead, consider something like the Reynold's number of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more reading to help you get started:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node128.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heat Transfer From a Fin&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Convective Heat Transfer&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6488" LastEditorUserId="8055" LastEditDate="2017-09-05T22:11:16.900" LastActivityDate="2017-09-05T22:11:16.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11787" PostTypeId="2" ParentId="11776" CreationDate="2016-10-08T05:21:12.323" Score="1" Body="&lt;p&gt;Precision and Accuracy are independent. As Andy has stated, accuracy refers to the ability of the device to measure as close to the absolute value as possible, whereas, precision refers to the closeness of the values in repeated measurements. Precision depends on repeatability and reproducibility of both the device and the measurers, assuming all other conditions remain the same (environmental conditions, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you should check how the manufacturer is defining the precision and accuracy. Typically, accuracy or trueness is the tolerance. Precision, may actually be expressed as the Precision/Tolerance ratio, so it is important in determining how you are arriving at your own capability. I would think that what you are after, is either a calibration of your device, or your own reproducibility and repeatability, rather than your precision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your particular numbers confuse me. How did you calculate your Standard Deviation? I am a bit thrown off, because I am tempted to interpret your mean as 0.002 and your SD as 0.0006. Please clarify. Here is some guidance on pipette accuracy determination:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://bitesizebio.com/613/how-accurate-are-your-pipettes/&quot; rel=&quot;nofollow&quot;&gt;http://bitesizebio.com/613/how-accurate-are-your-pipettes/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7989" LastActivityDate="2016-10-08T05:21:12.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11789" PostTypeId="2" ParentId="11780" CreationDate="2016-10-08T11:16:02.067" Score="3" Body="&lt;p&gt;The reason LED's are blinking is because they use Pulse Width Modulation (PWM) power the LED. Using PWM, the LED are switched on for a specific period of time and switch off rest of the time. The reason as suggested is Efficiency. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Over a period of time LED used less power, or in other words less average current&lt;/li&gt;&#xA;&lt;li&gt;The LED dissipates less heat, thus will degrade slow.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There reason that you are notice the blinking is because the duty cycle in your flashlight might be pretty low. Chances are if the LED is switched on around 200Hz, most humans will not notice blinking. To learn more I suggest investigating &lt;a href=&quot;https://en.wikipedia.org/wiki/Persistence_of_vision&quot; rel=&quot;nofollow noreferrer&quot;&gt;Persistence of vision&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Flicker_fusion_threshold&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flicker fusion threshold&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of a 50% duty cycle PWM signal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vHP08.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vHP08.png&quot; alt=&quot;PWM Signal&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/19040/limit-of-human-eye-flicker-perception&quot;&gt;Limit of human eye flicker perception?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://embeddedgurus.com/stack-overflow/2009/09/lowering-power-consumption-tip-2-modulate-leds/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lowering power consumption tip&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/325/what-is-best-way-to-control-brightness-of-a-led&quot;&gt;What is best way to control brightness of a LED?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://laptopmedia.com/highlights/what-is-pwm-why-is-it-important-and-why-did-we-add-this-test-to-our-reviews/&quot; rel=&quot;nofollow noreferrer&quot;&gt;What is PWM, why is it important and why did we add this test to our reviews?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-10-08T11:21:46.263" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11790" PostTypeId="2" ParentId="11783" CreationDate="2016-10-08T11:50:21.590" Score="2" Body="&lt;p&gt;On Keil uVision goto Project &gt; Manage &gt; Run-Time Environment. You will get a screen like the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ujwsU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ujwsU.jpg&quot; alt=&quot;Keil Run-Time Environment&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your device is not properly configure as indicated by the Red box. Click on the Device + to expand. Then select the appropriate &quot;System and Start up file&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you may try the resolve button in bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is screen shot of when the error is fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1yjJU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1yjJU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2016-10-08T11:50:21.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11791" PostTypeId="1" AcceptedAnswerId="11801" CreationDate="2016-10-08T12:25:25.363" Score="3" ViewCount="407" Body="&lt;p&gt;Tension and compression re-bars in a concrete beam are connected by stirrups vertically tied together. If the space in between (this corresponds to web of an I-Beam)  carrying shear load predominantly why are stirrups not placed along $\pm 45^0$ principal stress direction for better material economy ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a design practice could be in aerospace engineering for lightweight construction but not found in the originating civil/structural construction practice ? &lt;/p&gt;&#xA;" OwnerUserId="1270" LastEditorUserId="1270" LastEditDate="2016-10-08T16:00:17.373" LastActivityDate="2018-11-28T04:18:30.507" Title="Stirrups are not shear bearing diagonals but just spacers?" Tags="|structural-engineering|stresses|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11792" PostTypeId="1" CreationDate="2016-10-08T15:55:43.997" Score="0" ViewCount="449" Body="&lt;p&gt;I'm looking for a way to mount a 1000W electric heater to a stainless base. Electric kettle heaters could fit my requirements, but I'm wondering how the Aluminium heater is mounted to the stainless base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pictures below show a kettle heater on a stainless base. The steel is SS304 and the element's metal is a manganese-aluminium alloy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it mounted and what is the maximum temperature it can sustain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/73czD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/73czD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yeOdj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yeOdj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/AQpCN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AQpCN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8313" LastActivityDate="2016-10-08T16:43:44.710" Title="How is this aluminum heater mounted to the stainless plate?" Tags="|metallurgy|metals|aluminum|welding|brazing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11793" PostTypeId="2" ParentId="11780" CreationDate="2016-10-08T16:16:33.263" Score="4" Body="&lt;p&gt;The basic problem is that the power source is not matched to produce exactly the right voltage and current that the LED takes to put out the light you want.  Some &lt;i&gt;conversion&lt;/i&gt; is required between the power source and the LED.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting DC at one voltage and current to another voltage and current is done most efficiently with pulses.  It isn't too hard to smooth out these pulses by the time they get to the LED, but it does take a few extra parts and cost a few extra cents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In high volume mass market products, those few extra cents add up.  People are used to LEDs flashing, so the manufacturers don't feel much push to prevent the flashing.  Put another way, way more people are going to buy on price than whether the light flashes a bit.  Most people don't even think of looking at the latter, and it's rarely featured, so it's not a consideration in purchasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extra components not only add a little cost, but also weight and they take space.  All around, it's just not worth it when selling to the average consumer.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-10-08T16:16:33.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11794" PostTypeId="2" ParentId="11792" CreationDate="2016-10-08T16:34:21.923" Score="1" Body="&lt;p&gt;It looks like the element itself sits inside a casting (presumably the mag/aluminium alloy) which itself sits on a disk of the same material which is screwed to the stainless base. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not entirely clear is the screws in the picture are holding the element housing to the base or are there for some other purpose, but that is certainly one way to do it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see slightly darker grey area around the joint which I suspect may be thermal paste/adhesive  of some description to ensure good thermal conductivity between the element casing and the stainless steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermal epoxies ect are typically good for for up to around 300 deg C which should be plenty for a kettle as the boiling water will never get hotter than 100 deg C and acts as a good heat sink, although of course the element itself will need to get somewhat hotter than this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not really possible to weld aluminium to stainless steel (certainly not with any general purpose production process) , they can be soldered together with the right filler and flux, although these tend to be relatively low strength and low temperate and so may not be that much better than a combination of mechanical fixings and thermal paste. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermal paste/adhesive is important because unless you have a very, very good mechanical fit between the element and the stainless steel there will be a significant gap over much of the contact area, greatly reducing the rate of heat transfer. The thermal paste is very conductive (usually aluminum or sliver particles in some semi-fluid medium) which fills the gaps and creates direct, conductive contact over the entire surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are processes which involve &lt;a href=&quot;http://www.esabna.com/us/en/education/blog/can-i-weld-aluminum-to-steel.cfm&quot; rel=&quot;nofollow&quot;&gt;bimetallic interface inserts&lt;/a&gt; to allow welding of steels to aluminium alloys but these definitely fall under the category of 'advanced' welding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the reasons for using the aluminium casting is that the thermal conductivity of stainless steel is  fairly poor so you need a heat spreader made of a conductive material to distribute the relatively high temperature of the heating element evenly across the stainless steel to water interface to get the best heat transfer rate from the element to the water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a copper housing for the heating element would give you more options for higher temperature bonds  as it's a lot easier to weld or braze copper to stainless than aluminium to stainless. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-08T16:43:44.710" LastActivityDate="2016-10-08T16:43:44.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11795" PostTypeId="2" ParentId="10488" CreationDate="2016-10-08T18:11:20.947" Score="0" Body="&lt;p&gt;Usually data sheets will give a qualitative indication of maintainability along the lines of excellent/good/poor/not possible etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However you may need to seek out more subjective advice for specific alloys and processes as there are a huge variety of possible applications and there isn't a generally applicable  metric of maintainability. For example lots of long stringy swarf may be a major problem for automated production but no more than a mild inconvenience for one-off manual operations. Equally some issues like work hardening may be reasonably easy to mitigate by correct procedures if you are aware of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In very general terms many grades of aluminium are at least reasonably machinable and you are much less likely to encounter the severe difficulties that you can get with some types of stainless or high alloy steels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly some alloys may be easy to rough out but very resistant to taking a fine finish. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-10-08T18:11:20.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11796" PostTypeId="1" CreationDate="2016-10-08T19:38:09.920" Score="2" ViewCount="2773" Body="&lt;p&gt;Usually, worm gears drive helical wheels. For my application (hand-driven orrery), I am studying a more cost effective approach: using a straight wheel. It is technically possible as you can see with this LEGO model (&lt;a href=&quot;http://cs.smith.edu/~100e/gears.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rjUkc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rjUkc.jpg&quot; alt=&quot;LEGO worm gear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you design this kind of gear train? Specifically, is there a relation linking the thickness of the wheel $b$, its number of teeth $Z_2$, its module $m$, the base diameter of the worm $d$, its number of teeth $Z_1$, and its helix angle $\gamma$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that the tooth profile as involute and a pressure angle ($\alpha$) of 20°.&lt;/p&gt;&#xA;" OwnerUserId="8315" LastActivityDate="2020-06-19T07:04:04.787" Title="Design a worm gear that works with a straight wheel" Tags="|design|gears|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11797" PostTypeId="2" ParentId="11796" CreationDate="2016-10-08T23:29:08.247" Score="4" Body="&lt;p&gt;The reason why it works with the LEGO is because the clearances (i.e. the gaps between the worm and the spur gear) are sufficiently large so that the lead angle of the worm is not significant in preventing the two from meshing. In low load capacity where smooth power transmission is not required, this is okay. However, it is worth noting this is a sort of &quot;illegal&quot; mesh, illegal in the sense that spur gears are not designed to be used with cylindrical worms, except if the worm is oriented at an angle to the spur gear equal to the lead angle. Instead helical gears (at the least, for simple cylindrical worm drives*) are to be used. The region of contact will be considerably smaller for the &quot;illegal&quot; mesh compared to normal worm drives, so the contact stresses will be much higher, and so this illegal mesh is very unsuitable for high load capacity scenarios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the unsuitability of spur gears with worm drives with orthogonal shafts, you will be hard pressed to find any design equations; it just isn't a suitable configuration to design with. However, one natural consequence that still arises, despite the illegal mesh, is that the transmission ratio (ratio of turns of the worm to turns of the spur gear) is given by the number of teeth on the spur gear divided by the number of starts (i.e. number of unique threads) of the worm. This is quite an intuitive result to come across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, if you are just using the worm drive for kinematic purposes (e.g. precision alignment), under light loads, and you want to keep things cheap, then this soft of illegal mesh is reasonable; just don't expect the working life of the drive to be very high. It will be a matter of trial and error to design an operating drive. If, however, the machine is subject to high loads or requires smooth power transmission, I would highly suggest sticking to conventional worm gearing with existing design standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*More advanced worm drives exist. &quot;Single envelope&quot; worm gearing has a worm gear (not worm) as a modified &quot;helical&quot; gear, where the worm gear wraps around the worm more in an attempt to increase the area of contact. &quot;Double envelope&quot; is similar, except the worm gets more narrow in the middle so that its shape wraps around the worm gear more, further increasing contact area. These are often favoured over even simple cylindrical worm drives due to their higher load capacity. The more you know!&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2016-10-08T23:29:08.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11798" PostTypeId="2" ParentId="11766" CreationDate="2016-10-09T00:03:08.527" Score="-1" Body="&lt;p&gt;With a little bit of searching, a good theoretical value of the spring rate/stiffness $k$ (i.e. ratio of force to extension) can be calculated using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k=\frac{G d^4}{8D^3 N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $G$ is the shear modulus of the material, $d$ is the wire diameter of the spring, $D$ is the diameter of the turns of the spring, and $N$ is the effective number of turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be calculated from scratch by equating the work done by the tensile/compressive force acting on the spring with the elastic torsional energy stored inside the wire of the spring. The formula assumes that the length of the wire of the spring is approximately equal to the number of turns multiplied by the circumference of a circle of diameter $D$, i.e. the turns of the spring are rather flat.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2016-10-09T00:03:08.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11799" PostTypeId="1" CreationDate="2016-10-09T02:12:32.540" Score="3" ViewCount="1466" Body="&lt;p&gt;As everyone learns in physics class, Hooke's law states that a spring's force is proportional to the distance it is stretched/compressed.  The constant of proportionality is usually referred to as $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently working on a project for which I need a spring and I found online that I should probably use a &quot;K25&quot; spring for my application.  I assumed that this K is the same $k$ that is commonly used to represent the spring constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I actually started shopping for said spring I have found that it seems to be different.  Springs are listed with &lt;em&gt;both&lt;/em&gt; a &quot;rate constant&quot; and a K-something value.  See &lt;a href=&quot;http://www.mcmaster.com/#compression-springs/=14ijc66&quot; rel=&quot;nofollow&quot;&gt;this catalog list&lt;/a&gt;* from McMaster (the K-value is at the end of the part number, i.e. 9637&lt;strong&gt;K25&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two numbers don't seem to be related.  For instance there are two K25 springs, one has a rate constant of 28.53 and the other has a rate constant of 83.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what's the difference between these two numbers, what do they represent physically, and what is their relation to the spring constant $k$ from Hooke's Law?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;*For some reason McMaster's website doesn't allow me to link to a subsection of springs, I am looking in the &quot;Cut-to-Length Compression Springs&quot; category.&lt;/p&gt;&#xA;" OwnerUserId="8316" LastActivityDate="2017-10-25T05:50:26.337" Title="Difference between a spring's k-value and rate constant?" Tags="|springs|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11801" PostTypeId="2" ParentId="11791" CreationDate="2016-10-09T10:26:51.230" Score="3" Body="&lt;p&gt;Unfortunately the situation is more complicated than dealing with the principal stresses (see &lt;a href=&quot;https://engineering.stackexchange.com/questions/8043/why-dont-we-use-principal-maximum-stresses-when-designing-a-structural-element/8067#8067&quot;&gt;this answer&lt;/a&gt;). The principal stress direction will only be $\pm 45^o$ when in a condition of pure shear. So for a beam this is at the neutral axis, away from the neutral axis the principal stress direction will change. This can be seen here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lVA1Mm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lVA1Mm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current &lt;a href=&quot;http://www.concretecentre.com/Concrete-Design/Design-Codes/Eurocode-2/Sheart.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;concrete shear design&lt;/a&gt; takes a &lt;a href=&quot;http://www.concretecentre.com/Concrete-Design/Design-Codes/Eurocode-2/Stut-and-Tie.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;strut-and-tie&lt;/a&gt; approach. For this we assume a truss made from steel tension ties (reinforcement) and concrete compression struts:&#xA;&lt;a href=&quot;http://manuelalvarezlopez.blogspot.co.uk/2012/12/strut-and-tie-to-z.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BeMib.png&quot; alt=&quot;enter image description here]&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this image the dashed lines are in compression (which can be carried by the concrete), while the solid lines are in tension (which are carried by the reinforcement). According to Eurocode 2, the inclination of the compression structs is allowed to vary between $21.8^o$ and $45^o$, and depends on the shear present in the beam. This will vary along the length of the beam, resulting in the stirrup spacing changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inclining the stirrups can achieve an increase in shear resistance. However, inclined stirrups are very unusual in practice. This is due to some of the disadvantages of inclined stirrups, for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;During load reversal (during an earthquake, for example), you would end up with effectively no shear reinforcement - which could result in failure.&lt;/li&gt;&#xA;&lt;li&gt;The extra expense/complication involved in placing inclined reinforcement is usually not worth the relatively minor increase in shear resistance. &lt;/li&gt;&#xA;&lt;li&gt;In most design situations the extra cost/time/risk involved in developing a non-standard design is discouraged except when absolutely necessary.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-10-09T10:26:51.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11802" PostTypeId="1" CreationDate="2016-10-09T11:02:07.857" Score="1" ViewCount="4715" Body="&lt;p&gt;Since globe valves are unidirectional and should be installed in one direction only I encountered a valves supplier that does not mark the globe valve body with flow direction, the valve eventually got welded (socket weld ends, not flanged) in a by-pass branch on a skid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might go wrong if the valve is installed incorrectly? Is there any way to check this other than the flow direction on the body (in the as-built situation)?  (the body is symmetrical for both inlet and outlet sides).&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2016-10-10T09:57:04.407" Title="What would happen to a misinstalled globe valve?" Tags="|mechanical-engineering|piping|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11803" PostTypeId="1" CreationDate="2016-10-09T22:43:58.223" Score="1" ViewCount="719" Body="&lt;p&gt;If water goes through a pipe and ends up at higher elevation, what does it mean for the water to have friction head loss?&lt;/p&gt;&#xA;" OwnerUserId="8325" LastEditorUserId="8325" LastEditDate="2016-10-10T03:51:23.063" LastActivityDate="2016-10-10T04:49:31.203" Title="What is friction head loss through a pipe mean?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11804" PostTypeId="1" AcceptedAnswerId="25406" CreationDate="2016-10-10T00:34:28.137" Score="4" ViewCount="256" Body="&lt;p&gt;&lt;strong&gt;How to deal with thermal expansion in a city wide heating system?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fptiK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fptiK.jpg&quot; alt=&quot;Picture of the pipes in street&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As part of the program to combat pollution in and around Beijing, my adopted &lt;a href=&quot;https://en.wikipedia.org/wiki/Zhuozhou&quot; rel=&quot;nofollow noreferrer&quot;&gt;city of Zhuozhou&lt;/a&gt; (population, over half million, about 90 minutes Southwest of central Beijing, inhabited for at least 5,000 years) is installing a hot water heating system. In the last 3 weeks, they have torn up every major street in town, and they are in the process of installing pipes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two pipes because the water will be conserved; according to my father in law, the hot water will flow through one pipe, they enter nearby buildings, flow through the radiators in the building, then return back to the central heating facility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, everyone is happy as this will stop a lot of coal burning, and vastly improve the air quality. Nobody seems too bothered that every street in town has been dug up at once. While traffic is horrendous, it seems like they want centralized heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have seen thus far is that they put the pipes close to each other, then weld them together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think when the heat is running, the length and size of these pipes will change drastically. &lt;strong&gt;How do you make sure the street above the pipes won't buckle? When the pipes get longer, how to prevent them from leaking? Won't the welds break? Also, are they a special type of pipe?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M8anh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M8anh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8328" LastEditorUserId="6489" LastEditDate="2018-03-30T14:43:03.780" LastActivityDate="2019-01-11T10:37:41.833" Title="How to deal with thermal expansion in a city wide heating system?" Tags="|civil-engineering|temperature|piping|heating-systems|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11805" PostTypeId="1" CreationDate="2016-10-10T02:10:22.567" Score="0" ViewCount="153" Body="&lt;p&gt;The given problem asks for the Volume of the given figure below&lt;a href=&quot;https://i.stack.imgur.com/9ww3Q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ww3Q.jpg&quot; alt=&quot;The Problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a plane figure that somewhat forms a Two-circular base Paraboloid with radii 20 and 12 cms respectively, and an altitude of 16 cm when revolved about the y-axis. According to the book, the answer is 15,388 cubic centimeters and my answer on my first attempt was 4352pi cubic centimeters. I would like to know if this is an author's mistake or mine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used &lt;strong&gt;Pappus-Guldinus Theorem  II - The Solid of Revolution&lt;/strong&gt; to compute manually and used the formula of a two-circular base Paraboloid to verify the answer. But I still ended up getting 4352pi cu cm.&lt;/p&gt;&#xA;" OwnerUserId="8329" LastActivityDate="2016-10-12T17:38:57.127" Title="Statics - Volume of a Revolution" Tags="|statics|" AnswerCount="1" CommentCount="1" ClosedDate="2016-10-12T17:17:53.897" ContentLicense="CC BY-SA 3.0" />
  <row Id="11806" PostTypeId="1" CreationDate="2016-10-10T02:33:32.647" Score="1" ViewCount="298" Body="&lt;p&gt;I'm recently under industrial training at a plywood mill. My supervisor is from foreign country so there are some difficulties for me to understand what he is trying to say.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the power plant, the turbine is speed is maintain at 5600 rpm, no matter how much is the generator output is. So, why require more steam for the turbine to rotate at the speed of 5600 rpm during high electricity usage than low usage even the rpm is the same?&lt;/p&gt;&#xA;" OwnerUserId="8330" LastEditorUserId="33" LastEditDate="2019-05-09T20:36:20.707" LastActivityDate="2019-05-09T23:09:06.490" Title="Why is more steam required to turn a steam turbine when electric power consumption is higher" Tags="|electrical-engineering|turbines|steam|boilers|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11807" PostTypeId="1" AcceptedAnswerId="11808" CreationDate="2016-10-10T03:25:58.230" Score="3" ViewCount="341" Body="&lt;p&gt;I'm a Junior computer engineer working on a project involving a 3d printer. It is a mobile printer with the print head attached to wheels which move along the x,y axis of the surface it is printing on. The wheel movements may be off by less than a millimeter as the printer is moving. This causes a serious problem after printing a few layers as each layer does not land on top of the other.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am in the process of implementing a sensor to detect the location of the print head with respect to some arbitrary reference point. My task is to implement a feedback control loop to correct the printer's movements and have the object printed properly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an idea of how I want to solve this; using the sensor to detect the current location, then read the code to check what is the desired location, I then implement code to move the difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is where does the concept of PID control even fit in? The following equations were taken from wikipedia for time/frequency domains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://wikimedia.org/api/rest_v1/media/math/render/svg/7e13a0cb7bc543e507ae58384edcebcdcace5d27&quot; alt=&quot;equation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://wikimedia.org/api/rest_v1/media/math/render/svg/842e237127cf16ee7e961c84116df3946c887b99&quot; alt=&quot;equation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused about the values of t and s in the above equations. I have numeric values for input, output, error and the K constants. None of the values are functions so I don't understand what happens with the derivative and integration involved. Sorry if this seems like a dumb question but I am a complete novice to control theory and have been learning from online articles.&lt;/p&gt;&#xA;" OwnerUserId="8327" LastActivityDate="2016-11-24T23:20:24.980" Title="Implementing PID feedback control" Tags="|control-theory|3d-printing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11808" PostTypeId="2" ParentId="11807" CreationDate="2016-10-10T04:38:53.767" Score="1" Body="&lt;p&gt;You do in fact have a function. You have a value that changes, $e(t)$ is the error in your system (error in function of time). It will take you a while to learn to read math notation, don't worry took me ages to learn. But basically the formulation reads as follows, term by term:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Proportional term:  $K_P * (v_m - v_d)$, where $v_m$ is the measured value and $v_d$ is what you desire the value to be. $K_P$ is just a multiplier on how aggressively you want to maneuver the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially the proportional term works by adjusting the system towards to desired value. But may never reach it due to needing to overcome forces and reaches equilibrium before hitting target.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The integral term: Is just the sum of errors over time. So what you do is you calculate $ v_m-v_d $ and add it to a stored value from the previous time you calculated this same thing multiplying that by $K_I$. Again the $K_I$ is just a multiplier how much you want this effect to affect the maneuvering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of the integral term is to gradually overcome the persistent error and account for any changes in the system (hydraulic fluid leak etc).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The derivative term is the error from last time minus the error now. Its not needed in most systems, but it has a slowing effect on the other values. It is also somewhat noisy so you may want to calculate it with a more sophisticated method than just from two previous samples.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Its best not to look too much on the mathematical formulation. Best look up some code that does it, its much easier.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Like this article on &lt;a href=&quot;http://brettbeauregard.com/blog/2011/04/improving-the-beginners-pid-introduction/&quot; rel=&quot;nofollow&quot;&gt;PID and code&lt;/a&gt; and avoiding small problems.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2016-10-10T07:15:56.437" LastActivityDate="2016-10-10T07:15:56.437" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="11809" PostTypeId="2" ParentId="11803" CreationDate="2016-10-10T04:49:31.203" Score="1" Body="&lt;p&gt;In simplest terms to convey a volume of fluid through a pipe some amount of energy is required. A part of this energy is lost to the resistance to flow. When a fluid flows through the pipe, it encounters resistance and head loss due to shear stresses (within the fluid) and turbulence at boundary with pipe wall( depending on surface roughness). This resistance measured in meters (head) of the fluid is the frictional head loss.&#xA;You can use the Darcy-Weisbach equation in conjunction with Moody chart to find friction factor and head loss due to friction in pipe run. You also can account for pressure drop and frictional losses due to changes in direction in pipe bends and fittings (using manufacturer's k factors)&lt;/p&gt;&#xA;" OwnerUserId="3946" LastActivityDate="2016-10-10T04:49:31.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11810" PostTypeId="2" ParentId="11802" CreationDate="2016-10-10T05:04:49.727" Score="1" Body="&lt;p&gt;Globe valves are installed in such a direction that the pressure side is under the disc to protect the packing from pressure in the off position. If installed backward the pressure will be on the top side and this reduces ability to throttle the valve at near closed position. Maintenance of spindle gland cannot be done in this orientation and there is more chance of leaks developing due to pressure on the packing which cannot be serviced as well.&#xA;To check the direction of installation you can take help of embossed orientation arrow,if the valve is of asymmetrical construction the end under the disc is the inlet, if the valve is of symmetrical make and you have already installed it you cannot peer into it to observe which end has the mechanism you might have to settle for identifying the end by using the vendor drawings and direction of rotation of hand wheel as a reference (CW or CCW)&lt;/p&gt;&#xA;" OwnerUserId="3946" LastEditorUserId="3946" LastEditDate="2016-10-10T09:57:04.407" LastActivityDate="2016-10-10T09:57:04.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11812" PostTypeId="2" ParentId="11806" CreationDate="2016-10-10T08:08:09.217" Score="4" Body="&lt;p&gt;Because you can't get something for nothing?&#xA;If the generator is generating more electrical power then that energy has to come from somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more power you try to pull from a generator the harder it becomes to turn the generator. If it's harder to turn the generator then you need to push harder on the turbine driving it. The way you push harder on the turbine is by feeding it more steam.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-10-10T08:08:09.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11813" PostTypeId="1" CreationDate="2016-10-10T11:11:05.113" Score="2" ViewCount="168" Body="&lt;p&gt;Classically, if you have free flow (high level to low, no pump) between two vessels you want to control you use some restricting orifice, or flow over a variable height weir. Both dissipate energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you could have a turbine in the flow, attached to a generator. &lt;em&gt;If&lt;/em&gt; you adjust the torque applied by the generator to the turbine, you can control the flow over some range and extract power. I think this is an old idea, I just don't know the corect term to properly google it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many considerations why this may not be feasible or worth the trouble in a given instant (e.g. hydraulic efficency of turbine over the flow ranges required). The applications I'm thinking of would be wastewater treatment plants with flows 1-10m³/s and heads &amp;lt;1m - so single digit kW most likely. At these heads, Banki or axial flow type turbines would probably be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about the possibility - Can you actually control flow by appying breaking torque to a turbine, what does this do to hydraulic efficiency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I'll ask about the electrical side in a different question, while certainly not trivial it's out of scope for now.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2022-11-29T17:30:56.493" Title="Does this exist - Water flow control with energy recovery" Tags="|mechanical-engineering|fluid-mechanics|turbines|hydraulics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11814" PostTypeId="1" CreationDate="2016-10-10T13:42:53.190" Score="1" ViewCount="98" Body="&lt;p&gt;I'm coming from the software engineering world (please don't rip me to shreds here), and I'm somewhat unfamiliar with the manufacturing process when it comes to cut and bend steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To provide broader details, we're making a pretty simple steel enclosure (basically a box), that will house some electronic, and need to build a relatively low number of them (a few hundred).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So my question is:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are there CAD tools specifically created to allow for the design of a folded steel enclosure?&lt;/li&gt;&#xA;&lt;li&gt;If I needed to provide a standard design document to a variety of manufacturing places to get quotes, which should my document look like? What info should it have about the enclosure?&lt;/li&gt;&#xA;&lt;li&gt;What are my best resources to find a vendor to help us build something like this?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sorry about how noobish this all sounds, I'd love to learn more about this process, any help is very appreciated.&lt;/p&gt;&#xA;" OwnerUserId="8334" LastActivityDate="2016-12-11T16:48:26.090" Title="Folded/bent sheet steel design" Tags="|steel|cad|industrial-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11815" PostTypeId="1" CreationDate="2016-10-10T17:55:59.210" Score="1" ViewCount="11734" Body="&lt;p&gt;I have spent two days searching the web but haven't found out how to define the following parameters for weep holes in retaining walls.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;horizontal and vertical (if staggered situation) spacing between each weep hole;&lt;/li&gt;&#xA;&lt;li&gt;slope degree;&lt;/li&gt;&#xA;&lt;li&gt;diameter;&lt;/li&gt;&#xA;&lt;li&gt;the space between bottom of the ground and the weep hole also form the top as well.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8338" LastEditorUserId="1832" LastEditDate="2016-10-10T18:01:29.237" LastActivityDate="2017-10-07T02:14:53.960" Title="weep hole in retaining wall" Tags="|structural-engineering|retaining-wall|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11817" PostTypeId="1" CreationDate="2016-10-10T20:14:04.333" Score="1" ViewCount="1663" Body="&lt;p&gt;I'm trying to use LED's as a point source of light for a project I have.&#xA;a crucial idea is that the intensity of the light of the led falls off as 1 over the distance squared as we move away from it. i.e. it should obey the inverse square law.&#xA;I ran some measurements on an LED and tried to fit the results on a curve but did not get that result,(the intensity was more proportional to $1/d^3$).&#xA;my measurements could be non-accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't found any helpful sites that point out that LEDs actually follow the inverse square law, but I'm still skeptic that it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so my question is, can an led be used to some extent as a point source? how does an LED light vary with distance and with direction? if unfit, what other light emitting circuit components can be used instead if any?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate any kind of help. &lt;/p&gt;&#xA;" OwnerUserId="8341" LastActivityDate="2016-10-27T11:59:40.493" Title="Variation of LED light as a function of distance" Tags="|electrical-engineering|materials|mathematics|experimental-physics|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11818" PostTypeId="1" AcceptedAnswerId="38221" CreationDate="2016-10-10T20:16:38.460" Score="7" ViewCount="168" Body="&lt;p&gt;I'm looking for a control method for a production process with the following characteristics:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 control variable&lt;/li&gt;&#xA;&lt;li&gt;many process parameters (+-50)&lt;/li&gt;&#xA;&lt;li&gt;1 resulting variable&lt;/li&gt;&#xA;&lt;li&gt;continuous measurement of resulting variable with 30-90 second delay&lt;/li&gt;&#xA;&lt;li&gt;complex physics that govern the process and numerous factors that determine the (non-linear) relation between control variable and resulting variable&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With constant process/input parameters, the distribution of the resulting variable is normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control method should ensure that the resulting variable should be above a set minimum and that the average of the resulting value should be as close to the limit as possible. Being above the lower limit makes the product more expensive, being below the lower limit however is more expensive: the product is scrapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to try the following: I have want to make a statistical model using some process parameters as input predicting the control variable that leads to the desired value for the resulting variable. Next to the value of the control variable, I also want to determine the 'confidence' or distribution of that value. Then I want to state a portion of the time the control variable must be above the lower limit (i.e. 3 sigma). Consecutively I'd like to set the setpoint for the control variable at the such a value that I achieve the given portion above the lower limit. To illustrate what I envision I made the illustration below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E4uRa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E4uRa.png&quot; alt=&quot;Feed forward Control Optimization based on prediction confidence&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think this setup makes any sense? Am I missing any obvious alternative solution that is much simpler? Is this a known approach? &lt;/p&gt;&#xA;" OwnerUserId="6099" LastEditorUserId="18163" LastEditDate="2018-11-04T19:14:50.010" LastActivityDate="2020-10-16T06:21:39.280" Title="How to optimise feed-forward control of a process based on a prediction using the prediction confidence?" Tags="|control-engineering|steel|process-engineering|optimal-control|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11819" PostTypeId="2" ParentId="11813" CreationDate="2016-10-10T20:28:02.893" Score="1" Body="&lt;p&gt;Flow can be controlled using the various &lt;strong&gt;governing mechanisms&lt;/strong&gt; of the turbines.&lt;br /&gt;A generator generally does not have anything to do with the control system of turbine.&lt;br /&gt;Usually the flow is controlled using the gate opening with the help of governing mechanisms which nowadays are automated.&lt;br /&gt;Efficiency of the turbine if operated under constant head has maximum value at certain speed at certain gate opening.&lt;br /&gt;Graphs have been plotted that give you a fair idea of the values of these parameters.&#xA;&lt;br /&gt;&lt;strong&gt;Efficiency increases with increasing flow rate&lt;/strong&gt; but then it remains constant after a certain value of flow rate.&lt;br /&gt;According to your application, space, budget and arrangements, you can go with &lt;em&gt;kaplan&lt;/em&gt; or &lt;em&gt;propeller&lt;/em&gt;.&lt;br /&gt;Since you want to generate power you will probably run the turbine at a constant speed.&lt;br /&gt;Try searching for &lt;strong&gt;operating characteristics&lt;/strong&gt; and &lt;strong&gt;main characteristics&lt;/strong&gt; of turbines.&lt;/p&gt;&#xA;" OwnerUserId="8340" LastActivityDate="2016-10-10T20:28:02.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11820" PostTypeId="2" ParentId="11778" CreationDate="2016-10-10T23:08:20.057" Score="0" Body="&lt;p&gt;You need to check with your local building codes. California has special codes for treating elevated water tanks and pools and containers. Not only they are treated with dynamic load factor but other considerations in structural assembly, detailing and use of facility apply.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively in case of an earthquake after part of pool water is spilled out, after a few seconds the left over will start to swing back and forth and collide to the walls of the pool and break on them in an explosive impact, much like the shore line retaining walls and piers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to be mindful of local seismic spectra and avoid resonance and in the absence of dynamic lateral analysis follow the prescribes scheme advised by you local building department. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-10-10T23:08:20.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11822" PostTypeId="1" CreationDate="2016-10-11T00:15:34.807" Score="3" ViewCount="72" Body="&lt;p&gt;I'm working with a spectrometer that collects raw data as a 16-bit integer value, so the readout I get from a measurement always varies between 0 and 65,535.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to avoid implementing data with false precision in subsequent analysis of the data. In order to quantify the level of instrument precision / repeatability, I took multiple measurements of the same target with consistent illumination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest and lowest values measured for the consistent target varied by 1250 values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would one use these results to properly round the data collected by the spectrometer in order to avoid implementing false precision? Is there a set of 'best practices' for rounding measured values? Would I want to round my output to the nearest 1250 values? To the nearest 1000? The nearest 2000?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the output from the instrument for a measurement on a completely different target is 59279. Knowing the instrument is imprecise to 1250 values, what should the output be rounded to in order to avoid false precision?&lt;/p&gt;&#xA;" OwnerUserId="7169" LastActivityDate="2017-01-09T16:01:48.173" Title="Quantifying Instrument Precision Using Multiple Measurements: Rounding" Tags="|measurements|spectrometry|calibration|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11823" PostTypeId="2" ParentId="8903" CreationDate="2016-10-11T02:33:48.477" Score="3" Body="&lt;p&gt;One state space representation of $G(z)$ is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{array}{l c r}&#xA;A = \begin{bmatrix}&#xA;0 &amp;amp; 0 &amp;amp; -1 \\&#xA;1 &amp;amp; 0 &amp;amp; 5 \\&#xA;0 &amp;amp; 1 &amp;amp; 1&#xA;\end{bmatrix} &amp;amp;&#xA;B = \begin{bmatrix}&#xA;1 \\ 2 \\ 1&#xA;\end{bmatrix} &amp;amp; &#xA;C = \begin{bmatrix}&#xA;0 &amp;amp; 0 &amp;amp; 1&#xA;\end{bmatrix}&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with dynamics,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x(t+1) = A\,x(t) + B\,u(t)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y(t) = C\,x(t)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to find a $L$ and $K$, such that both $A-L\,C$ and $A-B\,K$ have of its eigenvalues at zero. Since I already put the system in the observable canonical form, finding $L$ will be easy, namely the last column of $A$. In order to find $K$ it might be easier to transform the system to the controllable canonical form, find $K$ there and transform it back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now estimate the internal states of the system $x(t)$ with $\hat{x}(t)$, using information from $y(t)$, $u(t)$ and the dynamics of the system,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\hat{x}(t+1) = L\,y(t) + \left(A - L\,C\right) \hat{x}(t) + B\,u(t)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose the input $u(t)$ as follows,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;u(t) = -K\,\hat{x}(t)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and define $e(t)$ as the error between the actual state $x(t)$ and its prediction $\hat{x}(t)$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;e(t) = x(t) - \hat{x}(t)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the total dynamics of the system can be written as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;x(t+1) \\ e(t+1)&#xA;\end{bmatrix} = &#xA;\begin{bmatrix}&#xA;A - B\,K &amp;amp; B\,K \\&#xA;0 &amp;amp; A - L\,C&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;x(t) \\ e(t)&#xA;\end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However you have chosen the eigenvalues of each matrix on the diagonal to have all zero eigenvalues, so after six step the system will always go to zero (independent of how good your initial guess is for $\hat{x}(t)$).&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2016-10-11T02:33:48.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11824" PostTypeId="1" AcceptedAnswerId="13349" CreationDate="2016-10-11T05:56:17.237" Score="3" ViewCount="13328" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XUpGx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XUpGx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;For the connection above, I think that once the beam  is loaded, their is no room for rotation of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as per British Standard this is a pinned connection. So I have been trying to determine how a connection is defined as Pinned or fixed. After reviewing American and British codes, I have come to the conclusion that steel structure designers need to follow their Standard connections for design (for which they have done experiments) or conduct experiments on connections to determine to evaluate whether its a pinned or connections based on certain criteria. Is my understanding correct?&lt;/p&gt;&#xA;" OwnerUserId="1693" LastActivityDate="2017-01-20T19:45:31.633" Title="Steel Structure connection pin or fixed" Tags="|steel|structures|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11825" PostTypeId="2" ParentId="11817" CreationDate="2016-10-11T11:48:52.187" Score="7" Body="&lt;p&gt;All EM radiation from a point source falls off in intensity with the square of the distance.  LEDs can't magically change this basic physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LEDs can be quite directional in their emission pattern.  One possibility is that the orientation of the LED relative to the receiver wasn't exactly the same in all your measurements.  Another is that you were fairly close to the LED so that the point source approximation was not valid.  Yet another possibility is that scattering and reflection from surrounding objects was contributing a significant fraction of the received light.  Or, the receiver could just be calibrated badly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many possible explanations, but light intensity falling off with the cube of the distance from a point source is not one of them.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2016-10-27T11:59:40.493" LastActivityDate="2016-10-27T11:59:40.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11826" PostTypeId="2" ParentId="11822" CreationDate="2016-10-11T13:52:26.263" Score="1" Body="&lt;p&gt;It's more than merely a question of absolute max and min.  I recommend you read Bevington's  &quot;Data Reduction and Error Analysis&quot; as a good, simple start.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precision can only be estimated based on the statistics of your data, and that depends on your entire setup:  temporal and spatial uniformity of the light source, for example.  First, calculate the mean value and the standard deviation. Then wander over to formalisms such as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Student%27s_t-test&quot; rel=&quot;nofollow&quot;&gt;Student's T-test&lt;/a&gt; and the general area of &quot;confidence intervals.&quot;  These algorithms allow you to use the measured parameters (mean, standard deviation, etc) to estimate the uncertainty in your measurement.&lt;br&gt;&#xA;Roughly speaking, if your uncertainty is &lt;code&gt;x&lt;/code&gt; then you shouldn't claim more precision than &lt;code&gt;x&lt;/code&gt; . (Strictly speaking, there's nothing wrong with reporting a mean value of  $3.1415926536 \pm 0.003$ but it's not particularly useful.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-10-11T13:52:26.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11827" PostTypeId="2" ParentId="11817" CreationDate="2016-10-11T13:56:39.210" Score="2" Body="&lt;p&gt;As Olin said, LEDs, even without their lens cap, are quite directional.  If you want a decent point source, the typical approach is to use an external lens to refocus the LED output as best you can onto a pinhole (20 microns if possible). This will cause some light loss but the output of the pinhole should be much closer to an ideal point source. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-10-11T13:56:39.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11829" PostTypeId="2" ParentId="11824" CreationDate="2016-10-11T16:18:13.413" Score="1" Body="&lt;p&gt;A fixed connection is a connection that would develop at least full plastic limit of the beam before yielding in rotation. Any connection less than that is considered pin connection.&lt;br&gt;&#xA;All the members joining at the connection have to be reinforced by stiffeners and connection details to bear the moments and stresses and transfer them per approved load tested methods. The joint in your diagram has a couple of grave weaknesses: the flanges are not continued into the connection, the connection is into the web of the column which has the least stiffness and will warp immediately allowing rotation of joint in all directions, the bolts will pop under high stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joint design is a very complex process, not only you have to consider service loads you have to mind buckling, torsion, stresses beyond rupture point and delayed critical failure!&#xA;Many codes provide for predictable ductile mode of failure of the joint in case of a catastrophic event such as explosion or earthquake.&lt;br&gt;&#xA;I saw this on youtube, it may be helpful. I do not know them and can not vouch for them.&lt;a href=&quot;https://www.youtube.com/watch?v=SmNJk51R-RM/%22See%20Here%22&quot; rel=&quot;nofollow&quot;&gt;see this clip&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2016-10-11T21:21:34.413" LastActivityDate="2016-10-11T21:21:34.413" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11830" PostTypeId="2" ParentId="11817" CreationDate="2016-10-11T17:34:48.170" Score="0" Body="&lt;p&gt;Technically, any EM radiation that has a reflecting surface can fall off with distance cubed.  It is the same thing that happens with radio transmission and ground bounce.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, I would highly recommend checking your test equipment.  Make sure it is linear.&lt;/p&gt;&#xA;" OwnerUserId="4403" LastActivityDate="2016-10-11T17:34:48.170" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11831" PostTypeId="2" ParentId="3648" CreationDate="2016-10-11T18:42:10.120" Score="0" Body="&lt;p&gt;Another option to consider is that many phones have a high intensity LED built in near the camera lens which works as a flash. It may be possible to use mirrors or a light pipe to make use of this. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-10-11T18:42:10.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11832" PostTypeId="1" CreationDate="2016-10-11T21:01:54.577" Score="2" ViewCount="1127" Body="&lt;p&gt;In the book &quot;&lt;em&gt;Against the Tide: Rickover's Leadership Principles and the Rise of the Nuclear Navy&quot;&lt;/em&gt;, Rear Admiral Dave Oliver (Ret.) narrates some of his experiences in the USS &lt;em&gt;George Washington Carver&lt;/em&gt;, SSBN-656, launched in 1966, and in which he was hired as reactor controls officer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, he mentions the problem of hydrogen generation and build-up which occurs during the battery charging process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Wikipedia entry for &lt;a href=&quot;https://en.wikipedia.org/wiki/Oxyhydrogen&quot; rel=&quot;nofollow&quot;&gt;oxyhydrogen&lt;/a&gt;, also called &quot;Knallgas&quot; for obvious onomatopoeic reasons, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At standard temperature and pressure, oxyhydrogen can burn when it is&#xA;  between about 4% and 95% hydrogen by volume&quot;. So one would really like&#xA;  to keep hydrogen concentration in an underwater closed space below the&#xA;  4%-by-volume mark. Incidentally, the most bang for the mix would be&#xA;  reached at the 2:1 hydrogen:oxygen ratio (I think this means that we&#xA;  would have a 66%-by-volume hydrogen concentration, not sure).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a diesel submarine, hydrogen generation is not a problem. Batteries are charged by running the diesel engine while on surface or schnorkeling. The diesel engine is air-breathing and is configured to suck air through the length of the submarine hull. Air circulation is high enough that free hydrogen never reaches dangerous values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In nuclear submarines however, batteries can be charged while fully submerged. Thus free hydrogen can become a problem quickly. To mitigate the risk, air is made to circulate throughout the hull and the hydrogen is captured (likely, the air is made to circulate through a catalyzer). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author narrates an event &quot;in early 1968&quot; whereby &lt;em&gt;George Washington Carver&lt;/em&gt; was put into &quot;Condition Baker&quot; while battery charging was ongoing. In Condition Baker, the boat is compartmentalized and airflow thus evidently stops. Hydrogen concentrations in the compartments skyrocket immediately to dangerous levels. It turns out that &quot;Condition Baker&quot; was meant to make a collision between a diesel submarine operating at periscope depth near commercial vessel lanes and a surface ship survivable. Nuclear submarines would operate far from commercial lanes and setting them into &quot;Condition Baker&quot; was just a procedure fossil, now dangerous in its out-of-context application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on page 42, the author says the following which is not entirely in accordance with known oxyhdrogen behaviour:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...the process of charging batteries, even on board a nuclear&#xA;  submarine, is one of the most dangerous routine evolutions the&#xA;  submarine performs. The procedure produces hydrogen, especially in the&#xA;  last hour of the charge, and hydrogen is not welcome on board a&#xA;  submarine. As most high school chemists have learned, if the hydrogen&#xA;  in a flask reaches the value of 8 percent, it spontaneously combines&#xA;  with the oxygen in the flask. The chemical reaction produces a lost of&#xA;  heat - lots and lots - which on board a submarine is enough to peel&#xA;  the hull back as if it were an overripe banana.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Were could the &quot;8% hydrogen will kill your boat&quot; rule come from? Is it a rule of thumb? Would it not be better to keep the concentration below 4% instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 8% are serious business and have Hollywood precision. The author is monitoring hydrogen concentration using two measuring devices (how does one measure hydrogen concentration btw?). During the Condition Baker event:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...the hydrogen needles had leaped to the red area of the gauge. The&#xA;  were now indicating 3 1/2 and 4 percent. One gauge, two needles, one&#xA;  indicative of the conditions in the forward part of the battery well,&#xA;  the second reading hydrogen concentrations in another sector. No-one&#xA;  in the manoeuvering said anything for the next minute except &quot;The&#xA;  charge is off, Sir.&quot; Every eye was focused on the climbing hydrogen&#xA;  needles. Everyone was silently willing them to stop. Both quivering&#xA;  indicators had passed 6 percent, and their thin points were ever&#xA;  creeping upward toward the next number.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When we built &lt;em&gt;Carver&lt;/em&gt;, I had helped install the sensors that fed&#xA;  those needles. An electrician and I had tried to optimally position&#xA;  them in the well, but our efforts had really just relied on a guess.&#xA;  What if I had been been wrong by just a few inches -- a few tenths of&#xA;  a percent? The announcement of Baker had set in process an evolution&#xA;  that completely isolated the battery well from any cleansing airflow.&#xA;  We were not going to restore the diluting flow until we reached the&#xA;  surface. Were the gauges stabilizing around 7 1/2 percent, or was I&#xA;  engaging in wishful thinking?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I do not want to impugn the author's recollection, but I feel he is going a bit overboard with the description here. Would anyone in the know be able to elucidate?&lt;/p&gt;&#xA;" OwnerUserId="8349" LastActivityDate="2016-10-11T21:01:54.577" Title="Hydrogen in submarines: &quot;keep below 8%&quot; rule of thumb?" Tags="|measurements|marine-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11833" PostTypeId="1" AcceptedAnswerId="11834" CreationDate="2016-10-12T00:25:06.803" Score="-1" ViewCount="435" Body="&lt;p&gt;I am currently tinkering with a railgun experiment, and I was thinking of using a liquid nitrogen loop to cool the rails down to very low resistance, and inject some nitrogen into the barrel between the rails, to remove some oxygen and prevent corrosion, and excessive arcing. I would run the liquid nitrogen through a hole down the length of the rail, and into the barrel through much smaller holes connecting to the main line in the rail. The nitrogen would go from the tank through a pump, through one rail, into a tubing to the other rail and back to the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I know nitrogen is magnetically transparent, and being a mostly inert gas doesn't run the chance of reacting unexpectedly. Some things I am unsure about:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I have looked around but it seems there are not a lot of places that sell tubing and fittings for liquid nitrogen. Is there some place I can browse to look for standard size fittings and tubing, and a small pump to carry liquid nitrogen through my system? The other difficulty is that I need to keep fittings and tubings to magnetically transparent non conductive materials, otherwise the magnetic field while firing (let's say 10&amp;nbsp;kJ) will mangle them. I am guessing there will be some custom fabrication involved, but still an idea of the basic sizes will help me make some design decisions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;At what pressure is it safe to carry liquid nitrogen through a system like this? Obviously the pressure will not be very high, but since I need to bleed some gas into the barrel, I need to set a target main line pressure to calculate injector channels size.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there an inherent danger I am overlooking with running an inert gas through a rail that will conduct a 1&amp;nbsp;ms pulse of 10&amp;nbsp;kJ? Obviously it won't ignite or explode, but is there a chance it will ionize at those relatively low energies?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some more looking, it seems PTFE (Teflon) is the right material for the job, but its stability under a high magnetic field is unclear. Is there a better material for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a crude cross section of the proposed arrangement. Orange is the copper rails, dark grey is PTFE, light grey is payload.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E2OSG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E2OSG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 3:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has been concern about the heat dissipation in the system, and I should mention that the system is fed power through some large inductors that should dissipate the bulk of the heat. They are in place to stretch out the pulse in the first place. The dissipation in the rails should be very, very low, considering oxygen-less copper at 77K has an abysmally low resistance. The coils will probably also get the nitrogen treatment.&lt;/p&gt;&#xA;" OwnerUserId="8356" LastEditorUserId="8356" LastEditDate="2016-10-12T04:03:57.550" LastActivityDate="2016-10-12T04:03:57.550" Title="Some questions about liquid nitrogen cooling" Tags="|pressure|power-electronics|hydraulics|cooling|electromagnetism|" AnswerCount="1" CommentCount="4" ClosedDate="2016-10-19T01:25:01.443" ContentLicense="CC BY-SA 3.0" />
  <row Id="11834" PostTypeId="2" ParentId="11833" CreationDate="2016-10-12T03:10:14.110" Score="2" Body="&lt;p&gt;Regarding questions 2 and 3, I think you need to look deeper in the chemistry, reactivity, and heat transfer characteristics of Nitrogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) It is not a noble gas. It is reactive, and you benefit from this reactivity each time you eat a plant protein.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I have not designed a railgun and I don't have much sense of the thermal inefficiencies here. But I think it would be a good idea to consider what kind of heat your launch pulse is going to transfer to the liquid nitrogen in the system. Depending on the temperature of the nitrogen (presumably just at -196 C, i.e. boiling point) then (some?) nitrogen in the system is likely to &quot;flash&quot; into its gaseous form during the launch, and pressure will go up considerably. (Assuming atmospheric pressure and ideal gas behaviour, it will take up ~700x the volume as soon as it gasifies.) Obviously you'd want to make sure the sudden gas pressure doesn't immediately blow up all the tubing, or the barrel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some aspects of this system will be similar to a water-cooling system in a foundry or a power plant, and there will be much more engineering experience with that coolant. You're just dealing with something 300 C colder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possibly you've already modelled out this thermal stuff but if so, mention it in the question to prevent us from worrying. :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your project sounds like fun, good luck! &lt;/p&gt;&#xA;" OwnerUserId="796" LastActivityDate="2016-10-12T03:10:14.110" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11836" PostTypeId="2" ParentId="11815" CreationDate="2016-10-12T04:37:48.070" Score="2" Body="&lt;p&gt;retaining walls are usually under 10-foot high if they are designed based on gravity with regular toe and heel foundation. These walls are treated on the backside with a waterproofing chemical or a plastic membrane which is backfilled with gravel to let the water run down to a 4-inch diameter PVC pipe perforated on top every 6-inches installed with a slope of 2% to 3% on the bottom of the wall draining out into storm drain system. There is weep holes as a back up every 4-feet by either 1-inch PVC or ornate cast iron pipes.&lt;br&gt;&#xA;Retaining wall shorter than 8-feet are usually made by concrete blocks and the same general detail of drainage applies to them except the weep holes are created by leaving every 4th joint on the first row ungrouted.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taller retaining walls or the new thin retaining walls anchored by horizontal epoxy coated tension cables (nails) have elaborate drainage systems engineered by the geotechnical soils engineer as an integral part of the backfill drainage network.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-10-12T04:37:48.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11837" PostTypeId="1" CreationDate="2016-10-12T04:38:37.307" Score="0" ViewCount="74" Body="&lt;p&gt;I need help figuring out the most efficient way (NOT electronically) to manually compress plastic in a small confined space. I was thinking there were small compression shocks that would help to triple my human body force but have been unsuccessful in finding something that works.&lt;/p&gt;&#xA;" OwnerUserId="8359" LastEditorUserId="163" LastEditDate="2016-10-12T06:07:22.103" LastActivityDate="2016-10-13T10:17:19.200" Title="multiplying up human compression power!" Tags="|waste-disposal|crushing|" AnswerCount="3" CommentCount="1" ClosedDate="2016-10-14T16:47:46.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="11838" PostTypeId="2" ParentId="11837" CreationDate="2016-10-12T06:06:42.633" Score="5" Body="&lt;p&gt;Levers are your friends: a lever allows you to multiply up the force of your own strength. Here's a manual can crusher, demonstrating this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lcs4b.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lcs4b.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;sub&gt;&lt;a href=&quot;http://www.childrenscornerbk.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your hand moves three or four times the distance that the crushing part does; as a result, the crushing part moves with three or four times the force that your hand does.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2016-10-12T06:06:42.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11839" PostTypeId="1" AcceptedAnswerId="11841" CreationDate="2016-10-12T06:16:28.417" Score="1" ViewCount="263" Body="&lt;p&gt;Is there a word for the tendency of some metals to make crayon-like marks on other metals?  For example, a lead bullet makes faint marks on some steels, but not on others with a similar surface finish.&lt;/p&gt;&#xA;" OwnerUserId="2140" LastActivityDate="2016-10-12T07:44:47.723" Title="Word for the tendency of one metal to rub off on another?" Tags="|metallurgy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11841" PostTypeId="2" ParentId="11839" CreationDate="2016-10-12T07:44:47.723" Score="6" Body="&lt;p&gt;&quot;Galling&quot; is probably the word you want here. It's the tendency of a (usually) soft metal to break down under pressure and adhere to the harder material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's common with soft metals like aluminium, so that's why I think this is what you've noticed on lead too. However harder materials like stainless steel can gall too in certain circumstances, so it's not a hard and fast rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The authoritative place to look would be to read up on Tribology by the way - that's the field of surface wear and other effects.)&lt;/p&gt;&#xA;" OwnerUserId="6665" LastActivityDate="2016-10-12T07:44:47.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11842" PostTypeId="1" AcceptedAnswerId="11855" CreationDate="2016-10-12T07:58:38.240" Score="2" ViewCount="142" Body="&lt;p&gt;Here's the setup, going through those four valves will be some fluid, probably water, but I need some fishing line going through this cylinder to be sealed.  The line is anchored at two points, the top and bottom, and will be used to draw two points together.  This apparatus is to heat and cool wound up fishing line, as that can be used as a muscle fiber.  When hot water is applied, it contracts, when cool water is applied, it expands.  This does mean that the line will be moving some percentage of the length, which may also mean that the diameter will change slightly.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://spectrum.ieee.org/tech-talk/robotics/robotics-hardware/fishing-line-makes-superhuman-artificial-muscles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fishing Line Makes for Superhuman Artificial Muscles&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UOUsd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UOUsdm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8361" LastEditorUserId="1632" LastEditDate="2016-10-13T08:04:19.877" LastActivityDate="2023-12-08T07:40:10.803" Title="Can I make a fishing line going through a small hole waterproof?" Tags="|fluid-mechanics|heat-transfer|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11843" PostTypeId="2" ParentId="11837" CreationDate="2016-10-12T10:54:09.920" Score="2" Body="&lt;p&gt;A bench press or arbor press, combined with a suitable die for horizontal containment, will allow you to produce rather high forces and pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bqs0M.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bqs0M.jpg&quot; alt=&quot;typical arbor press&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any numbers &amp;mdash; volume, pressure, etc. &amp;mdash; regarding exactly what it is you're trying to achieve?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have to confess that I missed the &lt;a href=&quot;/questions/tagged/waste-disposal&quot; class=&quot;post-tag&quot; title=&quot;show questions tagged &amp;#39;waste-disposal&amp;#39;&quot; rel=&quot;tag&quot;&gt;waste-disposal&lt;/a&gt; tag on my first reading, so my first suggestion isn't very practical in that application. Probably the most compact mechanism that can provide both the required force (about a ton) and the travel distance (18&amp;quot; to 24&amp;quot;) is the &quot;scissors jack&quot; &amp;mdash; but it requires a lot of cranking to operate that might not sit well with users.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5G2H0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5G2H0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2016-10-13T10:17:19.200" LastActivityDate="2016-10-13T10:17:19.200" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11845" PostTypeId="2" ParentId="11837" CreationDate="2016-10-12T14:20:34.663" Score="1" Body="&lt;p&gt;I'm not clear from your question if you are trying to crush a plastic object, or manipulate hot plastic (eg for injection molding.) Lever systems are certainly a good option as has been mentioned in other posts, but depending on your application, a compact lever may not provide enough force for you..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more force, hand-pumped hydraulic jacks will let you get a serious amount of force in a small space. For example, Enerpac makes a &lt;a href=&quot;http://www.enerpac.com/en-us/industrial-tools-imperial/hydraulic-cylinders-jacks-lifting-products-and-systems/general-purpose-hydraulic-cylinders/rsm-rcs-series-low-height-hydraulic-cylinders&quot; rel=&quot;nofollow noreferrer&quot;&gt;2.25&quot;x1.5&quot; cylinder that generates 5 tons of force&lt;/a&gt; or a  &lt;a href=&quot;http://www.enerpac.com/en-us/industrial-tools-imperial/hydraulic-cylinders-jacks-lifting-products-and-systems/general-purpose-hydraulic-cylinders/clp-series-pancake-lock-nut-cylinders&quot; rel=&quot;nofollow noreferrer&quot;&gt;5&quot;x5&quot; cylinder that generates 60 tons of force&lt;/a&gt;. Hydraulic hand pumps are readily available although the cylinder will move very slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0wUwu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0wUwu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-10-12T14:20:34.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11846" PostTypeId="2" ParentId="11814" CreationDate="2016-10-12T14:42:24.807" Score="0" Body="&lt;p&gt;I've done some research and I think I've learned enough to throw some novice advice up here regarding my question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The tool doesn't matter that much, as long as it can produce CAD drawings that a machine shop (depending on the complexity and precision of the part) can follow.&lt;/li&gt;&#xA;&lt;li&gt;I went with Solidworks (using a friend's to see if I want to actually spring for the subscription, it's not cheap)&lt;/li&gt;&#xA;&lt;li&gt;As far as community goes, I posted a similar question on Element14, but nobody has responded (2 days), so I'm still not sure about where to go if you're a noob in this space. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8334" LastActivityDate="2016-10-12T14:42:24.807" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11847" PostTypeId="2" ParentId="11814" CreationDate="2016-10-12T15:17:40.150" Score="2" Body="&lt;p&gt;Many CAD packages do have tools for sheet metal design which can plan folds, tabs etc for manufacturing. However you do need to have at least some understanding of the practicalities of the process for these to be of any real use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that the best advice is, if you are going to a manufacturer in any case then make use of their expertise by concentrating your design efforts on what you actually want to achieve. For example work out what dimensions are important and where any brackets, mounting points etc need to go. This will give them the information they need to manufacture what you actually &lt;em&gt;need&lt;/em&gt; and let them work out how to achieve it in a practical fashion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sort of situation it it a common problem for a client to show up with a very specific solution which, while it isn't 'wrong' as such isn't the most effective way to solve their underlying problem with a specific manufacturing process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that doing a CAD model will certainly be very helpful as it will allow you to clearly articulate your requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed it may be useful to do a simple model of the part(s) that you want to put in the enclosure, this will help you work out how much space you need and is also useful in assessing if the whole thing will be reasonable easy to assemble and maintain. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-12T15:23:39.987" LastActivityDate="2016-10-12T15:23:39.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11848" PostTypeId="2" ParentId="11805" CreationDate="2016-10-12T17:38:57.127" Score="2" Body="&lt;p&gt;tl;dr: I got the same answer as you. It appears the book is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we need to solve the parabola, assuming the vertex is at the origin: $$f(x)=y=a(x-h)^2 +k$$&#xA;with $h=k=0$: &#xA;$$f(20)=y+16=a(20)^2$$&#xA;$$f(12)=y=a(12)^2$$&#xA;$f(20) - f(12)$ would give:&#xA;$$a(400)-a(144)=y+16-y$$&#xA;$$a(256)=16$$&#xA;$$a=\frac{16}{256}=\frac{1}{16}$$&#xA;therefore $y=\frac{1}{16}x^2$ is the equation driving the parabola.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the centroid of the composite shape. After careful analysis of the image from the problem and our parabolic equation we see that the shape is the area inside a parabola defined by our formula above and bounded by the $y$ axis and between the lines $y=9$ and $y=25$. We also know that the centroid of a composite shape about the $y$ axis ($\bar{X}$) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{X}=\frac{\sum\bar{x}A}{\sum{A}}$$ &#xA;where $\bar{x}$ is the centroid about the $y$ axis of the component shape and $A$ is the area of the component shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the following to find the centroid of a semi-parabolic area:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4MPRC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4MPRC.png&quot; alt=&quot;Centroid of a Parabola&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and using $y=\frac{1}{16}x^2$ to solve for $y$ (or $h$) we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{X}=\frac{\left(\frac{3(a_1)}{8}\right)\left(\frac{2(a_1)(h_1)}{3}\right)-\left(\frac{3(a_2)}{8}\right)\left(\frac{2(a_2)(h_2)}{3}\right)}{\left(\frac{2(a_1)(h_1)}{3}\right)-\left(\frac{2(a_2)(h_2)}{3}\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{X}=\frac{\left(\frac{3(20)}{8}\right)\left(\frac{2(20)(25)}{3}\right)-\left(\frac{3(12)}{8}\right)\left(\frac{2(12)(9)}{3}\right)}{\left(\frac{2(20)(25)}{3}\right)-\left(\frac{2(12)(9)}{3}\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{X}=\frac{\left(7.5\right)\left(333.333\right)-\left(4.5\right)\left(72\right)}{\left(333.333\right)-\left(72\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{X}=8.326$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we know that the volume of an area swept around an axis is equivalent to the volume of that same area extruded along a length equal to the circumfrence of the circle scribed by the centroid as it rotates about said axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area to be swept is $333.333-72 = 261.333$, and the circumference of the circle scribed by the centroid rotated about the y axis is $2\pi(8.326)\approx52.313$, giving a volume of $13,671.3$ or $4352\pi$ $cm^3$.&lt;/p&gt;&#xA;" OwnerUserId="8370" LastActivityDate="2016-10-12T17:38:57.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11850" PostTypeId="1" AcceptedAnswerId="11854" CreationDate="2016-10-12T19:38:37.030" Score="2" ViewCount="171" Body="&lt;p&gt;I am working with a peristaltic pump to move liquid to/from a reservoir through a closed loop of tubing. The motor can rotate clockwise or counter-clockwise when set with either of the directional buttons below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Azc9e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Azc9e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direction of rotation can also be controlled remotely using contacts #1/2 in the green barrier strip below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cHAli.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cHAli.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have little experience with this kind of work. Would this be a simple enough task to connect, program and use a controller to alternate the direction of the motor for some time interval? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if someone could describe the process for me in simple terms, it might be helpful in my understanding of how to approach the challenge. &lt;/p&gt;&#xA;" OwnerUserId="8371" LastEditorUserId="282" LastEditDate="2016-10-13T08:04:31.473" LastActivityDate="2016-10-13T08:04:31.473" Title="Using a controller to remotely alternate the direction of a peristaltic pump" Tags="|control-engineering|pumps|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11851" PostTypeId="1" CreationDate="2016-10-12T20:27:33.083" Score="4" ViewCount="227" Body="&lt;p&gt;I have solved the Rayleigh-Plesset equation for a time varying bubble of a super critical fluid. I am trying to find if the impact pressure caused by SCF bubbles is lesser as compared to that caused by water bubbles. Here is the equation I arrived at:&#xA;$$&#xA; x x'' + \frac{3}{2}(x')^2 + A\frac{x'}{x} = W&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $A$ and $W$ are constants that depend on the viscosity, temperature and specific heat of the fluid continuum. I have used Weierstrass' elliptic function to find an analytical solution for this second order, nonlinear equation. I have used the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Chapman%E2%80%93Enskog_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chapman-Enskog theory&lt;/a&gt; for finding the viscosity of the SCF and its distribution.This is the equation I arrived at:&#xA;$$&#xA;  \boxed{&#xA;    R \ddot{R} + \frac{3}{2} (\dot{R})^2 + \frac{4\mu_0}{\rho_L}\left[1 + b_0 \rho_L \left(\frac{1}{y} + 0.8 + 0.7614\right)\right]\frac{\dot{R}}{R} = W&#xA;  }&#xA;$$&#xA;where&#xA;$$&#xA;  W = \frac{\exp(\gamma)\left(\frac{p_2}{4} - 1\right) + \frac{p_2}{4} + \frac{p_c}{2}}{\rho_L} \,.&#xA;$$&#xA;My question is how do I go about finding the impact pressure of such a system? And how do I model their distribution by a mathematical distribution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impact pressure in general would be $$\frac{P_{t}}{P}=\left(1+M^{2}\frac{\gamma-1}{2}\right)^{\frac{\gamma}{\gamma-1}}$$, for isentropic flow. $P_{t}$ is the total pressure and $P$ is the static pressure; the impact pressure is $$P_{I}=P_{t}-P$$. M is the Mach number and in this case it would be &amp;lt;0.8 as the bursting of a bubble in a fluid would be subsonic.$\gamma$ is the ratio of the specific heats.&lt;/p&gt;&#xA;" OwnerUserId="8375" LastEditorUserId="8375" LastEditDate="2016-11-24T16:59:52.623" LastActivityDate="2016-11-24T16:59:52.623" Title="How do I find the impact pressure of cavitation due to implosion of a super critical fluid bubble?" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|pressure|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11852" PostTypeId="1" CreationDate="2016-10-12T20:35:40.787" Score="4" ViewCount="575" Body="&lt;p&gt;&lt;strong&gt;Some extensive background information to start&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm attempting to reduce water consumption at 25+ production facilities, and I've identified that the largest consumption of water (60+%) comes from evaporation cooling for all the unit operations in the process. This is done at the facilities via an evaporative cooling tower. The towers are all the same base design, and relatively similar;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2, 3 or 4 cells&lt;/li&gt;&#xA;&lt;li&gt;counter flow induced draft (see picture)&lt;/li&gt;&#xA;&lt;li&gt;design capacity 45-85 MW heat load, with most being 65-75.&lt;/li&gt;&#xA;&lt;li&gt;the largest discrepancy is that each tower was designed for 10 degree Fahrenheit (5.5 C) Delta-T, but with a wet bulb reference temp that encompasses 350/365 days statistically per year for that respective latitude / longitude (facilities are spread out across the US Midwest, so that number varies). This is considerable, but shouldn't matter for what I'm about to ask (I include on the off chance it does)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Gqcis.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gqcis.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I understand the basic workings of a cooling tower (wet bulb temp, efficiency, cycles of concentration, make-up, blow down, drift etc) fairly well. With that, I understand that the amount of evaporation is proportional and fixed to the heat load. A such, ones first opportunity to reduce water is to recover/minimize blow down and drift. The facilities already have 0.1% drift on average, and run anywhere from 10-25 cycles of concentration in addition to recovering the blowdown for process water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite this, the cooling towers still represent over 60% of water usage, so the next step is looking at the evaporation side. During the summer, the evaporative load is fixed and cannot be reduced. However, during the winter we already see reduction in water consumption. This stems from a combination of latent + sensible heat changes during the winter, as opposed to exclusively latent heat changes during the summer. As much as 25% of the heat load at certain facilities is dissipated via sensible heat change during the deep winter (January February). This is a commonly referred to phenomenon in the cooling tower industry and even has it's own correction factor in some simplified equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has led me to believe that there is potential to control this effect and reduce water consumption even further. Adding to this evidence is the graph below. In this graph, one can see that for a substantial portion of the year, the basin temperature goes as low as 55 degrees F. The cooling water for the process &lt;strong&gt;does not need to be any colder than 70-75 degrees F.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bO8Ho.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bO8Ho.png&quot; alt=&quot;Cooling tower basin temperature&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that the basin temperature is consistently so much colder than necessary, my gut reaction says &quot;we have more cooling taking place than necessary&quot;. While we don't want to reduce total cooling MW's, one would want to minimize mass transfer driving force (to stop evaporation), and maximize temperature driving force (to increase cold air to hot water heat transfer). Doing this would shift more of the heat load from latent to sensible heat. I was thinking this could be achieved by slowing the circulation fans down with VFD's. Lower air flow rate for the same water flow rate, air gets saturated sooner to reduce evaporation, and the water circulating would be warmer to &lt;em&gt;hopefully&lt;/em&gt; increase heat exchange between air/water. In effect, what I'm hoping will be the case is being able to operate the cooling towers as a &quot;wet cooling&quot; unit during the summer, but utilize it as a &quot;partial dry cooling&quot; unit when possible to save water, but without spending the capital to purchase an actual dry cooling unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now my questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I cannot find ANYTHING online about doing this to reduce water consumption. Am I missing something blatantly obvious that shoots down the idea of trying to control the heat load towards sensible vs latent heat? (effectively trying to &quot;dry cool&quot; instead of &quot;wet cool&quot;, but using a cooling tower designed for wet cooling)&lt;/li&gt;&#xA;&lt;li&gt;If the idea isn't completely bunk, how would I go about modeling this with mass &amp;amp; energy balances &amp;amp;/or a psychrometric chart to get a ballpark number for how much water could be saved? Given that I know: Historical wet bulb temp &amp;amp; temp, Hot in temp, cold out temp, fan amperage (er go fan speed), water re-circulation rate, blow down, make-up and cycles of concentration&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8369" LastActivityDate="2018-03-25T15:39:01.770" Title="Can I control latent/sensible heat changes in a counter-flow, induced draft cooling tower using climate?" Tags="|thermodynamics|chemical-engineering|process-engineering|cooling|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11854" PostTypeId="2" ParentId="11850" CreationDate="2016-10-13T03:24:50.380" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;Find the manufacturers spec sheet if possible. This will detail the voltages, pin configuration, and more control details. &lt;/li&gt;&#xA;&lt;li&gt;If the spec sheet specifies serial commands like 706Astor mentioned, you may be able to use a usb-to-serial converter and a serial console like &lt;a href=&quot;http://www.chiark.greenend.org.uk/~sgtatham/putty/download.html&quot; rel=&quot;nofollow&quot;&gt;putty&lt;/a&gt; for testing it. You could then send serial commands via &lt;a href=&quot;https://pythonhosted.org/pyserial/&quot; rel=&quot;nofollow&quot;&gt;python&lt;/a&gt; or other programming language.&lt;/li&gt;&#xA;&lt;li&gt;If serial is not an option, test the inputs with an &lt;a href=&quot;http://www.apogeekits.com/images/alligator_leads_red_black.jpg&quot; rel=&quot;nofollow&quot;&gt;alligator clip&lt;/a&gt; or other testing leads. I assume based on the diagram that connecting 1 and 2 makes it run counter-clockwise, leaving them disconnected is clockwise. Connecting 3 and 4 makes it run, disconnecting makes it stop. Connect a 1.5v AA or AAA battery up to 5(+side) and 6(-side) to test its speed at 1.5/5 = 30% speed.&lt;/li&gt;&#xA;&lt;li&gt;If your tests work out and you now understand the operation, and all you want to do is change the direction at some interval, you might be able to use a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B012FSL2GK&quot; rel=&quot;nofollow&quot;&gt;cheap dc timer relay&lt;/a&gt; to connect contacts 1 and 2.&lt;/li&gt;&#xA;&lt;li&gt;If you need more control or more adjustment on the timing, you will want to get an &lt;a href=&quot;https://www.arduino.cc/en/Main/ArduinoBoardUno&quot; rel=&quot;nofollow&quot;&gt;Arduino UNO&lt;/a&gt;. They are very popular with lots of tutorials and internet community support. There are also some good tutorials on &lt;a href=&quot;https://learn.adafruit.com/category/learn-arduino&quot; rel=&quot;nofollow&quot;&gt;Adafruit&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-10-13T03:24:50.380" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11855" PostTypeId="2" ParentId="11842" CreationDate="2016-10-13T03:47:20.263" Score="2" Body="&lt;p&gt;A high-quality seal that diameter would be very difficult to fabricate. You always want to avoid seals in designs when you can help it. For example, you could fix one end inside the cylinder so that you only have one moving end to worry about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow for some leakage and run the water at low pressure (less than 12&quot; of head) you could probably get away with melting a small cup of wax around the line. Have the wax be as tall as the diameter of your primary tube. The longer distance will reduce water leakage. Pull the line taught while it cools so the path is straight. It should pull free with very little force, and with little movement during operation, it should provide a fair amount of operating life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more robust solution would be to put a diaphragm (as mentioned in the comments) on the end cap instead of a seal. The diaphragm will act like a soft spring, and can be compensated for with other spring components if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A design that you may or may not have considered; would be to use a single tube as your tension member and run hot and cold fluid through this tube. The design would be simpler, the volume of fluid per tension force would be lower, and the heat transfer would be quicker. Something to think about.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-10-13T03:47:20.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11856" PostTypeId="1" AcceptedAnswerId="11858" CreationDate="2016-10-13T03:53:42.183" Score="0" ViewCount="63" Body="&lt;p&gt;I was wondering, in laymen terms, what is the exact definition of renewable energy and what are its various forms, which ones are better and in what situations can they / can they not be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you weigh the cost of setting up this technology with its payoff, and how do you set these benefits apart from those derived by investing in other areas such add healthcare, research, and development in other areas such add food and disease fighting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sends like there are manpower-based investments and research based investments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this is a very broad question, but tried Wikipedia and could not find a laymen answer to my broad question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: If my question is to broad, then please just pinpoint this very basic question: how do you define renewable energy?&lt;/p&gt;&#xA;" OwnerUserId="7542" LastEditorUserId="163" LastEditDate="2016-10-13T06:55:47.817" LastActivityDate="2016-10-13T07:01:13.503" Title="What is the exact definition of renewable energy" Tags="|renewable-energy|" AnswerCount="1" CommentCount="2" ClosedDate="2016-10-15T10:12:42.077" ContentLicense="CC BY-SA 3.0" />
  <row Id="11858" PostTypeId="2" ParentId="11856" CreationDate="2016-10-13T06:16:32.023" Score="3" Body="&lt;h3&gt;overview:&lt;/h3&gt;&#xA;&lt;p&gt;Renewable energy is one for which the input energy is renewed on human timescales.&lt;/p&gt;&#xA;&lt;p&gt;Remember, energy is always conserved. It doesn't come from nowhere. Pretty much all the energy we use on Earth comes from the stars - and almost all of it comes from one star in particular, the Sun.&lt;/p&gt;&#xA;&lt;h3&gt;not renewables:&lt;/h3&gt;&#xA;&lt;p&gt;Uranium (for &lt;strong&gt;nuclear&lt;/strong&gt; power) is in the earth after being manufactured in generations of stars. This takes billions of years, and is thus not renewable. &lt;strong&gt;Coal&lt;/strong&gt;, &lt;strong&gt;gas&lt;/strong&gt; and &lt;strong&gt;oil&lt;/strong&gt; are in the earth after having been living organisms (plants, animals, etc) millions of year ago, and are thus not renewable.&lt;/p&gt;&#xA;&lt;h3&gt;renewables:&lt;/h3&gt;&#xA;&lt;p&gt;&lt;strong&gt;Wind&lt;/strong&gt;, &lt;strong&gt;solar&lt;/strong&gt;, &lt;strong&gt;hydro&lt;/strong&gt;, &lt;strong&gt;biomass&lt;/strong&gt; and &lt;strong&gt;waves&lt;/strong&gt; are all powered by incoming solar energy every second of the day, and so are constantly replenished at human timescales. The &lt;strong&gt;tides&lt;/strong&gt; are renewed every day by the relative motion of the Earth and the Moon (and to a much lesser extent, the sun too). &lt;strong&gt;Geothermal&lt;/strong&gt; energy is renewed every second of the day by slow nuclear reactions deep in the ground, and in some cases by being replenished constantly by the heat emitted from magma.&lt;/p&gt;&#xA;&lt;h3&gt;endnote:&lt;/h3&gt;&#xA;&lt;p&gt;You've asked a very broad range of questions. Many of them, &lt;em&gt;when refined&lt;/em&gt;, would be best handled on our Sustainability Stack, where we already have quite a few  &lt;a href=&quot;https://sustainability.stackexchange.com/questions/tagged/renewables&quot;&gt;questions on renewable energy&lt;/a&gt; - the economics, the engineering, and so on. You'll also find questions on the &lt;a href=&quot;https://physics.stackexchange.com/questions/tagged/renewable-energy&quot;&gt;physics concepts behind renewable energy, on our Physics site&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-10-13T07:01:13.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11860" PostTypeId="1" CreationDate="2016-10-13T18:31:26.900" Score="1" ViewCount="457" Body="&lt;p&gt;I am analyzing results of my turbine simulation. I have created one line from hub to casing at downstream of the flow...please see the following figures. I want to investigate along this line how the axial velocity changes. In one of my results, my new blade(VT in last Figure or blue line in last figure) has the lower axial velocity than to base blade(CT) at hub and tip. (Please see attachment.) What does this mean? I think that this is due to low separation in downstream of flow of new blade(VT). Am I right?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uqjYK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uqjYK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Kmre.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Kmre.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AE8qQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AE8qQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gVk8G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gVk8G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I6L7D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I6L7D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/apFj6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/apFj6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the following picture, Is axial velocity equal to velocity V in my case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8wUxX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8wUxX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2016-10-13T18:37:14.837" LastActivityDate="2023-12-09T12:01:30.140" Title="What is reason of low axial velocity at tip and hub of an axial turbine?" Tags="|mechanical-engineering|fluid-mechanics|turbines|cfd|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="11864" PostTypeId="2" ParentId="1743" CreationDate="2016-10-14T06:33:24.583" Score="1" Body="&lt;p&gt;I don't know the US situation. Codes over here (Germany) demand, among other things, that a room with a funrace or boiler in it is not at an underpressure re. the outside, which in many cases larger than a domestic water heater mandates active ventilation. This serves two goals I think:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;underpressure would make it hard to open doors (have to open outwards, because escape route)&lt;/li&gt;&#xA;&lt;li&gt;Makes sure enough combustion air is supplied&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I've only ever seen one sizing calculation for a ventilation system for a boiler room and the combustion air was definitly included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OTOH, th furnace/boiler whatever has it's own emission control scheme - be it an active lambda sensor or the tunig of a nozzle - that has to meet, among other things , CO requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you, as the civil engineer planning or delegating the ventilation system have to make sure that the ventilation satisfies certain conditions and the manufacturer of the furnace has to make sure the furnace satisfies emission limits in 'normal' conditions. Over here, both (ventilation &amp;amp; safety of boiler room, actual emissions) are controlled by the chimney sweep. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the two ways to specify it you mentioned:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;volume of the room that contains the equipment  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;would work if (big if!) the room is ventilated to achieve certain airchanges per hour, then the flow is proportional to the volume&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;an area of free-flowing vent from the outside in a ratio to the BTUs or Watts&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Air consumption is proportional to Watts or BTU, just not exactly the same for each fuel. So sizing a free vent to wattage makes sense, if the code contains a large enough margin of safety.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-10-14T10:18:34.360" LastActivityDate="2016-10-14T10:18:34.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11866" PostTypeId="2" ParentId="11860" CreationDate="2016-10-14T08:54:41.407" Score="0" Body="&lt;p&gt;I searched about the pitch and I deserve a detailed description  about the pitch because i am new in turbine. I am grateful that tell me that i am right or wrong. Please be careful that performance of Wells turbine is similar to Wind turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Section 1:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According Brian Drummond  stated, I think that the pitch is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bk7Ik.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bk7Ik.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mpoweruk.com/wind_power.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;enter link description here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact(according to above picture), for a given wind speed, &lt;strong&gt;to maintain a constant angle of attack along the length of the blade&lt;/strong&gt;, the pitch of the blade must be increased as it approaches the bade root. Am i right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Section 2:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sample of twisted blade:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uxWyb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uxWyb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.see.murdoch.edu.au/resources/info/NSWTC/install.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;enter link description here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, &lt;strong&gt;i think that a twisted blade has different pitch along the height of blade in order to maintain a constant angle of attack along the height of the blade&lt;/strong&gt; . Am i right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;section 3:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here, i have one question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any deference between &lt;strong&gt;the pitch&lt;/strong&gt; and &lt;strong&gt;pitch angle&lt;/strong&gt; in turbines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Section 4(Major section for me):&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the following figure, VT blade(my new blade designed) has better performance than to CT blade because there is lower axial velocity(lower AOA) at hub and tip in VT blade.  Am i right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;In fact, I think that because of unchanging the pitch in wells turbine(my case), We should change the axial velocity distribution in order to having better AOAs along the height of blade. Am i right?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aH70x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aH70x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Please be careful that My means is downstream of the turbine rotor not downstream of the flow.&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The turbine rotational direction and axial flow are in attachment are as follows:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/42IR7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/42IR7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow is bi directional in this turbine. the turbine is named Wells turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have another question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have created one line from hub to casing &lt;strong&gt;&lt;em&gt;at upstream of the turbine rotor&lt;/em&gt;&lt;/strong&gt; ..please see attachment. I plotted the axial velocity distribution on the line according to attachment. I think that the axial velocity distribution has not changed in these two blade because of unchanging flow blockage but variation of before axial velocity distribution is due to efferent thickness along of the blade height. Am i right? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GMPuQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GMPuQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SFbnp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SFbnp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i2nbJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i2nbJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FufHv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FufHv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am grateful that guide me more about these sections.&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2016-10-15T10:16:13.820" LastActivityDate="2016-10-15T10:16:13.820" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11867" PostTypeId="2" ParentId="11860" CreationDate="2016-10-14T11:25:50.343" Score="0" Body="&lt;p&gt;Regarding section 4, i designed a blade that thickness changes from hub to tip. In other words, thickness increases from hub to tip along of blade height(variable thickness blade). In high flow coefficients, this design makes lower separation at tip in variable thickness blade than to constant thickness blade because thicker profile has lower separation and names trailing edge stall. therefore, performance of improved blade is better in higher flow coefficients. The turbine tuque is better in variable thickness blade in high flow coefficient....I think that this event is due to more uniform axial velocity distribution along of blade height in variable thickness blade at downstream of the turbine rotor .  What is your opinion?&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2016-10-14T11:31:12.510" LastActivityDate="2016-10-14T11:31:12.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11869" PostTypeId="1" CreationDate="2016-10-14T13:23:45.660" Score="0" ViewCount="55" Body="&lt;p&gt;I need to build a structure like the one in the picture below. Basically, it is a pendulum that holds a measurement hammer and lets it oscillate without friction. I don't know how to find the two necessary components (the one that holds the hammer and the one that is fixed on the shaft). How are they called?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ovT7i.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ovT7i.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8392" LastActivityDate="2016-10-14T23:37:36.253" Title="How are these components called?" Tags="|structures|" AnswerCount="1" CommentCount="1" ClosedDate="2016-10-17T00:58:05.057" ContentLicense="CC BY-SA 3.0" />
  <row Id="11870" PostTypeId="1" CreationDate="2016-10-14T14:23:15.570" Score="2" ViewCount="9561" Body="&lt;p&gt;Micrometer calipers measure thickness. Vernier calipers can measure both thickness and inside diameter. For the same least count and accuracy, both are similarly priced. Do micrometers have any advantage over verniers (besides size- micrometers being compact)?&lt;/p&gt;&#xA;" OwnerUserId="6614" LastEditorUserId="368" LastEditDate="2016-10-14T16:59:57.627" LastActivityDate="2017-02-07T09:15:38.497" Title="Do micrometer calipers have any advantage over vernier calipers?" Tags="|measurements|tools|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11871" PostTypeId="2" ParentId="11870" CreationDate="2016-10-14T19:03:17.207" Score="2" Body="&lt;p&gt;Micrometers tend to be used for higher precision measurements and are usually specific to one particular type of measurement (thickness, depth, bore etc) they are defined by their vernier screw mechanism which can be adapted to various functions (the fine feeds on lathes use a similar principal). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vernier calipers (and their dial and digital equivalents) are a bit more convenient to use and are more versatile, having a larger range of measurement and being able to measure, thickness, length,  diameter, bore and depth in the same instrument.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's common  to use vernier calipers for general measurements where reasonable precision is required (say to around 0.1 mm) as they can be fairly inexpensive and are plenty accurate enough for a lot of tasks eg checking if a drill is 10 or 10.5 mm  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Micrometers and the like tend to be reserved for higher precision tasks like fine machining and grinding where accuracy better than 0.1 mm is required, especially for repetitive measurements over a narrow range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously precision will vary with quality but typically a micrometer will have better precision than a vernier caliper in the same sort of price range. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-10-14T19:03:17.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11872" PostTypeId="1" CreationDate="2016-10-14T23:01:48.660" Score="0" ViewCount="1011" Body="&lt;p&gt;Do gears on the same axle share both speed and torque? So if a large gear and a small gear were in the same axle the small gear would have the torque of the large gear?&lt;/p&gt;&#xA;" OwnerUserId="8395" LastActivityDate="2016-10-15T01:49:35.977" Title="Gears on the same axle question" Tags="|gears|" AnswerCount="1" CommentCount="0" ClosedDate="2016-10-19T01:25:31.840" ContentLicense="CC BY-SA 3.0" />
  <row Id="11873" PostTypeId="2" ParentId="11869" CreationDate="2016-10-14T23:37:36.253" Score="2" Body="&lt;p&gt;There are a few identifiable things in the picture &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The large P shaped object would be called a 'bearing housing', this looks like it may have been custom manufactured for this application, it is certainly not  a very common component although it is broadly similar to a &lt;a href=&quot;http://www.bestbearingsonline.co.uk/np1116--ucwp203-11-budget-two-bolt-cast-iron-1116-borelummer-illow-block-housed-unit-with-grub-screw-5321.html&quot; rel=&quot;nofollow&quot;&gt;pillow block&lt;/a&gt; which are readily available. The one in the example is probably machined from billet of steel or aluminium (possibly based on some other component but I'm not sure what). &lt;/li&gt;&#xA;&lt;li&gt;the o shaped component holding the horizontal shaft is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Needle_roller_bearing&quot; rel=&quot;nofollow&quot;&gt;needle roller bearing&lt;/a&gt;. This class of bearing has the rolling elements bearing directly onto the shaft they support rather than having an inner shell. &lt;/li&gt;&#xA;&lt;li&gt;The horizontal shaft is probably &lt;a href=&quot;https://en.wikipedia.org/wiki/Silver_steel&quot; rel=&quot;nofollow&quot;&gt;silver steel&lt;/a&gt; or similar, this is convenient for shafts fitted to bearings as it is available precision ground to standard sizes and has good wear and mechanical properties. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This would be a fairy simple job for any machine shop as the important operations are just drilling and tapping holes. Machining the P shape is a bit more involved but that is not strictly necessary for the basic functionality. at  a push you could do it with a pillar drill and a hacksaw.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-10-14T23:37:36.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11874" PostTypeId="2" ParentId="11872" CreationDate="2016-10-15T01:49:35.977" Score="3" Body="&lt;p&gt;Yes, of course. Or more accurately, one gear would have the &lt;em&gt;opposite&lt;/em&gt; torque of the other. Torques, like forces, need to balance.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-10-15T01:49:35.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11875" PostTypeId="1" CreationDate="2016-10-15T07:47:44.947" Score="1" ViewCount="2161" Body="&lt;p&gt;I am a computer engineering student and currently having my undergraduate thesis. My thesis is about temperature control system of a chicken egg incubator. I searched over the internet and found &lt;a href=&quot;http://www.idpublications.org/wp-content/uploads/2016/02/Full-Paper-DEVELOPMENT-AND-TEMPERATURE-CONTROL-OF-SMART-EGG-INCUBATOR.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this research&lt;/a&gt;. Please verify if i can use this transfer function for my temperature control system of an chicken egg incubation system with desired temperature of &lt;strong&gt;37.8 °C&lt;/strong&gt;.&lt;br&gt;&lt;br&gt;Inside the research, the researcher calculated the transfer function &#xA;from the equation Q = CT and the netwon's law of cooling and arrived at this transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mug3k.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mug3k.png&quot; alt=&quot;Modeling the incubation temperature control system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1&lt;/strong&gt;. Is this transfer function applicable to my chicken egg incubation system?&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Question 2&lt;/strong&gt;. Is there other things to be added or considered for calculating the transfer function of a thermal system aside from what the researcher calculated?&lt;br&gt;&lt;br&gt;&#xA;&lt;strong&gt;Question 3&lt;/strong&gt;. The paper tells us that Tau is the time it takes to achieve the maximum temperature of the incubator. I turned on the 200W heat rod in my incubator and the time it took to reach 37.8 °C is 240s. Isn't that very fast? Cause i did a simulink of this transfer function and the transient response(rise time, overshoot, settling time) is very fast compared to the real scenario. I tried 940s and its closer to the real system than 240s. &lt;/p&gt;&#xA;" OwnerUserId="8398" LastActivityDate="2016-10-15T17:10:41.300" Title="Transfer function of a temperature controlled incubation system" Tags="|control-engineering|heat-transfer|transfer-function|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11877" PostTypeId="1" AcceptedAnswerId="11886" CreationDate="2016-10-15T10:41:08.093" Score="0" ViewCount="235" Body="&lt;p&gt;I am a computer engineering student and currently having my undergraduate thesis. My thesis is about a chicken egg incubator and i need to control the humidity inside the incubator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, i dont know how to calculate the transfer function of the humidity control system. Please someone point me to the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to get the transfer function if i can give a step input (turn on the humidifier) and then get the s-domain of the output, then divide the output over the input because transfer function = output/input? or is this method difficult considering im only an undergraduate computer engineering student?&lt;/p&gt;&#xA;" OwnerUserId="8398" LastActivityDate="2016-10-19T11:18:08.943" Title="How to model a humidity control system inside a box?" Tags="|thermodynamics|control-engineering|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11878" PostTypeId="2" ParentId="11775" CreationDate="2016-10-15T13:42:00.760" Score="2" Body="&lt;p&gt;It sounds like you're inquiring about the development of rebar at RC beam-column connections. Based on my (largely academic) experience, you're likely looking at something similar to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0q7gl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0q7gl.jpg&quot; alt=&quot;Rebar Detailing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key attribute is extending the beam bars to the far face of the column. Realistically, all connections are somewhere between fixed and pinned, but extending the bars straight to the far column face rather than the near face more closely approximates a fixed condition. This detailing approach also gives the joint the best ductility - critical in seismic applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why bend the bars at all? Because the full tension strength of the beam bar needs to be developed at the near face of the column and this requires a certain length of rebar to be embedded in the concrete. In general, the only way to accomplish this is to hook the bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why bend the beam bars so that they enclose the beam rather than having them stick up and down along the column length? I assume it's to do with avoiding rebar congestion but I don't know for sure. So long as the hooked development length is achieved, it seems like a qualitative preference to me.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2016-10-15T13:42:00.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11879" PostTypeId="2" ParentId="11875" CreationDate="2016-10-15T17:10:41.300" Score="2" Body="&lt;p&gt;The transfer function you provide is merly a model of the egg incubation system it is not a controller. In addition a lot of assumptions are made, for instance, heat does spread evenly through the egg incubator. Heat is lost evenly throughout the egg incubator, et cetera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model is derived from basic heat equations, this is described in the article. Tau is the so called time constant of the system, read: &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_constant&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Time_constant&lt;/a&gt;. What the time constant is depends on the thermal resistance and the heat capacity or your egg incubation system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can derive the time constant by of your egg incubation system by an experiment. Put a temperature sensor inside the incubation chamber and change the heat from (lets say) 20 degrees to 38.5 degrees. Monitor how the temperature changes. The time at which the temperature has reached 66% of its end value (w.r.t. to its starting value) is your time constant. In this case, it would reach 66% = 32.21 degrees ((38.5-20) * 0.66 + 20).&lt;/p&gt;&#xA;" OwnerUserId="3357" LastActivityDate="2016-10-15T17:10:41.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11880" PostTypeId="1" CreationDate="2016-10-15T18:07:12.263" Score="0" ViewCount="52" Body="&lt;p&gt;I am interested in buying electric motor components separately so I can add my own bearings, not use a motor with sealed OEM bearings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know the correct terminology for this type of application, so I do not know how to search for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find motors like this?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2016-10-15T21:20:08.803" Title="Terminology for motors with custom bearings" Tags="|motors|bearings|" AnswerCount="2" CommentCount="3" ClosedDate="2016-10-17T11:31:33.610" ContentLicense="CC BY-SA 3.0" />
  <row Id="11882" PostTypeId="2" ParentId="11880" CreationDate="2016-10-15T20:22:24.687" Score="1" Body="&lt;p&gt;In the case of industrial type AC motors it is generally not that difficult to just change the bearings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of building from scratch you can get rotor and stator assemblies as replacement parts. The difficulty is that the housing is usually an aluminium casting which is made to fit the rotor and stator and is usually made in two pieces which also incorporates the bearing housing so if you want a significantly different bearing housing you have a fairly major fabrication and machining job on your hands. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, while it may be desirable to replace a generic bearing with a branded one this is usually a like for like swap as you will presumably just want better quality rather than a radically different size and is not usually a hugely difficult task on larger AC motors such as you find on lathes etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say AC because they are usually capable of fairly easy  disassembly but the same applies to larger DC motors as well. The smaller ones tend to be a bit more difficult to get into and are less likely to have separate spares readily available. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-15T21:00:55.267" LastActivityDate="2016-10-15T21:00:55.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11883" PostTypeId="2" ParentId="11880" CreationDate="2016-10-15T21:20:08.803" Score="1" Body="&lt;p&gt;In addition to Chris Johns' answer I have done further research to answer this question and have discovered the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) In general, there are no special customizable motors made; motors are made in such a way that they can be taken apart with the right tools, so every motor more or less is customizable with the right tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Some manufacturers sell &quot;modifiable&quot; motors that allow a buyer to alter the way the motor is configured to a limited degree, but usually you must be making a bulk order to take advantage of this. Note that such &quot;modifiable&quot; motors only allow the customer to switch out components for fixed alternatives that the motor manufacturer has pre-determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) To change the bearings on a motor requires equipment such as bearing pullers and snap ring pliers. It may also require an arbor press for the larger motors or even a hydraulic press for really big motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) In many cases you can get different bearings with the same form factor, so you can change your bearings to a different type with no further modification. For example, you could change ball bearings with roller bearings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) If your wanted bearing is in a different form factor, you may need to modify or fabricate the plate that holds the bearing to the frame, or alternatively create an adapter that fits to the existing plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(6) If the motor does not have enough room to accommodate your bearing, you can remove the motor's bearings entirely and place your bearings outside the frame. This will probably require removing the motor's &lt;em&gt;bells&lt;/em&gt;, the end caps. Your subframes for holding your custom bearings must be aligned so that the spacing between the rotor and stator is exactly equal all around. (If you can improve this spacing, you may be able to improve the performance of the motor over the factory build.) You will probably need a new, longer shaft to place bearings external to the existing frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7) The shaft of nearly all motors is joined to the rotor with a simple interference fit, or a force fit in the case of very small motors. To remove the shaft warm the rotor to approximately 200F using a heating strip or other mechanism and cool the shaft using freeze spray, a cooling jacket, dry ice, or some similar mechanism. The shaft should now be removable with an arbor press. You can try to force the shaft out cold with a 10 ton hydraulic press, but this risks damaging the rotor, so using a thermal differential is recommended. A new shaft can be inserted into the motor by the same method.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2016-10-15T21:20:08.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11884" PostTypeId="1" CreationDate="2016-10-16T05:09:04.373" Score="1" ViewCount="428" Body="&lt;p&gt;I'm having trouble understanding the 4th step to the Buckingham Pi theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My variables are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;weight ($\text{N}$) : $\text{MLT}^{-2}$&lt;/li&gt;&#xA;&lt;li&gt;h ($\text{m}$): $\text{L}$&lt;/li&gt;&#xA;&lt;li&gt;Diameter ($\text{m}$): $\text{L}$&lt;/li&gt;&#xA;&lt;li&gt;Orifice diameter ($\text{m}$): $\text{L}$&lt;/li&gt;&#xA;&lt;li&gt;flow rate Q ($\text{m}^3/\text{s}$) : $\text{L}^3\text{T}^{-1}$&lt;/li&gt;&#xA;&lt;li&gt;density ($\text{kg/m}^3$) : $\text{ML}^{-3}$&lt;/li&gt;&#xA;&lt;li&gt;fluid viscosity ($\text{Ns/m}^2$) : $\text{ML}^{-2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From this $k = 7$, $r = 3$, $\text{number of Pi groups} = 4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the criteria for repeating variables, but I'm not 100% sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if i picked weight, height and flow rate will that have fulfilled all the criteria? I don't understand the dependent variable criteria.&lt;/p&gt;&#xA;" OwnerUserId="2994" LastEditorUserId="1832" LastEditDate="2016-10-18T00:31:03.517" LastActivityDate="2016-10-18T00:31:03.517" Title="Buckingham Pi theorem, repeating variables" Tags="|fluid-mechanics|civil-engineering|mathematics|fluid|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11886" PostTypeId="2" ParentId="11877" CreationDate="2016-10-16T17:25:27.573" Score="0" Body="&lt;p&gt;If a system can be approximated as a (stable) linear time invariant system, than any bounded input which contains all frequencies (for digital signals limited to the Nyquist frequency) can be used to  identify the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Laplace transform of a step function is $s^{-1}$, so high frequency behavior might be more prone to be hidden below a noise floor. So you would need more measurements in order to average away this noise, if you are interested in those high frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have multiple measurements of input data, which contain &quot;all frequencies&quot;, and output/system-response data, then you can approximate the frequency response function of the system, denoted with $P(s)$, using,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P(s) \approx \frac{\sum Y_i(s)\,\bar{U}_i(s)}{\sum U_i(s)\,\bar{U}_i(s)},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Y_i(s)$ is the &lt;a href=&quot;https://nl.wikipedia.org/wiki/Fast_Fourier_transform&quot; rel=&quot;nofollow&quot;&gt;fast fourier transform&lt;/a&gt; (FFT) of the $i$th measurement of the output, $U_i(s)$ the FFT of the $i$th input data, and $\bar{U}_i(s)$ means taking the &lt;a href=&quot;https://en.wikipedia.org/wiki/Complex_conjugate&quot; rel=&quot;nofollow&quot;&gt;complex conjugate&lt;/a&gt; of $U_i(s)$ (which will help reduce the amount of noise in $P(s)$ the more measurements you use). Usually you would also apply a &lt;a href=&quot;https://en.wikipedia.org/wiki/Window_function&quot; rel=&quot;nofollow&quot;&gt;window function&lt;/a&gt; to each time domain signal, before calculating the FFT. A common choice is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Window_function#Hann_.28Hanning.29_window&quot; rel=&quot;nofollow&quot;&gt;Hanning window&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastEditorUserId="1765" LastEditDate="2016-10-19T11:18:08.943" LastActivityDate="2016-10-19T11:18:08.943" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11888" PostTypeId="2" ParentId="8769" CreationDate="2016-10-16T23:07:05.530" Score="4" Body="&lt;p&gt;Short answer: No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The code equations (e.g. ACI) relating modulus of rupture with compressive strength cannot be mathematically derived. It is an empirical relationship based on curve fitting to extensive experimental data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ASTM test for flexural strength (i.e. Modulus of rupture) is &lt;a href=&quot;http://www.astm.org/Standards/C78&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASTM C78/C78M&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, the most common test consists of loading an unreinforced simply-supported concrete beam at third points and determining the load at which fracture occurs. From this data, the modulus of rupture can be &lt;a href=&quot;https://ceen.et.byu.edu/sites/default/files/snrprojects/26-kirtikumar_k_shah-1969-kwk.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;calculated&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LCGAZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LCGAZ.jpg&quot; alt=&quot;Flexural strength test&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is generally a large amount of scatter in modulus of rupture data because a number of factors affect $f_r$ (mix design, aggregate size, rate of loading, specimen size, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many researchers have studied the relationship between modulus of rupture and compressive strength. Relating $f_r$ to $\sqrt f'_c$ seems to provide good correlation. Data from one such researcher is shown below. (Admittedly, it's not the definitive reference, but even Google has its limits when it comes to academic publications and paywalls.) I think some of the more recent research has proposed a 2/3 power relationship, though that may only be for high strength concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2DIXP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2DIXP.jpg&quot; alt=&quot;Relationship between modulus of rupture and compressive strength&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2016-10-16T23:07:05.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11890" PostTypeId="2" ParentId="6064" CreationDate="2016-10-17T02:46:41.303" Score="1" Body="&lt;p&gt;To start off, I thought I'd just throw up a conceptual sketch of an interaction diagram as a point of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PqP5x.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PqP5x.jpg&quot; alt=&quot;Interaction Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re: Question #1 ---&#xA;My best guess is that you're referring to a location in the compression controlled region where the strain profile is as in Case A, below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you state in your question: By definition, $c$ is the distance from the extreme fiber (at the assumed compressive strain of 0.003) to the neutral axis. To produce the interaction diagram, we assume values of $c$ and then use equilibrium equations to calculate $P$ and $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, there &lt;em&gt;is&lt;/em&gt; a point on the interaction curve where $c$ is the distance from the extreme fiber to the lowest layer of steel (Case B, below). Now the concrete below the steel is in tension, the concrete is assumed to be cracked, and the strength contribution of that concrete will be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MA0UO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MA0UO.jpg&quot; alt=&quot;Strain Profiles in Compression Region&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose theoretically, there is a small zone on the interaction curve where the concrete below the steel layer is in such a small amount of tension that the concrete could be uncracked. But the convention in practice is to assume concrete has no tension capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re: Question #2 ---&#xA;You are correct that in practice we assume elastic-perfectly-plastic material properties for the steel (as shown below) and therefore, while the strain may continue to increase, the stress in the steel cannot exceed the yield stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y8Me5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y8Me5.jpg&quot; alt=&quot;Steel Stress Strain Curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that in creating the interaction diagram, we're starting with a &lt;em&gt;strain&lt;/em&gt; profile. The tension strain in the steel simply is whatever it is for a given neutral axis location. Then you're probably using Young's Modulus ($E$) to calculate the corresponding stress in the steel. By doing this, you're assuming a &lt;em&gt;perfectly elastic&lt;/em&gt; material. Since we know we're actually elastoplastic, whenever $E\epsilon$ is greater than the yield stress, we can say that the steel is actually just at the yield stress.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="1832" LastEditDate="2016-10-18T00:26:57.600" LastActivityDate="2016-10-18T00:26:57.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11891" PostTypeId="1" CreationDate="2016-10-17T08:25:30.587" Score="1" ViewCount="107" Body="&lt;p&gt;I have a vertical wall which faces the sun (North in my case as I am in New Zealand) and it gets hot in Summer.I should like to take some of that heat and transfer it into a nearby swimming pool as it seems a shame to waste it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My plan is to fix 20mm HDPE pipe to the wall on the sunny side and pump water from the pool through it, then back to the pool.Most of the internet based plans have big spirals of exposed black pipe, but this is ugly (though effective) and wastes pipe.It should be possible to apply the pipe to the wall, spaced appropriately then render over the pipe so that the pipe is protected, concealed and also able to extract the energy from the bits of the wall between the pipes. 100% efficiency is not needed but from an engineering perspective the issues are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The less pipe I can use the better as this will limit costs, reduce flow resistance and of course make it easier to apply the render however the cost of the render is quite high so I need to extract as near as possible all the solar energy from the wall.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I will use electronics to detect when the temperature of the wall is higher than that of the pool so there is no need to plan for winter or not so sunny days - it will just run when it's going to heat the water rather than cool it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is no realistic prospect of the pipe freezing - it's a sunny place with gentle winters and the water is salty.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My plan would be to render 25mm deep - the depth of the pipe plus 5mm over the top.  chicken wire over the pipe will hold the pipe and the render in place nicely and should deal with expansion in the pipe well enough.  Fewer pipes will make this less of an issue of course.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the pipes are placed too far apart, they will not collect all the energy from the sun.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So what is the question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far apart should I place the pipes to have most of the solar energy absorbed by the wall enter the pipe, and how should they be configured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be possible to make the render deeper or the pipes narrower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two configurations spring to mind also:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a.  A spiral.  This has the advantage of simplicity and low material cost.  No leaks either, but is a very long pipe so resistance is high and airlocking might be a problem.  (not likely as I can just use a more powerful pump of course as there will be no joins)  I favour this option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b.  A parallel array with large infeed pipe along the bottom of the wall, similar atop it and vertically placed risers between.  Lots of joints in this case but much lower pump force needed and it is easy to bleed any air out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will this all just fall apart due to differential expansion in heat?&lt;/p&gt;&#xA;" OwnerUserId="8418" LastEditorUserId="7083" LastEditDate="2016-10-25T10:17:30.853" LastActivityDate="2016-10-25T10:17:30.853" Title="How to extract solar energy from a wall?" Tags="|solar-energy|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11892" PostTypeId="2" ParentId="10690" CreationDate="2016-10-17T09:18:52.337" Score="0" Body="&lt;p&gt;These batteries are designed for quite high current applications so can provide quite a punch.  It won't hurt you but if you short circuit it, it will get very hot very quickly.  Short-circuits aside, it should be fine.  I used to set fire to steel wool using one of these...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not waste time or money on the DC DC converter as the arduino will handle 6V very effectively.&#xA;Also, if you look at the voltage curves on the datasheet, you will see that the voltage will rapidly decline to 5V anyway so a converter would just be draining power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at &lt;a href=&quot;http://gadgetmakersblog.com/arduino-power-consumption/&quot; rel=&quot;nofollow&quot;&gt;http://gadgetmakersblog.com/arduino-power-consumption/&lt;/a&gt; which discusses arduino power consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have suggested try to transmit your information infrequently to save power.  Unless you really need real-time data, you could consider storing the data on an SD card and retrieving it when the project is over.  Of course someone might run off with it and the internet is so much more fun!&lt;/p&gt;&#xA;" OwnerUserId="8418" LastActivityDate="2016-10-17T09:18:52.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11893" PostTypeId="2" ParentId="11634" CreationDate="2016-10-17T12:01:27.357" Score="0" Body="&lt;p&gt;I found 2 possible approaches:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mayergoyz describes in &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/B9780124808737500037&quot; rel=&quot;nofollow&quot;&gt;chapter 2 of his book&lt;/a&gt; &quot;Mathematical Models of Hysteresis and Their Applications&quot; a Preisach model of magnetostricitve material. He considers a system with two cross-coupled inputs, which suits the problem prefectly.&lt;/li&gt;&#xA;&lt;li&gt;If the hysteresis of the disturbance value $z$ is negligible, one might try the &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.463.7198&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow&quot;&gt;modified Prandtl-Ishlinksi&lt;/a&gt; approach, in which the influence of $z$ on the $x$-$u$ hysteresis is modelled by an unambiguous characteristic. Furthermore this approach is highly suited for compensation purposes, since the model can be analytically inverted.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8170" LastActivityDate="2016-10-17T12:01:27.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11894" PostTypeId="1" AcceptedAnswerId="11897" CreationDate="2016-10-17T12:05:58.090" Score="2" ViewCount="1211" Body="&lt;p&gt;I am trying to derive the isentropic flow equations for a compressible gas by myself and at the end, I have different formulation than the one in the literature. Can you please tell me what am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have a nozzle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YE0vB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YE0vB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do an energy balance and consider that the kinetic energy and the potential energy in point 1 is negligible, one ends up with this relation:&#xA;$h_1=h_2+\dfrac{v_2^2}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using the ideal gas relation $h = CpT$, and if we then divide the equation by Cp and then by $T_2$, we end up with something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{T_1}{T_2}=1 + \dfrac{v_2^2}{2T_2Cp}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally using $Cp = Cv + R$, denoting $Cp/Cv = k$, using the formula for the mach number $M = v/c$ and the speed of sound in gas $c = \sqrt{(T_2Rk)}$, we end up with this relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{T_1}{T_2}=1 + \dfrac{k-1}{2}M^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from the &lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/isentrop.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;literature&lt;/a&gt; one can find that this formula is written like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dfrac{T_t}{T}=1 + \dfrac{k-1}{2}M^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is... Is the total temperature $T_t$ in their equation the same as $T_1$ in our case? And if the nozzle discharges to the ambient, is $T$ in their formulation the same as $T_2$ in ours? I'm a bit confused with the symbols and the meanings and I want to learn how this works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This is a copy of a question from &lt;a href=&quot;https://physics.stackexchange.com/questions/286940/isentropic-flow-equation-derivation-vs-literature&quot;&gt;enter link description here&lt;/a&gt;. I have been suggested to try and ask here.&lt;/p&gt;&#xA;" OwnerUserId="8419" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2016-11-16T16:37:36.340" Title="Isentropic flow equation derivation" Tags="|fluid-mechanics|thermodynamics|compressed-gases|compressible-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11895" PostTypeId="2" ParentId="11891" CreationDate="2016-10-17T12:31:09.180" Score="2" Body="&lt;p&gt;You might want to take a look at commercial designs -- for example there are solar panels with fluid loops on the underside to extract waste heat exactly as you want to do.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, on a technical note: ideally you'd like to maximize the heat transfer into your fluid. If you could cover the tubing (and/or the wall) with a material whose absorptance is high in the visible wavelength range but low in the infrared, then you will minimize radiative heat loss &amp;amp; thus improve the conductive heat transfer to the fluid.&lt;br&gt;&#xA;The other thing do to is to use tubing on the wall that is rectangular or elliptical in cross-section, so the tubing maximizes solar absorption and minimizes the physical distance from the surface to all the fluid. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-10-17T12:31:09.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11896" PostTypeId="1" CreationDate="2016-10-17T12:33:27.583" Score="1" ViewCount="98" Body="&lt;p&gt;I was reading about piezo elements and its frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found on the internet that : The composition of the ceramic material and the shape and volume of the element determine the resonance frequency -- generally, a thicker element has a lower resonance frequency than a thinner element of the same shape. &lt;a href=&quot;https://www.americanpiezo.com/knowledge-center/piezo-theory/determining-resonance-frequency.html&quot; rel=&quot;nofollow&quot;&gt;Determining Resonance Frequency&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found also that : At ultrasonic frequencies large surface area parts draw considerable current and resistive heating of the electrodes becomes the limiting factor. &lt;a href=&quot;http://www.piezo.com/tech3faq.html&quot; rel=&quot;nofollow&quot;&gt;Piezo Systems - Frequently Asked Questions&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First question, does that mean that if the piezo element is circular has a different frequency as a rectangular piezo element plate when they have the same thickness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that frequency = velocity/(2*thickness)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second question, what is the relation between frequency of piezo element and its shape?&lt;/p&gt;&#xA;" OwnerUserId="8420" LastEditorUserId="7083" LastEditDate="2016-10-25T09:17:36.653" LastActivityDate="2016-10-25T09:17:36.653" Title="What is the relation between frequency of piezo element and its shape?" Tags="|electrical-engineering|sensors|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11897" PostTypeId="2" ParentId="11894" CreationDate="2016-10-17T15:49:09.513" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the total temperature $T_t$ in their equation the same as $T_1$ in our case? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$T_t$ is called the stagnation temperature, which is the temperature of a flow when it's brought to rest ($v=0$), the $T_1$ you defined in the first equation is called the stagnation temperature since you ignored kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_pT_t = C_pT + \frac{v^2}{2}$$&#xA;$$T_t = T + \frac{v^2}{2C_p}$$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And if the nozzle discharges to the ambient, is $T$ in their formulation the same as $T_2$ in ours?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, the arbitary $T$ in the equation is for any location in the nozzle including the discharge.&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2016-10-17T15:49:09.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11898" PostTypeId="2" ParentId="11817" CreationDate="2016-10-17T17:32:49.683" Score="2" Body="&lt;p&gt;As Olin mentioned, the emission pattern is quite anisotropic. It depends on the LED. Most have primary lenses which all have unique emission patterns. One example of such can be seen in the Cree X-Lamp XP-G Datasheet (p.10)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cree.com/sitecore-modules/web/~/media/Files/Cree/LED-Components-and-Modules/XLamp/Data-and-Binning/XLampXPG.pdf#search=%22xp-g%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cree® XLamp® XP-G LEDs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2xbD8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2xbD8.png&quot; alt=&quot;Emission pattern and LED&quot;&gt;&lt;/a&gt;&#xA;On the left is the 'typical emission pattern' given by the manufacturer. On the right is the type of LED I'm assuming you are talking about when talking about a 'point light source'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best case scenario in my mind would be measuring it by yourself if you have access to a goniometer. The second best (and the easiest) would be searching the manufacturers datasheet for information, keeping in mind that reality is often quite different from manufacturer datasheet figures. If you have nothing else and must move forward a Gaussian distribution would be my best guess. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this only applies to LEDs such as illustrated. I do not know how COB -type LEDs radiate.&lt;/p&gt;&#xA;" OwnerUserId="8424" LastEditorUserId="7083" LastEditDate="2016-10-25T05:39:07.833" LastActivityDate="2016-10-25T05:39:07.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11900" PostTypeId="2" ParentId="11527" CreationDate="2016-10-17T18:54:31.620" Score="1" Body="&lt;p&gt;Flow rate by itself doesn't tell you the potential for extracting power.  You need to know the pressure drop, or if this water is open to the atmosphere, the height drop (which is a special case for determining pressure drop).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's say you have 1.5 m&lt;sup&gt;3&lt;/sup&gt;/s water flow that is dropping 2 m in height.  1.5 m&lt;sup&gt;3&lt;/sup&gt; of water has a mass of 1500 kg, which on earth weighs 14.7 kN.  That times the drop distance of 2 m yields 29.4 kJ/s, which is better expressed as 29.4 kW.  Of course any real device won't actually get all of that due to some inevitable inefficiency.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2016-10-18T11:36:01.847" LastActivityDate="2016-10-18T11:36:01.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11903" PostTypeId="1" AcceptedAnswerId="11946" CreationDate="2016-10-17T21:45:08.487" Score="5" ViewCount="1073" Body="&lt;p&gt;Consider a simple structure such as a beam, plate, ... Assume that you know how to determine the different modes of vibration of the structure. You are given an external input, such as a force or moment at a certain location. How can you then determine which modes will be excited how much? (Just a general question, I am not considering a specific case.)&lt;/p&gt;&#xA;" OwnerUserId="3633" LastActivityDate="2016-10-21T02:32:18.250" Title="How to know which modes are excited by a given input in modal analysis?" Tags="|structural-analysis|modal-analysis|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11904" PostTypeId="1" CreationDate="2016-10-17T23:51:33.410" Score="1" ViewCount="138" Body="&lt;p&gt;I would like to learn what type of &lt;a href=&quot;https://en.wikipedia.org/wiki/Pump#Positive_displacement_pumps&quot; rel=&quot;nofollow noreferrer&quot;&gt;positive displacement&lt;/a&gt; (PD) pumps are used to remove liquids from vacuum applications. I have been searching for an example application, but it seems to be a rare case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, imagine you have a liquid trap in a large vacuum system. The flow of water accumulation is so high that it warrants an extraction pump to remove the water. The pressure is about 90% vacuum. The goal is to remove the water to dump so it only need be pumped back up to ambient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a type of PD pump that is better at this application? From what I have found, very few pump manufacterers publish performance for vacuum operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have an example of a vacuum application where a PD pump is used to extract liquid?&lt;/p&gt;&#xA;" OwnerUserId="2780" LastEditorUserId="6790" LastEditDate="2020-05-04T11:28:51.783" LastActivityDate="2020-05-04T11:28:51.783" Title="What type of positive displacement pump is used for vacuum liquid extraction?" Tags="|pumps|vacuum|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11905" PostTypeId="1" AcceptedAnswerId="11914" CreationDate="2016-10-18T01:32:23.593" Score="3" ViewCount="5396" Body="&lt;p&gt;I'm trying to reverse engineer a set of spur gears that I have on a model RC car, but they are tiny, so there's a limit to what I can measure with my calipers.  My plan is to create a new gear design with a few important changes that will mesh with these existing gears, then have it professionally 3d printed in nylon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used this image as a reference for what data I might want to obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.unf.edu/~aschonni/Research/design%20analys%20fabr%20verific/gear%20tooth/Gear_design_analysis_files/image005.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's an image of the gears in question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ViFFm.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have these measurements for each gear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gear 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Teeth:  10&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pitch Diameter: 5mm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Outside Diameter: 6mm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Root Diameter: 4mm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gear 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Teeth: 22&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pitch Diameter: 11mm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Outside Diameter: 12mm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Root Diameter: 9.5mm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note:  I don't know how to obtain the base diameter, only the root diameter, since the base diameter doesn't seem to be directly measurable.  Also, the pitch diameters are a guess, but I'm assuming for simplicity the people who designed them probably picked 5mm and 11mm to go with 10 and 22 teeth respectively.  It's the sort of thing I'd probably do.  Also it matches with the centre-to-centre measurement of the meshed gears being 8mm, which is the same as (5mm + 11mm)/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of using this OpenSCAD script to replicate the gear designs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/openscad/MCAD/blob/dev/gears/involute_gears.scad&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/openscad/MCAD/blob/dev/gears/involute_gears.scad&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I have enough data to make a replica of these gears using that script?  If not, what else do I need?  The script's comments seem to indicate that I need to know the pressure angle, but looking at the diagram I linked, I can only figure that out if I have the base diameters.&lt;/p&gt;&#xA;" OwnerUserId="8428" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.343" LastActivityDate="2016-10-18T11:20:17.110" Title="Reverse-engineering spur gears - do I have enough data to make a gear that will mesh correctly?" Tags="|mechanical-engineering|design|gears|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11908" PostTypeId="2" ParentId="11891" CreationDate="2016-10-18T05:08:27.147" Score="2" Body="&lt;p&gt;While black pipe in any configuration will collect heat from the sun, I recommend not reinventing the wheel. There are many companies around the world that manufacture low-cost black-vinyl solar collectors specifically for pool heating. If you do some price hunting you can likely get them cheaper than you could get the equivalent surface area of tubing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.heliocol.com/heliocol-is/our-products&quot; rel=&quot;nofollow&quot;&gt;Heliocol Solar Pool Panel Manufaturer&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.umasolar.com/solar-products/#pool-heating&quot; rel=&quot;nofollow&quot;&gt;UMA Solar Pool Heating Distributor&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.heliocol.com/heliocol-vs/fafco&quot; rel=&quot;nofollow&quot;&gt;Heliocol vs Fafco&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you insist on making your own collector I recommend looking at manufactured panels and following them as a guide for achieving low cost and best efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while putting them on a wall will work, to get the most energy per solar collector you should put them at an angle that depends on your latitue. This &lt;a href=&quot;http://www.gogreensolar.com/pages/solar-panel-tilt-calculator&quot; rel=&quot;nofollow&quot;&gt;solar angle calculator&lt;/a&gt; has New Zeland. Choose the angle for the time of year you need the most heat. This will probably be the spring/autum angle if you do not plan on using the pool in the winter.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-10-18T05:08:27.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11909" PostTypeId="1" CreationDate="2016-10-18T08:48:32.237" Score="0" ViewCount="49" Body="&lt;p&gt;Assuming, I've got a machine, that runs with pressurized air and provides the following measurements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\dot m$ - mass-flow through the machine (same for input and output)&lt;/li&gt;&#xA;&lt;li&gt;$p_{1}$ - input pressure&lt;/li&gt;&#xA;&lt;li&gt;$T_1$ - input temperature&lt;/li&gt;&#xA;&lt;li&gt;$p_{2}$ - output pressure&lt;/li&gt;&#xA;&lt;li&gt;$T_2$ - output temperature&lt;/li&gt;&#xA;&lt;li&gt;$P_{mech}$ - mechanical output power&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the power $P_{fluid}$ the air gives to the machine?&#xA;In a first step, the heat-flow through the housing of the machine can be neglected, yet its influence to the process is interesting to me.&lt;/p&gt;&#xA;" OwnerUserId="8431" LastActivityDate="2016-10-24T06:03:46.740" Title="Power/Efficiency of expanding air-flow" Tags="|fluid-mechanics|thermodynamics|energy-efficiency|turbomachinery|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11911" PostTypeId="1" CreationDate="2016-10-18T10:48:50.857" Score="0" ViewCount="36" Body="&lt;p&gt;Materion sells 4 types of amorphous metal alloys,it cost huge amount of money,1000 dollar per kilogram.My question is,why would anybody choose it over lets say titanium 6al-4v?&lt;/p&gt;&#xA;" OwnerUserId="8433" LastActivityDate="2016-10-18T11:35:03.070" Title="Materion Bulk Metalic Glass" Tags="|materials|metals|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11912" PostTypeId="1" CreationDate="2016-10-18T10:59:22.537" Score="4" ViewCount="2436" Body="&lt;p&gt;I am going through a literature where these terms are regularly used and I am confused are they different and if yes then what does each term mean?&#xA;I also want to know relation of plastic strength of a rock (say granite) with pressure. I mean which equation describes this relation?&lt;/p&gt;&#xA;" OwnerUserId="8434" LastEditorUserId="3010" LastEditDate="2016-10-18T14:02:34.223" LastActivityDate="2016-10-21T19:35:28.827" Title="Difference between 'plastic yielding' , 'plastic yield strength', plastic strength' and 'plastic stress'?" Tags="|stresses|geotechnical-engineering|plastic|deformation|yield-point|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="11913" PostTypeId="2" ParentId="11904" CreationDate="2016-10-18T11:11:13.983" Score="2" Body="&lt;p&gt;Condensors in steam plant are a common application - about 90% vacuum and you have to extract the condensate at the rate the engine is using steam - something like 9 lbs per hp/hour if you'll excuse the non-SI units common in steam days. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you may be looking for a &lt;a href=&quot;http://en.wikipedia.org/wiki/Boiler_feedwater_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;boiler feedwater pump&lt;/a&gt; which not only works against the vacuum, but into a high pressure boiler. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LKwEy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LKwEy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2016-10-18T11:11:13.983" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11914" PostTypeId="2" ParentId="11905" CreationDate="2016-10-18T11:20:17.110" Score="5" Body="&lt;p&gt;Given these measurements I think these are standard Module 0.5mm (Mod 0.5) gears, and that alone plus a little web search should let you design for them to mesh correctly, and &lt;a href=&quot;http://www.technobotsonline.com/mod-0.5-gear-range.html&quot;&gt;buy&lt;/a&gt;,  &lt;a href=&quot;http://www.thingiverse.com/thing:683030&quot;&gt;design&lt;/a&gt; or &lt;a href=&quot;http://www.chronos.ltd.uk/acatalog/Gear_Cutters_and_Arbors.html&quot;&gt;manufacture&lt;/a&gt; any further gears that will mesh correctly with them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Easiest to measure:&lt;br&gt;&#xA;Outside diameter / (Tooth count + 2) = 0.5mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Useful cross check:&lt;br&gt;&#xA;Pitch diameter / (Tooth count) = 0.5mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically you may have to consider the possibility of incompatible Pressure Angles, but for modern (&amp;lt; 50 year old) injection moulded gears, I suspect that's such a remote possibility as to be non-existent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I deal with 14.5 degree pressure angle Imperial gears on pre-WW2 lathes but as far as I know everything recent uses 20 degree pressure angles.&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2016-10-18T11:20:17.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11915" PostTypeId="2" ParentId="11911" CreationDate="2016-10-18T11:35:03.070" Score="2" Body="&lt;p&gt;I assume you're talking about their bulk metallic glass (BMG), aka amorphous glass forming alloys.&lt;/p&gt;&#xA;&lt;p&gt;To quote &lt;a href=&quot;https://materion.com/Products/Alloys/BulkMetallicGlass.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;their site&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;HIGH PERFORMANCE ALTERNATIVE TO TITANIUM&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;BMGs have strength-to-weight&#xA;ratios more than twice those of titanium, magnesium and aluminum&#xA;alloys.  Amorphous alloys typically have elastic limits that can be&#xA;three times higher than crystalline metallic alloys and exhibit&#xA;outstanding corrosion resistance due to their unique atomic structure&#xA;and composition.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;EASY TO FABRICATE&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;These amorphous alloys have the additional benefit&#xA;of near net shape fabrication by injection molding and by a&#xA;thermoplastic forming process, similar to plastics.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Whether or not that's worth the extra cost depends on your circumstance.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-10-18T11:35:03.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11916" PostTypeId="1" CreationDate="2016-10-18T14:13:25.433" Score="-1" ViewCount="48" Body="&lt;p&gt;I want to make a knife, but I'm not shure what kind of steel should I use? &#xA;I want to form it through blacksmithing, and normalize it later.&#xA;Any suggestion?&lt;/p&gt;&#xA;" OwnerUserId="7760" LastActivityDate="2016-10-19T11:38:02.000" Title="Knife makeing, choose the steel" Tags="|steel|" AnswerCount="1" CommentCount="3" ClosedDate="2016-10-18T19:22:24.603" ContentLicense="CC BY-SA 3.0" />
  <row Id="11918" PostTypeId="2" ParentId="11916" CreationDate="2016-10-18T18:48:28.823" Score="3" Body="&lt;p&gt;For a beginner the best choice is a low alloy carbon steel as these are relatively simple to heat treat and more forgiving to work with than more complex alloys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular high alloy steels are often 'red short' this means that they have a relatively narrow temperature window in which they can be worked and can crumble if worked at too low a temperature and may require complex heat treating to make the most of their properties.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a knife you typically want a carbon content of at least 0.6% to be able to harden it adequately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spring steels are generally pretty good and &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Silver_steel&quot; rel=&quot;nofollow&quot;&gt;silver steel&lt;/a&gt;&lt;/em&gt; is probably the most readily available of forgeable knife steels as it can be purchased in small quantities from model engineering suppliers.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-19T11:38:02.000" LastActivityDate="2016-10-19T11:38:02.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11922" PostTypeId="1" AcceptedAnswerId="11924" CreationDate="2016-10-19T10:35:33.580" Score="3" ViewCount="854" Body="&lt;p&gt;I want to do some experiments on the dynamic behavior of the beam, plate and so on. Four boundary conditions I want to simulate: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Free&lt;/li&gt;&#xA;&lt;li&gt;Clamped&lt;/li&gt;&#xA;&lt;li&gt;Simply supported&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For free boundary condition, I can use a string to tie the beam or plates up  and hang them on some place. For clamped boundary, I can use some bolts to fix the beam or plate on a base. But how can I simulate the simply supported boundary conditions?&lt;/p&gt;&#xA;" OwnerUserId="8441" LastEditorUserId="1832" LastEditDate="2016-10-19T10:45:09.433" LastActivityDate="2019-02-23T01:08:50.157" Title="How to simulate the simply supported boundary condition in experiments?" Tags="|mechanical-engineering|applied-mechanics|dynamics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11924" PostTypeId="2" ParentId="11922" CreationDate="2016-10-19T10:53:32.700" Score="2" Body="&lt;p&gt;For steel beams, these are relatively simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simply supported beam requires two different supports: one to withstand vertical and horizontal forces (pinned support) and one to withstand only vertical forces (roller support).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are examples of each (click the links to see more examples and different types):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=pinned+support&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pinned supports&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gj9We.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gj9We.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://web.mit.edu/4.441/1_lectures/1_lecture13/pinned_1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;mit.edu&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=roller+support&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roller supports&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/brHTV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/brHTV.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;http://ceephotos.karcor.com/wordpress/wp-content/uploads/20110923-kjans-Bridge-Roller-Support-Manistee-MI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;karcor.com&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="6767" LastEditDate="2019-02-23T01:08:50.157" LastActivityDate="2019-02-23T01:08:50.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="11925" PostTypeId="2" ParentId="2125" CreationDate="2016-10-19T12:34:19.797" Score="1" Body="&lt;p&gt;Not sure how you've gotten on but I'm in the same boat and have gotten stumped...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;it will entail gears, linkages and actuators&quot; I believe is very difficult for the hobbyist.  This level of modelling requires professional design software costing ££££.  If you can get that, then all is well after you've engaged with the 3D modelling way of thinking.  It's not like 2D drafting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I've been testing FreeCAD 0.16 (the current stable[?] version) and been disappointed.  It's not really at a usable point yet.  I'm think that it might never reach usability as a lot of open source packages don't, such as the 2D LibreCAD.  It also can't deal with more than one part, so can only try to model a link, not linkages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you're academically linked, might I suggest getting a student /academic license for something like SolidWorks or SolidEdge.  There might also be Fusion 360.  They work well and are of a suitable stature for linkages and such.  They also allow the range of motion to be modelled.  Unfortunately for me as a Linux user, there are no free usable 3D modelling packages at this time.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="2947" LastEditDate="2016-10-19T12:42:55.270" LastActivityDate="2016-10-19T12:42:55.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11926" PostTypeId="1" AcceptedAnswerId="11941" CreationDate="2016-10-19T15:05:04.573" Score="4" ViewCount="933" Body="&lt;p&gt;I am trying to make like a &quot;periscope&quot; that i can use with a mobile camera.&#xA;I have made a very quick prototype with two mirrors and it &quot;works&quot; but you can see the interior of the walls so it is not a good solution. It currently looks like this &lt;a href=&quot;https://i.stack.imgur.com/MKPRT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MKPRT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vzUuv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vzUuv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as I said the problem is that you can see the &quot;walls&quot; on the interior of the periscope when i put the camera in there.&#xA;Is there anyway I can adjust this or do I need to look at an optional solution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i want to achieve is to make the backcamera of the phone to be like a frontcamera and have it under the camera that is vertical. (just like my current model). Can I somehow adjust the periscope (build it better, add more mirrors, maybe add a magnifying glass?) or do I need to build something else completely in order to make it work flawlessly without interior walls etc being in the way making the quality bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any tips, tricks, links is very much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="8445" LastEditorUserId="1832" LastEditDate="2016-10-19T17:38:35.770" LastActivityDate="2017-01-31T12:07:45.657" Title="Periscope camera with a mobile phone, how to make walls not visible on the camera/alternative solution?" Tags="|optics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11927" PostTypeId="1" AcceptedAnswerId="11934" CreationDate="2016-10-19T17:55:39.077" Score="1" ViewCount="656" Body="&lt;p&gt;The problem given is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Carnot cycle has a heat engine fluid efficiency of 30%. The heat transfer to the fluid in the boiler &lt;em&gt;Qh&lt;/em&gt; happens at 270 centigrade. At 270 centigrade, the entropy of saturated vapour and liquid water are 6.001KJ/(kgK) and 3.067KJ/(kgK), respectively. Determine the net work produced by the Carnot engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure to solve it you need to use the Carnot efficiency equation so 0.30 = &lt;em&gt;W&lt;/em&gt; / &lt;em&gt;Qh&lt;/em&gt; but I am not too sure how to get &lt;em&gt;Qh&lt;/em&gt; from the information provided.  Please explain or put me on the right track if you can!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="8264" LastActivityDate="2016-10-19T22:44:07.077" Title="How do you calculate the work of a Carnot Heat Engine using entropy?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11928" PostTypeId="1" AcceptedAnswerId="11929" CreationDate="2016-10-19T19:35:50.790" Score="1" ViewCount="112" Body="&lt;p&gt;I am currently trying to attach a small flat piece of plastic (no bigger than 2&quot;, could be smaller) to a 1.9&quot; hollow aluminum tube bent into a 2 foot radius torus, spinning at 2Hz.&#xA;My first and normal approach would be to drill a tapped hole in the hollow tube; however this is not an option in this situation, as the inside of the cylindrical tube is kept in a vacuum, and has to be in order for the project we are working on to work.&#xA;My second approach would be to mill the plastic to match the diameter of the tube, but the problem remains; how would I attach the resulting plastic to the tube?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have access to a machine shop, so I can do some basic mill and lathe stuff in order to accomplish this.&#xA;How should I do this?&lt;/p&gt;&#xA;" OwnerUserId="7156" LastEditorUserId="7156" LastEditDate="2016-10-19T20:48:18.097" LastActivityDate="2016-10-19T20:48:18.097" Title="Ways to attach stuff to a cylindrical tube that is in vacuum?" Tags="|mechanical-engineering|design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11929" PostTypeId="2" ParentId="11928" CreationDate="2016-10-19T20:16:51.057" Score="0" Body="&lt;p&gt;One way would be to clamp the outside of the pipe with a two-piece shaft collar. Properly tightened, a shaft collar will resist pretty heavy loads before it slips. You could tap the shaft collar and attach your plastic part as you normally would before attaching the collar to the pipe under vacuum. Note that a shaft collar will compress the pipe, so there is some risk of crushing the pipe, especially since it is already under inwards pressure. I doubt this will be a practical concern, but it is worth doing at least a rough check if you are using a thinner pipe wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem might be that shaft collars are typically manufactured for shaft sizes not pipe sizes, so you probably won't be able to find a perfectly sized collar. Since you have access to a shop, you could machine your own from scratch just for the pipe, or you could buy an 1 7/8&quot; collar (like McMaster Carr 6436K26) and machine the ID slightly larger. In a pinch you could open up the ID with a grinder too, but turning or milling would be a much better option.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-10-19T20:16:51.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11930" PostTypeId="1" CreationDate="2016-10-19T21:18:44.733" Score="0" ViewCount="67" Body="&lt;p&gt;I bought a little compressor that does 2.6 SCFM at 90 PSI. But I have a tool that needs 10 SCFM at 90 PSI. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you imagine PSI as voltage, and flow rate as amperage, then it might follow that the only thing limiting the flow rate is resistance. If so, why don't all air compressors reduce air resistance to achieve the same high flow rate?&lt;/p&gt;&#xA;" OwnerUserId="8450" LastActivityDate="2016-10-21T23:06:12.097" Title="Why do different air compressors have different flow rates?" Tags="|airflow|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11934" PostTypeId="2" ParentId="11927" CreationDate="2016-10-19T22:10:24.073" Score="2" Body="&lt;p&gt;Assuming reversible heat addition process:&#xA;$$q = \int_{s_1}^{s_2} T \,ds = T \int_{s_1}^{s_2}ds=T\Delta s $$&#xA;$$ w = \eta_{carnot} \,q$$&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2016-10-19T22:10:24.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11935" PostTypeId="2" ParentId="11851" CreationDate="2016-10-20T01:45:21.933" Score="2" Body="&lt;p&gt;The quantity that you are interested in is related to the momentum transfer from the bubble to an object.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a solution for the rate of change of the radius as a function of time.  Convert that into a velocity and compute the momentum of the bubble mass as a function of time.  Use Newton's equation (momentum equation) to compute the resulting force.  Convert that force into a pressure by dividing by the appropriate surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That will give you an estimate of the impact pressure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the equations you have solved assume that the bubble is isolated and in an infinite fluid.  Therefore, your estimates will only be qualitative.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2016-10-20T01:45:21.933" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11938" PostTypeId="2" ParentId="11926" CreationDate="2016-10-20T13:23:26.553" Score="3" Body="&lt;p&gt;The basic answer is that the field of view (FOV) of your camera is much wider than the FOV of the periscope.  Unless you add a lens system between the periscope and the phone to limit the effective FOV of the camera, there is no way to avoid seeing the walls of the periscope.  (Well, you &lt;em&gt;could&lt;/em&gt; build a periscope large enough that the walls are outside the camera's FOV, but that would be ugly :-) )&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-10-20T13:23:26.553" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="11939" PostTypeId="2" ParentId="11903" CreationDate="2016-10-20T17:18:15.863" Score="3" Body="&lt;p&gt;Firstly, applying a constant load will not excite any modes. You need a time-varying load to induce vibration (e.g. wind, seismic, blast). Which modes are excited will depend on the frequency content of the applied loading.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2016-10-20T17:18:15.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11940" PostTypeId="2" ParentId="10333" CreationDate="2016-10-20T18:14:35.690" Score="1" Body="&lt;p&gt;You are correct that the response for each mode must be calculated before applying some modal combination methodology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of thinking about the structure as being discretized into numerous elements, I like to think of it in terms of the lumped masses and how many DOFs there are. A more complex structure will simply have more DOFs (and therefore more modes) than a simple 'lollipop' model. But you still come up with modes for the structure as a whole - not modes for individual elements within the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you mention RSA, I'm assuming you're approaching all this from the perspective of structures subjected to earthquakes. During an earthquake, multiple modes will be excited, depending on the frequency content of the earthquake. How then, do you determine the overall structure response? RSA is a simplified tool for seismic design -- allowing you to first determine the peak response of the individual modes and then combine those responses using SRSS, CQC, etc. This is opposed to running time-history analyses (very computationally intensive, but sometimes done for complex and/or important structures).&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2016-10-20T18:14:35.690" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11941" PostTypeId="2" ParentId="11926" CreationDate="2016-10-20T19:14:20.833" Score="5" Body="&lt;p&gt;Ok let me draw some pictures you right now have the following situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0o3Pr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0o3Pr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add a convex lens to this setup you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EoJNf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EoJNf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which does work. However you may want to add a concave lens after the convex lens so that you can more readily adjust the beam focus and shape giving you more freedom to shape your periscope.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2016-10-20T19:14:20.833" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11942" PostTypeId="1" CreationDate="2016-10-20T22:09:19.490" Score="1" ViewCount="99" Body="&lt;p&gt;I am designing a potato peeler which is practically a bucket that has a rotating disk (figure below) at the bottom driven by a motor. Water and potatoes are put inside the bucket and the disk rotates creating a swirling motion and peeling the potatoes by rubbing them against an abrasive surface. How do I calculate the power required by the disk to rotate a given weight of water plus potatoes (in kgs) at a given rotational speed (in rpm)?&lt;a href=&quot;https://i.stack.imgur.com/Qnoc2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qnoc2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="8461" LastActivityDate="2016-10-21T08:16:33.120" Title="How do you calculate the required power of a disk turbine (Rushton turbine) to rotate a fluid at a given speed?" Tags="|mechanical-engineering|fluid-mechanics|process-engineering|product-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11944" PostTypeId="2" ParentId="11912" CreationDate="2016-10-20T23:13:47.310" Score="6" Body="&lt;p&gt;All these terms refer to the effect of loading on the deformation of materials.  Let us assume that we start with zero load and zero deformation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Elastic deformation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase the load you get an increase in deformation.  During the process of &lt;em&gt;elastic&lt;/em&gt; deformation, if you decrease the load to zero you will not have any residual non-zero deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Plastic yielding&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;process&lt;/em&gt; during which the material deforms to such an extent that when the load is removed the deformation does not return to its initial state.  The material appears to flow like a fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Plastic yield strength&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load beyond which elastic deformations cease to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Plastic strength&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This term is not specific enough and typically not used when technical accuracy is desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Plastic stress&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is another term that does not mean much and should be avoided.  An alternative is the &lt;em&gt;yield stress&lt;/em&gt; which means the same thing as &lt;em&gt;plastic yield strength&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Relation between yield stress and pressure&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term pressure can have multiple meanings depending on the context. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are thinking of a load applied to the surface of a rock sample, there is no straightforward way of relating the yield stress to the applied load.  There are numerous models for rocks and other granular material. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Yield_surface&quot; rel=&quot;nofollow&quot;&gt;https://en.wikipedia.org/wiki/Yield_surface&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if you are thinking of the change in hydrostatic yield stress with plastic strain, &lt;em&gt;crush curves&lt;/em&gt; are typically used to model the evolution of caps on yield surfaces.  See &lt;a href=&quot;http://www.iaea.org/inis/collection/NCLCollectionStore/_Public/15/061/15061729.pdf#page=43&quot; rel=&quot;nofollow&quot;&gt;http://www.iaea.org/inis/collection/NCLCollectionStore/_Public/15/061/15061729.pdf#page=43&lt;/a&gt; for an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rate-dependent plasticity&lt;/strong&gt;&#xA;If you load a rock sample in uniaxial strain compression at various speeds, the load at which it yields will appear to increase with the speed of loading.  If you define the yield stress based on a low speed (quasistatic) test, the stress will appear to exceed the quasistatic yield stress.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="3488" LastEditDate="2016-10-21T19:35:28.827" LastActivityDate="2016-10-21T19:35:28.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11945" PostTypeId="2" ParentId="11903" CreationDate="2016-10-21T00:59:35.843" Score="1" Body="&lt;p&gt;For a static problem, you can determine the displacements of a structure using only forces and stiffness of the structure using static equations for text books:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{normal} = K \Delta \\ M_{bending} =  EI \phi \\ T_{torsion} = GJ \psi $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that none of these equations depend on time. For the purpose of most problems in civil engineering, load application is slow enough that inertia forces are very small compared to other forces, so they are neglected (appart from wind and earthquakes loads which often will require dynamic analysis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modes of vibration of a structure require dynamic analysis, so they do not depend only on the stiffness of the structure, they also depend on its weight distribution and to a lesser extent to it's damping. To determine the modes of vibration of a structure, you need to know &lt;strong&gt;at least&lt;/strong&gt; its stiffness and its mass (and how it is distributed). The simplest example is a mass M on wheels tied to a horizontal spring of stiffness K. The natural frequency of this system would be :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f = 2 \pi\sqrt\frac{K}M $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note here that to obtain the natural frequency of the system, no external loads had to be applied. Since our case in unidirectional, it's quite simple to figure out which way an external load would have to go to excite our structure (at the calculated frequency of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more complex case the mass is usually distributed (rather than punctual) and the stiffness is often in bending, shear and torsion in addition to axial. For these more complex cases, we will usually also be working with more than one degree of freedom, so you will have multiple equations with multiple variables to solve (usually using finite element analysis). In using finite element analysis, your mass will be represented as a matrix M of n by n elements and your stiffness by a matrix K of n by n elements, where n is your number of degrees of freedom. Using eigenvalues, we can solve an equation very similar to the one above to obtain n modes of vibration and n natural frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each mode of vibration coupled with it's natural frequency gives you information on how to load each degree of freedom if you want to achieve resonance. Any other loading with any other frequency will not excite one mode alone, so you can only know the behaviour of any other system with more complex calculations.&lt;/p&gt;&#xA;" OwnerUserId="5691" LastActivityDate="2016-10-21T00:59:35.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11946" PostTypeId="2" ParentId="11903" CreationDate="2016-10-21T02:25:32.097" Score="5" Body="&lt;p&gt;They way to answer this question is transform the dynamics model into modal coordinates, and see what happens to the force term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we can describe the stiffness and mass properties of the structure as matrices $\mathbf K $ and $\mathbf M$, and its displacement as a vector $\mathbf x$, in a physical coordinate system, and we apply a vector of forces that vary sinusoidally in time, $\mathbf F e^{i\omega t}$ to the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation of motion of the system is then $$(\mathbf K - \omega^2 \mathbf M) \mathbf x e^{i\omega t} = \mathbf Fe^{i\omega t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can cancel the $e^{i\omega t}$ terms and reduce this to $$(\mathbf K - \omega^2 \mathbf M) \mathbf x = \mathbf F$$&#xA;(Note, I'm ignoring damping, for the sake of making things a bit simpler - doing that doesn't affect the final conclusion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can find the normal modes of the system and write the eigenvectors as a matrix $\mathbf \Phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write the displacements $\mathbf X$ as a linear combination of the eigenvalues, i.e. $\mathbf X = \mathbf \Phi \xi$ where $\xi$ is a vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute that in the equation of motion, and pre-multiply both sides by $\mathbf \Phi^T$ and we get $$\mathbf \Phi^T(\mathbf K - \omega^2 \mathbf M) \mathbf \Phi \xi = \mathbf \Phi^T \mathbf F$$&#xA;or $$(\mathbf \Phi^T\mathbf K \mathbf \Phi - \omega^2 \mathbf \Phi^T\mathbf M \mathbf \Phi) \xi = \mathbf \Phi^T \mathbf F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we use &lt;em&gt;mass-normalized&lt;/em&gt; eigenvectors, we know that $\mathbf \Phi^T\mathbf M \mathbf \Phi$ is a unit matrix, and $\mathbf \Phi^T\mathbf K \mathbf \Phi$ is a diagonal matrix of the eigenvalues squared. So the matrix equation becomes a set of scalar equations, and for the $i$th mode we have&#xA;$$(\omega_i^2 - \omega^2)\mathbf \xi_i = \mathbf \Phi_i^T \mathbf F$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Translating that equation back into words answers the OP's question: For each mode, you take the scalar product of each eigenvalue with the applied forces to find the &quot;modal component of the force&quot; (i.e. right hand side term of the final equation), and from that you can find the relative amplitude of that mode (i.e. $\xi_i$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases we only apply a force to a single degree of freedom of the structure. Then, the result is fairly intuitive - there are two effects which are relevant, when they are taken together: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Looking at the right hand side of the final equation, &lt;em&gt;The modes that will be excited most are those with the biggest displacements at the point where the force is applied&lt;/em&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Looking at the left hand side of the same equation, &lt;em&gt;The modes that will be excited most are those whose natural frequencies are close to the forcing frequency&lt;/em&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-10-21T02:32:18.250" LastActivityDate="2016-10-21T02:32:18.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11947" PostTypeId="1" AcceptedAnswerId="11949" CreationDate="2016-10-21T06:59:42.180" Score="5" ViewCount="342" Body="&lt;p&gt;I'm was looking into the &lt;code&gt;ASME B16.5&lt;/code&gt;Pipe flanges and pipe fittings and it has 2 dimension tables metric and inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I found was rather strange for me. I was looking at a 2 inch welding neck flange, and the hub diameter must match the pipe size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the metric table this is &lt;code&gt;60.3mm&lt;/code&gt; and it is the same in all pipe sizes charts that I have available. When I look at this dimension in the Inch table, it is &lt;code&gt;2.38in&lt;/code&gt; when I convert this to &lt;code&gt;mm&lt;/code&gt; this varies from the metric table..&#xA;&lt;code&gt;2.38in * 25,4 = 60,452mm&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a &lt;code&gt;,152&lt;/code&gt; difference between these values. &lt;strong&gt;What is correct?&lt;/strong&gt; These differences are to be found on more flanges then just the 2&quot; flange size.  &lt;/p&gt;&#xA;" OwnerUserId="8464" LastActivityDate="2017-01-22T20:45:23.597" Title="What is the correct pipe size for a 2inch pipe?" Tags="|process-engineering|piping|pipelines|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11948" PostTypeId="2" ParentId="11942" CreationDate="2016-10-21T08:09:52.040" Score="2" Body="&lt;p&gt;We have a fair share of questions about how to calculate maximum speed, but the fact remains that this is extremely hard calculation to make. See a mass will continue to accelerate until the losses are equal to the gains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Estimating the losses is hard, if not impossible in practice. That is because friction of your system is extremely hard to predict. Friction depends on nearly everything, from tolerances to how the actual micro surface geometry, chemical reactions at the surfaces and total energy in system etc.. So the easiest answer is build and see. Bear in mind that you have to win at least known forces by some margin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can however make a simplified model and estimate each interface by finding a somewhat corresponding case for each thing. Its still just a guess though. Without the resources of a big engineering company its most likely easier to exaggerate engine size then build and see. But do measure the frictions and estimate them this will come handy next time you do something similar.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2016-10-21T08:16:33.120" LastActivityDate="2016-10-21T08:16:33.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11949" PostTypeId="2" ParentId="11947" CreationDate="2016-10-21T12:08:34.460" Score="6" Body="&lt;p&gt;I would guess that the discrepancy is down to rounding fractional inch sizes to decimal inches ie 2.38 is not exactly 2 3/8&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that 2.375 (two and three &lt;s&gt;quarter&lt;/s&gt; eighth inches) is 60.325mm &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-26T23:35:23.253" LastActivityDate="2016-10-26T23:35:23.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11950" PostTypeId="2" ParentId="11947" CreationDate="2016-10-21T13:54:54.920" Score="9" Body="&lt;p&gt;As of the 2013 edition of ASME B16.5, the scope section explains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This Standard states values in both SI (Metric) and&#xA;  U.S. Customary units. As an exception, diameter of bolts&#xA;  and flange bolt holes are expressed in inch units only.&#xA;  These systems of units are to be regarded separately as&#xA;  standard. Within the text, the U.S. Customary units are&#xA;  shown in parentheses or in separate tables that appear&#xA;  in Mandatory Appendix II. The values stated in each&#xA;  system are not exact equivalents; therefore, it is required&#xA;  that each system of units be used independently of the&#xA;  other. Except for diameter of bolts and flange bolt holes,&#xA;  combining values from the two systems constitutes nonconformance&#xA;  with the Standard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is actually a pretty common practice where metric and inch systems are regarded separately. While the numbers may not match exactly, the intention is that they work as a system within each unit of measure. Of course, as far as I'm aware, you would not buy a metric flange for an NPS or DN size pipe separately, they will be the same product manufactured to some company's internal drawings, so in essence the standard is accepting that any rounding errors are within tolerances intended by the standard. One reason they don't allow intermixing when interpreting the standard is that they don't want you to be able to cherry-pick the most permissive set of dimensions and tolerances for each feature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forward also notes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The 2003 Edition included metric units as the primary reference units while maintaining U.S.&#xA;  Customary units in either parenthetical or separate forms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it would be reasonable to consider the metric information in the standard to be slightly more authoritative than the inch information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forward to ASME B36.10M-2004, which covers the dimensions of the actual pipe, explains&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The standard was revised in 1978 to include SI metric dimensions. The outside diameter and&#xA;  wall thicknesses were converted to millimeters by multiplying the inch dimensions by 25.4.&#xA;  Outside diameters larger than 16 in. were rounded to the nearest millimeter, and outside diameters&#xA;  16 in. and smaller were rounded to the nearest 0.1 mm. Wall thicknesses were rounded to the&#xA;  nearest 0.01 mm. These converted and rounded SI metric dimensions were added to Table 2. A&#xA;  formula to calculate the SI metric plain end mass, in kilograms per meter, using SI metric diameters&#xA;  and thicknesses was added to section 5. The SI metric plain end mass was calculated and was&#xA;  added to Table 2. These changes in the standard were approved and it was designated an American&#xA;  National Standard on July 18, 1979.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So they do acknowledge some use of rounding. For 2&quot; NPS pipe they list the OD as  2.375 inches and 60.3mm (DN 50 size.) There there is only an error of .025mm which is consistent with their rounding philosophy. While your observed error of .152mm is definitely larger than I can directly justify, it is unlikely to cause a serviceability issue in a welded joint - .006&quot; in below the manufacturing tolerances of most metal parts that are to be welded. Tolerances that small are usually achieved by precision machining. I would not be concerned about the actual function of the joint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address your specific question in the title, the pipe should be sized according to its own spec. That standard may have its own dimension tables, or it may reference ASME B36.10M and then specify manufacturing tolerances on top of it. In the US, if you'r ejust looking at plain black pipe, the relevant standard would be ASTM A36, but a multitude of standards exist for other materials, tighter tolerances, and application-specific pipes.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2016-10-21T13:54:54.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11951" PostTypeId="1" CreationDate="2016-10-21T15:52:57.503" Score="1" ViewCount="198" Body="&lt;p&gt;A viscometer of the Redwood type has an oil-containing cylinder 4.75cm in diam and an agate tube 0.17&amp;nbsp;cm in diameter and 1.2&amp;nbsp;cm long. The oil surface, when flow starts, is 9&amp;nbsp;cm above the outlet from the agate tube. To allow for the sudden contractions at entry to the tube, the effective length of the tube may be taken as the actual length plus the tube radius. Making allowance for the decreasing head of oil, the viscous resistance through the tube, and the KE (kinetic energy) of discharge, calculate using arithmetical integration, the time required for 50&amp;nbsp;cm&lt;sup&gt;3&lt;/sup&gt; of an oil of viscosity 0.5&amp;nbsp;poise and specific gravity 0.92 to flow through the viscometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you find the time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this question, I have tried to use quadratic equation to solve for $v$, but it is proving more difficult than I thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far this is what I have: $$h=\frac{32\mu (l+d/2)v}{\rho g d^2}+\frac{v^2}{2g}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$av^2 + bv + c = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a = 1/ 2g$,  $b=\frac{32\mu (l+d/2)v}{\rho g d^2} and  $c = $-h$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $v = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I eventually got $v = - \frac{32\mu (l+d/2)v}{\rho d^2} \pm (\frac{32\mu (l+d/2)v}{\rho d^2} + \sqrt{2gh})$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this expression $-b + b = 0$, so how do I deal with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the fall in level be $dh$ in a small interval of time $\text{d}t$, water area of cylinder be $A$ and area of tube be $a$.&#xA;Therefore, continuity equation of tank and pipe yields quantity discharge in $\text{d}t$ seconds = $A \text{d}h$ = $a v \text{d}t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you integrate the expression for this time for the head level to fall from $H1$ to $H2$ for this question?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any ideas that can help me, so that I can solve this question?&lt;/p&gt;&#xA;" OwnerUserId="8013" LastEditorUserId="8013" LastEditDate="2016-10-22T16:16:12.550" LastActivityDate="2016-10-22T16:16:12.550" Title="Time for oil to flow through Redwood Viscometer" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11953" PostTypeId="1" AcceptedAnswerId="11955" CreationDate="2016-10-21T17:21:51.757" Score="0" ViewCount="188" Body="&lt;p&gt;Is there something similar to the Section Modulus in the Flexure Formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_{max} = \dfrac{M_{max}}{S}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $S = \dfrac{I}{c}$ for shear stress formula?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau_{max} = \dfrac{VQ}{IT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then there should be a Shear Section Modulus published which would be $Q(\text{of centroid}) / IT$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can get $\tau_{max} = V * \text{Shear Section Modulus}$.&lt;/p&gt;&#xA;" OwnerUserId="5967" LastEditorUserId="1832" LastEditDate="2016-10-21T17:40:40.347" LastActivityDate="2016-10-22T03:52:56.223" Title="Shear Stress / Flexure Stress Formulas / Tables" Tags="|stresses|moments|shear|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11955" PostTypeId="2" ParentId="11953" CreationDate="2016-10-21T23:01:58.543" Score="1" Body="&lt;p&gt;What you're looking for is commonly referred to as the &lt;strong&gt;shear area&lt;/strong&gt; of the section (see &lt;a href=&quot;https://knowledge.autodesk.com/support/robot-structural-analysis-products/learn-explore/caas/CloudHelp/cloudhelp/2015/ENU/Robot/files/GUID-50986085-D4AB-41C5-9B21-CF9619B8EB8B-htm.html&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt; for a list of shear areas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a solid rectangle with depth 20mm and width 10mm. From &lt;a href=&quot;https://skyciv.com/free-moment-of-inertia-calculator&quot; rel=&quot;nofollow&quot;&gt;this calculator&lt;/a&gt; we can see that the section properties are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = 6667 \text{ mm}^4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = 500 \text{ mm}^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t = 10 \text{ mm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the shear area is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_{s} =\dfrac{It}{Q}=\dfrac{(6667)(10)}{500} = 133.3 \text{ mm}^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the shear stress is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau_{max}=\dfrac{VQ}{It}=\dfrac{V}{A_s} = \dfrac{V}{133.3 \text{ mm}^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare this to the &lt;a href=&quot;https://knowledge.autodesk.com/support/robot-structural-analysis-products/learn-explore/caas/CloudHelp/cloudhelp/2015/ENU/Robot/files/GUID-50986085-D4AB-41C5-9B21-CF9619B8EB8B-htm.html&quot; rel=&quot;nofollow&quot;&gt;resource&lt;/a&gt; I shared earlier and you can see that for a thick walled rectangular section the shear area (denoted by W in the resource) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A_s=\dfrac{2}{3}hb = \dfrac{2}{3}(20)(10) = 133.3 \text{ mm}^2$&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2016-10-22T03:52:56.223" LastActivityDate="2016-10-22T03:52:56.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11956" PostTypeId="2" ParentId="11930" CreationDate="2016-10-21T23:06:12.097" Score="2" Body="&lt;p&gt;The assumption you made is intuitive and correct, however this is practically already accomplished. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resistance is analogous to water (or air) flow in a pipe, and the resistance from the walls of the pipe are effectively our predominant resistance. However, when you are measuring resistance in a pipe, the end effect is a drop in pressure from one end to the other, which is similar to voltage potential. To make electricity flow from one end to another, you have to have a voltage potential. Well, flow through a pipe is no different, and you require a pressure potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to measure the pressure of the air at the exit of the compressor, it would be 90 PSI, 2.6 SCFM. If you measured it at the inlet of the piping to your tool, it might be 89.99 PSI, 2.6 SCFM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, at THAT point, you need to actually increase the speed of your compressor to put more air through at the same pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you care to understand that concept:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compress air, we have to do something called Pressure-Volume work (PV work). In this case, you'll be doing PV work &lt;em&gt;onto&lt;/em&gt; the system, or the air you want to compress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks like:&#xA;$$work=\int_{V_1}^{V_2}PdV$$&#xA;Note that normally there's a negative sign in front of the integral, but in this case we're doing work &lt;em&gt;on&lt;/em&gt; the system, so the sign is positive indicating that work is being applied to the air we're talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you change the $V$ to $V/second$ or a flow rate (such as 2.6 SCFM), then work (joules) becomes power (joules/second or watts). That is effectively limited by the motor on your compressor, or any mechanical limitations of a compressor (such as a max volume per cycle, and a maximum cycles per unit time etc) &lt;/p&gt;&#xA;" OwnerUserId="8369" LastActivityDate="2016-10-21T23:06:12.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11958" PostTypeId="1" AcceptedAnswerId="11961" CreationDate="2016-10-22T07:03:41.303" Score="1" ViewCount="431" Body="&lt;p&gt;I am trying to solve the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A masonry chimney having the shape of a conical frustrum is 25 m high.&#xA;  The external diameter at the top and internal diameter at the bottom&#xA;  is 2 m. The chimney is 0.5 m thick at its base. If the weight of the&#xA;  chimney is 1800 kN, find the uniform horizontal wind pressure that may&#xA;  act on the chimney per unit projected area of the chimney in order for&#xA;  tension at the base to be just avoided.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am using the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathrm{Centroid} (\bar{y}) = \dfrac {A_{1}y_{1} + A_2y_{2}} {A_{1} + A_{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A1$ and $A2$ are the shaded areas in the figure below. Observe that the object is a chimney, so it is hollow in the centre portion; we are not adding up the area of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tisSD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tisSD.png&quot; alt=&quot;Cross-sectional diagram of the chimney with the area on the left marked A1 and the area on the right marked A2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The centroid becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar{y} = \dfrac{ (\dfrac{1}2  *0.5 *25) * \dfrac{25}3   + (\dfrac{1}2  *0.5 *25) * \dfrac{25}3}{(\dfrac{1}2  *0.5 *25) +(\dfrac{1}2  *0.5 *25) }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{y} = 8.333$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, I created a 3D model of the solid chimney in CAD and found the centroid to be 8.035, which does not match my calculated centroid. It is close to 8.333 but not exactly the same as my $\bar{y}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I calculate the total wind pressure, which centroid should I use? The one I calculated by hand from the 2D cross-section, or the one I generated in CAD from the 3D model?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CAD generated centroid is for a 3D solid and $\bar{y}$ calculated is for that central section.&lt;/p&gt;&#xA;" OwnerUserId="2116" LastEditorUserId="368" LastEditDate="2016-11-21T17:17:56.000" LastActivityDate="2016-11-21T17:17:56.000" Title="How should I calculate the centroid of a hollow structure when analyzing a wind load?" Tags="|structural-engineering|geometry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11960" PostTypeId="1" CreationDate="2016-10-22T08:47:39.210" Score="3" ViewCount="556" Body="&lt;p&gt;AFAIK, all high speed trains (200+ kmh) run on electric track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this? Is there something about a diesel-electric engine that makes high-speed trains more difficult?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some other data, and correct me if I'm wrong:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric track is high-maintenance. Using a pantograph to pick up power from overhead wires is also non-trivial at speeds of 200 or 300 kmh. The electric system is much more prone to failure in bad weather (storms, snow/ice, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've often heard the claim that electrified rail is more energy-efficient than a diesel-electric. However, I've never seen numbers showing &lt;strong&gt;exactly how much more efficient&lt;/strong&gt; it is compared to diesel-electric. Maybe someone knows a source on that.&lt;/p&gt;&#xA;" OwnerUserId="3510" LastActivityDate="2024-02-01T07:06:12.133" Title="Why do all high-speed trains run on electric track (as opposed to normal track with diesel-electric trains)?" Tags="|mechanical-engineering|electrical-engineering|civil-engineering|rail|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11961" PostTypeId="2" ParentId="11958" CreationDate="2016-10-22T11:20:14.640" Score="2" Body="&lt;p&gt;I believe that you are confusing yourself by introducing the 3D aspect when building your CAD model, and yet trying to determine the answer by hand using a 2D calculation.&#xA;Due to the volume being different to an area, this will skew your answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps nudge you in the right direction&lt;/p&gt;&#xA;" OwnerUserId="8212" LastActivityDate="2016-10-22T11:20:14.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11962" PostTypeId="1" CreationDate="2016-10-22T14:54:18.637" Score="6" ViewCount="60020" Body="&lt;p&gt;I have been reading a lot about guns in Wikipedia. I know much of the terminology of firearms. However, I was wondering why some firearms such as shotguns and rifles have very long barrels opposed to small handguns such as revolvers and pistols, that normally have shorter barrels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, there are functional differences between the types of weapons I described: Shotguns normally shot shogun shells with pellets, Rifles have &quot;rifling&quot; inside their barrel, pistols are made to be held with the hand without shoulder support, revolvers have a cilynder of cartridges that rotates. (This descriptions is rough and doesn't intend to be a very accurate defintion of every gun type).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, shotguns rifles and snipers normally have longer barrels than pistols and revolvers &lt;a href=&quot;https://www.google.com.ar/search?q=pistol%20long%20barrel&amp;amp;espv=2&amp;amp;biw=1366&amp;amp;bih=638&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwj7g4H20u7PAhVKipAKHa7HBjUQ_AUIBigB#imgrc=ipF0t51__picwM%3A&quot; rel=&quot;noreferrer&quot;&gt;(althought there are handguns with long barrels)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.backwoodshome.com/how-long-should-your-guns-barrel-length-be/&quot; rel=&quot;noreferrer&quot;&gt;Here there is an article&lt;/a&gt; about the subject&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is:&#xA;What advantages and disadvantages come with differences in length of the barrel? And does a longer barrel improve accuracy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made up some hypotheses. I list them here. :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Longer barrels are used for higher explosive ammo because the barrel can contain the expanded gases and made them push the bullet over a longer barrel, giving the bullet more energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Longer barrel means the sight is further and thus one can make more precise shots because the two aligning marks are further, and thus in order to seem aligned to the eye they have to be in &quot;more aligned&quot; than with a shorter barrel.&lt;/p&gt;&#xA;" OwnerUserId="5033" LastEditorUserId="7590" LastEditDate="2016-10-28T10:50:28.523" LastActivityDate="2023-10-18T22:06:21.237" Title="Do longer gun's barrel improve accuracy?" Tags="|design|" AnswerCount="7" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11963" PostTypeId="1" AcceptedAnswerId="12038" CreationDate="2016-10-22T23:55:50.420" Score="11" ViewCount="4581" Body="&lt;p&gt;Cavitation and boiling are names for phenomenon that both involve sudden appearance of bubbles of vapor within a liquid, and in both cases they happen when the local hydrostatic pressure is lower than the vapor pressure of the fluid, but that doesn't necessarily mean they are the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://youtu.be/Lwk9Bi3j58o?t=105&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video of an electrical heating element in water&lt;/a&gt;, between 01:00 and 02:00 the sound produced by rapid bubble collapse gets louder and louder but there are few visible bubbles. Is the process that produces this sound considered &lt;em&gt;boiling&lt;/em&gt; or &lt;em&gt;cavitation&lt;/em&gt;? What is the distinction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've left a &lt;a href=&quot;https://space.stackexchange.com/a/18573/12102&quot;&gt;provisional answer&lt;/a&gt; to the related question on another SE site: &lt;a href=&quot;https://space.stackexchange.com/q/18486/12102&quot;&gt;How (actually) do sub-cooled propellants reduce cavitation within turbo pumps and make feed easier?&lt;/a&gt; I couldn't bring myself accept &lt;a href=&quot;https://space.stackexchange.com/a/18496/12102&quot;&gt;the answer&lt;/a&gt; on that question that starts with the claim, &quot;cavitation is boiling.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although they are related, &lt;em&gt;what fundamentally distinguishes cavitation and boiling as different phenomena?&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="6264" LastEditorUserId="6264" LastEditDate="2020-06-02T00:32:49.680" LastActivityDate="2022-09-19T07:43:21.990" Title="What fundamentally distinguishes cavitation and boiling as different phenomena?" Tags="|fluid-mechanics|thermodynamics|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="11964" PostTypeId="2" ParentId="11963" CreationDate="2016-10-23T00:51:40.557" Score="10" Body="&lt;p&gt;I think this is more about language than physics. The basic physical phenomenon - the phase change from liquid to gas when the vapor pressure equals the hydrostatic pressure in the fluid - is the same for boiling and cavitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In common (non-scientific) usage, &quot;boiling&quot; means &lt;em&gt;heating&lt;/em&gt; the liquid until its vapor pressure is equal to the internal pressure of the fluid. In most &quot;non-scientific&quot; cases, the heating is done at (approximately) constant pressure with an interface between the liquid and a gas (e.g. water and air), and the vaporized liquid (steam) leaves the liquid and mixes with the gas, transferring heat from the liquid to the gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand &quot;cavitation&quot; is a local reduction of the &lt;em&gt;pressure&lt;/em&gt; in the liquid, at (approximately) constant temperature. As with boiling, some of the liquid vaporizes when the liquid pressure equals the vapor pressure, but the vapor can't escape anywhere because the surrounding liquid is at a higher pressure. If a bubble of vapor starts to move through the fluid, it soon reaches a point where the fluid pressure is higher, and it collapses back into a liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sudden pressure waves in the liquid, which are created when the bubbles collapse, can cause damage to metal components like propellors, water turbines, etc.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-10-23T00:51:40.557" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="11965" PostTypeId="2" ParentId="7586" CreationDate="2016-10-23T03:22:04.403" Score="0" Body="&lt;p&gt;small diameter of bar  center of gravity is near to the concrete surface. larger diameter of bar center of gravity is far from the concrete surface.&#xA;small diameter of bar easy to handla then the larger diameter of bar.&#xA;The larger the diameter of the rebar, the larger its bending radius.&lt;/p&gt;&#xA;" OwnerUserId="8485" LastActivityDate="2016-10-23T03:22:04.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11968" PostTypeId="1" CreationDate="2016-10-23T07:33:56.763" Score="0" ViewCount="2173" Body="&lt;p&gt;I want to understand how in process of design of a controller I have to meet requirements using root locus method. I know that for the given damping ratio I need to estimate the arccos of it and everything below that angle is the desired region of my poles and zeros. But how about the settling time when it's said that it should be less than 1 second, how do I find the region (the vertical line) that corresponds to that requirement?&lt;/p&gt;&#xA;" OwnerUserId="8490" LastActivityDate="2017-02-21T16:49:55.877" Title="How do I define settling time vertical line on root locus?" Tags="|control-theory|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11969" PostTypeId="2" ParentId="11909" CreationDate="2016-10-23T08:37:09.470" Score="2" Body="&lt;p&gt;Well here goes.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given that there is no heat transfer to the surroundings, indicated by the neglection of heat to the machine, this means that we can assume the system to be adiabatic - a fancy term for no heat exchange. This is important because it makes the equations that we can use much simpler.&lt;/li&gt;&#xA;&lt;li&gt;We want to calculate power, which is work over time. $$P = \frac{W}{t}$$&lt;/li&gt;&#xA;&lt;li&gt;Work done in a thermodynamics setting is the integral of pressure multiplied by the change in volume$$W = \int_a^b PdV$$&lt;/li&gt;&#xA;&lt;li&gt;Since our system is adiabatic, we can use the equation $$PV^{\gamma} = K (constant)$$ which is derived &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiabc.html#c1&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Where gamma is derived from the constant pressure and constant volume specific heats for the fluid which can be looked up $$\gamma = \frac{C_p}{C_v}$$&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This gives us $$W = K\int_{Vi}^{Vf}\frac{dV}{V^{\gamma}}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Integrating gives $$W = \frac{K(V_f^{1-{\gamma}}-V_i^{1-{\gamma}})}{1-{\gamma}}$$ &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Substituting K with Pressure and volume as before gives $$W = \frac{PV^{\gamma}*(V_f^{1-{\gamma}}-V_i^{1-{\gamma}})}{1-{\gamma}}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rearranging the equation gives $$W = \frac{PV*V^{\gamma-1}*(V_f^{1-{\gamma}}-V_i^{1-{\gamma}})}{1-{\gamma}}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since volume equals mass times specific volume, which can be obtained from the fluid tables based on your temperature and pressure, $$V = m\nu $$ we can insert this into the previous equation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Which gives $$W = \frac{PV^{\gamma}*(m\nu_f^{1-{\gamma}}-m\nu_i^{1-{\gamma}})}{1-{\gamma}}$$ The values of $\nu_i$ and $\nu_f$ can be found from the property tables of your fluid in question (air) based on your initial and final conditions for pressure and temperature&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;All of a sudden, we can replace m with $\dot{m}$ which is mass / time and after pulling it out of the brackets we get $$\frac{\dot{m}PV^{\gamma}*(\nu_f^{1-{\gamma}}-\nu_i^{1-{\gamma}})}{1-{\gamma}}$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Which is the same as $$\frac{W}{t} = P$$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And there is your answer.  Much of this comes from the help of &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html&quot; rel=&quot;nofollow&quot;&gt;hyperphysics&lt;/a&gt;, as I'm a bit rusty from my thermodynamics courses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="8487" LastEditorUserId="8487" LastEditDate="2016-10-24T06:03:46.740" LastActivityDate="2016-10-24T06:03:46.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11970" PostTypeId="1" CreationDate="2016-10-23T14:41:00.580" Score="0" ViewCount="228" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AaJ7R.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AaJ7R.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I did is that I draw a Gaussian contour with radius p&gt;a. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric flux outside the cylinder is Ps/2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric flux inside the cylinder is PL/(2*pi*p)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I add them it is not the same answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is wrong , can someone show me the step of solving this problem &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you &lt;/p&gt;&#xA;" OwnerUserId="8494" LastActivityDate="2017-04-23T07:21:05.567" Title="Electric flux density outside cylindrical surface?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11972" PostTypeId="1" CreationDate="2016-10-23T17:28:58.750" Score="1" ViewCount="1885" Body="&lt;p&gt;I wanted to calculate the required thickness of a circular plate of radius 0.06m made of A-36 steel. The plate is supported by a rod at its center welded together with the circular weld of radius 6mm. A uniform load of 981N is applied on the top of the plat. I wanted to calculate the minimum thickness of the plate so that the plate may not bend. Any help would be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="8496" LastEditorUserId="8055" LastEditDate="2016-10-24T14:22:21.260" LastActivityDate="2016-10-24T19:03:18.220" Title="How to calculate the thickness of a circular plate centrally supported?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2016-10-25T21:37:07.067" ContentLicense="CC BY-SA 3.0" />
  <row Id="11974" PostTypeId="1" CreationDate="2016-10-23T20:19:13.647" Score="2" ViewCount="2077" Body="&lt;p&gt;Which type of support should I use when trying to simulate stress in structure that is lifted. Lifting is done by using four lifting lugs and chain angle is 45 degrees. What type of support to use on this surface:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nnJcR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nnJcR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lifting chains (ropes) are not modeled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structure is loaded with force of 20 000 [N] acting on plate shown on picture. Lug will be welded to plate.&lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="5855" LastEditDate="2016-10-24T15:06:48.807" LastActivityDate="2016-10-24T18:50:45.807" Title="ANSYS Workbench - Support type when lifting" Tags="|mechanical-engineering|structural-engineering|structural-analysis|finite-element-method|ansys|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11975" PostTypeId="2" ParentId="11962" CreationDate="2016-10-23T22:13:24.550" Score="7" Body="&lt;p&gt;The short answer: &lt;strong&gt;Yes, a longer gun barrel improves accuracy.&lt;/strong&gt;  Theoretical accuracy of handguns and rifles is driven by rifling, barrel length, and bullet mass. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Rifling: When the bullet travels through the barrel, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rifling&quot; rel=&quot;nofollow noreferrer&quot;&gt;rifling&lt;/a&gt; (helical grooves in the barrel) forces the bullet to spin by converting some of its forward kinetic energy to angular momentum.  This gives the bullet gyroscopic stability.  Remember, a gun without rifling is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Musket&quot; rel=&quot;nofollow noreferrer&quot;&gt;musket&lt;/a&gt; - without spin, the bullet trajectory is unstable, like a &lt;a href=&quot;http://baseball.physics.illinois.edu/knuckleball.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;'knuckle ball'&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Barrel length: A longer barrel extends the time time interval where chamber pressure acts on the bullet's mass.  Therefore, a longer barrel increases the exit velocity of the bullet and the effective range of the bullet.  Friction limits the allowable barrel length.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bullet mass:  Increased bullet mass increases the bullets inertia (resistance to external forces).  Therefore, the bullets trajectory is less effected by external forces such as cross-winds or contact with obstacles (like brush, etc.).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that these factors are interrelated.  Rifling is a function of barrel length (Twist per Length).  So a longer barrel (rifle) imparts more spin to the bullet than a short barrel (handgun).  But exit velocity also depends on bullet mass and barrel length.  The bullets exit velocity and stability (its effective accuracy) depends on rifling, barrel length, and bullet mass.  (Common calibers and their respective rates of twist are listed &lt;a href=&quot;http://www.chuckhawks.com/rifling_twist_rate.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Standard shotguns do not have rifling.  Their accuracy is more relative because they shoot multiple projectiles.  Longer barrels create a more focused scatter shot than shorter barrels.  The focus of the scatter is commonly modified with &lt;a href=&quot;http://www.rem870.com/2012/05/06/shotgun-chokes-explained-cylinder-improved-cylinder-modified-full/&quot; rel=&quot;nofollow noreferrer&quot;&gt;chokes&lt;/a&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dfyf7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dfyf7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A few practical considerations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Sight radius: For a gun without an optical scope, a longer sight base increases accuracy.  Misalignment errors by the shooter are reduced from trigonometry between the front and rear sights.  However, long distance accuracy demands minute angle corrections only possible with optical scopes - the length of sight base is irrelevant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat strain: Gun barrels often get very hot from high energy rounds and high rates of fire.  As a result of temperature differences, the barrel deforms/bends - therefore, the accuracy changes from the first shot to the last.  To minimize heat strain and deflections, &lt;a href=&quot;https://en.wikipedia.org/wiki/Free-floating_barrel&quot; rel=&quot;nofollow noreferrer&quot;&gt;free floating&lt;/a&gt; barrels, and/or larger diameter (or &lt;a href=&quot;http://www.snipercountry.com/articles/realbenefitsbarrelfluting.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;fluted&lt;/a&gt;) barrels are used to maintain accuracy.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ultimately, there are advantages to each type.  Hand guns excel at close ranges where maneuverability and rate of fire is critical.  Shotguns dont require absolute accuracy to hit the target (an advantage when shooting at moving targets like birds) but have a very limited range.  Rifles offer the best accuracy and range but reduced manueverability and often reduced rates of fire (depending on the action: Bolt vs Semi-Auto vs Full-Auto).&lt;/p&gt;&#xA;" OwnerUserId="2538" LastEditorUserId="-1" LastEditDate="2020-06-12T14:04:37.610" LastActivityDate="2020-06-12T14:04:37.610" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="11976" PostTypeId="1" AcceptedAnswerId="12023" CreationDate="2016-10-23T22:21:47.787" Score="0" ViewCount="792" Body="&lt;p&gt;I'd like to understand the new SRI &quot;Abacus&quot; transmission, recently reported at&#xA;&lt;a href=&quot;http://spectrum.ieee.org/automaton/robotics/robotics-hardware/sri-demonstrates-abacus-rotary-transmission&quot; rel=&quot;nofollow&quot;&gt;http://spectrum.ieee.org/automaton/robotics/robotics-hardware/sri-demonstrates-abacus-rotary-transmission&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to find any information on it that doesn't seem derivative of the IEEE story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious about how one determines the reduction ratio, and what the &quot;grooves&quot; look like mathematically (for example, the function from angle to width).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Judging from the pictures, this one seems to have &#xA;    6 setscrews on the inner race, &#xA;    7 &quot;beads&quot;, and&#xA;    8 setscrews on the outer race.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to tell the number of cycles for the grooves, but the ones in the outer race seem to be aligned with the setscrews.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They say the beads can have any profile you like; it seems that must then determine the groove profiles, but how would you get from one to the other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to reason out the relationship between number of groove cycles on each race, diameter profile of the beads, offset of the driveshaft's attachment to the inner cam, and the resulting reduction ratio. But my intuition is failing even after watching the video many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain the relationships involved here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!!!&lt;/p&gt;&#xA;" OwnerUserId="8497" LastActivityDate="2016-10-26T03:30:19.487" Title="How does the SRI &quot;Abacus&quot; rotary transmission work?" Tags="|mechanical-engineering|gears|bearings|mechanisms|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11977" PostTypeId="1" CreationDate="2016-10-23T22:27:58.857" Score="-3" ViewCount="1202" Body="&lt;p&gt;How to convert 10 VDC to VAC at a high frequency, say 10 MHz? VACpp can be +-10V or +-5V.&lt;/p&gt;&#xA;" OwnerUserId="8498" LastEditorUserId="7083" LastEditDate="2016-10-25T08:34:52.303" LastActivityDate="2016-10-25T08:34:52.303" Title="How to convert VDC to VAC at a high frequency?" Tags="|electrical-engineering|ac|" AnswerCount="1" CommentCount="3" ClosedDate="2016-10-26T16:10:56.817" ContentLicense="CC BY-SA 3.0" />
  <row Id="11978" PostTypeId="2" ParentId="11597" CreationDate="2016-10-23T22:53:07.467" Score="1" Body="&lt;p&gt;Yes you are correct in that the book assumes the initial start position to be either open or stationery after being stopped during previous opening and it would require two presses to get it off the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both your answer and the book forget to make a loop in state 6, and also forget the 5 minute timer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your solution does not account for initial starting position other that either closed or open, ie, if there was a power failure during closing/opening, the door will not move again unless you press the button whilst also pressing a top/bottom limit switch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are going to describe or create something with a flow- or state chart which is safe and fully/consistently functional according to specification you have to account for all possible positions/states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was a pain to go to that external site for reference. Please paste pics in question.&lt;/p&gt;&#xA;" OwnerUserId="8499" LastActivityDate="2016-10-23T22:53:07.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11979" PostTypeId="1" AcceptedAnswerId="12078" CreationDate="2016-10-24T00:45:02.897" Score="3" ViewCount="2325" Body="&lt;p&gt;I'm stuck with how to work out the total pressure and centre of pressure in this example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At an outdoor boundary between two properties there is a stepped change in ground level of the clay soil. The ground abruptly changes level by H (roughly 0.9 - 1.2m if it matters). A vertical retaining wall is present, whose design is not relevant here. The soil's depth is much greater than the wall height so there are no relevant local boundaries or property changes in the soil below the retaining wall (however deeply it is embedded in the soil). There are no buildings or other sources of pressure on the soil other than its own weight and soil mechanics, and hydrostatic effects of any water it contains. There is considerable rainfall at times, so the soil will periodically become saturated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm ignoring for now, any other issues, such as bearing, shear and other failure modes, but for safety I should probably assume active rather than passive pressure, I think. I want to find the maximum design lateral pressure such a wall must resist in the presence or absence of built-up hydrostatic pressure, and the height above the lower ground level at which that pressure should be taken as being exerted for calculation purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically I think what I'm trying to find is the maximum design lateral pressure exerted by the soil and any ground-water on the wall, and the point at which that pressure is exerted for calculation purposes (a) if there are ample weepholes/drainage, (b) if there are no weepholes/drainage. I'm expecting the answer to depend on the characteristics of the specific clay to an extent, but I can't work out which characteristics are critical and how much they affect the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt; Sketch as requested:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZSc0P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZSc0P.png&quot; alt=&quot;Sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2016-10-26T05:47:26.450" LastActivityDate="2016-10-30T13:25:12.227" Title="Total lateral pressure from clay soil when hydrostatic pressure is also present" Tags="|structural-engineering|civil-engineering|soil|retaining-wall|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11980" PostTypeId="1" AcceptedAnswerId="11981" CreationDate="2016-10-24T04:09:06.450" Score="1" ViewCount="215" Body="&lt;p&gt;I'd like to know how do the authors of Mechanical Engineering books(books of engineering dynamics, mechanics of materials, fluid mechanics...) and some university professors of Mechanical Engineering formulate those beautiful problems and exercises ? Especially those involving real situations. That is, where do they get those problem situations ? Almost always involving a real mechanical system, thermal system, fluid system...&lt;/p&gt;&#xA;" OwnerUserId="7129" LastActivityDate="2016-10-24T08:27:13.060" Title="Formulation of problems and exercises in Mechanical Engineering" Tags="|mechanical-engineering|education|" AnswerCount="1" CommentCount="0" ClosedDate="2016-10-26T10:04:56.397" ContentLicense="CC BY-SA 3.0" />
  <row Id="11981" PostTypeId="2" ParentId="11980" CreationDate="2016-10-24T05:34:17.743" Score="1" Body="&lt;p&gt;They look them up in books, articles, standards, catalogues or they observe their surroundings. Mechanical engineering has been around for a while so many of the problems are known; Looking up a existing solution is quite feasible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example back when i was teaching mechanism design, many of my exercises were from things that one could be observed at the premises of the building itself. The benefit of this was that students could actually go and observe for themselves. Door closers, cupboard hinges even the cafeteria washing machine hood were all pretty good sources of exercises.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Catalogs are also great as they usually include drawings for the things. Ikea for example has extremely well done vector isometric pictures of everything they sell. Same applies to industrial vendors such as robot manufacturers and heavy construction gear producers. Who often even give out full 3D models of their outer geometry for design purposes (which allows you to repose and build your own highly accurate images). Once you have an idea for a source and some vendor names you can usually find something relevant pretty quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zY5k2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zY5k2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: A image of Kuka KR6 SIXX and a overall instructions on how image was made plus model download can be found &lt;a href=&quot;https://graphicdesign.stackexchange.com/questions/59032/from-a-cad-model-to-an-illustration-drawing-how-to/59071#59071&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Articles are great as they can include all the relevant math needed, although they tend to be very niche stuff and may or may not have good images. A usually better source is some of the more comprehensive mechanics books that exist (like &lt;a href=&quot;https://accessengineeringlibrary.com/browse/mechanisms-and-mechanical-devices-sourcebook-fifth-edition#Preface01&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;). These books are wast catalogs of interesting mechanisms and have information about relevant design problems related to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, drawing images with CAD/hand is, or at least used to be part of the education of mechanical engineers. So taking what you have seen and sketching it up should not be a overwhelming task. But then I am also an illustrator so i tend to draw stuff just to keep my skills up.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="-1" LastEditDate="2017-04-13T12:46:19.207" LastActivityDate="2016-10-24T08:27:13.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11982" PostTypeId="1" CreationDate="2016-10-24T07:07:54.877" Score="0" ViewCount="70" Body="&lt;p&gt;I try to simulate flow through a manifold with multiple inlets &amp;amp; outlets but I'm facing convergence issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that it may be due to poor quality mesh, but my mesh statistics are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Minimum Orthogonal Quality =  0.115844&lt;/li&gt;&#xA;&lt;li&gt;Maximum Ortho Skew =  0.853992&lt;/li&gt;&#xA;&lt;li&gt;Maximum Aspect Ratio =  31.8148e&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which doesn't look that bad to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point could be due to wrong BCs and I'm not sure wether my BCs are good. For the inlets I calculated mass flow in the domain (since the manifold is a physical part from a measurement system, I know roughly what the flow field looks like and what the flow in and out my domain is) and set it as &lt;strong&gt;mass-flow-inlet&lt;/strong&gt; BC. For the outlets I also calculated mass flow and set it as &lt;strong&gt;pressure-outlet&lt;/strong&gt; with a good guess for the initial gauge pressure and a &lt;strong&gt;target-mass-flow&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as I said, I don't get a converged solution, is it due to these BCs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attach some residual plots.&#xA;(I wanted to add more but I can only post 2 links...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more information let me know, I tried to keep it short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iKgcv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iKgcv.png&quot; alt=&quot;residuals&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/V2dnb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V2dnb.png&quot; alt=&quot;mass-flow outlet-1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8282" LastActivityDate="2016-10-24T07:07:54.877" Title="Flow through manifold: convergence issues" Tags="|cfd|fluent|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11984" PostTypeId="1" CreationDate="2016-10-24T11:09:18.703" Score="1" ViewCount="383" Body="&lt;p&gt;I am an M.Tech student from Manipal Institute of Technology, Manipal. I want to deduce the mathematical relationship between Forward Voltage and Forward Current of an LED from the characteristics given in the datasheet. How to derive equation from the graph?&lt;a href=&quot;https://i.stack.imgur.com/GLeWE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GLeWE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8505" LastActivityDate="2016-10-24T12:51:03.647" Title="Derive equation from graph" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11985" PostTypeId="2" ParentId="11984" CreationDate="2016-10-24T11:28:17.290" Score="0" Body="&lt;p&gt;The term you would be requesting for is for a curve fit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of ways to do this. A linear, or non-linear one. &#xA;In this particular case you would berequire a non-linear. &#xA;or specifically a quadratic one. &#xA;Typically, you choose the model order by the number of bends you need in your line. Each increase in the exponent produces one more bend in the curved fitted line. It’s very rare to use more than a cubic term. &lt;a href=&quot;https://i.stack.imgur.com/a7oB4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would require a sample space. Which means x and y positions for the graph. You can use excel/minitab/spss/R/python/matlab to get your final curve out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quadratic linear model can be done by going to the option Stat &gt; Fitted Line Plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are more examples in the link I shared&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a7oB4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a7oB4.png&quot; alt=&quot;Example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a7oB4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; &lt;a href=&quot;http://blog.minitab.com/blog/adventures-in-statistics/curve-fitting-with-linear-and-nonlinear-regression&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://blog.minitab.com/blog/adventures-in-statistics/curve-fitting-with-linear-and-nonlinear-regression&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6215" LastActivityDate="2016-10-24T11:28:17.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11986" PostTypeId="2" ParentId="11984" CreationDate="2016-10-24T11:31:44.533" Score="0" Body="&lt;p&gt;There are models already out there for the current thru a diode as a function of its forward voltage.  Go find one, then tweak its constants to match the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a case where a polynomial approximation is not going to work well unless you use a reasonably high order.  Note the logarithmic scale for current in the graph.  Diode current as a function of voltage is predominantly a exponential function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the graph carefully and you can see that for currents up to 100 mA or so, the graph is linear, meaning the exponential term dominates to the point you can ignore the others.  It should be easy to write the exponential for that range of current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At higher currents, other effect become more relevant, so the equation gets more complicated.  Again, there are models out there that take all sorts of details into account.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-10-24T11:31:44.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11987" PostTypeId="1" CreationDate="2016-10-24T11:33:21.027" Score="0" ViewCount="89" Body="&lt;p&gt;I'm not exactly sure where this question would go, so I've decided to post it here. Hopefully you can help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a smaller computing device (i.e. an Arduino or Raspberry Pi) and other outside tools, would it be possible to construct a magnetic stripe device with an extremely low coercivity so that its code could be changed on the fly by a computer and an encoding device? If this is possible, what devices would be best for this kind of project?&lt;/p&gt;&#xA;" OwnerUserId="8506" LastEditorUserId="1832" LastEditDate="2016-10-24T13:17:43.760" LastActivityDate="2017-03-23T16:41:57.243" Title="Magnetic Stripe Encoding on the fly?" Tags="|design|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11988" PostTypeId="2" ParentId="11987" CreationDate="2016-10-24T11:41:35.120" Score="1" Body="&lt;p&gt;Go to the dusty archives and find some stone tablets that mention &lt;i&gt;magnetic tape&lt;/i&gt;.  That's how analog and then later even digital data was stored back in the Pleistocene, before steel tools, indoor plumbing, flash memory, and DVDs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need something called a &lt;i&gt;tape head&lt;/i&gt; to read and write data from/to a magnetic strip.  This is a coil wrapped around magnetic material with the ends brought together in a very small gap.  The gap is swept along the strip.  As the magnetic orientation stored on the strip changes, a small voltage is induced in the coil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process also works in reverse.  You drive the coil as the gap is swept along the strip to set the magnetic orientation of the small piece of the strip that is against the gap.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-10-24T11:41:35.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="11989" PostTypeId="2" ParentId="11977" CreationDate="2016-10-24T12:33:29.700" Score="1" Body="&lt;p&gt;You want something called a &lt;i&gt;oscillator&lt;/i&gt;.  It is powered from DC, and produces AC output, usually a sine or square wave.  For the case of 10 MHz, you can use a resonant design since the components can be physically small.  RF transmitters are examples of circuits that take DC power in and AC at a specific frequency at substantial power.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-10-24T12:33:29.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11990" PostTypeId="2" ParentId="11984" CreationDate="2016-10-24T12:51:03.647" Score="2" Body="&lt;p&gt;I may be misunderstanding the main thrust of your question, but my interpretation is that you want to extract data from the graph.  My favorite tool is &lt;a href=&quot;http://datathief.org/&quot; rel=&quot;nofollow&quot;&gt;DataThief&lt;/a&gt;. It's very easy to use and quite robust.  You load the graph image and DataThief generates an array of ordered points (x,y).  At that point you can use whatever nonlinear curve fitting tools you desire, e.g. polynomial, log, exponential, and so on. &#xA;My choice there is the &lt;code&gt;R&lt;/code&gt; language and perhaps the tool &lt;code&gt;nls&lt;/code&gt; for nonlinear fitting, but there are similar tools in any decent analysis application.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-10-24T12:51:03.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11991" PostTypeId="1" CreationDate="2016-10-24T13:18:03.767" Score="0" ViewCount="41" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B1qfF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B1qfF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;To simplify the problem, if we just imagine a rectangle resting vertically, and on the bottom face (the one closer to the 'ground') there are slits on that face. On the opposite face to that (the one closer to the 'sky'), there are also slits. In this case I am assuming no q in or out through the other 4 faces. The reason I put 'ground'  in quotes, is because I am imagining the rectangle is not actually resting on the ground, but is instead raised off the ground through some mechanism that doesn't affect heat transfer or fluid flow of the air. Also assume the are is some T1 below the bottom face, and T2 above the top face. T2&gt;T1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how I would go about calculating the temperature gradient within the house? The aim is to see if this is a viable way of cooling. To be clear, I am not asking for someone to work this out in detail, but rather an approach to the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spoke with a professor and he suggested using density differences to calculate the fluid flow, but I do not know how to get a temperature gradient from that information. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="5469" LastEditorUserId="5469" LastEditDate="2016-10-24T18:40:42.417" LastActivityDate="2016-10-24T18:40:42.417" Title="What method can I use to calculate the temperature within a house being cooled by vents on the floor?" Tags="|heat-transfer|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="11992" PostTypeId="1" AcceptedAnswerId="12036" CreationDate="2016-10-24T13:04:02.103" Score="1" ViewCount="1419" Body="&lt;p&gt;I have a Li-Po battery charger that comes with its own cooling fan which, like most of the cooling fans, blows hot air out of the charger to cool the charger down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a spare fan and I was wondering whether it should blow ambient air towards the charger or draw the hot air away from it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be an external fan outside the casing of the charger. Charger itself is roughly the size of a normal Wi-Fi Adapter and the auxiliary fan has ben taken from a computer casing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it does not really make much of a difference but was just curious as to which setup will perform better in order to dissipate the heat generated a little better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supplementary question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I had 2 extra fans would it be better if one was blowing air towards the charger and other was drawing away or both should be doing the same thing?&lt;/p&gt;&#xA;" OwnerUserId="5607" OwnerDisplayName="Hanky Panky" LastActivityDate="2016-10-26T20:25:20.673" Title="Should an auxiliary cooling fan blow air or draw it?" Tags="|thermodynamics|cooling|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="11993" PostTypeId="2" ParentId="11992" CreationDate="2016-10-24T13:09:28.580" Score="3" Body="&lt;p&gt;I think it would depend on the enclosure. In an open environment, it would be better to blow cool air at the target. For closed cases, it seems more common to blow hot air out if there are not fans in both directions.&lt;/p&gt;&#xA;" OwnerUserId="8514" OwnerDisplayName="EL_DON" LastActivityDate="2016-10-24T13:09:28.580" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="11994" PostTypeId="2" ParentId="11968" CreationDate="2016-10-24T14:21:48.040" Score="1" Body="&lt;p&gt;The settling time $T_s$ is related to the damping ratio $\zeta$ and the natural frequency $\omega$ by $$ T_s = \frac{4}{\zeta \omega}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $T_s&amp;lt;1$ this means that $\zeta \omega &amp;gt;4$. To plot it on the complex plane we need to first solve for $\zeta$ and $\omega$ in terms of $x$ and $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relationship is  $$ x=-\zeta  \omega,y=\omega \sqrt{1-\zeta ^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we solve for $\zeta$ and $\omega$, we need to find the region where $\zeta \omega &amp;gt;4$. These calculations are tedious, so I used Mathematica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VOv3C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VOv3C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2016-10-24T14:34:30.787" LastActivityDate="2016-10-24T14:34:30.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11995" PostTypeId="2" ParentId="5763" CreationDate="2016-10-24T15:01:26.147" Score="0" Body="&lt;p&gt;Actually, scipy has a function that does exactly this. I used it on each of the 70 arrays that make up the time series, and points generate a smooth line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://docs.scipy.org/doc/scipy-0.16.1/reference/generated/scipy.ndimage.measurements.center_of_mass.html&quot; rel=&quot;nofollow&quot;&gt;https://docs.scipy.org/doc/scipy-0.16.1/reference/generated/scipy.ndimage.measurements.center_of_mass.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3343" LastActivityDate="2016-10-24T15:01:26.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11996" PostTypeId="1" CreationDate="2016-10-24T16:30:01.270" Score="1" ViewCount="111" Body="&lt;p&gt;Are there standardized ways of dealing with precipitation in chemical engineering continuous flow processes? I have an application in which I need to cool a fluid via a heat exchanger and the fluid will precipitate some solid on cooling. The two specific issues I predict are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I need to select a type of heat exchanger that won't clog from the precipitate. Probably the best cases are something where the precipitate can just fall out by gravity or it is carried out by the fluid. What types of heat exchangers are good for this? Single pass shell-and-tube?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I need an automated way to remove the precipitate from the fluid. What pieces of equipment are used for this? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2016-10-27T20:45:39.247" Title="Managing precipitation in cooling process" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="11998" PostTypeId="2" ParentId="11974" CreationDate="2016-10-24T18:44:47.337" Score="1" Body="&lt;p&gt;I would put a pin through the hole with the proper clearances. The include a contact condition with friction between the pin and the hole. Otherwise the stresses on the lifter will be &lt;em&gt;way&lt;/em&gt; off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then apply the load on the ends of the pin. If you don't care for stresses on the pin, make it rigid so the bending does not cause edge stress on the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to simulate lifting you need to include standard gravity and constrain the ends of the pin instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't want to constrain the lugs because then you are over-constraining the structure as it won't be allowed to flex properly. As a last resort you might try a remote displacement in the center of the holes with deformation on. If the lifting is done with a bolt, then a coupled condition on the remote displacement would do the trick also.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastEditorUserId="1051" LastEditDate="2016-10-24T18:50:45.807" LastActivityDate="2016-10-24T18:50:45.807" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="11999" PostTypeId="2" ParentId="11972" CreationDate="2016-10-24T19:03:18.220" Score="1" Body="&lt;p&gt;The governing equation for a circular ring is&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{{\rm d}}{{\rm d}r}\left[\frac{1}{r}\frac{{\rm d}}{{\rm d}r}\left(r\frac{{\rm d}y}{{\rm d}r}\right)\right]=\mbox{-}\frac{Q(r)}{D}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;sub&gt;where &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; is the axial deflection, &lt;span class=&quot;math-container&quot;&gt;$D=\frac{E\,t^{3}}{12(1-\nu^{2})}$&lt;/span&gt; is a material constant with thickness &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$Q(r)$&lt;/span&gt; is the shear force rate at a particular radius. It is defined as &lt;span class=&quot;math-container&quot;&gt;$Q(r)=\frac{\mbox{load inside radius } {r}}{\mbox{perimeter at }{r}}$&lt;/span&gt; &lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;For a uniform pressure &lt;span class=&quot;math-container&quot;&gt;$q = \frac{F}{\pi R^2}$&lt;/span&gt; on the top surface the shear rate is thus &lt;span class=&quot;math-container&quot;&gt;$Q = \frac{F \;r}{2 \pi R^2}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If only the center is fixed, then the deflection curve is solved by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ y(r) = \frac{3F\,r^{2}(1-\nu)}{16\:\pi R^{2}\,E\,t^{3}}\left(2R^{2}(\nu+3)-r^{2}(\nu+1)\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Finally the deflection on the outer edge is &lt;span class=&quot;math-container&quot;&gt;$$\boxed{ \delta=\frac{3F\,R^{2}(1-\nu)(\nu+5)}{16\,\pi\,E\,t^{3}} }$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1051" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-10-24T19:03:18.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12001" PostTypeId="2" ParentId="11963" CreationDate="2016-10-25T00:11:36.807" Score="1" Body="&lt;p&gt;I'm a little confused, if you ask in relation about pump cavitation then my ans : cavitation emphasized on reverse boiling (when vapor back to liquid, the collapsing bubble). That's when destructive action done, surface impingement.&#xA;Thats why impeller parts get damaged is near exit.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2016-10-25T00:11:36.807" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12003" PostTypeId="2" ParentId="7586" CreationDate="2016-10-25T01:00:38.600" Score="-2" Body="&lt;p&gt;I think that 4 bars D1 will better than 1 bar D2 (same tensile area, but more bonding surface).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that, bigger bars are more brittle while smaller bars have dominant surface yield (more elastic and plastic than brittle). It adds safety when failure happens.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2016-10-25T01:00:38.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12005" PostTypeId="2" ParentId="11970" CreationDate="2016-10-25T04:03:04.130" Score="1" Body="&lt;p&gt;Consider a length $L$ of the cylindrical surface and the line. Draw a cylindrical contour with radius $\rho&amp;gt;a$ and length $L$, centered about the charged cylinder and line charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charge contained within the contour is equal to the linear charge density of the line of charge times its length, plus the surface charge density of the cylindrical surface times its surface area, that is:&#xA;$$Q = (\rho_L)(L)+(\rho_S)(A)$$&#xA;The surface area of the charged cylinder is the circumference of the cylinder times its length: $A=2\pi a L$. Thus, the charge contained within the Gaussian contour is:&#xA;$$Q = (\rho_L)(L)+(\rho_S)(2\pi a L)$$&#xA;The flux density is equal to the flux passing through the Gaussian contour divided by the area of the contour:&#xA;$$|\vec{D}|=\frac{Q}{A}=\frac{(\rho_L)(L)+(\rho_S)(2\pi a L)}{2\pi\rho L}$$&#xA;Note that the length $L$ can be canceled from each term. Due to the symmetry of the problem, the electric flux acts in the radial direction:&#xA;$$\vec{D}=\frac{\rho_L+2\pi a \rho_S}{2\pi\rho}\vec{a}_\rho$$&lt;/p&gt;&#xA;" OwnerUserId="8518" LastActivityDate="2016-10-25T04:03:04.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12006" PostTypeId="2" ParentId="11992" CreationDate="2016-10-25T05:27:01.693" Score="3" Body="&lt;p&gt;There is a big difference in the resulting flow during blowing and suction. Blowing creates a reasonably well directed jet of air, while suction pulls air from all around. Therefore blowing is better if you want to cool a small region or object (which is your case), because you can direct the flow there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of two fans, I am not sure what the best configuration would be. Again since the object to be cooled is small and in the open, it makes sense to blow on them. Perhaps keeping the fans side by side and blowing them in the same direction is a good idea, but I can't be sure.&lt;/p&gt;&#xA;" OwnerUserId="8519" LastActivityDate="2016-10-25T05:27:01.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12020" PostTypeId="1" CreationDate="2016-10-25T21:43:26.430" Score="0" ViewCount="88" Body="&lt;p&gt;So, there is this teacher who wants to test how an chemical compound acts in the velocity of drying in concrete. Basically, i want to measure continuously (it would be great if i could do it with an arduino) the humidity level in a block of concrete.&#xA;What is the best option? To use a air humidity sensor (DHT22, for example) or an ground humidity sensor? Or any other better option? &#xA;The main purpose here is to store every measure of humidity along approximately 12 ou 13 hours. Any ideas? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="8537" LastActivityDate="2016-10-26T21:33:10.663" Title="Is there any good way to measure humidity constantly in a concrete block?" Tags="|civil-engineering|control-engineering|sensors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12021" PostTypeId="1" AcceptedAnswerId="12030" CreationDate="2016-10-25T22:05:20.053" Score="1" ViewCount="100" Body="&lt;p&gt;A certain truck has 5 (per side) *2 sides= 10 bolts holding the body to the frame. The bolts are marked 10.9 and the area of the shank is 153 sq mm. From my math I got 3e8 lbs for all 10 bolts together.&lt;/p&gt;&#xA;&lt;p&gt;From a mechanical engineering point of view, how would you describe the mechanical characteristics of these bolts?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&lt;p&gt;Thank you for the discussion. I was dividing by the bolt section area; I should have been multiplying. Now the answers make much more sense.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Each bolt has a tensile strength of about 32,000 lbs * 10 bolts = 320k lbs.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Of course the more interesting tests in a vehicle application are the shear loads you would encounter in a collision, but that is also a MUCH more complicated question.&lt;/p&gt;&#xA;" OwnerUserId="8538" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-10-26T20:58:23.707" Title="How much do these bolts hold? Marked 10.9, with shank area of 153 sq mm" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12023" PostTypeId="2" ParentId="11976" CreationDate="2016-10-26T03:30:19.487" Score="3" Body="&lt;p&gt;It's just a fancy rolling cycloidal drive (not that that being simple makes it any less cool). It really bothers me that all the available articles only compare it to harmonic drives when it only bears a passing resemblance to them. Probably because it's owned by Harmonic Drive LLC. The bearing races can be abstracted as wire ring races that have been bent sinusoidally. The reason the metal one has a conical bearing is so that the races are easier to manufacture, since they're just sinusoidal rather than cycloidal, like they'd almost certainly have to be for the round bearing, and good luck milling that. What also bothers me about the video posted about it is that the guy claims it is a 1:3 drive. Well, the formula for the gear ratio of a cycloidal drive is P-L/L where P is the number of outer ring pins, in this case the number of periods of the outer sinusoidal wave, and L is the number of lobes on the cycloidal disk, in this case the number of bearings. (I think you're right about the set screws) (8-7)/7 = 1/7, not 1/3. &lt;/p&gt;&#xA;" OwnerUserId="8541" LastActivityDate="2016-10-26T03:30:19.487" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12025" PostTypeId="1" AcceptedAnswerId="12031" CreationDate="2016-10-26T05:59:40.567" Score="0" ViewCount="102" Body="&lt;p&gt;I have an engineering application where I would like to use a single 24v DC Stepping Motor to power two different axles of opposing direction. Basically when the motor turns clockwise, Axel A will spin and Axel B will be stationary. When the motor turns counter-clockwise, Axel A will be stationary and Axel B will spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need it to be as cheap and simple as possible. So far, the best solution that I could come up with to accomplish this is by using two gears similar to the bicycle Freewheel / Freehub. The problem is, I cannot for the life of me find a gear that is similar to that application, but for uses outside of bicycles (possibly because I don't know that specific gear's name). Can anyone recommend a gear or something that is similar to the application that I need?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a general idea of the function that I need, but I don't know the name of this type of ratcheting gear.&lt;/p&gt;&#xA;" OwnerUserId="8543" LastEditorUserId="8543" LastEditDate="2017-07-31T02:20:22.507" LastActivityDate="2017-07-31T02:20:22.507" Title="Looking for a gear similar to bicycle freewheel" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12026" PostTypeId="2" ParentId="12020" CreationDate="2016-10-26T07:27:37.547" Score="1" Body="&lt;p&gt;If this is a test block (and not a part of a building), the easiest solution is to place it on a scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scale will indicate the amount of water that has evaporated from the block. From this it is easy to calculate the moisture content, either by volume or by mass. There are also scales that output the weight through e.g. serial port, which you can then log.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air or ground humidity sensors would probably give you values that are related to the concrete water content, but they would need multi-point calibration if you need accurate readings. For example, temperature would affect the air moisture readings a lot.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-10-26T07:27:37.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12028" PostTypeId="2" ParentId="12025" CreationDate="2016-10-26T08:40:05.360" Score="1" Body="&lt;p&gt;The type of clutch is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Freewheel&quot; rel=&quot;nofollow&quot;&gt;&quot;freewheel&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though if the forces and speeds aren't too big, you can use a psuedo-planetary gear configuration without the ring gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the sun spins in one direction the planet gear will move along until it hits the output gear for the first shaft and will mesh with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the sun then spins in the other direction the planet will move away from the first shaft and move to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has the downside that it requires meshing the gears as you change directions which may cause extra wear and tear.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2016-10-26T08:40:05.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12029" PostTypeId="2" ParentId="12021" CreationDate="2016-10-26T10:06:36.823" Score="0" Body="&lt;p&gt;See &lt;a href=&quot;https://www.boltdepot.com/fastener-information/materials-and-grades/bolt-grade-chart.aspx&quot; rel=&quot;nofollow&quot;&gt;https://www.boltdepot.com/fastener-information/materials-and-grades/bolt-grade-chart.aspx&lt;/a&gt; for the meanign of the 10.9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence F = stress x area. Excluding safety factors e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are meany characteristic we can discuss, please clarify first.&lt;/p&gt;&#xA;" OwnerUserId="1077" LastActivityDate="2016-10-26T10:06:36.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12030" PostTypeId="2" ParentId="12021" CreationDate="2016-10-26T12:35:29.843" Score="1" Body="&lt;p&gt;The nominal capacity of a high  tensile bolt are defined by its designation eg 10.9 here the first figure (10) is the ultimate tensile stress in hundreds of MegaPascals and the second figure (after the point) is the yield stress as a fraction of the UTS ie 9 represents 9 tenths or 90% &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conveniently if you multiply the yield stress (MPa) by the bolt section (mm) you get an answer for the yield load in tension of the bolt in Newtons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that the static tensile strength of the bolt is only part of the capacity of a bolted joint and you need to consider any loads in other direction as well as the effect of dynamic loading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the preload on a bolt will have a significant effect on how it behaves, not all joints are preloaded but when they are external shear loads are carried by friction between the mating surfaces. It is also possible for bolted joints to fail in ways other than the bolt breaking eg by pulling through the member they are attached to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally the design load for a structure may incorporate a factor of safety which puts the rated safe working load at some fraction of the theoretical maximum and where  you have multiple bolts you can't always assume that they are equally loaded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is further complicated in something like a vehicle chassis in that the members are also carrying other loads as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I would add that if you are working from metric units your answer should be metric as well, convert your final answer &lt;em&gt;as well&lt;/em&gt; if necessary but having the problem and the answer in mixed units makes errors more likely and more difficult to identify. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-26T12:41:24.567" LastActivityDate="2016-10-26T12:41:24.567" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12031" PostTypeId="2" ParentId="12025" CreationDate="2016-10-26T13:12:03.923" Score="0" Body="&lt;p&gt;You can get &lt;a href=&quot;http://www.bocabearings.com/bearing-types/one-way-bearings&quot; rel=&quot;nofollow&quot;&gt;one way bearings&lt;/a&gt; which rotate freely on one direction and lock in the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you use one to attach a gear or pulley to your motor output shaft then the pulley will be coupled to the shaft in one direction and not in the other so with two mounted in opposite orientation each with a pulley driving a separate shaft (A and B)  you should be able to achieve what you want.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-26T13:28:03.103" LastActivityDate="2016-10-26T13:28:03.103" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12032" PostTypeId="1" AcceptedAnswerId="12034" CreationDate="2016-10-26T13:31:09.983" Score="2" ViewCount="453" Body="&lt;p&gt;I attend to engineering design of a new irrigation system, comprising 4 pump stations and 3 lengthy water mains between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definitions by ISO 12006-2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construction Entity is an independent construction result of significant scale serving at least one user activity or function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construction Complex means two or more adjacent construction entities collectively serving one or more user activity or function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, the irrigation system must be identified as a construction complex, comprising 4 pump stations + 3 pipe mains = 7 'construction entities'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I collectively and separately name the mentioned parts of the project? (In my native language those all are 'objects', however, definition for 'construction object' by ISO is something different).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please advise, can I use the term 'entity' in English-written documents (e.g. 'list of newly built entities', 'entity # 1'), or it is better to use such terms as works, structure, facility, unit, site, plant? So that it be understandable for English-speakers in construction environment. (Note that the words used do not need to conform to ISO 12006-2; I have just used it as a reference.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What terms have you really met and used for such cases?&lt;/p&gt;&#xA;" OwnerUserId="8547" LastEditorUserId="255" LastEditDate="2016-10-26T15:39:56.150" LastActivityDate="2016-10-26T15:47:57.700" Title="Construction Entity and Construction Complex" Tags="|civil-engineering|construction-management|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12034" PostTypeId="2" ParentId="12032" CreationDate="2016-10-26T15:41:54.120" Score="3" Body="&lt;p&gt;I am a native English speaker and practising civil engineer who had never heard of ISO 12006-2 until this question, and if you had talked about &quot;Construction Entities&quot; I wouldn't have had a clue what you were on about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider using &quot;4 pump stations and 3 water mains&quot;, as this is clear.      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a word which covers all 7 &quot;construction entities&quot; I could comprehend:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;7 separate &lt;strong&gt;works&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;7 separate &lt;strong&gt;components&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;7 separate &lt;strong&gt;subsystems&lt;/strong&gt; (&lt;em&gt;though this may be more suitable for relatively complicated mains)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't think &quot;structure&quot; would be appropriate for a water main, and your other suggestions sound even less appropriate.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2016-10-26T15:47:57.700" LastActivityDate="2016-10-26T15:47:57.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12036" PostTypeId="2" ParentId="11992" CreationDate="2016-10-26T19:11:42.000" Score="4" Body="&lt;p&gt;The air coming out of the fan will be highly turbulent, wheras the air being sucked into the fan is less so. Hence blowing at the target will increase heat transfer. On the other hand, if you are looking at a casing, a fan blowing into the case will also blow dust into the case. A fan blowing out of the case will typically suck in less dust, as the fluid velocity at the opening of the case will typically be lower than directly next to the fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the choice of two fans in serial (one blowing, one sucking) or parallel (both blowing or sucking) depends on what you are trying to achieve (i.e. the &lt;a href=&quot;http://www.greenheck.com/library/articles/42&quot; rel=&quot;nofollow&quot;&gt;system curve&lt;/a&gt; of the device to be cooled). Two fans in series will increase the total pressure you can build up, and is a good idea if you need to overcome a high pressure loss. Two fans in parallel will increase your flow rate if the pressure loss of your system is relatively low.&lt;/p&gt;&#xA;" OwnerUserId="8552" LastEditorUserId="8552" LastEditDate="2016-10-26T20:25:20.673" LastActivityDate="2016-10-26T20:25:20.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12037" PostTypeId="2" ParentId="11963" CreationDate="2016-10-26T20:03:52.433" Score="4" Body="&lt;p&gt;The short answer is that cavitation and boiling both refer to a phase change from liquid to gas that causes bubbles to form, where cavitation is &lt;em&gt;driven by&lt;/em&gt; a drop in pressure and boiling is &lt;em&gt;driven by&lt;/em&gt; an increase in temperature. For a citation, see &lt;a href=&quot;https://books.google.com/books?hl=en&amp;amp;id=yRhaAQAAQBAJ&amp;amp;oi=fnd&amp;amp;pg=PR11&amp;amp;ots=O6wQyNqbc6&amp;amp;sig=VNf3KwQPtB9OacnAuE7kp4zrtyw#v=snippet&amp;amp;q=distinguishing&amp;amp;f=false&quot; rel=&quot;nofollow&quot;&gt;Cavitation and Bubble Dynamics&lt;/a&gt;, page 1:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A rough but useful way of distinguishing these two processes is to&#xA;  define cavitation as the process of nucleation in a liquid when the&#xA;  pressure falls below the vapor pressure, while boiling is the process&#xA;  of nucleation that ocurs [sic] when the temperature is raised above the&#xA;  saturated vapor/liquid temperature. Of course, from a basic physical&#xA;  point of view, there is little difference between the two processes...&#xA;  The differences in the two processes occur because of the different&#xA;  complicating factors that occur in a cavitating flow on the one hand&#xA;  and in the temperature gradients and wall effects that occur in&#xA;  boiling on the other hand.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want to know &lt;em&gt;how&lt;/em&gt; this distinction can be useful, &lt;a href=&quot;http://resolver.caltech.edu/CaltechBOOK:1995.001&quot; rel=&quot;nofollow&quot;&gt;the full text of an older edition of the book is available via the Caltech library web site.&lt;/a&gt; Finding a more recent edition at the library shouldn't be difficult, considering the work has been cited nearly 3,000 times according to Google Scholar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long answer starts by noting that this quote doesn't pretend to give the &lt;strong&gt;only&lt;/strong&gt; definitions of cavitation and boiling; it explicitly proposes &lt;strong&gt;one way&lt;/strong&gt; of defining them as two processes that is &quot;rough but useful.&quot; I expect Dr. Brennen would agree that there exist contexts in which other definitions are more useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a very general sense, &quot;cavitation&quot; can mean the spontaneous appearance of cavities (also known as voids or bubbles) within a liquid. If you're researching how different materials or surface geometries promote or suppress nucleation, this might be a more useful definition to you than one that excludes heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a more restrictive sense, &quot;cavitation&quot; can mean only that subset of the former that occurs at relatively constant temperature, in the presence of a solid interface, which later implodes and contributes to wear on mechanical components. If you're building a propulsion system for a submarine, this might be a more useful definition than either of the previous two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.etymonline.com/index.php?term=boil&quot; rel=&quot;nofollow&quot;&gt;The word &quot;boiling&quot; pre-dates modern thermodynamics&lt;/a&gt; so we shouldn't be surprised if it's hard to pin down. We typically think of boiling as a process involving bubbles but &lt;a href=&quot;http://cdn.intechopen.com/pdfs-wm/13191.pdf&quot; rel=&quot;nofollow&quot;&gt;film boiling&lt;/a&gt; is an exception—clearly the folks researching what happens when you apply a ton of heat at a solid/liquid interface thought it was useful to put this phenomenon in the same category as nucleate boiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, liquids are also said to &quot;boil&quot; in a vacuum (and &lt;a href=&quot;https://www.youtube.com/watch?v=9q5gEZGoBnk&quot; rel=&quot;nofollow&quot;&gt;here's a video of that&lt;/a&gt;, if you're curious—try to make out where the nucleation occurs!). Do you think the folks at NASA care whether boiling requires heat when they're working to mitigate risks associated with explosive decompression? I don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You gain very little by expecting, or giving the expectation of, objectively correct terminology. If you're doing some technical writing on the subject and intend to distinguish between cavitation and boiling, just make your definitions explicit. Do your due diligence to ensure that your definitions aren't a significant departure from consensus, or else build a very strong argument to support them.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-10-27T16:27:03.563" LastActivityDate="2016-10-27T16:27:03.563" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12038" PostTypeId="2" ParentId="11963" CreationDate="2016-10-26T21:22:31.353" Score="16" Body="&lt;p&gt;Mechanical engineer here, former US Navy nuke. The textbook definition of cavitation is, from my nuclear training:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The formation &lt;strong&gt;and subsequent collapse&lt;/strong&gt; of vapor bubbles as the suction pressure falls below and then rises above saturation pressure.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This definition is referring to the suction pressure as in a pump, but I would say more generally and apparently against most of the other posters here that cavitation refers more to the &lt;strong&gt;formation and subsequent collapse of vapor bubbles&lt;/strong&gt; than it does about how those vapor bubbles occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I understand that the cavitation effect generally happens (or is most discussed when it happens) in pumps and propellers, but it also happens in boiling water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you bring water to a boil, initially it is quiet and there are no bubbles. At some transition point (nucleate boiling), bubbles form on the bottom of the pan, break away, but collapse &lt;em&gt;before they reach the surface&lt;/em&gt;. This kind of boiling (referred to as a simmer in culinary terms) can correctly be referred to as cavitation. This is also a very noisy phase in the boiling process - this is the &quot;noisy&quot; period in OP's video. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After cavitation comes (for cooking, at least) the final boiling phase, in which the bulk fluid boils and bubbles reach the surface of the water (departure from nucleate boiling). Despite boiling appearing to be more vigorous, this is actually much quieter &lt;em&gt;because cavitation is no longer occurring&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cavitation is the pinging sound a pot of water makes before a full boil. Once a full boil is achieved, steam bubbles reach the surface and the quality of the sound changes from a pinging to more of a gurgle. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;All of that said, there has been a lot of talk on other posts about boiling being the application of heat and cavitation being about the reduction of pressure. Again, the reduction of pressure (below saturation pressure) is a &lt;strong&gt;cause&lt;/strong&gt; of cavitation, but reducing pressure is not the &lt;strong&gt;definition&lt;/strong&gt; of cavitation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term for &lt;strong&gt;creating&lt;/strong&gt; vapor bubbles by reducing pressure is called &lt;strong&gt;flash distillation&lt;/strong&gt; or &lt;strong&gt;flash evaporation&lt;/strong&gt;. The term for &lt;strong&gt;creating&lt;/strong&gt; vapor bubbles by increasing heat is called &lt;strong&gt;boiling&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term cavitation refers to the formation &lt;strong&gt;and subsequent collapse&lt;/strong&gt; of the vapor bubbles. Cavitation happens in pumps, in a pot of spaghetti water, in a submarine propeller, etc. It is not restricted to either mode of creation (pressure or heat). The video in OP's post shows cavitation during a boiling process.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;:EDIT:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I felt challenged by &lt;a href=&quot;https://engineering.stackexchange.com/questions/11963/what-fundamentally-distinguishes-cavitation-and-boiling-as-different-phenomena/12038?noredirect=1#comment21330_12038&quot;&gt;Air's comment&lt;/a&gt; to produce a source for the definition of cavitation that I provided here. The line I quoted above is as-memorized from about 15 years ago now. I have (on a bookshelf at home) a condensed technical handout of non-classified information we were given at the conclusion of the nuclear training courses for personal reference. In trying to find this manual online, I found &lt;a href=&quot;http://nuclearpowertraining.tpub.com/&quot; rel=&quot;noreferrer&quot;&gt;a technical publications website&lt;/a&gt; that appears to reproduce some of the content we were taught in the nuclear power training program. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nuclearpowertraining.tpub.com/h1018v1/&quot; rel=&quot;noreferrer&quot;&gt;The first mechanical science volume&lt;/a&gt; has a &lt;a href=&quot;http://nuclearpowertraining.tpub.com/h1018v1/css/Cavitation-106.htm&quot; rel=&quot;noreferrer&quot;&gt;section on cavitation&lt;/a&gt; that states,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If  the pressure  drop is large enough, or if the temperature is high enough, the pressure drop may be sufficient to cause the liquid to flash to vapor when the local pressure falls below the saturation pressure for the fluid being pumped.   Any vapor bubbles formed by the pressure drop at the eye of the impeller are swept along the impeller vanes by the flow of the fluid.   When the bubbles enter a region where local pressure is greater than saturation pressure farther out the impeller vane, the vapor bubbles  abruptly  collapse.    &lt;strong&gt;This  process  of  the  formation  and  subsequent  collapse  of  vapor bubbles in a pump is called cavitation.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Emphasis added) The definition we were instructed to memorize (as I quoted at the top) is the condensed version of this statement for reproduction on exams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is no source &lt;em&gt;on this particular website&lt;/em&gt;, where the reference volumes are broken out by section, as to where this material originates, but at the top of the page is given the DOE document &quot;DOE-HDBK-1018/1&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can look that number up and find the document posted in full on the &lt;a href=&quot;http://energy.gov/sites/prod/files/2013/06/f2/h1018v1.pdf&quot; rel=&quot;noreferrer&quot;&gt;Department of Energy's website&lt;/a&gt;, where that passage can be found on page 12. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, regarding the comment about &quot;industry not toeing the line of the US Navy,&quot; the copy hosted on the DOE website includes a foreword and an overview that state the material was prepared with input from the nuclear industry and is intended for use in training nuclear operators. So, maybe some industries do not use the definition of cavitation I have provided, but the nuclear industry &lt;em&gt;does&lt;/em&gt;, and it seems like (from &lt;a href=&quot;https://engineering.stackexchange.com/questions/11963/what-fundamentally-distinguishes-cavitation-and-boiling-as-different-phenomena/12038?noredirect=1#comment21297_12038&quot;&gt;Bryon Wall's comment&lt;/a&gt;) that the chemical industry does as well. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-10-27T18:33:14.250" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12039" PostTypeId="2" ParentId="12020" CreationDate="2016-10-26T21:33:10.663" Score="0" Body="&lt;p&gt;You can also think of the concrete as being similar to wood ( a porous, composite solid) and use a wood moisture meter. They come in a wide variety of styles, choose the one that suits your application. You might have to drill holes in the concrete to insert the moisture meters that have probes, so if you choose this method you will have to drill precise diameter holes so that the probes will be in intimate contact with the concrete.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2016-10-26T21:33:10.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12040" PostTypeId="1" AcceptedAnswerId="12046" CreationDate="2016-10-26T22:24:51.253" Score="0" ViewCount="49" Body="&lt;p&gt;I was reading a paper about a &quot;Controller Design for Temperature Control of Heat Exchanger&quot; and I couldn't understand how to model the valve used in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No need to read the paper as I will wrap up the whole issue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;capacity of control valve is 1.6&amp;nbsp;kg/sec&lt;/li&gt;&#xA;&lt;li&gt;time constant is 3&amp;nbsp;sec&lt;/li&gt;&#xA;&lt;li&gt;valve input is pressure varying from 3 to 15&amp;nbsp;psi&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The resulting transfer function is:&#xA;$$G(s)=\dfrac{0.13}{3s+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that they considered the valve as a first order transfer function.The gain $K_p$ was calculated by: $\dfrac{1.6\text{ kg/s}}{15-3\text{ psi}}=1.6/12=0.133$. Time constant = 3&amp;nbsp;seconds and that's it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem is: shouldn't a step input of 15&amp;nbsp;psi output a signal (1-exponential) with a final value of 1.6&amp;nbsp;kg/sec. But that transfer function won't actually output that. It still needs some kind of a shift. So am I missing something? Is the model wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also simulated the transfer function response with MATLAB SIMULINK along with a suggested alternative that probably solve the offset thing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Abl1d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Abl1d.png&quot; alt=&quot;SIMULINK MODEL&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting waveform:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QxOOv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QxOOv.png&quot; alt=&quot;Waveforms&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yellow waveform is the input step function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue waveform is the output of the transfer function of the valve according to the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yellow waveform is the waveform with an offset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem with the blue waveform: wrong final value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem with the yellow waveform: starts from a negative value (-0.4&amp;nbsp;kg/sec)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the correct model?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If necessary, &lt;a href=&quot;http://www.wseas.org/multimedia/journals/control/2014/a125703-288.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here's the pdf file&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8554" LastEditorUserId="1627" LastEditDate="2016-10-27T13:36:54.887" LastActivityDate="2016-10-27T13:36:54.887" Title="Control System: offset issue with a simple valve model" Tags="|modeling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12046" PostTypeId="2" ParentId="12040" CreationDate="2016-10-27T13:33:56.853" Score="0" Body="&lt;p&gt;The offset needs to be in the input and it's value needs to be 3.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/RWRRT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RWRRT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then with the 15 psi input, the final value is 1.6 kg/sec.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/T9YUL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T9YUL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-10-27T13:33:56.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12048" PostTypeId="2" ParentId="11996" CreationDate="2016-10-27T20:45:39.247" Score="2" Body="&lt;p&gt;My apologies for answering both 1) and 2) in the same answer, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This sounds a like a perfect application for a crystallization process.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crystallizers use an external cooling loop, as opposed to the unit op being done inside of a heat exchanger. This often looks like a large vessel with a sloped bottom, and some sort of external cooling loop with tubes running inside of the tank. The tubes have a cold fluid flowing through that applies the cooling load, and can often act as the nucleation point for crystallization as well (be careful though, solids/crystal buildup could cause scaling/loss of heat transfer). After precipitation / crystallization, the solids will (usually) sink, and then get funneled into the bottom. From there they can be slurried/pumped to a solids/liquids separation unit, such as a classifier, gravity screen, mesh screen, filtration screen, etc all dependant on the physical properties of your material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fw16e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fw16e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option one could try is doing these two unit ops separately if possible. Often one can do a simple chemical adjustment to instigate the precipitation such as adjusting pH, ORP, adding a reagent etc. Separate with a settling tank and then as imilar method to the solid-liquid separation in the above diagram. Once separated, the fluid could then be sent to a traditional heat exchanger. &lt;/p&gt;&#xA;" OwnerUserId="8369" LastActivityDate="2016-10-27T20:45:39.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12049" PostTypeId="1" AcceptedAnswerId="12061" CreationDate="2016-10-28T01:01:49.823" Score="0" ViewCount="2940" Body="&lt;p&gt;What factors would degrade a spring's life span other than stress applied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will high cycles (e.g. 100k load-unload continuous cycles) and high speed (100 ms/cycle), affect the spring constant (k)? What does it means if the spring constant decreases or increases after 1 million cycles (does it means the spring has reached its lifespan, or degrade, or...)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the spring is under applied appropriate force.&lt;/p&gt;&#xA;" OwnerUserId="8563" LastEditorUserId="5899" LastEditDate="2016-10-29T00:16:45.097" LastActivityDate="2016-10-29T11:06:35.953" Title="What affects the spring constant or spring life spans?" Tags="|mechanical-engineering|springs|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12050" PostTypeId="1" AcceptedAnswerId="12062" CreationDate="2016-10-28T08:26:27.260" Score="1" ViewCount="2003" Body="&lt;p&gt;I have a structure with a lot of interconnecting column and beam frame elements. I want to compute the displacement, member forces/moments via Finite Element Method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem now is that I don't know how to properly incorporate moment release at beam end. I understand that moment release means that at the beam's end, the bending moment is $0$, but I don't know how I can incorporate this piece of information ( as a boundary condition? Or as a form of penalty function, or...?) into my FEM model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to incorporate support into my FEM model though; I can just make the relevant appropriate stiffness terms very big. But not moment release.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea how moment release can be formulated in finite element terms to achieve the desired end results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is different from the one that I asked &lt;a href=&quot;https://engineering.stackexchange.com/q/10864/3353&quot;&gt;here&lt;/a&gt;, because that one asked about mathematical formulation ( more on conceptual level), whereas this one asks about explicit FEM formulation. These two maybe related, but to go from previous question to this one is by no means trivial, and the connection not easy to see. &lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2022-07-28T06:33:00.767" Title="How to incorporate moment release at beam end in FEM formulation?" Tags="|structural-engineering|structural-analysis|finite-element-method|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12051" PostTypeId="1" CreationDate="2016-10-28T08:36:23.730" Score="3" ViewCount="1032" Body="&lt;p&gt;I am trying to plot the efficiency graph of a rotary compressor.&#xA;I already calculated the volumetric &amp;amp; polytropic efficiencies and the graph versus evaporating temperature is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4VG7L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4VG7L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now i would like to find the volumetric efficiency of the compressor versus pressure ratio, but what i see is not the single graph is 4 series of curves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rhnEp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rhnEp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have always seen in datasheet they show just one curve as the volumetric efficiency of the compressor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you please tell me what is the approach to calculate the voleumetric efficiency graph of the compressor?&lt;/p&gt;&#xA;" OwnerUserId="8568" LastActivityDate="2018-01-22T09:07:33.623" Title="Volumetric efficiency versus pressure ratio for rotary compressor" Tags="|compressors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12053" PostTypeId="1" AcceptedAnswerId="12054" CreationDate="2016-10-28T13:47:20.667" Score="0" ViewCount="132" Body="&lt;p&gt;We are looking to connect a non spinning coolant pipe to a spinning drill pipe. The connector needs to spin on the end with the drill pipe but be stationary on the end with the coolant pipe. Is there some type of bearing you know of or some other connector?&lt;/p&gt;&#xA;" OwnerUserId="3629" LastActivityDate="2016-10-28T14:30:00.607" Title="What kind of connection allows rotation at only one end?" Tags="|statics|dynamics|bearings|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12054" PostTypeId="2" ParentId="12053" CreationDate="2016-10-28T14:30:00.607" Score="2" Body="&lt;p&gt;I think OP is asking about transferring fluid (coolant) through a rotating joint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kinds of joints are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotary_union&quot; rel=&quot;nofollow&quot;&gt;rotary unions&lt;/a&gt; or swivel joints. One end of the rotary union is stationary, the other end may rotate continuously, and fluid passes through the joint, which is sealed even when the joint is in motion. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-10-28T14:30:00.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12055" PostTypeId="1" AcceptedAnswerId="12060" CreationDate="2016-10-28T15:02:58.190" Score="1" ViewCount="1540" Body="&lt;p&gt;What would be some common applications involving high friction and low wear rates? for example, gear-driven mechanisms which require high friction between the gear teeth but ideally low wear between them.&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2017-04-01T20:10:02.300" Title="Applications with combination of high friction and low wear?" Tags="|mechanical-engineering|applied-mechanics|friction|mechanisms|tribology|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12056" PostTypeId="1" AcceptedAnswerId="12063" CreationDate="2016-10-28T16:41:08.630" Score="0" ViewCount="105" Body="&lt;p&gt;I have 3 programable logic controllers and other one which isn't part of my solution but I can access usefull data via modbus. I've setup windows service on a remote server to gather data and store it in a SQL Server.&#xA;But the problem is if the Internet fails? So I'm searching for a hardware data logger solution.&#xA;I saw a schneider scadapack but I think it doesn't support modbus as client.&#xA;Everything I'm seeing is hardware with I/O for digital and analog input/outputs but I have already that in PLCs. What I need now is to gather data from the PLCs via modbus RTU/TCP/UDP. After the service detects connection errors it registers the begin date and end date. After it detects PLCs are reachable again it downloads the usefull data that it missed from the hardware data logger solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which solutions support this?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Yokogawa MW100&lt;/li&gt;&#xA;&lt;li&gt;Schneider scadapack models&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both this are data acquisition systems via I/O and probably not via modbus.&#xA;I'm lost. Help?&lt;/p&gt;&#xA;" OwnerUserId="8570" LastActivityDate="2016-10-30T08:53:25.130" Title="Data logger via modbus hardware solution" Tags="|data|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12059" PostTypeId="2" ParentId="6653" CreationDate="2016-10-28T17:16:06.270" Score="1" Body="&lt;p&gt;C15 equivalent in IS standard is 15C4. You can check reference IS standards for 15C4 to confirm, or see &lt;a href=&quot;https://books.google.com/books?id=SpcF9IrYKRoC&amp;amp;pg=PA147&amp;amp;lpg=PA147&amp;amp;dq=15c4%20steel&amp;amp;source=bl&amp;amp;ots=CfBATl65Lg&amp;amp;sig=pQaQhKlEwJyQyxNZgswPcCA2jpQ&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwi_8Mjsz5XRAhUW8mMKHSJCAN4Q6AEIGjAA#v=onepage&amp;amp;q=15c4%20steel&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;page 147 of Worldwide Guide to Equivalent Irons and Steels&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8573" LastEditorUserId="368" LastEditDate="2016-12-28T00:27:28.213" LastActivityDate="2016-12-28T00:27:28.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12060" PostTypeId="2" ParentId="12055" CreationDate="2016-10-28T18:31:14.840" Score="3" Body="&lt;p&gt;As mentioned in the comments you don't actually want high friction between gear teeth, the tooth profiles are designed to roll over each other indeed the ideal is  to minimise any sliding friction between the teeth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obvious examples of devices which use sliding friction are brakes and friction clutches, these will have a limited wear life depending on the material used and in general there is a compromise between performance and life, especially for clutches which tend to be relatively difficult to replace. These devices may also also include some sort of self adjusting mechanism to compensate for wear of the friction surfaces. Often two different materials will be used with one designed to wear faster  but be easier to replace  eg bolt or clip on pads working against a steel or cast iron friction plate (as in disk brakes). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is tyres which need to have low rolling resistance but high static friction to provide grip and traction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friction is also very relevant in screw threads, wear is significant for things like lead screws on lathes etc  which are subject to repeated sliding friction and where it is usually desirable to keep backlash to a minimum. Also on threaded fasteners there is a conflict between wanting low sliding friction to allow consistent torque setting and reduce the risk of galling but having adequate static friction to prevent fasteners from coming undone through vibration etc while still allowing reasonable manufacturing tolerances and there are various retaining devices such as nyloc nuts, spring and serrated washers and locking compounds used to achieve this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also say that cutting and abrasive operations fall broadly into this category although this is normally considered separately from conventional friction per-se. Although of course the analogy if pretty good for things like bonded abrasives and diamond coated cutting tools. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-29T16:59:04.397" LastActivityDate="2016-10-29T16:59:04.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12061" PostTypeId="2" ParentId="12049" CreationDate="2016-10-28T18:53:42.230" Score="2" Body="&lt;p&gt;As mentioned in the comments spring designs and materials can vary widely to it is difficult to generalise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if we consider steel springs which are very common then there are a few factors to consider. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly if properly specified a spring should never exceed its elastic limit, however if this does happen then its rate may very well change due to work hardening and early failure is quite likely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a spring operating within specification then &lt;a href=&quot;https://en.wikipedia.org/wiki/Fatigue_(material)&quot; rel=&quot;nofollow&quot;&gt;fatigue&lt;/a&gt; is one possible cause of failure this occurs over a large number of cycles within the elastic limit as small local stress concentrations which cause local overloading and gradual crack growth untill the crack reaches a critical length and then propagates very rapidly. The gradual crack growth will have a small effect on the performance of a part as it will reduce its cross section slightly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific materials will have a fatigue limit, this is stress at which a large number of cycles will lead to failure. Some materials have a fatigue limit close enough to the yield stress that fatigue does not need to be considered, in others the limit is so low that any practical component will be subject to fatigue eventually. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fatigue failures are predicted statistically so you might say that a particular spring has a 99% chance of not failing before 100k cycles and a 50% chance of failing after 500k cycles based on experimental data and empirical formulae so the nominal design life before replacement is a compromise between the consequences of premature failure and the cost of replacement.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is creep this is plastic type deformation which occurs below the elastic limit when constant loads are applied for long periods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both creep and fatigue are generally calculated from empirical data specific to a specific material, manufacturing process and component class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another consideration is environmental corrosion and degradation. Metal springs can suffer from corrosion in certain conditions and polymers may break down over time due to UV light and reactions with environmental chemicals.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-10-29T11:06:35.953" LastActivityDate="2016-10-29T11:06:35.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12062" PostTypeId="2" ParentId="12050" CreationDate="2016-10-28T22:09:41.490" Score="2" Body="&lt;p&gt;Conceptually, all you have to do is &lt;em&gt;not&lt;/em&gt; assemble the &quot;released&quot; rotation variable to anything else in the system stiffness and mass matrices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from keeping track of the &quot;extra&quot; variables at some nodes in the model, everything else in the FE method &quot;just works&quot; without needing any special math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to eliminate the &quot;released&quot; variable(s) from the &lt;em&gt;element&lt;/em&gt; matrices before you assemble them into the global matrices. In that case, you need to keep track of what you did, so you can undo it when you want to recover the element stresses, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the general situation, you might also need to create a local coordinate system at the grid (node) point, so you have a single rotation variable in each orientation that you want to &quot;release&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, the same technique is sometimes useful with translation variables as well as rotations - for example, if you are modeling a spline joint where the two parts can move relative to each other along the line of the joint, but the other two translations and all three rotations can transmit forces and moments across the joint.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-10-28T22:14:49.550" LastActivityDate="2016-10-28T22:14:49.550" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12063" PostTypeId="2" ParentId="12056" CreationDate="2016-10-28T22:55:35.670" Score="0" Body="&lt;p&gt;It seems the easiest solution would be to set up a local SQL server with the same setup you have going now on the remote server. I don't see a &quot;hardware&quot; solution being easier, cheaper, or more reliable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also a hardware logger usually means nasty csv files to parse afterward. If price is an issue, you could adapt it to run on a smaller linux server or an RPi with MySQL. Put that sever some place clean and on a UPS and you will be just as robust as any &quot;hardware&quot; based solutions that may exist.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2016-10-28T22:55:35.670" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12064" PostTypeId="1" CreationDate="2016-10-28T23:20:25.130" Score="2" ViewCount="1038" Body="&lt;p&gt;I have a project in mind where I want to measure the pulling force of each finger while doing a climbing exercise. My original plan was to build a wooden climbing hold that used 4 load cells under a hold that looks more like 4 piano levers with the cells mounted under each lever. After a bit of research I found these &lt;a href=&quot;http://www.tml.jp/e/product/strain_gauge/gauge_list/wood_gypsume_use.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;strain gauges&lt;/a&gt; and was wondering if it's possible to mount these on the wooden levers (approx 20x80mm surface area) to make the the piece of wood itself into a load cell and get good enough data. My criteria for each load cell is a max load of 75-100kg and accuracy to be within +/-1kg. Is this possible with wood using strain gauges or should I go for a more complex build using movable parts pressing down on a load cell?&lt;/p&gt;&#xA;&lt;p&gt;Short description: Is it possible to turn a wooden stick 20mm wide and 80mm long into a load cell capable to measure force up to 100kg using a strain gauge?&lt;/p&gt;&#xA;&lt;p&gt;Update:&#xA;Here is a image of what I had in mind:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gXcpF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gXcpF.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As alephzero said, its better to use plastic, but how can I find a type of plastic and calculate the correct thickness/height that can hold 100kg without breaking or deform over time?&lt;/p&gt;&#xA;" OwnerUserId="7467" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-10-29T08:58:29.000" Title="Strain gauge for wood" Tags="|sensors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12065" PostTypeId="2" ParentId="12064" CreationDate="2016-10-29T01:14:05.957" Score="1" Body="&lt;p&gt;Strain gauges on wood are useful for detecting failure conditions like cracks developing, by picking up &lt;em&gt;changes&lt;/em&gt; in strain over time when the loading is constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want to convert the strain into a force or stress measurement, you need to know the elastic properties of the wood. Wood is anisotropic, and the properties will vary depending on the exact position of the gauge relative to the wood grain and any &quot;defects&quot; like knots, etc. They will also be affected by humidity, unless you seal the wood against moisture - but doing that may make it difficult to bond the gauges to the wood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want &quot;accurate&quot; force measurements (i.e. to within say 5% rather than 50% accuracy) a load cell attached to a statically determinate structure (so the load path from your hand to the load cell does not depend on the stiffness of the material) would be a much better design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use strain gauges, you would be better to use a man-made material where the elastic properties are known and consistent - for example some type of plastic, if you don't want to use metal.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-10-29T01:19:44.050" LastActivityDate="2016-10-29T01:19:44.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12066" PostTypeId="2" ParentId="11772" CreationDate="2016-10-29T08:11:21.543" Score="0" Body="&lt;p&gt;Not sure how much this might help, but I recently finished a similar project, designing a fixture to spin a 350g mass at a radius of about 80mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used an arduino, but also created a circuit to switch 12v into the stepper motor using pwm.  One of the issues I faced was reading a correct value from the encoder.  It turned out that the large current through the motor,  ~1 amp, was affecting the signal on my encoder line. I had wires crossing close by each other, so when I designed a new pcb with noise consideration in mind,  separating the power from the signal as much as possible,  I was able to get it working perfectly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of the large inductive load from stepping a motor,  this is something one of the electrical engineers was telling me might be a concern for my board.  One of the solutions which we added was a Schottky Diode across the P-FET, but I am not familiar with why exactly that would help.  (mechanical engineering student)&lt;/p&gt;&#xA;" OwnerUserId="8487" LastEditorUserId="8487" LastEditDate="2016-10-29T08:17:20.937" LastActivityDate="2016-10-29T08:17:20.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12069" PostTypeId="1" AcceptedAnswerId="12070" CreationDate="2016-10-29T18:50:19.300" Score="1" ViewCount="113" Body="&lt;p&gt;Generally, in a pressure drop control valve that uses a piston and a spring (for example, in jet engines to control fuel flow), there is always potential for error in terms of the height of the spring. As we know, a shim is usually used to fix this. However, I was wondering if there are alternatives to such a procedure. Are there other ways to correct the height on a valve spring, or perhaps produce the spring in a way that reduces the need for a correction method like shimming?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking maybe laser correction could be a viable method for correcting a spring that is off by a few thousands of an inch? It is extremely accurate, but then the main disadvantage is cost. Are there other methods or technologies that exist or are perhaps being worked on at the moment?&lt;/p&gt;&#xA;" OwnerUserId="7802" LastActivityDate="2016-10-29T19:04:06.250" Title="Alternatives to shimming valve springs" Tags="|pressure|measurements|aerospace-engineering|flow-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12070" PostTypeId="2" ParentId="12069" CreationDate="2016-10-29T19:04:06.250" Score="0" Body="&lt;p&gt;Shimming is always attractive for fit adjustment as it is fairly straightforward to do ie you can measure any gap with a feeler gauge and select the appropriate thickness of shim to suit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The downside of manufacturing parts to fit exactly is that you then need to match individual parts to their location throughout their life and you then need a different system if you want to compensate for wear. Equally if you need a replacement part you need to report the exact size you need and get it made to fit as opposed to just ordering a standard part and having an existing known method to compensate for any variation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again if you know that the springs will always be a bit short you know you will always be filling some gap and its just the gauge of the shim which varies however if you are aiming for an acceptable fit out of the box there is always a risk that it will be slightly too long in which case a shim will no longer fix the problem.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-10-29T19:04:06.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12071" PostTypeId="1" AcceptedAnswerId="12110" CreationDate="2016-10-29T19:07:29.040" Score="0" ViewCount="116" Body="&lt;p&gt;Assume I have a system comprising of an airfoil &lt;strong&gt;A&lt;/strong&gt; immersed in a fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further assume, that the airfoil is affixed via two bolts at one end, and is untethered at the other end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system can be parametised as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;S&lt;/em&gt; is the span of the airfoil&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;C&lt;/em&gt; is the chord of the airfoil&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;V&lt;/em&gt; (vega) is the relative velocity of the fluid passing over the airfoil&lt;/li&gt;&#xA;&lt;li&gt;$\rho$ is the density of the fluid in which the airfoil is is immersed&lt;/li&gt;&#xA;&lt;li&gt;$\theta$ is the current angle of attack of the airfoil (in radians)&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;L&lt;/em&gt; the distance between two bolts affixing the airfoil at one end (&amp;lt; &lt;em&gt;C&lt;/em&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[[&lt;em&gt;Question&lt;/em&gt;]]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Given a new angle $\theta_i$, what would be the formula for calculating the force required to be exerted on the two bolts to change the current angle of attack of the airfoil from $\theta$ to $\theta_i$? (assuming all other variables held constant)&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[[Notes]]&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the lift characteristics of an airfoil, I think its fair to assume that a greater force will be required to be exerted as the angle of attack increases (up until the stall angle)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, since the force required is likely to be monotonically increasing (up until stall angle), I would prefer if the function actually returned the supremum of the the forces required to increase the angle of attack from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta$ $\rightarrow$ $\theta$ + $\delta$$\omega$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\delta$$\omega$ is the change in radians divided into an infinitesimal number of steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, the formula should derived from first principles (or the answer be provided as pseudocode for an algorithm), so that I can follow the logic, and apply it to an airfoil of non-rectangular shape.&lt;/p&gt;&#xA;" OwnerUserId="932" LastEditorUserId="932" LastEditDate="2016-10-29T21:37:35.677" LastActivityDate="2018-05-02T15:29:30.237" Title="Formula (or heuristic) for calculating force required for incremental change in angle of attack of an airfoil" Tags="|mechanical-engineering|fluid-mechanics|applied-mechanics|aerodynamics|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12072" PostTypeId="2" ParentId="11184" CreationDate="2016-10-29T20:56:59.570" Score="2" Body="&lt;p&gt;If a retaining wall is immobile, it incurs lateral soil pressure from both sides, and that is the so called resting pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the retaining wall, due to load from the retained soil massive, which tends to slide down (wedge of failure), starts moving towards the soil in front of the wall (and thus off the retained soil), then the resting pressure from the retained soil becomes active pressure; while the pressure from the soil in front of the wall, resisting against the wall moving towards it, becomes passive pressure (displacement wedge pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Active pressure is less than resting pressure, while passive pressure always exceeds resting pressure. Practically, some centimeters movement is sufficient to completely realize passive pressure; as well as practically not any retaining wall undergone active pressure remains completely immobile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the wall so moves, the shifting force (due to active pressure) is reducing, while resisting force (due to passive pressure) is increasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As soon as growing passive pressure load becomes equal to active pressure load, the wall stops moving. As a rule, if the retaining wall is designed adequately, the passive pressure does not reach its maximally possible value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: active pressure acts actively, it holds, let's say, 'first-mover right' trying to 'push off' the retaining wall. While the passive pressure will not 'respond', until it is 'attacked' by wall pushed by active pressure.&lt;/p&gt;&#xA;" OwnerUserId="8547" LastActivityDate="2016-10-29T20:56:59.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12073" PostTypeId="1" AcceptedAnswerId="12074" CreationDate="2016-10-29T22:38:26.630" Score="3" ViewCount="190" Body="&lt;p&gt;I am having issues building a recreation of Focaults Pendulum (basic, inexpensive, undergrad project). At this point, I am going to purchase aircraft cable or suspension cable unless someone can recommend a slightly flexible rod.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to use a 32oz plumb bob because I can't find a symmetrical spherical bob unless someone has suggestions. Last, if anyone has any recomendations on a frictionless or low friction pivot (360 degrees and pendulus) that would be amazing.&lt;/p&gt;&#xA;" OwnerUserId="8581" LastEditorDisplayName="user16" LastEditDate="2016-10-29T23:44:51.783" LastActivityDate="2016-11-04T19:19:33.393" Title="Recreating Focaults Pendulum" Tags="|mechanical-engineering|structural-engineering|friction|experimental-physics|building-physics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12074" PostTypeId="2" ParentId="12073" CreationDate="2016-10-29T22:59:35.547" Score="2" Body="&lt;p&gt;There are two simple options for the bearing &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is to suspend the bob from a flexible cord passing through a fairly tight grommet so it will maintain a constant point of contact in all directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to get a large ball bearing, ideally plastic or unhardened (or at least not through hardened) steel and then drill and tap it for an attachment point for your pendulum (or weld a nut to it in which case 316 stainless is a good option. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then mount this in a plate with a hole of something like 95% of the diameter of the bearing, for this slip-on pipe flanges are convenient. To get the best possible bearing surface fit  you can lap the ball into the join with valve grinding paste or similar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example below is a slightly more complex version and can be locked made from a 50mm diameter 316 stainless ball bearing and a 1/1/2&quot; bsp pipe flange. These flanges are failry thick but if you chamfer the top edge of the hole it gives you a bit more clearance and it can act as a reservoir for grease. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nt9ek.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nt9ek.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit to address questions in the commenst : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For a grommet or collet type support monofilament rather than braided wire might be best as the tension in braided wire might mean it behaves slightly differently in differnt directions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For the ball bearing option it might be easiest to weld a nut to the bearing (or drill and tap for a stud) and thread the end of a round bar to fit. Idally you want  the connection to the ball to be fairly rigid or there is little point using the ball at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) There is probably no advantage in going for a very heavy ball as more weight will only increase the friction at the bearing surface having said that a larger ball give syou a larger range of movement. In my example a 50mm ball in a 48mm beveled hole gives around 45 degrees of motion in all directions from vertical. With this approach you are not relying on the felxibility of a 'string' of provide the range of mevement so you have more options in how you attach the bob to the bearing whihc in turn should allow you to use a heavier bob while keeping bearing friction and damping to a minimum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) I would guess that the magnetic donut is providing an impulse to the pendulum to compensate for friction losses and keep it swinging indefintely. This requires a coil and a modified ocsillator circuit...but that probably merits a new question. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-11-04T19:19:33.393" LastActivityDate="2016-11-04T19:19:33.393" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12075" PostTypeId="2" ParentId="12056" CreationDate="2016-10-30T08:53:25.130" Score="0" Body="&lt;p&gt;Ice got experience with horner apg and mt-1xx/2xx by inventia.pl. Personally I've did data collecting from twido (modbus) on mt-151 (inventia, sdcard). You can did same thing on any horner device.&lt;/p&gt;&#xA;" OwnerUserId="8583" LastActivityDate="2016-10-30T08:53:25.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12076" PostTypeId="2" ParentId="6251" CreationDate="2016-10-30T10:50:23.817" Score="1" Body="&lt;p&gt;Directionally, yes. Treat it exactly like a compressibility term in the material balance equation (pressure drop and temperature rise both cause the grain material to expand). Second-order effects mentioned by the last contributor (e.g. permeability reduction from pore throat contraction) would need to be taken into account, but that's a productivity term, not a volumetric term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://accuratus.com/fused.html&quot; rel=&quot;nofollow&quot;&gt;This reference&lt;/a&gt; gives the bulk modulus of Fused Silica (kinda-sorta like sand grains?) as 6e6psi so the compressibility is 0.16e-6 psi. The coefficient of thermal expansion is 0.55e-6 per degree Celsius, so tripling that (for bulk coefficient) we get about 1.5e-6 per Celsius. So each degree Celsius produces (approximately) the same volumetric effect as a 10psi pressure change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, however, there is so much physics and physical chemistry going on in a thermal flood (passive or active) that I would be very surprised to see an interpretable response at the reservoir scale.&lt;/p&gt;&#xA;" OwnerUserId="8584" LastActivityDate="2016-10-30T10:50:23.817" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12077" PostTypeId="2" ParentId="5585" CreationDate="2016-10-30T11:02:14.287" Score="1" Body="&lt;p&gt;Think in order of magnitude terms of inter-well spacing (500 metres in an offshore waterflood?) divided by water breakthrough time (say, 3 years). 500m/1000 days = 0.5 metres per day. Yes, I know a saturation contour isn't a fluid particle (phase vs. group velocity), but the relevant multiplier will be of order unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want particle velocity, another approach is to do the calculation (flowrate * formation volume factor) / (linear dimension * net thickness * porosity). Say (100Mb/d * 1.3)/(10km*100m*0.15) = 0.14 metres per day, so same close order of magnitude. Of course things will speed up when you get into convergent flow near the wells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you using a commercial reservoir simulator? They don't normally concern themselves with fluid velocities, at least not in terms of what goes in and out of the code.&lt;/p&gt;&#xA;" OwnerUserId="8584" LastEditorUserId="8584" LastEditDate="2016-10-30T12:13:09.177" LastActivityDate="2016-10-30T12:13:09.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12078" PostTypeId="2" ParentId="11979" CreationDate="2016-10-30T13:25:12.227" Score="3" Body="&lt;p&gt;Usually since hydrostatic pressure along with freezing and thawing are such a pain (quite hard to figure out the pressure from freezing water in soil), in practice I try and keep my retaining walls 100% water free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/04CxQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/04CxQ.png&quot; alt=&quot;Typical retaining wall construction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you were to design the wall for lateral earth pressure of clay + water table pressure, I found in my old notes  some active, rest and passive values for clay which are around :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$k_a = 0.6 \\ k_o = 0.7 \\ k_p = 1.7$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually you will start with a hypothesis (which side is active or passive or at rest), and then confirm that your had chose correctly with maximum displacements. Since your aren't analysing the wall yet, we will keep that for later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For safety, you should probably at least consider at rest pressure on both side (use $k_o$ for both sides) and verify your wall displacements afterwards (If you consider active pressure on the right hand side, this means your wall is relatively flexible and the soil can mobilise some of its shear stiffness).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually for safety, we also put the water table all the way up to the level of the ground, since this can easily happen after a storm in non-draining clay. You then apply your basic soil equations, first dry-soil pressure and add water pressure (both triangular loads). In this case we assume there is no external load on top of the right hand side of the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h4Bnf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h4Bnf.png&quot; alt=&quot;Retaining wall pressures&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations for the maximal pressure are the same for each side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ q_{clay} = k * \gamma_{clay} * H \\ q_{water} = k * \gamma_{water} * H $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ever you decide to design this wall, you will see that your critical moment and shear will be at the base, even if you have some soil supporting on the left hand side. If you have enough movement to mobilise active pressure on the right and passive pressure on the left, then you can simply use the same equations with $k_a$ on the right and $k_p$ on the left.&lt;/p&gt;&#xA;" OwnerUserId="5691" LastActivityDate="2016-10-30T13:25:12.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12079" PostTypeId="1" CreationDate="2016-10-30T14:15:21.913" Score="0" ViewCount="396" Body="&lt;p&gt;I studied a paper about turbine and i seen that there is negative static pressure in contour. Why does this happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HjvXL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HjvXL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1877705815019244&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wave Energy Harvesting Turbine: Performance Enhancement&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2016-10-31T03:42:27.160" LastActivityDate="2016-10-31T03:42:27.160" Title="Why is there negative static pressure in contour?" Tags="|mechanical-engineering|pressure|turbines|cfd|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12080" PostTypeId="2" ParentId="12071" CreationDate="2016-10-30T15:54:10.927" Score="0" Body="&lt;p&gt;Without knowing the exact geometry of the system, such as the shape of the foil, it would be hard to come up with an exact model. You might be better of doing an experiment, applying some force and record what kind of angle you get in response for different relative fluid speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This force for a given angle will only be the force required in steady state, but during the transition from one angle to another you would also need some extra force to initially speed up and slow down the rotational speed of the airfoil. So you would need some time varying force, which can be achieved using some feedback control. For example a PID controller would be a possibility. The integral term of such PID controller should also help bring the airfoil angle to the desired angle in steady state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also include the model from your experiment in the form of &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Feed_forward_(control)&quot; rel=&quot;nofollow&quot;&gt;feed forward control&lt;/a&gt;, which could be used to improve the performance of the system.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2016-10-30T15:54:10.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12081" PostTypeId="1" CreationDate="2016-10-30T16:47:29.940" Score="3" ViewCount="72" Body="&lt;p&gt;Recently a certain 2 lane exit ramp changed from merge right to merge left, but the entrance ramp immediately preceding it is merge right. Why was this changed? In a given circumstance, what dictates a merge left rather than a merge right? &lt;/p&gt;&#xA;" OwnerUserId="8587" LastEditorUserId="1832" LastEditDate="2016-10-30T16:50:05.040" LastActivityDate="2016-10-30T16:50:05.040" Title="How is a merge left or a merge right chosen?" Tags="|civil-engineering|highway-engineering|" AnswerCount="0" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12084" PostTypeId="1" CreationDate="2016-10-30T20:40:22.940" Score="1" ViewCount="135" Body="&lt;p&gt;I'm currently working on the thermal analysis of a pipe, but i do not know how to define a third type boundary condition on the inner surface.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;finish&#xA;/out&#xA;/CLEAR,ALL&#xA;&#xA;/PNUM,KP,1  &#xA;/PNUM,LINE,1&#xA;/PNUM,AREA,1&#xA;&#xA;/PREP7&#xA;!Constants used in the simulation&#xA;a=0.05&#xA;b=0.2&#xA;!third type boundary condition is used on the inner surface&#xA;Tb=30&#xA;qb=260&#xA;p0=3.1e5&#xA;!second type boundary condtion on the outer surface&#xA;hk=120&#xA;Tk=15&#xA;qk=490&#xA;hb=10&#xA;rho=7600&#xA;E=210e9&#xA;nu=0.3&#xA;alpha=1.2e-5&#xA;k=45&#xA;Tinf=20&#xA;!!!!!!!!!!!!!!!!!!!!&#xA;ET,1,PLANE77&#xA;!Material props&#xA;TOFFST,273&#xA;MPTEMP,1,Tinf&#xA;MP,EX,1,E&#xA;MP,REFT,1,Tinf&#xA;MP,PRXY,1,nu&#xA;MP,KXX,1,k&#xA;&#xA;!!!!!!&#xA;K,1,0,0&#xA;K,2,a,0&#xA;K,3,b,0&#xA;K,4,0,a&#xA;K,5,0,b&#xA;LARC,2,4,1,a&#xA;LARC,3,5,1,b&#xA;L,4,5&#xA;L,2,3&#xA;AL,1,2,3,4&#xA;&#xA;Lsel,s,LINE,,1,2&#xA;LESIZE,ALL,,,50&#xA;LSEL,inve&#xA;LESIZE,ALL,,,30&#xA;mshkey,1&#xA;AMesh,all&#xA;/solu&#xA;lsel,all&#xA;SFL,3,HFLUX,0&#xA;SFL,4,HFLUX,0&#xA;SFL,2,HFLUX,490&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The problem is that i do not know how to apply convection boundary condition on the inner surface, where the inner radius is a and the temperature of the fluid is Tb.&#xA;Thanks for your help in advance.&#xA;yours sincerly,&#xA;David&lt;/p&gt;&#xA;" OwnerUserId="8592" LastEditorUserId="1832" LastEditDate="2016-10-31T13:11:48.547" LastActivityDate="2016-10-31T13:11:48.547" Title="How can I define third type boundary condition in ANSYS Mechanical APDL?" Tags="|thermodynamics|temperature|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12085" PostTypeId="1" AcceptedAnswerId="12095" CreationDate="2016-10-30T21:14:05.357" Score="2" ViewCount="989" Body="&lt;p&gt;Please help, what is the correct term for:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The bottom/foot part of building or structure which carries loads from the building/structure and transfers the loads to the soil/earth/ground beneath it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The soil massive to which said load from the structure is transferred.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In my native tongue, the two meanings have different terms, however, in dictionaries they are named both base and/or foundation. The question is to distinguish the part of structure from the soil.&lt;/p&gt;&#xA;" OwnerUserId="8547" LastEditorUserId="1832" LastEditDate="2016-10-30T22:44:59.413" LastActivityDate="2017-01-25T01:37:51.393" Title="Base or foundation?" Tags="|structural-engineering|civil-engineering|structures|geotechnical-engineering|terminology|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12086" PostTypeId="2" ParentId="12055" CreationDate="2016-10-30T21:22:11.397" Score="0" Body="&lt;p&gt;Grinders, crushers, shredders, pulverisers ...&lt;/p&gt;&#xA;" OwnerUserId="8584" LastActivityDate="2016-10-30T21:22:11.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12087" PostTypeId="1" AcceptedAnswerId="12088" CreationDate="2016-10-31T00:42:38.170" Score="2" ViewCount="122" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5g9sS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5g9sS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everytime I try this question I keep getting 93.75&amp;nbsp;mm. I was just wondering where I'm going wrong. Here's my work: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;BC=\dfrac{FL}{AE}=\dfrac{1500000\text{ N}\cdot 1250\text{ mm}}{100000\text{ mm}^2*200\text{ MPa}}=93.75\text{ mm}&#xA;$$&#xA;For deflection to be in $\text{mm}$, all units have to be in $\text{N}$ and $\text{mm}$, right?&lt;/p&gt;&#xA;" OwnerUserId="8594" LastEditorUserId="1832" LastEditDate="2016-10-31T00:59:13.377" LastActivityDate="2016-10-31T01:21:42.643" Title="Deflection problem confusing units?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12088" PostTypeId="2" ParentId="12087" CreationDate="2016-10-31T01:21:42.643" Score="1" Body="&lt;p&gt;You are correct. The book probably meant $E = 200\text{ GPa}$, which is frequently adopted for steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To confirm, I made a simple little model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wDrYN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wDrYN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're probably aware of this, but just to make it explicit: your math only worked because $200\text{ MPa} = 200\text{ N/mm}^2$.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-10-31T01:21:42.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12089" PostTypeId="1" AcceptedAnswerId="12090" CreationDate="2016-10-31T08:55:30.163" Score="0" ViewCount="1082" Body="&lt;p&gt;I was just wondering for this question if you can see anything wrong with it. The answer in the textbook is 220MPa but i get 193MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bw6tV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bw6tV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bffaa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bffaa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8594" LastActivityDate="2016-10-31T09:51:45.570" Title="Maximum Bending Stress" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12090" PostTypeId="2" ParentId="12089" CreationDate="2016-10-31T09:51:45.570" Score="2" Body="&lt;p&gt;The correct answer is 193 MPa as you have written. I have used &lt;a href=&quot;https://skyciv.com/beam-calculator-features&quot; rel=&quot;nofollow noreferrer&quot;&gt;SkyCiv Beam&lt;/a&gt; software to confirm this. It's highly unlikely you are both wrong. The shaft is essentially a beam so it's fine to use beam software for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HPmxh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HPmxh.png&quot; alt=&quot;SkyCiv Beam used to determine bending stress of shaft&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/21GKh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/21GKh.png&quot; alt=&quot;Bending stress profile in SkyCiv&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1848" LastActivityDate="2016-10-31T09:51:45.570" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12091" PostTypeId="1" AcceptedAnswerId="12094" CreationDate="2016-10-31T10:11:31.097" Score="1" ViewCount="81" Body="&lt;p&gt;I am thinking of building a prototype for a solar generator using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fresnel_lens&quot; rel=&quot;nofollow&quot;&gt;fresnel&lt;/a&gt; lens to concentrate sun rays to a relatively small area, for the purpose of harnessing the thermal energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this is that the temperature at the concentrated point, &lt;a href=&quot;https://www.quora.com/Thermodynamics-What-is-the-maximum-temperature-a-1-meter-square-fresnel-lens-can-reach-at-its-focal-point&quot; rel=&quot;nofollow&quot;&gt;temperatures as high as 3000F&lt;/a&gt; are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What (preferably inexpensive) material can be placed at the focal point of the lens, to transfer the heat away to do some useful work?&lt;/p&gt;&#xA;" OwnerUserId="932" LastEditorUserId="1832" LastEditDate="2016-10-31T11:08:10.883" LastActivityDate="2016-10-31T15:28:07.950" Title="What material (preferably relatively inexpensive), can I use for absorber in a fresnel solar concentrator" Tags="|materials|thermodynamics|heat-transfer|solar-energy|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12092" PostTypeId="2" ParentId="12091" CreationDate="2016-10-31T14:37:41.867" Score="1" Body="&lt;p&gt;It's not that easy. You're combining several questions (unintentionally, I guess). The first question is how to make sure you absorb as many photons as possible.  This is typically achieved both by using materials which are &quot;black&quot; in the wavelengths of interest &lt;em&gt;and&lt;/em&gt; by making the surface as rough as possible to increase the chances of absorbing photons which reflect on first impact.&lt;br&gt;&#xA;Next, you want a material with high thermal conductivity so that you can pull the heat (energy) into a sink. If you don't, then much of the energy will be re-radiated in accordance with black-body radiation theory.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're just building a &quot;proof of concept&quot; setup,  a thin layer of carbon dust in good contact with a thin-wall metal pipe might do.  The thin layer enhances heat transfer to the pipe;  pump water or ethylene glycol (automotive radiator fluid) thru the pipe to transfer the heat to something else (perhaps a &lt;a href=&quot;https://en.wikipedia.org/wiki/Stirling_engine&quot; rel=&quot;nofollow&quot;&gt;Stirling engine&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-10-31T14:37:41.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12094" PostTypeId="2" ParentId="12091" CreationDate="2016-10-31T15:28:07.950" Score="2" Body="&lt;p&gt;Crucial to this is how you intend to exploit this energy as this will be very significant in how quickly and uniformly heat is extracted from the target. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3000F is outside the working range of most readily available engineering materials and if you want something which can withstand that sort of temperature outright you are in the realms of things like engineering ceramics, carbon composites and tungsten. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that as long as you are removing heat quickly enough you should be able to keep the temperature in a sensible range indeed many heat engines have components which are exposed to flame temperatures in excess of their melting points. A gasoline engine will break pretty quickly if the cooling system fails. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a logical approach is to use a heat engine with a working fluid with good heat transfer properties here steam power is an obvious choice and &lt;a href=&quot;https://youtu.be/uJGpbvvJA2I&quot; rel=&quot;nofollow&quot;&gt;solar steam engines&lt;/a&gt; do indeed exist (note that this is an illustration rather than a reccomendation of how to do it in detail).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously any sort of boiler or steam power is fraught with hazards and not something which should be attempted without taking detailed and  specialist engineering advice. There may also be leagal requirements for inspection and certification of boilers whcih apply.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the video answer simple balck steel pipe should be adequate as long as you have enought coolant flow and has the advanatge that there are existing standards for steam pipe and the associated fittings and it is raltiavely inexpensive and straightforward to work with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach has the advantage that you are using the working fluid to directly take heat from the target of the lense with minimal intermediate steps between the heat source and the engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also steam power is a mature and well developed technology so if you are more interested in eveluating the potential output of the lense you could just run it as an open system and measure the energy trasnfer to the water, avoiding the hazrds of a pressurised steam system. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-10-31T15:28:07.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12095" PostTypeId="2" ParentId="12085" CreationDate="2016-10-31T17:40:58.467" Score="2" Body="&lt;p&gt;The part of the building that is in contact with the soil is the &lt;strong&gt;&lt;a href=&quot;http://www.dictionary.com/browse/foundation?s=t&quot; rel=&quot;nofollow&quot;&gt;foundation&lt;/a&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;the lowest division of a building, wall, or the like, usually of masonry and partly or wholly below the surface of the ground. &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you'll see from the same link, that &lt;em&gt;might&lt;/em&gt; also be used for the ground itself:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;the natural or prepared ground or base on which some structure rests. &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I would suggest that definition 2 (referring to the ground as the foundation) is not in use as a technical engineering definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest using &quot;foundation&quot; or &quot;base&quot; for the part of the structure. I am struggling to think of a good term for the soil/found on which the foundation sits. You might need to stick to simple terms such as &quot;supporting ground&quot;. Possibly &quot;bearing strata&quot; or &quot;founding strata&quot; might be better (though maybe not applicable in all situations).&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2016-10-31T17:40:58.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12096" PostTypeId="2" ParentId="12071" CreationDate="2016-10-31T17:44:07.507" Score="0" Body="&lt;p&gt;For a given angle of attack and velocity you can compute a pitching moment about the cg (Or the normal force on the airfoil and the center of pressure), this moment can then be resolved onto the bolts to determine the loading on the bolts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll need a controller (like a PID) to exert the loads on the bolts correctly to stop at the correct new angle of attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;We need to first define the center of pressure where by definition the pitching moment is zero. Some math will show us that the center of pressure can be found from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h_{cp}$ = $h_{ac}$ - $Cm_{ac}/C_l$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $C_l$ can be found from the lift curve slope for your airfoil in Theory of Wing Sections and $Cm_{ac}$ (Moment coefficient about the aerodynamic center) which is constant can also be obtained from Theory of Wing Sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for your back of the envelope calculation you can just assume that the aerodynamic center is roughly quarter chord from the leading edge ($h_{ac} = 0.25$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we know where the Lift force is acting ($h_{cp}$) as well as its value ($l=0.5*C_l*\rho*V^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we just need to resolve that onto the bolts using simple statics. I'm going to use a pin-roller beam for the example. Let's call the distance to bolt 1 from the leading edge $x_1$ and the distance to the second bolt $x_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing the moments about the 1st bolt and setting equal to zero we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_1 = l * x_{cp} / x_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then summing the forces in the Lift direction and setting equal to zero we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_2 = l - F_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x_{cp}$ and $x_2$ are the dimensional distances to the center of pressure and second bolt, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now bear in mind this is a VERY rough estimate. This doesn't take into account 3D effects (A whole topic on its own), drag contribution to the forces on the bolts, and the fact that the beam is actually a pin-pin which is indeterminate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For references I used &lt;a href=&quot;http://www.dept.aoe.vt.edu/~lutze/AOE3104/airfoilwings.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.dept.aoe.vt.edu/~lutze/AOE3104/airfoilwings.pdf&lt;/a&gt; and &lt;a href=&quot;http://bendingmomentdiagram.com/tutorials/calculating-reactions-at-supports/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://bendingmomentdiagram.com/tutorials/calculating-reactions-at-supports/&lt;/a&gt; as well as some experience ;)&lt;/p&gt;&#xA;" OwnerUserId="8225" LastEditorUserId="8225" LastEditDate="2016-11-15T04:03:33.787" LastActivityDate="2016-11-15T04:03:33.787" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12098" PostTypeId="1" AcceptedAnswerId="12099" CreationDate="2016-10-31T18:38:36.077" Score="0" ViewCount="34" Body="&lt;p&gt;I am doing research on objects frozen with liquid nitrogen then seeing how they thaw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to monitor objects temperatures as they thaw. Ideally getting the variance of the external and internal temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What strategy or equipment would you recommend to get this data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="8605" LastActivityDate="2016-10-31T22:01:52.077" Title="Lab Equipment for Monitoring Sub Zero Temperatures" Tags="|temperature|" AnswerCount="1" CommentCount="2" ClosedDate="2016-11-05T01:56:04.437" ContentLicense="CC BY-SA 3.0" />
  <row Id="12099" PostTypeId="2" ParentId="12098" CreationDate="2016-10-31T22:01:52.077" Score="1" Body="&lt;p&gt;You can get &lt;a href=&quot;http://www.thermometricscorp.com/thermocouple.html&quot; rel=&quot;nofollow&quot;&gt;thermocouples&lt;/a&gt; which are rated for cryogenic temperatures. These produce a voltage whcih can be calibrated to temperature and so are faily easy to connect to digiotal or analogue data logging equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are essentially just a wire so they are easy to embed in things and they can be fitted to various probes and pads for differnt mounting configurations.   &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-10-31T22:01:52.077" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12100" PostTypeId="2" ParentId="12085" CreationDate="2016-11-01T00:52:47.450" Score="1" Body="&lt;p&gt;In USA the part of structure that is in contact with earth and transfers all the live and dead load of the super-structure to earth is called &lt;strong&gt;foundation&lt;/strong&gt;.  Foundation is responsible to support the design loads with design settlement parameters:   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normal loads.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear loads.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment loads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torsion loads&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the earth under this foundation or encasing it if the foundation is set of pylons or deep caissons is called &lt;strong&gt;soil&lt;/strong&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Foundation engineer design foundation by coordinating input received from soils engineer and structural engineer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each site has to be studied by a geologist to determine its properties with respect to its location and type of underlying earth strata and its distance from known seismic faults or expected stresses and other large scale characteristics. The geologist will give his report to the soils engineer.    Soils engineer will excavate a number of test pits to measure the properties and strength of soil and gives his recommendations to foundation or in some cases to structural engineer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In small projects on areas that have competent known soil and are flat and not near any fault line or stress line the building departments publish simplified standards that replace the job if soils engineer and foundation engineer with prescribed tables of soils load bearing and foundation's details.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-11-01T00:52:47.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12101" PostTypeId="1" AcceptedAnswerId="12107" CreationDate="2016-11-01T04:41:52.447" Score="3" ViewCount="183" Body="&lt;p&gt;I am trying to do this question and I have everything but one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XRQIE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XRQIE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for $A_x$, $x$ force at $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;F_x = 6\cdot20 + A_x \\&#xA;A_x= -120\text{ kN}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Sign normally doesn't matter, if you're consistent, but it's causing an error in my working out the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaluating it, I get -1852.5 which is incorrect. The answer is 1132.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I change the sign of $6\cdot20$ to $-6\cdot20$, I get the ‘correct’ answer, -1132.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;F_x = 20\cdot6 + A_x = 0 \\&#xA;A_x = -120 \\&#xA;M_D = -30\cdot7\cdot3.5+2\cdot E_y = 0 \\&#xA;E_y = 367.5 \\&#xA;F_y = D_y + E_y - 30\cdot7 = 0 \\&#xA;D_y = -157.5 \\&#xA;F_y = -D_y - 60 + A_y = 0 \\&#xA;A_y = 60-157.5 = -97.5 \\&#xA;\sum M_A = M_A + 6\cdot20\cdot3-60\cdot4-D_y\cdot11=0&#xA;\end{gather}$$&lt;/p&gt;&#xA;" OwnerUserId="8607" LastEditorUserId="1832" LastEditDate="2016-11-01T14:14:03.167" LastActivityDate="2016-11-01T14:14:03.167" Title="Is a force to the right positive?" Tags="|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12104" PostTypeId="2" ParentId="12101" CreationDate="2016-11-01T10:52:39.900" Score="1" Body="&lt;p&gt;You are correct that sign doesn't matter as long as you are consistent.&lt;br&gt;&#xA;In that final stage you are calculating the torque about point A, you need to be consistent as to the direction of that torque, is it clockwise or counterclockwise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking at point A if you use positive for horizontal forces to the right then you need to use positive for vertical forces that are down. Mixing it and using positive for left and down or for right and up will give the wrong answer because you're being inconsistent as to the direction of the torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does mean that for torque calculations you may have to change the sign for a given force depending upon it's position relative to the point you are analyzing.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastEditorUserId="7659" LastEditDate="2016-11-01T11:05:29.343" LastActivityDate="2016-11-01T11:05:29.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12105" PostTypeId="1" AcceptedAnswerId="12126" CreationDate="2016-11-01T11:33:02.207" Score="1" ViewCount="281" Body="&lt;p&gt;I'm trying to simulate a compound action potential for calibrating research instruments. The image below is a small overview of the design. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/GW1Dm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GW1Dm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already completed the data acquisition from a live animal, and processed the data in MATLAB to make a nice, noise-less signal, with 789 values in 12-bit format. I then cloned the repository to the Pi using Git. Below is the Python script I've written on the RPi. You can skip to def main in the script to see functionality.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/python&#xA;&#xA;import spidev&#xA;from time import sleep&#xA;import RPi.GPIO as GPIO&#xA;import csv&#xA;import sys&#xA;import math&#xA;&#xA;DEBUG = False&#xA;spi_max_speed = 20 * 1000000&#xA;V_Ref = 5000&#xA;Resolution = 2**12&#xA;CE = 0&#xA;&#xA;spi = spidev.SpiDev()&#xA;spi.open(0,CE)&#xA;spi.max_speed_hz = spi_max_speed&#xA;&#xA;LDAQ = 22&#xA;GPIO.setmode(GPIO.BOARD)&#xA;GPIO.setup(LDAQ, GPIO.OUT)&#xA;GPIO.output(LDAQ,GPIO.LOW)&#xA;&#xA;def setOutput(val):&#xA;    lowByte = val &amp;amp; 0b11111111 #Make bytes using MCP4921 data sheet info&#xA;    highByte = ((val &amp;gt;&amp;gt; 8) &amp;amp; 0xff) | 0b0 &amp;lt;&amp;lt; 7 | 0b0 &amp;lt;&amp;lt; 6 | 0b1 &amp;lt;&amp;lt; 5 | 0b1 &amp;lt;&amp;lt; 4&#xA;    if DEBUG :&#xA;        print(&quot;Highbyte = {0:8b}&quot;.format(highByte))&#xA;        print(&quot;Lowbyte =  {0:8b}&quot;.format(lowByte))&#xA;    spi.xfer2([highByte, lowByte])&#xA;&#xA;def main():&#xA;    with open('signal12bit.csv') as signal:&#xA;        signal_length = float(raw_input(&quot;Please input signal length in ms: &quot;))&#xA;        delay = float(raw_input(&quot;Please input delay after signal in ms: &quot;))&#xA;        amplitude = float(raw_input(&quot;Please input signal amplitude in mV: &quot;))&#xA;        print &quot;Starting Simulant with signal length %.1f ms, delay %.1f ms and amplitude %.1f mV.&quot; % (signal_length, delay, amplitude)&#xA;        if not DEBUG : print &quot;Press ctrl+c to close.&quot;&#xA;        sleep (1) #Wait a sec before starting&#xA;        read = csv.reader(signal, delimiter=' ', quotechar='|')&#xA;        try:&#xA;            while(True):&#xA;                signal.seek(0)&#xA;                for row in read: #Loop csv file rows&#xA;                    if DEBUG : print ', '.join(row)&#xA;                    setOutput(int(row)/int((V_Ref/amplitude))) #Adjust amplitude, not super necessary to do in software&#xA;                    sleep (signal_length/(data_points*1000) #Divide by 1000 to make into ms, divide by length of data&#xA;                sleep (delay/1000)&#xA;        except (KeyboardInterrupt, Exception) as e:&#xA;            print(e)&#xA;            print &quot;Closing SPI channel&quot;&#xA;            setOutput(0)&#xA;            GPIO.cleanup()&#xA;            spi.close()&#xA;&#xA;if __name__ == '__main__':&#xA;    main()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This script almost works as intended. Connecting the output pin of the MCP4921 to an oscilloscope shows that it reproduces the signal very well, and it outputs the subsequent delay correctly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the data points are seperated much further than I need them to be. The shortest time I can cram the signal into is about 79 ms. This is due to dividing by 789000 in the sleep function, which I know is too much to ask from Python and from the Pi, because reading the csv file takes time. However, if I try making an array manually, and putting those values out instead of reading the csv file, I can achieve a frequency over 6 kHz with no loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is this&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I get this signal to appear at a frequency of 250 Hz, and decrease it reliably from the user's input? I've thought about manually writing the 789 values into an array in the script, and then changing the SPI speed to whatever value fits with 250 Hz. This would eliminate the slow csv reader function, but then you can't reduce the frequency from user input. In any case, eliminating the need for csv.read would help a lot. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="8614" LastEditorUserId="8055" LastEditDate="2016-11-01T13:26:24.387" LastActivityDate="2016-11-02T19:49:30.313" Title="Python delays on Raspberry Pi" Tags="|electrical-engineering|digital-communication|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12106" PostTypeId="2" ParentId="12105" CreationDate="2016-11-01T13:45:47.380" Score="0" Body="&lt;p&gt;I posted a comment to the question asking you to provide more information about what your problem is exactly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can kind of side-step that issue by acknowledging that it seems you have some kind of question about timing with Python on the Raspberry Pi, to which I would redirect you to &lt;a href=&quot;https://stackoverflow.com/a/1133888/5171120&quot;&gt;this answer on StackOverflow&lt;/a&gt;. In summary, for Linux operating systems (like Raspian), &quot;non-realtime Linux kernels have minimum sleep interval much closer to 1ms then 10ms but it varies in a non-deterministic manner.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a rate of 250Hz, you're asking for a 4ms delay between samples. A side/coding comment of mine at this point would be - instead of asking the user for a signal duration, why not ask for the sample rate? Then your sampling delay is the sample period, which is 1/sampleRate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any event, the sample period timing you're asking for doesn't appear to be outside the limits of what's achievable with Python's timing. This would seem to me to imply, then, that your hunch may be correct - things you are doing outside of the &lt;code&gt;sleep&lt;/code&gt; itself are impacting the timing of your program. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the question is - &lt;em&gt;why are you doing those things?&lt;/em&gt; Let's take a look at the essence of what you're doing:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;while(True):&#xA;    signal.seek(0)&#xA;    for row in read: #Loop csv file rows&#xA;        if DEBUG : print ', '.join(row)&#xA;        setOutput(int(row)/int((V_Ref/amplitude))) #Adjust amplitude, not super necessary to do in software&#xA;        sleep (signal_length/(data_points*1000) #Divide by 1000 to make into ms, divide by length of data&#xA;    sleep (delay/1000)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Your signal never changes - it's hard-coded into the source code. Despite this, you still have to manually input the signal parameters by hand every time the code is run (not sure why you chose to do this). But all that's beside the point - what I mean to say is, &lt;em&gt;once you get to the loop&lt;/em&gt;, you never swap out a different signal. As described, your signal isn't very big either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advice to you would be to move all of your data processing outside of your loop so the &lt;em&gt;only things you have to do&lt;/em&gt; are transmit the signal and sleep. No math. For example, consider the following changes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;signal.seek(0)&#xA;signalValues = [int(row)/int((V_ref/amplitude)) for row in read]&#xA;samplePeriod = signal_length/(data_points*1000)&#xA;signalDelay = delay/1000&#xA;while(True):&#xA;    for currentVal in signalValues:&#xA;        if DEBUG : print ', '.join(currentVal)&#xA;        setOutput(currentVal)&#xA;        sleep (samplePeriod)&#xA;    sleep (signalDelay)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Hopefully you see here that all of the floating point division and CSV reading has been moved outside of the loop. &lt;em&gt;None of your parameters change during the loop so there's no point in waiting to modify them on-the-fly.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Load the entire CSV (all 789 values) into a list that's held in memory. Scale them however they need to be scaled (warning on that later), then just iterate your way through the list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About scaling your values: You ask for the inputs as float, but then skew your result by casting them to &lt;code&gt;int&lt;/code&gt; mid-division. Specifically, you have the line:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;setOutput(int(row)/int((V_Ref/amplitude)))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Maybe you are thinking that an &lt;code&gt;int&lt;/code&gt; divided by and &lt;code&gt;int&lt;/code&gt; forces integer division - this is &lt;a href=&quot;https://stackoverflow.com/a/183870/5171120&quot;&gt;not correct for Python 3.0&lt;/a&gt;. If you are using Python 3.0 or later, then you're very likely getting a floating point number and then doing binary bit-wise operations on it when you attempt to send it. This could result in Bad Things happening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More to the point, if your desired output is an &lt;code&gt;int&lt;/code&gt;, then I would suggest you avoid prematurely truncating your numbers by casting the &lt;em&gt;entire result&lt;/em&gt; to &lt;code&gt;int&lt;/code&gt; as opposed to casting the individual operands. Specifically, consider this alternative:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;setOutput(int(row/(V_Ref/amplitude)))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here the output is guaranteed to be an &lt;code&gt;int&lt;/code&gt;, because you cast it that way, and you get better numeric accuracy in &lt;em&gt;which&lt;/em&gt; integer results from your math because you waited until the end of the operation to truncate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want integer division, you can get that with the &lt;code&gt;//&lt;/code&gt; integer division operator - &lt;code&gt;int(row)//int(V_Ref/amplitude)&lt;/code&gt; - but as with my comment about waiting to the end to cast as &lt;code&gt;int&lt;/code&gt;, here the integer division operator guarantees an integer output; &lt;em&gt;it accepts floating point inputs&lt;/em&gt;. So you could also have &lt;code&gt;row // (V_Ref/amplitude)&lt;/code&gt; and should give the same result as &lt;code&gt;int(row/(V_Ref/amplitude))&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For speed, consider converting your CSV to a list. &lt;/li&gt;&#xA;&lt;li&gt;For speed, consider pulling your static math operations (those whose operands do not change during your loops) outside of the loop. &lt;/li&gt;&#xA;&lt;li&gt;For accuracy, consider waiting to cast values to integer until after all math operations are complete.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="-1" LastEditDate="2017-05-23T12:38:59.030" LastActivityDate="2016-11-01T13:45:47.380" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12107" PostTypeId="2" ParentId="12101" CreationDate="2016-11-01T13:50:27.703" Score="1" Body="&lt;p&gt;A moment is defined as the cross-product of distance and force vectors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;M &amp;amp;= r\times F \\&#xA;&amp;amp;= \det\left(\left|\begin{matrix}&#xA;i &amp;amp; j &amp;amp; k \\&#xA;r_x &amp;amp; r_y &amp;amp; r_z \\&#xA;F_x &amp;amp; F_y &amp;amp; F_z \\&#xA;\end{matrix}\right|\right) \\&#xA;&amp;amp;= (r_yF_z-r_zF_y)\hat{i} -(r_xF_z-r_zF_x)\hat{j} + (r_xF_y-r_yF_x)\hat{k}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a 2D frame, $r_z = F_z = 0$, which simplifies this to $M = (r_xF_y-r_yF_x)\hat{k}$. So, a positive vertical force at a positive horizontal distance (to the right of the studied node) results in a positive moment. Meanwhile, a positive horizontal force at a positive vertical distance (above the studied node) results in a negative moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple way of thinking about it is that force components are positive to the right or upwards and distances are positive if a positive force would generate a positive (counter-clockwise) rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your last equation, you made the moment due to the distributed load along $\overline{AB}$ positive, when it should be negative. After all, a positive horizontal force at a positive vertical distance (above $A$) results in a negative moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should, however, be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum M_A =&amp;amp; M_A \\&#xA;&amp;amp; - (3)\cdot(6\cdot20) \\&#xA;&amp;amp; + (4)\cdot(-60) \\&#xA;&amp;amp; + (14.5)\cdot(-7\cdot30) \\&#xA;&amp;amp; + (13)\cdot (F_{E,y})= 0 \\&#xA;\therefore M_A =&amp;amp; -1132.5\text {kN}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally hate writing equations like this, with signs in the middle of the multiplications, but it makes all the variables explicit. So, in this calculation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;all the distances are positive (everything is either above or to the right of $A$)&lt;/li&gt;&#xA;&lt;li&gt;the horizontal distributed load is positive&lt;/li&gt;&#xA;&lt;li&gt;the vertical loads are negative&lt;/li&gt;&#xA;&lt;li&gt;$F_{E,y}$ is whatever it will be (positive, in our case)&lt;/li&gt;&#xA;&lt;li&gt;When calculating the moment due to vertical forces, the sign is positive&lt;/li&gt;&#xA;&lt;li&gt;When calculating the moment due to horizontal forces, the sign is negative&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-11-01T13:50:27.703" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12108" PostTypeId="1" AcceptedAnswerId="12109" CreationDate="2016-11-01T15:15:46.360" Score="2" ViewCount="2009" Body="&lt;p&gt;I am currently designing a prototype. I have asked to use low boiling point liquid to run a steam turbine. So i am thinking of using liquid ammonia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the complications involved when using vapor ammonia in a steam turbine. Is it possible to run a small steam turbine using ammonia vapor?&lt;/p&gt;&#xA;" OwnerUserId="8615" LastActivityDate="2022-02-27T10:39:42.580" Title="Is it possible to use Ammonia vapor in steam turbine" Tags="|turbines|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12109" PostTypeId="2" ParentId="12108" CreationDate="2016-11-01T15:40:49.180" Score="1" Body="&lt;p&gt;Of course it's possible, but the real question is, is there any advantage to justify using anything other than water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just run your steam plant off water, and if you want to use low boiling point liquid, just set the boiling point of water to about 40C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, you need a condenser, cooled to the point where the low pressure turbine stages exhaust into a pretty good vacuum (once air has been purged out of the system) and you'll find that allows you to use the energy in the steam down to about 40C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exact numbers relating temperature and pressure were traditionally published as &quot;steam tables&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using ammonia instead, you certainly wouldn't want to exhaust it to atmosphere, so either approach requires a condenser. Having added that, there is not much to be gained from moving away from water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may still be a thermodynamic advantage to ammonia (or diethyl ether) if it has much less latent heat of vaporisation than water; I can't comment on that.&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="435" LastEditDate="2016-11-01T15:49:54.383" LastActivityDate="2016-11-01T15:49:54.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12110" PostTypeId="2" ParentId="12071" CreationDate="2016-11-02T00:45:12.560" Score="1" Body="&lt;p&gt;As per &quot;NASA Aeronautics And Space Administration&quot; the thin foil lift equation is $$L = Cl * A * .5 * r * V^2  $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cl is the lift coefficient and in small angle range is directly related to angle of attack, multiplied by other factors. They have a java app here &lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/foil3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;FoilSim app&lt;/a&gt; Which is similar to what you seem to be asking. You have to set the security of your computer Java to let this app run. They offer help to set it up.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as your model of the wing and its attachment to support via two bolts, it is not practical and minimum number of bolts required would be three non-collinear bolts to turn yor hinge connection into a cantilever connection. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2016-11-02T00:45:12.560" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12111" PostTypeId="1" CreationDate="2016-11-02T00:46:19.330" Score="5" ViewCount="455" Body="&lt;p&gt;I'm a computer science student doing research in wireless sensor networks (WSN) optimization. Based on past research works, sensors are capable of failing due to a number of factors such as harsh environmental conditions, communication errors, and malicious attacks. According to &lt;a href=&quot;https://books.google.com.ph/books?id=aR0bAAAAQBAJ&amp;amp;pg=PA183&amp;amp;lpg=PA183&amp;amp;dq=sensor%20failure%20probability&amp;amp;source=bl&amp;amp;ots=wsZPlN3MKq&amp;amp;sig=gmynyYOd3UuaQfudwqKA-9PnqoA&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjuhYKg6IjQAhWMerwKHe8YBqEQ6AEILzAD#v=onepage&amp;amp;q=sensor%20failure%20probability&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source&lt;/a&gt;, it is possible to assign a &lt;em&gt;failure probability&lt;/em&gt; to each sensor. The source states that initially, it can be set by the manufacturer; however, many factors such as weather, accidents, interferences can affect the sensor failure probability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the network topology is known (the location and therefore environment is known) and that I have access to some hazard maps, how do I assign failure probabilities to each sensor? For example, if sensors in flood-prone areas are more likely to fail, how do I set the exact values of the failure probabilities? Is it acceptable to just assign arbitrarily higher values to sensors in harsher environments? Is there a way quantify this from an engineering perspective? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks and any insight would be extremely helpful.&lt;/p&gt;&#xA;" OwnerUserId="8621" LastEditorUserId="90" LastEditDate="2016-11-02T13:36:05.643" LastActivityDate="2016-11-04T21:08:02.267" Title="How to obtain sensor failure probabilities?" Tags="|sensors|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12112" PostTypeId="1" AcceptedAnswerId="12123" CreationDate="2016-11-02T02:32:21.293" Score="0" ViewCount="313" Body="&lt;p&gt;I'm trying to find a battery that I can use for a project. Let's be hypothetical for a moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 5-8 of &lt;a href=&quot;https://www.adafruit.com/products/1142&quot; rel=&quot;nofollow noreferrer&quot;&gt;these Adafruit Servos&lt;/a&gt;, connected to this &lt;a href=&quot;https://www.adafruit.com/products/1411&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adafruit Servo Shield&lt;/a&gt;. Leaving the Arduino itself out of the equation, how can I calculate the amperage draw that these motors and this shield will pull? And additionally, how can I use that with the Ah of a battery to calculate the time that the circuit could be on and in motion with the battery in question?&lt;/p&gt;&#xA;" OwnerUserId="8623" LastEditorUserId="8055" LastEditDate="2016-11-02T13:51:16.520" LastActivityDate="2016-11-02T14:20:13.537" Title="Amperage draw for Servo" Tags="|electrical-engineering|motors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12114" PostTypeId="1" AcceptedAnswerId="12124" CreationDate="2016-11-02T06:04:46.167" Score="2" ViewCount="59" Body="&lt;p&gt;When a body travels through any fluid(like aircrafts and cars), due to viscosity of the fluid a boundary layer is formed around the surface of the body which separates the velocity of the fluid inside the boundary layer(near the surface of object) to velocity of the fluid.The boundary layer consists of three zones laminar, transient and turbulent zones respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it compulsory for all the three zones to exist in a boundary layer? If we consider a small smooth surface, is it possible that the surface ends at the laminar zone of the boundary layer and there is no turbulent zone in boundary layer over the surface? And if it is possible will there be a low pressure zone at the end of the surface or something?&lt;/p&gt;&#xA;" OwnerUserId="8625" LastActivityDate="2016-11-04T19:16:56.193" Title="What happens when the surface ends at the laminar zone of the boundary layer?" Tags="|fluid-mechanics|aerodynamics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12115" PostTypeId="2" ParentId="12108" CreationDate="2016-11-02T07:01:23.060" Score="1" Body="&lt;p&gt;An established approach when you want a low boiling point in a steam engine is to use another fluid, just not ammonia - &lt;a href=&quot;https://en.wikipedia.org/wiki/Organic_Rankine_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Organic Rankine cycle engines&lt;/a&gt; use flueorcarbons, hydrofluorcarbons (HFC) and hydrocarbons. Last time I talked to ORC producers, HFC seemed to be pretty much the norm since those are also used in refridgerators.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2016-11-02T07:01:23.060" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12116" PostTypeId="1" AcceptedAnswerId="12120" CreationDate="2016-11-02T07:21:41.393" Score="1" ViewCount="119" Body="&lt;p&gt;I'm doing a project which simulates a car accident, with a cube block which has sensors on it crashing into a wall. &#xA;This object is on a 3&amp;nbsp;m track mounted on a linear bearing to reduce friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this object needs to be travelling at a speed and it impacts a solid wall (with about 40&amp;nbsp;G of deceleration in that impact). This impact is what I am recording.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can I get this object to travel at that speed? (which is about 35 to 45&amp;nbsp;mph, if my math isn't wrong that is)&lt;/p&gt;&#xA;" OwnerUserId="8626" LastEditorUserId="1832" LastEditDate="2016-11-02T10:48:49.713" LastActivityDate="2016-11-03T14:11:32.680" Title="Getting an object to travel at 35 to 45 mph" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12117" PostTypeId="2" ParentId="12112" CreationDate="2016-11-02T10:12:25.657" Score="0" Body="&lt;p&gt;The most practical way current draw is to power up the arduino board with a power supply capable of measuring current draw. Each time the Servos are activated the additional current draw can be observer on the &lt;a href=&quot;http://www.newegg.com/Product/Product.aspx?Item=9SIA1HE0ZR4821&quot; rel=&quot;nofollow noreferrer&quot;&gt;power supply&lt;/a&gt; with a digital read out like the one below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JPqxv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JPqxv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For much more accurate calculation, find a power supply capable of logging current draw. Calculate the current draw in relation to time to obtain the Ah. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you can you use a multimeter on supply line to record the current draw. A multimeter similar to &lt;a href=&quot;http://www.keysight.com/en/pd-692834-pn-34410A/digital-multimeter-6-digit-high-performance?cc=US&amp;amp;lc=eng&quot; rel=&quot;nofollow noreferrer&quot;&gt;34410A Digital Multimeter, 6½ Digit&lt;/a&gt; is capable of data logging the current draw. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VebmW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VebmW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-11-02T10:27:27.133" LastActivityDate="2016-11-02T10:27:27.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12118" PostTypeId="1" AcceptedAnswerId="13342" CreationDate="2016-11-02T11:31:46.443" Score="3" ViewCount="7998" Body="&lt;p&gt;Three common terms used in leadscrews are lead, # of starts, pitch. There definitions are usually as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lead usually refers to the amount of linear travel when the leadscrew&#xA;turns one revolution. &lt;/li&gt;&#xA;&lt;li&gt;# of Starts refers to the number of independent threads along the length of the leadscrew. &lt;/li&gt;&#xA;&lt;li&gt;Pitch refers to the distance between threads. In leadscrew with one start, pitch = lead. With multiple starts pitch = lead / # starts. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In many leadscrew catalogues, the equation listed for translating a load's linear mass into angular inertia is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ J_{load} = m\times({\frac{1}{2\pi p}})^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$ is the mass of the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p$ is the pitch in &lt;strong&gt;threads per inch&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main issue here is that the units are &lt;strong&gt;threads per unit length&lt;/strong&gt; instead of &lt;strong&gt;length per thread&lt;/strong&gt;. The mix in nomenclature here is very confusing. In a leadscrew with one start, I can easily do $\frac{1}{lead}$ and plug that as $p$ into the equation. But what about when there are multiple starts? Do I still use $\frac{1}{lead}$, or $\frac{1}{\#ofstarts\times lead}$? What is the derivation of this equation?&lt;/p&gt;&#xA;" OwnerUserId="8629" LastEditorUserId="8629" LastEditDate="2016-11-02T11:47:39.227" LastActivityDate="2019-02-07T19:38:43.330" Title="Calculating Load Inertia in a Leadscrew System" Tags="|mechanical-engineering|threads|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12119" PostTypeId="2" ParentId="12116" CreationDate="2016-11-02T11:41:52.303" Score="2" Body="&lt;p&gt;First, start by using consistent units.  40 miles/hour is 18 m/s.  For steady acceleration, the average speed when starting at 0 is therefore 9 m/s.  You have 3 m of acceleration distance, so the acceleration time is &amp;frac13; second.  That means the acceleration is 54 m/s&lt;sup&gt;2&lt;/sup&gt;.  That's about 5&amp;frac12; g.  How much force and energy that requires depends on the mass that is being accelerated.  Note that this includes anything the accelerating mass is riding on, like a sled on the track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You haven't said anything about the mass of the object, so I'll just pick 10 kg to see what that requires.  The final kinetic energy is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp; E = &amp;frac12; m V&lt;sup&gt;2&lt;/sup&gt; = &amp;frac12; (10 kg)(18 m/s)&lt;sup&gt;2&lt;/sup&gt; = 1.6 kJ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since that needs to imparted over 1/3 s, the power required is 4.9 kW.  The right electric motor with the right power source could do this, but something like a pneumatic piston is probably easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to let gravity do the acceleration if you're not absolutely stuck with 3 m of level track.  In 1 g gravity as here on earth, a 16.5 m drop results in 18 m/s terminal speed and takes 1.8 seconds.  This 16.5 meters drop doesn't need to be straight down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can arrange for a track that slopes down with a total drop of 17 m with maybe a small straight run-out section at the bottom, a lot of things get simpler.  The acceleration is independent of the mass.  Nothing needs to be adjusted as more or less gets piled onto the sled.  One way or another, you still have to supply the energy.  However, now you're supplying it by hoisting the sled up to the top.  This doesn't have to be done as fast, so you can trade off power with how long you're willing to wait.  It's also mechanically easier since the hoisting system won't be envolved in the crash.  It could be as simple as a cable and a crank, with some sort of release mechanism at the top.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-11-02T11:41:52.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12120" PostTypeId="2" ParentId="12116" CreationDate="2016-11-02T12:33:49.630" Score="3" Body="&lt;p&gt;Lets say we want 40 MPH (since I don't even have enough information to verify that it will give you the 40 G's).  As Olin mentioned, $40 MPH \approx 18 m/s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like in this situation, a spring would be your best bet.  We want it to be travelling at 18 m/s at the end of the 3 m path.  Potential energy in a spring is: $$ PE = \frac{1}{2} K x^2 $$&#xA;where K is the spring constant and x is the spring displacement.&#xA;  We want to compare that to the kinetic energy required to get 18 m/s.  This is given by: $$ KE = \frac{1}{2} m v^2$$&#xA;where m is the mass of the object and v is the velocity of the object.  If we want to use all the potential energy in the spring to move the block, we can equate them:$$\frac{1}{2}K x^2 = \frac{1}{2}m v^2$$&#xA;Then cancelling some terms and rearranging:$$\frac{v^2}{x^2} = \frac{K}{m}$$ Using v = 18 m/s and x = 3 m: $$\left(\frac{18}{3}\left[\frac{1}{s}\right]\right)^2 = \frac{K}{m}$$ &#xA;Simplifying and rearranging a little more: $$36 \left[\frac{1}{s^2}\right]\cdot m = K$$ &#xA;So if your mass was 10 kg for example, then you need a spring with a spring constant $$K = 36 \left[\frac{1}{s^2} \right] \cdot 10 kg = 360 \left[\frac{N}{m}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can play around with the length of spring if you don't want it to travel the whole 3 m on a spring.  Obviously you can also adjust your mass accordingly depending on what it really is.  You could choose a tension or compression spring; whatever you feel would work best in your setup.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2016-11-02T12:33:49.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12121" PostTypeId="2" ParentId="12111" CreationDate="2016-11-02T13:50:04.633" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Probabilistic_risk_assessment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Probabilistic Risk Assessment&lt;/a&gt; (PRA) is a big deal in the Nuclear industry, basically it is a mathematical approach to quantify how safe/unsafe a power plant is. Without going too deep into PRA it is a method to combine risk factors from many different failure possibilities and their interactions with everything else; everything from major weather events like earthquakes and hurricanes/tsunamis to human error in the face of stressful environments to sensor and actuator failures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PRA has to deal with two major problems, epistemic and stochastic uncertainties in failure events. For example, having a new sensor for which no one knows the failure probability (i.e. #failures/time) is a stochastic uncertainty, there is simply not enough data to provide a believable failure probability. The best way to overcome stochastic uncertainty is to get some data. Epistemic uncertainties are when the model used to predict failures is not accurate, like having inaccurate weather records to predict how often hurricanes will occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question, if you know your system layout and you have failure probabilities you can use the methods in PRA to properly combine all the failure modes (like combine flooding probability with other ways the sensor could fail). In the cases where you don't have data you should either collect/find some or consult an expert who can give you an idea of how conservative to be. No matter what you choose to do, there will always be scrutiny on values you make a &quot;guess&quot; for. &lt;/p&gt;&#xA;" OwnerUserId="1174" LastActivityDate="2016-11-02T13:50:04.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12122" PostTypeId="2" ParentId="23" CreationDate="2016-11-02T13:58:59.830" Score="2" Body="&lt;p&gt;You might want to look up the the &quot;Bond Work Index&quot;. It's an empirical method for understanding the energy required to achieve certain amounts of comminution (think P80) for a given method and mineral. &lt;/p&gt;&#xA;" OwnerUserId="8635" LastActivityDate="2016-11-02T13:58:59.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12123" PostTypeId="2" ParentId="12112" CreationDate="2016-11-02T14:20:13.537" Score="1" Body="&lt;p&gt;The &lt;a href=&quot;http://www.electronicoscaldas.com/datasheet/MG996R_Tower-Pro.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;datasheet for the device&lt;/a&gt; should always list a maximum or &quot;stall&quot; current. For this device, it's 2.5A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use that for a worst-case estimate of battery life by multiplying it by the amount of time you want to run. For example, if you want 12 minutes of run time, then the battery needs to have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(2.5\mbox{A})*(12\mbox{min})/(60\mbox{min/hr}) = 0.5\mbox{Ahr}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Per servo&lt;/strong&gt; that you have on your device. That is, if you have 5 servos, then you need a battery with a capacity of 2.5 Ah to run for 12 minutes. If you have 8, then you need a 4 Ah battery to run for 12 minutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this is a worst-case scenario, if you try to run your servo at stall current for a long duration then you'll probably wind up burning them out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much capacity you actually need depends on the the &lt;strong&gt;duty cycle&lt;/strong&gt; at which you're running the servos. If you're running an excavator then you might actually be running them close to stall current frequently. If you're trying to position a camera then you won't be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advice to you would be to go get a battery pack, any battery pack, and use that to power your system. Then, run the system as hard as you think it would normally be run. Compare the run time you achieved with the run time you desire. Then, adjust the capacity of the battery pack accordingly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mbox{Required Capacity} = (\mbox{Test Capacity})*\frac{\mbox{Desired Run Time}}{\mbox{Actual Run Time}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you had a 1.75 Ah battery and it ran for 5 minutes, but you want it to run for 20 minutes, then you need (1.75)*(20/5) = 7 Ah battery pack. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-11-02T14:20:13.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12124" PostTypeId="2" ParentId="12114" CreationDate="2016-11-02T17:09:05.697" Score="2" Body="&lt;p&gt;It is certainly possible to achieve this in certain circumstances for example &lt;a href=&quot;http://www.aviation-history.com/theory/lam-flow.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.aviation-history.com/theory/lam-flow.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically it is a lot easier to claanly accelerate a flow arround the leading edge of a surface than to keep it attached as the flow converges again. So the best chance of achieveing this is a long tapering teardrop shape as in sprint cycling helmets and some solar powerd endurance vehicles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in practice achieving completely laminar flow tends to compromise practical design to the extent that it is not useful and most practical aerodynamics is more about managing turbulence and vortex generation rather than attempting to eliminate it entirely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in F1 cars a lot of &lt;a href=&quot;http://en.f1i.com/magazine/77473-f1-fechnical-snapshot-austin.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;engineering effort&lt;/a&gt; goes into making vortices do a useful job (particularly) in separating the turbulent wash from the tyres from the laminar flow under the floor/diffuser) &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-11-04T19:16:56.193" LastActivityDate="2016-11-04T19:16:56.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12125" PostTypeId="2" ParentId="12114" CreationDate="2016-11-02T17:45:32.280" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it compulsory for all the three zones to exist in a boundary layer?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;in practice, yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If we consider a small smooth surface, is it possible that the surface&#xA;  ends at the laminar zone of the boundary layer and there is no&#xA;  turbulent zone in boundary layer over the surface?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IF you have a sufficiently smooth surface, and sufficiently slow velocity, the NSE simplify (I use the term loosely) into a PDE that can be solved analytically. (stokes flow around a cylinder or sphere, for example)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that they can be solved analytically indicate that the flow is not turbulent! (again, this is theoretical)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And if it is possible will there be a low pressure zone at the end of&#xA;  the surface or something?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'd have to crack open my fluid mechanics books before I can give a definite answer&lt;/p&gt;&#xA;" OwnerUserId="6488" LastActivityDate="2016-11-02T17:45:32.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12126" PostTypeId="2" ParentId="12105" CreationDate="2016-11-02T19:49:30.313" Score="0" Body="&lt;p&gt;Figured it out earlier today, so I thought I'd post an answer here, in case someone comes upon a similar problem in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the internal delay between data points cannot be solved with sleep(), for several reasons. What I ended up doing was the following&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Move &lt;strong&gt;all&lt;/strong&gt; math and function calling out of the critical loop&lt;/li&gt;&#xA;&lt;li&gt;Do a linear regression analysis on the time it takes to transfer the values with no delay&lt;/li&gt;&#xA;&lt;li&gt;Increase the number of datapoints in the CSV file to &quot;plenty&quot; (9600) in MATLAB&lt;/li&gt;&#xA;&lt;li&gt;Calculate the number of points needed to meet the user's wanted signal length&lt;/li&gt;&#xA;&lt;li&gt;Take evenly seperated entries from the now bigger CSV file to fit that number of points as closely as possible.&lt;/li&gt;&#xA;&lt;li&gt;Calculate these values and then calculate the SPI bytes explicitly&lt;/li&gt;&#xA;&lt;li&gt;Save the two byte lists, and output them &lt;strong&gt;directly&lt;/strong&gt; in the critical loop&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The new code, with a bit of input checking, is below&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#!/usr/bin/python&#xA;&#xA;import spidev&#xA;from time import sleep&#xA;import RPi.GPIO as GPIO&#xA;import sys&#xA;import csv&#xA;import ast&#xA;&#xA;spi_max_speed = 16 * 1000000 # 16 MHz&#xA;V_Ref = 5000 # 5V in mV&#xA;Resolution = 2**12 # 12 bits for the MCP 4921&#xA;CE = 0 # CE0 or CE1, select SPI device on bus&#xA;total_data_points = 9600 #CSV file length&#xA;&#xA;spi = spidev.SpiDev()&#xA;spi.open(0,CE)&#xA;spi.max_speed_hz = spi_max_speed&#xA;&#xA;LDAQ=22&#xA;GPIO.setmode(GPIO.BOARD)&#xA;GPIO.setup(LDAQ, GPIO.OUT)&#xA;GPIO.output(LDAQ,GPIO.LOW)&#xA;&#xA;def main():&#xA;&#xA;    #User inputs and checking for digits&#xA;    signalLengthU = raw_input(&quot;Input signal length in ms, minimum 4: &quot;)&#xA;    if signalLengthU.isdigit():&#xA;        signalLength = signalLengthU&#xA;    else:&#xA;        signalLength = 4&#xA;&#xA;    delayU = raw_input(&quot;Input delay after signal in ms: &quot;)&#xA;    if delayU.isdigit():&#xA;        delay = delayU&#xA;    else:&#xA;        delay = 0&#xA;&#xA;    amplitudeU = raw_input(&quot;Input signal amplitude in mV, between 1 and 5000: &quot;)&#xA;    if amplitudeU.isdigit():&#xA;        amplitude = amplitudeU&#xA;    else:&#xA;        amplitude = 5000&#xA;&#xA;    #Calculate data points, delay, and amplitude&#xA;    data_points = int((1000*float(signalLength)-24.6418)/12.3291)&#xA;    signalDelay = float(delay)/1000&#xA;    setAmplitude = V_Ref/float(amplitude)&#xA;&#xA;    #Load and save CSV file&#xA;    datain = open('signal12bit.csv')&#xA;    read = csv.reader(datain, delimiter=' ', quotechar='|')&#xA;    signal = []&#xA;    for row in read:&#xA;        signal.append(ast.literal_eval(row[0]))&#xA;&#xA;    #Downsampling to achieve desired signal length&#xA;    downsampling = int(round(total_data_points/data_points))&#xA;    signalSpeed = signal[0::downsampling]&#xA;    listlen = len(signalSpeed)&#xA;&#xA;    #Construction of SPI bytes, to avoid calling functions in critical loop&#xA;    lowByte = []&#xA;    highByte = []&#xA;    for i in signalSpeed:&#xA;        lowByte.append(int(i/setAmplitude) &amp;amp; 0b11111111)&#xA;        highByte.append(((int(i/setAmplitude) &amp;gt;&amp;gt; 8) &amp;amp; 0xff) | 0b0 &amp;lt;&amp;lt; 7 | 0b0 &amp;lt;&amp;lt; 6 | 0b1 &amp;lt;&amp;lt; 5 | 0b1 &amp;lt;&amp;lt; 4)&#xA;&#xA;    print &quot;Starting Simulant with signal length %s ms, delay %s ms and amplitude %s mV.&quot; % (signalLength, delay, amplitude)&#xA;    print &quot;Press ctrl+c to stop.&quot;&#xA;    sleep (1)&#xA;&#xA;    try:&#xA;        while(True): #Main loop&#xA;            for i in range(listlen):&#xA;                spi.xfer2([highByte[i],lowByte[i]]) #Critical loop, no delay!&#xA;            sleep (signalDelay)&#xA;    except (KeyboardInterrupt, Exception) as e:&#xA;        print e&#xA;        print &quot;Closing SPI channel&quot;&#xA;        lowByte = 0 &amp;amp; 0b11111111&#xA;        highByte = ((0 &amp;gt;&amp;gt; 8) &amp;amp; 0xff) | 0b0 &amp;lt;&amp;lt; 7 | 0b0 &amp;lt;&amp;lt; 6 | 0b1 &amp;lt;&amp;lt; 5 | 0b1 &amp;lt;&amp;lt; 4&#xA;        spi.xfer2([highByte, lowByte])&#xA;        GPIO.cleanup()&#xA;        spi.close()&#xA;&#xA;if __name__ == '__main__':&#xA;    main()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The result is exactly what I wanted. Below is seen an example from the oscilloscope with a signal length of 5 ms; 200 Hz. Thanks for your help, guys!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZQNEU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZQNEU.png&quot; alt=&quot;Oscilloscope reading&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8614" LastActivityDate="2016-11-02T19:49:30.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12127" PostTypeId="2" ParentId="7033" CreationDate="2016-11-02T20:19:16.780" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If, for any reason, the fatigue process becomes time-dependent, then it also becomes frequency-dependent.  Under normal conditions, fatigue failure is independent of frequency.  But when corrosion or high temperatures, or both, are encountered, the cyclic rate becomes important.  The slower the frequency and the higher the temperature, the higher the crack propagation rate and the shorter the life at a given stress level (1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This quote is taken from &lt;em&gt;Mechanical Engineering Design&lt;/em&gt;, 7th Ed. in reference to metallic parts.  Please note - metallic fatigue stresses are taken at high cycle times.  The standard R.R. Moore rotational testing machine has capacities up to &lt;a href=&quot;http://www.asminternational.org/documents/10192/1849770/06156G_Sample.pdf/6202c5f3-9ae8-479b-a8bd-d2050d37d534&quot; rel=&quot;nofollow noreferrer&quot;&gt;10,000 rpm&lt;/a&gt;, and thus the heat generated from fast cycling is usually already present in the sample data.  Instead the quote states time dependency for fatigue,  and a failure of the Wöhler curve to recognize low frequency crack propagation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time dependency means external, time dependent factors that increase the stresses slowly over time.  For example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Corrosion removes thickness at a fixed rate with time.  A part that is designed for endurance limit stress levels placed in a corrosive environment will only handle a finite number of cycles before the corrosion reduces the load bearing cross-section to a level below the endurance limit.&lt;/li&gt;&#xA;&lt;li&gt;At elevated temperature, the ultimate strength is reduced.  A part that is near its endurance limit at room temperature could cycle into and out of high temperatures (for example, automobile parts that are stressed for a few hours per day).  In this case, the frequency of the loading will become a factor, as the number of cycles taken during the times of high temperature will count towards fatigue and thus reduce life at low temperature simultaneously.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A different model is used for low-frequency, high stress loading. (So severe that plastic deformation / crack propogation begins almost immediately) This method, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fatigue_(material)#Low-cycle_fatigue&quot; rel=&quot;nofollow noreferrer&quot;&gt;Manson-Coffin Relationship&lt;/a&gt;, attempts to find the remaining cycles left after a crack has begun.  If $\Delta \epsilon$ is the strain from loading, $\sigma^{'}_F$ is the &lt;a href=&quot;http://www.engineeringarchives.com/les_mom_truestresstruestrainengstressengstrain.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;true stress&lt;/a&gt; for first fracture, $\epsilon^{'}_F$ is the &lt;a href=&quot;http://www.engineeringarchives.com/les_mom_truestresstruestrainengstressengstrain.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;true strain&lt;/a&gt; for first fracture, then the estimated reversals (half-cycles) $(N)$ after first fracture can be predicted for all frequencies as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\Delta \epsilon}{2} &amp;gt; \frac{\sigma^{'}_F}{E}(2N)^b + \epsilon^{'}_F(2N)^c$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b and c are empirically determined coefficients.  b is in the -0.055 to -0.09 range, while c is in the -1.0 to -0.66 range.  More data can be gathered from (2) and (3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, while internal heat generation may very well be a key at high cycles speeds for plastics, metallic testing is taken at high frequencies already and the Wöhler curve actually presents problems at low frequencies, which is better covered by the Manson-Coffin model.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Mishke, Charles R., and Richard G. Budynas. &quot;Endurance Limit Modifying Factors, Cyclic Frequency.&quot; &lt;em&gt;Mechanical Engineering Design&lt;/em&gt;. By Joseph E. Shigley. 7th ed. New York: McGraw-Hill, 2004. 335. Print. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;J.F. Tavernelli and L.F. Coffin, Jr., &quot;Experimental Support for Generalized Equation Predicting Low Cycle Fatigue,&quot; and S. S. Manson, discussion, &lt;em&gt;Trans. ASME, J. Basic Eng.&lt;/em&gt; vol. 84, no. 4, pp. 533-537. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;N.E. Dowling's &lt;em&gt;Mechanical Behavior of Materials&lt;/em&gt;, 2nd ed., Prentice-Hall, Englewood Cliffs, N.J., 1999, Chap. 14.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2016-11-02T23:58:43.933" LastActivityDate="2016-11-02T23:58:43.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12129" PostTypeId="1" AcceptedAnswerId="12141" CreationDate="2016-11-03T03:45:22.140" Score="2" ViewCount="330" Body="&lt;p&gt;On the image below is a &lt;a href=&quot;http://www.kopemash.ru/products/2/76.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;mining machine produced&lt;/a&gt; by the Kopeysk Machine-Building Plant. I'm not sure how to call it in English. The Russian name is &quot;prokhodchesko-ochistnoy kombain&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multitran offers the translation &quot;&lt;a href=&quot;http://www.multitran.ru/c/m.exe?l1=2&amp;amp;l2=1&amp;amp;s=%EF%F0%EE%F5%EE%E4%F7%E5%F1%EA%EE-%EE%F7%E8%F1%F2%ED%EE%E9%20%EA%EE%EC%E1%E0%E9%ED&quot; rel=&quot;nofollow noreferrer&quot;&gt;heading and winning machine&lt;/a&gt;&quot;, but I'm not sure: when I enter this term in Google, it finds predominantly Russian-based websites. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine is used for underground mining. This particular model is mentioned in a text I'm translating about an overpricing row between the Russian potash producer Uralkali and the machine-building plant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wcnH8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wcnH8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;http://www.kopemash.ru/en/products/68/367.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the company's description&lt;/a&gt; of the machine:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The «Ural-20R-00» heading-and-winning machine is designed for longwall operation in сhambers and driving of oval-arch shaped workings in potash ore seams with thickness of 3.0-3.7 m at dip angles of ± 12° and rock resistance to cutting of up to Аp=450 H/mm. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="349" LastEditorUserId="6790" LastEditDate="2020-09-24T11:15:36.690" LastActivityDate="2020-09-24T11:15:36.690" Title="What would you call this mining vehicle? (Ural-20R, used in mining)" Tags="|terminology|mining-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12131" PostTypeId="2" ParentId="8976" CreationDate="2016-11-03T06:11:14.100" Score="2" Body="&lt;p&gt;The confusion arises in terminology.  What instrumentation engineers call a &lt;em&gt;gyroscope&lt;/em&gt; is actually an &lt;em&gt;integrating gyroscope&lt;/em&gt; or &lt;em&gt;rate gyro&lt;/em&gt;, and what they call a &lt;em&gt;gimbal&lt;/em&gt; in &lt;em&gt;gimbal lock&lt;/em&gt; isn't the gimbal used to mount the gyroscope itself - the gimbals are used to keep a &lt;a href=&quot;http://inertialnavigations.blogspot.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;platform housing three gyroscopes level&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.tpub.com/neets/book15/63a.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard gyroscope&lt;/a&gt; will point in the same direction at all times, while a &lt;a href=&quot;http://www.tpub.com/neets/book15/63e.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;rate gyroscope&lt;/a&gt; is used for measuring angular velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wzFNw.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wzFNw.gif&quot; alt=&quot;small and large tilts on a rate gyro&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate gyro will deflect the springs in proportion to how fast the frame is spinning - the faster the spin, the larger the springs deflect, which via strain gauge or similar measurements engineering, can be turned into an electrical signal.  Measure the change in rate to derive the angular acceleration.  Have a computer integrate the rate over time to derive current angular position.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the pivoting gimbals for this device though - they are in the exact configuration of a gimbal lock.  A rate gyro's housing can only measure changes in a single axis.  If the frame housing the gimbal moves away from the axis, then the device wouldn't measure properly.  So, three &lt;em&gt;rate gyroscopes&lt;/em&gt; are required to measure angular movement in all three angles.  These are mounted on a platform that maintains a fixed orientation in space, so the rate gyroscopes can function properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SzjeV.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SzjeV.gif&quot; alt=&quot;Gyroscopes on a platform.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the rate gyroscope feeds a porportional signal to the instrumentation panel, it also feeds this signal to three motors.  These motors are the &lt;em&gt;torque motors&lt;/em&gt; in the quote. They turn the platform in response to the signal, keeping the platform level.  When these gimbals keeping the platform level lock, then there is the problem, and the goal is trying to keep the feedback loop away from turning the gimbals until they lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plan to turn the pitch gimbal 180 degrees would essentially turn the platform upside down, which would revese two of the three accelerometers - and all of the rate gyros.  Specifically, the pitch gyro would now be spinning in opposite orientation after the platform was upside down (counterclockwise instead of clockwise).  This would mean that it reads a 0.05 rpm counterclockwise turn as instead a 0.05 rpm clockwise turn - and the feedback loop to the motors would similarly be reversed, pulling the platform out of gimbal lock.  This could be compared to jumping over the discontinuity at 90 degrees for the tangent function and continuing to perform the integration - in theory.  In practice, the system never worked.  &lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2016-11-03T06:36:23.057" LastActivityDate="2016-11-03T06:36:23.057" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12134" PostTypeId="1" AcceptedAnswerId="14856" CreationDate="2016-11-03T12:44:56.823" Score="1" ViewCount="534" Body="&lt;p&gt;The gas $(A)$ is diffusing into the solid sphere $(B)$, The sphere phase is assumed to be stagnant so we have no bulk movement, $(A)$ is being consumed in the sphere by a rate of $KC_A$ which $K$ is a kinetic constant and $C_A$ is concentration of $A$ in the sphere, $C_A$ in sphere as a function of $r$ (radius) is desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the species continuity Eqn. we have :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{r^2} \frac{d}{dr}(r^2N_A)=-KC_A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By definition of $N_A$ we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_A=y_AN_t-D_{AB}\frac{dC_A}{dr}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which $N_t$ is zero by the stagnant assumption. By substituting $N_A$ in the first Equation and differentiating we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_A''+\frac{2}{r}C_A'-\frac{K}{D_{AB}}C_A$=0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just don't know how to solve this differential equation, maybe there's a need for a simplifier assumption that I didn't make, anyway I welcome any advice from you guys helping me find the answer.&lt;/p&gt;&#xA;" OwnerUserId="5452" LastEditorUserId="5452" LastEditDate="2016-11-03T15:21:01.467" LastActivityDate="2017-04-19T13:27:42.160" Title="Molecular diffusion into sphere" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12135" PostTypeId="2" ParentId="12129" CreationDate="2016-11-03T13:28:19.397" Score="3" Body="&lt;p&gt;Generally speaking, I would say it looks like a &quot;boring machine,&quot; but you might clarify that it's a &quot;coal&quot; boring machine or a &quot;potash&quot; boring machine. Specifically, it looks like some kind of a hybrid between a &lt;a href=&quot;https://www.google.com/search?q=Longwall+Miner&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;longwall miner&lt;/a&gt; and a &lt;a href=&quot;https://www.google.com/search?q=Continuous+Miners&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;continuous miner&lt;/a&gt;, the longwall having a toothed drum whose face is in-line with the direction of the track and the continuous miner having a toothed drum perpendicular to the track.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-11-03T13:28:19.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12136" PostTypeId="2" ParentId="12116" CreationDate="2016-11-03T14:11:32.680" Score="1" Body="&lt;p&gt;You can't really calculate the G-force for two rigid objects, as the collision is nearly a delta function.  If you want to simulate the force on, say, the driver inside a car, then you have to model the rate of deceleration as the front of the car collapses -- which is strongly dependent on the total vehicle design.  As an historical note, Volvo back in the 60s and 70s ran ads showing their car being dropped nose-first off a roof, and retaining the basic frame shape after hitting the ground.  That's great for rigidity, but then everyone realized this just transferred the peak impulse deceleration straight to the occupants. After that, cars' designs were modified to dissipate as much energy as possible by completely destroying the front end up to the firewall. This greatly reduces the peak G-force on the passenger compartment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, back to your block-to-wall experiment: unless your block is collapsible, you're going to be doing the 1970s Volvo thing.  You may want to modify your experiment in light of this.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-11-03T14:11:32.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12138" PostTypeId="1" CreationDate="2016-11-03T14:51:40.323" Score="1" ViewCount="492" Body="&lt;p&gt;I'm working on project and after a good amount of research I've ended up with a final design that looks roughly like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/OSCZW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OSCZW.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The design is inspired by load cells used in bathroom scales, but I found them to be too clunky for my application so I have to make a custom set of load cells. Each T-shaped arm will have a strain gauge on top of the section that gets narrower where hopefully most of the strain will be. I've kept the 8 load cells on one piece of metal to gain rigidity.&#xA;On top of this plate I will mount four 80x20mm bars of wood So I can measure four different forces at the same time and also the where the force is being applied on the bar.&lt;/p&gt;&#xA;&lt;p&gt;My main obstacle is to calculate the optimal thickness of the metal, width of the t-shaped arm and find the right strain gauge so I can measure weights up to 100kg per load cell without breaking. Lets say I use steel or stainless steel for this. Should I use a strain gauge with 1000ohm?&lt;/p&gt;&#xA;&lt;p&gt;Updated: &lt;a href=&quot;https://i.stack.imgur.com/UkUcq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UkUcq.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This whole part will be bolted down flat on a thick wooden surface and will be used to measure weight distribution on my fingers when I do climbing related excercises like pullups. The part is mounted completely horizontal.&lt;/p&gt;&#xA;" OwnerUserId="7467" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-01-02T18:21:18.033" Title="Selecting the correct strain gauge and metal thickness for custom load cell" Tags="|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12139" PostTypeId="2" ParentId="12138" CreationDate="2016-11-03T15:32:26.870" Score="1" Body="&lt;p&gt;Mechanical designer here - how do you get an inside cut on a sharp angle like that? What tool makes perfect corners? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You said, &quot;On top of this plate I will mount four... bars of wood.&quot; So, if the bars bolt to the top, how does this bolt to anything else? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You said, &quot;I've kept the 8 load cells on one piece of metal to gain rigidity,&quot; and, &quot;...I can measure four different forces at the same time.&quot; You get rigidity by interactions between the individual members - this means one &quot;load cell&quot; will skew the others because they're all coupled together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start out asking about the metal thickness and end up asking if you should use a 1$\mbox{ k}\Omega$ strain gauge. All of this leads me to ask (no offense intended) - &lt;em&gt;do you understand how a strain gauge works?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strain gauge measures &lt;em&gt;strain&lt;/em&gt; - elongation/deformation in the material. The change in resistance of the strain gauge is proportional to the amount of deformation in the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you need to determine how much force you are expecting to experience. Then, via part geometry, determine how much stress will exist in the part at a particular region. Then, via part material, determine how much strain you expect at the region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Gauge_factor&quot; rel=&quot;nofollow noreferrer&quot;&gt;gauge factor&lt;/a&gt; with the expected strain and strain gauge resistance to determine the anticipated range of resistances. Setup your &lt;a href=&quot;https://en.wikipedia.org/wiki/Wheatstone_bridge&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wheatstone bridge&lt;/a&gt; to control the expected voltage swing based on the expected resistance swing and/or use an op-amp to buffer/scale the output voltage. Read the output with an analog to digital converter that has a resolution capable of meeting your project needs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing to note about the strain gauge is that it only measures the strain in the member to which it is attached. If you want to measure &lt;strong&gt;all&lt;/strong&gt; of the applied force, then you need to make sure that &lt;strong&gt;all&lt;/strong&gt; of the applied force gets transmitted through the part to which the gauge is attached, &lt;strong&gt;and&lt;/strong&gt; that it is all applied in a &lt;em&gt;direction&lt;/em&gt; that the gauge is capable of reading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How accurate are you trying to be? Where will this be setup? Have you taken temperature effects into account? As metal changes temperature it expands and contracts - this will register as a &quot;phantom&quot; stress. Is the member to which the strain gauge attached experiencing &lt;em&gt;pure&lt;/em&gt; tension/compression, or &lt;em&gt;pure&lt;/em&gt; bending, or a combination?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now you don't provide any information about how this will be used beyond being attached to wood and subjected to a 100kg load. Can you provide a drawing of where you're planning to attach the load cell? Of how the wood attaches? Of how the assembly mounts to a structure? Is the load distributed or a point load? Cantilevered or centered? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without more information about the nature of the load and how your assembly goes together it's not really possible to comment on how the part should be sized. I will say, though, that you would want the strain gauge in an area of uniform stress, which means that you should consider an &lt;a href=&quot;https://en.wikipedia.org/wiki/Load_cell#/media/File:ZugDruckstab_WZ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;area of uniform cross-section&lt;/a&gt;, which doesn't really appear to exist anywhere in your drawing.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-11-03T15:32:26.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12141" PostTypeId="2" ParentId="12129" CreationDate="2016-11-03T15:59:20.280" Score="4" Body="&lt;p&gt;The machine is nothing like a longwall miner. Longwall mining machines are limited in how they operate. A wall of shields protects the face being cut while a shearing head runs along the face cutting coal, dropping coal onto a short conveyor belt running parallel to the face. That conveyor belt then drops the coal onto another conveyor belt running perpendicular to the face being cut. As the longwall face is cut, the shields slowly advance forward. Also, longwall miners cannot produce curved tops to excavation; they can only produce flat topped excavations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine in the question is more like a &lt;a href=&quot;http://alpinecutter.com/en/products/roadheaders.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;road header&lt;/a&gt;. They are used to &lt;a href=&quot;https://www.linkedin.com/pulse/sany-developed-ebs630-roadheader-potash-mining-%E6%89%AC-%E6%9D%8E&quot; rel=&quot;nofollow noreferrer&quot;&gt;mine postash&lt;/a&gt; and other soft stratified deposits like salt. If needed, they can produce excavations with curved profiles. The machine in question may have purposely been designed &amp;amp; manufactured to mine a particular type of potash.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2016-11-04T06:16:58.070" LastActivityDate="2016-11-04T06:16:58.070" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12142" PostTypeId="1" CreationDate="2016-11-03T20:27:40.363" Score="1" ViewCount="1108" Body="&lt;p&gt;I have two abs pieces that I would like to create a waterproof seal between via compressive force via fasteners like screws. How much do o-ring/gasket compress? Certainly it is dependent on the material and how much pressure is applied, but are there any resources on this? There is a pocket on one of the abs pieces that will be mating with the other piece on the bottom with very tight tolerances, as I plan to put something in the pocket and seal it with the second abs piece. &lt;/p&gt;&#xA;" OwnerUserId="8644" LastActivityDate="2017-03-06T10:49:40.440" Title="o-ring/gasket compressibility" Tags="|seals|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12143" PostTypeId="1" AcceptedAnswerId="12144" CreationDate="2016-11-04T01:42:04.383" Score="4" ViewCount="2388" Body="&lt;p&gt;Given a 3.25&quot; long, 0.125&quot; diameter rubber rod with tensile strength 1200 psi supported on both ends, how much weight can you hung from the middle of it without breaking it? (This is not homework I am trying to design a PC case and I have no idea what I am doing.) How does this change if we use a 0.25&quot; diameter rod instead?&lt;/p&gt;&#xA;" OwnerUserId="8645" LastEditorUserId="2015" LastEditDate="2016-11-04T04:39:00.993" LastActivityDate="2016-11-05T14:13:17.607" Title="Given the tensile strength of a rod, how do you calculate the max weight that can be hung from it?" Tags="|mechanical-engineering|structural-engineering|structures|beam|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12144" PostTypeId="2" ParentId="12143" CreationDate="2016-11-04T02:31:35.920" Score="7" Body="&lt;p&gt;You're hanging it from a rod supported by both ends - and need to use the &lt;a href=&quot;http://www-classes.usc.edu/engr/ce/457/moment_table.pdf&quot;&gt;bending equations&lt;/a&gt;.  For this case (case 7 in the link), the max weight is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W_{max} = \frac{\pi d^3 \sigma}{8L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma$ is the tensile stress, $L$ is the rod length, $d$ is the rod diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sag is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta = \frac{4W_{max}L^3}{3E \pi d^4}$$ where $W_{max}$ is the weight, $L$ is the rod length, $d$ is the rod diameter, and $E$ is a mechanical property called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Young%27s_modulus&quot;&gt;Youngs modulus&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2016-11-04T04:38:06.790" LastActivityDate="2016-11-04T04:38:06.790" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12147" PostTypeId="1" AcceptedAnswerId="12150" CreationDate="2016-11-04T10:52:22.373" Score="6" ViewCount="1710" Body="&lt;p&gt;There is this device I saw when I was a child at a big music store, that makes me curious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was meant to listen some CDs to try them out before the purchase.&#xA;You stand under a bell-shape kind of &quot;speaker&quot; to hear the music, but as soon as you step out you aren't able to hear anything anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been searching the net, but probably I haven't used the appropriate keywords because I didn't find anything.&#xA;What's this device called, and how is does it work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in particular about the sound and how it's possible to transmit it only in a very limited area, and being able to hear it well only inside that little &quot;circle&quot;&lt;/p&gt;&#xA;" OwnerUserId="8651" LastActivityDate="2016-11-04T13:03:20.563" Title="What device is this, and how does it work?" Tags="|power-electronics|sound-isolation|waves|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12148" PostTypeId="2" ParentId="12147" CreationDate="2016-11-04T11:15:00.000" Score="8" Body="&lt;p&gt;With a description as vague as that, it was probably a parabolic reflector with a speaker at its focus. Pointed straight down, the &quot;beam&quot; of acoustic energy would be reflected from the floor right back into the &quot;bell&quot;, with remarkably little leakage out the sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reflector with a diameter of 4-5 feet would be directional for frequencies down in the hundreds of Hz, which would be fine for auditioning music. They probably rolled off the bass response deliberately in order to prevent problems from the lack of directionality at those frequencies.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-11-04T11:15:00.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12150" PostTypeId="2" ParentId="12147" CreationDate="2016-11-04T13:03:20.563" Score="9" Body="&lt;p&gt;Something like this? &#xA;&lt;a href=&quot;http://www.browninnovations.com/sound-dome/single-localizer-sound-dome&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.browninnovations.com/sound-dome/single-localizer-sound-dome&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As others have said, it is a speaker with a parabolic reflector to direct the sound down at the listener. You can make your own fairly easily:  &lt;a href=&quot;http://www.instructables.com/id/umbrelAudio-Sound-Dome-Umbrella/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.instructables.com/id/umbrelAudio-Sound-Dome-Umbrella/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-11-04T13:03:20.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12151" PostTypeId="2" ParentId="12143" CreationDate="2016-11-04T19:37:53.143" Score="2" Body="&lt;p&gt;I searched and could not find any rubber material which has the same modulus of elasticity for compression and tension. Also rubbers swell under stress and creep( plastic behavior over time).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question does not have an analytical answer. However as a hypothetical case and just for illustration one can try to stablish a range of I, (second moment of area) by testing a short sample of the material under different compression stresses and set a lose curve fitting graph for the I by stablishing a new neutral axis and integrating y^2.da separately for top and bottom parts over and below the axis. And go from there.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2016-11-05T14:13:17.607" LastActivityDate="2016-11-05T14:13:17.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12153" PostTypeId="2" ParentId="12111" CreationDate="2016-11-04T21:01:37.930" Score="0" Body="&lt;p&gt;Your question is very general and it is difficult to give specific answers, but I will try to offer some pointers.  For example I cannot determine whether you are interested in &lt;em&gt;time to failure&lt;/em&gt;, &lt;em&gt;general system reliability&lt;/em&gt; or &lt;em&gt;both&lt;/em&gt;.  Also you indicated that the failure could be driven by a number of factors, such as &lt;em&gt;weather&lt;/em&gt;, &lt;em&gt;accidents&lt;/em&gt; or &lt;em&gt;interferences&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of these potential causes of failure is different.  For example, &lt;em&gt;weather&lt;/em&gt;, as a natural hazard, may be considered as time dependent - being related to the risk of a storms over a period of time.  The others could simply be due to random or unpredictable causes; therefore each of these factors may require a different probability density form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the first step is to determine the form of mass or density function that will be used to model the failure of each sensor.  For time to failure, a density function often used is the &lt;em&gt;Hazard&lt;/em&gt; function or one of its derivatives.  In simple terms the &lt;em&gt;Hazard&lt;/em&gt; function is also called a &lt;em&gt;bathtub&lt;/em&gt; function as the failure risks are typically much higher in early-life, decrease in mid-life and increase again at end-of-life. Human life can be modelled with this form of a risk profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manufacturer may offer useful data as the &lt;em&gt;Hazard&lt;/em&gt; function is commonly used in product reliability assessment.  For example, the risk of failure of light bulbs is modelled well with a &lt;em&gt;Hazard&lt;/em&gt; function. In your case, a paper illustrating the application of the &lt;em&gt;Hazard&lt;/em&gt; function to natural hazards is found &lt;a href=&quot;http://engineering.tufts.edu/cee/people/vogel/documents/2016_hazardFunctionTheory.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proper assessment of the risk of flooding due to &lt;em&gt;weather&lt;/em&gt; requires consideration of the rainfall intensity associated with a, say, 100-year storm, for example, and the associated flooding levels from the characteristics of the watershed and it's unit hydrograph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consideration of the forms associated with &lt;em&gt;accidents&lt;/em&gt; or &lt;em&gt;interferences&lt;/em&gt; and how they may be combined is important.  If the events are statistically independent then the overall result may be computed as a product of the probabilities.  Otherwise dependency effects must be considered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to assess the reliability of the entire system - this is again another rather large subject.  I have set out several alternative techniques in &lt;a href=&quot;https://engineering.stackexchange.com/a/5867/3266&quot;&gt;this thread&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="3266" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-11-04T21:08:02.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12155" PostTypeId="1" CreationDate="2016-11-05T02:26:12.637" Score="3" ViewCount="197" Body="&lt;p&gt;I am an engineering student who is working on a project. One of the things that will make the project a success is I need to know how to join a very thin type 304 stainless steel wire loop onto a slightly bigger high carbon steel pole in a very efficient way and can do it in an industrial scale. I know high carbon steel has low weldability and it is the interest of this project that the strength of the carbon steel and the ductility of the stainless steel(its softness) are not undermined.The mechanical shop in my school doesn't know anyone that is involved with these processes. So could you guys help me out here?&lt;/p&gt;&#xA;" OwnerUserId="8659" LastEditorUserId="9" LastEditDate="2018-10-27T15:18:15.797" LastActivityDate="2018-10-27T15:18:15.797" Title="How can high carbon steel and stainless steel be connected firmly without using nuts and bolts?" Tags="|materials|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="12157" PostTypeId="1" AcceptedAnswerId="12207" CreationDate="2016-11-05T02:46:39.877" Score="0" ViewCount="94" Body="&lt;p&gt;So I have the following circuit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8VzpZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8VzpZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the first question is, if DV1 is pushed all the way to the right, what speed will the clamp cylinder attain and as it extends what will be the pressure at P1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly Once the clamp cylinder has become fully extended, if the button on DV2 is pressed, what will be the speed of the piston of the lift cylinder and the pressure at P1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specifications are as such:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Clamp cylinder -&gt; Bore = 20 mm diameter, rod of 15mm diameter and a stroke of 1m.&lt;/li&gt;&#xA;&lt;li&gt;Lift Cylinder -&gt; Bore = 200mm diameter, rod = 15mm diameter, stroke = 1m&lt;/li&gt;&#xA;&lt;li&gt;Clamp cylinder has a resistive force of 2.5kN that opposes motion&lt;/li&gt;&#xA;&lt;li&gt;Lift Cylinder moves a mass of 50kg and no resistive forces&lt;/li&gt;&#xA;&lt;li&gt;The relief valve is set to 20 MPa&lt;/li&gt;&#xA;&lt;li&gt;Pressure valve, PV2 is set to 10MPa&lt;/li&gt;&#xA;&lt;li&gt;Pump -&gt; 5mL/rev and 1440 RPM&lt;/li&gt;&#xA;&lt;li&gt;no pressure losses in line or components or internal leakage&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So my thoughts are;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pump is pumping at 5 *1440 = 7200mL/minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to just ignore PV1 as it's just a pressure relief valve, so it will just limit the system to whatever it is set to (20MPa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think we can ignore PV2 and everything beyond for the first question too. Or since this flow goes back to tank, I think it may mean that the system can't be more than 10MPa?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I think we calculate the Area of the head of the cylinder which is A = (0.2/2)^2 * pi = 0.0314 m^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from here I think we can just put it into the formula Vext = Qp / A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 7.2*10^-3 / 3.14*10^-2 = 0.2293 m/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct, or do I have to subtract some flow that would go to the other parts of the system (towards flow 2 and PV2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the pressure at head should be P = F/A. This is my main problem, as 2.5*10^3/0.0314 = 80MPa which is way higher than any of the possible answer choices. Not to mention higher than what PV1 is set for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where have I gone wrong and how can I solve these problems?&lt;/p&gt;&#xA;" OwnerUserId="8155" LastActivityDate="2016-11-08T00:31:08.270" Title="Calculating speed, forces and pressures when extending a cylinder" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12158" PostTypeId="1" CreationDate="2016-11-05T03:12:38.173" Score="0" ViewCount="379" Body="&lt;p&gt;I am working on a project that need to redesign a sewing needle. Because I am planning to use a stainless steel loop to replace a needle's eye, using high carbon steel, which make all modern needles, would not be industrially profitable as it is hard to weld and might take a lot of time. I am wondering is there other material that can replace high carbon steel as the raw material for a needle? Or, is there other materials soft enough to be bented into any shape and can be joint with high carbon steel or whatever material that can replace it? Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="8659" LastActivityDate="2016-11-05T13:16:49.730" Title="what can replace high carbon steel as the raw material for sewing needle" Tags="|materials|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12160" PostTypeId="1" CreationDate="2016-11-05T08:15:46.880" Score="1" ViewCount="2377" Body="&lt;p&gt;I m going to find the shear stress at section n-n . &#xA;So , shear stress in beam has formula of &amp;tau; = V(A)(y) / It, where y = distance from the centroid of particular area to neutral axis &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get the correct $I$ (second moment of inertia), but I'm having problem of finding the shear stress at $A$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V$ (shear force) is a constant 40&amp;nbsp;kN from 0&amp;nbsp;m to 0.4&amp;nbsp;m (in my SFD diagram).&#xA;For $A_y$, I have $100\times10^{-3}\cdot12\times10^{-3}\cdot(94-50)\times10^{-3} = 5.28\times10^{-5}\text{ mm}^4$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my Area (green coloured part) correct ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so $\tau = \dfrac{40\times10^3\cdot5.28\times10^{-5}}{5553152 \cdot 12\times10^{-3}} = 3.81\times10^{-6}$, but the given ans is $15.85\times10^6$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the section n-n mean when we cut off the beam at n-n , the side view of n-n look like in the diagram?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kXX7L.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HKTDH.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: sorry , it should be &amp;tau; = V(A)(y) / It , where y = distance from the centorid of particular area to neutral axis &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;strong&gt;&lt;em&gt;How  to  get  the 21.05 Mpa ?&lt;/em&gt;&lt;/strong&gt;  I only  get  the  15.8Mpa  after  correction.... $$\tau$$ = (40x10^3)(5.28x10^-5)/ [(5553152x10^-12)(24x10^-3)] = (15.8x10^6)MPa&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="8664" LastEditDate="2016-11-05T23:35:07.767" LastActivityDate="2016-11-05T23:35:07.767" Title="shear stress in hollow beam" Tags="|structural-engineering|" AnswerCount="0" CommentCount="4" ClosedDate="2016-11-14T17:23:17.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="12162" PostTypeId="2" ParentId="12158" CreationDate="2016-11-05T13:16:49.730" Score="1" Body="&lt;p&gt;There are a large number of stainless steels which are hardenable and can be used as a more or less direct replacement for high carbon steel. Common grades like 304 and 316 would be possible candidates and 440 is widely used for cutting tools. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Needles are typically made by drawing form wire stock and punching the eye so welding or other jointing processes are not required unless there is a very specific reason why you want it made from two separate materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth mentioning that without more details of what you are planning to achieve it is hard to give more specific recommendations. Obviously needles have been around for a very long time and as such there is limited scope for improvement unless you have a new and different application. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-11-05T13:16:49.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12163" PostTypeId="2" ParentId="12142" CreationDate="2016-11-05T13:32:53.530" Score="0" Body="&lt;p&gt;Here is a GENERAL guideline on O-Ring &lt;em&gt;Gland&lt;/em&gt; design that you are describing:&#xA;&lt;a href=&quot;http://allorings.com/gland_static_axial.htm&quot; rel=&quot;nofollow&quot; title=&quot;face seal gland design&quot;&gt;Oring Face Seal Guide&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several resources online(google search) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure, size, fluid type, serviceability, safety(redundancy) etc will influence your design. If you have more details, we could provide a more detailed answer, but this should get you started. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2016-11-05T13:32:53.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12165" PostTypeId="1" AcceptedAnswerId="12167" CreationDate="2016-11-05T14:28:07.443" Score="3" ViewCount="2171" Body="&lt;p&gt;In this question my bending moment diagram (BMD) doesnt return to 0, what's wrong with it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the bending stress $\sigma = My / I$ , so i have to get the M (maximum moment) first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the maximum bending moment, I have to get it from the graph of bending moment. To get bending moment, i have to get the shear force from the shear force diagram (SFD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/81GVb.jpg&quot; alt=&quot;&quot;&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pCsqB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pCsqB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="255" LastEditDate="2016-11-07T12:44:07.723" LastActivityDate="2019-12-28T18:10:28.010" Title="Bending moment diagram from shear force diagram" Tags="|structural-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12167" PostTypeId="2" ParentId="12165" CreationDate="2016-11-05T14:49:20.483" Score="3" Body="&lt;p&gt;Your shear diagram is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bending moment is equal to the integral of shear. Therefore, the point of maximum bending moment is the point of zero shear. I can see you calculated that point as 4.38, which is correct (in fact, it's 4.375, but rounding is fine). That is found by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$6\cdot\dfrac{175}{175+65} = 4.375\text{ m}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you know the point of maximum bending moment. You chose to calculate the value by integrating the shear diagram. That is equivalent to calculating the total area of the shear diagram up to that point, which is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$175\cdot\dfrac{4.375}{2} = 382.8\text{ kNm}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your mistake was that you calculated the area of the rectangle &lt;span class=&quot;math-container&quot;&gt;$(175\cdot4.375$&lt;/span&gt;) instead of the triangle that describes the shear diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adopting the correct bending moment value, you can then correct the rest of it as well:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;M_B &amp;amp;= 382.8 - 65\cdot\dfrac{6-4.375}{2} = 330.0\text{ kNm} \\&#xA;M_C &amp;amp;= 330.0 - \dfrac{125+205}{2}\cdot2 = 0\text{ kNm}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now, to check our work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K0W8C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K0W8C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Diagram obtained with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis program.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:10:28.010" LastActivityDate="2019-12-28T18:10:28.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12170" PostTypeId="1" CreationDate="2016-11-05T16:13:33.643" Score="0" ViewCount="1952" Body="&lt;p&gt;For this question, I am having problem for part b. We know that the bending stress $$ \sigma = My / I $$&#xA;For part (b), the author only stated he wanted the bending stress at a section 2&amp;nbsp;m from A, but he didnt specify at which point of the beam (cross-sectional view)?&#xA;So, how to do this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ChWYT.jpg&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Aavtv.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="255" LastEditDate="2016-11-07T14:26:07.043" LastActivityDate="2016-11-07T14:30:38.477" Title="Maximum bending stress at particular section of beam" Tags="|structural-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12171" PostTypeId="2" ParentId="6909" CreationDate="2016-11-05T17:19:48.303" Score="1" Body="&lt;p&gt;Anderson PowerPole connectors are used in high vibration critical systems to make reliable connections. At the board use a ring terminal and screw with a Nylock locknut. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always solder the cables (no crimp) and use Silicone high strand count wires, the type used by radio control model battery power systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strain relief may be needed if there is a large differential movement between parts - otherwise leave the cables over-length and allow them to flex naturally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arbutus&lt;/p&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="7021" LastEditDate="2016-11-05T17:30:26.753" LastActivityDate="2016-11-05T17:30:26.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12172" PostTypeId="2" ParentId="12170" CreationDate="2016-11-05T17:36:31.353" Score="0" Body="&lt;p&gt;You have all of the necessary information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the BM diagram to determine the magnitude and direction of the moment. The question is asking for the maximum stress in the section. (Hint S=0 at NA).&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2016-11-05T17:36:31.353" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12173" PostTypeId="2" ParentId="12155" CreationDate="2016-11-05T19:52:50.380" Score="2" Body="&lt;p&gt;Taking the question in isolation resistance welding and forge welding are potential techniques for joining dissimilar materials in mass production and for very small items vacuum brazing is often attractive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that if your main consideration is to achieve a malleable eye  then joining two materials may not be the best option. Firstly stainless steel isn't especially ductile compared to carbon steels and if anything it tends to be more difficult to cold work ( I speak from experience here). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably a better solution is to deferentially harden the needle as a plain carbon steel can provide both high strength and hardness and reasonable ductility depending o how it is worked and heat treated and it is certainly possible to have a needle which is hard at the point and soft at the eye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple way to achieve this is to start with annealed stock (which you would anyway as needles are generally drawn to shape) and either heat the whole thing and just quench the point or just heat the point and quench the while thing. either approach is reasonably viable for mass production. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also consider that different steel alloys have variable hardenability, ineed the &lt;a href=&quot;http://www.industrialheating.com/articles/92224-understanding-the-jominy-end-quench-test&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard test of hardenability&lt;/a&gt; is quite close to what you want to achive. So you might want to focus some of your research efort on selecting a steel grade with hardening cahractersistics which best suits this apllication. With care you shodl be able to arrive at a method which goves you a very hard point, a tough shaft and a ductile eye. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also add that even if your design concept is a bit flaky (which it is) this sort of research will still make the project worthwhile and stand you in good stead for your future career as you will learn much about steel alloys and their processing and heat treating. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-11-05T19:52:50.380" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12174" PostTypeId="1" CreationDate="2016-11-05T21:51:25.377" Score="2" ViewCount="566" Body="&lt;p&gt;I'm new to this site so I apologize if I'm asking a previously answered question, but I couldn't find anything anywhere. I was trying to derive the pressure coefficient over a cylinder without using inviscid potential flow derivations. So for the usual potential flow equations, we obtain, for flow on the surface of the cylinder,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at $r = R, V_{\theta} = -U(1+R^2/r^2)\sin \theta = -2U\sin \theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where U is the free flow velocity. After plugging into Bernoulli's equation and rearranging,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C_p = 1 - 4\sin^2\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, let's assume we don't want to use potential flow derivations, but just resolve the free flow vector into $V_r$ and $V_{\theta}$. In that case, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_r = U\cos \theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_{\theta} = -U\sin \theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where does the discrepancy come from? Why does the potential flow solution give double that of a resolved vector? Where does the extra $R^2/r^2$ term come into play when we try to resolve it directly?&lt;/p&gt;&#xA;" OwnerUserId="8672" LastEditorUserId="1832" LastEditDate="2017-04-18T12:37:45.577" LastActivityDate="2018-03-14T17:21:17.353" Title="Pressure coefficient over a cylinder without using potential flow equations" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12175" PostTypeId="1" AcceptedAnswerId="12195" CreationDate="2016-11-05T23:48:20.657" Score="0" ViewCount="136" Body="&lt;p&gt;In &lt;a href=&quot;https://i.stack.imgur.com/AxBsb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this example&lt;/a&gt;, I suspect the selected A' (area) is wrong... From &lt;a href=&quot;https://i.stack.imgur.com/bRnF0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;the notes&lt;/a&gt;, A' is the &quot;portion of the member's cross sectional area above the section where t is measured&quot;. But, in the example, we could see that the selected area is located to the right of the point where shear stress is calculated. Are the notes wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I believe the selected A' should look like &lt;a href=&quot;https://i.stack.imgur.com/AfHQe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt; (refer to green shaded part). Am I right?&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="8664" LastEditDate="2016-11-08T05:04:44.040" LastActivityDate="2016-11-08T05:04:44.040" Title="shear stress at particular point in beam" Tags="|structural-engineering|" AnswerCount="1" CommentCount="5" ClosedDate="2016-11-07T21:56:28.650" ContentLicense="CC BY-SA 3.0" />
  <row Id="12176" PostTypeId="1" AcceptedAnswerId="12242" CreationDate="2016-11-06T01:21:10.557" Score="2" ViewCount="6962" Body="&lt;p&gt;For a isosceles triangle with base &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; and height &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;, the surface moment of inertia around the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-axis is &lt;span class=&quot;math-container&quot;&gt;$\frac{bh^3}{36}$&lt;/span&gt; (considering that our coordinate system has &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; in the horizontal and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; in the vertical axis, and has its origin on the triangle's center of mass (which is at &lt;span class=&quot;math-container&quot;&gt;$\left\{\frac{b}{2},-\frac{h}{3}\right\} $&lt;/span&gt; if you put your coordinate system's origin at the bottom left corner if the triangle).&lt;/p&gt;&#xA;&lt;p&gt;I know that the formula for the moment of inertia around the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-axis is &lt;span class=&quot;math-container&quot;&gt;$I_z = \int_A{y^2 dA}$&lt;/span&gt;, but I cannot figure out how to derive the formula from that. How is it done?&lt;/p&gt;&#xA;" OwnerUserId="8674" LastEditorUserId="90" LastEditDate="2023-11-30T09:13:11.117" LastActivityDate="2023-11-30T09:13:11.117" Title="Derivation of the formula for the surface moment of inertia of an isosceles triangle" Tags="|structural-engineering|applied-mechanics|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12179" PostTypeId="1" CreationDate="2016-11-06T06:26:49.037" Score="2" ViewCount="648" Body="&lt;p&gt;I've been thinking about hybrid drivetrains. One thing that is bugging me is operation when one motor is off while the other is on. In a scenario in which the motors are connected with all the rotors on the same axis, I can easily see how the velocities would add up. But if one motor was turned off, wouldn't the motor just cause the other one to spin in reverse and not put out much power to the shaft (assuming there is a large load on the shaft)?  The whole equal and opposite thing. If the load is much bigger than the drag of the motor running now backward, then all your effort would be going into spinning the off motor backward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the motors were hooked up in parallel through some gear and one was turned off, I can see how the power output would be decreased by at least half. The single motor that is on is losing power because of back EMF from the off motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the only way to overcome this is through some sort of clutch? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bfpzJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bfpzJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8676" LastEditorUserId="8676" LastEditDate="2016-11-06T17:30:30.070" LastActivityDate="2016-11-07T09:15:30.153" Title="2 motors on same shaft but 1 is off. How does that work?" Tags="|mechanical-engineering|design|motors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12184" PostTypeId="1" CreationDate="2016-11-06T14:17:36.253" Score="8" ViewCount="4855" Body="&lt;p&gt;I keep reading all over the internet that a &lt;a href=&quot;https://en.wikipedia.org/wiki/Torque_converter&quot; rel=&quot;noreferrer&quot;&gt;torque converter&lt;/a&gt; &lt;strong&gt;multiplies torque&lt;/strong&gt; at stall speed. Every explanation I have found sounded more like hand-waving than quantitative physical explanation. Essentially, the argument goes that the stator redirects flow which makes the torque converter &quot;more efficient&quot;, thereby multiplying torque. This would not qualify as a demonstration to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand it, the efficient slippage a torque converter allows lets the engine revolve faster and that alone can produce more torque due to the internal combustion engine output torque curve. But this is probably not the whole story: people and manufacturers insist that a torque converter does multiply torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give a quantitative explanation of how torque can get multiplied in this context ? I assume this is physically possible since only energy conservation must be met in such a system. I expect the explanation to resemble somewhat the operation of a boost converter that can increase voltage to the expense of current.&lt;/p&gt;&#xA;" OwnerUserId="8678" LastActivityDate="2022-06-28T03:55:56.030" Title="Torque converter torque multiplication" Tags="|mechanical-engineering|fluid-mechanics|automotive-engineering|torque|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12186" PostTypeId="1" CreationDate="2016-11-06T22:39:59.373" Score="2" ViewCount="106" Body="&lt;p&gt;I have an N-scale model train (N-scale is about half the size of HO scale). It's so small I could set up an entire layout on something the size of a coffee table. Put a glass top over it and I could use it as a coffee table. Problem is that people bump into coffee tables and whenever they do, the train will jump off the track, and I'll have to take the lid back off in order to put the train back on the track. So I'm wondering if I could make a shockproof mounting for this small model train layout.&lt;/p&gt;&#xA;" OwnerUserId="8682" LastActivityDate="2016-12-08T13:26:19.323" Title="I want a shockproof table for a model train. That is, protection from physical jolts, not electrical volts" Tags="|vibration|shock|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12187" PostTypeId="1" CreationDate="2016-11-07T01:42:30.990" Score="1" ViewCount="155" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Flettner_rotor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flettner rotor&lt;/a&gt; can be used to provide propulsion for ship from the wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it make sense to fit a Flettner rotor to a full displacement expedition trawler in the 55-70 foot class like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Nordhavn_(yacht)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nordhavn&lt;/a&gt; or Bering? These vessels are deep draft ships that move slow, about 8 knots typically, and are designed for stable ocean travel over long distances. The idea is that the rotor would augment the diesel propulsion to make the ship faster and more fuel efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a photo of a 65-foot Bering with a rectangle showing the imaginary location of a Flettner rotor. A 4 inch drive shaft would go through the decks to the engine room and be coupled to the engine or main drive shaft via a gear box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4R40Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4R40Z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawbacks I can see are that it could potentially decrease the stability of the ship and would be added weight. Also there would be a 6 inch column going through the center of the lounge and a loss of space on the top deck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question would be how much benefit would the rotor bring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the rotor could increase cruising speed by 20% and decrease fuel usage by 30% on a transoceanic crossing, that would be significant benefit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question would be how would the rotor affect the roll stability of the ship? Would it tend to dampen roll or increase it?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2017-01-10T01:16:28.213" Title="Would a Flettner rotor make sense for an ocean cruising trawler?" Tags="|naval-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12192" PostTypeId="2" ParentId="12179" CreationDate="2016-11-07T09:15:30.153" Score="1" Body="&lt;p&gt;It depends on the type of motor and how they are wired but as a first approximation a motor that's not being driven electrically will act like a generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no electrical connection to that motor then it can spin relativity freely and won't put much additional load onto the system. If however there is a heavy electrical load attached to the motor (e.g. a short circuit) then the motor will become hard to turn. By varying how much power you try to pull from the motor you can effect how well the motor acts as a brake. By changing the connections to the motors the car can vary to behavior of each motor as needed for the current conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said I would expect a car to include a clutch system to disengage any motor that isn't currently needed. Spinning a disconnected motor may not take much power but it will take some and there is no point in throwing away battery range for no reason.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-11-07T09:15:30.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12194" PostTypeId="1" CreationDate="2016-11-07T10:40:10.273" Score="1" ViewCount="1224" Body="&lt;p&gt;The Panama Canal has these huge gates. They close so the locks can be flooded or drained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2zr23.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2zr23.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're swing gates, so 3 questions come to mind:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How are they water-tight at the hinges?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How are they water-right at the middle, where the two gates meet?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How are they water-tight at the bottom, where the gate drags across the floor?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Okay, they don't need to be perfectly water-tight. We can call them &quot;decently sealed&quot;.&lt;/p&gt;&#xA;" OwnerUserId="3510" LastEditorUserId="3510" LastEditDate="2016-11-08T04:23:52.220" LastActivityDate="2016-11-08T04:23:52.220" Title="In the Panama Canal, how are the lock gates kept water-tight (at the hinges, middle, and bottom)?" Tags="|waterproofing|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12195" PostTypeId="2" ParentId="12175" CreationDate="2016-11-07T12:34:07.020" Score="1" Body="&lt;p&gt;The area A' is the excluded area for the cut under consideration. To get the shear stress at a location you &quot;cut&quot; the section there, and look at the area beyond the cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the shear stress in the web at the top of the web, the whole of the flange is excluded (your third picture). For the shear stress in the flange adjacent to the web, only the part of the flange beyond that cut is excluded (your first picture).&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2016-11-07T14:19:43.210" LastActivityDate="2016-11-07T14:19:43.210" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="12196" PostTypeId="1" CreationDate="2016-11-07T12:47:38.470" Score="1" ViewCount="3580" Body="&lt;p&gt;I'm working on a Home-Made Automatic Grappling Hook of sorts, as a project for university. I've designed a Blueprint accounting for everything but (embarrassingly) the motor, as I'm unsure how to calculate the minimum motor power required to lift me. Is there an equation/simple solution to work this out?&lt;/p&gt;&#xA;" OwnerUserId="8689" LastActivityDate="2016-11-07T21:01:03.340" Title="Calculate the minimum motor power required to lift a human?" Tags="|mechanical-engineering|motors|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12198" PostTypeId="2" ParentId="12170" CreationDate="2016-11-07T14:30:38.477" Score="2" Body="&lt;p&gt;As you've said, $$ \sigma = My / I $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a particular location (i.e. 2m from A), &lt;strong&gt;M&lt;/strong&gt; and &lt;strong&gt;I&lt;/strong&gt; are constant. The question asks for &lt;strong&gt;maximum&lt;/strong&gt; bending stress at that location, so you need maximum &lt;strong&gt;y&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you need to work out where the neutral axis is, and whether top face or bottom face is furthest from it, and use &lt;strong&gt;y&lt;/strong&gt; for that face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Maximum bending stress is very likely to mean maximum &lt;em&gt;magnitude&lt;/em&gt; of bending stress, so while one face will give you a positive y and the other a negative y, just use the one with the largest magnitude. Keep the sign in order to determine whether it is compressive or tensile stress.)&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2016-11-07T14:30:38.477" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12199" PostTypeId="2" ParentId="12196" CreationDate="2016-11-07T14:33:19.633" Score="1" Body="&lt;p&gt;The minimum force required to lift an object vertically  &gt; it's weight (mass.g) however some excess force will be required to produce a useful acceleration according to F = ma &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you consider an appropriate acceleration is up to you to specify but there is obviously an upper limit which is reasonably safe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how your design works you will need to convert this force to a torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of argument you decide that you want to be able to &lt;em&gt;just&lt;/em&gt; support a mass of 100kg (ie this is the load at which the motor stalls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F=mg gives us 100kg x 9.8m/s^2 = 9800N &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you connect the motor directly to a pulley of radius (r)  0.01m then the torque requirement is Fr = 98 Nm in practice this will require a fairly high reduction ratio gearbox but that's fine as it will keep the speed of the cable feed in a sensible range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With appropriate gearing this is probably equivalent to about a 250-500 watt motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an extra reality check &lt;a href=&quot;http://www.safetyliftingear.com/products/electric-wire-rope-hoist-125kg--240volt-x-18mtr-hol/wrh-yt125-250&quot; rel=&quot;nofollow noreferrer&quot;&gt;this electric hoist&lt;/a&gt; has a capacity of 125kg with a 500W 240V motor. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-11-07T14:33:19.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12200" PostTypeId="2" ParentId="12196" CreationDate="2016-11-07T14:56:21.410" Score="3" Body="&lt;p&gt;Power is work divided by the time interval required to complete the work, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P = \frac{W}{\Delta t} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work is force applied for a particular distance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;W = F\Delta x \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force is mass times acceleration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F = ma \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And acceleration is change in speed divided by the time interval required to complete the speed change:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;a = \frac{\Delta v}{\Delta t} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you're moving up, you need to counteract gravity in addition to the acceleration required to change speeds, so you would find the total force required to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F_{\mbox{peak}} = m\left(\frac{\Delta v}{\Delta t} + g\right) \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak force you need will occur during this acceleration segment, but the peak power you will need will occur during the constant-speed hoist. It's important to check both portions of the drive cycle for numbers - the acceleration requires a large force (torque) capability, the hoist requires slightly less force, but requires that force to be delivered at a higher speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant-speed portion of the drive cycle (assuming accelerate-constant speed-decelerate) requires a force of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F = mg \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the motion you will decelerate, so the power required will be even less. I'll point out that these delta values - the definition given for work, acceleration, and power - all depend on constant application of even force applied along a linear path. If this isn't true then you need to &lt;em&gt;integrate&lt;/em&gt; those values instead of &lt;em&gt;averaging&lt;/em&gt; them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example time! Say you weigh 165lb (75kg). You want to go from a stop to moving up at a speed of 2.2mph (1 m/s), and you want to reach that speed in 2 seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, first find the peak force:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F_{\mbox{peak}} = 75\mbox{kg}\left( \frac{1\mbox{m/s}}{2\mbox{s}} + 9.81 \mbox{m/s}^2 \right) \\&#xA;\boxed{F_{\mbox{peak}} = 773\mbox{N}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Work is force applied for a particular distance. &lt;strong&gt;Because I'm assuming constant acceleration&lt;/strong&gt;, the average speed is the average of the start and end speed values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xA;v_{\mbox{average}} = \frac{0 + 1}{2} \mbox{m/s} \\&#xA;\boxed{v_{\mbox{average}} = 0.5 \mbox{m/s}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance is then speed times time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = v_{\mbox{average}} \Delta t \\&#xA;x = 0.5 \mbox{m/s} 2 \mbox{s} \\&#xA;\boxed{x = 1 \mbox{m}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the work applied is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;W = F\Delta x \\&#xA;W = 773 \mbox{N} 1 \mbox{m} \\&#xA;\boxed{W = 773 \mbox{Nm}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the power you need for this section is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P = \frac{W}{\Delta t} \\&#xA;P = \frac{773 \mbox{Nm}}{2 \mbox{s}} \\&#xA;\boxed{P = 387 \mbox{W}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare this to the power required to go straight up 10 meters, &lt;em&gt;after&lt;/em&gt; the initial stopped-to-speed transient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;F = 75\mbox{kg}(9.81 \mbox{m/s}^2) \\&#xA;F = 736 \mbox{N} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = 10 \mbox{m} \mbox{(From the problem statement)} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;W = F\Delta x \\&#xA;W = 736 \mbox{N} (10 \mbox{m}) \\&#xA;W = 7360 \mbox{Nm} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x = v\Delta t \\&#xA;\Delta t = \frac{x}{v} \\ &#xA;\Delta t=  \frac{10 \mbox{m}}{1 \mbox{m/s}} \\&#xA;\Delta t = 10 \mbox{s} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xA;P = \frac{W}{\Delta t} \\&#xA;P = \frac{7360 \mbox{Nm}}{10 \mbox{s}} \\&#xA;\boxed{P = 736 \mbox{W}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here you can see that, for this example, acceleration needs 5% more force (torque), but the constant-speed lift requires &lt;strong&gt;90%&lt;/strong&gt; more power. You can tweak those numbers to see how the various parameters affect the results; in particular you should notice the impact that a shorter acceleration time makes on the acceleration power requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did all this math with linear numbers because you didn't provide any drawings of what your device looks like or how anything's attached (read: gearing, drive radii, etc.), so I can't give you anything specific on your &lt;em&gt;motor&lt;/em&gt; needs. I will say, though, that there is a very straightforward way to convert linear motion to rotational motion: $s = r\theta$. Arc length is drum distance times the total rotational angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can gear the motor however you want, but power is power - you either provide lower torque at a higher speed, or higher torque at a lower speed, but either way you're going to provide the power required as given above or you're not going to get the performance you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the example I gave, I'll point out that a 750W motor is about &lt;a href=&quot;https://electricbikereview.com/8fun/bbs02/&quot; rel=&quot;nofollow noreferrer&quot;&gt;8 pounds&lt;/a&gt;, which happens to be the same weight as a gallon of milk. This doesn't include the frame to hold the motor, the battery pack, the winch drum, 10 meters of cable, the gearbox, the grappling hook, or the grappling hook ejector. One gallon of milk, for &lt;strong&gt;just the motor&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as a final comment, none of the math here took friction into account. The power I gave was the &lt;em&gt;output mechanical power&lt;/em&gt; required to do the work. I'd probably estimate the efficiency of the mechanical drive system at 80 percent, so you could very likely wind up needing a motor 25% larger to achieve the desired performance - over 900 Watts. You can reduce weight or lower the top speed to enable use of a smaller motor, but then you're looking more at speeds of a winch like &lt;a href=&quot;https://engineering.stackexchange.com/a/12199/1633&quot;&gt;Chris Johns referenced&lt;/a&gt;, which operates at &lt;a href=&quot;http://www.safetyliftingear.com/products/electric-wire-rope-hoist-125kg--240volt-x-18mtr-hol/wrh-yt125-250&quot; rel=&quot;nofollow noreferrer&quot;&gt;8 meters per &lt;strong&gt;minute&lt;/strong&gt;&lt;/a&gt;, meaning it would elevate about one story every thirty seconds. Not exactly breathtaking.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-11-07T14:56:21.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12201" PostTypeId="1" CreationDate="2016-11-07T16:07:28.827" Score="1" ViewCount="105" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/feMdU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/feMdU.jpg&quot; alt=&quot;22&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/12165/bending-moment-diagram-from-shear-force-diagram&quot;&gt;Bending moment diagram from shear force diagram&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We all know that $$\tau = VQ/ It $$&#xA;how to determine the shear stress at &lt;strong&gt;G&lt;/strong&gt; ?&#xA;I'm having problem of finding Ay &#xA;centroid if the solid that i found earlier is y = 98.5mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so , my working is &#xA;Q = Ay = (250x10^-3)(50x10^-3)( 50+125-98)(10^-3) = 9.63x10^-4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering the area above the G .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my concept correct ? &#xA;edited : There's a dashed line from G to the bottom . It's making me confused . for Q , should i consider the area either to the left or right of the G ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5TfAp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5TfAp.png&quot; alt=&quot;jj&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2019-01-09T22:34:14.430" Title="shear force at boundary" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12202" PostTypeId="2" ParentId="12186" CreationDate="2016-11-07T16:34:01.773" Score="1" Body="&lt;p&gt;How much money do you have?  You could buy, for example, an air-suspension table such as those sold for optical lab work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could make the table out of half-inch forged iron and bolt it thru the floor to the foundation, and then isolate the train/track setup with some rubber pads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want a &quot;sensible&quot; coffeetable and don't want to mortgage your house to do the isolation, there's not much you can do.  Us old folks remember going thru this to keep our Thorens or KLH turntables from skipping when someone jumped up and down. The problem is that stereo turntable arms, like HO-gauge railroad cars, are extremely light and have a very small acceptable range of motion before jumping the groove or track.  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-11-07T16:34:01.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12203" PostTypeId="1" AcceptedAnswerId="12211" CreationDate="2016-11-07T16:50:53.833" Score="1" ViewCount="2564" Body="&lt;p&gt;I believed that the flame propagates over the place where fuel contacts the oxygen. If it contacts a fule+O2 mixture, the mixture inflames as well, and so on and on and on until there is no more inflammable material available around. If there is a gas comes to the surface, the flame stops and starts burning at the gas-air contact zone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, &lt;a href=&quot;https://diy.stackexchange.com/questions/102195/&quot;&gt;as it was explained yesterday&lt;/a&gt;, I can have the gas burning in my gas baking chamber because it gets mixed with air prior to getting into the chamber. This explain the mystery of having the flame in a sort of isolated chamber, but, what they refused to explain is that how it is possible that mixture does not blow up in the pipe as soon as you mix gas + air? Why don't you have the flame there, how do you prevent the mixture ignition until after it crosses the designed burner? How does the gas know how the burner/nozzle look like in order to inflame in the right place? My belief was that flame propagates in all directions over the fuel+oxygen mixture, so, it should propagate from the burner down the pipe until the gas-air mixer, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, is it right that &lt;a href=&quot;https://diy.stackexchange.com/questions/21504/why-is-there-an-opening-where-the-burner-connects-to-the-gas-line-on-a-gas-oven&quot;&gt;a hole in a gas pipe sucks the air despite gas&lt;/a&gt; in the pipe has pressure &gt; 1 atm (it must be greater than 1 atm in order to be pushed out of the gas pipe instead of sucking the air into it)? How is it possible that gas flows from lower pressuire into the higher pressure area without violation of entropy principle?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have noticed that there is a similar question on Quora. It is even more infomred than mine assumption -- it asks why no flahback arrestors are installed in the kitchen ovens? There should be absolutely no chance for the flame to go upstream over the explosive mixture! Why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I keep asking my question because highly appreciated &lt;a href=&quot;https://www.quora.com/Why-dont-we-provide-flash-back-arrestors-for-cooking-gas-cylinders&quot; rel=&quot;nofollow noreferrer&quot;&gt;answer there&lt;/a&gt; is unsatisfactory. It just says that we use propane cooking cylinders&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in our kitchens. These cillinders have much higher pressure than the air oxygen, and, therefore, backpropagation is not possible, says the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it prevents the &lt;strong&gt;backward propagation of GAS&lt;/strong&gt; whereas safety issue is propagation of &lt;strong&gt;flame&lt;/strong&gt;.  how does it make the flame speed necessarily lower than the gas flow speed, what happens on the last days of cylinder exhaustion when pressure goes down and what about the centralized (natural) gas kitchens?&lt;/p&gt;&#xA;" OwnerUserId="8691" LastEditorUserId="-1" LastEditDate="2017-04-13T12:22:17.377" LastActivityDate="2023-09-23T18:32:54.313" Title="How do you prevent ignition of the gas-air mixture before it reaches the burner nozzle?" Tags="|gas|safety|combustion|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12204" PostTypeId="2" ParentId="12203" CreationDate="2016-11-07T18:27:30.403" Score="2" Body="&lt;p&gt;No fuel (in the conventional sense) will burn without being mixed with oxygen.  However it is entirely possible for air to get mixed with fuel in unexpected ways, especially if you have leaks in the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context is is important to remember that gas pressure measured at a regulator is &lt;em&gt;static pressure&lt;/em&gt; but if you use that pressure to move it through a pipe and nozzle it is subject to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bernoulli%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;bernoulli principal&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also very relevant is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Venturi_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;venturi effect&lt;/a&gt;, indeed gas burners without forced air induction rely on this to work. Here the high pressure gas flow is accelerated through a small nozzle whcih casues a pressure drop thus sucking in air through holes in the back of the burner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another complication is that if you have a flame in a contained space the convection caused by heating can cause upredicatable pressure fluctuations and even further variations in combustion rate can cause rapid ignition of unburned gas reaulting in shockwaves which can propogate back into the gas line potentially sucking in air further compicated by the fact that transsonic shickwves follow entirely differnt rules. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tyhe short answer is that this is why we have &lt;a href=&quot;https://en.wikipedia.org/wiki/Flashback_arrestor&quot; rel=&quot;nofollow noreferrer&quot;&gt;flashback arrestors&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-11-07T18:41:25.400" LastActivityDate="2016-11-07T18:41:25.400" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12205" PostTypeId="2" ParentId="12196" CreationDate="2016-11-07T20:54:07.127" Score="1" Body="&lt;p&gt;The power can be as small as you want. It is just a matter of how slow you find to be acceptable. Disconsidering friction:&#xA;$$&#xA;P[W] = m[kg]\cdot g[m/s^2]\cdot v[m/s]&#xA;$$where&lt;br&gt;&#xA;P = required power&lt;br&gt;&#xA;m = your mass&lt;br&gt;&#xA;g = acceleration of gravity = 9.80665m/s2&lt;br&gt;&#xA;v = your body elevation speedy&lt;/p&gt;&#xA;" OwnerUserId="8696" LastEditorUserId="8696" LastEditDate="2016-11-07T21:01:03.340" LastActivityDate="2016-11-07T21:01:03.340" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12207" PostTypeId="2" ParentId="12157" CreationDate="2016-11-08T00:31:08.270" Score="0" Body="&lt;p&gt;So when DV1 is pushed to the right the flow will be acting on the head of the cylinder so we work out the head Area:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = PI * r^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you said however you used the radius of the lift clamp not the clamp cylinder so it should be 0.02 (20mm) divided by 2 squared times pi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give A = 0.000314 m^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is, as you said, using the formula:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v = q/A&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However your units are a bit messed up. If your &lt;code&gt;q&lt;/code&gt; is in L/s (q = 5 * 10^-3 * 1440/60 = 0.12) then you need to convert it to m^3 by multiplying it by 10^-3 and then dividing it by the area:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v = 0.12 L/s * 10^-3 / 0.000314 m^2&#xA;v = 0.382 m/s&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now the pressure will be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P = F/A&#xA;P = 2500/0.000314&#xA;P = 7.9 * 10^6 Pa =~ 8 MPa&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since this is less than the 10 for the sequence valve we won't have to go back and adjust the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your second question, since the lift cylinder is fully extended the flow cannot go to the clamp cylinder anymore so all of it goes to the lift cylinder so the process is the same:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;v = q/A&#xA;v = 0.12 * 10^-3 / 0.0314&#xA;v = 0.00382 m/s =~ 4 mm/s&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and same for pressure (with a slight difference)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P = F/A&#xA;P = 50 * 9.8 / 0.0314&#xA;P = 15.6 kPa&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since P here is 15.6 kPa, which is less than the 10 MPa of the sequence valve, the pressure at P1 will be 10 MPa.&lt;/p&gt;&#xA;" OwnerUserId="8155" LastActivityDate="2016-11-08T00:31:08.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12208" PostTypeId="2" ParentId="12184" CreationDate="2016-11-08T01:08:06.310" Score="0" Body="&lt;p&gt;I think the torque multiplier effect is related to the size and shape of the impellers like it is with gears with teeth (in a gearbox).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;" OwnerUserId="8698" LastEditorUserId="7461" LastEditDate="2017-04-06T12:08:00.713" LastActivityDate="2017-04-06T12:08:00.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12209" PostTypeId="1" CreationDate="2016-11-08T03:26:28.317" Score="1" ViewCount="57" Body="&lt;p&gt;First post in the engineering forum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Problem Statement&lt;/strong&gt; &#xA;I have designed a prototype agitator tank - its a circular square duct like pipe filled with incompressible liquid (acetone/water) which spins on a vertical axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/joIh4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/joIh4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had intended to build fluid momentum thru viscous action (achieving something akin to solid body rotation) then rapidly stop the pipe body hoping the fluid momentum will continue to circulate a flow for a time until viscosity degrades it &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated all the minor and major losses in the 'circuit' and they indicate that viscosity should degrade quite slowly but experimenting it appears that the fluid stops flowing almost immediately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I miscalculated the losses as the K-values I have used as they appear derived from steady flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h_{Minor loss}= K.\frac{v^2}{2g}\ $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a difference in how the relative velocity $v^2$ to the pipe is developed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;NB: Interestingly it does not require any more force to stop the full tank than the empty one so I'm really confused as to where this fluid momentum is going.&#xA;Additionally no matter how well I fill it (displacing all air , putting under pressure) I can still hear 'sloshing' when I stop the body, is there some sort of cavitation going on ?&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="8700" LastActivityDate="2016-11-08T03:26:28.317" Title="Minor Losses Pipeflow Momentum vs Steady Flow" Tags="|fluid-mechanics|pipelines|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12211" PostTypeId="2" ParentId="12203" CreationDate="2016-11-08T08:01:44.690" Score="4" Body="&lt;p&gt;Depending on fuel, a flame has a certain propagation speed - the speed with which the flame wanders through the medium. If the fuel - oxygen mixture moves faster through the pipe than the propagation speed of the flame, the flame can't 'walk'back. Like trying to run on a threadmill that's to fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of the fuel-air mixture depends on the area of the pipe, the pressure and the mass flow. The smaller the pipe, the higher the speed. Once the mixture leaves the pipe it will slow down. Look at the picture of a burning oil well a bit down on the &lt;a href=&quot;http://www.eng-tips.com/viewthread.cfm?qid=408099&quot; rel=&quot;nofollow noreferrer&quot;&gt;linked page&lt;/a&gt;: If you look carefully, you will see a cone shaped area between the end of the pipe and the flame. This is the spray of oild widening after leaving the pipe, slowing down. The flame starts where the speed of flame propagation is the same as the speed of the oil spray. Of course, a pre-mixed flame would have a higher flame propagation speed than a diffusion flame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a little graph of flame propagation speeds for pre-mixed stochiometric propane-air mixtures:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/g9m4K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g9m4K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;em&gt;Image source: Vorlesungsskript Technische Verbrennung, N. Peters&lt;/em&gt;&lt;br&gt;&#xA;The temperature refers to the unburnt part. As you see with higher temperatures the flame propagation is much faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that when the pressure in the bottle falls, the speed of fuel will fall also and it can become possible for the flame to 'walk' back to the bottle - so you should always have a properly designed system with a flame arrestor and if you have to ask you can't design it yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also asked how high pressure gas can 'suck' air through a whole. That's the &lt;a href=&quot;https://en.wikipedia.org/wiki/Venturi_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;venturi effect&lt;/a&gt;, fast flowing fluids have lower pressure. In all gas burners that used this for mixing in air I have seen, the mixing was after the pressure regulator where the gas is close to atmospheric pressure allready.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2016-11-08T09:03:07.257" LastActivityDate="2016-11-08T09:03:07.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12212" PostTypeId="1" AcceptedAnswerId="12220" CreationDate="2016-11-08T09:09:54.240" Score="3" ViewCount="1037" Body="&lt;p&gt;Since diesel runs lean, why is soot produced in diesel exhaust? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soot is carbon particles due to incomplete combustion. But, since diesel runs lean with plenty of oxygen, why does this happen?&lt;/p&gt;&#xA;" OwnerUserId="4846" LastEditorUserId="6790" LastEditDate="2020-05-03T18:12:48.310" LastActivityDate="2020-05-03T18:12:48.310" Title="Why is soot produced in diesel exhaust, since diesel runs lean?" Tags="|combustion|engines|diesel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12213" PostTypeId="1" CreationDate="2016-11-08T09:47:30.020" Score="1" ViewCount="5595" Body="&lt;p&gt;Hi how to calculate the effective section modulus of a class 4 section using Eurocode 1993? Can someone show me an example? I tried searching online but cannot find anything useful. My main problem is to determine the stress at both ends of the web plate. Is the stress ratio always -1 since the code mention pure bending? But if so, then the web is always fully effective.&lt;/p&gt;&#xA;" OwnerUserId="8705" LastEditorUserId="255" LastEditDate="2016-11-08T10:21:33.167" LastActivityDate="2016-11-08T11:32:11.540" Title="How to calculate effective section modulus of a slender section?" Tags="|structural-engineering|civil-engineering|steel|structures|building-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12214" PostTypeId="2" ParentId="12213" CreationDate="2016-11-08T10:17:55.470" Score="1" Body="&lt;p&gt;Class 4 sections are designed to &lt;a href=&quot;https://law.resource.org/pub/eu/eurocode/en.1993.1.5.2006.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;EC3-1-5&lt;/a&gt; (more properly known as EN1993-1-5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clause 4.3(4) states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The effective section modulus &lt;strong&gt;W&lt;sub&gt;eff&lt;/sub&gt;&lt;/strong&gt; should be determined assuming the cross section is subject only to bending stresses&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Clause 4.4(3) states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For web elements the stress ratio used in Table 4.1 should be obtained using a stress distribution based on the effective area of the compression flange&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;NB My interpretation is that the &quot;effective area&quot; of the compression flange is &lt;strong&gt;A&lt;sub&gt;eff&lt;/sub&gt;&lt;/strong&gt; from Clause 3.3(1), which allows for both shear lag and plate buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Table 4.1 shows the stress ratio to be the stress at one end of the internal compression element relative to the other. Hence for a section with equal flanges the stress ratio would be -1 for the web, &lt;em&gt;but only if the compression flange isn't reduced by plate buckling or shear lag&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A stress ratio of -1 doesn't guarantee that a web is always fully effective. Table 4.1 states that a stress ratio of -1 gives you a buckling factor of 23.9 (or 23,9 in European notation, as used in the code). This plugs into the equations in clause 4.4(2) for determining plate buckling. If your web is slender enough, it can still be reduced for plate buckling, whatever your buckling factor is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If neither your flange nor your web is reduced for either plate buckling or shear lag, then you don't have a Class 4 section.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Hence the entire process is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Plate buckling for compression flange to Clause 4.4(2)&lt;/li&gt;&#xA;&lt;li&gt;Shear lag of compression flange to Clause 3.3(1)&lt;/li&gt;&#xA;&lt;li&gt;Shear lag of tension flange to Clause 3.2.1&lt;/li&gt;&#xA;&lt;li&gt;Using the flanges from above, calculate stress ratio in web. Plate buckling for web to Clause 4.4(2)&lt;/li&gt;&#xA;&lt;li&gt;Using all the reduced elements above, calculate section modulus.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;You may wish to refer to &quot;&lt;a href=&quot;http://www.eurocodes.fi/1993/1993-1-5/background/EUR22898EN.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Commentary and Worked Examples to EN 1993-1-5&lt;/a&gt;&quot;. I have found it of some use, but not as clear as I would like it in some areas.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2016-11-08T11:32:11.540" LastActivityDate="2016-11-08T11:32:11.540" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12216" PostTypeId="2" ParentId="12212" CreationDate="2016-11-08T14:02:38.600" Score="5" Body="&lt;p&gt;From the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lean-burn#Diesel_engines&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;All diesel engines can be considered to be lean-burning with respect to the total volume, however the fuel and air is not well mixed before the combustion. Most of the combustion occurs in rich zones around small droplets of fuel. Locally rich combustion like this is a source of NOx and particles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-11-08T14:02:38.600" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12218" PostTypeId="2" ParentId="11852" CreationDate="2016-11-08T15:57:07.880" Score="0" Body="&lt;p&gt;The fundamental problem you're running into is that evaporation was originally selected over sensible heating because it is much more energy dense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you evaporate water, you can reject 2257 kJ/kg. This requires a water flow rate equal to the heat rejected, Q, divided by the enthalpy of fusion, h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When rejecting heat to water sensibly, however, you can only remove 4.18 kJ/kg-C. If we take a best case scenario where you can always reject heat to the winter water temperature, we can heat it by maybe 30°C before it starts evaporating really quickly. Here, the water flow rate must be equal to the heat rejected, Q, divided by the heat capacity times the delta T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing these two flow rates, you need 18 times more water to heat it sensibly. Then once all that water is hot, you need to cool it down again to reuse it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has been a lot of research into ways to do minimize the water consumption of water cooled condensers. In short, research has shown that you can't save a majority of the water through retrofits- you have to switch to dry cooling. That switch requires new equipment and hurts your system efficiency. If you want to look into the research, much of it was done in the concentrating solar power area since those plants are operating in the desert. &lt;a href=&quot;http://social.csptoday.com/technology/csp-cooling-options-workarounds-water-scarcity&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article isn't technical but may point you in some interesting directions.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2016-11-08T15:57:07.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12219" PostTypeId="1" AcceptedAnswerId="12221" CreationDate="2016-11-08T16:04:54.833" Score="0" ViewCount="1052" Body="&lt;p&gt;The wikipedia article for the Darcy Weisbach pressure drop equation states that &quot;In effect, the friction loss in the laminar regime is more accurately characterized as being proportional to flow velocity, rather than proportional to the square of that velocity: one could regard the Darcy–Weisbach equation as not truly applicable in the laminar flow regime.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What pressure drop equation would be more appropriate for flow with Reynolds numbers 1-100?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2016-11-08T19:07:14.903" Title="Pressure drop in very low reynolds number flows" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12220" PostTypeId="2" ParentId="12212" CreationDate="2016-11-08T17:55:36.000" Score="2" Body="&lt;p&gt;One important reason is that diesel fuel had a high molecular weight compared to gasoline this means that is is more difficult to disperse as it forms liquid droplets as opposed to vapour and even more importantly there are many more intermediate reactions involved in complete combustion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example Hydrogen, H2 burns very easily in oxygen as combustion is essentially one reaction process, H2 and O2 are separated and recombine as H20, simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you have a complex hydrocarbon fuel with long chain branched molecules there are lots of intermediate steps to go through to fully break it down into Carbon dioxide and water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an IC engine this combustion process has a very limited time to take place as an ignition cycle may take a fraction of a second &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-11-08T17:55:36.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12221" PostTypeId="2" ParentId="12219" CreationDate="2016-11-08T19:07:14.903" Score="0" Body="&lt;p&gt;Googling &quot;pressure drop laminar flow&quot; gives a Wikipedia link to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Hagen-Poiseuille&quot; equation&lt;/a&gt; for laminar flow through circular pipes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that answer your question?&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2016-11-08T19:07:14.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12222" PostTypeId="1" CreationDate="2016-11-08T23:35:00.597" Score="1" ViewCount="2666" Body="&lt;p&gt;Consider a beam that is simply supported at both ends by fixed supports and a point load is applied directly in the middle. The force (F), length of beam (L), Young's Modulus (E) and tensile and crushing strength (σ). The amount of deflection is not known. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say 1,000,000kN was applied to the beam, how would I calculate how large the cross-section would need to be in order for it to not fail and break?&lt;/p&gt;&#xA;" OwnerUserId="8713" LastActivityDate="2016-11-10T13:31:34.133" Title="How to calculate area of cross section needed to sustain a point load on a beam?" Tags="|civil-engineering|statics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12224" PostTypeId="2" ParentId="12222" CreationDate="2016-11-09T15:07:10.527" Score="3" Body="&lt;p&gt;$$\begin{gather}&#xA;M = \dfrac{FL}{4} \\&#xA;\sigma = \dfrac{My}{I} = \dfrac{FLy}{4I}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$F$ is your vertical force at midspan&lt;/li&gt;&#xA;&lt;li&gt;$L$ is your span length&lt;/li&gt;&#xA;&lt;li&gt;$y$ (distance from neutral axis to extreme fibre) and $I$ (second moment of area) depend on the shape and size of the section. For a rectangular section of width $b$ and depth $d$, $y = \dfrac{d}{2}$ and $I = \dfrac{bd^3}{12}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is assuming your material is linear elastic, and $\sigma$ is your elastic limit. It also ignores any safety factor(s).&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="1832" LastEditDate="2016-11-10T13:31:34.133" LastActivityDate="2016-11-10T13:31:34.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12225" PostTypeId="1" CreationDate="2016-11-09T16:08:21.200" Score="0" ViewCount="1047" Body="&lt;p&gt;In the notes , I dont understand that flange is thin , and the top and bottom surface are free of stress , can someone help to explain please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly , why the q' is assumed to be q '  throughout the flange is assumed to be 0 ? why are they stress free ? We could see that , the shear force V is applied to the top of the beam right ? &lt;a href=&quot;https://i.stack.imgur.com/k4FCn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k4FCn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2016-11-20T01:51:46.037" Title="shear flow in thin wall members" Tags="|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12228" PostTypeId="1" AcceptedAnswerId="12229" CreationDate="2016-11-09T22:49:58.057" Score="0" ViewCount="52" Body="&lt;p&gt;As we know, if the damping model only depends on the instantaneous velocity, it can be called viscous damping, for which the damping force $f = Cv$. However, if $C$ depends on the deformation as $C(u),$ is this still named viscous damping? And if the damping force is modeled as $f=C g(v),$ where $g$ is a function of $v$, do we call this nonlinear viscous damping? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="8721" LastActivityDate="2016-11-09T23:48:08.283" Title="deformation-dependent damping" Tags="|mechanical-engineering|structural-engineering|modeling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12229" PostTypeId="2" ParentId="12228" CreationDate="2016-11-09T23:48:08.283" Score="2" Body="&lt;p&gt;I think &quot;viscous&quot; refers to the physical effect that causes the damping - i.e. the viscosity of a fluid. The viscous forces might be a linear or nonlinear function of the velocity, or something more complicated which might include a function of the deformation. The terms &quot;linear viscous damping&quot; and &quot;nonlinear viscous damping&quot; are meaningful in that context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A damping force like of the $f = Cg(v)$ could model something like frictional damping, where for example $C$ was a (negative) constant and $g(v) = v/|v|$, (i.e. for linear motion $g(v) = \pm 1$ depending on the direction of the motion) but calling it &quot;nonlinear viscous damping&quot; instead of &quot;friction damping&quot; seems rather pointless.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-11-09T23:48:08.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12230" PostTypeId="1" AcceptedAnswerId="12239" CreationDate="2016-11-10T00:38:22.347" Score="2" ViewCount="179" Body="&lt;p&gt;This is about navigable or irrigation water ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Navigable channel or navigable canal? (it is about approach water ways to sea ports, or artificial inland water ways).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Channel dredging or canal dredging? (it is about navigable water ways).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Irrigation channel or irrigation canal? (it is about water ways within open irrigation systems).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Drainage canal or drainage channel? (it is about drainage water ways cut in earth).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Comment: in my native language (Russian), the terms canal and channel have the same translation in dictionaries.&lt;/p&gt;&#xA;" OwnerUserId="8547" LastEditorUserId="1832" LastEditDate="2016-11-10T13:27:43.243" LastActivityDate="2016-11-10T16:29:58.657" Title="Channel or Canal?" Tags="|civil-engineering|terminology|water-resources|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12231" PostTypeId="1" AcceptedAnswerId="12238" CreationDate="2016-11-09T05:49:18.947" Score="0" ViewCount="163" Body="&lt;p&gt;in tthe notes , It's stated that the shear flow will be varies lineraly prependicular to the direction of shear force , V . But , in the phooto, it shoed that the shear flow will be varies parabolically prependicular to the direction of shear force ...&#xA;Which is correct ? the notes or the example in the  photo ?&lt;a href=&quot;https://i.stack.imgur.com/5LkSu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5LkSu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Who9B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Who9B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" OwnerDisplayName="kelvinmacks" LastEditorUserId="1832" LastEditDate="2016-11-10T16:17:14.597" LastActivityDate="2016-11-10T16:17:14.597" Title="shear flow changes" Tags="|structural-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12232" PostTypeId="1" CreationDate="2016-11-10T09:21:24.637" Score="0" ViewCount="3890" Body="&lt;p&gt;In this notes , i dont understand how the Q=(Ay) of the is considered ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the both Q at part C and part B should be = (60)(15)(mm^2) ... i have redrawn the structure , i am not sure whether it is correct or not... &#xA;From the definition of shear flow , A is the cross section area of segment that is connected to the beam at juncture where the shear flow is calculated .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iLGii.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iLGii.png&quot; alt=&quot;shear flow formula&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/a1rSP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a1rSP.png&quot; alt=&quot;question&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XaWc1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XaWc1.png&quot; alt=&quot;sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2016-11-10T15:24:26.310" Title="shear stress at nail at different points" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2016-11-14T17:19:51.753" ContentLicense="CC BY-SA 3.0" />
  <row Id="12233" PostTypeId="1" CreationDate="2016-11-10T10:45:07.687" Score="0" ViewCount="939" Body="&lt;p&gt;In product catalogues, beams like UB 152x89x16 have dimensions written in mm to the nearest one decimal place. Are these beams cut and measured using an SI ruler or an imperial ruler. I am asking this question because I have trouble matching some of the section properties given in the table for some of the beams. So I presume that the dimensions given are rounded off dimensions and the actual dimension could be given in exact fractions of an inch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct? &lt;/p&gt;&#xA;" OwnerUserId="8726" LastActivityDate="2016-12-08T19:29:21.937" Title="Are British Universal Beams and Columns cut according to imperial or SI measurements?" Tags="|civil-engineering|steel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12234" PostTypeId="2" ParentId="12233" CreationDate="2016-11-10T12:19:41.717" Score="1" Body="&lt;p&gt;Universal Beams (UBs) are defined in millimetres, and are not based on the imperial rolled steel joists (RSJs). So no, the dimensions are not rounded off exact fractions of inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the designation of a section does not match its dimensions exactly. The &lt;a href=&quot;http://tsbluebook.steel-sci.org/EN/Browsers/Main.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;online blue book&lt;/a&gt; (navigate to &quot;Advance UKB&quot; for UBs) shows that for UB 914x419x388, the depth is actually 921mm, whereas for UB 914x419x343 the depth is 911.8mm - neither is actually 914mm deep!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, small errors in matching section properties may be caused by failing to take account of radii.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2016-11-10T15:20:21.640" LastActivityDate="2016-11-10T15:20:21.640" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12235" PostTypeId="1" AcceptedAnswerId="12236" CreationDate="2016-11-10T12:28:42.490" Score="0" ViewCount="79" Body="&lt;p&gt;How can we find it with the current technologies? Can it done without touching the haystack? Any cool stuffs like X-ray or some penetrating technologies ?&lt;/p&gt;&#xA;" OwnerUserId="8727" LastActivityDate="2016-11-10T15:22:39.543" Title="Find needle in a haystack?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="4" ClosedDate="2016-11-10T19:12:41.560" ContentLicense="CC BY-SA 3.0" />
  <row Id="12236" PostTypeId="2" ParentId="12235" CreationDate="2016-11-10T15:22:39.543" Score="0" Body="&lt;p&gt;I'd say X ray is the best bet. Or possibly ultrasound, metal detector (high powered). &#xA;Maybe since you never said the haystack has to remain intact, you could burn it down and search for the needle with a metal detector in the ashy remains. &lt;/p&gt;&#xA;" OwnerDisplayName="user8731" LastActivityDate="2016-11-10T15:22:39.543" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12237" PostTypeId="2" ParentId="12232" CreationDate="2016-11-10T15:24:26.310" Score="4" Body="&lt;p&gt;For starters, I'm a little confused by your units; so I'll explain my assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimensions you give are completely different from the dimensions in the picture.  It looks like you're taking inches and multiplying by 10 to get mm (e.g. your &quot;60 mm x 15 mm&quot; is 6&quot; and 1.5&quot; in the drawing; these units are not equivalent).  Maybe you just prefer metric and don't really care about the units or the answer; but still.  In North America it's good practice to be able to work in metric or imperial units. Regardless; for your reference and future information 1&quot; is 25.4 mm; and other conversion factors are usually found in the front or back of your textbook (or online).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, that's not really relevant to your question, I just wanted to make sure you aren't making that mistake when it matters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two cross sectional areas they have make sense.  Both the nails are supporting different loadings.  In the area you propose to use, you would be trying to account for both of the different loadings on the nails at the same time.  In their case, they split the problem into the cases where each nail is expected to experience the shear force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking B for example; the nail is in one side; but nail C isn't doing the same thing as nail B.  Instead, the shear is being resisted by the wood at B', so they take the area of 7.5&quot;x1.5&quot; knowing that the wood should be able to resist the shear as well.  The same idea is used for C and C'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they used your section, both Qb and Qc would be the same.  Just looking at the structure of the wooden box, that doesn't seem reasonable since the two nails are acting 90 degrees from each other while the applied loading is only in one direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this isn't clear let me know where it gets confusing.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2016-11-10T15:24:26.310" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12238" PostTypeId="2" ParentId="12231" CreationDate="2016-11-10T15:34:04.447" Score="4" Body="&lt;p&gt;You're getting something mixed up in your head here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear force is travelling in the vertical direction (up/down).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The segment perpendicular to that shear force is the horizontal component (left/ right) AKA the top and bottoms of the I beam.  In this section the shear flow is linear on the drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The centre section is going vertically, therefore it is parallel to the shear; and it shows the shear flow acting parabolically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize; both the notes and photo are correct; something in your interpretation was just misunderstood.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2016-11-10T15:34:04.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12239" PostTypeId="2" ParentId="12230" CreationDate="2016-11-10T16:29:58.657" Score="2" Body="&lt;p&gt;I thought this would be easily answerable by looking at English dictionaries. But they're not as useful as I'd hoped.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.dictionary.com/browse/channel?s=t&quot; rel=&quot;nofollow noreferrer&quot;&gt;Channel&lt;/a&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the bed of a stream, river, or other waterway.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Nautical.&lt;/em&gt; a navigable route between two bodies of water.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the deeper part of a waterway.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;http://www.dictionary.com/browse/canal?s=t&quot; rel=&quot;nofollow noreferrer&quot;&gt;Canal&lt;/a&gt;&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;an artificial waterway for navigation, irrigation, etc.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a long narrow arm of the sea penetrating far inland.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Well, I disagree with both of those.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A sea is not a &amp;quot;waterway&amp;quot;, but a deeper, navigable route through it is definitely a channel; possibly this is covered by definition 2? I'm not sure.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A long narrow arm of the sea penetrating far inland is a river, not a canal. In my experience (and, to be honest, backed up by the &amp;quot;British dictionary&amp;quot; further down the webpage), a canal can &lt;em&gt;only&lt;/em&gt; be a manmade / artificial waterway.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;So, to answer your questions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Navigable channel or navigable canal? (it is about approach water ways to sea ports, or artificial inland water ways). &lt;em&gt;If a navigable path through the sea it can only be a channel. If a navigable artificial inland waterway, it can be a navigable canal, but you can also call it a navigable channel. Navigable channel would therefore always be acceptable for this.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Channel dredging or canal dredging? (it is about navigable water ways). &lt;em&gt;As above&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Irrigation channel or irrigation canal? (it is about water ways within open irrigation systems). &lt;em&gt;Either is acceptable, assuming it is man-made (seems likely for an irrigation system). Personally, I would lean towards &amp;quot;irrigation channel&amp;quot;. According to &lt;a href=&quot;https://books.google.com/ngrams/graph?content=irrigation%20channel%2C%20irrigation%20canal%20&amp;amp;case_insensitive=on&amp;amp;year_start=1980&amp;amp;year_end=2008&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=t4%3B%2Cirrigation%20channel%3B%2Cc0%3B%2Cs0%3B%3Birrigation%20channel%3B%2Cc0%3B%3BIrrigation%20channel%3B%2Cc0%3B%3BIrrigation%20Channel%3B%2Cc0%3B.t4%3B%2Cirrigation%20canal%3B%2Cc0%3B%2Cs0%3B%3Birrigation%20canal%3B%2Cc0%3B%3BIrrigation%20Canal%3B%2Cc0%3B%3BIrrigation%20canal%3B%2Cc0&quot; rel=&quot;nofollow noreferrer&quot;&gt;google ngrams&lt;/a&gt;, &amp;quot;irrigation canal&amp;quot; is more common, but not massively so.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Drainage canal or drainage channel? (it is about drainage water ways cut in earth). &lt;em&gt;This is similar to the irrigation question above. Again, I would lean towards &amp;quot;channel&amp;quot;. For this one, &lt;a href=&quot;https://books.google.com/ngrams/graph?content=drainage%20channel%2Cdrainage%20canal&amp;amp;case_insensitive=on&amp;amp;year_start=1980&amp;amp;year_end=2008&amp;amp;corpus=15&amp;amp;smoothing=3&amp;amp;share=&amp;amp;direct_url=t4%3B%2Cdrainage%20channel%3B%2Cc0%3B%2Cs0%3B%3Bdrainage%20channel%3B%2Cc0%3B%3BDrainage%20channel%3B%2Cc0%3B%3BDrainage%20Channel%3B%2Cc0%3B%3BDRAINAGE%20CHANNEL%3B%2Cc0%3B.t4%3B%2Cdrainage%20canal%3B%2Cc0%3B%2Cs0%3B%3Bdrainage%20canal%3B%2Cc0%3B%3BDrainage%20Canal%3B%2Cc0%3B%3BDrainage%20canal%3B%2Cc0&quot; rel=&quot;nofollow noreferrer&quot;&gt;google ngrams&lt;/a&gt; agrees with me (just!).&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="255" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-11-10T16:29:58.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12240" PostTypeId="1" AcceptedAnswerId="12245" CreationDate="2016-11-10T17:09:39.270" Score="0" ViewCount="7898" Body="&lt;p&gt;I need to accurately measure (+/-0.25mm) the bolt hole circle drilled into a heavy metal plate.  There are three holes on one plate and nine holes on a second plate. The plates are approximately 300mm diameter. The holes are approximately 18mm diameter tapped M20 and the bolt hole circle is about 200mm diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When known, I will create a mating part on the CNC mill and lathe. The finished part rotates at around 2000 RPM, so the accuracy of the machining and fitment will determine the vibration of the finished system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I measure the hole circle?  Tools? Technique? Calculations?&lt;/p&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="7021" LastEditDate="2016-11-10T21:23:12.620" LastActivityDate="2016-11-10T21:23:12.620" Title="Accurately measure a bolt hole circle?" Tags="|mechanical-engineering|machining|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12241" PostTypeId="2" ParentId="5601" CreationDate="2016-11-10T17:33:17.213" Score="1" Body="&lt;p&gt;Also, the plastics used in most hobbyist printers are really not well suited to carrying any load and are usually too soft and flexible to machine properly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you can find a piece of tube, either metal or rigid plastic, to fit the axle accurately, design the wheel bore so that the tube can be pressed in. It should be a snug fit. Screws, glue, etc can attach the axle to the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say the wheel is &quot;wobbly&quot; do you mean that the bore is not aligned with the tire along the axle axis? Or is the bore offset from the centre somehow?  Can you can share your design so that we can understand the problem more clearly?&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2016-11-10T17:33:17.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12242" PostTypeId="2" ParentId="12176" CreationDate="2016-11-10T17:47:52.457" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ybr3h.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ybr3h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The infinitesimal area $\text{dA}$ is $2  z   \text{dy}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between $z$ and $y$ can be got from the slope.&#xA;$$\frac{z}{y-\frac{2 h}{3}}=\frac{0-\frac{b}{2}}{\frac{2&#xA;   h}{3}+\frac{h}{3}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving, we get&#xA;$$z=\frac{b (2 h-3 y)}{6 h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&#xA;$$I_{zz}=\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 dA&#xA;=\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 2 z  \, dy&#xA;=\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 2 \frac{b (2 h-3 y)}{6 h}  \, dy=\frac{b h^3}{36}$$&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-11-10T17:47:52.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12243" PostTypeId="2" ParentId="12240" CreationDate="2016-11-10T18:21:05.713" Score="1" Body="&lt;p&gt;One approach is to use a transfer punch to transfer the hole centres onto another plate. This is complicated a bit by the fact that the holes are threaded so you may need to machine one to the correct diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will certainly make it easier to measure the centre distances. A similar method would be to make up some short studs, centre drill them and measure the distances with dividers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With either method you should be able to construct the centre point of the pitch circle diameter using dividers &lt;a href=&quot;http://www.mathopenref.com/constcirclecenter.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mathopenref.com/constcirclecenter.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also get &lt;a href=&quot;https://www.wonkeedonkeetools.co.uk/centre-and-edge-finding-tools/what-is-a-centre-square/&quot; rel=&quot;nofollow noreferrer&quot;&gt;centre squares&lt;/a&gt; althoug hthis does depend on the outside edge of the plate being round and concentric with the PCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may also be that the CNC mill has the capability to be used as a coordinate measuring machine. This would certainly be the most straightforward way to get accurate measurements as you can just input the data to your CAD software and construct whatever dimensions you need. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth considering that you can improve the fit of bolted connections by using dowel pins to provide alignment while still allowing adequate clearance holes for the bolts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly it may be worth making your component slightly oversized then bolting  them together and machining both to the finished size at the same time . &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-11-10T18:21:05.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12244" PostTypeId="1" AcceptedAnswerId="12248" CreationDate="2016-11-10T18:41:02.703" Score="0" ViewCount="51" Body="&lt;p&gt;If I take a battery that is &quot;dead&quot; and completely replace the electrolyte and maybe the terminals, would it be a &quot;fresh&quot; battery?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So instead of pulling up to a charging station in your electric car, plugging in, and waiting half an hour or more to be able to continue, could you pull up, suction the electrolyte out, pump some fresh stuff in, and be back up to a fully charged state? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the electrolyte have to be charged before pumping it in?&lt;/p&gt;&#xA;" OwnerUserId="8734" LastActivityDate="2016-11-10T19:57:23.817" Title="How much of a battery needs to be replaced for it to be a fresh battery?" Tags="|battery|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12245" PostTypeId="2" ParentId="12240" CreationDate="2016-11-10T19:09:25.783" Score="3" Body="&lt;p&gt;The bolt pattern should be a standard; if there are three bolt holes then there should be 120 degrees between each hole. If there are nine bolt holes then there should be (360/9) = 40 degrees between each hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's talk for a second about vibration. Vibration will exist if the centerline of your flange does not coincide with the centerline of the existing flange. While your bolt pattern &lt;em&gt;does&lt;/em&gt; matter, it doesn't matter &lt;em&gt;that much&lt;/em&gt;. &lt;strong&gt;If you are relying on your bolt pattern to perfectly align two pieces of rotation machinery you are doing it wrong.&lt;/strong&gt; The machinist that attaches the two flanges together should have a means to shim one piece of equipment, or the other, or both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to rely on the shims to bring the two centerlines together, both coincident and parallel. Again, if you're relying on the bolt pattern to do this for you, then that implies that the flange needs to be &lt;em&gt;perfectly&lt;/em&gt; attached to the shaft, that the bearings have to be &lt;em&gt;perfectly&lt;/em&gt; sized with no room for dimensional tolerance, the body has to be perfect, the mounting feet have to be perfect, and the sub-structure to which both pieces of equipment attach has to be perfect. &lt;strong&gt;You need to have room for adjustment where the machinery bolts down.&lt;/strong&gt; You will &lt;strong&gt;never&lt;/strong&gt; attain the degree of perfection required to count on a bolt pattern to mate two pieces of rotation equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, that said, where vibration &lt;em&gt;does&lt;/em&gt; matter with regards to your bolt pattern is that your bolt pattern should be EVEN. If you have three bolts that are supposed to be 120 degrees apart, but two are 110 degrees apart and the others are 125 each, then you'll get vibration because the bolt pattern isn't rotationally symmetric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my advice to you would be to get the bolt pattern dimensions from the existing flange, then &lt;strong&gt;ignore minor variations&lt;/strong&gt; and make your new pattern as symmetric as you can achieve. If the existing flanges are threaded, then your holes are through holes. There needs to be some tolerance there for the bolt to be able to pass through, and the tolerance should also allow for the minor variations in hole placement in both parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clamping force of the bolts holds the flanges together. The shims under the mounting feet of the machinery brings the centerlines together. The location of the holes matters for vibration only if they're not rotationally symmetric. The fit of the the flange doesn't have anything to do with vibration provided you have adequately clamped the two flanges together and you aligned the shaft centerlines to be coincident, and again, that's done by moving the equipment, not by adjusting the flange bolt holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm harping on this to try to impress on you that a flange with through holes is not, and should not be, a piece of high precision equipment. You might need the &lt;em&gt;face&lt;/em&gt; to be very flat if you're using a gasket, and it should definitely be perpendicular to the shaft centerline, but the through holes are just through holes. If one flange is able to rotate relative to the other flange then you haven't torqued it well enough and/or you haven't used enough fasteners. If the flanges were able to have relative motion then you're &lt;a href=&quot;https://en.wikipedia.org/wiki/Clutch_control#Riding_the_clutch&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;riding the clutch&quot;&lt;/a&gt; so to speak and can expect (very) premature joint failure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, all that said, here's how to calculate what the parameters &lt;strong&gt;should&lt;/strong&gt; be for your circular bolt pattern. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Find the angular distance between bolts, $\theta$, by dividing 360 degrees by the number of bolts. A 3-bolt pattern is 120 degrees between fasteners, 4-bolt is 90, 9-bolt is 40, etc. &lt;/li&gt;&#xA;&lt;li&gt;For a threaded hole, thread a bolt into each of two adjacent holes. Not necessary if you can get the measurement tips of a pair of calipers into the holes, but it does make it easier. &lt;/li&gt;&#xA;&lt;li&gt;Measure the largest outside-outside distance between the two bolts. Measure the smallest inside-inside distance between the two bolts. Average those numbers (add together and divide by two) and you get the center-center distance. &lt;/li&gt;&#xA;&lt;li&gt;The bolting pattern's radius is given by:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;r = \frac{\mbox{average distance}/2}{\sin{(\theta /2)}} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can put your blank flange on a lathe, find the center, measure out a radius $r$, mark that circle, and locate your through hole centers at the appropriate angular positions on that circle. If you're doing it by hand you could open a compass to the $\mbox{average distance}$ between bolt holes that you found, put the pointy end anywhere on the circle and sweep it, then move the pointy end to any sweep-circle intersection and sweep again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a graphic if that'll make it any clearer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kHudn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kHudn.png&quot; alt=&quot;Bolt pattern geometry&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-11-10T19:09:25.783" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12246" PostTypeId="1" AcceptedAnswerId="12251" CreationDate="2016-11-10T19:44:58.423" Score="2" ViewCount="535" Body="&lt;p&gt;I am trying to see how much I can reduce the torque on a rudder post by changing the foil shape. I think I need a rudder with a centre-of-effort further ahead than average but I don't know what rudder series to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Balancing the rudder (in a spade rudder style) would be much simpler (and more efficient) - however, there is a design requirement for a sharply angled skeg for this particular project. The rudder will be attached/trailing directly behind a skeg.&lt;/p&gt;&#xA;" OwnerUserId="5871" LastEditorUserId="368" LastEditDate="2017-01-09T23:17:57.637" LastActivityDate="2017-01-09T23:17:57.637" Title="Hydrofoil for skeg-hung rudder with forward centre-of-effort" Tags="|mechanical-engineering|fluid-mechanics|aerodynamics|naval-engineering|airfoils|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12247" PostTypeId="2" ParentId="12244" CreationDate="2016-11-10T19:52:16.803" Score="2" Body="&lt;p&gt;In theory yes (although what you would actually need to change isn't the electrolyte as such and may very well not be a fluid depending on the battery type), however you are effectively changing most of the mass of the battery and you need to compromise the design and packaging to be able to practically achieve this, especially as many batteries need to be well sealed and cost effective manufacturing mass manufacturing processes tend no to lend themselves well to disassembly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes far more sense, in this context just to change the whole battery. This is certainly not fundamentally impractical as it makes reasonable sense to install it under the floor of the vehicle in any case (as fuel tanks tend to be in IC engine cars). &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-11-10T19:57:23.817" LastActivityDate="2016-11-10T19:57:23.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12248" PostTypeId="2" ParentId="12244" CreationDate="2016-11-10T19:55:37.900" Score="2" Body="&lt;p&gt;One plate is lead, one plate is lead dioxide. Hand-waving a big: sulfuric acid dissociates in water, making a lot of hydrogen sulfate ions. The lead plate reacts with the hydrogen sulfate to make lead sulfate, one hydrogen, and two electrons. The lead oxide plate takes a hydrogen sulfate, three hydrogens, and one electron to (also) make lead sulfate. This means there are three lead reactions required to make sufficient hydrogens for the lead oxide reaction, but the lead oxide reaction only consumes one of the six electrons generated. That is, the process releases 5 electrons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process also generates a considerable amount of lead sulfate. This forms as a powdery coating over all the plates in the battery. Interesting tidbit: any mechanical shock (bumps in the road, etc.) can knock that powdery coating off. As the coating is no longer attached to a particular plate, it's not going to get &quot;energized&quot; when the charge reaction happens, and thus will not convert from lead sulfate back to lead or lead oxide. Eventually, enough of this powder accumulates on the bottom of the battery and will short two (or more) plates, resulting in a higher self-discharge rate. Eventually, the self-discharge rate gets high enough that the battery won't &quot;hold a charge&quot; and then it gets replaced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, anyways, if all you do is pump out the electrolyte, then you haven't done anything to address the fact that the lead plates in the battery have corroded away (to form lead sulfate). Replacing just the electrolyte doesn't charge a battery. You would have to replace the electrolyte AND the lead plates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if you take a dead battery, and then replace the electrolyte and charge plates, then it's a new battery. The easier thing to do than cracking open the battery and replacing all the internals would be to just have an exchange program, like &lt;a href=&quot;http://www.bluerhino.com/Tank-Exchange.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;propane tank exchanges in the US&lt;/a&gt;. Users are charged a fee which equates to the energy (gas, or in this example electricity) added to the &quot;vessel&quot; as well as a small service fee to clean/inspect/recondition the &quot;vessel.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I think that's the way of the future for electric vehicles. It's just too hard to provide the electric power required to charge a battery at anywhere close to the rate gas vehicles refuel, which &quot;charge&quot; at a rate of megawatts - check energy density of gasoline and multiply it by 10 gallons per minute. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-11-10T19:55:37.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12249" PostTypeId="1" AcceptedAnswerId="12256" CreationDate="2016-11-10T22:46:45.373" Score="3" ViewCount="3853" Body="&lt;p&gt;Large steam engines seem to get pretty good efficiency, but there are other options such as combustion turbines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring all other considerations, what methodology can extract the most energy out of a given amount of fuel?&lt;/p&gt;&#xA;" OwnerUserId="8736" LastActivityDate="2017-04-18T15:28:03.977" Title="What is the most efficient way to convert a fuel into electrical energy?" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12250" PostTypeId="2" ParentId="12249" CreationDate="2016-11-10T23:09:44.197" Score="0" Body="&lt;p&gt;Scale matters as does the type of fuel. Steam power, in general, has the advantage that the boiler is physically separate from the working fluid and from the working parts of the engine itself and you can design the combustion and heat exchange process independently from the engine. So steam power lends itself well to using solid fuels which burn relatively slowly and leave a solid ash residue such as wood, coal and waste biomass and of course nuclear reactors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also steam power, using water as a working fluid with phase change means that there are lots of opportunities to recover waste heat as the working fluid is recirculated rather than being dumped to the atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas turbines are typically a bit less efficient but more flexible as they can achieve peak efficiency at smaller scales so in a grid system steam turbines may be used for the ongoing base load and gas turbines to pick up peak loads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously gas turbines also require specific types of fuel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course in the real world pure thermal efficiency matters less than economical considerations of fuel cost and availability, capital cost of plant and legislative constraints such as subsidies and penalties. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-11-10T23:09:44.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12251" PostTypeId="2" ParentId="12246" CreationDate="2016-11-10T23:10:36.980" Score="3" Body="&lt;p&gt;As a Naval Architect, this is a question that I am often asked.  The centre of effort is very close to the geometric centre of the spade as you know.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a fixed skeg or keel immediately in front of the rudder? That affects the hydrodynamics significantly because the hydrofoil would then have an effective length that is the skeg width (aligned with the fluid flow) plus the rudder chord length. While not precise, this approximation will allow a better calculation of Rn (Reynolds Number) to calculate lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rudder torque is the response to the hydrodynamic pressure on the moving part of the hydrofoil. A perfect spade rudder would be long and thin (think aircraft wing), with its axis approximately 1/3 to 1/2 chord from the leading edge at the centre of lift. Unfortunately with ships the rudder tends to be short and wide and much less efficient, so to achieve the same force, a larger profile area is needed ( F= 1/2 PAV^2) P=fluid density, A = Foil profile area, V=fluid velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding a skeg achieves exactly the same result as an aircraft wing with ailerons. The ailerons change the lift characteristic of the foil in two ways - firstly the fluid flow is diverted, changing the relationship of the pressure over the top and bottom surface; secondly the centre of lift moves forward slightly as the aerofoil deflection increases, and therefore increasing effort is needed to operate the aileron until stall is achieved. However the actual effort required with a skeg vs. a free spade rudder is significantly lower since the fixed skeg is part of the foil and carries most of the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question depends on the physical geometry required, the fluid flow rate, fluid density, and the aerofoil shape itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the low speed Selig or NACA series foils may be suitable - eg NACA 0015 with the movable portion of the foil (rudder) at 50-60% chord from the LE of the skeg. Obviously a leading edge radius on the rudder and a trailing edge radius on the skeg are required. The gap between should be minimised.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dHzvO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dHzvO.png&quot; alt=&quot;NACA0015&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Useful resources:-&#xA; &lt;a href=&quot;http://www.mh-aerotools.de&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mh-aerotools.de&lt;/a&gt;&#xA; &lt;a href=&quot;http://airfoiltools.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://airfoiltools.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2016-11-10T23:10:36.980" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12253" PostTypeId="2" ParentId="12187" CreationDate="2016-11-10T23:39:37.860" Score="1" Body="&lt;p&gt;It is very unlikely to create enough motive power for that vessel. Two fundamental issues are stability (heavy structure high above the waterline) and the absence of a keel (required to reduce lateral drift or leeway). Adding a rotor to that type of vessel would be IMHO inefficient and dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flettner and similar devices are not particularly efficient at capturing wind power when compared with a typical sail. High performance sailing vessels (e.g. Admirals Cup racers) have high aspect ratio, semi-rigid sails. A flettner would need to be very large indeed to be effective.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2016-11-10T23:39:37.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12255" PostTypeId="1" AcceptedAnswerId="12257" CreationDate="2016-11-11T05:26:19.467" Score="0" ViewCount="199" Body="&lt;p&gt;I recently had an idea about the design of an internal combustion engine. Normally, the spark must be timed using some kind of circuit, but if there was a &quot;spike&quot; on the piston:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c7FBS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c7FBS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then would there be no need for a specialized ignition system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there would have to be some kind of insulation between the spike and the casing, and between the spike and the piston.&lt;/p&gt;&#xA;" OwnerUserId="8740" LastActivityDate="2016-11-11T12:34:22.157" Title="IC engine where the piston is the spark plug" Tags="|mechanical-engineering|electrical-engineering|combustion|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12256" PostTypeId="2" ParentId="12249" CreationDate="2016-11-11T08:59:13.247" Score="6" Body="&lt;p&gt;In theory, &lt;strong&gt;direct&lt;/strong&gt; chemical to electrical energy offers the highest efficiency: it can deliver arbitrarily close to 100% efficiency. However, to date, fuel cells haven't delivered on anything close to that promise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, the worst way to do it, in efficiency terms, is to go via heat - e.g. by burning the fuel. That throws away a lot of the exergy (the ability of the fuel to do work) straight away. The amount that's irretrievably thrown away is given by Carnot's Law, and is determined by the absolute temperatures of the combusting fuel, and of the sink where the cooling happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hotter the combustion, and the colder the sink, the more efficient the process can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best-in-class combined-cycle gas turbines (CCGT) deliver around 60% efficiency, by combining a steam turbine with a gas turbine. And to date, that's about as good as it gets for fuel consumption.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2016-11-11T09:15:10.983" LastActivityDate="2016-11-11T09:15:10.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12257" PostTypeId="2" ParentId="12255" CreationDate="2016-11-11T12:26:23.853" Score="7" Body="&lt;p&gt;Well, you'd still need something to generate the high voltage for the spark, so there isn't much overall simplification. A magneto driven by points is about as simple as it gets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But more importantly, the spark does not always occur at TDC &amp;mdash; in fact, in most operating regimes for most engines, it &lt;em&gt;never&lt;/em&gt; does. In order to get the best efficiency, the ignition timing must be variable relative to the piston position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, at idle, ignition occurs a few degrees before TDC. But as the speed goes up, or under hard acceleration, the ignition might be advanced by as much as 40&amp;deg;-50&amp;deg;. Back when we used mechanical points for ignition timing, these were known as &quot;centrifugal advance&quot; and &quot;vacuum advance&quot;, respectively.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2016-11-11T12:34:22.157" LastActivityDate="2016-11-11T12:34:22.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12258" PostTypeId="1" CreationDate="2016-11-11T19:02:26.277" Score="2" ViewCount="153" Body="&lt;p&gt;I've recently came across Google's cartboard VR headset, which allows you to use your smartphone as a virtual reality device. &#xA;The headset basically splits the screen in your left and right eyes, which makes 3D vision possible.&#xA;This is a well known cognitive mechanism that I won't discuss here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what bothers me is actually the &lt;strong&gt;sensor&lt;/strong&gt; side of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Let us use the following frame of reference for the rest of this discussion:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c4Pue.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c4Pue.png&quot; alt=&quot;frame of reference&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been astonished by the precision with which your head's movement is detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sure, the integrated &lt;strong&gt;accelerometer&lt;/strong&gt; (which can detect gravity), can account for rotation around &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; (roll and pitch) since it knows the direction of the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;&lt;em&gt;how do you deal with rotations around z (yaw)&lt;/em&gt;&lt;/strong&gt;, which is obviously collinear with the direction of gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that the system uses the smartphone's &lt;strong&gt;magnetometer&lt;/strong&gt; as a compass (your orientation is then computed relatively to the Earth's magnetic north)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, many people argue that yaw is detected via the &lt;strong&gt;gyrometer&lt;/strong&gt;. Which doesn't seem right to me, for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First, even though it's made to detect a rotation, it will easily get lost in some cases. Indeed, if you rotate you head around &lt;em&gt;z&lt;/em&gt; at a constant rate (&lt;em&gt;i.e.&lt;/em&gt; without any angular acceleration), the device can not tell whether you're rotating clockwise or anti-clockwise (the tangential acceleration is non-existent). In practice, I agree that this problem could perhaps be fixed by looking at the previous time-steps (if an acceleration has stopped but no subsequent deceleration has been recorded, it means the user is still rotating at constant speed). But would it be precise? Since there is no absolute frame of reference, wouldn't the results diverge quickly? &lt;/li&gt;&#xA;&lt;li&gt;Second, the very fine precision of the measurement. I've tried to rotate the phone around its vertical axis as slow as I could, and the program was still sensing my moves. Can a smartphone's gyrometer measure such tiny accelerations?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For these reasons, I believe that the magnetometer (&lt;em&gt;i.e.&lt;/em&gt; the compass) is the way to go. Even if you turn and shake you phone very aggressively, the virtual reality reference frame doesn't seem to be disturbed. Which lets me think that the device is taking the external world as a reference, rather than relying on it's internal sense of position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, can a magnetometer achieve such a precision of measurement as well? How is the detection of the Earth's magnetic field not disturbed by surrounding sources (I've put a magnet next to my phone, but never managed to mislead the VR program).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or could the solution lie in a combination of different sensors? (comparing the measurements of the accelerometer, the gyroscope and the magnetometer to cancel most sources of error)&lt;/p&gt;&#xA;" OwnerUserId="8745" LastEditorUserId="8055" LastEditDate="2017-01-16T16:23:24.703" LastActivityDate="2017-01-16T16:23:24.703" Title="How does smartphone VR positioning work?" Tags="|electrical-engineering|sensors|signal-processing|" AnswerCount="0" CommentCount="4" ClosedDate="2017-01-17T17:04:24.880" ContentLicense="CC BY-SA 3.0" />
  <row Id="12259" PostTypeId="1" CreationDate="2016-11-11T19:03:20.187" Score="1" ViewCount="2696" Body="&lt;p&gt;I am trying to compelte this question, however when i run the simulation there is no voltage atall on the output votlage, and cahnging the capacitor value as requested in the question does not chane the output at all. &lt;a href=&quot;https://i.stack.imgur.com/qejx2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qejx2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Here is my version, the only thing i can see that is different is that the example circuit does not have AC listed in the parameters. however i cant find a votlage source like that one without the AC parameter.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8SCvn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8SCvn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could help me to get this working i would be extremely grateful.&lt;/p&gt;&#xA;" OwnerUserId="6104" LastEditorUserId="8055" LastEditDate="2017-01-11T18:33:09.153" LastActivityDate="2017-07-20T00:29:23.320" Title="No Voltage appearing on Pspice" Tags="|electrical-engineering|circuits|circuit-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12260" PostTypeId="2" ParentId="2934" CreationDate="2016-11-12T02:19:32.847" Score="1" Body="&lt;p&gt;You need a variable-reluctance stepper motor.  They have no permanent magnet and virtually no cogging torque.  When unenergized, they will rotate freely in either direction.  Available from 24 steps/revolution to 200 steps/revolution.&lt;/p&gt;&#xA;" OwnerUserId="8748" LastActivityDate="2016-11-12T02:19:32.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12262" PostTypeId="1" CreationDate="2016-11-12T06:16:49.690" Score="0" ViewCount="515" Body="&lt;p&gt;Why do power plants that use the Rankine cycle use the working fluids:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;sodium&lt;/li&gt;&#xA;&lt;li&gt;mercury&lt;/li&gt;&#xA;&lt;li&gt;serine&lt;/li&gt;&#xA;&lt;li&gt;benzene&lt;/li&gt;&#xA;&lt;li&gt;potassium&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can such plants use other fluids?&lt;/p&gt;&#xA;" OwnerUserId="8751" LastEditorUserId="90" LastEditDate="2016-11-12T08:37:08.520" LastActivityDate="2023-01-16T14:35:41.990" Title="Other working fluids in Rankine cycle power plants" Tags="|thermodynamics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12267" PostTypeId="2" ParentId="12259" CreationDate="2016-11-12T15:46:35.537" Score="0" Body="&lt;p&gt;Well, for one thing, you have two different nodes in your circuit connected to ground. Obviously, that can't work! Consider what happens to D8 when the AC supply goes negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For another thing, you have added &lt;code&gt;AC = 0&lt;/code&gt; to your AC source, which forces the RMS value to zero, and this apparently overrides the peak specification &lt;code&gt;VAMPL = 100V&lt;/code&gt;. Specify one or the other, but not both.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-11-12T15:46:35.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12268" PostTypeId="2" ParentId="7394" CreationDate="2016-11-12T15:53:07.150" Score="0" Body="&lt;p&gt;Chuk almost got it to the end. But from the point of view to answer the question, (now that is what I am told) ASME B&amp;amp;PV code restricts heating rate to 30 degrees centigrade per hour. Normal plants works at around 300 degrees centigrade. This will give you minimum theoritical heat up rate of plant. Secondly when a plant is tripped first cause of trip is found and its rectification. To heat secondry side, steam is required for which there are auxilary boilers that are started. Lastly the water chemistry of all plant is restored and this takes time.&lt;/p&gt;&#xA;" OwnerUserId="8757" LastActivityDate="2016-11-12T15:53:07.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12269" PostTypeId="2" ParentId="8240" CreationDate="2016-11-12T16:01:55.153" Score="0" Body="&lt;p&gt;To shield a reactor we have to take care of gamma rays and nuetrons. Neutrons are shielded easily if they are slow. We slow them through elastic collision with a light material and most economical one (water) is used. To shield fron gamma ray heavy material that is lead or iron is used. The answer to your question is that water is most probably used.&lt;/p&gt;&#xA;" OwnerUserId="8757" LastActivityDate="2016-11-12T16:01:55.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12272" PostTypeId="1" AcceptedAnswerId="12798" CreationDate="2016-11-12T21:14:36.873" Score="6" ViewCount="2131" Body="&lt;p&gt;Why does the ballast of train track need good drainage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read something about preventing &quot;fouling&quot; of the ballast, but don't know what that means or why it would be structurally bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could foundational support play a role, where soggy ground would give way under the track?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also trying to understand why ballastless track apparently doesn't need good drainage. That continuous slab of concrete surely is not permeable to water.&lt;/p&gt;&#xA;" OwnerUserId="3510" LastActivityDate="2020-09-15T13:38:02.167" Title="For train track, why does the ballast need &quot;good drainage&quot;?" Tags="|rail|transportation|foundations|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12274" PostTypeId="2" ParentId="12272" CreationDate="2016-11-13T00:16:36.540" Score="4" Body="&lt;p&gt;For both kinds of track, the answer is the same: to prevent the track from shifting as a result of freeze-thaw cycles. The idea with ballasted track is that water is not allowed to accumulate in the ballast if it has good drainage. The idea with unballasted track is that the concrete redirects the water so that the ground immediately below it stays dry, achieving the same result.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-11-13T00:16:36.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12277" PostTypeId="1" CreationDate="2016-11-13T07:45:35.843" Score="2" ViewCount="1505" Body="&lt;p&gt;I'm looking for, hopefully, a qualitative answer to how a compressor increases pressure. I understand the equations that predict this result but I can't picture it in my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rotor spins, deflecting air into the engine and increasing the air's velocity. Next, there are stationary fins that this relatively high speed flow collides with. Somehow this contact takes away velocity and adds pressure. This crucial step is the part I don't understand.&lt;/p&gt;&#xA;" OwnerUserId="8763" LastEditorUserId="282" LastEditDate="2016-12-01T07:52:47.540" LastActivityDate="2017-05-30T22:12:14.150" Title="How does a stator convert velocity into pressure?" Tags="|aerodynamics|compressors|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12279" PostTypeId="1" AcceptedAnswerId="12317" CreationDate="2016-11-13T14:34:53.690" Score="1" ViewCount="184" Body="&lt;p&gt;I have encounter this notion: &quot;Dimensionless Thickness&quot;. Does anyone have any idea how to interpret this? One can find such a term &lt;a href=&quot;https://www.sharcnet.ca/Software/Ansys/17.0/en-us/help/flu_am/th_bat_MSMD_abuse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interpretation is that the name comes from the fact that the thickness alone doesn't mean too much, but it means a lot if it is related to a reference number (e.g. $ratio = t/t_{reference}$, where t is the dimensionless thickness). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is true, can anyone provide me with an example of such a term in the fluid-dynamics, mechanics or thermodynamics field?&lt;/p&gt;&#xA;" OwnerUserId="8419" LastActivityDate="2016-11-15T20:02:08.190" Title="Dimensionless Thickness" Tags="|mechanical-engineering|chemical-engineering|building-physics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12280" PostTypeId="1" CreationDate="2016-11-13T14:37:50.047" Score="0" ViewCount="11348" Body="&lt;p&gt;I want to use a boat with an aluminium hull. The authorities wants me to prove by calculation that the hull strength (longitudinal, lateral and local strength) equals at least the strength that would result from the use of steel plate of a thickness of 0.11811 inches. So what I'm looking for is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A gross idea of the aluminium plate thickness I need;&lt;/li&gt;&#xA;&lt;li&gt;A method for showing that the aluminum plate is strong enough to be substituted for the steel one.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How can I calculate that thickness?&lt;/p&gt;&#xA;" OwnerUserId="8766" LastEditorUserId="368" LastEditDate="2016-11-14T18:58:24.873" LastActivityDate="2016-11-14T19:04:01.837" Title="What thickness of aluminium sheet do I need to achieve the same strength of a 1/8&quot; steel sheet?" Tags="|mechanical-engineering|steel|aluminum|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12281" PostTypeId="1" CreationDate="2016-11-13T17:20:26.960" Score="4" ViewCount="833" Body="&lt;p&gt;Solar panels can absorb sunlight and use it for energy, but can they absorb the heat that is generated from their use and use the heat for energy?&lt;/p&gt;&#xA;" OwnerUserId="6553" LastActivityDate="2017-04-26T14:57:29.570" Title="Can solar panels absorb heat for energy?" Tags="|mechanical-engineering|heat-transfer|environmental-engineering|solar-energy|photovoltaics|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12282" PostTypeId="2" ParentId="12281" CreationDate="2016-11-13T18:47:43.370" Score="11" Body="&lt;p&gt;Yes, there are indeed combined PV-T (photovoltaic-thermal) hybrid panels that turn some of the incident light into electricity, and have a circulating fluid and heat exchanger to put some of the heat into a heat store.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's very little take-up of them, because they're just not economic in most circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory looks great: PV panels are more efficient when they're colder, so extracting the heat from them should give a double win - useful heat is harvested, and more electricity is generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, it adds complexity. And PV is a mass consumer product as well as a mass utility-scale product - over a billion PV panels have been installed globally - so PV sees all the economies of scale and learning-by-doing, whereas PV-T remains an oddity and a niche.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In almost all circumstances, it's preferable just to install one or the other technology. In some circumstances, it might be appropriate to install both, but separately. There are some edge cases where space is very limited and it's appropriate to install both technologies - and those edge cases are the market for PV-T panels.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2016-11-13T18:47:43.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12283" PostTypeId="2" ParentId="12262" CreationDate="2016-11-13T20:03:48.827" Score="0" Body="&lt;p&gt;Yes, and the absolutely overwhelming majority of them do.  The number that use any of the fluids you listed is trivial.  Most use water.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2016-11-13T20:03:48.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12285" PostTypeId="2" ParentId="12280" CreationDate="2016-11-13T23:56:36.573" Score="0" Body="&lt;p&gt;For tensile, metals are usually used until their yield (start to have permanent deformation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A36 steel in plates, bars, and shapes with a thickness of less than 8 in (203 mm) has a minimum yield strength of 36,000 psi (250 MPa) and ultimate tensile strength of 58,000–80,000 psi (400–550 MPa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yield strength of pure aluminium is 7–11 MPa, while aluminium alloys have yield strengths ranging from 200 MPa to 600 MPa.[9]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes steel is stronger 22.7-35.7 times alumunium. So alumunium thk required is 22.7-35.7 times (2.84&quot; to 4.46&quot;). Depends on what steel and al-alloys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use tensile test to check.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2016-11-13T23:56:36.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12287" PostTypeId="2" ParentId="12280" CreationDate="2016-11-14T02:20:07.500" Score="6" Body="&lt;p&gt;Using your plate as boat hull implies a complex load profile probably involving compression, shear and bending, possibly even torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In such cases, it is effectively impossible to give a simple answer to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/12285/1832&quot;&gt;@RainerJ's answer&lt;/a&gt; is perfectly acceptable for tensile and shear loads: if you're replacing material X with material Y and Y's yield strength is half of X's, then you'll need twice the thickness to withstand the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For compressive loads, yield strength is probably irrelevant given that the hull would most likely buckle before reaching its compressive yield strength, and buckling is a purely elastic behavior. In this case, what matters is the effective span $\ell$ of the element under compression (which is a function of the true span and the boundary conditions), its moment of inertia $I$ and its Young's Modulus $E$. Perfect Euler buckling is given by&#xA;$$P = \dfrac{\pi^2EI}{\ell^2}$$&#xA;Actual designs don't use Euler buckling due to imperfections and other behaviors, but this at least gives you an idea of the variables at your disposal. If you're replacing material X with Y and Y's Young's Modulus is half of X's, then you'll need (assuming a simple sheet without stiffeners or whatnot) $\root3\of{2}$ times the thickness to withstand the compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're dealing with flexo-compression, then you're beyond the scope of a simple answer here, especially given the absolute lack of any sort of details as to your situation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And as &lt;a href=&quot;https://engineering.stackexchange.com/questions/12280/what-thickness-of-aluminium-sheet-do-i-need-to-achieve-the-same-strength-of-a-1/12285#comment21778_12280&quot;&gt;@alephzero mentioned in a comment&lt;/a&gt;, if you intend to actually build a boat (as it would seem you do, if you're needing to confer with the authorities), you'll need serious engineering calculations and designs, with considerations far more precise and detailed than anything any of us can offer here. You'll need to either truly learn the engineering principles behind boat design or you'll need to hire an engineer who can do it. This is not the sort of thing to be taken lightly.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-11-14T19:04:01.837" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12288" PostTypeId="1" CreationDate="2016-11-14T07:30:08.333" Score="1" ViewCount="2192" Body="&lt;p&gt;How to calculate flow rate of pump and velocity if only given length of pipe and volume of tanks?&lt;/p&gt;&#xA;" OwnerUserId="8781" LastEditorUserId="1293" LastEditDate="2016-11-14T09:33:59.373" LastActivityDate="2016-11-16T05:00:51.597" Title="Pump flow rate calculation" Tags="|mechanical-engineering|fluid-mechanics|pumps|" AnswerCount="2" CommentCount="1" ClosedDate="2016-11-14T21:11:23.793" ContentLicense="CC BY-SA 3.0" />
  <row Id="12289" PostTypeId="1" AcceptedAnswerId="12294" CreationDate="2016-11-14T09:25:05.447" Score="2" ViewCount="3556" Body="&lt;p&gt;I'm trying to find the best way to move a heavy weight around. Probably the best way is to put it on wheels but I don't know which physics concepts it involves so I can make a better research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found about rolling force and initial force but it's still unclear for me what can I do to decrease the force needed. What I found but not sure:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Increase diameter of the wheels&lt;/li&gt;&#xA;&lt;li&gt;Wheels/floor made of low rolling resistance coefficient&lt;/li&gt;&#xA;&lt;li&gt;Decrease the weight (can't be done)&lt;/li&gt;&#xA;&lt;li&gt;Good bearings&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this correct? What about the number of wheels, moment of inertia? Am I missing something else?&lt;/p&gt;&#xA;" OwnerUserId="8782" LastActivityDate="2016-11-14T15:50:47.373" Title="Force required to move object on wheels" Tags="|mechanical-engineering|wheels|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12290" PostTypeId="2" ParentId="12288" CreationDate="2016-11-14T09:40:59.043" Score="0" Body="&lt;p&gt;The volume of the tank shall give you an indication (if filled) of the level of the fluid stored inside which means that you can calculate the potential energy stored in the fluid which is converted to kinetic energy at tank delivery nozzle (meaning that velocity can be calculated simply using Bernoulli's equation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll then use the length of the pipe to calculate the friction losses (head losses $\triangle h_{loss}$) that is added to the total head requirement of the piping system $\triangle h_{total}$ and finally the power of the pump can be calculated as: $P = \rho gQ \triangle h_{total}$&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2016-11-14T09:40:59.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12291" PostTypeId="2" ParentId="12289" CreationDate="2016-11-14T09:52:03.673" Score="1" Body="&lt;p&gt;If you have a smooth and level floor without drain holes you can instead use an air cushion to move the weight around (think hovercraft).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air-pressure you need can be calculated with the relation between the footprint of the object and its weight. The flow rate required from the air-compressors will depend on the leakage between the floor and the skirt.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2016-11-14T12:37:32.433" LastActivityDate="2016-11-14T12:37:32.433" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12293" PostTypeId="2" ParentId="12288" CreationDate="2016-11-14T10:31:18.587" Score="0" Body="&lt;p&gt;The problem is not providing enough data, but it happens most of the time. I usually assumed pressure drop of 0.1 bar for every 100m pipe length.&#xA;From there and pipe size, you have velocity and flowrate.&#xA;Tank size is not relevant info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On second thought, you can just run the pump and stopwatch to empty half volume of tanks. Flowrate = (Volume of half of tanks)/Time.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastEditorUserId="4630" LastEditDate="2016-11-16T05:00:51.597" LastActivityDate="2016-11-16T05:00:51.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12294" PostTypeId="2" ParentId="12289" CreationDate="2016-11-14T12:55:25.837" Score="1" Body="&lt;p&gt;As you say wheels are usually the first option considered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the simplest option is to look at an industrial wheel/castor supplier. For example &lt;a href=&quot;http://www.rosscastors.co.uk/?gclid=CIDLnLajqNACFQ06Gwody4sJTw&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt; has a huge range of wheels castors and other parts for moving and handling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wheels are typiacly rated with a maximum load, note that if you use 4 castors it is usual to make sure that they are rated so that 3 are adequately rated for the total load ie if the load is 1000kg you need castors rated to at least 333kg. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Castors are generlaly pretty good for reasonably smooth surfaces like a decent concrete floor. Soft or uneven surfaces may require pneumatic wheels which tend to have significantly larger diameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Castors have the advantages of being self contained, vesatile and compact and they can often be fixed directly to th estructure you want to move but only work well on reasonably smooth, hard surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically harder wheels give less rolling resistance on smooth surfaces but are more affected by uneven floors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inertia is a fundamental property of matter and there is nothing you can do to decrease it. But when moving things by hand the ability to control heavy loads on wheels becomes a problem before they are too massive to get moving. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-11-14T15:50:47.373" LastActivityDate="2016-11-14T15:50:47.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12296" PostTypeId="1" CreationDate="2016-11-14T14:54:00.433" Score="1" ViewCount="379" Body="&lt;p&gt;I am trying to understand the physical locking mechanism used in bicycle docks, as in the image below, to secure the bike to the dock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7uHT7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7uHT7m.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want an in-depth understanding of the mechanism for a design project I am doing on bike sharing systems. I most interested in the type of docks &lt;a href=&quot;https://en.wikipedia.org/wiki/Citi_Bike&quot; rel=&quot;nofollow noreferrer&quot;&gt;Citi Bike&lt;/a&gt; uses.&lt;/p&gt;&#xA;" OwnerUserId="8783" LastEditorUserId="368" LastEditDate="2016-11-14T18:47:21.227" LastActivityDate="2016-11-14T18:47:21.227" Title="How does the locking mechanism of a secure bicycle dock work?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2016-11-18T00:22:25.330" ContentLicense="CC BY-SA 3.0" />
  <row Id="12297" PostTypeId="1" CreationDate="2016-11-14T15:12:57.293" Score="2" ViewCount="1178" Body="&lt;p&gt;I try to understand what the PID controller of a gimbal stabilizer boards&#xA;(e.g &lt;a href=&quot;http://www.olliw.eu/storm32bgc-wiki/STorM32_Boards&quot; rel=&quot;nofollow noreferrer&quot;&gt;Storm32BGC&lt;/a&gt;) does influence? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the IMU we get position and velocity after sensor funsion. The input variable of the plant seems to be the speed of the gimbal motor (since its driven by 3 sine waves), not the voltage as usual in a brushed dc motor setup. So one can not just implement a torque control for holding the position. When I investigated the Storm32BGC, it looks like the PID does influence the position and the speed at the same time: Fast movements leads to fast speed corrections, but also the deviation from the stationary position leads to a given speed. This two speed signals seems to add up. Has anyone an Idea how this is done? I could not find any papers about this topic.&lt;/p&gt;&#xA;" OwnerUserId="8784" LastEditorUserId="8055" LastEditDate="2019-01-01T15:56:19.157" LastActivityDate="2019-01-01T15:56:19.157" Title="Suggestion to control a gimbal bldc motor for positioning" Tags="|control-engineering|motors|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12301" PostTypeId="1" CreationDate="2016-11-14T17:14:26.453" Score="0" ViewCount="945" Body="&lt;p&gt;This question is going to have an application in aviation (to avoid doubts and relocation to other forum) but let's concentrate on theoretical mechanics here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to allow a shaft to rotate in one direction and not the other, then a ratchet would be a solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what I need is a base/platform with shaft M (master) and shaft S (slave) connected to each other the way that if I rotate shaft M (both clockwise or counter-clockwise) the shaft S will rotate with the same speed and direction. But if I try to rotate shaft S in any direction it should not be possible the same way as if I tried to rotate a ratchet in blocked direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the name of a device I described?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Shaft M --&amp;gt; ===[Device in]=== &amp;lt;-- Shaft S&#xA;_______________[question ]________ &amp;lt;-- platform&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="8786" LastEditorUserId="368" LastEditDate="2016-11-14T18:42:30.527" LastActivityDate="2016-11-15T15:44:39.910" Title="How can I couple two shafts along the same axis so that they rotate together, but only one shaft is allowed to drive the rotation?" Tags="|mechanical-engineering|gears|applied-mechanics|torque|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12303" PostTypeId="2" ParentId="12301" CreationDate="2016-11-14T17:38:07.453" Score="2" Body="&lt;p&gt;One basic class of mechanism which can achieve this is a &lt;strong&gt;non-backdrivable&lt;/strong&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Worm_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;worm drive&lt;/a&gt;, however this does come with a failry large reduction ration and so may not be suitable for your application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would also be achievable, in some circumstances with an automatically engaging clutch such as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Centrifugal_clutch&quot; rel=&quot;nofollow noreferrer&quot;&gt;centrifugal clutch&lt;/a&gt; or you could envisage a sytem where the master shaft has a cam lug which engages in a spiral groove to pull the shaft fowrard and engage with a dog clutch with a return spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively  a clutch mechnism which is engaged by a governor a bit like an automatic gearbox, this could be achieved by mechanical, hydraullic or electrical means. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course a lot depends on the sorts of torques and angluar speeds you are looking to deliver as well as the response, stiffness and backlash that you require I've seen very simple inertial engagement mechanisms in toys etc which achieve this but wouldn't be up to precision or heavy duty applications. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-11-15T15:44:39.910" LastActivityDate="2016-11-15T15:44:39.910" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12304" PostTypeId="2" ParentId="12296" CreationDate="2016-11-14T17:38:16.033" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://nymag.com/guides/summer/2012/communal-bikes/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This nymag page&lt;/a&gt; says &quot;The mechanical titanium lock sends pins into the docked bike&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://citibikeblog.tumblr.com/post/91458138556/dock-dynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;This citibikeblog page&lt;/a&gt; shows a view inside the dock and states that when the lock hasn't worked properly, then without a bike in there may still be a &quot;metal bar across the slot&quot; - this is presumably the titanium lock pin (not that I can see it in that page's photo).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.uk/search?site=&amp;amp;tbm=isch&amp;amp;source=hp&amp;amp;biw=825&amp;amp;bih=839&amp;amp;q=front+of+a+citibike&amp;amp;oq=front+of+a+citibike&amp;amp;gs_l=img.3...310.3798.0.4053.19.13.0.6.0.0.172.1444.7j6.13.0....0...1ac.1.64.img..0.12.1356...0.ZjpB6HqI6nc#tbm=isch&amp;amp;q=citibike+front+triangle&amp;amp;imgrc=90PAzxws2cjzaM%3A&quot; rel=&quot;nofollow noreferrer&quot;&gt;This google image search&lt;/a&gt; shows the goldish triangle at the front of the bike which goes into the dock for locking. You can see a hole at the front of it, and some dimples part way down - presumably the pins are supposed to engage one or either of these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm afraid I don't know how the pin is sent across.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2016-11-14T17:38:16.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12306" PostTypeId="1" AcceptedAnswerId="12310" CreationDate="2016-11-15T00:12:34.990" Score="2" ViewCount="251" Body="&lt;p&gt;I need to build a device that has motorized a door with a hinge on top, similar to cat doors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ruIaJs.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between these options, which one makes the most sense:&#xA;- use an actuator on the side to push the door up&#xA;- put a pulley on the door hinge and drive it with a stepper motor through a belt&#xA;- put an anchor above the door, on the same plane, and pull a string on a reel with a stepper motor&#xA;- or, any other idea :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What makes the most sense from a mechanical point of view? it's for a small device where the door is roughly 7x7&quot; and weights about 0.6LBS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actuator seems the simplest so far, but before I order parts I would like to have opinions from people that understand this stuff :) &lt;/p&gt;&#xA;" OwnerUserId="8789" LastEditorUserId="1832" LastEditDate="2016-11-15T02:45:20.583" LastActivityDate="2016-11-15T07:15:00.510" Title="Automatic door, linear actuator or stepper motor" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12309" PostTypeId="2" ParentId="10767" CreationDate="2016-11-15T06:04:55.710" Score="1" Body="&lt;p&gt;Pipe diameter and material of construction is also a factor.  One must take the volume of water into account.  With smaller diameters, such as factory reticulation, 1 meter per second is usually the upper limit to prevent hydraulic shock (water hammer), which can destroy some valves, such as butterfly valves.&lt;br&gt;&#xA;With larger diameter pipes, such as for wide area reticulation, even 1 meter per second can damage or destroy concrete pipes, so a bypass should be installed to protect valves and piping.&lt;/p&gt;&#xA;" OwnerUserId="8792" LastActivityDate="2016-11-15T06:04:55.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12310" PostTypeId="2" ParentId="12306" CreationDate="2016-11-15T07:15:00.510" Score="2" Body="&lt;p&gt;All of your suggested methods seem workable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it seems that the major differences are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Whether it resists opening the door by external force. A linear actuator would, while a string wouldn't. It depends on your use case which is more desirable, or whether it matters at all.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Amount of force produced. A linear actuator is the strongest, but if there is a possibility of someone being stuck in between the door you might actually prefer a pulley that can slip without causing damage.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How to find the zero position. After power is cut out and restored, your device must somehow know what is the current position of the door. Typically this would be done with a microswitch at one of the end positions. This would probably be easiest to place in the closed position for the linear actuator or pulley system, and in the open position with the string system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-11-15T07:15:00.510" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12311" PostTypeId="1" CreationDate="2016-11-15T07:18:33.033" Score="9" ViewCount="4847" Body="&lt;p&gt;It is known that Bluetooth Low Energy transmits data only during short time intervals called Connection Events. Connection events occur regullary with predefined period. The rest of the time Bluetooth LE device is does not transmit nor receive data. That is how low energy consumption is reached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, classical Bluetooth BR/EDR has Park state. In this mode the parked slave wakes up at regular intervals to listen to the channel in order to re-synchronize and to check for broadcast messages. The rest of the time Bluetooth BR/EDR device is does not transmit nor receive data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why Bluetooth LE consumes less energy?&lt;/p&gt;&#xA;" OwnerUserId="8785" LastEditorUserId="8055" LastEditDate="2017-01-16T15:29:20.377" LastActivityDate="2019-08-21T05:11:14.733" Title="What is the difference between Bluetooth Low Energy and Bluetooth BR/EDR in Park mode?" Tags="|electrical-engineering|telecommunication|wireless-communication|power|energy-efficiency|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12312" PostTypeId="2" ParentId="12297" CreationDate="2016-11-15T07:30:42.587" Score="1" Body="&lt;p&gt;I have not looked at the specific implementation, but many of the more advanced BLDC control schemes offer torque control:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Direct_torque_control&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Direct_torque_control&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Field_oriented_control&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Field_oriented_control&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The torque of a BLDC motor is directly related to the total current passing through the coils. A typical Field Oriented Control (FOC) gimbal system would have two cascaded PID controllers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Outer PID, which takes input angle from IMU and setpoint for the desired angle, and outputs torque signal.&lt;/li&gt;&#xA;&lt;li&gt;Inner PID, which takes current measurements from motor coils, and setpoint for the desired current (calculated from torque), and outputs the voltage signals. This inner PID usually operates in a rotating coordinate system and a mathematical transform is used to calculate the 3-phase output voltages.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For general information on BLDCs and field-oriented control schemes in particular, &lt;a href=&quot;http://krex.k-state.edu/dspace/bitstream/handle/2097/1507/JamesMevey2009.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this thesis&lt;/a&gt; by James Mevey is a great source.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastEditorUserId="1290" LastEditDate="2016-11-15T14:21:33.137" LastActivityDate="2016-11-15T14:21:33.137" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12313" PostTypeId="2" ParentId="8428" CreationDate="2016-11-15T12:56:52.093" Score="0" Body="&lt;p&gt;Band brakes around each disk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you do to the disk itself depends on the use case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if it's unloaded, like an indicator like a digit in a mechanical computer, friction couple it to the shaft (which must then be capable of driving 80 times the friction torque of each disk).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it has to transmit power, the band brake can hold the outer ring of an epicyclic gear. That'll allow efficient slippage and transfer of the motion to something else (in the 2mm gap between disks).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each band brake is just a length of spring steel like a watch spring, wrapped round part or even most of the disk depending on the torque required ... one end is fixed, the other is either tightened or slackened by some actuator ... the keyboard from a mechanical typewriter?&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2016-11-15T12:56:52.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12315" PostTypeId="1" CreationDate="2016-11-15T16:25:07.997" Score="4" ViewCount="355" Body="&lt;p&gt;I have a system consisting of 4 magnets rotating at speeds between 200 and 6,000rpm, which are facing 4 other magnets with a 0.5mm gap between the two sets. The set-up looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wi4Rk.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wi4Rk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aZeHw.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aZeHw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both sets of magnets have their noth pole facing out so that they repel each other (the 0.5mm gap is maintained by other means). The resulting effect is that when the bottom set starts spinning, the top set follows because the top magnets naturally fall (radially) between the gaps of the bottom ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to do is work out what the decoupling torque provided by this set up is. In other words, if the bottom part is spinning, how much torque do I need to provide to stop the top part from rotating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter between the centres of the magnets is 17mm, they are 90 degrees apart, and I have the following (very basic) information on the magnets themselves:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Diameter: 6mm&lt;/li&gt;&#xA;&lt;li&gt;Performance: 5,900 Gauss&lt;/li&gt;&#xA;&lt;li&gt;Vertical pull: 1.4 kg&lt;/li&gt;&#xA;&lt;li&gt;Slide resistance: 0.28kg&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can anybody point me in the right direction? I have made the following attempt, but I don't think that's right:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;T = 4 * (8.5e-3 * 0.28 * 9.81)  = 0.093 Nm&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2024-03-27T08:08:26.023" Title="Magnetic decoupling torque" Tags="|magnets|electromagnetism|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12316" PostTypeId="1" CreationDate="2016-11-15T17:38:55.823" Score="0" ViewCount="69" Body="&lt;p&gt;Preface:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roskam in book &quot;Airplane Aerodynamics and Performance&quot; explains (in paragraph &quot;1.1 ATMOSPHERIC FUNDAMENTALS&quot;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To determne how water vapor affects air density, consider a given volume filled with a dry air and water vapor mixture. Dalton's law of partial pressures states that the obvious pressure, $p$, of the mixture equals the sum of the dry air pressure, $p_a$ and the water vapor pressure, $p_v$: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$p = p_a + p_v \qquad \mbox{(1.2)}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Also, since the total mass of air inside the volume is equal to the sum of the dry air mass and the water vapor mass, the observed density of the mixture is given by:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\rho = \rho_a + \rho_v \qquad \mbox{(1.3)}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In Eqn. (1.3) it is also assumed that the distribution of masses of dry air and water vapor is uniform. By using the perfect gas law, $p=\rho g R T$, it is possible to show that the density of the mixture is given by:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\rho=\frac{p-p_v}{gRT}+\frac{p_v}{gR_vT} = \frac{1}{gRT}\left( p-p_v\left( 1 - \frac{R}{R_v}\right)\right) \qquad \mbox{(1.4)}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For water vapor or steam, $R_v$=85.89 ft/°R. It follows that $R &amp;lt; R_v$ so that second term in equation (1.4) is negative. Therefore, the density of a mixture of dry air and water vapor is less than that of dry air. Normally the effect of water vapor on air density is small. To illustrate this fact, assume that the observed temperature and pressure of a mixture of dry air and water vapor are 90 °F and 2116.2 lb/ft^2 respectively. If the relative humidity is 100%, the water vapor pressure can be found to be 100.6 lbs/ft^2. The density of the mixture then follows from Eqn (1.4) as: &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\rho =\frac{1}{53.35*32.17*549.7}(2116.2-100.6*0.379)=0.002203  \mbox{slug}/ft^3 \qquad \mbox{(1.5)}$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The corresponding dry air density is 0.002243 slugs/ft^3. The density reduction in this case is therefore only 1.8%.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Granted that in the equations above the term &quot;R&quot; is referred to dry air, in my opinion the sentence &quot;The corresponding dry air density is 0.002243 slugs/ft^3&quot; is not correct because in my calculation I found 0.002130 with the following calculation: &#xA;$$\begin{array}{}&#xA;\rho_a &amp;amp; =\frac{p_a}{gRT} \\&#xA;\rho_a &amp;amp; =\frac{p-p_v}{gRT} \\ &#xA;\rho_a &amp;amp; =\frac{2116.2-100.6}{32.17*53.5*549.7} \\&#xA;\rho_a &amp;amp; =0.002130 &#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, can someone explain if I'm wrong?&lt;/p&gt;&#xA;" OwnerUserId="4916" LastEditorUserId="4916" LastEditDate="2016-11-16T16:08:59.640" LastActivityDate="2016-11-16T16:08:59.640" Title="Roskam Book equation of dry air density is not clear. I need explanation" Tags="|thermodynamics|gas|compressed-gases|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12317" PostTypeId="2" ParentId="12279" CreationDate="2016-11-15T20:02:08.190" Score="1" Body="&lt;p&gt;What matters is not the dimensional value of the thickness but rather the ratio with a reference thickness like you stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples I can think of are when dealing with turbulence:&#xA;$U^+$ and $y^+$ which are the non-dimensional velocity and distance from the wall, respectively, when discussing the Law of the Wall in Turbulence.&lt;/p&gt;&#xA;" OwnerUserId="8225" LastActivityDate="2016-11-15T20:02:08.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12318" PostTypeId="1" CreationDate="2016-11-15T20:41:42.617" Score="2" ViewCount="92" Body="&lt;p&gt;so I am very interested in flying.  I don't have any formal education in the topic so take my questions for what they are.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about VTOL aircraft and the current state of such machines.  A thought crossed my mind as I was pondering the topic.  Would it be possible to produce a solid structure VTOL hybrid airship/airplane that is only slightly heavier than the air around it?  It seems like this type of craft would provide for very long flights and &quot;futuristic&quot; landings.  I understand this is no spaceship but the implications of building such a device seem like the &quot;henry ford&quot; moment of our generation.  An interesting hybrid called the solarship was recently developed by some private company.  That craft uses two different methods for generating lift (lifting gas and a wing).  There is another method that we use to generate lift on helicopters.  Could all three be combined with a craft slightly heavier than air?  The mechanical motion required for lifting such devices is very low (little RPM required).  The solarship is extremely small.  I don't suspect we will develop personal airships.  I was thinking more like larger commercial craft though both would be cool.  I understand the craft would need to be made up of something very light, maybe a metal microlattice or a carbon body or even 3D printed plastic.  It seems like such devices could benefit from hollow wings filled with lifting gas as well.  Additional concerns would deal with punctures to the hull.  Maybe a balloon inside the hull?  There are solid structure airships already but those craft are easily damaged by wind and other extreme weather.      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional question I have relates to the design of airships.  Some recently built craft have deviated from the traditional blimp design.  Why are there no vertical airships?  I assume this is because of buoyancy or just the fact nobody has thought to build one.  With airships the heaviest part will always be on the bottom.  Even if we put the hull on top it would just flip over right?  It seems like we could build some pretty cool designs with these considerations in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now don't get me wrong, I understand there are some major challenges associated with such an undertaking.  Hydrogen is dangerous, helium is in short supply and hot air probably won't provide the necessary lift.  Perhaps this will be more appealing once nuclear fusion is available.  Though I think people will be more concerned with creating new thrust devices once fusion is a reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any helpful information.  I do apologize if the question sounds stupid.  Again, I am not formally educated in aeronautical engineering. &lt;/p&gt;&#xA;" OwnerUserId="8802" LastActivityDate="2016-11-16T21:11:42.383" Title="Lift Requirements for Hybrid Airship" Tags="|aircraft-design|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12319" PostTypeId="2" ParentId="12318" CreationDate="2016-11-15T22:02:48.667" Score="1" Body="&lt;p&gt;In college the graduating class (Cal Poly San Luis Obispo 2012) before us did the AIAA design competition for hybrid airships. I know one of the biggest limitations is the fact that these vehicles are efficient only in large sizes. This then creates a limitation in terms of locations to land and take off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These vehicle are definitely feasible. I believe they used a Helium bladder for the buoyancy and then when compressed back down the ship was able to return to the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess for the reason they haven't been created yet is this size limitation which also incurs a large cost for not enough benefit.&lt;/p&gt;&#xA;" OwnerUserId="8225" LastActivityDate="2016-11-15T22:02:48.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12320" PostTypeId="2" ParentId="12318" CreationDate="2016-11-15T23:07:24.030" Score="3" Body="&lt;p&gt;As with any aircraft an airship need to be able to generate lift equal to its weight to remain airborne and also some additional lift to be able to gain height. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally important is the ability to manoeuvre well enough to avoid collisions with terrain or other aircraft in a variety of reasonably foreseeable weather conditions. It is the latter point which has hampered a lot of lighter than air craft in the past. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It actually makes a reasonable amount of sense to have a marginally heavier than air craft which carries the bulk of its payload via buoyancy and used direct thrust for low speed manoeuvring and body lift for cruising. Indeed there have been several reasonably serious concepts along these lines which are potentially attractive as they combine the heavy lift capacity and range of a fixed wing with the versatility of a helicopter. slight negative bouyancy also tends to make the problems accociated with ground handling and trim/ballast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's certainly not that difficult to come up with a reasonably credible concept for something like this, after all even in the 1930s German zeppelins worked pretty well and the Hindenburg crash was in many way more notable as extremely unfortunate PR than highlighting any fundamental engineering flaw. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the problems was that a the outbreak of WWII fixed wing aircraft had more clear cut military uses and so huge resources were put into the development and so at the end of the war there was a reasonably mature technology and manufacturing base available for adoption for civilian use at a time when there was little spare cash available for developing speculative technologies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the meantime the civil aviation industry has become very conservative in the face of commercial, cultural and regulatory pressure to deliver very very high standards of safety. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of lifting gas helium is indeed in short supply and likely to become more so as it is very important for cryogenic applications. Having said that hydrogen isn't necessarily that bad there are plenty of applications where flammable and explosive flues are routinely stored and transported and it is really just a case of mitigating the risk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For propulsion &lt;a href=&quot;https://en.wikipedia.org/wiki/Ducted_fan&quot; rel=&quot;nofollow noreferrer&quot;&gt;ducted fans&lt;/a&gt; are quite attractive as they are compact in size relative to the thrust they generate and can be vectored to provide manouverability and lift (specifically a vertical thrust component) even at low speeds. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-11-16T21:11:42.383" LastActivityDate="2016-11-16T21:11:42.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12321" PostTypeId="1" CreationDate="2016-11-16T00:55:57.117" Score="3" ViewCount="315" Body="&lt;p&gt;I am doing a process flow question and trying to calculate the grams of $\text{C}_{10}\text{H}_{14}$, $\text{C}_4\text{H}_6\text{O}_3$, and $\text{C}_{12}\text{H}_{16}\text{O}$ produced during a reaction. 134 grams of $\text{C}_{10}\text{H}_{14}$ and 134g of $\text{C}_4\text{H}_6\text{O}_3$ were used as reactants. The reaction also produced 68g of acetic acid. I defined my stream variables in terms of the elements ($\text{CHO}$), since the reaction is of unknown stoichiometry. I created a system of equations as follows, where $x$ is the grams of $\text{C}_{10}\text{H}_{14}$ produced, $y$ is the grams of $\text{C}_4\text{H}_6\text{O}_3$ produced, and $z$ is the grams of $\text{C}_{12}\text{H}_{16}\text{O}$ produced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\text{C}$: $12.98 = 10x/134 + 4y/102 + 12z/176$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\text{H}$: $17.347 = 14x/134 + 6y/102 + 16z/176$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\text{O}$: $1.67 = 0x + 3y/102 + z/176$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I combined these into linear matrix equation&#xA;$$\left[\begin{matrix}&#xA;\dfrac{10x}{134} &amp;amp; \dfrac{4y}{102} &amp;amp; \dfrac{12z}{176} \\&#xA;\dfrac{14x}{134} &amp;amp; \dfrac{6y}{102} &amp;amp; \dfrac{16z}{176} \\&#xA;0 &amp;amp; \dfrac{3y}{102} &amp;amp; \dfrac{1z}{176}&#xA;\end{matrix}\right]&#xA;\left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = &#xA;\left[\begin{matrix}12.98 \\ 17.347 \\ 1.67\end{matrix}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $Ax=b$, and $x=A^{-1}b$, I then took the inverse matrix of $A$ and multiplied it by the matrix $b$ (in my calculator). This resulted in the solution that $x = -16.46$, $y= 18.7$, and $z=197.64$. While this does satisfy the equations, the textbook gave the answers: $x=1.6$, $y=22.9$, and $z=175.5$, which also satisfies the equations, and makes sense unlike my answer where I have a negative gram amount produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe my error has something to do with setting a basis, though I'm not sure what that means. Also, the book has the same equations in its solution that I have, but the answer comes out differently (it does not explain how the three equations were solved, other than that it was simultaneous.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I get the correct answer and where is my mistake? Also if you could explain setting a basis that would be great as well. &lt;/p&gt;&#xA;" OwnerUserId="8803" LastEditorUserId="1832" LastEditDate="2017-04-20T11:06:45.013" LastActivityDate="2018-10-12T20:00:15.930" Title="Chemical Engineering Process Analysis - Process Flow Calculations using linear matrices" Tags="|chemical-engineering|process-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12322" PostTypeId="1" CreationDate="2016-11-16T02:11:56.653" Score="-1" ViewCount="885" Body="&lt;p&gt;Will Hyper loop succeed as a means of human transportation? What would be the adverse effect on the human body travelling at 1000km/h?&lt;/p&gt;&#xA;" OwnerUserId="8805" LastEditorUserId="33" LastEditDate="2016-11-16T16:35:24.657" LastActivityDate="2016-11-16T16:35:24.657" Title="What would be the adverse effect on the human body travelling at 1000km/h?" Tags="|safety|" AnswerCount="3" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12323" PostTypeId="1" AcceptedAnswerId="12333" CreationDate="2016-11-16T03:10:44.273" Score="4" ViewCount="20254" Body="&lt;p&gt;When I see electrical generators specified, they usually give the fuel consumption as at &quot;rated load&quot;. So,for example, a 5 kilowatt generator might consume 0.6 gallons per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what happens if the load is less than 5 kilowatts? Does the generator still run at the fixed rate and any excess electricity is just shorted to ground?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does the engine run slower and just generator less energy? How does it know what rate to run at? Does it have to manually adjusted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the engine is run at less than its maximum load, what would be a typical loss of efficiency? Can someone provide an example curve that shows how a generator loses efficiency at sub-optimal loads?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2022-06-05T11:19:51.680" Title="Do generators run slower if the load is lower?" Tags="|generator|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12325" PostTypeId="2" ParentId="12322" CreationDate="2016-11-16T04:51:49.983" Score="-1" Body="&lt;p&gt;Not in my opinion, due to more design cons.&#xA;Effect due to speed is no adverse effect, but in Hyperloop at least we may have acceleration (G-force) which have adverse effect to body.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2016-11-16T04:51:49.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12327" PostTypeId="2" ParentId="12322" CreationDate="2016-11-16T08:06:21.700" Score="0" Body="&lt;p&gt;Acceleration has an effect on bodies in classical mechanics, but not the speed itself.&#xA;Nevertheless, for friction to the air should be considered. I hope you do not have to consider friction to the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The human has (almost) no movement in the vehicule.&lt;/p&gt;&#xA;" OwnerUserId="8112" LastActivityDate="2016-11-16T08:06:21.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12329" PostTypeId="2" ParentId="12315" CreationDate="2016-11-16T08:36:00.873" Score="0" Body="&lt;p&gt;I think you are facing 2 different modes, depending on the difference of speed between both magnets. When both magnet are are at the same speed, the magnetic torque from one of the other is null.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's think on it first as a static problem, and suppose that your body1 is immobile, without any torque, the body2 make an angle of 45°. Let's rotate magnet2 from 45° to 0°. An opposite magnetic torque is created (in order to go back to the stqble position). Maximum torque would be reach at 0°. Continue to rotate, directly after position 0°, the repeal force of the magnet apply in the opposite direction (so do the torque). The magnets are pushing your system to the next stable position at -45°. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under this maximum torque, both bodies are lightly coupled, an rotate at almost the same speed.&#xA;The torque can be calculated from the repealling force between the magnets. You should calculate for one magnet the repealling effect to a second magnet depending to the distance between both magnets. &#xA;Then express the force of 4 magnets from 1 body on one magnet of the second body (then multiply per 4), and deduce the torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyong this torque, both bodies are decoupled, and the magnetic torque is alternatively positive and negative. As a result, speed of both bodies are not not significantly impacted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to stop one spinning body thanks to the second, I would say, you first have to couple both (same rotating speed), then apply a torque. This torque must be under the maximum one, otherwise you decouple the bodies anew.&lt;/p&gt;&#xA;" OwnerUserId="8112" LastEditorUserId="8112" LastEditDate="2016-11-16T08:43:23.283" LastActivityDate="2016-11-16T08:43:23.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12330" PostTypeId="2" ParentId="12323" CreationDate="2016-11-16T09:11:41.623" Score="4" Body="&lt;p&gt;At a lower load they run at the same speed but the generator becomes easier to turn which means the throttle to the engine driving it can be reduced to save fuel.&#xA;The throttle is controlled by a system designed to maintain the correct speed no matter the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The efficiency Vs load curve is going to be generator specific.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-11-16T09:11:41.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12331" PostTypeId="2" ParentId="12322" CreationDate="2016-11-16T09:25:45.053" Score="2" Body="&lt;p&gt;No impact at all as long as you don't open the window.&#xA;The effect of ~500 mph wind (~800km/h) on someones face can be seen here: &lt;a href=&quot;https://www.youtube.com/watch?v=IU4SDDNXuUA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=IU4SDDNXuUA&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone in the space station will be going a lot faster than that (about 27,600km/h) and they have a nice smooth ride (and even bigger problems if someone opened a window). The rough bits are getting there and back, they have to accelerate up to those speeds and then slow down again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assume that the air the person is breathing is moving with the person then speed isn't an issue but acceleration is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acceleration to get to 1000 km/h and slow down again could have an impact but it only needs 1g for 30 seconds. 1g sitting down and sitting facing forward isn't excessive. Slowing down would have to take longer unless you somehow rotated the seats so that you faced backwards for the braking. It's one thing to be pushed back into a chair, it's another to be thrown forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other big issue is bumps or corners in the track. At those speeds it doesn't take much of a bump to result in large accelerations, similarly any corner will have to have a massive turning circle to avoid excessive lateral forces. You need a very smooth track to avoid it feeling like the worlds largest rollercoaster.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastEditorUserId="7659" LastEditDate="2016-11-16T14:20:19.247" LastActivityDate="2016-11-16T14:20:19.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12332" PostTypeId="1" CreationDate="2016-11-16T11:56:09.370" Score="2" ViewCount="1328" Body="&lt;p&gt;I have encountered some issues with sketching the nyquist plot for the following loop transfer function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(s) = \frac{25K}{S^2+25}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that you first take:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(\omega=0) = K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L(\omega = infinity) = |0| e^{-180}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the larger part of the D contour, as for the smaller section of the D counter, I sub $\omega = \epsilon e^{\gamma}$ to represent a number with small magnitude and phase, if I substitute that into the transfer function, that is $L(\epsilon e^{\gamma}) = \frac{25K}{25} = K$ as $\epsilon e^{\gamma} + 25 \approx 25 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this information I can do a rough sketch of the nyquist plot, but when I plot it out on matlab I get something very different. Is my thought process on this correct?&lt;/p&gt;&#xA;" OwnerUserId="3637" LastActivityDate="2016-11-17T17:49:02.530" Title="Question on sketching nyquist plot of transfer function with poles on imaginary axis" Tags="|control-engineering|control-theory|matlab|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12333" PostTypeId="2" ParentId="12323" CreationDate="2016-11-16T12:40:19.140" Score="4" Body="&lt;p&gt;Generators don't use fuel at all.  You are apparently asking about a generator coupled to some sort of chemical engine, like a &quot;genset&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually these engine/generator combinations have a regulator that tries to keep the speed reasonably constant.  This supplies as much fuel to the engine as necessary to maintain that speed.  With lower electrical load on the generator, it puts less mechanical load on the engine, which then requires less fuel to maintain the same speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is no different from the cruise control in your car.  It maintains 55 miles/hour, for example, on a highway as you go up and down hills.  However, it opens the throttle more, and delivers more fuel to the engine, when going up hills than when going down.  Your milage (miles/gallon) is quite different going up a hill than down it, even though both are at the same speed.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-11-16T12:40:19.140" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12334" PostTypeId="2" ParentId="12323" CreationDate="2016-11-16T13:34:21.733" Score="1" Body="&lt;p&gt;If you are talking about gas electric generators designed to supply household ac power, there are two basic types.  These generators must supply a reasonably accurate 60hz frequency power, and there are two basic ways to achieve that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;most commonly the generator is wound to produce 60hz at a specific speed and the generator &lt;em&gt;must be&lt;/em&gt; run at that constant speed regardless of load (as noted the throttle is closed to reduce air/fuel under reduced load.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the alternate is to use an electronic device called an inverter to regulate the ac frequency. This permits the actual generator to be run at variable speed. (This is typical of small camping type generators).  Inverter type gensets do indeed drop speed to a minimal level under light load. Note this has as much to do with making them quiet as saving fuel.  Note also the inverter itself has an efficiency cost to it so these units may be less fuel efficient at high load than comparable fixed speed units.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2016-11-16T13:40:17.553" LastActivityDate="2016-11-16T13:40:17.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12338" PostTypeId="2" ParentId="12332" CreationDate="2016-11-16T14:55:40.017" Score="2" Body="&lt;p&gt;For the pole at $5 i$, the contour that has to be considered is $5 i+\epsilon e^{i \theta }$. Here $\epsilon \to 0$ and $\theta \in[-\frac{\pi }{2},\frac{\pi }{2}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The denominator of the transfer function becomes $\left(5 i+\epsilon e^{i \theta } \right)^2+25$. Expanding and neglecting higher order terms, we get $-10 \epsilon  \sin (\theta )+10 i \epsilon  \cos (\theta )$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transfer function becomes $$\frac{25}{-10 \epsilon  \sin (\theta )+10 i \epsilon  \cos (\theta )}$$ which after simplification is &#xA;$$-\frac{5 (\sin (\theta )+i \cos (\theta ))}{2 \epsilon }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically a semicircle of infinite radius because $\epsilon \to 0$ and it lies in the lower half plane because $\theta \in[-\frac{\pi }{2},\frac{\pi }{2}]$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GEovh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GEovh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar analysis can be done for the pole at $-5 i$, and the semicircle will lie in the upper half plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the complete Nyquist plot will loook like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qZWL1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qZWL1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2016-11-17T17:49:02.530" LastActivityDate="2016-11-17T17:49:02.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12340" PostTypeId="1" CreationDate="2016-11-16T16:40:03.253" Score="1" ViewCount="456" Body="&lt;p&gt;Hey guys I need some help with a fluid mechanics problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a capillary with the given dimensions $d=0,15 mm$ and $L=5 cm$ and the operating conditions  $p_{1}=4\cdot10^{5} Pa$,  $p_{2}=0.25\cdot10^{5} Pa$ &lt;br/&gt; &lt;br/&gt; &#xA;I want to calculate the velocity of oxygen flowing from 1 to 2. So I start with the formula for pressure drop with laminar flow &lt;br/&gt; &lt;br/&gt; &#xA;$\Delta p=\zeta \frac{L}{d}\frac{\rho u^{2}}{2}$ with $\zeta =\frac{64}{Re}$&#xA;&lt;br/&gt; &#xA;&lt;br/&gt; &#xA;which gets me $u=261 \frac{m}{s}$&lt;br/&gt; &lt;br/&gt; &#xA;After calculating Reynoldsnumber to check wether laminar assumption was right, I see that flow is turbulent ($Re=10175$)&lt;br/&gt; &lt;br/&gt; So instead I have to take $\zeta =\frac{0.3164}{\sqrt[4]{Re}}$ (Blasius)&lt;br/&gt;&lt;br/&gt;&#xA;But with a little help of wolfram alpha&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;solve p=(0.3164/((u&lt;em&gt;d&lt;/em&gt;rho)/eta)^(1/4))&lt;em&gt;l/d&lt;/em&gt;(rho*u^2)/2 for u&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I get a velocity of $u=142.9 \frac{m}{s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make any sense? I'm really not sure if I can calulate the velocity like this. &lt;/p&gt;&#xA;" OwnerUserId="8282" LastActivityDate="2018-10-21T23:00:34.207" Title="Calculating velocity in pipe flow" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12342" PostTypeId="1" AcceptedAnswerId="12343" CreationDate="2016-11-16T17:33:05.040" Score="1" ViewCount="4547" Body="&lt;p&gt;If an air pump needs to able to generate a particular pressure at a particular flow rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we size the electrical motor for this particular scenario? The electric motor would be coupled to the pump shaft in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean how do I calculate the mechanical power the motor shaft has to deliver to achieve this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the pump technology have any bearing on this I mean diaphragm, piston etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets take an example pressure of 1 psi at a flow rate of 10 liters / min.&#xA;I need to know the wattage of the motor that would be a suitable fit for this specification.&lt;/p&gt;&#xA;" OwnerUserId="8812" LastActivityDate="2016-11-17T16:18:51.620" Title="Relationship between output mechanical power and max pressure for a pump?" Tags="|pressure|pumps|power|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12343" PostTypeId="2" ParentId="12342" CreationDate="2016-11-16T19:50:08.870" Score="3" Body="&lt;p&gt;Generally speaking, for fluid systems the mechanical power can be computed as the product of pressure and volumetric flow rate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{E} = P\dot{V}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\dot{E}$ is the power, $P$ is the pressure and $\dot{V}$ is the flow rate. Therefore, for your application (with unit conversions to get the output in watts):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{E} = (1 \text{ psi})(\frac{6894.76 \text{ Pa}}{1 \text{ psi}})(10 \text{ L/min})(0.001 \text{ m}^3/\text{L})(\frac{1 \text{ min}}{60 \text{ s}}) \approx 1.15 \text{ W}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the required flow and pressure are not the only considerations. You need to consider all the sources of friction and other energy losses in the system: friction in the motor bearings, efficiency losses when converting mechanical energy in the impeller to fluid flow in the system, etc. Therefore you would need an electrical motor with more than just 1.15 W of power. The energy losses from the impeller to the fluid flow would be difficult to estimate if you're building your own pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider that an electric motor cannot output 1.15 W for any combination of speed and torque, so you need to check the motor's spec sheet to make sure that it can provide the power you need at the desired motor speed.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="6695" LastEditDate="2016-11-17T16:18:51.620" LastActivityDate="2016-11-17T16:18:51.620" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12345" PostTypeId="1" AcceptedAnswerId="12871" CreationDate="2016-11-17T04:31:33.917" Score="0" ViewCount="747" Body="&lt;p&gt;I'm designing a gear system which is 1 motor(drive gear) will drive the last gear(driven gear), and there will be a few gear between the motor and output gear. I understand that the gear ratio between motor and output gear didn't change even I add how many gear in between, and even I change the size of gears in between. As long as I keep the input and output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the link reference for above explanation &lt;a href=&quot;http://www.wikihow.com/Determine-Gear-Ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.wikihow.com/Determine-Gear-Ratio&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what happen if I add cluster gear in between, does it change the ratio between drive gear and driven gear? As what I understand, it will not change the gear ratio, but a few engineer's friend told me different thing. I cannot found solid explanation on this.&#xA;Please refer attached image for what I mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zUjhh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zUjhh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="8822" LastEditorUserId="8822" LastEditDate="2016-12-16T02:01:21.150" LastActivityDate="2016-12-16T09:53:45.150" Title="Does Cluster Gear Change The Gear Ratio" Tags="|motors|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12346" PostTypeId="1" CreationDate="2016-11-17T04:48:13.260" Score="-1" ViewCount="59" Body="&lt;p&gt;I am wondering whether a typical mechanic can break a bolt using a wrench.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's say the wrench is 15&quot; long and the force is applied at a radius of 13&quot;. The mechanic applies a force of 40 pounds. Imagine the thread of the bolt is 3/8&quot; in diameter and has 16 threads per inch. The bolt has an ultimate tensile strength of 150,000 PSI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the bolt fail and break apart?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="255" LastEditDate="2016-11-17T12:39:33.010" LastActivityDate="2016-11-17T13:00:36.127" Title="Can a mechanic pull apart a strong bolt by hand power alone with an ordinary wrench?" Tags="|mechanical-engineering|torque|mechanical-failure|" AnswerCount="1" CommentCount="4" ClosedDate="2016-11-17T22:16:33.413" ContentLicense="CC BY-SA 3.0" />
  <row Id="12348" PostTypeId="1" AcceptedAnswerId="12351" CreationDate="2016-11-17T08:38:46.820" Score="0" ViewCount="60" Body="&lt;p&gt;Where can I find (cite-able sources) properties (such as specific heat at constant volume, thermal conductivity, viscosity etc...) of air at different temperatures (Range of 200K to 2000 K)?&lt;/p&gt;&#xA;" OwnerUserId="7124" LastActivityDate="2016-11-17T12:17:58.727" Title="Standard properties of air" Tags="|standards|" AnswerCount="1" CommentCount="1" ClosedDate="2016-11-17T16:35:25.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="12349" PostTypeId="1" AcceptedAnswerId="12358" CreationDate="2016-11-17T08:44:32.970" Score="3" ViewCount="454" Body="&lt;p&gt;I have some questions about the enthalpies between two points during an isentropic process in a nozzle. By doing the energy balance between the two points, one gets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h_1 + \frac{v_1^2}{2}+gz_1=h_2+\frac{v_2^2}{2}+gz_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one neglects the potential energy and assuming that the velocity at the inlet is 0, one ends up with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$h_1 =h_2+\frac{v_2^2}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the part that raises some questions. If the process is isentropic, it means that the entropy doesn't change and therefore there is no heat added (or vice-versa) to the flow. But wouldn't that make $h_1 = h_2$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that because $h = u + Pv$, the Pv term is the one that makes them different, considering that the flow energy (Pv) is different. But this would make $u_1 = u_2$. Is this true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if this is true, why don't we just use the following relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_1v_1=P_2v_2+\frac{v_2^2}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, one can't really define $Cp$ or $k=Cp/Cv$, which would lead to the well-known isentropic flow equations. Am I getting something wrong? Can anyone help me to understand this? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="8419" LastActivityDate="2016-11-17T15:30:14.583" Title="Shouldn't enthalpies be equal in isentropic process?" Tags="|fluid-mechanics|thermodynamics|airflow|compressible-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12350" PostTypeId="1" CreationDate="2016-11-17T08:44:44.423" Score="0" ViewCount="1020" Body="&lt;p&gt;Wouldn't it be easier to just pump concrete over the reactor until it's all covered up with a solid concrete instead of hurting so many workers and putting so much effort and costs building a structure over it and cleaning up the area?&lt;/p&gt;&#xA;" OwnerUserId="8168" LastEditorUserId="8168" LastEditDate="2016-11-19T04:06:46.230" LastActivityDate="2016-11-19T04:06:46.230" Title="Why did they build a sarcophagus over Chernobyl instead of pouring concrete over the building?" Tags="|nuclear-technology|architecture|construction-management|" AnswerCount="2" CommentCount="7" ClosedDate="2016-11-17T22:15:41.910" ContentLicense="CC BY-SA 3.0" />
  <row Id="12351" PostTypeId="2" ParentId="12348" CreationDate="2016-11-17T12:08:18.217" Score="3" Body="&lt;p&gt;Cengel's Book: Yunus A. Cengel;John M. Cimbala;Robert H. Turner, Fundamentals of Thermal-Fluid Sciences. 4th. McGraw-Hill, 2012&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an extract of the property tables: &lt;a href=&quot;http://www.kostic.niu.edu/350/_350-posted/350Chengel7th/Appendix1Udated.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.kostic.niu.edu/350/_350-posted/350Chengel7th/Appendix1Udated.pdf&lt;/a&gt; (for air, see A-2, A-16 and A-17)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also: Yunus Cengel and Afshin Ghajar, Heat and Mass Transfer, Fundamentals and Applications, 4th McGraw-Hill, 2011 (for air, see A-15)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have all the properties of air at 1 atm ($\rho , C_p, k, \alpha , \mu , \nu, Pr$) from -150 $^{\circ}$C to 2000 $^{\circ}$C. See &lt;a href=&quot;https://www.scribd.com/doc/30589475/Heat-Transfer-Appendix-tables&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8419" LastEditorUserId="8419" LastEditDate="2016-11-17T12:17:58.727" LastActivityDate="2016-11-17T12:17:58.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12353" PostTypeId="2" ParentId="12350" CreationDate="2016-11-17T12:33:44.910" Score="5" Body="&lt;p&gt;The likely reason is that simply pouring concrete over it would not have contained it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The site is smoldering and emits gases of various types. It also has voids and shifts around, various areas collapsing into pockets. If it were covered with concrete, the concrete would develop cracks and radioactive particles would be emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sarcophagus structure is designed to be flexible and to seal everything in as tightly as possible without cracking, although it does have ventilation, so radioactive gases do escape. The structure also has viewing ports that allow the interior to be monitored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A large amount of concrete was used in the construction of the sarcophagus.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2016-11-17T17:52:04.660" LastActivityDate="2016-11-17T17:52:04.660" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12354" PostTypeId="2" ParentId="12346" CreationDate="2016-11-17T13:00:36.127" Score="1" Body="&lt;p&gt;torsion T = 40 pounds * 13 inches = 520 pound-inches&lt;/p&gt;&#xA;&#xA;&lt;p&gt;shear stress = Tr/J&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that threads would be ignored, and stress would only be calculated on the shank diameter, but seeing as you haven't provided that, I'll just use 3/8in as the available shaft to resist the torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torsional constant J = pi * D&lt;sup&gt;4&lt;/sup&gt; / 32 = 3.1415 * (3/8)&lt;sup&gt;4&lt;/sup&gt; / 32 = 1.941E-03 in&lt;sup&gt;4&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;radius r = D/2 = 3/16in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore stress = 520 * (3/16) / 1.941E-03 = &lt;strong&gt;50,220psi&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear strength is about 55% of tensile strength, giving you a limiting shear stress of 0.55 * 150,000 = &lt;strong&gt;82500psi&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on your numbers, the bolt survives. But I reckon I could supply 100 pounds of force (and I'm not exactly a strong guy), which would be enough to shear the bolt.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2016-11-17T13:00:36.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12355" PostTypeId="2" ParentId="12350" CreationDate="2016-11-17T14:37:29.063" Score="4" Body="&lt;p&gt;The original sarcophagus built around 1986 has some issues and it was estimated in 1988 that it would only last 20-30 years (here we are). Check out this &lt;a href=&quot;https://en.wikipedia.org/wiki/Chernobyl_Nuclear_Power_Plant_sarcophagus&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; that explains some of the main issues with the current sarcophagus; namely structural instability. Trying to repair the current sarcophagus is impractical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new confinement structure is designed to help shield the site from weather (e.g. rain water drains through the reactor picking up radiocative materials and then leaking into the groundwater). It also allows workers to partially dismantle and clean up what is there, which is a better long term solution. Here is a link about the new &lt;a href=&quot;https://en.wikipedia.org/wiki/Chernobyl_New_Safe_Confinement&quot; rel=&quot;nofollow noreferrer&quot;&gt;confinement structure&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1174" LastActivityDate="2016-11-17T14:37:29.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12356" PostTypeId="1" CreationDate="2016-11-17T14:46:08.903" Score="0" ViewCount="54" Body="&lt;p&gt;i am designing an experimental platform to test some 3D printed gears that i am designing. i want to test the durability and what i am looking for is a way to make them wear out fast.&#xA;i will basically put 2 or 3 gears in series and form a gear train, then run them until they &quot;fail&quot;, i am basically cehcing the vibration emitted by each gear, and i currently do it by placing an accelerometer above the shaft supporting the gear itself (above the shaft supporting bearing to be precise)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;however the gears reach a point where the vibration increases no more just after running for 15 min.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice on what kind of way to install the accelerometers or other sensors?&#xA;Or if i should change my DC motor, runs at 24V at 2000RPM but heating up too fast so i have to stop every 10 15 min of run.&lt;/p&gt;&#xA;" OwnerUserId="6682" LastActivityDate="2016-11-17T21:06:40.970" Title="How to drive a gear system to wear out fast" Tags="|motors|gears|experimental-physics|acceleration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12358" PostTypeId="2" ParentId="12349" CreationDate="2016-11-17T15:30:14.583" Score="4" Body="&lt;p&gt;Constant entropy doesn't mean constant temperature, the gas flow through nozzle is subject to change in pressure and velocity (continuity equation) due to the variation in cross section, and being compressible the change in pressure will be reflected as change in temperature and density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, since the flow is isentropic and there will be no energy dissipation due to friction and by applying the first law of thermodynamics to the nozzle you will (naturally!) find that the sum of energy across the nozzle is constant, meaning that the stagnation enthalpy ($h_1$) will equal the sum of enthalpy and kinetic energy at every other point in the nozzle (any increase in kinetic energy will be reflected as decrease in enthalpy and vice versa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your question: $h_1 \neq h_2$ and also the sensible temperature $T$ reflected in internal energy is not necessarily the same for the two point in comparison, meaning that also $u_1 \neq u_2$, you have to look at the big picture where the change in cross section will affect the velocity, pressure, temperature and density (that's why the analysis of compressible flow is far more complex than incompressible flow).&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2016-11-17T15:30:14.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12362" PostTypeId="2" ParentId="12356" CreationDate="2016-11-17T21:06:40.970" Score="1" Body="&lt;p&gt;The difficulty is that 'wearing out' is a bit of a vague concept in engineering and it is not always easy to accelerate wear in a way which is meaningful in the real world. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many applications running gears under a reasonable approximation of normal operating conditions continuously won't actually take that long to get to a substantial fraction of their design life, bearing in mind that most plastic gear applications tend to run fairly intermittently. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also as mentioned in the comments you really need to consider what failure mechanisms are most significant. In my experience plastic transmission parts tend to fail through overloading, especially shock and stall type loads long before they wear out as such. Often the failure point is the point at which the gear is attached to the shaft rather than the teeth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally what load do you have on the gear train ?  Apart form anything else the no-load torque generated by the gear train is an initial indicator of quality of terms of fit, finish and dimensional tolerances, which is always a concern with 3D printed parts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With any sort of testing you need to be quite focused on what you are actually trying to measure ad what you intend to learn from the test. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-11-17T21:06:40.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12366" PostTypeId="1" AcceptedAnswerId="12367" CreationDate="2016-11-17T23:49:05.767" Score="3" ViewCount="264" Body="&lt;p&gt;I'm intrigued by &lt;a href=&quot;https://www.youtube.com/watch?v=TsaES--OTzM&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; in which the roof camera rotates continuously. One reason could be to capture 360 degrees around, but then again two cameras or more could do the same job capturing different parts of the landscape. It seems to me a waste of energy in making the camera rotate, and also having mobile parts make it more vulnerable to breaking. I don't see why not using a system of many static cameras.&lt;/p&gt;&#xA;" OwnerUserId="5033" LastEditorUserId="6767" LastEditDate="2019-01-04T13:13:45.070" LastActivityDate="2019-01-04T13:13:45.070" Title="Why does &quot;Google self-driving car&quot;'s roof camera rotate?" Tags="|control-engineering|car|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12367" PostTypeId="2" ParentId="12366" CreationDate="2016-11-18T00:09:54.910" Score="7" Body="&lt;p&gt;It isn't an ordinary video camera. It's a high-resolution LIDAR that uses laser beams to measure the distance to thousands of points on every rotation. I believe the one used by the car in that video is the &lt;a href=&quot;http://velodynelidar.com/hdl-64e.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Velodyne HDL-64e&lt;/a&gt;, which I have used myself on a similar project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wtyjb.png&quot; alt=&quot;photo of Velodyne HDL-64e&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sends out a vertical array of 64 pulsed laser beams and uses time-of-flight to measure the distance to a reflective object. The physical rotation gives you the scanning in the horizontal direction. The data is in the form of a &quot;cloud&quot; of 3-D points that represent reflective surfaces that the sensor can see. It's fascinating to watch a 3-D plot of the raw point cloud change in real time as both the sensor and the objects in view move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes quite a bit of sophisticated software to connect the individual points together to represent actual surfaces and objects. It is the output of this software that you see in the video, representing vehicles, pedestrians, etc. as boxes that approximate their actual shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resolution of the sensor is specified as angles (0.4&amp;deg; vertical, 0.08&amp;deg; horizontal), so the actual spatial resolution depends on the range. For example, at a range of 10 m, the vertical resolution is about 6 cm and the horizontal resolution is about 1.3 cm. The range resolution is about 2 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular sensor is the &quot;gold standard&quot; for such devices, used in research projects where the accuracy of the data is paramount and cost is no object. It's far too expensive to be used in a mass-produced vehicle design. Indeed, it may require a breakthrough in 3-D (stereoscopic) image processing algorithms using cheap video sensors to get to the point where the data required for navigation can be acquired at a reasonable cost.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="6767" LastEditDate="2019-01-04T12:40:20.400" LastActivityDate="2019-01-04T12:40:20.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12368" PostTypeId="2" ParentId="12366" CreationDate="2016-11-18T00:11:03.543" Score="4" Body="&lt;p&gt;That's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lidar&quot; rel=&quot;nofollow noreferrer&quot;&gt;lidar&lt;/a&gt; sensor, which is making a continuous 360 degree sweep, rather than a camera which has a field of view as such. Rotating mirrors are a fundamental part of how the system works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in the video the system represents traffic as fairly crude boxes rather than rendered images, so it may be looking specifically at reflectors, which also helps it to identify signs and traffic cones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, they are talking very much about the software and control systems, so it seems likely that the sensor rig in the video is chosen to be convenient for testing, rather than a fully developed and refined consumer version. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, a fully autonomous car may well need a variety of different sensors to guarantee redundancy and effectiveness on a variety of conditions, but in good weather with a professional human driver supervising the vehicle, a bolt-on lidar sensor is a convenient way to fine tune the software. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="6767" LastEditDate="2019-01-04T13:12:53.253" LastActivityDate="2019-01-04T13:12:53.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12369" PostTypeId="1" CreationDate="2016-11-18T03:50:48.497" Score="3" ViewCount="94" Body="&lt;p&gt;I am an architect in need  help with an aerodynamics problem for a design that I am developing. I need  to know the  the potential velocity of air moving from one end of a pipe to the other if any with a constant  20° difference in temperature. The temperature at one end of the pipe would be uniformly 20°F colder than the temperature at the opposite end.  I'm assuming the temperature difference would cause warm air to move to the cold end. The pipe /tunnel would be 30 feet in diameter and 5000 feet in length. I am interested in using any potential work or force from this volume of air, moving due to temperature difference. Thank you for any help that you can provide.&lt;/p&gt;&#xA;" OwnerUserId="8839" LastActivityDate="2016-12-18T11:10:55.117" Title="Air velocity through a tunnel 5,000 feet long and 30 feet in diameter?" Tags="|aerodynamics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12370" PostTypeId="1" CreationDate="2016-11-18T06:22:05.643" Score="1" ViewCount="130" Body="&lt;p&gt;Suppose I have nine machines in total which require a continuous supply of air at 2.5-3 kg/cm2. So, can I get maximum safe working pressure of the compressor by working on the formula of circumferential and longitudinal stress for pressure vessels which requires parameters such as internal pressure, wall thickness and shell diameter. Secondly, how do I calculate the capacity and HP of compressor which would suit my requirement. Also, at some websites, the CFM(cubic feet per minute) is explained as mass of 1 cubic feet of air which a compressor breathes per minute. But isn't it volumetric flow rate ?&lt;/p&gt;&#xA;" OwnerUserId="739" LastActivityDate="2016-11-18T11:53:31.870" Title="How to know what capacity of a compressor will suit the demand of machines in a plant?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12372" PostTypeId="2" ParentId="12369" CreationDate="2016-11-18T10:40:56.463" Score="1" Body="&lt;p&gt;First of all, the flow of air is caused by the difference in pressure, and not in temperature, although they are related. So you need to know the pressure drop. You have to guess the pressure difference, or use the ideal gas law for finding the pressure at the ends:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=\rho RT$, where $\rho$ is the density of air at that specific temperature (see &lt;a href=&quot;http://www.engineeringtoolbox.com/air-density-specific-weight-d_600.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;), $T$ is the temperature that you must know and $R$ is a gas constant for air ($R$=53.35 ft lbf/lb/$^{\circ}$R).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to find the velocity, but that would require to know the roughness of the pipe, flow regime etc. I recommend that you do your calculations for velocity &lt;a href=&quot;http://www.endmemo.com/physics/poiseuille.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; once you know your pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And use the properties of air from &lt;a href=&quot;http://www.engineeringtoolbox.com/air-absolute-kinematic-viscosity-d_601.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can fine the conversion in velocity (from Imperial to SI) &lt;a href=&quot;http://www.unitconversion.org/unit_converter/velocity.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8419" LastActivityDate="2016-11-18T10:40:56.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12374" PostTypeId="2" ParentId="12370" CreationDate="2016-11-18T11:53:31.870" Score="3" Body="&lt;p&gt;You've got the 2.5-3 kg/cm^2 which is the pressure; approximately 2.5-3 bar. You still need to find the volumetric or mass flow too to fully determine your requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't try to go with stress calculations if you're shopping for a compressor, and not building one from scratch. Get one rated for pressure you require with a healthy margin - not a hard thing with 3 bar required; most available on the market can easily provide more. Since you need continuous supply, tank volume is secondary - you really need the flow to exceed the requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CFM measurement is expressed that way because the flow will change with pressure - if the compressor takes 100 cubic feet of air per minute in, it will output 50 cubic feet of air per minute at 2 bar, or 25 cubic feet per minute at 4 bar. By using mass equivalent you can easily convert your pressure and volume requirement to CFM - just multiply the volumetric flow by pressure in bar, to get the multiplier required on the intake. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2016-11-18T11:53:31.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12376" PostTypeId="2" ParentId="12225" CreationDate="2016-11-18T12:46:34.027" Score="2" Body="&lt;p&gt;If you are at the face of an element, there cannot be any &lt;strong&gt;internal&lt;/strong&gt; shear stress, because there is nothing to balance it out. The bottom horizontal force I added cannot be balanced by another force since there is no more material at the face of the flange. The thinner your element, the less it will resist in shear and more it will resist in bending.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IevTs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IevTs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5691" LastEditorUserId="5691" LastEditDate="2016-11-18T22:09:27.870" LastActivityDate="2016-11-18T22:09:27.870" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12378" PostTypeId="1" CreationDate="2016-11-18T18:50:34.547" Score="0" ViewCount="1067" Body="&lt;p&gt;I'm currently designing a project for a local museum through Purdue University's engineering class. My team has come down to three options: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Trying on a space suit with functioning parts&lt;/li&gt;&#xA;&lt;li&gt;Looking though a telescope that uses a VR-type experience that displays the Hubble telescope images in a 3D like manner on the end of the telescope, &lt;/li&gt;&#xA;&lt;li&gt;Or a mars rc rover that uses a QR reader to &quot;scan&quot; various objects on a &quot;martian&quot; surface and displays info to the user.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My team is looking for information on just how reliable QR code readers would be. We would only be using 2-3 simple designs (vertical lines, horizontal lines, and a circle). I could not find much information online other than &quot;best ios QR reader&quot; or &quot;best QR readers!&quot;, so if anyone could give some insight to the range of precision that the reader would have (i.e. angle from normal that it would still work, etc) that would be awesome. The reader and code itself would always be at the same height. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously budgets are a factor for this project, and each QR reader differs, but on average how does the QR code reader and software hold up?&lt;/p&gt;&#xA;" OwnerUserId="8852" LastEditorUserId="8852" LastEditDate="2016-11-21T00:59:31.603" LastActivityDate="2017-05-09T06:11:21.043" Title="How reliable is a QR code reader?" Tags="|electrical-engineering|sensors|software|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12379" PostTypeId="1" CreationDate="2016-11-18T19:11:20.337" Score="1" ViewCount="59" Body="&lt;p&gt;I read a manufacturer's web site which has the following statement:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some manufacturers promote their higher horsepower as a selling&#xA;  feature over competitors. However, it is critical to recognize that&#xA;  horsepower is not the determining criteria on saw performance, but&#xA;  instead it is how much torque is delivered to the band wheels. The&#xA;  amount of torque delivered to the wheels is based on the quality and&#xA;  ratio of the gearbox. Today, the higher technology saws can both run&#xA;  more efficiently and deliver more torque to the band wheels, with&#xA;  lower horsepower than on older saws.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand the importance of the gear ratio. Higher gear ratio, more torque. However, why would the &quot;quality&quot; of the gearbox make any ddifference to the torque. If the gearbox is 24:1, then it is same whether it is an expensive high-quality gearbox or a cheapo gearbox, I would think. Why is the manufacturer saying that the gearbox quality affects the saw performance?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2016-11-18T19:19:44.163" Title="Why would &quot;quality of the gearbox&quot; affect torque of a machine tool?" Tags="|gears|torque|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12380" PostTypeId="2" ParentId="12379" CreationDate="2016-11-18T19:19:44.163" Score="1" Body="&lt;p&gt;It is likely due to frictional losses, which can affect the torque output for a certain ratio. The gear material and design will also determine the maximum torque it can handle.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2016-11-18T19:19:44.163" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12381" PostTypeId="1" AcceptedAnswerId="13428" CreationDate="2016-11-18T22:18:43.940" Score="2" ViewCount="2852" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I sometimes design my retaining walls with longer toes and heels, this is because in my FEA program, when I pull my footing under my wall towards the toe, I get lower pressures on my soil (springs) that when I pull it towards the heel (even with the added benefit of the earth on top). This tends to happen when I don't have too much counter balancing soil on my heel. Anyone else get these results? I've never seen any long toe retaining walls in the literature.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When calulating over turning moment by hand, all of the literature I've seen takes the point of rotation at the toe of the wall. If I look at my FEA design, the wall seems to be rotating much more around the heel. Is it justifiable to calculate the overturning moment at the toe? This seems to be the point that's most favorable for design, but is it realistic?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KgCPY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KgCPY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5691" LastEditorUserId="1832" LastEditDate="2016-11-19T11:14:21.773" LastActivityDate="2017-07-12T03:25:50.747" Title="Postion of overturning moment for retaining wall" Tags="|structural-engineering|geotechnical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12382" PostTypeId="1" CreationDate="2016-11-19T01:21:54.567" Score="-7" ViewCount="426" Body="&lt;p&gt;I saw a high altitude jump where someone had exceeded the speed of sound before the air slowed him down enough to deploy a parachute. While inflated in space could a blimp enter the atmosphere while gradually increasing air pressure not to collapse from increased air pressure while descending to allow a soft non retrorocket landing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the air on Mars is carbon dioxide mostly and 1% of Earth's atmosphere could a hard body blimp use the atmosphere to slow and land like the space shuttle or glider?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes it would not be a blimp anymore on Mars but a glider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would change shape and size for higher speeds to use the lift of the wing to over the buoyancy of the helium. I added the ship from the movie Flight of the Navigator because it was a good example of a blimp shaped ship changing shape for different purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZYqD4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZYqD4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a0S32.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a0S32.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorUserId="33" LastEditDate="2018-09-25T18:18:23.407" LastActivityDate="2018-09-25T18:18:23.407" Title="Can a blimp enter Mar's atmosphere as a glider?" Tags="|structural-engineering|materials|aerospace-engineering|experimental-physics|aircraft-design|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="12383" PostTypeId="2" ParentId="8177" CreationDate="2016-11-19T01:40:21.633" Score="0" Body="&lt;p&gt;The key is to understand that you're not inputting the Pressure but the change in pressure ($\Delta P $) which is the difference between the upstream pressure and the back pressure downstream at the exit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise yet you can input the standard density and an appropriate friction factor $f $&lt;/p&gt;&#xA;" OwnerUserId="8225" LastActivityDate="2016-11-19T01:40:21.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12384" PostTypeId="1" CreationDate="2016-11-19T01:48:59.443" Score="0" ViewCount="126" Body="&lt;p&gt;Suppose I want to purchase a DC motor online, how can i determine how much torque my motor will need to run at a certain speed? I.e. 15mph&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for my electric skateboard project and the load on top of the board will be 140 lbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really having a hard time figuring this out and was hoping someone can share their knowledge.&lt;/p&gt;&#xA;" OwnerUserId="8832" LastActivityDate="2016-11-19T12:54:12.593" Title="Figuring out how much torque dc motor will need to run at a certain speed. (E-skateboard project)" Tags="|mechanical-engineering|automotive-engineering|motors|torque|" AnswerCount="1" CommentCount="1" ClosedDate="2016-11-30T23:22:34.523" ContentLicense="CC BY-SA 3.0" />
  <row Id="12385" PostTypeId="2" ParentId="12174" CreationDate="2016-11-19T02:18:04.937" Score="1" Body="&lt;p&gt;It seems that when you try to just resolve the freestream velocity onto the cylinder you neglect the fact that from potential theory flow over a cylinder includes a doublet. (Uniform flow + doublet = flow over a cylinder).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What those resolved components basically do is resolve a freestream uniform velocity in cylindrical coordinates at some distance $r = R$ from an arbitrary point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll need to add the contributions of the doublet at a distance $r$ from the center. That is where is $R^2/r$ comes into play.&lt;/p&gt;&#xA;" OwnerUserId="8225" LastActivityDate="2016-11-19T02:18:04.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12387" PostTypeId="2" ParentId="12384" CreationDate="2016-11-19T11:15:01.573" Score="1" Body="&lt;p&gt;For any powered vehicle the maximum speed on level ground is determined by the sum of the forces resisting its movement. These forces can be roughly broken down into: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rolling resistance : between the wheels and the road surface&lt;/li&gt;&#xA;&lt;li&gt;Internal friction from bearing, transmission etc&lt;/li&gt;&#xA;&lt;li&gt;Air resistance&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Typically the first two will be either more or less constant or proportional to speed, air resistance (drag) tends to increase with the square of speed and it is this which is generally the limiting factor on top speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you apply a constant torque to the wheels the vehicle will continue to accelerate untill the accelerating force is equal to the retarding force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The powered wheels convert torque to force with the relationship T = Fr , where T = Torque at the wheel axle, F = Force and r = wheel radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/drag-coefficient-d_627.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; details how to estimate drag forces and provides a coefficient for a standing person. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To estimate friction forces a simple method is to load up the board with the approriate weight and uses a spring balanc to determine the force required to just get it moving, this is a bit crude but should give you a ballpark estimate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of selecting a motor you will need to consult the &lt;a href=&quot;http://lancet.mit.edu/motors/motors3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;torque curve&lt;/a&gt; for the motor which describes teh relationship between torque and RPM will also influence what overall gear ratio you need. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course maximum speed is no the only thing you need to consider you also need enough torque to provide adequate acceleration and hill-climbing as well as taking bulk, weight and your power supply into account. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-11-19T12:54:12.593" LastActivityDate="2016-11-19T12:54:12.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12388" PostTypeId="1" CreationDate="2016-11-19T14:24:03.393" Score="1" ViewCount="204" Body="&lt;p&gt;As the title says, does a sinusoidal input voltage make a BLDC motor nearly into a PMSM?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.ti.com/lit/an/sprabq7a/sprabq7a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sensorless Trapezoidal Control of BLDC Motors&lt;/a&gt;, it follows that a BLDC motor has a trapezoidal back emf while a PMSM has sinusoidal back EMF. But if I use a sinusoidal commutation, will the back EMF of a BLDC motor still be trapezoidal?&lt;/p&gt;&#xA;" OwnerUserId="8784" LastEditorUserId="368" LastEditDate="2016-11-21T16:56:59.840" LastActivityDate="2016-12-21T17:17:09.623" Title="Does a sinusoidal input voltage make a BLDC motor into a PMSM?" Tags="|electrical-engineering|control-engineering|motors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12390" PostTypeId="2" ParentId="12277" CreationDate="2016-11-19T16:56:49.530" Score="0" Body="&lt;p&gt;As the fluid enters it first hits the compressor which increases the pressure by doing work (adding energy) to the flow. This pressure is dynamic pressure as the velocity increases at the compressor stage ( $q = 0.5 * \rho * V^2$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluid then enters the diffuser stage where the stators turn this dynamic pressure into static pressure by slowing down the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to summarize, the compressor adds velocity to the flow by doing work (adding energy) to the flow. This energy is then converted from dynamic pressure to static pressure by the diffuser as it slows down the fluid.&lt;/p&gt;&#xA;" OwnerUserId="8225" LastActivityDate="2016-11-19T16:56:49.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12391" PostTypeId="2" ParentId="12225" CreationDate="2016-11-20T01:51:46.037" Score="0" Body="&lt;p&gt;If you recall from practicing finding Vy when you make cuts along a beam, you have a shear force (Vy) on each side of the cut, each opposite and equal to the other. Now imagine cutting nothing off the surface of a body. What is the shear force of the surface? Nothing! Why? Because the shear force of the nothing you cut off is...nothing. Equal and opposite, remember? &lt;/p&gt;&#xA;" OwnerUserId="8862" LastActivityDate="2016-11-20T01:51:46.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12392" PostTypeId="1" CreationDate="2016-11-20T05:20:03.700" Score="0" ViewCount="42" Body="&lt;p&gt;I have an infrared thermometer that supposedly goes up to 1200F (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00IT0ODPQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/gp/product/B00IT0ODPQ/ref=oh_aui_search_detailpage?ie=UTF8&amp;amp;psc=1&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I used this on my electric stove when turned to max. The coils are cherry red which is normally 1500F. Nichrome is usually rated to go up to 1400F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, the thermometer never showed more than 900F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the problem? Are these types of thermometers just unreliable?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2016-11-20T08:53:06.457" Title="Infrared thermometer unreliable" Tags="|sensors|" AnswerCount="1" CommentCount="2" ClosedDate="2016-11-25T10:37:57.633" ContentLicense="CC BY-SA 3.0" />
  <row Id="12393" PostTypeId="2" ParentId="12392" CreationDate="2016-11-20T08:53:06.457" Score="1" Body="&lt;p&gt;Infrared meters are reliable. They are, however, indirect measurements. For you to measure temperature you need to know the emissivity of your material, and any intervening material, in the temperature range your measuring. This info must be input to the device for it to show correct values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a good &lt;a href=&quot;http://www.raytek.com/raytek/en-r0/ireducation/emissivity&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference&lt;/a&gt; for starting out. Measuring stuff is rarely just pointing a sensor at your problem and reading output values.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2016-11-20T08:53:06.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12394" PostTypeId="1" CreationDate="2016-11-20T09:37:02.790" Score="-1" ViewCount="25" Body="&lt;p&gt;What happened to the final power generation if  the hydraliuc turbine rotates discontinously for example  if a turbine runner rotates with 2000 rpm for 2 minutes and stop rotating for 4 seconds and again rotates for a minute and this cycle continues.. And then what is the difference  between  final power generated by normal hydraliuc  turbine and this discontinous turbine.. &lt;/p&gt;&#xA;" OwnerUserId="8865" LastActivityDate="2016-11-20T12:11:53.343" Title="Doubt on power generation by hydraliuc turbine" Tags="|power-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2016-11-22T17:43:54.177" ContentLicense="CC BY-SA 3.0" />
  <row Id="12395" PostTypeId="2" ParentId="12394" CreationDate="2016-11-20T12:11:53.343" Score="1" Body="&lt;p&gt;You need to consider the system as a whole any rotating turbine has a certain amount of angular momentum so it is not just going to stop dead and come back up to speed in in a few seconds without some substantial forces involved. 2000 rpm to 0 to 2000 rpm in 4 seconds is fairly substantial acceleration for anything. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of AC power distributed to a grid it is important that the angular speed of all the generators is synchronised to ensure a stable frequency so bringing individual generators on and off line needs to be done carefully and in response to the overall load on the system. With this in mind intermittent running of a generator could well cause problems downstream with power management. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-11-20T12:11:53.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12397" PostTypeId="2" ParentId="12388" CreationDate="2016-11-20T15:24:16.513" Score="1" Body="&lt;p&gt;The back EMF waveshape of a brushless DC motor is a function of the motor geometry.  It has nothing to do with how the motor is driven.  There may be other signals superimposed on the back EMF, but the back EMF, by definition, is how it acts as a generator.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-11-20T15:24:16.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12399" PostTypeId="1" CreationDate="2016-11-20T17:29:04.200" Score="3" ViewCount="84" Body="&lt;p&gt;Modelica touts as one of its advantages over Simulink that it is &quot;component-based&quot; whereas Simulink is &quot;block based&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly does this mean?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2017-05-31T17:55:01.350" Title="Component-based simulation versus block-based simulation" Tags="|simulation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12400" PostTypeId="1" CreationDate="2016-11-20T18:46:13.643" Score="1" ViewCount="76" Body="&lt;p&gt;I'm looking for a practical efficient way to open and close a 1/4 inch gravity fed hose. Initially I was thinking of using a Solenoid, but can't find anything suitable. My second thought was using a servo and valve. I was also looking into some sort of piezo valve but they seem very specialized and expensive.&lt;/p&gt;&#xA;" OwnerUserId="539" LastActivityDate="2017-01-19T20:01:19.703" Title="How can I build a electronically controlled valve for a 1/4 inch tube" Tags="|valves|servo|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12404" PostTypeId="2" ParentId="12400" CreationDate="2016-11-20T19:23:17.417" Score="1" Body="&lt;p&gt;How about using a &lt;a href=&quot;https://www.adafruit.com/products/997&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plastic Water Solenoid Valve - 12V - 1/2&quot; Nominal&lt;/a&gt; coupled with &lt;a href=&quot;https://www.fastenal.com/products/details/69254&quot; rel=&quot;nofollow noreferrer&quot;&gt;1/2&quot; to 1/4&quot; adapter&lt;/a&gt;. You could use simple arduino microcontoller to control the valve. Below are some example images of the hardware. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LiuLJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LiuLJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dyd8M.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dyd8M.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2016-11-20T19:23:17.417" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12406" PostTypeId="2" ParentId="12382" CreationDate="2016-11-20T20:25:15.533" Score="-1" Body="&lt;p&gt;It depends on the altitude it is dropped from, the aerodynamic drag of the blimp and the fact that the atmosphere rotates with the Earth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to say is that you would have to do some calculations to figure out the answer to your question.&lt;/p&gt;&#xA;" OwnerUserId="436" LastEditorUserId="436" LastEditDate="2016-11-21T07:46:19.437" LastActivityDate="2016-11-21T07:46:19.437" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12408" PostTypeId="1" AcceptedAnswerId="12565" CreationDate="2016-11-20T23:42:25.383" Score="0" ViewCount="1000" Body="&lt;p&gt;I was given the following equations and asked to design a controller for $u=-Kx$ using a pole placement method with the closed loop system having the damping and settling time $T_s$ given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I supposed to use the $\dot x_1$ and $\dot x_2$ equations or are they totally unrelevant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, which pole placement method is easier for the case, root locus or Bode and Nyquist plots?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;\dot x_1 = x_2 \\&#xA;\dot x_2 = -x_1 +\dfrac{1}{6}x_1^5-x_2+u \\&#xA;u = -Kx \quad \zeta=1.02 \quad T_s = 0.40&#xA;\end{gather}$$&lt;/p&gt;&#xA;" OwnerUserId="4088" LastEditorUserId="1627" LastEditDate="2016-11-29T15:43:27.993" LastActivityDate="2016-11-29T15:43:27.993" Title="Controller design with pole placement method with given damping and settling time" Tags="|control-engineering|control-theory|stability|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12409" PostTypeId="1" CreationDate="2016-11-21T00:31:13.667" Score="1" ViewCount="83" Body="&lt;p&gt;We have a system that heats up when voltage is applied, temperature is then measured against time. We have several different data plots of temperature versus time at varying voltages. We need to develop a PID controller to control the system. How would we find the system equation? &lt;/p&gt;&#xA;" OwnerUserId="8875" LastActivityDate="2017-04-05T14:55:41.633" Title="How to characterize a system given several curves" Tags="|electrical-engineering|control-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12410" PostTypeId="2" ParentId="12382" CreationDate="2016-11-21T01:00:20.373" Score="5" Body="&lt;p&gt;I assume that the person you are referring to is Felix Baumgartner, who achieved a maximum speed of Mach 1.25. While this speed is very fast, this is nowhere near the speeds achieved by meteorites, satellites, or various space junk. The space shuttle, when it was in operation, reentered the atmosphere between Mach 10 to Mach 25, which would destroy both Felix, and your blimp.&lt;a href=&quot;https://www.grc.nasa.gov/WWW/BGH/hihyper.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;(1)&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer your question, I think you may need to look into your question a little more. When you say re-enter, is that really what you mean? Is the blimp coming from outer space and going to Earth? From what altitude? It sounds like it isn't really in space if it used its Helium to get up there. This is the case in the Red Bull Stratos jump, where the capsule reached an altitude of 38.9 km (stratosphere). Space can loosely be argued to start at the Karman line (100km)&lt;a href=&quot;https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;(2)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, honestly, seems like the exchange to put this type of question. In my Aerospace Engineering undergrad, we dealt with a few problems where we ran calculations for balloons in the upper atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, if you are seeking further information on this subject, I would recommend looking into this subject more yourself. Do some research, and understand the subject better, so you can ask better questions. Then awesome people at stack exchange can give better answers!&lt;/p&gt;&#xA;" OwnerUserId="8862" LastEditorUserId="8862" LastEditDate="2017-05-07T01:58:50.483" LastActivityDate="2017-05-07T01:58:50.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12411" PostTypeId="1" CreationDate="2016-11-21T01:01:38.453" Score="4" ViewCount="722" Body="&lt;p&gt;I have an old peristaltic pump with a 9 pin male sub-d socket labeled Gilson Serial Input Output Channel (GSIOC) that connects a master device to devices with RS422A specifications, or RS485 specifications. The GSIOC implements RS422A/RS485 transmitters and receivers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cHAli.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cHAli.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could any computer serve as a master device by simply connecting the pump and computer directly through a serial to USB &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B002VY523I&quot; rel=&quot;nofollow noreferrer&quot;&gt;converter&lt;/a&gt; with proper specifications? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building from here could serial commands be written on and sent directly from the computer to the pump in order to control speed/rotor direction...etc? Or am I missing important conceptual understanding on how to remotely control the pump? &lt;/p&gt;&#xA;" OwnerUserId="8371" LastEditorUserId="8055" LastEditDate="2017-01-19T19:30:44.767" LastActivityDate="2020-07-29T15:47:23.720" Title="Using python to send automated commands through a serial-usb-cable adapter" Tags="|electrical-engineering|control-engineering|control-theory|computer-hardware|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12412" PostTypeId="2" ParentId="12411" CreationDate="2016-11-21T03:27:15.290" Score="1" Body="&lt;p&gt;Yes that will be feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears from the image that the green 6pin connector is for external manual controls. One of the signals is an analog voltage, presumably to control the motor speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the RS232/485 signal there should be a digital PWM (pulse width modulated) value to control the motor speed through a power driver (FET/transistor/SCR/etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other signals serve as on-off commands to manage the motor enable and direction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will need to find which DB9 pin serves as the speed control signal - (Analog V+) and feed a variable PWM source to that pin.  Now RS232/485 devices are not high current devices, so you will need to figure out how the power is provided to drive the motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that the computer/MCU will be able to set each pin on the DB9 to a LOW (0v) or HIGH (5V or 3.3v) voltage which is then passed through the RS232 &quot;converter&quot; to shift the signals to the correct levels (+/- 5V) to control the device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the wiring and power are known this is an easy job.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2016-11-21T03:27:15.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12415" PostTypeId="1" AcceptedAnswerId="12418" CreationDate="2016-11-21T13:13:32.990" Score="2" ViewCount="523" Body="&lt;p&gt;I need to buy a particular brand and model of stepper motor and I have found the product &lt;a href=&quot;http://motion.schneider-electric.com/products/mdrive/mdrive_plus_motion_control.php?nema=17&quot; rel=&quot;nofollow noreferrer&quot;&gt;on the manufacturer's web catalog&lt;/a&gt;, but with no price listed.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can I find the price of device? I tried writing to the manufacturer but they haven't responded. &lt;/li&gt;&#xA;&lt;li&gt;Why has the manufacturer hidden the price?   &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8880" LastEditorUserId="368" LastEditDate="2016-11-21T17:40:05.897" LastActivityDate="2016-11-21T17:58:40.513" Title="How can I find the price of a product if the manufacturer doesn't show it on their website?" Tags="|mechanical-engineering|equipment-selection|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12416" PostTypeId="1" AcceptedAnswerId="12422" CreationDate="2016-11-21T15:39:34.643" Score="2" ViewCount="147" Body="&lt;p&gt;I'm building a hovercraft, which transfer function i have to find for further control. The hovercraft can be found in the following &lt;a href=&quot;https://i.stack.imgur.com/g9xVG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g9xVG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the hovercraft has 2 motors for direction. I have been given an advice to apply a step to the system, via the direction motors, and determine the transfer function, from the stepresponse. For this several questions comes to my mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A step on the motor, meaning for example 5V on the motors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should the ouput of the transfer function represent? (should i monitor velocity, acceleration, position after x seconds?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards&lt;/p&gt;&#xA;" OwnerUserId="8881" LastActivityDate="2016-11-21T20:18:16.237" Title="Transfer function for hovercraft" Tags="|transfer-function|" AnswerCount="1" CommentCount="0" ClosedDate="2016-11-25T10:36:34.200" ContentLicense="CC BY-SA 3.0" />
  <row Id="12417" PostTypeId="1" CreationDate="2016-11-21T17:42:08.900" Score="0" ViewCount="47" Body="&lt;p&gt;Im working on efficienxy of heavy vehicle. Since i never had any experience with this kimd of things ( cars ) i dont have so much insight in what kind of efficiencys are important in vehicle. I have following insights: Slip efficiency of a tire Motion efficiency Feul efficiency etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have data on all kinds of forces gears slip and so on of a vehicle. What is there to maximize.&lt;/p&gt;&#xA;" OwnerUserId="8884" LastActivityDate="2016-11-22T11:10:14.480" Title="Efficiency of off road vehicles" Tags="|applied-mechanics|dynamics|energy-efficiency|optimal-control|" AnswerCount="1" CommentCount="0" ClosedDate="2016-11-21T23:01:25.750" ContentLicense="CC BY-SA 3.0" />
  <row Id="12418" PostTypeId="2" ParentId="12415" CreationDate="2016-11-21T17:58:40.513" Score="10" Body="&lt;p&gt;In industry, an equipment manufacturer hardly ever sells directly to the end user. They supply distributors in bulk at a negotiated price for each shipment of however many units the distributor orders. The price-per-unit varies accordingly, so there's no point in listing it the way you would for retail products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only need a single unit of a product, generally the manufacturer is not going to want to do business with you. They produce many units at a time, as required to supply their distributor(s), in order to use their facilities and personnel efficiently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to do is find a distributor or wholesaler who carries that manufacturer's products and is willing to sell directly to you (retail business). &lt;a href=&quot;https://en.wikipedia.org/wiki/W._W._Grainger&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grainger&lt;/a&gt; is one example of a company that does this, though you may have to work for a company that has a business account with them in order to just walk in and order a single item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can usually find the contact information for a local or regional distributor on the company's web site or in their catalog, provided they do business in your area. Sometimes, especially when dealing with a foreign supplier, they won't have an existing business relationship with anyone in your region, and then you may need to look for an alternative product or work with a foreign distributor as well, which can be a real headache for just one piece of equipment.&lt;/p&gt;&#xA;" OwnerUserId="368" LastActivityDate="2016-11-21T17:58:40.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12419" PostTypeId="1" CreationDate="2016-11-21T18:05:18.530" Score="0" ViewCount="71" Body="&lt;p&gt;I read on &lt;a href=&quot;http://www.itworld.com/article/2826935/networking/11-cutting-edge-car-tech-innovations.html?page=3&quot; rel=&quot;nofollow noreferrer&quot;&gt;itworld.com&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Companies such as Volvo and Mercedes are working on crash avoidance systems. Radar measures the distance to obstacles ahead of the car, and if one comes too close the system will warn the driver. If the driver fails to act in sufficient time to stop the car safely, the cars computer takes over and applies the brakes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is anyone working on making the cars respond sooner at colder temperatures, or when the road appears to be in slippery conditions? (car has less traction, etc.)&lt;/p&gt;&#xA;" OwnerUserId="8885" LastActivityDate="2016-12-22T04:26:36.907" Title="What do Rear-End Collision Avoidance System do on Ice?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12420" PostTypeId="2" ParentId="12417" CreationDate="2016-11-21T18:35:38.080" Score="2" Body="&lt;p&gt;There are a few issues to consider. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dedicated off road vehicles tend to have ladder or in some cases space-frame chassis. There are robust and make it easier to add specialist bodies but tend to be heavier than equivalent monocoque designs. Equally they may also need roll protection which adds further weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They &lt;em&gt;tend&lt;/em&gt; to have high displacement engines for low-end torque, again these are heavier and tend to have worse fuel economy compared to the trend in road vehicles which is towards smaller higher revving engines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four wheel drive transmission is inevitably heavier, bulkier and more complex than two wheel drive, especially front wheel drive. You need extra drive shafts, differentials and prop shafts. Dedicated off road vehicles may also have low range gearboxes and diff locking mechanisms all of which add weight, complexity and transmission losses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Off road tyres tend to have more rolling resistance than road tyres due to a more aggressive tread pattern and larger and heavier wheels and tyres overall.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They often have large frontal areas, increasing aerodynamic drag and combined with gear ratios optimised for traction  this tends to reduce fuel economy at road speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parts in general will tend to be heavier and more robust to cope better with shock loading and there will often be extra components such as sump guards and other under-body skid plates as well as snorkels, winches etc etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While not directly related to efficiency off road vehicles tend to have high centres of mass and soft suspension with a lot of travel which tends to compromise their handling in normal road conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of these issues can be overcome at least to some extent by the use of active suspension, lightweight materials etc etc all of course at the expense of increased cost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course 'off road' is a pretty broad category and can range from relatively lightweight utility vehicles to heavy plant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that an off-road vehicle has a 'hard' minimum performance requirement in order to fulfil its role, compared to road cars where performance is more of an incremental scale. So there is less design scope to make compromises between efficiency or fuel economy and performance. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From a control perspective key areas are traction control and suspension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traction control has the advantage that there is a lot you can do with existing hardware and can include&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;decent control&lt;/li&gt;&#xA;&lt;li&gt;ABS &lt;/li&gt;&#xA;&lt;li&gt;anti-wheel-spin / torque distribution &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Active suspension requires a bit more on terms of hardware design &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-11-22T11:10:14.480" LastActivityDate="2016-11-22T11:10:14.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12421" PostTypeId="1" CreationDate="2016-11-21T19:31:11.523" Score="9" ViewCount="526" Body="&lt;p&gt;The standard light petroleum distillate for vehicle engines, &quot;regular gasoline,&quot; is (or is equivalent to) some mix of heptane (C7) and octane (C8).  Higher proportions of C8 are more knock-resistant, which allow for higher compression ratios and thus more efficient energy utilization in internal combustion engines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't modern refineries produce pretty much any hydrocarbon mix they want through combinations of distillation, cracking, and alkylation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, then why is &quot;regular&quot; gasoline produced in such quantities, and why is a premium charged for more knock-resistant (&quot;higher octane&quot;) blends?  For example, if a refinery had to produce only one light fuel distillate, couldn't it just as easily and cheaply produce a &quot;100 octane&quot; blend as the current &quot;87 octane&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Or is it &lt;em&gt;actually&lt;/em&gt; cheaper to produce &quot;lower octane&quot; petrol?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that &quot;100 octane&quot; doesn't mean 100% octane, as the &quot;octane&quot; number of a hydrocarbon depends on its isomers, with &lt;a href=&quot;http://www.csgnetwork.com/octaneratecalc.html&quot; rel=&quot;noreferrer&quot;&gt;more highly-branched isomers having more knock resistance&lt;/a&gt;.  So 100 octane fuel can be produced with many blends of C7, C8, and even include lighter and heavier hydrocarbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A characterization of the isomeric constituents of crude oil might answer this question.  For example, if crude feedstock tends to have more linear isomers, then energy would have to be put into isomerization to produce more knock-resistant distillates.&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="77" LastEditDate="2016-11-25T20:04:30.527" LastActivityDate="2018-04-06T19:43:45.323" Title="Why is &quot;regular&quot; gasoline standard instead of something more knock-resistant?" Tags="|automotive-engineering|chemical-engineering|combustion|petroleum-engineering|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="12422" PostTypeId="2" ParentId="12416" CreationDate="2016-11-21T20:18:16.237" Score="1" Body="&lt;p&gt;If you want to determine the transfer function of a system by applying a step response, you'll need to decide in advance what kind of dynamic model you expect your system to have. If you're looking for a transfer function, that means you've already assumed your system is linear, so start considering linear models that would work for your case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you decide on a model, derive its transfer function. Apply a step response and use the response characteristics (such as time constant, natural frequency, steady state value, or any others that are relevant to your model) to solve for the unknown parameters of the model transfer function. &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Example&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let's say your hovercraft primarily has two forces acting on it: the force from the motors, and the force due to friction/drag. In that case you could approximate the dynamics using a classical linear, single degree-of-freedom model, like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m\dot{v} = K_m u(t) - bv$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $v$ is the velocity of the hovercraft, $\dot{v}$ is the acceleration, and $u$ is the voltage applied to the motors. The unknown parameters of the system are: $m$ (the mass of the hovercraft), $K_m$ (the conversion between applied motor voltage and thrust force), and $b$ (the drag/friction coefficient). You should be able to identify $m$ by simply weighing your hovercraft. That leaves $K_m$ and $b$ unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get the transfer function of the model by applying the Laplace transform:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$sV(s) = K_m U(s) - bV(s)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{V(s)}{U(s)} = \frac{K_m}{ms + b} = \frac{K_m/b}{(m/b) s + 1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can identify the different parameters by measuring the velocity of the hovercraft when you apply a step function to the input. The step function can have any value, but you want it to be large enough that the velocity is measurable, and small enough that the hovercraft doesn't speed out of control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are looking for two characteristics of the response:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The steady state velocity $v_{ss}$, which is the maximum speed that the hovercraft reaches for a given input, even if left for an infinite amount of time. Your hovercraft will probably stop accelerating after a few seconds if your input isn't too large, use that velocity as an approximation. Solve for the ratio of unknown parameters using $v_{ss} = (K_m/b)u_0$ if $u_0$ is the magnitude of the step input, maybe 2 V or something.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Time_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;time constant&lt;/a&gt; $\tau$ of the response, which is the time it takes for the system to go from 0 to approximately 63.2% of the steady state velocity (see the Wikipedia link for more info). Solve for $b$ using $\tau = m/b$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2016-11-21T20:18:16.237" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12424" PostTypeId="1" AcceptedAnswerId="14382" CreationDate="2016-11-21T20:58:16.100" Score="0" ViewCount="111" Body="&lt;p&gt;Okay, Hi there Engineers! I am a newbie to this site, and to engineering.&lt;/p&gt;&#xA;&lt;p&gt;So, I wanted to make this post to get all the help I can get. I have been set a task of creating a prototype of the following;&lt;/p&gt;&#xA;&lt;p&gt;I have been given a device (the device being a &lt;strong&gt;Rubidium Atom Clock/Standard&lt;/strong&gt;. Pictures are provided below.)&lt;/p&gt;&#xA;&lt;p&gt;The fault with this device is that different internal/external temperatures will affect the clock. So what I need to do is; somehow make the interior temperatures a constant, where ever the device is (so if it is in Iceland, the internal temperature must be the same as it will be if it was in the Sahara Desert.)&lt;/p&gt;&#xA;&lt;p&gt;Oh yeah, and the interior temperature must be within &lt;strong&gt;-20°C&lt;/strong&gt; to &lt;strong&gt;+65°C&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;If you need any information before you can help me, please go ahead and ask. Also, as I've stated above, I am new to engineering, so if you are trying to help me, please explain as much as you can so I can understand. And, I won't have advanced tools available, just assume that I have the tools in a workshop.&lt;/p&gt;&#xA;&lt;p&gt;Bear in mind, I have very little space to work with in the device, and watch out for voltage, I don't want any systems like fans or something to affect the power supply.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Specifications&lt;/strong&gt;&#xA;(Not exact to the device I am using but I will get that as soon as possible, they are pretty similar):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CbRc6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CbRc6.png&quot; alt=&quot;Specifications&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Device Images:&lt;/strong&gt;&#xA;...That is rather annoying. I can't upload more pictures due to my account not having 10 reputation. If you are looking for the device, just search up &amp;quot;Rubidium Oscillator PRS10B&amp;quot;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mbp13.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mbp13.jpg&quot; alt=&quot;Interior&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user8887" LastEditorUserId="33" LastEditorDisplayName="user8887" LastEditDate="2020-07-06T18:21:45.817" LastActivityDate="2020-07-06T18:21:45.817" Title="Creating a constant internal temperature that is negligible to the external environment" Tags="|electrical-engineering|power-electronics|temperature|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12426" PostTypeId="2" ParentId="12424" CreationDate="2016-11-21T23:01:28.740" Score="0" Body="&lt;p&gt;Not easily done I suspect. You'd need to carefully design an insulated container of some sort with some kind of active cooling/heating. I'd personally use peltier devices with a simple temperature monitoring circuit. You'd also need some form humidity control and some form of semi sealed automatic humidity controlling case vent. &lt;/p&gt;&#xA;" OwnerUserId="8890" LastActivityDate="2016-11-21T23:01:28.740" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12427" PostTypeId="1" AcceptedAnswerId="12438" CreationDate="2016-11-21T23:16:07.713" Score="11" ViewCount="421" Body="&lt;p&gt;My workplace has a policy to provide standing desks upon request, but no policy to provide chairs of matching height.  (I work for the government...)  We can buy our own, or build our own.  However they added a stipulation that the chairs must be bought or built with five or more wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since five is an awkward symmetry to measure and to sit in, I'd probably build my chair with six wheels.  Or perhaps eight because I can start with a square and lop off the corners to make an octagon base.  Or maybe make the base a perfect circle and line the bottom with innumerable tiny wheels?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This set me to wondering:  is there any situation in which a chair with N+1 wheels is strictly less stable than a chair with N wheels?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the question interesting and rule out trivial answers, assume that the wheels are more-or-less at the chair's perimeter and more-or-less evenly distributed.  Assume the floor is flat.&lt;/p&gt;&#xA;" OwnerUserId="7246" LastActivityDate="2016-11-23T13:01:16.213" Title="Is a chair with N+1 wheels ever less stable than a chair with N wheels?" Tags="|wheels|stability|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12428" PostTypeId="2" ParentId="11539" CreationDate="2016-11-21T23:27:37.733" Score="1" Body="&lt;p&gt;Generally speaking the filter is on the low pressure side of the system. That low pressure is either supplied by an in tank lift pump or a low pressure lift pump (usually integrated into the high pressure pump) sucks it from the tank. In 15 years working with an assortment of vehicle, generator, marine diesel engines I've never seen one burst a fuel filter. On automotive use the lift pumps operate at around 5 - 10 psi and flow around a gallon per minute. Worst that could happen in the event of a blocked filter would be the engine stopping, possibly some fuel leaks and you might get the lift pump hot or even cause it to fail if left long enough pumping against a blockage. On some designs the fuel return line from the high pressure pump sometimes goes back into the fuel filter head (to heat the incoming fuel by bleeding a little of the returning hot high pressure fuel back into the cold incoming fuel) before going back to the tank. I'd guess that an internal fault on the high pressure pump fuel return might cause a fuel filter to burst...? Never seen it personally though.  &lt;/p&gt;&#xA;" OwnerUserId="8890" LastActivityDate="2016-11-21T23:27:37.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12429" PostTypeId="1" CreationDate="2016-11-21T23:32:01.400" Score="1" ViewCount="1024" Body="&lt;p&gt;I would like to implement the Field Oriented Control (FOC) algorithm on my STM32 board for controlling a 3 phases sensorless BLDC motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By googling I found a very interesting &lt;a href=&quot;https://www.google.it/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;url=http://www.ti.com/lit/an/sprabq3/sprabq3.pdf&amp;amp;ved=0ahUKEwj3-OvG8brQAhUMKcAKHTLRBFEQFggiMAE&amp;amp;usg=AFQjCNEpbD4Rea0AH-HvNzs4VAFAIYH5wQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Application Note from Texas Instruments&lt;/a&gt; which explains very clearly how to apply the FOC algorithm for driving a sensorless 3 phase BLDC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On page 10 of the application note, there is a block diagram of the algorithm and there is a block called &lt;strong&gt;Sliding-Mode&#xA;Based&#xA;Rotor&#xA;Position Estimator&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1FeSY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1FeSY.jpg&quot; alt=&quot;Block diagram on page 10 of application note&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since in my application I have an IMU (Inertial Measurment Unit) fixed on the shaft of the BLDC motor, I can get both rate and acceleration data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once combined the data into my Kalman Filter (outputs a quaternion), I have both speed and position information of the rotor.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Can I use IMU's data for the rotor position, instead of estimating as shown?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Is it correct considering the rotor angle got from the IMU for closing the loop?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8888" LastEditorUserId="8055" LastEditDate="2017-02-03T07:01:51.357" LastActivityDate="2017-02-03T07:01:51.357" Title="FOC algorithm for BLDC motor and rotor position estimation via IMU" Tags="|electrical-engineering|control-engineering|motors|control-theory|pid-control|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12430" PostTypeId="1" CreationDate="2016-11-22T00:23:43.687" Score="0" ViewCount="155" Body="&lt;p&gt;If I have a linear quadratic regulator (quadratic objective function with linear dynamics) and use an open loop control - are there stability/controllability requirements on the system?  I know that there would be these requirements if I were to implement a closed loop control but I can't find any corresponding requirements for open loop. &lt;/p&gt;&#xA;" OwnerUserId="8891" LastActivityDate="2016-11-22T13:44:24.583" Title="stability/controllability of a finite time linear quadratic regulator" Tags="|stability|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12431" PostTypeId="1" AcceptedAnswerId="12439" CreationDate="2016-11-22T02:39:33.900" Score="4" ViewCount="869" Body="&lt;p&gt;I'm designing a 5 axis CNC, and preloading my lead screws has never been an issue for me,  but for the first time,  I find myself needing to preload a worm gear.  How is this done, and what's the best way to do it?  Thanks. &lt;/p&gt;&#xA;" OwnerUserId="8894" LastActivityDate="2016-11-22T13:25:01.867" Title="Preload a worm gear?" Tags="|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12432" PostTypeId="2" ParentId="12427" CreationDate="2016-11-22T03:02:55.447" Score="5" Body="&lt;p&gt;I think the &quot;health and safety&quot; regulation about 5 wheels is a compromise between stability and cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your weight is on the edge of the chair seat and the chair has only 3 wheels, it is much less stable if you are in line with one of the wheels than if you rotate through 60 degrees to be mid-way between two wheels. That might happen (1) because the occupant turns on the seat, (2) if the chair is moving and one wheel hits an obstruction which rotates the legs, or (3) the occupant is sitting centrally on the chair, but moves from &quot;leaning forward&quot; to &quot;leaning backward&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result could be that a stable sitting position suddenly becomes unstable. For a 3-legged chair, the minimum stable distance for a load offset from the center is only half the maximum stable distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bigger number of wheels reduces this issue, but increases the friction that has to be overcome to point all the wheels in the correct direction to move the chair. It is also more expensive to manufacture. In the worst case situation of tipping over, all the load on the chair is carried on just one wheel independent of how many wheels the chair has, so increasing the number of wheels doesn't allow the size of each leg and wheel to be reduced!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;min:max stability ratio&quot; of 0.5 for a 3-wheel chair increases to about 0.7 for 4 wheels, 0.8 for 5, and 0.9 for 7 wheels. IIRC, the safety regulations in the UK changed from 4 to 5 wheels some time during the 1970s.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-11-22T03:02:55.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12435" PostTypeId="2" ParentId="12419" CreationDate="2016-11-22T04:00:16.110" Score="1" Body="&lt;p&gt;That would either not be feasible, or make no sense. There is no effective, economical and safe way to test for &quot;slippery conditions&quot; before the fact. Once the car starts to brake, its ABS systems will automatically detect the exact road conditions, and do the best they can under those conditions. Cold temperatures by themselves do not significantly affect braking/stopping performance. Conversely, there are many road conditions that are not temperature-related that significantly affect braking performance also.&lt;/p&gt;&#xA;" OwnerUserId="8248" LastActivityDate="2016-11-22T04:00:16.110" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12438" PostTypeId="2" ParentId="12427" CreationDate="2016-11-22T13:04:04.530" Score="11" Body="&lt;p&gt;Assuming all the wheels are evenly spaced on the same circle, then more wheels is always more stable than less wheels.  However, there is diminishing return as the number of wheels gets large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metric of stability is how far from the center of the circle the center of mass can be before the chair tips over.  The chair is stable whenever the center of mass is inside the polygon formed by all the wheel points.  The worst case is in the center of one of the edges, since these are the closest points on the polygon to the center of the circle.  In the limit, with infinite number of support points, the minimum distance to instability is the radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can therefore quantify stability as the minimum distance to instability relative to the radius.  A value of 1 is the maximum, with infinite support points.  After a little trig, it's easy to see that this stability metric is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;S = cos(&amp;Pi;/N)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where N is the number of support points.  The stability metrics for values of N up to 20 are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;   N      S&#xA;----   ----&#xA;   2   0.00&#xA;   3   0.50&#xA;   4   0.71&#xA;   5   0.81&#xA;   6   0.87&#xA;   7   0.90&#xA;   8   0.92&#xA;   9   0.94&#xA;  10   0.95&#xA;  11   0.96&#xA;  12   0.97&#xA;  13   0.97&#xA;  14   0.97&#xA;  15   0.98&#xA;  16   0.98&#xA;  17   0.98&#xA;  18   0.98&#xA;  19   0.99&#xA;  20   0.99&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Office chairs commonly use N=5, which is a tradeoff between being good enough but not too expensive.  The extra 7% stability from adding a 6th wheel isn't worth the cost.  Or, put another way, you can achieve the same stability as 6 wheels by using 5 wheels but growing the base another 7% outward.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2016-11-23T13:01:16.213" LastActivityDate="2016-11-23T13:01:16.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12439" PostTypeId="2" ParentId="12431" CreationDate="2016-11-22T13:25:01.867" Score="2" Body="&lt;p&gt;For worm gears, the most common method to reduce backlash is by forcing the worm screw tightly against the gear. Many manual rotary tables used on milling machines have this adjustment.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2016-11-22T13:25:01.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12440" PostTypeId="2" ParentId="12430" CreationDate="2016-11-22T13:44:24.583" Score="1" Body="&lt;p&gt;How do you have an LQR with open loop control? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that there would be these requirements if I were to implement a closed loop control&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A linear quadratic regulator is basically a multi input version of a state feedback controller; your $Q$ and $R$ matrices determine the &quot;cost&quot; or priority each input should take. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the $Q$ and $R$ matrices, you solve the matrix Riccati equation to get the optimizing matrix $P$. That matrix then allows you to solve for the optimal control gain matrix $K$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Then $K$ is used to calculate the control input, $u = -K\hat{x}$, just like a state feedback controller.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The LQR controller is a &lt;em&gt;multi-input state feedback controller&lt;/em&gt;. To say that you want an &quot;open loop LQR controller&quot; is not technically correct, because again you &lt;em&gt;do&lt;/em&gt; have feedback: your current state estimate $\hat{x}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any criteria you have for &quot;closed loop control&quot; applies to an LQR controller. The only difference is that what you're calling &quot;closed loop&quot; refers to either direct measurement of the system states or the use of &lt;a href=&quot;https://en.wikipedia.org/wiki/State_observer&quot; rel=&quot;nofollow noreferrer&quot;&gt;observers&lt;/a&gt; to correct your estimate. Either way, you're either controlling based on an uncorrected state estimate or a corrected state estimate, but &lt;em&gt;you are always controlling based on a state estimate.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-11-22T13:44:24.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12441" PostTypeId="1" CreationDate="2016-11-22T13:53:57.787" Score="0" ViewCount="51" Body="&lt;p&gt;I want to get manufacturing done for custom designed copper connector which I am planning use it in my home automation project. I created design of the connector in Freecad . I have exported the designs in PDF and published to local manufacturers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be frank I am from software background and it is first time I am creating designs and getting them manufactured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After publishing my designs I got few replies. One of the reply asked me which production process to be followed extrusion or die casting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea of this, tried google to search for any clues which would help me deciding . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After browsing few pages I got to know only one difference one of process involves melting the metal and other  one include forcing cold metal to shape it in desired design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please help me deciding.how to chose the production process which would suite my application?&lt;/p&gt;&#xA;" OwnerUserId="8900" LastActivityDate="2016-11-22T18:53:43.110" Title="How to chose production process of copper electrical connector, Extrusion vs Die casting" Tags="|casting|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12442" PostTypeId="1" AcceptedAnswerId="12448" CreationDate="2016-11-22T14:48:02.600" Score="2" ViewCount="370" Body="&lt;p&gt;I'm sorry if the question is trivial but for me it is not. I'm wondering what's the principle behind the suction process in a piston pump or in a piston of an internal combustion engine. I mean when the piston goes down, pressure inside lowers and air/water enters. Is it the conservation of momentum?&lt;/p&gt;&#xA;" OwnerDisplayName="user6267" LastActivityDate="2019-01-07T05:15:02.220" Title="Physical principle behind suction of a piston" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12444" PostTypeId="2" ParentId="12441" CreationDate="2016-11-22T16:26:53.110" Score="3" Body="&lt;p&gt;Extrusion and die casting are both fairly high volume processes which allow relatively rapid and cheap costs per item at the expense of high setup costs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extrusion involves forcing a plastic material through a die, a bit like piping icing onto a cake this means that you are limited to producing prisms (in the geometric sense) ie a strip of arbitrary length with constant section. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Die casting gives a bit more flexibility in producing shapes with dimensions which vary in all three axes although there can be limitations compared to other casting processes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases once you have designed and manufactured the tooling you are very limited in what changes you can make. So if you are still at the prototyping stage it may be that neither is appropriate. Going into volume production without having produced and evaluated a physical prototype is pretty risky, especially if you don't have much design for manufacturing experience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the comments it is impossible to give any definite advice without seeing your design and specification in more detail and the best bet is just to ask your suppliers for prices for the volume you want. Note that the volume you will need can make a big difference to the price per unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider that for smaller volumes especially there are lots of other possible methods but again without seeing the design it is a bit difficult to make suggestions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to be careful about designing things which are unnecessarily difficult to manufacture which is easy to do if you don't have much experience in the manufacturing processes you are using. Again it can be helpful to have a proper dialogue with your suppliers about what it is that you actually want to achieve. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-11-22T18:53:43.110" LastActivityDate="2016-11-22T18:53:43.110" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12445" PostTypeId="2" ParentId="8289" CreationDate="2016-11-22T20:18:04.547" Score="3" Body="&lt;p&gt;Originally the spiral shaped devices were fitted to the earth wires on top of pylons as vibration dampers (the very top single wire on pylons is the earth wire). It was however noted that they also had the effect of discouraging birds from landing on the wires. Therefore the spiral bird diverters can be used as dual purpose devices. Keeping birds away and effectively managing vibration issues on lines in particularly windy areas. Although these days newer weighted vibration dampers are used on most problem lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They use spirals so that during periods of vibration the spiral will bounce around and hit the wire and disrupt any vibration patterns (specifically they were designed to prevent a vibration type called aeolian vibration) in the line. The weighted dampers can't do that. Generally they use the weighted dampers on thicker cables and spirals on thinner ones.&lt;/p&gt;&#xA;" OwnerUserId="8890" LastEditorUserId="255" LastEditDate="2016-11-24T09:44:51.290" LastActivityDate="2016-11-24T09:44:51.290" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12448" PostTypeId="2" ParentId="12442" CreationDate="2016-11-22T22:07:12.687" Score="4" Body="&lt;p&gt;The ideal gas law says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;PV = nRT \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P$ is pressure, $V$ is volume, $n$ is the number of moles (molecules, basically) of gas, $R$ is the ideal gas constant, and $T$ is the &lt;em&gt;absolute&lt;/em&gt; temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of molecules in the piston doesn't change and you assume that temperature stays relatively constant, then, because the ideal gas constant is a &lt;em&gt;constant&lt;/em&gt;, everything on the right hand side of that equation is constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that pressure and volume are linearly inversely related - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P = k\frac{1}{V} \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $k$ is a constant that affects how much of a change in pressure occurs for a particular change in volume. This is just an &quot;umbrella&quot; to capture the other terms that we know (are assuming) are constant. That is, $k = nRT$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can see that, if you make the volume bigger, the reciprocal of volume, $\frac{1}{V}$ is getting smaller, which means the pressure $P$ is getting smaller. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume gets bigger, pressure gets smaller, and vice-versa. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when you make the volume bigger, the pressure drops, and this means that the cylinder &quot;sucks&quot; more fluid into the chamber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say fluid - a gas or liquid - but the ideal gas law is for (not surprisingly): gas. There's a similar principle at work for liquids, but you have to make some more assumptions. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2016-11-22T22:07:12.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12450" PostTypeId="1" CreationDate="2016-11-23T02:10:10.520" Score="2" ViewCount="116" Body="&lt;p&gt;I'm currently working on resonance frequencies of buildings and have to do a practical work/experiment on which floor of the building is likely to collapse during the earthquake. I have the slight idea on how it will work but I have no idea how I should do an experiment to show and prove it. My guess is the top floor as it will move more distance in same time (therefore move faster) and it is higher than the center of the gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have quite a variety of apparatus such as the amplifier, signal generator, arbitrary function generator, vibration generator, sliding platforms and some metal plates for building structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if I can get some insights and tips on how I should go about doing the experiment. Some equations relating to thus would be great. &lt;/p&gt;&#xA;" OwnerUserId="8912" LastEditorDisplayName="user16" LastEditDate="2016-11-23T02:11:32.317" LastActivityDate="2016-11-24T09:34:09.057" Title="Which floor is likely to collapse during the earthquake?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="4" ClosedDate="2016-11-28T17:28:59.930" ContentLicense="CC BY-SA 3.0" />
  <row Id="12453" PostTypeId="1" AcceptedAnswerId="12456" CreationDate="2016-11-23T12:51:59.090" Score="20" ViewCount="5351" Body="&lt;p&gt;I have noticed that windmills are generally built in empty fields with no trees around, and I've been wondering why... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A windmill is generally way taller than trees and I can imagine that trees don't actually affect the flow (see picture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wnaZ7.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wnaZ7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is this the reason why there is nothing around or is there something else that makes them waste so much space around? And is the velocity profile from the first figure a realistic one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I search for the theory of external convection, the velocity profile looks like this for flow over a plateau:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wPFOW.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wPFOW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;In the second case, wouldn't the velocity be higher if there is an obstacle? Or is it because the obstacle is a porous medium that damps the flow? And lastly, is the velocity profile at earth level looking like that? Is there really an origin to it, or is it fully turbulent all the way?&lt;/p&gt;&#xA;" OwnerUserId="8419" LastEditorUserId="8419" LastEditDate="2016-11-25T13:09:20.007" LastActivityDate="2017-01-31T05:07:32.907" Title="Windmills in empty fields. Why no trees?" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|heat-transfer|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12454" PostTypeId="2" ParentId="12453" CreationDate="2016-11-23T13:05:43.760" Score="3" Body="&lt;p&gt;Look at how and why the &lt;a href=&quot;https://en.wikipedia.org/wiki/Betz&amp;#39;s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Betz limit&quot;&lt;/a&gt; works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As energy is lost in your first diagram to the trees, the airflow through and around the trees expands and slows down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That slowdown reduces the energy available to the turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(note. I am not the resident wind energy expert!)&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2016-11-23T13:05:43.760" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12455" PostTypeId="1" AcceptedAnswerId="13436" CreationDate="2016-11-23T14:03:28.143" Score="0" ViewCount="2520" Body="&lt;p&gt;Has anyone ever tried running buckling simulations in Nastran where inertia relief is invoked? I am just using a simple plate and I am getting &quot;funny&quot; modes (unrealistically low eigenvalues which correspond to in-plane deformations - no buckling). I went through the documents and they do not seem to suggest that it should be avoided or anything of the sort. In fact the following is stated under &lt;code&gt;PARAM,INREL,-2&lt;/code&gt; that &lt;em&gt;&quot;This method leads to indeterminate matrices which are not supported by buckling. If attempted the solution will fail.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i am using &lt;code&gt;PARAM,INREL,-1&lt;/code&gt; and the model runs through. The &lt;code&gt;*.f06&lt;/code&gt; files looks ok too (&lt;code&gt;OLOAD&lt;/code&gt;, &lt;code&gt;epsilon&lt;/code&gt;, &lt;code&gt;strain energy&lt;/code&gt; and so on..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody shed any light or suggestions on how to proceed with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a plate modeled with CQUAD4s referencing a PCOMP and i am trying to find its eigenvalues. The loads on the plate are a combination of all in plane loads (Nx, Ny, Nxy). Due to the existance of in-plane shear, the setting up of the boundary conditions is not a trivial task and that is the reason inertia relief was considered. Also, with inertia relief in the model and along with the &quot;funny&quot; modes, i also get realistic ones, the eigenvalues of which match those i get with ESAComp and ESDU 81047. &lt;em&gt;So to summarise:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If i could get the eigenvalues &lt;strong&gt;ESAComp&lt;/strong&gt; and &lt;strong&gt;ESDU 81047&lt;/strong&gt; gives without inertia relief i would gladly do it&lt;/li&gt;&#xA;&lt;li&gt;The fact that the first realistic eigenvalue i get with inertia relief in the model matches those from above suggests (imo) that it is somehow doable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8031" LastEditorUserId="8031" LastEditDate="2016-11-24T09:28:10.593" LastActivityDate="2019-05-20T12:00:05.880" Title="Inertia Relief and Buckling SOL105 in Nastran" Tags="|finite-element-method|buckling|nastran|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12456" PostTypeId="2" ParentId="12453" CreationDate="2016-11-23T14:03:52.640" Score="38" Body="&lt;p&gt;Trees increase the turbulence of the air that reaches the turbines. That creates all sorts of uneven, rapidly-shifting loads on the blades and structure. That increases the maintenance costs, decreases availability, decreases the capacity factor, and decreases the life expectancy of the turbine. So, higher costs, lower revenue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the ways we measure the impact is the surface roughness coefficient $z_0$. Here are the figures from the book &quot;&lt;a href=&quot;http://www.wind-energy-the-facts.org/fundamentals.html&quot;&gt;Wind Energy - The Facts&lt;/a&gt;&quot;. As you can see, forest and woodland has a much higher $z_0$ than open farmland - and that means higher turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open land also makes physical access to the site easier. That's particularly important during construction, when very very long vehicles carry each blade (40-60 metres long) in one piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iOFBH.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iOFBH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2016-11-23T14:11:42.480" LastActivityDate="2016-11-23T14:11:42.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12459" PostTypeId="2" ParentId="12453" CreationDate="2016-11-23T16:43:32.487" Score="7" Body="&lt;p&gt;There is a non-engineering reason as well: the people most open to placing turbines on their land are farmers who are interested in the extra income.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turbines are not (yet) placed near built-up areas, which generally leaves large farms, undeveloped lots and nature reserves. Getting a building permit for a commercial enterprise in a nature reserve is close to impossible, and undeveloped lots may be zoned for housing, which would also prohibit the erection of wind turbines.&lt;/p&gt;&#xA;" OwnerUserId="8923" LastActivityDate="2016-11-23T16:43:32.487" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12462" PostTypeId="1" CreationDate="2016-11-23T19:54:24.840" Score="1" ViewCount="837" Body="&lt;p&gt;Suppose you have a thin sheet of a material of size say (1mtr X 2mtr X 1mm). You do not know what material it is. How to find its resistivity? One way I can think of is to measure resistance across smallest cross section i.e. 1mtr X 1mm, and then using $\rho = R\frac{A}{l}$, one can calculate the resistivity. However, in case of metal alloys, resistivity can be very low resulting in very low values of resistance which may not be picked by general multimeter and one may have to use micro ohm meter. Will this method work given the unusual shape of the conductor? Is there any other way to calculate the resistivity?&lt;/p&gt;&#xA;" OwnerUserId="6614" LastEditorUserId="6614" LastEditDate="2016-11-26T06:20:37.160" LastActivityDate="2017-12-13T01:17:07.613" Title="How to measure resistivity of a given sheet of material?" Tags="|electrical-engineering|materials|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12463" PostTypeId="2" ParentId="12455" CreationDate="2016-11-23T22:19:54.070" Score="0" Body="&lt;p&gt;From the theory, I can't see how trying to use inertia relief makes any sense at all, except for (linear) stress analysis of a free body which is accelerating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you haven't given any information about what you are actually trying to model, it's hard to give any advice on &quot;how to proceed!&quot;&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-11-23T22:19:54.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12465" PostTypeId="2" ParentId="10537" CreationDate="2016-11-24T00:19:04.933" Score="1" Body="&lt;p&gt;Imagine that the engine is a person throwing a 1 kilogram ball backward at a speed of 1 meter per second. The reverse thrust system can be modeled as a wall attached to the plane that bounces the ball so it goes forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forward impulse caused by each ball is, obviously. 1 kg m/s. Then when it bounces, its velocity changes from 1 m/s backward to 1 m/s forward, for a net change of 2 m/s, applying a reverse impulse to the plane of 2 kg m/s. So the net total impulse is 1 kg m/s backwards.&lt;/p&gt;&#xA;" OwnerUserId="757" LastActivityDate="2016-11-24T00:19:04.933" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12466" PostTypeId="1" CreationDate="2016-11-24T01:52:40.510" Score="2" ViewCount="59" Body="&lt;p&gt;I'm looking for some advice on what study of engineering I should pursue. I want to be a PLC / SCADA programmer, but with all these engineering titles I am lost on what to even look for. I was thinking electrical engineering and then take some smaller programming courses. If anybody has any experience in the field or knows anybody and can guide me in the right direction it will be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="8936" LastActivityDate="2016-11-24T01:52:40.510" Title="PLC / Scada education requirements." Tags="|education|" AnswerCount="0" CommentCount="1" ClosedDate="2016-11-27T11:19:22.643" ContentLicense="CC BY-SA 3.0" />
  <row Id="12467" PostTypeId="1" CreationDate="2016-11-24T04:18:47.883" Score="1" ViewCount="363" Body="&lt;p&gt;I would like to model below mechanical system in Patran.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MKRbB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MKRbB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I don't know how to configure boundary condition of ball joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ball joint is spherically rotated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the described system is in gravity, the upper beam and supports would be deformed, then contact point of ball joint would be slightly rotated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my model, I just rigidly connected the ball joint by using RBE 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, with my model, I can not consider rotation of ball joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain how to model this kind of ball joint, or give me some sources about this issue?&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="4624" LastEditDate="2016-11-28T00:21:32.563" LastActivityDate="2016-11-28T00:21:32.563" Title="How to model ball joint in Patran/Nastran?" Tags="|mechanical-engineering|modeling|simulation|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12468" PostTypeId="1" CreationDate="2016-11-24T08:37:37.050" Score="3" ViewCount="9421" Body="&lt;p&gt;Can aluminum alloy be an alternative to steel for reinforcement in terms of the same forces encountered by reinforced concrete beams? Will it be economical and feasible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there organizations who already practices the use of aluminum alloy reinforcement as a replacement to steel?&lt;/p&gt;&#xA;" OwnerUserId="8939" LastEditorUserId="7371" LastEditDate="2016-12-12T23:53:44.997" LastActivityDate="2017-06-10T18:53:14.097" Title="Can steel be replaced with aluminum alloy in reinforced concrete?" Tags="|civil-engineering|aluminum|reinforced-concrete|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12469" PostTypeId="2" ParentId="12450" CreationDate="2016-11-24T09:34:09.057" Score="3" Body="&lt;p&gt;This is a pretty big topic but to point you in the right direction: start by having a read of the Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Seismic_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Seismic analysis&lt;/a&gt;. There you will see there are different methods of analysis with increasing complexity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answering your question&lt;/strong&gt;: to predict failure at a certain floor I would suggest carrying out a response spectrum analysis (modal analysis with lumped mass model) which will give you displacements at each floor for a given input signal frequency. Note that failure will not necessarily occur at the floor with the largest absolute displacement but at that with the largest inter-floor drift (the largest displacement relative to the floor beneath it). You can then re-run this analysis with a range of input frequencies until you find one where it is very clear which floor is most likely to fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as equations go, the above approach involves solving a system of harmonic equations treating each floor as a mass with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Harmonic_oscillator#Damped_harmonic_oscillator&quot; rel=&quot;nofollow noreferrer&quot;&gt;damped harmonic oscillator&lt;/a&gt;. The system is then excited by an arbitrary forcing function, $\mathbf{F}(t)$, from your signal generator:&#xA;$$&#xA;\mathbf{F}(t) = M\ddot{\mathbf{x}} + C\dot{\mathbf{x}} + K\mathbf{x}&#xA;$$&#xA;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M$ is your lumped mass matrix (one entry per line representing floor weight)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C$ is your damping matrix (represents damping force on each floor related to velocities of the floors either side)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K$ is the spring force matrix (representing spring force on each floor related to relative displacement of the floors either side)&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2016-11-24T09:34:09.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12470" PostTypeId="2" ParentId="12468" CreationDate="2016-11-24T10:42:52.177" Score="10" Body="&lt;p&gt;A very important property of steel used in reinforced concrete is that it has similar &lt;a href=&quot;http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;coefficient of thermal expansion&lt;/a&gt; as concrete:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Concrete: $14.5 \cdot 10^{-6} \frac{m}{m\,K}$&lt;/li&gt;&#xA;&lt;li&gt;Steel:    $12.0 \cdot 10^{-6} \frac{m}{m\,K}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Compare this with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Aluminum: $22.2 \cdot 10^{-6} \frac{m}{m\,K}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So a likely result of putting aluminum inside a concrete beam would be the concrete cracking when the aluminum expands more in the summer.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-11-24T10:42:52.177" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12474" PostTypeId="1" AcceptedAnswerId="12488" CreationDate="2016-11-24T21:42:50.263" Score="2" ViewCount="5261" Body="&lt;p&gt;I have an overall question that's been bothering me for years. Which is more dangerous for our health, a diesel car or a gasoline car?&lt;/p&gt;&#xA;&lt;p&gt;This is quite a cliche question and I can find a lot of newspapers reporting that diesel is way more dangerous, but I would like to hear a more engineering/scientific/economic approach to this. Can anyone help me with this? I have a series of other questions that would help me to answer this overall question. Here is goes...&lt;/p&gt;&#xA;&lt;p&gt;In &lt;a href=&quot;https://www.theguardian.com/uk/2013/jan/27/diesel-engine-fumes-worse-petrol&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;, they write:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Diesel fumes are significantly more damaging to health than those from petrol engines, according to research which shows that related air pollution contributes to lung disease, heart attacks, asthma and other respiratory problems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And they are referencing a nice report written by Department for Energy and Climate Change.&lt;/p&gt;&#xA;&lt;p&gt;So it is agreed by many that Diesel is more dangerous than gasoline. But at the same time, the diesel engine has a higher efficiency.&lt;/p&gt;&#xA;&lt;p&gt;Q1: Is the efficiency also reflected in the amount of gases getting out? In other words, an engine with a higher efficiency releases a lower amount of gases?&lt;/p&gt;&#xA;&lt;p&gt;Then there is a second question, which might sound a bit stupid but,&lt;/p&gt;&#xA;&lt;p&gt;Q2: Why are people so much for buying Diesel cars (especially in Europe)? They are way more expensive, and yet so popular in Europe.&lt;/p&gt;&#xA;&lt;p&gt;Q3: Is the lower price of Diesel fuel really making a difference? Or is it the power and reliability?&lt;/p&gt;&#xA;&lt;p&gt;At the same time, the regulations in US are more strict to Diesel cars.&lt;/p&gt;&#xA;&lt;p&gt;Q4: Is it because of the emissions or is there another reason for this?&lt;/p&gt;&#xA;&lt;p&gt;Q5: Is really NOx worse for our health than COx?&lt;/p&gt;&#xA;&lt;p&gt;Q6: What are the engineering limitations in reducing the NOx's?&lt;/p&gt;&#xA;&lt;p&gt;I know that some question are not strictly related to engineering, but I want to avoid posting the same questions in other places. However, I would appreciate if you could answer to any of the questions. I would highly appreciate any help in understanding this.&lt;/p&gt;&#xA;&lt;p&gt;I think this is a very important issue, because most of people (like me) are unaware of these issues.&lt;/p&gt;&#xA;" OwnerUserId="8419" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-02-21T04:40:17.147" Title="Diesel or Gasoline? Which one is more dangerous to our health?" Tags="|automotive-engineering|chemical-engineering|gas|petroleum-engineering|diesel|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12475" PostTypeId="1" AcceptedAnswerId="12482" CreationDate="2016-11-24T22:59:15.280" Score="1" ViewCount="48" Body="&lt;p&gt;I have created a self balancing robot. It uses PID control to control its pitch angle and gets the feedback from gyroscope and accelerometer. But I have not used any sensor to feedback the displacement to the controller. I have just approximated the voltage applying to the motor to the displacement through programming (let suppose 5v to the motor will make the motor to rotate for 1 rev/s and after integrating it , it will give the displacement covered). Through which I was able to maintain the robot at a single position. But now its time write the report and I do not know what to call this type of feedback of displacement.&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="8950" LastActivityDate="2016-11-25T06:51:05.290" Title="What is the name of my feedback procedure in control system that I am using?" Tags="|control-engineering|control-theory|pid-control|systems-engineering|feedback-loop|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12476" PostTypeId="2" ParentId="11807" CreationDate="2016-11-24T23:20:24.980" Score="1" Body="&lt;p&gt;e(t) and u(t) are functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e(t) shows the difference between required location and current location. &#xA;u(t) is the voltage input to the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means let suppose your printer head is at 4cm w.r.t reference point at t=1s but it must be at 5cm w.r.t reference point so the difference will be 1cm at t=1s. So the voltage input the motor will be u(1)=Kp(1)+Ki(1+e(t-1))+Kd(1-e(t-1))/1. So when t=1.1s e(1.1) will decrease due to previous effort on the motor by u(1). And soon error will reduce to zero. Hence u(t) and e(t) are varying with time they are functions. &lt;/p&gt;&#xA;" OwnerUserId="8950" LastActivityDate="2016-11-24T23:20:24.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12477" PostTypeId="1" AcceptedAnswerId="12478" CreationDate="2016-11-24T23:49:21.307" Score="1" ViewCount="90" Body="&lt;p&gt;How large of an area of the earth's surface would we need to cover in mirrors to negate current levels of global warming?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume that the reflective component of the mirror is made of aluminium foil and that it is kept clean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let's also forget about ocean acidification. I'm only really interested in negating the retained heat component of atmospheric carbon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question was initially asked &lt;a href=&quot;https://physics.stackexchange.com/questions/292765/how-large-of-an-area-of-the-earths-surface-would-we-need-to-cover-in-mirrors-to&quot;&gt;here&lt;/a&gt; on physics.stackexchange but was rejected because it &quot;was an engineering question&quot;.&lt;/p&gt;&#xA;" OwnerUserId="8951" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2023-03-01T12:53:21.890" Title="How large of an area of the earth's surface would we need to cover in mirrors to negate global warming?" Tags="|materials|temperature|environmental-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12478" PostTypeId="2" ParentId="12477" CreationDate="2016-11-25T02:29:25.623" Score="3" Body="&lt;p&gt;The idea is not new, but using mirrors on the Earth's surface is a particularly inefficient way to go about it.&lt;/p&gt;&#xA;&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Space_sunshade&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lagrange-point sunshade&lt;/a&gt; is actually much more practical, and the thermal analysis has been worked out in some detail. You basically need to block about 2% of the sun's light currently reaching the Earth.&lt;/p&gt;&#xA;&lt;p&gt;Beware of unintended consequences!&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; The topic of Earth's heat-flow balance and resulting surface temperature is surprisingly complex. A &lt;a href=&quot;https://www.youtube.com/watch?v=oqu5DjzOBF8&amp;amp;t=23s&quot; rel=&quot;nofollow noreferrer&quot;&gt;video by Sabine Hossenfelder&lt;/a&gt; is a good start at &amp;quot;peeling the onion&amp;quot;.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2023-03-01T12:53:21.890" LastActivityDate="2023-03-01T12:53:21.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12479" PostTypeId="1" CreationDate="2016-11-25T03:46:49.890" Score="1" ViewCount="11053" Body="&lt;p&gt;what is the difference between a PI and a PD controller? In which instance or for what kind of sensor(s)would you use one as opposed to the other?&lt;/p&gt;&#xA;" OwnerUserId="8911" LastActivityDate="2016-11-25T09:32:11.480" Title="Difference between PI and PD controller?" Tags="|sensors|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12480" PostTypeId="1" CreationDate="2016-11-25T03:58:15.707" Score="3" ViewCount="212" Body="&lt;p&gt;In structural analysis of a bridge using whatever software, such as MIDAS or SAP 2000, it is easy to obtain the deformation of individual nodes. However, for bearings, what we want is the relative displacement of top and bottom nodes. As we have thousands of load cases, we envelope them into load combinations to get the maximum and minimum forces and deformations. How does one derive the bearing movements with ease?&lt;/p&gt;&#xA;" OwnerUserId="8952" LastActivityDate="2016-11-25T03:58:15.707" Title="How to obtain relative movement and rotation of bridge bearings?" Tags="|civil-engineering|structural-analysis|modeling|bridges|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12482" PostTypeId="2" ParentId="12475" CreationDate="2016-11-25T06:51:05.290" Score="1" Body="&lt;p&gt;I would call this &lt;a href=&quot;https://en.wikipedia.org/wiki/Dead_reckoning&quot; rel=&quot;nofollow noreferrer&quot;&gt;dead reckoning&lt;/a&gt;, because you do not have any actual measurement of the position, but are just calculating it based on the estimated speed of the motor.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-11-25T06:51:05.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12483" PostTypeId="2" ParentId="12479" CreationDate="2016-11-25T09:32:11.480" Score="4" Body="&lt;p&gt;PI - Proportional - Integral&lt;br&gt;&#xA;The output is a combination of how far you are from the goal and the integral of your distance from the goal (total error over time). This means that it will track small changes well but in the event of a large change it will be prone to overshooting. Good for systems which are inherently heavily damped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PD - Proportional - Differential&lt;br&gt;&#xA;The output is a combination of how far you are from the goal and how fast you are moving towards the goal. The differential part is normally negative, this means that if you are rapidly approaching the goal then you start to slow down. It handles large changes well with minimal overshoot but isn't great for tracking small changes or errors. Good for systems which inherently have a lot of momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the third option is PID which is a little more complex since it has 3 parameters rather than 2 but provides the best of both systems.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-11-25T09:32:11.480" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12484" PostTypeId="1" AcceptedAnswerId="12487" CreationDate="2016-11-25T12:07:48.617" Score="0" ViewCount="822" Body="&lt;p&gt;Suppose I have a liquid in a container, the type of liquid is such that it expands when its temperature rises. The contrainer is marked with scale to measure the height of liquid expanded. There is a controller that maintains the temprature at it desired value. Temprature sensor feedback the temprature value to generate error signal into the controller. But the main purpose of the system is to maintain the height which does not have any feedback sensor. The only way to find out is through scale reading which is done by human observer. To maintain the height, the observer changes set point temprature. So first, what is the actual name of this kind of system, an open loop system for height control? Second is this block diagram correct in control theory?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/laZpe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/laZpe.png&quot; alt=&quot;Control sytem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8950" LastActivityDate="2016-11-25T16:48:10.153" Title="What type of control system is this? open loop?" Tags="|control-engineering|control-theory|temperature|pid-control|feedback-loop|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12485" PostTypeId="2" ParentId="12468" CreationDate="2016-11-25T13:18:44.927" Score="0" Body="&lt;p&gt;Al is more expensive than regular steel per unit strength. Plus, the reason steel is used in direct contact with concrete is the coincidence of the thermal expansion rate of the two materials at typical temperatures. &#xA;IOW, Aluminum would crack concrete when it expands if the Al were inside the concrete. &#xA;However, if you allow for movement between the two materials, say, design the structure so one can slide over the other, in certain applications, yes, Al can be used. For example, you might use an Al roof on a concrete-walled structure. &lt;/p&gt;&#xA;" OwnerUserId="8956" LastActivityDate="2016-11-25T13:18:44.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12487" PostTypeId="2" ParentId="12484" CreationDate="2016-11-25T16:48:10.153" Score="2" Body="&lt;p&gt;This system can be considered &quot;closed loop,&quot; since the control input is being determined by some sort of feedback loop (even though the mathematical expression of the feedback loop through the operator is unknown). Also, yes the diagram you've shown is valid. I often work with control systems that have a human operator in the control loop, such as robotic teleoperation systems, and they are often shown as some black box operating on a feedback signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it would be difficult to apply any sort of control theory to your system without some sort of model of how the human operator reacts to the height of the liquid. Often you can make an assumption, like maybe the operator behaves like a proportional controller with a bit of noise (you would need to have solid justifications for any assumptions you make). Either way, in this case if you want to control the height of the fluid with any amount of precision, replace that operator with an ultrasonic rangefinder (or some other sensor) plus some variation on a PID controller.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2016-11-25T16:48:10.153" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12488" PostTypeId="2" ParentId="12474" CreationDate="2016-11-25T16:58:50.467" Score="4" Body="&lt;p&gt;1) In an engineering context efficiency has a very specific meaning ie it is the ratio between the energy you put into a system and the useful work that you get out. In itself this doesn't tell you very much about the amount of pollution produced in the process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fuel economy is a bit different as it considers the overall energy(fuel) use of the car as a whole system and while engine efficiency is part of this the mass and drag of the vehicle are equally important. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pollution emissions is more closely related to fuel economy than efficiency but even then they are not identical. Clearly all else being equal if you burn more fuel you produce more emissions and equally an efficient engine is not necessarily clean, indeed there can be situations where emissions and efficiency are in conflict (especially in the case of NOx emissions).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 &amp;amp; 3) Fuel economy is the main selling point for diesel engines. Even if the fuel is not cheaper per litre (in fact it has been substantially more expensive than gasoline in the recent past) the overall savings in fuel uses can be significant, especially for high mileage users. In very general terms diesel engines provide more low rev  torque and are more robust and durable &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Diesel engines tend to produce more emissions which are immediately dangerous to humans, although this tends to be a local problem first and foremost, particularity in urban areas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) It depends, carbon dioxide is an asphyxiatiant in large enough concentrations but not actually toxic per-se, carbon monoxide is (sort of) toxic in that it binds to haemoglobin and can cause asphyxiation although this is only a real problem in moderate concentrations in enclosed spaces rather than being a long term cumulative toxin. Actually one of the biggest concerns with diesel is particulates especially nano-scale carbon particles (soot) which can accumulate in the lungs with the problems associated with any fine particle inhalation and may also be small enough to cross biological membranes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) In some cases reducing NOx emissions can be in direct conflict with efficiency as they are produced more readily in lean-burning, high compression conditions which are preferred for efficiency. There is also the fact that catalytic converters (which are intended, among other things, to Reduce NOx emissions) require fairly specific exhaust conditions to work properly and can be damaged and clogged by the particulates in diesel emissions. There are also diesel particulate filters but these are themselves fraught with engineering challenges and, in particular, tend not to work well in the low mileage, low speed conditions encountered in urban driving where particulate emissions are the biggest problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said all of this there is a difference between the health effects of local pollution and the global effects of more persistent emissions. CO2 and its effect as a greenhouse gas is the obvious one but there are also more general effects from petroleum extraction, handling and processing to consider. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally you can also look at the dangers inherent in the fuel itself. Gasoline is certainly more dangerous to store and handle and being more volatile can contribute to atmospheric pollution just by evaporating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another complication is that many fuels now have at least some bio-fuel component ethanol for gasoline and vegetable oils for diesel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in this context technology is changing rapidly in response to consumer demand and government regulation  so there is always a danger of comparing apples and oranges eg a 2 to 5 litre truck engine is a very different thing from a 1.2 litre hatchback regardless of petrol vs diesel. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-11-25T23:01:00.293" LastActivityDate="2016-11-25T23:01:00.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12490" PostTypeId="2" ParentId="12468" CreationDate="2016-11-25T18:49:49.157" Score="7" Body="&lt;p&gt;In short: you can't use aluminium to reinforce concrete. You shouldn't even embed uncoated aluminium in concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting &lt;a href=&quot;https://www.copper.org/applications/plumbing/techcorner/pdf/corrosion_nonferrous_metals_contact_concrete.pdf&quot; rel=&quot;noreferrer&quot;&gt;Corrosion of Non Ferrous Metals in Contact with Concrete&lt;/a&gt;, &quot;Aluminium suffers attack when embedded in concrete&quot;. Corrosion of aluminium embedded in concrete causes total destruction of aluminium bars, therefore it's dangerous and inadvisable to use aluminium as rebar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have the reference at hand (I think it was in &lt;a href=&quot;https://books.google.es/books?redir_esc=y&amp;amp;hl=ca&amp;amp;id=mJ0FAAAAMAAJ&amp;amp;focus=searchwithinvolume&amp;amp;q=aluminum&quot; rel=&quot;noreferrer&quot;&gt;this book&lt;/a&gt;) but I remember having read about some accident caused by use of aluminium as rebar during a shortage of steel caused by some strikes in the USA. Aluminium bars examined after accident had completely disintegrated leaving just some white powder.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; I got the book again. My memory stated above is inexact, although the conclusion doesn't change. The book is Feld, Jacob. &quot;Lessons from Failures of Concrete Structures&quot;. American Concrete Institute. Detroit 1964.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to it, during the steel strike in 1959 several buildings in New York substituted aluminium conduit for the conventional steel item. It resulted in distress with cracking ceilings along the conduits due to aluminium conduits expansively disintegrating into white powder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the reference was about aluminium conduits instead of aluminium rebar. However, aluminium rebar in concrete would result in the same effects: disintegrating bars, therefore losing all strength and breaking concrete due to expansion.&lt;/p&gt;&#xA;" OwnerUserId="7371" LastEditorUserId="7371" LastEditDate="2016-12-12T23:08:27.990" LastActivityDate="2016-12-12T23:08:27.990" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12492" PostTypeId="1" CreationDate="2016-11-25T21:31:45.970" Score="1" ViewCount="182" Body="&lt;p&gt;I want to build a proximity sensor that will detect if an object is within 100 cm of the sensor and then send a signal to a raspberry pi. What type of short range detection device should I use to build me sensor? I don't need to know the distance, just if the object is there or not. I have found many different types with a wide range in  prices. I want the parts to be low cost.I have been looking at the MuRata Proximity Sensor Pyro-Infrared Sensor, but the range doesn't appear to be enough. Any resources and/or part recommendations would also help.&lt;/p&gt;&#xA;" OwnerUserId="8961" LastActivityDate="2016-12-26T14:19:57.023" Title="How can I build a proximity sensor?" Tags="|electrical-engineering|sensors|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12494" PostTypeId="2" ParentId="12492" CreationDate="2016-11-26T04:13:12.380" Score="1" Body="&lt;p&gt;IMO Infrared Sensor are the most cost effective and currently most commonly used promiximity sensing technology used for detecting a object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also take a look at ultrasonic sensors. TDC1000 from Texas Instrument in combination with a piezo ceramic transducer is one that I am fairly familiar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time of Flight (TOF) methodology can be used with Ultrasonic and IR techolgy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another common technology is Capacitive Proximity Sensors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on object you are trying to detect Hall Effect Sensors might work as long as object your trying to detect can be embedded with a magnet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Proximity_sensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proximity sensor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Referances:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.st.com/content/ccc/resource/technical/document/datasheet/c4/11/28/86/e6/26/44/b3/DM00112632.pdf/files/DM00112632.pdf/jcr:content/translations/en.DM00112632.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Proximity and ambient light sensing (ALS) module&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lit/ds/symlink/tdc1000.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;TDC1000 Ultrasonic Sensing Analog Front End (AFE) for Proximity Sensing Applications&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://microchip.wikidot.com/touch:mgc3130&quot; rel=&quot;nofollow noreferrer&quot;&gt;MGC3130 - E-Field Based 3D Tracking and Gesture Controller&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-11-26T12:56:46.243" LastActivityDate="2016-11-26T12:56:46.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12496" PostTypeId="2" ParentId="12492" CreationDate="2016-11-26T06:47:21.100" Score="0" Body="&lt;p&gt;Usually the simplest system is to place a mirror on opposite end and then shoot lightbeam to the mirror then use a light sensor to see if the beam reaches back or not. This can be a infrared or a visible light beam and is called a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Photoelectric_sensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;photoelectric sensor&lt;/a&gt;. This type of sensor has the beneffit of working even in badly lit conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to do this is to use a motion sensor, that is in essence a very low resolution camera that senses change in color. This can also work in infrared range and use its own lightsource so it will work regardless of surrounding light conditions. The benefit of a motion sensor is that you can make the detection area quite large.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2016-11-26T06:47:21.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12498" PostTypeId="1" CreationDate="2016-11-26T09:10:15.867" Score="0" ViewCount="148" Body="&lt;p&gt;In the notes, the author stated that the shear stress is constant over the bottom face of the segment ... why in figure 7-4 , we can see that the shear stress varies ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2X52x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2X52x.png&quot; alt=&quot;shear formula 1 &quot;&gt;&lt;/a&gt;&#xA;Which is correct ? The red circled part or the figure ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3lOYw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3lOYw.png&quot; alt=&quot;shear formula 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2016-11-26T10:54:25.610" Title="shear formula confusion" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12499" PostTypeId="1" CreationDate="2016-11-26T10:24:52.923" Score="0" ViewCount="102" Body="&lt;p&gt;Biological process is the secondary stage in most of wastewater treatment plant. Is there any wastewater treatment plant without undergoing biological process? &lt;/p&gt;&#xA;" OwnerUserId="6422" LastActivityDate="2023-04-25T08:55:02.650" Title="Biological process in wastewater treatment" Tags="|waste-water-treatment|waste-disposal|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12500" PostTypeId="2" ParentId="12498" CreationDate="2016-11-26T10:54:25.610" Score="1" Body="&lt;p&gt;If you look at the 3D view of Figure 7.4c, it explains this quite clearly. The shear stress varies along the beam's &lt;strong&gt;height&lt;/strong&gt;. Along its &lt;strong&gt;width&lt;/strong&gt; (such as along the bottom face), the stress is indeed constant.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-11-26T10:54:25.610" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12501" PostTypeId="1" CreationDate="2016-11-26T12:31:32.323" Score="0" ViewCount="4536" Body="&lt;p&gt;Recently I started to study about &quot;Design of mechanical machinery&quot;. Since I am a person who is very sensitive about words I want to know the difference between these notions and their precise definitions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Linkage&lt;/li&gt;&#xA;&lt;li&gt;Kinematic chain&lt;/li&gt;&#xA;&lt;li&gt;Mechanism&lt;/li&gt;&#xA;&lt;li&gt;Mechanical element&lt;/li&gt;&#xA;&lt;li&gt;Mechanical machine&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It will be very good if you introduce some source textbooks which treat these concepts very carefully and accurate as well.&lt;/p&gt;&#xA;" OwnerUserId="8968" LastEditorUserId="1832" LastEditDate="2016-11-26T21:04:24.900" LastActivityDate="2017-03-05T08:45:00.843" Title="Difference between Linkage, Kinematic chain, Mechanism, Mechanical element, Mechanical machine and their precise definitions" Tags="|mechanical-engineering|terminology|mechanisms|linkage|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12502" PostTypeId="2" ParentId="12492" CreationDate="2016-11-26T13:23:57.000" Score="0" Body="&lt;p&gt;You can use ultrasonic sensor. It has range of about 3 meters and is cheap also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Daps&amp;amp;field-keywords=ultrasonic+sensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Daps&amp;amp;field-keywords=ultrasonic+sensor&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8950" LastActivityDate="2016-11-26T13:23:57.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12504" PostTypeId="2" ParentId="12462" CreationDate="2016-11-26T13:33:08.737" Score="1" Body="&lt;p&gt;No it will not work for unusual shape. Because resistance varies as area of the conductor varies. So to calculate resistivity you must pick a part of your sample whose area remains same between the two point at which you connected the multi-meter and then calculate the area and length between those two points to find the resistivity.&lt;/p&gt;&#xA;" OwnerUserId="8950" LastActivityDate="2016-11-26T13:33:08.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12505" PostTypeId="1" CreationDate="2016-11-26T13:43:13.353" Score="1" ViewCount="692" Body="&lt;p&gt;for shear flow at D , the Q = Ay , i tried another method , i gt 30000 , instead of 29750, why the ans is different from 29750? why cant i use this method( i break the shaded area into 3 parts&#xA; ( as shown in figure) my working is Q = Ay = (60x10x35) + (2x30x10x150) = 30000&#xA;&lt;a href=&quot;https://i.stack.imgur.com/zelth.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zelth.jpg&quot; alt=&quot;area&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g8T3O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g8T3O.png&quot; alt=&quot;image2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my working not acceptable ? &lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="8664" LastEditDate="2016-11-26T16:25:56.517" LastActivityDate="2016-11-26T21:40:05.383" Title="shear flow in hollow beam" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12506" PostTypeId="1" CreationDate="2016-11-26T14:33:50.567" Score="1" ViewCount="3266" Body="&lt;p&gt;I'm trying to work out a &amp;quot;rough and ready&amp;quot; calculation for the simple retaining wall below (about 1.00m high).&lt;/p&gt;&#xA;&lt;p&gt;The structural engineer has designed a cantilever with a large heel, as usual, but on checking, it's hard to excavate under the retained bank to put a &amp;quot;heel&amp;quot; in.  On the other hand, it already has a very large &amp;quot;toe&amp;quot; in the design, because there's a path going in front of the wall, so making the &amp;quot;toe&amp;quot; more substantial or adding a &amp;quot;key&amp;quot; beneath it to stop sliding would be much easier.&lt;/p&gt;&#xA;&lt;p&gt;I'd like to get an idea about load calcs for replacing the usual &amp;quot;T&amp;quot; shaped cantilever with an &amp;quot;L&amp;quot; shaped cantilever as I've seen elsewhere, like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/myVsz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/myVsz.png&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The physical dimensions we have so far (all in mm) are W1=200, H1 = 950-1050 varying along the length of the wall, W2 = 900, H2 = 300. The soil is London Clay (in London, UK), and the base is embedded below ground level which may be all that's needed to prevent sliding. I'm happy to do worst case assumptions and over-engineer for simplicity and durability, which would be fully saturated soil, active soil parameters, &amp;quot;worst case&amp;quot; clay, and poured foundation grade RC35 or similar concrete. Minor soil settlement isn't an issue, as it isn't being built on. The excavator can't quite reach to dig at the end of the toe, so if a key (if any) can be near the stem, that would be easier by far.&lt;/p&gt;&#xA;&lt;p&gt;The current design is for a base slab about 800-900 front-to-back and 300 high, a uniform stem 200 wide, and 2 x A393 welded grid in each of the base and the stem with right-angle 10mm rebar connecting them, so I've taken those values and would keep the same reinforcement (they might be the best or not, I can't tell, but they seem reasonable?).&lt;/p&gt;&#xA;&lt;p&gt;I'm hoping to find out what might be possible, and how to calculate any better options.&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2022-05-09T09:42:33.220" Title="Calculation for cantilever retaining wall, with small/no heel" Tags="|structural-engineering|retaining-wall|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12509" PostTypeId="1" CreationDate="2016-11-26T18:50:51.073" Score="4" ViewCount="75" Body="&lt;p&gt;We suppose thermodynamic equilibrium in a phase diagram, where cooling rates are infinitely long and diffusion in both phases can finish completely. So when we cross the solidus line and we are left with a crystal, will it have a monocrystalline structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take for example the iron-carbon phase diagram, when cooling a hypoeutectoid steel below its upper critical temperature, ferrite will start to form around the grain boundaries of the austenite. But if there are no grain boundaries where can the ferrite start to nucleate from?&lt;/p&gt;&#xA;" OwnerUserId="8973" LastEditorUserId="18132" LastEditDate="2021-06-22T08:18:56.007" LastActivityDate="2021-06-22T08:18:56.007" Title="What is the theoretical grain size in an ideal phase diagram of a metal?" Tags="|steel|metallurgy|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="12511" PostTypeId="1" CreationDate="2016-11-26T21:33:09.500" Score="5" ViewCount="4959" Body="&lt;p&gt;I am a high school physics student and I've been trying to simulate the acceleration of a car as it fully accelerates from rest, but I'm finding it's much harder than I thought. Here was my first attempt (without including air resistance):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the horsepower to calculate the kinetic energy of the car since power = $\frac{\Delta E}{\Delta t}$ and in this case $E$ is the kinetic energy of the car. I substituted this into $KE = \frac{1}{2}mv^2$ and after manipulating the equation I ended up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(t)=\sqrt\frac{2\dot{}hp\dot{}t}{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After taking air resistance into account, I substituted actual hp and mass of a Ferrari and took the limit as $t\rightarrow \infty $. My simulated top speed was very close to its actual top speed, but something wasn't right. The acceleration curve is decaying and at $t=0$ acceleration is infinite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/eUHOx-_1xZc?t=8m3s&quot; rel=&quot;nofollow noreferrer&quot;&gt;One source&lt;/a&gt; tested the acceleration of a few super cars from 0 to 60 and the acc vs time curve looks like this:&lt;a href=&quot;https://i.stack.imgur.com/vOe20.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vOe20.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; After more research, I figured I must take into account torque, gear ratios, and RPM, but it only confused me more. Does anyone have any insight to make a realistic simulation, what equations to use, or which properties of the car are relevant? One car I would like to use for the simulation would be the &lt;a href=&quot;https://en.wikipedia.org/wiki/Koenigsegg_Regera&quot; rel=&quot;nofollow noreferrer&quot;&gt;Koenigsegg Regera&lt;/a&gt; because it has one fixed gear, which I thought could make calculations more simple. Thanks for any help or advice.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Update)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to previous answers, I have found out that I can use a torque curve to relate the engine's RPM to torque, and thus solve for acceleration. The only problem that remains is how to find the rate of change of engine rpm. Consider a car at rest is running at 1000 rpm. If the driver then completely floors the throttle, how would I figure out how fast the rpm increases and approaches redline? To my current understanding, I need this information to find the torque at any given moment.&lt;/p&gt;&#xA;" OwnerUserId="8974" LastEditorUserId="8974" LastEditDate="2016-11-29T00:38:43.327" LastActivityDate="2018-06-02T22:58:10.223" Title="How to simulate car acceleration" Tags="|mechanical-engineering|automotive-engineering|gears|modeling|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12512" PostTypeId="2" ParentId="12505" CreationDate="2016-11-26T21:40:05.383" Score="2" Body="&lt;p&gt;Acceptable for what? For practical engineering purposes it would usually perfectly acceptable. The difference between the two values is less than 1%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book is making more &quot;precise&quot; assumptions (see the plot of the shear distribution in Fig. e) but the author might have &quot;forgotten&quot; that all of this mathematical theory is based on approximations (Euler-Bernoulli-Timoshenko beam theory) in any case! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;correct&quot; answer depends on exactly how the shear loads are applied to the beam and how it is supported - both of which are ignored in the book example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course in some situations you might need to do a full 3-D finite element analysis of the hollow structure to get even more details of the stress distribution - but that would be pointless if it is just being used as a structural beam and you don't know &lt;em&gt;exactly&lt;/em&gt; how it is loaded and restrained.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-11-26T21:40:05.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12515" PostTypeId="1" AcceptedAnswerId="12516" CreationDate="2016-11-27T03:32:21.413" Score="0" ViewCount="1352" Body="&lt;p&gt;In the notes , i noticed that the shear stress on the top plane and the front plane will 'meet together' at the edge . I dont understand it why the direction is like that .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for both the shear stress to become like this (opposite direction ) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WffLW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WffLW.png&quot; alt=&quot;image1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0GNkA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0GNkA.png&quot; alt=&quot;iamge2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ifJHj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ifJHj.png&quot; alt=&quot;image3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rGUv2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rGUv2.jpg&quot; alt=&quot;image4&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2016-11-27T04:30:38.793" Title="direction of shear stress" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12516" PostTypeId="2" ParentId="12515" CreationDate="2016-11-27T04:30:38.793" Score="2" Body="&lt;p&gt;Yes, it is. If you draw the shear force arrows on all four sides of a quadrilateral, they point &quot;towards&quot; two diagonally opposite corners, as in the picture in example 7.3(c) and &quot;away from&quot; the other two corners, as in your own picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about a very small quadrilateral element of the structure, small enough so that the shear and direct stresses can all be considered constant over the element. If the shear force arrows pointed &quot;head to tail&quot; around the sides of the quadrilateral, the forces would produce a moment about the center point, so there would have to be some other forces to counterbalance that moment if the structure was in equilibrium. But uniform &lt;em&gt;direct&lt;/em&gt; stresses on the edges of the quadrilateral can't produce a moment about the center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the shear forces must be arranged so that the two forces on one pair of opposite sides produce a moment in one direction, and the two forces on the other pair of sides produce an equal moment in the opposite direction.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-11-27T04:30:38.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12519" PostTypeId="1" CreationDate="2016-11-27T09:49:52.277" Score="2" ViewCount="191" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yVWnx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yVWnx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a newcomer to Plastic Injection Molding and would really appreciate some advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a prototype object that will be made from a type of polypropylene. I only need a maximum of 10 units initially to ensure that the product is satisfactory. The issue is that there are many undercuts in the design. Imagine if you will, a variety of different shaped trees growing randomly in a field. This will be the design. The trees and the field will need to be formed as a single object to maintain strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I printed a similar design which was aesthetically OK but due to material limitations was not suitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe Plastic Injection Molding is not the approach I should be taking? &lt;/p&gt;&#xA;" OwnerUserId="8979" LastEditorUserId="1832" LastEditDate="2016-11-27T10:42:29.387" LastActivityDate="2016-11-28T22:19:57.463" Title="Plastic Injection Moulding a Piece with Undercuts" Tags="|manufacturing-engineering|plastic|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12520" PostTypeId="1" AcceptedAnswerId="12528" CreationDate="2016-11-27T09:56:28.537" Score="-2" ViewCount="58" Body="&lt;p&gt;Are these right angle projections correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://postimg.org/image/64l4ibu25/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://postimg.org/image/64l4ibu25/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://postimg.org/image/pnpprostp/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://postimg.org/image/pnpprostp/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8980" LastActivityDate="2016-11-27T16:01:23.737" Title="Are these right angle projections correct?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ClosedDate="2016-11-29T23:07:44.777" ContentLicense="CC BY-SA 3.0" />
  <row Id="12521" PostTypeId="1" AcceptedAnswerId="12551" CreationDate="2016-11-27T11:13:40.840" Score="0" ViewCount="250" Body="&lt;p&gt;I want to know how to model and how to use an injection nut in SolidWorks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem starts with me not knowing the correct term in English for them thing I want to model (&lt;em&gt;Einpressmutter&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm referring to a metric nut (with metric thread, if course) which exposes a circular, radial notch at one end. This end is to be forcefully inserted into sheet metal with a hole slightly smaller in diameter than the nut's end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.pemnet.com/fastening-products/nuts-for-sheetmetal/&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Examples for those devices&quot;&gt;Penn Engineering nuts for sheet metal&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no problem for me in modeling such a part. What I want to have, is like a combination of a part and a forming tool. I.e. I want to drag this part into a closed sheet metal area and the holes for the nut should be created automagically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google didn't reveal anything useful. I found only tutorials in creating regular nuts, yawn. The SW tutorials weren't helpful either. But this, again might be caused by using the wrong search term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate &lt;strike&gt; a link to &lt;/strike&gt; a tutorial&lt;strike&gt;s&lt;/strike&gt; as well as a better search term or a basic outline of the method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;further clarification&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Yes, &quot;press nut&quot; was the word I'm searching for. &lt;/li&gt;&#xA;&lt;li&gt;Yes the nuts I want to use need holes drilled/punched/lasered into the sheet before insertion. &lt;/li&gt;&#xA;&lt;li&gt;I probably need different nuts of different sizes and threading, so I'm here for an abstract solution&lt;/li&gt;&#xA;&lt;li&gt;I know the hole wizard. I know how to place a part into an assembly and to align it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I want is this behaviour:&#xA;Upon manually placing a press nut into an assembly and aligning it in its recommended position on a metal sheet, the necessary hole shall be generated automatically in the metal sheet. &lt;/p&gt;&#xA;" OwnerUserId="3786" LastEditorUserId="3786" LastEditDate="2016-11-29T11:46:22.370" LastActivityDate="2016-11-29T11:46:22.370" Title="How to model a sheet metal nut in SolidWorks?" Tags="|modeling|solidworks|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12522" PostTypeId="2" ParentId="12519" CreationDate="2016-11-27T12:15:55.330" Score="1" Body="&lt;p&gt;I've worked on products designed by others which used injection moulding so while not an expert (and my terminology usage will be terrible) , do have a good feel for what may be attainable. I'd suggest that injection moulding for 10 prototypes is totally unsuitable and for volume production, it is probably unsuitable except if volumes are utterly vast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main issue is (I think) that tooling inserts or mould proper need a clear removal path once your product is cast. this could be achieved with removable inserts plus multi part split moulds. These will grow at a rate related to numbers of trees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'd be surprised if your assertion that &quot;The trees and the field will need to be formed as a single object to maintain strength.&quot; is correct.&lt;br&gt;&#xA;It  is easy to produce individual 'plantable trees' which can be inserted into a 'field' and then bonded in place by a number of means, in such a way that the retention forces exceed the strength of the material.&lt;br&gt;&#xA;These may include adhesive (although polypropylene is not a good material for this), ultrasonic welding, possibly &quot;stir welding&quot; with rotation of tree in hole to generate heat, and possibly mechanical retention by screws, pins etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also - why polypropylene? There may be a good reason related to the properties of the trees, but for injection moulding ABS is far more common, and AS may provide a superior result to AS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A far more extensive of the requirement  would help.&lt;br&gt;&#xA;Why one piece?&lt;br&gt;&#xA;Why PP?&lt;br&gt;&#xA;Why should multi partv strength be an issue?&lt;br&gt;&#xA;Why .....?&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2016-11-27T12:15:55.330" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12524" PostTypeId="2" ParentId="12519" CreationDate="2016-11-27T13:10:34.397" Score="2" Body="&lt;p&gt;For prototyping or small test series vacuum molding is the way to go. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses silicone casting moulds which can be easily split into several parts to feature undercuts. The moulds themselves can be cast from 3D printed parts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From one mould several up to tens of pieces can be cast before it has to be replaced.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Injection moulding on the contrary is suitable for high output numbers from thousands to hundred thousands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore massive parts like the top of your trees can't be injection moulded easily, because of the effects if thermal expansion. They won't keep their desired shape during cool down.&#xA;If you really want to produce a larger numbers of your meadow, you can resort to thermoplastic foam injection moulding. It is much better in keeping shape of thick parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;update based on your comments&lt;/strong&gt;&#xA;Some more words on injection moulding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tool for injection molding for a mentioned size of 100 * 80 * 20 mm may cost 5000,- € even with low quality surfaces. But for each tree you need probably two sliders to form the undercuts. A sliders may cost something between 500,- € and 3000,- € depending on size and complexity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be aware that as Russel already pointed out in his post a certain combination of trees on that meadow may be unmoldable at all because of trees being in the way of sliders needed for another tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially the room between a group of more than two trees is prone to be inaccessible to sliders. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMHO having more than 5 trees on your plate brings the probability that this can be injection moulded close to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tooling can be extremely expensive with injection moulding. And the restrictions are somewhat complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that when scaling a part you have to make a new tooling casting tens of thousands again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What can you do instead of injection moulding?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may think of using elastomeres instead of thermoplastic materials. They can be dispensed, which makes the tools much cheaper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also think of producing the trees separately. You can laser weld them to the base plate which makes it nearly as solid as moulded in one piece.&lt;/p&gt;&#xA;" OwnerUserId="3786" LastEditorUserId="3786" LastEditDate="2016-11-28T22:19:57.463" LastActivityDate="2016-11-28T22:19:57.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12525" PostTypeId="2" ParentId="12462" CreationDate="2016-11-27T13:22:34.287" Score="0" Body="&lt;p&gt;Given a homogenous Material with only scalar resistivity like in ordinary metals, you can derive the resistivity for any given shape by integration, after you have acquired one measurement of resistance between two planar contacts. This can be tedious with pen and paper, but it is at least possible. Depending on the complexity if the shape you can go straight ahead with integration or you have to make some assumptions and simplifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For inhomogenous and/or non-scalar resistivity as in crystals, it is very difficult. I think software like Ansys or Comsol will supply a sufficiently practical numerical algorithm. &lt;/p&gt;&#xA;" OwnerUserId="3786" LastEditorUserId="3786" LastEditDate="2016-11-29T09:30:28.057" LastActivityDate="2016-11-29T09:30:28.057" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12528" PostTypeId="2" ParentId="12520" CreationDate="2016-11-27T16:01:23.737" Score="0" Body="&lt;p&gt;Link 6414ibu25 is missing a dashed (hidden) line in the upper left projection on the first image.&#xA;The second image in 6414ibu25 appears to be correct&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link pnpprostp - the upper right image is projected from the wrong side. &#xA;The lower image in pnpprostp is correct.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2016-11-27T16:01:23.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12529" PostTypeId="2" ParentId="12511" CreationDate="2016-11-27T17:07:03.057" Score="3" Body="&lt;p&gt;The crucial thing here is that for IC engines torque is some function of RPM and so you need to consult the torque curve for a given engine to do any sort of meaningful modelling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also closely related to the way the gearbox works and indeed the primary function of the gearbox is to keep the RPM withing a particular range where the engine provides the most torque. Also engine torque characteristics can vary a lot for engine to engine and some engines may have quite a narrow rev band where they are providing peak torque which drops off sharply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth nothing that power is proportional to torque x rpm and peak power and peak torque are not necessarily the same and while power is a good guide to maximum speed torque is much more relevant to acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is further complicated by the fact that in accelerating form a standstill different limiting  factors come into play at different phases of acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the initial getaway traction dominates ie the ability to transfer torque through the tyres to the road without spinning the wheels. Most cars are capable of delivering enough torque to the wheels to spin them from a standstill in 1st gear so you can't just assume that the maximum torque the engine can provide is available at all times as an accelerating force. More powerful high performance cars may be able to spin the wheels on any gear-change. This initial traction phase is quite difficult to model without  a lot of extra information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also air resistance is proportional to the square of speed so it is exponentially more significant as speed increases and is indeed the limiting factor of top speed (for a given top gear ratio)  in most cases. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit in response to comment by OP &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming no slip (at either wheels or clutch) engine RPM is proportional to road speed, with the constant of proportionality being the overall gear ratio, including the current gear selected, the differential ratio and tyre diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases the best traction is achieved with a small slip angle, although you may want to ignore this for the sake of simplicity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the engine in question has a relatively flat torque curve it is probably OK to use the average torque between gear change points, otherwise you have to deal with the calculus relating instantaneous speed and torque. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-12-01T20:04:25.030" LastActivityDate="2016-12-01T20:04:25.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12531" PostTypeId="1" AcceptedAnswerId="12533" CreationDate="2016-11-27T19:46:33.890" Score="0" ViewCount="367" Body="&lt;p&gt;I'm looking to create something like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XSZiL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XSZiL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem with solid copper wires is it tends to spring back into it's original shape and this construction will flatten.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried creating a wooden cone and wrap the copper around it; however, after leaving it for 24 hrs, it still flattens when I take it off the wooden cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated, thanks!&lt;/p&gt;&#xA;" OwnerUserId="8986" LastActivityDate="2016-11-27T22:30:44.983" Title="How to create intersected conical copper coils?" Tags="|metallurgy|metals|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12533" PostTypeId="2" ParentId="12531" CreationDate="2016-11-27T22:24:39.253" Score="1" Body="&lt;p&gt;When you bend copper, it becomes much more &quot;springy&quot; than its original state. The wire will almost certainly have been bent at some stage before you use it, so the first thing you need to do it get it back to its original &quot;soft&quot; state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do that by &lt;em&gt;annealing&lt;/em&gt; the wire. Heat it with a gas torch till it is red hot, and allow it to cool in air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After it is annealed, make sure you only bend it &lt;strong&gt;&lt;em&gt;once&lt;/em&gt;&lt;/strong&gt;, into its final shape. You can repeat the annealing and bending as many times as you like, but you can only make one bend that will &quot;hold its shape&quot; accurately each time you  anneal the wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving the wire wrapped around the cone for 24 hours (or even for a year) won't change anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a some of videos of annealing copper and bending a spiral which might help. You could try bending a spiral, annealing the wire again, and then &quot;expanding&quot; the spiral to make the cone shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=3jSz11lz8MA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=3jSz11lz8MA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=n_p4bNZv1lM&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=n_p4bNZv1lM&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=zvg1CdbBQvI&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=zvg1CdbBQvI&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also get some ideas from websites and videos on bonsai - bent copper wire is used to &quot;train&quot; the tree branches into the desired shapes.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-11-27T22:30:44.983" LastActivityDate="2016-11-27T22:30:44.983" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12534" PostTypeId="2" ParentId="12511" CreationDate="2016-11-28T00:51:36.253" Score="2" Body="&lt;p&gt;&lt;strong&gt;Real-world torque at standstill:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not your main area of concern but needs addressing so that you can then properly deal with the main issue.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your statement   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;at t=0 acceleration is infinite.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indicates (as you are  aware :-) ) a basic misunderstanding or misapplication of formulae.&lt;br&gt;&#xA;Where one formula produces results that do not match known reality, it can help to do a &quot;sanity check&quot; using other formulae which should give the same answer but by using a different approach. If (when) they fail to do so, looking for the apparent contradictions can help you to identify incorrect assumptions and applications of formulae to reality.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very well known basic formula is f = m x a. - Force = mass x acceleration.&lt;br&gt;&#xA;Rearranging - a = f/m&lt;br&gt;&#xA;For acceleration to be infinite mass must be zero or force must be infinite.&lt;br&gt;&#xA;In this case the relationship between power and torque and rotational velocity are liable to be misleading you.&lt;br&gt;&#xA;In an ideal system torque is inversely proportional to power when both are plotted against rotational velocity.&lt;br&gt;&#xA;ie Torque = Power/RPM x k    (k adjusts to get units numerically correct).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This suggests that for a given supplied power Torque becomes infinite at zero RPM.&lt;br&gt;&#xA;This is true only if there are no losses and if the energy source is able to be properly matched to the load. To transfer power from a rotating energy source ('motor') to a stationary object (wheel, axle, driveshaft, gear, ...) requires an infinite ratio gearbox, 'which we have not got'.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest example in 'everyday' automotive reality is electric motor driven vehicles where maximum torque occurs &quot;at rest&quot; as the electric motor can deliver torque without rotation. One limitation to how fast a modern electric sportscar 'comes off the line' is providing a drivetrain that will not shatter under starting torque but which is not excessively oversized and expensive for its task across the rest of its operating range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real-world applications the motor (especially a petrol or diesel engine) power decreases at low rpm and is zero at 0 RPM so torque is also zero at 0 engine RPM. This is compensated for by initially &quot;slipping the clutch&quot; in a manual gearbox system and effectively &quot;slipping the torque converter&quot; in an &quot;automatic gearbox&quot; system. At standstill the motor is allowed to operate at  nonzero rpm, power and torque are produced and the torque is used to apply force to the drivetrain - either via a slipping clutch or via circulating transmission fluid exerting force but no power or motion at the instantaneous moment of application. Once there is some force available, f=ma can be applied and &quot;we're off&quot;.&#xA;In electric cars (with Tesla's easily coming to mind) it is positive with some types of electric motor to produce maximum torque at zero motor rpm and &quot;we're off&quot; faster per available engine power than with internal combustion engines that must produce start torque in a lossy manner, as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The real problem - air resistance:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your basic equations for power and energy re correct then it is likely that your comment  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... After taking air resistance into account ....  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;indicates a very significant misunderstanding of the effects and magnitude of energy consumed by air resistance.&lt;br&gt;&#xA;For cars of typical size and dimension, air resistance is THE major source of power loss at velocities above about 10 m/s.  As speed increases a proper modelling of drag and energy lost to air resistance is utterly essential to calulating acceleration profiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simplistic formula for air drag which is nevertheless extremely useful and which gives results close to those seen in reality is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force (drag) = 0.5 x d x Cd x A x V^2&lt;br&gt;&#xA;&amp;amp;&lt;br&gt;&#xA; Power = 0.5 x d x Cd x A x V^3&#xA;Where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;d = air density = 1.2 kg/m^3 at sea level  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cd = drag coefficient&lt;br&gt;&#xA;Proportion of flat-plate-drag which you can achieve by aerodynamic   streamlining. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A = projected frontal area&#xA;Area you would see viewing vehicle front on.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;V = velocity  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In SI units where A = m^2, V = m/2, d = kg/m^2, F = N (or F ~= kg x 10)&lt;br&gt;&#xA;Power is in Watts  where 1 HP ~= 760 Watts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At sealevel this reduces to ~= &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power = 0.6 x V^3 watts per square meter for a flat plate, reduced by whatever Cd you can achieve.  eg a modern blockish shaped saloon or SUV may achieve Cd = 0.6-0.8 and the very bet sportscars MAY achieve Cds of 0.2-0.3.  ie you can reduce power taken by drag by a factor of 3 or 4 compared to driving in a brick shaped vehicle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an illustration of how much power is involved.&lt;br&gt;&#xA;Assume 2 m^2 frontal area, Cd = 0.5, V = 27 m/s (= 60 mph / 96 kph) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power = 0.6  x Cd x A x V^3 = 0.6 x 0.5 x 2 x 27^3&lt;br&gt;&#xA;= 11,810 Watts&lt;br&gt;&#xA;~=  16 HP  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 100 mph that increases by the cube of the velocity so by a factor of (100/60)^3 = &#xA;4.6 times to about 74 HP. At 200 mph the power taken to oppose air resistance is 2^3 = 8 times larger again at 590 HP.  For vehicles with hundreds of rear wheel HP available the jump from  about 75 HP at 100 mph to about 600 HP at 200 mph means top speed will be &quot;somewhere in this range&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot (lot lot) more could be said on this topic, but applying the above drag and power formula as a starting point should greatly improve the match of your results to reality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;_____________________________&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Torque and power:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is commonly a very major misunderstanding of the very simple relationship of torque and power.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque = Power per RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's it. Tht IS the formal meaning (or one of the several different but consistent ones.  There will be a constant involved depending on the uits used but it's just a &quot;scaling factor&quot;. Using foot-pounds and Horsepower&lt;br&gt;&#xA;Torque = HP x 5252 / RPM  or&lt;br&gt;&#xA;HP = Torque x RPM / 5252 or&lt;br&gt;&#xA;RPM = HP x 5252 / Torque  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be that simple?&lt;br&gt;&#xA;Yep!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque is &quot;twisting force&quot; expressed in force x distance.&lt;br&gt;&#xA;If you lean on a 1 foot bar with 100 lbf force you produce 100 foot pounds of force. Double the force or the bar length and you double the torque.&lt;br&gt;&#xA;If the bar end does not move you are not delivering any power (despite what your muscles and brain may think). Only as you  apply rotation do you deliver power - see formulae above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A useful 'trick':&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &quot;almost right&quot; formula is Watts = kg.m.RPM.&lt;br&gt;&#xA;This almost works because various constants almost cancel. It's about 2% out but very useful for in-the-head estimations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;________________________________________________&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The point where I am stuck is figuring out what the rpms will be at any given point for a maximum accelerating car... from there I can find the force and then the acceleration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If&lt;/strong&gt; you have an automatic transmission you can ~~~ assume that the combination of the correct gear plus torque converter action*&lt;br&gt;&#xA;&lt;strong&gt;THEN&lt;/strong&gt; you can assume RPM is variable around engine optimum at all times.&lt;br&gt;&#xA;Optimum is (probably/usually) such that mean torque is maximised in that gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*A fluid drive torque converter all by itself is effectively an automatically variable gearbox across a limited range, using fluid flow to 'match' driving and driven surfaces. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;eg if the&#xA;  &quot;pump&quot; (engine) rotates faster than the driven surface then the system&#xA;  is designed so that fluid efficiently transfers energy to the slower&#xA;  moving driven &quot;plate&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have a manual gearbox and a competent driver then you can assume a driver will enter a gear and exit a gear at such revs that Rmax ~= redline and Rmin is the Redline /gear_ratio between this gear and the next lower. For VERY narrow power bands, redline may be above optimum speed for engine so driver will exit at such revs that overall revs in that gear produce maximum torque. If the above does not make sense then draw diagrams with rpm on and power curves and gear ratios nd play until it becomes clear (assuming no typos above :-). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very simplistic example: Assume a gearbox with 1.5:1 ratios between all gear changes (1 to 2, 2 to 3, 3 to 4, ...) - which is very high in most cases, and an engine that keeps on building power all the way to redline with power increasing fater than RPM and redline at 9,000 RPM.&lt;br&gt;&#xA;The optimum way to use this engine is to change up AT redline and to try to start it from rest such that when the clutch fully engages the engine is doing 6000 rpm (assuming your clutch can stand that). I'll leave out the 1st gear startup as that is somewhat arguable. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accelerate in 1st gear to 9000 RPM.&lt;br&gt;&#xA;Change 1-2&lt;br&gt;&#xA;2nd gear is now running at 9000/1.5 = 6000 RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accelerate in 2nd gear to 9000 RPM.&lt;br&gt;&#xA;Change 2-3&lt;br&gt;&#xA;3rd gear is now running at 9000/1.5 = 6000 RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accelerate in 3rd gear to 9000 RPM.&lt;br&gt;&#xA;Change 3-4&lt;br&gt;&#xA;4th gear is now running at 9000/1.5 = 6000 RPM.&#xA;Accelerate ... &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Comment only:&lt;br&gt;&#xA;I had a TY250 trials bike (still have but now sleeping in my basement waiting engine repair from long ago death).&lt;br&gt;&#xA;The ratio 1st:2nd is an immense 2:1&lt;br&gt;&#xA;The ratio 2nd:3rd is an immense 1.5:1&lt;br&gt;&#xA;4:3 and 5:4 are somewhat more sane.&#xA;Most fun and (just) possibly optimum way to accelerate was to pull wheelies in 1st 2nd 3rd and 4th successively staring with a near loop in 1st and successively lower ones as the bike desired and changing up as the front wheel kissed the ground after each wheelie. Hitting 5th usually did not produce a wheelie and you then chugged way at about 50 kph :-). &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2016-11-29T01:40:10.063" LastActivityDate="2016-11-29T01:40:10.063" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12535" PostTypeId="1" AcceptedAnswerId="12537" CreationDate="2016-11-28T01:06:09.933" Score="0" ViewCount="6213" Body="&lt;p&gt;In the notes , the author stated that when the force is applied through the centorid of cross section , the channel will bend and twist.&#xA;but , on the second page , the author stated that the shear center lies on an axis of symmetry of member's cross sectional area...&#xA;So, i am confused whether the member will twist or not when the force P is applied thru the centroid or shear center ....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why the shear center is located at O , which is located outside of the C channel ? &#xA;&lt;a href=&quot;https://i.stack.imgur.com/MEH1W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MEH1W.png&quot; alt=&quot;iamge2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HSc0B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HSc0B.png&quot; alt=&quot;image&quot;&gt;&lt;/a&gt;&#xA;Is there anything wrong with the notes ?&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2016-11-28T06:57:37.250" Title="twisting at shear center / centroid" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12536" PostTypeId="2" ParentId="12511" CreationDate="2016-11-28T06:17:17.083" Score="0" Body="&lt;p&gt;A quick shortcut for a vehicle of mass $m$ is the following procedure:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;At each vehicle speed $v$ calculate the engine rpm $n$.&lt;/li&gt;&#xA;&lt;li&gt;For that rpm calculate the engine power $P(n)$ using the torque/power curves.&lt;/li&gt;&#xA;&lt;li&gt;Assume a transmission loss of $\lambda=0.15$ for 15%&lt;/li&gt;&#xA;&lt;li&gt;Assume a drag coefficient $\beta$ (more on this later)&lt;/li&gt;&#xA;&lt;li&gt;The acceleration at this instant is $$ a(v) = \frac{ (1-\lambda) P}{m v} - \beta v^2$$&lt;/li&gt;&#xA;&lt;li&gt;If the coefficient of friction is $\mu$ and the % of weight on the driving wheels $\gamma$ then the traction limit is $$a_{limit} = \mu \gamma g$$&lt;/li&gt;&#xA;&lt;li&gt;Apply which ever is lowest between $a(v)$ and $a_{limit}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;NOTE: You can estimate the drag coefficient $\beta$ if you know the peak power $P_{max}$ and peak speed $v_{max}$ of the vehicle. $$ \beta = \frac{(1-\lambda) P_{max}}{m v_{max}^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need to simulate this using a numerical integration scheme. For quick and dirty use Euler's method where at each time frame, you apply a time step $h$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{aligned} &#xA;t &amp;amp; \leftarrow  t + h \\&#xA;x &amp;amp; \leftarrow  x + h\, v \\&#xA;v &amp;amp; \leftarrow  v + h\, a(v)&#xA;\end{aligned}$$&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2016-11-28T06:17:17.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12537" PostTypeId="2" ParentId="12535" CreationDate="2016-11-28T06:57:37.250" Score="2" Body="&lt;p&gt;Nothing wrong with the book or notes. This is just a little unintuitive at first until you take some time to internalize it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear center lies on an axis of symmetry. If two axis of symmetry exist, it will lie on both of them at the centroid. Otherwise, it is only constrained to lie on one of them. The location of the shear center on the unsymmetric axis is determined by the equation for e.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the centroid is located at the shear center, then it will not twist. However, if they are in different locations, twisting will happen at the centroid if a force is applied there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dVYiS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dVYiS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to McGraw Hill Companies inc, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear Center: Of any cross section of a beam, that point in the plane of the cross section through which a transverse load must be applied in order that there will be only bending of the section and no twisting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture shows how this works:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hGi1g.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hGi1g.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Y0272.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y0272.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IJ1SK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IJ1SK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the downwards force is applied along the screw at any point other than the shear center, twisting will occur. &lt;/p&gt;&#xA;" OwnerUserId="8862" LastActivityDate="2016-11-28T06:57:37.250" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12538" PostTypeId="1" AcceptedAnswerId="12539" CreationDate="2016-11-28T07:22:38.040" Score="1" ViewCount="2181" Body="&lt;p&gt;I am trying to solve the following beam deflection problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The cantilever beam is subjected to the point load at C.&lt;/p&gt;&#xA;  &#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Generate the equation for the elastic curve by using the double integration method.&lt;/li&gt;&#xA;  &lt;li&gt;Find the maximum deflection and slope if L = 3 m and P = 10 kN acted at 2 m from A.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K6SdQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K6SdQ.png&quot; alt=&quot;A cantilever beam is fixed on the left (point A). Force P is shown acting in the downward direction at the rightmost end of the beam (point C). The length of the beam is L.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Ans:&lt;/strong&gt; $y_{max} = -46.67/{EI}$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For part 1, my answer is $EI\dfrac{d^2y}{dx^2} = - Px$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part 2, my answer is:&#xA;$$\begin{gather}&#xA;EI\dfrac{dy}{dx} = -\dfrac{Px^2}{2} +c_1 \\&#xA;EIy = -\dfrac{Px^3}{6} +c_1x +c2&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;at $x= 0$, $y = 0$, so $c_2 = 0$&lt;/li&gt;&#xA;&lt;li&gt;at $x = 0$ , $\dfrac{dy}{dx} = 0$, so $c_1 = 0$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, $EIy = -\dfrac{Px^3}{6}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$EIy$ max occurs at $L=3$, so $EIy_{max} = -\dfrac{10\cdot3^3}{6} = -45$, but the answer is  $EIy_{max} =-46.67$. What have I done wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the slope at x = 2, my answer is $EI\dfrac{dy}{dx} = -\dfrac{Px^2}{2} =  -\dfrac{10\cdot2^2}{2} = -\dfrac{40}{EI}$.&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="368" LastEditDate="2016-12-01T17:51:38.667" LastActivityDate="2016-12-01T17:51:38.667" Title="How to calculate deflection of a cantilever beam subject to point loading" Tags="|structural-engineering|beam|deflection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12539" PostTypeId="2" ParentId="12538" CreationDate="2016-11-28T09:55:42.460" Score="3" Body="&lt;p&gt;Both answers are incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your answer to (a) would imply that the bending moment is equal to zero at $x=0$ and increases linearly until $x=L$. This is only true if you put $x=0$ at the free end of the beam and $x=L$ at the fixed support, in which case you'd have to change your constraints. What you actually need is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EI\dfrac{d^2y}{dx^2} = -PL + Px$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (b), you didn't notice that the behavior is different in such a beam. You can't simply use the equation above and be done with it. After all, this equation describes a beam under load, but the span from $x \in [2, 3]$ is not under load and therefore its beam equation is different (it follows a linear path).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easy way to solve it is to start by finding the slope and deflection at $x=2$. For this, we can use the equation above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;EI\dfrac{dy}{dx} = -PLx + \dfrac{Px^2}{2} + c_1 \\&#xA;\dfrac{dy}{dx}(x=0) = 0 = c_1 \\&#xA;EIy = -\dfrac{PLx}{2} + \dfrac{Px^3}{6} + c_2 \\&#xA;y(x=0) = 0 = c_2 \\&#xA;\therefore EI\dfrac{dy}{dx} = -PLx + \dfrac{Px^2}{2} \\&#xA;\therefore EIy = -\dfrac{PLx^2}{2} + \dfrac{Px^3}{6}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From which we gather that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\dfrac{dy}{dx}(2) &amp;amp;= -\dfrac{20}{EI} \\&#xA;y(2) &amp;amp;= -\dfrac{26.667}{EI}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $x = 2$ to $x = 3$, the deflection is linear, meaning that the slope is constant, so the result above for $\dfrac{dy}{dx}$ is the same at $x=3$. Knowing this, we can find the deflection at $x=3$ by finding the linear component and then adding it to $y(2)$ (where $\ell$ is the linear span):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\delta &amp;amp;= \dfrac{dy}{dx}(2)\cdot\ell = -\dfrac{26.667}{EI}\cdot1\\&#xA;\therefore y(3) &amp;amp;= y(2) + \delta = -\dfrac{46.667}{EI}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-11-28T09:55:42.460" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12541" PostTypeId="2" ParentId="11772" CreationDate="2016-11-28T12:55:49.647" Score="1" Body="&lt;p&gt;This would be best off asked in electronics.SE. But in short, there are two aspects that would be helpful to consider: DC resistance of the windings, and back-EMF. When the motor starts up (rotational velocity is zero), the current drawn will simply be determined by the DC resistance (If winding inductance is relatively low).  As the motor speeds up, it begins to act like a generator; the back-EMF opposes the voltage driving it. This difference reduces to zero with no load when the motor reaches full speed. As long as your circuit can handle the start-up current, there aren't any concerns. Otherwise you'll need to step the voltage up in increments. Torque will need to be considered. A general rule is torque is proportional to current, and voltage proportional to speed, but that is where my knowledge gets hazy. But that is why current is largest when first starting the motor, and why increasing voltage in steps reduces peak current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a vaguely similar process when slowing down but with added complications. If the driving circuit has infinite impedance when it removes power from the motor, the windings act as an inductor with stored energy. And since inductors oppose changes in current, it will create a very high voltage spike (theoretically infinite) to &quot;try&quot; to keep the current flowing: plus the effects of back-EMF. This can damage the circuitry, but for DC motors this is easily solved with diodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty of circuits in Google images to get ideas from. But without more information, that's about all I can say.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2016-11-28T12:55:49.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12542" PostTypeId="1" CreationDate="2016-11-28T13:36:45.957" Score="0" ViewCount="3387" Body="&lt;p&gt;I am working on ANSYS Mechanical. As i was reading the user's guide i saw a section about contact behavior. Symmetric and asymmetric. Can someone explain me on this? How to use these contacts?&lt;/p&gt;&#xA;" OwnerUserId="8990" LastActivityDate="2021-10-25T22:37:42.920" Title="ANSYS Mechanical Contact behavior" Tags="|mechanical-engineering|ansys|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12543" PostTypeId="1" CreationDate="2016-11-28T14:50:56.600" Score="5" ViewCount="288" Body="&lt;p&gt;I have a sample of aluminium which I'm interested in finding the rough composition of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sample is a part of a large number of samples which I sanded down and polished with a polishing table. I cleaned the samples with acetone, washed and rinsed in ultrapure water. Area spectrum SEM EDS analyses were then ran on all the samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some impurities were left on the surface of the sample as expected. All other samples show oxygen, silicon and carbon impurities assumedly left from the sanding process (silicon carbide paper), the carbon tape for the grounding and possibly carbon impurities in the SEM-chamber. The samples were left to dry overnight so assumedly the amounts of acetone and water on the surface are fairly small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the other samples show accuracies and composities that seem sensible to me. However, the aluminium sample that I have shows levels of oxygen much larger than I expect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YPIpv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YPIpv.png&quot; alt=&quot;EDS results&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, three measurements of four areal spectrums each. Carbon is left out. And a mean of 17.37-at% of oxygen? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that aluminium has a passive oxide layer and that EDS is a relatively surface active technique. However this &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0169433215011617&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; cites passive layer thicknesses in the order of nanometers, and I understand that the EDS has a penetration thickness in the order of micrometers. A comparable steel sample that I analysed has levels of oxygen that I expected (~~0.1%)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's going on? Can oxides precipitate inside bulk aluminium during manufacture or aging?&lt;/p&gt;&#xA;" OwnerUserId="8424" LastEditorUserId="8424" LastEditDate="2017-07-19T13:15:29.677" LastActivityDate="2017-07-19T13:15:29.677" Title="Anomalous oxygen in aluminium sample SEM EDS" Tags="|materials|measurements|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12544" PostTypeId="1" CreationDate="2016-11-28T14:53:41.590" Score="0" ViewCount="20" Body="&lt;p&gt;If I took the filament in the bulb and made it smaller, and also increase the thickness of it, what would happen?&lt;/p&gt;&#xA;" OwnerUserId="7041" LastActivityDate="2016-11-28T18:01:02.433" Title="If I made a incandescent light bulb filament smaller what would happen?" Tags="|power-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2016-11-30T23:23:18.257" ContentLicense="CC BY-SA 3.0" />
  <row Id="12546" PostTypeId="1" CreationDate="2016-11-28T15:22:43.120" Score="-1" ViewCount="234" Body="&lt;p&gt;Here's my ans , but the ans given is EIy = -4953 , but i gt EIy = -5078 , anything wrong with my answer ? for slope at A , i gt EI(dy/dx) = 0 instead of EI(dy/dx) =625&lt;a href=&quot;https://i.stack.imgur.com/WiznY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WiznY.png&quot; alt=&quot;1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UsFpZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UsFpZ.jpg&quot; alt=&quot;solution  &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="8664" LastEditDate="2016-11-28T15:29:11.237" LastActivityDate="2016-11-28T23:38:04.023" Title="slope and deflection of beam engineering" Tags="|structural-engineering|" AnswerCount="1" CommentCount="11" ClosedDate="2016-12-01T17:18:24.893" ContentLicense="CC BY-SA 3.0" />
  <row Id="12547" PostTypeId="2" ParentId="12546" CreationDate="2016-11-28T16:30:50.317" Score="4" Body="&lt;p&gt;You messed up your integration from $EI\dfrac{dy}{dx}$ to $EIy$. It should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EIy = \dfrac{75x^3}{6} - \dfrac{15x^4}{24} + c_1x + c_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but your denominator for the first term is $3$, not $6$. (In your calculations, however, you seem to have used 6).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, your solving for the values of $c_1$ and $c_2$ is incorrect. What you have is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;y(x=0)=0 \therefore c_2 = 0 \\&#xA;y(x=L)= \dfrac{75L^3}{6} - \dfrac{15L^4}{24} + c_1L = 0 \\&#xA;\therefore c_1 = \dfrac{15L^3}{24} - \dfrac{75L^2}{6} \\&#xA;\therefore EIy = \dfrac{75x^3}{6} - \dfrac{15x^4}{24} + \left(\dfrac{15L^3}{24} - \dfrac{75L^2}{6}\right)x \\&#xA;EIy(5) = -1953&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice my answer is different from the given answer, just as I said in a comment that the given answer is (almost) always right. You'll notice something, though: there's a single digit difference between my answer and the given one: change the $1$ in my answer to a $4$ and you get the given answer. Given that 1 and 4 are neighboring keys on the number pad, this could be easily explained as a typo in the book, instead of a calculation error on the author's part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not satisfied that this is right, I created a computer model to check my work and it checks out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1vFKW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1vFKW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2016-11-28T23:38:04.023" LastActivityDate="2016-11-28T23:38:04.023" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12548" PostTypeId="2" ParentId="12544" CreationDate="2016-11-28T18:01:02.433" Score="1" Body="&lt;p&gt;All other things being equal,  the resistance of a wire is inversely proportional to its cross-section and proportional to its length.   You could think of increasing the cross-section as analogous to resistors in parallel, and increasing the length as analogous to resistors in series. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, pick a given filament, find its resistance, then apply the factors I listed to find the new resistance.   Assuming a &quot;stiff&quot; voltage source, you have a constant voltage, so the power dissipated is $ \frac{V^2}{R} $ &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-11-28T18:01:02.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12549" PostTypeId="2" ParentId="12501" CreationDate="2016-11-28T18:12:42.467" Score="3" Body="&lt;p&gt;Here are some definitions for you, based on my experience with these terms:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;linkage&lt;/strong&gt; is a group of bodies (or &lt;strong&gt;links&lt;/strong&gt;) connected together by joints.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;kinematic chain&lt;/strong&gt; is a subset of linkages, specifically referring to linkages with rigid, ideal joints, and rigid links. Since the joints and links are rigid, this allows you to use geometry to relate the &lt;strong&gt;kinematics&lt;/strong&gt; (position and velocity) of each link to every other link in the chain.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;mechanism&lt;/strong&gt; is synonymous with &lt;strong&gt;mechanical machine&lt;/strong&gt; and refers to any collection of parts that can perform a mechanical action or transformation. For example, a gearbox is a mechanism that converts (transforms) an input torque and speed to a different output torque and speed without the addition of energy. The relationship between the input and output mechanical variables is known and quantifiable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;strong&gt;mechanical element&lt;/strong&gt; is the smallest subdivision of a machine that is still useful for analysis (similar usage as in chemistry, if you consider what happens beyond the scale of atoms to be the domain of physics). For example, if you are considering the gearbox from the previous example, and are interested in the motion of each shaft and gear within the box, then the individual gears and shafts would be considered as &lt;strong&gt;mechanical elements&lt;/strong&gt;. However, if you were analyzing the entire drive train of a car, it would be more convenient to consider the entire gearbox as a single &lt;strong&gt;mechanical element&lt;/strong&gt;, ignoring the internal motion and focusing on its input-output characteristics as it interacts with the rest of the system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Terminology like this can vary from place to place, so it is important to consider your audience. If you are writing an internal report or presentation for a group of engineers or in a course where the terminology is known to you, just use their terminology. Otherwise, it is important to include definitions for any terminology you might use.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="6695" LastEditDate="2016-11-28T20:35:20.657" LastActivityDate="2016-11-28T20:35:20.657" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12550" PostTypeId="1" CreationDate="2016-11-28T19:10:47.307" Score="1" ViewCount="1666" Body="&lt;p&gt;How do i figure out what the wing area/sqft needs to be for a functioning wing-flapping ornithopter? Lets say for example there is 500# of force moving the wings up and down, and the wings and passenger combined weighs 250#. What is the minimum needed wingspan/area to climb through the air?&lt;/p&gt;&#xA;" OwnerUserId="8994" LastEditorUserId="1832" LastEditDate="2016-11-28T19:24:59.937" LastActivityDate="2020-02-29T14:53:28.250" Title="Wingspan formula for an Ornithopter" Tags="|aerospace-engineering|airflow|aircraft-design|airfoils|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12551" PostTypeId="2" ParentId="12521" CreationDate="2016-11-28T19:14:34.293" Score="3" Body="&lt;p&gt;You might be interested in &lt;a href=&quot;http://help.solidworks.com/2016/English/SolidWorks/sldworks/c_Smart_Components.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Smart Components / Smart Features&lt;/a&gt;.  This technique will associate parts with features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the naming, we call those &quot;Press Nuts&quot;, or more generically &quot;Press Hardware&quot;&lt;/p&gt;&#xA;" OwnerUserId="4708" LastActivityDate="2016-11-28T19:14:34.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12552" PostTypeId="2" ParentId="12521" CreationDate="2016-11-28T19:37:41.973" Score="1" Body="&lt;p&gt;I can't tell what you want. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A combination of a part and forming tool&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I want to drag this part into a closed sheet metal area and the holes for the nut should be created automagically&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, you want the hole wizard? Go to the hole wizard, pick the size hole you want, then go to the &quot;Positions&quot; tab, click a face and you're able to drop holes of the spec you chose wherever you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You linked the entire PEM nut catalog for press nuts (as &lt;a href=&quot;https://engineering.stackexchange.com/a/12551/1633&quot;&gt;S Barry correctly called them&lt;/a&gt;), so I don't know which one you're talking about specifically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, as far as I know, all of those nuts have you drill one hole in the sheet. Specification varies by nut, but if you &lt;a href=&quot;http://catalog.pemnet.com/viewitems/floating-self-clinching-fasteners-a4-as-ac/floating-self-clinching-fasteners-a4-as-ac-unified&quot; rel=&quot;nofollow noreferrer&quot;&gt;look at any of the pages&lt;/a&gt; on the site you linked, for a specific nut, you can see the hole size and tolerance you need to specify to the sheet metal fabricator. It's that big column labeled, &quot;Hole Size in Sheet.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can specify a custom hole size and tolerance when you're in the hole wizard. If you want to re-use that hole wizard setting repeatedly without having to type the values in every time, then you could &lt;a href=&quot;http://blogs.solidworks.com/solidworksblog/2013/01/how-to-add-a-custom-hole-size-to-the-hole-wizard-database-in-solidworks.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;add the custom hole settings to the wizard&lt;/a&gt; to save those settings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this isn't the functionality you're looking for, or if you have a specific part you're referring to that couldn't use this method then please link to the specific part.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-11-28T19:37:41.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12553" PostTypeId="1" AcceptedAnswerId="12580" CreationDate="2016-11-28T23:29:59.473" Score="0" ViewCount="695" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uRKBT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uRKBT.png&quot; alt=&quot;image1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LIDbv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LIDbv.png&quot; alt=&quot;im2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the first photo , it's stated that the dy/dx at center = 0 , however , in the second photo (website) , it's stated that the dy/dx at the boundary = 0 ,which is correct ? I'm confused now&#xA;&lt;a href=&quot;http://www.geom.uiuc.edu/education/calc-init/static-beam/support.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.geom.uiuc.edu/education/calc-init/static-beam/support.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2016-11-29T23:55:16.993" Title="slope of deflection of beam at center" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12554" PostTypeId="1" CreationDate="2016-11-29T01:05:32.680" Score="4" ViewCount="2082" Body="&lt;p&gt;By flexible endoscope I mean colonoscope, gastroscope, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://youtu.be/nTIEXCE9BB4?t=56s&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;this video&lt;/strong&gt;&lt;/a&gt; I can see that an endoscope can swivel its tip while in place, kind of like a periscope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But while moving axially, i.e. &quot;telescoping&quot; (is that the right word?) or &quot;snaking its way in,&quot; or in other words, while being inserted, how does an endoscope turn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have to hit &lt;em&gt;something&lt;/em&gt; (e.g. the interior wall of the colon), and only due to the resistance of hitting that &lt;em&gt;something&lt;/em&gt; is it able to turn?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it designed in such a way that it can turn in midair, without having to come into contact with anything, while maintaining the turning point at the same point in space as it snakes further in, à-la the classic computer game &lt;a href=&quot;https://www.youtube.com/watch?v=GiOjqqWCsEQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;snake&lt;/strong&gt;&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="8997" LastEditorUserId="8997" LastEditDate="2016-11-29T05:40:31.317" LastActivityDate="2020-06-19T20:58:05.353" Title="How does a flexible endoscope turn?" Tags="|mechanical-engineering|biomedical-engineering|mechanisms|medical-devices|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12555" PostTypeId="2" ParentId="12554" CreationDate="2016-11-29T02:44:53.497" Score="3" Body="&lt;p&gt;The endoscope has pairs of cords running along the length, which are called &lt;em&gt;angulation wires&lt;/em&gt;.  The levers on the operator end pull the cords differentially, which makes the tip turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W5MNg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;( &lt;a href=&quot;http://fibertechmedical.com/support/how-it-works/how-it-works-the-angulation-system/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source; the article pretty much answers the original question.&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uBFlF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uBFlF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;( &lt;a href=&quot;http://clinicalgate.com/how-endoscopes-work/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source.&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/N5RlV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;( &lt;a href=&quot;http://www.gastrohep.com/ebooks/thumbnails.asp?book=1405120762&amp;amp;id=4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source.&lt;/a&gt; )&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2016-11-29T02:54:07.103" LastActivityDate="2016-11-29T02:54:07.103" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12556" PostTypeId="1" CreationDate="2016-11-29T03:24:33.870" Score="1" ViewCount="82" Body="&lt;p&gt;I have modulus of rupture (MOR) and maximum crushing strength of various species of wood. I am trying to understand how the thickness of wood a wood beam laid flat on the ground would affect both modulus of rupture and crushing strength. The practical application is this: We make &quot;crane mats&quot; which are essentially 6 pieces of long timbers bolted together. Each timber is 18 feet by 7.5&quot; x 7.5&quot;. The entire mat therefore is 18' x 7.5&quot; x 45&quot;. One species we use is eucalyptus grandis, and has a MOR and crushing strength of 118 and 60 MPa respectively. In contrast, red oak has an MOR and crushing strength of 75 and 42 respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my eucalyptus mat is, for example, 7.5&quot; thick, and it lies flat on the ground, is that equal in MOR or crushing strength to a 12&quot; thick mat of red oak?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be very appreciative for any tutoring. I have spent a full day trying to figure this out by myself but I'm not trained as an engineer. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="8998" LastActivityDate="2016-11-29T03:24:33.870" Title="how does thickness affect the maximum crushing strength and modulus of rupture of wood?" Tags="|wood|elastic-modulus|crushing|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12558" PostTypeId="1" AcceptedAnswerId="12561" CreationDate="2016-11-29T04:51:35.987" Score="2" ViewCount="17807" Body="&lt;p&gt;In Euler's bending theory, it's stated that the critical stress of a beam is always larger than the yield stress of the beam. I don't really undertstand the difference between them. Critical stress is defined as the stress that the maximum stress applied before the beam starts to buckle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the yield stress here mean the stress which the beam start to deform plastically? Why shouldn't the yield stress be smaller than the critical stress? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the object is subjected to stress, the beam will deform plastically before it starts to bend and break.  So I think that yield stress is smaller than the critical stress.&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="33" LastEditDate="2016-11-29T15:28:58.267" LastActivityDate="2016-12-05T07:32:28.073" Title="yield stress vs critical load stress" Tags="|structural-engineering|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12559" PostTypeId="2" ParentId="12550" CreationDate="2016-11-29T06:46:51.577" Score="2" Body="&lt;p&gt;&lt;strong&gt;Wing loading:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph below provides areas for living and man made flying machines - and this can be seen to be a quite good log-log fit of mass and wing loading. See his text for comment on the departure of Humming birds from the continuum mapped by most other birds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph suggests wing loadings of 30 to 80 kg/m&lt;sup&gt;2&lt;/sup&gt; as mass rises from 100 to 300 kg.&lt;br&gt;&#xA;Or about 3.3 m&lt;sup&gt;2&lt;/sup&gt; at 100 kg and 3.75 m&lt;sup&gt;2&lt;/sup&gt; at 300 kg.&lt;br&gt;&#xA;ie close to constant across your likely mass range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If correct, this answers your question regarding area. &#xA;The graph of wing loadings below is from the site referenced below.&lt;br&gt;&#xA;I have added the red lines and white text.&lt;br&gt;&#xA;Not that he works in Newton and I have roughly converted to kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aR7Nl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aR7Nl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Flapping rate - &quot;flat plate lift&quot;:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the drag equation, and miscellaneous assumptions (which I can explain if of enough interest) I estimate a required &quot;flapping rate&quot; of about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flaps per second ~~~~~&gt;= 4 x sqrt(Mass kg / Area m&lt;sup&gt;2&lt;/sup&gt;) / stroke&lt;sup&gt;2&lt;/sup&gt;&lt;br&gt;&#xA; = Fps = 4.sqrt(M/A)/stroke&lt;sup&gt;2&lt;/sup&gt;. &#xA;Stroke in meters.&lt;br&gt;&#xA;e.g. for 4 m&lt;sup&gt;2&lt;/sup&gt; wing area, 160 kg load, 1.0 metre stroke:&lt;br&gt;&#xA;Flap rate &gt;= 4 x sqrt(160/4)/1.0 = 25 flaps/second&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on the force on a flat plate when waved in air at sea level in one direction over a linear distance = 'stroke' art a rate of 'flap rate' per second with drag in one direction and mone in other, and high drive:return time ratio.&lt;br&gt;&#xA;(Using drag equation - Force ~~~= 0.6 x Rho x Cd x A x V&lt;sup&gt;2&lt;/sup&gt;. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;_______________________________________&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://asec.ir/%D8%A7%D9%84%DA%AF%D9%88%D8%B1%DB%8C%D8%AA%D9%85-%D8%B7%D8%B1%D8%A7%D8%AD%DB%8C-%D8%A8%D8%A7%D9%84-%D8%A8%D8%A7%D9%84%D8%B2%D9%86-10/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Ornithopter wing design algorithm:&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MATLAB code for the wing design is given &lt;a href=&quot;http://s7.picofile.com/d/8266613568/b2315634-d2a0-443f-9893-cbfe128c4de1/Code_Ornicopter_www_asec_ir_.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; on &lt;a href=&quot;http://s7.picofile.com/file/8266613568/Code_Ornicopter_www_asec_ir_.txt.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to 'potential navigation difficulties' due to language used I have appended the simple sample code at the end of this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;____________________________________&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://asec.ir/%D8%B3%D8%A7%D8%AE%D8%AA%D8%A7%D8%B1-%D8%A8%D8%A7%D9%84-%D9%BE%D8%B1%D9%86%D8%AF%D9%87-%D8%A8%D8%A7%D9%84%D8%B2%D9%86-2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;The structure of bird wings with applications to ornithopters&lt;/strong&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first animated GIF is of arguable usefulness but a great thought starter - &#xA;&lt;a href=&quot;http://cdn.instructables.com/FO7/186H/H9K2G6CN/FO7186HH9K2G6CN.LARGE.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wheeee!&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://asec.ir/%D8%A2%DB%8C%D8%B1%D9%88%D8%AF%DB%8C%D9%86%D8%A7%D9%85%DB%8C%DA%A9-%D8%A8%D8%A7%D9%84%D8%B2%D9%86-%D8%AF%D8%B1-%D8%AD%D8%A7%D9%84%D8%AA-%D8%A8%D8%A7%D9%84-%D8%AB%D8%A7%D8%A8%D8%AA-%D8%A8%D8%A7%D9%84/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Aerodynamic ornithopter fixed wing considerations&lt;/strong&gt;&lt;/a&gt; - all sorts of technical goodies. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Real world non-ornithopter but relevant man powered flying machine&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man powered flying machine in the videos below is not an ornithopter but maintains takeoff &amp;amp; flight lift based solely on &quot;man powered&quot; surfaces and the same basics can be applied. In the video example there are 8 unidirectionally driven lifting surfaces. An ornithopter could conceptually drive the same area at about the same mean forwards velocity but with powered downbeat and (probably) liftless upbeat to produce the same net effect. &#xA;For now, &lt;a href=&quot;http://bit.ly/2fHej8I&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;see this most lovely 1m30s video&lt;/strong&gt;&lt;/a&gt;&#xA;Also (slightly longer) &lt;a href=&quot;https://www.youtube.com/watch?v=syJq10EQkog&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=syJq10EQkog&lt;/a&gt;&lt;br&gt;&#xA;and &lt;a href=&quot;http://bit.ly/2fHej8I&quot; rel=&quot;nofollow noreferrer&quot;&gt;slightly longer 'making of'&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Some hard data and real models here.&lt;br&gt;&#xA;In Iranian (Persia)(Farsi?)- web translate works OK.&#xA;&lt;a href=&quot;http://asec.ir/%D8%B7%D8%B1%D8%A7%D8%AD%DB%8C-%D8%A2%DB%8C%D8%B1%D9%88%D8%AF%DB%8C%D9%86%D8%A7%D9%85%DB%8C%DA%A9%DB%8C-%D8%A8%D8%A7%D9%84-%D8%A8%D8%A7%D9%84%D8%B2%D9%86-9/0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aerodynamic design wing ornithopter &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3azbP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3azbP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://s6.picofile.com/file/8266611400/52.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://s6.picofile.com/file/8266611400/52.JPG&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His table of wing functions vs wingspan is liable to be of interest.&lt;br&gt;&#xA;This gives values for all birds for (as well as other hings) power, wingspan, wing area, power in terms of body mass &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Span = 1.17 x m&lt;sup&gt;0.39&lt;/sup&gt;&lt;br&gt;&#xA;Power = 10.9 x m&lt;sup&gt;0.19&lt;/sup&gt;&lt;br&gt;&#xA;Beat rate = 3.98 x m&lt;sup&gt;-0.33&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging ornithopter scale data into these suggests I'm missing something. TBD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://s6.picofile.com/file/8266611518/55.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://s6.picofile.com/file/8266611518/55.JPG&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UxNQM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UxNQM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MATLAB Ornithopter wing design code&lt;/strong&gt; &lt;a href=&quot;http://s7.picofile.com/file/8266613568/Code_Ornicopter_www_asec_ir_.txt.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;**from **&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pasted as is - tidyfy as desired or see source page.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;%% ---------------Ornithopter Wing design -------------------- %&lt;br&gt;&#xA;  --------------------- wwww.Asec.ir -------------------------  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;clc;clear;close all %% Fase ONE..... m=0.4; AR=3; W=m*10;&#xA;  W_S=30.6*W^(1/3) S=W/W_S span=sqrt(AR*S) a=span/2; b=S/pi/a&lt;br&gt;&#xA;  f=3.98*m^(-0.27)   x=0:a/20:a; y=b*sqrt(1-(x./a).^2); Xc=[0 a];Yc=[0&lt;br&gt;&#xA;  0]; X=[0 x];Y=[0 y]; fill(X,Y,'b','LineWidth',3);hold&lt;br&gt;&#xA;  on;plot(Xc,Yc,'r','LineWidth',3);axis equal text(a/2,3*a/4,'&lt;br&gt;&#xA;  \leftarrow MAC','FontSize',14)  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2020-02-29T14:53:28.250" LastActivityDate="2020-02-29T14:53:28.250" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="12560" PostTypeId="2" ParentId="12558" CreationDate="2016-11-29T09:33:03.867" Score="4" Body="&lt;p&gt;I think you mean Euler Buckling Theory right? Buckling is a different mode of failure to yielding. The plastic strength of a member will usually be constant as it depends on the material. However the &quot;buckling strength or Euler Strength&quot; of a member depends on its slenderness as well as the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below illustrates this very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h3yhK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h3yhK.jpg&quot; alt=&quot;Euler Strength and Buckling Strength&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1848" LastEditorUserId="1848" LastEditDate="2016-11-29T10:10:53.473" LastActivityDate="2016-11-29T10:10:53.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12561" PostTypeId="2" ParentId="12558" CreationDate="2016-11-29T10:51:07.357" Score="3" Body="&lt;p&gt;I was going to put this as a comment but it felt like more of an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buckling is a fairly unique failure mode.  When buckling, the compression force doesn't have to be as high as the yield strength.  If a beam is slender enough, the compressive force can cause substantial bending in the beam.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This bending goes beyond the elastic range and causes the beam to fail, even though the applied force usually wouldn't make it fail according to direct equations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest reason that critical stress is lower than yield stress is that it is an assumption for Euler buckling.  The equation might not work in the plastic range.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2016-11-29T10:51:07.357" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12562" PostTypeId="1" CreationDate="2016-11-29T11:40:32.350" Score="1" ViewCount="127" Body="&lt;p&gt;Recently I purchased a &lt;a href=&quot;http://www.tp-link.com/us/products/details/cat-5516_HS100.html#overview&quot; rel=&quot;nofollow noreferrer&quot;&gt;TP-Link HS100 smart plug&lt;/a&gt; which I configured using an app and all is working well. Now I am trying to understand the underlying software and hardware architecture for such implementation. Based on simple google searches following is my understanding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The headless smart plug makes a peer to peer connection to a device such as a mobile phone or computer. Using a device with a UI, network configurations are entered into the IOT device. The IOT device then connects to access point to make some type of authentication with a cloud service, and establishes a connection. The mobile app communicates the commands the cloud service which relays the command to the smart plug. This is pretty straight forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like to understand a little bit deeper in both software and hardware architecture for this type of design. It is fair say that there is some type of wifi device like &lt;a href=&quot;http://www.ti.com/product/CC3100&quot; rel=&quot;nofollow noreferrer&quot;&gt;CC3100&lt;/a&gt; or similar. There is some type of microcontroller like an ARM Cortex. It could be an integrated wifi controller such as a &lt;a href=&quot;http://www.ti.com/product/CC3200&quot; rel=&quot;nofollow noreferrer&quot;&gt;CC3200&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the underlying software and hardware architecture to implement similar microcontroller based design to setup and establish a secure communication link? What type of minimum security needs to implement to secure the device? I am hoping someone in this forum has experience in such an adaptation (2.4 GHz) and is willing to share their experience.&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2016-12-19T11:37:48.953" LastActivityDate="2016-12-19T11:37:48.953" Title="How to implement wifi for a headless IOT device?" Tags="|iot|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12563" PostTypeId="1" AcceptedAnswerId="12566" CreationDate="2016-11-29T14:28:20.993" Score="0" ViewCount="825" Body="&lt;p&gt;How can water be removed from a submarine that is submerged?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a submarine is opened, while it is submerged, water would get in because of the difference between the pressures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have seen that there is a diving room in submarines, these rooms are in direct contact with the sea but the level of the water remains steady.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain for me this phenomenon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If just the water level was steady, why don't we threw the water directly to the sea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MEWCe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MEWCe.jpg&quot; alt=&quot;explanation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9002" LastEditorUserId="90" LastEditDate="2016-11-30T02:52:59.833" LastActivityDate="2016-12-01T08:17:28.517" Title="How to get remove water from a submarine that is submerged" Tags="|fluid-mechanics|pressure|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="12564" PostTypeId="1" CreationDate="2016-11-29T15:34:09.263" Score="-2" ViewCount="81" Body="&lt;p&gt;How should I draw the semi cylinder in the sketch?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://postimg.org/image/lhl7x7t33/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://postimg.org/image/lhl7x7t33/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://postimg.org/image/6mwmj1ji7/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://postimg.org/image/6mwmj1ji7/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I don't know how to show that there is a semi cylinder in the horizontal view. (lower view)&lt;/p&gt;&#xA;" OwnerUserId="8980" LastEditorUserId="90" LastEditDate="2016-11-30T02:56:20.163" LastActivityDate="2016-11-30T02:56:20.163" Title="How should I draw the semi cylinder in the sketch?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ClosedDate="2016-12-15T14:21:27.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="12565" PostTypeId="2" ParentId="12408" CreationDate="2016-11-29T15:42:50.977" Score="3" Body="&lt;p&gt;&lt;strong&gt;Step 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to do is to determine the desired poles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural frequency can be computed using $\omega =\frac{4}{\zeta  T_s}$. This is a rule-of-thumb calculation for underdamped systems. The system here is slightly overdamped and is nonlinear as well. If the desired settling time is not obtained in the end, we have to come back and increase the constant 4 slightly. The design procedure is typically iterative. So we start with $\omega =9.80392$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the characteristic equation can be computed as $s^2+2 \zeta  s \omega +\omega ^2$, which after substituting values gives $s^2+20 s+96.1169$ and has roots $-11.9706$ and $-8.02944$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put the system in a linear from $$\dot{x}=\text{A}.x+\text{B}.v$$ where $$x=\left(&#xA;\begin{array}{cc}&#xA; x_1 &amp;amp; x_2 \\&#xA;\end{array}&#xA;\right)^T \ \ \ v=u+\frac{x_1^5}{6}$$&#xA;$$A=\left(&#xA;\begin{array}{cc}&#xA; 0 &amp;amp; 1 \\&#xA; -1 &amp;amp; -1 \\&#xA;\end{array}&#xA;\right) \ \ \ \ \ B=\left(&#xA;\begin{array}{c}&#xA; 0 \\&#xA; 1 \\&#xA;\end{array}&#xA;\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3&lt;/strong&gt;&#xA;Do the pole-placement design which gives $v=-k.x$ using Ackerman's formula.&#xA;$$k=\left(&#xA;\begin{array}{cc}&#xA; 0 &amp;amp; 1 \\&#xA;\end{array}&#xA;\right).&#xA;\left(&#xA;\begin{array}{cc}&#xA; \text{B} &amp;amp; \text{A}.\text{B} \\&#xA;\end{array}&#xA;\right)^{-1}.&#xA;(\text{A}^2+20 \text{A}+96.1169I)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting values, we get&#xA;$$k=\left(&#xA;\begin{array}{cc}&#xA; 95.1169 &amp;amp; 19 \\&#xA;\end{array}&#xA;\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the back transformation to get the value for $u$.&#xA;$$u+\frac{x_1^5}{6}=-95.1169 x_1-19 x_2$$&#xA;$$u=-95.1169 x_1-19 x_2-\frac{x_1^5}{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verification. We have to see if the design has met the requirements. (These simulations were done in Mathematica. The above calculations could also have been done there. I went through them manually above to explain things.) From the plot we see that the settling time constraint has been satisfied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y3MG7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y3MG7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-11-29T15:42:50.977" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12566" PostTypeId="2" ParentId="12563" CreationDate="2016-11-29T16:05:46.207" Score="1" Body="&lt;p&gt;No matter how you set it up, you'll need a two-door lock room.  For example, torpedos are placed in the firing tube in air, the rear hatch is locked, and the tube is flooded. Only then is the torpedo launched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divers can and do exit submarines which have a similar lock room for people. The same basic sequence is performed -- and reversed when the divers return.  Here's a great photo  &lt;a href=&quot;https://en.wikipedia.org/wiki/Airlock#/media/File:SEAL_Delivery_Vehicle_Team_(SDV)_2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Seal entry to sub&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-11-29T16:05:46.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12567" PostTypeId="2" ParentId="12564" CreationDate="2016-11-29T16:59:23.000" Score="0" Body="&lt;p&gt;The same way you would show it if it was a rectangle, dotted lines.  It's literally just a hidden feature, the view above it will show that it's a semicircle, this bottom view will show that extends all the way through the part.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2016-11-29T16:59:23.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12568" PostTypeId="2" ParentId="12564" CreationDate="2016-11-29T17:06:58.937" Score="2" Body="&lt;p&gt;The break out areas of the circle is what needs to be represented. It will be drawn with dotted line because it will be hidden in your view. No other call outs will need to be made. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your circle was off center from the break out points, and the side walls of the bore were intact, then you would draw the break out points AND the side walls, both in hidden dotted lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="9003" LastActivityDate="2016-11-29T17:06:58.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12569" PostTypeId="1" AcceptedAnswerId="12570" CreationDate="2016-11-29T17:43:59.080" Score="1" ViewCount="87" Body="&lt;p&gt;Consider $$\hat{G}(s) = \frac{1}{s^2+s}$$&#xA;than the Nyquist plot is&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gD1aj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gD1aj.jpg&quot; alt=&quot;nyquist plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the Bode plot is&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BYylp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BYylp.jpg&quot; alt=&quot;Bode Plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both plots it seems that the closed-loop system is stable even when the eigenvalues are {0, -1}. For a higher order closed-loop system one can see that a system like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{G}(s) = \frac{1}{s^3+s^2+s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is at its stability limit since ne Nquist plots hits the real axis at -1 and in the Bode plot there is no phase reserve when magnitude hits 0. Why dose the Nyquist and the Bode plot fail for a closed-loop system of 2nd order?&lt;/p&gt;&#xA;" OwnerUserId="7055" LastEditorUserId="7055" LastEditDate="2016-11-29T19:01:12.273" LastActivityDate="2016-11-29T19:01:12.273" Title="Why can’t one see for 2nd order system that it is at its stability limit neither in the Nyquist plot nor Bode plot?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12570" PostTypeId="2" ParentId="12569" CreationDate="2016-11-29T18:10:18.533" Score="3" Body="&lt;p&gt;I think you are mixing up the closed- and open-loop systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed-loop system is $$\frac{\frac{1}{s^2+s}}{1+\frac{1}{s^2+s}}=\frac{1}{s^2+s+1}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has poles at $-0.5\pm 0.866025 i$ which is stable. &lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-11-29T18:10:18.533" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12574" PostTypeId="2" ParentId="12542" CreationDate="2016-11-29T19:17:21.200" Score="1" Body="&lt;p&gt;Just look into the help pages (&lt;code&gt;C:\Program Files\ANSYS Inc\v172\commonfiles\help\en-us\help\wb_sim\ds_Contact_Definition.html&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A contact object in ANSYS consists of a contact and a target object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lDQpC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lDQpC.png&quot; alt=&quot;pic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a contact region is defined in Workbench two contact objects are created (symmetric contact), one as seen above, and the other with the contact/target geometries flipped. This is default (program controlled).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only when a flexible body is in contact with a rigid body the above behavior does not apply and an asymmetric behavior is needed. In this case there is only one target and one contact surface (I think target is on rigid body only).&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2016-11-29T19:17:21.200" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12576" PostTypeId="1" CreationDate="2016-11-29T20:01:55.673" Score="3" ViewCount="16327" Body="&lt;p&gt;A &quot;Pulley&quot; and a &quot;Wheel and axle&quot; seem much similar to each other. What is their actual difference and what are their precise definitions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of the two terms is much relied on the context of &lt;a href=&quot;https://en.wikipedia.org/wiki/Simple_machine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Simple machines&lt;/a&gt;. The wikipedia articles of the two terms provide a definition which makes them seem very similar.  But since simple machines are considered (by ancient greeks) as six distinct different basic machines then they must have an important difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure, but maybe modern treament of mechanical machines in terms of &quot;Linkage synthesis&quot; could give a better explanation. If it's possible, please explain the differences and the definitions both in the context of &quot;Simple machines&quot; and &quot;Linkage synthesis&quot; methods.&lt;/p&gt;&#xA;" OwnerUserId="8968" LastEditorDisplayName="user16" LastEditDate="2016-11-30T12:34:19.953" LastActivityDate="2021-12-24T13:35:29.553" Title="Difference between a &quot;Pulley&quot; and a &quot;Wheel and axle&quot;" Tags="|mechanical-engineering|wheels|pulleys|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12579" PostTypeId="2" ParentId="12558" CreationDate="2016-11-29T23:45:12.217" Score="2" Body="&lt;p&gt;Your statement &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In Euler's bending theory, it's stated that the critical stress of a&#xA;  beam is always larger than the yield stress of the beam&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;seems just wrong, or you misunderstood the source you got it from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euler buckling has nothing to do with the yield stress. The theory assumes the material behavior is linear elastic, at least up to the load at which the bucking failure &lt;em&gt;starts&lt;/em&gt;. (Euler's theory of buckling says nothing about what happens &lt;em&gt;after&lt;/em&gt; it has started).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily demonstrate that beam buckling can be purely elastic. Get a steel engineer's ruler (the longer the better, preferably 0.5 or 1 meter), or a thin wooden dowel or a sheet of thin plywood at least a meter long. Put one end on a flat surface and apply a load to the other end with your hands until it starts to buckle (i.e. bow outwards). Remove the load, and it will return to its initial straight configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if you continue to apply the load &lt;em&gt;after&lt;/em&gt; it starts to buckle, it will continue to bow outwards until the material yields or fractures, but that &lt;em&gt;post-buckling&lt;/em&gt; behaviour is not considered by Euler's theory.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-11-29T23:45:12.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12580" PostTypeId="2" ParentId="12553" CreationDate="2016-11-29T23:55:16.993" Score="2" Body="&lt;p&gt;If $\theta$ is supposed to be the same quantity as $dy/dx$, the picture is both drawn and labelled wrongly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations correctly state that at $x=2$, $[dy/dx]_\text{AC} = [dy/dx]_\text{BC}$, but that does not mean $[dy/dx]_\text{AC} = [dy/dx]_\text{BC} = 0$!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, $\theta$ might be some other quantity that isn't defined in your post - in which case we can't guess what it means.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-11-29T23:55:16.993" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12581" PostTypeId="1" CreationDate="2016-11-30T00:19:04.757" Score="0" ViewCount="55" Body="&lt;p&gt;I am looking for some guidance on calculations for a natural condenser project. I am a Computer Engineering student, so this stuff is out of my range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental question I am trying to answer is: &lt;strong&gt;what fan speed is required to push air through the following system to overcome head loss?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic mockup of the system is shown below. Some additional details of the design:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This will be installed in a greenhouse to reduce humidity and ideally collect some water for reuse.&lt;/li&gt;&#xA;&lt;li&gt;The long, narrow rectangles represent PVC pipes of diameter 20cm (0.2m). Both pipes are identical, aside from length.&lt;/li&gt;&#xA;&lt;li&gt;The storage tank on the bottom will collect any condensed water.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/agydm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/agydm.png&quot; alt=&quot;Condenser Schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Functional procedure of the design:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Humid air (90% relative humidity, 30 deg C) will enter the pipe at stage 1. A fan will be installed to move air down.&lt;/li&gt;&#xA;&lt;li&gt;The air will travel 2m down into the ground (denoted by the rectangle) to a storage tank at (12.7 deg C). The humid air will condense into liquid in the cooler temperature and collect in the storage tank.&lt;/li&gt;&#xA;&lt;li&gt;The air will travel 2.5m up and return into the greenhouse (unknown humidity and temperature). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In our small-scale testing, we were able to achieve condensation using a computer fan and 1.5&quot; diameter PVC pipe. However, we'd like to calculate the required fan speed for the full scale prototype. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We attempted to calculate head loss using equations in MATLAB, but weren't sure what the pressures would be at each stage. I understand this is a lot of info, but any guidance would be much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="2077" LastActivityDate="2016-12-01T23:01:28.647" Title="Force and Headloss Calculations for Three Stage Condenser" Tags="|thermodynamics|pressure|piping|vapor-pressure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12582" PostTypeId="1" CreationDate="2016-11-30T06:46:28.097" Score="3" ViewCount="388" Body="&lt;p&gt;If I have defined the size tolerance of a shaft, how is the maximum cylindricity tolerance automatically specified by rule-1 of GD&amp;amp;T?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean when a shaft shoulder diameter is defined (e.g. 33g6) then it is usually checked by measuring the diameters at multiple axial locations. What is the guarantee that each of the measured sections at various axial locations will lie between the coaxial cylinders as required by the cylindricity tolerance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="994" LastEditorUserId="282" LastEditDate="2016-12-01T06:14:55.523" LastActivityDate="2018-05-22T00:56:15.587" Title="How does Rule-1 guarantee cylindriccity in GD&amp;T?" Tags="|mechanical-engineering|metrology|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12583" PostTypeId="2" ParentId="12558" CreationDate="2016-11-30T09:50:53.760" Score="0" Body="&lt;p&gt;Buckling and yielding are completely different concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Buckling&lt;/em&gt; occurs when an elastic structure does not move back to its initial position as long as the load is applied. Buckling is a purely elastic phenomenon. When elastic material is subjected to a small load, the deformation is usually small as well. This is different with buckling. Buckling requires a significant load in one direction, but then a very, very small load in a perpendicular direction can lead to destructive deformation (depending on the amount of deformation allowed by the system). Buckling is a stability problem, and the sample geometry is essential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Yielding&lt;/em&gt; occurs when the behavior of the material itself changes (due to the high load). When a material yields, the relative position of the atoms change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; To realise the 'very, very small load' in perpendicular direction it is usually sufficient to apply the large load not perfectly in line with the stability axis. Since buckling is a stability problem, the large load leads to an instable system in the first place. In order to actually deform, some initial deformation in perpendicular direction is required which is then amplified. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also slightly changed the first sentence of the buckling paragraph to account for Wasabi's comment.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="4122" LastEditDate="2016-12-05T07:32:28.073" LastActivityDate="2016-12-05T07:32:28.073" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12588" PostTypeId="2" ParentId="12323" CreationDate="2016-11-30T21:44:05.297" Score="1" Body="&lt;p&gt;Most home gasoline-engine generators have a &quot;2-pole&quot; rotor which is spun at 3600 RPM by a governor on the engine.  This results in a pair of 60 Hz sine-wave volatges coming out of the stator winding...180 degrees out-of-phase with each other.  The losses in a typical gasoline engine at 3600 RPM are quite significant.  If you happen to have an automobile with a tachometer, you might note that the RPM rarely goes over 3600.  If the electrical load on the generator increases, the rotor current is increased to maintain 120 volts (normal line voltage).  That puts a higher load on the gasoline engine and the governor opens the throttle to maintain speed.  Very simple and inexpensive.&lt;/p&gt;&#xA;" OwnerUserId="8748" LastActivityDate="2016-11-30T21:44:05.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12590" PostTypeId="1" CreationDate="2016-12-01T04:41:46.030" Score="1" ViewCount="21" Body="&lt;p&gt;Which will require more force (kgf)? small or big scissors?&lt;/p&gt;&#xA;" OwnerUserId="9026" LastActivityDate="2016-12-01T04:41:46.030" Title="Which will require more force? small or big scissors?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="2" ClosedDate="2016-12-01T21:22:23.910" ContentLicense="CC BY-SA 3.0" />
  <row Id="12592" PostTypeId="1" AcceptedAnswerId="12593" CreationDate="2016-12-01T05:06:32.543" Score="0" ViewCount="46" Body="&lt;p&gt;I am an engineering student. I am working on a project where we need to use something called a vertical pipeline crawler. It is just a device that travels inside pipelines for inspection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a crawler online, and I wanted to calculate the amount of power it uses. In specifications (there is more specs than below):&#xA;Speeds up to 10 m/min&#xA;Mass: 10 kg&#xA;Maximum pull: 27 kg&#xA;Power: 600 W 115 / 230 VAC&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially what I did was to multiply the weight it can carry by the speed:&#xA;m&lt;em&gt;g&lt;/em&gt;v = 27*9.81*(10/60) = 44.145 Watts&#xA;But it also says 600 W for power in specs. So how much power does this device need? Is the input 600 and output 44.145, isn't that very inefficient for an electrical device? Is my thought process wrong, what should use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I didn't want to share the link for the product thinking it might be against the website rules, however its pretty easy to find online)&lt;/p&gt;&#xA;" OwnerUserId="9027" LastActivityDate="2016-12-01T08:07:32.070" Title="Calculation of power for a pipeline crawler" Tags="|mechanical-engineering|motors|power|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12593" PostTypeId="2" ParentId="12592" CreationDate="2016-12-01T08:07:32.070" Score="0" Body="&lt;p&gt;Your question &quot;power does this device need?&quot; is unclear. If you are concerned about how much electrical power it requires, thats 600W and the current it requires will be given by dividing thus with the line voltage. The diameter of the electrical power cable will be determined by this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you multiply speed with the downward force you are neglecting how much energy the device has to expend in operating the gripping mechanism. In addition there is the electrical power requirement for the inspection head. You need to consider all of these.&lt;/p&gt;&#xA;" OwnerUserId="994" LastActivityDate="2016-12-01T08:07:32.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12595" PostTypeId="2" ParentId="12563" CreationDate="2016-12-01T08:17:28.517" Score="0" Body="&lt;p&gt;The picture that you have drawn is not of submarine but a diving bell (check up wikipedia). In a diving bell the pressure inside the chamber balances the hydrostatic pressure of water and hence the water level inside the diving bell does not rise. You have see this if you push a glass tumbler into a tub. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a true submarine, the water in the balast tanks is vented by pushing in compressed air from reservoirs. However this is not the only way submarines can control their depth.&lt;/p&gt;&#xA;" OwnerUserId="994" LastActivityDate="2016-12-01T08:17:28.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12598" PostTypeId="2" ParentId="12576" CreationDate="2016-12-01T10:59:54.507" Score="1" Body="&lt;p&gt;The construction of a &quot;pulley&quot; and a &quot;wheel and axle&quot; can be similar or even same. But in the simple machines classification their purposes differ:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pulley&lt;/strong&gt; is used to change the direction of a rope or a chain. The actual rotation of the pulley wheel is a side effect.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Wheel and axle&lt;/strong&gt; are used to amplify rotational force. The force can be provided through a rope wrapped around the wheel, but the change of direction is a side effect.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example, a chain sprocket is an example of wheel and axle, while a chain idler would be an example of a pulley.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-12-01T10:59:54.507" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12599" PostTypeId="1" CreationDate="2016-12-01T11:23:45.993" Score="0" ViewCount="426" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5D2eL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5D2eL.png&quot; alt=&quot;1&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rguGy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rguGy.jpg&quot; alt=&quot;22&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the solution , it's clear that the top part of the beam undergo tension , while the lower part undergo compression .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But , i dont understand why it is so . When the moment is applied in the horizontal axis , the structure will slide become like this , right ? ( the top part will be slided into the book ) &#xA;I cant imagine how the moment applied causing the top part to experience tension , while the lower part experience compression .&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2016-12-01T12:37:16.397" Title="moment applied causing bending stress in beam" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12600" PostTypeId="2" ParentId="12599" CreationDate="2016-12-01T12:37:16.397" Score="2" Body="&lt;p&gt;You are correct that the given bending moment will create compression on top and tension in the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is where you see the book state that the tension is on top. If you look at the calculation of the centroid $\overline y$, you'll see it is 3.4&quot; from the top fiber. If you then look at the calculation of the stresses, you'll see that the $c$ value used in $\sigma_t$ is $(10.5-3.40)$. It is therefore equal to the distance from the centroid to the bottom fiber, meaning $\sigma_t$ is on the &lt;strong&gt;&lt;em&gt;bottom&lt;/em&gt;&lt;/strong&gt; fiber, as would be expected. For $\sigma_c$, $c = 3.40$, which means $\sigma_c$ is found on the &lt;strong&gt;&lt;em&gt;top&lt;/em&gt;&lt;/strong&gt; fiber.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-12-01T12:37:16.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12601" PostTypeId="1" CreationDate="2016-12-01T13:41:53.950" Score="0" ViewCount="353" Body="&lt;p&gt;In the diagram bottom of the image uploaded , it's clear that width of the section after and before the cut are the same , right , which is 25x3=75mm , why the author stated that the width of section after cut become 25mm ? I cant understand &lt;a href=&quot;https://i.stack.imgur.com/aAxKH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aAxKH.png&quot; alt=&quot;iamge1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2016-12-01T14:13:29.963" Title="shear stress before and after cut" Tags="|structural-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12602" PostTypeId="1" CreationDate="2016-12-01T13:49:38.270" Score="4" ViewCount="170" Body="&lt;p&gt;The definition of passivity for control systems is:&#xA;$\langle y,u \rangle _T \geq 0 \: \: \: \forall T &amp;gt;0 \: \: \: \forall u$.&#xA;Considering  signals $u,y :[0,T] \rightarrow R^m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, passivity is when the system is still stable and does not change the direction and lose stability. If we look at the vectors I understand that in the scalar multiplication $\cos\theta \geq 0$. Does that mean that the direction or phase doesn't change from positive to negative and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another definition using the storage function is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(x(T)) \leq \int_0^Ty(t)u(t)dt + V(x(0)), \: \: \forall T &amp;gt; 0 $$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Passivity idea: “Increase in stored energy ≤ Added energy”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I describe the passivity for a system in words without the storage function? I mean, what is passivity and why is it useful? In what way is it different from stability?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8884" LastEditorUserId="1832" LastEditDate="2016-12-01T14:05:29.023" LastActivityDate="2016-12-01T20:51:10.250" Title="Passivity in automatic control (Linear and nonlinear control)" Tags="|stability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12603" PostTypeId="2" ParentId="12601" CreationDate="2016-12-01T14:13:29.963" Score="3" Body="&lt;p&gt;As @JMac pointed out in a comment, the cut in the beam causes the following cross-section to occur (exaggerating the size of the cut):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/At9rH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/At9rH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally, the shear stress can flow from the top to the bottom along the three webs. Therefore, the effective width to calculate the shear stress was equal to the sum of the three webs. After the cut, however, the shear can only flow through the central web. This therefore triples the maximum shear stress.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-12-01T14:13:29.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12604" PostTypeId="1" AcceptedAnswerId="12607" CreationDate="2016-12-01T15:01:45.447" Score="2" ViewCount="361" Body="&lt;p&gt;I am a researcher on robotics and AI, and wonder if traditional manufacturing machines, such as lathe, CNC, or grind use any kind of machine learning technique or algorithm. This algos are widely used in mobile robots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e.: in a plant where there are some lathes, do they interact somehow between them? Do they adjust on the fly their parameters when machining a new piece? Or do lathes try to adjust to the parameters to the lathe that is performing the best? Kind of collaborative learning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not find this kind of examples in industry yet, but I wonder if industry 4.0 is about implementing this kind of techniques.&lt;/p&gt;&#xA;" OwnerUserId="5553" LastActivityDate="2020-06-16T05:25:16.063" Title="Artificial Intelligence in manufacturing machines" Tags="|control-engineering|manufacturing-engineering|robotics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12607" PostTypeId="2" ParentId="12604" CreationDate="2016-12-01T20:24:10.193" Score="3" Body="&lt;p&gt;In general terms what you really want from CNC machines is to repeat the same actions exactly the same every time, while you may need to monitor and account for things like tool wear, temperature changes, small variations in material properties  etc this is still in the realms of fairly straightforward feedback control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately lathes aren't the best example as they really only only really have 3 degrees of freedom, cutting speed, tool feed and tool travel. A better example would be 5 or 6 axis milling which is inherently a lot more complex. Here there may well be scope for AI applications, particularly in designing and developing tool path design. This is the process of specifying the sequence of cutting moves to machine a part as quickly and efficiently as possible to a specific quality standard and may require numerous tool changes as well as avoiding collisions, nesting parts, deciding on which order to perform operations, optimising part/billet setup and orientation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular there is a lot to be gained in the paths used for roughing cuts and finding the optimum tool path to move material quickly and effectively for example by taking a smoothed path around sharp corners rather than just using a fixed offset from the final geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Automated tool path design has been around for a while but it is certainly conceivable that AI could have a role to play and it is certainly an important field. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-01T20:24:10.193" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12608" PostTypeId="1" AcceptedAnswerId="12695" CreationDate="2016-12-01T20:26:14.687" Score="1" ViewCount="93" Body="&lt;p&gt;What is the difference between impulse, step and sum of sinusoids signals in terms of their spectra and their degrees of excitation ?&lt;/p&gt;&#xA;" OwnerUserId="9038" LastActivityDate="2016-12-06T00:33:55.710" Title="Control System Question - spectra and their degrees of excitation" Tags="|control-engineering|control-theory|systems-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12609" PostTypeId="2" ParentId="3863" CreationDate="2016-12-01T20:37:21.527" Score="0" Body="&lt;p&gt;The advent of machine tools more or less coincides with the adoption of high speed, high accuracy, high power gears as means of power transmission. For obvious reasons if you don't have powered machines you have limited need for high quality gears. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in another answer the gears for clocks and watches were originally made by hand with great care and skill using saws and files. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earliest gears were &lt;a href=&quot;http://www.instructables.com/id/Basic-Gear-Mechanisms/step14/Cage-and-Peg-Gears/&quot; rel=&quot;nofollow noreferrer&quot;&gt;peg and cage gears&lt;/a&gt; which could be made from wood and were used in watermills etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a specific process for gear cutting called &lt;a href=&quot;https://en.wikipedia.org/wiki/Hobbing&quot; rel=&quot;nofollow noreferrer&quot;&gt;hobbing&lt;/a&gt; which allows accurate gears to be made on fairly simple machine tools. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gears can also be cast from a pattern, requiring only moderate finishing to an acceptable surface finish and accuracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also be surprised what level of accuracy a skilled craftsman can achieve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed CNC technology has tended to improve overall productivity but doesn't do much to improve accuracy and there were automated production processes several centuries before computer controlled machining was a thing. &lt;a href=&quot;https://en.wikipedia.org/wiki/Jacquard_loom&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Jacquard_loom&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-03-29T20:54:21.430" LastActivityDate="2018-03-29T20:54:21.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12610" PostTypeId="2" ParentId="12602" CreationDate="2016-12-01T20:51:10.250" Score="4" Body="&lt;p&gt;A short answer for each of your questions in order:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;How do I describe passivity without the storage function?&lt;/strong&gt; Passivity analysis is based around the idea of energy flowing into and out of the system. When your system is passive it means it is not generating energy, only storing it or 'passing it through'.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Why is passivity useful?&lt;/strong&gt; When dealing with nonlinear systems it can be difficult if not impossible to prove that they are stable. It is much easier to prove that they are passive, and to design controllers that ensure passivity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;In what way is passivity different from stability?&lt;/strong&gt; It is important to understand that passivity is a sufficient condition for stability, in that a system that is passive is also stable. However, not all stable systems are passive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2016-12-01T20:51:10.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12612" PostTypeId="2" ParentId="12581" CreationDate="2016-12-01T23:01:28.647" Score="1" Body="&lt;p&gt;Not an expert with fluids but maybe this will help narrow down stuff-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Fan rpm should not be the first priority.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You should rather calculate the total pressure drop across your path (straight pipe, rectangular box which is approximately a U-bend and the return straight pipe). The fan you select should provide a static pressure which is greater than this total pressure drop. For example if you calculate the pressure drop for your air path to be 2 inch of water column (one of the common units for fan static pressure measurement also written as I.W.C), then you need to select a fan that is capable of providing its rated volume of air at 2 I.W.C or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Next, you need to find out how much airflow you need. This will be limited by the heat transfer happening in the rectangular box. Let's assume some values. If your box is transferring heat to the ground at 2.5 Btu/second (Btu is a unit for measuring heat energy), then you can move a maximum of 1 pound of air through the box in 1 second, so that it's temperature will drop from 30 degree celcius to 20 degree Celsius (these values are calculated using properties of air which are easily available). So now we know that we can move 1 lb of air per second, which will give us our airflow rate for the fan. This is measured in cubic foot per minute or cfm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) To summarize, your process will be-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Find out pressure drop in the air path and select fan static pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Find out maximum possible heat transfer in box and calculate maximum possible air flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Select a fan which will give you the required static pressure and airflow (you will have to select the fan from an available list and may not have a choice for rpm)&lt;/p&gt;&#xA;" OwnerUserId="8934" LastActivityDate="2016-12-01T23:01:28.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12614" PostTypeId="1" CreationDate="2016-12-02T05:57:41.640" Score="3" ViewCount="277" Body="&lt;p&gt;Breaking down water into H&lt;sub&gt;2&lt;/sub&gt; and O&lt;sub&gt;2&lt;/sub&gt; through electrolysis takes &lt;a href=&quot;https://chemistry.stackexchange.com/a/5859&quot;&gt;241.8 Kj/mol&lt;/a&gt;. &lt;a href=&quot;http://health.howstuffworks.com/human-body/systems/respiratory/question98.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to howstuffworks.com, the average person uses 550L of pure oxygen per day&lt;/a&gt;, which equates to about 24.6 moles of pure O&lt;sub&gt;2&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new Tesla car batteries are 85 kWh, which (in a perfect world) would be enough energy to extract about 1265 moles of O&lt;sub&gt;2&lt;/sub&gt; from water through electrolysis, enough oxygen to breathe for 51 days! Even a system at 1% efficiency would be able to provide enough oxygen to breathe for 12 hours. Current scuba tanks only last a fraction of that (using bottled air).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do scuba divers still use bottled air rather than generating their oxygen through electrolysis, in a world where batteries have enough capacity to generate plenty of oxygen through electrolysis?&lt;/p&gt;&#xA;" OwnerUserId="8974" LastEditorUserId="-1" LastEditDate="2017-04-13T12:57:06.907" LastActivityDate="2016-12-08T22:37:51.723" Title="Why do we still scuba dive with bottled air?" Tags="|mechanical-engineering|chemical-engineering|" AnswerCount="2" CommentCount="5" ClosedDate="2016-12-08T22:44:15.813" ContentLicense="CC BY-SA 3.0" />
  <row Id="12618" PostTypeId="5" CreationDate="2016-12-02T16:33:37.787" Score="0" Body="&lt;p&gt;Laser, which stands for &lt;em&gt;light amplification by stimulated emission of radiation&lt;/em&gt;, is a device that typically produces a well-defined narrow beam of light. Usually it is monochromatic, or single-colour, i.e. its radiation encompasses a very narrow spectral range. Lasers can be ultraviolet, visible and infrared, as well as continuous or pulsed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most common lasers are based in a semiconductor diode, but there are also gas-, liquid-, fibre- and solid-state lasers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lasers are used in a vast variety of applications nowadays, including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;telecommunication&lt;/li&gt;&#xA;&lt;li&gt;surgery&lt;/li&gt;&#xA;&lt;li&gt;remote sensing&lt;/li&gt;&#xA;&lt;li&gt;measuring distances &lt;/li&gt;&#xA;&lt;li&gt;data storage&lt;/li&gt;&#xA;&lt;li&gt;entertainment&lt;/li&gt;&#xA;&lt;li&gt;metal welding and cutting&lt;/li&gt;&#xA;&lt;li&gt;research&lt;/li&gt;&#xA;&lt;li&gt;security&lt;/li&gt;&#xA;&lt;li&gt;laser pointers&lt;/li&gt;&#xA;&lt;li&gt;and many more&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="315" LastEditorUserId="315" LastEditDate="2016-12-03T08:39:01.710" LastActivityDate="2016-12-03T08:39:01.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12619" PostTypeId="4" CreationDate="2016-12-02T16:33:37.787" Score="0" Body="Questions related to lasers, laser-based systems or devices, and their applications. " OwnerUserId="315" LastEditorUserId="315" LastEditDate="2016-12-02T20:16:43.480" LastActivityDate="2016-12-02T20:16:43.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12620" PostTypeId="5" CreationDate="2016-12-02T16:37:19.977" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-12-02T16:37:19.977" LastActivityDate="2016-12-02T16:37:19.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12621" PostTypeId="4" CreationDate="2016-12-02T16:37:19.977" Score="0" Body="Computer-Aided Design is the use of computers in the engineering design process. It involves drafting, modelling, simulation, optimisation, and creation of technical documentation. The term CAD is often used to denote software for computer-aided design." OwnerUserId="315" LastEditorUserId="315" LastEditDate="2016-12-02T20:17:05.120" LastActivityDate="2016-12-02T20:17:05.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12622" PostTypeId="1" CreationDate="2016-12-02T16:41:59.810" Score="1" ViewCount="898" Body="&lt;p&gt;My question is about differences between reservoir-hydropower and run-of-river power plants. Are run-of-river plants less efficient? what would be an approximate guess of the differences? (assuming they are placed in the same catchment). Would annual capacity factors (% of time in operation) differ? by how much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/6444/what-theoretical-efficiencies-can-wind-hydro-turbines-achieve&quot;&gt;This answer&lt;/a&gt; makes me think that reservoir hydropower would be more efficient because it can capture a larger % of the kinetic energy than wind power. Arguably the run-of-river plants would be in a similar situation, because they are less able to slow down the water flow in comparison with reservoir hydropower. Obviously water and air are very different fluids and I can't just translate the efficiencies from windpower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding capacity factors, in principle I have no reason to think ROR has lower capacity factors. Water throughput would depend on the catchment area and I guess both have similar maintenance requirements. However, all these are intuitions, I have not found real data to corroborate my hypothesis.&lt;/p&gt;&#xA;" OwnerUserId="9049" LastEditorUserId="9049" LastEditDate="2016-12-03T13:41:18.417" LastActivityDate="2017-01-02T16:21:02.183" Title="What is the difference (if any) in capacity factors and efficiency between reservoir hydropower and run-of-river power plants?" Tags="|thermodynamics|hydraulics|power|energy-efficiency|hydrology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12623" PostTypeId="2" ParentId="12622" CreationDate="2016-12-02T16:56:24.227" Score="1" Body="&lt;p&gt;Capacity factors over a plant's lifetime are largely a design decision. There is no standard one-size-fits-all answer. The situation in any one particular case will be determined by the local climate and the design decisions made that reflect the local economics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no &quot;same local conditions&quot;, it's not a meaningful question really. If there's a reservoir, one builds reservoir storage. But all other things being equal, they'd yield the same energy. (NB that's not the same as capacity factor or efficiency.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you've misunderstood the other answer. Hydro can be more efficient than wind. I didn't say that storage hydro was more efficient than run-of-river.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, efficiency and capacity factor are completely different things. Don't confuse the two.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2016-12-03T15:36:31.710" LastActivityDate="2016-12-03T15:36:31.710" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12624" PostTypeId="1" CreationDate="2016-12-02T19:42:43.260" Score="0" ViewCount="174" Body="&lt;p&gt;We take a cylindrical concrete sample and apply compression load on it, perpendicular to its head. To measure its strain we use 2 Lvdt(strain measurement devices)on both sides of the concrete sample . Surprisingly they give us different numbers (close but not the same). Why is that?&#xA;Firstly I think that it is because concrete sample might not be fully homogeneous (in preparation the mixture process might not have been perfectly mixed. The second reason is that it might have some bulk defects inside.&#xA;Let me know what you think   &lt;/p&gt;&#xA;" OwnerUserId="9053" LastActivityDate="2017-01-31T14:26:37.080" Title="Why do lvdt sensors give us different results?" Tags="|civil-engineering|concrete|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12629" PostTypeId="2" ParentId="203" CreationDate="2016-12-03T07:11:58.427" Score="1" Body="&lt;p&gt;I think for that I'd probably try using MasterBond EP21TCHT-1 for what your looking for, It has excellent performance on difficult to bond materials and is also excellent at cryogenic temperatures from minus 450 degrees F (4 degrees K) right up to +400 degrees F. The surfaces need to be spotless though, absolutely grease free and slightly roughened for best adhesion. &lt;/p&gt;&#xA;" OwnerUserId="8890" LastActivityDate="2016-12-03T07:11:58.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12630" PostTypeId="1" AcceptedAnswerId="12643" CreationDate="2016-12-03T14:41:24.447" Score="1" ViewCount="551" Body="&lt;p&gt;Hello I have a problem determining the value of K in this block diagram, to calculate this value, I applied this method, but the value at the end of K I is different from 216, you could explain to me where I'm wrong&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rRkmf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rRkmf.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/edZs5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/edZs5.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9061" LastEditorUserId="29410" LastEditDate="2021-02-28T10:14:35.293" LastActivityDate="2021-02-28T10:14:35.293" Title="Automation: steady state error calculation block diagram" Tags="|mechanical-engineering|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12631" PostTypeId="1" CreationDate="2016-12-03T16:12:15.917" Score="1" ViewCount="711" Body="&lt;p&gt;In statics we consider the force deflection constraints (i.e. assuming the structure to be flexible) to generate equations in order to solve the extra degrees of indeterminacy. So, basically if we consider a rigid body that does not produce any deflection how do we get the reactions at the supports? Can they be determined? If not, basically what does it mean from an intuitive perspective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lZf2g.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lZf2g.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, now all the 3 reactions are due to a pin support. This way, in statics we equate the deflection caused by R1 to the deflection caused by the pair of 100 pound loads. This gives us R1. But in the process we are assuming the structure to be elastic. If the body is rigid, in this case can R1, R2 and R3 assume any values as long as the body is in equilibrium? But how is it possible?&lt;/p&gt;&#xA;" OwnerUserId="4569" LastActivityDate="2016-12-04T02:33:14.433" Title="How do we solve a rigid body indeterminate structure?" Tags="|civil-engineering|statics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12632" PostTypeId="1" CreationDate="2016-12-03T16:22:31.977" Score="-2" ViewCount="5554" Body="&lt;p&gt;I am working the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The simply supported beam shown in Fig. 12-12a is subjected to the&#xA;  concentrated force &lt;strong&gt;P&lt;/strong&gt;. Determine the maximum deflection of the&#xA;  beam. &lt;em&gt;EI&lt;/em&gt; is constant.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NBZ8S.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NBZ8S.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The text later says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;By inspection of the elastic curve, Fig. 12-12b, the maximum deflection&#xA;  occurs at &lt;em&gt;D&lt;/em&gt;, somewhere within region &lt;em&gt;AB&lt;/em&gt;. Here the slope must be&#xA;  zero.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why does the author say that, by inspection, the maximum deflection occurs at &lt;em&gt;D&lt;/em&gt;? How do we know that? He didn't show any work or explain why it should be at &lt;em&gt;D&lt;/em&gt; and not somewhere else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I consider the maximum slope to occur at a point 2 m from &lt;em&gt;A&lt;/em&gt;, then I will take EI(dv2/dx2) = 0, then my answer is, -2((x2)^2) + 12x2 -44/3 = 0, then x = 4.29 m .... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;here's the full question . &#xA;in hree , we can notice that there are 2 sets of slope equation that we can use . Which is equation 5 &amp;amp; 7 . &#xA;In equation 5 , we will get 1.633 as in the working . ( the author use by 'inspection' the max deflection occur at region AB) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However , when as @Jmac stated , we dont know where is the position where the max deflection is located , how can we use equation 5 to solve ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;why We dont have to consider equation , which involve region DC ??? &lt;a href=&quot;https://i.stack.imgur.com/TpUak.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TpUak.png&quot; alt=&quot;22&quot;&gt;&lt;/a&gt;&#xA;in &#xA;&lt;a href=&quot;https://i.stack.imgur.com/aUDvd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aUDvd.png&quot; alt=&quot;33&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solving equation 7 = 0 , i have x = 5.23 , 3 and  0.763  , which is correct ? &lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="8664" LastEditDate="2017-01-04T14:12:00.203" LastActivityDate="2017-01-04T14:12:00.203" Title="How can we know &quot;by inspection&quot; the location of maximum deflection of a simply supported beam with a point load?" Tags="|structural-engineering|beam|deflection|" AnswerCount="1" CommentCount="22" ContentLicense="CC BY-SA 3.0" />
  <row Id="12633" PostTypeId="1" CreationDate="2016-12-03T17:44:50.907" Score="1" ViewCount="90" Body="&lt;p&gt;I am not a mechanical engineer, so just trying to figure this out. I want to attach a servo to my telescope focus, but the servo only has about 180 degrees motion, whereas you need to turn the knob about 540 degrees for the full range of focus. It has a rack and pinion gear, so my plan was to find a pinion gear with more teeth (maybe 48?) to get the full range in 180 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pinion gear has 16 teeth and an outer diameter of about 8mm. I measured 10 teeth on the rack which was about 15mm. The diameter of the shaft is about 5mm. My question is, what size gear would I need, how many teeth, what pitch? Does it exists? My best guess is 48 teeth would be needed, but I am not sure about my pitch. Also, if the gear exists, do they make them to attach directly to a servo? Or will I need to get a shaft and a shaft to servo coupler?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DUdmF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DUdmF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9064" LastEditorUserId="7083" LastEditDate="2016-12-15T23:21:42.063" LastActivityDate="2016-12-15T23:21:42.063" Title="Larger gear for telescope focus rack?" Tags="|motors|gears|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12634" PostTypeId="1" AcceptedAnswerId="12647" CreationDate="2016-12-03T17:55:30.227" Score="2" ViewCount="606" Body="&lt;p&gt;I hope I'm asking this in the right place.&#xA;I have the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jq2lf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jq2lf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and am to calculate A,B &amp;amp; C so that the system is in static equilibrium.&#xA;I calculated A without too many problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is with B &amp;amp; C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I have done so far:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realised that my placing my moment at the point ABF I can eliminate B as an unknown and solve for C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated the resultant for the diagonal part ($R_1=q_0\cdot l$) and for the vertical part ($R_2=\frac{q_0\cdot 2l}{2}=q_0\cdot l$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated the distances of $R_1$ and $R_2$ from my moment $M^{ABF}$ using the Pythagorean theory and set them both negative as they are turning my system in a mathematically negative direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force C I added to my moment as $C\cdot 2l$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My moment calculation is at this point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Sigma M^{ABF}=2l\cdot C-(\sqrt{(2l)^2+(2l)^2}+\frac{l}{2})\cdot R_1-\sqrt{(2l)^2+(\frac{2}{3}2l})^2\cdot R_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My problem:&lt;/strong&gt;&#xA;I don't know what to do with the moment $M$ which is at a distance of $l$ from $M^{ABF}$. I have tried adding as it is, adding it and multiplying by $l$, but both times the answer comes out wrong (I have a numerical answer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm obviously doing something wrong but I don't know what.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it helps to have values:&lt;br&gt;&#xA;$F=4q_0l$,  $M=2q_0l^2$,  $\alpha=45°$&lt;br&gt;&#xA;$q_0=3kN/m$, $l=2m$&lt;/p&gt;&#xA;" OwnerUserId="9065" LastActivityDate="2019-12-28T18:21:25.000" Title="Static equilibrium and moment" Tags="|statics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12635" PostTypeId="2" ParentId="12614" CreationDate="2016-12-03T18:10:39.710" Score="5" Body="&lt;p&gt;Bottled air is used for SCUBA because it is simple, reliable and relatively inexpensive and an open circuit system with a demand valve is pretty safe to use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are closed circuit systems (&lt;a href=&quot;https://en.wikipedia.org/wiki/Rebreather&quot; rel=&quot;nofollow noreferrer&quot;&gt;rebreathers&lt;/a&gt;) which use bottled pure oxygen and CO2 scrubbers, which give greatly extended endurance per weight, but they are significantly more complex than gas-mix SCUBA as the partial pressures of O2 and CO2 need to be carefully monitored and controlled and tend only to be used for specialist military and commercial applications as even small variations in gas mixture can be fatal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are breathing a gas mixture which already has the correct partial pressure of oxygen then there is much less to go wrong. In this case only a small fraction of the oxygen carried is actually absorbed but having an open circuit means that you are always breathing the correct partial pressure of oxygen and exhaled CO2 just escapes as bubbles with each breath. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need at enough pressure in the system to ensure that you can actually inhgale against the water pressure at whatever depth you are at, again a high pressure cylinder with a demand valve is a failry simple way to achieve this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you start using pure oxygen you need to use either a closed circuit system or prvide an inert gas to dilute it to the correct partial pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two key things to consider here : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The partial pressure of oxygen needs to be right, too little and it can't be absorbed by osmosis, too much and it becomes toxic. Equally you only inhale a small amount of oxygen for each lungfull as it is absored passively across a concentration gradient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) It is just as important to be able to get rid of carbon dioxide as it is to absorb oxygen, indeed the breathing mechanism is driven by CO2 concentration which is why it is such a powerful asphixiant. In a sealed room you will die from CO2 asphixiation beore you run out of oxygen. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-12-05T01:55:59.920" LastActivityDate="2016-12-05T01:55:59.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12636" PostTypeId="1" AcceptedAnswerId="12641" CreationDate="2016-12-03T18:38:49.653" Score="4" ViewCount="2439" Body="&lt;p&gt;It may be a simple or obvious question, but I'm confused what's changed between those. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fSDbM.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fSDbM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would be an example of a non structural rivet, does the bottom half of the mandrel steam inside the hollow rivet fall out, unlike structural where it stays, thus strengthening the rivet due to it not being hollow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bit confused between both.&lt;/p&gt;&#xA;" OwnerUserId="9066" LastEditorUserId="1832" LastEditDate="2016-12-03T19:29:30.230" LastActivityDate="2018-10-05T14:42:31.463" Title="What's the difference between structural pop rivets, and non structural?" Tags="|rivets|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12637" PostTypeId="1" CreationDate="2016-12-03T20:31:43.013" Score="2" ViewCount="115" Body="&lt;p&gt;Sorry if the question is confusing as such, here's an image that makes it hopefully clearer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tXnMI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tXnMI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's say this building consists of two elevated concrete slabs. The first elevated slab is supported by the 9 massive reinforced concrete piers arranged at even intervals of 5m x 5m, and connected with massive reinforced concrete beams. Thus the building is 10m x 10m in dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second concrete slab is supported by eight of the 9 massive concrete piers, the ones around the perimeter,  but the concrete pier in the middle is only one-storey high. Instead two less massive piers (labelled A &amp;amp; B) start from the height of the first elevated slab and support the second slab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, my questions are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;is this design potentially safe &amp;amp; practical as a way of supporting the top slab?&lt;/li&gt;&#xA;&lt;li&gt;does this  mean that the beam along the junction of A-B requires extra reinforcement compared to a design where the top slab is supported by the same (extended) pier as the bottom slab.&lt;/li&gt;&#xA;&lt;li&gt;are there any other major ways that the forces differ, e.g, in terms of the forces transmitted by the lower columns, or the loads experienced by the lower slab?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6518" LastEditorUserId="1832" LastEditDate="2016-12-04T02:13:46.823" LastActivityDate="2017-01-04T17:54:33.897" Title="Do beams in a structure transmit to piers?" Tags="|structural-engineering|concrete|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12638" PostTypeId="1" CreationDate="2016-12-03T22:25:01.317" Score="1" ViewCount="177" Body="&lt;p&gt;As part of an experiment, I'm verifying the existence of a linear relationship between curvature and bending moments for three beams using different materials. The plotted data can be seen below. &lt;a href=&quot;https://i.stack.imgur.com/KfpCV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KfpCV.jpg&quot; alt=&quot;graph&quot;&gt;&lt;/a&gt;Visually, there is a very strong positive linear relationship. Visually, the residuals are small. Of the three populations of data, the lowest Pearson coefficient is 0.9998, the highest is 0.9999.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of verifying this linear relationship in a lab report, need I discuss anything other than the regression lines being visually straight and the residuals being visually small? Is it necessary to calculate the Pearson coefficients? Should any other statistical tests be carried out such as the coefficient of determination?&lt;/p&gt;&#xA;" OwnerUserId="9070" LastEditorUserId="1832" LastEditDate="2017-03-05T10:41:58.943" LastActivityDate="2017-03-05T16:26:54.193" Title="How can the linear relationship between curvature and bending moments be verified?" Tags="|mechanical-engineering|statics|beam|modeling|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12639" PostTypeId="2" ParentId="12638" CreationDate="2016-12-03T22:35:09.720" Score="1" Body="&lt;p&gt;Not knowing &lt;em&gt;exactly&lt;/em&gt; how you did the experiment, I think you should explain carefully how you actually measured (and/or calculated) the curvature value, and how you know it really was &quot;the radius of curvature&quot; and not some other measure of &quot;how curvy the beam was&quot; which happens to give a linear plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question is whether your point at (0,0) is really &quot;measured data&quot; or just an assumption - and whether your test pieces really were &quot;perfectly straight&quot; before &lt;em&gt;and after&lt;/em&gt; the experiment. If it isn't measured data, should it be on the graphs at all?&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-12-03T22:35:09.720" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12641" PostTypeId="2" ParentId="12636" CreationDate="2016-12-04T00:26:21.297" Score="3" Body="&lt;p&gt;PopRivets (tm) are a simple example of non-structural rivets.  When assembled the steel core breaks off leaving a hollow rivet.  These cannot be considered to be structural since the forces needed to destroy the joint are acting on a very small cross sectional area and failure is possible in shear, tension and torsional modes.&#xA;They are appropriate for assembling a surface panel or decorative element to a design. Where significant loadings can occur, of course a denser rivet pattern will be needed, preferably following the distribution of load in the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A structural rivet is usually comprised of a single piece of material that is deformed to bind the parts together. Some hot riveting systems rely on the contraction of the cooling fastener to hold the parts together. Unlike PopRivets there are also cored and coreless versions available in a variety of materials with the necessary properties. Again the rivet pattern is important and some effort should be applied to understanding the optimum distribution.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2016-12-04T00:26:21.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12642" PostTypeId="2" ParentId="12633" CreationDate="2016-12-04T00:40:16.253" Score="2" Body="&lt;p&gt;I suggest that you use a stepper motor, not a servo for this application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steppers can be controlled in small rotational increments. The speed of rotation and the torque delivered by the stepper motor are easily determined and controlled.  Furthermore, a stepper with an encoder provides an accurate and repeatable method of precisely positioning a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given your gearing, the final answer depends on several factors with the torque and speed driving the design.  From the appearance of the mechanism a small NEMA 17 motor with 1.8 degree steps would be more than sufficient assuming that the mechanism is reasonably free-running.  If it is stiff, a larger motor with a larger current supply would improve performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stepper controllers are quite simple and easily available. For a telescope, a LiPo battery would make sense so I would consider a NEMA 17, 3S LiPo and a small controller with an external potentiometer/PWM used for servo testing to adjust the position, to be sufficient&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2016-12-04T00:40:16.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12643" PostTypeId="2" ParentId="12630" CreationDate="2016-12-04T01:16:17.377" Score="2" Body="&lt;p&gt;So, for this problem, you first need to find the transfer function from U to E. An easy way to do this to use the forward gain over one plus the loop gain. This simply mean $\frac{E(s)}{U(s)} = \frac{forward~gain}{1+loop~gain}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forward gain in this problem is 1 (U to E has no gains between it) while the loop gain is $\frac{K}{(s+2)(s+3)}*\frac{1}{s+4}$ (all gains around loop from E back to U).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steady state error is $e_{ss} = \lim_{s\to0} s\frac{E(s)}{U(s)} U(s)$. This simplifies to what you had above. If you substitute everything is, you have $0.1 = \lim_{s\to0}\frac{1}{1+\frac{K}{(s+2)(s+3)(s+4)}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply both top and bottom by $(s+2)(s+3)(s+4)$ and set s equal to 0. Multiply it all out and you should get 216.&lt;/p&gt;&#xA;" OwnerUserId="9073" LastActivityDate="2016-12-04T01:16:17.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12645" PostTypeId="2" ParentId="12631" CreationDate="2016-12-04T02:33:14.433" Score="1" Body="&lt;p&gt;For starters, as @alephzero mentioned in a comment, in the real world there are no rigid bodies, so this question is entirely theoretical. No matter how massive and rigid the beam, some infinitesimal deflection will always occur and that's all we need to calculate the beam as any other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, let's see how (or whether) we can solve this for rigid beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force method for statically indeterminate structures has us remove constraints until we get to an isostatic structure, at which point we replace the removed constraints with forces according to the deformation or rotation they would have inhibited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's remove $R_1$. What happens then? The beam is perfectly rigid, so it's deflection at $R_1$ is null. Since there's no deflection to counter, this means that $R_1$ is itself null. If $R_1$ is null, by symmetry we can tell that $R_3$ would also have to be null, meaning $R_2$ will absorb the entirety of the forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if instead we removed $R_2$, we'd get that $R_2$ is null and $R_1 = R_3 = 100$. A paradox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's see what we get with the stiffness matrix method. Actually, never mind. The stiffness matrix has an $EI$ (or $EA$) in the numerator of every item, which would throw the entire matrix into infinity, making any calculation nonsensical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the conclusion is clear: &lt;strong&gt;there is no numerical solution for an absolutely rigid structure&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But... what if the supports aren't rigid? Then the solution is trivial. Since the loading is symmetric, the beam will deform all spring supports equally (since it's infinitely rigid), meaning they will all have the same value: $R_1 = R_2 = R_3 = \dfrac{200}{3}$.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-12-04T02:33:14.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12647" PostTypeId="2" ParentId="12634" CreationDate="2016-12-04T03:29:35.727" Score="0" Body="&lt;p&gt;Bending moments due to forces and distances are found by the product of the force and the perpendicular distance of its projection to the point of study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the following structure (ignore the dimensions, they are just placeholders):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bKgP0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bKgP0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transforming the distributed loads into equivalent concentrated loads, you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jBQ8E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jBQ8E.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That figure already shows you the dimensions you need to consider for each force:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The linear load along the &lt;span class=&quot;math-container&quot;&gt;$2\ell$&lt;/span&gt; bar is horizontal. Therefore, the perpendicular distance to use is the vertical distance.&lt;/li&gt;&#xA;&lt;li&gt;The uniform load along the &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt; bar is at a 135 degree angle with the horizontal. Therefore, the distance must be at a 45 degree angle.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The concentrated bending moment &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; in the middle of the horizontal &lt;span class=&quot;math-container&quot;&gt;$2\ell$&lt;/span&gt; bar is simply added to your sum. Such loads are always just added to the equilibrium equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we can get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;M_{ABF} = 0 &amp;amp;= M + 2C\ell - \dfrac{2q_0\ell}{2}\cdot\dfrac{2}{3}2\ell - q_0\ell\left(2\ell\sqrt2 + \dfrac{\ell}{2}\right) \\&#xA;&amp;amp;= M + 2C\ell - q_0\ell^2\left(\dfrac{4}{3} + \left(2\sqrt2 + \dfrac{1}{2}\right)\right) \\&#xA;\therefore C &amp;amp;= \dfrac{q_0\ell^2\left(\dfrac{4}{3} + \left(2\sqrt2 + \dfrac{1}{2}\right)\right) - M}{2\ell} \\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing this, it is then trivial to find the value of &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; via the equilibrium of vertical forces &lt;span class=&quot;math-container&quot;&gt;$\left(\sum F_y = 0\right)$&lt;/span&gt; equation. This is left as an exercise for the reader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;All diagrams were done with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:21:25.000" LastActivityDate="2019-12-28T18:21:25.000" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12653" PostTypeId="1" AcceptedAnswerId="13538" CreationDate="2016-12-04T13:06:28.790" Score="1" ViewCount="111" Body="&lt;p&gt;I thought of an injector nozzle for a diesel engine with an alternate combustion chamber format (somewhat like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Heron_cylinder_head&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heron combustion chamber&lt;/a&gt;), and would now like to get an estimate of the size of the opening for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The injector nozzle is basically a tube with a pintle inside it; the pintle has a flange at its end, outside of the tube and in the combustion chamber, which extends past the radius of the tube. When the injector is spraying, the fuel flows between the tube wall and pintle and is deflected outwards by the flange into the combustion chamber, forming a planar 'disk' of fuel between the piston and the top of the cylinder. The problem is to calculate an estimate of the opening $l$ between the flange and the butt end of the tube from a simplified description of the intended spray front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qhk06.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qhk06.jpg&quot; alt=&quot;Stylized illustration of the nozzle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be using cylindrical coordinates $r,\theta,z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Design parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$Q_0$, the mass flow rate in the tube&lt;/li&gt;&#xA;&lt;li&gt;$r_0$, the flange radius (and thus of the nozzle)&lt;/li&gt;&#xA;&lt;li&gt;$r_c$, the radius of the combustion chamber&lt;/li&gt;&#xA;&lt;li&gt;$t_c$, the time allotted for fuel injection at max RPM.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Problem domain:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S(r)=2\pi(r-r_0) l$ (spray front)&lt;/li&gt;&#xA;&lt;li&gt;$V(r)=\pi(r^2-r_0^2) l$ (volume occupied)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The fuel occupies the whole opening of the nozzle, doesn't diverge vertically and distributes itself uniformly ($\frac{\partial\rho}{\partial r} = \frac{\partial\rho}{\partial\theta} =\frac{\partial\rho}{\partial z} = 0$)&lt;/li&gt;&#xA;&lt;li&gt;$Q_0$ is the flow not just in the nozzle, but also at the spray front ($Q_0=\iint_S \mathbf j\cdot d\mathbf S$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m(t_c)=\pi\rho(r_c^2-r_0^2) l=Q_0 t_c\\&#xA;\implies l=\frac{Q_0 t_c}{\pi\rho(r_c^2-r_0^2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get $l$ = 1.34×10&lt;sup&gt;-5&lt;/sup&gt; cm = 134 nm for an application in a high-speed two-stroke ($r_c$ = 2.2cm, $r_0$ = 1mm, $t_c$ = 2×10&lt;sup&gt;-4&lt;/sup&gt;s (more or less the time taken to traverse 10°, my estimate of the injection duration, at 8000 rpm), $Q_0$ = .8 g/s), which &lt;strong&gt;is&lt;/strong&gt; absurdly tiny.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though the order of magnitude of the result comes mainly from the injection duration, I don't trust my results and think that it comes from the assumptions I made. On top of that, I don't know how to take into account the end of injection prior to combustion, so I assumed the injector stops injecting at combustion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything wrong with the analysis or calculation that I did?&lt;/p&gt;&#xA;" OwnerUserId="6589" LastEditorUserId="163" LastEditDate="2017-02-01T18:32:26.007" LastActivityDate="2017-09-30T16:11:35.847" Title="Diesel injector nozzle geometry" Tags="|geometry|diesel|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12654" PostTypeId="1" AcceptedAnswerId="12682" CreationDate="2016-12-04T13:43:35.990" Score="3" ViewCount="521" Body="&lt;p&gt;I'm having trouble with this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OP64K.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OP64K.jpg&quot; alt=&quot;problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G and a are given, and we are asked to calculate A, B and C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I've Done:&lt;/strong&gt;&lt;br&gt;&#xA;The first thing I tried was a free body diagram cutting through A and C to create one body and C and B to create another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vvHTr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vvHTr.jpg&quot; alt=&quot;Body 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here I calculated $A_y$ and $C_y$ as being equal to $-G\frac{\sqrt{2}}{4}$&lt;br&gt;&#xA;Then I moved onto the second body:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oRfgQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oRfgQ.jpg&quot; alt=&quot;Body 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I ran into problems immediately. If I placed the moment at either B or C then it follows that the x component of the other must be zero, which it isn't (I have a numerical solution). So I've either made a mistake with my free body diagram or I'm doing something wrong with the second moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my equalibrium calculations:&lt;br&gt;&#xA;Body 1:&lt;br&gt;&#xA;$$\Sigma M^{A}=C_y\cdot 2a+G\frac{\sqrt{2}}{2}=0 \\&#xA;\Sigma F_x=A_x-C_x+G\frac{\sqrt{2}}{2}=0 \\&#xA;\Sigma F_y=A_y+C_y+G\frac{\sqrt{2}}{2}=0$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Body 2:&#xA;$$\Sigma M^{B}=3a\cdot C_x=0 \\&#xA;\Sigma F_x=C_x+B_x=0 \\&#xA;\Sigma F_y=B_y-C_y=0$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My results obviously make no sense. What am I doing wrong? I have also tried to calculate for the whole structure but there were too many unknows which I couldn't get rid of.&lt;/p&gt;&#xA;" OwnerUserId="9065" LastEditorUserId="9065" LastEditDate="2016-12-04T15:20:16.257" LastActivityDate="2019-12-28T18:21:13.477" Title="Free Body Diagram Problem" Tags="|statics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12655" PostTypeId="1" CreationDate="2016-12-04T13:55:29.933" Score="1" ViewCount="519" Body="&lt;p&gt;Consider:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9LZ5l.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9LZ5l.png&quot; alt=&quot;Enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because the 2nd and 3rd satellites' centres are collinear? Can a satellite overlap happen? If so then how is it avoided? If not avoided then what happen to the signal that is sent from such a satellite?&#xA;This is just a 2D representation. The same concept can be applied in 3D.&lt;/p&gt;&#xA;" OwnerUserId="9068" LastEditorUserId="9068" LastEditDate="2016-12-05T06:14:35.450" LastActivityDate="2016-12-05T13:15:01.163" Title="How are three satellites sufficient to find a unique point in GPS? What happens in case of an overlap?" Tags="|wireless-communication|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12656" PostTypeId="1" CreationDate="2016-12-04T13:57:26.237" Score="1" ViewCount="72" Body="&lt;p&gt;I know that pattern generator &lt;em&gt;device&lt;/em&gt; (no pattern generator &lt;em&gt;software&lt;/em&gt;) is needed to generate a model signal system that help you use electron diffraction for wafer surface printing by the use of the TEM. These signals are templates called &lt;em&gt;patterns&lt;/em&gt; and is generated by some electronic unknown components inside of pattern generator, maybe with an amplifier or oscillator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v6esm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v6esm.jpg&quot; alt=&quot;pattern generator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand &lt;strong&gt;what modulation system&lt;/strong&gt; allow this &quot;driving&quot; signal between this different devices (TEM, pattern generator, computer..) to generate correctly &lt;em&gt;template-pattern&lt;/em&gt; for final printing on wafer ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To &lt;em&gt;print&lt;/em&gt; a chip project from design (software) you need to generate a modulation to communicate with pattern generator device and this pattern generator device needs to communicate with TEM but you need still a modulation to &lt;em&gt;tell&lt;/em&gt; what TEM should print and &lt;em&gt;how&lt;/em&gt; to print &#xA;&lt;br&gt;What type of modulation or what system/component is used to modulate different signals between this devices ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Indeed&lt;/em&gt; we must &lt;em&gt;generate&lt;/em&gt; an interference &lt;em&gt;before&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Diffraction cannot be explained using a particle view of light (geometrical or ray optics); it requires a wave interpretation (physical optics), according to which any image is actually an interference pattern formed by light rays that take different paths to reach the same point in the image&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But &lt;em&gt;still before&lt;/em&gt; generating this interference to have correct diffraction electrons used to print a pattern on wafer (we use a TEM system) &lt;strong&gt;you need a &lt;em&gt;correct&lt;/em&gt; modulation&lt;/strong&gt;. &#xA;&lt;br&gt; What and how signal generating check if this modulation system is correct to ensure that the devices understand each other ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of technique use something like &lt;code&gt;Electron backscatter diffraction&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="9080" LastEditorUserId="7083" LastEditDate="2016-12-16T07:51:39.433" LastActivityDate="2016-12-16T07:51:39.433" Title="What system-electronic components or physic mechanism is used to modulate signals between different devices used in lithography?" Tags="|lithography|modulation|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12657" PostTypeId="2" ParentId="12655" CreationDate="2016-12-04T15:04:33.313" Score="6" Body="&lt;p&gt;You can't ignore the difference between 2D and 3D in this situation. To a first approximation, all GPS satellites are located on the surface of a sphere. There is no way that any three points on a sphere can be colinear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one situation that can be ambiguous, and that's when N satellites are equidistant from the receiver. However, in this case, one solution will be on the surface of the Earth, and the other solution will be thousands of miles straight up from there. It's relatively easy to pick the correct solution.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-12-04T15:04:33.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12658" PostTypeId="2" ParentId="12655" CreationDate="2016-12-04T15:14:34.947" Score="3" Body="&lt;blockquote&gt;how are three satellites sufficient to find a unique point in gps?&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are not.  It takes four satellites minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GPS needs to solve for the point in four dimensions, X, Y, Z and time.  Four unknowns requires at least four inputs to solve.  In the case of GPS these are the relative time offsets of the signals received at the coordinate in question.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-12-04T15:14:34.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12661" PostTypeId="2" ParentId="12632" CreationDate="2016-12-04T20:14:44.787" Score="1" Body="&lt;p&gt;Engineers, by design, are lazy. If you are looking at a beam, where you can see that under a given load it deflects most at a certain point, then you can say that it deflects most at that point. It's like when you have symmetry conditions to only have to figure out half of the numbers for a beam, and extrapolating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Textbook writers say 'by inspection' when they can use their engineering intuition and what they believe to be common sense in order to save labor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you only had the beam, some forces, and we knew that the beam would bend downwards, bottom out, and start to bend back up to the other end, it's intuitive that if you rolled a marble from either end of the beam, it would roll down and settle on the lowest point, where it had a max deflection and a slope of zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if there was a beam with multiple forces pointing up and down, with the beam looking like a ride at Disney Land, then that would be a different story. Then, you would have to go through more calculations and really get that engineering intuition into action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But most often, engineering intuition is knowing how to use the minimum computation for the required results.  The author simply said, there is only one global and local minimum, and said, &quot;close enough for government work.&quot; &lt;/p&gt;&#xA;" OwnerUserId="8862" LastActivityDate="2016-12-04T20:14:44.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12663" PostTypeId="1" AcceptedAnswerId="12671" CreationDate="2016-12-04T21:24:31.650" Score="0" ViewCount="162" Body="&lt;p&gt;What is the difference between 250W band heater and a 200W one? Can they reach the same temperature?&lt;/p&gt;&#xA;" OwnerUserId="9086" LastActivityDate="2016-12-05T04:33:00.973" Title="What is the difference between 250W band heater and a 200W one?" Tags="|heat-transfer|power-electronics|" AnswerCount="1" CommentCount="2" ClosedDate="2016-12-05T12:15:24.227" ContentLicense="CC BY-SA 3.0" />
  <row Id="12665" PostTypeId="2" ParentId="12655" CreationDate="2016-12-04T22:36:14.827" Score="0" Body="&lt;p&gt;First, a GPS satellite cannot actually tell the receiver how far away it is. The receiver &lt;em&gt;can&lt;/em&gt; detect the difference between the distance to one satellite A and another satellite B. As a result, four satellites are needed to find your position (however, if there are only three satellites then GPS can guess that you might be on the surface of the earth and calculate your location on that assumption. And if you are standing on the street, the location would be right. If you are on the tenth floor of a high building, you might be 100 metres off. If you are in a flying airplane, it's totally wrong). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's ignore that. If three satellites could tell you the distance from each satellite in 3D space, then there are usually two solutions to the equations. (Same in 2D space with two satellites). Typically, one solution will be close to the earth surface, and the other one won't. You pick the one close to the earth surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also do the same measurement again 100 ms later. So the two solutions move. The solution for the point where you really are moves with the speed of earth rotation. (When you sit still on your sofa, you are actually moving at about 1700 km/h or so; the number is from memory and may be wrong but it's roughly that). GPS can calculate how fast a solution moved relative to the earth surface; that speed will be low (say 190mph if you are travelling on Eurostar). The other incorrect solution will move at a massively different speed, mostly because the satellites themselves are moving at a high speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have another satellite (the third one in a 2D situation in your example), then the satellites are actually arranged in a pattern around the earth so the situation that you described will not happen. So all the circles will have one point in common. Actually, because all your data is slightly imprecise, all the circles will meet at slightly different points very close together. And you can mathematically find the point that is closest to all the circles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And interestingly, it is illegal to build a GPS system that could be used to guide ballistic missiles. If a GPS system detects that it is at high altitude and/or moving at very high speed, as it would be typical for a ballistic missile, then it is not allowed to tell its location. You can bet that the second solution will be far from the earth surface and moving fast, so &lt;em&gt;if&lt;/em&gt; it was the correct location it would be illegal for the GPS to report that location. &lt;/p&gt;&#xA;" OwnerUserId="9088" LastActivityDate="2016-12-04T22:36:14.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12667" PostTypeId="1" CreationDate="2016-12-05T02:30:54.730" Score="1" ViewCount="158" Body="&lt;p&gt;Hello I was wondering if you guys can help me figure out how to calculate the tension (or weight) or a rope in a given system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So A cylinder pipe that has outside diameter of 70cm (.7m) and inside of 60cm (.6) and ropes are split by a spreader bar at 45*. What is tension of AB and AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinda looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1k8rj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1k8rj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have $\pi(.7-.1)\cdot2320\cdot9.81= 9286\text{ N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then after that I get confused what to do.&lt;/p&gt;&#xA;" OwnerUserId="9089" LastEditorUserId="1832" LastEditDate="2016-12-05T11:17:53.467" LastActivityDate="2016-12-05T17:07:21.750" Title="Tension in a rope" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="6" ClosedDate="2016-12-09T15:42:52.797" ContentLicense="CC BY-SA 3.0" />
  <row Id="12668" PostTypeId="2" ParentId="11104" CreationDate="2016-12-05T02:32:43.593" Score="0" Body="&lt;p&gt;take a look at the following 2 patents for the information you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.patentsencyclopedia.com/app/20160070369&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.patentsencyclopedia.com/app/20160070369&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/patents/US20130100088&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/patents/US20130100088&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8890" LastActivityDate="2016-12-05T02:32:43.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12669" PostTypeId="1" CreationDate="2016-12-05T04:14:51.177" Score="3" ViewCount="906" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DDbJC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DDbJC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ive got the moments of :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;B:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vertical : $\sin(75)AB + \sin(45)BC = 33\text{ lbs}$&lt;/li&gt;&#xA;&lt;li&gt;Horizontal : $\cos(75)AB = DB + \cos(45)BC$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;D:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vertical: $\sin(60)DC + \sin(65)DE = 100\text{ lbs}$&lt;/li&gt;&#xA;&lt;li&gt;Horizontal: $\cos(60)DC + BD = \cos(65)DE$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vertical: $W_F = \sin(75)AB$&lt;/li&gt;&#xA;&lt;li&gt;Horizontal: $\cos(75)AB$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;E:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vertical: $W_B = \sin(65)DE$&lt;/li&gt;&#xA;&lt;li&gt;Horizontal: $CE = \cos(65)DE$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How do I solve this truss problem with multiple unknowns? It seems every possible manipulation leaves me with at least one unknown.&lt;/p&gt;&#xA;" OwnerUserId="9090" LastEditorUserId="1832" LastEditDate="2016-12-05T11:15:07.117" LastActivityDate="2017-05-05T10:08:29.077" Title="Truss problem with multiple unknowns" Tags="|structural-engineering|statics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12671" PostTypeId="2" ParentId="12663" CreationDate="2016-12-05T04:33:00.973" Score="1" Body="&lt;p&gt;Heating elements generally have a maximum operating temperature above which they'll start to degrade and eventually break. If these heaters are from the same product line they are likely made of the same materials and will have the same maximum operating temp therefore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Separately, you want to know how hot they can get in your application. The hotter they get, the faster they lose heat to the environment. Once they are losing heat as quickly as they generate it, their temperature will stop rising. From that perspective, the 250 W heater will reach a higher temperature if all other things are the same in your system. Alternatively, you may add insulation to the lower power heater to achieve the same final temp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last thing to consider, is that the higher power heater will heat things up more quickly.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2016-12-05T04:33:00.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12673" PostTypeId="1" AcceptedAnswerId="12693" CreationDate="2016-12-05T06:18:30.933" Score="0" ViewCount="1347" Body="&lt;p&gt;I am looking for a method changing the coordinate system of RBE2 or RBE3 elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the nastran description of RBE2 in Nastran user guide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z97bs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z97bs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no field for coordinate configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am wondering if coordinate change is possible or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone explain how to change coordinate of RBE2?&lt;/p&gt;&#xA;" OwnerUserId="4624" LastEditorUserId="4624" LastEditDate="2016-12-05T07:58:41.457" LastActivityDate="2016-12-05T22:16:22.343" Title="Nastran RBE2 coordinate change" Tags="|mechanical-engineering|modeling|simulation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12676" PostTypeId="2" ParentId="12655" CreationDate="2016-12-05T13:15:01.163" Score="0" Body="&lt;p&gt;3 satellites are sufficient for a 2D position fix where the receiver assumes that it is at a fixed height. Generally these are avoided whenever possible because a small error in this assumption can result in large position errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a true 3D position fix 4 satellites are needed. As others have said this is required to solve for 4 unknowns, you need to know the exact time in order to calculate your location. Generally far more SVs are tracked and a least squares or similar optimization is used to find the best solution to the measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't ever measure range to the satellites, you measure range differences between them, e.g. you know that you are nnn meters closer to SV 1 than SV 2. With 4 satellites you do actually end up with two possible solutions but one of them will be 10,000km above the earth and in a direction where none of the GPS signal is broadcast. The receivers are generally smart enough to ignore this possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all the satellites are at the same altitude and arranged into a fixed number of orbital planes I don't think the geometry you describe is possible in the real world. If it was it would only remain true for a tiny fraction of a second, the satellites are moving at high speeds rather than the static points always show in illustrations.&lt;br&gt;&#xA;The system is also tracking the satellites velocities relative to the receiver, even if the geometry you describe did somehow happen at the exact moment measurements were taken and only the 3/4 satellites involved were being tracked then you would still get the correct location, only one of the two positions would match both the measured ranges and the measured velocities.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-12-05T13:15:01.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12677" PostTypeId="1" CreationDate="2016-12-05T13:23:56.500" Score="0" ViewCount="46" Body="&lt;p&gt;I need to select some linear acturator or build it. I see it as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N8MYs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N8MYs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On green platform there is pink device. This device must be measured. For this purpose we need to choose linear acturator with carriage (dark blue on picture). Carriage must go along one direction. This carriage must to carry some electrical equipment. Mass of this equipment = 400 g (0,88 lb). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specification:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OFbFN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OFbFN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carriage (other moving element) must have range of motion = 50 mm (2 inch). Minimum motion (step) must equal 0,25 μm (or less). Allowable error of measurement = 7%. We plan to measure 100 points along and only ahead because repeatedly of measurements don't care. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What types of acturators may be appropriate for this task? What variant is better?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Additional explanation:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have considered two options: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Linear acturator on base ball screw rail with carriage and stepper motor. This solution does not satisfy this problem because has low precision.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Piezo acturator - useless too, because it has limitation on range of motion (max. 30 mm, when need 50 mm). Moreover piezo acturator very sensitive to&lt;br&gt;&#xA;payload (if that's devices can move 100 g - it's already good). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's acceptable for my objective if linear acturator will be move platform (green on picture) with electrical equipment. Blue stand with measurement equipment can fixed. In that case payload to platform = 2 kg. And step same = 0,25  μm. &lt;/p&gt;&#xA;" OwnerUserId="8880" LastActivityDate="2016-12-05T13:23:56.500" Title="What type of acturator with motor to choice for measurement with step = 0,25 μm?" Tags="|measurements|equipment-selection|actuator|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12679" PostTypeId="2" ParentId="12654" CreationDate="2016-12-05T15:02:30.973" Score="1" Body="&lt;p&gt;The balance equations would be the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Sigma F_x = A_x-C_x+\frac{G}{\sqrt{2}}=0$$&#xA;$$\Sigma F_y = A_y+C_y+\frac{G}{\sqrt{2}}=0$$&#xA;$$\Sigma M_P = A_y a-C_y a=0$$&#xA;$$\Sigma F_x = B_x+C_x-\frac{G}{\sqrt{2}}=0$$&#xA;$$\Sigma F_y = B_y-C_y-\left(\frac{G}{\sqrt{2}}+G\right)=0$$&#xA;$$\Sigma M_B = C_x (3 a)-\frac{G}{\sqrt{2}}(7 a)+&#xA;   \left(\frac{G}{\sqrt{2}}+G\right)(3 a)=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the solution turns out as&#xA;$$ A_x=\frac{1}{6} \left(\sqrt{2} G-6 G\right) \ \ A_y=-\frac{G}{2&#xA;   \sqrt{2}}$$&#xA;$$B_x=\frac{1}{6} \left(6 G-\sqrt{2} G\right)\ \ B_y=\frac{1}{4} \left(\sqrt{2}&#xA;   G+4 G\right)$$&#xA;$$C_x=\frac{1}{3} \left(2 \sqrt{2} G-3 G\right) \ \ C_y=-\frac{G}{2&#xA;   \sqrt{2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free-body diagrams&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h5OB8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h5OB8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2016-12-05T15:12:19.260" LastActivityDate="2016-12-05T15:12:19.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12680" PostTypeId="1" CreationDate="2016-12-05T15:03:07.290" Score="6" ViewCount="1033" Body="&lt;p&gt;In many application, the knowledge of the orientation of a body is essential for control purposes. There are several representations, where one is the Euler Angles and another is the Quaternion represenation, where the Euler Angles suffer from gimbal lock. My question is, how can one use the Quaternion represenation for control the angular position in space? If this would be possible, then why are the Euler Angles so widespread?&lt;/p&gt;&#xA;" OwnerUserId="8784" LastActivityDate="2022-03-31T17:40:14.063" Title="Quaternion in control applications" Tags="|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12681" PostTypeId="2" ParentId="12680" CreationDate="2016-12-05T15:25:44.250" Score="5" Body="&lt;p&gt;Euler angles are easier to understand and use. Imagine a airtrafic controller getting a aircraft heading info as quaternion data. Euler angles are significantly easier to understand interpret and interpolate.&lt;/p&gt;&#xA;&lt;p&gt;While quaternions do have benefits they are also conceptually more complex to work with. In many applications the downsides of euler angles and matrices are of no concern.&lt;/p&gt;&#xA;&lt;p&gt;In many mechanical cases you can not avoid gimbal lock anyway, having a control system that can navigate a problem area your physical counterpart can not is not very  useful. All you end up with is diverging from your real system such as a industrial robot.&lt;/p&gt;&#xA;&lt;p&gt;Similar thig applies to simulation where the euler-larange equation just sidesteps  the issue. The euler angle representation can represent all orientations just not interpolate well in this nonuniform space, but since the larangian is not operating in this space it matters not and the solution is well known. No gimbal lock concerns. Altough quats have a very small advantage in speed of normalisation. But then all formulas are usually in angular or matrix form.&lt;/p&gt;&#xA;&lt;p&gt;Third, while accumulating number of revolutions is usually futile, its viable to do in euler angles for the kinds of angles that dont come into question. While gimbal lock is nasty, it comes up quite rarely.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2022-03-31T17:40:14.063" LastActivityDate="2022-03-31T17:40:14.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12682" PostTypeId="2" ParentId="12654" CreationDate="2016-12-05T15:48:58.253" Score="2" Body="&lt;p&gt;Let's start by transforming the load &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; into the resultant forces applied to the structure. At point &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, we have a vertical component equal to &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; and an inclined component &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;. Since the inclined component is at 45°, we get a total force equal to &lt;span class=&quot;math-container&quot;&gt;$G\left(1 + \dfrac{1}{\sqrt2}\right)$&lt;/span&gt; downwards and &lt;span class=&quot;math-container&quot;&gt;$\dfrac{G}{\sqrt2}$&lt;/span&gt; to the left. Relatedly, at the end of the rope (between &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;), the concentrated force has upwards vertical and rightward horizontal components both equal to &lt;span class=&quot;math-container&quot;&gt;$\dfrac{G}{\sqrt2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model is therefore equal to (for &lt;span class=&quot;math-container&quot;&gt;$G = 100$&lt;/span&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QdReC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QdReC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you already found, we can isolate &lt;span class=&quot;math-container&quot;&gt;$AC$&lt;/span&gt;. It behaves like a simply supported beam with one horizontal constraint at &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;. Therefore, the horizontal load is fully absorbed by &lt;span class=&quot;math-container&quot;&gt;$A_x$&lt;/span&gt;. The vertical component is evenly split (due to being at the midspan of &lt;span class=&quot;math-container&quot;&gt;$AC$&lt;/span&gt;) between &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to deal with the forces at the top of the structure. The vertical force will be fully absorbed by &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; (but is slightly offset by the vertical component at the end of the rope discussed above which is transmitted to &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;). The horizontal force, however, we can't yet know. For that, we need to balance out the moment around &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; considering these forces and the horizontal force applied on/by &lt;span class=&quot;math-container&quot;&gt;$AC$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}\sum M_B = -G\left(1 + \dfrac{1}{\sqrt2}\right)\cdot3a + \dfrac{G}{\sqrt2}\cdot7a - A_x'\cdot3a = 0 \\&#xA;\therefore A_x' = G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3}&#xA;\end{gather}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$A_x'$&lt;/span&gt; is the horizontal reaction in &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; due to those forces, to be added to the reaction found previously of &lt;span class=&quot;math-container&quot;&gt;$\dfrac{G}{\sqrt2}$&lt;/span&gt;. So, &lt;span class=&quot;math-container&quot;&gt;$B_x = \dfrac{G}{\sqrt2} - A_x'$&lt;/span&gt; (in this case &lt;span class=&quot;math-container&quot;&gt;$A_x'$&lt;/span&gt; is negative, so this ends up being a sum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we end up with:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{alignat}{3}&#xA;A_x &amp;amp;= -\dfrac{G}{\sqrt2} + G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &amp;amp;&amp;amp;= G\left(\dfrac{1}{3\sqrt2} - 1\right) &amp;amp;&amp;amp;\approx -0.764G \\&#xA;A_y &amp;amp;= -\dfrac{G}{2\sqrt2} &amp;amp;&amp;amp;&amp;amp;&amp;amp;\approx -0.354G \\&#xA;B_x &amp;amp;= \dfrac{G}{\sqrt2} - G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &amp;amp;&amp;amp;= G\left(1 - \dfrac{1}{3\sqrt2}\right) &amp;amp;&amp;amp;\approx 0.764G \\&#xA;B_y &amp;amp;= G\left(1 + \dfrac{1}{\sqrt2}\right) - \dfrac{G}{2\sqrt2} &amp;amp;&amp;amp;= G\left(1 + \dfrac{1}{2\sqrt2}\right) &amp;amp;&amp;amp;\approx 1.354G&#xA;\\ C_x &amp;amp;= G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &amp;amp;&amp;amp;= G\left(\dfrac{4}{3\sqrt2} - 1\right) = &amp;amp;&amp;amp;\approx -0.057G \\&#xA;C_y &amp;amp;= -\dfrac{G}{2\sqrt2} &amp;amp;&amp;amp;&amp;amp;&amp;amp;\approx -0.354G \\&#xA;\end{alignat}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now, to check our work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W1Tg0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W1Tg0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alternatively, using your method, what you forgot to do was to consider the bending moment due to the forces at the top of the structure in your second cut (&lt;span class=&quot;math-container&quot;&gt;$BC$&lt;/span&gt;). That moment is equal to &lt;span class=&quot;math-container&quot;&gt;$G\left(1 + \dfrac{1}{\sqrt2}\right)3a - \dfrac{G}{\sqrt2}\cdot4a = Ga\left(3 - \dfrac{1}{\sqrt2}\right)$&lt;/span&gt;, which can only be balanced by the force binary generated by &lt;span class=&quot;math-container&quot;&gt;$A_x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B_x$&lt;/span&gt;. Dividing the moment by the &lt;span class=&quot;math-container&quot;&gt;$3a$&lt;/span&gt; lever arm between &lt;span class=&quot;math-container&quot;&gt;$A_x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B_x$&lt;/span&gt;, we get that each of those forces must be equal to &lt;span class=&quot;math-container&quot;&gt;$\pm\dfrac{Ga\left(3 - \dfrac{1}{\sqrt2}\right)}{3a} = \pm G\left(1 - \dfrac{1}{3\sqrt2}\right)$&lt;/span&gt;, which is precisely what we got for &lt;span class=&quot;math-container&quot;&gt;$A_x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B_x$&lt;/span&gt; above. Knowing this, you can then solve for the other variables as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;All diagrams given by &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis tool.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T18:21:13.477" LastActivityDate="2019-12-28T18:21:13.477" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="12683" PostTypeId="2" ParentId="12667" CreationDate="2016-12-05T17:07:21.750" Score="1" Body="&lt;p&gt;Solution:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Copper density = 8.96 g/cube cm. It means, that Volume =  π*250 cm *(35cm*35cm - 30cm*30cm) ~   255 125 cube cm. &#xA;Of cource R = 0,5 D, formula for founding square = π(r&lt;strong&gt;x&lt;/strong&gt;r);&#xA;Mass = 255 125 cube cm* 8,96 g ~  2 285 920 g = 2,285 tons. &#xA;Force of weight = 2,285 * 9,8 * 1000  ~ 22 850 N. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rPQfw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rPQfw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Tension = mg/4  ~ 5712 Н cause you have 2 rope and two side.&lt;/li&gt;&#xA;&lt;li&gt;Сhoose the axis like picture. &lt;strong&gt;OY&lt;/strong&gt;||mg; &lt;strong&gt;OX&lt;/strong&gt;⟂mg;&lt;/li&gt;&#xA;&lt;li&gt;In accordance to second law of Newton, all strength in picture must be compensated (equal zero) in point &lt;strong&gt;B.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Equations:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OY&lt;/strong&gt;: mg/4 = &lt;strong&gt;Tr&lt;/strong&gt; * sin(45);  - therefore &lt;strong&gt;Tr&lt;/strong&gt; = mg/(4*sin(45)) = 22 850/(4*0,7) ~ 8160 N ;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;OX&lt;/strong&gt;: Reactrion = Tr * cos(45) = 8 160 * 0,7 ~ 5712 N;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thats All. &lt;/p&gt;&#xA;" OwnerUserId="8880" LastActivityDate="2016-12-05T17:07:21.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12684" PostTypeId="2" ParentId="12680" CreationDate="2016-12-05T17:41:36.877" Score="3" Body="&lt;p&gt;I think joojaa summed it up well, Euler angles are fairly intuitive, given a set of angles you can quickly get an idea of what the object's orientation is. Gimbal lock does present a problem especially in video games and some applications in biomechanics/robotics. Quaternions are frequently used when gimbal lock is an issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily one can convert between quaterions and Euler angles, so you could have a control system that internally runs on quaternions but reports data as Euler angles if the user needs to look at what is happening.&lt;/p&gt;&#xA;" OwnerUserId="1174" LastActivityDate="2016-12-05T17:41:36.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12685" PostTypeId="2" ParentId="12637" CreationDate="2016-12-05T17:41:40.877" Score="3" Body="&lt;h2&gt;Questions 1 + 3:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Your first question is unanswerable. Any design is theoretically safe. Just create a massively thick slab with high-strength concrete and massive reinforcement and you can do just about anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, your two stories are completely different. The first floor can be best described as four different slabs, all of which are fixed-fixed-pinned-pinned. The second floor is a single slab which is pinned around its entire perimeter and at two points near the center. The second floor will certainly suffer larger stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The points where the slab is supported by the columns are especially critical. There will be massive shear stresses in that region, and specialized design will be necessary to handle those.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question 2:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Yes, the beams will certainly require additional reinforcement. After all, in this situation they will need to absorb the load transferred by the columns. This will lead to a massive shear load in the beam and probably massive bending moment as well. If the support system from the first floor was repeated in the second floor, then the beams in the first floor obviously wouldn't suffer those loads and therefore would require less reinforcement.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-12-05T17:41:40.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12686" PostTypeId="2" ParentId="12614" CreationDate="2016-12-05T18:19:10.687" Score="4" Body="&lt;p&gt;A scuba diver needs more air per minute than only 7 to 8 liters, because he is not at rest. A diver swimming with moderate speed needs about 20 to 25 liters per minute. But this is the consumption at surface, in ten meters depth the necessary mass flow is doubled and in 20 meters tripled. Breathing pure oxygen is safe only in very shallow water, about 3 m depth. Using only the minimal mass flow of oxygen is possible only if a scrubber is used to remove the exhaled carbon dioxide within the closed loop. If pure oxygen is used in a open loop the minimum flow is again about 20 to 25 liters per minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scuba diving with compressed air in bottles is the simplest and safest way to do it. Any kind of possible failure would be noticed by the diver immediately. In case of a failure, the diver is able to react and use an independent second system, share the air with his buddy or directly ascend to surface (if not to deep).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A closed loop oxygen rebreather may fail in several ways resulting in losing consciousness of the diver without warning before.&lt;/p&gt;&#xA;" OwnerUserId="950" LastEditorUserId="86" LastEditDate="2016-12-08T22:37:51.723" LastActivityDate="2016-12-08T22:37:51.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12687" PostTypeId="2" ParentId="12409" CreationDate="2016-12-05T18:43:16.893" Score="0" Body="&lt;p&gt;Whenever you want to model a physical system, you are faced with a choice of what model you are going to use. Your choice of model depends on how much accuracy you need, whether the fundamental physical relationships of the system are known, and what information you hope to get from your model.&lt;/p&gt;&#xA;&lt;h1&gt;What are your options?&lt;/h1&gt;&#xA;&lt;h2&gt;1. A naive model&lt;/h2&gt;&#xA;&lt;p&gt;You could decide that you have no idea what physical processes are happening to your system and try to come up with an input-output relationship that &amp;quot;fits&amp;quot; your data directly. In your case you seem to want the temperature change over time, so you would derive a relationship like so:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T = f(t,V)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the temperature, &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; is time, &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; is the input voltage, and &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; is some function. You have to decide what function best fits your data. For example, it could be linear, exponential, or polynomial. Often software packages like MATLAB, Microsoft Excel, and a multitude of others will automatically fit a curve to your data if you specify a type of function to use.&lt;/p&gt;&#xA;&lt;p&gt;The problem with this method is that you may be unaware of some of the parameters that affect your system. For example, do you need to include the surrounding air temperature? Also, you have no idea if your model will work outside of the temperature range you tested, in fact your model could be wildly divergent from the true system (this is especially true if you use a polynomial fit for your data).&lt;/p&gt;&#xA;&lt;h2&gt;2. A physics/systems based model&lt;/h2&gt;&#xA;&lt;p&gt;If you have an understanding of the underlying physical phenomena in your system, then you can construct a model based on the known underlying relationships between those phenomena. Use the elemental physical equations of each component in your system to create a set of differential equations describing the states of your entire system. Then adjust the parameters derived from your differential equation until the model provides the same output as the data you measured. There is a very wide variety of methods for doing that, which would be too huge to list and explain in a single answer, so I would suggest doing what was suggested in the comments and finding a textbook on system identification.&lt;/p&gt;&#xA;&lt;p&gt;The advantage to the physics-based model is that, if you've sufficiently accounted for all the variables in your system you can use it to estimate the output of your system in ranges that you haven't tested before. Another advantage is that you can usually derive a frequency domain expression of your system from the differential equation for use in control design. The disadvantage is that it can be more difficult to derive and identify all the relevant parameters.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-12-05T18:43:16.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12688" PostTypeId="1" CreationDate="2016-12-05T18:46:09.913" Score="0" ViewCount="193" Body="&lt;p&gt;I am trying to use plastic for a cheap and easy casting but I'm having trouble in molding it to any given shape. I'm trying to see if I can make a spark plug and want to cast out a model for measuring and getting a feel for the design. But I never worked with plastic. I'm using HDPE plastic.&lt;/p&gt;&#xA;" OwnerUserId="7041" LastActivityDate="2018-06-17T06:00:23.800" Title="How to cast and mold melt plastic to a shape and design you want?" Tags="|casting|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12689" PostTypeId="2" ParentId="8807" CreationDate="2016-12-05T19:12:28.897" Score="1" Body="&lt;p&gt;If you do it right, paper may be folded in half more than seven times:&#xA;&lt;a href=&quot;http://mathworld.wolfram.com/Folding.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mathworld.wolfram.com/Folding.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="950" LastActivityDate="2016-12-05T19:12:28.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12691" PostTypeId="1" CreationDate="2016-12-05T21:14:57.023" Score="2" ViewCount="249" Body="&lt;p&gt;I'm considering getting a SLA 3D printer, but the big uncertainty is whether I can build large things like propellers by printing them piecewise and fusing them together like you can with ABS and acetone (or butanone).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resins are proprietary, so I guess that's why I can hardly find any information on this. But according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Photopolymer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, photopolymers are thermoset, so that rules out melting them. But are there any solvents that can fuse them?&lt;/p&gt;&#xA;" OwnerUserId="4443" LastActivityDate="2024-01-23T14:45:49.940" Title="Can photopolymers from 3D printer be fused together like for thermo plastics?" Tags="|3d-printing|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12693" PostTypeId="2" ParentId="12673" CreationDate="2016-12-05T22:16:22.343" Score="1" Body="&lt;p&gt;The RBE2 element uses the output coordinate systems that are defined for the grid points (field CD on the GRID card).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use two different coordinate systems at the same point in your model, you can define two grid points with different output coordinate systems, and join them with a zero-length RBE2 or RBAR element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if a grid point is used in more than one rigid element, it can only be a &lt;em&gt;dependent&lt;/em&gt; grid point in &lt;em&gt;one&lt;/em&gt; of those elements. But if the rigid element connects all 6 degrees of freedom at each grid, it makes no difference which grid you choose as the independent one, so this isn't a &quot;real&quot; restriction on what you can do - it's only a feature of the way NASTRAN works internally!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-12-05T22:16:22.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12694" PostTypeId="1" CreationDate="2016-12-06T00:06:16.433" Score="1" ViewCount="57" Body="&lt;p&gt;I would like to know how the pressure of a blast is measured when using a piezoelectric gauge as well as if this is the best way to measure the reflective wave from an explosion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In better terms I guess would be maybe how does the sensor, sense, the passing pressure wave and what about it triggers it? Also I know there are other sensors out there that measures a different point of the wave and I would like to know how that differs from the pencil piezoelectric sensor.&lt;/p&gt;&#xA;" OwnerUserId="9104" LastEditorUserId="9104" LastEditDate="2016-12-07T21:27:36.047" LastActivityDate="2016-12-07T21:27:36.047" Title="Piezoelectric Pressure Gauge" Tags="|pressure|" AnswerCount="0" CommentCount="3" ClosedDate="2016-12-06T12:42:49.153" ContentLicense="CC BY-SA 3.0" />
  <row Id="12695" PostTypeId="2" ParentId="12608" CreationDate="2016-12-06T00:33:55.710" Score="2" Body="&lt;p&gt;An impulse has a flat/constant power density for all frequencies. The impulse response of a system can also be used to characterize a system, namely convolution of the input with impulse response, will yield the system response (however usually you would do this in the frequency domain, since then convolution becomes simple multiplication). The downsides of using an impulse as a way to excite a system is that the amount energy you can input into the system will be limited, since a too big impulse might damage a physical system. This might mean that you need a sensor with a higher accuracy, to measure lightly damped modes with (very) small gains, otherwise this might go unnoticed due to the resolution of the sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A step has a spectrum with a slope of minus one. So low frequency dynamics will get a higher gain than high frequency dynamics. And again a step can only supply a limited amount of energy into the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you actually mean just a sum of a couple of sinusoidal signals, then you will mainly see the system response of the system at those frequencies, but also some leakage to other frequencies, which could excite eigenmodes of the system. After these eigenmodes have died out (assuming the system is stable and sufficiently damped) then you should only see the response of the system at those frequencies. This can be useful if you are only interested in a specific frequency range or to see if the system can be modeled as a linear time independent system, because nonlinear or time variant systems will most likely also contain different frequencies in the response after the transient has died out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good options for signals are sine sweep, by applying a single &quot;sinusoidal&quot; signal whose frequency changes (relatively slowly) over time, and (white) noise. The sine sweep is similar to the sum of sinusoidal signals, but it might be harder to be used to detect non-LTI system behavior. White noise, similar to the impulse, also has a flat spectral power density, but has the advantage that it continuously adds energy to the system and keeps exciting all modes. But it would be even harder to detect non-LTI system behavior.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2016-12-06T00:33:55.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12696" PostTypeId="2" ParentId="12669" CreationDate="2016-12-06T03:36:59.793" Score="0" Body="&lt;p&gt;First just to clarify each of your equations at a point is looking at the $force$ not the $moment$. That being said, you can take the moment about any point as well. Hint: it may convenient to take the moment of the whole truss system about a point that you don't know the load at, i.e. A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma M_A = 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing so would yield the following (assuming clockwise positive):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma M_A = 33lbs * (4&quot;) + 100lbs * (4&quot; + 24&quot;) - WB * (4&quot; + 24&quot; + 7&quot;) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using your equations continue from there. For Statics problems remember your two main equations are $\Sigma M = 0$ and $\Sigma F = 0$&lt;/p&gt;&#xA;" OwnerUserId="8225" LastActivityDate="2016-12-06T03:36:59.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12697" PostTypeId="2" ParentId="10605" CreationDate="2016-12-06T03:43:16.907" Score="1" Body="&lt;p&gt;What software are you using?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're just calculating this in a spreadsheet, &quot;Flow in Open Channels&quot; by Subramanya (2009) provides a mathematical treatment of side-weirs which you may find helpful (Chapter 8.4).&lt;/p&gt;&#xA;" OwnerUserId="9107" LastActivityDate="2016-12-06T03:43:16.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12698" PostTypeId="1" AcceptedAnswerId="12701" CreationDate="2016-12-06T07:56:58.303" Score="0" ViewCount="950" Body="&lt;p&gt;A metallic cube of side 10cm, density 6.8gm/cc is floating in liquid mercury (density 13.6gm/cc), with 5cm height of cube exposed above mercury level. Water is filled over this to submerge the cube fully. What is the new height of the cube exposed above the mercury level?&lt;/p&gt;&#xA;" OwnerUserId="9110" LastEditorUserId="9" LastEditDate="2016-12-07T12:46:05.723" LastActivityDate="2016-12-07T12:46:05.723" Title="Body submerged in two liquids" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="4" ClosedDate="2016-12-30T00:08:25.127" ContentLicense="CC BY-SA 3.0" />
  <row Id="12701" PostTypeId="2" ParentId="12698" CreationDate="2016-12-06T11:19:04.407" Score="1" Body="&lt;p&gt;Question was really simple, &lt;strong&gt;We need to equate the total buoyancy force with the weight of the body&lt;/strong&gt;. I was stuck with before and after situation so was not able to solve it. Answer is the block will rise by 0.4cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total Buoyancy force for a body immersed in two fluid &#xA;= Density of water*g(S1*V1+S2*V2) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;V1 &amp;amp; V2 being volume of block in respective fluid. S1 and S2 specific gravity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this question 1-Mercury, 2-water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When water is filled over mercury, let us assume that the block will rise by a x units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buoyancy force= 1000*g(S1*(5-x)+S2(5+x))= Weight of block.&lt;/p&gt;&#xA;" OwnerUserId="9110" LastEditorUserId="9110" LastEditDate="2016-12-06T15:04:24.023" LastActivityDate="2016-12-06T15:04:24.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12702" PostTypeId="1" CreationDate="2016-12-06T11:27:28.710" Score="1" ViewCount="107" Body="&lt;p&gt;I notice bridges designed by bridge engineers in my company invariably uses two models. One model for seismic load case and another for other load cases. I was told that some bridge elements, such as the slab and the walls have to be modelled differently, using plate elements, to capture the actual stiffness. But because it is not easy to extract the output such as bending moment, these have to be modelled as beam grillage in the non-seismic model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the models are different, additional effort is required to extract and combine the outputs, before processing them for design. If you are a bridge engineer, I would like to know if you have found a simpler way to model a bridge.&lt;/p&gt;&#xA;" OwnerUserId="9114" LastActivityDate="2016-12-06T11:27:28.710" Title="What is the best way to model a beam and slab bridge?" Tags="|structural-engineering|civil-engineering|modeling|bridges|" AnswerCount="0" CommentCount="2" ClosedDate="2016-12-09T15:41:15.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="12703" PostTypeId="2" ParentId="12691" CreationDate="2016-12-06T15:00:37.790" Score="0" Body="&lt;p&gt;It will undoubtedly depend on the resins in use.  I've found several forums which suggest that isopropyl will dissolve cured residue -- but not necessarily allow bonding/welding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted, since photopolymers change their chemical composition, you can't weld them.  My strong suspicion is that the same holds for attempts at chemical dissolving, i.e. the cured material won't re-bond after a solvent destroys the molecular structure.&lt;br&gt;&#xA;While lack of evidence is not evidence of lack, the fact that manufacturers of photopolymer material and 3Dprinters which use the material don't mention or sell any post-production bonding compounds certainly suggests there's no straightforward way to do so. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2016-12-06T15:00:37.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12705" PostTypeId="1" AcceptedAnswerId="12717" CreationDate="2016-12-06T16:19:40.757" Score="7" ViewCount="18547" Body="&lt;p&gt;I have a friend who is German and he says that wire mesh is much stronger than steel bars (rebar) and is mandated in Germany for reinforcing concrete slabs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, is that true, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a typical concrete reinforcement here (in Indonesia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mjRdh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mjRdh.jpg&quot; alt=&quot;rebar&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are tied rebar beams, with rebar for floor reinforcement. Instead of the floor bars, we could instead install mesh between the beams. But what is the difference? Obviously we could buy 10.0&amp;nbsp;mm wire mesh @ 15&amp;nbsp;cm spacing, or we could install 10.0&amp;nbsp;mm bars at the same spacing (and let's say for the sake of argument that the same steel is used to make the mesh as we receive as bars). This would result in the same weight of steel per m&lt;sup&gt;3&lt;/sup&gt;, However whereas the wire mesh is welded, the rebars are only tied with wire, which I guess has trivial strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[note, let us assume, for the sake of argument that the mesh and bars are properly placed]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there, in any case, an overwhelming advantage from the welds in the mesh, or can such advantage as there is be eliminated at relatively low cost, say by increasing the numbers of bars used by say 10%.&lt;/p&gt;&#xA;" OwnerUserId="6518" LastEditorUserId="7371" LastEditDate="2017-03-05T10:42:13.163" LastActivityDate="2020-02-18T03:27:54.663" Title="Relative strength of wire mesh vs rebar for reinforcing a concrete slab" Tags="|structural-engineering|concrete|reinforced-concrete|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12707" PostTypeId="1" AcceptedAnswerId="12713" CreationDate="2016-12-06T19:03:40.323" Score="2" ViewCount="622" Body="&lt;p&gt;Why is it necessary for the moment of the forces on a rigid body in static equilibrium to be zero about any point, even if the position of that point is not on the body itself.&lt;/p&gt;&#xA;" OwnerUserId="3134" LastActivityDate="2016-12-06T22:20:15.700" Title="Why is the moment of forces on a rigid body in static equilibrium must be zero about any point, even if the position of that point is not on the body" Tags="|statics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12713" PostTypeId="2" ParentId="12707" CreationDate="2016-12-06T22:20:15.700" Score="5" Body="&lt;p&gt;Consider a body to be in equilibrium under the action of forces $F_i$, $i=1,\ldots,n$ which act at locations $r_i$, $i=1,\ldots,n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equilibrium of the body implies&#xA;$$\sum _{i=1}^n r_i \times F_i=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sum _{i=1}^n F_i=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about the moment about an arbitrary point p (that may or may not be on the body). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4fTKr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4fTKr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That can be computed as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum _{i=1}^n (r_i-p) \times F_i = \sum _{i=1}^n r_i \times F_i-p\times\sum _{i=1}^n F_i = 0-p\times 0 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the result. The moment is zero about any point, whether it is on the body or not.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-12-06T22:20:15.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12714" PostTypeId="2" ParentId="11852" CreationDate="2016-12-06T22:52:51.283" Score="1" Body="&lt;p&gt;I believe I have answered my own question after some time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Can one control latent/sensible heat changes using climate?&lt;/em&gt;&#xA;&lt;strong&gt;Yes, one can - but at a cost&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while searching for more information on the topic I ran across &lt;a href=&quot;https://www.ashrae.org/File%20Library/docLib/Events/Simbuild2016/Papers/C002.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a recent beautiful model for cooling towers that is easy to use&lt;/a&gt;. This model fairly well takes into account the behavior of water during the conditions input, while the only thing necessary is the operating conditions and a psycrometric chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two operating parameters that I was changing were return temperature (return to the cooling tower, before being cooled) and the air mass flow rate (which I have reduced to fan speed %). By changing these and finding the appropriate humidity, saturation humidity and average inlet temp &amp;amp; humidity, I can model the cooling tower water consumption. pairing the cooling tower water consumption alongside the chang in sensible heat of the air from inlet to outlet (assuming identical to water &lt;strong&gt;inlet&lt;/strong&gt;), I now have the total heat load on the cooling tower (sensible air + evaporative water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this, one can simply change the return temperature condition &amp;amp; input new humidity parameters, then run the model. The heat load is now not the same as the previous run, which is not comparable (heat load in the facilities is constant). This is corrected by adjusting the fan speed % (air mass flow rate) and running the model until I get the exact same heat load of sensible air + evaporative. I did this procedure for +/- 10 degrees F &lt;strong&gt;and I now have 3 sets of operating conditions modeled to compare.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TiODu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TiODu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing that jumps out is that at higher return temps, we have less fan usage and more water usage. At lower return temps, we have more fan usage and less water usage. &lt;strong&gt;This is the fundamental &quot;wall&quot; that you have to deal with when using evaporative cooling&lt;/strong&gt;. Now, these results make sense to me in hindsight for these reasons;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If one looks at a pscyrometric chart, it can be seen that Saturation humidity goes up exponentially with temperature. &lt;strong&gt;This means the air can hold more water, and the mass transfer driving force is increased the hotter we let the air get, whith the inverse also being true&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;This is analogous to dry cooling vs wet cooling - wet cooling uses less fan energy, but more water. Well, in this situation we're trading electric energy for water, just like wet vs dry cooling. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So in essence, if you allow the fans to run harder (and thus keep the cooling tower temperatures lower), you are maximizing the amount of dry cooling (sensible heat change), but using significantly more electrical energy. In contrast, if you slow the fans down, you use less electrical energy but increase your evaporation rate significantly (reduced sensible heat change). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some things I plan on looking into in the future when I run an actual experiment on some of these cooling towers. Most importantly, that changing the air flow rate linearly (fan speed %) changes HP usage of the fans as a cubic function, and the fact that the psycrometric chart saturation humidity increases exponentially with temperature; these things make me beleive there may be an optimization between electrical usage and water consumption here. Once I complete an experiment, I'll post the emperical results here to ensure the model conclusions match real world values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Empirical trials -&lt;/strong&gt; I had the opportunity to trial this idea at an industrial facility during the month of March. The trials were run at ~44 degrees north latitude. The setup consisted of 2 closely designed induced draft cooling towers. Both were 3 cells and designed to handle identical heat loads for the same weather conditions, although operations have changed and they were not handling even heat loads. Additionally, one tower is considerably newer than the other, and the fill media undoubtedly had an effect on the performance of the tower. To avoid this from detracting from the analysis, the older tower with less efficient fill was used as a &quot;control&quot; group. During two time periods that had average dry bulb temperatures of 33.7 F &amp;amp; 33.3 F (effectively controlling for temperature), the control tower was kept with a sump temperature set point of 65 F, while the variable tower was switched between 50 F &amp;amp; 65 F set points. The tower temperature controller was allowed to control fan speed to hit set point in both towers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cycles of concentration in each tower were controlled, as no changes to makeup water quality or conductivity set points were changed during the trials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat load on the newer, variable tower averaged 38.6 MW &amp;amp; SD of 3.2. Control tower averaged 28.6 MW &amp;amp; SD of 3.6. It should be noted that MW estimation of the control tower is a bit crude, as I had to estimate flowrate based on the number of active pumps (as opposed to a live flow transmitter). Regardless, the method I used of relative changes in each tower to ensure climate effects aren't interpreted as an actual system response should stand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final caveat is that the colder sump temperatures of 50 F were difficult for operators to maintain over long periods of time due to water freezing issues on the louvers. This required frequent fan reversals, which bottled up heat in the tower and altered the effect sump temperature. This data was disregarded and only non-transient data at the achieved set point of 50F was kept for analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On to the results: at the set point of 50 F, water consumption per MW of heat rejected was reduced by 16%, with the total fan kW consumption increasing by 168%. This falls right in line with the predictions of the model, and answers the question with a definitive &lt;strong&gt;Yes - one CAN control latent/sensible heat changes in a cooling tower by utilizing climate/weather effects&lt;/strong&gt;&lt;a href=&quot;https://i.stack.imgur.com/VEoAv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VEoAv.png&quot; alt=&quot;Results from trial&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the true economic optimum will be variable dependent on water and electrical costs. Below is a graph showing the break even point of water vs electrical costs. Under certain conditions of extreme water costs (grey water reclamation, dirty water sources that require considerable pretreatment and chemical costs etc) and cheap electricity, it may make more sense to operate at the colder temperature. It should also be noted that this curve is for ~33 F dry bulb temperatures, and this will not be recognized all year long. Additionally, I anticipate that a different dry bulb temperature would likely change the slope of this line. As we increase the heat capacitance of the incoming air per lb (more available Delta T), we would require less air for a given heat load. This would further reduce electrical consumption for a given heat load. The converse would likely also hold true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hnDEq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hnDEq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8369" LastEditorUserId="8369" LastEditDate="2018-03-25T15:39:01.770" LastActivityDate="2018-03-25T15:39:01.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12715" PostTypeId="1" CreationDate="2016-12-07T06:28:35.383" Score="2" ViewCount="213" Body="&lt;p&gt;I'm reading David Hutton's Finite Element Analysis, and recently programmed the truss analysis from scratch (in java). The figure shows the screenshot of a regular truss, with two ends fixed and a load at center. The hue and the stroke width indicate the axial stress, while the red-blue figure distinguish tension from compression. &#xA;I find the zero stress (marked with red circles) in the case of N=5, 9 (and 13, 17, etc.) interesting but confusing. Is there explanation for such pattern? or has my program mistakes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lZlDL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lZlDL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9125" LastActivityDate="2016-12-07T10:46:57.737" Title="Pattern of zero stress in regular truss structure" Tags="|structural-engineering|finite-element-method|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12716" PostTypeId="2" ParentId="12715" CreationDate="2016-12-07T10:46:57.737" Score="2" Body="&lt;p&gt;Without a better understanding of your program I can't be sure, but this looks like a modelling mistake I've already done countless times (so I recognize it from experience).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe you defined both supports as vertical &lt;strong&gt;&lt;em&gt;and&lt;/em&gt;&lt;/strong&gt; horizontal constraints. This is not the expected behavior for trusses. Instead, only one support should be RX and RY, while the other is only RY.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trusses can be broadly conceptualized as beams, with the top and lower chord representing the respective fiber of a beam and the diagonals represent the shear force. In your case, the truss should therefore behave like a simply supported beam, with compression at the top and tension at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you're getting zero stress on those spans is because they are the transition from tension to compression in your lower chord. Notice that the neighboring bars closer to the support are actually in compression. This is because a truss is meant to respond to load by deforming, which includes an outward horizontal expansion (due to the tension applied to the lower chord). By restraining horizontal displacements in both supports, you forbid this expansion and therefore generate a strong compression force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Releasing horizontal displacements in one of the supports should solve your problem.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-12-07T10:46:57.737" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12717" PostTypeId="2" ParentId="12705" CreationDate="2016-12-07T11:36:39.617" Score="9" Body="&lt;p&gt;This is going to depend a great deal on what you want to achieve with the reinforcement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not aware of any ultimate strength increase for using welded wire mesh allowed by Eurocode 2 or ACI318. I had a quick look and couldn't find anything. I'm also sceptical that welded wire mesh is mandated for all slabs in Germany. However, it is common in Germany. I suspect that is more to do with construction practice than legislation, but I haven't worked in Germany so don't know for sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other advantages for using Welded Wire Mesh:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Less labour intensive to place&lt;/li&gt;&#xA;&lt;li&gt;Lower possibility for placement errors (forgetting to place a bar - or not getting the spacing right) and bars don't move when concrete is poured.&lt;/li&gt;&#xA;&lt;li&gt;You could use smaller bars at a closer spacing to reduce cracking without increasing the labour costs significantly.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are also disadvantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Wire mesh is often produced by cold-drawing - which can decrease the ductility in comparison to standard steel bars.&lt;/li&gt;&#xA;&lt;li&gt;Lap splices can be more awkward due to the cross wires being welded in place.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://rebar.ecn.purdue.edu/wwr/Library%5CPotential%20Gains%20Through%20Welded-Wire%20Fabric%20Reinforcement.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; has a more detailed comparison of Welded Wire Mesh and conventional reinforcement&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2016-12-07T15:06:10.477" LastActivityDate="2016-12-07T15:06:10.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12718" PostTypeId="2" ParentId="12233" CreationDate="2016-12-07T13:48:21.977" Score="4" Body="&lt;p&gt;There is a misnomer in the question in that the beams are not cut to section at all, and not measured with any sort of ruler during manufacture.  The sections are rolled on a hot rolling mill where billets of hot (hot enough to have softened) steel is driven between a series of rollers that force it to the required shape.  See for example &lt;a href=&quot;https://www.youtube.com/watch?v=6xnKmt_gsLs&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=6xnKmt_gsLs&lt;/a&gt; (particularly around 2:09)&lt;/p&gt;&#xA;&lt;p&gt;(Other sections may be formed by cold rolling, but those to which the question relates - BS UB and UC sections - are hot rolled.  Cold rolling is used for relatively thin sections, such as purlins or cladding rails supporting the cladding common on industrial sheds.)&lt;/p&gt;&#xA;&lt;p&gt;The figures which designate a section are the 'serial size' (which is nominally in millimetres) and weight of the section.  The dimensions of the sections are now explicitly defined by reference to SI dimensions.  The current relevant standard is BS 4-1:2005 'Structural steel sections. Specification for hot-rolled sections'.  &lt;a href=&quot;http://shop.bsigroup.com/ProductDetail/?pid=000000000030128029&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://shop.bsigroup.com/ProductDetail/?pid=000000000030128029&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Section 3 of that standard states:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;3 Designation&lt;/p&gt;&#xA;&lt;p&gt;A section shall be designated by the serial size in millimetres (mm) and the mass per unit length in kilograms per metre (kg/m).&lt;/p&gt;&#xA;&lt;p&gt;This form of designation shall be used in any enquiry and order.&lt;/p&gt;&#xA;&lt;p&gt;EXAMPLE&lt;/p&gt;&#xA;&lt;p&gt;A beam of 920.5 mm × 420.5 mm at 388 kg/m is designated a 914 × 419 × 388 beam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However, the values of the units are not round numbers partly because they do originate as imperial values.  BS 4 is (not surprisingly, given its very low number) one of the earliest British standards.  The first version of BS 4 was BS 4:1903 &lt;a href=&quot;http://shop.bsigroup.com/ProductDetail?pid=000000000030304821&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://shop.bsigroup.com/ProductDetail?pid=000000000030304821&lt;/a&gt;  In 1903, Britain was imperial (digression - BSI originates when Sir John Wolfe-Barry instigated the Council of the Institution of Civil Engineers to form a committee to consider standardising iron and steel sections on 22 January 1901, the day Queen Victoria died - about when Britain's empire was near its peak).&lt;/p&gt;&#xA;&lt;p&gt;Sections very similar to BS 4:1903 sections are still in current use.  Some of the nominal or serial size sections match exactly.  For example, the BSB20 from BS 4:1903 was a 12&amp;quot; x 5&amp;quot; x 32 lbs/ft beam which had 0.55&amp;quot; thick flanges and 0.35&amp;quot; thick webs.  In precise conversion, 12&amp;quot;=304.8mm, 5&amp;quot;=127.0mm, 32lbs/ft=47.631...kg/m, 0.55&amp;quot;=13.97mm, 0.35&amp;quot;=8.89mm.  Now, 305 x 127 x 48 is a current section designation in BS 4-1:2005 (within table 2 in the standard) and has 14.0mm thick flanges and 9.0mm thick web.&lt;/p&gt;&#xA;&lt;p&gt;(Although even where the nominal figures match exactly, the sections are not a precise match - BSB beams had tapered flanges, early UB sections had somewhat less tapered flanges, and modern UB beams have parallel flanges.)&lt;/p&gt;&#xA;&lt;p&gt;However, if you actually put a ruler on this beam, it won't read as a round number in either imperial or SI units, because it isn't actually that size at all.  A 305 x 127 x 48 UB is actually 310.4mm deep and 125.2mm wide.  The designation is just a label, which reflects a nominal size termed the 'serial size'.  The various beams designated as 305 deep vary between 303.8mm and 312.7mm deep.&lt;/p&gt;&#xA;&lt;p&gt;This arises because you can roll all the beams from the same serial size from one set of rollers.  For example, 305 x 127 x 48 UB, 305 x 127 x 42 UB, and 305 x 127 x 37 UB can all be rolled from the same rollers, you 'simply' set the rollers slightly further apart.  You can see that from the section tables - those three sections are different depth and width, but the internal distance between flanges is the same (264.6mm) and the root radii are the same (8.9mm).  The rollers can physically be the same set for all those sizes, the rollers that form the outer faces are set further away from the inner ones and the inner ones are themselves spaced further apart, making the flanges, web and also the overall dimensions thicker and greater.&lt;/p&gt;&#xA;&lt;p&gt;That is, in summary:&lt;/p&gt;&#xA;&lt;p&gt;The 'actual' dimension is not either of the designation dimension in millimetres or the conversion of that dimension into inches.&lt;/p&gt;&#xA;&lt;p&gt;The actual dimensions are, however, defined in millimetres.&lt;/p&gt;&#xA;&lt;p&gt;Some of those metric dimensions are the same dimension as a round number in inches, rounded to the nearest 0.1mm (but not all of them).&lt;/p&gt;&#xA;&lt;p&gt;If you want to explore historical steel sizes, in addition to the BSs I've linked, there's a useful British Constructional Steelwork Association (BCSA) book - the Historical Structural Steelwork Handbook - &lt;a href=&quot;http://www.steelconstruction.info/index.php?title=Special:ImagePage&amp;amp;t=Historical+Steelwork+Handbook.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.steelconstruction.info/index.php?title=Special:ImagePage&amp;amp;t=Historical+Steelwork+Handbook.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="902" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-12-08T19:29:21.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12719" PostTypeId="1" CreationDate="2016-12-07T14:20:49.923" Score="3" ViewCount="1580" Body="&lt;p&gt;I am not a professional in engineering and was wondering what I could use to add traction to a wheel. (I have to make a small self-car for a school project) I was wondering if there was a cheap or almost free material I could use for it. The wheel is going to be made of wood, it is going to drive on ceramic tile and the car is not very heavy, maybe around a pound. I am not &lt;em&gt;necessarily&lt;/em&gt; looking for a product, just a way to increase traction.&lt;/p&gt;&#xA;" OwnerUserId="9130" LastEditorUserId="9130" LastEditDate="2016-12-08T14:12:41.513" LastActivityDate="2016-12-08T21:11:30.927" Title="How to add traction to a wheel?" Tags="|wheels|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12720" PostTypeId="1" AcceptedAnswerId="12724" CreationDate="2016-12-07T14:27:28.150" Score="0" ViewCount="138" Body="&lt;p&gt;I am looking for making a &lt;strong&gt;chip&lt;/strong&gt; or a &lt;strong&gt;microchip&lt;/strong&gt;, &lt;em&gt;which can be traced with the help of the satellites&lt;/em&gt;. Means it is merely a chip and when I tune my satellite to a particular frequency I would get to know that this particular chip is present here and so. But it should be a chip only means no battery and all i.e. &lt;em&gt;no other circuitry&lt;/em&gt;. I am thinking so if it is possible now and if not please suggest something similar to it so that I can use that in the similar way. &#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="9131" LastActivityDate="2016-12-07T17:28:52.560" Title="How can I make a GPS traceable microchip?" Tags="|electrical-engineering|telecommunication|wireless-communication|communication|" AnswerCount="2" CommentCount="3" ClosedDate="2016-12-09T14:15:44.893" ContentLicense="CC BY-SA 3.0" />
  <row Id="12721" PostTypeId="2" ParentId="12720" CreationDate="2016-12-07T16:26:58.047" Score="1" Body="&lt;p&gt;I second Pisda's point, but lets get hypothetical here:&#xA;RFID and related chip technology already uses a chip that is powered by the radio signal send to it. So in 'theory' you could use that to located the chip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HOWEVER, you are asking about orbital detection of a chip. Over more then 100 kilometers (let me Google this for you: &quot;distance to lowest satalite in orbit&quot;) comes up with a satellite at 705 km up in orbit. &#xA;If I do not count all the interference caused by: other radio sources (on the same frequency), space radiation, static created by (atmospheric) magnetic fields etc etc.&#xA;You still need to somehow send out radio waves that reach said chip. And lets keep that part more simple, use multiple coils and pick up charge from multiple frequencies.&#xA;You are still going to TRY to send a signal from such a weak power source all the way back to your satellite. I'll not try to come up with a proper calculation as this is not really my field and I hope I have already demonstrated why this question is a bit silly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I deem this endeavour unlikely with current technology unless there are something within quantum mechanics that could fix the distance problem. (I am thinking about quantum entanglement, but that field is still in early development and does not fix the direction problem. But that you could potentially fix by letting the chip locate its location with GPS signals and send the coordinates with the technology I just mentioned)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So allow me to propose something different:&#xA;A tracer element like a radioactive material. Preferably one that radiates gamma radiation in larger quantities (material with low half life) and detect said radiation.&#xA;Let me skip over the part that this creates a large health hazard :-)&#xA;and tell you that there is already a net of satellite detectors that can sniff out radioactive radiation and I'm pretty sure they can determine some specific characteristics about said source to point out your tracer.&lt;/p&gt;&#xA;" OwnerUserId="9133" LastActivityDate="2016-12-07T16:26:58.047" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12723" PostTypeId="1" AcceptedAnswerId="12736" CreationDate="2016-12-07T17:24:50.653" Score="2" ViewCount="76" Body="&lt;p&gt;According to ASME B31.8-2008 to calculate minimum thickness if a pipe under internal pressure you use:&#xA;$$t=\dfrac{PD}{2(SEW+PY)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen this formula re-written in terms of pressure of the fluid within the pipe as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p(psig)=\dfrac{2t \cdot S \cdot E}{D-2 \cdot t \cdot Y}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the reduction factor $W=1$, how do they get to the rest of the formula?&lt;/p&gt;&#xA;" OwnerUserId="9134" LastEditorUserId="1832" LastEditDate="2016-12-07T17:44:53.633" LastActivityDate="2016-12-08T10:34:28.500" Title="Max. Allowable pressure in process pipe" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12724" PostTypeId="2" ParentId="12720" CreationDate="2016-12-07T17:28:52.560" Score="4" Body="&lt;p&gt;Firstly you seem to be under the common misunderstanding that the GPS satellites somehow calculate where a GPS receiver is. They don't. The satellites are transmit only systems (ignoring the command and control radio links with their control stations) and a GPS receiver is a receive only system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As user114749 indicated RFID is the closest to what you want and even that doesn't quite fit the requirements, it still needs an antenna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 900MHz RFID tag doesn't actually transmit anything, it's not picking up anywhere near enough power to run a transmitter, it works by modulating the reflections of the readers signal bouncing off its antenna.&lt;br&gt;&#xA;A good tag can have a read range of ~20 meters when running the transmitter at 1W with a moderately directional antenna. Radio follows the inverse square rule, if you double the range you need 4 times the power (a 6 dB increase). But with RFID since the signal has to reflect back to the transmitter, you lose 6 dB getting to the tag and another 6 dB getting back to the transmitter meaning 12 dB of losses for every doubling of the range.&lt;br&gt;&#xA;The lowest possible practical orbit is around 350km (the fuel cost to maintain a stable orbit at this altitude is very high but it's possible). So when your satellite is directly over the tag you are trying to read we have log(350km/20m) / log(2) doublings of our 20 meter range. That works out as just over 14.&lt;br&gt;&#xA;Which means we need 14 x 12 dB = 168 dB more power needed at the transmitter over our 1 W RFID reader.&lt;br&gt;&#xA;So in order to pick up a tag that is directly under it we need our RFID reading satellite to broadcast with around 63,000 TW of power, about 2.5 times the total worlds power output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To locate a tag you'd need a range of several times this (add an extra 0 or two to the power number), you can't fly directly over every point on the earth, and a fleet of satellites to give reasonable coverage. But those are relatively small issues in comparison to getting it to work with a single satellite to start with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These numbers may be off by a little bit. In fact they could well be off by a lot. But unless I've gotten it wrong by a factor of a few million it's not even close to being practical.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-12-07T17:28:52.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12726" PostTypeId="1" CreationDate="2016-12-07T18:11:07.037" Score="1" ViewCount="42" Body="&lt;p&gt;As mentioned in the question above, I am working on a clinker silo, I want to know if the forces applied on the cone of the silo are exclusively coming from the weight of the material stocked or their any other source that I need to consider specifically in the emptying phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I stay available for any information required for you to answer my question.&lt;/p&gt;&#xA;" OwnerUserId="9136" LastEditorUserId="9136" LastEditDate="2016-12-07T19:03:21.820" LastActivityDate="2016-12-07T19:03:21.820" Title="Effort applied on the cement silo" Tags="|structural-engineering|civil-engineering|structural-analysis|prestressed-concrete|" AnswerCount="0" CommentCount="5" ClosedDate="2016-12-09T15:42:23.360" ContentLicense="CC BY-SA 3.0" />
  <row Id="12727" PostTypeId="1" CreationDate="2016-12-07T21:43:11.960" Score="2" ViewCount="530" Body="&lt;p&gt;Hey guys my first post here. I was planning on building a motorbike that you can also pedal like a bicycle. The idea is that you dont have to have a license if the IC is under 50cc. So the plan is I want to have a hybrid powertrain consisted of a 50cc 2stroke(7kw) IC and a 1kw electric motor. The quetion is will the torque add from the two sources or will one always pull while the other will be getting pulled struggling to keep up? Is it necessary to have the two sources connected to a differential for them to work together?&lt;/p&gt;&#xA;" OwnerUserId="9140" LastActivityDate="2016-12-09T14:56:06.410" Title="How can you sync an electric motor with an IC engine?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12730" PostTypeId="1" CreationDate="2016-12-08T03:24:47.687" Score="4" ViewCount="3977" Body="&lt;p&gt;does anyone know how does the central lock mechanism in articulated arm (magic arm) works? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vAOL1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vAOL1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magic arm usually used to hold camera, with only 1 knob, it can lock all sides. How?&lt;/p&gt;&#xA;" OwnerUserId="8563" LastActivityDate="2017-01-08T01:33:22.400" Title="How does the central lock mechanism in articulated arm (magic arm) works?" Tags="|mechanical-engineering|design|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12732" PostTypeId="1" CreationDate="2016-12-08T06:45:48.333" Score="3" ViewCount="748" Body="&lt;p&gt;Well, I'm having a hard time finding the forces through summation of moments in figure 4.14, the problem is, there's 3 hinges, which means there's 4 unknown forces. It's kinda hard to find the forces.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried researching about it but I can't find anything similar to this one.&lt;br&gt;&#xA;Do you have any advice or do you know the right process to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't need the specific answer to the image below, I just want to know how to find the 4 unknown forces.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZOhc5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZOhc5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9145" LastEditorUserId="9" LastEditDate="2016-12-09T03:04:39.360" LastActivityDate="2016-12-14T02:47:21.297" Title="Finding all members using method of joints" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12733" PostTypeId="2" ParentId="12730" CreationDate="2016-12-08T06:54:02.590" Score="0" Body="&lt;p&gt;I haven't used any, but from the image, it's clear that there are only two linked bars, so if the knob simply restricts the pin joint at the common vertex, the two links will be locked with respect to each other. The end points are ball joints and they won't be locked. &lt;/p&gt;&#xA;" OwnerUserId="8778" LastActivityDate="2016-12-08T06:54:02.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12736" PostTypeId="2" ParentId="12723" CreationDate="2016-12-08T10:34:28.500" Score="1" Body="&lt;p&gt;Getting from one form of the equation to the other involves a little bit of algebraic manipulation and factorisation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, assuming $W=1$, multiply both sides of the equation by $2(SE + PY)$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2tSE + 2tPY = PD$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, bring terms that have $P$ in them to one side and factorise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P(D-2tY) = 2tSE$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, divide by $(D-2tY)$ on both sides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \frac{2tSE}{D-2tSE}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for a given pressure, a minimum thickness is specified. So intuitively, for a given thickness, a maximum pressure should be specified. Therefore, minimum thickness $t$ can be substituted with &quot;given thickness&quot;, and simultaneously the given pressure $P$ can be substituted with &quot;maximum pressure&quot;, hence giving you the second equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very similar approach is used to obtain the first equation from the second.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2016-12-08T10:34:28.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12737" PostTypeId="1" CreationDate="2016-12-08T11:03:52.340" Score="1" ViewCount="133" Body="&lt;p&gt;In the following kinematic chain, I would like to get equation to go from&#xA;&lt;span class=&quot;math-container&quot;&gt;$F_{in}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$T_{out}$&lt;/span&gt;.&#xA;(excuse the poor drawing, nodes 3,4,5 are always aligned)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/T940J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T940J.jpg&quot; alt=&quot;Kinematic chain&quot; /&gt;&lt;/a&gt;&#xA;Here's my attempt :&lt;/p&gt;&#xA;&lt;p&gt;The torque at &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; is : &lt;span class=&quot;math-container&quot;&gt;$C_1 = F_{in} \times a$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So &lt;span class=&quot;math-container&quot;&gt;$F_2 = \frac{C_1}{b}$&lt;/span&gt; ( opposite direction of &lt;span class=&quot;math-container&quot;&gt;$F_{in}$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;But &lt;span class=&quot;math-container&quot;&gt;$F_3 = F_2 \times \cos{(\alpha_1)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Here's where I'm suspecting to go wrong :&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_5 = F_3 \times \cos{(\beta_1)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T_{out} = (F_5  \times \cos{(\theta_1)}) \times f$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;TL;DR; Can I chain force transmission property along a kinematic chain ?&lt;/p&gt;&#xA;" OwnerUserId="9149" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-12-08T11:03:52.340" Title="force transmission in kinematic chain" Tags="|torque|kinematics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12738" PostTypeId="2" ParentId="12719" CreationDate="2016-12-08T11:28:15.047" Score="3" Body="&lt;p&gt;Depends on a lot of factors, like &lt;strong&gt;what surfaces the car is going to go one, the size and weight of the car, if wheel-well dimensions are a constraint, etc...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general however, if you're trying to maximize traction on a flat driving surface, &lt;strong&gt;you'll want as wide a tire as possible&lt;/strong&gt;, maximizing contact and thus friction against the driving surface. In terms of materials, there's nothing wrong with a normal rubber compound or something comparable to what you would find on a car outside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cost is another issue. If you are truly trying to maximize the performance of your car, you'll need to spend an appropriate amount of money to get proper materials and components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extra details as to the nature of this car project would be helpful in giving a more accurate//applicable answer.&lt;/p&gt;&#xA;" OwnerUserId="9143" LastActivityDate="2016-12-08T11:28:15.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12739" PostTypeId="2" ParentId="12186" CreationDate="2016-12-08T13:26:19.323" Score="4" Body="&lt;p&gt;Build two tables. The outer table will have a hole in the middle and only hold the glass plate. There should be about 10 cm of free space between the tables. The train will be placed on the inner table:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  ================== &amp;lt;-- glass plate&#xA;   ||   ______   ||&#xA;   ||  ||    ||  ||  &amp;lt;-- train platform&#xA;   ||  ||    ||  ||&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This way people won't bump the inner table directly, and hitting the outer table will serve as an early warning. Then after a bump, you just have to move the outer table a bit to re-center it.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-12-08T13:26:19.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12740" PostTypeId="2" ParentId="10903" CreationDate="2016-12-08T17:04:54.670" Score="1" Body="&lt;p&gt;As I see it, there are a couple of ways to go about this. First is to change the existing truck bushing itself to add more damping. You can change the geometry of the bushing, the material or both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second would be adding some kind of secondary damping device to the truck. Rather than a complex fluid damping or magnetic system, you might consider something like the dampers used on washing machines which use a grease soaked compressed urethane foam between two sliding tubes. They're pretty simple to prototype and you can tune the design pretty easily. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2016-12-08T17:04:54.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12743" PostTypeId="1" CreationDate="2016-12-08T20:25:07.813" Score="1" ViewCount="509" Body="&lt;p&gt;Using Abaqus 6.12-1, I am trying to submit a restart job with an added step. I've followed exactly the instructions in section 19.6.5 of Abaqus 6.12-1 CAE user's manual, but receive this error:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A node set with the same name &quot;ASSEMBLY_SEM-2D-PART-1_Fixed Nodes&quot; defined in the original analysis has been redefined in the current analysis. Existing sets cannot be modified in a restart analysis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at the input file for job-2, I see that this is in fact true, Two Nsets used for BCs have been redefined. One is used for a displacement and the other for an encastre. I did not modify these sets in any way. I also checked the .rpy file and it didn't report any changes to sets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I intend to modify the value for the displacement in the added step. However, to isolate the problem, I have tried the following, none of which have worked.:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Not modifying the value for the displacement (propagating it).&lt;/li&gt;&#xA;&lt;li&gt;Deactivating the displacement in the added step.&lt;/li&gt;&#xA;&lt;li&gt;Adding a deactivated step in the original model and activating it upon restart.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Why does this happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, my end goal is to do this in a Python script. Modifying an input file (manually or programmatically) is not an option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can work around the problem as per &lt;a href=&quot;https://engineering.stackexchange.com/questions/12743/set-redefined-error-when-running-a-restart-job-in-abaqus-6-12-1#comment22605_12743&quot;&gt;agentp's comment&lt;/a&gt; and many helpful comments on &lt;a href=&quot;https://www.linkedin.com/groups/786907/786907-6211904027945508866&quot; rel=&quot;nofollow noreferrer&quot;&gt;this LinkedIn page&lt;/a&gt;. The problem, however, remains unsolved. I'll leave the question open for anyone with the same problem or with a solution to see.&lt;/p&gt;&#xA;" OwnerUserId="8855" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2016-12-15T19:24:08.607" Title="&quot;Set redefined&quot; Error When Running a Restart Job in Abaqus 6.12-1" Tags="|finite-element-method|software|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12744" PostTypeId="2" ParentId="12719" CreationDate="2016-12-08T21:11:30.927" Score="0" Body="&lt;p&gt;You'll need to pick a tire material that has a high coefficient of friction with respect to the surface(s) you'll be driving on. For the class of vehicle you're talking about, this generally means a fairly soft rubber compound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to a hobby store and look at the tires made for R/C model cars and trucks to get a &quot;feel&quot; for what works well on various surfaces.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-12-08T21:11:30.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12747" PostTypeId="1" CreationDate="2016-12-09T08:51:35.177" Score="-1" ViewCount="128" Body="&lt;p&gt;I would like to write software that can reliably get the &quot;relative physical time&quot;, with the highest resolution, and lowest overhead as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;relative physical time&quot;, I mean, for example, the actual amount of physical time that has passed in the physical world since the computer was turned on. It doesn't need to be relative to some external time system, calendar, or event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While there is a plethora of information about this, I think it is a topic this is misunderstood by SW people (including myself), and thus, lot's of misinformation. Hopefully an electrical or computer engineer is better informed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do computers (ex. iPhones, game consoles, personal PC's) come equipped with hardware to do this? Or, are there trade offs that need to be made: (i.e. if we want high res + low overhead, use a time step counter (ex. RDTSC), but no longer &quot;wall time&quot;).&lt;/p&gt;&#xA;" OwnerUserId="9159" LastEditorUserId="9159" LastEditDate="2016-12-12T08:04:40.117" LastActivityDate="2016-12-12T08:04:40.117" Title="Computer engineering: Physical time clock timing" Tags="|computer-engineering|computer-hardware|" AnswerCount="2" CommentCount="8" ClosedDate="2016-12-12T11:43:48.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="12749" PostTypeId="2" ParentId="12747" CreationDate="2016-12-09T10:43:46.030" Score="2" Body="&lt;p&gt;Most computers / computing devices will have a real time clock (RTC) built in. This is normally included in the processor or motherboard chip set rather than as a separate part. It will be designed such that when the system is on it uses the main system power to run but when the system is off it has a separate power pin that is connected to a small battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will keep track of time to a reasonable level of accuracy (say about 50 ppm (parts per million)) for years running off a watch battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The RTC is a low power part and so relatively slow and generally not very high resolution, when running most computer systems will run a separate faster and higher resolution clock that they initialize based on the time from the RTC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a software point of view unless you are working on smaller embedded systems then once the system is up and running the underlying OS will include a system for tracking and reporting time to applications. That is probably the best trade off between accuracy and speed that you will have available.&lt;br&gt;&#xA;In the absence of any other sources the OS time will be periodically synced in the background to the clock from the RTC. If a more trusted time source becomes available (e.g. the user, NTP, a GPS signal, the signal from a cell tower) then the system may switch to using that and update the value in the RTC.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-12-09T10:43:46.030" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12751" PostTypeId="2" ParentId="12747" CreationDate="2016-12-09T12:03:37.620" Score="1" Body="&lt;p&gt;There are many ways to track real time.  Most general purpose computers include a real time clock, with OS calls to get the time.  Small embedded processors can have RTC (real time clock) hardware built-in.  RTC chips are also available that can be connected to a processor via IIC and the like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also quite possible to perform the RTC function in firmware.  It is often useful for other reasons to have a periodic interrupt, like every 1 ms (1 kHz rate).  This can easily be the basis for a real time clock.  The accuracy is as good as the oscillator that the 1 ms events are derived from.  This can be a crystal, so down to 20 PPM or so is cheaply available.  Down to about 5 PPM can be possible with selected (extra money) crystals.  Lower requires some sort of stabilization, external oscillator module, or the like.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-12-09T12:03:37.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12754" PostTypeId="2" ParentId="12727" CreationDate="2016-12-09T12:43:13.947" Score="2" Body="&lt;p&gt;It might make more sense not to have the IC engine delivering power directly to the wheels at all. Rather just use it to generate electricity to power the motor and charge the batteries. This approach has  number of advantages.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You eliminate quite a bit of mechanical complexity in the transmission as you don't need a variable ratio gearbox or sophisticated  clutch and avoid coupling two different motors into the same drive train. &lt;/li&gt;&#xA;&lt;li&gt;The IC engine can either be running at its optimum RPM or off, improving efficiency. &lt;/li&gt;&#xA;&lt;li&gt;You will get better low speed torque which is good for a moderately heavy bicycle. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this case you might want to look at a less powerful IC engine and a more powerful electric motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example you might have various different conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Constant speed on the flat : ICE running powering the motor and charging the batteries EM providing motive power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Uphill/accelerating : ICE engine powering motor with extra power provided by the batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) downhill : ICE off and EM in regenerative mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) electric only : ICE off and EM in battery mode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This setup also give you a bit more flexibility as to how you package and place the main components. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-12-09T14:56:06.410" LastActivityDate="2016-12-09T14:56:06.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12755" PostTypeId="1" AcceptedAnswerId="12913" CreationDate="2016-12-09T12:54:29.687" Score="4" ViewCount="1263" Body="&lt;p&gt;In regulation, if we represent a system with a state space model, we end with three matrices A, B and C (without taking into account feed-forward part) that define a specific part of the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{x}=Ax+Bu$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y=Cx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it also possible to define the transfer function $G=\frac{N}{D}=C(sI-A)^{-1}B$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to find the poles, we search when the denominator ($D$) is equal 0 and for the zeroes when the numerator ($N$) is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, if we want to define which part of the state space model influence whether the poles or the zeroes. I would say by looking at the previous equation that $A$ is the only one to influence the poles and $B,C$ act on the zeroes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the corrections made to my exam gave a different answer: $C$ influences the poles and all the matrices the zeroes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand, why is that the case?&lt;/p&gt;&#xA;" OwnerUserId="1374" LastEditorUserId="368" LastEditDate="2016-12-09T18:53:30.553" LastActivityDate="2016-12-20T16:13:37.377" Title="Which equations of a state space model influence the poles and zeroes?" Tags="|control-engineering|transfer-function|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12756" PostTypeId="1" AcceptedAnswerId="12758" CreationDate="2016-12-09T14:28:38.143" Score="2" ViewCount="209" Body="&lt;p&gt;GPS receivers are able to operate at signal level of -150 dbm or less, but mobile phones need far more than -115 dbm level signal. Why GPS receivers are so much more sensitive than GSM or WiFi receivers? See these receivers &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.quectel.com/product/prodetail.aspx?id=77&quot; rel=&quot;nofollow noreferrer&quot;&gt;GNSS L86&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.quectel.com/product/prodetail.aspx?id=93&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wi-Fi FC10&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6614" LastEditorUserId="1586" LastEditDate="2016-12-09T18:00:19.047" LastActivityDate="2016-12-09T18:00:19.047" Title="Why GPS receivers are more sensitive than GSM or WiFi receivers?" Tags="|electrical-engineering|wireless-communication|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12757" PostTypeId="2" ParentId="12756" CreationDate="2016-12-09T14:35:12.053" Score="0" Body="&lt;p&gt;Signal bandwidth, for one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data that a GPS receiver gets from the satellites is far less than a streaming phone conversation over the same time.  This allows the RF bandwidth to be used to get higher signal to noise ratio.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-12-09T14:35:12.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12758" PostTypeId="2" ParentId="12756" CreationDate="2016-12-09T15:47:27.427" Score="2" Body="&lt;p&gt;There are a few factors at play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly as Olin indicated the signal bandwidth and data rates are different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fundamental rule of information theory that the maximum amount of data you can pass through a data channel depends upon the channel bandwidth and the signal to noise ratio. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Shannon-Hartley theory&lt;/a&gt; for details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if you have a lower data rate or a wider channel you can cope with a higher signal to noise ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The GPS navigation message sends data at 50 bits per second over a channel that is around 20MHz wide. WiFi 802.11n tries to get 72Mb/s or around 1 million times more data over a channel that is 20MHz wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would indicate that GPS can cope with significantly lower signal levels than WiFi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that only tells you what is theoretically possible, not how to achieve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A GPS signal is very predictable. Each satellite is broadcasting a pesudo random data sequence which repeats every millisecond, the data is then added on top of that every 20ms. The GPS reciever knows what this pattern is and looks for it in the background noise. It does this by making a guess at the timing of the signal and then moving this expected timing around until the correlation between the expected and recieved signals is maximised. By doing this the GPS reciever can effectivly find a signal that is weaker than the random background noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has the added advantage that since the signal is random it looks like noise unless you are looking for it. This means that all the satellites can broadcast on the same frequency at the same time without interfering with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage of this system is that the primary information of interest to a GPS reciever is the exact timing of the recieved signals. By maximising the  correlation between the expected and actual signals you get a very robust and accurate way of determining the exact signal timing.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-12-09T15:47:27.427" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12760" PostTypeId="5" CreationDate="2016-12-09T16:46:41.167" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2016-12-09T16:46:41.167" LastActivityDate="2016-12-09T16:46:41.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12761" PostTypeId="4" CreationDate="2016-12-09T16:46:41.167" Score="0" Body="IoT stands for Internet of Things, and refers to all manner of devices connected to the internet, and machine-to-machine communication in particular. For Internet of Things related questions that are not directly relevant to Engineering, consider asking at iot.stackexchange.com" OwnerUserId="255" LastEditorUserId="11227" LastEditDate="2017-05-08T07:59:54.573" LastActivityDate="2017-05-08T07:59:54.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12762" PostTypeId="1" AcceptedAnswerId="12780" CreationDate="2016-12-09T17:06:08.103" Score="4" ViewCount="512" Body="&lt;p&gt;I just finished my first controls course (on linear control systems) and we learned about 3 types of controllers: PID, State Space, and Frequency Response/Loopshaping. From what I understand, PID has the advantages of being really easy and it can be applied without knowing the dynamics of the system. State space provides really good responses. But loopshaping seemed very guess and check when we were doing it and kind of arbitrary. In addition, we got to actually implement the first two on a multirotor for our labs and didn't have time to do the last beyond homework. So when would it be advantageous to use loopshaping?&lt;/p&gt;&#xA;" OwnerUserId="9073" LastActivityDate="2020-05-10T07:14:02.547" Title="What benefits does loopshaping have over other control design methods?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12763" PostTypeId="5" CreationDate="2016-12-09T18:49:08.457" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-12-09T18:49:08.457" LastActivityDate="2016-12-09T18:49:08.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12764" PostTypeId="4" CreationDate="2016-12-09T18:49:08.457" Score="0" Body="questions about rules and regulations in the political sense; not to be used for questions about voltage regulation" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-12-09T18:49:08.457" LastActivityDate="2016-12-09T18:49:08.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12769" PostTypeId="1" AcceptedAnswerId="12810" CreationDate="2016-12-10T17:40:33.553" Score="2" ViewCount="352" Body="&lt;p&gt;Picture shows the electrical system. Analytically determine and draw system response $u_2(t)$ for the step input shown (use Laplace transform) (tp&gt;&gt;0).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fWTQe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fWTQe.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I know to solve RC system transfer function and get G(s)=1/(RCs+1), but I don't understand how to do this problem, we have never done it in class and professor often asks this question. If anyone could help me out, thank you!&lt;/p&gt;&#xA;" OwnerUserId="9175" LastEditorUserId="1627" LastEditDate="2016-12-12T17:35:54.160" LastActivityDate="2016-12-12T17:35:54.160" Title="Analytically determine and draw the system response to a piecewise step input for an electrical system" Tags="|electrical-engineering|control-engineering|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12770" PostTypeId="1" CreationDate="2016-12-10T20:28:15.327" Score="0" ViewCount="538" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VnM8y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VnM8y.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial strength for balsa wood at different densities&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand are the subscripts under the sigmas (strenghths.)&#xA;How do I know where the compression (crushing) and tensile strengths are? I'd like to know what represents what. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="9176" LastEditorUserId="9176" LastEditDate="2016-12-10T21:36:03.460" LastActivityDate="2016-12-11T05:48:11.400" Title="What do these sigma subscripts represent?" Tags="|mechanical-engineering|civil-engineering|modeling|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12772" PostTypeId="1" CreationDate="2016-12-10T23:00:19.183" Score="6" ViewCount="3265" Body="&lt;p&gt;I'm casting a 4m long triangular-profile reinforced gravity retaining wall against soil movement due to a 1 metre jump in ground level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lateral forces and moments on the formwork have been easy to handle, but I'm trying to find a good approach to the vertical uplift hydrostatic force on small-scale/DIY formwork while concrete is poured, which is a general problem for which I can't seem to find a good solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Details -&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using a very fluid self-consolidating (SCC) mix since access for compaction would be hard once poured (slump-flow ~ 550mm, full spec available if needed). The form has one vertical face and one face with a 2:1 slope; it's this second face that creates uplift forces. The actual height of the visible wall above ground is about 92 - 100 cm but I've dug down to about 1.2 - 1.4m to allow a &quot;heel&quot; and more mass under the &quot;soil&quot; side, and also because then it can be braced against slipping by the other concrete under the pathway itself.  The wall has about 0.5 m&lt;sup&gt;2&lt;/sup&gt; cross-section area. I've gone for an &quot;over engineered&quot; approach - it's also being laterally braced at both ends by perpendicular concrete works, and will have 10mm (A393) welded mesh and 12+16mm rebar and &quot;L&quot; bends added along the length and at the corners. The form is lined with polythene for easy removal of the formwork. I'm indifferent to surface finish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Cx1O.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Cx1O.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confident in the actual formwork - plywood backed by C16 2x4's (50x100 mm), which are in turn backed by 3x6's (70x150 mm) held together with heavy duty structural screws, 12mm s/steel threaded bar between the formwork timbers of the two faces, all the entire form ultimately being braced laterally against solid ground on both sides nearby. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concern is countering the vertical uplift forces from the hydrostatic pressure of the fluid concrete when it's poured, which will be about 4 - 5 tons equivalent (40 - 50kN), ie the weight of the &quot;missing&quot; concrete above the slanted face of the form. Basically, I want it not to be lifted off the ground by the fluid concrete. I've thought of several possible approaches to offset this force, but the reality is that I'm just not sure which is best or most foolproof, or if there are standard solutions used in civil engineering that I'm unaware of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible solutions + sketches I've come up with - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zXQuy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zXQuy.png&quot; alt=&quot;Possible solutions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Reduce the footprint and hence uplift, by making the wall slimmer.&lt;/strong&gt; Perhaps with strong lateral restraint, rebar, and concrete under an adjacent path, the usual 50% rule of thumb is excessive and I can reduce the width of the wall to ~ 40cm (and therefore the upward force by ~ 30-50%)?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Allow the concrete to fill the full vertical space of the wall's footprint, removing the excess pour when set.&lt;/strong&gt; Precisely zero hydrostatic pressure guaranteed as all concrete is at the same top level; the extra volume can be cast with plywood dividers so it comes away easily in sensible sized slices.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Cast in 2 parts, using a smaller 1st cast steel-linked to the form, as a deadweight for a larger 2nd cast.&lt;/strong&gt; Cast the bottom 1/4 only, but with steel rod embedded to link it to the form, and vertical rebar for the join. Then cast the top 3/4 about 48 hrs later.  The bottom 1/4, attached to the form itself, acts as a deadweight to hold down the form against upward pressure when casting the narrower but taller top 3/4. Extra weight still needed but only 1/4 as much; hopefully the vertical rebar linking the pours will ensure long term integrity.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Add a 'false floor' in and fixed to the sides of the formwork so the weight rests on the formwork not the ground; remove and underfill this space beneath the wall once the 1st pour is set.&lt;/strong&gt; 5 tons of concrete along a 4m length can easily be supported on 80 side-by-side 2x4 (50x100mm) timber (weight per 50mm 'slice' about 62kg which is small for a 0.8m clear span). Because the 'floor' is itself supported by the formwork, there is no net uplift force - it's replaced by vertical tension in the formwork which is much easier to handle. After the form has set, remove alternate 50cm sections of these 2x4 timbers, and pour concrete into the void below, to properly support the wall. When that's set, remove the other half of the timbers and pour under the rest of the wall. Needs 3 pours but again, guaranteed to work since there is zero net uplift force and the weight spread across many 50mm slices is small for each slice. Also simple to set up.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Brute force.&lt;/strong&gt; The simple solution. Add about 5 tons of weight. For stability it's added to the heavy members at the base of the form, not on top. problem is that 5 tons is a heck of a lot to move.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How can  I easiest ensure the formwork doesn't rise at the base and the pour escape, when it's poured?&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2016-12-11T07:49:01.657" LastActivityDate="2021-02-17T03:57:18.320" Title="Concrete formwork - best approach to counter uplift forces, for a freestanding pour?" Tags="|structural-engineering|civil-engineering|concrete|retaining-wall|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12773" PostTypeId="1" AcceptedAnswerId="12783" CreationDate="2016-12-11T01:06:45.827" Score="2" ViewCount="188" Body="&lt;p&gt;It seems that most bridge are composed of &lt;a href=&quot;http://www.nvfnorden.org/library/Files/Utskott-och-tema/Broar/Conference-2014/05_NVF%20Iceland%202014.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;concrete&lt;/a&gt;, e.g in Latvia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nabm8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nabm8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, concrete encompases a lot of bridge types:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most concretes used are lime-based concretes such as Portland cement&#xA;  concrete or concretes made with other hydraulic cements, such as&#xA;  ciment fondu. However, asphalt concrete, which is frequently used for&#xA;  road surfaces, is also a type of concrete, where the cement material&#xA;  is bitumen, and polymer concretes are sometimes used where the&#xA;  cementing material is a polymer.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which concrete material do modern civil engineers prefer when designing  bridges over motrways less than 30 meters above the underpass?&lt;/p&gt;&#xA;" OwnerUserId="2022" LastEditorUserId="6790" LastEditDate="2018-05-07T10:46:15.973" LastActivityDate="2018-05-07T10:46:15.973" Title="What is the preferred composition for modern concrete bridges?" Tags="|civil-engineering|bridges|concrete|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12774" PostTypeId="2" ParentId="12770" CreationDate="2016-12-11T05:48:11.400" Score="3" Body="&lt;p&gt;Wood is anisotropic. Its mechanical properties are different in different directions relative to the grain of the wood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The R, L and T suffixes probably refer to the radial, tangential, and longitudinal directions (defined relative to the tree, when the wood was still growing). See the diagram in &lt;a href=&quot;http://www.fpl.fs.fed.us/documnts/fplgtr/fplgtr113/ch04.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.fpl.fs.fed.us/documnts/fplgtr/fplgtr113/ch04.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea what the other subscripts might mean. I could guess some of them, but if you are making any serious use of this data, guessing is &lt;em&gt;always&lt;/em&gt; the wrong option!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-12-11T05:48:11.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12775" PostTypeId="1" CreationDate="2016-12-11T05:49:15.490" Score="1" ViewCount="47" Body="&lt;p&gt;I'm building a self balancing robot and would like to know how the wheel diameter and the torque provided by DC motors, would affect the overall stability and response of the robot. The balancing bot is based on the principle of an inverted pendulum.&lt;/p&gt;&#xA;" OwnerUserId="9181" LastActivityDate="2016-12-11T05:49:15.490" Title="What will be the effect of torque and wheel diameter on the stability of a robot?" Tags="|design|stability|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12776" PostTypeId="1" CreationDate="2016-12-11T11:44:12.293" Score="1" ViewCount="2048" Body="&lt;p&gt;Please forgive me if this isn't the right place for this question but I'm trying to understand the difference between AT (Automatic Transmission) and AMT (Automated Manual Transmission) as far as the driving experience is concerned.&lt;br&gt;&#xA;Is there any difference and if yes, what are these differences ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you trying move forward in a MT (Manual Transmission) car, when the car is parked uphill, then you will need to use half-clutch to prevent it from rolling backwards. I've heard that such a problem doesn't exist in an AT but I have no idea about AMT. This is just one aspect of the driving experience I'm trying to learn about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's any aspect where they do differ, please include them in your answer.&#xA;I'm not interested in the mechanical differences though.&lt;/p&gt;&#xA;" OwnerUserId="9183" LastActivityDate="2016-12-11T17:07:06.500" Title="Difference between AT and AMT transmissions as far driving experience is concerned" Tags="|automotive-engineering|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12777" PostTypeId="2" ParentId="12776" CreationDate="2016-12-11T11:53:09.677" Score="0" Body="&lt;p&gt;Well speeds. In a manual car you can go from 2 to 5 and keep performing, were as in a automatic you need to follow the numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And not all manual cars can jump gears some cars stops when you skip gears. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a manual cause you can accelerate and you'll be at the top of the hill safe. Half clutch is for people who want to go uphill relaxed. In my neighborhood you won't find people driving slow when they go uphill. We all accelerate &lt;/p&gt;&#xA;" OwnerUserId="9102" LastActivityDate="2016-12-11T11:53:09.677" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12779" PostTypeId="2" ParentId="12776" CreationDate="2016-12-11T13:09:33.113" Score="1" Body="&lt;p&gt;With manual transmissions every time the gears are changed, while the vehicle is in motion, the driver must engage and disengaged the clutch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &lt;a href=&quot;https://en.wikipedia.org/wiki/Automatic_transmission&quot; rel=&quot;nofollow noreferrer&quot;&gt;automatic transmissions&lt;/a&gt; the car automatically changes gears, when a vehicle is in motion, based on the torque requirements while the vehicle is moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Semi-automatic_transmission&quot; rel=&quot;nofollow noreferrer&quot;&gt;Automated manual transmission&lt;/a&gt; (also known as semi automatic transmissions, clutchless manual transmission &amp;amp; flappy paddle gear shift) allow gears to be changed without the driver activating the clutch. This is achieved using sensors, actuators, pneumatics to change the gears; making the clutch pedal redundant because the clutch is activated by electronic equipment.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-12-11T13:09:33.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12780" PostTypeId="2" ParentId="12762" CreationDate="2016-12-11T13:46:49.487" Score="2" Body="&lt;p&gt;When designing a controller the first priority should always be stability and then performance/robustness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know what method you used for tuning a PID controller, but I would suspect that it might not be easy to proof stability. Increasing performance would then probably involve increasing the gains. And it is probably hard to make a PID controller robust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For state space controller design you can look at the closed loop dynamics (matrix) and check whether it is stable (Hurwitz for continues time and Schur for discrete time). For performance you can try to place the closed loop poles as far as possible into the left half plane. I haven't looked at it, but there will probably be methods for robustness in state space. For disturbance rejection you can find a state estimator using a Kalman filter (by solving a Riccati equation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For loopshaping you can proof stability using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Nyquist_plot&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nyquist plot&lt;/a&gt; using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nyquist_stability_criterion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nyquist stability criterion&lt;/a&gt;. For performance you can just increase your bandwidth (sometimes defined at the point where the open loop crosses the 0 dB). Gain and phase margin can be used to see how stable/robust your closed loop system is. Closely related to this (but not often mentioned) is the modulus margin, which is basically the shortest distance to the minus one point in the Nyquist plot. This inversely relates to the highest peak of the sensitivity function of the closed loop system, which is a measure of the worst disturbance rejection. There are also other methods for robustness, such as H infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a PID controller might be the easiest, but it is much less advanced. The advantage of state space control is that it can also be applied to MIMO systems, which is harder to do for loopshaping. But the advantage of loopshaping is that you can just measure a frequency response function and design a controller for that directly, while for state space you first have to fit a model onto it first. Also if your system is continues, but also has delays it is harder to capture that into a state space model, but for loopshaping this does not change anything.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastEditorUserId="1765" LastEditDate="2020-05-10T07:14:02.547" LastActivityDate="2020-05-10T07:14:02.547" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="12783" PostTypeId="2" ParentId="12773" CreationDate="2016-12-11T15:07:33.733" Score="4" Body="&lt;p&gt;In structural situations, &lt;a href=&quot;https://en.wikipedia.org/wiki/Concrete&quot; rel=&quot;nofollow noreferrer&quot;&gt;portland cement concrete&lt;/a&gt; is what is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The properties of &lt;a href=&quot;https://en.wikipedia.org/wiki/Asphalt_concrete&quot; rel=&quot;nofollow noreferrer&quot;&gt;asphalt&lt;/a&gt; change with temperature and it never really &quot;sets&quot;. It is a viscous liquid that cannot support long-term loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There really shouldn't be any confusion as to what is being described in that chart as &quot;concrete&quot;.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2016-12-11T15:07:33.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12784" PostTypeId="2" ParentId="12772" CreationDate="2016-12-11T15:35:56.740" Score="2" Body="&lt;h2&gt;Solutions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Yes, this is a very real problem. In &quot;professional&quot; construction I have seen similar situations handled by either:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pouring in layers&lt;/li&gt;&#xA;&lt;li&gt;Placing the bucket of construction equipment on the form&lt;/li&gt;&#xA;&lt;li&gt;Tying (wire) the form down to something embedded in the ground (posts, columns, other concrete). &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As an example, when we specify pouring concrete (or flowable fill) around a pipe culvert, we typically specify pouring layers of 6&quot;-12&quot; at a time along with holding the pipe down.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Personal Use&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I would say for smaller scales, a combination of methods would be easiest. I would try to pour in thin layers if your concrete supplier can handle it. Also place whatever you can on the form to help weigh it down (concrete blocks, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To pour in layers, you shouldn't have to wait 48hrs. The concrete should set enough sooner than that, but you still may not be able to do multiple pours in one day. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;A Different Mix?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;This might be easier if you use a mix of concrete that isn't so fluid (not SCC). You will have to vibrate the concrete and you will likely have more honey combing on the sloped side, but that side will be buried anyway.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Final Words&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;You are thinking about the right concerns, but there isn't really a simple solution. You may need to combine as many of your methods as you can reasonably do to be safe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No matter what you do, be ready to stop pouring the instant that you see any movement in the form! You don't want to have to shovel a bunch of concrete by hand!&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2016-12-11T15:35:56.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12786" PostTypeId="2" ParentId="12776" CreationDate="2016-12-11T17:07:06.500" Score="0" Body="&lt;p&gt;AMT can cover a few different systems. Modern versions typically use a dual clutch system which pre-selects the next gear. The actual driving experience will depend on the control system and it's not uncommon to have selectable modes which work in concert with engine mapping, traction control, stability control ABS etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases they will still have a manual clutch, either as a hand paddle or foot pedal and may also have a neutral button as well as an automated anti-stall system. The clutch is not required for normal gear changes but may be used for pulling away from a standstill. Equally AMTs can emulate traditional type automatic transmissions and just use throttle and brake with no driver input to gear change or clutch operation at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, there is quite a lot of technical information on this subject in the context of F1 gear boxes. Until recently F1 cars had a dual paddle clutch system with one paddle used to blend in the clutch from a preset bite point  and one which was more of an on/off switch with respect to the initial preset bite point. The idea being to improve reaction time to the start lights as the first one is was released immediately in reaction to the lights and the second was released gradually to give fine control of torque delivery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary a modern semi-automatic gearbox has quite a lot of potential parameters available which can be adjusted to affect the driving experience which is determined by a combination of user settings and the software which controls the whole system. So in a sophisticated version you might have settings on a spectrum between convenience and ease of driving and maximum performance with a range of settings in between as well as modes for specific conditions and to maximise fuel economy. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-11T17:07:06.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12787" PostTypeId="1" CreationDate="2016-12-11T17:15:00.127" Score="1" ViewCount="44" Body="&lt;p&gt;I implemented the FOC (Field Oriented Control) algorithm to control a BLDC motor at low speed. A phenomena I observed is, that at different electrical angles, the torque producing current breaks down and the field producing current increases (mostly around +-90°). I use a simple PI controller and no cross coupling compensation. If I keep this angle, the field producing current stays then constant at a nonzero value. Does anyone know, what could producing this phenomena? Maybe someone of you already implemented a FOC algorithm for a BLDC motor and also noticed this problem. &lt;/p&gt;&#xA;" OwnerUserId="8784" LastEditorUserId="7083" LastEditDate="2016-12-16T01:18:09.057" LastActivityDate="2016-12-16T01:18:09.057" Title="FOC current ripples with angle dependencies" Tags="|control-engineering|motors|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12788" PostTypeId="1" AcceptedAnswerId="12790" CreationDate="2016-12-11T17:17:20.580" Score="-1" ViewCount="154" Body="&lt;p&gt;I'll soon get my engineering diploma (~MSc, from a &quot;renowned-but-not-in-the-top-3&quot; so-called french &lt;em&gt;grande école&lt;/em&gt;). The degree itself is not specialized, but I mainly studied mechanical and industrial engineering.&lt;br/&gt;&#xA;I aim to work in the industry – or, in other words, I don't want to make an academic career... even if I am still not sure, what position I will look for (R&amp;amp;D, project management, or maybe even entrepreneurship). I'd be ok to work for some years (4-12 years) abroad (Europe of further), but I think that I'd like to come back to France at some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm now wondering if doing a PhD in France (i.e. during 3 years) would be worth it, regarding my estimated career plan.&lt;br/&gt;&#xA;Question: What would then be the &lt;strong&gt;pros and cons  I should consider&lt;/strong&gt; before choosing to do (or not) a PhD?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;This question have been asked on Academia.SE. According to comments/advices there, I do ask it again here.&lt;br/&gt; I wanted to add these tags, but I don't have enough rep on this site to do it: &lt;code&gt;job&lt;/code&gt;, &lt;code&gt;career&lt;/code&gt;, &lt;code&gt;phd&lt;/code&gt;.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="9187" LastActivityDate="2016-12-11T19:22:12.777" Title="Young engineer wanting to work in the industry: is PhD worth it?" Tags="|mechanical-engineering|industrial-engineering|" AnswerCount="1" CommentCount="4" ClosedDate="2016-12-11T23:45:14.403" ContentLicense="CC BY-SA 3.0" />
  <row Id="12789" PostTypeId="1" CreationDate="2016-12-11T17:45:29.333" Score="1" ViewCount="487" Body="&lt;p&gt;I am planning a mechanical 40% keyboard build and are coincidentally on the home stretch of a homemade CNC project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing the CNC needs to do for the keyboard project is to drill 7*48 holes. So what I need to do now is layout those holes in SVG. Therein lyes the question. What resolution should I use for the SVG. I want to space the center of the keyboard switches 19mm apart. An online pixel to mm converter suggested that 72 pixels is exactly 19.05mm (which actually is what Cherry MX says should be their spacing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I do understand that this really doesn't matter, but I am curious as I am new on CNCs and was suspecting that there is a number that will &quot;just work&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="9188" LastEditorUserId="9188" LastEditDate="2016-12-11T20:07:54.520" LastActivityDate="2016-12-12T20:29:50.383" Title="What resolution (pixels/mm) should I use in SVGs for CNC" Tags="|cnc|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12790" PostTypeId="2" ParentId="12788" CreationDate="2016-12-11T18:23:14.143" Score="0" Body="&lt;p&gt;A PhD is specifically geared towards academic research and so if you are working in a more 'pure' field like maths or physics it is certainly a good thing. However engineering is not by any means a pure academic field and so there is a reasonable argument that industrial experience is at least as valuable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally PhD research tends to be quiet niche and you may be limited in what you can research by the facilities of your chosen institution and the interests of the available academic supervisors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are unsure it may not be bad advice to look at working industry for a year or so and then evaluating your options. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing about engineering is that it is always very useful to have some experience of getting your hands dirty at actually making things. Ultimately engineering is about pragmatic problem solving, something which academic qualifications are not necessarily all that great at teaching.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a potential problem that if you have a PhD you end up overqualified for junior positions but don't have the specific knowledge and experience required for more senior ones, especially if you don't have a very strong interest in a particular field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also an argument that a year spent working in a foundry, fabrication or machine shop will give you an insight into the real world practicalities of engineering which you will not get elsewhere. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-12-11T19:22:12.777" LastActivityDate="2016-12-11T19:22:12.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12791" PostTypeId="1" AcceptedAnswerId="12797" CreationDate="2016-12-11T18:28:36.543" Score="0" ViewCount="39" Body="&lt;p&gt;Can I inject plastic in a RTV mould?&#xA;Will the interior of the mould get deformed by pressure?&lt;/p&gt;&#xA;" OwnerUserId="9086" LastActivityDate="2016-12-12T00:06:54.537" Title="Can I inject plastic in a RTV mould?" Tags="|pressure|plastic|deformation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12792" PostTypeId="1" CreationDate="2016-12-08T16:41:08.923" Score="6" ViewCount="18813" Body="&lt;p&gt;I am designing gears in FreeCad, and printing then on a FlashForge Finder. I want to know if the pitch diameter is the same as the distance between the centers of the two gears. Both gears are external spur gears.&lt;/p&gt;&#xA;" OwnerDisplayName="Avagadro's Number" LastEditorUserId="5899" LastEditDate="2016-12-12T00:23:27.510" LastActivityDate="2016-12-14T23:17:44.773" Title="Is &quot;pitch diameter&quot; the same as the distance between the centers?" Tags="|gears|3d-printing|cad|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12793" PostTypeId="2" ParentId="12792" CreationDate="2016-12-11T18:53:16.550" Score="0" Body="&lt;p&gt;Yes they should, in theory, and for most practical purposes  be the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pitch circle diameter is the effective radius of a gear in terms of torque, although for some tooth profiles this may be a mean value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice gear design often uses the module system which defines gear ratios by number of teeth for a given tooth profile and ensures that you always have an integer number of teeth and eliminates pi, making calculations more convenient. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-11T18:53:16.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12794" PostTypeId="2" ParentId="12732" CreationDate="2016-12-11T19:29:31.400" Score="-1" Body="&lt;p&gt;What's going on here is that you have a what looks like&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Statically_indeterminate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hyperstatic structure&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Statically_indeterminate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Statically indeterminate&lt;/a&gt; structure, but isn't. Since you have a hinge in the middle, you get and extra equation. Cut your structure at the hinge and solve for your reactions. &lt;a href=&quot;https://en.wikipedia.org/wiki/Statically_indeterminate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is an example video&lt;/a&gt; from youtube. Start by the left hand side triangular truss and make your way right.&lt;/p&gt;&#xA;" OwnerUserId="5691" LastEditorUserId="5691" LastEditDate="2016-12-11T20:05:59.743" LastActivityDate="2016-12-11T20:05:59.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12795" PostTypeId="2" ParentId="12792" CreationDate="2016-12-11T21:24:28.100" Score="6" Body="&lt;p&gt;Pitch diameter is the diameter of the pitch circle described by the mid point of the length of the teeth around the gear, as shown in this diagram:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ti9uH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ti9uH.jpg&quot; alt=&quot;Pitch diameter&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The pitch circle defines the point where the teeth of two gears meet:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6ZVYk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6ZVYk.png&quot; alt=&quot;The pitch circles of two spur gears&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let's say you have two gears, each with a respective pitch circle diameter of &lt;span class=&quot;math-container&quot;&gt;$d_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$d_2$&lt;/span&gt;. The distance between the two gear centers, &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;, is given by,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$C = \frac{d_1 + d_2} { 2 }$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If the gears are identical (&lt;span class=&quot;math-container&quot;&gt;$d_1 = d_2$&lt;/span&gt;), then the pitch diameter is, indeed, the same as the distance between the two gear centers, &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;. Otherwise, is it not.&lt;/p&gt;&#xA;&lt;p&gt;Another method of calculating the gear center distance is,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$C = \frac{N_1 + N_2}{2P_d}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$P_d$&lt;/span&gt; is the diametrical pitch, and &lt;span class=&quot;math-container&quot;&gt;$N_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$N_2$&lt;/span&gt; are the number of teeth of the respective gears. From &lt;a href=&quot;http://www.engineersedge.com/gear_formula.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gear design equations and formula&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;You may find the following useful - from Wikipedia - &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_gear_nomenclature#Standard_pitch_diameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Standard Pitch Diameter&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The standard reference pitch diameter is the diameter of the standard pitch circle. In spur and helical gears, unless otherwise specified, the standard pitch diameter is related to the number of teeth and the standard transverse pitch. The diameter can be roughly estimated by taking the average of the diameter measuring the tips of the gear teeth and the base of the gear teeth.&lt;/p&gt;&#xA;&lt;p&gt;The pitch diameter is useful in determining the spacing between gear centers because proper spacing of gears implies tangent pitch circles. The pitch diameters of two gears may be used to calculate the gear ratio in the same way the number of teeth is used.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d = \frac{N}{ P_d } = \frac {pN}{\pi}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is the total number of teeth, &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the circular pitch, and &lt;span class=&quot;math-container&quot;&gt;$P_d$&lt;/span&gt; is the diametrical pitch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="5899" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-12-14T23:17:44.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12797" PostTypeId="2" ParentId="12791" CreationDate="2016-12-11T21:40:38.250" Score="0" Body="&lt;p&gt;RVT silicone is commonly rated by Shore Hardness. some grades are hard and heat resistant enough to be used for centrifugal casting of low temperature metal alloys (aka white metal). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in practice for plastic casting they are best suited to 'cold cast' resins such as Polyester, Epoxy and especially Polyurethane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally injection moulding is specified for high volume production using rigid steel dies and for low to medium volume production filled resins are more practical and can provide equivalent or better mechanical properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can provide more details of your application we can give you more specific advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can certainly do very good quality castings, in terms of both detail and mechanical strength with gravity casting of PU resins in RTV rubber moulds in moderate quantity without needing injection moulding. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-12-12T00:06:54.537" LastActivityDate="2016-12-12T00:06:54.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12798" PostTypeId="2" ParentId="12272" CreationDate="2016-12-11T23:07:10.833" Score="9" Body="&lt;p&gt;There are a number of good reasons why the ballast layer needs to be free draining. The main objective of the ballast layer being free draining is somewhat tautologically to keep the water out of the track structure.&lt;/p&gt;&#xA;&lt;p&gt;Why is this necessary? Well, there are are two main underlying reasons why you don't want water in your track structure:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It tends to lead to deterioration in the track geometry through various mechanisms, which leads to lower speeds, increased derailment risk and more expensive track resurfacing.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It tends to lead to deterioration of your ties (concrete, timber and steel alike), which means you have to replace them more often.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;There are a few key mechanisms that cause the deterioration in track geometry:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When the water stops moving through the ballast it stops washing out contamination of fines (through local sedimentation or from the breakdown of the ballast itself). The standing water then tends to bind with the fines to create a mud slurry which both erodes the ties and quickly rounds the ballast rocks.&lt;/p&gt;&#xA;&lt;p&gt;This is bad because ballast rocks rely on their irregular shape to provide a solid and stable layer; when new, the ballast rocks can interlock with one another and quickly reach a relatively stable lowest energy state with high strength characteristics. As the rocks become rounded and smaller they settle further and interlock less well, reducing the overall strength of the layer.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The pumping action of trains passing over the ballast layer when it is wet tends to drive water into cracks in the subballast, which can drive the water down into the track foundation. Either way, both the subballast and the foundation will tend to be weaker when wet, and the action of the water infiltration will also erode these layers, which only makes the problem worse.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In soils with a high expansiveness (in particular clay), they may vary volumetrically quite substantially when they are wet as opposed to being dry. So, when the formerly dry formation gets wet, it can heave, and then when it dries out, it shrinks again. These processes are not uniform and so lead to poor top geometry.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly, in areas where freezing temperatures are encountered the water in the soil can freeze; ice has a lower density than liquid water and so you get heave, and then shrinking again when it melts. Melt water is also a great eroder.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;For all of these reasons we like to keep water from entering the formation (so far as is reasonably possible) and drain it out quickly when it does get in there.&lt;/p&gt;&#xA;&lt;p&gt;I'm afraid I don't know very much about slab track but I suspect that as you speculate, it is relatively impermeable to water and also not especially susceptible to erosion, and so the formation stays dry even if the slab track above it is practically under water.&lt;/p&gt;&#xA;" OwnerUserId="1918" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-12-11T23:07:10.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12799" PostTypeId="2" ParentId="12732" CreationDate="2016-12-12T00:03:53.463" Score="1" Body="&lt;p&gt;This question is far simpler than it looks. Since it has four supports (RX and RY at A plus RY at D and E), it would seem to be statically indeterminate, but it actually isn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see this, let's count the number of unknowns and the number of equations we have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 axial force for 10 bars = 10 unknowns&lt;/li&gt;&#xA;&lt;li&gt;4 reactions ($A_x$, $A_y$, $D_y$, $E_y$) = 4 unknowns&lt;/li&gt;&#xA;&lt;li&gt;2 equations ($\sum F_x = 0$, $\sum F_y = 0$) for each of 7 nodes = 14 equations&lt;/li&gt;&#xA;&lt;li&gt;Number of equations = number of unknowns, therefore we have an isostatic (statically determinate structure).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One way of solving this would be to create a system of 14 equations and solving it. This would work, but &lt;a href=&quot;https://youtu.be/8cT_Ulmcrys?t=6s&quot; rel=&quot;nofollow noreferrer&quot;&gt;ain't nobody got time for that&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much simpler way of solving this is by observing that this is actually two trusses, with the one to the left being supported by the one to the right by the node at $C$ (much like a Gerber beam).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's start by solving the truss to the left, placing a support at $C$ representing the effect of the other truss. Also, since we know that the horizontal force at $G$ will be fully absorbed by $A_x$, let's apply that force at $C$. Due to the Gerber-beam-esque behavior of this truss, we don't have to worry about the bending moment generated by shifting the force this way. After all, that bending moment wouldn't have crossed over the hinge at $C$ and the force itself would indeed have been transmitted via $C$, so there's no issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lvYTo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lvYTo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very straightforward, so I won't go into how to calculate the internal forces in each bar, but just calculate the reactions. It is clear that the results are $A_x = -50\text{ kN}$, $A_y = 60\text{ kN}$ and $C_y = 180\text{ kN}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now move on to the rightmost truss, where we apply the fictitious value of $C_y$ (and notice that the horizontal force at $G$ is still there!):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v0QaU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v0QaU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also easy to solve for the reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum M_E &amp;amp;= -4D_y + 8\cdot180 - 3\cdot50 = 0 \\&#xA;\therefore D_y &amp;amp;= 322.5\text{ kN} \\&#xA;\sum F_y &amp;amp;= D_y + E_y - 180 = 0 \\&#xA;\therefore D_y &amp;amp;= -142.5\text{ kN}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the reactions, the structure is also easy to solve for the internal values, so that's left as an exercise for the reader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a computer model of the structure, as well as of the individual truss models. Notice I had to place a fictitious RX at $E$ for the isolated rightmost truss to be stable to horizontal forces, so ignore its horizontal reaction. This also alters the axial compression values for the bottom chord, so just subtract 50&amp;nbsp;kN from the compression in that model to make it align with the original structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bkixg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bkixg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-12-12T00:03:53.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12800" PostTypeId="1" AcceptedAnswerId="13361" CreationDate="2016-12-12T06:15:27.237" Score="2" ViewCount="541" Body="&lt;p&gt;I am wondering how to determine screw size for a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the attached pictures of an example system.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/xmNSz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xmNSz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/M3hC2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M3hC2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say the mass of object, m = 10kg&#xA;&lt;br&gt;distance, x = 0.5m, x2=0.5m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with moment = 0, &#xA;&lt;br&gt;0 = F*1.0m - 100N*0.5m&#xA;&lt;br&gt;F = 50N as the screw clamping force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the calculation to determine required clamping force correct?&#xA;&lt;br&gt;If its correct, how to determine the required size and number of screw?&#xA;(FYI, socket head cap screw is used) &lt;/p&gt;&#xA;" OwnerUserId="8563" LastActivityDate="2017-01-21T15:59:47.037" Title="How to determine which size of screw needed?" Tags="|mechanical-engineering|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12801" PostTypeId="2" ParentId="12789" CreationDate="2016-12-12T11:33:26.040" Score="0" Body="&lt;p&gt;You should probably seek out a program that lets you use millimeters as units directly. Atleast Inkscape supports this for .svg files (select &quot;default unit&quot; in document properties), and of course all proper CAD programs do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failing that, you can pick any value as long as you remember it and apply proper scaling when doing the CAM work. If you are ordering the pieces from somewhere, you'll want to include a dimension drawing to make sure they come out the proper size.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-12-12T11:33:26.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12802" PostTypeId="1" CreationDate="2016-12-12T11:46:38.983" Score="1" ViewCount="678" Body="&lt;p&gt;When designing a bearing plinth, should anti-bursting reinforcement be provided? Under BS 5400, it was only mentioned in the section for end block anchorages of tendons. How should the code be interpreted?&lt;/p&gt;&#xA;" OwnerUserId="9194" LastActivityDate="2017-01-11T12:58:34.437" Title="When should anti-bursting reinforcement be used?" Tags="|civil-engineering|bridges|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12803" PostTypeId="2" ParentId="12802" CreationDate="2016-12-12T12:10:08.577" Score="3" Body="&lt;p&gt;BS5400 &lt;a href=&quot;http://shop.bsigroup.com/ProductDetail/?pid=000000000030071551&quot; rel=&quot;nofollow noreferrer&quot;&gt;has been superseded and withdrawn&lt;/a&gt;, so you probably shouldn't be using it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, when I last used it, I did provide bursting reinforcement for bearing plinths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, just because something isn't explicitly stated, it doesn't mean you shouldn't use your engineering judgement / common sense. If you recognise the possibility of bursting failure, you need to do a design check for bursting failure. This will either tell you that reinforcement isn't needed, or give you the required reinforcement against said failure.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2016-12-12T12:10:08.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12807" PostTypeId="2" ParentId="12732" CreationDate="2016-12-12T15:06:46.320" Score="1" Body="&lt;p&gt;Assuming positive $x$ to be to the right, positive $y$ to be upwards, and all the bars to be in compression, we can write the force balance for each joint. In the end if a force turns out negative it means that the reaction force is in the negative $x$ or $y$ direction, or the bar is in tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joint A:&#xA;$$ A_x-F_{\text{AB}}-F_{\text{AF}} \cos (\theta ) = 0 $$&#xA;$$A_y-F_{\text{AF}} \sin (\theta ) = 0 $$&#xA;Joint B:&#xA;$$F_{\text{AB}}-F_{\text{BC}} = 0 $$&#xA;$$-F_{\text{FB}}-120000 = 0 $$&#xA;Joint F:&#xA;$$F_{\text{AF}} \cos (\theta )-F_{\text{FC}} \cos (\theta ) = 0 $$&#xA;$$F_{\text{AF}} \sin (\theta )+F_{\text{FB}}+F_{\text{FC}} \sin (\theta ) = 0 $$&#xA;Joint C:&#xA;$$F_{\text{BC}}-F_{\text{CD}}-F_{\text{CG}} \cos (\theta )+F_{\text{FC}} \cos (\theta ) = 0 $$&#xA;$$F_{\text{CG}} (-\sin (\theta ))-F_{\text{FC}} \sin (\theta )-120000 = 0 $$&#xA;Joint D:&#xA;$$F_{\text{CD}}-F_{\text{DE}} =0 $$&#xA;$$D_y-F_{\text{GD}} = 0 $$&#xA;Joint G:&#xA;$$F_{\text{CG}} \cos (\theta )-F_{\text{GE}} \cos (\theta ) +50000 = 0 $$&#xA;$$F_{\text{CG}} \sin (\theta )+F_{\text{GD}}+F_{\text{GE}} \sin (\theta ) = 0 $$&#xA;Joint E:&#xA;$$F_{\text{DE}}+F_{\text{GE}} \cos (\theta ) = 0 $$&#xA;$$E_y-F_{\text{GE}} \sin (\theta ) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I concur that nobody got time for solving these equations, so all my computations were done in &lt;a href=&quot;https://i.stack.imgur.com/k6Vrj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mathematica&lt;/a&gt; to get the following results:&#xA;$$ A_x=-50kN, \ \ \ A_y=60kN, \ \ \ D_y=322.5kN, \ \ \ E_y=-142.5kN$$&#xA;$$F_{\text{AB}}=-130kN, \ \ \ F_{\text{AF}}=100kN, \ \ \ F_{\text{BC}}=-130kN, \ \ \ F_{\text{CD}}=190kN, \ \ \ F_{\text{CG}}=-300kN$$&#xA;$$F_{\text{DE}}=190kN, \ \ \ F_{\text{FB}}=-120kN, \ \ \ F_{\text{FC}}=100kN, \ \ \ F_{\text{GD}}=322.5kN, \ \ \ F_{\text{GE}}=-237.5kN$$&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-12-12T15:06:46.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12808" PostTypeId="2" ParentId="12755" CreationDate="2016-12-12T15:52:43.333" Score="0" Body="&lt;p&gt;The concept of poles and zeros in terms of the roots of the numerator and denominator is very crude and will not work for multivariable systems. Also computing them using the transfer function model is problematic because of pole-zero cancellations that can occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many notions of poles and zero and they are computed using both the transfer function and system matrix model. The latter is a generalization of the state-space model. I will provide a brief summary, and for more details you can refer to [1] or [2].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Transmission zeros:&lt;/strong&gt;&#xA;These are frequencies at which transmission through the system is blocked. They are computed from the Smith-McMillian form of the transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Transmissison poles:&lt;/strong&gt;&#xA;These are frequencies at which transmission through the system blows up. They are also computed from the Smith-McMillian form of the transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Input-decoupling zeros:&lt;/strong&gt;&#xA;These are the modes that are uncontrollable. They are computed from the system matrix model, and do not involve the output equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Output-decoupling zeros:&lt;/strong&gt;&#xA;These are the modes that are unobservable. They are computed from the system matrix model, and do not involve the inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Input-output-decoupling zeros:&lt;/strong&gt;&#xA;These are the modes that are both uncontrollable and unobservable. They are computed from the system matrix model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Invariant zeros:&lt;/strong&gt;&#xA;They are computed from the system matrix model. They involve the transmission zeros and the some of the input- and output-decoupling zeros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;System zeros:&lt;/strong&gt;&#xA;This is defined as {input-decoupling zeros, output-decoupling zeros, transmission zeros}-{input-output-decoupling zeros}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;System poles:&lt;/strong&gt;&#xA;This is defined as {input-decoupling zeros, output-decoupling zeros, transmission poles}-{input-output-decoupling zeros}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom-line is that it needs to be clear what type of pole or zero you are interested in, and the resulting computations may or may not include the various matrices of the state-space model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refs:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A. G. J. MacFarlane and N. Karcanias, Poles and zeros of linear multivariable systems: A survey of the algebraic, geometric and complex-variable theory, International Journal of Control 24(1):33-74, July 1976&lt;/li&gt;&#xA;&lt;li&gt;H. H. Rosenbrock, State-space and multivariable theory. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-12-12T15:52:43.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12810" PostTypeId="2" ParentId="12769" CreationDate="2016-12-12T17:29:57.517" Score="2" Body="&lt;p&gt;Because this is a linear and time-invariant system, you can use the principle of superposition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the input to $u_i(t)$ is $y_i(t)$, then &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The response to $\alpha u_i(t) $ is $\alpha y_i(t) $&lt;/li&gt;&#xA;&lt;li&gt;The response of $u_i(t)+u_j(t)$ is $y_i(t)+y_j(t)$&lt;/li&gt;&#xA;&lt;li&gt;The response of $u(t-t_p)$ is $y(t-t_p)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The response of the system to a unit step input $\theta (t)$ is $1-e^{-\frac{t}{R C}}$. (I assume you have no problem with this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input to the system can be split into two. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$5 \theta (t)$ when $0\leq t\leq t_p$. The response of the system to this input is $5(1-e^{-\frac{t}{R C}})$.&lt;/li&gt;&#xA;&lt;li&gt;$5 \theta (t)-\theta(t-t_p)$ when $t_p &amp;lt; t$. The response is  $5(1-e^{-\frac{t}{R C}})-(1-e^{-\frac{t-\text{tp}}{R C}})$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The final response is a sum of these two piecewise functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have plotted the result for $R=1$, $C=1$ and various values of $t_p$. The dashed line shows the response if the second step was not applied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0E7rT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0E7rT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2016-12-12T17:29:57.517" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12814" PostTypeId="2" ParentId="12789" CreationDate="2016-12-12T20:13:35.460" Score="2" Body="&lt;p&gt;&lt;strike&gt;I found one of those printer things that puts ink on dead trees and tested to print a simple SVG file.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;svg xmlns=&quot;http://www.w3.org/2000/svg&quot;&#xA;     width=&quot;400px&quot; height=&quot;800px&quot;&amp;gt;&#xA;  &amp;lt;rect x=&quot;10&quot; y=&quot;10&quot; width=&quot;72&quot; height=&quot;72&quot; fill=&quot;#999999&quot; /&amp;gt;&#xA;  &amp;lt;rect x=&quot;10&quot; y=&quot;100&quot; width=&quot;378&quot; height=&quot;378&quot; fill=&quot;#999999&quot; /&amp;gt;&#xA;&amp;lt;/svg&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As I suspected 72 pixels came out pretty much exactly 19mm. (72/19.05)*100~=378 came out 100mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this I am going to assume that 72/19.05 is the de facto best pixel to mm ratio to use for CNC projects.&lt;/strike&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found this documentation: &lt;a href=&quot;http://w3.org/TR/SVG/coords.html#Units&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://w3.org/TR/SVG/coords.html#Units&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;lt;svg xmlns=&quot;http://www.w3.org/2000/svg&quot;&#xA;     width=&quot;400px&quot; height=&quot;800px&quot;&amp;gt;&#xA;  &amp;lt;rect x=&quot;10&quot; y=&quot;10&quot; width=&quot;19.05mm&quot; height=&quot;19.05mm&quot; fill=&quot;#999999&quot; /&amp;gt;&#xA;  &amp;lt;rect x=&quot;10&quot; y=&quot;100&quot; width=&quot;100mm&quot; height=&quot;100mm&quot; fill=&quot;#999999&quot; /&amp;gt;&#xA;&amp;lt;/svg&amp;gt; &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Much simpler to use mm as units right away &lt;/p&gt;&#xA;" OwnerUserId="9188" LastEditorUserId="9188" LastEditDate="2016-12-12T20:29:50.383" LastActivityDate="2016-12-12T20:29:50.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12816" PostTypeId="1" CreationDate="2016-12-13T05:19:52.863" Score="0" ViewCount="104" Body="&lt;p&gt;Suppose we have a isotropic homogeneous turbulent channel flow. We know that this flow is associated with many swirling eddies of different length and time scales. I need to know the direction of rotation of eddies.&#xA;More specifically, I need to know if most of the eddies are co-rotating, counter-rotating or there is no definite answer (randomly rotating)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, does your answer change if I remove the constraint of isotropy and homogeneity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance  &lt;/p&gt;&#xA;" OwnerUserId="9203" LastActivityDate="2017-03-13T08:39:03.330" Title="Direction of rotation of turbulent eddies in channel flow" Tags="|turbulence|fluid|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12817" PostTypeId="2" ParentId="12816" CreationDate="2016-12-13T07:00:18.250" Score="1" Body="&lt;p&gt;Non-laminar flow is often hard to characterize with much mathematical rigor (though in certain special cases and geometries may be done with a little limitation). The phenomena depends heavily on channel geometry. For example, A backflow eddy can develop from a sudden stepwise increase in channel height/width. The Reynolds number of the fluid flow under examination will determine the strength and formation of the eddy. The formation of the eddy results from the water lamina closest to the wall being deformed by the sudden channel widening due to friction. Viscous interactions between adjacent lamina propagate away from the wall and result in the curling of local flow lines into an eddy. The Reynold's number will determine if an eddy forms, or if there is a smooth lamellar flow into the new channel width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, the Reynolds number is a very useful quantity for characterizing different types of fluid flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the bulk of the fluid, most eddies will be locally co-rotating over a length scale that depends nontrivially on the fluid viscosity, as local lamina will tend to drag each other rather than slip. Of course, on some larger scale, this will appear as turbulent motion. Again, the Reynolds number is useful in this characterization. &lt;/p&gt;&#xA;" OwnerUserId="9204" LastActivityDate="2016-12-13T07:00:18.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12818" PostTypeId="2" ParentId="12792" CreationDate="2016-12-13T11:46:39.847" Score="5" Body="&lt;p&gt;As indeed answered by Greenonline, the centre distance is the average of the pitch diameters of the two meshing gears, but this is only strictly true when the gears are operating at standard centre distances, i.e. where the pitch circles are tangent to one another. There are indeed cases where two gears can be operating at non-standard centre distances, and therefore the standard pitch circles are no longer tangent to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes non-standard centre distances are accidental and occur due to difficulty in exactly aligning two gears to have a standard centre distance. Other times it may be a deliberate part of the design. For example, slightly increasing the centre distance above the standard value will implement some backlash (as shown below) which helps to prevent any jamming from occurring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bn8N0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bn8N0.jpg&quot; alt=&quot;Illustration of backlash&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two gears operating at standard centre distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5yEUq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5yEUq.jpg&quot; alt=&quot;two gears operating at standard centre distance&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And operating at non-standard centre distance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6ReFO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6ReFO.jpg&quot; alt=&quot;two gears operating at non-standard centre distance&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most gears in industry have teeth that are involute (the notable exception to this is in the clockmaking industry, where teeth are frequently cycloidal instead), and the advantage of involute teeth is that gears can still operate smoothly at non-standard centre distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dealing with non-standard centre distances, we have to be careful by what is meant by pitch circle and pitch diameter. The diameter of the pitch circle, the pitch diameter, is a dimension that belongs to a single gear on its own, regardless of how and with what it mesh. However, the pitch circles for a pair of meshing gears also represent the diameters of two equivalent disks that are rolling without slip; this means the pitch circles should be tangent. For operation at non-standard centre distance, there appears to be a contradiction in the definition of a pitch circle: therefore, the two following terms should be observed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Standard pitch diameter&lt;/strong&gt;: this is a dimension on a gear that is independent of how or with what it meshes, and defined for any two meshing gears as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_1 = N_1m = \frac{N_1}{P_D}$$&#xA;$$d_2 = N_2m = \frac{N_2}{P_D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $N$ is the number of teeth, $m$ is the module (a measure of tooth size in millimetres), $P_D$ is the diametral pitch (a measure of tooth fineness in teeth per inches) and the subscripts 1 and 2 indicate which gear the variable refers to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the dimension that is provided when buying gears &quot;off the shelf&quot;. Taking the average of these diameters for both gears will give you the standard centre distance, which may or may not be equal to the actual centre distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_1 = \frac{2C}{\frac{N_1}{N_2}+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_2 = \frac{2C}{\frac{N_2}{N_1}+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = \frac{d_1+d_2}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $C$ is the standard operating distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Operating Pitch Diameter&lt;/strong&gt;: this is a dimension that only exists when two gears are meshing together, and it represents the diameters of the equivalent rolling disks. The operating pitch circles are tangent to one another, and taking the average of these will indeed give you the actual centre distance, i.e. operating centre distance. If the centre distance is larger than standard, then the operating pitch circles will be larger than the standard pitch circles. The operating pitch diameters are defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d'_1 = \frac{2C'}{\frac{N_1}{N_2}+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d'_2 = \frac{2C'}{\frac{N_2}{N_1}+1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C' = \frac{d'_1+d'_2}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $d'$ is the operating pitch diameter, and $C'$ is the (actual) operating centre distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/imfxP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/imfxP.jpg&quot; alt=&quot;Operating at standard center distance&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UeBu9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UeBu9.jpg&quot; alt=&quot;Operating at non-standard center distance&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final point worth noting: there are limits to how much you can vary the centre distance from standard: there is a minimum centre distance below which the gear teeth will jam with one another, and there is a maximum above which the teeth will no longer reach and make contact with each other:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/om8g8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/om8g8.jpg&quot; alt=&quot;Two gears beyond maximum limit of non-standard center distance&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="5899" LastEditDate="2016-12-13T15:30:04.767" LastActivityDate="2016-12-13T15:30:04.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12820" PostTypeId="1" CreationDate="2016-12-13T16:45:40.363" Score="5" ViewCount="169" Body="&lt;p&gt;First of all, excuse my little knowledge, I'm a computer programmer by trade and unfortunately have very little knowledge of physics, engineering, DSP but I'm willing to learn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to measure the quality of a road surface when riding on it on a skateboard. Basically I'd be interested in the smoothness of the road. For &quot;smoother&quot; roads, there should be little vibration, for less smooth roads there should be more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideas I had so far is strapping the &lt;a href=&quot;http://processors.wiki.ti.com/index.php/CC2650_SensorTag_User&amp;#39;s_Guide&quot; rel=&quot;nofollow noreferrer&quot;&gt;TI SensorTag 2 (CC2650)&lt;/a&gt; to the skateboard and recording the accelerometer graph during a ride and then taking those frequencies (fourier transform). Unfortunately, the maximum sampling frequency I can get from the SensorTag seems to be &lt;a href=&quot;http://processors.wiki.ti.com/index.php/CC2650_SensorTag_User&amp;#39;s_Guide#Movement_Sensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;10 Hz&lt;/a&gt; which seems way too little. My uninformed back of the envelope calculation is that I kind of need to detect 1cm large &quot;potholes&quot; and that I'm going about 10 km/h. So the frequency I need detect needs to be at least 277 Hz (10 km/h / 1 cm). Also I believe to derive the &quot;degree of vibration&quot; from that I need the velocity that I'm traveling. The question is where I'd get that from... Sure, I could integrate the acceleration data but that seems like it'd be quite imprecise. Also I could use the GPS on my phone for that but that seems rather imprecise too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend of mine proposed a potentially simpler approach of recording the sounds the wheels create while riding. And then derive the degree of vibration from that. But I'd also need the travel velocity for that, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make any sense? And if yes, what could be a good approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="9211" LastEditorUserId="9211" LastEditDate="2016-12-13T23:03:43.393" LastActivityDate="2016-12-16T18:57:12.987" Title="Measuring the road surface quality" Tags="|electrical-engineering|sensors|measurements|vibration|acceleration|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12822" PostTypeId="1" AcceptedAnswerId="12862" CreationDate="2016-12-13T16:49:44.763" Score="5" ViewCount="2903" Body="&lt;p&gt;This is more of a general physics question to help me understand how to choose sports headsets in the future, however it is too specific to a certain use case (bluetooth headset) to belong in the Physics StackExchange.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Problem:&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;3 of my 4 sports bluetooth headsets suffer from choppy sound quality (signal interference?) the minute I step outside. It seems to be especially bad under a clear sky no matter the time of day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4th and the most expensive headset does not suffer from this, but it has a different (more compact) design. The worst performance was by the cheapest sports headset (from ebay) and it was unusable outdoors.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Observations:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In all cases my phone is located in my trouser pocket, but moving it closer to my head (the headset) seems to lessen signal interference, but not remedy the issue completely.&lt;/li&gt;&#xA;&lt;li&gt;In all tests the headsets have been fully charged.&lt;/li&gt;&#xA;&lt;li&gt;The speed of movement also seems to play a factor, but I am not able to reproduce that affect every time.&lt;/li&gt;&#xA;&lt;li&gt;I can say that clothes are not a factor, as the same clothing indoors produces no interference.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;Theories:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;background EM radiation coming from outer space?&lt;/li&gt;&#xA;&lt;li&gt;EM radiation coming from our sun? Although I have not observed if the moon reflecting the EM from the Sun plays a part or not.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3081" LastEditorUserId="8055" LastEditDate="2016-12-20T15:39:33.593" LastActivityDate="2017-01-17T23:33:15.387" Title="Why do bluetooth headsets get interference (choppy sound quality) outdoors?" Tags="|electrical-engineering|telecommunication|wireless-communication|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12823" PostTypeId="1" CreationDate="2016-12-13T18:18:20.930" Score="2" ViewCount="1737" Body="&lt;p&gt;I'm having some trouble understanding the proof of the Principle of Virtual Work for deformable bodies. I'll give below the proof that I've read, and, next, I'll remark what I'm not understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to remember before going through the proof is that the virtual work done by a system of virtual forces in equilibrium as a rigid body undergoes a small, compatible displacement is zero.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;PROOF:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Suppose that a deformable body is in static equilibrium under the external loads of a virtual $Q$-force system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the body as a whole is in equilibrium, any particular particle can be isolated and will be in equilibrium under the internal virtual $Q$ stresses developed by the external virtual $Q$ forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now suppose that the body is subjected to a small change in shape caused by some other source than the virtual $Q$-force system. Owing to this change in shape, any particle might be deformed as well as translated and rotated as a rigid particle. Hence, the boundaries of such a particle would move and hence do virtual work. Let the virtual work done by the $Q$ stresses on the boundaries of the differential particle be designated by $\text{d}W_s$. Part of this virtual work will be done because of the movements of the boundaries of the particle caused by the deformation of the particle itself; this part will be called $\text{d}W_d$. The remaining part of $\text{d}W_s$ will be the virtual work done by the $Q$ stresses during the remaining part of the displacement of the boundaries and will be equal to $\text{d}W_s-\text{d}W_d$. However, this remaining is caused by the translation and rotation of the particle as a rigid body, and, as reminded above, the virtual work done in such a case is equal to zero. Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{d}W_s=\text{d}W_d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the virtual work done by the $Q$ stresses on all particles of the body is now added, this equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W_s=W_d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To evaluate first $W_s$, we recognize that this term represents the total virtual work done by the virtual $Q$ stresses on all the boundaries of all the particles. However, for every internal boundary of a particle there is an adjoining particle whose adjacent boundary is actual the same line on the body as whole, and therefore these adjacent boundaries are displaced exactly the same amount. Since the forces acting on the two adjacent internal boundaries are numerically equal but opposite in direction, the total virtual work done on the pair of adjoining internal boundaries is zero. Hence, since all internal boundaries occur in pairs, there is no net virtual work done by the forces on all the internal boundaries. $W_s$ therefore consists only of the work done by the external $Q$ forces on the external boundaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W_d$, on the other hand, was obtained by integrating the virtual work associated with deformation of the element. This work includes the effects of all forces on the element, both stress resultants and external forces. However, when an element deforms, only the stress resultants perform any work. Thus, Wd represents the virtual work done by the stress resultants alone.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;END OF PROOF&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I understood the part concerning $W_s$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't understand why &quot;when an element deforms, only the stress resultants perform any work&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, it is said that &quot;$W_d$ represents the virtual work done by the stress resultants alone&quot;. But, according to the paragraph pertaining to $W_s$, this work should be zero, since all particle's internal boundaries occur in pairs and that one side of an internal boundary of a particle undergoes the same displacement as the common side of a neighboring particle and such two sides are subjected to equal and opposite stresses, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to apply this principle according to my (flawed) conclusion, all points of the deformable body would undergo no displacement at all: a complete nonsense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where's the flaw in my reasoning?&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2016-12-14T19:34:10.310" LastActivityDate="2022-04-22T22:41:21.273" Title="On the proof of the Principle of Virtual Work" Tags="|structural-analysis|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="12824" PostTypeId="1" CreationDate="2016-12-13T18:56:54.960" Score="1" ViewCount="2210" Body="&lt;p&gt;Draw the third view and the points on each view. In the first picture you have what the exercise has given us and in the second what I have drawn. I have turned the page so that the first view is the one on the left, the horizontal one is the one on the right. Is that OK? Should I use the one at the right as a lateral view?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tinypic.com/view.php?pic=qy7ltc&amp;amp;s=9&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hswMq.jpg&quot; alt=&quot;drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://tinypic.com/view.php?pic=9aznk8&amp;amp;s=9&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/liE0J.jpg&quot; alt=&quot;drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8980" LastEditorUserId="6767" LastEditDate="2019-01-05T19:52:57.087" LastActivityDate="2021-02-26T08:39:24.767" Title="Draw the third view and the points on each view" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12826" PostTypeId="2" ParentId="7586" CreationDate="2016-12-13T19:33:52.380" Score="2" Body="&lt;p&gt;In addition to Wasabi's answer - which is great - large diameter bars need insane &lt;a href=&quot;https://www.quora.com/What-is-development-length-Anchorage-length-and-lap-length-in-rcc&quot; rel=&quot;nofollow noreferrer&quot;&gt;development length and lap length&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rule of thumb estimate for development and lap length is 40 diameters. For 20 mm bars that is a large but reasonable 80 cm, but it becames 128 cm for 32 mm bars and 2 meters for 50 mm bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lap length problem can be avoided by using mechanical connectors, but 2 meters development length would waste a lot of steel and need a large space, very often unavailable.&lt;/p&gt;&#xA;" OwnerUserId="7371" LastEditorUserId="7371" LastEditDate="2016-12-14T19:33:17.477" LastActivityDate="2016-12-14T19:33:17.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12828" PostTypeId="2" ParentId="12820" CreationDate="2016-12-13T22:14:45.413" Score="3" Body="&lt;p&gt;While the acoustic analysis idea &lt;em&gt;may*&lt;/em&gt; work, I think the accellerometer data would be much easier to analyze. If you are looking for something that can give you better rates of data, try looking into other accellerometers. &lt;a href=&quot;https://www.adafruit.com/products/2019&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; one from Adafruit has a maximum rate of 800 Hz(&lt;a href=&quot;https://cdn-shop.adafruit.com/datasheets/MMA8451Q-1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;datasheet&lt;/a&gt;). According to your calculations, this should be enough.&lt;/p&gt;&#xA;&lt;p&gt;You could log this data with something like an arduino to be analyzed later (I haven't looked into how quickly that will be able to gather/process data). A module like the raspberry pi zero should have enough processing power to sort through the accellerometer data in real time.&lt;/p&gt;&#xA;&lt;p&gt;As for measuring the speed of the skateboard, a common, cheap, and fairly easy route is to use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hall_effect_sensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;hall effect sensor&lt;/a&gt; on a wheel. This will give you the angular velocity of the wheel (rpm/rps) and from there you use rpm times the circumference of the wheel (say in centimeters) to give you the speed in cm/min. Convert appropriately for your program.&lt;/p&gt;&#xA;&lt;p&gt;*I'm sure given enough time, trial and error, it could work, I am doubtful using sound to analyze the road would be worth the effort.&lt;/p&gt;&#xA;" OwnerUserId="8907" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-12-13T22:14:45.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12829" PostTypeId="1" CreationDate="2016-12-13T23:19:07.023" Score="2" ViewCount="74" Body="&lt;p&gt;We are designing a button panel at work, made out of glass and are running into some issues. Glass, of course, can shatter. But we noticed that when dropping a steel ball in the center of the glass it is far less likely to shatter than when the same experiment is performed near the corners. It almost always shatters. Why is that? Is there something we can do with the geometry of the panel (currently rectangle) that will alleviate this issue? Maybe rounding the corners to be circular? Does anyone have a link to some research on the topic (i.e. how thick the glass should be to survive a ball of mass $m$ being dropped on it from height $h$)?&lt;/p&gt;&#xA;" OwnerUserId="9215" LastEditorUserId="1832" LastEditDate="2016-12-13T23:34:20.573" LastActivityDate="2016-12-14T10:55:51.297" Title="Glass panel engineering at work" Tags="|mechanical-engineering|materials|applied-mechanics|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12830" PostTypeId="2" ParentId="12829" CreationDate="2016-12-13T23:40:41.570" Score="5" Body="&lt;p&gt;Glass tends to contain quite a lot of residual stresses from the manufacturing process and these will tend to be concentrated near corners (depending on how the sheet was manufactured). There is also the fact that for a panel supported at the edges the centre will be less stiff and thus can deform more before it fractures in the same way that a simple beam deforms most at the centre. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly mounting the panel in a shock absorbing frame such as using a rubber or foam casket or an elastomer adhesive can help as this will allow the whole assembly to defect smoothing out the energy transfer from the impactor to the panel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of calculation in glass is complicated by the fact that it is a brittle material thus any local stress concentration which exceeds its UTS can cause it to fail completely even if the average stress is quite low where an equivalent ductile material would just be scratched or dented. This is why you can easily break a window with an impact from a hard sharp object with much less effort than just pushing against it with something blunt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally any surface defects like pits or scratches can have a significant effect on the fracture toughness of glass, which is why scoring it with a diamond scribe is a very effective way to 'cut' it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some types of glass are designed to be much more impact resistant and to fail in a safe way, this include tempered glass which is heat treated to alleviate residual stresses and laminated glass which is composed of alternating layers of thin glass and a transparent polymer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like you really need to talk to a glass supplier to find the appropriate material for your requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no one generic formula for the impact toughness of glass (or really any material) any you would need to look at empirical data for a specific product and impact toughness metrics tend to only be applicable to quite specific conditions, especially for brittle materials. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-12-14T10:55:51.297" LastActivityDate="2016-12-14T10:55:51.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12832" PostTypeId="1" CreationDate="2016-12-14T00:08:19.207" Score="1" ViewCount="31" Body="&lt;p&gt;I'm trying to join two different parts together but it's hard trying to find the best method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot weld them because they must be able to be removed easily. I  can't use any nuts or bolts because then it won't be interchangeable and for what I'm trying to fit my part on, there are hundred of different variations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need out to have a very strong seal like I bolted or welded it on but without having to really do so. It also must be extremely air tight. I need a lot of creation on this one. I thought about making a suction effect but didn't know if it would be strong enough and hold it like bolts would. I'm trying to fit my own cylinder head to a little engine I have.&lt;/p&gt;&#xA;" OwnerUserId="7041" LastEditorUserId="90" LastEditDate="2016-12-14T02:38:57.180" LastActivityDate="2016-12-14T02:38:57.180" Title="What are some different sealing methods?" Tags="|seals|" AnswerCount="0" CommentCount="2" FavoriteCount="0" ClosedDate="2017-01-31T14:24:06.780" ContentLicense="CC BY-SA 3.0" />
  <row Id="12834" PostTypeId="2" ParentId="12820" CreationDate="2016-12-14T00:17:43.797" Score="2" Body="&lt;p&gt;As suggested I would try to use accelerometer. Below you will find a link to source code that might be helpful. I used a ADXL 345 accelormeter in combination with a TIVA ARM Cortex M4 Microcontroller. You are welcome to fork the source code&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/6ZFy4zMGLgg&quot; rel=&quot;nofollow noreferrer&quot;&gt;TIVA - ADXL345 Accelerometer Demo Using I2C&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://github.com/mahengunawardena/Tiva_I2C_Nokia_ADXL345&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tiva_I2C_Nokia_ADXL345 source code&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2016-12-14T00:17:43.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12835" PostTypeId="2" ParentId="12732" CreationDate="2016-12-14T02:47:21.297" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Consider the whole system: $$\sum x = H_A+50=0 $$ Thus: $$H_A=-50$$&lt;/li&gt;&#xA;&lt;li&gt;Consider the left part ABCF: $$\sum m_C=V_A\times 8-120\times 4=0$$ Thus: $$V_A=60$$&lt;/li&gt;&#xA;&lt;li&gt;Consider the right part CDEG: $$\sum m_D=V_E\times 4+(120+60)\times 4-50\times 3=0$$ Thus: $$V_E=-142.5$$&lt;/li&gt;&#xA;&lt;li&gt;Consider the whole system again:$$\sum y=V_D+V_A+V_E-120-120=0$$ Thus: $$V_D=322.5$$&lt;/li&gt;&#xA;&lt;li&gt;Do you only want to find the 4 unknown reactions? If yes, the 4 steps above might help. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8747" LastActivityDate="2016-12-14T02:47:21.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12836" PostTypeId="1" CreationDate="2016-12-14T04:22:14.647" Score="1" ViewCount="193" Body="&lt;p&gt;I have connected a 200 bar helium tank and a manometer to a pressure regulator. The end of the manometer's tube connected to the regulator is leaking. It's attached with anaerobic thread sealant, so I can't remove it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/gaZa_frCPKs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is the video of the connection&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It leaks between rotating and non rotating parts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aACFF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aACFF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using bath sealant, but it doesn't sustain 200 bar. I also attempted sealing it with melted soldering wire, but it doesn't stick, no matter how much flux I use. I need some kind of sealant, that is tough enough, but aerobic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tube is guaranteed to sustain up to 300 bar, but apparently it doesn't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I fix the leakage?&lt;/p&gt;&#xA;" OwnerUserId="9218" LastActivityDate="2016-12-14T10:43:40.310" Title="Helium is leaking through manometer" Tags="|pressure|gas|seals|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12837" PostTypeId="2" ParentId="12836" CreationDate="2016-12-14T10:43:40.310" Score="1" Body="&lt;p&gt;For high pressure connections it is common to use self sealing connectors with either tapered thread or dome shaped seats, in the second case it is not the thread itself which provides the seal. So the first thing is to verify what sort of connector you have as applying sealant to a seated connector can actually stop if from sealing properly and the mating surfaces need to be scrupulously clean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the role of the thread is to tighten the seats together and the fact that the whole thing can rotate implies that it isn't tight enough or that the fittings are incompatible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have a link to the original product ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way the chances are that you are going to have to disassemble the connection as you are unlikely to get a good seal now without cleaning it just applying more layers from the outside never works. You can usually remove anaerobic sealant by applying gentle heat.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-14T10:43:40.310" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12838" PostTypeId="1" CreationDate="2016-12-14T12:43:37.067" Score="2" ViewCount="6603" Body="&lt;p&gt;I have one of these slide-out trash cans:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0MYyx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0MYyx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the trash can to slide out with the door being opened and slide in with the door being closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of a method to convert the door opening motion (rotational?) into the linear motion of the trash can. Now I have tried searching multiple ways for solutions on converting rotational to linear motion, but came up empty for this seemingly simple problem. The door motion of course isn't purely rotational as we're not talking about converting the hinge-motion, but the motion of the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A method that works is just attaching a string to the side of the door and onto the trash can. That works for pulling out the trash can, but pushing in obviously doesn't work with a flexible string.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm lacking to tools to properly model this problem, right now I am stuck to fiddling around - I'm assuming there are better ways to solve this.&lt;/p&gt;&#xA;" OwnerUserId="9219" LastEditorUserId="1832" LastEditDate="2016-12-14T13:08:13.797" LastActivityDate="2020-04-27T20:40:08.670" Title="Converting rotary motion to linear motion (door for sliding trash can)" Tags="|mechanical-engineering|prototyping|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12839" PostTypeId="2" ParentId="12838" CreationDate="2016-12-14T13:14:07.333" Score="3" Body="&lt;p&gt;You have already arrived at half of the simplest solution. You need a dual tension system. When you open the door, the string pulls the slide out. When you close the door, you simply need something to pull it back. That something can be a spring mounted to the back of the slide and the back of the cabinet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It needs to be sized in length and spring stiffness to not overcome (and break) the string, but to have enough force to overcome the friction of the slide. &lt;/p&gt;&#xA;" OwnerUserId="5865" LastActivityDate="2016-12-14T13:14:07.333" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12840" PostTypeId="2" ParentId="3668" CreationDate="2016-12-14T14:38:17.313" Score="0" Body="&lt;p&gt;Failure of cylindrical shells under external pressure depends on length in addition to all the factors for internal pressure. If you can get your hands on a copy of the ASME boiler and pressure vessel code Section VIII division 1, and section II part D, you can calculate it yourself using the formulae in UG-28. Alternatively, there are a few pressure vessel software suites that offer free trials.&lt;/p&gt;&#xA;" OwnerUserId="9212" LastActivityDate="2016-12-14T14:38:17.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12842" PostTypeId="1" CreationDate="2016-12-14T16:43:04.607" Score="1" ViewCount="233" Body="&lt;p&gt;Many induction cooktops for sale have the option of controlling the temperature instead of setting the heat.  How does the temperature control typically work?  Is it an open loop or closed loop control system?  Is the temperature measured using an infrared sensor, or by some other means?&lt;/p&gt;&#xA;" OwnerUserId="9220" LastActivityDate="2016-12-14T16:43:04.607" Title="Temperature Control on and Induction Cooktop" Tags="|electrical-engineering|control-engineering|" AnswerCount="0" CommentCount="1" ClosedDate="2016-12-14T16:57:15.820" ContentLicense="CC BY-SA 3.0" />
  <row Id="12844" PostTypeId="1" AcceptedAnswerId="12901" CreationDate="2016-12-14T17:29:42.797" Score="1" ViewCount="204" Body="&lt;p&gt;As I see it, if you're going to calculate a twin-spool turbojet engine, you&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Determine expected max aircraft speed, and engine thrust at that speed and when stationary.&lt;/li&gt;&#xA;&lt;li&gt;Calculate peak mass of air flow&lt;/li&gt;&#xA;&lt;li&gt;Calculate energy needed to propel this mass of air&lt;/li&gt;&#xA;&lt;li&gt;Calculate bypass ratio&lt;/li&gt;&#xA;&lt;li&gt;Calculate compressor properties&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And from here I have no idea where to move on. Some books I've read just go through the engine from compressor to exhaust nozzle, but none explains which parameter of the engine is primary and where do calculations start and how they're performed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you give me a clue?&#xA;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="9221" LastActivityDate="2016-12-18T14:29:42.343" Title="In what order turbojet engine parameters are calculated?" Tags="|turbines|turbomachinery|propulsion|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12848" PostTypeId="1" CreationDate="2016-12-14T20:27:44.640" Score="1" ViewCount="153" Body="&lt;p&gt;I  want to use sensor for stretchable or bendable movements on knees, on fabric. But I am confused that which sensor is better to use and why?&lt;/p&gt;&#xA;" OwnerUserId="9222" LastEditorUserId="8055" LastEditDate="2017-01-16T16:23:34.973" LastActivityDate="2017-01-16T16:23:34.973" Title="Flex sensor vs stretch sensor" Tags="|sensors|" AnswerCount="0" CommentCount="1" ClosedDate="2017-01-20T23:14:01.060" ContentLicense="CC BY-SA 3.0" />
  <row Id="12849" PostTypeId="1" AcceptedAnswerId="13632" CreationDate="2016-12-14T21:16:27.537" Score="0" ViewCount="77" Body="&lt;p&gt;It's written that at constant pressure &amp;amp; temperature for Otto &amp;amp; Diesel cycles&#xA;the efficiency of diesel is more than efficiency of Otto.&lt;br /&gt;&#xA;Efficiency = Work / Heat added = 1- (Heat rejected / Heat added )&lt;/p&gt;&#xA;&lt;p&gt;from fig :&lt;br /&gt;&#xA;-(4-&amp;gt;1)  Heat rejected is same&lt;br /&gt;&#xA;-(2'-&amp;gt;3) Heat added for diesel is greater than (2-&amp;gt;3) )heat added for Otto&lt;/p&gt;&#xA;&lt;p&gt;so for diesel the term (Heat rejected / Heat added ) is less&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;so efficiency of diesel is more.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;What make it not clear for me is that&lt;br /&gt;&#xA;work ( Cv(T4-T3) ) (3-&amp;gt;4)  is same for both &amp;amp; heat added for diesel is more than that of Otto.&lt;br /&gt;&#xA;So if we use  eff= Work/(heat added)&lt;br /&gt;&#xA;&lt;strong&gt;Eff of otto will be more since less heat will be added and same work is done&lt;/strong&gt;,&lt;br /&gt;&#xA;What makes it wrong?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G3cb8.jpg&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;P-V For same pressure and temperature&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G3cb8.jpg&quot; alt=&quot;P-V For same pressure and temperature&quot; title=&quot;P-V For same pressure and temperature&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;source : &lt;a href=&quot;http://slideplayer.com/slide/10748942/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://slideplayer.com/slide/10748942/&lt;/a&gt;  slide 30&lt;/p&gt;&#xA;" OwnerUserId="9223" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-02-24T14:50:22.303" Title="cycle efficiency for (otto -diesel -dual) at constant Pr &amp; Temp" Tags="|mechanical-engineering|thermodynamics|energy-efficiency|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="12850" PostTypeId="1" CreationDate="2016-12-14T21:59:13.783" Score="2" ViewCount="2698" Body="&lt;p&gt;I found that &lt;em&gt;Resistance Temperature Detectors&lt;/em&gt; (RTD) are mostly used for bearings, but it's not easy to drill the casing and we need to take a signal out to PLC and LabVIEW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a sensor that can measure bearing temperature through the casing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I use an IR gun temperature and take a signal from it?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VrFdf.png&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Cross section of Bearing case&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VrFdf.png&quot; alt=&quot;Cross section of Bearing case&quot; title=&quot;Cross section of Bearing case&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9223" LastEditorUserId="5899" LastEditDate="2016-12-15T01:32:43.963" LastActivityDate="2016-12-15T07:26:22.560" Title="How to measure a bearing temperature in a pump?" Tags="|sensors|measurements|pumps|temperature|bearings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12851" PostTypeId="1" AcceptedAnswerId="12858" CreationDate="2016-12-15T01:21:19.937" Score="0" ViewCount="40" Body="&lt;p&gt;I try to understand how is possible to turn a photo receptive sensor into an emitter of light.&#xA;&lt;br&gt;&#xA;&lt;em&gt;Settings, tools, modulation&lt;/em&gt;..&lt;br&gt; &lt;/p&gt;&#xA;" OwnerUserId="9080" LastActivityDate="2016-12-15T12:38:00.673" Title="How to turn a photo receptive sensor into an emitter of light?" Tags="|sensors|modulation|" AnswerCount="1" CommentCount="1" ClosedDate="2016-12-15T14:21:48.403" ContentLicense="CC BY-SA 3.0" />
  <row Id="12852" PostTypeId="1" AcceptedAnswerId="12943" CreationDate="2016-12-15T02:11:15.073" Score="1" ViewCount="15403" Body="&lt;p&gt;Consider the two scenarios in this picture: &lt;img src=&quot;https://qph.ec.quoracdn.net/main-qimg-28763a636910916c63557f31d77183d8?convert_to_webp=true&quot; alt=&quot;Built Up Member Constructions&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the left, calculating Q=A'y[bar]' is easy. A is the area of the top board since the cross sectional area of that board corresponds to the area on the opposite side of the shear plane to the centroid of the member. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused about how to choose Q on the right hand side scenario. The two shear planes are perpendicular to the plane that separates A' from the rest of the member, meaning the definition I used for the first case on the left simply doesn't apply. Based on other sources I have consulted, the area I should choose is the area of the board in the center of the two upright members. I have no good justification for this choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: More information: Before I was introduced to the concept of shear flow, I was taught how to make shear stress calculations at any given height above or below the centroid of a beam cross section. To calculate this shear stress, a first moment of area (Q) needed to be calculated. This Q value was explained to me to be the area of the cross section of the beam on the opposite side of the desired horizontal shear plane.&#xA;&lt;img src=&quot;https://www.codecogs.com/users/23287/Shear-Stress-004.png&quot; alt=&quot;CSA diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the example above. In that diagram, the shear plane is a plane running along the longitudinal axis of the beam at a height $y$ above the neutral axis. The $Q$ value = [Area of the hatched region (which is all of the area on the opposite side of the shear plane from the neutral axis)]$\times$[$\bar{y}'$(height of the centroid of this area above the neutral axis)].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is how I understand Q. Now let's apply this understanding of Q to figure b in the first picture. The two shear planes are vertical planes that run along the longitudinal axis of the beam. Thus, a horizontal plane cannot be defined co-planer to the 2 vertical shear planes to denote that I want to calculate the shear flow along this horizontal plane. In short, my definition of Q has broken down spectacularly. How is Q defined for case (b)? &lt;/p&gt;&#xA;" OwnerUserId="9226" LastEditorUserId="-1" LastEditDate="2017-03-10T09:42:50.357" LastActivityDate="2016-12-22T23:19:00.390" Title="Determining the first moment of area (Q) when calculating the shear flow of a built up member" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12853" PostTypeId="1" CreationDate="2016-12-15T02:21:43.320" Score="3" ViewCount="51" Body="&lt;p&gt;This a not a homework or any similar task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding determine the determination of static nonlinearity, I know that one can draw two lines that contains the nonlinearity part. For instance for three cases below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t1zkCm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t1zkCm.jpg&quot; alt=&quot;three different non linearities&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition for sectors is&#xA;$ a \leq \frac{f(y)}{y} \leq b $ &#xA;Where $f(y)$ is shown in figure above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a) I draw a straight line with slipe of 1 and one line with slope of zero and for the other nonlinearities, the sectors are marked  as figures below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BuayAm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BuayAm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GmdX5m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GmdX5m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MNQNVm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MNQNVm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yellow area is the area that the sector should be. Regarding the sector area of the third curve. Isn't it misleading the the orange area could be accidentally interpreted as the sector area?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How would that look like for a backlash?&lt;/li&gt;&#xA;&lt;li&gt;How do I calclulate the gain of those nonlinearity? Assume that &#xA;$gain =Sup \frac{||f(y)||_2}{||y||_2}$&lt;/li&gt;&#xA;&lt;li&gt;When it comes to passivity, we know that $\int_0^T y^T u \geq 0$. But cant figure that out for those system and backlash. Whatever I  do I end up with all system being passive.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8884" LastEditorUserId="9" LastEditDate="2016-12-15T17:07:40.583" LastActivityDate="2016-12-15T17:07:40.583" Title="Sector determination of a nonlinearity and passivity" Tags="|control-engineering|automotive-engineering|control-theory|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12855" PostTypeId="2" ParentId="12838" CreationDate="2016-12-15T04:07:03.343" Score="2" Body="&lt;p&gt;A small plastic rack and pinion will work well and eliminate the need for strings. The rack will be on the drawer slider while the pinion at the hinge of the door. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x8GAA.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x8GAA.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8778" LastActivityDate="2016-12-15T04:07:03.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12857" PostTypeId="2" ParentId="12850" CreationDate="2016-12-15T07:26:22.560" Score="1" Body="&lt;p&gt;An infrared thermometer (when properly used) will give the temperature of the surface it sees. So if you measure the surface of the bearing casing, you are not seeing the internal bearing temperature. It depends on your application whether the casing temperature is close-enough to be useful in your measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it seems to me that if the infrared thermometer can see the surface, installing a regular RTD or other temperature sensor on the surface shouldn't be much of an issue. If drilling is too invasive, e.g. silicone-based glues offer good thermal transmission and are resistant to vibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify what I mean by &quot;when properly used&quot; above: it is relatively difficult to correctly apply infrared thermometers. The surface roughness and reflectivity affect the reading a lot. For example shiny metal surfaces may act as mirrors and cause the meter to accidentally measure the temperature of the reflected surroundings, instead of the surface you were aiming for. For paper and other translucent materials, it can read the temperature behind the object. However IR thermometers are not able to measure temperatures through metals.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-12-15T07:26:22.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12858" PostTypeId="2" ParentId="12851" CreationDate="2016-12-15T12:38:00.673" Score="1" Body="&lt;p&gt;In the general case, you can't.  Not all processes are reversible.  For example, a Cadmium-sulfide photoresistor won't emit light when you apply a voltage to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There really shouldn't be the expectation that this would work in general since there are so many every day examples.  Your car doesn't put gas back into the tank when going down a hill.  You don't get less hungry by walking down a mountain.  Etc, etc, etc.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-12-15T12:38:00.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12859" PostTypeId="1" CreationDate="2016-12-15T16:42:54.190" Score="1" ViewCount="15" Body="&lt;p&gt;We make plastic tensile bars per astm 638. The problem that I'm having is the bars are breaking outside the gauge marks in the reduced section of the samples. I'd like to know as to what I can do to prevent this ?&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="9231" LastActivityDate="2016-12-15T16:42:54.190" Title="Plastic tensile bars for Tye testing" Tags="|plastic|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12861" PostTypeId="1" CreationDate="2016-12-15T20:16:43.233" Score="2" ViewCount="115" Body="&lt;p&gt;I am looking for ideas for the following. I want to design a flexible plate where I can control the bending of the plate. In particular, a 5”x2”  plate will be used which can be bent  with some attached mechanisms.  The plate will be dragged in water so it should  not buckle under water pressure itself, rather its buckling should be controllable. I know there are certain shape memory alloys which can do that when powered (may be some piezoelectric actuator can do this also, I am not sure). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mtht6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mtht6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be designed with some actuating mechanism like this below:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LRl1c.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LRl1c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate your comments and suggestion. Thanks&lt;/p&gt;&#xA;" OwnerUserId="9237" LastEditorUserId="7083" LastEditDate="2016-12-16T00:01:58.530" LastActivityDate="2017-11-21T17:53:51.683" Title="designing a flexible plate" Tags="|mechanical-engineering|mechanisms|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12862" PostTypeId="2" ParentId="12822" CreationDate="2016-12-15T20:37:30.500" Score="6" Body="&lt;p&gt;As you might have already realized there is a good reason that some of Bluetooth headset are inexpensive. Poor design. High quality, well designed Bluetooth headset tend to be expensive. Also not all expensive Bluetooth headsets are high quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is difficult to root cause failures without physically analyzing the hardware, I will provide some possibilities. Since the device is a handset the device is using Bluetooth classic. Either way some understanding of Bluetooth technology might serve well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bluetooth operates in the 2.4 MHz frequency band, more specifically between 2402 and 2480 MHz frequency. If the 2 MHz bottom and 3.5 MHz upper guard band are included then the total bandwidth is 2400 to 2483.5 MHz.  Bluetooth classic uses 79 channels, of which each channel has a 1 MHz bandwidth. More importantly Bluetooth uses a radio technology called &lt;strong&gt;frequency hopping spread spectrum&lt;/strong&gt;, and if adaptive frequency hopping (AFH) is enabled it usually performs 800 hops a second. Bluetooth divides data into packets and each packet is transmitted on one of the 79 channels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LWdgo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LWdgo.png&quot; alt=&quot;Bluetooth Classic 80 channels&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When transmitting each of the data is subjected to &lt;a href=&quot;https://en.wikipedia.org/wiki/Path_loss&quot; rel=&quot;nofollow noreferrer&quot;&gt;path loss&lt;/a&gt; (or path attenuation). That is when the propagation of electromagnetic waves is subjected to a reduction in power density. In your case this path loss is influence by propagation medium such as dry or moist air as well as distance between the transmitter and receiver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your situation propagation losses would include absorption loss or penetration loss which a when the signal passes through a medium not transparent to electromagnetic waves or diffraction losses, which is when the radio signal is obstructed by an opaque obstacle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The signal transmitted by your device also takes multiple paths before arriving at the receiver. The signal received by the receiver might vary in strength, and propagation time etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A well designed system will account for all of the above and many more factors. In your case the inexpensive headset might have low &lt;strong&gt;Receive Signal Level&lt;/strong&gt; (RSL), or might not have accounted for &lt;strong&gt;fade margins&lt;/strong&gt;. A good design will include about 20 to 25 dbm fade margin. So if your headset has a RSL of 40 dbm, and a receive threshold is about 60 dbm. The design has accounted for 20 dbm of fade margin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that different environmental factors cause different level of signal attenuation. Example a water drop can cause about 5 dbm of attenuation. Not all environmental factors are present at all times. Depending on the application designer choose applicable fade margins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I have given you a good over view of Bluetooth, Radio Frequency and electromagnetic propagations help you make an informed technical decision on your next purchase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.aowireless.com/blog/bid/34066/Does-Weather-Effect-Wireless-The-5-Misconceptions-Part-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Does Weather Effect Wireless? The 5 Misconceptions - Part 1&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://smallbusiness.chron.com/signal-interference-issues-bluetooth-technology-58273.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;What Are Some Signal Interference Issues for the Bluetooth Technology?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://support.trangosys.com/hc/en-us/articles/216202223-Receive-Signal-Level-RSL-or-RSSI-is-Too-Low&quot; rel=&quot;nofollow noreferrer&quot;&gt;Receive Signal Level (RSL or RSSI) is Too Low&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/13292/110&quot;&gt;What is the difference between Bluetooth Low Energy and Bluetooth BR/EDR in Park mode?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-01-17T23:33:15.387" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12863" PostTypeId="1" AcceptedAnswerId="12887" CreationDate="2016-12-15T21:24:16.553" Score="2" ViewCount="435" Body="&lt;p&gt;I've been looking at using Aluminium 6082-T6 structures in impact energy absorption mechanisms. I planned to do some material testing, made specimens for testing then was distracted with another project for about a year. When I finally tested these specimens they broke in a surprisingly brittle manner with a much higher than expected proof stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering whether 6082-T6 is an alloy that age hardens at room temperature as one of the technicians suggested this as the cause. If so, does anyone know of any studies into this phenomenon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is would be rather important for impact energy absorption as increasing proof stress can result in higher risk of fatality. &lt;/p&gt;&#xA;" OwnerUserId="9233" LastActivityDate="2020-12-22T23:21:32.940" Title="Natural age hardening of aluminium 6082-t6" Tags="|materials|aluminum|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12864" PostTypeId="2" ParentId="12824" CreationDate="2016-12-15T22:27:30.823" Score="1" Body="&lt;p&gt;When drawing orthographic projections, it is convention to draw the &quot;front&quot; view in the middle of all the other views; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The &quot;top&quot; view is drawn above the front view,&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;the &quot;left side&quot; view is drawn to the left of the front view,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;the &quot;bottom&quot; view is drawn below the front view, and&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;the &quot;right side&quot; view is drawn to the right of the front view.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a &quot;back side&quot; view is desired, it is drawn either left of the &quot;left side&quot; view or right of the &quot;right side&quot; view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagonal areas of the page are used for isometric projections or renderings of the object in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the example below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://qph.ec.quoracdn.net/main-qimg-16a22851b6a38e07653d268a21fc73c3-c?convert_to_webp=true&quot; alt=&quot;Example Projections&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While all the views described above can be put onto a drawing to represent the object on each side. With the use of &lt;strong&gt;both object lines and hidden lines&lt;/strong&gt;, the whole object can be defined using exactly three of the views described above: one drawing for each dimension of space. Typically, the front and top view are drawn in conjunction with either the left or right side view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addressing your problem specifically: The problem you have been given seems to be a bit ambiguous as to which view is considered the front view and which view is considered the side view for this object. Based on your drawing, it seems you chose the left hand drawing to be the front view, and the right hand drawing to be the right side view. If you have been given no other instructions, this choice should have been fine; it was left for you to decide. However, those two drawings seem to be arranged on the page incorrectly. The left hand drawing seems to be the right side view of the right hand drawing but was placed on the left hand side for some reason (use the subtracted cylinder and different height &quot;shelves&quot; to verify this observation for yourself.) Because of this, I think the problem you have been given is poorly designed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we work off the assumption that the left hand drawing is the front view, then the 3rd view you have drawn should be the top view. However, based on the way you have drawn your object lines and hidden lines, it looks to me like you drew a bottom view. (You are trying to communicate all the geometry above the base with hidden lines.) If you did want to draw a bottom view, that view should be on the opposite side of the front view. Otherwise, you should redraw that top view, keeping in mind what object lines are visible, and which object lines are hidden.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: joojaa has enlightened me to the existence of the ISO 1st angle projection. OP, you seem to be drawing your views fine so long as you intended to use ISO 1st angle projection.&lt;/p&gt;&#xA;" OwnerUserId="9226" LastEditorUserId="9226" LastEditDate="2016-12-16T07:54:05.093" LastActivityDate="2016-12-16T07:54:05.093" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="12866" PostTypeId="2" ParentId="12823" CreationDate="2016-12-15T23:24:35.967" Score="2" Body="&lt;p&gt;Imagine a one dimensional beam with its left edge fixed at $x=0$ and its right edge at $x=L$. Now imagine you push on the right edge. What happens? The beam will contract. Let's concentrate on a infitissemal region at some point in the middle of the bar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens to this region? The proof you quote breaks the movement of this region into two pieces: a rigid motion, and a strain. The rigid motion cannot lead to any work over the little region, basically because the stress is uniform throughout the beam, so you have opposite forces on the left and right sides of the region, but the same displacement so total work is zero (this is the reminder before the proof).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the strain can cause work. The strain component of the motion corresponds to the center of the region remaining stationary and the ends moving in. In this case, the force is opposite at both ends, but the displacement is oppositie as well, so you get a net work. When we look at the work this way, the left boundary of our region moves to the right, while the right boundary of hte region to the left moves to the left, so there is no obvious cancellation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this explains why $W_d$ is non-zero, but to make things even more obvious, lets consider a beam of length $8L$ that gets compressed to a length of $4L$ upon application of a force $F$. Let's break this beam into four peices originally of length $2L$. In each piece's center of mass frame, both the ends move in by $L/2$, so the work done is $2 F L/2=FL$. and so the total work done is $4FL$. This is how you calculate things the $W_d$ way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of looking at things in the frame of each piece, you can look at things in the lab frame. For the leftmost piece, the left end doens't move, and the right end moves by $L$, so the work done on it is $FL$. For the next piece, the left end moves to the left by $L$, while the right end moves to the left by $2L$, so thet net work is $-FL+2FL$, which equals $FL$. Proceding in a similar way, the work ion the third and fourth pieces are $-2L+3L$ and $-3L + 4L$ respectively. So again we have that the work done on all pieces is $FL$.But we notice that the work at the interior interfaces cancel when we do the sum for the total work, $0 + FL - FL + 2FL-2FL+3FL-3FL+4FL$, thus the full work can be written just in terms of forces on the surface, and the interior contributions cancel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above answer was copied from the &lt;a href=&quot;https://physics.stackexchange.com/a/298822/23785&quot;&gt;physics stack exchange site&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9240" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2016-12-15T23:24:35.967" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12867" PostTypeId="1" CreationDate="2016-12-15T23:46:24.097" Score="2" ViewCount="423" Body="&lt;p&gt;I disassembled a usb wifi adapter. I was surprised to see a 40MHz oscillator, when wifi works at 2.4GHz. How the output of the oscillator is scaled to the required frequency?&lt;/p&gt;&#xA;" OwnerUserId="6614" LastEditorUserId="282" LastEditDate="2016-12-17T01:14:30.760" LastActivityDate="2016-12-17T01:14:30.760" Title="A 40MHz oscillator in usb wifi adapter" Tags="|electrical-engineering|telecommunication|wireless-communication|rf-electronics|wifi|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12868" PostTypeId="1" CreationDate="2016-12-16T06:23:30.723" Score="2" ViewCount="137" Body="&lt;p&gt;WiFi at 60 GHz provides a theoretical data-rate of 7 Gbit/s. The trade-off is limited range and penetration. The main reason for this awesome data-rate is a 9 GHz bandwidth, from 66 GHz to 57 GHz. I think this is country/regulation dependent, so 7 Gbit/s doesn't depend on the full 9 GHz bandwidth. Instead of focusing on increased frequency, why not scrap all the obsolete bands and reassign the 1 GHz - 6 GHz range to WiFi?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The range and penetration of the 1 GHz - 6 GHz bands limit suitability to home usage. Therefore, does it need to be regulated as part of a global spectrum? Regulations should ensure availability of some bandwidth for other &lt;em&gt;home&lt;/em&gt; uses only:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Punch out some small bands for bluetooth, cordless phone, TV remote, etc. (below the 1 GHz cutoff: cellular phone, NFC, etc.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Adaptive frequency hopping to avoid local interference (other routers, microwave, etc.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Backport latest technology from previous standards: QAM, MIMO, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Make the standard versioned so newer features can be tacked on without requiring a whole new range.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Right? Now tell me why this is a pipe dream.&lt;/p&gt;&#xA;" OwnerUserId="9242" LastEditorUserId="368" LastEditDate="2016-12-16T19:55:25.683" LastActivityDate="2016-12-19T09:38:56.047" Title="Why is WiFi increasing frequency?" Tags="|telecommunication|wifi|regulations|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12869" PostTypeId="1" CreationDate="2016-12-16T07:56:35.393" Score="4" ViewCount="2047" Body="&lt;p&gt;I would like to know how to synchronize two (nearly)-identical DC motors (approx. 1 Amp/12 VDC each) with permanent magnet, while they are driving on the same surface (railroad Track).&#xA;The problem is that 2 motors of the same brand and type always have somewhat different characteristics at different speeds, resulting in (unnecessary) heating of one of the engines and its bearings. &#xA;I'd like to sync them in an elect(ron)ic way as simple as possible.&#xA;Thanks for help. A.&lt;/p&gt;&#xA;" OwnerUserId="9243" LastEditorUserId="8055" LastEditDate="2016-12-16T14:44:47.357" LastActivityDate="2021-02-19T09:54:36.913" Title="DC Motors synchronization" Tags="|electrical-engineering|motors|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12870" PostTypeId="1" CreationDate="2016-12-16T09:17:49.177" Score="2" ViewCount="546" Body="&lt;p&gt;I'm making one project for my class, some kind of smart wristband, but not sure which type of battery should I use. I found a lot of good wristband online, I see that they usually have 60 - 100mAh batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone tell the type of batteries used in Smart wristbands?&lt;/p&gt;&#xA;" OwnerUserId="9244" LastEditorUserId="7083" LastEditDate="2016-12-17T01:14:24.137" LastActivityDate="2016-12-17T01:14:24.137" Title="What type of batteries are used in smart wristband?" Tags="|electrical-engineering|battery|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12871" PostTypeId="2" ParentId="12345" CreationDate="2016-12-16T09:53:45.150" Score="2" Body="&lt;p&gt;Having a sequence of gears connecting the driving and driven gears that all mesh in the same plane will produce a gear ratio independent of the intermediate gears: the gear ratio will be the number of teeth on the driven gear divided by the number of teeth on the driver gear. The following image shows this type of arrangement (not all teeth are shown):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xyrUd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xyrUd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is also the case for arrangement 2 because, even though there exists a gear out of plane, all of the gears involved in transmitting rotation from driver to driven all lie in plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, whenever you have rotation transmitted from one gear to a gear of a different size via a shaft, as is the case for the cluster gear in arrangement 1, the gear ratio does depend on the intermediate gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;IN-PLANE&quot; GEAR TRAIN&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why the gear ratio is independent of intermediate gears for a simple chain is because all the meshing gears have the same circumferential speed (to be more precise, the tangential speed of a gear at its pitch circle): any two meshing gears must have the same circumference speed, and this extends to all the gears because they all lie in plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H2pFK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H2pFK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above diagram, I use the following notation: $V$ is the circumferential speed, $\omega_D$, $\omega_d$ are the angular velocities and $r_D$, $r_d$ are the pitch circle radii for the driving (D) and driven (d) gears respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given $V$ is the same everywhere in an in-plane gear train, it can be shown that, according to kinematics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \omega_D r_D = \omega_d r_d$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the (reduction) gear ratio is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G = \frac{\omega_D}{\omega_d} = \frac{r_d}{r_D} \implies G = \frac{N_d}{N_D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_D$ and $N_d$ are the number of teeth on the driving and driven gears respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so it can be seen that the gear ratio is indeed independent of the intermediate gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;COMPOUND GEAR TRAIN&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For gear trains like in arrangement 1, where at least one pair of gears involved in transmission motion are linked by a shaft, the circumferential speed is not the same everywhere:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8kgAm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8kgAm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above, by kinematics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \frac{V_1}{r_1} = \frac{V_2}{r_2} \rightarrow r_1 \ne r_2 \implies V_1 \ne V_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, to determine the gear ratio of such a system, it helps to consider gear ratios of sub-systems consisting of only in-plane gear trains:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i5EQR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i5EQR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above diagram, there are two sub-systems of in-plane gears: the lower set and the upper set. Consider the lower set of gears having its own gear ratio, noting that all the gears in this set are in-plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_1=\frac{\omega_D}{\omega_1}=\frac{N_1}{N_D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, for the upper sub-system of gears:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G_2=\frac{\omega_1}{\omega_d}=\frac{N_d}{N_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the overall gear ratio can be determined:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G = \frac{\omega_D}{\omega_d}=\frac{\omega_D}{\omega_1} \frac{\omega_1}{\omega_d} = G_1 G_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, it can be shown that for a compound (out-of-plane) gear train, the gear ratio is equal to the gear ratios of in-plane sub-systems all multiplied together.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2016-12-16T09:53:45.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12872" PostTypeId="2" ParentId="12867" CreationDate="2016-12-16T10:34:07.853" Score="2" Body="&lt;p&gt;Generally when you need to scale up a clock you use a PLL (Phase lock loop).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very crude explanation is that you have a voltage controlled oscillator at roughly the required frequency. You divide the output of this down to your reference frequency (a factor of 60 in this case) and then compare it to your reference clock input. The difference between the two is then used to adjust the frequency of the voltage controlled oscillator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system will take a short time period to lock on and stabilize but after that you end up with a clock which is an exact multiple of the input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is glossing over a lot of the details but google for &quot;PLL theory of operation&quot; or similar and you'll find more details than you ever wanted to know.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-12-16T10:34:07.853" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12873" PostTypeId="1" CreationDate="2016-12-16T11:20:10.063" Score="6" ViewCount="184" Body="&lt;p&gt;I consider a spherical nanoparticle (NP) that is shrinking, for instance due to a melting process. I know that the pressure $p$ is proportional to $1/R(t)$, where $R(t)$ is the radius of the NP at time $t$. Conservation of energy states &#xA;\begin{equation}&#xA;\rho\frac{\partial u}{\partial t}=\nabla\cdot\vec q,&#xA;\end{equation}&#xA;where $\rho$ is the density, $u$ is the internal energy and $\vec q$ is the heat flux given by Fourier's law,&#xA;\begin{equation}&#xA;\vec q=-k\nabla T.&#xA;\end{equation}&#xA;The classical heat equation is obtained by assuming that the volume is constant along the whole process, which leads to $du=c_vdT$. Under constant pressure, a similar equation holds, derived from&#xA;\begin{equation}&#xA;dh=du+d(pv),&#xA;\end{equation}&#xA;where $h$ is the specific enthalpy.&#xA;In this case, where neither the volume nor the pressure is constant, in which way could the internal energy be related to the temperature to obtain an equation involving only $T$? The only quantity which is kept constant in this case is the density, but I don't know how to go on...&#xA;Any help would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="9247" LastActivityDate="2024-04-02T20:09:11.133" Title="Derive a heat equation for a shrinking nanoparticle" Tags="|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12874" PostTypeId="2" ParentId="12870" CreationDate="2016-12-16T11:28:01.923" Score="1" Body="&lt;p&gt;Looks like smart wristbands are using  lithium polymer batteries. Here are some examples.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Jawbone UP24 JL01-52M-US - Bluetooth 4.0 Wristband - &lt;strong&gt;32 mAh Li-ion polymer&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Xiaomi Mi Band XMSH01HM - Bluetooth 4.0 Fitness Wristband - &lt;strong&gt;3.7 V, 41 mAh Li-Polymer&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Atlas Wristband A101 - &lt;strong&gt;3.7 V, 120 mAh, Li-Polymer&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Amzer FitZer KA - Bluetooth Activity Tracker - &lt;strong&gt;3.7 V, 70 mAh, Li-Polymer&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://techinsights.com/reports-and-subscriptions/open-market-reports/Report-Profile/?ReportKey=10045&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jawbone UP24 JL01-52M-US - Bluetooth 4.0 Wristband&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://techinsights.com/reports-and-subscriptions/open-market-reports/Report-Profile/?ReportKey=10231&quot; rel=&quot;nofollow noreferrer&quot;&gt;Xiaomi Mi Band XMSH01HM - Bluetooth 4.0 Fitness Wristband&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.techinsights.com/reports-and-subscriptions/open-market-reports/Report-Profile/?ReportKey=11211&quot; rel=&quot;nofollow noreferrer&quot;&gt;Atlas Wristband A101 - Bluetooth Activity Tracker&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.techinsights.com/reports-and-subscriptions/open-market-reports/Report-Profile/?ReportKey=11259&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amzer FitZer KA - Bluetooth Activity Tracker&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.techinsights.com/teardown.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Teardown Reports&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2016-12-16T11:28:01.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12877" PostTypeId="2" ParentId="12870" CreationDate="2016-12-16T15:02:57.550" Score="1" Body="&lt;p&gt;As Mahendra's answer indicates just about all of the commercial smart bands use Li-something based batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if this is for a class project I'd recommend using something like a CR2032. If you buy a bulk pack then they are relatively cheap and hold about 200mAh each.&#xA;For testing and development you can easily replace the battery holder with a wired connection to two AA batteries to prevent burning through too many batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your end product won't be rechargeable but as a class project you probably don't want to spend half of your time designing battery charging and monitoring circuits.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2016-12-16T15:02:57.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12878" PostTypeId="1" AcceptedAnswerId="12886" CreationDate="2016-12-16T15:34:49.250" Score="1" ViewCount="601" Body="&lt;p&gt;I made a calculation for a pipe that needs to handle 1&amp;nbsp;MPa pressure. I based my calculation on &lt;code&gt;ASME VIII div1&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My calculation says that my pipe needs to be 3&amp;nbsp;mm thick. But do I also need to account for the tolerances pipe manufactures have on the thickness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This should be 12,5% according to a colleague of mine (sorry I don't have the source) but I am quite confident he is correct about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find a note in the &lt;code&gt;ASME VIII div1&lt;/code&gt; about this maybe I over-read this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is: do I need to add the pipe wall thickness tolerance to my calculated thickness to be theoretically correct according the &lt;code&gt;ASME VIII div1&lt;/code&gt;? &lt;/p&gt;&#xA;" OwnerUserId="8464" LastEditorUserId="1832" LastEditDate="2016-12-16T18:39:32.433" LastActivityDate="2016-12-17T01:23:53.873" Title="Pipe wall thickness and production tolerances" Tags="|mechanical-engineering|pipelines|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12879" PostTypeId="2" ParentId="12820" CreationDate="2016-12-16T18:57:12.987" Score="1" Body="&lt;p&gt;First, a bit of background. I spent 6 years running a packaging test lab here in the U.S. and part of my job was collecting shock and vibration data from the field for different transportation modes (semi tractor/trailers, TOFC, railcar, delivery trucks, ships...etc) around the world, then using that data to develop and/or validate lab test protocols. I've also done vibration measurement and analysis on rotating machinery and vehicles. So, not exactly what you're doing, but pretty close. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For measurements, I've typically used self contained &quot;black box&quot; data recorders from &lt;a href=&quot;http://isthq.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;IST&lt;/a&gt;, but these recorders cost about $10k each and are probably out of your budget. They do rent equipment, so it might be worth looking into and maybe there's an affordable equivalent wherever in the world you are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you decide you want to homebrew your own hardware, I would be cautious about mounting an Arduino or Ras-Pi to a skateboard without taking steps to protect it, they're simply not ruggedized for the dynamic environment. That, and you'll have real problems sampling data fast enough to avoid aliasing. The typical rule of thumb is that the sampling rate should be 10 the highest frequency of interest. Proceed at your own risk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were doing this, I'd look for a way to rent/borrow a real portable dynamic signal analyzer...something like this &lt;a href=&quot;http://www.crystalinstruments.com/wireless-analysis-anywhere&quot; rel=&quot;nofollow noreferrer&quot;&gt;Crystal Instruments Spyder-20&lt;/a&gt; . I'd setup a hall sensor or small optical sensor to use as a tach input (0-5V pulse train). Then a miniature accelerometer like &lt;a href=&quot;http://www.pcb.com/products.aspx?m=352A21&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; from PCB mounted to the skateboard truck to detect vibration vertically. Setup the DSA to trigger data collection when it senses input from the tach. Put the DSA in your backpack and go skate on different streets/surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your speed will vary, there's probably not much value in comparing acceleration vs time data or histograms between data collected on different surfaces except as maybe a gut check. I would take the data and process it into different waterfall plots the first plotting time vs frequency vs intensity like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lp7gn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lp7gn.jpg&quot; alt=&quot;Time v Freq v Intensity&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and another with speed vs frequency vs intensity, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2DqCx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2DqCx.jpg&quot; alt=&quot;Speed v Freq v Intensity&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to look at other things for the Z axis, but in general this would be a good start and you can narrow in on specific areas of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also look at the fundamental frequencies that will be created by the bearings in your skateboard wheels and apply some filtering to remove that from your data as well. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2016-12-16T18:57:12.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12881" PostTypeId="1" AcceptedAnswerId="13012" CreationDate="2016-12-16T21:13:10.593" Score="1" ViewCount="321" Body="&lt;p&gt;One of the cellular vendors wants to install a 180 foot monopole antenna in our neighborhood.  They claim that there is significant growth in LTE consumers. So in order to better serve the customers they need to install a new monopole. Following are two reasons that they provided recently.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The current towers are over capacity thus cannot reliably serve the&#xA;consumers&lt;/li&gt;&#xA;&lt;li&gt;The cellular signal is weak therefore consumers cannot make phone calls&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; What are the alternate ways to solve this problem beside installing a new monopole?  Are there new  LTE/Cellular technology that can help reduce cell tower capacity as well as permit the consumer to make phone calls reliably using cells phones?  What are any future technology that might address this issue?&lt;/p&gt;&#xA;" OwnerUserId="7083" LastEditorUserId="8055" LastEditDate="2016-12-28T22:43:14.557" LastActivityDate="2017-01-28T03:15:44.157" Title="What are alternative to installing LTE / Cell Tower to address capacity issue?" Tags="|electrical-engineering|wireless-communication|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12882" PostTypeId="1" AcceptedAnswerId="12968" CreationDate="2016-12-16T21:19:06.343" Score="1" ViewCount="71" Body="&lt;p&gt;I am currently designing a small weather station that will be suspended from two fiber optic cables and a power wire. You can imagine for now a three point attachment to each wire with a wind vane suspended below, like &lt;a href=&quot;https://www.dropbox.com/s/r0jf7pxd119rt79/20161218_095830.jpg?dl=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;so&lt;/a&gt;. I need some help figuring out the best way to support it while allowing rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I would like to be able to have the device adjust its pitch so that if one fiber gets pulled vertically, the wind vane will stay level. The yaw adjusted by the force from wind. I am not sure how important roll will be to add in, so for now I have not considered it too much.&#xA;The reason for controlling pitch is that as the fibers are pulled up in the air, one may be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My basic idea now is to have a rod go through a small hole in the top part of the device, where the device can rotate, adjusting the pitch. You can see the 3 holes for inserting that rod; &lt;a href=&quot;https://youtu.be/sNi7g8Lp--c&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; is a video of the device. I would choose one nearest the center of mass (to be determined). For a sense of scale, those holes are about 1.5 mm in diameter. The whole device is small, and should be about 100-200g total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plan to have a clevis structure attach to that rod; what my fingers do in the video. After that it will have to go to an assembly above that will allow yaw rotation (not designed yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I find with this design for pitch is there is no damping, so the device oscillates upon disturbances instead of finding an equilibrium point. Also, I have had some trouble finding rods and other parts at this size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions on how to best design this and parts that may be useful would be great. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(P.S. I tried putting the third image, but stack exchange limits me to 2 links until my reputation score is at least 10.)&lt;/p&gt;&#xA;" OwnerUserId="4819" LastEditorUserId="4819" LastEditDate="2016-12-18T18:05:23.843" LastActivityDate="2016-12-25T16:36:56.813" Title="Weather Station Gimbal Support" Tags="|mechanical-engineering|bearings|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12883" PostTypeId="1" CreationDate="2016-12-16T23:47:23.420" Score="2" ViewCount="393" Body="&lt;p&gt;I'm just an average person who questions the validity of my treadmill's claims about how many calories I've burned after an hour's worth of a cardio workout.  How can I get a more accurate measurement based on pre and post workout conditions?  I suspect the most accurate approach is to use rather sophisticated/expensive equipment (i.e. things I don't have/can't afford), but I'm also curious if there is a low cost/low tech approach to measure calories burned without relying solely on a heuristic formula? &lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="8055" LastEditDate="2016-12-19T17:23:33.297" LastActivityDate="2024-02-05T15:41:06.220" Title="How to measure calories actually burned from exercise?" Tags="|measurements|biomechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12884" PostTypeId="1" CreationDate="2016-12-17T00:20:42.453" Score="5" ViewCount="4757" Body="&lt;p&gt;I don't really understand the explanation in the circled part of the image below (from &lt;a href=&quot;https://en.wikipedia.org/wiki/Mohr&amp;#39;s_circle&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the diagram, we can see that the shear stress that cause the element to turn clockwise is positive, but in the text there, the author stated that the shear stress that cause the element that to turn clockwise is negative. So, which is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wRPn2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wRPn2.jpg&quot; alt=&quot;2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;img src=&quot;https://i.stack.imgur.com/Nejr6.jpg&quot; alt=&quot;image2[3]&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9254" LastEditorUserId="1832" LastEditDate="2016-12-17T01:07:13.013" LastActivityDate="2016-12-29T15:59:19.470" Title="Sign convention in Mohr's circle" Tags="|structural-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12885" PostTypeId="1" CreationDate="2016-12-17T00:56:49.720" Score="3" ViewCount="409" Body="&lt;p&gt;I read that the mercury vapour inside a fluorescent lamp is pressurized at 0.2% of atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can mercury escape undectably through a small crack in the glass yet still light up? Or will the pressure difference cause it to explode and/or stop working?&lt;/p&gt;&#xA;" OwnerUserId="9255" LastActivityDate="2016-12-17T19:37:25.920" Title="Can mercury escape undectectably from a CFL bulb? or does it have to shatter for the mercury to be liberated?" Tags="|electrical-engineering|pressure|gas|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12886" PostTypeId="2" ParentId="12878" CreationDate="2016-12-17T01:23:53.873" Score="1" Body="&lt;p&gt;The ASME calculation gives you a minimum allowable wall thickness.  You have to ensure that the actual thickness is equal to, or greater than the minimum allowable.  Therefore you have to account for the manufacturing tolerance by specifying a pipe schedule which is certain to be at least the minimum calculated value.  So yes, you have to add the tolerance to the calculated value.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2016-12-17T01:23:53.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12887" PostTypeId="2" ParentId="12863" CreationDate="2016-12-17T01:51:10.397" Score="2" Body="&lt;p&gt;Just leaving the alloy sitting on a shelf for a year at room temperature shouldn't have any effect on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, 6082-T6 &quot;as supplied&quot; has already been heat treated and artificially aged. That's what the -T6 temper designation means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted &quot;soft&quot; ductile material for energy absorption, maybe you should have used 6082-O (or some other temper designation) instead of -T6. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is that the manufacturing operations to make your test specimens affected the state of the material, and the finished specimens should  have been heat treated again. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-12-17T01:51:10.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12888" PostTypeId="1" CreationDate="2016-12-17T05:28:14.123" Score="1" ViewCount="157" Body="&lt;p&gt;I want to machine a spiral bevel gear on a manual universal milling machine using a involuter  gear cutter. I also have a differential indexing setup for this job. But the difficulty I am facing is that how to calculate lead for this gear so that I can link up indexing head with feed . All I could find online with some research was that mean bevel angle is 35~40 degree. I would appreciate some links to reliable sources for more research&lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2016-12-17T05:28:14.123" Title="Lead calcuation for spiral bevel gear" Tags="|mechanical-engineering|gears|machining|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12889" PostTypeId="1" CreationDate="2016-12-17T08:48:53.780" Score="1" ViewCount="33" Body="&lt;p&gt;I wonder which permeation properties are critical in concrete with and without surface treatment (e.g sodium silicates-based solution) , is it air permeability,water permeability or absorption. If u were to measure two of them, what would u choose? &lt;/p&gt;&#xA;" OwnerUserId="9259" LastActivityDate="2016-12-17T08:48:53.780" Title="permeation properties of concrete under surface treatment" Tags="|civil-engineering|chemical-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12890" PostTypeId="2" ParentId="12883" CreationDate="2016-12-17T10:58:57.903" Score="3" Body="&lt;p&gt;This is actually quite a difficult problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring the work done by a &lt;em&gt;specific&lt;/em&gt; exercise output, eg treadmill, cycling, rowing machine etc is easy enough. However this only measures what you might call 'shaft horsepower' and ignores any effort that doesn't go directly into the output being measured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also ignores the efficiency of the underlying metabolic processes ie how much food you use up to produce a specific amount of mechanical work and it is certainly reasonable to assume that the amount of energy consumed to perform a specific amount of work will vary somewhat from person to person.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of setting up an experiment the big difficulty is setting meaningful system boundaries such that you can measure energy input and output, especially since the human body has several different mechanisms for storing energy. For example a sensible way to measure the rate of metabolism is to measure the amount of carbon dioxide you exhale as that is a direct product of respiration and really you need to measure everything which goes in and out of the body during a period of exercise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weighting yourself doesn't help much either on its own as your mass can vary a lot due to fluctuations in water and waste products which aren't directly related to energy consumption and you will certainly lose a certain amount of water mass due to sweating and exhalation during exercise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that even knowing exactly how many calories you have used doesn't tell the whole story as you don't necessarily know how much usable energy is actually absorbed from a given quantity of food and unless you are eating precisely measured and processed food calorie counts are an average at best. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, measuring output work from an exercise machine is as good as you are realistically going to get without very specialist equipment, as that is at least measuring something specific and well defined and if nothing else you do know that you have expended at least that amount of energy. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-12-19T18:20:57.730" LastActivityDate="2016-12-19T18:20:57.730" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12891" PostTypeId="1" AcceptedAnswerId="12893" CreationDate="2016-12-17T14:46:56.897" Score="7" ViewCount="5240" Body="&lt;p&gt;I am working on product where we need pogo pins, as copper is better conductor of electricity I was searching for the copper pogo pins. After about a month of searching, it is come to my notice that most of the manufacturers are dealing with pogo pins not of copper but of brass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when I inquired about the pogo pin that suited voltage specification of 250V 15A. The manufacturers insisted on using Brass than copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from this I also seen many smart watches which were using pogo pins were using Brass instead of copper?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any basic issue with using copper? What are the scenarios when we should chose Brass over the copper while building electrical connectors?&lt;/p&gt;&#xA;" OwnerUserId="8900" LastActivityDate="2016-12-17T22:27:24.677" Title="Why manufacturers prefer Brass over Copper in electrical conductors?" Tags="|mechanical-engineering|electrical-engineering|materials|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12892" PostTypeId="2" ParentId="12885" CreationDate="2016-12-17T15:15:11.530" Score="4" Body="&lt;p&gt;If there were a crack in the glass, then the pressure difference would first cause air to flow into the lamp, which would cause it to stop working (both the mercury and the low pressure are necessary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there &lt;em&gt;were&lt;/em&gt; a way for mercury to escape from the lamp while it still worked, it would most likely be at the glass-to-metal seal around the filament/electrode terminals. But if that were a problem then lots of bulbs would be leaky and it would be fixed by design.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is &lt;em&gt;possible&lt;/em&gt; for a fluorescent lamp to break (and therefore eventually lose its mercury) without “exploding”. I once witnessed just such an event, which might be of interest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lamp was a spiral-style CFL lamp pointing sideways (in an above-the-mirror bathroom fixture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't remember at what point it stopped lighting or if the failure happened when it was switched on. But in any case, the filament at one end glowed brightly and broke. The hot filament fell down against the glass, which then cracked at that point, with a distinct sound, making a small hole. The rest of the glass remained intact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The overheated filament would have been caused by a failure in the starting circuit.)&lt;/p&gt;&#xA;" OwnerUserId="9262" LastEditorUserId="9262" LastEditDate="2016-12-17T19:37:25.920" LastActivityDate="2016-12-17T19:37:25.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12893" PostTypeId="2" ParentId="12891" CreationDate="2016-12-17T16:59:31.123" Score="9" Body="&lt;p&gt;Pure copper is a very soft metal and while it's ductility is good for wires and other flexible connectors it lacks the mechanical strength required for pins etc. Also brass can have somewhat better corrosion resistance and so may be preferred for mechanical contacts as while pure copper has good conductivity this advantage is lost as soon as you get a layer of surface oxides.  This doesn't really matter for something like a PCB where oxidation only affects the outside surface of the tracks but where you have metal to metal contact without something like a soldered joint surface conductivity becomes much more important &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brass is also somewhat easier to manufacture as it casts well and is easy to machine and the huge range of brasses available allow you to tailor its properties to the application to a great extent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally in many connectors the mechanical strength requirements mean that they may have quite a large cross-sectional area in any case, consider a UK 13 A plug, the pins are very chunky compared to the to the wires of equivalent rating. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2016-12-17T22:27:24.677" LastActivityDate="2016-12-17T22:27:24.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12894" PostTypeId="2" ParentId="11904" CreationDate="2016-12-17T18:06:52.037" Score="1" Body="&lt;p&gt;One approach is to use a gravity trap so that the liquid flows to the bottom of a sump. The sump can then be isolated from the rest of the system with a valve to allow it to drain under gravity a bit like an airlock. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-17T18:06:52.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12897" PostTypeId="2" ParentId="12891" CreationDate="2016-12-17T21:18:26.293" Score="3" Body="&lt;p&gt;For pogo pins to work correctly, there must be a good electrical connection across the sliding joint between the tip and base parts of the pin. If the joint develops a high resistance and current flows through the spring inside the pin, that introduces a random and potentially variable amount of inductance into the connection which can affect high frequency signals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper is a very &quot;soft&quot; material after it has been annealed, but any mechanical strain causes significant work hardening in the material which would lead to poorly-defined geometric tolerance and mechanical stiffness across the sliding joint. Brass doesn't have those disadvantages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create the required mechanical stiffness, copper pins would need to be thicker than brass, but thin brass pins already have good enough electrical conductivity for the application, so the mechanical disadvantage of using copper are more important than its theoretical electrical advantage.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-12-17T21:18:26.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12898" PostTypeId="1" CreationDate="2016-12-17T21:25:08.480" Score="2" ViewCount="195" Body="&lt;p&gt;I have the following SISO (Single Input, Single Output) system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;\dot{x}_1 \\ \dot{x}_2&#xA;\end{bmatrix}&#xA;=&#xA;\begin{bmatrix}&#xA;0 &amp;amp; 1 \\ -1 &amp;amp; -1&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;x_1 \\ x_2&#xA;\end{bmatrix}&#xA;+&#xA;\begin{bmatrix}&#xA;0 \\ 1&#xA;\end{bmatrix}&#xA;u&#xA;$$&#xA;$$&#xA;y = \begin{bmatrix}&#xA;1 &amp;amp; 0&#xA;\end{bmatrix} \begin{bmatrix} x_1 \\ x_2&#xA;\end{bmatrix}&#xA;$$&#xA;I know how to calculate the step response of that system (open loop and closed loop).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have the same system but another control variable:&#xA;$$&#xA;\begin{bmatrix}&#xA;\dot{x}_1 \\ \dot{x}_2&#xA;\end{bmatrix}&#xA;=&#xA;\begin{bmatrix}&#xA;0 &amp;amp; 1 \\ -1 &amp;amp; -1&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;x_1 \\ x_2&#xA;\end{bmatrix}&#xA;+&#xA;\begin{bmatrix}&#xA;0 \\ 1&#xA;\end{bmatrix}&#xA;u&#xA;+ &#xA;\begin{bmatrix}&#xA;-40 \\ 0&#xA;\end{bmatrix}&#xA;\omega&#xA;$$&#xA;$$&#xA;y = \begin{bmatrix}&#xA;1 &amp;amp; 0&#xA;\end{bmatrix} \begin{bmatrix} x_1 \\ x_2&#xA;\end{bmatrix}&#xA;$$&#xA;And I have to calculate the step response for open loop and closed loop (I suppose I have to use integral action, but maybe another technique is better). That's a MISO system and I don't know how to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I know that the system is linear so maybe I can calculate assuming that first B = [0 1] (transpose) and then B = [-40 0] (transpose) and then applying superposition or something like that&lt;/p&gt;&#xA;" OwnerUserId="8089" LastEditorUserId="8089" LastEditDate="2016-12-17T21:42:04.023" LastActivityDate="2017-02-27T06:15:44.590" Title="From Single to Multiple Input (from SISO to MISO)" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12901" PostTypeId="2" ParentId="12844" CreationDate="2016-12-18T14:29:42.343" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Determine expected max aircraft speed, and engine thrust at that speed and when stationary.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, sort of. At the very beginning, you may know the &lt;em&gt;desired&lt;/em&gt; thrust at a given condition, but you have to work all the way through the calculations to get the &lt;em&gt;actual&lt;/em&gt; thrust at that speed. If the actual thrust and desired thrust don't match, then you would need to change the design assumptions (e.g. change fan diameter, or compressor pressure ratio or something) in order to meet that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to say is, I think you may be combining the concept of how do I design a jet engine to meet a certain performance goal, with how do I evaluate the performance of a particular jet engine design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do the later, you start with upstream pressure, upstream temperature, and velocity, and the known design of your engine, and walk from front to back.  At the end of that process you will know thrust, mass flow, fuel consumption, etc for that given design and operating conditions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do the former, you start with a set of assumption, calculate the performance, and iterate until you have the desired design. For a large commercial turbofan, this process would typically be done to optimize the performance at cruise (since that is where the plane spends the majority of them), and then you would verify that other conditions (e.g. sea level static takeoff) are acceptable.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2016-12-18T14:29:42.343" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12903" PostTypeId="1" AcceptedAnswerId="12905" CreationDate="2016-12-18T21:40:30.803" Score="0" ViewCount="108" Body="&lt;p&gt;I noticed that blades of HP compressor are shorter in axial dimension next to its LP counterpart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand why it is smaller in radial dimension (this is because compressed air has smaller volume, and whole air passage is converging), but it seems to me that longer (again, in axial dimension) blades can transfer more energy to the air being compressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is obvious to be some good reason to make them small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why?&lt;/p&gt;&#xA;" OwnerUserId="9221" LastActivityDate="2016-12-19T16:05:50.393" Title="turbojet axial compressor: blades of HP compressor" Tags="|turbines|turbomachinery|propulsion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12904" PostTypeId="2" ParentId="12868" CreationDate="2016-12-19T09:38:56.047" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;why not scrap all the obsolete bands and reassign the 1 GHz - 6 GHz range to WiFi?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;First of all, these bands are not obsolete, there are many uses that might seem outdated, but won't go away easily. Check for example the use of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultra_high_frequency&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ultra high frequency&lt;/a&gt; band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice also that - yes - the use is country dependent, so if one of these bands have a critical application somewhere, it will not be available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the RF spectrum is a managed and finite resource. In just about every country the portions of the spectrum is auctioned for use. For mobile telephony there is a business that can pay for the use; for WiFi there is not. Don't expect authorities to give a spectrum for free. Perhaps they shouldn't, since part of the spectrum could better be used for other technologies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the free ISM bands the are utilised by WiFi are mainly free because they have been considered useless for other purposes. And since this is a physical property the band happens to be free in most of the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One common standard is a nice theoretical idea, but it requires the world to agree on spectrum use. It will also require a large semiconductor industry to redevelop their product lines, which is for sure less attractive than gradually tuning what they have. I think some better incentives than &quot;better WiFi&quot; is needed for this to fly.&lt;/p&gt;&#xA;" OwnerUserId="9278" LastActivityDate="2016-12-19T09:38:56.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12905" PostTypeId="2" ParentId="12903" CreationDate="2016-12-19T16:05:50.393" Score="1" Body="&lt;p&gt;There is a practical limit to the amount of compression that can be achieved in a single stage of an axial compressor. As a general principle, the rotor blades increase the velocity of the gas without much increase in pressure, and the stator blades increase the pressure while reducing the velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The changes in the gas velocity are mainly in the &lt;em&gt;circumferential&lt;/em&gt; direction. The maximum circumferential exit velocity of the gas that can be produced is related to the circumferential velocity of the rotor blades themselves (i.e. radius times angular velocity). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can design a blade with a short chord length which can do that (as in real-life compressors) there is no value in making the chord length of a single compressor stage longer. Even if you could get a higher compression ratio when the machine was running at one particular design condition of flow rate and gas input temperature and pressure, there would be a greater loss of performance and efficiency away from that design point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way you get a &quot;longer chord length,&quot; to give a higher compression ratio, is by making a multi-stage compressor with alternating rotors and stators. Jet engine compressors may have 10 or more stages in total, on two or three different rotors running at different speeds.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-12-19T16:05:50.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12907" PostTypeId="2" ParentId="11178" CreationDate="2016-12-19T21:46:01.467" Score="2" Body="&lt;p&gt;The required strain is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; (0.125 - 0.115)/0.115 = 0.087&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;the thermal expansion coefficient of aluminum is ~ 22 10^-6/C, so to expand that much you need to heat to&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 0.087/(22 10^-6) or ~4000C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;ie You will melt the aluminum long before it expands that much.&#xA;Note the drawing tolerance is -0/+0.01, so maybe buy a bunch and you will get lucky :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hunt around for a split tube..&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2016-12-19T21:46:01.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12909" PostTypeId="1" CreationDate="2016-12-20T03:12:31.233" Score="0" ViewCount="53" Body="&lt;p&gt;I want to design a leakage detection system along the pipeline that will use differences in pressure of flowing liquid (change in pressure between two points). Which sensor I should use?&lt;/p&gt;&#xA;" OwnerDisplayName="E Alphonce" LastEditorUserId="6790" LastEditDate="2020-08-01T15:07:13.360" LastActivityDate="2020-08-01T15:07:13.360" Title="Leakage detection system along the pipe" Tags="|design|sensors|pipelines|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="12911" PostTypeId="1" AcceptedAnswerId="12927" CreationDate="2016-12-20T15:49:12.553" Score="1" ViewCount="25" Body="&lt;p&gt;I need to conduct nano-indentation within a glovebox of a split-test cell for battery testing. I know that vibration may be a potential issue, and I should probably buy an isolation table. However, I do not know how to select the proper isolation table for this application. Does anyone have any advice? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2016-12-22T06:37:29.960" Title="Selection of an isolation table" Tags="|mechanical-engineering|vibration|sound-isolation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12913" PostTypeId="2" ParentId="12755" CreationDate="2016-12-20T16:13:37.377" Score="6" Body="&lt;p&gt;Half of your statement is correct: If we are considering the following system:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$G : \begin{aligned} \dot{x} &amp;amp;= Ax+Bu \\ y &amp;amp;= Cx \end{aligned}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;then &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; all influence the zeros of &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;, but &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; certainly does not affect the poles. The poles &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; are the eigenvalues of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This makes sense from a purely common-sense consideration: the matrix &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is the output matrix, it is only capable of scaling and adding the states to result in an output. It has no effect on the states themselves, therefore it won't affect things like stability of &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; (which is related to the poles).&lt;/p&gt;&#xA;&lt;p&gt;You can easily see this if you look at the transfer function equation you provided, with the inverse expressed as the determinant and the adjunct (or adjugate):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$G(s) = C(sI-A)^{-1}B = \frac{C \text{adj}(sI-A) B}{\text{det}(sI-A)}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, and the adjunct of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; all appear in the numerator, but only the determinant of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; appears in the denominator.&lt;/p&gt;&#xA;&lt;h1&gt;However...&lt;/h1&gt;&#xA;&lt;p&gt;Here's where your problem is likely coming from: If we consider &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; to be a system plant being controlled through some feedback element with a transfer function &lt;span class=&quot;math-container&quot;&gt;$H(s)$&lt;/span&gt; (like the following picture from &lt;em&gt;Modern Control Engineering&lt;/em&gt;, 5th ed. by Ogata):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rOUoE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rOUoE.png&quot; alt=&quot;Image of a feedback system&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Then the overall transfer function for the entire system, from reference input &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; to output &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{C(s)}{R(s)} = \frac{G(s)}{1+G(s)H(s)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, since &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt; is in both the numerator and the denominator, you can see that &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; affect both the poles and zeros of the &lt;strong&gt;closed-loop&lt;/strong&gt; transfer function.&lt;/p&gt;&#xA;&lt;p&gt;In conclusion:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Only &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; affects the poles of the system plant &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; affect the zeroes of the system plant &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; affect both the poles &lt;strong&gt;and&lt;/strong&gt; zeroes of the closed loop system&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Reference: &lt;em&gt;Modern Control Engineering&lt;/em&gt;, 5th ed. (2009) by Katsuhiko Ogata&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2016-12-20T16:13:37.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12914" PostTypeId="1" CreationDate="2016-12-20T16:55:09.540" Score="1" ViewCount="130" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nH3Dn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nH3Dn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if my calculations are correct for the above question.&#xA;$$&#xA;E = \frac{\sigma}{\epsilon} = \frac{\frac{My}{I}}{\frac{\delta}{L}} = k. \frac{1}{I}&#xA;$$&#xA;Where k is constant for both beams, therefore:&#xA;$$&#xA;E_1I_1 = E_2I_2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{I_1}{I_2} = \frac{1}{10}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\frac{W(H/n)^3}{12}}{\frac{WH^3}{12}} = \frac{1}{n^3} = \frac{1}{10}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;n = 10^{\frac{1}{3}}&#xA;$$&#xA;Is this correct? If so should I round up or down for the number of layers? Also any comments on part (b) are welcome :)&lt;/p&gt;&#xA;" OwnerUserId="8437" LastActivityDate="2016-12-20T18:26:00.870" Title="Number of layers in a laminated steel transplant, using Young's Modulus" Tags="|materials|steel|beam|biomechanics|elastic-modulus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12915" PostTypeId="2" ParentId="12914" CreationDate="2016-12-20T18:26:00.870" Score="2" Body="&lt;p&gt;This is quite a poorly-worded question. After all, how do the layers interact with one another? Can they be considered to be fully bonded to one another? I assume not, because then they've behave as one element, which kind of beats the purpose. Should we assume that the load is evenly distributed between each of the layers (so, for a given load $q$, if you have three layers then they'll behave like three beams each under $q/3$)? Something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to assume that the load is evenly distributed between each of the layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you have some sort of load $q$ being supported by $N$ layers of material. This means that the load being resisted by each layer is equal to $\dfrac{q}{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the inertia of a rectangular cross-section is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \dfrac{bh^3}{12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we have $N$ layers sharing the available height $h$, the inertia of each layer is equal to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \dfrac{b\left(\dfrac{h}{N}\right)^3}{12} = \dfrac{bh^3}{12N^3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, depending on the support conditions (simply supported, cantilever, fixed-and-pinned, etc), the deflection will be equal to (for simple loading conditions):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta =K\dfrac{qL^x}{EI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K$ depends on the boundary conditions and loading and $x=3$ for concentrated loads and $x=4$ for distributed loads. For example, a simply-supported beam under uniformly distributed loads has&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta =\dfrac{5}{384}\dfrac{qL^4}{EI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, lets call the deflection of the single beam $\delta_1$ (with $E_1=20\text{ GPa}$) and the deflection of one of the layers $\delta_2$ (with $E_2=200\text{ GPa}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\delta_1 &amp;amp;= \delta_2 \\&#xA;K\dfrac{qL^x}{E_1I_1} &amp;amp;= K\dfrac{\dfrac{q}{N}L^x}{E_2I_2} \\&#xA;\dfrac{1}{E_1I_1} &amp;amp;= \dfrac{1}{NE_2I_2} \\&#xA;E_1I_1 &amp;amp;= NE_2I_2 \\&#xA;E_1I_1 &amp;amp;= 10NE_1I_2 \\&#xA;I_1 &amp;amp;= 10NI_2 \\&#xA;\dfrac{bh^3}{12} &amp;amp;= 10N\dfrac{bh^3}{12N^3} \\&#xA;1 &amp;amp;= 10\dfrac{1}{N^2} \\&#xA;\therefore N &amp;amp;= \lceil\sqrt{10}\rfloor = 3&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With three layers, each layer has an inertia $\frac{1}{27}$ that of the solid beam. But they each only need to withstand a third of the load, meaning the they will deflect 9 times more than a solid beam of equal elastic modulus.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-12-20T18:26:00.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12917" PostTypeId="1" CreationDate="2016-12-20T20:41:18.213" Score="0" ViewCount="84" Body="&lt;p&gt;I am a senior in high school and have been accepted into multiple engineering schools, and the top engineering school I'm looking at is OSU(industrial and systems engineering). I have noticed that there is no Calculus 2 and the only programming class required is an intro to Java. I understand that OSU has an impressive engineering program, but I am unsure what courses will be necessary. Could someone please help me understand what mathematics and programming skills are needed in industrial engineering? &lt;/p&gt;&#xA;" OwnerUserId="9295" LastActivityDate="2016-12-21T18:40:49.557" Title="Industrial Engineering Program" Tags="|mathematics|industrial-engineering|" AnswerCount="1" CommentCount="4" ClosedDate="2016-12-21T21:49:51.787" ContentLicense="CC BY-SA 3.0" />
  <row Id="12923" PostTypeId="2" ParentId="12917" CreationDate="2016-12-21T18:40:49.557" Score="1" Body="&lt;p&gt;I am a senior mechanical engineering student at Wilkes University in Pennsylvania, so I know what you mean about trying to understand the mathematics and programming courses required for your degree, i'm still dealing with it now as it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are looking at a program, different schools have different names for different things, so for example at my school you have calc 1, calc 2, differential equations (practical applications of both calc 1 and 2), and then multi-variable calculus. OSU might however have my differential equations class as an engineering specific math course such as applications of differential equations in engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is the same case with the programming class, it might not say programming in &lt;em&gt;insert language here&lt;/em&gt;, but it might be like our schools linear systems class, which sounds like a math course, but was a MATLAB programming class on specifically how to solve calculus problems using a program that you design yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mathematics skills are important and if you feel that it's even more important for you to focus in something because you wan to go into a specific industry, you can take a minor in something, or just take a class because it never hurts. The most important thing that i have learned at my internship that I've been working at for the past 2 years is that you need to adapt and learn whatever material is relevant to your current problem. if there's a formula or a concept you don't know or don't understand, don't be afraid to ask your peers, supervisor or even someone on the internet (hi there). also you might be able to find a research study or a book dealing with that specific topic that could help you understand what your trying to do, and who knows you might even learn something else in the long run, which could help you in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's good that your thinking about this now because you've got your entire college career ahead of you, that way if you wan to deviate and possibly explore something you might be interested in, you can do it. I say good on you and good luck with everything you do, and just make sure to read up on the classes so you can understand the main premise of it before you go into it.&lt;/p&gt;&#xA;" OwnerUserId="9305" LastActivityDate="2016-12-21T18:40:49.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12924" PostTypeId="2" ParentId="6310" CreationDate="2016-12-21T19:20:25.177" Score="1" Body="&lt;p&gt;Have you considered a Industrial Plastics grinder? they are typically a series of metal plates that have teeth on them that spin in one axis with two sets in opposite directions. you can typically find them that take things down to i think 1/4 inch but you could probably find one that would take it down even smaller or even use a secondary meat grinder to take it further from your initial grinder size. i have a link to a place that sells them below and a video of an extremely large one crushing a car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.jordanreductionsolutions.com/plastic-grinder.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.jordanreductionsolutions.com/plastic-grinder.html&lt;/a&gt; : Plastics Grinder&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=tPDLX0koXFs&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=tPDLX0koXFs&lt;/a&gt; : Car Crush&lt;/p&gt;&#xA;" OwnerUserId="9305" LastActivityDate="2016-12-21T19:20:25.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12925" PostTypeId="1" CreationDate="2016-12-21T23:26:05.567" Score="3" ViewCount="2512" Body="&lt;p&gt;What kind of material(s) would make a good friction disk, or disks, for a friction hinge, or torque hinge? This will be for the hinge on an arm that needs to be adjustable and also hold it's position with 1 to 10 pounds on the end. The hinge may be a vertical tilt hinge or a horizontal swivel joint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two pieces that will swivel against each other are flat metal with 2 surfaces of 1.5 inch diameter of overlap where the friction disk will be. I plan to use a bolt with a spring washer to supply the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specifics above are to give an idea of the scale of the forces. My question is generally about what materials are used for friction disks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brass or nylon washers are sometimes used for hinges or joints when you don't want friction. I want the opposite of that - I want friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want something durable enough that won't wear out quickly or disintegrate, and grippy enough to hold without slipping. The hinge may be repositioned several times per day. The worst thing would be a hinge that slowly slips over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, the materials would slide without making noise, like scratching or grinding sounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edits: I reduced the weight from 40 down to 1-10 lbs. &lt;/p&gt;&#xA;" OwnerUserId="9307" LastEditorUserId="9307" LastEditDate="2016-12-22T15:04:17.337" LastActivityDate="2017-03-09T20:58:28.200" Title="Good material for friction disk for a torque hinge?" Tags="|mechanical-engineering|materials|friction|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12926" PostTypeId="2" ParentId="12925" CreationDate="2016-12-22T06:20:39.773" Score="0" Body="&lt;p&gt;The first problem I see is that static friction doesn't care about movement direction. So if you have e.g. 40 lbs worth of holding force in the &quot;downwards&quot; direction, you yourself must apply 80 lbs of force to move the screen upwards. This is because in upwards movement, you'll have both the weight of the screen and the friction working against you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is that your surfaces are made of aluminum, which is a very soft metal. To avoid wearing them out, you will have to at least mount a flange of more durable material around the hole. This mounting must be strong enough to transmit the 30 Nm torque to the aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general this kind of problems are solved by using a spring to take up most of the load, so the required amount of friction is much lower. You might try looking into mechanism that are used to hold windows and lids open; they usually have both a gas spring and the friction mechanism combined into one device.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-12-22T06:20:39.773" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12927" PostTypeId="2" ParentId="12911" CreationDate="2016-12-22T06:37:29.960" Score="0" Body="&lt;p&gt;It's not possible to give an exact answer, but here are a few things you may want to consider:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Do you want to place the isolation table inside the glovebox or outside it? On the outside, it will isolate building vibrations, but soundwaves hitting the glovebox surface and anyone touching it will still cause movements. On the inside, however, you are heavily limited by the size of the box.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How much space and weight do you have available? Very sturdy isolation tables may need reinforced floor and will be difficult to move into the room. However, making the table heavier is the easiest way to improve the vibration isolation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How much isolation do you need? This is the hardest part to estimate, and isolation table manufacturers use several ways to report attenuation performance. One way is just to try it out: if you need more isolation, add another isolation device on top of the previous one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2016-12-22T06:37:29.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12929" PostTypeId="1" CreationDate="2016-12-22T10:55:58.827" Score="1" ViewCount="175" Body="&lt;p&gt;Why is electrical energy most commonly stored in relatively expensive metal acid batteries, which have a short life, rather than hydrogen gas accumulated through electrolysis? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen is a versatile energy source itself and with low cost technology is a cheap, clean and reliable fuel.&lt;/p&gt;&#xA;" OwnerUserId="9314" LastEditorUserId="33" LastEditDate="2017-01-11T19:14:22.200" LastActivityDate="2017-01-11T19:14:22.200" Title="Why is electrical energy most commonly stored in metal acid batteries?" Tags="|electrical-engineering|battery|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12930" PostTypeId="1" AcceptedAnswerId="12932" CreationDate="2016-12-22T11:18:12.403" Score="0" ViewCount="644" Body="&lt;p&gt;Metal oxides are hard in nature, that's what my professor says. Technically  Iron oxide is a metal oxide and must be hard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;strong&gt;is it used in industries for surface finishing operations like buffing?&lt;/strong&gt; If not, why so?&lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2016-12-24T12:41:43.417" Title="Use of iron oxide as surface cleaning agent" Tags="|industrial-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12931" PostTypeId="2" ParentId="12929" CreationDate="2016-12-22T11:40:23.517" Score="3" Body="&lt;p&gt;One of the reasons why hydrogen isn't more widely used is that is extremely explosive requires great care when bein handled. It also needs to be stored away from naked flames or sparks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hydrogen atom is the smallest atom and as a H2 molecule it is still extremely small. Being so small, hydrogen molecules can pass through the walls of their containers which requires the walls of such containers to be thicker than for other gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once hydrogen has been stored in a container the main way to use its energy is to burn it, producing heat. If electricity is need to be generated using the hydrogen the heat of combustion needs to be converted by other means into electricity. By storing electricity from alternative sources in batteries, the energy is already in a more usable form.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2016-12-22T11:40:23.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12932" PostTypeId="2" ParentId="12930" CreationDate="2016-12-22T12:24:27.713" Score="1" Body="&lt;p&gt;Like metals and minerals, metal oxides have different grades of hardness depending on a lot more than just its chemical components. Crystal structure is the main factor here. Consider carbon, for example: both graphite and diamond are chemically carbon, but one is used as a lubricant and the other for cutting and grinding. Similarly, just the fact that something is a metal oxide doesn't mean it's well suited for use as an abrasive and for similar processes like buffing and polishing. Apart from crystal structure, particle size and shape are also factors to be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, iron oxide is, to the best of my knowledge, rarely or never used as an industrial abrasive. Aluminium oxide, on the other hand, is. Aluminum oxide comes in various forms, from alumina to carborundum.&lt;/p&gt;&#xA;" OwnerUserId="9316" LastActivityDate="2016-12-22T12:24:27.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12933" PostTypeId="2" ParentId="12929" CreationDate="2016-12-22T12:31:37.593" Score="3" Body="&lt;p&gt;Cost per watt-hour of storage. Other means are either more expensive to manufacture, more expensive to maintain, or have shorter total lifetime increasing the amortization costs. The one viable alternative currently is pumped storage power plant, but it doesn't scale nearly as neatly for smaller amounts - it needs to be &lt;em&gt;huge&lt;/em&gt; to be economically viable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrogen may seem like a very cheap option if you look only at the very essentials necessary to produce and utilize it, but it being highly explosive, prone to leak through any microfractures and porous materials nothing else can cross, difficult to store in large amounts for prolonged periods of time, and in general very dangerous in handling, necessitates security features which outweigh cost of the actual essential infrastructure costs by orders of magnitude - making the entire solution more expensive than metal-acid batteries, despite core functionality being cheaper.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2016-12-22T12:31:37.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12934" PostTypeId="2" ParentId="6529" CreationDate="2016-12-22T12:32:03.243" Score="0" Body="&lt;p&gt;Calculating permittivity as a function of frequency is extremely difficult at best, to the point of being considered so impractical as to be impossible. Permittivity as a function of frequency behaves in response to a many properties of the dielectricum, and does not follow a predictable curve. As you sweep across the frequency spectrum, permittivity of a given dielectricum may go up and down at seemingly random intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need accurate data of the permittivity of a given dielectricum under given circumstances (temperature, voltage, mechanical properties etc.) measurement is the only recommended option.&lt;/p&gt;&#xA;" OwnerUserId="9316" LastActivityDate="2016-12-22T12:32:03.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12936" PostTypeId="2" ParentId="12421" CreationDate="2016-12-22T12:49:09.760" Score="1" Body="&lt;p&gt;Depending on where you are, &quot;regular&quot; automotive fuel is anything between 85 and 95. However a number of, say, 95, does not mean it's a mixture of 95% octane and 5% heptane. It merely means that the fuel has knocking resistance properties similar to this mixture. The actual composition can be any mixture of isomeric hydrocarbons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the number that denotes the knocking resistance goes up, the mixture becomes progressively more expensive to manufacture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that the ideal fuel composition for a given internal combustion engine is subject to variables. Air pressure, for example, is a factor. Ideal fuel/air mixtures vary for different fuel compositions. At, say, 6000ft altitude (which is where I live) various fuel mixtures perform differently than they do at sea level. So local conditions may also be a consideration when it comes to formulating fuel mixtures.&lt;/p&gt;&#xA;" OwnerUserId="9316" LastActivityDate="2016-12-22T12:49:09.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12938" PostTypeId="2" ParentId="12929" CreationDate="2016-12-22T13:43:32.093" Score="6" Body="&lt;p&gt;Using hydrogen efficiently is difficult. If you burn it, you're typically back to IC engine efficiencies, i.e. catastrophically low. If you use it in a fuel cell, that usually involves rare metal catalysts, which are expensive and can be vulnerable to poisoning through contamination. By expensive, I mean platinum expensive, or even more exotic metals like rhodium or palladium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what a practical lifetime of a fuel cell catalyst is, in terms equivalent to charge cycles of a battery, but it's an important thing to know if you're looking at the economics of energy storage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is on top of the difficulties and expense of storing the hydrogen, as well as the energy losses in electrolysis and fuel cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's look at the competition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;metal acid&quot; batteries I presume you mean lead-acid, with a lifetime of several hundred cycles - (say, 200-500) thus adding the battery price divided by that number of cycles to the energy cost - indeed making it uncompetitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a 12V 160Ah battery stores 12*160 Wh, or 1920Wh - call it 1 kWh at 50% depth of discharge. And costs $200, and lasts 200-500 charge cycles. (Using more than 50% of its capacity will kill a lead-acid battery fast).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I pay 12.3p (about $$0.15 ) for that 1kWh but it would cost me $200/(200 to 500) or $0.40 to $1.00 to store it in a lead acid battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite this, lead acid are still popular by virtue of being initially fairly cheap, proven and reliable within their limitations, and practical to recycle. Thus they are an easy route avoiding the problems involved in any of the higher tech solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your H2 fuel cell, you need to find out how much in costs, and how many kWh you can convert through it before the catalyst is junk, in addition to the costs of actually storing hydrogen. I don't know that information, but I don't think you'll like the numbers. Why? because fuel cells have been around since at least the 1970s, and we aren't all using them already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And lead-acid isn't the only battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium-ion batteries have similar lifetimes, (again, say 200-500 cycles) but their weight makes them popular enough that their disadvantages can be tolerated in applications from phones to cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, LiFePo4 cells can achieve as many thousands of cycles, (say, 2000-5000 cycles) even up to 80% DOD, so their higher cost can be amortised over thousands of charge cycles, and this can bring down the cost of storage, especially if their costs come down with mass production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a LiFePo4 battery made of 4 of &lt;a href=&quot;https://www.ev-power.eu/Winston-40Ah-200Ah/WB-LYP100AHA-LiFeYPO4-3-2V-100Ah-WIDE.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;these cells&lt;/a&gt; (about 1.2kWh) would store about 1kWh (80% DOD) and cost about $$560. Then the cost of storing 1 kWh is $560/(2000 to 5000) or $0.28 to $0.11 which finally overlaps with the range of retail electricity prices. (I can't find lifetime specs for this model, so caveat emptor!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium Titanium Dioxide (often called LTO) can achieve tens of thousands of cycles, (maybe 20-50000) again at slightly higher cost and slightly lower energy storage density - still well above lead acid - this potentially reduces the cost of battery storage further. It also offers a very high charge efficiency : you get about 98% of the charge back out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could store 1 kWh in an array of &lt;a href=&quot;https://www.ev-power.eu/LTO-technology/Lithium-Titanate-Oxid-Battery-Cell-LTO-2-4V-30AH.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;these cells&lt;/a&gt; : 5 of them would make 12V, and 4 arrays would give 120Ah to give 1kWh at well under the rated 80% discharge - at about $$1200. Then, over 20000 to 50000 cycles, we are paying $0.06 to $0.024 to store 1 kWh.&lt;br&gt;&#xA;(But there's a problem with this analysis : if we're only getting 1 charge cycle per day, 20000 cycles is over 50 years! There may be other reasons the battery doesn't last that long! So use them for more frequent demands; braking/acceleration cycles in an EV, or running the fridge every hour, etc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there are batteries offering longer lived alternatives to lead acid, and that does reduce the price of storage, using what's available today. Given some momentum in either improving performance, or reducing prices, they could reduce the niche for hydrogen further. And there's probably a sweet spot for large capacity batteries as cheap as possible with a lifetime about 10,000 cycles...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT : As a footnote, &lt;a href=&quot;http://www.popsci.com/researchers-accidentally-make-batteries-last-400-times-longer?con=TrueAnthem&amp;amp;dom=fb&amp;amp;src=SOC&amp;amp;utm_campaign=&amp;amp;utm_content=587599d7b8a9fe0007b612a1&amp;amp;utm_medium=&amp;amp;utm_source=&quot; rel=&quot;nofollow noreferrer&quot;&gt;this story&lt;/a&gt; holds out the prospect of batteries lasting for hundreds of thousands of cycles.&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="435" LastEditDate="2017-01-11T13:12:47.500" LastActivityDate="2017-01-11T13:12:47.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12940" PostTypeId="1" AcceptedAnswerId="12942" CreationDate="2016-12-22T16:21:29.747" Score="1" ViewCount="84" Body="&lt;p&gt;I apologize for the confusing question title. I work with area rugs, and have almost never done measurements before. I am tasked with converting these measurements into the right unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have four measurements:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;19&lt;/p&gt;&#xA;&#xA;&lt;p&gt;39&lt;/p&gt;&#xA;&#xA;&lt;p&gt;81&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all in Square Inches I believe. On the spreadsheet it shows &quot;Square Footage (in.)&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to convert them to Square FT. I'm unfamiliar with measurements, but is it any different than regular inches? Would 6 square inches be 0.5 square ft?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Some were saying the question is confusing, I agree. If it helps, here is the product width for each of these measurements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 = 24 inches wide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;19 = 31 inches wide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;39 = 63 inches wide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;81 = 94 inches wide&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really wish I could provide more information, but I am just as confused on the entire thing as anyone else.&lt;/p&gt;&#xA;" OwnerUserId="9321" LastEditorUserId="9321" LastEditDate="2016-12-22T17:00:53.713" LastActivityDate="2016-12-22T17:02:59.693" Title="How to convert Square FT (in.) to Square FT (ft)?" Tags="|measurements|" AnswerCount="1" CommentCount="3" ClosedDate="2016-12-23T12:24:35.867" ContentLicense="CC BY-SA 3.0" />
  <row Id="12941" PostTypeId="1" AcceptedAnswerId="12952" CreationDate="2016-12-22T16:36:18.877" Score="5" ViewCount="941" Body="&lt;p&gt;I want to understand how a laser printers laser writes on a photosensitive drum. I am trying to understand this process.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the photosensitive cylinder is illuminated by a laser beam that projects above the negative of the image.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;After you illuminate with laser beam what happens exactly on the photosensitive drum? What is this projection?&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;negative image&lt;/em&gt;? Where does this come from?&lt;/li&gt;&#xA;&lt;li&gt;When the drum receives this laser light, the photosensitive drum material generates a current, voltage? How and what is handled this photosensitive drum variation due to the absorption of light?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I understand the process &lt;code&gt;drum-toner&lt;/code&gt; but I don't understand this step well:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;when you want to perform printing, it is loaded with positive electric energy from the corona wire, an electric current thread that runs through it in all its length.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9322" LastEditorUserId="7659" LastEditDate="2016-12-23T15:56:47.443" LastActivityDate="2018-03-25T07:25:45.567" Title="How does a laser printer write with light on the photosensitive drum?" Tags="|lasers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12942" PostTypeId="2" ParentId="12940" CreationDate="2016-12-22T16:53:48.693" Score="5" Body="&lt;p&gt;Let's think about how we work out the area in inches. What we are simply doing is multiplying the lengths (in inches) of the shape (let's say rectangle for simplicity) to get the square footage. ie. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a * b = area_{inches}$$ where $a$ and $b$ are sides of the rectange (measured in inches).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of using the length of the sides in feet, we could convert it into inches first. Given that there are 12 inches in a foot (or $1/12$ feet in an inch):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac a  {12} * \frac b  {12} = area_{feet}$$&#xA;$$ { a * b \over 144} = area_{feet}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This can also be derived from the units, we can treat the units as an equation themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ feet = {{inches} \over 12} $$&#xA;$$ feet^2 = ({{inches} \over 12})^2 $$&#xA;$$ feet^2 = {{inches}^2 \over 12^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of logic can be used for any unit conversations. Given the relationship of the base units (in this case inches and feet), more complicated relationships can be worked out.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For example, using your 81 square inch measurement. If we replace the ${inches}^2$ with $81$, we get 81 inches squared in square feet&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ feet^2 = {81 \over 12^2} $$&#xA;$$ feet^2 = 0.5625 $$&lt;/p&gt;&#xA;" OwnerUserId="9323" LastEditorUserId="9323" LastEditDate="2016-12-22T17:02:59.693" LastActivityDate="2016-12-22T17:02:59.693" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12943" PostTypeId="2" ParentId="12852" CreationDate="2016-12-22T17:11:51.160" Score="1" Body="&lt;p&gt;As preamble, I note that the equation for this shear stress / force / flow is something that seems to suffer from almost everyone adopting a different nomenclature.  I learnt it as longitudinal shear = Q A y[bar] / I.  In that form, the longitudinal shear is a force per unit length along the section, and Q is the shear force in the section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I will adopt the nomenclature that the question is adopting.  In this answer, I'm adopting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S = longitudinal shear force per unit length across a defined boundary of interest, that boundary being perpendicular to a plane section.  This boundary is normally a single plane on which you could slice off a part of the section, but it doesn't have to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V = vertical shear force in the plane section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = an area (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = a distance (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I = second moment of area of the whole plane section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation in question is then S = V A y / I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The question has an intermediate value Q = A y, but I'm going to ignore that - you can substitute it back in if you like, having calculated A and y, at any point in this discussion.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of A is the area of a 'chunk' of the section.  The value of y is the distance from the neutral axis of the whole section to the centroid of the sliced off part.  It doesn't actually matter if the chunk is the result of a single planar slice, or an arbitrary portion of the section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case b there's a chunk that's the top plank.  It's a rectangle d wide and b high.  The centroid of that part is exactly central, b/2 from the top and bottom faces of that plank.  The y value will be the distance from the centre of that plank to the neutral axis of the whole section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sake of discussion, suppose the neutral axis of the whole section is 0.75xc from the bottom edge.  Suppose also that a=d+2xb and e=b+c (ie, as it appears to be drawn).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case a):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = a * b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = 0.25 * c + 0.5 * b&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in case b):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = b * d&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = exactly the same as case a), because the centroid of area A in case a) is at the same place as the centroid of area A in case b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second half of the question contains an error about what y is.  The value y in the equation is not the distance from the centroid of the whole section to the cut plane.  That is, y is not what is labelled as y in the diagram, it should be the distance from the whole section neutral axis at NA to the centroid of the hatched area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that the S = V A y / I equation is only valid in the case of a prismatic section where the chunk is arranged such that A y and I do not vary along the length of the member.  Some thinking about the derivation of the equation is relevant (and explains the assertions made above):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axial bending fibre stress at any point on a section is calculated from stress = M y / I, where M is bending moment, y is now the distance from the neutral axis to the point where the stress is calculated and I is the second moment of area of the whole section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider any arbitrary chunk of the cross section, the axial force (F) on that whole chunk is the integration across the area of the chunk of (M y / I).  That is, F = M A y / I, if A is the area of the chunk and y is back to being the distance from neutral axis of the whole section to centroid of the chunk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, consider repeating that an increment along the member.  There's now a different axial force on the chunk, and the way that force got there was by longitudinal shear across the perimeter of the chunk.  Or, to put it another way, if x is position along the beam, the longitudinal shear flow across the chunk boundary = dF/dx = d(M A y / I)/dx.  However in that differentiation, A y and I are all constant with varying x, so dF/dx = dM/dx A y / I.  However, we know dM/dx (rate of change of bending moment along the beam) as something else - it's the vertical shear force in the section: dM/dx = V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence longitudinal shear across the chunk boundary S = dF/dx = V A y / I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important thing to note is that this derivation depends absolutely upon A y and I not varying with x - if they do vary, you can't simply recalculate S = V A y / I with different A values along the section, because with an A that varies with x (for example), d(M A y / I)/dx does NOT equal dM/dx A y / I.  In that case you need to apply the differentiation product rule, so d(M A y / I)/dx = (dM/dx A + M dA/dx) y / I.  Actually, of course, if A is varying with x it's almost certain that y and I are likewise varying with x, and you'll need to multiply out all four terms as varying-with-x.&lt;/p&gt;&#xA;" OwnerUserId="902" LastActivityDate="2016-12-22T17:11:51.160" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12944" PostTypeId="1" AcceptedAnswerId="12951" CreationDate="2016-12-22T18:28:01.437" Score="5" ViewCount="3004" Body="&lt;p&gt;I've written a program to generate gears and write them to dxf files. I thought all was well until I looked at the mesh of two gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hpLE2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hpLE2.jpg&quot; alt=&quot;Gear Mesh&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From everything I've read there should be no backlash when the pitch circles touch. So I'm ether reading wrong, or my calculations are skewed. If I drop the pressure angle to like 8, the teeth fit nicely. The image is with the angle set to 14.5 This is the actual code, I derived it from &lt;a href=&quot;http://faculty.tamucc.edu/psimionescu/PDFs/ASEE-2008.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;//angle of involute to outside circle&#xA;//see fetch.pdf, equ (6),(7) then from...&#xA;&#xA;//polar_radius = base_radius * sqrt( 1 + t^2 )&#xA;//where t is the arc AB&#xA;//the limit of polar_radius is the outer_radius&#xA;&#xA;//   so solve for t at the limit&#xA;// t= ( ( outside_diameter / base_diameter )^2 - 1 )^1/2&#xA;&#xA;&#xA;//for the end point of the involute arc&#xA;double Dop= gp.od / gp.bd; //ouside diameter / base diameter&#xA;double t= sqrt( Dop * Dop - 1 ); //angle of tangent to involute arc, at limit&#xA;double Ac= t - atan( t ); // angle of involute arc, at limit&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;then I create a line string for the involut curve as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for( size_t step= 0; step &amp;lt;= ITERS; ++step )&#xA;{&#xA;    double arc= t / ITERS * step;&#xA;    *ls+= bg_point( base_radius * ( cos( arc ) + arc * sin( arc ) ),&#xA;                    base_radius * ( sin( arc ) - arc * cos( arc ) ) );&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And thinking about it, how could you not have backlash when the teeth arc from the base circle, the point where the line starts at half a tooth/space distance. But I'm just not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm planing to put all the code on GetHub and will do it sooner than later if needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&#xA;The fix is in and now it looks like it should. This is with a 20 degree pitch.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rRlir.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rRlir.jpg&quot; alt=&quot;20 degree pitch meshed&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've uploaded to GetHub using the MIT license. If I think this useful, maybe others will also. The user interface needs a lot of work, gear params are still hard coded. Only these two files have any thing to do with gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/lakeweb/spur-gear-gen/blob/master/project/gear_gen.h&quot; rel=&quot;nofollow noreferrer&quot;&gt;gear_gen.h&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://github.com/lakeweb/spur-gear-gen/blob/master/project/gear_gen.cpp&quot; rel=&quot;nofollow noreferrer&quot;&gt;gear_gen.cpp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is a CAD drawing interface. Thanks again SprocketsAreNotGears!!&lt;/p&gt;&#xA;" OwnerUserId="9324" LastEditorUserId="9324" LastEditDate="2016-12-24T00:52:55.927" LastActivityDate="2016-12-24T00:52:55.927" Title="Involute Gear Calculation" Tags="|design|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12946" PostTypeId="1" CreationDate="2016-12-22T21:54:29.367" Score="1" ViewCount="172" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5XnQQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5XnQQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above diagram is misleading. In reality, both the length and thickness are equal to 8mm.  As this is from a past paper (one notorious for awkward questions), I'm not sure if it's a typo (an 8m length would be much more straightforward).  Is there any possible outcome/plane in which this pipe would indeed buckle?&lt;/p&gt;&#xA;" OwnerUserId="8437" LastActivityDate="2016-12-23T02:54:18.367" Title="Possibility of buckling in non-slender member?" Tags="|structural-engineering|materials|deformation|buckling|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12948" PostTypeId="2" ParentId="12946" CreationDate="2016-12-23T02:54:18.367" Score="2" Body="&lt;p&gt;It is not &lt;em&gt;necessarily&lt;/em&gt; a typo, but you wouldn't normally call a thin walled cylinder 8mm high and 100mm diameter a &quot;pipe&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the formulas for Euler buckling are most definitely &lt;em&gt;not&lt;/em&gt; valid for a structure with those dimensions. The appendix of this rather old report &lt;a href=&quot;http://www.dtic.mil/dtic/tr/fulltext/u2/a801283.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.dtic.mil/dtic/tr/fulltext/u2/a801283.pdf&lt;/a&gt; gives some basic theory and graphs for &quot;hand calculation&quot;. Such problems would now be solved by computer modeling, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real life, the critical load will be very sensitive to the exact boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the OP's diagram looks as it if came from a rather old textbook, most likely it &lt;em&gt;is&lt;/em&gt; a typo, but since the length as shown in the diagram is apparently less than the &quot;obvious&quot; typo of 8m it is not very clear what the &quot;correct&quot; dimensions for the numerical part of question should be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deriving a general formula for a pipe of length L, diameter d and wall thickness t is a sensible exercise, though.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-12-23T02:54:18.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12949" PostTypeId="1" CreationDate="2016-12-23T10:27:14.610" Score="-1" ViewCount="95" Body="&lt;p&gt;Here's is my problem I am trying to spinal object at high speeds&#xA;My motors watts is 242.42&#xA;My motor max speed is 70,000 rpm&#xA;My circular rotating object that will be right on motor is 60grams&#xA;My motor torque is 35 milli newton meter&#xA;Can anyone show me formula or figure out if this motor will spin a 60 gram circular object at any rpms?&#xA;This is a DC motor brushless.&lt;/p&gt;&#xA;" OwnerUserId="9318" LastEditorUserId="9318" LastEditDate="2016-12-23T10:51:19.600" LastActivityDate="2017-03-23T14:35:25.513" Title="Motor MnM torque max weight can spin circular object" Tags="|motors|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="12950" PostTypeId="2" ParentId="12949" CreationDate="2016-12-23T12:19:03.177" Score="1" Body="&lt;p&gt;You are missing some basic physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply spinning a mass at a fixed rate by itself takes no power at all.  It does take power to get it there, with more power required the faster you want to get it up to speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a fixed speed, the power to keep it at that speed is only to overcome friction and losses in the motor.  It has nothing to do with the mass being spun, other than the weight of that mass probably has some effect on the friction.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-12-23T12:19:03.177" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12951" PostTypeId="2" ParentId="12944" CreationDate="2016-12-23T12:31:53.430" Score="5" Body="&lt;p&gt;First of all, there should indeed ideally be no backlash between the teeth. There is a slight problem with the code, which stems from the misunderstanding that consecutive points of the involute curves that lie on the base circle occur a half-tooth's distance away from each other. If this were the case then, yes, there would be backlash. However, in fact, it is actually the points of the involute curves at the &lt;strong&gt;pitch&lt;/strong&gt; circle that are a half-tooth's distance apart. I'll illustrate this below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What your program is currently doing is as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/msM8w.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/msM8w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that each of the radial lines are one of the $2N$ division lines (for a gear with $N$ teeth) and each are separated by an equal angle of $\frac{\pi}{N}$ radians. Note the black dots are the points of the involute curves that lie on the base circle, and these lie on the equal division lines according to your program. This is not the case for standard gears with no deliberate backlash implemented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take note of the following diagram, which is basically a zoom-in of the previous:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RmyaT.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RmyaT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the pitch point on the involute curve, and take note of the angle $\theta_P$ that occurs between the radial lines passing through the base and pitch points. We will need this angle later on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to instead have the pitch points on consecutive involute curves separate by equal angles, and so the following diagram should be obtained: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QgBOj.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QgBOj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the black dots are the pitch points that are lying on the equal division lines. To get from the previous incorrect case to the correct one, this involves rotating the involute profiles about the centre of the gear by the angle $\theta_P$: the involutes that curl out clockwise from the base circle will be rotated by $\theta_P$ in an anticlockwise direction, and the ones that curl out anticlockwise will be rotated by $\theta_P$ clockwise. Now it is a matter of determining the value of $\theta_P$, and this can be done by looking at the polar parametric equation you were provided for the involute curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's determine the value of $t$ at the pitch circle, which we will call $t_P$. Using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = r_B \sqrt{1+t^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And noting that the pitch radius is $r_P = r_B/\cos{\phi}$, where $\phi$ is the pressure angle, then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{r_B}{\cos{\phi}} = r_B \sqrt{1+t_P^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, by algebraic manipulation, it can be shown that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t_P = \tan{\phi}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\phi$ is in radians. Now, $\theta_P$ can be determined using the second of the polar formulae:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = t - \arctan{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_P = \tan{\phi} -\phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotate your profile curves by this angle in the appropriate directions, and that should solve the issue! (Note how $\theta_P$ drops to zero as the pressure angle decreases, aligning with the observation you made earlier.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best of luck with your gear generator! :)&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2016-12-23T12:31:53.430" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12952" PostTypeId="2" ParentId="12941" CreationDate="2016-12-23T12:40:33.900" Score="3" Body="&lt;p&gt;You are apparently asking about the Xerographic process, such as used in original copy machines and now in laser printers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process works by charging up the surface of a cylinder (the drum) to a high voltage.  The drum then rotates so that any parts of it that are charged pick up black particles, called the &lt;i&gt;toner&lt;/i&gt;.  Then a piece of paper is pressed against the drum as it rotates further.  That causes the black stuff to stick to the paper.  The paper is then heated and squished to make the black stuff stick to the paper permanently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you end up with black on the paper wherever there was electric charge on the drum.  As I described things above, the drum was charged everywhere, so the paper ends up all black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If only we could selectively discharge parts of the drum, then we could end up with white areas on the paper.  Fortunately there is a way.  Instead of making the drum out of purely insulating material, we make it out of material that insulates unless it gets hit with light.  Early drums were made from selenium, for example.  If we connect the inside of the drum to low voltage (discharge potential), then all we have to do is shine light on the areas where we ultimately want the paper to be white.  The light discharges the high voltage on the surface of the drum, which causes that area to not pick up toner, which won't make the paper black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we have a way of exposing a image onto the drum surface, and ending up with that image on a piece of paper.  This is exactly what early copiers did.  A strip of bright light was shone on the original to be copied.  The image of that was optically focused onto the drum as it rotated and the bright light strip moved along.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laser printers use the same principle, except that a modulated laser &quot;writes&quot; onto the drum to discharge it, instead of the optical projection of the paper to be copied.  Since the modulation of the laser doesn't have to come from scanning a original, you now have a &quot;printer&quot; that can discharge any pattern on the drum, which means it can produce any pattern of black/white on the resulting piece of paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern copiers are really digital scanners with a laser printer back end.  There is no longer a direct optical connection between the original and the drum.  This allows for image processing, enhancements, storing the result, or even generating the pattern in a computer without ever scanning any original (printer versus copier).&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-12-23T12:40:33.900" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12953" PostTypeId="1" CreationDate="2016-12-23T15:04:42.517" Score="2" ViewCount="1450" Body="&lt;p&gt;For one axle with one wheel, can the following rules be met and if so, what would be the most simplistic implementation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rule #1. Axle rotates clockwise: Wheel rotates at a 1:1 ratio with the axle. The direction of the wheel compared to the axle does not matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rule #2. Axle rotates counter clockwise: Wheel rotates, but at less than a 1:1 ratio with the axle. The direction of the wheel compared to the axle does not matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this already exists as a part manufacturers produce, what is it called?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far have I gotten with this concept? Not very. Having a ratcheting mechanism sounds like it'd be a necessity or possibly some kind of planetary gear train.. but I'm coming up empty for actual solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of this would be to have a single motor drive a small toy car/robot forward/right when the motor rotates one way and forward/left when the motor rotates the other way. The motor would be attached to a bevel/mitre gearbox similar to the image below. The bottom is where the motor would go and the mechanism I'm describing should go on both the right and left sides along with the wheels. At least, this is what I'd envision so far to accomplish my goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I'm really just doing this for the novelty of it. I'm just getting into electronics and this was a challenge for myself (which I need help with it would seem, lol).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hsetv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hsetv.jpg&quot; alt=&quot;Mitre Gearbox&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9294" LastEditorUserId="9294" LastEditDate="2016-12-24T15:40:10.120" LastActivityDate="2019-05-11T22:09:15.863" Title="How to make wheel rotate slower in one direction than the other?" Tags="|gears|wheels|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="12954" PostTypeId="2" ParentId="3614" CreationDate="2016-12-23T16:08:53.520" Score="-2" Body="&lt;p&gt;yeah, there is a method to multiply both amperage and voltage at same time using motor-generator combination, but the current value is badly less for motor-generator combination to work out well. say like a 400 watt alternating current input can be amplified to 1200 watt direct current output. a 250 watt direct current input can be amplified to 24000 watt direct current output.&lt;/p&gt;&#xA;" OwnerUserId="9332" LastActivityDate="2016-12-23T16:08:53.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12956" PostTypeId="2" ParentId="3614" CreationDate="2016-12-23T21:30:13.937" Score="0" Body="&lt;p&gt;I get the impression that what you are really looking for is a way to achieve more current at the same voltage.  You cannot do this from a single generator.  But what you can do is add another generator in parallel to your circuit.  This should double the available current while keeping the same voltage.  Of course, this assumes that you can operate both generators at the same speed simultaneously, which can be quite remarkably more difficult than it sounds and may not be feasible for your application's needs.&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2016-12-24T00:37:12.097" LastActivityDate="2016-12-24T00:37:12.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12957" PostTypeId="2" ParentId="12930" CreationDate="2016-12-24T12:41:43.417" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.cooksongold.com/Jewellery-Tools/Jewellers-Rouge-Powder-120g4oz-prcode-998-193&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jeweller's rouge&lt;/a&gt; is fine Iron III oxide. It is generally used for fine polishing of relatively soft metals like gold and silver as well as some plastics and minerals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't aggressive enough  to be used on harder metals like steel except perhaps for very fine specialist polishing operations. Consider that aluminium oxide typically forms a hard, stable layer on metal whereas rust has very friable structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminium oxide is one of the most widely used general purpose abrasives (usually colour coded brown), zirconium oxide is used for more aggressive stock removal at larger grit sizes (usually colour coded blue) and other high performance 'ceramic' abrasives (colour coded red) are sometimes used for stock removal on hard to grind materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as their hardness the performance of different abrasives depends on how the particle wear for example aluminium oxide tend to become smoother as it is used which can be good for surface finishing whereas zirconium tends to fracture and so retains an aggressive cut for longer making it more effective for stock removal especially at larger grit sizes but at grits higher than about 120 the is no significant advantage over aluminium oxide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diamond grit may be used for some applications, especially lapping and honing of hard metals either as a paste or embedded in a metal substrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also various proprietary compounds, typically an abrasive powder blended with grease or wax used for specific buffing operations on specific metals. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-24T12:41:43.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12958" PostTypeId="1" CreationDate="2016-12-24T14:53:41.850" Score="0" ViewCount="1393" Body="&lt;p&gt;Can someone explain why the M is assigned to be anticlockwise here ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yIznY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yIznY.jpg&quot; alt=&quot;1&quot;&gt;&lt;/a&gt;&#xA;When i assign it as clockwise , i will get -P(δ -v) , which is different from the author ... &#xA;If i do so , then , all the sign will be different &#xA;Can i do so ? Why ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the second example here , i can understand that M = -Pv , since M+Pv = 0 at either end when it's in equlibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P/s : I know the the sign convention of the bending moment of beam is positive when the beam upwards as shown ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first example  , i gt M+P(∂-v) = 0 , so M = - P(∂-v) . &#xA;I am not sure whether is my concept correct or not . &#xA;&lt;a href=&quot;https://i.stack.imgur.com/0YHl2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0YHl2.jpg&quot; alt=&quot;2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bsuyp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bsuyp.jpg&quot; alt=&quot;3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DWPXg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DWPXg.jpg&quot; alt=&quot;4&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xJ36h.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xJ36h.jpg&quot; alt=&quot;5&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2016-12-31T10:07:06.627" Title="sign of moment in buckling of column" Tags="|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12959" PostTypeId="1" CreationDate="2016-12-24T16:24:09.127" Score="1" ViewCount="80" Body="&lt;p&gt;Can a signal below -176 dbm, which is 1.6 db below thermal noise at room temperature, be decoded irrespective of the modulation type? &lt;/p&gt;&#xA;" OwnerUserId="9337" LastEditorUserId="8055" LastEditDate="2016-12-27T19:23:49.253" LastActivityDate="2016-12-27T19:23:49.253" Title="Can a signal below -176 dbm be decoded?" Tags="|electrical-engineering|rf-electronics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12961" PostTypeId="1" AcceptedAnswerId="12966" CreationDate="2016-12-24T19:30:50.080" Score="1" ViewCount="372" Body="&lt;p&gt;I find this scheme &lt;a href=&quot;http://www.explainthatstuff.com/laserprinters.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qYuwI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qYuwI.png&quot; alt=&quot;printer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1&lt;/strong&gt; - is &lt;em&gt;data stream&lt;/em&gt; come from computer. This 'data' is an electric signal moudulated by the n°&lt;strong&gt;2&lt;/strong&gt;&lt;br&gt;&#xA;&lt;strong&gt;2&lt;/strong&gt; - this is the &lt;strong&gt;electronic circuit&lt;/strong&gt; inside printer that &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;a)&lt;/strong&gt; activates the corona wire &lt;code&gt;→&lt;/code&gt; &lt;strong&gt;b)&lt;/strong&gt; the corona wire charges up the photoreceptor drum&lt;br&gt;&#xA;&lt;strong&gt;a)&lt;/strong&gt; at the same time circuit activates the laser to make it draw the image of the page onto the drum. The laser beam doesn't actually move: it bounces off a moving mirror that scans it over the drum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;when data stream come &lt;em&gt;from&lt;/em&gt; computer to the printer &lt;em&gt;inside&lt;/em&gt; the electronic circuit what happens inside this electronic circuits ?&#xA;How mechanism 'tranform' this electric signal of &lt;em&gt;data&lt;/em&gt; into instruction for laser diode and moving mirror ?&lt;br&gt;&#xA;But this step is possible only if you have correct modulation between data stream and instruction laser otherwise printing should be wrong and electronic circuit may break or not work properly.&lt;br&gt;&#xA;What modulation technique is used for laser diode ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;which components inside electronic printer is needed to drive and control this ?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9322" LastEditorUserId="8055" LastEditDate="2016-12-27T14:09:54.400" LastActivityDate="2016-12-27T14:09:54.400" Title="How the electronic circuit control laser diode?" Tags="|electrical-engineering|lasers|circuits|modulation|" AnswerCount="1" CommentCount="0" ClosedDate="2016-12-27T14:08:11.507" ContentLicense="CC BY-SA 3.0" />
  <row Id="12962" PostTypeId="1" CreationDate="2016-12-24T20:24:01.600" Score="2" ViewCount="65" Body="&lt;p&gt;Problem: how much water (volume ratio) does it take on the exit side to maintain a siphon? Vertical lift 24&quot; from water level to top of an open container? Does it have to exceed the volume on the lift side to begin the process? And how far down the exit line must it remain &quot;solid&quot;, filling the tube, before it will start &quot;trickling&quot;, not filling the tube? Where does the air start mixing with the water on the low side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an engineer, so please refrain from equations or physics if you can. I am trying to solve a friends pumping from a 60 gallon container problem. He thought the water may be siphoning past the pump impellers after they had stopped and I felt pretty sure that could not happen then got to thinking about what it takes to siphon. Thank you to anyone who can answer the question in a simple fashion. &lt;/p&gt;&#xA;" OwnerUserId="9341" LastEditorUserId="4716" LastEditDate="2016-12-25T04:00:25.687" LastActivityDate="2016-12-25T04:00:25.687" Title="Concerning siphoning water" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="1" ClosedDate="2016-12-26T17:29:01.963" ContentLicense="CC BY-SA 3.0" />
  <row Id="12965" PostTypeId="2" ParentId="12958" CreationDate="2016-12-25T12:02:25.247" Score="1" Body="&lt;p&gt;Simple. The bending moment caused by $P\delta$ is clockwise, so the bending moment reaction must be counter-clockwise to resist it. The bending equilibrium equation should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;\sum M_{support} = M + (-P\delta) = 0 \\&#xA;\therefore M = P\delta&#xA;\end{gather}$$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2016-12-25T12:02:25.247" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="12966" PostTypeId="2" ParentId="12961" CreationDate="2016-12-25T14:56:13.683" Score="1" Body="&lt;p&gt;What you are asking in similar to how the beam in a CRT monitor is modulated to make lighter and darker areas on the screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lasers used in these devices can be electronically controller to vary their light output.  This is done in sync with the laser beam being mechanically swept across the drum.  As the laser hits each spot on the drum, the electronics modulates it to get the desired amount of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A moving mirror sweeps the laser beam across the drum to make one dimension of the picture, and the drum moves orthogonal to that to make the other dimension of the picture.  The result is a 2D image &quot;painted&quot; on the drum with laser light.  Again, this is much like the electron beam being swept across the face of a CRT and modulated per pixel to make a 2D image on the face of the CRT.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-12-25T14:56:13.683" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="12968" PostTypeId="2" ParentId="12882" CreationDate="2016-12-25T16:36:56.813" Score="1" Body="&lt;p&gt;First, the functional name of the asked mechanism is spherical bearing. Please have a look at &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Spherical_bearing&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Spherical_bearing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use a spherical bearing to allow continuous rotation (in yaw) while also allowing for some pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Damping can be enhanced by inserting some sort of friction to the assembly.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2016-12-25T16:36:56.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12970" PostTypeId="2" ParentId="12959" CreationDate="2016-12-25T20:35:42.820" Score="2" Body="&lt;p&gt;No, not irrespective of modulation type. If you know that you need to deal with signals like that, you pick a modulation scheme that provides &quot;gain&quot; in the demodulation process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, GPS uses &lt;a href=&quot;https://en.wikipedia.org/wiki/Direct-sequence_spread_spectrum&quot; rel=&quot;nofollow noreferrer&quot;&gt;direct-sequence spread spectrum&lt;/a&gt;, which allows the wideband energy of the signal to be &quot;refocused&quot; to a narrow bandwidth, providing a huge amount of gain over the thermal noise.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2016-12-25T20:35:42.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12972" PostTypeId="1" CreationDate="2016-12-26T15:16:45.877" Score="2" ViewCount="129" Body="&lt;p&gt;I'm looking at convective heat transfer for air flow over a solid at Ma=0.2.  At this speed, it is plausible to model the flow as incompressible because of small density variations in the air.  Normally, when modeling compressible flow, we include the kinetic energy term in the energy equation.  But for viscous incompressible flow at this speed, is it necessary?  Can I neglect kinetic energy as a component of the energy equation due to the incompressibility assumption?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="33" LastEditDate="2016-12-27T14:12:32.513" LastActivityDate="2016-12-27T14:12:32.513" Title="Can I neglect kinetic energy when modeling convective heat transfer for viscous, low mach airflow? Why or why not?" Tags="|fluid-mechanics|heat-transfer|convection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12975" PostTypeId="2" ParentId="12972" CreationDate="2016-12-26T23:48:32.413" Score="1" Body="&lt;p&gt;You should be fine to neglect kinetic energy from a thermal perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy per unit volume will be $1/2 \rho u^2$, which is about 2800 $J/m^3$ for your conditions. Given that the volumetric heat capacity is ~ 1200 $J/m^3.K$ The most you'd get out of your KE is a few degrees K (if you're lucky), due to viscosity, rather than compression. Unless you're dealing with such a small $\Delta T$ between your gas and solid, that effect is probably unimportant.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2016-12-26T23:48:32.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12982" PostTypeId="1" AcceptedAnswerId="13314" CreationDate="2016-12-27T15:17:51.353" Score="5" ViewCount="17808" Body="&lt;p&gt;I have an electronic device that uses black potting materiel similar the one below to I believe to protect the electronic components from elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y6WYR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y6WYR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to safely remove this material. This material might be some type of epoxy. There are plenty of suggestions how to remove potting material (&lt;a href=&quot;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=removing%20potting%20electronics&quot; rel=&quot;nofollow noreferrer&quot;&gt;removing potting electronics&lt;/a&gt;). Some indicate that the chemicals in the potting might be harmful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I safely remove the potting material? Through observation can I safely narrow down the potting material?    &lt;/p&gt;&#xA;" OwnerUserId="8055" LastEditorUserId="8055" LastEditDate="2017-01-17T14:27:46.437" LastActivityDate="2024-02-20T15:55:44.247" Title="How to remove potting material from an electronic package?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12983" PostTypeId="1" CreationDate="2016-12-27T15:20:51.820" Score="2" ViewCount="46" Body="&lt;p&gt;I have a pair of two-way switches at different floors to control a light bulb. I wish to put an Android-phone-controllable external switch over either switch to flip it up or down. I am familiar with the programming and circuit side, but the problem is I couldn't find any mechanic part to do the external switch's job like mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I'm not going wiring over the original switch is because my mother isn't familiar with smartphones, so I want to implement it without interfering with the basic usage.&lt;/p&gt;&#xA;" OwnerUserId="9360" LastEditorUserId="368" LastEditDate="2016-12-28T17:45:48.537" LastActivityDate="2016-12-28T17:45:48.537" Title="How can I modify a pair of two-way switches so that the circuit can also be opened or closed remotely using a device?" Tags="|electrical-engineering|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12984" PostTypeId="1" AcceptedAnswerId="12986" CreationDate="2016-12-27T17:46:39.277" Score="18" ViewCount="5869" Body="&lt;p&gt;I'd like to mount a shelf on the wall. I have two shelf supports to do so, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pluMo.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pluMo.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative to the shelf, where do I place the (blue) supports to achieve the best distribution of load?&lt;/p&gt;&#xA;" OwnerUserId="9362" LastEditorUserId="368" LastEditDate="2016-12-27T20:19:51.107" LastActivityDate="2021-06-11T01:19:59.827" Title="How should I position two shelf supports for the best distribution of load?" Tags="|mechanical-engineering|statics|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12985" PostTypeId="2" ParentId="12984" CreationDate="2016-12-27T18:00:05.077" Score="2" Body="&lt;p&gt;There are several criteria you need to consider:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;To distribute the load most evenly between the two supports, put them as far apart as possible.  That would be at the ends in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;The worst case heavy object placed at one end with nothing else on the shelf has to not flip the shelf over.  This puts a limit on how far from the ends the supports can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;The strength of the shelf needs to be higher to support the same load as the supports are moved farther apart.  In this case, consider the worst case load placed exactly at center.  Even if the shelf is strong enough to not be damaged by such a load, the resulting bend may be objectionable to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only you can decide how these various competing criteria trade off in your case.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-12-27T18:00:05.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12986" PostTypeId="2" ParentId="12984" CreationDate="2016-12-27T18:57:59.687" Score="25" Body="&lt;p&gt;As with all good things, it depends.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you can assume that your supports are totally stiff and that the loading on the shelf will be approximately uniform, then you basically have the following structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l79qK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l79qK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rectangular cross-section (such as a plank) will behave equally under positive or negative bending moment, so your objective should be to balance both. To do so, you want your main span to be &lt;span class=&quot;math-container&quot;&gt;$2\sqrt2 \approx 2.83$&lt;/span&gt; times the cantilevers. This is found by calculating the cantilever required to offset half of the bending moment due to a uniform load along a simply supported beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;M_{mid} &amp;amp;= \dfrac{qL_{mid}^2}{8} - M_{cant} \\&#xA;M_{cant} &amp;amp;= \dfrac{qL_{cant}^2}{2} = \dfrac{M_{mid}}{2} = \dfrac{qL_{mid}^2}{16} \\&#xA;\therefore 8L_{cant}^2 &amp;amp;= L_{mid}^2 \\&#xA;2\sqrt2 L_{cant} &amp;amp;= L_{mid}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If your shelf may be empty, except for a single concentrated load anywhere, then you have to find the points for the supports where the maximum bending moment due to a concentrated load at the edges will be equal to that of a load at midspan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yXenj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yXenj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;M_{mid} &amp;amp;= \dfrac{PL_{mid}}{4} \\&#xA;M_{cant} &amp;amp;= PL_{cant} = M_{mid} = \dfrac{PL_{mid}}{4} \\&#xA;\therefore 4L_{cant} &amp;amp;= L_{mid} \\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking an average of the two cases above, I'd adopt a main span some 3.5 times larger than the cantilevers as a first guess.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If your supports are loose, meaning they won't hold the shelf if it is raised, so there's the risk of it toppling over with too much load on one cantilever, then you need to calculate the maximum load the weight of your shelf can resist. There's too many variables to figure this out trivially, though (linear weight of the shelf, maximum adopted weight, etc). And, honestly, if your supports are loose, you really should just put the supports at the ends, eliminating this risk entirely (but therefore reducing the maximum load the shelf will resist).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Diagrams obtained with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis program.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-12-28T17:55:01.567" LastActivityDate="2019-12-28T17:55:01.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="12988" PostTypeId="1" AcceptedAnswerId="12993" CreationDate="2016-12-27T22:28:00.633" Score="3" ViewCount="3460" Body="&lt;p&gt;I am trying to design something that rotates 90 degrees based on a small (10mm or so) linear motion, using &lt;a href=&quot;https://www.youtube.com/watch?v=09jZsBDHalE&quot; rel=&quot;nofollow noreferrer&quot;&gt;this basic setup&lt;/a&gt; as an example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a lot of questions about how long the arms need to be, how far they need to anchor from the point of rotation, etc. that i need to know. I am assuming there is some formula that governs the relationship that someone can point me to?&lt;/p&gt;&#xA;" OwnerUserId="9364" LastActivityDate="2016-12-28T14:07:29.117" Title="Formula for converting linear to rotational motion" Tags="|mechanical-engineering|design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12990" PostTypeId="2" ParentId="12983" CreationDate="2016-12-28T00:14:56.700" Score="5" Body="&lt;p&gt;You don't really want an &quot;external case&quot; to mechanically switch one of the existing switches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you really want is a DPDT relay that your Arduino (or whatever) controls, and wire that into the circuit as a third switch. Here's a diagram that shows what I mean:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/njI3v.png&quot; alt=&quot;schematic&quot;&gt;&lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2016-12-28T00:24:21.223" LastActivityDate="2016-12-28T00:24:21.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="12993" PostTypeId="2" ParentId="12988" CreationDate="2016-12-28T08:52:22.390" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1B2xn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1B2xn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your given mechanism can be shown by figure above. the driving element is link DF, which moves linearly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;first of all calculate theta by,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CptEi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CptEi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where x is distance travelled by DF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now position of C can be given by, (origin at A)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EKvBA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EKvBA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have 2 equations,&#xA;both have common origin at A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oY2ZS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oY2ZS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;notice that both are equation of circle, one at origin with radius of AB, other at point C with radius BC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;solve the second equation for y in terms of x, then substitute it in equation 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you will arrive at quadratic equation where you can easily solve for x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also you know x = r cos alpha  (alpha is the angle you want to be 90)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the above information you can find your required term&lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2016-12-28T08:52:22.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12995" PostTypeId="1" CreationDate="2016-12-28T11:07:07.560" Score="1" ViewCount="2650" Body="&lt;p&gt;so for my project im designing a Tesla turbine and have calculated that the torque it will output is 38Nm. I would like to know what more information I would need that i could calculate theoretically to calculate how much power it would output from this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="9375" LastActivityDate="2016-12-28T13:00:27.387" Title="Calculating power from torque" Tags="|torque|power|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="12996" PostTypeId="2" ParentId="12995" CreationDate="2016-12-28T12:54:23.327" Score="0" Body="&lt;p&gt;Power is force times speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your shaft is turning at 1 Hz, for example, then the power is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(38 N)(2&amp;Pi; m/s) = 239 Nm/s = 239 W&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-12-28T12:54:23.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="12997" PostTypeId="2" ParentId="12995" CreationDate="2016-12-28T13:00:27.387" Score="1" Body="&lt;p&gt;general power formula is 2*3.14159 * N *T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you need to know revolutions per second for finding power.&#xA;there are various tachometers available. (stroboscope,aluminum cup,mechanical tachometer, etc) get the reading of revolution per min. convert it to rev per sec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then use it in the formula p = 2*3.1415*N*(38)&lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2016-12-28T13:00:27.387" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="12998" PostTypeId="2" ParentId="12988" CreationDate="2016-12-28T13:54:18.210" Score="3" Body="&lt;p&gt;&lt;strong&gt;There is a mechanism called:&lt;/strong&gt; &lt;a href=&quot;https://www.google.com.tr/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=slider%20crank%20mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;SLIDER CRANK MECHANISM&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can copy one of these mechanism for just converting linear to rotational motion. In most of the cases, this will do the trick. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for general, if you want to design a mechanism that will do your desired motion, get ready to do some work!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;First, here some links you will probably need for getting the dimensions for your particular shape:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://saltire.com/atlas/CrankRocker.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Web Atlas of Linkages&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.cedarville.edu/cf/engineering/kinematics/ccapdf/ccpagef.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Four-bar Coupler Curves and Centrodes&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.saltire.com/LinkageAtlas/Crank%20Rocker.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Atlas of Linkages - Crank Rocker&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.mekanizmalar.com/fourbar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Four-Bar Linkage and Coupler Curve&lt;/a&gt; (Especially this one!)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then you must start your motion (PVA) analysis, you can use some schema like that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JNUEu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JNUEu.png&quot; alt=&quot;by Virtual Labs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you will need it, if you only want to perform your motion. But if you want to have an analysis over your mechanism (for every position of crank), you can start with inspecting &lt;a href=&quot;http://facultad.bayamon.inter.edu/elay/mecn4110/ch04%20Position%20Analysis.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;these slides&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here it comes, there are some example of mine because the process is much or less same.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Four Bar Mechanism for Straight Line&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is the desired shape:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bj2EG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bj2EG.png&quot; alt=&quot;Straight Line&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the calculations for PVA analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HuKHY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HuKHY.png&quot; alt=&quot;Theta&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4UWoU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4UWoU.png&quot; alt=&quot;Alpha&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and some more of course...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all, to sum up:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Finding the dimensions of linkes for the desired shape&lt;/li&gt;&#xA;&lt;li&gt;Writing and solving Vector Loop&lt;/li&gt;&#xA;&lt;li&gt;PVA Analysis&lt;/li&gt;&#xA;&lt;li&gt;Test final mechanism in MATLAB or Solidworks.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here is MATLAB generated mechanism: (Also &lt;a href=&quot;https://www.youtube.com/watch?v=Ax-GNOOiQmc&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; @youtube with Solidworks)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nswNe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nswNe.png&quot; alt=&quot;Four Bar for Straight Line&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps, I wanted to help as much as I can because I've been experienced your suffer and it will be good to be informed when you are at the start.&lt;/p&gt;&#xA;" OwnerUserId="7468" LastEditorUserId="7468" LastEditDate="2016-12-28T14:07:29.117" LastActivityDate="2016-12-28T14:07:29.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13000" PostTypeId="1" CreationDate="2016-12-28T15:40:11.327" Score="0" ViewCount="61" Body="&lt;p&gt;There are many sensor fusion algorithms available like the complementary filter or the EKF. They all do a decent job for rotational movement attitude estimation, but what happen when I have a translational movement? For example the complementary filter does rely on the accelerometer alone when there is slow angular movement, which is the case for translational movement (no angular movement at all). So, the estimated angle is wrong and I get a deviation from the real position. Is there anything I can do to avoid this drift? Would a Kalmanfilter compensate for that? If we take as an example a quadcopter, where often a complementary filter is used. How does they avoid this drift? There is a lot of translational movement in this kind of application.&lt;/p&gt;&#xA;" OwnerUserId="8784" LastActivityDate="2016-12-29T13:46:09.673" Title="Sensor Fusion under translational movement" Tags="|control-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13001" PostTypeId="1" CreationDate="2016-12-28T15:46:36.243" Score="3" ViewCount="115" Body="&lt;p&gt;My client is aksing me if I can manufacture a small but long strip-shaped aluminum-alloy (7075 T651) part. The sectional dimension is about 15mm by 12 mm and the length is about 1000mm. Ideally, this part shall be made as extrusions. However, the client just needs one piece of the product, so he asks me if I can manufacture this product by milling.&#xA;The main problem is that there's a through slot in the lengthwise direction, and the wall is very thin, about 1mm. The process I come up with is that I mill the slot first when much stock is left around it and hence the wall is thick enough, and then &lt;strong&gt;fill something&lt;/strong&gt; into the slot so as to maintain/increase the rigidity of the part for the next milling process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: which material shall I use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, this material should adhere/stick to the metal part, and more importantly, it is easy to be removed from the part after I finish all processes. Maybe paraffin wax is a candidcate, but I suspect if it's strong enough. Someone suggested I use PVC soldering rod, but I tried and failed. It didn't stick to the part. I also considered expoxy resin, but it's not easy to be removed and cleared from the part.In thi case, I may need to polish the slot remove the resin. That would be annoying. Can someone here shed light on the material selection? Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="7460" LastActivityDate="2016-12-31T00:21:51.667" Title="Which material is suitable to be fiiled in a thin-walled workpiece to increase the rigidity during the machining process?" Tags="|materials|machining|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13002" PostTypeId="2" ParentId="13001" CreationDate="2016-12-28T16:53:57.063" Score="2" Body="&lt;p&gt;You can get machinable waxes which have a bit more mechanical strength than plain paraffin wax you can actually make it yourself by melting paraffin wax and mixing in shredded polythene at about 15-25% by mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with epoxies or similar resins is that it will either adhere and be hard to remove or id you use a release agent it won't adhere at all unless it is held in by other means. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-28T16:53:57.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13003" PostTypeId="1" CreationDate="2016-12-28T16:56:17.507" Score="-1" ViewCount="73" Body="&lt;p&gt;I'm not an engineer, but I'm doing some consulting work for a company that does this.  While I don't need to know the specifics of what they do for what I am doing, I'd like to be able to talk about it without sounding completely clueless.  My google searches haven't provided much insight.&lt;/p&gt;&#xA;" OwnerUserId="9381" LastActivityDate="2017-07-27T17:15:54.250" Title="What is someone talking about when they indicate a company is involved in assembling and sequencing of automobile wheels?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2017-07-28T00:39:58.693" ContentLicense="CC BY-SA 3.0" />
  <row Id="13004" PostTypeId="1" AcceptedAnswerId="13009" CreationDate="2016-12-28T17:33:45.707" Score="4" ViewCount="155" Body="&lt;p&gt;I'm trying to increase the Helium detection capabilities of a sniffer unit. Could a cyclonic separator, like that used in the oil/gas industry, with enough fluid velocity move the heavier gas molecules in air (O&lt;sub&gt;2&lt;/sub&gt;, N&lt;sub&gt;2&lt;/sub&gt;, Ar, CO&lt;sub&gt;2&lt;/sub&gt;) to the sides of it while keeping the lighter elements (He, H&lt;sub&gt;2&lt;/sub&gt;) in the center to be detected by a sniffer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're trying to detect leaks of helium from refrigeration systems on an assembly line. we're designing a positive air box to prevent any outside contamination from the filling and recovery stations. The box will have a blower attached to it to pull the volume of the box in 10 seconds. because most of the gases in the box are not our trace gas we have a very low mbar l/s level (allowable is 9e-7) that is on the borderline of the minimum of our sniffer unit we are looking to use. The sniffer unit has the precision of 1e-7 mbar l/s at 300 SCCM and 1e-6 mbar at 3000 SCCM. We are going to have an array of sniffer heads in the center of this conic that will sample the air at the center (with the thought that the lighter elements would be there) leading to a more accurate reading because it is sampling less of the non-trace gas. This process of sampling and detection along with the lifting/closing of doors and shifting of the equipment is done on a conveyor belt within a 40 second time period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current system is a person at the assembly line with a hand held sniffer going around the connections, but are getting readings that there are leaks even though there might not be, probably due to the helium charging station only 3 feet away from the leak testing area. we are looking to automate the process for less false readings and quicker turn over time in the process.&lt;/p&gt;&#xA;" OwnerUserId="9305" LastEditorUserId="9305" LastEditDate="2016-12-28T18:17:08.953" LastActivityDate="2016-12-28T23:17:47.430" Title="Can cyclonic separators be used to separate heavier and lighter gases?" Tags="|mechanical-engineering|fluid-mechanics|airflow|gas|cyclones|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13006" PostTypeId="2" ParentId="13004" CreationDate="2016-12-28T20:44:08.970" Score="0" Body="&lt;p&gt;Gas centrifuges (I assume that's what you mean by &lt;i&gt;cyclonic separators&lt;/i&gt;) certainly work.  They are used to separate Uranium-hexaflouride with U&lt;sup&gt;235&lt;/sup&gt; and U&lt;sup&gt;238&lt;/sup&gt;.  The relative densities of helium and components of air is &lt;i&gt;much&lt;/i&gt; greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting good information on gas centrifuges may be tricky because of their military use, and digging too hard may put you on some watch-list.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2016-12-28T20:44:08.970" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13009" PostTypeId="2" ParentId="13004" CreationDate="2016-12-28T23:17:47.430" Score="1" Body="&lt;p&gt;Cyclones are designed to quickly and cheaply remove contaminants that would tend to settle out of your process gas (or liquid) if it stopped flowing. They're very good at controlling coarse PM, not so good at controlling fine PM and really quite poor at controlling ultrafine PM. What you're looking for here is so far past ultrafine PM control on this spectrum that you would never achieve a useful result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you could slightly concentrate your heavier gases around the outside of the cyclone, where are they going to go? In a cyclone, the volumetric flow is the same at the inlet as it is at the outlet. The only thing that gets knocked out is the particulate, when it impacts the wall and settles down into the hopper. This works because that particle would tend to settle anyway, if it weren't entrained in your process flow, and near the wall of the cyclone the flow approaches zero (recall the &lt;a href=&quot;https://en.wikipedia.org/wiki/No-slip_condition&quot; rel=&quot;nofollow noreferrer&quot;&gt;no-slip condition&lt;/a&gt; of fluids 101). The same can't be said of a theoretical packet of slightly-heavier-than-average air.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We are going to have an array of sniffer heads in the center of this conic that will sample the air at the center (with the thought that the lighter elements would be there) leading to a more accurate reading because it is sampling less of the non-trace gas.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not sure this really ends up being more accurate, given that your sample is no longer representative of the mixture on which your leak detection threshold of 9 x 10&lt;sup&gt;-7&lt;/sup&gt; mbar·L/s is based. This isn't a situation where you want to set your leak detection threshold at zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think any clever gas stratification scheme would just over-complicate the process. Using the PPV box might be enough by itself to make your client happy. Otherwise, if you just need a little bit more sensitivity and you can't go with a more sensitive device, I'd keep thinking about optimizing the sampling conditions—can your client afford to spend 90 seconds per unit rather than 40, if it cuts down on false positives? Or maybe your client would be willing to perform some more reliable leak test on a sample, rather than testing every unit, if they can accept a non-zero defect rate.&lt;/p&gt;&#xA;" OwnerUserId="368" LastActivityDate="2016-12-28T23:17:47.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13010" PostTypeId="2" ParentId="12898" CreationDate="2016-12-29T00:19:56.430" Score="1" Body="&lt;p&gt;The given system is linear time invariant (LTI), therefore the superposition principle can be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is denoted as:&#xA;$\dot{x}(t) = Ax(t)+ B_1 u(t) + B_2 \omega (t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the superposition principle the stepresponce (depending on the initial condition and the input) is given by&#xA;$x(t) = e^{A t} x(0) + \int_0^t e^{A(t-\tau)} B_1 u(\tau)d\tau + \int_0^t e^{A(t-\tau)} B_2 \omega(\tau)d\tau$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed loop response can be found by substituting the (linear) feedback law in the above equation.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2016-12-29T00:19:56.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13012" PostTypeId="2" ParentId="12881" CreationDate="2016-12-29T00:31:24.387" Score="0" Body="&lt;p&gt;US base Cellular vendors (Such as Verizon, Sprint, AT&amp;amp;T) in combination with cell phone manufactures are starting to introduce a technology call Carrier-branded wifi calling. This is similar to Skype, Google Hangouts, Facebook Messenger and WhatsApp, except the phone dialer is integrated to the smartphone's wifi radio. So the cellular connection is weak, then the phone will automatically switch over to Wifi-Calling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note as I understand the LTE uses VoIP (VoLTE) to make the call. Skype like services use VoIP. This is an integration of both wifi and cellular technologies. Depending of the wifi connection the consumer is likely to experience better voice quality. Currently a VoIP call will requires about 1MB/minute of data for a voice call. The typical video call will take about 6-8 MB/minute of data for a video call. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As consumers move over to Wifi calling the need for cellular towers will decrease thus will be used for true mobile connectivity. Thus the residential consumer will less depend on cell phone towers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.verizonwireless.com/support/wifi-calling-faqs/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wi-Fi Calling FAQs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.cnet.com/news/what-you-need-to-know-about-wifi-calling/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Everything you need to know about Wi-Fi calling&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.gsma.com/network2020/technology/volte/&quot; rel=&quot;nofollow noreferrer&quot;&gt;VoLTE Realising revenue potential with Voice over LTE&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2017-01-28T03:15:44.157" LastActivityDate="2017-01-28T03:15:44.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13013" PostTypeId="1" AcceptedAnswerId="13041" CreationDate="2016-12-29T09:33:51.663" Score="8" ViewCount="588" Body="&lt;p&gt;Consider a 1D point-mass moving along an axis. A force $u$ is applied as control. There is no gravity or other forces involved. The system can be described in state space equations as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;A &amp;amp;= \begin{bmatrix} 0 &amp;amp; 1 &amp;amp; 0 \\&#xA;                     0 &amp;amp; 0 &amp;amp; 1 \\&#xA;                     0 &amp;amp; 0 &amp;amp; 0 \end{bmatrix} \\&#xA;B &amp;amp;= \begin{bmatrix} 0 \\&#xA;                     0 \\&#xA;                     \dfrac{1}{M} \end{bmatrix} \\&#xA;C &amp;amp;= \begin{bmatrix} 0 &amp;amp; 0 &amp;amp; 1 \end{bmatrix} \\&#xA;D &amp;amp;= [0]&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shown system is controllable, but not observable. Not even structurally observable and most certainly not fully observable. Thus, it should be impossible to construct an observer for this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if I know the initial state of the system, I can compute the full state at every time, i.e. by integrating the system's output. How does this fall in line with the concept of observability? How would I incorporate the initial state into the equations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find the error in my train of thought, but I am certain there is one. Do I misunderstand observability?&lt;/p&gt;&#xA;" OwnerUserId="9388" LastEditorUserId="6790" LastEditDate="2020-05-05T09:14:16.580" LastActivityDate="2020-05-05T09:14:16.580" Title="Why is it impossible to create an observer for this not fully observable system?" Tags="|control-engineering|control-theory|linear-control|linear-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13014" PostTypeId="2" ParentId="13003" CreationDate="2016-12-29T13:06:04.350" Score="1" Body="&lt;p&gt;It's not that common to talk about 'assembling' something like automotive wheels as they are usually cast, pressed or machined and as such tend to have relatively few parts and 'sequencing' could mean all sorts of things from the order of manufacturing operations to broader production scheduling. There is certainly no well defined general engineering which is obvious from the context. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest your best course is to just ask them to explain exactly what they do. If you are consulting about something other than the manufacturing and engineering side of the business then this would be an entirely normal thing to ask.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-29T13:06:04.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13015" PostTypeId="2" ParentId="13000" CreationDate="2016-12-29T13:46:09.673" Score="2" Body="&lt;p&gt;It is as @Brian Drummond said. 'Fusion' implies that you have multiple sources of data to combine. This isn't just limited to sensors. Anything that gives data about your system will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sensor fusion you always need a model, aka some assumption about your system. However, this only works if the desired state is &lt;strong&gt;observable&lt;/strong&gt;. Consider these two examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you attach an accelerometer to a 1d pendulum &#xA;$$m\ddot{x} = -kx+u~~\text{and}~~h(x,\dot{x},\ddot{x}) = \ddot{x}$$&#xA; you can reconstruct $x(t)$ based on $h(\cdot),u(t)$ readings, because all states are observable. In this case a Kalmanfilter can compensate and the position error $e_x(t)\to 0$ for $t\to\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you put the accelerometer on a freely movable 1D point-mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m\ddot{x} = u~~\text{and}~~h(x,\dot{x},\ddot{x}) = \ddot{x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can only get information about $\ddot{x}$, but $\dot{x}$ and $x$ are not observable. Now it is impossible for a Kalmanfilter (or any other observer) to reconstruct $x$. You can still integrate acceleration, but the error will only increase for $t\to\infty$. This is also known as &lt;strong&gt;dead reckoning&lt;/strong&gt;. If you wanted to track this system you need to directly measure $x$. If you do, you are able to stop measuring $\ddot{x}$ and still reconstruct the entire state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same principle holds for a quadcopter. Depending on the &lt;strong&gt;observability&lt;/strong&gt; of the position it may or may not be possible to get a precise position estimate. Thus, quadcopters often measure position directly and mainly use accelerometers to increase precision.&lt;/p&gt;&#xA;" OwnerUserId="9388" LastActivityDate="2016-12-29T13:46:09.673" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13016" PostTypeId="2" ParentId="12884" CreationDate="2016-12-29T15:53:13.710" Score="1" Body="&lt;p&gt;The sign convention for Mohr's Circle is dependent on the user. The direction of $\tau_n$ can be used in both ways, but you need to make sure that if you do it in the opposite method, that everything is opposite. Standard mohr's circle is what is used in Method 2 with $\tau_n$ pointing down and $\sigma_n$ pointing to the right ($\sigma_n$ will always point to the right no matter what method you use). In any of my classes that I use Mohr's Circle in, I never use method 1 as it has confused me too much with the change in direction of all of the elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are dealing with a situation that is similar to method 3, then you simply invert the direction of $\tau_n$ to get the proper sign convention. For example if you took a square bar of rubber that was mounted to a solid immovable wall and twisted it towards you with one hand, the positive $\tau_n$ ($\tau_y$) would be in the closest and farthest sides from you because those are in tension while the top and bottom are in compression meaning that $\tau_n$ is negative ($\tau_x$). This results in the picture seen in method 2. Now twist it the other way and the torsional stresses change in direction. This would result in the picture having $\tau_{xy}$ in the top left corner and bottom right corner rather than what is seen in method 2. They fix this method by inverting the signs resulting in the $\tau_{xy}$ being located where that are in method two, but the values of these stresses are now all opposite due to their change in direction. The Mohr's Circle used in method 3 depicts this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you were to use the values in method 3 in the Mohr's circle right away you would end up with something similar to method 1, there isn't anything wrong with that but because all the values are opposite it would be very easy to confuse signage during the calculations, so the $\tau_n$ is inverted so that you end up with something similar to method 2 that is easier to use. Then you just solve using the values you have and you should get your proper answer while making sure to pay attention to the direction of the values (positive/negative) that you get from your calculations.&lt;/p&gt;&#xA;" OwnerUserId="9305" LastEditorUserId="1832" LastEditDate="2016-12-29T15:59:19.470" LastActivityDate="2016-12-29T15:59:19.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13017" PostTypeId="1" CreationDate="2016-12-29T16:46:43.217" Score="1" ViewCount="397" Body="&lt;p&gt;We recently covered fundamental laws in our Computational Continuum Mechanics course last term. There are many forms of these laws and I am confused which one to use and when.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance in mass continuity, we can write the integral version (I am not gonna write it here) as usual. Then we can use the Jacobian to write it like this &#xA;$$ \int_{V_0} \rho_0 - J \rho \ \ dV $$&#xA;which can be said to be the weak form. Then by localisation, we can have&#xA;$$ J(\pmb{x},t) = \frac{\rho_0(t)}{\rho(\pmb{y},t)} $$&#xA;where $\pmb{y}(\pmb{x},t)$ is the deformed configuration position and $\pmb{x}$ is the undeformed configuration position (if that make sense) which is now the strong form. So isn't this good enough already? Why don't we use this in our code, because this is also point-wise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also derive the differential form of course which is&#xA;$$ \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \pmb{u})$$&#xA;where $u$ is velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I just wanted density with time, is it better to use the strong form? Would I just use the differential form if I want to include velocity or something? Also our lecturer mentioned that the Jacobian strong form is in Lagrangian form but differential is in Eulerian form. I know the meaning of both but I am not sure what it means in this context.&lt;/p&gt;&#xA;" OwnerUserId="7651" LastActivityDate="2016-12-29T17:24:49.903" Title="Difference between strong, weak and differential forms?" Tags="|mechanical-engineering|fluid-mechanics|cfd|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13018" PostTypeId="2" ParentId="13017" CreationDate="2016-12-29T17:24:49.903" Score="1" Body="&lt;p&gt;If your course started with an indigestible dose of multi-dimensional calculus without much discussion of how the math relates to the physics (and there are certainly textbooks on computational CFD which start that way!) the answers to these questions may become clearer when you start developing the actual numerical methods and seeing how they perform solving &quot;real&quot; problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two basic ways to devise a numerical solution scheme. One way is to consider the values of quantities defined at &lt;em&gt;specific points&lt;/em&gt; in space (e.g. the grid points of a finite difference mesh). The computational method ignores what happens to the quantities &quot;in between&quot; the mesh points, and you hope that as you increase the number of points, the computed values will converge to the solution of the underlying differential equations. (Sometimes, you can prove the solution will converge in this sense, but for most &quot;real world&quot; problems involving nonlinear differential equations you can't prove it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way is to consider the value of properties over &lt;em&gt;regions&lt;/em&gt; of space - for example the mass, energy and momentum of a &quot;unit cell&quot; defined by the grid points that surround it, and the flows into and out of the cell across its boundaries. This has the advantage that the solution method can guarantee that it satisfies some fundamental physical laws like conservation of mass and energy. Usually, a solution method based only on values at specific points will not automatically satisfy such conservation laws, though (if it is a useful solution method!) increasing the number of solution points should reduce the errors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But satisfying the global conservation laws doesn't necessarily mean that the grid-point values are &quot;accurate&quot; - for example there may be unphysical oscillations at adjacent grid points which don't affect the &quot;average&quot; energy, momentum, etc. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also our lecturer mentioned that the Jacobian strong form is in&#xA;  Lagrangian form but differential is in Eulerian form. I know the&#xA;  meaning of both but I am not sure what it means in this context.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Without knowing exactly what was said (or in the lecture notes) I'm not sure what that means either. You can certainly create four versions of the mathematical formulation, i.e. weak and strong forms of both the Lagrangian or Euleran equations - though some are more intuitively &quot;obvious&quot; than others. In fact some numerical methods use a combination of more than one of the four options.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-12-29T17:24:49.903" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13019" PostTypeId="2" ParentId="10889" CreationDate="2016-12-29T19:16:59.017" Score="0" Body="&lt;p&gt;Really pressure is the metric that  you are interested in as this will determine how hard or soft it feels to sit on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this may indeed vary you don't have to rely on instantaneous pressure and it is entirely possible to base your settings on an average over some period of time and there are all sorts of control systems which can cope with quite large perturbations and sensor noise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed there are existing similar applications, for example tyre pressure sensors in F1 cars which can be very accurate despite large variations in external load and changes in temperatures. Similarly there are systems for off road vehicles which allow tyre pressures to be adjusted on the fly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physical design of the system can also help to damp variations, for example by filling the cushion with open cell foam. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-29T19:16:59.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13020" PostTypeId="1" CreationDate="2016-12-29T21:01:37.160" Score="2" ViewCount="48" Body="&lt;p&gt;How do I implement scientific experiment to derive a equation relevant to subject of stress and fatigue called Gerber Line equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tHVzA.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tHVzA.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;May you also provide some historical context to that?&lt;/p&gt;&#xA;" OwnerUserId="252" LastActivityDate="2016-12-29T21:01:37.160" Title="How do I implement scientific experiment to derive a equation called Gerber Line?" Tags="|mechanical-engineering|stresses|machining|fatigue|" AnswerCount="0" CommentCount="1" ClosedDate="2016-12-30T00:01:04.343" ContentLicense="CC BY-SA 3.0" />
  <row Id="13021" PostTypeId="1" AcceptedAnswerId="13030" CreationDate="2016-12-29T22:41:19.893" Score="1" ViewCount="43" Body="&lt;p&gt;I have some high pressure and vacuum lines off of a tool from the 1980s and they are labeled as, for example, &quot;3/8 imperial - eastman,&quot; which seems to be different from how normal compression connections are labeled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the same as regular imperial units, or is there some difference I need to be aware of? Where can I find these parts?&lt;/p&gt;&#xA;" OwnerUserId="7273" LastEditorUserId="368" LastEditDate="2016-12-30T17:57:33.413" LastActivityDate="2016-12-30T17:57:33.413" Title="Are &quot;Imperial Eastman&quot; compression connections sized the same as regular imperial connections?" Tags="|pressure|compressed-air|vacuum|tools|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13022" PostTypeId="1" AcceptedAnswerId="13044" CreationDate="2016-12-30T02:53:26.033" Score="2" ViewCount="263" Body="&lt;p&gt;I have two basic questions about helicopters. Let's say the z-axis is the vertical axis from ground to sky. I know the tail rotor is there to counteract the angular momentum effects of the main rotor, and keep the total angular momentum about the z-axis at 0.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Say the tail rotor spins about the x-axis. What keeps the body of the helicopter spinning about the x-axis in the opposite direction? I assume it's a slight tilt of the main rotor forward to balance this effect. Am I right?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How are the speeds of the two rotors are kept coordinated? I assume there must be a sensor/feedback mechanism by which rotor speeds change to counteract spinning.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9396" LastActivityDate="2016-12-31T14:33:56.043" Title="Questions about helicopter rotors - conservation of angular momentum" Tags="|aerospace-engineering|aerodynamics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13023" PostTypeId="1" CreationDate="2016-12-30T13:27:34.163" Score="4" ViewCount="139" Body="&lt;p&gt;I have to pour outdoor concrete next week. The pour is in two main parts - a 100&amp;nbsp;mm cover / 2nd pour over a 6&amp;nbsp;x&amp;nbsp;3&amp;nbsp;m area of existing concrete from a 1st rough pour, and the bottom stub part of a retaining concrete wall profile about 500&amp;nbsp;mm wide x 750&amp;nbsp;mm high x 4&amp;nbsp;m long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the weather outlook is quite cold, getting up to 6-8°C (43-46°F) late next week daytime, but generally otherwise around 2-4°C (36-41°F) in daytime, which is below the 5°C minimum for concrete. The concrete can be poured in the morning so it'll have about 6 hours of daylight to start setting, but nights are forecast to be cold - close to zero (0 - 1°C / 32-34°F). Both are exposed outdoor pours, there isn't an issue with wind or rain, just temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't really afford to put off the pour (previous pour date didn't happen for unrelated reasons) especially with no guarantee that the week after (or the one after that) will be any better,but I do have several days to prepare, if preparations are needed to ensure success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nothing else comes up, I might have the capability, as a last resort, of &quot;tenting&quot; the entire poured areas under a lightweight wood frame overlaid with heavy polythene (DPM) and running some air /space heaters under it for a couple of days before or after, if there isn't a better option. But that's only a layperson's guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are my options for ensuring the concrete sets properly despite the forecast low temperatures?&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="368" LastEditDate="2016-12-30T17:14:15.553" LastActivityDate="2017-01-02T20:27:45.303" Title="Got to pour outdoor concrete next week, but forecast is cold. What are my options?" Tags="|structural-engineering|temperature|concrete|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13024" PostTypeId="2" ParentId="13023" CreationDate="2016-12-30T14:01:42.743" Score="3" Body="&lt;p&gt;The absolutely critical factor is that the concrete does not cool to 0C before it has developed enough strength to resist the expansion if any free water in the mix freezes. If it does freeze before setting, most likely you will have to remove all of it and start again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 5C, concrete should develop enough strength to resist freezing after about 48 hours, and faster at higher temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of the concrete itself when it is poured is an important factor in this. Your supplier may be able to deliver heated concrete - typically at 10C minimum when it is poured. If your supplier can't guarantee a pour temperature of at least 5C, then whatever you do it going to be risky. If you are mixing the concrete on site, consider using pre-heated water. Water taken straight from the supply pipes will be at a higher temperature than water from an uninsulated tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical reaction while the concrete is setting is exothermic, so good insulation against wind chill and radiation cooling at night is probably enough, so long as the ambient temperature stays above freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you do supply artificial heat, there is a risk of cracking the concrete when the heat source is removed, if the ambient temperature is low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the good news is that the final strength of the concrete will be higher if it sets at slowly at low temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small scale work (e.g. setting fence posts in concrete), you can get fast-setting concrete which will set within an hour or so of pouring, but that isn't a sensible option for your 18 square meter slab - unless you really need to build it &quot;right now&quot; even though the climate is permanently below freezing!&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2016-12-30T14:14:47.963" LastActivityDate="2016-12-30T14:14:47.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13025" PostTypeId="1" CreationDate="2016-12-30T14:56:02.030" Score="3" ViewCount="1161" Body="&lt;p&gt;&lt;strong&gt;How can I automatically compute and draw/display the &lt;a href=&quot;https://en.wikipedia.org/wiki/Voronoi_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voronoi diagram&lt;/a&gt; of a given set of vertices on a surface in AutoCAD?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;&lt;a href=&quot;https://apps.autodesk.com/ACD/en/Detail/HelpDoc?appId=1006119760063675415&amp;amp;appLang=en&amp;amp;os=Mac#inunininfo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voronoi Sketch Generator&lt;/a&gt;&quot; is the best I could come up with on a google search. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite close, but:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It doesn't seem to work for curved surfaces, only flat 2d planes&lt;/li&gt;&#xA;&lt;li&gt;It is a plugin for Fusion360, not AutoCAD&lt;/li&gt;&#xA;&lt;li&gt;It doesn't let me define the set of vertices of the diagram (it generates them randomly)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is my understanding that I can import an AutoCAD drawing to Fusion360 and then export it back to AutoCAD. Maybe tuning up this plugin for this task would be my best choice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being a telecommunications software engineer I have near zero experience with AutoCAD, so I apologize for my absolute ignorance on the topic.&lt;/p&gt;&#xA;" OwnerUserId="9403" LastEditorUserId="368" LastEditDate="2016-12-30T17:10:34.853" LastActivityDate="2019-02-14T13:46:13.120" Title="How can I draw Voronoi regions in curved surfaces in AutoCAD?" Tags="|autocad|geometry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13026" PostTypeId="2" ParentId="13022" CreationDate="2016-12-30T15:51:34.340" Score="2" Body="&lt;p&gt;You're right on Q1, but typically the blades are designed in such a way as to take advantage of Bernoullis principle. Notice the pitch of the blades: this animation is pretty cool:&#xA;&lt;a href=&quot;https://m.youtube.com/watch?v=C7N3ME5NTOs&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://m.youtube.com/watch?v=C7N3ME5NTOs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Control systems engineering. Feedback loops, pid control, lots of computational control of dynamic systems keeps them in order. &lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-12-30T15:51:34.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13027" PostTypeId="2" ParentId="8525" CreationDate="2016-12-30T16:46:06.437" Score="-2" Body="&lt;p&gt;Your safest best is to run a simple FEA with mesh analysis, but you can calculate shear stress and strain by hand and determine the factor of safety using $S=\frac{M\cdot y}{I}$ (max moment * distance over inertia).&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="368" LastEditDate="2016-12-30T17:39:04.267" LastActivityDate="2016-12-30T17:39:04.267" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13029" PostTypeId="1" AcceptedAnswerId="13031" CreationDate="2016-12-30T17:22:29.170" Score="3" ViewCount="728" Body="&lt;p&gt;I am trying to model the strain generated by inserting an electrode device into brain tissue. I've had some success in doing this in ANSYS Explicit Dynamics. &lt;a href=&quot;https://i.stack.imgur.com/VTTFB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's a link to a GIF showing the penetration&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My issue is that I had to do a very fast insertion (10 m/s) in order to have a faster computation time (~18 h). The actual insertion speed is on the order of 0.1 mm/s, which ends up being 5 seconds of simulation. 5 seconds of explicit dynamics simulation is days of computation on my computer! My question is, can I do this sort of slow penetration/ballistics model in ANSYS transient dynamics or some other implicit model? Or am I stuck with Explicit dynamics and the super long computation time?&lt;/p&gt;&#xA;" OwnerUserId="9405" LastEditorUserId="368" LastEditDate="2016-12-30T18:03:10.183" LastActivityDate="2016-12-31T22:16:14.900" Title="How can I reduce the computation time required to simulate a brain tissue penetration model?" Tags="|finite-element-method|ansys|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13030" PostTypeId="2" ParentId="13021" CreationDate="2016-12-30T17:37:02.633" Score="1" Body="&lt;p&gt;&quot;Imperial Eastman&quot; is the name of the now-defunct company that originally manufactured the tool. It exists today as &lt;a href=&quot;http://imperial-tools.com/about-us&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Imperial Tools brand of Stride Tool&lt;/a&gt;. The part you mention is probably just a standard 3/8&quot; fitting or hose but if you're concerned about it, you could take a look at their &lt;a href=&quot;http://imperial-tools.com/support/replacement-parts&quot; rel=&quot;nofollow noreferrer&quot;&gt;replacement parts guides&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get the parts elsewhere and you're worried about compatibility, you should be able to get equivalent parts from any large supplier. If you can find the actual part number on the fitting (not just the nominal dimension), you may be able to find the equivalent part in a document like &lt;a href=&quot;https://www.parker.com/parkerimages/Parker.com/ParkerStore2014/Global%20Assets/PDFs/Product%20Info/4400_B10_Imperial%20Eastman%20to%20Parker.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Equivalent Products List for Parker&lt;/a&gt; and not even have to get on the phone.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2016-12-30T17:54:38.723" LastActivityDate="2016-12-30T17:54:38.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13031" PostTypeId="2" ParentId="13029" CreationDate="2016-12-30T18:11:55.250" Score="3" Body="&lt;p&gt;Given the subject matter my gut tells me to take the time to compute once you've triple checked your conditions. What material properties are you using? You can alter the mesh to decrease cpu usage as well. Just food for thought, but that simulation reminds me of gelatin. If you can find an article that has investigated the hyperelastic material properties of the brain you may find that there is less displacement and therefore much less vibrational analytics. Then again It could be dead on. What are you using for boundary conditions? Crack propagation ?&lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-12-30T18:11:55.250" CommentCount="14" ContentLicense="CC BY-SA 3.0" />
  <row Id="13032" PostTypeId="2" ParentId="4244" CreationDate="2016-12-30T18:26:57.243" Score="2" Body="&lt;p&gt;Remember than large structures have to move or they will fracture. Think of the wear equation and the coefficient of friction in the two options. PTFE is an amazing polymer, goretex. Even if it is a dead load, there will be significant shear due to bending and environmental conditions (so rotational too). If it's welded the stress observed will be tensile and compressive with stress concentrations along the beads, less evenly distributed and more prone to fracture due to fatigue. &lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-12-30T18:26:57.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13033" PostTypeId="2" ParentId="13022" CreationDate="2016-12-30T20:04:33.533" Score="2" Body="&lt;p&gt;Considering that the helicopter's forward direction is the y-axis, the rear rotor should deliver torque, but the thrust created by the main rotor is connected to a swashplate so the thrust is not directly orthogonal. The main rotor can tilt into the y-axis in order to accelerate forward and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just my own speculation. Check out this page:&lt;/p&gt;&#xA;" OwnerUserId="9408" LastActivityDate="2016-12-30T20:04:33.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13034" PostTypeId="2" ParentId="11013" CreationDate="2016-12-30T20:17:41.733" Score="0" Body="&lt;p&gt;@Dopeybob435 have gave the answer that I need.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fans placed in series on a closed system will increase your pressure of the fan curve with little increase in your volume&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand the static pressure and airflow. I needed pressure so I bought a single radial fan that have 300ps pressure capacity, regarding the noice factor (50db). I still don't know the resistance of granular carbon that amount I am using. But it works just fine, for now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all.&lt;/p&gt;&#xA;" OwnerUserId="7567" LastActivityDate="2016-12-30T20:17:41.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13035" PostTypeId="2" ParentId="3887" CreationDate="2016-12-30T21:33:45.433" Score="1" Body="&lt;p&gt;I realize this is a bit late, but in option one you reference Datum G 4X individually. This is not allowed per Y14.5, you either recognize Datum G as the full pattern or you need 4 separate datums. In the second option you've just referenced a pattern of 4 diameters as two Datums at the same time. This also is not allowed per Y14.5&lt;/p&gt;&#xA;" OwnerUserId="9410" LastActivityDate="2016-12-30T21:33:45.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13036" PostTypeId="2" ParentId="13001" CreationDate="2016-12-31T00:21:51.667" Score="2" Body="&lt;p&gt;If for some reason the machinable waxes mentioned in previous answer don't work, you might also consider use of &lt;a href=&quot;https://en.wikipedia.org/wiki/Wood%27s_metal&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Wood's metal&lt;/strong&gt;&lt;/a&gt; as a filler while machining.  It's a low-melting point metal alloy.&#xA;As noted in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wood%27s_metal&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article about it, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wood's metal is commonly used as a filler when bending thin-walled metal tubes. For this use the tubing is filled with molten Wood's metal. After this filler solidifies the tubing is bent. The filler prevents the tube collapsing. The Wood's metal is then removed by heating, often by immersion in boiling water.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would be less likely (than wax) to shift or squeeze out during machining, but has some important drawbacks that make it not a panacea:&lt;br&gt;&#xA;• 10% cadmium content. &lt;a href=&quot;https://en.wikipedia.org/wiki/Cadmium&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cadmium&lt;/a&gt; is listed in the European Restriction of Hazardous Substances &lt;a href=&quot;https://en.wiktionary.org/wiki/bull#Noun_2&quot; rel=&quot;nofollow noreferrer&quot;&gt;bull&lt;/a&gt;.&lt;br&gt;&#xA;• Expensive.  For example, about $50 a pound at &lt;a href=&quot;http://www.flexbar.com/shop/pc/FIXTURING-ALLOY-1LB-CAKE-p4035.htm?gclid=CIqjiZGSndECFcGGaQodVscC6w&quot; rel=&quot;nofollow noreferrer&quot;&gt;flexbar.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wood%27s_metal&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article lists several variants that may cost less, or be less hazardous, mostly with higher melting points, but some, like Cerrolow, with lower MPs.&lt;/p&gt;&#xA;" OwnerUserId="9412" LastActivityDate="2016-12-31T00:21:51.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13037" PostTypeId="2" ParentId="12277" CreationDate="2016-12-31T09:18:54.697" Score="1" Body="&lt;p&gt;The main principle is the conversion of dynamic pressure to static pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pathway of the fluid is such that initially velocity decreases due to the mass flow rate equation (I'm not writing equations as you mentioned you knew them). The important idea here has to do with the energy lost in the ducts as it converts kinetic energy into internal energy and therein lies the pressure increase. Think of a stressor matrix in 3D with the sum of the principle components, which we know to be equal to hydrostatic pressure &lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="90" LastEditDate="2016-12-31T12:34:09.970" LastActivityDate="2016-12-31T12:34:09.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13038" PostTypeId="2" ParentId="3570" CreationDate="2016-12-31T09:52:12.113" Score="-1" Body="&lt;p&gt;Apologies that you didn't find the article informative, I figured the simplified equations for losses without stiffness matrices and polynomial derivations would be helpful as well as thinking of the structure as a string. I'm sure you're aware of metal beams being simplified to $f=Kx$ where $EA/L$ is the spring constant, however to incorporate a realistic model you have to take creep, elastic deformation, etc I'm sure you know all that. I suppose I'm confused as slippage occurs at the end connections. Simply $PL/EA = \Delta$ or slip of anchorage (P is prestress etc). In your case you need nodal functions. You are looking at a simple beam using two force diagrams, am I missing something? Use $f=\int B^T \tau \text{d}V$ to find the nodal forces which have to satisfy the basic conditions for elemental analysis. First you need the stress, which I'm sure you know. Your area calc is confusing as your $K$ matrix should be an integration of the derivatives of the area function at the very least i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_L E A(x)dx/L^2\begin{bmatrix} 1 &amp;amp; -1\\ -1 &amp;amp; 1\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a simple beam, then find $d^T f= F1x + F2x...$ etc. Remember $\text{strain} = Bd$, and for shape functions you'll most likely have to use at least polynomial expressions, but the derivative matrices make it easier. Also, if you want to deal with creep have fun with Maxwell's equations! Simple answer, simplify and ask any specifics you run into.&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="1832" LastEditDate="2017-01-02T10:24:58.460" LastActivityDate="2017-01-02T10:24:58.460" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13039" PostTypeId="2" ParentId="12958" CreationDate="2016-12-31T10:07:06.627" Score="2" Body="&lt;p&gt;You choose your coordinate system with a FBD, however the different pin supports have to counteract the exerted force to be static. Typically bending moments are defined using the rights hand rule. In example 1, M is negative by convention only because of the orientation typically used. Notice the author has x as the y axis. You can also ignore all that as the anti clockwise moment is opposing the bending moment caused by P (negative, aka clockwise). It's a cut section. &lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-12-31T10:07:06.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13040" PostTypeId="1" CreationDate="2016-12-31T11:45:54.147" Score="2" ViewCount="90" Body="&lt;p&gt;I am trying to find if making energy from human waste (excrement, sewage) is profitable? Lets say I live in a city of 16 million people which does not put waste into use. What could be the profitability?&lt;/p&gt;&#xA;" OwnerDisplayName="user5270" LastEditorUserId="90" LastEditDate="2017-01-01T01:35:41.133" LastActivityDate="2017-01-01T09:03:48.790" Title="Can we make significant energy from human waste?" Tags="|energy|" AnswerCount="2" CommentCount="2" ClosedDate="2017-01-19T18:08:58.860" ContentLicense="CC BY-SA 3.0" />
  <row Id="13041" PostTypeId="2" ParentId="13013" CreationDate="2016-12-31T12:02:27.810" Score="7" Body="&lt;p&gt;Observability means that you can estimate the complete state using only the output, without knowing the initial state. In other words, you have to figure out where you are without knowing where you were initially. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more practical reason why this rarely works is that when you are limited by non-perfect sensors and non-zero sampling time, taking the integral of the acceleration will cause growing errors in your estimates of positions and velocity. Thus, even if you know the initial state you will &quot;lose track&quot; of it over time. &lt;/p&gt;&#xA;" OwnerUserId="9418" LastActivityDate="2016-12-31T12:02:27.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13042" PostTypeId="2" ParentId="10889" CreationDate="2016-12-31T14:09:52.390" Score="2" Body="&lt;p&gt;I asked a few professors about similar hyperelastic deformation, and basically empirical testing is an imperative. If you take into account hysteresis (assuming testing of the product), strain history, etc, modeling in ANSYS or abaqus fe-safe is a personally advised necessity if you're creating a product for market. Localized displacement and strain energy density are also often dismissed as metals have a predictable stress strain relation. What I'm getting at is I can't really give you my ideas as that'd be foolish, but there are ways to obtain system values that you'd need for your feedback loop to be effective. Or go simple and use localized pressure and hope it works out! &lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2016-12-31T14:09:52.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13043" PostTypeId="2" ParentId="13040" CreationDate="2016-12-31T14:20:47.917" Score="2" Body="&lt;p&gt;Sounds like a thesis. The chemical reactions for decomposition of detritus and waste give energy to certain eukaryotes, so it's possible but out of the scope of a forum to determine financial factors without a process or business plan. That being said, carbon nanotubes (3D) can catalyze reactions unlike anything we've seen. Harness that and give me a call. &lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2017-01-01T09:03:48.790" LastActivityDate="2017-01-01T09:03:48.790" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13044" PostTypeId="2" ParentId="13022" CreationDate="2016-12-31T14:33:56.043" Score="1" Body="&lt;p&gt;A conventional helicopter has three main flight controls. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Torque pedals : these function is a similar way to the rudder on a fixed wing aircraft. They work by adjusting the pitch of the blades of the tail rotor to produce more or less thrust, allowing the pilot to balance the torque from the rotor blades and also control the yaw of the aircraft. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The cyclic lever : this controls the pitch of individual rotor blades at a specific point in their rotation, controlling pitch and yaw. This is analogous to tilting the rotor disk at about its centre. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Collective pitch :  this controls the pitch of all of the rotor blades simultaneously controlling overall lift. As the pitch is changed the throttle must also be used to adjust engine power to maintain a more or less constant rotational speed as the load on the blades changes. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is worth reiterating that it is the pitch of the blades which provides the direct control over thrust and lift and engine power is adjusted to keep their rotational speed withing an optimum range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a conventional helicopter the tail rotor and main rotor are directly coupled together via the gearbox and only their blade pitch rather than speed can be adjusted independently ie the tail rotor speed is proportional to the main rotor speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are designs which deviate from the conventional layout though for example there are tail designs which use ducted airflow rather then an external rotor and also various configurations of twin rotor which dispense with tail thrust altogether eg the Chinook which has fore and aft rotors and various &#xA;Russian military helicopters which have coaxial rotors. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-31T14:33:56.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13046" PostTypeId="2" ParentId="13040" CreationDate="2016-12-31T16:31:04.547" Score="2" Body="&lt;p&gt;There are certainly viable ways to generate energy from all sorts of biological waste. Typically involving bio-reactors which use microorganisms to produce eg. ethanol or methane. some types of waste can also be burned directly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Profitability is another matter and depends on a whole host of factors. On one hand there is the cost of building the initial infrastructure, it certainly seems likely that most urban sewerage systems would need a certain amount of modification and there are potential issues with separating non-biodegradable materials and industrial waste which can find their way into sewage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial outlay and running costs will depend on the scale and design of the plant as well as how centralised it is in respect to the overall layout of the city and existing infrastructure as well as the availability of suitable sites. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a reasonable possibility that any sort of novel human waste processing facility may meet with suspicion and resistance from residents in the planing stages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Economic and regulatory factors such as taxation, land prices and long term energy costs in terms of estimating financial viability. Depending on jurisdiction this sort of project may also have complex licensing requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed any sort of large scale energy project is financially risky as the capital costs tend to be large and it usually takes a significant amount of time to bring a return on investment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also a range of different technologies and infrastructure models to consider from fully centralised collection and processing to distributed systems which have micro-generators in individual households, buildings or city blocks all of which have their own pros and cons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly a distributed system is more of a retail business model like central heating or roof mounted solar panels where you are selling the plant while a centralised system is closer to the traditional utilities model where you are selling the energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While all of the above considerations are clearly highly relevant to the engineering side something like this goes far beyond being purely an engineering or technological problem. Indeed one of the difficulties with this field is that the engineering, economic, political and social aspects are so closely intertwined and it is hard to come up with a practical engineering solution without knowing what will work economically and vice versa. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-31T16:31:04.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13047" PostTypeId="1" CreationDate="2016-12-31T17:40:16.617" Score="4" ViewCount="625" Body="&lt;p&gt;I am planning to move from conventional grinding wheels to super-abrasive grinding wheels. I am unable to decide the right type of super-abrasive wheel for my manufacturing operations (I am into knife manufacturing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I select a the correct grinding wheel type for precision grinding of tool grade steels such as those used in knife manufacturing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically I would like to know what the differences are between resin bonded and vitrified CBN grinding wheel. I would also like to know which wheel can be used for precise and fine grinding applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankyou. &lt;/p&gt;&#xA;" OwnerUserId="9420" LastEditorUserId="7430" LastEditDate="2017-02-12T02:53:22.277" LastActivityDate="2017-09-28T12:13:22.557" Title="How to select the correct grinding wheel type for precision grinding of tool steels?" Tags="|mechanical-engineering|materials|manufacturing-engineering|grinding|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13049" PostTypeId="2" ParentId="13047" CreationDate="2016-12-31T20:21:34.503" Score="0" Body="&lt;p&gt;I have some experience of knife making. I would say that you might also look at abrasive belts rather than wheels these give rather more flexibility in terms of both range of abrasives for different operations like grinding vs sharpening of hardened steel and also allow a greater range of grind profiles. eg with a wheel you are stuck with hollow grinds whereas a belt grinder allows for both hollow grinds of varying radius and flat grinds of any arbitrary angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;certainly most professional knife makers I know of prefer belt grinders. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2016-12-31T20:21:34.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13050" PostTypeId="2" ParentId="13029" CreationDate="2016-12-31T22:16:14.900" Score="0" Body="&lt;p&gt;I don't know anything about the material properties of brains, but if you are applying some loads/displacements at 0.1mm/s, I don't understand why you are doing dynamics at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect the inertia forces caused by slowly deforming the material will be completely negligible compared with the elastic forces. A nonlinear stress analysis (probably with large displacement capability switched on) should work fine. Try it, and compare the results with the dynamics model you already ran.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2016-12-31T22:16:14.900" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13052" PostTypeId="2" ParentId="12624" CreationDate="2017-01-01T11:58:37.567" Score="-1" Body="&lt;p&gt;The reason is because of errors. &#xA;The first one is that the specimen might not be completely homogeneous(there might more cement on one side than the other for example). The second reason is that, the loading might not be in the center of the  specimen, this means that one side will have to carry more loading than the other side.&lt;/p&gt;&#xA;" OwnerUserId="9053" LastActivityDate="2017-01-01T11:58:37.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13053" PostTypeId="1" CreationDate="2017-01-01T12:05:26.660" Score="2" ViewCount="65" Body="&lt;p&gt;I have closed loop - discrete model. This is how it looks in matlab:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/su4D6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/su4D6.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I need to compute analitically output &lt;span class=&quot;math-container&quot;&gt;$y(t)$&lt;/span&gt; (same as in the Matlab's scope) having &lt;span class=&quot;math-container&quot;&gt;$Y(z)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Y(z)=U(z)*G(z)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Input is sine wave so Z transform for sine is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$U(z) = \frac{z\sin(\omega T)}{z^2-2zcos(\omega T)+1}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$G(z) = \frac{G_o(z)}{1+G_o(z)} = \frac{kz^3}{(a+k)z^3+bz^2+cz+d} = \frac{kz^3}{(a+k)(z-z_1)(z-z_2)(z-z_3)}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; - are given parameters.&lt;/p&gt;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Y(z) = \frac{z\sin(\omega T)kz^3}{(z^2-2zcos(\omega T)+1)(a+k)(z-z_2)(z-z_3)(z-z_4)}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I saw on Wolfram Alpha that roots of &lt;span class=&quot;math-container&quot;&gt;$z^2-2zcos(\omega T)+1$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;are&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$z_0 = cos(\omega T)-isin(\omega T)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$z_1 = cos(\omega T)+isin(\omega T)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;and rest of roots (&lt;span class=&quot;math-container&quot;&gt;$z_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$z_3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$z_4$&lt;/span&gt;) I compute using &lt;code&gt;roots&lt;/code&gt; function in matlab from &lt;span class=&quot;math-container&quot;&gt;$G(z)$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$z_2 = 0.9889+0.022i$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$z_3 = 0.9889-0.022i$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$z_4 = 0.8784$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The residue method is what I know but other would be good too.&lt;/p&gt;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$res_{z=z_0}=\lim_{z\to z_0}(z-z_0)Y(z)z^{n-1}=\lim_{z\to z_0}\frac{z\sin(\omega T)kz^3z^{n-1}}{(a+k)(z-z_1)(z-z_2)(z-z_3)(z-z_4)}=lim_{z\to z_0}\frac{\sin(\omega T)kz^{n+3}}{(a+k)(z-cos(\omega T)-isin(\omega T))(z-z_2)(z-z_3)(z-z_4)}=\frac{\sin(\omega T)k[cos(\omega T)-isin(\omega T)]^{n+3}}{(a+k)(-2isin(\omega T))(cos(\omega T)-isin(\omega T)-z_2)(cos(\omega T)-isin(\omega T)-z_3)(cos(\omega T)-isin(\omega T)-z_4)}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$y(t)$&lt;/span&gt; as you know is sum of residues for all poles.&lt;/p&gt;&#xA;&lt;p&gt;In order to plot graph from my function &lt;span class=&quot;math-container&quot;&gt;$y(t)$&lt;/span&gt; I have to get rid of imaginary parts of equation. How can I do this?&lt;/p&gt;&#xA;&lt;p&gt;Yeah it's boring because it's a homework.&lt;/p&gt;&#xA;" OwnerUserId="9425" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-01-01T16:07:54.993" Title="calculate output $y(t)$ from $Y(z)$ having discrete closed loop model with sine input - residue method" Tags="|control-engineering|mathematics|transfer-function|matlab|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13054" PostTypeId="2" ParentId="13025" CreationDate="2017-01-01T14:40:47.193" Score="3" Body="&lt;p&gt;Computing a Voronoi diagram on a surface is not an easy problem, and there is unlikely to be a solution within AutoCAD.&#xA;You might start with this paper, and work backward &amp;amp; forward through its references using Google Scholar:&#xA;&lt;hr /&gt;&#xA;&amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &amp;nbsp; &#xA;&lt;a href=&quot;https://i.stack.imgur.com/GN9Oy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GN9Oy.jpg&quot; alt=&quot;CatVD&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr /&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Yan, Dong‐Ming, Bruno Lévy, Yang Liu, Feng Sun, and Wenping Wang. &quot;Isotropic remeshing with fast and exact computation of restricted Voronoi diagram.&quot; In &lt;em&gt;Computer Graphics Forum&lt;/em&gt;, vol. 28, no. 5, pp. 1445-1454. Blackwell Publishing Ltd, 2009. (&lt;a href=&quot;http://i.cs.hku.hk/~gfxgroup/projects/remesh/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Authors' link&lt;/a&gt;.)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9427" LastEditorUserId="9427" LastEditDate="2017-01-01T17:29:21.387" LastActivityDate="2017-01-01T17:29:21.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13055" PostTypeId="2" ParentId="2253" CreationDate="2017-01-01T21:41:02.883" Score="1" Body="&lt;p&gt;Rather than trying to have sensors and receivers aligned perfectly, most light curtains pulse different lights and different times and then only look for signals from receivers at the moments when the corresponding lights are on.  This greatly eases the need for perfect alignment, but ensures that any object which gets between a sensor and its corresponding receiver can be detected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The practical limits to a light curtain's &quot;beam&quot; spacing stem from the need to have a certain amount of independent control and detection circuitry for each &quot;beam&quot;; having a beam every 2cm would cost about half as much as having a beam every 1cm, but about twice as much as having a beam every 4cm.  It would not be difficult to design a unit which had a beam every 5mm, but for most purposes the benefits of such tight spacing would be insufficient to justify the cost&lt;/p&gt;&#xA;" OwnerUserId="397" LastActivityDate="2017-01-01T21:41:02.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13056" PostTypeId="1" CreationDate="2017-01-02T05:43:23.693" Score="3" ViewCount="523" Body="&lt;p&gt;The soil in my area has a very thick clay content (apparently left behind by glaciers, damn ice age). Because of this drainage in my yard is horrible. I've put in some underground drainage (french drain) that empties into a ditch. But unfortunately the french drain is below the drain pipe, which mean the french drain system when really flooded just sits there full of water instead of draining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really want to stay away from sump pumps, or anything powered. I looked at creating a bell siphon, but that doesn't seem like it could pump above itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I somehow use a holding tank above ground to provide the necessary &quot;push&quot;of the water. I was thinking something similar to how a toilet works, that I would just have to &quot;flush&quot; when I wanted it to drain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fH2bQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fH2bQ.png&quot; alt=&quot;current drainage diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9432" LastActivityDate="2017-01-03T19:18:13.380" Title="Automatic siphon above max height" Tags="|pressure|valves|siphon|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13057" PostTypeId="1" CreationDate="2017-01-02T13:33:56.573" Score="0" ViewCount="588" Body="&lt;p&gt;In the figure , we can notice that the moment due to Pe is counterclokwise , but moment due to shear flow is also counterclockwise , How can the Pe prevent the twisting of the beam ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is the direction of the shear flow wrong ? Should it flow from the below to the top ?&lt;a href=&quot;https://i.stack.imgur.com/mKxH5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mKxH5.png&quot; alt=&quot;1&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/FSbwK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FSbwK.png&quot; alt=&quot;22&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2017-01-03T00:22:51.687" Title="direction of shear flow to cancel out the moment" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13058" PostTypeId="2" ParentId="13057" CreationDate="2017-01-02T17:45:42.013" Score="2" Body="&lt;p&gt;Any time you cut a beam to look at the shear flow distribution, the equilibrium condition will give you the correct orientation of the distribution. Additionally, the shear flow distribution is always equal and opposite when looking at the two faces of the cut. From Fig. 7-23a the loading from $P$ is causing a &lt;em&gt;clockwise&lt;/em&gt; moment and rotation of the beam. Which is indicated by the equilibrium shear flow in the non-cut image of Fig. 7-23a. When the beam is cut in Fig. 7.23-b, we are looking at the shear flow in the cross section in an equal and opposite manner as indicated by the equilibrium condition. &lt;/p&gt;&#xA;" OwnerUserId="9399" LastEditorUserId="9399" LastEditDate="2017-01-03T00:22:51.687" LastActivityDate="2017-01-03T00:22:51.687" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="13062" PostTypeId="2" ParentId="13023" CreationDate="2017-01-02T20:27:45.303" Score="0" Body="&lt;p&gt;There are complete specifications that you could find for your area and read. Department of Transportation specifications are usually available for free.  All of these specifications will have some sort of the following (which is what you can reasonably follow yourself):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Make sure that there is no snow or frozen material where the concrete will touch, e.g. reinforcing steel or soil.&lt;/li&gt;&#xA;&lt;li&gt;Make sure that concrete temperature stays above freezing. (This means concrete blankets and or plastic to keep wind away and heat in.)&lt;/li&gt;&#xA;&lt;li&gt;Sometimes you will see requirements for the maximum temperature difference between the outside of the concrete and the inside. All this mean is that you don't want to uncover the concrete before it reaches air temperature on its own (a couple of days usually).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Placing concrete is completely doable even when it is really cold, you just need to make sure that the air around the concrete stays above freezing.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2017-01-02T20:27:45.303" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13063" PostTypeId="1" AcceptedAnswerId="13074" CreationDate="2017-01-02T21:10:57.560" Score="4" ViewCount="51" Body="&lt;p&gt;I am working with silicone injection molding as a hobby and I am wanting to make a ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried with 100% silicone and the results are a clear ring that is not quite the consistency I want. I would like to experiment with making it more or less elastic if possible, as well as different swirled colors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I change the elasticity of a silicone ring and how can I introduce swirling colors into it as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched around for other existing things, and this silicone ring is close to what I want to make.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://groovelife.co/products/silicone_ring_breathable_groove?variant=13528608580&amp;amp;gclid=CjwKEAiAkajDBRCRq8Czmdj-yFgSJADikZggxMT5d91MuDELqOKU6eFGvRvJWkAbHWH1_Gxu9of58RoC9oDw_wcB&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://groovelife.co/products/silicone_ring_breathable_groove?variant=13528608580&amp;amp;gclid=CjwKEAiAkajDBRCRq8Czmdj-yFgSJADikZggxMT5d91MuDELqOKU6eFGvRvJWkAbHWH1_Gxu9of58RoC9oDw_wcB&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9436" LastEditorUserId="90" LastEditDate="2017-01-03T12:09:11.053" LastActivityDate="2017-01-03T12:23:43.943" Title="silicone injection molding - making a flexible ring" Tags="|structural-engineering|chemical-engineering|industrial-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13065" PostTypeId="2" ParentId="12378" CreationDate="2017-01-02T23:10:58.113" Score="0" Body="&lt;p&gt;Is a analog QR code reader necessary?  Couldn't you use a camera to take a photo, and then use something like OpenCV to write a program to crop the QR code from the photo, and then parse it using the extracted image?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://aishack.in/tutorials/sudoku-grabber-opencv-detection/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SuDoKu Grabber in OpenCV&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at this tutorial to see what I mean.&lt;/p&gt;&#xA;" OwnerUserId="9437" LastEditorUserId="7083" LastEditDate="2017-05-09T06:11:21.043" LastActivityDate="2017-05-09T06:11:21.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13066" PostTypeId="1" AcceptedAnswerId="13070" CreationDate="2017-01-02T23:16:05.640" Score="6" ViewCount="9863" Body="&lt;p&gt;I know that according to the thin airfoil theory all airfoils (symmetric or not) have lift curve slope of $2\pi$. I also know that finite wings should have less inclined lift curve because of the downwash. Then I saw this formula used for finite elliptic wings which is fine.&#xA;$$&#xA;\frac{dC_L}{d\alpha}=a=\frac{a_0}{1+a_0/\pi AR}&#xA;$$&#xA;The problem is $a_0$ is not $2\pi$ in examples that above formula is used. I am thinking that since non of the downwash effects are present for airfoils (infinite wings) $a_0$ should always be $2\pi$. What am I missing? Is it because not all airfoils are thin?&lt;/p&gt;&#xA;" OwnerUserId="9438" LastEditorUserId="90" LastEditDate="2017-01-03T12:16:02.537" LastActivityDate="2017-01-03T12:16:02.537" Title="Do all airfoils have a lift curve slope of $2\pi$?" Tags="|fluid-mechanics|aerospace-engineering|aircraft-design|aerodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13068" PostTypeId="1" CreationDate="2017-01-03T01:51:57.263" Score="0" ViewCount="50" Body="&lt;p&gt;If I use the forging process of clay tempering, can I take a thinner piece of steel and and make it as strong as a very thick piece of steel? Getting the same strength, without the extra weight? &lt;/p&gt;&#xA;" OwnerUserId="9441" LastEditorUserId="33" LastEditDate="2017-01-03T13:58:11.553" LastActivityDate="2017-01-03T13:58:11.553" Title="If I use clay tempering, can I take a thin piece of steel and make it hard as a very thick piece?" Tags="|heat-treatment|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2017-01-03T23:20:34.700" ContentLicense="CC BY-SA 3.0" />
  <row Id="13070" PostTypeId="2" ParentId="13066" CreationDate="2017-01-03T02:43:28.657" Score="6" Body="&lt;p&gt;In general, most airfoils only approximately display the $2\pi$ lift slope as predicted by &lt;em&gt;thin&lt;/em&gt; airfoil theory. That is because airfoils are not actually infinitely &lt;em&gt;thin&lt;/em&gt; in practice, and will deviate from thin airfoil theory by a small amount. $a_0$ is left in the above equation so one can predict the elliptical finite wing lift slope (effects of aspect ratio), $a$, given the best predicted/measured infinite wing lift slope $a_0$. The thin airfoil theory will predict a finite wing lift slope of $2\pi$ rad$^{-1}$ or 0.1097 deg$^{-1}$. It is common to measure airfoil lift slopes near the 0.1097 deg$^{-1}$ ($2\pi$) for the NACA 4 and 5 series airfoils with thickness ratios below 10%. However, once you go above this thickness, the measured lift slope begins approaching 0.1 deg$^{-1}$ or 5.73 rad$^{-1}$ for smooth airfoils with thickness ratios near the upper limit of 24%. Similarly, if roughness effects are taken into consideration, the measured lift slope is even lower and reaches 0.1 deg$^{-1}$ at only a thickness ratio of 14% and can reach as low as 0.085 deg$^{-1}$ or 4.87 rad$^{-1}$ near thickness ratios of 24%. This is all detailed in the plot below from the Theory of Wing Sections (excellent reference text).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JSnFK.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JSnFK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oddly enough, for the smooth NACA 63-series and 64-series, the lift slope will actually increase with increasing thickness ratios. However, the roughness effects keeps the measured lift slope around 0.11 for thickness ratios up to 24%. This is detailed in the following 2 plots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qScNg.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qScNg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the $a_0$ is left as a variable for the designer to be predicted or measured using vortex panel methods or experimental wind tunnel testing, to which the designer can use the above equation to estimate the effects of aspect ratio on the elliptical finite wing lift slope, $a$, given the predicted/measured $a_0$.&lt;/p&gt;&#xA;" OwnerUserId="9399" LastActivityDate="2017-01-03T02:43:28.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13071" PostTypeId="1" CreationDate="2017-01-03T07:21:53.360" Score="-2" ViewCount="558" Body="&lt;p&gt;why the mean area enclosed within the boundary of the centerline of tube thickness is (0.035)(0.057) ? Is it wrong ? Shouldnt it be (0.040-0.005-0.005) x ( 0.060 -0.003 -0.0030 = 0.054 ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question , why for the ds it's 2(57) and 2(35) respctively ? what is ds actually ? &#xA;&lt;a href=&quot;https://i.stack.imgur.com/R7yNh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R7yNh.jpg&quot; alt=&quot;500&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GwHpr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GwHpr.jpg&quot; alt=&quot;498&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5JMJf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5JMJf.jpg&quot; alt=&quot;499&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2017-01-03T10:27:28.750" Title="angle of twist / stress of thin wall having closed section" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13072" PostTypeId="2" ParentId="13071" CreationDate="2017-01-03T10:27:28.750" Score="2" Body="&lt;p&gt;The first question is trivial. The section's &lt;strong&gt;&lt;em&gt;centerline&lt;/em&gt;&lt;/strong&gt; is the line along each face's center. The right and left faces are 5&amp;nbsp;mm thick, so their centerlines are 2.5&amp;nbsp;mm from the exterior, and $40 - 2\times2.5 = 35\text{ mm}$. Likewise, the top and bottom faces are 3&amp;nbsp;mm thick, so their centerlines are 1.5&amp;nbsp;mm from the exterior, and $60 - 2\times1.5 = 57\text{ mm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second one, you need to observe that this is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Line_integral&quot; rel=&quot;nofollow noreferrer&quot;&gt;line integral&lt;/a&gt;, as demonstrated by the $\oint$ symbol. This is a simple one, so you just have to sum the ratio $s/\text{d}t$ for each side of the centerline, where $s$ and $t$ are the length of each centerline segment and the associated thickness, respectively. So that gives you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\oint \dfrac{\text{d}s}{t} = \dfrac{35}{3} + \dfrac{57}{5} + \dfrac{35}{3} + \dfrac{57}{5} = 2\left(\dfrac{57}{5} + \dfrac{35}{3}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-01-03T10:27:28.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13073" PostTypeId="2" ParentId="13086" CreationDate="2017-01-03T12:06:31.050" Score="2" Body="&lt;p&gt;It depends what you want to achieve and what you mean by 'strength'. . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat treating can be used to increase the yield stress of a hardenable steel (to a greater or lesser extent depending on the alloy and process used). But what it won't do is to increase the stiffness of the part as hardening and tempering have no significant effect on the Young's Modulus of steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So heat treating will increase the stress at which a part will fail but it won't change how it deflects under load. In the case of a simple beam under bending load it will bend further before it fails but will still bend the same amount under the same load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to increase the stiffness and reduce weight you need to change the section for example a hollow tube will be much stiffer in bending than a solid one of the same mass per unit length and the same goes for 'I' sections. In other loading conditions webs, ribs and swages can be used to increase stiffness significantly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of material properties there is usually a trade-off between hardness and toughness (ie the ability to tolerate impact loading) and in some cases it is desirable to sacrifice a bit of hardness and ultimate tensile strength for the sake of ductility in order to prevent brittle failures which tend to be sudden and catastrophic. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-11-28T19:31:35.673" LastActivityDate="2017-11-28T19:31:35.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13074" PostTypeId="2" ParentId="13063" CreationDate="2017-01-03T12:23:43.943" Score="3" Body="&lt;p&gt;If you are using RTV silicone of the type used for prosthetics and animatronics then you can get a variety of grades with different mechanical properties, commonly rated by Shore Hardness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also get additives known as 'deadeners' which create a more matte surface and reduce the slippery feel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create swirl colours you want to make up 2 or more batches of silicone of different colours mix in the catalyst , (pigments are available to add to clear silicone) and either mix them lightly once both area catalysed or inject  them into the mould at the same time. Different ratios and mixing methods will produce different results. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-01-03T12:23:43.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13077" PostTypeId="2" ParentId="13056" CreationDate="2017-01-03T19:18:13.380" Score="2" Body="&lt;p&gt;I think you can create a self-starting siphon if you are able to add pipe to both ends of the &quot;drain pipe.&quot;  You'd extend the output end downwards until the exit of the pipe is below the bottom level of the ditch. The inlet end of the drain pipe is extended down to the bottom of the ditch (with possible filters, etc).&lt;br&gt;&#xA;If I got this right, the ditch will fill until the water reaches the level of the existing drain pipe. Then the water will flow thru the drain pipe and initiate the siphoning in a manner (as you suggested) similar to a flush toilet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawback is that if there's only a little runoff from the French drain and the level in the ditch doesn't reach the level of the existing output drain pipe, the siphon won't start.  Not much you can do about that, I fear. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-01-03T19:18:13.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13079" PostTypeId="2" ParentId="11043" CreationDate="2017-01-03T20:41:06.477" Score="1" Body="&lt;p&gt;I believe there are &lt;strong&gt;3&lt;/strong&gt; mainly options for this case:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;use microfluidic devices that are made to be reusable;&lt;/li&gt;&#xA;&lt;li&gt;use microfluidic devices that are not are not meant to be reusable but cheaper;&lt;/li&gt;&#xA;&lt;li&gt;devise a protocol of gentle cleaning.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The main options for devices made to be &lt;strong&gt;reusable&lt;/strong&gt; are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;silicon;&lt;/li&gt;&#xA;&lt;li&gt;glass;&lt;/li&gt;&#xA;&lt;li&gt;low temperature co-fired ceramics (LTCC).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The main options for &lt;strong&gt;disposable devices&lt;/strong&gt; are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;thermosets;&lt;/li&gt;&#xA;&lt;li&gt;hydrogel;&lt;/li&gt;&#xA;&lt;li&gt;thermoplastics;&lt;/li&gt;&#xA;&lt;li&gt;paper;&lt;/li&gt;&#xA;&lt;li&gt;elastomer (includes PDMS, which is the most commom elastoemr used in microfluidics).[1]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which one to choose depends on your &lt;strong&gt;constraints of use and price&lt;/strong&gt;. And price depends on if you can &lt;strong&gt;fabricate&lt;/strong&gt; your devices or if you intend to &lt;strong&gt;buy&lt;/strong&gt; it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For imaging, probably the glass devices, which you already use, are the best choice. If you don't want to buy them, it is possible to do them, but the lack of characterisation might affect your reproducibility (about fabrication in [2]). If you are in a multidisciplinar research center there should be a &lt;strong&gt;microfluidics laboratory that can help you to make it&lt;/strong&gt;, but it can take time and distract you from the focus of your research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, maybe the best option is to learn a way to clean it gently. I don`t know if it can be done with with your particular device, but you can try some ideas from this document from Sigma-Aldrich for cleaning glassware: &lt;a href=&quot;https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Aldrich/Bulletin/al_techbull_al228.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Aldrich/Bulletin/al_techbull_al228.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] Materials for Microfluidic Chip Fabrication Kangning Ren, Jianhua Zhou, and Hongkai Wu. Accounts of Chemical Research 2013 46 (11), 2396-2406 doi: 10.1021/ar300314s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] Iliescu C, Taylor H, Avram M, Miao J, Franssila S. A practical guide for the fabrication of microfluidic devices using glass and silicon. Biomicrofluidics. 2012;6(1):016505-016505-16. doi:10.1063/1.3689939.&lt;/p&gt;&#xA;" OwnerUserId="9450" LastActivityDate="2017-01-03T20:41:06.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13080" PostTypeId="2" ParentId="3868" CreationDate="2017-01-03T21:10:42.013" Score="2" Body="&lt;p&gt;I believe some laser equipment does have the 20µm that you are looking for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sick.com/us/en/product-portfolio/distance-sensors/displacement-measurement-sensors/c/g150251&quot; rel=&quot;nofollow noreferrer&quot;&gt;Displacement measurement sensors&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This product for example has a 20µm precision, but it is said that the precision depends on the optical properties of the surface, so it would require a test for your solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sick.com/us/en/product-portfolio/distance-sensors/displacement-measurement-sensors/od-mini/c/g276251&quot; rel=&quot;nofollow noreferrer&quot;&gt;OD Mini - Compact, lightweight sensor for precise measurement&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, another set of options is to &lt;strong&gt;measure properties of the liquid that are proportional to the volume of the liquid&lt;/strong&gt; (supposing that the only variable of your experiment is the height and that the liquid composition, temperature, pressure or any other variable doesn't change). In that case you could, for example, measure the &lt;strong&gt;liquid resistance&lt;/strong&gt; from one side of the cylinder to the other. However this would require a custom apparatus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option that falls into this category (of using properties proportional to volume) is to use &lt;strong&gt;absorbance&lt;/strong&gt;, putting a light sensor at the bottom at the flask and a light source at the top (or the inverse). In this case the precision is only limited by the precision of your apparatus. However a precision measurement might require a sophisticated apparatus (I don't know how the 50µm precision translates into complexity in this case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, as it was already mentioned, &lt;strong&gt;mass&lt;/strong&gt; is another option. Coupling a reservoir to the cylinder which can be measured by a scale. If your experiment support this modification, I believe it is the easiest option, since precision scales are available at most labs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope my answer provides some inspiration and please keep us updated on the results!&lt;/p&gt;&#xA;" OwnerUserId="9450" LastEditorUserId="1832" LastEditDate="2017-03-06T20:37:02.747" LastActivityDate="2017-03-06T20:37:02.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13081" PostTypeId="2" ParentId="6031" CreationDate="2017-01-04T03:36:08.513" Score="3" Body="&lt;p&gt;A lot of the responses here are very interesting but the inverse relation of magnetic fields and magnetic strength must be considered. if two attracting magnets are a certain distance apart (say 1 inch) their attraction is for example 2lb. if the two magnets are 1/2 inch apart their attractive force is 4lbs and a quarter inch 16 pounds. all simplified numbers. In a magnetic lock, the two units are touching and therefore very strong. any external magnetic force applied to the outside of a door would be the thickness of the door (plus a little) away from the electromagnet. to overcome a mid-range magnet of 800lbf you would need a much much larger electromagnet (exponential). this seems to be as much of a physics question as an engineering question. If you have an hour, this guy is very engaging and demystifies magnets and electromagnets in one shot. it is filmed 10 years ago at UC Berkley but is meant to be understood by the likes of me so that is saying a lot...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=WxitGR-9qGA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=WxitGR-9qGA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9455" LastActivityDate="2017-01-04T03:36:08.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13082" PostTypeId="1" CreationDate="2017-01-04T12:59:27.183" Score="3" ViewCount="1845" Body="&lt;p&gt;So, where I work, we have this arcade game, which is a classic claw machine.  IN order to raise the claw up and down the machine makes use of a worm gear, like so:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rrX41.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rrX41.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(although for the meticulously curious, this is actually the drive that moves the trolley the claw rests on back and fourth but it is almost identical to the part in question.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that the gear seems to be on little grooves here.  On one of our machines , the gear has actually come off of the grooves on the horizontal shaft.  This has been causing the gears to slip whenever the motor tries to rise the claw.  I assume this can happen because the vertical gear puts friction on the other gear that makes it tend to slide in the direction of rotation along the shaft, so my question is this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this gear on the horizontal shaft held in place by friction (if I slide it in place until it is hard to move from pressure between the shaft and diameters of the gears, am I good) or do I need to clean the shaft and apply some adhesive in order to expect it to stay?  It works okay now, but I didn't exactly test it extensively.  &lt;/p&gt;&#xA;" OwnerUserId="1611" LastEditorUserId="90" LastEditDate="2021-04-17T06:04:24.783" LastActivityDate="2021-04-17T08:05:29.887" Title="Is a gear usually held to a shaft by friction or an adhesive?" Tags="|mechanical-engineering|gears|friction|mechanical-failure|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="13085" PostTypeId="1" AcceptedAnswerId="13105" CreationDate="2017-01-04T22:32:35.320" Score="1" ViewCount="84" Body="&lt;p&gt;Any idea on how this type of pneumatic solenoid works? I know that NC is normally closed, NO normally open, but not sure how air flows? Thanks! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uCcu6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uCcu6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7273" LastActivityDate="2017-01-05T19:23:39.667" Title="pneumatic solenoid operation?" Tags="|airflow|compressed-air|pneumatic|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13086" PostTypeId="1" AcceptedAnswerId="13073" CreationDate="2017-01-04T22:53:24.153" Score="1" ViewCount="1526" Body="&lt;p&gt;During the heat treatment process, people use clay to harden the steel in certain areas like the spine. Then they have to go through another process to make the steel tough. I need very hard steel and want to know will the hard steel be brittle and able to withstand large amounts of pressure? I ask because the steel is heavily stressed from the forging process. Also, can I make a thinner piece of steel as hard as a thicker piece with this method?&lt;/p&gt;&#xA;" OwnerUserId="9468" LastEditorUserId="368" LastEditDate="2017-01-05T16:41:08.177" LastActivityDate="2017-11-28T19:31:35.673" Title="Is hardened steel better than tough steel?" Tags="|steel|heat-treatment|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ClosedDate="2017-01-05T16:42:43.357" ContentLicense="CC BY-SA 3.0" />
  <row Id="13087" PostTypeId="2" ParentId="13086" CreationDate="2017-01-04T23:16:00.847" Score="3" Body="&lt;p&gt;In steel hardness is closely related to tensile strength so a hardened steel will have both high hardness and high tensile strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Toughness is a bit more difficult to quantify than strength but in general is the ability to resist impact and resistance to brittle fractures. Ductility plays a part in toughness as it tends to help a material to resist crack propagation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simple steels heat treating is usually a balance between hardness and toughness. Cooling a hardenable steel from critical temperature (around 850 C) creates a very hard but brittle material, tempering (heating to a lower temperature, around 200-300 C) partially reverses the hardening process and improves toughness and ductility. Controlling the tempering temperature allows the relative hardness/toughness to be controlled to an extent.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact properties you get from a given heat treating process will depend on the alloy used and different alloys will have different heat treating requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to hardening residual manufacturing stresses and grain growth can be mitigated by &lt;em&gt;annealing&lt;/em&gt; and/or &lt;em&gt;normalising&lt;/em&gt; if necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that heat treating steel is a complex subject on which many large and complex books have been written and it's not easy to summarise in a short answer. It is also important that you read up on the precise engineering definitions of the terms involved eg &lt;em&gt;hardness, toughness, tensile strength, ductility&lt;/em&gt; etc as they have very specific meanings in this context and talking about 'strength' generally can cause confusion. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-01-05T13:35:12.963" LastActivityDate="2017-01-05T13:35:12.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13089" PostTypeId="1" AcceptedAnswerId="13091" CreationDate="2017-01-05T01:53:25.077" Score="4" ViewCount="2853" Body="&lt;p&gt;Being a mechanical engineering novice (I stick closer to solid state electronics), when I look inside of any toy-grade RC car, I see a series of different-sized gears between the motor shaft and the drive axle. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why does a single-speed RC car require a gearbox? &lt;/li&gt;&#xA;&lt;li&gt;Instead, why can't a gear on the motor shaft connect to a gear on the axle?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So far, two of the plausible explanations I have come up with are that&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The motor turns at a different RPM than desired by the axle, thus&#xA;conversion is required.&lt;/li&gt;&#xA;&lt;li&gt;To put protection between the spinning of the axle and the&#xA;spinning of the motor.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I ask because I am setting out to make my own Lego car driven by a DC brush motor and have no idea what to do for a single speed gearbox other than hook one gear to the motor shaft and one gear to the axle and let it go.&lt;/p&gt;&#xA;" OwnerUserId="9470" LastEditorUserId="8055" LastEditDate="2017-01-07T07:08:04.080" LastActivityDate="2017-01-07T07:08:04.080" Title="Why do single speed RC cars use a gearbox?" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13090" PostTypeId="1" AcceptedAnswerId="13098" CreationDate="2017-01-05T03:04:44.633" Score="2" ViewCount="3223" Body="&lt;p&gt;Is it due to the PET material that it is made of or is it more the shape and structure of the bottle? If it is due to the shape/structure of the container, is it possible to achieve such pressure resistance using another plastic?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2017-01-05T14:43:43.027" Title="Why can soda bottles withstand so much pressure?" Tags="|materials|pressure|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13091" PostTypeId="2" ParentId="13089" CreationDate="2017-01-05T10:52:24.277" Score="4" Body="&lt;p&gt;Your first guess covers it fairly well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small DC motors work best at high speeds, the gearbox is needed to reduce the higher motor speed down to something more reasonable for the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why use a series of gears rather than a single one? Because there are practical limitations on gear ratios, especially when you need to fit them into a small space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gears don't offer any protection to the motor unless they are designed to break before the motor does. Since little bits of broken gear flying around is not a nice thing to happen this isn't normally the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a lego car that you want to protect against the motor getting overloaded in the event that it drives into a wall and tries to keep going then adding a belt drive stage is probably the easiest thing to do. If the load gets too high the belt will slip. This costs you a little performance and efficiency but is a simple way to protect the motor. &lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2017-01-05T10:52:24.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13092" PostTypeId="1" AcceptedAnswerId="13096" CreationDate="2017-01-05T11:17:48.473" Score="3" ViewCount="9341" Body="&lt;p&gt;Which is better to use in contructing ceiling frame? RHS or SHS? And why?&lt;/p&gt;&#xA;" OwnerUserId="9474" LastActivityDate="2017-01-06T09:17:21.633" Title="Rectangular hollow section vs square hollow section" Tags="|structural-engineering|structures|" AnswerCount="2" CommentCount="1" ClosedDate="2017-01-08T13:11:09.177" ContentLicense="CC BY-SA 3.0" />
  <row Id="13093" PostTypeId="1" CreationDate="2017-01-05T11:36:50.943" Score="1" ViewCount="217" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f2orq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f2orq.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With usage, old laser rods eventually will need maintenance by cleaning and refurbishing it. Does anyone know how this is done?&lt;/p&gt;&#xA;" OwnerUserId="9475" LastEditorUserId="90" LastEditDate="2017-01-06T02:23:52.373" LastActivityDate="2017-03-07T03:11:29.843" Title="Does anybody have information on how to refurbish an old laser rod (alexandrite or Nd Yag)? )" Tags="|biomedical-engineering|maintenance|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13094" PostTypeId="2" ParentId="13092" CreationDate="2017-01-05T11:54:58.987" Score="6" Body="&lt;p&gt;Use an RHS over an SHS when you want significantly more bending/shear capacity about/in one axis than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additional depth that an RHS has in one axis gives it greater shear capacity in this axis, and greater moment capacity about the perpendicular axis.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2017-01-06T09:17:21.633" LastActivityDate="2017-01-06T09:17:21.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13095" PostTypeId="1" AcceptedAnswerId="13097" CreationDate="2017-01-05T12:10:19.563" Score="0" ViewCount="598" Body="&lt;p&gt;i'm a chemical eng. student from Italy and i'm &quot;sizing&quot; (&quot;setting&quot;?) a heat exchanger to cool some octane with toluene.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that pressure drops shell side in a 1,2 pull-through floating head heat exchanger are lower than 0.001 atmosphere? I used really wide buffles and buffle cuts opened at 45%. The heat exchange is in range and the pressure drops tube side are &quot;normal&quot; (0.2 atm). I just find it weird that shell side they are so low. Can someone help me?&lt;/p&gt;&#xA;" OwnerUserId="9476" LastActivityDate="2017-01-05T13:23:46.850" Title="Can pressure drops in a heat exchanger shell side be super low?" Tags="|heat-transfer|chemical-engineering|heat-exchanger|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13096" PostTypeId="2" ParentId="13092" CreationDate="2017-01-05T12:24:34.397" Score="5" Body="&lt;p&gt;Obviously this depends on the relative size of the sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the rectangular section's smallest dimension is equal to (or greater than) the square section's side and their thicknesses are equal, then the rectangular section will always be stronger than the square, for any load in any orientation. This is because the rectangular section's geometric properties (namely area and moment of inertia) will always be greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same reasons, if the rectangular section's largest dimension is equal to (or lesser than) the square section's side, then the square section will always be stronger than the rectangular section for any load in any orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, the rectangular section's smallest dimension is lesser than the square section's side but it's largest dimension is greater, then you need to calculate the section properties and see which is best:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;A &amp;amp;= BH - bh \\&#xA;I &amp;amp;= \dfrac{BH^3 - bh^3}{12}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $B$ and $H$ are the section's width and height, respectively, and $b$ and $h$ are the hole's (so $b = B - 2t$, where $t$ is the section's thickness). The moment of inertia also depends on the rectangular section's orientation (&quot;standing up&quot; means $H&amp;gt;B$, &quot;sideways&quot; means $B&amp;gt;H$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible to get dimensions where the rectangular section has a greater moment of inertia but smaller area or vice-versa. In these cases, the best section also depends on the expected loading: moment of inertia is important for bending and torsion, while area is important for normal and shear loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there may be case-specific height restrictions or architectural/aesthetic reasons to choose one over the other, but that's obviously not in the scope of this answer.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-01-05T12:24:34.397" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13097" PostTypeId="2" ParentId="13095" CreationDate="2017-01-05T13:23:46.850" Score="1" Body="&lt;p&gt;This is a bit of a vague question but I can give you two pieces of information which may help:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In theory the pressure drop can be arbitrarily low. That is to say that if the flowrate is small enough relative to the size of the flow path, then the pressure drop could be 0.001 atm.&lt;/li&gt;&#xA;&lt;li&gt;In practice, for any heat exchanger design you would not have a pressure drop that low. If your pressure drop is that low I suggest the heat exchanger is over-sized and uneconomical to produce.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2017-01-05T13:23:46.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13098" PostTypeId="2" ParentId="13090" CreationDate="2017-01-05T13:36:39.260" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it due to the PET material that it is made of or is it more the shape and structure of the bottle? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Short answer&lt;/strong&gt;: &lt;em&gt;Partially because of both&lt;/em&gt;. With that I mean it would be possible to produce weaker bottles changing the material or changing the usual design of the bottles. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If it is due to the shape/structure of the container, is it possible to achieve such pressure resistance using another plastic&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Short answer&lt;/strong&gt;: It is because of both, being impossible to say what contributes more without a definition of contribution to resistance. &lt;em&gt;Yes&lt;/em&gt;, it would be possible to use other polymers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Additional info&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comment about the material:&#xA;The resistance to pressure of polymer containers usually depends on the combination of three tensil properties: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ultimate tensile strength (force needed to break it pulling from extremities);&lt;/li&gt;&#xA;&lt;li&gt;tensile elongation&#xA;(how much you can pull it apart until it rips);&lt;/li&gt;&#xA;&lt;li&gt;tensile modulus ( ratio of stress to elastic strain in tension).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The table below illustrates these values for some polymers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u0Rqq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u0Rqq.jpg&quot; alt=&quot;Tensile property polymers&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.matweb.com/reference/tensilestrength.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.matweb.com/reference/tensilestrength.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown in the table, PET doesn't have the highest values. However, it does withstand reasonable pressure and certainly within the limits of proposed use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comment about the shape&lt;/strong&gt;:&#xA;Intuitively, if standing the pressure was the only requirement, the ideal shape would be a sphere. Although a sphere would allow thinner walls without deforming, I believe it is not usually chosen because of ergonomics and transportation as others have commented ( I have seen one brand of water that has spherical bottles, but I think it was done for aesthetics). The most common designs just prioritise a shape that is easy to hold and without sharp edges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why PET was chosen even though it is not the most resistant polymer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although many other polymers could satisfy the pressure requirement, I believe the reason of why polyethylene terephthalate (PET, material of most soda bottles) is of widespread use is &lt;strong&gt;compatibility&lt;/strong&gt;. Even though PET is what we see everyday, polypropylene (PP), Ethylene-Vinyl Alcohol Copolymer (EVOH), Polyethylene Terephthalate (PET),&#xA;and other plant derived plastics are used for containers &lt;a href=&quot;https://i.stack.imgur.com/u0Rqq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;. Although I did not do extensive research on their applications, I believe come of blends do satisfy the pressure requirements of 200~700 KiloPascals inside a PET bottle &lt;a href=&quot;https://i.stack.imgur.com/3gHnI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;( The comments below are my personal justification on the compatibility argument)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Machinery&lt;/strong&gt;&#xA;The process of producing machinery for polymer molding usually is polymer specific, that is, not always a machine that molds one polymer can mold another and therefore it is cheaper for industry as a hole to adopt a convention.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Regulation&lt;/strong&gt;&#xA;To test and approve a polymer that stays in long and close contact to food is extremely expensive. Also, considering that the approval must happen for the health agency of every country where the product is sold, there is no reason to make a change for another polymer if you know one that attends your health requirements and that has been tested for decades. I believe one evidence for this is that most os the polymers that I cited are used for bottles of liquids (sometimes pressurised) but not for human consumption. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Recycling&lt;/strong&gt;&#xA;Just as for producing, recycling processes are usually polymer specific, for what they needed to be separated. I don't believe they thought of that by the time the industry adopted the PET as the standard for carbonated drinks, but today it is an extra reason. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I believe that because of these reasons and others (please suggest edits) the &lt;strong&gt;transition of PET to anything will take a long time&lt;/strong&gt; and would be very bothersome. For example, we still were not able to transition completely from glass. In many countries glass is still used for carbonated drinks.  This transition was proposed because PET is lighter, doesn't break, cheaper, resistant to microrganisms (which is not so good today as it takes a long time to degrade in nature) and I believe because it was developed in a time when it was believed that men were unable to consume all petroleum available on Earth. However PET is no perfect polymer, it can't for example hold bear for very long, since it has considerable permeability to oxygen. Below there is a table with some permeability data of common use polymers [3] (unfortunately it doesn't include the other polymers that I said can be used for bottles, but you can find the values online; note that the permeability depends on the gas)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3gHnI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3gHnI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment what is being done is research on different blends of PET that can fill the gaps. In the future, however, there might be a new polymer that takes its place, hopefully addressing the environmental concerns. &lt;/p&gt;&#xA;" OwnerUserId="9450" LastEditorUserId="9450" LastEditDate="2017-01-05T14:43:43.027" LastActivityDate="2017-01-05T14:43:43.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13099" PostTypeId="1" AcceptedAnswerId="14099" CreationDate="2017-01-05T14:03:34.870" Score="0" ViewCount="1091" Body="&lt;p&gt;I'm attempting to find the highest free stream velocity over a flat plat before a sensor placed 'delta' away from it experiences turbulent flow. When I plot this in Matlab it seems that the turbulent boundary layer thickness decreases with velocity. This seems pretty counter intuitive to me, am I thinking of this in the wrong way? Or am I making a silly coding error?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the governing equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\delta &amp;amp;\approx \dfrac{0.37x}{Re_x^{1/5}} \\&#xA;Re_x &amp;amp;= \dfrac{\rho u_0x}{\mu}&#xA;\end{align}$$&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Boundary_layer_thickness&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; : Do not include any 'clear' or 'delete' code when posting, because many people do not want their workspace cleared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Matlab Code&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;% clc&#xA;% clear all&#xA;rho = 1025;   % density of water&#xA;mu = 0.00108; % dynamic viscosity&#xA;x = 65e-3;    % distance from leading plate edge&#xA;U = linspace(0,10); % Free Stream Velocity&#xA;Re = (rho*U *x)/mu;&#xA;turb_lay = (0.382*x)./(Re).^0.2&#xA;lam_lay = (5*x)./(Re).^.5&#xA;plot(U,lam_lay,U,lam_lay2,U,turb_lay)&#xA;xlabel('Velocity (m/s)')&#xA;ylabel('Boundary Layer Height (m)')&#xA;legend('Laminar Boundary', 'Turbulent Boundary')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="9478" LastEditorUserId="615" LastEditDate="2017-01-05T14:49:48.017" LastActivityDate="2017-03-06T21:22:26.410" Title="Turbulent Boundary Layer Thickness as Velocity Increases" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13101" PostTypeId="1" AcceptedAnswerId="13120" CreationDate="2017-01-05T13:32:00.463" Score="3" ViewCount="113" Body="&lt;p&gt;These days I'm designing a 5-DOF parallel manipulator with Autocad. Though I know that some people might say that Autocad is not a proper software for robot designing but I only have access to this software at the moment. So here is my question (and I thank you in advance for your help) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\textbf{Parametric constraints:}$ Unfortunately I do not know how to put geometrical constraints on my parallel manipulator though I have searched a lot on net and watched some videos about it on youtube. By geometrical constraint I mean putting some restrictions on for example two 3d objects that they can make some restricted moves with respect to each other. For example I have a spherical joint in my robot which I want it to perform some rotations without actually leaving its socket (as depicted in the following basic picture). &#xA;&lt;a href=&quot;https://i.stack.imgur.com/3eO8T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3eO8T.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9480" OwnerDisplayName="Arvin Rasoulzadeh" LastEditorUserId="9480" LastEditDate="2017-01-05T15:07:32.860" LastActivityDate="2017-01-06T14:44:05.473" Title="Designing a Parallel Robot by Autocad" Tags="|design|autocad|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13102" PostTypeId="2" ParentId="13099" CreationDate="2017-01-05T14:51:11.743" Score="0" Body="&lt;p&gt;It would appear that your setup is OK. I would recommend reading a few documents or texts which discuss boundary layer behavior. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found possibly useful monographs via Google.&#xA;&lt;a href=&quot;http://www.mne.psu.edu/cimbala/me320web_Spring_2015/pdf/Flat_plate_turbulent_BL.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This PDF&lt;/a&gt; includes some tables such as &#xA;&lt;a href=&quot;https://i.stack.imgur.com/QMoSr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QMoSr.png&quot; alt=&quot;Table10.4&quot;&gt;&lt;/a&gt; .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is a PowerPoint lecture which may also help&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0ahUKEwikmsCSnqvRAhUDNSYKHRqLDZcQFggcMAA&amp;amp;url=http%3A%2F%2Fwww3.nd.edu%2F~msen%2FTeaching%2FIntHT%2FSlides%2F07A%2520-%2520Chapter%25207%2520-%2520Sec%25207.1-7.3%2CApp%2520F%2520-%2520Black.ppt&amp;amp;usg=AFQjCNHVRXdiJvniCPMhTytKD_jkmT3v6g&amp;amp;sig2=gEo0gSDeDhOXuV-50dq2WQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;GoogleLink&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-01-05T14:51:11.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13103" PostTypeId="2" ParentId="13093" CreationDate="2017-01-05T14:55:42.657" Score="1" Body="&lt;p&gt;Well, rather obviously you need to find an optics house which will take on the task of regrinding the ends to optical flatness while maintaining perpendicularity of the faces to the long axis.  Cleaning the barrel is easier, assuming there aren't any nicks or cracks.  If the latter, it's more than likely that the rod will not survive much longer in any case due to the thermal stresses involved.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the application, you then follow up by having new coatings (AR or possibly an end-mirror) applied.  This is a job that only a specialized optics coating house can perform.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-01-05T14:55:42.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13104" PostTypeId="2" ParentId="11104" CreationDate="2017-01-05T15:28:00.797" Score="1" Body="&lt;p&gt;&lt;strong&gt;Short answer&lt;/strong&gt;: Any conductive material with the approximate size of a human finger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Longer answer&lt;/strong&gt; &#xA;It might be difficult to make one that includes multiple technologies. For example, the stylus from the Samsung Note and the Ipad Pro rely on proprietary technology that might be hard to duplicate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, it is feasible to make one that imitates the touch of a finger. In fact there is an &lt;a href=&quot;http://www.instructables.com/id/DIY-Capacitive-Stylus/?ALLSTEPS&quot; rel=&quot;nofollow noreferrer&quot;&gt; insctructable that I will sum up here. &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Screens map changes in an electric field produced by a charged conductive layer. Whether one or more points is detectable depends on the capacitive technology used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When a change in the electric field is detected, it is checked if it has the size of a finger and if yes the position is tracked.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, any material that is not an insulator can be used, this includes things such as aluminium foil, a wet sponge or a wet piece of leather. I believe that also means that you could use other methods for modifying the electric field of the screen, such as charges or currents (one observation that I made accidentally is that the cord of my earphone has touching capabilities even though it is coated in plastic (please try with yours). Also, the electric currents of big monitors &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Capacitive_sensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;make it impossible to use capacitive technology in bigger screens. &lt;/a&gt; But if you plan to use currents or point charges to control a capacitive screen, it may not behave nicely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you can use any material that is not an insulator, I think &lt;strong&gt;your second worry should be damaging the screen&lt;/strong&gt; and, at the same time allowing it to glide on the screen (low friction coefficient). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;One very convenient idea is to add graphite&lt;/strong&gt; To a polymer of choice. This is a trick long knew in electronics: add graphite powder to anything you want to make conductive or with less friction. So you can choose a polymer with your desired physical properties and mix graphite during curing. &lt;a href=&quot;http://www.instructables.com/id/Make-Conductive-Rubber-Transparent-stylus-iPodiP/?ALLSTEPS&quot; rel=&quot;nofollow noreferrer&quot;&gt;There is an insctructable about that. &lt;/a&gt; Any conductive powder such as silver may also work, but it is more expensive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I assume you are making an stylus to make it somehow different from the existing ones, you can try different mixtures to see which ones produces better results. Good luck and please edit with the results. &lt;/p&gt;&#xA;" OwnerUserId="9450" LastEditorUserId="9450" LastEditDate="2017-01-05T15:39:16.020" LastActivityDate="2017-01-05T15:39:16.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13105" PostTypeId="2" ParentId="13085" CreationDate="2017-01-05T19:23:39.667" Score="0" Body="&lt;p&gt;When no voltage is applied to the solenoid, air will flow between port 2 'Common' and port 3 'Normally Open,' but not to port 1 'Normally Closed.' When voltage is applied, air will flow between port 2 and port 1, but not to port 3. We could likely assume that the direction of air flow is intended to be from port 2 to ports 1 &amp;amp; 3, but it isn't a guarantee and it may not matter to the valve. Note that in pneumatic and hydraulic circuits, normally closed and normally open have the opposite definition as in electrical circuits. For a valve, closed means &lt;em&gt;not passing&lt;/em&gt; air/oil, whereas for a relay closed means &lt;em&gt;passing&lt;/em&gt; electrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more standard diagram for this valve would be #3 in the image below:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6fRSa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6fRSa.jpg&quot; alt=&quot;Example valve diagrams&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although it's worth noting that the 'T' designation implies a hydraulic valve, and numeric port designations are more common for pneumatic valves.&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2017-01-05T19:23:39.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13110" PostTypeId="1" CreationDate="2017-01-06T04:02:51.050" Score="0" ViewCount="406" Body="&lt;p&gt;How can I measure stress in a thin- walled sphere containing a gas? Is there a way to predict what the stress will be without actually needing to fill a sphere with gas? I am talking about hoop stress/longitudinal stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thickness of the walls of the container is unknown at this point; I know the radius and the material (&lt;a href=&quot;https://en.wikipedia.org/wiki/Ethylene_vinyl_alcohol&quot; rel=&quot;nofollow noreferrer&quot;&gt;EVOH&lt;/a&gt;); the internal pressure will be 10 atm. The question then is how will I know how much stress is too much for the material to handle? Would it be the material's yield stress value? Once that is found, how will I know how much of a safety factor is acceptable?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="368" LastEditDate="2017-01-09T22:58:21.087" LastActivityDate="2017-01-09T22:58:21.087" Title="Measure stress of gas acting on a spherical container" Tags="|pressure|measurements|pressure-vessel|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13111" PostTypeId="1" CreationDate="2017-01-06T05:28:54.470" Score="0" ViewCount="1320" Body="&lt;p&gt;why we need to consider 2 moment caused by the 85Nm in this question ? The question ask for the determine the angle of twist of gear A relative to D . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why shouldnt the angle of twist = &lt;strong&gt;(85)(0.4)&lt;/strong&gt; / (      (0.5pi)( 0.015^4- 0.01^4)(75)(10^9)  )  + (85)(0.25) / (0.5pi)(0.02^4)(75)(10^9) only ? &lt;a href=&quot;https://i.stack.imgur.com/Fd0Uu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fd0Uu.png&quot; alt=&quot;c&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="33" LastEditDate="2017-01-06T15:12:39.943" LastActivityDate="2017-01-06T15:12:39.943" Title="angle of twist of gear relative to another" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ClosedDate="2017-01-14T17:36:24.040" ContentLicense="CC BY-SA 3.0" />
  <row Id="13114" PostTypeId="2" ParentId="13111" CreationDate="2017-01-06T10:48:18.607" Score="0" Body="&lt;p&gt;I almost had this in a comment but it might as well be an answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There should indeed be that 2 there.  You are summing the twist in each of the bars and adding them together.  Both the hollow sections attached to gears have the same torque and sizes, so they will have the same value.  Instead of writing out the twist for that bar separately they just multiply the first term by 2 since it will just be that again.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-01-06T10:48:18.607" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13115" PostTypeId="1" AcceptedAnswerId="13118" CreationDate="2017-01-06T10:53:06.257" Score="1" ViewCount="452" Body="&lt;p&gt;I'm having a presentation about dc generators and I wanted to demonstrate how it works. Anyone have an idea on how to build one. I left my electric motors at home because of Christmas break, now I can't find anything.&lt;/p&gt;&#xA;" OwnerUserId="9446" LastActivityDate="2017-01-06T13:50:52.100" Title="How do I build a dc generator at home for my presentaion?" Tags="|electrical-engineering|motors|generator|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13116" PostTypeId="1" AcceptedAnswerId="13117" CreationDate="2017-01-06T12:32:56.383" Score="0" ViewCount="3204" Body="&lt;p&gt;why the torque is negative in this case ? the question ask for determine the angle of twist of gear C with respect to B . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to 'fix' the B , and determine the twist of C ? So , imagine that we cut off the shaft and gear , now only left with gear B and gear C connected by a shaft . So , when B is fixed , the gear C rotate , with the direction of torques pointing to the left ( by using right hand grip rule )  , &#xA;So , the finger point towards B  , so it's negative ? &lt;a href=&quot;https://i.stack.imgur.com/Gmlah.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gmlah.png&quot; alt=&quot;11&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="33" LastEditDate="2017-01-06T15:12:18.997" LastActivityDate="2017-01-07T06:22:50.737" Title="angle of twist of gear relative to another gear" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13117" PostTypeId="2" ParentId="13116" CreationDate="2017-01-06T12:55:08.493" Score="2" Body="&lt;p&gt;Negative and positive are essentially arbitrary and are really just to differentiate the directions of the moments and forces.  You can chose your positive y direction to be facing down for example, if that's how you want to arrange your coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard orientation is right and up are both positive coordinates.  If we assume they are using standard coordinates; then it should be negative using the right hand rule.  You yourself said when using the right hand rule that your thumb pointed left.  That is generally considered the negative x direction, therefore the moment is considered negative.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-01-06T12:55:08.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13118" PostTypeId="2" ParentId="13115" CreationDate="2017-01-06T13:28:46.783" Score="1" Body="&lt;p&gt;For a presentation it might makes sense to reproduce one of &lt;a href=&quot;https://en.wikipedia.org/wiki/Faraday&amp;#39;s_law_of_induction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Faraday's&lt;/a&gt; original experiments which give more of a visible insight into what is going on than a modern generator which is essentially just a sealed box. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like the experiment in &lt;a href=&quot;https://www.youtube.com/watch?v=vwIdZjjd8fo&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; would be quick and easy to set up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously making a rotating armature which produces a steady current is a bit more difficult than just moving a magnet in and out of a coil but in some ways the simple version demonstrates the underlying concept better. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-01-06T13:50:52.100" LastActivityDate="2017-01-06T13:50:52.100" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13120" PostTypeId="2" ParentId="13101" CreationDate="2017-01-06T14:44:05.473" Score="1" Body="&lt;p&gt;I can only answer the question you asked about. You asked about a spherical joint, so I'll give advice about the spherical joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, are you using Autocad? That looks like Inventor. They are two separate software packages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two ways you could constrain (mate) the spherical joint would be to make a &quot;cup&quot; (spherical cut) on the socket and then constrain the ball to be tangent to the cup (cup is same radius as the ball) or concentric to the cup (not necessarily the same radius). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way would be to create a reference point on the cup and a reference point on the ball and constrain the two points to be coincident. That's probably the way I would do it, though a concentric mate would probably be the same thing behind the scenes. Tangent mates would be a lot more taxing on the software and would bog down your simulation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use driven constraints to get a joint to move to a particular parameter's value - &lt;em&gt;those&lt;/em&gt; would be parametric constraints. Check out constraint limits; I forget what they're called in Inventor off hand, but you can essentially make &quot;end stops&quot; for actuator min/max strokes, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're in academia then you might try to see if you can get an academic-use copy of Matlab. There is the SimMultibody package that lets you do probably a lot of what you want to do anyways, along with the Robotic System toolbox. Don't confuse the latter with the Robotic Toolbox - the Robotic System Toolbox is offered by Mathworks; the Robotic Toolbox was written by a third party and is out of date. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have any other &lt;em&gt;specific&lt;/em&gt; questions, please ask another. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2017-01-06T14:44:05.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13121" PostTypeId="2" ParentId="13089" CreationDate="2017-01-06T14:49:15.150" Score="1" Body="&lt;p&gt;While it is certainly possible to connect input and output shafts with gears there are a number of reasons why having a separate gearbox might be desirable. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For more complex compound gear systems the gearbox casing provides mounting points for shaft bearings for intermediate gears. For larger gear rations compound gears are generally more compact than a single pair of gears with a large difference in diameter. &lt;/li&gt;&#xA;&lt;li&gt;The gearbox casing allows lubricant (oil or grease) to be retained and helps prevent contamination by dust, debris, water etc from outside.&lt;/li&gt;&#xA;&lt;li&gt;Having a self contained modular gearbox can make assembly and maintenance easier and can allow for a standard gearbox to be used for a number of different applications. &lt;/li&gt;&#xA;&lt;li&gt;Gears and their shafts need to be accurately aligned to work effectively, a rigid casing helps with this. If one of the shafts needs to move or has large loads on it, as in the case of a drive shaft in a vehicle with suspension the gears can be separated from the loads on the shaft via constant velocity joints or other articulated couplings. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-01-06T14:49:15.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13122" PostTypeId="1" CreationDate="2017-01-06T15:46:08.380" Score="0" ViewCount="42" Body="&lt;p&gt;Trying to work out the closed curve circulation of a velocity profile. The final integral that has been formed is:&#xA;$$\int_0^{2\pi}r^3g(t)dt$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody help? Not sure whether g(t) can even be integrated?&lt;/p&gt;&#xA;" OwnerUserId="9498" LastEditorUserId="163" LastEditDate="2017-01-07T06:13:27.207" LastActivityDate="2017-01-07T06:56:33.217" Title="Can g(t) be integrated with respect to t? Been stuck on this one for a while" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="2" ClosedDate="2017-01-07T07:07:49.013" ContentLicense="CC BY-SA 3.0" />
  <row Id="13126" PostTypeId="2" ParentId="13110" CreationDate="2017-01-07T01:07:51.867" Score="2" Body="&lt;p&gt;Spherical vessels have a stress in the wall equal to,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma = \frac{p r}{2 t} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $p$ is the internal gage pressure, $r$ is the inner radius of the sphere, and $t$ is the thickness of the wall, where the only constraint is $r/t \geq 10$.&lt;/p&gt;&#xA;" OwnerUserId="9399" LastActivityDate="2017-01-07T01:07:51.867" CommentCount="26" ContentLicense="CC BY-SA 3.0" />
  <row Id="13127" PostTypeId="1" AcceptedAnswerId="13130" CreationDate="2017-01-07T03:41:05.567" Score="1" ViewCount="373" Body="&lt;p&gt;In this problem, why did the author use &lt;span class=&quot;math-container&quot;&gt;${(30×10^3)} {/s} =  {VQ} /{ I} $&lt;/span&gt; only?&#xA;Shouldn't it be &lt;span class=&quot;math-container&quot;&gt;$4(30×10^3) /s = VQ / I$&lt;/span&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I said so because the board at the centered is connected by 4 bolts , 2 on the left and 2 on the right . The faces on the right also resisted by the same arrangement of nails , right ?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pH5tw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pH5tw.png&quot; alt=&quot;22&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JK9uV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JK9uV.png&quot; alt=&quot;226&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="16486" LastEditDate="2019-08-04T04:20:41.417" LastActivityDate="2019-08-04T04:25:37.200" Title="shear stress resisted by bolt" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13128" PostTypeId="2" ParentId="13122" CreationDate="2017-01-07T06:09:12.930" Score="1" Body="&lt;p&gt;Use g(t) as its equivalent (derivative of velocity) and integrate using an abstract function unless the question is assuming g(t) is the constant function used on earth. I think using F = M&lt;em&gt;mg/r^2 = m&lt;/em&gt;dv\dt and substituting g(t) into your function would be a plausible option as g(t) =~a(t). If it's a periodic function then you have the square root relation of T but I'd need more info to confirm. As you are integrating over 2pi I believe g is the relation of a periodic function (root relation), which can be explicitly solved for and plugged into the formula. A picture would certainly clarify! :)&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="9404" LastEditDate="2017-01-07T06:56:33.217" LastActivityDate="2017-01-07T06:56:33.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13129" PostTypeId="2" ParentId="13116" CreationDate="2017-01-07T06:22:50.737" Score="1" Body="&lt;p&gt;they are not opposing. Their rotation is a ratio regarding the number of teeth n1/n2. If you fix b so that there is torsion I think you're moving in the wrong direction. The torque* down the shaft is constant. The torque is relative to the diameter of the gear i.e. F*r = T. The angle is simply geometric unless it is restricted by an outside torsional force. They both rotate in the same direction but with different torque. &lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2017-01-07T06:22:50.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13130" PostTypeId="2" ParentId="13127" CreationDate="2017-01-07T06:27:47.337" Score="1" Body="&lt;p&gt;The four connected boards are encompassed by the inertia in the shear equation. In the inertial formula they account for the four connections (the multiplier 4). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear force on the element in question is defined by the bodies and forces that are acting on it, and inertia is the resistance to rotational movement (simplification to the extreme but I think you get my point), so by using the inertial mass of the connected elements we can deduce the force on the main body by accounting for the connected masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go back to what shear force is on a body. In equilibrium you should be able to make a theoretical slice in the body at any point and have a balanced system (axial, shear, rotational), the 4 outside the brackets of the inertia equation (I) accounts for the four masses connected.&lt;/p&gt;&#xA;" OwnerUserId="9404" LastEditorUserId="1832" LastEditDate="2019-08-04T04:25:37.200" LastActivityDate="2019-08-04T04:25:37.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13131" PostTypeId="2" ParentId="13110" CreationDate="2017-01-07T06:49:25.243" Score="-1" Body="&lt;p&gt;Create a 3D stress tensor and calculate the known forces for a tube (easily googleable). You simply need to know the pressure inside the tube if it is in quasi-equilibrium. Given that there is no torsion it should be quite simple, but you must know the force exerted on the body. You can predict the pressure using PV=nRT assuming an ideal gas. If I'm missing something let me know and I'll modify my answer. &lt;/p&gt;&#xA;" OwnerUserId="9404" LastActivityDate="2017-01-07T06:49:25.243" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13132" PostTypeId="1" AcceptedAnswerId="13133" CreationDate="2017-01-07T08:51:38.030" Score="2" ViewCount="313" Body="&lt;p&gt;Our teacher gave us a project before Christmas break, it was to build a miniature house, she didn't state what specific materials should be used, so I built a cardboard house. She didn't accept it because it wasn't firm enough for her. Now I want to know if it is possible to build a miniature concrete house? &#xA;Considering I can't find any tutorials online I'm guessing it has a high chance of collapsing, but I don't know, that's why I came here.&lt;/p&gt;&#xA;" OwnerUserId="9446" LastEditorUserId="90" LastEditDate="2017-01-07T11:19:19.617" LastActivityDate="2017-01-09T03:23:47.367" Title="How do I build a miniature concrete house?" Tags="|civil-engineering|structures|architecture|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13133" PostTypeId="2" ParentId="13132" CreationDate="2017-01-07T13:26:43.863" Score="5" Body="&lt;p&gt;Concrete isn't a good material for such a small model. I would suggest foamboard. You can get it at craft stores. If you can't find half inch thick material, laminate two layers of quarter inch board. Architects use foamboard all the time. You can glue it with Elmers. If you really need heavier duty stuff, there is concrete wall board used in bathrooms. You can get it at stores like Home Depot. Much harder to cut and work with. &lt;/p&gt;&#xA;" OwnerUserId="9506" LastActivityDate="2017-01-07T13:26:43.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13134" PostTypeId="1" CreationDate="2017-01-07T16:23:51.007" Score="1" ViewCount="70" Body="&lt;p&gt;Don't worry guys i won't ask how to calculate the moment of inertia of the gate. I already have, but i have no experience in real life engineering, so i would be thankfull if any of you could review my result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Door wieghs about 90 kg Has a length of 3000 mm a height of 1115 mm and a width of 40 mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I Calculated the Moment of Inertia on the hinges and the result was 227.5 kgm². I added a Picture so you can get an idea of how the gate looks like. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/wBvsO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wBvsO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9507" LastEditorUserId="9507" LastEditDate="2017-01-08T11:42:46.733" LastActivityDate="2017-01-08T11:42:46.733" Title="Realistic Value for the Rotational Inertia of a Gate" Tags="|mechanical-engineering|torque|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13136" PostTypeId="2" ParentId="12142" CreationDate="2017-01-07T20:21:19.650" Score="1" Body="&lt;p&gt;The amount of o-ring compression or squeeze is dependent on the size of the ring and the specific application. The o-ring groove geometry likewise varies. The easiest way to design with o-rings is to use a design guide. I know it is a commercial website, but &lt;a href=&quot;http://www.applerubber.com/seal-design-guide/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Apple Rubber&lt;/a&gt; provides a lot of nice design information for using O-Rings. For the record, I have no affiliation with Apple Rubber.&lt;/p&gt;&#xA;" OwnerUserId="9506" LastEditorUserId="9506" LastEditDate="2017-02-04T09:58:47.783" LastActivityDate="2017-02-04T09:58:47.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13138" PostTypeId="1" AcceptedAnswerId="13139" CreationDate="2017-01-07T21:16:40.977" Score="1" ViewCount="33" Body="&lt;p&gt;How can the vertical wind direction be measured?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a project I have to design a wind turbine and I would like the blades to make slight adjustments based on the wind direction in order to achieve the most efficient energy conversion. Whilst I can use a wind vane to get the direction of the wind on a horizontal plane; I have been unable to find a method for finding the direction of the wind on a vertical plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate any help.&lt;/p&gt;&#xA;" OwnerUserId="6505" LastActivityDate="2017-01-07T21:51:24.083" Title="Vertical wind measurements" Tags="|mechanical-engineering|aerodynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13139" PostTypeId="2" ParentId="13138" CreationDate="2017-01-07T21:51:24.083" Score="1" Body="&lt;p&gt;Given the budget of up to £3m, then lidar scanning seems to be current state of the art for 3-d profiling of wind for active management of wind turbines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://www.dewi.de/dewi_res/fileadmin/pdf/publications/Magazin_37/08.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;sample business case&lt;/a&gt; for one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The WINDCUBE v2 is a vertical wind profiler that uses doppler pulsed lidar. &lt;a href=&quot;http://www.norcowe.no/doc/Documents%20for%20web/SMI%20S%20Presentasjonar/E%20Santiago,%20A.,%20Leosphere.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's a presentation on it&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, lidar has mostly been used in offshore wind. There are quite a few academic papers on its use in the wind industry. I don't know what, if anything, has been done about processing vertical-plane wind data, but there's an increasing number of conference= and journal- papers on yield optimisation for wind turbines, so there could be quite a bit of existing literature in the field on this already.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2017-01-07T21:51:24.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13140" PostTypeId="2" ParentId="13134" CreationDate="2017-01-07T22:09:14.817" Score="2" Body="&lt;p&gt;Your units are wrong. I think you meant $227.5\,\text{kg}\,\text{m}^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your calculations, the radius of gyration about the hinge is $$k = \sqrt{\frac I m} = \sqrt{\frac {227.5}{90}}\,\text{m} = 1590\,\text{mm}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a point mass at the middle of the gate, $k = 1500\,\text{mm}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a uniformly distributed mass in the horizontal direction, $k = 3000/\sqrt3 = 1732\,\text{mm}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design of your gate is somewhere in between those two limits, so $k = 1590\,\text{mm}$ is &quot;not obviously wrong&quot;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-01-07T22:09:14.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13141" PostTypeId="1" AcceptedAnswerId="13143" CreationDate="2017-01-07T23:40:57.997" Score="1" ViewCount="1676" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IhsLg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IhsLg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is a past exam question from an introductory structural analysis course, one in which although we have studied the Euler buckling load equation, we have just been given parameters for the equation based on the standard end support conditions (fixed/fixed, fixed/pinned, etc.).  I don't think the above example fits any of the standard end conditions and am therefore finding it difficult to derive an equation for $P_{cr}$&#xA;I have tried to use the bending moment equation by including torque (that opposes bending) from B and the net load (due to spring C) as follows:&#xA;$$&#xA;T_B = \beta_r \theta&#xA;$$&#xA;Taking a small deflection x in the vertical direction and y in the horizontal:&#xA;$$&#xA;tan\theta = \frac{dy}{dx} \approx \theta&#xA;$$&#xA;$$&#xA;\therefore T_B = \beta_r \frac{dy}{dx}&#xA;$$&#xA;Using the bending moment equation:&#xA;$$&#xA;M=-EI\frac{d^2y}{dx^2}&#xA;$$&#xA;$$&#xA;\therefore \sum M=-EI\frac{d^2y}{dx^2} + \beta_r \frac{dy}{dx} = Qy&#xA;$$&#xA;The net load $Q$ can be expressed in term of the force P and the reaction force due to the spring at C as:&#xA;$$&#xA;Q = P - \beta dx&#xA;$$&#xA;Leaving the final second order equation:&#xA;$$&#xA;EI\frac{d^2y}{dx^2} - \beta_r \frac{dy}{dx} + (P-\beta dx) y = 0&#xA;$$&#xA;&lt;p&gt;&#xA;This is beyond the scope of our course, so I have no idea if:&lt;br&gt;&#xA;a) The above equation is correct or solvable (and if so how one goes about solving it)&lt;br&gt;&#xA;b) There is a different approach using more basic techniques (which is more than likely the case)&lt;/p&gt;&#xA;" OwnerUserId="8437" LastActivityDate="2017-01-08T01:05:41.293" Title="Critical Buckling Load for a Spring Supported Bar" Tags="|structural-engineering|structures|buckling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13142" PostTypeId="1" CreationDate="2017-01-08T00:11:51.440" Score="1" ViewCount="120" Body="&lt;p&gt;I'm trying to figure out what kind of materials my disposable plastic syringe is made from. I have a &lt;a href=&quot;http://medefilinc.com/saline-iv-flush-syringes-.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Medefil MIS-1130&lt;/a&gt; syringe and when I visited the company website I couldn't find any information on what the disposable syringes were made from (although they did tell me a lot on what's NOT in them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find a regulation or something that could help me identify what kind of plastic and rubber are used to make the shill, plunger and seal that make up the disposable syringe. Can anyone on here please give me some guidance?&lt;/p&gt;&#xA;" OwnerUserId="9511" LastActivityDate="2017-01-08T00:11:51.440" Title="What kind of materials are plastic syringes made from?" Tags="|materials|biomedical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13143" PostTypeId="2" ParentId="13141" CreationDate="2017-01-08T01:05:41.293" Score="3" Body="&lt;p&gt;To answer the specific question: (a) No, that equation is irrelevant and (b) Yes. see below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question isn't about Euler buckling. It says the bar is rigid. Euler buckling only applies to a flexible bar - the Euler formulas include Young's modulus, and also the moment of inertia of the bar which is not mentioned in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to solve the question is to use the principle of virtual work, and compare the strain energy in the springs with the work done by the external force P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also notice that the condition $\beta_r = 3\beta L^2/2$ is related to the geometry of the triangle ABC.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-01-08T01:05:41.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13144" PostTypeId="2" ParentId="12730" CreationDate="2017-01-08T01:33:22.400" Score="5" Body="&lt;p&gt;&lt;strong&gt;Locking the ball joints:&lt;/strong&gt;&#xA;Rotating the handle also rotates two cylinders with a section cut out. (see photo) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When rotated to a locked position, each cylinder pushes on a rod in each arm, applying force to the ball joint, holding the ball joints in place by friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When unlocked, each rod slides into the cut out in the cylinder, releasing the force on the ball joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Locking the central pivot joint:&lt;/strong&gt;&#xA;Rotating the handle presses the two halves of the arm together, locking two rings of teeth together (pictured below). I am unsure of the exact mechanism to convert the rotational force from the handle into force pressing the halves together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model may work differently from the one pictured in the question.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uhcjZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uhcjZ.png&quot; alt=&quot;Magic Camera Arm Design Explanation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Images used:&#xA;Spare part pdf under the downloads tab on this Manfrotto product page&#xA;&lt;a href=&quot;https://www.manfrotto.us/magic-arm-kit-143a-003-035&quot; rel=&quot;noreferrer&quot;&gt;https://www.manfrotto.us/magic-arm-kit-143a-003-035&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Product comparison image from this product listing page. I have not idea if this is a reputable website or not, I just found my image there.&#xA;dx.com/p/aputure-a10-multifunctional-magic-arm-stabilizer-grip-for-camera-black-387217#.WHGSchsrLb1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't post more than 2 links yet, so you may need to add three &quot;w&quot; and a period before the third link to get it to work.&lt;/p&gt;&#xA;" OwnerUserId="9512" LastActivityDate="2017-01-08T01:33:22.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13145" PostTypeId="1" CreationDate="2017-01-08T01:37:43.037" Score="1" ViewCount="2056" Body="&lt;p&gt;I am interested in learning more about ESC/POS commands. I have an EPSON TM-T88IV printer connected to my PC via parallel cable at port LPT1. Connected to this printer is an APG Cash Drawer. The PC is running Windows 7 Professional. I would like to know if it is possible to send ESC/POS commands to this printer through the Windows command line. If not, I would like to know what steps to take to be able to set up some sort of test environment.&lt;/p&gt;&#xA;" OwnerUserId="2399" LastActivityDate="2017-06-07T07:50:52.743" Title="Is it possible to send ESC/POS commands to EPSON TM-T88IV Printer through Windows command line?" Tags="|electrical-engineering|computer-engineering|computer-hardware|computer|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13146" PostTypeId="2" ParentId="13145" CreationDate="2017-01-08T06:10:03.843" Score="1" Body="&lt;p&gt;In principle this should work fine. According to &lt;a href=&quot;https://reference.epson-biz.com/modules/ref_escpos/index.php?content_id=82&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://reference.epson-biz.com/modules/ref_escpos/index.php?content_id=82&lt;/a&gt; the commands to drive the printer are just combinations of ASCII control characters like ESC, FS, GS, etc, which can be transmitted to the printer exactly the same way as &quot;ordinary&quot; printable letters and numbers. A complete table of ASCII codes is here: &lt;a href=&quot;http://www.asciitable.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.asciitable.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you actually send an &quot;unprintable&quot; character like ESC etc will depend on the software you are using. You may be able to simply &quot;type&quot; the characters - for example to generate ESC (character number 27 in the ASCII table) hold down the ALT key while typing 027 on the numeric keypad, use ALT + 029 for GS, etc.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-01-08T06:10:03.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13148" PostTypeId="1" AcceptedAnswerId="16339" CreationDate="2017-01-08T13:14:21.553" Score="1" ViewCount="9219" Body="&lt;p&gt;Is there any way to display element coordinate system in ANSYS Mechanical APDL GUI (the same way the global coordinates system is displayed)?&lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="12198" LastEditDate="2017-07-18T12:33:05.863" LastActivityDate="2021-01-23T09:34:03.713" Title="Display element coordinate system using Ansys Mechcanical APDL" Tags="|ansys|ansys-apdl|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13149" PostTypeId="1" AcceptedAnswerId="13150" CreationDate="2017-01-08T14:33:57.323" Score="1" ViewCount="1016" Body="&lt;p&gt;My friend and I are currently working on a cloud chamber cooled by the compressor of a deep-freezer. We have built an isolated box that contains the cold pipes of the compressor. The pipes cool down to a whopping -40C. On the top of our box lies an aluminium plate, and on top of that a well isolated chamber. With the aluminium plate 5 or 6 cm away from the compressor pipes, the plate just gets -15C. Connecting the tubes and the plate with a folded fin heatsink gives us -20C. Moving the plate down onto the tubes results in a much better -35C, however this is not a very favorable option as this makes the plate harder to look at. Unfortunately, bending the pipes is not an option either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We think the air in the box is simply not thermally conductive enough. The heatsink was not doing a very good job either, presumably because the fins where too thin and were in contact with a lot of air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our question is: What would be the best way to transport the &quot;coolness&quot; of the pipes towards the plate. Would it be better to fill the box with some sort of liquid that would transport heat more efficient then air? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are using a computer fan to circulate the air in the box. The fan is pointed upward, so the airflow flows via the pipes onto the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Images:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UaXTa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UaXTa.jpg&quot; alt=&quot;Top of the box with the aluminium plate&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/OsQTf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OsQTf.jpg&quot; alt=&quot;Top of the box without the aluminium plate, you can see the pipes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9516" LastEditorDisplayName="user16" LastEditDate="2017-01-08T18:09:48.377" LastActivityDate="2017-01-08T18:09:48.377" Title="Cooling an aluminium plate" Tags="|heat-transfer|temperature|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13150" PostTypeId="2" ParentId="13149" CreationDate="2017-01-08T17:06:26.277" Score="0" Body="&lt;p&gt;The goal is to remove heat energy from the aluminum plate to reach a temperature of at least $-15\unicode{xb0}$F = $-26.1\unicode{x2103}$ by increasing heat transfer between the plate and the refrigeration tubing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One option is to try to increase convection. This would increase the &quot;h&quot; value in the heat transfer equation below. &#xA;$$Q = h*A_S*\Delta T$$&#xA;$Q$ is the Heat Transferred. $h$ is the heat transfer coefficient. $A_S$ is the surface area. $\Delta T$ is the temperature difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding liquid might work, but would be heavy and messy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest trying to increase conduction between the plate and the tubes. The distance between the plate and the tubes is too much for using heat transfer compound. An inexpensive option could be to lightly crumple up aluminum foil and put it under the plate in such a way where it contacts both the plate and tubing. &lt;/p&gt;&#xA;" OwnerUserId="3569" LastActivityDate="2017-01-08T17:06:26.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13151" PostTypeId="1" AcceptedAnswerId="13168" CreationDate="2017-01-08T17:24:06.170" Score="1" ViewCount="515" Body="&lt;p&gt;How I can calculate or find a value for the start-up cost of a natural gas boiler? The boiler will have a capacity of about 700 kW and is intended for heating the water for a greenhouse.&lt;/p&gt;&#xA;" OwnerDisplayName="user5026" LastEditorUserId="368" LastEditDate="2017-01-09T22:59:48.793" LastActivityDate="2017-01-09T23:03:06.137" Title="How to calculate the start-up cost of a natural gas boiler" Tags="|boilers|cost-estimation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13153" PostTypeId="1" AcceptedAnswerId="13157" CreationDate="2017-01-08T19:51:24.200" Score="0" ViewCount="1105" Body="&lt;p&gt;I'm using existing architectural designs and some mechanical designs too, all CAD files in formats *.dwg, *.dgn and the drawings have smart solid geometries including voids. Using CAD software what is a void space and why is it important to create voids as an object?&lt;/p&gt;&#xA;" OwnerUserId="9508" LastActivityDate="2017-01-09T09:10:18.170" Title="What is a void in CAD" Tags="|autocad|cad|geometry|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13155" PostTypeId="1" AcceptedAnswerId="13166" CreationDate="2017-01-09T02:06:58.160" Score="1" ViewCount="8549" Body="&lt;p&gt;I got this stand from a friend of mine years ago. Its unbranded. I want to use it to hold a 55 gallon fish tank which can weigh about 456 pounds to 622.5 pounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can i calculate a estimate of how much weight this stand will hold ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wood is 61 in by 17 in and about 3/4 in thick. The metal underneath it is 2 inches. The metal legs are 28 inches high and 1 inch on each side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, then is there any way i can easily modify the stand for more support? Perhaps stick some 2x4s or cinderblocks under it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/eRlRo.jpg&quot; alt=&quot;a busy cat&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/sNgeC.jpg&quot; alt=&quot;a busy cat&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/iHvqL.jpg&quot; alt=&quot;a busy cat&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Mx72H.jpg&quot; alt=&quot;a busy cat&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/eQnJ3.jpg&quot; alt=&quot;a busy cat&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/oYoaS.jpg&quot; alt=&quot;a busy cat&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9522" LastEditorUserId="9522" LastEditDate="2017-01-09T06:30:55.257" LastActivityDate="2017-01-09T15:33:42.900" Title="How much weight can this stand hold?" Tags="|structural-engineering|civil-engineering|structures|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13156" PostTypeId="2" ParentId="13132" CreationDate="2017-01-09T03:23:47.367" Score="-1" Body="&lt;p&gt;If you must use concrete, you could consider ShapeCrete -- it's a clay-like moldable concrete, more info could be found on their web site www.shapecrete.com . They also have a projects gallery that you can use to decide whether this material will suite your needs. It is available from Amazon.&lt;/p&gt;&#xA;" OwnerUserId="9518" LastActivityDate="2017-01-09T03:23:47.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13157" PostTypeId="2" ParentId="13153" CreationDate="2017-01-09T09:10:18.170" Score="0" Body="&lt;p&gt;You can add a void geometry in a window object in order to make a hole when you insert the window in a wall. Autodesk Revit works like this.&lt;/p&gt;&#xA;" OwnerUserId="4428" LastActivityDate="2017-01-09T09:10:18.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13160" PostTypeId="1" CreationDate="2017-01-09T12:46:11.947" Score="2" ViewCount="815" Body="&lt;p&gt;I have a question about intravenous flow rate controllers, which are controlled using microprocessors. For example: &lt;a href=&quot;https://i.stack.imgur.com/mkUI4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mkUI4.jpg&quot; alt=&quot;iv-flow-controller&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to control the IV flow rate, the flow is in range from 0 to 300. So, I just put a number between 0 drips/minute to 300 drips/minute in a 10 scale from the computer interface, and the microprocessor would control the flow rate to the desired number. For example, when I put 60 drips/minute on the computer interface, the microprocessor controls the mechanism until the flow rate reaches 60 drips/minute. The tubing used by the IV dispenser is a form of PVC plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to use a pinching mechanism like in the picture below. I'm using that kind of mechanism because it is easier to insert and remove the tube from the device when required. But, the mechanism still doesn't give the precise output that is specified, also it is unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yAkHH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yAkHH.jpg&quot; alt=&quot;pinch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can the flow rate be controlled just by using motors such as a servo motor or a motor DC and microprocessor?&lt;/p&gt;&#xA;" OwnerUserId="8972" LastEditorUserId="90" LastEditDate="2017-01-11T01:30:01.900" LastActivityDate="2017-01-11T01:30:01.900" Title="IV drip/flow rate controller design" Tags="|mechanical-engineering|electrical-engineering|applied-mechanics|mechanisms|medical-devices|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13162" PostTypeId="1" CreationDate="2017-01-09T13:47:48.870" Score="1" ViewCount="16605" Body="&lt;p&gt;As in the title, I need some indication about the optimal tightening torque to choose when screwing in a SS screw (A2-70, say) into an aluminium threaded hole. In particular, I am concerned with metrical dimensions (M8, M10 and M12).&lt;/p&gt;&#xA;" OwnerUserId="9528" LastActivityDate="2017-02-17T20:53:29.467" Title="Optimal tightening torque for a SS screw on an aluminium threaded hole" Tags="|mechanical-engineering|threads|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13163" PostTypeId="1" AcceptedAnswerId="13178" CreationDate="2017-01-09T14:16:29.247" Score="3" ViewCount="1878" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nlFcC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nlFcC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File:DieselElectricLocomotiveSchematic.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not just take the AC from the alternator to the motors? Why go AC to DC to AC again?&lt;/p&gt;&#xA;" OwnerUserId="9529" LastEditorUserId="33" LastEditDate="2017-05-02T18:02:24.567" LastActivityDate="2017-05-02T18:02:24.567" Title="Why use rectifier / inverter in locomotives?" Tags="|mechanical-engineering|electrical-engineering|rail|transportation|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13164" PostTypeId="1" AcceptedAnswerId="13165" CreationDate="2017-01-09T14:21:34.947" Score="3" ViewCount="325" Body="&lt;p&gt;I have a part made from 5mm alloy plate and I need to cut a large cutout in the middle of said part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3IW2o.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3IW2o.png&quot; alt=&quot;An alloy part requiring a large cutout&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As seen above I intend to use a small cutter to cut through the stock (not necessarily plunging as shown here) and cut around the inner diameter of the part instead of roughing out the entire area of the cutout which would take a lot longer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concern is that when the cut is complete the large piece of internal cutout could get caught by the cutter and thrown around the machine enclosure, wreaking death and destruction as it goes. Is this something I should be worried about and if so, what's the usual strategy for making this type of feature?&lt;/p&gt;&#xA;" OwnerUserId="1161" LastEditorUserId="1161" LastEditDate="2017-01-09T14:24:35.167" LastActivityDate="2017-05-13T16:07:37.643" Title="CNC milling of large cutouts" Tags="|machining|safety|cnc|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13165" PostTypeId="2" ParentId="13164" CreationDate="2017-01-09T14:45:03.747" Score="5" Body="&lt;p&gt;I'm assuming that you are cutting on a base sheet of some sort in my answer, if not, please do. In all my shop applications I have found that double sided taping the product to the base sheet is sufficient when my machine doesn't have adequate suction.&lt;/p&gt;&#xA;" OwnerUserId="9433" LastActivityDate="2017-01-09T14:45:03.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13166" PostTypeId="2" ParentId="13155" CreationDate="2017-01-09T15:33:42.900" Score="0" Body="&lt;p&gt;If you look carefully at most fish tanks, the bottom glass is actually not touching the table underneath.  Disclaimer: I've only dealt with tanks up to about 30 gal.  The bottom is held up a small distance because it sits inside the bottom rectangle of the frame.&lt;br&gt;&#xA;Assuming your tanks are built similarly, you can easily see that the only thing which matters for the table is the strength of the outer frame, not the top itself. Despite what some comments say, I'd be surprised if those two-inch frame parts could not handle considerable load.  The legs are fine, since they only take a compressive load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running a test load w/ cinderblocks or pig iron will always outrank theoretical analysis, of course.  Alternatively, take a look at aquarium stands designed for such tanks and see what the dimensions of the metal parts are (and look at the L- or I- shape of the beams). &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-01-09T15:33:42.900" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13167" PostTypeId="2" ParentId="13162" CreationDate="2017-01-09T17:33:45.590" Score="1" Body="&lt;p&gt;The most likely point of failure is tearing out the female thread so the strength of the aluminium will be the limiting factor. For more details see &lt;a href=&quot;http://www.calqlata.com/proddetail.asp?prod=00016&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual torque which equates to a certain pull-out force will depend on both the tensile strength of the aluminium (which will depend on the alloy and temper condition) and the coefficient of friction between the two, which itself will depend on the quality of the threads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that torque, on it's own, isn't a great way to prevent  self unscrewing and if this is an issue some sort of self locking fastener or locking adhesive are usually required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it is often better when using threaded fasteners into softer materials or into a cast or machined part to use studs secured with locking adhesive and apply the torque with a nut as this significantly reduces the risk of damaging the thread. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thread locking compound will also help to mitigate galvanic corrosion between dissimilar metals. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-01-09T17:33:45.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13168" PostTypeId="2" ParentId="13151" CreationDate="2017-01-09T19:25:24.980" Score="3" Body="&lt;p&gt;For accurate cost estimation you will need to find a vendor and have them supply you a &quot;budgetary quote&quot; for your 700 kW boiler. The vendor can usually get this type of quote to you more quickly than a firm quote because they are not held to that price or terms; though I find that frequently the price is the same when we go to purchase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I need an estimate in the next 10 minutes I try to find something on the internet with a price that I can scale. I find a unit closest to the type and capacity I need. Usually economy of scale ensures that I over estimate, but it doesnt hurt to give yourself some safety factor.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;http://www.supplyhouse.com/Lochinvar-KBN701-660100-BTU-Knight-XL-High-Efficiency-Boiler&quot; rel=&quot;nofollow noreferrer&quot;&gt;574,000 BTU Boiler&lt;/a&gt; for 14,000USD&lt;br&gt;&#xA;  700 kW is 2,390,000 BTU&lt;br&gt;&#xA;  2,390,000 BTU / 574,000 BTU  = 4.16&lt;br&gt;&#xA;  4.16 * 14,000 USD = 58,240 USD&lt;br&gt;&#xA;  58,240 USD * 1.2 safety factor = 70,000 USD&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Obviously you will have to do this for all your major components, piping, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For installation costs you typically estimate some number of labor hours and multiply by your &quot;shop rate.&quot; Accounting should be able to give you this number as it factors in all employee costs like benefits, vacation, training, tool overhead costs, etc.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="368" LastEditDate="2017-01-09T23:03:06.137" LastActivityDate="2017-01-09T23:03:06.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13170" PostTypeId="1" CreationDate="2017-01-09T21:57:25.823" Score="2" ViewCount="830" Body="&lt;p&gt;I need to join a half-inch motor shaft to a gearbox with a 14 mm input (the input is not a male shaft, but a female collar with a 14mm hole). I'm considering using a sleeve with a wall thickness of 0.026&quot; (0.6604 mm) but not sure it will be strong enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it &lt;em&gt;will&lt;/em&gt; work to sleeve this, what material should I make the sleeve out of?&lt;/p&gt;&#xA;" OwnerUserId="9534" LastEditorUserId="9534" LastEditDate="2017-01-10T15:31:26.847" LastActivityDate="2017-01-10T21:45:37.567" Title="How can I join a 1/2&quot; motor shaft to a gearbox with a 14 mm input?" Tags="|mechanical-engineering|materials|motors|design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13172" PostTypeId="1" AcceptedAnswerId="13176" CreationDate="2017-01-09T23:41:47.927" Score="0" ViewCount="200" Body="&lt;p&gt;I'm wondering what causes the price difference between a &lt;a href=&quot;http://www.cablewholesale.com/products/serial-modem-cable/modular-adaptors/product-31d3-37200.php?utm_source=GoogleShopping&amp;amp;utm_medium=cpc&amp;amp;utm_term=31D3-37200&amp;amp;utm_campaign=Modular%20Adapter%2C%20Gray%2C%20DB25%20Male%20to%20RJ45&amp;amp;gclid=Cj0KEQiAhs3DBRDmu-rVkuif0N8BEiQAWuUJr6FoZ_2-R_gjiAJBk-9abyF9pj5E38frxKYkp7pGCTwaAgqD8P8HAQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Serial-to-RJ45 adapter&lt;/a&gt; and a &lt;a href=&quot;https://gridconnect.com/uds1100.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Serial-to-Ethernet server&lt;/a&gt;. Given a serial-interfaced instrument, can one use either to connect the instrument to a computer? (Of course, the first must be wire correctly first).&lt;/p&gt;&#xA;" OwnerUserId="9535" LastActivityDate="2017-01-10T03:38:35.513" Title="Difference between Serial-to-RJ45 adapter and Serial-to-Ethernet server" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13173" PostTypeId="1" CreationDate="2017-01-10T01:59:19.270" Score="1" ViewCount="10430" Body="&lt;p&gt;Is this equation valid for a given mechanical system&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wgcwr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgcwr.png&quot; alt=&quot;moto&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Inertia of the motor is $J_m$,there is no loss in the system.then conservation of energy can be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the motor is revolving with speed $\omega$.and the gears ratio are equal.&#xA;$v $ is the velocity of the weight $w$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking that the can the total InertIa can be expressed as  $$\frac{1}{2}J'_m\omega^2=\frac{1}{2}J_m\omega^2+\frac{1}{2}mv^2$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;equivalent moment of inertia referred to the motor shaft is $J'_m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was told by my teacher I think there is something wrong here.&#xA;I am skeptical about this because we can't talk about adding the both terms and referring them to as total inertia referred to motor shaft.&lt;strong&gt;BOOK SOLUTION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pgzMl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pgzMl.png&quot; alt=&quot;jjl&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YFCKi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YFCKi.png&quot; alt=&quot;enterhere&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you might be saying is &lt;/p&gt;&#xA;" OwnerUserId="8808" LastEditorUserId="8808" LastEditDate="2017-01-10T13:46:30.500" LastActivityDate="2017-01-10T21:24:09.123" Title="inertia referred to motor shaft" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13175" PostTypeId="2" ParentId="13173" CreationDate="2017-01-10T03:11:27.627" Score="2" Body="&lt;p&gt;Your teacher is right. This is called equivalent mass moment of inertia, and it is often used in system dynamics problems to simplify things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case the equivalent inertia $J_m^\prime$ is given by the equation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J_m^\prime = \left[\frac{1}{2}J_m \omega^2 + \frac{1}{2} m v^2\right]\left[\frac{2}{\omega^2}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J_m^\prime = J_m + m \left(\frac{v}{\omega}\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are confused, think of it this way: inertia is really a measure of how much energy it would take to slow a moving object to a stop. Note that this only works if there is a lossless gearbox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are still skeptical, try determining the torque the motor must output for a given angular acceleration $\alpha$. You should find that the $T_m = J_m \alpha + rma/n$, where $r$ is the radius of the drum and $n$ is the gear ratio. Substituting $\alpha r/n$ for $a$ and dividing both sides by $\alpha$ yields $$\frac{T_m}{\alpha} = J_m + m \left(\frac{r}{n}\right)^2$$ which is equivalent to the expression we found above.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2017-01-10T03:11:27.627" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13176" PostTypeId="2" ParentId="13172" CreationDate="2017-01-10T03:38:35.513" Score="2" Body="&lt;p&gt;To answer your first question: the Serial-to-RJ45 adapter is just wires, the Serial-to-Ethernet server is a computer in a box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't answer your second question without knowing more about your equipment, but perhaps this information will help you.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Serial-to-RJ45 adapter&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This adapter to me looks like it is designed to permit users to use inexpensive ethernet cables in place of relatively rare serial cables. In order for it to work, you need to have a serial port on your computer. The ethernet jack on your PC will not output serial signals and this adapter will not turn them into serial signals.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Serial-to-Ethernet server&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is a small computer which connects to your local network and has a serial port which you can connect equipment to. It comes with software that allows you to use it as if your computer had the serial port itself.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Other notes&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If your computer doesn't have a serial port, USB-to-serial adapters are available very cheaply&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are a lot of different serial communication protocols. Generally &quot;serial&quot; on an instrument means RS-232 or RS-485, but could be something different. Make sure you understand the requirements of your equipment before buying something that may not work.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="709" LastActivityDate="2017-01-10T03:38:35.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13178" PostTypeId="2" ParentId="13163" CreationDate="2017-01-10T07:32:35.967" Score="3" Body="&lt;p&gt;To change varying AC to fixed AC .for example the alternator that is attached to the engine will  output varying AC voltage, whose frequency would be different from the operating frequency or rated values of the drive.so to get a rated values accordingly to drive AC-DC-AC converter are used. &lt;/p&gt;&#xA;" OwnerUserId="8808" LastActivityDate="2017-01-10T07:32:35.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13180" PostTypeId="2" ParentId="13164" CreationDate="2017-01-10T10:59:12.990" Score="2" Body="&lt;p&gt;If the alloy is magnetic then you can use a magnet as the cut is nearly complete to fix the cut out part in place as the cut finishes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the CNC doesn't allow that then you can instead let a few tabs remain and cut them out on another machine where the center part is properly held in place.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-01-10T10:59:12.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13181" PostTypeId="2" ParentId="13170" CreationDate="2017-01-10T14:40:29.903" Score="2" Body="&lt;p&gt;Why would you make anything? Shaft couplings are readily available off the shelf, including inch to metric versions? The exact variety of coupling will depend far more on how precisely the two shafts are aligned, speed, torque and tolerable backlash than on the small size mismatch you have.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edited in light of the comments below&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you need one of two things; a reducer bushing/bore reducer or a shaft adapter &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reducer bushing is a sleeve much as you propose with the exception that it has a &quot;C&quot; shaped cross section to allow for tolerances and the shaft key. I've only used them with mating a metric shaft to a inch bore, but I'm sure you can google something. Worst case is you have a custom one made. They're typically brass or aluminum. The torque should &lt;strong&gt;not&lt;/strong&gt; be transmitted through the adapter. If you have a keyed shaft, you'll need a special step key, which you can buy or if you have a mill, they're simple to make from key stock for the larger of the two slots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have axial space, you can get a 1/2&quot; to 14mm shaft adapter which has a female side that accepts your 1/2&quot; shaft and a short 14mm shaft on the other side. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2017-01-10T21:45:37.567" LastActivityDate="2017-01-10T21:45:37.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13182" PostTypeId="2" ParentId="13173" CreationDate="2017-01-10T18:23:54.890" Score="2" Body="&lt;p&gt;Ignoring the inertia in the gears and drum, assuming a gear ratio of 1, and denoting the radius of the drum by $r$, the total kinetic energy of the system is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T=\frac{1}{2}J_m\omega_m^2+\frac{1}{2}m v^2=\frac{1}{2}J_m\omega_m^2+\frac{1}{2}m r^2\omega_m^2=\frac{1}{2}(J_m+mr^2)\omega_m^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the equivalent moment of inertia is $J_m+mr^2$.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2017-01-10T21:24:09.123" LastActivityDate="2017-01-10T21:24:09.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13183" PostTypeId="1" CreationDate="2017-01-10T20:10:52.937" Score="5" ViewCount="1707" Body="&lt;p&gt;I want to build power meter for wifi signals. One way to do it is to use a diode to rectify AC signal and amplify that before measuring. But are there diodes which work at such high frequency? Also, are there other methods?&lt;/p&gt;&#xA;" OwnerUserId="9542" LastEditorUserId="8055" LastEditDate="2017-01-11T17:20:24.903" LastActivityDate="2020-04-26T07:03:50.370" Title="Building RF power meter" Tags="|electrical-engineering|rf-electronics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13185" PostTypeId="2" ParentId="10470" CreationDate="2017-01-10T21:58:00.270" Score="2" Body="&lt;p&gt;The signals mean exactly what you think they do:  They tell the drivers of trains when it is safe to proceed, and often, how fast is safe.  Everyone knows that trains are heavy and can take a long time to stop.  Trains must be warned in advance long before they bump into one another so they can stop before that happens.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Railroad signals vary in many ways, even on the same railroad or the same route.  There are different classes of signals which serve different purposes.  The two main classes are &lt;em&gt;absolute&lt;/em&gt; and &lt;em&gt;permissive&lt;/em&gt;.  Absolute signals require a train to stop and stay stopped when they are &quot;red.&quot;  Permissive signals allow a train to continue on at a slow, controlled speed.  In the United States, these signals are more-or-less identical in form and appearance, except that permissive signals have a &lt;em&gt;number plate&lt;/em&gt; indicating the track mile of the signal.  Railroad signals, especially in the United States, can have many varied appearances, even of the same class.  This is because signaling has evolved over the past 100+ years and different railroads and eras have held different views on how best to do it.  Because of this, signals have formed a part of many railroads' identities because are unique to that railroad.  Your description of the two different forms of signals near you sound just like you said, an implementation detail (of tradition).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am the author of the second link about signaling posted by alephzero above.  If you still have any more questions, feel free to ask.&lt;/p&gt;&#xA;" OwnerUserId="9545" LastActivityDate="2017-01-10T21:58:00.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13186" PostTypeId="1" CreationDate="2017-01-10T23:26:28.690" Score="1" ViewCount="62" Body="&lt;p&gt;Are there diodes which can switch at multiple GHz speed? Most of them are limited up to few hundreds of MHz.&lt;/p&gt;&#xA;" OwnerUserId="9546" LastActivityDate="2017-01-13T08:14:54.363" Title="Are there diodes which can switch at multiple GHz speed?" Tags="|rf-electronics|" AnswerCount="2" CommentCount="3" ClosedDate="2017-01-24T10:22:24.593" ContentLicense="CC BY-SA 3.0" />
  <row Id="13187" PostTypeId="1" CreationDate="2017-01-11T03:01:12.960" Score="0" ViewCount="76" Body="&lt;p&gt;There is no secondary containment to hold any potential shrapnel. How would this change the safety factor that is standard at 2.5? The consequences of this container exploding would be quite extreme as it is under high pressure.&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="90" LastEditDate="2017-01-11T11:12:40.147" LastActivityDate="2017-01-11T11:12:40.147" Title="Safety factor for pressurized container at 10 ATM" Tags="|pressure|measurements|pressure-vessel|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13188" PostTypeId="1" AcceptedAnswerId="13195" CreationDate="2017-01-11T04:32:09.900" Score="-1" ViewCount="2766" Body="&lt;p&gt;For this problem , i found that the internal force AB at point A and point B pointed in the same direction ( as shown) in my working , so , how they cancel off each other ???Since they cant cancel off each other , so they are not stable , right ? The structure is statically indeterminate ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/szLHe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/szLHe.jpg&quot; alt=&quot;22&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/OuMey.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OuMey.jpg&quot; alt=&quot;222&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ve1LE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ve1LE.jpg&quot; alt=&quot;2131&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="255" LastEditDate="2017-01-12T09:55:06.887" LastActivityDate="2017-01-12T09:55:06.887" Title="internal forces in truss" Tags="|structural-engineering|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13190" PostTypeId="2" ParentId="13183" CreationDate="2017-01-11T07:22:14.490" Score="1" Body="&lt;p&gt;The basic approach is correct, but it won't be as easy as just slapping a diode there, for several reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If you are receiving the signal with an antenna, the power level will be very small. A typical value might be -40dBm, which equals 0.1µW. In a 50 ohm load, this would be 2mV of voltage and 44µA of current.&lt;/li&gt;&#xA;&lt;li&gt;Diodes have forward voltage ranging from 200mV up to 1V, depending on diode type. You'd need to bias it somehow to avoid your 2mV signal getting completely cut off.&lt;/li&gt;&#xA;&lt;li&gt;Diodes have capacitance ranging from 1pF up to hundreds of pF. This capacitance will provide a path for the RF signal to get past your diode, no matter what polarity it is. So you won't get an exact cutoff but it will just conduct slightly better in one direction than the other.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So I suggest starting with a premade schematic which has these things thought out. Any such schematic would also indicate a suitable diode, for example this one uses MBD301:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FOjSt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FOjSt.png&quot; alt=&quot;Wattmeter&quot;&gt;&lt;/a&gt;&#xA;(Image from &lt;a href=&quot;http://www.qsl.net/n9zia/wireless/appendixF.html#10&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.qsl.net/n9zia/wireless/appendixF.html#10&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The MBD301 used there has a 15ps charge carrier lifetime, so in theory you could claim it can switch up to 66 GHz. But as I've explained above, for RF signals it is not so clear-cut. Therefore most datasheets won't give a figure for maximum switching frequency, as it depends on application.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2017-01-11T07:22:14.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13191" PostTypeId="1" CreationDate="2017-01-11T09:21:35.143" Score="1" ViewCount="650" Body="&lt;p&gt;I bought a 4&amp;nbsp;W Electromagnet to collect aluminium cans laying on the floor. A can weighs around 0.015&amp;nbsp;kg. I read aluminium is only paramagnetic. And playing manually with the current magnet, I cannot appreciate any force being exerted to the aluminium can. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How powerful should a magnet be to be able to pick the cans from the floor? &#xA;(let's say, from 10&amp;nbsp;cm away?)&lt;/p&gt;&#xA;" OwnerUserId="9551" LastEditorUserId="1832" LastEditDate="2017-01-11T10:41:29.043" LastActivityDate="2022-05-31T07:56:53.047" Title="How powerful should a magnet be to be able to pick the cans from the floor?" Tags="|electromagnetism|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13194" PostTypeId="1" CreationDate="2017-01-11T10:53:32.600" Score="1" ViewCount="86" Body="&lt;p&gt;I wonder, do fine and course screws have different diameter if they are the same standard size? For example, what is the outer diameter for 5/16″-18 and 5/16″-24 if I measure them with a caliper? Is it possible that a 5/16″-24 has a diameter of 0.3015inch diameter (which I measured)?&lt;/p&gt;&#xA;" OwnerUserId="8563" LastEditorUserId="2061" LastEditDate="2017-01-12T14:45:12.400" LastActivityDate="2017-01-12T21:19:32.430" Title="Do fine and coarse screws have different diameters?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13195" PostTypeId="2" ParentId="13188" CreationDate="2017-01-11T11:19:47.890" Score="3" Body="&lt;p&gt;As @AndyT stated, you need to add the horizontal reaction at A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You start by finding the reactions. Your global equilibrium equations are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= R_{A,x} + F = 0 \\&#xA;\therefore R_{A,x} &amp;amp;= -F = -4\text{ kN} \\&#xA;\sum F_y &amp;amp;= R_{A,y} + R_{B,y} = 0 \\&#xA;\therefore R_{A,y} &amp;amp;= -R_{B,y} \\&#xA;\sum M_A &amp;amp;= 8R_{B,y} - 3F = 0 \\&#xA;\therefore R_{B,y} &amp;amp;= \dfrac{3F}{8} = 1.5\text{ kN} \\&#xA;\therefore R_{A,y} &amp;amp;= -1.5\text{ kN}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, looking at the nodal equilibrium diagrams, we have for B:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7dHZy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7dHZy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before going into the math, it is worth noting that this truss is composed of two &lt;a href=&quot;https://en.wikipedia.org/wiki/Special_right_triangle#Common_Pythagorean_triples&quot; rel=&quot;nofollow noreferrer&quot;&gt;3-4-5 triangles&lt;/a&gt;, so we already know $\overline{AC}$ and $\overline{BC}$ are 5&amp;nbsp;m long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\overline{BC}$ is the only member that can resist the vertical reaction, and it does so by suffering compression equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum F_{B,y} &amp;amp;= R_{B,y} + \dfrac{3\overline{BC}}{5} = 0 \\&#xA;\therefore \overline{BC} &amp;amp;= -\dfrac{5R_{B,y}}{3} = -2.5\text{ kN (compression)} \\&#xA;\sum F_{B,x} &amp;amp;= \overline{AB} + \dfrac{4\overline{BC}}{5} = 0 \\&#xA;\therefore \overline{AB} &amp;amp;= -\dfrac{4\overline{BC}}{5} = 2\text{ kN (tension)} &#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then looking at A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G3NuB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G3NuB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum F_{A,y} &amp;amp;= R_{A,y} + \dfrac{3\overline{AC}}{5} = 0 \\&#xA;\therefore \overline{AC} &amp;amp;= -\dfrac{5R_{A,y}}{3} = 2.5\text{ kN (tension)} \\&#xA;\sum F_{A,x} &amp;amp;= R_{A,x} + \overline{AB} + \dfrac{4\overline{BC}}{5} = 0 \\&#xA;\therefore \overline{AB} &amp;amp;= -\dfrac{4\overline{BC}}{5} - R_{A,x} = 2\text{ kN (tension)} &#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both nodes agree that $\overline{AB}$ is in tension, so it checks out. Also, $\overline{AC}$ and $\overline{BC}$ have equal but opposite results, which is as it should be (and, if you actually do the work for node C, you'll see they are also correct).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-01-11T11:31:41.860" LastActivityDate="2017-01-11T11:31:41.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13196" PostTypeId="1" AcceptedAnswerId="13201" CreationDate="2017-01-11T14:05:31.257" Score="2" ViewCount="1156" Body="&lt;p&gt;For a beam supported at both ends with a mass in the middle, the flexural modulus is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_b = \frac{F L^3}{48I d}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with I as the area moment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \frac{1}{12}wh^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/87kWQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/87kWQ.png&quot; alt=&quot;Flexural modulus measurement&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the expression for modulus change if we have one fixed end as in the diagram below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kjN4R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kjN4R.png&quot; alt=&quot;scenario&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ja72 states &lt;a href=&quot;https://physics.stackexchange.com/a/279972/12568&quot;&gt;in this answer&lt;/a&gt; that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_b = \frac{F L^3}{3I d}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not clear on the derivation.&lt;/p&gt;&#xA;" OwnerUserId="9552" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2017-01-11T18:30:08.240" Title="Flexural modulus for a beam fixed at one end" Tags="|structural-engineering|structural-analysis|elastic-modulus|moments|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13197" PostTypeId="1" CreationDate="2017-01-11T14:24:04.600" Score="-1" ViewCount="4138" Body="&lt;p&gt;I am trying to solve the following problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;EXAMPLE 2.10&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;Determine the reactions on the beam in Fig. 2–30&lt;em&gt;a&lt;/em&gt;. Assume $B$&#xA;  is a pin and the support at $B$ is a roller (smooth surface).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I30zQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I30zQ.jpg&quot; alt=&quot;A bent beam is supported on the left at $A$. The beam extends 7 ft rightward and then bends sharply upward. The section to the right of the bend extends 3 ft horizontally and 4 ft vertically to a support at $B$. The horizontal section to the left of the bend is subject to a uniform load of 500 lb/ft in the downward direction.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;h3&gt;SOLUTION&lt;/h3&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Free-Body Diagram.&lt;/strong&gt; As shown in Fig. 2–30&lt;em&gt;b&lt;/em&gt;, the support (&quot;roller&quot;) at $B$ exerts a &lt;em&gt;normal force&lt;/em&gt; on the beam at its point of&#xA;  contact. The line of action of this force is defined by the 3-4-5&#xA;  triangle.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0tgjt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0tgjt.jpg&quot; alt=&quot;Free body diagram of the beam from previous image. The uniform load is replaced by a downward point load of 3500 lb, shown as acting 3.5 ft to the right of $A$. The reactions at $A$ are decomposed into components $\mathbf{A}_x$ and $\mathbf{A}_y$, acting rightward and upward, respectively. A single reaction $\mathbf{N}_B$ is shown acting from below-right at $B$, perpendicular to that section. The horizontal distance between the substituted point load and the reaction at $B$ is shown as 6.5 ft. The sharp bend in the beam is circled in red for emphasis.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Equations of Equilibrium.&lt;/strong&gt; Resolving $\mathbf{N}_B$ into &lt;em&gt;x&lt;/em&gt; and &lt;em&gt;y&lt;/em&gt; components and summing moments around $A$ yields a direct solution for $N_B$. Why? Using this result, we can then obtain $A_x$ and $A_y$.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$\begin {align} \\ \Sigma M_A = 0; \quad &amp;amp; -3500 (3.5) + (\frac{4}{5}) N_B (4) + (\frac{3}{5}) N_B (10) = 0 \\ &amp;amp; \qquad N_B = 1331.5\ \mathrm{lb} = 1.33\ \mathrm{k} \\ \Sigma F_x = 0; \quad &amp;amp; A_x - \frac{4}{5}(1331.5) = 0 &amp;amp; A_x = 1.07\ \mathrm{k}\\ \Sigma F_y = 0; \quad &amp;amp; A_y - 3500 + \frac{3}{5}(1331.5) = 0 &amp;amp; A_x = 2.70\&#xA;\mathrm{k}\\ \end {align}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For the moment about $A$, why does the reaction at $B$ produce a moment of $(\frac{4}{5})N_B(4)$? I think it's not necessary to include that moment because the moment arm $d = 0.4\ \mathrm{m}$ is not measured directly from point A, but from the bend in the beam, which I circled red in figure (b).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qRZPU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qRZPU.jpg&quot; alt=&quot;opp&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="8664" LastEditDate="2017-01-12T00:24:17.407" LastActivityDate="2017-01-12T00:57:43.077" Title="Calculating the moments due to reaction forces on a bent beam" Tags="|structural-engineering|statics|moments|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13198" PostTypeId="2" ParentId="13196" CreationDate="2017-01-11T14:38:43.433" Score="2" Body="&lt;p&gt;'I' doesn't change for the different situations.  'I' is a property of the cross-section of the beam - a rectangle of width w and depth h. The Eb equations are correct for the situations set out, with the following caveats:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The first situation is not for a beam that is fixed at both ends, it&#xA;is for a beam that is supported at both ends but not 'fixed'. &#xA;'Fixed' implies a moment continuity, in which case the '48' in the&#xA;equation would be '192'.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second situation needs the end condition to be fixed - if it is&#xA;supported but not fixed (which is what is shown in the diagram), it&#xA;is a mechanism.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="902" LastActivityDate="2017-01-11T14:38:43.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13199" PostTypeId="2" ParentId="13197" CreationDate="2017-01-11T15:59:53.763" Score="3" Body="&lt;p&gt;The reason is that both the horizontal and the vertical component of $N_B$ generate a moment around $A$. Since the moment due to a force is equal to the product of the force and the perpendicular lever-arm between the force's &lt;a href=&quot;https://en.wikipedia.org/wiki/Line_of_action&quot; rel=&quot;nofollow noreferrer&quot;&gt;line of action&lt;/a&gt; and the point being considered, this therefore means that the total moment around $A$ is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;N_{B,x} &amp;amp;= \dfrac{4}{5}N_B \text{ (horizontal component)} \\&#xA;N_{B,y} &amp;amp;= \dfrac{3}{5}N_B \text{ (vertical component)} \\&#xA;M_A &amp;amp;= -3500 \cdot 3.5 + 4N_{B,x} + 10N_{B,y} = 0&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all, the horizontal distance from $A$ to $B$ is 10&amp;nbsp;ft, so that's the lever arm for the vertical component, while the vertical distance is 4&amp;nbsp;ft, so that's used for the horizontal component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hoping to make this as clear as possible, here's a diagram of the forces, their lines of action and the respective perpendicular distances. They are color-coded: everything relevant for the horizontal component is green, everything for the vertical component is red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zyzFB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zyzFB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-01-12T00:57:43.077" LastActivityDate="2017-01-12T00:57:43.077" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13200" PostTypeId="1" AcceptedAnswerId="13202" CreationDate="2017-01-11T17:54:15.843" Score="1" ViewCount="308" Body="&lt;p&gt;Can tensile strength be the same as yield strength? Can Flexural Yield Strength be related to the yield strength of a material in any way?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2017-01-11T19:35:07.923" Title="Definitions of strength?" Tags="|strength|yield-point|unit|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13201" PostTypeId="2" ParentId="13196" CreationDate="2017-01-11T17:59:37.910" Score="3" Body="&lt;p&gt;The fundamental beam equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\text{d}^2}{\text{d}x^2}\left(EI\dfrac{\text{d}^2w}{\text{d}x^2}\right) = q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which basically translates to &quot;the fourth derivative of the deflection function is equal to the applied load&quot;. In fact&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the first derivative is the tangent of the deflection, which for small angles is approximately equal to the angle of deflection&lt;/li&gt;&#xA;&lt;li&gt;the second derivative is the bending moment&lt;/li&gt;&#xA;&lt;li&gt;the third derivative is the shear force&lt;/li&gt;&#xA;&lt;li&gt;the fourth derivative (to repeat myself), is the applied load.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All deflection results are obtained using this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To simplify this answer, I'm going to start off from the second derivative, the bending moment, since it (and the ones after it) is trivial to find by inspection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the simply-supported beam with a concentrated load at midspan, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;M &amp;amp;= \begin{cases}&#xA;\dfrac{Fx}{2} &amp;amp;\text{ for } x \in [0,\dfrac{L}{2}] \\&#xA;\dfrac{F(L-x)}{2} &amp;amp;\text{ for } x \in [\dfrac{L}{2}, L] \\&#xA;\end{cases} \\&#xA;EI\theta = \int M\text{d}x &amp;amp;= \begin{cases}&#xA;\dfrac{Fx^2}{4} + C_1 &amp;amp;\text{ for } x \in [0,\dfrac{L}{2}] \\&#xA;\dfrac{FLx}{2} - \dfrac{Fx^2}{4} + C_2 &amp;amp;\text{ for } x \in [\dfrac{L}{2}, L] \\&#xA;\end{cases} \\&#xA;EI\delta = EI\int \theta\text{d}x &amp;amp;= \begin{cases}&#xA;\dfrac{Fx^3}{12} + C_1x + C_3 &amp;amp;\text{ for } x \in [0,\dfrac{L}{2}] \\&#xA;\dfrac{FLx^2}{4} - \dfrac{Fx^3}{12} + C_2x + C_4 &amp;amp;\text{ for } x \in [\dfrac{L}{2}, L] \\&#xA;\end{cases}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $\delta(0) = \delta(L) = 0$ and that $\theta\left(\dfrac{L}{2}^+\right) = \theta\left(\dfrac{L}{2}^-\right)$ and $\delta\left(\dfrac{L}{2}^+\right) = \delta\left(\dfrac{L}{2}^-\right)$ (that is, the deflection and angle is continuous at $\dfrac{L}{2}$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you solve it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;\delta(0) = C_3 = 0 \\&#xA;\delta(L) = \dfrac{FL^3}{4} - \dfrac{FL^3}{12} + C_2L + C_4 = 0 \\&#xA;\therefore C_4 = -\dfrac{FL^3}{6} - C_2L \\&#xA;\theta\left(\dfrac{L}{2}^+\right) = \theta\left(\dfrac{L}{2}^-\right) \\&#xA;\therefore \dfrac{FL^2}{16} + C_1 = \dfrac{FL^2}{4} - \dfrac{FL^2}{16} + C_2 \\&#xA;\therefore C_1 = \dfrac{FL^2}{8} + C_2 \\&#xA;\delta\left(\dfrac{L}{2}^+\right) = \delta\left(\dfrac{L}{2}^-\right) \\&#xA;\therefore \dfrac{FL^3}{96} + \dfrac{FL^3}{16} + \dfrac{C_2L}{2} = \dfrac{FL^3}{16} - \dfrac{FL^3}{96} + \dfrac{C_2L}{2} - \dfrac{FL^3}{6} - C_2L \\&#xA;\therefore C_2 = -\dfrac{9FL^2}{48} \\&#xA;\therefore C_1 = -\dfrac{3FL^2}{48} \\&#xA;\therefore C_4 = \dfrac{FL^3}{48}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, by inspection we can easily see the deflection is at the midspan, so let's calculate it (doesn't matter which $\delta$ equation you choose).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta\left(\dfrac{L}{2}\right) = \dfrac{1}{EI}\left(\dfrac{FL^3}{96} - \dfrac{3FL^3}{96}\right) = -\dfrac{FL^3}{48EI}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The same process can be repeated for a cantilevered beam, only it's much simpler:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;M &amp;amp;= FL - Fx \\&#xA;EI\theta = \int M\text{d}x &amp;amp;= FLx - \dfrac{Fx^2}{2} + C_1 \\&#xA;EI\delta = EI\int \theta\text{d}x &amp;amp;= \dfrac{FLx^2}{2} - \dfrac{Fx^3}{6} + C_1x + C_2 \\&#xA;\theta(0) &amp;amp;= C_1 = 0 \\&#xA;\delta(0) &amp;amp;= C_2 = 0 \\&#xA;\therefore EI\delta &amp;amp;= \dfrac{FLx^2}{2} - \dfrac{Fx^3}{6} \\&#xA;\therefore \delta(L) &amp;amp;= \dfrac{FL^3}{3EI}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-01-11T17:59:37.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13202" PostTypeId="2" ParentId="13200" CreationDate="2017-01-11T18:59:09.950" Score="4" Body="&lt;p&gt;Strength can mean different things in different contexts and technical definitions can vary significantly from what is commonly understood by the word 'strength'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For structures where actual forces are more relevant you might talk about rated loads, safe working loads or design loads in conjunction with factors of safety but this usually needs to be qualified by specific details of how the load is applied and is much less generally applicable than stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of yield it is generally more precise to use the term yield stress. This is a numerical value which applies to a particular material. For example Ultimate Tensile Stress is a common indicator of the resistance of a material to failure. For many ductile materials eg steel UTS is somewhat higher than yield stress, usually by some concistant factor for a given material. For example metric bolt grades are defined by UTS and yield stress as a factor of UTS eg for an 8.8 bolt UTS is 800 MPa and yield stress is 80% of that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also crucial is how failure is defined in a specific context and there are quite a few possible modes of failure. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Yield : the point at which plastic deformation begins to occur. The material remains deformed after the load is removed but may still be capable of supporting a significant load. &lt;/li&gt;&#xA;&lt;li&gt;Fracture : the material 'breaks' and is no longer able to support any meaningful load. &lt;/li&gt;&#xA;&lt;li&gt;Excess strain : the material deflects elastically to the point where it is no longer able to carry out its intended function. This does not necessarily imply permanent damage or deformation to the material. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note also that we tend to use different terms for material properties and structural properties, ie rigidity for a material vs stiffness for a structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When discussing flexing it is important to maintain a clear distinction between stiffness/rigidity and strength. Where stiffness is  the degree to which a material deforms under a certain load expressed as a ratio (strain) and strength is the load it can resist per unit area without failing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally the stresses experienced by a part are as much a function of its geometry as its material properties. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For homogeneous materials (including most engineeriong metal alloys) yield stress is a pretty generally applicable figure, what is more important is the model used to determine stresses as even a simple applied load can result in complex internal induced stresses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However for certain other materials such as many natural materials, textiles and composites the situation becomes rather more complex as both strength and stiffness can be directional. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also nominal values for stress are generally based on static loads and apparent material properties can change significantly with strain rate or loads applied over long periods of time. Some materials may also be subject to fatigue failure under cyclic loads well below the static yield stress.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-01-11T19:35:07.923" LastActivityDate="2017-01-11T19:35:07.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13203" PostTypeId="1" AcceptedAnswerId="13213" CreationDate="2017-01-11T19:33:37.457" Score="4" ViewCount="854" Body="&lt;p&gt;Few months ago I opened the boards of an old, scrap radio-cum-tape-recorder (Philips DR183); whose tuning knob (that adjusts the gang); was very peculiar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;( * I did not found its datasheet on web)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Xklj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Xklj.png&quot; alt=&quot;inside&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The channel-indicator is a narrow rod at the tip of red plastic-thing. That indicator moves along a straight-linear segment when I turns the wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ytcxu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ytcxu.png&quot; alt=&quot;what its work look like&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is just a rough sketch.  &lt;em&gt;I could not figure out, what mathematical principle/ geometric theorem being used in this mechanic. It is neither like pulley and rope; and nor-like a piston-with-crankshaft. &lt;strong&gt;My question is&lt;/strong&gt;, on what-basis they have chosen the angles, distances and ratios to make such movements? &#xA; Or, For say if I try to make one with cardboard etc; what angles, lengths, ratios etc I have to choose, and in which logical basis?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I've provided some of its allowed conformations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0JrEJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0JrEJ.jpg&quot; alt=&quot;Steps&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please feel free to move this question if it is more appropriate for any other SE site, such as &lt;a href=&quot;https://math.stackexchange.com/&quot;&gt;mathematics.SE&lt;/a&gt; . &lt;/p&gt;&#xA;" OwnerDisplayName="user8831" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:37.633" LastActivityDate="2017-01-12T15:21:28.780" Title="How this radio knob converting a rotation into a straight-linear motion?" Tags="|mechanisms|geometry|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13204" PostTypeId="1" AcceptedAnswerId="13229" CreationDate="2017-01-11T19:36:04.810" Score="1" ViewCount="2792" Body="&lt;p&gt;I've been making technical drawings for various parts I've designed. Since these parts were meant to be 3D printed, this is mainly just to have them in my portfolio so I have something to show/talk about at the upcoming career fair (I'm an undergraduate ME student).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've encountered difficulty with one particular part, which is designed to hold the motor for a perpetually spinning top. Because the motor ran a little hot when turning the rather heavy disc it was attached to, I used the SolidWorks wrap tool to remove portions of the cylindrical casing in order to encourage heat dissipation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ko1FR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ko1FR.png&quot; alt=&quot;Technical drawing of a motor holder.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see in the image, SolidWorks is unable to annotate the resulting dimensions, at least not using the basic tools I'm familiar with. So my question is: how would the dimensions of this part be represented on a technical drawing? Relatedly, since it probably determines the answer: by what process could a part like this be machined, if 3D printing was not an option and the arguably unnecessary complexity could not be removed?&lt;/p&gt;&#xA;" OwnerUserId="9311" LastActivityDate="2017-01-18T20:15:55.917" Title="Technical drawing for a part with complex (non-line, non-arc) geometry" Tags="|solidworks|drawings|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13205" PostTypeId="2" ParentId="13194" CreationDate="2017-01-11T20:39:37.597" Score="2" Body="&lt;p&gt;The Basic Major Diameter for both 5/16&quot; -18 UNC and 5/16&quot;-24 UNF threads is 0.3125&quot;. The depth of the threads and thus the pitch diameters are different 0.2764&quot; for UNC and 0.2854&quot; for UNF. The limits for those diameters will depend on the tolerance class for the bolts in question. The &lt;strong&gt;minimum&lt;/strong&gt; Major Diameter for a  5/16&quot; UNC bolt varies with its tolerance class as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Min Maj Dia for 5/16 - 18 UNC Class 1A = 0.2982&quot; &lt;/li&gt;&#xA;&lt;li&gt;Min Maj Dia for 5/16 - 18 UNC Class 2A = 0.3026&quot;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Min Maj Dia for 5/16 - 18 UNC Class 3A = 0.3038&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Min Maj Dia for 5/16 - 24 UNF Class 1A = 0.3006&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Min Maj Dia for 5/16 - 24 UNF Class 2A = 0.3042&quot;&lt;/li&gt;&#xA;&lt;li&gt;Min Maj Dia for 5/16 - 24 UNF Class 3A = 0.3053&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, while the Basic Major Diameter of both coarse and fine thread 5/16&quot; fasteners is the same, in reality the diameters can certainly be different within the allowable tolerances. The diameter of the fastener which you measured at 0.3015&quot; fits within the acceptable range for Class 1A fasteners. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.efunda.com/designstandards/screws/unified.cfm?start=64&amp;amp;finish=147&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reference Table including Tolerance Classes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2017-01-12T21:19:32.430" LastActivityDate="2017-01-12T21:19:32.430" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13206" PostTypeId="1" AcceptedAnswerId="13226" CreationDate="2017-01-11T23:05:47.460" Score="1" ViewCount="2165" Body="&lt;p&gt;I need a pump that can remove water from certain volume (~300ml). Water mixed with washing agent (soap) is pumped into this volume using another pump ~5-20L/minute. This volume is not completely sealed - air can get inside (sometimes easily, sometimes in very limited quantities). I need to guarantee that water will not spill out of this volume. In my understanding I need a water transfer pump, but such that can run dry for a while and work as vacuum pump. I looked what market has to offer, but didn't find anything like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is - does such pump exist at all? If so, how it is called, so I can search for it online?&lt;/p&gt;&#xA;" OwnerUserId="9555" LastActivityDate="2017-02-14T23:06:37.870" Title="Transfer pump suitable for air, water and their mixture" Tags="|pumps|vacuum-pumps|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13208" PostTypeId="1" AcceptedAnswerId="13233" CreationDate="2017-01-12T01:45:53.300" Score="4" ViewCount="195" Body="&lt;p&gt;I imagine this is an easy concept but I do not know much about mechanical engineering design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to make a &quot;knob&quot; like a control knob in solidworks that turns and is locked in place after twisting.  I am thinking of a twisting knob in a labyrinth game or something like a foosball table knob that is locked in place after turning, or just a general control knob on an oven.  I want to figure out how to design the &quot;interface&quot; that lets the knob turn yet stay in place after turning but do not know enough vocabulary to point myself in the right direction.  Does the &quot;interface&quot; I am looking for have a name? How does one go about making it?&lt;/p&gt;&#xA;" OwnerUserId="9556" LastActivityDate="2017-01-13T08:46:22.750" Title="Make something that turns yet locked in place" Tags="|mechanical-engineering|solidworks|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13209" PostTypeId="2" ParentId="13204" CreationDate="2017-01-12T04:42:42.233" Score="2" Body="&lt;p&gt;You probably need some auxiliary sections as well as the three basic views.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if you draw a section at mid-height looking down the axis, you can easily dimension the &quot;straight&quot; ribs. If you draw a section normal to one of the curved ribs, you can dimension the cross section of that rib, and presumably they are all the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the dimensions of the ribs don't look very critical (presumably only the top part is actually load-carrying) just conveying the message that &quot;this rib follows the contours of the cylinder and its cross section is approximately $x$ inches square&quot; is probably as much information as anyone needs in real life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you really want it to look like that for aesthetic reasons, a way to machine something with the same functionality would be make to a hollow cylinder, and then drill a pattern of holes through it radially. Since the cylinder is thin, it might be easier to drill a pattern of radial blind holes into the solid cylinder before you bore out the interior. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to machine something that looks more like your 3D printed version, you could mill triangular holes with fillet radii in the corners, instead of drilling circular holes. (The fillets radii would be the same size of the milling tool, of course.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With either of those options, it's more obvious what you need to include on the drawing - the purpose of the auxiliary views and dimensions is to tell the operator how to set up the machine tool, and the tool itself will &quot;automatically dimension&quot; everything else.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-01-12T04:42:42.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13211" PostTypeId="1" CreationDate="2017-01-12T13:26:16.177" Score="2" ViewCount="222" Body="&lt;p&gt;I am trying to understand the difference between Latches and Flip-Flops, as an example: RS-Latch and RS-Flip-Flop. In my lecture script the only difference is that latches are level triggered and Flip-Flops are edge triggered, so far now problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one task we had to build an RS-Flip-Flop with Nor gates, could someone give me an example of it? Because in the Solutions there is a picture of an RS-Latch (the one you can see on wikipedia), but latches are edge-triggered, so we can't build an Flip-Flop by using just two nor gates, can we?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="9561" LastEditorUserId="8055" LastEditDate="2017-01-12T19:42:43.040" LastActivityDate="2017-01-12T19:42:54.980" Title="Difference Latches and Flip-Flops" Tags="|electrical-engineering|computer-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2017-01-16T13:42:03.833" ContentLicense="CC BY-SA 3.0" />
  <row Id="13212" PostTypeId="2" ParentId="13211" CreationDate="2017-01-12T13:47:00.183" Score="1" Body="&lt;p&gt;The main difference latches and flip-flops are that former are level triggered that is once the latch is enabled the change in inputs can show change in output after latch is disabled the values are fixed, the latches are edge triggered that is when the clock pulse start rising,or falling the output across the latch changes if there is change in input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For making a SR flip flop you should first make the table and draw the K map of the outputs and then you can use NOR gates to solve the problem.&#xA; &lt;a href=&quot;https://i.stack.imgur.com/jekFa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jekFa.png&quot; alt=&quot;vjh&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cs.ucr.edu/~ehwang/courses/cs120b/flipflops.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Latches and Flip-Flops&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8808" LastEditorUserId="8055" LastEditDate="2017-01-12T19:42:54.980" LastActivityDate="2017-01-12T19:42:54.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13213" PostTypeId="2" ParentId="13203" CreationDate="2017-01-12T14:14:47.500" Score="5" Body="&lt;p&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Chebyshev&amp;#39;s_Lambda_Mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chebyshev Link Motion&lt;/a&gt;. (cognate form known as the Lambda Motion) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a classic, also sometimes seen in its &lt;a href=&quot;https://en.wikipedia.org/wiki/Chebyshev_linkage&quot; rel=&quot;nofollow noreferrer&quot;&gt;original form&lt;/a&gt; using circular motion to provide a 6th order approximation to a straight line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crucial dimensions are in a 2:4:5 ratio (or 1:2:2.5 in the animation linked above)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an improvement on the earlier (by about a century!) James Watt &lt;a href=&quot;https://en.wikipedia.org/wiki/Watt&amp;#39;s_linkage&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parallel Link motion&lt;/a&gt;, which provides a fifth order approximation - and was the invention he was most proud of - as well as sparking the whole field of four-bar linkages as field of systematic study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some time ago I simulated these linkages ( in a spreadsheet) and plotted the errors - the Chebyshev, designed for 1 metre of linear travel, gives an equi-ripple error reaching 0.6mm off the straight line. (For other travels, the error scales accordingly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can actually tune the Watt linkage's error to be even lower than this, despite it being only 5th order - if you have space for slightly longer links.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysis with a spreadsheet isn't particularly difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start with a fixed point at some coordinate, X1 = (0,0) and a length, say, R=5 and use the equation of rotary motion (X^2 + Y^2 = R^2) to plot multiple values of Y1 at different angles of rotation. From a second fixed point, say, X2 = (4,0) and R2 = 5, find values of Y2 such that Y2 is also a fixed distance (R3 = 2) away from each Y position. This turns out to be solving quadratics, and not all angles have valid roots. Finally plot the midpoint of the Y-Y2 bar and if you've chosen the dimensions above you have the regular Chebyshev link motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll see that cancellation of two rotary motions is at the heart of it, but the rotation of the third link introduces complications and the devil is in the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the other answer says, synthesis is more &quot;involved&quot; - to the extent that some of the world's top mathematicians devoted decades of their lives to it, and as far as I know, synthesising any motion you might want is not yet a solved problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is related to the use of Chebyshev's polynomials in electronic filter design, where equi-ripple errors on a straight line passband (or stopband) are also important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally note that the straight line portion is also practically linear in distance per degree rotation (with a relatively fast transit through the curved part of the path). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to &quot;Tchebyshevs plantigrade machine&quot; which maintains a practically constant speed of perfectly linear motion. I am glad to see that animation - I &quot;invented&quot; it independently after seeing the above motions, and actually built a slightly simpler form in Meccano years ago, though I'm not surprised it already existed and had a name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EYGru.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EYGru.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If ever there was a steampunk walking robot, to an authentic 19th century design, this is it.&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="435" LastEditDate="2017-01-12T15:21:28.780" LastActivityDate="2017-01-12T15:21:28.780" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13214" PostTypeId="2" ParentId="13203" CreationDate="2017-01-12T14:21:17.450" Score="0" Body="&lt;p&gt;The mechanism shown in your picture is known as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Four-bar_linkage&quot; rel=&quot;nofollow noreferrer&quot;&gt;four-bar linkage&lt;/a&gt;. These mechanisms are used quite often in mechanical engineering to convert rotational motion into complex trajectories, and vice-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific mechanism you've shown, as BrianDrummond mentioned in his comment, is &lt;a href=&quot;https://en.wikipedia.org/wiki/Chebyshev&amp;#39;s_Lambda_Mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chebyshev's Lambda Mechanism&lt;/a&gt;. This four-bar linkage is often used to produce a linear motion from the rotational motion of a crank, and can be seen in some simple walking robots. There are other ways to do produce linear motion from rotational motion, such as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Four-bar_linkage#Slider-crank_linkage&quot; rel=&quot;nofollow noreferrer&quot;&gt;Slider-crank&lt;/a&gt;, another type of four-bar linkage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how to design such mechanisms, the methods are too involved to give an adequate answer on this site, but you can easily find resources online by searching &quot;four-bar linkage synthesis.&quot; Keep in mind that there are two different ways of going about the design: graphical methods and analytical methods. With graphical methods you place a series of anchor points and trajectories on a piece of paper, then using some simple geometrical tricks you can easily generate a four-bar that performs that task. Analytical methods involve having a mathematical expression of the trajectory and using that to solve for your four-bar.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2017-01-12T14:21:17.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13215" PostTypeId="2" ParentId="13206" CreationDate="2017-01-12T16:47:10.067" Score="0" Body="&lt;p&gt;I would recommend looking at Tesla turbines. If you use a motor to turn it, it acts as a pump and can pump anything from air to crude oil. If you can find one with different distances between the blades, that might help when trying to pump water and/or air. I'm just a student though, so look into it yourself to make sure you're getting what you need.&lt;/p&gt;&#xA;" OwnerUserId="9563" LastActivityDate="2017-01-12T16:47:10.067" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13216" PostTypeId="2" ParentId="13186" CreationDate="2017-01-12T16:50:52.513" Score="1" Body="&lt;p&gt;I think you're looking for Schottky diodes.&lt;/p&gt;&#xA;" OwnerUserId="9563" LastActivityDate="2017-01-12T16:50:52.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13217" PostTypeId="1" CreationDate="2017-01-12T17:23:28.423" Score="0" ViewCount="210" Body="&lt;p&gt;In the example A , Why the author stated the total reactions = 9 only ? How to determine the total reactions for this type of questions ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rDCXx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rDCXx.jpg&quot; alt=&quot;k;ok&quot;&gt;&lt;/a&gt;&#xA;we can see that there is a total of 3 reactions (vertical , horizontal and moment) at the cut of the 4 parts , right (circled part) , and 3 reactions below . So , the total reactions is 4x3 + 3 = 15.&lt;/p&gt;&#xA;" OwnerUserId="9254" LastActivityDate="2017-01-12T17:40:15.337" Title="number of total reactions in a structure" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13218" PostTypeId="2" ParentId="13217" CreationDate="2017-01-12T17:40:15.337" Score="2" Body="&lt;p&gt;The reason is simple: the axial and shear forces and bending moment on opposite sides of the cuts are the same. So you don't have $4\times3$ unknowns along the cut, but just $2\times3$. Add that to the three support reactions and you get nine unknowns.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-01-12T17:40:15.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13219" PostTypeId="1" CreationDate="2017-01-12T18:21:08.587" Score="1" ViewCount="69" Body="&lt;p&gt;I am planning to do a project in which I plan to mimic the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Laser_Interferometer_Space_Antenna&quot; rel=&quot;nofollow noreferrer&quot;&gt;LISA interferometer&lt;/a&gt;. For this I use the fabric material, Lycra to mimic space time and use spherical balls, forming a couple and connected to a rotating shaft thus mimicking binary stars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the balls rotate, they would cause waves in the fabric. Further away from the binary system is the triangular interferometer, this system comprises of three discs, each having a laser and a sensor. When the laser in aligned with the second discs' sensor, and the laser of the second to the third's sensor and lastly the third's laser to the primary's sensor, the three form a equilateral triangular system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three discs would be placed on individual corks and each disc would be connected to a vertical shaft with a shock absorber( so that the disc can oscillate about its mean position when the spherical balls start to spin and thus produce ripples). The three shafts are connected and hence form a system where the individual components can oscillate about their mean position when ever a disturbance is caused in its surrounding fabric. When the individual components oscillate, the alignment of the lasers would change. I need this system to be sensitive to minute translational displacements, so that when the ripples hit the system, the contraction and expansion of relative lengths between the lasers can be measured by some device connected to this system.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Below is an actual representation of LISA:&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/69JRq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/69JRq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is regarding the position instruments I am to use/make, what sort of instruments and/or devices would be able to detect the minute changes in the position of the interferometers and measure the curvature of the wave when the wave propagates in the elastic medium?&#xA;Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="8375" LastEditorUserId="8375" LastEditDate="2017-01-16T13:42:16.373" LastActivityDate="2017-01-16T13:42:16.373" Title="On construction of a triangular interferometer like LISA" Tags="|lasers|experimental-physics|building-design|" AnswerCount="1" CommentCount="3" ClosedDate="2017-01-16T13:04:44.613" ContentLicense="CC BY-SA 3.0" />
  <row Id="13220" PostTypeId="1" CreationDate="2017-01-12T19:23:28.417" Score="1" ViewCount="72" Body="&lt;p&gt;I came upon enough 3&quot;x3&quot; structural aluminum square tubing to build a countertop over the washer, dryer, and sink. I wanted to double check the sag/deflection of the tubing but am having trouble figuring it out. I think it should be strong enough but just wanted to double check. Where is a simple calculator that I can plug in the components and see if the aluminum will bend. The front piece is 3x3x7.5'- 1/4&quot; thick. There will be 3x3x24&quot; pieces going to the back wall to support as well. Really I'm concerned about the main span. I don't imagine it would get over 300 lbs across the whole weight. &#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1909" LastActivityDate="2017-01-12T19:23:28.417" Title="Countertop Span/sag Calculations" Tags="|aluminum|deflection|" AnswerCount="0" CommentCount="7" ClosedDate="2017-01-16T13:44:40.487" ContentLicense="CC BY-SA 3.0" />
  <row Id="13222" PostTypeId="1" CreationDate="2017-01-12T21:04:31.243" Score="1" ViewCount="27" Body="&lt;p&gt;Laptops have processors running on few GHz clock. Is there a way to generate electromagnetic signals in that range using some software or coding? Say generating 2.4 GHz signal if the CPU runs at 3 GHz.&lt;/p&gt;&#xA;" OwnerUserId="9566" LastActivityDate="2017-01-12T21:04:31.243" Title="Generating electromagnetic signal using CPU clock" Tags="|rf-electronics|computer-hardware|" AnswerCount="0" CommentCount="1" ClosedDate="2017-01-19T19:45:19.783" ContentLicense="CC BY-SA 3.0" />
  <row Id="13225" PostTypeId="1" CreationDate="2017-01-12T23:52:11.140" Score="3" ViewCount="2281" Body="&lt;p&gt;I have a structure which is essentially a curved beam (not an arch or anything simple) which I have modeled in microstran/spacegass. It looks something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kSWpQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kSWpQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or for something more simple, imagine I have a beam like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HEr36.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HEr36.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of stiffness as I know it is: &lt;strong&gt;&lt;em&gt;k = F/d&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ie the stiffness of a structure in a direction is proportional to the Force applied to the location divided by the deflection of the structure at that location due to that force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in the case of these structures, the stiffness will vary depending on the direction I apply the load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have applied a downward load of 1 N to the center of each of the 'arches' and they deflect accordingly. However, they obviously deflect differently (in magnitude) when I apply an upward load at the same location. &lt;strong&gt;&lt;em&gt;Note: I am doing a non-linear analysis.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which is the correct deflection that I need to use in my calculation of &lt;strong&gt;&lt;em&gt;k = F/d&lt;/em&gt;&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using this stiffness for structural dynamics purposes, I am trying to find the natural frequency at that location. The structure will therefore vibrate in both the upward and downward direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial thought was to apply a load of 0.5N in both directions and add the deflections. Would this be correct?&lt;/p&gt;&#xA;" OwnerUserId="9569" LastEditorUserId="9569" LastEditDate="2017-01-13T08:27:39.937" LastActivityDate="2017-01-13T09:41:26.473" Title="Definition of stiffness for structural dynamics: How would you find the stiffness of an unusual beam directly?" Tags="|civil-engineering|structural-analysis|dynamics|stiffness|frequency-response|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13226" PostTypeId="2" ParentId="13206" CreationDate="2017-01-13T00:27:34.260" Score="5" Body="&lt;p&gt;The terms you are looking for are &quot;self priming pump&quot; and/or &quot;run dry pump&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Centrifugal self priming pumps require some water to be in them to start, but can pull a vacuum including air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diaphragm pumps pump air or water very well. They can run dry and are self priming. Some are specifically designed to just pump air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vane pumps or any other positive displacement pump will be self priming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the reliability needs of your application a low cost centrifugal sump pump may work just fine. They are built for some abuse and many have integrated floats that control when they turn on and off.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-01-13T00:27:34.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13227" PostTypeId="1" CreationDate="2017-01-13T00:32:30.287" Score="1" ViewCount="229" Body="&lt;p&gt;I cannot wrap my head around this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made two simple models, a triangular (or curved or arched) beam that is pin ended and has a single point load of 1N downwards at it's center, and a flat beam with the same loading/supports. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The secondary effects response for the flat beam makes sense to me, It looks like this:&lt;a href=&quot;https://i.stack.imgur.com/VnErZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VnErZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Point Load (1N)&lt;br&gt;&#xA;b) Deflection using a non-linear (secondary) analysis (Light Green)&lt;br&gt;&#xA;c) Deflection using a linear analysis (Dark Green)&lt;br&gt;&#xA;d) Un-deformed structure&lt;br&gt;&#xA;e) Pin supports&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear analysis: The beam deflects, tension is formed in the beam&lt;br&gt;&#xA;Non-Linear Analysis: The tension in the beam &quot;pulls up&quot; the beam so it deflects LESS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then, why does the curved beam deflect MORE due to secondary effects? I would assume that since the load creates compression in the members, the secondary analysis would reduce deflection and push the beam upward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Deflection using a linear analysis (Dark Green)&lt;br&gt;&#xA;b) Deflection using a non-linear (secondary) analysis (Light Green)&lt;br&gt;&#xA;c) Un-deformed structure&lt;br&gt;&#xA;d) Point Load (1N)&lt;br&gt;&#xA;e) Pin supports  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/esGIf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/esGIf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9569" LastActivityDate="2017-01-20T01:50:15.913" Title="Why does a simple pin-ended triangular (arched/curved) beam deflect more due to secondary effects (non-linear analysis)?" Tags="|structural-engineering|civil-engineering|statics|deflection|stiffness|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13228" PostTypeId="2" ParentId="13227" CreationDate="2017-01-13T02:24:36.703" Score="6" Body="&lt;p&gt;Play around with a simple version of this structure, made from a sheet of paper fixed in a slight curve, and see what happens when you apply a load to the mid point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the first example, any deflection of the beam will increase its length, which creates more stiffness that is included in the nonlinear analysis but not in the linear one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second diagram, you can change the &lt;em&gt;shape&lt;/em&gt; of the beam by making it more curved at the ends and flatter in the centre, &lt;em&gt;without&lt;/em&gt; changing its length. In fact if you apply more load, the stiffness will become zero and the beam will suddenly deflect into a curved shape with the center &lt;em&gt;below&lt;/em&gt; the end points, not above them. After that has happened, the stiffness will start to increase, but if you remove the load the beam won't return to its original shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That type of behaviour is often called &quot;snap through buckling,&quot; and it is usually a bad thing if it happens to a real structure, for the obvious reasons. You often get this behaviour if the lid of a tin can is not quite flat. The lid has as two stable states, bent in opposite directions, and the sudden displacement of the air makes a loud noise if it flips from one state to the other.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-01-13T02:24:36.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13229" PostTypeId="2" ParentId="13204" CreationDate="2017-01-13T03:25:08.257" Score="5" Body="&lt;p&gt;You don't need to dimension it at all, because as you probably know, that part would never be manufactured by looking at the 2D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At my company, one of the standard notes we put on every drawing is &quot;Referring to solid model for geometry not dimensioned is permissible. All unspecified surfaces must match solid model within .020 in. (and we put the .020 in a &lt;a href=&quot;http://www.gdandtbasics.com/profile-of-a-surface/&quot;&gt;GD&amp;amp;T profile symbol&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You only need to put the necessary dimensions on a print so that the design intent is clearly conveyed and there is a standard by which the part can be inspected. Therefore, I would advise that you &lt;em&gt;don't&lt;/em&gt; dimension any features that aren't critical to the design intent. &lt;em&gt;Do&lt;/em&gt; dimension every feature that interfaces with your motor and other parts of your spinning top and &lt;em&gt;do&lt;/em&gt; dimension every feature that must be made correctly for part strength, but don't stress about the angles and curves of your ribs unless those dimensions are critical to your design.&lt;/p&gt;&#xA;" OwnerUserId="709" LastActivityDate="2017-01-13T03:25:08.257" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13231" PostTypeId="1" AcceptedAnswerId="13238" CreationDate="2017-01-13T06:40:21.540" Score="2" ViewCount="207" Body="&lt;p&gt;I have a laptop computer (Dell L701X) which requires an 11.1 V battery with a rating of 8400 mAh (Battery Type R795X, 9 cell, 11.1 V, 90 Wh, 8400 mAh, other battery ID are: R1122J-733 &amp;amp; B052R796-9021).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I place a new 11.1 V battery with the same physical dimensions in the computer, but with a rating of 5200 mAh the computer doesn't like the battery. On start-up the computer beeps &amp;amp; gives a text warning on the black start up screen. The on-keyboard battery indicator flashes between white &amp;amp; amber when both my old &amp;amp; new battery are in the computer. After boot up, the battery icon in the task bar acknowledges the 5200 mAh battery but the computer will not charge the battery - the battery is new and it only has a charge of 17%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the computer need a 8400 mAh battery or is there a fault with the battery charger in my computer?&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2017-01-13T11:51:24.637" LastActivityDate="2017-01-15T02:45:16.300" Title="Battery mAh ratings" Tags="|electrical-engineering|battery|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13232" PostTypeId="2" ParentId="13231" CreationDate="2017-01-13T07:24:55.443" Score="0" Body="&lt;p&gt;*first draft, which I will edit with more details&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The capacity of battery is usually not the parameter you look at to determine  battery compatibility.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes batteries with different capacities can be used by the same product with no problems, but often ignored parameters for compatibility would be the C and E rates. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;C- and E- rates – In describing batteries, discharge current is often expressed as a C-rate in order to normalize against battery capacity, which is often very different between batteries. A C-rate is a measure of the rate at which a battery is discharged relative to its maximum capacity. A 1C rate means that the discharge current will discharge the entire battery in 1 hour. For a battery with a capacity of 100 Amp-hrs, this equates to a discharge current of 100 Amps. A 5C rate for this battery would be 500 Amps, and a C/2 rate would be 50 Amps. Similarly, an E-rate describes the discharge power. A 1E rate is the discharge power to discharge the entire battery in 1 hour&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.jp/url?q=http://mit.edu/evt/summary_battery_specifications.pdf&amp;amp;sa=U&amp;amp;ved=0ahUKEwiroYHpyb7RAhWCEpQKHajFB6oQFggLMAA&amp;amp;sig2=ZzSuIP0Ktg0NL3DzIYtuJw&amp;amp;usg=AFQjCNFKs7fQH89Qg26GFIu3qjAPVtyTWQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9450" LastActivityDate="2017-01-13T07:24:55.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13233" PostTypeId="2" ParentId="13208" CreationDate="2017-01-13T07:49:37.557" Score="5" Body="&lt;p&gt;A cogged positioning system uses the concept of a &lt;em&gt;detent&lt;/em&gt; - a pin, ball, pawl or bump that fits against a mating cavity. Usually there is some kind of spring to allow the detent to disengage when enough force is applied. This gives the familiar click as a knob is rotated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer the question the interface is called a detent. Google search &lt;em&gt;detent ball&lt;/em&gt; for examples of how a detent works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wiki definition here &lt;a href=&quot;https://en.wikipedia.org/wiki/Detent&quot;&gt;https://en.wikipedia.org/wiki/Detent&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="7021" LastEditDate="2017-01-13T08:46:22.750" LastActivityDate="2017-01-13T08:46:22.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13234" PostTypeId="2" ParentId="13186" CreationDate="2017-01-13T08:14:54.363" Score="1" Body="&lt;p&gt;Ultra low capacitance diodes &amp;lt;1pF are capable of operating at GHz frequencies. They are low loss Schottky types.  BAS69xxx from STMicro is one example.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-01-13T08:14:54.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13235" PostTypeId="2" ParentId="13225" CreationDate="2017-01-13T09:41:26.473" Score="3" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;k=F/d&lt;/em&gt;&lt;/strong&gt; is a linear relationship. If you're doing a non-linear analysis, you shouldn't expect a linear response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this beam, a geometrically non-linear analysis is appropriate if your deflections are any way significant. This is because a downwards load (if it is big enough) will cause snap-through buckling behaviour, whereas an upwards load will not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you cannot use &lt;strong&gt;&lt;em&gt;k=F/d&lt;/em&gt;&lt;/strong&gt; as a) the downwards deflection is not directly proportional to the applied load, and b) deflection due to upwards loads behaves differently to deflection due to downwards loads.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My knowledge of dynamics is extremely limited. Possibly a linear analysis might be an appropriate approximation as long as deflections are small.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-01-13T09:41:26.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13236" PostTypeId="2" ParentId="12262" CreationDate="2017-01-13T10:29:10.283" Score="1" Body="&lt;p&gt;I doubt many Rankine cycles use any of these as working fluids, if working fluid means the fluid that undergoes the actual phase change etc. Rankine cycle engines want fluids with a convenient boiling point - within the temp. range the machine will operate in, at a managable pressure. Most of the times this is water. &lt;a href=&quot;https://en.wikipedia.org/wiki/Organic_Rankine_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Organic Rankine Cycle&lt;/a&gt; engines exist, these use fluids with lower boiling point to exploit lower temp. heat sources like industrial waste heat or geothermal sources. The wiki article I linked goes into the choice of the working fluid, with more explanation :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Isentropic saturation vapor curve&lt;/li&gt;&#xA;&lt;li&gt;Low freezing point, high stability temperature &lt;/li&gt;&#xA;&lt;li&gt;High heat of vaporisation and density &lt;/li&gt;&#xA;&lt;li&gt;Low environmental impact&lt;/li&gt;&#xA;&lt;li&gt;Safety&lt;/li&gt;&#xA;&lt;li&gt;Good availability and low cost&lt;/li&gt;&#xA;&lt;li&gt;Acceptable pressures&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-01-13T10:29:10.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13237" PostTypeId="1" CreationDate="2017-01-13T11:20:32.993" Score="1" ViewCount="89" Body="&lt;p&gt;In a calculation I have received a number of values in the range $1e4-1e6$. Now I would like to display the values with 3 significant figures according to the rules of using significant figures. But this causes some issues when the values vary in magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I have the values $34745$ and $211360$. It seems inconsistent to me to render these with 3 significant figures independently of each other, as $34700$ and $211000$, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, choosing either as a baseline, rendering $35000$ and $211000$ or $34700$ and $211400$ seems to break the rules as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the standard course of action in this situation?&lt;/p&gt;&#xA;" OwnerUserId="8944" LastEditorUserId="8944" LastEditDate="2017-01-13T13:12:14.073" LastActivityDate="2017-01-13T14:45:31.163" Title="Choosing significant figures when dealing with a wide range of values" Tags="|statistics|data|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13238" PostTypeId="2" ParentId="13231" CreationDate="2017-01-13T11:50:51.723" Score="0" Body="&lt;p&gt;If dell recommends the use of a 5200mAh battery then it should work. There is something else not in sync with the laptop. I would suggest that you update the BIOS firmware. All thought the computer recognizes the lower capacity battery, there might be some parameters that need to be updated. Below is link that might help you.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.dell.com/support/home/us/en/4/Products/?app=drivers&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dell Drivers &amp;amp; Downloads Identify your product&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Although highly unlikely your power adapter might not match in new battery. If you a getting error similar to the one below you might want to follow the direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i34Oj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i34Oj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might be some discrepancies between your laptop BIOS, power adapter and the new battery. Here is an example where there was inconsistency between the power adapter and the BIOS.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Before&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/An6Se.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/An6Se.jpg&quot; alt=&quot;Batter adapter not recognized&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After replacing the power adapter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0nmeA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0nmeA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have listed the links to help you further perform root cause analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.dell.com/support/Article/us/en/19/266588/EN&quot; rel=&quot;nofollow noreferrer&quot;&gt;AC Adapter Not Powering on the Computer or Charging the Battery&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.dell.com/support/article/us/en/19/SLN85590#Issue7&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to improve your laptop battery's performance?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://pcroger.com/dell-laptop-battery-will-not-charge/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dell Laptop Battery Will Not Charge&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://en.community.dell.com/support-forums/laptop/f/3518/t/19452828&quot; rel=&quot;nofollow noreferrer&quot;&gt;(XPS L701X) Battery not charging, computer shuts down instantly if unplugged&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2017-01-15T02:45:16.300" LastActivityDate="2017-01-15T02:45:16.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13239" PostTypeId="2" ParentId="13237" CreationDate="2017-01-13T12:01:06.403" Score="0" Body="&lt;p&gt;It depends on what these different values are. If these different values are for different entities, then I would have each to 3 significant figures based on itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. For the equation &lt;strong&gt;&lt;em&gt;d = FL/EA&lt;/em&gt;&lt;/strong&gt;, working in kN and m:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;F = 5412, I would report as 5410&lt;/li&gt;&#xA;&lt;li&gt;L = 6.911, I would report as 6.91&lt;/li&gt;&#xA;&lt;li&gt;E = 210265, I would report as 210000&lt;/li&gt;&#xA;&lt;li&gt;A = 0.043211, I would report as 0.0432&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, if I have a lot of deflections at different positions, then the &lt;em&gt;relative&lt;/em&gt; deflection is important, and I would report all values based on which values are significant for the biggest deflection:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If Nodes 1-6 have deflections (in mm) of 0.000114, 0.256, 1.54, 17.9, 9.5211 and 0.0187, I would report them as: 0.0, 0.3, 1.5, 17.9, 9.5, 0.0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this case, the accuracy of 0.000114mm for node 1 is unimportant: when node 4 moved 17.9mm, node 1 effectively didn't move at all.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-01-13T12:01:06.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13240" PostTypeId="1" CreationDate="2017-01-13T13:13:11.407" Score="1" ViewCount="56" Body="&lt;p&gt;I have a building to evenly split a water stream into two: Water flows into central chamber, over two weirs of equal height and length into two chambers and from there into pipelines. I'm looking for rules of the thumb to correctly size the chambers. Weir length is straightforward. But to guarantee even flow over both weirs I need to ensure even flow conditions in the central chamber. So I'm looking on rules of the thumb how wide and high my central chamber has to be, depending on inflow conditions. So far I had no luck with the DWA/ATK (German water and wastewater associaction) guidelines, maybe codes from other countries are helpful here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example drawing (top - layout, bottom - cut):&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GywNB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GywNB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-01-13T13:13:11.407" Title="Water distribution - sizing weir buildings" Tags="|civil-engineering|hydraulics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13241" PostTypeId="2" ParentId="13237" CreationDate="2017-01-13T14:45:31.163" Score="1" Body="&lt;p&gt;The fundamental problem here is that you've got a bunch of numeric values which are not in engineering/scientific notation, so you have to make up some arbitrary guesses as to how many sig figs really are there.&lt;br&gt;&#xA;Let's look at one of your values:  34745  .  Where did it come from?  If it's, say the reading on a digital display for a widget whose operating manual specifies an accuracy of +/-1 in the last digit, then you can use it as  $(3.4745 \pm 0.00005)E5 $.  But if it's the result of a calculator giving you a ton of digits after dividing two values which were accurate to, say, 1%, then you can't assign an accuracy to the result better than that of the inputs.  Skipping the exact error propagation formula, you'd roughly be limited to $(3.47 \pm 0.03)E5$ .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately,  the &quot;rule&quot; of sig figs depends on how you combined the source data to get the calculated result. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-01-13T14:45:31.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13242" PostTypeId="1" CreationDate="2017-01-13T15:04:01.300" Score="-4" ViewCount="759" Body="&lt;p&gt;why we need to consider 2 conditions , the beam curve downwards ( as in 11-3) and beam curve upwards ( as in 11-4) ?&#xA;What's the difference between 2 cases ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I notice that in both cases , the author assume clockwise as positive&#xA;here are the 3 consecutive pages of notes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the second cases where the curve curve upwards is unnecessary because we can see that the curve curve downwards (U shape) in the first photos .&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lk0p0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lk0p0.png&quot; alt=&quot;hh&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/a5P2F.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a5P2F.jpg&quot; alt=&quot;cc&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/j1c0l.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j1c0l.jpg&quot; alt=&quot;dgr&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;EDIT : ONE more question here , why for the relative linear dispalcement , we only need to displace B relative to A ? We dont have to displace A relative to B ?&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="8664" LastEditDate="2017-01-13T16:18:32.390" LastActivityDate="2017-01-13T16:27:49.860" Title="slope deflections method analysis of beam" Tags="|structural-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2017-01-16T13:13:14.550" ContentLicense="CC BY-SA 3.0" />
  <row Id="13243" PostTypeId="2" ParentId="13242" CreationDate="2017-01-13T15:27:09.290" Score="2" Body="&lt;p&gt;You should really try to read the book more carefully, it clearly explains what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method calculates the effect of each degree of freedom: what happens when you rotate A (and only A, so B stays fixed)? What happens when you rotate B (and A stays fixed)? What happens when you impose a deflection at one of the supports (but not the other)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate these effects, the rotations are all applied clockwise. $\theta_A$ is a clockwise rotation around A, which causes the beam to deflect downwards. $\theta_B$ is a clockwise rotation around B, which causes the beam to deflect upwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this method is for arbitrary loads, you don't know if $\theta_A$, $\theta_B$ or $\Delta$ will be positive or negative. But by calculating the forces generated by each of them, you can solve for the entire structure. With the generic loading shown in Figure 11-2 it looks like $\theta_A &amp;gt; 0$, $\theta_B &amp;lt; 0$ and $\Delta &amp;gt; 0$, but there's no way to mathematically know that for any load pattern until you do the work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding why $\Delta$ is only calculated for node B, again: read the book. Here, I'll transcribe it for you:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If far the far node B of the member is displaced relative to A, so that the cord of the member rotates clockwise (positive displacement) and yet both ends do not rotate, then &lt;strong&gt;equal but opposite&lt;/strong&gt; moment and shear reactions are developed in the member [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, who cares whether it was A or B that got displaced? The result is the same! That's why they only did it for B, they'd just be repeating themselves if they did it for both nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for finally posting a question about a concept instead of a specific exercise, but next time please try reading the concept &lt;strong&gt;&lt;em&gt;carefully&lt;/em&gt;&lt;/strong&gt; before asking.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-01-13T16:27:49.860" LastActivityDate="2017-01-13T16:27:49.860" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="13244" PostTypeId="1" CreationDate="2017-01-13T16:19:29.853" Score="1" ViewCount="127" Body="&lt;p&gt;I'm trying to figure out whether lenses for thermal (long-wave infrared, i.e.,  7-14µm) cameras are inherently more expensive or less capable than the lenses for visible cameras.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Production lenses for digital cameras often incorporate a dozen or more individual glass elements up to perhaps 50mm in diameter.  With surface coatings these systems still transmit over 90% of light in the visible spectrum, and are able to nearly eliminate all forms of distortion while providing a wide focal (zoom) range.  These systems can be mass produced for a few hundred dollars per unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the limit (i.e., assuming the same economies of scale were provided), is similar performance and/or price possible with current uncooled thermal lens technology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble determining the state of the art in thermal lenses.  I know that for production cameras at least the front element is typically made of coated Germanium, or Ge-As, and a single Ge element in real-world temperatures (i.e., below 50°C) transmits over 90% across the LWIR spectrum.  (My understanding is that Zn-Se and Zn-S are also very good in the thermal spectrum, but they are too soft to be exposed, so they could work as interior elements, but not the front element.)  But I don't know, for example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is transmission through these materials so low that using more than a couple of elements is not practical?&lt;/li&gt;&#xA;&lt;li&gt;Are these materials inherently significantly more expensive to manufacture than the glasses used in visible light lenses?&lt;/li&gt;&#xA;&lt;li&gt;Are the physics of refraction in these materials in the LWIR spectrum too variable to build lenses with many elements, or with &quot;zoom&quot; capabilities?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="77" LastActivityDate="2017-11-01T22:15:27.550" Title="Are thermal (LWIR) lenses as capable as visible-wavelength optics?" Tags="|optics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13245" PostTypeId="1" CreationDate="2017-01-13T16:24:55.227" Score="-2" ViewCount="1733" Body="&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi I was working for an engineering club and we needed to do this as soon as possible:&#xA;We wanted to make a pulley system with a mechanical advantage of 0.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that we have to do something with the string angles but I dont know how the pulleys should be oriented&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks!&lt;/p&gt;&#xA;" OwnerUserId="9572" LastActivityDate="2017-01-13T17:44:06.743" Title="How to make a pulley system with a mechanical advantage of 0.5" Tags="|mechanical-engineering|experimental-physics|pulleys|building-physics|" AnswerCount="2" CommentCount="2" ClosedDate="2017-01-16T12:58:38.930" ContentLicense="CC BY-SA 3.0" />
  <row Id="13247" PostTypeId="2" ParentId="13245" CreationDate="2017-01-13T17:22:48.457" Score="0" Body="&lt;p&gt;When the load moves up and down the angles are going to change so the mechanical advantage will also change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to use a winch, the load is attached to a big drum and the other string is attached to a drum half the size.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-01-13T17:22:48.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13248" PostTypeId="1" AcceptedAnswerId="13251" CreationDate="2017-01-13T17:25:23.947" Score="2" ViewCount="11278" Body="&lt;p&gt;Trying to design a 2:1 gear ratio planetary gear, and I'm not sure how to begin. In the past I've had luck using &lt;a href=&quot;http://geargenerator.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; site to create an illustrator file that I can extrude in Rhino. This does not generate planetary gears, and I'm also unclear as to how design a planetary gear with a 2:1 ratio.&lt;/p&gt;&#xA;" OwnerUserId="539" LastActivityDate="2017-01-14T11:54:03.807" Title="How to I design a planetary gear with a gear ratio of 2:1 ( doubling the torque)" Tags="|mechanical-engineering|gears|servo|" AnswerCount="1" CommentCount="1" ClosedDate="2017-01-16T12:58:12.223" ContentLicense="CC BY-SA 3.0" />
  <row Id="13249" PostTypeId="2" ParentId="13245" CreationDate="2017-01-13T17:44:06.743" Score="2" Body="&lt;p&gt;What you need is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Block_and_tackle&quot; rel=&quot;nofollow noreferrer&quot;&gt;block-and-tackle&lt;/a&gt;, namely a gun-tackle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HXMhg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HXMhg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;sub&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Block_and_tackle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general trick to determine advantage is how many verticals you have bearing the load (other than the end you are pulling from). Also, it must be a single rope that loops around the block. So, for example, a tackle with an advantage of 4 (or 0.25) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zBKuO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zBKuO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;sub&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Block_and_tackle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-01-13T17:44:06.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13250" PostTypeId="1" AcceptedAnswerId="13267" CreationDate="2017-01-14T00:51:53.653" Score="2" ViewCount="317" Body="&lt;p&gt;I am not sure if this is the best place to put this, but if I'm in the wrong place please tell me where I can go. So I know a peltier module is not the most efficient, however my goal is to make a hand warmer / cooler with a petlier module. My operating temperatures will be around 60f to 90f, nowhere near the max of most peltiers. My first question is can it be powered by a small battery? Second, would I need a heatsink with this kind of temperatures that I want to produce and if it's required, is there any passive cooling I can use without being thick? Third, is this idea practical?&lt;/p&gt;&#xA;" OwnerUserId="9575" LastEditorUserId="9575" LastEditDate="2017-01-14T00:59:05.280" LastActivityDate="2017-01-15T08:03:23.587" Title="Peltier Module to warm hands?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13251" PostTypeId="2" ParentId="13248" CreationDate="2017-01-14T01:09:48.090" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xSThp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xSThp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(Image from Wikipedia)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When designing planetary gears, there are three different types of gear present to be designed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Annulus gear, an internal gear surrounding the other gears (shown above in red)&lt;/li&gt;&#xA;&lt;li&gt;Sun gear, an external gear in the middle of the gear set (shown in yellow).&lt;/li&gt;&#xA;&lt;li&gt;Planet gears, external gears surrounding the sun gear, usually three or four of them are present (shown in blue).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;First of all, it's worth noting about internal gears. The sun and planet gears, being external gears (i.e. teeth point outward), are the exact same as any other kind of spur gear. However, the annulus gear is an internal gear, with the teeth pointing inward. An internal gear may look like a circle with an external gear subtracted from inside of the circle, and it almost is; there is just a slight difference in how far the teeth extend from the pitch circle. Not noting this difference in internal gears can lead to gear teeth undesirably colliding before properly meshing, causing the gear set to jam. Therefore, if you are fine with gears being output as .dxf files, then I suggest using &lt;a href=&quot;http://hessmer.org/gears/InvoluteSpurGearBuilder.html&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Dr. Rainer Hessmer&amp;#39;s gear generator&quot;&gt;Dr. Rainer Hessmer's gear generator&lt;/a&gt;, as it allows you to create both external and internal gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to actually designing these three different types of gears, the following equation, which I'll refer to as the &lt;em&gt;speed rule&lt;/em&gt;, should be observed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_C (1+\rho) = \omega_A \rho + \omega_S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_C$ is the rotation speed of the &lt;em&gt;carrier&lt;/em&gt;, which is shown in green in the above image. $\omega_A$ is the rotation speed of the annulus gear and $\omega_S$ is the rotation speed of the sun gear. Rotation speeds are all defined to be positive in the same direction of rotation. $\rho$ is the ratio of the number of teeth on the annulus gear to the number of teeth on the sun gear. It is worth noting that $\rho$ cannot be less than or equal to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, out of the carrier, the annulus gear and the sun gear, one is selected to be fixed in place ($\omega=0$), one is selected to be the input and one is selected to be the output (it is however possible to have two inputs or two outputs). By fixing one of the rotation speeds, a gear ratio can be determined between the input and output, given as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix the carrier:&#xA;$\omega_C = 0, \frac{\omega_S}{\omega_A} = -\rho$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix the annulus:&#xA;$\omega_A = 0, \frac{\omega_S}{\omega_C} = \rho+1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix the sun:&#xA;$\omega_S = 0, \frac{\omega_A}{\omega_C} = 1+\frac{1}{\rho}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is now a matter of selecting one of the above configurations that give you the appropriate gear ratio. Sometimes it requires a bit of trial and error: stick with a configuration and determine the value of $\rho$. If $\rho$ is less than or very close to one, try another setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your case, with a gear ratio of $2:1$, let's try the following configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fixed: carrier&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Input: sun&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Output: annulus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in this configuration, $\frac{\omega_S}{\omega_A} = -\rho$, so the input and output need to be rotation in opposition direction to get a positive value of $\rho$. Let's say the input is spinning in a &quot;negative&quot; direction and the output in a &quot;positive&quot; direction. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\omega_S}{\omega_A} = \frac{-2}{1} = -2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = 2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A value of $\rho$ which is greater than 1 is obtained: this configuration seems to work well. Now we enter the second phase of the planetary gear design: assigning the number of teeth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the planet gears are generally spaced equally apart from one another. To achieve this, the number of teeth of both the sun gear and the annulus should be a multiple of the number of planets. Otherwise, difficulties may arise when it comes to assembling the gears together. This is a process that may require some trial and error, and it may even required very slightly changing the value of $\rho$ to make it easier to make the teeth number multiples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our example let's assume that there are four planet gears. Therefore, it can be seen that a sun gear of 31 (not multiple of 4) teeth is not appropriate, but one of 32 (multiple of 4) teeth is. If the sun gear has 32 teeth, the annulus gear will have $32 \times \rho = 64$ teeth. 64 is also a multiple of 4 and is therefore appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the number of teeth for the sun and annulus gears are selected, the number of teeth for the planet gear is calculated using the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{planet teeth} = \frac{\text{annulus teeth} - \text{sun teeth}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the division by two: this means that the number of teeth on the annulus and sun gears must both be odd numbers, or both be even numbers. If this turns out not to be the case, doubling the number of teeth on the sun and annulus gears quickly resolves the issue. In our example, the number of teeth on each planet gear is 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the number of teeth selected, now select an appropriate module or diametral pitch, depending on how strong you want the gears, and that is the planetary gear set designed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use speed rule to select appropriate configuration &lt;/li&gt;&#xA;&lt;li&gt;Determine ratio of annulus teeth to sun teeth&lt;/li&gt;&#xA;&lt;li&gt;Determine number of teeth for each gear&lt;/li&gt;&#xA;&lt;li&gt;Select the module/diametral pitch&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2017-01-14T11:54:03.807" LastActivityDate="2017-01-14T11:54:03.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13252" PostTypeId="1" CreationDate="2017-01-14T07:59:17.223" Score="-1" ViewCount="5209" Body="&lt;p&gt;In the first photo, i was told that the fixed end moment is zero when the far end is pinned or roller supported . Why in the example in the second and third photo( both end are fixed)  , the angle is zero ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's wrong , because the fixed end must be able to resist shear force . In conjugate beam method , angle of rotation = shear force . So , as the fixed end can resist the moment , there must be some values for the angle if rotation , am i right ? &lt;a href=&quot;https://i.stack.imgur.com/Aa8Yx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Aa8Yx.jpg&quot; alt=&quot;iii&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8hjCZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8hjCZ.jpg&quot; alt=&quot;jjj&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9EfxX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9EfxX.jpg&quot; alt=&quot;khihi&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[![ee][4]][4]&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="8664" LastEditDate="2017-01-14T11:46:45.610" LastActivityDate="2017-01-14T16:18:20.100" Title="Angle of rotation in fixed end of beam" Tags="|structural-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13254" PostTypeId="2" ParentId="13252" CreationDate="2017-01-14T16:05:08.700" Score="3" Body="&lt;p&gt;Ok. Yeah.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question seems to be why a fixed support has zero rotation ($\theta=0$). The answer is simple: &lt;em&gt;that's the definition of a fixed support&lt;/em&gt;. Here is a table representing the constraints placed by different types of supports (for a 2D frame structure, of course):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Support       | dX | dY | rZ |&#xA;--------------+----+----+----+&#xA;Pinned        | X  | X  |    |&#xA;Fixed         | X  | X  | X  |&#xA;Plate         |    |    | X  |&#xA;Roller*       | *  | *  |    |&#xA;Fixed roller* | *  | *  | X  |&#xA;(* one or the other)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So, pinned supports do not allow any sort of translation (dX, dY), but do allow rotation (rZ). Fixed supports do not allow any translation or rotation. Plate supports (I don't know of a better name for them) allow translation but no rotation. Roller supports allow for rotation and for translation in one direction but not the other. Fixed rollers are the same, but don't allow rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first circle you drew states that a pinned or roller support does not resist bending moments and therefore allows rotations ($M_R = 0 \therefore \theta \neq 0 $). A fixed support, on the other hand, does not allow rotations and therefore generates a resisting bending moment ($\theta = 0 \therefore M_R \neq 0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a comment under your question you state that &quot;In the first post, i was told that at the pinned and roller support, then the theta is zero.&quot; I don't know if this was a reading mistake on your part (since the author clearly states that the &lt;strong&gt;bending moment&lt;/strong&gt; is zero, not $\theta$) or if you simply don't know the difference between them. Hopefully it is the former, because the latter would be such a fundamental misunderstanding that I simply won't be able to help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second circle you drew states that &quot;moments at A and D are zero&quot;. Looking at Figure 11-9, we can see that A and D are pinned supports at the extremities of the beam, so yeah, by definition we know that the moments at A and D are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third circle you drew states that &quot;$\theta_A = \theta_D = 0$ since A and D are fixed supports.&quot; Given the definition, no further explanation should be necessary.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-01-14T16:18:20.100" LastActivityDate="2017-01-14T16:18:20.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13259" PostTypeId="1" CreationDate="2017-01-14T21:53:50.667" Score="2" ViewCount="4509" Body="&lt;p&gt;Is there a non intrusive way to measure if water is flowing through a copper pipe or just standing still in it? The copper pipe belongs to my heating system and is always under pressure with water, flowing or not.&lt;/p&gt;&#xA;" OwnerUserId="9585" OwnerDisplayName="user2021331" LastEditorUserId="8055" LastEditDate="2017-01-16T15:28:17.163" LastActivityDate="2017-01-18T08:20:51.820" Title="How can I measure if water is flowing through a copper pipe?" Tags="|electrical-engineering|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13260" PostTypeId="2" ParentId="13259" CreationDate="2017-01-14T22:02:29.927" Score="7" Body="&lt;p&gt;Non Invasive flow meters typically use ultrasonic sensors clamped to the pipe. The keyword is simply non-invasive flow meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But regular flow meters are much cheaper. All you need is the appropriate size, cut the pipe and thread it, attach t he meter and done. It's done all the time.&lt;/p&gt;&#xA;" OwnerUserId="3895" OwnerDisplayName="cde" LastEditorUserId="3895" LastEditDate="2017-01-15T03:03:13.113" LastActivityDate="2017-01-15T03:03:13.113" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13262" PostTypeId="2" ParentId="13259" CreationDate="2017-01-15T02:54:25.953" Score="2" Body="&lt;p&gt;You can use two piezo ceramic transducers in combination with a TDC1000 AFE from TI to measure the flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O3KT7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O3KT7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lit/ds/symlink/tdc1000.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;TDC1000 Ultrasonic Sensing Analog Front End (AFE) for Flow Sensing Applications&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2017-01-15T02:54:25.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13263" PostTypeId="1" AcceptedAnswerId="13266" CreationDate="2017-01-15T04:23:29.840" Score="1" ViewCount="236" Body="&lt;p&gt;In a &lt;a href=&quot;https://en.wikipedia.org/wiki/Split-flap_display&quot; rel=&quot;nofollow noreferrer&quot;&gt;split-flap display&lt;/a&gt; there is a barrel which rotates by a small angle N times to get to some next display element. However I fail to understand, how it is possible that the barrel rotates &lt;strong&gt;exactly&lt;/strong&gt; by some desired angle? How is the rotation error after each flip is not accumulating? The only explanation I can think of is that it has sensors that sends signal back to the the motor to stop it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it possible that a motor can make exact rotations without error accumulation? I don't think it is possible unless the motor itself has built-in controller systems which resets the position each time to eliminate the error.&lt;/p&gt;&#xA;" OwnerUserId="4314" LastActivityDate="2017-01-15T05:05:12.403" Title="How split-flap display rotates by exact angle" Tags="|mechanical-engineering|electrical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13264" PostTypeId="1" AcceptedAnswerId="13265" CreationDate="2017-01-15T04:28:03.740" Score="0" ViewCount="96" Body="&lt;p&gt;Should standard SI units be used in this formula or are other units supposed to be used for the formula for stress in a thin-walled spherical pressure vessel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_{yield}=\frac{P_g r_i}{2t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\sigma_{yield}$ is the yield strength of the material, $P_g$ is the gauge pressure, $r_i$ is the inner radius of the sphere, and $t$ is the thickness of the sphere wall.&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="6695" LastEditDate="2017-01-18T17:20:23.440" LastActivityDate="2017-01-18T20:09:14.053" Title="Units to use in this formula?" Tags="|pressure|measurements|unit|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13265" PostTypeId="2" ParentId="13264" CreationDate="2017-01-15T04:41:45.077" Score="7" Body="&lt;p&gt;You can use any &lt;em&gt;consistent&lt;/em&gt; set of units. That includes SI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you try to use a pressure in ATM, a sphere radius in feet, a wall thickness in mils, and want the stress in tons per square inch, you will probably get the wrong answer that you deserve!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-01-15T04:41:45.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13266" PostTypeId="2" ParentId="13263" CreationDate="2017-01-15T05:05:12.403" Score="2" Body="&lt;p&gt;The barrel is usually rotated by a ratchet mechanism, not directly by a rotary motor. The ratchet might be driven by such a motor (e.g., a clock) or by a solenoid (e.g., railroad station status board).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the former case, the display is initialized manually. In the latter, there is usually a microswitch that indicates the index position of the barrel.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2017-01-15T05:05:12.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13267" PostTypeId="2" ParentId="13250" CreationDate="2017-01-15T08:03:23.587" Score="0" Body="&lt;p&gt;For heating, it would be practical. There are &lt;a href=&quot;https://www.milwaukeetool.com/power-tools/cordless/2345&quot; rel=&quot;nofollow noreferrer&quot;&gt;battery powered coats&lt;/a&gt; that are currently on the market. A peltier module pumps heat, plus makes a heat; so you get more heat than electricity you put in. So instead of getting 100% heat from electricity in a resistive heater, you are getting 150% heat (for example) from electricity in a heat pump (peltier module in this case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A heatsink will increase the efficiency by allowing more heat to be pumped. Lots of variables, but lets say an increase from 150% to 180%. Add a fan to that heatsink and say an increase from 180% to 200%. I would recommend starting with a heatsink at least, and increasing or decreasing the chosen voltage to achieve your desired temperature. If you have a 12v module, start with something around 6 volts. Remember not to drain lipo batteries too low and to watch battery chemistry or you could damage them or start a fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-01-15T08:03:23.587" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13268" PostTypeId="1" CreationDate="2017-01-15T10:04:34.823" Score="4" ViewCount="1611" Body="&lt;p&gt;I have many photos of devices which must be traced into AutoCAD. Is there any software for tracing object edges/outlines automatically so that a Primary Plan of the object can be made in AutoCAD?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: the photo below, shows an object I would like to have a line drawing of in AutoCad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pGHsg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pGHsg.jpg&quot; alt=&quot;main&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to trace the object's edge/outline from a top view and create a line drawing in AutoCAD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried Photoshop by using it's &lt;strong&gt;Magic Wand Tool&lt;/strong&gt; and made this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mx1zg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mx1zg.jpg&quot; alt=&quot;1st&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method I used was from &lt;a href=&quot;http://www.youtube.com/watch?v=ICxp0Puao1Q&quot; rel=&quot;nofollow noreferrer&quot;&gt;How To Transform an Image into Lines to AutoCAD by youtube&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this site and made this photo by using it's edge detection function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My photo is in www.photo-kako.com/en/edge.cgi:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uW4ZZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uW4ZZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after this how could I change the main edge/outline to a line drawing in AutoCAD automatically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or how could I do this better, any help or instruction for saving time and getting a better result would be appreciated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the lack of time, I have made that work by my hand  similar to this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gBCDm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gBCDm.jpg&quot; alt=&quot;hand line panting!!!&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2678" LastEditorDisplayName="user16" LastEditDate="2017-10-14T23:16:40.713" LastActivityDate="2017-10-14T23:16:40.713" Title="Is it Possible to Automatically Transform an Image of an Object into a Line Drawing in AutoCAD?" Tags="|measurements|autocad|cad|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13269" PostTypeId="2" ParentId="13268" CreationDate="2017-01-15T12:09:09.113" Score="3" Body="&lt;p&gt;One option is to import the jpeg directly into AutoCad and simply trace the lines you want over the top in  a new layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this isn't automatic as such I suspect that it may end up being less trouble than going through multiple pieces of software and a your example shows there is a good chance that you will need to do a bit of manual tidying up in any case especially as the photo in the example isn't very high contrast and you have a lot of shadows and perspective to deal with. So it is unlikely that an automatic edge detection tool is going to reliably do what you want, certainly with that sort of quality of photograph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plus be aware that close-up photos like this can distort the image quite a bit especially at the corners. Just the fact that you can see the part receding into the background indicates that it is not a true flat projection and unless you know the lens type and focal length used to take the picture this is not easy to correct. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-01-15T16:49:53.693" LastActivityDate="2017-01-15T16:49:53.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13271" PostTypeId="1" AcceptedAnswerId="13276" CreationDate="2017-01-15T14:24:06.860" Score="0" ViewCount="3222" Body="&lt;p&gt;In the solution , we notice that the 2823.6 is the result of carry over factor from BA to AB . Why we dont need to do the same thing for BC to CB ? &lt;a href=&quot;https://i.stack.imgur.com/lS69U.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lS69U.jpg&quot; alt=&quot;dd&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fSpNN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fSpNN.jpg&quot; alt=&quot;dd&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2017-01-16T00:04:43.840" Title="carry over factor in beam ( moment distribution )" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13273" PostTypeId="1" CreationDate="2017-01-15T15:50:57.967" Score="1" ViewCount="49" Body="&lt;p&gt;I've read through many PDF's on the web but can't seem to find the equations I'm after, I am trying to make a small model axial flow compressed jet engine, a little smaller then a shoe box. Any good online resources I can read up on? Much appreciated. &lt;/p&gt;&#xA;" OwnerUserId="8827" LastActivityDate="2017-01-15T15:50:57.967" Title="Equation for Axial flow compressor blade/stator vein geometry?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="5" ClosedDate="2017-01-19T19:43:29.470" ContentLicense="CC BY-SA 3.0" />
  <row Id="13275" PostTypeId="1" CreationDate="2017-01-15T19:57:32.770" Score="0" ViewCount="200" Body="&lt;p&gt;In an assignment we had to build a Clock D-Latch and give the boolean function and the truth table for the states $Q_1$/$Q_1^*$ and the following states $Q_2$/$Q_2^*$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latch looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tIANV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tIANV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got the functions but I don't know how I proceed with the truth table for the states $Q_2$ and $Q_2^*$.  At first I thought I could just replace the $Q_0$ in the boolean function with $Q_1$ and could calculate the states, but that doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Functions are: $Q_1 = Q_0^* (D+ {\sim}CK)$ and $Q_1^* = {\sim}Q_0^*({\sim}D + {\sim}CK)$ where $\sim$ means not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more specific: With $Q_0 = 0$, $D = CK = 1$ we get $Q_1 = Q_1^* = 0$. But the state two is $Q_2 = 1$ and $Q_2^* = 0$. I don't understand how we get from 0,0 to 1,0?&lt;/p&gt;&#xA;" OwnerUserId="9561" LastEditorUserId="8055" LastEditDate="2017-01-16T15:29:03.477" LastActivityDate="2024-03-20T11:05:15.600" Title="Latches - Question regarding states" Tags="|electrical-engineering|computer-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13276" PostTypeId="2" ParentId="13271" CreationDate="2017-01-15T23:43:18.493" Score="4" Body="&lt;p&gt;Because it is an end pin joint, it is assumed that no moment can occur in the joint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moment carried from BA to AB because AB is a fixed joint, and thus incur a moment as a result of the moment at joint B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the &lt;a href=&quot;https://en.wikipedia.org/wiki/Moment_distribution_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moment Distribution Method&lt;/a&gt; Wikipedia page and look at their worked example, it should help clear things up. &lt;/p&gt;&#xA;" OwnerUserId="9594" LastEditorUserId="1832" LastEditDate="2017-01-16T00:04:43.840" LastActivityDate="2017-01-16T00:04:43.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13277" PostTypeId="1" CreationDate="2017-01-16T00:51:40.327" Score="4" ViewCount="125" Body="&lt;p&gt;I was reading an article in the press this week about a project to lay an undersea cable across the Tasman Sea. The ship had had to return to port for repairs and return to complete the job later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This got me wondering: how does the ship secure the end of the laid cable, and how does it retrieve the cable to continue the cable-laying operation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this instance it was a fibre-optic cable for telecoms use - would there be differences for other cable types?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, I imagine that very long cable runs (e.g. trans-Pacific) would need to be laid in more than one piece. Would similar techniques be used, or would some different approach be used for planned breaks?&lt;/p&gt;&#xA;" OwnerUserId="9595" LastEditorUserId="6790" LastEditDate="2020-05-02T19:36:36.400" LastActivityDate="2020-05-02T19:36:36.400" Title="How does a cable-laying ship retrieve a partly-laid cable?" Tags="|marine-engineering|ships|infrastructure|fiber-optics|offshore|" AnswerCount="0" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13278" PostTypeId="2" ParentId="13268" CreationDate="2017-01-16T03:47:41.323" Score="1" Body="&lt;p&gt;You would want to use a feature in adobe illustrator called &quot;live trace&quot; you can &quot;place&quot; an image into an illustrator file and select live trace. You can tweak the settings and presets until you get the best result. Then, you &quot;expand&quot; the result which you can export to DXF. It's not perfect, but if your images are sharp and have good contrast, you may get good results. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-01-16T03:47:41.323" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13279" PostTypeId="2" ParentId="13219" CreationDate="2017-01-16T06:29:19.303" Score="2" Body="&lt;p&gt;If I'm understanding your question correctly, you are asking how you can build a functional laser interferometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is a pretty complex topic and impossible to cover in detail in a single answer. I'll however point you to Sam's laser FAQ, which is an excellent resource and has a few thoughts about building such from a CD-ROM drive:&#xA;&lt;a href=&quot;http://www.repairfaq.org/sam/laserlia.htm#liaint4&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.repairfaq.org/sam/laserlia.htm#liaint4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to consider using either time-of-flight or triangulation based laser distance sensors for your demonstration. They are nowhere near as accurate; they have a typical accuracy of 1 mm, as opposed to interferometry's 0.0001 mm. But their construction is significantly simpler and they are commonly available as premade modules.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2017-01-16T06:29:19.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13281" PostTypeId="1" CreationDate="2017-01-16T12:06:36.293" Score="1" ViewCount="52" Body="&lt;p&gt;I have a shifter kart (125ccm kz2) with 6 gears.&#xA;What I want to do is to change gears from the steering wheel. Like in the system mentioned in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the electrical and electronic system needed but I need a fast and powerful actuator (using 12VDC or 5-6VDC, 3 cm travel, going both ways, about 30Nm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about using just the actuator from one of these systems (don't have that much money to spend on it), but I don't know what it is.&#xA;This is the website: &lt;a href=&quot;http://me-shifter.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://me-shifter.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="9599" LastActivityDate="2017-01-16T12:06:36.293" Title="What actuators do the me-shifter gear shifting system uses?" Tags="|automotive-engineering|gears|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13283" PostTypeId="1" AcceptedAnswerId="13311" CreationDate="2017-01-16T14:50:49.170" Score="0" ViewCount="422" Body="&lt;p&gt;In a laminar flow along an Annular cross section with heating in one side and insulation in the other, what is the value of the Nusselt number in this flow? and is the diameter in the Nusselt expression Nu=h*D/k the hydraulic diameter?&lt;/p&gt;&#xA;" OwnerUserId="5886" LastActivityDate="2017-01-19T11:59:40.120" Title="What is the value of Nusselt number in laminar flow?" Tags="|fluid-mechanics|thermodynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13284" PostTypeId="1" CreationDate="2017-01-16T14:53:29.797" Score="-4" ViewCount="1559" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/my7IU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/my7IU.png&quot; alt=&quot;jjg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why at C . the shear force V change from -0.5 to 0.5 ? We wantg to determine the shear force at different location , and a unit force is applied at arbitary location ( measured from A) ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMO , there's no upwards force of 1 acting exactly on C , so the shear force will decrease from -0.5 at C to the cantilever end .&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2017-01-16T18:13:59.143" Title="shear force in influence line of beam" Tags="|structural-engineering|structural-analysis|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13285" PostTypeId="2" ParentId="13259" CreationDate="2017-01-16T15:02:23.940" Score="1" Body="&lt;p&gt;So what you're hoping is to keep circulating after the thermostat has shut down the furnace, in the hope of extracting as much heat as possible from the water?  Your question only asked about flow, not temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem, I fear, is that you're not looking at the entire system's gains/losses.  The best thing you can do, by far, to minimize waste heat, is to insulate the pipes everywhere except the radiators themselves.  The next largest waste heat source is the furnace &amp;amp; exhaust system, and unless you want to put in some passive air feed to send furnace-room air into the house proper, there's not much you can do about that.&lt;br&gt;&#xA;Compared with those two items, the proposed heat recovery from warm water in the pipes is miniscule at best. Yes, it is sensible to set the circulator to continue for a few minutes after the thermostat triggers, but you get diminishing returns in a hurry. For one thing, the rate of heat transfer in the radiators is a strong function of the temperature difference.  You're going to end up burning more electric power to run the pump than you'll ever get in room warmth. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-01-16T15:02:23.940" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13287" PostTypeId="1" AcceptedAnswerId="13293" CreationDate="2017-01-16T15:47:48.397" Score="1" ViewCount="3945" Body="&lt;p&gt;My research has found a &lt;a href=&quot;https://www.quora.com/What-is-the-difference-between-a-plunger-and-a-piston&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quora link&lt;/a&gt; and another &lt;a href=&quot;http://www.drucklufttechnik.de/www/temp/e/drucklfte.nsf/b741591d8029bb7dc1256633006a1729/959ffe76f432a534c125662500774784?OpenDocument&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt; giving two different, albeit not contrasting, definitions. Also, as far as I have seen, looking up &quot;plunger compressor&quot; also refers me to pumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking to understand the proper definitions, and see some animated GIFs should you have some, that exactly tell these two apart.&lt;/p&gt;&#xA;" OwnerUserId="7994" LastActivityDate="2019-08-07T18:07:37.910" Title="What is the difference between a piston and a plunger compressor/pump?" Tags="|mechanical-engineering|pumps|compressors|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13288" PostTypeId="2" ParentId="13284" CreationDate="2017-01-16T16:31:28.160" Score="3" Body="&lt;p&gt;Read the book more carefully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a plot of the &lt;em&gt;shear force along the beam for a fixed point load&lt;/em&gt;. That plot would &lt;em&gt;not&lt;/em&gt; have a step change at C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot in the book shows the &lt;em&gt;shear force at the fixed point C as the loading point moves along the beam&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-01-16T16:31:28.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13289" PostTypeId="1" CreationDate="2017-01-16T17:50:31.233" Score="1" ViewCount="222" Body="&lt;p&gt;A sawblade cuts a piece of steel which has a rectangular profile with height $h$ and width $b$. The length of this piece is given by $\ell$ (not relevant for the cutting process). I have a sawblade with length $L$ and sawtooths that have length $\Delta L$ (the sawblade therefore has $N = \frac{L}{\Delta L}$ sawtooths), peak height $f$ (the distance from minimum to maximum of a sawtooth) and thickness of the sawblate is $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sawblade is attached to the piece of metal with force $F$ and uses full length for sawing (velocity: $v$). A periodic motion front - back - front - back - ... is exerted by the sawblade. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1: Which material parameters of the steel and blade determine how much the saw will penetrate through the piece of metal during ONE saw motion with full length?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: How long will it take until the piece of metal is cut?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there formulas to compute such things?&lt;/p&gt;&#xA;" OwnerUserId="8941" LastEditorUserId="1832" LastEditDate="2017-01-16T18:28:35.480" LastActivityDate="2017-01-16T18:28:35.480" Title="How much time is needed to cut steel?" Tags="|mechanical-engineering|machining|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13290" PostTypeId="2" ParentId="13284" CreationDate="2017-01-16T18:13:59.143" Score="2" Body="&lt;p&gt;As &lt;a href=&quot;https://engineering.stackexchange.com/a/13288/1832&quot;&gt;@alephzero mentioned&lt;/a&gt;, you are misunderstanding how influence lines work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To quote &lt;a href=&quot;https://en.wikipedia.org/wiki/Influence_line&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, &quot;an influence line graphs the variation of a function (such as the shear felt in a structure member) at a specific point on a beam or truss caused by a unit load placed at any point along the structure.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in your case, the influence line shown describes how much shear will be felt at C when a concentrated unit load is applied at any given point along the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance if you applied a concentrated downward unit force:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;at A, what would the shear force be at C? Zero.&lt;/li&gt;&#xA;&lt;li&gt;at B, what would the shear force be at C? Zero.&lt;/li&gt;&#xA;&lt;li&gt;immediately to the right of C? 0.5&lt;/li&gt;&#xA;&lt;li&gt;immediately to the left of C? -0.5&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Or, more visually, here's the influence line at C (the red dot at midspan).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ma7Bc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ma7Bc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is the shear diagram for loads along the span. Notice that the shear value &lt;strong&gt;&lt;em&gt;at C&lt;/em&gt;&lt;/strong&gt; is always equal to the value shown in the influence line for the point where the load is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For C, I applied the load a tiny bit to the left and right of C, so the value is a bit off from the 0.5 it should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HeGZ1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HeGZ1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aEJnd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aEJnd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4rLFT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4rLFT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DYfJO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DYfJO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you see that the point where a concentrated load is applied is also a discontinuity in the shear diagram. The influence line is merely representing that: if you apply a load immediately to the left of C, the result is completely different than if you'd applied it immediately to the right of C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also notice that the difference between the left and right sides of the discontinuity point must be equal to the concentrated force (in this case, $-0.5 - 0.5 = -1$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If C were not at midspan, then the diagram still would go from negative to positive, with a gap of 1. So, let's say C is just two meters away from A: in this case, the discontinuity will go from -0.25 to 0.75.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-01-16T18:13:59.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13291" PostTypeId="1" AcceptedAnswerId="15599" CreationDate="2017-01-17T01:09:53.690" Score="3" ViewCount="237" Body="&lt;p&gt;I'm an EE student by trade, but I've started a project that requires some mechanical analysis, and I've run up against a bit of a wall in my calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system in question that I'm looking at is an actuator, connected via ball joint to a rod, which is then connected with a ball joint to a platform. During operation, the platform will be constrained to movement in either only the x direction, or only the y direction. I'm trying to find out, based on the force of the actuator and the geometry, what the X and Y components of theses forces will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gbX60.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gbX60.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Apologies for the cartoonish drawing). The actuator is on the bottom left. For the purposes of calculation, I'm assuming that the rod is rigid, has zero mass, and that the actuator does not need to move to apply any amount of force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using static equilibrium to calculate the X and Y components of the force on the upper right hand side, which will be the X and Y force that my actuator applies to the moving platform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actuator will apply an initial force Fa to the rod. Since both supported ends are ball joints, I know that there can be no moments on either end of the rod, only forces acting in-line with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of $\theta_M$ being equal to $\theta_R$, the solution is straightforward: all of the force is being applied in-line with the rod, so we set the sum of forces in X and Y equal to zero, and solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I'm having trouble wrapping my head around what happens if $\theta_M \neq \theta_R$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My naive approach to solving this would be as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dv3v5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dv3v5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Break down the initial applied force into components that are parallel with the rod, and components that are perpendicular to the rod. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the part I'm unsure about. Since the actuator can't move, the perpendicular force &lt;em&gt;should&lt;/em&gt; be cancelled by an equal and opposite normal force of the actuator pushing back against the rod, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However something in the back of my head is telling me that I'm missing something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, if we were to apply the actuator force not in-line with the axis of the rod, I'd think that there would be some kind of lever action present. An example would explain what I'm trying to say best I think. Can I not analyze a system like this statically? Or am I just completely off base?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: to make the point I'm trying to get at clear, I was able to find a mechanism that outlines what I'm talking about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IC8ae.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IC8ae.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wheel is driving the slider, won't the output force of the slider be at its peak value when  members 2 and 3 are near parallel to one another? I feel like the same effect should be at play in my system above, but I don't know how to mathematically articulate that. Or my intuition about when the force is at peak value could be totally wrong.&lt;/p&gt;&#xA;" OwnerUserId="3486" LastEditorUserId="3486" LastEditDate="2017-01-17T05:33:05.940" LastActivityDate="2017-05-31T16:17:56.620" Title="Equilibrium of a Rod with an Actuator" Tags="|statics|dynamics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13292" PostTypeId="2" ParentId="12311" CreationDate="2017-01-17T02:26:54.640" Score="4" Body="&lt;p&gt;There are many factors that make BLE low power, and I have attempted to address as many of them as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to better understand the power consumption differences between Bluetooth classic and BLE, it would helpful to look at some of the differences between the Bluetooth technologies. This would help appreciate the difference in power consumption. For starter Bluetooth classic consist of Bluetooth 1.0-3.0. These include Bluetooth BR (Basic Rate) round 1.2Mbit/sec, Bluetooth EDR (Enhance Data Rate) at 3Mbits/Sec and Bluetooth HS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bluetooth operates in 2.4 GHz ISM band, with Bluetooth classic uses 79 channels from 2.4GHz to 2.4835 GHz each spaced 1Hhz apart whereas BLE uses 40 channels from 2.402 GHz 2.480 GHz each spaced 2MHz apart. Of the 40 channels 3 of them are dedicated to advertising requests.  Initial parameters are exchanged using same channel used for the connection request.  Upon successful discover and connection, regular data channels are used for communication. Also note that advertising channels don’t overlap with Wifi &lt;a href=&quot;https://en.wikipedia.org/wiki/Direct-sequence_spread_spectrum&quot; rel=&quot;nofollow noreferrer&quot;&gt;Direct-sequence spread spectrum (DSSS)&lt;/a&gt; channels 1, 6, and 11. So Bluetooth uses a 2.4 GHz frequency band but implements a simpler &lt;a href=&quot;https://en.wikipedia.org/wiki/Frequency-shift_keying#Gaussian_frequency-shift_keying&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaussian Frequency Shifting Protocol&lt;/a&gt; to &lt;strong&gt;reduce power&lt;/strong&gt; as well as DSSS modulation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IER36.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LkvF8.jpg&quot; alt=&quot;rf spectrum br edr vs bluetooth smart&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2apaO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2apaO.png&quot; alt=&quot;advertising and data channels&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uQcI9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uQcI9.png&quot; alt=&quot;Bluetooth LE channel arrangements&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BLE has many different modes of which the main modes of operation are advertising mode, scanning mode, master device, and slave device. In the advertising mode the BLE base device will receive responses from other BLE devices for advertising events. On the scan mode the BLE device will scan for advertising request from other BLE devices and will respond with additional information depend on the status of the active scan status. There is also the passive mode, scanner only as well as advertiser only in which case the receiver and transmitter function of the RF module is required respectively. &#xA;Some understanding of the Link Layer State machine is beneficial to understanding management of &lt;strong&gt;power consumption&lt;/strong&gt;.  There are five states and they are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Standby&lt;/strong&gt;: Can be entered into from any other state and no transmitting or receiving packets&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Advertising&lt;/strong&gt;: This state can be entered from standby state. In this state the link layer will be transmitting advertising packets as well as responding to advertising related data exchanges&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Scanning&lt;/strong&gt;: Scanning state can be entered from standby state, which listens for advertising channel packets from devices&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Initiating&lt;/strong&gt;: The link layer in this state initiates a connection with another device responding to advertising channel packets from specific devices&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Connection&lt;/strong&gt;: The connection state has two defined roles, namely master and slave. A device in the master role will define timing for transmission&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kmapO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kmapO.jpg&quot; alt=&quot;State Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A connection is established by one device being in advertiser mode and other in initiator mode. The initiator becomes the master and the advertiser becomes the slave. This master slave data exchange defines critical connection parameters such as defining the channel and timing, which includes connection interval and salve latency.  The slave latency is important because this determines the number of connection intervals the slave can ignore without losing connection.  This helps the slave to optimize and &lt;strong&gt;preserve power consumption&lt;/strong&gt;. The slave can request to update the communication parameters to better fit the slave’s application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question you had referenced connection event.  The diagram below describes a connection event. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ORCV8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ORCV8.png&quot; alt=&quot;Connection Event&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power consumption during a connection event will be discussed later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A CONNECT_REQ PDU  is sent by the initiator or received  by the advertiser at which point the connection parameters are exchanged.  These parameters have a profound effect on power consumption. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Connection Interval&lt;/strong&gt; determines the time between two connections. This can be as low as 7.5 ms or as high as 4 sec. As one can imagine longer connection intervals mean low power consumption, but also means low data rates. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Slave Latency&lt;/strong&gt; defines the number of the consecutive connection events the slave can ignore from the master which once again have an effect on low power consumption&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Supervision Timeout&lt;/strong&gt; is the timeout between two received data packets before the connection is lost.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Bluetooth Low Energy framework too contributes to &lt;strong&gt;low power consumption&lt;/strong&gt;. The shortest packet transmitted can be 80 bits with a transmission time of 80usec. The longest packet can be 376 bits with a transmit time of around 0.3 mSec. These are very important for single mode BLE devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RvMMa.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RvMMa.gif&quot; alt=&quot;Packet Breakdown&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to manage power consumption as well as maintain legacy designs bluetooth 4.0 standards were developed.  Bluetooth 4.0 effectively has two modes, single and dual. The single mode supports &lt;strong&gt;low power&lt;/strong&gt; slave device using the standard better known as BLE. The dual mode as one might guess support both Bluetooth BR/EDR and BLE. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RNtSj.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RNtSj.gif&quot; alt=&quot;BLE 4.0 Configuration&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another power saving option is white lists. This allows the link layer to filter advertisers, initiators and scanners. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So BLE technology scans only 3 advertising channels bluetooth has to scan 32 channels. This is around 0.6 to 1.2 ms of discovery time for BLE as oppose to 22.5 ms of discovery time for Bluetooth, this the power savings for BLE. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also BLE devices in 3 ms can scan, connect, send data, confirm validate reception and terminate  where is Bluetooth take over 100 ms to accomplish he the same tasks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the BLE packets are much shorter than Bluetooth classic data packets which also contribute to power savings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To conclude this response below are scope measurements of connection events and related power consumption for bluetooth low energy which have been done on a TI CC2541.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lNgom.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lNgom.png&quot; alt=&quot;Power Consumption - 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6wcHB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6wcHB.png&quot; alt=&quot;Power Consumption - 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mF1eq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mF1eq.png&quot; alt=&quot;Power Consumption - 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=8uZzihQtV68&quot; rel=&quot;nofollow noreferrer&quot;&gt;MSP430F5529 CC2650 Bluetooth Low Energy Demonstration&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.digikey.com/en/articles/techzone/2011/dec/bluetooth-goes-ultra-low-power&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bluetooth Goes Ultra-Low-Power&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/12862/110&quot;&gt;Why do bluetooth headsets get interference (choppy sound quality) outdoors?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/258836/bluetooth-dual-mode-br-edr-le&quot;&gt;Bluetooth dual mode (BR/EDR + LE)&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://www.eetimes.com/document.asp?doc_id=1278927&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bluetooth 4.0: An introduction to Bluetooth Low Energy—Part I&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.eetimes.com/document.asp?doc_id=1278966&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bluetooth 4.0: An introduction to Bluetooth Low Energy—Part II&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.electronicdesign.com/industrial-automation/how-pick-best-bluetooth-protocol-your-application&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to Pick the Best Bluetooth Protocol for Your Application&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-07-27T11:06:17.593" LastActivityDate="2019-07-27T11:06:17.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13293" PostTypeId="2" ParentId="13287" CreationDate="2017-01-17T05:13:10.323" Score="3" Body="&lt;p&gt;Per the &lt;a href=&quot;https://en.wikipedia.org/wiki/Plunger_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plunger Pump Wiki&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A plunger pump is a type of positive displacement pump where the high-pressure seal is stationary and a smooth cylindrical plunger slides through the seal. This makes them different from piston pumps and allows them to be used at higher pressures.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R9OBi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R9OBi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-01-17T05:13:10.323" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13294" PostTypeId="1" CreationDate="2017-01-17T06:35:29.137" Score="0" ViewCount="47" Body="&lt;p&gt;Can this formula be used to evaluate the minimum thickness of a spherical pressure vessel that holds hydrogen gas at 10 ATM for a long time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma = \dfrac{p r}{2 t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told, &quot;As long as you don't need to hold the pressure for a long time or above room temperature, which will cause creep.&quot; It would seem that the person is trying to say that the container will fail over time due to creep. Is this true if the container will never go above room temperature? If yes, how should the design of the container change to address this problem?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="7538" LastEditDate="2017-01-17T16:22:15.857" LastActivityDate="2017-01-19T19:49:43.510" Title="Long-term storage of a gas?" Tags="|pressure|design|pressure-vessel|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13296" PostTypeId="1" CreationDate="2017-01-17T07:16:45.330" Score="1" ViewCount="49" Body="&lt;p&gt;If these exist, then I would imagine they are in research. Is an M.Eng. looked as favorably upon as an M.S.? &lt;/p&gt;&#xA;" OwnerUserId="7926" LastActivityDate="2017-01-17T07:16:45.330" Title="Are there ever job openings that say &quot;Only M.S. degrees apply&quot;, and if so, will an M.Eng. still cut it?" Tags="|mechanical-engineering|employment|" AnswerCount="0" CommentCount="11" ClosedDate="2017-01-17T14:27:29.593" ContentLicense="CC BY-SA 3.0" />
  <row Id="13297" PostTypeId="2" ParentId="13268" CreationDate="2017-01-17T08:29:49.727" Score="3" Body="&lt;h1&gt;TL;DR;&lt;/h1&gt;&#xA;&lt;p&gt;Possible but not nesseserily worth it.&lt;/p&gt;&#xA;&lt;h1&gt;Introduction&lt;/h1&gt;&#xA;&lt;p&gt;Its possible, but it is a bit questionable whether its actually practical or saves you any time. See when you photograph a section you get into measurement, and measurement is hard. Even just photographing that you get a easily traceable image needs some careful thinking about setup. In addition you get into trouble with noise, and distortions caused by lens and not entirely perpendicular camera shooting angle. You can handle all this but its quite much work. Also you need much more than one shape for this setup to be beneficial.&lt;/p&gt;&#xA;&lt;h1&gt;Photographing better&lt;/h1&gt;&#xA;&lt;p&gt;Even so there is quite much you can do for the photography session.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Using the scanner as suggested by user &lt;a href=&quot;https://engineering.stackexchange.com/users/4314/mikhail-v&quot;&gt;Mikhail V&lt;/a&gt; eliminates many of your problems. Its not prone to perspective distortions (except to one caused by uneven cut), and comes with its own light setup. The rubbing suggested can also be a good way to make the image flat for photography too. You get the scale.&lt;/li&gt;&#xA;&lt;li&gt;Use a lens with a flat focus plane. Most macro lenses do but most general purpose and portrait lenses do not.&lt;/li&gt;&#xA;&lt;li&gt;Use a lens with a long focal length, the longer focal length the more like a orthographic your projection will be and the more uniform your lens distortion. You can also photograph form further away and use less of the resolution area of the lens for same effect.&lt;/li&gt;&#xA;&lt;li&gt;Use a tripod.&lt;/li&gt;&#xA;&lt;li&gt;Use a marker/brush to make the surface you want to capture colored. This will make it easier to see and photograph the end. And makes a huge difference in the masking stage (and can in fact eliminate it entirely).&lt;/li&gt;&#xA;&lt;li&gt;Use a diffuser on your light, or use a soft box for photographing.&lt;/li&gt;&#xA;&lt;li&gt;Print a checkerboard on the background so that you can correct for distortions and perspective errors. Also would be good if the object is a flat slice but you can not have everything.&lt;/li&gt;&#xA;&lt;li&gt;Include a scale in your image on the plane your trying to capture. Images captured by a camera  lose scale information.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;In general its best to actually spend time on the photo shoot to get things correct as it saves you lot of time in the masking stage.&lt;/p&gt;&#xA;&lt;h1&gt;Digital conversion of image to&lt;/h1&gt;&#xA;&lt;p&gt;You can simply do this in Photoshop, the process is pretty easy. Once you have selected the area you want to capture, and have a selection tool enabled. Right click on your canvas and choose &lt;strong&gt;Make Work Path...&lt;/strong&gt;, this will simply trace the selection edge.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xCyZT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xCyZT.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: Before and after making selection a work path&lt;/p&gt;&#xA;&lt;p&gt;You can then use &lt;strong&gt;File → Export → Path to Illustrator&lt;/strong&gt; this will create an old AI file that is in postscript format. You can either open it in illustrator (or copy paste the path) and save it as DXF and then read it in form there. Or you can just read the postscript file in AutoCAD with many available importers. (or write your own the format is fairly trivial)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QNJmq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QNJmq.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Image 2&lt;/strong&gt;: Resulting quickly masked image path in AutoCAD, image lost scale and could have been done better in first place with better source images.&lt;/p&gt;&#xA;&lt;p&gt;Now the results aren't stellar because your source image is bad, one can do much better with more advanced masking techniques*. In general the results will not be super good as all your actual geometric flaws are in the image and round holes will no longer be round for example. And the design is supposed to depict the ideal object. But in general if you take precautions and work your photography session slightly more then you can use a camera as a cheap coordinate measuring device (look up stereo geometry).&lt;/p&gt;&#xA;&lt;p&gt;But in this case i think you'd be better off taking the object in one hand and calipers in the other. I mean I wouldn't scan this thing, but your mileage may differ.&lt;/p&gt;&#xA;&lt;p&gt;* Some of which use manual tracing which you wnated to avoid.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-01-21T20:27:23.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13298" PostTypeId="1" AcceptedAnswerId="13301" CreationDate="2017-01-17T09:39:06.980" Score="3" ViewCount="632" Body="&lt;p&gt;I am studying Gaudí and how does he created some of his work. I found that he used inverted weight models (as the shown in the picture below) to recreate the catenary. In many articles it is pointed as a good practice to simplify and find a better/more sophisticated and solid structure, but I can't see why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1467f.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1467f.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1262" LastActivityDate="2017-01-17T12:13:47.080" Title="Does the use of the catenary in an inverted weight model imply a more solid structure?" Tags="|structures|architecture|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13299" PostTypeId="1" CreationDate="2017-01-17T09:59:22.507" Score="2" ViewCount="423" Body="&lt;p&gt;Gears are used to transfer power between two &lt;em&gt;rotating&lt;/em&gt; shafts, either by increasing or decreasing the angular speed, with a corresponding inverse change in the torque. If I want to do the same for &lt;em&gt;linear&lt;/em&gt; motion, I see only two options in front of me:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Use Pascals law for liquids and connect pipe of varying diameters to multiply (or divide ) the force, as is done in car jacks, or, &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Connect the input linear drive to a rotating shaft, connect it to a step-up or step down gear, and then convert rotary motion back to linear. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hydraulic cylinders are kind of costly and keeping the fluid at a higher pressure is not quite convenient. Also using a linear-rotary-linear power transfer is easy to build but quite inefficient. Is there another way to do this?&lt;/p&gt;&#xA;" OwnerUserId="7057" LastActivityDate="2017-01-18T19:01:16.687" Title="Linear force multiplication (or division) without hydraulics" Tags="|gears|hydraulics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13300" PostTypeId="1" CreationDate="2017-01-17T10:14:25.543" Score="0" ViewCount="90" Body="&lt;p&gt;How can I calculate the cost of heating a building such as &lt;a href=&quot;http://group.canarywharf.com/estate-building/5_north_colonnade/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://group.canarywharf.com/estate-building/5_north_colonnade/&lt;/a&gt; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its size: 59,142 sq m / 636,600 sq ft&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature is in many places more than 27 degrees celsius. I would like to show, that an investment in better heating control would pay off quickly, if the temperature was reduced to 20 degrees celsius.&lt;/p&gt;&#xA;" OwnerUserId="9605" LastEditorUserId="9" LastEditDate="2017-01-17T12:23:26.923" LastActivityDate="2017-01-18T20:04:15.123" Title="What is the cost of heating of a 60k sqm building to 27 degrees celsius vs 20 degrees celsius?" Tags="|heating-systems|" AnswerCount="2" CommentCount="9" ClosedDate="2017-02-02T16:23:11.390" ContentLicense="CC BY-SA 3.0" />
  <row Id="13301" PostTypeId="2" ParentId="13298" CreationDate="2017-01-17T12:13:47.080" Score="2" Body="&lt;p&gt;A catenary is the shape a cable will hang in under self-weight. Cables only work in tension, not in bending. If you take this shape and flip it upside down, then (due to gravity now operating in the opposite direction relative to the structure) the structure is purely in compression, with no bending.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly this wouldn't work for a cable, as they don't work well under compressive loads. But it does work well for concrete structures, or for steel sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial force &lt;strong&gt;&lt;em&gt;P&lt;/em&gt;&lt;/strong&gt; over an area &lt;strong&gt;&lt;em&gt;A&lt;/em&gt;&lt;/strong&gt; causes a stress of &lt;strong&gt;&lt;em&gt;P/A&lt;/em&gt;&lt;/strong&gt;. Moment &lt;strong&gt;&lt;em&gt;M&lt;/em&gt;&lt;/strong&gt; on a section with an elastic section modulus &lt;strong&gt;&lt;em&gt;Z&lt;/em&gt;&lt;/strong&gt; causes a stress &lt;strong&gt;&lt;em&gt;+M/Z&lt;/em&gt;&lt;/strong&gt; on one edge, linearly decreasing to 0 at the neutral axis, and further decreasing to &lt;strong&gt;&lt;em&gt;-M/Z&lt;/em&gt;&lt;/strong&gt; on the other edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you have a section under compression and bending, then stress at one edge is equal to &lt;strong&gt;&lt;em&gt;P/A + M/Z&lt;/em&gt;&lt;/strong&gt;, and at the other edge &lt;strong&gt;&lt;em&gt;P/A - M/Z&lt;/em&gt;&lt;/strong&gt;. Clearly if you design so that your maximum stress (P/A + M/Z) is equal to your allowable stress, then most of your section is at less than the allowable stress. However, if you only have compression, then your whole section is at P/A, and if you design such that this is equal to your allowable stress then your whole section is fully utilised - much more efficient.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-01-17T12:13:47.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13302" PostTypeId="2" ParentId="13300" CreationDate="2017-01-17T14:08:33.977" Score="3" Body="&lt;p&gt;This is a heat transfer problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get a first order estimate by considering only conductive heat transfer, neglecting radiative and convective.  Simplifying it even more, (again we don't know anything about your building, so this is the best we can do)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;q = k*dT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where q is the RATE of heat transfer, dT is the temperature difference between outside and inside, and k is a constant that we determine experimentally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume it's 10 degrees C out.  The average difference between 20 and 10 is 5 (that is, dT changes every time the temperature changes, but we're simplifying, so let's set it to somewhere in the middle).  So we'll let dT = 5.  So, per hour (or minute, or second) you have 5*k units of heat exiting the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Versus 8.5*k units of heat per hour if the temperature is set to 27 C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how much do you pay per unit of heat?  Multiply that by 3.5, and there's the savings &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind this is a VERY rough estimate.  I've neglected a LOT of factors.  Your problem may be a matter of configuring the HVAC system for optimal air flow, or getting new insulation.  I'm neglecting the shape of your building vs how the wind/sun impact it all day/season/year long.&lt;/p&gt;&#xA;" OwnerUserId="6488" LastActivityDate="2017-01-17T14:08:33.977" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13303" PostTypeId="1" CreationDate="2017-01-17T14:27:19.807" Score="-1" ViewCount="144" Body="&lt;p&gt;I've worked in SolidWorks for 3 years now and have found it to be a highly productive and capable software. I recently heard in passing that Autodesk surface modelling is far superior to SolidWorks. His case was that SolidWorks is great fro mechanical design but that Autodesk tools are much better for surface modelling. My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In what ways are the Autodesk Suite tools superior to Solidworks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What is the difference between &quot;surface modelling&quot; and the mechanical design stuff SolidWorks facilitates?&lt;/p&gt;&#xA;" OwnerUserId="9143" LastActivityDate="2017-01-17T16:53:40.810" Title="Autodesk Suite vs SolidWorks comparison" Tags="|computer-aided-design|solidworks|" AnswerCount="2" CommentCount="0" ClosedDate="2017-01-18T05:16:29.017" ContentLicense="CC BY-SA 3.0" />
  <row Id="13304" PostTypeId="2" ParentId="13303" CreationDate="2017-01-17T16:10:44.583" Score="1" Body="&lt;p&gt;This may be the fact that the Autodesk brand covers quite a broad range of software including more graphics orientated products like Maya and Mudbox which are intended for things like animation whereas Solidworks is more narrowly focused on engineering and manufacturing applications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically the more 'artistic' surface modelling software is focused on manipulating complex meshes eg for creating realistic face and figure models and is much more geared towards creating complex organic surfaces rather than building up solids according to specific dimensions. For example Mudbox is more or less a digital equivalent of modelling in clay as opposed to assembling a kit from carefully dimensioned parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that the Autodesk stable does include things like Inventor which very capable for designing mechanical parts and assemblies and includes things like basic finite element analysis and specific environments for welded and sheet metal assemblies as well as tools for things like threaded fasteners, holes as well as standard section materials (I beams etc) as well as the ability to easily generate projected drawings, parts lists etc from models and assemblies. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-01-17T16:10:44.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13306" PostTypeId="2" ParentId="13303" CreationDate="2017-01-17T16:53:40.810" Score="2" Body="&lt;p&gt;First of all, the comparison you suggest isn't really valid since SolidWorks is really a parametric solid modeling software package, while The Autodesk Suite encompasses several different software packages, one of which, Inventor, is roughly the functional equivalent to Solidworks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an attept to answser your questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The surface modelling functions that Autodesk offers are primarily part of other Autodesk software packages and not Inventor. From my perspective, the choice between one or the other comes down to interoperability with other engineering software like Product Lifecycle Management/Product Data Management tools, how they deal with legacy data and, in many cases, interoperability with the customer's software. &lt;/li&gt;&#xA;&lt;li&gt;As others have mentioned, surface modeling is just that, modeling of just the surface. There's no material properties and the models aren't parametrically driven. Their real application is in things where apprearance is the primary goal (think artist's renderings, sales brochures or animation).  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I've used both Inventor and Solidworks (along with Catia and Pro/E), though never as part of my primary work function, mostly to revise and modify existing parts. I prefer Autodesk Inventor over Solidworks, but it's really just personal, I know others who feel exactly the opposite.  &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-01-17T16:53:40.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13308" PostTypeId="1" CreationDate="2017-01-17T18:46:20.113" Score="3" ViewCount="104" Body="&lt;p&gt;I have some rolls of coated textile I want to characterize, including some measure of how well the material conducts electricity in the plane of the textile so I can judge which type is least likely to produce electrostatic sparks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found two methods used for measuring surface resistivity or resistance in high-resistance materials.  The first is using a flat plate at either edge of a square sample, where sample size is irrelevant as long as it's square.  This gives resistivity in ohms per square.  This would require me to cut samples and would prohibit in-situ testing of the end-product that this textile is going to be used for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second method used a set of needle-point probes to pick random points on a sheet.  First a measurement was taken with 1cm between the probes, and then a second measurement was taken with a 10cm distance between the probes.  The results were not the same, but no explanation was given as to why two measurements were taken or what this technique is useful for.  This technique would be much easier to implement, but I'm not sure if the results are useful or how they should be interpreted.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the appropriate measurement method for my application?  Are there any standard procedures for performing the second method I described above?&lt;/p&gt;&#xA;" OwnerUserId="9602" LastActivityDate="2017-01-17T18:46:20.113" Title="Resistance measurement for electrically conductive textile sheet" Tags="|electrical-engineering|materials|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13310" PostTypeId="1" AcceptedAnswerId="13315" CreationDate="2017-01-17T23:35:25.920" Score="1" ViewCount="868" Body="&lt;p&gt;I'm trying to build a self-regenerating door lock as a side project in college and i'm having some difficulty accurately calculating the gear trains output (in rpm), the purpose of this gearing system is to operate a AC generator. Any help at all would be much appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've estimated that a torque of about 1.12 Nm will be used to turn the door handle down 50 degrees. This torque will be transmitted through the shaft which is attached internally to a one way gear. The one way gear is attached to both a fly wheel and the gear train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to get an output of about 100 rpm to operate the generator. My question is what factors must i include in the calculation? do i need to include things such as the weight of the gear train and how do i factor in the fly wheel? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9619" OwnerDisplayName="Tom" LastActivityDate="2017-01-18T10:46:14.827" Title="How to accurately calculate the rpm from a gear train" Tags="|generator|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13311" PostTypeId="2" ParentId="13283" CreationDate="2017-01-18T07:33:14.597" Score="2" Body="&lt;p&gt;For&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Re &amp;lt; 2300$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.1 \leq Pr \leq \infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0.1 \leq Pr \frac{d}{h} &amp;lt; 10^4$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the correlation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Nu_m = \left( 49.37 + (1.615\frac{Pe \cdot d}{h})^{\frac{1}{3}} -0.7)^3\right)^{\frac{1}{3}} K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d = hydraulic diameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h = length of pipe&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K = 1$ for gases&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K = \left(\frac{Pr_f}{Pr_w}\right)^{0.11}$ for liquids&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pr_f$ = Pr for fluid temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Pr_w$ = Pr for wall temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.schweizer-fn.de/waerme/waermeuebergang/waerme_uebergang.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a native speaker, so someone please feel free to add a source for that correlation in English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the geometry is not circular, you use the hydraulic diameter, as always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may derive it from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_h = \frac {4 A}{U}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A denotes the crosssectional area and U the circumference.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastEditorUserId="1632" LastEditDate="2017-01-19T11:59:40.120" LastActivityDate="2017-01-19T11:59:40.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13312" PostTypeId="2" ParentId="13259" CreationDate="2017-01-18T08:20:51.820" Score="1" Body="&lt;p&gt;Mount two surface temperature sensors (or the hot and cold junctions of a thermocouple arrangement) to the pipe say 100mm (4&quot;) apart.  Place a resistive heater between the two temperature sensors.  Add 5-20W of heat to the pipe with the heater.  Compare the temperatures.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When water is stationary you will measure the same temperature at both locations.    &lt;/li&gt;&#xA;&lt;li&gt;When water at above or below ambient flows (and stops) you will initially see large change in both sensor values and small differential due to additional heat indicating flow direction.      &lt;/li&gt;&#xA;&lt;li&gt;Water flowing at ambient temperature will only result in small change in one sensor indicating flow direction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3497" LastActivityDate="2017-01-18T08:20:51.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13314" PostTypeId="2" ParentId="12982" CreationDate="2017-01-18T09:46:50.257" Score="1" Body="&lt;p&gt;It's not easy to identify which kind of potting material it is. However, you can try the following common ways of de-potting, ordered from easiest to hardest:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Heating it up. Some compounds separate from the substrate when you just heat it to about 150 Celsius. Soaking it in water between heating sessions can help the separation.&lt;/li&gt;&#xA;&lt;li&gt;Hit it with a hammer. If the coating is only on one side of the PCB like in the image, a heavy blow can cause it to separate.&lt;/li&gt;&#xA;&lt;li&gt;Acetone bath. Leave it for an hour, scrape off any goo, repeat until all plastic is gone. Basic precautions like protective gloves and goggles are necessary.&lt;/li&gt;&#xA;&lt;li&gt;Heated nitric acid bath. This will need about 70% nitric acid and fume removal. I wouldn't attempt this without a proper lab setup.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Obviously all of these have significant risk of damaging the electronics.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastEditorUserId="8055" LastEditDate="2017-01-19T16:51:16.777" LastActivityDate="2017-01-19T16:51:16.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13315" PostTypeId="2" ParentId="13310" CreationDate="2017-01-18T10:24:53.660" Score="2" Body="&lt;p&gt;Obtain a cheap plastic hand crank generator or several and play.&lt;br&gt;&#xA;That will show you much of what you need to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you turn the handle 1/8th turn (45 degrees) in 1/4 second then its&lt;br&gt;&#xA;RPM is 60/time_to_turn x fraction of turn =&lt;br&gt;&#xA;60s/0.25s x 1/8 turn ~~=  30 RPM.&lt;br&gt;&#xA;In engineering 30 ~~~= 100 in many cases :-).&lt;br&gt;&#xA;But if you want closer to 100 RPM then a 3 or 4 to 1 step up gear ratio will give around 100 RPM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO coupling the shaft to the alternator with a one way clutch and a say 4:1 speed up gearbox gives you about what you want. One stage of the epicyclic gearbox in many battery electric drills may be about right. These may have multiple stages to get higher ratios.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just turning the chuck on a battery electric drill and measuring motor voltage produced may be worthwhile. Slightly tidier than &lt;a href=&quot;http://media.treehugger.com/assets/images/2013/02/macgyver-generator.jpg.650x0_q70_crop-smart.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; ideally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful rule of thumb is Power is ~~~= RPM x kgm &#xA;Here RPM = 100 and torque = torque in / gear ratio = 1.12 Nm/4 = 0.28 Nm ~= 0.028 kg.m&lt;br&gt;&#xA;So power = kg.m x RPM = 0.028 x 100 = 2.8 Watts.&#xA;This is produced only during 1/4 second of turning so energy&lt;br&gt;&#xA;= power x time = 2.8W x 0.25s = 0.9 Watt seconds = 0.9 Joule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A useful amount of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My experiments with cheap plastic squeeze type hand alternators shows they make about 1 Watt with very enthusiastic cranking so the above calculation is &quot;in the order of right&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flywheel does not alter energy delivered but can modify user handle-feel and rate of delivery - eg if a flywheel is spun up by a 1 way clutch it may spin down over 1+ seconds and deliver energy at a less peaky rate.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;______________________________________&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus:&lt;/strong&gt; Obtaining energy from the opening or closing door is liable to be rather more productive if suitable to purpose. The door can eg have a spring closer and drive the alternator on one or both halves of open/close cycle. Energy taken from this source is liable to less noticed and easier to obtain in greater quantity. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.nz/search?tbm=isch&amp;amp;sa=1&amp;amp;q=hand+crank+generator&amp;amp;oq=hand+crank+generator&amp;amp;gs_l=img.3..0l9.989370.994934.0.995430.20.15.0.5.5.0.358.2325.2-9j1.10.0....0...1c.1.64.img..5.15.2338...0i67k1j0i10k1.tJlPO4CMl9o&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Many&lt;/strong&gt; ideas here&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://www.google.co.nz/search?tbm=isch&amp;amp;sa=1&amp;amp;q=hand+squeeze+generator&amp;amp;oq=hand+squeeze+generator&amp;amp;gs_l=img.3...44995.47128.0.47963.8.8.0.0.0.0.226.1697.2-8.8.0....0...1c.1.64.img..0.1.225...0i7i30k1j0i8i7i30k1.dnq3vanXmNo&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;and here&lt;/strong&gt;&lt;/a&gt; - each image links to a related page.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Cheap plastic hand crank alternator.&#xA;Mechanical efficiency is usually poor.&#xA;Note gear ratio is probably higher than apparent as another gear is probably not shown. ie their alternator spins MUCH faster than yours. The plastic Z-ish shaped piece in the flywheel acts as a one way clutch/pawl.&lt;br&gt;&#xA;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:%D0%9C%D0%B5%D1%85%D0%B0%D0%BD%D0%B8%D1%87%D0%B5%D1%81%D0%BA%D0%B8%D0%B9_%D1%84%D0%BE%D0%BD%D0%B0%D1%80%D1%8C_inside_(1).jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;From Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/10fwc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/10fwc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://patentimages.storage.googleapis.com/US7893551B2/US07893551-20110222-D00000.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Patent imahge&lt;/a&gt;&lt;br&gt;&#xA;from here &lt;a href=&quot;https://www.google.com/patents/US7893551&quot; rel=&quot;nofollow noreferrer&quot;&gt;hand squeeze generator&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can buy &lt;a href=&quot;http://www.ebay.com/itm/3-Phase-AC-Micro-Mini-Wind-Hand-Brushless-Generator-Teaching-Model-DIY-Motor-24V-/272526120811&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt; for far far too much $ on ebay OR pull apart old CD or DVD drives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4CMEx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4CMEx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stepper motors also may serve. Old large ones from 5.25&quot; floppy drives (remember them :-) ) are usually better.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2017-01-18T10:46:14.827" LastActivityDate="2017-01-18T10:46:14.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13316" PostTypeId="2" ParentId="11357" CreationDate="2017-01-18T14:19:14.307" Score="2" Body="&lt;p&gt;The following sentence is not correct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the moment of inertia is given by the integer of an area times the square of the distance from its centroid to the axis&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have to add to that, the moment of inertia of the area around its own centroid. That is what the parrallel axis theorem is all about:&#xA;$$ I = I_o + A\cdot d^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&#xA; - Io the moment of inertia around centroid&#xA; - I is the moment of inertia around any parallel axis and &#xA; - d the distance between the two axes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So applying the above to your example, each half area (below and above centroidal axis) should have a moment of inertia equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I_{half} = \frac{b (h/2)^3}{12} + \frac{bh}{2}\cdot\left(\frac{h}{4}\right)^2 $$&#xA;$$ I_{half} = \frac{b h^3}{96} + \frac{b h^3}{32}  $$&#xA;$$ I_{half} = \frac{b h^3}{24} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, for the whole section, due to symmetry: &#xA;$$ I = 2 I_{half} = \frac{b h^3}{12} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Demonstrating the example with your numbers:&#xA;$$ I = 2\left(\frac{20 (100)^3}{12} + 20\cdot 100\cdot\left(50\right)^2 \right)\,mm^4$$&#xA;$$ I = 2\left(1666666.7 + 5000000 \right) \,mm^4 $$&#xA;$$ I = 13333333.3 \,mm^4 = 1333.33 cm^4 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually in enginnereing cross sections the parallel axis term $Ad^2$ is much bigger than the centroidal term $I_o$. It is rather acceptable to ignore the centroidal term for the flange of an I/H section for example, because d is big and flange thickness (the h in the above formulas) is quite small. In other circumstances however this is not accepteble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://calcresource.com/moment-of-inertia-parallel.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calculation of moment of inertia for parallel axes&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="9620" LastEditorUserId="9620" LastEditDate="2017-01-18T21:12:37.257" LastActivityDate="2017-01-18T21:12:37.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13319" PostTypeId="1" CreationDate="2017-01-18T16:47:39.790" Score="1" ViewCount="34" Body="&lt;p&gt;I plan to mimic binary stars and their orbital decay due to gravitational wave radiation. I want to make a rotating shaft with extended arms, each being able to hold cast iron balls of specific weight, hence mimicking a binary with a circular orbit, when rotated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This I can construct, my problem however is that I need the extended arms to vary in length such that they trace an ellipse with a specific semi-major axis and eccentricity. The arm must continue to vary in both of these parameters according to the ODE's which explain the time dependency as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$\frac{de}{dt}=-\frac{304}{15}\frac{G^{3}m_{1}m_{2}(m_{1}+m_{2})}{c^{5}a^{4}(1-e^{2})^{\frac{5}{2}}}\left[1+\frac{121}{304}e^{2}\right]$$&#xA;  $$\frac{da}{dt}=-\frac{64}{5}\frac{G^{3}m_{1}m_{2}(m_{1}+m_{2})}{c^{5}a^{4}(1-e^{2})^{2}}\left[1+\frac{73}{24}e^{2}+\frac{37}{96}e^{4}\right]$$&#xA;  Note: The value of 'c' and 'G' I would be taking are  1 and 2.226e-19, to simplify the calculations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would then measure the time taken via a sensor, this would be the life time of the binary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I build such a device? Is there any specific tech available that I can use for this specific issue that I have highlighted? Or can anyone provide some idea how to go ahead. Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="8375" LastEditorUserId="8375" LastEditDate="2017-01-18T18:13:48.630" LastActivityDate="2017-01-18T18:13:48.630" Title="Creating a rotating unit with the extended arms varying the trajectory in accordance to ODE's" Tags="|mechanical-engineering|electrical-engineering|materials|experimental-physics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13320" PostTypeId="2" ParentId="13299" CreationDate="2017-01-18T19:01:16.687" Score="2" Body="&lt;p&gt;Take into a count what you want to do. Hydraulics is used for a reason: it can transmit huge power. The idea of using gears and then using a linear one is good, but you need to know if it is going to be able to handle the weight and internal reactions on the gears. Also, take into a count the size of the gears, if they already exist inside a catalog and you can buy them easily, because if you have to manufacture them you will find it is really expensive.&lt;/p&gt;&#xA;" OwnerUserId="9623" LastActivityDate="2017-01-18T19:01:16.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13322" PostTypeId="2" ParentId="13300" CreationDate="2017-01-18T20:04:15.123" Score="-1" Body="&lt;p&gt;This webpage has a variety of equations including heat transfer/loss of windows, walls, and thermal properties of common building materials.  &lt;a href=&quot;http://www.engineersedge.com/heat_transfer/heat_transfer_table_content.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineersedge.com/heat_transfer/heat_transfer_table_content.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9624" LastActivityDate="2017-01-18T20:04:15.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13323" PostTypeId="2" ParentId="13264" CreationDate="2017-01-18T20:09:14.053" Score="-2" Body="&lt;p&gt;Yield should be in MPa, Pressure Pa, and R in mm t is thickness therefor mm&lt;/p&gt;&#xA;" OwnerUserId="9624" LastActivityDate="2017-01-18T20:09:14.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13324" PostTypeId="2" ParentId="13204" CreationDate="2017-01-18T20:15:55.917" Score="0" Body="&lt;p&gt;&quot;Try working on a collaborative project with half a dozen large companies all using different CAD software, and the probability that &quot;the solid model&quot; will look the same when they all import it into their own system is just about zero.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I disagree, step file translation and exchange is industry standard works great..  Undimensioned drawings are common practice... Partial dimensioned drawings are also called &quot;Critical Feature Drawings&quot; or &quot;Limited Dimension Drawings&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google it..&lt;/p&gt;&#xA;" OwnerUserId="9624" LastActivityDate="2017-01-18T20:15:55.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13325" PostTypeId="2" ParentId="13162" CreationDate="2017-01-18T20:23:01.777" Score="1" Body="&lt;p&gt;You need to determine the minimum thread engagement for full thread strength requirements of the weaker material in the assembly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equations and calculators for thread pullout pet industriy standards, See:  &lt;a href=&quot;http://www.engineersedge.com/fastener_thread_menu.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineersedge.com/fastener_thread_menu.shtml&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also evaluate the environment and corrosion risks of SS and Aluminum in your application. &lt;/p&gt;&#xA;" OwnerUserId="9624" LastActivityDate="2017-01-18T20:23:01.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13326" PostTypeId="1" AcceptedAnswerId="13330" CreationDate="2017-01-19T08:39:15.300" Score="3" ViewCount="1825" Body="&lt;p&gt;I need to calculate the plastic section modulus of a rectangular section with rounded corners. First I need to know the formula for first moment area of a quadrant. I can't find it anywhere on the internet. Does anyone know?&lt;/p&gt;&#xA;" OwnerUserId="9630" LastEditorUserId="9620" LastEditDate="2017-01-19T17:08:20.477" LastActivityDate="2017-01-22T02:23:34.253" Title="What is the first moment area of a rounded rectangle?" Tags="|civil-engineering|mathematics|" AnswerCount="1" CommentCount="2" ClosedDate="2017-01-19T19:44:22.020" ContentLicense="CC BY-SA 3.0" />
  <row Id="13327" PostTypeId="1" CreationDate="2017-01-19T11:56:41.803" Score="3" ViewCount="186" Body="&lt;p&gt;When I look at the types of &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_screw_drives&quot; rel=&quot;nofollow noreferrer&quot;&gt;screw drives&lt;/a&gt;, I see... a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;History probably explains why so many existed, but I fail to understand why some &lt;em&gt;still&lt;/em&gt; exist today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, compare Double Hex &lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/6/60/Screw_Head_-_Double_Hex.svg/40px-Screw_Head_-_Double_Hex.svg.png&quot; alt=&quot;&quot;&gt; with Spline &lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/f/f0/Screw_Head_-_Spline.svg/40px-Screw_Head_-_Spline.svg.png&quot; alt=&quot;&quot;&gt;.&lt;br&gt;&#xA;It seems &lt;em&gt;obviously clear&lt;/em&gt; to me that the Double Hex has a higher chance of wearing out or slipping, and it seems to have no advantage whatsoever.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or for example, compare the usual Hex Socket &lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/c/ce/Screw_Head_-_Hex_Socket.svg/40px-Screw_Head_-_Hex_Socket.svg.png&quot; alt=&quot;&quot;&gt; with the Robertson &lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/54/Screw_Head_-_Robertson.svg/40px-Screw_Head_-_Robertson.svg.png&quot; alt=&quot;&quot;&gt;.&lt;br&gt;&#xA;Again, the Hex Socket appears to have a far higher chance of eventually wearing out or slipping, and again, no apparent advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do these keep being used? Do they &lt;em&gt;really&lt;/em&gt; &lt;strong&gt;all&lt;/strong&gt; have at least one physical advantage over the others? Or is it a cost thing, or is it something else?&lt;/p&gt;&#xA;" OwnerUserId="4985" LastEditorUserId="615" LastEditDate="2017-01-19T14:41:16.913" LastActivityDate="2017-01-19T15:56:16.283" Title="Why are there so many types of screw head drives? Why aren't some strictly better than others?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13328" PostTypeId="1" CreationDate="2017-01-19T15:32:14.313" Score="1" ViewCount="483" Body="&lt;p&gt;I have a small drum that will be immersed half way in a water bath (drag). The drum will contain some 500ml of photo chemicals (weight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;is there an easy way to calculate the gear ratio to get between 35 rpm and 75 rpm?&lt;/li&gt;&#xA;&lt;li&gt;OR is there a gear ratio that will keep the rpm of the motor (variable speed motor)?&lt;/li&gt;&#xA;&lt;li&gt;considering drag/weight should I worry about stress on gears? Should I use bushes (brass/nylon) or small bearings around shafts?&lt;/li&gt;&#xA;&lt;li&gt;what kind of electric motor do I need (DC motor for toy or drill?)&lt;/li&gt;&#xA;&lt;li&gt;do I need a lot of torque to obtain durability?&lt;/li&gt;&#xA;&lt;li&gt;using gears, is there an easy way to calculate the necessary motor power/torque vs weight of drum (if I use a bigger drum)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;See drawing for general concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for helping me to learn!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xsgZF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xsgZF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9632" LastActivityDate="2017-02-18T18:12:34.620" Title="Gear ratio and motor power" Tags="|applied-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13329" PostTypeId="2" ParentId="13327" CreationDate="2017-01-19T15:46:47.497" Score="4" Body="&lt;p&gt;If you read the Wikipedia page you've linked to, it gives the reasoning behind the deveolpment of many of the types of screw drive. Some ARE better than others, just not at the same thing. There are design tradeoffs to be made when selecting fasteners. For example, some applications may require that a unique tool be used to limit access. (e.g. Apple's use of the pentalobe screw in the iPhone or the penta screw commonly used on municipal water systems) For other applications, it may be desirable to have a commonly available tool such as a phillips screwdriver. For other applications, it may be necessary to be able to drive a lot of screws easily and &quot;camming out&quot; may be an issue...the list of choices goes on and on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, don't underestimate the power of cost. Different heads will have a higher or lower cost of manufacture and thus a higher or lower cost to produce. The difference between a fastener that costs 0.03 USD vs. a fastener that costs 0.035 USD may not seem like much, but if you're designing something that uses 12 of them and has a production volume in the millions of units, that adds up. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2017-01-19T15:56:16.283" LastActivityDate="2017-01-19T15:56:16.283" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13330" PostTypeId="2" ParentId="13326" CreationDate="2017-01-19T16:08:12.447" Score="3" Body="&lt;p&gt;I don't have a simple formula but that is the way i would handle it. Assuming the four corners are circular with equal radius r, so that symmetry exists, the first moment of area of &lt;strong&gt;one half&lt;/strong&gt; of the section should be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{half} = S_{half\,rect}-2S_{corner}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$S_{half\,rect} = b(h/2)(h/4) = \frac{bh^2}{8} $ the 1st moment of&#xA;area of the half rectangle (with intact corners) around neutral axis and&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;$S_{corner}$ the 1st moment of area of the removed material from each&#xA;corner, around neutral axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GpzzT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GpzzT.png&quot; alt=&quot;basic shape&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;How to find $S_{corner}$&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We have to find its area and the distance of its centroid from the section neutral axis. It helps to consider the removed corner as a quarter circle subtracted from an r x r square (see figure below). The area is then simply found, by subtracting form the small r x r square, a circular quarter (see figure below):&#xA;$$A_{corner} = r^2 - \frac{πr^2}{4} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DMlOz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DMlOz.png&quot; alt=&quot;detail of the removed corner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The centroid of the removed corner area, relative to the top edge, is similarly found  considering the centroid of the rxr square (red dot) which is located $y=r/2$ below top edge and the centroid of the quarter circle (blue dot) which is located $y=r-\frac{4r}{3\pi}$ from the top edge. Combining them both, we get the distance of centroid of the removed corner from the top edge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ y_{corner} = \frac{1}{A_{corner}}\left(r^2\cdot\frac{r}{2} - \frac{\pi r^2}{4} \left(r-\frac{4r}{3\pi}\right)\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having found the above, the first moment of area of one removed corner around the section neutral axis is:&#xA;$$ S_{corner} = A_{corner}\left(\frac{h}{2} - y_{corner}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Finally&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Substituting $ S_{corner}$ to the 1st formula, we get the 1st moment of area of the half section $S_{half}$. Then, the plastic modulus of the total section, taking advantage of symmmetry, is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z=2S_{half}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For verification of the result &lt;a href=&quot;http://calcresource.com/cross-section-rectround.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;a rounded rectangle calculator&lt;/a&gt; may prove handy. &lt;/p&gt;&#xA;" OwnerUserId="9620" LastEditorUserId="9620" LastEditDate="2017-01-22T02:23:34.253" LastActivityDate="2017-01-22T02:23:34.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13332" PostTypeId="1" CreationDate="2017-01-19T16:46:08.237" Score="1" ViewCount="236" Body="&lt;p&gt;Assume two vertical pistons with their flat head facing upward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to couple them to one rotating shaft in a way that the position vs. time diagram gets like this? (This is one complete cycle)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iZbDJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iZbDJ.png&quot; alt=&quot;diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each color being representing a piston.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Red: $$2 \left|\cos(\pi t)\right| - \frac12$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blue: $$2 \left|\sin(\pi t)\right| - \frac12$$ (The first with a phase difference of $\pi \over 2$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think because of the Absolute values, this can not be a simple crankshaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually I want to get as close as possible to this diagram of motion, which is the ideal theoretical case. Maybe this can help you for a better approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zr5Do.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zr5Do.jpg&quot; alt=&quot;real thing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9633" LastEditorUserId="9633" LastEditDate="2017-01-19T18:23:26.327" LastActivityDate="2020-11-22T05:08:50.957" Title="How to couple two pistons to a rotating shaft so that the piston heads move like this?" Tags="|design|mechanisms|pistons|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13333" PostTypeId="2" ParentId="13328" CreationDate="2017-01-19T18:11:07.393" Score="1" Body="&lt;p&gt;Gearing gives a simple ratio between torque and angular speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have a motor which delivers torque T at 1000 rpm and a reduction ration of 10:1 then you will get an output at from the gearbox of 10T at 100rpm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key thing here is to estimate the torque required to rotate the drum, from what you have described, my educated guess would be that something in the couple of hundred watt range, typical of a small electric drill, should be fine once it is geared down to the relatively low speeds that you are talking about. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can buy DC motors off the shelf with integrated fairly high reduction ratio gearboxes relatively cheaply which should give the sort of performance you are looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also say that for this sort of application it is good to include a belt drive element in the transmission as this is a simple way to alter gear ratios if you need it and also has the advantage that it is simple and robust, being tolerant of minor misalignment and will tend to slip rather than break or stall the motor if something jams, which is also useful if you want to leave it unattended. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of bearings &lt;a href=&quot;https://en.wikipedia.org/wiki/Pillow_block_bearing&quot; rel=&quot;nofollow noreferrer&quot;&gt;pillow blocks&lt;/a&gt; are often the simplest solution for this sort of application as they are very simple to install, especially if you aren't very tight on space. and tend to be a good solution for low-volume and prototype production where you just want to support a shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also looks a lot like a &lt;a href=&quot;http://www.cooksongold.com/Jewellery-Tools/Metal-Barrelling-Starter-Kit-With--3lb-Machine-prcode-999-600K&quot; rel=&quot;nofollow noreferrer&quot;&gt;barrel polisher&lt;/a&gt; these tend to work by having the drum sit on a pair of rollers with one or both rotating. This makes it very easy to remove the drum itself to put what you want to agitate in and get it out again, plus you can prepare one drum while the other is working as they aren't physically attached to the rotation mechanism. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-01-19T18:11:07.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13336" PostTypeId="1" CreationDate="2017-01-19T18:44:26.083" Score="1" ViewCount="59" Body="&lt;p&gt;A unsolicited lighting installer went up on my roof and told me that I had had a lightning strike and that the cable along the roof line (going from one rod to another) was crystallized and that I needed to have all cables replaced. I sensed a scam and sent him away. Was I right? Can these cables 'crystallize'?&lt;/p&gt;&#xA;" OwnerUserId="9639" LastEditorUserId="1832" LastEditDate="2017-01-19T18:56:08.693" LastActivityDate="2017-01-19T19:07:02.857" Title="Can lightning rod cables crystalize?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13337" PostTypeId="2" ParentId="13336" CreationDate="2017-01-19T19:07:02.857" Score="2" Body="&lt;p&gt;If you look at this related question &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/1785/why-doesnt-a-lightning-strike-destroy-the-lightning-rod&quot;&gt;Why doesn&amp;#39;t a lightning strike destroy the lightning rod?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accepted answer explains that the actual energy dissipated in a lightning conductor can be expected to be really fairly modest and as such you would not normally expect a conductor to be damaged by a strike. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crystallisation of some metals at temperatures which are a substantial fraction of their melting point is a thing  but as discussed in the question I referred to this is far above the temperatures you would reasonably expect from a strike. Furthermore i can only imagine that &lt;em&gt;crystallisation&lt;/em&gt; really refers to excessive grain growth in a metal when hot working etc and really only applies to alloys like steel where you can get phase changes rather than the relatively pure copper which is typically used for electrical conductors. As far as I am aware pure copper doesn't suffer any adverse effects from heating apart from being annealed which is not a bad thing for a conductor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I am certainly not an expert in lightning rods this certainly sounds bogus. I'm sure that there are many potential things which can degrade lightning conductors over time but 'crystallisation' does sound made-up. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-01-19T19:07:02.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13338" PostTypeId="1" CreationDate="2017-01-19T19:12:47.523" Score="0" ViewCount="140" Body="&lt;p&gt;Consider a typical private parking garage for a condominium or apartment building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is currently a typical garage gate that opens and closes (slowly) to allow vehicles to pass safely through the gate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the optimal method for preventing access into this garage by pedestrians/foot traffic, but ensuring the same safe entry/exit for vehicles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to make sure people cannot enter/exit through the same gate that vehicles may enter/exit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically implementable solutions only, please.  Cost need not be a factor.&lt;/p&gt;&#xA;" OwnerUserId="9640" LastActivityDate="2017-01-20T18:48:57.900" Title="How can one reliably prevent pedestrian access to a parking garage, but allow vehicular access?" Tags="|electrical-engineering|automotive-engineering|design|building-design|infrastructure|" AnswerCount="4" CommentCount="8" ClosedDate="2017-01-21T01:54:57.260" ContentLicense="CC BY-SA 3.0" />
  <row Id="13339" PostTypeId="2" ParentId="4113" CreationDate="2017-01-19T20:16:57.007" Score="0" Body="&lt;p&gt;The critical compressibility factor Zc for the Peng Robinson equation of state is reported as 0.307 in Walas, Phase Equilibria in Chemical Engineering.  The coefficients were 0.45724 and 0.07780.  Using Excel, I found Zc = 0.3213.  Using 0.457235 and 0.077796 as reported in Stryjek and Vera's paper, Zc = 0.3111.  You can see that the solution is very sensitive to the values used.  Still, I can't get 0.307 either.&lt;/p&gt;&#xA;" OwnerUserId="9642" LastActivityDate="2017-01-19T20:16:57.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13340" PostTypeId="1" AcceptedAnswerId="13341" CreationDate="2017-01-19T20:28:27.133" Score="0" ViewCount="116" Body="&lt;p&gt;How can I make an elevator that goes up while twisting like in &lt;a href=&quot;https://youtu.be/KCElBEfwLfc&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt;? (The man that made this used an old printer)&lt;/p&gt;&#xA;" OwnerUserId="9643" LastEditorUserId="255" LastEditDate="2017-01-20T17:27:51.043" LastActivityDate="2017-01-20T17:27:51.043" Title="How to make a mechanical elevator with rotation?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13341" PostTypeId="2" ParentId="13340" CreationDate="2017-01-19T20:52:46.617" Score="2" Body="&lt;p&gt;In this case it looks like the lift and rotation elements are mechanically separate ie the lift mechanism sits on a rotating turntable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the key element is to be able to provide electrical power to a rotating assembly, I can't see in detail how this is done in this particular case but &lt;a href=&quot;https://en.wikipedia.org/wiki/Slip_ring&quot; rel=&quot;nofollow noreferrer&quot;&gt;slip rings&lt;/a&gt; would be the conventional solution. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-01-19T20:52:46.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13342" PostTypeId="2" ParentId="12118" CreationDate="2017-01-19T21:47:00.267" Score="3" Body="&lt;p&gt;Firstly, in reference to nomenclature: in &lt;em&gt;ISO&lt;/em&gt; (metric) standards the pitch of a leadscrew is given in units of distance/revolution. However, &lt;em&gt;Unified&lt;/em&gt; (inch series) standards, pitch is given in &lt;em&gt;TPI&lt;/em&gt; (threads per inch). As you mentioned in your question, lead is related to pitch by &lt;span class=&quot;math-container&quot;&gt;$p = L/N$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the pitch, &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the lead and &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; is the number of starts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Juvinall &amp;amp; Marshek, &lt;em&gt;Fundamentals of Machine Component Design, 5th Ed.&lt;/em&gt;, pp. 412.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a derivation of the equation.&#xA;The equation relating linear load to torque for a lead screw is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ T = \frac{W d_m}{2}\frac{f\pi d_m + L \cos{\alpha_n}}{\pi d_m \cos{\alpha_n} - f L} $$&lt;/span&gt;&#xA;Where &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the torque, &lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt; is the linear load, &lt;span class=&quot;math-container&quot;&gt;$d_m$&lt;/span&gt; is the mean diameter of thread contact, &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; is the coefficient of friction, &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the metric lead in distance/thread, and &lt;span class=&quot;math-container&quot;&gt;$\alpha_n$&lt;/span&gt; is the thread angle measured in the normal plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Juvinall &amp;amp; Marshek, &lt;em&gt;Fundamentals of Machine Component Design, 5th Ed.&lt;/em&gt;, pp. 418.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us examine the simplified case of a square thread (&lt;span class=&quot;math-container&quot;&gt;$\alpha_n = 0$&lt;/span&gt;). The above equation then simplifies to:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$T = \frac{W d_m}{2}\frac{f\pi d_m + L}{\pi d_m - f L} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Values of &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; tend to be around 0.08 to 0.20 for steel against cast iron and bronze, and can be even lower for ballscrews (same source as above). Therefore let us simplify further by assuming &lt;span class=&quot;math-container&quot;&gt;$f = 0$&lt;/span&gt;. Then we get the following:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ T = \frac{WL}{2\pi}  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in this case our load &lt;span class=&quot;math-container&quot;&gt;$W= ma$&lt;/span&gt; (mass times acceleration). Acceleration is related to angular acceleration by &lt;span class=&quot;math-container&quot;&gt;$a = \frac{L}{2\pi}\frac{d^2\theta}{dt^2}$&lt;/span&gt;. This gives:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ T = \frac{mL^2}{(2\pi)^2}\frac{d^2\theta}{dt^2} $$&lt;/span&gt;&#xA;From the definition of rotational inertia (&lt;span class=&quot;math-container&quot;&gt;$T = J\frac{d^2\theta}{dt^2}$&lt;/span&gt;) we arrive at the final answer:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ J = \left(\frac{L}{2\pi}\right)^2m$$&lt;/span&gt;&#xA;In other words, the torque you need to apply to move a linear load is independent of the number of starts of the leadscrew. For a single start, and using the &lt;em&gt;Unified&lt;/em&gt; standard definition of pitch we get:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ J = \frac{1}{(2\pi p)^2}m $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results of my derivation are supported by the catalogue I found that lists the calculation for the force-torque conversion for a lead screw (pp. 2, eq. 3 and the sample calculations on pp. 9) &lt;a href=&quot;https://www.sdp-si.com/D810/PDFS/Ball%20And%20Acme%20Lead%20Screw%20Technical%20Info.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&#xA;Note that they include the efficiency value &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; which is used to approximate the friction in the system while simplifying the calculations. You should double check the equation given by whatever catalogue you are using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="7430" LastEditorUserId="7430" LastEditDate="2019-02-06T22:38:16.663" LastActivityDate="2019-02-06T22:38:16.663" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="13343" PostTypeId="1" AcceptedAnswerId="13350" CreationDate="2017-01-19T23:17:42.650" Score="0" ViewCount="75" Body="&lt;p&gt;Does some material/metal exist that can be stretched and still transfer current? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;Say I have a rectangular item made of some metal or material(unknown to me at the time). On both ends, I have two small sensors attached. These sensors do nothing but want to establish communication between each other. For the purpose of this question, let's refer to them as sensor A and sensor B. Sensor A sends a message to Sensor B. Since the metal is not very stretched, it manages to transfer the message in electrical current to Sensor B in an elapsed time of .1 second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yG2xK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yG2xK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, imagine that we stretch the metal a bit more, where its width is lowered and its length is increased. Since the message now has to travel further, I would expect the delivery to be greater than .1 second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ebcB8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ebcB8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Numbers used are just an example.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What sort of material or metal can I use to pass current, that when stretched, would alter the total distance a current would have to travel?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ideally, the two sensors will at the minimum be separated by 3-4 inches, and the metal/material would only stretch up to 1 inch.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9644" LastEditorUserId="9644" LastEditDate="2017-01-19T23:22:48.500" LastActivityDate="2017-01-20T07:32:41.340" Title="Strechable object that can transfer current" Tags="|mechanical-engineering|electrical-engineering|metals|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13345" PostTypeId="2" ParentId="13343" CreationDate="2017-01-19T23:46:16.750" Score="3" Body="&lt;p&gt;Given the theoretical dimensions stated in the question, there would be no measurable difference in the transmission speed of the message. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible materials include various conductive rubber compounds, where a conductive filler, such as copper, aluminum or nickel particles are mixed with a flexible carrier such as silicone rubber, latex or other elastomers to suit requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google &quot;electrically conductive rubber&quot; for some commercial products.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-01-19T23:46:16.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13346" PostTypeId="1" CreationDate="2017-01-20T00:59:11.230" Score="0" ViewCount="94" Body="&lt;p&gt;I have a hearing condition and need to wear ear muffs on my scooter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My muffs are 3M Peltor with a dB rating of 35dB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When riding my bike, the combined noise from the engine and the wind is about 80dB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in theory that should get me into about 45dB, but it seems the wind buffering makes it louder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the muffs take off 35dB at rest, but with movement and wind they get louder from inside my head because of the wind buffering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are there any specifications or any research done on wind impact decibel levels on muffs for ear protection?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9645" LastEditorUserId="255" LastEditDate="2017-01-20T17:26:06.657" LastActivityDate="2017-01-20T21:46:47.387" Title="How many decibels does wind buffering reduce from ear muffs protection?" Tags="|medical-devices|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13347" PostTypeId="2" ParentId="13227" CreationDate="2017-01-20T01:36:25.917" Score="3" Body="&lt;p&gt;The answer of alephzero is spot on. I just want to mention that the arch structure is particularly sensitive to 2nd order effects when it is shallow (say depth to span ratio ~0.1) . It's not acceptable to model it using 1st order theory. The figure below demostrates the type of behaviour expected (load vs vertical displacement curve) for an arch with pinned supports and pinned apex. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The blue curve is the response when you apply displacement &lt;/li&gt;&#xA;&lt;li&gt;The orange   curve is the response when you apply load&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The small figures below the graph, indicate the deformed shape and applied load to the arch, at the respective curve points (of the blue curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BHCWl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BHCWl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the stiffness of the structure decreases initially as the bars tend to become horizontal. When they become horizontal there is no way to carry the vertical load (except for bending, but it's not allowed due to the pinned apex), therefore stiffens becomes zero and the structure snaps under load control to an opposite geometry which is more stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand when performing linear analysis, no change in stiffness is considered. Therefore, since initial stiffness is the highest one, it is expected to find lower displacements. &lt;/p&gt;&#xA;" OwnerUserId="9620" LastEditorUserId="9620" LastEditDate="2017-01-20T01:50:15.913" LastActivityDate="2017-01-20T01:50:15.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13348" PostTypeId="2" ParentId="13346" CreationDate="2017-01-20T02:44:05.020" Score="2" Body="&lt;p&gt;The amount of protection offered by ear muffs depends on a number of things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, how the muffs are constructed &amp;amp; the materials used and the sound attenuation properties of those materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing that is important is how the muffs are worn. You probably know that the cup of each muff must fully encapsulate each ear. For ear muffs to function properly there must be a seal between the cup of each muff &amp;amp; the skin around the side of the head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entrapment of hair by the muff cups will reduced the protection offered by the ear muffs because that can create a gap between the skin &amp;amp; the muffs which will allow air and sound to travel through the gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While riding your bike it is mostly likely that your head is moving, even though imperceptible vibrations, road roughness, jolts, whatever. Such movement can cause the seal between the muff cups and the skin to be broken, thus letting in more sound into the ear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if there is a tight fit between the muffs and a helmet any contact between the helmet and the muffs may cause the muffs to move and again breaking the seal between the muffs and the skin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To improve the noise reduction you receive while riding your bike you may need to also insert ear plugs in addition to using ear muffs.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-01-20T02:44:05.020" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13349" PostTypeId="2" ParentId="11824" CreationDate="2017-01-20T03:28:54.413" Score="5" Body="&lt;p&gt;Pinned connection in structural context doesn't mean free rotation/zero moment. Pinned connection means much bigger rotations and much lower strength compared to the connected members. According to Eurocode 3 (EN1993-1-8:2005 section:5.2), if the moment strength is below 0.25 of the connected beam and its stiffeness is below a limit (relative to the beam) then it should be considered nominally pinned, provided adequate rotational capacity is available (for example through ductile critical mechanism of failure). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally one &lt;strong&gt;should find the moment strength and stiffness of the connection and classify it against the limits&lt;/strong&gt;. A connection can be classified  either as full strength/rigid or pinned or semirigid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the connection in question, it is obvious that no transfer of force occurs through the beam flanges. The bending moment finds its way through the flanges mostly. So this connection is ineffective to transfer enough of the beam end moment. Also take into account that the connection is bolted to the web of the column which is quite flexible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, i want to emphasize that the following statement is not accurate:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;once the beam is loaded, their is no room for rotation of the beam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Don't assume that based on the undeformed geometry!&lt;/p&gt;&#xA;" OwnerUserId="9620" LastEditorUserId="9620" LastEditDate="2017-01-20T19:45:31.633" LastActivityDate="2017-01-20T19:45:31.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13350" PostTypeId="2" ParentId="13343" CreationDate="2017-01-20T07:32:41.340" Score="5" Body="&lt;p&gt;First of all, with your 4 inch distance, the propagation time wouldn't be 0.1 seconds. It would be 0.000&amp;nbsp;000&amp;nbsp;000&amp;nbsp;339 seconds. Measuring such short time differences is a significant challenge and there are no obvious off-the-shelf solutions capable of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But more than that, the propagation time doesn't depend only on the distance. The shape and surroundings of the conductor also affects it, due to &lt;a href=&quot;https://en.wikipedia.org/wiki/Transmission_line&quot;&gt;transmission line effects&lt;/a&gt;. Thus even if you had the propagation time, it would need a more accurate modelling of the situation to arrive at a distance reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may want to ask another question about what kind of distance sensors would be applicable to your problem, but you'll also have to give more details about the actual goal that you want to achieve.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2017-01-20T07:32:41.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13351" PostTypeId="2" ParentId="13338" CreationDate="2017-01-20T09:09:07.203" Score="2" Body="&lt;p&gt;Dual gates, with the outer one always closing before the inner one opens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For extra security, a computer vision system could check that no people are present in the intermediate space before opening the inner gate.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2017-01-20T09:09:07.203" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13352" PostTypeId="2" ParentId="13338" CreationDate="2017-01-20T09:26:55.303" Score="2" Body="&lt;p&gt;Pressure sensor that triggers alarm if something less than 1 ton but more than 10kg enters the garage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrify the floor, anyone walking in will be fried without a Faraday's cage protection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fill whole garage with carbon monoxide fumes after entering. Suck out the fumes after 5min.&lt;/p&gt;&#xA;" OwnerUserId="9630" LastActivityDate="2017-01-20T09:26:55.303" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13353" PostTypeId="1" CreationDate="2017-01-20T13:44:15.897" Score="1" ViewCount="98" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hC3YO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hC3YO.png&quot; alt=&quot;A shelf&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is the shelf that I'm intending to put up.  It will have two symmetrically placed supports. This is probably a simple question that really should be answered in all DIY shops but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a uniformly distributed load across the entire shelf, what should the value of &lt;strong&gt;x&lt;/strong&gt; be to obtain the minimum deflection anywhere along L?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2017-01-20T13:44:15.897" Title="How should I position two shelf supports for the minimum deflection?" Tags="|structures|" AnswerCount="0" CommentCount="3" FavoriteCount="0" ClosedDate="2017-01-20T16:16:01.477" ContentLicense="CC BY-SA 3.0" />
  <row Id="13354" PostTypeId="1" CreationDate="2017-01-20T14:32:22.890" Score="2" ViewCount="146" Body="&lt;p&gt;Suppose I have a rectangular plate (width $a$, height $b$, thickness $t$). By using a Finite-Element discretization I obtain the system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M \ddot{x} + Kx= f$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with mass matrix $M$, stiffness matrix $K$ and force vector $f$. Now I want to know whether the solution vector $x$ does change if I vary $a,b$ or $t$; this is called sensitivity analysis. I know the direct method, but the adjoint method is also known. How I obtain the sensitivities by the adjoint method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not understand this method well after I looked in literature.&lt;/p&gt;&#xA;" OwnerUserId="8941" LastActivityDate="2017-01-23T14:35:32.023" Title="Sensitivity analysis: How do I perform the adjoint method?" Tags="|structural-engineering|dynamics|vibration|" AnswerCount="0" CommentCount="1" ClosedDate="2017-02-01T00:12:16.480" ContentLicense="CC BY-SA 3.0" />
  <row Id="13355" PostTypeId="2" ParentId="13338" CreationDate="2017-01-20T16:07:29.343" Score="1" Body="&lt;p&gt;You can install a passive solution like a Cattle Grid for people that would heavily discourage anyone from trying to walk on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then you might have to deal with the idiots that might get stuck and/or break they legs trying to cross one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another passive solution is a ditch with about 6 inches of water, because I don't think anyone would want to soak their shoes and pants trying to use the car entrance.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2017-01-20T16:07:29.343" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13356" PostTypeId="2" ParentId="13338" CreationDate="2017-01-20T18:48:57.900" Score="1" Body="&lt;p&gt;Since cost is not an issue how about an armed guard. Not every problem is an engineering problem. &lt;/p&gt;&#xA;" OwnerUserId="9506" LastActivityDate="2017-01-20T18:48:57.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13357" PostTypeId="2" ParentId="3328" CreationDate="2017-01-20T19:23:06.820" Score="2" Body="&lt;p&gt;I don't know much about the operation of a Telsa Turbine but I can talk in general about what affects the stiffness of an object.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Short answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The geometry of the disc and the material of the disc both affect the stiffness of the disc. Smaller radius = stiffer. Thicker disc = stiffer. If you are looking for a trade-off between stiffness and weight then you can employ tricks like ribs or hollow pockets inside the disc. The material property you are looking for is the Young's modulus, $E$. Higher $E$ = stiffer material. You can find a list of Young's Moduli for various materials here:&lt;a href=&quot;http://www.engineeringtoolbox.com/young-modulus-d_417.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/young-modulus-d_417.html&lt;/a&gt;. Modern HDD's are made glass/ceramic composites apparently. If I had to suggest a material and money was no problem, I would say graphene or some other nano-engineered material where you could design the stiffness properties at the molecular level. In terms of affordable materials however, you probably won't be able to find anything better than the HDD platter. Of course you will also need to worry about corrosion resistance, temperature stability, etc. due to the environment in which you are operating.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Now the long answer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Stiffness is generally characterized in engineering as the relationship between the force applied to an object and the amount that object deflects:&#xA;$$\delta = f(F,K)$$&#xA;Where $\delta$ is the displacement of the point where the force is applied, $F$ is an applied force and $K$ is the stiffness. In the equation above, $K$ is not necessarily a constant with respect to time or deflection. In fact, it can be a function of many things, including but not limited to the deflection, the rate of deflection, the rate of load application, the temperature of the material, etc. The relationship between force and displacement is often simplified to a linear relationship, where $K$ is assumed to be constant. In this case, K is often called the &lt;em&gt;spring constant&lt;/em&gt;.&#xA;$$F = K\delta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ignore those more complicated effects, then at its most basic level the stiffness of an object is determined by two things: the material properties and the geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material property most relevant to stiffness is the Young's modulus, $E$. Higher Young's Modulus = stiffer material. From what I found (&lt;a href=&quot;https://en.wikipedia.org/wiki/Hard_disk_drive_platter&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Hard_disk_drive_platter&lt;/a&gt;) modern HDD platters are made of a glass/ceramic composite, which I'm guessing would be a proprietary, specially-engineered material for each manufacturer. The Young's modulus for these materials is probably not easily available online. As a comparison however, glass has a Young's modulus of 50-90 GPa. Aluminum is what they used to use for HDD platters, which has a Young's modulus of around 70 GPa. Stainless steel has a Young's modulus of around 180 GPa. If money wasn't an issue then you could make a specially engineered carbon-based structure, like graphene which has a Young's modulus of around 1000 GPa. At the end of the day, you're going to be hard pressed to find a material that is as thin as a HDD platter but stiffer. I don't have a source for this, but in my experience steel can't be machined that thinly (it would get seriously warped due to the heat of the machining process), so you'd need to use other techniques to produce the disc. That's why these things are made with glass and ceramics, which can be assembled with chemical techniques instead of mechanical ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometry part is where things get &lt;em&gt;really&lt;/em&gt; complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram of how I am picturing your situation:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/t3d85.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t3d85.png&quot; alt=&quot;Circular plate deflection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a circular disk pinned in the center by some kind of axle. The plate deflects to a new position (dotted line) when a force is applied. If I understand correctly, the stiffness that you are worried about is the &quot;floppy&quot; direction of the floppy disc, i.e. the bending stiffness, as I have drawn it and not the torsional stiffness or compression stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a case of plate bending and here is a Wikipedia article where you can read about how that works. &lt;a href=&quot;https://en.wikipedia.org/wiki/Bending_of_plates&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Bending_of_plates&lt;/a&gt;. Yes, the math is very nasty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a better idea of how geometry affects stiffness, I propose a simplified case of cantilever beam bending:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DAgg2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DAgg2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a known solution for this case (source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&lt;/a&gt;).&#xA;$$F = \frac{3EI}{L^3}\delta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So where does the &quot;thickness&quot;, $h$, come in? $I$ is called the &lt;em&gt;area moment of inertia&lt;/em&gt; and it is calculated from the cross-section of the beam. In the simple rectangular beam case that I have drawn, $I$ is equal to the following:&#xA;$$I = \frac{bh^3}{12}$$&#xA;Check out my source here: &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_second_moments_of_area&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_second_moments_of_area&lt;/a&gt;. That page contains many different cross section shapes. Some are more efficient than others. A cross section that is often used in the construction industry is the I-beam (&lt;a href=&quot;https://en.wikipedia.org/wiki/I-beam&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/I-beam&lt;/a&gt;) because it handles bending and shear loads quite well, while also conserving material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what can we conclude from the simpler case of cantilever beam bending for our plate bending problem? If you have a solid cross-section for a flat disc, the only ways to improve the stiffness geometrically is to reduce the radius of the disc (analogous to $L$) or increase its thickness (analogous to $h$). If you aren't constrained to solid cross-sections or flat discs then you can use ribs or pockets in your disc to reduce material usage while maintaining stiffness. Again, this is ignoring the more complicated effects that can occur with plate bending.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe somebody else can look into whether or not reducing the radius of the disc will affect the functionality of the turbine. Fluid dynamics isn't my specialty!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wow. Bit of an essay there. Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2017-01-20T19:23:06.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13358" PostTypeId="2" ParentId="3328" CreationDate="2017-01-20T21:07:58.283" Score="1" Body="&lt;p&gt;As was previously mentioned, a Tesla turbine is attractive from a simplicity standpoint, but it is not the most efficient solution given current material technology. There are lots of other efficiency considerations on a small project like this that need engineering attention. Don't set your hopes too high; if you get 20% efficiency you are doing very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you operate at high rpm; the blades will be in tension permitting very thin blades. This &lt;a href=&quot;https://www.youtube.com/watch?v=rYfkhdKcEiE&quot; rel=&quot;nofollow noreferrer&quot;&gt;video of a paper table saw blade&lt;/a&gt; qualitatively demonstrates how a very thin member can resist deflection. Composite carbon fiber sheets may be a good option. Laser cut thin steel may be your most cost effective. You may need to have the blades cleaned up on a lathe and/or have the finished rotor professionally balanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steam is a very dangerous form of energy. Always pressure test the system with water and have a properly sized pressure relief installed. If you ever plan on selling systems remember that steam is extensively regulated to protect the public; there are many codes you will need to follow. This is the primary reason you see very few small steam turbines.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-01-20T21:07:58.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13360" PostTypeId="2" ParentId="13346" CreationDate="2017-01-20T21:46:47.387" Score="1" Body="&lt;p&gt;Why not wear foam earplugs? The right earplug, properly used, have a NRR of 33dB, which is really close to the 35 dB rating of your earmuffs. 3dB is considered the minimum human detectable threashold for sound, so it's unlikely that you'd notice any difference even in ideal conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's the ideal conditions part that typically causes problems with earmuffs. To work properly, ear protection, plugs or muffs, has to block the path of the sound from reaching your eardrum. They achieve this by creating a sealed barrier between the sound and the eardrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While slightly more effective, the muffs have three disadvantages in achieving this seal. First, they're sealing against a much larger and flatter surface area and using a fairly weak spring to do so. This means the seal can be disturbed fairly easily and noise can intrude. Second, there are lots of common things that can get between the seal on the muff and the wearer's skin, typically eyeglass temples and hair are the big offenders here. Third, they're much more likely to become dislodged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that in your case, it's some combination of those factors coupled with airflow over your head pushing on and disrupting the seal between the muff and your head. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earplugs on the other hand achieve a good seal much easier and unless you have really hairy ear canals like Yoda, they also don't have problems with foreign object disrupting the seal. The biggest problem with earplugs is in teaching people how to wear them correctly, but it's easy to do. The other big plus is that earplugs are much more discreet than those big yellow Peltor muffs. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-01-20T21:46:47.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13361" PostTypeId="2" ParentId="12800" CreationDate="2017-01-20T22:42:21.817" Score="1" Body="&lt;p&gt;I believe your equillibrium is very wrong. This is like a rigid body and we have to consider all forces applied to it. The base shouldn't be part of the rigid body, since the screw force would be then an internal one, with no way to find it. So, an apporopriate equillibrium would be as shown in the figure below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p5Xa4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p5Xa4.png&quot; alt=&quot;rigid body equiilibrium&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forces are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weight F&lt;/li&gt;&#xA;&lt;li&gt;Reaction from base at rotation point R&lt;/li&gt;&#xA;&lt;li&gt;Screw tension T&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming small displacements, the equillibrium equations should be:&#xA;$$ F + T - R = 0 \\  F\cdot 0.5b - T \cdot 0.5d = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the second one we get:&#xA;$$ T = F\frac{b}{d}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If b/d is let's say 5, then T = 5F !!!&#xA; That means the screws are loaded with 5x the weight F. It can only get worse with a more slender post.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How to improve it&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Add some stiffeners between the base and the posts. The rotation point is moved away from the screws, offering a larger lever arm for T, and simultaneously reducing the lever arm of the weight F.&lt;/p&gt;&#xA;" OwnerUserId="9620" LastEditorUserId="9620" LastEditDate="2017-01-21T15:59:47.037" LastActivityDate="2017-01-21T15:59:47.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13362" PostTypeId="1" CreationDate="2017-01-20T22:56:17.983" Score="1" ViewCount="115" Body="&lt;p&gt;As we all know the industry is heading into PE pipelines in the gathering fields fast. Some applications use DN560-DN1000, and the spades is very heavy and in awkward positions. Is there PE spades/blanks in the market to ease the manual handling?&lt;/p&gt;&#xA;" OwnerUserId="9655" LastActivityDate="2017-01-20T22:56:17.983" Title="Spades used in pressure testing PE Pipe lines" Tags="|gas|" AnswerCount="0" CommentCount="2" ClosedDate="2017-01-24T13:47:20.837" ContentLicense="CC BY-SA 3.0" />
  <row Id="13364" PostTypeId="1" AcceptedAnswerId="13398" CreationDate="2017-01-21T01:52:27.167" Score="1" ViewCount="210" Body="&lt;p&gt;I was working on a project where I have to launch a ball. The current design has two wheels that are side by side with the ball going between them. I noticed that in almost every commercial ball launcher, the wheels are stacked on top of one another rather than side by side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what advantage this has over my current system. It seems a lot more accurate and precise, but I don't see why.&lt;/p&gt;&#xA;" OwnerUserId="2164" LastActivityDate="2017-01-23T14:16:07.627" Title="Why do baseball pitching machines have wheels stacked instead of side by side?" Tags="|wheels|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13366" PostTypeId="1" CreationDate="2017-01-21T17:08:23.847" Score="0" ViewCount="84" Body="&lt;p&gt;I need help with interpreting the kinetics of a scientific paper: &lt;a href=&quot;https://www.researchgate.net/publication/284014208_Kinetic_modelling_of_methanol_conversion_to_light_olefins_process_over_Silicoaluminophosphate_SAPO-34_Catalyst&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.researchgate.net/publication/284014208_Kinetic_modelling_of_methanol_conversion_to_light_olefins_process_over_Silicoaluminophosphate_SAPO-34_Catalyst&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to reproduce the reactions in table 1 using a pfr in hysys. However, I am at a loss how to interpret the kinetics data given in table 3. The units given are in (mol gCatalyst−1 atm−1 h−1) which do not make much sense to me. I thought a 2nd order rate constant should have units of atm-1 h-1. Hence, I am unsure how to input the rate kinetics data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="9660" LastEditorUserId="9660" LastEditDate="2017-01-22T02:28:26.160" LastActivityDate="2017-04-14T01:24:58.913" Title="Need help with interpreting units of kinetics data in a scientific paper" Tags="|chemical-engineering|modeling|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13367" PostTypeId="1" CreationDate="2017-01-21T19:57:57.060" Score="3" ViewCount="119" Body="&lt;p&gt;I am doing research on the impact of vehicles on roadways, specifically garbage trucks. In my research I am finding largely different &quot;answers.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example &quot;Using American Association of State Highway and transportation Officials (AASHTO) vehicle load factors (VLF), a full, residential garbage truck is equivalent to 11,700 passenger cars.&quot; &lt;a href=&quot;http://files.dep.state.pa.us/waste/Recycling/RecyclingPortalFiles/Doylestown_Township_504.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://files.dep.state.pa.us/waste/Recycling/RecyclingPortalFiles/Doylestown_Township_504.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when calculating based on EASL the number is 1,279 &lt;a href=&quot;https://www.lrrb.org/PDF/201432.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.lrrb.org/PDF/201432.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference and why are the answers so different?&lt;/p&gt;&#xA;" OwnerUserId="9664" LastEditorUserId="1832" LastEditDate="2017-01-21T22:11:03.317" LastActivityDate="2017-01-23T11:12:33.273" Title="Vehicle load factor discrepancies" Tags="|civil-engineering|transportation|roadway|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13369" PostTypeId="2" ParentId="10355" CreationDate="2017-01-21T20:36:08.667" Score="0" Body="&lt;p&gt;Basically the two types of pump that are appropriate for this application are the centrifugal pump and the displacement pump.  Both types can be submerged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The displacement pump uses small chambers built into a rotating disk. When operating, the action creates suction on one side, often allowing several feet of water to be LIFTED and ejected.  Displacement pumps such as the FloJet series do not &lt;em&gt;need&lt;/em&gt; to be submerged and can be located some distance from the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The centrifugal pump requires fluid to fill the vane chamber, without air. These work by spinning an impeller at high speed, creating a pressure difference between the fluid at the centre of the vane and the outer edge. The low pressure at the inlet draws fluid into the chamber, but any air present will disturb the vacuum.  Centrifugal pumps generally have a higher flow rate and operate at much lower pressures than displacement pumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the water dish, if you can guarantee that there will always be water for the pump to work with, the centrifugal pump would be least expensive, quieter and reasonably reliable. If the dish runs dry, the pump could become damaged, so a method of detecting the water level would be required.  &lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-01-21T20:36:08.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13370" PostTypeId="1" AcceptedAnswerId="13374" CreationDate="2017-01-22T00:02:00.843" Score="1" ViewCount="464" Body="&lt;p&gt;I'm a 3d artist and I'm trying to produce a type of scaffolding within a biodome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For lack of knowledge in technical terms, I'm having a hard time learning about the design pattern/structure that I must follow (I suspect just making a triangle-array would be ignorant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the technical term for the triangular scaffolding often used in future-esque projects&lt;/strong&gt;? I've seen it used to hold up giant solar arrays as well as biodomes, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I only knew what it was called I could go ahead and research it!&lt;/p&gt;&#xA;" OwnerUserId="9666" LastActivityDate="2017-01-22T15:41:22.560" Title="What do they call that elegant triangular scaffolding often used in future-oriented architecture?" Tags="|architecture|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13373" PostTypeId="2" ParentId="8627" CreationDate="2017-01-22T03:47:00.487" Score="2" Body="&lt;p&gt;A ground motion record (accelerogram) is a measurement with high specificity to the fault that caused the earthquake. The recorded peak acceleration, frequencies, duration etc depend on the fault geometry and movement. Of course there are not two identical faults and every time a fault is activated it doesn't produce the same ground motion. Therefore, the value of designing a structure by one &quot;as recorded&quot; ground motion is small, since it is highly unlikely to encounter it in the future. Typically the design with real ground motions requires a set of them (at least 5). &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why scale a ground motion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In an ideal world a structural engineer would have a vast amount of historical ground motions to select from, for a particular site, when designing a new structure. There would be no need to use any records from other parts of the world. Of course this is not happening because severe earthquakes are not an every-day phenomenon and recording ground accelerations started not to far in the past. There are sites where only one or two good records are available. So the solution is to use available records from all over the world, but then the problem is that the seismic risk is not the same all over the planet. Typically, the seismic code provides us with the expected intensity of the seismic excitation, in the form of peak ground acceleration. A solution is to scale an accelerogram to match the code requirements. A problem arises though, because a distrortion in the frequencies is unavoidable. Therefore, it is better to avoid too much scaling, or alternativly to use artificial accelerograms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, scaling of the ground motions is inherently needed for some advanced types of analyses (incremental dynamic analysis). &lt;/p&gt;&#xA;" OwnerUserId="9620" LastActivityDate="2017-01-22T03:47:00.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13374" PostTypeId="2" ParentId="13370" CreationDate="2017-01-22T04:18:07.773" Score="6" Body="&lt;p&gt;The general term for these structures in 3-d is a &lt;strong&gt;space frame&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Just making a triangle array&quot; is a pretty good starting point, because three bars forming a triangle make a rigid structure without any bracing at the corners, unlike four bars forming a rectangle which can be &quot;squashed&quot; out of shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another key to making a stiff lightweight structure is to make it curved in both directions (i.e. not like a cylinder or a plane), so that effectively you have groups of four bars forming tetrahedra. The curved structure works the same way as a &lt;strong&gt;dome&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to study designing something like this, you could start with the same idea in two dimensions, called a &lt;strong&gt;truss&lt;/strong&gt;. Trusses are a very common engineering design for the roof supports in buildings, and also for bridges, tower cranes, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make a space frame which is cheap and easy to assemble, you need to make the bars with a few standard lengths and meeting at the same angles, rather than every bar and every joint being different. Of course a computer aided design &lt;em&gt;can&lt;/em&gt; have every part different if you want, but the manufacturing and assembly costs will be higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to look at &lt;strong&gt;polyhedra&lt;/strong&gt; (including &quot;semi-regular&quot; polyhedra, not just the five &quot;Platonic regular solids&quot;) where all the edges are the same length, and &lt;strong&gt;buckyballs&lt;/strong&gt; which have similar properties. In particular, there are a few &lt;strong&gt;deltahedra&lt;/strong&gt; where all the faces are equilateral triangles but the complete object does not have a &quot;regular&quot; shape. The regular polyhedra were studied by the ancient Greek mathematicians, but the irregular shaped deltahedra were not discovered until nearly 2000 years later - most likely simply because nobody bothered to look to see if such a thing could exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google any of the terms in bold type for more information.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-01-22T04:18:07.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13375" PostTypeId="2" ParentId="625" CreationDate="2017-01-22T04:55:43.690" Score="0" Body="&lt;p&gt;The viscous resistance of air on an aircraft is really tiny compared to the turbulent resistance caused the formation of eddies in the air behind the aircraft. For that reason, hydrophobicity of the surface is not really what we care about. An object travelling through air is to either be very small or be moving through the air very slowly in order to be experiencing only viscous resistance. All objects going through air experience linear resistance from air's vescosity below a certain speed and quadratic and quadratic resistance from turbulence above that speed and for any shape, the lowest speed an object of that shape can go and experience quadratic resistance varies as the reciprocal of the size of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, a substance doesn't have to be hydrophobic in order for water to roll off it. Any amorphous or monocrystalline solid etched nanosmooth, if water doesn't completely wet it, it will roll off because its receding contact angle will be so close to its advancing contact angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Airplanes do have other problems that are very worth solving. One such problem is supercooled water droplets in the air going into the engine and freezing and building up a thick layer of ice in it. There's an entire Wikipedia article about ways to prevent that problem at &lt;a href=&quot;https://en.wikipedia.org/wiki/Ice_protection_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Ice_protection_system&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6419" LastActivityDate="2017-01-22T04:55:43.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13377" PostTypeId="1" CreationDate="2017-01-22T14:54:23.160" Score="0" ViewCount="250" Body="&lt;p&gt;In i ) I was told that for the pinned support , we can assume to replace it with a roller , while for the cantilever , we can assume it to be supported by double roller . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the cantilever , why we cant replace it with roller instead of double roller ? What 's the difference between double and single roller ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In iv ) I was told that we can use roller guide , so from the figure , it's single roller , right ? So , IMO , the ii) was wrong , for the cantilever , we should always use single roller&lt;a href=&quot;https://i.stack.imgur.com/UNjgd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UNjgd.jpg&quot; alt=&quot;22&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w6oWC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w6oWC.jpg&quot; alt=&quot;3232&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2017-01-23T12:25:16.560" Title="replace support in qualitative influence line analysis" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13378" PostTypeId="2" ParentId="13370" CreationDate="2017-01-22T15:24:26.330" Score="1" Body="&lt;p&gt;You may be thinking of geodesics, commonly seen in geodesic domes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Geodesics&quot; or geodetics refer to lines following a &lt;a href=&quot;https://en.wikipedia.org/wiki/Geodesic&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;great circle&quot;&lt;/a&gt; route across a surface. They can produce strong structures with no internal reinforcement, maximising usable space (in principle ... in practice, filling the space with cuboid furniture is problematic!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One early exploitation of this was in late 1920s airship designs, where it is associated with &lt;a href=&quot;https://en.wikipedia.org/wiki/Barnes_Wallis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Barnes Wallis&lt;/a&gt; and his successful R100 design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structure has a lot of redundancy and can be quite resistant to damage, a property that proved useful in a later Barnes Wallace design, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Vickers_Wellington&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vickers Wellington&lt;/a&gt; bomber from early WW2.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/n6Jpr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n6Jpr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The post-war Vickers Viking airliner followed the geodesic design, one survived a bomb exploding in the toilet but landed successfully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Post war geodesics are more commonly associated with another genius and dreamer, &lt;a href=&quot;https://en.wikipedia.org/wiki/Buckminster_Fuller#Geodesic_domes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Buckminster Fuller&lt;/a&gt;. He famously championed finding peaceful uses for weapons of war.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is now highly feasible to take care of everybody on Earth at a higher standard of living than any have ever known. It no longer has to be you or me. Selfishness is unnecessary. War is obsolete. It is a matter of converting the high technology from weaponry to livingry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether he knew of Wallis's weaponry or independently invented geodesics for &quot;livingry&quot; I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common variant is composed of pentagons and hexagons with slightly raised centres. using just two slightly different sizes of triangle, as can be seen here (not quite complete in the photo). This greenhouse, made of lightweight 2x1 timber and polycarbonate survived quite a few hurricanes in the Hebrides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/63qWB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/63qWB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the relatively recently discovered C60 atom mentioned in the other answer, which earned the name Buckyball, or Buckminsterfullerene for obvious reasons.&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="435" LastEditDate="2017-01-22T15:41:22.560" LastActivityDate="2017-01-22T15:41:22.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13380" PostTypeId="1" CreationDate="2017-01-22T17:53:12.570" Score="1" ViewCount="193" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EZQsR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EZQsR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I constructed a small building to protect irrigation valves from freezing. The roof is hinged and rests at a 110 degree angle. The goal is to lift it up to about 20 degrees. The roof is 4'x 8', and weighs 100#. I want counter balance the lid to assist in lifting. The hard part is the lift mechanism needs to be hidden inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea for lift is to mount a bar to lid and mount a bar to the building base, then connect the two rods, and have all 3 connection points hinged. Then use pulley's and a counter weight to pull the bars to near vertical alignment. So the bars will be in a tight v position when lid is closed. As lid opens the v would get larger until bars are almost I vertical alignment.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I improve this design?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hsg0F.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9674" LastEditorUserId="9" LastEditDate="2017-01-22T22:33:47.853" LastActivityDate="2017-01-23T14:17:36.497" Title="How to lift a roof using no electricity?" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13384" PostTypeId="2" ParentId="13380" CreationDate="2017-01-22T23:23:33.637" Score="1" Body="&lt;p&gt;Try this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/PxEpz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PxEpz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lifting arm would need to be assisted by something like spring struts (e.g. car tailgate/hood lifters) between the floor/back wall and the lid.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-01-22T23:23:33.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13385" PostTypeId="1" CreationDate="2017-01-23T02:56:08.397" Score="2" ViewCount="1120" Body="&lt;p&gt;How would I go about collimating the light from a smartphone's flashlight into a more focused beam? I'm working on a project where I need a focused beam of light to act as a light source so I can take pictures of the retina of a human patient using a smartphone. I'd rather use the built-in flash of the camera than build my own light source, but the light from the flash is very diffuse.&lt;/p&gt;&#xA;" OwnerUserId="9683" LastEditorUserId="9" LastEditDate="2017-01-23T15:00:56.580" LastActivityDate="2017-01-23T15:00:56.580" Title="How to focus the flashlight of a smartphone?" Tags="|electrical-engineering|optics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13386" PostTypeId="1" CreationDate="2017-01-23T04:44:44.860" Score="1" ViewCount="105" Body="&lt;p&gt;The main raw water has following parameters, water comes from a soap manufacturing factory&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;pH 7.1  &lt;/li&gt;&#xA;&lt;li&gt;BOD 520 mg/L  &lt;/li&gt;&#xA;&lt;li&gt;COD 4890 mg/L  &lt;/li&gt;&#xA;&lt;li&gt;TSS 360 mg/L  &lt;/li&gt;&#xA;&lt;li&gt;No Oil &amp;amp; Grease &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;if you need expected treated effluent data&#xA; - pH 6.5-8.5&lt;br&gt;&#xA; - BOD &amp;lt;50 mg/L&lt;br&gt;&#xA; - COD &amp;lt;30 mg/L&lt;br&gt;&#xA; - TSS &amp;lt;250 mg/L&lt;br&gt;&#xA; - Oil &amp;amp; Grease &amp;lt;10 mg/L &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is activated sludge process efficient for industrial waste water treatment of this kind of raw water? what are the reasons if it isn't a good way to treat such raw water. Any recommendations with reputable sources such as research papers etc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my proposed process is initial sieve filtration --&gt; airation --&gt; addition of floc (bacteria and protozoa) --&gt; settling tank --&gt; gravity filtration &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm really new to this and trying to understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;please letme know if I asked it in a incorrect way. thanks&lt;/p&gt;&#xA;" OwnerUserId="4273" LastEditorUserId="4273" LastEditDate="2017-01-23T15:37:45.093" LastActivityDate="2018-01-09T02:20:24.500" Title="Is activated sludge process efficient for industrial waste water treatment ? why if isn't?" Tags="|waste-water-treatment|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13388" PostTypeId="1" AcceptedAnswerId="13390" CreationDate="2017-01-23T08:56:46.110" Score="1" ViewCount="133" Body="&lt;p&gt;For the highlighted part , why we need to replace the support at A with a roller guide ? We could see at A , it's a pinned support , right ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pinned support will resist horizontal and vertical reaction , why the author said it will not resist vertical force ? Is the author wrong ?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gMhaW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gMhaW.png&quot; alt=&quot;fsf&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2017-01-23T10:43:32.817" Title="qualitative influence line of pinned support" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13389" PostTypeId="1" CreationDate="2017-01-23T10:10:34.007" Score="3" ViewCount="174" Body="&lt;p&gt;In a few days, there will be a power outage that will probably last for 8 hours. We had this issue before, but then only our part of de building was affected. It was temporarily resolved by using 2 industrial extension cords. But in this case, the whole street won't have power. So that's not an option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because we've done it before with extension cords, we where wondering if we can do it again using a power generator. I've counted about 15 computers. Assuming they use an estimated 250 watt average. But I'm aware this is a big guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the things to keep in mind when choosing a temporary backup generator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equipment we want to supply:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mail server, file server, firewall and all other primary network devices use an APC&lt;/li&gt;&#xA;&lt;li&gt;Other devices (8 computers, 8 27&quot; screens) will be powered off. They are both on 2 power loops (hope I'm using the correct terminology). Once these are connected to the temporary power supply, the devices can be switched on again.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9690" LastEditorDisplayName="user16" LastEditDate="2017-01-24T00:45:04.873" LastActivityDate="2018-09-25T14:22:25.033" Title="Planned power outage and backup generator" Tags="|electrical-engineering|power|generator|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13390" PostTypeId="2" ParentId="13388" CreationDate="2017-01-23T10:43:32.817" Score="2" Body="&lt;p&gt;The answer is in your quote/picture:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In order to draw the deflected shape properly, the capacity of the beam to resist the applied function must be removed so the beam can deflect when the function is applied.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Muller-Breslau principle is that the influence line is equal to: the deflected shape of the relevant force/moment at the relevant location &lt;em&gt;if the resistance to the effect at the location of interest is removed&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to find the influence line for a vertical reaction at A, we therefore want to find the deflected shape for a vertical force at A, when there is no vertical support at A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we had left the vertical support in at A, then a vertical force applied at A would give us no deflection, and hence the influence line would be zero everywhere. Clearly that's wrong, as we know a force applied just next to A would give us a reaction at A!&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-01-23T10:43:32.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13391" PostTypeId="2" ParentId="13367" CreationDate="2017-01-23T11:12:33.273" Score="5" Body="&lt;p&gt;I think you are possibly concentrating on the wrong thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your state.pa.us link:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The VLF of a passenger car (.0004) is so small that cumulative pavement impact is essentially moot&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, essentially, comparing a garbage truck to a car is pointless: the car may as well not be there. In other words: the car is so insignificant compared to a garbage truck that you can't use &quot;a car&quot; as your unit. It's like me looking out my window in London and saying &quot;That building is about 1000 times the height of a person&quot;, and someone in New York looking out their window and saying &quot;That building is about 10000 times the height of a person&quot;. Is the building in New York really 10 times higher? Or could I have been using a taller-than-average person, and rounded down, while the person in New York used a shorter-than-average person and rounded up. Maybe the difference in height is only 3 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, your state.pa.us link also says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Roadway damage caused by vehicles is exponential (x4). For example, doubling the load for a given axle increases the damage 16 times.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So to get a vehicle damage ratio of 11700/1279 = 9.15, you'd need a vehicle weight ratio of 1.74. Is it inconceivable that the Pennsylvania investigation used a garbage truck that was 1.2 times heavier, and a car that was 1.5 times lighter, than the Colorado investigation? I suggest not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary: I suggest you look at &quot;damage done by garbage trucks&quot; in terms of longevity of road surfaces used by them, and not in terms of relative damage done by cars.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-01-23T11:12:33.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13392" PostTypeId="2" ParentId="13385" CreationDate="2017-01-23T11:49:17.997" Score="2" Body="&lt;p&gt;Focusing and collimating are two different things. Focusing means focusing into a small point (which will show the image of the light source); collimating is analogous to focusing into infinity, so that the beam diameter stays constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either of these is readily done using a lens. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Lens_(optics)#Imaging_properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;basic formula&lt;/a&gt; for lenses is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{S_1} + \frac{1}{S_2} = \frac{1}{f}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6pVI2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6pVI2.png&quot; alt=&quot;From wikipedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $S_1$ would be the distance from light source to lens, $S_2$ would be either the focus distance or $\infty$ for collimating; and $f$ would be the lens focal length. You can choose two of these parameters quite freely, and then calculate the third one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However, a word of caution:&lt;/strong&gt; The flash light of a smartphone contains easily enough power to cause permanent damage to the eye. Especially if focused using a lens. Make sure you do the proper calculations and measurements for safety limits before trying this on anyone.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2017-01-23T11:49:17.997" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13393" PostTypeId="2" ParentId="13380" CreationDate="2017-01-23T11:58:05.770" Score="6" Body="&lt;p&gt;Try looking into gas spring based door openers and hatch lifts. They are commonly available for various weights, and can easily be installed hidden inside the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2xW74.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2xW74.jpg&quot; alt=&quot;door opener&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2017-01-23T11:58:05.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13394" PostTypeId="2" ParentId="13377" CreationDate="2017-01-23T12:25:16.560" Score="1" Body="&lt;p&gt;The answer is in your quote/picture:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the support A is cantilever type, a double roller guide must be used at A &lt;strong&gt;since this type of support will then transmit both a moment M&lt;sub&gt;A&lt;/sub&gt; at the fixed support&lt;/strong&gt; and axial load, R&lt;sub&gt;Ax&lt;/sub&gt; but will not transmit R&lt;sub&gt;Ay&lt;/sub&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In other words, we want the influence for the vertical reaction, so we release the vertical support degree of freedom only, &lt;em&gt;and leave the other supported degrees of freedom alone&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as iv) goes, I can see three rollers in the picture. This is analogous to a &quot;double roller&quot;: it will transmit axial load and moment but not shear load.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-01-23T12:25:16.560" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="13395" PostTypeId="1" AcceptedAnswerId="13402" CreationDate="2017-01-23T13:06:04.727" Score="1" ViewCount="255" Body="&lt;p&gt;The following picture is pretty self-explanatory. I need a connection for a beam that is coming in very close to a concrete wall edge (north/south beam in picture). The embedded plate with nelson studs in already in place. I was thinking a T type welded plate with bolts already fixed to plate (as drawn in purple), but they wont be able to weld both sides of the plate. Maybe with a full penetration backed weld? Any better ideas on how to fix this beam? Required shear for full beam capacity is about 175 kN and factored load is about 50 kN to give a quick idea.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/W8GPO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8GPO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5691" LastEditorUserId="1832" LastEditDate="2017-01-23T14:53:11.490" LastActivityDate="2017-01-23T15:38:59.673" Title="Connection for beam very close to concrete wall" Tags="|structural-engineering|shear|connections|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13396" PostTypeId="2" ParentId="10426" CreationDate="2017-01-23T13:46:57.520" Score="0" Body="&lt;p&gt;I think your best bet is to provide a far more porous &quot;interface&quot;: sew on a strip or whatever other shape you need of a porous cloth - cotton, flax, etc, then glue that to polyurethane - even using spray-on PUR foam - either as glue, or just as &lt;em&gt;the&lt;/em&gt; foam, shaping it as needed in a cast.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-01-23T13:46:57.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13397" PostTypeId="1" CreationDate="2017-01-23T13:51:17.563" Score="-1" ViewCount="949" Body="&lt;p&gt;For the maximum moment occur under the 1.5k load , why when we consider the left part of the beam, we just need to consider the forces 2k and 1.5k only ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about FR = 4.5k and also the 1k ? &#xA;Cant i just consider all of them ? That's means i make an iamginary cut after the 1k load ... &#xA;&lt;a href=&quot;https://i.stack.imgur.com/76S43.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/76S43.jpg&quot; alt=&quot;jj&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wvYfL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvYfL.jpg&quot; alt=&quot;njj&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2017-01-23T14:49:16.863" Title="absolute max shear and moment" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13398" PostTypeId="2" ParentId="13364" CreationDate="2017-01-23T14:16:07.627" Score="4" Body="&lt;p&gt;Quite a few of the machines allow to tilt the pair of wheels to arbitrary angle too. Some allow tuning speed of the two motors separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's all about curveballs and &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnus_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magnus Effect&lt;/a&gt;. One wheel going faster will give the ball a spin on top of the forward speed - and the ball trajectory will curve in direction perpendicular to rotation axis and current velocity vector. &quot;Stacked&quot; layout allows to pitch the ball a longer distance, by counter-acting gravitational drop, and keeping the trajectory more straight. Sideways placement will tend to curve the ball to a side. Which you want depends on how you want to train.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-01-23T14:16:07.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13399" PostTypeId="2" ParentId="13380" CreationDate="2017-01-23T14:17:36.497" Score="0" Body="&lt;p&gt;As has been suggested, gas springs are probably the way to go here, they are available in a wide range of stroke length, spring force and attachment methods. I have the most experience with products from &lt;a href=&quot;https://www.suspa.com/us/products/gas-springs/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Suspa&lt;/a&gt; here in the USA, but there are hundreds of manufacturers world wide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure you get a gas &lt;strong&gt;spring&lt;/strong&gt; and not a gas strut or a gas damper, they are not the same despite similar appearances. Don't be like me and have to learn this the hard way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Size the spring and select your mounting points such that the force of the spring offsets about 90% of the lid's weight. For your application, assuming that you'd use two springs, one on each side, I would look into a locking gas spring  on on side so that the lid stays in place once it is in the fully open position.  &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-01-23T14:17:36.497" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13400" PostTypeId="2" ParentId="13397" CreationDate="2017-01-23T14:49:16.863" Score="2" Body="&lt;p&gt;You can always find the bending moment at a given point by calculating the moment generated by all the forces to one side of the point. So, you can calculate it as $M = A_yD_A + 2D_2$ (left section) &lt;strong&gt;&lt;em&gt;OR&lt;/em&gt;&lt;/strong&gt; as $M = B_yD_B + 1D_1$ (right section), where the $D$'s are the lever arms, since both sections must give equal results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the authors chose to calculate from the left section, but that was just their personal choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resultant force of 4.5&amp;nbsp;kN is completely irrelevant for this and was only used to calculate the reactions $A_y$ and $B_y$.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-01-23T14:49:16.863" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13402" PostTypeId="2" ParentId="13395" CreationDate="2017-01-23T15:38:59.673" Score="2" Body="&lt;p&gt;Full penetration weld would be ok. I am more concerned about that bolts between the beam and the concrete edge. It could be difficult to tight them and inspection  would be problematic too. Also you loose a considerable part of the flanges, at the beam edge, and therefore you cause a severe drop of the moment strength. I don't know if this is acceptable for your design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more simple solution i came up, is shown in the figure below. The beam is cut a little shorter in order to be flush with the concrete face and thus avoid flange trimming. A plate (pink color) is fastened with fillet welds at the anchor plate. A filler plate (green) is used to fill the space between the pink plate and the beam web. With this design, access for inspection and repairs remains good. No installation problems too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage from placing the pink plate parallel to the edge, is that no bending is transferred to it. As a result, you avoid any additional bending of the anchor plate, tension of the anchors and compression stresses on the concrete. But you may induce more shear to the anchors (not certain if it is worst, because shear is induced by your design too).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qa1dX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qa1dX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What verifications are needed:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fillet Welds of the pink plate&lt;/li&gt;&#xA;&lt;li&gt;Anchors due to the additional shear (there will be a torsional component) &lt;/li&gt;&#xA;&lt;li&gt;Bolts (considering the filler plate)&lt;/li&gt;&#xA;&lt;li&gt;Pink plate in bending/shear&lt;/li&gt;&#xA;&lt;li&gt;Beam web&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9620" LastActivityDate="2017-01-23T15:38:59.673" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13403" PostTypeId="1" AcceptedAnswerId="13433" CreationDate="2017-01-23T17:25:55.897" Score="1" ViewCount="215" Body="&lt;p&gt;(Discrete systems for my question are further divided into periodic and event based)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By reading around I have a feeling that analog does not equal continuous and that digital does not equal discrete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in the case of a periodic discrete control the measurements may not be digitized, not even quantized and simply fed into a controller. Is this correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, does this mean that there can be 6 types of systems? Analog-continuous, analog-periodic, analog-event_based, digital-continuous, digital-periodic, digital-event_based?&lt;/p&gt;&#xA;" OwnerUserId="9696" LastEditorUserId="8055" LastEditDate="2017-01-23T18:28:21.330" LastActivityDate="2017-01-25T00:06:20.190" Title="Is analog|digital different from continuous|discrete?" Tags="|electrical-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13404" PostTypeId="1" CreationDate="2017-01-23T18:27:40.320" Score="0" ViewCount="79" Body="&lt;p&gt;I am asking for a categorization of the control mechanism that TCP uses for flow and congestion control. Obviously, it is closed-loop but would that be discrete? Event-based? Continuous?&lt;/p&gt;&#xA;" OwnerUserId="9696" LastEditorUserId="8055" LastEditDate="2017-01-23T19:19:18.590" LastActivityDate="2017-06-25T02:47:17.283" Title="What type of control does the TCP protocol offer?" Tags="|electrical-engineering|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13405" PostTypeId="1" CreationDate="2017-01-23T20:47:19.073" Score="1" ViewCount="1644" Body="&lt;p&gt;I have used a MATLAB finite difference code to solve a lid driven cavity flow, based on a Stream function-Vorticity formulation of the viscous, incompressible Navier Stokes equations. Details about the method can be found &lt;a href=&quot;http://www3.nd.edu/~gtryggva/CFD-Course/2013-Lecture-5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to change the code to simulate the flow around a square box in a rectangular domain, where the flow is uniform on the left side, and the flow is limited by the horisontal walls on the top and bottom. However, on the right side of the domain, where the flow exits, I have no idea how to express the boundary conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I express the boundary conditions on the right side? Is it even possible for this method?&lt;/p&gt;&#xA;" OwnerUserId="9697" LastActivityDate="2017-06-23T18:12:35.673" Title="Boundary conditions for Stream function-Vorticity method" Tags="|fluid-mechanics|cfd|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13406" PostTypeId="1" CreationDate="2017-01-23T21:17:14.540" Score="2" ViewCount="197" Body="&lt;p&gt;During &lt;a href=&quot;https://en.wikipedia.org/wiki/Northeast_blackout_of_2003&quot; rel=&quot;nofollow noreferrer&quot;&gt;the 2003 Northeast blackout&lt;/a&gt; I had the misfortune of being in Manhattan.  After about 12 hours municipal water pressure was gone &amp;ndash; I assume because the pressure is supplied from water towers or reservoirs that are fed by electric pumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In such a condition, what would happen if someone determined to get water from a tap connected a (perhaps gas-powered) pump to a municipal water supply line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that older municipal systems may often lack functional backflow check valves everywhere.  In such an event, it seems likely that the pump would suck air into the system through any other consumer taps that were left open.  So it would get whatever water is between it and the closest open taps, and then it would pull get air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, for purpose of this question let's assume that all outlets are either closed or else protected with working backflow valves.  Will the pump be able to pull water all the way from the raw water supply, through whatever treatment facilities and supply pumps normally feed the mains, and finally out the service tap to which it is connected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect not, but I have no idea of where it would stop.  (For example, if municipal water is always fed by gravity from a reservoir of treated water, then presumably when the reservoir runs dry the utility would lock it to prevent air from entering the supply system.  But I have no idea what facilities or mechanisms are upstream in a major water utility system.)&lt;/p&gt;&#xA;" OwnerUserId="77" LastEditorUserId="77" LastEditDate="2017-01-24T14:08:58.660" LastActivityDate="2017-01-24T14:08:58.660" Title="What happens if one pumps from municipal water supply during power outage?" Tags="|civil-engineering|water-resources|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13407" PostTypeId="2" ParentId="13275" CreationDate="2017-01-23T21:28:24.363" Score="0" Body="&lt;p&gt;Your question is very ambiguously worded making it hard to explain the behaviour that you do not understand. D-type latches are very common and their behaviour is &lt;a href=&quot;https://en.wikipedia.org/wiki/Flip-flop_(electronics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;straightforward&lt;/a&gt; once their operation is explained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please find a full truth table and explanation of your particular logic circuit &lt;a href=&quot;http://www.daenotes.com/electronics/digital-electronics/flip-flops-types-applications-woking&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $Q$ and $Q^*$ should always be opposites so I believe you have made a false statement when saying $ Q1 = Q1* = 0 $.&lt;/p&gt;&#xA;" OwnerUserId="8933" LastActivityDate="2017-01-23T21:28:24.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13409" PostTypeId="2" ParentId="13389" CreationDate="2017-01-23T23:00:26.883" Score="2" Body="&lt;p&gt;The most practical and simplest is to review the breaker box. Below is an example breaker box for a typical US single family house. Note: US base houses are power using 120 VAC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gv299.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gv299.jpg&quot; alt=&quot;Breaker box&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add up total current for circuits that are of interest, which essentially will help the estimate the total current. Then calculate the total wattage for the generator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember the startup current for most electrical equipment is higher than steady state current. So if the startup sequence is staged appropriately a smaller generator might work. This requires knowledge of the startup current and steady state current for the electrical devices. Below are some reference that might help further investigate the solution. Note this is for a US single family house, the theory should be same for a larger business. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: It is best to consult a professional before making the final decision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.twincityelectrical.com/Calculator/sizingcalculator.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Selecting the Correct Generator&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.norwall.com/power-expert/Sizing-Guide-for-Commercial-Generators/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sizing Guide for Commercial Generators&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2017-01-24T13:01:54.067" LastActivityDate="2017-01-24T13:01:54.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13410" PostTypeId="1" CreationDate="2017-01-23T23:33:47.853" Score="1" ViewCount="663" Body="&lt;p&gt;In my area traffic lights with bulbs are being gradually replaced with LED traffic lights. Sometimes these lights have no diffusion lens and the individual LEDs can be seen. Other times the lights have diffusion lenses which give a very smooth, consistent color.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do these lenses work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traffic light in Brookline, MA, with no diffuser:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KS1Ma.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KS1Ma.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="1672" LastEditDate="2017-01-24T03:32:16.167" LastActivityDate="2017-01-24T03:32:16.167" Title="How do LED traffic light diffusion lenses work?" Tags="|optics|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13413" PostTypeId="1" CreationDate="2017-01-24T06:15:29.973" Score="2" ViewCount="33" Body="&lt;p&gt;I am currently using a AI-6061-T6 plate measuring 750mmx370mmx6mm as a rack between two frames at about 3/4 meter above ground. The load on the plate would be about 8-12kg. How much load will be able to take before bending? And how would the plate react underwater at a depth of 3-4m? Please help.&lt;/p&gt;&#xA;" OwnerUserId="9700" LastActivityDate="2017-01-24T06:15:29.973" Title="Aluminium plate being used in an AUV" Tags="|aluminum|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13414" PostTypeId="1" CreationDate="2017-01-24T10:00:52.687" Score="0" ViewCount="39" Body="&lt;p&gt;I need to know, or estimate the physical characteristics of the alloy which consists 40% copper and 60% molybdenum. I know only:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Density: $9.62 \frac{g}{cm^3}$&lt;/li&gt;&#xA;&lt;li&gt;Thermal conductivity: $223 \frac{W}{m·K}$&lt;/li&gt;&#xA;&lt;li&gt;Coefficient of thermal expansion $9.3 \cdot 10^{-6} \frac{1}{°C}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How to calculate, or at least, estimate its boiling point, clectrical resistivity, Young's modulus, and Vickers hardness? Also partial answers are okay.&lt;/p&gt;&#xA;" OwnerUserId="8880" LastEditorUserId="9" LastEditDate="2017-01-25T19:59:58.893" LastActivityDate="2017-01-25T19:59:58.893" Title="How to estimate the physical parameters of 40% Cu + 60% Mo alloy?" Tags="|building-physics|alloys|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13416" PostTypeId="2" ParentId="13389" CreationDate="2017-01-24T12:28:15.487" Score="3" Body="&lt;p&gt;Electronics can be quite sensitive to the quality of the power supply, particularly in terms of providing a stable voltage and the right sort of AC waveform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two obvious solutions for backup power. The first is to use a generator powered by an IC engine. The complication here is that the engine output needs to be matched to the load by regulating the throttle which means that if the load is inconsistent the voltage and frequency  can vary. So a simple generator consisting of an engine driving an alternator may need additional regulation of the electrical output to provide a supply suitable for electronics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now get inverter generators which convert the AC generated by the alternator to DC and then use and inverter to convert it back to a more stable AC supply. A decent one will produce a stable voltage with a waveform very close to a pure sine wave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other alternative is to use inverters powered by batteries which can be charged by the normal mains supply when not required. This method has the advantage of providing a 'clean' power supply (with the right inverter) and can be a bit more modular. Deep cycle lead acid batteries are generally used for this sort of application as they are reasonably economical and can tolerate quite a lot of discharge without being damaged. Note that conventional car type batteries don't like being fully discharged and prefer to be have their charge regularly topped up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Batteries also tend to deal better with fluctuating demand as they can deliver fairly high currents eg to cope with surged on startup of equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to account for the fact that whatever supply you use will really need to be able to cope with the peak demand as well as the sustained average. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-01-24T13:28:56.233" LastActivityDate="2017-01-24T13:28:56.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13417" PostTypeId="2" ParentId="12323" CreationDate="2017-01-24T12:38:21.193" Score="1" Body="&lt;p&gt;Most generators will attempt to maintain a constant sped/frequency regardless of load. As far as the engine is concerned the load is the torque on the alternator which is itself dependant on the current being drawn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engine RPM is controlled by some sort of electronic or mechanical regulator ie some sort of feedback control system, which detects engine speed and adjusts the throttle to attempt to maintain a preset rpm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice this regulation is imperfect and there will often be a some lag with the engine speed slowing noticeably when a large load is applied, for example starting a large electric motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the power generated is proportional to the electrical impedance across the alternator ie the current that what ever is connected to it is trying to draw so less load simply means that the engine needs to deliver less torque to maintain the same rpm. For example if the engine is running with nothing connected to the electrical output then no electrical power is generated and the engine only has to overcome its own internal friction, as soon as a load is connected this creates a back emf in the alternator which imposes a torque on the engine which needs to be matched by opening the throttle to maintain a constant speed. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-01-24T12:38:21.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13418" PostTypeId="1" CreationDate="2017-01-24T13:03:58.887" Score="1" ViewCount="297" Body="&lt;p&gt;I'm making a corner shelf for spare toilet paper from a thin plywood sheet, say 5mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to connect the plywood to the corner walls with a bolt/nut joint like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/NVWAr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NVWAr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is -- how do I calculate the size of a cutout for such a joint so that a wrench would fit?&lt;/p&gt;&#xA;" OwnerUserId="5053" LastActivityDate="2017-01-24T13:51:55.007" Title="Design for CNC -cut sheet+bolt connection" Tags="|design|cnc|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13420" PostTypeId="2" ParentId="13418" CreationDate="2017-01-24T13:20:24.850" Score="1" Body="&lt;p&gt;The external size of wrenches can vary quite a lot. Ring spanners tend to be the slimmest but you need enough room to get them off the end of the bolt once it is tight. You can also get ring spanners with slots which fit over a bolt shaft but theses are a bit more chunky. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really you just need to measure the size of the wrench you want to use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that you might want to consider using a captive nut, this will make assembly easier and you may well find that it is more stable and distributes the load better for this sort of installation in wood. It doesn't take much for a hex nut to dig into plywood. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed you may find that 5mm plywood is really a bit thin to use this sort of construction method at all. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-01-24T13:27:25.573" LastActivityDate="2017-01-24T13:27:25.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13421" PostTypeId="2" ParentId="13406" CreationDate="2017-01-24T13:20:53.013" Score="3" Body="&lt;p&gt;It isn't just a question of air. When you put suction on a system that's normally under pressure, you're going to draw in contaminants through every available opening &amp;mdash; leaks in underground pipes, garden hoses left in buckets, heating systems, etc. Generally a very bad idea.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2017-01-24T13:20:53.013" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13422" PostTypeId="2" ParentId="13418" CreationDate="2017-01-24T13:51:55.007" Score="2" Body="&lt;p&gt;What you're looking for is called &quot;Wrench Clearance&quot;. There are tables for these in Machinery's Handbook (pp 1344-1355 in 24th ed). There are also resources online like &lt;a href=&quot;http://www.engineersedge.com/wrench_clearance.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this table&lt;/a&gt; for box end wrenches or &lt;a href=&quot;http://www.engineersedge.com/general_engineering/open-end-wrench_clearance.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; for open end wrenches&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will caution you that the clearances listed in these types of tables should be considered minimums and more is generally better within reason of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also think that you'll find this joinery marginal for plywood this thin. It's much better suited for 1/2&quot; - 3/4&quot; plywood.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-01-24T13:51:55.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13423" PostTypeId="2" ParentId="13403" CreationDate="2017-01-24T14:49:39.257" Score="0" Body="&lt;p&gt;Yes. {snarky answer}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details:  it's trivial to collect discrete analog samples, e.g., by checking the level in a capacitor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW, you cannot continuously collect digital samples, since the process of collecting any sample requires an integrator and thus a sample time-window. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of your proposed sub-categories are not really of interest.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-01-24T14:49:39.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13424" PostTypeId="2" ParentId="13405" CreationDate="2017-01-24T15:29:06.150" Score="-1" Body="&lt;p&gt;(In the following I assume twodimensional with main flowin positive x direction, the velocity vector is considered $\vec{u} = (u \quad v)^T$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary conditions you could apply to the outlet:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;homogeneous Neumann for the streamwise velocity and no tangential stress, i.e. $\frac{\partial u}{\partial x}$ and $v = 0$&lt;/li&gt;&#xA;&lt;li&gt;If zour flow is turbulent you might want a proper outflow condition based on the wave equation: $\frac{\vec{u}}{t} + \vec{c} \frac{\vec{u}}{\vec{x}} = 0$ with $\vec{c}$ the convective outlet velocity which should be in the order of your bulk velocity.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both these conditions ignore pressure. From a physical point of view your method then needs to generate the pressure gradient driving the flow. I think that formulation does allow that but I have never worked with it. Another issue that you have to keep in mind is that your domain needs to be long enough to allow the pressure gradient to develop behind the obstacle. Or in other words you need to simulate the whole wake (or the most part of it). If you cut through the wake your solver might not even converge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be possible to reformulate both boundary conditions to streamfunction and vorticity values. The first option is probably the simpler one as it only requires $\omega = 0$ and the streamlines to be horizontal. &lt;/p&gt;&#xA;" OwnerUserId="9531" LastActivityDate="2017-01-24T15:29:06.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13426" PostTypeId="2" ParentId="8246" CreationDate="2017-01-24T17:18:54.440" Score="3" Body="&lt;p&gt;I would rethink your design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cost and difficulty of producing your component with a conventional machine setup is likely outside your range of expectations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest the following approach to create a micro orifice.   Instead of a conventional printer nozzle, consider using a thin piece of sheet material, perhaps 1mm thick, which has been drilled by a water jet or laser to your specified bore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water jet could then cut out a small piece of the drilled sheet to fit the inside of a holder that attaches to the head.  This would produce a 2-piece nozzle with the conventional exterior and thread fittings, containing a small drilled disk with the orifice. The outer part of the nozzle can be produced in aluminum for example using a standard lathe setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ctBzG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ctBzG.png&quot; alt=&quot;2 part nozzle with tungsten micro-bore orifice&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GuzD9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GuzD9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Jf6KM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jf6KM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="7021" LastEditDate="2017-01-24T17:25:16.390" LastActivityDate="2017-01-24T17:25:16.390" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13428" PostTypeId="2" ParentId="12381" CreationDate="2017-01-24T22:39:57.703" Score="2" Body="&lt;p&gt;I have designed plenty of (probably several dozen) retaining walls with long toes in situations with low allowable bearing capacities. Usually designed with RetainPro (which is not FEA), but also using &quot;by hand&quot; calcs as well. So your results are not unusual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to rotation about the toe or heel: overturning and rotation are two different things. You actually want the wall to rotate about the heel just a bit so you can transition from at-rest to active soil conditions, and relieve the locked-in horizontal soil load (from compaction behind the wall). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the wall is going to catastrophically fail, which is what an OTM check is looking at, it WILL rotate about the toe eventually. At the point in time it fails, such a wall will probably have initially rotated around the heel, but the overturning failure will be about the toe. Given the presence of the soil underneath constraining movement into the soil mass, how could it not? But the walls you are analyzing don't seem to be in danger of catastrophically failing (or your FEA would not converge). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it this way: in order for the wall to fail due to the applied overturning moment about the heel, the footing toe would have to rotate INTO the soil. This cannot happen unless there is a slope failure, but that is a separate discussion topic. And that failure mode is not being analyzed in your FEM- that is, unless it's a geotechnical analysis package and you have some kind of decent soil modeling design parameters, which nobody ever has. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try this experiment in your FEM: put an X fixity at the footing toe (so it cannot slide). Then incrementally ramp up the horizontal load only (no corresponding increase in vertical load). The structure will eventually rotate about the toe. Checking this kind of load condition- high horizontal loading (usually due to live load surcharge BEHIND the wall heel) with high resistance to sliding- is the purpose of the OTM check. Under such a loading condition, it would make no sense to check the OTM about the heel. The wall simply will not fail that way unless there is a localized bearing capacity failure. &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2017-07-12T03:25:50.747" LastActivityDate="2017-07-12T03:25:50.747" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13429" PostTypeId="2" ParentId="12085" CreationDate="2017-01-24T22:53:48.880" Score="2" Body="&lt;p&gt;As was said in the other answers, the part in contact with the soil is the &quot;foundation&quot;, or perhaps &quot;footing&quot; (a specific type of foundation, but there are also piles, drilled shafts, rafts - &lt;em&gt;all&lt;/em&gt; of these are foundations). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The soil under a footing will usually be called either the &quot;bearing surface&quot;, or &quot;bearing strata&quot;. However, this would not apply in the case of a deep foundation such as a pile foundation unless you are specifically talking about end-bearing piles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other terminology you may hear and find useful in different situations: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;soil matrix, or soil profile&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This refers to the layers of soils at some location, such as you will see in a boring log. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;bearing capacity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a wildly variable term that very often means different things depending on who says it. When a structural engineer asks &quot;What is the bearing capacity?&quot;, they are almost always asking for an ALLOWABLE bearing capacity based on some limitation for settlement (often &quot;1 inch or less&quot;). However, bearing capacity also means the amount of stress a soil can experience before undergoing some kind of shear failure. These are related, but totally different things. Always be sure you understand what is being talked about when people start throwing around &quot;bearing capacity&quot;. &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2017-01-25T01:37:51.393" LastActivityDate="2017-01-25T01:37:51.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13430" PostTypeId="1" AcceptedAnswerId="13443" CreationDate="2017-01-24T22:58:21.500" Score="1" ViewCount="599" Body="&lt;p&gt;I paddle several different types of small craft in the ocean and bays near my home.  One phenomenon I've observed is beyond my understanding of drag on a narrow displacement hull.  When paddling in water less than maybe 6 feet in depth, there is a very noticeable increase in drag, which increases as the depth decreases.  This is true for all the types of boats I paddle, from a large 6-person Hawaiian outrigger canoe (L=40', width=2', draft=0.67') to a one-person racing kayak (L=20', width=1.5', draft=0.33').  I should add that hull speeds are generally in the 6-8 mph range.  I've read that rowing shells also encounter this same drag and they are longer, narrower, but with drafts in more or less the same range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read one explanation that the boundary layer on the hull makes contact with the bottom and the bottom increases the drag on the outer part of the boundary layer which in turn is transmitted to the hull.  This seems highly unlikely to me, especially when the water is more than a maybe a foot deep.  I don't believe the boundary layer from a 20 foot kayak extends that deep.  As I said, this is noticeable at water depths up to 6 feet or more.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also heard that pressure waves from the hull bounce off the bottom and reflect back up to the hull and cause drag but as an engineer that doesn't really sound very rigorous to me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me there might be some interaction between the bow wave of a boat and the bottom, since I know that the dynamics of surface waves do extend approximately as deep as their wavelength, and at these speeds the wavelength of a bow wave is certainly on par with the water depth.  Can anyone give me a properly defensible answer to this?&lt;/p&gt;&#xA;" OwnerUserId="9637" LastActivityDate="2022-07-02T21:58:49.970" Title="Boat Hull Drag in Shallow Water" Tags="|hydraulics|marine-engineering|drag|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13431" PostTypeId="2" ParentId="13332" CreationDate="2017-01-24T23:24:16.460" Score="1" Body="&lt;p&gt;You can (come close to) this motion profile with two cams on one shaft. the tops being the dwell portion of the cams, and the bottom being the peak between two accelerations.  As some one else mentioned since the bottom of your motion profile is an instantaneous acceleration change, you will never achieve it perfectly. this can be minimized by reducing mass, speed, and stroke length. This is going to be the closest mechanical way to do this. &lt;/p&gt;&#xA;" OwnerUserId="9712" LastActivityDate="2017-01-24T23:24:16.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13433" PostTypeId="2" ParentId="13403" CreationDate="2017-01-25T00:06:20.190" Score="3" Body="&lt;p&gt;Let's go back to the original fundamental definitions of the terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Analog&quot; mean that there is an &lt;em&gt;analogy&lt;/em&gt; between two things.  For example, in a mercury thermometer, there is a relationship between the height of the mercury in the thermometer tube and the outside temperature.  Or in a car tachometer, there is a relationship between the angle of the pointer on the dial and the speed of the engine. Or in an analog clock there is a relationship between the angle of the two hands and the time. A very common type of device is where there is an electrical voltage and there is a relationship between the voltage and something else.  e.g. the voltage is related to the temperature or the engine rpm. Bottom line: analog simply means that one thing is represented by analogy to another thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Digital&quot; means that a thing is not represented by another thing, but directly by a number.  So in a digital thermometer, we don't have a level of mercury in a tube, we have a number.  In a digital tachometer, we don't have a pointer on a dial, we have a number.  In a digital clock, we don't have hands we have a number.  So in digital electronics, we don't have a voltage that is proportional to a quantity, we have a number.  That number is usually commonly encoded in binary, although it could be something else, like balanced ternary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, &quot;continuous&quot; means that given any two values, there are an infinite number of intermediate value.  E.g. like the real numbers. In between 1.0 and 1.1 there are an infinite number of intermediate values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Discrete&quot; means that there are some values which do not have any values in between.  e.g. like the integers.  There is no integer between 1 and 2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, continuous and discrete may refer to either time or the quantity being measured.  e.g. you can have continuous time systems, discrete time systems, but you can also have a continuous valued waveform or a discrete quantized waveform.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, strictly speaking, analog does not equal continuous and digital does not equal discrete. However, in common usage in the context of digital signal processing, analog systems are usually continuous in time and value, and digital systems are usually discrete in time and value.  So the terminology does get abused somewhat.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-01-25T00:06:20.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13434" PostTypeId="1" AcceptedAnswerId="13493" CreationDate="2017-01-25T00:29:02.063" Score="0" ViewCount="91" Body="&lt;p&gt;I'm designing something with a vertical linear actuator. It is a quite large actuator lifting a load. I'm anticipating sway from the actuator as it moves up and down so I'm creating a support structure. What is the best way to determine how much this sway will be? I can figure out the weight of the load and the acceleration of the actuator &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help!&lt;/p&gt;&#xA;" OwnerUserId="9714" LastActivityDate="2017-02-28T14:42:41.740" Title="Vertical Linear Actuator Sway Movement" Tags="|mechanical-engineering|structures|torque|stability|actuator|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13435" PostTypeId="2" ParentId="13430" CreationDate="2017-01-25T01:06:13.137" Score="0" Body="&lt;p&gt;There is a phenomenon often observed when two vessels are travelling close together on a parallel course. The effect causes the hulls to draw closer together due to the increased water velocity between the hulls, due to reduction in dynamic water pressure.  This slight increase in water velocity on one side is enough to push the hull sideways. (Google 'Bernoulli' to learn more) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar effect is called squat (Google 'Squat effect'), where a vessel travelling in relatively shallow water causes a slight increase in the water velocity between the hull and the seabed. The result is a small increase in draft with a corresponding change in the wake profile, which slightly affects the resistance experienced by the hull. &lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-01-25T01:06:13.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13436" PostTypeId="2" ParentId="12455" CreationDate="2017-01-25T01:14:44.547" Score="1" Body="&lt;p&gt;Here is what inertia relief does:  Let's say that we have a flexible free body (e.g. satellite in space).  We apply some forces to it. The forces are going to do two things.  First, they are going to cause a rigid body acceleration according to F=ma.  i.e. the center of mass of the object is going to accelerate based on the sum of the forces.  But, because we have a flexible body, these forces are also going to cause local deformations and stresses.  We don't care about the rigid body acceleration of the center of mass.  We only care about the local stress.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we could solve a transient dynamic solution (i.e. sol 109 in NASTRAN), to get both the rigid body acceleration and the local deformation, and then afterward subtract off the rigid body acceleration to get just the local deformation that we care about.  The problem is that would be computationally expensive.  We are spending a lot of CPU resources to compute something that we are just going to through away anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we want to do is to solve just a static solution (i.e. sol 101 in NASTRAN).  But there is a problem.  Because the structure is unconstrained (free), the stiffness matrix is singular.  We cannot invert it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could add a fake support to the structure somewhere, just for the sake of constraining it to get an invertible stiffness matrix.  But the reaction forces at this constraint will cause local deformations and stresses that are not real and will mess up the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have a problem, we don't want to solve a transient dynamic problem because it would take too long, but a static solution doesn't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here is what inertia relief does.  First, we add a constraint. i.e. we make the structure fixed at the boundary instead of free.  Then, we add a distributed body force to the entire structure that exactly balances out the rigid body acceleration.  That means that the reaction force at our new dummy constraint is zero.  So, the system is now constrained, so the stiffness matrix is non-singular, so now we can invert it.  But, the reaction force at the dummy support is zero, so it doesn't have any effect on the local deformations and stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might help to illustrate: &lt;a href=&quot;https://www.simutechgroup.com/tips-and-tricks/fea-articles/251-fea-tips-tricks-ansys-static-analysis-inertia-relief&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.simutechgroup.com/tips-and-tricks/fea-articles/251-fea-tips-tricks-ansys-static-analysis-inertia-relief&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, having said all that, can you use inertia relief for buckling?  If you are trying to model buckling of a free body that is accelerating, then I don't see why not.  For example, if you have a satellite in space, and this satellite has some flexible structure with thruster attached to it, and you want to know what thrust it will take to buckle your structure, then inertia relief makes sense.  The added fake constraints are only going to constrain out rigid body motion which should not affect the buckling results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you have a constrained structure that is not accelerating, then I don't see how it makes sense. I guess what you are doing is deleting the boundary conditions entirely (because they are hard to model), so fictitiously unconstraining the model, and then using inertia relief to put them back in...  Inertia relief is going to try to generate a body force over your entire structure in order to enforce a &lt;em&gt;zero&lt;/em&gt; reaction at the fictitiously created constraints.  But your structure really does have real constraints, and so there should be a &lt;em&gt;non-zero&lt;/em&gt; reaction at the real constraints.  So the applied body force will be wrong.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not quite sure how NASTRAN handles that body force.  If it uses it with all of the other loads in generating the differential stiffness matrix, then you definitely get the wrong answer.  But if it ignores it when generating the differential stiffness matrix, then you might get the right answer, &lt;em&gt;if&lt;/em&gt; the supports that got added back in were equivalent to the ones that got deleted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that was a really long winded answer... bottom line, although it might work, unless there is something special about how NASTRAN does this that I'm not aware of, I would not recommend it. Spend some extra time to get the boundaries modeled correctly.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-01-25T01:14:44.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13437" PostTypeId="1" CreationDate="2017-01-25T04:45:40.163" Score="1" ViewCount="9569" Body="&lt;p&gt;What are the difference between these two ? Here's my notes , The author just stated the formula without explaining much . Can someone explain further what 's their difference ? &lt;a href=&quot;https://i.stack.imgur.com/ZA9No.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZA9No.jpg&quot; alt=&quot;gg&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2020-02-03T09:09:06.677" Title="External indeterminate vs internal indeterminate" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13438" PostTypeId="1" AcceptedAnswerId="13441" CreationDate="2017-01-25T08:33:20.517" Score="0" ViewCount="877" Body="&lt;p&gt;It's given that the applied force is 10kN at F and 25kN at E . I have trouble of finding the other real force in the member of the trusses . &#xA;So , i start by drawing the FBD at F . I have 10kN point downwards , so the reaction is 10kN upwards  , so it's in equilibrium . But , &lt;strong&gt;&lt;em&gt;i really have no idea how the author get the reaction 15kN ( in tension) pointing to the left and to the right .&lt;/em&gt;&lt;/strong&gt; Can someone explain about it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/G3f9x.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/roR5v.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="1832" LastEditDate="2017-01-25T14:00:35.270" LastActivityDate="2017-01-25T14:00:35.270" Title="finding real force in trusses" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13440" PostTypeId="2" ParentId="5732" CreationDate="2017-01-25T12:15:01.907" Score="1" Body="&lt;p&gt;This is an optimization problem, and specifically a topology optimization one. There are software packages that perform that task. However an optimal solution depends on the contraints of the problem, which sometimes are hard to quantify. Normally the objective is to minimize total cost, but other considerations can be valid (architectural style for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An insight about the factors that make a truss more cost effective is useful. These factors may contradict one another, so the optimal solution  depends on the specific project. The most profound factors are mentioned next.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Material cost&lt;/strong&gt;: Optimally should be minimized, but this can lead to a&#xA;more complicated design, and increased labor costs.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Labor cost&lt;/strong&gt;: Again optimally should be minimized, but this usually&#xA;means members are not varying much, which results in increased&#xA;material costs. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Joints&lt;/strong&gt;. Joints are a considerable part of the total cost, so an&#xA;optimal design should minimize them and keep them as simple. Also  depending    on the material some types of joints can be more difficult.&#xA;For wood, it is harder to joint tensile members compared to&#xA;compressive ones, so one should try to minimize such joints. These decisions can have an impact on material costs. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Sensitivity to buckling&lt;/strong&gt;: Steel members usually suffer from buckling when in compression, so an optimal design should try to drive compression through the short members. Consider for example the Pratt truss in the figure. For gravity loads, the diagonal braces, which are the most lengthy ones, are in tension while the vertical braces, which are the shorter ones, are in compression. This solution seems suitable for a steel truss, since it drives compressive forces through the shorter members.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aLArF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aLArF.png&quot; alt=&quot;Pratt truss&quot;&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare this to the Warren truss below, where all the braces have equal lengths and some of them are in compression. If the members are sensitive to buckling, then the increased length of the compression members (compared to the Pratt truss), may drive the material costs for them high. However, if buckling is not an issue (for wooden structure maybe) then this design is simpler and therefore more cost effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XRtYL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XRtYL.png&quot; alt=&quot;Warren truss&quot;&gt;&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Life cycle costs&lt;/strong&gt;. Maintenance costs, upgrade costs, demolition costs, environmental costs etc. can vary from one design to another. Often this cost is the bigger one but also the harder to predict.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9620" LastEditorUserId="9620" LastEditDate="2017-01-25T12:27:54.437" LastActivityDate="2017-01-25T12:27:54.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13441" PostTypeId="2" ParentId="13438" CreationDate="2017-01-25T13:22:04.620" Score="2" Body="&lt;p&gt;&lt;em&gt;Note: a &lt;strong&gt;reaction&lt;/strong&gt; is an external force, such as at A and D. The internal forces in members are not reactions; you could refer to them as &quot;axial loads&quot; in this case.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove unknowns one at a time. The easiest are the external reactions at A and D. Take moments about A to get the reaction at D, take moments about D to get the reaction at A. Resolve vertically to make sure they add up!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, draw free body diagrams at nodes with the fewest members; these are likely to allow you to eliminate unknowns. In this example, resolving vertically at A will give you the force in member AB, given that you know the reaction at A. Resolving horizontally at A will give you the force in member AF in terms of the force in member AB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Then&lt;/em&gt; you can resolve a free body diagram at B, which will give you that the force in member AF is equal to the force in member FE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue in this manner and you can get the forces in all members.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-01-25T13:22:04.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13442" PostTypeId="2" ParentId="13437" CreationDate="2017-01-25T15:27:52.800" Score="1" Body="&lt;p&gt;For starters, &lt;a href=&quot;https://engineering.stackexchange.com/questions/13437/external-indeterminate-vs-internal-indeterminate/13442#comment24109_13437&quot;&gt;@alephzero's comment&lt;/a&gt; is entirely true: there is, in real practice, no difference between them. Also, you seem to be using a garbage textbook. Do yourself a favor and throw it away. You can probably learn more from videos on Youtube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to actually answer your question, here's the basic difference between them: using only the global equilibrium equations, in externally indeterminate structures, you can't even solve for the support reactions, while in internally indeterminate structures you can find the reactions but not the internal forces in the members.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Externally indeterminate structures&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;These structures have too many supports. By this, I mean that you cannot calculate the reactions for such structures with just the global equilibrium equations. Take, for instance, the following continuous beam (with nodes $A$, $B$ and $C$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FrcEk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FrcEk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's try to calculate the reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= 0 \therefore A_x = 0 \\&#xA;\sum F_y &amp;amp;= A_y + B_y + C_y - 1\cdot(5+8) = 0 \\&#xA;\sum M_A &amp;amp;= 5B_y + (5+8)C_y - 1\cdot(5+8)\cdot\left(\frac{5+8}{2}\right) = 0&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we know that $A_x$ (the horizontal reaction from the first support) is equal to zero. But we still have three unknowns ($A_y$, $B_y$, and $C_y$) and only two equations left. We're stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve this, we need to create one more equation: the fact that the beam's rotation on the left side of $B$ (the middle support) is equal to the rotation on the right side of $B$. This is called a &lt;strong&gt;compatibility equation&lt;/strong&gt;, which is the sort of equation necessary to solve indeterminate structures.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Internally indeterminate structures&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;These structures have just enough supports, but the structure itself is too complex. You can therefore calculate the support reactions with the global equilibrium equations, but not the members' internal forces. Take, for instance, the following truss (with supports $A$ and $B$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6378b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6378b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can easily find the reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= 0 \therefore A_x = 0 \\&#xA;\sum M_A &amp;amp;= 5B_y - 1\cdot3 = 0 \therefore B_y = \frac{3}{5} \\&#xA;\sum F_y &amp;amp;= A_y + B_y - 1 = 0 \therefore A_y = \frac{2}{5} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's try to find the axial forces in each bar. Let's start with node $A$ (the bottom left node with the support), where $N_1$, $N_2$, and $N_3$ are the axial forces in the vertical, diagonal and horizontal bars, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum F_{x,A} &amp;amp;= N_2\cos45 + N_3 = 0 \\&#xA;\sum F_{y,A} &amp;amp;= N_1 + N_2\sin45 = 0&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three unknowns, but only two equations. Can't be done. And you can do this for all the nodes, but you'll still end up with more unknowns than equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you once again need to create additional compatibility equations, such as the fact that the deflection of all the bars at a given node must be equal (for instance, the vertical, diagonal and horizontal bars at $A$ must all end up with a deflection of zero).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-01-25T15:27:52.800" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13443" PostTypeId="2" ParentId="13430" CreationDate="2017-01-25T15:46:58.970" Score="1" Body="&lt;p&gt;I found some possibly applicable explanations &lt;a href=&quot;http://hydrocompinc.com/blog/time-trial-racing-%E2%80%93-deep-water-versus-shallow-water&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  The figure referred to in the quote below is  &lt;a href=&quot;https://i.stack.imgur.com/75c7u.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/75c7u.jpg&quot; alt=&quot;drag graph&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The attached plot [Hofman 2000] illustrates the basic principles of&#xA;  added drag due to shallow water. A non-dimensional depth-based speed&#xA;  parameter is on the X axis (usually noted as FNH, but shown here as&#xA;  FL), and a ratio of drag in shallow water to that in deep water is on&#xA;  the Y axis. You can see how shallow water drag increases rapidly as it&#xA;  approaches the “critical speed”, after whic h it just as rapidly drops&#xA;  off and produces negative added drag. Displacement hulls generally&#xA;  stay in the “sub-critical region” as they typically don’t have the&#xA;  necessary thrust-making “oomph” to get past the critical region drag&#xA;  hump. This is the mode that was described in the paper you read from&#xA;  our Knowledge Library. Planing hulls, however, typically can get past&#xA;  the critical hump, at which point shallow water becomes beneficial.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are a couple interesting-looking references at the bottom of that web page.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-01-25T15:46:58.970" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13444" PostTypeId="1" AcceptedAnswerId="13523" CreationDate="2017-01-25T16:08:33.420" Score="1" ViewCount="1779" Body="&lt;p&gt;I dont understand how the diagram of M/EI of conjugate beam drawn , can someone explain about it ? &#xA;According to conjugate beam theorem , i know that the M/EI represent the w(x) , which is force per unit length of the beam . I dont understand why the author draw in in 2 directions ( upwards and downwards) ... &#xA;&lt;a href=&quot;https://i.stack.imgur.com/Rh1yj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rh1yj.png&quot; alt=&quot;ddd&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2017-02-01T05:56:52.857" Title="Conjugate beam (M/EI)" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13445" PostTypeId="1" CreationDate="2017-01-25T16:10:07.280" Score="2" ViewCount="134" Body="&lt;p&gt;I'm trying to replace connections between a trunk cable and drop cables in a DeviceNet system. The system is in a high humidity environment, so the current clamps used now are not holding up with the harsh conditions. The plan is to replace the cable with a round cable with t-port taps. The taps are IP67 rating, which can handle full submersion in water for 30 mins. I'm wondering though if it will still hold up in the high humidity environment which is constant. My question is asking if IP ratings correlate with a products ability to tolerate humidity.&lt;/p&gt;&#xA;" OwnerUserId="9721" LastActivityDate="2021-03-31T10:49:37.897" Title="Question about IP Ratings" Tags="|electrical-engineering|materials|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13446" PostTypeId="1" CreationDate="2017-01-25T21:50:20.913" Score="3" ViewCount="81" Body="&lt;p&gt;Does the &quot;polymer&quot; in CFRP directly and unambiguously imply the use of an epoxy resin?&lt;/p&gt;&#xA;" OwnerUserId="9726" LastActivityDate="2017-01-27T16:12:58.090" Title="Can the phrases “carbon fibre reinforced polymer composites” (CFRP) and “carbon fibre-epoxy composites” be used interchangeably?" Tags="|composite|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13447" PostTypeId="2" ParentId="13404" CreationDate="2017-01-25T23:40:28.790" Score="1" Body="&lt;p&gt;I would say it is event based.  You send out some packets and then nothing happens until you either receive an ACK, a NACK, or a certain amount of time passes without receiving anything.  Once one of those events occurs, then you take some actions (e.g. send out more packets or re-send dropped packets, possibly more or less packets than you sent out the first time).  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-01-25T23:40:28.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13448" PostTypeId="2" ParentId="590" CreationDate="2017-01-26T00:13:07.087" Score="0" Body="&lt;p&gt;The answer (believe it or not) is no. Why? Because Protactinium is now being discussed as a &quot;additive&quot; to nuclear reactors to improve fuel burn. The cost of removing Protactinium is not necessary at all.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;238\text{Th} + 1\text{n}= 233\text{Pa} = 233\text{U} \\&#xA;233\text{Pa}+1\text{n}=\text{U}234+1\text{n}=\text{U}235&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both are fissionable. So, short answer is no. &lt;/p&gt;&#xA;" OwnerUserId="9728" LastEditorUserId="1832" LastEditDate="2017-01-26T00:59:07.287" LastActivityDate="2017-01-26T00:59:07.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13451" PostTypeId="2" ParentId="7328" CreationDate="2017-01-26T03:21:06.920" Score="4" Body="&lt;p&gt;It's of importance to note that historically, &quot;stone&quot;, &quot;gravel&quot;, and many other related terms have great regional variability. It is always advised to make certain what is meant by such terms when they are being used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the term &quot;gravel&quot; does have a specific engineering definition (aggregate of a certain size range), in design specifications the terms &quot;crushed stone&quot; and &quot;crushed gravel&quot; have come to have a slightly different and generally agreed upon meaning. Using these terms together is a way of specifying several things at once, including the &lt;em&gt;source&lt;/em&gt; of the material, and its size, as well as its shape characteristics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this context, &quot;stone&quot; is rock that is sourced- usually quarried- from some parent rock (such as granite, limestone, and dolomite). Stone has generally not been naturally created by weathering. On the other hand, &quot;gravel&quot; is rock fragments sourced from an existing deposit of weathered rock, often from rivers and streams, but also gravel pits. As such, gravel tends to be more rounded in shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modifier &lt;em&gt;crushed&lt;/em&gt; specifies two things at once. First, that the aggregate be angular in shape and not rounded. Second, that the aggregate consist of a variety of sizes. When required, these very general terms are often fleshed out a bit in the design specification by requiring the aggregate meet some kind of gradation curve, or a durability test (such as the LA Abrasion Test). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When these terms are listed such as in a design specification like the one quoted above, the specification is allowing for just about &lt;em&gt;any&lt;/em&gt; material &lt;em&gt;source&lt;/em&gt;, so long as the material meets the size and shape requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the material is high quality and meets specification, telling the difference between the two on site should be difficult. Look for smaller rounded pieces in the stockpile that were missed by the crusher to identify a gravel source. Rounded pieces from a quarried stone should be very rare. &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2017-02-09T19:17:49.550" LastActivityDate="2017-02-09T19:17:49.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13455" PostTypeId="2" ParentId="13445" CreationDate="2017-01-26T12:53:23.973" Score="1" Body="&lt;p&gt;Yes, most definitely. IP66 can withstand high pressure water jets from any direction and IP67 has a higher protection against moisture.&lt;/p&gt;&#xA;" OwnerUserId="9737" LastActivityDate="2017-01-26T12:53:23.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13456" PostTypeId="1" CreationDate="2017-01-26T15:34:21.787" Score="3" ViewCount="55" Body="&lt;p&gt;I am looking for a solution to monitor and log ambient He levels down to 50-100ppm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The detector is meant to be installed in an experimental hall where a large number of Photo Multiplier Tubes are in use. PMTs ageing speed increases with the He concentration. We consider 100-50ppm a safe limit for the PMTs and we want to make sure that the ambient level of He does not exceed that threshold.&#xA;Unfortunately there is a (normally sealed) device in the same area that could leak Helium so we need to actively monitor the He level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution I could think of is to look for a Thermal Conductivity Detector, such as those used in handheld gas sniffers. But a TCD will give only a reading related to the overall difference between sample and reference. Can you suggest any other solutions?&lt;/p&gt;&#xA;" OwnerUserId="9740" LastEditorUserId="9" LastEditDate="2017-01-27T20:18:26.473" LastActivityDate="2017-01-31T14:24:56.310" Title="Ambient helium level monitoring" Tags="|measurements|gas|instrumentation|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13457" PostTypeId="1" AcceptedAnswerId="13460" CreationDate="2017-01-26T17:18:19.023" Score="5" ViewCount="5419" Body="&lt;p&gt;When talking about roller supports in a statics problem, we say that the roller prevents only the vertical translation of the beam, but not the rotation or the horizontal translation of it. So there is only one vertical reaction applied on the beam by the roller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we know a beam is connected to the roller using a pin, and the reaction force is the result of the pin that acts on the beam, as shown in this picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9od1C.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9od1C.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, suppose that I have a beam that is connected to a roller and I am going to push the beam in the horizontal direction (rollers don't prevent translation in the horizontal direction). What reactions should I expect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the pin is going to apply a force in the horizontal direction (due to the contact between the pin and the beam), which means that there is a reaction force in the horizontal direction. Is that true? If it is true, then why don't we consider horizontal reactions when analyzing rollers?&lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="368" LastEditDate="2017-01-26T20:10:43.783" LastActivityDate="2022-10-26T12:19:20.033" Title="Why do we ignore horizontal reactions at roller supports?" Tags="|civil-engineering|statics|" AnswerCount="4" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13458" PostTypeId="1" AcceptedAnswerId="13462" CreationDate="2017-01-26T18:39:42.313" Score="6" ViewCount="1664" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FVm5U.png&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FVm5U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for yield point you can say that it is a point after which the material will have plastic (permanent) deformation. I'm wondering if there is any similar meaning to the tensile strength? And is there any importance to it?&#xA;The thing is, whenever we calculate tension of anything (in college), the tension in the material must always be lower than it's yield strength. So it makes me wonder why do we care about ultimate tensile strength?&lt;/p&gt;&#xA;" OwnerUserId="8851" LastEditorUserId="1832" LastEditDate="2017-01-26T19:28:37.810" LastActivityDate="2017-01-26T23:06:50.163" Title="Does the ultimate tensile strength of a material have a physical meaning?" Tags="|mechanical-engineering|materials|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13459" PostTypeId="2" ParentId="13458" CreationDate="2017-01-26T19:53:09.643" Score="2" Body="&lt;p&gt;Ultimate strength is the stress at which the member will truly collapse. Between the yield and ultimate strengths, the member will suffer plastic deformations. After the ultimate strength, the member will be no more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, well, we don't always design to the yield strength. There are cases where plastic deformation is accepted and taken into consideration, at which point the ultimate strength is certainly used, since it's the stress at which the member will truly collapse. Why we usually design to the yield instead of ultimate strength is another question entirely.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-01-26T19:53:09.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13460" PostTypeId="2" ParentId="13457" CreationDate="2017-01-26T19:58:35.157" Score="6" Body="&lt;p&gt;If a bar is only supported by roller supports (which allow for horizontal displacements) and a horizontal force is applied, the bar will not suffer any internal forces because the entire structure (bar and supports) will suffer a rigid body motion sliding along the ground. At most, there'll be a slight force due to the friction of the rollers on the ground, but that's usually assumed to be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you get into a dynamic system under a constant force (and therefore constant acceleration), then sure, you might get some internal inertial forces, but I don't think that's what you're asking about.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-01-26T19:58:35.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13461" PostTypeId="2" ParentId="13457" CreationDate="2017-01-26T20:04:35.450" Score="1" Body="&lt;p&gt;If you have a horizontal beam with a roller on the bottom and you give it a push in the horizontal direction, it will accelerate in that direction. That is no longer a statics problem and you can't analyze it using a static model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are told, &quot;here is a statics problem,&quot; then what that means is that you should &lt;strong&gt;assume&lt;/strong&gt; that the sum of forces and moments are zero in all directions. Can you make this assumption? Sure, we usually hope that the beams in our lives are not accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might ask, &quot;What if I have already pushed the beam and it's moving at a constant rate? Then it could still be a statics problem because there is no acceleration (sum of forces and moments can be zero).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if you stopped pushing it and it's moving at a constant rate, then the net horizontal force is zero. The only way that is possible is if there is no friction in the roller, which brings us back to the static model with zero horizontal reaction force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you add a horizontal reaction (friction force, in this scenario), then your beam has to accelerate opposite the direction of motion, and you're back to having a dynamics problem (not statics)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, unless you keep pushing it, to balance out the friction. Now you're chasing this beam, pushing it down the road for who knows what reason, and in this case we can probably still use the principles of statics to model this problem but we probably &lt;em&gt;wouldn't&lt;/em&gt; model it with a roller support when there is friction. Why? Because it would be even &lt;em&gt;simpler&lt;/em&gt; to model it as a beam sliding down a flat surface, and that would give us the same result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We care less about whether the actual beam has wheels or bearings or anything that rolls, than we care about choosing the simplest model that will allow us to make a correct decision.&lt;/p&gt;&#xA;" OwnerUserId="368" LastActivityDate="2017-01-26T20:04:35.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13462" PostTypeId="2" ParentId="13458" CreationDate="2017-01-26T23:01:00.180" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, for yield point you can say that it is a point after which the material will have plastic (permanent) deformation. I'm wondering if there is any similar meaning to the tensile strength?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes: &lt;strong&gt;tensile strength&lt;/strong&gt;, or &lt;strong&gt;ultimate tensile strength&lt;/strong&gt; (UTS), is the point at which, if the load is not reduced, the material ruptures or fractures and thereby becomes discontinuous. Rupture is commonly viewed as a form of material failure. Note that for brittle materials rupture occurs at the UTS. For ductile materials the UTS often does not coincide with rupture because the material is able to change shape to accommodate the strain. The shape change, or plastic deformation, is limited because the volume of the material is constant, hence why necking occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the yield stress, metals work harden, which is why required engineering stress increases with increasing strain. While work hardening is occurring the material is also necking, up to the UTS. With strain increasing beyond the UTS, voids and cracks begin to form within the material's microstructure and start to coalesce into successively larger voids, until they bridge and the material fails. Voids and cracks can not support loads, so the effective cross-sectional area of the material decreases, hence the decreasing engineering stress with increasing strain after the UTS. Once the voids have completely bridged the cross-section of the material, rupture has occurred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a sense the UTS can be viewed as a &quot;point-of-no-return&quot; for constant-stress applications. Once the UTS has been reached the material will inevitably rupture if the load is not sufficiently reduced in time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And is there any importance to it? The thing is, whenever we calculate tension of anything (in college), the tension in the material must always be lower than it's yield strength. So it makes me wonder why do we care about ultimate tensile strength?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Absolutely, it is quite important for shaping metals. As an example, consider the process of wire drawing. Wire drawing is the process of reducing a wire stock from a larger diameter to a smaller diameter through mechanical force. Wire drawing is a continuous process involving very large spools of material moving very quickly through dies, to produce many miles of wire per hour. Without going into mathematical details, the wire-drawing process requires precise control of stresses to achieve optimal plastic deformation. If the UTS is inadvertently reached anywhere in the wire, the wire will rupture and stop the entire continuous process. Most likely the spool will be ruined, because the wire can not be rejoined, and customers expect a certain continuous length when they make a purchase. Restarting the wire drawing process can be a time-consuming waste of materials and thus costly. Therefore, wire manufacturers rely on knowledge of the UTS of the metal they are working to avoid costly mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason most introductory materials and mechanics courses only involve the elastic regime, below the yield point, is because the calculations are less complicated. Plastic deformation requires a greater understanding of the mathematics behind material flow, as well as incorporation of strain hardening and other fairly involved phenomena. That said, plastic deformation and the UTS are of utmost importance for shaping of ductile materials in industrial processes.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2017-01-26T23:06:50.163" LastActivityDate="2017-01-26T23:06:50.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13463" PostTypeId="1" CreationDate="2017-01-26T23:08:20.023" Score="6" ViewCount="419" Body="&lt;p&gt;I have a staggered 2x4 wall on a 2x6 sill plate as illustrated below (16&amp;quot; on center on each side = 8&amp;quot; on center each stud). The wall is loaded from both sides by floor joists, so I would consider it to be loaded in the center of the axis of the &amp;quot;5.5&amp;quot; inch wall. My problem is that this causes an eccentricity of 1 inch with regards to the 2x4 studs centre line. Should I consider this to be equivalent to a 2x4 wall with studs @ 8&amp;quot; on center with a 1&amp;quot; eccentricity? Can the studs buckle in opposite directions?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y05FV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y05FV.png&quot; alt=&quot;Staggered 2x4 wall&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5691" LastEditorUserId="16486" LastEditDate="2022-11-01T07:48:36.493" LastActivityDate="2024-03-25T10:08:36.143" Title="Eccentricity for staggered 2x4 wall" Tags="|structural-engineering|wood|buckling|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="13466" PostTypeId="1" CreationDate="2017-01-27T02:46:32.047" Score="3" ViewCount="25333" Body="&lt;p&gt;In this question , I dont understand why the KBC is 4EI / L .... I was told that for the far end pinned or roller supported , the K = 3EI / L , so shouldn't the KBC = 3EI / L &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 3rd photo , we can see that for far end pinned supported , k = 3EI / L , not 4EI / L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the difference between the example in photo 2 and photo 3 ? &#xA;&lt;a href=&quot;https://i.stack.imgur.com/P861M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P861M.png&quot; alt=&quot;gg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vRd8Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vRd8Z.png&quot; alt=&quot;ggf&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bmSLr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bmSLr.png&quot; alt=&quot;efsdsfd&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="8664" LastEditDate="2017-02-11T06:49:20.520" LastActivityDate="2020-03-16T09:35:00.377" Title="stiffness factor for member in beam" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13467" PostTypeId="2" ParentId="13430" CreationDate="2017-01-27T09:15:03.670" Score="0" Body="&lt;p&gt;Note that the mean water depth around the vessel is reduced&#xA;when the depth-based Froude number is close to 1.&#xA;It is not just that the vessel sinks and trims, which would imply&#xA;that the wetted surface area would increase significantly.&#xA;Finite width effects further complicate the situation.&lt;/p&gt;&#xA;" OwnerUserId="1908" LastActivityDate="2017-01-27T09:15:03.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13469" PostTypeId="1" CreationDate="2017-01-27T12:27:22.450" Score="4" ViewCount="6090" Body="&lt;p&gt;In the picture shown : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/USqxz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/USqxz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that there is a man applying a force $F$ in order to support the box using a cord that wraps over a frictionless pulley (the system is in equilibrium and nothing is moving)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now , if we want to draw the free body diagram for the pulley, it will look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uOLZn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uOLZn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that $T_1=T_2$ because the pulley is frictionless, and we can check that this true using summation of moments about point A = zero (assume counterclockwise positive)&#xA;$$\begin{align}&#xA;T_2r-T_1r&amp;amp;=0 \\&#xA;T_2-T_1&amp;amp;=0 \\&#xA;\therefore T_2&amp;amp;=T_1&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if the pulley is not frictionless (there is friction between the pulley and the cord), then the man must apply a greater force in order to support the box (because you are supporting the box and some of your force is dissipated due to friction) (the system here is also in equilibrium). That means that $T_1$ is greater than $T_2$, but this situation will not satisfy the equilibrium equation (summation of moments about point A =zero) because $T_1 &amp;gt;T_2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is my mistake when analyzing both situations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a solved example in my statics book &#xA;&lt;a href=&quot;https://i.stack.imgur.com/dTKeL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dTKeL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="9742" LastEditDate="2017-01-27T15:28:11.490" LastActivityDate="2017-01-29T18:06:21.300" Title="How does a pulley (with friction) satisfy equilibrium?" Tags="|civil-engineering|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13470" PostTypeId="2" ParentId="13469" CreationDate="2017-01-27T12:41:41.037" Score="2" Body="&lt;p&gt;You're thinking about the friction wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friction opposes relative movement.  Why should the man have to pull harder if there's friction in the pulley?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With help from ratchet freak I've cleaned up the rest.  Friction is opposing motion.  If your applied force ($F_a$) is the same as the weight of your mass ($F_m = m g$) then there is no relative motion to try to oppose:$$T_1 = F_a = T_2 = F_m = mg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $F_a &amp;lt; F_m$ but $F_m - F_a \leq F_{f_{static}}$ then it can still balance as: $$T_1 = F_a = T_2 - F_{f_{static}} = mg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which means you actually need to apply less force if there is friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sketched a FBD to help illustrate how the moments and everything will balance.  Note that this is approximate as the friction would actually be spread out over the contact surface of the rope.  Also note that I don't think you would usually worry about friction in the rope-pulley interface.  I believe the bearing in the pulley would usually resist since ideally the pulley is actually rotating with the rope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KzIbV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KzIbV.jpg&quot; alt=&quot;Free Body Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7736" LastEditorUserId="7736" LastEditDate="2017-01-27T15:00:43.337" LastActivityDate="2017-01-27T15:00:43.337" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13471" PostTypeId="1" AcceptedAnswerId="13475" CreationDate="2017-01-27T12:44:25.847" Score="2" ViewCount="3173" Body="&lt;p&gt;I am trying to source the parts for a &lt;a href=&quot;http://reprap.org/wiki/Category:Wilson_II&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wilson II 3D printer&lt;/a&gt;, and the recommended struts are made from 2020 aluminium extrusions, like so:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cKLj5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cKLj5.png&quot; alt=&quot;2020 Aluminium extrusion&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, in Bangkok, I can only find this type:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y7mat.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y7mat.png&quot; alt=&quot;2020 Aluminium extrusion in Bangkok&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Note that the profile/shape of the slot is markedly different from the image above.&lt;/p&gt;&#xA;&lt;p&gt;Whilst looking about on the web, I have also found other types, like these:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KfVZg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KfVZg.png&quot; alt=&quot;Various types of 2020 Aluminium extrusion&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Again, the profile of the slot varies considerably.&lt;/p&gt;&#xA;&lt;p&gt;My question is whether standard full size&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gPhHI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gPhHI.png&quot; alt=&quot;Full T-nut&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;and half size T nuts&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2fMA0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2fMA0.png&quot; alt=&quot;Half T-nut&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;will still fit all of these different types? Or are there different T-nuts for the different types of extruded 2020 aluminium, depending on the profile/shape of the slot.&lt;/p&gt;&#xA;&lt;p&gt;Also, why are there the different designs, does each type have an advantage for particular applications (with respect to strength, weight, etc.)?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h3&gt;Update&lt;/h3&gt;&#xA;&lt;p&gt;For completeness and accuracy, the 2020 extrusion, that I obtained in Bangkok, actually has this profile:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MoVBE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MoVBE.jpg&quot; alt=&quot;2020 Aluminium Extrusion BKK#1&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;and here is a more in-focus photo, of four, in a 2 x 2 configuration:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vzce3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vzce3.jpg&quot; alt=&quot;2020 Aluminium Extrusion BKK#2&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;As ericnutsch states, in his answer, only regular hexagonal M5 nuts will fit.&lt;/p&gt;&#xA;" OwnerUserId="5899" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-06-01T03:17:00.227" Title="Will standard M5 T-slot nuts fit these different 2020 aluminium extrusions?" Tags="|aluminum|fasteners|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13472" PostTypeId="2" ParentId="13446" CreationDate="2017-01-27T16:12:58.090" Score="3" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use other polymers.  Polyesters, vinylesters, and epoxies are all common in fibre-polymer composites and could be used.  You can also use other special stuff if you need, eg, fire performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Epoxies are common because if you're using carbon fibres (rather than a variety of glass) you are evidently looking for a high performance component.  Of the commonly used polymers, epoxy is the family at the high performance end of the range.  Epoxies typically out-perform other resins both in mechanical properties terms and in environmental degradation.&lt;/p&gt;&#xA;" OwnerUserId="902" LastActivityDate="2017-01-27T16:12:58.090" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13473" PostTypeId="1" CreationDate="2017-01-27T18:28:10.407" Score="5" ViewCount="1124" Body="&lt;p&gt;In my propulsion notes I always see mention of the importance of minimizing stagnation pressure loss across the inlet shock of a ramjet/turbojet/turbofan. Could anyone explain further why this is the case, possibly with equations to explain certain points in more detail? Thanks very much.&lt;/p&gt;&#xA;" OwnerUserId="9755" LastEditorUserId="9755" LastEditDate="2017-01-29T01:46:14.913" LastActivityDate="2019-01-04T01:47:41.477" Title="Why is high stagnation pressure important for propulsion?" Tags="|pressure|propulsion|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13474" PostTypeId="2" ParentId="13463" CreationDate="2017-01-27T20:22:02.580" Score="0" Body="&lt;p&gt;I am not an SE, but I would think that with a significantly ridging top and bottom plate, your wall would be equivalently strong as a non-eccentric 2x4 wall with equivalent spacing. Calculating the whole wall with 1&quot; eccentricity like you mentioned would ensure some safety factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eccentricity occurs when you have a moment. If your top wall plate was thin plywood you would have to certainly figure eccentricity for each. Since your top plate has 1 or 2 laminated ridged 2x6 boards it will transfer majority of the moment between left and right members. To know how much you will probably have to figure in the geometry of your floor joists. It would be a good fea project, but at the end of the day you are likely going to have to follow code anyways so I would contact someone with code experience in your area and ask their thoughts. Looks like you are building a standard sound isolation wall; its likely someone may have already calculated and published an equivalent loading factor for this.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-01-27T20:22:02.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13475" PostTypeId="2" ParentId="13471" CreationDate="2017-01-27T20:32:22.320" Score="1" Body="&lt;p&gt;I don't think that any profile has any major advantage over any other. I use &lt;a href=&quot;http://openbuildspartstore.com/v-slot-linear-rail/&quot; rel=&quot;nofollow noreferrer&quot;&gt;openbuilds v-slot&lt;/a&gt; on a lot of my projects (its 2020 designed for perpendicular v rollers) and all the &lt;a href=&quot;https://www.aliexpress.com/item/100pcs-lot-M5-Nickel-Plated-Carbon-Steel-Hammer-Nut-Aluminum-Connector-T-Fastener-Sliding-Nut-For/32317449512.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;aliexpress t-nut hardware&lt;/a&gt; I have bought works fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Worst case, you can use regular nuts with the aluminum rail you have access to; they are not as convenient but will work fine.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-01-27T20:32:22.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13476" PostTypeId="1" CreationDate="2017-01-28T09:14:15.027" Score="0" ViewCount="193" Body="&lt;p&gt;I want to make a smaller version of a revolving stage with rotation and lift.&#xA;But how do you combine those mechanism that creates a turntable and that rotates and lifts up at the same time? I know that you can use a slip ring, but how do you put that together to make a lift? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there are lots of tutorials about the lazy susan but no lift... Does anyone has a simular tutorial on how to make this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you want to recycle an old printer which part could you use the best to make this rotating-lift up-turntable like a revolving stage?&lt;/p&gt;&#xA;" OwnerUserId="9643" LastActivityDate="2017-01-28T13:14:58.743" Title="How to make a lazy susan 'lift'?" Tags="|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13477" PostTypeId="1" AcceptedAnswerId="13566" CreationDate="2017-01-28T09:58:59.693" Score="1" ViewCount="42" Body="&lt;p&gt;I'm measuring thicknesses of ~15 $\mu$m nickel thin films electroplated on a silicon substrate. I only have access to a SEM for this task. I need to cut the samples to get a side view of the film. However, the nickel film is plastic and the silicon undergoes a brittle fracture. As a result the nickel film often detaches from the substrate near the fracture point before I have the chance to cut it, making imaging impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I cut through both the plastic thin film and the brittle silicon simultaneously? I'm getting access to liquid nitrogen in a week or so with the intention of freezing the sample. Is there a ~room temperature process which would allow me to prepare samples meanwhile? &lt;/p&gt;&#xA;" OwnerUserId="8424" LastActivityDate="2017-02-02T23:56:44.733" Title="Cutting silicon samples with plastic thin films" Tags="|materials|measurements|thin-films|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13478" PostTypeId="1" AcceptedAnswerId="13619" CreationDate="2017-01-28T11:49:04.247" Score="1" ViewCount="2067" Body="&lt;p&gt;I have an absolute value encoder with 14-bit resolution and want to implement a speed controller. For high speeds, the offset between the old and the new position value is large and so the accuracy of the estimation for velocity is high. If I target a low speed, then the accuracy of my estimation shrinks, since I have only a little difference between two steps of the encoder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see two solutions to increase the resolution. First, the time between two measurements can be increased, but then the speed controller begins to fail for very large speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trajectories for the position controller could be calculated to come up with a velocity controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think is the best solution for speed control to achieve a large dynamic of speed and how is this done normally for industrial applications?&lt;/p&gt;&#xA;" OwnerUserId="8784" LastActivityDate="2021-10-01T07:27:56.873" Title="Speed estimation with absolute encoder" Tags="|control-engineering|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13479" PostTypeId="2" ParentId="13476" CreationDate="2017-01-28T13:14:58.743" Score="2" Body="&lt;p&gt;In terms of simple lifting mechanisms there are a few good options. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://www.google.co.uk/search?q=scissor%20lift%20how%20it%20works&amp;amp;safe=off&amp;amp;rlz=1C1GIWA_enGB625GB625&amp;amp;espv=2&amp;amp;biw=1440&amp;amp;bih=770&amp;amp;tbm=isch&amp;amp;imgil=j8QJT7vf2ozm2M%253A%253ByZ3_C-JOaY28VM%253Bhttp%25253A%25252F%25252Fwww.globalspec.com%25252Flearnmore%25252Fmaterial_handling_packaging_equipment%25252Fmaterial_handling_equipment%25252Fscissor_lifts&amp;amp;source=iu&amp;amp;pf=m&amp;amp;fir=j8QJT7vf2ozm2M%253A%252CyZ3_C-JOaY28VM%252C_&amp;amp;usg=__o6xcmxa2T9vRs9m_FXGhP8T1Lxc%3D&amp;amp;ved=0ahUKEwj6q7Gr9OTRAhWmKMAKHZVtAmQQyjcIkQE&amp;amp;ei=rZeMWLqhI6bRgAaV24mgBg#safe=off&amp;amp;tbm=isch&amp;amp;q=scissor%20lift%20&quot; rel=&quot;nofollow noreferrer&quot;&gt;scissor lift&lt;/a&gt; type  mechanism is reasonably simple and can be powered b y a lead screw. Depending on the number of elements you have it can get quite a lot of lift for a fairly compact footprint and retracted volume. They also give quite a large mechanical advantage and so can be operated with a fairly small motor driving a lead-screw. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;another option is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rack_and_pinion&quot; rel=&quot;nofollow noreferrer&quot;&gt;rack and pinion&lt;/a&gt; type mechanism, also reasonably simple but it doesn't actually collapse as such so you need space underneath for the column to retract into. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-01-28T13:14:58.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13480" PostTypeId="1" AcceptedAnswerId="13482" CreationDate="2017-01-28T13:33:00.523" Score="2" ViewCount="82" Body="&lt;p&gt;Recently I found it essential to make my room soundproof and for this I have worked out the following way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The walls of my room are made of cement. I want to add two glass layers (of different thickness) inside them. To make it clear, think of my room for the time being a hollow cement cube. Now I want to add one hollow glass cube inside this cement cube (whose dimensions are intentionally kept less than the cement cube) and then do the same for the second glass cube (whose dimensions are also intentionally kept less than the first glass cube). Now what I want to do is to suck out the air between two consecutive layers by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Stopcock&quot; rel=&quot;nofollow noreferrer&quot;&gt;stopcock&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Will this procedure make my room soundproof? &lt;/li&gt;&#xA;&lt;li&gt;Will my room be durable?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user9761" LastEditorUserId="1832" LastEditDate="2017-01-28T17:18:05.473" LastActivityDate="2017-01-29T18:47:42.187" Title="Is my procedure applicable to make a room soundproof?" Tags="|civil-engineering|sound-isolation|" AnswerCount="1" CommentCount="4" ClosedDate="2017-01-29T20:03:10.200" ContentLicense="CC BY-SA 3.0" />
  <row Id="13482" PostTypeId="2" ParentId="13480" CreationDate="2017-01-28T17:16:02.207" Score="4" Body="&lt;p&gt;If I am understanding your question correctly you want to surround your room a vacuum so sound waves will not travel through your glass walls. In theory this would work very well but there are some major obstacles to overcome. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say your room is quite small, $12$ by $12$ by $8$. From the standpoint of a vacuum, earths atmosphere is at $14.7$ psi. $12$ feet times $12$ feet equals $152$ tons of force on your ceiling and floor due to atmospheric pressure. ($12\ \text{ft}\times 12\ \text{ft} = 20726\ \text{in}^2$, $20726\ \text{&quot;}^2 \times  14.7\ \text{lb/in}^2 = 304819\ \text{lb}$ or $152\ \text{tons}$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also at some point you are going to have to interface your inside box with your outside box for support and access. Any loud noises would vibrate the inside box which would vibrate your doorframe/supports, which would vibrate your outside box and the air as well. While the noise would probably be minimal it would not be 100% soundproof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer to your questions &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Theoretically yes but I doubt you have the time or money.&lt;/li&gt;&#xA;&lt;li&gt;No it would not be durable without very thick glass or some kind of internal and external support.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you want to soundproof your room I would look into getting some foam padding specifically made to cancel sound waves. These will absorb sound waves and deter sound waves from traveling through your walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.soundproofcow.com/soundproofing-101/soundproof-a-room/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a website that will help you get an idea of how to do what you want based on the reason you want to sound proof your room.&lt;/p&gt;&#xA;" OwnerUserId="9656" LastEditorDisplayName="user9761" LastEditDate="2017-01-29T18:47:42.187" LastActivityDate="2017-01-29T18:47:42.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13484" PostTypeId="2" ParentId="13469" CreationDate="2017-01-28T19:32:04.837" Score="1" Body="&lt;p&gt;There are two possible sources of friction, not one - if that's not clear, you may get confused. So let's start from basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the cord can slip freely, or experience friction, sliding on the pulley. Second (and I know your question kind of said &quot;ignore this&quot;) the pulley can turn freely or experience friction sliding on the rod that's supporting it. We will treat the two sources of friction as if they acted like one source &lt;em&gt;as far as the cord is concerned&lt;/em&gt;, but it's important to notice it could exist and might need careful consideration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A third point would be the angular momentum/moment of inertia of the pulley itself, if the pulley were heavy and needed significant energy to begin to turn as the cord moves on it, but we will ignore this as well, and assume a lightweight pulley.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have drawing software here, but your answer goes like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic equation:  &lt;em&gt;Net Force = mass x acceleration. ($F=mA$)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Forces on the box&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;There are 2 forces acting on the box. A force due to gravity (call it $W$) &lt;strong&gt;downwards&lt;/strong&gt;, and tension in the string (call it $T$) &lt;strong&gt;upwards&lt;/strong&gt;. The box is in equilibrium so $W = T$. The force due to gravity acting on a mass $m$ is $mg$, so $W$ is easily worked out as $W = mg$. Because the box is in equilibrium, $T$, the tension in the cord, is the same as this in size, so $T = W = mg$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Forces acting on the cord/tension in the cord&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The cord (slightly simplifying as is usual for questions at this level) is also in equilibrium, so from the &lt;em&gt;cord's&lt;/em&gt; perspective, it experiences three forces which also balance out. On one end, the force of the box, on the other end the force due to the man pulling and in the middle, any static frictional force from the contact with the pulley (that exists when the cord isn't moving). There may be some, or none. But if there is a frictional force, it's going to resist movement of the cord, so it'll act the opposite way to whichever way the cord  would otherwise move.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Condition for equilibrium&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Suppose the pulley can exert a force on the cord due to friction, up to an amount of $N$ newtons. Then what will happen is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The man pulls with force $F$.  But the cord is in equilibrium. The net force from pulling and from the weight of the box is $F-W$, and because it's in equilibrium, this has to be &quot;small enough&quot;, between $+N$ and $-N$, otherwise friction can't provide enough force to balance it and it won't stay static in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So remembering that $W=mg$, the condition will be that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-N \leq F - m.g \leq N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding $mg$ to all terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mg - N \leq F \leq mg + N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and splitting this into separate conditions and rearranging:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F \geq mg - N$ &lt;strong&gt;and&lt;/strong&gt; $F \leq mg + N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't do more because in the question, the force needed by the man to maintain equilibrium depends on 2 things - the mass of the box, and the maximum force possible due to friction, and we don't have any information to work either of these out any further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what this says in plain English is that the force the man has to apply, needs to be &quot;close enough&quot; to $mg$, that friction can supply the rest of the balancing force needed for equilibrium. If friction provided no force ($N=0$) then you'd get $F =mg$ which is the exact solution for a frictionless pulley.&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2017-01-29T18:06:21.300" LastActivityDate="2017-01-29T18:06:21.300" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13485" PostTypeId="1" CreationDate="2017-01-28T21:16:42.550" Score="0" ViewCount="104" Body="&lt;p&gt;Is it possible to generate power from fruit juice, using a device with a simple circuit which charges a battery using juice from fresh oranges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so what would be the ratio between emf emitted and juice from orange?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the device used rechargeable cells charged by the orange juice, would there be enough force to cut oranges automatically with a small blade connected to a battery powered motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am hoping to collect and store energy collected from the juice in a rechargeable battery, would this be possible? of so would be energy stored be enough to power a small motor to operate a knife which would cut through the fruit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the device would cut oranges into quarters at the press of a button, it would use a collection tray beneath the cutting area which collects juice residue from the cutting process and convert this into electrical energy stored in a battery. &lt;/p&gt;&#xA;" OwnerUserId="6104" LastEditorUserId="8055" LastEditDate="2017-01-30T17:50:36.570" LastActivityDate="2017-01-30T17:50:36.570" Title="Generating power from fruit" Tags="|electrical-engineering|design|power|renewable-energy|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13486" PostTypeId="2" ParentId="13485" CreationDate="2017-01-28T22:13:38.367" Score="5" Body="&lt;p&gt;You seem to be asking about using fruit juice as the electrolyte of a battery.  Yes, that's possible, but you don't get a lot of power that way.  The power also doesn't just come from the fruit juice.  That's only one of the chemical reactants.  One or both electrodes are the other side of the chemical reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The EMF is a function of the battery chemistry, which means the chemical makeup of the elctrodes and the electrolyte.  No matter how much juice is used, the EMF will be the same for the same electrodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of your question confuses force and power, so there is no sensible way to answer it.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-01-28T22:13:38.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13487" PostTypeId="2" ParentId="13457" CreationDate="2017-01-28T22:15:40.053" Score="0" Body="&lt;p&gt;Step one should always be to define what your free body is and what question you are trying to answer. If you are looking at the bar pictured as a whole then you aren't concerned with how the roller is attached to the bar (pin). You could draw the free body diagram to include the pin, but everything is going to balance out and muck up your solution with unnecessary forces. Since the pin is internal to the system, no need to include it. It's simply a bar attached the the ground or support with a roller (no horizontal reaction or moment).&lt;/p&gt;&#xA;&lt;p&gt;If you do care about the pin, draw that free body diagram separately to understand the forces. Simplify!&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="9448" LastEditDate="2022-10-26T12:19:20.033" LastActivityDate="2022-10-26T12:19:20.033" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="13489" PostTypeId="1" AcceptedAnswerId="13624" CreationDate="2017-01-29T02:55:32.770" Score="0" ViewCount="49" Body="&lt;p&gt;consider following diff equations&#xA;$$&#xA;\dot{v} = \frac{f_x}{m} = \mu g\\&#xA;\dot{\omega} = \frac{T}{J} - \frac{Rf_x}{J}\\&#xA;\dot{R} = \eta&#xA;$$&#xA;where the input is $T$. There is a help equation for slip ratio. The unknown variation is $\dot{R}$.where $\eta$ is a state and variable that I declared. $f_x$, $J$, $m$ and $g$ are known. $R$ is known at the beginning but then it varies.&#xA;$$&#xA;\lambda = 1-(\frac{R\omega}{v})^n, \:\: n=\left\{-1,1\right\} $$&#xA;What I have done is to define some state variable:&#xA;$$&#xA;x = [x_1,x_2,x_3]^T = [\omega, v, R]^T \\&#xA;u = T\\&#xA;y = [x_1, x_2]$$&#xA;When I build the system I end up with a singular matrix. What kind of observation approach does work on such system to observe $$x_3$$. I really want to use a slide mode observer. Otherwise everything else works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appriate it alot if someone gave me a hint.&lt;/p&gt;&#xA;" OwnerUserId="8884" LastEditorUserId="8884" LastEditDate="2017-01-29T22:44:23.750" LastActivityDate="2017-02-06T08:14:19.267" Title="Nonlinear observer of an unknown variation state" Tags="|control-engineering|control-theory|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13490" PostTypeId="1" CreationDate="2017-01-29T11:31:56.573" Score="1" ViewCount="366" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xz1KR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xz1KR.jpg&quot; alt=&quot;Image one.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved it using the logic that the vertical link of the beam is basically there to make the beam act like a cantilever.&#xA;So I think it is a cantilever problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&#xA;$$EI\frac{d^2y}{dx^2}=-px=M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and further integrating it we get&#xA;$$y_{max}=\frac{PL^3}{3EI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the book people say that the answer is wrong and the answer it 4 times a this answer. Why?&lt;/p&gt;&#xA;" OwnerUserId="3856" LastActivityDate="2017-01-31T14:14:52.110" Title="What is the logic to solve this cantilever problem?" Tags="|mechanical-engineering|civil-engineering|structures|beam|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13491" PostTypeId="2" ParentId="7828" CreationDate="2017-01-29T12:27:15.610" Score="1" Body="&lt;p&gt;This is a simple isometric projection&#xA;The angle is, by the picture 30.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QUHcV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QUHcV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3856" LastActivityDate="2017-01-29T12:27:15.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13493" PostTypeId="2" ParentId="13434" CreationDate="2017-01-29T14:33:42.360" Score="0" Body="&lt;p&gt;A sketch would be helpful here, but you should have an addition linear bearing to stabilize the load and then you decouple any moment loads on the actuator. If space/design allows, you could use your actuator in tension instead of compression which would reduce/remove any buckling conditions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding &quot;sway&quot;, you should start with a free body diagram, and then apply some simple statics to get deflections and stresses. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-01-29T14:33:42.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13494" PostTypeId="1" CreationDate="2017-01-29T16:03:11.877" Score="4" ViewCount="210" Body="&lt;p&gt;I have a CO2 production for my aquariums (currently 2, but the solution should be extenable to - let's say - 5) based on fermentation of bio mass (CO2 is used as fertilizer there) which consist of a bottle, a connector, valves and diffusors and flippers which help dissolving the CO2 underwater. They're connected with 4/6mm tubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the efficiency of the fermentation is irregular the presure in the system varies. Extranous CO2 is released into the water, but can't be dissolved and thus is waisted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to avoid that by adding a storage which collects CO2 if the pressure in the system is above 2 bar and releases it when it's below. What components can I use for such a system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plan:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CO2 source --&amp;gt;--|-- valve 1 -- Aquarium 1 (diffusor, underwater)&#xA;                |&#xA;                |-- valve 2 -- Aquarium 2 (diffusor, underwater)&#xA;                |&#xA;                |-- ...&#xA;                |&#xA;                |-- storage&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Functional sketch of my idea for a storage:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                                                      |---------|&#xA;--&amp;gt;- backstop valve opening &amp;gt; 2 bar before storage ---- Storage ---- pressure regulator opening &amp;lt; 2 bar after the storage --&#xA;                                                      |---------|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The storage isn't necessary in the scenario since the CO2 is cheap and over-pressure could easily be released in the room (1 1.5 l PET bottle with water and sugar can't cause a hazard). It's more of an engineering exercise.&lt;/p&gt;&#xA;" OwnerUserId="9780" LastEditorUserId="9780" LastEditDate="2017-02-07T01:02:15.983" LastActivityDate="2019-06-07T01:57:58.403" Title="Which components to use for a CO2 gas storage?" Tags="|fluid-mechanics|valves|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13495" PostTypeId="1" CreationDate="2017-01-29T16:08:53.570" Score="3" ViewCount="302" Body="&lt;p&gt;Sawdust has to be dried while in storage waiting to be processed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time window for the drying is 12 to 36 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The storage room is 20m x 20m x7m (L x W x H).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sawdust is stored on 300m2 of the 400m2 available. It is placed on tent vented steel sheets at 0.5m above ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives a sawdust volume of 100m3 and an aproximate weight of 21000kg. Sawdust humidity 50%. I would like to take it as low as possible in the available time window. Let's say 30% would be acceptable. 20% would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What airflow has to be provided from under the vented steel sheets in terms of volume, temperature and humidity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking to use tubes to distribute the air under the steel sheet. The air in will be matched with roof fans taking the air out. I would like to have a rough idea so I know with what equipment to start.&lt;/p&gt;&#xA;" OwnerUserId="9694" LastActivityDate="2018-02-25T21:16:15.273" Title="How to calculate the required airflow to dry a batch of sawdust?" Tags="|airflow|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13497" PostTypeId="1" AcceptedAnswerId="13502" CreationDate="2017-01-29T21:51:30.877" Score="4" ViewCount="205" Body="&lt;p&gt;When you buy plastic sheets such as acrylics, polycarbonate sheets etc, they come with a paper/plastic protective film. Like in this image : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/15YQX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/15YQX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://qdjnd.en.alibaba.com/productshowimg/50172628-200203123/Protective_Film_for_Acrylic_Sheet_PMMA_Panel.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Protective Film for Acrylic Sheet PMMA Panel&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You notice that there is no adhesive (at least not something tacky) that is present on the sheets, and there is no residue when the sheet is peeled off, I was wondering if anyone can shed some light on how these protective films are applied ?&lt;/p&gt;&#xA;" OwnerUserId="9784" LastEditorUserId="8055" LastEditDate="2017-01-30T17:50:52.913" LastActivityDate="2022-01-03T10:37:47.337" Title="How are protective sheets attached to plastics?" Tags="|plastic|adhesive|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13498" PostTypeId="2" ParentId="13495" CreationDate="2017-01-30T06:37:06.637" Score="0" Body="&lt;p&gt;The Volume of the entire room per hour of airflow at 105 degrees will take  16 hours at 25% humidity.&lt;br&gt;&#xA;If you want it done in 8 hours youll have to double the volume of air to pass through the room in 1 hour, this will start to look like a vacuum.&lt;/p&gt;&#xA;" OwnerUserId="9785" LastActivityDate="2017-01-30T06:37:06.637" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13499" PostTypeId="1" AcceptedAnswerId="13501" CreationDate="2017-01-30T09:49:14.383" Score="1" ViewCount="292" Body="&lt;p&gt;I have here a BLDC motor that is controllable with FOC and I already implemented a position controller. The next step is to implement a velocity controller, but I actually don't see how to control the velocity at very low speeds. The speed is proportional to the reference torque producing current i_d for FOC and so for low speeds, there is only a very low current and so nearly no power. This immediately can be seen, that the motor becomes bumpy for low speeds. The implemented speed controller looks like this and works fine for speeds around 50rpm or higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i_q_ref = i_q_ref_old + K_P&lt;em&gt;err_speed + K_I&lt;/em&gt;sum_err_speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed measurement is pretty accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of other control structures can be used to achieve a velocity controller for low speeds with FOC?&lt;/p&gt;&#xA;" OwnerUserId="8784" LastEditorUserId="8055" LastEditDate="2017-01-30T16:52:16.810" LastActivityDate="2017-01-30T16:52:16.810" Title="Speed control of a BLDCM at low speed" Tags="|control-engineering|motors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13501" PostTypeId="2" ParentId="13499" CreationDate="2017-01-30T11:08:15.947" Score="2" Body="&lt;p&gt;If you already have a functional position controller, you can just integrate the target speed to get a target position (which is moving). This should work relatively well for low speeds, but probably not when the motor starts moving a significant fraction of a revolution per control loop iteration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think the underlying problem is probably either latency or inaccuracy in the speed measurement. I assume it comes from some kind of encoder? At low speeds, it can take a long time for the motor to move to the next encoder tick, thus the velocity change is not immediately visible to the controller. And if you don't have any smoothing on the velocity reading, it can render PID D-term quite ineffective; and if you add smoothing, it will increase the latency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to increase the speed of the feedback loop is to form some kind of a feed-forward system. Consider having a Kalman filter with state variables (velocity, inertia, external_torque). Then you can update that Kalman filter by both your velocity measurements when you get them, and more often by torque values that you drive to the motor.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2017-01-30T11:08:15.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13502" PostTypeId="2" ParentId="13497" CreationDate="2017-01-30T11:20:44.943" Score="8" Body="&lt;p&gt;The material is called &quot;Electrostatic Protective Film&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is manufactured by letting the plastic film cool in a strong electrostatic field: there are high-voltage electrodes on top and bottom of the plastic. This causes the polar molecules in the plastic to align so that one side of the film has a positive charge and the other side has a negative one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This charge will persist quite long, and will keep the film adhered to the plastic sheet by electrostatic forces. However, because the electrostatic force is quickly reduced by distance, it is important that no dust gets between the protective film and the base plastic during manufacturing.&lt;/p&gt;&#xA;" OwnerUserId="1290" LastActivityDate="2017-01-30T11:20:44.943" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13503" PostTypeId="1" AcceptedAnswerId="13507" CreationDate="2017-01-30T16:54:14.747" Score="-1" ViewCount="444" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fUPEo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fUPEo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Does the left red wheel turn in the same direction as the leftmost orange wheel?If so P and S go up? If not, Q and S go up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasoning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Leftmost orange cc&lt;/li&gt;&#xA;&lt;li&gt;Left red cc (by assumption)&lt;/li&gt;&#xA;&lt;li&gt;Leftleast orange clock --&gt; P goes up&lt;/li&gt;&#xA;&lt;li&gt;Gray cc&lt;/li&gt;&#xA;&lt;li&gt;Right least orange clock&lt;/li&gt;&#xA;&lt;li&gt;Right most orange clock (similar to assumption)&lt;/li&gt;&#xA;&lt;li&gt;Right red cc --&gt; s goes up&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If not&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Leftmost orange cc&lt;/li&gt;&#xA;&lt;li&gt;Left red clock&lt;/li&gt;&#xA;&lt;li&gt;Leftleast orange cc --&gt; Q goes up&lt;/li&gt;&#xA;&lt;li&gt;Gray clock&lt;/li&gt;&#xA;&lt;li&gt;Right least orange cc&lt;/li&gt;&#xA;&lt;li&gt;Right most orange clock (similar to assumption)&lt;/li&gt;&#xA;&lt;li&gt;Right red cc --&gt; s goes up&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9791" LastActivityDate="2017-01-31T00:20:24.883" Title="How do Cogs and belts work? / Cogs and belts in an ICAS exam" Tags="|machining|pulleys|turbomachinery|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13504" PostTypeId="2" ParentId="13244" CreationDate="2017-01-30T18:46:28.040" Score="0" Body="&lt;p&gt;Got the following info from Edmunds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These lenses can have comparable performance to visible spectrum&#xA;  lenses given the proper coatings and arrangements within the optical&#xA;  system. As far as I know, the greatest limiting factor for high&#xA;  precision IR imaging lenses is manufacturing costs and substrate&#xA;  durability. Many of the IR substrates cannot handle as wide a range of&#xA;  conditions, whether that be moisture/weather related or temperature&#xA;  related.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Now ... LWIR is such a wide range of wavelengths that a given imaging&#xA;  system cannot possibly work across all of it. If you compare to the&#xA;  visible spectrum, most systems are accounting for about 300nm of&#xA;  variation in optical properties. Some systems cannot really handle&#xA;  that, but are used in more specific situations at precise design&#xA;  wavelengths so it is not really a problem. The LWIR has a range of&#xA;  several microns which means that some regions of the LWIR transmit&#xA;  through materials in a very different fashion than others. If LWIR&#xA;  applications are more specified to a specific region that has a&#xA;  similar bandwidth to the visible spectrum, I think designing lenses&#xA;  with great performance would be quite feasible, albeit still very&#xA;  expensive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="77" LastActivityDate="2017-01-30T18:46:28.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13505" PostTypeId="1" AcceptedAnswerId="13527" CreationDate="2017-01-30T23:23:46.437" Score="0" ViewCount="69" Body="&lt;p&gt;I would like to calculate or find the electrical resistance (in ohms) of aluminium oxide at or past its melting point, for the purposes of using it as an electrolyte in a molten state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies for the broadness I just can't seem to find anything specific to this. Even if you don't have an answer due to the nature, any direction for a conclusion to this question would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="9794" LastEditorUserId="9" LastEditDate="2017-02-01T12:35:27.073" LastActivityDate="2017-02-01T12:46:58.727" Title="Electrical Resistance of molten aluminium oxide" Tags="|electrical-engineering|chemical-engineering|metallurgy|" AnswerCount="1" CommentCount="4" ClosedDate="2017-02-01T13:51:14.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="13506" PostTypeId="1" CreationDate="2017-01-30T23:57:09.990" Score="1" ViewCount="156" Body="&lt;p&gt;Assume the horizontal member can be treated as a beam and assuming the diagonal member may be treated as a rod, what is its axial force as a function of the applied loading $P$ and length parameters $L_i$. Assume that the connection between the beam and the rod may be treated as as socket joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/21JWK.png&quot; alt=&quot;Diagram of problem&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, I have defined the vertical, horizontal and moment reactions at the top and bottom built-in ends as $V_1, H_1, M_1, V_2, H_2,$ and $M_2$ respectively. I have also defined the axial force (assumed tensile) in the diagonal member as $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vertical equilibrium of the beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P + V_1 - Tsin(\phi) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Horizontal equilibrium of the beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_1 - T cos(\phi) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment equilibrium of the beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_1 + P(L_1 + L_2) - T sin(\phi) L_1 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vertical equilibrium of the rod:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T \sin{\phi} + V_2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Horizontal equilibrium of the rod:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T cos(\phi) + H_2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment equilibrium of the rod:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Global vertical equilibrium: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P + V_1 + V_2 =0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Global horizontal equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_1 + H_2 = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I attempt to solve the simultaneous equation, I get an indeterminate problem and can not solve for the value of $T$.&lt;/p&gt;&#xA;" OwnerUserId="9726" LastEditorUserId="9726" LastEditDate="2017-01-31T08:00:58.557" LastActivityDate="2017-02-01T09:49:18.897" Title="How would the axial force in a rod of a doubly-built in structure be determined?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13507" PostTypeId="2" ParentId="13503" CreationDate="2017-01-31T00:20:24.883" Score="1" Body="&lt;p&gt;When two wheels/pulleys/cogs are connected by a belt, when one of the wheels turns the other one turns in the same direction as the first wheel that turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When two cogs/gears are positioned so they are touching, when one cog is turned in one direction the other cog turns in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for your example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The upper left yellow cog turns counter clockwise (ccw)&lt;/li&gt;&#xA;&lt;li&gt;Because its connect to the red cog via a belt, the red cog also&#xA;rotates ccw and the middle yellow cog turns clockwise (cw) causing&#xA;weight P to rise.&lt;/li&gt;&#xA;&lt;li&gt;Because the middle yellow cog rotates cw the black cog rotates the&#xA;opposite direction, ccw. This causes the upper right yellow cog to&#xA;turn cw.&lt;/li&gt;&#xA;&lt;li&gt;The upper right yellow cog is connected to the lower right yellow&#xA;cog via a belt, so the lower right yellow cog also turns cw.&lt;/li&gt;&#xA;&lt;li&gt;The lower right yellow cog touches the lower right red cog which&#xA;then turns the opposite direction, ccw.&lt;/li&gt;&#xA;&lt;li&gt;This results in weight S being raised.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-01-31T00:20:24.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13511" PostTypeId="2" ParentId="13506" CreationDate="2017-01-31T09:31:09.080" Score="1" Body="&lt;p&gt;You are correct that the problem is indeterminate. The relative axial stiffness of the rod and bending stiffness of the beam are required. In order to solve this problem you need to find the stiffnesses and use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Direct_stiffness_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;stiffness method&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2017-02-01T09:49:18.897" LastActivityDate="2017-02-01T09:49:18.897" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13516" PostTypeId="1" AcceptedAnswerId="13518" CreationDate="2017-01-31T13:15:53.430" Score="2" ViewCount="5256" Body="&lt;p&gt;I am currently integrating a C-Code (written by a third person) in a Simulink model. In the C-Code various sensor signals are filtered, using a PT1 filter with the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y(t) = \dfrac{u(t) + Cy(t-1)}{C + 1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $y(t)$, $u(t)$ and $C$ are the output (the filtered signal), input and a factor applied for each signal individually, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the step function response, I can see a typical PT1-behaviour with a gain factor of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot figure out the derivation of this function from the general formulation (the differential equation) of a PT1, which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Ty'(t) + y(t) = Ku(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any of you guys help me out?&lt;/p&gt;&#xA;" OwnerUserId="9808" LastEditorUserId="1832" LastEditDate="2017-01-31T13:54:43.563" LastActivityDate="2017-01-31T14:04:55.613" Title="PT1 Filter without Derivative" Tags="|control-engineering|electronic-filters|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13517" PostTypeId="2" ParentId="13490" CreationDate="2017-01-31T13:22:50.317" Score="2" Body="&lt;p&gt;I think your 'book people' are wrong, but so are you.  The support conditions are not well defined, but if I've interpreted them correctly, I think the tip vertical displacement is (b) $-\dfrac{2PL^3}{3EI}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming that the support points are intended to be fixed in position, but not providing rotational restraint.  If the restraint at the corner was supposed to be a moment restraint, it would just be represented as a built-in support and there would be no need of the vertical leg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers all only have $EI$ terms, so it is evident that the author of the question only wants to consider flexural effects (not shear or axial deformations - we'd need terms with shear or axial area in them if those effects were included).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question asks about logic, so this is my thought sequence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start at the loaded tip. From there all the way to the corner it's just a cantilever, so we'll get cantilever behaviour. If it were a rigid support, this would give the classic result tip $\delta=\dfrac{PL^3}{3EI}$. This is the result the questioner derives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at the corner we have a moment ($PL$) that must be resisted. There's no restraint at the corner support, but restraint is provided by the vertical leg.  In isolation, this member is just a pin-ended beam with a moment applied at one end. That's another standard result - in that situation the beam end rotation (at the loaded end) is $\dfrac{ML}{3EI}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we know $M = PL$, so our corner rotation is $\dfrac{PL^2}{3EI}$. As far as the horizontal leg is concerned, that will a rigid body rotation superimposed on the bending deformation of the horizontal leg. A rigid rotation of $\dfrac{PL^2}{3EI}$ about the corner moves the tip downwards $\dfrac{PL^3}{3EI}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have $\dfrac{PL^3}{3EI}$ bending deformation in the horizontal leg, plus $\dfrac{PL^3}{3EI}$ due to rotation about the corner, hence a total tip deflection of $\dfrac{2PL^3}{3EI}$, which is answer (b).&lt;/p&gt;&#xA;" OwnerUserId="902" LastEditorUserId="1832" LastEditDate="2017-01-31T14:14:52.110" LastActivityDate="2017-01-31T14:14:52.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13518" PostTypeId="2" ParentId="13516" CreationDate="2017-01-31T14:04:55.613" Score="2" Body="&lt;p&gt;The equation being used in code is a discrete-time implementation of the filter. The second equation you gave is a continuous-time expression of the filter dynamics. In order to create a discrete-time filter from the continuous-time dynamics the person who wrote the code used a first order approximation (backwards difference, specifically) of the derivative of $y$, like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{dy}{dt} \approx \frac{y(t)-y(t-\Delta t)}{\Delta t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we use the notation $k$ to denote a sample taken at time $t$ and $k-1$ to denote a sample taken at time $t-\Delta t$, then we can derive the equation used in code as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T\frac{dy}{dt} + y(t) = Ku(t)$$&#xA;$$T \frac{y(k)-y(k-1)}{\Delta t} + y(k) = Ku(k)$$&#xA;$$\left(\frac{T}{\Delta t} + 1\right) y(k) = K u(k) + \frac{T}{\Delta t} y(k-1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let $K=1$ and $C = \frac{T}{\Delta t}$, and you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(C + 1\right) y(k) = u(k) + C y(k-1)$$&#xA;$$ y(k) = \frac{u(k) + C y(k-1)}{C + 1}$$&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2017-01-31T14:04:55.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13519" PostTypeId="2" ParentId="12624" CreationDate="2017-01-31T14:26:37.080" Score="1" Body="&lt;p&gt;With any measurement system, there are errors. Based on what details have been given there are plenty of potential error sources. A few that I can think of off the top of my head are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Physical differences in the sensors&lt;/li&gt;&#xA;&lt;li&gt;Errors in their mounting locations&lt;/li&gt;&#xA;&lt;li&gt;Calibration errors&lt;/li&gt;&#xA;&lt;li&gt;Differences between the sig. cond. and/or A/D hardware for each sensor&lt;/li&gt;&#xA;&lt;li&gt;Differences in the dimensions of the test specimens&lt;/li&gt;&#xA;&lt;li&gt;Irregularities in the test specimens&lt;/li&gt;&#xA;&lt;li&gt;All manner of variations in setup&lt;/li&gt;&#xA;&lt;li&gt;Operator dependent error&lt;/li&gt;&#xA;&lt;li&gt;Etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you're really interested in understanding  what you're measuring, my recommendation  is to take the time to do a real &lt;a href=&quot;http://www.six-sigma-material.com/MSA.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Measurement System Analysis&lt;/a&gt; (MSA) to understand the limitations of the measurements you're taking and what the sources of your error really are. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-01-31T14:26:37.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13520" PostTypeId="1" CreationDate="2017-01-31T14:52:48.550" Score="1" ViewCount="275" Body="&lt;p&gt;I'm working on a solution to this problem and I would like some advice and help. A fixed-ring epicyclic gear train is required to have an overall ratio of 5.5:1. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the sun gear is to have between 10 to 15 teeth, suggest the most appropriate numbers of teeth for each of the constituent gears.&lt;/li&gt;&#xA;&lt;li&gt;How many planet wheels would you propose for this design?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9809" LastEditorUserId="1832" LastEditDate="2017-01-31T16:07:07.573" LastActivityDate="2017-01-31T17:55:31.737" Title="Epicyclic Gear Train" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="2" ClosedDate="2017-02-01T13:54:15.607" ContentLicense="CC BY-SA 3.0" />
  <row Id="13522" PostTypeId="2" ParentId="13520" CreationDate="2017-01-31T17:55:31.737" Score="2" Body="&lt;p&gt;Start with 3 planets. Calculate the contact stresses and bearing loads. If too high then try 4 planets and repeat. Each individual planet will go down with more planets as you are sharing the load between them. Keep adding planets until you can either take the loads or you run out of room to add more. At that point, if you still can't take the load you probably need to up size the entire gearbox. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-01-31T17:55:31.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13523" PostTypeId="2" ParentId="13444" CreationDate="2017-01-31T18:38:30.443" Score="1" Body="&lt;p&gt;The purpose of the conjugate beam method is to find deflection and slope of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method uses equilibrium equations in conjugate beam, instead of using moment area diagram in real beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you must know the support of both real and conjugate beams, and then you  &lt;a href=&quot;http://www.mathalino.com/sites/default/files/reviewer-strength/06-beam-deflections/conjugate_support_real_support.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;transform&lt;/a&gt; the support of the real beam to the support of the conjugate beam, and draw it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, you plot the M/EI of &quot;real beam&quot; on the conjugate beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you take the equilibrium equations and determine the reactions at the conjugate beams. The reactions of the conjugate beam, V and M, will equal to θ and Δ, respectively, for the real beam.&lt;/p&gt;&#xA;" OwnerUserId="9810" LastEditorUserId="9810" LastEditDate="2017-02-01T05:56:52.857" LastActivityDate="2017-02-01T05:56:52.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13526" PostTypeId="1" AcceptedAnswerId="13630" CreationDate="2017-02-01T08:50:01.280" Score="5" ViewCount="2193" Body="&lt;p&gt;I want to generate NACA 63824 airfoil coordinates to use them in my bachelor thesis. &#xA;I tried airfoil tools and java foil. But I can't generate 638xx series. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you recommend me a program or a code that generates these airfoil coordinates? &lt;/p&gt;&#xA;" OwnerUserId="9817" LastActivityDate="2017-02-06T14:29:01.253" Title="How to generate NACA 5 digit airfoil coordinates?" Tags="|airfoils|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13527" PostTypeId="2" ParentId="13505" CreationDate="2017-02-01T11:22:32.687" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://www.researchgate.net/publication/231422619_The_Electrical_Conductivity_of_Liquid_Al2O3_Molten_Corundum_and_Ruby&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; you can find the resistivity of the molten $Al_2 O_3$ at $2400 {}^o C$: $384 ± 5\% \Omega m$. I've found this link with lesser than 10 seconds of googling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its melting point is only $2072 {}^o C$ (&lt;a href=&quot;http://de.wikipedia.org/wiki/Aluminiumoxid&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;). Its resistivity is probably smaller as the referred data. Checking this isn't very simple, because you can't easily extrapolate from the available experimental data (parameters such this tends to varies greatly around the phase shifts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your actual case, it depends also on its shape. For a correct calculation, you could make a numerical simulation depending on its temperature distribution. But, as a fist rule, you can relative easily estimate it by simplifying the calculation to a constant-temperature tube.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditDate="2017-02-01T12:46:58.727" LastActivityDate="2017-02-01T12:46:58.727" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13528" PostTypeId="1" AcceptedAnswerId="13532" CreationDate="2017-02-01T12:20:11.330" Score="5" ViewCount="201" Body="&lt;p&gt;Let's say I have a large isotropic body in uniaxial tension, onto which a small rectangular piece of film is applied. Think of a strain gage on a big body, but the 'strain gage' is one single continuous isotropic layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let x- and y-direction be on the surface of the big body and z-direction the normal (piercing trough the strain gage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;em&gt;big body&lt;/em&gt; I get $\sigma_x = \sigma_0; \sigma_y=\sigma_z=0$.&#xA;Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\epsilon_x &amp;amp;= \frac{\sigma_x}{E} \\&#xA;\epsilon_y = \epsilon_z &amp;amp;= \frac{\sigma_x\nu}{E}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the strain in the small body?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking was: Assuming I have an ideal boundary layer and negligible thickness of the film, the strain should be the same as in the big body. This is consistent with literature for $\epsilon_x$ and $\epsilon_y$.&#xA;But I keep reading an expression for $\epsilon_z$ in the &lt;em&gt;small body&lt;/em&gt; involving $\nu_B$ of the big body and $\nu_s$ of the small body:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon_{z,smallbody}=-\frac{\nu_s(1-\nu_B)}{1-\nu_s}\sigma_x/E$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain how to get this result?&lt;/p&gt;&#xA;" OwnerUserId="9819" LastEditorUserId="9819" LastEditDate="2017-02-01T16:36:19.397" LastActivityDate="2017-02-01T16:36:19.397" Title="Poisson ratio boundary condition" Tags="|mechanical-engineering|structural-engineering|applied-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13529" PostTypeId="1" AcceptedAnswerId="13534" CreationDate="2017-02-01T14:04:17.380" Score="4" ViewCount="39" Body="&lt;p&gt;Security valves with two connectors open if the pressure exceed a value x +/- tolerance on one side. I'd like to know if there's a mechanical valve which opens if the pressure on one side sinks below a certain value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use case is a system with a desired pressure d (e.g. 2 bar), but irregular pressure during operation (because it's based on bio mass fermentation) and a gas storage which stores extranous gas if the pressure is above 2 bar. This storage needs a value which can release pressure from the storage into the system if the pressure in the system sinks below the desired pressure.&lt;/p&gt;&#xA;" OwnerUserId="9780" LastActivityDate="2017-02-01T15:37:48.343" Title="Is there a (sort of inverse security) valve which opens if one connected side has a pressure below a certain value?" Tags="|fluid-mechanics|valves|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13531" PostTypeId="1" CreationDate="2017-02-01T14:32:23.037" Score="0" ViewCount="41" Body="&lt;p&gt;Why we need to consider only 3 cases when we want to find the max positive shear ? IMO , We know that the max shear will only occur when the 67.5kN act on B , am i right ? There is no need to consider 3 different cases .....Or my concept is incorrect ? If so , can anyone show me the example where the max shear will occur at B although the max concentrated load doesn't act on the particular point ? &lt;a href=&quot;https://i.stack.imgur.com/k2fHW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k2fHW.jpg&quot; alt=&quot;ggg&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0V5d7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0V5d7.jpg&quot; alt=&quot;gggsfsf&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastActivityDate="2017-02-01T16:00:19.917" Title="max shear act on point" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13532" PostTypeId="2" ParentId="13528" CreationDate="2017-02-01T14:33:07.850" Score="1" Body="&lt;p&gt;I would like to start by stating that my answer is not exactly the same as the one you've found in the literature. It is almost the same, but one sign is switched. I don't know if you wrote it wrong or if I messed up somewhere (&lt;strong&gt;very possible&lt;/strong&gt;, please correct me if you find my mistake). Regardless, I am so close that I'm sure the mistake is somewhere on this page (and not in the literature, of course). And it is close enough to show you where that equation comes from, which is your actual question.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is worth noting a key difference between the small and big bodies: the big body is in uniaxial tension, but the small one is under biaxial stress. It is under tension in the x axis but under compression in the y axis. Now, the deformation equations for biaxial stress are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\epsilon_x &amp;amp;= \dfrac{1}{E}(\sigma_x - \nu\sigma_y) \\&#xA;\epsilon_y &amp;amp;= \dfrac{1}{E}(-\nu\sigma_x + \sigma_y) \\&#xA;\epsilon_z &amp;amp;= \dfrac{-\nu}{E}(\sigma_x + \sigma_y) \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your small body is under known $\epsilon_x$ and $\epsilon_y$ (equal to those of the big body). From that we can calculate $\sigma_{x,s}$ and $\sigma_{y,s}$. (All variables with index $s$ are for the small body, $b$ for the big body)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\epsilon_{x,s} = \epsilon_{x,b} &amp;amp;= \dfrac{1}{E_s}(\sigma_{x,s} - \nu_s\sigma_{y,s}) \\&#xA;\dfrac{\sigma_{x,b}}{E_b} &amp;amp;= \dfrac{1}{E_s}(\sigma_{x,s} - \nu_s\sigma_{y,s}) \\&#xA;\therefore \sigma_{x,s} &amp;amp;= \dfrac{E_s\sigma_{x,b}}{E_b} + \nu_s\sigma_{y,s}\\&#xA;\epsilon_{y,s} = \epsilon_{y,b} &amp;amp;= \dfrac{1}{E_s}(-\nu_s\sigma_{x,s} + \sigma_{y,s}) \\&#xA;\dfrac{\nu_b\sigma_{x,b}}{E_b} &amp;amp;= \dfrac{1}{E_s}(-\nu_s\sigma_{x,s} + \sigma_{y,s}) \\&#xA;\therefore \sigma_{y,s} &amp;amp;= \dfrac{E_s\nu_b\sigma_{x,b}}{E_b} + \nu_s\sigma_{x,s} \\&#xA;\therefore \sigma_{x,s} &amp;amp;= \dfrac{E_s\sigma_{x,b}}{E_b} + \nu_s\left(\dfrac{E_s\nu_b\sigma_{x,b}}{E_b} + \nu_s\sigma_{x,s}\right) \\&#xA;\sigma_{x,s} &amp;amp;= \dfrac{E_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} \\&#xA;\therefore \sigma_{y,s} &amp;amp;= \dfrac{E_s\nu_b\sigma_{x,b}}{E_b} + \nu_s\left(\dfrac{E_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right) \\&#xA;\sigma_{y,s} &amp;amp;= \dfrac{E_s\sigma_{x,b}}{E_b}\left(\nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we just plug these in the $\epsilon_z$ equation. There's a lot to simplify here, so I'll show each of the steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\epsilon_z &amp;amp;= -\dfrac{\nu_s}{E_s}\left(\dfrac{E_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} + \dfrac{E_s\sigma_{x,b}}{E_b}\left(\nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right)\right) \\&#xA;\epsilon_z &amp;amp;= -\nu_s\left(\dfrac{\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} + \dfrac{\sigma_{x,b}}{E_b}\left(\nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right)\right) \\&#xA;\epsilon_z &amp;amp;= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} + \nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right) \\&#xA;\epsilon_z &amp;amp;= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}(1+\nu_s) + \nu_b\right) \\&#xA;\epsilon_z &amp;amp;= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s} + \nu_b\right) \\&#xA;\epsilon_z &amp;amp;= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s} + \dfrac{(1-\nu_s)\nu_b}{1-\nu_s}\right) \\&#xA;\epsilon_z &amp;amp;= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b + \nu_b-\nu_s\nu_b}{1-\nu_s}\right) \\&#xA;\epsilon_z &amp;amp;= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_b}{1-\nu_s} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is almost identical to the equation you've given from the literature, with the exception that my equation has $1 + \nu_b$, while the literature has $1 - \nu_b$. I can't figure out where I went wrong (or if you wrote it wrong), but this is so close that I believe it should demonstrate where the literature got this equation from.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-02-01T15:12:37.813" LastActivityDate="2017-02-01T15:12:37.813" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13534" PostTypeId="2" ParentId="13529" CreationDate="2017-02-01T15:37:48.343" Score="3" Body="&lt;p&gt;You're looking for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pressure_regulator&quot; rel=&quot;nofollow noreferrer&quot;&gt;pressure regulator&lt;/a&gt; (and this is indeed what you need in your use case, but that is another question.) The regulator works so that over a range of flows, the pressure on the outlet side is kept constant.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-02-01T15:37:48.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13535" PostTypeId="1" AcceptedAnswerId="13540" CreationDate="2017-02-01T15:59:09.647" Score="1" ViewCount="92" Body="&lt;p&gt;Here's the setup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QBIKv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QBIKv.jpg&quot; alt=&quot;Cantilever set-up&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem asks to find the relationship between the force $P$ and the displacement of B ($u_B$). I have done this correctly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \left(\dfrac{E_1A_1}{L_1}+\dfrac{E_2A_2}{L_2}\right)u_B$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine that members AB and BC both have yield stresses ($\sigma_{Y_1}$ and $\sigma_{Y_2}$) and are perfectly plastic after yield. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find the relationship between $u_B$ and $P$ in the following cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;No member yields - ($0 &amp;lt; u_B &amp;lt; \sigma_{Y_1}$) &amp;amp; ($0 &amp;lt; P &amp;lt; P_{y_1}$) &lt;/li&gt;&#xA;&lt;li&gt;Member AB yields, but not member BC - ($u_{Y_1} &amp;lt; u_B &amp;lt; u_{Y_2}$) &amp;amp; ($P_{y,1} &amp;lt; P &amp;lt; P_{y,2}$)&lt;/li&gt;&#xA;&lt;li&gt;Both members yield - ($u_B &amp;gt; u_{Y_2}$) &amp;amp; ($P &amp;gt; P_{y,2}$)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble finding explicit expressions for $P_{y_{1/2}}$ and $u_{Y_{1/2}}$. I'm not quite sure how to relate yield stress to elongation and displacement of the members. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest, this problem has tied my brain in a knot and I believe my conceptual understanding (or misunderstanding) of the problem is limiting my ability to solve it mathematically. Any help at all would be greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here is the exact problem&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6cUI4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6cUI4.png&quot; alt=&quot;Exact Problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9822" LastEditorUserId="1832" LastEditDate="2017-02-01T23:03:00.573" LastActivityDate="2017-02-01T23:03:00.573" Title="Deformation of a fixed-fixed beam with perfectly plastic yield stresses" Tags="|structural-engineering|statics|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13537" PostTypeId="2" ParentId="13531" CreationDate="2017-02-01T16:00:19.917" Score="3" Body="&lt;p&gt;Sure. Change your wheel loads to (from left to right): &lt;strong&gt;99kN, 99kN, 100kN, 1kN&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the first wheel has just reached B (Case 1), you have its 99kN at max influence, the second wheel's 99kN at a high positive influence, and the third 100kN wheel at a small positive influence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare to Case 3 when the third (and maximum) wheel reaches B: the 100kN wheel at max influence only provides slightly more positive effect than the 99kN wheel did in Case 1, but instead of having two big wheels with positive influence we instead have two big wheels with negative influence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's clear from inspection, without running the numbers, that Case 1 is more onerous. If you change the wheel loads a bit, you can make it less obvious, and need to run the numbers to confirm which is more onerous.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-02-01T16:00:19.917" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13538" PostTypeId="2" ParentId="12653" CreationDate="2017-02-01T16:12:31.173" Score="3" Body="&lt;p&gt;Revisiting the problem anew, it turns out that the units were interacting in a way I wasn't taking into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I asked this question, I had input $r_c$ and $r_0$ in cm. However, $\rho$ was in &lt;em&gt;kg/m&lt;sup&gt;3&lt;/sup&gt;&lt;/em&gt;, and $Q_0$ was in &lt;em&gt;g/s&lt;/em&gt;; additionally, my $Q_0$ was too large (an &lt;a href=&quot;https://en.wikipedia.org/wiki/Brake_specific_fuel_consumption#Examples_of_values_of_BSFC_for_shaft_engines&quot; rel=&quot;nofollow noreferrer&quot;&gt;Audi 3.3L V8 TDI&lt;/a&gt; achieves 0.014 g/s per injector).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely enough, putting $r_c$ and $r_0$ into meters, then putting $\rho$ in g/m&lt;sup&gt;3&lt;/sup&gt; and finally correcting $Q_0$ gave me $l$ = 2.9342×10&lt;sup&gt;-6&lt;/sup&gt;m &amp;#x2248; 2.93 &amp;mu;m, which is more reasonable given that $2\pi r_0 l$ = 9.2181×10&lt;sup&gt;-8&lt;/sup&gt; m&lt;sup&gt;2&lt;/sup&gt; &amp;#x2248; 9.22 &amp;mu;m&lt;sup&gt;2&lt;/sup&gt;, comparable to the injection area per injector for a &lt;a href=&quot;https://fr.wikipedia.org/wiki/HDi#Syst.C3.A8me_d.27injection&quot; rel=&quot;nofollow noreferrer&quot;&gt;PSA HDi DW10D&lt;/a&gt; with 8×110-micrometer nozzles of 7.6 &amp;mu;m&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's a lesson to be learned here, it's to double check not just your dimensions, but your units... especially the ones indirectly related to the result.&lt;/p&gt;&#xA;" OwnerUserId="6589" LastEditorUserId="6589" LastEditDate="2017-09-30T16:11:35.847" LastActivityDate="2017-09-30T16:11:35.847" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13539" PostTypeId="1" CreationDate="2017-02-01T17:28:21.040" Score="2" ViewCount="5224" Body="&lt;p&gt;I'm trying to design a two-stage &quot;arm&quot; which articulates from an attached base. The arm uses a pulley and cable at the &quot;elbow&quot; and a hydraulic cylinder mounted on the lower stage to lift the upper stage (similar to a boom truck).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given arm length $L$, pulley diameter $D$, and maximum weight $W$ at the end of the arm (obviously the arm itself has some weight too), how can I determine the maximum required hydraulic cylinder pull force ($F$) required to lift the arm from a horizontal orientation?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's a formula where I can plug in the values, I can experiment with different arm lengths, pulley circumferences, and weights to hopefully find a sweet spot in relation to available hydraulic cylinders and pulley sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a rough drawing. Thanks for any pointers, links, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kdj47.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kdj47.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I found &lt;a href=&quot;http://www2.mae.ufl.edu/designlab/motors/Useful%20Equations.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this document&lt;/a&gt; which clearly expresses the formulas used to calculate forces required to lift a weight on an arm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this document, I understand that $F = ma$, and $T = Fd$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So am I doing this right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume 225&amp;nbsp;kg weight to be lifted, and an arm length of 6&amp;nbsp;m, and a 30&amp;nbsp;cm pulley diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;F &amp;amp;= 225 \times 9.81 &amp;amp;&amp;amp;= 2207.25\text{ N} \\&#xA;T &amp;amp;= 2207.25 \times 6 &amp;amp;&amp;amp;= 13243.5\text{ Nm}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So pulley radius actually becomes the length of a lever, right?  So the same torque would need to be generated by applying a linear force to a 15cm lever?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = 13243.5\text{ Nm}$ and $d = 0.15\text{ m}$, so $N = 88290 N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this example I would need a cylinder that would be able to generate 88,290&amp;nbsp;N pull force (or roughly 20,000&amp;nbsp;lb pull), and a cable that could also withstand more than 20,000&amp;nbsp;lb strain?&lt;/p&gt;&#xA;" OwnerUserId="1558" LastEditorUserId="1832" LastEditDate="2017-02-02T12:32:15.320" LastActivityDate="2017-06-21T04:51:08.803" Title="Calculating the force and/or torque required to pivot an arm from one end" Tags="|mechanical-engineering|mathematics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13540" PostTypeId="2" ParentId="13535" CreationDate="2017-02-01T18:10:25.550" Score="1" Body="&lt;p&gt;From Hooke's Law, we know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon = \dfrac{\sigma}{E}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is valid up until the yield point. Considering a perfectly plastic material, the deformation $\epsilon$ is constant after the yield point, which gives us the following graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pU5DH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pU5DH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to answer your question, what is the value of $u_{Y_1}$? Well, it is equal to the displacement under the yield stress:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{Y_1} = \dfrac{\epsilon}{L} = \dfrac{\sigma_{Y_1}}{E_1L_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise for $P_{Y_1}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{Y_1} = \sigma_{Y_1}A_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the rest of the question, I think it is worth expanding your equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \dfrac{E_1A_1u_B}{L_1} + \dfrac{E_2A_2u_B}{L_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \sigma_1A_1 + \sigma_2A_2 = P_1 + P_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(where $\sigma_1$ and $\sigma_2$ are variables and proportional to $u_B$) which means that the force applied is equal to the sum of the forces applied by each member, which each are equal to the product of the stress and the area of the member. Keep that in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the question states that we can assume that AB will become plastic first. So we know that the situation will change when $u_B = u_1$. In this case, the equation above will change to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \sigma_{Y_1}A_1 + \sigma_2A_2 = P_{Y_1} + \dfrac{E_2A_2u_B}{L_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as of that point, member AB will no longer increase how much it resists deformation; it is no longer proportional to $u_B$, but is instead a constant equal to $P_{Y_1} = \sigma_{Y_1}$. Member BC, however, has yet to yield, and therefore remains proportional to $u_B$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the force increases, $u_B$ will eventually reach $u_2$, at which point BC's collaboration will also become a constant and equal to $P_{Y_2}$, so we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \sigma_{Y_1}A_1 + \sigma_{Y_2}A_2 = P_{Y_1} + P_{Y_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us that once both members cross into the plastic region, the deformation will tend to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there's your answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \begin{cases}&#xA;\begin{aligned}&#xA;&amp;amp;\left(\dfrac{E_1A_1}{L_1} + \dfrac{E_2A_2}{L_2}\right)u_B &amp;amp;&amp;amp;\text{ for }u_B \in [0, u_1] \\&#xA;&amp;amp;\sigma_{Y_1}A_1 + \dfrac{E_2A_2}{L_2}u_B &amp;amp;&amp;amp;\text{ for }u_B \in [u_1, u_2] \\&#xA;&amp;amp;\sigma_{Y_1}A_1 + \sigma_{Y_2}A_2 &amp;amp;&amp;amp;\text{ for }u_B \geq u_2&#xA;\end{aligned}&#xA;\end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives you the following behavior (the red line shows AB's contribution and the green, BC's; the black is the entire structure's resistance and is equal to the sum of both lines):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QlPdO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QlPdO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-02-01T18:10:25.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13541" PostTypeId="1" CreationDate="2017-02-01T19:26:51.663" Score="0" ViewCount="52" Body="&lt;p&gt;I'm working on a system with non-deterministic pressure (because it's based on bio mass fermentation) where CO2 is generated and input into the system at maximal pressure &lt;code&gt;x&lt;/code&gt; (e.g. 2 bar). If additional CO2 is produced it ought to be stored in a container with pressure &lt;code&gt;&amp;gt; x&lt;/code&gt; and released back into the system through a pressure regulator if the pressure in the system falls &lt;code&gt;&amp;lt; x&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that the input side of the container/storage has to be a backstop valve in order to keep the pressure in the storage. Based on technical terms (which I only know superficially) it has to be a security valve, because it has the function of releasing pressure &lt;code&gt;&amp;gt; x&lt;/code&gt;. Usually security valves do that into a space with less pressure (in my use case the pressure will be higher at the second occurance of over-production of CO2). Is there a backstop-security value which allows to redirect all pressure &lt;code&gt;y&lt;/code&gt; with &lt;code&gt;y &amp;gt; x&lt;/code&gt; into a container with pressure &lt;code&gt;&amp;gt; y&lt;/code&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The container/storage has to contain a (different) security valve which releases pressure into the room if it exceed the maximum of the container.&lt;/p&gt;&#xA;" OwnerUserId="9780" LastActivityDate="2017-04-03T11:28:10.823" Title="Can I increase pressure on the output side of a backstop/security valve?" Tags="|fluid-mechanics|valves|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13543" PostTypeId="1" AcceptedAnswerId="13544" CreationDate="2017-02-01T20:31:33.050" Score="2" ViewCount="1641" Body="&lt;p&gt;I can't find the name for this kind of hole. It allows the release of the plate by just loosening the bolt a bit, without having to remove it completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searches using &quot;teardrop&quot; or &quot;hole with oblong&quot; do not return the desired results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZfK5m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZfK5m.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6517" LastEditorUserId="368" LastEditDate="2017-02-02T16:51:19.190" LastActivityDate="2017-02-02T16:51:19.190" Title="What is the name for a bolt hole with a slot that allows the bolt to be inserted or removed when the nut is loose?" Tags="|bolting|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13544" PostTypeId="2" ParentId="13543" CreationDate="2017-02-01T21:21:32.823" Score="9" Body="&lt;p&gt;I would call that a 'key hole'.&lt;/p&gt;&#xA;" OwnerUserId="9827" LastActivityDate="2017-02-01T21:21:32.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13545" PostTypeId="1" AcceptedAnswerId="13548" CreationDate="2017-02-01T22:18:53.947" Score="2" ViewCount="519" Body="&lt;p&gt;Say I make a fiberglass/epoxy part on a plastic mold, and it cures at room temperature. The epoxy system calls for a postcure at 250F, but the mold cannot withstand 250F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I take off the part and postcure it separately, am I likely to see dimensional changes in the cured part? Is there any way to postcure the part off-the-mold without losing accuracy?&lt;/p&gt;&#xA;" OwnerUserId="9831" LastActivityDate="2017-08-09T23:32:53.290" Title="Can you postcure epoxy composites outside the mold?" Tags="|mechanical-engineering|materials|aerospace-engineering|composite|epoxy|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13547" PostTypeId="1" CreationDate="2017-02-02T00:23:53.807" Score="1" ViewCount="167" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zui07.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zui07.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is a screw press used to hmm disconnect the clamp connection (sorry if these are not the correct terms I really don't know how to translate it properly). Anyway, I think it's obvious what this is used for to anyone familiar with the matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to draw the detail &lt;strong&gt;A&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a big discussion among my friends on how it's supposed to be drawn. There are two options shown on the picture below, which one is correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/roKjX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/roKjX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, I used &lt;strong&gt;green color&lt;/strong&gt; for the screw and &lt;strong&gt;red color&lt;/strong&gt; for the nut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those who think &lt;strong&gt;case no 1&lt;/strong&gt; is correct, think that that screw teeth press upwards because of the reactive force that the clamp connection is giving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those who support the &lt;strong&gt;case no 2&lt;/strong&gt; think that since you are 'pushing' the inside of the clamp connection downward, then the teeth of the screw also push down on the nut teeth.&lt;/p&gt;&#xA;" OwnerUserId="8851" LastEditorUserId="8851" LastEditDate="2017-02-02T09:44:30.957" LastActivityDate="2017-04-03T09:00:58.950" Title="Where is the contact point between screw spindle teeth and nut teeth?" Tags="|mechanical-engineering|civil-engineering|pressure|experimental-physics|tools|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13548" PostTypeId="2" ParentId="13545" CreationDate="2017-02-02T03:20:01.247" Score="4" Body="&lt;p&gt;Yes it can be post cured - however, without support there is a risk of sagging or warping, particularly if it is a large and heavy part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermal curing is simply a way of speeding up the chemical hardening process, which in the case of typical epoxies may normally take place over several days, even weeks. The second benefit of heating the material is the removal of dissolved gasses in the plastic (outgassing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no need to cure at 250F - any temperature above ambient will hasten the curing process, so I would suggest soaking at say 125F for 12 hours instead of 250F for 4 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warping and distortions are always a possibility but in my experience if the temperature is managed and some support is supplied, (e.g. a sand bed) any distortions are imperceptible, at least in my applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very good reference for those interested in the process: &lt;a href=&quot;http://www.masterbond.com/interview/q-how-properly-cure-epoxy-aerospace-industry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Masterbond.com&#xA;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="7021" LastEditDate="2017-02-02T03:42:31.087" LastActivityDate="2017-02-02T03:42:31.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13549" PostTypeId="2" ParentId="13526" CreationDate="2017-02-02T03:30:34.577" Score="1" Body="&lt;p&gt;NASA Technical Memorandum 4741 describes the calculations for Series 6 and 6A foils on page 8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19970008124.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19970008124.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-02-02T03:30:34.577" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13551" PostTypeId="2" ParentId="13545" CreationDate="2017-02-02T09:30:06.863" Score="2" Body="&lt;p&gt;Curing conditions depend on the resin sytem. Some systems require a minimum temperature for a certain duration to cure, others cure completely at room temperature. If in doubt, you will need to ask the manufacturer of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also depends on the requirements you have for the final part. If the final properties of the part are somehow safety relevant, you should stick to the manufacturer's instructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the tolerable distortion one solution may be the sand bed proposed by Donald Gibson. Another possibility could be to cure at lower temperature in the mold first and post-cure in a sand bed at higher temperatures later.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2017-02-02T09:30:06.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13553" PostTypeId="2" ParentId="13541" CreationDate="2017-02-02T09:52:49.023" Score="1" Body="&lt;p&gt;Usually gases only flow from higher to lower pressure. Thus if the container shall contain CO2 at a greater pressure than the generator, you will need a pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it differently: A container that shall passively &quot;collect&quot; all CO2 above 2 bar needs to operate at a pressure below 2 bar. If it is operating below 2 bar, it cannot passively feed gas into a system operating at 2 bar. Therefore for your idea to work, you need a pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having read your other posts here (especially &lt;a href=&quot;https://engineering.stackexchange.com/questions/13494/which-components-to-use-for-a-co2-gas-storage&quot;&gt;the first one&lt;/a&gt;), your solution is to add a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pressure_regulator&quot; rel=&quot;nofollow noreferrer&quot;&gt;pressure regulator&lt;/a&gt; to the outlet of your bio mass fermenter as pointed out by @mart's answer to your &lt;a href=&quot;https://engineering.stackexchange.com/questions/13529/is-there-a-sort-of-inverse-security-valve-which-opens-if-one-connected-side-ha&quot;&gt;second post&lt;/a&gt;. You need a tolerable pressure difference between your generator and your consumers to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one problem is that the pressure in the fermenter rises above its maximum tolerable pressure, one possible solution may be to increase the volume of your fermenter. Pressure energy is usually stored in volume. However, there will be some loss, since you can only use the energy of the pressure difference between the maximum tolerable pressure of the fermenter and the minimum pressure of your output.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-02-02T10:26:03.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13554" PostTypeId="1" CreationDate="2017-02-02T11:22:51.770" Score="4" ViewCount="650" Body="&lt;p&gt;Drop (suspended) ceiling tiles are of course quite common, but I am in the process of trying to evaluate the long term safety of a specific kind of suspended tile made of fired clay. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/hMZLR.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hMZLR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When installed it looks like this&#xA;&lt;a href=&quot;https://i.stack.imgur.com/naMQF.jpg&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/naMQF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tile measures about 12 inches x 9 inches and weighs 2 kilograms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the important point is that &lt;strong&gt;the tile is supported only on two ends and therefore a single breadthwise fracture causes it to lose all support&lt;/strong&gt;. In such cases it has been observed that it may sometimes &quot;hang in&quot; for a few days before it falls out, but I don't think that is necessarily true. This type of tile has been widespread in its use in India for over a century now, and also in parts of Europe. Experienced engineers and architects of repute think this is as safe as any other type of suspended tile ceiling. From its design it seems that is not true. However its widespread usage seems to indicate that I am missing something. Does any structural engineer on the forum have an opinion or observation? Is this really as safe as any other suspended tile? If so what makes it safe despite the observations above? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="9837" LastActivityDate="2020-01-31T09:49:34.973" Title="Safety of suspended (drop) clay ceiling tiles" Tags="|structural-engineering|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13555" PostTypeId="1" AcceptedAnswerId="13556" CreationDate="2017-02-02T11:26:11.323" Score="0" ViewCount="90" Body="&lt;p&gt;For this problem , why we dont have to consider the max shear occur under the concentrated load just like the case in &lt;a href=&quot;https://engineering.stackexchange.com/questions/13531/max-shear-act-on-point/13537?noredirect=1#comment24328_13537&quot;&gt;max shear act on point&lt;/a&gt; ?&#xA;In the previous question , the author 'move the load' so that the the &lt;strong&gt;&lt;em&gt;center of the beam is between the FR (resultant force ) and the load&lt;/em&gt;&lt;/strong&gt; ... &#xA;Why in this case , we just need to consider 2 supports only ? Or the question is wrong ? &lt;a href=&quot;https://i.stack.imgur.com/Z05fd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z05fd.jpg&quot; alt=&quot;fds&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mdCBh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mdCBh.jpg&quot; alt=&quot;fdf&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8664" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-02-02T12:04:08.113" Title="absolute shear occur at support?" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13556" PostTypeId="2" ParentId="13555" CreationDate="2017-02-02T12:04:08.113" Score="3" Body="&lt;p&gt;This question is looking for the &lt;strong&gt;absolute&lt;/strong&gt; largest shear, the previous question was looking for the largest shear &lt;strong&gt;at midspan&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw the shear force diagram for a simply supported beam, subjected to loads in one direction only, you will notice that the largest shear is &lt;em&gt;always&lt;/em&gt; at one of the supports. Hence there was no need to consider anything other than the supports in this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a note: in my opinion the author of this question missed one possibility they should have considered: what happens if you move the load far enough to the left that the 5kN load drops off the beam, and move the first 20kN load to be directly over A? Answer: you get a shear at A of 60.5kN. Still not quite as big as found by the author at B, but bigger than what they found at A! &lt;em&gt;Of course, this only applies if its possible to move the loads off the beam. As I normally design bridges, I expect loads to be able to move off the end.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-02-02T12:04:08.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13558" PostTypeId="1" AcceptedAnswerId="13560" CreationDate="2017-02-02T16:40:51.350" Score="3" ViewCount="3410" Body="&lt;p&gt;Given an equation of motion of an undamped system $\mathbf{M}\mathbf{\ddot{q}} + \mathbf{K}\mathbf{q} = \mathbf{f}$, $\mathbf{M}$ indicates the mass-matrix, $\mathbf{K}$ the stiffness matrix, $\mathbf{q}$ the time-dependent displacement, and $\mathbf{f}$ the applied force. The roots, $\omega^2$, of the equation $det(\mathbf{K} - \omega^2\mathbf{M}) = 0$ indicate the vibration frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what is the physical interpretation of the &lt;strong&gt;eigenvalues&lt;/strong&gt; of the stiffness matrix $\mathbf{K}$?&lt;/p&gt;&#xA;" OwnerUserId="9839" LastEditorUserId="9839" LastEditDate="2017-02-08T09:37:10.397" LastActivityDate="2017-02-08T09:37:10.397" Title="What is the physical interpretation of the eigenvalues of the stiffness matrix in the finite element method?" Tags="|structural-engineering|finite-element-method|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13560" PostTypeId="2" ParentId="13558" CreationDate="2017-02-02T20:07:24.007" Score="9" Body="&lt;p&gt;Assume a 1d version of the equation. Then the K matrix becomes a &lt;em&gt;k&lt;/em&gt;, spring constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 1-D The equation reduces to : $m \ddot{q} + kq = f $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The q matrix also becomes x vector in one dimension. This is the differential equation for a forced mass-spring system. (@Jmac added the 1-d equation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, The physical meaning of the matrix eigenvalues is how stiff the system is in the corresponding eigen vector direction. And as such, this determines how much elastic position/movement the force causes in the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most interesting eigenvalues/vectors of K are the zero eigenvalues. These correspond to rigid body motions of the structure. These eigenvectors can be a useful check for structures that contain moving joints, etc, where the number of rigid body motions is not expected to be 6. They can also show problems with the FE model, for example &quot;hourglassing&quot; modes caused by inappropriate choice of elements. (Thanks to @alephzero for this paragraph)&lt;/p&gt;&#xA;" OwnerUserId="1097" LastEditorUserId="1097" LastEditDate="2017-02-03T17:11:47.917" LastActivityDate="2017-02-03T17:11:47.917" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13562" PostTypeId="2" ParentId="13547" CreationDate="2017-02-02T20:53:21.790" Score="1" Body="&lt;p&gt;Screws are effectively a spiral  wedge so when the screw is exerting a downward force case 1 is more representative. Case 2 would only be correct if there was an external force pushing on the back of the screw shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that it is not the convention to draw screw threads in this way. The gap and contact area between threads is defined by the thread standard and tolerances which are better indicated by notation rather than trying to draw them. In many cases it is simply not possible to draw tolerances to scale. Usually threads are &lt;a href=&quot;http://www.blinn.edu/brazos/matheng/jcoffelt/graphics/files/ege/fast/fast_page5.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;drawn using some sort of abbreviation&lt;/a&gt;, of which there are various levels of detail. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-02-02T20:53:21.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13563" PostTypeId="1" AcceptedAnswerId="13568" CreationDate="2017-02-02T23:16:31.107" Score="1" ViewCount="80" Body="&lt;p&gt;This is a noob question, please don`t be too harsh on me. I would like to use a motor for turning a mixer. The shaft of the mixer should be removable and I though of 3d printing a case which holds the motor and a gearbox in place. I made a very simple drawing of how it should look like (or how i can imagine it). I was wondering if such a thing already exist but I do not find it because I do not know the name. It should be easy to remove the shaft. Some sort if quick release/fix. The motor will run at 300 to 1000 rpm at low torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lKmq6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lKmq6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9847" LastActivityDate="2017-05-04T05:43:02.893" Title="some kind of gearbox for driving a removable shaft; is there a name for it?" Tags="|motors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13564" PostTypeId="2" ParentId="13494" CreationDate="2017-02-02T23:17:58.207" Score="0" Body="&lt;p&gt;I guess you want to keep fermentation process anaerobic. Just snap rubber/latex balloons on top of bottle, bleed of excess gas automatically by poking a needle hole on the balloon. The hole will release excess CO2 but could withstand water ingress.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastEditorUserId="4630" LastEditDate="2019-06-07T01:57:58.403" LastActivityDate="2019-06-07T01:57:58.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13566" PostTypeId="2" ParentId="13477" CreationDate="2017-02-02T23:56:44.733" Score="0" Body="&lt;p&gt;Liquid nitrogen performed poorly in this case. The change in temperature and tension popped the relatively thick film straight off the substrate!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best results were achieved with encasing the small square samples in epoxy. It allowed the samples to be cut with a diamond saw without even fracturing the silicon. Tool marks were apparent and the film suffered, but in this case it was enough to estimate the thickness of the film.&lt;/p&gt;&#xA;" OwnerUserId="8424" LastActivityDate="2017-02-02T23:56:44.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13568" PostTypeId="2" ParentId="13563" CreationDate="2017-02-03T02:46:49.163" Score="1" Body="&lt;p&gt;Most general purpose gearboxes will have a fairly short stub shaft which is connected to the main shaft by some sort of coupling. Exactly how this is implemented will depend on the application.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light duty couplings might just be a collar held on with grub screws or cotter pins heavier duty applications may use keys or splines. Splines are often preferred if the shaft needs to accommodate some axial movement and for very heavy duty applications such as vehicle transmission. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The version you have drawn has the added complication that a completely removable shaft needs to accommodate both the final drive gear and a set of bearings with a pretty good fit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution would be to have the final gear running on a hollow, internally splines shaft but this would be rather complex and expensive to manufacture. There is also the consideration of keeping the gearbox internals reasonably well sealed to keep the grease in and any dirt and grit out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice many mixers with removable blades are similar to electric drills, using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Chuck_(engineering)&quot; rel=&quot;nofollow noreferrer&quot;&gt;chuck&lt;/a&gt; or similar quick release mechanism to allow easy fitting and removal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic 3 jaw drill chuck will transmit moderate torque to a round shaft and somewhat higher torque to a hexagonal shaft and they have the advantage of being cheap and easily available. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-02-03T02:46:49.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13572" PostTypeId="1" AcceptedAnswerId="13784" CreationDate="2017-02-03T09:16:46.857" Score="3" ViewCount="217" Body="&lt;p&gt;Doing a fun project to reconnect with my engineering training: I would like to design a control scheme for hovering in-place at a set altitude with a vertically-oriented propeller, rocket, etc. Right now I'm not even worried about lateral motion, just the vertical component. My plan is to use PID control and tune with the Ziegler-Nichols method for &lt;a href=&quot;https://en.wikipedia.org/wiki/Ziegler%E2%80%93Nichols_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;no-overshoot&lt;/a&gt;. The system is so simple I can write down a model of the P-only controlled system rather than tune experimentally. Just using Newton's 2nd law and ignoring aerodynamic effects:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;m x^{\prime\prime} &amp;amp; = g m + T \\&#xA;x^{\prime\prime} &amp;amp; = g + \frac{K(x - x_{target})}{m}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $g$ is gravity, $m$ is mass, $x$ is height, and $T = K(x-x_{target})$ because my thrust is proportional to the error times the gain $K$. Using $sin(\omega t)$ as a trial solution I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\omega &amp;amp; = \textstyle\sqrt{\frac{K}{m}} \\&#xA;T_u &amp;amp; = 2\pi\textstyle\sqrt{\frac{m}{K}}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it seems like I get free choice of the ultimate gain $K$, since any choice of $K$ causes the system to oscillate, as far as I can tell. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="9850" LastActivityDate="2017-02-16T16:32:43.857" Title="Model-based Ziegler-Nichols tuning of a hover PID controller" Tags="|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13574" PostTypeId="1" AcceptedAnswerId="13580" CreationDate="2017-02-03T15:15:27.450" Score="2" ViewCount="2361" Body="&lt;p&gt;As is well known, a safety factor is often applied for engineering structures as a measure of precaution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to ask for which loads (if not all) does the safety factor apply for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, take an aircraft wing. The safety factor will need to be applied for the aerodynamic loads but does it need to be applied for any other loads such as the self-weight of the wing or the weight of the powerplant? &lt;/p&gt;&#xA;" OwnerUserId="9726" LastActivityDate="2017-02-05T12:19:07.530" Title="What loads does the safety factor apply for?" Tags="|structures|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13575" PostTypeId="1" AcceptedAnswerId="13598" CreationDate="2017-02-03T15:16:45.803" Score="3" ViewCount="1512" Body="&lt;p&gt;I am not an engineering student, but can understand if explained, so excuse me if this is too basic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a personal DIY project I envisioned, I need to make an object, attached to a set of gears, fall at a certain speed - 6ft / 24 hours. This object will rotate the first driving gear and the resultant driven gear at 3000-5000 RPM. I estimate the ratio to be 1:100000.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically my question is whether the speed at which the object falls (from gravity) be regulated? The weight of the object doesn't matter, since I can set it to whatever is required for my rate of descent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The setup will be similar to: &lt;a href=&quot;http://s.hswstatic.com/gif/clock-weight-train.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://s.hswstatic.com/gif/clock-weight-train.gif&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9856" LastEditorUserId="9856" LastEditDate="2017-02-04T13:10:11.087" LastActivityDate="2017-02-05T10:22:47.003" Title="Can the speed of a falling object be regulated using gears?" Tags="|mechanical-engineering|gears|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13576" PostTypeId="2" ParentId="12142" CreationDate="2017-02-03T15:19:00.023" Score="2" Body="&lt;p&gt;With any O-ring seal, static or dynamic, the key parameter is &quot;O-ring Squeeze&quot; which is defined as the percentage reduction in cross sectional area when compressed from the free state. Since the o-ring acts like a spring, this squeeze is directly related to the sealing force. The greater the squeeze, the greater the sealing force, and the greater resistance to leakage...at least in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with most things, more is not always better, too much squeeze can result in a poor joint since it can cause excess stress in the mating parts, excess friction (in the case of a dynamic seal), and O ring damage since there's a greater risk of the o-ring getting pinched between the two mating surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In designing O-ring seals in stainless steel with good surface finish, I would try to stick to &amp;lt; 15% squeeze for dynamic seals and 15%-30% for static seals using a standard NBR O-ring from Parker, but it really depends on the type of O-Ring, the materials involved and all of the other specifics of the application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot that goes into the design of a good O-ring joint. The best resource I've found is the &lt;a href=&quot;https://www.parker.com/literature/ORD%205700%20Parker_O-Ring_Handbook.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parker O-ring Handbook&lt;/a&gt; (link goes to the pdf). &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-02-03T15:19:00.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13577" PostTypeId="1" AcceptedAnswerId="13592" CreationDate="2017-02-03T15:38:20.683" Score="5" ViewCount="259" Body="&lt;p&gt;I always wondered how it is possible to have electromagnets so strong that they keep a heavy maglev-train levitating without forcing anything metallic to be flinged away or to be pressed against the ceiling of the vehicle-cabin?&#xA;And how come a person inside that train which has a pacemaker implanted is not affected? With a pacemaker someone cannot have a MRI scan but still ride a maglev train?&lt;/p&gt;&#xA;" OwnerUserId="9857" LastActivityDate="2017-02-14T13:49:41.707" Title="How is it possible electromagnets keep a heavy maglev train levitating without forcing anything metallic to be flinged away?" Tags="|electromagnetism|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13578" PostTypeId="1" CreationDate="2017-02-03T15:45:37.903" Score="2" ViewCount="60" Body="&lt;p&gt;I want to make a radar system which needs to send EM wave and receive the reflected wave from the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a ready to use RF signal processing hardware but can't be sure which is applicable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I found that, the required frequency range for us is approx. 700 - 1000 MHz. So, we need to process this frequency of signal. I talk to somebody who told be that we need ADC having double the frequency we use which is 2GHz or something. I am not sure what we need else and what should be looked for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to visualize beneath the ground with spatial image created from the signal read. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched for articles but couldn't find sufficient data about the hardware that we can use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it is ok to have some ready to use evaluation boards from IC companies if they fit. Such as &lt;a href=&quot;http://www.analog.com/en/products/analog-to-digital-converters/high-speed-ad-10msps/ad9625.html#product-evaluationkit&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="1287" LastActivityDate="2017-02-03T15:45:37.903" Title="Is there any hardware to use in GPR applications?" Tags="|rf-electronics|circuit-design|" AnswerCount="0" CommentCount="3" ClosedDate="2017-02-05T16:14:08.333" ContentLicense="CC BY-SA 3.0" />
  <row Id="13579" PostTypeId="1" AcceptedAnswerId="13596" CreationDate="2017-02-03T15:56:57.847" Score="0" ViewCount="73" Body="&lt;p&gt;What type of machines are used to paint a bulk of small objects all over... things like buttons, keys, ping pong balls, paperclips etc.&lt;/p&gt;&#xA;" OwnerUserId="9396" LastActivityDate="2017-02-04T16:17:45.303" Title="Machine that can paint a bulk of small objects all over" Tags="|machining|" AnswerCount="1" CommentCount="3" ClosedDate="2017-02-06T11:25:51.380" ContentLicense="CC BY-SA 3.0" />
  <row Id="13580" PostTypeId="2" ParentId="13574" CreationDate="2017-02-03T16:41:18.207" Score="1" Body="&lt;p&gt;I'm a civil engineer, but you seem to be interested in aeronautical engineering (as far as I can tell from the example you gave), so my answer might not be applicable to your case (but it probably is).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, any and all loading is calculated with a safety factor for ultimate limit state checks. However, the value of that safety factor may be different for different types of loads. The Brazilian code for reinforced concrete bridges, for example, defines that permanent loads (self-weight and dead loads) must have a safety factor of 1.35, while live loads need 1.5, and wind needs 1.4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are secondary safety factors, such as if you are testing the summation of different non-permanent load types: for instance, heavy traffic probably won't cross your bridge during hurricane-strength winds, so you might make two tests: one with heavy traffic as the primary load and wind as a secondary load (with a &quot;reduced safety factor&quot;), and another with wind as the primary load and traffic as secondary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some checks under exceptional conditions, certain load types might even be given a safety factor of 1.0, that is, no safety factor. Also, if doing serviceability checks, there usually are no factors.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-02-03T16:41:18.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13581" PostTypeId="2" ParentId="13574" CreationDate="2017-02-03T17:02:29.933" Score="1" Body="&lt;p&gt;The choice of safety factor is typically driven by certification requirements. The largest body overseeing these certifications is the Underwriters Laboratory, commonly UL. Check stuff around you- most of it will have their logo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ksWo6s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ksWo6s.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can obtain (for a non-trivial amount of money), the UL standards for the product you're designing. These will list the required safety factor and typically also a test procedure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These standards will often be aligned with an industry organization. For example, the UL standards for pressure vessels rely on the American Society of Mechanical Engineers (ASME) standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other standards organizations and may be more applicable for your field. I'm not familiar with the aerospace industry standards but many components will be UL listed.&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2017-02-03T17:02:29.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13582" PostTypeId="2" ParentId="13574" CreationDate="2017-02-03T17:04:54.220" Score="1" Body="&lt;p&gt;From a structural engineering perspective - different loads have different partial factors, reflecting different levels of uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Live loads are generally less certain than dead loads, and hence carry a higher factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under British Standards concrete had a higher factor than steel, because the weight of concrete was harder to predict. Road surfacing had a very high factor, to allow for the possibility of surfacing thickness increasing in future.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-02-03T17:04:54.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13583" PostTypeId="1" AcceptedAnswerId="13585" CreationDate="2017-02-03T17:12:27.117" Score="10" ViewCount="27488" Body="&lt;p&gt;This is a stress-strain diagram &#xA;&lt;a href=&quot;https://i.stack.imgur.com/405ro.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/405ro.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have noticed that the curve decreases between points C &amp;amp; D and between E &amp;amp; F. &#xA;Why does that happen? I expected the curve to always increase since the tension force in a tensile test is always increasing and the cross sectional area of the specimen decreases, which means that the stress is always increasing.&lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="1832" LastEditDate="2017-02-03T17:57:28.303" LastActivityDate="2019-05-13T08:08:15.743" Title="Why does the stress-strain curve decrease?" Tags="|mechanical-engineering|materials|civil-engineering|industrial-engineering|strength|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13585" PostTypeId="2" ParentId="13583" CreationDate="2017-02-03T17:55:45.743" Score="12" Body="&lt;p&gt;For the explanation of the loss of stress from E to F, I'll refer you to &lt;a href=&quot;https://engineering.stackexchange.com/a/8490/1832&quot;&gt;my answer to another question&lt;/a&gt;. Basically, the drop in stress seen in the diagram is a consequence of the fact that we use engineering stress (not taking the loss of cross-sectional area into consideration) instead of true stress. In a true stress diagram, there is no such drop. See &lt;a href=&quot;https://engineering.stackexchange.com/q/6020/1832&quot;&gt;this question&lt;/a&gt; for answers to why we use engineering stress instead of true stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as for why there's a drop between C and D, that has nothing to do with our use of engineering stress (though the drop is smaller in a true stress diagram). Instead, that occurs due to the steel's microstructure. Since I'm not a materials engineer, I apologize for the &quot;loose&quot; vocabulary I will use to explain this phenomenon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens due to the fact that steel contains carbon which occupies interstitials and effectively impedes the &quot;flow&quot; of the surrounding material that wants to yield at the lower yield value. Additional stress is required to overcome the carbon's resistance, at which point everything can move uninterrupted, which means that the material can then continue to yield at a lower stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the upper yield stress is momentary. Once that is overcome, the yield stress drops to the lower value.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-02-03T17:55:45.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13587" PostTypeId="2" ParentId="13558" CreationDate="2017-02-03T20:27:22.563" Score="1" Body="&lt;p&gt;The matrix $\mathbf{K}$ simply represents the force response to a unit displacement on each of the degrees of freedom of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a 2D cantilever beam of length $\ell$ with two degrees of freedom. The end displacement $\delta$ and the end slope $\theta$. You can assemble a stiffness matrix of the form $\mathbf{f} = \mathbf{K} \mathbf{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{vmatrix} F \\ M \end{vmatrix} = \begin{bmatrix} \frac{48 E I}{\ell^3} &amp;amp; -\frac{18 E I}{\ell^2} \\ -\frac{12 E I}{\ell^2} &amp;amp; \frac{6 E I}{\ell} \end{bmatrix} \begin{vmatrix} \delta \\ \theta \end{vmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interpretation is the force/moment needed to achieve unit deflections $[1\;0]$ or $[0\;1]$ between the two columns of the matrix.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-02-03T20:27:22.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13588" PostTypeId="2" ParentId="7408" CreationDate="2017-02-03T20:43:03.583" Score="1" Body="&lt;p&gt;Plastic is not used for windows mainly because it is not durable, in fact:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;it is degraded by UV radiation, which breaks atomic bonds, yielding a matt and thinned (and fragile) surface, not so desirable for a window;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;because of the previous point, plastic generates micro- and nanoparticles which diffuse inside the house and inhaling them is not healthy;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;it is less hard than glass, then it can be easily scratched with wider variety of materials;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;it is flammable(!) and can produce dioxin as it is burnt (especially if it contains chlorine atoms in its structure), an extremely toxic and carcinogenic compound.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Moreover, glass is able to shield approximately 10% of solar radiation and almost 100% of infrared radiation (emitted by human body as heat), so it is a powerful insulating material, which helps keep your house relatively fresh in summer and warm in winter.&lt;/p&gt;&#xA;" OwnerUserId="2541" LastActivityDate="2017-02-03T20:43:03.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13589" PostTypeId="1" CreationDate="2017-02-03T21:41:24.270" Score="1" ViewCount="233" Body="&lt;p&gt;Im in daily basis see a lot of controller. Sliding mode PID. Fuzzy, adapttive. Controller based on lyapunov etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how many are they that are actually in use in industry and when do we use them? If you want to drive a state to a certain set points what do you use? And output? &lt;/p&gt;&#xA;" OwnerUserId="8884" LastActivityDate="2017-02-03T21:41:24.270" Title="How many controllers of output or state are available?" Tags="|control-engineering|control-theory|pid-control|systems-engineering|optimal-control|" AnswerCount="0" CommentCount="4" ClosedDate="2017-02-04T21:08:34.533" ContentLicense="CC BY-SA 3.0" />
  <row Id="13590" PostTypeId="1" CreationDate="2017-02-03T22:02:54.360" Score="1" ViewCount="187" Body="&lt;p&gt;I have a situation that I have a combination of free and forced convection over a cylinder. Remembering back from heat transfer, to combine these I use the formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Nu_{\mathrm{comb}}^n = Nu_{\mathrm{forced}}^n + Nu_{\mathrm{free}}^n$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's fine. However, what dimension do I use to find $h_{\mathrm{comb}}$? Because&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Nu_{\mathrm{forced}} = \frac{h_{\mathrm{forced}} k}{D}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Nu_{\mathrm{free}} = \frac{h_{\mathrm{free}} k}{L}$$&lt;/p&gt;&#xA;" OwnerUserId="6885" LastActivityDate="2017-05-06T14:30:38.607" Title="Characteristic dimension for combining free and forced convection?" Tags="|heat-transfer|convection|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13591" PostTypeId="2" ParentId="13590" CreationDate="2017-02-03T22:43:03.153" Score="1" Body="&lt;p&gt;Okay, I'm not 100% sure if I got the question right but, to combine natural and forced convection for flow over a cylinder (I assumed horizontal) you have the following formula:&#xA;$$Nu_{comb}=(Nu_{free}^n+Nu_{forced}^n)^{1/n}\tag{1}$$&#xA;where $n$ depends on your cylinder type (horizontal or vertical). Next you need to calculate the Nu for each regime (free and forced). For the natural convection over a horizontal cylinder, you have this correlation:&#xA;$$Nu_{free}=\left(0.6+\frac{0.387Ra^{1/6}}{[1+(0.559/Pr)^{9/16}]^{8/27}}\right)^2\tag{2}$$&#xA;You can find whatever correlation you find better for your situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the forced convection, you have this correlation:&#xA;$$Nu_{forced} = 0.989Re^{0.330}Pr^{1/3}\tag{3}$$ (for 0.4 &amp;lt; Re &amp;lt; 4) For other type of flows you can find different Nusselt correlations. You can find them on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you replace them in Eq. 1 and finally you get:&#xA;$$h_{comb}=\frac{k_{fluid}\cdot Nu_{comb}}{D}\tag{4}$$&#xA;where D is the diameter of your cylinder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find some Nusselt correlations &lt;a href=&quot;http://machineryequipmentonline.com/hydraulics-and-pneumatics/wp-content/uploads/2016/03/NATURAL-CONVECTION-0489.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://machineryequipmentonline.com/hydraulics-and-pneumatics/wp-content/uploads/2016/03/FORCED-CONVECTION-0414.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8419" LastEditorUserId="8419" LastEditDate="2017-05-06T14:30:38.607" LastActivityDate="2017-05-06T14:30:38.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13592" PostTypeId="2" ParentId="13577" CreationDate="2017-02-03T22:46:28.983" Score="5" Body="&lt;p&gt;First off there are many different types of maglev prototypes out there each with their pros and cons. In fact some designs have the drawback of needing shielded compartments for people with pacemakers! However, instead of going into much detail about all the different kinds of magnets that can be used on maglevs, I will try to explain how magnetic fields can be manipulated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, some maglevs use an array of magnets called a Halbach array.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lMz87.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lMz87.png&quot; alt=&quot;Halbach Aray&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the left you can see that the magnetic field lines are somewhat uniform and on the right the maganetic field is not only on the top but also &lt;em&gt;much more condensed&lt;/em&gt;  than the image on the left. This is just one of the ways that enginerers can mitigate the problems caused by strong magnetic fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it is important to note that magnets will pull or repel each other with much more strength that when they are close together than when they are far apart. For example dipole magnetic fields decay at a rate of $1/d^3$ which means that as the distance $d$ increases magnetic field gets weak very fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maglev trains exploit these properties of magnetic fields by fitting powerful magnets on the ground and on the bottom of the train. As the magnets approach each other the fields repel with a force strong enough to keep the train afloat while keeping the passengers from experiencing a strong magnetic field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this has shown you that magnetic fields can be shaped and that magnetic fields tend to decay quickly as you move away from them. As a final note consider how little magnetic material humans tend to carry around! Most metals in common use are not strongly attracted to magnetic fields like aluminum. So unless you are moving you weight set or you cast iron skillet you can ride the maglev in safety without prying your stuff from the ceiling!&lt;/p&gt;&#xA;" OwnerUserId="9656" LastEditorUserId="9656" LastEditDate="2017-02-14T13:49:41.707" LastActivityDate="2017-02-14T13:49:41.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13593" PostTypeId="1" CreationDate="2017-02-03T23:12:51.920" Score="1" ViewCount="903" Body="&lt;p&gt;Here's an image of my structural plan that my group and I are designing for a university assignment. It is a 7-story steel structure with bracing at the two ends and at the verticle core near the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4a87s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4a87s.png&quot; alt=&quot;Structural plan with bracing at ends and center&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our professors have suggested that we use continuous primary and secondary beams however we are unsure what they are. All of the structural calculations (bending moment, shear, deflection check) will be done with a computer program however we must first understand the structural diagram of a continuous beam in order to enter the proper information into the program. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the continuous beam structural diagram look like for the primary and secondary beams of this structure? Is the diagram below correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4wKnV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4wKnV.jpg&quot; alt=&quot;Structural diagram with two fixed ends and nine hinged connections in between&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any explanation of how a steel continuous beam works in a multi-level structure is greatly appreciated. Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="9860" LastActivityDate="2017-02-06T23:43:44.783" Title="Steel Continuous Beam Structural Diagram" Tags="|structural-engineering|civil-engineering|structures|steel|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13594" PostTypeId="2" ParentId="13575" CreationDate="2017-02-04T00:03:08.163" Score="2" Body="&lt;p&gt;In theory this is completely possible. In practice you may have problems. Remember that if you have a 100000:1 ratio of speed, you also have a 1:100000 ratio of torque. At that ratio, a tiny torque at the high speed shafts becomes huge at the low speed shaft. E.g let's say the bearing on your high speed shaft has a drag torque of 0.05 in-lbf. That becomes 5000 in-lbf at the low speed shaft. If your low speed shaft is 1&quot; radius, then your object needs to be 5000 pounds just to overcome the friction needed to turn the output shaft. With very high precision bearings with very low drag torque, might be possible. But I suspect those would be expensive. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-02-04T00:03:08.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13595" PostTypeId="2" ParentId="13583" CreationDate="2017-02-04T06:41:29.467" Score="2" Body="&lt;p&gt;Further adding to Wasabi answer,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yield point phenomenon occurs due to the segregation of impurity solute atoms(C and/ or N in Fe) around dislocations so as to reduce the strain energy associated with the distorted atomic arrangement (maximum in bcc metals, less in hcp metals and least in fcc metals). This additional stress required to free the dislocations and set them in motion needed for plastic deformation is called the upper yield point. Once dislocations have been freed, the stress needed for their motion drops abruptly and is called the lower yield point. There comes slight variation in its value due to the interaction of moving dislocations with the impurity solute atoms obstructing their paths.&#xA;&lt;a href=&quot;https://www.researchgate.net/post/What_are_the_reasons_for_the_phenomenon_of_upper_and_lower_yielding_behavior_in_mechanical_testing&quot; rel=&quot;nofollow noreferrer&quot;&gt;Further Reference&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9789" LastActivityDate="2017-02-04T06:41:29.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13596" PostTypeId="2" ParentId="13579" CreationDate="2017-02-04T11:52:36.210" Score="2" Body="&lt;p&gt;In terms of general coatings there are a few options for bulk treatment of small objects. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Powder coating : the part is electrostatically coated with a polymer powder which is then baked on. This produces quite a thick and tough coating which can cover complex surfaces well and doesn't require volatile solvents. &lt;/li&gt;&#xA;&lt;li&gt;Dipping : the part is dipped is a rubber based medium which may also be foamed,  coating the surface. This leaves a very thick, soft coating which is good for things like tool handles and lever arms. &lt;/li&gt;&#xA;&lt;li&gt;Spray painting : can be mechanised by spraying from different directions on a conveyor type system, often going on to drying ovens. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are also various chemical techniques eg plating, passivating, anodising, phosphating, bluing which induce the formation of a relatively inert layer on a metal surface. For example most steel fasteners are bright zinc passivated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nickel, chrome, brass or bright zinc plating are fairly common on small metal parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sheet metal parts may also be made from material which is pre-coated with a plastic film often over galvanising. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminium parts are often anodised, an electrolytic process which encourages the formation of an even and stable layer of aluminium oxide which can also be dyed. tending to produce a semi-transparent colour. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-02-04T16:17:45.303" LastActivityDate="2017-02-04T16:17:45.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13597" PostTypeId="2" ParentId="13575" CreationDate="2017-02-04T14:56:43.767" Score="1" Body="&lt;p&gt;The 100k to 1 overall gear ratio you want makes the realization difficult, as Danial Kiracofe has already pointed out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to answer your question, yes it is possible to regulate the speed.  There are quite a few different ways.  A conceptually simple one is a centrifugal brake.  You have a vertical rotating shaft with weights hinged from it so that the weights move outward due to centrifugal force as the shaft speed goes up.  When the weights move a certain distance from the shaft, they touch something that causes friction.  This slows down the rotation speed of the shaft.  The net effect is that the frictional drag on the shaft is little up to some set speed, then increases rapidly for higher speeds.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-02-04T14:56:43.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13598" PostTypeId="2" ParentId="13575" CreationDate="2017-02-04T16:37:05.553" Score="2" Body="&lt;p&gt;A gearbox on its own won't really regulate speed, apart from the fact that it will have some internal friction but this is difficult to regulate. For example just having a 1000:1 gear ration won't make the object fall 1000 times slower unless there is some external torque applied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple solution is to use viscous drag for example having the gearbox drive a fan or paddle wheel. The works because viscous drag increases with the square of speed so you will get to a point where the drag force equals the weight of the object at some reasonably constant speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also various types of mechanical governor which can be used to adjust the force on a mechanical brake in response to speed. A step further is to use an electronic system with speed sensors and actuators on a brake, this potentially allows a lot of flexibility and the ability to self adjust on the fly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to use an escapement of they type used in clocks. This doesn't give true constant speed as the motion becomes a series of discrete steps but for the very low speeds you are talking about this may not be a problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that the gear ratios you are talking about are huge, to put it into perspective if you did this with a single pair of gears and the driving gear was 5cm in diameter then the driven gear would need to be 5km in diameter &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit in response to comment: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use this arrangement to generate power then you need to bear in mind that gravitation potential energy has pretty poor energy density compared to other small to medium scale energy storage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically the total gravitational potential energy is 9.8J per kg per metre of drop. 24 hours is 86400 seconds so for a 2m drop this translates as 1.1e-4 Watts ie you would need 5000kg to provide 1W over 24 hours before considering losses. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-02-05T10:22:47.003" LastActivityDate="2017-02-05T10:22:47.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13601" PostTypeId="1" AcceptedAnswerId="13611" CreationDate="2017-02-04T17:55:45.560" Score="2" ViewCount="153" Body="&lt;p&gt;A spec characterizes the vibration stability by acceleration noise density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Na &amp;lt; \dfrac{1\text{ μg}}{\sqrt{Hz}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for 1-100&amp;nbsp;Hz, where g=9.8&amp;nbsp;m/s&lt;sup&gt;2&lt;/sup&gt;. But usually what I see is something like 3.5&amp;nbsp;nm pk-pk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how to convert these two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried this as below, but not sure if it's correct, any comment is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Displacement noise density (is this correct?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Nx = \frac{Na}{4\pi^2f^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA; x^2 &amp;amp;= \int (Nx)^2 df = \int \frac{(Na)^2}{16\pi^4f^4} df \\&#xA; &amp;amp;= \frac{(1\mu g)^2}{16\pi^4} \int_{1}^{100} \frac{1}{f^4} df \\&#xA; &amp;amp;\approx 2 \times 10^{-14} m^2 &#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="9875" LastEditorUserId="1832" LastEditDate="2017-02-05T00:15:29.147" LastActivityDate="2017-02-05T13:39:34.387" Title="Vibration characterization: how to convert acceleration noise density to displacement pk-pk value?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13602" PostTypeId="1" AcceptedAnswerId="13606" CreationDate="2017-02-04T18:25:14.397" Score="2" ViewCount="372" Body="&lt;p&gt;In the old days there was a type of sign/light you would find in bars and restaurants which was animated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be a lit scene usually of water, like a lake with a waterfall, and by some means it would animate and look like the water was flowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They would be used to advertise cigarettes or beer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What was this technology called? How did it work?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2022-07-11T21:51:27.287" Title="Animated light box signs" Tags="|lighting|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13603" PostTypeId="2" ParentId="13583" CreationDate="2017-02-04T20:05:44.777" Score="9" Body="&lt;p&gt;To add to the other answers: interstitial solutes effectively &quot;lock&quot; dislocations in BCC metals (such as with carbon and nitrogen dissolved in iron), by forming what are called Cottrell atmospheres (&lt;a href=&quot;https://en.wikipedia.org/wiki/Cottrell_atmosphere&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;). The lowest free energy is achieved when the interstitial atoms are in the dislocation, because the dislocation has a larger interatomic spacing due to the missing atomic plane. Thus, over time, the interstitials will gradually migrate to, and remain at, the dislocations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interstitials locally strain the lattice around the dislocation, increasing the stress required to move the dislocation. The effect of Cottrell atmospheres is to increase the yield strength of the material, creating an upper yield point at C in the diagram. If the carbon is unable to diffuse as quickly as the dislocations glide, then when point C is reached, the dislocations move away from the carbon and become &quot;unlocked.&quot; When unlocked, the stress required to move the dislocations decreases, resulting in a lower yield point at D in the diagram. While the dislocations are moving due to the increasing strain, the density of dislocations is increasing, i.e. work hardening. Between C and D, the decrease in strength from dislocation unlocking is more rapid than the strength increase caused by work hardening. After D, the strength can no longer decrease, so the normal work hardening curve resumes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you were to strain the material to somewhere between C and D, stop, and then allow the material to sit for a few days. After allowing the material to age, you then resumed straining the material, it will have regained its yield strength, and to a point higher than C. The reason is that by aging the material, the carbon has had time to diffuse to the dislocations again. Additionally, the increased dislocation density and thus work hardening would have also increased the strength of the material. This effect is called static strain aging: the material is allowed to age after extension to a static (constant) strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that at or above the temperature where interstitial diffusion is as fast as the dislocation motion, the upper and lower yield point phenomena change into a serration on the stress-strain curve. This effect is known as dynamic strain aging, in contrast with static strain aging, because the material is being aged as fast as, and while, it is being strained. The effect is shown in the diagram below. One may also change decrease the strain rate to achieve a similar effect to increasing the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram, at X the sample is unloaded and then immediately reloaded. At Y the sample is unloaded and allowed to age for sufficient time. Note the recovery of the yield point to a higher value than originally: static strain aging. Note that there are serrations after the upper yield point, which indicates the material is being strained sufficiently slowly to allow the carbon to catch up to the dislocations, i.e. dynamic strain aging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FjJmu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FjJmu.png&quot; alt=&quot;Stress-strain diagram showin static and dynamic strain aging.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from &lt;a href=&quot;https://gatesolution-metallurgicalengineering.blogspot.com/2011/07/yield-point-and-strain-aging.html&quot; rel=&quot;noreferrer&quot;&gt;GATE Metallurgical Engineering&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2017-02-04T20:05:44.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13605" PostTypeId="2" ParentId="13602" CreationDate="2017-02-05T01:32:25.060" Score="-1" Body="&lt;p&gt;I think you may be referring to Kinegrams:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fSc8X.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fSc8X.gif&quot; alt=&quot;windmill&quot;&gt;&lt;/a&gt;&#xA;Here are some links:&#xA;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Kinegram&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Kinegram&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://thinkzone.wlonk.com/Kinegram/MakingKinegrams.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://thinkzone.wlonk.com/Kinegram/MakingKinegrams.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2017-02-05T01:38:56.743" LastActivityDate="2017-02-05T01:38:56.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13606" PostTypeId="2" ParentId="13602" CreationDate="2017-02-05T04:06:07.997" Score="2" Body="&lt;p&gt;After some more research, I found it: &lt;a href=&quot;http://www.collectorsweekly.com/articles/hamms-scene-o-rama-beer-sign-bliss/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scene-o-rama&lt;/a&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These animated water motion signs from the 1950s and 1960s work by a plastic overlayer and roll film on the underside that constantly rotates in a loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scene-o-Rama was first widely used by Hamm's beer, but many other manufacturers adopted them as an advertising methodology. I remember seeing Michelob scene-o-ramas myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you watch one on YouTube you can see how cool it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ekSoO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ekSoO.jpg&quot; alt=&quot;Scene-o-Rama&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2017-02-05T04:06:07.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13607" PostTypeId="1" AcceptedAnswerId="15400" CreationDate="2017-02-05T04:46:13.427" Score="0" ViewCount="192" Body="&lt;p&gt;I know that the 4G network is only for data and doesn't support voice calls. Most of phones switch to 2G/3G when receiving a voice call. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my problem is how does the phone gets to know about the reception of the voice call when it is in 4G network. We can think, a signal is sent over 4G data to the phone to switch from 4G to 2G/3G. But what will happen if the mobile data is switched off in the device, when receiving a phone call.&lt;/p&gt;&#xA;" OwnerUserId="9879" LastActivityDate="2017-05-19T08:47:56.800" Title="How does 4G know the reception of a voice call?" Tags="|telecommunication|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13608" PostTypeId="2" ParentId="13478" CreationDate="2017-02-05T05:02:29.073" Score="0" Body="&lt;p&gt;Depends on the critically of application. Your system may have a speed limit specification, defined by encoder resolution. This specification shall comply with your operational, security and reliability requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After defined this, you can count the time between two encoder signals, for determine speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of you have the control of a motor, you may determine the rotation frequency of the motor, through PWM, or in many cases, determine the power and time of operation. &lt;/p&gt;&#xA;" OwnerUserId="9864" LastActivityDate="2017-02-05T05:02:29.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13610" PostTypeId="2" ParentId="13574" CreationDate="2017-02-05T12:19:07.530" Score="0" Body="&lt;p&gt;Yes, you have to multiply all the loads with safety factor, I think It is already based on the regulation worldwide in engineering field in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose is to avoid the underestimation in calculation process and to human-error, and any other technical mistake in the field.&lt;/p&gt;&#xA;" OwnerDisplayName="user8531" LastActivityDate="2017-02-05T12:19:07.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13611" PostTypeId="2" ParentId="13601" CreationDate="2017-02-05T13:39:34.387" Score="2" Body="&lt;p&gt;Conversion between $g/\sqrt{hz} $ and $m/\sqrt {hz} $ is possible. However, converting from that to m pk-pk is usually not possible. The reason is that spectral density is a unit used to quantify &lt;em&gt;random&lt;/em&gt; vibration, and m pk-pk is a unit used to quantify &lt;em&gt;sinusoidal&lt;/em&gt; or periodic vibration. Random and periodic vibration simply are not the same thing. For specific applications, there are ways to come up with equivalences between them. But in the most general sense it is apples and oranges. I have some tutorials on my webite that may help explain this concept in more detail &lt;a href=&quot;http://mechanicalvibration.com/Power_spectral_density.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mechanicalvibration.com/Power_spectral_density.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-02-05T13:39:34.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13614" PostTypeId="1" CreationDate="2017-02-05T15:10:42.047" Score="1" ViewCount="148" Body="&lt;p&gt;I would assume that different materials absorb sound differently (maybe not?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this called?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I compare the sound absorbent properties of different materials with the same surface geometry, such as plaster and gypsum board?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2017-02-06T19:10:36.203" Title="Acoustic absorption of different materials? (gypsum board versus plaster)" Tags="|acoustics|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13615" PostTypeId="2" ParentId="13593" CreationDate="2017-02-05T16:19:21.617" Score="1" Body="&lt;p&gt;For multiple stories, you usually have continuous columns going up and simple span beams in between. Having continuous beams would require putting the top level column on top of the beam, so you would have an extra beam bearing calculation which would become more difficult as you have more stories. There is also the question of stability of the beam with a top flange load which would have to be addressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your diagram is about right, but catches my eye as maybe a bit long for a beam. Usually steel rolled shapes will be max. 60 feet long, so if you want to do multi-span, you will be limited by this. Id suggest you take a look at &quot;Gerber beam systems&quot; which are sometimes used for economical multi span beams going over columns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also there is no need to use fixed ends at both ends, there are quite a few column connections that can be considered flexible enough to be designed as pins (Shear tabs, Single and double angle, beam face plate). While putting a rigid (fixed) connection in steel is possible, it is rarely done outside of framed construction because the connection is more complex and you have to design your column for the transferred moment. If I understand your drawing correctly, you have X braces here so no need for steel frames.&lt;/p&gt;&#xA;" OwnerUserId="5691" LastActivityDate="2017-02-05T16:19:21.617" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13616" PostTypeId="1" CreationDate="2017-02-05T19:57:00.957" Score="1" ViewCount="579" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IXiLc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw this plastic on a baby changing station and was wondering what sort of plastic it was. It seems like injection molding plastic but I'm not sure which one and also what pepper color scheme that is. &lt;/p&gt;&#xA;" OwnerDisplayName="user9826" LastActivityDate="2017-03-08T03:35:48.760" Title="What sort of plastic is this? What would you call this color scheme?" Tags="|plastic|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13617" PostTypeId="2" ParentId="13602" CreationDate="2017-02-05T21:40:34.537" Score="0" Body="&lt;p&gt;At least some of such signs were animated using polarized light. The water or whatever would be made as a mosaic of little pieces of polarizing plastic, and the whole thing would be illuminated from behind by a lamp shining through a rotating polarizer. This causes different segments to brighten and dim at different times.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2017-02-05T21:40:34.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13618" PostTypeId="1" CreationDate="2017-02-05T21:57:04.947" Score="1" ViewCount="35" Body="&lt;p&gt;Is there any application or circuit design for RF radar systems to use directly the transmitter frequency for the receiver side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've seen, a mixer is always used to get the frequency difference to calculate the distance. Is that still applicable to use for Ground Penetrating Radar systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What determines the frequency required for scanning depth and size of the objects to be detected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the use of higher frequency results in more finer details but it's difficult to penetrate deep into ground. I guess the maximum of 1GHz is enough to penetrate about 1 meter into the ground, of course depending on the conductivity of the soil. But I am not sure which specifications I need to process that much frequency. How much bandwidth do I need to use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One said that you need to have a bigger bandwidth to have a higher resolution for depth information. But I am not sure if I use 1GHz do I need some sort of bandwidth about 2 GHz or sth. That much bandwidth supplied to ADCs are about 3000 USD and I am not sure if I am looking to correct direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you guide me a bit?&lt;/p&gt;&#xA;" OwnerUserId="1287" LastEditorUserId="90" LastEditDate="2017-02-06T11:27:01.747" LastActivityDate="2017-02-06T11:27:01.747" Title="Are there any negative aspects to using baseband frequencies in radar systems?" Tags="|rf-electronics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13619" PostTypeId="2" ParentId="13478" CreationDate="2017-02-06T01:28:37.527" Score="4" Body="&lt;p&gt;You have stumbled across the fundamental problem of using an encoder to measure speed: they are not great for measuring low speeds. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why is using an encoder to measure speed an issue?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The encoder has a certain angular resolution $\theta_r$ in radians and the number of counts detected by your digital counter $N$ is reported after a certain sampling time $\Delta t$. Your angular velocity is then determined for each sample time by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega = \frac{N\times \theta_r}{\Delta t} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest number of counts that can be recorded (other than zero counts, of course) is 1 count over the sample time. This means that your angular velocity resolution $\omega_r$ is determined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\omega_r = \frac{\theta_r}{\Delta t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets say you try to measure a speed that is 1/4 of your resolution. Your measurement device will indicate $\omega = 0$ for $3\times\Delta t$ and then your minimum speed $\omega=\omega_r$ for $1\times \Delta t$ and repeat every ten sample time steps. For a speed that is 1/10 of your resolution it will measure zero for $9\times \Delta t$ and then $\omega_r$ for $1\times \Delta t$ and so on. See the figure below for an example when the speed is 1/10 of your speed resolution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gli8x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gli8x.png&quot; alt=&quot;Encoder position and velocity in normalized time&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you fix it?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;My recommendation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Buy a different sensor. If you are operating an encoder near its speed resolution limit then you just won't be getting a very usable control signal. Look instead for a &lt;em&gt;tachometer&lt;/em&gt;. A tachometer is a sensor that is specifically designed to measure speeds. There are many different varieties of this type of sensor available so it is hard to recommend a specific type without knowing more about your application. However, if you need electrical output for some kind of controller then I think your best bet would be a &lt;em&gt;variable reluctance sensor&lt;/em&gt; tachometer or a &lt;em&gt;hall effect sensor&lt;/em&gt; tachometer.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Otherwise&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If buying a good sensor is simply not an option then, as you stated in your question, you can increase your sampling time (decrease your sampling rate) to improve your velocity resolution. The main problem with this solution is that you could increase your sampling time to the point where your system is no longer controllable, or is not controllable in a practical sense. Again, without knowing more about your application it is difficult to say what a reasonable sampling time would be. I would not suggest getting better encoder resolution because 14-bit is already very fine angular resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: Beckwith, Marangoni and Lienhard. &lt;em&gt;Mechanical Measurements, 6th Ed.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2017-02-06T01:28:37.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13621" PostTypeId="1" AcceptedAnswerId="13655" CreationDate="2017-02-06T02:31:14.130" Score="8" ViewCount="3219" Body="&lt;p&gt;Why do some high speed trains like the Shinkansen E5 and E6, have &lt;em&gt;really&lt;/em&gt; long noses while other trains, like the Eurostar and Javelin, have shorter noses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the shape of the nose affect the train?&lt;/p&gt;&#xA;" OwnerUserId="9893" LastEditorUserId="368" LastEditDate="2017-06-06T16:45:11.783" LastActivityDate="2017-06-13T11:43:22.113" Title="Why do some high speed trains have long noses, while others do not?" Tags="|mechanical-engineering|aerodynamics|rail|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13622" PostTypeId="1" CreationDate="2017-02-06T02:34:56.333" Score="1" ViewCount="41" Body="&lt;p&gt;I wish to calculate the amount of energy required for electrolysis through induction, instead of electrodes. If there is a crucible metal core &#xA;with solution inside and the coil wrapped around it. Can there be the required voltage for the electrolysis reaction and enough current to transport the ions? &#xA;If I have the resistance of the solution inside the crucible (derived from its resistivity of an area) and the voltage produced from the inductor will this tell me how much current is flowing through the solution and then inturn allow me to calculate Faradays laws? &#xA;Not really asking for a solution more the correct steps to get there.&#xA;Thank you for taking the time to read the question.  &lt;/p&gt;&#xA;" OwnerUserId="9794" LastActivityDate="2017-02-06T02:34:56.333" Title="using a coil for an electrochemical reaction" Tags="|electrical-engineering|chemical-engineering|electromagnetism|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13623" PostTypeId="2" ParentId="13616" CreationDate="2017-02-06T02:37:27.403" Score="2" Body="&lt;p&gt;You might be looking for a speckle additive or as this company(link below) refers to it as stone/granite/marble effect...Looks like a Granite type effect to me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a link:&#xA;&lt;a href=&quot;https://www.rtpcompany.com/products/color/special-effects/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plastic special effects&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what grade of plastic, we could only make some educated guesses based on its use. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-02-06T02:37:27.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13624" PostTypeId="2" ParentId="13489" CreationDate="2017-02-06T08:14:19.267" Score="0" Body="&lt;p&gt;Its not a complete answer. But I found out that the dynamics need redefining such that a transformation of the state is performed. A system can be generically observable if the systes can be expressed by output and input and finite numbers of their derivatives i.e&#xA;$$x=X(y, \dot{y},..., y^j , u , \dot{u}, ..., u^i)$$&#xA;Where &#xA;$$j,i \in \mathbb{N}_+$$&#xA;Denote the number of derivation.&#xA;When the dynamics are transformed to a simpler and standard form you can take the unkowns as uncertainties and handle the system using desired observer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More about this can be found in the book Advances and Applications of Nonlinear Control. &lt;/p&gt;&#xA;" OwnerUserId="8884" LastActivityDate="2017-02-06T08:14:19.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13625" PostTypeId="2" ParentId="13621" CreationDate="2017-02-06T08:26:46.290" Score="-4" Body="&lt;p&gt;In high speeds the impact of aerodynamic forces need to be included in analysis. Recall from the formula for aerodynamics  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_a = 1/2 \rho v^2 C_d A$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\rho , v, C_d,A$ are air density, speed of the vehicle , drag coefficient and contact area with air. The drag coeffient has other components. The viscous effect has an impact on the aerodynamics forces.&lt;/p&gt;&#xA;" OwnerUserId="8884" LastEditorUserId="288" LastEditDate="2017-02-06T09:24:13.553" LastActivityDate="2017-02-06T09:24:13.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13626" PostTypeId="2" ParentId="13494" CreationDate="2017-02-06T09:46:03.130" Score="1" Body="&lt;p&gt;Your setup as with the pressure regulator could work. Before buying one, you need to know the approximate pressure you regulate to: Static pressure + dynamic loss pipe + dynamic loss diffusor&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Static pressure: 1 cm water column above the diffusor equals 1 mbar&lt;/li&gt;&#xA;&lt;li&gt;Dynamic loss pipe - dependant on flow rate, use &lt;a href=&quot;https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Darcy Weissbach&lt;/a&gt;. &lt;/li&gt;&#xA;&lt;li&gt;Dynamic loss diffusor - dependant on flow rate, impossible to know without knowing the diffusor.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The gas from the fermentation will be at saturation moisture, condensation inside the regulator can happen. I would position that regulator so that drops can leave and see what happens, though this may damage the regulator. Alternativly preheat the gas somehow upstream of the regulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avoid low points in the tubing, put tubing with a consistent grade downward towards containers so that condensate can flow along.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't really go into safety considerations here, you have small amounts of gas (I assume?) and PET containers. Make sure that the whole setup is in a well ventilated place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure the include vents with manual valves so you can de-pressurize your system, before opening.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-02-06T09:46:03.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13627" PostTypeId="1" CreationDate="2017-02-06T11:17:07.657" Score="4" ViewCount="175" Body="&lt;p&gt;I have created the following PCB design:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wCAUS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wCAUS.png&quot; alt=&quot;PCB layout&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the right is a USB port, on the left are a number of pads for connecting a round multi-core cable. There will be between 3 and 10 connections, depending on what is being connected. The cable will most likely be soldered directly to the PCB, although a connector could be used. The large number of pads is to allow for different configurations, only 7 to 12 will be used at any one time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The board is limited to 50x50mm for cost reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My plan is to use a low cost &quot;sandwich&quot; enclosure, basically two acrylic laser cut parts, a top and a bottom, with screws and spacers to hold the PCB in the middle. The case can be larger than the PCB if required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I add effective but cheap strain relief to this design?&lt;/strong&gt; I was considering extending the case by 10mm to the left, and adding a couple of holes to thread a cable tie though. Alternately I could thread more screws through and bend the cable around them, but that would require more space. Perhaps some kind of slot in the bottom and top case parts would allow a rubber strain relief grommet to be fitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a hobby project, doesn't need to be commercial grade.&lt;/p&gt;&#xA;" OwnerUserId="9670" LastEditorUserId="9670" LastEditDate="2017-02-07T15:57:20.250" LastActivityDate="2018-03-05T00:42:55.500" Title="Low cost, compact cable strain relief for laser cut sandwich case" Tags="|mechanical-engineering|design|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13630" PostTypeId="2" ParentId="13526" CreationDate="2017-02-06T14:29:01.253" Score="1" Body="&lt;p&gt;If you are a MATLAB or Oculus user, there is the &lt;a href=&quot;https://www.mathworks.com/matlabcentral/fileexchange/23241-naca-5-digit-airfoil-generator/content/naca5gen.m&quot; rel=&quot;nofollow noreferrer&quot;&gt;NACA 5 digit Airfoil Generator&lt;/a&gt;&#xA;program you can use. It works very well, considering it calculates the plot, as well as other main features of the airfoil, given the airfoil type. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am unaware of a program that gives plots based on the formula, but if you ever DO find out, please do not hesitate to post the link here.&lt;/p&gt;&#xA;" OwnerUserId="8862" LastActivityDate="2017-02-06T14:29:01.253" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13631" PostTypeId="1" CreationDate="2017-02-06T14:44:34.503" Score="1" ViewCount="183" Body="&lt;p&gt;I need help with a question regarding diffraction gratings. It goes something like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that you have a diffraction grating. When you shine light of different wavelength through this grating it will give different amounts of maxima. Let's say the grating has to satisfy the following: If we shine light of wavelength $\lambda_1$ we get $3$ maxima and if we shine light with wavelength $\lambda_2$ we get $5$ maxima. What values of the grating constant are possible given the criteria stated above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to solve this using the grating equation, $d\sin a = k\lambda$ where I set $k = 1$ for 3 maxims and $k = 2$ for $5$ maxima and then look at what different values for the angle $a$ give. With $90^{\circ}$ as the maximum angle. However, using this method I find that $d &amp;gt; \lambda_1$ for $3$ maxima, and $d &amp;gt; 2\lambda_2$ for $5$ maxima which seems wrong to me. Is this answer wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate all help!! &lt;/p&gt;&#xA;" OwnerUserId="9900" LastActivityDate="2017-02-06T14:44:34.503" Title="Possible values of grating constant" Tags="|waves|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13632" PostTypeId="2" ParentId="12849" CreationDate="2017-02-06T15:01:56.613" Score="-1" Body="&lt;p&gt;As  @Brian Drummond said ,i overlooked the work done between 2' and 3&lt;/p&gt;&#xA;" OwnerUserId="9223" LastActivityDate="2017-02-06T15:01:56.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13633" PostTypeId="1" AcceptedAnswerId="13657" CreationDate="2017-02-06T15:25:31.517" Score="7" ViewCount="6956" Body="&lt;p&gt;I'm having some trouble figuring out how to design the corresponding internal gear for any particular spur gear. With spur gears, meshing is accomplished by having two gears with the same modulus (MOD) or, in the US, the same diametral pitch. The spacing between the two gears is such that their pitch circles touch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at an internal gear the tooth profile for an internal involute gear is the inverse of the external involute. However if I try to create such a system (in Fusion 360), I get some interference issues:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ETImA.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ETImA.png&quot; alt=&quot;Internal to Spur gear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've done is create two spur gears, both MOD = 1, one with 24 teeth which I've just simply inverted to create an internal ring gear and the other with 12 teeth. My questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The pitch circles are marked in green. Is the pitch circle for the internal gear the same as the spur gear that I used in inverse to create it? Or is it moved somewhat. The proper relative gear positions should be where the pitch circles touch so that's a main issue.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Although it doesn't look too bad, the left and right teeth in the image are interfering. I think if the addendum of the internal gear's teeth was reduced this would help matters. It seems that the larger the ratios (eg: if the spur gear was 8 teeth instead) the more the gears would crash and perhaps the more the addendum should be reduced.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In addition I could add backslash, would this help the interference issue?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are there specific rules to design an internal involute gear for a particular MOD?&lt;/p&gt;&#xA;" OwnerUserId="2555" LastEditorUserId="2555" LastEditDate="2017-02-06T15:36:26.180" LastActivityDate="2023-12-04T08:43:51.983" Title="Internal teeth involute gear pitch circles" Tags="|gears|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13637" PostTypeId="2" ParentId="13614" CreationDate="2017-02-06T19:10:36.203" Score="3" Body="&lt;p&gt;What you're looking for is called the Acoustic Absorption Coefficient. Which is a measure of how much sound a material absorbs. Unfortunately, it's not as simple as that, mostly because the coefficient is dependent on frequency meaning that it will absorb some frequencies better than others. It's also dependent on the angle of incidence. Google will bring up a variety of tables, but use these with caution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind as well that when sound is incident upon a surface, a portion is reflected, a portion is absorbed and a portion is transmitted. Since part of the sound is transmitted, it's not just the material in question that's important, but the material behind it as well. As with the absorption, reflection and transmission are dependent upon both frequency and angle of incidence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any tabulated data of absorption coefficients for different materials has to be taken with a huge grain of salt and an understanding of what those numbers are and how they were measured.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-02-06T19:10:36.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13639" PostTypeId="1" AcceptedAnswerId="13653" CreationDate="2017-02-06T23:41:12.447" Score="0" ViewCount="73" Body="&lt;p&gt;I am aware that it is easier to run a static timing analysis on a CPLD than a FPGA, but why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any information would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6104" LastEditorUserId="8055" LastEditDate="2017-02-07T18:53:14.833" LastActivityDate="2017-02-07T18:53:14.833" Title="Why is it easier to run an STA on a CPLD than an FPGA?" Tags="|electrical-engineering|computer-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13640" PostTypeId="2" ParentId="13593" CreationDate="2017-02-06T23:43:44.783" Score="0" Body="&lt;p&gt;In real life you would simulate the whole building as a 3D frame. Particularly for lateral loads (such as wind or eartquakes) it is almost necessary. You could omit secondary beams from such a model though.With an assignment things can be more simplistic though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A beam can be considered continuous if it can transfer its full end moments and forces to the next span. This, doesn't neccessarily mean that the beam is physically continuous, as a non-split member. It rather means, that an appropriate connection exists at beam end that can carry the full moments and other forces (aka full-strength, rigid connection). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your structural model seems appropriate, except for the end supports. Making them fully fixed you assume that the columns at that points are infinitely less flexible than the beam, which is not correct. A detailed approach would be to envelope the beam response between the two extremes of fully fixed and pinned end supports. However you may skip this if the beam cross-section remains unchanged and loads are also similar for all spans. In that case, it would be ok if you considered only the pinned ended model, because this would give you simultaneously worst sagging moments (at the end spans) and almost the same hogging moments (at the interior supports). Thus, you cover both extremes simultaneously. &lt;/p&gt;&#xA;" OwnerUserId="9620" LastActivityDate="2017-02-06T23:43:44.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13641" PostTypeId="2" ParentId="13494" CreationDate="2017-02-07T00:09:34.477" Score="1" Body="&lt;p&gt;If I am understanding your design correctly, I think you are going to need a little more than a regulator to operate efficiently. When your storage pressure is only a little higher than your use pressure in a system like this, your storage volume goes up, as does your waste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend a &lt;a href=&quot;https://www.zoro.com/shurflo-pump-diaphragm-115-vac-8000-813-238/i/G0489711/&quot; rel=&quot;nofollow noreferrer&quot;&gt;gas/water diaphragm pump&lt;/a&gt; with a pressure switch or level switch. This would let you store at a higher pressure and capture all of your waste gas instead of just what is above 2 bar. Returning the gas could be done with a regulator or control valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the sugar production industry, a device used to mix CO2 with water is called a carbonator. That might help you locate some other references. &lt;a href=&quot;http://www.liquicel.com/applications/carbonation.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Membrane carbonators&lt;/a&gt; might also be an option. There are lots of brands out there, but they basically allow you to add CO2 in a more controlled manner without the large footprint or gas recirculation of a bubbling type device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you give some more details about what your are trying to achieve with your setup I may be able to provide more specific information.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-02-07T00:09:34.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13642" PostTypeId="2" ParentId="13554" CreationDate="2017-02-07T00:22:43.837" Score="2" Body="&lt;p&gt;If it is tried for so many years with no incidents (or small number of them, that may come from wrong installation) then how can we say it is unsafe? Of course improvements can be made to any design and i will try my best to offer some. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of the tile seems simple and convenient for easy installation and also can be stacked one upon the other for less transportation costs. As with any structure, we want it to be simple and economic. Since this is made of clay, I think a major improvement would be if its resistance to brittle fructure is enhanced, so that even if a crack initiates at some point it doesn't propagate too fast and even if a breadthwise fracture occurs, the tile doesn't break apart and fall down. This can be achieved with some sort of reinforcement, like concrete, where steel rebars reinforce and restore its lacking tensile strength. What sort of reinforcement? I can think of small synthetic fibers, or straw, that has been used since ancient times to reinforce clay bricks. But these fibers can be visible to the surface of the tile. Another possibility is a net of wires, placed in the mid-thickness of the tile. The goal is to prevent the tile from breaking apart and fall down, even if a serious crack develops. The size and type of reinforcement should be based on this requirement, Of course experimental testing should verify these claims.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach could be to fasten the tiles at the wooden members, they rest on. This would be easy if the tiles were made with some holes at their ends (along the little step)&lt;/p&gt;&#xA;" OwnerUserId="9620" LastEditorUserId="9620" LastEditDate="2017-02-07T11:07:53.093" LastActivityDate="2017-02-07T11:07:53.093" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13643" PostTypeId="2" ParentId="13627" CreationDate="2017-02-07T03:08:38.647" Score="1" Body="&lt;p&gt;I would suggest using a right angle shrouded IDC Male connector. You will have to modify the PCB layout. Based on my understanding of your question you can seat the connector between PCB board the acrylic part. Here is an example of a shrouded IDC connector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/62AHz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/62AHz.jpg&quot; alt=&quot;Shrouded IDC Connector&quot;&gt;&lt;/a&gt;   &lt;a href=&quot;https://i.stack.imgur.com/i4i2I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i4i2I.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/h5BDq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h5BDq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2017-02-07T03:08:38.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13645" PostTypeId="2" ParentId="11870" CreationDate="2017-02-07T09:15:38.497" Score="0" Body="&lt;p&gt;~A micrometer screw gauge has a smaller scale compared to vanier calipers       therefore  measuring short distances&#xA;~a micrometer calipers measures thickness while a vanier calipers measures both thickness and diameter&lt;/p&gt;&#xA;" OwnerUserId="9910" LastActivityDate="2017-02-07T09:15:38.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13647" PostTypeId="1" CreationDate="2017-02-07T11:14:50.530" Score="1" ViewCount="43" Body="&lt;p&gt;How to obtain a constant pressure by a mini water pump to 9 bar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ex given a common engine for the aquarium, how to obtain at the output a pressure that is stable to 9 or 10 bar?&lt;/p&gt;&#xA;" OwnerUserId="9912" LastEditorUserId="1832" LastEditDate="2017-02-07T11:24:26.637" LastActivityDate="2017-02-20T11:32:45.327" Title="Stable liquid pressure" Tags="|mechanical-engineering|pressure|" AnswerCount="1" CommentCount="6" ClosedDate="2017-02-20T21:39:13.220" ContentLicense="CC BY-SA 3.0" />
  <row Id="13648" PostTypeId="2" ParentId="13554" CreationDate="2017-02-07T11:48:01.217" Score="0" Body="&lt;p&gt;Even if they cracked how would those tiles fall out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would take a wide crack for the tile to actually move.  A hairline fracture would still have both the cracked faces leaning on each other.  I feel like you'd notice significant failure before it fell down(or you're throwing heavy things at your ceiling anyways; so tiles falling down is only one of your worries).&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-02-07T11:48:01.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13653" PostTypeId="2" ParentId="13639" CreationDate="2017-02-07T14:49:25.327" Score="1" Body="&lt;p&gt;Basically, a CPLD has far fewer choices in terms of the number of different &lt;em&gt;kinds&lt;/em&gt; of paths through which a signal can flow from one register to the next, which means that the job of analyzing all of the different &lt;em&gt;combinations&lt;/em&gt; of paths is reduced by many orders of magnitude.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2017-02-07T14:49:25.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13654" PostTypeId="1" AcceptedAnswerId="13719" CreationDate="2017-02-07T16:25:09.080" Score="0" ViewCount="96" Body="&lt;p&gt;I'm playing around with a radio receiver on 433MHz and I am receiving signals from a temperature / humidity sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already figured out what bytes represent humidity, but I'm stuck with the temperature as I don't know how it is represented in the signal.&#xA;Here are some example messages I receive:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;C1 | 75.2F | 70%&#xA;0001010001010011010000000100100000000100011011010111000011&#xA;C1 | 24.1C | 68%&#xA;0001010001010100100000000100100000010100010000000001000011&#xA;C1 | 24.1C | 56%&#xA;0001010001010100100000000100100000010011100001010110000011&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The temperature sensor has the following readouts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Temperature&lt;/li&gt;&#xA;&lt;li&gt;Humidity&lt;/li&gt;&#xA;&lt;li&gt;Unit (Celsius / Fahrenheit)&lt;/li&gt;&#xA;&lt;li&gt;Channel (1-8)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I already figured out what the &lt;strong&gt;humidity&lt;/strong&gt; is as well as what represents the &lt;strong&gt;channel&lt;/strong&gt;. I also figured out a &lt;strong&gt;temperature range(?)&lt;/strong&gt; [1 = ..-11 | 2= -1..-10 | 3= 0..15 | 4= 15..30 | 5=31..40]. But I have a hard time figuring out where I can find the exact temperature or the unit (F/C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prepared this public Google Sheet with my findings and raw data so far (sorry I don't know a better way to share on Stackexchange):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://docs.google.com/spreadsheets/d/1q2sk5bK6Bv3gi8ahTLIUM4DaMR7AvnTVPY-NvmGMHoQ/edit?usp=sharing&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://docs.google.com/spreadsheets/d/1q2sk5bK6Bv3gi8ahTLIUM4DaMR7AvnTVPY-NvmGMHoQ/edit?usp=sharing&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vrhIM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vrhIM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(the channel with a zero based index I already figured out is violet, the humidity is in blue, some bits always stay the same hence I made them grey)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is my best course of action now? I have been trying around with all the bits and bytes, adding them together, adding the one to the other, multiplying the temperature times 10 to get a decimal... nothing.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Yeah... Plotting is an idea, but it doesn't really help me:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UE91o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UE91o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Column &lt;strong&gt;I&lt;/strong&gt; seems to follow the temperature in a way (negative temperature/high temperatures and the values just don't relate, look to the end of the graph), but looking at the decimal representation of I can see values such as 7,8,9 and wildly changing temperatures. I am assuming there is some kind of correlation between the temperature range and this byte but I haven't figured out yet how to achieve the calculation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Alright I found something:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w0bdj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w0bdj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature ranges in correlation with I &amp;amp; J seem to be the key. I &amp;amp; J corresponds to a value between 0-255 (I guess) for each range and represents a particular temperature value. The I &amp;amp; J scale seems to be logarithmic looking at the difference between 17,8C (17) and 20,8C (71) and the difference between 20,8C (71) and 22,8C (106). But that's just a guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I somehow need to correlate the range (15 degrees C wide) with the 0-255 index to get a temperature. Crazy.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Converted H &amp;amp; I &amp;amp; J 12 bit number to decimal and get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lnjtn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lnjtn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does look very related, but I can't figure out a formula.&lt;/p&gt;&#xA;" OwnerUserId="9219" LastEditorUserId="9219" LastEditDate="2017-02-11T09:03:16.900" LastActivityDate="2017-02-12T13:36:46.980" Title="How to decode radio thermometer / hygrometer bytes to floats?" Tags="|wireless-communication|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13655" PostTypeId="2" ParentId="13621" CreationDate="2017-02-07T18:50:59.873" Score="10" Body="&lt;p&gt;It all depends on the level of efficiency you seek. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A train, given its size, has a ridiculously small cross section. This small frontal area footprint is being 'pushed' by the inertia of hundreds of tonnes of metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All high speed train have aerodynamic noses, but some will tend to be pedantic about how many thousandths of a percent efficiency they can squeeze out of an aerodynamic profile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a couple of other factors that come into play with the design of the nose and that is the pressure pulse that is generated as a train passes another object, generally another train going in the opposite direction or a tunnel. It's why passengers are told to stay away from the platform edge as fast trains pass by, its not that you'll be blown away, but may sucked towards the train by the low pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two high speed trains passing each other can generate a huge amount of suction between them at the nose, because of the low pressures there, so having a longer nose helps spread this out a bit and reduce the area of minimum pressure. If not the trains could be pulled towards each other, or some instability set up by the rapid sideways pressure pulse. &#xA;The nose of the E5 Series, at 15 metres, is a massive 9 metres longer than the previous incarnation of the bullet train (Shinkansen), the E2 Series. This, according to its designers at JR East, will help eliminate the phenomena of &quot;tunnel boom&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Japan's rail tunnels are somewhat narrower than their European counterparts, so when the Shinkansen enters a tunnel at speeds above 200 kilometres per hour, the sudden increase in air pressure can cause a loud &quot;boom&quot; at the other end of the tunnel. In some cases, such shock waves are thought to have damaged tunnels in Japan, ripping chunks of material from tunnel ceilings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of the front car has evolved gradually to combat this danger, and the striking &quot;Long Nose&quot; design of the E5 Series is the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference &lt;a href=&quot;http://www.jreast.co.jp/e/development/tech/pdf_1/22_28tecrev.pdf&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;,&#xA;&lt;a href=&quot;https://www.newscientist.com/blogs/onepercent/2011/03/japans-long-nosed-bullet-train-1.html&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9789" LastEditorUserId="90" LastEditDate="2017-06-13T11:43:22.113" LastActivityDate="2017-06-13T11:43:22.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13656" PostTypeId="1" AcceptedAnswerId="13682" CreationDate="2017-02-07T18:59:17.533" Score="1" ViewCount="970" Body="&lt;p&gt;As an assignment for my Automotive Engineering study, I have to practice manual FEM calculations. I'm free to choose whichever existing construction, and with that I have to make FEM analyses of both the original and the improved construction. I chose the chassis of the '57 MGA i'm restoring, which is shown below:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J3GWa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J3GWa.jpg&quot; alt=&quot;mga chassis&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose there are several ways to improve the driveability, but I expect increased torsional stiffness to have the most impact. The rear bridge seems rather wobbly to me, but I can't really tell; I haven't driven an MGA before.&lt;br&gt;&#xA;(If you have better ideas how to improve the chassis please do suggest.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be a good approach to improve the torsional stiffness of the chassis?&lt;br&gt;&#xA;With that, I mean what kind of construction/structure would be effective; a triangular structure between the two side beams for instance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Please note:&lt;/strong&gt; it doesn't have to be practically feasible or cost effective. It's just for practicing.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="9920" LastEditDate="2017-02-07T22:26:28.203" LastActivityDate="2017-02-09T15:29:17.467" Title="How could an MGA's frame/chassis best be improved?" Tags="|structural-engineering|finite-element-method|reinforcement|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13657" PostTypeId="2" ParentId="13633" CreationDate="2017-02-07T19:07:15.403" Score="5" Body="&lt;p&gt;First of all, yes the pitch circle is the same for an internal gear as it is for the equivalent external gear. Second, note that indeed the addendum refers to how much the teeth extend inward of the pitch circle for an internal gear (instead of outward, as is the case for external gears). Vice versa for the dedendum. For standard gears, both internal and external, the addendum $a$ and dedendum $d$ are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=m=\frac{1}{D_p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d=1.25m=\frac{1.25}{D_p}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$ is the module (in millimetres) and $D_p$ is the diametral pitch (in inches$^{-1}$). The units of the addendum and dedendum are consistent with that of the module or diametral pitch, depending what you chose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, it looks like you will need to increase the dedendum for the internal gear to $1.25m$, and make sure the addendum is only $m$. This will then prevent these inference issues, and will generally be fine as long as any external gears are not significantly larger than half the diameter of the internal gear they are inside of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k3HeO.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k3HeO.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2017-02-07T19:27:18.530" LastActivityDate="2017-02-07T19:27:18.530" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13658" PostTypeId="1" AcceptedAnswerId="13664" CreationDate="2017-02-07T22:02:25.827" Score="5" ViewCount="998" Body="&lt;p&gt;I am fascinated by construction. My university is building a new facility, and they are using 2 tower cranes in the construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would a contractor choose to use 2 cranes on one construction project? Is it simply a matter of saving time by having the two cranes work simultaneously? What are some other considerations that might be taken into account? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: after checking your answers, I agree that it is most likely that using two cranes in this situation was mostly done to increase productivity/throughput, and the positioning of the cranes depends on the building design and ease of loading materials to be transported by the crane. &lt;img src=&quot;https://i.stack.imgur.com/sSpAw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2N1u9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9922" LastEditorUserId="9922" LastEditDate="2017-02-09T22:40:01.467" LastActivityDate="2021-08-17T14:01:21.103" Title="Why use two tower cranes on one construction site?" Tags="|civil-engineering|construction-management|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13659" PostTypeId="1" CreationDate="2017-02-07T23:33:33.047" Score="1" ViewCount="175" Body="&lt;p&gt;I am measuring BLDC motor vibrations with ADXL345. I use 2 propellers, one perfectly balanced and other bad, which vibrates a lot. The problem is I get similar results from accelerometer with both props. I tried ODRs from 200 to 3200 and results look similar with both props (1000-5000 RPM). I get real time data from MCU which reads raw data from accelerometer, then my program draws a real time chart (amplitude vs time, no processing just raw accelerometer data), so I visually check the results while trying different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be the problem here? I want to be able to detect vibrations on the motor.&lt;/p&gt;&#xA;" OwnerUserId="9325" LastEditorUserId="8055" LastEditDate="2017-02-13T17:16:22.380" LastActivityDate="2021-10-26T11:12:03.507" Title="Measuring motor vibrations" Tags="|electrical-engineering|motors|vibration|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="13660" PostTypeId="2" ParentId="13658" CreationDate="2017-02-07T23:57:10.003" Score="2" Body="&lt;p&gt;I'd say it's most likely a time saving operation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure a project manager looked at the situation and decided it was beneficial to the schedule and productivity to have a second crane operating as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the area and height of the building there's only so much you can do with 1 crane; everything takes time.  If you can run two cranes at once and keep them both busy it's bound to benefit you from time savings.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-02-07T23:57:10.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13661" PostTypeId="2" ParentId="2478" CreationDate="2017-02-08T00:52:15.473" Score="0" Body="&lt;p&gt;There are other approaches. HVOF, cold-spray, etc. can build up very thick coatings of a large number of materials (such as titanium, platinum, aluminum, stainless, plastics, etc) on almost any substrate - including paper, plastics, ceramics, and metals. The term coating is not expressive enough as these processes are also used to deposit tool steel to repair worn bearing surfaces, which are subsequently machined. Such techniques are not anisotropic as they rely on supersonic ballistic splatting of extremely fine powders.&lt;/p&gt;&#xA;" OwnerUserId="9923" LastActivityDate="2017-02-08T00:52:15.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13662" PostTypeId="1" CreationDate="2017-02-08T03:41:42.417" Score="1" ViewCount="111" Body="&lt;p&gt;If the material is manufactured by 3D printing (laser sintering) then what will be change in its mechanical properties than the properties by casting and machined part? What would be appropriate failure criteria for it?&lt;/p&gt;&#xA;" OwnerUserId="9925" LastActivityDate="2019-12-17T21:01:35.407" Title="mechanical Strength of 3D printed part" Tags="|materials|mechanical-failure|failure-analysis|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13663" PostTypeId="1" CreationDate="2017-02-08T04:53:37.973" Score="5" ViewCount="535" Body="&lt;p&gt;I'm assuming here that all materials will have a change in elasticity due to thermal expansion/contraction and a resultant reduction/increase in the strength of the bonds of the atoms leading to a varying young's modulus. But how significant is it (at normal non-extreme &lt;strong&gt;ambient temperatures&lt;/strong&gt; -  maybe from -10°C to 50°C)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone know's of any research into this. Or even a table of values like the following, that would be very useful: &lt;a href=&quot;http://www.engineeringtoolbox.com/young-modulus-d_773.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/young-modulus-d_773.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I am asking is I am interested in Structural Health Monitoring, and I've read a lot of researchers claim that temperature has a major impact upon structural stiffness due to changes in material stiffness. I'm yet to find substantial evidence of this. &lt;/p&gt;&#xA;" OwnerUserId="9569" LastEditorUserId="1832" LastEditDate="2017-02-08T10:26:42.097" LastActivityDate="2021-01-26T06:02:15.197" Title="How much does the elasticity (Young's Modulus) of concrete and asphalt change with ambient temperature?" Tags="|structural-engineering|materials|concrete|stiffness|elastic-modulus|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13664" PostTypeId="2" ParentId="13658" CreationDate="2017-02-08T05:08:02.870" Score="7" Body="&lt;p&gt;Tower cranes are a very significant cost on any construction site, and so builders will usually only use as many cranes as is necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I work for a construction scheduling company, and the one of the biggest factors for deciding on the number of cranes and their location is access. Obviously the crane needs to be able to access all parts of the building during construction, but the placement of the crane is also often determined by access to roads and materials on the ground.  Other big restrictions such as site boundaries and roads also often come into play in urban areas. If the access limitations mean you need two cranes, then two cranes will be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, as you noted, is time. Time is money on a site, and if having a second crane can save you a couple of months off your finish date, then it is usually a cost effective idea. Other times, there may be a hard deadline for a project finish, in these cases more than one crane may be used to accelerate the program at the request of the client despite it being more expensive. &lt;/p&gt;&#xA;" OwnerUserId="9569" LastActivityDate="2017-02-08T05:08:02.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13665" PostTypeId="1" AcceptedAnswerId="14179" CreationDate="2017-02-08T07:55:57.670" Score="2" ViewCount="351" Body="&lt;p&gt;I am planning on buying a kitchen stand mixer with planetary action.&#xA;&lt;br/&gt;One important factor to consider is the type of gears the mixer uses, they can be plastic, composite or metal.&#xA;&lt;br/&gt;As far as I know metal gears are the best and plastic the worst in terms of durability and handling stress.&#xA;&lt;br/&gt;However I don't know &lt;em&gt;what&lt;/em&gt; are composite gears and how they fare, compared to metal or plastic.&lt;/p&gt;&#xA;" OwnerUserId="9927" LastActivityDate="2017-03-10T10:23:34.770" Title="Composite vs metal gearing" Tags="|materials|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13666" PostTypeId="1" AcceptedAnswerId="13667" CreationDate="2017-02-08T08:22:38.177" Score="-1" ViewCount="148" Body="&lt;p&gt;Ohms law defines R = V / I, so according to this equation, when resistance increases, voltage and current decreases.&#xA;Now i have two questions.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Ain't there somekind of &quot;pressure&quot; buildup in front of a resistor? because the resistor limits the current flow, I would expect something like a &quot;traffic jam&quot;.&lt;/li&gt;&#xA;&lt;li&gt;How is it that a resistor reduces the voltage? Is it converted to heat by electrons &quot;bumping&quot; into more massive particles or does something else happens to the voltage?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Wasabi&lt;/em&gt;:&#xA;Sorry if this question doesn't have the correct tag, didn't know which tag to give. If you have any suggestions for tags i'll put that in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;hazzey&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to visualize for myself how Volt\Amp\Resistance work togather. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that Voltage is a measurement of potential energy expressed in Jouls / Coulomb and the Current is the number of Coulombs per second passing a certain point, the only part that isn't quite clear to me is the resistance. How resistance physically works to reduce current and in the process uses some energy (voltage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand the answer correctly, the following is true.&#xA;The voltage that is used by a resistor is always converted to heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I have a circuit with a single resistor of 1 Ohms with 1 Amp of current, the resulting 1 volt is converted to heat?&lt;/p&gt;&#xA;" OwnerUserId="9929" LastEditorUserId="9929" LastEditDate="2017-02-08T14:58:37.190" LastActivityDate="2017-02-08T14:58:37.190" Title="What happens to the voltage?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2017-02-09T17:51:56.110" ContentLicense="CC BY-SA 3.0" />
  <row Id="13667" PostTypeId="2" ParentId="13666" CreationDate="2017-02-08T08:29:21.600" Score="1" Body="&lt;p&gt;No. If resistance increases, voltage increases and current decreases. The extra energy is spent thermally, which is why the resistor gets hot when current flows through it, and the higher the resistance, the hotter it becomes for the same current. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to remember that voltage is a differential measurement, like pressure or altitude: you always measure a voltage relative to something or a reference (like earth). Current, on the other hand, is measured when flowing through something, like flow rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good analogy would be water flowing through a dam or similar flow restriction. The more you restrict the water flow (i.e. the higher the resistance becomes), the lower the water flow through the dam, but the higher the pressure upstream of the dam.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2017-02-08T08:29:21.600" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13668" PostTypeId="2" ParentId="13665" CreationDate="2017-02-08T08:47:26.713" Score="0" Body="&lt;p&gt;I am not sure if it is a thing to buy. There is a composite made of thermoset plastic (hard plastic, e.g. epoxy resin...) with fibers embedded in it to give it more strength. I have personally had such a gear in my hands and it's hard I can tell you that.&lt;/p&gt;&#xA;" OwnerUserId="9911" LastActivityDate="2017-02-08T08:47:26.713" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13670" PostTypeId="1" CreationDate="2017-02-08T00:43:35.107" Score="1" ViewCount="2641" Body="&lt;p&gt;I'm building a robot that will weigh less than 75kg, and be capable of carrying people, so I need to select motors that are powerful enough to move about 166kg at a speed of 12mph, with standard 18in golf-cart wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 20s acceleration was arbitrarily chosen because it seemed reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are my calculations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;200 lbs person is 91 kg.&#xA;Total Robot Weight: 75 kg + 91 kg = 166 kg&#xA;&#xA;Calculate Required Torque:&#xA;&#xA;12mph / 20s is 0.268224 m/s^2&#xA;0.268224 m/s^2 * 166 kg is 44.5N of Force&#xA;44.5N * 18in (Wheel Diameter) is 20.3 J (aka N*m) of Torque&#xA;&#xA;Potential Motor:&#xA;&#xA;ZY6812 250W24V (Razor E300 Motor)&#xA;Specs: 3350RPM No Load, 2750RPM Load, 0.9 N*m, 8.9A (15.4A Stall)&#xA;&#xA;Motor Torque:&#xA;&#xA;4x motors = 1000W @ 3.6 N*m of Torque total.&#xA;1:8 Gear Reduction -&amp;gt; 22.8 N*m Torque @ 418.75 RPM.&#xA;&#xA;Motor Speed:&#xA;&#xA;12mph with 18in wheels = 224 RPM Required Speed.&#xA;250W / 20.3J = 12.32 Hz = 739.2 RPM, so speed is possible with desired torque.&#xA;22.8 &amp;gt;= 20.3 and 418.75 &amp;gt;= 224, the motors can move 166 kg at 12mph or more and accelerate in 20s or less.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Supposedly, 4 of these 250W motors should be enough to carry the weight at twice the speed, given a 1:8 gear ratio. Are these calculations valid, or did I mess something up horribly?&lt;/p&gt;&#xA;" OwnerDisplayName="Pecacheu" LastActivityDate="2017-02-08T13:02:24.633" Title="Calculating Required Motor Power for Robot" Tags="|robotics|torque|" AnswerCount="0" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="13671" PostTypeId="2" ParentId="13654" CreationDate="2017-02-08T13:22:22.813" Score="1" Body="&lt;p&gt;This is not a full answer but I can't comment yet due to too less reputation, sorry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I+J seems to have a strong correlation with the temperature. I did a scatter plot of I+J vs. Temp and I can see at least one strong linear relationship (did not do the fit yet, though). &lt;a href=&quot;https://i.stack.imgur.com/ZcAu0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZcAu0.png&quot; alt=&quot;See plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also see some outliers but they seem to be parallel. You could think of a sawtooth waveform. This implies different measurement ranges or that more bits are needed to see the complete result.&lt;/p&gt;&#xA;" OwnerUserId="9819" LastEditorUserId="9819" LastEditDate="2017-02-08T13:46:14.830" LastActivityDate="2017-02-08T13:46:14.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13672" PostTypeId="1" CreationDate="2017-02-08T14:29:16.287" Score="2" ViewCount="138" Body="&lt;p&gt;I am working on a design for a small air pump that uses a single piston.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When choosing a piston I have two parameters diameter and stroke. So, I can make it fatter and shorter, or skinnier and longer, for the given volume of air that I want to move per stroke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any rules of thumb concerning how to make this choice?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="2819" LastEditDate="2017-02-09T19:21:47.297" LastActivityDate="2018-03-07T02:48:23.603" Title="Rules of thumb for sizing an air pump piston?" Tags="|mechanical-engineering|pneumatic|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13673" PostTypeId="2" ParentId="13659" CreationDate="2017-02-08T15:19:27.407" Score="3" Body="&lt;p&gt;Typically, motor vibration analyses and monitoring are done in the frequency domain, taking the time data and processing it into the frequency domain via subsequent FFT's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, the approach would go as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine the frequencies of interest. In this case, you'll probably be interested primarily in the rotational frequency and harmonics, so ~16.7Hz - ~83Hz. This is where you would anticipate finding peak amplitudes in the frequency domian.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Generate a baseline frequency/amplitude profile. Typically this would be done with the ideal case, for you, this is probably the balanced prop. and identify the frequencies and amplitudes that correspond to the frequencies of interest. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Once that's done, the effect of any change in vibration will show up as a shift in the amplitude and frequency of the corresponding peak.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-02-08T15:19:27.407" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13674" PostTypeId="2" ParentId="13672" CreationDate="2017-02-08T18:52:13.913" Score="1" Body="&lt;p&gt;Depends on a several factors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stroke length should just be application dependent.  You should know how far you require it to actuate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diameter is a bit more complicated.  Your $Force = \frac{Pressure}{Area}$ so if you know the required actuation force, you just need one more parameter.  If space is an issue but air pressure isn't, then you would want smaller diameter (area) and higher pressure.  If you want lower operating pressures and size isn't as much of a constraint; you would want a larger diameter/area to minimize pressure required.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-02-08T18:52:13.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13675" PostTypeId="2" ParentId="13656" CreationDate="2017-02-08T20:36:44.400" Score="2" Body="&lt;p&gt;There are really two aspects to this: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Will increasing roll stiffness improve handling to an extent which offsets any increase in weight by adding extra chassis members ?&lt;/li&gt;&#xA;&lt;li&gt;If yes, what is the best strategy for improving the stiffness of the chassis? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The issue here is that this is a fairly simple ladder chassis so we are really looking at creating some sort of ladder/spaceframe hybrid. However we also need to consider the fact that the rear suspension is pretty basic in the form of longitudinal leaf springs so there is no guarantee that stiffer is better without identifying specific characteristics which you need to improve. Eg increasing rear roll stiffness isn't necessarily better unless this is identified as a limiting factor on performance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some ways it might be a more productive exercise to investigate where you can lose weight by removing material from the existing chassis by identifying areas where it is under-stressed. Apart from anything else this should &lt;em&gt;definitely&lt;/em&gt; give some easily quantifiable benefit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of an assignment it may also make more sense to demonstrate how a space-frame chassis could give the same torsional stiffness for less mass. Again this gives your project a much more well defined objective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if you manage to find a way to make the existing chassis 10% stiffer it is then very hard for you to show what real world performance benefits this gives without actually carrying out the modifications and testing them. On the other hand if you can make it 10% lighter for the same stiffness that is a very obvious benefit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also add that having a project with a well defined strategy for improving a design is fundamentally better than just changing some arbitrary design parameter without any clear idea of how it will improve performance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So project titles like 'A worked example of the weight advantages of a space-frame vs ladder chassis' or 'Reducing the weight of an MGA chassis' will make much more sense to an engineering professor than 'Showing that adding more mass to a structure can make it stiffer'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg say you make it 10% stiffer and 10% heavier... is that better? Unless you can &lt;em&gt;prove&lt;/em&gt; it is better the whole exercise is a bit pointless. If it was an experimental exercise things would be a bit different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point here is that in any scientific and engineering experiment you need to be able to reach some conclusion. If your intended approach didn't work that is fine as long as you can say that it didn't work and why what is bad is when there is no way to tell within the scope of the project whether it was worth doing or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg 'I investigated whether it was practically possible to lighten the chassis of an MGA using the following strategies but none of the approaches I used were effective' is useful information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However 'I made the chassis of an MGA 10% stiffer by increasing its mass by 10%' doesn't really tell you anything useful, unless you can add further information how this improves its performance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside there is also the fact that for classic cars (in the UK at least)  any marginal increase in performance has much less value attached to it than preserving its authenticity and originality.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="1832" LastEditDate="2017-02-08T21:49:09.037" LastActivityDate="2017-02-08T21:49:09.037" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13677" PostTypeId="1" CreationDate="2017-02-09T08:01:14.973" Score="1" ViewCount="64" Body="&lt;p&gt;I have done experiments on a grinding machine and have collected the data of grinding wheel profile (using a laser sensor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The raw data consists of info on grinding wheel profile, waviness and &#xA;roughness (because of abrasive grains). I would like to process the collected raw data. I want to filter out the high frequency and low frequency data. (I want to apply a simple band pass filter). One cycle data is about 200,000 data units and i have them in an excel file (csv data) and would like &#xA;to perform digital filtering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I filter out the waviness data separately and analyze, from this? &lt;/p&gt;&#xA;" OwnerUserId="9949" LastEditorUserId="90" LastEditDate="2017-02-09T11:36:27.347" LastActivityDate="2017-02-12T23:45:26.043" Title="Measurement of grinding wheel waviness" Tags="|electronic-filters|grinding|" AnswerCount="0" CommentCount="7" ClosedDate="2017-02-13T13:02:22.583" ContentLicense="CC BY-SA 3.0" />
  <row Id="13679" PostTypeId="1" AcceptedAnswerId="13681" CreationDate="2017-02-09T10:29:48.567" Score="2" ViewCount="253" Body="&lt;p&gt;Hi there (from Europe),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am completing what should be a relatively quick assessment for the client, but I have started to second guess my self and would like some reassurance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially the job involves removing a portion of an existing 2.6m x 1.22m concrete slab in order to install a manhole (1.2m x 0.75m). I've taken the fairly rudimentary (and conservative imo) approach of simply reducing the existing slabs moment capacity by reducing it's width. For example, for assessing the moment capacity of the current slab I took a width of 1.22m, but it's reduced capacity would be based on a width of 1.22-0.75=0.47m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an original moment capacity (89kNm), and reduced moment capacity(37kNm), we know the difference between these two (&lt;strong&gt;52kNm&lt;/strong&gt;). My idea was to simply take two Square Hollow Sections and attach them to the underside of the slab, on either side of the manhole, and size them such that each was capable of holding half the difference in moment, &lt;strong&gt;26kNm&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gaBkq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gaBkq.png&quot; alt=&quot;Sketch of situation and proposed solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fairly simple approach but the more I think about it the more I'm beginning to reconsider and instead consider assessing the steel and concrete as a single composite section to Eurocode 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a small scheme with a low budget and I lack experience of Eurocode 4 to quickly assess this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is thus:&#xA;&lt;strong&gt;Is my approach conservative enough to safely size steel members or is there something fundamental that I am missing in composite design that makes my solution inaccurate and dangerous?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="3093" LastActivityDate="2017-02-10T08:22:12.067" Title="Assessment of a concrete beam being strengthened by an additional steel member" Tags="|civil-engineering|steel|concrete|composite|eurocodes|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13681" PostTypeId="2" ParentId="13679" CreationDate="2017-02-09T14:11:51.507" Score="2" Body="&lt;p&gt;It isn't clear from your sketch whether the end supports are fixed or pinned, so I'm going to assume they were pinned. However, this is going to be a qualitative answer, so that doesn't actually matter too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming pinned supports, the original slab was therefore basically a simply-supported beam, with no moment at the ends and positive longitudinal moment at the midspan equal to $\frac{qL^2}{8}$. Transversal moment would be effectively null.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, with your suggested reinforcement, you will be completely changing the slabs structural arrangement. Indeed, it would no longer be &quot;the slab&quot;, but instead eight different slabs (for manual analytical purposes):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ViyNI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ViyNI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, slabs S1, S3, S6 and S8 are effectively equivalent, as is S2 equal to S7, and S4 to S5. So now, when discussing the different slabs, I'll only mention slabs S1, S2 and S4, each representing all their equivalent slabs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, slab S1 needs to be calculated like a fixed-pinned, fixed-free slab; S2 as a fixed-fixed, pinned-free slab and (out of an abundance of caution) S4 as both a pinned-free, fixed-fixed slab and as a pinned-free, pinned-pinned slab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The division of the unmodified strips of the original slab (S1-S2-S3 and S6-S7-S8) into three slabs may seem a bit odd. After all, this method would imply in a negative bending moment between S1 and S2 even though they are actually the same slab with no beam beneath them. But that is in fact correct, since S4 will now resist any rotation around the S1-S2 interface. Meanwhile, between S1 and S4 there is now the reinforcing beam, which adds stiffness to that interface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you want to be very conservative, you should calculate the slab with all its loading as if it were originally constructed with that hole and see what is the required steel reinforcement (rebar) in both the longitudinal and transversal directions. Check to see whether the built slab satisfies all of these requirements. If it does, you're fine. If it doesn't, then the slab itself may need to be reinforced, and simply adding steel beams may not be enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, my concern is with the negative reinforcement that will be necessary in the slab. The positive will probably be fine since you are effectively reducing the effective span of each span area and whatever load is transferred from S4 and S5 will go to the beams and not the neighboring slabs. The original slab was probably not designed with any significant negative reinforcement, so I would not be surprised if a problem occurs there, especially on the S1-S4 edge, where you're dealing with negative transversal reinforcement which is probably almost nonexistent in the original slab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An argument could be made that considering the entirety of the load when calculating the reinforcement would be excessive. After all, some portion of the self-weight and dead load will probably be unmolested by the hole (for instance, whichever part of the load which already goes from S1 to S3) and therefore shouldn't be considered when calculating the new configuration. This would be especially important when considering the effect of the beams, since if the slab doesn't redistribute much load to them, they won't be of much use. However, the hole is so massive (my drawing is more to scale than your sketch) that considering a full redistribution of loads seems entirely reasonable to me. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another question is how to fix the beams to the slab in such a way as to certify that they will indeed work in tandem. As &lt;a href=&quot;https://engineering.stackexchange.com/questions/13679/assessment-of-a-concrete-beam-being-strengthened-by-an-additional-steel-member#comment24519_13679&quot;&gt;@AndyT commented&lt;/a&gt;, this is usually done by taking special measures when the slab is poured. Obviously, in this case that is not possible. My suggestion would therefore be to use anchor bolts. This, however, would not be possible with SHS profiles since you wouldn't be able to access the bolts along the beam's span. You'd therefore need to change to L-, T- or I-shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as described above, you may need to add some negative reinforcement to your slab. If that is the case, then you can make a slightly larger intervention:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Get SHS sections with shear studs.&lt;/li&gt;&#xA;&lt;li&gt;Drill two lines of large holes in the slab along each beam's span.&lt;/li&gt;&#xA;&lt;li&gt;Place the SHS sections under the slab, with adequate space between the beam and the slab for the studs. Also add some shear reinforcement passing through the holes.&lt;/li&gt;&#xA;&lt;li&gt;When concreting the slab with the negative reinforcement, also concrete the joint between the slab and the beam. Also add some rebar going through the shear reinforcement placed in the previous step. Given the small access, special treatment may be necessary for the concrete.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/08Pkl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/08Pkl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-02-09T14:11:51.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13682" PostTypeId="2" ParentId="13656" CreationDate="2017-02-09T15:29:17.467" Score="1" Body="&lt;p&gt;The overall stiffness of the frame is determined by the stiffness of the side beams augmented primarily by the three tubular cross members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting academic exercise would be to evaluate how changing the diameter of the cross members affects the overall stiffness.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2017-02-09T15:29:17.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13683" PostTypeId="2" ParentId="13672" CreationDate="2017-02-09T19:29:07.210" Score="0" Body="&lt;p&gt;When making a piston pump, it will depend on the torque and RPM of the motor you have selected, head pressure the pump is pumping against, and flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller diameter means less volume per stroke and less motor torque required.&#xA;Shorter stroke means less volume per stroke, and a shorter camshaft lever arm resulting in less torque required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will not be able to help much beyond those concepts without nailing down some variables like motor rpm, motor torque, flowrate, head pressure, etc.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-02-09T19:29:07.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13685" PostTypeId="1" CreationDate="2017-02-10T00:44:58.043" Score="0" ViewCount="2678" Body="&lt;p&gt;It has been a while since I had thermodynamics and am attempting to solve this problem for work. I need to figure out the other two mass flow rates. I worked it out with specific heats, but don't really trust it. Seems like there should be a simple way to do it with enthalpy; the process just escapes me. Thanks for the help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I9rFk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I9rFk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2781" LastEditDate="2017-02-10T16:08:32.850" LastActivityDate="2017-02-10T16:08:32.850" Title="How to calculate saturated steam flow rates?" Tags="|thermodynamics|steam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13686" PostTypeId="2" ParentId="13679" CreationDate="2017-02-10T00:53:49.810" Score="0" Body="&lt;p&gt;Regarding composite action, it will be minimal  therefore you are doing well to ignore it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the design is adequate (except maybe for point 4 below)but i want to mention some missing elements from your analysis:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You are restoring the slab strength for sagging moments but not for hogging ones.&lt;/li&gt;&#xA;&lt;li&gt;It is important to know when the shs members are installed. If after the hole is opened, then a verification is needed for the reduced slab under the dead loads plus any live before the shs installation.&lt;/li&gt;&#xA;&lt;li&gt;Verification of deflections of the reduced slab.&lt;/li&gt;&#xA;&lt;li&gt;The hole surface should probably get strengthened, particularly its corners, due to stress concentrations.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9620" LastEditorUserId="9620" LastEditDate="2017-02-10T08:22:12.067" LastActivityDate="2017-02-10T08:22:12.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13688" PostTypeId="2" ParentId="13685" CreationDate="2017-02-10T03:06:09.303" Score="2" Body="&lt;p&gt;With steam in general and absolutely with any phase changes and pressure changes, you need to use enthalpies instead of specific heat. If there are phase changes, specific heat neglects the enthalpy of vaporization. Even if there aren't, the specific heat of steam changes too much with temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just need to do mass and energy balances (isn't it always mass and energy balances?). The mass balance is $\dot m_h + \dot m_c = \dot m_l$ and the energy balance is $\dot m_h h_h + \dot m_c h_c = \dot m_l h_l$. The enthalpies should be available in any thermodynamics textbook or various software packages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll let you solve the whole thing, but if you have more issues, leave a comment or update the question.&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2017-02-10T03:06:09.303" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13692" PostTypeId="1" CreationDate="2017-02-10T16:13:45.993" Score="1" ViewCount="590" Body="&lt;p&gt;I am making a project where i need to put a water valve below the tank and then need to open it automatically but the problem is that already available brass electric valves aren't good enough to handle the 120C temp besides i heard that brass isnt good for something that will be consumed by Humans later.&#xA;And that overtime brass get mixed up in material and it will be consumed.&#xA;what's the best solution for this problem ?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oUIEs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oUIEs.jpg&quot; alt=&quot;i was thinking maybe is there anyway that i can put a motor or something on this silver valve and remove this handle and then open it with that motor or something&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i was thinking maybe is there anyway that i can put a motor or something on this steel valve and remove this handle and then open it with that motor or something&lt;/p&gt;&#xA;" OwnerUserId="9965" LastActivityDate="2017-02-10T16:44:18.667" Title="Water valve for over 120C temperature" Tags="|mechanical-engineering|electrical-engineering|valves|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13693" PostTypeId="2" ParentId="13692" CreationDate="2017-02-10T16:29:27.947" Score="0" Body="&lt;p&gt;120C water is steam at standard conditions.  You should look into steam valves or vales designed for a steam/water mixture.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modifying the valve opening mechanism won't make the actual valve any stronger, so I don't suggest that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK a lot of residential pipe fittings are brass, though it may only be safe in specific temperature ranges.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-02-10T16:29:27.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13694" PostTypeId="2" ParentId="13692" CreationDate="2017-02-10T16:33:25.280" Score="1" Body="&lt;p&gt;I suggest an air actuated stainless steel ball valve. There are literally tens of thousands(if not more) of available options, including kits that will refit an existing manual ball valve to be air actuated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My preferred source for this type of valve and accessorie is Parker. Start &lt;a href=&quot;http://ph.parker.com/us/17054/en/industrial-ball-valves-stainless-steel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; and you can use their configurator to get exactly what you want. If you can't find it, call them, they can make it. Sometimes that can get expensive though. There are other vendors and as with many things, you can order such things from &lt;a href=&quot;https://www.mcmaster.com/#air-actuated-ball-valves/=16amu4s&quot; rel=&quot;nofollow noreferrer&quot;&gt;McMaster-Carr&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solenoid valves are another option, but they are typically limited in tubing size. Again, McMaster-Carr has a wide &lt;a href=&quot;https://www.mcmaster.com/#standard-solenoid-valves/=16amv6v&quot; rel=&quot;nofollow noreferrer&quot;&gt;selection&lt;/a&gt; and a tool that lets you search by your particular requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the concern over brass contaminating water, it's used all over the place in domestic faucets and fittings. That said, brass does contain small amounts of lead that &lt;a href=&quot;https://www.sciencenews.org/blog/science-public/newfound-water-risk-lead-leaching-valves&quot; rel=&quot;nofollow noreferrer&quot;&gt;can leach into water even when the material is compliant&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stainless steel doesn't have that problem, but is more expensive.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2017-02-10T16:44:18.667" LastActivityDate="2017-02-10T16:44:18.667" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13699" PostTypeId="2" ParentId="13047" CreationDate="2017-02-11T00:37:41.977" Score="2" Body="&lt;p&gt;Here is what you need to know when selecting a grinding wheel. My experience is in manufacturing research and I do not know much about the specifics of knife making, but there are some general rules you can look at when selecting a grinding wheel. Always check the specs provided by suppliers for specific information on the correct wheel to order from them.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR: for precision grinding/finishing of a stainless steel workpiece, you should use a cBN abrasive type, metal or vitrified bond type, small (fine) grain size, soft grade wheel.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Grinding wheel specification&lt;/h2&gt;&#xA;&lt;p&gt;In order to choose a grinding wheel you need to understand what the wheel markings mean. Superabrasive grinding wheels are often specified using the ANSI Standard B74.13-1977 marking system:&lt;/p&gt;&#xA;&lt;p&gt;XX - B - 150 - P - YY - V - ZZ - 3&lt;/p&gt;&#xA;&lt;p&gt;The parameters are, in order:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;XX, Manufacturer's prefix: not important&lt;/li&gt;&#xA;&lt;li&gt;B, Abrasive type: D = diamond, B = Cubic boron nitride (cBN)&lt;/li&gt;&#xA;&lt;li&gt;150, Grain size: 8 to 24 = Coarse, 30 to 60 = Medium, 70 to 180 = Fine, 220 to 600 = Very Fine&lt;/li&gt;&#xA;&lt;li&gt;P, Wheel grade: Value from A to Z. A = soft, M = medium, Z = hard&lt;/li&gt;&#xA;&lt;li&gt;YY, Concentration: Varies between manufacturers, check the datasheet for whatever company you are looking at.&lt;/li&gt;&#xA;&lt;li&gt;V, Bond type: B = Resin, M = Metal, V = Vitrified&lt;/li&gt;&#xA;&lt;li&gt;ZZ, Bond modification: Varies between manufacturers&lt;/li&gt;&#xA;&lt;li&gt;3, Depth of abrasive: Working depth of the abrasive in inches or mm, check the manufacturer's datasheet (unlike conventional wheels, subperabrasives are often only applied in a thin layer to the outside of a wheel rim because they are rather expensive).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The above was somewhat paraphrased from [1].&lt;/p&gt;&#xA;&lt;p&gt;I will now comment in more detail on the parameters that you are interested in.&lt;/p&gt;&#xA;&lt;h2&gt;Bond types&lt;/h2&gt;&#xA;&lt;p&gt;The difference between resin bonded and vitrified bonded grinding wheels is the material used to bond the grains. Different bond types have different hardness, toughness and temperature resistance properties.&lt;/p&gt;&#xA;&lt;p&gt;In general:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;[Vitrified bonds consist] chiefly of baked clay and ceramic materials. Most grinding wheels in common use are vitrified bonded wheels. They are strong and rigid, resistant to elevated temperatures, and relatively unaffected by water and oil that might be used in grinding fluids.&amp;quot; [1]&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;[Resinoid bonds consist] of various thermosetting resin materials, such as phenol-formaldehyde. They have very high strength and are used for rough grinding and cutoff operations.&amp;quot; [1]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Vitrified bonds are better for precise finishing operations than resinoid bonds. I work in a manufacturing research lab and most of the wheels we use are vitrified bond grinding wheels, for the reasons stated above.&lt;/p&gt;&#xA;&lt;p&gt;There is, however, another option. Often cBN and diamond superabrasive wheels use &lt;em&gt;metallic&lt;/em&gt; bonds (usually bronze). Grains are attached in a very thin layer to the metal substrate with metal bonded wheels, which tends to save significant amounts of abrasive and therefore money.&lt;/p&gt;&#xA;&lt;h2&gt;Abrasive types&lt;/h2&gt;&#xA;&lt;p&gt;Good idea moving to superabrasives for your application. Good quality knives to my knowledge are made of tool grade stainless steels [2], and that's what cBN (cubic boron nitride) superabrasives are good for [1].&lt;/p&gt;&#xA;&lt;p&gt;I've ground stainless steel with conventional aluminum oxide wheels. They work alright, but it takes a long time with lots of re-dressing of the wheel. Diamond superabrasive would also work, but would probably be overkill for your application. Silicon carbide can work for stainless steels but is &lt;em&gt;not&lt;/em&gt; compatible with most other steels due to chemical reactions between the wheel and workpiece.&lt;/p&gt;&#xA;&lt;h2&gt;Other things you should note&lt;/h2&gt;&#xA;&lt;p&gt;For a precision grinding application, assuming that the abrasive you have chosen is compatible with the workpiece material, the most important parameter that you need to select is the grain size.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Small grit sizes produce better ﬁnishes, whereas larger grain sizes permit larger material removal rates. Thus, a choice must be made between these two objectives when selecting abrasive grain size. The selection of grit size also depends to some extent on the hardness of the work material. Harder work materials require smaller grain sizes to cut effectively, whereas softer materials require larger grit sizes. [1]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So because you are cutting a relatively hard material (tool grade stainless steel) and you want a fine surface finish you will want to select a small (fine) grain size. Note that grit/grain sizes are numbered &amp;quot;backwards&amp;quot; to the way you might expect: small grains have high grit numbers and large grains have small grit numbers.&lt;/p&gt;&#xA;&lt;p&gt;You also want a soft wheel grade. Soft wheels are better for hard materials and low material removal rates than hard wheels. However, the tradeoff is that hard wheels are better for removing lots of material quickly (for a grinding process anyway).&lt;/p&gt;&#xA;&lt;p&gt;[1] Mikell P. Groover, &lt;em&gt;Fundamentals of Modern Manufacturing, 5th Ed.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;[2] &lt;a href=&quot;http://www.buckknives.com/about-knives/choosing-the-right-knife/#types-of-steel&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.buckknives.com/about-knives/choosing-the-right-knife/#types-of-steel&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7430" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-02-14T20:27:11.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13701" PostTypeId="1" CreationDate="2017-02-11T00:57:19.773" Score="0" ViewCount="76" Body="&lt;p&gt;I have sections of steel tube I want to set up for use as zip line anchors and was wondering what the allowable load would be on the vertically mounted tubes. I read that a zip line could have momentary tensions of up to 8,000 lbs put on the anchoring structures. The tubes are rolled, welded steel, 0.83 inch wall thickness (14 ga.). The end anchor point I was planning on was 10' long buried 2 feet in the ground in an 18&quot; diameter concrete pier. The anchor point for the line would be 4 inches from the top so 7 ft 8 in. above ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The start side anchor post would be the same steel tube but buried 3 feet down in 15&quot; a pier. The anchor point will either be at 9 feet above ground or 15.5 ft. Depending on my ability to brake the trolley. The start side tube will have another one directly behind it at 22&quot; center to center that I could hook up a guy wire between to beef it up since there will obviously be more deflection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your thoughts are muchappreciated!&lt;/p&gt;&#xA;" OwnerUserId="9975" LastActivityDate="2017-02-11T12:57:48.683" Title="Max horizontal momentary load of vertical steel tube" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2017-02-12T17:18:58.680" ContentLicense="CC BY-SA 3.0" />
  <row Id="13703" PostTypeId="1" AcceptedAnswerId="13736" CreationDate="2017-02-11T09:06:43.680" Score="0" ViewCount="158" Body="&lt;p&gt;I'm a software engineer whose main concentration has been data science with some minor electrical engineering on the side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to delve into some mechanical engineering by building a proper mini vending machine but am having trouble knowing which materials to use as a consequence of not knowing where to find good documentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is corrosion resistance important for a vending machine? Do I use aluminum or steel? What factors would I need to take into account if it were a 2 meters tall, 1 meter deep machine, for example? Or another size?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apologies if this is mistagged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="9982" LastEditorUserId="9982" LastEditDate="2017-02-11T09:30:00.210" LastActivityDate="2017-02-13T11:53:47.157" Title="What factors are needed to be taken into account when building a vending machine?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="8" ClosedDate="2017-02-13T16:34:35.507" ContentLicense="CC BY-SA 3.0" />
  <row Id="13704" PostTypeId="1" CreationDate="2017-02-11T09:13:53.693" Score="2" ViewCount="471" Body="&lt;p&gt;I recently asked a friend how I should recycle my used deep frying oil because I just don't have enough oil to justify taking my car, drive 20 minutes away to the waste collection site for few liters of oil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my friend advised me to just throw it within the garbage. The oil has to be burn anyway so it will be burned at the recycle station and generate electricity and heat from it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other solution, is to simply throw the oil in the sink. When it reaches the wastewater treatment plant, settling basins will separate the solid, the liquid and the floating oil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I propose here three solutions, and I would like to know which one is the most adapted for someone that want to eliminate about 5 liters used cooking oil per year. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Store it in the basement then go to the waste collection site located 20 minutes away by car and get rid of the oil once a Year.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Save that storing space by throwing 1 liter within the regular garbage 5 times a year in a sealed container. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Throw the used oil in the toilet and throw away the shrank empty plastic container in the prepaid garbage bags (I have to pay 3$ per 35l bag).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These propositions have to come with goals: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Most energetically viable solution&lt;/li&gt;&#xA;&lt;li&gt;Less environment impact&lt;/li&gt;&#xA;&lt;li&gt;Most convenient solution &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2070" LastActivityDate="2017-02-13T19:38:17.400" Title="Is it better to throw used domestic cooking oil in toilet or within the garbage?" Tags="|environmental-engineering|waste-water-treatment|waste-disposal|recycling|" AnswerCount="2" CommentCount="4" ClosedDate="2017-02-13T19:43:41.760" ContentLicense="CC BY-SA 3.0" />
  <row Id="13706" PostTypeId="2" ParentId="13701" CreationDate="2017-02-11T12:57:48.683" Score="1" Body="&lt;p&gt;Although I agree with Wasabi that this is a very broad question, here is the 10,000 ft view of the process&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Draw a free body diagram of the tube.&lt;/li&gt;&#xA;&lt;li&gt;Solve the equations of static equilibrium to determine the reaction forces and moments on the tube&lt;/li&gt;&#xA;&lt;li&gt;Draw the shear/moment diagram to determine the loads at each point on the tube&lt;/li&gt;&#xA;&lt;li&gt;Determine the stress at the most highly loaded point&lt;/li&gt;&#xA;&lt;li&gt;Compare that to the yield stress for your material&lt;/li&gt;&#xA;&lt;li&gt;apply an appropriate factor of safety&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you are not sure what I mean by some of these, you'll want to pick up a textbook on Statics and a textbook on Strength of Materials (typically these are semester long sophomore level classes for Mechanical Engineering students)&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-02-11T12:57:48.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13707" PostTypeId="2" ParentId="13704" CreationDate="2017-02-11T14:11:29.543" Score="0" Body="&lt;p&gt;Storage space for a 5-litre liquid shouldn't be that much, I think. So, if it doesn't smell, I'd go for option 1.&lt;/p&gt;&#xA;" OwnerUserId="5408" LastActivityDate="2017-02-11T14:11:29.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13708" PostTypeId="1" AcceptedAnswerId="15495" CreationDate="2017-02-11T16:43:42.620" Score="1" ViewCount="297" Body="&lt;p&gt;Suppose that there is a piston cylinder device . It contains a gas with pressure P. The external pressure is Pext . In order for the gas to expand , I think that P must be greater than P ext , and it will expand until the pressure of the gas is equal to the external pressure. &#xA;The thing that I cannot understand is that when we calculate the boundary work we multiply the external pressure by the change in volume. Since the gas is expanding , then the gas is DOING the work ON the surrounding , then I expect that when calculating the boundary work is to consider the gas pressure. What is wrong with that ?&lt;/p&gt;&#xA;" OwnerUserId="9742" LastActivityDate="2017-05-24T16:28:37.137" Title="why do we consider the external pressure when calculating the boundary work" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13709" PostTypeId="2" ParentId="13708" CreationDate="2017-02-11T18:07:59.653" Score="0" Body="&lt;p&gt;The surrounding is only ever resisting with external pressure; that is the only work done on the piston (neglecting it's mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider it that way, the work is just a constant pressure with a change in volume.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-02-11T18:07:59.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13710" PostTypeId="1" CreationDate="2017-02-11T22:15:03.423" Score="1" ViewCount="122" Body="&lt;p&gt;Efficiency = Actual Mass displaced by the compressor divided by the theoretical mass displaced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we continuously increase the pressure in the compressor tank, will the efficiency stay roughly constant or is there another relationship between Pressure in the tank and efficiency of the compressor?&lt;/p&gt;&#xA;" OwnerUserId="9991" LastEditorUserId="9991" LastEditDate="2017-02-11T22:36:26.200" LastActivityDate="2017-05-13T00:24:58.533" Title="Is the efficiency of a reciprocating air compressor constant for all allowable values of Pressure?" Tags="|compressors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13712" PostTypeId="2" ParentId="13710" CreationDate="2017-02-11T22:57:19.447" Score="1" Body="&lt;p&gt;I am not familiar with measuring efficiency based on mass the way you described it.  However, there is a thing called volumetric efficiency which is the volume of air actually drawn into the cylinder divided by the piston volume.  And yes, this value will drop as the discharge pressure increases. At some pressure, the efficiency will become zero, which means the compressor cannot compress air to any higher pressure than that.  Here's a page with some formulas, &lt;a href=&quot;http://petrowiki.org/Reciprocating_compressor&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://petrowiki.org/Reciprocating_compressor&lt;/a&gt; (look for volumetric efficiency halfway down the page).  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-02-11T22:57:19.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13713" PostTypeId="2" ParentId="49" CreationDate="2017-02-12T00:45:19.470" Score="1" Body="&lt;p&gt;You have two options: pass light through a circular aperture or (inversely) block light with a circular mask. Either way, you create edge shadows without focusing the light.  Light sensors centered the edge of the shadow (on both axes) will balance when the device is properly aligned.  If the distance from the aperture to the sensors is less than the aperture diameter, the device will tell you which way to rotate to achieve alignment (balance) even for very large error angles.  You may need diffusers over the aperture or sensors to soften the transition from light to dark.&lt;/p&gt;&#xA;" OwnerUserId="9992" LastActivityDate="2017-02-12T00:45:19.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13714" PostTypeId="1" AcceptedAnswerId="13717" CreationDate="2017-02-12T01:05:31.573" Score="0" ViewCount="729" Body="&lt;p&gt;I am trying to create a simulation for car suspension and tyre physics and to do the tyre physics I need to know how much weight is over the friction point that is under the spring.&lt;/p&gt;&#xA;" OwnerUserId="9993" LastActivityDate="2017-03-06T23:59:19.223" Title="How would I calculate a force on a spring? Or more specifically the weight of an object that spring is supporting" Tags="|springs|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13715" PostTypeId="1" AcceptedAnswerId="13740" CreationDate="2017-02-12T01:25:01.190" Score="0" ViewCount="95" Body="&lt;p&gt;How to finish part (a) and (b) of this dynamics problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aIQGb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aIQGb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no solution or answer in the book...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor say that this problem is very tricky, don't even try by yourself.&lt;/p&gt;&#xA;" OwnerUserId="252" LastEditorUserId="1627" LastEditDate="2017-02-16T14:36:32.480" LastActivityDate="2017-02-16T14:36:32.480" Title="How to compute the mass and max speed of a ball moving on a circular track under the influence of gravity and a spring" Tags="|mechanical-engineering|dynamics|" AnswerCount="1" CommentCount="3" ClosedDate="2017-02-20T21:39:58.130" ContentLicense="CC BY-SA 3.0" />
  <row Id="13716" PostTypeId="1" AcceptedAnswerId="13894" CreationDate="2017-02-12T05:19:31.233" Score="1" ViewCount="69" Body="&lt;p&gt;I'm trying to find PSIG from other variables that I have values for. I'll try my best to explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have Helium flowing through a 1&quot; copper pipe at 65 psig and at 40 deg F to a 1&quot; ball valve and power control valve. There is a flow meter connected after the power control valve, with the ball valve throttled to achieve 40 CFM. I need to find the PSIG after the throttled ball valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for the formula or a web calculator to solve this equation. I don't need a precise answer, just a rounded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be appreciated.&#xA;Thanks, Steve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P3WEE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P3WEE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8936" LastActivityDate="2017-02-22T18:46:01.117" Title="Need help solving for PSIG" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13717" PostTypeId="2" ParentId="13714" CreationDate="2017-02-12T05:20:01.523" Score="2" Body="&lt;p&gt;The force associated with springs is well described by Hooke's law. The spring force is given by,&#xA;$$ F = k x $$ &#xA;where $F$ is the spring force, $x$ is the displacement of the spring from its original resting position, and $k$ is the spring constant in units of force per length.&lt;/p&gt;&#xA;" OwnerUserId="9399" LastEditorUserId="1324" LastEditDate="2017-02-12T08:14:52.587" LastActivityDate="2017-02-12T08:14:52.587" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13718" PostTypeId="1" CreationDate="2017-02-12T10:52:23.767" Score="0" ViewCount="75" Body="&lt;p&gt;I'm currently trying to replace a broken battery for an electric moped, but the old one had two power ports,  one for the engine and one for the lights/horn.  I haven't been able to find anything that seems suited for splitting a 48volt 30amp connection into one 25 amp and one 5 amp. &#xA;If anyone could share a blueprint and a quick explanation it would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="10000" LastEditorUserId="8055" LastEditDate="2017-02-13T17:16:12.667" LastActivityDate="2017-02-14T12:20:21.643" Title="Uneven split of amperage from battery" Tags="|electrical-engineering|battery|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13719" PostTypeId="2" ParentId="13654" CreationDate="2017-02-12T13:36:46.980" Score="0" Body="&lt;p&gt;Figured it out!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature is always being send in Fahrenheit not in Celsius, see the following table:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ygrGR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ygrGR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An increase of one in the 12bit (columns H&amp;amp;I&amp;amp;J together) column leads to 0,1 degree Fahrenheit increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting the 12bit to decimal, subtracting 400 and dividing by 10 actually leads to the exact value in Fahrenheit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(H&amp;amp;I&amp;amp;J-400)/10 = Temperature in Fahrenheit&#xA;((H&amp;amp;I&amp;amp;J-400/10)-32)/1,8 = Temperature in Celsius&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The formulas hold for all values!&lt;/p&gt;&#xA;" OwnerUserId="9219" LastActivityDate="2017-02-12T13:36:46.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13723" PostTypeId="2" ParentId="13718" CreationDate="2017-02-12T18:57:15.930" Score="1" Body="&lt;p&gt;The two different circuits in parallel will draw current from the battery depending on their relative resistances you don't need to 'supply' a specific current to specific circuits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the components in each circuit need to be rated for the maximum  current which that circuit will draw. &lt;/li&gt;&#xA;&lt;li&gt;Each outlet should be protected by a fuse or circuit breaker of an appropriate rating, ideally as close to the battery as possible. &lt;/li&gt;&#xA;&lt;li&gt;It is entirely likely that some ancillaries work off a lower voltage than the motor (probably 12v) in this case you will need a transformer or voltage converter to provide the appropriate voltage. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-02-14T12:20:21.643" LastActivityDate="2017-02-14T12:20:21.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13724" PostTypeId="1" AcceptedAnswerId="13739" CreationDate="2017-02-12T20:51:06.747" Score="0" ViewCount="107" Body="&lt;p&gt;I am trying to make a device that can go up or down a screw by twisting a knob, kind of like a wing nut, but the wing would be sideways (I suppose). To clarify what I mean by this, please see a simplified version in a diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dcw10.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dcw10.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this type of engineering has already been used in various devices, such as a microscope, where the course adjustment knob will raise or lower the &quot;stage&quot; of the microscope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know where a part/device like this could be found, or how I would go about making something similar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much for your time and kindness,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kelsey Nealon&lt;/p&gt;&#xA;" OwnerUserId="10006" LastActivityDate="2017-02-13T14:31:59.613" Title="How to Make an Object Able to be Screwed In/Out With a Sideways Knob" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13726" PostTypeId="2" ParentId="2300" CreationDate="2017-02-13T00:18:58.807" Score="4" Body="&lt;h1&gt;lbf has two definitions and a friend called Poundal&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1) EE System&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force required to accelerate 1 lbm 32.174049 ft/s^2 (i.e., acceleration due to gravity) However, the problem with this is that it &lt;strong&gt;MUST&lt;/strong&gt; retain 32.174049 in its units! Which is not ideal, Consider F = ma, which means ma will always have to be divided by 32.174049 making this equation F =(ma)/32.174049 however, this approach has 1 added convenience, your mass is equal to the force you exert on the surface of Earth (i.e., the magnitude of lbm and lbf are equal and interchangeable IFF considering your force on Earth due to acceleration caused by gravity at 32.174049ft/s^2)&#xA;$$&#xA;lbf := \frac{lbm*32.174049ft}{s^2} &#xA;$$&#xA;&lt;strong&gt;(2) BG System&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, it is in units of slugs. The force required to accelerate 1 slug 1 ft/s^2, where 1 slug is conveniently defined as 32.174048 lbm (i.e. the same value as the acceleration due to gravity) this approach also has the same added convenience as (1), your mass is equal to the force you exert on the surface of Earth (i.e., the magnitude of lbm and lbf are equal and interchangeable IFF considering your force on Earth due to acceleration caused by gravity at 32.174049ft/s^2))&#xA;$$&#xA;lbf = \frac{1slug}{32.174049lbm}\frac{1lbm*32.174049ft}{s^2} &#xA;$$&#xA;$$&#xA; := \frac{slug*ft}{s^2} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Know the base units of the unit system you are working in for ANY final solution to be applied appropriately. Both forms are correct!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0mmk2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0mmk2.png&quot; alt=&quot;British Gravitational v. English Engineering v. Absolute English&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3)AE System&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Poundal, the force required to accelerate 1 lbm 1 ft/s^2. Similar in approach to (2), except it is multiplied by a normalizing factor instead of a unit conversion, therefore retaining lbm ft/s^2 units:&#xA;$$&#xA;pdl = \frac{1}{32.174049}\frac{lbm*32.174049ft}{s^2}&#xA;$$&#xA;$$&#xA;:=\frac{lbm*ft}{s^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, (1),(2) and (3) are all dividing by 32.174049, however, it is when and how that makes all the difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Know the base units of your system, lbf will always be an ambiguity problem as long as it exists in its current symbolic form. &lt;strong&gt;I'd suggest adopting &lt;em&gt;sdl&lt;/em&gt; for (2) lbf with unit slug&lt;/strong&gt;, the ambiguity of pound is an unusual punishment lb, lbs, lbm, lbf, lbf...&lt;/p&gt;&#xA;" OwnerUserId="10008" LastEditorUserId="10008" LastEditDate="2017-02-13T16:46:42.460" LastActivityDate="2017-02-13T16:46:42.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13727" PostTypeId="1" CreationDate="2017-02-13T01:28:23.563" Score="3" ViewCount="1989" Body="&lt;p&gt;I'm asked to analytically calculate the composite torsional constant of the following structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UJ9KO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UJ9KO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No other information is given, such as the applied torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to calculate the torsional constant for each component (an equilateral triangle and a semicircle), but I don't think I can just add them weighted to the areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything else I can try?&lt;/p&gt;&#xA;" OwnerUserId="5218" LastActivityDate="2017-11-10T16:55:15.973" Title="Torsional Constant of Thin-Walled Multi-Cell Structures" Tags="|structures|aerospace-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13728" PostTypeId="1" CreationDate="2017-02-13T04:22:48.110" Score="2" ViewCount="158" Body="&lt;p&gt;What is the equation to determine the inside surface temperature along the length of the counter flow heat exchanger?&lt;/p&gt;&#xA;" OwnerUserId="10012" LastEditorUserId="9" LastEditDate="2017-02-13T11:00:46.737" LastActivityDate="2017-04-17T21:40:59.943" Title="How to determine heat exchanger equation?" Tags="|heat-transfer|chemical-engineering|heat-exchanger|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13729" PostTypeId="1" AcceptedAnswerId="13735" CreationDate="2017-02-13T04:30:07.313" Score="3" ViewCount="1060" Body="&lt;p&gt;With folding barriers like the one below, how does the orange bar get turned by the green mechanism? What's inside the yellow triangular part?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2gqpQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2gqpQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=LF8kSTCZIxw&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=LF8kSTCZIxw&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4302" LastActivityDate="2020-06-11T21:12:35.463" Title="How do folding barriers work?" Tags="|mechanical-engineering|gears|linkage|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13733" PostTypeId="2" ParentId="13729" CreationDate="2017-02-13T10:06:05.010" Score="1" Body="&lt;p&gt;Think of the orange bar as the humerus bone in your upper arm, the blue bar as your radius, and the pink one as your ulna. The gate folds up in much the same way your arm folds up, and opens up much the same too.&lt;/p&gt;&#xA;" OwnerUserId="5408" LastActivityDate="2017-02-13T10:06:05.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13735" PostTypeId="2" ParentId="13729" CreationDate="2017-02-13T11:44:03.403" Score="4" Body="&lt;p&gt;What you see here is two sets of &lt;em&gt;four-bar&lt;/em&gt; linkages connected to each other to form a &lt;em&gt;six-bar&lt;/em&gt; linkage. To be more precise this is a &lt;strong&gt;Watt type 1&lt;/strong&gt; six-bar linkage. This particular configuration is known for its ability to fold together by rotating only one member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z4GwC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z4GwC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: The classification of some planar 6 bar linkages&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing inside the yellow part although in parallel to that there is a straight linkage bar. The way you analyze these is just draw a simplified image contain just joints and apparent connections.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-02-13T11:54:31.480" LastActivityDate="2017-02-13T11:54:31.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13736" PostTypeId="2" ParentId="13703" CreationDate="2017-02-13T11:48:21.110" Score="2" Body="&lt;p&gt;Before you start thinking about technical solutions like materials and mechanisms you really need to come up with a reasonably detailed design specification, this should address the problem in terms of what you want to achieve rather than how you will achieve it. This also involves precisely setting out the problem that you want to solve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly one crucial element will be the product you are dispensing for example :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Shelf life : how will you ensure efficient stock rotation ?&lt;/li&gt;&#xA;&lt;li&gt;Volume : how much of the product should the machine hold ?&lt;/li&gt;&#xA;&lt;li&gt;Storage requirements : temperature, humidity etc &lt;/li&gt;&#xA;&lt;li&gt;Is the product individually packaged or does it need to be dispensed into containers at the point of delivery ? Does it have specific handling requirements ?&lt;/li&gt;&#xA;&lt;li&gt;How many different types of product will the machine stock ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other important elements might include :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Security &lt;/li&gt;&#xA;&lt;li&gt;Safety &lt;/li&gt;&#xA;&lt;li&gt;Branding and aesthetic appearance &lt;/li&gt;&#xA;&lt;li&gt;How payment will be taken &lt;/li&gt;&#xA;&lt;li&gt;User interface &lt;/li&gt;&#xA;&lt;li&gt;Reliability &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-02-13T11:53:47.157" LastActivityDate="2017-02-13T11:53:47.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13737" PostTypeId="2" ParentId="13727" CreationDate="2017-02-13T11:56:43.127" Score="1" Body="&lt;p&gt;You need to find the shear flows in each individual cell, and then apply the usual torsion constant formula to the cells and add the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get one equation from the equilibrium of the &quot;corner points&quot; where the sections join (there are two points, at the ends of the vertical rib, but by symmetry you get the same equation for each point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get another equation, you can to assume each cell has the same amount of twist - i.e. the cross-section does not warp under torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works through the details: &lt;a href=&quot;http://web.aeromech.usyd.edu.au//structures/acs1-p83.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://web.aeromech.usyd.edu.au//structures/acs1-p83.html&lt;/a&gt;. Google &quot;torsion multicell beams&quot; for other references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to include warping of the cross section, the best way to deal with this type of structure is model the correct geometry in a finite element analysis, rather than trying to reduce it to a &quot;beam element&quot; Aside from the assumptions made in calculating a single torsion constant, the actual torsional stiffness is very sensitive to the way the ends of the &quot;beam&quot; are connected to the rest of the structure, which may or may not prevent warping or other deformations of the cross section.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-02-13T11:56:43.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13738" PostTypeId="1" CreationDate="2017-02-13T13:27:15.200" Score="0" ViewCount="1834" Body="&lt;p&gt;For patent and security issues I want to know if SolidWorks (relevant versions are 2012, 2015 and 2016) stores user, license and further metadata inside .sldprt, .sldasm and .slddrw files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so: &#xA;Are there measures of evaluating the metadata and/or tampering it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to circumvent the delivery of metadata by e.g. exporting as STEP?&lt;/p&gt;&#xA;" OwnerUserId="3786" LastActivityDate="2017-03-26T15:49:10.210" Title="Does solidworks store license and user information inside files?" Tags="|solidworks|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13739" PostTypeId="2" ParentId="13724" CreationDate="2017-02-13T14:31:59.613" Score="3" Body="&lt;p&gt;I can't see a worm gear working in this instance as it would be trying to reverse drive the worm - there is too much friction to achieve this. A rack and pinion would not stay in place after releasing the knob. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use a 90 degree bevel gear with an internal thread in the output gear like the one in the photo. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DZozr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DZozr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4189" LastActivityDate="2017-02-13T14:31:59.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13740" PostTypeId="2" ParentId="13715" CreationDate="2017-02-13T16:45:17.533" Score="2" Body="&lt;p&gt;Starting with equating the total energies at the starting and max velocity point. The energy at the starting point comes from the potential energies of the mass and spring. The energy at the other point also has the kinetic energy component of the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;m g (R \tan (\alpha )+R)+\frac{1}{2} k \left(\sqrt{R^2+(R \tan (\alpha&#xA;   )+R)^2}-\ell \right)^2 = \\&#xA;m g (R-R \sin (\beta ))+\frac{1}{2} k \left(\sqrt{(R-R&#xA;   \sin (\beta ))^2+(2 R-R \cos (\beta ))^2}-\ell \right)^2+\frac{1}{2}m v^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($\ell$ is the unstretched spring length)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can solve for $v^2$ (or $v$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1}{m}(-4. \sin ^2(\beta )-4. \cos ^2(\beta )+16. \cos (\beta )+40. \sqrt{-0.08 \sin&#xA;   (\beta )-0.16 \cos (\beta )+0.24}+(3.92 m+8.) \sin (\beta )-12 +\\&#xA;4. \tan ^2(\alpha )-40. \sqrt{0.04 \tan ^2(\alpha )+0.08 \tan (\alpha )+0.08}+8.&#xA;   \tan (\alpha )+3.92 m \tan (\alpha ))&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at this expression we see that part of it depends on $\beta$ and the other part on $\alpha$, and there are no cross variables. The other variable is the mass $m$. Given the information that the max velocity is at $\beta=34 {}^{\circ}$, we can equate the derivative of $v^2$ to 0 and solve for $m$. This gives $m = 0.143 kg$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is as far as we can get with the information provided. The actual velocity is a function of $\alpha$ and as expected higher values produce larger velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;2.64447 \sqrt{4. \tan ^2(\alpha )-40. \sqrt{0.04 \tan ^2(\alpha )+0.08 \tan&#xA;   (\alpha )+0.08}+8.56054 \tan (\alpha )+12.0611}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/487UA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/487UA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;[Update]&lt;/strong&gt; If $\alpha=45{}^{\circ}$ then the velocity is $6.86192\ m/s$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All the computations were done in Mathematica and I am attaching the screenshot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NtN37.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NtN37.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2017-02-16T14:22:13.873" LastActivityDate="2017-02-16T14:22:13.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13741" PostTypeId="1" CreationDate="2017-02-13T17:08:53.243" Score="2" ViewCount="42" Body="&lt;p&gt;I'm trying to model a capacitive force sensor with a bilayer configuration of polymer - Si on Mechanical APDL 16.2 .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 0.2 mm layer of Si deposited on 0.005 mm of polyimide polymer. My goal is to find out the the length change in polyimide for an exerted force of 3 mN. For that purpose, I plot the displacement curve along x direction (origin to the edge). However, I obtain a nonlinear result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand the origin of this nonlinearity. In my opinion, the polymer has to obey Hooke's law. What could be its reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://gist.github.com/anonymous/f3ad42a0824657714d818ac81cc20838&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is the model code.&lt;/p&gt;&#xA;" OwnerUserId="10021" LastEditorUserId="9" LastEditDate="2017-02-13T17:33:26.540" LastActivityDate="2017-02-13T17:33:26.540" Title="Convergence problem in polymer - silicon structural analysis" Tags="|sensors|ansys-apdl|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13742" PostTypeId="1" CreationDate="2017-02-13T17:42:37.433" Score="4" ViewCount="134" Body="&lt;p&gt;I noticed &lt;a href=&quot;https://twitter.com/NWSSacramento/status/830961883708608513?ref_src=twsrc%5Etfw&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Tweet&lt;/a&gt; about the &lt;a href=&quot;http://www.latimes.com/local/california/la-live-updates-oroville-dam-20170212-htmlstory.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oroville Dam concerns&lt;/a&gt; this week:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CBngQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CBngQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems most media is not emphasizing/noting the distinction.  Is there a difference in the consequences of a spillway failure?  Seems intuitively that the release of water might be slower due to what may well be a smaller gap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is that true?  And would unmitigated spillway failures generally add concern that additional stress/erosion can result in full dam failure shortly thereafter?&lt;/p&gt;&#xA;" OwnerUserId="7655" LastEditorUserId="6790" LastEditDate="2020-07-31T21:03:13.907" LastActivityDate="2020-07-31T21:03:13.907" Title="Does a spillway failure have different consequences versus dam failure?" Tags="|civil-engineering|hydroelectric-dams|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13743" PostTypeId="2" ParentId="13728" CreationDate="2017-02-13T18:01:44.647" Score="1" Body="&lt;p&gt;The temperature equations for a counter flow exchanger are the same for a parallel flow exchanger as long as steady-state is a valid assumption, so the flow must be fully developed and the temperature profile must be fully developed. For a differential element where the flow travels along $x$ we have the conservation of heat. &#xA;Properties are known about substances and geometry and we analyze only the averages across the pipe's cross sections, $P$ is perimeter, $h$ is convection coefficient, $c_p$ fluid specific heat at average temperature, $T_m$ and constant (from steady state) average mass flow rate, $\dot{m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_{in,pipe}+Q_{in,conv} = Q_{out}$$&#xA;$$\dot{m} c_p T_m(x) + h P dx (T_m(x) - T_s(x)) = \dot{m} c_p T_m(x+dx)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the definition of the derivative and some algebra&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\dot{m} c_p}{h P dx} \frac{dT_m(x)}{dx} + T_m(x) = T_s(x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the mean temperature profile of the fluid, then you can evaluate this expression. The mean temperature is often identified with the inlet and outlet temperatures and it may be the case that you will need a set of boundary conditions that describe heat transfer from surroundings into the pipe material and then into the fluid. While these may be closed form, they might be series solutions.&lt;/p&gt;&#xA;" OwnerUserId="10020" LastActivityDate="2017-02-13T18:01:44.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13745" PostTypeId="2" ParentId="13704" CreationDate="2017-02-13T19:38:17.400" Score="2" Body="&lt;p&gt;Option 4: Put it in the container, leave it on your doorstep and offer it for free on &lt;a href=&quot;https://www.craigslist.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;craigslist&lt;/a&gt;. You will likely find someone that regularly needs it and you can just email/text message them any time you set some out. People use waste vegetable oil to burn for heat, bio-diesel, soap making, animal feed, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that disposing oil down the toilet is very irresponsible. Oil water separation is not so trivial when it is spread across millions of gallons of sewage. Oil also makes the water underneath go aerobic; creating additional smell issues. When in doubt about a type of waste; it should go to the landfill. Remember that the water from your toilet or storm drain has to be processed and cleaned to a level that can enter a river or ocean.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-02-13T19:38:17.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13746" PostTypeId="2" ParentId="13742" CreationDate="2017-02-13T19:52:10.883" Score="2" Body="&lt;p&gt;Spillways are designed to control the water flow (and excess waterflow) past a dam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spillway failure could mean a variety of things, but the primary difference is that the spillway is most always located at the upper level of the dam. Water that escapes this area may be damaging, but unlike dam failure, not all the water in the reservoir is going to be released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/2017_Oroville_Dam_crisis&quot; rel=&quot;nofollow noreferrer&quot;&gt;2017 Oroville Dam Crisis Wiki&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-02-13T19:52:10.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13747" PostTypeId="2" ParentId="7394" CreationDate="2017-02-13T22:03:52.110" Score="1" Body="&lt;p&gt;They are many reasons for the time it takes to startup or return to full power operation in commercial nuclear power plants. In the US there are two main types of plants, Boiling Water Reactors(BWRs) and Pressurized Water Reactors (PWRs).  Answers will differ according to the type of reactor and even which version of the type. A common explanation that I did not see mentioned is that all commercial nuclear power plants avoid making &gt;15% thermal power changes in any 4 hour period. This is to protect the integrity of the fuel cladding.  I worked in the commercial nuclear power industry for almost 20 years - and have been out of it for over 20 years -so maybe they have improved the fuel cladding and this is no longer an issue - but it was a mandatory constraint in my day.&lt;/p&gt;&#xA;" OwnerUserId="10025" LastActivityDate="2017-02-13T22:03:52.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13750" PostTypeId="1" AcceptedAnswerId="13754" CreationDate="2017-02-14T00:19:56.200" Score="2" ViewCount="2655" Body="&lt;p&gt;I would appreciate a quick peer check on the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently enrolled in a class where, I think, the professor is repeatedly confusing &lt;em&gt;percent increase&lt;/em&gt; and factor by which something increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There &lt;em&gt;is&lt;/em&gt; a difference isn't there? I'm not going crazy here am I? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, to calculate the percent increase I would do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;IV = initial \; value\\&#xA;FV = final \; value\\&#xA;Assume \; FV &amp;gt; IV\\&#xA;\% \, Inc. = \left( \frac{FV - IV}{IV} \right) * 100 = \left( \frac{FV}{IV} - 1 \right) * 100&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want the factor by which the value increased with respect to the initial value I would take the ratio of the final value to the initial:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Inc. \; Factor = \frac{FV}{IV}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reiterate, these are not the same things. I just want to make sure I'm not confusing anything before I point it out to him and ask for clarification.&lt;/p&gt;&#xA;" OwnerUserId="9732" LastActivityDate="2017-02-14T12:07:58.013" Title="Percent Increase vs. Factor By Which Something Increases" Tags="|heat-transfer|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13751" PostTypeId="2" ParentId="13047" CreationDate="2017-02-14T02:16:41.330" Score="1" Body="&lt;p&gt;CBN grinding wheel provides you with great grinding ability : low grinding wheel wear, longer wheel life and good work material removal, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can opt of &lt;strong&gt;softer grade&lt;/strong&gt; CBN grinding wheels with a &lt;strong&gt;fine abrasive grain size&lt;/strong&gt; for your high precision application. I would suggest you to go for &lt;strong&gt;vitrified type bond&lt;/strong&gt; for higher durability, reliability and high resistance to corrosion due to coolant used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can call up or meet up with any grinding wheel manufacturer/retailer near you and seek their help in selecting the best grinding wheel for your application. They know the best, and thats what they do for a living. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps. &lt;/p&gt;&#xA;" OwnerUserId="9949" LastActivityDate="2017-02-14T02:16:41.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13752" PostTypeId="2" ParentId="13206" CreationDate="2017-02-14T08:16:59.650" Score="0" Body="&lt;p&gt;Look at &lt;a href=&quot;https://www.dekkervacuum.com/resource-library/knowledge-database/principles-of-operation/two-stage-liquid-ring-vacuum-pump/&quot; rel=&quot;nofollow noreferrer&quot;&gt;liquid ring vacuum pumps&lt;/a&gt; (&lt;a href=&quot;http://www.graham-mfg.com/usr/pdf/TechLibVacuum/212.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;sizing tips&lt;/a&gt;) (really a version of the vane pumps mentioned by ericnutsch). Any fluid they suck will add to the liquid ring, excess fluid will be discharged. Talk to an applications engineer if the pump will have a problem with soap. These pumps exist at flowrates around 10-20 l/min. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your situation, however, I would look at diaphragm pumps first and at other solutions if the pulsed flow is a problem.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-02-14T08:16:59.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13753" PostTypeId="2" ParentId="13148" CreationDate="2017-02-14T09:31:59.560" Score="1" Body="&lt;p&gt;I think the answer for yor question can be the following command.&#xA;/PSYMB,ESYS,1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find further information in the software, through Help&gt;Mechanical APDL&gt;Command Reference&gt;XVII. P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tamás&lt;/p&gt;&#xA;" OwnerUserId="10032" LastActivityDate="2017-02-14T09:31:59.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13754" PostTypeId="2" ParentId="13750" CreationDate="2017-02-14T12:07:58.013" Score="0" Body="&lt;p&gt;Yes, what you show is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a value going from 8 to 12 is a 50% increase.  You can also say it was increased by a factor of 1.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People sometimes get a little sloppy with this.  If it's clear enough from context what the professor really means, I would let it go.  If you're the professor, it's OK to insist your students use the terms correctly.  When you're the student, it's better to reserve picking a fight for something more important.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-02-14T12:07:58.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13755" PostTypeId="1" CreationDate="2017-02-14T12:44:11.233" Score="1" ViewCount="235" Body="&lt;p&gt;We have a project which requires Ground Penetrating Radar(GPR). We got a student license of NI AWR software and Ansys HFSS (high frequency structural simulator). Are those the correct software for this types of system simulation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we mainly want to achieve is finding a proper ready to use circuit to connect antennas if applicable, determine the system requirements by using the simulation outputs and implementing algorithm to read and analyze data gathered form the circuit in MATLAB. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If those software are OK, can you suggest a reference/source/tutorial or whatever to model the system. As far as I understand, AWR works like a 2D environment which makes me confused how to simulate especially the ground side. HFSS seems more appropriate to model the ground layers, voids or objects with different material properties to find EM characteristics. But I am not sure how to combine this with a antenna system.&lt;/p&gt;&#xA;" OwnerUserId="1287" LastEditorUserId="8055" LastEditDate="2017-02-16T05:30:37.270" LastActivityDate="2017-02-16T05:30:37.270" Title="How to simulate Ground Penetrating Radar system?" Tags="|electrical-engineering|finite-element-method|rf-electronics|electromagnetism|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13756" PostTypeId="1" CreationDate="2017-02-14T13:15:39.523" Score="1" ViewCount="116" Body="&lt;p&gt;Switching speed of PN diode is limited by reverse recovery time. Can Schotkky diodes rectify signals of GHz frequency? If yes, which ones?&lt;/p&gt;&#xA;" OwnerUserId="10033" LastEditorUserId="8055" LastEditDate="2017-02-17T23:19:00.367" LastActivityDate="2017-02-17T23:19:00.367" Title="Can Schotkky diodes rectify signals of GHz frequency?" Tags="|electrical-engineering|rf-electronics|" AnswerCount="1" CommentCount="2" ClosedDate="2017-02-15T01:47:47.940" ContentLicense="CC BY-SA 3.0" />
  <row Id="13757" PostTypeId="1" CreationDate="2017-02-14T14:11:58.643" Score="-3" ViewCount="339" Body="&lt;p&gt;I am an earth scientist, I have a presentation about rare-earth elements. &#xA;I want to get a review of exhaust pipes system. Also, what kind of REE are used in their manufacturing?&#xA;I have no knowledge of engine's mechanics or automobile parts, so let's avoid technical jargon :p.&lt;/p&gt;&#xA;" OwnerUserId="10036" LastEditorUserId="10036" LastEditDate="2017-02-14T16:53:29.537" LastActivityDate="2017-02-14T16:53:29.537" Title="What rare earth metal is used in exhaust pipes?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="3" ClosedDate="2017-02-14T18:05:54.123" ContentLicense="CC BY-SA 3.0" />
  <row Id="13758" PostTypeId="1" AcceptedAnswerId="13765" CreationDate="2017-02-14T14:40:20.667" Score="2" ViewCount="302" Body="&lt;p&gt;As far as I know, propane with oxygen will burn at 2800 °C and steel will melt at ~1500 °C. Brass is even less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am trying to do is a rocket nozzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way I can built this with steel or is it simply too hot? And for the burning temperature of propane, 2800 °C, is it the temperature &lt;strong&gt;IN&lt;/strong&gt; the nozzle, &lt;strong&gt;OUT&lt;/strong&gt; of it, or &lt;strong&gt;IN&lt;/strong&gt; the combustion chamber?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3nDU4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3nDU4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2017-02-16T16:26:18.770" Title="Propane rocket nozzle material" Tags="|cad|propulsion|autodesk-inventor|rocketry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13759" PostTypeId="2" ParentId="13742" CreationDate="2017-02-14T14:44:13.587" Score="2" Body="&lt;p&gt;If a dam collapses catastrophically, the entire volume of the lake is released in one 'pulse.'  This means the downstream area will be subjected not only to a wave of great height but also of great peak power (energy per unit time).  That's a BadThing(TM) . You could look up info on what happened eons ago when Ice-Age ice dams gave way in the Northwest United States.  Canyons were carved out in hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a spillway gives out, then, depending on just how it fails,  there'll be a gradually increasing flow rate as the new escape channel is widened.  To a limited extent this means the same total water volume is released, but over a longer time period -- again unless at some point the spillway hillside blows out catastrophically.  So the peak wave height and the peak power will be smaller, albeit applied for a longer period of time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way,  you will not be happy if you're downstream.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-02-14T14:44:13.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13760" PostTypeId="2" ParentId="13757" CreationDate="2017-02-14T15:09:53.563" Score="1" Body="&lt;p&gt;&lt;strong&gt;Cerium or Ce&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cerium is a silvery-white metal that easily oxidizes in the air. It is the most abundant of the rare earth elements and has many uses. For instance, cerium oxide is used as a catalyst in catalytic converters in automotive exhaust systems to reduce emissions, and is highly desirable for precision glass polishing. Cerium can also be used in iron, magnesium and aluminum alloys, magnets, certain types of electrodes, and carbon-arc lighting.&lt;/p&gt;&#xA;" OwnerUserId="10028" LastActivityDate="2017-02-14T15:09:53.563" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13761" PostTypeId="2" ParentId="7015" CreationDate="2017-02-14T15:28:14.697" Score="1" Body="&lt;p&gt;This link specifies that Alloy 20, specifically CARPENTER® alloy 20Cb-3, has an amazing resistance to chemicals containing chlorides and sulfuric, phosphoric, and nitric acids. It also resists pitting, crevice corrosion, SCC, and intergranular attack. It can be used for pickling tanks, gas scrubbers, piping, heat exchangers, pumps, shafts and valves. Synthetic rubber, pharmaceutical production and other process equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While Ti 6-4 (Grade 5) is significantly stronger than commercially pure titanium while having the same stiffness and thermal properties ( excluding thermal conductivity, which is about 60% lower in Grade 5 than in CP Ti). Among its many advantages, it is heat treatable. This grade is an excellent combination of strength, corrosion resistance, weld and fabric-ability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing it to alloy C-276, specifically HASTELLOY® alloy C-276, possesses an outstanding corrosion resistance in reducing and oxidizing environments. It maintains corrosion resistance in welded condition, and is excellent resistance to pitting and stress-corrosion cracking (SCC). &#xA;It is widely used in severest environments in chemical processing, pollution control, pulp and paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not too sure about the comparison with alloy 316, but the latter possesses an improved resistance to pitting by solutions containing chlorides and other halides. In addition, it provides excellent elevated temperature tensile, creep and stress - rupture strengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your last concern, obtaining and availability of Alloy 20 that I am not too sure about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hpalloy.com/Alloys/corrosionResistant.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Corrosion Resistant Materials&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10028" LastActivityDate="2017-02-14T15:28:14.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13762" PostTypeId="2" ParentId="7492" CreationDate="2017-02-14T15:51:28.657" Score="2" Body="&lt;p&gt;It is rare in the mechanical engineering field to have true stress and strain curve diagrams on &lt;em&gt;compression&lt;/em&gt; but some of my laboratories were done on it. Here is a diagram of true stress and strain curve in tension, &lt;a href=&quot;https://i.stack.imgur.com/kqjiz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kqjiz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; and here is a diagram of true stress and strain curve in compression &lt;a href=&quot;https://i.stack.imgur.com/aN202.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aN202.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compression will still rise above tension curve in the plastic region, according to the pictures provided; but you cannot have a general talk about simply &quot;typical metals&quot;. Each material has its' own properties and might be completely different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you mentioned about the extra rise in force for compression with respect to tension is indeed compensated by the fact that for compression the area increases and for tension it decreases, due to necking. Necking is a type of tensile deformation where relatively large amounts of strain localize disproportionately in a small region of the material, therefore causing the tension to decrease.&lt;/p&gt;&#xA;" OwnerUserId="10028" LastActivityDate="2017-02-14T15:51:28.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13763" PostTypeId="1" AcceptedAnswerId="13768" CreationDate="2017-02-14T15:57:32.287" Score="1" ViewCount="131" Body="&lt;p&gt;Ok, I wish to plot the speed time curve of the my urban service . The travel includes the curvature ( as shown in image ( just for reference )).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BPMSw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BPMSw.jpg&quot; alt=&quot;curvature in the track&quot;&gt;&lt;/a&gt;&#xA; The train speed is decreased while it runs along the arc ( i am pretty sure ) . The question is whether this decrease in train speed is considered while plotting a speed time curve or is it considered to be negiligible. I have to plot one . &lt;a href=&quot;https://i.stack.imgur.com/KNjF3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KNjF3.jpg&quot; alt=&quot;urban service speed time curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10040" OwnerDisplayName="Kaushal" LastEditorUserId="86" LastEditDate="2017-02-14T17:59:31.770" LastActivityDate="2017-02-14T22:40:15.690" Title="Does the decrease in speed of the train at the curvature changes the properties of Speed Time Curve Of Urban Service" Tags="|rail|transportation|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13764" PostTypeId="2" ParentId="13757" CreationDate="2017-02-14T16:26:12.880" Score="0" Body="&lt;p&gt;The structure of an exhaust system is usually made from steel or stainless steel (which gives better corrosion resistance for increased cost). Mild steel parts may be plated with chrome (especially if they are visible eg in motorcycles) or coated with aluminium to manage corrosion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exhaust system may also include catalytic converters and/or particle filters which may incorporate more exotic materials depending on the technology used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Catalytic converters are generally focused on educing emissions of nitrogen compounds (NOX), metal catalysts, especially platinum are commonly used, although this is an area of ongoing research and development, especially to find cheaper and more easily available catalysts as a catalytic converter can now represent a significant fraction of the cost of a car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diesel engines in particular also use physical filter to reduce emissions of very fine particulates which contribute to smog and are hazardous to human health, especially in urban areas with high traffic. These are often made from ceramic fibres of matrices and need to be able to tolerate temperatures high enough to burn off accumulated soot. So these engineering ceramics are quite a likely use of rare earth metals. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-02-14T16:26:12.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13765" PostTypeId="2" ParentId="13758" CreationDate="2017-02-14T17:43:15.527" Score="4" Body="&lt;p&gt;The answer is that in order to survive and function, the rocket nozzles need to be cooled. How that's done varies with the application, but includes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Liquid Cooling Jacket: The liquid propergols and/or fuels ar circulated through a jacket around the nozzle before being directed to the combustion chamber. This is how the Saturn V F-1 engines and the Space Shuttle Main Engine operated. In other applications, th fuel is circulated through a jacket but is dumped rather than sent to the combustion chamber&lt;/li&gt;&#xA;&lt;li&gt;Ablative cooling: The combustion chamber and/or nozzles are designed to wear away and be ejected along the exhaust stream. This is typically how solid fuel rockets operate, but some liquid fueled engines work this way as well. One problem is that as the material is worn away, the geometry of the nozzle and combustion chamber change, possibly affecting engine performance&lt;/li&gt;&#xA;&lt;li&gt;Radiative/convective cooling: The heat is radiated through the material at a fast enough rate that the material never gets too hot. This only works on relatively small engines.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-02-14T17:43:15.527" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13766" PostTypeId="2" ParentId="13756" CreationDate="2017-02-14T18:00:54.530" Score="2" Body="&lt;p&gt;Ultra low capacitance diodes &amp;lt;1pF are capable of operating at GHz frequencies. They are low loss Schottky types. BAS69xxx from STMicro is one example.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-02-14T18:00:54.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13767" PostTypeId="2" ParentId="7749" CreationDate="2017-02-14T22:33:00.550" Score="1" Body="&lt;p&gt;It appears that you are trying to derive the &lt;em&gt;transfer function&lt;/em&gt; of the system, that is, the input/output relationship in the frequency domain for zero initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ G(s) = \frac{Y(s)}{R(s)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $G(s)$ is the transfer function, $Y(s)$ is the system output and $R(s)$ is the input to the system and can be a summation of any number of inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R(s) = R_1(s) + R_2(s) + ... + R_n(s)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note: normally in systems and controls we use $U$ for inputs, but you have already defined it as a unit step at time = 0 in your question so I will avoid that notation and use $R$ instead.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How to treat the constant gravitational force&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In a translational mass system, all external forces are inputs to the system. &lt;strong&gt;The gravity is an external force and therefore must be included in your input term when you derive the transfer function.&lt;/strong&gt; Therefore you should have something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m\ddot{y} = x(t) + mg\dot{}u(t) = r_1(t) + r_2(t) = r(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the Laplace transform of both sides yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ms^2Y(s) = X(s) + \frac{1}{s}mg$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ms^2Y(s) = R_1(s) + R_2(s) = R(s)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(s) = \frac{Y(s)}{R(s)} = \frac{1}{ms^2}$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;How to derive Y(s)/X(s)?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I am guessing you still want know how to derive the transfer function for $Y(s)/X(s)$. As you stated, it is not possible to separate out the $X(s)$ term, but you can instead define your input thrust as having a variable (controlled) component $x_c$ and a constant (operating point) component $x_{o}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(t) = x_c(t) + x_{o}\dot{}u(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then all you have to do is set your operating point to be equal to the gravitational force:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_{o} = -mg$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This then cancels with the gravitational force, allowing you to derive a transfer function between the variable component of your thrust and the altitude of the hovercraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m\ddot{y} = x_c(t) - mg\dot{}u(t) + mg\dot{}u(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{Y(s)}{X_c(s)} = \frac{1}{ms^2}$&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2017-02-14T22:33:00.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13768" PostTypeId="2" ParentId="13763" CreationDate="2017-02-14T22:40:15.690" Score="1" Body="&lt;p&gt;As is often the case the answer is that it depends. Urban train services tend to operate at fairly modest speeds and it may well be that they are run at an optimal constant safe speed for a given stretch of track, bearing in mind also that there are a lot of energy losses associated with accelerating and braking a heavy train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally there may be no real advantage in running at the maximum possible speed if there are other bottlenecks in the system associated with signalling and spacing of trains, especially in an urban system where there are likely to be lots of stops and large volumes of passengers getting on and off. For example there is little point expending fuel getting to between stations 5 minutes faster if that just means you need to wait an extra five minutes for the previous train to clear that section of track. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-02-14T22:40:15.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13769" PostTypeId="2" ParentId="7492" CreationDate="2017-02-14T22:54:12.847" Score="2" Body="&lt;p&gt;One real-life complication is that once you get into the plastic region the behaviour can diverge wildly depending on the geometry and work hardening characteristics of the sample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a cylinder of a fairly ductile material may just get flatter and flatter indefinitely untill you get a flat disk or it exceeds the bounds of the ram compressing it clearly in this case the effective area will increase so true stress will decrease relative to the applied force. This is not dissimilar to some real world forming processes like setting solid rivets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you could get relatively unpredictable failure mechanisms like buckling, mushrooming or rupture at some point especially if work hardening is significant equally if the sample is not dead square relative to the test rig  or the sample is not perfectly homogeneous you could start to see bending forces develop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary tension is pretty much just tension regardless of the dimensions of the sample but response to compression is much more dependant on specific test conditions so some care is needed in making direct comparisons. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-02-14T22:54:12.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13770" PostTypeId="1" AcceptedAnswerId="13777" CreationDate="2017-02-14T23:56:14.687" Score="2" ViewCount="139" Body="&lt;p&gt;Is there a device that acts like a spring (i.e. $F=kx$) but the spring constant, $k$,  has different values in compression and tension?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if $k_c &amp;gt; k_t$ (where c and t refer to compression and tension), then the spring would quickly compress und a compressive force but would slowly expand under a tensile force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine you could design a hydraulic system like this with some check valves and control valves. Is there a single device that does this? Or alternatively, are there practical engineered systems that do this?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2017-02-18T20:34:56.407" Title="Spring-like device with anisotropic k value" Tags="|springs|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13772" PostTypeId="1" AcceptedAnswerId="13775" CreationDate="2017-02-15T10:08:39.507" Score="0" ViewCount="461" Body="&lt;p&gt;I want to engrave some letters from a 2D dxf file with single stroke fonts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this easy to do in most CAM software?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not decided what CAM software to use yet, because the ability to do single stroke fonts is very important to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read on forums of different CAM software, that people often find this difficult. Maybe it is because single stroke fonts are not included in the software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I already have the font as a dxf, so all I am asking is for the machine to follow a single line, where the centre of the cutting tool follows the line.&lt;/p&gt;&#xA;" OwnerUserId="7879" LastActivityDate="2017-02-15T18:15:37.280" Title="CNC engraving single line fonts" Tags="|cnc|" AnswerCount="1" CommentCount="3" ClosedDate="2017-02-16T14:11:26.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="13773" PostTypeId="2" ParentId="13770" CreationDate="2017-02-15T15:31:37.013" Score="1" Body="&lt;p&gt;One limitation is that a spring, by definition, exerts a force proportional to its displacement so once you compress it the returning force is what it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From  your description it actually sounds like what you  really want its a &lt;strong&gt;damper&lt;/strong&gt; which provides a resistance proportional to the speed of movement and doesn't provide a returning force when it isn't being displaced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can certainly get dampers off the shelf which provide damping in one direction only which sounds like what you want this is one example picked at random : &lt;a href=&quot;http://www.slamproof.co.uk/epages/BT3287.sf/en_GB/?ObjectPath=/Shops/BT3287/Products/%22HD22/100%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.slamproof.co.uk/epages/BT3287.sf/en_GB/?ObjectPath=/Shops/BT3287/Products/%22HD22/100%22&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also have a compression and extension dampers of different rates  in parallel if you still want some damping in one direction. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-02-15T15:31:37.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13774" PostTypeId="1" CreationDate="2017-02-15T17:35:21.603" Score="1" ViewCount="110" Body="&lt;p&gt;I want to know something about the mechanical properties of the MAX phases. How ductile can $\text{Ti}_3\text{SiC}_2$ for example be at elevated temperatures? Can it be as ductile as Zinc or Aluminium?&lt;/p&gt;&#xA;" OwnerUserId="6487" LastEditorUserId="1832" LastEditDate="2017-02-15T18:29:58.567" LastActivityDate="2019-09-16T11:55:56.517" Title="How much ductile can be the MAX phases?" Tags="|mechanical-engineering|materials|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13775" PostTypeId="2" ParentId="13772" CreationDate="2017-02-15T18:15:37.280" Score="0" Body="&lt;p&gt;Yes this is a simple operation.  The only real consideration is the shape/size of the cutter. Obviously a tapered cutter will carve a path where the width is proportional to the depth. Therefore some understanding of the toolpath and interference with other objects or cuts is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is easily visualized and controlled by most CAM applications once the machine setup is understood. The DXF files may also need to be cleaned up if they were produced from a raster conversion.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-02-15T18:15:37.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13776" PostTypeId="1" CreationDate="2017-02-15T19:22:13.690" Score="0" ViewCount="897" Body="&lt;p&gt;How would you determine the flow rate of a liquid that flows through a narrow microneedle channel and then through the opening of a funnel, i.e. pouring liquid through the other end of a funnel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've chosen arbitrary diameters but the pressures are from literature.&#xA;The diameter of the lumen/tip is 50 $\mu$m and the base diameter is 1 mm. The pressure at the tip is 25 mmHg and the other end is open to the air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to split the problem into two parts: first determine the flow rate through the straight part of the needle, and then determine the flow rate through the funnel. I calculated the flow rate in the first part using Poiseville's equation but I think that there must be another model for narrow channels or fluids with capillary action coupled with initial pressure. The Reynold's number is clearly in the turbulent range for the first part and I'm not sure if the numbers should be this high for narrow channels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a fluid model that's better suited for microneedles?&lt;/p&gt;&#xA;" OwnerUserId="10057" LastEditorUserId="10057" LastEditDate="2017-02-15T21:36:57.700" LastActivityDate="2017-11-01T23:07:46.040" Title="How to calculate flow rate of viscous fluids through a funnel-shaped needle?" Tags="|fluid-mechanics|microfluidics|fluid|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13777" PostTypeId="2" ParentId="13770" CreationDate="2017-02-16T02:12:42.000" Score="2" Body="&lt;p&gt;You can get something very similar to this with belleville washers. A stack of belleville washers takes load only in compression. It has zero stiffness in tension. I cannot find any good illustrations of this on google, but what you can do is to set up two separate stacks of washers, one for each direction. In ASCII art it looks kinda like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;|&gt;&gt;&gt;&gt;&gt;&gt;|&amp;lt;&amp;lt;&amp;lt;&amp;lt;&amp;lt;&amp;lt;&amp;lt;|&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the middle bar is moved to the left, the left stack is in compression and the right stack is in tension.  Since the belleville washers have no stiffness in compression, the force on the bar is controlled only by the left stack.  If you move the middle bar to the right, the right stack is in compression and the left stack is in tension.  Then, to get an anisotropic spring, just put a different number (or size) of washers in the left stack versus the right.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-02-16T02:12:42.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13779" PostTypeId="2" ParentId="13774" CreationDate="2017-02-16T14:53:24.230" Score="1" Body="&lt;p&gt;Relating to this link &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1359645401000118&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mechanical properties of polycrystalline Ti3SiC2&lt;/a&gt;, I found that dense polycrystalline Ti3SiC2 samples were fabricated by reactively hot-isostatic pressing (HIPing) a mixture of elemental Ti, Si and C powders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanical properties, including load–strain response, bending strength, fracture toughness and crack propagation, were investigated from ambient temperature to 1573 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Non-linear stress–strain responses were observed in the polycrystalline Ti3SiC2 materials at ambient temperature. It is conceivable that the inelastic deformation is attributable to micro-deformations that consist of slip between micro-lamellae within individual grains and the formation of microcracks between grains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polycrystalline Ti3SiC2 exhibited a brittle-to-ductile transition at about 1473 K; above this the Ti3SiC2 samples deformed plastically and exhibited high strains (&gt;1.5%), whereas below 1373 K only limited inelastic deformation was observed prior to fracture. The mode I fracture toughness, $K_{ic}$, was measured by the single-edge notched beam (SENB) method to be 4.52 MPa $m^2$ at ambient temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both fracture strength and fracture toughness decrease only slightly with increasing temperature up to 1273 K, above which they decrease more rapidly and reach half of their room-temperature values by 1473 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in regards of the comparison with Zinc and Aluminum, I am not certain about it but you can always search the properties of Zinc and Aluminum and compare them with the mechanical properties found in this link &lt;a href=&quot;http://www.azom.com/article.aspx?ArticleID=2363&quot; rel=&quot;nofollow noreferrer&quot;&gt;Titanium Silicocarbide ( Ti3SiC2 )- Properties and Applications&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10028" LastActivityDate="2017-02-16T14:53:24.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13780" PostTypeId="1" AcceptedAnswerId="13782" CreationDate="2017-02-16T15:11:21.757" Score="2" ViewCount="56" Body="&lt;p&gt;I need to understand some basic concepts about the cooling characteristics of a black &quot;polymer&quot;-and-zinc-alloy handgun (specifically, &lt;a href=&quot;http://www.hi-pointfirearms.com/Hi-Point-handguns/9MM_handgun.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this gun was left on the ground under daytime sunlight, I assume that both the ground and the gun would get warmer.  I assume, also, that after the sun went down, both the ground and the gun would begin to cool.  My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) would the gun warm under sunlight to a higher temperature than the ground, because it is black and/or because of what it's made of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)  after sundown, would the gun cool at the same rate as the ground?  Or, would the rate of cooling be different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  I should clarify that the surrounding ground is essentially brown dirt/mud.&lt;/p&gt;&#xA;" OwnerUserId="10071" LastActivityDate="2017-02-16T16:59:19.130" Title="Rate of Cooling of a Black &quot;Polymer&quot; Handgun" Tags="|heat-transfer|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13781" PostTypeId="2" ParentId="13780" CreationDate="2017-02-16T15:53:22.193" Score="0" Body="&lt;p&gt;It would likely be a higher temperature than the ground, as it would be heated by the suns radiation, and black generally absorbs better (if the surface finish were too smooth it may change this).  The material may also &lt;em&gt;feel&lt;/em&gt; warmer than the ground, even if they were at the same temperature.  This relates to the conductivity of the materials and how much heat they would let off into your hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rates of cooling would be very different.  The black of the gun will also make it emit it's radiated heat faster once the surroundings drop below the surface temperature due to radiation.  It will also cool down a lot faster than the ground because it has a much smaller mass, and higher relative surface area.  The ground is a big thermal mass which slowly gains and loses heat.  A small object generally gains and loses heat fast.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastEditorUserId="7736" LastEditDate="2017-02-16T16:59:19.130" LastActivityDate="2017-02-16T16:59:19.130" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13782" PostTypeId="2" ParentId="13780" CreationDate="2017-02-16T16:16:20.737" Score="2" Body="&lt;p&gt;There are three main heat transfer mechanisms:  conductive, convective, and radiative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conductive transfer means direct transfer from one item to another it's in contact with. The transfer rate depends on the empirically determined thermal conductivity of both materials.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convective refers to heat transfer between an object and a moving fluid such as wind.  This depends heavily on the total surface area available (such as fins on a radiator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiative means absorption or emission of photons.  The energy transfer rate depends on the surface properties, called &quot;emissivity&quot; or equivalently &quot;absorptivity&quot; and is wavelength-dependent for any given material.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, so with that in  mind, yes a black object generally will absorb more sunlight than a lighter shade of ground/dirt.  This is a generality, as the absorptivity in the infrared may be different than in the visible.   At the same time, the conductive heat loss/gain to the ground depends on the total contact area, and further will be much greater if the ground is damp, because water is good at conducting a lot of heat 'fast.'  After dark, the radiative processes are sort of reversed.  It's not an exact reversal because the spectral radiation rates depend on the material AND on the temperature of the object. (actually the radiation rates apply all the time; it's just that in daylight each object absorbs more light than it radiates until it reaches a thermal equilibrium).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sum, the black object probably will get cold faster, and get colder than the ground.  As a practical example, come to New England in the late Fall.  On cool nights, black stuff such as roof shingles often frost over when other things like white walls don't, because of radiative cooling. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-02-16T16:16:20.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13783" PostTypeId="2" ParentId="13758" CreationDate="2017-02-16T16:26:18.770" Score="2" Body="&lt;p&gt;To answer the other part of your question, the temperature is that in the chamber, but note that that figure assumes perfect combustion, which may not be the optimum point for specific impulse in your geometry. Many rockets run fuel rich as a reducing atmosphere in the hot bits is easier to cope with then an oxidising one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest thermal flux is usually at the nozzle throat, as that is where the combination of high pressure, high temperature and high mass flow come together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nozzle is a complex thermodynamic machine, which trades hot and high pressure, for cooler, a pressure ideally just above external atmospheric, and high velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ablative nozzles, graphite is a magic material. &lt;/p&gt;&#xA;" OwnerUserId="10072" LastActivityDate="2017-02-16T16:26:18.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13784" PostTypeId="2" ParentId="13572" CreationDate="2017-02-16T16:32:43.857" Score="0" Body="&lt;p&gt;Ziegler-Nichols tuning will not work for this system because there is no energy loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that your system oscillates regardless of the value of $K$ that you choose is because there is no energy loss in your model. All of the energy is stored as potential gravitational energy (which can be ignored for now), kinetic energy in the mass, and &quot;potential&quot; energy in the proportional control (it acts like a spring in your system connecting the mass to the desired position). So if your system starts where $x \neq x_{target}$ you can think of it as pulling and releasing a spring (without any friction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally a real system will have some amount of air resistance, which I believe is proportional to the square of velocity. That will give you a non-linear system, which will require non-linear control techniques, so for a fun exercise I would suggest instead using the following equation to approximate the damping in your system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$mx'' + bx' = gm + T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for what value to pick for $b$, who knows? It will depend largely on the aerodynamics of your aircraft. Ziegler-Nichols will work for this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I would recommend using more mathematical methods for tuning the system (like having a target closed-loop overshoot and rise time, for example) rather than Ziegler-Nichols, which is primarily an experimental tuning method for when we do not have an accurate model of the system plant.&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2017-02-16T16:32:43.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13787" PostTypeId="1" CreationDate="2017-02-16T21:14:17.243" Score="1" ViewCount="41" Body="&lt;p&gt;I am a CEO of New Beginning Homes. I am looking into building a waterless non composting toilet. This toilet will be run using gears and a handle that turns the gears to push the waste through a decomposing process where the waste will then be able to be used again for agricultural purposes. I am already aware of incinerating toilets. Any ideas on what could be used to break down the waste and on what types of gears would be best for this type of thing?&lt;/p&gt;&#xA;" OwnerUserId="10078" LastEditorUserId="10078" LastEditDate="2017-02-17T19:08:45.057" LastActivityDate="2017-02-17T19:08:45.057" Title="How to Build waterless non composting toilets?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="5" ClosedDate="2017-02-17T15:18:34.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="13789" PostTypeId="1" CreationDate="2017-02-17T00:19:00.807" Score="2" ViewCount="243" Body="&lt;p&gt;I am looking for plastic film in the 10-50 micron range. The property I value most is high strength to weight ratio. Is there some material that is superior to BoPET/Mylar?&lt;/p&gt;&#xA;" OwnerUserId="10080" LastEditorUserId="421" LastEditDate="2017-02-17T02:47:38.357" LastActivityDate="2017-02-17T02:47:38.357" Title="Is there any plastic film material stronger than Mylar?" Tags="|materials|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13790" PostTypeId="1" CreationDate="2017-02-17T03:06:46.863" Score="-4" ViewCount="740" Body="&lt;p&gt;I see some track race cars, especially old ones made in the 1960s-70s, are designed with the heavy components of the car sitting between the rear wheels. In fact, some of old rear-/mid-engine cars use transmissions where the differential goes between gearbox and the engine to provide even closer engine placement to the center of rear axle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the advantages of this layout, when the center of gravity is placed over the center of the rear axle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my understanding front wheel drive provides stability, so if the engine is placed at the rear axle does it providing better grip?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see some modern track vehicles have the engine placed even farther back. Does this mean the older configuration is not as effective as they thought back in old days?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned that a car with instantaneous center of rotation goes further with slip from the rear wheels. Is this reason why they placed the engine at the geometric center of car in modern track cars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do cars with smaller engines perform better with the engines placed over the rear axle, or is this an illusion of longer wheelbases and smaller engines in modern cars?&lt;/p&gt;&#xA;" OwnerUserId="8826" LastEditorUserId="368" LastEditDate="2017-02-17T16:17:42.613" LastActivityDate="2017-02-17T16:17:42.613" Title="Effects of center of gravity very close to rear wheels of a car" Tags="|automotive-engineering|engines|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ClosedDate="2017-02-19T02:14:00.387" ContentLicense="CC BY-SA 3.0" />
  <row Id="13791" PostTypeId="1" CreationDate="2017-02-17T04:00:47.587" Score="1" ViewCount="665" Body="&lt;p&gt;The adhesion between two solids is characterized by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Adhesion#Surface_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;adhesion energy&lt;/a&gt;. If you have two solids adhered at an interface and you want to pull them apart, I understand that you have to supply the energy described by the surface energy (J/m^2) times the area (m^2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, it seems like there should also be a minimum adhesive force. If I try to separate two materials at an interface using a tiny force, it is never going to act over a distance so it will never do any work. There is then, presumably, some minimum amount of force required to separate them. This force would of course be specific to the angle at which it was applied (pulling vs peeling, for example). Is this adhesive force a property used for analysis? Is it generally extensible, ie, if I double the surface area of an interface does the specific adhesive force (N/m^2) stay the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondarily, what is the relationship between the force and work required to de-adhere to materials at an interface? If the energy required is characterized by a adhesive energy, $E$, and I pull them apart with force $F$, what is the characteristic length over which the work is done? $\int dE/dF$? Is the adhesive energy you might look up in a table not actually appropriate for finding the required force?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2017-11-16T02:49:38.920" Title="Adhesion energy vs force" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13792" PostTypeId="1" AcceptedAnswerId="13797" CreationDate="2017-02-17T04:13:21.307" Score="4" ViewCount="165" Body="&lt;p&gt;I am working on a DIY project, I am trying to replicate a bar compressor which for some reason is only sold in Europe so I thought it would be fun to make my own without having to waste so much material. I simply want to make sure the force going down with the plate somewhat replaces a person to jump in the trashcan. Somewhere above 80+ lbs, since I am no engineer I was wondering how would one go about finding the force applied going down to the trash if about 50lbs is pulling down the lever. I understand that materials are key to this but as an amateur I just wanted to know how to get started. Sorry for posting here but I thought it would suit this page the most for this question. Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8CRBu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8CRBu.jpg&quot; alt=&quot;Drawing of beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aZwty.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aZwty.png&quot; alt=&quot;Actual use of bar&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8165" LastActivityDate="2017-02-18T15:00:33.097" Title="Finding downward force of lever bar" Tags="|mechanical-engineering|civil-engineering|applied-mechanics|statics|dynamics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13796" PostTypeId="2" ParentId="13792" CreationDate="2017-02-17T09:09:49.787" Score="-3" Body="&lt;p&gt;The force at the middle would be like 2 times bigger than the input. I calculated for this exact situation and it would be around 94lbs.&lt;/p&gt;&#xA;" OwnerUserId="9911" LastActivityDate="2017-02-17T09:09:49.787" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13797" PostTypeId="2" ParentId="13792" CreationDate="2017-02-17T09:54:35.357" Score="4" Body="&lt;p&gt;The input force is 18.5+20.5 ft away from the fulcrum and the load is 18.5 away from the fulcrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the force applied to the load is $\frac{(18.5+20.5)}{18.5} = 2.1$ times greater than the input force or for an input force of 50 lbs you get an output force of &lt;strong&gt;105 lbs&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this assumes that all forces are applied &lt;em&gt;perpendicular&lt;/em&gt; to the line between point and fulcrum. If this is not the case then you need to decompose the forces into tangential and normal components.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-02-17T09:54:35.357" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13798" PostTypeId="1" AcceptedAnswerId="13805" CreationDate="2017-02-17T09:55:55.440" Score="1" ViewCount="5821" Body="&lt;p&gt;I have a rectangular shape with uniform mass, approx 6m x 1m. It rotates around an axis in the centre of its width but 1m from one end. The axis has some friction but this is not known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need to work out is how much force is required to start rotating this object from different positions from the axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fear I may be plunged out of my depth but if anyone could point me in the right direction it would be really appreciated!&lt;/p&gt;&#xA;" OwnerUserId="10086" LastActivityDate="2017-02-17T20:22:08.220" Title="Force required to rotate an object on a plane - moment of inertia?" Tags="|mechanical-engineering|applied-mechanics|moments|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13800" PostTypeId="2" ParentId="13798" CreationDate="2017-02-17T13:13:24.253" Score="5" Body="&lt;p&gt;Since the friction is not known, the torque to overcome that friction can't be known either.  You therefore can't do what you want since the problem is under-constrained.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-02-17T13:13:24.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13802" PostTypeId="2" ParentId="13790" CreationDate="2017-02-17T14:40:19.490" Score="3" Body="&lt;p&gt;One possible reason for placing the engine and most of the weight at the back of the car would be to improve rear wheel tire friction. This would be useful if the car has a rear-wheel drive, since the car's ability to accelerate is limited by the amount of torque the wheels can support before slipping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding weight to the back of the car would increase the normal force exerted by the rear wheels on the track. Since friction is proportional to the normal force, increasing the weight in the rear will increase the friction, thus improving the acceleration of the car.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2017-02-17T14:40:19.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13805" PostTypeId="2" ParentId="13798" CreationDate="2017-02-17T20:22:08.220" Score="1" Body="&lt;p&gt;If we neglect friction and the object is rotating about its centre of mass then it will experience an &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_acceleration&quot; rel=&quot;nofollow noreferrer&quot;&gt;angular acceleration&lt;/a&gt; proportional to the torque applied and its moment of inertia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is a direct analogue of the linear acceleration equation &lt;em&gt;F = ma&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But instead of force we use torque (force x radius) and instead of mass we use &lt;a href=&quot;https://en.wikipedia.org/wiki/Moment_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;moment of inertia&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as with linear acceleration, if we ingnore friction any force greater than zero will produce some acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is slightly different if the axis of rotation is offset from the centre of mass. In this case the weight of the body will exert a torque on it ie it will behave as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pendulum&quot; rel=&quot;nofollow noreferrer&quot;&gt;pendulum&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for your example with a 1m x 6m plank rotating around a point 1m from one end the centre of mass will lie 3m from each end so it can be simplified to a point mass connected to a pivot by a mass-less rigid bar 2m long. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be a constant force acting on the point mass of Mg acting vertically downwards. This will create a  torque when it is displaced at some angle x from  the horizontal so, again any torque greater than zero will produce some displacement but in order to get it horizontal you need to support the its full wieght so :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the minimum constant torque to rotate it in a full circle is given by :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T &gt; M.g.r&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where M is the mass of the object, g is acceleration due to gravity and r is the distance from centre of mass to the pivot point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking this is the torque required to hold it stationary in the horizontal position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To work out the holding torque for other positions you need to use trigonometry to work out the resultant toque from the vertical force due gravity. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-02-17T20:22:08.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13806" PostTypeId="1" CreationDate="2017-02-17T20:39:52.517" Score="0" ViewCount="45" Body="&lt;p&gt;I'd like to design a minimal thin light switch which holds an illuminant with standardized socket, e.g. E14, E27, and has itself such a socket to be screwed in a support and a switch which will be very small and distant from the thread and socket. Apart from the switch the object should be 3d printable with PLA for the thread and socket and the conductive parts will be printed as well or mounted after the print.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there's the question about the thickness of the PLA support at the buttom where the ground and phase are isolated against each other. I found no information which material is used in average sockets and which criteria it has to fulfill.&lt;/p&gt;&#xA;" OwnerUserId="9780" LastActivityDate="2017-02-18T14:16:55.980" Title="Thickness of a 3d print general purpose plastic/PLA 230 V ~ isolator?" Tags="|electrical-engineering|3d-printing|lighting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13807" PostTypeId="1" CreationDate="2017-02-17T20:50:42.860" Score="1" ViewCount="184" Body="&lt;p&gt;I use standard end mills for cutting carbon fibre sheets. Generally the cutters are 2 to 3 mm diameter, with a flute length of about 8 mm. I use 3- and 4-flute end mills. These work quite well if they are surfaced (TiN) or solid carbide. Uncoated or HSS tooling becomes dull very quickly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I need to work with an aramid composite, where the core is carbon fibre and the outer layers are Kevlar. Overall thickness is 2.5 to 5.0 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface layers are not cutting cleanly - leaving a fine fuzz that is a pain to remove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cutter is cooled by airjet, spindle speed is at the maximum setting of 10,000 RPM. I have tried feed rates from 5 mm/sec to 25 mm/sec at various cutting depths without any significant improvement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I improve the quality of the cut? I would like to try alternative types of cutters or machining strategies.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="368" LastEditDate="2017-03-06T17:17:23.997" LastActivityDate="2021-01-01T15:42:37.727" Title="Milling aramid composites produces unwanted fuzz at surface" Tags="|machining|cnc|composite|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13808" PostTypeId="1" CreationDate="2017-02-18T08:19:42.627" Score="2" ViewCount="4453" Body="&lt;p&gt;I have an MDOF system with 4 degrees of freedom with forced input at one of the dof (4th dof)(it is a sinusoidal wave with a particular frequency expected to be generated using a shaker); displacement response is measured for this system. I would like to extract mode shapes for this system experimentally, but in order to achieve this objective I took up a theoretical case for understanding various complications and problems in it. So, now my system can be represented as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;\begin{equation}&#xA;G_{YY}=H G_{YX}&#xA;\end{equation}&#xA;  \begin{Bmatrix}&#xA;Y_{1}\\ &#xA;Y_{2}\\ &#xA;Y_{3}\\ &#xA;Y_{4}&#xA;\end{Bmatrix}=\begin{bmatrix}&#xA;H_{11} &amp;amp; H_{12} &amp;amp; H_{13} &amp;amp; H_{14}\\ &#xA;H_{21} &amp;amp; H_{22} &amp;amp; H_{23} &amp;amp; H_{24}\\ &#xA;H_{31} &amp;amp; H_{32} &amp;amp; H_{33} &amp;amp; H_{34}\\ &#xA;H_{41} &amp;amp; H_{42} &amp;amp; H_{43} &amp;amp; H_{44}&amp;amp; &#xA;\end{bmatrix}&#xA;  \begin{Bmatrix}&#xA;0\\ &#xA;0\\ &#xA;0\\ &#xA;X_{4}&#xA;\end{Bmatrix}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here H is the Frequency Response Function (FRF) and it can be seen that only one column would be needed to obtain the mode shapes of the system. Now, writing a code on matlab give me:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; FRF(i,:)=GYY./GYX&#xA;&amp;gt; plot(fr,abs(real(GYY./GYX)))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The amplitudes and shapes could be extracted from the above expressions, but my concern is on obtaining mass normalized mode shapes, i.e. &#xA;\begin{equation}&#xA;\phi^{T}\left[M_A\right]\phi=I&#xA;\end{equation}&#xA;If I am not wrong, it should be possible in the case of input + output experimental modal analysis. I know that there would be many complications but at least in the case of a theoretical case it should come right. I generated a code for the above case by using mdof equations of motion for a  4dof MDOF system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I am supposed to get, as verified with eigen analysis of the mass M and stiffness K matrices:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   -0.1442    0.3651    0.4152   -0.2710&#xA;   -0.2710    0.3651   -0.1442    0.4152&#xA;   -0.3651   -0.0000   -0.3651   -0.3651&#xA;   -0.4152   -0.3651    0.2710    0.1442&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But from the frf the the first mode shape is extracted to be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.0098&#xA;0.0186&#xA;0.0254&#xA;0.0296&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It looks porportional with the actual one but is not scaled to the real values to satisfy mass normalization.&#xA;Where am I going wrong?&lt;/p&gt;&#xA;" OwnerUserId="10097" LastActivityDate="2017-02-18T08:19:42.627" Title="How to get mass normalized mode shapes?" Tags="|dynamics|vibration|modal-analysis|eigenvalue-analysis|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13809" PostTypeId="1" CreationDate="2017-02-18T11:49:57.010" Score="0" ViewCount="367" Body="&lt;p&gt;what do we mean by lightest? I mean, 10 g of aluminum is just as light 10 g of steel. So what's the point? &lt;a href=&quot;https://i.stack.imgur.com/vchK5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vchK5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10098" LastActivityDate="2017-02-18T19:35:49.990" Title="What is the lightest known metal?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13811" PostTypeId="1" AcceptedAnswerId="13812" CreationDate="2017-02-18T12:35:42.947" Score="12" ViewCount="2666" Body="&lt;p&gt;I've heard that in the past, spray cans (aerosol cans) contained Freon. I found that the Montreal Protocol made the usage of Freon illegal due to its effect on the ozone layer, but I can't find the reason for why Freon was used in the first place. I understand its role in refrigeration systems, but I fail to see a functionality for spray cans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why would anyone use Freon in spray cans, what was its contribution? I assume that it affected the pressure inside somehow, but it's just a guess.&lt;/p&gt;&#xA;" OwnerUserId="8419" LastEditorUserId="90" LastEditDate="2017-02-18T15:24:06.283" LastActivityDate="2017-02-18T23:17:23.070" Title="Why was Freon used in spray cans before it was declared a threat to ozone?" Tags="|thermodynamics|pressure|compressed-air|refrigeration|compressed-gases|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13812" PostTypeId="2" ParentId="13811" CreationDate="2017-02-18T12:48:12.793" Score="18" Body="&lt;p&gt;Spray cans work by containing a propellant which has a boiling point a small amount below room temperature. When the can is sealed the liquid and vapour reach equilibrium at a relatively modest pressure. When the spray valve is opened this pressure forces out the contents as an aerosol and more of the propellant evaporates as the internal pressure drops. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has two advantages firstly the can doesn't need to be anywhere near as strong as if the propellant was a fully pressurised gas and it provides a fairly constant pressure untill the propellant is exhausted without needing mechanical regulators which would add cost and complexity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is pretty much the same principal as a  butane lighter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you need in a propellant is something with a very specific boiling point which provides the required vapour pressure at the expected operating temperature. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the comments butane and/or propane are now often used as propellants in aerosol type cans indeed 'canned air' dusters used for cleaning computers etc is often just butane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One downside of butane is that there is an inherent cooling effect as the liquid evaporates and the gas expends which can cause a significant cooling effect and, as the boiling point of butane is a bit on the low side this can cause significant cooling of the contents of the can and a consequent drop-off in pressure. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-02-18T23:17:23.070" LastActivityDate="2017-02-18T23:17:23.070" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13813" PostTypeId="2" ParentId="13806" CreationDate="2017-02-18T14:16:55.980" Score="1" Body="&lt;p&gt;Your major gotcha in this is probably thermal rating which is not going to be exactly wonderful with a PLA part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a lamp does not connect between ground a phase (except in weird circumstances), it connects between phase and neutral both of which are viewed as live current carrying conductors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a reason most electrical accessories are made from a thermosetting material!  &lt;/p&gt;&#xA;" OwnerUserId="10072" LastActivityDate="2017-02-18T14:16:55.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13814" PostTypeId="2" ParentId="13792" CreationDate="2017-02-18T15:00:33.097" Score="1" Body="&lt;p&gt;Balance the moments acting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EffortForce x 39 x cos 30 = 18.5 cos 30 x resulting force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or more accurately,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pBX46.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pBX46.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sin(90+ theta - alpha ) x 39 x effortForce = cos(theta) x 18.5 x requiredForce&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation will give you answer for any angle of level. &lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2017-02-18T15:00:33.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13815" PostTypeId="1" CreationDate="2017-02-18T18:09:51.400" Score="0" ViewCount="276" Body="&lt;p&gt;Is it possible to improve strength of a hollow cylinder against radial load by making holes in it or converting it into scaffolding like structure without changing material, internal dia, external dia and length?&lt;/p&gt;&#xA;" OwnerUserId="10100" LastActivityDate="2017-02-18T19:27:35.810" Title="Strength optimization of a hollow cylinder" Tags="|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13816" PostTypeId="2" ParentId="13815" CreationDate="2017-02-18T19:03:51.620" Score="4" Body="&lt;p&gt;Generally speaking, the strength of a structure for a static load cannot be improved by removing material. Usually material is removed in low-stress regions to improve the strength-to-weight ratio, but that does not improve the overall strength of the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some cases where removing material could help improve strength would be in rounding a sharp edge to remove a stress concentration, or reducing the mass of a part to make it less susceptible to dynamic loads of a certain frequency (by changing its resonant frequencies).&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2017-02-18T19:03:51.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13817" PostTypeId="2" ParentId="7947" CreationDate="2017-02-18T19:13:07.260" Score="2" Body="&lt;p&gt;The elephant in the room is that while gasses are compressible, hydraulic oil is only negligibly so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This difference has two major effects:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, the output volume flow is limited by the intake volumetric flow divided by the compression ratio. Let's say we have a hypothetical compressor that pulls in 10 L of air per second at STP, and raises the pressure to 300 bar. Once the output gasses have cooled down from temperatures best described as &quot;red hot&quot;, the output volume will be 1/30th of a liter per second. All that work, and we only have 30 mL of compressed gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second issue is that as you compress a gas it heats up. This is an effect that is inherent to the physics of compression. The heat (motion) already present in gasses doesn't go away when they are compressed. When our example 10 L (10,000 mL) of air is smashed into a much smaller volume of 33 mL, all the heat energy the air already contained remains, but in a much smaller volume. The air has been compressed to occupy a space that is 0.3% of its original volume - a size reduction of 99.7%. All of that heat has to be absorbed and dissipated by the compressor mechanism somehow before we will end up with cool high pressure gasses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that air at 300 bar has an O2 partial pressure of around 63 Bar. High pressure O2 is extremely dangerous around oils or anything else flammable. For example, the oils like those used in and are essential to seal hydraulic systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your energy source doesn't matter, whether hydraulic, electric, or steam. They are just how you apply the energy to the system, and make very little difference to the complexity of the compressor mechanism itself.&lt;/p&gt;&#xA;" OwnerUserId="10072" LastEditorUserId="20783" LastEditDate="2019-06-07T21:10:30.720" LastActivityDate="2019-06-07T21:10:30.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13818" PostTypeId="2" ParentId="13815" CreationDate="2017-02-18T19:27:35.810" Score="0" Body="&lt;p&gt;Use thicker wall perforated cylinders and add hoop/bulkhead rings or helical reinforcements. They increase radial resistance against tendency to buckle (as in underground gas tanks). A slight curving like hypar also would help but at a cost.&lt;/p&gt;&#xA;" OwnerUserId="1270" LastActivityDate="2017-02-18T19:27:35.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13819" PostTypeId="2" ParentId="13809" CreationDate="2017-02-18T19:35:49.990" Score="2" Body="&lt;p&gt;This is not a very well worded question. One problem is that the definition of a 'metal' is not entirely unambiguous depending on context, for example a cosmologist might consider anything which isn't hydrogen or helium to be a metal and chemists might disagree amongst themselves about exactly what is and is not a metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In engineering there is also the complication that the majority of metals are really metal alloys, normally this difference is trifling but of you are talking about 'known metals' that implies a chemistry type definition. A physicist may argue that hydrogen has metallic phases under certain conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in the context using 'lightest' is actually ok as chemists and physicists  will talk of heavy and light elements, although in an engineering context using density is less ambiguous as engineers are more interested in ratios between density and strength etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much less ambiguous way to phrase the question would be 'which of the following materials has the lowest density ?'. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-02-18T19:35:49.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13821" PostTypeId="2" ParentId="13770" CreationDate="2017-02-18T20:34:56.407" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vgRc5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vgRc5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially a bi-modulus or bilinear characteristic? In this suggestion two strips are pinned together and one end is fixed at the other. When compressed the buckling resistance is low. In tension strip stresses in metal are high, so acts as a stiff spring.&lt;/p&gt;&#xA;" OwnerUserId="1270" LastActivityDate="2017-02-18T20:34:56.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13822" PostTypeId="2" ParentId="13791" CreationDate="2017-02-18T20:43:56.920" Score="1" Body="&lt;p&gt;Mechanical interlocking due to surface roughness plays a significant role in practical adhesion (e.g. the mean peeling force required to de-adhere an interface).  This is in addition to the surface energy values of the materials as they appear in the literature, if they were measured using a surface force apparatus with atomically smooth material coupons.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contact areas and lines of force for the purposes of mechanical analysis will vary depending on the size and shape of asperities, presence of contaminants, and wettability by the adhesive—possibly by orders of magnitude (cf. stress risers).   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two very rough surfaces with large surface areas, effectively wet by an adhesive, or having few interfacial voids through plastic deformation of each other's asperities, will have a higher adhesive energy as a result.  Surfaces of the same materials, but smoother, will have lower adhesive energy.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Pocius (2012), Adhesion and Adhesives Technology, 3e, § 6.5 for a concise discussion on the topic.  Most textbooks on tribology have a section on adhesion.&lt;/p&gt;&#xA;" OwnerUserId="10105" LastActivityDate="2017-02-18T20:43:56.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13823" PostTypeId="2" ParentId="4275" CreationDate="2017-02-18T22:09:31.490" Score="1" Body="&lt;p&gt;With a 20 ms pulse width you really should not be using interrupts directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual thing is to have some timer generating periodic interrupts at maybe 250Hz or so, then if on two consecutive timer events you read the input as active you can be reasonably sure you have a pulse lasting at least 1/Timer frequency, similar logic can be applied to filtering the falling edge.  &lt;/p&gt;&#xA;" OwnerUserId="10072" LastActivityDate="2017-02-18T22:09:31.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13824" PostTypeId="1" CreationDate="2017-02-19T00:53:37.193" Score="3" ViewCount="426" Body="&lt;p&gt;When I was little I would often take apart my toys and found that if they contained any gears or motors they'd always be coated in a white lubricant. What is this substance?&lt;/p&gt;&#xA;" OwnerUserId="10108" LastEditorUserId="368" LastEditDate="2017-02-21T16:41:00.830" LastActivityDate="2017-02-21T16:41:00.830" Title="What's the name of the white lubricant that children's toys often use?" Tags="|mechanical-engineering|tribology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13825" PostTypeId="2" ParentId="13824" CreationDate="2017-02-19T01:19:15.273" Score="8" Body="&lt;p&gt;It is &lt;em&gt;lithium grease&lt;/em&gt; aka &lt;em&gt;white grease&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often sold in aerosol can packaging. It adheres well to moving parts and tends to work well for long term lubrication of low to moderate load mechanisms. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-02-19T01:19:15.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13826" PostTypeId="1" AcceptedAnswerId="13955" CreationDate="2017-02-19T03:10:11.253" Score="1" ViewCount="786" Body="&lt;p&gt;I encountered a problem in measuring electrical energy using a multi ratio current transformer (CT). The catalogue of CTs is unavailable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have little information about the type of CT I am using. I have several questions about this matter. Firstly, the CT I work with has 7 connections:&#xA;X7,100,200,300,400,500&amp;amp;600.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The connection for ratio 100/5 was on X7 &amp;amp; 100. Something that is confusing  me is that X7 a common connector and if connection X7 to 100  measures 100/5, are other windings from 200 to 600 in circuit or open? If they are in circuit how does it produce a 100/5 ratio, and if they are open how it that possible for a CT?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, the power transformer has delta connections and the CT connection is a Star. After I have connected the meter to the CT I get an error in energy metering. Is it possible that this error is due to the star connection of the CT and should it be a delta connection, as for the power transformer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its impossible to change the CT connection to delta. Can I define any coefficient factor to remove this error with out changing the circuits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power transformer ratio is 14400/3300, with a delta connection and the &#xA;CT's ratio is 200/5, with star connection.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/jKvz5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4411" LastEditorUserId="4411" LastEditDate="2017-02-24T19:44:42.033" LastActivityDate="2017-03-01T02:15:01.060" Title="Multi Ratio Current Transformers" Tags="|electrical-engineering|sensors|measurements|circuits|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13828" PostTypeId="1" CreationDate="2017-02-19T13:52:01.237" Score="1" ViewCount="526" Body="&lt;p&gt;Usually, the frequency of a computer's CPU is expressed in GHz. &#xA;But does this mean that the individual transistors etched on the chip also switch on and off billions of times per second? Or are the high frequencies of CPUs achieved thanks to some &quot;group effect&quot; ? (since there are billions of transistors in a CPU, they may not need to switch at gigahertz frequencies themselves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: For sure, this kind of information can easily be found for &quot;macroscopic&quot; scale MOSFETs. But I am absolutely not sure that they can be compared at all to their nanometric counterpart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="8745" LastActivityDate="2017-02-19T22:36:08.893" Title="Switching speed of a (nano) transistor" Tags="|electrical-engineering|cmos|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13829" PostTypeId="1" CreationDate="2017-02-19T16:53:53.900" Score="0" ViewCount="120" Body="&lt;p&gt;I have a line in my text that states the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;For systems with low boiling points (e.g. Ethane-Ethylene, propane-propylene etc.), high pressures are used so the vapour leaving the column at the top is cooled by water.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand the utility of high pressure over here. Isn't usage of water in the condenser already the case?&lt;/p&gt;&#xA;" OwnerUserId="2569" LastActivityDate="2017-02-20T12:12:53.940" Title="Boiling Point, Pressure and Distillation" Tags="|pressure|chemical-engineering|process-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13830" PostTypeId="1" CreationDate="2017-02-19T17:51:15.810" Score="1" ViewCount="282" Body="&lt;p&gt;It's one of questions for my exam, and I couldn't find anything about inversion related with architecture on the internet. Thanks for help!&lt;/p&gt;&#xA;" OwnerUserId="10113" LastActivityDate="2017-02-19T19:51:31.527" Title="What's the definition of concept of inversion in architecture? Can you also give some example for this and explanation?" Tags="|architecture|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13832" PostTypeId="1" AcceptedAnswerId="13850" CreationDate="2017-02-19T19:00:17.193" Score="1" ViewCount="222" Body="&lt;p&gt;I am building a free flight helicopter for a Science Olympiad competition, and the general rules are as such: &lt;a href=&quot;https://www.soinc.org/sites/default/files/uploaded_files/Helicopters_2017_Diagram.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.soinc.org/sites/default/files/uploaded_files/Helicopters_2017_Diagram.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves a lot of opportunity for math that is well beyond my ability, so my questions are as such:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You get scored based on how long the craft stays in the air, but you can get 25% added (up to 75% total) for each single bladed rotor, is a single bladed rotor a worthwhile investment?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have tried using an airfoil shaped wing at the regional competition, and it failed because of how heavy it was, but people have been suggesting a helical blade shape, what are the mathematical advantages of this over a flat rotor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The central shaft has no given dimensional restriction, what is the optimum length of the core? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="10116" LastActivityDate="2017-02-20T12:16:34.600" Title="Rubber band Powered Free Flight Helicopter" Tags="|mathematics|building-physics|" AnswerCount="1" CommentCount="1" ClosedDate="2017-02-24T13:00:00.453" ContentLicense="CC BY-SA 3.0" />
  <row Id="13833" PostTypeId="1" AcceptedAnswerId="14076" CreationDate="2017-02-19T19:45:36.843" Score="3" ViewCount="1339" Body="&lt;h1&gt;Background&lt;/h1&gt;&#xA;&lt;p&gt;Looking to build a table for a dryer to fit inside a laundry closet. The table must be assembled inside the closet, as such it has three parts: two legs and a metal sheet for the table top. Here's a picture of the table outside the closet:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vzyFI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vzyFI.png&quot; alt=&quot;laundry closet washer&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here's a picture of the dryer on top of the table:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JZbG9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JZbG9.png&quot; alt=&quot;laundry closet washer and dryer&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Weights, Measures, and Materials&lt;/h1&gt;&#xA;&lt;h2&gt;Dryer&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dimensions: 65.5cm x 60cm&lt;/li&gt;&#xA;&lt;li&gt;Weight: 50kg&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Table&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dimensions: 106cm x 63.5cm&lt;/li&gt;&#xA;&lt;li&gt;Material: Steel alloy&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XdBHr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XdBHr.png&quot; alt=&quot;Table Dimensions Wireframe&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Same, but without the wireframe:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JR9qI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JR9qI.png&quot; alt=&quot;Table Dimensions&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Without the table top to show the supporting flaps:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nS1UI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nS1UI.png&quot; alt=&quot;Table Flaps&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Problem&lt;/h1&gt;&#xA;&lt;p&gt;I'm concerned that the table thickness will be insufficient to support the load. A &lt;a href=&quot;https://www.engineersedge.com/calculators/flat-plate-deflection.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;flat plane deflector&lt;/a&gt; calculator is useful for calculating the total load applied to center of plate, but doesn't quite tell me if the distributed load of the dryer (not a point load) will be structurally sound.&lt;/p&gt;&#xA;&lt;p&gt;In comparison, commercial &lt;a href=&quot;http://www.globalindustrial.ca/g/work-benches/tables-stands/machine-tables/welded-12-gauge-machine-tables&quot; rel=&quot;nofollow noreferrer&quot;&gt;heavy gauge machine tables&lt;/a&gt; support over 900kg, so it looks like the design is sound.&lt;/p&gt;&#xA;&lt;h1&gt;Question&lt;/h1&gt;&#xA;&lt;p&gt;Are these calculations reasonable? (Especially pertaining to the forces imparted by the dryer?) That is, will a 1.25cm (approx. ½&amp;quot;) steel alloy table top outlast the dryer?&lt;/p&gt;&#xA;&lt;h1&gt;Related Links&lt;/h1&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.engineersedge.com/beam_bending/beam_bending3.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineersedge.com/beam_bending/beam_bending3.htm&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.engineersedge.com/calculators/flat-plate-deflection.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineersedge.com/calculators/flat-plate-deflection.htm&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.engineeringtoolbox.com/young-modulus-d_417.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineeringtoolbox.com/young-modulus-d_417.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.engineeringtoolbox.com/poissons-ratio-d_1224.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineeringtoolbox.com/poissons-ratio-d_1224.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10117" LastEditorUserId="5411" LastEditDate="2021-02-17T08:40:50.800" LastActivityDate="2021-02-17T16:57:51.890" Title="Custom heavy gauge machine table to support clothes dryer" Tags="|structural-engineering|materials|stresses|mathematics|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="13834" PostTypeId="2" ParentId="13830" CreationDate="2017-02-19T19:51:31.527" Score="2" Body="&lt;p&gt;Hard to say without knowing the material covered in the course. However, it could be referring to Robert Hooke: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As hangs a flexible cable so, inverted, stands a rigid arch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which describes a way to find the funicular form for an arch by inverting the shape of a hanging cable under the same loading. In other words, &lt;a href=&quot;http://dataphys.org/list/gaudis-hanging-chain-models/&quot; rel=&quot;nofollow noreferrer&quot;&gt;experimental form finding&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2017-02-19T19:51:31.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13835" PostTypeId="2" ParentId="13828" CreationDate="2017-02-19T22:36:08.893" Score="2" Body="&lt;p&gt;Transistors can switch incredibly fast, yes billions of times per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CPU speed is the rate at which the clock path, which interacts with transistors in clocking elements like flip-flops and latches, oscillates between high and low. And transistors must be able to respond to the clock appropriately!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure if you look hard enough on google you might be able to find information about the rise/fall and delay times of fets from various technology nodes from various fabs/foundries. I hope you aren't too surprised if the units for those delays and rise/fall times are in picoseconds!&lt;/p&gt;&#xA;" OwnerUserId="10120" LastActivityDate="2017-02-19T22:36:08.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13836" PostTypeId="2" ParentId="12259" CreationDate="2017-02-19T22:44:37.213" Score="1" Body="&lt;p&gt;I believe the voltage source you are looking for is Vsin, which is just a sin wave output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from that as the other answer mentions only use one ground!&lt;/p&gt;&#xA;" OwnerUserId="10120" LastActivityDate="2017-02-19T22:44:37.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13837" PostTypeId="1" AcceptedAnswerId="13863" CreationDate="2017-02-19T22:45:38.343" Score="4" ViewCount="1763" Body="&lt;p&gt;I'm building a server type computer consisting of integrated motherboard, CPU, RAM and hard disk.  There will also be a standard power supply unit.  This is to be a Steam Punked computer, so...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the feasibility of passively cooling the server using only a chimney effect.  So the electronics would be enclosed, with a chimney at the top and air intake(s) at the bottom.  All the components will be custom mounted so there is no typical computer case and therefore no particular size constraint.  How would I begin to evaluate /design this?  I believe that there are relationships between chimney proportions, temperatures and exhaust velocities.  Perhaps like this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LiGi9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LiGi9.png&quot; alt=&quot;Computer cooled by chimney effect&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mini computers (ARM /Celeron) exist that have no fans, so perhaps it isn't really a big deal and anything at all will work?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="2947" LastEditDate="2017-02-21T21:50:55.503" LastActivityDate="2017-02-21T21:50:55.503" Title="How to design a cooling chimney for a computer server?" Tags="|thermodynamics|cooling|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13838" PostTypeId="1" AcceptedAnswerId="13855" CreationDate="2017-02-20T04:23:03.620" Score="3" ViewCount="354" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Combined_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Combined Cycle Turbines&lt;/a&gt; are a gas turbine whose exhaust heat powers a second turbine running on steam. These can get thermal efficiencies greater than 50%, whereas a plain turbine would get 30 to 35%. Large power plants tend to use combined cycle for obvious reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I want to know if this can be small enough to power a train?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dreamworksrc.com/catalog/Jetcat-P200Rx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Example&lt;/a&gt; of a very small jet engine, which is very similar to a gas turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Desired specs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Fuel: either natural gas or kerosene&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Horsepower: ~4500 (taken from a typical diesel train engine (&lt;a href=&quot;https://en.wikipedia.org/wiki/GE_Dash_9-44CW&quot; rel=&quot;nofollow noreferrer&quot;&gt;GE Dash 9&lt;/a&gt;))&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Max volume: 25 x 3 x 6 meters (about the volume of a train car)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Max weight: 100 metric tons&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm assuming a fuel car can follow behind the engine car. The turbines would generate electric power sent to the car wheels, which is what a modern diesel-electric train does. I was imagining combined cycle turbo trains for freight and passenger services, but at &quot;normal&quot; speeds (100 - 150 kmh).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can this be done with today's technology? The engine would still need to be insulated enough and efficient enough to retain 50+ % thermal efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, I'm aware of &lt;a href=&quot;https://en.wikipedia.org/wiki/Turboliner&quot; rel=&quot;nofollow noreferrer&quot;&gt;this failed example&lt;/a&gt; from Amtrack starting in the 1970's. I don't think those were combined cycle turbines, and I could not find an explanation for why &quot;The high cost of operating the trains led to their withdrawal from the Midwest in 1981.&quot; But I did notice those were tiny trains, 5 cars total (2 engine cars, 2 coaches, and a food car). That is probably far too small to get economy of scale.&lt;/p&gt;&#xA;" OwnerUserId="3510" LastEditorUserId="3510" LastEditDate="2017-02-20T12:17:56.817" LastActivityDate="2018-05-04T06:09:03.717" Title="Can combined cycle turbines be small enough for trains?" Tags="|civil-engineering|turbines|energy-efficiency|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13839" PostTypeId="1" CreationDate="2017-02-20T05:22:56.207" Score="0" ViewCount="83" Body="&lt;p&gt;From a pie in the sky approach, similar to the question of whether a jetliner could be theoretically solar powered, what are the limiting factors on the magnitude of lifting capability of a controllable blimp? At what point would the control surfaces be so large that they would be impractically massive? What size would be too large for the fabric? Does a blimp with slightly heavier than air properties have the capacity to be much larger than its lighter than air brother? Anything you can supply would be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="8862" LastEditorUserId="8862" LastEditDate="2017-02-20T13:32:54.290" LastActivityDate="2017-02-20T13:32:54.290" Title="Largest possible navigable blimp" Tags="|aerospace-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13840" PostTypeId="1" AcceptedAnswerId="13842" CreationDate="2017-02-20T06:29:44.617" Score="3" ViewCount="17356" Body="&lt;p&gt;I recently purchased a &lt;a href=&quot;http://www.ebay.com/itm/282070870420&quot; rel=&quot;nofollow noreferrer&quot;&gt;digital scale from eBay&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the listing it can weigh in the following units:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Multiple Weighting Units: g, oz, tl, ct.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Screenshot of that page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mkdxa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mkdxa.png&quot; alt=&quot;Units of weighing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I know that &quot;g&quot; is grams, &quot;oz&quot; is ounces, and &quot;ct&quot; is carats (and the figures displayed agree with that) however what is &quot;tl&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I measure a 10g reference weight it shows as 0.853 tl, thus 1g is 0.0853 tl. Allowing for slight measurement errors (the last significant figure changes a bit with different measurements) I can't find any unit of weight that &quot;tl&quot; could correspond to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest in name I could find is &quot;tael&quot; but the conversion factor doesn't agree. (10g is 0.264555 tael).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jRd7O.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jRd7O.jpg&quot; alt=&quot;10g in tl&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The unit also measures in grains (gn) although that is not advertised.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The device only measure up to 200g, so I imagine that whatever &quot;tl&quot; is will be some sort of measurement of jewellery or similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is: &lt;strong&gt;What is tl?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10124" LastActivityDate="2023-11-21T15:50:14.737" Title="What is the &quot;tl&quot; unit of weight measurement?" Tags="|measurements|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13841" PostTypeId="2" ParentId="13839" CreationDate="2017-02-20T07:58:10.907" Score="2" Body="&lt;p&gt;To remain aloft in air physics of floatation requires total weight to volume ratio should be less than air density as in a LTA blimp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dfrac{W}{V} &amp;lt; \rho_{air}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which also means that percentage increments of weight added should be less than percentage increment of volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dfrac{\Delta W}{\Delta V} &amp;lt; \rho_{air}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a neutral balloon floating equilibrium &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dfrac{\Delta W/W}{\Delta V/V} =1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot understand heavier than air blimps.&lt;/p&gt;&#xA;" OwnerUserId="1270" LastActivityDate="2017-02-20T07:58:10.907" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13842" PostTypeId="2" ParentId="13840" CreationDate="2017-02-20T08:13:31.860" Score="9" Body="&lt;p&gt;It could be a &lt;a href=&quot;https://en.wikipedia.org/wiki/Tola_(unit)&quot; rel=&quot;noreferrer&quot; title=&quot;tola&quot;&gt;tola&lt;/a&gt;. 1 tola = 3/8 troy ounce (exactly) = 11.66g (approx.), so 10g = 0.857 tolas. If the scale has an accuracy of 0.01g (it's not very clear from the ebay description!) that would be near enough what is displayed in the picture.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-02-20T08:13:31.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13843" PostTypeId="2" ParentId="3328" CreationDate="2017-02-20T09:21:16.227" Score="0" Body="&lt;p&gt;I am surprised it had not been done earlier. Constant stress design which is also not far away from stiffest designs are  Gauss Bell shaped plates/ shims that thin out at periphery varying according to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{t}{t_1} = e^{-(r/r_m)^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t$ is thickness and $r$ radius. However changing air gap width effect for boundary layer has to be a proper trade-off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Composite design with Kevlar, Carbon cloth /epoxy cut as circular disks omni-directionally placed at shaft center and together cured would produce very high speed rotors. Layer balancing is a tricky affair though.&lt;/p&gt;&#xA;" OwnerUserId="1270" LastEditorUserId="1270" LastEditDate="2017-02-20T09:34:24.910" LastActivityDate="2017-02-20T09:34:24.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13844" PostTypeId="1" CreationDate="2017-02-20T10:31:52.070" Score="1" ViewCount="47" Body="&lt;p&gt;How to calculate the pressure needed for a pump to ensure a specific temperature in a pipe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there is a pipe 100m long. A booster pump is needed to pump water with supply temperature 70 deg C. I need to ensure a temperature of 65 deg C at the end of the pipe. How can I calculate the pump pressure needed to ensure 65 deg C at the outflow?&lt;/p&gt;&#xA;" OwnerUserId="8686" LastEditorUserId="90" LastEditDate="2017-02-20T11:34:31.670" LastActivityDate="2017-02-20T11:43:36.243" Title="What is the pressure to secure certain temperature?" Tags="|fluid-mechanics|thermodynamics|pumps|pipelines|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13845" PostTypeId="1" AcceptedAnswerId="13849" CreationDate="2017-02-20T11:18:24.717" Score="1" ViewCount="571" Body="&lt;p&gt;I have asked this question in 3DPrinting.stackexchange forum but got no answer answer. Please take a minute to see if you can help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am building a simple RWD vehicle using 3d printer parts. See the picture below. &lt;a href=&quot;https://i.stack.imgur.com/FQIfX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FQIfX.jpg&quot; alt=&quot;RWD vehicle with steering&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see from the picture, I am using 3d printed linkages for steering mechanism and I used regular bolt and lock nuts to secure them together (thats what I had laying around the house at the moment). The problem here is that, holes in the linkages are not exactly the size of the bolt, and because the bolts are threaded, it generates a lot of friction. Other than that, the bolt seems to sit loose inside the hole and creates an unstable pivot. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/pTxK4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pTxK4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I tighten the nut more, it just presses them together making it hard for the servo to operate. I am looking for a solution that will use simple bolts, nuts and washers which can be bought easily off the shelf (or even with the bolts and screws everyone has around there house). What can be done to make this joint stable in xy plane, rotating freely around the bolt. By the way, even though I used nylon lock nuts, it seems to loosen after couple of repetitive rotations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help in advance and I hope ya'll have a great day. &lt;/p&gt;&#xA;" OwnerUserId="10126" LastActivityDate="2017-02-20T11:35:07.350" Title="How to build a pivot linkage for a robot." Tags="|3d-printing|bolting|linkage|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13846" PostTypeId="2" ParentId="13844" CreationDate="2017-02-20T11:20:35.950" Score="4" Body="&lt;p&gt;Pressure adds only a to the temperature of water. Pump pressure allows water to be moved, it is not used to maintain the temperature of the fluid being pumped. To minimize thermal loss of fluid in a pipe, the pipe need to be sufficiently insulated.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastEditorUserId="90" LastEditDate="2017-02-20T11:43:36.243" LastActivityDate="2017-02-20T11:43:36.243" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13847" PostTypeId="2" ParentId="13716" CreationDate="2017-02-20T11:29:40.597" Score="0" Body="&lt;p&gt;The pressure downstream valve can be any value less than 65psig.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If downstream pressure at 60psig, valve maybe fully open. If downstream pressure at 0psig, valve maybe only slightly open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, pressure downstream valve will be 65psig - pressure drop between Pressure Gauge to inlet valve - pressure drop of valve at current opening.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2017-02-20T11:29:40.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13848" PostTypeId="2" ParentId="13647" CreationDate="2017-02-20T11:32:45.327" Score="0" Body="&lt;p&gt;Use pneumatic pump (diapraghm or piston type) that has 1:1 pressure ratio. Feed the pump using compressed air that have pressure regulator set at 9 bar.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2017-02-20T11:32:45.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13849" PostTypeId="2" ParentId="13845" CreationDate="2017-02-20T11:35:07.350" Score="0" Body="&lt;p&gt;Simple, easy but lower-quality approach: two nuts and a tension washer between them. Turn the nut until the linkages are secured in position with correct amount of play. Put the tension washer and a second nut, then holding the first nut immobile with a wrench, tighten the second nut very firmly. That way the pair of nuts will remain in place on the bolt without moving either way.&#xA;Put a flat washer between the linkages; friction of 3d-printed plastic against 3d-printed plastic is much higher than plastic against metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better quality, harder approach: nut-based bearings. Drill in the holes of the linkages, to size a little smaller than the nut. Heat the nut to temperature sufficient to melt the plastic, and place it in the hole, fusing it with the linkage. Screw the bolt in; leave some room between the linkages - about 2 turns worth. Secure the bolt by tightening it to one (top-most) nut, so that it's immobile relative to one of the linkages; the nut in the other acts as a bearing; it will introduce some half a millimeter of travel up or down as it turns, but that should be well within your tolerances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the best approach would involve actual ball bearings, but it seems for your application that might be an overkill. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-02-20T11:35:07.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13850" PostTypeId="2" ParentId="13832" CreationDate="2017-02-20T12:03:12.593" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, even through a simple parallelization of a single-rotor design. If there's no other payload than the helicopter itself, and the challenge is for staying airborne, you gain about nothing performance-wise by adding more blades, but hey, +75% score. Go with quadcopter and four bands. Driving four rotors through a single shaft in line would be a hell of mechanics.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Single-bladed airfoil will mean the blade is way off-center, introducing hopeless torque. If your single blade is a full turn of a helix, its center of mass is dead in the axis of rotation. It's not really beneficial aerodynamically but it's a way to cheat center of mass into rotation axis with the contrived &amp;quot;single blade&amp;quot; restriction through making the blade big.&lt;/p&gt;&#xA;&lt;p&gt;The airfoil gives a decent lift to mass ratio at higher airspeeds, but you won't get high airspeeds from a rubber band. You must buy your lift with wing surface area - and the helical blade will give you that. The blade doesn't need to be THAT big for that, but that's one way to make center of mass and axis the same.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It will likely be dictated by many factors - keeping the rubber band tension, minimizing mass, keeping center of mass low enough. I don't think you can find a good &lt;em&gt;formula&lt;/em&gt; for it. It's something that should be found experimentally.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="203" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-02-20T12:16:34.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13851" PostTypeId="2" ParentId="13829" CreationDate="2017-02-20T12:12:53.940" Score="3" Body="&lt;p&gt;In ambient pressure, ethane boiling point is -89 °C; ethylene boiling point is -103.7 °C. Water, to act as efficient coolant needs to be liquid, so above 0 °C. You're not going to condense ethane vapor at ambient pressure with coolant at 89 degrees above its boiling point!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your two options are to either use a much colder - and more expensive and difficult to handle coolant (say, liquid nitrogen) or increase the pressure of the distillate so that it condenses at temperature of the cheap, ubiquitous coolant like water.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-02-20T12:12:53.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13852" PostTypeId="1" AcceptedAnswerId="13868" CreationDate="2017-02-20T13:03:40.970" Score="9" ViewCount="8703" Body="&lt;p&gt;The answer to a question on &lt;a href=&quot;https://engineering.stackexchange.com/a/12951/2555&quot;&gt;involute gear&lt;/a&gt; curve calculations shows how the gear tooth curve proceeds as an involute calculation from the base diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the &lt;em&gt;pitch diameter&lt;/em&gt; is D = m.N where N is the number of teeth and m is the modulus (for metric gears), then the &lt;em&gt;base circle diameter&lt;/em&gt; for a pressure angle &amp;alpha;, most usually 20 degrees, is D.cos &amp;alpha; and the &lt;em&gt;root diameter&lt;/em&gt; is given by the amount the tooth falls below the pitch circle. This amount is the dedendum and is usually 1.25m. Thus the root circle diameter is D - 2.5m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The involute is calculated in a standard way and looks like the unwinding of a string from a circle, the base circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YafhH.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YafhH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gear tooth is then described by the intersection of the involute with the root diameter (the bottom of the tooth) and the outer diameter (the top of the tooth). An issue arises if the root diameter is less than the base diameter as the involute is not defined below the base diameter (you get a math domain error in Python if you try it!).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pJUHq.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pJUHq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These tooth are described as &quot;undercut&quot;. I cannot find a description of how the curve proceeds once you &quot;run out of involute&quot; - ie cross the base diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standard way of describing the rest of the gear tooth curve?&lt;/p&gt;&#xA;" OwnerUserId="2555" LastEditorUserId="8055" LastEditDate="2017-07-21T22:28:57.380" LastActivityDate="2020-02-18T03:05:55.293" Title="Involute gear curve when root diameter falls below base diameter" Tags="|design|gears|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13853" PostTypeId="2" ParentId="13839" CreationDate="2017-02-20T13:17:01.197" Score="1" Body="&lt;p&gt;This is unanswerable unless acceptable meteorological conditions are defined, and even then it would be exceedingly difficult to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can increase the size of the blimp arbitrarily through bundling separate balloons indefinitely, and it will work perfectly well in still air, at any size that doesn't exceed the atmosphere thickness vertically and planet surface horizontally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as soon as you add &lt;em&gt;weather&lt;/em&gt; it falls apart, as masses of air acting on such object will make it behave more like liquid or gas than a single solid blimp. You need to reduce size before its breaking point. And calculating what kind of winds it can withstand? Obviously it won't be a rigid structure due to the mass overhead restricting the size; it will behave more like one of these huge soap bubbles:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TR1sS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TR1sS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;and good luck readily calculating breaking point of these!&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-02-20T13:17:01.197" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13855" PostTypeId="2" ParentId="13838" CreationDate="2017-02-20T15:19:01.657" Score="4" Body="&lt;p&gt;We can try to answer this question by looking at existing equipment in terms of size and weight and make an educated guess from there if all this fits on a train. Our CC will consist of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gas turbine&lt;/li&gt;&#xA;&lt;li&gt;Exhaust HX/Boiler&lt;/li&gt;&#xA;&lt;li&gt;Steam turbine with condensor and other aux. parts&lt;/li&gt;&#xA;&lt;li&gt;Transmission&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the following, I will not match components exactly, but pick stuff that sits in the right power range, just to get a feel if the parts could fit into the design envelope given above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Basic sizing&lt;/strong&gt;&lt;br&gt;&#xA;You give 4500hp, that's a bit more than 3MW. Looking at wikipedia, the gas turbine is typically twice the power of the steam turbine. Soe we look for a 2MW gas turbine and a 1MW steam plant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Gas turbine&lt;/strong&gt;&lt;br&gt;&#xA;What are some turbine around 2MW?  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.kawasaki-gasturbine.de/files/Gas_Turbine_Generator_Set_GPB17D_englisch_rev2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kawasaki has one at 1.7 MW&lt;/a&gt;, it seems the 'naked' turbine fits into a 6.5m container + there's another one stacked on top for air vents. No mass given, does not seem to include exhaust HX.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.opraturbines.com/en/OPRA-WEB---Products---Products/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This 2MW Opra engine&lt;/a&gt; again fits into a 6.5 m container (without air supply), the naked engine and weighs maybe 13.5 t.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For the following, I'll take the stats of the Kawasaki engine and assume the weight of the Opra engine is not too far off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kawasaki:&lt;br&gt;&#xA;Electrical Power 1.690 kW&lt;br&gt;&#xA;Fuel input 6.360 kW&lt;br&gt;&#xA;Efficiency 26,9 %&lt;br&gt;&#xA;Exhaust gas mass flow 8,00 kg/s&lt;br&gt;&#xA;Exhaust gas temperature 521 °C  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Boiler/Exhaust HX&lt;/strong&gt;&lt;br&gt;&#xA;If we take the Kawasaki engine ebove, we have about 480kg/min exhaust gas. Looking at &lt;a href=&quot;http://www.ejbowman.co.uk/pdf/Exhaust%20Gas%20R-angle%20Iss%201.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this table of exhaust HX&lt;/a&gt;, we would need 7 of the largest, each about 600kg and length x diameter 2.6 x 0.7 m. Roundabouts. At up to 700kW heat recovery each, we should get north of the 2MW worth of steam we need for our steam turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Steam turbine&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.energy.siemens.com/hq/pool/hq/power-generation/steam-turbines/Pre-Designed_Steam_Turbines_en.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Siemens has one that covers the 800kW-6MW range&lt;/a&gt;. Without condensor it fits into a LxWxH evelope of 1.5 x 2.5 x 2.5 m. No weight is given, I guess a few tons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Condensor&lt;/strong&gt;&lt;br&gt;&#xA; We need to shed about 1MW of heat for our 1MW steam engine. Let's take some measurements from this &lt;a href=&quot;http://www.guentner.eu/fileadmin/literature/europe/condensers_drycoolers/GVW/Guentner_GVW_Data_sheet_DE_EN.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;product sheet from Güntner&lt;/a&gt;. Note that these a condensors for refridgerant. I guess the bottleneck is rather the airflow and the heat transfer towards air, then these should be ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest have a nominal capacity of 500kW, weigh 2t and are about LxWxH 10x1.2x1.7 m. Two of those side by side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Transmission&lt;/strong&gt;&#xA;My guess is that an electrical trasnmission would be used, then you'd have to add a 3MW motor and a 1MW and 2MW alternator. Because by now I'm lazy I'd say that these add maybe 2-3 m to our 'length budget'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;br&gt;&#xA;We don't seem to really reach the mass budget of 100t (which seems way large for just the engine, OTOH).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our train could look like, from from to back:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;6m length dedicted to air intake, filters etc.&lt;/li&gt;&#xA;&lt;li&gt;6.5 m for the gas turbine (13t)&lt;/li&gt;&#xA;&lt;li&gt;3m for a stack of HX (3.5t)&lt;/li&gt;&#xA;&lt;li&gt;3 m for the steam engine (3t?)&lt;/li&gt;&#xA;&lt;li&gt;10m for the condensors (4t)&lt;/li&gt;&#xA;&lt;li&gt;2-3 m for alternators and motor (10t?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a total of 31m. This is larger than the size envelope given by in the OP, but not by much. So my conclusion is: &lt;em&gt;You can fit 3MW worth of combined cycle plant on a train car.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's lots of room to improve the whole thing at the HX and condensors, OTOH we need to fit large electrical systems on the train. In summary I believe you could build a combined cycle powered train &amp;amp; you would not have to invent too many components to do it. &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2018-05-04T06:09:03.717" LastActivityDate="2018-05-04T06:09:03.717" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="13856" PostTypeId="1" CreationDate="2017-02-20T15:27:33.377" Score="1" ViewCount="3945" Body="&lt;p&gt;Could an air compressor be used to compress steam? What would be practical problems if any? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have saturated steam at 1 barg and I need to compress it to 2 barg. In theory could I just pipe to the inlet of a conventional air compressor (reciprocating type). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds dicey to me but when someone came up with this idea during brainstorming for a DIY project I couldn't come up with cogent reasons why this would be a bad idea. &lt;/p&gt;&#xA;" OwnerUserId="6277" LastActivityDate="2021-04-12T16:50:45.537" Title="Could an air compressor be used to compress steam?" Tags="|compressors|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13857" PostTypeId="2" ParentId="8473" CreationDate="2017-02-20T15:31:26.537" Score="2" Body="&lt;p&gt;Answering your first question: In the control field of mechanical engineering, a spring can indeed be a link. In the book Modern Control Engineering by Ogata, Katsuhiko the spring is frequently used as a link!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering your second question: &quot;Reuleaux called the ideal connections between components that form a machine, &quot;kinematic pairs&quot;. He distinguished between higher pairs which were said to have line contact between the two links and lower pairs that have area contact between the links. J. Phillips shows that there are many ways to construct pairs that do not fit this simple classification.&quot;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Kinematics#Kinematic_pairs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kinematics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I would say in a broad sense they're called kinematic pairs but I think there might be a lot more different types of joints that might have different names. It really depends in which circumstance the subject is being discussed, but in Kinematics they are called indeed both names.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering your third question: &quot;Rigid bodies (&quot;links&quot;) connected by kinematic pairs (&quot;joints&quot;) are known as kinematic chains. Mechanisms and robots are examples of kinematic chains. The degree of freedom of a kinematic chain is computed from the number of links and the number and type of joints using the mobility formula. This formula can also be used to enumerate the topologies of kinematic chains that have a given degree of freedom, which is known as type synthesis in machine design.&quot;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Kinematics#Kinematic_pairs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kinematics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that being said, if there is no fixed link connected by kinematic pairs, then there are neither kinematic chains nor mechanisms.&#xA;Mechanism is simply an example of kinematic chains, which makes mechanism a subset of kinematic chains.&lt;/p&gt;&#xA;" OwnerUserId="10028" LastActivityDate="2017-02-20T15:31:26.537" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13860" PostTypeId="2" ParentId="13856" CreationDate="2017-02-20T19:45:30.803" Score="1" Body="&lt;p&gt;Practically speaking, it would be very inefficient. Just thinking about its application, the pressure would not increase since the steam will just turn into water in a liquid state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have found this &lt;a href=&quot;http://www.tlv.com/global/TI/products/210000.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; that might be helpful, depending on your specific application, and they mention that the compressor is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&quot;Capable of recovering steam even at atmospheric pressure.**Requires the use of a COS or CV-COS control valve.&quot;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also mentioned that the different application areas include&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Removing vapor clouds above tanks of boiling water.&lt;/li&gt;&#xA;&lt;li&gt;Pressurizing and reutilizing flash steam from tanks.&lt;/li&gt;&#xA;&lt;li&gt;Pressurizing and reutilizing excess steam used in power generation.&lt;/li&gt;&#xA;&lt;li&gt;Pressurizing and reutilizing low pressure steam generated after use in processes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm assuming the third application is the one suiting the description and question you asked.&lt;/p&gt;&#xA;" OwnerUserId="10028" LastActivityDate="2017-02-20T19:45:30.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13861" PostTypeId="1" AcceptedAnswerId="14373" CreationDate="2017-02-20T20:33:13.083" Score="0" ViewCount="71" Body="&lt;p&gt;How are these springed ball connectors called? Can not find them without right name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OJmBV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OJmBV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9325" LastActivityDate="2017-03-23T05:52:38.603" Title="Name or type of these connectors" Tags="|connections|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13862" PostTypeId="2" ParentId="4035" CreationDate="2017-02-20T20:38:22.693" Score="1" Body="&lt;p&gt;There are quite a few options when it comes to computational electromagnetics - both free and commercial. ANSYS has various options to do this sort of analysis, as well as COMSOL Multiphysics, but both are commercial and quite costly. There are some free options as well - openEMS comes to mind, as well as getdp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like you're asking for a coupled simulation where the force from the field is used to impart some motion to the projectile. It may honestly be simpler to do that math by hand if you're new to modeling. All of the above options will require you to construct some geometry - ANSYS and COMSOL can import things like Solidworks as well as having their own methods of defining geometry, but this is often one of the most complex steps of creating a useful simulation. There are free methods of constructing geometry - Gmsh, Salome, FreeCAD. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simulation and modeling takes a lot of time. You may be better served by seeking an analytical treatment of the force or acceleration produced (which, for the parameters you're talking about in solenoids, almost certainly exist in the literature).&lt;/p&gt;&#xA;" OwnerUserId="3583" LastActivityDate="2017-02-20T20:38:22.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13863" PostTypeId="2" ParentId="13837" CreationDate="2017-02-20T21:09:23.007" Score="3" Body="&lt;p&gt;The key technical term you're looking for here is convective cooling. Either you allow the air to rise through the chimney on its own accord (natural convection cooling) or you pump it through with a fan (forced convection cooling). The second is obviously more effective at pulling heat out of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of heat removed will be a factor of the flow rate of air through the system as well as the heat transfer coefficient (HTC) between the air and the CPU. You can get a good estimate by grabbing an HTC function from engineering toolbox or a text book and using Newton's Law of Cooling:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;q = hc A dT&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where q is heat transfer (in watts), hc is the heat transfer coefficient (W/m^2 K), A is the surface area to be cooled (m^2), and dT is the temperature difference between the air and the surface (in K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HTC for air is (for v between 2-20 m/s):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;hc = 10.45 - v + 10 v^1/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where v is your air flow speed in m/s. The faster the air is moving, the higher your HTC (and the more heat you can pull out of the system). &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Broad estimates of HTCs:&#xA;Free convection: 0.5-1000 W/(m^2 K)&#xA;Forced convection: 10-1000 W/(m^2 K)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You should know from the specs of the machine roughly how much heat is produced in watts. Then your air flow will dictate your HTC - and try to maximize it so that you can pull more heat out of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept of convective cooling is pretty widely discussed; any general engineering thermodynamics textbook will have more detail (&quot;Thermodynamics: An Engineering Approach&quot; by Cengel or &quot;Fundamentals of Heat and Mass Transfer&quot; by Incropera are good resources). There's quite a lot of stuff about general engineering approaches online, too: where to locate the object to be cooled in your enclosure, what sort of fans to use, etc. These are the sorts of problems that can be analyzed at a really detailed level with computational fluid dynamics. To check the feasibility, I would start with Newton's Law of Cooling, though.&lt;/p&gt;&#xA;" OwnerUserId="3583" LastActivityDate="2017-02-20T21:09:23.007" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13865" PostTypeId="1" AcceptedAnswerId="13867" CreationDate="2017-02-20T22:49:07.947" Score="4" ViewCount="10252" Body="&lt;p&gt;After some research, I found that diesel is my choice of fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I need a way to pump it to the engine. My main constraint is price.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is can I use a water pump like one of &lt;a href=&quot;https://www.amazon.ca/SODIAL-Priming-Diaphragm-Spray-Dispenser/dp/B00U8N07HU/ref=sr_1_8?ie=UTF8&amp;amp;qid=1487630446&amp;amp;sr=8-8&amp;amp;keywords=water%20pump%20sodial&quot; rel=&quot;noreferrer&quot;&gt;these&lt;/a&gt;, to pump the diesel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VXuZu.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VXuZu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2023-04-02T10:27:18.080" Title="Can a water-pump pump diesel?" Tags="|pumps|diesel|rocketry|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13866" PostTypeId="1" AcceptedAnswerId="13876" CreationDate="2017-02-20T23:28:46.850" Score="1" ViewCount="135" Body="&lt;p&gt;Most specs for light bulbs list luminous efficiency (or, more accurately sometimes, luminous efficacy) but not energy efficiency. Luminous efficacy tells you how bright the light is as observed by a human eye per unit of input electricity. Unfortunately, the human eye is terrible at measuring power and its very hard to convert between lumens/spectrum/human eye/radiant power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious as to which types of lights have the highest energy efficiency, which I'm defining the amount of energy that would be absorbed by a perfect absorber surrounding the light bulb divided by the electric input. In this metric, the light is just a one form that the energy takes between electricity and its absorbed form (eg, heat). Obviously the exergy is degraded in this process, but I'm not worried about that here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect that incandescents are near 100% efficient by this metric since the only losses should be conduction/convection from the element through the near-vacuum and wires. &lt;a href=&quot;https://www.dial.de/en/blog/article/efficiency-of-ledsthe-highest-luminous-efficacy-of-a-white-led/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This weakly scientific analysis&lt;/a&gt; suggests that LEDs are probably only 40% efficient by this metric because all their luminous inefficiencies are thermal losses in the cells. I think fluorescent lights might be extremely efficient by this metric but I can't really find any data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the phrase &quot;radiant efficiency&quot; somewhat useful but still can't find much in manufacturers data sheets. Is there a better name for the metric I've described or a good database for what values might be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to comments: I am interested in the power radiated away from the light bulb in the form of electromagnetic radiation divided by the electric power that goes in to the lightbulb. I don't care about the wavelength of the light (even though I understand that it changes both how much you can see and how much can be recaptured in another form later). I do care about thermal losses inside the bulb that don't radiate power across a distance.&lt;/p&gt;&#xA;" OwnerUserId="2781" LastEditorUserId="2781" LastEditDate="2017-02-21T15:04:18.193" LastActivityDate="2017-02-21T15:04:18.193" Title="Light bulbs with highest energy efficiency" Tags="|energy|lighting|" AnswerCount="2" CommentCount="6" ClosedDate="2017-02-24T12:55:07.257" ContentLicense="CC BY-SA 3.0" />
  <row Id="13867" PostTypeId="2" ParentId="13865" CreationDate="2017-02-20T23:38:42.733" Score="15" Body="&lt;p&gt;With pumping diesel, your main constraint should be safety, not cost. There are very good reasons diesel pumps aren't for sale for $5 on amazon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main things you want to look at are 1) whether the wetted materials are &lt;a href=&quot;https://www.coleparmer.com/Chemical-Resistance&quot; rel=&quot;nofollow noreferrer&quot;&gt;compatible with diesel&lt;/a&gt; (EDPM and silicone for example will dissolve) and 2) how it is grounded. Many pumps can build up static charge which can cause the diesel to explode in the pump. You'll also need to be careful to avoid cavitation which could easily lead to explosions also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Explosion proof motors (and intermediate ratings like totally enclosed) are also worth investigating. If your plumbing leaks, is the vapor going to find its way to the motor and explode?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also recommend reading more about explosion-proof pumps and reading up more generally on what types of pumps are used for diesel.&lt;/p&gt;&#xA;" OwnerUserId="2781" LastEditorUserId="2781" LastEditDate="2017-02-21T14:29:32.640" LastActivityDate="2017-02-21T14:29:32.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13868" PostTypeId="2" ParentId="13852" CreationDate="2017-02-21T00:30:22.650" Score="18" Body="&lt;p&gt;In general, each tooth of an involute gear is made up of two curves: the involute curve and the root fillet. The involute curve is essential for gears to operate properly, and allows power to be smoothly transmitted from one gear to the other. The root fillet on the other hand is not involved directly with power transmission or the kinematics of the gears (i.e. how the motion of one gear influences the motions of a meshing gear.). This means that the root fillet curve's shape is reasonably arbitrary, provided, of course, that the root fillet on one gear does not at any point make contact with the teeth of the other gear. In addition, the root fillet also plays a roll in determining how much bending stress develops at the root of the teeth during operation: if the root fillet has a low radius of curvature, this will result in a large stress concentration at the root.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite the root fillet curve having the potential to be almost any shape, there are a few shapes that occur commonly, the two which I believe to be most common are as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Connecting involute curve and root circle with radial lines&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest root fillet to determine mathematically is simply a straight radial line that connects the start of the involute curve to the root circle. This is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RdPj3.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RdPj3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantages of the root &quot;fillet&quot; (a sharp corner is not much of a fillet I guess) is the geometry is very simple. However, there are two big disadvantages: 1. The sharp corner produces a very large stress concentration at the root of the gear, meaning larger root bending stresses. This means that these gears are just not suitable for applications that subject them to strong torques. 2. In cases where &lt;em&gt;undercut interference&lt;/em&gt; becomes an issue, these radial lines are going to get in the way of the teeth, causing gears to jam. For gears with pressure angles at the standard value of 20°, undercut interference usually occurs for gears with teeth less than 18 (if gear is meshing with a rack gear) or 13 (if gear is meshing with another gear of equal number of teeth). For these reasons, you will rarely see gears with this kind of root fillet in industrial application, but are typically fine in areas where teeth numbers are quite high and the gears are not subjected to high torques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Trochoidal root fillet&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most common processes for generating spur gears and helical gears is &lt;a href=&quot;https://en.wikipedia.org/wiki/Hobbing&quot; rel=&quot;noreferrer&quot;&gt;hobbing&lt;/a&gt;, and the root fillet that is naturally produced as a result of hobbing is the trochoidal curve. An example of a gear with trochoidal root fillets is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9H7cD.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9H7cD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The individual trochoidal curve and involute curve are displayed along with the root circle and outer circle as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ueewg.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ueewg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing to notice is that the trochoidal fillet is considerably more rounded than the previous case, so less bending stress will develop at the root of the tooth and so the gear is more suitable in high-torque scenarios. Also, notice how the trochoidal curve intersects with the involute curve at a point slightly to the right of the start of the involute: some of the involute curve has been inadvertently lost by the hobbing process! The smaller the number of teeth, the greater this involute loss will be, and it becomes apparent as undercut interference starts to occur: in such a case, the gears are said to be &lt;em&gt;undercut&lt;/em&gt;. While for radial-line fillet curves the undercut interference would have caused the gears to jam, gears with trochoidal fillets will not jam in such circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever gear teeth are generated by hobbing, it is as though a rack gear is rolling along a gear blank (circle with diameter equal to outer diameter of gear tooth), material being subtracted wherever the rack gear has been. The below animation should illustrate how the rack gear can generate the shape of a spur gear with trochoidal fillets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3wnsT.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3wnsT.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how the teeth on the rack gear are trapezia shapes (and also that the rack gear here, being a cutter, has an addendum of $1.25m$ and a dedendum of $m$, unlike usual rack gears), and the cutting of these trapezium teeth into a gear blank manages to magically create two curves, the involute and the trochoidal curves. More precisely, the involute curve is generated by either of the diagonal lines of the trapezium, and the trochoidal curve is generated by either of the two points on the end of the trapezium. With this information, it is possible to express the trochoidal fillet as a parametric equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next section of this question will look at the mathematics behind deriving such a parametric equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Deriving the trochoidal curve&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, let's consider a single trapezium of the rack gear, and see how a point on the end of a trapezium moves relative to the gear being cut. The curve that this point follows will be the trochoidal curve. The pitch point of the trapezium, $P$, is the point on the trapezium that will initially lie on the gear's pitch circle, as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wUrQj.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wUrQj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the pitch circle radius is $R$, and that there is a general point $Q$ that lies on the trapezium, offset from $P$ by $x_0$ and $y_0$ in the $x$ and $y$ directions (later, we will set this point to be one of the points on the end of the trapezium). Also, note the solid black triangle: this is to help identify the orientation of the gear. The current position of $Q$ can be expressed as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;        \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} R+x_0 \\ y_0 \end{array} \right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's rotate the gear anticlockwise by an angle $t$, which then pushes the trapezium up by distance $Rt$ (no slip between pitch circle and pitch line). Illustration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E9wP1.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E9wP1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how the solid black triangle has moved in accordance with the gear rotation. The position of $Q$ is now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;        \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} R+x_0 \\ Rt+y_0 \end{array} \right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trapezium has moved, but so has the gear: we are interested in the motion of $Q$ &lt;em&gt;relative&lt;/em&gt; to the gear. Therefore, we need to rotate the entire system about the origin by angle $t$ clockwise. This simply rotates the perspective so that it returns the gear to its default position while retaining the new movement of the trapezium, i.e. we get the motion of a trapezium as the rack rolls around the stationary gear, similar to the animation above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uyWKN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uyWKN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the position of point $Q$, we need to multiply the previous position by a clockwise rotation matrix, so the position is expressed as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;        \left[ \begin{array}{c} x \\ y \end{array} \right] = \begin{bmatrix} \cos{t} &amp;amp; \sin{t} \\ -\sin{t} &amp;amp; \cos{t} \end{bmatrix}\left[ \begin{array}{c} R+x_0 \\ Rt+y_0 \end{array} \right]= \left[ \begin{array}{c} (R+x_0)\cos{t}+(Rt+y_0)\sin{t} \\ -(R+x_0)\sin{t}+(Rt+y_0)\cos{t} \end{array} \right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, by setting $Q$ to be the &quot;upper left&quot; point on the trapezium, i.e. $x_0 = -1.25m$ and $y_0 = \frac{1}{4}\pi m - 1.25m\tan{\phi}$, where $\phi$ is pressure angle, and also noting that $R=\frac{1}{2}Nm$, where $N$ is number of gear teeth, you get the parametric equation for the trochoidal fillet curve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;        \left[ \begin{array}{c} x(t) \\ y(t) \end{array} \right] = \left[ \begin{array}{c} (\frac{1}{2}Nm-1.25m)\cos{t}+(\frac{1}{2}Nmt+\frac{1}{4}\pi m - 1.25m\tan{\phi})\sin{t} \\ -(\frac{1}{2}Nm-1.25m)\sin{t}+(\frac{1}{2}Nmt+\frac{1}{4}\pi m - 1.25m\tan{\phi})\cos{t} \end{array} \right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the parameter $t$ will start at $t=-\frac{y_0}{R}=-\frac{\frac{1}{4}\pi  - 1.25\tan{\phi}}{\frac{1}{2}N}$ and increases up until a particular value that represents the intersection of the trochoid with the involute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that answers your question :)&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2017-02-21T00:30:22.650" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13869" PostTypeId="2" ParentId="13861" CreationDate="2017-02-21T01:01:27.853" Score="0" Body="&lt;p&gt;&lt;a href=&quot;http://www.woodencamera.com/tool-less-lcd-evf-adapter-weapon-p/233800.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;I found this link that is from the seller.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a 26 pins adapter that they call:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tool-less LCD/EVF Adapter&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;believe&lt;/em&gt; this is a unique product to the brand.&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2017-02-21T01:01:27.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13871" PostTypeId="2" ParentId="12474" CreationDate="2017-02-21T04:40:17.147" Score="2" Body="&lt;p&gt;My understanding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Petrol engines produce far more carbon monoxide. Carbon monoxide is a colourless, odourless and highly toxic gas. This makes running petrol engines indoors far more dangerous than running diesel engines indoors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outdoors however carbon monoxide is not so bad. It's lighter than air so it won't tend to persist near ground level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diesel engines produce more particulates and NOx, these things aren't going to kill you immediately but chronic exposure is not healthy and since they are heavy they are more likely to remain near ground level where people can breathe them. &lt;/p&gt;&#xA;" OwnerUserId="3377" LastActivityDate="2017-02-21T04:40:17.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13872" PostTypeId="1" CreationDate="2017-02-21T06:27:29.837" Score="1" ViewCount="81" Body="&lt;p&gt;I am trying to simulate multiphase fluid flow (gas-particle) through a electrostatic field. Are there any ways to do this in ANSYS?&lt;/p&gt;&#xA;" OwnerUserId="10139" LastEditorUserId="8055" LastEditDate="2017-02-21T14:57:14.667" LastActivityDate="2017-02-21T14:57:14.667" Title="Multiphase-Electrostatic ANSYS Workbench" Tags="|mechanical-engineering|fluid-mechanics|ansys|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13873" PostTypeId="1" CreationDate="2017-02-21T09:44:18.307" Score="0" ViewCount="52" Body="&lt;p&gt;Electric hoists such as this one &lt;a href=&quot;http://www.builderauthority.com/product/6013242/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.builderauthority.com/product/6013242/&lt;/a&gt; commonly have two load capacities depending on whether a single-line or double-line configuration is used. What is it about the double-line configuration that allows the load capacity to be doubled?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="8835" LastActivityDate="2017-02-21T10:52:40.053" Title="What is the principle behind electric hoists having two load capacities?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13874" PostTypeId="2" ParentId="13873" CreationDate="2017-02-21T10:52:40.053" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What is it about the double-line configuration that allows the load capacity to be doubled?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because the lifting force is literally split between 2 cables. One to the static anchor and one to the hoist's winch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U1AuG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U1AuG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;(image source: &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Pulley1.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-02-21T10:52:40.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13875" PostTypeId="2" ParentId="13866" CreationDate="2017-02-21T11:46:18.517" Score="1" Body="&lt;p&gt;I don't think your metric is particularly useful.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You even elude to the flaw yourself, incandescent bulbs should be more efficient than LEDS in your system, because they create more heat than light.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are thinking of efficiency wrong in this case.  It isn't just the ratio of energy out to energy in.   It's the ratio of useful energy out to energy in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of light bulbs the main factor you want is light, not heat.  That means a good measure of efficiency is the amount of light vs. Energy input (not heat  vs. Energy input).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incandescents are more efficient than LEDS when it comes to home heating, but then a lot of things heat better than lights anyways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry I couldn't directly answer your question, I'm not sure where exactly you would find the information  on bulb radiation.  Energy management texts would likely be a good start, I know lighting and heat management come up in that quite a bit.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-02-21T11:46:18.517" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13876" PostTypeId="2" ParentId="13866" CreationDate="2017-02-21T12:30:44.710" Score="2" Body="&lt;p&gt;By your metric, a electrically-heated black body radiator in vacuum would be very efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incandescent bulbs are close, but run at high temperatures to make (at least some) visible light.  However, the filament is not in vacuum, so incandescent bulbs have significant heat conduction losses.  A typical bulb gets too hot to touch after running for a few minutes.  The filament is run so hot that if it were in a vacuum, it would evaporate too quickly and coat the inside of the bulb with the filament material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A modified incandescent that runs a bit cooler with the filament in vacuum should work nicely.  At dimly glowing red temperatures, the evaporation of the filament is greatly reduced, and you should get useful run times in vacuum.  The spectrum would be shifted even lower than typical incandescent light bulbs, but your spec says nothing about wavelength restrictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only losses would be a little heat conduction thru the filament connections to the outside, and transmission losses thru the bulb at the wavelengths actually emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to do this is to start with a standard incandescent light bulb and place it within another bulb that is evacuated.  Now there can't be conduction losses from the incandescent bulb, except the little bit thru the feed wires.  The problem with this setup is that the inner bulb will likely overheat at full power, since it was probably designed with some air cooling assumptions.  You could compensate by running the inner bulb at lower than rated power.  That will shift the light more toward longer wavelength, but probably doesn't need to be as far as in the first example.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2017-02-21T12:35:46.503" LastActivityDate="2017-02-21T12:35:46.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13879" PostTypeId="1" CreationDate="2017-02-21T16:26:25.713" Score="1" ViewCount="2712" Body="&lt;p&gt;I'm struggling to confirm the Velocity in an open channel flow with 0 percent slope.  Water enters the channel from an open pipe.  The rectangular channel dimensions (66' Long x 3.33' W x 6.75' D) and Q (15 MGD) are known.  The channel is concrete and empties into a larger basin via free fall (i.e. no weir at end).  I would use mannings; however with no slope I'm stumped.  Any help would be appreciated.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This is a real world problem at a waste water treatment plant I'm working on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&#xA;C&lt;/p&gt;&#xA;" OwnerUserId="10149" LastActivityDate="2017-02-27T11:03:41.353" Title="Open Channel Flow with no slope" Tags="|water-resources|waste-water-treatment|open-channel-flow|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13882" PostTypeId="1" CreationDate="2017-02-22T03:58:05.283" Score="3" ViewCount="271" Body="&lt;p&gt;The question title says it all. I'm looking for the opposite of a UPS, ie it won't turn back on if there is a power outage. I know they exist but, I can't remember what they are called or find them on Google.&lt;/p&gt;&#xA;" OwnerUserId="6885" LastActivityDate="2017-02-22T21:38:52.380" Title="What is the opposite of an uninterruptible power supply called?" Tags="|equipment-selection|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="13883" PostTypeId="1" AcceptedAnswerId="13888" CreationDate="2017-02-22T07:13:50.710" Score="0" ViewCount="183" Body="&lt;p&gt;I am working at a construction site and came accross this issue. a borehole needs to be expanded in length inside a container-office. Working space is as little as 6m x 3m (19ft by 9.5ft). Height of the container is about 2m (6'6&quot;). Heigt can somehow be altered up to 5 meters (16'5&quot;). Borehole diameter is 150mm (6&quot;). Current depth of the borehole is 30 m (98 ft). The depth should be increased to 42m (138 ft). See the attached sketch for details. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ddjJf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ddjJf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The borehole was bored before the containers were placed. It is used to measure water levels with piezometric sensors inside them. Water levels have dropped lower than expected and the borehole needs to be expanded in depth to reach lower groundwater deposits. Containers cannot be moved, and the location cannot be changed since this specific borehole is in a sensitive location, so please do not offer boring a new borehole outside the container. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What method or which equipment can be used to work in a small container space and can reach such depth. &lt;/p&gt;&#xA;" OwnerUserId="10126" LastActivityDate="2017-02-22T12:16:26.560" Title="Equipment or method for borehole" Tags="|civil-engineering|geotechnical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13885" PostTypeId="1" CreationDate="2017-02-22T08:31:49.247" Score="2" ViewCount="1028" Body="&lt;p&gt;I am currently coding a Forward and Inverse Kinematics solver for a PUMA 560 robot. For the Inverse Kinematics part I am using the closed for solution given in this &lt;a href=&quot;https://deepblue.lib.umich.edu/bitstream/handle/2027.42/6192/bac6709.0001.001.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt;. But my issue is, my solution for IK for a given set of (x,y,z) does not return the same values returned by my FK values. The reason I am doing this is to verify my code accurately computes the FK and IK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the DH parameters for my robot (These are Python code, since I was testing my algorithm on Spyder IDE before implementing on C++).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;DH Parameters &lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Link lengths &lt;br&gt;&#xA;   &lt;code&gt;a = [0, 650, 0, 0, 0, 0]&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Link offsets&lt;br&gt;&#xA;  &lt;code&gt;d = [0, 190, 0, 600, 0, 125]&lt;/code&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Link twist angle&lt;br&gt;&#xA;  &lt;code&gt;alpha = [-pi/2, 0, pi/2, -pi/2, pi/2, 0]&lt;/code&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So basically I finding the &lt;code&gt;T&lt;/code&gt; transformation matrix for each link from the the &lt;code&gt;base frame {B}&lt;/code&gt; to &lt;code&gt;wrist frame {W}&lt;/code&gt;. This is my code;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt; Function to compute Forward Kinematics&lt;/b&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def forwardK(q):&#xA;&#xA;#T06 is the location of Wrist frame, {W}, relative to Base frame, {B} &#xA;T01 = genT(q[0],0,d[0],0)&#xA;T12 = genT(q[1],a[0],d[1],alpha[0])&#xA;T23 = genT(q[2],a[1],d[2],alpha[1])&#xA;T34 = genT(q[3],a[2],d[3],alpha[2])&#xA;T45 = genT(q[4],a[3],d[4],alpha[3])&#xA;T56 = genT(q[5],a[4],d[5],alpha[4])&#xA;&#xA;#Tool frame {T}&#xA;#T67 = genT(0,0,d[5],0)&#xA;&#xA;T03 = matmul(T01,T12,T23)&#xA;T36 = matmul(T34,T45,T56)&#xA;T06 = matmul(T03,T36)    &#xA;#T07 = matmul(T06,T67)&#xA;&#xA;x = T[0][3]&#xA;y = T[1][3]&#xA;z = T[2][3]&#xA;&#xA;print(&quot;X: &quot;,x)&#xA;print(&quot;Y: &quot;,y)&#xA;print(&quot;Z: &quot;,z,&quot;\n&quot;)&#xA;print(&quot;T: &quot;,T,&quot;\n&quot;)&#xA;&#xA;return T06  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt; The function to compute &lt;/b&gt; &lt;code&gt;T Matrix&lt;/code&gt; &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;def genT(theta, a, d, alpha):&#xA;T =  array([[cos(theta), (-sin(theta)), 0, a],&#xA;    [sin(theta)*cos(alpha), (cos(theta)*cos(alpha)), -sin(alpha), (-   d*sin(alpha))],&#xA;    [sin(theta)*sin(alpha), cos(theta)*sin(alpha), cos(alpha), cos(alpha)*d],&#xA;    [0, 0, 0, 1]])&#xA;&#xA;return T&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;from the &lt;code&gt;T Matrix&lt;/code&gt; relating the &lt;code&gt;{B} frame to the {W} frame&lt;/code&gt; position vector of the &lt;code&gt;{w}&lt;/code&gt; &lt;code&gt;[x y z]&lt;/code&gt; is extracted. &lt;code&gt;R Matrix&lt;/code&gt; (orientation) of the &lt;code&gt;{W}&lt;/code&gt; relative to the &lt;code&gt;{B}&lt;/code&gt; is obtained by the following piece of code;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T = forwardK([30,-110,-30,0,0,0])&#xA;x = T[0][3]&#xA;y = T[1][3]&#xA;z = T[2][3]&#xA;R = T[0:3,0:3]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where &lt;code&gt;T&lt;/code&gt; is the transformation matrix relating &lt;code&gt;{W}&lt;/code&gt; to &lt;code&gt;{B}&lt;/code&gt;. Then this information is fed in to the &lt;code&gt;invK(x,y,z,R,ARM,ELOBOW,WRIST)&lt;/code&gt; function to check if the algorithm returns the same set of angles fed to the &lt;code&gt;forwardK(q1,q2,q3,q4,q5,q6)&lt;/code&gt; function. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the &lt;code&gt;invK(x,y,z,R,ARM,ELOBOW,WRIST)&lt;/code&gt;&#xA;      - &lt;code&gt;ARM, ELBOW, WRIST&lt;/code&gt; are orientation specifiers to describe various possible configurations of the manipulator. Each of these parameters are either &lt;code&gt;{+1,-1}&lt;/code&gt;. These values are then used in the closed form geometrical solution presented by the afore-mentioned paper.&lt;br&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I did not post the code for the&lt;/em&gt;&lt;/strong&gt; &lt;code&gt;invK(x,y,z,R,ARM,ELOBOW,WRIST)&lt;/code&gt; &lt;strong&gt;&lt;em&gt;since it is a direct implementation of the closed form solution presented in the paper and also it significantly long hence making it highly unreadable.&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think I am doing wrong? I am quite sure the way I am computing the FK is correct but I could be wrong. The matrix multiplications of my &lt;code&gt;Python&lt;/code&gt; code are correct since I double checked them with &lt;code&gt;Matlab&lt;/code&gt;. Any advice is appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;I already asked this question on &lt;a href=&quot;https://robotics.stackexchange.com/questions/11653/programming-forward-and-inverse-kinematics-of-puma-560&quot;&gt;robotics.stackexchange.com&lt;/a&gt;, but could not get a proper answer. That's why I am here. Thanks&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="1940" LastEditorUserId="-1" LastEditDate="2017-04-13T12:49:17.810" LastActivityDate="2017-02-22T08:31:49.247" Title="Forward and Inverse Kinematics of PUMA 560" Tags="|kinematics|robotics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13886" PostTypeId="1" CreationDate="2017-02-22T09:48:20.297" Score="0" ViewCount="672" Body="&lt;p&gt;How to convert uniform linear motion(mm/s) to uniform rotary motion(deg/s). I.e. table saw blade tilt mechanism. Something like reversed Scotch yoke mechnism. &lt;/p&gt;&#xA;" OwnerUserId="10159" LastActivityDate="2017-02-23T21:14:40.283" Title="Uniform linear movement to uniform rotary movement" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ClosedDate="2017-02-24T12:58:25.613" ContentLicense="CC BY-SA 3.0" />
  <row Id="13887" PostTypeId="2" ParentId="13886" CreationDate="2017-02-22T10:11:54.163" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Rack_and_pinion&quot; rel=&quot;nofollow noreferrer&quot;&gt;rack and pinion&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in a table saw tilt mechanism the rack is actually just a segment of a much) larger gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Howecer the scotch yoke is not uniform rotary to uniform linear, it is actually uniform rotary to sine linear. You can get rotary motion out of reciprocal motion using a crankshaft. You may need to add a flywheel to maintain the rotary momentum.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-02-22T10:11:54.163" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13888" PostTypeId="2" ParentId="13883" CreationDate="2017-02-22T11:52:54.483" Score="1" Body="&lt;p&gt;I would recommend you consider a low profile diamond drill rig; something similar to the &lt;a href=&quot;http://www.miningandconstruction.sandvik.com/sandvik/1181/Internet/CIS/S000924.nsf/Alldocs/Products*5CDrill*rigs*and*rock*drills*5CUnderground*exploration*drill*rigs*2ADE140/$File/DE140_080814_low.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sandvik DE140&lt;/a&gt; (I'm not advertising). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such rigs would definitely be able to drill the length of hole you require. Your space restrictions are the complicating factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given you space restrictions, I also recommend you consider the underground drill rig version, as opposed to the surface drill rig version, as equipment designed to operate underground tends to be smaller than equipment designed to operate on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will most likely have to raise the ceiling of the container where the hole is to be extended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram of the rig on page 7 shows it be 4250 mm tall when drilling vertical holes. It is advisable you have about 1 metre clearance above the rig, when it is operating; the more clearance, the better. For such a rig you will need to raise the ceiling of the container to the 5 metres mentioned in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the type of access and the dimensions of the access available into the container, where the hole is located, it may be possible to used a vehicle mounted rig similar to the one in the first picture on &lt;a href=&quot;http://www.resemin.com/en/pdf/agxyzq-drillmeters.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;. However, re-positioning the drill directly over the hole every time the rig has to be moved to add more drill steels, as the hole gets longer, will be a problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more expert advice, contact an underground mining drilling contractor who has low profile drilling rigs.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2017-02-22T12:16:26.560" LastActivityDate="2017-02-22T12:16:26.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13889" PostTypeId="2" ParentId="13882" CreationDate="2017-02-22T12:45:23.707" Score="5" Body="&lt;p&gt;I don't know what such a thing is called, but you can easily construct one by putting this relay circuit in-line with the power to the supply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xth1I.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xth1I.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This construct is called a &lt;i&gt;electrically latching relay&lt;/i&gt;.  When the relay is energized, pins 1 and 3 are connected, which keeps the incoming AC power applied to the relay coil.  This keeps the relay on and SW-HOT on as long as the incoming power stays on.  If the incoming power goes off, the relay turns off.  Once off, it stays off even if power is re-applied.  After power is re-applied, SW1 needs to be pressed for a short time to turn the relay on again.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-02-22T12:45:23.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13890" PostTypeId="2" ParentId="8473" CreationDate="2017-02-22T12:58:18.640" Score="0" Body="&lt;p&gt;@katipra I am adding this as another answer since it was too long for the comment section. But it's a reply to your aforementioned comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good news. Based on the actual definitions of Merriam-Webster of machine vs. machinery:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Machine: 1) an assemblage of parts that transmit forces, motion, and energy one to another in a predetermined manner.&#xA;             2) an instrument (as a lever) designed to transmit or modify the application of power, force, or motion.&#xA;             3)  a mechanically, electrically, or electronically operated device for performing a task.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Mechanism: 1) a process, technique, or system for achieving a result &#xA;                      2) mechanical operation or action.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the best conclusion would be that machine would be the most suitable word for an IC engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I have found a reliable link in NASA's database that mentions &quot;When discussing engines, we must consider both the mechanical operation of the machine and the thermodynamic processes that enable the machine to produce useful work.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;a href=&quot;https://www.grc.nasa.gov/WWW/K-12/airplane/icengine.html]&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.grc.nasa.gov/WWW/K-12/airplane/icengine.html]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10028" LastActivityDate="2017-02-22T12:58:18.640" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13893" PostTypeId="2" ParentId="6529" CreationDate="2017-02-22T18:21:36.270" Score="2" Body="&lt;p&gt;There are a lot of methods for determining permittivities at high frequency, but they are uniformly experimentally-based. One of the easy methods is a resonant cavity. You create a resonant cavity that has a known resonant frequency. Then insert the material inside the cavity by some holder and see how much the resonant frequency is perturbed (using a vector network analyzer or similar equipment). From this you can figure out the permittivity. This is a pretty accurate method but an open probe test can do a reasonable job with a lot less manufacturing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From experience, there typically is good data on common dielectric materials around the heating frequency (2.45 GHz). It will not likely differ meaningfully for 1-4 GHz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tested microwave cavities with dielectric materials at the 18-20 GHz range and usually published values are fairly accurate for 10 GHz even applied at 20 GHz. It is certainly still a meaningful thing to have!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for specific data, I'd look to data sheets from manufacturers of a given material. Matweb is also a good resource - you can sign up for a free account to get full results, I believe: &lt;a href=&quot;http://www.matweb.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.matweb.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also - just as an aside, doing antenna + dielectric simulations may be pretty easy to test the viability of dielectric materials, depending on the complexity of your antenna geometry. I've used COMSOL and a bunch of other packages to determine applicable relative permittivity / loss tangent ranges for microwave antennas / end launches. It sure beats purchasing expensive materials &amp;amp; manufacturing only to find out your frequency response is poor.&lt;/p&gt;&#xA;" OwnerUserId="3583" LastActivityDate="2017-02-22T18:21:36.270" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13894" PostTypeId="2" ParentId="13716" CreationDate="2017-02-22T18:46:01.117" Score="1" Body="&lt;p&gt;Unfortunately for all of us who have to deal with fluid mechanics, pressure drop through a valve from a flow rate can't be computed without knowing the Cv of the valve. You actually need the Cv of both valves here. From that, and knowing your target flow rate, you can compute how much the pressure will need to drop to achieve the flow rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can work the problem the other way, too: have a fixed pressure at the outlet and work backwards to find what your flow rate will be. Either way, you need the Cv or the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a million different references out there on valve sizing and the relevant equations: my personal favorite is &quot;Fluid Flow Basics of Throttling Valves&quot;: &lt;a href=&quot;http://www.controlglobal.com/assets/Media/MediaManager/RefBook_Cashco_Fluid.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.controlglobal.com/assets/Media/MediaManager/RefBook_Cashco_Fluid.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation you want, for gas flow, is: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Cv = Qdot / (N7 * Fp * P1 * Y) * sqrt(Mw * T1 * Z1 / X)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Qdot - volumetric flow rate in SCFH&#xA;N7 - Unit correlation constant = 1360&#xA;Fp - Piping geometry factor (can be taken as 1 for your case)&#xA;P1 - Absolute upstream pressure in psia&#xA;Y - Expansion factor (maximum of 1, minimum of 0.667 - plug both in to get a range of possible pressures)&#xA;Mw - molecular weight in lb/lb mole&#xA;T1 - Temperature in Rankine&#xA;Z1 - compressibility factor &#xA;X - Pressure drop ratio (i.e., (UpstreamP - DownstreamP) / UpstreamP)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So clearly here you need to either solve for X given Cv and Qdot, or Cv given X and Qdot...and probably now you can see why this can end up being someone's full time job.&lt;/p&gt;&#xA;" OwnerUserId="3583" LastActivityDate="2017-02-22T18:46:01.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13895" PostTypeId="2" ParentId="13882" CreationDate="2017-02-22T20:36:19.520" Score="1" Body="&lt;p&gt;In the UK  this is called an &lt;em&gt;NVR (no-volt release) switch&lt;/em&gt; they are compulsory in the EU for most static machine tools but not domestic appliances. Typically these are machines with exposed moving parts such as saws, drills, grinders, lathes ect &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have an 'on' button (marked I and green) and 'off' button  (marked O and red). The on button needs to be pressed for the machine to be powered on and hitting the off button immediately cuts power. In some cases there is a separate switch to actually operate the machine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the power is interrupted the on button needs to be pressed again to restart the machine ie if the power is interrupted the machine will not restart untill the on button is pressed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They often also have a mushroom switch type cover for easier manual shutoff in an emergency. This is just a mechanical means of pressing the off button. In some cases it also can be padlocked closed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also get rotary switches which perform a similar function, these are more common on large, especially 3 phase, machines and for remote isolation of an entire electric circuit, for example where a supervisor needs to be present for machines to be operated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.axminster.co.uk/kedu-kjd12-nvr-switch-230v-1ph-c-w-box-600072?gclid=CM3JmIXHpNICFfgW0wodtyUHBA&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; is one example &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-02-22T21:38:52.380" LastActivityDate="2017-02-22T21:38:52.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13896" PostTypeId="1" AcceptedAnswerId="13916" CreationDate="2017-02-22T22:07:40.227" Score="2" ViewCount="97" Body="&lt;p&gt;According to the &lt;a href=&quot;http://voyager.jpl.nasa.gov/spacecraft/spacecraftlife.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA Website&lt;/a&gt;, the Voyager probes will begin powering down electrical components soon with all electronic functions ceasing some time around 2025. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With no ability to transmit back to earth and thus of 0 value in the Interstellar Medium...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...could either one (or both) of the Voyagers be &lt;strong&gt;directed&lt;/strong&gt; back to Earth at all before electronic shut down?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not; do aerospace engineers have any idea where either (or both) of the probes may end up?&lt;/p&gt;&#xA;" OwnerUserId="10169" LastActivityDate="2017-02-23T23:28:05.523" Title="Can Voyager 1 or 2, theoretically, return to earth if given instructions before their electronic instruments shut down in 2025?" Tags="|aerospace-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13897" PostTypeId="1" CreationDate="2017-02-23T05:03:50.457" Score="0" ViewCount="1162" Body="&lt;p&gt;I have a face mask that I'm planning on making, a cosplay of Wrench from Watchdogs 2. Not satisfied with existing ones, I want to go for a more realistic one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, this requires that I make the entire part of the mask covering the eyes opaque. I'd also like to be able to look around with my eyes instead of moving my entire head around to view detail. Thid is also critical to the function of the mask. Basically high FOV covering my entire range of vision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might I implement this? I've thought about using stripped down VR goggles with a camera, mirrors to bounce light around from another point on the mask, or pinholes in the mask with lenses to allow you to see through them. (If that's possible, I'm not well acquainted with lenses.)&lt;/p&gt;&#xA;" OwnerUserId="10154" LastActivityDate="2017-02-23T13:37:22.277" Title="How to see through an opaque mask?" Tags="|optics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13899" PostTypeId="1" CreationDate="2017-02-23T07:39:40.873" Score="6" ViewCount="1386" Body="&lt;p&gt;I've been a civil/structural engineer for 20 years now. My HP 48GX has recently done its last calculation. So I looked for a replacement and it seems that the market for scientific/graphing calculators has changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: Could it be that smartphones are replacing calculators in the engineering world and all one needs to do is buy an &quot;graphing calculator&quot; app? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am not referring to the open sourced HP48 app)&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="23945" LastEditDate="2019-12-19T06:25:26.720" LastActivityDate="2019-12-19T06:25:26.720" Title="Are Scientific Calculators still current?" Tags="|engineering-general|" AnswerCount="4" CommentCount="10" ClosedDate="2017-02-24T12:55:47.580" ContentLicense="CC BY-SA 4.0" />
  <row Id="13900" PostTypeId="2" ParentId="13897" CreationDate="2017-02-23T08:48:33.903" Score="4" Body="&lt;p&gt;Just a lot of pinholes (without lenses) will be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a high enough density Your eyes and brain will be able to compensate the lack of complete vision pretty dang well. It'll be darker and you shouldn't go driving while wearing it but you'll be able to see enough to enjoy a party.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method could be to stretch a piece of thin black fabric over a frame. The weave will have a lot of hole for the light to come in.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2017-02-23T13:37:22.277" LastActivityDate="2017-02-23T13:37:22.277" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13901" PostTypeId="1" AcceptedAnswerId="13908" CreationDate="2017-02-23T09:07:35.813" Score="3" ViewCount="3029" Body="&lt;p&gt;I made a mathematical model (transfer function) of a rlc circuit. I simulated it in matlab simulink with step input and square wave input and observed the result. I can also calculate this system with step input since its transfer function is simple 1/s. But i do not know how to calculated square wave transfer function. Is there a way to find square wave transfer function or just i have to break the signal in to high or low form.&#xA;THanks &lt;/p&gt;&#xA;" OwnerUserId="8950" LastActivityDate="2017-02-23T20:23:43.827" Title="Square wave transfer function?" Tags="|control-theory|mathematics|transfer-function|systems-engineering|circuit-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13902" PostTypeId="2" ParentId="13901" CreationDate="2017-02-23T09:28:00.600" Score="2" Body="&lt;p&gt;You want to look for the &lt;a href=&quot;https://math.stackexchange.com/questions/1782935/laplace-transform-of-a-square-wave-function&quot;&gt;Laplace transform of a square wave&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The transfer function $H(s)$ is the ratio of the Laplace transforms of output $Y(s)$ and input $U(s)$:&#xA;    $Y(s)=H(s)\cdot U(s)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the Laplace transform of a function (input our output), it is not called a transfer function.&lt;/p&gt;&#xA;" OwnerUserId="3633" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:37.633" LastActivityDate="2017-02-23T09:28:00.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13903" PostTypeId="2" ParentId="13899" CreationDate="2017-02-23T11:21:49.420" Score="3" Body="&lt;p&gt;IMHO an app such as wolfram alpha can easily replace a calculator.&#xA;Some additional advantages:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Possibility of integration with other apps. E.g. Save results and&#xA;send them to your computer for further analysis.&lt;/li&gt;&#xA;&lt;li&gt;Extended functionality, such as 3d plotting.&lt;/li&gt;&#xA;&lt;li&gt;you do not need an extra device, since you always have your phone with you.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some reasons to stay using a calculator:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It is faster to use. The HMI is tailor made for entering equations.&lt;/li&gt;&#xA;&lt;li&gt;You do not need an internet connection.&lt;/li&gt;&#xA;&lt;li&gt;Battery life is better.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7590" LastActivityDate="2017-02-23T11:21:49.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13904" PostTypeId="2" ParentId="13899" CreationDate="2017-02-23T12:31:42.173" Score="4" Body="&lt;p&gt;When I'm at my desk I have a desktop computer connected to the internet in front of me and a smartphone with a scientific calculator app installed in my pocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I need to do some quick calculations I will still reach for my old calculator that I've had for around 24 years now (it still has the sticker on the back saying it's OK to use in university exams).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find the physical keys quicker and less error prone to use and it just feels more natural.&#xA;And while the actual processing speed is a tiny fraction of the other devices I have available it's good enough for the job and the battery life is far better (So far it's averaging 1 AAA every 10 years.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it died I'd certainly look for a replacement but I suspect the market is significantly smaller these days, if I didn't use it regularly I'd probably make do with the smart phone app myself.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2017-02-23T12:31:42.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13906" PostTypeId="1" CreationDate="2017-02-23T14:48:01.123" Score="2" ViewCount="53" Body="&lt;p&gt;I am working as TV repairer but some &quot;IC jungles for TV's&quot; are from China without almost information and often they are impossible to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is : Is it possible to do a IC jungle for TV, LCD, plasma, etc by aplying FPGA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know by microcontroller I can't do it because they are not so flexibles as FPGA's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example look this IC is from a TV and it's impossible to find out because is from China:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xrcnh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xrcnh.jpg&quot; alt=&quot;https://i.stack.imgur.com/xrcnh.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is the diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oThq4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oThq4.png&quot; alt=&quot;https://i.stack.imgur.com/oThq4.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to do most IC's impossible to find (because they are from china mostly) by FPGA?&lt;/p&gt;&#xA;" OwnerUserId="10179" LastEditorUserId="9" LastEditDate="2017-03-14T08:13:19.533" LastActivityDate="2017-03-14T08:13:19.533" Title="Is it possible to make commercial (or not) IC´s with FPGA?" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13908" PostTypeId="2" ParentId="13901" CreationDate="2017-02-23T20:23:43.827" Score="5" Body="&lt;p&gt;I thought I would expand a little on the answer offered by Karlo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Long story short, I would not try to calculate the analytical time response of a system to a square wave. That would be a serious pure-maths exercise, and not necessary for most engineering applications.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, I would suggest using the step response analytical vs. simulated to validate your model, and then trust that your MATLAB Simulink model is working properly for the square wave case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you feel like trying it out, here is what you need to know:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transfer function $G(s)$ is a description of the relationship between the input $U(s)$ and output $Y(s)$ of your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G(s) = \frac{Y(s)}{U(s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you want to determine the time response of your system to a specified input analyticalally, you rearrange for $Y(s)$ and then use the inverse Laplace Transform:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Y(s) = G(s)U(s)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y(t) = \mathcal{L}^{-1}\{G(s)U(s)\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated in the Question, for a simple step input $U(s) = \frac{1}{s}$ the time response is then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y(t) = \mathcal{L}^{-1}\{\frac{G(s)}{s}\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gets a lot more difficult for a square wave. A square wave is a series of time-shifted step functions (or Heaviside functions) $H(t-T)$ where $T$ is the time at which the step occurs. The derivation for the Laplace transform of a square wave is given in the answer to &lt;a href=&quot;https://math.stackexchange.com/q/1782998&quot; title=&quot;this question&quot;&gt;this question&lt;/a&gt; by &lt;a href=&quot;https://math.stackexchange.com/users/103399/alexjo&quot;&gt;alexjo&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$u(t) = A\sum_{k=0}^{\infty} \left[H\left(t-kT\right)-2H\left(t-\frac{2k+1}{2}T\right)+H\left(t-(k+1)T\right)\right]$$&#xA;  $$U(s) =\frac{A}{s}\frac{1-\mathrm e^{-\frac{Ts}{2}}}{1+\mathrm e^{-\frac{Ts}{2}}}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where $A$ is the amplitude of the square wave. You would then multiply this expression by your transfer function and then take the inverse Laplace transform of the entire mess. I am not aware of any simple way of doing this. Good luck.&lt;/p&gt;&#xA;" OwnerUserId="7430" LastEditorUserId="-1" LastEditDate="2017-04-13T12:19:37.633" LastActivityDate="2017-02-23T20:23:43.827" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13909" PostTypeId="1" AcceptedAnswerId="13911" CreationDate="2017-02-23T20:44:36.997" Score="2" ViewCount="1024" Body="&lt;p&gt;What is the best way to design a circuit based on the following transfer function: (0.0364*s)/(0.0002*s + 1) ? I am trying to design the derivative stage of a PIDF controller using op-amps&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: This is part of a bigger problem where i have to design a PIDF controller to achieve the minimum possible settling time for the plant function: $$\ G(s)= \frac{100*20*100}{(s+10)(s+20)(s+100)}$$&lt;/p&gt;&#xA;" OwnerUserId="10182" LastEditorUserId="10182" LastEditDate="2017-02-23T23:06:11.723" LastActivityDate="2017-02-23T23:06:11.723" Title="Design op-amp circuit from transfer function?" Tags="|pid-control|matlab|" AnswerCount="1" CommentCount="1" ClosedDate="2017-02-24T14:16:40.947" ContentLicense="CC BY-SA 3.0" />
  <row Id="13910" PostTypeId="2" ParentId="13886" CreationDate="2017-02-23T21:14:40.283" Score="0" Body="&lt;p&gt;Well, to answer the question directly, a belt and pulley system precisely converts linear to rotary motion. Sprockets &amp;amp; chains, toothed belts, etc can drive in either direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ezmRl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ezmRl.jpg&quot; alt=&quot;Pulley and belt diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-02-23T21:14:40.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13911" PostTypeId="2" ParentId="13909" CreationDate="2017-02-23T21:25:41.100" Score="4" Body="&lt;p&gt;To my knowledge the simplest way to accomplish the transfer function...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ G(s) = \frac{E_o(s)}{E_i(s)} =  \frac{0.0364s}{0.0002s + 1} = \frac{KTs}{Ts +1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...is the following high-pass filter circuit:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Xc24a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xc24a.png&quot; alt=&quot;High pass filter with opamp gain&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transfer function of this circuit is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{E_o(s)}{E_i(s)} = \frac{R_2+R_3}{R_3}\frac{R_1 C s}{R_1 C s + 1} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let $R_1C = T = 0.0002$ and $\frac{R_2+R_3}{R_3} = K = 0.0364/0.0002$. Then solve for $R_1$, $R_2$, $R_3$ and $C$. You'll have some wiggle room as to what values you pick, because you have two equations and four parameters. There are multiple solutions and you will have to decide what combination of values work best for your application.&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2017-02-23T21:25:41.100" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13913" PostTypeId="2" ParentId="11583" CreationDate="2017-02-23T22:33:00.393" Score="0" Body="&lt;p&gt;The torque required by the motor is the sum of different actors, for instance it can be the friction in the motor ($T=K_f \dot \phi$), the inertia of the rotor itself ($T=J\ddot \phi$) and then all the loads connected to the motor shaft. You have to include the inertia as well as the damping and stiffness between the different loads.&lt;/p&gt;&#xA;" OwnerUserId="1374" LastActivityDate="2017-02-23T22:33:00.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13914" PostTypeId="2" ParentId="8655" CreationDate="2017-02-23T22:46:01.877" Score="1" Body="&lt;p&gt;You essentially have a fluid mechanics problem here that can be approximated fairly well without heat transfer during the process of mixing. One approach (not too difficult) is to simply discretize the problem into steps of a fraction of a second performing the following computations:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Compute the mass flow rate from the upstream pressure, valve flow coefficient (Cv), and downstream pressure. Because this is a fairly high pressure problem, there will be some pressure drop from the tubing itself (but much smaller than the valve). You can choose whether to incorporate that loss into this step. See valve sizing equations from Crane Technical Paper 410, among many other references, for further details. You'll need to split the C and O into separate masses and assume they are evenly mixed for your own sanity (so you deduct 90% of the mass flow from C, 10% from O at each step).&lt;/li&gt;&#xA;&lt;li&gt;Deduct the mass flow that would exit the first tank and add it to the second tank (taking into account your time step).&lt;/li&gt;&#xA;&lt;li&gt;Recompute your upstream and downstream pressure based on the new masses.&lt;/li&gt;&#xA;&lt;li&gt;Repeat until you reach steady state (i.e. your pressures aren't changing anymore). Note that in a real world scenario the valve may have a minimum pressure to result in flow (&quot;cracking pressure&quot; for check valves, for ex) - so steady state may mean P1 != P2. You can then back out the temperature in each cylinder from the ideal gas law.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you need to deal with heat exchange during the process, this turns into a fully-fledged 1D compressible flow solver. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the problem was meant to be constructed as some sort of Bernoulli Equation example where you handwave and say the completely open valve acts exactly as a pipe, rather than the dominant factor (since they give you pipe dimensions but no valve details). &lt;/p&gt;&#xA;" OwnerUserId="3583" LastActivityDate="2017-02-23T22:46:01.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13915" PostTypeId="2" ParentId="8557" CreationDate="2017-02-23T23:10:59.517" Score="1" Body="&lt;p&gt;The question is not answerable using only the equations and constraints given. However, we can get close to the answer.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Power out of the motor&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Mechanical power is calculated by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = Fv = mgv$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $m$ is the mass, $g$ is the gravitational constant and $v$ is the speed at which the mass is being lifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the string wraps around some kind of barrel with a radius $r_b$ attached to a gear with radius $r_2$, then the torque $T_2$ applied to the gear by the load is:&#xA;$$T_2 = mgr_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the speed $\omega_2$ at which the gear rotates is&#xA;$$\omega_2 = \frac{v}{r_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let the radius of the gear attached to the motor be $r_1$ and let the gear ratio be defined as:&#xA;$$G = \frac{r_1}{r_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the torque applied on the motor is $T = GT_2$ and the speed of the motor is $\omega = \omega_2/G$. You can see that the power out of the motor (i.e. the power to lift the load) is independent of G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{out} = T\omega = GT_2\frac{1}{G}\omega_2 = T_2\omega_2$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Power into the motor&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assuming a DC motor, it will be powered by a power supply which outputs a voltage $V$ and a current $i$. The power consumed by the motor is then whatever is being supplied by the power source:&#xA;$$P_{in} = Vi$$&#xA;In the question you gave the following equations:&#xA;$$V = iR + K\omega$$&#xA;$$T = (i-i_o)K$$&#xA;If we rearrange the second equation for $i$ and then insert both equations into the power equation we get (after a lot of algebra):&#xA;$$P_{in} = \frac{(T+Ki_o)^2}{K^2}\left(R+\frac{K^2\omega}{T+Ki_o}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can substitute the expressions for $T_2$ and $w_2$ to get the gear ratio into the equation:&#xA;$$P_{in} = \frac{(GT_2+Ki_o)^2}{K^2}\left(R+\frac{K^2\omega_2}{G(GT_2+Ki_o)}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a little rearranging again we get:&#xA;$$P_{in} = \frac{(GT_2+Ki_o)^2}{K^2(G^2T_2+GKi_o)}\left(R+K^2\omega_2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this can be simplified a lot by grouping all of the constant terms so that we can see what effect G has on the power required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{in} = a_1\frac{(G+a_2)^2}{G(G+a_3)}(G(G+a_4)+a_5)$$&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Great algebra, but what does it all mean?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you take the limits as $G$ approaches infinity and $G$ approaches zero you will notice that in both cases the $P_{in}$ goes to infinity. This means that there is likely some optimal value or several optimal values of $G$ between zero and infinity that minimize the consumed power for a given output power. Those values will depend on the values of every other variable in the equation (how fast are you raising the load, what is the mass of the load, what are the values of $R$ and $K$, etc.?).&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2017-02-23T23:10:59.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13916" PostTypeId="2" ParentId="13896" CreationDate="2017-02-23T23:28:05.523" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;...could either one (or both) of the Voyagers be directed back to Earth at all before electronic shut down?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. It would be impossible since these satellites do not carry any type of course changing propellant. According to Wikipedia the Voyager satellites carry 16 hydrazine thrusters that are only used to reorient the satellite. This satellite was made to be like a bullet. Once it left the rocket that got it off earth it was only affected by gravity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;do aerospace engineers have any idea where either (or both) of the probes may end up?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes we do know where the probes are going. The short answer is nowhere in particular. Space is much more empty than many people can even fathom. The long answer is that in 300 years it will reach the Oort cloud which itself is theoretical and after that 40,000 years till it passes relatively close to a star. A mere 1.6 light years will separate them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These probes were not made to return to earth. They were designed to capture and return data to earth as they traveled out of our solar system.&lt;/p&gt;&#xA;" OwnerUserId="9656" LastActivityDate="2017-02-23T23:28:05.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13917" PostTypeId="2" ParentId="13899" CreationDate="2017-02-24T05:19:13.110" Score="1" Body="&lt;p&gt;The market has definitely changed: HP only has 8 calculators on their web site, and the hp50g (the last descendant of the hp48 series - or of the hp28 series if you look even farther back) was discontinued in 2015.   &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;all one needs to do is buy an &quot;graphing calculator&quot; app?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have and use an hp48 emulator on my phone, but it's not the same as a real 48, so I have two of them kicking around at all times - one mostly at the office, one at home.  I have an hp48sx that I bought in college, so it's more than 20 years old, and a 48G I bought off of craigslist about 5 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even on a smartphone with a large screen, the interface on a real hp48 is better - the positive feel to button presses and the lack of swiping issues are the two things I find the most beneficial about the real calculator.  Just turning it on is easier on the calculator - push the on button, no unlocking, swiping to the right screen, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even for a simple 4-function calculator, I'd rather use an old hp than a phone.  I have a bunch of old calculators and at different times I've used an hp34c or an hp12c - even with LEDs, they last a long time on a set of AAs.  (I'd fire up my 28S or 41CV except that N cell batteries are a pain.)&lt;/p&gt;&#xA;" OwnerUserId="3445" LastActivityDate="2017-02-24T05:19:13.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13919" PostTypeId="2" ParentId="13899" CreationDate="2017-02-24T07:01:12.573" Score="2" Body="&lt;p&gt;A calculator, today, is more about giving school exams a sandbox than about calculating. Any general purpose computer can do the same with fewer resrictions, including your phone. When you are sitting on your desktop you should be using the desktop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While its true that a calculator can have a better tactile user interface than your phone. Its still a marginal benefit when your phone can integrate into all your systems better more easily. You can even find, FEM, statics and other specialized apps. Even when you can not, you can network with your phone and use matlab, mathematica, ansys, abaqus... that is on your work computer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you can save calculations for reuse, for documentation nicely typeset etc. You use less time switching between systems for less transfer of errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to the question is: No. There is only a marginal use for a calculator, many reach for a calculator out of habit more than benefit.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-02-24T07:09:50.760" LastActivityDate="2017-02-24T07:09:50.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13922" PostTypeId="1" AcceptedAnswerId="13925" CreationDate="2017-02-24T08:06:26.997" Score="2" ViewCount="8290" Body="&lt;p&gt;In Ansys workbench , structural module, in modelling, what is the exact difference between pressure and surface force application? Is surface force applied in same manner as that of pressure application or different?&#xA;In more detail, I want to know how the total force is applied on the surface equally?&lt;/p&gt;&#xA;" OwnerUserId="9925" LastActivityDate="2019-01-08T17:59:56.767" Title="ansys workbech structural- difference between pressure and surface force application" Tags="|ansys|ansys-workbench|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13923" PostTypeId="1" CreationDate="2017-02-24T13:19:22.110" Score="2" ViewCount="87" Body="&lt;p&gt;I see figures claiming that the front brakes provides 75% of the stopping power for a bike. So why is the front tire not as wide as the rear, especially since the engine and rider are closer to the front? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The front tire has bigger demands on grip and supporting weight, but is given less.&lt;/p&gt;&#xA;" OwnerUserId="444" LastActivityDate="2017-02-24T14:18:37.310" Title="Why do bikes have different sized tires?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13924" PostTypeId="2" ParentId="13923" CreationDate="2017-02-24T14:18:37.310" Score="4" Body="&lt;p&gt;Braking is only one of several factors affecting tire and wheel design.  The rear wheer, for one thing, is the drive wheel.  That means not only does it experience far more stress at all times, but maintaining a larger contact area with the ground reduces the chance of breaking traction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to braking: while it's true that the front wheel ends up handling a lot of the stopping power due to basic mechanical physics, it's also true that the rear brake is critical to avoiding a somersault crash.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rear wheel handles far more weight at all times other than a hard brake. Most of the bike &amp;amp; operator mass is nearer the rear wheel than the front one. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-02-24T14:18:37.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13925" PostTypeId="2" ParentId="13922" CreationDate="2017-02-24T15:10:02.863" Score="3" Body="&lt;p&gt;&quot;Pressure&quot; is always normal to the surface. &quot;Surface force&quot; can be applied in any direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As example of a surface force that is &lt;em&gt;not&lt;/em&gt; normal to the surface would be snow loading on a sloping roof - the weight of the snow acts vertically downwards, not perpendicular to the slope of the roof. (Of course there must be some friction between the snow and the roof, otherwise the snow would just slide down the roof.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are doing a nonlinear analysis, a pressure load can &lt;em&gt;change direction&lt;/em&gt; as the structure deforms, to stay normal to the surface. A surface force (like the snow loading) may always act in a fixed direction.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-02-24T15:10:02.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13927" PostTypeId="2" ParentId="11583" CreationDate="2017-02-24T18:54:43.080" Score="0" Body="&lt;p&gt;Yes, motor torque depends on load. More the load torque more the torque produced by motor. Yes, as previously mentioned answers every motor have some losses, that may be friction losses etc, and at no load the motor will produce a torque equal to the torque due to loss. So the some of Torque produced = Load torque + Torque (friction loss). The torque is not a thing that you can calculate that it will produce. There is a torque-speed charaterstic curve of different motors(dc series,parallel etc) in which it shows how the motor speed changes as you add load on it.&lt;/p&gt;&#xA;" OwnerUserId="8950" LastActivityDate="2017-02-24T18:54:43.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13929" PostTypeId="2" ParentId="217" CreationDate="2017-02-25T02:33:59.313" Score="-2" Body="&lt;p&gt;If you can get to like 3000 psi, you could store a lot more energy in a a smaller space. What if you had a several solar-powered, high pressure compressors and a fleet of high pressure tanks? If one were to get the numbers right, they could basically have an air-powered house, no?&lt;/p&gt;&#xA;" OwnerUserId="10208" LastActivityDate="2017-02-25T02:33:59.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13931" PostTypeId="1" CreationDate="2017-02-25T13:17:54.380" Score="1" ViewCount="31" Body="&lt;p&gt;How do I hold a end milling tool in a rod which is smaller than the tool(smaller efers to the diameter of the rod).&lt;/p&gt;&#xA;" OwnerUserId="10214" LastActivityDate="2017-02-25T13:17:54.380" Title="Tool fixture and holding" Tags="|mechanical-engineering|manufacturing-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13933" PostTypeId="2" ParentId="217" CreationDate="2017-02-25T16:33:37.343" Score="2" Body="&lt;p&gt;The existing answers do a pretty good job of explaining the sizing and the economics, but surprisingly for an engineering site, skip over the equipment. And that's where this becomes totally infeasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lightsail.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;LightSail&lt;/a&gt; is arguably the closest to making compressed air energy storage (CAES) economically feasible at any scale. &lt;a href=&quot;https://www.greentechmedia.com/articles/read/Layoffs-at-Lightsail-Khosla-Funded-Compressed-Air-Energy-Storage-Startup&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to LightSail's CEO, the problem is the efficiency.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Efficiency of compressors/expanders can be hard to pin down, so let's just look at the motor/generators first. &lt;a href=&quot;http://www.engineeringtoolbox.com/electrical-motor-efficiency-d_655.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The NEMA standards for motors are that large ones need to be mid-90's percent efficient.&lt;/a&gt; So let's assume Lightsail is using a 94% efficient motor/generator. If you are using the &lt;a href=&quot;http://www.homedepot.com/p/DEWALT-15-Gal-Portable-Electric-Air-Compressor-D55168/100497924&quot; rel=&quot;nofollow noreferrer&quot;&gt;biggest compressor at the local home store&lt;/a&gt;, you're looking at at most 2 hp. NEMA suggests you'll be looking at 78% efficiency on that motor. So just running the motor to generate the compressed air and running it in reverse to get the electricity back out, you've already lost $1-.78^2=0.39$ of your energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the compressor/expander. These numbers are not regulated and are generally trade secrets (although sometimes you can back them out from datasheets). This website suggests though that a small low speed reciprocating compressor might also be 80% efficient (my experience is actually that that is much too high for a product like the one I linked above, but I can't find any sources to cite). Now let's also assume you can use that same device as an expander at the same efficiency (highly unlikely). That's another $1-0.8^2=0.34$ loss. So in total, we're looking at $.78^2-0.8^2=0.39$ efficiency of the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the economics, if electricity costs you 5.4 p/kWh and you can store it at 39% efficiency, your recovered price is $5.4/0.39=13.9$p/kWh. Bad news! That's more expensive than your daytime price, so no, you unfortunately can't do this cost effectively at home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also look at the this heuristically. If Lightsail, and no one else, can do it economically at large scales with &lt;a href=&quot;https://www.crunchbase.com/organization/lightsail-energy#/entity&quot; rel=&quot;nofollow noreferrer&quot;&gt;$37M research dollars&lt;/a&gt;, its almost impossible that you could do it at small scales with smaller off the shelf equipment.&lt;/p&gt;&#xA;" OwnerUserId="2781" LastEditorUserId="2781" LastEditDate="2017-02-25T17:03:47.357" LastActivityDate="2017-02-25T17:03:47.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13936" PostTypeId="1" CreationDate="2017-02-25T19:05:24.197" Score="0" ViewCount="93" Body="&lt;p&gt;I have a solid plate of steel I need to see through, as a mask. I need a wide FOV (at least 210 degrees). I cannot have any obstructions in the image, so no pinhole meshes or anything like that.&#xA;Furthermore, I can drill no more than 4 small (pinhole) holes in it. It only covers my eyes and parts of my nose. Think ski goggle coverage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not well versed in optics, so I'd like to know how I could somehow &quot;re-route&quot; my vision, like a periscope, while maintaining a reasonable field of view. I've also looked into using VR headsets as a way to do this but none currently released have wide enough FOVs for my tastes. &lt;/p&gt;&#xA;" OwnerUserId="10154" LastActivityDate="2017-02-26T08:02:46.207" Title="How might I see through a completely solid plate of steel worn as a mask?" Tags="|optics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13937" PostTypeId="2" ParentId="13936" CreationDate="2017-02-25T20:10:59.507" Score="0" Body="&lt;p&gt;The problem with any sort of camera or periscope type VR is that it is likely to be quite disorientating as your point of view will be changes plus camera lenses withe the required field of view will have a significantly different focal length to your eyes which will seriously alter your depth perception. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle through which your own eyes can clearly register images is actually quite narrow and a lot of the human filed of vision is achieved by rapid scanning eye movements and the brains ability to manufacture a composite image in fill in details as well as being able to process and detect points of interest and focus on them. All of this is very difficult to achieve with cameras trying to give real time information about the real world. In fact a wholly virtual reality is much easier to achieve than a replicating a real environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't believe me try walking around or picking things up just looking through the view finder of a camera. There is also the problem of the minimum focal distance of the human eye, you simply won't be able to focus on a screen close enough to your eyes to fit in a goggle type setup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind that that the F35 fighter jet uses a similar system to what you want  and that has had persistent problems despite the eye watering sums of money spent on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get materials which appear opaque from one side but allow pretty decent transparency form the other. Mirrored sunglasses are one obvious example pl;us there are various other coated plastics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have discounted meshes, fine mesh and perforated sheet can give pretty good vision, they are, for example used for eye protection in air-soft and mesh face I have personally used mesh face shields in foundry work where I can assure you you need excellent depth perception and peripheral vision.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-02-25T20:10:59.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13939" PostTypeId="1" AcceptedAnswerId="13945" CreationDate="2017-02-26T00:16:07.463" Score="1" ViewCount="262" Body="&lt;p&gt;I have a small flat surface (~80mm x 80mm) that needs to adhere to a round surface (radius = 100mm). Does anyone know how to mate the two surfaces together using tape to give a good enough bond?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thought appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The part with the flat surface (~250g) is supposed to be hung on the round part (a lamp shade). Both parts are plastic. I'm hoping to avoid causing damage to either part. The whole assembly will be kept indoor.&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="4302" LastEditDate="2017-02-26T02:43:58.200" LastActivityDate="2017-02-26T04:32:14.483" Title="Mating a round surface with a flat surface" Tags="|adhesive|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13941" PostTypeId="1" AcceptedAnswerId="13947" CreationDate="2017-02-26T01:27:21.390" Score="3" ViewCount="3821" Body="&lt;p&gt;Arch bridges operate according to the following (I am assuming arch dams are the same):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Tensional force in arch bridges are virtually negligible. The natural curve of the arch and its ability to dissipate the force outward greatly reduces the effects of tension on the underside of the arch.&lt;/p&gt;&#xA;&lt;p&gt;The greater the degree of curvature (the larger the semicircle of the arch), the greater the effects of tension on the underside of the bridge.&#xA;(&lt;a href=&quot;http://science.howstuffworks.com/engineering/civil/bridge5.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7cuBs.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;WHY does a greater degree of curvature results in more tension?&lt;/strong&gt; A diagram would be useful if possible.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wvRJN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvRJN.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;2. How does the changing the radius of curvature affect the tension and compression in the dam, and ultimately the strength of it.&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GhRjD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GhRjD.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7745" LastEditorUserId="6790" LastEditDate="2020-07-31T21:04:01.907" LastActivityDate="2020-07-31T21:04:01.907" Title="How and WHY does the tension in an arch dam or bridge change as the degree of curvature and radius of curvature change?" Tags="|civil-engineering|bridges|hydroelectric-dams|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="13942" PostTypeId="1" CreationDate="2017-02-26T01:53:39.063" Score="1" ViewCount="439" Body="&lt;p&gt;Does anyone know of any type of small, refillable oxygen tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to build a rocket and the smallest cylinder I could find is BernzOmatic's 1.4 oz. cylinder.&#xA; A big cylinder isn't really safe for my purpose and I really need it to be refillable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there any DIY solutions? Safety is very important for me, and putting a big tank on a rocket ain't really a good idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI, the tank has to be around 2&quot; - 2.5&quot; in diameter, and it's very &lt;strong&gt;important that it is REFILLABLE!!!&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2017-02-27T14:38:32.920" Title="Small oxygen tanks" Tags="|rocketry|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13943" PostTypeId="1" AcceptedAnswerId="13965" CreationDate="2017-02-26T03:05:44.277" Score="1" ViewCount="159" Body="&lt;p&gt;Could someone tell me if there's a general name for the kind of spring clips that slide onto a rod and set a position for some other attached element. I've seen them on sinks and toilet tanks -- here are pictures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vvz3K.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vvz3K.jpg&quot; alt=&quot;toilet float level setter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Burj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Burj.jpg&quot; alt=&quot;Sink stopper lever position adjustment&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10221" LastActivityDate="2017-02-27T13:36:38.400" Title="What's the name of position-setting spring clips used in plumbing?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13945" PostTypeId="2" ParentId="13939" CreationDate="2017-02-26T04:32:14.483" Score="1" Body="&lt;p&gt;Without knowing more about your requirements here are some options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Permanent&quot; options:&#xA;Try a two-part epoxy designated for plastics. Using sandpaper, like 120 grit, to roughen the surfaces helps with adhesion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, maybe a double-sided foam tape would work. Specifically VHB tape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're using acrylic, you could use the special acrylic glue solvent which makes a strong bond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Non-permanent&quot; options:&#xA; zip ties/cable ties or a hose clamp to hold the parts together&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-02-26T04:32:14.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13946" PostTypeId="2" ParentId="13936" CreationDate="2017-02-26T08:02:46.207" Score="1" Body="&lt;p&gt;I looked up the character you mentioned. Actually the grid looks quite coarse. I counted about 20 pixels over the height of the mask. About 10 pixels per inch. I think you could easily drill a small (1/64&quot;) hole between every pixel, and then maybe cover the back with a black gauze. That should allow you to see out, but no one could see in. Seems to me that is the only real option&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-02-26T08:02:46.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13947" PostTypeId="2" ParentId="13941" CreationDate="2017-02-26T11:36:04.667" Score="6" Body="&lt;p&gt;The presence of tension in an arch is not really dependent on the curvature but on how well the arch matches a &lt;a href=&quot;https://en.wikipedia.org/wiki/Catenary&quot; rel=&quot;noreferrer&quot;&gt;catenary shape&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Robert Hooke famously stated:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As hangs a flexible cable so, inverted, stand the touching pieces of an arch.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So a catenary arch will have only compression stresses (since a flexible cable can only have tension stresses). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GKhAq.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GKhAq.jpg&quot; alt=&quot;Catenary Arch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how the line of action of the compression in the arch runs through the centre of the arch. If the arch shape is not a catenary, the line of action will still follow a catenary shape but will not follow the centre line of the arch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CLcKTm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CLcKTm.jpg&quot; alt=&quot;Non-catenary&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see how this can cause tension by examining the distribution of stress as a result of the position of the line of action (green is tension):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aZWXF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aZWXF.png&quot; alt=&quot;Stress Distribution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general the rule for arches is the cantenary line must be within the middle $\frac{1}{3}$ of the arch to avoid tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Form finding can be used to find the catenary shape for a particular loading. This can be done computationally, or even by the use of &lt;a href=&quot;http://dataphys.org/list/gaudis-hanging-chain-models/&quot; rel=&quot;noreferrer&quot;&gt;hanging models&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arch dams behave in the same way. However, since the main loading on the dam (from water) is a pressure normal to the dam surface the equivalent of the catenary shape is a circle. In this case the compressive stress is calculated in the same way as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cylinder_stress#Hoop_stress&quot; rel=&quot;noreferrer&quot;&gt;hoop stress&lt;/a&gt; of a pressure vessel (where I have assumed r &gt;&gt; t):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma = \frac{Pr}{t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\sigma$ is the stress P is the applied pressure, r is the radius and t is the thickness. Since the pressure P increases with the depth of water dams sometimes decrease radius with depth (variable-radius dams) or change the thickness (constant-radius dams).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally the design of a dam has many other complicating factors which affect the design.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2017-02-26T17:23:48.323" LastActivityDate="2017-02-26T17:23:48.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13950" PostTypeId="1" CreationDate="2017-02-26T12:45:35.443" Score="1" ViewCount="70" Body="&lt;p&gt;I have to build an electro-mechanical device that has a few sliding and rotating parts controlled by servos and actuators from Phidgets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made the prototype with wood; it was crude, imprecise, and looked like it was built from scrap from a junkyard, but it proved the concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we are going to need to make the real thing. I spoke with a local company that does CNC work and I tried to explain to them the concept. But I suck at drawing, and the original prototype covers just a part of what we need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have several options that I am contemplating:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Building something out of wood again to show them what I need. But it was time consuming and not very good last time&lt;/li&gt;&#xA;&lt;li&gt;Build something, without the motors, out of legos to demonstrate the moving parts.&lt;/li&gt;&#xA;&lt;li&gt;Make a proper model in SketchUp, but it won't be able to demonstrate the moving parts. The CNC company told me they'd be happy to see SketchUp/Autocad files, but I'm afraid this won't convey how things move during the operation.&lt;/li&gt;&#xA;&lt;li&gt;Writing on Stack Exchange to get some fresh ideas :)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What would be the best way to describe a device to people foreign with the concept? They need to understand how it operates, the constraints for the moving parts, etc and eventually come up with the final schematics. So I need to get the point across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never done this before, and this is domain that is foreign to me.&lt;/p&gt;&#xA;" OwnerUserId="8789" LastEditorDisplayName="user16" LastEditDate="2017-02-26T14:48:31.177" LastActivityDate="2017-05-27T18:50:22.390" Title="Having to build an apparatus, trying to show the concept" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13951" PostTypeId="1" AcceptedAnswerId="13956" CreationDate="2017-02-26T13:24:25.177" Score="-1" ViewCount="147" Body="&lt;p&gt;If one searches for the different levels of CO2 emissions in cars, one can find two very different levels. For instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: &lt;a href=&quot;https://ec.europa.eu/clima/policies/transport/vehicles/cars_en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; it is written that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The law requires that the new cars registered in the EU do not emit more than an average of &lt;strong&gt;130 grams of CO2 per kilometer&lt;/strong&gt; (g CO2/km) by 2015.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So it can be seen that the mass of CO2 is described as being over 100 grams/km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: &lt;a href=&quot;https://www.dieselnet.com/standards/eu/ld.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt;, one can see that the level of &lt;strong&gt;CO is around 1 g/km&lt;/strong&gt;. So way less than 100 g/km, which is a huge difference. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, in the second reference, the table shows that those are the values for CO and not CO2, as in the first case. But in the second one, I can't find the CO2 levels and I assumed that the CO is a typo, and it's missing the &quot;2&quot;. However, something tells me that it is not a typo and that they read the levels of CO and not CO2. Now, I have two questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a typo in the second reference? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Yes, &lt;strong&gt;Q1:&lt;/strong&gt; Why is the difference so huge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If No, &lt;strong&gt;Q2:&lt;/strong&gt; Why do they give the values in CO and not in CO$_2$?&lt;/p&gt;&#xA;" OwnerUserId="8419" LastEditorUserId="8419" LastEditDate="2017-02-26T13:30:12.343" LastActivityDate="2017-02-26T15:20:24.507" Title="What is the real CO2 emission for cars? 1g/km or &gt;100g/km?" Tags="|manufacturing-engineering|combustion|emissions|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13952" PostTypeId="1" AcceptedAnswerId="13958" CreationDate="2017-02-26T13:30:14.400" Score="0" ViewCount="3257" Body="&lt;p&gt;This question is similar to a question I am stuck on, but I have changed it to get an understanding how to works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spring is being made by pulling a 12cm long cylinder of material in tension with a desired stiffness of 6200kN/m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has a density of 2.9 g/cm^3, an ultimate tensile strength of 375 MPa and a Young's modulus of 70 GPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What should the diameter (in mm) be for the metal &quot;spring&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is how can I go about solving this question?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at the Young's Modulus and Hooke Law's formulas.  But keep getting stuck with material displacement.  Do I need to know the amount of length the object changes to work out this solution, or am I on the wrong track?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I assume: $\Delta L = 1$ if I used the following formula to find the $A_0$, which I can then calculate the diameter from:&#xA;$$&#xA;F = \frac{E A_0 \Delta L}{L_0} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be most appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10223" LastEditorUserId="10223" LastEditDate="2017-02-26T16:26:33.070" LastActivityDate="2017-02-26T17:01:44.770" Title="Finding the diameter of a material using tension stiffness and Young's Modulus" Tags="|springs|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13953" PostTypeId="2" ParentId="13950" CreationDate="2017-02-26T14:45:32.837" Score="1" Body="&lt;p&gt;I think that it depends upon what your purposes are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking to pitch the concept to investors, then there are a couple of routes to consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious route, as you mentioned, would be to build a second, more refined prototype.  It may be worth giving this another consideration despite the investment of your time.  Nothing sells a concept to an investor quite as well as an actual, functioning prototype.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For perspective, Dyson claims that he &lt;a href=&quot;http://www.dyson.com/community/aboutdyson.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;went through over 5,000 prototypes&lt;/a&gt; in designing his first bagless vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another route to consider is that &lt;a href=&quot;https://help.sketchup.com/en/article/3000147&quot; rel=&quot;nofollow noreferrer&quot;&gt;sketchup appears to support creating animations&lt;/a&gt;.  If you eventually want to ship this concept off to a machine shop, then you'll need to invest the time to create digitized plans.  This is likely the fastest path to having a refined presentation of the concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other concept to consider is to create a video recording of the existing prototype.  That will both demonstrate the moving portions of the concept as well as digitizing the concept so it is more easily shared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rebuilding the prototype in a non-functional way (leaving out motors) with a material that you never intend to use for production doesn't seem like a great approach.  That seems like a step backwards from where you're currently at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option to consider is to contract with a mechanical design shop.  Not all CNC shops have mechanical design capabilities, so you may need to shift who you are talking with at this point.  A design shop should have an engineer or draftsman that can work with you on the concept and then build a specification.  It's worth noting that this can appear to be an expensive option, but you are paying for their expertise in creating schematics.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2017-02-26T14:51:55.377" LastActivityDate="2017-02-26T14:51:55.377" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="13954" PostTypeId="2" ParentId="13879" CreationDate="2017-02-26T15:16:53.773" Score="-1" Body="&lt;p&gt;Assume a slope of 1:500 it is impossible in reality to build a completly flat pipe. Construction tolerances accept that a flat pipe is in a gradient of 1:500. Flow calculations to this will prove adequately precise.&lt;/p&gt;&#xA;" OwnerUserId="8863" LastActivityDate="2017-02-26T15:16:53.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13955" PostTypeId="2" ParentId="13826" CreationDate="2017-02-26T15:19:06.550" Score="0" Body="&lt;p&gt;I've encountered this problem before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem didn't depend on the connection of your power transformer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem happens when meter coefficients (depend on your CT's and PT's ratios)are inappropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The note is this:Check your circuits node and use KCL law to calculate the powers in node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards&lt;/p&gt;&#xA;" OwnerUserId="4412" LastEditorUserId="4411" LastEditDate="2017-03-01T02:15:01.060" LastActivityDate="2017-03-01T02:15:01.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13956" PostTypeId="2" ParentId="13951" CreationDate="2017-02-26T15:20:24.507" Score="2" Body="&lt;p&gt;Q1.&#xA;The CO is Carbon Monoxide Level, and CO2 is the Carbon Dioxide Level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is huge because the potent levels for both of them are very different.  Generally for clean fresh air there will be between 0-0.5ppm CO, and between 250-350ppm CO2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exposing ourselves to around 800ppm of CO will kill us in 2-3 hours, where as we would need to be exposed to over 40,000ppm of CO2 for it to become fatal for an average person dur to oxygen deprivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2.&#xA;The reason for some datasets using CO while other use CO2 is usually down to standards within the Governments, Councils, Companies parameters.  Certain datasets require CO levels where as others use CO2 levels.  Theis could also be due to the questions that are being asked.&lt;/p&gt;&#xA;" OwnerUserId="10223" LastActivityDate="2017-02-26T15:20:24.507" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13958" PostTypeId="2" ParentId="13952" CreationDate="2017-02-26T17:01:44.770" Score="1" Body="&lt;p&gt;The axial stiffness of an isotropic material with a uniform cross-section is a fundamental part of most engineering mechanics of materials concepts. We can derive it as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force-displacement relation of a spring is described by Hooke's Law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = k \Delta L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $F$ is the force exerted on the spring, $\Delta L$ is the change in length or displacement, and $k$ is the stiffness or spring constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can rearrange to get an expression of the stiffness&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = \frac{F}{\Delta L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we want to express $k$ exclusively in terms of the geometry and material properties. For that, we need to make use of the engineering stress $\sigma = F/A_0$ where $A_0$ is the cross-sectional area of the material before deformation. We can rearrange and substitute this back into the previous equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$k = \frac{\sigma A_0}{\Delta L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We still need to get rid of $\sigma$, and to do that we can use $\sigma = E \epsilon$ where $E$ is the Young's modulus of the material and $\epsilon$ is the engineering strain. Substituting that and using the definition of engineering strain $\epsilon = \Delta L/L_0$ where $L_0$ is the length of the material before deformation and we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ k = \frac{E(\Delta L/L_0)A_0}{\Delta L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ k = \frac{E A_0}{L_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $\Delta L$ cancels, so we don't need to know the change in length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you want to figure out the necessary dimensions for a block of material to have a given stiffness, just substitute the appropriate equation for $A_0$ and rearrange. For example, a cylinder has cross-sectional area $A_0 = \frac{\pi}{4} D_0^2$, sub that in and you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ D_0 = \sqrt{\frac{4 k L_0}{\pi E}}$$&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2017-02-26T17:01:44.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13959" PostTypeId="2" ParentId="13942" CreationDate="2017-02-26T18:23:24.620" Score="3" Body="&lt;p&gt;Why are you trying to use O2 stored as a gas in a rocket? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get any real quantity of the stuff you would need enormous pressures which argues directly against a light weight tank, the tank structure would weigh more then the gas it contains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oxygen using rockets all start with liquid oxygen, which has all sorts of fun issues of its own in a small rocket (Just don't go there!).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is wrong with Nitrous oxide instead? It liquifies at reasonable pressures at room temperature, so it is much easier to get a sane propellant mass fraction in a small rocket, it is &lt;strong&gt;MUCH&lt;/strong&gt; safer to handle then pure O2 (or heaven forbid, liquid oxygen). &lt;/p&gt;&#xA;" OwnerUserId="10072" LastActivityDate="2017-02-26T18:23:24.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13960" PostTypeId="1" AcceptedAnswerId="13988" CreationDate="2017-02-26T18:36:51.930" Score="2" ViewCount="4229" Body="&lt;p&gt;I've been working on simulating drone flight in Gazebo (an open source robotics simulator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I'm trying to control the drone's pitch / roll rotation using PIDs. It's important to remember that a drone moves linearly only by angular rotation (forward / backward requires pitch rotation, left / right requires roll).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I'm testing PID pitch control.  The goal is to get the drone to rotate to a specified angle and hold that position - perfect job for a PID.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the physics implementation complicates that.  Essentially, the PID takes the setpoint - the desired pitch rotation angle (in radians), and the current pitch rotation angle (which I get directly form the simulation state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PID output is then used to calculate the resultant torque.  The torque equation is t = I * a (torque = Intertia * angular acceleration).  So, essentially, I'm using a PID to control the angular acceleration of the drone given the current and desired angles of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem, of course, is that a PID output scales linearly with time, so although it decreases to zero as the drone approaches the setpoint, it has been applying positive accelerations through the entire time.  The result is that the drone has a lot of angular momentum, and so it naturally overshoots.  Then, of course, the PID responds by reversing the rotation, but with the same problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, what happens is that the drone starts with a horizontal attitude, rotates to the setpoint, then rotates back to horizontal, then back to the setpoint.  If I add any I or D gain it becomes unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is really simple.  It's like controlling a vehicle in space.  Thrust is applied in pulses in order to give fine control over position / rotation.  The PID however, provides a continuous output.  Which means I need some sort of physical system that behaves as depicted here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://vignette4.wikia.nocookie.net...imation.gif/revision/latest?cb=20141124024926&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://vignette4.wikia.nocookie.net...imation.gif/revision/latest?cb=20141124024926&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or I need to rewrite the physics so that the PID output scales appropriately to bring angular velocity to zero by the time it reaches the desired angle of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this is proving more difficult than I thought - probably because I'm not a strong physics or controls guy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've posted this problem already in another forum.  You can read the thread here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.physicsforums.com/threads/pid-control-for-drone-rotations.905431/#post-5701849&quot; rel=&quot;nofollow noreferrer&quot;&gt;PID control for drone rotations&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where I get really specific.  I'm afraid I wasn't able to articulate the problem well enough, partly because the guy who tried to help wasn't really familiar with PIDs (or C++).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, Any thoughts from someone who has dealt with this sort of thing (or thinks they know more about it than I do...) would be really welcome. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  You can see the behavior in the following animated gif:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://gph.is/2lKf9aT&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://gph.is/2lKf9aT&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10225" LastEditorUserId="6790" LastEditDate="2020-05-04T08:03:09.563" LastActivityDate="2020-05-04T08:03:09.563" Title="Using a PID to control rotation" Tags="|control-engineering|aerospace-engineering|pid-control|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="13961" PostTypeId="1" CreationDate="2017-02-26T19:55:10.413" Score="1" ViewCount="27" Body="&lt;p&gt;Why the circuit does not freeze/lock when the signals SUM ? But the electrical signals are SUM together in what sense?&#xA;&lt;br&gt;This question is about neural network circuits because I don't understand what exactly at hardware level node is it&lt;br&gt;&#xA;Is a line ?&#xA;&lt;br&gt;I have a wrong image in my head about node concept because I translate like &lt;em&gt;point&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;if 10 machines meet at the same point or node occurs the incident, then how is adjusting the time within a neural network? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What happens inside a 'node' in artificial neural network ? Can you show me an example of hardware electronic circuit implementation about a neural network?&#xA;a neural network?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another suggest me this image to understand but I don't see circuit&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a circuit with input terminals 10 to 10 independent signals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But this is a &lt;em&gt;line&lt;/em&gt; not a 'node'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I try to find a circuit for what you call 'node' about an artificial neural network. i don't understand how is implemented a 'node' with an electronic circuit&lt;/p&gt;&#xA;" OwnerUserId="10226" LastEditorUserId="10226" LastEditDate="2017-02-26T22:47:02.010" LastActivityDate="2017-02-27T11:53:43.950" Title="How do they get more signals simultaneously inside a node of a neural network?" Tags="|circuits|circuit-design|" AnswerCount="1" CommentCount="3" ClosedDate="2017-03-05T10:43:35.773" ContentLicense="CC BY-SA 3.0" />
  <row Id="13963" PostTypeId="1" CreationDate="2017-02-26T20:30:55.840" Score="0" ViewCount="78" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/biQo3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/biQo3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two motors are mounted in parallel as you can see in the picture and each one of them is connected to a gearbox through an elastic coupling. My question is : how do they operate together ? is one of them supposed to be off ? if you have any book or any website that discuss such transmissions I'd be thankful.&lt;/p&gt;&#xA;" OwnerUserId="9832" LastActivityDate="2017-02-27T11:23:21.447" Title="How does the second Motor operate in this diagram?" Tags="|mechanical-engineering|electrical-engineering|motors|gears|power|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13964" PostTypeId="2" ParentId="13963" CreationDate="2017-02-26T22:04:21.390" Score="2" Body="&lt;p&gt;Three gears (and gearhousing body) in that configuration is a differential. Differentials work in both directions, connected this way they simply add the outputs together to one shaft. That is simply one of the uses of the differential. Although the reverse of this is more commonly known and is used in cars to spread the forces of one shaft to two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically a differential works as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A + B = C&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since you have motors connected to A and B their combined input is the output  C. Also note this means the other motor can be not running it still works perfectly fine (since A + 0 = A = C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, before you ask, differentials can be used to do mechanical arithmetical calculations of adding or subtracting. It was used in all kinds of targetting devices in WW2.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-02-27T06:30:51.943" LastActivityDate="2017-02-27T06:30:51.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13965" PostTypeId="2" ParentId="13943" CreationDate="2017-02-26T23:22:26.813" Score="3" Body="&lt;p&gt;After doing some searching it seems that style of clip is limited to toilet hardware like you have shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A google reverse image search found the &lt;a href=&quot;http://www.prettyhandygirl.com/repairing-pop-up-sink-drain/&quot; rel=&quot;nofollow noreferrer&quot;&gt;originating site&lt;/a&gt; of your first image. They call it a &quot;joint clamp clip&quot; but a &lt;a href=&quot;https://en.wikipedia.org/wiki/Joint_clip&quot; rel=&quot;nofollow noreferrer&quot;&gt;join clamp clip on wiki&lt;/a&gt; says &quot;Joint clips are used to prevent a joint from separating during a reaction process&quot; so that name does not fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many niche industries and unique devices do not have names even if the device is very old. I'm sure various plumbers refer to this clip in many different ways. From my experience in construction, the names are often unintelligent, named after a specific brand of that product, racist, sexist, or overlap with another industry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I am creating engineering documentation and I see a new device that does not have an obvious, appropriate, or searchable name; I will often create a name for it. If the name is logical and valuable, searchable documentation is spread with that name; it may very well become the industry standard name for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here potential names for it. If you like one of them set it as the alt text for your image; and you may have just named it ;-)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;rod retention clasp&lt;/li&gt;&#xA;&lt;li&gt;adjustable rod retention clasp &lt;/li&gt;&#xA;&lt;li&gt;rod retention clip&lt;/li&gt;&#xA;&lt;li&gt;adjustable rod retention clip&lt;/li&gt;&#xA;&lt;li&gt;float retention clip&lt;/li&gt;&#xA;&lt;li&gt;adjustable float retention clip&lt;/li&gt;&#xA;&lt;li&gt;float retention clasp&lt;/li&gt;&#xA;&lt;li&gt;adjustable float retention clasp&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2017-02-27T13:36:38.400" LastActivityDate="2017-02-27T13:36:38.400" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="13969" PostTypeId="2" ParentId="13963" CreationDate="2017-02-27T07:47:55.093" Score="0" Body="&lt;p&gt;Although I'm sure there are exceptions, I think it is common for the two motors to share the load 50-50. Adding the gearbox adds cost and complexity to the system (versus just one motor), so you are only going to do it if you get a big benefit from the second motor. If it's a 90-10 split, you might as well get just rid of the second motor and save yourself the cost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, the split may deviate from 50-50 due to slight differences between the two motors. In this case a master-slave arrangement might be used to get an exactly even split. E.g. run one motor on speed control, measure the output torque, and the run the other motor on  torque control. For a dc motor, volarge is proportional to speed and current is proportional to torque. So you run one motor on a constant voltage and then match the other motor to the same current. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-02-27T07:47:55.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13970" PostTypeId="1" AcceptedAnswerId="13975" CreationDate="2017-02-27T08:46:03.323" Score="1" ViewCount="101" Body="&lt;p&gt;Apparently, during a world fair (presumably Paris in 1889&lt;sup&gt;[Don't trust that fact]&lt;/sup&gt;), an engineer showed a perpetual machine that could not be stopped. For months, he challenged visitors to stop it, but that machine always recovered the energy that people tried to take from it. Everyone was convinced it worked!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;trick&quot; was so simple: the action of trying to stop it would give the machine extra energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I could have heard that from a highly unreliable source, but whether the story is true or not doesn't change my question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would that machine look like and how exactly would it work?&lt;/p&gt;&#xA;" OwnerUserId="10237" LastActivityDate="2017-02-27T11:46:04.743" Title="How does this obviously fake but convincing perpetual machine work?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2017-02-27T14:13:02.477" ContentLicense="CC BY-SA 3.0" />
  <row Id="13971" PostTypeId="1" AcceptedAnswerId="14012" CreationDate="2017-02-27T09:42:47.523" Score="3" ViewCount="599" Body="&lt;p&gt;Note - This is a duplicate of the question I asked over on &lt;a href=&quot;https://physics.stackexchange.com/questions/314474/two-dimensional-plasticity-with-radial-return&quot;&gt;Physics&lt;/a&gt; which hasn't received any attention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the proper method to scale deviatoric stresses back to the yield surface in plane stress 2D?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot get the &lt;a href=&quot;http://www.dynasupport.com/tutorial/computational-plasticity/radial-return&quot; rel=&quot;nofollow noreferrer&quot;&gt;radial return&lt;/a&gt; method to produce the desired results under plain stress conditions in 2D. I'll show this using an example in both 3D and 2D in the hope that my mistake might become apparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Three Dimensions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume an arbitrary trial &lt;a href=&quot;https://en.wikipedia.org/wiki/Cauchy_stress_tensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cauchy Stress tensor&lt;/a&gt; of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_{tr} = \begin{bmatrix}&#xA;    90 &amp;amp; 20 &amp;amp; 0 \\&#xA;    20 &amp;amp; 90&amp;amp; 0 \\&#xA;    0 &amp;amp; 0&amp;amp; 0&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the &lt;a href=&quot;http://www.continuummechanics.org/hydrodeviatoricstress.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrostatic component&lt;/a&gt; is given by:&#xA;$\sigma_{tr,hyd} = \begin{bmatrix}&#xA;    50 &amp;amp; 0 &amp;amp; 0 \\&#xA;    0 &amp;amp; 50&amp;amp; 0 \\&#xA;    0 &amp;amp; 0&amp;amp; 50&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cauchy_stress_tensor#Stress_deviator_tensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;deviatoric component&lt;/a&gt; by:&#xA;$\sigma_{tr,dev} = \begin{bmatrix}&#xA;    40 &amp;amp; 20 &amp;amp; 0 \\&#xA;    20 &amp;amp; 10&amp;amp; 0 \\&#xA;    0 &amp;amp; 0&amp;amp; -50&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Von_Mises_yield_criterion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Von Mises criterion&lt;/a&gt; to determine whether the stresses are still in the yield surface. This calculated as (for general plane stress):  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_v = \sqrt{\sigma_{11}^2-\sigma_{11}\sigma_{22}+\sigma_{22}^2+3\sigma_{12}^2} = 86.6025$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of argument let's assume a yield stress of $\sigma_y=80$. In which case the deviatoric stresses need to be scaled back by a factor of $\alpha={\sigma_y}/{\sigma_v}=0.9238$. So the new deviatoric stress is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_{dev}=\alpha*\sigma_{tr,dev}&#xA;= \begin{bmatrix}&#xA;    36.95 &amp;amp; 18.48 &amp;amp; 0 \\&#xA;    18.48 &amp;amp; 9.24&amp;amp; 0 \\&#xA;    0 &amp;amp; 0&amp;amp; -46.19&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the scaled back &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma=\sigma_{tr,hyd}+\sigma_{dev}&#xA;= \begin{bmatrix}&#xA;    86.95&amp;amp; 18.48 &amp;amp; 0 \\&#xA;    18.48 &amp;amp; 59.24&amp;amp; 0 \\&#xA;    0 &amp;amp; 0&amp;amp; 3.81&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we calculate the corresponding Von Mises stress we get $\sigma_{v}=80$ indicating that the scaling was correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Two Dimensions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't repeated a lot of the text here for brevity. Check page 157 &lt;a href=&quot;https://www.wou.edu/las/physci/taylor/es406_structure/TM_chap8_stress.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for confirmation of hydrostatic stress in 2D, it's just $(\sigma_{11}+\sigma_{22})/2$. Assuming plane stress then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_{tr} = \begin{bmatrix}&#xA;    90 &amp;amp; 20 \\&#xA;    20 &amp;amp; 90 \\&#xA;\end{bmatrix}$&#xA;$\sigma_{tr,hyd} = \begin{bmatrix}&#xA;    75 &amp;amp; 0 \\&#xA;    0 &amp;amp; 75 \\&#xA;\end{bmatrix}$&#xA;$\sigma_{tr,dev} = \begin{bmatrix}&#xA;    15 &amp;amp; 20 \\&#xA;    20 &amp;amp; -15 \\&#xA;\end{bmatrix}$&#xA;$\sigma_v = 86.6025$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that scaling the deviatoric stresses ($\alpha$ is the same as before) results in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_{dev}=\alpha*\sigma_{tr,dev}&#xA;= \begin{bmatrix}&#xA;    13.86 &amp;amp; 18.48 \\&#xA;    18.48 &amp;amp; -13.86 \\&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma=\sigma_{tr,hyd}+\sigma_{dev}&#xA;= \begin{bmatrix}&#xA;    88.86&amp;amp; 18.48  \\&#xA;    18.48 &amp;amp; 61.14 \\&#xA;\end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the Von Mises stress gives a value of $\sigma_{v}=85$, which is outside the yield surface. What is the problem here? What is the proper method to scale the deviatoric stresses back to the yield surface in 2D?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&#xA;I am slowly coming around to the ideal that this radial return method is only appropriate for full 3D or plane strain (in which case the $\sigma_{33}\neq0$). Page 25 of &lt;a href=&quot;http://publications.jrc.ec.europa.eu/repository/bitstream/JRC102336/vpjcfin.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; seems to also indicate that this is the case, but I am still unsure. &lt;/p&gt;&#xA;" OwnerUserId="4002" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2017-03-08T10:50:26.543" Title="Two dimensional plasticity with radial return" Tags="|modeling|stresses|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13972" PostTypeId="1" AcceptedAnswerId="23725" CreationDate="2017-02-27T10:23:38.380" Score="7" ViewCount="1780" Body="&lt;p&gt;I'm being annoyed by a pulsating hum which comes from my neighbor's HVAC system. I read up a little on the design of HVAC systems and I came to a tentative conclusion that what I'm hearing is a compressor being driven by an AC induction motor. A frequency spectrum shows a spike at 120 Hz and another at 117 Hz; which of course produces a 3 Hz beat. The 117 Hz spike actually moves around between say 117 and 118 Hz. I guess what I'm hearing is called a &quot;slip frequency beat&quot;? I'm in the US, where the mains frequency is 60 Hz; so 120 Hz is the audible hum produced from magnetic fields in motors, transformers, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://www.bksv.com/doc/BO0269.pdf&quot; rel=&quot;noreferrer&quot;&gt;this document&lt;/a&gt; (p. 3), the &quot;slip frequency&quot;, defined as &quot;per-unit slip times line frequency&quot; is the &quot;actual frequency of current in the rotor conductors&quot;. Per-unit slip is proportional to torque and is commonly 0.03 - 0.05. This means 3 Hz could be a typical slip frequency, and I guess 117 Hz could be a slip-frequency sideband of 120 Hz. But there is no 123 Hz spike, and in fact the rest of the spectrum is pretty quiet except for spikes at 59 Hz and 29 Hz. There are harmonics at 234 Hz and 240 Hz, but the harmonic at 352 Hz is stronger than that at 234 Hz. All the harmonics have much less energy than the 117 Hz and 120 Hz fundamentals (say &gt;20 times less). The 29 Hz and 59 Hz spikes don't seem to vary much when the compressor turns on and off, and neither does the 120 Hz spike. Only the 117 Hz spike and its harmonics show an &quot;on-off&quot; square wave in a time-series plot, corresponding to compressor activation at 10-minute intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't provide decibel-calibrated sound pressure levels (and since I don't have access to the equipment, I can't measure the actual vibration intensity). I would say the sound is not loud enough to be annoying unless you have to live next to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated that the sound amplitude in the 117 Hz &quot;spike&quot; is 3 times &lt;em&gt;higher&lt;/em&gt; than that in the 120 Hz spike. I don't know if this reveals anything significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read that the speed of an AC motor is 120 Hz / (number of poles), and that the minimum number of poles is 2, so I'm concluding that the motor cannot be turning at just under 120 Hz (7200 RPM) but rather must be going at some slower speed like 60 Hz (3600 RPM). This seems to imply that the 117 Hz vibration is not mechanical vibration from the rotor - but I'm not sure what else it could be. The linked document mentions certain faults producing vibrations at 2 x RPM, but it seems these should be concurrent with vibrations at 1 x RPM. But, as I said, the 59 Hz spike doesn't vary with compressor activation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a time-series plot of various spectral amplitudes, taken over 6 hours at night. The spectral bands shown here are 58-59 Hz, 116-118 Hz, 174-177 Hz, etc. You can see that compressor activation doesn't seem to affect the 59 Hz spike, but there is a clear signal at 117 Hz and 3x117 = 351 Hz. (I also appended a spectrum plot)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tYhpV.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tYhpV.png&quot; alt=&quot;Harmonics of 59 Hz, 6-hour time-series, plus spectrum&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the linked document, some amount of vibration is normal and the surest way to diagnose a developing fault is to take spectral readings over time and see if anything is getting worse. The neighbor's HVAC system is 20 years old, but I only have readings for the past six months, and there seems to be a lot of fluctuation just from weather conditions and diurnal temperature variations influencing the propagation of sound across the property line. Because of this I can't say that there has been an obvious trend upwards in sound amplitude over this time period.&lt;/p&gt;&#xA;" OwnerUserId="10236" LastActivityDate="2018-09-14T18:12:53.690" Title="Is a 3 Hz beat frequency from my neighbor's HVAC a sign of a malfunction?" Tags="|vibration|hvac|compressors|ac|frequency-response|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="13973" PostTypeId="2" ParentId="13879" CreationDate="2017-02-27T11:03:41.353" Score="2" Body="&lt;p&gt;My open channel hydraulics are a little rusty having ended up in project management for a few years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming sub-critical flow, i.e. the Froude number is less than 1.  This is reasonable unless the inlet to your channel is horizontal and the water is entering at high velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The depth of flow is governed by a control structure at the down stream end.  Two conditions could exist - either the water free falls into the basin, or the discharge is flooded and the downstream level is the same as the basin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming free-fall, calculate the critical depth of the flow.  This is the same as a weir equation, setting the height of the weir above the invert of the channel as zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal equations for flow in an open channel fall apart when the depth is critical.  To overcome this, go say 3 depths back upstream and add 10% to the depth.  Now split the remaining length of channel into small segments, and working upstream calculate the frictional losses in the next segment, hence the change in total head, hence the change in depth.  Use the new depth for the start of your next segment.  Do this all the way upstream to the inlet.&lt;/p&gt;&#xA;" OwnerUserId="10240" LastActivityDate="2017-02-27T11:03:41.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13974" PostTypeId="2" ParentId="13963" CreationDate="2017-02-27T11:23:21.447" Score="0" Body="&lt;p&gt;Elastic couplings are going to eliminate/reduce shock loading in the gear train. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does it work? Either the second motor is for redundancy/backup and/or additional power/torque. It's the same concept as adding more cylinders to an internal combustion engine. They apply more power to the crank shaft. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-02-27T11:23:21.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13975" PostTypeId="2" ParentId="13970" CreationDate="2017-02-27T11:46:04.743" Score="2" Body="&lt;p&gt;There are many possible approaches - let's take a simple one; electrostatic. A static charge based &lt;a href=&quot;http://physics.kenyon.edu/EarlyApparatus/Static_Electricity/Electric_Whirl/Electric_Whirl.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;whirl&lt;/a&gt; connected to a capacitor for storing more charge, and a fluffy carpet around. Stopping the whirl with your fingers charges it (and the capacitor) with new static charge picked through your feet from the carpet.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-02-27T11:46:04.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13976" PostTypeId="2" ParentId="13961" CreationDate="2017-02-27T11:53:43.950" Score="2" Body="&lt;p&gt;First off, neural networks are only very rarely done in hardware, as actual summation of electric signal. Much more often they are just programs that run on standard computers, simulating a network with numerical inputs and outputs, and when you want superior speed and size, you implement it in hardware of FPGA, but still with numerical input and output. Analog neural networks, while possible and present in academic research, are not used in practice due to analog circuitry being considerably more expensive and bulkier than digital circuitry capable of simulating the same effects with adequate speed and precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the actual summation takes form of $ax+by = z$ where $x, y$ are inputs, $z$ is output, and $a,b$ are parameters which are adjusted through the process of teaching the neural network, and since they can take negative values, $z$ may just as well come out as $x-y$ for $a=1; b=-1$.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-02-27T11:53:43.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13977" PostTypeId="2" ParentId="13942" CreationDate="2017-02-27T13:00:08.813" Score="4" Body="&lt;p&gt;First, if you plan to use the same tank for flight and storage, that's not a good approach, because the tank to last without leaking, corroding, or being at risk of damage, needs to be quite thick. Use a dedicated oxygen storage tank (like a gas welding tank), and a small 'flight' tank for the rocket, filled right before launch. It doesn't need to be up to quite high standards, as both the amount and duration of exposure to oxygen won't cause a significant risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to adapt a tanks used for Airsoft gas pistols - there's a wide range of sizes, they are refillable, can hold a respectable pressure, and you can get all the &quot;plumbing&quot; as spare parts for ASG pistols.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although, if I were you, I'd go for nitrous oxide. The cartridges &lt;a href=&quot;https://en.wikipedia.org/wiki/Whipped-cream_charger&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;whippets&quot;&lt;/a&gt;, while not refillable, are cheap enough to treat them as disposables, safe, light, and you can get all the &quot;plumbing&quot; from a whipped cream dispenser, for which these cartridges are made. Never mind they hold 8 grams of nitrous oxide for 22g of dry mass. You'll never get this weight ratio out of $O_2$ unless you go with liquid oxygen.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-02-27T13:00:08.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13978" PostTypeId="1" AcceptedAnswerId="13981" CreationDate="2017-02-27T14:30:28.327" Score="0" ViewCount="2145" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yqMly.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yqMly.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1QX6o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1QX6o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lmZxe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lmZxe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the closed-loop transfer function is equal to &#xA;Y/Ysp = (P controller * DC motor) /(1+(P controller * DC motor)) then I am struck.  Can anyone give me some hints on how to proceed?&lt;/p&gt;&#xA;" OwnerUserId="5359" LastActivityDate="2017-02-28T03:11:42.643" Title="Find the differential equation to the closed-loop system" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13979" PostTypeId="2" ParentId="13942" CreationDate="2017-02-27T14:33:08.057" Score="0" Body="&lt;p&gt;Norco, Airgas, or your local oxygen supplier is only going to refill cylinder with a &lt;a href=&quot;http://www.c-f-c.com/gaslink/docs/dot_cylinder.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;DOT stamp and a current inspection date.&lt;/a&gt; Oxygen may also have NFPA requirements because it is an oxidizer. So for purchasing oxygen or storing it you need to use a purchased cylinder that is rated and consequently has a large safety factor. The oxygen will not be liquid in this container. To get the energy density, you will probably be looking at a cryogenic system. With the proper arrangements you may be able to purchase liquid oxygen in a dewar at atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For amateur rocketry, a 1.5 safety factor may be sufficient. That is with small volumes, safety glasses, and filled at the pad minutes before launch. Selecting a volume, pressure rating, pressure relief valve, and interface plumbing with your storage will take a lot of engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were not aware, liquid oxygen or any (&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Liquid_rocket_propellant?wprov=sfla1&quot; rel=&quot;nofollow noreferrer&quot;&gt;liquid propellant system&lt;/a&gt;) is expert level amateur rocketry; it will require strong engineering skills or access to those skills. Pre-manufactured solid fuel is the best to start with then work up to a non-cryogenic or cryogenic liquid fuel system&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2017-02-27T14:38:32.920" LastActivityDate="2017-02-27T14:38:32.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13980" PostTypeId="1" CreationDate="2017-02-27T14:51:02.127" Score="4" ViewCount="324" Body="&lt;p&gt;In &lt;em&gt;internal&lt;/em&gt; combustion heat engines, there is no heat transfer through walls required between  the source of heat and the expanding medium. With an &lt;em&gt;external&lt;/em&gt; combustion system, we heat the expanding medium by transferring heat through the walls of some type of enclosure such as a boiler or cylinder.  The boundary layer of a gas doesn't transfer heat very fast. But for the external combustion case, consider what happens if we build larger or smaller versions of the same engine design.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume a scale factor $\alpha$ times each length dimension. It would appear that the ratio of surface area divided by enclosed volume varies by $\frac{1}{\alpha}$.  Then the time required to heat a given volume of expanding medium is changed by a  factor alpha. Take $\alpha = 0.1$, then you have ten times the area heating a volume of medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can picture an empty saucepan on a hot stove. Dump in a quart of water. After awhile it will  heat up.  Compare to the case of dropping a single drop of water which sizzles into steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, bear with me a moment while I picture the time required to heat the gases in an IC engine  at time of combustion as being fast, something like the width dimension divided by the speed of sound in the compressed gas.  If I take alpha as smaller and smaller, the external combustion  case (boiler) gets faster and faster.  How small do we have to get before we are heating our volume of gas through the wall at speeds comparable to the IC case? Yes, I know about choking in compressible fluid flow, but small pipes may not  be needed in our external engine design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #1: If I need a very small engine, should I consider  external combustion?  Let's say I'm willing to trade  some power to mass ratio to get quiet operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #2: Should I consider ganging multiple scaled-down external combustion engines  to get the advantages like quiet operation, self-starting, etc. to get any required power?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a background comment, note the following regarding IC engine size scales:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In their book &quot;On Size and Life&quot;, 1983, Scientific American Books, Inc., pp. 60-65, Thomas A. McMahon and John Tyler Bonner compare commercial IC engines from the smallest to the largest.  They note that larger power users such as planes and ships use multiple cylinders per engine and multiple engines per vehicle, partly because &quot;the requirements of cooling apparatus are relatively more expensive, both in subtracted power and added mass, the larger the engine becomes&quot;.  I believe this is because of the difficulty of cooling the cylinders having smaller and smaller surface area available per unit volume as size increases.  They note that 757,778 Webra Speedys have the same mass as one huge marine engine, yet would have about 12 times the power if you could somehow link them together. It's 3.33 hp/kg for the model airplane engine, 0.27 hp/kg for the giant diesel.&lt;/p&gt;&#xA;" OwnerUserId="10241" LastEditorUserId="9" LastEditDate="2017-03-02T16:58:31.233" LastActivityDate="2017-03-02T16:58:31.233" Title="If I need a very small heat engine, should I consider external combustion?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="13981" PostTypeId="2" ParentId="13978" CreationDate="2017-02-27T15:46:21.147" Score="3" Body="&lt;p&gt;You don't need to use the transfer function for this problem, just substitute the known equations in the time domain to get a differential equation for the closed loop. You have&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ V(t) = K_p e(t) \qquad e = Y_{sp} - Y(t) \qquad \tau_L = 0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ T \frac{d^2Y(t)}{dt^2} + \frac{dY(t)}{dt} = K V(t) - K_B \tau_L(t)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Just insert the first three equations into the fourth and you have your closed loop differential equation.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T \frac{d^2Y(t)}{dt^2} + \frac{dY(t)}{dt} = K K_p (Y_{sp} - Y(t))$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Rearrange to get into the standard differential equation form with your states on the left-hand-side and your inputs on the right-hand-side:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T \frac{d^2Y(t)}{dt^2} + \frac{dY(t)}{dt} +  K K_p Y(t) = K K_p Y_{sp}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now you can analyze the control system as a simple second-order DE with a constant input.&lt;/p&gt;&#xA;&lt;h1&gt;Edit: Steady state value&lt;/h1&gt;&#xA;&lt;p&gt;The steady state value can easily be derived from the differential equation. We know that when the system reaches steady state, by definition, &lt;span class=&quot;math-container&quot;&gt;$\frac{d^2Y(t)}{dt^2} = 0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\frac{dY(t)}{dt} = 0$&lt;/span&gt;. Therefore, substituting those values in the DE and setting &lt;span class=&quot;math-container&quot;&gt;$Y(t)=Y_{ss}$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T \cdot 0 + 0 +  K K_p Y_{ss} = K K_p Y_{sp}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Y_{ss} = Y_{sp}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-02-28T03:11:42.643" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="13982" PostTypeId="2" ParentId="13980" CreationDate="2017-02-27T16:32:17.800" Score="3" Body="&lt;p&gt;Most small external combustion engines have abysmal efficiency compared to internal combustion ones - simply, most of heat is vented without ever getting transferred to the working gas/liquid. They are so inferior, you'll be considerably better off with an electric engine instead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You only use a tiny combustion engine if you really need considerable torque and excellent mass efficiency - and even that to a degree; model airplane engines are about as small as you can reasonably get (and they are awfully loud), and slowly losing to electric anyway. Anything smaller will lose to electric due to overhead of structural/mechanics/friction. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-02-27T16:32:17.800" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13983" PostTypeId="1" CreationDate="2017-02-27T17:12:17.397" Score="1" ViewCount="694" Body="&lt;p&gt;Imagine two objects are being dropped from an arbitrary height. One object is prismatic, and the other object is spherical. Both are isotropic and made of the same material (perhaps even the same volume of material). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force upon impact is a straightforward calculation for an arbitrary mass. This can be done by utilizing the energy imparted on the object through the complete transfer of potential to kinetic energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in knowing how geometry affects the resulting distribution of forces, and how one can take this knowledge to better design objects braced for impact. Is this something that can be easily calculated by hand, or does it require a more thorough simulation (via ANSYS, et al) in practice? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: My original post lacks assumptions &amp;amp; needs clarification. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I have no velocity or material in mind for the dropped objects. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, by force distribution, I mean stress distribution. I would like to calculate resultant peak stresses from impact, then compare these values to various material yield strengths.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The objects fall on a flat surface. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The object can rebound and experiences elastic shock waves. Further, consider that the flat impact surface has infinite stiffness. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &quot;back of the envelope&quot; estimate should suffice. This information is needed to better inform prototyping design. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10242" LastEditorUserId="10242" LastEditDate="2017-02-27T21:21:06.290" LastActivityDate="2017-05-29T21:46:01.840" Title="Impact energy absorption with objects of varying geometry" Tags="|structural-engineering|materials|structural-analysis|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="13985" PostTypeId="2" ParentId="13980" CreationDate="2017-02-27T19:56:51.010" Score="5" Body="&lt;p&gt;I have done a lot of work with small sterling heat engines. While decreasing the size generally does have a favorable effect on your heat transfer, your energy density doesn't increase substantially over the feasible range. Other issues come into play, like being able to fabricate an engine that small while keeping internal friction low. Also your cylinder head area to cylinder ring circumference ratio decreases; yielding a lower efficiency cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For increasing sterling engine power density, I usually added fins to the inside and outside of the heat transfer wall to increase the surface area without shrinking the parts to impractical sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A heat engine is not going to match the power density of an internal combustion engine (unless you switch to exotic materials and run very high temperatures perhaps). And it is certainly not the match the power density of an electric motor. A Stirling engine's niche is low-maintenance waste-heat to electrical conversion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For portable power tools, electric is the way to go. Energy density of lithium ion is quite high, and battery technology increases every day. Electric motors have highly controllable speeds and torques, especially with brushless dc motors (variable frequency). Electric systems are also much much lower maintenance.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-02-27T19:56:51.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13988" PostTypeId="2" ParentId="13960" CreationDate="2017-02-27T20:50:45.263" Score="2" Body="&lt;p&gt;So, I'm going to base my answer off a simulation I just finished in a flight dynamics class I'm taking. The book is called Small Unmanned Aircraft Theory and Practice by Beard and McLain. There are slides available at &lt;a href=&quot;http://uavbook.byu.edu/doku.php?id=lecture&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://uavbook.byu.edu/doku.php?id=lecture&lt;/a&gt; (check out chapter 6). While we are simulating fixed-wing UAVs, I think the concepts should be similar. For our pitch and roll loops (which are kind of the same thing in a quadrotor), we use a basic PD loop (no integral term because each is an inner loop). The equation for a commanded control input is $y=k_pE - k_d\dot{E}$. For convenience, we used $q$ (pitch rate) instead of $\dot{E}$ because we have measurements of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your PID loop, you add $k_d\dot{E}$ instead of subtracting it. By subtracting, it reduces the input as you increase angular velocity. For instance, if I am $\frac{\pi}{2}$ radians off and not rotating, my input will be large. As I increase in angular velocity, my input decreases from both the decrease in error and the increase in angular velocity. Eventually, there must be 0 input and it will happen at some point before I get to my position. The UAV keeps rotating and error decreases. However, $\dot{E}$ remains at its previous level and the input becomes negative, decreasing my angular velocity. If, somehow, the angular velocity, got to 0 before achieving the setpoint, this whole process would be back at the beginning. This means you should only have to tune your gains. &lt;/p&gt;&#xA;" OwnerUserId="9073" LastActivityDate="2017-02-27T20:50:45.263" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13989" PostTypeId="2" ParentId="13983" CreationDate="2017-02-27T21:07:49.887" Score="2" Body="&lt;p&gt;Making the material as general as possible makes the problem very hard. E.g. rubber falling on an infinitely stiff surface is completely different than steel. To be completely general with respect to both material and velocity makes this problem into a PhD dissertation. At the very least, you will need to restrict yourself to a certain class of materials to have a tractable problem. The easiest is to stick with linear elastic materials. E.g. this would include most metals if the velocity is not too high, but would exclude polymers (i.e. plastics) and very high velocities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: adding how I would approach the problem if it were limited to linear elastic materials :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer: use a transient dynamics simulation in ANSYS with the appropriate contact elements.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long answer (disclaimer, although this isn’t a PhD. Dissertation level problem, it still may require a lot of time to pick up all of the necessary knowledge)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the first thing you’ll need to think about is contact mechanics.  For the case of the sphere, the problem is that the area of contact will change over the time of the impact.  Initially, only a small area of the sphere is in contact.  Then, the sphere deforms, and a larger area is in contact.  As the sphere rebounds and moves back up again, the area of contact shrinks.  The video that ja72 referenced gives some idea of that concept, an here is one of a tennis ball that is a little lower speed (&lt;a href=&quot;https://www.youtube.com/watch?v=YTwDH-9rM7c&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=YTwDH-9rM7c&lt;/a&gt;) . If you stick with relatively stiff metals on a stiff surface, the applicable theory is called Hertz contact. For a sphere, it works out that the total force exerted on the sphere is proportional to the indentation to the 3/2 power.  The pressure will vary over the region of contact, with the highest pressure in the center, tapering off to zero at the edge.  Wikipedia has a pretty good page on contact mechanics &lt;a href=&quot;https://en.wikipedia.org/wiki/Contact_mechanics&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Contact_mechanics&lt;/a&gt;.  If you go with something like rubber, you can’t use Hertz contact, and probably need to use something more involved, like JKR contact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider a prismatic object, like a cube, you could get away without using contact mechanics if you are willing to assume that the object always lands perfectly flat.  I.e. either the entire surface area is in contact, or none of it is in contact.  The pressure is just the force divided by the area and is the same over the whole area.  In practice, you will probably never get your object to land perfectly flat, but you said you wanted back of the envelope, so let’s just go with this.  So ignore spheres for now and focus only on prismatic objects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve for the internal stresses due to the contact forces, we will want to use the theory of linear elasticity (&lt;a href=&quot;https://en.wikipedia.org/wiki/Linear_elasticity&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Linear_elasticity&lt;/a&gt;).  However, it’s pretty complicated.  If we are willing to make some assumptions on the geometry, we can simplify the problem greatly.  Start by considering a rod or bar.  i.e. the height of the object is much greater than the other two dimensions.  Think of the aspect ratio of a pencil.  With that assumption, we can simplify the three dimension theory of linear elasticity down to a single one dimension.  In other words, the stresses will vary greatly along the length of the pencil, but won’t vary too much from the inside to the outside, so we can just assume that the stress is essentially identical throughout the cross section. We can write down a partial differential equation describing the motion of the bar. See equation 5 in this course handout from Texas A&amp;amp;M &lt;a href=&quot;https://oaktrust.library.tamu.edu/bitstream/handle/1969.1/93279/HD14%20vibes%20continuous%20systems%202008.pdf?sequence=1&amp;amp;isAllowed=y&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://oaktrust.library.tamu.edu/bitstream/handle/1969.1/93279/HD14%20vibes%20continuous%20systems%202008.pdf?sequence=1&amp;amp;isAllowed=y&lt;/a&gt;  it’s a little too much for me to retype here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, that’s the vibration of the bar.  We will want to set that up with free-free boundary conditions, a gravity load across the entire bar, and finally a non-linear force at the end of the bar that represents the contact.  i.e. the force is zero if the end of the bar is not in contact with the surface, and some non-zero value if it is contact.  the easiest assumption is just a force that varies proportional with the indentation.  i.e. don’t assume a perfectly rigid surface, just give it some fairly high stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you’ll need to solve that partial differential equation as a function of both time and space.  Unfortunately, even this fairly simple 1-D equation is going to be pretty hard to solve analytically.  What we need to do is to reduce the partial differential equation into a series of ordinary differential equations.  If I were doing this, I’d use a Galerkin discretization.  However, that is probably way more math that you are really looking to do.  Most people would use the finite element method (i.e. ANSYS) to discretize the structure spatially. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of the Galerkin discretization is that it is analytical.  So variables like the geometry of the bar or the Young’s modulus of the material will appear in the final answer.  But the disadvantage is that it’s a lot of math.  The advantage of a commercial finite element pacakge is that the computer does all the work for you.  But, to study the effect of different variables you will have to run many different cases (i.e. a numerical experiement).  Another advantage of ANSYS is that extending from a 1-D rod to a 2-D or even 3-D structure is pretty straightforward.  Just add more elements. An analytical study will get much harder in 2-D.  Also Ansys can automate the contact mechanics, so you can study spheres or other contacts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only trick in ANSYS is that you will need to make sure you have a fairly small time step to capture the dynamics.  Try some initial guess, and then run it again with the time step cut in half.  If the answer changes, then your time step is too big.  Keep making it smaller until you get the same answer even when you cut the step in half.  Do the same thing for mesh density.  Try a mesh, then refine it.  If you get a different answer, you need more element.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The really serious impact guys, like those that simulate automotive crash tests, will probably use LS-DYNA instead of ANSYS.  But LS-DYNA is not for the faint of heart and probably too much of a learning curve for what you want to do. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="9275" LastEditDate="2017-02-28T19:57:31.483" LastActivityDate="2017-02-28T19:57:31.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13990" PostTypeId="1" CreationDate="2017-02-27T22:10:41.933" Score="1" ViewCount="525" Body="&lt;p&gt;$1m^3$ of gas undergoes an adiabatic compression from 1 bar to 10 bar. Determine the work required per kg, if its ratio of specific heats is 1.4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that 1 bar is equal $1$x$10^5$Pa. I also know that $PV^k=constant$. I also know that through the adiabatic compression process that work, $w=[m*R*(T2-T1)]/(1-k)$ .&#xA;I tried using this equation. However how can I calculate the work using this equation if no mass has been specified and no change in temperature?&lt;/p&gt;&#xA;" OwnerUserId="10247" LastEditorUserId="9" LastEditDate="2017-03-02T17:01:05.613" LastActivityDate="2017-03-02T17:01:05.613" Title="How to determine the work using volume and pressure in an adiabatic process?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13991" PostTypeId="2" ParentId="13990" CreationDate="2017-02-27T23:06:12.967" Score="1" Body="&lt;p&gt;Adiabatic expansion is the exchange of pressure for volume WITHOUT heat exchange. Therefore any temperature differential is zero. Furthermore there is no loss of material during the expansion/compression, so the mass of the fluid is irrelevant. The work done on compression and during expansion is ideally equal, and is simply the integral of the volume and pressure.  Pretty much any thermo textbook will cover the principles.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/PLAsI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PLAsI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-02-27T23:06:12.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13992" PostTypeId="1" AcceptedAnswerId="13993" CreationDate="2017-02-28T02:12:42.857" Score="4" ViewCount="1410" Body="&lt;p&gt;Internal Combustion Engine powered cars need multi-speed gear shifts since their power is highly dependent on speed and engines have narrow usable power/speed bands. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some electric vehicles are also coming out with gear shifts. We have Antonov already working on a 3-speed gearbox for electric motor drive, and the new Exagon Furtive eGT is said to be coming with a 4-speed gearbox. Why multiple speeds on an electric vehicle? The reasoning is the same as for an engine. It is to allow lots of torque in first gear to start but also to eliminate the need for the motor to run insanely high RPM's when you're going 80 mph on the freeway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metal band saws also come with some form of mechanical speed control (CVT or* gear shift) since you will want to cut softer stuff using high gear, but you will need low gear to cut iron, steel, etc. There's Clausing Industrial who have put two xor THREE speed controls: a belt-drive CVT, 4-speed gearbox, and possibly an infinitely-variable** inverter motor, in their V4014H vertical band saw! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what's odd is that many tower cranes also have gear shifts (on the hoist winch) even though they're electric! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the hoist winch on the Liebherr 357 HC-L 12/24 luffing boom tower crane appears to have a 3-speed gearbox immediately after the motor. Using a 215-horsepower (160 kW) motor and 1-fall operation, the maximum no-load lifting speeds are (capacities based on halving the capacities of 2-fall operation): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;330 m/min (3rd gear) maintainable up to 1,500(?) kgf &lt;/li&gt;&#xA;&lt;li&gt;185 m/min (2nd gear) &lt;/li&gt;&#xA;&lt;li&gt;113 m/min (1st gear) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And the maximum full-load line speed at 12,000 kgf (even on third gear, I presume) is 68 m/min. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller 147-horsepower (110 kW) hoist winch also, apparently, has 3 speeds (these are again on 1-fall operation): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;261 m/min (3rd gear) maintainable up to 1,100(?) kgf &lt;/li&gt;&#xA;&lt;li&gt;157 m/min (2nd gear) &lt;/li&gt;&#xA;&lt;li&gt;91 m/min (1st gear) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And the maximum full-load lifting speed (again, on any gear) is 45 m/min. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTES: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;No-Load Speed&quot; is at zero hook load (when the crane is lifting only the hook-block) &lt;/li&gt;&#xA;&lt;li&gt;kgf = Kilogram-force&lt;/li&gt;&#xA;&lt;li&gt;*Logical OR&lt;/li&gt;&#xA;&lt;li&gt;** &quot;Infinitely-variable&quot; = &quot;Continuously variable down to zero speed&quot; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My questions are: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Despite the fast lifting speeds often encountered and the high load capacities, shouldn't the working range of a tower crane be smaller than that of a car? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If electric motors do so well on cars without a gear shift, why would you need such a thing on a winch? In this case, a crane hoist) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user10249" LastEditorDisplayName="user10249" LastEditDate="2017-04-05T00:09:22.010" LastActivityDate="2021-03-16T18:36:42.327" Title="Why do tower cranes have gear shifts?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="13993" PostTypeId="2" ParentId="13992" CreationDate="2017-02-28T02:52:51.377" Score="11" Body="&lt;p&gt;You need very slow speeds to make precise movements of the load. At high speed, the cables will stretch as the load accelerates and decelerates and then oscillate longitudinally, &quot;bouncing&quot; the load up and down (and bouncing the whole crane boom as well). That is not a good idea if you want to raise or lower the load accurately through a small distance (say 10mm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, moving a large load a small precise distance &lt;em&gt;vertically&lt;/em&gt; is different problem from moving a car a small precise distance &lt;em&gt;horizontally&lt;/em&gt;, because the crane motor has to provide the force to counter-balance the full weight of the load all the time that the brakes are off, but the car engine does not. A lower gear ratio means the torque on the motor is less, and therefore the motor generates less heat while it is effectively stalled holding the load stationary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can easily move a car 10mm horizontally by pushing it with one hand, but you can't do the same thing to a 12-tonne weight hanging from a cable!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to avoid any misunderstanding, you don't &quot;change gear&quot; on the crane to accelerate up to the maximum speed &lt;em&gt;while the load is moving&lt;/em&gt;. You select the appropriate gear for the next step in the procedure while the load is stationary and the crane brakes are on.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-02-28T03:05:33.533" LastActivityDate="2017-02-28T03:05:33.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="13994" PostTypeId="1" CreationDate="2017-02-28T04:05:55.103" Score="1" ViewCount="101" Body="&lt;p&gt;So I was thinking that its not good to let a car engine sit. Since the oil and all moves down to the pan and all that other fun stuff, so what about small engines? I was thinking about whether it was good to start, for example, my 3 small engines. Being the push mower (4-cycle), riding lawnmower (4-cycle), and weed wacker/trimmer/eater (2-cycle). It seems to me to be a good idea since the oil and stuff must, like a car, sit someplace. And for 2 cycle all that lubrication would eventually dry up, and that would seize the engine, right? Now I am going to add that there is no fuel left in these engines. Well the rider does have a half tank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So simplifying, is it a good idea to start these smaller engines like once or twice a month? Just to keep everything lubricated. Also adding, if this is benifitial, which one (4-cycle/2-cycle) would get the most benefit? Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note: I hope this is fine for this site, but there is no place else to ask (that I know of).&lt;/p&gt;&#xA;" OwnerUserId="8540" LastActivityDate="2017-09-26T18:09:02.003" Title="Is it a good idea to start any engine once in a while?" Tags="|automotive-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="13995" PostTypeId="1" CreationDate="2017-02-28T06:43:19.630" Score="1" ViewCount="80" Body="&lt;p&gt;In my experience, many companies do not use the recommended cutting speeds for materials.  In general, it seems they opt for lower cutting speeds.  Is there a reason they typically do this?&lt;/p&gt;&#xA;" OwnerUserId="10251" LastEditorUserId="2538" LastEditDate="2017-03-01T11:29:55.587" LastActivityDate="2017-03-01T11:29:55.587" Title="What are reasons to use cutting speeds different from recommended?" Tags="|mechanical-engineering|materials|pumps|machining|" AnswerCount="1" CommentCount="1" ClosedDate="2017-03-02T23:45:18.167" ContentLicense="CC BY-SA 3.0" />
  <row Id="13999" PostTypeId="1" CreationDate="2017-02-28T08:54:15.880" Score="1" ViewCount="3114" Body="&lt;p&gt;Is it possible to сalculate discharge / flow rate on the pipe diameter and the pressure (for constant pressure)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V1tPx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V1tPx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Pressure_internal = 5 bar&lt;/code&gt;, &lt;code&gt;Diameter_pipe = 36 mm&lt;/code&gt;, &lt;code&gt;Density_water = 1000 kg/m^3&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bernoulli's Formula &lt;a href=&quot;https://www.wolframalpha.com/input/?i=sqrt%28%285-1%29%5Bbar%5D+*+2+%2F+%281000+%5Bkg%2Fm%5E3%5D%29%29+to+%5Bm%2Fs%5D&quot; rel=&quot;nofollow noreferrer&quot;&gt;(online calculation)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\rho v^2}{2} + \rho g h + p = const$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v \approx \sqrt{(p_i-p_a)\frac{2}{\rho_w}} = \sqrt{(5 \mathrm{bar}-1 \mathrm{bar})\frac{2}{1000 \frac{\mathrm{kg}}{\mathrm{m}^3} }}=28.284 \frac{\mathrm{m}}{\mathrm{s}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the discharge should be &lt;a href=&quot;https://www.wolframalpha.com/input/?i=pi%20*%20%2818%5Bmm%5D%20%29%5E2%20*%2028.284%5Bm%2Fs%5D%20to%20%5Bl%2Fmin%5D&quot; rel=&quot;nofollow noreferrer&quot;&gt;(online calculation)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Q = Sv = (\pi r^2) v = \pi * (18\mathrm{mm})^2 *28.284\frac{\mathrm{m}}{\mathrm{s}} = 1717.81 \frac{\mathrm{l}}{\mathrm{min}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these values are very far from reality. How can I perform a calculation correct?&lt;/p&gt;&#xA;" OwnerUserId="10253" LastEditorUserId="10253" LastEditDate="2020-04-10T10:55:03.967" LastActivityDate="2020-04-10T11:27:29.617" Title="Calculating discharge on the pipe diameter and the pressure?" Tags="|pressure|hydraulics|piping|pipelines|" AnswerCount="0" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="14001" PostTypeId="1" AcceptedAnswerId="14003" CreationDate="2017-02-28T13:12:05.403" Score="2" ViewCount="108" Body="&lt;p&gt;None of my fellow commuters know what this white panel with, what appears to be a red slider attached, is called and what it is for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular one is on a platform edge which happens to be over a bridge (Maidenhead in Berkshire).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know? The track is a straight section at this point. So I don't think it's any indication of cant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nGKzR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nGKzR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10259" LastEditorUserId="10259" LastEditDate="2017-02-28T13:38:56.850" LastActivityDate="2017-02-28T14:29:17.473" Title="What is this indicator called?" Tags="|rail|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14003" PostTypeId="2" ParentId="14001" CreationDate="2017-02-28T14:29:17.473" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://www.google.co.uk/search?q=indicator%20track%20side%20at%20station&amp;amp;sourceid=ie7&amp;amp;rls=com.microsoft:en-GB:IE-Address&amp;amp;ie=&amp;amp;oe=&amp;amp;gfe_rd=cr&amp;amp;ei=Zoi1WKKHF4Om8we5mK2gBA&amp;amp;gws_rd=ssl&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google&lt;/a&gt; has the answer (&lt;a href=&quot;https://groups.google.com/forum/#!topic/uk.railway/_z2B3F_8jyM&quot; rel=&quot;nofollow noreferrer&quot;&gt;second result&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;They mark the height of the nearest rail. &lt;/li&gt;&#xA;  &lt;li&gt;Height with respect to the little slider thing, so the track position is &#xA;  usually an amount below the indicator.   They are also found on some pieces &#xA;  of trackside equipment such as OHLE or signal gantries - but they aren't so &#xA;  readily seen, so mostly go without comment.   Anywhere where the track &#xA;  position cannot vary too much away from the 'as installed' position &#xA;  basically. &lt;/li&gt;&#xA;  &lt;li&gt;There are datum plates, when track is relaid or worked on it allows &#xA;  measurement to ensure the vehicle clearance is maintained &lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So basically it's there to mark what height the rail should be installed to, presumably in your case to ensure sufficient clearance to the bridge above.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-02-28T14:29:17.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14004" PostTypeId="2" ParentId="13994" CreationDate="2017-02-28T14:45:15.727" Score="0" Body="&lt;p&gt;Gasoline has a relatively short &quot;shelf life&quot; before sufficient volatiles have evaporated, leaving a highly flammable but not particularly aerosolizable remainder.  That's why owner's manuals (the &lt;strong&gt;first&lt;/strong&gt; thing you should have read before posting) recommend running the engine until the tank is dry before storing.&lt;br&gt;&#xA;Lubricating oils have a much lower evaporation rate, and can pretty much sit around forever without loss of functionality.  The reason to change the oil is to remove &quot;sludge,&quot; which is a mixture of foreign particles and partially oxidised oil.  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Stern  Reprimand&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Just because there's no automotive.stackexchange site hardly means there's &quot;No place to ask&quot; on the net.  There's approximately googol-7 websites dedicated to ICE design and maintenance.  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-02-28T14:45:15.727" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14005" PostTypeId="1" AcceptedAnswerId="14006" CreationDate="2017-02-28T15:44:59.690" Score="2" ViewCount="4732" Body="&lt;p&gt;I'm using airfoil data that is generated by a program. The generated airfoil has a chord length of 1 unit. I want to know how to scale up this airfoil to required chord length. My airfoil series number 638xx. NACA five digit series.&lt;/p&gt;&#xA;" OwnerUserId="9817" LastActivityDate="2017-02-28T20:39:28.557" Title="How to scale airfoil coordinates to achieve the required size?" Tags="|aerospace-engineering|aerodynamics|airfoils|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14006" PostTypeId="2" ParentId="14005" CreationDate="2017-02-28T16:00:32.637" Score="3" Body="&lt;p&gt;This is a simple math question. Multiply your x and y coordinates by the scaling factor you want. Say your chord length is 100mm. To make it 150mm, multiply all your values by 1.5. If you have a picture of an airfoil, just scale the image by 150%. &lt;/p&gt;&#xA;" OwnerUserId="8862" LastEditorUserId="8862" LastEditDate="2017-02-28T16:47:42.630" LastActivityDate="2017-02-28T16:47:42.630" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14007" PostTypeId="1" CreationDate="2017-02-28T17:31:40.570" Score="0" ViewCount="4565" Body="&lt;p&gt;I don't see what the difference is between turbo wastegate and variable geometry turbo. Both seem to allow some of the gas flow, without turning the turbine, ie. bypassing the turbine, have I understood correctly? The only difference I see, is that the wastegate allows the gas to flow before the turbine, while the variable geometry, after the turbine. Any idea? &lt;/p&gt;&#xA;" OwnerUserId="4846" LastActivityDate="2017-05-30T03:50:01.590" Title="What is the difference between turbo wastegate and variable geometry turbo?" Tags="|turbomachinery|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14008" PostTypeId="2" ParentId="13995" CreationDate="2017-02-28T18:27:21.160" Score="4" Body="&lt;p&gt;By necessity recommended speeds and feed rates are generalisations and, in practice the optimum for a given job can vary a lot. Recommended speeds are typically intended to be a reasonable compromise between material removal rate, quality of finish and tool life and it is entirely possible that in certain circumstances it is preferable to sacrifice one in favour of the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one machine has to perform a number of different operations it may not be practical to change the speed for each operation (especially with pulley driven pillar drills with manual ratio changes) and in this case a compromise towards the lower end of the speed range as too high a speed is likely to damage tooling whereas too low a speed will just reduce productivity and perhaps reduce finish quality a bit (although this isn't the case in all circumstances). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or it may be a strategy to prolong tool life and reduce down-time due to tool changes and refurbishment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most speed tables only cover a few generic materials but the machinability of materials can vary significantly even for ostensibly similar materials and even between batches of nominally identical alloys so a lower speed may be perceived as a safer default to achieve consistent quality. It is not necessarily even very easy predict or generalise what will work best for a given material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact that for any drilling operation with solid drills the speed will be a compromise between the relative surface speeds across the radius of the tool which obviously is exaggerated as the hole diameter gets bigger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important factor is that practical drill speeds depend greatly on the cooling, lubrication and chip flushing regime used, which can vary hugely according to context. For example a basic hand held pr pillar drill may have no cooling or lubrication but even something as simple as a manual oil dropper may significantly improve performance and at the most sophisticated end of the spectrum there is fully automated flood, mist or air jet cooling not to mention a variety of oils, cutting fluids and emulsions as well as tool coatings. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-03-01T02:22:41.060" LastActivityDate="2017-03-01T02:22:41.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14009" PostTypeId="2" ParentId="14005" CreationDate="2017-02-28T20:39:28.557" Score="2" Body="&lt;p&gt;MountainClimberi is correct. Multiplying both x and y coordinates by the same factor will scale your airfoil correctly. Using a different factor for each will result in a different aspect ratio and a different(stretched) version of that airfoil that will not match its published properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember that after scaling the data, your reference location will have moved if it was not (0,0). This may require a translation operation if you need to adjust it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per &lt;a href=&quot;https://en.wikipedia.org/wiki/Scaling_(geometry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scaling(geometry) wiki&lt;/a&gt;: &quot;The scaling is uniform if and only if the scaling factors are equal (vx = vy = vz)&quot;&#xA;&lt;br&gt;Willamate.edu has a good article on &lt;a href=&quot;https://www.willamette.edu/~gorr/classes/GeneralGraphics/Transforms/transforms2d.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;2D transformations&lt;/a&gt; including scaling.&#xA;&lt;br&gt;Python would be a good language to manipulate this data with. Here is an &lt;a href=&quot;https://stackoverflow.com/questions/2450035/scale-2d-coordinates-and-keep-their-relative-euclidean-distances-intact&quot;&gt;SE python scaling question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="-1" LastEditDate="2017-05-23T12:38:59.030" LastActivityDate="2017-02-28T20:39:28.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14010" PostTypeId="1" CreationDate="2017-02-28T22:03:14.323" Score="2" ViewCount="481" Body="&lt;p&gt;I'm working on a problem where I need to design and 3D print bevel gears to fit in the inside of a box and transmit a gear train around the inside corner, as roughly illustrated by the cross section image below. The box is composed of 3mm thick material and the distance from each gear axis hole to the edge of the box is 15mm. How can I design these bevel gears? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a fairly competent user of Solidworks and Autodesk Fusion 360 and I've attempted using the Toolbox design libraries and plugins respectively but both generate gears based on specific parameters (which do not include the outside diameter, or pitch diameter). I'm working under the assumption that I'll need to draw them based on the outside diameter so that they fit nicely into the corner of the box - is there another way of doing this, or is there a way of designing them to these constraints?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit - the constraints available in Solidworks to design the gear are the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Module&lt;/li&gt;&#xA;&lt;li&gt;No. of teeth&lt;/li&gt;&#xA;&lt;li&gt;Pinion's no. of teeth&lt;/li&gt;&#xA;&lt;li&gt;Pressure Angle &lt;/li&gt;&#xA;&lt;li&gt;Face Width&lt;/li&gt;&#xA;&lt;li&gt;Hub Diameter&lt;/li&gt;&#xA;&lt;li&gt;Mounting Distance&lt;/li&gt;&#xA;&lt;li&gt;Nominal Shaft Diameter&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It's also worth noting that some of these parameters are restricted to certain values by Solidworks i.e. I can't enter any random value, only pick from a drop-down list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yfGdH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yfGdH.jpg&quot; alt=&quot;Drawing of box/bevel gears&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10261" LastActivityDate="2017-02-28T22:03:14.323" Title="How to design CAD bevel gears from known outside diameter?" Tags="|gears|cad|3d-printing|solidworks|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14012" PostTypeId="2" ParentId="13971" CreationDate="2017-03-01T02:43:31.450" Score="3" Body="&lt;p&gt;Let's look at this problem mathematically a la Simo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let stress tensor ($\boldsymbol{\sigma}$) and its deviator ($\boldsymbol{s}$) in plane stress space be&#xA;$$&#xA;  \boldsymbol{\sigma} = \left[\sigma_{11}\, \sigma_{22}\, \sigma_{12}\right]^T &#xA;  \quad \text{and} \quad&#xA;  \boldsymbol{s} = \left[s_{11}\, s_{22}\, s_{12}\right]^T&#xA;$$&#xA;Note that $\sigma_{33} = \sigma_{13} = \sigma_{23} = s_{23} = s_{13} = 0$.  Even though $s_{33} \ne 0$, we can ignore $s_{33}$ because $s_{11} + s_{22} + s_{33} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $\bar{\mathsf{T}}$ be the matrix that takes $\boldsymbol{\sigma}$ to $\boldsymbol{s}$, i.e., $\boldsymbol{s} = \bar{\mathsf{T}}\,\boldsymbol{\sigma}$.&#xA;Also recall the convention for writing the plane stress strains ($\boldsymbol{\varepsilon}$) and their deviators ($\boldsymbol{e}$):&#xA;$$&#xA;  \boldsymbol{\varepsilon} = \left[\varepsilon_{11}\,\varepsilon_{22}\, 2\varepsilon_{12}\right]^T \quad \text{and} \quad&#xA;  \boldsymbol{e} = \left[e_{11}\, e_{22}\, 2\varepsilon_{12}\right]^T&#xA;$$&#xA;You can show, after some algebra and making sure that the convention of the strain tensors is respected, that if we use a slightly modified version of $\bar{\mathsf{T}}$ (let's call it $\mathsf{T}$):&#xA;$$&#xA;  \mathsf{T} = \frac{1}{3}\begin{bmatrix} 2 &amp;amp; -1 &amp;amp; 0 \\ -1 &amp;amp; 2 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 6 \end{bmatrix}&#xA;$$&#xA;we can get relatively simple expressions for the yield condition and flow rule for plane stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yield function for von Mises plasticity in 3D is (in tensor notation)&#xA;$$&#xA;  f = \sqrt{\frac{3\, \boldsymbol{s}:\boldsymbol{s}}{2}} - \sigma_Y&#xA;$$&#xA;In plane stress, and using the matrices we have defined earlier (and a bit of algebra), you can show that the yield function can be written as&#xA;$$&#xA;  f = \sqrt{\frac{3\, \boldsymbol{\sigma}^T\,\mathsf{T}\,\boldsymbol{\sigma}}{2}} - \sigma_Y&#xA;$$&#xA;You will get yielding at your trial stress values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have to compute the return using the plane stress flow rule:&#xA;$$&#xA;  \dot{\boldsymbol{\varepsilon}}^p = \dot{\lambda}\,\mathsf{T}\,\boldsymbol{\sigma}&#xA;\quad \text{where} \quad&#xA;\boldsymbol{\varepsilon}^p = \left[\varepsilon_{11}^p\,\varepsilon_{22}^p\, 2\varepsilon_{12}^p\right]^T &#xA;$$&#xA;These modifications are needed for you to reach the right point on the yield surface in your return process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Update&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the linear elastic relation is&#xA;$$&#xA;  \boldsymbol{\sigma} = \mathsf{C}\,\boldsymbol{\varepsilon}^e&#xA;$$&#xA;where $\mathsf{C}$ is the plane-stress $3\times 3$ stiffness matrix, we can use the flow rule to show that&#xA;$$&#xA;  \boldsymbol{\sigma_{n+1}} = \left[\mathsf{C}^{-1} + \Delta\lambda\mathsf{T}\right]^{-1} \mathsf{C}^{-1} \boldsymbol{\sigma_{n+1}}^{\text{trial}}&#xA;$$&#xA;We solve for $\Delta\lambda = \dot{\lambda} \Delta t$ by using the consistency condition $f(\sigma_{n+1}) = 0$.  The resulting equation is a quadratic in $\Delta\lambda$ and therefore a simple scaling relation cannot be used for the plane stress radial return.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="3488" LastEditDate="2017-03-02T05:26:03.040" LastActivityDate="2017-03-02T05:26:03.040" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14013" PostTypeId="2" ParentId="14007" CreationDate="2017-03-01T03:02:35.680" Score="1" Body="&lt;p&gt;The real difference is that a wastegate or dump valve  is a way of limiting the maximum boost pressure in order to prevent damage to the turbo itself of downstream parts on the other hand variable geometry is intended to improve performance across a range of engine speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One disadvantage of simple turbos is that they tend to have a fairly narrow effective RPM range which can adversely affect drivability and make it difficult to predict and control torque response. Specifically a wide aperture tends to make the turbo less effective at lower engine speeds and increases lag while a narrow aperture can choke the exhaust flow at higher speeds and limit peak power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A variable geometry outlet is intended to allow the turbine to be effective across a wider range of engine speeds.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-03-01T03:02:35.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14014" PostTypeId="1" CreationDate="2017-03-01T03:10:53.553" Score="8" ViewCount="1224" Body="&lt;p&gt;Besides satisfying nominal bar provision in the code, is there any special requirement in the provision of &lt;em&gt;transverse reinforcement&lt;/em&gt; to ensure that the longitudinal reinforcement is fully effective in a wide beam design? That is, can I use the full reinforcement area (all bottom bars) to calculate the bending moment capacity of a wide RC beam, provided that all detailing and curtailment requirement is satisfied? Do I have to calculate the transverse reinforcement necessary to confine the longitudinal bars and make them fully effective? Or do I have to consider a reduction using some spread angle from edge of support? For example, in the design of a pile cap with wide pile spacing, using Eurocode or British standard.&lt;/p&gt;&#xA;" OwnerUserId="10264" LastActivityDate="2023-12-31T02:03:05.443" Title="Design of transverse reinforcement in a wide beam" Tags="|beam|reinforced-concrete|building-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14015" PostTypeId="1" AcceptedAnswerId="14023" CreationDate="2017-03-01T08:53:36.003" Score="4" ViewCount="9542" Body="&lt;p&gt;Internet is full of twisted belt examples (30, 60, 90 degrees), however, the allowed misalignment tolerances claimed by the manufacturers are less than several degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The very term misalignment seems to impose the two pulleys are out of single plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, would it be correct stating that?:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;180 / 360 degree twist is ok&lt;/li&gt;&#xA;&lt;li&gt;any other twist is a misalignment and therefore is a bad practice&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Apart from timing belts, are there other belt types which would enable me driving i.e. 30kW load at 30 degree belt twist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/92mQk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/92mQk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10268" LastActivityDate="2020-06-13T01:32:07.040" Title="Is twisted belt configuration a good practice?" Tags="|torque|pulleys|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14016" PostTypeId="1" CreationDate="2017-03-01T09:06:42.793" Score="3" ViewCount="5453" Body="&lt;p&gt;I was given the compiled program from an automotive ECU by my colleagues in HEX-format, as well as the corresponding A2L-file. For my project I need to read the applied parameters from that HEX-file. &#xA;I want to do this automatically, via a MATLAB function, as the parameters are used as inputs in a Simulink model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as an example the HEX-file looks like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[…]&#xA;:2001C00000000000000000000000000000000000000000000000000000000000000000001F&#xA;:2001E0000000000000000000000000000000000000000000000000000000000000000000FF&#xA;:2002000054710780809A0780809A0780809A07808C530680000011A0140011A000000000D4&#xA;:200220000000000000000000000000000000000000000000000000000000000000000000BE&#xA;:2002400000000000000000000000000000000000000000000000000000000000000000009E&#xA;:2002600000000000000000000000000000000000000000000000000000000000000000007E&#xA;[…]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(please note that it is just an exemplary excerpt)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The address for a certain parameter can be read from the A2L for example as &lt;code&gt;‘0xA0143B0B’&lt;/code&gt;.&#xA;From the specification on the XCP-protocol I know that the address in the A2L is 40-bit long  (32-bit address and 8-bit extension). &#xA;As you can see the address I read is 64-bit. &#xA;Now if I got it right in my exemplary address &lt;code&gt;‘A01’&lt;/code&gt; is reserved, &lt;code&gt;‘4’&lt;/code&gt; is the address extension and &lt;code&gt;‘3B0B’&lt;/code&gt; is the actual 32-bit address.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone confirm this or correct me if I am mistaken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading the data from the HEX-file in position &lt;code&gt;‘3B0B’&lt;/code&gt; is incorrect. My colleagues exported the data for me in a text file to verify my results, and they clearly differ.&#xA;Are the ‘:’ counted when the position is addressed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I just found out that &lt;code&gt;'A01'&lt;/code&gt; defines the RAM-Segment the data is written in. But how can I find this Segment within the HEX-File?&lt;/p&gt;&#xA;" OwnerUserId="9808" LastEditorUserId="9808" LastEditDate="2017-03-01T09:28:57.080" LastActivityDate="2018-02-21T14:37:04.410" Title="Accessing parameters in HEX file (ECU program) using A2L file" Tags="|electrical-engineering|automotive-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14017" PostTypeId="1" CreationDate="2017-03-01T09:51:01.290" Score="0" ViewCount="1706" Body="&lt;p&gt;It seems that in low loads the engine efficiency is very low, while in heavier loads the engine efficiency or fuel efficiency/economy is much better. What are the reasons for this?&lt;/p&gt;&#xA;" OwnerUserId="4846" LastActivityDate="2022-03-16T16:59:53.060" Title="What are the reasons that engine efficiency drops in low loads?" Tags="|automotive-engineering|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14018" PostTypeId="2" ParentId="14015" CreationDate="2017-03-01T11:55:03.320" Score="2" Body="&lt;p&gt;Twisting the belts increases the &lt;a href=&quot;http://www.statics.marcks.cc/friction/pdf/belt_friction.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;angle of wrap&lt;/a&gt;, or &lt;a href=&quot;https://www.coursehero.com/file/13619565/Ch-5-Beltspdf/&quot; rel=&quot;nofollow noreferrer&quot;&gt;arc of contact&lt;/a&gt;, that the belt has with the pulleys - ie the belt has a greater contact area with each pulley.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, the greater the angle of wrap, the greater the amount of power that can be transferred. &lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-03-01T11:55:03.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14019" PostTypeId="1" AcceptedAnswerId="14032" CreationDate="2017-03-01T12:17:43.237" Score="1" ViewCount="232" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NpVA8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NpVA8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is my daughter's tricycle that I plan to make run off some spare Lithium-ion 18650s. I was just thinking of mounting a small DC motor to where I am pointing, and it will use a small tyre to make contact with the wheel, and rotate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few questions / concerns:&#xA;1) How do I calculate the Power (Watt) and/or Torque required from the motor, given the total weight of the tri-cycle, motor, and my daughter... in order to make it go at 5Kmph on a flat surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Both the rear wheels are free rotating independent of each other. So, if I rotate only one of them using the motor, will that push the cycle straight and forward, or in circles/curved path?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) If she pedals instead of using the electric mode, I am guessing it will charge the battery, although I doubt she'd pedal fast and enough to overcharge it. Would I need to do something to prevent this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Is there anything else that I haven't considered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the tricycle will not be used for more than an hour in a day, I can make a relatively small battery pack with less mAh capacity.&lt;/p&gt;&#xA;" OwnerUserId="9856" LastActivityDate="2017-03-01T20:01:21.617" Title="Evaluate my electric tricycle project" Tags="|torque|electric-vehicles|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14020" PostTypeId="2" ParentId="217" CreationDate="2017-03-01T12:53:58.183" Score="-1" Body="&lt;p&gt;If you are looking for house size scale, some panels would help: PV for compression &amp;amp; or termal to boost heat exchange for decompression,you could make also a puffer (water tank)with 2 coils, 1 at the bottom &amp;amp; 1 at the top,throw the bottom one will pass &amp;amp; cool the compressed air and on reverse on the top coil so you could recuperate some of the heat energy.&#xA;Depending on where you purchase the materials or where you gather (junk) plus your work you maybe come close to a sustenability .With all the Green advantages and long life span(20-30 years) i think that beat the batteries storage price at least(5-8 years)    &lt;/p&gt;&#xA;" OwnerUserId="10271" LastActivityDate="2017-03-01T12:53:58.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14021" PostTypeId="1" CreationDate="2017-03-01T13:04:33.970" Score="0" ViewCount="55" Body="&lt;p&gt;I am new to control modeling. Given the following problem: I want to control the heating of a house. Therefore, a top unit should be switched on (nominal load, no ramp up) as soon as the production of the main unit is above a certain level. How would you do this? Is it possible to set the output of a p-controller to a fixed value? &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if main_unit &amp;lt; level:&#xA;  top_unit = 0&#xA;else:&#xA;  top_unit = nominal_load&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your help&lt;/p&gt;&#xA;" OwnerUserId="5906" LastEditorUserId="8687" LastEditDate="2017-07-03T16:50:07.593" LastActivityDate="2017-07-03T16:50:07.593" Title="Using a controller for a simple heating system" Tags="|control-engineering|modeling|simulink|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14023" PostTypeId="2" ParentId="14015" CreationDate="2017-03-01T13:33:35.663" Score="4" Body="&lt;p&gt;The twist introduces a sideways torque, making the belt align non-uniformly, stretch and compress non-uniformly, and rub against any stops that prevent it from falling off (or be prone to fall off in absence of such stops). This increases load and wear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stick to manufacturer-specified misalignment, you can fully expect the belt's life time will be as in specifications at nominal load. If you apply the twist, the wear will progress much faster, and the load non-uniformity reduces the loads the belt will able to handle, as nominal load isn't distributed evenly throughout the whole width, but strains one edge above nominal, while leaving the other slack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And on top of that, all that extra compression/expansion, misalignment and friction cause a significant performance loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply put, your device will still work, but belt will need to be replaced more often, and you'll need to use a belt rated for higher load than typical for this application in straight configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That makes the twisted belt configuration non-optimal for main transmission; continuous operation. But if your loads and usage are occasional - e.g. driving a regulation mechanism that is pre-set for beginning of a production batch and set in place for the remainder of the batch, these disadvantages vanish completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course disadvantages of belt transmission in general, like large play, stretching under load etc remain, but the twist becomes non-issue.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2017-03-02T13:18:02.127" LastActivityDate="2017-03-02T13:18:02.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14024" PostTypeId="2" ParentId="14021" CreationDate="2017-03-01T13:42:57.823" Score="2" Body="&lt;p&gt;You shouldn't use a proportional control. The type of controller you describe (switching from 0 to full load and back to 0 almost instantaneously) is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bang%E2%80%93bang_control&quot; rel=&quot;nofollow noreferrer&quot;&gt;bang-bang controller&lt;/a&gt;. This controller is often used as a simplified model of a furnace heating system in a house, which can only be on or off. It is important to include a dead band (a range where the control stays on if it's currently on and off if it's currently off) into this type of controller to prevent high frequency switching near the setpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your question is tagged with Simulink, I would suggest looking into the 'Relay' block as a way of implementing this type of controller. &lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="6695" LastEditDate="2017-03-01T14:10:27.893" LastActivityDate="2017-03-01T14:10:27.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14025" PostTypeId="2" ParentId="14016" CreationDate="2017-03-01T14:01:09.030" Score="1" Body="&lt;p&gt;This is &lt;a href=&quot;https://en.wikipedia.org/wiki/Intel_HEX&quot; rel=&quot;nofollow noreferrer&quot;&gt;Intel hex&lt;/a&gt; format, not your raw hex dump - so no, position in the file won't be of much help. These are records, with length, address, record type, data and checksum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7BiZm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7BiZm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, as result of that overhead, the byte at position 3B0B in your hex file will be completely different. You need to decode the Intel Hex file into raw binary (apply offsets/addresses, strip headers and so on - the data in .hex is not guaranteed to represent continuous segment of memory) and only there your 3B0B will point at the right location. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-03-01T14:01:09.030" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14026" PostTypeId="2" ParentId="13994" CreationDate="2017-03-01T15:57:23.317" Score="2" Body="&lt;p&gt;Gas tanks are not completely sealed. As gas goes to your engine, air must enter the tank to equalize the pressure. This vent is typically located in the gas cap. As the this tank sits, it gets hot and cold with daily temperature swings. Every hot cold cycle, it pulls a little moisture in and condenses. This results in water accumulation in the tank that causes rust and can prevent startup. Modern 10% ethanol gasoline will absorb some amount of water and will go through your system as normal. This is why people recommend running engines every once in a while. Burning this gas is one method of removing that moisture. Realistically though, you are probably not going to go run a tank of gas through your weedeater/lawn mower once a month in the middle of winter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend filling the tank full of gas for storage and adding fuel stabilizer. I don't know all the chemistry of fuel stabilizers, but here are some links:&#xA;&lt;br&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Gasoline#Fuel_stabilizers_.28antioxidants_and_metal_deactivators.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiki Fuel Stabilizer Section&lt;/a&gt;&#xA;&lt;br&gt;&lt;a href=&quot;https://www.briggsandstratton.com/na/en_us/support/maintenance-how-to/browse/the-importance-of-fuel-stabilizer-by-briggsandstratton-stratton.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brigs &amp;amp; Stratton Fuel Stabilizer Article&lt;/a&gt;&#xA;&lt;br&gt;&lt;a href=&quot;http://fuelandfriction.com/weekend-warrior/why-use-fuel-stabilizer/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fuel and Friction Article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember, an empty tank can collect moisture in the same way. Removing all the gas may remove the stale gas related issues, but with a larger empty volume, it will actually make the water and rust problem worse.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-03-01T15:57:23.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14027" PostTypeId="2" ParentId="12277" CreationDate="2017-03-01T16:20:01.530" Score="0" Body="&lt;p&gt;In terms of a &quot;qualitative&quot; answer, I think the best way to think about a compressor is in kneading or shaping dough (like in baking).  Imagine you have long strip of bread dough on a working surface.  Using your hands as paddles, if you were to take turns pushing the dough, alternating with each hand, the dough would turn into a smaller and smaller strip with each pass.  This is similar to rotors and stators (&quot;paddles&quot;) compressing air in a compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a compressor with multiple stages, you have many stages of rotors/stators that gradually increase the static pressure, but the &lt;em&gt;way&lt;/em&gt; in which it increases the pressure is dynamic.  With a rotor, you are imparting momentum (and with it velocity) to the air.  Next, when the air encounters the stator,  the momentum and veclocity &quot;crash&quot; into the stator. Because the air is a compressible fluid, this action increases the density.  An increase in density is the definition of compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual mechanics are a little more elegant than &quot;crashing&quot; (think of paddling the dough),  but I think you get the picture.  Stators are designed to change the direction of the flow, and this causes a &quot;build up&quot; of air as the air switches direction.  In addition, the exit area (or &quot;throat&quot;) of the stator further restricts free flow and causes a pressure to build.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="10276" LastActivityDate="2017-03-01T16:20:01.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14029" PostTypeId="1" CreationDate="2017-03-01T16:27:46.457" Score="1" ViewCount="53" Body="&lt;p&gt;For a school scientific experiment to be flown onboard the ISS in a few months, I need to generate mechanical waves in a cylindrical capsule made of polycarbonate filled with oil (height about 9 cm, radius about 2 cm, wall thickness a few mm) and electrically heated on one side. Objective is to study heat conduction in oil in microgravity, with and without mechanical vibrations propagating in the oil. Frequency needs to be varied and in the range 3 kHz to, say, 100 kHz. Amplitude of vibrations should be high enough to, hopefully, generate standing waves in the oil. Thought of using a ceramic heater to heat the oil, and a piezo transducer to generate the vibrations but really don't know about possible issues with these ideas. Any advice highly welcome !! Thanks !&lt;/p&gt;&#xA;" OwnerUserId="10277" LastActivityDate="2017-03-01T16:27:46.457" Title="Need to generate mechanical vibrations for experiment in ISS" Tags="|mechanical-engineering|electrical-engineering|materials|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14030" PostTypeId="2" ParentId="14017" CreationDate="2017-03-01T16:52:36.843" Score="1" Body="&lt;p&gt;I think the simplest answer is, generally, it's all about the design of the engine.  An engine is generally designed with the load in mind first, and the torque and RPMs second, but the torque is the driving factor for efficiency.  Since:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P=\omega*\tau=rpm*torque$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the torque and RPMS are integral variables in achieving your desired goal for the engine performance. It is important to realize, however, that &lt;strong&gt;maximum efficiency&lt;/strong&gt; of the combustion cycle corresponds with the &lt;strong&gt;maximum torque&lt;/strong&gt;, whereas the &lt;strong&gt;maximum power&lt;/strong&gt; corresponds to the &lt;strong&gt;maximum product of torque and RPMs&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum torque (and hence maximum efficiency) is designed around a specific rpm where the engine is likely to operate.  Moving away from this point in either direction (i.e. faster or slower rpms) will allow you to change the power to meet requirements, but generally cause a lower overall efficiency.&lt;/p&gt;&#xA;" OwnerUserId="10276" LastActivityDate="2017-03-01T16:52:36.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14031" PostTypeId="2" ParentId="14017" CreationDate="2017-03-01T17:17:28.143" Score="1" Body="&lt;p&gt;Well, at zero load the efficiency is zero by definition. If the curve of efficiency versus load is continuous and smooth, it must turn toward the zero-zero point. I don't need to know anything about engines to say that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then again, power falls off below a peak located at high engine speed, because less fuel/air (energy source) is brought in and burned per second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that were the only factor, then power output would be proportional to angular speed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P∼ω$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and torque would remain constant down to idle. This is based on the simple relation between torque, power and angular speed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$τ=\frac{P}{ω}$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact passenger car engines have a fairly broad and flat torque curve. This can been seen in the curves at &lt;a href=&quot;http://en.wikipedia.org/wiki/Power_band&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Power_band&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However engine efficiency drops at low speeds since combustion chamber shape, bore/stroke ratio, manifold runner shape and length, valve lift and intake/exhaust valve overlap, to name just a few factors, are tuned for best performance at higher engine speeds. Thus torque eventually falls. In racing cars, the tuning is &quot;peakier,&quot; that is, they produce far more peak power but only over a narrow RPM range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you might expect, the torque curve isn't as flat in this case, and it falls off more rapidly. See Fig. 3 here &lt;a href=&quot;http://www.corvetteactioncenter.com/tech/hp_torque.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.corvetteactioncenter.com/tech/hp_torque.html&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vehicles that are optimized for very high torque at very low vehicle speed either have no high end to speak of (road graders, bulldozers) or, if they need both, use different systems (diesel-electric locomotives).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.physicsforums.com/threads/why-does-the-torque-curve-drop-off-at-low-rpm-in-a-typical-piston-engine.244434/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reference&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10028" LastActivityDate="2017-03-01T17:17:28.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14032" PostTypeId="2" ParentId="14019" CreationDate="2017-03-01T20:01:21.617" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;To calculate required torque, we are going to need a lot more information. Probably easiest to just pick a power like say 100 Watts. Then calculate the motor rpm you need based on the tire diameter you select.&lt;/li&gt;&#xA;&lt;li&gt;Powering one is actually better. If you powered both wheels you would need a differential. The steering of the front wheel should easily overpower the offset driving wheel.&lt;/li&gt;&#xA;&lt;li&gt;Don't mess around with charging the battery. Just install a double pull double throw (DPDT) switch (if using a dc motor) to give you forward and reverse options. Then install a on/off momentary push button.&lt;/li&gt;&#xA;&lt;li&gt;Design the motor interface such that fingers wont be pinched (remember kids are always trying to get injured). A tire type drive might not be your best option. Getting a gearbox that can direct drive the wheel (might have to modify its axle) might be a better option.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is going to be a pretty large project when you get into it, so feel free to post new questions specific to gearing, battery calculations, etc. Good Luck!&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-03-01T20:01:21.617" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14033" PostTypeId="2" ParentId="12277" CreationDate="2017-03-01T20:05:52.837" Score="0" Body="&lt;p&gt;A stator vane is nothing but a little wing, so it is probably best to first explain how a wing works. Please follow the &lt;a href=&quot;https://aviation.stackexchange.com/questions/16193/how-do-wings-generate-lift/16202#16202&quot;&gt;links&lt;/a&gt; over to &lt;a href=&quot;https://aviation.stackexchange.com/questions/8877/how-does-an-aircraft-form-wake-turbulence/8882#8882&quot;&gt;Aviation SE&lt;/a&gt; for a detailed explanation. Here is the gist of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'A wing creates lift by accelerating the flow of air downwards. This is called downwash and is also the effect which creates the slipstream behind a spinning propeller. The stator vane is not spinning itself, but is placed into a rotating flow, such that the flow around it is very similar to that of a wing or a propeller blade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotational flow component is transformed into an axial acceleration which in turn is converted into ram pressure by the narrowing flow path in a compressor. If you look at it at an energy level, kinetic (rotational) energy is converted into potential (pressure) energy.'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the stator's purpose is to redirect the flow into straight lines parallel to the center shaft. The stators are not responsible for pressure increases in that sense but still do have some responsibility. But the narrowing compressor is the main reason of the pressure increase. The stators just help to force the flow uniformly into the smaller space. Although, the stators still do have some responsibility in increasing pressure: since air is compressible, when the fluid's particles hit onto the stators, density increases and therefore compression does as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stator's purpose is to redirect flow to the next set of turbine blades by redirecting the downward flow from the turbine blades, back to the top of the next turbine. This is what as seen as an S-flow.&lt;/p&gt;&#xA;" OwnerUserId="10028" LastEditorUserId="-1" LastEditDate="2017-04-13T13:00:01.440" LastActivityDate="2017-03-01T20:05:52.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14034" PostTypeId="1" CreationDate="2017-03-01T20:28:20.323" Score="2" ViewCount="89" Body="&lt;p&gt;Is there some regulatory body/inspection that verifies the AISC specified Moment(s) of Inertia for steel I-Beams?  So for example, if a mill produces a W8x21 and says hey, it meets the specs, it's 8 in Deep and is 21 lbs/ft, and because I say so, it has Moment Of 75 in^4.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody actually go in a caliper the flange width/thickness web thickness/depth and fillet profile to get some idea whether this manufacturer is within a percent +/- of the spec? Like particularly if their fillet building process ends up putting an extra half pound per foot of material?&lt;/p&gt;&#xA;" OwnerUserId="5967" LastActivityDate="2017-03-01T20:28:20.323" Title="Moment Of Inertia Specifications AISC" Tags="|beam|specifications|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14035" PostTypeId="1" AcceptedAnswerId="14043" CreationDate="2017-03-01T20:30:22.070" Score="0" ViewCount="52" Body="&lt;p&gt;The sentence is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a closed loop system when the loop gain grows typically we have an instability&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10280" LastActivityDate="2017-03-02T12:24:36.243" Title="Can someone explain my this sentence about feedback systems?" Tags="|control-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14037" PostTypeId="1" AcceptedAnswerId="14057" CreationDate="2017-03-01T23:06:29.800" Score="-1" ViewCount="112" Body="&lt;p&gt;The high-pressure oxygen in the cabin is supplied through a system of regulators and valves from a high-pressure oxygen tank. There are two pairs of the sub-systems of check valves, shut-off valves and nonreturn automatic shut-off valves in the system. The function of these valves is to stop the reverse flow of air from the cabin to the gas tank in the case of pressure drop and to close the line supply if there is some sudden pressure drop in header line or cabin in order to avoid the waste of the gas. Each pair of the valve systems consists of two alternative branches. One consists of a non-return automatic emergency shut-off valve, and the other consists of a check valve and a shut-off valve in series. Any branch of the two pairs is capable of supplying sufficient gas to the cabin. There are three alternative paths between the oxygen tank and the pair of valves. Oxygen can be transmitted to the cabin through either of the two regulators and the pair of valves connected to the regulator. It can also be transmitted to the cabin through a selector valve and either of the two pairs of valves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DHUx4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DHUx4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9331" LastEditorUserId="9" LastEditDate="2017-03-02T17:12:17.957" LastActivityDate="2017-03-02T23:16:38.840" Title="Where is the source of the high pressure oxygen supply system in this spacecraft life support system?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2017-03-05T14:08:18.287" ContentLicense="CC BY-SA 3.0" />
  <row Id="14038" PostTypeId="1" CreationDate="2017-03-02T03:28:05.647" Score="0" ViewCount="135" Body="&lt;p&gt;In my fluid mechanics class, I've learned that a fluid traveling in a pipe will generate a force when the exit area is smaller then the entrance area. Suppose a pipe is attached to a car that will use the kinetic energy of the air passing over the vehicle. Assuming that the car is moving at a constant velocity and the pipe is straight and the exit area is 1/4 the size as the frontal area, is it feasible for a moving vehicle to generate enough force from the air to reduce energy use?&lt;/p&gt;&#xA;" OwnerUserId="10282" LastActivityDate="2017-04-05T23:53:11.370" Title="Can you use the kinetic energy of air to power a car?" Tags="|fluid-mechanics|automotive-engineering|airflow|energy-efficiency|renewable-energy|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14041" PostTypeId="1" AcceptedAnswerId="14042" CreationDate="2017-03-02T10:22:36.677" Score="0" ViewCount="36" Body="&lt;p&gt;Does work behave the same way for all temperature range while in a refrigeration setup. When quality of heat changes with the temperature, Does it apply to work also. IF we consider refrigerator Source-sink of 1000-500k and 300-250k, Does 1KJ of work input have the same effect or the effect is different. My doubt may be a very silly, i might be missing something. Any input is appreciated. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="9110" LastActivityDate="2017-03-02T10:45:20.657" Title="How does Work behave in two different refrigeration setup?" Tags="|thermodynamics|refrigeration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14042" PostTypeId="2" ParentId="14041" CreationDate="2017-03-02T10:45:20.657" Score="1" Body="&lt;p&gt;In theoretical, ideal setup, work will be approximately the same for the same temperature deltas between original temperature of the body, target temperature, and heat sink temperature. Cooling something from 200K to 100K given 300K heatsink should take the same work as cooling it from 1000K to 900K given 1100K heatsink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, obtaining efficient heat sinks for low temperatures is difficult. You can just leave something sit in open air for a couple minutes to bring it from 1000K to 900K while you need to jump through hoops and spend a lot of energy to get something from 200 to 100K - simply because your heatsink will be at ~300K in both cases. If you can provide a heat sink of temperature that is the same offset from your target temperature, you have comparable work - but usually, you can't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also operational temperatures of the coolant. Most heat is stored/extracted at phase change, gas&amp;lt;-&gt;liquid. This needs to correspond to boiling point of the coolant, so you'll use e.g. sodium to cool a nuclear reactor, but helium to cool superconducting magnets. Since they have different enthalpy of evaporation, they will provide different performance - not due to difference in temperature, but due to being different substances, that happen to boil at different temperatures needed for their respective processes - affecting performance of the refrigerators.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-03-02T10:45:20.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14043" PostTypeId="2" ParentId="14035" CreationDate="2017-03-02T12:24:36.243" Score="1" Body="&lt;p&gt;I can explain the very basic sentence with a very basic idea what happens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your gain grows, that means you're amplifying the signal &lt;strong&gt;more&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By loop gain I assume they mean the total gain from the loop.  If that is growing constantly, and the loop is closed, every time you go around the loop your signal strength should increase.  A constantly increasing signal isn't stable behaviour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Positive_feedback&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia's article on positive feedback&lt;/a&gt; has a lot of information and more links to more information (where I'm sure you could find actual academic sources).&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-03-02T12:24:36.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14045" PostTypeId="1" CreationDate="2017-03-02T14:09:02.313" Score="1" ViewCount="2465" Body="&lt;p&gt;I am trying to do some calculations to figure out how much power output would be produced from a water powered generator that uses horizontal current flow to move the generators coils rather than vertical water drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula to calculate power is: &#xA;$$\text{Power} = \rho gh\,Q\eta  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = \text{density (kg/m3) (~ 1000 kg/m3 for water)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q = \text{water flow (m3/s)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g = \text{acceleration of gravity}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h = \text{falling height, head (m)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta = \text{efficiency}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am unsure how to calculate when gravity and falling head are out of the equation, as setting them to 1 or 0 will give incorrect answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the water flow is $3.1m^3/second$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the efficiency is $0.78$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could explain how to calculate the power output I would be extremely grateful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a link to a similar device to what I am describing: &lt;a href=&quot;https://www.youtube.com/watch?v=hVREafgPcOs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Waterwheel Documentary- Harnessing Electricity on the Zambezi River&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6104" LastEditorUserId="8055" LastEditDate="2017-03-03T18:01:51.633" LastActivityDate="2022-04-19T17:58:17.373" Title="How to calculate hydropower in kW without water head" Tags="|power|power-engineering|water-resources|generator|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14046" PostTypeId="2" ParentId="14045" CreationDate="2017-03-02T14:37:28.970" Score="-5" Body="&lt;p&gt;Just measure what comes out of the generator and take an average ,peak and low.&lt;/p&gt;&#xA;" OwnerUserId="10289" LastActivityDate="2017-03-02T14:37:28.970" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14047" PostTypeId="1" CreationDate="2017-03-02T15:03:57.490" Score="1" ViewCount="170" Body="&lt;p&gt;I need a 1 to 1 planetary gear box that torque increases&#xA;I was told torque in planetary gear sets are because of the number of teeth moving over the sun gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was told the more spur gears with teeth means the more torque regardless of the increase .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="5770" LastActivityDate="2017-03-02T15:03:57.490" Title="1 to 1 planetary gear" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="5" ClosedDate="2017-03-03T02:47:39.307" ContentLicense="CC BY-SA 3.0" />
  <row Id="14048" PostTypeId="2" ParentId="14045" CreationDate="2017-03-02T15:11:43.667" Score="4" Body="&lt;p&gt;If you have horizontal water flow (no head) you are extracting its kinetic energy (from its velocity) not its potential energy (from its head). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to know the velocity of that flow as well as the flow rate to calculate the energy available. (You can calculate it from the flow rate and channel dimensions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot extract all the available energy - that would imply the water stopping downstream of the turbine! For wind power (compressible fluid), the maximum energy you can extract is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Betz&amp;#39;s_law&quot; rel=&quot;noreferrer&quot;&gt;Betz limit&lt;/a&gt; or 59% of the original kinetic energy. Something similar will apply to incompressible fluids - apparently Betz itself, as it applies to Newtonian fluids, which do include incompressible fluids - and it implies slower flow in a larger (wider,deeper) channel after the turbine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice if you extract half the Betz limit (between 25 and 30% of the available energy) you're doing pretty well, but given you have a figure for efficiency, you are probably meant to calculate power as efficiency * Betz * available kinetic energy.&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="435" LastEditDate="2017-03-02T15:22:20.513" LastActivityDate="2017-03-02T15:22:20.513" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14049" PostTypeId="1" CreationDate="2017-03-02T15:33:42.357" Score="0" ViewCount="55" Body="&lt;p&gt;Using a hydro generator which gives an output of 4.838KW, I am trying to figure out how many batteries I can charge and how long it would take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am doing the calculations based on 12V, 17Ah batteries. this is for a theoretical large scale operation so i am therefore looking to charge between 25 and 100 batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an extra generator is necessary this can be factored in, however the core of this question is to figure out the ration between power output it watts and charging of multiple batteries in V &amp;amp; Ah.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone could help it would be very much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="6104" LastEditorUserId="8055" LastEditDate="2017-03-03T16:31:33.287" LastActivityDate="2017-03-03T16:31:33.287" Title="How many batteries can I charge and how long would it take?" Tags="|power|power-engineering|battery|renewable-energy|water-resources|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14051" PostTypeId="2" ParentId="14049" CreationDate="2017-03-02T17:11:49.480" Score="1" Body="&lt;p&gt;P = VI&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a 4838 joule/second input, and a system that holds 12 * 17 Ah = 12*17*3600 =734400 joules of energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, if you need 7.344 K joules of energy, and your input is 4.838 J/s that's 734400/4838 = 151.798263 seconds to get that energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the real challenge is that you probably can't charge a battery that fast- I'm a mechanical engineer, so I can't give you specifics, but you need a charge controller that limits the amount of current going in to your battery, otherwise they could overheat and start a fire&lt;/p&gt;&#xA;" OwnerUserId="6488" LastActivityDate="2017-03-02T17:11:49.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14053" PostTypeId="1" CreationDate="2017-03-02T19:29:09.297" Score="1" ViewCount="2144" Body="&lt;p&gt;This is a question one of my students brought up in class and I thought it was worth posting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When considering the stability of a linear time-invariant system it is often useful to look at the pole positions on the pole-zero map. Usually one of the first things that is taught in an undergraduate systems engineering course is that poles located on the LHS s-plane are stable, while poles located on the RHS s-plane are unstable. Furthermore, poles located on the imaginary axis are considered &quot;critically stable&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of such critically stable systems would be:&#xA;$$G_1 = \frac{Y_1}{U_1} = \frac{1}{s^2+\omega_n^2} $$&#xA;$$G_2 = \frac{Y_2}{U_2} = \frac{1}{s}$$&#xA;The exact value of $\omega_n$ is not important for this question. The poles of $G_1$ and $G_2$ are plotted on the pole-zero map below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yVPwV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yVPwV.png&quot; alt=&quot;Pole zero map showing some critically stable poles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets take a look at the behaviour of these critically stable poles. If subjected to a finite step input, $G_1$ will produce a bounded oscillation with constant amplitude:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vxdLZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vxdLZ.png&quot; alt=&quot;Step response of a critically stable system, bounded oscillation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If subjected to a finite step input, $G_2$ will produce an unbounded response (linear response approaching infinity as time approaches infinity):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HM8WA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HM8WA.png&quot; alt=&quot;Step response of a critically stable system, unbounded response&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that only unstable systems had unbounded finite step responses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the finite step response of a LTI system with a pole at the origin is unbounded, why do we say that the system is critically stable?&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2017-03-02T19:29:09.297" Title="The finite step response of a system pole at the origin is unbounded, how is this critically stable?" Tags="|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14054" PostTypeId="2" ParentId="14053" CreationDate="2017-03-02T19:29:09.297" Score="2" Body="&lt;p&gt;The main issue here is a misunderstanding of how stability is defined. Stability is not defined by the step response, but is instead defined by the system response to initial conditions (or an impulse response). From Katsuhiko Ogata &lt;em&gt;Modern Control Engineering, 5th Ed.&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A linear time-invariant control system is stable if the output eventually comes back to its equilibrium state when the system is subjected to an initial condition. A linear time-invariant control system is critically stable if oscillations of the output continue forever. It is unstable if the output diverges without bound from its equilibrium state &lt;strong&gt;when the system is subjected to an initial condition&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[Emphasis mine].&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;From a purely physical perspective&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Another way of phrasing this is: in the presence of zero input and non-zero initial conditions, does the energy of the system increase, decrease or remain constant over time? If the energy decreases over time, the system is stable. If the energy increases over time, the system is unstable. If the energy of the system remains constant over time, the system is critically stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us analyze the previous systems from a purely physical perspective and not worry too much about the systems engineering side of things. An example of a critically stable system that is equivalent to $G_1$ would be a spring-mass system with no damping:&#xA;$$m\ddot{y}_1(t) + ky_1(t) = u_1(t) $$&#xA;Where $y_1$ is the position of the spring, $u_1$ is an external force applied to the spring, $k$ is the spring constant, and $m$ is the mass. If you pull a spring away from its resting position or if it has some initial velocity (or both), and there is no friction or any kind of energy dissipation, and there is no external force performing work on the system ($u_1=0$) then it is clear that the system will oscillate freely forever. As it oscillates, it transfers energy between kinetic and potential forms (energy stored in the mass vs. energy stored in the spring). No energy is being created or dissipated, therefore critically stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets look at the $G_2$ case. This system could be represented as a mass sliding on a frictionless surface:&#xA;$$m\dot{y}_2(t)=u_2(t)$$&#xA;Where $y_2$ is the velocity of the mass, $m$ is the mass, and $u_2$ is an external force applied to the mass. If you apply some initial velocity to the mass, and there is no friction or any kind of energy dissipation, and there is no external force performing work on the system, then it will remain at that velocity forever. The energy in the system is all kinetic, and is stored in the mass. No energy is being created or dissipated, therefore critically stable.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;From a systems engineering perspective&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You can represent non-zero initial conditions using the impulse response of the system. So for the purposes of systems engineering, if the impulse response of the system is bounded, then the system is at least critically stable. If the impulse response of the system approaches zero, then the system is stable. If the impulse response of the system is unbounded, then it is unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example showing how the impulse response is equivalent to non-zero IC's:&#xA;$$m\dot{y}(t) = u(t)$$&#xA;Taking the Laplace transform:&#xA;$$m(sY(s) - y(0)) = U(s)$$&#xA;$$Y(s) = \frac{U(s)/m + y(0)}{s}$$&#xA;For zero input and non-zero initial conditions:&#xA;$$Y(s) = \frac{y(0)}{s}$$&#xA;If $y(0) = 1$ then:&#xA;$$Y(s) = \frac{1}{s}$$&#xA;Similarly, for zero initial conditions and a impulse scaled to $m$ (i.e. $u(t) = m\delta(t)$, where $\delta(t)$ is the dirac delta function):&#xA;$$Y(s) = \frac{1}{s}$$&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2017-03-02T19:29:09.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14055" PostTypeId="5" CreationDate="2017-03-02T20:11:34.820" Score="0" Body="" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2017-03-02T20:11:34.820" LastActivityDate="2017-03-02T20:11:34.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14056" PostTypeId="4" CreationDate="2017-03-02T20:11:34.820" Score="0" Body="Quality Function Deployment method of developing requirements based on client needs." OwnerUserId="368" LastEditorUserId="368" LastEditDate="2017-03-02T20:11:34.820" LastActivityDate="2017-03-02T20:11:34.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14057" PostTypeId="2" ParentId="14037" CreationDate="2017-03-02T23:16:38.840" Score="-1" Body="&lt;p&gt;From Oxygen tanks or H2O electrolysis (back to fuel cell as H2 and O2)&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2017-03-02T23:16:38.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14058" PostTypeId="1" AcceptedAnswerId="14059" CreationDate="2017-03-03T07:46:01.813" Score="0" ViewCount="108" Body="&lt;p&gt;What will be the best thermal insulation material for insulation of pipe lines having temperature of 500℃. Mineral wool and Calcium Silicate can be used but here we have a constraint of both weight and space.&lt;/p&gt;&#xA;" OwnerUserId="10297" LastActivityDate="2017-03-03T10:59:41.300" Title="Best insulation material for pipes @ 500℃" Tags="|thermal-insulation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14059" PostTypeId="2" ParentId="14058" CreationDate="2017-03-03T10:59:41.300" Score="0" Body="&lt;p&gt;There is a well known but counter-intuitive result that for a given temperature, below a certain diameter, any insulation at all is worse than no insulation, because it conducts heat (poorly but better than air) to a larger external radiating surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details should be in any good book on heat flow, but for common steam pipe temperatures, the critical diameter is of the order of 5mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, below a certain pipe diameter, leave the pipe exposed to keep the radiating area small. We don't know from your question if your pipe dia can be small enough to allow this.&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2017-03-03T10:59:41.300" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14060" PostTypeId="1" CreationDate="2017-03-03T13:53:18.870" Score="1" ViewCount="230" Body="&lt;p&gt;Beer, Johnston, and DeWolf's &lt;em&gt;Mechanics of Materials&lt;/em&gt; includes the following image of the distribution of shear stresses in a bending beam:&lt;a href=&quot;https://i.stack.imgur.com/jNVFF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jNVFF.jpg&quot; alt=&quot;Beer Jonston and Dewolf 6.12&quot;&gt;&lt;/a&gt; (Axes and load vector carried forward from Fig 6.1 in the text by me for clarity in this question) This image represents the bottom part of a beam below a slice in an offset xy plane so the analysis of the shear stresses is being conducted in the top shown surface. The beam is being bent by the load V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text mentions that the shear stress along $D_1'-D_2'$ are a parabolic function of x. However, they never mention why. So my question is, why? I don't understand where this would pop out of the math or why this should be the case intuitively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this example. I have a book in front of me so that I'm looking at the cover. I fix the bound edge and push the opposite edge with a force perpendicular to the cover. The book will bend in a uniform profile at any slice parallel to the ground. This would seem to be to demonstrate that the stresses are independent of the height of the slice. Why is this wrong?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2017-03-03T13:53:18.870" Title="Distribution of shear stress in beam bending" Tags="|beam|stresses|shear|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14062" PostTypeId="1" AcceptedAnswerId="14079" CreationDate="2017-03-03T19:12:57.023" Score="2" ViewCount="1017" Body="&lt;p&gt;I am on the task of analysing the load capacity of a warren-ish truss bridge similar to &lt;a href=&quot;https://images.duckduckgo.com/iu/?u=https%3A%2F%2Fs-media-cache-ak0.pinimg.com%2F736x%2F7f%2Ff1%2Fa1%2F7ff1a1719eadfdc3567338d9b659284b.jpg&amp;amp;f=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;. I do not, however, have the software or skill to model and analyse a 3D structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering wether or not i could analyse a &lt;a href=&quot;https://images.duckduckgo.com/iu/?u=http%3A%2F%2Fwww.mathalino.com%2Fsites%2Fdefault%2Ffiles%2Freviewer-mechanics%2F418-warren-truss.gif&amp;amp;f=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;2D&lt;/a&gt; simplification and still come to a somewhat credible conclusion. What considerations should be made when approximating a 3D truss as a 2D truss? What would be the limitations of such an approach? Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;I do not have to know the exact maximum load. My only goal is to figure out where the truss will fail first.&lt;/p&gt;&#xA;" OwnerUserId="10302" LastEditorUserId="10302" LastEditDate="2017-03-04T12:29:20.083" LastActivityDate="2017-03-05T15:57:29.730" Title="Drawing conclusions about 3d truss from 2d model" Tags="|structural-engineering|statics|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14063" PostTypeId="1" CreationDate="2017-03-03T19:26:41.710" Score="0" ViewCount="909" Body="&lt;p&gt;I want to model an ideal truss in ANSYS 16.1 and by ideal I mean all of the joints are pinned joints and so the momentum can not be transferred at joints(like attached image). The problem is that I don't know how to model the pin joints for the truss structure. I would be appreciated if some body can help me by providing solutions or ideas.&#xA;Cheers&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2fPMn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2fPMn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10304" LastEditorUserId="10304" LastEditDate="2017-03-03T23:22:14.947" LastActivityDate="2019-08-29T14:55:46.347" Title="modelling truss in ANSYS" Tags="|ansys|ansys-workbench|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14066" PostTypeId="2" ParentId="14038" CreationDate="2017-03-04T06:27:28.160" Score="2" Body="&lt;p&gt;No, you can not create energy or a force from nothing. It would break the laws of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically addressing your question in a simplified way; the large area on the front multiplied velocity equals the small area on the back multiplied by the higher velocity. In practice, the losses are quite high and it would waste lots of energy.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-03-04T06:27:28.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14067" PostTypeId="1" CreationDate="2017-03-03T16:46:06.430" Score="3" ViewCount="251" Body="&lt;p&gt;Hi I am trying to make a CAD model of something like the scoop sampler shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.rolls-royce.com/~/media/Files/R/Rolls-Royce/documents/customers/nuclear/scoop-sampling-tcm92-50918.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.rolls-royce.com/~/media/Files/R/Rolls-Royce/documents/customers/nuclear/scoop-sampling-tcm92-50918.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a university coursework that requires design of bespoke machines. I am trying to design a machine that can literally scoop out a 1 mm thick sample from any aerospace component. My intention is to use a motor to rotate the shaft which is connected to the cutter (with abrasive particles to cut through metal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/diIAa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/diIAa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;See this would initiate the cutting and only cut at the controlled depth (based on the dimension of the cutter) but how can I possibly make the cutter to eventually go up? (lets say after 35 - 40mm length I want the cutter to make a vertical movement so that it scoops out). To clarify I want the final metal scoop to be roughly 40mm long, 20mm wide and max thickness of 1mm. So once it reaches around 35mm how can I make it go up vertically while the shaft and cutter is rotating in the horizontal axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally what if I don't have a flat surface. Imagine if I have a concave surface. I have some ideas but I can't seem to get through these initial problems.&lt;/p&gt;&#xA;" OwnerDisplayName="Arun John" LastActivityDate="2024-01-12T20:01:05.143" Title="Rotational Motion Kinematics - Not sure how to fix axis of rotation" Tags="|kinematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14068" PostTypeId="2" ParentId="14014" CreationDate="2017-03-04T10:47:54.040" Score="0" Body="&lt;p&gt;I can't answer with reference to Eurocode or British Standards, but there is some guidance in official comments to Spanish norm EHE-08. This answer follows  &lt;a href=&quot;http://web.archive.org/web/20120709002749/http://www.fomento.gob.es/NR/rdonlyres/029BEBA6-A895-40E4-BA9F-FD0D75E3B865/107241/5EHE2008ultimo.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;comments to article 42.3.1 in page 194&lt;/a&gt;, where the following image is taken from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V0RmF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V0RmF.png&quot; alt=&quot;Comments to article 42.3.1 EHE-08&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rules are for compressed bars and aimed to prevent buckling and therefore they should only apply to the compressed face wide beams. However, bar distribution in wide beams is usually quite symmetrical and then all longitudinal bars get the same confinement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, each compressed longitudinal bar needs to be confined by transversal reinforcement, but if distance between longitudinal bars is less than 15 cm only one on two bars needs to be confined.&lt;/p&gt;&#xA;" OwnerUserId="7371" LastEditorUserId="7371" LastEditDate="2017-03-05T16:42:14.523" LastActivityDate="2017-03-05T16:42:14.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14069" PostTypeId="2" ParentId="12638" CreationDate="2017-03-04T11:16:09.813" Score="0" Body="&lt;p&gt;The question of how to verify that there is a linear relationship between two variables in your data is more an statistical question than an engineering one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pearson's correlation and coefficient of determination measure if your points are aligned. However, that's just a part of the answer because but they doesn't take in account the number of points. For example, if you only had two data points, Pearson's correlation would be 1 because there is one straight line that fits perfectly two points, whatever relation were between the involved variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more complete answer would perform an hypothesis test in addition to correlation and coefficient of determination. The hypothesis test would compare the slope of the fitted line with the residuals to tell if that slope is significant. Unless you are interested in doing the maths by hand, I suggest using any statistical package to perform regression analysis - for example, R and RCommander are free, but even Excel has capabilities for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more advanced test would be to adjust a non linear model - for example a second degree polynomial - and test if the second degree component is significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, your lines are so straight that you can be sure that the relation between your data is linear. At least, you can be sure that if there was a non linear component it would be so small that it would be unlikely to be found using just four data points.&lt;/p&gt;&#xA;" OwnerUserId="7371" LastActivityDate="2017-03-04T11:16:09.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14070" PostTypeId="1" AcceptedAnswerId="14074" CreationDate="2017-03-04T13:16:34.493" Score="0" ViewCount="154" Body="&lt;p&gt;I have a choice of two hydraulic power units with specs as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A. 2.2kW motor, 8lpm flow rate, 120 Bar pressure rating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B. 1.1kW motor, 2lpm flow rate, 200 Bar pressure rating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using F=PA I assume pump B is capable of producing more force in the same system (albeit a lot slower!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, hearing the phrase &quot;resistance to flow produces the pressure not the pump&quot; many times gives me a nagging feeling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct or is there something I'm missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="10086" LastActivityDate="2017-03-04T17:12:34.340" Title="Hydraulic pump choice - which produces most force" Tags="|fluid-mechanics|hydraulics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14073" PostTypeId="1" CreationDate="2017-03-04T17:04:35.933" Score="1" ViewCount="356" Body="&lt;p&gt;I am currently calculating various quantities for the circuit shown below. However, I am stuck on how to calculate the current (Ix) and the input resistance (0 output current). The circuit is an ideal op amp circuit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There should be a virtual short between the two opamp inputs so that in the active region, the are at nearly identical voltages. I am still unsure how to calculate the voltage at the non-inverting input.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fe3e0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fe3e0.jpg&quot; alt=&quot;the circuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10311" LastEditorUserId="10311" LastEditDate="2017-03-04T18:34:31.220" LastActivityDate="2021-07-26T19:42:16.673" Title="Ideal op amp circuit with variable voltage gain, how do I calculate the current?" Tags="|electrical-engineering|circuits|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14074" PostTypeId="2" ParentId="14070" CreationDate="2017-03-04T17:12:34.340" Score="0" Body="&lt;p&gt;For a given cylinder, B has higher pressure and will produce a larger force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could select a larger diameter cylinder for A such that it would produce the same or larger force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a cylinder is moving with zero resistance, the pressure in the cylinder will be very low (just enough to overcome its own friction).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are likely looking for maximum force (pump max pressure x cylinder head area) which occurs when the cylinder blocked or fully extended (zero flow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At flowrates and pressures in between full flow zero pressure and full pressure zero flow will require a pump curve, plumbing information, and more involved calculations.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-03-04T17:12:34.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14076" PostTypeId="2" ParentId="13833" CreationDate="2017-03-04T22:29:11.697" Score="0" Body="&lt;p&gt;Based on the numerous helpful comments, I've settled on the following design.&lt;/p&gt;&#xA;&lt;p&gt;After researching structural integrity of steel alloys, my understanding is that fatigue failure due to cyclic stresses can be eliminated by ensuring that the endurance limit of the alloy is higher than the cyclic forces the dryer will impose. I've also read that some steel alloys will not fatigue regardless of the number of cycles.&lt;/p&gt;&#xA;&lt;p&gt;Here are my assumptions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Point load = 35kN (estimated)&lt;/li&gt;&#xA;&lt;li&gt;(x, y) offset to the right of the table top&lt;/li&gt;&#xA;&lt;li&gt;Dimensions = 64cm x 106.5cm x 1cm&lt;/li&gt;&#xA;&lt;li&gt;Young's modulus = 200GPa&lt;/li&gt;&#xA;&lt;li&gt;Poisson's ratio = 0.287&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The (x, y) stresses are therefore (193MPa, 113MPa).&lt;/p&gt;&#xA;&lt;p&gt;Structural steel (ASTM-A36) has a yield strength of 250MPa and tensile strength of 400MPa. AISI 8620 is 385MPa/530MPa, respectively. At 2cm thickness, the point load would have to exceed 140kN for the same stresses as 1cm, thus a ½&amp;quot; thickness for the table top should not have its endurance limit exceeded by the dryer's cyclic stresses.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.homedepot.ca/en/home/p.laundry-anti-vibration-pads.1000672766.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laundry pads&lt;/a&gt; prevent walking and reduce vibration. The end result will resemble (pads not rendered):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/weEZw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/weEZw.jpg&quot; alt=&quot;Table w/Dryer&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Instead of an X brace, I've opted for two horizontal braces.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dQOQr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dQOQr.jpg&quot; alt=&quot;Table without Dryer&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Without the top, it'll look like:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uQuuU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uQuuU.jpg&quot; alt=&quot;Table without Top&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The metalworker decided simple was best and basically welded a table together inside the closet:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZM8Bq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZM8Bq.jpg&quot; alt=&quot;Finished table&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10117" LastEditorUserId="10117" LastEditDate="2021-02-17T16:57:51.890" LastActivityDate="2021-02-17T16:57:51.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14077" PostTypeId="2" ParentId="12501" CreationDate="2017-03-05T08:45:00.843" Score="0" Body="&lt;p&gt;Kinematic chain is an assembly of link in which the relative motion of the links is possible and the motion of each relative to the other is definite. A linkage is obtained if one of the link of kinematic chain is fixed to the ground. If motion of any of the movable links results in definite motions of the others,the linkage is known as mechanism.&lt;/p&gt;&#xA;" OwnerUserId="10318" LastActivityDate="2017-03-05T08:45:00.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14079" PostTypeId="2" ParentId="14062" CreationDate="2017-03-05T11:26:50.480" Score="2" Body="&lt;p&gt;The answer to your question &quot;whether or not I could analyse a 2D simplification and still come to a somewhat credible conclusion&quot; is obviously &lt;strong&gt;yes&lt;/strong&gt;. After all, how do you think engineers calculated complex structures before the advent of the computer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real question, of course, is how to do this &lt;strong&gt;well&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first question is how to determine the loading applied to the truss. Each of the trusses must obviously support its own self-weight. The division of the remaining loads among the trusses, however, is a bit more circumstantial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reasonable first hypothesis would be to get the total load and divide it evenly between both trusses. If the structure is transversally symmetrical (that is, there is nothing that'd draw more loads to one truss than the other), then that is a perfectly valid assumption for all the dead loads (paving, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the live loads, however, a bit more care may be required if discrete moving loads are considered. If a car is driving along your bridge, will it be perfectly centered (or close enough)? Then sure, divide the live load evenly between both trusses. Otherwise, if (for example) the bridge is wide enough that the car may be closer to one truss than the other, then more of the live load will go to the near truss. Assuming a transversally symmetric bridge, if the live load will be at a distance $a$ from one truss and $b$ from the other, and $a&amp;gt;b$, then the closer truss will have to withstand a fraction $\dfrac{a}{a+b}$ of the total live load. Obviously, if you're just considering uniformly distributed live loads, then the worst case scenario is to load the entire bridge and divide it evenly between both sides (assuming a transversally symmetric bridge).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know the loads, you can then move on to checking the trusses themselves. Trusses are composed of members that are either in tension or compression. For those in tension, the 3D reality of the structure is irrelevant. For those in compression, it is very relevant. That is because compression can lead to buckling, and so we need to know how the members are braced. If we were dealing with a real 2D truss, the upper chord (under compression) would have no out-of-plane bracing, meaning its effective length would be equal to almost the entire span of the bridge, dramatically reducing its buckling strength. However, we aren't dealing with a 2D truss, but a 3D bridge. And the 3D reality of the truss is that the transversal bracing is restraining the out-of-plane displacement of all the nodes. This has absolutely no effect on the structural analysis calculation of the 2D truss (calculating the internal forces of each member), but it is essential when calculating the truss' design strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other issues such as wind and other non-vertical loads, but I have a feeling that these are out-of-scope for your question, so I'll leave them out of this answer. Let me know in a comment if I'm mistaken.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-03-05T11:26:50.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14080" PostTypeId="1" CreationDate="2017-03-05T12:22:53.577" Score="1" ViewCount="6067" Body="&lt;p&gt;Can anyone please explain to me the difference in the levels of static water and water table? &lt;/p&gt;&#xA;" OwnerUserId="10319" LastActivityDate="2018-05-31T08:05:31.990" Title="Difference between static water level and water table level" Tags="|civil-engineering|water-resources|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14082" PostTypeId="2" ParentId="14062" CreationDate="2017-03-05T15:57:29.730" Score="0" Body="&lt;p&gt;It depends on the loads applied to it. If there are only vertical loads (or the lateral loads are insignificant in comparison) then of you can assume each side to behave in similar manner to a 2D truss. Of course you have to consider that the vertical loads in such a bridge reach to the side trusses through bending of the transverse members, so you must consider the possibility of failure of these members too. Furthermore even if you analyse each truss as a 2D structure you should propably consider the presence of the the transverse members when you calculate the resistances of the truss, for lateral buckling failure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other circumstances, such as when severe lateral loads exist, or the dynamic response is wanted (which is typical true for bridges) then 3D modeling is required. &lt;/p&gt;&#xA;" OwnerUserId="9620" LastActivityDate="2017-03-05T15:57:29.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14083" PostTypeId="2" ParentId="12638" CreationDate="2017-03-05T16:26:54.193" Score="0" Body="&lt;p&gt;Taken into account all the uncertainties of experimental testing of actual specimens, these curves look prefectly linear to me. Discussing the results, you can report the excellent Person coefficients $\rho$ or the coefficients of determination $\rho^2$ to defend your experimental procedure, that sould be detailed too in your report (including curvature calculation as alephzero mentioned). Both coefficents should prove the strong correlation of the data, so you point is perfectly made. As far as i am concerned that would be enough to convince me. &lt;/p&gt;&#xA;" OwnerUserId="9620" LastActivityDate="2017-03-05T16:26:54.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14085" PostTypeId="1" CreationDate="2017-03-05T18:44:51.703" Score="1" ViewCount="319" Body="&lt;p&gt;Let's say that a fluid is undergoing a one dimensional motion with the following velocity field: $\mathbf{U} = V(x,t)\mathbf{i}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem asks to find $V(x,t)$ given that the position of a particle initially at coordinates $(x_0,t_0)$ is $x=(x_0/t_0)t^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first wrote the langrangian derivative of $\mathbf{U}$ and equated it with the acceleration of a particle $2(x_0/t_0)$. This led me to a transport equation and ultimately to a wrong result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I've been thinking about it for hours and I'm really getting nowhere. Any help is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10325" LastEditorUserId="1832" LastEditDate="2017-03-06T03:59:38.847" LastActivityDate="2017-04-05T20:43:45.577" Title="How to go from a Lagrangian to an Eulerian description of the velocity of a particle" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14087" PostTypeId="1" AcceptedAnswerId="16229" CreationDate="2017-03-05T23:33:33.563" Score="1" ViewCount="6491" Body="&lt;p&gt;I have been looking at the location of shear flow centers for profiles for my mechanics of materials course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to figure out how to determine the direction of the shear flow in a given beam section with a shear force which is either directed up or down (aligned with y-axis). In my lectures it has been said that 'the shear flow in the web of the beam section must always follow the same direction as the shear force'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found some examples where it appears this rule does not work though:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=zs_dEtwgxSM&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=zs_dEtwgxSM&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/6601/interpretation-of-shear-flow-shear-center-of-l-profile&quot;&gt;Interpretation of shear flow/shear center of L-profile&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In these 2 examples the shear flow direction in the web of the beam section is drawn in opposite direction of the shear force. I do not know why the aforementioned rule does not apply to these examples. Is this rule (shear flow direction in web must follow shear force direction) correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I determine the direction of shear flow for these 2 examples/in general?&lt;/p&gt;&#xA;" OwnerUserId="10328" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-07-10T05:47:00.070" Title="Determine shear flow direction in profile/beam section" Tags="|beam|shear|cross-section|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14088" PostTypeId="1" CreationDate="2017-03-06T01:50:06.940" Score="2" ViewCount="248" Body="&lt;p&gt;For my electronic project, I have been searching the internet for a non invasive flow meter that does not require disassembly. I have found an ultrasonic measurement technique that use the clamp on method (just fitting on to the pipe) for flow measurement of liquid though a pipe. But when I search through internet I found that ultrasound can behave like light as it is reflected back from a solid surface. Is this true? Or it is like ordinary sound that can pass through a medium?&lt;/p&gt;&#xA;" OwnerUserId="8950" LastEditorUserId="3633" LastEditDate="2017-03-06T11:44:06.223" LastActivityDate="2017-03-07T17:32:51.187" Title="Can ultrasonic sound pass through solid medium?" Tags="|control-engineering|flow-control|sound-isolation|ultrasound|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14089" PostTypeId="2" ParentId="14088" CreationDate="2017-03-06T02:25:00.607" Score="5" Body="&lt;p&gt;Ultrasound partially reflects when it encounters a medium with different acoustic impedance.  It behaves like light and obeys &lt;a href=&quot;https://en.wikipedia.org/wiki/Fresnel_equations&quot; rel=&quot;noreferrer&quot;&gt;Fresnel equations&lt;/a&gt;.  If the angle of incidence is smaller than the total internal reflection angle, then part of the ultrasound beam will go forward (with some refraction) and part will reflect back from the interface between the media.  The greater the difference between acoustic impedance of the two, the more the ultrasound will bounce back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ki7sW.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="90" LastEditDate="2017-03-06T06:50:16.657" LastActivityDate="2017-03-06T06:50:16.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14090" PostTypeId="2" ParentId="14067" CreationDate="2017-03-06T02:56:21.317" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;I would move the cutter closer to the motor. &lt;/li&gt;&#xA;&lt;li&gt;Design the shaft so that it passes through the motor and through the cutter so that you have some length of shaft extending out both ends.  Then, you can attach some type of arm that can lift the cutter assembly.  Maybe some type of lever, or something more complicated like a power screw that can be controlled to move the cutter + motor vertically.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10329" LastActivityDate="2017-03-06T02:56:21.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14091" PostTypeId="1" CreationDate="2017-03-06T04:22:46.253" Score="0" ViewCount="2916" Body="&lt;p&gt;I have many M2 flush-mount, counter-sunk hex screws ordered from a variety of sources, including McMaster Carr, Ebay, etc. The Wiha 263 1.5x50 screwdriver I have does not drive any of these properly, which means I have to step down to a 1.3mm screwdriver which is too loose and causes stripping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me identify the proper screwdriver to drive M2 Hex screws? Numerous google searches don't seem to shed any light on this, although pages like &lt;a href=&quot;https://www.fastenal.com/content/product_specifications/M.SHCS.A2-70.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; seem to suggest a 1.5mm hex driver SHOULD fit. &lt;/p&gt;&#xA;" OwnerUserId="10330" LastActivityDate="2020-07-25T07:05:25.607" Title="What is the proper screwdriver for a M2 counter-sunk hex screw?" Tags="|mechanical-engineering|metals|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14093" PostTypeId="2" ParentId="13807" CreationDate="2017-03-06T13:19:29.763" Score="-1" Body="&lt;p&gt;I would suggest trying a diamond-coated end mill. We use those to mill graphite and carbon-fiber products at my job.&lt;/p&gt;&#xA;" OwnerUserId="10336" LastEditorUserId="368" LastEditDate="2017-03-06T17:13:14.523" LastActivityDate="2017-03-06T17:13:14.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14095" PostTypeId="2" ParentId="14091" CreationDate="2017-03-06T16:13:51.323" Score="1" Body="&lt;p&gt;A couple pages , &lt;a href=&quot;http://www.csgnetwork.com/screwsochdcaptable.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;csgnetwork&lt;/a&gt;  and &lt;a href=&quot;http://www.ruland.com/ti_hardware_tool_sizes.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;ruland&lt;/a&gt;  specify 1.5 mm.  Perhaps your Wiha is a bad unit, or perhaps you would be happier with a ball-end driver which is much easier to insert.   At these sizes, it only takes a little bit of edge roughness to cause great difficulty inserting a flat-end driver. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-03-06T16:13:51.323" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14097" PostTypeId="2" ParentId="12691" CreationDate="2017-03-06T19:48:52.513" Score="1" Body="&lt;p&gt;No such luck: the UV-curing resins used in SLA and polyjet printers aren't thermoplastics.  These are cross-linked polymers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, polyjet materials and SLA can be bonded with epoxy or cyanoacrylate glues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, I wouldn't try to make a functional propeller out of SLA material, because it's fragile.  You could, however, use the SLA print as a master for a mold.  A cast urethane propeller would be considerably stronger than an SLA.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2017-03-06T19:48:52.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14099" PostTypeId="2" ParentId="13099" CreationDate="2017-03-06T21:22:26.410" Score="1" Body="&lt;p&gt;As the velocity increases, the fluid will not be able to form a cohesive boundary layer because the fluid particles will not have the needed residence time near the boundary layer to add to it. &lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2017-03-06T21:22:26.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14100" PostTypeId="2" ParentId="14038" CreationDate="2017-03-06T23:46:06.433" Score="0" Body="&lt;p&gt;Imagine holding a cup out the window with a small hole cut in the bottom.  With the top of the cup facing the direction of travel, the air traveling into the cup would act to force the cup rearward, so it would actually increase the drag on the vehicle.  In your example, the force acting on the pipe would add to the drag not decrease it.  &lt;/p&gt;&#xA;" OwnerUserId="10329" LastActivityDate="2017-03-06T23:46:06.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14101" PostTypeId="2" ParentId="13714" CreationDate="2017-03-06T23:59:19.223" Score="0" Body="&lt;p&gt;The weight supported by each spring will depend on how the weight is distributed by the vehicle.  Most vehicles are slightly heavier at the front so, a 60:40% or 55:45% distribution would be a good approximation for most cars.  Sports cars get closer to an even split.  Then, you would subtract the weight of the tyre and supporting axle, brake, etc. since that weight is not supported by the spring.  &lt;/p&gt;&#xA;" OwnerUserId="10329" LastActivityDate="2017-03-06T23:59:19.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14102" PostTypeId="1" AcceptedAnswerId="14112" CreationDate="2017-03-07T02:32:54.750" Score="-1" ViewCount="388" Body="&lt;p&gt;I have been trying to find out the significance of the usual streetlight design - Why is there curvature of the arm holding the bulb of the light? I have been unable to any reason as to why it may be. Here is an picture of the &quot;usual&quot; of which I am referring to: &lt;a href=&quot;https://i.stack.imgur.com/KsFqw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KsFqw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10348" LastEditorUserId="255" LastEditDate="2017-03-07T10:36:52.477" LastActivityDate="2017-03-14T17:49:53.183" Title="What is the significance of curvature in the arm of a streetlight?" Tags="|design|lighting|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14105" PostTypeId="2" ParentId="1861" CreationDate="2017-03-07T03:39:51.983" Score="4" Body="&lt;p&gt;The reason why a hydroelectric generator is fundamentally different to a gas turbine is because water under pressure is not a gas, and does not change size significantly as energy is extracted from it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gas engine has to account for considerable thermal and volume changes of the gases inside the engine, so multiple parts and multiple materials are generally required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydroelectric turbines have different challenges, and have to tolerate items such as leaves and branches passing through them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design schemes of the rotating elements of hydroelectric turbines are substantially different to gas engines: archimedes screws, kaplan fans, Pelton wheels, crossflow turbines and water wheels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multi-stage designs are employed under some circumstances.&lt;/p&gt;&#xA;" OwnerUserId="10351" LastActivityDate="2017-03-07T03:39:51.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14108" PostTypeId="1" CreationDate="2017-03-07T10:10:13.223" Score="2" ViewCount="305" Body="&lt;p&gt;I have an unknown DC motor with specifications I was testing. I used a Prony Brake setup and scale to evaluate torque, a tachometer for RPM, and multimeters for voltage and current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get a linear graph relating torque and RPM. However, when I physically stalled the motor and checked the stall torque, it was much lower than what the linear graph expected! Almost 1/2 the expected value!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a physical reason to explain this? This outlier is pretty consistent through 4 trials. Is it possible it's due to the Prony Brake? Or perhaps something within the motor? Back EMF or something due to friction? I'm at a loss as to why such a data occurred, and not only once but multiple times!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="10342" LastActivityDate="2017-03-07T12:43:23.173" Title="Stall Torque of a DC Motor Less than Theoretical Values" Tags="|motors|torque|current|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14109" PostTypeId="5" CreationDate="2017-03-07T10:46:41.997" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-03-07T10:46:41.997" LastActivityDate="2017-03-07T10:46:41.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14110" PostTypeId="4" CreationDate="2017-03-07T10:46:41.997" Score="0" Body="Traffic lights are coloured lights that are used to control traffic flow. They instruct traffic when it must stop (usually a red light) or may proceed, or continue to proceed (usually a green light). In some countries intermediate lights (usually amber) may also be used when changing from Stop to Go or from Go to Stop." OwnerUserId="255" LastEditorUserId="90" LastEditDate="2017-03-07T13:16:44.633" LastActivityDate="2017-03-07T13:16:44.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14111" PostTypeId="2" ParentId="14108" CreationDate="2017-03-07T12:43:23.173" Score="1" Body="&lt;p&gt;Torque should be proportional to current, whether stalled or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measure the current at stall.  Is it what you expected?  It should be just the applied voltage divided by the DC resistance of the motor coils that are switched in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider that there can be significant ripple on the torque over a rotation.  This is especially true for mechanically commutated motors.  Try testing the stall torque and current at multiple points over one magnetic rotation.  Note that there may be multiple magnetic rotations in one physical rotation.  The multiple is the number of &quot;poles&quot; the motor has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back EMF has nothing to do with your observations.  This is effectively a voltage in series with the motor that is proportional to speed.  When stalled, the speed is 0, so the back EMF is 0.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-03-07T12:43:23.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14112" PostTypeId="2" ParentId="14102" CreationDate="2017-03-07T13:34:11.103" Score="-4" Body="&lt;p&gt;One possible reason is that it reduces the radius of the moment arm a bit.  So, the downward force is the same, but the moment arm is smaller.  The fixed joint will be under less stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;clarification: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;remember, the weight acts vertically&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so to find the moment arm we should draw a line perpendicular from the vertical pole and another line from the weight that goes straight up and down (incidentally parallel to the main pole)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the two intersect, that's how we can find the length of the moment arm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another clarification, my explanation assumes a shorter distance between the two&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zJqyw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zJqyw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6488" LastEditorUserId="6488" LastEditDate="2017-03-07T15:02:28.763" LastActivityDate="2017-03-07T15:02:28.763" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14113" PostTypeId="1" CreationDate="2017-03-07T16:28:50.980" Score="1" ViewCount="141" Body="&lt;p&gt;Atmel's &lt;a href=&quot;https://www.microchip.com/wwwproducts/en/ATTINY20&quot; rel=&quot;nofollow noreferrer&quot;&gt;ATtiny20&lt;/a&gt; is 1.55x1.4x0.53mm in size. It is one of the smallest microcontrollers. What is the actual chip size inside it (i.e. without the packaging),like what is described &lt;a href=&quot;https://electronics.stackexchange.com/questions/56649/what-is-a-die-package&quot;&gt;here&lt;/a&gt;? &lt;/p&gt;&#xA;" OwnerUserId="10359" LastActivityDate="2017-03-08T11:44:03.647" Title="Actual chip size of Atmel ATtiny20" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14114" PostTypeId="1" CreationDate="2017-03-07T16:49:51.557" Score="1" ViewCount="34" Body="&lt;p&gt;Typically a chip is packaged to protect it and to make it easier to handle. Is it possible  to look at exposed  chip (see images  here &lt;a href=&quot;https://electronics.stackexchange.com/questions/56649/what-is-a-die-package&quot;&gt;https://electronics.stackexchange.com/questions/56649/what-is-a-die-package&lt;/a&gt;) under microscope and know the components and the circuit involved?&lt;/p&gt;&#xA;" OwnerUserId="10359" LastActivityDate="2017-03-07T18:57:50.287" Title="Is it possible to look at exposed chip and know the components and the circuit involved?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14116" PostTypeId="1" CreationDate="2017-03-07T17:39:36.853" Score="1" ViewCount="2405" Body="&lt;p&gt;What is the correct formula to calculate the effective strain under &lt;a href=&quot;https://en.wikipedia.org/wiki/Plane_stress&quot; rel=&quot;nofollow noreferrer&quot;&gt;plane stress&lt;/a&gt; conditions?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In plane stress it is assumed that the out of plane stresses are zero. The out of plane strains are not zero but these tend to be ignored. Wikipedia gives the equation for &lt;a href=&quot;https://en.wikipedia.org/wiki/Infinitesimal_strain_theory#Equivalent_strain&quot; rel=&quot;nofollow noreferrer&quot;&gt;equivalent strain&lt;/a&gt; as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\epsilon_{eq} = \sqrt{\frac{2}{3}{\epsilon}^{dev}_{ij}:{\epsilon}^{dev}_{ij}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from the 2/3 in the equation above it is clear that this was derived from the full strain tensor. How is the effective strain calculated if the out of plain strains are not calculated in the plane stress approximation? Do these have to be calculated to then calculate the effective stress or is it common practice to just neglect these in the formula above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am using the terms effective and equivalent here interchangeably, please correct me if this is wrong as I can't seem to clear up what the difference is if plasticity is not involved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my current thinking. If stresses are calculated from strains using a plane stress simplification then in vector notation (&lt;a href=&quot;https://en.wikipedia.org/wiki/Hooke&amp;#39;s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;from wiki&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;\begin{bmatrix}&#xA;   \sigma_{11} &amp;amp; \sigma_{12} \\&#xA;   \sigma_{12} &amp;amp; \sigma_{22}&#xA;\end{bmatrix}&#xA;=\frac{E}{1-v^2}\left((1-v)&#xA;\begin{bmatrix}&#xA;   \epsilon_{11} &amp;amp; \epsilon_{12} \\&#xA;   \epsilon_{12} &amp;amp; \epsilon_{22}&#xA;\end{bmatrix}&#xA;+v\mathbf{I}(\epsilon_{11}+\epsilon_{22})\right)&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $\epsilon_{33}$ can be calculated from $\frac{-v}{E}(\sigma_{11}+\sigma_{22})$. Considering that all of the strain components have now been calculated it is possible to then calculate the deviatoric strain using the conventional 3D formulas for the hydrostatic and deviatoric components. Does this seem correct?&lt;/p&gt;&#xA;" OwnerUserId="4002" LastEditorUserId="4002" LastEditDate="2017-03-08T14:28:07.367" LastActivityDate="2017-03-08T14:28:07.367" Title="How to calculate effective strain in plane stress?" Tags="|stresses|numerical-methods|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14117" PostTypeId="1" AcceptedAnswerId="14118" CreationDate="2017-03-07T18:23:14.700" Score="-2" ViewCount="82" Body="&lt;p&gt;If you were aiming for 300mph then what would need to change and would it be possible. There is less heat so less drag which must be a huge advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kWh battery pack would be needed?&lt;/p&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="10361" LastEditDate="2017-03-07T19:41:27.987" LastActivityDate="2017-03-07T22:49:52.433" Title="What would be needed to make a tesla fastest car in the world?" Tags="|automotive-engineering|motors|electric-vehicles|" AnswerCount="1" CommentCount="6" ClosedDate="2017-03-08T23:31:41.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="14118" PostTypeId="2" ParentId="14117" CreationDate="2017-03-07T18:46:11.367" Score="5" Body="&lt;p&gt;Maximum speed in a straight line is largely determined by the vehicles maximum power output compared to its drag although once you get past about 200 mph dynamic stability can start to become a serious issue as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current Land speed Record is held by &lt;a href=&quot;https://en.wikipedia.org/wiki/ThrustSSC&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thrust SSC&lt;/a&gt; which was powered by two jet engines, reached supersonic speeds and has as much in common with a plane as a conventional car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crucial point is that aerodynamic drag is proportional to the square of speed so all else being equal to double the maximum speed you need to quadruple the thrust/torque available and 8x the power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any car, even a high  performance one there is a trade off between maximum speed and handling and even very high end performance car manufacturers have pretty much given up chasing headline top speed as if becomes a bit meaningless once you get cars which can top 200mph and acceleration, braking and handling become more meaningful measures of performance. Even current F1 cars don't get anywhere near 300 mph simply because adding aerodynamic down-force increased overall average speeds around a track at the expense of increased drag. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it is possible that you could use the electric drive train from a Tesla to attempt to get to 300mph but it would for all practical purposed be a completely different vehicle as you would need totally different chassis, body shape and gear ratios. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a point of reference the current &lt;a href=&quot;http://www.autocar.co.uk/car-news/motorsport-formula-e/electric-land-speed-record-set-venturi&quot; rel=&quot;nofollow noreferrer&quot;&gt;Land Speed Record for an electric car&lt;/a&gt; is 341 mph. As you can see from the article this is a very different vehicle from a Tesla. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-03-07T22:49:52.433" LastActivityDate="2017-03-07T22:49:52.433" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14119" PostTypeId="2" ParentId="14114" CreationDate="2017-03-07T18:57:50.287" Score="1" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A experienced chip designer can make some educated guesses, but can't know for sure.  A highly regular array will be some memory, for example.  Larger structures near pin pads will be the output drivers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't tell the dopant levels just by looking, and there are many layers to modern chips.  You can only see the top interconnect layer directly.  That will obscure other layers.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-03-07T18:57:50.287" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14120" PostTypeId="1" AcceptedAnswerId="14170" CreationDate="2017-03-07T23:57:58.887" Score="3" ViewCount="216" Body="&lt;p&gt;I have a dynamical model of a planetary gearbox of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{M{\ddot q}+{\Omega_c}G{\dot q}+{[K_b+K_m-\Omega_c^2K_\Omega]q}=T(t)+F(t)} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_b$: bearing stiffness symmetrical matrix&#xA;$K_m$: mess stiffness matrix&#xA;$K_\Omega$: diagonal centripetal stiffness matrix&#xA;$T$: applied external torque&#xA;$F$: static transmission error excitation&#xA;$q$: the vector with the traslation and rotational coordinates for the sun,ring and carrier&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the free vibrations I have the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{M{\ddot q}+{[K_b+K_m]q}=0} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{-\omega_i^2M{\phi_i}+{[K_b+K_m]\phi_i}=0} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $\phi_i$ is the vibration modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{\omega_i^2M{\phi_i}={[K_b+K_m]\phi_i}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the $\Omega_c$ term for the gyroscopic effect is neglected at first and there's no excitation input because it's free vibration. But then I don't understand where the $\phi_i$ comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have basic knowledge of vibration theory for a single degree system, but not for multi DOF which is this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to determine the natural frequencies and vibrational modes of the system from this equation but I don't know how or where to start. I have initial numerical values for Stiffness, mass, preassure angle of the carrier, sun and planet gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerically how would I know how many modes of vibration I have and how to determine them with their natural frequencies? Like I said this is the first time I deal with a MDOF system and I've read a lot but this model is far to complicated and the math is confusing me.&lt;/p&gt;&#xA;" OwnerUserId="4088" LastEditorUserId="4088" LastEditDate="2017-03-25T20:51:35.567" LastActivityDate="2017-03-25T20:51:35.567" Title="How to simulate the Natural Frequencies of a Planetary Gearbox?" Tags="|mechanical-engineering|gears|dynamics|vibration|frequency-response|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14121" PostTypeId="1" CreationDate="2017-03-08T00:41:21.357" Score="0" ViewCount="5215" Body="&lt;p&gt;I will be performing some tests and measuring the vibrations of a planetary gearbox. The accelerometer I will be using stays attached to the gearbox and saves the data into a log, which is available after the test is finished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main objective is to extract the natural frequencies of the accelerometer data and compare them with a simulation derived from an analytical model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best method for the signal conditioning of the accelerometer data? &#xA;I don't understand if the steps between, do I need to have an Integrator for determining the velocity and displacement also, or will a direct fourier analysys like the fft will yield the natural frequencies?&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2017-03-08T05:30:57.930" Title="How to get natural frequencies from accelerometer measurements?" Tags="|measurements|vibration|signal-processing|instrumentation|frequency-response|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14122" PostTypeId="1" AcceptedAnswerId="14232" CreationDate="2017-03-08T00:59:43.463" Score="4" ViewCount="374" Body="&lt;p&gt;I am trying to build a low power output generator with 12 pole pairs and 9 stator coils. The design will be similar to the one described in this link &lt;a href=&quot;http://www.sparweb.ca/2_Gen_Ax/AXIAL_FLUX_HowItWorks_V4.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Basic Principles Of The Homebrew Axial Flux Alternator&lt;/a&gt;  , but at a different scale/size. I am trying to find some way of estimating the number of turns I will need in the coils to produce an output voltage of 6.5V. The generator should produce this voltage at 100rpm. I am getting very confused about how to use Faraday's law to do my calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\varepsilon = N\frac{d(BA)}{dt}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.6pie.com/faradayslaw.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Faraday's Law&lt;/a&gt; as explained in this example where the area used is that of the magnet, while &lt;a href=&quot;https://www.electrical4u.com/faraday-law-of-electromagnetic-induction/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Faraday Law of Electromagnetic Induction&lt;/a&gt;  states at the bottom that A is the area of the coil. Does this difference have to do with when the magnetic field is stationary while the coil is moving versus when the magnetic field is moving relative to a stationary coil? Also, should I be using $$\varepsilon = N\frac{d(BAcos(\theta))}{dt}$$ instead? Thanks very much in advance, and apologies if I am missing something very obvious here.&lt;/p&gt;&#xA;" OwnerUserId="5469" LastEditorUserId="1586" LastEditDate="2017-03-13T21:54:17.847" LastActivityDate="2017-03-14T02:37:38.573" Title="What area should I use when calculating induced voltage?" Tags="|generator|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14124" PostTypeId="1" CreationDate="2017-03-08T02:11:36.437" Score="1" ViewCount="239" Body="&lt;p&gt;I'm doing an optional review sheet before a test and I'm stuck on this one problem. I really don't know what it's asking. Can someone help lead me in the right way? Thank you&lt;a href=&quot;https://i.stack.imgur.com/bfUma.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bfUma.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10366" LastActivityDate="2017-03-08T02:11:36.437" Title="Fictional AB-Type flip-flop" Tags="|control-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14125" PostTypeId="2" ParentId="14121" CreationDate="2017-03-08T04:37:38.087" Score="2" Body="&lt;p&gt;It depends on what your data looks like. I recently got some accelerometer data that looked like this after running it through &lt;a href=&quot;https://docs.scipy.org/doc/scipy-0.18.1/reference/generated/scipy.fftpack.fft.html#scipy.fftpack.fft&quot; rel=&quot;nofollow noreferrer&quot;&gt;Python's Scipy fft function&lt;/a&gt;:  &lt;a href=&quot;https://docs.scipy.org/doc/scipy-0.18.1/reference/generated/scipy.fftpack.fft.html#scipy.fftpack.fft&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t0q4K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; The results are pretty straightforward- natural frequency at about 5.5 Hz. In that particular case, the analysis was basically trivial- 20 lines of python, most of which was importing and plotting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the frequency domain data looks nastier, there are lots of post-processing tools like &lt;a href=&quot;https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.signal.find_peaks_cwt.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;scipy's find_peaks_cwt&lt;/a&gt;. The details of which of those are best is more a question for SO, and you'll find lots of questions (including my own) on the topic over there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the first step should be trying to shove it through fft. If that doesn't work well, you can either follow up here with more details on how to improve the hardware/data collection, or with SO to clean the data with existing software tools&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2017-03-08T04:37:38.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14126" PostTypeId="2" ParentId="14121" CreationDate="2017-03-08T05:30:57.930" Score="1" Body="&lt;p&gt;I've done this before.  I had to be pretty careful to generate the FRF.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It helped me to know the input (force vs. time) and the output (acceleration vs. time).  I have done it without knowing the input.  I typically plot the FFT using a logarithmic scale.  When you know the input, you divide the output by the input (in the frequency domain) and the peaks are much more obvious.  Furthermore, you can look for a 180 deg phase shift and that typically identifies the peaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good resources include:  &lt;strong&gt;Modal Testing&lt;/strong&gt; by Ewins or &lt;strong&gt;Mechanical Vibrations&lt;/strong&gt; by Rao&lt;/p&gt;&#xA;" OwnerUserId="7569" LastActivityDate="2017-03-08T05:30:57.930" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14127" PostTypeId="1" CreationDate="2017-03-08T08:01:16.090" Score="3" ViewCount="153" Body="&lt;p&gt;I want to create a diagram. It will consist of a plate with a uniform heat load, a heat sink, and an adiabatic boundary condition. I thought of displaying the sink as a rigid attachment (as in mechanical problems). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how do I display the adiabatic boundary conditions?&lt;/p&gt;&#xA;" OwnerUserId="9839" LastEditorUserId="90" LastEditDate="2017-05-20T11:55:34.007" LastActivityDate="2017-05-20T11:55:34.007" Title="How to display an adiabatic boundary condition in a diagram?" Tags="|thermodynamics|technical-drawing|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14128" PostTypeId="1" AcceptedAnswerId="14130" CreationDate="2017-03-08T08:21:39.617" Score="1" ViewCount="67" Body="&lt;p&gt;Is it possible to define any camera without &quot;flash&quot; (sensible to any band of radio wave) as a passive radar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, where could I find references to that? I've been looking all over the place and nothing has come up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has any sources or comments, please let me know.&#xA;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The idea of camera is merely a device that captures light in any range (visible light, infrared, ultraviolet). &#xA;This is close in meaning to a passive radar, which is sensible to the electromagnetic waves coming from a target.&lt;/p&gt;&#xA;" OwnerUserId="10368" LastEditorUserId="10368" LastEditDate="2017-03-08T11:30:22.000" LastActivityDate="2017-03-08T12:41:17.613" Title="Defining camera as a passive radar" Tags="|electromagnetism|waves|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14129" PostTypeId="1" CreationDate="2017-03-08T10:45:25.507" Score="3" ViewCount="4916" Body="&lt;p&gt;According to Matlab documentation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;sigma&lt;/em&gt; plots the singular values of the frequency response of a dynamic system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;bode&lt;/em&gt; creates a Bode plot of the frequency response of a dynamic system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am reasonably familiar with Bode plots and dynamic systems but I don't understand what the singular values of the system are or how they are calculated. Are they related to the magnitudes of the system (which can be outputted by &lt;em&gt;bode&lt;/em&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, I have a discrete state-space system which I am attempting to model as a reduced order system. One of the papers I have been using to accomplish this mentions using the sigma function to compare the two models. The paper is 'Dynamic Mode Decomposition with Control' by J.L.Proctor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help with understanding singular values or this &lt;em&gt;sigma&lt;/em&gt; function would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10362" LastActivityDate="2018-05-21T18:55:48.817" Title="What are the Singular Values of a dynamic system and how are they calculated in the *sigma* function in Matlab?" Tags="|control-engineering|transfer-function|systems-engineering|matlab|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14130" PostTypeId="2" ParentId="14128" CreationDate="2017-03-08T11:39:42.837" Score="3" Body="&lt;p&gt;Not really. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly the RA in RADAR stands for RAdio which covers significantly longer wavelengths than the visible light (an in some cases near infra-red) which cameras can detect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While light and radio are both part of the electromagnetism spectrum they are &lt;em&gt;different&lt;/em&gt; parts of it and not interchangeable. While they are all part of a spectrum in terms of physics, from an engineering perspective they might as well not be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally there are laser based systems analogous to radar, usuallly called &lt;a href=&quot;https://en.wikipedia.org/wiki/Lidar&quot; rel=&quot;nofollow noreferrer&quot;&gt;LIDAR&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also radar implies a device which is capable of deducing information about the range and possibly speed and direction of a target object, most cameras can't do this, or if they can it is through a separate system. In fact the auto-focus on most cameras is more akin to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Coincidence_rangefinder&quot; rel=&quot;nofollow noreferrer&quot;&gt;optical rangefinders&lt;/a&gt; widely used for military purposes before the advent of radar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A radar system consists not just of an emitter and receiver but also the processing equipment to deduce the required information while a camera is intended to capture and store a 2D image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest thing to a 'radio camera' is a &lt;a href=&quot;http://abyss.uoregon.edu/~js/glossary/radio_telescope.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;radio telescope&lt;/a&gt;. A key point here is that the differences in wavelength between radio and visible light means that a radio telescope is very different from any camera in terms of both scale and structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While radar and cameras could both be described as devices for detecting electromagnetic radiation that about as far as the similarity goes and is not a complete definition of either. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-03-08T12:41:17.613" LastActivityDate="2017-03-08T12:41:17.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14131" PostTypeId="2" ParentId="14113" CreationDate="2017-03-08T11:44:03.647" Score="1" Body="&lt;p&gt;ATtiny20 is available WLCSP (Wafer level chip scale package) form factor, which is the bare chip with absolute minimum of padding required to mount it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The summary &lt;a href=&quot;http://ww1.microchip.com/downloads/en/DeviceDoc/Atmel-8235-8-bit-AVR-Microcontroller-ATtiny20_Datasheet_Summary.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;datasheet&lt;/a&gt; on page 12 specifies for the WLCSP:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.540 x 1.388mm Body, 0.433 mm thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Width and length will be exact match, but thickness includes the pads. A technical drawing on page 13 gives the chip thickness, although with tolerances:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;min: 0.228mm, nom: 0.305, max: 0.330.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming nominal values then, &lt;strong&gt;1.540 x 1.388 x 0.305 mm&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-03-08T11:44:03.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14132" PostTypeId="1" CreationDate="2017-03-08T13:19:00.787" Score="1" ViewCount="42" Body="&lt;p&gt;Firstly, I'm unsure about the scope of this stack, but let's try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing electron beam lithography on semi-insulating GaAs wafer pieces. And I have a hard time getting the PMMA to stick onto the GaAs. It detaches from the surface, and makes EBL possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial approach was cleaning in U/S acetone, then rinse with IPA. Secondly, I tried rather long cleaning steps (30 minutes of sonication time), and longer solvent bake of the PMMA (solvent: chlorobenzene). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've had limited success with 60s etch in ammonia, followed by a 3 hour solvent bake, after spin coating. Limited success means that the thin film did not entirely peel off, in the vacuum of the SEM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aiming for a 100nm PMMA layer, and I need to produce features of down to 200nm in linewidth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any pointers, or ideas for surface cleaning/etching? I'm contemplating the standard RCA cleaning procedure, but it's also a rather significant step up in complexity and work safety considerations. &lt;/p&gt;&#xA;" OwnerUserId="10373" LastEditorUserId="8055" LastEditDate="2017-03-08T22:03:24.107" LastActivityDate="2017-03-08T22:03:24.107" Title="Improving PMMA adhesion on Gallium Arsenide wafer pieces" Tags="|electrical-engineering|thin-films|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14133" PostTypeId="1" AcceptedAnswerId="14140" CreationDate="2017-03-08T14:25:30.510" Score="2" ViewCount="602" Body="&lt;p&gt;Today, I was thinking about seat belts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On that note, why do the ones in cars and planes differ? Planes have only belts on the waist, but cars have waist and diagonal. Is there a reason that this is so, or are the difference just... [Insert suitable word]?&lt;/p&gt;&#xA;" OwnerUserId="10374" LastEditorUserId="2061" LastEditDate="2017-03-08T16:11:42.057" LastActivityDate="2017-03-09T03:16:52.280" Title="Difference between seat belts" Tags="|automotive-engineering|aerospace-engineering|stresses|safety|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14134" PostTypeId="1" CreationDate="2017-03-08T14:30:42.130" Score="0" ViewCount="35" Body="&lt;p&gt;So let's say you set your thermostat to 20 degrees centigrade, and the current room temperature is 10. A flow of water will gush into the radiators. What I'm wondering, is this flow constant the whole time, or does it slowly descend as the room temperature approaches 20 degrees?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to put it, in the same scenario as above, will the room temperature get to 20 degrees faster if I put the thermostat on 30 degrees? Or will it take the same time?&lt;/p&gt;&#xA;" OwnerUserId="8851" LastActivityDate="2017-03-08T14:38:17.867" Title="When heating a house, does the water flow slowly decrease as the difference in temperatures approaches 0?" Tags="|thermodynamics|heating-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14135" PostTypeId="2" ParentId="14134" CreationDate="2017-03-08T14:38:17.867" Score="2" Body="&lt;p&gt;A hot water based central heating system is fed by a pump and the only signal from the thermostat is heat/don't heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature regulation is done by duty cycling the pump so that when the temperature is low the pump will be started and when it's high the pump will be stopped. There will be some hysteresis of a few degrees to reduce the number of times it will cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, changing the temperature only changes how long the pump will be active until it turns off again.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-03-08T14:38:17.867" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14137" PostTypeId="2" ParentId="13366" CreationDate="2017-03-08T14:56:20.577" Score="0" Body="&lt;p&gt;My guess is (mol catalyst)^-1^-1(hr)^-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;(mol catalyst)^-1&quot; meaning : per gram-mol (which is just a regular mole, not to be confused with a pound mole) of catalyst material.&lt;/p&gt;&#xA;" OwnerUserId="8264" LastActivityDate="2017-03-08T14:56:20.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14139" PostTypeId="2" ParentId="7477" CreationDate="2017-03-08T15:28:01.983" Score="2" Body="&lt;p&gt;There are two main types of polyurethane : Aliphatic and Aromatic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aliphatic offer better UV resistance (so it can retain its gloss and color really well). They're common for exterior coating formulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aromatic have better chemical resistance but are not as resistant to UV. But they are usually applied in a fairly thin film (Commonly used in marine and industrial applications)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of chemical resistance, that depends on the specific coating system but here's a rough list : &lt;a href=&quot;http://www.engineeringtoolbox.com/polyurethane-chemical-resistance-d_1878.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/polyurethane-chemical-resistance-d_1878.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for glue or foam I'm not too sure.&lt;/p&gt;&#xA;" OwnerUserId="8264" LastActivityDate="2017-03-08T15:28:01.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14140" PostTypeId="2" ParentId="14133" CreationDate="2017-03-08T16:10:13.710" Score="3" Body="&lt;p&gt;There's no doubt that lap and shoulder belts are safer than lap belts alone in both cars and planes. The FAA already requires shoulder/lap belts in small general aviation craft, but not (or not yet) in large passenger aircraft. If you notice, the crew on those planes wear 4 point harnesses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, there are multiple challenges to implementing safer passenger restraints on commercial aircraft. First is just the engineering challenge of where to place the upper attachment point. In a car, the frame of the vehicle is close and easily accessible while in a plane it's not. That means adding the required structure to the seat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding that structure to the seat will inevitably result in a seat that is more expensive, heavier and physically larger. Just replacing all of the seats on pasenger aircraft would be cost-prohibitive on its own, but a larger seat means there can be fewer seats/plane which drives the actual cost even higher. Finally, the seats would be heavier. Heavier seats mean greater fuel consumption also resulting in higher costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, there's the behavioral aspect of shoulder belts. Passengers won't like them, especially on long flights. Flight attendants already have their hands full trying to get people to wear lap belts, a more restrictive belt will only aggravate that problem.   &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-03-08T16:10:13.710" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14142" PostTypeId="2" ParentId="14133" CreationDate="2017-03-08T16:18:07.273" Score="2" Body="&lt;p&gt;Different types of seat belts are designed for different purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main function of car seat belts is to stop you being thrown &lt;em&gt;forwards&lt;/em&gt; during an &lt;em&gt;unexpected&lt;/em&gt; accident. A lap-only belt would not prevent your head hitting the windscreen. The lap belt prevents you sliding forwards feet-first, which could cause injury if the front of the car is crushed, or it could cause problems if you need to get out of the car quickly, for example because the accident started a fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main function of an aircraft seat belt is to stop you being thrown &lt;em&gt;vertically&lt;/em&gt; by unexpected turbulence, or simply to keep you sitting in your seat during takeoff and landing. The chance of a &lt;em&gt;horizontal&lt;/em&gt; impact in a plane crash is usually predictable, and injuries can be avoided by the &quot;brace position&quot; without the need to wear a diagonal belt all the time.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-03-08T16:18:07.273" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14143" PostTypeId="1" CreationDate="2017-03-08T17:52:05.300" Score="2" ViewCount="919" Body="&lt;p&gt;I am studying the application of Pade approximations on a system with a dead time. In most cases, we consider a transfer function with a dead time and apply the Pade approximation on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is my question: can we introduce a delay in &lt;strong&gt;state space&lt;/strong&gt;? If yes, how one can apply the Pade approximation on it? If no, why is it not possible?&lt;/p&gt;&#xA;" OwnerUserId="7847" LastEditorUserId="1832" LastEditDate="2018-04-23T14:22:50.663" LastActivityDate="2018-04-23T14:22:50.663" Title="Pade approximations for state space systems" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14144" PostTypeId="1" CreationDate="2017-03-08T18:02:51.477" Score="-4" ViewCount="90" Body="&lt;p&gt;Let's say I want to measure the weight of a bag of clothes (or some other object), in a situation where I don't have a weighing machine (&quot;scales&quot;?) available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Objects that are available:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A broom.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The specialty of a broomstick is that its weight is evenly distributed.&lt;/li&gt;&#xA;&lt;li&gt;If the broom end cannot be removed, it means there is an extra weight on one end.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;A chair or something to balance the broom on.&lt;/li&gt;&#xA;&lt;li&gt;A counterweight with a known weight of, say, 1kg.&lt;br&gt;&#xA;This could be a packaged food item, e.g. a milk package, placed in a plastic bag so it can be hanged on the stick.&lt;/li&gt;&#xA;&lt;li&gt;A pencil or some other thing to make marks on the broomstick.&lt;/li&gt;&#xA;&lt;li&gt;A smartphone with calculator.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Objects that &lt;em&gt;might&lt;/em&gt; be available:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A (wooden) stick with even distribution of weight, e.g. a broomstick with the broom end removed.&lt;/li&gt;&#xA;&lt;li&gt;A tape measure.&lt;/li&gt;&#xA;&lt;li&gt;A smartphone with internet and/or an app that does the math for me (if such an app exists).&lt;/li&gt;&#xA;&lt;li&gt;If you think you desperately need another object that is commonly available and does not radically change the setup of the scenario, you may include it in the answer.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;None of these objects should be damaged in the process! E.g. we don't know if the owner of the broom would approve of removing the broom end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a method and a formula to determine the weight of the bag of clothes. So in other words, a primitive / improvised weighing machine / scales.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: The difference to traditional scales is that we don't have arbitrary counterweights available. It is hard enough to find one object with a known weight.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A perfect answer would have&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Simple and easy-to-follow instructions.&lt;/li&gt;&#xA;&lt;li&gt;Different versions of the instructions for the &quot;optional&quot; objects.&lt;/li&gt;&#xA;&lt;li&gt;Some background or proof or derivation of the formula, but separate from the instructions, so this part is optional for the reader.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;P.S.: Yes, I might figure this out on my own if I think really hard. But even if I do, I still think this should go into a stackexchange Q/A. The best would be macgyver.stackexchange.com, but engineering is the closest fit.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;About the &quot;on hold&quot; / &quot;too broad&quot;.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I seriously don't get what is wrong with the question. It is a simple problem with a simple answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the time of asking, I was sure that a weighing machine can be built with a stick balanced on something, and stuff hanging on the stick. I just was not sure about the exact setup (which is embarrassing, since this is very basic stuff), and which physical properties are required from the stick object. This is why I tried to provide details without narrowing it down unnecessarily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal of this Q/A is that someone with the same problem finds the information that is relevant to the solution, so they learn how to build the thing and why it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could have asked &quot;How can I build an improvised weighing machine from commonly available household materials?&quot;, but this would be too vague because &quot;commonly available household materials&quot; is open to interpretation, and would possibly result in a wide range of different apparatuses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it was clear that it does not matter if it is a broom or another stick-like object, so there was no benefit in insisting on this. Naming it a broom-stick makes it easier to imagine, so I kept it as part of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally I was not sure if additional items are required for the task, so I left this open.&lt;/p&gt;&#xA;" OwnerUserId="2640" LastEditorUserId="2640" LastEditDate="2017-03-09T22:14:28.857" LastActivityDate="2017-03-09T22:14:28.857" Title="Improvised weighing machine from broomstick and 1kg counterweight" Tags="|applied-mechanics|" AnswerCount="1" CommentCount="7" ClosedDate="2017-03-09T10:34:10.330" ContentLicense="CC BY-SA 3.0" />
  <row Id="14145" PostTypeId="1" CreationDate="2017-03-08T18:21:05.780" Score="3" ViewCount="93" Body="&lt;p&gt;I am working with a tdc1000-tdc7200 evm board for water level detection. I am trying to mount a &lt;a href=&quot;https://www.steminc.com/PZT/en/piezo-ceramic-disc-10x2mm-r-215-khz-wire-leads-smd10t2r111wl&quot; rel=&quot;nofollow noreferrer&quot;&gt;steminc 1MHz ultrasonic transducer&lt;/a&gt; under a polycarbonate jar to measure the water level in the jar. According to TI's recommendation (&lt;a href=&quot;http://www.ti.com/lit/an/snaa266a/snaa266a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;) I am using Cyanoacrylate for transducer mounting. I am wondering is there other options of adhesive that I can use for mounting? I am looking for an adhesive that does not react with water. Also TI recommended using hot glue over sensor which also absorbs. Is there a better alternative that does not hinder transducer performance?&lt;/p&gt;&#xA;" OwnerUserId="10379" LastEditorUserId="1586" LastEditDate="2017-03-13T21:55:21.573" LastActivityDate="2017-03-13T21:55:21.573" Title="mounting a transducer under a polycarbonate jar for fluid level detection" Tags="|electrical-engineering|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14146" PostTypeId="2" ParentId="14145" CreationDate="2017-03-08T21:30:43.677" Score="2" Body="&lt;p&gt;Water improves superglue's bonding, I'm not sure I see the problem with using it (and the hot glue) in your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, I suspect that they specify the CA glue and hot glue because they provide a good bond, are inexpensive and reversible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I have used a variety of 2 part epoxies for mounting sensors to a variety of materials. Make sure you've got it where you want it because once it sets, there's no do-overs.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-03-08T21:30:43.677" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14150" PostTypeId="2" ParentId="12184" CreationDate="2017-03-09T00:58:01.907" Score="1" Body="&lt;p&gt;A torque converter has a pump driven by the engine and a turbine, which drives the remaining powertrain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fluid leaving the pump has velocity components in the axial and rotational reference frames (it's swirling).  Increased swirling means that the fluid impacts the turbine with a higher angle of attack.  The effect is similar to  pitch angle vs. torque for a wind turbine, or attack angle vs lift for an airplane wing.  When the fluid's angle of attack increases, the torque multiplier at the turbine increases(not absolute torque, but the torque ratio).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stators are used in an attempt to control the rotation of the flow to achieve some effect. For example, to increase compression ratios in jet engines. &#xA;See this page to view the torque ratio vs speed ratio/ efficiency tradeoff for heavy duty applications, where torque converters are very popular:&#xA;&lt;a href=&quot;http://jmclutch.com/site/book/export/html/7&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://jmclutch.com/site/book/export/html/7&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-03-09T00:58:01.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14151" PostTypeId="2" ParentId="14087" CreationDate="2017-03-09T02:22:48.743" Score="1" Body="&lt;p&gt;Propably the confusion comes from different conventions used for the internal forces. If we make a cut at any section of a member, two internal actions are defined, at opposite directions, each applied to one of the two faces of the cut. Consider as an example the following cantilever beam problem, with two cuts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KMuJj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KMuJj.jpg&quot; alt=&quot;cantilever beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear force, and the shear stresses have alternate directions, depending on the cut face you look at. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one may consider the following convention, where shear stresses match the direction of &lt;strong&gt;internal shear force&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/weWFP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/weWFP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or this convention, where shear stresses have opposite direction to &lt;strong&gt;external shear force&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vKtet.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vKtet.jpg&quot; alt=&quot;opposite direction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear flow is just shear stress multiplied by section thickness and therefore both shear flow and shear stresses (and internal shear force) have the same direction on the same face. &lt;/p&gt;&#xA;" OwnerUserId="9620" LastEditorUserId="9620" LastEditDate="2017-03-09T12:57:30.000" LastActivityDate="2017-03-09T12:57:30.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14152" PostTypeId="2" ParentId="14144" CreationDate="2017-03-09T02:47:52.417" Score="1" Body="&lt;p&gt;There is more than one possible configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to describe one that seems to be the most simple and is painfully obvious. It is really quite &quot;duh&quot;, I don't know why I did not come up with this myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Balance the broom over the chair, without any weights on either side. Mark the center of gravity (the point where it touches the chair), to make sure it does not move around.&lt;/li&gt;&#xA;&lt;li&gt;Hang on the stick: the bag of clothes on one side of the chair, the counterweight on the other side of the chair, in a way that the stick stays balanced. (The point where the broom touches the chair must remain the same as before!)&lt;/li&gt;&#xA;&lt;li&gt;Compare (*) the distance of each item to the center of gravity (the point where the broom touches the chair). The ratio of these two distances is the inverse ratio of the two weights.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This means if the distance of the 1kg counter-weight is 3x the distance of the bag of clothes, then the bag of clothes weighs 3x1kg = 3kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight and shape of the broom are really irrelevant in this scenario. So it does not matter if the broom end is removed or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) To measure the ratio of the two distances, one could of course use a tape measure. But I am sure there are other ways to do this. Remember, we don't need any absolute length, only the ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One part of this can be to have a picture of a tape measure (or some kind of pixel grid with numbers on it), on your smartphone. Since absolute distances are irrelevant, it does not matter if these are real centimeters or inches.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Explanation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The two objects add torque in opposite directions, proportional to distance x weight (gravitational force). The sum of all torques (negative plus positive) must be zero, if the broom is in balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of torque of the broom itself is already zero, because it already was in balance before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Torque&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Torque&lt;/a&gt; (torque = rotational force)&lt;/p&gt;&#xA;" OwnerUserId="2640" LastEditorUserId="2640" LastEditDate="2017-03-09T02:55:42.410" LastActivityDate="2017-03-09T02:55:42.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14153" PostTypeId="2" ParentId="14133" CreationDate="2017-03-09T03:16:52.280" Score="1" Body="&lt;p&gt;Originally car seat belts were like the airline &lt;a href=&quot;https://en.wikipedia.org/wiki/Seat_belt&quot; rel=&quot;nofollow noreferrer&quot;&gt;seat belts&lt;/a&gt;. In 1959 the Swedish car marker &lt;a href=&quot;http://jalopnik.com/volvo-gave-away-their-most-important-invention-to-save-1069825878&quot; rel=&quot;nofollow noreferrer&quot;&gt;Volvo invented the three point seat belt&lt;/a&gt; and determined it to be the safest type of seat belt, by the way it restrains vehicle occupants. Volvo let other car makers use the belt for a fee of just one dollar ($1) to save lives around the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Airline type seat belts are simple to use &amp;amp; simpler to get out of in an emergency in a bulk passenger situation. The seat belt ends stay on the seat. If airlines used the triple point seat belts that are used in cars some passengers may get injured by the belt buckles as they fly back up to there resting positions.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-03-09T03:16:52.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14156" PostTypeId="1" CreationDate="2017-03-09T09:08:14.463" Score="1" ViewCount="2874" Body="&lt;p&gt;I simulated a VLC system for an AWGN channel and plotted the BER against Ebno. I used the LED impulse response and convolved it with the signal produced to transmitting. But the results are not OK. It means that Ber is fixed or 0.5. It is a PAM-DMT system which the transmitter is a 50*50 array of LEDs. The input is a binary sequence that are modulated by PAM modulator. They are used to make OFDM symbols by PAM-DMT algorithm. OFDM symbols are modulated with LED intensity and transmitted into the channel. At the receiver side, after the demodulating the received symbols, the received binary sequence is compared with the input binary sequence to calculate BER. In each OFDM symbol, there are some data carriers, pilot symbol carriers and null carriers.&#xA;I think the problem is about the scheme I use to cut the extended length (made by convolving the LED input with the LED impulse response) of the transmitted signal at the receiver side. Could anyone help me if I am wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my Matlab code: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; clear,clc;&#xA;&#xA; close all;&#xA;&#xA;%%&#xA;&#xA;k=4;M=2^k;%4 bits per PAM symbol%&#xA;&#xA;c=3*1e8;&#xA;&#xA;f_3dB=15e6;&#xA;&#xA;N=32;&#xA;&#xA;R=0.54;%Responsivity at A/W%&#xA;&#xA;Ar=1e-4;&#xA;&#xA;index=1.5;&#xA;&#xA;carier_freqSpac=f_3dB/N;&#xA;&#xA;T_sym=1/f_3dB;&#xA;&#xA;Tb=T_sym/k;&#xA;&#xA;Rb=1/Tb;&#xA;&#xA;T_ifft=1/carier_freqSpac;&#xA;&#xA;l_cp=N/4;&#xA;&#xA;T_cp=T_sym*l_cp;&#xA;&#xA;T_ofdmsym=T_ifft+T_cp; %OFDM symbol duration is 4us%&#xA;&#xA;l_ofdm_sym=N+l_cp;&#xA;&#xA;n_pilot_carrier=N/16;&#xA;&#xA;n_null=N/16;&#xA;&#xA;n_data_carrier=N-n_null-n_pilot_carrier;&#xA;&#xA;nofdm=2e4;&#xA;&#xA;TTran_ofdm_sym=((n_data_carrier+n_pilot_carrier)*T_sym)+T_cp;&#xA;&#xA;Rb_ofdm=((n_data_carrier+n_pilot_carrier)*k)/TTran_ofdm_sym;% at b/s&#xA;&#xA;&#xA;hpammod=comm.PAMModulator(M,'BitInput',true);&#xA;&#xA;hpam_demod=comm.PAMDemodulator(M,'BitOutput',true);&#xA;&#xA;%Pilots in the 1st half&#xA;&#xA;if N==32;&#xA;&#xA;    PS=[1];&#xA;&#xA;elseif N==64;&#xA;&#xA;    PS=[-1 1];&#xA;&#xA;elseif N==128;&#xA;&#xA;    PS=[-1 -1 1 1];&#xA;end;&#xA;&#xA;&#xA;input_b=randi([0,1],nofdm*(n_data_carrier/2)*k,1);%1000036,1160&#xA;&#xA;n_bit=length(input_b);&#xA;&#xA;PAM_sym=step(hpammod,input_b);&#xA;&#xA;n_PAMsym=n_bit/k;&#xA;&#xA;PAM_sym=reshape(PAM_sym,n_data_carrier/2,nofdm);&#xA;&#xA; PAM_sym_PS_1st=[];&#xA;&#xA;PAM_sym_PS_2nd=[];&#xA;&#xA;if N==32;&#xA;&#xA;for i=1:nofdm;&#xA;&#xA;   PAM_sym_ps_1st=1j.*[0;PAM_sym(1:14,i);PS(1)];&#xA;&#xA;   PAM_sym_PS_1st=[PAM_sym_PS_1st PAM_sym_ps_1st];&#xA;&#xA;   PAM_sym_ps_2nd=circshift(conj(flipud(PAM_sym_ps_1st)),1); &#xA;&#xA;   PAM_sym_PS_2nd=[PAM_sym_PS_2nd PAM_sym_ps_2nd];&#xA;&#xA;end;&#xA;&#xA;elseif N==64;&#xA;&#xA;    for i=1:nofdm;&#xA;&#xA;   PAM_sym_ps_1st=1j.*[0;PAM_sym(1:14,i);PS(1);0;PAM_sym(15:28,i);PS(2)];&#xA;&#xA;   PAM_sym_PS_1st=[PAM_sym_PS_1st PAM_sym_ps_1st];&#xA;&#xA;   PAM_sym_ps_2nd=circshift(conj(flipud(PAM_sym_ps_1st)),1); &#xA;&#xA;   PAM_sym_PS_2nd=[PAM_sym_PS_2nd PAM_sym_ps_2nd];&#xA;&#xA;end;&#xA;&#xA;elseif N==128;&#xA;&#xA;    for i=1:nofdm;&#xA;&#xA;   PAM_sym_ps_1st=1j.*&#xA;[0;PAM_sym(1:14,i);PS(1);0;PAM_sym(15:28,i);PS(2);0;PAM_sym(29:42,i);PS(3);0;PAM_sym(43:56,i);PS(4)];&#xA;&#xA;   PAM_sym_PS_1st=[PAM_sym_PS_1st PAM_sym_ps_1st];&#xA;&#xA;   PAM_sym_ps_2nd=circshift(conj(flipud(PAM_sym_ps_1st)),1); &#xA;&#xA;   PAM_sym_PS_2nd=[PAM_sym_PS_2nd PAM_sym_ps_2nd];&#xA;&#xA;end;&#xA;&#xA;end;&#xA;&#xA;ifft_in=[PAM_sym_PS_1st;PAM_sym_PS_2nd];&#xA;&#xA;ifft_out=ifft(ifft_in,N);&#xA;&#xA;cp=ifft_out(N-l_cp+1:N,:);&#xA;&#xA;ofdm_sym=[cp;ifft_out];&#xA;&#xA;ofdm_sym1=reshape(ofdm_sym,l_ofdm_sym*nofdm,1);% P/S%&#xA;&#xA;uni_ofdm_sym=(ofdm_sym1+abs(ofdm_sym1))./2;&#xA;&#xA;&#xA;t=0:T_sym:100e-9;%ns&#xA;&#xA;n_led=2500;&#xA;&#xA;hled=(n_led).*exp((-2*pi*f_3dB).*t);&#xA;&#xA;hled=hled./sum(hled);&#xA;&#xA;ledin=48e-6*n_led+uni_ofdm_sym;&#xA;&#xA;optsig=R.*conv(ledin,hled);&#xA;&#xA;l=length(optsig)-length(ofdm_sym1);&#xA;&#xA;Ebno_dB=0:20;&#xA;&#xA;SNR_db=Ebno_dB+10*log10((N/N+l_cp))...&#xA; +10*log10(((n_data_carrier+n_pilot_carrier)/N))+10*log10(k);&#xA;&#xA;&#xA; Ber=zeros(1,length(SNR_db));&#xA;&#xA;for rr=1:length(SNR_db);&#xA;&#xA;inoisy=awgn(optsig,SNR_db(rr),'measured');&#xA;&#xA;inoisy1=inoisy((1+floor(l/2)):(floor(l/2)+length(ofdm_sym1)));&#xA;&#xA;inoisy22=reshape(inoisy1,l_ofdm_sym,nofdm);&#xA;&#xA;fft_in=inoisy22(l_cp+1:N+l_cp,:);&#xA;&#xA;fft_out=fft(fft_in,N);&#xA;&#xA;if N==32;&#xA;&#xA;    fft_out2=fft_out(2:15,:);&#xA;&#xA;elseif N==64;&#xA;&#xA;    f1=fft_out(2:15,:);&#xA;&#xA;    f2=fft_out(18:31,:);&#xA;&#xA;    fft_out2=[f1;f2];&#xA;&#xA;elseif N==128;&#xA;&#xA;    f1=fft_out(2:15,:);&#xA;&#xA;    f2=fft_out(18:31,:);&#xA;&#xA;    f3=fft_out(34:47,:);&#xA;&#xA;    f4=fft_out(50:63,:);&#xA;&#xA;    fft_out2=[f1;f2;f3;f4];&#xA;&#xA;end;&#xA;&#xA;fft_out3=2.*imag(fft_out2);&#xA;&#xA;fft_out4=reshape(fft_out3,n_data_carrier*0.5*nofdm,1);&#xA;&#xA;demoded_sig=step(hpam_demod,fft_out4);&#xA;&#xA;ber1=xor(input_b,demoded_sig);&#xA;&#xA;Ber(1,rr)=sum(ber1)/length(ber1)&#xA;&#xA;end;&#xA;&#xA;semilogy(Ebno_dB,Ber,'g-s','LineWidth',2.5,'MarkerSize',8);grid on;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10390" LastEditorUserId="10390" LastEditDate="2017-03-10T12:30:09.583" LastActivityDate="2017-03-13T17:20:58.133" Title="Matlab code for a visible light communication system" Tags="|optics|communication|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14162" PostTypeId="1" AcceptedAnswerId="14167" CreationDate="2017-03-09T15:33:03.303" Score="1" ViewCount="4091" Body="&lt;p&gt;Is it possible to build a two storey building just using bricks (no wooden beams, steel beams or steel reinforcement) to suspend the floor of the second storey?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that using wooden beams is an option, however because of the environment of the build (tropical rains, humidity, termites, ants, beetles and millions of other insects), I would much prefer to use only brick - if this is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally there will be no intermediate supports; the floor structure will span from one external wall to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an intuitive feeling that an efficient structure can be built using just bricks - but I can't quite articulate it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has this been done before - is this even possible?&lt;/p&gt;&#xA;" OwnerUserId="932" LastEditorUserId="255" LastEditDate="2017-03-09T17:10:53.143" LastActivityDate="2017-03-09T17:20:51.523" Title="Two storey brick building without using wood or steel" Tags="|structural-engineering|civil-engineering|structures|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14163" PostTypeId="2" ParentId="14162" CreationDate="2017-03-09T15:58:11.693" Score="0" Body="&lt;p&gt;Yes it can be done, a 2 or 3 story brick house with no steel is the standard home construction method in plenty of places. However wood beams are normally used for the upper floor and roof supporting structures. Avoiding any use of structural wood  would get tricky, for the roof you could dome it in some way but I have no idea how you would create the upper floors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And don't try this anywhere with a risk of earthquakes.&lt;/p&gt;&#xA;" OwnerUserId="7659" LastActivityDate="2017-03-09T15:58:11.693" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14164" PostTypeId="1" AcceptedAnswerId="14182" CreationDate="2017-03-09T16:11:49.880" Score="2" ViewCount="113" Body="&lt;p&gt;I'm working on a seaplane and find myself doubting about the specific nautical term for a keel placed between the &lt;em&gt;bilge keel&lt;/em&gt; and the &lt;em&gt;main keel&lt;/em&gt;. This might be a bit academic in nature since I have a working way of naming it, but I'd rather take the chance to find if it's correct or just a fossilized mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My working term is &quot;sister keelson&quot;, which is commonplace in seaplane design, but something tells me it's not really right, as in naval engineering a keelson is the &lt;strong&gt;&lt;em&gt;inner&lt;/em&gt;&lt;/strong&gt; structural element that holds ribs against a keel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second nearest term would be &quot;docking keel&quot;, because of its location, but that implies it can be used to beach the craft, which it may not be necessarily designed for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't add images right now, but a google image search for &quot;sister keelson seaplane&quot; will give you pictures of what this is, while a search for &quot;sister keelson&quot; will point you to the inner elements that would be used in a wooden hull boat.&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2017-03-10T13:07:40.790" Title="Name of a keel between the bilge keel and the main keel" Tags="|aerospace-engineering|terminology|aircraft-design|naval-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14165" PostTypeId="1" CreationDate="2017-03-09T16:34:18.577" Score="1" ViewCount="1891" Body="&lt;p&gt;I have read that corrugations give strength to pipes and the same strength can be achieved with thinner wall when compared to plain pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't understand why (how) the corrugations gave strength to the pipe. How do they work?&lt;/p&gt;&#xA;" OwnerUserId="10401" LastEditorUserId="421" LastEditDate="2017-03-13T21:55:10.233" LastActivityDate="2018-09-02T13:52:07.107" Title="Why do corrugations give strength to pipes?" Tags="|civil-engineering|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14166" PostTypeId="2" ParentId="14165" CreationDate="2017-03-09T17:20:08.677" Score="1" Body="&lt;p&gt;The corrugations will increase the moment of inertia of the pipe, making it more resistant to deformation out of its circular shape.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you slit the pipe long ways and unroll it into a corrugated sheet.  It is easy to understand that the corrugations will make the sheet more rigid and resistant to bending along that direction.  The same principle will apply to the pipe.&lt;/p&gt;&#xA;" OwnerUserId="4708" LastActivityDate="2017-03-09T17:20:08.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14167" PostTypeId="2" ParentId="14162" CreationDate="2017-03-09T17:20:51.523" Score="3" Body="&lt;p&gt;Of course it can be done - &lt;a href=&quot;https://www.google.com/search?site=&amp;amp;tbm=isch&amp;amp;source=hp&amp;amp;biw=784&amp;amp;bih=797&amp;amp;q=masonry%20bridge&amp;amp;oq=masonry%20bridge&amp;amp;gs_l=img.3..0j0i8i30k1j0i24k1l2.187.3369.0.3672.16.14.1.1.1.0.157.1193.11j3.14.0....0...1ac.1.64.img..0.16.1201.2LIkfpgDGSE#tbm=isch&amp;amp;q=masonry%20arch%20bridge&amp;amp;*&quot; rel=&quot;nofollow noreferrer&quot;&gt;masonry bridges&lt;/a&gt; suspend a bridge deck above a void, you could use the same technique to support the floor for your second storey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, in order to avoid tension in your masonry (masonry is not good at tension) you need an arch that is close to a catenary in shape. This means that the crown (top) of your arch is going to be a fairly big height above your springing (bottom of the arch). My intuition is that your &quot;two storey&quot; house will be the height of a normal three storey house, with the arch taking up the height of a storey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the floor of your second storey to be flat, you'll need fill over the arch. This adds weight and hence causes more thrust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a masonry arch bridge there's a big thrust resisted at the abutments. This is resisted by the mass of soil. For your house to resist it you're going to need very thick walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So: technically feasible but not very practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are also plenty of other materials out there which aren't steel or wood. Fibre reinforced polymers would work, but they're not going to be cheap!&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-03-09T17:20:51.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14168" PostTypeId="2" ParentId="14165" CreationDate="2017-03-09T17:55:21.597" Score="4" Body="&lt;p&gt;For a thin wall structure, corrugations behave like a frame, the outer shell rests upon. Imagine the fuselage of an airplane or the hull of a ship. They are typically consist of a strong framing, made by bars, both in the cross-sectional plane and longitudinal ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7EcNh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7EcNh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross-sectional bars (the circular rings in the image above) provide stiffness and resistance against the normal forces towards the fuselage (either due to internal or external high pressure), while the longitudinal bars support the rings against lateral buckling and provide longitudinal stiffness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without such framing, a shell structure should rely on it's own thickness to resist normal wall pressure (through bending) as well as lontidutinal forces. Not an efficient use of material.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;For pipes, internal corrugations are typically only longitudinal ones, since otherwise flow would be severly inhibited. In that case, the resitance against normal wall pressure isn't as much improved by the longitudinal corrugations alone, but they provide an improvement for local phenomena, and for short pipes, or for pipes with external reinforcement rings. In such cases they function like beams for the thin shell. &lt;/p&gt;&#xA;" OwnerUserId="9620" LastEditorUserId="9620" LastEditDate="2017-03-09T20:30:30.130" LastActivityDate="2017-03-09T20:30:30.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14169" PostTypeId="2" ParentId="12873" CreationDate="2017-03-09T18:33:08.957" Score="0" Body="&lt;p&gt;Have you considered an elastic control volume around the solid?  The boundary shrinks as the particle melts, and calculations can be made with extrinsic (total internal energy) rather than intrinsic variables (specific internal energy). If you draw your control volume only around the solid phase, you'll have to estimate the mass leaving the control volume as liquid:&#xA;$$ \left. \frac{dm}{dt} \right|_{NP} = -\dot{m}_{liq} $$&#xA;and estimate the internal energy leaving the control volume as liquid as well as the internal energy used to melt the particle:&#xA;$$\left. \frac{dU}{dt} \right|_{NP} = -\dot{m}_{liq}(u_{ls} +u_{l}) + \int_{A} \nabla \cdot \vec{q} $$ &#xA;Where $u_{l}$ is the specific internal energy at which the NP melts and $u_{ls}$ is the latent heat of melting  I integrated the heat flux in this case, over the surface area of the particle.  The equations can be given in terms of temperature and mass if you know the specific heats ,$C_{s}$ and $C_{l}$ for solid and liquid, and .&#xA;$$\left. c_{s}\left( m\frac{dT}{dt} + T\frac{dm}{dt} \right) \right|_{NP} = \int_{A}\nabla \cdot \vec{q} -\dot{m}_{liq}\left(T_{sat}c_{l} + u_{ls} \right) $$&lt;br&gt;&#xA;I'm sorry to say, there's still a lot of loose ends.  How fast does liquid flow off the particle surface?  How will you model the change in surface area of the particle?  I assumed that liquid leaves the control volume at exactly the melting temperature, and that there was no material in the control volume which was &quot;mushy&quot;(saturated solid-liquid).  best of luck!    &lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-03-09T18:33:08.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14170" PostTypeId="2" ParentId="14120" CreationDate="2017-03-09T19:48:52.600" Score="5" Body="&lt;p&gt;To obtain the modes shapes and resonant frequencies, you start from your equation of motion with no externally applied forces, which is indeed as you've stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf M \mathbf{\ddot q} + \mathbf K \mathbf q = \mathbf 0 \qquad (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For brevity, I've let $\mathbf K = \mathbf K_b + \mathbf K_m$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, $\mathbf q(t)$ is a function of time. If the system is vibrating in the $i$th mode, everywhere will be vibrating at frequency $\omega_i$ in the same phase, assuming there is no damping. Therefore, we can assume the following:&#xA;$$\mathbf q(t) = \mathbf u^{(i)}\exp{(j\omega_i t)}\qquad(2)$$&#xA;, where $j$ is the imaginary number defined as $\sqrt{-1}$ and $\mathbf u^{(i)}$ is the $i$th mode shape, which is independent of time. (This assumption may seem bizarre as $\exp{(j\omega_i t)}$ is a complex number, but this does not matter in this scenario, and is used just to show that $\mathbf q(t)$ is sinusoidal in time with frequency $\omega_i$. Equivalently, you could instead use $\mathbf q(t) = \mathbf u^{(i)} (A \sin{\omega_i t} + B \cos{\omega_i t})$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf {\ddot q}(t) = \frac{d}{dt}\left(\mathbf u^{(i)}\exp{(j\omega_i t)}\right)=-\omega_i^2\mathbf u^{(i)}\exp{(j\omega_i t)}\qquad (3)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting equations (2) and (3) into (1) will give you the following, noting that the exponential functions cancel out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(-\omega_i^2 \mathbf M + \mathbf K \right) \mathbf u^{(i)}=0 \qquad (4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation 4 is a type of problem called the &lt;em&gt;generalised&lt;/em&gt; eigenvalue problem, which involves solving equations of the form $\mathbf A \mathbf x = \lambda \mathbf B \mathbf x$ (compare with the &lt;em&gt;standard&lt;/em&gt; eigenvalue problem, which involves solving $\mathbf A \mathbf x = \lambda \mathbf x$). In this case, the eigenvalue is $\omega_i^2$ and the eigenvector is $\mathbf u^{(i)}$. To obtain the eigenvalues and eigenvectors involves solving the generalised eigenvalue problem. There are two main approaches: hand calculation or numerical computation. For your problem, the latter will definitely be the recommended option, however, the former should help with smaller problems and lends some insight to how generalised eigenvalue problems are solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;HAND CALCULATION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how equation (4) also has the form $\mathbf A \mathbf u^{(i)} = \mathbf 0$. Obviously, $\mathbf u^{(i)}=\mathbf 0$ would satisfy such an equation, but this is a trivial result that corresponds to no actual vibration. Instead, we are interested in values of $\mathbf u^{(i)}\ne\mathbf 0$ that solve the equation. In order for such a non-zero answer to exist, $\mathbf A$ must have no inverse, i.e. $\mathbf A$ is a singular matrix. (If an inverse did exist, then you could premultiply both sides of the equation with the inverse matrix, getting $\mathbf u^{(i)} = \mathbf A^{-1} \mathbf 0 = \mathbf 0$.) In order for a matrix to be singular, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Determinant&quot; rel=&quot;noreferrer&quot;&gt;determinant&lt;/a&gt; of the matrix must be zero. i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left| \mathbf A \right| = \left| -\omega_i^2 \mathbf M + \mathbf K \right| = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will produce a polynomial in $\omega_i^2$, the roots of which will be each of the resonant frequencies squared. Take care to note that some modes will have the same resonant frequencies, in the form of repeated roots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf M = \left[\begin{matrix}&#xA;        m &amp;amp; 0\\&#xA;        0 &amp;amp; m\\&#xA;        \end{matrix}\right]&#xA;\qquad&#xA;\mathbf K = \left[\begin{matrix}&#xA;        2k &amp;amp; -k\\&#xA;        -k &amp;amp; 2k\\&#xA;        \end{matrix}\right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The determinant of interest would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left|\begin{matrix}&#xA;        2k-\omega_i^2 m &amp;amp; -k\\&#xA;        -k &amp;amp; 2k-\omega_i^2 m\\&#xA;        \end{matrix}\right|&#xA;=0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By expanding out the determinant expression, we obtain the polynomial:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(2k-\omega_i^2 m)^2-k^2=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two roots of which are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_i^2 = \frac{k}{m},\frac{3k}{m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the two resonant frequencies are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_1 = \sqrt{\frac{k}{m}} \qquad \omega_2 = \sqrt{\frac{3k}{m}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you should obtain a number of (not necessarily distinct) eigenvalues equal to the number of rows or columns of your $\mathbf M$ or $\mathbf K$ matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain the eigenvector corresponding to a particular eigenvalue, substitute back into equation (4), to obtain a series of simultaneous equation, the solving of which should determine the eigenvector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our example, finding the eigenvector corresponding to the eigenvalue $\omega_1^2$, substitute $\omega_i=\omega_1$ into equation (4)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left[\begin{matrix}&#xA;        2k-k &amp;amp; -k\\&#xA;        -k &amp;amp; 2k-k\\&#xA;        \end{matrix}\right]&#xA;\left[\begin{matrix}&#xA;        u^{(1)}_1\\&#xA;        u^{(1)}_2\\&#xA;        \end{matrix}\right]&#xA;=&#xA;\left[\begin{matrix}&#xA;        k u^{(1)}_1-k u^{(1)}_2\\&#xA;        -k u^{(1)}_1+k u^{(1)}_2\\&#xA;        \end{matrix}\right]&#xA;=&#xA;\left[\begin{matrix}&#xA;        0\\&#xA;        0\\&#xA;        \end{matrix}\right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, we can see that $u^{(1)}_1 = u^{(1)}_2$, and so the eigenvector is therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf u^{(1)} = &#xA;\left[\begin{matrix}&#xA;        1\\&#xA;        1\\&#xA;        \end{matrix}\right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NUMERICAL COMPUTATION&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your problem does not seem practical to do by hand. Thankfully, there are a variety of different eigen-solver algorithms out there that are dedicated to solving this very problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of these include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Power Iteration method&lt;/li&gt;&#xA;&lt;li&gt;Lanczos method&lt;/li&gt;&#xA;&lt;li&gt;Krylov-Schur method&lt;/li&gt;&#xA;&lt;li&gt;Jacobi-Davidson method&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To explain how each of these algorithms work would take more time than I currently have, however I would definitely recommend tracking down these algorithms, reading how they work and using them to solver the generalised eigenvalue problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2017-03-09T19:48:52.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14172" PostTypeId="1" CreationDate="2017-03-09T20:16:16.743" Score="1" ViewCount="110" Body="&lt;p&gt;Ultrasonic transducer is used to detect flaws. What is the smallest flaw that can be detected by ultrasonic technology? One thing I know is that higher the frequency, higher is the sensitivity of the transducer i.e. smaller is the size of the flaw it can detect.&lt;/p&gt;&#xA;" OwnerUserId="10406" LastActivityDate="2017-03-09T20:16:16.743" Title="What is the smallest flaw that can be detected by ultrasonic technology?" Tags="|ultrasound|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14174" PostTypeId="2" ParentId="12925" CreationDate="2017-03-09T20:58:28.200" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=8Jl6MzjVbHI&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gem Products&lt;/a&gt; friction hinges use a sandwich of different metal washers, for example, phosphor bronze, stainless steel, and phosphor bronze. &lt;a href=&quot;https://www.google.com/patents/US20140090204&quot; rel=&quot;nofollow noreferrer&quot;&gt;Their patent&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The washers 135 of the present application can provide advantageous&#xA;  properties relative to conventional friction hinges. For example, the&#xA;  washers 135 may be made from phosphorus bronze, beryllium copper,&#xA;  brass, stainless steel and 17-4 PH hardened stainless steel, for&#xA;  example. The inventors of the present application discovered that the&#xA;  above materials are advantageous when used in washers for a friction&#xA;  hinge configuration like that shown in the figures. For example, the&#xA;  inventors discovered that a three washer configuration including a&#xA;  first washer made of phosphor bronze, a second washer made of&#xA;  stainless steel, and a third washer made of phosphor bronze,&#xA;  frictionally coupled together with advantageous wear properties. That&#xA;  is, the three washer configuration discussed above provides sufficient&#xA;  friction to ensure the desired angular position of the hinge 100 even&#xA;  after many rotations. Likewise, the inventors discovered that washers&#xA;  135 made of beryllium copper, brass, 17-4 PH hardened stainless steel&#xA;  or any combination of the above materials, are advantageous for wear&#xA;  properties.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9307" LastActivityDate="2017-03-09T20:58:28.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14175" PostTypeId="2" ParentId="3181" CreationDate="2017-03-09T23:04:02.120" Score="1" Body="&lt;p&gt;The photos are of exhaust fans.  They may be exhausting restrooms, or could be drawing from other spaces, such as janitor closets, storerooms, etc. The exhaust from these fans may have excessive moisture and/or odors (restrooms), or could contain dilute cleaning chemical fumes (janitor closets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are not venting the sanitary waste piping.  Plumbing vents will look like open pipes stubbed up through the roof.  Those connect to the sanitary sewer and can vent harmful gases (such as hydrogen sulfide)- they are required by the Uniform Plumbing Code and the International Plumbing Code to be located at least 10 feet horizontally from air intakes, doors, and windows (or 3 feet above).&lt;/p&gt;&#xA;" OwnerUserId="7763" LastActivityDate="2017-03-09T23:04:02.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14176" PostTypeId="2" ParentId="14165" CreationDate="2017-03-09T23:50:11.540" Score="3" Body="&lt;p&gt;The increase in strength in terms of moment of interia is small since the change in overall size is small.  This also doesn't address the different properties of the metal when it goes through the corrugated process.  When the corrugated metal is made it is cold-worked which takes that portion of the metal beyond its yield point.  This makes the bent portion of the metal stronger and more force is required to reach its new yield point.  So, when the corrugated sheet is made, it is the bending process that creates a new geometry that adds strength but you're also changing the material property of the metal making it stronger.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information on the cold-working principle look at stress-strain curves if you want to read more about it.  It's also called strain hardening.  &lt;/p&gt;&#xA;" OwnerUserId="10329" LastActivityDate="2017-03-09T23:50:11.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14178" PostTypeId="2" ParentId="13494" CreationDate="2017-03-10T09:30:12.407" Score="2" Body="&lt;p&gt;Your option 1, done &quot;according to specs&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two check valves and a pressure regulator with separate output and regulation input (these are usually joint; output pressure being the regulating/feedback pressure; you need a variant with these separate).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The regulator has regulation input open to air, throughput input in your circuit, output to the tank, through a check valve. Set to two bars, allows pressure in only when pressure in circuitry exceeds 2 bar, filling the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A separate, parallel connection through the second check valve depletes the tank whenever the pressure in the circuitry drops below the pressure in the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion this - while meeting the specs - is absolutely unnecessarily complex. What normally is used in your situation is a buffer tank:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option 2: Adjusted specs. Extra storage for CO2 filling when pressure rises, depleting when pressure drops, no hysteresis on filling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply connect the tank directly to the circuitry. No check valves, no extra parts. Fill it with CO2 at ambient pressure initially. As pressure grows, it accepts more gas, in proportion to the pressure. As gas is spent, the tank releases its storage. Simple, robust, no extra complexity or cost, and does its job roughly as well as the original design.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-03-10T09:30:12.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14179" PostTypeId="2" ParentId="13665" CreationDate="2017-03-10T10:23:34.770" Score="1" Body="&lt;p&gt;Composite is a very broad term in engineering - it's essentially any material that is made of two or more substances merged at macroscopic level - so not a mixture or alloy. This may be just a mix of larger grains, or it can be a complex structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One most common composite is reinforced concrete, where structure of rebar is embedded in concrete. Another is carbon fiber reinforced polymer (CFRP), where a texture/fabric of carbon fiber is embedded in a resin. Even plain concrete, even if rarely defined as such, is a composite - cement binder plus gravel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mechanisms, composite is usually the middle ground between plain polymer and metal. This may be metal-reinforced polymer (thin sheet metal gears embedded in plastic for stability), or it may be a CFRP cut-out, or a gear with metal axis mount and plastic teeth. Or it may be plain plastic with some waste organic fiber (cotton) added, increasing its strength insignificantly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to say without knowing the exact device in question, and while that may be something of performance quite comparable to metal, the marketing department can get away with the buzzword to peddle a total BS that way - any additive that won't dissolve in plastic will make the plastic &quot;a composite&quot; regardless of any performance gains - or losses.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-03-10T10:23:34.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14180" PostTypeId="1" AcceptedAnswerId="14181" CreationDate="2017-03-10T10:56:39.777" Score="1" ViewCount="89" Body="&lt;p&gt;I am finding different opinion, but I can not find a mathematical demonstration about this topic.&#xA;for example here is described as non homokinetic:&#xA;&lt;a href=&quot;https://engineering.stackexchange.com/questions/7898/oldham-coupling-vs-universal-joint/7935?newreg=9adda9c4d4144e99af298e1434ee12d2&quot;&gt;Oldham coupling vs Universal joint&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is homokinetic:&#xA;&lt;a href=&quot;https://books.google.it/books?id=X0AHKxwWTsYC&amp;amp;pg=PA120&amp;amp;lpg=PA120&amp;amp;dq=constant%20velocity%20oldham&amp;amp;source=bl&amp;amp;ots=KtiVJpXE_U&amp;amp;sig=5SYjn7HEgm2Yr5_mKvx1Y0H7i3Y&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ved=0ahUKEwiDtYCV4svSAhWKXBoKHWBfAzQ4ChDoAQg_MAY#v=onepage&amp;amp;q=constant%20velocity%20oldham&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Book&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10413" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-03-10T12:13:10.780" Title="Is the oldham coupling homokinetic?" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14181" PostTypeId="2" ParentId="14180" CreationDate="2017-03-10T12:13:10.780" Score="1" Body="&lt;p&gt;It's kinda trivial, the middle disk is homokinetically and rotationally coupled to the outer disks. Because the slot and tab in the connections are always parallel (not counting play in the fit).&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-03-10T12:13:10.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14182" PostTypeId="2" ParentId="14164" CreationDate="2017-03-10T13:07:40.790" Score="2" Body="&lt;p&gt;Sister keelson definitely does not translate to conventional marine usage. There are a couple of terms that apply to similar structural elements.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Strake. A longitudinal hull element that may be backed by an internal longitidinal structural element. In planing hulls the strake at the turn of the hard chine may be referred to as a lifting strake. Lifting strakes may also be fitted on the flat of bottom to improve hydrodynamic efficiency.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Which leads to the other possible term. &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Chine_(boating)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chine&lt;/a&gt;. In a multi chine vessel the hard angle between strakes is referred to as a chine. There will typically be a chine log or composite chine log backing the chine longitidinally.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is a reasonable comparative illustration &lt;a href=&quot;https://www.google.com.au/search?q=strakes%20on%20a%20boat&amp;amp;client=ms-android-samsung&amp;amp;sa=X&amp;amp;prmd=ivsn&amp;amp;tbm=isch&amp;amp;tbo=u&amp;amp;source=univ&amp;amp;fir=ITCNuPhUtqLjfM%253A%252CYvGOzp8PjEj-eM%252C_%253B3ziv3qLqc8OgzM%253A%252CtoQs-4812V7NFM%252C_%253BftJ94Ujy71allM%253A%252C3EhS_rE2FDYqtM%252C_%253Ba0Cpu-pY-NvReM%253A%252CvWfBJMOajLkyYM%252C_%253BLla2SrsZht-k2M%253A%252CeS5eMb-SYdn4VM%252C_%253By2HpFf6GcfqsCM%253A%252CnEW36Lbel-ALkM%252C_%253BSTg5Hs9ZasXUuM%253A%252CkDr7cxOdGcNgFM%252C_%253BnJsghWHEciCu-M%253A%252CSexDWV_dTQ8ZZM%252C_%253B4ld3NXuttglZxM%253A%252CBL47r_DZrW1jpM%252C_%253BwST5lZoHWdr6bM%253A%252Co1vxoqgWu_tAFM%252C_&amp;amp;usg=__oKx9GkfEpZ6QQ_WDPd3Z5tm2ydw%3D&amp;amp;ved=0ahUKEwiSuf-m_cvSAhVE3mMKHSpnDDUQsAQIGA&amp;amp;biw=360&amp;amp;bih=560#imgrc=nJsghWHEciCu-M:&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This usage is primarily restricted to planing hulls (there are no chines on faired hulls) and on a faired hull a strake is a longitudinal run of planking / plating extending from the garboard strake adjacent to the keel to the sheer strake at the hull/deck join.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I guess your seaplane float is a planing hull &amp;amp; so the terminology applies!&lt;/p&gt;&#xA;" OwnerUserId="10415" LastActivityDate="2017-03-10T13:07:40.790" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14183" PostTypeId="2" ParentId="13708" CreationDate="2017-03-10T15:08:03.517" Score="0" Body="&lt;p&gt;Per this article: &lt;a href=&quot;https://en.wikipedia.org/wiki/Working_fluid&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Working_fluid&lt;/a&gt; &lt;br&gt;&#xA;Work done by a fluid during the process $1 \to 2$ is given by the equation:&#xA;$$W = -\int\limits_{V_{1}}^{V_{2}} PdV$$&#xA;Unless pressure ($P$) varies linearly with volume, this integral can't be solved analytically.  So, people make an assumption, they either choose a constant pressure to multiply by the volume change:&#xA;$$W = -P\int\limits_{V_{1}}^{V_{2}} dV$$&#xA; or a constant volume to multiply by the pressure change:&#xA;$$ W = -V\int\limits_{P_{1}}^{P_{2}} dP$$&#xA;There's nothing so wrong with assuming that the process takes place at a pressure equal to the external pressure, if the pressure change is much smaller than the volume change, like it usually is when air escapes from a tank.  &lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-03-10T15:08:03.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14184" PostTypeId="1" CreationDate="2017-03-10T15:50:00.017" Score="2" ViewCount="2974" Body="&lt;p&gt;Draw the FBD of the entire frame, assuming that friction and the weights of the members are negligible. How many unknowns appear on this FBD?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cTavi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cTavi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I've tried so far. I have four unknowns. They are F&lt;sub&gt;1&lt;/sub&gt;, F&lt;sub&gt;2&lt;/sub&gt;, F&lt;sub&gt;3&lt;/sub&gt;, F&lt;sub&gt;4&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sQhIT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sQhIT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10416" LastEditorUserId="10416" LastEditDate="2017-03-11T01:29:01.750" LastActivityDate="2021-03-20T23:03:12.277" Title="Draw the FBD of the entire frame, assuming that friction and the weights of the members are negligible. How many unknowns appear on this FBD?" Tags="|statics|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14186" PostTypeId="1" CreationDate="2017-03-10T18:05:37.543" Score="1" ViewCount="715" Body="&lt;p&gt;In a textbook on process controls, I read that when tuning a controller there must be a balance between performance and excessive control element cycling. My question is, what corresponds excessive cycling?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To narrow down the question, assume we have a control valve with linear characteristics. What is the minimum time the controller should take to move the valve 5% of the controller output? For a typical set point change, how many cycles (up and down) of the valve are too much prior to reaching the final valve position?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect there is a process specific part of the answer (minimize disturbance, smooth operation criteria), but what I am after is: what is too much cycling to avoid damaging the valve itself or to prolong the valves lifetime?&lt;/p&gt;&#xA;" OwnerUserId="1031" LastActivityDate="2017-03-11T07:32:16.923" Title="How much is too much control valve cycling?" Tags="|control-engineering|valves|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14187" PostTypeId="2" ParentId="14186" CreationDate="2017-03-10T19:52:53.390" Score="1" Body="&lt;p&gt;Assuming you have correctly selected/sized the control valve and actuator assembly, you could expect valve motion to act within 10s of seconds (you could make it open slower by fiddling with the actuating mechanism). Cycling usually isn't a problem with respect to the valve lifetime (typically stated in the years timeframe) when the valve is properly sized, but it can become an issue if the valve's operating temperature or pressure is exceeded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any experience with valves that have internal packing, so I won't comment on them further than saying cycling may affect the lifetime of valve packing. It would be worth talking with a controls equipment company to get a more detailed understanding on this area.&lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2017-03-10T19:52:53.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14188" PostTypeId="2" ParentId="2478" CreationDate="2017-03-10T20:07:05.633" Score="0" Body="&lt;p&gt;Someone mentioned allclad plating with stainless steel. That is not quite correct. They take a pair of thin stainless steel pots, then position them to have a 1/8 inch gap, and pour aluminum into the void. Since the melting point of aluminum is so low compared to stainless steel, this works quite well, and the stainless pots do not deform.&lt;/p&gt;&#xA;" OwnerUserId="10420" LastActivityDate="2017-03-10T20:07:05.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14189" PostTypeId="1" CreationDate="2017-03-10T22:06:48.337" Score="1" ViewCount="30" Body="&lt;p&gt;I have a CNC that does not understand the G81 code for drilling. G-Simple looks to be configurable but I am finding it difficult to find information on it. It looks to have a &quot;USE&quot; command with a &quot;G81&quot; parameter but no idea how to make it not use this command and just drill a hole by constructing the movements itself. Just a simple CNC so not too bothered about the timing etc. &lt;/p&gt;&#xA;" OwnerUserId="10421" LastActivityDate="2017-03-10T22:06:48.337" Title="How do I configure g-simple so it does not use G81 code?" Tags="|machining|computer-aided-design|cnc|drilling|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14192" PostTypeId="1" CreationDate="2017-03-11T01:19:53.933" Score="5" ViewCount="6541" Body="&lt;p&gt;Many of us think FEA is used for structural problems and CFD is used for fluid dynamics problems. But I have read that the above statement is wrong. So please explain the exact usage of those two.Also please give the basic difference  between them.&lt;/p&gt;&#xA;" OwnerUserId="10401" LastEditorUserId="1586" LastEditDate="2017-03-13T21:55:23.967" LastActivityDate="2019-05-22T20:05:49.673" Title="What is the difference between FEA and CFD?" Tags="|finite-element-method|cfd|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14193" PostTypeId="2" ParentId="14192" CreationDate="2017-03-11T02:29:51.550" Score="19" Body="&lt;p&gt;CFD (computational fluid dynamics) includes any numerical method used to solve fluid flow problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FEA (finite element analysis) is one numerical method for solving partial differential equations, independent of what the equations are modelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that FEA is the most popular method for solving computational mechanics problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several different approaches used to solve CFD problems, one of which is FEA - though the CFD community often describe FEA as an &quot;unstructured grid method&quot; instead of &quot;FEA&quot;, in contrast to the regular &quot;structured grids&quot; used in finite &lt;em&gt;difference&lt;/em&gt; solution methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many partial differential equations can be classified as &quot;elliptic&quot;, &quot;parabolic&quot;, or &quot;hyperbolic&quot; depending on how the solution propagates to different parts of the domain over time. Most structural analysis problems are elliptic, though problems involving time-dependent heat transfer are parabolic. Even for differential equations which strictly speaking are hyperbolic, if the speed of sound in the material is large compared with the speed of the response of the structure, it is often useful to use an approximate parabolic or elliptic equation instead (for example by assuming the speed of sound is infinite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions of differential equations for fluid flow often don't fit into this neat mathematical classification, and have a mixture of elliptic and hyperbolic behaviour in different parts of the domain (and the boundary between the different types of behaviour is unknown before the solution has been found). So, even when finite element methods are used in CFD, much of the detail of the numerical algorithms is very different from the FEA methods used for structural analysis problems.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-03-11T02:29:51.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14195" PostTypeId="2" ParentId="14186" CreationDate="2017-03-11T07:32:16.923" Score="1" Body="&lt;p&gt;I would assume that most of the time a feedback gain is limited by how big of an input you can give, so when it starts to saturate. You would like to avoid is breaking the system, for example if a too big of a force is applied, or the system might overheat if too much energy is supplied in a too short of a time frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your sensor sampling rate also limits the fastest possible rate at which you can control your system. So applying a feedback gain which would result is an analytically faster response (when modeling it as an continues system) would also not improve the performance.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2017-03-11T07:32:16.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14196" PostTypeId="1" CreationDate="2017-03-11T11:59:36.167" Score="6" ViewCount="635" Body="&lt;p&gt;Regarding crosshead bearing failure, I have read that at slow speed it is difficult to establish a lubricating film because there is an imbalance between forces due to combustion and forces due to inertia. This information is from a maker of big low speed two stroke engines used for the propulsion of ships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't quite understand the explanation. How are these forces imbalanced and how does this contribute to eventual failure of the bearing?&lt;/p&gt;&#xA;" OwnerUserId="5825" LastEditorUserId="368" LastEditDate="2017-03-16T16:08:24.657" LastActivityDate="2017-03-20T22:37:50.157" Title="What makes crosshead bearings hard to lubricate?" Tags="|mechanical-engineering|bearings|engines|tribology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14198" PostTypeId="1" CreationDate="2017-03-11T19:13:13.790" Score="1" ViewCount="135" Body="&lt;p&gt;A patent that I read that stated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;  Coating tungsten with nickel then with aluminum, will stop oxidation in open air environments. And even pass the melting points of both the tungsten and nickel, the metals doesn't seem to melt&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will this hold up in a real engineering application, even if it goes past the melting points of nickel and aluminum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heres the patent: &#xA;&lt;a href=&quot;https://www.google.com/patents/US2682101&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/patents/US2682101&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9468" LastEditorUserId="90" LastEditDate="2017-03-12T03:02:27.573" LastActivityDate="2017-03-12T03:44:36.610" Title="Can coating tungsten to stop oxidation?" Tags="|metals|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14200" PostTypeId="1" AcceptedAnswerId="14204" CreationDate="2017-03-11T19:38:19.000" Score="0" ViewCount="111" Body="&lt;p&gt;I read that a human&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.co.uk/search?client=ms-android-samsung&amp;amp;ei=l0_EWNjsHKXEgAafuIUg&amp;amp;q=john+stapp+45+g&amp;amp;oq=john+stapp+45&amp;amp;gs_l=mobile-gws-serp.1.0.0.6460.9902.0.10829.15.14.1.8.8.0.185.2085.0j13.13.0....0...1c.1j4.64.mobile-gws-serp..1.14.1112.3..35i39k1j0i67k1j0i10k1.tC-NnmcrLLw&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.co.uk/search?client=ms-android-samsung&amp;amp;ei=l0_EWNjsHKXEgAafuIUg&amp;amp;q=john+stapp+45+g&amp;amp;oq=john+stapp+45&amp;amp;gs_l=mobile-gws-serp.1.0.0.6460.9902.0.10829.15.14.1.8.8.0.185.2085.0j13.13.0....0...1c.1j4.64.mobile-gws-serp..1.14.1112.3..35i39k1j0i67k1j0i10k1.tC-NnmcrLLw&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can survive 46g force so what type of rocket and how much fuel would be best to stop a human falling at terminal velocity without exceeding this g force limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also read spacex use rockets not parashutes for the weight advantage so would a short (i think 1 second) burst rocket be better in anyway to a parashute (despite such deceleration causing injury(yet non permanent)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would the rocket and fuel be far too big.&lt;/p&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="10361" LastEditDate="2017-04-11T21:57:08.263" LastActivityDate="2017-05-11T22:04:12.890" Title="Rocket to slow freefall" Tags="|aerospace-engineering|rocketry|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14202" PostTypeId="2" ParentId="14200" CreationDate="2017-03-11T20:13:44.840" Score="0" Body="&lt;p&gt;Your question does not provide a definition of &lt;em&gt;better&lt;/em&gt;, so I'll try to answer as generally as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parachutes have a considerable advantage over a rocket in that they are passive once deployed and self-stabilize, whereas a rocket needs a fairly sophisticated control system to point it the right way (and remain so) during the burn. Additionally, pilots can only endure high accelerations like those involved with rocket engines thanks to flight suits, training and careful orientation to minimize the effects, which may not be feasible in an emergency scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the only criteria under which a rocket engine could outperform a parachute as a means of slowing a human from free-fall to a safe speed is the vertical space required for the deceleration to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your SpaceX example, they already have a rocket engine, fuel tank and control system, thus if they can make those elements continue to perform after the launch to decelerate the rocket, it makes sense to drop the parachute system, especially since they want precise landings to reduce recovery costs (no fishing the boosters out of the ocean like with the shuttle missions). Hard to say if the added fuel weight and complexity would be worthwhile in any other scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your fuel question, you'd need to apply Tsiolkovsky's equation for a desired $\Delta V$ of 53m/s (if you mean atmospheric freefall), but more details would be needed to give an actual number, like what type of rocket engine you envision (to estimate an Isp) and how much can we assume your human+engine system to weight. Also, I'm pretty sure you've got a typo&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2017-03-11T20:13:44.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14203" PostTypeId="1" AcceptedAnswerId="14224" CreationDate="2017-03-11T20:25:35.910" Score="0" ViewCount="2666" Body="&lt;p&gt;For a cantilever beam of circular cross section, loaded at the end, how does the deflection of the free end vary as a function of beam length and diameter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect the answer will be something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{L^2}{D^3}$$&lt;/p&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="255" LastEditDate="2017-03-13T17:06:39.957" LastActivityDate="2017-03-13T17:06:39.957" Title="What is the formula for deflection of an end-loaded cantilever rod beam?" Tags="|beam|deflection|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14204" PostTypeId="2" ParentId="14200" CreationDate="2017-03-11T21:47:53.430" Score="0" Body="&lt;p&gt;There are human portable &lt;a href=&quot;https://en.wikipedia.org/wiki/Jet_pack&quot; rel=&quot;nofollow noreferrer&quot;&gt;rocket packs&lt;/a&gt; which do work, albeit to a fairly limited extent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum terminal velocity in a stable 'belly to earth' position is about 50 metres per second so even at a fairly modest 10 g that is only about 5 seconds to slow down to a stop. So it's certainly not impossible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However parachutes work pretty well at what they do. They are very simple and reliable and light and compact enough that they don't encumber the wearer much and you can carry a second one as a reserve which further improves their reliability. They don't require much maintenance beyond proper storage and for a safety-critical item they aren't really that expensive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally importantly they are pretty easy to use. If you jumped out of a plane wearing one you would have a pretty good chance of surviving, even with minimal instruction. Obviously a moderate amount of training is desirable for dealing with emergencies and reducing the risk of injury on landing but even so there are a significant number of competent amateur parachutists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also once a parachute has deployed successfully there is little left to go mechanically wrong &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand rocket packs require significant skill to fly at all they usually launch from a standing position and the pilot stays upright throughout the flight whereas free-fall parachutes are usually deployed from a belly down spreadeagled position which is inherently stable and works nicely with a backpack type parachute harness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a limited fuel supply you need to be able to time the deployment of thrust pretty accurately so you have a fairly narrow window to get it right, as opposed to a parachute where any altitude above the minimum safe deployment is basically fine (obviously jumping from very high altitudes has a whole separate set of problems). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally a rocket pack is mechanically complex, heavy and bulky and most rockets use fuel which is hazardous and requires special handling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really you are talking about replacing a simple, inexpensive and proven system with something which has never really got beyond the prototype stage, doesn't offer any obvious advantages and has several inherent drawbacks. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-03-11T21:47:53.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14210" PostTypeId="1" CreationDate="2017-03-12T11:54:44.777" Score="0" ViewCount="2771" Body="&lt;p&gt;While learning about normal and tangential stress, we had a task, where I had to calculate the cross sectional area of a beam that could withstand the force applied to it, if the beam was sliced in the middle on an angle (Like two 90 deg trapezoids glued together).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes sense.. having a language barrier here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was given the maximum allowed normal and tangential stresses under which it would not break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know, Maximum tangential stress = 0.5 maximum normal stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was given two values. For example, the max normal stress was 7.6 MPa and the tangential stress was 4.1 MPa and according to the rule, if tangential stress is 4.1 then the max normal stress can be 8.2 MPa. So the calculations are done based on the lowest value, since it can't hold any higher...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if I'm only basing on one of the values and they're both connected to the equation, why does the task give me both? Is it just for the task, or in real life there are there materials that somehow have specifications for the tangential stress and normal stress boarder values which are different when using the equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope any of this makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question basically is-  &lt;strong&gt;How can the same material have two different values of stress it can withstand?&lt;/strong&gt; Is it just for the task or..?&lt;/p&gt;&#xA;" OwnerUserId="10438" LastEditorUserId="1832" LastEditDate="2017-03-15T16:35:37.273" LastActivityDate="2017-07-23T18:16:11.427" Title="Normal and tangential stresses" Tags="|structural-engineering|civil-engineering|stresses|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14213" PostTypeId="1" CreationDate="2017-03-12T14:22:38.280" Score="1" ViewCount="442" Body="&lt;p&gt;I bought an old turning lathe ( &lt;strong&gt;Żebrak TOS S32&lt;/strong&gt;) but there is problem with the &lt;strong&gt;o-ring&lt;/strong&gt; of it's &lt;strong&gt;spindle&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have opened the lathe and don't know how to take out the spindle shaft and replace the damaged o-ring with a new o-ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following pictures show the part in question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tkAUx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tkAUx.jpg&quot; alt=&quot; Żebrak TOS S32&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as you can see from the pictures, how could I separate the spindle shaft from it's gear wheel to remove the  spindle shaft and change it's o-ring?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manual for the lathe can be found here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tumlikovo.cz/wp-content/uploads/2010/manualy/manS32.zip&quot; rel=&quot;nofollow noreferrer&quot;&gt;Żebrak TOS S32 manual&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.4shared.com/zip/p8PFn9Dtca/manS32.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Żebrak TOS S32 manual in 4shared&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, I also found this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.4shared.com/zip/p8PFn9Dtca/manS32.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DEFoF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2678" LastEditorUserId="90" LastEditDate="2017-03-13T01:59:27.030" LastActivityDate="2024-03-12T17:03:10.480" Title="How can I change a turning lathe's spindle o-ring?" Tags="|lathe|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14216" PostTypeId="2" ParentId="11491" CreationDate="2017-03-12T17:34:31.030" Score="0" Body="&lt;p&gt;(a)  Head lost in friction $H_f = 4fLv^2/ D2g$, &#xA;where L = Length of hose, D = Hose diameter, v = hose velocity and f = friction coefficient&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If H = Head at inlet to pump and hoses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Head at nozzle $H_p = H - 4fLv^2/ D2g$&lt;br&gt;&#xA;or $H_p = Head at inlet to hoses - frictional head loss in each hose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If jet velocity = V and $C_v = 0.97$&#xA;Jet Velocity $V = C_v \sqrt {2g * Hp}$, where $V = 24m/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$H_p = V^2/ {2g * C_v^2}$ = (24)^2 / $2g * (0.97)^2, so Hp = 31.208 m &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For continuity of flow, flow from hoses = flow from nozzle, so &#xA;1/4 Pi*D^2*v = 1/4 Pi*d^2*V, where d = jet diameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v = V (d^2/ D^2)$ = 24 (0.0375)^2 / D^2 = $v = 0.03375/ D^2$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have then put this in the frictional head loss formula &#xA;$4fLV^2/ D2g$ = 4*0.007*45*(0.03375D^-2)^2 / (D * 19.62) = $1.43521875*10^-3*D^-4/ D * 19.62$  = $1.43521875 * 10^-3/ (D^5 * 19.62)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note $2g = 19.62$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If power lost in overcoming friction in hoses is not to exceed 20 per cent of hydraulic power available at inlet end of hoses,&#xA;then H = 5 * 4fLv^2/ D2g&lt;br&gt;&#xA;so $H_p = 5 * (4fLv^2/ D2g)$ - (4fLv^2/ D2g) = 4 * (4fLv^2/ D2g) = 4*1.43521875 * 10^-3/ (D^5 * 19.62)&#xA;Hp = 5.740875*10^-3/ (D^5*2g) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 2g * Hp = 5.740875*10^-3/ D^5 = 19.62*31.2018 = 612.1798m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D^5 = 5.740875*10^-3 / 612.179, so D = (5.740875*10^-3/612.1798)^0.2 = 0.09872m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so Hose Diameter = 98.72 mm &#xA;IN MY TEXT BOOK ANSWER WAS GIVEN AS 98.5mm, WOULD THIS BE DUE TO ROUNDING?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Power from pump P = wQHp, where w = unit weight of water, Q = flow through pipes and Hp = Head from pump&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to use Bernoulli equation with pump head term in it as well as head loss&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z1 + P1/pg + V1^2/2g + Hp = Z2 + P2/pg + V2^2/2g + Hf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this formula we assume that P1 ands P2 are atmospheric pressures at entry to the pump and exit of the nozzle, so P1 = P2 = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z1 = 0 and Z2 = 3m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V1 = velocity in hoses and V2 = jet velocity from nozzle = 24m/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to calculate velocity in hoses and then frictional head loss &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From before $v = V (d^2/ D^2)$&#xA;so v = 24 * (0.0375^2)/ (0.09872^2) = 3.463m/s&#xA;so V1 = 3.463m/s &#xA;Now we have to put V1 into frictional head loss formula $H_f = 4fLv^2/ D2g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so Hf = 4*0.007*45*(3.463)^2 / (0.09872 * 19.62) = 7.80m &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to put values in Bernoulli equation to find head from pump&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so using Z1 + P1/pg + V1^2/2g + Hp = Z2 + P2/pg + V2^2/2g + Hf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 + 0 + (3.463^2)/2g + Hp = 3 + 0 + (24^2)/2g + 7.8&#xA;Hp = 3 + (24^2)/2g + 7.8 - (3.463^2)/2g = 3 + 29.3578 + 7.8 - 0.611&#xA;Hp = 39.5468m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow from one hose Q = 1/4 Pi*D^2*v = 1/4 Pi*0.09872^2* 3.463 = 0.026507m^3/s    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power from one hose P = wQHp = 9810 * 0.026507 * 39.5468 = 10284W = 10.284kW&#xA;as there are 2 hoses, total Power from hoses P = 10.284 * 2 = 20.567kW&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as the pump has an efficiency of 70 per cent &#xA;then Power from pump = 20.567/ 0.7 = 29.382kW&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IN MY TEXT BOOK THE ANSWER WAS GIVEN AS 31.1kW, IS THERE SOMETHING ELSE I SHOULD TAKE INTO CONSIDERATION? &lt;/p&gt;&#xA;" OwnerUserId="8013" LastEditorUserId="8013" LastEditDate="2017-03-12T17:47:18.493" LastActivityDate="2017-03-12T17:47:18.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14218" PostTypeId="1" CreationDate="2017-03-13T00:10:06.397" Score="1" ViewCount="140" Body="&lt;p&gt;I have a system in which there is a screw with an electrical contact on it. When the screw is in all the way, one electrical connection is made. When it is out all the way, another connection is made. If it is in-between, no connection exists and my circuit doesn't work. I can adjust the endpoints which determine what is &amp;quot;in&amp;quot; and what is &amp;quot;out&amp;quot;. The piece is from a garage opener that I am trying to repurpose. Here is a picture of the system.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TWYLa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TWYLa.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I &lt;em&gt;can&lt;/em&gt; control my unit with this setup, however I would like to use a light switch, since turning a dial really isn't as easy as flipping a switch. The lightswitch, unfortunately toggles between on and off. I would like to control between connections 1 and 2 instead, since another switch gives power to the unit. Is there a way I can do this or am I just going about this entirely the wrong way?&lt;/p&gt;&#xA;" OwnerUserId="3806" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-03-13T02:04:36.623" Title="Converting a light switch to a Boolean toggle switch" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14219" PostTypeId="2" ParentId="14218" CreationDate="2017-03-13T02:04:36.623" Score="2" Body="&lt;p&gt;The light switch is a single-throw switch; you need a double-throw switch for what you're doing. Most electronics suppliers should have one for the voltage and current you need.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2017-03-13T02:04:36.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14220" PostTypeId="2" ParentId="14184" CreationDate="2017-03-13T03:31:01.817" Score="0" Body="&lt;p&gt;General planar joints have 3 reactions: two force reactions (one for each independent direction) and one moment reaction. Pin joints are free to rotate in-plane, which means the moment reaction is null and there are only two reactions. Pin slots are free to rotate like a pin joint, and free to slide along only one direction. Consider how many reactions a pin slot would have. Each joint thus has a number of reactions depending on its type. From that you can figure out the total number of reactions each member has individually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But consider: where each member is joined to another member, the reactions of one member are counteracted by an equal-but-opposite reaction in the other member. If this weren't true, the members would not be static because they would be moving, due to the imbalance of forces. Recall that a net force causes acceleration. Since the free-body diagram requested is for the frame as a whole unit, consider which joints attach the frame to something that isn't part of the frame. Consider what type of joints those are. Consider how many reactions each of those joints has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you've answered those questions, drawing the free-body diagram should be simple, and should appear similarly to how you've drawn it, perhaps with some of the force arrows drawn differently. I would say that two of the ones you've drawn are correct, and two are incorrect.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2017-03-13T03:31:01.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14221" PostTypeId="1" CreationDate="2017-03-13T08:37:53.020" Score="1" ViewCount="85" Body="&lt;p&gt;I need to measure the level of the liquid in a tank. The tank is opened and closed frequently. Measurements are made in milliseconds. I tried to measure the level of the liquid by using an ultrasonic sensor, but the measurement was not accurate enough. I need to measure accurate to one milliliter. Measurements are made from the top of the tank. Can anyone suggest some sensor that will accurately measure the liquid level?&lt;/p&gt;&#xA;" OwnerUserId="10441" LastEditorUserId="421" LastEditDate="2017-03-13T21:53:50.060" LastActivityDate="2017-03-14T11:07:54.260" Title="Precision measurement of liquid level in a tank which is continuously opened and closed" Tags="|sensors|distance-measurement|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14222" PostTypeId="2" ParentId="14184" CreationDate="2017-03-13T09:17:39.153" Score="3" Body="&lt;p&gt;A free body diagram has typically &lt;strong&gt;each&lt;/strong&gt; body drawn free of each other*. So in you case 3 separate free body diagrams is a good start.  In this specific case you must do so, as otherwise you do not capture the joints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a common mistake not to draw the free body diagrams, in fact many students do not seem to know what a free body diagram is. The purpose of a free body diagram is to enable you to enumerate all forces in a systematic way, yes each body is separate form each other. Sometimes you see people claiming that its tedious to make the free body diagrams, but doing so makes this child's play. So the free body diagrams are (in reading order left to right top down):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2yTSo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2yTSo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QLatX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QLatX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O6jHq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O6jHq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inages 1-3&lt;/strong&gt; The three bodies and their reaction forces (free body diagrams)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The direction of force arrows is a pure guess (if I am worng i just get negative results). The meaningful thing to note is that the equivalent force on the opposing member of a link is reverse in direction. I have color coded the forces so that it is possible to see what force is the same as the other. You would probably need to name the forces somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that many of these forces are 0 like the cyan sideways force on body 2-3 interface, due to the need of each part to be in static equilibrium. Which in turn means the yellow arrow is also 0 and so on... From these diagrams its easy to see that body 3 is Statically determined. Now all you need to do is solve the equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA; \sum F = 0 \tag{1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA; \sum M = 0 \tag{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for each member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* One might split a body too but in this case its not really very constructive.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-03-13T09:35:01.160" LastActivityDate="2017-03-13T09:35:01.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14224" PostTypeId="2" ParentId="14203" CreationDate="2017-03-13T11:01:51.787" Score="2" Body="&lt;p&gt;For a cantilever with a point load at the free end, the deflection at that free end is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{WL^3}{3EI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Handbook of Structural Steelwork - 3rd edition&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$W$ = point load&lt;/li&gt;&#xA;&lt;li&gt;$L$ = beam length&lt;/li&gt;&#xA;&lt;li&gt;$E$ = Young's modulus&lt;/li&gt;&#xA;&lt;li&gt;$I$ = second moment of area&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Young's modulus depends on the material, see &lt;a href=&quot;https://en.wikipedia.org/wiki/Young&amp;#39;s_modulus#Approximate_values&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; for various examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second moment of area for a circle is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{\pi r^4}{4} = \dfrac{\pi d^4}{64}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your overall equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{64WL^3}{3\pi Ed^4}$$&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="1832" LastEditDate="2017-03-13T11:46:52.540" LastActivityDate="2017-03-13T11:46:52.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14226" PostTypeId="1" AcceptedAnswerId="14233" CreationDate="2017-03-13T16:53:57.200" Score="1" ViewCount="281" Body="&lt;p&gt;So I got a new belt, and it was the right length but the wrong thickness. I needed a 3/8 belt but it was a 1/2. I did go to get it but the guy looked it up and told me I needed this one. So I got it. But when I installed, as I figured, there was a small issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the belt is running up on the &quot;brake&quot; as I call it. It is along the idler and when not engaged this flat metal holds it in place, so the belt does not run when I do not want it to I guess. But the belt is so thick that it is rubbing against this metal when engaged, when the metal is far from the pulley (I think it is called). So it has been burning the belt from friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is my thought to solving...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let it burn. I am sure it wouldn't take time to burn down to the right thickness. My only problem is that the 'strings' could show. If this is a good idea in general. &lt;/li&gt;&#xA;&lt;li&gt;Cut that metal off. All I see it doing is keeping the belt from spinning when not engaged. But the belt would be loose were I do not think that would be a problem to begin with. So unless it does have another purpose than to just keep the belt from moving. Couldn't I just cut that off? Then it would not contact the belt and problem solved. Thing I do not like on this is when I do get the proper belt will it be needed then. Side thought: Does this keep the belt from falling off?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;2nd Update: Another thought is, since the belt is too thick, how could I thin it while it is running? I would have to remove a whole 1/8 inch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not want to return the belt, it was only $7 anyway. So I hope this was understandable and there is a answer to this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I again, am not sure were to ask a question like this so I wold hope this is fine.&lt;/p&gt;&#xA;" OwnerUserId="8540" LastEditorUserId="8055" LastEditDate="2017-03-13T21:55:19.083" LastActivityDate="2017-03-14T08:08:54.027" Title="Belt is to thick for my snowblower, how to thin or leave?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14227" PostTypeId="1" AcceptedAnswerId="15223" CreationDate="2017-03-13T17:32:27.717" Score="2" ViewCount="2203" Body="&lt;p&gt;I have mounted a piezo-ceramic ultrasonic transducer (MODEL # SMD10T2R111WL, Vendor Website: &lt;a href=&quot;https://www.steminc.com/PZT/en/piezo-ceramic-disc-10x2mm-r-215-khz-wire-leads-smd10t2r111wl&quot; rel=&quot;nofollow noreferrer&quot;&gt;PIEZO CERAMIC DISC 10X2MM R 215 KHZ WIRE LEADS&lt;/a&gt;) under a plastic tank using epoxy. Please see the below picture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FlGC6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FlGC6.jpg&quot; alt=&quot;[1]: https://i.stack.imgur.com/sN7Sj.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to confirm that the wire did  not get detached from the transducer while mounting. Is there a way to test it. The two wires that are soldered to the transducer are not connected so there is no way to check the continuity of the wires using multi meter short circuit test. &lt;/p&gt;&#xA;" OwnerUserId="10379" LastEditorUserId="282" LastEditDate="2017-05-26T02:13:26.987" LastActivityDate="2020-08-01T22:33:41.757" Title="How to test connectivity between wire and piezo-ceramic transducer" Tags="|electrical-engineering|sensors|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14229" PostTypeId="2" ParentId="14221" CreationDate="2017-03-13T22:35:31.600" Score="1" Body="&lt;p&gt;Maybe you could use one of the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA; &lt;li&gt;level glass&#xA; &lt;ul&gt;&#xA;  &lt;li&gt;allows you to read the height of the liquid outside the tank.&lt;/li&gt;&#xA;  &lt;li&gt; The width of the tube would have to be large enough for capillary action to be neglectable &lt;/li&gt;&#xA;  &lt;li&gt;If there is a lot of movement inside the tank the water level in the tube might oscillate as well.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA; &lt;/li&gt;&#xA; &lt;li&gt;Electronic hydrostatic manometer&#xA;  &lt;ul&gt;&#xA;   &lt;li&gt;Attached to the bottom of the tank you could easily read the pressure and calculate the height of the liquid&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="8632" LastEditDate="2017-03-14T11:07:54.260" LastActivityDate="2017-03-14T11:07:54.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14231" PostTypeId="2" ParentId="14210" CreationDate="2017-03-13T23:00:59.623" Score="1" Body="&lt;p&gt;I'm not quite sure what you mean by tangential stress. If you mean shear stress, then &lt;em&gt;shear strength&lt;/em&gt; does not necessarily have to be $0.5\cdot yield strength$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between &lt;em&gt;ultimate tensile strength&lt;/em&gt; and &lt;em&gt;ultimate shear strength&lt;/em&gt; depend on what kind of material is used.&#xA;A rule of thumb is, that ductile materials fail due to shear and brittle materials due to tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; you can find a table with some values for shear/yield relationships.&#xA;E.g. for cast iron the ultimate shear strength is about 1.3 times the ultimate yield strength, for steels only about 0.75.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2017-03-13T23:00:59.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14232" PostTypeId="2" ParentId="14122" CreationDate="2017-03-14T02:37:38.573" Score="1" Body="&lt;p&gt;No expert, but the problem is interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Generate $6.5V_{RMS}$ at 100 rpm with 9 stator coils.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$V_{MAX} = 9.2V$&lt;/li&gt;&#xA;&lt;li&gt;100rpm = 1.67rps&lt;/li&gt;&#xA;&lt;li&gt;if radius to center of magnets/coils = 0.1m&lt;/li&gt;&#xA;&lt;li&gt;$C = 2 \pi r = 2 \pi \times 0.1m = 0.628m$&lt;/li&gt;&#xA;&lt;li&gt;at 1.67 rps, v = 1.05m/s&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;N48 Neodymium Bar Magnets 1 in x 1/2 in x 1/4 in&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;L = 0.0254m, W = 0.0127m, D = 0.00635m.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Separation z = 0.00635m.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Biggest problem is separation between magnets and windings.  Closer they are the better.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From: &lt;a href=&quot;https://www.supermagnete.de/eng/faq/How-do-you-calculate-the-magnetic-flux-density&quot; rel=&quot;nofollow noreferrer&quot;&gt;How do you calculate the magnetic flux density?&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.supermagnete.de/eng/faq/How-do-you-calculate-the-magnetic-flux-density&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KMaqh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From: &lt;a href=&quot;http://www.cy-magnetics.com/Mag_Pros_NdFeB.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magnetic Properties of Sintered NdFeB Magnets&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N48 has a Remanence field of $B_r$ = 1.48T, which gives a flux density B = 0.164T, when you process the above equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{Ind} = N B l v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N = \frac {V_{Ind}} {B l v} = \frac {9.2V} {0.164T \times 0.0254m \times 1.05m/s} = 2108 turns$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that assumes you have one coil.  Divide that by 3 (in series) gives 703 turns/coil (which is a lot).  This coil has to be within length of magnet of 0.0254m.  Whole area of coil has to fit within area of magnet for greatest effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The math is a bit distorted, but it should give you a ballpark to do your initial calculations.  I'd guess, by the second or third prototype, you will have what you want.  Not all three coils in series will generate maximum voltage at the same time, so some experimentation will have to come into play (i.e. more turns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realize that the coil will not experience a constant magnetic field.  The further away windings are from the magnet the less the voltage induced.  As already stated, biggest problem is air gap between windings and magnets.  Air is a poor conductor of flux.  I'd use a steel plate to attach your magnets.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2017-03-14T02:37:38.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14233" PostTypeId="2" ParentId="14226" CreationDate="2017-03-14T08:08:54.027" Score="3" Body="&lt;p&gt;A big item to note is a belt is not just a piece of rubber.  In fact, typical v-belts and other engineered belting systems have steel wires running through them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H7pGE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H7pGE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These steel wires do all the real work, but the steel has a low coefficient of friction.  It would take a lot of area of wire allow the belt to operate with just the wire alone, without welding the wire in place.  The rubber has a high coefficient of friction, and allows the sheaves to grip the belt and let the steel do the work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line is, don't modify a belt because you could damage the wires inside.  That's why they're sized differently instead of simply being modifiable.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-03-14T08:08:54.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14235" PostTypeId="1" CreationDate="2017-03-14T10:32:18.883" Score="0" ViewCount="3814" Body="&lt;p&gt;Given the spring-damper system below where $x_i$, $x_i$, and $y_i$ are position values, how do I find the transfer function $\frac{X_o(s)}{X_i(s)}$? Wouldn't it require mass? I tried neglecting the masses in the equation but it zeros out the acceleration term and does not work. Any diea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hfIxy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfIxy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10457" LastActivityDate="2020-09-04T15:50:44.180" Title="Transfer Function of Spring-Damper System" Tags="|control-engineering|control-theory|springs|transfer-function|stability|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14236" PostTypeId="1" AcceptedAnswerId="15321" CreationDate="2017-03-14T10:35:12.160" Score="0" ViewCount="2287" Body="&lt;p&gt;I am struggling with the ASME B16.5 code and the specified stud-bolt length that is taken into account in the tables (7, 10, 13, 15, 17, 19 and 21).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you calculate the stud-bolt length according the &quot;guidelines for stud-bolt calculation&quot; for a NPS 24 Class 150 welding neck flange, the stud-bolt length should be &lt;strong&gt;7.45in&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I look in the ASME B16.5 Table7 the length is &lt;strong&gt;6.75in&lt;/strong&gt; this is &lt;strong&gt;0.7in&lt;/strong&gt; difference with the ASME B16.5.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rL8wX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rL8wX.png&quot; alt=&quot;Studbolt length image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What are the required number of free threads? As far as I am aware from bolt size 1in and up the number of threads per inch is the same and size &lt;code&gt;S&lt;/code&gt; is fixed. If you keep calculating with &lt;code&gt;S = 1/3 bolt diameter&lt;/code&gt; the number of free threads will only increase with no use?&lt;/p&gt;&#xA;" OwnerUserId="8464" LastActivityDate="2023-10-31T17:46:20.293" Title="Flanged connection studbolt length calculation" Tags="|mechanical-engineering|bolting|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14239" PostTypeId="1" CreationDate="2017-03-14T13:11:16.293" Score="0" ViewCount="559" Body="&lt;p&gt;Can we use kevlar, a composite material in leaf springs.  I mean does it have the properties which are essential for a leaf spring (like resistance to bending, elasticity)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if it has, are(properties)  they sufficient in case of heavy vehicle leaf springs? &lt;/p&gt;&#xA;" OwnerUserId="10401" LastActivityDate="2017-03-15T01:34:41.400" Title="Does kevlar (composite materials ) have elasticity?" Tags="|springs|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14240" PostTypeId="1" CreationDate="2017-03-14T13:51:02.917" Score="0" ViewCount="541" Body="&lt;p&gt;Can an event on the critical path have non zero slack time in critical path method?&lt;/p&gt;&#xA;" OwnerUserId="10319" LastActivityDate="2017-03-15T09:30:30.947" Title="Slack time in critical path method" Tags="|civil-engineering|construction-management|project-management|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14245" PostTypeId="2" ParentId="14240" CreationDate="2017-03-14T16:06:57.203" Score="2" Body="&lt;p&gt;The critical path is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;the longest path of planned activities [...] to the end of the project&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Critical_path_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In what I would call &quot;purely theoretical&quot; Critical Path Analysis, the length of the project is the length of the critical path. Therefore tasks on the critical path have no slack by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the real world there may be external dates setting deadlines. Therefore although the critical path might take 15 months, if the deadline is not for 16 months, there would be 1 month of slack to tasks on the critical path. Similarly, if the deadline is in 14 months there is negative slack on the critical path.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2017-03-15T09:30:30.947" LastActivityDate="2017-03-15T09:30:30.947" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14248" PostTypeId="1" CreationDate="2017-03-14T17:12:49.950" Score="1" ViewCount="803" Body="&lt;p&gt;I am trying to use the Curve Number Method to calculate the peak flow over an area. The precipitation is provided in a table, with the first column being the time (hr), second column being the incremental rainfall at each time (inch.), and the third being the cumulative rainfall (inch).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am calculating the peak flow, should I just go with the highest incremental rainfall during the period (which is 0.75 inch at 1 hr), or the highest cumulative rainfall (which is 2.25 inch.)?&lt;/p&gt;&#xA;" OwnerUserId="10463" LastActivityDate="2022-11-11T15:55:03.057" Title="Peak flow calculation using rainfall data" Tags="|hydrology|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14250" PostTypeId="2" ParentId="14102" CreationDate="2017-03-14T17:43:08.767" Score="0" Body="&lt;p&gt;okay heres a thought. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have had similar thoughts about sign boards and their life.&#xA;We know what happens when there is a cantilever beam. There is more stress at the ends (SFD)&#xA;So think of the deflection to be maximum at the ends with respect to the original reference or starting point of the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we keep a curved pipe, the deflection with respect to the starting point will be lesser than the first case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The life and the amount of stress the pipe can handle would be more by using the curved pipe&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of this example of a camber in rollers, we change the shape a bit to accommodate the pressure the rollers face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qPX3E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qPX3E.png&quot; alt=&quot;Roll Bending&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6215" LastEditorUserId="6215" LastEditDate="2017-03-14T17:49:53.183" LastActivityDate="2017-03-14T17:49:53.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14251" PostTypeId="1" CreationDate="2017-03-14T18:02:31.500" Score="-1" ViewCount="55" Body="&lt;p&gt;As you can see inside this following link: $$$$&#xA; &lt;a href=&quot;https://books.google.it/books?id=bSq-cEf0EWsC&amp;amp;pg=PA19&amp;amp;lpg=PA19&amp;amp;dq=Aerodynamic+2.3.2+The+speed+of+sound&amp;amp;source=bl&amp;amp;ots=iFRByWKpwG&amp;amp;sig=MpcRIKUA6tsAwsEzG5lwOnRuySQ&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ved=0ahUKEwjfjb_o0dXSAhWLOxQKHRjIB40Q6AEIITAA#v=onepage&amp;amp;q=Aerodynamic%202.3.2%20The%20speed%20of%20sound&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://books.google.it/books?id=bSq-cEf0EWsC&amp;amp;pg=PA19&amp;amp;lpg=PA19&amp;amp;dq=Aerodynamic+2.3.2+The+speed+of+sound&amp;amp;source=bl&amp;amp;ots=iFRByWKpwG&amp;amp;sig=MpcRIKUA6tsAwsEzG5lwOnRuySQ&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ved=0ahUKEwjfjb_o0dXSAhWLOxQKHRjIB40Q6AEIITAA#v=onepage&amp;amp;q=Aerodynamic%202.3.2%20The%20speed%20of%20sound&amp;amp;f=false&lt;/a&gt; $$$$&#xA;Jan Roskam's book  : Airplane Aerodynamics and Performance, Chapter 2 Basic Aerodynamic Principles and Application, Subparagraph 2.3.2 THE SPEED OF SOUND,  uses continuity equation:&#xA; (2.2) $dm/dt= ρAV= cost$ , and,  differentianting it for constant area it yields $ V_ad ρ+ ρd V_a=0$ (2.28).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore the book uses The Euler equation of motion, Eq (2.9) $p+{1/2}ρ{V^2}=cost$ in order to yielding $dp+ρV_adV_a=0$.  (2.29)&#xA;In the continuation it eliminates $dV_a$ from Equations (2.28) and (2.29)&#xA;But equation (2.9) is the form of equation in incompressibiity hypothesis. Why does that book use mass continuity equation in condition of density not changing (incompressibility) (i.e. disregadind density variation term) in conjunction of another equation where it uses compressibility variation? furtheremore compressibility variation (in form of derivative) is used in the yielded equation (2.30) too: ${V_a}^2=dp/dρ$ &lt;/p&gt;&#xA;" OwnerUserId="4916" LastActivityDate="2017-04-14T00:22:32.887" Title="Roskam book uses Eulero motion eq. and mass continuity eq. to find speed of sound but (in my opinion) not in a coherent use (about density variation)" Tags="|fluid-mechanics|acoustics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14252" PostTypeId="2" ParentId="14248" CreationDate="2017-03-14T18:29:06.090" Score="1" Body="&lt;p&gt;According to TR-55, &lt;a href=&quot;https://www.nrcs.usda.gov/Internet/FSE_DOCUMENTS/stelprdb1044171.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, you use total rainfall. This requires a bit more information about the watershed, namely the time of concentration (which affects the unit peak discharge). Then, from TR-55:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}q_p &amp;amp;= q_uA_mQF_p\\&#xA;\text{where:}\\&#xA;q_p &amp;amp;= \text{peak discharge (cfs)} \\&#xA;q_u &amp;amp;= \text{unit peak discharge (csm/in)} \\&#xA;A_m &amp;amp;= \text{drainage area (mi$^2$)} \\&#xA;Q &amp;amp;= \text{runoff (in)} \\&#xA;F_p &amp;amp;= \text{pond and swamp adjustment factor}\end{align}$$ &lt;/p&gt;&#xA;" OwnerUserId="10465" LastActivityDate="2017-03-14T18:29:06.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14256" PostTypeId="2" ParentId="14251" CreationDate="2017-03-14T19:54:37.523" Score="0" Body="&lt;p&gt;It's a typo and he's actually changing the form of Eq. $2.8$, thus not applying the $\rho = constant$ constraint. So he says he's rewriting $2.9$, but actually rewrites $2.8$&lt;/p&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="10400" LastEditDate="2017-03-14T23:51:22.190" LastActivityDate="2017-03-14T23:51:22.190" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14257" PostTypeId="2" ParentId="13386" CreationDate="2017-03-14T20:29:02.860" Score="1" Body="&lt;p&gt;The BOD/COD ratio of this wastewater is low enough that it would be impractical and most likely toxic to the microbiology of an activated sludge process. &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.261.2969&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a good review of the general characterization of wastewater based on BOD, COD, and BOD/COD ratio. &lt;/p&gt;&#xA;" OwnerUserId="10465" LastActivityDate="2017-03-14T20:29:02.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14259" PostTypeId="1" CreationDate="2017-03-14T22:38:44.597" Score="0" ViewCount="63" Body="&lt;p&gt;I am not sure if I am on the right site for this, but here it goes. &#xA;I have a friend who makes a hot sauce for a company he runs. It is very pasty (made of pepper flakes in an oil).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, when he makes a batch, he's got to continually stir up the batch while trying to pour an equal (and evenly distributed) amount of the paste into the bottles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He wants system that makes it easy and predictable to do this faster. Ideally, he would make the paste, put it in something that keeps it stirred, and from there fill each bottle with the same amount of paste. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of contraption would do this?&lt;/p&gt;&#xA;" OwnerUserId="10470" LastActivityDate="2017-03-15T16:11:44.093" Title="How to pump same amount of paste every time?" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14260" PostTypeId="2" ParentId="14239" CreationDate="2017-03-15T01:14:27.193" Score="1" Body="&lt;p&gt;Of course. kevlar has high stiffness per weight so you would use it for a spring that needs to be stiff and light. It has somewhat less strain capability than glass fiber, which I suppose, along with cost, is why glass fiber composites seem to be selected for vehicle leaf springs.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2017-03-15T01:34:41.400" LastActivityDate="2017-03-15T01:34:41.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14261" PostTypeId="2" ParentId="14259" CreationDate="2017-03-15T02:29:16.233" Score="1" Body="&lt;p&gt;It's difficult to give the type of answer you require because the batch sizes &amp;amp; viscosity of the sauce mixes is unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try researching mixing bowls with agitators. These are just two I came across while doing a quick search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://pms-group.net/PDF_catalog/high_speed_mixer2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mixing bowl type 1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://my.hobartcorp.com/resourcecenter/ProductDocumentation/F19128%20Rev%20D.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mixing bowl type 2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-03-15T02:29:16.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14267" PostTypeId="1" CreationDate="2017-03-15T13:10:28.477" Score="3" ViewCount="73" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b2bvT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b2bvT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for my bad english in advance I am not a native speaker.&#xA;I have a system similar to the one above and we are supposed to statically and then dynamically balance it. To statically balance it we are told to set two of the masses to arbitrary angles, we chose 90 degrees away from each other. We were then told to create a vector diagram, where the lengths of the vectors correspond to the mass*radius values for each of the 4 masses. By creating a complete shape we managed to find the angles of the other two masses for which to statically balance. I understand why we do this and theory behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To dynamically balance it we are given these instructions &#xA;&lt;a href=&quot;https://i.stack.imgur.com/2NoEO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2NoEO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can follow the instructions but I do not understand why this should be done and how this relates to dynamic balancing. Would someone please be able to explain why?&lt;/p&gt;&#xA;" OwnerUserId="10480" LastActivityDate="2017-03-15T13:10:28.477" Title="What is the theory that makes this method work when dynamically balancing this system" Tags="|mechanical-engineering|dynamics|" AnswerCount="0" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14268" PostTypeId="1" AcceptedAnswerId="14271" CreationDate="2017-03-15T14:36:50.733" Score="1" ViewCount="99" Body="&lt;p&gt;Is there a way to use a radioactive elements (Americium in my case, but for demo, any radioactive material) to create a usable amount of energy (Power small LED?) AT HOME! Using materials that can be bought from eBay/amazon and the store?&lt;/p&gt;&#xA;" OwnerUserId="10482" LastActivityDate="2017-04-12T22:58:26.417" Title="Radioactive Materials to Energy AT HOME" Tags="|power|energy-efficiency|energy|thermal-conduction|radiation|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14270" PostTypeId="2" ParentId="14259" CreationDate="2017-03-15T15:41:05.693" Score="3" Body="&lt;p&gt;You want to achieve two things: Keep the paste homogenous, in this case keep the flakes from settling, and dose reliably into bottles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first task, you need a mixing bowl or hopper. At a skim, the links provided in Freds answer seem to go in the right direction (but you will need lower speed mixers, since you just want to suspend, not disperge). Since you gave no information on viscosity, amount produced in one batch, other considerations (heating? cooling?), we can't guide you any further for now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second step, you need a dosing pump. A dosing pump is any &lt;a href=&quot;https://en.wikipedia.org/wiki/Pump#Positive_displacement_pumps&quot; rel=&quot;nofollow noreferrer&quot;&gt;positive displacement pump&lt;/a&gt; that can be controlled in a way that it delivers reliable batches. Typically piston, progressing cavity or peristaltic pumps are used. Again, we can't guide you any further as we lack important information. However, given that you likely have small batches and a small budget my first look would be these two types of pumps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Peristaltic pumps: Only one part (the tube) touches the medium and get a pretty exact dosing result by controling the no. of revolutions per batch. These are likely to be affordable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Diaphragm pumps: If you manage to find one that lets you fill one bottle with round number of strokes, you will again have a relible dosing result at a lowish cost. Cleaning could be a bigger hassle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So you want to look for food grade mixing hopper or mixing bowl with a (food grade!) dosing pump. I don't work in food processing, but my hunch is that the exact application of your friend is boringly common and you can get a mixing hopper with dosing pump as an integrated unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before going on a shopping spree, your friend should educate themselves on the important regulations. Likely they will have to clean and thus dissasemble the device every working day &amp;amp; thousand other considerations. &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2017-03-15T16:11:44.093" LastActivityDate="2017-03-15T16:11:44.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14271" PostTypeId="2" ParentId="14268" CreationDate="2017-03-15T16:32:53.760" Score="1" Body="&lt;p&gt;Depending on how much nuclear material you have, you can use a system based on  &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermoelectric_materials#Materials_of_interest&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermoelectric materials&lt;/a&gt; to directly use the heat from the decay of the material to generate electricity. Following from Hazzey's comment, the safety of the experiment will quickly go out of the household realm the more power you try to generate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit for OP comment 1:&#xA;In theory Americium is okay to use for a thermoelectric device because it has an alpha particle decay route (easier to produce heat with alpha particles). Wikipedia says &lt;a href=&quot;https://en.wikipedia.org/wiki/Americium&quot; rel=&quot;nofollow noreferrer&quot;&gt;Americium based smoke detectors&lt;/a&gt; have 0.3 micrograms per unit so you would have to open many thousands of detectors to get a usable amount.  &lt;/p&gt;&#xA;" OwnerUserId="10344" LastEditorUserId="10344" LastEditDate="2017-03-15T16:59:14.207" LastActivityDate="2017-03-15T16:59:14.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14273" PostTypeId="1" AcceptedAnswerId="14274" CreationDate="2017-03-15T17:19:01.940" Score="1" ViewCount="238" Body="&lt;p&gt;I have begun to study control engineering by several books and awesome &lt;a href=&quot;https://www.youtube.com/playlist?list=PLUMWjy5jgHK1NC52DXXrriwihVrYZKqjk&quot; rel=&quot;nofollow noreferrer&quot;&gt;youtube videos by Brian Douglas&lt;/a&gt;. I love them because they give me new understanding of math that I learned before but I still can not get the point of what actually control engineers do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is meant by systems here? I mean all examples always are just physical systems like springs and etc, but all of them are governed by physical laws and no matter how dissatisfied with output we are, we are not able to modify physical laws. And in the case of, for instance, mobile robots what do control engineers do? Deriving PID coefficients? I feel like missing the most crucial part of it. &lt;/p&gt;&#xA;" OwnerUserId="10483" LastActivityDate="2017-03-15T17:43:16.313" Title="What do actually control engineers do?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14274" PostTypeId="2" ParentId="14273" CreationDate="2017-03-15T17:43:16.313" Score="4" Body="&lt;p&gt;In short, 'systems' refers to a combination of components that act together and perform a certain objective. A system can span across different physical and virtual domains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Controls engineers primarily seek to change these inputs to achieve an optimal response based on their objectives/requirements through means of some process (usually electrical or mechanical actuation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, in a commercial setting due to inherent complexities associated with non-isolated systems, engineers usually specialize in a specific process or task (eg. Tuning PID coefficients for different plants). Also controls engineers are sometimes called systems engineers in their designation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: Modern Control Engineering (K. Ogata) and personal experience.    &lt;/p&gt;&#xA;" OwnerUserId="7339" LastActivityDate="2017-03-15T17:43:16.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14275" PostTypeId="1" CreationDate="2017-03-15T18:47:21.663" Score="-1" ViewCount="232" Body="&lt;p&gt;So, I have this matlab problem, and I'm totally lost. Below, you will find my attempt. We are supposed to fit a polynomial curve to the data, but I'm a bit confused as to how the to get a polynomial to match the known relationship. I've got polyfit solving for the form of y = 1 / (ax + b), but I don't know where to go from there. Any help would be greatly appreciated.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4tPgJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4tPgJ.jpg&quot; alt=&quot;problem&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;x= 5:5:50;&lt;br /&gt;&#xA;y= [15 25 32 33 37 35 38 39 41 42];&lt;br /&gt;&#xA;p= polyfit(x,1 ./ y,1)&lt;br /&gt;&#xA;a= p(1);&lt;br /&gt;&#xA;b= p(2);&lt;/p&gt;&#xA;&lt;p&gt;xFit= 5:50;&lt;br /&gt;&#xA;yFit= xFit ./ (a * xFit + b)&lt;br /&gt;&#xA;semilogy(x,y,'o',xFit,yFit,'-');&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10485" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-03-16T02:01:43.660" Title="Matlab Polynomial Curve Fitting with Known Relationship" Tags="|matlab|" AnswerCount="1" CommentCount="1" ClosedDate="2017-03-17T19:08:50.317" ContentLicense="CC BY-SA 3.0" />
  <row Id="14276" PostTypeId="1" CreationDate="2017-03-15T19:46:14.510" Score="4" ViewCount="197" Body="&lt;p&gt;Can a new foundation be dug alongside an existing house with a deep foundation or cellar without compromising the structural integrity of the existing house?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subquestion: should I leave rebar or sth sticking out from the existing foundation to join it with the new foundation?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;existing house (not built yet):&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;foundation 1.8m deep (wall)&lt;/li&gt;&#xA;&lt;li&gt;foundation is on a pad of compacted crushed rock that creates a 1m perimeter around the house. The perimeter above that is filled with sand up to ground level&lt;/li&gt;&#xA;&lt;li&gt;wood framed, some brick walls&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h3&gt;new addition:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;walls and pipes will be joined to the existing house&lt;/li&gt;&#xA;&lt;li&gt;will have the same type of foundation and frame&lt;/li&gt;&#xA;&lt;li&gt;dug into ground, that is fresh - that is never compacted or dug into before.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3081" LastActivityDate="2017-11-03T07:14:58.313" Title="can a new foundation be dug alongside an existing house with a deep foundation without compromising the structural integrity of the existing house?" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14277" PostTypeId="2" ParentId="10634" CreationDate="2017-03-15T20:55:26.867" Score="1" Body="&lt;p&gt;No, no, no, guys. Vapor compression recovers the latent heat of condensation that is normally lost to cooling water in a condenser. It does this by taking vapors that would normally go to a condenser and compressing them, raising their temp and pressure, then reinjecting them into the process.  Since the energy in latent heat for water/steam is 5 or 6 times as much energy as it takes to raise water from freezing to boiling, we are talking about a LOT of energy. Ideally, a vapor compression system can operate on about 15% of the energy required to run the same process with a boiler and condenser.  The dirty trick is that the 15% is mechanical energy, not thermal, so it is penalized by the cost of converting thermal to mechanical.&lt;/p&gt;&#xA;" OwnerUserId="10487" LastActivityDate="2017-03-15T20:55:26.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14278" PostTypeId="2" ParentId="14210" CreationDate="2017-03-15T21:38:00.033" Score="0" Body="&lt;p&gt;There are two separate considerations: applied stress and allowable stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allowable normal stress and allowable shear stress are a function of the material. They are not functions of each other. This is why the problem would need to give both values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for applied stress. I assume you're looking at something like a bar under axial load and considering the state of stress for an inclined plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fZ1iL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fZ1iL.jpg&quot; alt=&quot;bar under axial load&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3QeoU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3QeoU.jpg&quot; alt=&quot;stress on inclined plane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear stress and normal stress for the inclined plane are both a function of the pure axial stress (P/A) and the angle of the inclined plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/idwiR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/idwiR.jpg&quot; alt=&quot;normal and shear stress&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by virtue of statics, maximum shear stress occurs on a plane inclined at 45 degrees and is half the pure axial stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum normal stress occurs on a plane inclined at 0 degrees and is equal to the pure axial stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to your numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allowable normal stress = 7.6 MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allowable shear stress = 4.1 MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial stress associated with allowable shear stress = 8.2 MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Which is greater than the allowable normal stress, so the 7.6MPa controls and should be used to calculate the required cross sectional area.)&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2017-03-15T21:38:00.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14279" PostTypeId="2" ParentId="14275" CreationDate="2017-03-15T22:44:25.057" Score="3" Body="&lt;p&gt;The trick to this one is at the end there. You took the reciprocal of your y data but what you need is the reciprocal of the entire equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original: $y = a\frac{x}{b+x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reciprocal: $\frac{1}{y}=\frac{1}{a}\frac{b+x}{x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when you simplify that a little, you'll get: $y^{-1}=\frac{b}{a}x^{-1}+\frac{1}{a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully, you can see that this is a linear equation but your variables are the reciprocals of x and y. So, when you get over to MATLAB, make the same vectors you did, but then get the reciprocals of those. Use polyfit() on those reciprocals and you'll get coefficients for $\frac{b}{a}$ and $\frac{1}{a}$. Solve those for $a$ and $b$ and check your work; you'll see a fit! &lt;/p&gt;&#xA;" OwnerUserId="9073" LastActivityDate="2017-03-15T22:44:25.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14280" PostTypeId="1" CreationDate="2017-03-15T23:23:58.083" Score="1" ViewCount="496" Body="&lt;p&gt;An experimental study to compare the stability of different plate systems made of titanium to fix fractures of the lower jaw (mandible) made of resin material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load displacement curve I have obtained was step wise both in the elastic region and in the post yield region. Is a step wise load displacement curve normal? How can it be explained, are there similar cases, and what is the references?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IIovJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IIovJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; explained&lt;/p&gt;&#xA;" OwnerUserId="10489" LastEditorUserId="90" LastEditDate="2017-03-16T03:04:52.300" LastActivityDate="2017-03-17T19:42:13.207" Title="Explaination for a stepwise load displacement curve" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14281" PostTypeId="2" ParentId="14280" CreationDate="2017-03-16T03:16:02.603" Score="3" Body="&lt;p&gt;From your graph, you will notice the stepped segments occur in the sloping or curved portions of the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the steps are most likely the result of the resolution/sensitivity of the equipment you are using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same issue arises on computer screens when diagonal lines are drawn on the screen. The pixels in the screens make it impossible to have a true sloping or curving line. The lines will only be straight in horizontal &amp;amp; vertical directions when the line is perfectly aligned with the pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also partially the result of using digital recording equipment, as analogue equipment tends to smooth out such steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To mitigate such stepped regions in your data, use equipment that has a higher resolution/sensitivity.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-03-16T03:16:02.603" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14282" PostTypeId="2" ParentId="14268" CreationDate="2017-03-16T07:30:14.490" Score="2" Body="&lt;p&gt;Attempting this as a home experiment is unlikely to be feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator#241Am&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator#241Am&lt;/a&gt; gives the efficiency of professional prototype designs using Americum as about &lt;strong&gt;2 watts / kilogram of Am&lt;/strong&gt;. The electrical power you will get from a few &lt;em&gt;micrograms&lt;/em&gt; of material will be undetectable (of the order of picowatts) without some &lt;em&gt;very&lt;/em&gt; sophisticated measuring equipment, and most likely a million times too small to &quot;power an LED&quot; in the sense of &quot;make the LED light up&quot;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-03-16T07:30:14.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14283" PostTypeId="1" AcceptedAnswerId="14288" CreationDate="2017-03-16T10:50:39.227" Score="1" ViewCount="214" Body="&lt;p&gt;I am making a vehicle simulation, but just noticed a big issue - there are a lot of cars with no clutch in them, but I haven't driven a car with automatic/clutchless-manual gearbox, nor been even a passenger in such a car. Consequently, I have no idea how such vehicle drive or operate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a vehicle with a clutch, I know how it drives and I also know the physics behind it operation  - torque from engine gets transmitted through the clutch to the transmission, then to differential and then it reaches wheels, the wheels generate torque from ground friction which goes the other way around and slows down the engine. If the clutch is disengaged, then the engine won't transfer torque. If the clutch is partially engaged, then some of the engine torque will be transmitted and the rest of it will be lost in some form of heat during clutch slippage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have no idea how vehicles with an automatic gearbox or a semi-automatic gearbox (e.g. sequential) work. As far as I know, they don't have a clutch. However, I've seen that such vehicles can have their engine revving while not transmitting any torque to the differential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it that they're directly connected to the transmission or is there something else there? What I am asking for is some sort of guidelines for me to figure out how to calculate their maths (plus, if possible, some equations would help).&lt;/p&gt;&#xA;" OwnerUserId="5281" LastEditorUserId="90" LastEditDate="2017-03-17T07:31:31.317" LastActivityDate="2017-08-17T11:13:56.907" Title="Non-Clutch Vehicle Drivetrain - How Does it Work?" Tags="|automotive-engineering|motors|torque|simulation|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14285" PostTypeId="2" ParentId="14283" CreationDate="2017-03-16T13:55:33.173" Score="1" Body="&lt;p&gt;There are 2 important human functions that an automatic transmission must replace:&lt;br&gt;&#xA;1.The driver chooses a gear that's appropriate for the torque and speed requirements of the vehicle.  In an automatic transmission, shift schedules are stored as breakpoints for upward and downward shifting, usually based on engine torque and speed.&lt;br&gt;&#xA;2.  The driver &lt;em&gt;feathers&lt;/em&gt; the clutch to shift smoothly between gears.  The automatic transmission incorporates a hydraulic torque converter, a fluid coupling between the engine and transmission, which damps any jolts in engine torque or speed.  This animation is very useful: &lt;br&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=z5G2zQ_3xTc&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=z5G2zQ_3xTc&lt;/a&gt;   &lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-03-16T13:55:33.173" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14287" PostTypeId="2" ParentId="14196" CreationDate="2017-03-16T16:03:48.920" Score="3" Body="&lt;p&gt;The crosshead bearing slides within a track and connects to the conrod and then to the crankshaft.  In a vertically oriented engine the pressure on the bearing is always downwards, resulting in a depleted lubrication film on the lower contact surfaces of the bearing shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To supply lubricant to the entire bearing, high pressure oil is injected into the bearing and distributed by a pattern of channels to guide the oil uniformly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oil pressures can be as high as 20bar in some designs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This article (search &quot;crosshead bearings&quot;) provides a good introduction to the topic &lt;a href=&quot;http://www.marinediesels.info/2_stroke_engine_parts/Other_info/xhead_lub.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Operational Information, Crosshead Lubrication&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="8055" LastEditDate="2017-03-20T22:37:50.157" LastActivityDate="2017-03-20T22:37:50.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14288" PostTypeId="2" ParentId="14283" CreationDate="2017-03-16T18:22:34.897" Score="3" Body="&lt;p&gt;Conventional automatic transmissions have a torque converter which uses a viscous rather than a rigid coupling to connect input and output. This performs a similar function to a friction clutch but doesn't require manual engagement. Also automatic gearboxes are based on epicyclic gear trains so all the gears are permanently meshed with each other and ratios are selected by locking different combinations of the sun, planet and annular gears so there is no need to match different shaft speeds to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequential gearboxes use a combination of straight cut (as opposed to helical) gears and dog clutches with gears engaging by sliding along the input and/or output shaft so the gears never go fully out of mesh with each other but rather are either locked to or freely spinning on their respective shaft. Again this means that it is not necessary to match input and output speeds, however a clutch is still required to start from a standstill. Motorcycles, for example typically have sequential gearboxes and hand operated clutches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more modern type of auto/semi-auto gearbox typically uses twin wet plate clutches to change between pre-selected gears in this case the gear selection and clutch engagement is done by actuators which are controlled by a computer. In many cases they can be operated as fully automatic or semi automatics where the driver shifts up or down by operating a paddle or lever. So while they do have a clutch is it not manually operated to change gear. Although, as in a sequential gearbox, manual clutch operation &lt;em&gt;may&lt;/em&gt; still be required to start from a standstill often this will be an additional paddle or paddles. Although many semi-auto transmission will also incorporate an anti-stall system which automatically releases the clutch if the revs drop below a certain level at low speed. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So from the perspective of the driver there are 3 main cases. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Full auto (either conventional auto or modern semi auto in full auto mode) : press accelerator pedal to go and the brake to stop and the clutch (or equivalent) sorts itself out automatically eg with automatic transmission if you brake to a stop the engine won't stall so no manual clutch operation required. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sequential/semi auto : manual clutch operation is required to start/stop but not to change gear. Often this will be a hand clutch. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Manual : manual (usually foot pedal) clutch required to start and to change gear. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These cover the most common systems, although there are various other transmission types around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sequential gearboxes are most common on motorcycles and some classes of competition cars (especially those heavily adapted from road cars (eg rallying). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most conventional road cars have either conventional automatic or manual  transmissions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computer controlled semi-auto/auto transmissions are mostly found on high-end sports and racing cars eg F1 and GT formulae. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also automatic transmissions which allow for an approximation of manual gear selection eg the Porche Tiptronic system and it is also &lt;em&gt;possible&lt;/em&gt; to automate a conventional manual transmission, although this is a lot of complexity for no real gain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of completeness there are also centrifugal clutches which only engage above a certain engine rpm but these are only really found on things like chainsaws and self propelled mowers. Also steam piston engines don't &lt;em&gt;require&lt;/em&gt; a clutch as they can develop torque from zero rpm. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is also worth noting that for a driving simulation, unless you have really good haptic feedback in the control hardware it will be very difficult to simulate the feel of a manual clutch.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="9718" LastEditDate="2017-08-17T11:13:56.907" LastActivityDate="2017-08-17T11:13:56.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14290" PostTypeId="1" CreationDate="2017-03-17T05:36:27.410" Score="0" ViewCount="383" Body="&lt;p&gt;By using, transformation of the stress tensor property, the stresses can be calculated for transformed orientations. The properties are unique in case of isotropic in all directions. &#xA;But what will be the case for which the orthotropic material is transformed? How the transformes stress matrix can be calculated for it?&lt;/p&gt;&#xA;" OwnerUserId="9925" LastActivityDate="2017-08-14T20:36:33.430" Title="transformation of orthotropic materials" Tags="|stresses|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14291" PostTypeId="1" AcceptedAnswerId="14434" CreationDate="2017-03-17T10:58:26.410" Score="6" ViewCount="193" Body="&lt;p&gt;I have a general question regarding energy coupling in multi phase flow.  My question comes is based on the text from this book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.co.uk/books?id=CioXotlGMiYC&amp;amp;pg=PA33&amp;amp;lpg=PA33&amp;amp;dq=thermal+coupling+parameter+continuous+dispersed&amp;amp;source=bl&amp;amp;ots=s9uWpatRmM&amp;amp;sig=-aRSlKoQXHc7gTlZTsl30ROncfo&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjh6-DfhdzSAhVELMAKHfKOBHAQ6AEIGjAA#v=onepage&amp;amp;q=thermal&quot; rel=&quot;noreferrer&quot;&gt;https://books.google.co.uk/books?id=CioXotlGMiYC&amp;amp;pg=PA33&amp;amp;lpg=PA33&amp;amp;dq=thermal+coupling+parameter+continuous+dispersed&amp;amp;source=bl&amp;amp;ots=s9uWpatRmM&amp;amp;sig=-aRSlKoQXHc7gTlZTsl30ROncfo&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjh6-DfhdzSAhVELMAKHfKOBHAQ6AEIGjAA#v=onepage&amp;amp;q=thermal&lt;/a&gt; coupling parameter continuous dispersed&amp;amp;f=false&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(page 33)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is kind of obscure so I'll provide a quick background:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the premise of the question is that there is a flow of fluid, the fluid has a continuous phase, and within the continuous there exists a dispersed phase, so something like water flowing with little droplets of oil inside the water. The idea is that, suppose the water is flowing through a pipe and into a control volume. The coupling parameter will compare the effect of the dispersed phase on the control volume to the continuous phases effect. So, a flow of fluid with a continuous phase containing n drops per volume flows into a cubic control volume of length L (a volume of $L^3$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy coupling is related to temperature. So, in energy coupling the two things that are to be compared are:&#xA;The energy (heat) released by the droplets (the convective heat transfer from droplets to water)&#xA;The energy provided by the continuous phase (the heat energy of the water)&#xA;Thus, the coupling parameter will be:&#xA;$$&#xA;\Pi_{\text{Energy}} = \frac{\text{Heat Released by Droplets}}{\text{Enthalpy Flux from Water}}&#xA;$$&#xA;The heat released by the droplets comes from Newtons law of cooling:&#xA;$$&#xA;mC_d\frac{dT_d}{dt} = Nu k_c \pi D (T_c - T_d)&#xA;$$&#xA;D is the diameter of the droplet, Tc and Td are the temperatures of the continuous/dispersed phases. Nu is the Nusselt number. Again, there are n droplets per unit volume therefore the total heat released from the droplets is:&#xA;$$&#xA;nL^3Nu k_c \pi D (T_c - T_d)&#xA;$$&#xA;The heat flux due to the continuous will just be:&#xA;$$&#xA;\rho_c u C_c T_c L^2&#xA;$$&#xA;Which is the density of the continuous phase multiplied by the volume flowing per unit time multiplied by specific heat capacity and it's temperature. The coupling parameter is therefore:&#xA;$$&#xA;\Pi_{\text{Energy}} = \frac{nL^3Nu k_c \pi D (T_c - T_d)}{\rho_c u C_c T_c L^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so that's all fine and dandy. The issue is when they simplify the above formula. So there exists a thermal response time constant (I do not believe it matters where it comes from, this is basically an algebraic exercise at this point):&#xA;$$&#xA;\tau_T = \frac{\rho_dC_dD^2}{12k_c}&#xA;$$&#xA;This can be included in the above equation such that :&#xA;$$&#xA;\Pi_{\text{Energy}} = \frac{nL^3Nu k_c \pi D (T_c - T_d) \rho_d C_d D^2 \tau_T}{\rho_c u C_c T_c L^2 \tau_T }&#xA;$$&#xA;This can be cleaned up, Nu/2 can be approximated to 1 at low Reynolds numbers. piD^2/6 multiplied by density of a droplet (spherical) gives the mass of that droplet, multiply that by n and you end up with the total droplet density per unit volume. The above equation simplifies to:&#xA;$$&#xA;\Pi_{\text{Energy}}= \frac{T_dCLC_d}{uC_c T_C \tau_T } \bigg(1-\frac{T_c}{T_d}\bigg)&#xA;$$&#xA;Where C is the ratio of total dispersed phase density and continuous phase density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the problem, the book quotes the final result as:&#xA;$$&#xA;\Pi_{\text{Energy}}= \frac{CL}{u\tau_T } \bigg(1-\frac{T_c}{T_d}\bigg)&#xA;$$&#xA;Which implies that:&#xA;$$&#xA;\frac{C_dT_d}{C_c T_c} = 1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see any justification for that though...does anyone know where the extra terms went?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4530" LastActivityDate="2017-03-26T17:08:13.083" Title="Multiphase Flow - Energy Coupling Parameter" Tags="|fluid-mechanics|heat-transfer|multiphase-flow|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14292" PostTypeId="1" CreationDate="2017-03-17T12:59:49.813" Score="7" ViewCount="4941" Body="&lt;p&gt;Aerospace engineering textbooks make frequent use of non-dimensional coefficients like &lt;span class=&quot;math-container&quot;&gt;$C_L$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$C_D$&lt;/span&gt; and a whole host of others. Usually, for equations of motion, it is preferred to replace variables with non-dimensional coefficients. However, fundamentally I do not understand why.&lt;/p&gt;&#xA;&lt;p&gt;I have two questions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Why do we make aircraft parameters non-dimensional (&lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$C_L$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$C_m$&lt;/span&gt;)?&lt;/p&gt;&#xA;&lt;p&gt;An often-heard response is you need the coefficients in order to compare different aircraft; &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; are really dependent on aircraft dimensions and flight conditions. However, as far as I know, coefficients &lt;span class=&quot;math-container&quot;&gt;$C_L$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_D$&lt;/span&gt; are also dependent on aircraft dimensions and flight conditions! Where is the inherent superiority of one over the other?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does non-dimensionalizing make the coefficient independent of anything (weight, aircraft dimensions, dynamic pressure, Mach number etc.)?&lt;/p&gt;&#xA;&lt;p&gt;Some textbooks state that non-dimensional coefficients are independent of weight and aircraft dimensions but do not give an argument for why that should be the case.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10512" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-03-23T00:16:37.007" Title="Why use non-dimensional coefficients?" Tags="|mechanical-engineering|aerospace-engineering|aerodynamics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14294" PostTypeId="1" CreationDate="2017-03-17T15:50:56.360" Score="1" ViewCount="1855" Body="&lt;p&gt;A vacuum evaporation system works keeping warm water inside a sealed vessel, which is evacuated with a reversed compressor (or vacuum pump) to keep an inside pressure lower than atmospheric pressure, with the objective of lowering evaporation temperature and increase overall process efficiency. Next picture shows a highly optimized vacuum evaporator system schematic (for the subject of this question we can ignore all heat recovery subsystems):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NB1qb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NB1qb.jpg&quot; alt=&quot;vacuum evaporation system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in an ambient temperature vacuum evaporator (i.e. 25 ºC). Vapor pressure of water at this temperature is 3.17 kPa, so we need an inside pressure lower than or equal to this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two main sources of energy expenses. First is enthalpy of evaporation, which at this conditions is 2440 kJ/kg, so we need to add this amount of energy to maintain water temperature at 25 ºC during process. Second is compressor work to keep process pressure inside vacuum vessel. We can approximate it as a reversible adiabatic thermodynamic process (i.e. isentropic process).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Entropy of saturated water vapor at 25 ºC and 3.17 kPa (the one that compressor evacuates because liquid water stays in the bottom of the tank) is  8.56 kJ/kg. Internal energy at same conditions is 2410 kJ/kg. Because process is isentropic, entropy of discharge steam is the same, so its thermodynamic state is 400 ºC 101 kPa and internal energy is 2970 kJ/kg. Because process is adiabatic, compressor work is equal as change in internal energy: 2970 - 2410 = 560 kJ/kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy needed to run an idealized system as the one described will be the sum of the enthalpy of evaporation and the compressor work: 2440 + 560 = 3000 kJ/kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this estimation correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;400 ºC compressor discharge steam looks quite hot to me. This can explain why most common industrial vacuum evaporation systems work at higher water temperatures and lower vacuum levels. It is right?&lt;/p&gt;&#xA;" OwnerUserId="1616" LastActivityDate="2017-03-18T22:24:11.577" Title="Water vacuum evaporation thermodynamics" Tags="|thermodynamics|waste-water-treatment|evaporation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14295" PostTypeId="2" ParentId="14292" CreationDate="2017-03-17T17:31:34.667" Score="4" Body="&lt;p&gt;There are two major benefits to dimensional analysis (non-dimensional coefficients) according to Frank M. White, &lt;em&gt;Fluid Mechanics&lt;/em&gt;, 2nd Ed. My answer is heavily paraphrased from this source.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They allow scientists/engineers to reduce the number of experiments required to explore a given phenomenon.&lt;/li&gt;&#xA;&lt;li&gt;They provide &lt;em&gt;scaling laws&lt;/em&gt; allowing experiments to be performed on small, cheap scale models which can then be extended to full-size applications.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Reducing the number of experiments&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If we wanted to design an experiment to measure the force on a cylinder immersed in a moving fluid, and we knew it depends on the following parameters:&#xA;$$ F = f(L, V, \rho, \mu) $$&#xA;Then if we want to properly identify the phenomenon we would need to explore a rather large parameter space. You would want to be able to fit a curve to the data, which is going to require at least 10 data points for each variable. So we would need to design an experiment with 10 values of $L$, and for each value of $L$ we would also have 10 values of $V$, and for each value of $V$ 10 values of $\rho$... and so on. This would be time consuming and expensive. However, the equation can be reduced using dimensional analysis to get:&#xA;$$ \frac{F}{\rho V^2 L^2} = g\left( \frac{\rho VL}{\mu}\right)$$&#xA;$$ C_F = g\left( Re\right) $$&#xA;Where $C_F$ is the dimensionless &lt;em&gt;force coefficient&lt;/em&gt; and $Re$ is the &lt;em&gt;Reynolds number&lt;/em&gt;.&#xA;Now we only need to explore around 10 values of $Re$ to start to get a basic characterization of the phenomenon. This is a greatly simplified example (there is more to fluid forces than the &lt;em&gt;Reynolds number&lt;/em&gt;) but it serves to illustrate the point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, we can use non-dimensional coefficients to compare different geometries. Consider, for example, the &lt;em&gt;Reynolds number&lt;/em&gt;. The &lt;em&gt;Reynolds number&lt;/em&gt; can be calculated for many geometries (inside a pipe, on an aircraft wing, around the body of a submarine), different flow conditions (internal vs. external flow) and different scales, yet it can tell us important information about the nature of the flow (laminar vs. turbulent vs. transition).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Scaling laws&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Non-dimensional coefficients are also useful because they allow easy comparison between engineering cases at different scales. They allow us to establish a condition of &lt;em&gt;similarity&lt;/em&gt; between a model and a full-scale prototype.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to this property (independence from scale), non-dimensional coefficients are used to design scale-model tests. If you make a 1:10 model of an airplane and put it in a wind tunnel, the layperson might think you would just scale the design air velocity by the same factor (and they would be wrong). Similarly if you want to design an RC model airplane, you can't just scale down a Boeing 747 and have it fly! Here is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Model_aircraft#Model_aerodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the example function $f$ given above, we can achieve what is called &lt;em&gt;similarity&lt;/em&gt; if the &lt;em&gt;Reynolds numbers&lt;/em&gt; of a model and a full-size prototype are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Re_m = Re_p$ then $C_{Fm} = C_{Fp}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the definition of the force coefficient from before, we get the following scaling law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{F_p}{F_m} = \frac{\rho_p}{\rho_m}\left(\frac{V_p}{V_m}\right)^2\left( \frac{L_p}{L_m}\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you were to make a model at 1:10 scale, $L_p/L_m = 10$. Assuming that the velocity of the fluid $V$ and the density of the fluid $\rho$ are identical for both, then the prototype will see 100 times the force compared to the model when subjected to these conditions.&lt;/p&gt;&#xA;" OwnerUserId="7430" LastEditorUserId="7430" LastEditDate="2017-03-23T00:16:37.007" LastActivityDate="2017-03-23T00:16:37.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14296" PostTypeId="2" ParentId="14290" CreationDate="2017-03-17T18:21:42.163" Score="1" Body="&lt;p&gt;I assume you want to know what the stress tensor given in one cartesian coordinate system looks like in another cartesian coordinate system. If not, please be more specific in your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, tensor coordinates transform with an orthogonal matrix. So if you have an order 1 tensor (a vector) $\underline{a}$ with coordinates $a_i$ in one cartesian coordinate system, and a transformation matrix $\alpha_{ij}$ to a second cartesian coordinate system, then the coordinates of $\underline{a}$ in the second cartesian coordinate system are $\tilde a_i = \alpha_{ji} a_j$ (assuming &lt;a href=&quot;https://en.wikipedia.org/wiki/Einstein_notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Einstein summation convention&lt;/a&gt;, which is usually applied in tensor algebra).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an order 2 tensor (like the stress tensor), every index transforms individually: $\tilde a_{ij} = \alpha_{mi} \alpha_{nj} a_{mn}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Tensor coordinates transform with the transposed matrix, while base vectors transform with the original matrix: $\underline{\tilde e}_i = \alpha_{ij} \underline{e}_j$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the transform matrix is &lt;a href=&quot;https://en.wikipedia.org/wiki/Orthogonal_matrix&quot; rel=&quot;nofollow noreferrer&quot;&gt;orthogonal&lt;/a&gt;, its inverse is identical to its transposed. All this can be deduced. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Tensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia article on tensors&lt;/a&gt; for more details.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The 3D orthogonal matrix that rotates the 1,2 plane (or $x$,$y$ plane) is $\begin{bmatrix} \cos \theta &amp;amp; -\sin \theta &amp;amp; 0 \\ \sin \theta &amp;amp; \cos \theta &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 1 \end{bmatrix}$ with rotation angle $\theta$.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2017-03-17T18:21:42.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14297" PostTypeId="2" ParentId="14210" CreationDate="2017-03-17T18:57:47.630" Score="-2" Body="&lt;p&gt;Think of it like this. When you place a load on a bar, the bar will want to bend. The bending puts the top in compression and the bottom in tension, these are the tangential stresses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can bend a bar and it will spring back, but if you bend it too far it will stay bent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the bend will become permanent is from when you exceed the strength of the tangential stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you understand the concept, the rest is just math!&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:16:11.427" LastActivityDate="2017-07-23T18:16:11.427" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14298" PostTypeId="2" ParentId="12421" CreationDate="2017-03-17T19:08:20.407" Score="0" Body="&lt;p&gt;The simple answer is profits. They can charge more for the higher numbers even if the cost is not relative to the increase. I can't tell you how much more it costs to produce high octane, but am confident there is additional expense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The winter blends as mentioned above are more for evaporation rates. If you consider a large city has many hundreds of petrol stations the total gasoline evaporated from all of those tanks is significant. Since the evaporation rate of fuel directly correlates to the temperature they can use a more evaporative fuel in the winter than they can in the summer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also have zones where they have to use the heavier gasoline even in the winter due to the concentration of petrol stations there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned all this while working for a pipeline company that stored and transferred fuel for all the major oil companies from their refineries to the terminals where they were loaded on trucks. Interesting trivia, the gasoline in my pipelines had an octane rating in the 60's when we tested it. The octane rating you see at the pump is from the ethanol they add at the terminal while loading the trucks. This is mostly due to the fact that you don't want alcohol in bulk storage tanks because it would soak up a tremendous quantity of water from the tanks.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:18:40.887" LastActivityDate="2017-07-23T18:18:40.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14299" PostTypeId="2" ParentId="14280" CreationDate="2017-03-17T19:34:59.967" Score="0" Body="&lt;p&gt;The step behavior of the data is almost certainly an artifact of the test setup and not of the material itself.  As @agentp mentioned in a comment, it looks like there is a mismatch between the output voltage span of your sensor and the expected input voltage span of the analog-to-digital converter.  Most oscilloscopes can be adjusted to match the span of the sensor they are reading from - you should check the documentation for whatever equipment you're using.  If the voltage of the sensor is too small for the ADC to deal with, you can use an amplifier to boost the sensor signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the following figure from &lt;a href=&quot;http://www.ti.com/lit/an/slyt173/slyt173.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt; that illustrates span mismatch pretty well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qjhbo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qjhbo.png&quot; alt=&quot;Span mismatch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In figure (a), the span of the sensor and ADC match, but there is a 1-volt offset that will result in lost data.  In figure (b), the sensor is only using half of the available span from the ADC.  This will effectively reduce the bit-depth of the sampled data, which I suspect is what happened to your data.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2017-03-17T19:42:13.207" LastActivityDate="2017-03-17T19:42:13.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14300" PostTypeId="1" CreationDate="2017-03-18T02:38:49.450" Score="2" ViewCount="1743" Body="&lt;p&gt;I used to use Solidworks a few years back and am now trying to learn Autodesk Fusion 360. So far I'm pretty impressed but one thing I've not found is something like the &lt;code&gt;equal&lt;/code&gt; constraint that Solidworks offered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts were that it would clutter a sketch less to make a load of lines constrained to be the same length and then specify the dimension once to lock those constraints down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that possible or is there an equivalent?&lt;/p&gt;&#xA;" OwnerUserId="10517" LastActivityDate="2019-07-12T20:40:33.180" Title="Is there an equivalent to the 'equal' constraint from solidworks in fusion 360?" Tags="|cad|solidworks|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14303" PostTypeId="1" CreationDate="2017-03-18T07:54:33.787" Score="1" ViewCount="54" Body="&lt;p&gt;What manufacturing process should be adopted to manufacture a LEGO® compatible strip made by material which is rubber like?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UKqvP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UKqvP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10521" LastEditorUserId="10521" LastEditDate="2017-03-18T08:00:45.157" LastActivityDate="2017-06-16T19:08:07.937" Title="What manufacturing process is suited for production of the elastomer strip?" Tags="|materials|manufacturing-engineering|product-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14304" PostTypeId="1" AcceptedAnswerId="15265" CreationDate="2017-03-18T14:00:22.883" Score="6" ViewCount="742" Body="&lt;p&gt;I am considering a regenerative braking system implemented with a AC permeant magnet motor with an efficiently of ~96%. It is easy to determine via energy equations the kinetic energy of the car and therefore how much energy is possible at max efficiently but regenerative systems also rely on the brakes of the car to slow it down to a stop. Is there a particular rule or reference to how much conventional braking is required for a given scenario? I wonder if it is possible to determine how much the motor would actually decelerate the car and then work out energies from there. Could anyone offer some advice?&lt;/p&gt;&#xA;" OwnerUserId="10522" LastActivityDate="2017-06-11T14:36:36.727" Title="How to determine how much energy recovered from regenerative braking" Tags="|mechanical-engineering|electrical-engineering|automotive-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14305" PostTypeId="2" ParentId="14303" CreationDate="2017-03-18T17:38:47.937" Score="1" Body="&lt;p&gt;For an arbitrary length, I believe hot-forming. Feed a rubber band into a system that runs it along a roller heated past the $T_g$ of the rubber that has the desired imprint etched into it, then cool it down as it leaves the forming roller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual temperatures, tension values for the band and pressure required are up to you to figure out once you settle on a specific material.&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2017-03-18T17:38:47.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14306" PostTypeId="1" AcceptedAnswerId="14320" CreationDate="2017-03-18T18:53:00.597" Score="1" ViewCount="1435" Body="&lt;p&gt;I am not sure where else to post this, so here goes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a velocity profile ($v_{\phi}$) for a radial distance ($r$). From this velocity profile I am trying calculate the shear rate ($y$) which can be calculated by ( Eq 3 in Image):&#xA;$$&#xA;y = \frac{\partial v_{\phi}}{ \partial r}  - \frac{v_{\phi}}{r}  = r  \frac{\partial}{\partial r} \left( \frac{v_{\phi}}{  r}\right )&#xA;$$&#xA;Please find the image of the equation from the papers attached,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/swMBh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/swMBh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Screenshot of the equation for the paper for your kind consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have implemented the equation as such,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for i = 1: size(v_phi,1)&#xA;&#xA;    y (i) = ( r(i) ) * (( ( v_phi (i) / r(i) ) - ( v_phi (i+1) / r(i+1) ) ) / ( r(i+1) - r(i)) );&#xA;&#xA;    y1 (i) = ( ( ( v_phi(i) / r(i) ) - ( v_phi(i+1) / r(i+1) ) ) * ( v_phi (i) / r(i) ) );&#xA;&#xA;&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However the y and y1 values calculated are very different. I am just wondering, whether I am doing something wrong in the implementation and looking for help if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Data &amp;gt;&amp;gt;&amp;gt;&amp;gt;&#xA;&#xA;v_phi = [36.99 37.00 35.78 31.43 26.91 23.32 19.46 16.97 14.49 12.12 10.42 8.52 7.25 6.06 5.11 3.92 3.30 2.54 2.11 1.42 1.46 1.46 0.97 0.63 0.63 0.51 0.40 0.91 0.24 0.36 0.63 0.71 0.45 0.26 0.74 ];&#xA;&#xA;r = [3.59 3.76 3.94 4.12 4.29 4.47 4.65 4.82 5.00 5.18 5.35 5.53 5.71 5.88 6.06 6.24 6.41 6.59 6.76 6.94 7.12 7.29 7.47 7.65 7.82 8.00 8.18 8.35 8.53 8.71 8.88 9.06 9.24 9.41 9.50 ];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="10524" LastEditorUserId="3488" LastEditDate="2017-03-20T11:08:09.767" LastActivityDate="2017-03-20T11:08:09.767" Title="Calculating a derivative in a loop?" Tags="|matlab|feedback-loop|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14307" PostTypeId="1" CreationDate="2017-03-18T20:39:21.727" Score="1" ViewCount="228" Body="&lt;p&gt;In the figure below we have that the gear $A$ is has an fixed axis and has the radius of $R_a = 0,1m$ and, in the instant represented in the figure spins clock-wise with angular velocity $\omega_a = 4\text{rad}/s$. The gear $C$  has radius $R_c = 0,1m$ and gear $D$ has radius $R_d =0,2m$ and they has been firmly weld each other. An arm of length $0,2m$ connects the rotation axis of $A$ and $C$ and $D$. The $B$ gear spins around $A$ but without any connection with $A$ and has radius $R_b = 0,4m$. In the instant presented by the figure $B$ spins in the counter clock-wise and has angular velocity of $\omega_b = 1\text{rad}/s$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PiJaG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PiJaG.png&quot; alt=&quot;Image of &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I have is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;What is the velocity of the center of gears $C$ and $D$ expressed in $m/s$?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt:&lt;/strong&gt; I've tried to derive the velocity of the point $v_P$ where $P$ is the point of connection between $A$ and $C$ and I've tried to encounter the angular velocity of $C$ and $D$ using that for a point $Q$ of $B$ must have the velocity equal &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_q = \omega_a(L/2) = 0,4m/s$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Witch leads us to the angular velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_c = \omega_d = v_q/L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But still I couldn't find the velocity using that. Someone have a hint?&lt;/p&gt;&#xA;" OwnerUserId="10526" LastEditorUserId="10526" LastEditDate="2017-03-18T21:42:42.703" LastActivityDate="2017-03-29T01:51:40.703" Title="Compound gear system: Finding velocity" Tags="|applied-mechanics|kinematics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14308" PostTypeId="1" CreationDate="2017-03-18T21:06:34.623" Score="3" ViewCount="118" Body="&lt;p&gt;I'm automating a non-thermostatic household radiator valve, by replacing the knob with a DC motor and adding some support structure. You can see the system at work here: &lt;a href=&quot;https://www.instagram.com/p/BRQ8xS9gPPN/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.instagram.com/p/BRQ8xS9gPPN/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motor applies a constant torque and stops automatically when it stalls, i.e. when the valve can't turn anymore. The water flowing in the valve has a temperature of 60-62 degrees Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a repeated stress test (fully open, wait 5s, then fully close, wait 5s, repeat), starting from a closed position and the radiator at ambient temperature, I noticed that the time required to open and close the valve had increased over time. It seemed like the valve is getting harder and harder to turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My strongest suspect is an effect of heat. If so, what could be happening? Is the valve opposing more resistance, or the motor providing less torque? Is it something that I have to take into account during regular use? What other causes could lead to this phenomenon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I couldn't do other tests because some parts broke due to mechanical stress. Will update this post when I'll have more results.&lt;/p&gt;&#xA;" OwnerUserId="10527" LastActivityDate="2017-04-11T20:38:11.360" Title="Why does a radiator valve require more torque after a few turns?" Tags="|mechanical-engineering|thermodynamics|heating-systems|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14309" PostTypeId="2" ParentId="14300" CreationDate="2017-03-18T21:28:24.073" Score="3" Body="&lt;p&gt;I'm not familiar with the Solidworks equal constraint but it sounds like this should be done parametrically in Fusion. Dimension the first line to a value. Then, dimension the second one, but instead of typing in a value, click on the dimension of the first line. It will set it to something like 'd1' and then if you change your first dimension, the second will also change. You can also set them all to a named variable.&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2017-03-18T21:28:24.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14310" PostTypeId="2" ParentId="11533" CreationDate="2017-03-18T21:42:50.880" Score="1" Body="&lt;p&gt;A stepper controlled proportional valve has the characteristics you're looking for. &lt;a href=&quot;http://www.clippard.com/downloads/PDF_Documents/Product%20Data%20Sheets/Clippard%20SCPV%20Proportional%20Valve.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This one&lt;/a&gt;, for example, only consumes 4J to open or close. Then, they consume no power to stay in position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawbacks are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The valves are much slower than a solenoid (1s to open or close for the linked one)&lt;/li&gt;&#xA;&lt;li&gt;They'll cost more because they're giving you functionality you don't apparently need (its a proportional control valve, not just on/off)&lt;/li&gt;&#xA;&lt;li&gt;They're typically harder to control. You either need to supply a chopper drive or you have to pay for them to supply one.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2017-03-18T21:42:50.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14312" PostTypeId="2" ParentId="14294" CreationDate="2017-03-18T22:24:11.577" Score="1" Body="&lt;p&gt;The temperature sounds correct, the energy balance does not. I get 405°C for the compressor discharge but the enthalpy increases from 2547 kJ/kg to 3289 kJ/kg, for a work input to the gas of 743 kJ/kg, not 560 kJ/kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't remember why you need to use enthalpy instead of internal energy in a compressor. But it is always correct and just as easy to look up, so you might as well always use enthalpy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your final question, compressing low density gases at high pressure ratios typically results in very high discharge temperatures, so this result makes sense. This doesn't even account for the fact that realistically this compressor is going to be very inefficient and will require significant cooling or the outlet will be even hotter.&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2017-03-18T22:24:11.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14314" PostTypeId="2" ParentId="14292" CreationDate="2017-03-19T07:48:39.213" Score="3" Body="&lt;p&gt;The qualitative behaviour of a physical system often depends on the relative size of two (or more) different effects. For example, in fluid flow, turbulence tends to develop when the force to accelerate the fluid (i.e. its inertia) is big compared with the viscous forces in the flow. In the opposite situation where the inertia forces are small relative to the viscous forces, the flow tends to remain laminar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a similar situation in heat transfer problems, depending on the amount of heat carried by convection in the fluid (which depends on the fluid physically moving around), compared with the amount moved by conduction (which doesn't depend on the fluid motion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we invent a simple but &quot;reasonable&quot; way to measure these different effects, the &lt;em&gt;ratio&lt;/em&gt; of the two measures will be a non-dimensional number. For the Reynolds number example, a reasonable way to measure the inertia forces is simply mass$\times$acceleration of an element of the fluid, i.e., $\rho L^3 v/t$ where $\rho$ is the density, $L$ is some &quot;average&quot; dimension of the fluid element, $v$ is the velocity, and $t$ some typical time scale in the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly a reasonable way to measure the viscous forces is $\mu(v/L)L^2$, where $\mu$ is the dynamic viscosity, $v/L$ represents the velocity gradient creating the viscous forces, and $L^2$ is the cross sectional area of the fluid element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing one measure by the other gives $$\frac{\rho\, (L/t)\, L}{\mu}$$ or (since $L/t$ is measuring some sort of velocity - the mean flow velocity through the pipe, for example) $$\text{Re} = \frac{\rho v L}{\mu}$$ where Re = Reynolds' number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a particular situation like flow through a pipe, the transition from laminar to turbulent flow typically happens when Re is about $3000$, if we take $L$ to mean the diameter of the pipe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is useful, because &lt;em&gt;the result is true in a wide range of flow situations&lt;/em&gt;. it doesn't matter if we are talking about low-pressure gas flowing through an $0.1\,\text{mm}$ diameter capillary tube in a vacuum pumping system, or crude oil being pumped through a $1.2\,\text{m}$ diameter long-distance pipeline - just plug in the relevant material properties, diameter, and flow velocity to find Re, and for the same value of Re the flow pattern will be basically the same.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-03-19T07:54:02.250" LastActivityDate="2017-03-19T07:54:02.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14315" PostTypeId="2" ParentId="11699" CreationDate="2017-03-19T09:00:50.380" Score="0" Body="&lt;p&gt;Thin shells can by modeled using Kirchhoff-Love or Mindlin approaches; the geometrical relations would be different in those cases. You will find exact geometrical relations in the following book, devoted to analysis of shells (including conical) based on the Kirchhoff-Love approach; please see the reference below. The reason why I recommend the following book is that it also contains a CD-ROM with complete source codes for the analysis of shells, including conical shells. In particular ,the example on the webpage stated below (also presented in the book)consists of several shells, including a conical component:&#xA;&lt;a href=&quot;http://members.ozemail.com.au/~comecau/quad_shell.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://members.ozemail.com.au/~comecau/quad_shell.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5992" LastActivityDate="2017-03-19T09:00:50.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14316" PostTypeId="2" ParentId="449" CreationDate="2017-03-19T09:14:57.767" Score="0" Body="&lt;p&gt;Yes, there are meshing software programs, allowing fully automatic meshing. If you are interested in meshing planar or curved surfaces, there are several products that provide completely automatic meshing, delivering 100% quadrilateral meshes on surfaces of any degree of complexity. I would suggest that you visit the following webpage and choose one of the programs that suits your needs as close as possible (some of those programs are best for structural engineering applications, other - for modeling of printed circuit boards, etc.)&#xA;&lt;a href=&quot;http://members.ozemail.com.au/~comecau/products.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://members.ozemail.com.au/~comecau/products.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5992" LastActivityDate="2017-03-19T09:14:57.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14320" PostTypeId="2" ParentId="14306" CreationDate="2017-03-19T22:31:38.187" Score="2" Body="&lt;p&gt;Your implementation is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for i = 1: size(v_phi,1)&#xA;  y(i) = r(i)*((v_phi(i)/r(i)  - v_phi(i+1)/r(i+1)) / (r(i+1) - r(i)));&#xA;  y1(i) = (v_phi(i)/r(i)  - v_phi(i+1)/r(i+1)) * (v_phi(i)/r(i));&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That means that the discretization you are using is&#xA;$$&#xA;  y_i = r_i \left(\frac{v_{\phi,i}}{r_i} - \frac{v_{\phi, i+1}}{r_{i+1}}\right) \left(\frac{1}{r_{i+1} - r_i}\right) \quad \equiv \quad y =  r \frac{d}{d r}\left(\frac{v_\phi}{r}\right)&#xA;$$&#xA;and&#xA;$$&#xA;  y_{1,i} = \left(\frac{v_{\phi,i}}{r_i} - \frac{v_{\phi, i+1}}{r_{i+1}}\right) \left(\frac{v_{\phi,i}}{r_i}\right) \quad \equiv \quad y_1 = \frac{d v_\phi}{d r} - \frac{v_\phi}{r}&#xA;$$&#xA;From the above we notice that the discretization may be inaccurate and there is at least one bug in your implementation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the ratio of the two values is&#xA;$$&#xA;  \frac{y_i}{y_{1,i}} = \frac{r_i^2}{v_{\phi,i} (r_{i+1} - r_i)}&#xA;$$&#xA;There is no reason why this ratio should be 1, indicating that there is an error in your implementation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fix the finite difference accuracy, you have to go to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Finite_difference&quot; rel=&quot;nofollow noreferrer&quot;&gt;definition of the  discretization scheme&lt;/a&gt;. In your case you appear to be using a forward difference scheme.  For that situation, as $h \rightarrow 0$, &#xA;$$&#xA;  \frac{df(x)}{dx} = \frac{f(x+h) - f(x)}{h} + O(h)&#xA;$$&#xA;The accuracy of your solution depends on the value of $h$.  To increase accuracy you should use a central difference scheme at the minimum.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2017-03-19T22:31:38.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14321" PostTypeId="2" ParentId="14306" CreationDate="2017-03-20T03:00:36.627" Score="1" Body="&lt;p&gt;Biswajit Banerjee already indicated that your code has bugs, which is why your results are so far off.  I will explain in more detail where these bugs are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;v_phi&lt;/code&gt; is a row vector, so &lt;code&gt;size(v_phi, 1)&lt;/code&gt; always returns &lt;code&gt;1&lt;/code&gt; because there is only one row.  You want &lt;code&gt;size(v_phi, 2)&lt;/code&gt; or better yet, &lt;code&gt;numel(v_phi)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you're using forward differences to compute the derivatives, you can only iterate up to the &lt;em&gt;second-to-last&lt;/em&gt; element in each of the input vectors, so the loop declaration should start like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for i = 1:size(v_phi, 2) - 1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;After that, you mix up the order of subtraction at least once or twice.  That is, where you should be doing &lt;code&gt;y(i+1) - y(i)&lt;/code&gt;, you instead do &lt;code&gt;y(i) - y(i+1)&lt;/code&gt;, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, as a matter of style, loops can often be replaced by vector operations in Matlab.  For example, the following one-liner computes $\frac{dv_{\phi}}{dr}$ over the whole data set:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dvdr = diff(v_phi) ./ diff(r);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Your equation can actually be solved without using any loops.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2017-03-20T03:00:36.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14323" PostTypeId="1" CreationDate="2017-03-20T08:09:09.553" Score="4" ViewCount="1442" Body="&lt;p&gt;I understand that to calculate the indeterminacy of a pin-jointed frame, the unknowns have to be considered i.e. the number of elements corresponding to the number of internal forces, plus the number of reaction forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations that can be written to solve for these unknowns are vertical and horizontal equilibrium at each node i.e. 2 equations per node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the number of unknowns minus the number of equations results in the indeterminacy of the pin-jointed frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, why do we not include the possibility of being able to write moment equilibrium as one of our equations? For the analysis of beams, the moment equilibrium equation is considered for calculating the indeterminacy of the beam.&lt;/p&gt;&#xA;" OwnerUserId="10539" LastEditorUserId="2015" LastEditDate="2018-08-08T20:08:58.520" LastActivityDate="2018-08-08T20:23:32.043" Title="How to calculate indeterminacy of pin-jointed frame?" Tags="|mechanical-engineering|structural-engineering|structures|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14324" PostTypeId="1" CreationDate="2017-03-20T08:51:26.650" Score="1" ViewCount="61" Body="&lt;p&gt;Is it possible to detect specific UVB wavelength ranges using just a UVM-30A UV sensor or similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensor in consideration is &lt;a href=&quot;http://image.dfrobot.com/image/data/TOY0044/guva-s12sd.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;GUVA-S12SD&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems to indicate that it generates a specific current level when sensing a specific wavelength, but because its sensing a range of wavelengths I guess you get the summation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to use any techniques to perform a UV spectral analysis on the output so as to detect the presence of a specific wavelength band.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also important to determine relative intensity over time for the band. Is this anyway possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not an engineer, but I can do any programming necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help!&lt;/p&gt;&#xA;" OwnerUserId="10534" LastEditorUserId="7083" LastEditDate="2017-05-09T06:10:49.487" LastActivityDate="2017-09-06T08:27:59.517" Title="Is it possible to detect specific UVB wavelength ranges using just a UVM-30A UV sensor or similar" Tags="|electrical-engineering|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14325" PostTypeId="1" AcceptedAnswerId="14328" CreationDate="2017-03-20T09:44:42.580" Score="0" ViewCount="840" Body="&lt;p&gt;I want to mount two patio heaters on my terrace in front of a window wall.  I cannot attach anything directly to the window wall so I am going to run an aluminium box beam between two brick columns.  The beam will be 5000mm long.  The beam dimensions are 80mm x 40mm with a 4mm thick wall, 13kg in weight.  The heaters weigh 6.5kg each and are 1000mm long.  They will be 1000mm from each end and therefore 1000 from each other.  I need to drill 8 x 8mm holes for mounting brackets (2 brackets per heater, 2 holes per bracket).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure the beam will deflect a little - any serious risk of collapse?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/xNptP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xNptP.jpg&quot; alt=&quot;Patio Heater Attachment Points&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/AHzvy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AHzvy.jpg&quot; alt=&quot;Proposed Positioning&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10540" LastEditorUserId="10540" LastEditDate="2017-03-22T14:56:53.570" LastActivityDate="2017-03-22T14:56:53.570" Title="Aluminium Beam - is it fit for purpose? Can it easily support this load?" Tags="|mechanical-engineering|building-design|construction-management|" AnswerCount="1" CommentCount="6" ClosedDate="2017-03-22T01:12:48.160" ContentLicense="CC BY-SA 3.0" />
  <row Id="14326" PostTypeId="1" CreationDate="2017-03-20T11:29:26.870" Score="-1" ViewCount="82" Body="&lt;p&gt;As you can see inside this following link:&#xA;&lt;a href=&quot;https://books.google.it/books?id=bSq-cEf0EWsC&amp;amp;pg=PA19&amp;amp;lpg=PA19&amp;amp;dq=Aerodynamic+2.3.2+The+speed+of+sound&amp;amp;source=bl&amp;amp;ots=iFRByWKpwG&amp;amp;sig=MpcRIKUA6tsAwsEzG5lwOnRuySQ&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ved=0ahUKEwjfjb_o0dXSAhWLOxQKHRjIB40Q6AEIITAA#v=onepage&amp;amp;q=Aerodynamic%202.3.2%20The%20speed%20of%20sound&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://books.google.it/books?id=bSq-cEf0EWsC&amp;amp;pg=PA19&amp;amp;lpg=PA19&amp;amp;dq=Aerodynamic+2.3.2+The+speed+of+sound&amp;amp;source=bl&amp;amp;ots=iFRByWKpwG&amp;amp;sig=MpcRIKUA6tsAwsEzG5lwOnRuySQ&amp;amp;hl=it&amp;amp;sa=X&amp;amp;ved=0ahUKEwjfjb_o0dXSAhWLOxQKHRjIB40Q6AEIITAA#v=onepage&amp;amp;q=Aerodynamic%202.3.2%20The%20speed%20of%20sound&amp;amp;f=false&lt;/a&gt;&#xA;Jan Roskam's book : Airplane Aerodynamics and Performance, Chapter 2 Basic Aerodynamic Principles and Application, Subparagraph 2.3.2 THE SPEED OF SOUND, I don't understand completely the assumptions, the hypothesi taken by book author: in particular the book says: &quot; To derive an expression for the speed of sound c, consider a one-dimensional duct with an infinitesimal pressure wave generated inside the gas at rest. If a coordinate system is chosen to be fixed with the disturbance, the gas velocity, $V_a$ will be the propagation speed of the disturbance before being affected by it. &quot; Please explain better if possible. Only what I understand is that the book with this assumptions permit to identify  &quot;the gas velocity, $V_a$&quot; with speed of sound that is a velocity (speed) of disturbance, so with those hypothesis it identifyies (at least about speed calculation) at the end liquid material motion with disturbance motion. With the good intention for the results I still have to undertsand better the assumptions I quoted above.&lt;/p&gt;&#xA;" OwnerUserId="4916" LastEditorUserId="4916" LastEditDate="2017-03-20T11:34:57.730" LastActivityDate="2017-03-20T15:46:48.530" Title="Roskam assumptions in order to calculate speed of sound formula, need explanation" Tags="|fluid-mechanics|acoustics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14327" PostTypeId="2" ParentId="14292" CreationDate="2017-03-20T12:04:38.207" Score="0" Body="&lt;p&gt;The reason for using non-dimensional coefficients is to be able to manipulate the different possible cases needed to be studied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By example, to be able to compare the real life scaled airplane to a prototype scaled model. It is extremely useful since it allows for companies and researchers to be able to simply do all the testing on prototypes and evaluate different possible scenarios that could occur in real life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By doing that, these companies or researchers reduce the amount of real life scaled airplanes for testing to possibly zero, and go straight to manufacturing (in the ideal case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in this &lt;a href=&quot;http://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/fprops/dimension/node6.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The importance of experiments in Fluid Mechanics needs no additional emphasis. Experiments are required in design and testing of vehicles such as aeroplanes, ships and automobiles, pumps, turbines, fans and other equipment. We also have experiments which are carried out from the point of view of understanding a flow and fundamental phenomena such as turbulence. Needless to say that the experiments have to be planned and executed methodically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question we must ask ourselves is :&quot;What are the parameters involved in the given flow? What is their relative significance?&quot; These are some of the questions facing one that conducts any experiment. At the same time, presentation of results in a concise fashion also requires thought and planning. A thoughtless experiment (or for that matter a modern computation) will generate a large body of numbers. It may be difficult to interpret them. A thoughtful presentation will help to render them in a concise and understandable manner. It is here that Dimensional Analysis helps us.&quot;&lt;/p&gt;&#xA;" OwnerUserId="10028" LastActivityDate="2017-03-20T12:04:38.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14328" PostTypeId="2" ParentId="14325" CreationDate="2017-03-20T12:53:32.203" Score="-1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2rUzK.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2rUzK.gif&quot; alt=&quot;Load equations&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the 5th equation from the bottom the displacement is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;disp = Wa^3b^3/3EIL^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can superimpose it for each of your loads. Essentially your beam is loaded with 3.25kg (~33N) at 4 points (1,2,3,4m along).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. disp(total) = disp(1st load) + disp(2nd load) ... + disp(4rth load)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all the loads W = 33N &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the first and fourth a = 1 and b = 4&#xA;for the second and third a = 2 and b = 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E = 69 x 10^9 (Elastic Modulus of Aluminium)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I = 7.11x10^-7 (Second moment of Area)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ends up being: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;disp = 0.001m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the middle x = 2.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So displacement = 0.0025m &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds like it should be fine.  &lt;/p&gt;&#xA;" OwnerUserId="2768" LastActivityDate="2017-03-20T12:53:32.203" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="14329" PostTypeId="1" AcceptedAnswerId="35070" CreationDate="2017-03-20T14:06:22.553" Score="0" ViewCount="4455" Body="&lt;p&gt;I accidentally inserted a mesh edit item and tried moving a node, but now can't remove it. The context menu doesn't have any helpful options and I don't see any relating tools in toolbars. How to remove it?&lt;/p&gt;&#xA;" OwnerUserId="2740" LastActivityDate="2020-04-08T17:42:17.907" Title="How to remove a 'mesh edit' item in ANSYS Workbench?" Tags="|ansys|ansys-workbench|meshing|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14331" PostTypeId="2" ParentId="14326" CreationDate="2017-03-20T15:46:48.530" Score="0" Body="&lt;p&gt;That is a very poor explanation IMO. What he is trying to do is consider a pulse of pressure moving along the tube and find its velocity, but he doesn't describe the assumptions clearly, and some diagrams would be useful! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This derivation follows the same idea, but with more explanation: &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/e/eb/Introduction_to_Physical_Chemistry_Lecture_5_Supplement.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://upload.wikimedia.org/wikipedia/commons/e/eb/Introduction_to_Physical_Chemistry_Lecture_5_Supplement.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally I prefer a different approach that sets up a differential equation (the wave equation) and then solves it, as in the Feynman lectures. It's longer, but doesn't need so many &quot;imaginative jumps&quot; to get from one step to the next. See &lt;a href=&quot;http://www.feynmanlectures.caltech.edu/I_47.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.feynmanlectures.caltech.edu/I_47.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-03-20T15:46:48.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14333" PostTypeId="2" ParentId="14324" CreationDate="2017-03-20T18:03:53.570" Score="1" Body="&lt;p&gt;Your spec sheet lists the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Spectral Detection Range λ 10% of R 240 - 370 nm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That means: over the range 240 to 370 nm, the responsivity is at least 10% of the peak responsivity (R).  So look at the spec for R:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Responsivity R λ = 300 nm, UR = 0 V - 0.14 - A/W&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The responsivity at other wavelengths is shown to be quite small in the following graph.   A brief ride on the InfoSuperHighway suggests GaN detectors in general have that sort of shortwave (and longwave) limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your other questions: rather obviously you cannot determine what the input wavelength is from the electrical output, as there is only one output.  You would need a set of narrowbandpass UV filters to measure a restricted band with this device. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-03-20T18:03:53.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14336" PostTypeId="2" ParentId="12421" CreationDate="2017-03-20T21:06:11.470" Score="3" Body="&lt;p&gt;Well first and foremost, refiners are a business so they must make the decisions that make them the most money. Considering this, lets think about cracking and isomerization via catalysis. The refiner is going to require additional reactors, piping and pumping, most likely additional distillation columns, catalyst loading etc. The energy alone to do distillation is generally very burdensome and if you're dealing with isomerization, you could likely be dealing with similar chemicals that don't separate easily (meaning huge energy and thus monetary operating costs). Catalysts can often be cheap, but usually are made from zeolite bases doped with rare or heavy metals, and can cost anywhere from $5 to thousands of dollars per kilogram, and they don't last forever (sintering, poisoning, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So even without numbers, you can see where it could become very involved for an entity to produce the higher octane materials. But if they can get a premium for it and make money they'll do it right (some do)? Here's another factor: Biofuels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biofuels are currently mandated through the Renewable Fuel Standard for blending into the US fuel supply. This includes many different renewable and biofuels, but the most prominent is fuel ethanol (from starch currently, cellulosic later). The link below provides an overview of the RFS mandate. It is important to note that the parties obligated to blend these biofuels are &lt;strong&gt;Refiners and importers of crude or finished petroleum fuel products&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.epa.gov/renewable-fuel-standard-program/final-renewable-fuel-standards-2017-and-biomass-based-diesel-volume&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.epa.gov/renewable-fuel-standard-program/final-renewable-fuel-standards-2017-and-biomass-based-diesel-volume&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because these entities are required by law (effectively under the clean air act) to blend these fuels, it makes the most sense to utilize them in the best way possible. Ethanol provides a very specific advantage to refiners, as it can drastically impact the final RON, MON, or combined Octane Number of lower quality gasoline fuels. &lt;a href=&quot;https://i.stack.imgur.com/kBD3G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kBD3G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from a business perspective, it makes the most economic sense to invest little to no capital or operating expense in octane enhancing process equipment, meet the mandated obligation and at the same time blend the ethanol with lower quality cuts of fuel but still make the same octane grade to the rack for the distributor / consumer. &lt;/p&gt;&#xA;" OwnerUserId="8369" LastActivityDate="2017-03-20T21:06:11.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14337" PostTypeId="1" AcceptedAnswerId="14347" CreationDate="2017-03-21T00:27:27.837" Score="2" ViewCount="399" Body="&lt;p&gt;I'm doing some vibration tests on planetary gear sets and a lot of the information I´ve read mentions a lot about mode shapes and multiplicities of the natural frequencies. I understand how one can derive the natural frequencies from the equations of motion of the gears and even the eigenvalues, but I don't understand what the multiplicity of the natural frequencies mean. Does it refer to the harmonics of each natural frecuency? If for example there's a fault it will probably be a multiple of the 1st harmonic? Also, does the eigenvalues have to do with the modal modes?&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2017-03-21T13:51:46.713" Title="What does multiplicity of a multiple natural frequency mean?" Tags="|mechanical-engineering|gears|vibration|modal-analysis|frequency-response|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14340" PostTypeId="1" CreationDate="2017-03-21T11:15:53.407" Score="1" ViewCount="402" Body="&lt;p&gt;I'm working on my university project but I can't find a proper  synthesis method for 5 bar mechanism. I'm working on underactuated mechanism. It's a slider crank with the connecting rod divided into 2 links with a pin joint. Thanks!!!!&lt;img src=&quot;https://i.stack.imgur.com/9DCdE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10548" LastEditorUserId="10548" LastEditDate="2017-03-22T05:00:32.037" LastActivityDate="2017-03-22T05:00:32.037" Title="Synthesis of 5 bar mechanism" Tags="|mechanical-engineering|mechanisms|linkage|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14341" PostTypeId="1" AcceptedAnswerId="14405" CreationDate="2017-03-21T11:33:21.197" Score="1" ViewCount="5726" Body="&lt;p&gt;I am trying to recreate in Microsoft Visio a diagram of a photovoltaic power plant that would look like the one below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have some issues for doing so, in particular : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I don't know how to create a &quot;grid&quot; symbol like the one that is at the bottom right of this picture&lt;/li&gt;&#xA;&lt;li&gt;I don't know how to make an &quot;abbreviated line&quot; with tildes like the ones you can see Inside the PV modules at the extreme left of the picture.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can someone give me some help with this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fhtft.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fhtft.png&quot; alt=&quot;Diagram of a photovoltaic power plant&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6497" LastActivityDate="2020-12-07T09:32:56.360" Title="Electrical flowchart in Microsoft Visio : issues" Tags="|photovoltaics|diagram|electrical|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14343" PostTypeId="1" AcceptedAnswerId="14349" CreationDate="2017-03-21T13:04:48.787" Score="5" ViewCount="20258" Body="&lt;p&gt;I want to make a &quot;gearbox&quot; that will transform input rotational movement to an output rotational movement with the same speed but in the different direction (clockwise / counterclockwise) on the same axis. How can this be done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F4FBr.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F4FBr.png&quot; alt=&quot;mystery gearbox&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10552" LastActivityDate="2018-05-30T09:27:52.650" Title="How to reverse the direction of rotational movement on the same axis?" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14345" PostTypeId="2" ParentId="14343" CreationDate="2017-03-21T13:16:59.400" Score="2" Body="&lt;p&gt;Bevel gears to each of the axles and one meshing with both of them attached perpendicular to the gearbox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially a differential where the input shaft is locked.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-03-21T13:16:59.400" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14346" PostTypeId="1" AcceptedAnswerId="14350" CreationDate="2017-03-21T13:49:07.710" Score="0" ViewCount="1141" Body="&lt;p&gt;How do i calculate the load of a glass facade and is it placed uniformly on the edge of the slab or should i place a point load on each edge of the panel of the facade? &#xA;And Can anyone give me a link on an example on how to calculate the load and how to place it.&#xA;Thank you in advance,&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2017-03-21T14:18:59.530" Title="Glass Facade Load on edge of slab?" Tags="|structures|glass|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14347" PostTypeId="2" ParentId="14337" CreationDate="2017-03-21T13:51:46.713" Score="3" Body="&lt;p&gt;&quot;Multiplicity&quot; usually refers to several different modes with the same natural frequency. For example, ignoring any fault conditions and tolerances, if there are several identical planet gears, each one will have the same mode shapes and frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are measuring the vibration response, you may not be able to distinguish these modes - you will measure one &quot;mode&quot; which is a combination of all of them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a fault involving just one of the nominally identical parts, then you may be able to identify different modes for that part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This situation can occur in any structure which is symmetrical - it's not restricted to planetary gearboxes.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-03-21T13:51:46.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14348" PostTypeId="2" ParentId="14307" CreationDate="2017-03-21T13:57:30.710" Score="1" Body="&lt;p&gt;first, I'm not sure that the figure is labelled correctly: it looks like D is colored orange and C is colored yellow. If this is the case, I would assume that $R_C = 0.2m,\; R_D = 0.1m$.  After changing those numbers, I found 2 equations on wikipedia:&#xA;$$N_{s}\omega_{s} + N_{p}\omega_{p} - \left( N_{s} + N_{p} \right) \omega_{c}= 0$$&#xA;$$N_{r}\omega_{r} + N_{p}\omega_{p} - \left( N_{r} - N_{p} \right) \omega_{c}= 0$$&#xA;Where $\omega$ is speed and N is the number of teeth on each gear.  The subscripts are: $s$ for sun (A), $p$ for planet (C,D), $r$ for ring (B), and $c$ for carriage (bar).  I'm confident that radius is an acceptable substitute for number of teeth in this case.  Hope this helps.  &lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-03-21T13:57:30.710" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14349" PostTypeId="2" ParentId="14343" CreationDate="2017-03-21T14:10:32.910" Score="8" Body="&lt;p&gt;Option 1 - just reverse direction without affecting the speed: Bevel gears in differential gear setup, with the carrier (the axle on which the intermediate gears are located) fixed to the chassis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cdcnH.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cdcnH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option 2: reverse direction and change speed by a large amount. Planetary/epicyclic gear, fixed carrier (the green part in the image below) fixed to the chassis with the second shaft connecting to the ring gear (pink). Changing gear ratio between the ring gear and the sun gear (yellow) changes the speed and torque on top of reversing the direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2syd1.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2syd1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option 3: Use two universal joints to offset the shafts relative to each other, so that they are no longer in line, then use a standard gear to reverse the direction, possibly changing speed a little in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1PJVo.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1PJVo.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2017-03-21T14:15:50.247" LastActivityDate="2017-03-21T14:15:50.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14350" PostTypeId="2" ParentId="14346" CreationDate="2017-03-21T14:18:59.530" Score="2" Body="&lt;p&gt;The value of the distributed load is trivial to determine: it is equal to the linear weight of the glass. The linear weight is itself equal to the product of the thickness, height and specific weight of the glass (depends on the type of glass, usually between 24-28 kN/m&lt;sup&gt;3&lt;/sup&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How this load should be placed, however, depends on how the glass is mounted on the structure. Does it rest on a support structure directly on the slab? In this case, apply it as a uniformly distributed load. If, however, it is actually supported by vertical structures which themselves rest on1 the slab, then the load should be concentrated at the supports. In this case, the value of the concentrated force at each support is equal to the distributed load times half of the span between supports (the other half is handled by the other support).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-03-21T14:18:59.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14352" PostTypeId="1" AcceptedAnswerId="14378" CreationDate="2017-03-21T16:11:53.370" Score="1" ViewCount="135" Body="&lt;p&gt;I have used the Equivalent Lateral Load Method from the IBC code to distribute forces and moments on each storey (level) but what I need to know is how to distribute the moment on each shear wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This maybe a stupid question, but I'm just a student. Do I distribute the force &amp;amp; moment on each storey to each shear wall individually or do I distribute shear &amp;amp; &lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="90" LastEditDate="2017-03-22T07:54:46.773" LastActivityDate="2020-04-23T14:00:17.737" Title="How to distribute moment from Lateral Load Method (Static Analysis IBC for example) to Each shear wall individually?" Tags="|design|seismic|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14353" PostTypeId="2" ParentId="14307" CreationDate="2017-03-21T16:37:54.247" Score="1" Body="&lt;p&gt;In solving many kinematic problems for gears, there is a key rule to key in mind: When two gears are meshing, the tangential velocities of both gears at the point of meshing are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, as said in EMiller's answer, gear C seems to be the smaller orange gear, and gear D the larger yellow gear, contrary to the diagram's labelling).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rule will help provide the equations we need to solve this problem. Note that there are 2 pairs of meshing gears (A meshes with C and B meshes with D). The points of meshing for each pair are $P$ and $Q$ respectively. Therefore, this gives us 2 equations to work with. It is then a matter of relating the tangential velocities at the point of meshing with the velocity of the centre of gear C and D. I will define the following terms:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $v_{a,p}, v_{c,p}$ be the tangential velocities of gears A and C, respectively, at point P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $v_{b,q}, v_{d,q}$ be the tangential velocities of gears A and C, respectively, at point P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $V$ be the velocity of the centre of gear C and D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The following diagrams represents gears A and C, and then B and D, as circles that are in pure rolling with one another:)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RNuz2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RNuz2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RLPAV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RLPAV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the rule above, we get the following two equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{a,p}=v_{c,p} \qquad (Eq.1)$$&#xA;$$v_{b,q}=v_{d,q} \qquad (Eq.2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{a,p}$ and $v_{b,q}$ are simple enough to determine since the centres of gears A and B don't move:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{a,p}=\omega_a R_a = 0.4\text{m/s} \qquad v_{b,q}=\omega_b R_b = 0.4\text{m/s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $Eq.1$ and $Eq.2$, we now know the values for $v_{c,p}$ and $v_{d,q}$. Now, all that remains is to determine the velocity $V$ using the values of $v_{c,p}$ and $v_{d,q}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a9KXX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a9KXX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that since gears C and D are welded together, they together form a single rigid body. In general, if you know the velocities at two different points on the body, it is possible to determine the velocity at the centre of the body. There are a few ways of determining this, but it may help to determine to centre of rotation and then the angular velocity of gear C/D in order to find the centre velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps! :)&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2017-03-21T16:37:54.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14355" PostTypeId="1" AcceptedAnswerId="14356" CreationDate="2017-03-21T20:29:30.993" Score="2" ViewCount="1427" Body="&lt;p&gt;I understand that coils have properties over a single wire in producing a magnetic field.  But I'm a hobbyist not an expert.  I know electricity was discovered from a wire carrying a charge when presented in a magnetic field.  So I wonder if that is why we use wire and wire coils to pass through the magnetic field.  (Because that's what was used at the time of discovery.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid copper presents a massive magnetic field, as seen when dropping a strong magnet through the inside of a length of copper pipe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone tried using solid copper plates to generate magnetic fields instead of coiled wire?  Would that work or not?  Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f70E3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f70E3.png&quot; alt=&quot;Coiled Wire vs. Copper Plate electrical generation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10559" LastEditorUserId="615" LastEditDate="2017-03-22T19:22:23.687" LastActivityDate="2019-11-09T07:44:37.533" Title="Why are coils used to generate magnetic fields" Tags="|generator|electrical|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14356" PostTypeId="2" ParentId="14355" CreationDate="2017-03-21T21:06:21.823" Score="6" Body="&lt;p&gt;The current generated by a conductor moving through a magnetic field is related to the length of the conductor and the magnetic field strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A coil is simply a convenient way to get a longer length of conductor within a given space. The cross sectional area doesn't actually matter in terms of generating power except insofar as it determines the current carrying capacity of the wire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like you've slightly misunderstood the configuration of the winding of a motor or generator as one complete loop from N to S is one turn of the coil, you don't just make one leg of the loop into a coil. See &lt;a href=&quot;https://electronics.stackexchange.com/questions/277602/where-does-back-emf-appear-in-motor&quot;&gt;this question&lt;/a&gt; on the electrical site for a good diagram. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2017-03-21T21:14:27.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14357" PostTypeId="1" AcceptedAnswerId="14371" CreationDate="2017-03-21T21:55:15.587" Score="0" ViewCount="265" Body="&lt;p&gt;I understand the Transmission Error is the displacement error of the gear shafts, but if measured how would the signal over time show and why? if its a sinusoid would that mean that only a few times per cycle the displacement occurs or is it always present?&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2017-03-23T01:11:55.457" Title="What is the Gear Transmission Error and How would a Signal Analysis look?" Tags="|mechanical-engineering|gears|dynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14359" PostTypeId="1" AcceptedAnswerId="14361" CreationDate="2017-03-22T02:15:31.103" Score="3" ViewCount="338" Body="&lt;p&gt;I notice from &lt;a href=&quot;https://engineering.stackexchange.com/q/10789/3353&quot;&gt;this question&lt;/a&gt; that the moments at joint connection are somehow &quot;conserved&quot;, or being &quot;transferred&quot; from one element to the others. In other words, there are no &quot;missing moment&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/np5DW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/np5DW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that at joint for left image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2.767\text{ kNm}=(2.37+0.397)\text{ kNm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for bending moment diagram that I examined so far (using structural software), this seems to be &lt;em&gt;always&lt;/em&gt; the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wonder whether this relationship is true for all kinds of joint connection? Or is there a deeper physics principle behind it (such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Conservation_of_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;the conservation of energy&lt;/a&gt;)? How can we derive it from the first physics principle, if it there is a first principle?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-03-22T08:01:24.260" Title="What are the principles (if there are any) behind the conservation of bending moment in frame analysis?" Tags="|structural-engineering|structural-analysis|moments|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14360" PostTypeId="1" AcceptedAnswerId="14369" CreationDate="2017-03-22T02:40:14.760" Score="2" ViewCount="87" Body="&lt;p&gt;If I have a planetary gear set consisting of one sun and N planets, what does it mean if it says that typical vibration modes of 6 natural frequencies always have multiplicity m = 1 for different N, except for the zero natural frequency? How would this multiplicities look in a frequency domain plot?&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2017-03-22T23:43:28.733" Title="What does natural frequencies with multiplicities m=1,2 or n mean?" Tags="|mechanical-engineering|gears|dynamics|vibration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14361" PostTypeId="2" ParentId="14359" CreationDate="2017-03-22T03:13:23.883" Score="8" Body="&lt;p&gt;Well, such an equilibrium (be it of moment, or shear or axial force) is necessary for any static system, and can be trivially demonstrated with Newton's second law. For forces, that is the classic $F=ma$ (force equals mass times acceleration), while for moments it is equal to $M = I\alpha$, where $M$ is moment (or torque), $I$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Moment_of_inertia&quot; rel=&quot;noreferrer&quot;&gt;rotational inertia&lt;/a&gt; (also known as the moment of inertia with dimensions $ML^2$, as opposed to the moment of inertia we commonly discuss in structural engineering, with dimensions $L^4$, which is usually called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Second_moment_of_area&quot; rel=&quot;noreferrer&quot;&gt;second moment of area&lt;/a&gt;), and $\alpha$ is the rotational acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you could calculate the force at a given point as equal to anything other than zero, that would mean that the entire system would accelerate in the net force's direction according to $F=ma$ (if $F\neq0$, then $a\neq0$). Equivalently, if the total moment around a point is non-null, then the entire system will have a rigid-body rotational acceleration (if $M\neq0$, then $\alpha\neq0$). This happens all the time in the real world, of course, but we call those mechanisms, not structures.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-03-22T03:13:23.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14362" PostTypeId="2" ParentId="10675" CreationDate="2017-03-22T10:59:04.253" Score="1" Body="&lt;p&gt;I solved this by converting the stick input to a Nz value. I then subtracted alpha in a certain range from that Nz input. Then converted that to elevator deflection using a integrator comparing actual Nz to the input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found out that in the real aircraft I am simulating a certain stick input correspond to certain commanded G-force, so in all, its a good solution.&lt;/p&gt;&#xA;" OwnerUserId="7236" LastActivityDate="2017-03-22T10:59:04.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14363" PostTypeId="2" ParentId="4170" CreationDate="2017-03-22T14:33:03.150" Score="0" Body="&lt;p&gt;I agree with the answer given by Murenrb but for one thing.  Engines aren't in danger of breaking as much as they're in danger of losing lubrication.  Engine oil in the piston rings forms a bead that seals the gap between the piston and the wall.  Hotter oil has lower viscosity, which means that the bead may blow out, breaking the seal.  This is why engines are liquid cooled, and why towing capacity and radiator size are so strongly correlated.&lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-03-22T14:33:03.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14364" PostTypeId="1" CreationDate="2017-03-22T16:55:57.310" Score="1" ViewCount="58" Body="&lt;p&gt;I am building a device to heat and measure the temperature of a small vessel (double boiler) of gasoline over a range of temperatures say -10F to 120F. I will be measuring the capacitance across a homemade probe to figure out the dielectric constant over this temperature range. What would be the best way to accurately log the temperature while simultaneously measuring the capacitance. Currently, I will be using a 555 timer and microcontroller to measure the capacitance so the same microcontroller could take an analog input from a temperature probe. Would a thermistor work best in this situation? Can I put a thermistor directly into gasoline safely? Would it be better to consider a thermal imaging approach? &lt;/p&gt;&#xA;" OwnerUserId="3551" LastActivityDate="2017-05-09T06:11:33.543" Title="Measuring the temperature of gasoline while conducting experiments over a temperature range" Tags="|electrical-engineering|sensors|petroleum-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14365" PostTypeId="2" ParentId="14364" CreationDate="2017-03-22T17:05:17.743" Score="1" Body="&lt;p&gt;I would use a thermocouple in the gasoline to a data-logger. You can get simple data-logger from omega (&lt;a href=&quot;http://www.omega.com/subsection/temperature-humidity-data-loggers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Temperature &amp;amp; Humidity Data Loggers&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be safe to put in the gasoline, just be sure to use a good fume extraction hood that is suited for combustable vapours and keep all metal in contact with the gas grounded and bonded to prevent sparks.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorUserId="7083" LastEditDate="2017-05-09T06:11:33.543" LastActivityDate="2017-05-09T06:11:33.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14366" PostTypeId="2" ParentId="14364" CreationDate="2017-03-22T18:41:49.343" Score="1" Body="&lt;p&gt;A thermistor will work fine. Most are epoxy dipped, and your test is for a relatively short period of time. A theromcouple would work well also, but they take more supporting electronics for signal amplification. As long as your wires running to your thermistor are not to long, it will work great for this temperature range.&#xA;&lt;br&gt;&lt;a href=&quot;http://www.digikey.com/product-detail/en/vishay-bc-components/NTCALUG03A103H/BC2386-ND/2230714&quot; rel=&quot;nofollow noreferrer&quot;&gt;10k NTC thermistor with lug&lt;/a&gt;&#xA;&lt;br&gt;&lt;a href=&quot;http://www.digikey.com/product-detail/en/murata-electronics-north-america/NXRT15XH103FA1B030/490-8601-ND/3788625&quot; rel=&quot;nofollow noreferrer&quot;&gt;10k NTC thermistor with 1% tolerance&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A data acquisition unit is typically the way engineers do data measurement since using a mictrocontroller requires extra work. If you have it already set up because you will be using this capacitance sensor on said microcontroller in the future that would be a good way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you go the data acquisition (DAQ) route, I recommend &lt;a href=&quot;https://labjack.com/products&quot; rel=&quot;nofollow noreferrer&quot;&gt;Labjack&lt;/a&gt; as it has &lt;a href=&quot;https://labjack.com/support/software/examples/ljm/python&quot; rel=&quot;nofollow noreferrer&quot;&gt;drivers for opensource python&lt;/a&gt;. You would feed both the capacitance (probably a circuit required or even output from your microcontroller), and your temperature to the DAQ unit. National Instruments makes good DAQ units too, but plan on running those with LabView.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the temperature loggers Gwydionforge recommended in his answer would work, but it is very difficult to correlate data captured on two different systems. The data will have to be combined in a spread sheet or scripting program. If the time stamps are not identical you will have to resample (interpolate) to get a capacitance and temperature at each timestamp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Soapbox safety speech: Gasoline vapors are very flammable; do your test outside or in a hood. Make sure to wear safety glasses and a face shield.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-03-22T18:41:49.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14367" PostTypeId="1" AcceptedAnswerId="14380" CreationDate="2017-03-22T19:31:06.240" Score="1" ViewCount="4197" Body="&lt;p&gt;If I have a transfer slab in the basement level how to find the tributary area for the columns because the columns in the upper floors have different location and tributary area than that of the basements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please if possible to show an example &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NbDlS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NbDlS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="8055" LastEditDate="2017-03-24T14:11:11.270" LastActivityDate="2017-03-24T14:11:41.240" Title="How to find tributary area for a column when I have a transfer slab?" Tags="|structural-engineering|columns|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14369" PostTypeId="2" ParentId="14360" CreationDate="2017-03-22T23:43:28.733" Score="2" Body="&lt;p&gt;I cannot find a freely available PDF of it, but if you have access to ASME journals, then this paper lays out the basics of planetary gear natural frequencies (at least in the most basic case, there are many later papers with different variations in assumptions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lin J, Parker RG. Analytical Characterization of the Unique Properties of Planetary Gear Free Vibration. ASME. J. Vib. Acoust. 1999;121(3):316-321. doi:10.1115/1.2893982.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As outlined in this paper, planetary gears have three types of modes: rotational modes, translational modes, and planet modes.  Rotational modes always have multiplicity 1, translational modes always have multiplicity 2, and planet modes have multiplicity N-3.  What the multiplicity means in that there is more than one vibration mode (eigenvector) with the same natural frequency (eigenvalues).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For translational modes, here is how to think about the multiplicity: there could be one mode where everything is translating vertical (e.g. just up down... in reality it might not be exactly up/down, but let's just assume it is for now).  There is a completely separate mode where everything is translating laterally (just left right).  Now, let's call the stiffness in the vertical direction $k_v$ and the stiffness in the horizontal direction $k_h$.  The mass is the same for both cases, let's call it $m$.  So the vertical natural frequency is $\sqrt{k_v/m}$ and the horizontal natural frequency is $\sqrt{k_h/m}$.  But because planetary gears are axisymmetric, $k_v = k_h$.  So the two natural frequencies are identical.  They are two completely separate modes, that just happen to have the same frequency.  That is what is meant by multiplicity 2.  If you were to make the bearings or some other part of the structure slightly stiffer in one direction, such that $k_v&amp;gt;k_h$, then it's no longer symmetric and you would have two different frequencies each of multiplicity 1.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-03-22T23:43:28.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14371" PostTypeId="2" ParentId="14357" CreationDate="2017-03-23T01:11:55.457" Score="2" Body="&lt;p&gt;Transmission error would usually be at the tooth mesh frequency. i.e. if you have N teeth on the gear, then you would see N cycles of oscillation over one revolution.  It is probably not a pure sinusoid, depending on your type of gear, but it is periodic.  If you just google &quot;dynamic transmission error&quot;, you can find many papers with plots of time histories.  e.g. &lt;a href=&quot;https://core.ac.uk/download/pdf/337497.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://core.ac.uk/download/pdf/337497.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-03-23T01:11:55.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14372" PostTypeId="1" CreationDate="2017-03-23T05:14:04.780" Score="1" ViewCount="61" Body="&lt;p&gt;EMA and OMA are two methods used for determining the mode shapes of the natural frequencies of mechanical systems. I understand it until that point. I know that for EMA we need an artificial excitation like a hammer and measure the response.&#xA;But what is the final objective of these methods? Is it to model the dynamical system or to obtain the parameters? Or maybe it is a vibration analysis or measurement method? It´s not quite clear what the answer is.&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2021-08-11T13:23:49.447" Title="What is Experimental and Operational MODAL ANAYLIS?" Tags="|mechanical-engineering|dynamics|modal-analysis|frequency-response|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14373" PostTypeId="2" ParentId="13861" CreationDate="2017-03-23T05:52:38.603" Score="0" Body="&lt;p&gt;Spring pin connector&lt;br&gt;&#xA;Spring loaded connector&lt;br&gt;&#xA;Spring-loaded connector&lt;br&gt;&#xA;Spring loaded pin connector&lt;br&gt;&#xA;Magnetic spring connector (obviously only if magnetic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=spring%20pin%20connector&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google image search &quot;spring pin connector&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mill-max.com/new_products/detail/74&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mill-Max spring-loaded connector&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-03-23T05:52:38.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14375" PostTypeId="1" CreationDate="2017-03-23T07:36:25.183" Score="0" ViewCount="51" Body="&lt;p&gt;The deflection of the sector is assumed in the following form&#xA;&lt;a href=&quot;https://i.stack.imgur.com/670X7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/670X7.png&quot; alt=&quot;deflection&quot;&gt;&lt;/a&gt;&#xA;But I don't know why they assumed G(θ) in the following form &lt;a href=&quot;https://i.stack.imgur.com/AMiXa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AMiXa.png&quot; alt=&quot;G(θ)&quot;&gt;&lt;/a&gt; &#xA;and how to compute η.Can somebody help me? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="10574" LastActivityDate="2017-03-23T23:24:08.727" Title="Vibration of annular sector plate" Tags="|vibration|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14376" PostTypeId="1" CreationDate="2017-03-23T08:06:39.880" Score="0" ViewCount="83" Body="&lt;p&gt;I have cylinder model that I control with 4/3 directional valve. When 4/3DV control input is 10V, cylinder moves forward and when valves control input is -10V, then cylinder moves backward. I can get any data from cylinder(position, speed, acceleration,force). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to make control system for this. I would like to make control system that I can set target position and target speed for system. I could use PID controller, but what parameter to control? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just asking what is the best way to make this kind of control system, be fore I start to make it. &lt;/p&gt;&#xA;" OwnerUserId="6971" LastActivityDate="2017-08-18T20:16:27.140" Title="Cylinder position and speed control" Tags="|control-engineering|pid-control|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14377" PostTypeId="1" CreationDate="2017-03-23T13:45:13.783" Score="0" ViewCount="72" Body="&lt;p&gt;I was wondering where the negative sign comes from in the following simplification for diffusion in 1 dimension.  The initial equation is given as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial C_i}{\partial t} A \Delta x= (N_{ix}|_{x} - N_{ix}|_{x+\Delta x}A + R_iA \Delta x )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial C_i}{\partial t} = -\frac{\partial N_{ix}}{\partial x} + R_i$$&lt;/p&gt;&#xA;" OwnerUserId="4781" LastEditorUserId="7627" LastEditDate="2017-04-21T16:04:44.407" LastActivityDate="2017-04-21T16:04:44.407" Title="Diffusion in one dimension" Tags="|chemical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14378" PostTypeId="2" ParentId="14352" CreationDate="2017-03-23T14:15:16.963" Score="1" Body="&lt;p&gt;Your question seems like it is incomplete so it's difficult to provide a complete answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the moment computation, what you'll need to do is compute the centroid of rotational stiffness, which is determined by the relative stiffness of each shear wall in the X and Y directions. Then you'll compute the moment arm based on the distance between the stiffness and loading centroids, which will give you the applied torque (or moment). The torque is then distributed to each shear wall in relation to its relative stiffness and its distance from the stiffness centroid. For pure shear (ie, ignoring rotation), you'll do a similar procedure but using forces instead of moments/torque and ignore the contribution of the walls perpendicular to the force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to do any of this, the first thing you have to do is compute the relative stiffness of each shear wall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't know how to do this, for a quick primer I recommend Seismic Design of Building Structures by Lindeburg. &lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2017-03-23T14:22:34.847" LastActivityDate="2017-03-23T14:22:34.847" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14379" PostTypeId="2" ParentId="14377" CreationDate="2017-03-23T14:46:08.470" Score="0" Body="&lt;p&gt;because you go from high to low.&#xA;For example, if you want to go from 10 to 1, you &quot;move forward -9 steps&quot;&lt;/p&gt;&#xA;" OwnerUserId="6488" LastActivityDate="2017-03-23T14:46:08.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14380" PostTypeId="2" ParentId="14367" CreationDate="2017-03-23T15:14:30.617" Score="1" Body="&lt;p&gt;As I understand your question, you are asking how to determine how the loads from the upper floors are divided between the lower columns. For instance, column C14 is going to transmit its loads to C12, C17, C18 and the unnamed wall near the core (assuming its load-bearing). How much of the load will go to each of the columns?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Curiously enough, the answer to this question depends on the chosen foundation for your structure and whether they allow rotations (such as shallow foundations) or don't (such as multi-pile foundations).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;&quot;Pinned&quot; Foundations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In the case of foundations which allow rotations, then you can roughly approximate the answer in this case (where the load is relatively centered between the four surrounding columns) by dividing the load equally between the four surrounding columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a rough estimate, because the load actually isn't centered between the four columns. Also, the core wall's vertical stiffness is actually much larger than that of the other columns, so it will certainly pull in more of the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simply supported beam with a concentrated load $P$ anywhere on the span (at a distance $a$ from the left support and $b$ from the right), the reactions are equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;R_{left} &amp;amp;= P\dfrac{b}{a+b} \\&#xA;R_{right} &amp;amp;= P\dfrac{a}{a+b}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps one could use something similar to better estimate the distribution of the load given that it isn't centered in the slab, but I don't know what that equation would be. It also would do nothing to solve the problem (which I believe might be significant) of the core wall's significantly larger vertical stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve that problem, you'd need to use a FEM program to model the structure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;&quot;Fixed&quot; Foundations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Tributary areas aren't really adequate in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orientations of the supporting columns aren't the same. This means that their rotational stiffnesses around the x and y axes will be different: column C12 will be far more rigid than C17 and C18 for rotations around the y-axis, while C17 and C18 will be more rigid around the x-axis. And this doesn't even mention the fact that the core structure will be insanely stiff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, let's look at how the load would be distributed between C17 and C18 if it were perfectly centered between them. Each of the columns can be represented by a rotational spring which represents their rotational stiffness around the relevant axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nmexH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nmexH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values themselves don't matter, but it is clear that the equal stiffness of the columns around the relevant axis means that the load is equally distributed between them (assuming it is centered, which it actually isn't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the distribution between C18 and C12, however, the former will be far stiffer (I'd guess some 10 times stiffer):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nVccE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nVccE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C18 therefore absorbs some 50% more of the load than C12, even though the load is centered between them. The same concept applies for C17 and the core, though in this case, the core would be approximately infinitely stiff:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GIREq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GIREq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The core'll therefore absorb more of the load than C17.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering just these local effects already shows that the distribution is non-trivial. However it isn't adequate to consider the isolated structure around each of the upper columns; one must consider global effects. The edge columns will always pull less of a reaction than the central ones due to continuity effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, let's look at how the load from C14 is distributed between C18 and C12, but remembering that there are also C4 and C9 in the distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5i6Tp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5i6Tp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The continuity effects of C4 and C9 mean that C12 absorbs 30% more load than we thought by looking at only C18 and C12 in isolation, while C12's reaction is slightly reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this is to show that this is very non-trivial; the models I've shown here are obscenely simplified, since they look at the load distribution along a line, without taking into consideration the influence of nearby structural elements. For instance, the continuity effect from C18-C12-C9-C4 is probably even stronger, since the nearby presence of the massively stiff core will probably make the span from C12-C9 even stiffer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reasonable recommendation in my opinion is to use a FEM program to model the entire structure, including the slab, at least floor-by-floor. Any &quot;manual&quot; method will be highly simplified and inadequate by modern standards.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Why the difference?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The easiest way to demonstrate this difference is with another model, now actually showing the support columns and the &quot;foundations&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sHtPY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sHtPY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the model to the left, the pinned foundations mean that the load is equally distributed, even though the columns have different stiffnesses (but equal area). To the right, however, the fixed foundations cause the load distribution to be uneven, with more load being absorbed by the stiffer column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can also be seen in the difference of the deflection of the models:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AryvE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AryvE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pinned supports allow the columns to rotate freely, eliminating any potential rotational stiffness. With the fixed supports, however, the columns do resist the beam's attempt to rotate the nodes.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-03-24T14:11:41.240" LastActivityDate="2017-03-24T14:11:41.240" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14381" PostTypeId="2" ParentId="14341" CreationDate="2017-03-23T18:54:33.813" Score="3" Body="&lt;p&gt;I had to draw electrical and hydraulic schematics in Microsoft Visio at a past job. I'm glad that is behind me because that software is barely good enough for making flow charts. I would never recommend anyone use it for engineering diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are expensive professional software programs that solve this problem much better, but if you are on a budget I would recommend &lt;a href=&quot;https://inkscape.org/en/about/screenshots/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Inkscape&lt;/a&gt;. It is much more stable, predictable, and has community/forum support. There is also &lt;a href=&quot;http://www.digikey.com/schemeit/project/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Digikey Scheme-it&lt;/a&gt;. It looks good aside from being proprietary, but I have not used it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addressing your question: Visio supports grouping. You can also add snap points inside of groups. To draw your hashed box add a bunch of snap points around the border, connect them with your hash lines, then group all of them. You will have to make an abbreviated line symbol as a separate object with two snap points. Its irritating, but is possible. Welcome to Visio; I wish you better software in the future ;-)&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-03-23T18:54:33.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14382" PostTypeId="2" ParentId="12424" CreationDate="2017-03-23T19:34:08.450" Score="0" Body="&lt;p&gt;You first need to set your operational criteria:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the minimum environment temperature:&lt;/li&gt;&#xA;&lt;li&gt;What is the maximum environment temperature:&lt;/li&gt;&#xA;&lt;li&gt;What are the physical size constraints?&lt;/li&gt;&#xA;&lt;li&gt;How much additional power can I use to maintain this temperature?&lt;/li&gt;&#xA;&lt;li&gt;How constant does the temperature need to be? (ex +/-1degC)&lt;/li&gt;&#xA;&lt;li&gt;Does only a portion of the unit need to have a constant temperature?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would not recommend messing around with cooling if you want the temperature to be very constant. I would operate the unit at the high end above your highest environmental temp. Perhaps something like 40C as your constant temp and use a heating element surrounding the unit to hold the temperature up in cold environments. Heating also takes less power and equipment than cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your space constraints are not too tight, some high quality insulation would be the easiest start. Closed cell foam or ceramic fiber would work well. Vacuum insulated like a thermos beverage container might be good too look at too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you cant increase the form factor you will need to add a uniform heating element around the outside of the unit and control it with microcontroller and thermistor using a pid loop. This will obviously cost some additional power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, most temperature dependent sensors and systems don't attempt to modify their environmental temperature, but instead use a micocontroller to digitally measure the temperature and compensate the value for that temperature.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-03-23T19:34:08.450" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14384" PostTypeId="2" ParentId="14375" CreationDate="2017-03-23T21:21:36.773" Score="1" Body="&lt;p&gt;The governing equation for free vibrations of a circular plate is &lt;a href=&quot;https://en.wikipedia.org/wiki/Vibration_of_plates#Circular_plates&quot; rel=&quot;nofollow noreferrer&quot;&gt;(see Wikipedia article on vibrating Kirchhoff plates)&lt;/a&gt;&#xA;$$&#xA;\frac{1}{r}\frac{\partial }{\partial r}\left[r \frac{\partial }{\partial r}\left\{\frac{1}{r}\frac{\partial }{\partial r}\left(r \frac{\partial w}{\partial r}\right)\right\}\right] = -\frac{2\rho h}{D}\frac{\partial^2 w}{\partial t^2}\,.&#xA;$$&#xA;Separation of variables leads to a general solution of the form&#xA;$$ &#xA;w(r,t) = \sum_{n=1}^\infty C_n\left[J_0(\lambda_n r) - \frac{J_0(\lambda_n a)}{I_0(\lambda_n a)}I_0(\lambda_n r)\right]&#xA;    [A_n e^{i\omega_n t} + B_n e^{-i\omega_n t}] \,.&#xA;$$&#xA;The term&#xA;$$&#xA;  A_n e^{i\omega_n t} + B_n e^{-i\omega_n t}&#xA;$$&#xA;can be expressed as a sum of sine, cosines, and hyperbolic sines and cosines when we take the real part of $w$.  That's why the expression for $G(\theta)$ in your book has that form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a annular sector of a circular plate, things are a bit more complicated and you will have to do the algebra yourself (or look it up from a book) to find out what $\eta$ is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  The correct term for these orthogonal functions is &lt;em&gt;basis&lt;/em&gt; functions (and not &lt;em&gt;basic&lt;/em&gt; functions) - see &lt;a href=&quot;https://en.wikipedia.org/wiki/Basis_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia basis function&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2017-03-23T21:21:36.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14385" PostTypeId="2" ParentId="14375" CreationDate="2017-03-23T23:24:08.727" Score="0" Body="&lt;p&gt;Biswajit covered most of it, but to build on his answer a little: when you are solving these types of problems, your answer has to satisfy both the partial differential equation and the boundary conditions.  $G(\theta)$ as given in your question is the most general solution to the PDE.  But it will not satisfy the boundary condition for all values of $\eta$, A, B, C, and D.  e.g. a clamped boundary at 0 would be expressed as $G(0)=0$, $G'(0)=0$. If you picked $A=B=D=0$, and $C=1$, you can see that this does not satisfy the boundary conditions, because $cos(0)=1$.  But there is some other combination of parameters that will satisfy it.  To solve for the 4 unknowns, you will need to apply 4 boundary conditions, 2 at each end of your sector.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-03-23T23:24:08.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14386" PostTypeId="1" AcceptedAnswerId="14402" CreationDate="2017-03-23T23:27:57.097" Score="1" ViewCount="2595" Body="&lt;p&gt;Here's the excercise: A tensile test is carried out on a mild steel specimen of gauge length 40mm and cross-sectional area 100mm&lt;sup&gt;2&lt;/sup&gt;. The results obtained for the specimen up to it's yield point are given below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Load (kN)     | 0 |   8   |  19   |  29   |  36&#xA;Extension(mm) | 0 | 0.015 | 0.038 | 0.060 | 0.072&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need to determine the gradient of the load / extension graph.&#xA;When the data is plotted (in the book) on a graph it gives a straight line therefore the gradient should be any load (of the given values) divided by the &#xA;corresponding extension gradient = load/(extension/1000) except e.g. the first one (8kN) gives 533*10&lt;sup&gt;6&lt;/sup&gt; while the third one (19kN) gives 500*10&lt;sup&gt;6&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why and which one do I use for the calculations?&lt;/p&gt;&#xA;" OwnerUserId="10583" LastEditorUserId="1832" LastEditDate="2017-03-23T23:40:45.120" LastActivityDate="2017-03-24T15:10:35.870" Title="How to determine gradient of load/extension graph?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14387" PostTypeId="1" AcceptedAnswerId="14391" CreationDate="2017-03-23T23:33:01.417" Score="0" ViewCount="61" Body="&lt;p&gt;Could the rotor be shaped like a fan inside the stator of an AC induction motor. And could the stator act as a duct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this lead to a high airflow:materials used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="10361" LastActivityDate="2017-03-24T01:57:23.950" Title="AC induction motor" Tags="|electrical-engineering|motors|ac|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14388" PostTypeId="1" CreationDate="2017-03-24T00:15:19.037" Score="0" ViewCount="89" Body="&lt;p&gt;For moment about D , how could the FGF sin26.6 generate moment about D ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FGF sin26.6 is located above point D , right ? How could FGF sin26.6 generate moment about D ?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yPEgM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yPEgM.png&quot; alt=&quot;fsdfsf&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fy9xU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fy9xU.png&quot; alt=&quot;ddada&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10127" LastActivityDate="2017-03-24T15:21:59.417" Title="cut section method in truss" Tags="|structural-engineering|structural-analysis|statics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14389" PostTypeId="2" ParentId="14372" CreationDate="2017-03-24T00:17:58.070" Score="2" Body="&lt;p&gt;Typically you are not doing modal analysis for its own sake.  Typically you have a problem and you are using modal analysis as a tool to solve that problem. The goal of the modal analysis will depend on what exactly your problem is. Here are some examples&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) You have a part which is exposed to vibration, and is failing (cracking) due to high cycle fatigue. Often in these cases it is because the part has a natural frequency that is close to the excitation frequency, and so the stresses in the part are being amplified by resonance. One possible fix is to redesign the part to raise the natural frequency away from the excitation frequency. Understanding the mode shape will help you understand the best way to raise the natural frequency to avoid the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Your customers are complaining that your device is too loud / generating too much noise.  A modal analysis will help you understand how the vibration is being transmitted / radiated into sound, and thus help you redesign the part to reduce the noise.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-03-24T00:17:58.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14390" PostTypeId="2" ParentId="14372" CreationDate="2017-03-24T01:53:30.003" Score="2" Body="&lt;p&gt;The most basic purpose of the measurements is &lt;em&gt;system identification&lt;/em&gt; - i.e. to determine the modal frequencies, mode shapes, and damping factors for the modes in some frequency range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use that information for many different purposes. If you measure it in sufficient detail, you can construct a mathematical model of the structure that can be used in the same way as finite element model. Alternatively, you can use the measurements to verify that a FE model is accurate, or fine-tune the FE model to improve its accuracy. Creating a FE model &quot;from first princples&quot; may be inaccurate if the structure contains joints with unknown flexibility, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually there is no way to model &lt;em&gt;damping&lt;/em&gt; accurately from first principles, but measured damping ratios can be included in a finite element model to improve on &quot;guesses&quot; of the damping level based on general experience of how  similar structures behave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that with modern measurement techniques, you can sometimes measure in &lt;em&gt;more&lt;/em&gt; detail than a finite element model will provide. For example, using a scanning laser Doppler vibrometer, you can quickly measure the vibration response of literally thousands of points on a machine such as an engine or gearbox while it is operating - though there are some limitations in what is measureable, for example the laser beam has to be able to &quot;see&quot; the part being measured.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-03-24T01:53:30.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14391" PostTypeId="2" ParentId="14387" CreationDate="2017-03-24T01:57:23.950" Score="2" Body="&lt;p&gt;It could, yes. But the tradeoff would be worse performance. The strength magnetic field is a function of distance from the coil, so increasing the volume by turning your rotor into a fan shape would reduce the efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Internal_fan-cooled_electric_motor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Typically, this is dealt with by close mounting a fan, sometimes inside the motor case&lt;/a&gt;. The motor stays compact and efficient while the cooling is provided by a fan specially built for the purpose.&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2017-03-24T01:57:23.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14393" PostTypeId="1" CreationDate="2017-03-24T03:37:41.010" Score="3" ViewCount="200" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I am building an engine from scratch for the sole purpose of becoming a better mechanical engineer--I'm currently a second year undergrad. The engine I'm designing is a single piston 4-stroke gasoline engine. I have no intention of using it on anything yet and only want to make it run reliably.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I'm using fuel injection and I've heard I need to control the pressure on the line fairly accurately. I'd like to avoid using a fuel pump in favor of something I can actuate with a microcontroller. One of the harder things I've found in mechanical engineering is that it is difficult to find what parts exist that can solve your problem. So my question is, what are some ways that I can do this if I don't use a fuel pump? I really want to become a good engineer so any insight is greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="7503" LastEditorUserId="7503" LastEditDate="2017-03-24T05:41:55.973" LastActivityDate="2017-07-23T18:18:30.600" Title="How can I build and maintain an accurate pressure on a fuel line without using a fuel pump?" Tags="|mechanical-engineering|pressure|automotive-engineering|pumps|actuator|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14395" PostTypeId="1" AcceptedAnswerId="14398" CreationDate="2017-03-24T05:59:31.990" Score="0" ViewCount="47" Body="&lt;p&gt;I was wondering why there isn't a 9/20&quot; diameter bolt and then realized the bolt sizes start at what ever size, the next size is + 39/1250 of an inch. Is there any reason for this? Is there any place to find a 9/20&quot; hardware?&lt;/p&gt;&#xA;" OwnerDisplayName="user10585" LastEditorUserId="90" LastEditDate="2017-03-24T07:33:47.590" LastActivityDate="2017-03-24T09:17:29.937" Title="why does sae increment by .0312" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14396" PostTypeId="1" CreationDate="2017-03-24T06:14:05.930" Score="0" ViewCount="919" Body="&lt;p&gt;I have a residential Tower (36 story) and i want to find the design loads (Live Loads and Sustained Dead Loads in t/m&lt;sup&gt;2&lt;/sup&gt;) for Technical and Parking floors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently using the ASCE (Table 4-1 Minimum Uniformly Distributed Live Loads, Lo, and Minimum Concentrated Live Loads) as a reference to find the loads for the typical floors but it doesn't refer to either technical Floors nor Parking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone know any reference that I could use to find these loads I would appreciate it.&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="1832" LastEditDate="2017-03-24T10:44:46.107" LastActivityDate="2017-03-24T10:44:46.107" Title="What should I take Live Load and Sustained Dead Load for Parking and Technical Floors?" Tags="|structural-engineering|civil-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14397" PostTypeId="2" ParentId="14393" CreationDate="2017-03-24T07:22:42.910" Score="2" Body="&lt;p&gt;&lt;strong&gt;Edit to clarify&lt;/strong&gt;&lt;br&gt;&#xA;As of now, The question is a bit unclear. I imagine you want a largely pressureless fuel system, each cylinder has it's own solenoid pump that replaces the fuel valve on this cylinder and is controlled by your microcontroller. So the fuel supply is not done via supplying a constant pressure to a &lt;a href=&quot;https://en.wikipedia.org/wiki/Carburetor&quot; rel=&quot;nofollow noreferrer&quot;&gt;carburetor&lt;/a&gt; that does the mixing, but by directly dosing fuel into each cylinder. Since you are building a sort of &lt;a href=&quot;https://en.wikipedia.org/wiki/Gasoline_direct_injection&quot; rel=&quot;nofollow noreferrer&quot;&gt;direct injection engine&lt;/a&gt;, you would likely not control the air flow, but just the fuel flow between lean and stochiometric. In &lt;a href=&quot;https://en.wikipedia.org/wiki/Common_rail&quot; rel=&quot;nofollow noreferrer&quot;&gt;common rail&lt;/a&gt; you'd have one high pressure fule lin and solenoid valves, instead the OP wants to have a low pressure line and solenoid pumps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the assumption I base my answer on, if this is not the engine the OP wants to build everything below is moot.&lt;br&gt;&#xA;&lt;strong&gt;End edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are looking for (search terms) solenoid driven diaphragm pumps or solenoid driven piston pumps. These exist as metering pumps. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Diaphragm_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;working principle of diaphragm pumps&lt;/a&gt; is explained on wikipedia, in metering pumps with solenoids there is only one chamber and the solenoid works in tandem with a spring. AFAICT these are mostly metering pumps for pharmaceutical or chemical industry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of these pumps each stroke will deliver the same volume. Is this acceptable? If not, &lt;a href=&quot;http://www.lmipumps.co.uk/products/c1-adjustable-stroke--speed&quot; rel=&quot;nofollow noreferrer&quot;&gt;variable stroke solenoid pumps exist&lt;/a&gt;, but I'm not sure you'll find ones where you can adjust the stroke online, on the fly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found &lt;a href=&quot;https://www.knf.com/products/oem-pumps/product/products/solenoid-diaphragm-dosing-pumps/solenoid-dosing-pumps/&quot; rel=&quot;nofollow noreferrer&quot;&gt;one supplier where the volume per stroke is in the µl range&lt;/a&gt;. Is this acceptable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supplier linked above states a frequency of 0-20Hz for it's pumps, so in a 4 cylinder engine you get max 80 Hz = 4800 rpm a most. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will likely need some pressure on your fuel line upstream of the pumps or they won'T fill reliably. Maybe it's sufficient to mount the fuel tank a few m up, and have a rather fat pipe and manifold upstream the pumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit to clarify:&lt;/em&gt; For fuel injection, you need a precise control over the volume. In common rail injection, this is achieved by maintaining a constant pressure upstram of tightly timed solenoid valves. When replacing the valves with pumps, you wouldnt need a constant pressure to achieve control over volume flow, 'just' enough pressure to overcome static pressure in cylinder and transport losses. Control over fuel amount would be as god or bad as your pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, I think what you need exists but it will be likely expensive and may not work very well in the context of your engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if your goal is to become a better mechanical engineer, you invent/design a variable stroke diaphragm metering pump where the stroke volume can be adjusted online, qickly, via analogous signal and the individual strokes are controlled by a digital signal. &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2017-03-27T06:56:03.850" LastActivityDate="2017-03-27T06:56:03.850" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14398" PostTypeId="2" ParentId="14395" CreationDate="2017-03-24T09:17:29.937" Score="4" Body="&lt;p&gt;0.03125 = 1/32 (exactly). The USA (and the UK before metrication) had a long tradition of subdividing basic measurements by factors of 2. For example the standard sizes of drills are usually state in 32ths or 64ths of an inch. This principle was used for other measures as well as lengths - e.g. gallons, quarts, pints - and sometimes for larger measures as well as fractions - e.g. pecks (4 gallons) and bushels (8 gallons), or weight units like hundredweights (8 stones).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The decimal conversions are more convenient to use with modern measuring devices than fractions, but anything less than 0.1 thousandth of an inch is not of much practical significance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The list of standard sizes has small enough increments that there would be no particular use for an &quot;odd size&quot; like 9/20&quot; = 0.45 in, when there is already a standard size 7/16&quot; = 0.4375 in. Most likely, this would just lead to confusion and accidental use of the incorrect size bolts, which could cause problems if the wrong replacements parts, or tools like wrenches, were under or over size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course in some very specialized applications, bolts and other fasteners might be designed to &lt;em&gt;any&lt;/em&gt; size to meet the required specification, if the cost of the parts is not a significant issue. Such parts are likely to be made from special materials, as well as in special sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard metric bolt sizes are more straightforward, because the bolt diameters and head sizes are all integer numbers of millimeters.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-03-24T09:17:29.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14399" PostTypeId="2" ParentId="6865" CreationDate="2017-03-24T11:37:53.107" Score="0" Body="&lt;p&gt;What you are looking for is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Variable_displacement_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;Variable displacement pump&lt;/a&gt;. These often take the shape of &lt;a href=&quot;https://en.wikipedia.org/wiki/Axial_piston_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;axial piston pumps&lt;/a&gt; with a (somewhat) movable swashplate. I've seen us in use in hydraulic systems (low volume flow, high pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://axleaddict.com/auto-repair/Variable-Displacement-Compressors&quot; rel=&quot;nofollow noreferrer&quot;&gt;Variable displacement compressors&lt;/a&gt; also exist:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J3sFf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J3sFf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be modified with a control scheme that controls displacement for a constant power at the shaft.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-03-24T11:37:53.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14401" PostTypeId="2" ParentId="14393" CreationDate="2017-03-24T14:14:17.250" Score="2" Body="&lt;p&gt;Without a pump you will have trouble maintaining flow. The pressure gauge on a fuel regulator I looked for maxed out at 140 psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can build pressure with an actuator but it will only run till the cylinder you make is empty. If you use a large actuator and a large cylinder (the size of a gas can) you might get it to work but that is not an efficient way to do it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are insistent on not using a pump, consider using a fuel bladder or making a cylinder and just putting a weight on top until you get sufficient pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your concern for the system is that you worry about the pump running but not moving much or enough fuel to keep it cool (which can be a problem) you should install a return line to the tank with a pressure regulator. That way you can keep constant pressure and keep the pump cool. This will also limit the amount of pressure change as the fuel flow to the engine changes.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:18:30.600" LastActivityDate="2017-07-23T18:18:30.600" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14402" PostTypeId="2" ParentId="14386" CreationDate="2017-03-24T15:10:35.870" Score="2" Body="&lt;p&gt;Theoretically, the elastic range of a material is linear. But as Yogi Berra said, &quot;In theory there is no difference between theory and practice. In practice there is.&quot; This data is obtained from a physical experiment, so errors are natural.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the piece or the strain gauge was incorrectly placed. Or, even more likely (given the data), you were dealing with a specimen which simply wasn't perfect (none ever is). Slight microimperfections meant that even in the elastic regime you might have a result which isn't perfectly linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to find a best estimate of what the &quot;theoretical&quot; Young's modulus is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the results for each of the points, take your pick:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;disp (mm) | F (kN)  | F/d (kN/mm)&#xA;----------+---------+------------&#xA;0         | 0       | -&#xA;0.015     | 8       | 533.3&#xA;0.038     | 19      | 500&#xA;0.060     | 29      | 483.3&#xA;0.072     | 36      | 500&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It is visually clear that the result is hovering around 500.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unsure of how serious the statistical analysis needs to be.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You can simply say 500 looks reasonable.&lt;/li&gt;&#xA;&lt;li&gt;You can take an average, which gives you 504&lt;/li&gt;&#xA;&lt;li&gt;You can do a least-squares analysis, which gives you 490.6, but with a line crossing the y-intercept at 0.25, as opposed to zero, as would be expected.&lt;/li&gt;&#xA;&lt;li&gt;You can force the least-squares to cross through zero, which gives you 495.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They all give you the following results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rTzzS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rTzzS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kpix6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kpix6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pqEFS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pqEFS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-03-24T15:10:35.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14403" PostTypeId="2" ParentId="14388" CreationDate="2017-03-24T15:21:59.417" Score="1" Body="&lt;p&gt;F.GF sin (26.6) is the vertical component of the force in the top member, but you can't simply shift the point of application of that force.  You can shift it where you like along the dotted line (ie, the line of action of the force), so you can shift it to O, at which time the moment of that force about D is indeed F.GFxsin(26.6)x6m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a singularly clumsy way of doing it though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For moment of F.GF about D I would do F.GF x 3m x cos(26.6), ie full F.GF force times the perpendicular distance between D and the line of action of the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, 3 cos(atan(0.5)) is exactly the same number as 6 sin(atan(0.5)), so it makes no difference to the answer.&lt;/p&gt;&#xA;" OwnerUserId="902" LastActivityDate="2017-03-24T15:21:59.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14404" PostTypeId="1" AcceptedAnswerId="14409" CreationDate="2017-03-24T15:56:29.570" Score="1" ViewCount="339" Body="&lt;p&gt;I have a tube with a circular cross-section, this tube protrudes from the ground vertically for a section, then arcs over to one side. At the end of the tube, a flat plate is mounted to the end of it. The arc is of a constant radius but is not necessarily a 90 degree bend. (i.e. the angle of the flat plate to the ground plane is not necessarily 90 degrees). How would you go about calculating, by hand,  how much load the flat plate could hold before the tube bends beyond it's yield point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming all geometries are known and the material properties are known and constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been working on a solution using a combination of formulas from Roarks book. But, I'd like to see how others would approach this problem.  &quot;Solve using FEA&quot; is not an answer i'm looking for, but rather a reasonably accurate hand calculation approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;-dimensions and loading direction added&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximate tube dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vertical straight section: 1m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;'Bent' section arc radius: 125mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flat end, 10 degrees to floor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tube OD - 25mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tube ID - 23mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loading direction will be in the negative z-axis i.e. gravitational loading only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GIgrr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GIgrr.png&quot; alt=&quot;Tube Sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10590" LastEditorUserId="10590" LastEditDate="2017-03-24T19:07:42.873" LastActivityDate="2018-01-21T04:00:39.153" Title="Circular tube stress loading" Tags="|structural-analysis|stresses|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14405" PostTypeId="2" ParentId="14341" CreationDate="2017-03-24T19:23:46.083" Score="0" Body="&lt;p&gt;Eventually I finished my flowchart with Visio. Here's how it looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ceIfv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ceIfv.png&quot; alt=&quot;Plant scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The PV modules were a template already present in Visio.&lt;/li&gt;&#xA;&lt;li&gt;For the inverters, the square box with a diagonal line was an existing template. I added the signs for &quot;DC&quot; and &quot;AC&quot; by finding them in vectorial on Google images.&lt;/li&gt;&#xA;&lt;li&gt;The transformers were an existing template.&lt;/li&gt;&#xA;&lt;li&gt;I used the option &quot;pattern&quot; to have a grid-pattern for the rectangle that represents the grid.&lt;/li&gt;&#xA;&lt;li&gt;The &quot;waves&quot; that means &quot;abbreviation of a line&quot; were made by hand.&lt;/li&gt;&#xA;&lt;li&gt;The battery sign was an existing template.&lt;/li&gt;&#xA;&lt;li&gt;All the rest is only basic geometrical shapes, lines, arrows.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6497" LastEditorUserId="6497" LastEditDate="2017-03-26T18:53:45.277" LastActivityDate="2017-03-26T18:53:45.277" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14406" PostTypeId="1" AcceptedAnswerId="14512" CreationDate="2017-03-24T19:43:45.777" Score="1" ViewCount="83" Body="&lt;p&gt;I want to build a professional CNC router for my garage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use the residential split phase configuration for power to my VFD and industrial power supplies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find any straight answer regarding connecting 240V 1ph (L/N/E input) apparatus to 120V/120V (L1/L2/E).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I feed the other line to neutral terminal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step-up transformer is out of question since it's high current draw.&lt;/p&gt;&#xA;" OwnerUserId="10591" LastActivityDate="2017-03-30T17:41:16.107" Title="Split phase on industrial devices" Tags="|electrical-engineering|power-electronics|power|electrical|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14407" PostTypeId="1" CreationDate="2017-03-24T20:03:03.687" Score="0" ViewCount="91" Body="&lt;p&gt;Whenever I happen to try to insert the card into the slot in the ATM wrong side - be it upside down, or backwards - it goes maybe 2cm in, barely past the plastic shield, and is mechanically stopped. The ATM only &quot;swallows&quot; it when inserted the right way in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is way before the chip interface enters the slot. Maybe a centimeter of the magnetic strip enters as well - if there even is one. NFC/RFID might work with determining the end of the card, but seems unlikely to help with the side, never mind older cards with no &quot;contactless payments&quot; feature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HOW does the ATM determine the orientation of the card?&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="8055" LastEditDate="2017-03-25T14:10:47.407" LastActivityDate="2017-03-25T14:10:47.407" Title="How does ATM determine the card is inserted the right side?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14408" PostTypeId="2" ParentId="489" CreationDate="2017-03-24T20:26:46.497" Score="1" Body="&lt;p&gt;Assuming money is not an option, Iridum is absurdly high density, ~22500 kg/m^3, Tungsten is close at ~19300 kg/m^3. both of which are extremely tough materials in compression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Headline asks for ceramics, but actual question asks for just material, hence the metal response).&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-03-24T20:26:46.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14409" PostTypeId="2" ParentId="14404" CreationDate="2017-03-25T01:37:45.167" Score="0" Body="&lt;p&gt;I'd check for a maximum overturning moment which will cause yield stress at the base of the member where it is attached to a base plate or enters the foundation as a flag pole, of course with any required safety. Then It is a simple arithmetic $M=\sigma\cdot s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ s=\pi\frac{d_o^4-d_i^4}{32\cdot d_o^2},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which $d_i$ is the inner diameter and $d_o$ is the outer diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more detailed diagram as to the attachment of the post to the base plate or the configuration of the plate at the top would allow more detailed observation and checking for local buckling, shear, normal loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am adding this clarification as to mode of failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pipe will start to rotate counter clockwise and undergo elastic defleection which will add to the moment arm, but as we add the vertical loading it will yield from the base which already has the greatest bending moment of approximately m = P.16 cm.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-01-21T04:00:39.153" LastActivityDate="2018-01-21T04:00:39.153" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14410" PostTypeId="1" CreationDate="2017-03-25T06:58:18.347" Score="1" ViewCount="11659" Body="&lt;p&gt;When preforming a grain size distribution on a sample of Soil using Mechanical Sieves (e.g. Sieve #4 and sieve #10 and sieve #40) one of the necessary calculation need for drawing the particle size distribution curve is the Cumulative Passing percentage of Soil Particles (aka Percent Finer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulas are pretty straight forward, but what is the physical meaning of Percent Finer? What does it imply ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a reference table: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b0udI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b0udI.jpg&quot; alt=&quot;Calculation Sheet&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10597" LastEditorUserId="90" LastEditDate="2022-08-03T00:36:05.270" LastActivityDate="2023-04-12T09:32:44.090" Title="What is the physical meaning of Cumulative Passing% in Grain Size distribution?" Tags="|geotechnical-engineering|soil|soil-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14412" PostTypeId="1" CreationDate="2017-03-25T12:04:47.890" Score="0" ViewCount="103" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SVnDY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SVnDY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a crucible with molten metal flowing under its own head. As per torricelli's law the velocity of the fluid would be $\sqrt{2gh}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any mechanism with which I can maintain a constant flow rate of my choice?&lt;/p&gt;&#xA;" OwnerUserId="6705" LastActivityDate="2017-04-05T20:15:51.680" Title="Best way to maintain flow of fluid under head?" Tags="|mechanical-engineering|fluid-mechanics|valves|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14413" PostTypeId="1" AcceptedAnswerId="14415" CreationDate="2017-03-25T12:52:38.153" Score="2" ViewCount="255" Body="&lt;p&gt;I have to determinate the power (Watts) for an oven to reach the Temperature Reference. And I have been given only the function to apply power to the oven and read the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've reading the Ziegler-Nichols method. And the first step is to obtain the tangent in the inflection point, and then check the values of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;L : The point where the tangent intersect the x axis&lt;/li&gt;&#xA;&lt;li&gt;T : The point where the tangent intersect the reference line&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But I do not understand one thing, the range of the power I can apply power from 0 to 10,000(W).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I have to calc this tangent for a static random Power? But which value, because it will be the same in &quot;production mode&quot;, when the power is vary because of the controller?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xKaXH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xKaXH.png&quot; alt=&quot;Calculation of L &amp;amp; T for Ziegler-Nichols PID method&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10598" LastEditorUserId="1586" LastEditDate="2017-08-21T23:57:08.000" LastActivityDate="2017-08-21T23:57:08.000" Title="Tuning PID Controller" Tags="|electrical-engineering|embedded-systems|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14415" PostTypeId="2" ParentId="14413" CreationDate="2017-03-25T13:39:52.137" Score="2" Body="&lt;p&gt;There are two answers here.  First answer is the theoretical one. It doesn't matter what static power you apply. You could apply any number. The reason is that the plot of y(t) that you end up drawing will be normalized to an input of 1.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, compare slides 12 and 13 in this presentation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hh.se/download/18.7502f6fd13ccb4fa8cb2f6/1360676928481/PID.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PID control - Simple tuning methods&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you apply an input power of 1, then the output plot you get gives you the value $K_p$ (you called it just K in your plot above).  But if you input a power of $\Delta$u, then the output plot gives you the value $\Delta$u$K_p$.  So to find the true value of $K_p$ you will divide by whatever power you input.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second answer is the practical one.  In reality you will never measure your parameters perfectly.  There will always be some error or noise in the measurement.  If you use a step of 1W on a 10,000W oven, you aren't going to get a very good measurement.  So higher power will always be a better measurement. But, on the other hand, if you are unfamiliar with this system, you might not want to try full power right away. You might want to have a reasonable idea of the open loop system behavior before going to full power.  If it were me, I would probably start by doing a 1000W (i.e. 10% full scale) step, making sure that I understood how the system was behaving and that I liked the answer. Then I would work up to full power in a few steps.  I would take the final Z-N parameters from a run at full power.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="7083" LastEditDate="2017-04-02T23:47:59.893" LastActivityDate="2017-04-02T23:47:59.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14416" PostTypeId="1" AcceptedAnswerId="14417" CreationDate="2017-03-25T15:04:54.873" Score="4" ViewCount="304" Body="&lt;p&gt;If this is the wrong place for this question, please migrate it to someplace where it's more suitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one of my old teachers from my previous school sent me a contraption to try and figure out what it was and how it worked. Pictures &lt;a href=&quot;https://i.stack.imgur.com/bPpP2.jpg&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've played around with it for a while. I tested to see if it did something when spinning really, really quickly by attaching a motor, tried taking it apart numerous ways, and plenty of things in-between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I've gathered, here's what I think it's doing. The propeller drives a shaft that is attached to something that converts the circular motion into linear motion. At this point the glass is frosted or there's some kind of glue on the inside that makes it very hard to see clearly what's going on. The top is connected to the bottle via two pipes that end in a hole in the side of the bottle. The cap itself has a hole in the top to which the top thing is screwed to, possibly also leaving space for water to go through -- it's unclear. The purple part is sticky on one side, suggesting that the whole thing sticks to the wall or maybe to something that moves like a car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe the propeller isn't what does the driving at all. Maybe it's filled with water and stuck to a wall or something facing downwards, and gravity would cause the water to cause the whole thing to leap to life! But, I can't find any way of filling it up -- there's no existing holes as far as I can tell. I may be able to take off the whole red cap but that's made difficult with the yellow pipes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm stuck. Does anyone have any idea about what this mysterious thing is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The propellers have arrows on them indicating the direction they should turn. This suggests that the propeller is indeed the driver and not the drivee.&lt;/p&gt;&#xA;" OwnerUserId="10600" LastEditorUserId="10600" LastEditDate="2017-03-25T15:58:38.330" LastActivityDate="2017-03-25T15:58:38.330" Title="What is this contraption?" Tags="|mechanical-engineering|fluid-mechanics|motors|mechanisms|linear-motors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14417" PostTypeId="2" ParentId="14416" CreationDate="2017-03-25T15:34:36.417" Score="5" Body="&lt;p&gt;It looks like it might be a air-powered propeller.  You pump up the bottle with air, and the high pressure is used to run a air motor that spins the propeller.  This might have been part of a toy airplane, with the wings and tail now gone.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-03-25T15:34:36.417" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14418" PostTypeId="1" CreationDate="2017-03-25T18:03:56.317" Score="6" ViewCount="345" Body="&lt;p&gt;In bridge design normally base supports are strong,fixed/rigid. Why did he choose thicker arch sections at middle above mid-river and narrow sections at abutment supports? If he chose a  simply supported condition, then for what structural advantage?&lt;/p&gt;&#xA;&lt;p&gt;EDIT1:&lt;/p&gt;&#xA;&lt;p&gt;Could it be some combination of arch and beam design?&lt;/p&gt;&#xA;&lt;p&gt;Between arch abutments it spans 5 pillar span lengths.Treating main structure as a long span Beam bridge simply supported at ends bending moment at mid-span is maximum so midsection should be designed deeply trussed as done here.&lt;/p&gt;&#xA;&lt;p&gt;However does it not compromise benefits of pure membrane arch action ( like St. Louis Gateway arch, Missouri design as a catenary) creating stress concentration by lower section property utilization?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Maria_Pia_Bridge&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eiffel Rail-Road Bridge Portugal&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;In 1876 they designed the 1,158-foot steel railroad Ponte Maria Pia (Maria Pia Bridge) over the Douro River in Oporto, Portugal&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1d9yL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1d9yL.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1270" LastEditorUserId="1270" LastEditDate="2023-02-08T19:21:19.137" LastActivityDate="2023-02-08T19:21:19.137" Title="Eiffel Bridge Design" Tags="|structural-engineering|bridges|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="14422" PostTypeId="1" AcceptedAnswerId="14440" CreationDate="2017-03-25T23:53:05.430" Score="-1" ViewCount="52" Body="&lt;p&gt;If you've been watching Bones series, there is an episode in which Bones and Hodgins were trapped beneath the ground, and they managed to send an SMS by wiring the phone to the car honk. Is this possible? What is the exact mechanism used?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate it if someone points out if the question is in the wrong site. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://vimeo.com/210057258&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://vimeo.com/210057258&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10608" LastActivityDate="2017-03-26T20:49:06.767" Title="Sending SMS using car horn" Tags="|mechanical-engineering|electrical-engineering|telecommunication|" AnswerCount="1" CommentCount="3" ClosedDate="2017-03-26T21:03:03.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="14423" PostTypeId="1" CreationDate="2017-03-26T01:40:16.880" Score="1" ViewCount="34" Body="&lt;p&gt;I need to conduct a vibration experiment with 3 accelerometers for a gear box. I'm reading the theory of mainly how to perform the vibration analysis but I don't understand why EMA needs a reference signal, and what it means by it? Also, is it the same for operational modal analysis?&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2017-03-26T01:40:16.880" Title="Why does Experimental Modal Analysis need a reference signal?" Tags="|mechanical-engineering|vibration|signal-processing|modal-analysis|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14427" PostTypeId="1" CreationDate="2017-03-26T13:58:37.463" Score="5" ViewCount="1133" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Hb5l.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Hb5l.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that flow seperation occurs when the fluid adjacent to a body deviates from the contours of the body . This occurs because the pressure increases in the direction of flow and velocity decreases until it reaches zero , and then , the flow reverses and a region of circulating flow develops which causes the fluid to deviate from the countours of the body ,, Is that true ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's true , my text book says :&quot; seperating flow is undesirable because it causes low pressure zones ...&quot; how does it cause low pressure zones and the main reason for the flow seperation to occur is that the pressure increases in the direction of flow ?  &lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="9742" LastEditDate="2017-03-26T15:02:31.360" LastActivityDate="2017-03-30T11:51:42.697" Title="Why does flow seperation cause low pressure zone?" Tags="|mechanical-engineering|fluid-mechanics|aerospace-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14428" PostTypeId="1" CreationDate="2017-03-26T14:45:30.840" Score="3" ViewCount="376" Body="&lt;p&gt;I am modellign some examples using the modelling language Modelica. Unfortunately, I am not an electrical engineer, so it is hard to understand the physical behavior of the following example (DC motor):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Iz3gs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iz3gs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where emf &quot;transforms electrical energy into rotational mechanical energy&quot;. &#xA;The model &quot;emf&quot; looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XJlF5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XJlF5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input voltage is plotted in the following graphic (top). &#xA;I dont understand the following things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why is the current zero in steady state?&lt;/li&gt;&#xA;&lt;li&gt;My interpretation: In steady state, the voltage input is equal to the voltage drop at the emf. The angular velocity is proportional to the voltage drop at emf.&lt;/li&gt;&#xA;&lt;li&gt;What is the &quot;emf model&quot; in a real DC-motor?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kSoAM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kSoAM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your help&lt;/p&gt;&#xA;" OwnerUserId="5906" LastActivityDate="2017-12-19T05:27:27.567" Title="Modelling a DC motor" Tags="|mechanical-engineering|electrical-engineering|motors|modeling|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14429" PostTypeId="1" CreationDate="2017-03-26T15:04:48.180" Score="0" ViewCount="814" Body="&lt;p&gt;I'm trying to build a device that, on receiving a signal, lifts the mattress from one side (in a &quot;slowly kick you out of bed&quot; manner). Since linear actuators are generally expensive and i'm on a tight budget, I though of using a cheap scissor jack (&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B004Y09L66&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this one &lt;/a&gt;), and move it with a simple electric motor, or most likely a gear reducer motor, but i'm no engineer (well technically a software engineer but that doesn't help here) so i don't know where to even begin to do the calculus. ¿Could anyone give me some pointers? I can do the math but i'm lost on the physics part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PD: This is also my first time building any device, so if i'm missing some painfully obvious solution to this problem please tell.&lt;/p&gt;&#xA;" OwnerUserId="10616" LastActivityDate="2017-03-26T17:59:43.930" Title="Electric motor power needed to move an scissor jack?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14430" PostTypeId="1" CreationDate="2017-03-26T15:14:46.283" Score="2" ViewCount="209" Body="&lt;p&gt;I am having difficulties trying to work out one problem. I've attached a picture explaining the situation - the task is to determine internal forces in &lt;strong&gt;C&lt;/strong&gt; but the bending moment &lt;strong&gt;M(x)&lt;/strong&gt; doesn't seem to get me the correct results (I used equivalance and went from LHS). I would greatly appreciate if somebody helped me construct a correct equation or find the mistake, thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h0kQy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h0kQy.jpg&quot; alt=&quot;picture of the problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10617" LastActivityDate="2017-06-25T19:23:35.863" Title="Structural mechanics - computing bending moment" Tags="|civil-engineering|moments|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14432" PostTypeId="2" ParentId="13738" CreationDate="2017-03-26T15:49:10.210" Score="1" Body="&lt;p&gt;You won't know for sure without contacting the developers directly but I've examined some .sldprt files with a HEX editor and there doesn't seem to important metadata.&#xA;Unlike other files such as mp3 where the metadata is displayed in plain text at the top of the file, sldprt files seem only to contain a file declaration and data.&#xA;However I would want to make sure if I were you as this sounds like it could be an  important thing to know.&lt;/p&gt;&#xA;" OwnerDisplayName="user10618" LastActivityDate="2017-03-26T15:49:10.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14434" PostTypeId="2" ParentId="14291" CreationDate="2017-03-26T17:08:13.083" Score="0" Body="&lt;p&gt;No answers, I agree with Carlton in the comments, some very large liberties have been taken.&lt;/p&gt;&#xA;" OwnerUserId="4530" LastActivityDate="2017-03-26T17:08:13.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14435" PostTypeId="2" ParentId="14429" CreationDate="2017-03-26T17:59:43.930" Score="1" Body="&lt;p&gt;This sort of thing comes up a lot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a low speed high load application like this the specification you are most interested in in torque. Essentially this is the sum of two components 1) overcoming the friction forces in the system 2) accelerating the mass you want to move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice it's not that easy to calculate this sort of thing from first principals and it is often best to start with a ballpark estimate based on something similar which already exists. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a scissor jack the fact that it is, by default manually operated helps a bit as you can easily set up a practical test eg set up an equivalent load situation and use a spring balance etc to find the minimum force which needs to be applied to the crank (and thus torque) required to get it moving. Anything more than that will give you extra margin as I don't imagine you require particularly snappy acceleration for this application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also suggest that you might want to consider over-specifying the motor a bit and using speed control as it is going to be tricky to hit the exact speed you want just from calculations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an initial estimate it is possible to jack up a car with a screw jack and a cordless drill so you are looking at something in the 500W sort of range. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-03-26T17:59:43.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14436" PostTypeId="1" AcceptedAnswerId="14438" CreationDate="2017-03-26T18:03:03.327" Score="0" ViewCount="1484" Body="&lt;p&gt;There is a steel plate with length of 26.5cm (10.4 inch)  and width of 4cm (1.57 inch). the steel plate is attached on one end, at the other end there is a load of 30kg, what thickness should be the plate so it will not bend ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8Gh6z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Gh6z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10619" LastActivityDate="2017-03-26T20:10:23.160" Title="Steel plate thickness for load" Tags="|steel|stresses|" AnswerCount="1" CommentCount="2" ClosedDate="2017-03-27T13:14:17.223" ContentLicense="CC BY-SA 3.0" />
  <row Id="14438" PostTypeId="2" ParentId="14436" CreationDate="2017-03-26T18:28:13.183" Score="1" Body="&lt;p&gt;It will always bend as steel has finite stiffness. You need to decide how much deflection is acceptable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineersedge.com/beam_calc_menu.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; lists the bending equations for a variety of loading situations. This case is a &lt;a href=&quot;http://www.engineersedge.com/beam_bending/beam_bending9.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;cantilever with load at one end&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations do rely on some assumptions but should be fine for this situation, be aware though that they apply to static loads only, if the load moves the stresses involved can be significantly higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations also assume that stresses are within the elastic limit of the material, if the yield stress is exceeded then the beam will be permanently deformed and may fail completely. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rough guide I would say that you are looking at about 8mm thick or more &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth nothing that a flat plate is a very inefficient way to support a cantilevered load and unless there is a strong reason why the support needs to be very slender you are much better off either adding an arch type support web below the beam or using a hollow section. For example a 25x25mm square tube with wall thickness of 2mm would support 30 kg no trouble with very small deflection at 27cm cantilever length and weigh much less than an equivalent 4cm wide strip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you also need to consider the fact that there will be a large torsion load where the plate meets the thing it is attached to which is a common point of failure.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-03-26T20:10:23.160" LastActivityDate="2017-03-26T20:10:23.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14440" PostTypeId="2" ParentId="14422" CreationDate="2017-03-26T20:49:06.767" Score="-1" Body="&lt;p&gt;From the clip you gave (which I glossed over the first time), I think they needed to power the phone so they wired into the circuit that draws power from the car battery for the car horn. That circuit doesn't require the car to be on in any way so that would be the reason to use that circuit.&lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2017-03-26T20:49:06.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14442" PostTypeId="1" CreationDate="2017-03-26T23:34:21.117" Score="0" ViewCount="175" Body="&lt;p&gt;What's the name for a kind of gear that slips over the top of another gear and can drive/turn the gear it's surrounding?&lt;/p&gt;&#xA;&lt;p&gt;I'm making a handcranked flashlight with my 7 year old. We're thinking of using a stepper motor with a gear on the end from an old printer, something like the one pictured. I've seen this kind of crank flashlight made with just a 'female' gear embedded in the crank handle, that slots right over the 'male' gear (as pictured here). That's the best way I can describe it. What's the name of this kind of gear? One that would slot right over the top of the gear in the picture below and be able to drive it?  I can't seem to find what I am looking for anywhere, any ideas? If I knew it's name then I could search for it. Thank you!&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/03nmA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/03nmA.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Edit** What I looking for is kind of like this internal spur gear (the image below), but with that outer ring FITTING REALLY TIGHT around the inner spur gear, with the number of cogs matching exactly (not like this picture, where the outer ring is really big compared to the inner gear).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wEIT8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wEIT8.jpg&quot; alt=&quot;enter preformatted text here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10622" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-03-31T09:59:41.980" Title="What's the name for a kind of gear that slips tightly over the top of another gear and can drive the gear it's surrounding?" Tags="|gears|prototyping|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14443" PostTypeId="1" CreationDate="2017-03-27T01:37:17.620" Score="1" ViewCount="39" Body="&lt;p&gt;I need some help to a design problem at my university. The design work is about designing a prefabricated wall panel. I calculated the U Value of the external wall and it is 0.0136&amp;nbsp;W/m&lt;sup&gt;2&lt;/sup&gt;K. The thickness of the wall is 0.353&amp;nbsp;m and total area equal to 3&amp;nbsp;m x 6&amp;nbsp;m = 18&amp;nbsp;m2 (assumed no openings).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need to do now is to estimate how much energy do I need to maintain the inside temperature. I know that the maximum summer outside temperature is 30°C and the minimum in winter -20°C. How can I calculate the additional cooling and heating I need by having this wall?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is basically a whole design project with the purpose of creating a prefabricated panel that can meet the needs of a specific country that was given. It was entirely flexible, nothing was given, just a city to make it fit there!  I divided my work as I thought it would be better and one part of it was the thermal properties.I tried to choose materials that can meet a rigorous U value of 0.13 and now I would like to show how much better it is in terms of operational energy (cooling and heating). After a research I did I found that probably I could calculate the heat loss and heat gain (the simplest I found with the data I have is by this Ht = A U (ti - to) and then according to my results propose a heating system or cooling etc. But I am not sure if this is the right way to do it.. this is why I asked some help.&lt;/p&gt;&#xA;" OwnerUserId="10624" LastEditorUserId="10624" LastEditDate="2017-03-27T23:39:12.973" LastActivityDate="2017-03-27T23:39:12.973" Title="Find additional heating and cooling needed by having a specific wall" Tags="|materials|civil-engineering|heat-transfer|" AnswerCount="0" CommentCount="5" ClosedDate="2017-03-29T14:53:39.483" ContentLicense="CC BY-SA 3.0" />
  <row Id="14444" PostTypeId="2" ParentId="14442" CreationDate="2017-03-27T01:44:58.860" Score="0" Body="&lt;p&gt;Internal spur gear is a pretty good name for it, actually.  In the context of a planetary gear, this would be called the ring gear.  But since this is not a planetary, I would just call it an internal spur gear. Boston gear sells them &lt;a href=&quot;http://www.bostongear.com/products/open-gearing/stock-gears/spur-gears/internal-gears&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.bostongear.com/products/open-gearing/stock-gears/spur-gears/internal-gears&lt;/a&gt; and I'm sure a number of other places do too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this sounds like a take-stuff-apart and re-use it kind of project, you might consider using an old cordless drill.  Those almost always have a small planetary gear transmission in them.  You can reuse the motor and transmission as-is to make a small generator out of it. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-03-27T01:44:58.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14446" PostTypeId="1" CreationDate="2017-03-27T03:42:17.843" Score="1" ViewCount="60" Body="&lt;p&gt;I am designing an automation machine. I remember that I read a document that the machine is stable if some certain amount of force to push on the top corner of the machine and the machine does not tile more than X degree then the machine is stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not find the document again. Do you see something like that online?&lt;/p&gt;&#xA;" OwnerUserId="8198" LastActivityDate="2017-03-27T03:42:17.843" Title="Machine design tip over safety requirement?" Tags="|design|safety|" AnswerCount="0" CommentCount="4" ClosedDate="2017-03-29T18:07:15.823" ContentLicense="CC BY-SA 3.0" />
  <row Id="14449" PostTypeId="2" ParentId="14080" CreationDate="2017-03-27T08:09:18.290" Score="3" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Static water level refers to the level of water (in a well for example) under normal, undisturbed, no-pumping conditions. Static water level is best determined when the well has not been pumped for several hours prior to measuring. You may get a false reading if the well was pumped just before the static water level is measured.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Water Table Level is The level below which the ground is completely saturated with water Also called water level.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They mean the same just in geotechnical engineering they use the name &quot;Water Table Level&quot; because they are interested in the soil but when you talk about a well or dams you name it &quot;Static water level&quot; but they are the same they mean the level of water.&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2017-03-27T08:09:18.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14450" PostTypeId="1" CreationDate="2017-03-27T11:16:18.737" Score="2" ViewCount="3924" Body="&lt;p&gt;I'd been having difficulty with a problem from Cengel and Boles, Thermodynamics An engineering approach. The problem is about the deviated Rankine cycle from Idealised ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I would like to ask is how to obtain the enthalpy at state 3 and at state 6. The image attached shows the data for the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7EC4c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7EC4c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to know is there a need to use the T-S diagram in obtaining the enthalpies in state 3 and state 6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I got so far:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The enthalpy at state 1 is obtained from the steam table and is h1 = 182.586 kJ/kg.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The actual work done by the pump is calculated using the isentropic efficiency value for pump and the value is w(pump, actual) = 18.986 kJ/kg&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To calculate the heat input, the enthalpy at state 3 and state 4 must be known. The problem is I do not know how to get the enthalpy at state 3.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Similarly for the work done by the turbine, w(turbine,actual) the enthalpy at state 5 and 6 needs to be know. The problem is that I do not know how to obtain for enthalpy at state 6.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Updates:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I managed to obtain the work done by actual turbine with the method described by @algo.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The problem that I still have is how to obtain the enthalpy at state 3. I need it to obtain the heat input to the boiler. Can considering irreversibilities in the pipes (State 2 to 3) actually help in obtaining the enthalpy at state 3? Currently, I have no clue on how to start.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any helpful reply would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="10633" LastEditorUserId="10633" LastEditDate="2017-03-28T14:28:55.953" LastActivityDate="2024-03-24T11:00:45.090" Title="Deviation of Ideal Rankine Cycle from Actual Rankine Cycle" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14451" PostTypeId="1" CreationDate="2017-03-27T11:31:48.757" Score="3" ViewCount="710" Body="&lt;p&gt;I have been repeatedly told when dealing with some data for Navigation purposes that I can assume it follows a &quot;bicycle model&quot;. I am not very familiar with vehicle dynamics so I consulted online. I saw many different models varying on the level of sophistication but most of them seems to consist in planar Dynamics similar to the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\begin{bmatrix}&#xA;x_{k+1} \\&#xA;y_{k+1} \\&#xA;\theta_{k+1}&#xA;\end{bmatrix} = \begin{bmatrix}&#xA;x_{k} \\&#xA;y_{k} \\&#xA;\theta_{k}&#xA;\end{bmatrix} + \begin{bmatrix}&#xA;v\cdot\Delta t\cdot\cos{\theta_k} \\&#xA;v\cdot\Delta t\cdot\sin{\theta_k} \\&#xA;f(\gamma,v)&#xA;\end{bmatrix} &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma$, and $v$  are the steering angle and the velocity, both inputs from the vehicle sensors, and $\theta$ is the angle respect to the origin of a fixed frame. $f(\gamma,v)$ is some function relating both angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My doubt arises from the fact that the data I have been granted contains only the following fields: $x$,$y$,$\theta$. My assumption is that, if we know it must follow a bicycle model it should hold a correlation like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\begin{bmatrix}&#xA;x_{k+1} \\&#xA;y_{k+1} \\&#xA;\theta_{k+1}&#xA;\end{bmatrix} = \begin{bmatrix}&#xA;x_{k} \\&#xA;y_{k} \\&#xA;\theta_{k}&#xA;\end{bmatrix} + \begin{bmatrix}&#xA;d_k\cdot\cos{\theta_k} \\&#xA;d_k\cdot\sin{\theta_k} \\&#xA;\Delta\theta_k&#xA;\end{bmatrix} &#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $d_k$ and $\Delta\theta_k$ are known at every time step and are the distance traveled and the change in the angle respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I generate data this way, starting from an initial point $(x_0,y_0)$ and building up, it doesn't seem to correlate well with the actual data I have been granted.  My question is if the deduction exposed has any conceptual mistake, so I can clear out if the problem is that the data given is very noisy and deviates after a while or that what I think is a bicycle model is not&lt;/p&gt;&#xA;" OwnerUserId="10632" LastActivityDate="2017-06-28T11:51:11.170" Title="&quot;Bicycle model&quot; in vehicle dynamics" Tags="|mechanical-engineering|dynamics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14452" PostTypeId="1" CreationDate="2017-03-27T13:10:47.227" Score="1" ViewCount="9434" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A2brT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A2brT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stucked at trying to find the vertical reaction at top and bottom constraint, i know how to do the rest. Can you show me how?&lt;/p&gt;&#xA;" OwnerUserId="10533" LastActivityDate="2017-04-28T09:38:01.013" Title="How to find reaction forces in trusses?" Tags="|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14453" PostTypeId="2" ParentId="14452" CreationDate="2017-03-27T13:35:49.843" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;No load can be in the far left member as it is supported at both ends. Axial force is directly proportional to change in length; if there can be no change in length (due to both ends being fixed in position) then there can be no axial force.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore resolving vertically at the top support, the vertical reaction is equal to the shear in the horizontal member. If we're assuming axial loads only for truss behaviour, then there is no shear and hence no vertical reaction&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore the whole vertical reaction is taken by the lower support&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2017-03-27T14:27:21.857" LastActivityDate="2017-03-27T14:27:21.857" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14454" PostTypeId="1" AcceptedAnswerId="14457" CreationDate="2017-03-27T14:44:03.420" Score="2" ViewCount="1705" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vYP02.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vYP02.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If i first assume the member(far left) to not be zero and use method of joint at both points of constraint, with both constraints having unknown vertical reaction, it cannot be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Load and lengths are known, horizontal reaction can be found.&lt;/p&gt;&#xA;" OwnerUserId="10533" LastActivityDate="2017-07-23T18:18:23.680" Title="Why is the truss member between two fixed supports zero?" Tags="|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14455" PostTypeId="2" ParentId="14454" CreationDate="2017-03-27T15:24:45.223" Score="0" Body="&lt;p&gt; If both constrained points are pinned, the model is over-constrained and cannot be determined without more information.&#xA;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both the constrained points are fixed there can be no load on the member between because the fixed constraints take that load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the bottom constraint is only constrained by a bearing (like a wheel) then there will still be no load on that member. The only way I can see to put a load on it would be to remove the bottom horizontal member and make the top vertex a fixed point or place the load there. This would make a downward force along the diagonal that is fixed to the bottom constraint and put a downward force on the leftmost member. But this is only if that bottom constraint is a roller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, there is no load on that member because there are no forces acting on it, the same could be said for the other vertical member, unless the bottom horizontal is made of two sections and pinned, then that other vertical only serves to keep the bottom from bowing.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:18:23.680" LastActivityDate="2017-07-23T18:18:23.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14456" PostTypeId="1" AcceptedAnswerId="14634" CreationDate="2017-03-27T15:31:17.183" Score="-1" ViewCount="4753" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dnPjL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dnPjL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to answer the following question which is (image shown above)&#xA;&quot;what is the force to keep the rectangular gate closed? in N/M&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all the information I have to solve the question. I understand there will be use of moments. I just require help understanding the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GZ4Sw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GZ4Sw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10468" LastEditorUserId="10468" LastEditDate="2017-03-29T16:31:51.620" LastActivityDate="2017-04-06T00:00:05.330" Title="Fluid mechanics- Force required to keep the gate closed" Tags="|fluid-mechanics|moments|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14457" PostTypeId="2" ParentId="14454" CreationDate="2017-03-27T15:43:07.907" Score="1" Body="&lt;p&gt;In this situations, it's worth remembering &lt;a href=&quot;https://en.wikipedia.org/wiki/Hooke%27s_law#For_linear_springs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hooke's Law&lt;/a&gt;, which tells us that internal forces and deformations are related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A beam can deform either because a load is directly applied to it or because one (or both) of its extremities is being deformed by a load on another beam somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, a beam has no load applied directly to it and both of its ends are constrained (can't move in any direction), then it will never deform in any way shape or form. And if it doesn't deform, then we know by Hooke's Law that it won't suffer any internal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the case for your truss: there are no loads applied to the member between both supports, and since it is between two supports, it will never deform and therefore never suffer any axial forces. And since this is a truss, any rotations suffered by the surrounding members won't be transmitted to the member either, so neither will there be any bending moment (or shear).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the more visually inclined, imagine the deformed configuration of the structure: all the other members will be deformed as the load pulls the cantilever down. This member, however, won't be deformed at all. And therefore, neither will it suffer any internal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MriYd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MriYd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-03-27T15:43:07.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14459" PostTypeId="1" AcceptedAnswerId="14482" CreationDate="2017-03-27T16:37:07.063" Score="2" ViewCount="118" Body="&lt;p&gt;(I posted this in the Sound-Design section but got advised to post here instead).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just purchased a 3d printer and I am working on a product with a box that has a hole in the bottom and the top. Inside this box I have a few electronics that has a pretty loud sound. What I now look to do is to kill that sound as much as possible so it is not possible to hear it as loud. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At home I have a few acoustic panels looking like this and as they are very lightweight it seems like a very good solution for this design: &lt;a href=&quot;https://www.thomann.de/se/eq_acoustics_classic_wedge_foam_tiles_grey.htm?sid=dbe7318446923736e3fdee98e5ba1174&quot; rel=&quot;nofollow noreferrer&quot;&gt;EQ Acoustics Classic Wedge 30 Tile grey&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I cover the interior of the box on all the walls would this make the sound not leave the box as much which would result in a lower sound? I am very novice when it comes to this area so any help/tips is very appreciated! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZJD3x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZJD3x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Width and length is around 30cm.&lt;/p&gt;&#xA;" OwnerUserId="10630" LastEditorUserId="8055" LastEditDate="2017-04-04T16:30:28.750" LastActivityDate="2017-04-21T01:35:29.127" Title="How can I kill the sound as much as possible in a rounded box? Acoustic panels/other option?" Tags="|civil-engineering|sound-isolation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14460" PostTypeId="2" ParentId="14450" CreationDate="2017-03-27T16:46:07.610" Score="0" Body="&lt;p&gt;Actually, you don't need $h_6$ to calculate the specific work done by the turbine. By definition isentropic efficiency for the turbine is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta _{is} = \frac {\text{Turbine actual work}} {\text{Turbine isentropic work}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The turbine isentropic work in your case can be calculated as $h_5 - h_{6s}$, calculating $h_{6s}$ is the tricky part here. For the state $6s$ you have two properties: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The pressure (10 kPa).  &lt;/li&gt;&#xA;&lt;li&gt;The entropy (which equals the entropy of&#xA;state 5 $(s_5 = s_{6s})$, for an isentropic turbine).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can obtain $h_{6s}$ by looking on your steam table with some painful interpolations using those two properties,or you can use a Mollier chart, or even better an EES code. Once you've obtained $h_{6s}$ you can calculate turbine isentropic work and finally the actual work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a quick EES code for your problem to make thing more clear:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&quot;Pump inlet and outlet states&quot;&#xA;h_1 = enthalpy(water, p = 9 [kPa] , t = 38)&#xA;s_1 = entropy(water, p = 9 [kPa] , t = 38)&#xA;eta_pump_is = p_work_isentropic / p_work_actual&#xA;p_work_isentropic = h_2s - h_1&#xA;h_2s = enthalpy(water, p =16 * 1000, s = s_1)&#xA;eta_pump_is = 0.85&#xA;&#xA;&quot;Boiler inlet and outlet states&quot;&#xA;h_2 = enthalpy(water, p=16 * 1000, t=35)&#xA;h_4 = enthalpy(water, p=15.2 * 1000, t=625)&#xA;q_boiler = h_4 - h_2&#xA;&#xA;&quot;Turbine&quot;&#xA;h_5 = enthalpy(water, p=15 * 1000, t=600)&#xA;s_5 = entropy(water, p=15 * 1000, t=600)&#xA;h_6s = enthalpy(water, p=10, s=s_5)&#xA;eta_turb_is = t_work_actual / t_work_isentropic&#xA;eta_turb_is = 0.87&#xA;t_work_isentropic = h_5 - h_6s&#xA;t_work_actual = h_5 - h_6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2017-03-27T16:46:07.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14461" PostTypeId="2" ParentId="14430" CreationDate="2017-03-27T16:52:46.760" Score="1" Body="&lt;p&gt;Let's start by simplifying this for ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In such a simple beam, horizontal loads don't have any influence in anything other than the axial loads. You apparently need to solve for those as well, but you're only asking for the bending moment, so I'm going to ignore them, so let's see how the load looks considering just the vertical component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;q_0 &amp;amp;= 3\sin{48} = 2.23\text{ kN/m} \\&#xA;q_1 &amp;amp;= 9\sin{48} = 6.69\text{ kN/m}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's also already calculate some intermediate values for the load at the left support ($A$) and at $C$, which we'll need later on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;q_A &amp;amp;= q_0 + (q_1-q_0)\dfrac{1.2}{10.2} = 2.75\text{ kN/m} \\&#xA;q_C &amp;amp;= q_0 + (q_1-q_0)\dfrac{2.2}{10.2} = 3.19\text{ kN/m}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By moment, we can see that the reaction at the left support ($A$) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;M_B = -9R_A + \dfrac{q_0 \cdot 10.2^2}{2} + \dfrac{(q_1-q_0) \cdot 10.2^2}{2\cdot3} = 0 \\&#xA;\therefore R_A = \dfrac{10.2^2}{9}\left(\dfrac{q_0}{3} + \dfrac{q_1}{6}\right) = 21.48\text{ kN}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then calculate the bending moment at $C$:&#xA;$$\begin{align}&#xA;M_C &amp;amp;= -\dfrac{q_0 \cdot 2.2^2}{2} - \dfrac{(q_C-q_0) \cdot 2.2^2}{2\cdot3} + R_A \cdot 1 \\&#xA;&amp;amp;= -2.2^2\left(\dfrac{q_0}{3} + \dfrac{q_C}{6}\right) + R_A = 15.31\text{ kNm}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you'd rather do this by finding the bending moment equation, you'll need to use the beam equation. In the main span, the distributed load has the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q(x) = q_A + (q_1-q_A)\dfrac{x}{9}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating that, we get the shear equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;Q(x) &amp;amp;= \int q\text{d}x = -q_Ax - (q_1-q_A)\dfrac{x^2}{18} + C_1 \\&#xA;Q(0) &amp;amp;= C_1 = \text{shear at }A^+ = -\dfrac{(q_0 + q_A)}{2}\cdot1.2+R_A = 18.492\text{ kN} \\&#xA;\therefore Q(x) &amp;amp;= -q_Ax - (q_1-q_A)\dfrac{x^2}{18} + 18.492 \\&#xA;\therefore Q_C &amp;amp;= Q(1) = 15.523\text{ kN}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating again, we get the bending moment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;M(x) &amp;amp;= \int Q\text{d}x = -\dfrac{q_Ax^2}{2} - (q_1-q_A)\dfrac{x^3}{54} + 18.492x + C_2 \\&#xA;M(0) &amp;amp;= C_2 = \text{moment at }A = -1.2^2\left(\dfrac{q_0}{3} + \dfrac{q_A}{6}\right) = -1.73\text{ kNm} \\&#xA;\therefore M(x) &amp;amp;= -\dfrac{q_Ax^2}{2} - (q_1-q_A)\dfrac{x^3}{54} + 18.492x - 1.73 \\&#xA;\therefore M_C &amp;amp;= M(1) = 15.31\text{ kNm}&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-03-27T16:52:46.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14463" PostTypeId="1" CreationDate="2017-03-27T19:47:46.343" Score="0" ViewCount="44" Body="&lt;p&gt;I am having an issue with caking in a process.  I am already trying to figure out ways to prevent the caking in the first place but I need a short term solution that will allow the line to continue running.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea is to take two filters and have them piped to run in parallel with one another.  However, I want a valve on the inlet of each individual filter.  I want this so while in production I can have Valve A open (allowing fluid through Filter A) while keeping Valve B closed (no fluid through Filter B).  Then when Filter A's mesh becomes clogged open Valve B, close Valve A, and remove and clean Filter A.  This way the fluid is still being constantly filtered and we do not have to stop the line for a filter change.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am struggling with a name for this arrangement of filters as in my mind its a simple enough set up that a commercial version should already exist (I could be wrong about that).  All I am able to find on line are filter arrangements that allow series or parallel flow but not the ability to turn off flow to individual filters.  It could also be an option while ordering that maybe I am not catching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any advice.&lt;/p&gt;&#xA;" OwnerUserId="10641" LastActivityDate="2017-07-23T18:18:16.157" Title="Trying to determine the name of a filter set up" Tags="|fluid-mechanics|fluid-filtration|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14465" PostTypeId="2" ParentId="14442" CreationDate="2017-03-28T05:52:01.130" Score="3" Body="&lt;p&gt;I would call it a &lt;a href=&quot;http://echo.bike/uploads/2/5/4/9/25498541/8121891.jpg?1398403185&quot; rel=&quot;nofollow noreferrer&quot;&gt;splined crank&lt;/a&gt; (i.e. a crank/handle that has a splined socket on one end -- like on some higher-end bicycle cranks).&lt;br&gt;&#xA;In your case, it seems that you want to place the splined end of your crank/handle onto the spur gear that is mounted on the stepper motor shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another application of such a lever is a handle that is used on a water valve  (e.g. an outside water &lt;a href=&quot;http://s7.cdn.hardwareandtools.net/is/image/HardwareandTools/Product/037155804177.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;spigot handle&lt;/a&gt; or some shut-off valve handles that are used under sinks for supply lines or for water heater supply lines).  Of course, those handles are round &amp;amp; not in a crank shape like your application seems to require.&lt;/p&gt;&#xA;" OwnerUserId="5778" LastEditorUserId="5778" LastEditDate="2017-03-31T09:59:41.980" LastActivityDate="2017-03-31T09:59:41.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14466" PostTypeId="2" ParentId="54" CreationDate="2017-03-28T06:53:42.807" Score="2" Body="&lt;p&gt;&lt;strong&gt;For reverse Osmosis&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://urila.tripod.com/desalination.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;This site gives the minimal required energy for sea water desalination by RO as 2.78 kJ/l (fresh water)&lt;/a&gt;, this is if you consider only the reversible process. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Reverse_osmosis#Desalination&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia, the best RO desalination plants operate at 3kWh/m³&lt;/a&gt; which trasnlates to 10.8 kJ/l.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK, energy losses are pressure loss through the membrane (in addition to the osmotic pressure, a membrane introduces irreversible pressure losses), water pretreatment and energy (in the form of pressure) in the brine. Also a lot of water simply needs to be moved about, there's pre treatment steps etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.iwapublishing.com/sites/default/files/ebooks/IWA_GlobalTrendReport2016.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to this IWA trend report&lt;/a&gt;, two areas within the wide field of membranes where more research is done are better membranes in terms of pressure loss and fouling resistance (fouling directly influences pressure losses). Recentish developments in RO desalination like &lt;a href=&quot;https://en.wikipedia.org/wiki/Forward_osmosis#Desalination&quot; rel=&quot;nofollow noreferrer&quot;&gt;forward osmosis&lt;/a&gt; mostly benefit from better fouling resistance compared to RO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For thermal desalination&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;crickets&lt;/em&gt;&lt;br&gt;&#xA; (will update when I find more info)&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-03-28T06:53:42.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14467" PostTypeId="1" AcceptedAnswerId="14480" CreationDate="2017-03-28T06:55:07.067" Score="0" ViewCount="47" Body="&lt;p&gt;I have a wall with the bottom end of it fixed to the ground. I want to derive from first principles the frequencies and mode shapes of the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I will need to start with the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K-\omega^2M=0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I know how to do that for a cantilever column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wall is a different thing than column because it is an area element instead of a line element. And I don't know how to generate the correct expression for $K$ and $M$ for a continuum area element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to proceed with the analytical solution for wall eigenvalue analysis? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I am not looking for a FEM solution. &lt;/p&gt;&#xA;" OwnerUserId="3353" LastActivityDate="2017-03-28T22:45:58.393" Title="Analytical solution for wall eigenvalue analysis" Tags="|structural-analysis|modal-analysis|eigenvalue-analysis|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14468" PostTypeId="1" CreationDate="2017-03-28T08:51:37.460" Score="0" ViewCount="704" Body="&lt;p&gt;How to calculate the gas constant of saturated moist air at 16 degree Celsius and 1.013 bar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where to go with this. I just know that the saturated vapor pressure from steam tables, at 16 degree Celsius is 0.018173 bar. Is this data even useful for calculating the gas constant?&lt;/p&gt;&#xA;" OwnerUserId="10643" LastEditorUserId="90" LastEditDate="2017-03-28T11:15:56.263" LastActivityDate="2017-03-28T12:52:25.700" Title="Calculating the gas constant of saturated moist air" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14469" PostTypeId="1" CreationDate="2017-03-28T10:14:47.357" Score="1" ViewCount="34" Body="&lt;p&gt;I am currently trying to create a small product with my 3D printer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product consists of a flat wide stick first, which then is connected to a 45 degrees angled stick. What I now want to achieve is that the 45 degree angled plate gets to the same level as the flat plate when pressure gets added. So it goes from 45 degrees to 0 looking something like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QXj5H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QXj5H.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been doing a bit of googling and came across something called a torsion spring which I assume could work pretty well looking like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Wi3Lk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wi3Lk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not currently own a torsion spring at home and I assume torsion springs cannot be made using a 3D printer. I was wondering if there is an optional solution for this task? &lt;/p&gt;&#xA;" OwnerUserId="8445" LastEditorUserId="90" LastEditDate="2017-03-28T11:20:21.590" LastActivityDate="2017-03-28T11:20:21.590" Title="How to make stick change angle after pressure is added?" Tags="|structural-engineering|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="14472" PostTypeId="2" ParentId="14463" CreationDate="2017-03-28T12:38:45.577" Score="1" Body="&lt;p&gt;I think this is the kind of thing you are looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://m.eaton.com/mobile/ProductsandSolutions/Filtration/ManualPipelineStrainers/DuplexBasketStrainers/Model505-8inch/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://m.eaton.com/mobile/ProductsandSolutions/Filtration/ManualPipelineStrainers/DuplexBasketStrainers/Model505-8inch/index.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are common in fuel systems where you need to be able to clean one strainer while maintaining flow. You could build your own system using any sort of bypass valve and piping. Install a Y or T in your pipe and a valve on both legs, install your method of filtration then a second valve after the filter / strainer. to isolate that section for cleaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You didn't mention the fluid or contaminant you are removing nor the rate of flow so I don't know exactly what you need.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:18:16.157" LastActivityDate="2017-07-23T18:18:16.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14473" PostTypeId="2" ParentId="14468" CreationDate="2017-03-28T12:52:25.700" Score="3" Body="&lt;p&gt;$$R=\frac{N_A \, k_B}{M}$$&#xA;with &lt;a href=&quot;https://en.wikipedia.org/wiki/Gas_constant#Specific_gas_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;specific gas constant&lt;/a&gt; $R$, &lt;a href=&quot;https://en.wikipedia.org/wiki/Avogadro_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Avogadro number&lt;/a&gt; $N_A$, &lt;a href=&quot;https://en.wikipedia.org/wiki/Boltzmann_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boltzmann constant&lt;/a&gt; $k_B$ and &lt;a href=&quot;https://en.wikipedia.org/wiki/Molar_mass&quot; rel=&quot;nofollow noreferrer&quot;&gt;molar mass&lt;/a&gt; $M$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a mixture of ideal gases (remembering $pV = N k_B T$):&#xA;$$x_i = \frac{N_i}{N} = \frac{p_i}{p}$$&#xA;with pressure $p$, volume $V$, number of molecules $N$, temperature $T$ and &lt;a href=&quot;https://en.wikipedia.org/wiki/Mole_fraction&quot; rel=&quot;nofollow noreferrer&quot;&gt;mole fraction&lt;/a&gt; $x$. Indices refer to individual species, non-indexed values are for the mixture: $p_{\mathrm{Water}}$ = water partial gas pressure, $p$ = total pressure. Remember that $V_i=V$ and $T_i=T$ for mixtures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally (see &lt;a href=&quot;https://de.wikipedia.org/wiki/Gasgemisch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;),&#xA;$$M =\sum_j \, x_j \, M_j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have $p_{\mathrm{Water}}$ and $p$, which allows you to calculate $p_{\mathrm{Air}}$ (the difference), all $x_i$, then $M$ and finally $R$.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2017-03-28T12:52:25.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14475" PostTypeId="1" AcceptedAnswerId="14478" CreationDate="2017-03-28T16:02:13.370" Score="3" ViewCount="1331" Body="&lt;p&gt;This may be a dumb question, but is a solid block 100% of the time stronger than a 'strategically hollow' part. Strategically hollow meaning, the internal structure is design in such a way to support the loading direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking is originating from both 3D printing internal supports but also from bio-mimicry designs. Looking at the internal structure of bones, they aren't solid, would they definitely always be stronger if they were solid? Intuitively I would say 'obviously' however could an internal structure is designed in such a way that the surface energy, and load paths create a stronger structure than a single solid block?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a way to design the internal structure, how would you go about designing/analyzing this, if not is there a way to prove that it's not?&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-03-28T17:36:48.463" Title="Solid versus strategically hollow" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14476" PostTypeId="2" ParentId="13950" CreationDate="2017-03-28T17:30:02.033" Score="0" Body="&lt;p&gt;The best route in this situation is a simple feedback loop:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qSSDe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qSSDe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact is, you don't need to have amazing drafting skills to have good ideas, nor do you need to have amazing machining skills to have good ideas.  You do need persistence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a video of your device (if you still have it), and have someone make the CAD file.  It will come out wrong.  Redo, over and over, until it comes out right.  Why CAD? It's cheaper to make CAD files then it is to make actual prototypes.  Then make prototypes, over and over, until it comes out right.  Then make production parts.  Test them over and over, with QC checks, until you find better ways to make them come out right without defects.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at it like &quot;I need a definitive solution for it to come out right the first time&quot; is how scientists approach things.  That's what makes engineers different.  Here, we modify and redo over and over until it comes out correct, not just settle for the &quot;one answer&quot;.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-03-28T17:30:02.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14477" PostTypeId="2" ParentId="14456" CreationDate="2017-03-28T17:33:05.783" Score="0" Body="&lt;p&gt;Is there more information from the question, there is enough to solve but it never hurts to have more info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I see this problem is you have to balance the applied force F with the distributed load from the water, this gets a little complicated because the water is fighting itself, the water at heights (from the bottom) 1.2m - 1.35m wants to close the door and the water between 1.55m and 4.2m wants to open the door. This is simply 3 forces acting on a lever, where 2 are distributed over a length. The distributed forces are not hard to find though, a quick google of &quot;distributed load moments&quot; turned up this gem&#xA;&lt;a href=&quot;http://engineering-references.sbainvent.com/strength_of_materials/pictures/strength_of_materials_shear_moment_diagram_cantilever_triangle_distribute_increase.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; This is for a beam but the principles and equations are the similar, just put the pivot in a different place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck if you haven't already solved it &lt;/p&gt;&#xA;" OwnerUserId="10652" LastActivityDate="2017-03-28T17:33:05.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14478" PostTypeId="2" ParentId="14475" CreationDate="2017-03-28T17:36:48.463" Score="8" Body="&lt;p&gt;Seems like you might not be considering the importance of the &lt;strong&gt;strength to weight ratio&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solid bar will be stronger, but it will also weigh more.  This additional weight will put extra stress on other components, and also have it's own internal stresses associated with supporting itself (imagine a long solid cantilever beam vs. a hollow one).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often times engineers will use hollow members or designs with webbing.  The general purpose of the designs is to put material where it needs to be for strength, but try not to overdo it, or else then you have weight that makes it harder to move (or causes more internal stress between the members).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often this involves making things hollow, especially since things like bending and torsion get the most resistance from the furthest out points.  You maximize your material where you would see the most stresses (think of I beams for bending and hollow poles for twisting).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nature has this down to a pretty good science.  Bones and such have strength where they need them, but there are enough cavities and such to fit other vital parts of the anatomy instead of being full of bone without as much structural benefit.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-03-28T17:36:48.463" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14480" PostTypeId="2" ParentId="14467" CreationDate="2017-03-28T22:45:58.393" Score="1" Body="&lt;p&gt;Your &quot;wall&quot; would be referred as a &quot;plate&quot; in the technical literature. The eigenvalue problem for a plate is significantly more complicated than a beam, but similar ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest that you look for a book on &quot;vibrations of continuous systems&quot;, for example: &lt;a href=&quot;http://www.wiley.com/WileyCDA/WileyTitle/productCd-0471771716.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.wiley.com/WileyCDA/WileyTitle/productCd-0471771716.html&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also specific books dedicated entirely to plates, e.g. &lt;a href=&quot;https://www.crcpress.com/Vibrations-of-Shells-and-Plates-Third-Edition/Soedel/p/book/9780824756291&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.crcpress.com/Vibrations-of-Shells-and-Plates-Third-Edition/Soedel/p/book/9780824756291&lt;/a&gt;  however they may be hard to understand without sufficient background.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-03-28T22:45:58.393" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14481" PostTypeId="2" ParentId="14307" CreationDate="2017-03-29T01:51:40.703" Score="0" Body="&lt;p&gt;I think I've solved. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is not an simple exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the problem we must first find the angular velocity of the gears $C$ and $D$. Note that both of them have the same angular velocity because they are strongly fixed one another. For that we use the contact point of $B$ and $D$ that will be (if denoting $L = 0,2m$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{\text{Contact point between B and D}} = \omega_D L = \omega_B (2L)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that  $\omega_D = \omega_C = 2\text{rad}/s$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we analise the gears sistem $A$ and $C$. In this sistem we need to calculate the center of rotation, wich obviously is in the center of $A$. With that we calculate the velocity of the rotation center of the gears $C$ and $D$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{\text{Center of  C of D}} = \omega_{\text{Bar}}L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can find the value of  $\omega_{\text{Bar}}$ using the contact point&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{\text{Contact point between A and C}} = \omega_{\text{Bar}}\frac{3L}{2} = \omega_C\frac{L}{2} \implies \omega_{\text{Bar}} = \frac{\omega_C}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that we will have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_{\text{Center of  C and D}} = \frac{\omega_C L}{3} = \frac{0,4}{3}m/s \approx 0,13m/s$$&lt;/p&gt;&#xA;" OwnerUserId="10526" LastActivityDate="2017-03-29T01:51:40.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14482" PostTypeId="2" ParentId="14459" CreationDate="2017-03-29T07:05:26.193" Score="0" Body="&lt;p&gt;Acoustic tiles reduce reflected sound. They are not so good at reducing sound penetrative.  And their design requires a large structure, so it won't work shrunk down to 30cm. This is down to the frequency of sound, and the speed of sound. You could use the principles to reduce ultrasonic emissions, or very high pitch sounds. To reduce lower pitch emissions, you need to focus on what's causing the noise in the first place. Perhaps start with the noisiest component and explore how you can change it.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2017-03-29T07:05:26.193" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14483" PostTypeId="1" CreationDate="2017-03-29T11:34:39.177" Score="3" ViewCount="169" Body="&lt;p&gt;I am using a psychophysical model to calculate the probability (P) of detecting an electrical stimulus to the skin (amplitude = x) over time (t):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;P(x;\alpha(t),\beta) = \dfrac{1}{1+e^{\beta(\alpha(t)-x)}}, \text{ } {\small \alpha \text{= threshold &amp;amp; } \beta \text{ = slope}}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the model parameters I would like to use a logistic regression function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;logit(\pi) = b_0 +b_1X +b_2T&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how to implement this logistic regression function in LabVIEW? As inputs I have a timevector, an amplitude vector and a response (0 or 1) vector. &lt;/p&gt;&#xA;" OwnerUserId="2509" LastEditorUserId="8055" LastEditDate="2017-03-30T01:26:33.020" LastActivityDate="2017-03-30T01:26:33.020" Title="Logistic regression in LabVIEW" Tags="|electrical-engineering|biomedical-engineering|labview|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14484" PostTypeId="1" AcceptedAnswerId="14504" CreationDate="2017-03-29T12:51:28.420" Score="1" ViewCount="125" Body="&lt;p&gt;I am currently making software to interact with a conductive rubber cord. I am very interested in figuring out the percentage of a stretch of the cord, to determine how the resistance changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QIK3M.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QIK3M.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stretch the cord 1 cm, see the difference in the resistance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stretch the cord 3 cm, see the difference in resistance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Programmatically, I can tell this change. When I use my hands, I can see the resistance flow change as my pressure increases on stretching the cord. However, as my hands are probably not the best instrument for accurate stretching, I am in need of an alternate machine to do this for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a such device exist, which can hold one end of this material for me, and pull the other end 1 cm? 3 cm? etc.&#xA;Since I am a software engineer, and certainly lack in the field of mechanical engineering, is a good approach to take here to use motors and a raspberry pi to possibly create my own stretch machine?&lt;/p&gt;&#xA;" OwnerUserId="9644" LastActivityDate="2017-03-30T14:17:10.897" Title="Stretching material accurately with a machine" Tags="|measurements|distance-measurement|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14485" PostTypeId="2" ParentId="14450" CreationDate="2017-03-29T14:51:45.017" Score="0" Body="&lt;p&gt;A good rule of thumb is that you only need to know 2 properties of a single-phase fluid to know all remaining thermodynamic properties.  As shown in the answer from @Algo, lookup tables or EES can provide specific enthalpy from temperature and pressure.  The utility of the T-S diagram is that it shows all points but 6 (and maybe 1) are outside the saturation curve, meaning that they are either pure liquid or pure vapor.  If the fluid was saturated, temperature and pressure would not be sufficient to obtain specific enthalpy.    &lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-03-29T14:51:45.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14490" PostTypeId="1" AcceptedAnswerId="14500" CreationDate="2017-03-29T23:19:36.893" Score="2" ViewCount="1630" Body="&lt;p&gt;Can we bring engineering to the paper in paper mache?  It's a mix of a binder and a reinforcing matrix.  PVA is common for the binder and is strong.  That's fine as I think it's strong enough and readily available.  Can we do better  for the traditional news paper?  I think that glass /carbon fibres would be incompatible as PVA doesn't readily adhere.  Is there another filler?  I read somewhere that they used to make air plane drop tanks and boats from some sort of paper mache, demonstrating it's engineering potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise that this is more materials science than pure mechanical engineering, but there are commonalities and it's all we have on SE.  I'm deliberately avoiding Arts &amp;amp; Crafts as I'm not looking for a child friendly, mix a long with mom type of material.  I'm looking for the dirtiest, leanest meanest (probably cancer causing) reinforcement that's as strong as possible drawing on modern technology.  The only stipulation is that it should be available to the domestic purchaser (so bags of nano tubes are out).  Is there something stronger than newspaper?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="2947" LastEditDate="2017-03-30T10:40:06.273" LastActivityDate="2017-07-23T18:18:09.287" Title="Can we make engineering grade paper mache by improving on the paper?" Tags="|materials|composite|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14491" PostTypeId="1" CreationDate="2017-03-30T06:03:15.420" Score="-5" ViewCount="103" Body="&lt;p&gt;I got a question on levelling. But I don't know how to do that. My question is in the picture below, with the values&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions are in the picture. Can you guys please help me to do this question? I get same value for one position of the equipment in one side. But in the other side the height is different. I don't understand why is that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find the error of the equipment and the height difference between x &amp;amp; y. I uploaded my progress on comments. And here is my rough work. &lt;a href=&quot;https://i.stack.imgur.com/JEtKT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JEtKT.jpg&quot; alt=&quot;my rough work is in the picture&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10668" LastEditorUserId="10668" LastEditDate="2017-03-30T17:59:44.023" LastActivityDate="2017-03-30T17:59:44.023" Title="How to do a calculation on levelling?" Tags="|civil-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2017-03-31T22:32:14.537" ContentLicense="CC BY-SA 3.0" />
  <row Id="14492" PostTypeId="2" ParentId="14484" CreationDate="2017-03-30T08:40:32.387" Score="1" Body="&lt;p&gt;If you are at a university, you can consider using a universal tensile testing machine to stretch the rubber cord at specific stretch rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you may develop a simple setup using stepper motor to also stretch it at fixed stretch rates.&lt;/p&gt;&#xA;" OwnerUserId="10586" LastActivityDate="2017-03-30T08:40:32.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14493" PostTypeId="1" AcceptedAnswerId="14502" CreationDate="2017-03-30T09:07:10.117" Score="1" ViewCount="181" Body="&lt;p&gt;I am asked to find the member force of AB and DE &#xA;&lt;a href=&quot;https://i.stack.imgur.com/4qeDo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4qeDo.png&quot; alt=&quot;fgfg&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Orhvk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Orhvk.png&quot; alt=&quot;32323&quot;&gt;&lt;/a&gt;&#xA;I make an imaginary cut of the question and consider the upper part of the cut . Bur , i didnt get the same ans provided . However , when i consider the lower section of the cut , i get the same ans with the book . Anything wrong with my working ? &#xA;Taking moment about A = 20(4) +30cos45 (3) + DE(4) = 0 , DE = -35.9 upwards &lt;/p&gt;&#xA;&#xA;&lt;p&gt;btw , the vertical reaction at a is 51.3 kN upwards , at E , vertical reaction = 35.9 kN upwards , and 21.2kN to the left ...&lt;/p&gt;&#xA;" OwnerUserId="10127" LastActivityDate="2017-03-30T14:14:59.003" Title="cut section method" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14494" PostTypeId="2" ParentId="14451" CreationDate="2017-03-30T09:07:41.950" Score="-1" Body="&lt;p&gt;Your problem may be simplified to this:&#xA;You have an initial position and velocity (including angular displacement) for discrete time points. And, you wish to find the position and direction at any given time instant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bicycle model you have discussed here is basically a first order approximation solution. It means you are ignoring the change in velocity between one time-step and the other. This is still acceptable if the time-step is small. If it is large, the approximation may no longer be accurate. If the time-step is too... small, there may be truncation errors on account of the precision of your computer (you may not possibly worry about this error).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conclusion can be that either the resolution of your data is low or else, as you have pointed out, the data may be too noisy.&lt;/p&gt;&#xA;" OwnerUserId="10586" LastActivityDate="2017-03-30T09:07:41.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14495" PostTypeId="2" ParentId="12462" CreationDate="2017-03-30T09:12:56.453" Score="2" Body="&lt;p&gt;Generally, for thin sheets, we measure sheet resistance by using a three point or four point probe method. Then, you can find the resistivity of the material by using a relation between sheet resistance, resistivity and thickness of the sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://four-point-probes.com/sheet-resistance-and-the-calculation-of-resistivity-or-thickness-relative-to-semiconductor-applications/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sheet Resistance and the Calculation of Resistivity or Thickness Relative to Semiconductor Applications&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10586" LastEditorUserId="8055" LastEditDate="2017-12-13T01:17:07.613" LastActivityDate="2017-12-13T01:17:07.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14496" PostTypeId="1" CreationDate="2017-03-30T10:21:54.663" Score="1" ViewCount="247" Body="&lt;p&gt;I noticed that there are a growing number of abandoned stacks of this type: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zbjg9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zbjg9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Rancho-Seco-power-plant-California_new.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if they are physically strong enough to hold water, for use in pumped storage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xRRWx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xRRWx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10670" LastActivityDate="2017-03-31T15:48:45.610" Title="Are abandoned nuclear smoke stacks strong enough to be filled with water (for pumped storage)?" Tags="|structural-engineering|structural-analysis|renewable-energy|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14497" PostTypeId="2" ParentId="14427" CreationDate="2017-03-30T11:51:42.697" Score="2" Body="&lt;p&gt;From wikipedia: &quot;The flow reversal is primarily caused by an adverse pressure gradient imposed on the boundary layer by the outer potential flow&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this website &lt;a href=&quot;http://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/fprops/introvisc/node9.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/fprops/introvisc/node9.html&lt;/a&gt;: &quot;One of the severe effects of an adverse pressure gradient is to separate the flow.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Fundamentals of Aerodynamics by John D Anderson Jr., page 795: &quot;Figure 15.3 Separated flow induced by an adverse pressure gradient&quot;. Figure 15.3 is very similar to the image on the left side of your post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand your question correctly, you are asking if separated flow causes low pressure zones, and I believe the answer is technically no. An adverse pressure gradient &lt;em&gt;causes&lt;/em&gt; the separated flow, and where we see separated flow there is lower pressure. However, the separation of flow is not the instigator, it is the adverse pressure gradient in viscous fluids due to shear stress and the retarding forces acting on the fluid as it flows past a body that causes the flow separation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the setting up of an adverse pressure gradient will depend on the geometry of the body that the fluid is flowing over, and at what angle the fluid is flowing over the body. If pressure increases along the body, then &lt;em&gt;I believe&lt;/em&gt;, an adverse pressure gradient will be set up.&lt;/p&gt;&#xA;" OwnerUserId="5469" LastActivityDate="2017-03-30T11:51:42.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14498" PostTypeId="2" ParentId="14496" CreationDate="2017-03-30T11:53:41.257" Score="7" Body="&lt;p&gt;Those aren't &quot;smoke&quot; stacks but cooling towers designed to let hot air rise and expand optimally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These things are typically 200 m high. Static pressure of 200 m of water is pretty high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They were never required to be water tight so they will leak. In fact towers are open at the bottom to let air in and rise inside. So to start you would need to seal that up. Sealing a retrofit patch to an adequate level that it'll keep that water in is very difficult. Getting waterproof concrete is usually done by poring the the entire slab in a single go to avoid leaky seams.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-03-30T11:53:41.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14499" PostTypeId="2" ParentId="14496" CreationDate="2017-03-30T12:50:50.083" Score="1" Body="&lt;p&gt;Cooling towers are designed for forced /induced draft cooling  purpose, no big loading other than self weight and laeral wind forces.They would not be efficient pressel vessels holding water for likes of 200 meter water column height of Seco Rancho/CA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If utilization of the twin towers is the main aim then alternate new big project  features can be meaningfully defined i.e, if at all there is some scope for its implementation. Structural engineering  strength,stiffness and stability for building a water reservoir &lt;em&gt;around&lt;/em&gt; the towers is required as it is  possible to design/model test/proto build. Much higher augmented capacity with efficiency and low unit power generation cost is possible with the stacks as &lt;em&gt;external&lt;/em&gt; pressure vessels. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Cooling towers of single sheet hyperboloid shape and negative Gauss curvature have &lt;em&gt;extra-ordinary stability against lateral bending loads&lt;/em&gt; and can resist not just wind but even heavy water loads. Please refer to &#xA;Handbook of Structural Engineering by Gould &amp;amp; Krätzig. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.crcnetbase.com/doi/abs/10.1201/9781420039931.ch27&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.crcnetbase.com/doi/abs/10.1201/9781420039931.ch27&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/opfoh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/opfoh.png&quot; alt=&quot;HdBk StrEngg G&amp;amp;K &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If water pressure acts on all four sides of stack, shell bending can be significantly reduced due to cancellation of equilibrated surrounding symmetrical forces.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Second point is much  more important. A reticulated structure ( that leaves much of external surface between reticulations/net beams open) around the stack could be cost effectively constructed to take up all membrane loads induced by water pressure. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The volume of water contained in the reservoir  would be very high compared to inner stack  volume. The turbines could be placed &lt;em&gt;inside&lt;/em&gt; the stacks for  potential power augmentation benefits.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;A similar concrete filled &lt;em&gt;very thin&lt;/em&gt; shell reticulated structure can be built around as a surrounding external wall /fort/dam/reservoir at a &lt;em&gt;very low&lt;/em&gt; cost, not as a dam structure whose weight need not fall in the middle third for civil dam stability.The shape of meridian and generators can be further optimized by  &lt;em&gt;Form-Finding&lt;/em&gt; design principles. This is the principal &lt;em&gt;new&lt;/em&gt; cost element here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The present sketch/description is inadequate to explain all features due to novelty of design proposal made here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1270" LastEditorUserId="1270" LastEditDate="2017-03-31T05:34:29.103" LastActivityDate="2017-03-31T05:34:29.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14500" PostTypeId="2" ParentId="14490" CreationDate="2017-03-30T13:25:56.667" Score="1" Body="&lt;p&gt;My favorite is papercrete, below is a link to the wikipedia article on it. This can be reinforced with simple fiberglass to make it stronger, unfortunately finite analysis would not be terribly effective on determining the strength of the material. You would have to make dozens of samples and test them individually to work on optimizing the mixture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Papercrete&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Papercrete&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:18:09.287" LastActivityDate="2017-07-23T18:18:09.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14501" PostTypeId="2" ParentId="14490" CreationDate="2017-03-30T14:08:05.000" Score="1" Body="&lt;p&gt;Absolutely. Paper mache at it's core is an early predecessor of carbon fiber. The paper is the fiber structure and the paste (not sure of the technical term) is the equivalent to the epoxy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase the 'strength' of the fibers you'll eventually reach a point that the paste is the weak point, beyond which increasing the strength of the paper won't do much. If you increase the strength of the paste to a better adhesive you'll then be able to increase the strength of the final part.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-03-30T14:08:05.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14502" PostTypeId="2" ParentId="14493" CreationDate="2017-03-30T14:14:59.003" Score="1" Body="&lt;p&gt;As noted in the comments to your question, you forgot to include the moment due to the internal force in BE.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-03-30T14:14:59.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14504" PostTypeId="2" ParentId="14484" CreationDate="2017-03-30T14:17:10.897" Score="1" Body="&lt;p&gt;Yup, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use something as simple as a bar clamp. Most hardware stores will carry some version. The type you'd need is the type that can be reversed to a spreader (95% of them can do this just double check). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With one of these you'd need something to connect your cord to it. But still fairly straight forward (more clamps for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This wont be super accurate but it'll work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For something more accurate you're looking at something more like a load frame. Most engineering universities will have a few of these (mechanical or civil departments). Very easy to use, but they take some set up time (a half hour ish). Every university I've worked with is usually pretty open to doing external testing, though you'll be limited to their schedule. There are other material testing places that will have them but they charge oodles usually.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-03-30T14:17:10.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14505" PostTypeId="1" CreationDate="2017-03-30T14:30:37.273" Score="1" ViewCount="30" Body="&lt;p&gt;I'm trying to solve this problem in elementary Carnot cycle study:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two (hot and cold) heat fountains of a motor, in air, at temperatures $T_a = 1500^0$C and $T_c = 10^0$C, respectively. The work done by this motor is employed to activate a heat pump, the cold fountain of the heat pump is at $T_c = 10^0$C and the hot fountain is at $T_b = 40^0$C. Both are suppose reversible machines. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;The fraction between the heat that the motor receives of the fountain and the heat that the heat pump provides to its hot fountain is?&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My attempt:&lt;/strong&gt; We have that both, the motor and the heat pump, can be modeled by a Carnot Cicle model. The question asks: What is $Q_a^M/Q_c^T$? So we have that the work done by the fountains is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau^M = Q_a^M-Q_c^M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is the work that it is used to activate the heat pump, so we have that $$\tau^M = Q_b^T$$ &#xA;Then we can manipulate this results to obtain, using that $Q_b^T/T_b = Q^T_c/T_c$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_c^T\frac{T_b}{T_c} = Q_a^M-Q_c^M = Q_a^M - Q_a^M\frac{T_c}{T_a} = Q_a^M\left(1 - \frac{T_c}{T_a}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_a^M/Q_c^T = \frac{T_b}{T_c}\left(\frac{1}{1 - T_c/T_a}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this gives the wrong answer. The correct one is $Q_a^M/Q_c^T \approx 0,1137 =  11.37$%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notations: $Q_{a,b,c}$ is the heat in the isotherm given by the temperatures $T_a$ or $T_b$ or $T_c$. The above indices $^M$ means that this is the heat motor and the above indices $^T$ means that it is the heat of the heat pump. &lt;/p&gt;&#xA;" OwnerUserId="10526" LastActivityDate="2017-03-30T14:30:37.273" Title="Carnot cicle problem" Tags="|thermodynamics|motors|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14506" PostTypeId="1" CreationDate="2017-03-30T14:59:55.837" Score="1" ViewCount="509" Body="&lt;p&gt;Good morning/afternoon/evening,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on a project in my spare time, and was wondering how this effect is created.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/EPbXf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EPbXf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QAEAZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QAEAZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to assume this is more than just a cutout with an LED behind it, but if I wanted to do a simple LED indicator, not shapes, how would that be done? My initial guess is that a piece of black translucent polycarbonate with an LED behind it would do the trick. Would this produce a nice little circle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="10672" LastEditorUserId="8055" LastEditDate="2017-03-30T18:34:00.977" LastActivityDate="2017-07-29T08:46:33.133" Title="Diffuse uniform LED back lighting" Tags="|electrical-engineering|lighting|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14507" PostTypeId="2" ParentId="14506" CreationDate="2017-03-30T15:41:48.807" Score="2" Body="&lt;p&gt;On first inspection this looks to be done with diffuser placed above an LED light source. That will expand the cone of light generated by the LED to cover a larger area. As for the shape, it looks to be clear polycarbonate or acrylic that is painted on the inside with the image cutout. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways to do this, but if you're working on something for simple hobbyist level development this is by far the most practical. Large scale operations may embed the image shape into the plastic. That will last longer but is more difficult and expensive to do&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-03-30T15:41:48.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14508" PostTypeId="1" AcceptedAnswerId="23441" CreationDate="2017-03-30T16:04:47.427" Score="1" ViewCount="628" Body="&lt;p&gt;Which dimensions of columns supporting flat slabs do not require a punching shear check (according to ACI code)?&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="33" LastEditDate="2018-08-24T19:59:09.787" LastActivityDate="2018-08-24T19:59:13.100" Title="Punching Shear in Slab?" Tags="|structural-engineering|civil-engineering|reinforced-concrete|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14509" PostTypeId="1" AcceptedAnswerId="14520" CreationDate="2017-03-30T16:16:01.880" Score="4" ViewCount="395" Body="&lt;p&gt;Here is the question, I am stuck on part B:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RAxBr.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RAxBr.png&quot; alt=&quot;Question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, this is my thought process so far, as $x_a \rightarrow 0$, the Henry's law line becomes tangential to the actual partial pressure curve.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tVoz0.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tVoz0.png&quot; alt=&quot;Graphs&quot;&gt;&lt;/a&gt; Henry's law states that:&lt;br&gt;&#xA;$$&#xA;p_{\alpha} = Hx_{\alpha}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, using the activity coefficient model:&#xA;$$&#xA;p_{\alpha} = \gamma_{\alpha}x_{\alpha}p^*_{\alpha}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where H is Henry's constant and $p^*_{\alpha}$ is the vapour pressure of the pure component. So at small $x_{\alpha}$ values one can assume:&#xA;$$&#xA;H = \gamma_{\alpha}p^*_{\alpha}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this I can get values for the $\gamma$ (I've been given values for pure component pressure). Then I can get values for $\ln \gamma_1, \ln \gamma_2$. My next step would be to substitute in values of $x_2$ and $x_1$ into the given correlations. So, for $\gamma_1$ to be correct, $x_1$ must be very small, assume it is zero therefore $x_2 \approx 1$, therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\ln \gamma_1 = \frac{A \times 1^2}{\bigg( (1-1)^2\frac{A}{B} + 1 \bigg)^2} \implies \ln \gamma_1 = A&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, this is wrong because if I run through the same process for $\ln \gamma_2$, I get a different A value. Can someone please advise, thanks.&lt;/p&gt;&#xA;" OwnerUserId="4530" LastActivityDate="2017-03-31T10:54:03.333" Title="Calculation of Activity Coefficient Model parameters" Tags="|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14510" PostTypeId="2" ParentId="14491" CreationDate="2017-03-30T16:50:22.067" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The first thing to note is that we are not provided with the level of the stations at A and B. We should therefore not assume that they are at the same height.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We know (from question (a)) that there is an error in the measurement in units of mm/m. Therefore we can assume that all levels are wrong, &lt;em&gt;but that they are wrong by a consistent error&lt;/em&gt;, and the size of the error is proportional to the distance from the instrument. Let us call this error &lt;strong&gt;e&lt;/strong&gt; mm/m&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore, working from A: the true height of Y from instrument A is 3.46 + 24e. The true height of Z from A is 3.46 + 42e. Hence the difference in height between Y and Z is 18e.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Working from B: the true height of Y from B is 3.92 + 36e. The true height of Z from B is 3.96 + 18e. Hence the difference in height between Y and Z is 0.04 - 18e.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Therefore 18e = 0.04 - 18e, e = 0.04m in 36m = 1/900 m/m = 1000/900 mm/m ~ 1.111 mm/m&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The methodology shown above can be extended to give the actual height differences between X and Y and between Y and Z. As this is a homework question, I suggest the OP attempts this themselves, and comes back with any problems they have.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2017-03-30T16:56:33.577" LastActivityDate="2017-03-30T16:56:33.577" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14511" PostTypeId="2" ParentId="14496" CreationDate="2017-03-30T17:04:27.557" Score="-2" Body="&lt;p&gt;Ironically, these could probably hold quite a bit of water.  Probably not to the top, or even to the narrow part, but quite a bit.  They're designed this way so they will work even without a fan at the bottom, the heat alone moves the water along and up the tower.  They're also not made with curved members, but a bunch of straight beams inside the concrete:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cD3zJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cD3zJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they're so freakishly tall, those towers need a lot of beams with a lot of strength to hold up the high height and massive weight of the large towers.  Cutting the towers to a shorter height, the beams wouldn't have to hold a lot of unsupported concrete weight on top, and that extra strength could be used to ... hold water.  Depending on the size of the beams inside (the concrete can hold a lot of compression, but the actual pressure holding would be from the beams inside), quite a bit of pressure could be retained.  You'd need to get a professional to review the actual tower, but the idea has merit.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-03-30T17:04:27.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14512" PostTypeId="2" ParentId="14406" CreationDate="2017-03-30T17:41:16.107" Score="1" Body="&lt;p&gt;I have bugged people who claim to know about this, and am no expert but here are my 2 cents:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Residential 110 voltage is taken from the 240 supply. The transformer is middle tapped to make 2 110 legs. Look at it as a single long coil that is 240 volts from top to bottom but if you connect at the middle you can get 2 legs at 110.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For alternating current that changes polarity with a sine wave, you are getting half the wave in each leg. If you look at it on a graph, I think you get the positive half in one leg and the negative in the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with 3 phase is you need the third phase to not be in sync with the others. So the only way to get the frequency right is to generate your third leg. You can run a motor on 2 phase that generates the third phase by coupling that motor with a generator. I think it might be possible to use a variable frequency device instead of the motor-generator couple but I now less about VFD.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastActivityDate="2017-03-30T17:41:16.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14513" PostTypeId="2" ParentId="2651" CreationDate="2017-03-30T18:36:57.903" Score="1" Body="&lt;p&gt;This approach has been used in many, many scientific studies, and is an active area of research in the materials engineering community, but with a different material.  FRP (&lt;a href=&quot;https://en.wikipedia.org/wiki/Fibre-reinforced_plastic&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fiber Reinforced Plastic&lt;/a&gt;) is used to wrap around an existing concrete column (or using an FRP pipe filled with concrete), with very good results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: &lt;a href=&quot;http://dx.doi.org/10.1016/j.conbuildmat.2008.07.008&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0263822305000231&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt; &lt;a href=&quot;http://www.mdpi.com/2073-4360/6/4/1040/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt; &lt;a href=&quot;http://ieeexplore.ieee.org/document/4737811/?reload=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;4&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-03-30T18:36:57.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14514" PostTypeId="1" AcceptedAnswerId="14560" CreationDate="2017-03-30T18:37:33.663" Score="2" ViewCount="1776" Body="&lt;p&gt;As I know, a thinner mat may be possible by providing a nominal amount of vertical reinforcement as compared to a mat without vertical reinforcement, the vertical reinforcement is used to resist punching shear at d/2 from columns or shear walls at foundation level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is to what distance (or offset) from the column to stop placing these vertical reinforcement is it at a distance d/2 ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please if possible provide reference in ACI Code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lz4bi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lz4bi.png&quot; alt=&quot;critical section outside slab shear reinforcement&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i need is the critical section outside slab shear reinforcement from which distance from the column ?&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="10554" LastEditDate="2017-04-02T09:51:55.063" LastActivityDate="2017-04-02T17:07:13.303" Title="To Which distance Punching Shear Reinforcement is placed in Mat Foundation?" Tags="|structural-engineering|civil-engineering|geotechnical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14517" PostTypeId="1" AcceptedAnswerId="14899" CreationDate="2017-03-31T07:34:07.857" Score="4" ViewCount="462" Body="&lt;p&gt;In electrical domain, we have resistors, capacitors, and inductors as fundamental constitutive elements. Similarly, in mechanical domain, we have masses, springs and dampers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most of the literature, we can immediately see that they have the one-to-one relation, perse i.e., we can go from one form of representation to other, or even mix them up (mechatronics). However, the problem I see is, in the electrical domain, we have the inductor which is an energy storage element, representing an integrator, collectively speaking. In contrast, when we look it up in the mechanical domain, all we have are mass-spring-damper equivalent, and there is no explicit representation for an inductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is this, does there exists a fundamental equivalent for an inductor in mechanical domain i.e., a mechanical equivalent of an integrator with a physical meaning in itself. If so, how it is called?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried a lot of literature to find one but failed in the quest. So, all your help and suggestions are welcome.&lt;/p&gt;&#xA;" OwnerUserId="10681" LastEditorUserId="10902" LastEditDate="2017-06-23T05:57:37.040" LastActivityDate="2017-06-23T05:57:37.040" Title="Does a Mechanical equivalent of an Integrator exist" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14518" PostTypeId="1" CreationDate="2017-03-31T07:47:08.687" Score="0" ViewCount="219" Body="&lt;p&gt;I'm looking for the easiest way to simulate an array of cubes that is being pushed. Therefor I need to model the motion/collision effects, the friction with the floor and between blocks and also thermal expansion of the cubes, although that is secondary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What libraries/packages/software do you recommend to have this simulation in place a.s.a.p.? I am familiar with python and js, therefor I've looked at pybullet &amp;amp; physicjs, however I'd like to hear your opinion on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The approach I have got in mind is to use some kind of 3D library that supports physics, which gives visualization and rapid development speed. Any other fast approach would be ok, though I will need to run many different combinations (block sizes, friction, etc) so it needs to be scriptable in some way.&lt;/p&gt;&#xA;" OwnerUserId="6099" LastEditorUserId="6099" LastEditDate="2017-03-31T08:15:08.130" LastActivityDate="2018-05-09T12:10:30.040" Title="Simulation of motion, collision, friction and thermal expansion in 3D" Tags="|simulation|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14519" PostTypeId="2" ParentId="14517" CreationDate="2017-03-31T08:22:31.037" Score="5" Body="&lt;p&gt;If position is what is being manipulated then it's pretty simple, convert the input position to speed using a friction wheel on a spinning disk where the input position decides where the wheel is in relation to the axis of the spinning disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/s1i-dnAH9Y4?t=30m52s&quot; rel=&quot;noreferrer&quot;&gt;This video with a description a mechanical computer's components&lt;/a&gt; includes a description of an integrator at 30:52&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-03-31T08:22:31.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14520" PostTypeId="2" ParentId="14509" CreationDate="2017-03-31T10:54:03.333" Score="3" Body="&lt;p&gt;There was an error in the question, the model in the question is the van Laar activity model, the coefficient for $\ln \gamma_2$ on the numerator should be B, in this case, it is easily solvable.&lt;/p&gt;&#xA;" OwnerUserId="4530" LastActivityDate="2017-03-31T10:54:03.333" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14522" PostTypeId="2" ParentId="14518" CreationDate="2017-03-31T12:43:38.683" Score="2" Body="&lt;p&gt;This is a multi-physics analysis, there are some open source packages though i'm not familiar with them (OpenFOAM is open source CFD software, and COMSOL is example of an commercial multi-physics solver). If you're looking at writing your own, &quot;a.s.a.p.&quot; (as soon as possible) is going to be not very soon as that takes a huge amount of work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best solution is using a prebuilt software designed for that, ANSYS and NASTRAN are my go to solutions. NASTRAN I think has a better multi-physics integration but that's an opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know there is no such thing as an &quot;easy&quot; way to do multi-physics analysis. Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;Given the relative simplicity of the system you're looking at analyzing you may want to consider just one aspect at a time as the multi-physics aspect might be minimal. I.e. thermal expansion of the cubes sliding on a surface likely wont effect the contact. A good first approximation is to do individual analyses and then manually check to see what type of effect they would have on each other. If thermal expansion changes the temperature by a tenth of a degree thermal analysis in multi-phys would be well below the threshold for simulation error and could justifiably be neglected. It really will depend on what answers you're hoping to get from the solution.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastEditorUserId="561" LastEditDate="2018-05-09T12:10:30.040" LastActivityDate="2018-05-09T12:10:30.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="14523" PostTypeId="2" ParentId="14496" CreationDate="2017-03-31T15:48:45.610" Score="1" Body="&lt;p&gt;Typically the walls on these stacks are very thin, less than 6 inches and their geometry provides the strength. They also from what i remember are typically build using wire mesh, not rebar on their structure though I could be wrong. If that's accurate, they would not be particularly stable for hoop pressure. They're designed for gravitational loading primarily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer, not likely.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-03-31T15:48:45.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14524" PostTypeId="1" CreationDate="2017-03-31T17:25:08.817" Score="0" ViewCount="1925" Body="&lt;p&gt;I.e. can I save a model in Fusion 360 that can then be opened in SolidWorks?&lt;/p&gt;&#xA;" OwnerUserId="10688" LastActivityDate="2017-07-23T18:18:00.973" Title="Is Fusion 360 compatible with SolidWorks?" Tags="|solidworks|software|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14525" PostTypeId="1" AcceptedAnswerId="14536" CreationDate="2017-03-31T17:43:05.230" Score="2" ViewCount="727" Body="&lt;p&gt;I am trying to use an Arduino Uno R3 as a tachometer (It is going to measure the RPM of my motorcycle). Basically it only has to count how many times an input pin goes HIGH in one second.&#xA;The problem is pulseIn is too slow for it...&#xA;The pin will be on for up to 230 times/ second (14.000 RPM).&#xA;So that would be 1000/230 (approx. 4 ms per rotation, i believe).&#xA;What is the easiest way to do the job? I can't figure any solution yet...&lt;/p&gt;&#xA;" OwnerUserId="10614" LastEditorUserId="8055" LastEditDate="2017-04-04T15:45:42.340" LastActivityDate="2017-04-04T15:45:42.340" Title="How to measure signal period with Arduino?" Tags="|mechanical-engineering|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14527" PostTypeId="2" ParentId="14524" CreationDate="2017-03-31T18:18:53.447" Score="0" Body="&lt;p&gt;I use Inventor, but 360 should allow you to export as a step file that you can import into solid works.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:18:00.973" LastActivityDate="2017-07-23T18:18:00.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14528" PostTypeId="1" CreationDate="2017-03-31T21:41:34.863" Score="1" ViewCount="601" Body="&lt;p&gt;In most (maybe all?) automobiles, the drum or disc brakes are positioned at the end of the axles, inside the wheel. This increases &lt;a href=&quot;https://en.wikipedia.org/wiki/Unsprung_mass&quot; rel=&quot;nofollow noreferrer&quot;&gt;unsprung mass&lt;/a&gt;, which reduces handling performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it not be just as effective to move the brakes inboard, positioning them on the axles themselves, somewhere under the chassis in between the suspension arms? It seems that this would have the same effect, but would eliminate the brakes' contribution to unsprung mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen it done this way on go-karts with a solid rear axle. Does the fact that automobiles have sprung suspension systems necessitate the brakes' placement inside the wheels? Are there other engineering tradeoffs at play?&lt;/p&gt;&#xA;" OwnerUserId="10691" LastActivityDate="2017-04-01T04:54:41.403" Title="Why are automotive brakes set inside the wheels, and not on the axles?" Tags="|automotive-engineering|springs|wheels|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14530" PostTypeId="2" ParentId="14528" CreationDate="2017-03-31T23:09:03.657" Score="1" Body="&lt;p&gt;This idea isn't even feasible on a typical front-wheel-drive car. There is no &quot;front axle&quot; inboard of the steering joints which the front brakes could be fitted to, and the only rotating parts of the rear axle are the stub shafts holding the wheels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably on a rear-wheel-drive go-kart, in principle you could put the rear brakes anywhere along the axle between the differential and the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the reduction of unsprung mass would be limited. If one wheel on an axle goes over a vertical bump, the vertical movement of the mid-point on the axle is half as much as the wheel. If both wheels go over a similar-shaped vertical bump, the whole axle moves the same amount as the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a car, the brakes would also occupy space and require a bigger clearance between the axle and the floor pan of the car body - and would probably be harder to access for maintenance as well.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-03-31T23:09:03.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14531" PostTypeId="2" ParentId="14528" CreationDate="2017-04-01T00:10:33.077" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;It would be more difficult to service/repair&lt;/li&gt;&#xA;&lt;li&gt;The axle of a typical car resides within a outer tube and the surface area of the axel is much smaller than the surface area of a typical rotor or a drum.&lt;/li&gt;&#xA;&lt;li&gt;The heat dissipation rate would likely be lower than conventional brakes.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5778" LastActivityDate="2017-04-01T00:10:33.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14533" PostTypeId="2" ParentId="14528" CreationDate="2017-04-01T04:54:41.403" Score="1" Body="&lt;p&gt;I did some more research, and it turns out that this is in fact a tried &amp;amp; true concept:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Inboard_brake&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Inboard_brake&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other answerers were accurate in their assessments, particularly in the difficulty of servicing, which seems to be the main reason why these aren't more popular. Accessing the brakes is more difficult and requires a lift or jack, and replacing a rotor requires disassembly of the axle (a lot harder than removing a wheel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from that, some pros:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;reduces unsprung weight&lt;/li&gt;&#xA;&lt;li&gt;reduces exposure to environmental contaminants&lt;/li&gt;&#xA;&lt;li&gt;eliminates torsion forces on suspension components while braking&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and cons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;harder to cool&lt;/li&gt;&#xA;&lt;li&gt;increases complexity; non-driven wheels require axles dedicated to braking, and cooling channels/ducts are generally required&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The cons seem to outweigh the pros, since the primary advantages are performance-related, but the disadvantages make this approach mostly unfeasible for performance (i.e., motorsports) use. Racing teams tend to work on brakes frequently and may need to change brake components quickly during practice or races, which isn't possible with inboard brakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do wonder if these may make a comeback with electric vehicle (EV) racing due to the decreased demand on friction brakes from the use of regenerative braking. I can imagine a set of pads lasting an entire race, in which case a racing team could swap them between races and eliminate the need for quick access.&lt;/p&gt;&#xA;" OwnerUserId="10691" LastActivityDate="2017-04-01T04:54:41.403" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14535" PostTypeId="2" ParentId="14525" CreationDate="2017-04-01T09:59:15.210" Score="1" Body="&lt;p&gt;I'd comment instead, but I don't have the rep so I'll have to answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &lt;a href=&quot;https://www.arduino.cc/en/Reference/attachInterrupt&quot; rel=&quot;nofollow noreferrer&quot;&gt;attaching an interrupt&lt;/a&gt; to the rising edge of your signal you can increment a counter. You read the counter every X milliseconds and calculate the rpm from that. Of course, the counter needs to be reset each time it is read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, there's a frequency meter example &lt;a href=&quot;https://www.arduino.cc/en/Tutorial/SimpleAudioFrequencyMeter&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; that looks promising (Arduino Zero only).&lt;/p&gt;&#xA;" OwnerUserId="10699" LastEditorUserId="10699" LastEditDate="2017-04-01T10:16:01.047" LastActivityDate="2017-04-01T10:16:01.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14536" PostTypeId="2" ParentId="14525" CreationDate="2017-04-01T12:08:55.167" Score="0" Body="&lt;p&gt;Similar to &lt;a href=&quot;https://engineering.stackexchange.com/a/14535/110&quot;&gt;this&lt;/a&gt; response, I would suggest using an interrupt, but also start a timer, and then capture the time elapsed on the next interrupt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you could start a timer on a interrupt, and count the number of rising edges in for the duration of the timer. So of the timer was set to 1 sec, and there was x rising edges, the RPM can be now calculated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the automotive's are magnet and hall effect sensor is used to calculate RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-04-01T12:08:55.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14539" PostTypeId="2" ParentId="14524" CreationDate="2017-04-01T15:27:10.203" Score="1" Body="&lt;p&gt;You can export the geometry yes by exporting a STEP (or IGES file  which is slightly worse). But not much else, the object will be a black box as it has no construction history. Whether this counts as the model moved over is up to you.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-04-01T18:19:52.237" LastActivityDate="2017-04-01T18:19:52.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14542" PostTypeId="1" CreationDate="2017-04-01T19:36:29.897" Score="2" ViewCount="1183" Body="&lt;p&gt;I want to have water source on the 2'nd floor of our house so I can wash stuff there without going down to our 1'st floor... Is there a way to make water rise from 1'st to 2'nd floor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KhD3a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KhD3a.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3252" LastActivityDate="2017-05-01T22:10:36.253" Title="Is there a way to make water rise from 1'st to 2'nd floor?" Tags="|water-resources|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14543" PostTypeId="2" ParentId="12055" CreationDate="2017-04-01T20:10:02.300" Score="1" Body="&lt;p&gt;As mentioned in the comments you don't want high friction between gear teeth, but if you're asking about the materials who exhibits higher friction with low wear rate, it depends on several parameters, such as, the nature of the two counterfaces, the environment, the applied load, the hardness, velocity, etc&#xA;In general there's cubic crystal oxides, with a high hardness above 10GPA whom with sliding against certain material, exhibits highr friction coefficient with a loaw wear rate, we take an example of AL2O3.&#xA;Best regards&lt;/p&gt;&#xA;" OwnerUserId="6487" LastActivityDate="2017-04-01T20:10:02.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14544" PostTypeId="1" CreationDate="2017-04-01T20:14:46.490" Score="0" ViewCount="461" Body="&lt;p&gt;I'm trying to understand what the ideal sampling time for an input change represented in the form of a timing diagram should look like, for a PLC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the relationship between the sampling rate and input change to an extent. The sampling rate must be faster (at a higher frequency) than the rate of the input changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I consider an example of an input that stays on for 25ms then switches off (when the input changes to a high), what is the ideal sampling rate to always measure this change and why?&lt;/p&gt;&#xA;" OwnerUserId="10184" LastEditorUserId="90" LastEditDate="2017-04-02T01:51:55.967" LastActivityDate="2017-04-04T13:45:39.987" Title="What must the ideal sampling rate be, to sample an input in regards to PLC's?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14545" PostTypeId="1" CreationDate="2017-04-01T20:28:39.423" Score="1" ViewCount="450" Body="&lt;p&gt;I know at least 3 different approaches to solve inverse kinematics problem. They are pseudo inverse jacobian, cyclic coordinate descent and ANFIS networks. I would like to know advantages and disadvantages of these methods comparing to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that CCD suffers from local minima but as I heard this problem seldom arises in practice. However it is more intuitive and easy to implement comparing to pseudo inverse jacobian. Therefore I wonder why would one use inverse jacobian instead of CCD?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While these two methods I have found to be the most popular there is another approach called ANFIS networks which requires gathering training data from forward kinematics equations and training a neural network. It seems a bit overcomplicated to me. Why would one prefer ANFIS to CCD or pseudo inverse jacobian methods? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="10483" LastEditorUserId="10483" LastEditDate="2017-04-06T19:23:01.083" LastActivityDate="2017-04-06T19:23:01.083" Title="Inverse kinematics solutions pros and cons" Tags="|control-engineering|control-theory|kinematics|robotics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14546" PostTypeId="2" ParentId="14542" CreationDate="2017-04-01T21:37:45.193" Score="2" Body="&lt;p&gt;Add a water pump on the main floor of your house that pumps it to the second floor. Or, assuming your house is connected to a city plumbing and water system have an extension line added to the second floor, using the city main pressure.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-04-01T21:37:45.193" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14549" PostTypeId="2" ParentId="11960" CreationDate="2017-04-02T04:45:28.480" Score="7" Body="&lt;p&gt;&lt;strong&gt;Diesel locomotives can't deliver enough power.&lt;/strong&gt; The power rating of the diesel generator on most locomotives ranges from &lt;a href=&quot;https://web.archive.org/web/20010821074000/www.getransportation.com/general/apps/specsheets/Amtrak_Genesis_2/Genesis_2.htm&quot; rel=&quot;noreferrer&quot; title=&quot;GE Genesis 2.4 MW&quot;&gt;2&lt;/a&gt; to &lt;a href=&quot;https://en.m.wikipedia.org/wiki/EMD_DDA40X&quot; rel=&quot;noreferrer&quot; title=&quot;EMD DDA40X 5 MW, though with two engines.&quot;&gt;4&lt;/a&gt; megawatts. The combined power rating of motors on high-speed trainsets range from &lt;a href=&quot;https://web.archive.org/web/20120515115623/http://www.traintesting.com/Acela_0029_Nov01_en.pdf&quot; rel=&quot;noreferrer&quot; title=&quot;Acela 4.6 MW&quot;&gt;5&lt;/a&gt; to &lt;a href=&quot;https://en.m.wikipedia.org/wiki/China_Railways_CRH380A&quot; rel=&quot;noreferrer&quot; title=&quot;CRH380AL 20.44 MW&quot;&gt;20&lt;/a&gt; megawatts, depending on size and design speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diesel engines are rated for maximum power, while motors are rated for continuous power. Motors can deliver much more than their continuous-rated power for a few seconds. This means much greater acceleration, and much greater power requirements, than any engine could provide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that aerodynamic drag, which is the dominant resistance on all but the slowest trains, is proportional to the square of speed.&lt;/p&gt;&#xA;" OwnerUserId="10706" LastEditorUserId="10706" LastEditDate="2017-04-02T08:34:59.373" LastActivityDate="2017-04-02T08:34:59.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14550" PostTypeId="1" CreationDate="2017-04-02T05:44:23.080" Score="4" ViewCount="359" Body="&lt;p&gt;Here is a photo of a railroad crossing near a road intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://alamedasun.com/sites/default/files/styles/nodeview__smartphone_landscape/public/primary-images/3-3-front-robin.gif?itok=8nCaeJIv&quot; rel=&quot;nofollow noreferrer&quot;&gt;Traffic, Rail Lights out of Sync at Fruitvale Avenue Crossing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The road signals are mounted to slender poles (or even suspended from a cable). Maintenance workers can access the road signals with a truck-mounted boom lift. In contrast, the poles that support the rail crossing signals are built much sturdier and also support a ladder and catwalk. (Ladders and catwalks seem to be common on railway operating signals, too.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are many more signals than workers, it seems like the total cost to outfit the workers with boom lifts would be cheaper than making every signal easily accessible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize that road and rail signals developed independently, but their common engineering criteria (to hold a light up in the air) should have resulted in road and rail converging on a single lowest-cost solution.&lt;/p&gt;&#xA;" OwnerUserId="10706" LastActivityDate="2017-05-02T08:21:10.873" Title="Why are rail signals overbuilt compared to road signals?" Tags="|rail|transportation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14552" PostTypeId="2" ParentId="14550" CreationDate="2017-04-02T08:13:31.067" Score="2" Body="&lt;p&gt;For a start, not all rail line traffic signals are like the ones in your picture - there is no uniformity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over-the-road signals will tend to be more common in high road traffic situations and in more populous regions than in low traffic or rural areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By having traffic lights over the road like that, at rail intersections, makes the intersection more noticeable to road users.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collisions between trains and road vehicles tend to be more devastating and disruptive than collisions between just road vehicles. Making the rail intersection more obvious &amp;amp; noticeable to road traffic is an attempt to mitigate the possibility of a collision between a train &amp;amp; road vehicles.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-04-02T08:13:31.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14553" PostTypeId="1" CreationDate="2017-04-02T08:33:39.813" Score="2" ViewCount="8177" Body="&lt;p&gt;Can anyone please explain to me the difference between surface loading rate and surface overflow rate when used during the design calculations of sedimentation tanks in water treatment plants? &lt;/p&gt;&#xA;" OwnerUserId="10319" LastEditorUserId="368" LastEditDate="2017-05-04T19:19:54.143" LastActivityDate="2017-05-04T19:19:54.143" Title="Difference between surface loading rate and surface overflow rate of sedimentation tanks" Tags="|civil-engineering|environmental-engineering|waste-water-treatment|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14555" PostTypeId="1" CreationDate="2017-04-02T10:49:49.843" Score="0" ViewCount="67" Body="&lt;p&gt;I'm working my way through learning vacuum technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 10l vessel which I want to evacuate to &amp;lt; 10mbar. I'll use &lt;a href=&quot;https://www.pfeiffer-vacuum.com/en/products/diaphragm-pumps/mvp-006/?detailPdoId=4223&quot; rel=&quot;nofollow noreferrer&quot;&gt;this diaphragm pump&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to add a (manual) venting valve to the system, or will switching off the pump do the job?&lt;/p&gt;&#xA;" OwnerUserId="10709" LastActivityDate="2017-05-02T19:24:02.863" Title="Vacuum vessel at 10mbar with a diaphragm pump: do I need a venting valve?" Tags="|vacuum|vacuum-pumps|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14556" PostTypeId="2" ParentId="14555" CreationDate="2017-04-02T12:38:46.933" Score="2" Body="&lt;p&gt;I am not a vacuum expert, but if you look at the technical data for your pump, you'll find this &quot;Leak rate 6x10-3 Pa m³/s&quot;. I take this to mean that your 10 liter = 0.01 m$^3$ vessel will leak at a rate of 0.6 Pa/s.  That's pretty slow.  You'll want a manual venting value.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-04-02T12:38:46.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14558" PostTypeId="1" CreationDate="2017-04-02T15:01:49.323" Score="1" ViewCount="597" Body="&lt;p&gt;How does &lt;a href=&quot;https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcQNg1Wpyk_7N98QZnPONEhLjVotl2s-7MUPgBq8jQY9mtS08xBu&quot; rel=&quot;nofollow noreferrer&quot;&gt;water sink&lt;/a&gt; works?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sr2WE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sr2WE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to do the same thing but the water just stayed:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IeVgH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IeVgH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3252" LastActivityDate="2017-04-02T23:16:26.767" Title="How does water sink works?" Tags="|water-resources|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14559" PostTypeId="2" ParentId="14558" CreationDate="2017-04-02T15:36:59.847" Score="0" Body="&lt;p&gt;The purpose of the kink in the line is to keep some water in the dip. The reason being that any smell in the rest of the line will get trapped in the top part and not flow back through into the house.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-04-02T15:36:59.847" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14560" PostTypeId="2" ParentId="14514" CreationDate="2017-04-02T16:56:40.570" Score="3" Body="&lt;p&gt;For a more complete discussion see: &lt;a href=&quot;https://www.concrete.org/store/productdetail.aspx?ItemID=421108&quot; rel=&quot;nofollow noreferrer&quot;&gt;ACI 421.1R-08 - Guide to Shear Reinforcement for Slabs&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in principle you need to continue to provide punching shear reinforcement until the shear capacity of the slab is sufficient to prevent punching shear at that perimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From ACI 421.1R-08 this means:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to check at $d/2$ from the column face to determine if shear reinforcement is required. If it is then you need to continue to check perimeters at a distance of $\alpha d$ from the column face until you find that (in SI units) &#xA;$$\frac{\nu_u}{\phi} \leq \frac{0.17\lambda\sqrt{{f'}_c}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the outermost perimeter of shear reinforcement must be at least $\alpha d - \frac{d}{2}$ from the column face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nu_u$ is maximum shear stress due to factored forces. $\phi$ is the strength reduction factor = 0.75. ${f'}_c$ is the specified concrete compressive strength. $\lambda$ is the modification factor for the reduced mechanical properties of lightweight concrete (=1 for normal weight concrete). $d$ is the effective depth of slab.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2017-04-02T17:07:13.303" LastActivityDate="2017-04-02T17:07:13.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14561" PostTypeId="1" AcceptedAnswerId="14588" CreationDate="2017-04-02T17:41:49.210" Score="1" ViewCount="1180" Body="&lt;p&gt;&quot;By mounting a gyro on a set of gimbal rings, the gyro is able to rotate freely in any direction. Even though the gimbal rings are tiled, twisted, or otherwise moved, the gyro remains in the plane in which it was originally spinning&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've seen that statement in several different textbooks, but I'm having trouble understanding the terminology used. It's stating that the gyro is able to rotate freely in any direction, but then it also says that the gyro remains in the plane it's rotating. If it can rotate freely in any direction, then it isn't in it's original plane of rotation. So if anyone could simplify that, it'd be helpful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the gimbals allow things to rotate around it while the gimbal doesn't rotate, but then wouldn't you need the gyro to move for precession to take effect? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an example if I said that I applied a force with regards to precession taking place, would the gyro maintain it's plane of rotation but that force would take affect on the gimbal? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you! &lt;/p&gt;&#xA;" OwnerUserId="9066" LastEditorUserId="9066" LastEditDate="2017-04-02T17:47:25.310" LastActivityDate="2017-04-03T18:55:27.520" Title="How do gimbals affect gyros?" Tags="|instrumentation|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14562" PostTypeId="2" ParentId="14561" CreationDate="2017-04-02T17:52:10.663" Score="0" Body="&lt;p&gt;The point of a gyroscope is that, because it has a substantial amount of angular momentum is will tend to stay in the same plane that it is rotating in unless a fairly significant amount of force is applied to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What actually happens with a gimbal mounting is that the gyroscope stays still and whatever it is attached to rotates around it (although this is of course relative to your frame of reference) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way the gimbals isolate the gyroscope from external forces. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-04-02T17:52:10.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14563" PostTypeId="1" AcceptedAnswerId="14583" CreationDate="2017-04-02T20:06:20.020" Score="0" ViewCount="1008" Body="&lt;p&gt;What is the simplest way if you have a shaft rotating clockwise for it to have an extension that rotates anti clockwise. Least loss of energy is best so i think a mechanical mechanism would be best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking is there a way you could use the main shaft in a turbo fan engine to drive another fan in the opposing direction in front of the original fan.&lt;/p&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="10361" LastEditDate="2017-04-02T21:12:08.390" LastActivityDate="2017-07-23T18:17:44.720" Title="Shaft to swap direction of rotation" Tags="|gears|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14564" PostTypeId="1" CreationDate="2017-04-02T20:20:41.340" Score="-1" ViewCount="164" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9WF4F.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9WF4F.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reasoning for this question comes out of pure curiosity.&#xA;say if there were, for example 4 bugatti veyron's ( represented by the rectangles on the diagram) all starting from the middle and all attached by rope, and they accelerate out at 90 degrees to each other, reaching a top speed of say 200mph (89.4m/s) before the rope becomes taught (diagram in top right corner) what kind of force will the centre point experience and also the force the driver. &#xA;assumptions can be made regarding the distance and time, also assuming there are no material limits and ideal situations meaning in the end the centre point does not move and rope joints are infinitely strong.&#xA;id just like to see how the newtons laws could be applied to a question like this.&lt;/p&gt;&#xA;" OwnerUserId="10468" LastActivityDate="2017-04-04T12:22:14.807" Title="4 Way Bugatti Tug of War" Tags="|mechanical-engineering|energy|acceleration|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14565" PostTypeId="1" CreationDate="2017-04-02T20:48:44.320" Score="3" ViewCount="97" Body="&lt;p&gt;Consider a &quot;cartridge&quot; of sorts. Is it possible to construct a &quot;pinhole&quot; in the cartridge such that only one molecule will pass through at a time? If so, what sort of equipment would I need to construct such a pinhole, or what sort of company would I need to contact? (The cheaper the method, the better.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, the molecules in question are &lt;a href=&quot;https://en.wikipedia.org/wiki/Deoxyribonucleotide&quot; rel=&quot;nofollow noreferrer&quot;&gt;deoxribonucleotides&lt;/a&gt;, which are about 13 angstroms or 1.3 nanometers in diameter. The molecules would I believe be in the state of a gas.&lt;/p&gt;&#xA;" OwnerUserId="10715" LastEditorUserId="10715" LastEditDate="2017-06-20T00:06:21.110" LastActivityDate="2017-06-20T00:06:21.110" Title="Creating a &quot;pinhole&quot; small enough to allow only one molecule to pass through" Tags="|mechanical-engineering|gas|biomechanics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14566" PostTypeId="1" CreationDate="2017-04-02T21:52:06.607" Score="-1" ViewCount="75" Body="&lt;p&gt;Greeting, I have two pipes A ,B same diameter, joined to be one pipe C with double diameter of A ,  water flow in A with pressure 90bar and in B with pressure 40 bar    . &#xA;The question is how much will be the water pressure in pipe c ? &#xA;Regards &lt;/p&gt;&#xA;" OwnerUserId="10692" LastActivityDate="2017-07-23T18:17:52.523" Title="Net water pressure in pipes" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ClosedDate="2017-04-04T15:44:27.407" ContentLicense="CC BY-SA 3.0" />
  <row Id="14567" PostTypeId="2" ParentId="14558" CreationDate="2017-04-02T23:16:26.767" Score="2" Body="&lt;p&gt;This is a simple case of &lt;a href=&quot;https://en.wikipedia.org/wiki/Communicating_vessels&quot; rel=&quot;nofollow noreferrer&quot;&gt;communicating vessels&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Communicating_vessels&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/2/20/ANIMvasicomunicanti.gif&quot; alt=&quot;example of communicating vessels&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When water enters your sink, it raises the water level on the right side of the kink all the way up to the sink. This means that the water pressure on the right side of the pipe is greater than on the left side of the pipe. This imbalance is &quot;not allowed&quot;, so the water level on the left side rises (therefore increasing the pressure on that side). If it could, it would rise up to the level of the water on the right side of the pipe. However, since the left side connects to the sewage column, the water merely drains away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you close the faucet, the water on the right hand side will drop until it reaches an equilibrium with the left side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you tried with the flexible tubing didn't work because both the entrance and exit levels are the same. If you are putting water in from the left (as would seem to be the case), then lower the right side just a bit. When you add the water, it will fill the tubing up on both sides until it reaches an exit, which will be the now-lowered right side.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-04-02T23:16:26.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14569" PostTypeId="2" ParentId="14544" CreationDate="2017-04-02T23:46:40.820" Score="1" Body="&lt;p&gt;The short answer to your question is 13 ms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should review the Nyquist-Shannon theorem &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. For a bandlimited signal the sampling frequency needs to be &gt;2f. If your input signal is digital then (technically) it is not bandlimited but practically there will be some limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step is to understand the signal you are sampling. I'd start with a 13ms sampling rate &amp;amp; run some tests to ensure you are catching all the transitions. If some are missed you can then ramp up the sampling rate until you are satisfied.&lt;/p&gt;&#xA;" OwnerUserId="10415" LastEditorUserId="10415" LastEditDate="2017-04-04T13:45:39.987" LastActivityDate="2017-04-04T13:45:39.987" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14572" PostTypeId="1" AcceptedAnswerId="14581" CreationDate="2017-04-03T02:31:04.823" Score="3" ViewCount="762" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OGull.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OGull.png&quot; alt=&quot;Image showing Ackermann steering where all of the projected wheel axles meet at the center of rotation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Ackermann_steering_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Ackermann_steering_geometry&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image above shows a vehicle with &lt;a href=&quot;https://en.wikipedia.org/wiki/Ackermann_steering_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ackermann steering&lt;/a&gt;. The front wheels are connected by a mechanism so that they turn at different angles such that the projected axles of the wheels meet at the same point, the center of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KYvue.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KYvue.png&quot; alt=&quot;Image showing non-Ackerman steering where the projected wheel axles do not all meet at the same location&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.google.com/patents/US6634109&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/patents/US6634109&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image shows a case where are four wheels do not have the same center of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Ackerman condition is not satisfied as shown in the figure for a car taking a turn, then what would be the equivalent centre of rotation for the vehicle? &lt;/p&gt;&#xA;" OwnerUserId="10075" LastEditorUserId="7430" LastEditDate="2017-04-05T13:07:52.460" LastActivityDate="2017-04-10T09:13:23.937" Title="Ackerman Conundrum" Tags="|automotive-engineering|dynamics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14574" PostTypeId="1" CreationDate="2017-04-03T05:44:48.050" Score="1" ViewCount="213" Body="&lt;p&gt;I am trying to design a science experiment involving lasers and want to know if what I am planning to build it technically feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the experiment requires that I make tiny nanometer changes to the apparatus in order to lengthen the time it takes for a beam of light to travel a particular distance.  I need to be able to control the distance very accurately because I have to make calculations based on how far the light is travelling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing the weight of the apparatus is roughly a kilogram.  What's the smallest reliable distance I can move the apparatus with the specifications I have?&lt;/p&gt;&#xA;" OwnerUserId="10720" LastActivityDate="2017-04-03T23:58:28.150" Title="What is smallest distance you can reliably move an object?" Tags="|mechanical-engineering|experimental-physics|research|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="14575" PostTypeId="2" ParentId="13547" CreationDate="2017-04-03T08:54:39.417" Score="0" Body="&lt;p&gt;It depends on:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The damping your system (if it is rusted or not)&lt;/li&gt;&#xA;&lt;li&gt;The speed wants to bring it down &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An experimental way to verify which of the case 1 or 2 apply could consist in removing the 'nut' and then let the system freely falls:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every movement slower than this falling speed, the case 1 will apply (gravitation is the main force in the system)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For every movement faster than this falling speed, the case 2 will apply (Your system does not goes down naturally and therefore you are pushing it downwards)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Except if your system is rusted or incudes safeties to prevent free fall, the case 1 is most likely correct&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="10684" LastEditDate="2017-04-03T09:00:58.950" LastActivityDate="2017-04-03T09:00:58.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14577" PostTypeId="2" ParentId="14564" CreationDate="2017-04-03T10:43:55.923" Score="-2" Body="&lt;p&gt;With the information and assumptions you have given, I would say that the forces act in a hyperbolic fashion toward each other, creating a saddle region at the center point as in your figure. To have a rough look see the image below, in which the red arrows indicates the forces acting in a hyperbolic fashion around the center location creating the above-mentioned saddle region.&lt;a href=&quot;https://i.stack.imgur.com/lk3iX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lk3iX.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Hope this helps!!&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt;: As J.Ari mentioned, only the net force at (0,0) will be 0N, all the others will behave hyperbolically.&lt;/p&gt;&#xA;" OwnerUserId="10681" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-04-03T13:07:37.067" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="14578" PostTypeId="2" ParentId="14574" CreationDate="2017-04-03T13:34:17.507" Score="1" Body="&lt;p&gt;Positioners with resolution on the order of 1-10 nm are totally doable.  Here's one example: &lt;a href=&quot;https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=2386&amp;amp;pn=MAX313D&quot; rel=&quot;nofollow noreferrer&quot;&gt;3-Axis NanoMax Flexure Stages&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you buy this with the closed loop piezo option, 5 nm resolution is possible and it has a 1 kg load capacity. Only $3000 too. Getting down to 0.1 nm resolution is certainly possible, and is in fact routinely done in the world of atomic force microscopy. However, you may have to sacrifice total travel, load capacity, or cost in order to get there.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="8055" LastEditDate="2017-04-03T23:58:28.150" LastActivityDate="2017-04-03T23:58:28.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14579" PostTypeId="2" ParentId="14564" CreationDate="2017-04-03T13:45:09.837" Score="4" Body="&lt;p&gt;The force on each car is purely longitudinal because the restraining force of necessity runs thru the central point (origin).  For example, the East and West forces cancel so far as the origin is concerned, so no lateral force exists from the viewpoint of the North or South car.&lt;br&gt;&#xA;The restraining force on each car, then, is the same as if it were tied to an unmoveable tiedown point-- the force thus is the torque currently being applied via each car's engine and drivetrain.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-04-03T13:45:09.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14581" PostTypeId="2" ParentId="14572" CreationDate="2017-04-03T15:15:13.133" Score="5" Body="&lt;p&gt;This is quite easy:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Choose one of the wheels&lt;/li&gt;&#xA;&lt;li&gt;build a second wheel so it would satisfy Ackerman condition&lt;/li&gt;&#xA;&lt;li&gt;Get the First rotation point ( O&lt;SUB&gt;1&lt;/SUB&gt; )&lt;/li&gt;&#xA;&lt;li&gt;Do the same with the second wheel, and get a second rotation point ( O&lt;SUB&gt;1&lt;/SUB&gt;' )&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The actual rotation point can be anything between these two points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose, for instance, the first point as rotation point, it means that the first wheel has no issue but the second wheel therefore has an additional damping to its movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the real rotation center you could compute including the damping on each wheel, but In a driving situation equal damping is quite unlikely, more over it can change in function of the directions. On top of this, this Damping changes in function of the object dynamics. So you may most likely end up with an irrelevant model.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="10684" LastEditDate="2017-04-10T09:13:23.937" LastActivityDate="2017-04-10T09:13:23.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14582" PostTypeId="2" ParentId="14566" CreationDate="2017-04-03T15:40:55.873" Score="0" Body="&lt;p&gt;Pressure will be 90 bar. Also, you will get reverse flow in pipe B because the pressure is greater. The water in B will not add to the overall pressure of the source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider A and B are connected to tanks with water level deep enough to make 90 and 40 bar pressure. Tank A will fill tank B until the level is nearly identical then both will contribute equal pressure to C and have near equal flow rates.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:17:52.523" LastActivityDate="2017-07-23T18:17:52.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14583" PostTypeId="2" ParentId="14563" CreationDate="2017-04-03T15:53:41.303" Score="1" Body="&lt;p&gt;Least loss of energy would be using one gear with adequate lubrication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prefer planetary gears because they last much longer and distribute the load better. Plus you can do it with a hollow shaft. The center shaft turns clockwise, the planet gears turn counterclockwise as does the outer shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link that should show you a gif of the gear in action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://giphy.com/gifs/yXQrz0IscQeWs&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://giphy.com/gifs/yXQrz0IscQeWs&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:17:44.720" LastActivityDate="2017-07-23T18:17:44.720" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14584" PostTypeId="2" ParentId="14574" CreationDate="2017-04-03T16:01:50.943" Score="0" Body="&lt;p&gt;This will depend on the setup of your system and how easily you want to be able to make the movement. Off the top of my head though this sounds feasible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The setup I would imagine, or start from is moving your mass laterally (parallel to the ground). Having it sit on low friction bearings or bushings of some sort. Then having that position controlled by a fine pitch lead screw. There several options available for nanometer precision lead screw  type tools (I found Nano-Gauge off a quick google search, but there are several dozen others as well). Cost will likely not be that cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could design and control your own system which would be cheaper, but calibration and the time to build it will be much much higher than buying a COTS (COTS - commercial off the shelf) part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nanometer precision is difficult but doable. Verification and calibration will be tricky. Temperature control and air pressure of  your system will be of a concern at this scale, doable, just tricky.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-04-03T16:01:50.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14585" PostTypeId="1" CreationDate="2017-04-03T16:24:36.520" Score="1" ViewCount="111" Body="&lt;p&gt;I'm looking into possible failures of a cast zinc part. I'm wondering if the Zinc-oxide that develops after a time would have a thickness great enough to cause dimensional tolerancing issues? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tolerances on my parts are plus-minus 0.1 mm. Assuming the part is made within spec could an oxide (or other zinc corrosion) cause the part to fall out of spec?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Environmental conditions are standard air pressure, room temperature and normal air composition.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastEditorUserId="368" LastEditDate="2017-12-14T16:45:14.837" LastActivityDate="2017-12-14T16:45:14.837" Title="How thick of an oxide might form on a cast zinc part?" Tags="|materials|mechanical-failure|corrosion|tolerance|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14587" PostTypeId="1" AcceptedAnswerId="14591" CreationDate="2017-04-03T18:24:11.433" Score="2" ViewCount="246" Body="&lt;p&gt;The heat of a lightning bolt causes the surrounding air to rapidly expand causing thunder in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reciprocating combustion engine ignites an air fuel mix to use the rapid expansion created in the process to build up pressure and use that to move pistons in order to create rotating motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a reciprocating engine that uses only air heated by a very high powered electric arc work? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It surely would be very ineffective, hard to supply with enough power without cables and would probably melt the engine block, but could it function in principle?&lt;/p&gt;&#xA;" OwnerUserId="10725" LastEditorUserId="368" LastEditDate="2017-04-03T22:45:36.193" LastActivityDate="2017-04-03T23:30:58.400" Title="Could a reciprocating engine be powered by electric arc instead of combustion?" Tags="|mechanical-engineering|automotive-engineering|electric-vehicles|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14588" PostTypeId="2" ParentId="14561" CreationDate="2017-04-03T18:55:27.520" Score="1" Body="&lt;p&gt;The confusion arises in terminology.  What instrumentation engineers call a &lt;em&gt;gyroscope&lt;/em&gt; is actually an &lt;em&gt;integrating gyroscope&lt;/em&gt; or &lt;em&gt;rate gyro&lt;/em&gt;, and what they call a &lt;em&gt;gimbal&lt;/em&gt; in &lt;em&gt;gimbal lock&lt;/em&gt; isn't the gimbal used to mount the gyroscope itself - the gimbals are used to keep a &lt;a href=&quot;http://inertialnavigations.blogspot.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;platform housing three gyroscopes level&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.tpub.com/neets/book15/63a.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard gyroscope&lt;/a&gt; will point in the same direction at all times, while a &lt;a href=&quot;http://www.tpub.com/neets/book15/63e.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;rate gyroscope&lt;/a&gt; is used for measuring angular velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wzFNw.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wzFNw.gif&quot; alt=&quot;small and large tilts on a rate gyro&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate gyro will deflect the springs in proportion to how fast the frame is spinning - the faster the spin, the larger the springs deflect, which via strain gauge or similar measurements engineering, can be turned into an electrical signal.  Measure the change in rate to derive the angular acceleration.  Have a computer integrate the rate over time to derive current angular position.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the pivoting gimbals for this device though - they are in the exact configuration of a gimbal lock.  A rate gyro's housing can only measure changes in a single axis.  If the frame housing the gimbal moves away from the axis, then the device wouldn't measure properly.  So, three &lt;em&gt;rate gyroscopes&lt;/em&gt; are required to measure angular movement in all three angles.  These are mounted on a platform that maintains a fixed orientation in space, so the rate gyroscopes can function properly.  &lt;strong&gt;In other words, motors specifically move the platform that the rate gyroscopes are mounted onto so it stays on the same plane.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SzjeV.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SzjeV.gif&quot; alt=&quot;Gyroscopes on a platform.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the rate gyroscope feeds a porportional signal to the instrumentation panel, it also feeds this signal to three motors.  They turn the platform in response to the signal, keeping the platform level.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-04-03T18:55:27.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14589" PostTypeId="2" ParentId="14566" CreationDate="2017-04-03T21:30:42.357" Score="0" Body="&lt;p&gt;I think we don't have enough information to solve this problem.&#xA;Are the given pressure values hydrostatic or hydrodynamic?&#xA;Hydrostatic would basically mean the height of the water level, e.g. $90\mbox{bar} \hat{=} 900\mbox{m}$, hydrodynamic is the difference of static pressure and pressure losses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a hydraulic system like this we can make some assumptions &lt;em&gt;at the intersection&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;flow continuity:&lt;/strong&gt; $Q_A+Q_B+Q_C=0$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This holds true, as water is almost incompressible, thus all that comes in must come out somewhere.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;energy equilibrium:&lt;/strong&gt; $H_A=H_B=H_C$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At the intersection the hydraulic head of all three flows must be equal.&#xA;In this case $H=\frac{p}{\rho g}+\frac{2Q^2}{\pi D g}$ &#xA;The pressures may indeed vary, but two pressure values do not suffice to calculate the third value.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2017-04-03T21:30:42.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14590" PostTypeId="2" ParentId="14544" CreationDate="2017-04-03T22:54:07.877" Score="0" Body="&lt;p&gt;My short answer is 24 ms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your pulse has a width of 25 ms.  Therefore if you poll at 42 Hz (1 over 24 ms) you will always be certain of catching the pulse.  Now it's possible that you'll only overlap the input pulse by 1 ms, leading to a maximum delay of 24 ms from the falling edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sampling theorem does not apply in this situation.  Shannon only applies if you are trying to exactly replicate the input signal, say a sine wave of frequency f.  That would require a minimum sample rate &gt; 2f.  In this case all you want is to catch the pulse passing, not replicate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PLCs have hardware interrupts.  If you create one of those on the appropriate I/O pin, you'll catch the pulse within the PLCs minimum specified propagation delay.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2017-04-03T22:54:07.877" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14591" PostTypeId="2" ParentId="14587" CreationDate="2017-04-03T23:23:11.467" Score="3" Body="&lt;p&gt;Yes, but it would probably be a very poor engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted in your question, there are huge practical issues with using the arc alone as your power source, compared to using it to ignite the fuel mixture in a conventional spark-ignition engine. But you are correct to observe that both the electric arc and the combustion process cause a rapid expansion inside the chamber, and that this rapid expansion could drive a piston.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you propose is to convert electrical energy to thermal energy to kinetic energy and useful work (you want to move the car). Implicit in this is some form of energy storage, which we can assume to be chemical for the sake of comparison with conventional technologies. So, chemical -&gt; electrical -&gt; thermal -&gt; mechanical. You have to go through at least three conversions of energy from one form to another in order to make this technology work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrast that with electric motor technology, which stores energy in batteries (or fuel cells) and produces work relatively efficiently. Chemical -&gt; electrical -&gt; mechanical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrast it also with the conventional internal combustion engine, which stores energy relatively densely in hydrocarbon fuels and produces relatively more work per refueling cycle. Chemical -&gt; thermal -&gt; mechanical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your proposed solution adds another conversion step to the electric motor solution. It seems to combine the worst characteristics of both established solutions, since you have the same storage challenges to work with as the electric car does. So I wouldn't call it a &lt;em&gt;practical&lt;/em&gt; technology; but you might be able to produce a working prototype of this sort of engine, for use as a novelty or teaching aid.&lt;/p&gt;&#xA;" OwnerUserId="368" LastEditorUserId="368" LastEditDate="2017-04-03T23:30:58.400" LastActivityDate="2017-04-03T23:30:58.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14593" PostTypeId="1" CreationDate="2017-04-04T06:22:27.700" Score="0" ViewCount="946" Body="&lt;p&gt;My question points that why the auto-ignition temperature of petrol is more than diesel even though petrol is more volatile and got least flash point than diesel.&lt;/p&gt;&#xA;" OwnerUserId="10734" LastActivityDate="2017-04-04T08:34:35.357" Title="why is the auto ignition temperature of petrol is more than the diesel even though petrol is more volatile than diesel?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2017-04-05T16:51:25.117" ContentLicense="CC BY-SA 3.0" />
  <row Id="14594" PostTypeId="2" ParentId="14553" CreationDate="2017-04-04T07:45:19.113" Score="2" Body="&lt;p&gt;I'm not 100% familiar with english terms of art. However, &lt;a href=&quot;http://www.deq.state.ms.us/MDEQ.nsf/8f3cca80758d46df86256c080014bab8/557702f42378ab3b862573220073b407/$FILE/QP7498-DEQ-Ch%204%20correction.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;according to this document&lt;/a&gt;, they are the same:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The &lt;em&gt;surface loading rate&lt;/em&gt; is a hydraulic loading factor expressed in terms of flow per surface area. This factor is also referred to as the &lt;em&gt;“surface settling rate”&lt;/em&gt; or &lt;em&gt;“surface overflow rate.”&lt;/em&gt; [...]&#xA;  Typical values of surface loading rates for primary clarifiers are 600 to 1,000 gallons per day per&#xA;  square foot.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-04-04T07:45:19.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14595" PostTypeId="2" ParentId="14593" CreationDate="2017-04-04T08:34:35.357" Score="3" Body="&lt;p&gt;I'd comment, but I don't have the rep, so I'll have to answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because they're a measure two &lt;em&gt;different&lt;/em&gt; properties.&#xA;Just because something is volatile (high vapor pressure) doesn't necessarily make it more flammable (low auto-ignition temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could think of volatility as more related to boiling point.&#xA;Auto-ignition temperature is related more to the minimum amount of energy required to begin a reaction.&lt;/p&gt;&#xA;" OwnerUserId="10699" LastActivityDate="2017-04-04T08:34:35.357" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14597" PostTypeId="1" CreationDate="2017-04-04T08:56:37.040" Score="2" ViewCount="43" Body="&lt;p&gt;Roskam, book &quot;Airplane Aerodynamichs and Performance&quot; at subparagraph: &quot;2.5.2 HIGH-SPEED AIRSPEED INDICATORS (COMPRESSIBLE FLOW)&quot; uses the following three equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) $\left(\frac{\gamma}{\gamma-1}\right)\frac{p}{\rho}+\frac{1}{2}V^2=\left(\frac{\gamma}{\gamma-1}\right)\frac{p_t}{\rho_t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) $\frac{p}{\rho^\gamma}=\frac{p_t}{\rho^\gamma_t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c)  $V_a^2=\gamma\frac{p}{\rho}$ ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and, after some steps, it reachs $M^2=\frac{2}{\gamma-1}\left[\left(\frac{p_t-p}{p}+1\right)^\frac{\gamma-1}{\gamma}-1\right]$  (called eq.n (2.44) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point the book says: &quot;...From this equation it can be seen that when $(p_t-p)/p$ is measured, the Mach number can be calculated. An instrument which measures both $(p_t-p)$ and the static pressure, $p$, to indicate the Mach number through Eqn (2.44) is called a Mach meter. Note, that a conventional airspeed indicator will only measure $(p_t-p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When, in the calibration of an airspeed indicator, $(p_t-p)/p$ is replaced by $(p_t-p)/p_0$ and $V_a$ is replaced by $V_{a_0}$ , the resulting airspeed is called the calibrated airspeed, $V_c$ ....&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some steps it yelds an equation with $M^2$ depending by variables $\gamma$, $\delta$ * , $V_c$ , $V_{a_0}$ and it tells that Fig 2.6, obtained with this last equation, represents a plot of true Mach numbers versus calibrated airspeed for constant pressure altitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My QUESTION is: what is the actual meaning, role and aim of $p_0$, $V_{a_0}$ and $V_c$ ? I suppose Roskam considers they play an experimental role to set the MachMeter before it will fitted to use, but I have to understand exactly how to use those variable. How to obtain them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that equation with calibrated variables is not something realistic about the flight because for example we have $\frac{(p_t-p)}{p_0}$, so we have in the same formula $p$ and $p_0$, something is changed and something is not changed, so it's not represent a specific physical situation but only something artificial in order to calibrate the Mach-meter, I suppose. But ... can someone explain me the details?&#xA;Maybe that Machmeter is measuring $V_c$ every moment during flight?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\delta$ is $p/p_0$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4916" LastEditorUserId="4916" LastEditDate="2017-04-13T16:32:12.393" LastActivityDate="2017-04-13T16:32:12.393" Title="what is the actual meaning, role and aim of $p_0$, $V_{a_0}$ and $V_c$ for Mach-meter?" Tags="|fluid-mechanics|thermodynamics|aerodynamics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14598" PostTypeId="1" AcceptedAnswerId="14606" CreationDate="2017-04-04T09:08:01.143" Score="5" ViewCount="4296" Body="&lt;p&gt;So today when I was taking a taxi and looked at the vehicles, I observed that the majority of the light vehicles (i.e. 16-person minibus/ 4-6 person cars) use 5 (or less frequently, 6) wheels studs for each wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, 6 evenly distributed components are more stable as it has 3 axes of symmetry, as compared to 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any mechanical reason for that? &lt;/p&gt;&#xA;" OwnerUserId="10736" LastEditorUserId="10736" LastEditDate="2017-04-04T09:19:26.420" LastActivityDate="2021-01-03T21:18:34.137" Title="Reason of Using exactly 5 Wheel Studs per Wheel" Tags="|automotive-engineering|wheels|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14600" PostTypeId="2" ParentId="14585" CreationDate="2017-04-04T11:07:21.283" Score="1" Body="&lt;p&gt;One possible issue is galvanic corrosion. In certain conditions, if dissimilar metals are in contact with each other the more electronegative metal will corrode at a greatly accelerated rate, effectively the same situation a a metal/metal battery. This can be a particular issue if you have steel studs or other interests in a non-ferrous casting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed sacrificial zinc anodes are used to protect thing like boat hulls and submerged piles from corrosion and this is why galvanising is such effective corrosion protection, even if the zinc surface layer is scratched through to the base metal the galvanic effect will continue to provide protection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally zinc is a fairly reactive metal which relies on a passive oxide layer for corrosion resistance so anything which disrupts this layer can cause very rapid corrosion. Even quite  weak acid can strip galvanising in a few hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also with any cast part you should consider shrinkage as a potential source of tolerating issues although this should be apparent from new. Equally inadequate design of a casting can generate high residual stresses and other defects which can cause failure over time. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-04-04T11:07:21.283" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14601" PostTypeId="2" ParentId="4113" CreationDate="2017-04-04T11:18:10.817" Score="0" Body="&lt;p&gt;It is important to understand that the critical point of a pure substance is characterized by certain properties. For example, the first and second partial derivative of the pressure with respect to the molar volume at constant temperature vanish. Secondly, it is important to understand that the critical point given by an equation of state (EoS) is not necessarily the same point as determined by measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the critical point of the Peng-Robinson (PR) EoS one can calculate the first and second partial derivative with respect to the molar volume, and look for a given volume/temperature pair where both partial derivatives vanish. Doing this with a solver for non-linear equation systems (e.g. a Levenberg-Marquardt method) I found the critical point of CO2 to be close to $1.0T_c$ and $0.89\rho_c$, giving a critical pressure of $1.0p_c$, which yields a critical compressibility of $Z_c=0.3074$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get good results I recommend the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you use iterative solvers be sure to check residuals and convergence of the method.&lt;/li&gt;&#xA;&lt;li&gt;Use the same constants and critical state variables everywhere, as the solution is quite sensitive.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10739" LastEditorUserId="10739" LastEditDate="2017-04-05T18:50:11.420" LastActivityDate="2017-04-05T18:50:11.420" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14602" PostTypeId="1" AcceptedAnswerId="14617" CreationDate="2017-04-04T11:31:59.130" Score="3" ViewCount="1594" Body="&lt;p&gt;I'm having trouble drawing an isometric view of this shape:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UPdLj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UPdLj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've only just started an engineering course and so far it's involved a lot of drawing which I usually can do but I can't seem to figure out how to 'place' this object to make an isometric view. Am I able to make the base of the object in my drawing whichever face I want?, as I feel perhaps if I rotate it so that the face with the hole is facing the ground might make it easier to draw, or do I have to use the 20x20 square as the base like it is in the current view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also kind of confusing me as well in regards to drawing it in parallel oblique view but that's another issue I guess which I'll probably be able to figure out once I wrap my head around how to attempt the isometric view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other shapes I've drawn (see &lt;a href=&quot;https://imgur.com/a/xsm2r&quot; rel=&quot;nofollow noreferrer&quot;&gt;example&lt;/a&gt;) felt a lot easier to simply start and don't mess with my brain like this particular shape does. It isn't being marked but I still want to understand how to do it since I'll probably have to deal with these types of shapes later on anyway and would rather have the basics down now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry if my explanation doesn't make sense but I'm not really sure how to word my issue at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated, thanks.&lt;/p&gt;&#xA;" OwnerUserId="10740" LastEditorUserId="7083" LastEditDate="2017-04-07T21:17:40.340" LastActivityDate="2017-04-07T21:17:40.340" Title="How to draw isometric view of this shape?" Tags="|mechanical-engineering|drawings|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14604" PostTypeId="1" CreationDate="2017-04-04T11:54:05.877" Score="3" ViewCount="1398" Body="&lt;p&gt;I'm trying to design a mobile robot which has to climb over flat 60° inclined wet glass surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking to use a large vacuum system composed by a large fan connected to a brushless motor. &lt;em&gt;By looking on google&lt;/em&gt;, I found this photo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DdEt4.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DdEt4.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is possible to see, they use a brushless motor with a fan to generete some kind of vacuum to keep the robot on the wall. This system is very similar to the RC wall car toys which are able to vertically move on flat walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea is to &lt;strong&gt;create a vacuum chamber&lt;/strong&gt; and to make it connect to the surface by using a &lt;strong&gt;soft rubber gasket&lt;/strong&gt; (to avoid leakage), but I do not know how to design the vacuum generator.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;How many RPMs do I need?&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;What shape should the fan have?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The robot has a weight of about 30-35Kg and its size is 1000x800 mm.&#xA;The surface which can be used for the vacuum generator is the space between the tracks and it is about 800x600 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you can help me!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you a lot!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&#xA;Information after @ericnutsch answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my ideal design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to use a cylider as case for the fan/impeller, use a rubber gasket for the cylider borders in order to avoid leakages (when possible). The cylinder/case will be suspended by springs in order to make it touch the surface by using the weight of the chassis. I do not know if the drawing is clear, but this is what I would like to acquire.&#xA;The dimensions showed in the figures are realistic and the weight of the robot is about 30-35kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wM5ZM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wM5ZM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LJxrV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LJxrV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on #Mark answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found this impeller which can be customized and can have a diameter of 550 mm. It comes with some specifications:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8pm2n.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8pm2n.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT1&lt;/strong&gt;:&#xA;This is what I tried so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During last months, I did several tests with different solutions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Using rubber tracks with Linatex which offers a great grip over flat surfaces&lt;/li&gt;&#xA;&lt;li&gt;Multiple suction cups with a vacuum generator&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fZDJF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fZDJF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Linatex works very great up to 60° if the surface is not wet, if it is wet, they works fine up to 18°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kPwF8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kPwF8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4 suction cups works fine only if the robot moves very very slowly (0.1 m/s) otherwise they lose the vacuum.&lt;/p&gt;&#xA;" OwnerUserId="10741" LastEditorUserId="10741" LastEditDate="2017-04-05T18:29:37.773" LastActivityDate="2017-04-07T06:20:23.190" Title="How to design a vacuum climbing robot" Tags="|automotive-engineering|robotics|vacuum|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14605" PostTypeId="2" ParentId="14564" CreationDate="2017-04-04T12:22:14.807" Score="0" Body="&lt;p&gt;The question assumes no material limits, but specifies the very fine Bugatti Veyron, which is made of specific materials.  Let's just take two cars driving apart at high speed, joined by an infinitely strong rope.  Assuming 89.4m/s and a mass per car of 1,888kg, the total energy of 2*1/2 mv^2 = 15MJ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the cars were to come to a halt, this is the amount of energy which would have to be dissipated.  This energy could be go into the elongation of the ropes, or more likely into the deformation of the cars.  Depending upon what the rope was attached to, either the cars would be ripped apart until the requisite amount of energy has been absorbed by the structure of the car, or the ropes will detach themselves from the car due to local failure of the connection point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is more about energy than balancing forces.&lt;/p&gt;&#xA;" OwnerUserId="10240" LastActivityDate="2017-04-04T12:22:14.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14606" PostTypeId="2" ParentId="14598" CreationDate="2017-04-04T12:36:33.050" Score="6" Body="&lt;p&gt;On reason is that die cast (alloy wheels) generally have an odd number of spokes. This is because having directly opposed spokes causes problems with residual stress distribution  as the casting cools and shrinks (this is also why cast iron hand wheels often have S shaped spokes). So 5 studs tends to be a more convenient number, if only for aesthetic reasons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wheels with an even number of spokes do exist but these are generally machined from a solid billet in several parts and are substantially more expensive than die cast ones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays most vehicles are designed on a common platform which is used across a wide range and even things like vans often have alloys wheels as an option so there is little point having different hub designs for steel and alloy wheels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the issue of redundancy. If one of five or six studs fails or is loose then you still have a reasonable distribution of functional nuts ie covering more than 180 degrees of the hub. If you only have 3 out of 4 present then it is entirely possible for the wheel to not sit flat on the hub. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-04-04T13:12:15.330" LastActivityDate="2017-04-04T13:12:15.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14607" PostTypeId="1" AcceptedAnswerId="14621" CreationDate="2017-04-04T12:41:40.323" Score="1" ViewCount="4161" Body="&lt;p&gt;I am currently building a box filled with electronics and I now look to isolate the sound a bit with the lightest and best material out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am looking for is a thin, lightweight and very soundproof material. I came across a material called Sortbothane and it seems like a very good fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ebay.com/itm/Isolate-It-Sorbothane-Acoustic-Vibration-Damping-Film-40-Duro-0-04-x-4-x-/172548443510?hash=item282cb04d76:g:EqcAAOSwZVlXnC9J&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sorbothane Acoustic &amp;amp; Vibration Damping Film 40 Duro&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;0.04&quot; of Sorbothane Will Absorb up to 4 dB&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That number is pretty impressive I must say and the material seems to be reasonable light and thin as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to my question, do you think Sortobothane is the way to go or is there any other materials out there with the same/better soundabsorbing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also thinking to combine a few materials to make it very solid. For instance I am planning to use a small layer of silicone as well.&lt;/p&gt;&#xA;" OwnerUserId="10630" LastEditorUserId="8055" LastEditDate="2017-04-04T16:30:31.753" LastActivityDate="2017-04-05T06:20:25.663" Title="What is the best thin/light material to kill/absorb sound? Sorbothane?" Tags="|sound-isolation|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14609" PostTypeId="2" ParentId="14598" CreationDate="2017-04-04T12:56:14.673" Score="1" Body="&lt;p&gt;Most probably cost optimization. 6 might be stronger, but 5 is strong enough and when you make thousands of vehicles a year those 4 nuts, bolts and assembly time add up to a significant enough value. 4 lug wheel mounting systems are more common in Europe where the cars are smaller and 10 inch rims are common in 'mini' style cars.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:17:36.947" LastActivityDate="2017-07-23T18:17:36.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14610" PostTypeId="2" ParentId="14598" CreationDate="2017-04-04T13:19:52.150" Score="1" Body="&lt;p&gt;To some extent, the number of bolt studs depends on the anticipated torque on the wheel.  That's why small cars may have 4, larger cars 6, and semitrailer trucks have 11 or 13.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's certainly a cosmetic contribution when considering sporty or luxury cars, but &quot;form follows function&quot; here as well.  The designer might add a bolt or two for looks but would never under-design the fitting. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-04-04T13:19:52.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14611" PostTypeId="1" CreationDate="2017-04-04T15:02:46.413" Score="7" ViewCount="4459" Body="&lt;p&gt;I had a look online and into some literature however I did not seem to find a good comparison for the two different methods. They are both used for determining displacements and slopes (rotation by theta) at a point in a continuum.&#xA;Former uses a virtual unit force which equals to the strain energy in the element (when multiplied by displacement of interest) and latter uses differential with respect to a virtual force that tends to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one is more efficient and which one is more accurate? Why would one choose Virtual Work over Castigliano's or vice versa? &lt;/p&gt;&#xA;" OwnerUserId="7651" LastEditorUserId="7651" LastEditDate="2017-04-04T16:47:26.443" LastActivityDate="2018-12-17T22:01:10.223" Title="Principle of Virtual Work vs Castigliano's (Second) Theorem" Tags="|mechanical-engineering|structural-engineering|structural-analysis|applied-mechanics|stresses|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14613" PostTypeId="2" ParentId="14604" CreationDate="2017-04-04T15:22:58.310" Score="1" Body="&lt;p&gt;Well if you would like to create your own vacuum, I would not recommend designing the actual pump yourself. You would need to design the turbomachinery components like the impeller (fyi an impeller is not a simple fan, it is much more complicated in terms of it's geomtery and the material used to make it), casing and various seals from scratch and those things are NOT easy to analyse, design and test in terms of the fluid dynamics, stress analysis and design engineering that goes into them.&#xA;I say go on a distributor like RS Components and find one that suits your needs. Of course the part that you would design is the way you would channel the air, create the suction chamber, so on and so forth. That will depend on your engineering/DIY knowledge and your specifications.&lt;/p&gt;&#xA;" OwnerUserId="7651" LastActivityDate="2017-04-04T15:22:58.310" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="14614" PostTypeId="1" AcceptedAnswerId="14623" CreationDate="2017-04-04T15:46:29.610" Score="3" ViewCount="1578" Body="&lt;p&gt;I want to make a shaft rotate in 21 discrete steps, from a motor that is turning smoothly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking of some kind of simplified clock escapement type of mechanism. Or non circular gears. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it could be done by programming a stepper motor. But it will be easier and more fun for me to do it mechanically with 3D printed nylon parts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a low torque low speed application. About 1 rpm and less than 0.1 Nm torque. &lt;/p&gt;&#xA;" OwnerUserId="7879" LastActivityDate="2017-04-05T07:05:28.337" Title="What mechanism can make a shaft rotate in steps" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14615" PostTypeId="1" AcceptedAnswerId="14973" CreationDate="2017-04-04T17:54:56.897" Score="0" ViewCount="1671" Body="&lt;p&gt;I just modeled a High-rise building of 43 stories with an area of 700 square meter, Using ETABS as a software to model the building I got the Static and Dynamic Base Shear as the image below, my Question is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it Reasonable that the dynamic base shear (Specx and Specy) is bigger than the static base shear (Ex and Ey) ? How can i know if this result is logical or not ?&#xA;If the result is Correct i don't need to scale the dynamic base shear ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kw63A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kw63A.png&quot; alt=&quot;ETABS Dynamic and Static Base Shear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2018-10-11T11:39:26.117" Title="Can Dynamic Base shear be greater than Static Base shear?" Tags="|structural-engineering|civil-engineering|structural-analysis|dynamics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14617" PostTypeId="2" ParentId="14602" CreationDate="2017-04-04T20:50:26.577" Score="3" Body="&lt;p&gt;Don't think, look at the numbers! Drawing isometric, or any projective images for that matter is very straightforward. Just as long as you do not try to find a trick to make it easier. Just work out the distance numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by sketching your primary axes. These are the directions you are going to measure against in all parallel projections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ehqfv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ehqfv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then start measuring lines along those directions according to your measurements. Do not try to eye them just measure one axis at a time. So for example from the bottom back corner to the top corner there is 40 units, to the beginning if the miter theres another 40 in the -30 degree direction (axis toward right)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2RTWX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2RTWX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just continue mechanically form there to measure all known points. First one plane then from that plane and so on. Don't worry about the circle at first, just find its center and radius as if it were a box. Just faithfully measure units on your ruler (or whatever your drawing software if any is using)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lWVqZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lWVqZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't worry about hiding lines, you can do that later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0wokO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0wokO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then once you have one or a few solids done. Hide some lines to make things clearer to see. If you missed any lines it will become apparent at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ob1Ip.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ob1Ip.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then block the next solid. Now you need to use some draftsman's assumptions in the circle, a draftsman would assume the circle is in middle of the shape at least vertically form that image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LOvmI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LOvmI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OuImW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OuImW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-04-04T20:56:02.530" LastActivityDate="2017-04-04T20:56:02.530" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14619" PostTypeId="1" AcceptedAnswerId="14995" CreationDate="2017-04-04T23:14:48.497" Score="1" ViewCount="601" Body="&lt;p&gt;Is there any way to know whether or not an unknown system has positive or negative zeros just from looking at the step response? This is the step response: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/DcQVy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DcQVy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10182" LastActivityDate="2017-04-26T22:18:47.480" Title="Extract information about zeros of system from step response" Tags="|control-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14621" PostTypeId="2" ParentId="14607" CreationDate="2017-04-05T06:20:25.663" Score="0" Body="&lt;p&gt;Isolate and absorb are two different things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Vibration_isolation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vibration isolation&lt;/a&gt; means let one side of the material vibrate but keep the vibrations from transmitting through.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Absorption_(acoustics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Acoustic absorption&lt;/a&gt; means to dissipate vibration energy into heat at the audible frequencies.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When used in the proper geometric configuration, a thin layer of &lt;a href=&quot;https://en.wikipedia.org/wiki/Sorbothane&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sorbothane&lt;/a&gt; could work well for either of these applications. Its just important to approach your problem with an intent. It will be best used at the interface between two masses, like the enclosure and the mounting location or the door of the enclosure and the enclosure, etc. Selecting this interface, and changing the amount of mass can direct toward isolating or absorbing the concerned frequency. This will take some math and/or some experimentation. Other means of attachment across this interface like bolts, screws, adhesives, etc will influence how well the absorbs/isolates behaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower-energy and higher frequency sounds that are already airborne (like a buzzing electronic component) may be better absorbed with &lt;a href=&quot;https://en.wikipedia.org/wiki/Acoustic_foam&quot; rel=&quot;nofollow noreferrer&quot;&gt;open celled acoustic foam&lt;/a&gt; inside the enclosure.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-04-05T06:20:25.663" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14623" PostTypeId="2" ParentId="14614" CreationDate="2017-04-05T07:05:28.337" Score="7" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Geneva_drive&quot; rel=&quot;noreferrer&quot;&gt;Geneva drive&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.thingiverse.com/thing:192448&quot; rel=&quot;noreferrer&quot;&gt;3D Printed Geneva Drive&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BXJqM.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BXJqM.gif&quot; alt=&quot;Geneva drive&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/5RB9m.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5RB9m.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Anchor_escapement&quot; rel=&quot;noreferrer&quot;&gt;Anchor Escapement (Clock Mechanism)&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.thingiverse.com/thing:1128317&quot; rel=&quot;noreferrer&quot;&gt;3D Printed Anchor Escapement&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E8SQZ.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E8SQZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-04-05T07:05:28.337" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14627" PostTypeId="2" ParentId="14604" CreationDate="2017-04-05T08:44:32.630" Score="4" Body="&lt;p&gt;You do not need full vacuum to accomplish your goal. A small amount of vacuum pressure adds up really quick over a large surface area. Try to make the entire footprint of your robot under vacuum; your sealing perimeter increases linearly with diameter, but area increases with diameter squared. Basically this means the problem is easier to solve the larger you go. And a circle is the best geometry because it has the highest area to perimeter ratio. Its really the design problem of a hovercraft, just in reverse. You will have to plan on some air leakage for it to operate on non-perfect surfaces in the real world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some existing equipment that may meet your needs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A hobby &lt;a href=&quot;https://hobbyking.com/en_us/alloy-dps-64mm-10-blade-electric-ducted-fan-assembley-3300kv.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;rc brushless ducted fan&lt;/a&gt;. They do not publish head pressure, but might supply -3.5kPa gauge (-0.5psig) or so at a very high flowrate. You may be able to find one rated for high pressure lower flow. The nice part about these is they are easily battery operated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;a href=&quot;https://www.zoro.com/metrovac-critical-area-vacuums-65-cfm-34-hp-foam-mdv-1esd/i/G2100975/&quot; rel=&quot;nofollow noreferrer&quot;&gt;vacuum blower&lt;/a&gt;. High flow rate and reasonably high pressure. This one has 60 inches of water max, which is approximately 14kPa. They are likely rating in positive pressure. Its vacuum abilities would be less, so lets say -7kPa gauge (-1psig).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A &lt;a href=&quot;https://www.zoro.com/ebm-papst-motorized-impeller-115v-530-cfm-r2e220-ra44-14/i/G5069285/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;motorized impeller&quot;&lt;/a&gt; (the internals of a shop vac). The website does not rate the vacuum pressure but i would estimate it would be -40kPa (-6psig).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;An &lt;a href=&quot;https://www.aliexpress.com/item/Hot-Sale-12V-13L-M-DC-Diaphragm-Micro-Air-Vacuum-Pump-For-massage-Chair-Home-Appliance/32707334920.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;electric diaphragm pump&lt;/a&gt;. These are typically low flowrate so you would be back to clean smooth surfaces, but the vacuum pressure on this one is -50kPa gauge. Might be more flow than you have on your current vacuum pump if you stack a few in parallel.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think you mean 80mm in your image not 80cm. 8cm results in an area of 0.005m^3 (7.8in^2). So at perfect vacuum 101kPa (-14.7psig), you are getting 500N (115 pounds of force), which is great until you lose that perfect vacuum. The force drops quickly because the vacuum pump you are using can not maintain the flow rate required by the imperfect surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say a motorized impeller would be your best bet. I don't know your scale here, but lets say putting a shroud around your footprint puts you at a 0.4m (16 inch) diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PT22a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PT22a.png&quot; alt=&quot;climbing robot footprint&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding area would be 0.126m^2 (200in^2). At -40kPa (-6psig) you now have roughly 5000N (1,200 pounds of force). And since this fan can handle a high flow rate, even large leaks are not a problem. You will want to design the shroud such that it is strong enough to not collapse under this force and design it such that it does not apply too much load to your climbing tracks.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2017-04-07T06:20:23.190" LastActivityDate="2017-04-07T06:20:23.190" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14628" PostTypeId="1" CreationDate="2017-04-05T09:30:01.617" Score="2" ViewCount="105" Body="&lt;p&gt;I am designing a 3V battery-powered simple micro-controller to collect sensor data in RAM. How can I avoid a conflict with the 5V of a USB when it comes time to upload the data? Disconnecting the battery would lose the RAM. The solution needs to be very simple as the PCB is very small. &lt;/p&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="8055" LastEditDate="2017-04-05T16:43:08.140" LastActivityDate="2017-04-05T16:43:08.140" Title="How to transition from battery to USB in a microcontroller?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14631" PostTypeId="2" ParentId="14628" CreationDate="2017-04-05T12:55:55.717" Score="4" Body="&lt;p&gt;There are two possible options based on the limited infomation you have provided&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Consider storing data in Non-Volatile Memory.&lt;/li&gt;&#xA;&lt;li&gt;Use a LDO such as TPS73633 that can support wide input voltage range TPS73633 support a range of 3.5V and 5.5V. The output is 3.3V. An provide both the supply voltage options. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vUl1c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vUl1c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2017-04-05T12:55:55.717" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="14632" PostTypeId="2" ParentId="12409" CreationDate="2017-04-05T14:55:41.633" Score="1" Body="&lt;p&gt;If you're completely stuck and all you have is plenty of (good) data, plug it into a program called &lt;code&gt;Eureqa&lt;/code&gt; from &lt;a href=&quot;http://www.nutonian.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;nutonian.com&lt;/a&gt;. Eureqa uses symbolic regression to search for an appropriate model. If your data isn't super nonlinear or discontinuous it'll find a fit pretty quickly. &lt;/p&gt;&#xA;" OwnerUserId="10699" LastActivityDate="2017-04-05T14:55:41.633" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14633" PostTypeId="2" ParentId="14604" CreationDate="2017-04-05T17:33:27.333" Score="3" Body="&lt;p&gt;An important point in the fluid dynamics design is about control points.  A simple experiment - hold your hand over a hose on the vacuum cleaner, you'll notice a light vacuum pulling on it, but hold your hand right on the top of the hose, and it feels like it is pulling significantly harder.  This is because fans operate on a load response curve, and when there is less air flow, it pulls with more pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few different fan designs.  Forward, Radial and Backwards curved blades:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/35S6J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/35S6J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These lead to different kinds of performance curves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Lqxx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Lqxx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As expected, with no flow, you have nearly the highest pressure, and with a very low pressure, you have almost no pressure.  The problem is you need to operate in a portion of the curve with a negative slope.  The area with the positive slope is called the surge region:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zrWHH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zrWHH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While forward curved fans have a lot more pressure available, they typically have very large surge areas because of the large area with a positive slope, but there is usually the small area with a negative slope right before the surge area.  In this application, a forward curved blade design could work so long as it stays above the local maxima in the mid-flow range, However,a radial blade design would be more conservative, and usually more cost effective, but the unit will necessarily have to be bigger to obtain large pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comprehensive stability analysis also needs to look at the power requirements.  For power design, we'd need to look at the power curve for the fan.  Notice the power requirements for the motor increase drastically as the flow rate goes up.  Since a user can always peel the robot off the wall while in operation, the design will need to account for the fact that a constant RPM fan will demand more torque as more air is moved while the system stabilizes to the lower pressure requirements.  Potential solutions could include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Electronics could detect this torque and turn the motor off&lt;/li&gt;&#xA;&lt;li&gt;Size the motor to handle the maximum airflow scenario&lt;/li&gt;&#xA;&lt;li&gt;Something new and creative that's mechanically governed to limit torque&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-04-05T17:33:27.333" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14634" PostTypeId="2" ParentId="14456" CreationDate="2017-04-05T19:54:36.980" Score="-5" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f3Eap.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8jCQT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope you understand!&#xA;If any questions, feel free to ask.&lt;/p&gt;&#xA;" OwnerUserId="5414" LastEditorDisplayName="user16" LastEditDate="2017-04-06T00:00:05.330" LastActivityDate="2017-04-06T00:00:05.330" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14636" PostTypeId="1" AcceptedAnswerId="22482" CreationDate="2017-04-05T20:01:00.530" Score="3" ViewCount="4011" Body="&lt;p&gt;I'm finding myself doing quite a lot of thread cutting in stainless steel lately, particularly 304 and 316, often these are moderately deep threads ~ 25 mm or so and at M12 it is becoming a bit of a chore. I'm using a decent tap wench and cutting oil with carbon steel taps (taper, second cut and finishing/plug) but I was wondering if anybody has any experience or knowledge of any ways to make this a bit easier eg by a different tap material or coating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not doing enough volume to justify any major investment in tooling as this is all prototype work but it's that difficult area of a dozen or so at a time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it is specifically stainless which is the problem and these are stressed welded components so free machining stainless isn't really an option (as far as I know) and I do appreciate that this is a bit of a 'how do I have my cake and eat it' type question. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The main issue is the taps getting progressively tighter, I haven't broken any yet but they often feel like they are getting to the point where they might. I generally start the taps off in a lathe (not under power) with a sliding chuck in the tail-stock so I'm fairly confident that they are going in pretty square. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-04-15T20:33:33.857" LastActivityDate="2020-07-24T22:07:01.110" Title="Hand thread taps for stainless steel" Tags="|machining|threads|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14637" PostTypeId="2" ParentId="14412" CreationDate="2017-04-05T20:15:51.680" Score="1" Body="&lt;p&gt;To maintain flow rate (Q), velocity (v) must be constant. To keep velocity constant, potential head(h) must be constant. To keep potential head constant either a large crucible (like a reservoir) can be used OR flow rate OUT of crucible must be equal to flow rate IN to the crucible.&lt;/p&gt;&#xA;" OwnerUserId="5414" LastActivityDate="2017-04-05T20:15:51.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14638" PostTypeId="2" ParentId="14085" CreationDate="2017-04-05T20:43:45.577" Score="1" Body="&lt;p&gt;U=dx/dt&#xA;  =2t*(x0/t0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now eliminate (x0/t0) from the above equation;&#xA;x=(x0/t0)*t^2&#xA;(X0/t0)=x/t^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &#xA;U=2x/t  which is in eulerian form.&lt;/p&gt;&#xA;" OwnerUserId="5414" LastActivityDate="2017-04-05T20:43:45.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14639" PostTypeId="2" ParentId="13776" CreationDate="2017-04-05T20:52:15.553" Score="1" Body="&lt;p&gt;Apply Bernoulli's equation across the two ends. You have pressure values at both ends. Elevation level is same. And substitute velocity(v)=flow rate (Q)/area of flow.&#xA;Assume head loss to be negligible &lt;/p&gt;&#xA;" OwnerUserId="5414" LastActivityDate="2017-04-05T20:52:15.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14641" PostTypeId="1" CreationDate="2017-04-06T02:36:04.533" Score="1" ViewCount="34" Body="&lt;p&gt;I have a wave propagation question that regards two materials impacting each other. The question asks to find the impact pressure of the impact with a tungsten carbide flyer and a steel target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated my two particle velocities from the left and right going Hugoniot equations however I always have trouble choosing which one is the correct one to use. I know if one is negative that the other is the obvious choice but for this specific problem I have two positive particle velocities. Up to this point I have simply done the calculations with both and decided to choose a reasonable pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping there was maybe a check I could use to make sure I am using the correct particle velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="9104" LastActivityDate="2017-04-06T02:36:04.533" Title="Hugoniot Equations and Impact Pressure" Tags="|pressure|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14642" PostTypeId="1" AcceptedAnswerId="14648" CreationDate="2017-04-06T08:37:29.457" Score="2" ViewCount="182" Body="&lt;p&gt;In Windows is it possible to write a simple program to read the binary value of a USB port without going thru a USB driver? &lt;/p&gt;&#xA;" OwnerUserId="10433" LastEditorUserId="8055" LastEditDate="2017-04-06T16:30:12.440" LastActivityDate="2017-04-06T16:30:12.440" Title="Is it possible to read a USB port directly?" Tags="|electrical-engineering|embedded-systems|" AnswerCount="2" CommentCount="2" ClosedDate="2017-04-08T02:11:42.057" ContentLicense="CC BY-SA 3.0" />
  <row Id="14644" PostTypeId="1" AcceptedAnswerId="15352" CreationDate="2017-04-06T09:16:10.287" Score="2" ViewCount="381" Body="&lt;p&gt;If you spin the stator of an AC induction motor than would the rotation of the rotor increase? Or would it have a different affect such as moving with a higher force?&lt;/p&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="10361" LastEditDate="2017-05-15T14:03:37.610" LastActivityDate="2017-05-16T20:53:14.767" Title="Mechanically increasing RPM of ac induction motor" Tags="|electrical-engineering|motors|ac|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14648" PostTypeId="2" ParentId="14642" CreationDate="2017-04-06T12:18:52.577" Score="5" Body="&lt;p&gt;No, at least not in any remotely modern version of Windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accessing hardware directly requires kernel privilege level, which ordinary application programs don't have.  &lt;i&gt;Think&lt;/i&gt; about it.  It has to be this way else one rogue process could cause all kinds of harm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drivers not only deal with the hardware details for you, but they also act as gate keeper to what operations applications can perform on the hardware.  Actually, applications don't perform hardware operations.  They send requests to drivers to perform hardware operations for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, you need a driver to access any hardware, including a USB.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-04-06T12:18:52.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14649" PostTypeId="2" ParentId="14642" CreationDate="2017-04-06T12:20:33.590" Score="4" Body="&lt;p&gt;It assumed that this question is a continuation from &lt;a href=&quot;https://engineering.stackexchange.com/questions/14628/how-to-transition-from-battery-to-usb-in-a-microcontroller&quot;&gt;How to transition from battery to USB in a microcontroller?&lt;/a&gt;. It is also assumed that your goal is to collect data using a PIC micro controller and transfer the data to a Windows base computer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate is to use configure a UART port on the PIC micro controller to transfer data serially. Then use a FTDI based USB serial cable such as the one in the image to transfer data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V3wWo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V3wWo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data can be accessed using a serial COM port on the PC. The USB port will be enumerated as a RS232 standard COM port. This will also help you create simple C program too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This alternate solution will solve your 3V and 5V predicament from the other question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sparkfun.com/products/9717&quot; rel=&quot;nofollow noreferrer&quot;&gt;FTDI Cable 5V VCC-3.3V I/O&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-04-06T12:53:05.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14652" PostTypeId="1" CreationDate="2017-04-06T13:13:58.943" Score="0" ViewCount="4895" Body="&lt;p&gt;I'd like to know if I can use a gear train to drive a pinion, and thereby increase the torque of the rack. If so, how does the gear ratio of the gear train affect the torque of the rack relative to the gear ratio of the rack and pinion on its own?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, and this should perhaps be another question but I'll go ahead and ask it: If I have two separate racks separated by a distance of X, connected so that when one rack moves the other moves also, and use a pinion to drive only one rack, should I expect a decrease in the torque of the secondary rack, and if so to what degree?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to help ensure I get an answer to my specific question, I'd like to state upfront that I'm aware that the gear ratio of the rack is determined by the distance the rack travels relative to each rotation of the pinion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question relates to whether or not the mechanical advantage created by a gear train is entirely lost if the gear train is used to drive a pinion which drives a rack, and how the torque of the rack scales when using a gear train to turn the pinion relative to the gear ratio of the rack and pinion themselves.&lt;/p&gt;&#xA;" OwnerUserId="10765" LastEditorUserId="78" LastEditDate="2017-12-23T13:00:19.113" LastActivityDate="2019-11-03T17:29:15.887" Title="Gear train with rack and pinion" Tags="|mechanical-engineering|gears|torque|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14653" PostTypeId="2" ParentId="14652" CreationDate="2017-04-06T13:30:18.993" Score="0" Body="&lt;p&gt;There is a direct correlation between any gear system input and output. The rack and pinion is just another gear. The gearing can be modeled as one gear against the rack if you do the math to make that gear an equivalent ratio to the track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one gear is twice the size of the next it has a 2 to one ratio. If your rack and pinion also have a 2 to one ratio then adding the gear to the rack and pinion will result in a 4 to one ratio (forgive me if i just made stupid error in math, i haven't had coffee yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you add gears to your pinion gear you will amplify torque or speed to your rack. If you use a second rack and pinion tied to the first the and the load is shared between the two you will only see a difference due to the losses of friction and normal gear transmission losses.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:17:29.877" LastActivityDate="2017-07-23T18:17:29.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14655" PostTypeId="1" CreationDate="2017-04-06T14:55:31.673" Score="1" ViewCount="7894" Body="&lt;p&gt;I have a door that will hang on a vertical hinge.  The door will be 20 feet wide and 10 feet tall, there will be one hinge at each end of the door and the hinges will be 11.75 inches from the top of the door, and will have an amount of counterweight above the hinge.  The counterweight will be a series of weights along the 20' width of the door.  The goal will be to open the door by creating torque at the hinge in the counterclockwise direction at either end, so I am trying to create an equation to determine the torque necessary so that I can play with the weight of the door (this is my first post, please let me know if there is any more information that you need):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ejZYY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ejZYY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10759" LastActivityDate="2017-04-06T18:03:33.537" Title="How to calculate force required to open a door on a horizontal hinge" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14658" PostTypeId="1" CreationDate="2017-04-06T15:37:14.247" Score="4" ViewCount="253" Body="&lt;p&gt;I have an arduino powered CNC machine with the Protoneer v3.00 shield. I have a probe setup for auto levelling which works correctly, I get a depth map for the board. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the issue is that this depth map is created to the point at which the bit touches the board, but the board has some flex in it such that it is raised above the table in the centre. This therefore means that when it comes to milling, any flex in the board gets pushed down by the bit before any milling occurs and so the depth map is not correct as it has not taken account of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realise therefore I need to ensure the board is firmly clamped to the table at all points and so I think a vacuum table is the best option. I have found one which is 210mm x 120mm and has a 10mm x 10mm grid for the vacuum holes (vacuumtablesuk SEAL2 VT2012).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be needing to clamp standard FR4 fibreglass PCB board, and I am wondering what vacuum pump would be suitable? I am lost as to how to choose a vacuum pump so any advice and links to suitable pumps would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="10772" LastActivityDate="2019-12-26T07:02:35.453" Title="Arduino CNC - which vacuum pump?" Tags="|vacuum|vacuum-pumps|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14659" PostTypeId="1" CreationDate="2017-04-06T16:19:00.303" Score="2" ViewCount="88" Body="&lt;p&gt;Is $p\cdot V^γ= \mathrm{const.}$ valid for only adiabatic reversible process and not for adiabatic irreversible process if not why?&lt;br&gt;&#xA;Today I was doing an objective question as mentioned above and chose option &quot;any adiabatic process&quot; but the ans was &quot;only reversible adiabatic process&quot; but i couldn't understand why. All i know is that path followed by both process will be different but how will it affect the adiabatic equation i.e $p\cdot V^γ= \mathrm{const}$.&lt;/p&gt;&#xA;" OwnerUserId="10643" LastEditorUserId="1339" LastEditDate="2018-01-19T22:48:44.750" LastActivityDate="2018-02-18T23:57:17.270" Title="validity of adiabatic equation" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14660" PostTypeId="1" CreationDate="2017-04-06T16:42:55.840" Score="1" ViewCount="139" Body="&lt;p&gt;I know that torque is proportional to RPM and vice versa, but id like to know why we cant just decrease the circumference of a generator shaft in order to compensate for increases in torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if the power input is held constant, will decreasing the circumference of a generator shaft increase RPM without decreasing torque? If not, why not and what happens? Im looking for reasons other than the laws of thermodynamics.&lt;/p&gt;&#xA;" OwnerUserId="10766" LastEditorUserId="10766" LastEditDate="2017-04-06T17:38:55.020" LastActivityDate="2017-07-23T18:17:19.307" Title="Decreasing the circumference of a generator shaft" Tags="|electrical-engineering|torque|generator|" AnswerCount="1" CommentCount="10" ClosedDate="2017-04-07T23:12:59.623" ContentLicense="CC BY-SA 3.0" />
  <row Id="14662" PostTypeId="2" ParentId="14655" CreationDate="2017-04-06T18:03:33.537" Score="1" Body="&lt;p&gt;This is a simple matter of balancing moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have a force $F$ applied at a distance $x_F$ from the hinge. This generates a moment of $M=Fx_F$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have a distributed load $q$ applied along the door (of height $h$ and width $w$). The total resultant force of this load is $Q=qhw$, applied at an average position equal to $\frac{h}{2}$, for a moment of $M=qwh\frac{h}{2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equalizing both moments, you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;Fx_F = \frac{qwh^2}{2} \\&#xA;\therefore F = \frac{qwh^2}{2x_F}&#xA;\end{gather}$$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-04-06T18:03:33.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14663" PostTypeId="2" ParentId="14660" CreationDate="2017-04-06T18:18:54.923" Score="1" Body="&lt;p&gt;Just like gear ratios, if you change the diameter you change the torque. The power will be the same, but if you decrease the primary gear the RPM will go up and the torque will increase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I always remember it from riding a bicycle.When you change from the larger sprocket on the pedal gear to the smaller it is easier for you to pedal but you go slower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In industrial application the shaft diameter is not often changed except to compensate for strength or weight. What is changed is the output shaft gear or pulley. To increase torque you decrease the drive pulley / gear or increase the loaded pulley / gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is often done with constant speed motors to change the speed and torque of the output. If you look at the top of a drill press you will often see a bunch of stacked pulleys connected by a drive belt. Scroll down about half way on the page I linked below to see images of a drill press pulley system. They use 3 sets of pulleys and 2 belts to give more control over the speed, it works kinda like changing gears in a car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.thewoodnerd.com/reviews/delta_18-900L.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.thewoodnerd.com/reviews/delta_18-900L.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:17:19.307" LastActivityDate="2017-07-23T18:17:19.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14664" PostTypeId="1" AcceptedAnswerId="14667" CreationDate="2017-04-06T18:48:58.757" Score="2" ViewCount="650" Body="&lt;p&gt;I have the following problem most of which I have calculated but have a difficulty with the beam under angle. The angle is 56,30 degress. I just cannot figure out how the forces there are acting - the equations I am making have three unknown variables in two equations, which cannot be calculated, so apparently there are forces I &quot;cannot see&quot; there as well and the equations are wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disributed loads have been replaced with concentrated loads: the point where they act is indicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jRKti.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jRKti.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for advice.&lt;/p&gt;&#xA;" OwnerUserId="10780" LastEditorUserId="255" LastEditDate="2017-04-07T08:43:33.483" LastActivityDate="2017-04-07T08:43:33.483" Title="Reaction force on a beam" Tags="|structural-analysis|applied-mechanics|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14665" PostTypeId="1" AcceptedAnswerId="14668" CreationDate="2017-04-06T19:19:31.900" Score="0" ViewCount="51" Body="&lt;p&gt;is there anyway of converting this machine into giving me digital readings?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bought it on ebay the other day and I am planning on cleaning it up and getting it looking good as new.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/A4EUU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A4EUU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10781" LastActivityDate="2017-04-07T21:18:35.600" Title="Anyway of converting this to outputting digital readings?" Tags="|electrical-engineering|design|" AnswerCount="1" CommentCount="2" ClosedDate="2017-04-07T23:13:43.530" ContentLicense="CC BY-SA 3.0" />
  <row Id="14667" PostTypeId="2" ParentId="14664" CreationDate="2017-04-06T20:40:14.910" Score="2" Body="&lt;p&gt;It is worth noting that this can be treated as two separate structures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zPU8R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zPU8R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that's how you solved the reactions of support A. Given that, you can ignore the entire left-hand structure and act like you only have the right-hand side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, let's use the standard equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= B_x + C_x + 29.4 = 0 \\&#xA;\sum F_z &amp;amp;= B_z + C_z - 24.5 - 7 = 0 \\&#xA;\sum M_B &amp;amp;= -29.4 \cdot 6 + 6C_z = 0 \\&#xA;\therefore C_z &amp;amp;= 29.4\text{ kN} \\&#xA;\therefore B_z &amp;amp;= 24.5 + 7 - C_z = 2.1\text{ kN}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might seem like you're now stuck without solving for $F_x$, but you can notice that the diagonal member is a truss, with only axial loads. That means that the vertical and horizontal forces applied to it must be proportional to its tangent. Therefore we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;C_x &amp;amp;= -C_z \cdot \dfrac{6}{9} = -19.6\text{ kN} \\&#xA;\therefore B_x &amp;amp;= -29.4 - C_x = -9.8\text{ kN}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check our work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1vkPV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1vkPV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-04-06T20:40:14.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14668" PostTypeId="2" ParentId="14665" CreationDate="2017-04-06T21:27:49.620" Score="2" Body="&lt;p&gt;I can think of a few different ways of doing something like this.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I believe that type of dial uses a magnetic field to move the needle. A well placed Hall effect sensor could do the trick. This is not invasive and also relatively hidden.&lt;/li&gt;&#xA;&lt;li&gt;A hall effect sensor may be a bit complex to place and calibrate. Another option would be to tap into the reading and pull the voltage generating the current for the needle. Link this to an ADC in a micro-controller (Arduino, MSP430, etc.) and you should get an decent result. &lt;a href=&quot;https://www.youtube.com/watch?v=ruuxn2u3yao&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electron microscope image capture via microcontroller (with drill bit animation)&lt;/a&gt;, I highly recommend this video, very similar to your project.&lt;/li&gt;&#xA;&lt;li&gt;Use a Raspberry-Pi and a Pi-camera. If you are more of a fan of programming as opposed to circuits you can have a pi-camera attached to device and read out where it is pointing with some relatively simple code.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10785" LastEditorUserId="7083" LastEditDate="2017-04-07T21:18:35.600" LastActivityDate="2017-04-07T21:18:35.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14671" PostTypeId="1" AcceptedAnswerId="14679" CreationDate="2017-04-07T05:53:54.157" Score="0" ViewCount="91" Body="&lt;p&gt;I want to build an electric motorcycle ( more of an electric bike, actually), but i don't want to use a conversion kit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to take out the pedals and put a motor there (either 12v DC or 220v AC).&#xA;My problem is how to transfer the power from the engine to the rear wheel:&#xA;Do I need some sort of gearbox?&#xA;Let's say I would use a 0.7kw, 3000rpm 220v AC motor:&#xA;How do I connect the spinning rod of the motor to a chain or belt (it doesn't necessarily have to be chain driven€ &lt;a href=&quot;https://i.stack.imgur.com/Xd8TJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xd8TJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a bad idea? Can the motor drive the wheel directly? The whole thing (including the rider) would weigh less than 150kg. 0.7kw is enough for the speed I desire. (I am interested in building a dirt bike actually...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excuse my English, I am not a native speaker.&lt;/p&gt;&#xA;" OwnerUserId="10614" LastEditorUserId="7083" LastEditDate="2017-04-07T21:18:40.320" LastActivityDate="2017-04-07T21:18:40.320" Title="Transmission for electric motor" Tags="|motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14672" PostTypeId="2" ParentId="14658" CreationDate="2017-04-07T06:17:26.953" Score="0" Body="&lt;p&gt;I would recommend starting with a wet/dry vacuum (shop-vac). They pull a fairly stong vacuum and have a high flowrate that will be forgiving for misalignment and leaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The down side is these vacuum cleaners is that they typically have a low dutycycle and are not designed to run for 8 hours straight. You may run into reliability problems depending on the quality of vacuum cleaner and how hard you push it. But regardless this is a very low cost way to test your design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have it working you can measure the pressure with a u-tube manometer and the flowrate by placing a trashbag of known volume over the exhaust and timing its fill time with a stopwatch. With those two values you can select an efficient and effective vacuum blower for your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.zoro.com/ebm-papst-motorized-impeller-115v-530-cfm-r2e220-ra44-14/i/G5069285/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Motorized Impeller&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-04-07T06:17:26.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14673" PostTypeId="1" CreationDate="2017-04-07T06:46:29.253" Score="1" ViewCount="56" Body="&lt;p&gt;For the sum of forces about x axis = 0  , how do we know there is a 1.73kN force and also the 0.81kN force ? Where are they coming from ? Can someone explain about it ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly , where are the fixed moments of 100kNm come from ? &#xA;&lt;a href=&quot;https://i.stack.imgur.com/RFXqn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RFXqn.png&quot; alt=&quot;rwr&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p6d2Q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p6d2Q.png&quot; alt=&quot;fsf&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10127" LastActivityDate="2017-04-07T06:46:29.253" Title="Sum of forces in frame" Tags="|structural-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14675" PostTypeId="1" CreationDate="2017-04-07T08:30:59.890" Score="2" ViewCount="69" Body="&lt;p&gt;In NOx sensors there are oxygen pumps that pump the oxygen out of the chamber. My question is, the oxygen is obviously pumped through the Zirconia zeolite, but how does it get released from at other side? Does it go through the electrode? &lt;/p&gt;&#xA;" OwnerUserId="4846" LastActivityDate="2017-04-07T08:30:59.890" Title="How does oxygen pump works?" Tags="|materials|automotive-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14676" PostTypeId="1" CreationDate="2017-04-07T08:49:10.237" Score="1" ViewCount="156" Body="&lt;p&gt;I have an old German caliper. It works and look like any other caliper but on the other side it has a series of numbers ranging from 0 to 61653. What are these? &lt;img src=&quot;https://i.stack.imgur.com/hIAsi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10796" LastActivityDate="2017-11-04T23:39:12.050" Title="What are the numbers on the other side of vernier caliper" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="7" ClosedDate="2017-11-09T22:53:13.210" ContentLicense="CC BY-SA 3.0" />
  <row Id="14677" PostTypeId="2" ParentId="589" CreationDate="2017-04-07T10:42:09.093" Score="1" Body="&lt;p&gt;I had the same question recently and was doing some research a wile ago. Sorry that I don't have all the literature prepared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NIL is for very small structures, the mold is created with a    lithography process. Depth are usually not very large. Can be very fast implemented in R2R. There are two basic    types:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Thermal NIL: Mold is places on substrated, mold is heated at first, then cooled down and separated from substrate with finished imprint.&lt;/li&gt;&#xA;&lt;li&gt;UV NIL: Same as above but instead of a heating source you have a UV source.&#xA;[Source: N. Kooy et al. A review of roll-to-roll NIL Nanoscale Research Letters 2014, 9:320]&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hot embossing is also used for small structures but also for bigger ones, small structures with great depth can be achieved.&#xA;In the process the substrate is preheated and in liquid phase. The mold is pressed on the substrate. Temperature and pressure are controlled. During cooling the mold is removed. Speeds are slower, depending on how big the structure is can be very slow. The mold can also be created by lithography but not necessarily as they often exceed the size possibilities of lithography. The molds are bigger than the imprinted structure depending on the flow characteristics of the material imprinted. Delamination happens during cooling.&#xA;[Source: M. Heckle et al. Hot embossing - The molding technique for plastic microstructures - Microsystem Technologies 4 (1998) 122-124]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differences:&#xA;Hot embossing is: slower, for bigger structures, or small but great depth (holes, lenses, deep channels...), mold is bigger than structure, substrate is preheated, temp. and pressure are always controlled, mold isn´t necessary made with lithography. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sure there are things I missed but I hope it helps someone.&lt;/p&gt;&#xA;" OwnerUserId="10799" LastActivityDate="2017-04-07T10:42:09.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14679" PostTypeId="2" ParentId="14671" CreationDate="2017-04-07T16:40:38.537" Score="2" Body="&lt;p&gt;A transmissions main job is to manage torque. What you have to do is look at the motor spec-sheet and see how much torque it generates.&#xA;Ensure it is sufficient to get you moving, calculate the friction coefficient with your weight and make sure the torque isn't so high you get wheel spin. &#xA;Finally tune it to make it comfortable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But really, this would be more for a &quot;finely tuned machine&quot;. Sounds like you want something fun and easy!!!  In reality you should just get a motor with a metal rotor(part that spins), weld/bolt it onto the sprocket, and pump some current through it. If your bike already has multiple gears I would say leave them on as you can adjust the torque ratio to accommodate a wider range of electric motors.&lt;/p&gt;&#xA;" OwnerUserId="10802" LastActivityDate="2017-04-07T16:40:38.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14681" PostTypeId="2" ParentId="12691" CreationDate="2017-04-07T20:12:27.080" Score="0" Body="&lt;p&gt;&lt;strong&gt;Yes you can... theoretically&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are absolutely correct that most 3D printed materials will not work for this application but I did recently come across a 3D printable PVC that I believe you can melt together. I was looking into this for a project of mine because I wanted to 3D print parts and ultrasonically weld (melt) them together. I was in contact with a rep from &lt;a href=&quot;http://www.chemson.com/en/loesungen/3d-filament/&quot; rel=&quot;nofollow noreferrer&quot;&gt;3D vinyl&lt;/a&gt; and they made it sound like a possibility, unfortunately for me my project changed directions and this rep disappeared but depending on your resources and exact needs I would say it is worth looking into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a very new product and I don't know what stage it is at but it looked promising to me&lt;/p&gt;&#xA;" OwnerUserId="10652" LastActivityDate="2017-04-07T20:12:27.080" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14682" PostTypeId="1" CreationDate="2017-04-07T20:29:17.157" Score="1" ViewCount="1021" Body="&lt;p&gt;I have the following structure, in which I am failing to calculate the two reactions: Cz (apparently vertical) and Dz, in the roller support of the beam. The beam length is 7.5 m, the reactions Az, Cx, and the momentum in C are calculated correctly. The distributed load replaced with concentrated on the first beam. Thank you in advance for advice and help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1m6zB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1m6zB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10780" LastActivityDate="2017-05-08T04:27:56.507" Title="Reaction forces in a structure" Tags="|structures|applied-mechanics|beam|building-physics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14684" PostTypeId="1" AcceptedAnswerId="14690" CreationDate="2017-04-07T21:47:13.767" Score="3" ViewCount="115" Body="&lt;p&gt;I am an occasional hobbyist &quot;DIYer&quot; (not an engineer) who needs help choosing a steel plate for a small home improvement project:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will use it to create a 20&quot; circle, suspended from a bolt at the center. It should be able to hold 350lbs pretty much evenly distributed over its surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is already a generous safety margin built into the 350lbs estimate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is a reasonable choice of steel plate (steel type + thickness) for me to go with?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be clear, this is not a critical application -- ie, no one will be hurt (fall/get squashed) if this fails.&lt;/p&gt;&#xA;" OwnerUserId="10805" LastEditorUserId="10805" LastEditDate="2017-04-09T01:58:36.487" LastActivityDate="2017-05-26T23:13:32.393" Title="choice of steel plate for simple application" Tags="|structural-engineering|materials|structures|steel|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14686" PostTypeId="1" CreationDate="2017-04-08T01:29:26.283" Score="1" ViewCount="96" Body="&lt;h1&gt;The exact question&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Assuming you have outsourced a FEA of a structural design (steel tubes and sheet metal, bolts and welds design):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would you look for in the report to have confidence in the outsourced work findings&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What questions would you ask them about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything that would set off a &lt;em&gt;red flag&lt;/em&gt; that the outsourced work did a poor job or you would simply not trust their findings?&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-04-08T01:29:26.283" Title="External FEA confirmation" Tags="|mechanical-engineering|finite-element-method|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14687" PostTypeId="2" ParentId="14682" CreationDate="2017-04-08T04:19:07.300" Score="1" Body="&lt;p&gt;The reasoning here is very similar to that in &lt;a href=&quot;https://engineering.stackexchange.com/q/14664/1832&quot;&gt;your previous question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by noting that the structure can be separated into three different isostatic structures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ENllr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ENllr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $F$ is obviously the sum of the reactions of the imaginary supports on the beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left-hand side you've already solved. All that's left is the right hand side, and that's just as easy: $R = \frac{1}{2} \cdot 5.1 \sqrt{4.5^2+6^2} = 19.13\text{ kN}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then with that you obviously know that the vertical reaction on the central column is equal to $19.13 + 14.28 = 33.43\text{ kN}$. The horizontal and moment reactions on that column you've already solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check our work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OUxNx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OUxNx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2017-04-13T12:51:48.920" LastActivityDate="2017-04-08T04:19:07.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14688" PostTypeId="1" AcceptedAnswerId="14691" CreationDate="2017-04-08T05:59:33.897" Score="0" ViewCount="118" Body="&lt;p&gt;I saw the following extract from Hibbeler's 'Mechanical Engineering - Statics, 13th Ed';&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GIxbu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GIxbu.png&quot; alt=&quot;relation between distributed load and shear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;general&lt;/em&gt; result $\frac{dV}{dx} = w(x)$ seems to rely on the shear forces at the interface acting in opposite directions. If my force distribution function along the top of the beam is linear with gradient 0, then my $\Delta F$ would act in the centre of $\Delta x$ and the shear force at the right and left interface of $\Delta x$ would be identical vectors, both acting downwards. My vertical force equilibrium would then be described as follows;&#xA;$$&#xA;\Delta F - V - (V + \Delta V)=0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This then breaks the general relationship described by (7-1). What am I misunderstanding?&lt;/p&gt;&#xA;" OwnerUserId="1992" LastActivityDate="2017-04-08T12:35:02.913" Title="Relation Between Distributed Load and Shear" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14689" PostTypeId="1" AcceptedAnswerId="14713" CreationDate="2017-04-08T12:15:53.317" Score="3" ViewCount="258" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LgMH3m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LgMH3m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/IVLO9m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IVLO9m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think, one could split a mercury drop into two drops with the magnetic force? Mercury is a liquid metal, hence if I put a current through it, it will become a electromagnet (a wire with a current induces a magnetic field). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/je8hhm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/je8hhm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I then place two magnets on each side, but sufficiently separated, I thought that the fluid &quot;electromagnet&quot; will split and both sides will travel to the magnets.&lt;/p&gt;&#xA;" OwnerUserId="9038" LastEditorUserId="10684" LastEditDate="2017-04-10T09:00:29.847" LastActivityDate="2017-04-10T15:19:32.927" Title="How to split mercury with a magnetic force?" Tags="|magnets|electromagnetism|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14690" PostTypeId="2" ParentId="14684" CreationDate="2017-04-08T12:22:02.203" Score="1" Body="&lt;p&gt;By 20&quot;x20&quot; circle, I assume you mean 20 inches diameter.  That puts the surface area at 314 in^2.  So 350 lbf evenly distributed is just a bit over 1 lbf/in^2.  That's a pretty small load. &lt;em&gt;Although I've not done any calculations here&lt;/em&gt;, my intuition is that you can probably use 1/4&quot; or 3/8&quot; plate. If your project will be outside or otherwise exposed to moisture, stainless steel would be recommended.  Otherwise I think any generic low carbon steel would be fine.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-04-08T12:22:02.203" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14691" PostTypeId="2" ParentId="14688" CreationDate="2017-04-08T12:35:02.913" Score="1" Body="&lt;p&gt;I think the misunderstanding is because the book carelessly ignores the difference between the shear &lt;em&gt;stress&lt;/em&gt; in the beam, and the shear &lt;em&gt;force&lt;/em&gt; at a point where you cut the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear &lt;em&gt;stress&lt;/em&gt; at the two cutting planes in Fig 1-13(b) has a consistent sign convention. But since the outward normal to the two planes are in opposite directions (one in the positive X direction, the other in the negative X direction), that flips the sign of one of the &lt;em&gt;forces&lt;/em&gt; on the cutting planes as shown in the figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig 1-13(b) show the shear &lt;em&gt;forces&lt;/em&gt; on the two cuts, but &quot;shear diagrams&quot; show the shear &lt;em&gt;stress&lt;/em&gt; - hence your confusion about the signs.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-04-08T12:35:02.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14695" PostTypeId="1" AcceptedAnswerId="14697" CreationDate="2017-04-08T21:51:46.627" Score="1" ViewCount="135" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dHflo.jpg&quot; alt=&quot;Vacuum Tube&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the current flow from the bottom left pin to the bottom right pin, or from the bottom pins to the cap on top?&lt;/p&gt;&#xA;" OwnerUserId="10810" LastActivityDate="2017-04-08T22:26:10.250" Title="How does electricity flow through this kind of vacuum tube?" Tags="|electrical-engineering|vacuum|audio-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14696" PostTypeId="2" ParentId="14695" CreationDate="2017-04-08T22:16:01.057" Score="1" Body="&lt;p&gt;I believe that's a 4 pin triode. The camera angle makes it look like 2 pins. The top connector is usually a connector for the plate/anode. You would need three pins for the heater/cathode and one pin for the grid, similar to &lt;a href=&quot;http://www.hamradioschool.com/g6b10-vacuum-tubes/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Operation is (simplified) heat up the negatively charged cathode to boil off electrons, which are attracted to the positively charged plate. Modulate the stream of electrons by varying the voltage on the grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P5Pby.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P5Pby.jpg&quot; alt=&quot;Diagram of a triode&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9682" LastEditorUserId="9682" LastEditDate="2017-04-08T22:21:32.490" LastActivityDate="2017-04-08T22:21:32.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14697" PostTypeId="2" ParentId="14695" CreationDate="2017-04-08T22:20:40.800" Score="0" Body="&lt;p&gt;This may be short, but might answer the question:-&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HHEiJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HHEiJ.png&quot; alt=&quot;pinout&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an &lt;a href=&quot;http://www.radiotechnika.hu/images/811A.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;811&lt;/a&gt; power triode.  There is no cathode per se as it's also the heater filament.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="2947" LastEditDate="2017-04-08T22:26:10.250" LastActivityDate="2017-04-08T22:26:10.250" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14698" PostTypeId="1" CreationDate="2017-04-09T01:15:21.507" Score="0" ViewCount="316" Body="&lt;p&gt;The problem is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the equations of motion for the simple pendulum in cartesian coordinates (Eq.(3.7)), numarically integrate the trajectory for the initial conditions &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(\[Theta][0], \[Theta]'[0]) = (0.1, 0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Plot the pendulum angle theta[t] and the pendulum length l[t]=sqrt[x[t]^2+y[t]^2] as a function of time.&#xA;The equation (3.7) is &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x''[t] == (-x[t] x'[t]^2 - x[t] y'[t]^2 + x[t] y[t] g)/l[t]^2&#xA;y''[t] == (-y[t] x'[t]^2 - y[t] y'[t]^2 - x[t]^2 g)/l[t]^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At first i try to make EOM(equation of motion) only express for x[t] and y[t]. So use l[t]=sqrt[x[t]^2+y[t]^2]]. And in this problem the initial boundary condition expressed polar coordinate, sine i try to change IBC(Initial Boundary Condition) as&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;\[Theta][t] == ArcTan[y[t]/x[t]]; &#xA;Tan[\[Theta][t]] == Sin[\[Theta][t]]/Cos[\[Theta][t]] == y[t]/x[t]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;y[0]=Sin[\[Theta][0]]&#xA;x[0]=Cos[\[Theta][0]]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Actually I don't know that another two IBC correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is my process by using Mathematica and solutions what i want to plot. Actually I just try to make to plot similarly to solution. So i was assumed the one of IBC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X2WoA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X2WoA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, I want to know method of changing IBC, and i want to plot the graph in solution.&lt;/p&gt;&#xA;" OwnerUserId="10811" LastActivityDate="2017-05-12T04:12:46.940" Title="In simple pendulum problem, how to change Initial boundary conditions polar into cartesian coordinate" Tags="|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14699" PostTypeId="1" CreationDate="2017-04-09T01:43:06.393" Score="1" ViewCount="40" Body="&lt;p&gt;(Simple Mechanics of Materials) Is there a sign convention for where x=0 is and x=L is on a simple horizontal axially-loaded rod? It seems that no matter where the fixed end is, the x=0 end is always on the left end of the rod, so is that a defined convention or can we choose either end to be where x=0? &lt;/p&gt;&#xA;" OwnerUserId="10814" LastActivityDate="2017-06-08T02:12:59.440" Title="X Axis Sign Convention for Axially-Loaded Rods" Tags="|mechanical-engineering|civil-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14700" PostTypeId="2" ParentId="14699" CreationDate="2017-04-09T01:48:46.610" Score="1" Body="&lt;p&gt;You can choose whatever you want. Or, more importantly, whatever is best to help solve your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a natural bias to put $x=0$ on the left side simply because, well, we're used to having the positive number line (from 0 to $L$) go to the right. But you could obviously just mirror the structure in any case to better suit your needs, or simply go with the positive number line to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatever works.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-04-09T01:48:46.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14701" PostTypeId="1" CreationDate="2017-04-09T06:07:28.820" Score="0" ViewCount="100" Body="&lt;p&gt;If a water pipe has a large diameter at the start and a small diameter at the end what part of the pipe would the water flow faster?&lt;/p&gt;&#xA;" OwnerUserId="10816" LastActivityDate="2017-04-11T14:43:19.297" Title="Velocity of Fluid Flow" Tags="|fluid-mechanics|fluid|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14703" PostTypeId="2" ParentId="14701" CreationDate="2017-04-09T09:33:20.860" Score="3" Body="&lt;p&gt;If there are no losses and no storing between the beginning $A$ and the end $B$, you can calculate the velocities based on the principle of continuity.&#xA;$$ Q_A= Q_B $$&#xA;whereas in a homogeneous flow $Q$ is equal to the cross-sectional area times the flow velocity:&#xA;$$ Q=A\cdot v $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the proportions of the two cross-sectional areas you can now determine, which part should have the higher velocity.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2017-04-09T09:33:20.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14708" PostTypeId="1" CreationDate="2017-04-09T17:18:05.857" Score="3" ViewCount="4852" Body="&lt;p&gt;I am required to create a Bill of Material for a steering system I am designing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am not entirely sure how to structure it. For example, what type of order are the parts listed as I have seen some that involve &lt;code&gt;levels&lt;/code&gt; (I have posted below)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just require some knowledge from someone who has created a full list on how to write one up clearly. Also the parts circled are what I like help explaining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rJ6uP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rJ6uP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10468" LastEditorUserId="8055" LastEditDate="2017-09-05T23:37:24.157" LastActivityDate="2017-09-05T23:37:24.157" Title="How to create a clear bill of materials list?" Tags="|materials|cad|product-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14710" PostTypeId="1" CreationDate="2017-04-09T20:58:14.247" Score="2" ViewCount="128" Body="&lt;p&gt;Say I have a line &lt;code&gt;(0,0)&amp;lt;---&amp;gt;(1,0)&lt;/code&gt; and I want to make a rounded &quot;buffer&quot; along it (or any curve, just keeping it simple) in such a way that I now have a closed curve with radius 0.5 so that I end up with a rectangle with semi-circles on each end that is 1 unit in height, and 2 units in width.&lt;/p&gt;&#xA;" OwnerUserId="10828" LastActivityDate="2017-04-09T20:58:14.247" Title="How to create a buffer along a curve in Rhinoceros 5?" Tags="|autocad|" AnswerCount="0" CommentCount="2" ClosedDate="2017-04-10T15:42:53.900" ContentLicense="CC BY-SA 3.0" />
  <row Id="14713" PostTypeId="2" ParentId="14689" CreationDate="2017-04-10T01:22:49.830" Score="2" Body="&lt;p&gt;It looks like it should work. It is a little more geometry specific than you are thinking, so I have modified your design below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will require a reasonable amount of experimentation to perfect the correct permanent magnet strength, required current, and distances to match the surface tension of mercury, etc. Other things that would need to be tested include containing the mercury droplet and bringing the droplets back together when current is reversed (if necessary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TmHOm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TmHOm.png&quot; alt=&quot;Mercury Drop Separating Apparatus&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Diagram Notes:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Green Arrows = Desired electromagnetic applied force&lt;/li&gt;&#xA;&lt;li&gt;Cyan Arrows = Surface tension force that must be overcome (just a generalization; it is much more complex than this)&lt;/li&gt;&#xA;&lt;li&gt;Yellow Arrows = Electric current (note that this is &quot;hole current&quot;; electrons move the opposite direction). Don't get too hung up on direction; you just reverse the polarity of the electrical current if you have it backwards.&lt;/li&gt;&#xA;&lt;li&gt;Purple Arrows = Magnetic field (out of and into the page respectively). Putting an iron back plate on these magnets will increase their strength. Leaving some gap between them will also increase their strength; this gap can be filled with any non-ferrous material.&#xA;&lt;br&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Design Considerations:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You will need to have long electrodes (as opposed to points in your drawing) so both droplets can stay in contact with the electrodes as they are pulled apart. If the current stops, so does the separating force.&lt;/li&gt;&#xA;&lt;li&gt;Surface tension is the force you are working against. Bringing the electrodes close together so the mercury droplet is squished will decrease the current required to separate it. If the droplet is not squished it may try to move the whole droplet one way or the other instead of separating it.&lt;/li&gt;&#xA;&lt;li&gt;You may need to experiment with different metals for the electrodes; some may have better &quot;wetting&quot; or angle of contact with the mercury droplet because of surface tension properties.&lt;/li&gt;&#xA;&lt;li&gt;The magnetic field through the center will not be into or out of the page, so electrical current in this area will not provide a force in the directions of concern. Possibly putting an insulator in this middle zone may improve the design.&#xA;&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;HyperPhysics - Magnetic Force on a Current-Carrying Wire&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-04-10T01:22:49.830" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14714" PostTypeId="2" ParentId="8692" CreationDate="2017-04-10T11:49:02.100" Score="1" Body="&lt;p&gt;In most of the residences, the garage door is constructed in a limited space and due to this counterweights are not used more. As this requires huge space, it is not applicable for those who are living in a small area. If we use counterweights, sometimes there can be jamming which can affect the garage door. Now, most of the local garage door installer doesn’t offer a counterweight system for a residential garage because of the disadvantages and limitations.  &lt;/p&gt;&#xA;" OwnerUserId="10839" LastEditorDisplayName="user16" LastEditDate="2017-04-10T12:23:56.123" LastActivityDate="2017-04-10T12:23:56.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14718" PostTypeId="2" ParentId="14689" CreationDate="2017-04-10T15:19:32.927" Score="0" Body="&lt;p&gt;To add to Eric's Answer, you can see a visual demonstration from &lt;a href=&quot;https://www.youtube.com/watch?v=pB-qAwkgfFQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Physics Girl&lt;/a&gt; - this really does work with another magnetic liquid - Oxygen&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-04-10T15:19:32.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14719" PostTypeId="1" CreationDate="2017-04-10T15:25:42.047" Score="1" ViewCount="56" Body="&lt;p&gt;I have a device that produces heat via Nichrome coils. I want to heat another small piece of metal (in this case stainless steel) with those Nichrome coils. I need to reach 175 degrees C, and so far I haven't found a readily available thermal adhesive that will work in my temperature range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a high temperature thermal adhesive available to consumers that I could use? Any suggestions on another way to effectively transfer the heat from the coils to my small metal component?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help!&lt;/p&gt;&#xA;" OwnerUserId="10845" LastEditorUserId="2819" LastEditDate="2017-04-11T00:13:30.813" LastActivityDate="2017-04-11T00:13:30.813" Title="Heat Transfer From Resistance Element to Metal Component" Tags="|thermodynamics|heat-transfer|thermal-conduction|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14720" PostTypeId="1" AcceptedAnswerId="15310" CreationDate="2017-04-10T17:16:33.633" Score="0" ViewCount="527" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XUnp1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XUnp1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part I'm stuck on is the last part. Basically, the question is to obtain the following equation for the entropy of vaporisation using the Redlich-Kwong equation:&#xA;$$&#xA;\Delta S = R\Bigg[ \ln \frac{V_2 -b}{V_1 - b} \Bigg] + \frac{0.5a}{bT^{1.5}}\ln \Bigg[ \frac{V_2(V_1-b)}{V_1(V_2-b)} \Bigg]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution Attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I should start by using the known fact that at equilibrium, the free energy of both phases must be the same. Using Gibbs free energy:&#xA;$$&#xA;dG = -SdT + VdP \implies -S_1dT + V_1dP = -S_2dT + V_2dP&#xA;$$&#xA;Therefore, I can write an equation for change in entropy due to vaporisation:&#xA;$$&#xA;S_1 - S_2 = (V_1 - V_2)\frac{dP}{dT}&#xA;$$&#xA;However when I simply differentiate the given EoS and multiply by $V_1 - V_2$ I don't get the same result. Clearly, in the answer they have integrated wrt v at some point but I just don't get why or how. Any help will be appreciated, thank you.&lt;/p&gt;&#xA;" OwnerUserId="4530" LastActivityDate="2017-05-17T12:16:47.183" Title="Derivation: Entropy of Vaporisation using Redlich-Kwong EoS" Tags="|mechanical-engineering|thermodynamics|chemical-engineering|energy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14721" PostTypeId="1" CreationDate="2017-04-10T17:22:55.963" Score="0" ViewCount="45" Body="&lt;p&gt;Say we got this circuit: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/4F4O2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4F4O2.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially we had the X resistance, but then I opened the circuit in the terminals of resistance X, and I moved the resistance X to the equivalent circuit (my teacher says this is a shortcut of how doing it). My doubt is about calculating the equivalent resistance: it is obvious that R1 is in series with R3, and then they are in parallel with R2. &#xA;But, since I opened the circuit, the current that are going to those terminals is going to be zero, so there are no nodes, because the current doesn't divide. So, why couldn't I say that R1 and R2 and R3 are in series, or R1 and R2 are in series and these are in parallel with R3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's missing me? Doesn't the current in terminals is zero? &lt;/p&gt;&#xA;" OwnerUserId="7947" LastActivityDate="2017-04-10T20:35:41.847" Title="Is the current on the terminals of an opened circuit is zero?" Tags="|electrical-engineering|circuits|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14722" PostTypeId="1" AcceptedAnswerId="14725" CreationDate="2017-04-10T18:10:11.917" Score="0" ViewCount="148" Body="&lt;p&gt;Why does the Ultimaker 3D Printer has a Heater + Heater transfer plate (aluminium) + Glass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder why a glass plate, and if is possible to remove the glass and print directly in the aluminium plate adjusting the heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ultimaker.com/en/products/ultimaker-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link to the ultimaker&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pictures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hBsTd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hBsTd.jpg&quot; alt=&quot;Ultimaker view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uf2vc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uf2vc.jpg&quot; alt=&quot;PCB Heater view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1262" LastActivityDate="2017-04-10T22:09:26.810" Title="Why does the Ultimaker 3D Printer has a Heater + Heater transfer plate (aluminium) + Glass?" Tags="|aluminum|3d-printing|heating-systems|glass|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14724" PostTypeId="2" ParentId="14721" CreationDate="2017-04-10T20:35:41.847" Score="1" Body="&lt;p&gt;First it would be really helpful to see the question you are trying to solve so I can give you a better explanation, but I think I see your confusion. The terminals are not actually an open circuit as you are assuming, in an example like this you can imagine that the two terminals connected to some unknown circuit and that they are providing a potential across the circuit you have drawn. Basically the undrawn portion of the circuit is the equivalent of a battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You asked about R1, R2, and R3 being in series, this is correct if and only if your terminals really are not connected to any voltage source. But this would result in a loop with no inputs or outputs, in short a very useless circuit not worth examining. S I doubt this is where your question is leading&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is most likely is you are looking at a voltage or current divider question. If that is the case please provide some more context to your circuit and I will show you how to solve it.&lt;/p&gt;&#xA;" OwnerUserId="10652" LastActivityDate="2017-04-10T20:35:41.847" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14725" PostTypeId="2" ParentId="14722" CreationDate="2017-04-10T22:09:26.810" Score="1" Body="&lt;p&gt;The heater PCB is not suitable for printing on for multiple reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A PCB is not as durable as the supplied glass is, nor is it as flat. The surface of a PCB held up by the corners with no other supports would droop in the middle, and no current Ultimaker has a way to compensate for that i.e. automatic mesh bed leveling. A PCB is also not as versatile as a swappable build plate. It's a bit silly to swap out a heater every time you scratch the PCB, eh? (You also probably don't want to scratch and expose the heater leads in the PCB.) With Ultimaker's system, you simply unclip the glass plate and take it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because most PCBs are not flat, as discussed earlier, the Ultimaker uses the aluminum transfer plate that you mentioned. If you simply place the glass on top of the PCB, the two will not make full contact at the center, resulting in uneven heating of the build plate. Assuming the glass and alu plates are both reasonably flat, if you set them on top of each other, there won't be an air gap between them. If you use adhesive or something to adhere the PCB to the alu plate, then you have an evenly heated build surface. And you can swap the glass out without having to rip off all that adhesive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for printing directly onto the aluminum plate, that's a question more appropriate for the &lt;a href=&quot;https://3dprinting.stackexchange.com/questions/3838/why-does-the-ultimaker-3d-printer-has-a-heater-heater-transfer-plate-aluminiu&quot;&gt;3D printing exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10852" LastEditorUserId="-1" LastEditDate="2017-04-13T13:00:58.317" LastActivityDate="2017-04-10T22:09:26.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14727" PostTypeId="2" ParentId="14719" CreationDate="2017-04-11T00:11:58.537" Score="1" Body="&lt;p&gt;You could use &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B000BOAZM8&quot; rel=&quot;nofollow noreferrer&quot;&gt;muffler cement&lt;/a&gt; from your local automotive shop. It would probably be good to keep the wire temp below 500C even though suppliers boast higher. Remember that if you want the metal to get to 175C in a reasonable amount of time, the wire temp will have to be higher than the 175C regardless of how well it is thermally connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most existing products that I have repaired or torn down use a high temp ceramic resistor, fiberglass insulated resistor or fiberglass insulated resistance wire. The wire or resistor is then physically clamped against the surface to be heated. Higher end products will use &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B004Z9XG0I&quot; rel=&quot;nofollow noreferrer&quot;&gt;low cost thermal grease&lt;/a&gt; at the interface. &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B0087X728K&quot; rel=&quot;nofollow noreferrer&quot;&gt;Silver filled thermal grease&lt;/a&gt; is even better, but probably unnecessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://reachelectrical.com/index.php?page=shop.product_details&amp;amp;category_id=209&amp;amp;flypage=flypage.tpl&amp;amp;product_id=866&amp;amp;option=com_virtuemart&amp;amp;Itemid=177&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fiberglass insulated resistance wire 750C&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.reachelectrical.com/index.php?page=shop.product_details&amp;amp;category_id=209&amp;amp;flypage=flypage.tpl&amp;amp;product_id=703&amp;amp;option=com_virtuemart&amp;amp;Itemid=177&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fiberglass insulated resistance wire 350C&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://www.digikey.com/products/en/resistors/chassis-mount-resistors/54&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chassis mount aluminum housed resistors rated 275C&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://www.digikey.com/product-detail/en/te-connectivity-passive-product/SBCHE1510KJ/SBCHE1510KJ-ND/2365138&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wire-wound Ceramic resistor 350C&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-04-11T00:11:58.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14730" PostTypeId="1" CreationDate="2017-04-11T07:13:55.730" Score="3" ViewCount="99" Body="&lt;p&gt;I see that 4 stroke engines use hard gaskets at the cylinder heads, whereas two stroke slow speed engines that are used for the propulsion of ships or for production of electricity on land have soft gaskets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What determines which gasket should be used?&lt;/p&gt;&#xA;" OwnerUserId="5825" LastEditorUserId="90" LastEditDate="2017-04-11T13:52:27.090" LastActivityDate="2017-07-23T18:17:11.793" Title="In internal combustion engines, what determines when soft or hard gaskets are used?" Tags="|mechanical-engineering|design|engines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14731" PostTypeId="1" CreationDate="2017-04-11T07:16:12.460" Score="4" ViewCount="75" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rbXua.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rbXua.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;practice calligraphy&quot; paper has a special coating that amplifies the contrast between light/dark when a wet brush is applied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physics behind this effect has been explained:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/30366/why-do-wet-objects-become-darker&quot;&gt;Why do wet objects become darker?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/30365/why-does-paper-become-translucent-when-smeared-with-oil-but-not-so-much-with-w&quot;&gt;Why does paper become translucent when smeared with oil but not (so much) with water?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From my understanding the refractive index of the liquid and roughened surface should be as close as possible (Eg: Glycerol 1.4729, and PMMA 1.4896) in order to maximize this effect. But this Pilot paper works better than any of my trials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest how it's made, or what materials/chemistry would work best?&lt;/p&gt;&#xA;" OwnerUserId="1864" LastEditorUserId="-1" LastEditDate="2017-04-13T12:40:16.027" LastActivityDate="2018-12-28T22:59:40.793" Title="Which coating best exploits &quot;darker when wet&quot; phenomena" Tags="|mechanical-engineering|chemical-engineering|plastic|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14732" PostTypeId="2" ParentId="6230" CreationDate="2017-04-11T07:24:52.493" Score="2" Body="&lt;p&gt;I live in a big old 1920s built house.  The bedroom is one floor above the router, and one room adjacent.  The floors in my house are wooden, the walls are stone, the cielings are high and the doors are solid wood.  If I try to stream video on Chromecast in my bedroom with the bedroom door closed, it intermittently buffers.  Chromecast even grumbled that the wifi signal was weak and recommended using a different connection. (side note, but my phone picks up full wifi signal).  If I open the bedroom door, and the kitchen door where the router lives, it NEVER buffers.  I have watched whole films and shows without any issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it is just a coincidence? I thought.  But I have repeated this experiment on several different days, at different times of day, and it always yields the same result.  Video can be streaming in HD beautifuly, if I go downstairs and close the kitchen door and then the bedroom door, within 5 seconds the video is buffering.  I have even tried it with the doors closed over a good 30 minute period.  I get small bursts of video followed by minutes of buffering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the numbers shown above, although probably quite accurate and scientifically measured ....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Generally, the amount of impedance added by the door is a negligible amount and would not be sufficient to noticeably improve the quality of the signal.&quot;&#xA;In my case it is not negligible, and makes the difference between watching video, and watching a spinning icon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite often a real world experiment will give different results to the science backing the accepted answer. This happens all the time, isn't engineering great.&lt;/p&gt;&#xA;" OwnerUserId="10860" LastActivityDate="2017-04-11T07:24:52.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14734" PostTypeId="2" ParentId="7446" CreationDate="2017-04-11T11:09:50.927" Score="1" Body="&lt;p&gt;Pulleys have a lot of play. This option is off. You don't have a lot of space, so a single 1:50 gear, while optimal for play, would take a lot of space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High ratios are easily obtainable with worm drive, but they turn the axis 90 degrees, so if the motor must be parallel, you'd need two sets of these (or one set + bevel gears to return to original direction). Still, that would probably create least play while keeping the motor orientation. Although if you could install the motor 90 degrees to the cylinder axis, a single worm drive, with the worm directly on the motor axis, and firmly held with in a bearing would provide your 1:50 ratio with minimum backlash quite easily.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-04-11T11:09:50.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14735" PostTypeId="1" CreationDate="2017-04-11T11:33:30.097" Score="1" ViewCount="2576" Body="&lt;p&gt;I'm designing a frame for supporting some agricultural equipment that will be composed of welded square hollow section members. The top of the frame will be a rectangle of SHS members, with four vertical SHS legs - one in each corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to the corner joints, I'm not sure which is the best way to go: (A) weld the top members together with 45° mitred corners, then weld it on top of the vertical members, or (B) weld the horizontal members in between the vertical members, and cap the ends of the vertical members. See the following image for a comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gibDg.png&quot; alt=&quot;Image of corner joints&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My feeling is that B (on the right) would be stronger, but I'm not sure why. I'd like some information - perhaps what terms I should be searching for - so that I can come to an informed decision based on the requirements of my application.&lt;/p&gt;&#xA;" OwnerUserId="10863" LastEditorUserId="1832" LastEditDate="2017-04-11T12:09:59.517" LastActivityDate="2017-05-11T17:04:18.790" Title="Advice on suitability of corner joints in welded frame" Tags="|mechanical-engineering|welding|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14736" PostTypeId="1" CreationDate="2017-04-11T11:37:30.910" Score="2" ViewCount="37" Body="&lt;p&gt;The question says, &quot;Perform RM scheduling for the following task set: T1=(2, 0.5, 4), T2=(3, 1, 10), T3=(5, 1, 10) and T4=(7, 0.45, 15)&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the standard notation (Period, execution time, relative deadline) is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the priority will be as follows: T1, T2, T3, T4&#xA;&lt;br&gt;&lt;br&gt;&#xA;In this case, job T1 is released consuming time 0 to 0.5 and then job T2 is released consuming time 0.5 to 1.5. At 2, job 1 needs to be released again and hence I cannot release job T3 from 1.5 to 2.5. Can I skip priorities and release job T4 from 1.5 to 1.95? &lt;/p&gt;&#xA;" OwnerUserId="10865" LastEditorUserId="7083" LastEditDate="2018-01-29T00:05:16.570" LastActivityDate="2018-01-29T00:05:16.570" Title="How do I perform RM scheduling?" Tags="|electrical-engineering|embedded-systems|systems-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14738" PostTypeId="2" ParentId="14730" CreationDate="2017-04-11T14:20:46.050" Score="3" Body="&lt;p&gt;Gaskets are selected based upon the mating surface materials, pressure, surface area, surface condition and what they are sealing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, harder gaskets are used in harsher locations. But there are many factors involved with modern to do with making them lighter and stronger. A strong steel gasket may be applied to a cylinder head because it adds strength to an aluminum block. The large engines used for power generation or ships have heavy steel or cast iron blocks that are machined to a high tolerance that only need a thin film or soft gasket to make the seal tight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many conditions that determine what gasket is best, but some of it is just what the designers choose.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:17:11.793" LastActivityDate="2017-07-23T18:17:11.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14739" PostTypeId="2" ParentId="14735" CreationDate="2017-04-11T14:31:10.013" Score="0" Body="&lt;p&gt;All else being equal the isn't a huge amount of difference as long as in case B the open end of the vertical tube is capped (ie has a plate welded on the end rather than just a plastic bung). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice the best method will really depend on the fit you can achieve. Often it is easier to achieve a consistent fit with square cut ends as it is both easier to cut the ends in the first place and easier to adjust the fit as necessary. Bearing in mind also that SHS is not always quite as straight as we might ideally like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact that SHS, especially thicker walled stuff often has quite rounded edges which can make cutting mitres to accurate length quite difficult although this does of course depend on exactly what process you have available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the short answer is that you'll get the biggest advantage from whatever will give you the best and most consistent fit to start with. Which depends on the processes you intend to use for cutting, finishing and welding the tubes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may also be production consideration as to what is the most convenient way to assemble and weld the various parts and in what order. Typically with these sorts of box section frames there is quite a lot of effort involved in getting everything held together and in getting at all of the welds which often involves turning the assembly several times or welding in awkward positions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally weld order matters in terms of minimising distortion which can be hard to control in this sort of space-frame construction without a very substantial jig. It is hard to eliminate this entirely but you can at least mitigate its effects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall I would say that for thin walled small section tube (say 25x25mm x 2mm) I would probably tend to go for mitred joints, favouring capped ends more as the section and wall thickness increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally if the actual manufacturing is going to be sub-contracted it may be worth discussing with them what they think the best approach would be. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-04-11T14:31:10.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14740" PostTypeId="2" ParentId="14701" CreationDate="2017-04-11T14:43:19.297" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If a water pipe has a &lt;strong&gt;large diameter at the start&lt;/strong&gt; and a &lt;strong&gt;small diameter at the end&lt;/strong&gt; what part of the pipe would the water flow faster?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If there are no losses of the fluid between the wide point and the narrow point, then the flow will be faster at the narrow point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because if you take the flow rates: $$\dot{Q}_{start} =A_{start} \times V_{start}  = \dot{Q}_{finish} =A_{finish} \times V_{finish}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final velocity is given then by the ratio of the areas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{finish} = \frac{A_{start}}{A_{finish}} \times V_{start}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If then,  $A_{start} &amp;gt; A_{finish}$ like in your case, then the fraction $\frac{A_{start}}{A_{finish}}&amp;gt;1$ and  $ V_{finish} &amp;gt;V_{start}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the opposite case the finishing velocity is less than the start, naturally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This behavior has all kinds of interesting applications such as in diffusers and nozzles and jet engines, although things are less straightforward once compressibility becomes an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For low Reynolds number and low mach number cases this kind of thing can be explored with tools such as Bernoulli's principle and can be readily seen in the Venturi effect.&lt;/p&gt;&#xA;" OwnerUserId="1104" LastActivityDate="2017-04-11T14:43:19.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14741" PostTypeId="1" CreationDate="2017-04-11T16:02:01.500" Score="3" ViewCount="303" Body="&lt;p&gt;I am trying to get my head around the behaviour and the effectivity of the the filaments movement of a toothbrush inside your mouth. I want to know the mechanical formulas and/or properties that I need to consider in order to understand how the filament moves and bends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ultimate goal is to know how much it will move and bend over the time and the relation of the parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JvClC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JvClC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1262" LastEditorUserId="90" LastEditDate="2017-04-12T02:43:12.230" LastActivityDate="2018-02-06T09:18:01.827" Title="How to calculate filament deflection and force applied to teeth by toothbrush bristles?" Tags="|mechanical-engineering|deflection|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14742" PostTypeId="2" ParentId="14741" CreationDate="2017-04-11T17:11:55.100" Score="1" Body="&lt;p&gt;The best approach to model this, since we will pass well beyond small-deflection beam theory, is to use the &lt;a href=&quot;https://www2.eecs.berkeley.edu/Pubs/TechRpts/2008/EECS-2008-103.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;elastica.&lt;/a&gt;  When you plug in the various sloped constraints of the tooth, and compare to the &quot;soft&quot; moment and force resistance of the individual filament, the elastica will yield the most accurate results.  Having said that, it's a heavily numerical method model, and subsequently has minimal actual useful results.  If you're attempting to define measurable quantities (such as how hard should you push the brush on your mouth?), the use of standard buckling theory will be the best model, but realize that all of your deflections will be invalid after the filament buckles (which will be a very small force).&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-04-11T17:11:55.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14743" PostTypeId="1" CreationDate="2017-04-11T17:31:46.343" Score="0" ViewCount="43" Body="&lt;p&gt;A common practice in naval research is to construct brass models of ships to study the reflectivity of a vessel. This can (supposedly) be used to understand the way it reacts to radar and the way antennas on the ship might be affected by the ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems strange to me though because the wavelength of radar usually has an intimate relationship with the dimensions of the object being sensed. For example, if you have a radar with a 2 foot wavelength, it will not even see a 1 foot model model at all. So, how could a brass model possibly be of use? It would not seem to scale.&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2017-04-11T17:31:46.343" Title="How do brass models work (radar)?" Tags="|electromagnetism|radar|radio|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14746" PostTypeId="1" AcceptedAnswerId="14758" CreationDate="2017-04-11T19:29:16.190" Score="3" ViewCount="2727" Body="&lt;p&gt;I have the following robot arm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zaa5O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zaa5O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how do I obtain the equation of motion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M \ddot\theta + \vec{b} + \vec{g} = \vec{\tau} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am familiar with forces, torque, velocity and so on. I know that this equation follows from force and torque equations. However, I don't know how to establish these force and torque equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque is applied by motors in the joints, so that the arm does not move. No payload is attached to the end-effector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understood, $\vec{\tau}$ contains the two scalar torques for each of the two joints. $\vec{g}$ should be the force vector in negative y-direction, although I am not quite sure. I have no idea what $\vec{b}$ might be.&lt;/p&gt;&#xA;" OwnerUserId="10870" LastEditorUserId="1832" LastEditDate="2017-04-11T20:18:09.067" LastActivityDate="2017-04-13T17:01:25.870" Title="Equation of motion for 2-link arm" Tags="|mechanical-engineering|torque|robotics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14748" PostTypeId="1" AcceptedAnswerId="14964" CreationDate="2017-04-12T00:07:24.663" Score="0" ViewCount="87" Body="&lt;p&gt;If you wanted highest efficientcy and speed from an AC motor what would you do?&#xA;Could the rate of alternation of current be optimizes?&#xA;More windings?&#xA;Materials?&#xA;More lamination?&#xA;Other factors I haven't thought of?&lt;/p&gt;&#xA;" OwnerUserId="10361" LastActivityDate="2018-07-03T14:37:21.243" Title="Increasing speed and efficiency of ac induction motor" Tags="|electrical-engineering|motors|ac|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14750" PostTypeId="2" ParentId="14698" CreationDate="2017-04-12T02:12:52.130" Score="1" Body="&lt;p&gt;You have two second order differential equations.  Thus, you need 4 initial conditions.  You already have half of them:&#xA;$x(t)=l \sin(\theta(t))$ and $\theta(0)=0.1$ has given you $x(0)$, and similarly $y(0)$.  Now you just need to figure out $x'(0)$.  This is actually fairly easy.  Just calculate the derivative of $x$ and evaluate at $t=0$.  You already have the expression for $x(t)$.  Just take derivative using chain rule.  i.e. $\frac{d}{dt}x(t) = \frac{d}{dt} ( l \sin(\theta(t))) = l \frac{d}{dt}\sin(\theta(t)) = l \frac{d}{d\theta}\sin(\theta)\frac{d}{dt} \theta(t) $.  I think you should be able to work this out.  Once you have that, just evaluate at $t=0$ and then do the same for $y$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(as an aside, I would not call these &quot;initial boundary conditions&quot;, I would just call them &quot;initial conditions&quot;. Usually the word &quot;boundary&quot; is reserved for partial differential equations where the variables are spatial.  This is just an ordinary differential equation where the variable is time.  So I'd say &quot;initial condition&quot;)&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-04-12T02:12:52.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14755" PostTypeId="1" AcceptedAnswerId="14771" CreationDate="2017-04-12T14:58:11.200" Score="0" ViewCount="67" Body="&lt;p&gt;I am trying to find the COP of a VCR system. I can reach the COP with using the EES(Engineering Equation Solver), without dealing with the U&lt;em&gt;A values, i.e. just with the energy balances for the cycle. How the COP can differ, if I incorporate the U&lt;/em&gt;A values of the Heat Exchangers into the model? Is there any need for this?&lt;/p&gt;&#xA;" OwnerDisplayName="user5026" LastActivityDate="2017-04-13T13:48:57.670" Title="How the COP differs for a Vapor Compression Chiller without the U*A values of Evaporator/Condenser?" Tags="|thermodynamics|energy-efficiency|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14756" PostTypeId="1" CreationDate="2017-04-12T16:11:32.263" Score="1" ViewCount="95" Body="&lt;p&gt;I've been send to do a project about the construction of a building. Throughout the project I had a doubt I couldn't solve and wish you could help me with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we are working with foundations, we can apply Terzaghi's formulas to get the bearing capacity of the ground. However, in everything I've read they always use a single column. How can I apply the formulas having the weight of the building and a more than one column?&lt;/p&gt;&#xA;" OwnerUserId="10880" LastEditorUserId="1832" LastEditDate="2017-04-12T18:29:25.613" LastActivityDate="2017-11-22T14:35:21.113" Title="Terzaghi's Equation with more than one column" Tags="|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14757" PostTypeId="2" ParentId="659" CreationDate="2017-04-12T16:30:43.740" Score="0" Body="&lt;p&gt;Wire T/C's require a bead or stress relief if separate at the surface to be measured, both methods work well on metal. Peening a place for your bead is best if surface is not delicate. Attach with either epoxy (without metal in it), or attach with aluminum tape covered with rtv for stress relief. If attaching bare wires, spot weld the wires with stress relief loops in the wires (this is so the wires don't break upon thermal expansion) then cover with epoxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spot welded wires are more accurate than the beads.&lt;/p&gt;&#xA;" OwnerUserId="10881" LastEditorUserId="1832" LastEditDate="2017-04-12T18:27:52.603" LastActivityDate="2017-04-12T18:27:52.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14758" PostTypeId="2" ParentId="14746" CreationDate="2017-04-12T16:51:42.910" Score="1" Body="&lt;p&gt;Define $\vec{\tau}_1$ on the first joint and $\vec{\tau}_2$ on the second joint.  The first arm is well defined.  The traditional $\sum \vec{F} = m\vec{a}$ and $\sum \vec{M} = I\vec{\alpha}$, valid only when looking at the center of mass, can be applied.  However, the moment equation is only valid relative to the center of mass.  The general formula for motion when described by another point P on the body (travelling with the body) is $\sum \vec{M}_P = I_p\vec{\alpha} + \vec{p} \times m\vec{a}_P$, where $\vec{p}$ is the distance vector from point P to the center of mass, and $\vec{a}_P$ is the acceleration at P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the pinned point doesn't have an acceleration, it can be established for pinned bodies that, with P being the point of the pin, $\sum \vec{M}_P = I_P\vec{\alpha}$. The second arm doesn't have this luxury, and we must establish an equation of motion for the body.  A major note is that we need an equation to describe $\vec{p} \times m\vec{a}_J$.  While $\vec{a}_J$ the acceleration of the joint, in terms of $\theta_1$, in Cartesian coordinates, is quite complex:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{a}_J = L_1\left[\left(-\ddot{\theta_1}\sin(\theta_1)-\dot{\theta_1}^2\cos(\theta_1)\right)\hat{i} + \left(\ddot{\theta_1}\cos(\theta_1)-\dot{\theta_1}^2\sin(\theta_1)\right)\hat{j}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $\theta_2$ is defined by the relative angle, this can be simplified as we evaluate across the cross product of the two vectors:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{p} \times m\vec{a}_J = \frac{mL_1L_2}{2}\left(\ddot{\theta_1}\cos(\theta_2)-\dot{\theta_1}^2\sin(\theta_2)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the equation of motion for the second body would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{\tau_2} = I_J\ddot{\theta_2}+ \frac{mL_1L_2}{2}\left(\ddot{\theta_1}\cos(\theta_2)-\dot{\theta_1}^2\sin(\theta_2)\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $I_J$ is the inertia of the second body relative to that joint.  Since $\vec{\tau_1(t)} = I_P\ddot{\theta_1}$, which implies that $\dot{\theta_1} = I_P^{-1}\int_0^t\vec{\tau_1(t)}dt$, we can simplify:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\vec{\tau_2(t)} = I_J\ddot{\theta_2}+ \frac{mL_1L_2}{2I_P}\left[\vec{\tau_1(t)}\cos(\theta_2)-\left(\int^t_0\vec{\tau_1(t)}dt\right)^2\frac{\sin(\theta_2)}{I_P}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simplification allows for solving in $\theta_2$ only, without solving for $\theta_1$ simultaneously.  Assuming the input function $\tau_1(t)$ is known, the resulting equation can be simple or vastly complex.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2017-04-13T17:01:25.870" LastActivityDate="2017-04-13T17:01:25.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14760" PostTypeId="1" CreationDate="2017-04-12T20:30:08.570" Score="0" ViewCount="95" Body="&lt;p&gt;I have a simulation based research vs. practical experience type question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a number of academic research papers available which claim 20% to 40% cooling energy savings by increasing the cooling setpoint by 5°F. The savings change by location. Some research papers also provide thumb rules like (approx.) 10% energy savings for 1°C rise in cooling setpoint. The studies I have come across are based on energy simulations.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, some 'on the field' engineers (retro-commissioning/energy engineer) I have spoken to suggest, these savings are inflated or at least not achievable with just change in the cooling setpoint. To achieve these savings you have to make changes in VAV controls (basically reduce the minimum airflow) along with changing the cooling setpoint. But these changes (reduction in VAV min airflow during reheat) can not be applied in actual buildings for various practical reasons. These reasons vary from burning up the electric reheat coil, high diffused air temperature leading to stagnation, not enough diffusion due to low velocity, IAQ issues in conference rooms, supply fan stalling, low air throw length etc.. There are savings but of the range 2-5% for changing just the setpoint by 3F (no VAV controls change as it is impractical for reasons mentioned earlier).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However my sample size for 'on-field experience' is small (a couple of engineers), and there is at least half a dozen of research papers based on simulations I could find which give results as I mentioned earlier, which suggest a good potential in energy savings by increasing the cooling setpoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone actually implemented these retrofits (changing the cooling setpoints) in regular existing office buildings (20,000 sqft to 70,000 sqft)? If yes, which camp do you agree with?  &lt;/p&gt;&#xA;" OwnerUserId="10882" LastActivityDate="2017-04-13T07:33:39.953" Title="Has anyone achieved significant energy savings by increasing the cooling setpoints in retrofit office buildings, in practice?" Tags="|mechanical-engineering|energy-efficiency|hvac|building-physics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14761" PostTypeId="1" CreationDate="2017-04-12T21:59:50.983" Score="2" ViewCount="44" Body="&lt;p&gt;I've got an elastomer, NinjaFlex, that I have been tensile testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I 3D printed the tensile test specimens. I made some specimens with 100% fill and no voids and some specimens with an 80% infill (with intentional voids).&#xA;When I put a side crack in both specimens, the 80% infill specimens failed later than the 100% fill specimens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping anyone could point me in the right direction of some theory which may explain why this is happening? My initial thoughts are the crack growth dissipates into the voids for the 80% whereas the 100% has no stopping mechanism and continues straight through the specimen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards!&lt;/p&gt;&#xA;" OwnerUserId="10825" LastActivityDate="2017-04-13T16:44:22.037" Title="Crack Growth in Elastomers" Tags="|materials|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14762" PostTypeId="2" ParentId="14268" CreationDate="2017-04-12T22:58:26.417" Score="0" Body="&lt;p&gt;Take the americium as one lump ball.  Surround them with a conductor with a decent (1-2 cm) air gap.  Connect the conductor to one side of a capacitor, and the Americium to the the other side of the capacitor.  Then connect to a &lt;a href=&quot;https://electronics.stackexchange.com/questions/103930/automatic-capacitor-charge-discharge-circuit&quot;&gt;circuit&lt;/a&gt;.  Each alpha particle that decays will increase the negative charge on the Americium side, and will increase the positive charge on the conductor's side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the charge has a sufficient voltage, it should discharge through the capacitor to generate the experiment.  Connect to a device that emits a &lt;a href=&quot;https://www.google.com/webhp?sourceid=chrome-instant&amp;amp;ion=1&amp;amp;espv=2&amp;amp;ie=UTF-8#q=creepy%20sound%20youtube&quot; rel=&quot;nofollow noreferrer&quot;&gt;creepy sound&lt;/a&gt;, and leave in the bedroom of your worst enemy.  Of course, if you really wanted measurable nuclear power at home, I'd suggest &lt;a href=&quot;https://www.youtube.com/watch?v=KKdzhPiOqqg&quot; rel=&quot;nofollow noreferrer&quot;&gt;tritium&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="-1" LastEditDate="2017-04-13T12:33:10.067" LastActivityDate="2017-04-12T22:58:26.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14763" PostTypeId="1" AcceptedAnswerId="14770" CreationDate="2017-04-13T00:41:10.803" Score="0" ViewCount="54" Body="&lt;p&gt;speed measured by pitot tubes is that of the wind speed relative to the aircrafts. Now, this itself depends on weather conditions which can affect wind speeds. So, then why employ pitot tubes at all if they can give deceiving readings? Can not satellite imaging be used rather for measuring accurate speeds?&lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2017-04-13T11:27:21.507" Title="Deceptional Pitot Tube" Tags="|aerodynamics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14764" PostTypeId="1" AcceptedAnswerId="14765" CreationDate="2017-04-13T02:36:00.443" Score="1" ViewCount="332" Body="&lt;p&gt;I'm making a double pendulum for an art installation, and I would like to add energy to the system to compensate for friction and air drag losses so it can run more or less indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design so far has the top pendulum connecting to an axle at the top which is parallel to the ground and can rotate freely. So far I've sketched out a design which uses a rotational position sensor and a microcontroller to determine the pendulum shaft's current speed and position. A small, reversible electric motor will connect to the axle with a somewhat &quot;limp&quot; drive belt around the consistency of a rubber band (to lessen the impact of rapid chaotic changes mismatching with power input direction). As the axle rotates unpredictably, the control software will reverse or advance the motor to add a slight force in the current direction of rotation. Once tuned, the hope is that it will not perceptibly interfere with the natural motion of the pendulum while adding enough energy to keep the pendulum swinging with the same total energy as it started at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&#xA;1) Are there mechanically simpler ways of adding energy to such a chaotic system?&#xA;2) Will energy added to the top pendulum distribute to the bottom pendulum in a way that would leave a casual observer unaware, or would the system develop obviously unnatural motion?&#xA;3) How can I roughly estimate (+-50%, to size the drive motor) the average energy input required to overcome friction and drag losses?&lt;/p&gt;&#xA;" OwnerUserId="10885" LastActivityDate="2017-04-14T05:55:27.247" Title="Power assist mechanism for double pendulum?" Tags="|power-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14765" PostTypeId="2" ParentId="14764" CreationDate="2017-04-13T04:51:53.950" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.lightinthebox.com/kinetic-orbital-desk-decoration_p407368.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Perpetual motion pendulum desk toys&lt;/a&gt; have a magnetic coil in the base that is pulsed at the correct pendulum displacement. This applies a small assisting force to the permanent magnet in the pendulum each pass. A &lt;a href=&quot;https://electronics.stackexchange.com/questions/145612/electronic-pendulum&quot;&gt;question on Electrical Engineering SE&lt;/a&gt; explains the driving circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If visually you can not accept a magnet passing near a coil, a DC motor would be an option too. I would probably not interface with a rubber band as that would be a very sloppy system with backlash to address. I would connect it direct drive or with a toothed belt. You could develop a circuit or micro-controller program that would provide positive feedback to the motor. Basically reading the voltage from the motor and adding a percentage to that. The polarity would change clockwise vs counter clockwise as would the assisting current. &lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2017-04-14T00:51:02.850" LastActivityDate="2017-04-14T00:51:02.850" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14767" PostTypeId="2" ParentId="14760" CreationDate="2017-04-13T07:33:39.953" Score="1" Body="&lt;p&gt;Any time you are given rules of thumb you have to understand the variables that were eliminated to make that simple relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt; 30% energy savings at 5 degree F reduction in office temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Variables:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Office temperature&lt;/li&gt;&#xA;&lt;li&gt;Outdoor temperature&lt;/li&gt;&#xA;&lt;li&gt;Building geometry&lt;/li&gt;&#xA;&lt;li&gt;Insulation R-value of walls&lt;/li&gt;&#xA;&lt;li&gt;Insulation R-value of roof&lt;/li&gt;&#xA;&lt;li&gt;Insulation R-value of floor&lt;/li&gt;&#xA;&lt;li&gt;Thermal mass of contents&lt;/li&gt;&#xA;&lt;li&gt;Solar radiation&lt;/li&gt;&#xA;&lt;li&gt;Shading&lt;/li&gt;&#xA;&lt;li&gt;Outside air exchange rate&lt;/li&gt;&#xA;&lt;li&gt;$/kWhr&lt;/li&gt;&#xA;&lt;li&gt;Cooling system efficiency&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The calculations would be quite involved and vary greatly based on the data you input and the assumptions you make. The biggest variable here is that the rule is stating a percentage rather than an amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the claim would work for the following scenario:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;65F office temperature&lt;/li&gt;&#xA;&lt;li&gt;70F outdoor temperature&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you changed the office temp to 70F you would have zero heat transfer and you would save 100% of cooling costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To defeat the claim consider another scenario:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;65F office temperature&lt;/li&gt;&#xA;&lt;li&gt;110F outdoor temperature&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this scenario if you changed the office temp to 70F you would save the same amount of energy as in the previous scenario, but the savings would not necessarily be 30% of the total cooling energy used.&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;It is best practice to not trust rules of thumb that you have not created or tested. In this case a better rule of thumb metric might be W/sqftF. This isn't an industry standard; I made it up just now. It still takes out the massive amount of heat transfer calculations, but gives us a much more meaningful answer than a percentage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say my house is 1500sqft and takes a continuous 3kW to keep it 5F below ambient. So the savings metric would be 3000/(1500*5) = 0.4 W/sqftF&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now I know that for a similar house, dropping the thermostat 1 degree F will roughly save 0.4 watts per square foot. Multiplying by the house square footage gives me the total savings. I could then take that number and divide it by that buildings actually cooling energy use to give a simple (and much more accurate) percentage.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-04-13T07:33:39.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14769" PostTypeId="2" ParentId="14763" CreationDate="2017-04-13T10:25:58.500" Score="0" Body="&lt;p&gt;The speed of the aircraft relative to the surrounding air is actually much more important to a pilot than the ground speed as this is what affects the aerodynamic behaviour of the aircraft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider landing, here the pilot needs to maintain a speed low enough to land safely (determined by ground speed) but high enough that the airflow over the wings and control surfaces provides enough lift to support the weight of the aircraft and control it which is determined by airspeed and this effected by head or tail winds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Airspeed also affects fuel consumption which is an important consideration for any aircraft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course knowing ground speed and position is very useful for navigation but doesn't replace the need to know airspeed as well. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-04-13T10:25:58.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14770" PostTypeId="2" ParentId="14763" CreationDate="2017-04-13T11:27:21.507" Score="3" Body="&lt;p&gt;There are two speeds that the pilot is concerned about. First is airspeed, and the second is ground speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ground speed (speed of the aircraft relative to the ground) can and is checked by GPS. This speed tells the pilot how long the flight will be and if they are staying on track with their fuel calculations for the flight, crucial, but less critical compared to airspeed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Airspeed, (Speed of aircraft relative to the wind that is passing over the wing) is checked by the pitot tube. This speed is a more critical measurement for actual flight. The aircraft dynamics require that the air passing over the wing to be within certain ranges for different types of flight maneuvers (take off and landing have their own critical speeds, cruising flight another, stall speed is another). These can only be measured directly through a pitot tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the challenge (conceptually) is, is in understanding the difference between the two speeds. If your aircraft is flying into the wind at 60 knots (airspeed) and the wind speed is also 60 knots, then your GROUND speed, will be zero. Your aircraft is moving forward relative to the wind at the same speed as the wind is pushing it back. If you've ever seen a bird on a windy day flapping really hard, but not moving anywhere relative to you standing on the ground, this is what is going on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the pilot turns 180 degrees and is now flying with the wind. Once the plane is at a steady flight speed, his airspeed is 60 knots, but his ground speed will be 120 knots! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, do not fly in 60 knot winds if your cruising speed is 60, this is ridiculously extremely insanely dangerous to do. &lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-04-13T11:27:21.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14771" PostTypeId="2" ParentId="14755" CreationDate="2017-04-13T13:48:57.670" Score="0" Body="&lt;p&gt;My opinion is, no, there's not any need to incorporate UA into your calculations. Think about it this way:&#xA;The cooling provided by the evaporator really only comes from boiling refrigerant. The heat absorbed by vapor is almost negligible.&#xA;To absorb more heat, you need to boil &lt;strong&gt;more&lt;/strong&gt; refrigerant rather than getting refrigerant to boil &lt;strong&gt;sooner&lt;/strong&gt;.&#xA;To boil more refrigerant, you need to run your compressor at a higher speed, which means adding more work.  Since: $$COP = \frac{Q_{evaporator}}{W_{compressor}}$$, you won't gain much clarity adding UA to the model.&lt;br&gt;&#xA;Caveat: if refrigerant is not exiting the condenser as pure liquid, efficiency improvements can be made by increasing UA of the condenser.&lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-04-13T13:48:57.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14772" PostTypeId="1" CreationDate="2017-04-13T14:11:51.953" Score="3" ViewCount="267" Body="&lt;p&gt;So I understand the principles of two-point time correlations (covariance) and how you can just normalise that to get the autocorrelation. I also understand the integral time scale which is quite intuitive really and tells us the largest scales in the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now Taylor's microscale is sightly trickier it seems. We look at the slope of how quickly the autocorrelation decays from 0-seperation which seems to be related to a short-time microscale $\lambda$ by an inverse-square law!? For clarity I am of course talking about&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;-\frac{2}{{\lambda}^2} = \frac{d^2\sigma}{d{\tau}^2}&#xA;$$&#xA;where $\tau$ is the time seperation and $\lambda$ is the Taylor's microscale. So can anyone explain what Taylor was thinking and how he got to this? Was it just a hypotheses which seemed to work or did he justify in theoretical ground? (probably a combination)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I am of course a mechanical engineering student and not a mathematician however still interested in these details so long as it could be explained simply.&lt;/p&gt;&#xA;" OwnerUserId="7651" LastActivityDate="2017-04-13T14:11:51.953" Title="Why Taylor's microscale gives us information about the smallest scales of the flow?" Tags="|fluid-mechanics|statistics|turbulence|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14773" PostTypeId="1" AcceptedAnswerId="14775" CreationDate="2017-04-13T14:13:38.350" Score="3" ViewCount="487" Body="&lt;p&gt;I am working on a building that would like to used compressed air for some pneumatic actuators.  They would like to bring the air in from an adjacent building that has more than enough capacity, but the air quality coming over is not sufficient and I need to add a dryer and filters to bring it to the correct quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the usage in the other building, I could feasibly see air that has run through the dryer and filters in the new building being effectively siphoned back to the other building.  Is there a reason that I could/should not install an inline check valve where the piping comes into the new building right before the filters and dryer?&#xA;Check valves are used extensively on the compressors and receivers, but a brief look online didn't turn up anywhere where people were using them as I propose.&lt;/p&gt;&#xA;" OwnerUserId="9770" LastEditorUserId="9770" LastEditDate="2017-04-13T15:03:56.940" LastActivityDate="2017-04-13T15:09:22.433" Title="Check valve in compressed air pipe network?" Tags="|compressed-air|industrial-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14774" PostTypeId="2" ParentId="14773" CreationDate="2017-04-13T14:33:20.657" Score="1" Body="&lt;p&gt;If you know that the check valve won't contribute too much frictional loss and that mixing the air from building 2 (higher quality air building) with air in building 1 (supply building) cannot happen, a check valve will be fine. The only other consideration I would have is to give a way to vent the air in building 2 because if there is some air stuck between the check valve and use point, there is no way to depressurize it safely for any future work. Usually the filter bank can be opened to atmosphere and you can bleed the pressure that way.&lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2017-04-13T14:33:20.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14775" PostTypeId="2" ParentId="14773" CreationDate="2017-04-13T15:09:22.433" Score="1" Body="&lt;p&gt;The problem comes down to pressure.  If your supply building (building 1), which has capacity, is far enough away, the pressure at the end of that line may be significantly lower than the line in new building, (building 2), especially before the filters.  If this is the case, the new line won't have enough force to open the check valve against the old line, and none of the new air will get into the new line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, a careful consideration and monitoring of the pressures should be done before beginning.  Monitor the use of building 2 during slow times, during peak usage, and then model if for the new higher usage.  Ideally use 3-4 points, knowing the flow rates and pressures for each.  Plot them on a curve in excel, using a &lt;a href=&quot;https://www.youtube.com/watch?v=kFasmbrDG5I&quot; rel=&quot;nofollow noreferrer&quot;&gt;quadratic regression,&lt;/a&gt; and figure out where the pressure will be at the new flow rate at the new use rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should note that the compressor's pressure should drop on that curve.  Then measure building 1's usage as well, ideally at similar flow rates and pressures if possible.  Finally, use some basic fluid dynamics losses to find the pressure loss after routing from so far away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $Q_{Overcapacity}$, is the new flow rate, $P(Q_{Overcapacity})$ is the pressure at the new flow rate, and $P(0)$ is the pressure at no flow rate, the system will only provide new air if the following equation holds true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_1(0) &amp;gt; P_2(Q_{Overcapacity})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this equation does hold true, then the actual flow rate of the new line can be determined by when the following equation is true:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_1(Q_{Makeup}) = P_2(Q_{Remainder}) + P_{losses}(Q_{Makeup})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scenario, $Q_{Makeup}$ is the flow rate of the makeup line, $Q_{Remainder}$ is the remainder to meet the complete flow demand.  If this pressure is to low for the equipment to operate, then the entire system needs to be scrapped and a higher pressure source is required anyways. &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-04-13T15:09:22.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14776" PostTypeId="2" ParentId="14761" CreationDate="2017-04-13T15:45:16.680" Score="2" Body="&lt;p&gt;One of the reasons for the difference that you're seeing is that both actually have voids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When 3D printing your extruder prints lines which don't fully bond in the XY plane, they're close but not perfectly solid. Remember, voids are typically measured in mincrometers or nanometers so you won't easily be able to see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With your 80% fill sample each hole you print has a single complete path around it preventing the crack formation in it, and then the lines to will in the rest of the layer span between these printed circles (or hexes in the case with 3D printed parts). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the printer you're using you may be able to change the infill pattern, with the same 80% fill. I would expect that different types of fill will have different loading results. Hex would likely be stronger than diamond, diamond stronger than Cat fill.. etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd love to see some results if you run more tests! This is an area I've been meaning to do some more formal research in for a while. &lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-04-13T15:45:16.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14777" PostTypeId="2" ParentId="14761" CreationDate="2017-04-13T16:44:22.037" Score="0" Body="&lt;p&gt;Printing samples will lead to fiber effects.  In effect, you are dealing with an anisotropic material.  After a quick &lt;a href=&quot;http://www2.lbl.gov/MicroWorlds/Kevlar/KevClue2Act1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;experiment&lt;/a&gt; with paper, you can see what is happening.  However, when you intentionally put voids, the fibers now twist and turn in a different pattern, resulting in potentially increased strength.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-04-13T16:44:22.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14778" PostTypeId="1" AcceptedAnswerId="14780" CreationDate="2017-04-13T16:45:43.217" Score="1" ViewCount="103" Body="&lt;p&gt;A gauge pressure sensor measures a pressure at a port against ambient pressure.&lt;br&gt;&#xA;A unidirectional differential pressure sensor measures a pressure at a &quot;high&quot; port against a &quot;low&quot; port.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a gauge pressure sensor the same thing as a unidirectional differential pressure sensor (simply using ambient pressure as the &quot;low&quot; side)?  Is this just a matter of where your reference point is?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing you could turn a differential sensor into a gauge sensor by using ambient as your &quot;low&quot; side.  Can you turn a gauge sensor into a differential sensor by using a pressure port rather than ambient?&lt;/p&gt;&#xA;" OwnerUserId="10894" LastActivityDate="2022-08-11T07:04:23.387" Title="Pressure sensors - gauge vs unidirectional differential" Tags="|pressure|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14779" PostTypeId="1" CreationDate="2017-04-13T17:25:16.193" Score="1" ViewCount="87" Body="&lt;p&gt;It looks like the digging of this ISIL tunnel introduced periodic waviness in the sidewalls of the tunnel. Similar (related?) waviness can sometimes be observed on dirt roads. What is the origin of that periodic waviness?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C4lBl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C4lBl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;(&lt;a href=&quot;http://www.reuters.com/article/us-mideast-crisis-syria-tunnels-idUSKBN16Y1UG&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="602" LastActivityDate="2017-04-14T17:20:11.493" Title="What casued periodic waviness in the sidewalls of this tunnel?" Tags="|tunnels|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14780" PostTypeId="2" ParentId="14778" CreationDate="2017-04-13T17:28:43.730" Score="1" Body="&lt;p&gt;A differential pressure sensor can be used as a gauge pressure sensor by leaving the low pressure port open to the atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gauge pressure sensor is unlikely to be effective as a differential pressure sensor.  The atmospheric side of the sensor has no need to be air tight, so if you did connect it to a low pressure point, it would probably leak air in or out, causing an error in the reading. &lt;/p&gt;&#xA;" OwnerUserId="8482" LastActivityDate="2017-04-13T17:28:43.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14782" PostTypeId="1" AcceptedAnswerId="50723" CreationDate="2017-04-13T21:45:20.537" Score="1" ViewCount="230" Body="&lt;p&gt;There are many new cars now that automatically turn off their engines when stopped at traffic lights.  They then automatically turn them on when moving off.  These systems have many names along the lines of &lt;em&gt;active&lt;/em&gt; or &lt;em&gt;eco&lt;/em&gt; .  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manufacturers claim that these systems are to save fuel, reduce running costs and save polar bears.  At the same time there are no published figures detailing the economies involved.  Is this just marketing hyperbole?  It seems to me that switching your engine off when it's at a warmed up idle is absolutely the last place that would generate any fuel saving worth printing.  Most cars idle at ~ 0.3 gallons per hour anyway which is nil in the context of a typical trip.  These systems also require a larger alternator, battery, and starter motor. It's a con init?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="2947" LastEditDate="2017-04-14T20:44:57.540" LastActivityDate="2022-04-25T01:30:32.750" Title="Do cars that automatically turn off engines at idle save any significant fuel?" Tags="|automotive-engineering|environmental-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14783" PostTypeId="1" AcceptedAnswerId="14787" CreationDate="2017-04-13T22:30:43.940" Score="1" ViewCount="1048" Body="&lt;p&gt;It is easy to find a very flat surface, &quot;perfect&quot; straight edge, or squares for ensuring things are perpendicular. The machines that make these need to be flat/orthogonal to do so, but how were the first squares and straight edges that were fairly accurate made?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Fairly accurate&quot; is subjective, and I don't know exactly how to define it, but in my mind I am imagining measurements made in a time before the age of heavy machinery and modern manufacturing.&lt;/p&gt;&#xA;" OwnerUserId="10897" LastActivityDate="2017-04-14T13:56:27.267" Title="How were 90 degree angles made?" Tags="|measurements|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14784" PostTypeId="2" ParentId="14782" CreationDate="2017-04-13T22:42:04.937" Score="1" Body="&lt;p&gt;From a study conducted by Oak Ridge National Labs:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Light-Duty Idling Summary:&lt;br&gt;  •&#xA;  Idling beyond 30 seconds is excessive and should be avoided&lt;br&gt;&#xA;  • Extended idling wastes fuel and causes reduction in fuel economy&lt;br&gt; &#xA;  • Excessive idling can create engine wear and carbon soot buildup in the engine and components&lt;br&gt;&#xA;  • Excessive idling can affect the life of engine oil&lt;br&gt;&#xA;  • At start-up: idle for 0-25 seconds and drive gently to warm the engine&#xA;  (as recommended in manual)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The full report is here: &lt;a href=&quot;https://sustainability-ornl.org/documents/ORNL%20Idle%20Reduction%20Guide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://sustainabilityornl.org/documents/ORNL%20Idle%20Reduction%20Guide.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-04-13T22:42:04.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14785" PostTypeId="2" ParentId="14783" CreationDate="2017-04-14T00:57:56.837" Score="3" Body="&lt;p&gt;It is fairly easy to construct a pretty accurate right angle using a divider to bisect a straight line with the limiting factors on accuracy being the straightness of your lines and the consistency of the arc length that your dividers can scribe (the accuracy of the arc length doesn't matter as long as it is consistent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use the angle between a horizontal line (achieved with a spirit level) and a vertical line (from a pendulum bob) to give you a right angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For flat surfaces and straight lines there are various methods a taunt wire being the simplest way to achieve a straight line. For flatness &lt;a href=&quot;https://en.wikipedia.org/wiki/Lapping&quot; rel=&quot;nofollow noreferrer&quot;&gt;lapping&lt;/a&gt; can achieve better flatness than the tools used in the process. Similarly any liquid will form surface parallel to the surface of the earth under gravity which is actually pretty flat for most practical purposes, for example float glass is generally pretty flat and parallel just as a product of the manufacturing process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key thing here is that when you are talking about geometric tolerances eg flatness, straightness, parallelism, perpendicularity ect there are often various tricks you can exploit to get decent accuracy with quite simple tools with a bit of creativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;another useful example is &lt;a href=&quot;https://en.wikipedia.org/wiki/Honing_(metalworking)&quot; rel=&quot;nofollow noreferrer&quot;&gt;honing&lt;/a&gt; which essentially uses free floating abrasive blocks to 'average out' irregularities in a surface and can produce flatness/roundness much better than the tolerances of the tool itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also don't discount the importance of the skill of individual craftsmen, especially in the contest of pre and proto industrial technology, medieval stone masons are quiet a good example of this. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-04-14T13:56:27.267" LastActivityDate="2017-04-14T13:56:27.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14786" PostTypeId="2" ParentId="13366" CreationDate="2017-04-14T01:24:58.913" Score="1" Body="&lt;p&gt;The 'mol gCatalyst^-1' (or &quot;per gmol catalyst&quot;) comes from the phenomenon (in this case) of the rate constant being a function of the catalyst they're using. That is, the rate constant of the given reaction is affected by the mass of catalyst present. In undergraduate-level CME reactor design courses, this typically comes up when designing packed bed reactors (&lt;a href=&quot;http://umich.edu/~elements/asyLearn/bits/pbr/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://umich.edu/~elements/asyLearn/bits/pbr/index.htm&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it's raised to the -1 power here, it suggests that the rate law is some form of a typical second-order rate law multiplied by the catalyst mass raised to the 1 power. See the link for how this would relate to a PBR (PFR with catalyst). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without looking too deeply into the article,  it looks like they're using a bed of catalyst particles, but instead of specifying a bulk density, they specify the particle size and the particle surface area per mass catalyst (which can easily be used to determine the catalyst concentration of the particles by using conc=[mass catalyst/SA]*[SA of sphere/Volume of sphere]), but I'm not sure whether or not they used that directly in their modeling or modified it to get some sort of bulk density value.&lt;/p&gt;&#xA;" OwnerUserId="10898" LastActivityDate="2017-04-14T01:24:58.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14787" PostTypeId="2" ParentId="14783" CreationDate="2017-04-14T01:33:24.493" Score="4" Body="&lt;p&gt;Perpendicularity can be achieved by using any item of constant length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Knotted_cord&quot; rel=&quot;nofollow noreferrer&quot;&gt;Knotted cord&lt;/a&gt; was used by the Greeks, Romans, and Ancient Egyptians. Sticks work well too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to establish perpendicularity (and one I frequently use in fabrication) is to measure diagonals and confirm they are equal. Even though I own a large steel square, this method has much less error on projects that are larger than the square. As joojaa pointed out in the comments, this only works if opposing sides are also equal as shown by the congruency symbols on the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Iub9n.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iub9n.gif&quot; alt=&quot;perpendicularity via equal diagonals&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 3:4:5 triangle also has a 90 degree corner, and can be more convenient that constructing a whole rectangle if it doesn't already exist in the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3THP0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3THP0.png&quot; alt=&quot;3 4 5 triangle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Perpendicular#Construction_of_the_perpendicular&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki &quot;construction of the perpendicular&quot; section&lt;/a&gt; and the links below have the compass method and a few others I have not used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IaJtM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IaJtM.png&quot; alt=&quot;perpendicularity via compass&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.instructables.com/id/handy-tricks-to-find-square/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Instructables, tricks to find square&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.bbc.co.uk/schools/gcsebitesize/maths/geometry/polygonsrev2.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quadrilaterals&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://matheducators.stackexchange.com/questions/9847/given-a-3-4-5-triangle-how-do-you-know-that-it-is-a-right-triangle&quot;&gt;Math Educators SE 3:4:5 Triangle Question&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://www.mathsisfun.com/geometry/triangle-3-4-5.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;MathIsFun 3:4:5 Triangle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2017-04-14T05:51:56.230" LastActivityDate="2017-04-14T05:51:56.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14788" PostTypeId="2" ParentId="14764" CreationDate="2017-04-14T05:55:27.247" Score="0" Body="&lt;p&gt;To show a complete free falling double pendulum, a clutch separating your motor would work sufficiently. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd still recommend a position and angular velocity sensor - when the magnitude of rotation slows down below a specified set point, have the motor begin speeding up the clutch to a speed beyond the angular rotation of the motor. When the angular rotation is in the correct direction, engage the clutch. Then have a random number generator dictate how long the clutch stays engaged, to ensure every second of the demonstration is new. During this energization, this will be noticeable to the eye, but it will make for a perpetual double pendulum that acts truly and completely random.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-04-14T05:55:27.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14790" PostTypeId="2" ParentId="14779" CreationDate="2017-04-14T17:20:11.493" Score="1" Body="&lt;p&gt;The most likely explanation for the waviness pattern is improper drilling and blasting practices leaving toe (the ridges) behind, particularly in the floor of the tunnel. In some case the ridging of the toe continues up the walls. It looks like the bottom row of blast holes (the lifters) weren't low enough &amp;amp; they weren't angled downwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the limited information provided by the picture, the rock through which the tunnel was excavated looks competent. Using manual pick against such rock would be very difficult. The most likely way the tunnel was excavated was to use drill and blast techniques.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the height &amp;amp; width of the tunnel, the most likely method of drilling would have been to use a jack leg (also known as an air leg) drill, similar to the one pictured below. Such drills are pneumatically operated. They can be powered using common air compressors, with holes linking the jack leg drill to the air compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compressed air not only provides power for the drill itself, but also extends the telescopic leg which continually forces the drill and drill steel into the rock face as drilling advances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is relatively easy to do for holes bored into the rock face from waist height upwards. Below waist height the telescopic jack leg can have difficultly getting a good enough grip to provide the force required to keep the drill steel drilling efficiently. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom row of holes, technically referred to as the lifter holes, need to be drilled as close to the bottom of the face (tunnel heading) as possible – preferably in the corner of the face and the floor. For this to happen, the jack leg drill needs to be placed horizontally on the floor of the tunnel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that position, the jack leg won’t have anything to push against. Typically, to drill lifter holes with a jack leg drill the miners/tunnelers would secure a steel ladder to the floor of the tunnel and place the bottom of the jack leg against a rung of the ladder to give the jack leg something to push against so the lifter holes can be drilled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the height of the ridges left in the floor of the tunnel I am speculating the people who excavated the tunnel didn’t use a steel ladder bolted to the floor to get the lifter holes in the correct location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also assuming that ammonium nitrate fuel oil (ANFO) was used as the main explosive for blasting. With the correct equipment it can be easily made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, ANFO is a blasting agent, not an explosive, because it can’t be detonated using a blasting cap, it needs a high energy explosive (like dynamite or another type of “stick” explosive) to set it off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When charging tunneling blast holes, the high energy explosive, with a blasting cap inserted, is placed in the toe of the hole (far end of the hole). ANFO is then pneumatically forced into the rest of the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The highest energy part of the blast is always at the end of the drill holes. If the holes weren’t properly, or fully filled with ANFO, the collar end of the hole (at the face) may receive substantially  less blast energy and this may not be enough to break the rock at the bottom of the face, thus potentially leaving a ridge in the floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spacing of the ridges, in the picture, appears to be relatively uniform, which would confirm a certain length of drill steel was used in drill all the holes for each of the faces during tunnel advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tunneling via drill and blast rarely produces a smooth tunnel throughout. There are always gouges throughout because of inconsistent or poor drilling and blasting practices or the nature of the rock the tunnel is being developed through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing the tunnelers didn’t do was to backfill the floor of tunnel with road base or gravel, some form of smaller rock during excavation. That would have produced a smoother floor in the tunnel.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-04-14T17:20:11.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14792" PostTypeId="1" AcceptedAnswerId="14798" CreationDate="2017-04-14T17:53:39.287" Score="0" ViewCount="47" Body="&lt;p&gt;I have a vessel that has a pressure measuring device. The value it shows as the pressure inside the vessel is 6 inches of water column. Is inches of water column meant to always be used as gauge pressure, or does one have to specify gauge vs atmospheric? &lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2017-04-15T03:16:59.277" Title="I need clarification on how pressure is measured" Tags="|pressure|unit|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14793" PostTypeId="1" AcceptedAnswerId="14794" CreationDate="2017-04-14T17:59:55.733" Score="1" ViewCount="58" Body="&lt;p&gt;I need someone to confirm or deny my understanding in the following situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an empty vessel that is in equilibrium with the atmosphere and I have a pressure gauge on the vessel that displays gauge pressure. The gauge should read 0 atm, correct? If I then close this vessel, pull vacuum on the vessel to completely evacuate it (assume this is possible), then fill it with it 2 atmospheres of nitrogen, the gauge will show 1 atm, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT -&lt;/strong&gt; &#xA;There is some specific volume of nitrogen that at ambient temperature will exert 2 atmospheres of pressure within my vessel. In my scenario, I mean to say I have a supply line that can deliver as much nitrogen as I need to the vessel at ambient temperature to meet this condition.&lt;/p&gt;&#xA;" OwnerUserId="10344" LastEditorUserId="10344" LastEditDate="2017-04-14T19:32:08.817" LastActivityDate="2017-04-14T20:04:13.437" Title="Measuring vessel pressure scenario" Tags="|pressure|measurements|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14794" PostTypeId="2" ParentId="14793" CreationDate="2017-04-14T19:15:40.263" Score="1" Body="&lt;p&gt;There's no such quantity as 2 atmospheres of nitrogen.  If you pulled a vacuum to -1 atm, then connected the tank to a nitrogen tank of the same size at 2 atm,  you'd have 2 tanks of nitrogen at 1 atm pressure (assuming that nitrogen is an ideal gas)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT-&lt;/strong&gt; from the ideal gas law: $$m \sim \frac{PV}{T} $$, assuming volume, $V &amp;gt; 0$, temperature, $T &amp;lt; \infty $.  Once the vacuum is pulled, $P =0\; atm (abs), m = 0$. By adding some nitrogen such that $P = 2\; atm (abs)$, $ m = 0 + \frac{2V}{T}$, then $P = 2\;atm(abs) = 1\;atm(gauge) $.  I think you're ok.  &lt;/p&gt;&#xA;" OwnerUserId="4058" LastEditorUserId="4058" LastEditDate="2017-04-14T20:04:13.437" LastActivityDate="2017-04-14T20:04:13.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14795" PostTypeId="1" AcceptedAnswerId="14808" CreationDate="2017-04-14T19:46:24.843" Score="1" ViewCount="92" Body="&lt;p&gt;For example,&#xA;How does one compare 100 g's for 6 milliseconds to 45 g's for 11 milliseconds. These are two different shock scenarios with different magnitudes and times, so how can I compare them? If I have something that is shock rated at 100 g's for 6 ms does this imply that it will withstand 45 g's for 11 ms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried finding a &quot;shock equivalence&quot; chart or a metric for comparing two shock scenarios but I haven't found anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking it may depend on the mass of the equipment that is being shock rated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v=at$$&#xA;$$F=ma$$&#xA;$$Power=Fv$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An energy metric like power may give me a reasonable estimate to compare, but I am just guessing here. Is there a standard way to compare two scenarios? A chart or equation would be nice..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for the help!&lt;/p&gt;&#xA;" OwnerUserId="6844" LastActivityDate="2017-04-15T20:48:02.977" Title="Comparing 2 shock scenarios with different magnitudes and times" Tags="|mechanical-engineering|shock|specifications|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14796" PostTypeId="2" ParentId="4065" CreationDate="2017-04-15T01:28:06.157" Score="1" Body="&lt;p&gt;The outflow case could be considered assuming a jet confined on one side. That should give a reasonable analytical solution for the ocean side.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inflow case is trickier, but you should be able to find some research on flow fields upstream of an orifice.  The bed confines the inflow, so that's a bit of a unique case.  Try looking for studies on an orifice intake with the bed/floor confining the inflow.  I would think someone's published flow fields for that case.&lt;/p&gt;&#xA;" OwnerUserId="10904" LastActivityDate="2017-04-15T01:28:06.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14797" PostTypeId="1" AcceptedAnswerId="14805" CreationDate="2017-04-15T01:35:15.480" Score="2" ViewCount="256" Body="&lt;p&gt;Under-steering is caused when, at a corner, the front wheels lose traction and as a result the car does not take the corner as quick as it should. So basically what we should do is try and get both the tyres to have a better grip on a corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what an anti-roll bar does is it allows both tyres to remain on ground during  a turn so that we get maximised traction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why don't we increase the stiffness of the anti-roll bar when faced with the problem of under-steering?&lt;/p&gt;&#xA;" OwnerUserId="10905" LastActivityDate="2017-04-15T18:57:36.210" Title="Why do we remove (or reduce stiffness of) anti-roll bars from the front side to prevent under-steering?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14798" PostTypeId="2" ParentId="14792" CreationDate="2017-04-15T03:16:59.277" Score="0" Body="&lt;p&gt;Inches of water is conventionally used as gauge pressure, not absolute.  You could conceivably specify it as inches of water absolute, but that would be an unusual use - sub-atmospheric pressures are more commonly measured in inches or mm of mercury.  &lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2017-04-15T03:16:59.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14799" PostTypeId="2" ParentId="14797" CreationDate="2017-04-15T05:02:06.807" Score="0" Body="&lt;p&gt;Because it reduces the coupling between the front wheels allowing both to follow the irregularities in the road surface better but sacrificing the roll control.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-04-15T05:02:06.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14801" PostTypeId="1" AcceptedAnswerId="14809" CreationDate="2017-03-25T21:05:49.670" Score="0" ViewCount="81" Body="&lt;p&gt;I'm currently building a sonar running off an arduino. My radar includes an ultrasonic sensor that will detect the distance an object is sitting at, and a servo motor that the sensor will sit on top of. I was wondering if it was at all practical to use a PIR sensor with this setup so that the radar will be able to detect people instead of just inanimate objects. I'm worried that while rotating, the sensor might just constantly stay high if it sees any moving person. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no prior PIR sensor experience and that is why I was hoping someone with some experience could let me know if this is worth pursuing or if will be too much of a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The range of this sonar isn't too important. Although I would like it if it could scan several meters, under 1 meter would still be fine.&lt;/p&gt;&#xA;" OwnerUserId="12810" OwnerDisplayName="Sheriff" LastEditorUserId="7083" LastEditDate="2017-06-29T11:51:14.303" LastActivityDate="2017-08-28T14:54:37.857" Title="Can I use a PIR sensor on a home made sonar?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14802" PostTypeId="2" ParentId="9092" CreationDate="2017-04-15T10:29:55.650" Score="-1" Body="&lt;p&gt;We made a car that works on acetylene.. which is obtained from reacting calcium carbide and water.. the alterations made on a Maruti 800 car.. as an engineering degree project&lt;/p&gt;&#xA;" OwnerUserId="10909" LastActivityDate="2017-04-15T10:29:55.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14803" PostTypeId="1" CreationDate="2017-04-15T11:34:19.667" Score="0" ViewCount="53" Body="&lt;p&gt;How can we validate the structural integrity of a small lightweight chassis manufactured from 6063-T6 aluminium that has been welded and then heat treated (back to t6 condition)? A simple FEA model has been created but does not consider any thermal analysis and just gives a basic approximation. We are setting up strain gauges and will perform the test today at around 5pm UK time. I am wondering, is it possible that while performing the strain gauging, could we also measure deflection somehow? Even using a string as a baseline then crudely get a deflection measurement etc. Any suggestions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is our first time strain gauging so using it as a sole method to validate our FEA will maybe not be wise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for all replies!&lt;/p&gt;&#xA;" OwnerUserId="7476" LastActivityDate="2017-04-15T16:24:28.373" Title="Validation of Lightweight Chassis (measuring deflection of structures/strain gauging)" Tags="|structural-analysis|finite-element-method|deflection|product-testing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14804" PostTypeId="2" ParentId="14803" CreationDate="2017-04-15T16:24:28.373" Score="1" Body="&lt;p&gt;Just seen this, you should set some fixed points, from the floor for example and then you can measure the deflections / changes as you apply the loads to specific points - you don't mention how you are applying loads but setting up strain gauges and performing the test leads me to believe you will be loading the chassis.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-04-15T16:24:28.373" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14805" PostTypeId="2" ParentId="14797" CreationDate="2017-04-15T18:49:58.077" Score="2" Body="&lt;p&gt;&quot;Now what an anti-roll bar does is it allows both tyres to remain on ground during a turn so that we get maximised traction.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually just the opposite. The roll bar shifts load from the inside wheel to outside, effectively lifting the inside wheel. With a sufficiently stiff roll bar the inside wheel can be lifted off the ground (Watch most any front wheel drive race event and you will see the inside rear wheels being routinely lifted in the air.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In less extreme case you have increased the contact pressure on the outer wheel and correspondingly lowered it on the inside. You might expect the net available cornering force to be a wash but it turns out the gain on the outside is less than the loss on the inside, so increasing roll stiffness results in a net loss of available traction.  The extreme case is useful to think about as those FWD racers are deliberately tending to oversteer cornering on three wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are competing effects related to real would suspensions geometry and how the body roll effects keeping the tire flat to the ground, so its not quite so simple.  This is why folks typically increase the rear stiffness to reduce understeer, rather than lowering the front stiffness.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2017-04-15T18:57:36.210" LastActivityDate="2017-04-15T18:57:36.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14806" PostTypeId="1" AcceptedAnswerId="14812" CreationDate="2017-04-15T19:36:29.217" Score="2" ViewCount="7366" Body="&lt;p&gt;How do you calculate the torque on the cam, with a force placed on the follower?&lt;/p&gt;&#xA;&lt;p&gt;Two possibilities:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Linear cam (schematic)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NdLTU.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NdLTU.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Radial cam (schematic)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sxwX1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sxwX1.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9038" LastEditorUserId="30442" LastEditDate="2021-01-26T07:57:10.997" LastActivityDate="2021-01-26T07:57:10.997" Title="Calculate torque for a cam" Tags="|torque|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="14808" PostTypeId="2" ParentId="14795" CreationDate="2017-04-15T20:48:02.977" Score="2" Body="&lt;p&gt;For car exidents there is a meassure, which is called Head Injury Criterion. It is a measure to estimate the possibility of an head injury during a car accident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HIC for the time interval $t_1,t_2$ is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$HIC_{t_1,t_2} = (t_2-t_1)^{-1.5}\int_{t_1}^{t_2}a(t)dt.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final HIC is given by the maximum value of the previuos expression, which can be obtained by sliding the interval over the total crash time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But depending on your application other similar measures can be used.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-04-15T20:48:02.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14809" PostTypeId="2" ParentId="14801" CreationDate="2017-04-15T20:51:59.250" Score="0" Body="&lt;p&gt;Perhaps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hobbyist level PIR sensor is actually a digital output device.  It &lt;em&gt;sees&lt;/em&gt; parts of 2D space split into various patterns and detects thermal transitions /movement between them.  If the thermal emitter moves from zone a to zone b within a certain time, a positive detection event is output.  Just like you see with your PIR burglar alarm. If the person doesn't move, there is no transition with a  static sensor.  But your sensor is rotating.  However, this might even happen with no one around as the thermal background radiation will be constantly changing wrt the sensor.  Again these rules govern the siting of PIR burglar alarm detectors.  One thing.  You might have to screen the sensor as they have very wide fields of &lt;em&gt;view&lt;/em&gt;.  That might increase angular accuracy.  They only cost a few quid.  Try it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mentioned that the detector is 2D which seems counter intuitive, but it has no ability to judge distance as far as you're concerned.  All it sees is a map of zones spread out over the x and y axes.  There is no depth perception that you can use.  It's not analogue like the ultrasonics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note.  If you have access to a scientific thermal sensor or camera you could judge distance as well so get 3D scanning as with ultrasonics, but I'm assuming that this is out of scope here.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastEditorUserId="2947" LastEditDate="2017-04-16T14:17:45.523" LastActivityDate="2017-04-16T14:17:45.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14812" PostTypeId="2" ParentId="14806" CreationDate="2017-04-16T13:55:40.340" Score="4" Body="&lt;p&gt;Does this help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k7syl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k7syl.png&quot; alt=&quot;vectors&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the [Angle] that's most prejudicial to opposing F, and consider the lever arm to the camshaft centre.  It looks like my screen grab is the worst case scenario, but with the changing lever arm to point of contact, you should really repeat this for sever angles of cam rotation.  I'm sure there's software for this, but it can be done graphically by hand by plotting a surface development of the cam profile against shaft rotation, say 0 - 180 degrees.  Repeat the calculation every 15 degrees and Fanny's your Aunt.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing.  You can't possibly have a fully algebraic solution unless you have an equation for the profile of the cam as you won't be able to determine [Angle] other than by numerical /graphical means.  I might  be tempted to scan the cam profile and use a graphics package to segment and develop the cam profile.  Or use a rule.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2017-04-16T13:55:40.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14816" PostTypeId="1" AcceptedAnswerId="14820" CreationDate="2017-04-17T04:08:12.613" Score="1" ViewCount="2206" Body="&lt;p&gt;I have a 120mmx120mm brushless fan with an Aluminum frame, which as you can imagine, isn't very thick. I want to thread the base, so I can mount it with a fairly large screw. For that I need to add more Aluminum so the screw has more to  grip on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of creating a casting mold around the exterior side of the Aluminum frame (proof sealed) and then pouring molten Aluminum in it. I've seen videos of people pouring Aluminum and it seems to have a relatively low Heat Capacity compared to other metals (i.e. it will cool down and solidify before it melts the surface of Aluminum frame), as I've seen Anthill Art being made without so much as scorching the earth. Also, as Aluminum is a good Thermal Conductor, the heat would dissipate throughout the frame pretty quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I alleviate the two problems of low Heat Capacity and being a good Thermal Conductor? I know that one always have to pour excess molten metal to the block being added to, but do you guys have better ideas??&lt;/p&gt;&#xA;" OwnerUserId="1915" LastEditorUserId="1915" LastEditDate="2017-04-17T04:16:26.113" LastActivityDate="2017-04-17T15:29:29.643" Title="Pouring aluminum into another block of Aluminum" Tags="|metals|aluminum|molding|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14817" PostTypeId="2" ParentId="6981" CreationDate="2017-04-17T08:36:37.943" Score="-1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Aluminum is a metal that can be used in die casting to make parts easily. Many die casting techniques like permanent mold casting and sand casting are used for casting aluminum. Permanent mold casting is highly repeatable and helps to decrease the spin balancing processes and secondary machining operations. On the other hand, sand casting uses temporary molds that are made from wood or metal, making the tooling investment very low. However, per part prices are higher in the case of sand casting than permanent mold castings.&lt;/p&gt;&#xA;&lt;p&gt;In the case of permanent mold casting techniques, the cycle times are short and this will decrease the per part price. However, tooling costs can be very high in this method. When compared to sand castings, permanent mold castings cool faster, offering a finer and uniform microstructure to the parts. This will help to boost the mechanical properties of the product by about twenty percent.&lt;/p&gt;&#xA;&lt;p&gt;Gravity fed permanent metal molds is used these days to manufacture near net shaped parts from different alloys of aluminum. However, it is the duty of the die cast designer to ensure that it is profitable and possible to use permanent molds to manufacture the part. The die cast company should know the limits of the casting method to take full advantage of this die casting technique.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Source : &lt;a href=&quot;http://pacdiecast.com/aluminum-die-casting/permanent-mold-casting-for-casting-aluminum-alloys/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://pacdiecast.com/aluminum-die-casting/permanent-mold-casting-for-casting-aluminum-alloys/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10917" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-04-17T11:36:02.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14819" PostTypeId="1" CreationDate="2017-04-17T11:10:40.840" Score="1" ViewCount="265" Body="&lt;p&gt;I am designing a pipe strut connected to a UB waler using a 100mm diameter pin through a series of 8 interlocking plates. Each plate is about 40mm thick. To be able to weld the plates to the waler and strut end, I need to leave sufficient space between consecutive plates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the plates are 300mm wide and long, what should be the minimum gap between the plates to allow butt welding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should there be a maximum gap between each pair of plates if the pin can resist the bending moment induced by the bigger lever arm?&lt;/p&gt;&#xA;" OwnerUserId="10920" LastActivityDate="2017-04-17T11:10:40.840" Title="Designing a pin connection" Tags="|structural-engineering|design|steel|connections|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14820" PostTypeId="2" ParentId="14816" CreationDate="2017-04-17T12:57:45.130" Score="3" Body="&lt;p&gt;This is unlikely to work very well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest issue is that you won't get good adhesion between the two surfaces. aluminium has a particularly resilient and inert surface oxide layer so molten aluminium won't just stick to the surface of solid aluminium. Welding it requires an inert atmosphere and reverse polarity current or alternating current to clean the surface. Pre-cleaning won't work either as the oxide layer forms very quickly. Equally the aluminium will shrink substantially as it cools so anything you cast onto the surface will in all likelihood just pop off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, casting a new part from scratch using a known castable alloy is far more likely to work and not likely to be significantly more effort. Note also that cast aluminium is prone to porosity and needs to be de-gassed to produce reliable mechanical parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly without knowing the alloy you are working with and its initial condition there is no way of knowing how the heat will affect its mechanical properties. Aluminium anneals as a fairly low temperature and if the existing part reaches that temperature it will be drastically weakened and there is a good chance that it will distort. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better solution would be to fabricate the extra part you need and either get it TIG welded on (even then there is some risk of distortion and not all aluminium alloys are readily weldable) or to bond it on with adhesive, which is likely to be the only practical solution which is reasonably likely to be problem free. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-04-17T15:29:29.643" LastActivityDate="2017-04-17T15:29:29.643" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14821" PostTypeId="2" ParentId="2039" CreationDate="2017-04-17T13:11:15.793" Score="1" Body="&lt;p&gt;As Herold has stated Nitrogen has a rather high ionization potential.  However the voltage needed to ionize air depends on alot of other factors:&#xA;density &#xA;altitude&#xA;humidity&#xA;mean temperature&#xA;presence of dust&#xA;etc&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-04-17T13:11:15.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14822" PostTypeId="1" AcceptedAnswerId="14843" CreationDate="2017-04-17T13:34:21.253" Score="2" ViewCount="2989" Body="&lt;p&gt;The problem is that when I try to solve the following equations of J2 perturbed gravity:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function xdot = Gravity( ~ , x )&#xA;&#xA;Re  = 6378.137;&#xA;J2  = 1.08262668e-3;&#xA;mu  = 398600.4418;&#xA;&#xA;r   = norm(x(1:3));&#xA;&#xA;% xdot = [x(4:6)&#xA;%         -mu/r^3*x(1:3)];&#xA;&#xA;xdot    = [ x(4);&#xA;            x(5);&#xA;            x(6);&#xA;            -mu*x(1)/r^3*(1+1.5*J2*(Re/r)^2*(1-5*(x(3)/r)^2));&#xA;            -mu*x(2)/r^3*(1+1.5*J2*(Re/r)^2*(1-5*(x(3)/r)^2));&#xA;            -mu*x(3)/r^3*(1+1.5*J2*(Re/r)^2*(3-5*(x(3)/r)^2))];&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;using any of the solvers (ode45, ode113, etc.), the propagated position and velocity has a considerable error compared to a trusted commercial software (STK), and the error keeps growing. I've tried various settings of tolerances, initial steps, and max steps, but the problem persists!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be surprising, but when I wrote a fixed step 4th order Runge-Kutta code as below, it didn't get much better.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function [t,x] = RK4( fn,tspan,x0 )&#xA;% An implementation of fixed step 4th order Runge-Kutta method&#xA;&#xA;f = str2func(fn);&#xA;t = tspan;&#xA;h = tspan(2)-tspan(1);&#xA;x(:,1) = x0;&#xA;&#xA;for i = 1:size(t,2)-1&#xA;    k1 = f(t(i),x(:,i));&#xA;    k2 = f(t(i)+h/2,x(:,i)+k1*h/2);&#xA;    k3 = f(t(i)+h/2,x(:,i)+k2*h/2);&#xA;    k4 = f(t(i)+h,x(:,i)+k3*h);&#xA;    x(:,i+1) = x(:,i) + (k1 + 2*k2 + 2*k3 + k4)*h/6;&#xA;end&#xA;&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;any idea of what the problem might be? I also have to mention that to a great surprise, I did get a result using RK4, but the same code with the same conditions misbehaved later!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sample result:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Cnwsx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cnwsx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10922" LastActivityDate="2022-03-10T08:43:59.393" Title="J2 orbit propagation problem in MATLAB (solver instability)" Tags="|mechanical-engineering|aerospace-engineering|dynamics|numerical-methods|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14823" PostTypeId="1" AcceptedAnswerId="14830" CreationDate="2017-04-17T13:51:24.903" Score="0" ViewCount="77" Body="&lt;p&gt;I am trying to understand how to make generation of electrical power more efficient by using Faraday's law to directly convert the kinetic energy of the ionized or easily ionizable exhaust gases of combustion (oxidation of hydrocarbons or to make things simpler just hydrogen) into electrical energy.  In this system, the only real moving parts would be the motion of the ionized or plasma exhaust gas medium through the surrounding faraday coils. We may need a fan to coax the pre-combustion gases in a linear flow. The usable current would be induced in the Faraday coils from the accelerating or decelerating ionized gas from the exhaust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main question is how to completely ionize the exhaust gases at the point of combustion.  My understanding is that these are already ionized at the instant of combustion but rapidly de-ionize when the oxidation reduction process is complete.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2020-10-07T06:06:10.573" Title="Magnetohydrodynamics, ionizing combustion exhaust then converting its kinetic energy to electricity" Tags="|electrical-engineering|fluid-mechanics|thermodynamics|combustion|experimental-physics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14824" PostTypeId="1" CreationDate="2017-04-17T13:59:01.507" Score="0" ViewCount="65" Body="&lt;p&gt;In YouTube I have seen various videos where someone cut with a sword into hard objects and has cut that into pieces. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=EVCWGwvctt4&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=EVCWGwvctt4&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the mass, the velocity and the material of the cutting blade, how I can compute whether some material is cut by it completely or not? Which basic equations should I use? Are there theories that model mathematically such things?&lt;/p&gt;&#xA;" OwnerUserId="8941" LastActivityDate="2017-04-19T17:14:04.163" Title="How I can compute cutting processes with continuum mechanics?" Tags="|applied-mechanics|modeling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14825" PostTypeId="2" ParentId="14816" CreationDate="2017-04-17T15:21:09.337" Score="0" Body="&lt;p&gt;Your ultimate goal is not clear - you've asked how to do a task rather than how to achieve a result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: if your fan is not terribly massive, then what matters is the surface area of contact between the screw head and the frame. This is more easily achieved with a wide flat washer or two (under the screw head and perhaps under the matching nut on the far side). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is to follow ChrisJ's answer and bond a piece on, using an epoxy rated for metal-to-metal. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-04-17T15:21:09.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14826" PostTypeId="1" AcceptedAnswerId="14874" CreationDate="2017-04-17T15:32:15.047" Score="1" ViewCount="6051" Body="&lt;p&gt;I'm trying to find online the derivation for the 2 parameter margules equation but cant find it anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;basically need to prove&#xA;$$&#xA;\frac{d}{dn1} (x_{1}x_{2}(A_{21}x_{1} + A_{12}x_{2})) = x_2^2(A_{12} + 2(A_{21}-A_{12})x_{1})&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but everytime I try just end up with a page full of rubbish. I've searched online for the derivation but cant find it anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: the A values are functions of temperature only and can be considered constants&lt;/p&gt;&#xA;" OwnerUserId="10923" LastEditorUserId="10923" LastEditDate="2017-04-17T15:40:51.263" LastActivityDate="2017-04-20T13:52:49.093" Title="2 Parameter Margules derivation" Tags="|chemical-engineering|vapor-pressure|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14827" PostTypeId="1" CreationDate="2017-04-17T15:34:04.043" Score="0" ViewCount="331" Body="&lt;p&gt;I am working on a building project that requires a high ventilation rate for fire safety: 6 air changes per hour (ACH).  The winter design temperature is -14, so moving that much air through the system will obviously impact the total building heat load significantly with a 200,000 ft&lt;sup&gt;3&lt;/sup&gt;area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I run the analysis with the Revit loads tool before setting the ACH, it returns a heat load of 103,000 Btu/hr.  If I set the zone ventilation to 6 ACH, it returns a heat load of 1,662,000 Btu/hr.&#xA;From a simple heat transfer standpoint, it would make sense that the heating load would skyrocket when you are pumping that much cold air into the building.  However, during the winter, part of that makeup air will be infiltrating the building &lt;em&gt;through&lt;/em&gt; a heater, so the delta T will not be the difference between the -14 degrees F, and the 50 degree F setpoint, but, rather the heater output temperature and the setpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know of a way to account for this within Revit?&lt;/p&gt;&#xA;" OwnerUserId="9770" LastEditorUserId="9770" LastEditDate="2020-01-04T16:59:40.760" LastActivityDate="2020-01-04T16:59:40.760" Title="Heat load calculations with heated air changes" Tags="|heat-transfer|software|bim|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14828" PostTypeId="2" ParentId="14827" CreationDate="2017-04-17T21:34:30.627" Score="1" Body="&lt;p&gt;I know it is not what you asked, but you could take the intake air below ground, 2m down for example, and benefit from the ground temperature compared to the ambient air temperature - called a &quot;puit canadien&quot; and i did this for a 4 bedroom house in Switzerland - needed 30m of pipe 2m down plus a heat exchanger as @Brian Drummond . Helps in summer to reduce air T and winter to increase air T. Several references can be found online - just make sure that you have a constant slope and get rid of any moisture...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-04-17T21:34:30.627" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14830" PostTypeId="2" ParentId="14823" CreationDate="2017-04-17T22:48:11.730" Score="1" Body="&lt;p&gt;Hydrocarbon combustion does produce ions, due to chemiionization mechanisms. This does make flames more highly ionized than other gases at the same temperature. That having been said, the concentration of ions is still very small (rarely more than $10^{-6}$ by volume). You can augment this somewhat with metal salts but even that has limits. Approaching complete ionization is unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not saying you can't generate some amount of electricity this way, it just may not be very efficient.&lt;/p&gt;&#xA;" OwnerUserId="68" LastActivityDate="2017-04-17T22:48:11.730" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14831" PostTypeId="1" AcceptedAnswerId="14832" CreationDate="2017-04-18T00:42:13.250" Score="0" ViewCount="2307" Body="&lt;p&gt;I know the reasons why old cars backfire, but I've noticed that new fuel injected supercars do this as well in the video game Driveclub. I know a video game doesn't seem like a reliable source, but I researched and found that the game creators simulate each car and the in-game driving experience very closely to their real life counterparts, even claiming they recorded the engine sounds of every car in real life to make sure that the in-game car audio reflects the sounds of the actual engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That given, I noticed that lots of the supercars/hypercars backfire a lot. Sometimes this happens during acceleration, but never during pure linear acceleration, mostly when coming out of a turn. This is usually indicated by somewhat subtle flames coming out of the exhaust and sometimes a soft popping noise. More prominently however, I've noticed it during deceleration, making two loud pops in quick succession accompanied by two quick bursts of intense flames shooting out the exhaust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've researched this online myself but I haven't found a complete explanation. I have also seen some explanations as why this happens in drag racing cars, but I wouldn't suspect the same explanation would hold for street legal cars since they are constructed and purposed differently. So, why do supercars backfire during acceleration, but not during linear acceleration? And why is the backfiring so much different during deceleration (specifically the two pops, louder noise, and bigger fire)? &lt;/p&gt;&#xA;" OwnerUserId="8974" LastActivityDate="2020-05-31T06:27:01.107" Title="Why do fuel injected supercars still backfire?" Tags="|automotive-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14832" PostTypeId="2" ParentId="14831" CreationDate="2017-04-18T01:27:37.920" Score="2" Body="&lt;p&gt;While fuel injected cars have the ability to cut off fuel when the throttle is closed they don't have to and this could happen either as a deliberate decision in designing the engine map or as a side effect of something else and while digital ECUs can respond very quickly they still aren't instantaneous an in a high revving engine an ignition cycle happens very quickly indeed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing which can cause this is if the engine mapping deliberately introduces fuel into the cylinders even when the throttle is closed. One reason to do this is to maintain a high velocity exhaust gas flow to maintain turbo speed or because the exhaust gasses are being used for some aerodynamic function as in a &lt;a href=&quot;http://www.formula1-dictionary.net/diffuser_blown.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;blown diffuser&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also high pressure turbos can make a variety of noises either from the dump valve or flutter when the turbine stalls under high boost or closed throttle. turbocharged engines may also prefer to run rich at high boost pressures to mitigate knocking.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is the rev limiter, these can limit revs either by reducing fuel input or cutting the ignition if the latter is the case then this can put unburnt fuel into the exhaust. This is less common in production cars as it is bad for emissions and can damage catalytic converters but may have advantages in racing where hitting the limiter is more normal. As well as a 'red line' limit variable rev limits may also be used as part of a traction control system especially form standing starts or pulling away from a corner which tend to be traction limited situations. Some systems may even cut the ignition to some but not all cylinders to control torque in traction limited situations.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-04-18T11:31:54.600" LastActivityDate="2017-04-18T11:31:54.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14834" PostTypeId="2" ParentId="14822" CreationDate="2017-04-18T07:23:31.220" Score="1" Body="&lt;p&gt;As far as I can see, you are trying to integrate a nonlinear ODE of a conservative system. It is very likely that your system exhibits chaotic behavior. RK4 is a fourth order numerical scheme. These numerical integrators tend to change the total energy of the system, which contradicts the conservativeness of the system but explains the differences between your solution and the solution of the program you are talking about (btw I am not sure if the program does integrate the equations correctly, but let's assume it does. You could look up the user manual or contact the support in order to find out if the software is suited for your problem). In order to get better numerical integration, you have different possibilities.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Use higher order approximation scheme (e.g. 8th order numerical scheme): This system will still show a drift in the total energy, but the drift will be less rapidly than for RK4. So your simulation will follow the &quot;real&quot; trajectory for a longer period. But as you can imagine at some point your trajectory will not follow the &quot;real&quot; trajectory anymore.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use custom integrators like &lt;a href=&quot;http://www.unige.ch/~hairer/poly_geoint/week2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;symplectic integrators&lt;/a&gt; (for Hamiltonian systems: keeps Hamiltonian constant) or &lt;a href=&quot;http://www.math.ucsd.edu/~mleok/pdf/LeLeMc2005_lgvi.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;variational integrators&lt;/a&gt; (for Lagrangian systems: keeps the Lagrangian constant). The hard part about these integrators is that you need to create them problem dependent, but they are usually much better than just increasing the order of the numerical scheme.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-04-18T07:30:04.283" LastActivityDate="2017-04-18T07:30:04.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14836" PostTypeId="1" CreationDate="2017-04-18T10:28:57.840" Score="0" ViewCount="293" Body="&lt;p&gt;Am I correct in saying that the residual stress of a component, that has undergone plastic deformation and then unloaded, is determined fully by the elastic component of the residual strain? And if so, is this because the plastic component is permanently deformed and hence is in a state of equilibrium unlike the elastic strain which is held out if equilibrium by the plastic deformation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense in my head but the notes I am using don't strictly state this but instead say: &quot;the residual stress is related to the residual elastic strain&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="7476" LastActivityDate="2017-06-17T14:13:20.790" Title="Residual Stress Theory" Tags="|mechanical-engineering|materials|stresses|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14837" PostTypeId="2" ParentId="11699" CreationDate="2017-04-18T11:32:29.163" Score="1" Body="&lt;p&gt;There is no way to connect rotations and translations. That is the reason why we need the kinematic differential equation to relate velocity and angular velocity with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only apply infinitesimal rotations, then this can be represented by infinitesimal translations. It even turns out that for infinitesimal rotations do commute, which is not true for general rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotation matrix $\mathbf{R}$ for an angle $\beta$ around the direction $\mathbf{u}$ is given by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;formula&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{R}(\mathbf{u},\beta)=\cos \beta \mathbf{I}+\sin \beta \tilde{\mathbf{u}}+(1-\cos \beta)\mathbf{u}\mathbf{u}^T.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Skew-symmetric_matrix#Cross_product&quot; rel=&quot;nofollow noreferrer&quot;&gt;skew-symmetric matrix&lt;/a&gt; is represented by $\tilde{\mathbf{u}}$, this is used to express the cross product as a matrix product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assume that we apply an infinitesimal rotation with the angle $\beta \rightarrow d \beta.$ Using small angle approximations we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{R}(\mathbf{u},d \beta)=\mathbf{I}+d\beta \tilde{\mathbf{u}}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this expression, it is easy to see that two rotations indeed do commute (after neglecting higher order terms), hence infinitesimal rotations do behave like vectors, which are commutative, and not like matrices, which in general do not commute.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-04-18T11:38:12.293" LastActivityDate="2017-04-18T11:38:12.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14838" PostTypeId="2" ParentId="14836" CreationDate="2017-04-18T11:55:11.793" Score="1" Body="&lt;p&gt;I think this is really a question about the terminology. The concept of &quot;the elastic component of the residual strain&quot; may be confusing, until you take it apart to see what it really means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;total&lt;/em&gt; strain in the material is well defined - it is what a strain gauge would measure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the structure has undergone plastic deformation, the total strain may be non-zero even when there are no applied loads - in other words, the structure has been permanently deformed from its initial shape. That is what &lt;em&gt;residual&lt;/em&gt; strain means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For continuum mechanics, it is useful to split up the total strain into different components, such as plastic strain, creep strain, thermal strain, elastic strain, etc. These are &lt;em&gt;theoretical&lt;/em&gt; concepts, and &lt;em&gt;by definition&lt;/em&gt; the elastic strain is the only component which is related to the elastic stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example that doesn't involve any nonlinear or time-dependent behaviour, consider a bar of material with both ends rigidly fixed, where the bar is then heated. The total strain is zero, because the length of the bar remains constant. However there is a non-zero &lt;em&gt;thermal&lt;/em&gt; strain (equal to what the thermal expansion would be if the bar was not fixed) and an equal-and-opposite &lt;em&gt;elastic&lt;/em&gt; strain which makes the total strain zero. The stress in the bar depends only on the elastic strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to summarize the answer, &quot;the elastic component of the residual strain&quot; and &quot;the residual elastic strain&quot; are two names for exactly the same thing.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-04-18T13:46:10.520" LastActivityDate="2017-04-18T13:46:10.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14840" PostTypeId="1" CreationDate="2017-04-18T13:01:13.113" Score="5" ViewCount="426" Body="&lt;p&gt;Given the following system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x} = \begin{bmatrix}1 &amp;amp; 1 &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 0 \\ 0 &amp;amp; 1 &amp;amp; 1 \end{bmatrix}x + \begin{bmatrix}0 &amp;amp; 1 \\ 1 &amp;amp; 0 \\ 0 &amp;amp; 1 \end{bmatrix} u$$&#xA;$$y = \begin{bmatrix}1 &amp;amp; 1 &amp;amp; 1 \end{bmatrix} x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find the minimal realization of the system. I know that the minimal realization means controllable and observable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system given is not controllable and not observable.&lt;/p&gt;&#xA;" OwnerUserId="8089" LastEditorUserId="9" LastEditDate="2020-05-04T08:03:41.530" LastActivityDate="2020-05-04T08:03:41.530" Title="Minimal realization of a MISO system" Tags="|control-engineering|control-theory|linear-control|linear-systems|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14841" PostTypeId="1" CreationDate="2017-04-18T14:49:16.077" Score="0" ViewCount="71" Body="&lt;p&gt;I'm very clumsy person so I like to buy tools that are just short of bomb proof. I want a pair of digital calipers. So I was wondering is there an equivalent of ip rating for being drop proof to say a 1 or 2 meters.&lt;/p&gt;&#xA;" OwnerUserId="2054" LastEditorUserId="2819" LastEditDate="2017-04-18T18:23:14.363" LastActivityDate="2017-04-18T18:23:14.363" Title="What is the equivalent of ip rating for mechanical protection?" Tags="|measurements|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14842" PostTypeId="2" ParentId="12249" CreationDate="2017-04-18T15:28:03.977" Score="2" Body="&lt;p&gt;We need to go directly from chemical energy to electrical energy.  Electrical energy can be attained by accelerating electrically charged particles through a coil or magnetic field, which causes electromotive force (voltage) and electrical current through a load to do work to neutralize the voltage. &#xA;The chemical reaction of fuel with oxygen is known as oxidation reduction and is exothermic.  That means that the end products of the reaction are more stable in terms of the resulting molecule's outermost electron sub-orbitals and pairing of electron spin defined by Pauli's exclusion principle and quantum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the end product is more stable, energy is released (exothermic).  This energy is manifested as photon emission as the electrons settle into the lower energy level orbitals, and as increase of the average kinetic energy of the molecules (heat) in this exhaust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Boyle's law, the increased temperature of the exhaust gas causes expansion of said gas to fill a greater volume (less dense).  This thermal expansion of the exhaust gas is what drives the turbines in a gas turbine electric power generation setup, jet engine, rocket engine, and what pushes the pistons inside the cylinders in an internal combustion engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the trick to converting directly to electrical energy is to ionize the molecules of the exhaust gas at the point of combustion and keep these molecules ionized as they accelerate through a magnetic field producing electrical energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The efficiency of this setup would be determined by how much of the exhaust gas can be ionized and how much of the thermal energy of this ionized exhaust could be extracted.  As the charged gas goes through the magnetic field it would slow down (cool) and/or possibly lose charge.  The greater the temperature difference between the point of combustion and the end of the magnetic field pipe, the greater the efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are converting the thermal kinetic energy of the ions directly to electrical energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, we can extend (stretch) the entire flame, or ionized process of combustion through the entire insulated tube which would be surrounded by magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A straight linear tube would avoid x-ray photon emission caused by rotational acceleration if the tube were curved, but this x-ray emission could be useful in keeping the gas ionized.  So a circular or spiraling setup may also work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a pulsed gas setup may work better than a continual stream.  The frequency of pulsation could be fine tuned to the physics of the system for optimal efficiency.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-04-18T15:28:03.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14843" PostTypeId="2" ParentId="14822" CreationDate="2017-04-18T17:03:10.237" Score="1" Body="&lt;p&gt;Solved it, thanks to a colleague. The problem is, I used &quot;J2 Propagator&quot; in STK, which in fact uses secular rates of orbital precession. Comparing that result with the real integration that MATLAB does, led to this result. When I used HPOP propagator with degree 2 and order 0 (which is the same as just using J2 perturbation), the results got much better. HPOP does indeed integrate the differential equations. There was a growing error at a rate of 500 meters per day, and that is due to the fact that HPOP considers earth's axis precession, nutation, and polar motion, which those equations above do not contain these phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also need to mention, to get a correct solution of J2 perturbed orbit, use a relative tolerance of 1e-8 or finer.&lt;/p&gt;&#xA;" OwnerUserId="10922" LastActivityDate="2017-04-18T17:03:10.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14844" PostTypeId="2" ParentId="14841" CreationDate="2017-04-18T18:21:15.227" Score="1" Body="&lt;p&gt;Mil-Spec, or &lt;a href=&quot;https://en.wikipedia.org/wiki/United_States_Military_Standard&quot; rel=&quot;nofollow noreferrer&quot;&gt;US Military Standards&lt;/a&gt; is one type I know of. You frequently see cell phones marketed with mil-spec drop tests, etc. &lt;a href=&quot;https://en.wikipedia.org/wiki/MIL-STD-810#MIL-STD-810G.2C_Part_two_-_Laboratory_test_methods&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mil-Specs&lt;/a&gt; come in all types, so they don't necessarily cover what you were hoping they did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main initial test for if a tool is reliable is its &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B000GSLKIW&quot; rel=&quot;nofollow noreferrer&quot;&gt;online user review rating&lt;/a&gt;. People use tools in a variety of situations at different expectation levels. A high user approval rating means that even some of those people that abuse it were happy with the performance. I find this to be a much more reliable metric than any lab preformed tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative to taking good care of measurement instruments is to calibrate them often or seek an accuracy 10 times less than the designed accuracy of the device. I have a pair of &lt;a href=&quot;http://www.harborfreight.com/6-inch-digital-caliper-47257.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;harbor freight digital calipers&lt;/a&gt; that gets all the dirty work and occasionally gets dropped. They hold up well. They measure out 3 decimal places, but just I round to two decimal places and get out the nicer calipers out when I need the additional accuracy.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-04-18T18:21:15.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14846" PostTypeId="1" AcceptedAnswerId="14848" CreationDate="2017-04-18T20:46:49.103" Score="3" ViewCount="2275" Body="&lt;h1&gt;3D Positioner&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Recently I used a &lt;a href=&quot;http://www.pmk.de/en/products/3d_positionierer&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;3D positioner&quot; made by PMK Mess&lt;/a&gt;.  It is pretty awesome -- basically you move it to any position, turn the locking knob, and it stays rock solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kQ5Gl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kQ5Gl.jpg&quot; alt=&quot;3D positioner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;How might it work internally?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I want to know how it works.  I would like to be able to answer the following questions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How can the joints be locked in place so well using one knob in the center?  &lt;/li&gt;&#xA;&lt;li&gt;Are there canonical mechanisms that solve this sort of issue?  &lt;/li&gt;&#xA;&lt;li&gt;What are some good resources or references for designing something like this?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It consists of 2 ball joints and 1 DOF rotational joint.  Pictures of each joint are shown below.  I would take it apart, but it's not mine.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Pictures of the joints&lt;/h1&gt;&#xA;&#xA;&lt;h3&gt;Bottom ball joint&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PybNu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PybNu.jpg&quot; alt=&quot;Bottom ball joint&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Center 1 DOF rotational joint&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1IGwh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1IGwh.jpg&quot; alt=&quot;Center 1 DOF rotational joint&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Top ball joint&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iA0pB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iA0pB.jpg&quot; alt=&quot;Top ball joint&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10938" LastActivityDate="2017-04-19T02:01:16.427" Title="How can multiple rotational joints be locked with one mechanism?" Tags="|mechanical-engineering|design|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14847" PostTypeId="1" CreationDate="2017-04-19T00:25:19.697" Score="1" ViewCount="614" Body="&lt;p&gt;I am trying to build a simple DC motor for a project. I used a 9 V battery, 70 loops of copper wire, aluminum foil for the brushes of the commutator, and the commutator is wrapped by Aluminium foil and I used neodymium magnets with a pull of 16 Kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My motor worked for one minute and stopped and it is not working now anymore. What can be the potential problem that occurred?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b7pKZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b7pKZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The copper wire has a thin layer of insulation, the brushes are touching the commutator, magnets are 6 cm close) Is it possible that a more powerful battery is required or can the brushes of the commutator be the problem? When the brushes are touching the commutator, I can feel that the motor vibrating but even if I try to give it a push it doesn't turn.&lt;/p&gt;&#xA;" OwnerUserId="10939" LastEditorUserId="282" LastEditDate="2017-05-20T17:10:59.320" LastActivityDate="2017-05-21T12:35:33.590" Title="Building a simple DC motor" Tags="|electrical-engineering|materials|design|motors|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14848" PostTypeId="2" ParentId="14846" CreationDate="2017-04-19T02:01:16.427" Score="3" Body="&lt;p&gt;They could be locked by friction with part of the socket (circled) pressed against the ball. That force could be applied via a push rod by a cam on the knob's shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CFMPT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CFMPT.png&quot; alt=&quot;friction surface&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6eMXp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6eMXp.png&quot; alt=&quot;cam mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force might also be applied by a toggle mechanism, which is two linkages that snap into a straight line, producing a very high axial force from a relatively small lateral force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same force that locks the ball joints could also lock (by friction) the each arm onto the knob's shaft simply because that's where the reaction force to it is.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2017-04-19T02:01:16.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14849" PostTypeId="1" AcceptedAnswerId="14925" CreationDate="2017-04-19T03:46:31.500" Score="0" ViewCount="647" Body="&lt;p&gt;If we apply a simple push to rotate a system of gears to generate electricity, what is the maximum practical and theoretical power efficiency of the system, assuming the push as the only source of energy? What if the gears are to be rotated continuously?&lt;/p&gt;&#xA;" OwnerUserId="5159" LastActivityDate="2018-04-30T07:56:06.417" Title="What is the maximum efficiency of a mechanical electric generator?" Tags="|mechanical-engineering|electrical-engineering|power|energy-efficiency|power-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14850" PostTypeId="2" ParentId="14847" CreationDate="2017-04-19T05:11:02.520" Score="0" Body="&lt;p&gt;Possibly poor contact at the commutator / brushes due to burning, loose or poor contact. But without more information that is only an educated guess.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-04-19T05:11:02.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14851" PostTypeId="1" CreationDate="2017-04-19T06:05:10.550" Score="1" ViewCount="51" Body="&lt;p&gt;This is a basic question but for some reason it has me stumped. I am modeling a continuous load on a soil layer on a commercial FEA package. This is a plane strain problem where the layer is modelled 20m wide by 10m deep. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying 2 variations with 0.2 m  thickness and 1 m thickness. The load the software is applying is 1 kN/m/m for the 1m thickness variation. I would like to find an equivalent loading (i.e one which will give me equivalent stresses and strains throughout the layer) for the 0.2 m thickness case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am hesitating between using 0.2 kN/m/m and 5 kN/m/m. The reasoning behind these two numbers is either 1/0.2 = 5 or 1*0.2 = 0.2. Can anyone please explain what 1kN/m/m actually means? and how do we convert it for different thicknesses?&lt;/p&gt;&#xA;" OwnerUserId="4808" LastActivityDate="2017-04-19T12:22:21.673" Title="Load conversion for geotech problem" Tags="|structural-engineering|structural-analysis|stresses|geotechnical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14852" PostTypeId="1" AcceptedAnswerId="14866" CreationDate="2017-04-19T09:17:00.713" Score="1" ViewCount="326" Body="&lt;p&gt;I need a way to move air which is somewhere in between a compressor and a fan: something along the lines of 20-40 cfm at 2-3 psi (500-1000 liters/minute at 10-20 kPa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical small compressors (of the right power rating) seem to be approx 0.5 cfm at 30 psi.  Typical centrifugal blowers are approx 60 cfm free flow but only 0.05 psi max static pressure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble visualizing what I need, let alone sourcing it.  Leaf blower?  Shopvac?  Supercharger?  Do I want a piston pump, diaphragm pump, screw compressor, turbopump, some type of higher pressure centrifugal fan, or ???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this application energy efficiency at the described operating point is most important; size/cost/noise is much less important.  &lt;strong&gt;UPDATE&lt;/strong&gt; It also needs to be oil-free.&lt;/p&gt;&#xA;" OwnerUserId="2335" LastEditorUserId="2335" LastEditDate="2017-04-19T22:05:20.560" LastActivityDate="2017-05-02T23:09:37.357" Title="What type of air blower/compressor to use?" Tags="|pumps|compressed-air|compressors|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14853" PostTypeId="2" ParentId="13856" CreationDate="2017-04-19T10:53:08.097" Score="2" Body="&lt;p&gt;You are describing &lt;a href=&quot;https://en.wikipedia.org/wiki/Vapor-compression_evaporation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vapor-compression Evaporation&lt;/a&gt;. According to wikipedia, since any real compression will also heat the gas &quot;&lt;em&gt;The compression work will deliver the steam superheated if compared to the theoretical pressure/temperature equilibrium.&lt;/em&gt;&quot; IMO this should be verified with the actual figures from your process design etc., you don't want liquid water in a compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.gea.com/de/binaries/GEA_Evaporation-Technology-Using-Mechanical-Vapour-Recompression_brochure_EN_tcm24-22946.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This document by GEA gives an overview over the process and the compressors types usually used&lt;/a&gt;. Not that this helpful selection diagram is marked in degree temp. diff, not pressure, since this is what you ulimately want:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z1Pga.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z1Pga.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most likely you will want a rotary fan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to sum it up, yes, you can pressurize vapor mechanically and this is an established process. &lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-04-19T10:53:08.097" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14854" PostTypeId="2" ParentId="14851" CreationDate="2017-04-19T12:22:21.673" Score="2" Body="&lt;p&gt;In the comments you seem to already understand that kN/m/m means &quot;kN per meter along your screen, per meter into your screen&quot;. This can be simplified as kN/m&lt;sup&gt;2&lt;/sup&gt;: kN per meter in each direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beauty of units is that they follow all the same rules of algebra as numbers. That's because kN/m/m actually means:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1\text{ kN}}{1\text{ m}\cdot1\text{ m}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we can do some cross-multiplication:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{1\text{ kN}}{1\text{ m}\cdot1\text{ m}} = \dfrac{X\text{ kN}}{1\text{ m}\cdot0.2\text{ m}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this it is clear that $X = 0.2$. If you are reducing the area where the total force is applied, you need to reduce the force in order to maintain the same overall stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, it depends on how exactly you are applying this load. Are you applying it already as concentrated nodal forces? In this case, the above description is valid. It is also valid if you are applying linearly distributed loads (kN/m) where the described distance is along the screen (meaning that you need to consider the &quot;depth&quot; of the applied load).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're using a software package that can handle distributed loads (and then internally transforms them into nodal forces), those are usually defined with area-load units (such as kN/m/m). In this case, you just keep applying the same 1&amp;nbsp;kN/m/m and let the program handle the math.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-04-19T12:22:21.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14856" PostTypeId="2" ParentId="12134" CreationDate="2017-04-19T13:27:42.160" Score="1" Body="&lt;p&gt;Your ODE is: $$c''+\frac 2r c'-\frac KDc=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming $r\neq 0$, use the substitution $c = \frac {f(r)}{r}$. After substitution you will obtain $f''-\frac KD f=0$. This ODE can be solved by hyperbolic sine and hyperbolic cosine functions or simple exponential functions.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-04-19T13:27:42.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14857" PostTypeId="1" AcceptedAnswerId="14862" CreationDate="2017-04-19T14:26:06.287" Score="0" ViewCount="1139" Body="&lt;p&gt;My background is in electronics, not mechanical engineering and fluid dynamics, so please keep that in mind. Also, this is not a homework or academic assignment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: A dewar is partially filled with liquid nitrogen (LN2) and is open to the atmosphere. If a container, open on the top, has a hole in the bottom and is lowered into the liquid, how quickly will the liquid fill the container?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifics:&#xA;The dewar has an inside diameter of 14&quot; and has 8&quot; of LN2 ( volume ~1200 cubic inch).&#xA;Second container is 4&quot; outside diameter, 3.75&quot; inside diameter, 7&quot; height.&#xA;A small hole, ~40 mils, is drilled in the bottom of the container.&#xA;Container will be lowered into the LN2 to a depth of 6&quot;.&#xA;Atmospheric pressure: Sea level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to determine the rate of flow into the secondary container so I know how quickly it will fill to a 6&quot; depth (~66 cubic inch).&lt;/p&gt;&#xA;" OwnerUserId="10948" LastEditorUserId="10948" LastEditDate="2017-04-19T16:03:56.953" LastActivityDate="2017-07-23T18:16:56.653" Title="Rate of flow of liquid nitrogen from a static tank through an orifice in a submerged vessel" Tags="|fluid-mechanics|cryogenics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14859" PostTypeId="2" ParentId="14852" CreationDate="2017-04-19T15:44:54.097" Score="1" Body="&lt;p&gt;If efficiency is important, I would not compress to a higher pressure, then expand the air.  I think the pressure ratio is in the right range for a positive-displacement pump, like the ones that are used in superchargers.  There's an efficiency map here: &lt;a href=&quot;https://en.wikipedia.org/wiki/Roots-type_supercharger&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Roots-type_supercharger&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-04-19T15:44:54.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14860" PostTypeId="2" ParentId="2690" CreationDate="2017-04-19T17:07:13.383" Score="0" Body="&lt;p&gt;The greater the photon density at the frequency of interest, the more the power as the photons excite the semiconductor's electrons to higher energy orbitals to the band-gap and beyond.  That being said,as Olin stated, the increase is not linear.  Eventually as temperature increases, the increased photon intensity results in ever decreasing increments in power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion would be to use lens filters and other methods to reject the photon wavelengths that are not of benefit.  We only want the photons of wavelength that are tuned to excite the electrons across the band-gap for that particular semi-conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any photons that don't do this only cause increase in temperature.  So you want to increase the incident photon density of only the photons of relevance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can actually cool the solar arrays by aluminum heat sinks beneath them that run water through them for water heater purposes.  I saw such a setup at a trade show.  It was by a Spanish company but I don't recall the name.  The setup combined solar power for both electricity and convective water heating. &lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-04-19T17:07:13.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14861" PostTypeId="2" ParentId="14824" CreationDate="2017-04-19T17:14:04.163" Score="4" Body="&lt;h2&gt;Real-world engineering example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To my knowledge there are no simple analytical methods of determining whether an object such as a sword can cut cleanly through any arbitrary object. In manufacturing engineering we would rather cut many small pieces of material than take one big cut (i.e. we generally use saws to cut material instead of giant guillotines). The exception would be for thin objects like sheet metal, which is cut with large punches or shears, as pictured below.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/NtYDL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NtYDL.png&quot; alt=&quot;Shearing operation: (a) side view of the shearing operation; (b) front view of power shears equipped with inclined upper cutting blade.&quot;&gt;&lt;/a&gt;&#xA;Image Source: Mikell P. Groover, &lt;em&gt;Fundamentals of Modern Manufacturing, 5th Ed.&lt;/em&gt; pp. 483&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you will notice about the shearing operation is that it requires an &quot;ideal&quot; shearing case. The blade has a very small amount of clearance between it and the die (light grey support) to ensure that the material fails in shear and not in bending. Furthermore, the blade is angled so that it minimizes the cutting force by not cutting the entire stock material at once. So the chances of encountering this sort of situation with a sword is unlikely. However, if you had this situation, here are the equations you would use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The recommended clearance, $c$ is calculated using:&#xA;$$c = A_ct $$&#xA;Where $A_c$ is the clearance allowance and $t$ is the stock thickness (thickness of the cut material). $A_c$ can vary from 0.045 for very soft materials (i.e. aluminum) to 0.075 for relatively hard materials (i.e. stainless steels).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cutting force, $F$, can be calculated using:&#xA;$$F = StL $$&#xA;Where $S$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;shear strength&lt;/a&gt; of the material and $L$ is the length of the cut edge. This equation assumes you will cut the entire length all at once, so you can apply a factor to reduce this amount if you have angled cutting like in Figure 19.3 above.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Unfortunately it's not that simple for swords&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In these types of operations we would generally apply high, constant forces at low speeds. This is very different from a sword, where you would likely be moving at higher speeds and applying time-varying forces (i.e. impact). Furthermore, the blades used in shearing operations are usually much thicker than the objects that they cut, so we can mostly ignore the stresses and deflections in the blade. This would not be the case for, say, a sword cutting through another sword, or armour. The relative hardness and fracture toughness of the materials, crack propagation, etc. would likely be relevant as well.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;If you're looking to do calculations for fun anyway&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;If you're just doing these calculations for fun, not for serious engineering design, here's my advice. Make some assumptions (i.e. ideal shearing case) and use the cutting force equation above to check what the load on the sword would be. Assume that the force is a point load placed at the middle of the sword, apply a safety factor of 5 and check whether the sword would fail due to bending. If it doesn't fail then you could say the sword would &quot;likely&quot; be able to chop cleanly through the material. Then check whether or not a human could actually apply that force at that point on the sword. If not, then you are free to imagine the sword being wielded by some kind of super-strong android samurai.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...Wait a sec, you aren't building a super-strong android samurai are you? XD&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2017-04-19T17:14:04.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14862" PostTypeId="2" ParentId="14857" CreationDate="2017-04-19T18:36:34.913" Score="0" Body="&lt;p&gt;I am at work and cannot perform the calculations for you, but if you solve the following equation you will find the velocity $v$ being the square root of 2 times Earth gravity times depth $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v = \sqrt{2 \cdot 9.81 \, \mathrm{m/s^2} \cdot x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I remember correctly, from the velocity you should be able to calculate the volume rate $\dot{V}$ by multiplying the area of the opening $A$ with the velocity $v$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{V} = A \cdot v$$&#xA;$$\mathrm{\frac{m^3}{s} = m^2 \cdot \frac{m}{s}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then just convert to the volume units of your choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/bernouilli-equation-d_183.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/bernouilli-equation-d_183.html&lt;/a&gt;&#xA;Read the example on Bernoulli's equation and the vented tank example.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:16:56.653" LastActivityDate="2017-07-23T18:16:56.653" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14863" PostTypeId="1" CreationDate="2017-04-19T23:18:40.637" Score="2" ViewCount="287" Body="&lt;p&gt;I want to build a small device (beacon) that can send its location (&lt;em&gt;inside&lt;/em&gt; of a building) and can receive signals in order to output information on an LED screen. The beacon should communicate with a central unit.&lt;br&gt;&#xA;So the device would be pretty similar to a &lt;a href=&quot;https://www.thetileapp.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tile&lt;/a&gt; in many ways. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture to illustrate the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pmXg8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pmXg8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The beacon is currently located in area 1. When pinged, it should either send back its exact coordinates (so that the central unit can calculate that the beacon is positioned in area 1) or the current area id.&lt;br&gt;&#xA;Once the beacon is moved, it should of course update its position.&lt;br&gt;&#xA;The areas are inside of a building, always rectangular and I can install hardware on site (e.g. on the four emitters in the corners of the rectangles). The side length of the areas is between 1 and 20 meters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I am evaluating different possible technologies. The biggest challenge is that I need 2-way communication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other requirements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The beacon should have a battery life of several &lt;em&gt;months&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;The beacon should update its location on a regular basis (e.g. once a minute) and on demand (when triggered from the central unit)&lt;/li&gt;&#xA;&lt;li&gt;There can be up to 5,000 beacons in one area at a time. They all have to work simultaneously &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RFID&lt;/strong&gt;&lt;br&gt;&#xA;+ Passive (no need for batteries)&lt;br&gt;&#xA;+ Cheap&lt;br&gt;&#xA;+ Location detection should be fairly easy with &quot;gates&quot;&lt;br&gt;&#xA;- Range probably too short for my scenario&lt;br&gt;&#xA;- No 2-way communication possible (would need additional module)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bluetooth&lt;/strong&gt;&lt;br&gt;&#xA;+ Battery life with low energy BT&lt;br&gt;&#xA;+ Range should suffice&lt;br&gt;&#xA;+ 2-way communication possible&lt;br&gt;&#xA;- Not sure how many BT beacons can be connected to the central unit at once (probably not 5,000)&lt;br&gt;&#xA;- Not sure how locating the beacon would work (signal strength?) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Wifi&lt;/strong&gt;&lt;br&gt;&#xA;+ Range&lt;br&gt;&#xA;+ 2-way communication&lt;br&gt;&#xA;+ Connection to thousands of beacons at the same time possible&lt;br&gt;&#xA;- Not wure how locating the beacon would work (signal strength?)&lt;br&gt;&#xA;- Battery life?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly appreciate any recommendations, tips and improvements. Right now, I don't really know which way to take and it would end in a lot of trial and error :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much.&lt;/p&gt;&#xA;" OwnerUserId="10956" LastEditorUserId="10956" LastEditDate="2017-04-19T23:26:01.287" LastActivityDate="2017-04-20T11:51:26.510" Title="RFID vs. Bluetooth vs. Wifi beacon" Tags="|electrical-engineering|design|wifi|" AnswerCount="1" CommentCount="0" ClosedDate="2017-04-22T18:07:29.157" ContentLicense="CC BY-SA 3.0" />
  <row Id="14864" PostTypeId="2" ParentId="12321" CreationDate="2017-04-20T00:48:50.693" Score="1" Body="&lt;p&gt;Your question has issues. First you say &quot;$\text{C}_{10}H_{14}$, $\text{C}_4\text{H}_6\text{O}_3$, and $\text{C}_{12}\text{H}_{16}\text{O}$ produced during a reaction&quot;, then you say &quot;134 grams of $\text{C}_{10}H_{14}$ and 134g of $\text{C}_4\text{H}_6\text{O}_3$&quot;, so $\text{C}_{10}H_{14}$, and $\text{C}_4\text{H}_6\text{O}_3$ are reactants or products?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume they are reactants, you say also acetic acid is produced so the reaction would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\text{C}_{10}H_{14}+\text{C}_4\text{H}_6\text{O}_3 \rightarrow \text{C}_{12}\text{H}_{16}\text{O}+\text{C}_2\text{H}_4\text{O}_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All stochiometric coefficients are equal to 1. So if you assume 68 grams of &#xA;acetic acid where produced, because $\text{MW}(\text{C}_2\text{H}_4\text{O}_2)=60$, more than one mole was produced. But because 134 g of $\text{C}_{10}H_{14}$ were used as reactant, and $\text{MW}(\text{C}_{10}H_{14})=134$, the maximum extent would be 1 mol, so the data are inconsistent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in your original equations, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $C$: $12.98 = 10x/134 + 4y/102 + 12z/176$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You didn't specify the units. Based on the rhs it appear there are mol quantities, if that's the case, the third term would need to have a negative side. And if lhs are mols produced (from acetic acid), the data don't match,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mol C produced (acetic acid) = $2\cdot68/60=2.26$&lt;/p&gt;&#xA;" OwnerUserId="10955" LastEditorUserId="1832" LastEditDate="2017-04-20T11:11:17.973" LastActivityDate="2017-04-20T11:11:17.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14866" PostTypeId="2" ParentId="14852" CreationDate="2017-04-20T05:44:50.580" Score="1" Body="&lt;p&gt;I think fans are out, as you won't get your combination of flow rate and pressure. The only thing you might want to look at is multistage fans. For this answer I'll concentrate on PD compressors. All of the following should be available in oil-free versions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side channel compressors come in the flow rate and pressure region you want. You will need to talk to an application engineer of a vendor to see if they fit your particular aplication. I don't know how they compare in terms of efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All rotary lobe compressors and sliding vane compressors I found with quick googling where designed for higher pressures or flow rates. If you find one in the right pressure range, talk to the vendor if the flow rate can be adjusted with a VFD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Membrane pumps are built for lower flow rates. If all else fails, get ten or twelve, pumping in paralell into the same pressure manifold.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-04-20T05:44:50.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14867" PostTypeId="1" CreationDate="2017-04-20T05:54:19.180" Score="2" ViewCount="2869" Body="&lt;p&gt;When doing section classification to calculate buckling of a plate subjected to some bending and compression on its edges, do we consider the plate width as outstand of flanges &quot;c&quot;? What are the limiting values for plate and bar in &lt;a href=&quot;http://www.steelconstruction.info/Design_codes_and_standards#Section_classification&quot; rel=&quot;nofollow noreferrer&quot;&gt;section classification&lt;/a&gt;?&lt;/p&gt;&#xA;" OwnerUserId="10920" LastActivityDate="2018-11-14T12:12:28.573" Title="How to do section classification for a plate or flat bar?" Tags="|structural-engineering|design|steel|eurocodes|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14868" PostTypeId="2" ParentId="14867" CreationDate="2017-04-20T08:26:47.233" Score="2" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For a flat bar - there is no local buckling, only global buckling.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a plate you follow the rules for plated sections. In Europe that would be EC3-1-5.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The local design code should define the limits at which a flat bar becomes a plate. (I can't remember off the top of my head where in the eurocodes this would be found.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-04-20T08:26:47.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14869" PostTypeId="1" CreationDate="2017-04-20T08:29:27.590" Score="-1" ViewCount="75" Body="&lt;p&gt;If the fluid moved by a pump is changed by another different one, what will be the effect of such kind of operation for the pump ? &#xA;Specifically, a pump moving 60 m3/hr of &quot;liquid nr. 1&quot; (sp.grav.: 0.800 / visc.: 0.65) should now move  53.5 m3/hr of &quot;liquid nr. 1&quot; + 6,5 m3/hr of    &quot;liquid nr. 2&quot; with sp. grav. 0.56 and a viscosity of 0.17. &#xA;Is there a way to locate to operation point of the pump on the performance curves delivered py the manufaturer and has somebody experience with a situation like this ? Considering that the viscosity as well as the specific gravity of the mixture will be lower than at the previous operating point, what problems could occcure ? (cavitation, erosion, motor overload etc. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards,&#xA;Marcus&lt;/p&gt;&#xA;" OwnerUserId="10960" LastActivityDate="2018-09-21T22:55:01.563" Title="Change of fluid of Pump" Tags="|mechanical-engineering|pumps|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14870" PostTypeId="2" ParentId="14869" CreationDate="2017-04-20T08:42:37.283" Score="0" Body="&lt;p&gt;I don't have access to my library at the moment but the link can give you some information : &lt;a href=&quot;https://neutrium.net/equipment/pump-power-calculation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://neutrium.net/equipment/pump-power-calculation/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-04-20T08:42:37.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14872" PostTypeId="2" ParentId="14863" CreationDate="2017-04-20T11:45:50.523" Score="2" Body="&lt;p&gt;Some thoughts on your options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RFID&lt;/strong&gt;, as you stated, likely won't work, most certainly not over 20 meters as low power devices, that's not what it is designed for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bluetooth&lt;/strong&gt; and &lt;strong&gt;WiFi&lt;/strong&gt; would work with signal strength measurement (proximity) and triangulation. The signal strength measurement could lack the desired accuracy, especially if you're in the 1m range. But if accuracy isn't very important, you probably should choose one of those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an alternative, I once worked with &lt;a href=&quot;http://www.decawave.com/products/dwm1000-module&quot; rel=&quot;nofollow noreferrer&quot;&gt;these RF Modules&lt;/a&gt; (low-power, good range, good precision), but 5'000 connections are not possible (if I remember it correctly). Maybe there are other solutions in this direction..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, sounds like an interesting project. (What do you need 5'000 beacons for?)&lt;/p&gt;&#xA;" OwnerUserId="10862" LastEditorUserId="10862" LastEditDate="2017-04-20T11:51:26.510" LastActivityDate="2017-04-20T11:51:26.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14873" PostTypeId="2" ParentId="14377" CreationDate="2017-04-20T13:35:35.653" Score="2" Body="&lt;p&gt;Because the definition of derivative is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{\Delta x\rightarrow0}\frac{f_{x+\Delta x}-f_x}{\Delta x}=\frac{\partial f}{\partial x}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applied to this case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\lim_{\Delta x\rightarrow0}\frac{N_{ix}|_{x}-N_{ix}|_{x+\Delta x}}{\Delta x}=-lim_{\Delta x\rightarrow0}{\Delta x}\frac{N_{ix}|_{x+\Delta x}-N_{ix}|_{x}}{\Delta x}=-\frac{\partial N}{\partial x}$&lt;/p&gt;&#xA;" OwnerUserId="10955" LastActivityDate="2017-04-20T13:35:35.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14874" PostTypeId="2" ParentId="14826" CreationDate="2017-04-20T13:52:49.093" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JSqS2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JSqS2.jpg&quot; alt=&quot;Hope this helps!&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope it makes a lil more sense. Be careful to multiply your excess gibbs by n &amp;amp; xlx2 from the activity coefficient relationship with gibbye! Also, be super grateful you didn't have to do the same thing for the Margules 3 parameter that I literally just worked out like 15 minutes before I saw this!&lt;/p&gt;&#xA;" OwnerUserId="10961" LastActivityDate="2017-04-20T13:52:49.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14876" PostTypeId="1" AcceptedAnswerId="14895" CreationDate="2017-04-20T15:02:48.300" Score="1" ViewCount="146" Body="&lt;p&gt;I am tasked with finding the radius of a hole at the bottom of a tank to obtain the desired outflow rate when the tank is full.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tank is a cylinder with dimensions Height: 4.8m and Radius: 1.79m. It holds roughly 48316.6897 liters of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already determined $C_d$ (The coefficient of discharge) to be 0.6372.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outflow rate I desire is 3500 liters per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that $\frac{dh}{dt}= \frac{a_0C_d\sqrt{2gH}}{A}$. where $a_0$ is the radius i need to obtain, and $A$ being the surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From chain rule I can get $\frac{dh}{dt}$ in terms of radius and $\frac{dv}{dt}$(The desired outflow rate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{dh}{dt} = \frac{\Delta{V}}{\pi r^2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I let them equal each other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\Delta{V}}{\pi r^2} = \frac{a_0 C_d\sqrt{2gH}}{A}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives me the equation&#xA;$$a0=\frac{\Delta{V}}{C_d\sqrt{2gH}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, substitute in what we know&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3500$ Litres per hour is $\frac{35}{36000} \frac{m^3}{s} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a0 = \frac{\frac{35}{36000}}{0.6372\sqrt{2\cdot 9.8 \cdot 4.8}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this $a_0$ is $0.157304491$ millimeters in radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am doubting my calculations because $3500$ liters going past a hole of radius $0.157304491$ millimeters every hour seems absurd. &lt;/p&gt;&#xA;" OwnerUserId="10963" LastEditorUserId="7627" LastEditDate="2017-04-22T14:08:21.873" LastActivityDate="2017-05-03T22:49:21.260" Title="Calculating the radius of a hole for a desired outflow rate" Tags="|mechanical-engineering|fluid-mechanics|fluid|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14877" PostTypeId="1" CreationDate="2017-04-20T15:17:24.827" Score="-4" ViewCount="83" Body="&lt;p&gt;How doe one design a circuit.  I know all the components but don't know where to place the components. It would be really helpful if one can explain a simple complex rc circuit. How does the current flow and why the components are there.i am trying to learn by example.&lt;/p&gt;&#xA;" OwnerUserId="10964" LastActivityDate="2017-07-23T18:16:47.430" Title="Circuit designing?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="2" ClosedDate="2017-04-20T15:35:44.283" ContentLicense="CC BY-SA 3.0" />
  <row Id="14879" PostTypeId="2" ParentId="14877" CreationDate="2017-04-20T15:26:16.833" Score="1" Body="&lt;p&gt;I would recommend the use of MultiSim. It's an electronic schematic capture and simulation program which is part of a suite of circuit design programs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will give you a great idea on where to place everything and you can even test it out in a simulation, to make sure everything will work out!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple complex circuit seems contradictory, but I have completed rc circuit setups and you need, obviously a resistor, capacitor, circuit board to set up everything on it, circuit cables to connect everything together, and a power source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current flows from the positive terminal of the power source, towards the rest of the components, to finally come back to the negative terminal of the power source.&lt;/p&gt;&#xA;" OwnerUserId="10028" LastActivityDate="2017-04-20T15:26:16.833" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14880" PostTypeId="2" ParentId="14877" CreationDate="2017-04-20T15:30:25.870" Score="0" Body="&lt;p&gt;A simple place to start is to find a Radio Shack that is still open and purchase a breadboard and some simple arduino circuit boards, then just experiment. There are loads of examples with devices made with arduino components and custom circuit boards made using bread boards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ebay.com/p/?iid=261740468056&amp;amp;lpid=82&amp;amp;&amp;amp;&amp;amp;ul_noapp=true&amp;amp;chn=ps&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ebay.com/p/?iid=261740468056&amp;amp;lpid=82&amp;amp;&amp;amp;&amp;amp;ul_noapp=true&amp;amp;chn=ps&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:16:47.430" LastActivityDate="2017-07-23T18:16:47.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14881" PostTypeId="1" AcceptedAnswerId="14883" CreationDate="2017-04-20T15:51:54.193" Score="0" ViewCount="173" Body="&lt;p&gt;So I am making a tech demo demonstrating fairly realistic penetration physics. I Have run into some confusion due to my lack of knowledge. If I have the resistance of the armor in MPa, that means I have the amount of joules acting in a square meter, so i divide by the area of the cross section of the shell, in meters (assuming it's a cylinder) then multiply by the thickness of the armor, also in meters, to get the required joules to punch through the armor of the given thickness with a round of the given cross sectional area?&lt;/p&gt;&#xA;" OwnerUserId="10635" LastActivityDate="2017-04-20T17:36:32.860" Title="penetration of rolled homogeneous steel, issues with finding the joules required to punch through" Tags="|materials|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14882" PostTypeId="2" ParentId="14876" CreationDate="2017-04-20T15:55:01.480" Score="0" Body="&lt;p&gt;I think the problem is that there's something wrong with your first differential equation &#xA;$$ \frac{dh}{dt}=\frac{a_0C_d\sqrt{2gH}}{A} $$ &#xA;the lefthand side has the unit $ \frac{m}{s} $, while the righthand side leads to the units $\frac{1}{s}$.&#xA;If you only need to know the outflow rate or needed radius for the instant where the tank is full, I would use Bernoullis principle:&#xA;$$ H=\frac{v^2}{2g} \qquad \mbox{and}\qquad v=\frac{Q_t}{\pi r^2} $$&#xA;$$ C_d=\frac{Q_{r}}{{Q_t}} $$, where $t =$ theoretical and $r=$ real.&#xA;When solving for $r$ you get:&#xA;$$ r = \left( \frac{\left(\frac{Q_r}{C_d}\right)^2}{2\pi^2Hg} \right)^{1/4} \qquad\qquad \star $$&#xA;If I put in the given values it returns &#xA;$$ r=6.88mm $$&lt;/p&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="8632" LastEditDate="2017-04-20T20:54:10.433" LastActivityDate="2017-04-20T20:54:10.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14883" PostTypeId="2" ParentId="14881" CreationDate="2017-04-20T16:51:59.707" Score="2" Body="&lt;p&gt;MPa is pressure, giving you measures of $\frac{N}{m ^2} \times 10^9$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying it by cross section ($m^2$) gives you Force, measured in Newtons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you multiply that by thickness, the value you get will be in Joules ($N \times m$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So great, the units seem to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is finding the values you should use.  I'm assuming you found the ultimate tensile stress or some similar property for the material and took that for the MPa value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot just multiply it by the volume removed to get required energy, even though the units work.  Consider the situation.  The cross sectional area shouldn't directly matter.  Regardless of the area, you are only really punching through the perimeter over a specific thickness, not an area through a thickness (you can keep the centre of what you punch intact).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality the equations are much more involved than the ultimate material strength and the amount of material removed.  It includes factors like the area vs. perimeter relationship and would also require information on what you are using for a punch.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-04-20T16:51:59.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14884" PostTypeId="2" ParentId="14881" CreationDate="2017-04-20T17:36:32.860" Score="0" Body="&lt;p&gt;A value in MPa is likely to be a yield stress or ultimate tensile stress. This is fine for static forces and conditions where strain rates are low eg if you are punching a hole in a plate with a hydraulic press. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in the context of armour we re often looking at very high strain rates and in this case material properties for static conditions may not apply. For example with kinetic energy penetrators you can see viscous flow as well as shockwave induced effects like spalling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So realistic models tend to be complex, based on empirical data and highly contextual. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might be better is too look at empirical standard for armour plating eg &lt;a href=&quot;http://www.ssab.co.uk/products/brands/armox/recommended-plate-thickness-for-different-protection-levels&quot; rel=&quot;nofollow noreferrer&quot;&gt;this graphic for Armox&lt;/a&gt; Obviously a lot of the fine detail of this sort of data is proprietary but you should at least be able to get order of magnitude figures. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-04-20T17:36:32.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14885" PostTypeId="1" CreationDate="2017-04-21T00:19:18.437" Score="1" ViewCount="29" Body="&lt;p&gt;If I Design this slab as a flat 1500mm thick slab throughout ignoring the step will the design be conservative. I'm worried the extra stiffness created by the step, will attract the moment into the step as it is acting like a beam and therefore my model may not be conservative and there could be possible failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PMo1Z.jpg&quot; alt=&quot;Slab Section&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10969" LastActivityDate="2017-04-21T00:19:18.437" Title="Is it possible that a considerably thicker portion of slab would decrease the moment capacity of a raft slab" Tags="|structural-engineering|structural-analysis|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14886" PostTypeId="1" CreationDate="2017-04-21T00:33:40.233" Score="1" ViewCount="27" Body="&lt;p&gt;I have been wondering, would it be possible to make an 8’x4’x’5 tank that can be put up and taken down as needed, where you would have a base, four removable corner posts, more removable posts as needed to support the walls, and a top connector.  Use either water pressure or expanders to hold the walls against the seals, and only one wall would need to be clear.  For even more extreme have the base be in multiple sections so it can fit in the back of a car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because many of my friends and myself are mermaids/mermen and we don't always have the funds to buy a massive 1000 gallon tank and the trailer to move it.  The tanks allow us to set up and entertain kids in dry environments, like a building or a fair.  I know water has the pressure to keep a car door from opening when a car goes into a lake, and I would imagine having water push against a wall will make a fairly good seal, and the bottom could be lip connector based.  I don't think it has to be 100% leak tight as it might only be set up for a few hours and something like a tarp could be placed under the tank to collect the water and pump it back into the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How insane is this idea?&lt;/p&gt;&#xA;" OwnerUserId="5848" LastActivityDate="2017-04-21T01:52:48.433" Title="Collapsable Water Tank" Tags="|pressure-vessel|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14887" PostTypeId="2" ParentId="6175" CreationDate="2017-04-21T00:45:33.623" Score="0" Body="&lt;p&gt;The advantages outweigh the disadvantages, so the we need to consolidate the political will to make the necessary investments.  The ideal locations in sunny arid areas would be next to the seas for powering desalination plants.  The CSPs can be used together with PV for these applications.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By desalinating and electrolyzing sea water, we get hydrogen in addition to fresh water in addition to electric power.  Through the electrolysis we can also obtain hydroxides to remove the carbon dioxide from the exhaust from hydrocarbon burning facilities.  The hydroxides, primarily sodium hydroxide, would react with the carbon dioxide to form carbonates which are solid precipitates.  The carbonate can be cashed in for carbon credits on the carbon sequestration market.&#xA;We also get chlorine from the electrolysis of seawater which is also very useful product especially in the third world where it can be used to sterilize drinking water.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-04-21T00:45:33.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14888" PostTypeId="1" CreationDate="2017-04-21T01:33:55.533" Score="1" ViewCount="1142" Body="&lt;p&gt;In the book (photo 1 )  , it's stated at the left span, the fixed end moment is clockwise , while on the right span , the fixed end moment is anticlockwise . But , accroding to wiki (photo2) , it's stated that at the left span of the beam , the fixed end moment is anticlockwise , while on the right span , it's clockwise ... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases , clockwise moment is assigned as positive . Which case is correct ? I'm confused . &#xA;ht//en.wikipedia.org/wiki/Fixed_end_moment&#xA;&lt;a href=&quot;https://i.stack.imgur.com/m96bT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m96bT.png&quot; alt=&quot;hhf&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/C7H6m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C7H6m.png&quot; alt=&quot;sfsf&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10127" LastActivityDate="2017-04-21T03:21:11.537" Title="fixed end moment" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14889" PostTypeId="2" ParentId="14459" CreationDate="2017-04-21T01:35:29.127" Score="0" Body="&lt;p&gt;If you can remove or reduce the openings you will have the most success in reducing sound. Failing that, if you need airflow, baffles will help to reduce the sound (at a cost of reducing the airflow). Once you've addressed the design of the holes, adding mass will be a good way to reduce sound transmitted through the material. For a good example, look at the vinyl sound deadening liners used in the automotive industry. In fact, you could try applying these to the inside of this product. They are much thinner than acoustic panels, albeit heavier. One popular aftermarket brand name is Dynamat.&lt;/p&gt;&#xA;" OwnerUserId="2606" LastActivityDate="2017-04-21T01:35:29.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14891" PostTypeId="2" ParentId="14886" CreationDate="2017-04-21T01:52:48.433" Score="1" Body="&lt;p&gt;Most temporay tanks are not rigid because of the increased weight and trouble to transport. &lt;a href=&quot;http://www.thefirestore.com/mobile/store/product.aspx/productId/11248/Husky-Aluminum-Frame-Folding-Water-Tank/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flexible colapsible tanks&lt;/a&gt; are commonly used for rangeland firefighting. You are probably not going to find any at your desired dimensions but you might get close. They wont have a clear window either. For this you will probably want to use a clear vinyl bonded to the fabric with an appropriate adhesive and lots of overlap. Even if you built one from scratch this would be a much easier way to go than a dissembleable rigid construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Constructing a rigid tank from plywood in 2x6 timbers would not be too difficult. The timbers would be mounted in the horizontal direction to counter the hoop stress of the filled tank. As far as the standard construction goes there looks to be lots of DIY support on fish tank forums.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.monsterfishkeepers.com/forums/attachments/pond-jpg.1230104/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Large diy windowed fish tank&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the tank disassembleable will definitely make things more difficult. The weight of each panel will be considerable; definitely a two-man lift. The corners will need lots of large bolts. Setup time might not be as quick as you are thinking without some careful design and an impact wrench. You will want to paint the plywood even if the use will be tempotary. You will have to get creative with the design to make it seal; this will be the most difficult part. While pumping it back in might be an option; it may leak faster than you can pump it back in ;-) Dont underestimate the difficulty of designing seals and containing water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better budget approach may be to get a low cost trailer from harborfreight or other and building a &quot;permanent&quot; tank on that fully sealed. You would use the trailer to transport the bulky tank 100% empty. Then use &lt;strong&gt;lots&lt;/strong&gt; of jacks to hold the weight of the tank before you fill it. Then get creative with some decorations to cover up the trailer beneath.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-04-21T01:52:48.433" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14894" PostTypeId="2" ParentId="14888" CreationDate="2017-04-21T03:21:11.537" Score="2" Body="&lt;p&gt;Everything if fine. You can use any one convention. Don't change conventions while doing a project. Keep it same all the time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I would add:&#xA;Your book shows moment formed &lt;strong&gt;inside the beam&lt;/strong&gt;. Wiki shows &lt;strong&gt;moment formed on the end walls.&lt;/strong&gt; Both are correct &lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2017-04-21T03:21:11.537" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14895" PostTypeId="2" ParentId="14876" CreationDate="2017-04-21T03:34:57.593" Score="0" Body="&lt;p&gt;&lt;em&gt;According to this a0 is 0.157304491 millimetres in radius.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am doubting my calculations because 3500 litres going past a hole of radius 0.157304491 millimetres every hour seems absurd.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is area. You need diameter. Multiply it by 4 then divide it by pi  then take root of the result. That will be diameter of hole&lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2017-04-21T03:34:57.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14897" PostTypeId="1" AcceptedAnswerId="14898" CreationDate="2017-04-21T08:04:44.737" Score="-1" ViewCount="867" Body="&lt;p&gt;Titanium is more durable and stronger than steel. I checked the price of titanium and it is not too high. Despite that, I could find only one car made of titanium costing, and it cost 2 million. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it have anything to do with aerodynamics?&lt;/p&gt;&#xA;" OwnerUserId="10977" LastEditorUserId="6790" LastEditDate="2020-05-02T20:17:33.577" LastActivityDate="2023-11-16T06:10:58.597" Title="Why are car bodies not made of titanium?" Tags="|mechanical-engineering|materials|automotive-engineering|metals|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="14898" PostTypeId="2" ParentId="14897" CreationDate="2017-04-21T08:48:11.537" Score="6" Body="&lt;p&gt;The bulk price of titanium doesn't tell the whole story. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A big chunk of the cost of titanium parts is down to the fact that it is significantly more costly to make things out of titanium than steel. Even before you start manufacturing parts there are costs associated with processing stock to the correct dimensions and specifications (in this case thin sheet). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass produced car bodies are generally made by cold pressing processes and then welded together and it is the cost of tooling rather than the bulk material price which is the biggest contributor to the manufacturing cost per vehicle. Titanium alloys are significantly more difficult to form and weld than steel. Especially as car design has tended towards deeper forms, compound curved surfaces and tighter radii in pressings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Welds in titanium are also much more prone to failure due to contamination and thus require much more tightly controlled welding processes and inspection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing is that stiffness is at least as important as tensile strength in a car body and the stiffness to weight ratio of titanium isn't much different to steel so if you are designing for stiffness you may not even save any weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the other end of the spectrum carbon and Kevlar composites are generally superior to titanium for body shells in almost all respects for low-volume high performance applications and a few carbon parts are starting to filter through to the higher end of production cars. So there isn't even any filter down from high performance applications which tends to be one of the drivers for new technology in production cars. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-04-21T17:55:51.813" LastActivityDate="2017-04-21T17:55:51.813" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14899" PostTypeId="2" ParentId="14517" CreationDate="2017-04-21T12:20:21.183" Score="6" Body="&lt;p&gt;Consider an analogy where voltage and force are &quot;efforts&quot; and current and velocity are &quot;flows&quot;. We can choose either the effort or flow as the input, and the other as the output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A spring is a mechanical integrator in the sense that it integrates its velocity to generate a proportional force. (It integrates a flow input to generate an effort output). Alternatively, a mass can be considered a mechanical integrator in the sense that it integrates the applied force to generate a change in velocity. (It integrates an effort input to generate a flow output).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogy to the electrical domain should be clear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider &quot;flow&quot; as the input and &quot;effort&quot; as the output, then a capacitor integrates electrical flow (current) to generate an electrical effort (voltage), i.e., $e = \frac{1}{C}\int i\,dt$. Similarly, a spring integrates mechanical flow (velocity) to generate mechanical effort (force), i.e., $F = k\int v\,dt$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if we consider the &quot;effort&quot; as the input and the &quot;flow&quot; as the output, then an inductor integrates electrical effort (voltage) to generate a change in electrical flow (current), i.e., $i = \frac{1}{L} \int e\,dt$. Similarly, a mass integrates mechanical effort (force) to generate a change in mechanical flow (velocity), i.e., $v = \frac{1}{m} \int F\,dt$. &lt;/p&gt;&#xA;" OwnerUserId="3232" LastEditorUserId="3232" LastEditDate="2017-06-22T19:41:27.607" LastActivityDate="2017-06-22T19:41:27.607" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14906" PostTypeId="1" CreationDate="2017-04-21T16:19:46.567" Score="-2" ViewCount="1349" Body="&lt;p&gt;I am facing problem with this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my book it is written that throughput increases with the increase in the number of stages. But I think this is wrong. I think throughput decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can anyone please help me?&lt;/p&gt;&#xA;" OwnerUserId="10988" LastEditorUserId="6790" LastEditDate="2020-08-01T15:07:27.820" LastActivityDate="2020-08-01T15:07:27.820" Title="Does increasing the number of pipeline stages increase the throughput of the pipeline?" Tags="|computer-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14913" PostTypeId="2" ParentId="14897" CreationDate="2017-04-21T22:16:29.033" Score="1" Body="&lt;p&gt;Aside from the manufacturing cost issues in another answer, titanium is much less ductile than steel. In a crash situation, it would not absorb energy permanently (by plastic deformation) but simply &quot;rebound,&quot; or possibly vibrate with a large amplitude (several inches peak-to-peak) for a relatively long time span. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The above is based on practical experience, trying to replace steel by titanium in a containment system to reduce weight - the idea just didn't work).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, titanium is quite a reactive metal - titanium fires can be started by friction, and can be as spectacular as burning magnesium ribbon.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-04-21T22:16:29.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14915" PostTypeId="1" CreationDate="2017-04-22T07:56:14.743" Score="0" ViewCount="48" Body="&lt;p&gt;I was reading about the new Intel Xeon Phi processor on Intel's site &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://software.intel.com/xeon-phi/x200-processor/remote-access&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://software.intel.com/xeon-phi/x200-processor/remote-access&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and ones of its features is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Highly-parallel performance: Features up to 72 cores with deep out-of-order buffers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking for a basic description of these buffers that highlights maybe their purpose (ie parallelism) and behavior. Are these buffers used in the cache?&lt;/p&gt;&#xA;" OwnerUserId="10996" LastActivityDate="2017-04-24T02:04:16.493" Title="Deep out-of-order buffers" Tags="|computer-engineering|architecture|computer-hardware|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14916" PostTypeId="1" CreationDate="2017-04-22T08:17:58.330" Score="0" ViewCount="120" Body="&lt;p&gt;A wi-fi router works at 2.4GHz. Can a wireless signal of 2.4 GHz or any other frequency, of sufficiently high energy, damage the device? &lt;/p&gt;&#xA;" OwnerUserId="10997" LastActivityDate="2017-04-25T02:39:20.653" Title="Can a wireless signal of sufficiently high energy damage a wireless transceiver?" Tags="|electrical-engineering|rf-electronics|electromagnetism|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14917" PostTypeId="1" CreationDate="2017-04-22T11:00:20.937" Score="1" ViewCount="116" Body="&lt;p&gt;Attenuation of radio frequency (rf) signals in a medium depends upon medium characteristics and frequency of the signal. How much does human tissue attenuate rf signal of a given frequency?&lt;/p&gt;&#xA;" OwnerUserId="11000" LastEditorUserId="90" LastEditDate="2017-04-22T12:06:21.603" LastActivityDate="2017-07-26T12:48:18.670" Title="How much does human tissue attenuate radio frequency signal of a given frequency?" Tags="|electromagnetism|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14918" PostTypeId="1" CreationDate="2017-04-22T12:05:59.790" Score="0" ViewCount="60" Body="&lt;p&gt;Given the following fictious scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A construction project requires that a big swampy landscape has to be cleared. This swamp landscape has various trees (also trees with a big diameter), a wet region that is a swamp, a lake and some rocks. A construction company has to clear all that things to build a theme park with big coasters and roads on this. Now I am interested in how this company would do this in a cost-saving and also fast way. How costs can be minimized without being too slow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Tree felling with chainsaw and then stump removal with excavator or pure removal of trees with excavators and/or bulldozers. The second version (version without felling trees before) is faster, but may require stronger machines (more costs?)&lt;/li&gt;&#xA;&lt;li&gt;Rocks can be removed by excavators if rocks are small. Larger rocks must be blasted.&lt;/li&gt;&#xA;&lt;li&gt;Excavators begin to dig holes around the swamp and the lake. &lt;/li&gt;&#xA;&lt;li&gt;Now water can be removed from the lake and the swamp by a pipe inserted between the wet region and the holes.&lt;/li&gt;&#xA;&lt;li&gt;The land must be made flat by bulldozers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is this procedure plausible in this fictious (it is not real) scenario?   &lt;/p&gt;&#xA;" OwnerUserId="8941" LastActivityDate="2017-04-22T12:18:59.940" Title="How land clearing would be done in this theoretical situation?" Tags="|civil-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2017-04-28T20:05:52.460" ContentLicense="CC BY-SA 3.0" />
  <row Id="14919" PostTypeId="2" ParentId="14918" CreationDate="2017-04-22T12:18:59.940" Score="1" Body="&lt;p&gt;You plan appears plausible, but does the theme park really need to be located where there is now a lake?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of a lake indicates that there is localized low elevation point on the landscape. Could this pose problems in the future with water inundation, or do you plan to have excellent water drainage infrastructure for such an event?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relocating a lake seems expensive when dry land nearby can be used, if it is available.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-04-22T12:18:59.940" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14921" PostTypeId="1" AcceptedAnswerId="14924" CreationDate="2017-04-22T15:59:03.613" Score="0" ViewCount="92" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YGTY6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YGTY6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4gBX8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4gBX8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made my code like this,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function DK = pendulum_cartesian(t,K)    &#xA;g = 9.8; l = sqrt(K(1)^2 + K(2)^2);  &#xA;DK = zeros(4,1);&#xA;DK(1) = K(3);&#xA;DK(2) = K(4);&#xA;DK(3) = ( (-K(1)*K(3)^2) - (K(1)*K(4)^2) + K(1)*K(2)*g )/(l^2);&#xA;DK(4) = ( (-K(2)*K(3)^2) - (K(2)*K(4)^2) - (K(1)^2)*g )/(l^2);&#xA;&#xA;theta = atan(K(2)/K(1));&#xA;&#xA;plot(t,theta);&#xA;&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and i typed,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[t,K] = ode45(@pendulum_cartesian, [0,10], [0.1,0.00017,0.1,0])&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but there is no value in theta, and plots nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where is a problem?&lt;/p&gt;&#xA;" OwnerUserId="11006" LastActivityDate="2017-04-22T17:31:43.350" Title="i have problem with cartesian coordinates pendulum EOM in matlab" Tags="|matlab|" AnswerCount="1" CommentCount="3" ClosedDate="2017-04-23T17:04:52.313" ContentLicense="CC BY-SA 3.0" />
  <row Id="14922" PostTypeId="2" ParentId="14916" CreationDate="2017-04-22T16:03:43.147" Score="2" Body="&lt;p&gt;Yes, but it's unlikely. Damage from other frequencies will be limited because typical receivers have a pre-select filter after the antenna to filter out other frequencies. In band signals pass through a Low Noise Amplifier (LNA) directly after the antenna, which can saturate if the signal level is high, and fail if the signal level exceeds its maximum input power spec. The most common approach is to put a limiter in front of the LNA to handle any signals that may damage the LNA. However this degrades the Noise Figure of the receiver by the limiter's insertion loss. Another way to prevent overloading is to detect the input power using a power detector and routing it past the sensitive electronics if it exceeds some threshold. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short answer: yes, any signal can damage the device if sufficiently high energy. However, wifi routers are not usually operating in environments where high power survivability is desired - the focus is more on range and sensitivity. Other sensors such as military radars are designed with robust power survivability specs to deal with jammers, self-reflections, and other electronic countermeasures.&lt;/p&gt;&#xA;" OwnerUserId="302" LastActivityDate="2017-04-22T16:03:43.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14924" PostTypeId="2" ParentId="14921" CreationDate="2017-04-22T17:31:43.350" Score="0" Body="&lt;p&gt;Assuming your equations and your code is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I used for the function:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function DK = pendulum_cartesian(t,K)    &#xA;DK = zeros(4,1);&#xA;DK(1) = K(3);&#xA;DK(2) = K(4);&#xA;DK(3) = ( (-K(1)*K(3)^2) - (K(1)*K(4)^2) + K(1)*K(2)*9.81 )/(K(1)^2 + K(2)^2);&#xA;DK(4) = ( (-K(2)*K(3)^2) - (K(2)*K(4)^2) - (K(1)^2)*9.81 )/(K(1)^2 + K(2)^2);&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Then run with:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[t,K] = ode45(@pendulum_cartesian, [0,10], [0.1;0.00017;0.1;0]);&#xA;theta = atan(K(:,2)/K(:,1));&#xA;plot(t,theta);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-04-22T17:31:43.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14925" PostTypeId="2" ParentId="14849" CreationDate="2017-04-22T17:50:36.733" Score="1" Body="&lt;p&gt;In theory, we can get arbitrarily close to 100%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, the higher the efficiency you want, the more the generator will cost.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2017-04-22T17:50:36.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14928" PostTypeId="1" AcceptedAnswerId="14936" CreationDate="2017-04-23T00:34:21.947" Score="-1" ViewCount="122" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rag1B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rag1B.png&quot; alt=&quot;Question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) I dont understand how to take the integration like its integral of ft from 0 to 2 pi = integral of velocity, and the integral of the first is the area under the graph= 8j and then I dont know how to get the answer fron there cause  dont get what to plug in ?&#xA;Any help is greatly appresiated. &lt;/p&gt;&#xA;" OwnerUserId="11012" LastEditorUserId="11012" LastEditDate="2017-04-24T05:36:04.267" LastActivityDate="2017-04-24T05:36:04.267" Title="Theory on tangential acceleration" Tags="|dynamics|acceleration|" AnswerCount="1" CommentCount="4" ClosedDate="2017-04-24T01:27:51.063" ContentLicense="CC BY-SA 3.0" />
  <row Id="14929" PostTypeId="1" AcceptedAnswerId="14953" CreationDate="2017-04-23T04:31:44.683" Score="3" ViewCount="446" Body="&lt;p&gt;I'm trying to use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Impedance_analogy&quot; rel=&quot;nofollow noreferrer&quot;&gt;mechanical impedance analogy&lt;/a&gt; to describe a spring mass damper system as an electrical circuit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanical Impedances for linear motion are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Z_{lin}(s) = \frac{F(s)}{\dot X(s)} = Ms + B + \frac{K}{s} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where M has units of kg, B has units of N-s/m and K has units N/m. The units of mechanical impedance then are N-s/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I translate this to the rotational domain, I start having problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotational impedance will again be the ratio of an effort (torque) to a flow (rotational velocity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Z_{rot}(s) = \frac{\tau(s)}{\omega(s)} = Js + B_{rot} + \frac{K_{rot}}{s} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on dividing torque by angular velocity, the units of rotational impedance are [N-m-s/rad], or $\frac{ML^2}{T}$. When I try to verify that the units of rotational impedance are consistent, I get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J\cdot s = \frac{kg\cdot m^2\cdot rad}{s} = N\cdot m\cdot s \cdot rad = \frac{ML^2}{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B = N\cdot m\cdot\frac{s}{rad} =\frac{ML^2}{T} $$ &#xA;and &#xA;$$\frac{K}{s} = \frac{N\cdot m\cdot s}{rad^2}=\frac{ML^2}{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't sleep with all those radians being all over the place! It seems awfully inconsistent to have them in different place and I haven't been able to find any sources on the web or in my textbooks that adequately deal with this since mechanical impedance is such a niche topic.   In doing this, I've assumed that the $s$ has units of [rad/s], which some people dislike and just use $s= \frac{1}{T}$ Is there a way for me to make this a little more consistent, possibly by using radians in the units for rotational inertia and the spring constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also looked at the discussions here:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/36079/unit-of-torque-with-radians&quot;&gt;https://physics.stackexchange.com/questions/36079/unit-of-torque-with-radians&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/11500/simple-harmonic-motion-what-are-the-units-for-omega-0&quot;&gt;https://physics.stackexchange.com/questions/11500/simple-harmonic-motion-what-are-the-units-for-omega-0&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/33542/why-are-radians-more-natural-than-any-other-angle-unit&quot;&gt;https://physics.stackexchange.com/questions/33542/why-are-radians-more-natural-than-any-other-angle-unit&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11013" LastActivityDate="2017-04-24T13:56:51.473" Title="Units for rotational impedance" Tags="|mechanical-engineering|control-engineering|robotics|circuit-design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14930" PostTypeId="1" AcceptedAnswerId="14934" CreationDate="2017-04-23T05:30:47.860" Score="2" ViewCount="105" Body="&lt;p&gt;I am looking at Hibbeler's section on a `Cable Subjected to a Distributed Load' in Engineering Mechanics - Statics, Ed13, as shown in the attached images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GlUhl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GlUhl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EiveO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EiveO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am getting stuck deriving equation (7-7). I derive the horizontal equilibrium equation just fine;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;|\vec{T} + \vec{\Delta T}| \cdot \cos(\theta + \Delta \theta) - |\vec{T}|\cdot \cos(\theta) = 0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I rearrange to describe horizontal tension component as a function of $\Delta x$ (given $\Delta T$ and $\Delta \theta$ are ultimately functions of $\Delta x$);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;|\vec{T}|\cdot \cos(\theta) = |\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I express the differential;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{\Delta(|\vec{T}|\cdot \cos(\theta))}{\Delta x}&#xA;=&#xA;\frac{|\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)}{\Delta x}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then when I go to express the derivative, I come up with a limit which does not exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d(|\vec{T}|\cdot \cos(\theta))}{dx}&#xA;=&#xA;\lim_{\Delta x \to 0}\frac{|\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)}{\Delta x}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further to this, it is not obvious to me how the resultant force from the distributed load acting over $\Delta x$ can be defined as $w(x)\Delta x$ (in line 8) as the author suggests. I feel it should be defined as;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\Delta F = \int_{x}^{x+\Delta x}w(x)dx&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;becuase the loading intensity function is not constant. Any pointers as to where my errors are? Appreciate any advice offered. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="1992" LastEditorUserId="1992" LastEditDate="2017-04-23T06:00:47.950" LastActivityDate="2017-04-23T13:28:22.620" Title="Expressing Derivative of Cable Tension Component" Tags="|structural-engineering|structural-analysis|structures|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14934" PostTypeId="2" ParentId="14930" CreationDate="2017-04-23T10:04:25.820" Score="2" Body="&lt;p&gt;You have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-T\cos \theta + (T+\Delta T) \cos( \theta + \Delta \theta)=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to note that $T$ and $\theta$ are both depending on $x$. Dividing the previous expression by $\Delta x$ and rewriting $T+\Delta T = T(x+\Delta x)$ and $\theta + \Delta \theta = \theta(x+\Delta x)$, will lead to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{T(x+\Delta x)\cos [\theta(x+\Delta x)]-T(x)\cos[\theta(x+\Delta x)]}{\Delta x}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take the limit $\Delta x \to 0$. As the above is the definition of the derivative of $T(x)\cos[\theta(x)]$ we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{d}{d x}\left[T(x)\cos[\theta (x)]\right]=0 \implies T(x)\cos[\theta (x)] = \text{const.}$$&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-04-23T13:28:22.620" LastActivityDate="2017-04-23T13:28:22.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14935" PostTypeId="1" CreationDate="2017-04-23T10:27:02.827" Score="-1" ViewCount="1667" Body="&lt;p&gt;Will an electronic device,say a smartphone or wifi router, get damaged if there is an arc welding going on in the vicinity? I am not talking about damage from direct contact with the fire/spark, but the electromagnetic pulse.&lt;/p&gt;&#xA;" OwnerUserId="11018" LastActivityDate="2017-04-23T13:24:46.363" Title="Effect of arc welding on nearby electronic devices" Tags="|electromagnetism|" AnswerCount="1" CommentCount="4" ClosedDate="2017-05-11T16:33:29.333" ContentLicense="CC BY-SA 3.0" />
  <row Id="14936" PostTypeId="2" ParentId="14928" CreationDate="2017-04-23T11:01:57.123" Score="1" Body="&lt;p&gt;For the first question, you can split up the equations of motion into normal and tangential direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ma_n=F_n$$&#xA;$$ma_t=F_t.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tangential force $F_t=12/\pi \text{ N}$. Hence, $a_t=\frac{12}{10\pi} \frac{\text{m}}{\text{s}^2}$. For the normal force (centripetal force) you have to remember that $F_n=m\frac{v^2}{r}$. Hence, $a_n=\frac{v^2}{r}=1\frac{\text{m}}{\text{s}^2}$. The total acceleration is given by $a=\sqrt{a_t^2+a_n^2}=1.07\frac{\text{m}}{\text{s}^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, now it's up to you. Try the other questions and tell us what you have tried or where you got stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;For b) we note that $$ma_t=F_t\implies \frac{F_t}{m}=a_t=\dfrac{dv}{dt}.$$&#xA;This can be rewritten by the chain rule (Note, that $ds/dt=v$) as:&#xA;$$\frac{F_t}{m}=\dfrac{dv}{ds}\dfrac{ds}{dt}=v\dfrac{dv}{ds}\implies \int_{s=0}^{2\pi}F_t ds=m\int_{v_0}^{v_{2\pi}}vdv.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral of on the right-hand side is just the area in your $s-F_t$-diagram. It turns out to be $8 \text{ J}$. Can you continue from here?&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-04-23T17:22:29.507" LastActivityDate="2017-04-23T17:22:29.507" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="14939" PostTypeId="2" ParentId="14935" CreationDate="2017-04-23T13:24:46.363" Score="1" Body="&lt;p&gt;The arc does produce a certain amount of RF interference which can certainly disrupt electronic devices in some circumstances but won't usually cause any permanent damage. Some welding machines have additional earthing points on the chassis which can be used to mitigate this if it is a problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TIG welders often use a high frequency pulse to make it easier to start the arc this can be a problem for sensitive equipment nearby and some machines have the facility to disable the HF start. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you are dealing with equipment which is especially sensitive you are only likely to cause actual damage if the device is electrically connected to the structure that is being welded (eg the ecu of a vehicle). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Welders, especially transformer based ones can also induce noise and voltage spikes in the circuit they are connected to. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-04-23T13:24:46.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14944" PostTypeId="1" AcceptedAnswerId="14960" CreationDate="2017-04-23T19:32:15.217" Score="1" ViewCount="1279" Body="&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following 2D rigid body in the $x$-$y$-plane, which is rigidly rotated around the $z$-axis with the angle $\beta$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hzwQ2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hzwQ2.png&quot; alt=&quot;Rigid Body Rotation - Kinematics&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(Image Source: &lt;a href=&quot;http://www.continuummechanics.org/smallstrain.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Continuum Mechanics - Small Scale Strains - Limitations of Small Strain Equations - Effect of Rotations on Strains&lt;/a&gt;)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A point $[X, Y]^T$ will then be rotated to a new point&#xA;$$[x, y]^T = \ldots$$&#xA;$$\ldots = [X \cdot \cos(\beta) - Y \cdot \sin(\beta), \ldots$$&#xA;$$\ldots X \cdot \sin(\beta) + Y \cdot \cos(\beta)]^T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engineering strains for this rigid body rotations can thus be given as&#xA;$$\epsilon_{xx} = \cos(\beta)-1$$&#xA;$$\epsilon_{yy} = \cos(\beta)-1$$&#xA;$$\epsilon_{xy} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that normal strains will be induced for (large) rigid body deformations. For $\beta=90°$, they will be $\epsilon_{xx} = \epsilon_{yy} = -1$. For $\beta=180°$, they will be $\epsilon_{xx} = \epsilon_{yy} = -2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a text book, I found a corresponding picture, which shows a FEM simulation based on engineering strains. As a result of the rigid body rotation, the body grew in size:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3HVUx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3HVUx.png&quot; alt=&quot;Rigid Body Rotation - FEM&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to correspond to a &lt;a href=&quot;https://de.wikipedia.org/wiki/Datei:Drehung.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;gif&lt;/a&gt; I found on the German Wikipedia site (here a static version) showing the von Mises stresses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DCe6o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DCe6o.png&quot; alt=&quot;Rigid Body Rotation - von Mises stresses&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does the body grow in size instead of getting smaller? The normal strains are always smaller than zero, which means that the stresses are also smaller than zero (for a linear stress-strain-relationship). Would that not entail compression?&lt;br&gt;&#xA;What do I miss?&lt;/p&gt;&#xA;" OwnerUserId="11025" LastEditorUserId="11025" LastEditDate="2017-04-24T05:40:51.720" LastActivityDate="2017-04-24T23:22:04.427" Title="Rigid Body Rotations and Engineering Strain" Tags="|stresses|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14945" PostTypeId="2" ParentId="14915" CreationDate="2017-04-24T02:04:16.493" Score="1" Body="&lt;p&gt;The term &quot;out-of-order&quot; here refers to &quot;out of order execution&quot;.  e.g. if your program has instruction A, then instruction B, then instruction C, it some situations it may turn out that if you execute them in the order A, C, B (instead of A, B, C), you get the exact same result but the program executes quicker.  An example of when this would happen is if instruction B requires data to be fetched from RAM but it will take some time to go fetch the data.  Instead of sitting around doing nothing while waiting for the data to arrive, the process can save time by executing instruction C while it is waiting (assuming, of course, that instruction C does not need the result of instruction B). Keeping track of what instructions have been re-ordered in what ways requires some memory space. i.e. the processor needs to remember that it put B on hold while waiting for data, so that it can start back up again when the data is ready. That is the buffers that this is referring to.   Note: this is a &lt;em&gt;very&lt;/em&gt; simplified explanation.  The reality of keeping track of everything is more complicated.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Out-of-order_execution&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Out-of-order_execution&lt;/a&gt; for some more details.   &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-04-24T02:04:16.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14946" PostTypeId="2" ParentId="14906" CreationDate="2017-04-24T02:25:34.150" Score="0" Body="&lt;p&gt;I agree with the commenters that this would be better asked on a different forum.  But here is what I know: Wikipedia basically agrees with your textbook. &lt;a href=&quot;https://en.wikipedia.org/wiki/Instruction_pipelining&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Instruction_pipelining&lt;/a&gt;  The total amount of work to complete each instruction is always the same, regardless of the number of stages. As you add stages to the pipeline, each indidvidual stage becomes simpler. Therefore, it can run quicker. Therefore the clock speed of the processor can be increased.  So the overall processor is faster.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-04-24T02:25:34.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14951" PostTypeId="1" AcceptedAnswerId="14954" CreationDate="2017-04-24T13:12:31.937" Score="2" ViewCount="2002" Body="&lt;p&gt;While reading about airplane engines on the internet, it is common to find figures of their thrust, in kilo Newton. However, in the case of ships and naval vessels, when talking about the engine, the only parameter mentioned is the horsepower of the power plant driving the propeller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is, what is the typical thrust generated by a large naval ship (such as the nuclear-powered USS Enterprise aircraft carrier, with a power of 210 MW)? How can this be calculated?&lt;/p&gt;&#xA;" OwnerUserId="11036" LastActivityDate="2017-04-24T18:29:27.710" Title="Thrust generated by a ship" Tags="|ships|thrust|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14953" PostTypeId="2" ParentId="14929" CreationDate="2017-04-24T13:56:51.473" Score="9" Body="&lt;h2&gt;Radians and unit analysis&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The comment by &lt;a href=&quot;https://engineering.stackexchange.com/users/7627/mryoumath&quot;&gt;MrYouMath&lt;/a&gt; is correct. For the purposes of dimensional analysis, the radian is considered a dimensionless quantity and therefore does not need to be included. This is because the radian is defined by an arc length (L) divided by a radius (also L) to give (L/L). The units of the Laplace domain complex number, $s$, are technically rad/s which, as the OP states, are considered to be $1/T$ for dimensional analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ J\cdot s = \left[\frac{N\cdot m \cdot s \cdot rad}{rad}\right] = \frac{ML^2}{T}\left(\frac{L}{L}\right)\left(\frac{L}{L}\right) = \frac{ML^2}{T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from a dimensional analysis perspective we are fine. You can double check the units for the other terms as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, radians are often ignored/assumed because they are so convenient (you don't end up with a bunch of constant factors in your equations). In engineering, radians serve mostly as a reminder that we are using them instead of degrees! So you can ignore them until you are figuring out what the units of your final answer are.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Mechanical impedance to blame?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The reason that the units don't work out has nothing to due with the fact that this is a mechanical system. Check out the units for the transfer function of an RLC circuit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{V}{I} = Ls + R + \left(\frac{1}{C}\right)\frac{1}{s} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[ \frac{kg\cdot m^2}{s^3 \cdot A^2}\right] = \left[\frac{kg \cdot m^2}{s^2 \cdot A^2}\right]\cdot\left[\frac{rad}{s}\right] + \left[\frac{kg\cdot m^2}{s^3\cdot A^2}\right] + \left[\frac{kg\cdot m^2}{s^3\cdot A^2}\right]\cdot\left[\frac{s}{rad}\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice once again those pesky radians sticking their noses where they don't belong! In all seriousness though, you will see this &quot;problem&quot; regardless of what domain you work in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact of the matter is simply this: when doing unit analysis in the Laplace domain (transfer functions, etc.), ignore the radian units for the complex number $s$. If you do this, the units &lt;em&gt;will&lt;/em&gt; work out as long as you use consistent units for your physical constants $J$, $B$ and $K$ (you'll have to check it for the mechanical system yourself, I've wasted far too much time at work already).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final note: Just remember that when you do your frequency analysis the units of frequency are still rad/s (because we use $\sin{\omega t}$ instead of $\sin{2\pi ft}$).&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2017-04-24T13:56:51.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14954" PostTypeId="2" ParentId="14951" CreationDate="2017-04-24T16:04:38.507" Score="1" Body="&lt;p&gt;If you know the Froude efficiency $\eta_{\text{Fr}}$ and assume a constant velocity $u$ for the ship. Then you can calculate the thrust $F_{\text{T}}$ for given power at the shaft $P_{\text{S}}$ as $$F_{\text{T}}=\eta_{\text{Fr}}\frac{P_{\text{S}}}{u}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a dimensionless thrust coefficient $C_{\text{T}}$ which is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{\text{T}}=\frac{F_{\text{T}}}{1/2\rho v^2 (D/2)^2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which $\rho$ is the density of the fluid (here: water) and $D$ is the diameter of the screw. $v = u(1-w)$ in which $w$ is the so-called &lt;a href=&quot;https://www.quora.com/What-is-the-wake-fraction-of-a-propeller-in-a-ship?share=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;wake fraction&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-04-24T16:30:24.250" LastActivityDate="2017-04-24T16:30:24.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14955" PostTypeId="1" CreationDate="2017-04-24T16:11:31.463" Score="1" ViewCount="277" Body="&lt;p&gt;Our professor has given us a review for a upcoming exam. But he has never done anything like this question and it has stumped a quite few of us. We are hoping someone can walk us through this so we can understand, as he will not assist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A modular two amplifier system consists of a first stage amplifier with an open loop gain of 100, an input impedance of 25kΩ and an output impedance of 2kΩ, and a second stage with an input impedance of 2kΩ and an open loop gain of 50 and an output impedance of 500Ω. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the gain of the overall system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a single op-amp, specify a third modular amplifier, using the original two in the order or otherwise, such that the input impedance of the overall system is 20kΩ, the output impedance is 2kΩ, and the magnitude of the overall voltage gain is 2000.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought you could just multiply the gains (gain = - Z2/Z1) of each op amp together, but some classmates disagree with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I approach it with op amps or signal circuit equivalents? Does a voltage divider have anything to do with it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can attach screenshots of circuit diagrams we have tried.&lt;/p&gt;&#xA;" OwnerUserId="11039" LastEditorUserId="8055" LastEditDate="2017-05-01T14:52:12.053" LastActivityDate="2017-05-01T14:52:12.053" Title="Calculating gain of an overall system. Analogue electronics" Tags="|electrical-engineering|circuits|circuit-design|electrical|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14956" PostTypeId="2" ParentId="14944" CreationDate="2017-04-24T16:18:34.827" Score="1" Body="&lt;p&gt;Engineering strains are only valid mathematically for &lt;em&gt;infinitesimally small rotations&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to work with &lt;em&gt;finite&lt;/em&gt; rotations (of arbitrary size) you need to use a different strain measure, such as &lt;a href=&quot;http://www.continuummechanics.org/greenstrain.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Green strain&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-04-24T16:18:34.827" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="14957" PostTypeId="2" ParentId="14951" CreationDate="2017-04-24T18:29:27.710" Score="3" Body="&lt;p&gt;This gives a complete analysis of how to calculate : &lt;a href=&quot;http://web.mit.edu/13.012/www/handouts/propellers_reading.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://web.mit.edu/13.012/www/handouts/propellers_reading.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-04-24T18:29:27.710" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14960" PostTypeId="2" ParentId="14944" CreationDate="2017-04-24T23:22:04.427" Score="1" Body="&lt;p&gt;The question seems to be &quot;why does the FEM seemingly produce results that are contradictory to theory?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a square with nodes &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;id  x    y&#xA;1 -1.0 -1.0&#xA;2  1.0 -1.0&#xA;3  1.0  1.0&#xA;4 -1.0  1.0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The FE displacement field is&#xA;$$&#xA;  u_x(x,y) = \sum_{j=1}^4 u_x^j N_j(x,y) ~,~~&#xA;  u_y(x,y) = \sum_{j=1}^4 u_y^j N_j(x,y)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strain field is&#xA;$$&#xA;  \varepsilon_{xx}(x,y) = \sum_{j=1}^4 u_x^j \frac{\partial N_j(x,y)}{\partial x} ~,~~&#xA;  \varepsilon_{yy}(x,y) = \sum_{j=1}^4 u_y^j \frac{\partial N_j(x,y)}{\partial y}&#xA;$$&#xA;and&#xA;$$&#xA;  \varepsilon_{xy}(x,y) = \tfrac{1}{2}\sum_{j=1}^4 \left[u_x^j \frac{\partial N_j(x,y)}{\partial y} + u_y^j \frac{\partial N_j(x,y)}{\partial x}\right]&#xA;$$&#xA;The nodal shape functions are&#xA;$$&#xA;  N_1(x,y) = \frac{(1-x)(1-y)}{4} ~,~~&#xA;  N_2(x,y) = \frac{(1+x)(1-y)}{4}&#xA;$$&#xA;$$&#xA;  N_3(x,y) = \frac{(1+x)(1+y)}{4} ~,~~&#xA;  N_4(x,y) = \frac{(1-x)(1+y)}{4}&#xA;$$&#xA;The gradients of the shape functions are&#xA;$$&#xA;  \begin{align}&#xA;   G_{1x} := \frac{\partial N_1(x,y)}{\partial x} &amp;amp; = -\frac{(1-y)}{4} ~,~~&#xA;   G_{1y} := \frac{\partial N_1(x,y)}{\partial y} = -\frac{(1-x)}{4} \\&#xA;   G_{2x} := \frac{\partial N_2(x,y)}{\partial x} &amp;amp; = \frac{(1-y)}{4} ~,~~&#xA;   G_{2y} := \frac{\partial N_2(x,y)}{\partial y} = -\frac{(1+x)}{4} \\&#xA;   G_{3x} := \frac{\partial N_3(x,y)}{\partial x} &amp;amp; = \frac{(1+y)}{4} ~,~~&#xA;   G_{3y} := \frac{\partial N_3(x,y)}{\partial y} = \frac{(1+x)}{4} \\&#xA;   G_{4x} := \frac{\partial N_4(x,y)}{\partial x} &amp;amp; = -\frac{(1+y)}{4} ~,~~&#xA;   G_{4y} := \frac{\partial N_4(x,y)}{\partial y} = \frac{(1-x)}{4}&#xA;  \end{align}&#xA;$$&#xA;Therefore the strain field is&#xA;$$&#xA;  \begin{bmatrix}  &#xA;   \varepsilon_{xx}(x,y) \\ \varepsilon_{yy}(x,y) \\ 2\varepsilon_{xy}(x,y) &#xA;  \end{bmatrix}&#xA; = \begin{bmatrix} &#xA; G_{1x} &amp;amp; G_{2x} &amp;amp; G_{3x} &amp;amp; G_{4x} &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xA; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; G_{1y} &amp;amp; G_{2y} &amp;amp; G_{3y} &amp;amp; G_{4y} \\&#xA; G_{1y} &amp;amp; G_{2y} &amp;amp; G_{3y} &amp;amp; G_{4y} &amp;amp; G_{1x} &amp;amp; G_{2x} &amp;amp; G_{3x} &amp;amp; G_{4x}&#xA;   \end{bmatrix} \mathbf{u}&#xA; $$&#xA; where&#xA; $$&#xA;   \mathbf{u} = \begin{bmatrix}&#xA; u_x^1 &amp;amp; u_x^2 &amp;amp; u_x^3 &amp;amp; u_x^4 &amp;amp; u_y^1 &amp;amp; u_y^2 &amp;amp; u_y^3 &amp;amp; u_y^4&#xA;   \end{bmatrix}^T&#xA; $$&#xA;Consider the case where the square is rotated by 90 degrees clockwise around node 1 so that the new positions of the nodes are&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;id  x    y&#xA;1 -1.0 -1.0&#xA;2 -1.0 -3.0&#xA;3  1.0 -3.0&#xA;4  1.0 -1.0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note that we are not assuming any deformation of the square.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the displacements are&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;id  u_x  u_y&#xA;1  0.0 0.0&#xA;2 -2.0 -2.0&#xA;3  0.0 -4.0&#xA;4  2.0 -2.0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If we plug in the values, at node 1 we get&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;B1 = [[G1x(1) G2x(1) G3x(1) G4x(1) 0 0 0 0];...&#xA;     [0 0 0 0 G1y(1) G2y(1) G3y(1) G4y(1)];...&#xA;     [G1y(1) G2y(1) G3y(1) G4y(1) G1x(1) G2x(1) G3x(1) G4x(1)]]&#xA;   =   -0.50000   0.50000   0.00000  -0.00000   0.00000   0.00000 0.00000   0.00000&#xA;       0.00000   0.00000   0.00000   0.00000  -0.50000  -0.00000 0.00000   0.50000&#xA;       -0.50000  -0.00000   0.00000   0.50000  -0.50000   0.50000 0.00000  -0.00000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and the nodal displacement vector is&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;u = [0 -2 0 2 0 -2 -4 -2]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The strain at node 1 is then &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;eps1 = B1*u' = [-1 -1 0]'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;strong&gt;the finite element solution is identical to your solution&lt;/strong&gt; and just says that &lt;strong&gt;stresses will develop in the element due to pure rigid body rotation even if the element does not deform&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finite element method is typically implemented in displacement form. So you specify the displacements and then find the stresses in the element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative would be to specify moments that would rotate the element.  That's tricky to do.  I'm more inclined to believe that the figure in the textbook is just meant to be an illustration.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2017-04-24T23:22:04.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14962" PostTypeId="2" ParentId="14916" CreationDate="2017-04-25T02:39:20.653" Score="0" Body="&lt;p&gt;of course.  Ronald Reagan's &quot;Star Wars&quot; missile defense is designed to damage the electronic navigation systems on board these missiles.  The electromagnetic pulse from a nuclear explosion will certainly deliver enough energy to damage much of the electronic devices within its radius. Satellites, which are transceivers, can be damaged by solar flares and coronal mass ejections.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-04-25T02:39:20.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14964" PostTypeId="2" ParentId="14748" CreationDate="2017-04-25T03:27:19.033" Score="1" Body="&lt;p&gt;You really can't change the frequency of alternating current.  This is fixed at 60 hertz (cycles/sec) for USA and 50 hertz most everywhere else.  by laminating the permanent magnets you reduce losses to hall effect currents on the surface. Most commercial motors have already made the most efficient design possible.  You can always improve but usually only getting marginal improvements for increased costs.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-04-25T03:27:19.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14965" PostTypeId="2" ParentId="14748" CreationDate="2017-04-25T05:06:42.123" Score="1" Body="&lt;p&gt;A possible avenue you could look at is the motor controllers that match power to load.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-04-25T05:06:42.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14967" PostTypeId="1" CreationDate="2017-04-25T17:51:20.723" Score="2" ViewCount="51" Body="&lt;p&gt;I am working on a burner management system for a gas/coal fired pulp dryer. We are providing a thermocouple temperature switch on the roof of the firebox for an Excess Temperature Limit Interlock (per NFPA86 8.18.2). The code is ambiguous and the manufacturer of this old dryer has long been out of business. Some sources are recommending the temperature be measured in the refractory, while others are recommending direct measurement. Below the surface increases the reliability of the sensor while exposed gives you quicker response and slightly more accurate data.&lt;/p&gt;&#xA;&lt;p&gt;Is it industry best practice (and/or code compliant) to install the thermocouple just below the fire side of the refractory surface, and calibrate for the small temperature reduction or should the sensor extend into the firebox?&#xA;&lt;br&gt;&lt;br&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;NFPA86 excerpts:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;8.18.2.6* The temperature-sensing components of the excess temperature limit interlock shall be rated for the temperature and atmosphere to which they are exposed.&lt;/p&gt;&#xA;&lt;p&gt;8.18.2.7* The temperature-sensing element of the excess temperature limit interlock shall be located where recommended by the oven manufacturer or designer.&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&amp;quot;A.8.19.8 The sensing element should be positioned where&#xA;the difference between the temperature control sensor and&#xA;the excess temperature limit sensor is minimized. The&#xA;temperature-sensing element of the excess temperature limit&#xA;interlock should be located where it will sense the excess temperature&#xA;condition that will cause the first damage to the furnace&#xA;or work as temperatures within the furnace rise above the&#xA;maximum operating set point most critical to safe operation.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-04-25T17:51:20.723" Title="Placement of a refractory wall excess temperature sensor?" Tags="|control-engineering|heat-transfer|sensors|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14968" PostTypeId="1" CreationDate="2017-04-25T17:51:44.047" Score="0" ViewCount="143" Body="&lt;p&gt;I have lived overseas for many years and purchased your standard appliances.  Anyway moving back to the US, I had to purchase transformers to bring the voltage down from 240V to 120V.  I know that there also issues with cycles (50cycles verse 60) but that not an issue yet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is I am trying to figure out the electrical cost of running these appliances so I can make a decision of when the return on investment is right to replace these appliances with ones that run off 120V.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. I have a 3000W transformer that runs an appliance - which uses 900W for two heaters and 600W for a pump.  I am paying about \$0.06 per kW-hour.  A new appliance is about \$325.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have about 6 240V various appliances so it could drop my electrical bill.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks &lt;/p&gt;&#xA;" OwnerUserId="11048" LastEditorUserId="163" LastEditDate="2017-05-28T08:05:01.113" LastActivityDate="2018-07-22T16:13:24.617" Title="Cost Value of Changing appliances that run on 240 down to 120" Tags="|electrical-engineering|electrical|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14969" PostTypeId="2" ParentId="14968" CreationDate="2017-04-25T18:37:11.713" Score="1" Body="&lt;p&gt;First, I would like to mention that you can find 240V outlets in the U.S. intended for residential use by electric stoves and dryers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, it is sometimes possible to run a 240V appliance using 120V at the expense of cooking time (given that it is purely an analog device). For example, a 240V water kettle with a given resistance heating element will perform the same job using 120V, but at 1/4 power output. This means it will also take 4 times longer to boil the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, its difficult to answer this question without knowing more details about the transformers and appliance loads. All transformers have some losses and its possible to calculate these if you have the parameters of the transformer. Otherwise, its best to approximate the efficiency of the transformer using the labeled rating (output power / input power). Some hardware tools like the kill-a-watt exist that may allow you to empirically determine the efficiency. I suggest you look into that to calculate the ROI of eliminating the transformers.&lt;/p&gt;&#xA;" OwnerUserId="11049" LastActivityDate="2017-04-25T18:37:11.713" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14971" PostTypeId="1" CreationDate="2017-04-25T20:56:38.513" Score="3" ViewCount="108" Body="&lt;p&gt;I have read the Brunel's shield directly exposed ground which was then excavated by the workers. But the construction was done just a few meters under the seabed. How wasn't the tunnel flooded immediately?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think of small pits that children dig on beaches in the sand near the water. As soon as water level is reached, the hole is flooded very quickly. Shouldn't the same have happened to the tunnel? Was the soil water-tight enough so that the water didn't seep so fast?&lt;/p&gt;&#xA;" OwnerUserId="11053" LastActivityDate="2017-04-26T13:04:07.193" Title="How wasn't Thames Tunnel flooded during construction?" Tags="|waterproofing|tunnels|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14973" PostTypeId="2" ParentId="14615" CreationDate="2017-04-26T05:55:00.613" Score="-1" Body="&lt;p&gt;yes it can be greater than the static base shear, and in this case dont scale &lt;/p&gt;&#xA;" OwnerUserId="11057" LastActivityDate="2017-04-26T05:55:00.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14974" PostTypeId="1" CreationDate="2017-04-26T06:06:48.263" Score="3" ViewCount="1627" Body="&lt;p&gt;According to ACI the stiffness modifier for moment of inertia for flat plate and flat slabs is 0.25 (ACI318/318R-02 10.11.1) to mitigate for the cracked section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in a prestressed slab (post-tensioned) the crack is controlled can the stiffness modifiers be taken higher than 0.25?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor suggests that it could be taken between 0.6-0.8, can anyone suggest any article or document or even a discussion about this topic to be sure about this problem ?&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="1832" LastEditDate="2017-04-26T11:35:08.583" LastActivityDate="2017-09-29T22:30:00.497" Title="What is the stiffness modifiers for a prestressed slab?" Tags="|structural-engineering|civil-engineering|structural-analysis|prestressed-concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14975" PostTypeId="1" AcceptedAnswerId="14976" CreationDate="2017-04-26T10:57:48.177" Score="0" ViewCount="127" Body="&lt;p&gt;I currently have a ditch that fills with water draining out into an area  lower than the water (via a hose), but the hose for most of the travel extend above the hight of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the hose stays filled with water, which it should assuming the water doesn't evaporate enough, this system works. Unfortunately, the water doesn't drain as quickly as I'd like after a rain storm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does the piping at all effect (e.g. the volume it holds) at what level the the basin will drain to?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If I increase the diameter of the piping will the volume of water flowing out per minute increase?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus points for providing the physics principles being applied here. I'm not familiar with the names.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BOyBv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BOyBv.png&quot; alt=&quot;basin drainage diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9432" LastActivityDate="2017-04-29T21:55:33.590" Title="Will water level of remain level between two points" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14976" PostTypeId="2" ParentId="14975" CreationDate="2017-04-26T13:01:58.907" Score="5" Body="&lt;p&gt;&lt;strong&gt;Does the piping at all effect (e.g. the volume it holds) at what level the basin will drain to?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outlet level of the pipe work determines the minimum level the basin will drain to.  At this point, the water is static and reaches its natural level, which will be to just fill the discharge (any excess will overflow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If I increase the diameter of the piping will the volume of water flowing out per minute increase?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes.  The higher the velocity, the greater the head loss through the pipe.  The head loss is fixed by the difference between the water levels in the basin and the discharge pipe.  So for a fixed, head loss, you want to reduce the velocity to increase the flow rate.  To do this, you need to increase the diameter of the pipe.&lt;/p&gt;&#xA;" OwnerUserId="10240" LastActivityDate="2017-04-26T13:01:58.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14977" PostTypeId="2" ParentId="14971" CreationDate="2017-04-26T13:04:07.193" Score="3" Body="&lt;p&gt;At the risk of performing a pure lmgtfy, take a look at &lt;a href=&quot;http://www.ikbrunel.org.uk/thames-tunnel&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting quote from that page:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The operation was started in 1825 and was halted a number of times&#xA;  when the river burst in and flooded the workings.  Isambard Brunel was&#xA;  put in charge of operations, and at one time nearly lost his life in a&#xA;  flood.  In November 1827 Marc’s son, Isambard Kingdom Brunel,&#xA;  organised a lavish underwater banquet in the tunnel – to help convince&#xA;  people that it was safe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And as to the basic construction method,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Thames Tunnel, from Wapping to Rotherhithe was the first tunnel&#xA;  under a river anywhere in the world.  It was designed by Marc Brunel&#xA;  (Isambard's father) using a revolutionary tunnel shield which he&#xA;  invented, comprising 36 cells in which a workman was engaged working&#xA;  independently of the others. Propulsion for the device was a screw&#xA;  which drove the device forward in 114mm/4.5in steps(the width of a&#xA;  brick)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-04-26T13:04:07.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14978" PostTypeId="1" CreationDate="2017-04-26T13:40:22.323" Score="1" ViewCount="489" Body="&lt;p&gt;Much of the currently used techniques of scrubbing the carbon dioxide from power plant exhaust uses sophisticated amine and other organic compounds.  These compounds are then forced to release the CO2 with electric current or heat so that the concentrated CO2 can be pumped underground or otherwise sequestered.&#xA;Why not just spray the power plant exhaust with a finely atomized Sodium Hydroxide mist to directly precipitate out the CO2 as Sodium Carbonate or Sodium BiCarbonate?  The carbonates can be sold as carbon credits and/or be useful products.  Sodium Hydroxide can be obtained in quantity cheaply from electrolysis of sea water and other methods. &lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-04-26T14:17:32.373" Title="Why not use Sodium Hydroxide mist to precipitate carbon dioxide from hydrocarbon combustion exhaust" Tags="|mechanical-engineering|fluid-mechanics|chemical-engineering|environmental-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14980" PostTypeId="2" ParentId="14978" CreationDate="2017-04-26T14:17:20.253" Score="2" Body="&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt; we use stupendous amounts of fossile fuels, capturing the carbon would require even more stupendous amounts of sodium hydroxide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of amines is that removing CO2 from amines only requires heat input and the amines can be recycled (though some degradation does occur).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You describe a &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbon_dioxide_scrubber#Sodium_hydroxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;caustic scrubber, chemistry as outlined here&lt;/a&gt;. you would need about 2 mol NaOH / 1 mol CO2 or about 6 kg NaOH per kg Carbon in the fuel. So imagine a coal plant, and for every train car full of coal six train cars of NaOH prills arrive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.researchgate.net/publication/235969094_Carbon_dioxide_equivalent_CO_2_-e_emissions_A_comparison_between_geopolymer_and_OPC_cement_concrete&quot; rel=&quot;nofollow noreferrer&quot;&gt;This paper&lt;/a&gt; (that I only skimmed) estimates the carbon footprint of NaOh tp be 1.9 kg CO2 equivalents (per kg). Caustic scrubbing will not provide a net carbon benefit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that someone, somewhere built a bicarbonate factory using exhaust gas as a CO2 source. The driving consideration re. sizing this plant will be: How much bicarbonate can I sell, not how much electricit do I need =&gt; how much coal do i burn etc.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-04-26T14:17:20.253" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="14981" PostTypeId="2" ParentId="14978" CreationDate="2017-04-26T14:17:32.373" Score="0" Body="&lt;p&gt;I would guess it's because the amines are able to be regenerated, so they don't need to buy more that often. Where as with the sodium hydroxide, it would not be regenerated because the CO2 took it's oxygen. Basically, the amines do not react with the CO2, but rather they absorb it and then are sent to a desorber where the amines and CO2 are separated and the pure amines get reused.&lt;/p&gt;&#xA;" OwnerUserId="11062" LastActivityDate="2017-04-26T14:17:32.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14982" PostTypeId="1" AcceptedAnswerId="14986" CreationDate="2017-04-26T14:49:38.753" Score="1" ViewCount="659" Body="&lt;p&gt;I have a 10 x 10cm square plate of metal I would like to insert into a hole in the wall of plastic water tank (cube shape), I'm planning on 3D printing the tank itself (only 1-2 litres). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the best design to stop water from leaking between the plate and the plastic body? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far all I can think of is to make the hole in the tank smaller (i.e. 8 x 8cm) than the metal sheet and perhaps using a sealant and affix it using screws possibly, but as there's live wiring around it I need to be sure it has the best possible seal. Any design ideas would be much appreciated &lt;/p&gt;&#xA;" OwnerUserId="11064" LastActivityDate="2020-02-27T01:06:10.100" Title="How to water seal a metal plate into the wall of a plastic water tank?" Tags="|materials|design|3d-printing|waterproofing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14983" PostTypeId="1" CreationDate="2017-04-26T14:51:06.937" Score="1" ViewCount="1188" Body="&lt;p&gt;I'll be teaching vector calculus to mechatronics engineers, and I'd like to provide them with industrially relevant examples, especially of the use of vector fields.  Can anyone suggest either an example application that they have personally used or a text that I can use to glean some examples from?&lt;/p&gt;&#xA;" OwnerUserId="11063" LastActivityDate="2017-04-26T14:51:06.937" Title="Industrial applications of vector calculus" Tags="|mathematics|" AnswerCount="0" CommentCount="6" ClosedDate="2017-05-06T22:48:15.563" ContentLicense="CC BY-SA 3.0" />
  <row Id="14984" PostTypeId="2" ParentId="12281" CreationDate="2017-04-26T14:57:29.570" Score="0" Body="&lt;p&gt;Both thermo-electric and photo-electric cells work on the same physical principles.  A photon is absorbed by an electron in the outermost orbital which causes ionization and voltage potential.  The photo-cell is tuned with with doping to semi-conductor substrate to specific wavelength photons for that of light in the visible or UV spectrum whereas thermo-electric cells would be tuned for absorbing photons in the infrared wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this type of hybrid system to work efficiently, there would have to be filters set up to not pass through photons that are not in the wavelengths of interest.  Otherwise we would have photons that are not useful for the electric potential generation striking the cells and just heating them up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will often find filters on photo-cell arrays to increase their efficiency.  Of course different materials and thus different filters will be required for the different wavelengths.    &lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-04-26T14:57:29.570" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14985" PostTypeId="2" ParentId="14840" CreationDate="2017-04-26T15:40:28.427" Score="8" Body="&lt;p&gt;One way of doing this is using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kalman_decomposition&quot; rel=&quot;noreferrer&quot;&gt;Kalman decomposition&lt;/a&gt;. For this you need the reachable and unobservable subspaces. These subspaces can be constructed using the image of the controllability matrix and the kernel of the observability matrix respectively. The controllability matrix and its image, the observability matrix and its kernel are,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathcal{C} = \begin{bmatrix} B &amp;amp; A\,B &amp;amp; A^2 B \end{bmatrix} = &#xA;\begin{bmatrix}&#xA;0 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 2 &amp;amp; 1 \\&#xA;1 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \\&#xA;0 &amp;amp; 1 &amp;amp; 1 &amp;amp; 1 &amp;amp; 2 &amp;amp; 1&#xA;\end{bmatrix},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{im}(\mathcal{C}) = \text{span}\left\{\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}1\\0\\1\end{bmatrix}\right\},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathcal{O} = \begin{bmatrix} C \\ C\,A \\ C\,A^2 \end{bmatrix} = &#xA;\begin{bmatrix}&#xA;1 &amp;amp; 1 &amp;amp; 1 \\&#xA;1 &amp;amp; 3 &amp;amp; 1 \\&#xA;1 &amp;amp; 5 &amp;amp; 1&#xA;\end{bmatrix},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\text{ker}(\mathcal{O}) = \text{span}\left\{\begin{bmatrix}1\\0\\-1\end{bmatrix}\right\}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kalman decomposition can now be found by constructing a similarity transformation $M\, z = x$, such that,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{z} = \underbrace{M^{-1} A\, M}_{A^*}\, z + \underbrace{M^{-1} B}_{B^*}\, u,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y = \underbrace{C\, M}_{C^*}\, z&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $M$ defined as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M = &#xA;\begin{bmatrix}&#xA;\text{im}(\mathcal{C}) \cap \text{ker}(\mathcal{O}) &amp;amp; &#xA;\text{im}(\mathcal{C}) \cap \text{ker}(\mathcal{O})^\complement &amp;amp;&#xA;\text{im}(\mathcal{C})^\complement \cap \text{ker}(\mathcal{O}) &amp;amp;&#xA;\text{im}(\mathcal{C})^\complement \cap \text{ker}(\mathcal{O})^\complement&#xA;\end{bmatrix}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case $\text{im}(\mathcal{C}) \cap \text{ker}(\mathcal{O})$ and $\text{im}(\mathcal{C})^\complement \cap \text{ker}(\mathcal{O})^\complement$ are empty sets, $\text{im}(\mathcal{C})^\complement \cap \text{ker}(\mathcal{O})=\text{ker}(\mathcal{O})$ and $\text{im}(\mathcal{C}) \cap \text{ker}(\mathcal{O})^\complement=\text{im}(\mathcal{C})$, so a possible transformation would be,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M = &#xA;\begin{bmatrix}&#xA;0 &amp;amp; 1 &amp;amp; 1 \\&#xA;1 &amp;amp; 0 &amp;amp; 0 \\&#xA;0 &amp;amp; 1 &amp;amp; -1&#xA;\end{bmatrix}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this yields,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left[\begin{array}{c|c}&#xA;A^* &amp;amp; B^* \\ \hline C^*&#xA;\end{array}\right] = &#xA;\left[\begin{array}{ccc|cc}&#xA;1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \\&#xA;1 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 \\&#xA;0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\&#xA;\hline&#xA;1 &amp;amp; 2 &amp;amp; 0&#xA;\end{array}\right].&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimal representation can be obtained by eliminating the rows and columns corresponding to any of the uncontrollable or unobservable parts so containing $\text{im}(\mathcal{C})^\complement$ or $\text{ker}(\mathcal{O})$: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\text{im}(\mathcal{C}) \cap \text{ker}(\mathcal{O})$ is empty so nothing has be be removed due to this.&lt;/li&gt;&#xA;&lt;li&gt;$\text{im}(\mathcal{C}) \cap \text{ker}(\mathcal{O})^\complement$ is the part we want, so the next two rows and columns should be skipped.&lt;/li&gt;&#xA;&lt;li&gt;$\text{im}(\mathcal{C})^\complement \cap \text{ker}(\mathcal{O})$ has a dimension of one, so the third row and column should be removed.&lt;/li&gt;&#xA;&lt;li&gt;$\text{im}(\mathcal{C})^\complement \cap \text{ker}(\mathcal{O})^\complement$ is empty so nothing has be be removed due to this.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the minimal representation could look like this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left[\begin{array}{c|c}&#xA;A^*_m &amp;amp; B^*_m \\ \hline C^*_m&#xA;\end{array}\right] = &#xA;\left[\begin{array}{cc|cc}&#xA;1 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \\&#xA;1 &amp;amp; 1 &amp;amp; 0 &amp;amp; 1 \\&#xA;\hline&#xA;1 &amp;amp; 2&#xA;\end{array}\right].&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A way to check whether you did not make any mistakes along the way is to calculate the corresponding transfer function, which should be the same as the one of the original system,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;G(s) = C (s\,I - A)^{-1} B = C^*_m (s\,I - A^*_m)^{-1} B^*_m = \begin{bmatrix}\frac{s + 1}{(s - 1)^2} &amp;amp; \frac{2}{s - 1}\end{bmatrix}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="1765" LastEditorUserId="1765" LastEditDate="2017-10-27T18:49:52.833" LastActivityDate="2017-10-27T18:49:52.833" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="14986" PostTypeId="2" ParentId="14982" CreationDate="2017-04-26T15:44:24.600" Score="2" Body="&lt;p&gt;A common way to do this is to use mechanical fixings (bolts, rivets etc) and a flexible or semi-flexible gasket to provide sealing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the mechanical fasteners physically hold the plate on and also compresses the gasket between the two materials to form a good seal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bolts are ideal for this as they allow quite fine adjustment of the clamping force so that you can get even pressure around the seal. If the tank material is relatively flexible it may also help to sandwich it between two plates to event and warping of the joint, although the second plate can be reduced to a frame or collar. If the tank is more or less as rigid as the plate washers may be adequate. Of course if you are 3D printing it you also have the option to thicken the material around the hole. you may also want a bit of extra material to allow you to grind/sand it flat to get a flat, smooth mating surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a variety of gasket materials including sheet, O-rings and compounds which can be applied as a paste, putty or liquid. For example neoprene sheet is fairly commonly used for water tank fittings. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-04-26T15:44:24.600" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14987" PostTypeId="1" CreationDate="2017-04-26T16:12:24.800" Score="0" ViewCount="1547" Body="&lt;p&gt;I have a project to convert wind energy directly to electrical energy with no moving parts.  The wind will enter a reducing tunnel for concentration, then go through an unobstructing de-humidifier, then through a multi-plate sandwich type electrode setup like that of a capacitor where the voltage would be very high to ionize the air passing through.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This ionized air molecule would then pass through a magnetic field to generate electricity.  A fraction of the electricity would be used to ionize the incoming stream again. I realize that the less the spacing between the capacitor plates, and the greater the voltage, the better the  ionization.  I was wondering what would be optimal plate spacing and voltage for dry air at STP, if a cylindrical plate arrangement would be better than rectangular, etc.  Other advise also appreciated. &lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2023-02-20T10:16:21.463" Title="What is the most efficient way to ionize air?" Tags="|electrical-engineering|fluid-mechanics|thermodynamics|experimental-physics|building-physics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="14988" PostTypeId="2" ParentId="14975" CreationDate="2017-04-26T16:48:31.327" Score="0" Body="&lt;p&gt;This is answering the second part of your question to get a mathematical relationship for the parameters that are involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We write down the continuity equation in integral form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{V}\dfrac{\partial \rho}{\partial t}dV+\oint_S\rho \mathbf{u}\cdot\mathbf{n}dS=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Greek letter $\rho$ is the density of your fluid. $\mathbf{u}\cdot\mathbf{n}$ is the dot product of the velocity at the surface $\mathbf{u}$ and the outward normal unit vector $\mathbf{n}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For stationary flow, the volume integral vanishes. At the inlet side, we assume a uniform velocity $u_{\text{i}}$, which is parallel with the unit vector of the surface, and a surface area of $A_{\text{i}}$. Same procedure at the outlet with $u_{\text{o}}$ and $A_{\text{o}}$. Evaluation of the surface integral is only necessary at the inlet and outlet. We get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\rho u_{\text{i}}A_{\text{i}}+\rho u_{\text{o}}A_{\text{o}}=0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that $u_{\text{i}}A_{\text{i}}$ is the volumetric flowrate at the inlet and $\rho u_{\text{i}}A_{\text{i}}$ is representing the massflowrate at the inlet. Assuming circular inlet and outlet geometries we obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{\text{i}}\frac{\pi}{4} d^2_{\text{i}}=u_{\text{o}}\frac{\pi}{4} d^2_{\text{o}} \implies u_{\text{i}} d^2_{\text{i}}=u_{\text{o}}d^2_{\text{o}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you know the flow rate at the inlet $Q_{\text{i}}=u_{\text{i}}A_{\text{i}}$ (can be determined by measuring the volume of output in a certain time). You can easily infer the velocity at the outlet for a given outlet $d_{\text{o}}$ and vice versa. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get more information about the system behavior it is possible to apply Bernoulli's equation with empirical pressure loss $\Delta p_{\text{loss}}$ due to dissipative effects (can be looked up in tables for different pipings):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{\text{i}}+\frac 12\rho u^2_{\text{i}}+\rho g h_{\text{i}}=p_{\text{o}}+\frac 12 \rho u^2_{\text{o}}+\rho g h_{\text{o}}+\Delta p_{\text{loss}}.$$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now assume that the pressure at the outlet and the inlet is the same, which is a good assumption for slowly moving air as surrounding &quot;fluid&quot;. Hence, we can drop the pressure terms. Subtract $\frac 12 \rho u^2_{\text{i}}$ and use the continuity equation to replace $u_{\text{i}}=\frac {A_{\text{o}}}{A_{\text{i}}}u_{\text{o}}$. Additionally, we subtract $\rho g h_{\text{o}}$ and $\Delta p_{\text{loss}}$. We get the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho g(h_{\text{i}}-h_{\text{o}}) -\Delta p_{\text{loss}}=\frac 12 \rho u^2_{\text{o}}\left(1-\frac{A_{\text{o}}}{A_{\text{i}}} \right).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this equation for $u_{\text{o}}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{\text{o}}=\sqrt{\frac{2g(h_{\text{i}}-h_{\text{o}}) -2\frac{\Delta p_{\text{loss}}}{\rho}}{1-\frac{A_{\text{o}}}{A_{\text{i}}}}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you additionally assume the following simplifications&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Assume that there is no pressure loss: $\Delta p_{\text{loss}}=0$&lt;/li&gt;&#xA;&lt;li&gt;Assume that the inlet area is much larger than the outlet area: $\frac{A_{\text{o}}}{A_{\text{i}}} \ll 1$ &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;you will obtain Torricelli's law as a crude estimate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u_{\text{o}} \approx \sqrt{2g(h_{\text{i}}-h_{\text{o}})}.$$&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-04-29T21:55:33.590" LastActivityDate="2017-04-29T21:55:33.590" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="14989" PostTypeId="2" ParentId="14968" CreationDate="2017-04-26T17:10:46.417" Score="0" Body="&lt;p&gt;It depends on the exact loss factor of the transformer and how much you use the appliance. E.g let's just assume the loss factor is 50%  it's probably much much less that, but just see what happens. Let's assume you use the appliance 1 hour a day every day. Then you are paying about ${$}72$ a year to run the appliance and with out the transformer you'd be paying ${$}36$. So it would take about 10 years to pay back the investment. Probably not worth it. Now if you use the appliance 24 hours a day every day might be a different story. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="302" LastEditDate="2017-04-27T08:55:12.800" LastActivityDate="2017-04-27T08:55:12.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14991" PostTypeId="1" CreationDate="2017-04-26T20:16:00.857" Score="2" ViewCount="42" Body="&lt;p&gt;Over the years I've encountered a number of motor controllers (e.g., controllers in the NI Motion series) that incorporate what I usually refer to as &quot;retargeting&quot; behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, if the controller is told to move a motor to position A and, while moving there, is told to move the motor to position B, it will stop moving toward position A and move to position B instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This is opposed to behavior where the controller simply ignores movement commands issued while the motor is moving or maintains a &quot;target position&quot; that can be updated during motion and gets checked upon the completion of every movement to determine if a correction movement is required.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standard/common name for this retargeting behaviour, particularly in the context of motor control?&lt;/p&gt;&#xA;" OwnerUserId="11066" LastActivityDate="2017-04-26T20:16:00.857" Title="Standard name for &quot;retargeting&quot; behaviour in motor controller" Tags="|motors|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14993" PostTypeId="2" ParentId="14975" CreationDate="2017-04-26T20:47:01.810" Score="1" Body="&lt;p&gt;While the other answers give a relationship, there are factors that need to be included such as friction factor, inlet &amp;amp; outlet coefficients and length, so this link gives a more practical approach : &lt;a href=&quot;http://files.engineering.com/getfile.aspx?folder=23da7aca-8762-4e14-b029-09&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://files.engineering.com/getfile.aspx?folder=23da7aca-8762-4e14-b029-09&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-04-26T20:47:01.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14994" PostTypeId="2" ParentId="14987" CreationDate="2017-04-26T21:29:30.610" Score="4" Body="&lt;p&gt;It will depend on the degree of ionization you require, as well as the pressure of the air passing through the ionization region. Your gas will not likely be at STP if the air is moving through a reducing tunnel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ionization and breakdown (the common name for the process of forming a plasma by ionization) occurs more easily at lower pressures (so a quickly moving flow will be easier to break down) and can be initiated by DC voltage, RF, microwave, laser power, or sufficient heat. In terms of efficiency, microwave frequency power will cause breakdown at lower power levels than RF or DC, but the equipment is in general more expensive. It can also be a complex engineering problem to deliver sufficient microwave power in the right configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DC discharges are less efficiently generated but are relatively simple: apply 3 kV across a 1 mm gap and you should be near the threshold of breakdown. You can get very high temperature gas this way, but your anode and cathode will get ablated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more strongly your gas is ionized, the more power will get generated by passing through the magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general you're going to spend a lot of power ionizing the air. For large flow rates, you will require thousands of watts. You will likely recapture a very small portion of it from running it through a magnetic field. I think it is important to realize you will need to pump more electrical power into the system than you will capture from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wikipedia page for Paschen's law, which shows the curve of applied electric field to pressure required to break down a gas (for DC), is a good starting point: &lt;a href=&quot;https://en.wikipedia.org/wiki/Paschen%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Paschen%27s_law&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in reading more, I recommend &quot;Gas Discharge Physics&quot; (back in print!) by Yuri Raizer. Sanborn Brown also has a good book, likely in most large university libraries, &quot;High Frequency Gas Discharge Breakdown,&quot; which is pretty readable. &lt;/p&gt;&#xA;" OwnerUserId="3583" LastEditorUserId="3583" LastEditDate="2017-08-28T04:59:40.707" LastActivityDate="2017-08-28T04:59:40.707" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14995" PostTypeId="2" ParentId="14619" CreationDate="2017-04-26T22:18:47.480" Score="1" Body="&lt;p&gt;Here is my approach to identifying this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the steady-state response to a step input is a nonzero constant, we can infer that the system must not have a zero at the origin (because the steady-state response to a step would be zero), and it must not have a pole at the origin (because the steady-state response to a step would be a ramp). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, when systems start off in the &quot;wrong&quot; direction, we can infer the presence of a right-half-plane (or &quot;nonminimum phase&quot;) zero. The location of the zero determines the degree of &quot;undershoot&quot;. I started off by guessing the zero was at +1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, I estimated the location of the poles. The dominant time constant appears to be about a second. Playing around with the transfer function in MATLAB suggested that a better estimate would be s = -1.5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nonzero initial condition suggested to me that there might be &lt;em&gt;another&lt;/em&gt; zero. To keep the transfer function proper, I increased the order of the pole to 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that I got a reasonable fit with the transfer function $G_p(s) = \frac{(s-1)^2}{(s+1.5)^2}$, though you could obviously do better by tweaking parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YX2rP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YX2rP.png&quot; alt=&quot;Step response.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3232" LastActivityDate="2017-04-26T22:18:47.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="14996" PostTypeId="1" CreationDate="2017-04-27T06:54:27.523" Score="0" ViewCount="128" Body="&lt;p&gt;assume we have a multi-dof system and a force is applied with same frequency as one of the natural frequencies. will this system be resonant in any circumstances?&#xA;(system is undamped).&lt;/p&gt;&#xA;" OwnerUserId="11072" LastActivityDate="2017-04-27T23:56:40.833" Title="will a force with same frequency as natural frequency make system resonance (in multi-dof)?" Tags="|mechanical-engineering|vibration|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="14997" PostTypeId="2" ParentId="14996" CreationDate="2017-04-27T07:12:59.097" Score="1" Body="&lt;p&gt;That resonance is exactly why most systems are designed to avoid vibrating at the natural frequency - there are many examples of the consequences - see the (millenium I think...) bridge in London people walked, bridge moved, people started to walk in step, bridge moved even more...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-04-27T07:12:59.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15001" PostTypeId="2" ParentId="14452" CreationDate="2017-04-27T09:46:08.720" Score="2" Body="&lt;p&gt;First you should check if the truss is statically determinate. In one of your comments you said, that only the bottom support has a vertical reaction.&#xA;I assumed that all members are joint-connected, thus are not able to transfer any moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IinJH.png&quot; width=&quot;400&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a formula to determine, if a truss is statically determinate:&#xA;$$ n= r+s-2k $$&#xA;$r$ … number of support reactions $(3)$&lt;br&gt;&#xA;$s$ … number of members $(7)$&lt;br&gt;&#xA;$k$ … number of nodes $(5)$&lt;br&gt;&#xA;$$ \to n=3+7-2\cdot5 =0$$ &#xA;thus the truss is statically determinate, which means you can find the support reactions.&#xA;$$ \sum H = A_H+B_H =0 $$&#xA;$$ \sum V= B_V =Q $$&#xA;$$ \sum M(B)=A_H\cdot a + Q\cdot b =0 \qquad \star$$&#xA;The $\star$-equation is the moment equation, which can be defined for any point, I chose to establish the moment equation with respect to point $B$, because the forces $B_H$ and $B_V$ do not produce any moment wrt point $B$, thus they cancel out. With these three equations you can solve for all three support reactions and then find the member reactions.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="8632" LastEditDate="2017-04-28T09:38:01.013" LastActivityDate="2017-04-28T09:38:01.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15004" PostTypeId="2" ParentId="14996" CreationDate="2017-04-27T10:25:32.647" Score="1" Body="&lt;p&gt;A Multi-DoF (system with multiple degrees of freedom) can be linear or nonlinear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For linear systems in most cases, the system will enter resonance. The application of so-called mass dampers can be used to avoid the resonance peak. The theoretical advantage of linear systems is that we know that there is no dependence of amplitude and frequency from the external excitement for the resonance frequencies and modes of resonance of the system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the nonlinear system, we observe that the resonance frequencies are depending on the amplitude and frequency of the external oscillations. In general, the term mode of resonance is not applicable for nonlinear systems. Additionally, there are a lot of effects that make vibrations of nonlinear systems very difficult to describe. &lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-04-27T23:56:40.833" LastActivityDate="2017-04-27T23:56:40.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15005" PostTypeId="2" ParentId="14917" CreationDate="2017-04-27T10:48:30.127" Score="1" Body="&lt;p&gt;The intensity $I_0$ of an electromagnetic wave gets attenuated to the intensity $I$. The law describing the relationship between $I_0$ and $I$ is given by the Beer–Lambert law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I(d) = I_0\exp(-\alpha d)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The depth of penetration is given by $d$ and $\alpha$ is the coefficient of absorbance(must be given, e.g. &lt;a href=&quot;http://biomedicaloptics.spiedigitallibrary.org/article.aspx?articleid=1362680&quot; rel=&quot;nofollow noreferrer&quot;&gt;this publication&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intensity $I$ is linked to the maximal amplitude of the electric field $E_0$ or the maximal aplitude of the magnetic field by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I=\frac{c\varepsilon_0}{2}E_{0}^{2}=\frac {c}{2\mu_{0}}B_{0}^{2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of light is represented by $c$. The Greek letters $\varepsilon_0$ and $\mu_0$ are the vacuum permittivity and vacuum permeability.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-04-27T10:48:30.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15006" PostTypeId="1" AcceptedAnswerId="15015" CreationDate="2017-04-27T11:50:03.547" Score="1" ViewCount="411" Body="&lt;p&gt;Could anyone tell me what the variables in this formula are? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau=\frac{S\;a\; \bar{v}}{I\; t}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe it is used for shear stress and that $I$ is moment of inertia but unsure on the other variables.&lt;/p&gt;&#xA;" OwnerUserId="11075" LastEditorUserId="3010" LastEditDate="2017-04-27T13:56:58.860" LastActivityDate="2017-04-27T16:24:44.963" Title="Shear Stress Formula" Tags="|mechanical-engineering|stresses|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15007" PostTypeId="1" CreationDate="2017-04-27T13:42:40.477" Score="2" ViewCount="230" Body="&lt;p&gt;&lt;em&gt;Maybe this is topic for English board, but it's terminus technicus&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm seeking a name for the shape which secures two parts in a defined mutual position. The literal translation from my language is &quot;lock&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: I want to make two shafts concentric, so I make ??? on their ends. On the first shaft, I reduce its outer diameter from 100 mm to 80 mm for last 10 mm of its length. On the second shaft, I made the negative shape. So these parts fit in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IXQue.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IXQue.png&quot; alt=&quot;Result of previous paragraph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What word can replace ???&lt;/p&gt;&#xA;" OwnerUserId="4180" LastEditorUserId="12754" LastEditDate="2017-09-26T01:51:11.347" LastActivityDate="2018-02-23T12:04:47.467" Title="What is proper name of shaped shaft ends for ensuring their mutual position?" Tags="|mechanical-engineering|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15008" PostTypeId="2" ParentId="15007" CreationDate="2017-04-27T14:01:08.113" Score="0" Body="&lt;p&gt;I can't think of a good name for the &quot;complete concept&quot; here (and this doesn't seem to be a complete drawing of the structure anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several names for parts of this type of joint, for example boss, bush, fixture, flange, lug, pad, shoulder, etc. This &lt;a href=&quot;http://homepages.cae.wisc.edu/~me349/resources/engineering_terms.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://homepages.cae.wisc.edu/~me349/resources/engineering_terms.pdf&lt;/a&gt; might help you choose the best term.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-04-27T14:01:08.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15010" PostTypeId="1" CreationDate="2017-04-27T14:19:22.440" Score="3" ViewCount="708" Body="&lt;p&gt;When using modal analysis to determine the building period do we need to use the stiffness modifiers (or in another word use the cracked section) to calculate the building period ?&#xA;Can Anyone refer to any article or ACI reference that indicate anything related to this topic ?&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2017-04-27T16:33:41.237" Title="Do we need to use cracked section in Fundamental period calculation for a building?" Tags="|structural-engineering|civil-engineering|design|structural-analysis|building-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15013" PostTypeId="1" CreationDate="2017-04-27T15:51:47.290" Score="0" ViewCount="1756" Body="&lt;p&gt;I don't understand the fixed end moment AE and also fixed end moment EA ? shouldn't they be the same ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the fixed moment AE / fixed moment EA , shouldn't it = -10(2) +5(2) = -10 , for fixed end moment , we always assume clockwise as positive and anticlockwise as negative , right ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S2hCH.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b85zJ.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11080" LastEditorUserId="1832" LastEditDate="2017-04-27T16:24:57.243" LastActivityDate="2023-12-24T06:04:50.667" Title="why fixed end moment in beam confusion" Tags="|structural-engineering|structural-analysis|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15014" PostTypeId="2" ParentId="15007" CreationDate="2017-04-27T16:21:36.440" Score="0" Body="&lt;p&gt;I'm not sure if this is the term you're really trying to describe, but my first thought was a &lt;a href=&quot;https://en.wikipedia.org/wiki/Spline_(mechanical)&quot; rel=&quot;nofollow noreferrer&quot;&gt;spline&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a couple issues suggesting that though, mainly:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It is for a rotating shaft (this is not specified in your question, and AFAIK it's only a spline if it transmits rotation)&lt;/li&gt;&#xA;&lt;li&gt;I believe splines are specifically on the mating concentric edges, where as your picture seems to suggest it is the mating parallel edges that are transmitting the motion (again assuming these shafts rotate)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the shafts rotate and it's the parallel edges that mate, &lt;a href=&quot;https://en.wikipedia.org/wiki/Coupling&quot; rel=&quot;nofollow noreferrer&quot;&gt;coupling&lt;/a&gt; might be a better term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are not rotating this answer should be disregarded and you can sort through the terms in the other answer.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-04-27T16:21:36.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15015" PostTypeId="2" ParentId="15006" CreationDate="2017-04-27T16:24:44.963" Score="1" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/SzMtI.png&quot; width=&quot;400&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign convention I currently use is as shown ($x$ &quot;out of the display&quot; towards you, $y$ to the left, $z$ down)&#xA;The way I was taught the formula is:&#xA;$$ \tau=\frac{V_z\cdot Q_y}{I_y \cdot t} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_z$ … shear force at position $x$ in $z$-direction &lt;br&gt;&#xA;$Q_y$ … first moment of area ($\int zdA$) of the blue area wrt neutral axis&lt;br&gt;&#xA;$I_z$ … second moment of area of ($\int z^2dA$) entire cross section wrt neutral axis&lt;br&gt;&#xA;$t$ … thickness of cross section at point where $\tau$ has to be determined&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q_y$ can also be written as $A\bar{z}$, where $A$ is the blue area and $\bar{z}$ the $z$-coordinate of its centroid.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2017-04-27T16:24:44.963" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15016" PostTypeId="2" ParentId="15010" CreationDate="2017-04-27T16:33:41.237" Score="2" Body="&lt;p&gt;Yes, you need to consider the effect of cracking on the stiffness of the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Eurocode 8 (clause 4.3.1 (6)):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In concrete buildings, in composite steel-concrete buildings and in masonry buildings the stiffness of the load bearing elements should, in general, be evaluated taking into account the effect of cracking.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't have an ACI reference at the moment, but this &lt;a href=&quot;http://nvlpubs.nist.gov/nistpubs/gcr/2016/NIST.GCR.16-917-40.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;NEHRP Seismic Design Technical Brief&lt;/a&gt; says in section 4.2:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When analyzing a special moment frame, it is important to&#xA;  appropriately model the cracked stiffness of the beams, columns,&#xA;  and joints, as this stiffness determines the resulting&#xA;  building periods, base shear, story drifts, and internal force&#xA;  distributions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2017-04-27T16:33:41.237" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15018" PostTypeId="2" ParentId="13163" CreationDate="2017-04-27T18:57:12.910" Score="1" Body="&lt;p&gt;To expand on this, controlling the speed of the motors requires feeding it AC and varying the frequency. The generator puts out AC at a fixed frequency. So you need to convert fixed-frequency AC to variable-frequency AC. The most straightforward way of doing that is to rectify the AC to DC, then invert it back to AC again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A variable-frequency drive is the standard way of accomplishing this. Most take AC input into a rectifier, then have an inverter stage feeding the motor. Occasionally it makes more sense to have a single large rectifier creating a large DC bus, and then feeding several inverters off that single bus.&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2017-04-27T18:57:12.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15019" PostTypeId="1" CreationDate="2017-04-27T18:59:13.117" Score="1" ViewCount="109" Body="&lt;p&gt;Some plastics will sustain a flame. Others won't. Not spreading a fire is obviously preferable in almost every circumstance. What advantages do combustible plastics have over non-combustible plastics, and in what circumstances?&lt;/p&gt;&#xA;" OwnerUserId="18" LastActivityDate="2017-10-20T17:54:03.133" Title="Why not use fire-retardant plastics everywhere?" Tags="|plastic|safety|fire|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15020" PostTypeId="1" CreationDate="2017-04-27T18:52:04.060" Score="1" ViewCount="130" Body="&lt;p&gt;First time designing with pneumatics here. I want to build a continuous use device (with ~30 back and forth actuation per minute)  but do not currently have access to a large compressor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to manually test my device as I build it, and then use a real compressor later?   I calculated that my device (a simple cylinder that lifts a 500g mass) will use 5 cubic feet of air per minute (cfm).  I will need a compressor capable of 15 cfm and cannot afford this cost at the moment (need to invest it on other things). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am guessing the pressures I will be working with are around 90 psi, but maybe up to 120 or 180 depending on the cylinder I use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way I can rig something up with say a bicycle pump?  Surely someone has been in this situation of wanting to engineer/prototype without a big compressor available. There's math and calculations of course, but I am more of a hands-on engineer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for suggestions. &lt;/p&gt;&#xA;" OwnerUserId="21120" OwnerDisplayName="user391339" LastEditorUserId="2819" LastEditDate="2017-04-28T18:44:06.943" LastActivityDate="2017-05-15T10:55:52.683" Title="Air flowrate required for testing a pneumatic cylinder?" Tags="|systems-design|pneumatic|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15023" PostTypeId="2" ParentId="15013" CreationDate="2017-04-28T01:50:06.690" Score="0" Body="&lt;p&gt;What is the total distributed load applied between &lt;em&gt;AE&lt;/em&gt;? It is $w\,\ell_{AE} = 10(2) =20$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the load applied? Half way across $\frac{1}{2} \ell_{AE} = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply the two to get the moment of the distributed load $10(2)(1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end load of $5$ is applied at a distance of $\ell_{AE} =2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment of the end load is $5(2)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combined moment on &lt;em&gt;AE&lt;/em&gt; is $10(2)(1)-5(2)$&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-04-28T01:50:06.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15025" PostTypeId="1" CreationDate="2017-04-28T02:40:45.490" Score="1" ViewCount="53" Body="&lt;p&gt;So I am trying to plot variation in Heave (&lt;strong&gt;Z-direction&lt;/strong&gt;) of three ships with a change in its &lt;strong&gt;Speed&lt;/strong&gt; (which is in Froude number i.e. non-dimensionalized) and change in its &lt;strong&gt;Draft&lt;/strong&gt; (mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the best way to visualized this in one graph/plot? This is for my research work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H6Gbc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H6Gbc.png&quot; alt=&quot;Graph this&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11087" LastActivityDate="2017-09-28T20:44:17.177" Title="Plot variation in heave v/s speed v/s change in mass in one graph?" Tags="|fluid-mechanics|dynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15026" PostTypeId="2" ParentId="11685" CreationDate="2017-04-28T05:43:33.133" Score="1" Body="&lt;p&gt;A rocket engine is impractical for 2 reasons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As other answers have stated, you must carry the oxidizer on board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly though, a rocket accelerates much faster, burns much more fuel, generates much more thrust because the target velocity is orbital velocity, 9.4 km/sec or mach 27.5  for LEO, or escape velocity 11.2 km/sec or mach 33.  A supersonic airliner would have great difficulty going faster than mach 4 through the atmosphere for a number of reasons, primarily friction or drag and the thinness of the air for power requirements of the jet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At mach 4 you can use a scram-jet instead of conventional jet.  The scram-jet does not need the compressor turbine but rather the speed of the airflow at mach4 and higher would be sufficient enough for the compression, thus less moving parts.  This would allow the supersonic aircraft to fly high in the stratosphere and avoid the weather.  A hybrid system that may work well is rocket propulsion for a quick 2 minute or less boost to reach the stratosphere at mach4 and then the scram-jets taking over.  You could theoretically go up to mach 15 with a scram-jet, which while not orbital velocity, would certainly please any traveler.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-04-28T05:43:33.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15027" PostTypeId="2" ParentId="2046" CreationDate="2017-04-28T09:08:38.573" Score="-1" Body="&lt;p&gt;You could try to analyze this problem with doing very complicated fluid dynamics analytically or numerically. The problem is nonstationary and the convective terms do not vanish, hence very difficult to treat analytically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inviscid approximation by Olin Lathrop seems do be a good model for this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative way and in my opinion the most reliable way would be just to use a simple experiment. Fix the syringe vertically so that the needle shows downward. Then use small weights as force and measure the time it takes to drain the syringe. Then vary the weight until you achieve the drain time of $ 20 \text{ s}$. If you don't quite get to this time, because you don't have this specific weight, you can interpolate your measurements (for example with: Excel, R, MATLAB) to estimate the appropriate weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might need to add an additional plate on the pushing side so that you can place your weights. If it is not too heavy then you don't have to take this also into account. The piston weight should also be negligible. &lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-04-28T09:08:38.573" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15028" PostTypeId="2" ParentId="8228" CreationDate="2017-04-28T10:28:52.547" Score="-1" Body="&lt;p&gt;Dear madam ,&#xA;             pls make beam with keypoints and then do meshing with mesh tool and use smisc3 and 16 for bending moment and smisc 6 &amp;amp; 19 for shear force&lt;/p&gt;&#xA;" OwnerUserId="11091" LastActivityDate="2017-04-28T10:28:52.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15029" PostTypeId="2" ParentId="15007" CreationDate="2017-04-28T10:44:33.787" Score="1" Body="&lt;p&gt;One specific form of such shapings is a &quot;taper&quot; - such as the well known Morse tapers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of a term for the general feature you're looking for, but tapers are adopted for many instances of it, probably because of their self-aligning nature. Consider the consequence of a machining error in the ID or OD of your diagram above - it will result in either imprecise alignment or no fit at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar error on a taper would translate to perfect centring with a slight displacement in the axial direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Morse tapers for example not only provide centring but can also provide drive force. Morse tapers are self-locking due to their shallow angle, combined with material properties such as friction and elasticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other tapers which are self-releasing (I believe the R8 tapers used in industrial tooling would come into this category) - they require some other means to hold the parts together - often a collet closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, many systems of collets use tapers to guarantee high accuracy in centring (low runout).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lathes.co.uk/latheparts/page7.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fitting a backplate to a lathe chuck&lt;/a&gt; is precisely this operation without a taper, and I see this instruction page uses the term &quot;spigot&quot; for the male part, but the female part (in the rear of the chuck) is simply called a &quot;recess&quot;.&lt;/p&gt;&#xA;" OwnerUserId="435" LastEditorUserId="435" LastEditDate="2017-04-28T11:02:21.510" LastActivityDate="2017-04-28T11:02:21.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15030" PostTypeId="1" CreationDate="2017-04-28T12:34:30.683" Score="0" ViewCount="109" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PCFEK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PCFEK.jpg&quot; alt=&quot;Schematic for idea&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi there, When I was very little I came up with this idea for a power station and I recently remembered it and was curious about the viability. The way it works theoretically is that a voltage supply is temporarily provided to an electric motor to get it spinning up to speed. The motor has a magnet on its armature that spins in a coil to generate electricity, this will generate output electricity. This output is taken from the generator and split, some will go as an output from the power station and some will be fed back in to the system. The portion that is fed back runs through a rectifier to convert it from ac to dc voltage and then through a dc amplifier to get it up to whatever the voltage supply to the motor initially was. The output of the amplifier then becomes the new power supply for the motor as the temporary voltage supply is cut off using a switch. This could potentially leave the system self sufficient. I understand that a small amount of electricity would still have to be used to power the amplifier and it won't be greatly efficient but it was just an idea that I had when I was little. I highly doubt this design will work but if anyone could explain to me why or provide improvements to the design then I would be very grateful.&lt;/p&gt;&#xA;" OwnerUserId="11092" LastActivityDate="2021-07-07T09:40:20.190" Title="Feedback on idea for a power station" Tags="|electrical-engineering|power|renewable-energy|electrical|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15032" PostTypeId="2" ParentId="15030" CreationDate="2017-04-28T13:19:39.157" Score="3" Body="&lt;p&gt;For pretty fundamental physical reasons any power generator is only a way of converting one sort of energy to another ie converting the energy you have available to the sort of energy that you want. For example :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chemical energy to mechanical power eg an internal combustion engine &lt;/li&gt;&#xA;&lt;li&gt;Mechanical power to electrical power eg a generator &lt;/li&gt;&#xA;&lt;li&gt;Chemical energy to electrical power eg a battery &lt;/li&gt;&#xA;&lt;li&gt;Electromagnetic radiation to electrical power eg a solar panel &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Even for sustainable energy sources you still need an ultimate energy source (in many cases this ends up being the sun). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can never do is get more energy out than goes in to the system. Again there are fundamental and well proven general reasons why this is the case and finding a 'loophole' in these basic laws of nature, while technically not entirely incredible, would require extraordinary evidence to be credible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact these conversion processes will always waste some energy and even if you can make a system with very tiny losses any energy you take out needs to be replace otherwise you are just storing energy not generating it. For it is certainly possible to design a flywheel with very few frictional losses eg by using magnetic bearings and running it in a vacuum and this may indeed be a useful way to store energy but you can still get out no more than you put in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular case the thing you are missing is that an amplifier doesn't &lt;em&gt;multiply&lt;/em&gt; electrical power, essentially it allows a low powered signal to switch/modulate a high powered output but you still need to put in as least as much power as you get out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Laws_of_thermodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laws of thermodynamics&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Perpetual_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Perpetual Motion&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-04-28T15:37:54.560" LastActivityDate="2017-04-28T15:37:54.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15033" PostTypeId="1" CreationDate="2017-04-28T14:14:40.627" Score="0" ViewCount="371" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JHRDS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JHRDS.png&quot; alt=&quot;Liquid exiting a nozzle and exerting a pressure force on an inclined plate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm attempting a fluid mechanics questions that requires me to sketch a pressure distribution along the surface of the plate and explain why it shaped the way it is. I'm struggling to comprehend how to sketch this distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it vary linearly/non-linearly? Is it continuous or constant along the plate? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My general thoughts are that it would be it triangular shaped and would decrease linearly as you went further away from where the water jet strikes the plate.&lt;/p&gt;&#xA;" OwnerUserId="11095" LastEditorDisplayName="user16" LastEditDate="2017-05-06T22:47:09.003" LastActivityDate="2017-05-06T22:47:09.003" Title="Pressure distribution on an inclined plate" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15034" PostTypeId="2" ParentId="11818" CreationDate="2017-04-28T14:22:51.830" Score="0" Body="&lt;p&gt;This is more a comment than an answer but I don't have enough reputation to comment yet:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;I understand that deriving a first-principles model may be too complex, but be aware that a good statistical model for your case will surely need a lot of good data. It is not enough with having a lot of data, the input variables must be excited in order to cover the input space. Normally this is not the case with industrial real data, where most of the online measurements are close to the same region in the input space&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why not feedback control? You don't need a model for feedback control (although it helps) and 30-90 seconds is not such a big delay for most chemical processes (it depends on the process and it may not be a chemical process at all)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Otherwise, if you have a lot of data and you manage to set up a good data driven model, your approach can work but you will need to recalibrate your model often  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3991" LastActivityDate="2017-04-28T14:22:51.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15036" PostTypeId="2" ParentId="15020" CreationDate="2017-04-28T18:30:50.450" Score="1" Body="&lt;p&gt;It depends on what you are trying to test; if you are trying to test leakage a bike pump will work fine, but if you are trying to test ring wear or operation at full power there is no way around supplying the required amount of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double check your volume calculations and force calculations. That seems like way to much power to do the task required.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0.5kg *9.81m/s^2 = 620kpa * Area &lt;br&gt;&#xA;  Area= 7.9E-6m^2 = PI()*R^2 &lt;br&gt;&#xA;  R=0.0016m &lt;br&gt;&#xA;  Diameter= 3.3mm at the 90psi you are proposing&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I don't know your stroke length but lets say 1 meter (~39inches)&lt;br&gt;&#xA;  Volume = 7.9E-6m^2 * 1m = 7.9E-6m^3 = 0.00028ft^3&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;30cycles/min * 2(double acting, assuming same volume both directions) * 0.00028ft^3 = &lt;strong&gt;0.0168CFM required @&#xA;  90psi&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You will need to run though these calculations again with a more reasonable cylinder diameter and pressure combination, but as you can see the cfm requirement is going to be much less than you estimated.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-04-28T18:30:50.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15037" PostTypeId="2" ParentId="15019" CreationDate="2017-04-28T19:18:43.993" Score="5" Body="&lt;p&gt;Fire retardant fillers are pretty cheap; often cheaper than the base plastic (Alumina Trihydrate for example). Like anything in engineering though, there are always exceptions and functional reasons why something is never the solution 100% of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some instances where I think fire retardant fillers would not be preferred:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Filler Capacity - A plastic can only accept so much filler. A fire retardant filler would take up this capacity that may be better used for a UV filler or other.&lt;/li&gt;&#xA;&lt;li&gt;Elastic Modulus (resistance to deformation)- Fire retardant fillers typically make a plastic more rigid.&lt;/li&gt;&#xA;&lt;li&gt;Transparency - Unless you have identical indexes of refraction, any filler will make the plastic opaque.&lt;/li&gt;&#xA;&lt;li&gt;Weight - Fire retardant fillers are typically heavier than the plastic.&lt;/li&gt;&#xA;&lt;li&gt;Disposal - Fillers may make it more difficult to recycle a plastic or burn it in municipal gassifier or boiler systems.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also, just because a plastic doesn't sustain combustion by itself, it does not mean that it is safe in a fire. Often there is additional sources of energy that bring the plastic to its smoke point. Smoke inhalation kills as many or more people than direct fire does in situations like plane crashes.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="13321" LastEditDate="2017-10-20T17:54:03.133" LastActivityDate="2017-10-20T17:54:03.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15038" PostTypeId="2" ParentId="15019" CreationDate="2017-04-28T19:20:34.030" Score="0" Body="&lt;p&gt;It depends on the plastic and its intended use. &#xA;Most, but not all plastics will burn, giving off noxious fumes. &#xA;To make them &quot;flame retardant&quot; increases the cost and in many cases reduces the physical performance properties of the material.&#xA;Many applications require flame retardant plastic materials, such as aircraft interior parts made of plastics. &#xA;Look at Underwriters Labs. 94-V ratings tests. &lt;/p&gt;&#xA;" OwnerUserId="11068" LastActivityDate="2017-04-28T19:20:34.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15039" PostTypeId="1" AcceptedAnswerId="15328" CreationDate="2017-04-28T22:38:27.050" Score="-1" ViewCount="1482" Body="&lt;p&gt;I read that a Tesla Model S P100D achieved 730 horsepower on a dynamometer. Would increasing battery size increase the horsepower? If not, what would?&lt;/p&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="368" LastEditDate="2017-06-12T16:48:20.687" LastActivityDate="2017-06-12T16:48:20.687" Title="How do you increase the horsepower of an electric motor?" Tags="|motors|automotive-engineering|battery|electric-vehicles|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15040" PostTypeId="1" CreationDate="2017-04-28T23:44:37.673" Score="4" ViewCount="182042" Body="&lt;p&gt;Why the Fixed End Moment (FEM) for  BC is 3PL / 16?&#xA;It's clear in the first figure that  that when one end is fixed, while the another end is pinned, then the fixed end moment is 3PL /16 ... But for the span BC, we could see that B is the roller and C is the pinned connection, there's no fixed support in the span BC&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gLH80.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gLH80.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ywnpr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ywnpr.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kdgZF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kdgZF.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11080" LastEditorUserId="19906" LastEditDate="2020-12-14T14:15:14.140" LastActivityDate="2020-12-14T14:15:14.140" Title="How to determine fixed end moment in beam?" Tags="|structural-engineering|structural-analysis|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15041" PostTypeId="2" ParentId="15039" CreationDate="2017-04-29T02:05:51.513" Score="1" Body="&lt;p&gt;Increasing the battery size would increase the range of the vehicle, not the horsepower.&#xA;Increasing the mechanical output of the motor would increase horsepower, but reduce range.&lt;/p&gt;&#xA;" OwnerUserId="11068" LastActivityDate="2017-04-29T02:05:51.513" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15042" PostTypeId="1" CreationDate="2017-04-29T05:38:44.767" Score="2" ViewCount="2171" Body="&lt;p&gt;In ASCE 7-10 Table 12.2-1 two types of shear walls can be used ordinary reinforced concrete shear walls and special reinforced concrete shear walls what is it the difference between these walls ? (Reinforcement, Shape, Usage) and when do i use each type ?&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2018-07-02T19:27:34.247" Title="What is the difference between Ordinary reinforced concrete shear walls and special reinforced concrete shear walls?" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15043" PostTypeId="2" ParentId="2964" CreationDate="2017-04-29T06:36:30.760" Score="-1" Body="&lt;p&gt;At slip=1 and speed of rotor N=0 rpm ,motor produces a torque called starting torque &lt;/p&gt;&#xA;" OwnerUserId="11102" LastActivityDate="2017-04-29T06:36:30.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15044" PostTypeId="1" AcceptedAnswerId="15045" CreationDate="2017-04-29T11:46:39.323" Score="1" ViewCount="272" Body="&lt;p&gt;Where is L measured from in the formula for the Reynolds Number. I know it is a linear dimension of the system measured in metres but unsure of how to measure it.&lt;/p&gt;&#xA;" OwnerUserId="11075" LastActivityDate="2017-04-29T16:42:56.120" Title="Reynolds Number" Tags="|fluid-mechanics|thermodynamics|" AnswerCount="2" CommentCount="0" ClosedDate="2017-04-30T09:40:06.847" ContentLicense="CC BY-SA 3.0" />
  <row Id="15045" PostTypeId="2" ParentId="15044" CreationDate="2017-04-29T12:12:45.520" Score="4" Body="&lt;p&gt;L is described as a &lt;em&gt;characteristic dimension&lt;/em&gt;, it's function is to indicate the overall scale of the system. For example in circular pipes it would be the pipe diameter and for a wing you would generally use the chord.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that the main function of Reynolds number is comparative, for example comparing a scale model in a wind tunnel to the real thing so it doesn't necessarily matter exactly what you pick as long as it is consistent. But there will be conventions for particular circumstances. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-04-29T12:12:45.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15047" PostTypeId="2" ParentId="15044" CreationDate="2017-04-29T16:42:56.120" Score="0" Body="&lt;p&gt;Reynolds number is a dimensionless constant used to describe the flow characteristics of a flow system. It is a tool for ccomparative analysis. So the absolute value of Reynolds number does not hold any physical meaning unless the flow conditions are specified. &#xA;The L is a parameter used to quantify the extent or size of flow.It can be choosen to be any length which gives an idea of the extent of the flow.But once we compute Reynolds number of a flow by choosing a particular length(say, the chord for a wing), it should be made sure that we have to use the same length in computing the Reynolds number of another flow if both are intended to be compared.&#xA;Hope you got it.&lt;/p&gt;&#xA;" OwnerUserId="11109" LastActivityDate="2017-04-29T16:42:56.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15048" PostTypeId="1" CreationDate="2017-04-29T18:23:55.553" Score="0" ViewCount="5367" Body="&lt;p&gt;I am planning to build an e-bike and I realized that a car starter motor would be a good motor, as it has enough power (1kW, enough for my bike) and the voltage is practical enough.&#xA;Will it handle the stress?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mechanic told me that the coil in it would be destroyed after 30 seconds, as it isn't meant to be used a lot.&#xA;I am only interested in the motor, not the solenoid.&#xA;I will be using it at 75% of its power, at most, so not too much stress.&#xA;Will it work?&#xA;It's going to replace a 50cc motor of a piaggio ciao.&lt;/p&gt;&#xA;" OwnerUserId="10614" LastEditorUserId="10344" LastEditDate="2017-05-01T20:50:14.423" LastActivityDate="2021-10-29T14:55:13.957" Title="Use car starter as motor" Tags="|motors|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15049" PostTypeId="1" AcceptedAnswerId="15616" CreationDate="2017-04-29T18:42:40.083" Score="0" ViewCount="1936" Body="&lt;p&gt;I'm doing a gear design project, and I'm using a multivariate mixed integer non-linear optimization suite known as &lt;code&gt;Couenne&lt;/code&gt; distributed by COIN-OR. Within Python, I use this solver to solve for all aspects of the gears in a 2 ratio reverted gear-train given 2 target gear train ratios (4.3 fwd, 9.1 rev). My classmates are telling me that the diametral pitch for each gear pair must be an integer value, or a value with a well known fraction in decimal form. Is this true? I've found that placing this constraint makes it much harder to approach the 2 target ratios. Perhaps it was a consideration about manufacturing, because It would be exceedingly hard to machine a diameter that is an irrational number. However, couldn't the same be said for any diameter? Manufacturing shouldn't depend on the dimension because for any given manufacturing precision we will never know what is after the decimal place. For both irrational and integer diametral pitches, the manufacturing error would be the same. So why would gear diametral pitches have to be integer values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without this constraint, I'm able to achieve a total gear train ratio error of 1.3*10^-8 percent from the target ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: when I refer to &quot;diametral pitch&quot; I am referring to the gear module, number of teeth per inch&lt;/p&gt;&#xA;" OwnerUserId="11110" LastEditorUserId="11110" LastEditDate="2017-04-29T20:31:47.783" LastActivityDate="2017-05-31T13:23:34.120" Title="Does gear diametral pitch have to be an integer? (teeth/diameter)" Tags="|gears|torque|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15050" PostTypeId="2" ParentId="15049" CreationDate="2017-04-29T18:57:29.943" Score="0" Body="&lt;p&gt;Obviously gears need an integer number of teeth and spur gears need to have the same tooth profile for all meshing gears and this does impose some practical constraints on what ratios can be achieved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many gears and most 'off the shelf' gears are designed with the &lt;em&gt;module&lt;/em&gt; system. which defines a ratio between tooth pitch and pitch diameter which also eliminates pi from the calculation. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-04-29T18:57:29.943" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15051" PostTypeId="1" CreationDate="2017-04-29T21:50:38.857" Score="5" ViewCount="59" Body="&lt;p&gt;My understanding is, given an optimal control problem, one can show that the optimal cost satisfies a Hamilton-Jacobi PDE and use dynamic programming to figure out the optimal control. However, sometimes this PDE has no strong solution, and the deep theory of viscosity solution was invented to make sense of a &amp;quot;solution&amp;quot; in this situation.&lt;/p&gt;&#xA;&lt;p&gt;Does such &amp;quot;optimal&amp;quot; cost have any practical meaning? Phrased slightly differently, can one still figure out what the control should be, and implement it?&lt;/p&gt;&#xA;&lt;p&gt;I feel like it has to be a function, not a distribution, in order to be implementable.&lt;/p&gt;&#xA;" OwnerUserId="11111" LastEditorUserId="6790" LastEditDate="2020-08-01T06:54:18.010" LastActivityDate="2020-08-01T06:54:18.010" Title="Is the viscosity solution of Hamilton-Jacobi equation of practical use in optimal control?" Tags="|control-engineering|control-theory|mathematics|optimal-control|partial-differential-equations|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="15052" PostTypeId="1" CreationDate="2017-04-29T23:54:51.177" Score="0" ViewCount="41" Body="&lt;p&gt;In this problem(attached as a picture), the authors have chosen the horizontal element to calculate the shear and I understand the solution. But, the thing I don't understand is: Why  don't they take the vertical element (shaded by me, in the blue color) if we take the horizontal element we find a smaller value for shear, which is in our interest cause we are looking for the smallest shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ROJwj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ROJwj.png&quot; alt=&quot;retrieved from Mechanics of materials P.Beer &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9053" LastActivityDate="2017-04-30T08:57:38.473" Title="shear on a beam element" Tags="|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15053" PostTypeId="1" CreationDate="2017-04-29T23:59:15.030" Score="1" ViewCount="38" Body="&lt;p&gt;In non-recirculating hydroponic systems, we want to fill containers with nutrient rich water  let's say up to an inch, let the plants take up the water , give the plants some air time for a few hours, and then restart a filling cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking a sponge could be used as a sensor and actuator.  The sponge valve would be dry and open.  As the water reached an inch the sponge would get wet and expand in a cylinder and push something that blocks the flow of water in a 4mm irrigation tube.  As the water gets absorbed by the plants, the sponge would dry out and the pressure lowers and eventually it becomes low enough that the water starts flowing again, for a new cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a cellulose sponge realistic for this. In particular would the wetting and drying state of the sponge be sufficiently stable or would it rapidly deteriorate?  Is there another material that would be better suited?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="11112" LastEditorUserId="11112" LastEditDate="2017-04-30T00:41:15.533" LastActivityDate="2017-04-30T00:41:15.533" Title="Can a cellulose sponge be used as an actuator to shut off flow of water temporarily?" Tags="|actuator|fluid|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15054" PostTypeId="1" AcceptedAnswerId="15164" CreationDate="2017-04-30T05:28:36.507" Score="3" ViewCount="3495" Body="&lt;p&gt;I want to build a home automation system by hooking up various sensors (e.g., passive IR motion sensors, magnetic door contact switches, vibration detectors, etc.) around the house to detect things then send the sensor signals to a central command center / control station.&lt;/p&gt;&#xA;&#xA;PIR Motion Sensor&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gfZxW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gfZxW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to do this &lt;em&gt;wirelessly&lt;/em&gt; and &lt;em&gt;solderlessly&lt;/em&gt; with a budget of roughly $10 per sensor. So, for example, I could hook up a breadboard circuit that receives the sensor signal and sends it to a Wifi network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple circuit component or IC that can convert the analog sensor signal to a wireless signal then transmit that signal to the network via wifi without, say, hooking up an entire microprocessor (e.g., Raspberry Pi) to each sensor? A Raspberry Pi for each sensor would blow my budget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking to install a simple component part in, say, a solderless breadboard circuit at the sensor level. Then read all the signals off the network at the control station using a Raspberry Pi or laptop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of components and configuration would anyone here recommend?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Edit 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Apparently, a USB wireless adapter or a network adapter might be the component I'm looking for. But would it be possible to build up a separate breadboard circuit for each sensor and pipe all the signals to the wifi network? If so, how? I want to keep my budget to about $10 per sensor.&lt;/p&gt;&#xA;&#xA;USB Wireless Adaptor&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/23etr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/23etr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Edit 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.electronicszone.eu/index.php/en/12-english/199-infrared-motion-detector&quot; rel=&quot;nofollow noreferrer&quot;&gt;I found this article&lt;/a&gt; which reflects what I'm trying to do with the sensor. But it says nothing about a wireless (analog-to-digital) conversion.&lt;/p&gt;&#xA;&#xA;PIR Motion Sensor Schematic&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E2xw7.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E2xw7.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;Breadboard Hookup&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NShw9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NShw9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Edit 3&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://openhomeautomation.net/getting-started-esp8266/&quot; rel=&quot;nofollow noreferrer&quot;&gt;I like the suggested ESP8266 chip described here&lt;/a&gt;. (Could this really be $ 5-10 per sensor?)&lt;/p&gt;&#xA;&#xA;ESP8266 chip&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YFYPl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YFYPl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;Breadboard configuration&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5c3yo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5c3yo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;Schematic&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FvrLI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FvrLI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2017-05-09T02:58:17.883" LastActivityDate="2018-02-20T06:09:48.827" Title="Send sensor signals to WiFi network" Tags="|electrical-engineering|sensors|embedded-systems|wifi|iot|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15055" PostTypeId="2" ParentId="15048" CreationDate="2017-04-30T06:03:08.243" Score="0" Body="&lt;p&gt;Starter motors are designed to be most efficient at starting engines : getting a large mass moving to a particular speed - but not for extended long period running. This will affect the design of the bearings etc The design of e-bike motors are optimised for that duty. So, it may work, and you may be happy with the performance, but other types of motor may give you more range or speed.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-04-30T06:03:08.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15056" PostTypeId="1" CreationDate="2017-04-30T06:31:42.447" Score="2" ViewCount="48" Body="&lt;p&gt;Could someone please explain some of the advantages/disadvantages of the wallace multiplier circuit vs a shifting and adding multiplier circuit vs an shift register multiplying circuit? We went over them briefly in class but didn't get to in depth and I'm trying to sort out which is better in what situation or even overall of applicable, and why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wallace multiplier is pretty straight forward and simple design (to design, the physical thing seems quite tedious) but quite huge... The shifting and adding one seems better sized and quite simple but it does have an output going back in which I guess could slow things down? But then with all the things going on it the The shift register I guess seems like a nice median but not sure if I'm missing anything there. &lt;/p&gt;&#xA;" OwnerUserId="8733" LastEditorUserId="8055" LastEditDate="2017-05-01T14:31:56.807" LastActivityDate="2017-05-01T14:31:56.807" Title="Comparisons of multiplier circuits" Tags="|electrical-engineering|circuits|circuit-design|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15058" PostTypeId="2" ParentId="15052" CreationDate="2017-04-30T08:57:38.473" Score="1" Body="&lt;p&gt;First of all let it be clarified that the smallest shear force is not our point of interest here. Here the question is to find the shear acting at the two joints,A and B.You can conclude by the first glance itself that the shear on these two joints are of different magnitudes as A and B are not symmetrical about any axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by the method followed in the given text, the average shear stress along a given plane is calculated using the Q of the element on one side of an axis parallel to our plane of interest, ie V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in the given problem the joint A is horizontal. So we should consider a line parallel to this and passing through A, and calculate the Q of the element on either side of the line(in this case the horizontal element AB). Now by applying the formula we can find out the average shear stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as we come to the second part of the question, the joint B is vertical. Hence we should choose a vertical element. You might have done it that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope everything is clear.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2017-04-30T08:57:38.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15059" PostTypeId="1" AcceptedAnswerId="15062" CreationDate="2017-04-30T10:56:15.203" Score="1" ViewCount="255" Body="&lt;p&gt;In my text book , it is written &#xA;&quot;Power generated by the turbine can be increased by using a gradual expansion at the turbine outlet&quot; &#xA;The picture below explains the meaning of gradual expansion at the turbine outlet &lt;a href=&quot;https://i.stack.imgur.com/09X1l.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/09X1l.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e the cross sectional area of the pipe at the turbine outlet is increasing gradually m, causing gradual expansion.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this thing cause the power generated by the turbine to increase ? &lt;/p&gt;&#xA;" OwnerUserId="9742" LastActivityDate="2017-04-30T15:30:48.480" Title="Does the turbine outlet affect the power generated by the turbine ?" Tags="|mechanical-engineering|fluid-mechanics|turbines|turbomachinery|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15060" PostTypeId="2" ParentId="15048" CreationDate="2017-04-30T12:42:17.107" Score="2" Body="&lt;p&gt;It might not burn out in 30 seconds, but I suspect your mechanic is probably on the right track. All motors are going to generate heat as they run. They differ in how quickly they heat up (and what temperatures they can withstand). For a given load rating (say 1 kW), a smaller motor will heat up quicker and a larger motor will heat up slowly. Your starter motor is probably relatively small for a 1 kW motor. It's going to heat up quickly with operation.  But that doesn't matter for your car, because it's only intended to work for a short period of time.  It's on for 10s or so, starts to heats up a little, but before it gets close to its maximum temperature, it gets turned off and given a chance to cool down for a long time (hours maybe) before being operated again. That's very different from what you are going to be doing with your bike.  You'll be running it for a lot longer periods.  The motor is going to get very hot very quickly, and you won't be giving it any time to cool down. It might not burn up in 30s, but I don't think it will last for 10 years either. You'll want a motor that is designed to be operated at 1 kW continuously.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.groschopp.com/blog/duty-cycle-need-know/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Duty Cycle: What you need to know&lt;/a&gt; page might help.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="8055" LastEditDate="2017-05-01T17:29:36.437" LastActivityDate="2017-05-01T17:29:36.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15061" PostTypeId="2" ParentId="15048" CreationDate="2017-04-30T13:33:02.183" Score="0" Body="&lt;p&gt;Starter motors are optimized for high torque and to survive the rough environment of a car.  They are not designed for sustained use, and are not particularly efficient.  Both those latter points make them unsuitable for a bicycle.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-04-30T13:33:02.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15062" PostTypeId="2" ParentId="15059" CreationDate="2017-04-30T15:10:02.980" Score="2" Body="&lt;p&gt;The turbine referred to, in your text book must be a specific type of turbine, namely reaction turbine. Reaction turbines utilise mainly the pressure energy of flowing water to produce Power. The flowing fluids consist of both pressure energy(a function of fluid pressure) and kinetic energy(a function of flow velocity). The reaction turbines are rotated by the pressure difference of water on both sides of turbine blades. The reaction turbines convert most of the pressure energy to mechanical power, but they are unable to convert much of the kinetic energy in such a way. So the kinetic energy brought in by the water is left to flow away through the outlet&#xA;But if a pipe whose diameter is small at first and then increasing slowly is used at outlet, then the pressure at the pipe beginning will be lowered significantly. This happens due continuity conditions and Bernoulli's principle.&#xA;According to continuity equation if the cross sectional area of a flow decreases the flow velocity increases. But according to Bernoulli's principle if flow velocity increases the pressure decreases, given there is no change in elevation, losses, external work etc. &#xA;Thus the pressure at outlet side of turbine decreases and the pressure difference between both sides of the turbine increases. Thus the driving force increases and hence the power. So such A pipe actually converts some of the kinetic energy into pressure energy for better utilisation in reaction turbines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the water cannot flow outside if its pressure remains below atmospheric pressure. So in order to increase the pressure the diameter is increased slowly. In this case also Bernoulli's principle is utilised. The increase in diameter of the outlet pipe is kept gradual to avoid losses (eg: Eddy formation due to sudden expansion). Such pipes with gradually increasing diameters used in reaction turbines are called Draft Tubes.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastEditorUserId="11114" LastEditDate="2017-04-30T15:30:48.480" LastActivityDate="2017-04-30T15:30:48.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15063" PostTypeId="2" ParentId="15054" CreationDate="2017-04-30T17:00:37.030" Score="2" Body="&lt;p&gt;I think you won't get this done without some kind of processor and customized firmware to convert the sensor signal to a digital signal for WiFi transmisson. The dongle is a good idea, but it provides only the transmission, not the convertion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more accurate advices, you should provide more details on the targeted sensors and the network you want to create.&lt;/p&gt;&#xA;" OwnerUserId="10862" LastActivityDate="2017-04-30T17:00:37.030" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15064" PostTypeId="1" CreationDate="2017-04-30T17:06:52.847" Score="4" ViewCount="861" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZPdma.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZPdma.jpg&quot; alt=&quot;Basel wood screw press&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to calculate the tonnage a medieval wood screw press could generate. In particular I’m speaking of a press used to squeeze water from newly formed sheets paper in-between layers of felt (see picture) however medieval wood screw presses used to extract olive oil or juice from grapes are identical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an engineer but I will be giving a PechaKucha (a short 6 minute 40 second lecture) to a group of hand papermakers in a few months from now in which I will include the concepts of simple machines like the lever and the inclined plane to gain mechanical advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such presses exist in museums throughout Europe however being over 500 years old we are unable to use these to test the actual force. Here’s what I already know. The formula for calculating the ideal mechanical advantage on jack screws is straight forward which includes the advantage gained from length of the lever and the pitch of the thread. The much more difficult part is allowing for friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve read that even well-lubricated jack screws have efficiencies of only 15% - 20% but that’s for a machined iron screw. The threads on these wooden screws were chiseled by hand with the most difficult part being the inside threads of the nut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is – is it possible to calculate the efficiency of a hand carved wooden screw? Presumably the upper limit of efficiency would be less than the 15% noted above (and is this 15% figure is correct in the first place)? But what might be the lower end of efficiency be?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oYLEz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oYLEz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the dimensions of the wood screw&lt;/p&gt;&#xA;" OwnerUserId="11116" LastEditorUserId="2819" LastEditDate="2017-09-22T23:35:27.560" LastActivityDate="2017-09-22T23:35:27.560" Title="Medieval wooden screw press tonnage" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15065" PostTypeId="1" CreationDate="2017-04-30T20:37:06.453" Score="2" ViewCount="1468" Body="&lt;p&gt;I want to make a durable lightweight container that maintains it's shape, I am aware that a hexagon shape has the best compression ratio, but does that translate into a stronger more robust structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to build a container that will be used to store refuse,soil, feed, fertilizer etc. The containers need to take a lot of abuse as currently the plastic round containers seem to not last at all.&lt;/p&gt;&#xA;" OwnerUserId="11119" LastEditorUserId="11119" LastEditDate="2017-05-03T12:22:26.517" LastActivityDate="2017-05-03T12:22:26.517" Title="round vs hexagon" Tags="|materials|structural-analysis|stresses|" AnswerCount="1" CommentCount="1" ClosedDate="2017-05-03T00:01:54.570" ContentLicense="CC BY-SA 3.0" />
  <row Id="15066" PostTypeId="1" CreationDate="2017-04-30T20:49:34.763" Score="3" ViewCount="170" Body="&lt;p&gt;I am trying to build a robot which would plant seeds instead of humans doing it (lettuce, radish and so on). The only problem i have so far is picking an individual seed. &#xA;How could i do it? I've seen some discs designed especially for such appliances, but those are really expensive and i think that there has to be an alternative.&lt;/p&gt;&#xA;" OwnerUserId="10614" LastActivityDate="2017-05-01T18:31:02.177" Title="How to make an arduino pick an individual plant seed?" Tags="|mechanisms|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15067" PostTypeId="1" CreationDate="2017-05-01T00:56:07.757" Score="2" ViewCount="314" Body="&lt;p&gt;What is the process in design a Compound Elliptical Concentrator and Compound Parabolic Concentrator if I wanted to achieve 100% optimal concentration ? Let's assume random numbers of: A light source being 14 meters wide, the aperture is 4 meters wide and the distance between them being 12 meters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was doing random searches on solar energy and came across solar concentrators which I've never heard of before, but saw you can design them based on knowing information above. I can't find anything easy enough to understand so a newbie like me can follow clearly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for the CPC. The equation is for the maximum concentration: $$C=\frac{1}{sin^2(\theta)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I wanted a 100% concenctration would have a 1 (100%) in place of the C then solve for the theta to get the half acceptance angle ? Would I go backwards from this and adjust the base of the concetrator to have an aperture of 4 meters and a base of whatever gives me an angle of theta?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image of what I meant (found randomly online with no explanation): &lt;a href=&quot;https://i.stack.imgur.com/0pIJe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0pIJe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11123" LastEditorUserId="1832" LastEditDate="2017-05-01T15:19:42.310" LastActivityDate="2018-08-25T01:01:06.837" Title="Design of Compound Elliptical Concentrator and Compound Parabolic Concentrator" Tags="|optics|solar-energy|photovoltaics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15069" PostTypeId="2" ParentId="15065" CreationDate="2017-05-01T02:08:26.840" Score="1" Body="&lt;p&gt;A circle (cylinder or sphere) is the strongest shape for a given wall thickness for a tank, pressure vessel, or reducing localized wall deflection. A circle has the highest area to perimeter ratio. Without knowing more details, this is likely your best solution. This is why most containers are round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hexagon shape works well as internal trussing structure in sandwich structured composites. As a vessel it is stronger than a rectangular box of similar volume because the wall spans are shorter, but it will not be as strong as a circle for a given wall thickness.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-05-01T02:08:26.840" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15070" PostTypeId="1" CreationDate="2017-05-01T05:48:17.150" Score="-1" ViewCount="26" Body="&lt;p&gt;I think that the varying load in roads affect the efficiency of IC engines in large amounts. I am planning to do a mechanism which stores the power produced in IC engines mechanically, and then release it as per requirement to the drive. Would such a system be feasible. Can anyone suggest an idea for this.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2017-05-01T07:29:35.287" Title="Improving efficiency in automobileIC Engines" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ClosedDate="2017-05-01T20:02:18.193" ContentLicense="CC BY-SA 3.0" />
  <row Id="15071" PostTypeId="2" ParentId="15070" CreationDate="2017-05-01T07:29:35.287" Score="1" Body="&lt;p&gt;Yes, research flywheel energy storage - and it has been used, IIRC on busses / coaches in the past, but some systems had issues with catastrophic failure.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-01T07:29:35.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15073" PostTypeId="1" CreationDate="2017-05-01T10:19:17.260" Score="2" ViewCount="406" Body="&lt;p&gt;I need some assistance understanding moments of inertia. I am doing some review for an upcoming exam, however I am slightly stumped by this question. I have already tried googling &quot;moments of inertia&quot; to try and understand the concept better, but I am having issues knowing when to apply which formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attached my specific question and the &quot;answer&quot; to the question, hopefully someone can help me understand how to get the answer.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Two identical slender rods of length l and mass m are linked together at 90deg, as shown in figure Q4a (image attached), to form a link in a mechanism. Point B is midway between A and C, and points B, D, E, F and G are equally spaced along the lower link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) Determine the position of the center of gravity of the link&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) Find the moment of inertia of the link about the point B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iii) Find the moment of inertia of the link about the point D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iv) Find the radius of gyration about the point G&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fxfGv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fxfGv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ANSWERS:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;i) Center of gravity is at D&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ii) I&lt;sub&gt;B&lt;/sub&gt;=(5/12)ml&lt;sup&gt;2&lt;/sup&gt; kg m&lt;sup&gt;2&lt;/sup&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;iii) I&lt;sub&gt;D&lt;/sub&gt;=(7/24)ml&lt;sup&gt;2&lt;/sup&gt; kg m&lt;sup&gt;2&lt;/sup&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;iv) k= sqrt(17/24) l m&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;My attempts&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;i)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know inherently that the centre of gravity is at D. How do i prove it mathematically?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;ii)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get the moment of inertial about B like so: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I&lt;sub&gt;b&lt;/sub&gt; = (ml&lt;sup&gt;2&lt;/sup&gt;)/3 + ((ml&lt;sup&gt;2&lt;/sup&gt;/3) - (ml&lt;sup&gt;2&lt;/sup&gt;)/4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i don't understand why I am supposed to subtract. Is it because the centre of gravity is l/2 below B?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;iii)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For I&lt;sub&gt;d&lt;/sub&gt; I get the answer by doing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I&lt;sub&gt;d&lt;/sub&gt; = ((ml&lt;sup&gt;2&lt;/sup&gt;)/3 - (ml&lt;sup&gt;2&lt;/sup&gt;)/4) + ((ml&lt;sup&gt;2&lt;/sup&gt;)/(3*16)) + (9ml&lt;sup&gt;2&lt;/sup&gt;)/(16*3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do I subtract (ml&lt;sup&gt;2&lt;/sup&gt;)/4) if that rod is above D?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;iv)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to use I&lt;sub&gt;G&lt;/sub&gt; to find the radius of gyration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for I&lt;sub&gt;G&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I&lt;sub&gt;g&lt;/sub&gt; = (7/24)ml&lt;sup&gt;2&lt;/sup&gt; - ml&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I am using the parallel axis theorem, but why do I subtract the (md&lt;sup&gt;2&lt;/sup&gt;) of the AC rod from the Moment of inertia about D?&lt;/p&gt;&#xA;" OwnerUserId="11039" LastActivityDate="2017-05-03T14:00:26.773" Title="Moments of inertia, rods mechanism" Tags="|mathematics|statistics|moments|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15074" PostTypeId="1" AcceptedAnswerId="15077" CreationDate="2017-05-01T10:44:09.803" Score="0" ViewCount="85" Body="&lt;p&gt;I am working on a project using springs. What is the maximum load per deflection of springs available in market?&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2017-05-01T15:08:43.030" Title="Load per deflection of springs" Tags="|mechanical-engineering|springs|" AnswerCount="1" CommentCount="4" ClosedDate="2017-05-01T17:28:48.167" ContentLicense="CC BY-SA 3.0" />
  <row Id="15075" PostTypeId="1" AcceptedAnswerId="15098" CreationDate="2017-05-01T12:17:38.737" Score="4" ViewCount="274" Body="&lt;p&gt;When brazing HVAC linesets, the correct method is to flow an inert gas through the line in order to prevent interior oxidation of the copper and brazing products.  However, as a specifier, I can say that this needs to be done until I'm blue in the face, but contractors (in my experience) are notorious for skipping this step.  It adds cost, time, and a system where this is not done will not fail immediately, but it will contribute to the premature failure of the system.  In the long term, this simply looks like a system that ran it's course, rather than the product of poor installation practices.&#xA;Is there a way to tell by inspection after installation that this practice was followed, or are you at the mercy of the contractor if someone is not watching?&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2019-11-26T02:43:33.390" Title="Inert gas use during HVAC line installation" Tags="|hvac|specifications|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15077" PostTypeId="2" ParentId="15074" CreationDate="2017-05-01T15:08:43.030" Score="2" Body="&lt;p&gt;There is no sensible answer to this. Springs are designed to have a relatively high deflection relative to their  to ultimate strength. Once you want very stiff springs you can just use solid bars of steel.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick with spring design is to end up with a product which has a &lt;em&gt;specific&lt;/em&gt; set of properties and it is possible to make arbitrarily stiff or light springs. What may be more difficult is combining a desired stiffness and capacity with a particular set of dimensions and lifetime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally any answer may also depend on what you define as the difference between a spring and a structure as all solid materials are essentially springs for a given load range. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-05-01T15:08:43.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15078" PostTypeId="2" ParentId="15025" CreationDate="2017-05-01T15:13:30.923" Score="1" Body="&lt;p&gt;Looks like you 3D data (draft, speed, heave) that you have to repeat for each ship. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F7tb8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F7tb8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2017-05-01T15:13:30.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15079" PostTypeId="2" ParentId="15073" CreationDate="2017-05-01T15:58:53.380" Score="1" Body="&lt;p&gt;To compute the c.g location, assume that is is at a distance Y below point B. Use the fact that the mass moment about that point should be zero to determine Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_{Y-L}^Y \frac{m}{l}y \, dy = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above equation has one unknown Y. Solving, we get $ Y = l/4 $. This is point D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding $I_D$ is straightforward integration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I_D=\int_{-\frac{l}{2}}^{\frac{l}{2}} \frac{m&#xA;   }{l} \left(\frac{l}{4}+x\right)^2\, dx+\int_{\frac{l}{4}-l}^{\frac{l}{4}} \frac{m}{l} y^2 \, dy$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving (using Mathematica), I get $$ I_D= \frac{7 l^2 m}{24}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I_B$ can then get obtained using the parallel axis theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I_B = I_D + 2 m \left(\frac{l}{4}\right)^2 = \frac{5 l^2 m}{12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can similarly find $I_G$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_G= I_D + 2 m \left(\frac{3 l}{4}\right)^2 = \frac{17 l^2 m}{12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of gyration follows from this value.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2017-05-01T15:58:53.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15080" PostTypeId="2" ParentId="15033" CreationDate="2017-05-01T16:37:06.983" Score="1" Body="&lt;p&gt;This task is not trivial at all. In order to get a crude estimate of the pressure distribution, we will assume inviscid flow. The flow is separated by a streamline in the middle (red line in the picture, sp = stagnation point). We will try to get solutions for the right half of the plate and for the left half of the plate separately by applying potential theory. To be more precise we will use the power law for the potential flow. The formula that we will apply is given by:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\phi = |A|\mathrm{e}^{\varphi_A}r^n \cos n\theta.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The radius &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is given as the distance to the stagnation point (sp) and the angle &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the angle in radians with respect to the horizontal &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-axis. The parameter &lt;span class=&quot;math-container&quot;&gt;$|A|$&lt;/span&gt; needs to be obtained by additional data (e.g. by assuming uniform flow for &lt;span class=&quot;math-container&quot;&gt;$Q=Au$&lt;/span&gt;). The angle &lt;span class=&quot;math-container&quot;&gt;$\varphi_A$&lt;/span&gt; is the angle in radians with which we rotate the base flow to get the same orientation as in the problem.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tVayH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tVayH.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The right half of the flow is a &lt;span class=&quot;math-container&quot;&gt;$60°$&lt;/span&gt;-corner flow rotated (&lt;span class=&quot;math-container&quot;&gt;$n=3$&lt;/span&gt;) with an additional rotation of &lt;span class=&quot;math-container&quot;&gt;$30°$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$\varphi_A=\frac{\pi}{6}$&lt;/span&gt;). This will result in a potential &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\phi = |A|\mathrm{e}^{i\frac{\pi}{6}}r^3 \cos3\theta.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The velocities can be obtained by using the relationships:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u_r = \frac{\partial \phi}{\partial r} \qquad u_{\theta}=\frac{1}{r}\frac{\partial \phi}{\partial \theta}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Analogous, we will use &lt;span class=&quot;math-container&quot;&gt;$n=3/2$&lt;/span&gt; for the left-hand side and &lt;span class=&quot;math-container&quot;&gt;$\varphi_A = \frac{\pi}{2}$&lt;/span&gt; got get:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\phi = |A|\mathrm{e}^{i\frac{\pi}{2}}r^3 \cos(3\theta/2).$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The velocities can be obtained by the previous relationships.&lt;/p&gt;&#xA;&lt;p&gt;Finally, to get our pressure distribution, we need to apply Bernoulli's equation to the right side of the flow and for the left side of the flow separately.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$p_1+\frac 12 \rho u^2_1+\rho g h_1 = p(r,\theta) + \frac{1}{2}\rho \left[u^2_r+u^2_{\varphi} \right]+\rho g h$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$p(r,\theta)-p_1=\frac 12 \rho \left[u^2_1-u^2_r-u^2_{\varphi} \right]+\rho g (h_1-h)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The height &lt;span class=&quot;math-container&quot;&gt;$h_1=4.85 \text{ m}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u_1= 12.2 \frac {\text{m}}{\text{s}} $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$p_1$&lt;/span&gt; is the pressure at the exit of the nozzle. If they are not given take any positions for which you know &lt;span class=&quot;math-container&quot;&gt;$h_1, p_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u_1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;On the inclined plate the height &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is given by &lt;span class=&quot;math-container&quot;&gt;$h=r\sin(30°)$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-05-01T22:22:41.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15081" PostTypeId="2" ParentId="15067" CreationDate="2017-05-01T17:48:40.443" Score="1" Body="&lt;p&gt;Per the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Nonimaging_optics&quot; rel=&quot;nofollow noreferrer&quot;&gt;non-imaging optics wiki&lt;/a&gt;, C is the &quot;ratio of input and output aperture areas&quot;. It is not efficiency like it is in some other equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So putting a 45 degree half angle into the equation gives C = 2. This means that if the input area is 1m^2 then the output will be 0.5m^2. A solar radiation intensity would increase from say 1000w/m^2 to 2000w/m^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a 30 degree half angle the equation gives C = 4. So solar radiaton could be increased to 4000w/m^2. This would be more favorable in a solar heating application but the trade off is the smaller acceptance angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a 90 degree half angle (light from 180 degrees) the equation gives C = 1 meaning no concentration is possible.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-05-01T17:48:40.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15083" PostTypeId="2" ParentId="15066" CreationDate="2017-05-01T18:31:02.177" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://genesis.farmbot.io/v1.2/docs/seed-injector&quot; rel=&quot;nofollow noreferrer&quot;&gt;farmbot&lt;/a&gt; uses a small diameter tube and a vacuum pump to pick up a single seed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NzHUi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NzHUi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-05-01T18:31:02.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15086" PostTypeId="1" CreationDate="2017-05-01T21:49:50.603" Score="1" ViewCount="4520" Body="&lt;p&gt;There is a back EMF in a motor, due to current flow in a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in a generator, is there a reverse rotation (opposed to original rotation) due to a current flow in magnetic field?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is, and this reverse rotation is zero when there is no load connected to generator (no current flow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the load connects to a generator, reverse rotation will arise, this will lead to a rotation speed down (slow down output frequency and voltage)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct?&lt;/p&gt;&#xA;" OwnerUserId="11137" LastEditorUserId="282" LastEditDate="2017-05-20T17:11:06.670" LastActivityDate="2017-08-18T18:46:41.063" Title="Are there reversed rotation direction in generators (as back EMF in MOTORS)?" Tags="|electrical-engineering|generator|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15088" PostTypeId="1" AcceptedAnswerId="15090" CreationDate="2017-05-02T08:49:26.620" Score="2" ViewCount="1503" Body="&lt;p&gt;At what height do we splice the vertical bars in columns and shear walls is it at the middle of the element or at the bottom ( Slab level) and why ?&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2017-05-02T14:54:11.313" Title="At What height do we splice vertical bars in Columns and Shear walls?" Tags="|structural-engineering|civil-engineering|design|structural-analysis|detailing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15089" PostTypeId="1" CreationDate="2017-05-02T14:46:49.047" Score="2" ViewCount="208" Body="&lt;p&gt;I have been attempting this question for a while now. Could anyone give me a hand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question and the work I have done so far is below. My final &quot;answer&quot; is &#xA;P (force required to keep the gate closed) = 109,100.1526w + 346.84 (where w is the width of the gate). Since i cannot rearrange my answer in the form &quot;P/w&quot;, I am assuming I have made a mistake somewhere, but I cannot identify my mistake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xv4mN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xv4mN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DEOkY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DEOkY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11147" LastEditorUserId="11147" LastEditDate="2017-05-03T12:08:28.763" LastActivityDate="2017-05-03T12:08:28.763" Title="Fluid Statics Problem" Tags="|fluid-mechanics|statics|water-resources|fluid|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15090" PostTypeId="2" ParentId="15088" CreationDate="2017-05-02T14:54:11.313" Score="3" Body="&lt;p&gt;In theory, a splice behaves exactly like a single continuous bar, so there should be no effect as to where you actually do the splice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in the real world it depends.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you're dealing with a compression-only column or wall, then I'd suggest splicing near (but not &lt;strong&gt;&lt;em&gt;at&lt;/em&gt;&lt;/strong&gt;) the slab level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason is simple ease of construction. For example, if you must splice between floors 2 and 3 of a building and you do so just above floor 2, then the construction will look like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pouring of column from floor 1 to 2 without the splicing rebar (but with protruding rebar awaiting the splice).&lt;/li&gt;&#xA;&lt;li&gt;Pouring of floor 2's slab&lt;/li&gt;&#xA;&lt;li&gt;Working over floor 2's slab, the workers now have easy access to the splicing area, without needing to use scaffolding or anything of the sort to reach it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Obviously, the splicing shouldn't occur too close to the slab otherwise you might end up with a jumble of rebar: you have to fit twice the column reinforcement due to the splice and you have to deal with the anchorage of the slab's reinforcement. So make sure the splicing doesn't affect interfere with anything else.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, if you're dealing with a moment-resisting column, then the best splicing location is wherever the moment is minimal, which frequently is near the midspan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, for some odd reason it is near one of the slabs, remember to check for interferences.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An important exception is if you have to deal with earthquake loads. In this case, one should always try to splice column reinforcement at the midspan. This is because the inertial loads at the nodes are huge, and you don't want to rely on your splice in this delicate area.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I also found &lt;a href=&quot;https://www.quora.com/Which-is-the-best-zone-to-provide-lapping-in-columns-and-beams&quot; rel=&quot;nofollow noreferrer&quot;&gt;this discussion&lt;/a&gt; on the subject useful.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-05-02T14:54:11.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15091" PostTypeId="1" AcceptedAnswerId="15092" CreationDate="2017-05-02T16:57:55.270" Score="2" ViewCount="3299" Body="&lt;p&gt;If we have a series of columns as shown in the picture below and to reduce deflection of the slab between two columns that are 11.7 m apart we placed a drop beam 30 x 100 cm (the slab is 35 cm thick) do we need to check the punching shear for these two columns ? and if we need to check the punching shear do we use the beam thickness (100 cm) or we use the slab thickness (35 cm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking this particular question because in SAFE for columns C5 through C12 does calculate the punching shear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PMVMt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PMVMt.png&quot; alt=&quot;Punching Shear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="1832" LastEditDate="2017-05-02T17:49:13.413" LastActivityDate="2018-03-16T10:54:02.977" Title="Do we need to check punching shear when a drop beam is connecting columns?" Tags="|structural-engineering|civil-engineering|design|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15092" PostTypeId="2" ParentId="15091" CreationDate="2017-05-02T18:43:36.583" Score="2" Body="&lt;p&gt;Absolutely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be clearly seen when you consider the load path for everything other than the loads in the region with the beams. Where will a load between C1 and C5 go? Straight to C1 and C5, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A load between C5, C6, C9 and C10 will probably go to the beams and then the columns, so there's no need for punching shear reinforcement in that region, sure, but the loads outside of that area will concentrate right on the columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only question is what the load should be. When calculating the punching shear for column C6, what load should you use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One option to think about is to just use the load applied in the rectangle bounded by C1-C3-C7-C5 (well, the portion of it that goes to C6). This seems reasonable: this load goes straight to the column (requiring punching shear reinforcement), while the load around C5-C7-C11-C9 first goes to the beams, avoiding the need for punching shear reinforcement. Basically (where full lines represent loads that go straight to the columns and dashed lines go to the beams):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c5T1v.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c5T1v.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how sure can you be that some fraction of the load in C5-C7-C11-C9 won't go straight to the column? After all, the column is far more stiff than the beam, so shouldn't it absorb more of the load than'd be trivially expected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A really conservative calculation would ignore the beams and calculate the punching shear as if all the load went to the columns. Worst case, you're considering around double the real load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another, slightly less conservative calculation would use a 2/3 slope to define the load paths that go straight to the column:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o2pcw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o2pcw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, you actually have a FEM program that can accurately calculate the slab-beam-column interaction, there's an easier way to know the punching force: get the column's axial load and then subtract the shear force of the beam at the column. With this, you'll get the total force that was directly absorbed by the column, which is the punching shear force. Just obviously make sure to model the structure accurately.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-05-02T18:43:36.583" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15093" PostTypeId="2" ParentId="569" CreationDate="2017-05-02T18:44:41.947" Score="1" Body="&lt;p&gt;If the other pipes are carrying liquids with a greater specific heat coefficient then they would not be expected to experience the same temperature ranges, i.e. the water pipes will be at some average of the outside temp and the water temp. &#xA;A low-pressure gas line would be made of lighter material and the gas inside would equalize with the external temp pretty quickly.  So the code for a gas line and a liquid line (or a light pipe and a heavy pipe) might be different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expansion joints are often used for exposed liquids pipelines, like the Alaska pipeline, that are subject to temperature extremes and cover long distances.  Buried pipelines don't experience the same range of temperature and don't normally employ expansion loops.&lt;/p&gt;&#xA;" OwnerUserId="11149" LastActivityDate="2017-05-02T18:44:41.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15094" PostTypeId="1" CreationDate="2017-05-02T19:01:23.040" Score="2" ViewCount="3788" Body="&lt;p&gt;The two towers of the Verrazano Bridge are not parallel: they are [???] to account for the curvature of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4nQGJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4nQGJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is that word?&lt;/p&gt;&#xA;" OwnerUserId="3740" LastActivityDate="2019-11-15T13:02:46.597" Title="What is the opposite of &quot;parallel&quot; in engineering?" Tags="|terminology|" AnswerCount="7" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15095" PostTypeId="2" ParentId="15094" CreationDate="2017-05-02T19:06:56.843" Score="5" Body="&lt;p&gt;angled, divergent, splayed... I'd probably go with angled depending how I was phrasing the description.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-02T19:06:56.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15096" PostTypeId="2" ParentId="14555" CreationDate="2017-05-02T19:24:02.863" Score="1" Body="&lt;p&gt;You will likely want more than just a vent valve - to avoid contamination of a vacuum chamber, typically you re-pressurize with an inert gas like nitrogen to avoid sucking condensates and water vapor into the chamber. Every time you add some condensate to the chamber, they'll need to desorb - which is an extremely slow process compared to pumpdown.&lt;/p&gt;&#xA;" OwnerUserId="3583" LastActivityDate="2017-05-02T19:24:02.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15098" PostTypeId="2" ParentId="15075" CreationDate="2017-05-02T19:40:05.960" Score="1" Body="&lt;p&gt;You could do a visual inspection with a pipe camera on the interior. Poorly-brazed copper will have discolorations / flaking due to oxidation (just like the exterior would). You can also just pull one of their brazes and slice it in half to inspect it. This PDF (&quot;Brazing Best Practices: 12 Tips for HVAC Technicians&quot; from Nov. 2014 issue of Welding Journal) has a good photo of what you'd see from a bad braze: &lt;a href=&quot;https://app.aws.org/bsmc/WJ_201411_62.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://app.aws.org/bsmc/WJ_201411_62.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pipe camera obviously is only cost-effective if you've got a lot of subcontractors doing brazing work that you need to check (or if you already have one). Slicing is probably easiest. You could also supply them with the nitrogen and check the tank pressure afterwards - obviously they could just bleed it for a little bit, but at some point fooling you is going to be more of a hassle than just doing the job correctly.&lt;/p&gt;&#xA;" OwnerUserId="3583" LastActivityDate="2017-05-02T19:40:05.960" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15099" PostTypeId="1" CreationDate="2017-05-02T20:21:50.337" Score="1" ViewCount="711" Body="&lt;p&gt;What is the minimum for vertical bars in a column according to ACI and I am looking for a particular condition where diameter must be greater than 0.625 in.&#xA;(when using S-concrete for column design it gives a warning that diameter must be 0.625 in. but I couldn't find anywhere in the ACI that include this condition) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any article or in any clause in ACI refer to this condition please share.&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2017-05-02T20:21:50.337" Title="What is the minimum diameter for vertical bars in a column according to ACI?" Tags="|structural-engineering|design|structural-analysis|building-design|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15101" PostTypeId="2" ParentId="15094" CreationDate="2017-05-02T22:14:18.343" Score="5" Body="&lt;p&gt;In general, we can distinguish four cases.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Case: parallel --&gt; If you take the curvature into account, the pillars cannot be treated as parallel. While I would guess that the error is pretty small on this scale.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Case: orthogonal --&gt; Obviously not the case. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Case: intersecting lines: If we assume that the axis of these pillars meets at the middle of the earth, we could call them intersecting lines. Note that for this case the lines should not be parallel and not be orthogonal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Case: skew lines: This case requires that the lines are not parallel, orthogonal and do not intersect. If you look at the axis of these pillars with high precision then it will be obvious the axis of these pillars do actually form skew lines. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Summary: Depending on your accuracy you would call the pillars by the following categories. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;low accuracy: parallel&lt;/li&gt;&#xA;&lt;li&gt;mid accuracy: intersecting lines or intersecting&lt;/li&gt;&#xA;&lt;li&gt;very high accuracy: skew lines or skew&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-05-02T22:14:18.343" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15102" PostTypeId="2" ParentId="2390" CreationDate="2017-05-02T22:18:25.910" Score="0" Body="&lt;p&gt;Generally, the pressure drop across a filter will follow the second fan law where:&#xA;$$\Delta P=C_v Q^2$$&#xA;Therefore, you only need one data point for the filter ($\Delta P_1, Q_1)$ to calculate the proportionality constant of that filter material. Then you can extrapolate for a continuous range of CFM. Assuming the filter is the biggest source of resistance, that should give you a pretty good approximation of the system curve ($Q$ vs. $\Delta P$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there you'll need to overlay your fan curves to find the operating point. That will be difficult if the fan specs don't include them. The operating point is where the fan curve intersects the system curve for a particular RPM, which you get to choose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To one of your other questions, if the fan is rated at 500 CFM without giving a static pressure, I think you would have to assume that it's for the free-running condition. In other words, it's the maximum CFM without any imposed resistance to the flow. In more words - no pressure drop, or 0 static pressure. I'm not sure how useful that is because you already know there's going to be resistance to flow in the form of the filter.&lt;/p&gt;&#xA;" OwnerUserId="11151" LastActivityDate="2017-05-02T22:18:25.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15103" PostTypeId="2" ParentId="7228" CreationDate="2017-05-02T22:35:48.627" Score="0" Body="&lt;p&gt;I can think of two ways you might increase airflow in this case - reduce the resistance to airflow or increase the speed of your fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first case, I'd look at the resistance in the ducts of your system. The 1&quot; exit pipe seems small; most commercial blowers are about 4&quot; in diameter. It should be noted that the resistance to flow in a pipe is proportional to $\frac{1}{D^4}$ of the pipe (&lt;a href=&quot;https://en.wikipedia.org/wiki/Airway_resistance&quot; rel=&quot;nofollow noreferrer&quot;&gt;airway resistance&lt;/a&gt;), so the performance does not scale directly. Small reductions in diameter can have big effects on the flow resistance, AKA back pressure or pressure drop.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also speed up your fan. However, there are certain combinations of flow and resistance where a particular fan likes to operate. If you change the speed, you could be improving or reducing the fan efficiency. &lt;/p&gt;&#xA;" OwnerUserId="11151" LastEditorUserId="90" LastEditDate="2017-05-03T00:55:55.057" LastActivityDate="2017-05-03T00:55:55.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15104" PostTypeId="2" ParentId="14852" CreationDate="2017-05-02T22:46:17.110" Score="1" Body="&lt;p&gt;To perform the amount of mechanical work described here, it would take about $330W$ of power, if you consider the maximum pressure and airflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Power = \Delta P \cdot Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is assuming 100% efficiency, which will never be the case. Does your power budget allow this?&lt;/p&gt;&#xA;" OwnerUserId="11151" LastActivityDate="2017-05-02T22:46:17.110" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15106" PostTypeId="2" ParentId="15086" CreationDate="2017-05-03T11:37:26.333" Score="2" Body="&lt;p&gt;Kind of, there is a torque produced proportional to the applied load that counters the rotation, as clearly must be the case to cause mechanical power to flow from the prime mover to the machine rotor, but in a grid connected machine the speed is set by the grid frequency, not the load on the individual machine (Ok, there can be some fun dynamics involved, but assume steady state).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dynamics are made complex by the fact that rotor speed (and thus frequency) is the integral wrt time of the net torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an ideal three phase machine connected to a large grid, if we start off with everything in sync and no net power flow then the angle between the rotors field and the grid connected stator induced field is zero and no torque is generated. If we now open the throttles what happens? The prime mover tries to speed up, which causes the rotor to start to lead on the fields produced by the grid connected stator and as this angle starts to increase, a torque is produced as the magnetic field of the stator pulls on the rotor, eventually this torque equals the torque being produced by the prime mover and we end up in a (hopefully) steady state where the rotor is leading the grid by exactly enough to match the torque from the prime mover. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting thing is that if the grid is massive compared to the generator, the speed has not changed (as that is set by the grid frequency), only the rotor angle has changed but we are now generating power.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is insufficient power input in total, the grid frequency falls (And all of the generators slow down), if there is excess power input in total the grid frequency rises, but in all cases the speed of all the machines matches the grid frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there has been no mention of voltage, which is usually controlled to set the reactive power, not the real power.&lt;/p&gt;&#xA;" OwnerUserId="10072" LastActivityDate="2017-05-03T11:37:26.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15108" PostTypeId="2" ParentId="15073" CreationDate="2017-05-03T14:00:26.773" Score="2" Body="&lt;p&gt;The point where center of mass lies so the center of gravity.&#xA;So, for center of mass&#xA;For rod AC.&#xA;B is the mid point of AC.So,let us take A as origin So coordinates of B($\frac{l}{2}$,0).Similarly for midpoint of BG We say there are four divisions which are equally space hence the length of BD,DE,EF,FG are$\frac{l}{4}$ So the mid point is E the mid point of BG. So coordinates of E($\frac{l}{2}$,$\frac{l}{2}$) are (according to the origin taken).&#xA;$$\Rightarrow X_{com}=\frac{m_{1}x_{1}+m_{2}x_{2}+.............+m_{n}x_{n}}{m_{1}+m_{2}+...........m_{n}}$$&#xA;So here mass of the mass of both the rods are same and we have worked out their individual center of masses.&#xA;$$\Rightarrow \frac{m\frac{l}{2}+m\frac{l}{2}}{m+m}$$&#xA;$$\Rightarrow \frac{2m\frac{l}{2}}{2m}$$&#xA;$$X_{com}=\frac{l}{2}$$&#xA;Now same for $Y_{com}$&#xA;$$Y_{com}=\frac{m(0)+m\frac{l}{2}}{2m}$$&#xA;$$Y_{com}=\frac{l}{4}$$&#xA;So, coordinates of center of mass of the system is ($\frac{l}{2}$,$\frac{l}{4}$)&#xA;Now coordinates of point G($\frac{l}{2}$,$l$),Point F($\frac{l}{2}$,$\frac{3l}{4}$),Point E($\frac{l}{2}$,$\frac{l}{2}$)And D($\frac{l}{2}$,$\frac{l}{4}$)which satisfies the $X_{com}$ and $Y_{com}$.Hence its is the center of gravity of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) moment of inertia about the point B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Due to AC rod  for which B is the mid point And for its the moment of intertia is as follows:&#xA;Assume the frame from  B is perpendicular to the rod AC passing through the point B.&#xA;So only considering AC rod &#xA;The mass about the B point is $\frac{m}{l}$ mass of rod is divided among the lenght l Now imagining a point x distance from B now the moment of inertia about the imagined point is $$\Rightarrow \frac{m}{l}x^2$$&#xA;now for the whole we integrate &#xA;so,$$\Rightarrow \int\limits_{\frac{-L}{2}}^{\frac{L}{2}}\frac{m}{l}x^2\,dx$$&#xA;$$\Rightarrow \frac{m}{l}\int\limits_{\frac{-L}{2}}^{\frac{L}{2}}\frac{x^3}{3}$$&#xA;$$\Rightarrow \frac{m}{3l}\left(\frac{L^3}{8}-\left(-\frac{L^3}{8}\right)\right)$$&#xA;now finally we get $$\Rightarrow \frac{m L^2}{12}$$&#xA;2)Due to BG rod As B is at B corner so just change the limits and put it 0 to l and we get $$\frac{m L^2}{3}$$.&#xA;So total moment of inertia about the point is &#xA;$$\Rightarrow \frac{m L^2}{12}+\frac{m L^2}{3}$$&#xA;which is &#xA;$$\Rightarrow \frac{5m L^2}{12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iii)just put the limits in above integral $\frac{-L}{4}$ to $\frac{3L}{4}$&#xA;We get moment of inertia  for BG to D is $$\frac{7m L^2}{48}$$&#xA;Hence we use the parallel axis theorem for AC so&#xA;$$\Rightarrow \frac{mL^2}{12}+m\left(\frac{L}{4}\right)^2$$&#xA;$$\Rightarrow \frac{7mL^2}{48}$$&#xA;Again adding the both of the moment of interia that is about BG and AC&#xA;$$\Rightarrow \frac{7mL^2}{48}+\frac{7mL^2}{48}$$&#xA;which is $$\frac{7mL^2}{24}$$.&lt;/p&gt;&#xA;" OwnerUserId="11154" LastActivityDate="2017-05-03T14:00:26.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15110" PostTypeId="2" ParentId="11005" CreationDate="2017-05-03T16:26:42.773" Score="1" Body="&lt;p&gt;Your question is stability. Will the design turn over? You don't need any software for this, not even algebra. Only a ruler (ok, you could do this in CAD). What you need to do, is make a Free Body Diagram (FBD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The FBD has:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a center of gravity,&lt;/li&gt;&#xA;&lt;li&gt;several pivot points,&lt;/li&gt;&#xA;&lt;li&gt;a human messing with your design.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The center of gravity is where the weight of your design applies (red). The pivot points is the point around which your design would turn over, I think these would be the wheels of your design (pawns). The human is the force trying to turn over your design (blue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gRehX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gRehX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue and red force cross at a certain point. That's where you can add them up. If you draw a triangle to the pivots from that point, and the resulting force stays within the triangle, it's stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add up forces by putting them head-to-tail, starting from the crossing point. As you can see, here it's just unstable. But that's enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to make this work for you, you have to draw both the dimensions and forces scale. It doesn't matter how big the forces are in relation to the dimensions, just draw them in a practical size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this for slightly more complicated designs too! Here is an example of mine: a lifting crane with several weights, loads, working under an angle, and fully parametric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VjwdK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VjwdK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11163" LastActivityDate="2017-05-03T16:26:42.773" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15111" PostTypeId="2" ParentId="6074" CreationDate="2017-05-03T17:30:20.150" Score="1" Body="&lt;p&gt;If you are looking for an individually-addressable mirror array, a Digital Micromirror Device (DMD) is what you are looking for. It is used to adjust the brightness of pixels inside a DLP projector by adjusting how much light is display per pixel via pulse-width modulation. By adjusting how fast the mirrors reflect from on to off, the pixels can be adjusted for brightness. The downside is that these are generally very small devices do to their application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're simply looking for a way to turn a clear pane of glass into a mirror, 'Smart Glass' may be what you're looking for. Electrochromic panels allow you to pass a current through them to adjust opacity, but not reflectance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please edit your comment to reflect (see what I did there?) your intended application and we can further help you identify the component/material you are looking for.&lt;/p&gt;&#xA;" OwnerUserId="11169" LastActivityDate="2017-05-03T17:30:20.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15112" PostTypeId="2" ParentId="12378" CreationDate="2017-05-03T17:41:24.017" Score="1" Body="&lt;p&gt;QR codes are &lt;em&gt;very&lt;/em&gt; accurate. QR Code was originally invented in automotive assembly environments to allow for high-speed scanning of components and parts. To allow for them to be scanned at high speeds, they must be able to be scanned very accurately to ensure no errors during the assembly process which can lead to Quality Assurance and Quality Control issues.&lt;/p&gt;&#xA;&lt;p&gt;So how accurate are they? Well that depends on the code design. Depending on the error-correction level and QR Code version, some labels allow for more errors than others. To quote &lt;a href=&quot;https://en.wikipedia.org/wiki/QR_code#Error_correction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Codewords are 8 bits long and use the Reed–Solomon error correction algorithm with four error correction levels. The higher the error correction level, the less storage capacity. The following table lists the approximate error correction capability at each of the four levels:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Level L (Low)   7% of codewords can be restored.&lt;/li&gt;&#xA;&lt;li&gt;Level M (Medium)    15% of codewords can be restored.&lt;/li&gt;&#xA;&lt;li&gt;Level Q (Quartile)  25% of codewords can be restored.&lt;/li&gt;&#xA;&lt;li&gt;Level H (High)  30% of codewords can be restored.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;In larger QR symbols, the message is broken up into several Reed–Solomon code blocks. The block size is chosen so that at most 15 errors can be corrected in each block; this limits the complexity of the decoding algorithm. The code blocks are then interleaved together, making it less likely that localized damage to a QR symbol will overwhelm the capacity of any single block.&lt;/p&gt;&#xA;&lt;p&gt;Due to error correction, it is possible to create artistic QR codes that still scan correctly, but contain intentional errors to make them more readable or attractive to the human eye, as well as to incorporate colors, logos, and other features into the QR code block.&lt;/p&gt;&#xA;&lt;p&gt;It is also possible to design artistic QR codes without reducing the error correction capacity by manipulating the underlying mathematical constructs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11169" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-05-04T02:49:22.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15114" PostTypeId="2" ParentId="15064" CreationDate="2017-05-03T19:42:08.840" Score="1" Body="&lt;p&gt;I can only try to give you an approximate answer. The principle is simple. We give rotational energy $E_{\text{rot}}$ into the system and we get axial force $F_{\text{ax}}$ times displacement $\Delta x$, taking into account that the system is not ideal we also need to multiply the rotational energy by the efficiency $\eta$. Assuming the axial force is constant we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta E_{\text{rot}}=F_{\text{ax}} \Delta x.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for a screw with thread pitch $\tan(\alpha)$ and mean diameter $d$ the variable $\Delta x$ can be calculated for $N$ rotations as $\Delta x = N\pi d\tan(\alpha)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to determine $E_{\text{rot}}$ we assume a radial input force $F_1$ and a lever of length $r$. Hence, the rotational energy for $N$ rotations is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ E_{\text{rot}}=N(2\pi r F_1).$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging these results into the first equation and solving for $F_{\text{ax}}$ yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{\text{ax}}=\eta \frac{E_{\text{rot}}}{\Delta x}=\eta \frac{N(2\pi r F_1)}{N\pi d \tan(\alpha)}=\eta\frac{2rF_1}{d\tan(\alpha)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this system is nothing else than a gearbox. Depending on your input lever $r$ (can be measured or estimated from the picture), mean diameter of the screw $d$ (can be measured or estimated from the picture), the efficiency $\eta$ and the pitch $\tan(\alpha)$ (can be measured or estimated from the picture) it is possible to get almost every axial force that you like. The Force $F_1$ can also be estimated by pushing against a balance and measuring the weight. The problem is, that at some point the stresses inside the wooden screw will reach a limit and the wooden screw will begin to bulge. I don't know the exact value for which wood will start to bulge under compression, but this will dictate the practical limit for your axial force.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Estimation of efficiency&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get the efficiency you simply solve the last equation for $\eta$. The other measures can be obtained by measurements. You need to do measurements as it is very unlikely that you will have an analytic solution for the efficiency without using any kind of measurements. One thing is for sure, the lower limit is an efficiency of zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta = d\frac{\tan(\alpha)}{2r}\frac{F_{\text{ax}}}{F_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only unknown in the system is the axial force $F_{\text{ax}}$. Just take the screw and deform a metal with given Young's modulus $E$ (the law should almost be the same for compression and strain for &quot;isotropic&quot; materials). Measure the strain $\varepsilon$ and use Hook's law to get the stress $\sigma=E\varepsilon$. Then $F_{\text{ax}}=\sigma A_{\text{cross section}}.$ &lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-05-03T20:04:20.567" LastActivityDate="2017-05-03T20:04:20.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15115" PostTypeId="1" CreationDate="2017-05-03T20:08:50.597" Score="2" ViewCount="8104" Body="&lt;p&gt;Are electronic components like diodes,transistors, micro-controllers, microprocessors, wireless transceivers etc are damaged by strong static or dynamic magnetic fields? If yes, at what strength the field becomes dangerous? Also, what are the ways to protect the components against such fields?&lt;/p&gt;&#xA;" OwnerUserId="11171" LastEditorUserId="10684" LastEditDate="2017-07-03T13:07:44.327" LastActivityDate="2023-02-21T23:12:55.980" Title="Effect of magnetic field on electronic components" Tags="|magnets|rf-electronics|microelectronics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15116" PostTypeId="2" ParentId="8290" CreationDate="2017-05-03T21:13:32.103" Score="1" Body="&lt;p&gt;According to my reference book on thermodynamics, the efficiency is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\eta = 1-\frac{1}{\varepsilon^{\varkappa-1}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat capacity ratio is given by $\varkappa$ and $\varepsilon$ is the compression ratio. The compression ratio is given by $\varepsilon =\frac{V_k+V_h}{V_k}$, in which $V_k$ is the volume after the isentropic compression and $V_k+V_h$ is the initial volume. Assuming $\varepsilon = r$ and $\varkappa = 1.4$ gives exactly your result. But $\varkappa$ does depend on temperature and pressure. So it might be that in the exercise they used another $\varkappa$ value or another fluid at all. It would help us if you could provide the full exercise statement and the solution, with which you compared your solution.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-05-03T21:13:32.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15117" PostTypeId="1" AcceptedAnswerId="15147" CreationDate="2017-05-03T22:43:38.133" Score="2" ViewCount="95" Body="&lt;p&gt;I'm wondering if the structural integrity of a solid cylindrical pylon is just as sturdy as a cubical one of the same volume given that they're both the same height.&lt;/p&gt;&#xA;" OwnerUserId="11173" LastEditorUserId="11173" LastEditDate="2017-05-04T23:01:23.793" LastActivityDate="2017-05-05T10:43:18.407" Title="is a cylindrical pylon just as sturdy as a cubical one of the same volume?" Tags="|structural-engineering|structural-analysis|structures|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15118" PostTypeId="2" ParentId="14876" CreationDate="2017-05-03T22:49:21.260" Score="0" Body="&lt;p&gt;To expand on previous answers, you're solving Bernoulli's equation which describes the conservation of mechanical fluid energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_1+\frac{\rho V_1^2}{2}+\rho g h_1 = P_2+\frac{\rho V_2^2}{2}+\rho g h_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming atmospheric pressure at the exit is 0, fluid in the tank has zero velocity, $h$ is the depth of the tank at the outlet and change in potential energy is neglected, the formula simplifies to:&#xA;$$P_1 = \frac{\rho V_2^2}{2}$$ where $P_1$ is equal to $\rho g h$ and simplifies to:&#xA;$$g h=\frac{V_2^2}{2}$$&#xA;You can solve for the exit velocity and you already know your flow rate:&#xA;$$V_2 = \sqrt{2gh}{}$$&#xA;$$Q_{actual}=C_d A V_2=C_d\frac{\pi D^2}{4}\sqrt{2gh}$$&#xA;Now you can solve for the diameter of the outlet:&#xA;$$D=\sqrt{\frac{4Q_{actual}}{C_D \pi\sqrt{2gh}}}$$&lt;/p&gt;&#xA;" OwnerUserId="11151" LastActivityDate="2017-05-03T22:49:21.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15120" PostTypeId="2" ParentId="2719" CreationDate="2017-05-03T23:41:29.350" Score="0" Body="&lt;p&gt;If the two loads are applied at the same point and in the same direction, they can be added together and represented as a single load in the same direction. I can't think if any exception to this case for constant loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding distributed loads, you can represent them with a point load for the purposes of determining reaction forces, but not for calculating internal moment or shear.&lt;/p&gt;&#xA;" OwnerUserId="11151" LastActivityDate="2017-05-03T23:41:29.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15121" PostTypeId="1" AcceptedAnswerId="15122" CreationDate="2017-05-04T05:15:01.103" Score="2" ViewCount="1330" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VbThd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VbThd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every train car I can remember has its axles arranged like the photo. Two are put very close together in one chassey, and two more the same way at the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why do that? If you want 4 axles, isn't it better to spread them out evenly like I drew? Or if you just want axles at the ends, why not just have 2 total instead of 4 unevenly spaced ones?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So why are there unevenly spaced axles?&lt;/strong&gt; If you want only 2 pivot points, why not just have 2 axles total (which would be spread evenly if they are equal distances from the ends of the car)? If you want 4 axles (presumably for load bearing), why not spread the axles evenly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not find this answer via googling. It might help if I knew the exact terminology of that chassey holding two axles.&lt;/p&gt;&#xA;" OwnerUserId="3510" LastEditorUserId="3510" LastEditDate="2017-05-05T01:42:24.917" LastActivityDate="2021-05-03T19:43:57.520" Title="Why do train cars have unevenly spaced axles?" Tags="|rail|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15122" PostTypeId="2" ParentId="15121" CreationDate="2017-05-04T05:40:51.457" Score="6" Body="&lt;p&gt;If you spread them out evenly then taking curves is more difficult - it's the turning circle : as the carriage goes round a curve the axles in the middle will need to move sideways in your diagram.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-04T05:40:51.457" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15124" PostTypeId="2" ParentId="15115" CreationDate="2017-05-04T06:02:05.360" Score="0" Body="&lt;p&gt;For protection : a Faraday cage.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-04T06:02:05.360" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15125" PostTypeId="2" ParentId="15121" CreationDate="2017-05-04T08:32:50.250" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bogie&quot; rel=&quot;noreferrer&quot;&gt;Bogies&lt;/a&gt;.  Each par of axles is on a small articulated platform, which can self steer, and helps smooth out bumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a wheel hits a bump. the other one dips, both their springs share the load. The central pivot point halfway between them doesn't move much, giving a very smooth ride.&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2017-05-04T08:32:50.250" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15127" PostTypeId="2" ParentId="15094" CreationDate="2017-05-04T12:46:21.933" Score="3" Body="&lt;p&gt;If you say two lines (towers) are parallel, you state a relationship between, and only between, the two lines. The relationship you are looking for is not just between the two lines. Therefore, there is no word to describe that relationship.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You need to include, for instance, the curvature of the earth. You could state: &lt;strong&gt;the two towers are &lt;em&gt;both&lt;/em&gt; perpendicular to the curvature of the earth.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Instead of &lt;em&gt;curvature&lt;/em&gt;, how about &lt;em&gt;center of gravity&lt;/em&gt; of the earth. The forces would like that. Then you could state: &lt;strong&gt;the two towers intersect each other at the &lt;em&gt;center of gravity&lt;/em&gt; of the earth&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is just no way to describe a three-way relationship with just two of the three parts.&lt;/p&gt;&#xA;" OwnerUserId="11163" LastActivityDate="2017-05-04T12:46:21.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15128" PostTypeId="1" CreationDate="2017-05-04T13:36:42.320" Score="1" ViewCount="30" Body="&lt;p&gt;I'm working on a P&amp;amp;ID drawing and need to add a symbol for an &quot;Atmospheric Break&quot;. Is there a symbol for this that I can use to represent this? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="11181" LastActivityDate="2017-05-04T13:36:42.320" Title="Atmospheric Break" Tags="|pi-diagram|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15130" PostTypeId="1" CreationDate="2017-05-04T15:55:54.783" Score="2" ViewCount="784" Body="&lt;p&gt;I'm designing a truss which has 9 elements in &lt;code&gt;ANSYS Workbench&lt;/code&gt;. since it is a truss, all of the elements are truss elements and can not tolerate momentum. I used &lt;code&gt;LINK 188&lt;/code&gt; element type for them. now, all I want is to define 2 of those elements as beams so that momentum can be transmitted between them(so we want 8 truss members and 2 beam members). but the important point is that only momentum about y axis must be transmitted, not momentum about z axis. I used &lt;code&gt;BEAM 188&lt;/code&gt; element type for these 2 members but it constraints all DOFs. any suggestion?!&lt;/p&gt;&#xA;" OwnerUserId="10304" LastActivityDate="2017-05-04T15:55:54.783" Title="how to release specific degree of freedom in ANSYS Workbench?" Tags="|structural-analysis|finite-element-method|ansys|ansys-workbench|ansys-apdl|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15133" PostTypeId="2" ParentId="15121" CreationDate="2017-05-04T18:24:08.073" Score="1" Body="&lt;p&gt;It is all about the turning radius of the train. To understand it you must look from above. Scroll down a bit on the link below to see a model train on a curving track. If there were wheels in the center then they would have to slide way off center when the train is rounding a sharp bend or turning onto a siding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ngaugelayouts.com/n-scale-layouts.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ngaugelayouts.com/n-scale-layouts.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wheels are positioned close to the coupling between trains to allow for the curving track. If you look closely at modern trains they have reduced the number of wheel trucks by pivoting 2 rail cars on one wheel truck. If you look closely there are 6 axles on this section (yes I realize this is lego but it's the best I could find). There are 2 axles at the front and back, but only 2 instead of 4 between the cars. I have seen groupings like this that were 5 or 6 cars long, reducing the weight of the train by one whole 2 axle wheel truck assembly for each car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.eurobricks.com/forum/index.php?/forums/topic/82085-moc-ttx-intermodal-for-7939-container-size/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.eurobricks.com/forum/index.php?/forums/topic/82085-moc-ttx-intermodal-for-7939-container-size/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:16:39.207" LastActivityDate="2017-07-23T18:16:39.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15134" PostTypeId="1" AcceptedAnswerId="15137" CreationDate="2017-05-04T20:19:24.827" Score="3" ViewCount="46" Body="&lt;p&gt;I am currently working on creating a tool to calculate the maximum heating rate on a re-entry vehicle given starting conditions and coefficients. I am using &lt;a href=&quot;https://arc.aiaa.org/doi/book/10.2514/4.862342&quot; rel=&quot;nofollow noreferrer&quot;&gt;Re-Entry Aerodynamics by Wilbur L. Hankey&lt;/a&gt; as my current source. On page 30, he introduces the following equation as a way to calculate the heating rate:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://chart.googleapis.com/chart?cht=tx&amp;amp;chl=%5Cdot%7Bq%7D%20%3D%20K%20%5Crho%5Em%20V%5E3&quot; alt=&quot;Equation&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The problem is that I do not understand the equation. The way I interpret it is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://chart.googleapis.com/chart?cht=tx&amp;amp;chl=%5Ctext%7BRate%20Of%20Change%20Of%20Heat%7D%20%3D%20%5Ctext%7BThermal%20Condcutivity%7D%20*%20%5Ctext%7BDensity%20Of%20Air%7D%5E%7B%5Ctext%7BMass%20OfObject%7D%7D*%5Ctext%7BVelocity%7D%5E3&quot; alt=&quot;Equation&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;What doesn't make sense to me is the ρ^m in the equation. If you attempt to solve for the units of ρ^m, you get it as K s^3/m^4, which makes no sense to me.&lt;/p&gt;&#xA;&lt;p&gt;My researching myself has been unable to lead me to an origin or explanation for the equation, and neither my coworkers or my student friends have ever seen it before. If someone is able to give me some idea what I'm looking at, that would be great.&lt;/p&gt;&#xA;" OwnerUserId="11188" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-05-04T23:47:51.967" Title="Finding Source of Heating Rate Equation" Tags="|fluid-mechanics|aerospace-engineering|aerodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15135" PostTypeId="1" CreationDate="2017-05-04T22:04:28.173" Score="1" ViewCount="516" Body="&lt;p&gt;I have a question regarding the sizing of a proper transformer. I have a source voltage and load. See the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Primary Source: 480VAC 100A (3phase)&#xA;Secondary: 415 VAC (3phase) Load: 2 parallel AC-DC units 10kW, Power Factor = 0.8 (ea). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VBC34.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VBC34.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you go about sizing a proper transformer to interface between the source and the load?&lt;/p&gt;&#xA;" OwnerUserId="11189" LastEditorDisplayName="user16" LastEditDate="2017-05-05T01:01:36.177" LastActivityDate="2018-12-18T16:08:37.293" Title="How to size proper Transformer (3phase quesiton)" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15137" PostTypeId="2" ParentId="15134" CreationDate="2017-05-04T23:47:51.967" Score="1" Body="&lt;p&gt;$m$ is an empirical curve-fitted constant, depending on the composition of the atmosphere, not a mass. For earth and mars, $m = 0.5$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually the &quot;3&quot; is also a curve-fitted constant. Your reference to Hankey is behind a paywall so I don't know why he fixed that at 3 but left $m$ variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://tfaws.nasa.gov/TFAWS12/Proceedings/Aerothermodynamics%20Course.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://tfaws.nasa.gov/TFAWS12/Proceedings/Aerothermodynamics%20Course.pdf&lt;/a&gt; page 19.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-05-04T23:47:51.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15138" PostTypeId="1" CreationDate="2017-05-05T02:01:32.653" Score="2" ViewCount="95" Body="&lt;p&gt;I have a big button &lt;a href=&quot;http://store.2xlnetworks.com/thebigbutton/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://store.2xlnetworks.com/thebigbutton/&lt;/a&gt; and a display counter Microframe Model 265 &lt;a href=&quot;https://www.microframecorp.com/d0265-6-digit-rugged-computer-controlled-display&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.microframecorp.com/d0265-6-digit-rugged-computer-controlled-display&lt;/a&gt; . &#xA;I want the push button to make the display counter go up one digit every time its pushed. Any idea how to go about this problem ?&lt;/p&gt;&#xA;" OwnerUserId="11190" LastActivityDate="2017-09-06T02:25:29.823" Title="How do I hook these components together to get my desired functionality?" Tags="|electrical-engineering|power-electronics|computer-engineering|systems-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15140" PostTypeId="1" CreationDate="2017-05-05T04:23:18.187" Score="3" ViewCount="301" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ypFij.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ypFij.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering about this specific type of truss in particular, any links to further source information and images regarding this particular geometry would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="11193" LastActivityDate="2017-05-05T09:48:08.517" Title="Is there a specific name for diagonally braced truss members like those on the Queensboro Bridge?" Tags="|structural-engineering|structures|steel|bridges|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15141" PostTypeId="1" AcceptedAnswerId="15153" CreationDate="2017-05-05T04:51:45.443" Score="0" ViewCount="177" Body="&lt;p&gt;I am doing mechatronics engineering and i am to choose an elective module. We have the choice of taking up Digital Signal Processing or Power Electronics and Drive. Majority of my friends are taking DSP cause our lecturer says its more useful for mechatronic student. I personally feel that PED has a better application for robotics. Could i get some advice on which is more reasonable for robotics.&lt;/p&gt;&#xA;" OwnerUserId="11192" LastActivityDate="2017-05-05T13:40:18.687" Title="PED or DSP for mechatronic student" Tags="|electrical-engineering|power-electronics|signal-processing|" AnswerCount="3" CommentCount="3" ClosedDate="2017-05-05T22:47:34.167" ContentLicense="CC BY-SA 3.0" />
  <row Id="15142" PostTypeId="2" ParentId="15141" CreationDate="2017-05-05T04:59:53.200" Score="1" Body="&lt;p&gt;In my view you should pick what you are interested in - that's why courses are designed to be modular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you have the choice to follow all the others : DSP or take PED and end up doing something different. Think about the robots that are being built at the moment that move heavy weights...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really it's only my opinion : I started with an apprenticeship as a vehicle electrician, now I lecture in Facilities...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-05T04:59:53.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15143" PostTypeId="1" CreationDate="2017-05-05T05:14:49.133" Score="4" ViewCount="898" Body="&lt;p&gt;I have just purchased a 45 W 12 V transformer with an integral photocell for landscape lights. I measured the voltage while it was plugged in and the photocell closed the circuit (off) and there was still about 3 volts across the terminals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, since the transformer is delivering 3 volts to the terminals, how much energy is it consuming just by being plugged in with no lights attached? Is this normal or bad design?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Input 120VAC 60Hz 52VA&#xA;OUTPUT 12VAC 3.75A 45VA&lt;/p&gt;&#xA;" OwnerUserId="11195" LastEditorUserId="11195" LastEditDate="2017-05-05T14:34:40.363" LastActivityDate="2017-05-05T14:34:40.363" Title="Does a transformer use power when devices aren't attached?" Tags="|electrical-engineering|consumer-electronics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15144" PostTypeId="2" ParentId="15143" CreationDate="2017-05-05T05:37:09.567" Score="1" Body="&lt;p&gt;Yes, it does use power, but only a small amount compared to its capacity. It's one of the issues that is called &quot;standby load&quot; and all who leave phone chargers plugged in when not charging their phones contribute to it. One solution is to switch it off - the extension leads with a supply switch work well - phone, laptop, tablet all off when not needed.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-05T05:37:09.567" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15145" PostTypeId="2" ParentId="15140" CreationDate="2017-05-05T05:39:20.327" Score="0" Body="&lt;p&gt;Trusses, struts, braces...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engineering question is usually is it in tension or compression .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for examples you could check out roofing designs as well as some of the old wooden ships - in some countries ship builders did roofs : all they did was invert the structure.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-05T05:39:20.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15146" PostTypeId="2" ParentId="15141" CreationDate="2017-05-05T07:11:58.013" Score="1" Body="&lt;p&gt;As Solar Mike already stated, pick what fits your personal interest (the topic, not any potential future advantages).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've experienced in my studies (electronics) and the bunch of elective modules I took, is that you won't become an expert anyways in the (usually) short time. As for potential future upsides, since you probably cannot tell for sure, what direction (or company) you will take after your studies, neither can you say which one is the better for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the two topics, I think both of them are reasonable choices for robotics. Some other criteria I considered:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Interest&lt;/li&gt;&#xA;&lt;li&gt;Lecturer&lt;/li&gt;&#xA;&lt;li&gt;Number of students&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10862" LastActivityDate="2017-05-05T07:11:58.013" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15147" PostTypeId="2" ParentId="15117" CreationDate="2017-05-05T08:46:47.587" Score="3" Body="&lt;p&gt;I assume that both pylons are composed of the same material. If you have a cylindrical pylon with height $h$ and volume $V$ then the radius can be calculated by $V = hA_{\text{base}} =h\pi r^2 \implies r = \sqrt{\frac{V}{h\pi}}$. For the cuboid pylon with a square cross-section with a length $a$ for the edge we can calculate $a$ by $V = h A_{\text{base}}=ha^2 \implies a=\sqrt{\frac{V}{h}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what you mean by sturdy we can distinguish different kinds of loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Short summary&lt;/strong&gt;: Circular structures are stiffer but more expensive to manufacture and sometimes not suited for the usage (e.g. as floor structure in buildings, where commonly I-beams are used). Read the part below for more detailed answer.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Axial load (compressive and tensile in elastic regime) $F$: As $F=\sigma A_{\text{base}}$ both structures have same &quot;sturdyness&quot; because $A_{\text{base}}$ is the same for both cases, which implies $\sigma$ to be the same. But this is only true in theory, because we neglected the effects of the geometry differences and the finite length of the structures.&lt;/li&gt;&#xA;&lt;li&gt;Torsional load $T$: For this case we have $\tau_{\text{max}}=\frac{T}{W_T}$, in which $W_T$ is the torsional resistance torque. For a circular cross-section, we have $W_T=\frac{\pi}{2}r^3$ and for a square cross section we have $W_T=0.208 a^3$. If you calculate &#xA;$$\frac{\tau_{\text{max,r}}}{\tau_{\text{max,a}}}=2\cdot 0.208 \sqrt{\pi}\approx 0.737$$ you will see that the maximal shear stress for the circular structure is preferable, as it is less. Similarly by using $\dfrac{d \varphi}{dx}=\frac{T}{GI_T}$, in which $G$ is the shear modulus (is same for both cases) and $I_T$ is the torsion constant. The integral of the derivative $\dfrac{d \varphi}{dx}$ gives the amount of rotation at length $x$.  $I_{T,r}=\frac{\pi}{2}r^4$ and $I_{T,a}=0.141a^4$. Again consider the fraction $$\frac{(d \varphi/dx)_{r}}{(d \varphi/dx)_{a}}=\frac{I_{T,a}}{I_{T,r}}=2\cdot 0.141 \pi\approx 0.886.$$ This implies that the circular cross section will show less rotational deflection unter the same torsional load $T$.&lt;/li&gt;&#xA;&lt;li&gt;Consider the bending torque $T_y$. The bending line $w(x)$ can be determined for Bernoulli-Euler-beams by solving the following differential equation $$EI_yw''(x)=-T_y.$$ For circular cross-section $I_{y,r}=\frac{\pi}{4}r^4$ and for square cross-section $I_{y,a}=\frac{1}{12}a^4$. If you integrate the differential equation and apply boundary conditions you can calculate the maximal bending / deflection. You could also consider the stress due to $T_y$ which is given by &#xA;$$\sigma = \frac{T_y}{I_y}z_{\text{height in beam cross section}}.$$&#xA;The fraction $\sigma_{r}$ by $\sigma_{a}$ is given as:&#xA;$$\frac{\sigma_r}{\sigma_a}=\frac{I_{y,r}}{I_{y,a}}=\frac{3}{\pi}\approx 0.955.$$ Which indicates that the stress in the circular structure is more favorable.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-05-05T10:43:18.407" LastActivityDate="2017-05-05T10:43:18.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15148" PostTypeId="2" ParentId="15140" CreationDate="2017-05-05T09:13:37.330" Score="3" Body="&lt;p&gt;There are a variety of names for these, but they all have a similar theme:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Laced strut/tie &lt;/li&gt;&#xA;&lt;li&gt;Lattice or Batten bracing &lt;/li&gt;&#xA;&lt;li&gt;Lattice truss &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lattice_girder&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lattice Girder&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm sure you get the idea. These were popular in the 19th/early 20th Century before the development of more efficient/cost effective structural sections such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Hollow_structural_section&quot; rel=&quot;nofollow noreferrer&quot;&gt;Structural Hollow Sections&lt;/a&gt;, castellated beams, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One paper which discusses the analysis of such elements is available &lt;a href=&quot;https://www.istructe.org/journal/volumes/volume-30-(published-in-1952)/issues/issue-5/articles/a-theory-for-struts-with-lattice-or-batten-bracing&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2017-05-05T09:48:08.517" LastActivityDate="2017-05-05T09:48:08.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15149" PostTypeId="2" ParentId="15140" CreationDate="2017-05-05T09:14:23.750" Score="0" Body="&lt;p&gt;Quadrangular Trough Warren. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I googled &quot;names for truss frame configurations&quot; for you: &lt;a href=&quot;https://s-media-cache-ak0.pinimg.com/originals/f7/3a/8c/f73a8c855f1f4669903fe9d866ddc0d8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://s-media-cache-ak0.pinimg.com/originals/f7/3a/8c/f73a8c855f1f4669903fe9d866ddc0d8.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11163" LastActivityDate="2017-05-05T09:14:23.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15150" PostTypeId="2" ParentId="12669" CreationDate="2017-05-05T10:00:25.090" Score="1" Body="&lt;p&gt;The problem is best understood with the text at top - that this is a frame for a bicycle, not a structure.  Statics is not the correct method for approaching bicycle frames for a variety of reasons, but problem makers need to keep it interesting.  Because it is a frame, none of the joints are truly pinned, which is why the welds are introduced.  At a minimum, arm AB is welded in place on the frame to minimize the front wheel from pulling away from the front, or too close to the pedals.  With this information, the problem could be approached several ways, from simple to insane, depending on your desired level of realism, and profit margins from building this bicycle:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Add a single moment $B_z$ onto B.  Summing moments about B should show how much arm AB is bending while the bike is moving.  To reduce the problem to a truss problem, we will make the assumption that only member AB experiences bending.  The bending is simple (see below).  This isn't a good model for a variety of reasons, but it solves the problem on paper.&lt;/li&gt;&#xA;&lt;li&gt;As described in the paper add a reaction $B_x$ and $C_x$, which makes no intuitive sense.  However, this is actually equivalent to answer 1 - which is why this type of problem is used frequently at end-of-term, it begins developing the concepts of strengths of materials.  This method allows the use of statics techniques to solve it, but won't require a full strengths analysis.  Summing forces in x directions will show $B_x$ = $-C_x$, and that the couple formed by these two reactions will have a moment equal to $B_z$.&lt;/li&gt;&#xA;&lt;li&gt;Realizing that the frame is still going to bend anyways (you already added a moment onto one member), turn this into a strengths problem, with all welds forming reactionary moments ($M = k*\theta$, k to be determined based upon weld length and material).  An FEA would expedite the process.  Note that deflection of A and E in y should be 0, as both tires should rest parallel to the ground.  &lt;/li&gt;&#xA;&lt;li&gt;Realizing that the loading of a human onto the bicycle frame is probably more complex and distributed then the point loads shown (and more than likely has dynamic road loads, climbing loads, accelerations within and without of the plane, impact loadings, and that human pedal pushing will almost always necessitate the need for loading on node C.), build a bicycle, cover it with strain gauges, and measure performance with real humans to ensure all assumptions are accurate.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All of this goes back to the first method as a bare bones basis though.  To begin, split $W_f$ into two components - one parallel to the member, and one perpendicular.  Acting in the truss equations, the parallel reaction will carry through to the rest of the equations, while the perpendicular component is the bending force that will act to bend AB.  Moment $B_z$ will need to exactly cancel this moment to ensure that only AB is experiencing bending, and the remaining members will undergo static deformations only.  This will ensure that the weld load will make AB act in pure bending alone, otherwise all of the equations would need to be solved as bending equations, not as truss equations (method 3 above). Therefore: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B_z = W_f*\cos(75)*\frac{4in}{\cos(75)} = 4&quot; * W_f$$&#xA;Then, summing moments about B: &#xA;$$B_z - 4in * W_f - 24in * 100 +31in * W_b = 0 \to W_b = \frac{100lb*24in}{31in} = 77.42lb$$&#xA;And summing forces in Y: &#xA;$$W_f+W_b-100lb-33lb = 0 \to W_f = \frac{133lb*31in - 100lb*24in}{31in} = 55.58lb$$&#xA;Which means $B_z = 55.58lb*4in = 222.32in*lb$, and the parallel portion of AB is simply $W_f*\sin(75) = 53.69lb$, which can then be put onto Node B, tilted at a 75 degree angle.  The rest of the equations follow simple statics. &lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2017-05-05T10:08:29.077" LastActivityDate="2017-05-05T10:08:29.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15152" PostTypeId="2" ParentId="15143" CreationDate="2017-05-05T12:30:06.293" Score="1" Body="&lt;p&gt;The loss in a transformer is due to a variety of real-world factors such as non-zero DC resistance in the wiring.  Quoting from (what else!) the &lt;a href=&quot;https://en.wikipedia.org/wiki/Transformer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; page,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For simplification or approximation purposes, it is very common to&#xA;  analyze the transformer as an ideal transformer model as presented in&#xA;  the two images.[5] An ideal transformer is a theoretical, linear&#xA;  transformer that is lossless and perfectly coupled; that is, there are&#xA;  no energy losses and flux is completely confined within the magnetic&#xA;  core. Perfect coupling implies infinitely high core magnetic&#xA;  permeability and winding inductances and zero net magnetomotive&#xA;  force.[6][c]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Deviations from ideal The ideal transformer model neglects the&#xA;  following basic linear aspects in real transformers:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;a) Core losses, collectively called magnetizing current losses,&#xA;  consisting of[17]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Hysteresis losses due to nonlinear application of the voltage applied&#xA;  in the transformer core, and Eddy current losses due to joule heating&#xA;  in the core that are proportional to the square of the transformer's&#xA;  applied voltage. b) Whereas windings in the ideal model have no&#xA;  resistances and infinite inductances, the windings in a real&#xA;  transformer have finite non-zero resistances and inductances&#xA;  associated with:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Joule losses due to resistance in the primary and secondary&#xA;  windings[17] Leakage flux that escapes from the core and passes&#xA;  through one winding only resulting in primary and secondary reactive&#xA;  impedance.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-05-05T12:30:06.293" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15153" PostTypeId="2" ParentId="15141" CreationDate="2017-05-05T13:40:18.687" Score="2" Body="&lt;p&gt;DSP is relevant to any engineering field that involves gathering data and/or controlling something. That includes almost all robotics applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not quite clear just from the title what &quot;Power electronics and drive&quot; means. &quot;Power electronics&quot; could be anything from a 20W amplifier driving a motor, through a few MW for transport applications (e.g. electric trains), up to GW of power and systems operating at hundreds of kV for electricity generation and distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that working in industry you will almost certainly end up knowing something about both fields, so choosing which to study right now might not be too important in the long run. But I would suggest that DSP would be easier to learn from an academic course, and power electronics easier to learn &quot;on the job&quot; as and when you need it, so long as your degree course has covered the basics of (low power) analog electronic circuit design and analysis.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-05-05T13:40:18.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15154" PostTypeId="2" ParentId="15135" CreationDate="2017-05-05T14:17:25.727" Score="0" Body="&lt;p&gt;So from what I remember  &lt;span class=&quot;math-container&quot;&gt;$P = \sqrt3 V\ I\ PF\ $&lt;/span&gt; and this gives just under 18A for each of your loads.&#xA;This means there is sufficient supply for only one running at any time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choosing the transformer will also need considerations as to starting load - usually different to running load and other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could start with this link : &lt;a href=&quot;http://www.schneider-electric.us/en/faqs/FA101694/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.schneider-electric.us/en/faqs/FA101694/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorDisplayName="user14407" LastEditDate="2018-12-18T16:08:37.293" LastActivityDate="2018-12-18T16:08:37.293" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15155" PostTypeId="2" ParentId="15094" CreationDate="2017-05-05T16:09:59.347" Score="2" Body="&lt;p&gt;Orthogonal to imaginary surface of Earth. &lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2017-05-05T16:09:59.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15158" PostTypeId="1" CreationDate="2017-05-05T18:57:52.033" Score="0" ViewCount="645" Body="&lt;p&gt;What is the best resolution obtainable by ultrasonic non destructive testing? I understand it depends on frequency. &lt;/p&gt;&#xA;" OwnerUserId="11203" LastActivityDate="2017-05-05T20:30:10.317" Title="What is the best resolution obtainable by ultrasonic non destructive testing?" Tags="|ultrasound|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15159" PostTypeId="2" ParentId="15094" CreationDate="2017-05-05T19:40:36.000" Score="2" Body="&lt;p&gt;Radial from the center of the earth?&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2017-05-05T19:40:36.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15160" PostTypeId="2" ParentId="15158" CreationDate="2017-05-05T20:30:10.317" Score="1" Body="&lt;p&gt;The resolution depends on the material you are using and the frequency $f$ applied. An approximation for the resolution is given by half of the wavelength $$\lambda = \frac{c}{f}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sonic speed $c$, in meters per second, for some materials, is given by the following list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Steel: 5900&lt;/li&gt;&#xA;&lt;li&gt;Aluminum: 6320&lt;/li&gt;&#xA;&lt;li&gt;Brass: 4430&lt;/li&gt;&#xA;&lt;li&gt;Lead: 2160 &lt;/li&gt;&#xA;&lt;li&gt;Water: 1480 &lt;/li&gt;&#xA;&lt;li&gt;Air: 330.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Depending on the mechanism that is used to generate the ultrasonic sound waves (piezoelectric effect: $f=200 \text{ kHz} \ldots 20 \text{ MHz}$ or megnetorestricitve effect: $f=10 \text{ kHz} \ldots 200 \text{ kHz}$) you will get different resolutions. A common frequency range for metals is between $0.2 \text{ MHz}$ up to $20 \text{ MHz}$. &lt;strong&gt;The resulting resolution is then $0.1 \text{ mm}$ for steel.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-05-05T20:30:10.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15161" PostTypeId="1" CreationDate="2017-05-05T21:21:25.673" Score="4" ViewCount="63" Body="&lt;p&gt;I have seen some resources on the Internet that claim to be able to make sheets of graphene (single atom thick lattice of hex-bonding carbon) from commercial off the shelf materials and a modest lab setup.  I have great interest in graphene and would like to attempt the manufacture but I don't want to waste my time with bunk.  Can anyone recommend a tried and proven way to manufacture without very elaborate and expensive lab?&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-10-26T02:15:03.403" Title="Can graphene really be made in a modest lab at home?" Tags="|materials|manufacturing-engineering|experimental-physics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15164" PostTypeId="2" ParentId="15054" CreationDate="2017-05-05T23:04:25.150" Score="3" Body="&lt;p&gt;I would keep the sensors passive as transponders to a house wide rfid radar scanner.  Changes in the sensor would change the code on the rfid of the sensor, thus reflecting back a different transponder code to the central scanner. See &lt;a href=&quot;http://www.harting-rfid.com/en/applications/rfid-transponders-with-external-sensors/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Intelligent RFID transponder takes over monitoring function&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example would be a temperature sensor where the rfid code would vary linearly with the temperature.  The central rfid scanner would then pick up the new sensor data from its next scan. No wires, no sensor power requirement, done deal except for maybe cost.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastEditorUserId="7083" LastEditDate="2017-05-09T06:10:54.957" LastActivityDate="2017-05-09T06:10:54.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15167" PostTypeId="1" CreationDate="2017-05-06T15:30:39.283" Score="2" ViewCount="1226" Body="&lt;p&gt;When explaining to someone what mechatronics is, I ususally end up stating that it is synonymous with robotics. While I think it is mostly true, are there any subtle differences between the two?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obligatory google search suggested that robotics is a subset of mechatronics involving autonomy. I'm not sure I completely agree. Is there any qualifier that would for sure separate a system into one or the other? Or is it just a different label?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also noticed some schools will have a robotics undergraduate degree but mechatronics graduate degree. Does this make any difference?&lt;/p&gt;&#xA;" OwnerUserId="161" LastActivityDate="2020-08-17T13:52:19.120" Title="What is the difference between mechatronics and robotics, if any?" Tags="|robotics|education|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15168" PostTypeId="1" CreationDate="2017-05-06T17:54:50.197" Score="-1" ViewCount="580" Body="&lt;p&gt;Before purchasing &lt;strong&gt;Wire cut cnc edm&lt;/strong&gt; I just wants to know about disadvantages of it over using mechanical lathe machine. &lt;/p&gt;&#xA;" OwnerUserId="11212" LastActivityDate="2018-06-14T21:08:42.963" Title="Disadvantages of wire cut cnc edm" Tags="|mechanical-engineering|machining|cnc|cutting|lathe|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15169" PostTypeId="1" AcceptedAnswerId="15692" CreationDate="2017-05-06T18:16:46.177" Score="0" ViewCount="400" Body="&lt;p&gt;I'm a non-engineer doing a little design project.&lt;/p&gt;&#xA;&lt;p&gt;I'm going to be running some M5-ish (have some flexibility) setscrews into a shaft of about 6mm. There's no reason why the shaft can't be flatted off a bit so I'm presuming I'll do that.&lt;/p&gt;&#xA;&lt;p&gt;There's no motors or electronics involved and the ends of the shafts are getting a pretty minor (I don't have numbers, sorry, I know I'll need them at some point and will need to do testing to get there) amount of force applied from a distance 20-30mm away. The setscrew connection must be highly resistant to slipping or loosening over extended (could be effectively permanent) use but also must be re-settable occasionally. When it does get reset, it could be right in the area of prior settings on the shaft, and it will have to be done with high accuracy (say down to a resolution of .02mm ideally), so I want to avoid marring as much as possible. (If this isn't possible then the whole thing will have to use a pinch bolt instead of a setscrew.) The shaft can be any material, within reason.&lt;/p&gt;&#xA;&lt;p&gt;The part will have some exposure to vibration but I don't intuit that it will be more than blue loctite will protect against.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How do I do the math to determine what exact materials to use and the torque on the setscrew to avoid the screw loosening, slipping, or significantly marring the surface of the shaft?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Any recommendations for reading on the topic would be great.&lt;/p&gt;&#xA;" OwnerUserId="11211" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-06-05T20:43:12.020" Title="How to determine materials and torque for a flat setscrew that won't loosen, slip, or mar" Tags="|torque|bolting|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15170" PostTypeId="2" ParentId="15169" CreationDate="2017-05-06T18:51:31.040" Score="0" Body="&lt;p&gt;A few thoughts :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In terms of protecting the surface finish of the shaft, hardness is the most important property so ideally you want a relatively hard shaft and a relatively soft screw (at least at the point). Of reasonably easily available materials something like silver steel may be a good bet as it has decent hardness even without heat treatment and also comes precision ground and so is good for shafts. With heat treatment you can significantly improve this at the expense of toughness. Possibly the best solution might be to surface harden it eg by case hardening. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the screw itself you really want some thing with a dog point as this should prevent the thread getting damaged and making it difficult to remove. Having the screw in a relatively oft material will obviously help prevent damage to the shaft but conversely will reduce the torque and thus the clamping force which can be applied. One way around this is to use a brass insert between the screw and the shaft, as long as there is enough space. If this is impossible then at least makes sure that the tip of the screw has a good surface finish and is free from burrs etc. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is no obvious way of actually calculating what you need as these sorts of wear situations really tend to be empirical and there is no simple way of calculating exactly what you need beyond the fact that having the shaft surface as hard as possible will certainty improve matters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of torque you will probably want to use the maximum torque that the screw will take as the greater the preload in the screw the less the chance of the interface flexing or slipping which is what is most likely to cause damage. This needs to be balanced against the fact that higher strength screws will also be harder, although if you can put something like a brass insert between the screw tip and the shaft this will help a lot. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-05-06T18:51:31.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15171" PostTypeId="2" ParentId="15169" CreationDate="2017-05-06T19:24:26.453" Score="0" Body="&lt;p&gt;If you have the space why not use a cylindrical clamp which slides over the ends and has its own clamping screws so that the clamping force is on the whole shaft not just provided by one or two setscrews.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-06T19:24:26.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15172" PostTypeId="2" ParentId="15168" CreationDate="2017-05-06T20:38:17.700" Score="1" Body="&lt;p&gt;Wire EDM and a lathe aren't really directly comparable. Wire EDM is typically used for profile cutting of very hard materials such as hardened tool steel with good dimensional accuracy and may also be useful for producing very small holes or slots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result it tends to be quite a specialist process and compared to conventional machining processes on materials with good machinability it will tend to have a low rate of material removal and may be less versatile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm tempted to suggest that if you don't already know &lt;em&gt;why&lt;/em&gt; you need an EDM machine you probably don't need one. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-05-06T20:38:17.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15174" PostTypeId="2" ParentId="15167" CreationDate="2017-05-06T21:39:34.950" Score="1" Body="&lt;p&gt;Mechatronics is a reaction to the fact that one can not design good next gen products without knowing both the electronical and mechanical realms. Having kickass electronics can not overcome a bad mechanical design. Conversely a badly done electrical system can not control the mechanical part so that you gain benefits over traditional mechanical engineering. This is not to say you can not have specialists, but the overal design needs people with a broader view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two or possibly three schools of mechatronics engineers. Some are electrical engineers, and some are mechanical enginers or possibly control enginers. So the fact that somebody is a mechatronics engineer tells quite little of what their education entails.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now robotics is quite clearly a subfield of mechatronics because mechatronics includes all kinds of things that arent entirely robotic in nature. Things like active vibration damping, electroactuated hydraulics, and many industrial automation things fall into this category.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Robotics is just more specific moniker. Its still not very specific but more specific.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2017-05-06T21:39:34.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15176" PostTypeId="1" CreationDate="2017-05-07T02:19:29.443" Score="4" ViewCount="167" Body="&lt;p&gt;If so, what would it be its official name? it might have another name than Torsion test because I googled it and nothing was found.&#xA;pdta: I know there is a torsion test for metals , but that is not what I am looking for. thank you.&lt;/p&gt;&#xA;" OwnerUserId="11215" LastActivityDate="2017-07-06T13:18:28.700" Title="Is there a torsion test in rocks?" Tags="|civil-engineering|geotechnical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15178" PostTypeId="1" CreationDate="2017-05-07T04:19:43.203" Score="5" ViewCount="8186" Body="&lt;p&gt;For RC beams, which one is preferable to be designed for construction? Is it multiple layer of reinforcements or one single layer? and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg. If the beam required 3T20, is it preferable one layer of 3T20 or multiple layers: 2T16+3t16?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="11219" LastActivityDate="2021-03-14T14:39:37.600" Title="Which one is preferable? Multiple layers of reinforcements or one layer of reinforcement?" Tags="|structural-engineering|reinforced-concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15179" PostTypeId="2" ParentId="14276" CreationDate="2017-05-07T04:24:27.000" Score="1" Body="&lt;p&gt;From what I know, you no need to stick out or drill in the new reinforcement of new footing to your existing footing. If you do that, some of new loadings will go into your existing footing which the previous designer does not take into account in their design.&lt;/p&gt;&#xA;" OwnerUserId="11219" LastActivityDate="2017-05-07T04:24:27.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15180" PostTypeId="1" CreationDate="2017-05-07T10:43:13.513" Score="0" ViewCount="621" Body="&lt;p&gt;I've come across two forms of the Darcy-Weisbach Equation, and each one seems to yield a different result when I am solving for $Q$ (flow). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_{f}=\frac{\lambda Lv^{2}}{2gD}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_{f}=\frac{f LQ^{2}}{3D^{5}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do these equations differ, and what (if any) is the relationship between $f$ and  $\lambda$ ?&lt;/p&gt;&#xA;" OwnerUserId="11147" LastEditorUserId="11147" LastEditDate="2017-05-07T12:11:15.620" LastActivityDate="2017-05-07T12:41:43.250" Title="Darcy-Weisbach Equation" Tags="|fluid-mechanics|hydraulics|pipelines|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15181" PostTypeId="2" ParentId="15176" CreationDate="2017-05-07T12:19:15.607" Score="3" Body="&lt;p&gt;I am unaware of a torsion test for rock. Shear, compression &amp;amp; &lt;a href=&quot;http://www.geotechdata.info/geotest/brazilian-test.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;tensile&lt;/a&gt; strength tests yes, but not torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the civil engineers, mining engineers and geotechnical engineers I have worked with over the years have used shear, compression and occasionally tensile strength of rock. No-one ever mentioned torsion strength.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-05-07T12:19:15.607" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15183" PostTypeId="2" ParentId="15180" CreationDate="2017-05-07T12:40:15.070" Score="1" Body="&lt;p&gt;I think &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; explains it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lambda is also called the Darcy friction factor, and equals 4 times the Fanning friction factor, f.  The two equations are equivalent, with the first written in the &quot;pressure loss&quot; form, and the second in the &quot;head loss&quot; form.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2017-05-07T12:40:15.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15184" PostTypeId="2" ParentId="15180" CreationDate="2017-05-07T12:41:43.250" Score="0" Body="&lt;p&gt;The first equation uses the &lt;em&gt;Darcy friction factor&lt;/em&gt; $\lambda$. The second one I don't know. You could rewrite the first equation.&#xA;$$ v=\frac{Q}{A}=\frac{4Q}{\pi D^2}$$&#xA;$$ \to h_f=\lambda\frac{8Q^2L}{\pi^2gD^5}$$&#xA;If you equate the two $h_f$ and solve for $f$:&#xA;$$ f=\frac{8\lambda}{3\pi^2g} $$&#xA;$f$ &lt;em&gt;might&lt;/em&gt; be a different friction coefficient, that holds true in certain hydraulic conditions, as $\pi$, $g$ and a constant factor $\frac{8}{3}$ are included in $f$. But I have no clue, where exactly $\frac{1}{3}$ comes from.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2017-05-07T12:41:43.250" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15185" PostTypeId="2" ParentId="11962" CreationDate="2017-05-07T12:58:55.650" Score="0" Body="&lt;p&gt;Also, for accuracy, the quality of the bullet or projectile - if it is &quot;out of balance&quot; ie the centre of gravity does not coincide with the axis of rotation then it will &quot;fly off&quot; or miss its intended target. And if you want a reference try &quot;The bullet's flight by F. W. Mann&quot; long out of print though.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-07T12:58:55.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15187" PostTypeId="2" ParentId="15178" CreationDate="2017-05-07T13:50:39.793" Score="5" Body="&lt;p&gt;On average, it's best to limit the number of layers of rebar. There are a few reasons for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, the more layers you have, the more complex the construction of the beam becomes. The first layer of rebar is easy to place: it simply rests on the stirrups and is held in place with some tie wire. All other layers, however, require the placement of plastic supports and more tie wire. So, in terms of ease of construction, the fewer layers you have, the better (with an especially big difference between one and two layers).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another constructibility difference is in pouring the concrete. The more layers of rebar you have, the harder it becomes to properly vibrate the concrete. Theoretically, so long as the horizontal spacing is sufficient, the number of layers shouldn't be much of a problem, but the more layers you have, the greater the odds that the vibrator will accidentally touch the rebar (which it shouldn't).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Also, the more layers you have, the more you risk segregating your concrete's aggregates. This is because the rebar will behave like a &quot;&lt;a href=&quot;https://s-media-cache-ak0.pinimg.com/736x/87/ec/ec/87ececd38b1d75cfded9322a0388b624.jpg&quot; rel=&quot;noreferrer&quot;&gt;Plinko board&lt;/a&gt;&quot;. This allows smaller aggregates to reach the bottom of the beam faster than the coarse one, as can be seen &lt;a href=&quot;https://www.youtube.com/watch?v=JNvMhdetjdQ&quot; rel=&quot;noreferrer&quot;&gt;in this video&lt;/a&gt; (though to a lesser extent, since the rebar will hopefully be vertically aligned). Obviously, this effect is relatively small unless you have many, many layers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And then there's the difference in strength efficiency between many or few layers. The more layers you have, the less efficient your reinforcement becomes. That's because the beam's strength is proportional to its effective depth, which is the distance from the steel's center of gravity to the opposite fiber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a single layer of rebar, then the effective depth $d$ is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d = h - c - \phi_s -\frac{1}{2}\phi_l$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $h$ is the beam's height, $c$, the concrete cover, $\phi_s$, the stirrup's diameter, and $\phi_s$, the longitudinal steel's diameter. If you have more layers, then the effective depth becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d = h - c - \phi_s - \dfrac{n}{2}\phi_l - \dfrac{n-1}{2}d_l $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $n$ is the number of layers and $d_l$ is the vertical distance between faces of rebar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the fewer layers you have, the more efficient your reinforcement becomes (requiring less area to get the same strength).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Having more layers, however, does offer some advantages. It implies in using more bars of smaller diameter. This usually means your adopted steel area is closer to the necessary steel area (making your layout more efficient). For instance, if you calculate that you need 7.8&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt; of steel, you can choose between one layer of $4\phi16$ (8.04&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;) or two layers of $10\phi10$ (7.85&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;) (5 bars per layer). The smaller step increment in area when working with smaller diameters can be an advantage. In the specific case you mentioned, though ($3\phi20$ or $5\phi16$), this doesn't apply, since $3\phi20$ has a smaller total area (9.42&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;) than $5\phi16$ (10.05&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while your design may be more efficient with more bars of smaller diameter, these efficiency measures are really just financial measures: &quot;how much money are you wasting by using more steel than is needed?&quot; One question which must be raised is whether the labor costs of placing so many more smaller bars (which must each be placed individually) will overrun any benefit from using less material. That's a question I can't answer since it is entirely dependent on the local market conditions (and I wouldn't know how to calculate it even if I had the information).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another advantage of using more bars of smaller diameter is crack control. Using more bars of smaller diameter increases the surface area available to transfer force between the steel and the concrete, therefore reducing the tensile stresses in those regions and therefore reducing the cracking. For example, $28\phi10$ (21.98&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;) and $11\phi16$ (22.11&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;) have basically the same cross-sectional area (error of 0.5%). However, $28\phi10$ has a total perimeter of 87.96&amp;nbsp;cm, while $11\phi16$ has only 55.29&amp;nbsp;cm, almost 40% less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if crack control is essential to your design, using more bars of smaller diameter may be interesting.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It should be noted that the two advantages I just gave for using smaller diameter bars have nothing to do with the number of layers, however. So, even if one wants to make use of these advantages, the bars should be placed so as to minimize the number of layers.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-05-07T13:50:39.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15188" PostTypeId="1" CreationDate="2017-05-07T13:54:13.297" Score="1" ViewCount="383" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vDTlY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vDTlY.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The above pic is describing the system how the car will be lifted if car is placed on the larger piston. But if the small piston is moved through 1m (say) ,then the piston is merely moved by millimetres. Then how, in practical life,small piston can actually be used to lift a car in a carwash station. If helpful, please suggest books, E-books or a few good articles.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The following portion is edited area after the question was declared answered&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;My question was very nicely answered and now my confusions have been eradicated and I'm grateful for that. I have hit upon a conclusion in the light of answer given and want the experts to confirm my following statement: If small multiple displacements &amp;quot;h&amp;quot; are not considered(see the answer with figure) and only stroke of the smaller piston is considered,&lt;/p&gt;&#xA;&lt;p&gt;THEN more fluid enters the large piston than fluid lost by by smaller piston and vice versa (on pushing small piston) i.e more fluid will come out of the larger piston than fluid gained by smaller piston upon pulling the smaller piston upwards.&lt;/p&gt;&#xA;&lt;p&gt;Note: I also wrote similar text in comments section but now I edited question thinking I followed wrong approach&lt;/p&gt;&#xA;" OwnerUserId="7625" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-05-17T23:10:50.823" Title="How can a car can be lifted using hydraulic press?(Pascal's Law)" Tags="|hydraulics|pistons|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15189" PostTypeId="1" AcceptedAnswerId="15205" CreationDate="2017-05-07T14:07:03.157" Score="1" ViewCount="174" Body="&lt;p&gt;I am given a step response of magnitude of 3 and the root locus and I have to find the transfer function of the system. The function I find gives me the step response(magnitude of 3 again) of the last diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oh183.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oh183.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a beginner at this so I've done something stupid probably but I have trouble finding answers regarding control engineering on the internet. This is what I tried doing:&#xA;I found the poles and the zeros from the root locus. z=-5,+4 p=-6,-10,-3&#xA;I think my transfer function is given from this formula but I'm not sure if we have an H(s) in the feedback and it is not stated : $$ T(s)=  \frac{KG(s)}{1+KG(s)} $$ &#xA;From the poles and the zeros my open-loop transfer function G(s) is :&#xA;$$ G(s)= \frac{(s+5)(s-4)}{(s+10)(s+6)(s+3)}$$ Doing the calculations I find : $$ T(s)= \frac{Ks^2+Ks-20K}{s^3+(K+19)s^2+(108+K)+180-20K}$$&#xA;From the step response(final value is 4) and the final value theorem I find $\frac{-20K}{180-20K}=-4/3=&amp;gt;K=5.14$  I divided 4 by 3 because the first step response is of magnitude of 3. With this K the step response is the one in the third diagram.It's close to the first one but it's not the one I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here? &lt;/p&gt;&#xA;" OwnerUserId="9603" LastActivityDate="2017-05-08T06:07:02.413" Title="Find transfer function from step response and root locus?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15190" PostTypeId="2" ParentId="15188" CreationDate="2017-05-07T14:09:37.557" Score="1" Body="&lt;p&gt;The main idea of the hydraulic press is that we use a piston for which we have to apply a small force $F_1$ and small area $A_1$ to displace a specific volume of the incompressible fluid. Imagine we want to displace a volume $V=A_{\text{base}}h.$ For the piston with a small area, we have large height displacement $h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the displaced volume in both pistons has to be approximately the same because water is in very good approximation incompressible, as long as velocities are small. So,&#xA;$$V_1=V_2 \implies A_{1}h_1=A_{2}h_2 \implies \frac{h_1}{h_2}=\frac{A_2}{A_1} \implies h_1=\frac{A_2}{A_1}h_2.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your choice of $A_2$ and $A_1$ you can create different transmission ratios. Using an additional valve which is connected to the reservoir we suck fluid back into the circuit in the backstroke. You can use repetitive motion with small $h$ displacements per cycle and accumulate them for a large total displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following picture shows the working principle. Note the both valves. The left one opens in the work stroke and closes in the backstroke. The right valve is operating in an inverted fashion.&lt;a href=&quot;https://i.stack.imgur.com/IFxAY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IFxAY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-05-08T12:41:20.957" LastActivityDate="2017-05-08T12:41:20.957" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15191" PostTypeId="2" ParentId="15188" CreationDate="2017-05-07T15:13:43.303" Score="0" Body="&lt;p&gt;To put the answer simply : the volume delivered by the small piston is delivered many times - this then moves the large piston accordingly. That is what the maths above shows.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-07T15:13:43.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15192" PostTypeId="1" CreationDate="2017-05-07T15:37:34.667" Score="1" ViewCount="67" Body="&lt;p&gt;In my text book it's written that head losses in elbows is produced by secondary flows..&lt;br&gt;&#xA;I have looked for the meaning of secondary flow,  and I found that :&#xA;The primary flow is parallel to the main direction of fluid motion and the secondary flow is perpendicular to this.. &#xA;But I didn't find why does it occur.. Can you help me?  &lt;/p&gt;&#xA;" OwnerUserId="9742" LastActivityDate="2017-05-07T17:45:27.880" Title="Why does secondary flow occur?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15194" PostTypeId="1" CreationDate="2017-05-07T16:40:12.220" Score="0" ViewCount="374" Body="&lt;p&gt;To maximize efficiency and reduce cost I am contemplating using the power from solar cells directly as dc without conversion to ac.  I plan to wire and connect the solar panels in series and parallel to charge batteries and super capacitors.  And then arrange the batteries and super-capacitors to yield 115 volts dc.  This arrangement will be completely off-grid.  By skipping the inverters, I hope to gain much efficiency and lower cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know direct dc wiring is used in rvs but the voltage is lower.  The 12 gauge and 14 gauge wiring in the house is already designed for 115 volts albeit rms ac.  The dc should be ok for all loads except ac motors and compressors for which I will either keep a dedicated line from the grid or use inverters just for these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides possible code violations, is there any other reason to not do this?&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-05-07T18:05:25.080" Title="Why not skip inverters and use solar power directly as dc?" Tags="|electrical-engineering|power-electronics|building-design|solar-energy|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15195" PostTypeId="2" ParentId="15194" CreationDate="2017-05-07T16:57:44.413" Score="0" Body="&lt;p&gt;The solar panel output voltage varies with the solar energy hitting it, ie varies with time of day : angle and cloud cover, part of the function of a solar controller can be to control the voltage output to the batteries.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-07T16:57:44.413" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15196" PostTypeId="1" CreationDate="2017-05-07T17:19:43.910" Score="0" ViewCount="68" Body="&lt;p&gt;(I'm not really sure what SE to post this on, so if this is inappropriate, my apologies)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to be able to remotely control the amount of chicken feed in the feeder. Here's how I envision the setup:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; _____________________________&#xA;| Huge bucket of chicken feed | &#xA; =============================&#xA;           |      | &amp;lt;--- Pipe (PVC?)&#xA;           ==Valve=---------Arduino&#xA;           |      |&#xA;          |        |&#xA;           oooooooo &amp;lt;---- Chicken feeder&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I would guess each granule of chicken feed is about 1/3 cm cubed. I would appreciate any suggestions on the setup as well as what kind of valve to use. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="11228" LastActivityDate="2017-05-07T18:12:01.527" Title="Chicken feeder design and valve" Tags="|mechanical-engineering|design|valves|" AnswerCount="1" CommentCount="4" ClosedDate="2017-05-07T21:28:54.640" ContentLicense="CC BY-SA 3.0" />
  <row Id="15197" PostTypeId="2" ParentId="15192" CreationDate="2017-05-07T17:45:27.880" Score="1" Body="&lt;p&gt;Secondary flow can result from obstacles, viscosity effects and because of changes in the size and orientation of the cross section. In general these causes lead to nonvanishing convective terms in the Navier-Stokes equation whicht lead to more complex flow patterns because of nonlinear effects.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-05-07T17:45:27.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15199" PostTypeId="2" ParentId="15194" CreationDate="2017-05-07T18:05:25.080" Score="1" Body="&lt;p&gt;Does your &quot;reduce cost&quot; objective include the cost of installing a duplicated house wiring system (including safety devices like breakers - not just stringing a few wires around!), and then ensuring that you never destroy an appliance by connecting it to the wrong supply? To do that reliably, you would need a different (and probably &quot;non-standard&quot;) type of plugs and sockets on the DC supply, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your inverter is &quot;only&quot; 90% efficient, just adding 10% more battery and/or solar panel capacity might be both cheaper and simpler.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-05-07T18:05:25.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15200" PostTypeId="2" ParentId="15196" CreationDate="2017-05-07T18:12:01.527" Score="1" Body="&lt;p&gt;One option is to use an Archimedes screw to actively draw the pellets out of a hopper that way some number of turns of the screw corresponds to a certain dose of pellets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also use a tipping bucket and counterweight or spring to dispense an consistent weight of pellets, this has the advantage that it doesn't depend on a consistent flow rate. A conveyor with buckets or ridges may also work in conjunction with this method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the difficulty here is that with large irregular granules they may not flow in a particularly consistent way just under gravity and can potentially wedge in mechanisms with close tolerances. So actively lifting them out of a hopper is often a useful approach in this sort of situation. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-05-07T18:12:01.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15201" PostTypeId="2" ParentId="15167" CreationDate="2017-05-07T18:33:58.310" Score="1" Body="&lt;p&gt;I don't think that they are fundamentally that different. Possibly mechatronics would lean more towards industrial automation in a process and manufacturing type context whereas robotics implies more of an experimental and product design type approach ie robotics is product and mechatronics is plant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly robotics tends to imply a more self contained and packaged system and also has some implication that its function is analogous in some way to the form of a human of animal, perhaps also including autonomous or remotely controlled vehicles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of an undergraduate degree the basic syllabus is likely to be fairly similar although you might expect a small difference in the focus on application and the background of the academic staff. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-05-07T18:33:58.310" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15204" PostTypeId="1" CreationDate="2017-05-07T20:12:08.270" Score="1" ViewCount="7101" Body="&lt;p&gt;The beam is fixed on both ends and there is a link in the middle. There is a force F on the link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach to solve this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A free body diagram is cut at the link and made into 2 systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Horizontal forces are ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum of force in vertical direction for system 1 on the left:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A+L(link)-F = 0$ (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of moments about A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ Ma+L*l-F*l=0  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the Right side I do the same as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is I keep getting nowhere and running in circles. I calculated the displacement for the left side and also the right side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I equaled them to tr and solve for unknown variables, as the displacement at the link (length l) has to be same for both the sides, but still can't cancel out the variables to solve for the reaction of the forces. My goal is to solve in terms of F for the reaction forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V2U25.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V2U25.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11230" LastEditorUserId="9" LastEditDate="2017-05-08T00:31:39.493" LastActivityDate="2018-01-12T12:51:44.947" Title="How to find the reaction forces, moments and the displacement of the fixed beam with a link?" Tags="|mechanical-engineering|structural-engineering|structures|statics|beam|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15205" PostTypeId="2" ParentId="15189" CreationDate="2017-05-07T22:30:55.190" Score="0" Body="&lt;p&gt;I think you did mix this up with proportional feedback. The transfer function is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G(s)=K\frac{(s+5)(s-4)}{(s+10)(s+6)(s+3)},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which $K$ is a parameter that needs to be determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $G(s)$ is a stable plant (all poles are in the left half plane) we can determine the DC gain by the final value theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G(s=0)=K\frac{5\cdot (-4)}{10\cdot 6 \cdot 3}=-\frac{K}{9}$$&#xA;And $G(s=0)=y(s=0)/u(s=0)=\frac{-4}{3}\implies a=12.$&#xA;Hence, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G(s)=12\frac{(s+5)(s-4)}{(s+10)(s+6)(s+3)}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MATLAB testing:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = tf('s');&#xA;G = 12*(s+5)*(s-4)/((s+10)*(s+6)*(s+3));&#xA;step(3*G); % 3 to scale the unit step response&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-05-08T06:07:02.413" LastActivityDate="2017-05-08T06:07:02.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15206" PostTypeId="1" CreationDate="2017-05-08T01:59:45.597" Score="3" ViewCount="195" Body="&lt;p&gt;Section 3.12.6 of CSA-S6-06 mentions that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Ice accretion load shall be taken on all exposed surface of superstructure members, structural supports, traffic signals, luminairies and&#xA;  railings. In case of sign panels, bridge girders, and solid barriers,&#xA;  ice accretion shall be considered to occur on one side only&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why in the latter part of the above statement, ice accretion is considered to occur only on one side?&lt;/p&gt;&#xA;" OwnerUserId="1693" LastEditorUserId="1693" LastEditDate="2017-05-10T02:58:13.440" LastActivityDate="2017-05-10T02:58:13.440" Title="Ice accretion on Structures in Canada" Tags="|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15207" PostTypeId="2" ParentId="15206" CreationDate="2017-05-08T05:43:24.083" Score="3" Body="&lt;p&gt;Because if ice accretion occurred equally on both sides there would not be a turning moment or side load on one side compared to the other. Having seen ice accretion on trees, lamp posts it is on one side only, normally due to the wind prevailing at that time.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-08T05:43:24.083" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15208" PostTypeId="1" AcceptedAnswerId="15211" CreationDate="2017-05-08T07:03:52.653" Score="0" ViewCount="6905" Body="&lt;p&gt;I am trying to build a &quot;mule&quot;: it s basically a go kart chassis with a 12v dc motor (I figured it would be easier to use). Now I need to determine how powerful should the motor be...    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is going to have 4 small wheels (15 cm in diameter - maximum), and be rear wheel driven, and it's going to do some soft kind of off roading (backyard terrain)&lt;br&gt;&#xA;It is going to carry up to 150 kg at low speeds (5km/h max). Should I use a torque converter? I was thinking of using 2 cordless drills to do the job...   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need torque or power in this case? I'm thinking I'd need plenty of torque as it is a pretty heavy load and the project requires low speeds...   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine the needed power / torque for future projects? Is there a formula (considering air friction 0).&lt;/p&gt;&#xA;" OwnerUserId="10614" LastEditorUserId="282" LastEditDate="2017-05-21T11:18:58.510" LastActivityDate="2017-05-21T11:18:58.510" Title="How to calculate the power/torque/force needed to move a 4 wheeled vehicle?" Tags="|electrical-engineering|motors|torque|power|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15209" PostTypeId="1" AcceptedAnswerId="15210" CreationDate="2017-05-08T08:34:46.840" Score="1" ViewCount="177" Body="&lt;p&gt;I have a beam as shown below that have a span having 60 cm in thickness and the other spans 30 cm do I consider this beam as continuous or does the changing in thickness affect that ? if there is anything like an article about these criteria (especially in ACI code) &#xA;&lt;a href=&quot;https://i.stack.imgur.com/EyuPj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EyuPj.png&quot; alt=&quot;Span 3 with changing thickness&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2017-05-08T09:41:32.257" Title="What is the thickness criteria to consider a beam with changing thickness Continuous?" Tags="|structural-engineering|civil-engineering|design|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15210" PostTypeId="2" ParentId="15209" CreationDate="2017-05-08T09:41:32.257" Score="2" Body="&lt;p&gt;Yes it is still continuous; all beams are still capable of producing a hogging moment above the columns.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-05-08T09:41:32.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15211" PostTypeId="2" ParentId="15208" CreationDate="2017-05-08T10:28:59.353" Score="3" Body="&lt;p&gt;The force required $F_{\text{r}}$ for driving a wheeled vehicle with mass $m$ is given the following formula (I will neglect slip):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{\text{r}}=F_{\text{tire}}+F_{\text{aero}}+F_{\text{acc.}}+F_{\text{slope}}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frictional force of the tires is given by $F_{\text{tire}}=c_{\text{tire}}mg.$ The dimensionless tire friction coefficient $c_{\text{tire}}$ is in general between $0.005$ and $0.010$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aerodynamic resistance is given by $F_{\text{aero}}=\frac{1}{2}\rho_{\text{air}}c_{\text{D}}A_{\text{ref}}v^2_{\text{rel}}.$ The dimensionless drag coefficient is between $0.28$ (good aerodynamics) and $0.80$ (bad aerodynamics e.g. for trucks). The reference area $A_{\text{ref}}$ is the projected area of the front face of the vehicle. For very small relative  velocities $v_{\text{rel}}=v+v_{\text{wind}}$ aerodynamical resistance can be neglected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force needed for a given translative acceleration $\ddot{x}$ is given by $F_{\text{acc.}}=e_{\text{m}}m\ddot{x}$. The dimensionless coefficient $e_{\text{m}}$ is there to account that it is also necessary to accelerate the components of the motor, gearbox and so forth. In most cases it is between $1.05$ and $1.40$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force $F_{\text{slope}}=mg\sin{\alpha}$ is necessary to overcome a slope of $\alpha$ (in degrees, so remember to calculate $\sin{\alpha}$ in degrees and not in rad). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in total we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{\text{r}}=c_{\text{tire}}mg+\frac{1}{2}\rho_{\text{air}}c_{\text{D}}A_{\text{ref}}v^2_{\text{rel}}+e_{\text{m}}m\ddot{x}+mg\sin{\alpha}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in order to get the required power $P_{\text{r}}$ you simply multiply $F_{\text{r}}$ with the velocity $v$ of the vehicle. Note that in general, when considering the wind this velocity $v$ will not be equal to the relative velocity $v_{\text{rel}}$. &lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-05-08T12:36:52.590" LastActivityDate="2017-05-08T12:36:52.590" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15212" PostTypeId="2" ParentId="14235" CreationDate="2017-05-08T13:12:37.990" Score="2" Body="&lt;p&gt;First, create the free body diagram for this system. If you cut through the spring $k_1$ and the damper $b_1$ you will get two forces $F_{k_1}=k_1(x_i-x_0)$ and $F_{b_1}=b_1(\dot{x}_i-\dot{x}_0)$ opposing the direction of $x_i$. Writing down Newton's second law of motion for the mass $m_i$ will result in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_i\ddot{x}_i=-b_1(\dot{x}_i-\dot{x}_0)-k_1(x_i-x_0) \implies m_i\ddot{x}_i+b_1\dot{x}_i+k_1x_i=b_1\dot{x}_0+k_1x_0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming zero initial conditions we can transform this into the Laplace domain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[m_is^2+b_1s+k_1\right]x_i(s)=\left[b_1s+k_1\right]x_0 \implies \frac{x_0(s)}{x_i(s)}=\frac{m_is^2+b_1s+k_1}{b_1s+k_1}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the point $x_i$ is attached to a massless position, then we simply set $m_i=0$ in the previous expression.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-05-08T13:12:37.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15216" PostTypeId="2" ParentId="8492" CreationDate="2017-05-08T16:55:30.347" Score="0" Body="&lt;p&gt;In the following, I will assume that the only thing that changed, was the number of blades of your test turbines.&#xA;Thus, all other parameter of your turbines such as swap area, blade geometry, height, generator, etc. will stay constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blades of wind turbines use the same aerodynamic principle as airplane wings do: By generating a pressure difference on both side, lift is caused.&#xA;Since the blades are forced to move on a circular path, lift is transformed into torque.&#xA;Hence, incrementing the number of blades, also increments the torque $M$ and the power $P$ since:&#xA;$$ P = 2 \pi n M \quad,$$&#xA;where $n$ denotes the number of revolutions.&#xA;As stated in the comments above, the more blade a wind turbine has, the less is its efficiency, but this may be neglected for small numbers of blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resistance of the model turbines (As the only difference is the number of blades) will only increase with an increase in revolutions until equilibrium is reached, explaining the increase in revolutions as it was observed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies for &lt;em&gt;real&lt;/em&gt; wind turbines.&#xA;However, for wind turbines it may be favorable to keep its number of revolutions constant or even reduce them as excitation frequencies are couple to them by&#xA;$$ f_{1P} = n \qquad \text{and} \qquad f_{3P} = 3n $$&#xA;due to imbalances in the rotor ($f_{1P}$) and tower shading ($f_{3P}$).&#xA;Eigenfrequencies of wind turbines are typically at 0.2 Hz to 0.3 Hz, what exactly is between those two frequencies for three-bladed wind turbines with a rated rotational speed of 10 rpm.&#xA;In order to avoid resonance, the number of revolutions should stay roughly the same or even decrease.&#xA;Said that, on increasing the number of blades you may also increase the generators torque in order to keep the number of revolutions constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other possibility could be the fact, that by keeping the power constant (for various reasons) you can reduce the number of revolutions by the same amount you increase the torque.&#xA;Despite the effect of reducing the excitation frequency with the number of revolutions, noise pollution (blade tip speed) and load on mechanical parts such as bearings and gears is reduced as well.&lt;/p&gt;&#xA;" OwnerUserId="11233" LastActivityDate="2017-05-08T16:55:30.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15217" PostTypeId="2" ParentId="2690" CreationDate="2017-05-08T17:39:07.457" Score="0" Body="&lt;p&gt;A new way to convert solar power has been discovered at the University of Michigan.  Please check out :&#xA;&lt;a href=&quot;https://phys.org/news/2011-04-solar-power-cells-hidden-magnetic.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://phys.org/news/2011-04-solar-power-cells-hidden-magnetic.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses the magnetic component of light which manifested itself when high intensity light goes through a transparent but non-electrically conductive material, glass for example. The light must be focused to an intensity of 10 million watts per square centimeter. Sunlight isn’t this intense on its own, but new materials are being sought that would work at lower intensities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your concentration of light using lenses and mirrors has limited potential for extracting more energy from the conventional solar cell, but it would certainly increase electrical energy with this new method.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-05-08T17:39:07.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15218" PostTypeId="1" CreationDate="2017-05-08T19:35:23.030" Score="1" ViewCount="73" Body="&lt;p&gt;I have essentially a hollow tube created from 3D splines that define the inner and outer surfaces.  How can I extrude a flange from the exterior surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not mention, this is Solidworks 2012.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done by wrapping onto a cylinder, but my thinking is that a 3D extrude would be more appropriate that would extrude from one inner surface to an outer surface ... like a boundary extrude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TJCvx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wrapping onto a cylinder&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a spline surface, and I can't figure out how to do it appropriately.  In my real world example, this is actually a tube, so I can't blind extrude, because it will also fill the inside void.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oOqiU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oOqiU.png&quot; alt=&quot;Spline surface&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11243" LastEditorUserId="11243" LastEditDate="2017-05-09T14:24:52.753" LastActivityDate="2017-05-09T14:24:52.753" Title="Creating flange from 3D surface" Tags="|solidworks|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15219" PostTypeId="1" CreationDate="2017-05-08T19:49:14.473" Score="0" ViewCount="982" Body="&lt;p&gt;Also, what does this excerpt from Wikipedia mean&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;However, since with open channel flow, the water surface is open to the atmosphere, the pressure term between two points has the same value and is therefore ignored.&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="11244" LastActivityDate="2017-07-23T18:16:30.273" Title="Why is pressure head ignored in open channel flow?" Tags="|hydraulics|open-channel-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15220" PostTypeId="2" ParentId="15219" CreationDate="2017-05-08T20:23:48.087" Score="2" Body="&lt;p&gt;Because head pressure is the weight of the water standing on top of the region being measured. Even though you can visualize an open channel as a slope you will not see a difference in pressure if you measure one foot deep at any point on the slope; this is because you are only measuring the weight of the water above the measuring point.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:16:30.273" LastActivityDate="2017-07-23T18:16:30.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15222" PostTypeId="2" ParentId="14227" CreationDate="2017-05-08T21:31:40.520" Score="1" Body="&lt;p&gt;If you can't do a continuity test then test the capacitance - 450pF at 1kHz as per the specs you refer to.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-08T21:31:40.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15223" PostTypeId="2" ParentId="14227" CreationDate="2017-05-09T00:05:14.727" Score="1" Body="&lt;p&gt;I suggest you use circuit similar to the below. The test circuit shown as below may be used to measure the resonant, anti-resonant, and impedance &#xA;characteristics of ultrasonic transducers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K3czK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K3czK.jpg&quot; alt=&quot;Piezo Ceramic Transducer Test Circuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Adjust frequency to obtain &lt;strong&gt;maximum&lt;/strong&gt; EOUT. Switch in VR and adjust to obtain the same output. VR now equals the minimum series impedance. &lt;/li&gt;&#xA;&lt;li&gt;Adjust frequency to obtain &lt;strong&gt;minimum&lt;/strong&gt; EOUT. Switch in VR and adjust to obtain same output. VR now equals the maximum series impedance at the antiresonant frequency. &lt;/li&gt;&#xA;&lt;li&gt;Measure phase between voltage EU and EOUT to determine impedance characteristics.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I suggest reviewing &lt;a href=&quot;http://www.senscomp.com/pdfs/using-piezo-sensors.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Application Notes on Ultrasonic Ceramic Transducers&lt;/a&gt; for additional details. Additionally take a look at &lt;a href=&quot;http://www.prowave.com.tw/pdf/an0508309.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Selection and use of Ultrasonic Ceramic Transducers&lt;/a&gt;. This almost identical to the first but has some additional details. &lt;/p&gt;&#xA;" OwnerUserId="7083" LastActivityDate="2017-05-09T00:05:14.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15224" PostTypeId="2" ParentId="15138" CreationDate="2017-05-09T00:35:06.510" Score="1" Body="&lt;p&gt;You will need a computer between the two devices to count the number of spaces sent by the button, keep a running total, then assemble a string of digits to send to the display. Neither the button nor the display is smart enough to understand the other, or to count the number of times the button is pushed. Also, the electrical port on the display (RS-232) and the port on the button (USB) have different electrical signal levels that need to be translated. A board like the one shown here found on-line has both a USB port and an RS-232 connector, with a microcontroller connected between them for under $30. While I have not used this particular board, the basic features appear to be a good foundation to bridge the button to the display. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technical hurdle is writing the program and loading it into the microcontroller to perform the task of counting up the button presses and sending a human readable character string out to the LED display. The ease or difficulty of the programming job is completely dependent your experience level. For someone who has used microcontrollers already the task should be trivial. It could be as few as a dozen lines of code. For someone who has never done a project with a microcontroller it will be a challenge, but totally doable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your reference, the particular microcontroller on the board shown is from a family of microcontrollers commonly referred to as PIC microcontrollers sold by Microchip. Development tools are available from their website, many for free. If you are interested in the PIC series of microcontrollers you may want to learn a bit about a special programming cable referred to as a PICKIT3 which is extremely affordable and can be used to transfer programs from a PC into the microcontroller when it is already in a circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dwc2g.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dwc2g.jpg&quot; alt=&quot;USB port, PIC processor, RS-232 port&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Microchip-PIC18F14K50-Nano-Development-Board-USB-Boot-Loader-w-Serial-Cable &lt;/p&gt;&#xA;" OwnerUserId="11222" LastEditorUserId="11222" LastEditDate="2017-05-09T01:01:40.453" LastActivityDate="2017-05-09T01:01:40.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15226" PostTypeId="1" CreationDate="2017-05-09T01:16:29.143" Score="2" ViewCount="422" Body="&lt;p&gt;I need to model the planetary gear-train with transformation matrices of  1 sun and 2 planets system, with no ring and an &amp;quot;arm&amp;quot; or carrier holding the gears ,like in the second picture. I already have a dynamical model of the simple system derived analytically, but I need to model it dimensionally to be able to specify the spatial relationships—the relative positions and orientations of the epicyclic geartrain to simulate it.&lt;/p&gt;&#xA;&lt;p&gt;I only have experience deriving homogeneous transformation matrices of robot joints but I don't know where to star for a gear train system or what method to use.&lt;/p&gt;&#xA;&lt;p&gt;I'm not sure about which frames of reference to use and how to set up the coordinate system between the gears correctly, for example; what would be the difference between considering a fixed frame of reference for the sun vs a rotating frame of reference? Also, I don't know which and how to include variables like mass, stiffness and damping, like shown in the first picture (for example).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cs9nl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cs9nl.jpg&quot; alt=&quot;Lumped parameter planetary model&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vy1xP.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vy1xP.gif&quot; alt=&quot;1 sun , two planets gear train&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4088" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-05-09T22:55:57.513" Title="How to model a planetary gear system with transformation matrices?" Tags="|mechanical-engineering|gears|modeling|dynamics|mechanisms|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15227" PostTypeId="2" ParentId="7500" CreationDate="2017-05-09T01:18:18.127" Score="1" Body="&lt;p&gt;Early TTL outputs (those typical of 74xx00 series logic gates) were capable of sinking (creating a current path to ground) more current than they could source (creating a current path to VCC). If you wanted a predictable current flow  which you could set by your choice of external current limiting resistors then the output transistor path to ground did a better job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know this is still true.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-05-09T01:18:18.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15228" PostTypeId="1" CreationDate="2017-05-09T05:07:40.037" Score="1" ViewCount="325" Body="&lt;p&gt;I saw some formula involving integration. Is there a straightforward formula to calculate this instead given b, d, t and alpha angle between x-x and u-u axis?&lt;/p&gt;&#xA;" OwnerUserId="11248" LastActivityDate="2017-05-09T05:07:40.037" Title="How to calculate monosymmetry index of an unequal angle?" Tags="|structural-engineering|civil-engineering|structural-analysis|steel|" AnswerCount="0" CommentCount="1" ClosedDate="2017-05-09T13:32:52.380" ContentLicense="CC BY-SA 3.0" />
  <row Id="15232" PostTypeId="1" AcceptedAnswerId="15241" CreationDate="2017-05-09T06:29:55.813" Score="1" ViewCount="2318" Body="&lt;p&gt;In the below picture I want to design B4 and B6 but I am not sure which one does support the other and how the detailing of the area of intersection between these two beams?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it better that B6 supports B4 because B6 have the shortest span?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XULmT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XULmT.png&quot; alt=&quot;Area intersection between two beams&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="1832" LastEditDate="2017-05-09T14:05:42.537" LastActivityDate="2017-05-09T18:04:19.007" Title="How is the detailing of the intersection between two RC beams perpendicular to each other?" Tags="|structural-engineering|civil-engineering|design|structural-analysis|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15235" PostTypeId="1" CreationDate="2017-05-09T10:08:27.260" Score="5" ViewCount="113" Body="&lt;p&gt;I'm modelling a vehicle as a multibody system in Matlab/Simulink.&#xA;My goals are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Have a realistic simulation model so I don't have to do prototype measurements all the time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Solve specific parameter estimation tasks and apply driver assistance control mechanisms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now for the first task I created a detailed multibody model that imitates the real vehicle's suspension kinematics, has a tire model and so forth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second task, this model would of course be taking too much computational effort for a real-time application on a vehicle's ECU.&#xA;That is why I created additional models with reduced complexity for different purposes (e.g. a single-track model).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I would like to know not only how &quot;well&quot; the complex model matches the real vehicle, but also to what degree the &quot;interesting&quot; dynamics of the complex model correspond to the simplified ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to avoid is just using test signals and looking at graphs to see if they are &quot;kind of similar&quot;. A more in-depth approach that gives me mathematically more or less sound results would be nice.&#xA;Numerical solutions are perfectly fine though, I don't expect to find nice analytical approaches for complex nonlinear systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I am looking for literature or some guidance on this kind of problem.&lt;/strong&gt; I am sure sensitivity analysis is an important part of the solution and I already found some resources on that. But it doesn't help too much when I want to compare 2 models (or model and reality).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thankful for every hint!&lt;/p&gt;&#xA;" OwnerUserId="11250" LastActivityDate="2020-04-19T06:02:45.260" Title="How to compare dynamic systems of different degree of abstraction?" Tags="|control-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15236" PostTypeId="1" CreationDate="2017-05-09T11:33:40.810" Score="2" ViewCount="277" Body="&lt;p&gt;Example: I want to use an M8x20 bolt in Inventor and edit the data of the file to get it right in BOM. I was able to do it in Inventor 2017 and earlier but not any more. I have tried to untick &quot;read only&quot; in the content center properties etc. Any suggestions? &#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZiUT3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZiUT3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9142" LastEditorUserId="9142" LastEditDate="2017-05-09T14:46:35.950" LastActivityDate="2017-05-09T14:46:35.950" Title="How to edit content center data (Inventor 2018)?" Tags="|autodesk-inventor|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15237" PostTypeId="2" ParentId="14227" CreationDate="2017-05-09T11:57:07.267" Score="1" Body="&lt;p&gt;Apply a small voltage to the piezo from a signal generator, maybe around 1 kHz.  If you can hear something, then the wires are connected.  A few Volts should be good enough, and won't damage the piezo, especially one that size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, you can connect the wires to a scope and see if there are any voltage spikes as you jar the device the piezo is attached to.  Be careful that this doesn't cause the piezo to touch anything other what it is already mounted on.  Otherwise, the voltage spikes could be due to the leads picking up static charges from different objects.  You want to make sure you are getting voltage spikes only from mechanical stress on the piezo.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-05-09T11:57:07.267" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15238" PostTypeId="1" CreationDate="2017-05-09T12:13:09.650" Score="0" ViewCount="227" Body="&lt;p&gt;Is 0.0 the appropriate external static pressure to use when sizing a high flow exhaust fan that is not ducted on either the inlet or outlet?  I can't find any specific guidance on this practice, yet I'm reluctant to assume 0.0 due to friction losses. On the other hand, I can't think what values to use to even begin trying to quantify the pressure.&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2017-05-09T12:13:09.650" Title="Static Pressure of non-ducted fan" Tags="|hvac|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15239" PostTypeId="1" AcceptedAnswerId="15248" CreationDate="2017-05-09T15:43:53.630" Score="0" ViewCount="80" Body="&lt;p&gt;I'm working on a project trying to combine a &lt;a href=&quot;http://www.mekanizmalar.com/pantograph.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;pantograph linkage&lt;/a&gt; with a &lt;a href=&quot;http://www.mekanizmalar.com/four-bar-infinity-curve.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;four-bar linkage&lt;/a&gt;. I'd like to build it out of relatively at-hand materials (AKA, nothing requiring a machine shop), and I'd like to do some educational projects in the future involving students building linkages, too, so low-cost would be great. Relatively stiff and able to run unattended as a display are also desirable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that bolts and nuts are going to be problematic after it runs for too long. Is there a particular type of bearing that would be easy to use/acquire? ...a good material for the rods? I have a Carvey CNC machine, so HDPE and wood are easy to machine for me. It's mostly the pivots/bearings that I'm having trouble with.&lt;/p&gt;&#xA;" OwnerUserId="11252" LastActivityDate="2017-05-10T03:02:52.603" Title="What are a few low-cost options for building linkages?" Tags="|bearings|linkage|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15241" PostTypeId="2" ParentId="15232" CreationDate="2017-05-09T18:04:19.007" Score="2" Body="&lt;p&gt;You decide which beam supports which by how you detail the steel reinforcement at the intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, one chooses to make the stiffest beam support the other. In your case, that means that yes, B6 would support B4. However, that decision isn't taken merely by looking at the beams' spans, but also at their $EI$. Since they seem to be of same (or very similar) cross-section, then the difference in span becomes the only relevant variable. However, if B6 were actually half as wide (or tall, which would be even worse), then it would probably be far less stiff than B4, in which case one'd define that B4 supports B6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you prefer, you can also let an analytical model make that decision for you. Look at the shear diagrams of both beams at the intersection. It should be clear that one &quot;lost&quot; shear force and the other &quot;gained&quot; it, meaning the latter is supporting the former.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, once you've defined which beam supports which (in this case, B6 supports B4), we need to inform our decision to the universe by detailing that intersection accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not merely a matter of which beam's rebar goes beneath the other's. After all, it is common to assume that reinforced concrete transmits its load from the bottom of the beam, at the end of the compression strut:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ylo4t.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ylo4t.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that, as far as the supporting beam is concerned, it just received a concentrated load near its bottom fiber. Therefore, additional suspension reinforcement in the form of stirrups is necessary to &quot;raise&quot; this load, allowing the full height of the supporting beam to work. This reinforcement is found by obtaining the support reaction between the beams and calculating&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_s = \frac{R}{f_{yd}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(when the supported beam's bottom fiber is above the supporting beam's bottom fiber, this reinforcement can be reduced somewhat). This reinforcement should be added to the normal stirrups and doesn't need to be exclusively within the intersection volume. Item 9.2.5 of Eurocode 2 allows the suspension stirrups to be placed in the following region:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zdtO7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zdtO7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-05-09T18:04:19.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15242" PostTypeId="2" ParentId="15135" CreationDate="2017-05-09T19:52:53.257" Score="1" Body="&lt;p&gt;You haven't said if your loads are motors or whether your voltages are line or phase.  I'll assume voltages are line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P_T = 20kW$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_T = \sqrt {3}\ V_{Line}\ I_{Line}\ cos \ \theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{Line}  = \frac {P_T}{\sqrt {3}\ V_{Line}\ cos \ \theta}= \frac {20kW}{\sqrt {3} \times 415V \times 0.8} = 34.78A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S = kVA  = \sqrt {3}\ V_{Line}\ I_{Line} = \sqrt {3}\times 415V \times 34.78A = 25kVA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need a 25kVA three-phase transformer or slightly larger.  If your 10kW loads are motors, add 20% for starting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference sites say &lt;em&gt;If motors are started more than once per hour, increase minimum transformer KVA by 20%&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://jeffersonelectric.com/wp-content/uploads/catalog/basics.pdf#page=3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Transformer Basics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://dongan.com/wp-content/uploads/SizingTransformers.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sizing Transformers&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2017-05-09T19:52:53.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15243" PostTypeId="1" AcceptedAnswerId="15267" CreationDate="2017-05-09T20:15:15.403" Score="1" ViewCount="3300" Body="&lt;p&gt;If I have a rectangular embedded beam in a slab supporting hollow core slab how is the connection and detailing between these two elements ? (any sketch would help)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the beam parallel to the hollow core how is the connection between these two ?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WHGqL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WHGqL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know how the connection between the hollow core and the beam how is the bars between these two are connected ?&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="10554" LastEditDate="2017-05-10T05:52:19.520" LastActivityDate="2017-05-11T16:26:04.653" Title="How the connection is between a Hollow core slab and an embedded beam?" Tags="|structural-engineering|civil-engineering|design|structural-analysis|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15244" PostTypeId="1" CreationDate="2017-05-09T22:02:42.637" Score="3" ViewCount="121" Body="&lt;p&gt;I need to design a water to air heat exchanger with a blower for a small greenhouse as an engineering project. There aren't much design specifications - only that it must maintain a temperature of between 15 and 25 degrees celsius during winter nights (-4 degrees celsius approx) the material of the box it will be placed in is polycarbonate with dimensions 1x1x1,5m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a good source or a textbook to work from for the design of the heat exchanger as well as  determining all the variables. it is a semester project and have just started with the design, if somebody could guide me in a direction I would appreciate the help. &lt;/p&gt;&#xA;" OwnerUserId="7092" LastEditorDisplayName="user16" LastEditDate="2017-05-10T02:10:56.537" LastActivityDate="2017-05-11T13:40:44.953" Title="Heat exchanger design guidance" Tags="|mechanical-engineering|heat-transfer|heat-exchanger|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15245" PostTypeId="2" ParentId="15226" CreationDate="2017-05-09T22:19:04.097" Score="2" Body="&lt;p&gt;(edit: didn't read your comment fully... here's a better answer based on your comment)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;in studying the vibrations caused by the meshing of the gears&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I take that to mean you are interested in a forced response vibration problem. Well, actually because it's gear system, the real driver of the response is the time varying mesh stiffness which leads to &quot;parametric excitation&quot;, and you will also have to contend with the non-linear behavior of the gear tooth contact.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by building a linear model for free vibration, i.e. natural frequencies. This paper &lt;a href=&quot;http://vibrationacoustics.asmedigitalcollection.asme.org/article.aspx?articleid=1470160&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Analytical Characterization of the Unique Properties of Planetary Gear Free Vibration by Jian Lin and R. G. Parker&quot;&gt;Analytical Characterization of the Unique Properties of Planetary Gear Free Vibration by Jian Lin and R. G. Parker&lt;/a&gt; is one of the classic references on planetary gear free vibration. If you follow Parker's modeling approach, the answer won't necessary be physically meaningful, since you have no ring gear, but it will get you in the right direction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have that basic free vibration model, then you will want to move on to the forced response / parametric instability modeling.  Here are a few references &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0022460X01938489&quot; rel=&quot;nofollow noreferrer&quot;&gt;PLANETARY GEAR PARAMETRIC INSTABILITY CAUSED BY MESH STIFFNESS VARIATION by J. LIN. and R.G. PARKER.&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0022460X06009023&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nonlinear dynamics of planetary gears using analytical and finite element models by Vijaya Kumar Ambarisha, and Robert G. Parker&lt;/a&gt;&#xA;&lt;a href=&quot;http://vibrationacoustics.asmedigitalcollection.asme.org/article.aspx?articleid=1661272&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vibration Properties of High-Speed Planetary Gears With Gyroscopic Effects by Christopher G. Cooley and Robert G. Parker&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty of other references out there.  These are just some of the ones I'm most familiar with.  Google scholar and the bibliography of these papers should help.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not familiar with parametric excitation and non-linear dynamics, you'll probably want pick up a basic non-linear dynamics textbook.  I recommend the one by Jordan and Smith.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="9275" LastEditDate="2017-05-09T22:55:57.513" LastActivityDate="2017-05-09T22:55:57.513" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15248" PostTypeId="2" ParentId="15239" CreationDate="2017-05-10T03:02:52.603" Score="0" Body="&lt;p&gt;I have tested linkage models using wood strips with small bearing recesses milled into the wood. Skate bearings 8mm ID x 22mm OD x 7mm width are widely available for about 60 cents if you buy 20 or so. I just milled the bearing recess a bit tight then pressed them in. If those are too large then smaller bearings are also available and may be cheaper in bulk. Be aware that skate bearings are available in some exotic ceramics and other materials that get very expensive, so don't be scared off if you see some for $10 per bearing. I have paid about 40 cents for them when I bought them in bulk.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-05-10T03:02:52.603" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15250" PostTypeId="2" ParentId="2300" CreationDate="2017-05-10T03:18:13.807" Score="1" Body="&lt;p&gt;I will try to make it as simple as possible and will provide an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-First of all ignore the word slug... I know it is the standard unit for mass and so is lbm. you will see lbm used in your text and in real life 99% of the time. Once you understand this concept well you can go on to familiarize yourself to using slugs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Think of newton as the force required to move a mass of 1kg by 1m/s^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Think of pound-force (lbf) as the force required to move a mass of 1lbm by 32.2ft/s^s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the last two points above, it is obvious that the newton is very different that the lbf&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;On the surface of earth, 1kg exerts a force of 9.81N... or 9.81kgm/s^2&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the surface of earth, 1lbm exerts a force of 1lbf... or 32.2lbft/s^2&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Makes sense?... lets try an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;QUESTION&lt;/strong&gt;: An astronaut has a mass of 100kg (220lbs) what is his weight (force) if he is on earth? what if he was on a planet with the gravity of 5m/s^2 (16.4ft/s^2)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ANSWER&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Earth&lt;/em&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;SI units --&gt; 100kg*9.81m/s^2= 981kgm/s^2= 981N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imperial units --&gt; 220lbs*32.2ft/s^2= 7084 lbmft/s^2 = 220lbf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Random planet&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SI units--&gt; 100kg*5m/s^2 = 500kgft/s^2 = 500N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;imperial units --&gt; 220lbs*16.4ft/s^2= 3608 lbmft/s^2 = 3608/32.2 = 112lbf&lt;/p&gt;&#xA;" OwnerUserId="11259" LastActivityDate="2017-05-10T03:18:13.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15251" PostTypeId="1" CreationDate="2017-05-10T05:17:51.200" Score="0" ViewCount="46" Body="&lt;p&gt;Which case is used to check the drift and P-delta for a building according to ASCE ?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Static lateral force procedure (with a period Comparing T the real fundamental period with CuTa)&lt;/li&gt;&#xA;&lt;li&gt;Static lateral force procedure (without comparing T with CuTa)&lt;/li&gt;&#xA;&lt;li&gt;Modal response spectrum analysis (without scaling)&lt;/li&gt;&#xA;&lt;li&gt;Modal response spectrum analysis (with scaling)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2017-05-10T05:17:51.200" Title="Which case is used to check Drift and P-delta for a building subject to seismic forces?" Tags="|structural-engineering|civil-engineering|design|structural-analysis|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15252" PostTypeId="2" ParentId="15244" CreationDate="2017-05-10T05:43:00.480" Score="0" Body="&lt;p&gt;I found one called Engineering Heat Transfer by Simonson J. R, but it was well written and had examples. Used it to size a heat exchanger for buildings which came out well. But any of the heat transfer books or heat exchanger books will give you the basic design parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to one : &lt;a href=&quot;http://www.ebay.co.uk/itm/Engineering-Heat-Transfer-by-Simonson-J-R-Author-9780333459997-/172390036831?hash=item28233f355f:g:0EwAAOSwB09YExgD&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ebay.co.uk/itm/Engineering-Heat-Transfer-by-Simonson-J-R-Author-9780333459997-/172390036831?hash=item28233f355f:g:0EwAAOSwB09YExgD&lt;/a&gt; , not cheap though - any second-hand bookshops near you?&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-05-10T05:48:05.147" LastActivityDate="2017-05-10T05:48:05.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15253" PostTypeId="1" CreationDate="2017-05-10T07:12:35.190" Score="1" ViewCount="251" Body="&lt;p&gt;I have known that the view factor reciprosity is that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ai x Fij = Aj x Fji&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i can't intuitively explain the logic behind this, i do know that the view factor Fij is the portion of the radiation emitted by surface i that is recieved by surface j&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked up for the proof but that doesn't seem to help much&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="11262" LastActivityDate="2017-05-10T22:42:16.610" Title="What is the inuition behind the radiation view factor reciprocity?" Tags="|heat-transfer|radiation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15255" PostTypeId="1" CreationDate="2017-05-10T09:40:56.770" Score="2" ViewCount="83" Body="&lt;p&gt;There are electronic components like microprocessors which work at 1-2V.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why do some semiconductor devices have a minimum voltage requirement?&lt;/li&gt;&#xA;&lt;li&gt;Are there (simple) microcontrollers which work at few tens of millivolts or less? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11265" LastEditorUserId="33" LastEditDate="2017-05-15T15:04:13.380" LastActivityDate="2017-05-15T15:04:13.380" Title="Why do some semiconductor devices have a minimum voltage requirement?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15257" PostTypeId="2" ParentId="15255" CreationDate="2017-05-10T12:47:15.243" Score="3" Body="&lt;p&gt;BJTs (bipolar junction transistors) require forward-biasing a diode junction to operate.  That inherently requires a minimum voltage.  That voltage depends on the material and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just being able to turn transistors on and off isn't good enough by itself.  The result of those on/off control signals have to actually do something.  That's going to require some additional voltage swing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often the point is to produce a voltage signal.  That signal has to have some swing to be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MOSFETs (metal oxide semiconductor field effect transistors) work on a principle that is somewhat &quot;soft&quot;.  There is no hard transition at a particular gate voltage.  If there isn't enough voltage range to swing the gates over, then the devices won't be solidly on at one end and solidly off at the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower supply voltage reduces the charge that has to be transferred each time the gate is switched, but also increases quiescent current due to higher off-state leakage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern processors do run at low voltages, like as low as 1.2 V, but it's a constant tradeoff between switching charge and leakage.  There was one Intel processor some years ago where about half the average power was due to leakage current when the MOSFETs were in the off state.  The leakage of one transistor might be very small, but this adds up due to lots of transistors.  A nano-Amp here, a nano-Amp there, and pretty soon you're talking about real current.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-05-10T12:47:15.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15258" PostTypeId="2" ParentId="15253" CreationDate="2017-05-10T22:42:16.610" Score="-1" Body="&lt;p&gt;It follows from the zeroth law of thermodynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two surfaces are at the same temperature, there is no net heat transfer between them, otherwise the temperature of one would rise and the other would fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node137.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node137.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-05-10T22:42:16.610" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15259" PostTypeId="2" ParentId="3358" CreationDate="2017-05-11T01:29:51.997" Score="1" Body="&lt;p&gt;It is too small a surface area to be an effective microwave reflector.  The shape is wrong too, it must be parabolic with respect to the transmitting fractal antenna of the router.  Yours is perpendicular to it.&lt;br&gt;&#xA;&lt;a href=&quot;http://www.instructables.com/id/Easy-to-Build-WIFI-24GHz-Yagi-Antenna/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.instructables.com/id/Easy-to-Build-WIFI-24GHz-Yagi-Antenna/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the advent of embedded fractal antennas, we no longer need external antennas sized according to the wavelength of the carrier frequency.  A good development or we would have to be carrying around antennas for our cell phones.  I'm not saying you can't shape an aluminum can to enhance directional transmission and reception.  It just takes some design and calculation.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-05-11T01:29:51.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15260" PostTypeId="2" ParentId="417" CreationDate="2017-05-11T04:22:51.973" Score="2" Body="&lt;p&gt;I am making one right now. These posts are all old and the problem is usually the same. I am using a camping pump, some tubes, and a vaporizer. I use no nicotine juice that I make myself just using glycerin diluted with a little bit of distilled water. The pipe is plugged at the end and as several holes to watch the airflow at different heights. Works great, but I only get smoke for 10 seconds at a time or it will burn out the coil. Wait 3 seconds and then fire it up again.&lt;/p&gt;&#xA;" OwnerUserId="11273" LastActivityDate="2017-05-11T04:22:51.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15261" PostTypeId="1" CreationDate="2017-05-11T12:46:03.333" Score="0" ViewCount="95" Body="&lt;p&gt;What I am after is the center line of this wave, which is 0. How can I calculate that value though? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, run the code below for a simple example in Matlab. This will plot a simple sine wave with a couple of periods.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = linspace(0,7.5*pi,100); y = sin(x); plot(x,y); mean(y)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SHnlb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SHnlb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result (and the problem) is that &lt;code&gt;mean(y)&lt;/code&gt; is 0.0368, and not 0.&#xA;But I am plotting a sine wave oscillating around 0 of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate these kind of averages correctly (without me having to plot and actually examine each signal manually to get the periods right)?&lt;/p&gt;&#xA;" OwnerUserId="11276" LastEditorUserId="11276" LastEditDate="2017-05-11T18:05:58.470" LastActivityDate="2017-06-10T19:17:15.497" Title="Center line of cut-off periodic signal" Tags="|mathematics|signal|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15262" PostTypeId="1" AcceptedAnswerId="15281" CreationDate="2017-05-11T12:53:28.977" Score="3" ViewCount="106" Body="&lt;p&gt;I am trying to translate the name of a special kind of beam from a set of Russian-language drawings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dictionary translates the name as simply supported beam, but it's not that simple. There is a slight difference in two kinds of beams. First - what is called a &lt;strong&gt;simply supported beam&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A beam that lies freely on two supports and covers one span. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Second - &lt;strong&gt;my beam&lt;/strong&gt;. This one is different from the previous that it lies on multiple supports and covers many spans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Important note: both of the beams are not continuous; i.e., they are parts of one long beam.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, what would I call the second beam?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0WkKu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0WkKu.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11277" LastEditorUserId="368" LastEditDate="2017-05-12T16:12:51.680" LastActivityDate="2017-05-12T16:53:24.630" Title="Translating the name of a type of beam from Russian to English" Tags="|structures|beam|terminology|building-design|" AnswerCount="1" CommentCount="7" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15264" PostTypeId="2" ParentId="15244" CreationDate="2017-05-11T13:40:44.953" Score="1" Body="&lt;p&gt;I don't do much heat tranfer stuff now, but I used my textbook from a graduate heat exchanger design course I took, &lt;a href=&quot;http://www.dl.begellhouse.com/buynow/book/7daba87f6fb65d65.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Process Heat Transfer by Hewitt, Shires and Bott&lt;/a&gt;. It has all of the info you need.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-05-11T13:40:44.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15265" PostTypeId="2" ParentId="14304" CreationDate="2017-05-11T14:17:12.833" Score="2" Body="&lt;p&gt;It would depend on how much energy you can store. If you are simply storing in a battery you can calculate the max charging rate and subtract that from the kinetic energy of the moving vehicle. Locomotives use dynamic braking and almost all of them do not attempt to store the energy, they convert it to heat and vent it out the top of the train. A civil Engineer by the name of Travis Painter did a thesis on attempting to store that energy. He selected a flywheel energy storage system due to the fact that it can store a high volume of energy rapidly. Given the other options in his paper that was the best choice but he determined that the losses in generation, storage and retrieval was too high to make it a viable means of regenerative braking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the link below is to a copy of his thesis, if not search for Travis Painter Dynamic Brake Energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.semanticscholar.org/paper/Recovering-Railroad-Diesel-electric-Locomotive-PAINTER-DIESEL-ELECTRIC/3c9f11f47e0ae8484a29248e4086e271dc16a667&quot; rel=&quot;nofollow noreferrer&quot;&gt;Travis Painter Thesis&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10516" LastActivityDate="2017-05-11T14:17:12.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15266" PostTypeId="2" ParentId="8110" CreationDate="2017-05-11T14:20:40.400" Score="0" Body="&lt;p&gt;What about using a sintered metal? Powdered metals can be mixed and an acid might be able to remove one metal in specific spots allowing the light to shine through. &lt;/p&gt;&#xA;" OwnerUserId="10516" LastActivityDate="2017-05-11T14:20:40.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15267" PostTypeId="2" ParentId="15243" CreationDate="2017-05-11T15:01:12.230" Score="2" Body="&lt;p&gt;This depends on what connection behavior you would like to achieve. The hollow core slab manufacturer will typically have standard connection details or assist you with the connection design. Usually beams/columns are designed such that the prefabricated hollow core slabs can be lifted into place. Therefore, usually there are corbels or ledges for the slabs to sit on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an idea, Hollow Core Concrete Pty Ltd (Australia) has &lt;a href=&quot;http://www.hollowcore.com.au/floor_slab_tech.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hollow Core Slab Technical Sheets&lt;/a&gt; on their website showing some typical connection details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have taken the detail for an &lt;a href=&quot;http://www.hollowcore.com.au/pdf_files/Internal%20Beam%20to%20Hollow%20Core%20Connection.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;internal beam to hollow core connection&lt;/a&gt; from their website: &lt;a href=&quot;https://i.stack.imgur.com/Us41j.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Us41j.png&quot; alt=&quot;internal beam to hollow core connection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth discussing the design with the hollow core manufacturer to get an expert opinion. I always recommend using the experts instead of trying to re-invent something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Wasabi pointed out I didn't address the situation you mention in the comments with the slab and beam having the same depth. This is a non-standard detail with regards to hollow core slabs. As I mentioned earlier, hollow core slabs are typically used in pre-cast construction. If you are casting integral beams, this removes much of the advantage of using hollow core slabs in the first place. You could just cast a &lt;a href=&quot;http://www.concreteconstruction.net/how-to/voided-slab-flat-plate-floor-construction_o&quot; rel=&quot;nofollow noreferrer&quot;&gt;voided slab&lt;/a&gt; in-situ for example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to use hollow core slabs in this situation there are potentially a few ways this could be achieved, for example: breaking out part of the hollow-core slab, adding the additional reinforcement then casting new concrete to fill the space; or using a &lt;a href=&quot;http://www-civ.eng.cam.ac.uk/struct/jml/CCSG_topic_prhd2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;concrete half-joint&lt;/a&gt; detail. The manufacturer may also be able to incorporate this beam into the design of the slab. Any of these non-standard connections would depend on your loading and other design and construction constraints and would likely need to be designed in collaboration with the hollow core slab manufacturer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are willing to use steel-concrete composite construction you could use a typical slimfloor approach:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OsT6jm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OsT6jm.jpg&quot; alt=&quot;slimfloor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The space around the steel beam would be filled with concrete, and if you wanted to get very clever you could even add rebar in this space.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2017-05-11T16:26:04.653" LastActivityDate="2017-05-11T16:26:04.653" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15268" PostTypeId="2" ParentId="15261" CreationDate="2017-05-11T15:27:32.377" Score="1" Body="&lt;p&gt;The problem that you describe comes from the fact that you calculating the average of the vector with numerical values. In order to calculate the exact mean of $\sin(x)$ or other functions, you should use computer algebra systems like (MATLAB symolic toolbox which is not always exact, Mupad in MATLAB, Maple, Mathematica, sympy package for Python).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is that you want to calculate the mean of $\sin(x)$ on the interval from $0$ to $7.5 \pi$. This should acutally not give you a value of $0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mean $&amp;lt;f(x)&amp;gt;$ of a function $f(x)$ on the interval $a$ to $b$ is given as:&#xA;$$&amp;lt;f(x)&amp;gt;_{[a,b]} = \frac{1}{b-a}\int_{a}^{b}f(x)dx.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your problem this leads to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&amp;lt;\sin(x)&amp;gt;_{[0,7.5\pi]}=\frac{1}{7.5\pi-0}\int_{0}^{7.5\pi}\sin(x)dx=\frac{1}{7.5\pi}\left[\cos(7.5\pi)-\cos(0)\right]\approx \frac{12}{90\pi}\approx 0.04244...$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the MATLAB code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;syms x;&#xA;1/(7.5*pi-0)*int(sin(x), x, 0, 7.5*pi)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that the mean is not exactly correct because MATLAB is using some weird fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not care about 100% precision, then you could also use more points:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mean(sin(linspace(0,7.5*pi,100000)))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-05-11T15:27:32.377" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15269" PostTypeId="2" ParentId="15261" CreationDate="2017-05-11T16:24:02.253" Score="0" Body="&lt;p&gt;You could keep the entire computation symbolic and defer computing for specific ranges.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mean=Integrate[Sin[x],{x,xmin,xmax}]/(xmax-xmin)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;(-Cos[xmax]+Cos[xmin])/(xmax-xmin)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From $0$ to $7.5 \pi$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mean/.{xmin-&amp;gt;0,xmax-&amp;gt;7.5 \[Pi]}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;0.0424413&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pygQA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pygQA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $1.5 \pi$ to $ 11.85 \pi$:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mean /. {xmin -&amp;gt; 1.5 \[Pi], xmax -&amp;gt; 11.85 \[Pi]}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;-0.0274025&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pQ8NY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pQ8NY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2017-05-11T16:24:02.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15270" PostTypeId="2" ParentId="14735" CreationDate="2017-05-11T17:04:18.790" Score="1" Body="&lt;p&gt;Typically, your welded joins are stronger than your base material itself. Looking at your two joint configurations, I would approach it from an ease of assembly/build perspective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joint A Wins in my book. You could fabricate the upper rectangle as a sub-assembly. Ensure that it's built square and correct and then weld the &quot;legs&quot;(the vertically oriented section) to it. Joint A also inherently caps the end of the section which is probably preferable. Most saws should be able to cut your 45 deg mitre. It's a pretty typical cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joint B would have you eventually fabricating in 3 planes at the same time. You would need more fixturing, manipulating, and checking as your welding to ensure you're positioning and alignment is correct.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-05-11T17:04:18.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15272" PostTypeId="1" AcceptedAnswerId="15275" CreationDate="2017-05-12T00:02:54.910" Score="4" ViewCount="1623" Body="&lt;p&gt;I am trying to transform any arbitrary vector $\bf{v_{1}}$ in a Cartesian frame $\mathscr{F_{1}}$ into a different Cartesian frame $\mathscr{F_{2}}$. Both $\mathscr{F_{1}}$ and $\mathscr{F_{2}}$ have the same origin. I have consulted &lt;em&gt;Aircraft Flight Dynamics and Control&lt;/em&gt; by Wayne Durham, and understand that book's solution as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the standard yaw-pitch-roll $ z,y,x$ $/$ {$\theta_{z},\phantom{s} \theta_{y},\phantom{s} \theta_{x}$} (usually called {$\psi, \theta, \phi$} respectively) / $3 2 1$ order. There exist three transformation matrices, one for each axis. First, rotate about the z axis of $\mathscr{F_{1}}$. Call the transformation matrix $T_{\mathscr{F'},1}$ since it goes from $\mathscr{F_{1}}$ to an intermediate frame $\mathscr{F'}$. $T_{\mathscr{F'},1}$ is given as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{\mathscr{F'},1}=\begin{bmatrix}&#xA;    cos(\theta_{z})       &amp;amp; sin(\theta_{z}) &amp;amp; 0 \\&#xA;   -sin(\theta_{z})       &amp;amp; cos(\theta_{z}) &amp;amp; 0 \\&#xA;    0                     &amp;amp; 0               &amp;amp; 1 \\         &#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go from $\mathscr{F'}$ to $\mathscr{F''}$, you then rotate about the y' axis using &#xA;$$T_{\mathscr{F''},\mathscr{F'}}=\begin{bmatrix}&#xA;    cos(\theta_{y})       &amp;amp; 0 &amp;amp; -sin(\theta_{y})              \\&#xA;    0                     &amp;amp; 1                &amp;amp; 0              \\&#xA;    sin(\theta_{y})       &amp;amp; 0               &amp;amp; cos(\theta_{y}) \\         &#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, to go from $\mathscr{F''}$ to $\mathscr{F_{2}}$, you then rotate about the x'' axis using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{\mathscr{F''},\mathscr{F'}}=\begin{bmatrix}&#xA;    1       &amp;amp; 0 &amp;amp; 0             \\&#xA;    0                     &amp;amp; cos(\theta_{x})       &amp;amp; sin(\theta_{x})   \\&#xA;    0       &amp;amp; -sin(\theta_{x})               &amp;amp; cos(\theta_{x}) \\         &#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A composite transformation matrix $\bf{T}$ can be made by multiplying these transforms together. Individual transform steps on a generic vector $\bf{v}$ are given in the book (and I reproduced them below) to show order determination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bf{v'}=T_{\mathscr{F'},1} \bf{v}$$&#xA;$$\bf{v''}=T_{\mathscr{F''},\mathscr{F'}} \bf{v'}$$&#xA;$$\bf{v_{2}}=T_{\mathscr{F_{2}},\mathscr{F''}} \bf{v''}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bf{v_{2}}=T_{\mathscr{F_{2}},\mathscr{F''}} T_{\mathscr{F''},\mathscr{F'}} T_{\mathscr{F'},\mathscr{F_{1}}} \bf{v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bf{v_{2}}=\bf{T}v_{1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $\bf{T} = T_{\mathscr{F_{2}},\mathscr{F''}} T_{\mathscr{F''},\mathscr{F'}} T_{\mathscr{F'},\mathscr{F_{1}}}$ (strictly in this order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem arose when I coded this transformation into MATLAB and tested some examples which I could verify qualitatively (with some coordinate axes I made out of wood).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transformation matrix $\bf{T}$ that works is $\bf{T} =  (T_{\mathscr{F'},\mathscr{F_{1}}}   T_{\mathscr{F''},\mathscr{F'}}  T_{\mathscr{F_{2}},\mathscr{F''}})^{T}$. Thus, in order for the transformation to work the way I expect it should, I need to multiply the matrices in reverse order and take their transpose. What has gone wrong?&lt;/p&gt;&#xA;" OwnerUserId="9226" LastActivityDate="2017-05-12T10:04:48.870" Title="3D Transformation Between Two Cartesian Coordinate Systems Using Euler Angles" Tags="|control-engineering|aerospace-engineering|mathematics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15274" PostTypeId="1" CreationDate="2017-05-12T06:47:26.803" Score="0" ViewCount="555" Body="&lt;p&gt;In Abaqus I print the problem's geometry of a plate 2D with a crack at the edge, bound on the left side in the x-direction and subjected to uniform effort in y-direction (uniform pressure - perpendicular to the two opposing edges) as shown in the figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8VRUT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8VRUT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I apply the tangential forces on the boundary as shown in the figure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QDSVh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QDSVh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerDisplayName="user11283" LastEditorDisplayName="user11283" LastEditDate="2017-05-13T19:09:24.893" LastActivityDate="2017-05-13T19:09:24.893" Title="Tangential forces on the boundary in Abaqus" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15275" PostTypeId="2" ParentId="15272" CreationDate="2017-05-12T07:55:05.030" Score="4" Body="&lt;p&gt;Suppose Frame 1 is a world coordinate frame and Frame 2 local robot frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that's a little confusing here is that when you transform from Frame 1 to Frame 2, you are saying that you want to... transform frame 1 to frame 2. You want to take all of frame 1's points, and put them in frame 2. You want measure where points in frame 1 are located, but in frame 2's coordinate system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a problem because what you probably want to do is measure/control the robot's frame (position) in the world frame coordinate system. You care if your vehicle is 10 degrees &lt;em&gt;above the horizon&lt;/em&gt;, or the global coordinate of an end effector, etc. This means that (counter-intuitively for me at least), if you want to see what something reads in frame 1's coordinates, then you need to import it into frame 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you want to use frame 1's coordinates to measure an object in frame 2, you need to take the transform from frame 1 to frame 2. To harp on this one more time, your math gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;v_2 = Tv_1 \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that your input is $v_1$, you transform the inputs by $T$, and then your outputs are &lt;em&gt;measured in frame 2's coordinates&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you &lt;em&gt;instead&lt;/em&gt; want to do is to take the &lt;em&gt;inverse&lt;/em&gt; transform. This gives you the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;v_2 = Tv_1 \\&#xA;T^{-1} \left(v_2\right) = T^{-1}\left(Tv_1 \right) \\&#xA;T^{-1}v_2 = v_1 \\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll point out briefly that you're using $T$ for a transform matrix, but the way you've defined it it's really a &lt;em&gt;rotation&lt;/em&gt; matrix; all of your $T_F$ functions are 3x3 matrices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either event, transform or rotation, it doesn't matter! The rotation matrix is the upper-left 3x3 section of a 4x4 transform matrix, and taking the &lt;em&gt;transpose&lt;/em&gt; of a rotation matrix is the same as taking the &lt;em&gt;inverse&lt;/em&gt; of the same rotation matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your coordinates are coming out wrong because you're not measured the points you provided in the frame you think you are. It's &quot;working&quot; when you take the transpose because that is actually the same as the &lt;em&gt;inverse&lt;/em&gt;, which means you are reversing the directionality of the transform. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2017-05-12T07:55:05.030" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15276" PostTypeId="2" ParentId="15272" CreationDate="2017-05-12T09:06:40.887" Score="1" Body="&lt;p&gt;You have actually not done anything wrong. Its just that Matrix algebra, much to the chagrin of some, does not actually define how you need to pack your data into the matrix. Your choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is do you consider that vectors inside your matrix to be columns or rows. This is called column major or row major. What you simply get is a transposed solution, and a inverse calculation order. It happens, some literature is in row major and some is in column major notation. Its all a question of whether you you want to look form object out or form outside in, your choice. But you need to be aware of this or you end up with problems like this, or well they aren't problems just definitinons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using qaternions or any other method does get you the same issue.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-05-12T10:04:48.870" LastActivityDate="2017-05-12T10:04:48.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15277" PostTypeId="2" ParentId="15274" CreationDate="2017-05-12T14:03:13.883" Score="1" Body="&lt;p&gt;In Abaqus to apply a shear constant shear force on a surface you need to construct a &lt;strong&gt;Surface Traction&lt;/strong&gt; load condition. This is an option in the &lt;em&gt;Load&lt;/em&gt; module. You will need to supply the direction and magnitude of the shear traction vector.&lt;/p&gt;&#xA;" OwnerUserId="7056" LastActivityDate="2017-05-12T14:03:13.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15278" PostTypeId="1" AcceptedAnswerId="15282" CreationDate="2017-05-12T14:07:18.353" Score="2" ViewCount="615" Body="&lt;p&gt;All,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm designing a minor slab to Eurocode 2. It's purpose is more for show than any real structural purpose however I need to justify it's ability to resist it's own self weight and possibly the accidental loading of a person climbing upon it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slab itself is 680mm wide, 160mm deep (100mm effective depth to rebar) and has a maximum span of 1200mm, so there is little room for shear reinforcement. Ideally, this slab should be kept as small and simple as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a small slab and acts as a cope for a sheepile wall and so spans between the 'zig-zag' of the piles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eurocode 2 states the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1IcM7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1IcM7.png&quot; alt=&quot;Clause 1, Section 6.2.2, Eurocode 2 (BS EN 1992-1-1)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be seen, for slabs with an effective depth (&lt;em&gt;d&lt;/em&gt;) of less than 200mm, the equations do not hold up as the inequality K &amp;lt;= 2.0 must be satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is very little explanatory text around these clauses and I am looking for guidance. Does this mean that a solution is impossible using these equations? If so, what other guidance can I refer too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what significance does the k value have? Does a k value &gt; 2.0 give me a conservative result that I can continue to use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately; How can the shear capacity of a concrete slab be determined for effective depths under 200mm within Eurocode 2? &lt;/p&gt;&#xA;" OwnerUserId="3093" LastEditorUserId="3093" LastEditDate="2017-05-12T14:22:20.147" LastActivityDate="2017-05-12T18:50:01.447" Title="How can the shear capacity of a concrete slab be determined for effective depths under 200mm? [Eurocodes]" Tags="|structural-engineering|civil-engineering|concrete|standards|eurocodes|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15279" PostTypeId="1" AcceptedAnswerId="15351" CreationDate="2017-05-12T14:50:00.817" Score="3" ViewCount="1257" Body="&lt;p&gt;I've run across a term in some grinding manufacturing research papers called the &lt;em&gt;specific material removal rate&lt;/em&gt;. The units given for this term are [mm$^3$/mm$\cdot$s] (I don't really like this style of formatting because it makes it unclear as to whether the [s] is in the numerator or denominator, but that is the style that has been used in all of the papers that I have seen with this term). I am familiar with the general machining term &lt;em&gt;material removal rate&lt;/em&gt; (or $MRR$), which is the volume of material removed per second from a workpiece in [mm$^3$/s], but what is the &quot;specific&quot; part of the &quot;specific material removal rate&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually in machining when we say &quot;specific&quot; we're talking about per unit volume removed. So the &lt;em&gt;specific energy of machining&lt;/em&gt; is normally expressed as unit energy per unit volume removed (e.g. [J/mm$^3$]). But for the &lt;em&gt;specific&lt;/em&gt; $MRR$ it is a volume per second per unit length (after all, it wouldn't make any sense to have volume removed per unit volume removed). So I guess the real question is, what is this length that we are dividing the $MRR$ by to get the &lt;em&gt;specific&lt;/em&gt; $MRR$? Why is this term useful?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems like a pretty basic question, but I have been unable to find a definition for the specific material removal rate anywhere. I've looked at several papers and from context in the papers it is rather unclear. I've checked my textbook, &lt;em&gt;Fundamentals of Modern Manufacturing 5th Ed.&lt;/em&gt; by Mikell P. Groover, no answer there. It may be a term that is specific to grinding operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate it if any answers could be backed up with a source.&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2021-09-14T18:13:59.353" Title="Definition of the &quot;specific material removal rate&quot;?" Tags="|manufacturing-engineering|grinding|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15280" PostTypeId="1" CreationDate="2017-05-12T15:00:33.347" Score="1" ViewCount="143" Body="&lt;p&gt;I am using Ansys Fluent for my simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a series of connected U pipes surrounded by a hot block (343K) with an inflow of cold water (298K) which is discharging the block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to simulate only one U pipe with the block around and run the simulation for 720 seconds. Then use the output temperatures as inlet for the simulation using UDF to simulate the second block and U pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outputs from the first section are a table of temperature with respect to time. It is an Excel file with two columns. I would like to use the exact values from the table (not an approximated function). How can I import these values into Ansys?&lt;/p&gt;&#xA;" OwnerUserId="11287" LastActivityDate="2017-05-12T15:00:33.347" Title="Importing a time-temperature list into Ansys Fluent for simulation" Tags="|ansys|fluent|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15281" PostTypeId="2" ParentId="15262" CreationDate="2017-05-12T16:53:24.630" Score="3" Body="&lt;p&gt;This is also considered a simply-supported beam. If you want to be more descriptive, you can call it a simply-supported beam with overhangs (the cantilevers beyond the main span).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, any beam which rests only on two hinged supports (allowing rotations) is considered simply supported. Beams which rest on more than two supports are considered continuous (so long as all the spans are covered by the same beam):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QM7oY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QM7oY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-05-12T16:53:24.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15282" PostTypeId="2" ParentId="15278" CreationDate="2017-05-12T18:30:44.983" Score="3" Body="&lt;p&gt;As @Wasabi and @AndyT mentioned in the comments, this means the maximum value $k$ can have is $2$, so if $d&amp;lt;200$ then $k=2$. This can be confirmed in the section on shear design for slabs from the document &lt;a href=&quot;http://www.concretecentre.com/Publications-Software/Publications/How-to-Design-Concrete-Structures-to-Eurocode-2.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;'How to Design Concrete Structures to Eurocode 2'&lt;/a&gt;. Table 7 from this document shows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yXJln.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yXJln.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has $V_{Rd,c}$ unchanged for effective depths $d&amp;lt;200$ and $k=2$ at the bottom of the $d\leq 200$ column. You can confirm this by doing the calculation by hand. Using $k&amp;gt;2$ would be non-conservative as it would predict a higher shear capacity than allowed by the code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in a more complete discussion of where this formula comes from you could refer to the &lt;a href=&quot;http://www.europeanconcrete.eu/publications/eurocodes/114-commentarytoeurocode2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Commentary to Eurocode 2&lt;/a&gt;, essentially it is a fit to experimental data.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2017-05-12T18:50:01.447" LastActivityDate="2017-05-12T18:50:01.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15286" PostTypeId="2" ParentId="13164" CreationDate="2017-05-13T03:47:11.957" Score="1" Body="&lt;p&gt;Some CAM software offers the option of generating &lt;em&gt;tabs&lt;/em&gt; or &lt;em&gt;bridges&lt;/em&gt; that look like this:&lt;a href=&quot;https://i.stack.imgur.com/V4k3Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V4k3Z.jpg&quot; alt=&quot;CAM Tabs&quot;&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position, qty and size of the tabs are configurable in the CAM software. Here's an example configuration dialog for Fusion 360:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZadXm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZadXm.png&quot; alt=&quot;Fusion 360 dialog with &amp;#39;tab&amp;#39; configuration&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tabs can be 'cleaned up' in a secondary operation.&lt;/p&gt;&#xA;" OwnerUserId="11292" LastEditorUserId="11294" LastEditDate="2017-05-13T16:07:37.643" LastActivityDate="2017-05-13T16:07:37.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15288" PostTypeId="1" AcceptedAnswerId="15297" CreationDate="2017-05-13T05:47:35.947" Score="1" ViewCount="279" Body="&lt;p&gt;I'm working on a process flow currently and am using carbon dioxide as the working fluid because of its heat properties and because it's an inert gas. In the reactor that I am designing I'm looking to operate at 4 atmospheres (~58.78 psia) and at around 977°F (~500°C). I know it would be a supercritical fluid at this P-T relation but can not find any information on the properties of carbon dioxide at these conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about finding or calculating for the thermodynamic properties at these conditions for carbon dioxide and then be able to continue this calculation through the entire process I'm designing?&lt;/p&gt;&#xA;" OwnerUserId="9305" LastEditorUserId="10902" LastEditDate="2019-10-04T05:39:18.547" LastActivityDate="2019-10-04T05:39:18.547" Title="How to find properties of supercritical CO2 as a working fluid in a system" Tags="|fluid-mechanics|thermodynamics|pressure|temperature|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15291" PostTypeId="1" CreationDate="2017-05-13T08:48:32.780" Score="13" ViewCount="1405" Body="&lt;p&gt;Typically materials which are good conductor of electricity are also good conductor of heat, and vice versa. Are there notable exceptions?&lt;/p&gt;&#xA;" OwnerUserId="11296" LastEditorUserId="-1" LastEditDate="2017-05-14T13:24:31.110" LastActivityDate="2017-05-14T18:04:25.020" Title="Does good electrical conductivity imply good thermal conductivity?" Tags="|thermal-conduction|electrical|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15292" PostTypeId="2" ParentId="15291" CreationDate="2017-05-13T10:06:06.767" Score="11" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Beryllium_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;Beryllium oxide&lt;/a&gt; is a very good electrical insulator but at the same time the best non-metal (except diamond) thermal conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to summarize. In general, good thermal conductivity is correlated with good electrical conductivity, but it is not a strict relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there is the empirical &lt;a href=&quot;https://en.wikipedia.org/wiki/Wiedemann%E2%80%93Franz_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiedemann-Franz law&lt;/a&gt; for metals which states that the ratio of thermal conductivity $k$ and electrical conductivity $\sigma$ is proportional to temperature $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{k}{\sigma} =c_0T $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proportionality constant is $c_0$. But as the wiki article points out there are exceptions to this empirical relationship.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-05-14T12:08:37.670" LastActivityDate="2017-05-14T12:08:37.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15293" PostTypeId="2" ParentId="15291" CreationDate="2017-05-13T10:19:07.857" Score="7" Body="&lt;p&gt;For metals, good electrical conductivity does indeed imply good thermal conductivity. This is known from the Wiedemann–Franz law, which gives the ratio between electronic contribution of thermal conductivity ($\lambda$) and electrical conductivity ($\sigma$) and is proportional to the temperature ($T$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\lambda}{\sigma} = LT $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the empirical constant $L$ known as the Lorenz number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ L = \frac{\lambda}{\sigma T} = \frac{\pi^2}{3}(\frac{\lambda_B}{e})^2 = 2.44 \times 10^{-8} W\Omega K^{-2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated, this law applies to metals. Diamond, for example, is an excellent thermal conductor because of its structure, but at the same time it has a very high resistance to electrical current.&lt;/p&gt;&#xA;" OwnerUserId="10979" LastEditorUserId="10979" LastEditDate="2017-05-13T10:43:27.407" LastActivityDate="2017-05-13T10:43:27.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15295" PostTypeId="1" CreationDate="2017-05-13T12:20:22.083" Score="1" ViewCount="34" Body="&lt;p&gt;The Beckhoff TwinCAT system allows me to put a box on a DIN rail which can act as  a realtime PLC controller (programmed in Ladder or SF) while allowing a Windows instance to run in the hardware at the same time.  Hence a Windows program (eg a UI written in .NET, or a comple high level algorithm) can make API calls to exchange data with the realtime PLC program.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other platforms out there which offer a similar sort of facility?  And how do they compare? &lt;/p&gt;&#xA;" OwnerUserId="11299" LastActivityDate="2017-05-13T12:20:22.083" Title="What &quot;Soft PLC&quot; platforms are there out there" Tags="|control-engineering|" AnswerCount="0" CommentCount="2" ClosedDate="2017-05-15T11:28:28.023" ContentLicense="CC BY-SA 3.0" />
  <row Id="15296" PostTypeId="1" CreationDate="2017-05-13T13:12:03.340" Score="1" ViewCount="342" Body="&lt;p&gt;How the tendon profile is calculated for banded-distributed disposition? In the following figure why is the tendons profile one in the top fiber and the second in the bottom fiber? If the location is at the middle of the slab why do we place them opposite to each other doesn't that cancel the effect of the lower tendon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WgbfJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WgbfJ.jpg&quot; alt=&quot;tendon opposite to each others&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;maybe this picture will clear my question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JSrUj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JSrUj.jpg&quot; alt=&quot;load in interior panel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="10554" LastEditDate="2017-05-14T14:41:02.780" LastActivityDate="2017-05-14T14:41:02.780" Title="What are the general consideration for tendon profile in a post tension slab?" Tags="|structural-engineering|civil-engineering|structural-analysis|prestressed-concrete|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15297" PostTypeId="2" ParentId="15288" CreationDate="2017-05-13T16:09:01.067" Score="1" Body="&lt;p&gt;Coolprop is a free API that's very similar to RefProp (not free):&#xA;&lt;a href=&quot;http://www.coolprop.org/contents.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.coolprop.org/contents.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-05-13T16:09:01.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15298" PostTypeId="1" CreationDate="2017-05-13T16:18:06.017" Score="2" ViewCount="54" Body="&lt;p&gt;The short version is: should I be worried about bass transmission through water into my neighbors' hulls?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am renovating a small houseboat (12x30 ft) to live on. It is one large room, and I plan on putting my reasonably large hifi system into it (260W/channel at 8Ω). One side of the boat is glass windows, the other is a plywood wall with an ~8&quot; cavity with studs. The floor is a plywood floor, and the hull is a fiberglass shell with marine ply inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am familiar with audio transmission/reflectance/resonance in, say, an apartment context, which can be roughly modeled as adjacent resonant cavities. How will transmission differ through water? Will I have higher transmission at bass frequencies? The houseboats adjacent to mine have thick concrete hulls (say, 6-12&quot;), so maybe the point is moot and I should be only worried about transmission over the surface of the water, which I know is highly reflective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tl;dr am I going to pwn my neighbors with bass through the water or should I be more worried about the glass wall and transmission through the 12 ft of air separating the boats? What other factors may contribute that I may be missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="11302" LastActivityDate="2017-05-13T16:50:10.827" Title="Acoustic transmission through houseboat hulls" Tags="|structures|acoustics|audio-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15299" PostTypeId="2" ParentId="15298" CreationDate="2017-05-13T16:42:04.183" Score="1" Body="&lt;p&gt;As @SolarMike pointed out the acoustic waves might harm the fish because the sonic speed in water is approximate $1484 \frac{\text{m}}{\text{s}}$ (at $20° \text{ C}$), this is quite high compared to the sonic speed in air which is approximate $330 \frac{\text{m}}{\text{s}}$. This might lead to the disorientation of the fish population close to your boat. The worst case scenario would be that your hi-fi system destroys the lateral line system (something like the ear of humans but not quite the same) of the fish population, which will lead to a permanent loss of the ability to orient and possibly to their death.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when underwater drilling is carried out it is necessary to create a curtain of air bubbles, which locally reduces the sonic speed, in order to protect the fish from taking harm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other questions, you can simply replace the sonic speed of air with the sonic speed of water and the physics should not change.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-05-13T16:50:10.827" LastActivityDate="2017-05-13T16:50:10.827" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15304" PostTypeId="2" ParentId="15291" CreationDate="2017-05-14T15:30:14.317" Score="2" Body="&lt;p&gt;As alluded to in the other answers, a most notable exception is diamond. Diamond is an &lt;a href=&quot;https://en.wikipedia.org/wiki/Material_properties_of_diamond&quot; rel=&quot;nofollow noreferrer&quot;&gt;excellent thermal conductor&lt;/a&gt;. The thermal conductivity of natural diamond is around 22 W/(cm·K), which makes diamond five times better than copper at conducting heat. At the same time, the electrical resistivity of most diamonds is on the order of 10E11 to 10E18 Ω·m.&lt;/p&gt;&#xA;" OwnerUserId="9506" LastEditorUserId="9506" LastEditDate="2017-05-14T18:04:25.020" LastActivityDate="2017-05-14T18:04:25.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15307" PostTypeId="1" AcceptedAnswerId="22711" CreationDate="2017-05-14T20:58:09.760" Score="2" ViewCount="202" Body="&lt;p&gt;If we are able to create a fusion reactor so there is a net gain in energy how can we turn the heat into electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how it is done normally for fission, steam, but how could this be adapted for a fusion reactor where all of the heat is confined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any concepts for the stage after fusion becomes net positive?&lt;/p&gt;&#xA;" OwnerUserId="10361" LastActivityDate="2019-09-19T10:54:09.270" Title="Extracting power from fusion?" Tags="|electrical-engineering|power-engineering|generator|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15310" PostTypeId="2" ParentId="14720" CreationDate="2017-05-14T23:16:02.127" Score="1" Body="&lt;p&gt;I arrived to an answer very similar to what it says it should end up being except for a difference in the sign of a term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how I proceeded:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total differential of a two-variable function can be written as&#xA;$$ dS(T,V) = \left( \frac{\partial S}{\partial T} \right) _V dT + \left( \frac{\partial S}{\partial V} \right) _T dV  $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a phase change occurs at constant temperature, $ dT = 0 $. Hence, the equation becomes simply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xA;dS = \left( \frac{\partial S}{\partial V} \right) _T dV&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing the right hand side of the above equation using other thermodynamic relations for the partial derivative in terms of $ P $ and $ T $, we get &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xA; dS = \left( \frac{\partial P}{\partial T} \right) _V dV&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we differentiate $ P $ with respect to $T$ at constant $V$:&#xA;$$&#xA; \left( \frac{\partial P}{\partial T} \right) _V = \frac{R}{V-b} + \frac{a}{2T^{3/2}} \frac{1}{V(V+b)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Noting that $ \Delta S = \int dS $,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA; \Delta S = S_2 - S_1 &#xA;= \int_{V_1}^{V_2}  \left( \frac{\partial P}{\partial T} \right) _V dV \\&#xA;= \left[  R \ln (V-b) +  \frac{a}{2b T^{3/2}} \ln  \left( \frac{V}{V+b} \right) \right]_{V_1}^{V_2} \\&#xA;=  R \ln \frac{(V_2-b)}{(V_1 - b)} +  \frac{a}{2b T^{3/2}} \ln  &#xA;\left[&#xA;\frac{V_2}{V_1} \left( \frac{V_1+b}{V_2+b} \right) \right]&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either the question is wrong in their signs or I made a mistake with the signs in my solution. Either way, this should hopefully guide you in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="9567" LastEditorUserId="10123" LastEditDate="2017-05-17T12:16:47.183" LastActivityDate="2017-05-17T12:16:47.183" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15311" PostTypeId="1" CreationDate="2017-05-15T01:53:03.293" Score="1" ViewCount="1406" Body="&lt;p&gt;Hey I was wondering what kind of chassis, wheels, gears are for a robot with a lot of torque. Please help. I have decided on a trapezoidal chassis, and I know that the servo connects to the small gear and the wheel connects to the bigger gear. But I do not know what the optimal gear ratio should be and size of the wheel. Also should I have 4 wheels (two in the back and 2 small ones in the front. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The competition is to see if you can pull the other robot in a tug of war. It is going to be happening on a typical gym floor at a school. We are provided with only two servos, and we can make any design that can move and work for the tug of war. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The robot has to be 8&quot; x 8&quot; x 8&quot; and should not exceed 2 lbs. &#xA;There is no budget, and I am building this for real, not fiction. I'm planning to laser cut the chassis but I don't know how long and wide it should be. We also can laser cut wheels (I don't know if I should use 4 wheels or 3). If I am going to do the tracks (like on a tank), how would I do that? Does it require 4 servos and 4 different gear boxes? &#xA;Also, for the type of gears, which one would be the best? &lt;/p&gt;&#xA;" OwnerUserId="11318" LastEditorUserId="11318" LastEditDate="2017-05-16T22:50:07.313" LastActivityDate="2017-05-16T22:50:07.313" Title="Tug of war robot w/ max torque" Tags="|gears|torque|robotics|" AnswerCount="3" CommentCount="2" ClosedDate="2017-05-16T16:24:01.847" ContentLicense="CC BY-SA 3.0" />
  <row Id="15313" PostTypeId="1" CreationDate="2017-05-15T06:29:41.600" Score="1" ViewCount="91" Body="&lt;p&gt;Heatseeker when locked on target(say a jet), follows the target until the it is hit, what happens when target is moving and changing it's direction.how does the heatseeker it's changes direction at such high speeds?&lt;/p&gt;&#xA;" OwnerUserId="11320" LastActivityDate="2017-07-23T18:16:19.857" Title="How heatseekers change their course?" Tags="|aerodynamics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15316" PostTypeId="1" CreationDate="2017-05-15T09:16:33.800" Score="1" ViewCount="83" Body="&lt;p&gt;Considering the situation where:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A non-waterproof motor is used inside a waterproof box&lt;/li&gt;&#xA;&lt;li&gt;Something must be rotated by that motor in a wet environment outside the box&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What is the correct way to transmit the rotational power from the motor to the exterior environment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rotation speed 1-10 rpm&lt;/li&gt;&#xA;&lt;li&gt;Exterior environment has constant low power water spray&lt;/li&gt;&#xA;&lt;li&gt;Low cost is a priority&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;NB : Apologies for using incorrect engineering language. I do not have an engineering background. I invite anyone to correct my words as appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&#xA;Elliot&lt;/p&gt;&#xA;" OwnerUserId="11321" LastActivityDate="2017-05-21T00:52:18.200" Title="Transmission of rotational power across a waterproof boundary" Tags="|motors|waterproofing|" AnswerCount="3" CommentCount="3" ClosedDate="2017-05-22T19:31:19.017" ContentLicense="CC BY-SA 3.0" />
  <row Id="15317" PostTypeId="2" ParentId="15316" CreationDate="2017-05-15T10:28:39.720" Score="2" Body="&lt;p&gt;A lot of bearings are sealed anyway, at least to some extent, to keep the grease in and water and grit etc out. Equally the inner and outer shells are typically a press or adhesive bonded  fit to the shaft/hosing so tend not to pose any inherent difficulties for sealing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the greater the pressure difference between the inside and outside the greater the demands on the seal but this is more a question of degree. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There also exist rotary seals which can seal a rotating shaft to a stationary housing, they are commonly used on things like the dive shafts of FWD cars where they exit the gearbox and fit in a similar way to a bearing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For just waster spray rather than something submerged a standard sealed bearing may well be fine &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-05-15T10:28:39.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15318" PostTypeId="2" ParentId="15316" CreationDate="2017-05-15T10:48:32.903" Score="1" Body="&lt;p&gt;You could also consider a magnetic coupling if the box material is suitable.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-15T10:48:32.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15319" PostTypeId="2" ParentId="15311" CreationDate="2017-05-15T10:51:53.120" Score="1" Body="&lt;p&gt;For maximum drag force, I recommend to have beam (long) chasis. Wheels on both farthest ends. This will maximize topple resistance, at every weight (more weight is better).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wheels, if possible is gear wheel on linear gear track affixed to the floor. This will enable drag force exceed friction weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gear ratio should be as big as possible (if no limitation, maybe 1:9999 or 1:99999 or more). You can use planetary gear for compact size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another tweak from me, pull point should be lower than wheel centerline. So that pull force will help to push the wheel more to the ground (I should draw this).&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2017-05-15T10:51:53.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15320" PostTypeId="2" ParentId="15020" CreationDate="2017-05-15T10:55:52.683" Score="0" Body="&lt;p&gt;With 1/3 air compressor flowrate, just test your actuator at ~10 strokes per minute.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2017-05-15T10:55:52.683" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15321" PostTypeId="2" ParentId="14236" CreationDate="2017-05-15T11:21:32.553" Score="2" Body="&lt;p&gt;In small bolts, threads is 1D but in larger size is less than 1D.&#xA;Free threads in best practice is 2-3 threads.&#xA;Don't forget to take gasket thickness (SPWD or RTJ?) in calc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional free threads of 1D may be required if hydraulic tension fastening method to be used. Hydraulic puller will need enough (1D) thread engagement to pull the studbolt.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastEditorUserId="4630" LastEditDate="2018-07-25T04:54:43.643" LastActivityDate="2018-07-25T04:54:43.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15322" PostTypeId="1" AcceptedAnswerId="15349" CreationDate="2017-05-15T11:25:38.043" Score="2" ViewCount="556" Body="&lt;p&gt;I'm currently using 400ml &lt;a href=&quot;http://www.farnell.com/datasheets/316910.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dust Off 67&lt;/a&gt; gas dusters and want to replace them with compressed air and a blow-out gun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to select the right equipment and get a better understanding I would like to compare the amount of air the gas duster contains to the compressed air solution. I know the gas duster does not actually contain compressed air, but for me it's all about the result. I know how much I can work with the gas duster and want to have a comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So considerung the 400ml Dust Off 67 (see linked datasheet for additional details, I don't know what might be relevant there). If I had a gas container which I can fill up with a pressure of up to 8 bar, how much volume (in liters) would the gas container need to get the same amount of air out of it as I would get the with the gas duster?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zi0B9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zi0B9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11323" LastActivityDate="2017-05-17T13:15:18.050" Title="volume comparison of compressed air and gas duster" Tags="|compressed-air|compressed-gases|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15325" PostTypeId="2" ParentId="15311" CreationDate="2017-05-15T13:42:34.740" Score="2" Body="&lt;p&gt;I would like to commented this on the question (but I can't due to the 50rep requirement)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if wheels are obligated or there a other options?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tracks should give more friction to the ground, so you will be able to have a smaller chasis with more traction than a long one with wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the design must be wheel based, this is also a option.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yDVmE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yDVmE.jpg&quot; alt=&quot;Wheel tracks&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also keep in mind for your design to have a low as possible center of gravity. The higher it is, the faster it tips.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Additional question(s)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To answer your additional question in the comments I would like some more info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you really building this in real life, are we talking about a scale model or is this just fiction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a budget limit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various ways to build &quot;tracks&quot; or something that gives a similar effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if there is a large budget for this, buying or making tracks would be the number one solution. But in many cases as in hobby projects there are no big budgets available. A cheaper solution that would give you more friction are snow chains, this will also work in mud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This technique is often used in by logging companies (link in swamp loggers discovery channel)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2Femj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Femj.jpg&quot; alt=&quot;Tire chains&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8464" LastEditorUserId="8464" LastEditDate="2017-05-16T13:54:15.820" LastActivityDate="2017-05-16T13:54:15.820" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15326" PostTypeId="2" ParentId="15313" CreationDate="2017-05-15T13:44:07.170" Score="2" Body="&lt;p&gt;There are fins on the missile that can control it's flight. Guidance is controlled by a thermal imaging camera that locks onto a heat source and computers keep the heat source in dead center of what you would seen on a view screen were you silly enough to ride the rocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other missiles use similar systems with radar or a laser 'painting' the target. The laser system uses a guidance laser targeting from usually a separate location, either the aircraft or sometimes an operator on the ground in the case of a moving vehicle. This can be watched in some modern military movies (but unfortunately I don't recall which one).&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:16:19.857" LastActivityDate="2017-07-23T18:16:19.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15327" PostTypeId="2" ParentId="15311" CreationDate="2017-05-15T13:58:15.770" Score="0" Body="&lt;p&gt;If this is for a robotics competition, you might keep in mind your design vs the opponent. If they put their weight on the rear of their vehicle and you can move your tie on point you can lift their rear wheels off the ground by tying off at an elevated location at the front of your vehicle. The downward force of the cable would actually add downforce to your wheels if they are close enough. You need to draw a tri-angle from the tie off point to the 2 wheels, then draw a straight line to the tie off point of the other robot. If the slope of the line is  shallow enough it will add sufficient downforce to your rear wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch some videos of tow trucks pulling vehicles out of ditches on you tube to get a better understanding of what will help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are required to use the same tie off point as your opponents Ranier J above has good advice on a long chassis and weight placement to minimize forces that might tip your robot over...although, thinking outside the box I wonder if a single axle robot might have more traction on the single axle and be 'un-tippable'. Look at the recent fad of 'hover boards' that have a platform located at the centerline of the axles.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastActivityDate="2017-05-15T13:58:15.770" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15328" PostTypeId="2" ParentId="15039" CreationDate="2017-05-15T14:02:20.150" Score="-4" Body="&lt;p&gt;A Tesla battery is made from many cells, there is a max current rating for each cell, a larger battery has more cells and therefore give out a larger peak current. (Ian ringrose)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also linked to the naximum output of the electric motor. Which produces the rotation.&lt;/p&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="10361" LastEditDate="2017-05-18T11:46:45.977" LastActivityDate="2017-05-18T11:46:45.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15329" PostTypeId="1" CreationDate="2017-05-15T14:07:35.563" Score="1" ViewCount="67" Body="&lt;p&gt;By no curvature I mean not following the earth's curvature. Would it be easier to tunnel the middle through the earth or raise the ends out of the earth? Is there anywhere that this kind of construction has been made and if so how much did it cost?&lt;/p&gt;&#xA;" OwnerUserId="11324" LastActivityDate="2017-05-15T14:07:35.563" Title="How feasible is a 200km long tube with no curvature?" Tags="|civil-engineering|construction-management|" AnswerCount="0" CommentCount="18" ClosedDate="2017-05-15T16:07:05.037" ContentLicense="CC BY-SA 3.0" />
  <row Id="15331" PostTypeId="2" ParentId="15313" CreationDate="2017-05-15T14:58:08.937" Score="0" Body="&lt;p&gt;Most heat seeking missiles (IR homing) have sensors fitted into GIMPAL(a kind of gyroscope) which help it to point towards the target even if missile is not.In a chase these missiles first need to have enough speed for its fins to have good affects.This is a considerable problem. &#xA;Mordern missiles consist of proportional guidance systems which help missiles to intelligently sort out a more easy and efficient way rather than chasing in original path.some of the modern IR homing missiles are AIM-9 Sidewinder(USA),ASRAAM(UK),IRIS-T(GER),MICA(FRA),PL-9(CHINA),Python-5(ISRAEL),R-73(RUSSIA),AAM-1(JAPAN).&lt;/p&gt;&#xA;" OwnerUserId="11325" LastActivityDate="2017-05-15T14:58:08.937" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15332" PostTypeId="2" ParentId="2599" CreationDate="2017-05-15T15:57:43.273" Score="1" Body="&lt;p&gt;I think this is a question better for the bean counters instead of Engineering. I worked as an Intern at a natural gas fired power plant (about 500 MW total capacity) and profits were the critical factor in determining what happens at a plant. Fuel was a pass-thru cost but capital and maintenance projects were different. Capital was considered re-investment but maintenance was an expense. Capital projects were approved if they invested enough value to the plant to be worth it, unplanned repairs counted against the plant. As Ericnutsch stated above, minor repairs and temp fixes and inspections are what make overhauls and nuke refueling take longer than you might think. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things like a leaking tube in a heat exchanger are a good example. You can isolate the heat exchanger to keep the unit operational but at a loss of efficiency so you may downgrade or limit the power output until an overhaul can repair the tubes. In this example what they usually do is isolate the heat exchanger, plug the tubes then once a set percentage of tubes are plugged you have to downgrade the unit until the entire tube rack can be replaced.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:15:25.483" LastActivityDate="2017-07-23T18:15:25.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15334" PostTypeId="1" AcceptedAnswerId="15356" CreationDate="2017-05-15T18:04:45.413" Score="2" ViewCount="386" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W6Xz4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W6Xz4.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Default Hydraulic solenoid valve issue&lt;/p&gt;&#xA;&lt;p&gt;I'm having a weird issue with some solenoid valves in a simple hydraulic circuit, let me try and describe the situation as best as I could.&lt;/p&gt;&#xA;&lt;p&gt;Coming out of a 2 station manifold, I have 2 4-way spring return center drain valves. 120VAC, controlled through relay outputs by a PLC. Each valve controls a set of cylinders, A position down, B position up. Wired up such that 1 leg of each coil jumped together to my 120VAC control, the other 4 legs to their corresponding PLC output.&lt;/p&gt;&#xA;&lt;p&gt;Initially, I had valves with a center position, A&amp;amp;B drain to T, P blocked. Everything worked correctly from a controls standpoint, only problem was when both valves were &amp;quot;off&amp;quot;, I was holding at full pressure (~1700 psi)and building up a lot of heat.&lt;/p&gt;&#xA;&lt;p&gt;So I swapped out the valves for center position, A to B blocked, P drain to T. Wired up exactly the same as before. Now when I actuate, let's say, valve 1, side A, nothing happens...until I actuate valve 2 side A, then cylinder on valve 1 actuates, as well as the cylinder on valve 2. Electrically everything is right when I check with a meter, program is running as it should as well. So that leads me to believe that something's going on hydraulically.&lt;/p&gt;&#xA;&lt;p&gt;Anyone have any idea what's going on? It seems like when I have one valve, P going to A and the other &amp;quot;off&amp;quot; (P to T) that I might have to much pressure going to tank, preventing B to T thus keeping my cylinder from moving?&lt;/p&gt;&#xA;&lt;p&gt;Let's assume that's the issue, how can I modify my design to remedy the problem? I'm thinking, change from one, 2 station manifold to 2, independent station blocks and add check valves to the T outputs to prevent pressure from backing into the T line and preventing the oil from draining from the exhaust of the cylinder.&lt;/p&gt;&#xA;&lt;p&gt;Solved: A &amp;quot;dump&amp;quot; or relief valve was added to the circuit, right before the solenoid manifold, plumbed into the tank line, right before the radiator. In the instances when the pump is running, but the machine/solenoid valves are not operating, the dump valve is opened, draining all pressure to tank through radiator.&lt;/p&gt;&#xA;" OwnerUserId="11327" LastEditorUserId="9448" LastEditDate="2022-10-04T18:01:17.230" LastActivityDate="2022-10-04T18:01:17.230" Title="Hydraulic System design help/ directional valves" Tags="|mechanical-engineering|hydraulics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="15339" PostTypeId="1" CreationDate="2017-05-16T01:12:25.747" Score="2" ViewCount="44" Body="&lt;p&gt;OK, let's be clear: I'm a &lt;em&gt;software&lt;/em&gt; engineer (retired), not a physicist, hydrologist, or hydraulics person, so please be gentle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Back Story:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the process of developing an open source system for monitoring water column &#xA;height in our well and of house / irrigation water usage (we're in N. CA), I discovered we had a substantial problem with sand and silt in our well water. We hadn't noticed it before because other components of the system (e.g. our 3,000 gal holding tank, ozone system, and calcite filter) were removing most of it before it came out of the kitchen tap. But that's &lt;em&gt;not&lt;/em&gt; what they were meant to do. (Well, the ozone is sort of meant to help with clumping, but not the other 2.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After going through several variations on filters -- paper, steel mesh, etc. -- it became clear that they &lt;em&gt;all&lt;/em&gt; clogged up, requiring frequent flushing / cleaning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading a lot of stuff which is &lt;em&gt;waay&lt;/em&gt; outside my area of expertise, it is clear what we need is a hydrocyclonic filter. User &lt;a href=&quot;https://engineering.stackexchange.com/users/1339/rul30&quot;&gt;rul30&lt;/a&gt; mentioned &lt;a href=&quot;http://nigc.ir/portal/Images/Images_Traning/files/files/chemist%20book%20cd2/chemical%20eng/SVAROVSKY,%20L.%20(2000)/Solid_Liquid_Separation_4E.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Solid-Liquid Seperation&lt;/em&gt;&lt;/a&gt; by Ladislav Svarovsky. Chapter 6, &lt;em&gt;Hydrocyclones&lt;/em&gt; appears to be exactly what I need. Except I don't know how to go from the formulas and academic description to a specific set of dimensions I could use with a 3D printer or other fabrication method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem Statement:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a well pump that is throttled to 3GPM to minimize silt migration into the well. This flow level is substantially below what all commercial hydrocyclonic filters require (mostly 10GPM and higher). We have a mix of heavy sand (which is easy to get rid of with even a half-assed cyclonic filter), and a fair volume of material that is as small as 50 &amp;micro;m. I have plenty of pressure available, so we can create significant velocity, but the &lt;em&gt;specific&lt;/em&gt; dimensions of the upper cylinder, the lower cone shape, and the depth of the vortex finder are based on equations that are beyond me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Help?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="1609" LastActivityDate="2017-05-16T01:12:25.747" Title="Determing the optimal dimensions for hydrocyclone water/sediment separation" Tags="|water-resources|fluid-filtration|cyclones|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15340" PostTypeId="1" CreationDate="2017-05-16T03:01:33.257" Score="2" ViewCount="48" Body="&lt;p&gt;Is there a universal rule for servo drive gear placement? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm planning to use &lt;a href=&quot;http://www.goteckrc.com/Download/GS-9025MG.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this servo&lt;/a&gt; for a robotics project, but I need to know exactly where the drive gear is located relative to the body. The technical sketch provides the other dimensions very nicely, but they fail to provide that &lt;em&gt;important&lt;/em&gt; measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume that one dimension is easily found by taking half the width, but the other dimension can't be eyeballed; although it appears to be roughly 3/4 of the length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone knows why they didn't provide this dimension and/or how I could find it, that would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:&#xA;If this isn't the right S.E. to ask this, would someone kindly point me in the direction of the correct one?&lt;/p&gt;&#xA;" OwnerUserId="3487" LastActivityDate="2017-05-16T03:01:33.257" Title="Servo drive gear placement relative to the body" Tags="|motors|servo|technical-drawing|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15341" PostTypeId="1" CreationDate="2017-05-16T09:11:03.040" Score="3" ViewCount="65" Body="&lt;p&gt;I am looking for a benchmark as a reference regarding the human Kinesiology that specifies the quickest movement a human can do on basic everyday movements like computer mouse clicking.&#xA;As far as I know, there is an standard called MTM (Measurement Time Methods) in the field of industrial engineering and ergonomics which specifies the required time for each basic movement to be done by human. This MTM standard time is provided on the basis of not putting the human (worker) in an overload or underload working situation. So I assume MTM cannot provide me with the shortest possible duration (time) a basic movement can be performed. Do you have any suggestion in this regard?&lt;/p&gt;&#xA;" OwnerUserId="4503" LastEditorUserId="4503" LastEditDate="2017-05-18T07:34:52.223" LastActivityDate="2017-08-16T11:02:35.267" Title="How to specify the quickest movement a human can do based on MTM standards?" Tags="|industrial-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15342" PostTypeId="2" ParentId="417" CreationDate="2017-05-16T10:31:32.320" Score="1" Body="&lt;p&gt;I'm thinking of using dry ice fog in water, or by heat up kerosene until smoking.&#xA;Dry ice produced fog when put in water, and kerosene will get gasification when heated.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastEditorUserId="4630" LastEditDate="2017-05-16T11:23:54.480" LastActivityDate="2017-05-16T11:23:54.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15343" PostTypeId="2" ParentId="666" CreationDate="2017-05-16T10:55:24.657" Score="1" Body="&lt;p&gt;Use Flange &amp;amp; Blind Flange ANSI 2&quot;-2500# WNRTJ ASME B16.5 A-182 F304. With a gasket RTJ SS304 and bolting.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2017-05-16T10:55:24.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15345" PostTypeId="1" AcceptedAnswerId="15348" CreationDate="2017-05-16T13:55:45.070" Score="0" ViewCount="144" Body="&lt;p&gt;I'd like to ask you if a C mex s-function model in Simulink can get the dSpace execution time lower, rather than having a model with blocks and functions.&lt;/p&gt;&#xA;" OwnerUserId="11338" LastEditorUserId="90" LastEditDate="2017-05-16T17:05:39.753" LastActivityDate="2017-05-16T17:05:39.753" Title="dSpace overrun issue" Tags="|electrical-engineering|control-engineering|matlab|simulink|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15347" PostTypeId="1" CreationDate="2017-05-16T15:36:00.923" Score="9" ViewCount="315" Body="&lt;p&gt;When a metal wire moves across a magnetic field, voltage is induced across it, which is equal to $BLV$, where $B$ is strength of magnetic field, $L$ is the length of wire and $V$ is its velocity. Will a carbon nanotube also exhibit this phenomenon? I am asking this because carbon nanotube is although a good conductor of electricity, but probably that is because of some quantum phenomenon (bulk carbon is bad conductor of electricity).&lt;/p&gt;&#xA;" OwnerUserId="11341" LastEditorUserId="10684" LastEditDate="2017-05-17T12:16:56.330" LastActivityDate="2017-05-19T06:54:39.393" Title="Moving carbon nanotube in a magnetic field" Tags="|electromagnetism|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15348" PostTypeId="2" ParentId="15345" CreationDate="2017-05-16T16:21:24.973" Score="0" Body="&lt;p&gt;&lt;strong&gt;Short Answer:&lt;/strong&gt; It depends on what blocks you're replacing, and even then the effects might be negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interconnected set of standard Simulink blocks (gain, sum, integral, derivative, etc.) will definitely be equally-fast or faster than any custom function you write, CMEX or not. On the other hand, replacing a MATLAB Function Block with a CMEX S-Function might have some effect, but it's very difficult to determine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects of using a CMEX S-Function will depend on your skill at writing efficient C code, and even then the effect could be small compared to the execution time of other parts of your model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also not as simple as asking which block is more efficient in 'Normal mode' Simulink. dSpace uses a combination of the Simulink Coder and their own proprietary software to translate your Simulink model into C code that is then compiled into an executable. That executable is then run in real time by the dSpace hardware. Therefore, during the translation process from Simulink to C code the benefits of using the CMEX S-Function might be negated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Note: I have not worked with dSpace but I do use a very similar software package in my research called QUARC, which is basically the Canadian equivalent.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2017-05-16T16:21:24.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15349" PostTypeId="2" ParentId="15322" CreationDate="2017-05-16T16:24:21.903" Score="3" Body="&lt;p&gt;According to to the SDS found &lt;a href=&quot;http://www.tme.eu/en/Document/bbd0a19d0233a00349282ad342bbaaed/DUST_OFF_67_Aerosol.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;here&lt;/strong&gt;&lt;/a&gt; the product contains tetrafluoroethane as an ingredient. I looked around and found other similar products with the same ingredient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sake of simplicity lets say that all the $400 ml$ are tetrafluoroethane and the compressed density is $1.01 g/cm^3$ according to your table. With this information we know that one can contains $404 g$ of tetrafluoroethane by multiplying the density by the volume. ( $1 ml = 1 cm^3$ ) According to &lt;a href=&quot;https://en.wikipedia.org/wiki/1,1,1,2-Tetrafluoroethane&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Wikipedia&lt;/strong&gt;&lt;/a&gt; the molar mass of tetrafluoroethane is $102.03 g/mol$. With this information we know that bottle contains 3.96 moles of tetrafluoroethane. ( $404 g / 102.03 g/mol = 3.96 mol$ )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can use the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/idegas.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;Ideal Gas Law&lt;/strong&gt;&lt;/a&gt; to solve for volume at &lt;a href=&quot;https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;STP&lt;/strong&gt;&lt;/a&gt;. In our case a handy rule of thumb is 1 mole = $22.4 L$ at STP. So you canister contains 88.7 liters of tetrafluoroethane ( $3.96 * 22.4$ ) if it were not compressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we reverse this process for air. This time around the handy rule is PV( initial ) = PV ( final ). 8 bar = 7.9 atmmospheres. $1 atm * 88.7 L = 7.9 atm * 11.2L$. So the answer to your question is that you will need 11.2 liters of compressed air at 8 bar to equal to your .4 liters of Dust Off! The reason that Dust Off is so efficient is that it liquefies when compressed where air does not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the process seems confusing read up on the Ideal Gas Law. It is very powerful when dealing with gasses at different conditions.&lt;/p&gt;&#xA;" OwnerUserId="9656" LastEditorUserId="9656" LastEditDate="2017-05-17T13:15:18.050" LastActivityDate="2017-05-17T13:15:18.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15350" PostTypeId="2" ParentId="417" CreationDate="2017-05-16T19:18:48.593" Score="6" Body="&lt;p&gt;In automotive testing we use baby oil. When heated it produces a thick white smoke that is non toxic in low volumes and doesn't stick to surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I built my own smoke tester using a metal pressure fed wax-oil/underseal sprayer, I drilled a hole and inserted an automotive glow plug to heat the oil. You can then experiment with a pressure that works best for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have successfully filled car interiors with smoke under pressure and it hasn't left any residue.&lt;/p&gt;&#xA;" OwnerUserId="11135" LastActivityDate="2017-05-16T19:18:48.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15351" PostTypeId="2" ParentId="15279" CreationDate="2017-05-16T20:46:40.170" Score="3" Body="&lt;p&gt;The specific material removal rate appears to be the product of the depth of cut and the feed rate. &quot;Specific&quot; refers to normalization by the width of the cut. From a quick search at Google Books, this definition appears, for example, in:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Li's &quot;Efficiency of Manufacturing Processes&quot;&lt;/li&gt;&#xA;&lt;li&gt;Klingelnberg's &quot;Bevel Gear: Fundamentals and Applications&quot;&lt;/li&gt;&#xA;&lt;li&gt;Bach et al.'s &quot;Properties of Optical Glass&quot; &lt;/li&gt;&#xA;&lt;li&gt;Kimber's &quot;Quality Management Handbook&quot;&lt;/li&gt;&#xA;&lt;li&gt;Shen's &quot;Minimum Quantity Lubrication Grinding Using Nanofluids&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2017-05-16T20:46:40.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15352" PostTypeId="2" ParentId="14644" CreationDate="2017-05-16T20:53:14.767" Score="1" Body="&lt;p&gt;A standard induction motor has to turn at the speed determined by the current frequency.  If you attempt to apply external torque to drive it faster it will act as a brake resisting that torque. &#xA;If you apply sufficient torque to force the rotor out of phase with the driving current it would effectively stall ( the &quot;braking force&quot; dropping way off and or becoming erratic )&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2017-05-16T20:53:14.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15353" PostTypeId="1" AcceptedAnswerId="15354" CreationDate="2017-05-16T22:50:05.560" Score="0" ViewCount="80" Body="&lt;p&gt;We have large scale aircraft with long endurance and much higher speed than birds. But it seems that aircraft of comparable size to birds (i.e. drones) have much lower endurance, top speed and flight range. What technologies are currently limiting us from achieving this?&lt;/p&gt;&#xA;" OwnerUserId="11241" LastEditorUserId="11241" LastEditDate="2017-05-17T02:26:38.097" LastActivityDate="2017-05-17T02:26:38.097" Title="What technologies prevent drones from being as efficient as birds?" Tags="|mechanical-engineering|aircraft-design|" AnswerCount="1" CommentCount="0" ClosedDate="2017-05-17T18:32:57.643" ContentLicense="CC BY-SA 3.0" />
  <row Id="15354" PostTypeId="2" ParentId="15353" CreationDate="2017-05-16T23:43:35.743" Score="6" Body="&lt;p&gt;Somebody needs to design a chocolate-powered drone. Yes, seriously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total energy stored in a 40AH 12V battery is about the same as the calorific value of five 100g chocolate bars from your nearest supermarket. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: from &lt;a href=&quot;https://www.tesco.com/groceries/product/details/?id=254381873&quot; rel=&quot;noreferrer&quot;&gt;https://www.tesco.com/groceries/product/details/?id=254381873&lt;/a&gt;, &quot;Tesco Everyday Value Milk Chocolate&quot; provides 3840 kJ/kg, and the fully charged battery holds 40x12x3600/1000 = 1728 kJ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a way to pack that much energy density into drone (and also consider that unlike LIPO batteries, chocolate bars don't spontaneously combust if they are mishandled!) and the comparison between drones and birds would come out rather differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, soaring birds (and sailplanes) can stooge around all day and night, getting their energy from thermals and the wave airflows over mountain ridges - but they still need &lt;em&gt;some&lt;/em&gt; external power to reposition themselves to take advantage of those &quot;free&quot; energy sources.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-05-16T23:50:52.213" LastActivityDate="2017-05-16T23:50:52.213" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15355" PostTypeId="1" CreationDate="2017-05-17T00:06:36.850" Score="1" ViewCount="1278" Body="&lt;p&gt;My next door neighbor's car backed into my wooden fence and knocked down a section.  I want to protect the fence from a future accident.  A steel pipe filled with concrete would cost more than PVC.  A pressure-treated 6x6 could also work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to do it up right.&lt;/p&gt;&#xA;" OwnerUserId="11345" LastActivityDate="2017-05-17T00:06:36.850" Title="Cement-Filled 6&quot; PVC Bollard" Tags="|steel|strength|" AnswerCount="0" CommentCount="4" ClosedDate="2017-05-17T10:27:43.227" ContentLicense="CC BY-SA 3.0" />
  <row Id="15356" PostTypeId="2" ParentId="15334" CreationDate="2017-05-17T04:18:13.643" Score="1" Body="&lt;p&gt;Yes. You are losing your pump pressure through the center (off) position of the the other valve. Once you do some work through the other valve, so as not to drain your pressure to the tank, then both cylinders get some pressure to operate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two approaches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Connect the T output from one valve to the P input of the other valve. The fluid can return to the tank easily when both valves are in the center position, but both valves will receive pressure from the pump. The pressure and volume going to the second valve will not be constant when the valve that is first in line is being switched, but both will still work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I have seen this approach, but I don't know the exact circuit. Go back to your earlier approach, and add a pressure sensing pilot controlled relief valve that opens the return path to the tank whenever neither of your control valves is drawing fluid. So your full 1700 psi drops to some much lower pressure to reduce heat when the pressure is not being drawn by either valve.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastEditorUserId="11222" LastEditDate="2017-05-17T16:34:17.807" LastActivityDate="2017-05-17T16:34:17.807" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15357" PostTypeId="2" ParentId="15347" CreationDate="2017-05-17T11:28:25.870" Score="1" Body="&lt;p&gt;A carbon nanotube is a conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is somehow a 'rolled graphene sheet' and the material graphite (stack of graphene layers) is quite a famous conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What characterize the carbon nanotube is its 'fairly' good strength, together with a ballistic conduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore as for any conductor, the equation &lt;code&gt;emf = B.L.V&lt;/code&gt; is perfectly valid (the equation reflect an accumulation of charges on both ends of the wire)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously if your wire is too short (let say &amp;lt;0.5um long) then you may have to start considering a bit more your system but with other conductors it would be the same.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="10684" LastEditDate="2017-05-19T06:54:39.393" LastActivityDate="2017-05-19T06:54:39.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15358" PostTypeId="2" ParentId="1905" CreationDate="2017-05-17T13:50:10.530" Score="0" Body="&lt;p&gt;I made a fast drawing with a bit more proper electronic:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iOoRr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iOoRr.png&quot; alt=&quot;voltage switcher&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You cannot put two electrical sources in parallel, this will break your system imediatelly or on the long term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using Not gates because they are the most common, also the simplest to build.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connect the green with the green ($12V$) and orange with orange ($6V$) at the moment the circuit is on $12V$, press the switch and it will work on $6V$&#xA;(you need to work a bit the switch so what is not connected is then connected to the ground)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also indicated that you can charge your batteries &lt;em&gt;in series please&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="10684" LastEditDate="2017-05-17T14:02:17.840" LastActivityDate="2017-05-17T14:02:17.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15360" PostTypeId="1" AcceptedAnswerId="15362" CreationDate="2017-05-17T18:45:27.497" Score="1" ViewCount="219" Body="&lt;p&gt;I was reading &lt;a href=&quot;http://www.control.tu-berlin.de/images/2/22/seel2012_msc.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; called &quot;&lt;em&gt;Joint Axis and Position Estimation from Inertial Measurement Data by Exploiting Kinematic Constraints&lt;/em&gt;&quot;, but I'm not understanding one part, which doesn't let me proceed in the reading of the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the physical point of view, we have the following situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IxNHu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IxNHu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two segments (one orange and the other green) connected by a &lt;em&gt;hinge joint&lt;/em&gt;. The 3-axis coordinate systems that you see in the picture (the one on segment orange and the other on the segment green) are the local coordinate systems of two 3-dimensional gyroscopes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dashed line that you see in the hinge should be the coordinate axis of the same joint. Since the joint can only move in one dimension, we have just one axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given this setting, the authors say (first page, column at the right, at the end of the same):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... let the angular velocities of the gyroscopes, in the coordinates of&#xA;  their local frames, be $g_1(t)$ and $g_2(t)$ for the first and the second segment, respectively. Then it is a geometrical fact, that $g_1(t)$ and $g_2(t)$ differ only by the joint angle velocity and a (time-variant) rotation matrix&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h3&gt;Questions&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the joint angle velocity?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is this time-variant rotation matrix?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Why do the angular velocities differ only by the joint angle velocity and a time-variant rotation matrix? What does it even mean, actually?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user11356" LastEditorDisplayName="user11356" LastEditDate="2017-05-17T22:18:50.530" LastActivityDate="2017-05-17T22:18:50.530" Title="Two angular velocities on two segments connected by a hinge joint differ only by the joint angle velocity" Tags="|mathematics|terminology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15361" PostTypeId="1" CreationDate="2017-05-17T20:04:52.720" Score="1" ViewCount="41" Body="&lt;p&gt;I'm designing a push button made from most likely ABS but possibly TPE, it presses a surface mount button on a PCB with 0.9mm travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ABS button is 8x8mm Square and travels through 4mm of polycarbonate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want there to be very little (if any) wobble and as little gap around the edges as possible for aesthetic reasons. But I cannot have the button snagging or jamming as the product wears and the inevitable dirt build up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big should the cut out be? &lt;/p&gt;&#xA;" OwnerUserId="11135" LastActivityDate="2017-05-17T20:04:52.720" Title="Required clearance for moving part (Push Button)" Tags="|plastic|industrial-engineering|product-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15362" PostTypeId="2" ParentId="15360" CreationDate="2017-05-17T21:04:38.523" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CLLFS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CLLFS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;What is the joint angle velocity?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The joint angle velocity is the angular velocity of you joint, because we are acutally dealing with a 2D problem. This will become clear if you look at your system from the $z_0,z_1$ axis direction. In the picture this angular velocity is given by $\dot{\beta}$.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;What is this time-variant rotation matrix?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You have to consider four coordinate frames. The first coordinate frame is fixed at your joint axis $K_0=(x_0,y_0,z_0)$. The second frame is frame 1 but now rotating around the $z_0=z_1$ axis. The other two frames are there in your inital image. If you tranform from frame 1 to the other frames you will see that the rotation matrix will be time dependent. This is obvious, because depending on the time dependent motion of $\beta$ the coordinate frames will also move with time dependence.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;Why do the angular velocities differ only by the joint angle velocity and a time-variant rotation matrix? What does it even mean,&#xA;  actually?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This means that the angular velocity of the left coordinate frame $\omega_1$ and the angular velocity of the right coordinate frame $\omega_2$ are related by the angular velocity of the joint $\dot{\beta}$. The relationship is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega_2-\omega_1=\dot{\beta}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that this simple equation is only true because we only have a 2D rotation. Imagine $\dot{\beta}=0$, then the joint is acting like a fixed link, hence the left and right coordinate frame have the same angular velocity. If $\dot{\beta}$ is positive then the angular velocity of the right frame $\omega_2$ will be the sum of the angular velocity of the left frame $\omega_1$ and the angular velocity of the joint $\dot{\beta}$.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-05-17T22:02:11.473" LastActivityDate="2017-05-17T22:02:11.473" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15364" PostTypeId="2" ParentId="14067" CreationDate="2017-05-17T22:40:32.073" Score="0" Body="&lt;p&gt;A suggestion: if the shaft has a reduced diameter section on the end, and that section is cut eccentrically to the scoop's axis of rotation, then the scoop will lift up and down as it rotates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB abrasive particles each cut their own little sample, so I think you might need to look for another idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the size of cut-out piece you mention is pretty big in &quot;cutting metal&quot; terms. That means high forces. You'll need to find some way to anchor the machine... especially if you're going to sample a turbine blade. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully it's just the mechanism you need. All the best&lt;/p&gt;&#xA;" OwnerUserId="11360" LastEditorUserId="17070" LastEditDate="2019-09-05T10:04:54.940" LastActivityDate="2019-09-05T10:04:54.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15365" PostTypeId="2" ParentId="15188" CreationDate="2017-05-17T23:10:50.823" Score="0" Body="&lt;p&gt;The 2 keys to a hydraulic jack (which is what a hydraulic car lift is) are:  the often quoted Pascals's law.. AND valves, a non return one to &quot;ratchet&quot; the big piston up, and a controllable one to let it down. Brilliant simplicity.&lt;/p&gt;&#xA;" OwnerUserId="11360" LastActivityDate="2017-05-17T23:10:50.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15367" PostTypeId="1" AcceptedAnswerId="15376" CreationDate="2017-05-18T04:12:55.570" Score="1" ViewCount="1394" Body="&lt;p&gt;I need design a trolley with four casters, the four casters are installed on the four corners of a 460mmX405mm metal plane. And the load will be a cubic 380mmX225mmX350mm (Height is 350mm), weight about 150Kg, site at the center of the metal plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimensions of the casters is as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hYlKp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hYlKp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see the support point of the caster and the wheel are not on a vertical line, so I'm afraid when all four wheels in the same direction, the trolley will not be stable. So I go here for some suggestions(or some formulas are better), if the trolley can works stable? Or I must change my design, such as enlarge the metal plane? &lt;/p&gt;&#xA;" OwnerUserId="11362" LastEditorUserId="11362" LastEditDate="2017-05-18T11:04:37.223" LastActivityDate="2017-05-18T11:18:01.587" Title="Design a trolley with four casters" Tags="|applied-mechanics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15369" PostTypeId="1" CreationDate="2017-05-18T04:24:25.790" Score="2" ViewCount="34" Body="&lt;p&gt;I am trying to model WCDMA MAC, but I cannot figure out how to demultiplex HS-DSCH unto DCCH, DTCH.&#xA;Also, since I am using E-DCH/HS-DSCH transport channel scheme, how does the data of DCCH  and DTCH mapped to E-DCH on transmitter side would be demapped on the receiver side?&#xA;The respective MAC headers do not seem to provide a clue to tell just by looking at the message which part belongs to DTCH and which part belongs to DCCH.&lt;/p&gt;&#xA;" OwnerUserId="11364" LastActivityDate="2017-05-18T04:24:25.790" Title="How does the demapping/demultiplexing of transport channels to logical channels work?" Tags="|telecommunication|networking|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15370" PostTypeId="2" ParentId="15367" CreationDate="2017-05-18T05:58:27.190" Score="1" Body="&lt;p&gt;Consider using two of the trailing ones you show at one end and find two similar that are fixed ie not trailing at the other end. This will give stability and manouverabilty as well.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-18T05:58:27.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15371" PostTypeId="1" CreationDate="2017-05-18T07:08:26.413" Score="1" ViewCount="1615" Body="&lt;p&gt;From the book &quot;Design of Liquid Propellant Rocket Engines&quot;. Under stress analysis related pages, the author used an equation for finding the thickness of the wall needed for sphere under stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Thickness of sphere wall} = \dfrac{\text{Yield pressure}\times\text{diameter}}{4\times\text{yield strength at 300°F}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know the equation for tubes.&lt;/p&gt;&#xA;" OwnerUserId="9679" LastEditorUserId="9679" LastEditDate="2017-05-20T06:04:41.543" LastActivityDate="2020-08-11T05:10:16.903" Title="Equation for finding the thickness of material required for tubes under radial stress" Tags="|stresses|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15372" PostTypeId="1" AcceptedAnswerId="15395" CreationDate="2017-05-18T07:40:37.413" Score="1" ViewCount="41" Body="&lt;p&gt;I have a small pump that does 6L/min flow-rate. With a pipe size of 0.5 inches and 3 nozzles spaced at a distance of about 1m each, is it possible to do effective misting? My total pipe length will be 4m. I have been trying out calculations from various references but it keeps driving me nuts. What am wondering is, is the pump fit for that job? I have seen similar pumps online with equal flow-rates and pressure ratings of about 130psi. What matters most?&lt;/p&gt;&#xA;" OwnerUserId="11368" LastActivityDate="2017-05-19T04:42:56.107" Title="Can a small pump be good for atomizing" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15373" PostTypeId="1" CreationDate="2017-05-18T08:27:22.533" Score="1" ViewCount="795" Body="&lt;p&gt;I'm having a question regarding the sensors and the techniques in fluid mechanics. Since my skills are amazing in this field, here I'm asking you questions xD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to measure the air pressure of this tube/pipe (P1,P2,P3). This pipe will be flushed with water and then with compressed air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[![enter image description here][1]][1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, it should be just pressure sensor in each pipe and problem solved! &lt;/p&gt;&#xA;" OwnerUserId="11369" LastEditorUserId="11369" LastEditDate="2017-09-06T13:50:33.650" LastActivityDate="2017-09-06T13:50:33.650" Title="How to measure the air pressure of a pipe without having water resistant sensor" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="2" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="15374" PostTypeId="2" ParentId="15341" CreationDate="2017-05-18T09:15:40.103" Score="1" Body="&lt;p&gt;I would suggest that sports science is probably a good place to look for this sort of information as reaction times are critical in many sports, for example the reaction times of sprinters to the starting gun are often quoted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For very simple operations like a button press then the time taken for the brain to process the stimulus, make a decision and react becomes significant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also reasonable to suppose that there will be a difference between making an expected reaction to a cue and tasks where there is some element of decision making. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally the time taken to complete the action will depend on the action itself eg if you are but actual reaction times are typically i the order of a few hundreds ms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also &lt;a href=&quot;https://www.humanbenchmark.com/tests/reactiontime&quot; rel=&quot;nofollow noreferrer&quot;&gt;reaction time testing game&lt;/a&gt;s online if you want to try it for yourself.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-05-18T09:15:40.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15375" PostTypeId="2" ParentId="15371" CreationDate="2017-05-18T10:15:07.663" Score="1" Body="&lt;p&gt;Assuming a very long cylinder (to neglect effects at the top and bottom of the cylinder) with a small wall thickness $t$ (to assume that the stress is constant across the thickness of the wall) and with mean diameter $d_{\text{m}}=d_{\text{inner}}+t$ you can easily derive the so-called Barlow's formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_{\text{tangetial}}=\frac{d_m}{2t}(p_{\text{inner}}-p_{\text{outer}})$$&#xA;$$\sigma_{\text{axial}}=\frac{\sigma_{\text{tangetial}}}{2}$$&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-05-18T10:21:25.450" LastActivityDate="2017-05-18T10:21:25.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15376" PostTypeId="2" ParentId="15367" CreationDate="2017-05-18T11:18:01.587" Score="1" Body="&lt;p&gt;As long as the worst case (all wheels pointing to the center) is still stable then you have no issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To look at the footprint first create it as if you didn't use castors. Then for cornet with a castor move the line so it's tangential to the inside of the circle the contact point makes when swiveling. Note that this may mean that the apparent contact point is outside that circle. For a rectangular base you can subtract twice the radius of the swivel's offset on each side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can use that footprint for your stability calculations.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-05-18T11:18:01.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15377" PostTypeId="2" ParentId="15373" CreationDate="2017-05-18T11:49:44.533" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/171Je.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/171Je.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps you could use a small extension tube with a loop in it... Mounted vertically not horizontally, so as to act as a liquid trap... As the would be no flow in that tube then water should not get to the pressure sensor - but you will have to make sure you will get the required accuracy. Fit the pressure sensor to the top end.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-18T11:49:44.533" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15379" PostTypeId="2" ParentId="15373" CreationDate="2017-05-18T13:32:39.937" Score="0" Body="&lt;p&gt;There are many solutions for this. The simplest is to use a standard pressure gauge on a vertical pipe sticking out of your pipes. It will be exposed to minimal water vapour and unless you are using a fancy delicate gauge you will be fine. Below is something I found on how gauges work and why you will be fine if water does get into the gauge:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Pressure gauges commonly use a Bourdon tube. Bourdon tubes are a coiled metal tube and as the pressure increases, it forces the coil straight which then moves the needle. (think blowing a party horn, which causes it to unroll) The type of fluid (water/air) makes no difference to the tube, only the pressure exerted against the tube does. And since most of the tubes are made from bronze, copper or stainless water will not corrode it any faster than moist air from the compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all said, the air compressor gauge will most likely work without any issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you really want to get a water gauge, go to a pool supply store. They have pressure gauges for filters that are designed for water and use a 1/4 npt. They also have a nifty little ring on the outside where you can mark your normal pressure so it is easy to see if your pressure is low or high.&#xA;&lt;a href=&quot;https://diy.stackexchange.com/questions/74354/can-i-use-a-gauge-designed-for-measuring-air-pressure-to-measure-water-pressure&quot;&gt;LINK&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Pipelines that are drained and filled often have surges of water that can damage piping and gauges by what is called 'water hammer' when a slug of water is propelled at high velocity by air behind it. If your pipe is designed to fill and drain it should be able to handle these surges but I would monitor the gauge for a few cycles to ensure excessive pressures are not observed.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastActivityDate="2017-05-18T13:32:39.937" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15380" PostTypeId="1" CreationDate="2017-05-18T13:37:17.490" Score="0" ViewCount="82" Body="&lt;p&gt;In order to make a neuron fire via an external electrode, the electrode must input some electrons in the extracellular space(just outside of the neuron). Imagine a straight conducting wire touching a neuron and other end free in the air. If you move a strong magnet near the wire, such that the wire cuts the magnetic flux, then voltage will be generated at the two ends of the wire. In this case will the neuron fire? or equivalently, will there be flow of electrons to the neuron? Note that the wire is in open circuit i.e. other end is in the air. If there will not be any flow of electrons, can some simple modification on this setup will make it happen?&lt;/p&gt;&#xA;" OwnerUserId="11375" LastEditorUserId="8055" LastEditDate="2017-05-19T16:30:52.560" LastActivityDate="2017-05-19T16:30:52.560" Title="Using electromagnetic induction to fire a neuron" Tags="|electrical-engineering|electromagnetism|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15381" PostTypeId="2" ParentId="3410" CreationDate="2017-05-18T13:51:28.233" Score="1" Body="&lt;p&gt;Maintaining a co-linear hole/feature with the axis of rotation of the part. Automatic lathes with bar feeders have also been around for multiple decades. These machines can be &quot;programmed&quot; to make that particular part, in high speed, without any need for an operator. Milling or drilling that feature in a round part (which is more difficult to hold in such a set up) would require much more to set up square and maintain concentric to the outside diameter. These machines can also be set up to run autonomously (CNC) however they would require a robotic arm to load and un load parts into the machine. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2017-05-18T13:51:28.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15383" PostTypeId="2" ParentId="3410" CreationDate="2017-05-18T15:04:58.570" Score="0" Body="&lt;p&gt;Another possibility to consider is that the part is on a moving track and can then proceed to the next station / operation.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-18T15:04:58.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15384" PostTypeId="1" CreationDate="2017-05-18T17:32:54.833" Score="1" ViewCount="175" Body="&lt;p&gt;Previously, I was told by someone I trust well that a column can avoid buckling at the first critical load by applying the load &quot;instantaneously&quot; with a magnitude between the first and second critical loads. Of course nothing is instantaneous, so an idealization would be applying the load as a step function. I am an undergraduate mechanical engineer with no practical background with buckling failures and was hoping someone can shed some light on this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry I have few details or sources backing this question up, I have been trying to confirm it and have been unable to find it online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this true? Can buckling be avoided at the first critical load by instantaneously applying a load between the first and second critical loads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="11377" LastActivityDate="2017-05-18T18:17:39.923" Title="Can buckling failure at the first critical load be avoided by &quot;instantaneous&quot; loading?" Tags="|mechanical-engineering|applied-mechanics|mechanical-failure|buckling|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15386" PostTypeId="2" ParentId="15380" CreationDate="2017-05-18T18:05:46.590" Score="1" Body="&lt;p&gt;Your goal is to move the electrons in a neuron with a magnetic field. If we assume a homogeneous magnetic field $\mathbf{B}$, then this will generate an electrodynamic force &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{F}=q\mathbf{v}\times\mathbf{B},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that acts on a charge $q$ which moves with the velocity $\mathbf{v}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we move the magnet from orthogonal to the direction of the neuron-axis, then this is equivalent to moving the electors in the reverse orthogonal direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that $\mathbf{v}=-v\mathbf{e}_y$ points in the negative $y$-direction. The magnetic field is pointing in the negative $z$-direction $\mathbf{B}=-B\mathbf{e}_z$ and we look at one electron with the charge $q=-1.609\cdot 10^{-19} \text{C}$. Then using $\mathbf{F}=q\mathbf{v}\times\mathbf{B}=-qvB\mathbf{e}_x$ (see picture).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that your electron will try to stay on a circular trajectory with radius $r$. This is obvious because your force is orthogonal to your velocity. The radius can be determined by assuming an equilibrium of the electrodynamic force and the centripetal force (both with magnitudes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$qvB=m\frac{v^2}{r}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would rotate the neuron-axis according to this knowledge, then you can move the electron from one side to the other side. But it is quite complicated compared to a standard electric field. Another more theoretical way would be to use a very small magnetic field so that the radius becomes very large, but that will lead to very slow accelerations because $\mathbf{F}$, will be very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WF4WF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WF4WF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-05-18T18:05:46.590" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15387" PostTypeId="2" ParentId="15384" CreationDate="2017-05-18T18:11:16.627" Score="2" Body="&lt;p&gt;This is just wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The notion of a &quot;critical load&quot; in Euler buckling is a very simplified mathematical idealization of the situation. In the real world, buckling behaviour always involved nonlinear dynamics, and if you apply a big enough load, the structure will always buckle - though if you apply the load fast enough, it may buckle into a &lt;em&gt;different shape&lt;/em&gt; from what would happen with a smaller load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only situation I can think of where this idea is &quot;sort of correct&quot; is shaft whirling in rotordynamics, which vaguely analogous to buckling. In that case, if you accelerate or decelerate the rotor fast enough through a critical speed, there isn't enough time for the displacement to get big enough to cause failure, and continuous operation at speeds &lt;em&gt;above&lt;/em&gt; the critical speed is stable. This idea &lt;em&gt;is&lt;/em&gt; used in the design of some types of rotating machines.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-05-18T18:17:39.923" LastActivityDate="2017-05-18T18:17:39.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15388" PostTypeId="2" ParentId="15384" CreationDate="2017-05-18T18:16:44.973" Score="1" Body="&lt;p&gt;Euler's buckling is the behavior of static loads and not time-dependent loads. You would need to look at nonstatic beam deflections but I doubt that instantaneous loading will prevent buckling. In contrast, it is more likely that a step function loading will excite some critical frequency of the system which might lead to instability and hence buckling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it might be useful to you to understand what buckling means. Buckling is an instability of an equation. Differential equations permit many solutions. Sometimes these solutions are stable and sometimes they are unstable (like in buckling). When a solution is unstable then little perturbations of the actual state will lead to divergent behavior of the state and you might get a totally different behavior (like buckling).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So actually it is possible to observe no buckling (which is always a possible solution for the bar under compression) of your system even for very large loads. But small perturbations from this state will lead to deviations from this solution, hence it is unlikely that you will observe this case in real life. &lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-05-18T18:16:44.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15389" PostTypeId="1" CreationDate="2017-05-18T19:04:48.267" Score="1" ViewCount="34" Body="&lt;p&gt;Attempt to edit into a more specific question as suggested by moderator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like define the shape and material properties of a 3D thin-walled shell/membrane-like structure (for instance a baloon). Then I would like to approximate the pressure volume-relation (PVR) of this structure - what pressure will we obtain when increasing the internal volume? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm planning to create a mesh of 2D triangular elements for a 3D shell with volume V0 and internal pressure zero. After defining the stiffness matrix for the material I should be able to compute the nodal forces, geometric configuration and internal volume for a certain internal pressure (the plan is to use an iterative approach searching for a configuration where all resulting nodal forces equals zero, that is where expanding forces arising from the internal pressure equals constricting forces arising from displacement). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this I need to calculate the nodal forces created by 1) displacement from the original V0 configuration and 2) from the internal pressure. I think I can figure out the first one, but I do not know how to transform the internal pressure (acting perpendicular on the triangular elements) into nodal forces. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="11380" LastEditorUserId="11380" LastEditDate="2017-05-21T19:32:39.210" LastActivityDate="2017-05-21T19:32:39.210" Title="Pressure-volume relation of thin-walled shell" Tags="|structural-engineering|structures|stresses|finite-element-method|pressure-vessel|" AnswerCount="0" CommentCount="4" ClosedDate="2017-05-19T12:44:59.467" ContentLicense="CC BY-SA 3.0" />
  <row Id="15390" PostTypeId="2" ParentId="15371" CreationDate="2017-05-18T20:38:46.153" Score="1" Body="&lt;p&gt;Had to post this as an answer:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YtVAM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YtVAM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which deals with thick-walled cylinders :&#xA;Source : &lt;a href=&quot;http://www.engineeringtoolbox.com/stress-thick-walled-tube-d_949.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/stress-thick-walled-tube-d_949.html&lt;/a&gt;&#xA;Or here, from Lamé's equations :&#xA;&lt;a href=&quot;http://www.mydatabook.org/solid-mechanics/stress-for-thick-walled-cylinders-and-spheres-using-lames-equations/#cylinder_axial_stress&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mydatabook.org/solid-mechanics/stress-for-thick-walled-cylinders-and-spheres-using-lames-equations/#cylinder_axial_stress&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-18T20:38:46.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15391" PostTypeId="2" ParentId="8596" CreationDate="2017-05-18T22:00:26.873" Score="1" Body="&lt;p&gt;I would use a low viscosity epoxy dip on the tube end and going/polish with an air flow and mask, when cured with repeated operations as required for surface finish.&lt;/p&gt;&#xA;" OwnerUserId="8847" LastActivityDate="2017-05-18T22:00:26.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15392" PostTypeId="1" AcceptedAnswerId="15397" CreationDate="2017-05-18T22:20:48.887" Score="0" ViewCount="468" Body="&lt;p&gt;How do I calculate the thrust of ducted contrarotating propellers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any formula or website calculator or would it have to be done through experimenting with scale replicas?&lt;/p&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="9679" LastEditDate="2017-05-18T22:36:04.490" LastActivityDate="2017-05-19T05:06:37.937" Title="Calculating Thrust for Contra-Rotating Propellers" Tags="|propulsion|thrust|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15393" PostTypeId="1" AcceptedAnswerId="15406" CreationDate="2017-05-19T04:08:41.570" Score="4" ViewCount="394" Body="&lt;p&gt;Consider this hypothetical ideal scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have a test coupon in a tensile test that has a constant cross-sectional area throughout its length (i.e. along the tensile loading direction) and there are no flaws or imperfections at all, where does necking occur?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case stress is uniform throughout the volume of the coupon (no stress concentrations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="11382" LastActivityDate="2019-05-30T06:23:05.163" Title="Tensile Test Necking Location" Tags="|materials|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15394" PostTypeId="1" CreationDate="2017-05-19T04:21:34.807" Score="5" ViewCount="1327" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uFJE1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uFJE1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Anemometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Anemometer&lt;/a&gt; measures wind speed. It's shaped with little cups---hollow half-spheres. That looks very logical as the cups grab the wind and the other side is rather streamligned to cut through the air as it spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So why are there no wind turbines designed like this?&lt;/strong&gt; (At least, I've never seen them.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could go further and make the cups ellipsoidal (more pointy). You could even add a inner shield, maybe like half a torus, around the inside to further streamlign the cups and help them grab wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be a vertical axis wind turbine (VAWT). Been looking around at VAWTs on Youtube for a while and found a lot of interesting designs, but never seen one with cups. Why not? I know I've seen anemometers spinning around &lt;strong&gt;very rapidly in good winds, so its seems logical.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="3510" LastEditorUserId="33" LastEditDate="2019-01-16T13:50:34.403" LastActivityDate="2019-01-16T13:50:34.403" Title="Why not design a Wind Turbine like an Anemometer?" Tags="|turbines|wind-power|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15395" PostTypeId="2" ParentId="15372" CreationDate="2017-05-19T04:42:56.107" Score="3" Body="&lt;p&gt;Droplet size through a spray nozzle is a strong function of pressure drop. The higher the pressure drop, the smaller the drop size. So for an atomizer you need high pressure (probably &gt;15 psi). Coarser droplets can be produced down to about 3 psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out the data sheets for you nozzles. They likely specify acceptable pressure drops and may also rate the droplet size (coarse to fine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, as is usual, the pressure drop across the nozzle is related to flow rate. Increasing one increases the other (check out &quot;flow coefficient&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, a small centrifugal pump probably won't let you atomize/mist, although that's not a fixed rule. Try positive displacement pumps like diaphragm or gear pumps to get higher pressures at low flows.&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2017-05-19T04:42:56.107" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15396" PostTypeId="2" ParentId="15394" CreationDate="2017-05-19T04:44:01.813" Score="3" Body="&lt;p&gt;Because the back of the cups create a lot of drag thereby reducing the power produced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blade profile used in classic 3-bladed HAWTs is also used in VAWTs and they look like &quot;egg beaters&quot; needing top and bottom bearings. The engineering complications of VAWTs make the 3-bladed HAWT much more popular also due to the power output, rotating mass and visual impact.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-05-19T05:12:12.923" LastActivityDate="2017-05-19T05:12:12.923" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15397" PostTypeId="2" ParentId="15392" CreationDate="2017-05-19T05:06:37.937" Score="1" Body="&lt;p&gt;This is one analysis that gives you information .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dULrf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dULrf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source : &lt;a href=&quot;http://naca.central.cranfield.ac.uk/reports/arc/rm/2218.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naca.central.cranfield.ac.uk/reports/arc/rm/2218.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-19T05:06:37.937" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15398" PostTypeId="1" CreationDate="2017-05-19T05:12:41.303" Score="-3" ViewCount="120" Body="&lt;p&gt;When unfolding the DJI Mavic Pro, the extensions with propellers seem to lock into place? How is it that when it is folded, it appears to be locked into place as well? How can I replicate it?&lt;/p&gt;&#xA;" OwnerUserId="11383" LastActivityDate="2017-05-19T11:20:08.747" Title="How is it that when the DJI Mavic Pro is unfolded, the extension and propellor locks into place?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15400" PostTypeId="2" ParentId="13607" CreationDate="2017-05-19T08:47:56.800" Score="0" Body="&lt;p&gt;First things first. 4G is not only for data, it is a only -packet switched network as opposed to both circuit switched and packet switched in 3G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as per your question as to how does a phone(user equipment or better UE) knows the reception of a voice call, it is done by control plane signalling over the network. So, yes, a signal is sent before the reception of call. But what is important to note that this procedure is not exclusive to 4G, 3G and lower networks also employ similar mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as per your doubt if call would be received with data switched off, yes it would be received. This is because signalling messages are transferred over LTE and not data( they are two separate things). In fact event before you can access data on LTE networks a process similar to voice call is carried out, and it is technically known as data calls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I had mentioned in my comment, VoLTE is used to carry voice calls. This is a new standard for transmitting voice messages. The need for this method arose because LTE supports only Packet Switching,  while conventionally voice was carried through Circuit Switched networks.&lt;/p&gt;&#xA;" OwnerUserId="11364" LastActivityDate="2017-05-19T08:47:56.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15401" PostTypeId="1" CreationDate="2017-05-19T10:01:02.863" Score="3" ViewCount="3831" Body="&lt;p&gt;I have a vehicle (I bought it and it proprietary and I have no information about any internals) which I want to integrate into my simulation environment. So far I have a physical model of it which I gained by driving around recording data and building a physical model from it. Now I'm at the point where I want to model the internal controller. I though that it might use a PID but I think now that something else is used instead.&#xA;First I ran (around 30k) simulation with the physical model as basis to come up with the PID constants. In the simulations I let the vehicle move around the way as I did with the real vehicle and compared the response of my controller with the measured response of the real controller. In the image below you see the best PID which my brute force parameter search came up with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N1Dtg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N1Dtg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see my PID (red line) does not respond as fast as the real controller (green line) So my question is now: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What kind of controller could have been used here ? And how can I model it ?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is definitive not a PID. Ten year ago I had lectures on control and system theory but since then I did not need to build a controller so I have not much experience. I looked into the literature and found the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lead%E2%80%93lag_compensator&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lead–lag compensator&lt;/a&gt; which look promising but I have no idea where to start or how to implement it. To get the constants here I again can run simulations in a try an error manor. Maybe another different approach has been used but I have no idea what.&lt;/p&gt;&#xA;" OwnerUserId="11388" LastEditorUserId="11388" LastEditDate="2017-05-19T10:06:32.680" LastActivityDate="2018-04-17T21:06:19.347" Title="Replacing PID with Lead–lag compensator?" Tags="|control-engineering|control-theory|pid-control|optimal-control|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15402" PostTypeId="2" ParentId="15398" CreationDate="2017-05-19T11:20:08.747" Score="0" Body="&lt;p&gt;According to the manual, this is how the extensions move:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X4Fgi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X4Fgi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know precisely which type of hinge they used, but it's probably follows the same general concept as this one used for folding tables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GwMIW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GwMIW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hinge is slotted, so once the extension reaches a certain point, it &quot;falls&quot; into the slot and can't rotate back to its original location (unless the tabletop is raised, removing it from the slot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many variations on this design, of course. Some have &quot;buttons&quot; which can be pressed to release the lock, for example. But the overall concept behind them (at a certain point, something mechanically impedes further motion) is the same.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-05-19T11:20:08.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15403" PostTypeId="2" ParentId="14847" CreationDate="2017-05-19T11:47:58.533" Score="2" Body="&lt;p&gt;Read the battery datasheet.  Those little 9 V consumer batteries with the snap on leads are not good for delivering this kind of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measure the battery voltage when connected to your motor, and you will probably find it is close to 0.  This may be the case even if the battery measures about right when open circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a benchtop power supply.  Something that can put out a few amps at 12 V should be good.  Don't use a car battery because those aren't current-limited.  If you cause a short, the car battery will deliver enough current to melt or explode something.  A benchtop supply with current limiting at a couple of amps should provide enough power, but be safe when the inevitable short happens.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Added in response to picture&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It is amazing that thing worked at all.  Obvious problems:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;The large diameter of the commutator.  Actually &lt;i&gt;think&lt;/i&gt; about it.  The brushes touching the commutator cause friction.  The bigger the diameter, the lager the friction force due to higher speed, &lt;i&gt;and&lt;/i&gt; that force is applied at a larger radius, so causes more torque.  You want the commutator diameter to be as small as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Single coil rotor.  This means there are two positions at which there is never any torque.  This is when the plane of the coil is vertical in your picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;The spacing between the magnets.  You want the magnetic field around the rotor to be as strong as possible.  This means placing the magnets as close together as possible.  They should be as close to the rotor without touching it as you can reliably manage.  The bottom magnet can certainly be closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;The coil is a clear mismatch to the little toy battery you are trying to power this motor with.  Use a bench power supply, as described above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2017-05-21T12:35:33.590" LastActivityDate="2017-05-21T12:35:33.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15406" PostTypeId="2" ParentId="15393" CreationDate="2017-05-19T12:05:42.193" Score="4" Body="&lt;p&gt;According to theories of failure, necking begins as small voids near the largest imperfection present in the specimen. But, as far as the case you proposed is concerned, the specimen has no imperfections at all. Then the clamping used to fix the specimen will induce stresses perpendicular to the axis of elongation. This will cause formation of imperfections near the clamps. Thus the necking will occur in close proximity of one of the clamps.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2017-05-19T12:05:42.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15407" PostTypeId="1" AcceptedAnswerId="15450" CreationDate="2017-05-19T12:48:42.123" Score="0" ViewCount="539" Body="&lt;p&gt;I want to measure the vibration of a roller coaster cart while it is riding down the tracks. Handheld data loggers with a pin on them won't work. I'd need something I can attach or place inside of the roller coaster track. I am worried that conventional data loggers won't be well suited for this because the acceleration of the cart influences the vibration readings when using an accelerometer. I have found exactly 0 results on Google for &quot;measuring vibration of moving objects&quot;-like queries. I want to measure 3-axis G-force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything you could recommend me to use?&lt;/p&gt;&#xA;" OwnerUserId="11389" LastEditorUserId="1832" LastEditDate="2018-11-28T18:52:45.250" LastActivityDate="2018-11-28T18:52:45.250" Title="How do I measure vibration of a moving object?" Tags="|vibration|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="15408" PostTypeId="2" ParentId="15407" CreationDate="2017-05-19T13:59:51.893" Score="0" Body="&lt;p&gt;And a google search for &quot;vibration datalogger&quot; gave : &lt;a href=&quot;https://www.crystalinstruments.com/coco-90-dynamic-signal-analyzer-and-data-collector/?gclid=CPz8w9KO_NMCFSEW0wodzs4FHQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.crystalinstruments.com/coco-90-dynamic-signal-analyzer-and-data-collector/?gclid=CPz8w9KO_NMCFSEW0wodzs4FHQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the first result....&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-19T13:59:51.893" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15411" PostTypeId="2" ParentId="15407" CreationDate="2017-05-19T15:27:52.030" Score="1" Body="&lt;p&gt;I'm sure there is more than one way to do what you want, but based on what you have posted and my experience measuring vibration in and out of vehicles, this is basically how I would do it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used an earlier version of &lt;a href=&quot;http://www.isthq.com/IST-Products/EDR-6DOF-6-Axis-Acceleration-Angular-Rate-Recorder&quot; rel=&quot;nofollow noreferrer&quot;&gt;these data recorders&lt;/a&gt; from IST to measure the vibration environment inside shipping containers and truck trailers as they travelled around the world. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this application, I would mount 1 inside the cart using the built in sensors, and a second unit mounted near the tracks, using external accelerometers. Triggering both systems to record at the same time would be tricky, but I would set up optical sensors to trigger both or use one optical sensor to trigger a radio module which would in turn trigger both recorders. They won't be perfectly synced, but probably good enough. I'm sure that the folks at IST would be more than happy to help you with getting this set up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another nice thing about IST is that they &lt;a href=&quot;http://www.isthq.com/Product-Rentals&quot; rel=&quot;nofollow noreferrer&quot;&gt;rent equipment&lt;/a&gt; so you don't necessarily have to make a huge investment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I am not associated with IST in any way aside from being a satisfied customer.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-05-19T15:27:52.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15412" PostTypeId="1" CreationDate="2017-05-19T15:31:46.160" Score="6" ViewCount="5825" Body="&lt;p&gt;Is there any material that blocks a static magnetic field? So that if I put it between a magnet and a piece of metal, they won't interact with each other?&lt;/p&gt;&#xA;" OwnerUserId="4846" LastEditorUserId="368" LastEditDate="2017-05-19T16:23:59.487" LastActivityDate="2017-05-19T20:08:17.947" Title="Is there any material that blocks a static magnetic field?" Tags="|electrical-engineering|electromagnetism|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15414" PostTypeId="2" ParentId="15412" CreationDate="2017-05-19T16:31:20.040" Score="13" Body="&lt;p&gt;Yes, there are several different ways to manipulate magnetic fields. One way to shield an area from a magnetic field is to place a highly ferrous material like iron next to the source of the magnetic field. The iron will absorb the field in such a way that the magnetic field would be much smaller on the other side of the iron material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ECgzp.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ECgzp.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice how in this simulation the iron sheet is absorbing the field and barely letting anything through. &lt;a href=&quot;https://www.kjmagnetics.com/blog.asp?p=shielding-materials&quot; rel=&quot;noreferrer&quot;&gt;This website&lt;/a&gt; goes into deeper detail on how materials absorb magnetic flux.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Superconductors can be used for magnetic field shielding as well. Superconductors repel magnetic fields much more efficiently than say steel but is much more expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EQLaP.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EQLaP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this picture you can see that the magnetic field lines are repelled from the sheet of superconducting material. If you would like to learn more about superconducting shielding take a gander at &lt;a href=&quot;http://www.coolmagnetman.com/magshield.htm&quot; rel=&quot;noreferrer&quot;&gt;this website&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also important to note that while both these cases will keep two magnets from interacting with each other that does not guarantee that they will not interact with the shield itself. In the case of an iron based shield magnets will be attracted to the shield and in the case of a superconducting shield it will be repelled.&lt;/p&gt;&#xA;" OwnerUserId="9656" LastEditorUserId="9656" LastEditDate="2017-05-19T19:19:00.947" LastActivityDate="2017-05-19T19:19:00.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15415" PostTypeId="1" CreationDate="2017-05-19T17:35:59.100" Score="0" ViewCount="667" Body="&lt;p&gt;It is my understanding that basic latching solenoid consist of coil, magnet, spring, two electrical wire, electrical connector, enclosure and armature plunger. My assumptions are based on &lt;a href=&quot;http://www.tlxtech.com/understanding-solenoids/theory-operation/latching-solenoid-theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Latching Solenoid Theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KYVSp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KYVSp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic 4 wire stepper motor consist of 2 additional coils, magnet, four electrical wire, enclosure and electrical connector. My assumptions are based on &lt;a href=&quot;http://www.eng.utah.edu/~cs5789/2012/Slides/Steppersx2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Intro to Stepper Motors&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4pPzz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4pPzz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above are my basic understanding with regard to internal differences and similarities between the two components. I understand vendors will include addition feature to differentiate themselves from competition.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt; Are there any other standard subcompacts that are missing in this basic analysis?   &lt;/p&gt;&#xA;" OwnerUserId="8055" LastEditorUserId="282" LastEditDate="2017-05-22T13:19:40.323" LastActivityDate="2017-05-22T13:19:40.323" Title="Internal differences between latching solenoid and Stepper Motor" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15419" PostTypeId="2" ParentId="15412" CreationDate="2017-05-19T20:08:17.947" Score="-2" Body="&lt;p&gt;Yeah, you can use what is known as a  Faraday cage or Faraday shield, which is an enclousure of iron used to block electromagnetic fields.&lt;/p&gt;&#xA;" OwnerUserId="11399" LastActivityDate="2017-05-19T20:08:17.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15420" PostTypeId="1" AcceptedAnswerId="15421" CreationDate="2017-05-19T20:28:31.923" Score="0" ViewCount="107" Body="&lt;p&gt;Why does chilled water emerge from the water pipe of air conditioner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that instead, lukewarm water should come out, the reason being the air conditioner air from the room and also, electricity produces heat energy so as a result warm water should come out. So what is the correct reason for this?&lt;/p&gt;&#xA;" OwnerUserId="11361" LastEditorUserId="90" LastEditDate="2017-05-20T12:02:25.230" LastActivityDate="2017-05-20T12:02:25.230" Title="The air conditioner mechanism" Tags="|electrical-engineering|heat-transfer|heating-systems|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15421" PostTypeId="2" ParentId="15420" CreationDate="2017-05-19T20:55:28.780" Score="2" Body="&lt;p&gt;An air conditioning system is basically a refrigerator, it passes air over a heat exchanger to cool it ie it injects cold air into a space, it doesn't 'suck' anything out'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water which comes out of the waste pipe is just condensed water vapour from the incoming warm air, the waste heat form the exchanger goes elsewhere. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-05-19T20:55:28.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15423" PostTypeId="1" CreationDate="2017-05-20T00:37:45.820" Score="0" ViewCount="213" Body="&lt;p&gt;I would load a sample measuring 130 mm long in the 4-point bending test (ASTM C393)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the standard, the sample should be 150 mm long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top layers consist of laminates (laminates are Fiber-reinforced composite )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I do the test on the 130 mm sample?&lt;/p&gt;&#xA;" OwnerUserId="11393" LastEditorUserId="33" LastEditDate="2017-05-20T12:24:32.290" LastActivityDate="2017-05-20T13:16:50.117" Title="Short sample in a 4-point bending test ( ASTM C 393 )" Tags="|product-testing|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15424" PostTypeId="2" ParentId="3367" CreationDate="2017-05-20T01:55:29.350" Score="0" Body="&lt;p&gt;If the cost and workspace are not the problems, you should take 2 separate machines. Otherwise, a combo machine will save the money and workspace&lt;/p&gt;&#xA;" OwnerUserId="11401" LastActivityDate="2017-05-20T01:55:29.350" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15425" PostTypeId="1" CreationDate="2017-05-20T02:01:30.493" Score="0" ViewCount="285" Body="&lt;p&gt;Currently I am powering a worm to drive a worm gear that is pinned together on the same shaft to a spur gear to drive another spur gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to simplify the process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to power a worm that drives a worm gear to mesh with another worm gear as spur gears mesh together?&lt;/p&gt;&#xA;" OwnerUserId="11402" LastActivityDate="2018-06-14T13:08:31.770" Title="Worm to worm gear to worm gear" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15426" PostTypeId="2" ParentId="15425" CreationDate="2017-05-20T03:08:50.633" Score="0" Body="&lt;p&gt;As far as I know, no.  Worm gears can only mesh with worms.  Worm gears cannot mesh with other worm gears.  There may very well be a way to simplify your process, but I don't think the idea you suggested is feasible.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-05-20T03:08:50.633" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15427" PostTypeId="2" ParentId="15420" CreationDate="2017-05-20T03:35:13.313" Score="0" Body="&lt;p&gt;I'm not quite sure of the context of the question or what the water pipeline you are referring to is since most air conditioners that are in homes do not use water, but rather refrigerants. If you are talking about the condensate line from the coil as Chris thought, then yes that is water vapor from the air. Warmer air is able to hold more moisture per volume than cold air and when warm moist air meets the cold heat exchanger (coil) the water vapor in the air will condense on the coil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the sucking out of odors and dust, that is the air return for the air handler that is sucking the air and at the most basic level it is basically pulling air in so that the fan in the air handler has something to blow out. The odors and dust are actually trapped in the filter which is there for the purposes of both filtering out particles for your benefit, but more importantly to protect the coil from all of that crap getting stuck between the fins and making it less effective at heat transfer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the electricity, the electricity going into your a/c system is not being deliberately used as a resistive heating element, so the heat generated is only coming off of the motor as a byproduct and is comparatively minimal in magnitude compared to the cooling capacity of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully that helps and if part of your question is still unanswered, please edit your original question to clarify what it is that you are specifically trying to figure out. &lt;/p&gt;&#xA;" OwnerUserId="9770" LastEditorUserId="1832" LastEditDate="2017-05-20T04:34:09.763" LastActivityDate="2017-05-20T04:34:09.763" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15428" PostTypeId="1" AcceptedAnswerId="15502" CreationDate="2017-05-20T03:52:53.960" Score="0" ViewCount="51" Body="&lt;p&gt;I often select makeup air handling units for buildings using Greenheck's CAPS tool. When selecting a model, you first select the unit based on your airflow needs. Then it will provide some choices for burners which it gives in units of Btus per hour. It is my understanding that this is a function of the unit to deliver that amount of heat over a period of time based on the volumetric flow that was selected. This would generally follow the rule of thumb for air of approximately: $$\dot{Q}_{Btu/hr}=1.08\cdot\dot{V}_{CFM}\cdot\Delta T_{^{\circ}F}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that when looking at the burner selection you would have to adjust for a lower air flow rate and subsequently less output or does the temperature delta change accordingly that the heat output for a given burner would still be effectively the same even at the reduced airflow?  On any of the software packages, I can't change the flow rate AFTER selecting a unit to see how the numbers change so that I could back into the answer to check for myself and if I simply pick a lower air flow rate, it gives me entirely different burner choices. &lt;/p&gt;&#xA;" OwnerUserId="9770" LastEditorUserId="9770" LastEditDate="2017-05-25T05:03:33.660" LastActivityDate="2017-05-25T05:03:33.660" Title="Makeup Air Gas Burner" Tags="|airflow|heating-systems|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15429" PostTypeId="2" ParentId="15401" CreationDate="2017-05-20T06:59:23.520" Score="1" Body="&lt;p&gt;Without knowing the transfer function of the plant you try to control it might be that a PID-controller would suffice, but for now I will assume this is not the case. Assuming that your graph is a step response; I am however a little confused by the drop back to zero, since there all systems seem to exert the same behavior. For the initial step the system with your PID-controller has a lower &lt;a href=&quot;https://en.wikipedia.org/wiki/Step_response#Time_domain_versus_frequency_domain&quot; rel=&quot;nofollow noreferrer&quot;&gt;rise-time and settling-time&lt;/a&gt; which would indicate that the real system has a higher bandwidth, which can be achieved using a higher gain. Also the system with your PID-controller has a slight oscillation, which would indicate that your system does not have the same amount of phase/gain/modulus margin. You can use a lead–lag compensator for this by placing the pole and zero of the lead–lag compensator at 3 times the bandwidth and 1/3 times the bandwidth respectively. This has the affect that you get more phase lead near the bandwidth without change the magnitude of the openloop much and thus increasing your phase margin. Increasing your phase margin usually also increases your modulus margin and depending on the situation also the gain margin (if you already have sufficient gain margin it won't change much).&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2017-05-20T06:59:23.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15430" PostTypeId="1" CreationDate="2017-05-20T09:35:51.630" Score="1" ViewCount="110" Body="&lt;p&gt;It is well known that the viscous force (or drag force) is directionally proportional to the square of the velocity in many cases. For example the torque on motors driving a fan-type load (pump) is proportional (approximately) to the square of the rotational speed. The air drag force on a car or a plane has a similar relation between linear drag force and straight line velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in a mass-spring-damper system common in mechanics and control theory, the viscous force &lt;em&gt;F&lt;/em&gt; is &lt;strong&gt;proportional to the velocity itself&lt;/strong&gt;, as in the relation is &lt;em&gt;F = -Bv&lt;/em&gt;, where &lt;em&gt;v&lt;/em&gt; is the velocity and &lt;em&gt;B&lt;/em&gt; is the viscous damping constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for such an anomaly? Are they for different conditions/situations? If so, how? Or am I missing something entirely here?&lt;/p&gt;&#xA;" OwnerUserId="7057" LastActivityDate="2017-05-21T02:15:07.500" Title="Anomaly in viscous force - vs - velocity relationships" Tags="|fluid-mechanics|control-theory|drag|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15431" PostTypeId="1" CreationDate="2017-05-20T11:09:20.940" Score="1" ViewCount="95" Body="&lt;p&gt;I'm working through an FEA problem and I'm stuck. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How can I determine the strain in member 3 from displacements known at node 3? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/16zNH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/16zNH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's at 45 degrees, I know I should be using some $cos\phi$ and $sin\phi$, and given the displacements I suspect there to be a negative strain (indicating compression), however, I'm not entirely sure how to find the strain in the inclined member using both displacements.&lt;/p&gt;&#xA;" OwnerUserId="11405" LastActivityDate="2017-11-17T02:52:51.177" Title="How to find strain in an inclined member of a struss" Tags="|mechanical-engineering|structures|finite-element-method|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15432" PostTypeId="1" CreationDate="2017-05-20T11:15:11.760" Score="0" ViewCount="1820" Body="&lt;p&gt;I have a pressure reducer calibrated for Argon and CO$_2$ output. If I connected it to a gas tank with compressed air, how can I convert the output information &quot;liters of Argon / CO$_2$ per minute&quot; to &quot;pressure of compressed air in bar&quot;?&lt;/p&gt;&#xA;" OwnerUserId="11323" LastActivityDate="2017-07-19T12:25:36.250" Title="compare l/min output of CO2 pressure reducer to bar (compressed air)" Tags="|compressed-air|compressed-gases|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15433" PostTypeId="2" ParentId="15432" CreationDate="2017-05-20T11:41:44.127" Score="1" Body="&lt;p&gt;The units of &quot;liters of Argon / CO2 per minute&quot; is a flow rate (volume delivered in time) and is not equivalent to &quot;pressure of compressed air in Bar&quot; which is pressure (force / area).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you get what you need if you connected a pressure gauge to the output side?&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-20T11:41:44.127" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15434" PostTypeId="2" ParentId="15423" CreationDate="2017-05-20T12:11:09.670" Score="2" Body="&lt;p&gt;Standards are established for a reason: so that tests are conducted to an established procedure so that the results are comparable. If the standard calls for a length of 150 mm then that is what is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A non-standard test can be conducted on a different length sample, but you then have to make adjustments to the results and the results may not be accepted.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-05-20T12:11:09.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15435" PostTypeId="2" ParentId="15423" CreationDate="2017-05-20T13:16:50.117" Score="0" Body="&lt;p&gt;if performing a standard compliant test is important be sure to read and understand the actual standards document. There are typically formula for parameters the give ranges of permissible dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you do that, is 150 the minimum? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can not meet the standard, you can of course do the test and report the data with an asterisk, but it is important you should understand why the parameters limits are what they are. The reasoning is explained in the standards document.  In some cases you are exceeding the accuracy bounds of the beam theory used for stress calculation (for example). In other cases you may risk generating a completely different mode of failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry to be general about it, but I think Its to you to go study the standard and ask more specific questions if you need help.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2017-05-20T13:16:50.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15436" PostTypeId="2" ParentId="15430" CreationDate="2017-05-20T14:38:21.223" Score="4" Body="&lt;p&gt;The examples you mentioned both occur due to turbulent flow of a gas. However in a lot of mechanical systems damping often occurs due visco-elastic deformation or due to shearing of the lubricant inside a bearing, which due to different scales can often be described as a laminar flow. In these cases a good approximation for the &quot;drag forces&quot; would be a linear model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason linear models are often used is that the differential equations  always have algebraic solutions and thus easy to calculate.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2017-05-20T14:38:21.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15438" PostTypeId="1" CreationDate="2017-05-20T18:28:43.830" Score="1" ViewCount="2142" Body="&lt;p&gt;In this &lt;a href=&quot;https://translate.google.com/translate?sl=auto&amp;amp;tl=en&amp;amp;js=y&amp;amp;prev=_t&amp;amp;hl=en&amp;amp;ie=UTF-8&amp;amp;u=http%3A%2F%2Fwww.channelionline.com%2F%25E0%25A6%25AA%25E0%25A7%258D%25E0%25A6%25B0%25E0%25A6%2595%25E0%25A7%258C%25E0%25A6%25B6%25E0%25A6%25B2%25E0%25A7%2580-%25E0%25A6%25AC%25E0%25A7%258D%25E0%25A6%25B2%25E0%25A6%2597%25E0%25A6%25BE%25E0%25A6%25B0-%25E0%25A6%2589%25E0%25A7%258E%25E0%25A6%25AA%25E0%25A6%25B2-%25E0%25A6%259A%25E0%25A6%2595%25E0%25A7%258D%2F&amp;amp;edit-text=&quot; rel=&quot;nofollow noreferrer&quot;&gt;piece of news&lt;/a&gt; it is said that an Engineer and Blogger named Utpal Chakrabarti has died after explosion of air conditioner compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how dangerous air conditioners are? Which is more dangerous: Window Type ACs or Split Type ACs?&lt;/p&gt;&#xA;" OwnerUserId="11410" LastActivityDate="2017-05-21T02:44:29.910" Title="Air Conditioners: Compressor Explosions?" Tags="|compressors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15439" PostTypeId="2" ParentId="15431" CreationDate="2017-05-20T19:19:11.627" Score="1" Body="&lt;p&gt;Resolve the nodal displacements into the &lt;em&gt;element coordinate system.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most likely, one of the element coordinate directions lies along the element. Since you didn't tell us whether the element is a rod or a beam, we don't know if the other two directions are important or irrelevant for finding the strains.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-05-20T19:19:11.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15440" PostTypeId="2" ParentId="8792" CreationDate="2017-05-20T20:39:54.563" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Andre, I assume you've got an answer to this question by now; please tell us how you solved it! – NauticalMile&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I didn't found any information regarding the high temperature values for the viscosity nor the conductivity for those gases. I used a very crude method to get around the issue. Basically, I interpolated the available data using some functions up to the upper temperature limit of the available data (usually around 1000K), and then used the obtained functions to extrapolate up to 2000K. The functions where selected on the basis that &quot;they maintained the general tendency on the behavior of the available data up to the maximum limit of 2000K, not allowing for any sharp inflexions or other oscillatory phenomenon&quot;. This method is extremely crude, but my application did not require very high precision. You can read it further in my Master's thesis (the work where I used this information). When the document is available to the general public I'll post the link here. Bellow is a figure of the used equations and the available data vs. interpolating/extrapolating functions. Note some discontinuity where the data from NIST (full lines) meet the extraplations (dashed lines):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6DiGs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6DiGs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6364" LastEditorUserId="6364" LastEditDate="2017-05-20T20:51:13.557" LastActivityDate="2017-05-20T20:51:13.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15441" PostTypeId="1" CreationDate="2017-05-20T22:23:50.793" Score="2" ViewCount="100" Body="&lt;p&gt;Suppose there is a rod, on the wall or at the centre, in a pressurized vessel, the pressure inside the vessel being P. The rod has a shear strength of S. Is it true that the rod will break, in shear, if P&gt;S, otherwise it will not? To me it seems so, but I am not able to get my head around it.&lt;/p&gt;&#xA;" OwnerUserId="11413" LastEditorUserId="90" LastEditDate="2017-05-21T03:00:41.673" LastActivityDate="2017-05-21T03:00:41.673" Title="A question on mechanical failure" Tags="|mechanical-engineering|stresses|pressure-vessel|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15442" PostTypeId="1" CreationDate="2017-05-20T23:08:02.383" Score="1" ViewCount="6949" Body="&lt;p&gt;Melting point of copper is around 1000 C. If you put a thin copper wire say 50 micron or so over a gas stove flame, it will break immediately. Has its melting point been reached? or, some other phenomenon is at work?&lt;/p&gt;&#xA;" OwnerUserId="11414" LastActivityDate="2017-05-20T23:48:46.833" Title="Melting point of a thin copper wire" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15443" PostTypeId="2" ParentId="15442" CreationDate="2017-05-20T23:32:27.883" Score="1" Body="&lt;p&gt;The melting point of copper = 1,085°C (1,984 °F).&#xA;Methane flame temp = ~1950 °C (3542 °F).&#xA;Therefore, your thin filament copper reaches it's melting point very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put a penny in the same flame &amp;amp; watch how long it takes for the penny to melt.&#xA;It may not even melt if the flame isn't applied properly because copper is a great heat dissipator.&lt;/p&gt;&#xA;" OwnerUserId="5778" LastActivityDate="2017-05-20T23:32:27.883" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15444" PostTypeId="2" ParentId="15442" CreationDate="2017-05-20T23:48:46.833" Score="1" Body="&lt;p&gt;Yes, the butane flame is melting the copper wire. According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Butane_torch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; butane torches can easily reach temperatures of $1430\unicode{xb0}C$. As you noted in your question this is well above the melting point of copper. In fact that maximum temperature that a butane flame can reach is nearly double the melting point of copper though this is hard to reproduce in real world applications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it can seem odd that the wire would break so easily is that metals are good conductors of heat. If the wire were much thicker the heat from the flame would wick away down the wire and be dissipated into the atmosphere. However with a thin wire the surface area to mass ratio is very large so the wire can heat up to its melting point before the heat has a chance to travel down the wire. &lt;/p&gt;&#xA;" OwnerUserId="9656" LastActivityDate="2017-05-20T23:48:46.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15445" PostTypeId="2" ParentId="15441" CreationDate="2017-05-21T00:13:52.093" Score="4" Body="&lt;p&gt;It won't. Solid objects don't break in hydrostatic compression. The pressure is pushing all parts of the rod together, so there's nowhere for them to go except stay together. You can imagine an object on the ocean floor. Pressure could easily exceed shear strength but even soft animals remain intact, as do rocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The von Mises failure criterion says a material that follows it can sustain infinite hydrostatic pressure or tension without reaching the failure surface.&#xA;See the diagram here. The yield surface is a tube that extends out to infinity in the hydrostatic compression direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Von_Mises_yield_criterion&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Von_Mises_yield_criterion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be other failure modes, like phase change. Or it could collapse if it contains air pockets. But that's not determined by the pressure exceeding the shear strength.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="4437" LastEditDate="2017-05-21T00:28:48.460" LastActivityDate="2017-05-21T00:28:48.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15446" PostTypeId="2" ParentId="15431" CreationDate="2017-05-21T00:25:45.170" Score="0" Body="&lt;p&gt;Strain = (deformed length - L3) / L3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the deformed length from the deformed position of node 3 and Pythagoras' theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the deformed position of node 3 by adding the displacement vector to the coordinates of node 3.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2017-05-21T00:25:45.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15447" PostTypeId="2" ParentId="15316" CreationDate="2017-05-21T00:52:18.200" Score="0" Body="&lt;p&gt;For low speeds like what you have, you can use a flexible diaphram between a pair of rotating rollers. A bit like how a diaphragm pump works. Here's one called a &quot;ball-diaphragm coupling&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/patents/US3157054&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/patents/US3157054&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also describes using a crank instead of balls.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2017-05-21T00:52:18.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15448" PostTypeId="2" ParentId="15430" CreationDate="2017-05-21T02:15:07.500" Score="1" Body="&lt;p&gt;So first, there things known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Dashpot&quot; rel=&quot;nofollow noreferrer&quot;&gt;dashpots&lt;/a&gt;, for which the force really is proportional to velocity. These are the things that intended to be represented by the damping term in the classical mass-spring-damper systems that you find in the textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is a much bigger picture idea here. The mass-spring-damper systems in textbook are just mathematical models. The important thing to realize about models is this: &lt;em&gt;All models are wrong, some models are useful&lt;/em&gt;.  We use the mass-spring-damper model not because it is exactly correct for real physical systems, but because it is simple to understand and is close enough for many purposes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically regarding damping, there are some systems for which the damping really is proportional to velocity squared (or maybe some other relation such as dry friction damping, aka coloumb frction, for which the force is proportional to the sign of the velocity). But including a velocity squared term can make the resulting equations much more difficult to solve. e.g. it could go from 30s to solve the equation to 3 hours to solve the equation. If we absolutely must know the exact answer to within 0.1%, we will spend that 3 hours.  But if we only want an approximate answer to within, say, 10%, we will take the easier route and just model the damping as if it was a dashpot.  We should remember that neither model is exactly correct, both of them are wrong, just one is closer to the truth than the other.  We pick the one that is most useful for the given situation.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-05-21T02:15:07.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15449" PostTypeId="2" ParentId="15438" CreationDate="2017-05-21T02:44:29.910" Score="3" Body="&lt;p&gt;That's a pretty qualitative question, but by and large, air conditioner compressors are not really inherently dangerous to be around. Danger comes in the form of unqualified people working on or installing systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a direct comparison, split systems are going to be larger than window units and will, therefore, have potential for more spectacular failures, but there's a reason you don't hear about death-by-air-conditioner all the time and that's because they are not prone to blowing up at random. &lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2017-05-21T02:44:29.910" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15450" PostTypeId="2" ParentId="15407" CreationDate="2017-05-21T02:58:41.130" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The exact problem is how do I measure vibration in a moving object like a rollercoaster vehicle. Because it accelerates and decelerates the G-forces measured by a vibration sensor using an accelerometer are off since they also include the acceleration of the vehicle. This in contrast to a stationary machine which does not move&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To do this, you do in fact want an accelerometer, and then you want to add to it a thing called a &quot;high pass filter&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this will work is that the &quot;rigid body acceleration&quot; you refer to is at a fairly low frequency compared to the vibration you want to measure. Consider your basic roller coaster ride.  Going up the first your accelerometer is going to measure 1g in the vertical direction. Then going down the hill, you measure much less than one g in vertical direction, say 0.2g for a few seconds (this is assuming that your accelerometer always stays in vertical direction and doesn't change with the direction of the car). Then at the bottom of the hill, you are measuring more than 1g, say 2.0g, again for a few seconds.  So our signal will have a 1 -&gt; 0.2 -&gt; 2.0 g waveform lasting maybe 5 seconds, or effectively 0.2 Hz.  Superimposed on top of that is the vibration you want to measure.  That will be at a higher frequency, perhaps 10 - 1000 Hz.  So if you apply a high pass filter with a 1 Hz cutoff frequency, you'll get rid of the 0.2 Hz rigid body acceleration and keep the 10 - 1000 Hz vibration.  The filtering could be done before the recording is made, or you could just record the whole thing and then filter it out later.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another entirely different option, would be to use a displacement sensor that measures the distance between the car and the track, which will change very slightly as the car vibrates.  These types of sensors can be made with a resolution of better than 0.001&quot;, so should be able capture to the vibration of interest.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-05-21T02:58:41.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15451" PostTypeId="1" CreationDate="2017-05-21T06:09:56.603" Score="0" ViewCount="112" Body="&lt;p&gt;Consider a mechanical setup, in which there are two concentric tubes, both to rotate counter to each other ie in a contra rotating fashion. Inner tube runs clockwise, and outer tube runs counter clockwise. Is there any kind of bearing that I can use between the tubes to ensure smooth functioning of mechanism. Or just a plain bearing is effective to put in between the tubes, for smooth running. &#xA;The rpm of inner tube is 1000 rpm, and outer is about 10000 rpm. So can I put any &#xA;bearing with capacity of about 11000 rpm, work in this scenario, ie can i install &#xA;any bearing in between the tubes, or are there specific application bearings to be used for the concentric contra rotating tubes.  &lt;/p&gt;&#xA;" OwnerUserId="11416" LastActivityDate="2017-05-22T05:08:02.237" Title="Bearings for a contra rotating tubes" Tags="|bearings|propulsion|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15453" PostTypeId="1" AcceptedAnswerId="15480" CreationDate="2017-05-21T09:33:55.267" Score="1" ViewCount="2033" Body="&lt;p&gt;I would like to determine the massflow rate of (a) steam (b) air flowing through a cylindrical pipe knowing that p_inlet = 5 bars p_outlet = 1 bar. Internal diameter is 0.5m. Length is 30m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface roughness is assumed equal to 5E-05 m. Only steady state situation is necessary. Temperature at the inlet is assumed in both cases to be equal to 150°C. For derivation, one can assume adiabatic flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the steps to determine the massflow rate ? &#xA;I assume one must first check whether the flow is compressible or not ? Then whether the flow is subsonic, sonic, supersonic ? Then also if laminar or turbulent ?&#xA;And then eventually, what would be the equation to be used to determine the massflow rate ?&lt;/p&gt;&#xA;" OwnerUserId="8261" LastEditorUserId="8261" LastEditDate="2017-05-21T11:20:21.757" LastActivityDate="2017-05-23T15:58:46.023" Title="Mass flow rate (air and steam) through a cylindrical pipe" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15454" PostTypeId="2" ParentId="5421" CreationDate="2017-05-21T17:57:49.767" Score="2" Body="&lt;p&gt;Hydraulic separation in hydronic systems is generally more important with high flow resistance heat sources such as condensing boilers.  If you don't hydraulically separate the heat source from the rest of the system, then  you will have unnecessarily large pressure drops through your boiler.  In order to have &quot;complete&quot; hydraulic separation, it is not necessary to decouple both fluid streams, however, as your question seems to indicate.  By using either a hydraulic separator or closely spaced tees (perform the same basic task as a hydraulic separator without the expensive specialty equipment price tag), the two different circuits (one for your boiler loop and one for your primary loop) can each perform their functions separately.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use case for a heat exchanger decoupling the fluid stream completely would be better usually only when one fluid stream cannot be mixed with another due to additives or contaminants, but the heat transfer will not be as efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caleffi (an Italian hydronics brand) has some excellent resources and explanations in their Idronics magazines that they release every so often.&lt;a href=&quot;https://www.caleffi.com/usa/en-us/technical-magazine&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;  Of particular interest might be #19: Proven Hydronics Systems. &lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2017-05-21T17:57:49.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15455" PostTypeId="1" AcceptedAnswerId="15456" CreationDate="2017-05-21T20:21:28.723" Score="6" ViewCount="2078" Body="&lt;p&gt;I know what pattern draft is and it's usefulness in removing castings from rigid moulds. That's the definition you can find anywhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is when you sand cast. Sand casting is the most common amateur metal casting technique, with lots of YouTube examples. Yet all the decent examples have a draft incorporated. Since you destroy the sand mould when extracting the casting, I can't see any benefit from a draft. It just adds an unnecessary level of complexity when designing and making the pattern. Any shape could be successfully removed, so what's the point of pattern draft in sand casting?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2017-05-21T20:44:31.287" Title="What's the point of draft in sand casting" Tags="|casting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15456" PostTypeId="2" ParentId="15455" CreationDate="2017-05-21T20:44:31.287" Score="10" Body="&lt;p&gt;You need pattern draft to be able to remove the pattern from the sand before pouring. If you don't have any / sufficient draft then edges can be weakened or damaged / broken when you remove the pattern from the sand or plaster - whatever material it is. It is not to help remove the casting as the sand mould is destroyed.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-21T20:44:31.287" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15458" PostTypeId="1" CreationDate="2017-05-21T22:03:16.357" Score="0" ViewCount="331" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tq9Fe.jpg&quot; alt=&quot;Problem 2.9&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the following question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[![This is how i tried, but I know it is wrong.][2]][2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is our attempt but We can't seem to solve it with inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Tq9Fe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tq9Fe.jpg&quot; alt=&quot;New attempt according to your approach&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11421" LastEditorUserId="11421" LastEditDate="2017-05-25T17:14:59.000" LastActivityDate="2017-05-25T17:14:59.000" Title="Two-way pulley system considering inertia of pulleys" Tags="|mechanical-engineering|applied-mechanics|dynamics|mathematics|pulleys|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15459" PostTypeId="1" AcceptedAnswerId="15465" CreationDate="2017-05-22T01:17:33.503" Score="1" ViewCount="127" Body="&lt;p&gt;I've designed a whole lot of different parts and products in my time: Plastic, metal, lighting products, electrical products and automotive parts.  I generally can look at a product and figure out why did they do that.  Here's one that has me baffled.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its the bottom of a consumer grade vacuum cleaner.  Molded in the plastic is a slot (which I'd generally refer to as a bayonet mount).  I can see that the mount definitely is out of normal die draw (large undercut) and requires a slide (and that tooling is definitely not cheap...)  I can't figure out what that mount is for.  It doesn't seem to match the cord or any of the accessories that I can see.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bJGX0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bJGX0.jpg&quot; alt=&quot;Vacuum cleaner mystery mount&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on what that 'bayonet mount' is for?  Why would the manufacturer spend real investment dollars for tooling something so obtuse?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Something they use in their manufacturing process? &lt;/li&gt;&#xA;&lt;li&gt;Something to ensure the part control during the molding ejection process from the die? Note: I've seen molded parts get stuck in the wrong half of the tool and that can be quite ugly...  The other side of the part is heavily engineered, with lots of ribs and other details.&lt;/li&gt;&#xA;&lt;li&gt;Could it be for planting a sales/marketing sign when the product is sitting in a sales display?  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any ideas? &lt;/p&gt;&#xA;" OwnerUserId="1567" LastActivityDate="2017-05-22T08:09:50.827" Title="Plastic Vacuum Cleaner Part -- Why did they design it this way?" Tags="|plastic|molding|injection|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15460" PostTypeId="2" ParentId="15451" CreationDate="2017-05-22T01:22:30.377" Score="1" Body="&lt;p&gt;Yes, pretty much any bearing (rolling element bearing or journal bearing) that can take 11,000 rpm should be able to handle being placed in a differential configuration. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-05-22T01:22:30.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15461" PostTypeId="2" ParentId="15451" CreationDate="2017-05-22T05:08:02.237" Score="0" Body="&lt;p&gt;My original comment as an answer : Check out roller bearings, other questions : will they need forced lubrication - is this possible ?  What is the required life? Will they be easy to change? Or replace complete unit. What are the loads that need to be supported end or side loading etc.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-22T05:08:02.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15462" PostTypeId="2" ParentId="15394" CreationDate="2017-05-22T05:08:16.650" Score="3" Body="&lt;p&gt;One other reason is that the propeller-type turbine allows it to be rotated to avoid damage in high winds. When the air is flowing transverse to the axis of the propeller it encounters relatively little drag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be hard to imagine an orientation that would allow the anemometer-type design to avoid this.&lt;/p&gt;&#xA;" OwnerUserId="11424" LastActivityDate="2017-05-22T05:08:16.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15465" PostTypeId="2" ParentId="15459" CreationDate="2017-05-22T08:09:50.827" Score="2" Body="&lt;p&gt;This was an unexpected surprise. The slot is for a mating element on the lower tube extension.  I totally missed this.  When stowed this way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the storage footprint is small and stable&lt;/li&gt;&#xA;&lt;li&gt;you can easily move the entire assembly in and out of a storage closet with one hand (by grabbing the handle on the base unit.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks to Solar Mike and Zimano for their responses.  Separate answer posted here to enable a photograph of the assembly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MgMm8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MgMm8.jpg&quot; alt=&quot;vacuum cleaner hose storage slot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1567" LastActivityDate="2017-05-22T08:09:50.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15467" PostTypeId="2" ParentId="15458" CreationDate="2017-05-22T11:27:38.450" Score="2" Body="&lt;p&gt;The first step in a problem like this is to draw the relevant free body diagrams.  You sort of did this in the top left diagram, but not really.  What you want to do is to split each mass and inertia up into a &lt;strong&gt;&lt;em&gt;separate&lt;/em&gt;&lt;/strong&gt; free-body diagram.  i.e. one FBD for mass 1, one FBD for pulley 1, one FBD for mass 2, one FBD for pulley 2.  Then, for each FBD, write down the equations of motion: $\sum F = m a$, $\sum M = \Theta \alpha$.  That would get you 8 equations (although 2 will be trivial as the 2 masses have no applied torque and no rotations, so the equation will be just 0=0)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram you drew, with everything all together, is going to get you confused.  For example, your second equation looks like $\sum F = T_1 - m_1 g +T_2 + T_3=0$.  That's probably wrong, because $T_3$ does not act upon the 1st mass.  If you draw out separate diagrams it will be easier to set up the correct equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the diagrams and drawn and you have the correct 8 equations, solving the equations is straightforward.    &lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="9275" LastEditDate="2017-05-22T11:52:00.177" LastActivityDate="2017-05-22T11:52:00.177" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15472" PostTypeId="1" AcceptedAnswerId="15478" CreationDate="2017-05-22T23:45:09.687" Score="-1" ViewCount="95" Body="&lt;p&gt;How much graphene is required to coat a 5 inch (diameter) sphere if the graphene layer must be 35 nanometers thick?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="7538" LastEditDate="2017-05-22T23:50:54.313" LastActivityDate="2017-05-23T07:30:14.643" Title="Graphene distribution over a sphere" Tags="|materials|mathematics|geometry|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15473" PostTypeId="1" AcceptedAnswerId="15479" CreationDate="2017-05-23T03:11:25.613" Score="1" ViewCount="5758" Body="&lt;p&gt;How can permeation units in Darcy be converted to the SI form. For example, how would 9.0 x 10-6 darcy be converted to SI units of permeation? How would it also be converted to the permeation units: (cm^3 . mm)/m^2 . d . atm ?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="7538" LastEditDate="2017-05-23T15:53:49.353" LastActivityDate="2019-03-22T09:55:54.630" Title="Darcy to SI permeation units" Tags="|measurements|mathematics|unit|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15474" PostTypeId="2" ParentId="15472" CreationDate="2017-05-23T04:03:40.537" Score="1" Body="&lt;p&gt;Surface area of the sphere is given by: $$A=\pi d^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In $mm$, the diameter is equal to: $$d=5\times25.4=127 mm$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the surface area is equal to about 50671 $mm^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply this by the thickness of the graphene layer (with a suitable conversion from $nm$ to $mm$) to get a volume of 0.0177 $mm^3$.&lt;/p&gt;&#xA;" OwnerUserId="11424" LastActivityDate="2017-05-23T04:03:40.537" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15477" PostTypeId="2" ParentId="15472" CreationDate="2017-05-23T07:16:21.653" Score="1" Body="&lt;p&gt;$$ A = 4\pi r^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ r = \frac{5 inch}{2} = \frac{127000000 nm}{2} = 63500000nm $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ V = \int_{63500000}^{63500035} 4\pi r^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This turns out to be around 1.773 mm³.&lt;/p&gt;&#xA;" OwnerUserId="11436" LastActivityDate="2017-05-23T07:16:21.653" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15478" PostTypeId="2" ParentId="15472" CreationDate="2017-05-23T07:19:14.940" Score="2" Body="&lt;p&gt;The volume $V$ needed to cover the surface is equal to the spherical shell volume with $d_{\text{inner}}=5 \text{ inch}=0.127\text{ m}$ and $d_{\text{outer}}=d_{\text{inner}}+2\cdot 35 \cdot 10^{-9}\text{ m}$. Remember to use everything in SI-units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V=\frac{4}{3}\pi\left[\left(\frac{d_{\text{outer}}}{2}\right)^3-\left(\frac{d_{\text{inner}}}{2}\right)^3\right]\approx1.7734\cdot 10^{-9} \text{ m}^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the density of graphene is given by $\rho$, then the necessary mass $m$ of graphene is given by $m=\rho V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, that the answer is very similar to the result obtained by Angus Murray, as the outer diameter is almost the same as the inner diameter. But I would not use approximative formulas as long as there is an exact formula because it will not be necessary to estimate the error that you have introduced by using this approximation.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-05-23T07:30:14.643" LastActivityDate="2017-05-23T07:30:14.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15479" PostTypeId="2" ParentId="15473" CreationDate="2017-05-23T09:32:57.493" Score="1" Body="&lt;p&gt;To convert &lt;a href=&quot;https://en.wikipedia.org/wiki/Darcy_(unit)&quot; rel=&quot;nofollow noreferrer&quot;&gt;darcy units&lt;/a&gt; to SI units&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;... 1 darcy is equivalent to 9.869233×10−13 m² or 0.9869233 (µm)². This conversion is usually approximated as 1 (µm)².[3] Note that this is the reciprocal of 1.013250—the conversion factor from atmospheres to bars.&lt;/p&gt;&#xA;&lt;p&gt;Specifically in the hydrology domain, permeability of soil or rock may also be defined as the flux of water under hydrostatic pressure (~ 0.1 bar/m) at a temperature of 20°C. In this specific setup, 1 darcy is equivalent to 0.831 m/day.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="90" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-05-23T09:32:57.493" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15480" PostTypeId="2" ParentId="15453" CreationDate="2017-05-23T15:58:46.023" Score="0" Body="&lt;p&gt;In pipe flow, we don't typically talk about subsonic / supersonic / transonic flow. We instead differentiate between choked and non-choked flow. Choking occurs when the upstream pressure is greater than the downstream pressure by some factor (dictated by the gas or fluid). That factor is usually ~1.5, depending (roughly) on the ratio of specific heats. You are clearly dealing with choked flow in your case. Choking means the flow speed at the exit is the speed of sound. Any excess pressure past the point of choking does not increase the flow speed. It will merely increase the density of the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, it does not matter whether your fluid is compressible or gas or liquid phase, because this is a relatively uncomplicated problem. In others, where you may have a restrictor plate or valve, it would. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for the condition of choked flow:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p* / p0 = (2 / (gamma + 1)) ^ (gamma / (gamma - 1))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where p*  is the downstream pressure, p0 is the upstream pressure, and gamma is the ratio of specific heats. If your downstream pressure is lower than p*, then your flow is choked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a straight cylindrical pipe we can (almost) always assume flow is laminar, but there are equations to check that for us. The Reynolds number for a pipe flow can be found by:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Re = rho V  D / mu&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with rho being density (kg / m3), V being flow speed (m/s), D being the pipe diameter (m), and mu being the dynamic viscosity (N s / m^2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For choked flow, we can simply use a mass flow rate equation based on pressure differential. If you were looking for mass flow rate without already knowing the downstream pressure, or looking for the pressure drop caused by the friction, we'd have to take account the surface roughness. Similarly more work is required with non-choked flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass flow rate equation for choked flow:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;mdot = Cd A sqrt(gamma rho0 p0 (2 / (gamma +1)) ^ (gamma + 1) / (gamma - 1))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cd - the coefficient of discharge, defined next&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A - pipe exit area&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gamma - ratio of specific heats&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rho0 - upstream density (which you should compute based on T, p0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p0 - upstream pressure&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Cd = mdot / (A sqrt(2 rho deltaP))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with deltaP being the pressure drop across the exit (here, 5 bars - 1 bar = 4 bars). You'll need to convert units here as appropriate (i.e. to Pascals).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coefficient of discharge basically is a refinement to the mass flow rate; it takes into account  that the cross section of the flow in the orifice at its maximum speed is not the same size as the total area of the orifice. If you're bored and want to learn more about that phenomena, look up vena contracta. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a wide array of pipe flow, valve, orifice, etc. problems out there. If you're looking for more information, Crane's TP-410 publication is the de facto reference for these sort of standard pipe flow problems. I also really like Robert Blevin's Applied Fluid Dynamics Handbook, as it has a host of information not available anywhere on the internet (for estimating pressure losses due to all sorts of pipe configurations). &lt;/p&gt;&#xA;" OwnerUserId="3583" LastActivityDate="2017-05-23T15:58:46.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15481" PostTypeId="1" CreationDate="2017-05-23T20:38:28.163" Score="1" ViewCount="230" Body="&lt;p&gt;I would like to know if there is a standard way of calculating the equivalent to a &lt;em&gt;capacity factor&lt;/em&gt; but for heating equipment. That is the ratio between the amount of heat delivered during one year divided by the amount of heat that could be delivered by the equipment (capacity x time in a year). Or otherwise, the equivalent full load heating hours divided by the number of hours of the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought of an approximation, dividing the average annual heating degree days by the maximum annual heating degree days of the city where the building is situated.  I would like to compare it with other ways of estimating the same factor.&lt;/p&gt;&#xA;" OwnerUserId="9049" LastEditorUserId="9049" LastEditDate="2017-05-23T23:13:52.240" LastActivityDate="2017-09-23T09:27:47.407" Title="equivalent full hours of heating equipment" Tags="|heat-transfer|heating-systems|building-physics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15482" PostTypeId="1" CreationDate="2017-05-23T21:14:29.070" Score="2" ViewCount="2127" Body="&lt;p&gt;I thought the term &lt;em&gt;sample period&lt;/em&gt; referred to the &lt;em&gt;time interval&lt;/em&gt; between one measurement and another. But I've come across &lt;a href=&quot;http://www.mdpi.com/1424-8220/14/4/6891/htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; which says&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;About every tenth of a second, a dataset, &lt;span class=&quot;math-container&quot;&gt;$S(i)$&lt;/span&gt;, of the form:&lt;/p&gt;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&lt;p&gt;is recorded (of course, the time between taking two datasets must be a multiple of the sample period, i.e., &lt;span class=&quot;math-container&quot;&gt;$\Delta t \mid (t_{i+1}- t_i), \forall i$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Now, the way I interpret this is that &lt;span class=&quot;math-container&quot;&gt;$t_{i+1}- t_i$&lt;/span&gt; is the time interval between a recording of a data set and the successive. So, if that's correct and &lt;span class=&quot;math-container&quot;&gt;$\Delta t$&lt;/span&gt; (i.e. the sample period) is the &lt;em&gt;time interval&lt;/em&gt; between one measurement and another, then &lt;span class=&quot;math-container&quot;&gt;$\Delta t = t_{i+1}- t_i$&lt;/span&gt;, but clearly some definitions must be wrong, otherwise the original author of that article would not have introduced two terms to identify the same thing, and would not have said that &lt;span class=&quot;math-container&quot;&gt;$t_{i+1}- t_i$&lt;/span&gt; must be a multiple of &lt;span class=&quot;math-container&quot;&gt;$\Delta t$&lt;/span&gt;?&lt;/p&gt;&#xA;&lt;p&gt;What exactly is the sample period &lt;span class=&quot;math-container&quot;&gt;$\Delta t$&lt;/span&gt;?  How is it different from the interval between one measurement and the other (i.e. &lt;span class=&quot;math-container&quot;&gt;$t_{i+1}- t_i$&lt;/span&gt;)? What exactly does it mean that &lt;span class=&quot;math-container&quot;&gt;$t_{i+1}- t_i$&lt;/span&gt; must be a multiple of &lt;span class=&quot;math-container&quot;&gt;$\Delta t$&lt;/span&gt; (I guess this will be answered once I know what's exactly the sample period)?&lt;/p&gt;&#xA;" OwnerDisplayName="user11356" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-05-24T11:55:55.780" Title="What exactly is the sample period?" Tags="|terminology|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15483" PostTypeId="1" CreationDate="2017-05-24T01:41:16.527" Score="2" ViewCount="88" Body="&lt;p&gt;Many text book literature on braced excavation mention the necessity to consider additional load due to temperature effect on struts. But nothing is mentioned about waling. Why is this so? Wouldn't axial load in waling be significant for waling with large cross sections? Is the consideration any different from struts? How would such effect be considered for waling, if temperature is a necessary consideration?&lt;/p&gt;&#xA;" OwnerUserId="11443" LastActivityDate="2018-04-26T13:24:49.890" Title="Is it necessary to consider temperature effect on waling?" Tags="|civil-engineering|structural-analysis|steel|temperature|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15484" PostTypeId="1" CreationDate="2017-05-24T06:17:10.590" Score="1" ViewCount="474" Body="&lt;p&gt;On a lawn mower there is a non-removeable metal shaft on which a pulley stack sits. The problem is that worn out bushings caused the shaft to wear in two places and this is causing the drive belt for the blades to be out of alignment and thus throw the belts. My question is, how can I engineer a solution to either shim up or machine the shaft on the deck when turning it on a lathe is not an option?&lt;/p&gt;&#xA;" OwnerUserId="11449" LastActivityDate="2021-08-02T09:05:50.693" Title="How to machine a metal shaft that cannot be turned on a lathe" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15485" PostTypeId="1" AcceptedAnswerId="15488" CreationDate="2017-05-24T07:14:55.997" Score="0" ViewCount="1927" Body="&lt;p&gt;I am currently doing a control systems course and I was lead to believe that if a pole lies on the imaginary axis in the s-plane (i.e s=0), then the system was marginally/critically stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I come across this transfer function and plotted it in Simulink. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MgLQF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MgLQF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysis shows that there are 3 poles at s=-1, s=-3 and s=0. So because there is a pole at s=0, the system should be marginally stable right? But the output of the transfer function shows this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d2Bpf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d2Bpf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, that is showing the system is unstable because the output is getting bigger and bigger as time goes on. Is this correct? Or is my understanding still flawed? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the system is unstable, then why aren't any of the poles in the right half of the s-plane? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="11451" LastActivityDate="2017-05-24T10:27:18.373" Title="Control Systems: Why is this system unstable?" Tags="|control-engineering|stability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15486" PostTypeId="1" CreationDate="2017-05-24T09:07:05.720" Score="1" ViewCount="76" Body="&lt;p&gt;When pouring copper onto another copper block and I want to have a good solid welding of the poured copper onto the block, do I need to use a flux? Or, if the block is freshly cleaned and free of most oxidation, would that be enough?&lt;/p&gt;&#xA;" OwnerUserId="1915" LastActivityDate="2017-05-25T21:58:12.917" Title="Pouring copper onto another copper block" Tags="|welding|casting|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15487" PostTypeId="2" ParentId="15482" CreationDate="2017-05-24T10:09:30.023" Score="0" Body="&lt;p&gt;I suggest a grammar problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read properly the article $\Delta t$ is the total time of the measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the word 'Period' is here not to describe the wavelength of a repetitive behaviour but the &lt;strong&gt;Duration&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is wrong: $\Delta t = t_{i+1}- t_i$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is correct $\Delta t = N.(t_{i+1}- t_i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;($N$ is defined in your article and also 'the time between taking two datasets')&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again there is no repetition in the time period. To make it less confusing, the author should have called it 'time interval' or 'duration'&lt;/p&gt;&#xA;" OwnerUserId="10684" LastActivityDate="2017-05-24T10:09:30.023" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15488" PostTypeId="2" ParentId="15485" CreationDate="2017-05-24T10:20:16.753" Score="2" Body="&lt;p&gt;Your system is open loop stable as the poles are at $s=-1$, $s=-3$ and $s=0$. Note, that if the order of the pole at $s=0$ is greater then 1, then the open loop system is also unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But closing the loop changes the poles of the system. If $F(s)$ is your transfer function of the open loop system, then the transfer function of the closed loop system is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G(s)=\frac{F(s)}{1+F(s)}=\frac{21}{s^3+4s^2+3s+21}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.wolframalpha.com/input/?i=s%5E3%2B4*s%5E2%2B3*s%2B21%3D0&quot; rel=&quot;nofollow noreferrer&quot;&gt;poles of $G(s)$&lt;/a&gt; are $s \approx -4.4022$, $s \approx 0.2011 - 2.1748 i$ and $s \approx 0.2011 + 2.1748 i$. Hence, the closed loop system is unstable.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-05-24T10:27:18.373" LastActivityDate="2017-05-24T10:27:18.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15489" PostTypeId="2" ParentId="15482" CreationDate="2017-05-24T10:32:42.990" Score="1" Body="&lt;p&gt;Your definition of the &lt;em&gt;sample period&lt;/em&gt; is correct but is misused here. In my opinion, this originates from the notations used in the referred article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, let me firstly introduce the following notation&#xA;$$f[n] \triangleq f(n\Delta t)$$&#xA;used to refer to discrete samples from a continuous function $f$ which are regularly distant (in the continuous domain) by an interval $\Delta t$ called the &lt;em&gt;sample period&lt;/em&gt; (you already know that). Here, we sample a continuous-time function and obtain a discrete-time function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, consider sampling $f[n]$, for example&#xA;$$f_s[n] \triangleq f[n\ N]$$&#xA;where $N$ is also a &lt;em&gt;sample period&lt;/em&gt;! But $\Delta t$ and $N$ are different because of having been defined in different contexts. Particularly, notice that $\Delta t$ is a time duration (in seconds) while $N$ is dimensionless. Furthermore, $\Delta t \in \mathbb{R}^+$ while $N \in \mathbb{N}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let us translate $f_s[n]$ into a continous-time function&#xA;$$f_s(t)&#xA;\begin{cases}&#xA;  \triangleq f_s\Big[\frac{t}{N\Delta t}\Big] &amp;amp; \text{if } \frac{t_n}{N\Delta t} \in \mathbb{Z}\\&#xA;  \text{undefined} &amp;amp; \text{otherwise}&#xA;\end{cases}&#xA;$$&#xA;As $f_s$ is defined over $\mathbb{R}$, when we have no corresponding sample for $t$, $f_s(t)$ can not be defined. Actually, you could define it as 0 or something else but this does not matter here because the interesting part is when it is defined: notice that the times at which the function is defined are expressed as $t_n \triangleq n\ N\Delta t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to the article, $\Delta t$ is the sample time for the sensors while $S(i)$ is a sample of a set of (already sampled) functions (or a set of samples, depending on the way you look at it) that they call a &lt;em&gt;dataset&lt;/em&gt;. When they talk about a &lt;em&gt;sample period&lt;/em&gt; for $S$:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;About every tenth of a second&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;they refer to $N\Delta t$, not $\Delta t$! And, when you understand this distinction,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;of course, the time between taking two datasets must be a multiple of the sample period, i.e., $\Delta t \mid (t_{i+1}- t_i), \forall i$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;as $(t_{i+1}- t_i)$ is $N\Delta t$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After another quick look at the article, I noticed that they use the symbol $N$ as well. It should be obvious that the meaning of $N$ here is different from what it expresses in the article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, as I have less than 50 reputation, I am not allowed to post comments so I will add this to my answer: Another answer states that $\Delta t$ is the total duration (in the article) while my previous definition of it comes from Section 3 &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Assume that two inertial sensors, one attached to the upper leg and the other attached to the lower leg, measure the accelerations, $a1(t),a2(t) \in \mathbb{R}^3$, and angular rates, $g1(t),g2(t) \in \mathbb{R}^3$, at some sample period, $\Delta t$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which I assumed to be the definition of $\Delta t$. I would appreciate any clarification of this.&lt;/p&gt;&#xA;" OwnerUserId="11450" LastEditorUserId="11450" LastEditDate="2017-05-24T10:51:46.770" LastActivityDate="2017-05-24T10:51:46.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15491" PostTypeId="2" ParentId="15486" CreationDate="2017-05-24T10:47:05.543" Score="4" Body="&lt;p&gt;There are a couple of issues here. To get a proper fusion weld both surfaces need to melt and whether or not this happens depends on the details of the heat transfer for the whole system. It is certainly possible that heat could be conducted way from the interface too quickly for the surface to melt well enough to get good fusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this situation the cleaner the surface the better, if metals are clean enough and have a good enough surface finish they can fuse cols in some circumstances and it is certainly true that a welded joint can never be too clean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with using flux is that it need a certain amount of heat to activate and it works by dissolving metal oxides from the surface so you still need to prevent flux/slag from being included in the joint (which in this case may be difficult.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It certainly is &lt;em&gt;possible&lt;/em&gt; to get decent fusion between solid copper and molten cast iron in a mould because I have done it, although here the mass of iron was significantly greater than that of the copper and it is very sensitive to the poured metal having a decent amount of super-heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It really depends on exactly what you want to achieve and without more context it is difficult to be more helpful. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-05-24T13:27:30.180" LastActivityDate="2017-05-24T13:27:30.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15492" PostTypeId="2" ParentId="15482" CreationDate="2017-05-24T11:55:55.780" Score="0" Body="&lt;p&gt;Collection of the dataset is performed less frequently than sampling - it is performed at certain (unnamed, nor given a symbol) intervals, which are the multiples of the sample period, described by symbol $Δt$, defined as $Δt = t_{i+1}−t_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is pretty standard - in case of simple sensors, like accelerometers, measurements are performed very frequently (very short sample period), because &lt;em&gt;why not?&lt;/em&gt; - their controller can do it, doesn't have anything else to do, and in case step of some analysis needs to be decreased, the higher density measurements are there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterwards, complex analysis of data ignores/discards part (often a major part) of the measurements, and only picks every &lt;em&gt;n&lt;/em&gt;th sample for the dataset - for reasons of performance, storage space, lack of accuracy of other inputs, and a range of other reasons - the sample period of the dataset is consciously made a multiple of sample period of the sensor. We have one dataset sample every $n$ samples, and complete dataset comprised of $N$ dataset samples. The whole dataset covers therefore a period equal $N \cdot n \cdot (t_{i+1}−t_i)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The authors deemed detailing how many sensor samples are dropped unimportant, but the gist of the paragraph (with N &gt;&gt; 1) is &quot;the sensor provides overwhelming amount of data, so we drop some of it regularly, but we still collect a lot.&quot;&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-05-24T11:55:55.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15493" PostTypeId="2" ParentId="15484" CreationDate="2017-05-24T13:20:53.423" Score="1" Body="&lt;p&gt;This could be done several ways that cost more than a new deck would. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best solution I can think of would be to grind off the existing shaft, if it was pressed in you can grind / press out the remaining shaft then weld across the hole (not easy but possible). Then grind the weld flat and drill a new hole. To make it easier you could weld on a .25&quot; or .125&quot; plate instead of filling in and re-drilling the hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would probably use a bolted shaft instead of a pressed in one so you can replace it in the future; a good shoulder bolt may work.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:14:16.003" LastActivityDate="2017-07-23T18:14:16.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15494" PostTypeId="1" AcceptedAnswerId="15499" CreationDate="2017-05-24T15:24:00.093" Score="2" ViewCount="501" Body="&lt;p&gt;What is the logic behind using a wall tap/piezometer to measure the static pressure. Is it that near the walls, the fluid velocity is zero, because of the no slip condition and hence the pressure measurement near the wall would only be static pressure. There is this point in the book that i am consulting, it says &quot;there is no pressure variation normal to straight streamlines, this fact makes it possible to measure static pressure in a flowing fluid using a wall pressure tap&quot;. Does this point mean that because there is non-variance of pressure in normal direction to the straight streamlines, thus giving a single value for static pressure, whereas in regions of curved streamlines, we would have progressively increasing values as we go away from the center of curvature of the family of curved streamlines. Also, why do we require a separate static pressure measuring instrument alongside a pitot tube (for fluids passing through a pipe) in order to measure velocity through the Bernoulli eq. Why can't we calculate velocity directly as is done, for free streams of fluids, such as river water. &lt;/p&gt;&#xA;" OwnerUserId="9537" LastEditorUserId="10902" LastEditDate="2017-05-24T17:26:30.600" LastActivityDate="2017-05-24T19:42:05.473" Title="Measurement of static pressure" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15495" PostTypeId="2" ParentId="13708" CreationDate="2017-05-24T16:02:29.173" Score="4" Body="&lt;p&gt;In case of a Quasi-static expansion, the internal pressure ($P_{in}$) would at all times be just infinitesimally greater than the external pressure($P_{ex}$). Mathematically, $P_{in} = P_{ex} + &#xA;\text{d}p$. Now work done as you said is, $\int P_{in} \text{d}v$. Therefore, using $P_{in} = P_{ex} + \text{d}p$, then after multiplication inside the integral you would have $\int P_{ex} \text{d}v + \text{d}p\text{d}v$. Since, $\text{d}p$ and $\text{d}v$ are both infinitesimally small, therefore their product can be neglected and you end up with an approximate value of work, which is $\int P_{ex} \text{d}v$.&lt;/p&gt;&#xA;" OwnerUserId="9537" LastEditorUserId="1832" LastEditDate="2017-05-24T16:28:37.137" LastActivityDate="2017-05-24T16:28:37.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15496" PostTypeId="1" CreationDate="2017-05-24T16:26:00.517" Score="1" ViewCount="407" Body="&lt;p&gt;I would like to know how to interpret the roots of the Peng Robinson equation of state for vapor-liquid-liquid equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that if two phases are present, the largest root is taken as the compressibility factor of the vapor and the smallest root is taken as the compressibility factor of the liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the smallest root represent if two liquid phases are present (in addition to the vapor)?  Is it meaningful?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And how can I determine the compressibility or molar volume of each liquid phase?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately my goal is to be able to calculate the fugacity coefficients for components in the vapor phase, liquid phase 1, and liquid phase 2 in order to calculate the K factors for use in a multicomponent equilibrium algorithm - but I believe that I need the molar volume of all three phases in order to do this.&lt;/p&gt;&#xA;" OwnerUserId="11458" LastActivityDate="2018-03-21T20:45:59.777" Title="Interpreting roots of a cubic EOS with 3 phases (VLLE)" Tags="|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15497" PostTypeId="1" CreationDate="2017-05-24T17:12:09.670" Score="1" ViewCount="144" Body="&lt;p&gt;Is there a way to use gyroscopes to control the motion of a pen-like device? I am trying to build a device that will be able to take in a radius input and draw shapes without physically moving my hand in the shape of the shape. An example of this type of device would be the spyro-gyro toy pen.&lt;/p&gt;&#xA;" OwnerUserId="11459" LastEditorUserId="9" LastEditDate="2017-05-24T21:13:33.987" LastActivityDate="2017-09-22T12:04:06.260" Title="How are gyroscopes used for motion control?" Tags="|mechanical-engineering|torque|robotics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15498" PostTypeId="1" CreationDate="2017-05-24T17:59:09.063" Score="1" ViewCount="1069" Body="&lt;p&gt;I am working on a project for an amazing hot springs project in CA. As part of this project, the county is requiring that we provide our own fire suppression water. We have 3 tanks of 40,000gal each. These tanks will equally feed into a 6&quot; PVC pipe. From the tanks, the 6&quot; pipe will drop 172' of elevation, and run approx. 1400'  with several bends and a few 90degree turns, terminating at a fire hydrant. The county has given us certain flow requirements which will be physically tested once the system is in place. I need to estimate the flow at that hydrant before we install this expensive pipe. If the estimate is considerably over our requirements, then I feel confident enough to begin the installation. If the estimate is close or below....then I have a problem.&#xA;I understand the equations for determining flow, BUT they always need numbers that I don't have- specifically velocity! I know my pipe diameter (6&quot;), but how can I estimate velocity? To find velocity I need flow, lol. And so on. &#xA;Any help would be immensely appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tanks- 3 tanks x 40,000gal = 120,000gal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tank height to top= 20'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pipe Interior Diameter= 6&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elevation from Tanks outlet to Hydrant= 172'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pipe length= 1400' (including approx. 5 x 45deg turns, 2 x 90deg turns)&lt;/p&gt;&#xA;" OwnerUserId="11460" LastActivityDate="2017-05-25T05:46:23.873" Title="Estimating Potential Flow through a pipe, from tanks to outlet" Tags="|fluid-mechanics|pipelines|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15499" PostTypeId="2" ParentId="15494" CreationDate="2017-05-24T19:42:05.473" Score="2" Body="&lt;p&gt;A piezometer is a device which senses pressure by producing electric signals proportional to the applied pressure. It is attached on the walls of the pipe so that the fluid pressure acts normal to it. Since the fluid flow is parallel to the walls, the fluid flow is neither obstructed by the piezometer, nor does it experience any momentum transfer. The sole force acting on the piezometer will be that caused by the hydrostatic pressure. So the need for measuring velocity of flow does not arise. This has nothing to do with no slip condition. We can place the piezometer at any point in the direction of streamline. It is more convenient to fix it on the walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of the type of gauges mentioned in your second question, they are placed normal to the flow so that they experience both hydrostatic pressure force and force due to momentum transfer(since they obstruct the flow). So we need to subtract the hydrostatic pressure to find the flow velocity.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2017-05-24T19:42:05.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15500" PostTypeId="2" ParentId="15498" CreationDate="2017-05-25T01:11:36.147" Score="2" Body="&lt;p&gt;I'm pretty sure that you have all of the info that you should need to determine flow velocity using Bernoulli's equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically you can calculate the hydraulic pressure based on the elevation (don't forget to add what's in the tank), use that to determine your final velocity, and then use that to determine your flow rate based on your pipe size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=z_2-z_1+\frac{P_2-P_1}{\rho*g}+\frac{V_2^2-V_1^2}{2g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I threw it into some hydraulic software and included the accounting for the fittings you listed.  I made a couple of assumptions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The water exits the tank at 0 ft in the tank.&lt;/li&gt;&#xA;&lt;li&gt;The tanks are all at the same elevation (so it doesn't really matter how many 40,000 gallon tanks you have, that just determines how long the water will flow.&lt;/li&gt;&#xA;&lt;li&gt;The tanks are essentially manifolded together and there is virtually no distance between the bottom of the tank and where you start counting the pipe run length.&lt;/li&gt;&#xA;&lt;li&gt;I used the default spec for 6&quot; Schedule 40 PVC pipe which actually has an ID of 6.065&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It came up with 1676 gpm flow at the pipe outlet at 8.65 psig.  Remember that as the tanks lose water, that pressure will drop slightly, though that is only 20 feet of the total 192 in elevation change.  &lt;/p&gt;&#xA;" OwnerUserId="9770" LastEditorUserId="9770" LastEditDate="2017-05-25T01:26:14.787" LastActivityDate="2017-05-25T01:26:14.787" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15502" PostTypeId="2" ParentId="15428" CreationDate="2017-05-25T04:54:19.133" Score="0" Body="&lt;p&gt;I went back and worked through it, and I was somewhat right.  If you request the manufacturer to maximize the burner, you may get a different size burner even when the base unit is the same because typical units are temperature limited.  So by reducing the airflow in half, you would double your delta T for a given input, to balance the equations, but the actual selections will not match simply because of manufacturer limits in place to prevent damage to the units.&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2017-05-25T04:54:19.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15503" PostTypeId="2" ParentId="15498" CreationDate="2017-05-25T05:40:56.463" Score="2" Body="&lt;p&gt;This link : &lt;a href=&quot;http://www.calctool.org/CALC/eng/civil/hazen-williams_g&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.calctool.org/CALC/eng/civil/hazen-williams_g&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives you a calculator BUT does not include any losses due to junctions, joints or bends. Losses due to all the junctions, joints, bends or any change in section will reduce this result which will / should be a maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note : I did not include the height of the tanks for this first run - you can enjoy tweaking the input !&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have their expected flow rate then you can work backwards using continuity of mass etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have included a picture of the results:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HzqiE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HzqiE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-05-25T05:46:23.873" LastActivityDate="2017-05-25T05:46:23.873" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15504" PostTypeId="2" ParentId="15497" CreationDate="2017-05-25T07:42:00.573" Score="1" Body="&lt;p&gt;&quot;Control&quot; as in &quot;monitor&quot; or &quot;drive&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gyroscopes can monitor motion of a pen-like device - gyroscopic sensors on chip are available and small enough to fit in one. I don't think you could drive motion of one though - that requires considerable torque, actuators in multiple axis, and a flywheel of mass no less than two orders of magnitude the mass of what is driven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of gyroscopes driving motion of an object, you can take International Space Station's &lt;a href=&quot;https://en.wikipedia.org/wiki/Control_moment_gyroscope&quot; rel=&quot;nofollow noreferrer&quot;&gt;CMG&lt;/a&gt;s. ISS has several of these, they weigh something of order of a ton each, and they take a couple hours to turn the station. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK there are no pen-sized CMGs in production, building one on your own, due to miniaturization required is out of reach of anyone without access to micromechanics workshop, and even then they only control angle - rotation, not translation, so your pen tip could draw nice circles in the air, but as soon as you push it to paper, with just rotation and no translation control the best you can get is an ellipse.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-05-25T07:42:00.573" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15505" PostTypeId="2" ParentId="13275" CreationDate="2017-05-25T08:36:59.107" Score="0" Body="&lt;p&gt;The truth table for &lt;em&gt;moment i&lt;/em&gt; would be:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CK_i| D_i|Q_(i-1)|| Q_i&#xA; ------------------------&#xA;  0 | 0  | 0     || 0 &#xA;  0 | 0  | 1     || 1 &#xA;  0 | 1  | 0     || 0 &#xA;  0 | 1  | 1     || 1 &#xA;  1 | 0  | 0     || 0 &#xA;  1 | 0  | 1     || 0 &#xA;  1 | 1  | 0     || 1 &#xA;  1 | 1  | 1     || 1 &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;$Q^*_i$ is &lt;strong&gt;always&lt;/strong&gt; trivially $\sim Q_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Describing this simply: for $CK_i=0: Q_i=Q_{i-1}.$ - Q is retained from previous state. For $CK_i=1 : Q_i=D_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to rewrite that into a function, $Q_i = ( CK_i \cdot D_i ) + (\sim CK_i \cdot Q_{i-1})$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your question didn't tag CK or D with their moments. I'll assume they are not changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_0=0$$&#xA;$$ D_1=D_2=CK_1=CK_2=1$$&#xA;$$ Q_1=?; Q_2 = ? $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have write enabled at all times, previous state is not retained and input is directly rewritten to output, Q=D. $Q_1 = 1; Q_2 = 1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And no,  $Q_1=Q^∗_1=0$ only if the mains switch is off or the circuit is on fire. $Q_1 = 1; Q^*_1 = 0$, period.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-05-25T08:36:59.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15506" PostTypeId="1" AcceptedAnswerId="15545" CreationDate="2017-05-25T10:46:20.190" Score="1" ViewCount="129" Body="&lt;p&gt;For a mechanical design, I would need a small/light axle, having high precision (as less loose as possible and reduced drag). With precision, there is not a very strict definition, however, I target ~0.2º of radial rotation (axial movement is not that an issue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the axle is around 24mm, on which I would attach a few elements (~50g).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The first idea is to use a bearing on each side of the axle, but I am not sure what is the good practice for attaching the bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following is my current sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/snj03.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/snj03.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest example of what I am doing is the axle in the actuator of a Hard-Disk, however, the bearing there is internally tapped for fixation. Most of the bearing I found on the market are not tapped. My feeling is that tapping the bearing is not a good idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kjyoV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kjyoV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are the commonly use pieces, practices, terms and fixations for this type of problem?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand is a quite broad question, but either I miss terminologies and in any case I did not found answers online. Also, I do not need a deep answer, but some indications of where to look further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others possible tags: axle, matched-set&lt;/p&gt;&#xA;" OwnerUserId="11129" LastEditorUserId="11129" LastEditDate="2017-05-25T14:16:42.377" LastActivityDate="2017-05-27T03:59:12.330" Title="How to design a precision (small) axle?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15508" PostTypeId="1" CreationDate="2017-05-25T11:34:47.050" Score="1" ViewCount="609" Body="&lt;p&gt;Has anyone tried to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two GPS receivers, and when they are near each other, they calculate different positions.&#xA;The difference is about 5-10 meters, but in reality it is several centimeters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm trying to do this, connect two receivers to one antenna. And again I get different positions. The difference is about 1-2 meters.&#xA;The receivers are the same model and on the same boards.&#xA;The other receivers work well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, has anyone dealt with something similar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I think, if they are connected to the same antenna, the calculated position should be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be somebody has two GPS receivers that are the same and can test this...?&lt;/p&gt;&#xA;" OwnerUserId="11466" LastEditorUserId="90" LastEditDate="2017-05-27T11:49:51.520" LastActivityDate="2017-05-27T11:49:51.520" Title="Two GPS receivers on one antenna?" Tags="|electrical-engineering|rf-electronics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15510" PostTypeId="2" ParentId="15496" CreationDate="2017-05-25T13:39:40.690" Score="1" Body="&lt;p&gt;Not anywhere near my area of expertise, but Wikipedia says of all cubic-equation-based equations of state that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For Pr&amp;lt;1 and Tr&amp;lt;1, the system is in a state of vapor–liquid&#xA;  equilibrium. The reduced cubic equation of state yields in that case 3&#xA;  solutions. The largest and the lowest solution are the gas and liquid&#xA;  reduced volume.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not clear on how or what single material can demonstrate two different liquid phases(other than superfluid).  Are you thinking of &lt;em&gt;immiscible&lt;/em&gt; liquid phases?  It may simply be that P-R cannot describe a system that complicated.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-05-25T13:39:40.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15511" PostTypeId="1" CreationDate="2017-05-25T14:57:21.117" Score="5" ViewCount="167" Body="&lt;p&gt;So I'm doing an experiment to verify the contact duration when a steel rod (Material: Mild Steel) is allowed to free fall longitudinally on a steel plate. The arrangement is real simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HkA2Q.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HkA2Q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rod is allowed to fall longitudinally on the plate, resulting in a number of impacts. &lt;strong&gt;I will only be considering the first impact&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Now here's the problem I'm facing&lt;/em&gt;: During the first contact and off-contact, the oscilloscope reads a much higher than expected value for the contact time. Same procedure is repeated for three lengths of the rod (200mm, 350mm and 475mm), all resulting in the same anomaly. Here are the results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hwoIt.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hwoIt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As clear from the plot, &lt;strong&gt;the experimental values are similar to the theoretical values except for an offset (found to be about 121 µs)&lt;/strong&gt;. Any idea why this is happening?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: &lt;em&gt;The theory used is that of elastic waves produced on spherical contact where the impact results in the creation of a compressive wave that travels to the end of the rod and returns as a tensile wave which pulls the rod off contact. As such the contact time t=  2l/(wave speed). Wave speed = 5000 m/s.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="11470" LastActivityDate="2020-08-05T18:18:08.647" Title="Longitudinal impact of steel rod on a steel plate to verify the contact duration" Tags="|mechanical-engineering|materials|applied-mechanics|beam|dynamics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15512" PostTypeId="2" ParentId="15508" CreationDate="2017-05-25T14:57:53.683" Score="1" Body="&lt;p&gt;Even when connected to the same antenna, two identical receivers might come up with different navigation solutions. Their analog front ends have different sensitivity and noise figures as a result of normal manufacturing tolerances. Their internal clocks are not synchronized, and among other things, these effects might cause them to choose different subsets of the visible satellites for their solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precision applications (e.g., GPS interferometry) need to take this into account. They typically work with the raw pseudorange and range-rate data associated with each satellite, making sure to select a common subset of satellites from both receivers. There is also usually a provision to have them running on a common internal clock.&lt;/p&gt;&#xA;" OwnerUserId="86" LastEditorUserId="86" LastEditDate="2017-05-26T02:02:44.557" LastActivityDate="2017-05-26T02:02:44.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15513" PostTypeId="2" ParentId="15511" CreationDate="2017-05-25T15:35:19.980" Score="0" Body="&lt;p&gt;Above my electrical skills, but as the rod bounces up and opens the contact, is the resistor discharging through the oscilloscope giving a false time? Would a capacitor help this?&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-25T15:35:19.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15514" PostTypeId="1" AcceptedAnswerId="15518" CreationDate="2017-05-25T16:50:27.427" Score="0" ViewCount="539" Body="&lt;p&gt;Is there a formula for the burst working pressure of a spherical pressure vessel that takes into account its material peoperties? It should be noted that this is not a concentric sphere design (it has only one radius) .This is a formula for a pipe: &lt;a href=&quot;http://www.engineersedge.com/calculators/pipe_bust_calc.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineersedge.com/calculators/pipe_bust_calc.htm&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="7538" LastEditDate="2017-05-27T02:50:11.150" LastActivityDate="2017-05-27T02:50:11.150" Title="Burst pressure of a sphere" Tags="|pressure|statics|stresses|pressure-vessel|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15515" PostTypeId="2" ParentId="8057" CreationDate="2017-05-25T17:15:14.197" Score="1" Body="&lt;p&gt;Your stationary domain can't incorporate the time/spatial dependency that the turbine blades should induce in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose there's a region of high velocity flow that trails a turbine blade by 10 degrees. When that high velocity flow crosses into the stationary domain, it should cross at a location that is 10 degrees trailing the blade, but that location is always moving in a stationary reference frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only time you can have a stationary and moving domain share a boundary is if you expect all of the free variables to be uniform in the direction of relative movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend making your entire domain rotating and applying a fixed velocity to the walls corresponding to zero velocity in the non-rotating frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Far from the blades the flow should become uniform across the circumferential direction, at that point if you wanted to have a boundary to a stationary domain you could. However, you would still get kinks in your streamlines at the border, as if the air is not rotating in the stationary reference frame the streamlines would be aligned with the turbine axis whereas in the rotating reference frame they would form a helix, so there would be a sharp transition between the two. &lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2017-05-25T17:15:14.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15516" PostTypeId="1" CreationDate="2017-05-25T17:16:25.217" Score="1" ViewCount="136" Body="&lt;p&gt;I've been given this bode diagram :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GStR5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GStR5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I've been asked to find the transfer function only by using the bode diagram.&#xA;It's the first time I'm doing this so here's what I thought. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The starting value is zero so the multiplying constant of the the transfer function has to be 1 and it is positive since the phase diagram starts from 0 degrees( it would start from 180 had it been negative)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the fact that the starting degrees are 0 we derive that we have no pole or zero that is at the origin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the final degrees are -270 and the dBs are decreasing so I must have 3 poles.Am I correct?&lt;br&gt;&#xA;Also ,I know that I can look at the curve , the dBs per decade at any point to see how many poles or zeros are acting. I can't get a good estimation though.It seems like it's 10 dBs per decade until the frequency reaches 10 rad/sec while I know it has to be a factor of 20 dBs. 60 dBs per dec mean 3 poles/zeros,40 dBs per dec mean 2 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I know where the poles are if I can't tell where the angle of the curve changes? Is it clear and I can't see it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, I've been given the answer to check if I do it right in the end and it's as follows: $$ G=\frac{500}{s^3+20s^2+200s+500} $$ Also p =&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-8.2390 + 8.6072i,&#xA;  -8.2390 - 8.6072i,&#xA;  -3.5220 + 0.0000i&lt;/p&gt;&#xA;" OwnerUserId="9603" LastActivityDate="2017-05-25T17:16:25.217" Title="Find transfer function from bode diagram?" Tags="|control-engineering|control-theory|transfer-function|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15518" PostTypeId="2" ParentId="15514" CreationDate="2017-05-25T19:06:18.780" Score="2" Body="&lt;p&gt;For spheres,stresses in the material  is same in all directions.So ,hoop stress and longitudinal stresses are the same. &#xA;Using similar abbreviations, as P for pressure(gauge) inside sphere ,FS for factor of safety,S for allowable stress and additionly&#xA;Ri for inner radius,Ro for outer radius&#xA;        P(gauge pressure)=((Ro^2-Ri^2)×S)÷((Ri^2)×FS)&#xA;For calculation of brusting pressure,take S as ultimate &#xA;stress for a given material and put FS=1&lt;/p&gt;&#xA;" OwnerUserId="10639" LastActivityDate="2017-05-25T19:06:18.780" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15519" PostTypeId="1" CreationDate="2017-05-25T19:35:14.777" Score="1" ViewCount="85" Body="&lt;p&gt;I have mounted a piezo-ceramic ultrasonic transducer (MODEL # SMD10T2R111WL, Vendor Website: PIEZO CERAMIC DISC 10X2MM R 215 KHZ WIRE LEADS) under a plastic tank using epoxy. Please see the picture below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q6Fpv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q6Fpv.jpg&quot; alt=&quot;Ultrasonic transducer mounted under a water tank&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to measure the electro-mechanical coupling of the transducer under the tank? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the specs of the transducer, the electro-mechanical coupling coefficient is &gt;= 60%. What is the meaning of it? Is there a way to measure it?&lt;/p&gt;&#xA;" OwnerUserId="10379" LastEditorUserId="282" LastEditDate="2017-06-02T14:03:46.313" LastActivityDate="2017-08-01T14:52:32.413" Title="How to measure electro-mechanical coupling?" Tags="|electrical-engineering|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15520" PostTypeId="2" ParentId="15519" CreationDate="2017-05-25T20:23:32.473" Score="1" Body="&lt;p&gt;I am new to the concept myself but according to search engine findings it is the efficiency of conversion from electrical energy to mechanical energy. A crude but educational approach might be to attach 2 to opposite ends of a very short liquid filled cylinder. Put a signal into 1 and see what comes out of the other. I am really winging it here. But if you are curious about it that might be a place to start.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-05-25T20:23:32.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15521" PostTypeId="2" ParentId="15486" CreationDate="2017-05-25T21:58:12.917" Score="0" Body="&lt;p&gt;I doubt if pouring liquid Cu onto solid , of similar mass, will produce a weld. At one time I poured various copper alloys into a simple mold. The mold was about twice the mass of the liquid alloys ; there was never a trace of bonding. However the alloys were at temperatures of 100 to 200 F below melting point of Cu.&lt;/p&gt;&#xA;" OwnerUserId="11475" LastActivityDate="2017-05-25T21:58:12.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15522" PostTypeId="2" ParentId="15393" CreationDate="2017-05-25T22:13:08.110" Score="0" Body="&lt;p&gt;If you have a tensile test bar with shoulder fillets, the grips will not be a problem. Then as the metal plastically deforms ( assuming it is ductile) , dislocations will pile-up - work harden. This will eventually produce a local stress concentration &gt; necking.  &lt;/p&gt;&#xA;" OwnerUserId="11476" LastActivityDate="2017-05-25T22:13:08.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15523" PostTypeId="1" CreationDate="2017-05-25T22:40:26.323" Score="2" ViewCount="363" Body="&lt;p&gt;The field of human anatomy seems to have standardized on a number of useful terms to unambiguously describe the human body.  For example the three &lt;a href=&quot;https://en.wikipedia.org/wiki/Anatomical_plane#Human_anatomy&quot; rel=&quot;nofollow noreferrer&quot;&gt;planes&lt;/a&gt; and &lt;a href=&quot;http://www.physio-pedia.com/Cardinal_Planes_and_Axes_of_Movement#Axes&quot; rel=&quot;nofollow noreferrer&quot;&gt;axes&lt;/a&gt; as well as &lt;a href=&quot;https://en.wikipedia.org/wiki/Anatomical_terminology#Standard_terms&quot; rel=&quot;nofollow noreferrer&quot;&gt;various directions&lt;/a&gt; seem to be unambiguously defined and those definitions universally accepted.  Those terms also seem indispensable for clear verbal and written communication.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were using the anatomical terms, I'd be able to refer to the plane made by the side of a book case as a sagittal plane, and refer to a dimension measured from left to right as measured along the frontal axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there corresponding standardized terms in mechanical or structural engineering that refer to axes, planes, and directions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, how are such axes, planes, and directions customarily communicated verbally and in writing?&lt;/p&gt;&#xA;" OwnerUserId="11477" LastActivityDate="2017-05-26T15:12:56.967" Title="Are there standard terms that correspond to the anatomical axes, planes, and directions?" Tags="|mechanical-engineering|structural-engineering|terminology|standards|architecture|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15525" PostTypeId="2" ParentId="15523" CreationDate="2017-05-26T00:37:46.240" Score="4" Body="&lt;p&gt;This answer comes from my experience in mechanical engineering, a structural engineer might have a different perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is difficult to come up with a standard set of terms for planes or axes in mechanical and structural engineering since each there are so many different types of work within each discipline. There are probably some specialist-specific terms out there for people who frequently deal with the same kind of geometry repeatedly. There are some terms that do get used across disciplines, for example the &quot;longitudinal axis&quot; would generally be the axis along the primary length of something. Could be along the length of a pipe or a beam, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the real answer to this question is: rather than relying on standardized terminology, &lt;strong&gt;engineers use diagrams&lt;/strong&gt;. Whenever you start discussing reference frames, planes or whatever else, drawing a schematic or diagram is standard practice. The advantage to this is that you aren't relying on everyone who comes across your report or design calculations to speak the same engineering jargon. Assuming you keep your diagram or drawing mostly symbolic you also have the advantage that even people who don't speak your language can understand your coordinate system.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2017-05-26T00:37:46.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15528" PostTypeId="2" ParentId="15523" CreationDate="2017-05-26T02:27:51.893" Score="2" Body="&lt;p&gt;In engineering there are four views, or axes that I can think of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;longitudinal - which is parallel to the longest vertical side of an&#xA;object,&lt;/li&gt;&#xA;&lt;li&gt;cross-sectional - which is parallel to the shortest vertical side of&#xA;an object &amp;amp; is at right angles to the longitudinal view/axis.&lt;/li&gt;&#xA;&lt;li&gt;plan - which is usually the horizontal view or axis, as viewed from&#xA;above the object, and&lt;/li&gt;&#xA;&lt;li&gt;oblique - which is at an angle between cross-sectional and&#xA;longitudinal.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-05-26T02:27:51.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15530" PostTypeId="2" ParentId="15523" CreationDate="2017-05-26T08:40:02.267" Score="2" Body="&lt;p&gt;For aviation, aerospace, and naval (and to a lesser degree automotive) engineering, the three axes are &lt;strong&gt;pitch, yaw and roll axes&lt;/strong&gt; - with the names &lt;strong&gt;pitch, yaw and roll&lt;/strong&gt; corresponding to rotational motion around these, and &lt;strong&gt;sway, heave and surge&lt;/strong&gt; for translational motion along them. Velocities in these are &lt;strong&gt;groundspeed&lt;/strong&gt;/&lt;strong&gt;speed over ground(SOG)&lt;/strong&gt; (measured along travel axis; airspace/naval terms respectively), &lt;strong&gt;slip rate&lt;/strong&gt; (sideways) and &lt;strong&gt;climb/sink rate&lt;/strong&gt; (vertical). (angular speeds are plain &lt;strong&gt;yaw, pitch, roll rate&lt;/strong&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, faces/directions would be &lt;strong&gt;fore/aft&lt;/strong&gt; (+/- along roll axis; front/back), &lt;strong&gt;ventral/dorsal&lt;/strong&gt; (+/- along yaw axis; bottom/top; sometimes yaw axis is oriented upwards - naval and space tends towards bottom orientation, aviation often uses upwards oriented yaw axis); and &lt;strong&gt;star/port or starboard/portboard&lt;/strong&gt; (+/-pitch axis; right, left.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complete the image, to locate these axis in space, we have directions: &lt;strong&gt;magnetic north&lt;/strong&gt; or &lt;strong&gt;geographic north&lt;/strong&gt; (obvious), &lt;strong&gt;zenith&lt;/strong&gt; and &lt;strong&gt;nadir&lt;/strong&gt; (directions relative to Earth, up and down respectively), location: with geographical coordinates &lt;strong&gt;latitude&lt;/strong&gt; and &lt;strong&gt;longitude&lt;/strong&gt;, and &lt;strong&gt;height&lt;/strong&gt; (relative to surface) or &lt;strong&gt;altitude&lt;/strong&gt; (sea level). Angle between magnetic north and roll axis is called &lt;strong&gt;heading&lt;/strong&gt; (the other two angles are just pitch and roll). To add a bit to the confusion, there's also &lt;strong&gt;track&lt;/strong&gt; which is the direction of travel (groundspeed + slip + climb/sink), and &lt;strong&gt;airspeed&lt;/strong&gt; (speed relative to surrounding air; couldn't find the related naval term)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xof6J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xof6J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2017-05-26T15:12:56.967" LastActivityDate="2017-05-26T15:12:56.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15531" PostTypeId="1" AcceptedAnswerId="15533" CreationDate="2017-05-26T11:06:28.477" Score="2" ViewCount="85" Body="&lt;p&gt;I was curious about this. If i have a tube which has water flowing in it. and mounted vertically a small extension tube with a loop. The water will go horizantally and only a few drops will go vertically. Why is that? and how can u determine how much water will go vertically? I tried to search for a scientic explaination but didn't find anything. What if I want to design a tube with this loop but I don't want any water to go vertically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram if you don't understand: &lt;a href=&quot;https://i.stack.imgur.com/M6ofE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M6ofE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anyway to prevent the water from going verically if the tube is blocked? &lt;/p&gt;&#xA;" OwnerUserId="11369" LastActivityDate="2017-05-26T11:41:39.900" Title="Why would the flow go horizantally not vertically in the tube?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15532" PostTypeId="2" ParentId="15531" CreationDate="2017-05-26T11:35:02.293" Score="0" Body="&lt;p&gt;Depends on the internal and external pressures. Apply the Bernoulli equation to the situation. Also check out coefficient of discharge and the Coanda effect. To stop the flow going vertically block it same as the tube.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-26T11:35:02.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15533" PostTypeId="2" ParentId="15531" CreationDate="2017-05-26T11:41:39.900" Score="1" Body="&lt;p&gt;Water likes to take the path of least resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scenario, you have 2 paths for the water to take when it reaches your T-junction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we compare the two paths, we can see why it doesn't want to travel through the vertical tube.  First, to go through the vertical tube it would need to overcome gravity.  The tube seems thinner, so it would also have greater frictional losses as it travels through it.  To add to that, it has a bend, which is another source of friction loss compared to a straight pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're familiar with basic DC electric circuits, you may know that if you have current travelling through parallel resistors, the higher current will flow through the lower resistance branch.  This is an analogous situation.  The current is like the flow rate, the resistance is like the frictional and gravity losses, and the voltage is like the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the equations of &lt;a href=&quot;https://en.wikipedia.org/wiki/Bernoulli%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bernoulli's Principle&lt;/a&gt; to get some idea on how these losses will effect the flowrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to prevent water flow vertically when the tube is blocked, you need to make sure the pressure losses in the vertical tube due to gravity and friction are greater than the maximum pressure you can build up. This will depend on how you are getting your flow, if it's a pump, the pump may stall or throttle itself once it reaches max operating pressure.  If it's flowing from a higher point, the safe thing to do would be to make sure your tube is higher than the high-point you feed in from.  (you can probably get away with slightly below the high point due to friction losses on the way there, especially depending on how long the pipes are).&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-05-26T11:41:39.900" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15534" PostTypeId="2" ParentId="8097" CreationDate="2017-05-26T12:41:25.903" Score="1" Body="&lt;p&gt;The horizontal tension remains constant. The vertical tension integrates the load. The ratio determines the direction of the cable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f(x)=\frac{dT_y(x)}{dx}$$&#xA;$$\frac{dy(x)}{dx}=\frac{T_y(x)}{T_x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y(x)=\frac1{T_x}\iint f(x) \, dx + C_0 +C_1 x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determining $T_x$ is the difficult part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the cable path, must equal the length of the correctly stretched cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$path=\int^b_a\sqrt{1+\left(\frac{dy(x)}{dx}\right)^2} \,dx$$&#xA;$$path=\int^b_a\sqrt{1+\left(\frac{T_y(x)}{T_x}\right)^2} \,dx$$&#xA;$$cable=relaxed+\int^b_a\epsilon(x)\sqrt{1+\left(\frac{dy(x)}{dx}\right)^2}dx$$&#xA;$$cable=relaxed+\int^b_a\frac{\sigma(x)}{E(x)}\sqrt{1+\left(\frac{T_y(x)}{T_x}\right)^2}dx$$&#xA;$$cable=relaxed+\int^b_a\frac{\sqrt{T_y(x)^2+{T_x}^2}}{A(x)E(x)}\frac{\sqrt{T_y(x)^2+{T_x}^2}}{T_x}dx$$&#xA;$$cable=relaxed+\int^b_a\frac{T_y(x)^2+{T_x}^2}{A(x)E(x)T_x}dx$$&#xA;$$\int^b_a\sqrt{1+\left(\frac{T_y(x)}{T_x}\right)^2} \,dx=relaxed+\int^b_a\frac{T_y(x)^2+{T_x}^2}{A(x)E(x)T_x}dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that's about as simplified as I can make it without know the form of your area, stiffness, and load curves.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2017-05-26T12:41:25.903" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15535" PostTypeId="2" ParentId="8823" CreationDate="2017-05-26T15:17:02.750" Score="2" Body="&lt;p&gt;The coefficient of moment is related to how much the wing makes the plane want to pitch up. This torque is taken about the center of gravity and thus moving the center of gravity changes the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you measure the torque about an arbitrary point, the value will usually depend on the angle of attack. However, there are some points where it remains constant. The point that lies on the cord of the wing where the torque remains constant across angles of attack is the aerodynamic center of the wing.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2017-05-26T15:17:02.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15536" PostTypeId="2" ParentId="8608" CreationDate="2017-05-26T16:02:30.890" Score="1" Body="&lt;p&gt;There are currently houses that are built with heat pumps to heat or cool the house using the ground a thermal reservoir. It's usually much more efficient than using the outside air as their thermal reservoir. However, as far as I know none of the heat pumps used approach the theoretical carnot efficiency as the heat transfer would be too slow without massive heat exchangers. Instead, they relinquish some of their thermodynamic efficiency to make their cold side colder than their cold reservoir, and their hot side hotter than their hot reservoir. this allows for the heat transfer rates to be greatly increased, making the heat exchanger sizes reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, trying the same tactic won't work when trying to increase the heat transfer rate of a stirling machine. Increasing, the difference between the cold side of the engine and the cold reservoir just cuts into energy available. To be effective, you'd have to have giant heat exchangers. Turns out it would be way cheaper to install solar panels, or just buy the electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you have a really large temperature difference. For example if you live near yellowstone national park, the temperature deep underground can be hundreds of degrees farenheit. Then it's &quot;easy&quot; to build a geothermal generator.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2017-05-26T16:02:30.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15537" PostTypeId="2" ParentId="8692" CreationDate="2017-05-26T16:39:01.137" Score="0" Body="&lt;p&gt;My 2 cents: Springs are easily adjustable , counter weights difficult. I remember  counter weighted windows; when the rope broke , the repair was a mess - getting into the frame. I have had many doors with both spring types and never had a spring failure ; I have had a couple cables break. And you need cables for a counter weight. I really have a hard time thinking of an advantage for weights. &lt;/p&gt;&#xA;" OwnerUserId="11486" LastActivityDate="2017-05-26T16:39:01.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15538" PostTypeId="2" ParentId="8247" CreationDate="2017-05-26T18:34:37.600" Score="0" Body="&lt;p&gt;It looks like you're trying to use area to determine how much back-flow there will be. The flow will be depended on the pressure gradients, which are dependent not just on cross sectional area but also length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like you're trying to model something like a 3D printer filament being pushed through a heated nozzle and want to determine how tight of a tolerance you'll need to prevent melted plastic from flowing back out the entrance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a pretty complicated problem but I'll try to give the basics of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we need to establish some pressures, and additional dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nT4pV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nT4pV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can see the melted piston in green can flow in two directions back around the solid piston, or forward through the exit orifice. First let's look at the forward flow rate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For fairly viscous media (like melted plastic) through thin pipes flow rate is calculated by the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hagen–Poiseuille equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q= \frac{\pi\, R^4\,\Delta P}{8\,\mu\, L}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $Q$ is your volumetric flow rate, $L$ is the length of your tube ($L_0$), $R$ is your tube radius ($r_0$), $\mu$ is your viscosity (going to be highly temperature dependent), and $\Delta P$ is the difference in pressure across the tube ($P_1-P_2$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So plugging in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_0= \frac{\pi\, r_0^4\,(P_1-P_0)}{8\,\mu\, L_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow around a piston can be modeled like flow between two parallel plates. Except the plates are wrapped into a circle to form cylinders. When one plate is moving relative to the other this is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Couette_flow&quot; rel=&quot;nofollow noreferrer&quot;&gt;Couette Flow&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=\frac{\Delta P\,h^3 \, w }{12 \mu\,L} - \frac{U\,h\,w}2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $h$ is the height of our chanel $r-r_1$, $w$ is the width of our channel (in this case circumference of the piston) $2\pi r_1$, $U$ is the velocity of our piston, $\Delta P$ is $P_1-P_2$, and $L$ is $L_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_1=\frac{\pi(P_1-P_2)\,(r-r_1)^3 \, r_1 }{6 \mu\,L_1} - \pi\,U\,(r-r_1)\,r_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we're looking for this second Q to be zero. This means all of the melted material gets pushed out the front hole. If the flow rate is positive, the edge of the melted plastic would flow further up the channel, increasing $L_1$ which would in turn decrease the flow rate. This is good news, it means as long as the sleeve is long enough the flow will stabilize around zero flow. Also since we are now assuming all the newly melted plastic is going to go out the exit, this allows us to relate the flow rate out the front hole with the velocity of the piston:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_0=U\, \pi {r_1}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=Q_1=\frac{\pi(P_1-P_2)\,(r-r_1)^3 \, r_1 }{6 \mu\,L_1} - \frac{Q_0\,(r-r_1)}{r_1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0=\frac{\pi(P_1-P_2)\,(r-r_1)^3 \, r_1 }{6 \mu\,L_1} - \frac{\pi\, r_0^4\,(P_1-P_0)\,(r-r_1)}{8\,\mu\,r_1 \, L_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, while viscosity will probably be different in the different regions due to temperature, it's super convenient if we ignore that because then we can just multiply it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{(P_1-P_2)\,(r-r_1)^3 \, r_1 }{3\,L_1} = \frac{r_0^4\,(P_1-P_0)\,(r-r_1)}{4\,r_1 \, L_0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further rearranging:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4(P_1-P_2)\,(r-r_1)^2 \, {r_1}^2 \,L_0 = 3\,{r_0}^4\,(P_1-P_0)\,L_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly if ($P_1=P_2$) we can cancel that out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$4\,(r-r_1)^2 \, {r_1}^2 \,L_0 = 3\,{r_0}^4\,L_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r=r_1+\frac{{r_0}^2}{r_1}\sqrt{\frac{3\,L_1}{4\,L_0}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give you the maximum size bore you can use with the given lengths and other sizes of your hole.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="1122" LastEditDate="2017-05-26T18:56:36.150" LastActivityDate="2017-05-26T18:56:36.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15539" PostTypeId="2" ParentId="8049" CreationDate="2017-05-26T19:39:19.640" Score="1" Body="&lt;p&gt;Even if the CPU cooler could make the patch of aluminum it was attached to exactly equal to atmospheric temperature, your plate would be much worse than a fan based cooler. The 1/8 inch air gap forms a very good insulator, so getting air moving along the surface of the laptop bottom is much better at cooling it than providing a heat sink 1/8 inch away.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2017-05-26T19:39:19.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15540" PostTypeId="2" ParentId="8049" CreationDate="2017-05-26T20:28:04.863" Score="1" Body="&lt;p&gt;Your best bet may be to make a hole in the aluminium sheet and mount the fan to either blow or suck the air from between the laptop base and the aluminium sheet ie using the 1/8&quot; gap...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-26T20:28:04.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15541" PostTypeId="2" ParentId="8247" CreationDate="2017-05-26T21:03:53.900" Score="1" Body="&lt;p&gt;I don't know about the math, but the answer to the stated question depends on how the bit is driven ( solid drill press , hand held , etc) and the type of drill/point. The point is generally not in the exact center so normally makes an oversize hole . To correct this, a &quot;gun drill&quot; only cuts on one side; and there are other types of points.&lt;/p&gt;&#xA;" OwnerUserId="11489" LastActivityDate="2017-05-26T21:03:53.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15542" PostTypeId="2" ParentId="14684" CreationDate="2017-05-26T21:19:54.760" Score="0" Body="&lt;p&gt;That is not much load. If it must be steel , I would find a local weld shop and go to see what leftovers they have; I suspect 1/8&quot; would be good. Otherwise plywood could do the stated  job.&lt;/p&gt;&#xA;" OwnerUserId="11491" LastActivityDate="2017-05-26T21:19:54.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15543" PostTypeId="2" ParentId="14684" CreationDate="2017-05-26T23:13:32.393" Score="0" Body="&lt;p&gt;The weight on the bolt will be 350 lbs plus the weight of the plate. Depending on what size washer and bolt head you use the force at the hub/bolt overlap could be 700 psi. 1/4&quot; plate is probably ok for the full disk, but I would double it with 1/4&quot; thick by  3&quot; diameter inner disk/washer.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-05-26T23:13:32.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15545" PostTypeId="2" ParentId="15506" CreationDate="2017-05-27T03:59:12.330" Score="1" Body="&lt;p&gt;Axle type rod is available under different names on-line. Try searching for &quot;8mm linear shaft&quot;, &quot;linear rail shaft&quot;, &quot;ground polished shaft&quot; just to start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bearings that fit snug don't require much of a fastener to keep them from walking off of the shaft. You could machine a hole through the end, or a groove around the end, and use an R-clip or c-clip snapped onto the end. It would probably be overkill, but you could drill and tap a hole straight into the center of each end for a screw and washer.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-05-27T03:59:12.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15547" PostTypeId="1" AcceptedAnswerId="15550" CreationDate="2017-05-27T08:46:39.313" Score="6" ViewCount="11342" Body="&lt;p&gt;To save on material, I am requested to design a continuously supported beam with a maximum bending capacity that is half of its maximum bending moment and to strengthen the beam at certain intervals where its capacity is exceeded. What is a feasible strengthening scheme without changing the beam's overall dimensions significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This beam is going to be used as a waling where stiffeners will be added at strut-waling connection. This is also the location where moment capacity will be exceeded.&lt;/p&gt;&#xA;" OwnerUserId="11498" LastActivityDate="2019-04-11T14:57:45.593" Title="How to strengthen an I beam" Tags="|structural-engineering|civil-engineering|steel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15549" PostTypeId="1" CreationDate="2017-05-27T12:31:16.023" Score="0" ViewCount="519" Body="&lt;p&gt;I have a project where I need to cut a typical 5 point star shape (i.e. ★) out of brass sheet. The stars are small (~8mm across) and in high quantities. I was thinking of making a punch and die of the appropriate size but have stumbled on the best way to do this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm stuck on how would I mill the sharp corners of the punch and dies - I was thinking possibly a gear cutter but the angle may be wrong? Would there be a better way to approach this entirely?&lt;/p&gt;&#xA;" OwnerUserId="10454" LastActivityDate="2018-02-24T11:31:23.350" Title="Cutting Small Stars From Sheet Metal" Tags="|machining|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15550" PostTypeId="2" ParentId="15547" CreationDate="2017-05-27T14:50:15.060" Score="8" Body="&lt;p&gt;If your beam design is governed by yielding in bending (not lateral-torsional buckling/plate buckling, etc) then you need to increase the second moment of area (I) to increase the bending capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually this is done by fastening additional plates to the beam, typically onto the flanges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some examples of such a strengthening strategy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bolting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1tQ6d.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1tQ6d.png&quot; alt=&quot;Bolting&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Welding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xf85e.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xf85e.png&quot; alt=&quot;Welding&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.newsteelconstruction.com/wp/strengthening-existing-steelwork/&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; on strengthening existing steel beams for more information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more elaborate and expensive method would be to use variable depth beams as is sometimes done with bridges:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jyEJQ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jyEJQ.jpg&quot; alt=&quot;Variable Depth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2017-05-27T14:50:15.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15551" PostTypeId="1" AcceptedAnswerId="15553" CreationDate="2017-05-27T17:35:34.510" Score="0" ViewCount="231" Body="&lt;p&gt;I have a beam. I have its Young's modulus. I don't have its maximum acceptable deflection. How can I determine the maximum acceptable deflection without breaking the beam? &lt;/p&gt;&#xA;" OwnerUserId="4214" LastActivityDate="2017-05-27T21:52:20.890" Title="How to find maximum acceptable deflection?" Tags="|beam|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15552" PostTypeId="1" CreationDate="2017-05-27T18:47:28.883" Score="1" ViewCount="76" Body="&lt;p&gt;I am interested in, perhaps, going in the direction of bio-plastics. However there might be some other options I haven't yet thought of, so that's the reason I'm opening this dialogue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wSh0j.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wSh0j.jpg&quot; alt=&quot;Conceptual prototype made from polyethylene and scotch tape&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before securing a manufacturer, I want to know what limitations, or options, I may have with regard to food safety, durability of product, look and feel of said materials, and implementation. Some compostable materials might not meet certain food-safety needs, perhaps exposure to acids or heat might result in types of unwanted degradation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Iutfg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iutfg.jpg&quot; alt=&quot;3D model&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am, basically, asking to collectively weigh the pros and cons of each material (e.g. Bio polyethylene, corn plastics) against the elements. &#xA;Which would be the absolute best material to use? Not only for functionality, but for the sake of being eco friendly also.&lt;/p&gt;&#xA;" OwnerUserId="11503" OwnerDisplayName="Joseph Pérez" LastEditorUserId="5899" LastEditDate="2017-05-28T17:28:56.257" LastActivityDate="2017-05-28T17:28:56.257" Title="Prototyping disposable protective tong sleeves - materials?" Tags="|materials|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15553" PostTypeId="2" ParentId="15551" CreationDate="2017-05-27T21:52:20.890" Score="2" Body="&lt;p&gt;Yield strength of steels is generally defined by 0.2% deflection . So if you do not exceed 0.2% strain under load you will not permanently deform it .The specification of the steel will give the minimum Yield Strength. Then if your load does not cause a stress greater than this specified yield strength it will not &quot;break&quot;. &lt;/p&gt;&#xA;" OwnerUserId="11502" LastActivityDate="2017-05-27T21:52:20.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15555" PostTypeId="1" CreationDate="2017-05-28T02:25:16.060" Score="1" ViewCount="29" Body="&lt;p&gt;Suppose there is a given piezoelectric material and under a given force it produces voltage $V$. If an external electric circuit is attached with load $R$, the current will be $I= V/R$. Also power generated will be $V^2/R$. By choosing $R$ very low, high energy can be obtained. Is this understanding correct?&lt;/p&gt;&#xA;" OwnerUserId="11504" LastEditorUserId="1832" LastEditDate="2017-05-28T02:55:46.143" LastActivityDate="2017-05-28T14:51:51.560" Title="Power from piezo material" Tags="|energy|ceramics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15556" PostTypeId="1" AcceptedAnswerId="15560" CreationDate="2017-05-28T04:34:07.040" Score="2" ViewCount="209" Body="&lt;p&gt;I recently bumbped into a rather basic but interesting question on aeroelasticity. I've learned to derive the fluttering critical speed from Pines's theory but it involves some spring stiffness like for normal spring $K_h$ and torsion spring $K_t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I want to see if I can derive the same thing if the wing is now treated as a continuous beam-rod model (i.e. torsion and bending). The equations of motion are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{\partial^2}{\partial y^2} \left( EI \frac{\partial^2 h}{\partial y^2} \right) + m \frac{\partial^2 h}{\partial t^2} + m x_\alpha \frac{\partial^2 \alpha}{\partial t^2}+L=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ -\frac{\partial}{\partial y} \left( GJ \frac{\partial \alpha}{\partial y} \right) + I_\alpha \frac{\partial^2 \alpha}{\partial t^2} + mx_\alpha \frac{\partial^2 h}{\partial t^2} - M = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, let $h(y,t)=0$ and $\alpha=s(y)e^{pt}$ so we can focus entirely on the torsion dynamic response. Let $L=qca_0(\alpha+\alpha_0)$ and $M=qcea_0(\alpha+\alpha_0)$. But now things get out of my control as I don't know how to solve for $p$ (to let its real part positive), given $s(y)$ unsolved too. &lt;/p&gt;&#xA;" OwnerUserId="11505" LastActivityDate="2017-05-28T13:17:43.587" Title="Wing as cantilever beam (beam-rod) model for aeroelasticity fluttering analysis" Tags="|beam|mathematics|aerodynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15557" PostTypeId="1" CreationDate="2017-05-28T05:56:19.200" Score="3" ViewCount="60" Body="&lt;p&gt;I am designing a device in which I want to lock each drawer until it's time to open them. &#xA;Let's say that there are 4 compartments in the device. Each compartment needs to be programmed to open at a pre-programed time.&#xA;For example, at T1 time, compartment 1 opens. Then at T1 time, none of the other compartments should be able to open even if pulled by the user. The same should be true for all the other compartments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My earlier approach included using a solenoidal actuator for this. So, the holding current of the actuator would keep it magnetised and hence latched. And whichever compartment needs to be opened, that actuator is set to low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am unable to find any solenoidal actuator below 1Amp rating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to hear if you have any other solutions for this sort of locking mechanism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.   &lt;/p&gt;&#xA;" OwnerUserId="11506" LastActivityDate="2017-05-28T13:57:29.907" Title="How to provide conditional locking to drawers?" Tags="|electrical-engineering|circuit-design|actuator|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="15559" PostTypeId="1" CreationDate="2017-05-28T08:17:19.760" Score="4" ViewCount="293" Body="&lt;p&gt;Hi I have this block diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qJpJA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qJpJA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I want to reduce it to such a block diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sdCdz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sdCdz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I find  H ?&lt;/strong&gt;&#xA;I am looking for a general method, of how to approach such problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what H should be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8v46f.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8v46f.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9038" LastActivityDate="2017-07-21T08:43:55.337" Title="Finding Transfer Function between specific points" Tags="|control-engineering|transfer-function|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15560" PostTypeId="2" ParentId="15556" CreationDate="2017-05-28T13:17:43.587" Score="0" Body="&lt;p&gt;By writing $\alpha=s(y) e^{pt}$, you've assumed that the solution is separable, i.e. that it can be separated into a part that depends only on y and another part that depends only on t.  This is pretty common for this type of equation and it should work.  The typical next step here would be to separate the variables in the equation. i.e. move everything that depends on s(y) (and its derivatives) to one side, and everything that depends on $e^{pt}$ (and its derivatives) to the other side. Then, since the left depends only on y, and the right only on t, the only way this can work if is both sides are equal to the same constant. You then just need to find that constant. Typically you'll proceed by assuming something like $s(y) = A \cos(kpy)+B \sin(kpy)$, where A and B are unknown constants and k is the wavenumber.  Then just plug in and solve.  There will be an infinite number of solutions to the equation. Each solution is a combination of k, A, B that solve the equation (and the boundary conditions), and represents an eigenmode of vibration. Typically, you'll keep only the first few (or maybe only first one) eigenmodes, and then find the time response of the modes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This page might help, &lt;a href=&quot;http://www1.aucegypt.edu/faculty/mharafa/MENG%20475/Continuous%20Systems%20Fall%202010.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www1.aucegypt.edu/faculty/mharafa/MENG%20475/Continuous%20Systems%20Fall%202010.pdf&lt;/a&gt; , or pick up any textbook on &quot;vibrations of continuous systems&quot;.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-05-28T13:17:43.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15561" PostTypeId="2" ParentId="15557" CreationDate="2017-05-28T13:57:29.907" Score="1" Body="&lt;p&gt;Try a bi-stable lock/unlock mechanism per drawer.  A solenoid or perhaps two flip the state of the lock.  No power is needed to stay in any one state, only to flip states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the solenoids only need to be run for a short time, like maybe 50 ms, the large current spike can be supplied locally by a capacitor.  That capacitor can then charge up more slowly between actuations.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-05-28T13:57:29.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15563" PostTypeId="2" ParentId="15555" CreationDate="2017-05-28T14:51:51.560" Score="1" Body="&lt;p&gt;This is wrong for at least two reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if you apply a &lt;em&gt;constant&lt;/em&gt; force to the piezoelectric material, you will produce a voltage across it but &lt;em&gt;no&lt;/em&gt; current will flow. You can only generate a current by doing &lt;em&gt;mechanical work&lt;/em&gt; on the material. That means you only get a current while you are &lt;em&gt;changing&lt;/em&gt; the force, and therefore changing the amount that the material bends or stretches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, you are ignoring the internal electrical impedance of the material. If the impedance was a constant resistance, it is fairly easy to show that the maximum power in the load occurs when the load $R$ is the same as the internal resistance. See the &lt;a href=&quot;https://en.wikipedia.org/wiki/Maximum_power_transfer_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maximum Power Transfer theorem&lt;/a&gt;. If $R$ is smaller than this, most of the power is &quot;wasted&quot; within the piezo material instead of being transferred to the load. If $R$ is bigger, the higher impedance of the complete circuit reduces the power that is generated, in a similar way to your attempt at analysing what happens using Ohm's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same idea applies to a piezoelectric material, but the internal impedance is more like a capacitance than a resistance, so you need to analyze the behaviour of the circuit for an alternating current, and the amount of power transferred depends on the &lt;em&gt;frequency&lt;/em&gt; of the force applied to the material, as well as on the internal and load impedances. Basically, you get more power by moving the piezo material faster.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-05-28T14:51:51.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15565" PostTypeId="5" CreationDate="2017-05-28T16:41:13.807" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-05-28T16:41:13.807" LastActivityDate="2017-05-28T16:41:13.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15566" PostTypeId="4" CreationDate="2017-05-28T16:41:13.807" Score="0" Body="Questions for this tag may be for improving processes or finding an efficient way to complete a task. " OwnerUserId="11512" LastEditorUserId="33" LastEditDate="2017-05-28T20:21:56.560" LastActivityDate="2017-05-28T20:21:56.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15567" PostTypeId="1" CreationDate="2017-05-28T20:06:47.710" Score="2" ViewCount="299" Body="&lt;p&gt;I saw that the rpm of a synchronous motor is proportional to the frequency of current alteration and I was wondering if there is a theoretical maximum frequency (regardless of magnet strength)? Also could changing the frequency be used to control the speed of a synchronous motor?&lt;/p&gt;&#xA;" OwnerUserId="10361" LastActivityDate="2017-05-28T21:24:07.563" Title="What is the maximum frequency of an AC current?" Tags="|stepper-motor|ac|frequency-response|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15569" PostTypeId="1" AcceptedAnswerId="15582" CreationDate="2017-05-28T22:14:22.737" Score="2" ViewCount="50" Body="&lt;p&gt;By this I mean can any DC to AC variable frequency drives invert digitally (no moving parts) and if so how do these work? Is this possible for both sine and square wave inverters?&lt;/p&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="8055" LastEditDate="2017-05-31T17:02:10.957" LastActivityDate="2017-05-31T17:02:10.957" Title="Are there any variable frequency drives that work digitally?" Tags="|electrical-engineering|frequency-response|ac|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15570" PostTypeId="1" CreationDate="2017-05-28T23:17:30.370" Score="1" ViewCount="1355" Body="&lt;p&gt;If a plate is elongated in the x and y direction, why is the stress sigma z considered negligible with respect to sigma x and sigma y?&lt;/p&gt;&#xA;" OwnerUserId="1248" LastActivityDate="2017-06-10T13:18:32.190" Title="Why is the normal stress assumed to be 0 in plane stress?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15572" PostTypeId="1" CreationDate="2017-05-29T00:35:58.137" Score="2" ViewCount="34" Body="&lt;p&gt;I'm trying to build a lamp which has about 20 identical layers stacked on top of each other. There is a central spine and, on each level, a pair of arcs attached to the spine that are hinged and which meet each other (this looks like a big rib cage, where the ribs are free to rotate in and out). I'd like to create some gearing such that moving one side of one layer will move the other one symmetrically, keeping each layer symmetrical. I don't need any automation to move them, but I just want them to be constrained to symmetry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general gear plan could be something like this: &lt;a href=&quot;http://www.gearsket.ch/#c74a934963cc29f8&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.gearsket.ch/#c74a934963cc29f8&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...but I'm not able to attach the outside gears (which would be attached to the ribs) at the rotation axes of the ribs, because of the hinges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may be able to get that to work, but I'm getting worried that the gears (to be CNC machined out of wood) are getting to be pretty big (each layer, when closed, is a circle about 8 inches in diameter), and are going to take over the look of the lamp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any alternate gear or linkage designs that would help to constrain these to symmetrical opening/closing motion?&lt;/p&gt;&#xA;" OwnerUserId="11252" LastActivityDate="2017-05-29T00:35:58.137" Title="Gear design for symmetrical opening of two parts" Tags="|gears|linkage|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15573" PostTypeId="1" AcceptedAnswerId="15577" CreationDate="2017-05-29T01:15:01.803" Score="5" ViewCount="492" Body="&lt;p&gt;I saw crane on very tall building. How do they put such heavy item on such high building? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4bd3P.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4bd3P.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11516" LastActivityDate="2017-05-30T10:26:29.383" Title="How they put heavy cranes on top of the building?" Tags="|civil-engineering|construction-management|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15575" PostTypeId="1" CreationDate="2017-05-29T04:40:25.337" Score="1" ViewCount="88" Body="&lt;p&gt;I would like to double the shear capacity of an I beam by welding some steel plates. What strengthening schemes are available? How would you define the shear area for the strengthening scheme proposed?&lt;/p&gt;&#xA;" OwnerUserId="11498" LastEditorUserId="11498" LastEditDate="2017-05-29T05:40:37.877" LastActivityDate="2018-11-07T22:47:59.293" Title="What shear strengthening schemes are available for an I beam?" Tags="|structural-engineering|civil-engineering|steel|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15576" PostTypeId="1" CreationDate="2017-05-29T07:48:50.297" Score="1" ViewCount="3542" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Sommerfeld_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sommerfeld&lt;/a&gt; number can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ S = \frac{6 \ \mu \ U \ b}{p \ c^2} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;U: Velocity&lt;/li&gt;&#xA;&lt;li&gt;b: Width&lt;/li&gt;&#xA;&lt;li&gt;p: Pressure &lt;/li&gt;&#xA;&lt;li&gt;c: Minimal Clearing&lt;/li&gt;&#xA;&lt;li&gt;$\mu$: Absolute Viscosity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As far as I'm aware, it may be used for both thrust and journal bearings. It is also known as the &lt;a href=&quot;https://www.google.de/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;ved=0ahUKEwjyiZ3ix5TUAhWBhiwKHRfQDUIQFggiMAA&amp;amp;url=https%3A%2F%2Fntrs.nasa.gov%2Farchive%2Fnasa%2Fcasi.ntrs.nasa.gov%2F19700026623.pdf&amp;amp;usg=AFQjCNHZ5Zkacd3bZO6T1RVNEWpUfeaWRw&amp;amp;cad=rja&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dimensionless Bearing Number&lt;/a&gt;&#xA;or dimensionless surface pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand or differently stated, I couldn't find any resources on the significance of that dimensionless number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance the Reynolds number is the ratio of the inertia and viscous forces. Depending on the situation, any value, approximately, above 1000 or 3000 is considered to be turbulent flow.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What is the ballpark for the Sommerfeld number? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When is it too small or too large to disrupt operation or damage the bearing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do several &quot;zones&quot; of values exist that exhibit similar behaviour? (such as between 1 and 3 it behaves as ....?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case there are huge differences between journal and thrust bearings in regard to the Sommerfeld number, I would be more interested in the thrust bearing case.&lt;/p&gt;&#xA;" OwnerUserId="11517" LastEditorUserId="11517" LastEditDate="2017-05-29T08:28:48.307" LastActivityDate="2017-05-29T08:28:48.307" Title="Significance of the Sommerfeld number" Tags="|mechanical-engineering|hydraulics|tribology|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15577" PostTypeId="2" ParentId="15573" CreationDate="2017-05-29T09:34:50.537" Score="7" Body="&lt;p&gt;Quite simple, lift it up there with another crane that is more expensive once it's up there the expensive crane is sent away and they can use the &quot;cheap&quot; crane for construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterwards the expensive crane returns to get the construction crane down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but most of the times they are from the ground up and attached to the building (see image)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ePn6S.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ePn6S.jpg&quot; alt=&quot;crane erection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crane can lift itself higher when required (see video in link)&lt;a href=&quot;https://www.youtube.com/watch?v=y39152VWCPk&quot; rel=&quot;noreferrer&quot;&gt;video&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8464" LastActivityDate="2017-05-29T09:34:50.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15580" PostTypeId="1" CreationDate="2017-05-29T11:36:56.840" Score="2" ViewCount="6184" Body="&lt;p&gt;I have the following issue:&lt;/p&gt;&#xA;&lt;p&gt;One year ago or so I bought a toy for my kid. Worked fine, then the battery got low (still working though but under performing). I checked the battery type, see is an AA (3 of them). Went to the store and bought some.&lt;/p&gt;&#xA;&lt;p&gt;Replaced the batteries and the toy won't start at all. Put the old back, toy worked so I checked what went wrong.&lt;/p&gt;&#xA;&lt;p&gt;Noticed the toy had Industrial Panasonic R6 batteries and I bought Normal Duracell LR6. So I have to questions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Are industrial batteries safe for children (I think yes their main purpose being stability afaik but I must ask :) - kind of like my kid)?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Considering the fact that both type of batteries are AA, both have same nominal voltage (1.5) and according to &lt;a href=&quot;https://en.wikipedia.org/wiki/AA_battery&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/AA_battery&lt;/a&gt; the LR6 has actually higher capacity... why is the toy not starting?&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11520" LastEditorUserId="110" LastEditDate="2021-03-06T19:09:30.763" LastActivityDate="2021-03-06T19:09:30.763" Title="Batteries AA - industrial R6 vs normal LR6" Tags="|electrical-engineering|battery|" AnswerCount="0" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="15582" PostTypeId="2" ParentId="15569" CreationDate="2017-05-29T13:59:15.860" Score="1" Body="&lt;p&gt;This question is rather confused, but I'll take it as &lt;i&gt;&quot;Can AC power be produced from DC power without mechanically moving parts&quot;&lt;/i&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Of course&lt;/b&gt;.  That's what devices called &quot;inverters&quot; do.  Most inverters are designed to work at a single frequency, but they certainly don't have to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of class D amplifiers as examples.  Clearly they can produce a wide range of frequencies.  In the case of amplifiers, the frequencies they put out are the frequencies in a small control signal.  However, that control signal could just as well be produced internally if you have a way of describing it, like a sine of a particular frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is not much difference between a class D amplifier and a modern sine wave inverter (except for resonant design, beyond the scope of this answer).  In the case of the inverter, the control signal is generated internally, but the power output electronics are quite similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another obvious existence proof of such things are uninterruptable power supplies (UPSs).  When the input power goes away, the power source is the internal battery.  That is inherently DC, yet the output power is AC at a regulated frequency and amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ye olde days, this kind of conversion was done with motor-generators.  There is still a niche for those today, although that niche is small and getting smaller as power electronics advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even utility-scale DC to AC conversion is done with solid state electronics nowadays.  For example, here is a picture from Google Earth of such a plant in Ayer Massachusetts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nGVlS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nGVlS.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DC power lines from dams nearly 1000 miles away in northern Quebec come in at top left.  The big square site at lower right converts that to AC at 60 Hz, and feeds it onto the large AC power lines shown horizontally at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transforming that much power at such high voltages and currents takes a large facility, but it is ultimately done with semiconductors.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2017-05-29T14:13:28.043" LastActivityDate="2017-05-29T14:13:28.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15583" PostTypeId="1" AcceptedAnswerId="15584" CreationDate="2017-05-29T14:26:28.013" Score="2" ViewCount="1620" Body="&lt;p&gt;In the &lt;strong&gt;real world&lt;/strong&gt;, would this pinned frame exhibit a deflection in the fixed, left hand side column?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arguments from peers largely surround:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) There &lt;strong&gt;wouldn't&lt;/strong&gt; be left-column horizontal deflection because the moment cannot apply through the pin to the column. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) There &lt;strong&gt;would&lt;/strong&gt; be left-column horizontal deflection because although the moment doesn't travel through the left-column pin, in reality the right side will sag and pull the left-column towards the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xjYd0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xjYd0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exceptional answers will explain if points 1) &amp;amp; 2) are correct; provides additional information we haven't considered; and the inclusion of visual diagrams if relevant. &lt;/p&gt;&#xA;" OwnerUserId="11523" LastEditorUserId="11523" LastEditDate="2017-05-29T17:01:09.647" LastActivityDate="2017-07-21T09:55:59.517" Title="Structural Frame Real World Deflection" Tags="|structural-engineering|deflection|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15584" PostTypeId="2" ParentId="15583" CreationDate="2017-05-29T14:59:20.890" Score="4" Body="&lt;p&gt;Yes the left column would deflect in the real world. The reasons are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Eccentricity of the connection / column:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality the load will not be transferred perfectly to the center of the column nor will the column be perfectly straight. This will cause a small amount of bending in the column. In most cases this will be extremely small. However, if the column is slender and the load is large this will cause buckling to occur significantly below the theoretical &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%27s_critical_load&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler buckling load&lt;/a&gt;. The variation in cross-section is accounted for in the building code material factors. Geometric imperfections must be explicitly checked. For example: Clause 5.2 in Eurocode 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The connection is not a true, friction-free hinge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture of a typical steel 'pinned' beam-column connection:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fjXLYm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fjXLYm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, this is not a friction free hinge and some bending will be transferred to the column. This bending transfer will be small assuming the deflections are small (which they typically are in structures). There is usually some play in the connection as well which will allow some movement. But the bending transfer will not be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) There is no such thing as a perfectly fixed support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foundation has some (non-infinite) stiffness, therefore the fixed support could rotate a small amount. An interesting exercise for the keen reader would be to consider a concentrated moment on an &lt;a href=&quot;https://www.brown.edu/Departments/Engineering/Courses/EN224/halfspace/halfspace.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;infinite half space&lt;/a&gt; made from sold steel (or another material) to get the foundation stiffness then see how this affects the deflection of a cantilever. But I digress...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, as with many things in engineering these deflections are very small in typical design situations and are therefore not usually explicitly calculated/considered. However, they do exist. We should not forget the idealisations we make for calculation convenience are simplifications which could have real consequences if not understood.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2017-05-29T15:11:08.207" LastActivityDate="2017-05-29T15:11:08.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15586" PostTypeId="2" ParentId="15583" CreationDate="2017-05-30T00:00:36.470" Score="0" Body="&lt;p&gt;Although all of mg4w's points are good, let me add another one: let's say you set up an experiment to determine experimentally whether there was a deflection or not. How would you know if there was a deflection or not?  You would have to use some measuring instrument, and that instrument must have a minimum range.  E.g. if you used a tape measure, you could detect motion of 1/4&quot;, but a deflection of 0.001&quot; might not be noticed.  With a dial indicator or LVDT, you could detect 0.001&quot;, but maybe not 1e-6 inches.  With a laser, you could probably measure 1e-6 inches, but not 1e-12. So experimentally in the real world, the question &quot;Is there a deflection?&quot; is ill-posed, because it cannot be answered definitely. If your instrument reads zero, it could be because there really is a deflection, but it's smaller than your instrument can read.  The best you can say is &quot;Is the deflection larger than X?&quot;, where X is the smallest resolution of your measurement device.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically though, all real pin joints will transmit some non-zero moment. There is always a non-zero coefficent of friction. It might be a very small amount of moment, but it is not zero.  Therefore, yes, there will be some deflection.  It is probably so small that you can ignore it for many practical purposes, but it is not zero.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-05-30T00:00:36.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15587" PostTypeId="1" CreationDate="2017-05-30T02:16:38.213" Score="4" ViewCount="5044" Body="&lt;p&gt;Is there a formula to calculate the moment of inertia of an unequal angle? I am not talking about x or y direction but the maximum and minimum moment of inertia about its rotated axes (usually labelled u and v) through its centroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can get this using Massprop function in AutoCAD, but I would like to have a formula so that I can use it in excel.&lt;/p&gt;&#xA;" OwnerUserId="11527" LastActivityDate="2017-06-29T08:06:02.717" Title="How to calculate principal moment of inertia of an unequal angle?" Tags="|civil-engineering|structural-analysis|steel|cross-section|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15588" PostTypeId="1" CreationDate="2017-05-30T05:23:19.390" Score="3" ViewCount="248" Body="&lt;p&gt;My first post here. I'm curious about how power is balanced in multiple diesel-electric locomotives. I understand to that two locos can be ganged in terms of controls, to set speed etc, so that one set of driver controls manages the whole pulling group. (sorry, I don't know the correct vernacular) How is it possible to ensure that each is pulling it's weight evenly compared to the others in the group. Or, put another way, how can I avoid one loco pulling at 5000Nm and a second pulling at 6000Nm. Is it possible to read the tow force in the couplings?&lt;/p&gt;&#xA;" OwnerUserId="11529" LastEditorUserId="11529" LastEditDate="2017-06-02T02:00:43.310" LastActivityDate="2022-11-17T12:30:25.370" Title="Balancing Locomotive Power" Tags="|electrical-engineering|rail|diesel|electrical|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15589" PostTypeId="2" ParentId="15587" CreationDate="2017-05-30T06:30:53.897" Score="0" Body="&lt;p&gt;In principle, it involves figuring out the orientation at which the product second moment of area is equal to zero. This orientation gives the principal second moments of area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This link explains more:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nguyen.hong.hai.free.fr/EBOOKS/SCIENCE%20AND%20ENGINEERING/MECANIQUE/MATERIAUX/Mechanics%20of%20Materials.rar_FILES/Mechanics%20of%20Materials/Volume%202/32666_01.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://nguyen.hong.hai.free.fr/EBOOKS/SCIENCE%20AND%20ENGINEERING/MECANIQUE/MATERIAUX/Mechanics%20of%20Materials.rar_FILES/Mechanics%20of%20Materials/Volume%202/32666_01.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11424" LastActivityDate="2017-05-30T06:30:53.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15590" PostTypeId="1" CreationDate="2017-05-30T06:52:16.400" Score="0" ViewCount="195" Body="&lt;p&gt;Can we see a nanowire, whose dia is say 1 nm but length is 10mm, with naked eye? To see an object does at least two out of three dimensions must be greater than the minimum size one's eye can see?&lt;/p&gt;&#xA;" OwnerUserId="11530" LastActivityDate="2017-05-30T15:47:52.553" Title="Can we see a nanowire?" Tags="|optics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15591" PostTypeId="2" ParentId="6108" CreationDate="2017-05-30T07:20:15.247" Score="0" Body="&lt;p&gt;A two phase chemistry that makes pressure using heat is necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pressure cooker with only air makes a lot less pressure than with a liter of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water is in effect potential pressure stored in a cold state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supercritical fluids are actually more efficient than steam but require higher pressure vessels, and a lot of ice CO2. and other exotic substances.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2017-05-30T07:20:15.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15592" PostTypeId="2" ParentId="14506" CreationDate="2017-05-30T07:32:46.557" Score="0" Body="&lt;p&gt;that's probably just a milky plastic sheet called a led diffuser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The led has a directionality, which spreads light over an angle by distance, it can be guided with optical guides, mirrors, reflectors. grooved notched transparent sheets than deflect light into an acrylic plate, and then diffused using a led diffuser. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.fr/search?q=led+diffuser&amp;amp;num=100&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwiHxsXbi5fUAhVDNhoKHdTKAJ4Q_AUICigB&amp;amp;biw=1004&amp;amp;bih=606&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.fr/search?q=led+diffuser&amp;amp;num=100&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwiHxsXbi5fUAhVDNhoKHdTKAJ4Q_AUICigB&amp;amp;biw=1004&amp;amp;bih=606&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2017-05-30T07:32:46.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15593" PostTypeId="1" CreationDate="2017-05-30T09:14:43.853" Score="0" ViewCount="102" Body="&lt;p&gt;I want to build a small 1x1x1 m insulated (wooden) box, with the purpose of testing an internal coating (the idea is to buy 5 square meters of it instead of 60 for a full scale room). My idea was to place an electrical heating element and a small fan inside, and raise the temperature from approximately 20 to 40 degrees Celsius. Given the air conditions inside, I used the Carrier diagram to find the thermal power I would need to heat up the humid air volume in 0.5 hours. I then tried to consider transient heat losses through the walls of the chamber, of which I know the transmittance: to do so, given the relatively small temperature differences between inside (from 20 to 40 C) and outside (20 C), I figured I would reduce it to a steady state case with a constant internal temperature of 30 C (arithmetic mean between initial and final temperature).&#xA;The thermal power lost through the walls and calculated this way is then, in my ideas, to be summed to the power needed to heat up the air volume, thus obtaining the power I need from the heating element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Is that of the average temperature a good assumption?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- Apart from that, am I doing anything wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reckon this is quite basic stuff, I'm sorry, but I never faced this kind of issues and second opinions would be of great help.&lt;/p&gt;&#xA;" OwnerUserId="11531" LastActivityDate="2017-07-19T22:29:23.207" Title="transient heat losses approximation, closed chamber" Tags="|mechanical-engineering|heat-transfer|thermal-insulation|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15594" PostTypeId="2" ParentId="5755" CreationDate="2017-05-30T09:48:59.580" Score="1" Body="&lt;p&gt;cartesian 2 axis robot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 axis robot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cartesian coordinate robot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;xz axis robot&lt;/p&gt;&#xA;&#xA;&lt;p&gt;xz gantry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your purposes, compare the required precision, the load weight, and the control board and display, pc link, if it uses GCode or some other control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this kind of cartesian robot system has been popularized by 3d printers and there are masses of fan info's for steppers and gantry's and 3d printer related control displays.&lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2017-05-30T09:48:59.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15595" PostTypeId="2" ParentId="15573" CreationDate="2017-05-30T10:26:29.383" Score="1" Body="&lt;p&gt;They lift it up in pieces and assemble the crane in place using a second mobile crane they bring in to build it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heaviest parts will be the winch assembly and the counter weights.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-05-30T10:26:29.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15596" PostTypeId="2" ParentId="15590" CreationDate="2017-05-30T10:51:28.597" Score="2" Body="&lt;p&gt;no, a nanometer is much smaller than the wavelength of light, so you can not even see it with an optical microscope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be worth a note the &quot;nano-&quot; prefix is often abused in the literature. Things described as &quot;nano&quot; are often &quot;really&quot; micron sized. Additionally such small stuff often exists as agglomerates/mat/powder , etc. which then of course you can see if there is enough of it&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2017-05-30T15:47:52.553" LastActivityDate="2017-05-30T15:47:52.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15597" PostTypeId="2" ParentId="15590" CreationDate="2017-05-30T11:55:20.420" Score="0" Body="&lt;p&gt;No.  Absolutely not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;i&gt;Think&lt;/i&gt; about it.  Visible light has wavelengths roughly from 400 to 700 nm.  You are talking about feature sizes many times smaller than the wavelength of anything your eyes can see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that, you certainly can't see things that small with a &lt;i&gt;naked&lt;/i&gt;, as you asked about.  Again, actually &lt;i&gt;think&lt;/i&gt; about it.  If you could see things even remotely that small with a naked eye, there would be no need for optical microscopes.  Of course anything using visible light, which includes optical microscopes, are limited by the wavelength of light they use.  Even the best optical microscope can't resolve 10 nm features, not even close.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-05-30T11:55:20.420" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15598" PostTypeId="2" ParentId="7842" CreationDate="2017-05-30T12:47:59.980" Score="1" Body="&lt;p&gt;In a clutch, the torque transmitted can be calculated by integrating the torque contribution of the shear pressure over the contact area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau=\int \sigma\,r \; dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear pressure will be equal to the friction coefficient multiplied by the contact pressure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma=\mu\,p$$&#xA;$$\tau=\int \mu\,p\,r \; dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;usually the contact pressure is considered constant, as areas of higher pressure tend to wear faster, reducing the pressure. Coefficient of friction is also usually considered constant. Thus those can be brought outside the integral:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau=\mu\,p \int r \; dA$$&#xA;$$\tau=\mu\,\frac{F}{A} \int r \; dA$$&#xA;$$\tau=\mu\,F\frac{\int r \; dA}{A} $$&#xA;So it's this mean radius that indeed you care about:&#xA;$$\frac{\int r \; dA}{A} $$&#xA;For a solid circle this gives:&#xA;$$\frac{\iint r\,d\theta\,dr}{\iint 1\,d\theta\,dr}$$&#xA;$$\frac{\int 2\,\pi\,r^2\,dr}{\int 2\,\pi\,r\,dr}$$&#xA;$$\frac{\frac23 \pi\,r^3}{\pi\,r^2}$$&#xA;$$\frac23 r$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a ring with outside radius $R_2$ and inside radius $R_1$ then you can just subtract the integrals on top and bottom:&#xA;$$\frac{\iint r\,d\theta\,dr_2 - \iint r\,d\theta\,dr_1}{\iint 1\,d\theta\,dr-\iint 1\,d\theta\,dr_1}$$&#xA;$$\frac{\frac23 \,\pi\, {r_2}^3 - \frac23 \,\pi\, {r_1}^3}{\pi\, {r_2}^2 - \pi\, {r_1}^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac23\left(r_1+r_2-\frac{r_1\,r_2}{r_1+r_2}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is close to the commonly sighted approximation $\frac{r_1+r_2}2$:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0p74S.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0p74S.png&quot; alt=&quot;Comparing mean effective radii graphs &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This graph assumes an outside radius of 1 and plots the mean effective radius varying the inner radius. It shows that the approximation is close when the mean radius is more than about half the size of the outer radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like to account for holes in your clutch plate you can use a similar method to the above. However, given most people just use the mean of the radii approximation, I don't think the added accuracy is likely necessary. &lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2017-05-30T12:47:59.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15599" PostTypeId="2" ParentId="13291" CreationDate="2017-05-30T13:27:58.433" Score="1" Body="&lt;p&gt;Your equations are mostly correct, but you're missing the tangential load on the actuator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{array}{l}&#xA;{F_a}_p  &amp;amp; = &amp;amp; \text{Force parallel to the actuator arm at the joint} \\&#xA;{F_a}_t  &amp;amp; = &amp;amp; \text{Force tangent to the actuator arm movement at the joint} \\&#xA;F_p  &amp;amp; = &amp;amp; \text{Force parallel to the rod} \\&#xA;F_t  &amp;amp; = &amp;amp; \text{Force perpendicular to the rod} &#xA;\end{array} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_p={F_a}_{p} \cos(\theta_m - \theta_r) - {F_a}_{t}sin(\theta_m - \theta_r)$$&#xA;$$0=F_t={F_a}_{p} \sin(\theta_m - \theta_r) + {F_a}_{t}cos(\theta_m - \theta_r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These equations can be rearranged to give:&#xA;$${F_a}_p=F_p \cos(\theta_m - \theta_r)$$&#xA;$${F_a}_t=F_p \sin(\theta_m - \theta_r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or rearranged again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_p = {F_a}_t \csc(\theta_m - \theta_r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does indeed give you a multiplying effect as the difference in angles approaches 0.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="1122" LastEditDate="2017-05-31T16:17:56.620" LastActivityDate="2017-05-31T16:17:56.620" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15601" PostTypeId="1" CreationDate="2017-05-30T16:20:18.030" Score="1" ViewCount="50" Body="&lt;p&gt;I have a centrifugal pump (Q: 64 m3/hr, H: 30 m) connected to a balance tank but it can't provide enough flow for my purpose. One quick option is to add a firefighting truck's piston pump to the circuit. I'm not sure about the features of the pump but I'm sure it delivers more flow and pressure than the centrifugal pump. My question is, do I have to worry about back flow when I'm not using none-return valves? &lt;/p&gt;&#xA;" OwnerUserId="11538" LastActivityDate="2017-09-28T15:39:24.023" Title="Setting up a centrifugal pump with a piston pump in parallel" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15602" PostTypeId="2" ParentId="15601" CreationDate="2017-05-30T16:42:06.863" Score="0" Body="&lt;p&gt;Yes, you would - the more powerful pump will discharge or stall it through the other wasting energy and time.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-05-30T16:59:53.243" LastActivityDate="2017-05-30T16:59:53.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15603" PostTypeId="1" CreationDate="2017-05-30T17:45:43.630" Score="2" ViewCount="541" Body="&lt;p&gt;I am a student who is interning at a company this summer, and I've been put in charge of its first R&amp;amp;D project which will require me to collect a lot of data.  The machine with which I am working is called a Tunnel Finisher.  It is essentially a long tunnel through which clothes are pulled in order to dry them and remove wrinkles after a wash cycle.  Hot air is blown down on the clothes through louvers from a shaft above the tunnel which leads from a fan that maintains the flow of air.  What I would like to do is measure air flow, above the clothes, at the level of the clothes, and below the clothes, at multiple points in the tunnel.  Because the air in the tunnel is hot and clothes move through it, this measurement cannot be done manually; I need sensors that can be placed in the tunnel, preferably stuck to the walls, that will provide data while the machine is operating normally.  The goal of this endeavor is to reveal dead spots of significantly less air flow in the machine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked at some different kinds of sensors, but I don't really know what I'm looking for, and I assume what I end up using shouldn't be too expensive.  I was hoping someone here could provide some insight as to which air flow sensors, or other type of sensor, I should be looking at for this purpose.&lt;/p&gt;&#xA;" OwnerUserId="11540" LastActivityDate="2017-10-28T18:56:24.697" Title="How to Measure Air Flow in a Large Tunnel at Various Different Points?" Tags="|mechanical-engineering|sensors|airflow|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15605" PostTypeId="2" ParentId="7989" CreationDate="2017-05-30T17:57:27.377" Score="2" Body="&lt;p&gt;You can calculate the shear stiffness directly and then get an effective area after the fact if you'd like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{shear}=\frac{V}{\gamma_{ave}\,G}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate average shear by averaging over the height (not area):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gamma_{ave}=\frac1h \int \gamma\;dz$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate shear strain from the shear stress:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\gamma=\frac{\tau}G$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And shear stress from shear flow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau=\frac{g}t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$g(z)=\frac{V\,Q(z)}{I}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now a lot of the variables cancel if you plug in at this point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{shear}=\frac{h\,I}{\int \frac{Q}{t}\;dz}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a beam with shear center in the middle (for example symmetric):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_{shear}=\frac{h\,\int^\frac{h}2_{-\frac{h}2}t\,z^2 dz}{\int^\frac{h}2_{-\frac{h}2} \frac{\int^\frac{h}2_zt\,z\; dz}{t}\;dz}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That will give you the effective shear area for any symmetric beam given the thickness/width as a function of z.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2017-05-30T17:57:27.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15607" PostTypeId="2" ParentId="15603" CreationDate="2017-05-30T18:43:57.037" Score="0" Body="&lt;p&gt;How about a pitot tube that you can draw across the width of the tunnel - had an experiment where the velocity profile was measured before and after a fan - smaller scale (pipe was 6 inches in diameter) but you could even have one tube that moves or make a &quot;comb&quot; of 20 or 30 which would then be stationary.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-05-30T18:43:57.037" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15608" PostTypeId="1" AcceptedAnswerId="15609" CreationDate="2017-05-30T21:40:41.113" Score="4" ViewCount="7868" Body="&lt;p&gt;Our small radio-control model cars have shims, spacers and washers. I am tasked with clarifying the manual and getting parts naming consistent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have these items from less than .5mm thin up to 1/8&quot; thick. I'm thinking of calling everything under .5mm a shim, and everything .5mm and thicker that is available off the shelf a washer. And spacers will be anything .5mm and thicker that we have manufactured ourselves for specific needs on the car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do engineers define the difference between these terms?&lt;/p&gt;&#xA;" OwnerUserId="11288" LastEditorUserId="11288" LastEditDate="2017-05-31T20:36:59.943" LastActivityDate="2018-10-23T09:32:13.440" Title="How do we specify among a shim, spacer, and washer?" Tags="|mechanical-engineering|computer-hardware|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15609" PostTypeId="2" ParentId="15608" CreationDate="2017-05-30T21:58:48.770" Score="7" Body="&lt;p&gt;The three terms specify the function of the component, not its size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;shim&lt;/strong&gt; is a component used to fill up an unwanted gap. Shims may be flat or tapered, and are often specified as &quot;select to fit&quot; parts, rather than as a part with a predefined thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;spacer&lt;/strong&gt; or &lt;strong&gt;standoff&lt;/strong&gt; is a fixed size component designed to separate two other parts by a known amount. It does not necessarily fill &lt;em&gt;all&lt;/em&gt; the space between the components - for example the purpose might be to separate them for electrical insulation, or to allow cooling air to flow between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shims are usually fairly thin, but spacers can be thicker (inches or even several feet thick, in large structures). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;washer&lt;/strong&gt; is a part used to spread the load between the two components, e.g., between a bolt- or screw-head and whatever the bolt or screw is fixing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Washers are not necessarily flat - for example, Belleville washers, split washers, star washers, wave washers, etc. The purpose of a non-flat washer is often to create a well-defined force between the parts on either side when it is installed by squashing it flat. That force may be used to resist unwanted rotation of a screw or bolt if the structure is vibrating, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A particular part may have more than one of these functions - for example a shim or spacer may also act as a flat washer. In that case, the naming of a particular part is a bit arbitrary, or a pedantic name like &quot;spacer-washer&quot; might be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All this might lead to some problem situations. For example if you are selling a self-build kit, you might be including a packet of &quot;20 washers&quot; (all the same size and bought pre-packed from a hardware supplier) but in fact some are used in the kit as washers, some as spacers, and some as shims. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you deal that might depend on what engineering knowledge you expect your customers to have. An experienced engineer might prefer to see the parts named &quot;correctly&quot; on an assembly drawing which follows international standard drawing conventions, while a beginner might prefer a more pictorial sketch which is mainly intended to show &quot;what part goes where&quot; when building the kit.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-05-30T22:15:23.157" LastActivityDate="2017-05-30T22:15:23.157" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="15612" PostTypeId="1" AcceptedAnswerId="15623" CreationDate="2017-05-31T08:01:01.820" Score="1" ViewCount="200" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zYDFY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zYDFY.png&quot; alt=&quot;system&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I need to model the system in figure, where turbine, shaft1 and gear1 are a unique rigid body, while shaft2 can be modeled like a torsional spring. Using as coordinates Theta1, clockwise angular displacement of the turbine, and Theta2, anticlockwise displacement of the generator, i obtain these equations of motion&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vpnje.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vpnje.gif&quot; alt=&quot;differential equation&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;where tau is the transmission ratio of the gears and J0 the equivalent moment of inertia of the turbine-shaft-gears block.&lt;/p&gt;&#xA;&lt;p&gt;The stiffness matrix is not symmetric and it should be impossible, where am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-05-31T23:37:30.047" Title="Asymettric stiffness matrix for a generator-gearbox-turbine system" Tags="|mechanical-engineering|stiffness|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15613" PostTypeId="1" CreationDate="2017-05-31T09:04:47.033" Score="1" ViewCount="48" Body="&lt;p&gt;I would like to design a buck converter to use a PV panel to charge a &lt;a href=&quot;https://www.amazon.co.uk/FLOUREON-3000mAh-Rechargeable-Airplane-Helicopter/dp/B01ITCMZ44&quot; rel=&quot;nofollow noreferrer&quot;&gt;11.1V 3000mAh 3S 30C Lipo battery&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to choose the minimum inductance for a CCM ($L=\dfrac{V_{out}T_s}{2I_{out}}$), what is the output current that I need to charge this battery? should $V_{out}=11.1\text{ V}$ and $I_{out}=3000\text{ mAh}$?&lt;/p&gt;&#xA;" OwnerUserId="11555" LastEditorUserId="1832" LastEditDate="2017-05-31T11:17:36.560" LastActivityDate="2017-05-31T11:17:36.560" Title="Charging a Li-po battery (battery output current)" Tags="|electrical-engineering|power-electronics|power|battery|solar-energy|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15615" PostTypeId="2" ParentId="13495" CreationDate="2017-05-31T12:32:58.077" Score="1" Body="&lt;p&gt;You may come up with various equations that will suggest various more or less (usually more) inaccurate value because the number of obscure factors, like sawdust grain shape, its layout, porosity, ambient air humidity and so on, will make any calculation such a distant approximation it will be useless. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, take a 20x20cm sheet of the same kind that's to be used, with air vent that can be partially closed, sprinkle 21 gram of the same wet sawdust that's to be dried on 3/4 of its surface, set vent to half diameter, blast air or 2 hours, check if sawdust is dry. If not, open vent more. If yes, close it more. Repeat as needed - in 3-4 iterations you'll have a throughput that gets the work done in desired time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then measure the air flow (even as simple as known volume plastic bag, check time needed to inflate) and multiply by 10,000. The result may be ~10-20% away from optimal, but certainly better than a simplistic equation would yield.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-05-31T12:32:58.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15616" PostTypeId="2" ParentId="15049" CreationDate="2017-05-31T13:23:34.120" Score="2" Body="&lt;p&gt;That's why it's a STANDARD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using reasonably available tools you can measure part X, round the results to nearest standard-allowed values, and pick a matching part from catalogue or manufacture following a simple set of standard-defined parameters, and it will fit. With weirdo sizes you have a weirdo system where every element needs to be custom-calculated and custom-made because it fits nothing in the world except what it was made for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's an arbitrary restriction to curb anarchy of a billion custom-purpose standards that don't match each other, allow various manufacturers to provide standarized parts that match each other and fulfill all reasonable expectations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the diametral pitch doesn't have to be integer. Engineers will hate you for it, and people will criticize you for vendor lock-in practices, plus you'll need to manufacture all the gears on your own, but there's no law (legal, or of physics) that forbids it.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-05-31T13:23:34.120" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15617" PostTypeId="2" ParentId="13473" CreationDate="2017-05-31T13:46:25.237" Score="3" Body="&lt;p&gt;The amount of useful work you can extract out of gas  in an engine is proportional to its stagnation pressure. Energy and be stored as thermal, kinetic, or as pressure. The stagnation pressure is the pressure the gas would be at if all the kinetic energy was converted to thermal and pressure storage with 100% efficiency ie. isentropic process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To generate thrust you want the highest exit velocity, which means you want to convert as much energy as possible from pressure and temperature. This processes is limited by atmospheric pressure (or the size of your nozzle). The pressure of your exit gas can't drop below atmospheric or it will stop exhausting/ there will be a shock to build pressure again so you can keep exhausting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you want to keep your static pressure as high as possible to get the largest transition to atmospheric, giving you the largest velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A shockwave, while it &lt;em&gt;increases&lt;/em&gt; static pressure, &lt;em&gt;decreases&lt;/em&gt; stagnation pressure. The discrepancy comes from an irreversible transfer of kinetic energy to thermal energy. While a shock wave increases pressure, it doesn't increase it as much as a smooth isentropic transition to the same velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some relevant equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between static pressure, pressure, and Mach number during an isentropic process:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{p}{p_0}=\left(1+\frac{\gamma-1}2M^2\right)^{\frac{-\gamma}{\gamma-1}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increase in pressure vs. strength of shock:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{p_2}{p_1} = \frac{2\gamma M_1^2}{\gamma + 1} - \frac{\gamma - 1}{\gamma + 1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Decrease in static pressure vs. strength of shock:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{p_{02}}{p_{01}} = \left(\frac{\frac{\gamma + 1}{2}M_1^2}{1 + \frac{\gamma - 1}{2}M_1^2}\right)^\frac{\gamma}{\gamma - 1}\left(\frac{1}{\frac{2\gamma}{\gamma + 1}M_1^2 - \frac{\gamma - 1}{\gamma + 1}}\right)^\frac{1}{\gamma - 1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.grc.nasa.gov/www/K-12/airplane/isentrop.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Isentropic equations are from Nasa&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Normal_shock_tables&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shock equations are from Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="18837" LastEditDate="2019-01-03T20:17:27.587" LastActivityDate="2019-01-03T20:17:27.587" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15618" PostTypeId="2" ParentId="2405" CreationDate="2017-05-31T14:04:03.637" Score="0" Body="&lt;p&gt;Pressure is the average outward force that molecules exert on their surrounding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take the case of air molecules bouncing around hitting everything they push outwards to the sides equally, but as you mentioned their weight means that they do indeed push downward harder than they push upwards. As the weight of air in a small space is very small, this difference can usually be neglected. However, without this difference balloons wouldn't float. This tiny difference adds up in the atmosphere till the pressures down here on the surface are actually pretty significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the cars would squish you is that when the car pushes down with a high pressure, it will move your surface inward until you push back with the same pressure. Unfortunately for you, as your internal pressure increases that makes your sides at higher pressure than the air around you, so your sides squish out since the air doesn't push back as hard. So, it's not enough to just be pushed from the top and bottom, or even from four sides. You have to be pushed from all directions, including up your nose and inside your lungs in order for your internal pressure to be able to comfortably push back against the high pressure.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2017-05-31T14:04:03.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15620" PostTypeId="2" ParentId="2405" CreationDate="2017-05-31T16:13:24.480" Score="0" Body="&lt;p&gt;Both statements are correct. The best way to understand how these two statements can coexist is to understand the concept of a gas pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to understand pressure we look at a container full of gas molecules. Gas molecules do not behave at all like solids or liquids. In a gas the molecules are not attracted to one another so they fly around at extreme speeds bouncing into objects and other gas molecules. These collisions are &lt;a href=&quot;https://en.wikipedia.org/wiki/Elastic_collision&quot; rel=&quot;nofollow noreferrer&quot;&gt;elastic&lt;/a&gt; so no energy is lost during collisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lclsZ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lclsZ.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time a collision occurs some kind of energy transfer takes place between molecules. However, on a macroscopic level, there are so many collision taking place that they average out to zero energy transferred. Imagine that a gas molecule is about to hit the wall off the above container. We know that when the molecule hits it will bounce off and head in the other direction just like a bouncy ball. The wall will also feel a force due to &#xA;&lt;a href=&quot;http://www.physicsclassroom.com/class/newtlaws/Lesson-3/Newton-s-Second-Law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton's second law&lt;/a&gt;. However on the other side of the container the exact same thing is happening. In fact the same thing is happening on the outside of the container as well. All these collisions exert a force but they all cancel each other out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let us apply this to your first deffinition. As you stated air pressure is caused by the weight of the air molecules above. Gas molecules are attracted by gravity towards the earths surface. As a gas molecule is pulled toward the surface of the earth chances are it will hit another gas molecule and bounce off of it in another direction. Now lets say that in this particular collision the first molecule hits the top of the second molecule. This causes the second molecule to travel downwards even faster than the first molecule. This happens again and again until the molecule bounces of the surface of the earth. This is how your first definition is derived. The key is to remember that this is a gas pressure and thus is from all sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the hardest concept to grasp because when someone hears that there are hundreds of pounds of air above them they imagine hundreds of pounds of steel plates on their shoulders. Do not think of it like that. If a bouncy ball falls on your head it pushes you down. However if it misses, hits the floor, bounces up and hits you the two forces cancel each other out. The trick is to realize that so many collisions are occurring at such a minute scale that you do not &quot;feel&quot; the pressure of the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid objects are very good at resisting an even force from all directions. Have you ever heard that you can't crush an egg if you squeeze it from all directions? The same concept applies to to your body. The atmosphere is pushing very hard from all directions (even from inside your lungs!) but they all cancel out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To contrast this imagine a steel drum with just a few gas molecules in it what would happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ba5t6.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ba5t6.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now despite this being cool notice that the sides of the barrel collapse as well. This means that air molecules were pushing from the side but there was nothing to push back from the inside. We can see from the imploding barrel that the atmosphere is compressing us with enough force to crumple a steel drum. However because this pressure is exerted from every direction the forces are cancel out and we don't feel a thing.&lt;/p&gt;&#xA;" OwnerUserId="9656" LastActivityDate="2017-05-31T16:13:24.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15621" PostTypeId="2" ParentId="12399" CreationDate="2017-05-31T17:55:01.350" Score="2" Body="&lt;p&gt;Components have connections that can be either inputs or outputs, while blocks have defined inputs and defined outputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the components you can have bidirectional causality while if you want to do something similar with blocks you have to figure out which way the causality will flow and explicitly code it that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least that's my very very limited understanding of it.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2017-05-31T17:55:01.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15622" PostTypeId="1" CreationDate="2017-05-31T18:27:53.903" Score="1" ViewCount="288" Body="&lt;p&gt;I have a fresh feed (stream 1) entering a reactor . This fresh feed consists 225kg/h of iso-butane , 225kg/h of butylenes and &#xA;50kg/h of n-butane . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the stream table I need to add in a flowrate value . How do I calculate this ? Do I take the average ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also the composition (wt%) how do I calculate them ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I calculate the percentage ? &#xA;This means iso-butane and butylenes have 45% each and 10% for the n-butane in the mixed fresh feed .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a question I have to solve . I'm not given wt% . So can I calculate the wt% by knowing the flowrate of all the products in the fresh feed ? &lt;/p&gt;&#xA;" OwnerUserId="11561" LastActivityDate="2017-05-31T21:12:01.243" Title="Flow rate into a reactor from feed" Tags="|chemical-engineering|flow-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15623" PostTypeId="2" ParentId="15612" CreationDate="2017-05-31T19:22:36.940" Score="1" Body="&lt;p&gt;One way to get this right is to start by ignoring the gearbox and modelling the system with three DOFs, i.e. the angular position of Shaft 1, and the positions of the two ends of Shaft 2. That gives the equations of motion as&#xA;$$\begin{bmatrix}J_0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; J_1 \end{bmatrix}&#xA;\begin{bmatrix} \ddot\theta_1 \\ \ddot\theta_g \\ \ddot\theta_2 \end{bmatrix} + &#xA;\begin{bmatrix} 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; k_{t2} &amp;amp; -k_{t2} \\ 0 &amp;amp; -k_{t2} &amp;amp; k_{t2} \end{bmatrix} &#xA;\begin{bmatrix} \theta_1 \\ \theta_g \\ \theta_2 \end{bmatrix} = 0$$ &#xA;or $$M_{3\times3}\,\ddot\theta_{3\times3} + K_{3\times3}\,\theta_{3\times3} = 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, apply the gearbox constraint, i.e. $\theta_g = -\tau\theta_1$. That is equivalent to the transformation matrix &#xA;$$\begin{bmatrix} \theta_1 \\ \theta_g  \\ \theta_2 \end{bmatrix} = &#xA;\begin{bmatrix} 1 &amp;amp; 0 \\ -\tau &amp;amp; 0 \\ 0 &amp;amp; 1 \end{bmatrix} &#xA;\begin{bmatrix} \theta_1 \\ \theta_2 \end{bmatrix}$$ &#xA;or &#xA;$$\theta_{3\times3} = L\,\theta_{2\times2}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the $2\times2$ stiffness and mass matrices are &#xA;$$\begin{align}&#xA;M_{2\times2} &amp;amp;= L^T M_{3\times 3}\, L \\&#xA;K_{2\times2} &amp;amp;= L^T K_{3\times 3}\, L &#xA;\end{align}$$&#xA;which are symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following a systematic procedure might seem more longer than trying to write the answer down &quot;by inspection,&quot; but it's more likely to avoid mistakes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one of the diagonal terms in the &lt;em&gt;correct&lt;/em&gt; stiffness matrix is $\tau^2 k_{t2}$, which doesn't appear anywhere in the OP's equations of motion. That suggests the OP's mistake was getting confused about the units being used to measure the various torques and rotations in the system in terms of $\theta_1$, $\theta_2$, and $\tau$.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-05-31T23:37:30.047" LastActivityDate="2017-05-31T23:37:30.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15624" PostTypeId="2" ParentId="15622" CreationDate="2017-05-31T21:12:01.243" Score="1" Body="&lt;p&gt;Yes, weight percent of a component is the mass (flow rate) of that component divided by the total mass (total flow rate). Total flow rate is equal to the sum of the flowrates of every component.&lt;/p&gt;&#xA;" OwnerUserId="10955" LastActivityDate="2017-05-31T21:12:01.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15625" PostTypeId="1" AcceptedAnswerId="15634" CreationDate="2017-06-01T01:05:22.337" Score="4" ViewCount="1962" Body="&lt;p&gt;Basic Googling on the mechanical generation of oscillating motion (moving something back and forth in the manner of a windshield wiper or a flapping wing) from a source of circular motion comes up with a concept known as a &quot;four-bar linkage&quot; or &quot;crank rocker&quot;.  For example, this video shows very clearly a simple way to create a windshield-wiper motion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/4tIo3AQQiU8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Crank-Rocker Four-Bar Linkage&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The problem is, &lt;em&gt;this doesn't look anything like a windshield wiper.&lt;/em&gt;  It has a bar attached to the tip of the &quot;wiper&quot; pulling it back and forth along its arc.  But in an actual windshield wiper (or wing, for that matter) the moving force comes from the &lt;em&gt;base,&lt;/em&gt; not the tip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you go about generating oscillating motion from the base of the oscillating device, windshield-wiper style?  Assume you're starting with a motor providing circular motion, as in the video.&lt;/p&gt;&#xA;" OwnerUserId="2181" LastEditorUserId="8055" LastEditDate="2017-06-02T11:54:22.090" LastActivityDate="2018-04-17T13:17:18.747" Title="How does a windshield wiper work?" Tags="|motors|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15626" PostTypeId="1" CreationDate="2017-06-01T02:16:32.573" Score="0" ViewCount="220" Body="&lt;p&gt;I'm not sure if this is the right place to ask, but I'm curious about what contributes to the asymmetry of force-distance graphs provided by keyboard switch manufacturers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, it is common to see graphs that look like this online:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/oSrrR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accompanied by gifs of the mechanism like this:&#xA;&lt;img src=&quot;https://www.pcper.com/files/fixed/MX-Red.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism here and what I've seen of actual keys look symmetrical in that there's no reason for any difference between pressing the key and the return of the key. So why is there the reduced amount of force on the return?&lt;/p&gt;&#xA;" OwnerUserId="11563" LastEditorUserId="6767" LastEditDate="2019-01-08T19:29:31.223" LastActivityDate="2019-01-08T19:29:31.223" Title="How do mechanical keyboard switch mechanisms relate to the force graphs?" Tags="|computer-hardware|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="15628" PostTypeId="1" AcceptedAnswerId="15707" CreationDate="2017-06-01T03:16:49.607" Score="0" ViewCount="214" Body="&lt;p&gt;This is a follow on from my previous question, &lt;a href=&quot;https://engineering.stackexchange.com/questions/13471/will-standard-m5-t-slot-nuts-fit-these-different-2020-aluminium-extrusions&quot;&gt;Will standard M5 T-slot nuts fit these different 2020 aluminium extrusions?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extrusion has this profile:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MoVBE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MoVBE.jpg&quot; alt=&quot;2020 Aluminium Extrusion BKK#1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here is a more in-focus photo, of four, in a 2 x 2 configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vzce3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vzce3.jpg&quot; alt=&quot;2020 Aluminium Extrusion BKK#2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gap, between the bottom of the flange and the top of the circle, is about 4.4 mm high. The M5 hexagonal nuts are 3.85 mm thick, and 7.85 mm wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been using these 2020 extrusions with regular M5 hexagonal nuts (and M5 bolts, obviously).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have recently found a need to use M3 bolts as well (to secure PCB mounted mechanical switches and endstop holders to the extrusion, for a 3D printer, see &lt;a href=&quot;https://3dprinting.stackexchange.com/questions/4173/complete-list-of-fasteners-for-the-sintron-kossel&quot;&gt;Complete list of fasteners for the Sintron Kossel&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The width of the slot is 5.4&amp;nbsp;mm and regular hexagonal M3 nuts are also around the same width (5.38 mm), and hence slip out of the slot. I guess that I &lt;em&gt;could&lt;/em&gt; use a wider washer to hold the nut in place, but does anyone know of a &quot;fat&quot; M3 nut that is available, preferably one that is the same size as an M5 nut, but with a smaller M3 bore?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem has arisen because I should be using &quot;European&quot; 2020 aluminium, which takes T-nuts, and those come in a range of bores (M3-M6).  However, I have been unable to obtain any, in Bangkok.&lt;/p&gt;&#xA;" OwnerUserId="5899" LastEditorUserId="5899" LastEditDate="2017-06-01T06:50:46.050" LastActivityDate="2017-06-06T05:22:58.073" Title="Fat hexagonal M3 nuts, with outer diameter greater than 6 mm" Tags="|aluminum|fasteners|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15630" PostTypeId="2" ParentId="15625" CreationDate="2017-06-01T05:31:49.340" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tA65X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tA65X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;You have to imagine an extra bar on the right hand side. Well 2 actually, one connecting to the swinging pivot joint that is depicted with the coloured arc and its length is the distance between the wipers then the second bar which will be attached to another fixed point. This means you have 2 fixed points, 2 swinging arms, 2 linking bars and 1 rotating bar . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wipers, in a car, are attached to the swinging arms (the one shown with the coloured arc) and extended past the joint with the linking bar as necessary.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-06-01T05:40:09.367" LastActivityDate="2017-06-01T05:40:09.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15632" PostTypeId="2" ParentId="15628" CreationDate="2017-06-01T06:49:37.360" Score="0" Body="&lt;p&gt;The solution would appear to be to use flanged hexagonal nuts&lt;sup&gt;1&lt;/sup&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OdBbX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OdBbX.jpg&quot; alt=&quot;Dimensions of flanged hexagonal nut&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The height, or thickness, &lt;em&gt;m&lt;/em&gt;, of the flanged M3 nut is slightly less than that of an M5 nut and the width, &lt;em&gt;d&lt;sub&gt;c&lt;/sub&gt;&lt;/em&gt;, is comparable to that of an M5 nut.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; However, whilst they fit in the slot, being able to tighten them is another matter altogether, and they will just spin around, as they have no grip on the sides of the slot...&lt;/p&gt;&#xA;" OwnerUserId="5899" LastEditorUserId="5899" LastEditDate="2017-06-01T08:33:07.567" LastActivityDate="2017-06-01T08:33:07.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15634" PostTypeId="2" ParentId="15625" CreationDate="2017-06-01T08:58:25.843" Score="6" Body="&lt;p&gt;For having dismantled a wiper motor once,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic principle is exactly as shown on the video, but the wiper is then linked to the to the rotation point of the rocking bar and not to the bar itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a schematic on how it works:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ucFlZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ucFlZ.png&quot; alt=&quot;Wiper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this system is quite standard, but there are many type of wipers and I am quite sure that a few constructors have also comes with interesting alternative mechanics.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="10684" LastEditDate="2017-06-01T09:19:52.790" LastActivityDate="2017-06-01T09:19:52.790" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15636" PostTypeId="1" CreationDate="2017-06-01T10:30:35.467" Score="1" ViewCount="182" Body="&lt;p&gt;My question is to calculate the electric field of a point in the Z axis of a empty hemisphere with a superficial density $\sigma$, but I have to do it from the formula of the electric field of a ring: $\vec{E}= \frac{\rho_l z a}{2 \epsilon_0 (z^2+a^2)} \hat{k}$, where $\rho_l$ is the lineal density, $z$ the z coordinate, $a$ the radius of the ring and $\epsilon_0$ is the electric constant. I have tried to express $\rho_l=\frac{dq}{dl}=\frac{\sigma dA}{dl}=\sigma dz'$ but if I change $a$ with $z'$ (because the radium of each ring changes) and I do the integral, I don't get the right answer. I know that I'm doing something wrong, and I'd appreciate if someone can help me.&lt;/p&gt;&#xA;" OwnerUserId="11568" LastEditorUserId="11568" LastEditDate="2017-06-02T07:42:39.580" LastActivityDate="2017-06-02T07:42:39.580" Title="Electric field of a hemisphere" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15639" PostTypeId="2" ParentId="2601" CreationDate="2017-06-01T17:18:50.810" Score="1" Body="&lt;p&gt;The coefficient that you need to use depends on what kind of heat source you're using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What units is that 280 in? Watts? Celsius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are in watts then you need a coefficient relating joules (watts*dt) to temperature change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would look something like:&#xA;$$k_{heating}=\frac1{dx \,dy \, \rho \, c_p \, t}$$&#xA;Where $t$ is the thickness of your metal&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2017-06-01T17:18:50.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15643" PostTypeId="1" CreationDate="2017-06-02T08:10:40.533" Score="3" ViewCount="3188" Body="&lt;p&gt;A transistor, say $NPN$, can be used as an amplifier. It amplifies base current by a factor beta. Is there a minimum base current threshold below which it cannot amplify? Say, are there transistors which can amplify femtoampere?&lt;/p&gt;&#xA;" OwnerUserId="11583" LastEditorUserId="8055" LastEditDate="2017-06-05T17:23:23.053" LastActivityDate="2018-07-31T08:14:46.647" Title="Minimum base current for transistor amplifier" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15644" PostTypeId="2" ParentId="15643" CreationDate="2017-06-02T08:36:47.620" Score="0" Body="&lt;p&gt;To answer the question, we can consider Ebers-Moll's ecuations:&#xA;$$I_E=I_{ES}(e^{V_{EB}/V_T}-1) - \alpha_RI_{CS}(e^{V_{CB}/V_T}-1)$$&#xA;$$I_C=\alpha_FI_{ES}(e^{V_{EB}/V_T}-1) - I_{CS}(e^{V_{CB}/V_T}-1)$$&#xA;$$I_B=I_E-I_C$$&#xA;where $I_{ES}$ and $I_{CS}$ are the reverse saturation currents, $V_T=\frac{KT}{q}$ ($K$ Bolzmann constant) and $\alpha_F$ and $\alpha_R$ the forward and the reverse constants. As we can see, unless both PN junctions are in reverse direction (it is said, $V_{EB}=V_{CB}=0)$, which depends on the circuit and the input voltage, the bipolar junction will amplifiy the current of the emitter.&lt;/p&gt;&#xA;" OwnerUserId="11568" LastActivityDate="2017-06-02T08:36:47.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15645" PostTypeId="1" AcceptedAnswerId="15646" CreationDate="2017-06-02T10:53:44.943" Score="0" ViewCount="31" Body="&lt;p&gt;i am first year engineer and this is my first class on electronics. &#xA;Can please someone explain why ix' = 0.2A ? &lt;img src=&quot;https://i.stack.imgur.com/kMlA1.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** I mean it is 0.2 if we add the resistors, but wouldn't it be 0.5 after the 6Ω? &lt;/p&gt;&#xA;" OwnerUserId="11586" LastEditorUserId="11586" LastEditDate="2017-06-02T11:01:54.350" LastActivityDate="2017-06-02T11:25:25.030" Title="Current calculation" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15646" PostTypeId="2" ParentId="15645" CreationDate="2017-06-02T11:14:39.407" Score="1" Body="&lt;p&gt;In that circuit the resistors are in series, so the current through them is the same. In addition, to calculate the current, you can sum the values of the resistors (because they are in series) and then apply Ohm's law: $I=\frac{V}{R}=\frac{3V}{(9+6)\Omega}=0.2A$&lt;/p&gt;&#xA;" OwnerUserId="11568" LastEditorUserId="10902" LastEditDate="2017-06-02T11:25:25.030" LastActivityDate="2017-06-02T11:25:25.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15650" PostTypeId="2" ParentId="14073" CreationDate="2017-06-02T14:40:37.260" Score="1" Body="&lt;p&gt;The key to doing these problems the &quot;easy way&quot; is to think how &quot;cause and effect&quot; works in the circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potentiometer is connected directly across the input voltage. An ideal op amp draws no input current, so the current in both parts of the potentiometer is the same. Therefore $V_x$ is just $\alpha V_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now look at R1. For an ideal op amp operating in the linear region, the two inputs are at the same voltage, so one end of R1 is at $V_x$. The other end is at $V_i$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can calculate $I_x$ by Ohm's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the ideal op amp draws no input current, there is only one place for the current $I_x$ to go after it has gone through R1, and that is through R2. So you can use Ohm's Law again to find $V_o$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;hard way&quot; to do the problem is set up all the circuit equations using KCL and KVL and then solve them. That's a good way for a computer program to simulate the circuit, but it's not what human engineers should be doing IMO - a bit of insight and understanding always beats a lot of grunt work!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-06-02T14:40:37.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15654" PostTypeId="2" ParentId="15626" CreationDate="2017-06-03T06:02:52.947" Score="0" Body="&lt;p&gt;Because on the &quot;pressing&quot; stroke you are compressing the spring requiring your energy / force and on the return the spring is now assisting the return.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-03T06:02:52.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15657" PostTypeId="1" CreationDate="2017-06-03T09:41:40.917" Score="2" ViewCount="2317" Body="&lt;p&gt;We are using one dental scanner (Vitality Scanner)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CTXHu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CTXHu.jpg&quot; alt=&quot;Vitality Scanner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the output format is in 3doxz Format,So we need to change it to stl for some analyzing on it, So i need to find out any software for this CAD format Converting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example one of it's 3DOXZ Format is here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.4shared.com/get/YjEgn_8Qei/51832_170524101342.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;3DOXZ CAD Example&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="2678" LastActivityDate="2017-06-03T09:41:40.917" Title="How to convert *.3OXZ CAD files to *.STL?" Tags="|cad|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15663" PostTypeId="1" AcceptedAnswerId="16812" CreationDate="2017-06-04T02:50:38.047" Score="0" ViewCount="4948" Body="&lt;p&gt;So I tried equating the sum of forces in the x,y and torque about $C$ to zero and found $B_x$ and $C_x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how to find $B_y$ and $C_y$? I have 3 equations and 4 unknowns? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gxu1T.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11605" LastEditorUserId="1832" LastEditDate="2017-06-05T02:56:27.947" LastActivityDate="2017-08-19T03:35:08.897" Title="Euler buckling formula,triangular structure" Tags="|structural-engineering|civil-engineering|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15664" PostTypeId="1" AcceptedAnswerId="15666" CreationDate="2017-06-04T04:40:54.390" Score="0" ViewCount="77" Body="&lt;p&gt;In the making of an inflatable toy ball (think of a basketball), a check valve is used to assist the user in the inflation of the ball by preventing backflow of inserted air. Is this check valve made of any metalic parts (such as conventional check valves) or is it all plastic? Also, what type of valve is it? Is there a maximum pressure this type of check valve can sustain before becoming damaged?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="7538" LastEditDate="2017-06-04T14:01:55.393" LastActivityDate="2018-02-12T04:38:14.327" Title="Toy ball check valve functionality and details" Tags="|pressure|valves|pressure-vessel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15665" PostTypeId="2" ParentId="15663" CreationDate="2017-06-04T06:48:38.300" Score="0" Body="&lt;p&gt;Bars AB and AC are pin-jointed and can only carry axial forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore looking at joint C we can say that $R_{cy}=0$ since the axial force transferred from bar AC will only be in the x direction. This leaves only three reactions left in the two pin-joints which can be found using three equations of statics: $\sum F_x=0, \sum F_y=0, \sum M=0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These will find the three reactions which will lead to the axial forces in the bars which can then be used in the axial stress and euler-buckling formulas respectively. Bending stress doesn't need to be checked since no bending occurs in the pin-jointed bars due to the first statement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On closer inspection, you can take sum of forces in X and Y at A which would give two equations to solve the two unknowns in the bars at that point $F_{ac} , F_{ab}.$ This is a quicker method to get the information you need but less information on the whole structure. &lt;/p&gt;&#xA;" OwnerUserId="11607" LastActivityDate="2017-06-04T06:48:38.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15666" PostTypeId="2" ParentId="15664" CreationDate="2017-06-04T06:53:16.497" Score="1" Body="&lt;p&gt;Depends on the design - some just use a soft plastic tube that collapses in on itself to seal. That is why some inflating kits have long needle adaptors amongst others in them tapered or parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum pressure a check valve could withstand IMHO is probably far greater than the material used for that ball itself. So, if you over pressurise it then it bursts, usually from the weakest point, and from what i have seen near the seam but not the valve failing first.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-06-04T06:59:08.537" LastActivityDate="2017-06-04T06:59:08.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15670" PostTypeId="1" CreationDate="2017-06-04T17:03:18.013" Score="1" ViewCount="497" Body="&lt;p&gt;I've got a glass box that measures 900mm x 450mm x 450mm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The width of the glass is 6mm thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I drill, cut, or grind a small hole in/through one of the faces; will it significantly affect the structural integrity of that piece of glass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm concerned it might become prone to cracking, if for instance, I pick it up to move it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought there might be some kind of ratio, or formula I can use to determine a safe size for a hole, a safe distance from the edge, and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xky1M.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xky1M.jpg&quot; alt=&quot;glass box schematic diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2829" LastActivityDate="2017-06-05T21:43:23.313" Title="Will drilling a hole in a pane of glass reduce it's structural integrity?" Tags="|structural-engineering|stresses|strength|glass|drain-waste-vent|" AnswerCount="0" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="15671" PostTypeId="1" AcceptedAnswerId="15672" CreationDate="2017-06-04T17:42:56.103" Score="0" ViewCount="2413" Body="&lt;p&gt;Suppose that I want to transport gases in a tube in only one direction. Can a liquid check valve be used in place of a gas check valve or is there a specific reason to use only a gas check valve?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2017-06-06T20:34:25.890" Title="Liquid vs. gas check valves" Tags="|pressure|valves|pressure-vessel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15672" PostTypeId="2" ParentId="15671" CreationDate="2017-06-04T20:53:23.913" Score="2" Body="&lt;p&gt;No, liquid check valves function on density, viscosity or velocity or a combination and the mass is taken into account - gas check valves are designed to deal with a fluid 700 or more times less dense than a liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some gas check valves will function with a liquid - a check valve designed for blowing up party baloons for example will work with a liquid at low pressure but not high pressure.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-04T20:53:23.913" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15673" PostTypeId="1" CreationDate="2017-06-04T21:39:57.123" Score="1" ViewCount="95" Body="&lt;p&gt;I'm trying to come up with a way to describe deformation on a body at each point on that body. My problem can be expressed and solved in two dimensions. Given two triangles, one being the initial shape, the other being the initial shape transformed arbitrarily :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XmJFD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XmJFD.png&quot; alt=&quot;Deformation of a triangle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's evident by looking at this that points along the left edge are now more distant to each other, points along the right edge are closer, points along the bottom edge have the same distance as before, and any point inside that triangle is blend of those three states. Is there any engineering standard to describe this, for any position on the body? I've looked at stress and strain, but I'm not sure if either describe what I want. In the end, I'm trying to find a value (or possibly two? Maybe a vector?) that describe, for a tiny area, if that area is compressed or stretched after the transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS.: I'm not a mechanical engineer if that wasn't already obvious. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="11615" LastEditorUserId="11615" LastEditDate="2017-06-04T22:15:41.767" LastActivityDate="2017-11-03T12:18:57.380" Title="Is there an standard description for deformation at each point on a body?" Tags="|mechanical-engineering|materials|deformation|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15674" PostTypeId="1" AcceptedAnswerId="15679" CreationDate="2017-06-05T00:48:41.710" Score="1" ViewCount="706" Body="&lt;p&gt;Recently, I heard that there is a law/ OSHA/ ASME piece of regulation dictating that hydrogen should be transferred (suppose from a tank of hydrogen to another location) by using specially approved tubing. Is this true?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2018-02-20T00:11:53.943" Title="Tubing for explosive gas or hydrogen transfer" Tags="|pressure|safety|compressed-air|pressure-vessel|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15676" PostTypeId="1" AcceptedAnswerId="15698" CreationDate="2017-06-05T02:19:39.920" Score="4" ViewCount="1720" Body="&lt;p&gt;Damascus steel is a multilayered hand-forged steel used in the past for arabic and japanese weapons and now for kitchen knives and various connoisseur blades. Usually, two alloys are used alternatively to stack layers having different properties and shades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if it still has a mechanical superiority over modern carbon steels, especially with &lt;a href=&quot;https://www.crucible.com/eselector/grades&amp;amp;names/0%7Egrades&amp;amp;namestitle.html&quot; rel=&quot;noreferrer&quot;&gt;&quot;super steels&quot; and tools steels&lt;/a&gt; like CPM 3V. By technical superiority, I mean better properties regarding fatigue resistance, impact resistance, tear propagation, edge durability, corrosion resistance, ultimate average tensile strength and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any studies about that ?&lt;/p&gt;&#xA;" OwnerUserId="11617" LastEditorUserId="11617" LastEditDate="2017-06-05T02:25:15.620" LastActivityDate="2017-06-06T10:30:34.160" Title="Steel : is Damascus steel still relevant compared to modern alloys?" Tags="|mechanical-engineering|steel|alloys|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15677" PostTypeId="1" CreationDate="2017-06-05T04:34:01.547" Score="0" ViewCount="51" Body="&lt;p&gt;I've been using 3d printers for prototyping for a long time, and I'll admit it's incredibly convenient to be able to throw together a CAD model, send it to the printer, and have it ready to assemble in a few hours. But that's just it, anything complicated with many moving parts often requires many prints and subsequent assembly which burns a huge amount of my prototyping time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was watching the first Iron Man movie the other day and noticed Jarvis say the line &quot;commencing automated assembly,&quot; which got me thinking, for something as complex as, say, a motorcycle, what's the current state of the art as far as general purpose automated assembly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; that by &quot;general purpose&quot; I mean able to construct essentially arbitrary parts using a mixture of printed components of a given material (say sintered metal) and standard parts (servos, wires, gears, PCBs, etc.). Feel free to ignore the difficulty in entirely automated PCB reflow.&lt;/p&gt;&#xA;" OwnerUserId="11618" LastActivityDate="2017-06-05T04:52:58.560" Title="General purpose automated assembly?" Tags="|manufacturing-engineering|cad|3d-printing|prototyping|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15678" PostTypeId="2" ParentId="15677" CreationDate="2017-06-05T04:52:58.560" Score="1" Body="&lt;p&gt;Do a search for the automation of car assembly : there was a joke about Fiat : designed on computer, built by robot, driven by monkeys. Robots exist that build engines, assemble bodies and also spray cars - and they use less paint.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-05T04:52:58.560" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15679" PostTypeId="2" ParentId="15674" CreationDate="2017-06-05T05:08:09.293" Score="3" Body="&lt;p&gt;I would say that it is true and for that case there will be many relevant codes / regulations that need to be respected. Make sure you read them all before designing / altering the system.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-05T05:08:09.293" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15681" PostTypeId="1" CreationDate="2017-06-04T13:05:36.677" Score="2" ViewCount="155" Body="&lt;p&gt;I have a problem when I'm extruding a trimmed surface, it's look alright in isometric view, but in 2d view it shows unrelated line due to the trimming of the circle as show in the photo below, any idea to remove it so that the 2d view looks normal which how it should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sample image unwanted line&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EYfjy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EYfjy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="tea" LastActivityDate="2018-04-01T16:51:51.990" Title="How to remove unwanted line from 2d trimming?" Tags="|autocad|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15682" PostTypeId="1" CreationDate="2017-06-05T10:06:33.620" Score="3" ViewCount="2669" Body="&lt;p&gt;Can some please explain why is the elastic modulus of aluminium and&#xA;work hardened aluminium the same?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it only the same for pure metals?&lt;/p&gt;&#xA;" OwnerUserId="11605" LastActivityDate="2017-06-05T19:01:05.097" Title="Work hardening?" Tags="|materials|manufacturing-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15683" PostTypeId="2" ParentId="15681" CreationDate="2017-06-05T11:35:37.520" Score="1" Body="&lt;p&gt;Find and delete the object(s) causing the lines.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-05T11:35:37.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15684" PostTypeId="2" ParentId="15682" CreationDate="2017-06-05T11:56:34.273" Score="6" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Work_hardening&quot; rel=&quot;noreferrer&quot;&gt;Work hardening&lt;/a&gt; is done to increase the &lt;em&gt;strength&lt;/em&gt; of the material, not the stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You change the yield stress to be closer to the failure stress of the material.  If I understand it correctly it also increases the elastic range and takes away from the plastic range of deformation (which seems reasonable as it requires plastic deformation to work harden in the first place).&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-06-05T11:56:34.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15685" PostTypeId="2" ParentId="15674" CreationDate="2017-06-05T13:13:57.077" Score="1" Body="&lt;p&gt;This is a complex topic, if you are responsible for designing/altering such a system, I would say you should consult an expert in hydrogen transport. You will need input on the appropriate materials for the pipe (e.g. MOC of pipe for hydrogen embrittlement) and the types of joints required at the minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting reading regarding the MECHANICAL design of hydrogen piping (note the scope) - &#xA;NFPA 50A: Standard for Gaseous Hydrogen Systems at Consumer Sites, 1999 Edition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the PROCESS design for hydrogen usage, it would be the same as any other gas transport application, using the appropriate materials and developing the proper process parameters. &lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2017-06-05T13:13:57.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15686" PostTypeId="1" CreationDate="2017-06-05T13:18:51.323" Score="1" ViewCount="61" Body="&lt;p&gt;Need to design 3D model of photodiode and calculate it (interaction light with semiconductor). What software product would be better for this objective and why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Important criteria:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Big library semiconductor elements.&lt;/li&gt;&#xA;&lt;li&gt;Comfortable interface&lt;/li&gt;&#xA;&lt;li&gt;Small expensive CPU\GPU (high speed of calculate).&lt;/li&gt;&#xA;&lt;li&gt;Price of product.&lt;/li&gt;&#xA;&lt;li&gt;Ability to import geometry from autocad\solidworks etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8880" LastActivityDate="2017-06-05T13:18:51.323" Title="Best software for 3D modeling and simulation semiconductors?" Tags="|modeling|finite-element-method|simulation|software|" AnswerCount="0" CommentCount="1" ClosedDate="2017-06-06T02:01:07.207" ContentLicense="CC BY-SA 3.0" />
  <row Id="15687" PostTypeId="1" CreationDate="2017-06-05T14:56:52.990" Score="0" ViewCount="145" Body="&lt;p&gt;Okay, I have bought an electric motor, it has a 14mm bore, unfortunately the part I need to affix has an internal diameter of 16mm (5/8&quot;).&#xA;So, to fix this problem I call upon your experience and expertise. Is their a way to perhaps sleeve the bore with a (tiny) grub screw holding it in position? Or perhaps a better alternative that will not create vibration at around 2800 rpm?&#xA;I should let you know that I do not have a milling machine or metal lathe at my disposal unfortunate as that is, but if I did I wouldn't be asking this most likely.&#xA;I probably should mention that this project is an abrasive belt sander, of which I have already built fromantic scratch. This is in fact the last hurdle.&#xA;Thank you for your time,&#xA;Regards,&#xA;Peter &lt;/p&gt;&#xA;" OwnerUserId="11105" LastActivityDate="2017-08-05T19:34:09.170" Title="Electric motor bore sizing problem" Tags="|mechanical-engineering|motors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15688" PostTypeId="2" ParentId="15687" CreationDate="2017-06-05T15:48:15.787" Score="1" Body="&lt;p&gt;Brass shim should do the trick. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically you need a strip of 1mm thick shim stock and wrap it most of the way around the motor shaft. It may help to make some sort of simple former to help get it the right shape and indeed maybe wrap it with 1.5 turns and trim off the excess as the ends tend to be harder to bend. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shim stock has pretty accurate dimensions so it should end up being adequately concentric. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the grub screw, it sounds like this is a moderately high torque application so I would suggest a decent sized screw, maybe M8 with a pointed tip. If you drill through the mounting bore with the shaft in place and carry on a bit so you make a small dimple in the shaft maybe a couple of mm deep that will provide a more positive grab for the screw and reduce the risk that it will slip and chew up the shaft and/or wear the shim. Then tap the hole and use loctite on the grub screw to stop it vibrating loose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want the grub screw to locate in the gap between the ends of the shim (hence not wrapping it all the way around the shaft). The reason for this is that brass can creep a bit so you may find it starts to get slack after a while if the screw is bearing on the brass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used exactly this solution to fit a polishing mop holder to a motor shaft (about 900W)  and it has worked for years with no problems. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-08-05T19:34:09.170" LastActivityDate="2017-08-05T19:34:09.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15689" PostTypeId="2" ParentId="15682" CreationDate="2017-06-05T15:55:11.767" Score="2" Body="&lt;p&gt;In general hardness has very little effect on the elastic modulus of metals. Hardness and tensile strength are closely related but elasticity is a pretty much independent property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is somewhat counter intuitive and often causes some confusion especially in the context of tempering steel. Contrary to popular belief springs are not heat treated to make them more elastic but to improve their strength, this increases the extent to which they can deflect without deforming but not their elastic modulus ans so called 'spring temper' is a balance between ultimate strength and toughness and has no bearing on the spring constant itself. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-06-05T15:55:11.767" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15690" PostTypeId="2" ParentId="15682" CreationDate="2017-06-05T19:01:05.097" Score="3" Body="&lt;p&gt;Elasticity of materials depends only on the strength of the atomic bonds. Atoms are pushed apart from each other by their negatively charged electron clouds. They are pulled closer together by the negative electrons of one atom attracting the positively charged nuclei of neighboring atoms, and vice-versa. At some distance the attractive and repulsive forces balance, and the atoms are in equilibrium. To separate the atoms (or bring them closer together) takes application of an outside force. Application of an outside force shifts atoms from their natural equilibrium positions, which is resisted by the changing electrostatic conditions. The differential resistance to differential change in position is measurable at large scales as elastic modulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast to elastic modulus, strength depends mostly on the microstructure of a material, and not relative atomic position. Most practical metals, like aluminum, are composed of many differently-oriented crystalline grains. Where the grains meet are grain boundaries. Plastic deformation of metals occurs by slip of planes in the crystalline grains, one line of atoms at a time. The lines of atoms that are actively moving are called dislocations. The more plastic deformation, or cold work, experienced by a metal sample, the greater the density of dislocations. Dislocations eventually get blocked by grain boundaries and by each other, increasing the resistance to further plastic deformation, and this effect is measurable as an increase in strength. As the dislocation density increases, these blockages occur with greater frequency and with greater resistance, and strength increases in proportion to cold work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the stresses that caused plastic deformation are removed, the material retains its increase in strength. However, the atoms haven't gotten closer together or further apart from each other on average. They've only been moved around relative to one another by crystal slip and dislocation motion. Thus the elastic modulus hasn't changed as a result of cold work.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2017-06-05T19:01:05.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15691" PostTypeId="1" AcceptedAnswerId="15704" CreationDate="2017-06-05T20:12:26.233" Score="3" ViewCount="5417" Body="&lt;p&gt;I have a building that is 7 story height (which 5 stories are basement floors which means they are underground) My water level is at 1.5 m from the surface. the stories have typical rectangular shape of 30 m x 36.2 m with a grid of columns as shown in the picture below. &#xA;When modeling the raft foundation that is subjected to gravity loads and water pressure (opposite to gravity loads) and since the raft is at a level of 19.5 m from the ground surface it is subjected to a lot of uplift the overall gravity load is of 108858 KN as for the uplift is (19.5-1.5)*10*30*36.2 = 195480 KN which is greater than the gravity loads so their will be an uplift of the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to solve this problem ? I mean how many solutions that are available today for this problem and what is the most economical one ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qOKY6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qOKY6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I1AE1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I1AE1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ajdd4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ajdd4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="1832" LastEditDate="2017-11-12T03:00:32.873" LastActivityDate="2020-03-27T15:49:40.973" Title="Building Uplift due to water pressure" Tags="|structural-engineering|civil-engineering|structural-analysis|foundations|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15692" PostTypeId="2" ParentId="15169" CreationDate="2017-06-05T20:36:17.103" Score="1" Body="&lt;p&gt;While 6mm is at the small end of shaft sizes, why not use a standard keyed shaft and hub? The standard key is 2mmx2mm for a 6mm shaft. For a normal fit, the shaft tolerance should be (-0.004/-0029) and the hub (+/-0.012). The setscrew tightens into the key so damaging the shaft becomes a non-issue. A keyed shaft/hub isn't going to slip unless you shear the entire key and if that happens, you've got much bigger problems. The alignment is repeatable to within the tolerance between the width of the key and the keyslots on the shaft and hub.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2017-06-05T20:43:12.020" LastActivityDate="2017-06-05T20:43:12.020" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15693" PostTypeId="1" AcceptedAnswerId="15697" CreationDate="2017-06-05T21:07:20.047" Score="1" ViewCount="47" Body="&lt;p&gt;I am not sure if this is the right place to ask this but. I have an offroad skateboard from &lt;a href=&quot;http://mbs.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;MBS&lt;/a&gt;. I bought their &lt;a href=&quot;https://www.mbs.com/parts/15005-mbs-v5-brake-system&quot; rel=&quot;nofollow noreferrer&quot;&gt;brake system&lt;/a&gt; which comes with a circular piece of aluminum. Pictures attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that on the circular aluminum brake part (don't know the proper name), it has 5 holes drilled. I called the company and they said they have CAD files and it is done in mass. I don't have those CAD files and I am not a CAD expert, nor do I have a machine to cut it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the best way to measure (with basic relatively inexpensive tools, if needed I'll buy more), and replicate those holes with high precision on the sprocket. As you can tell, this is going to spin, so any measurement that is a little bit off is going to make the sprocket wobble. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sprocket is not the same exact size as the aluminum brake part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not looking for measurements. I am looking for how to measure. Not the fish itself, but I want to learn how to fish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VgdT1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VgdT1.jpg&quot; alt=&quot;sprocket&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZJPQn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZJPQn.jpg&quot; alt=&quot;wheel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fb3zO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fb3zO.jpg&quot; alt=&quot;inside wheel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11626" LastActivityDate="2017-07-23T18:14:56.883" Title="How to duplicate measurements from one piece of material to another?" Tags="|steel|measurements|mathematics|aluminum|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15694" PostTypeId="2" ParentId="15674" CreationDate="2017-06-05T21:12:53.123" Score="0" Body="&lt;p&gt;Assuming the question pertains to smaller quantities as may be used in a laboratory or specialized heat-treating : H can easily be handled with ordinary materials = carbon steel, copper and stainless steel. A major concern with H is that if there is any leakage it will rise and collect at the top of any enclosure. So design needs to include things like  explosion proof/ gas tight light fixtures.&lt;/p&gt;&#xA;" OwnerUserId="11627" LastActivityDate="2017-06-05T21:12:53.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15695" PostTypeId="2" ParentId="15671" CreationDate="2017-06-05T21:29:10.363" Score="0" Body="&lt;p&gt;Industrial valves are ( or were ) marked W O G . Meaning they are suitable for water, oil, and gas, ( not gasoline .). It is up to the user to determine if the valve will function in his particular application. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2017-06-06T20:34:25.890" LastActivityDate="2017-06-06T20:34:25.890" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15697" PostTypeId="2" ParentId="15693" CreationDate="2017-06-05T21:48:24.817" Score="2" Body="&lt;p&gt;I would go either of two ways : one to mount both pieces rigid on a shaft and line centre punch the sprocket or two find or borrow (machine shop...) a pillar drill to get perpendicular and mount both parts so they rotate and centre punch again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the (old school) possibility of using compass and dividers.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-05T21:48:24.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15698" PostTypeId="2" ParentId="15676" CreationDate="2017-06-05T22:08:01.957" Score="4" Body="&lt;p&gt;Damascus and Samari blades are artisan made , so have variations. They were a very clever way to make medium carbon steel before the science was understood. They are a combination for wrought iron ( low carbon , soft, ductile) and a high carbon similar to cast iron ( high carbon). The two materials are layered together , forged and reheated a number of times. The carbon does diffuse to some degree to produce a medium carbon product , roughly 0.4% . This has a blend of toughness, strength, hardness ( depending on other elements like Mn ,Si and others, and heatreatment). I can't imagine them being made in high quality, quantity or at competitive cost.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-06-05T22:08:01.957" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15699" PostTypeId="2" ParentId="15693" CreationDate="2017-06-05T22:56:42.280" Score="0" Body="&lt;p&gt;I am confused about why you have the gear? The center of that gear does not look like it will fit on the axle where the brake disk is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are planning on cutting the center out of the gear and the 5 holes, I might mount it in a drill and clamp the 2 together. Spin the part with the drill in a vice or something until you an see that they are lined up correctly then you can spray paint through the holes in the brake disk to mark the gear. Depending upon the precision you need you can drill press then mill until the paint is gone. &lt;/p&gt;&#xA;" OwnerUserId="10516" LastEditorDisplayName="user16" LastEditDate="2017-07-23T18:14:56.883" LastActivityDate="2017-07-23T18:14:56.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15700" PostTypeId="2" ParentId="7746" CreationDate="2017-06-05T23:51:23.180" Score="0" Body="&lt;p&gt;The answer with 5/8 radius is irrelevant.  Well not completely.  Could help find answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The leg has a 1 1/4 in hole in it.  Therefore the leg with a 1 1/4 radius will be a 2.5 inch dia leg half way rounded.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now.  How long and wide is your actual part?&lt;/p&gt;&#xA;" OwnerUserId="11630" LastActivityDate="2017-06-05T23:51:23.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15702" PostTypeId="1" CreationDate="2017-06-06T00:58:24.547" Score="7" ViewCount="614" Body="&lt;p&gt;My wife and I are expecting our first child, and therefore are expecting to purchase a car seats. My wife and I have discovered that car seats have &quot;expiration dates&quot;: a car seat, no matter how lightly used, is deemed unsuitable for use after a particular date. We wonder why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A car seat which has been in an accident could be compromised in ways that simple visual inspection could not confirm. &lt;strong&gt;&lt;em&gt;I am not asking about these car seats.&lt;/em&gt;&lt;/strong&gt; I am asking about the car seats which have experienced the repeated loading/unloading which comes from placing a small human child in and out of the seat and driving them about in a driving style that could only be described as &quot;not aggressive.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempting to look up answers for this question have not yielded any satisfactory results! So I ask you, engineering stack exchange, &lt;strong&gt;why do infant and child car seats have expiration dates?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few possibilities, but I have issues with all of them:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The car seat is designed to withstand a number of cyclic loads, and critical car-seat parts have a fatigue life which the manufacturers base the expiration date on. This seems the most likely, but babies are really small and the materials are meant to protect them against high-speed crashes for the entire product life!&lt;/li&gt;&#xA;&lt;li&gt;The car-seat manufacturers are greedy, using social pressures and the label of &quot;unsafe&quot; to force parents (as a group) to buy car seats at regular intervals and deny the second-hand market any 'viable' material.&lt;/li&gt;&#xA;&lt;li&gt;Something in the car seats really does degrade with use, heat, or sunlight and therefore &lt;em&gt;actually expires.&lt;/em&gt; (This seems unlikely, but terrifying!)&lt;/li&gt;&#xA;&lt;li&gt;Safety standards are expected to update after some amount of time. This particular explanation seems very odd to me: safety standards are set be governments and industry groups, and I doubt car-seat needs have evolved so much that a regular update every few years is entirely needed, or that those updated industry safety standards will always render old products unsafe.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Or maybe it's something else!&lt;/em&gt; Sources are welcome!&lt;/p&gt;&#xA;" OwnerUserId="11631" LastEditorUserId="11631" LastEditDate="2017-06-06T15:22:52.920" LastActivityDate="2017-06-07T21:09:00.373" Title="Why do car-seats have an expiration date?" Tags="|materials|design|safety|fatigue|" AnswerCount="2" CommentCount="16" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15703" PostTypeId="1" CreationDate="2017-06-06T01:51:12.067" Score="1" ViewCount="196" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/msZS1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/msZS1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can someone tell me please the name of it or any info about it &lt;/p&gt;&#xA;" OwnerUserId="11633" LastActivityDate="2017-06-06T01:51:12.067" Title="What is the name of this mechanism" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="2" ClosedDate="2017-06-06T15:37:51.937" ContentLicense="CC BY-SA 3.0" />
  <row Id="15704" PostTypeId="2" ParentId="15691" CreationDate="2017-06-06T02:20:38.677" Score="4" Body="&lt;p&gt;There are a few possible solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use pile foundations, where the piles dig deep into the soil and resist the uplift. They'll work entirely in tension, instead of the usual compressive load, so point bearing capacity is non-existent, only consider lateral capacity. Depending on the soil type (and remembering it is saturated), this may or may not be viable, especially if you need to handle seismic loads. Regardless, it is probably not the best idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The usual solution is mass. Lots and lots of mass. Your walls and bottom basement slab are under massive loads, so just go crazy and make them monstrously thick. This will lower your steel reinforcement and reduce cracking. And, if you put in enough (including a safety factor), your structure will weigh more than the water it'll displace (weight &gt; uplift).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-06-06T02:20:38.677" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15705" PostTypeId="1" CreationDate="2017-06-06T02:25:27.773" Score="3" ViewCount="57" Body="&lt;p&gt;I want to be able to detect very small torques placed on the end of a cantilever beam. These torques are very small - 1mg by 10cm. I am trying to make this on the cheap and use a load cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plan to amplify the torque by running it through a self made very high stage planetary gear box, say like 250:1. I would apply the resulting torque to the sensor by having a rigid beam of known length hit the load cell. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is essentially a static test. I will have the apparatus so that the torque on the output would be resisted so it could be measured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will friction be a problem in terms of accuracy? Is there an initial threshold I would need to meet for the torque to make it through the gear box?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would I be better off with maybe using a worm gear setup? &lt;/p&gt;&#xA;" OwnerUserId="11634" LastEditorUserId="90" LastEditDate="2017-06-06T08:54:50.957" LastActivityDate="2017-06-06T11:16:30.650" Title="Would friction forces impact a torque measurement transmitted through a gear box" Tags="|gears|statics|torque|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15706" PostTypeId="2" ParentId="15628" CreationDate="2017-06-06T03:24:21.463" Score="1" Body="&lt;p&gt;For the M3 t-nuts, could you use this &lt;a href=&quot;https://www.thingiverse.com/thing:1573410&quot; rel=&quot;nofollow noreferrer&quot;&gt;printed solution&lt;/a&gt;.&#xA;If you are just attaching a PCB, you wouldn't need that much strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another excuse to use your 3D printer. ;)&lt;/p&gt;&#xA;" OwnerUserId="3925" LastEditorUserId="3925" LastEditDate="2017-06-06T05:22:58.073" LastActivityDate="2017-06-06T05:22:58.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15707" PostTypeId="2" ParentId="15628" CreationDate="2017-06-06T03:39:44.857" Score="1" Body="&lt;p&gt;Another option would be to make your own t-nut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a mill/CNC, here are some instructions to &lt;a href=&quot;http://rick.sparber.org/tn.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Make a T-Nut&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have access to a mill, you could make a pretty good one out of some 4mm x 5mm &lt;a href=&quot;https://www.mcmaster.com/#standard-aluminum-sheets&quot; rel=&quot;nofollow noreferrer&quot;&gt;bar stock&lt;/a&gt; and tap your own holes. All you would need is a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00HQONFVE&quot; rel=&quot;nofollow noreferrer&quot;&gt;drill press&lt;/a&gt; and an &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B01FTFX40S&quot; rel=&quot;nofollow noreferrer&quot;&gt;M3 Tap&lt;/a&gt;. The nut may not be as rigid as a t-nut that precisely matches you channel and you can't put as much torque on 4mm of bar stock aluminum as you could on a steel t-nut; but, for most application it would probably be sufficient.&lt;/p&gt;&#xA;" OwnerUserId="3925" LastActivityDate="2017-06-06T03:39:44.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15709" PostTypeId="2" ParentId="15705" CreationDate="2017-06-06T04:30:05.133" Score="2" Body="&lt;p&gt;Yes, friction will be a serious problem. So will backlash in the gears, the clearance in the gearbox bearings, the general flexibility of all the gearbox parts (including the casing), etc, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A worm gear will probably be even worse. Unless it is designed to operate in either direction (i.e. if you push on the rack, the gear will turn) the friction is intentionally large to hold the gear in position with no external loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would probably be better making a flexible cantilever and measuring its deformation with a strain gauge or an accelerometer if you want a &quot;cheap&quot; solutions - or use more sensitive measurement device like a laser based system if you want high accuracy and can afford it. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-06-06T04:30:05.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15711" PostTypeId="1" AcceptedAnswerId="15718" CreationDate="2017-06-06T08:41:44.173" Score="0" ViewCount="603" Body="&lt;p&gt;I have a mat foundation that is 3 m thick (44 story building), I know how to design the upper layer and bottom layer but since at the top layer the reinforcement is too dense I want to make a third layer of reinforcement but my problem is I don't know how to design the third layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for example in total in the top layer I need 55.8 cm2/m so I place T25@15 cm (32.8 cm2/m)in the top layer, for the remaining reinforcement area need I want to place them in a different layer just below the top layer bearing in mind that the effective depth will change so I can't just say 55.8-32.8 = 23 cm2/m and place them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the remaining reinforcement needed ?    &lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2017-06-06T13:02:53.463" Title="Mat foundation Reinforcement with 3 layers" Tags="|structural-engineering|civil-engineering|design|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15712" PostTypeId="2" ParentId="15676" CreationDate="2017-06-06T08:48:54.833" Score="2" Body="&lt;p&gt;No. The only application is historically accurate replicas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While damascus steel has parameters quite comparable to modern hardened steels, its manufacturing process is lengthy, expensive and ill-suited for mechanization and mass production. While its parameters would make it perfectly adequate for many applications, alternatives exist that provide equivalent performance at significantly lower manufacturing cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=mTsfPPR91qc&quot; rel=&quot;nofollow noreferrer&quot;&gt;This video&lt;/a&gt;, while focusing on the Japanese Katana and its manufacturing process, also discusses damascus steel, and points out similar shortcomings of both versus modern steels. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-06T08:48:54.833" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15713" PostTypeId="2" ParentId="15673" CreationDate="2017-06-06T09:31:38.937" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Displacement_field_(mechanics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Displacement field&lt;/a&gt; - a vector field, that assigns a displacement vector to each infinitesimal particle of the body over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object, in this case will be usually defined by a density field (scalar field) with zero density being &quot;outside the object.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This description works on all kinds of matter - gas, fluid mechanics, and non-rigid solids, their behavior changing with the kind of interaction between neighboring infinitesimal particles (stress/strain, pressure, viscosity etc.) which all interact with 2nd derivative of position over time (=acceleration) resulting in different equations of motion/displacement, but the underlying description of object changing remains a displacement field of one kind of another.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-06T09:31:38.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15714" PostTypeId="2" ParentId="15626" CreationDate="2017-06-06T09:57:30.097" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Friction is the cause of &quot;hysteresis-like&quot; difference over the whole stroke. Friction always acts in direction opposite than movement, so on the way &quot;down&quot; you act against spring strength + friction, while on the way &quot;up&quot; you act against spring strength - friction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;force hysteresis coming from the switch construction. The schematics don't show the switch accurately - there's a point where the switch &quot;latches&quot; requiring more force in the negative direction to disengage, than in the positive direction, to engage. In simpler keys, like typical microswitches, this is usually achieved by a small sheet of metal pressed to buckle in the middle; pressing upon it flips the bulge to the opposite side - a bit like a &quot;&lt;a href=&quot;https://www.cuteness.com/article/make-clicker-dog-training&quot; rel=&quot;nofollow noreferrer&quot;&gt;clicker lid&lt;/a&gt;&quot; used on some juices etc, but the amount of engineering that goes into these things when designing keyboards of pleasant key action goes far beyond solutions this simple. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-06T09:57:30.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15715" PostTypeId="2" ParentId="15676" CreationDate="2017-06-06T10:18:39.477" Score="-1" Body="&lt;p&gt;Wootz steel/ Damascus steel is an incredibly beautifully made alloying process and for this alone yes it is still relevant. I personally am working on all sorts of variables in the process of making said alloys.&#xA;It has many uses, perhaps not in the industrial scene, but yes it has its place.&#xA;Now, on to the &quot;super steels&quot; yes there are super steels that are stronger, hold a keener edge for longer etc etc etc, however as a knife maker I enjoy making wootz/damascus steels. And I am just guessing that you do not know how much some of these super steels cost?&#xA;Besides, as I forge my own blades, I am not one of these guys that just removes material and considers themself a knife maker! I have a healthy business, so yes, I believe damascus has its place and will continue to for a long long time.   &lt;/p&gt;&#xA;" OwnerUserId="11105" LastEditorUserId="11105" LastEditDate="2017-06-06T10:30:34.160" LastActivityDate="2017-06-06T10:30:34.160" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15716" PostTypeId="2" ParentId="15705" CreationDate="2017-06-06T11:16:30.650" Score="3" Body="&lt;p&gt;The forces associated with this torque will be of order of 10 micronewtons. It takes NASA level equipment to measure such forces (look up EM Drive tests). Your breath at a meter distance will exert this sort of disturbances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general approach to measuring forces of this order is using a mirror and a laser beam to detect displacement/rotation: a tiny mirror attached on a long arm, a laser beam aimed at it, and reflecting onto a scale placed a considerable distance from the test rig; minuscule displacements will make the point move a measurable distance. This is the simple measurement rig, but the real difficulty is in eliminating external disturbances - even minor air movement, vibrations from distant vehicles, steps of operators, dynamic pressure from audible noise - that all can introduce forces of that order and spoil your measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't even begin to dream torque of $10^{-4}$ Nm can be transmitted by any kind of macroscopic gear mechanism (it's different with nanomechanics, but can you make your planetary gear with sub-millimeter gears?)&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-06T11:16:30.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15717" PostTypeId="1" AcceptedAnswerId="15729" CreationDate="2017-06-06T11:37:08.053" Score="0" ViewCount="1271" Body="&lt;p&gt;Although I have seen a lot of literature online about how internal pressure resistance is calculated (i.e the pressure in a gas tank) I haven't found anything about the opposite scenario (i.e the pressures on a deep sea submarine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would one calculate the thickness of material needed and stresses involved at high pressures externally? Please assume a very basic level of physics knowledge (I have GCSE knowledge for sure and can probably work out and search up some of the more basic stuff, but I'm certainly no engineer).&lt;/p&gt;&#xA;" OwnerUserId="11637" LastActivityDate="2017-06-06T16:55:42.467" Title="How to calculate external pressure resistance in spheres and cylinders?" Tags="|structural-engineering|stresses|pressure-vessel|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15718" PostTypeId="2" ParentId="15711" CreationDate="2017-06-06T13:02:53.463" Score="1" Body="&lt;p&gt;Actually, you can just subtract the area from the top layer to calculate the subsequent layers. Obviously this is an approximation, since you're losing effective depth in the other layers, but it's a good start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check that the approximation is valid, calculate the center of gravity of the total steel in each face and then calculate the requisite steel area. If it is lower than your adopted value, you're fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, if you need 55.8&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;/m, you can adopt one layer of $\phi20/10$ (31.42&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;/m) and another $\phi25/20$ (24.54&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;/m), for a total of 55.96&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a concrete cover of 2&amp;nbsp;cm and a distance of 3&amp;nbsp;cm between rebar faces, your CoG moves from 3&amp;nbsp;cm (2&amp;nbsp;cm cover plus half the diameter of the $\phi20$ rebar) to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2 + \frac{\frac{2}{2}\cdot31.42 + \left(3 + \frac{2.5}{2}\right)\cdot24.54}{55.96} = 4.425\text{ cm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then recalculate your mat adopting the new effective depth. If the required steel area remains below 55.96&amp;nbsp;cm&lt;sup&gt;2&lt;/sup&gt;/m, you're fine. If it goes above it, just repeat the process. This works for any (reasonable) number of layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular case, where you have a mat that's three meters thick, you'll be fine: the change in the effective depth (of 1.425&amp;nbsp;cm) will be little more than a rounding error.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-06-06T13:02:53.463" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15719" PostTypeId="2" ParentId="15702" CreationDate="2017-06-06T13:16:57.043" Score="8" Body="&lt;p&gt;It's a combination of 1, 3, 4, the product development cycle, and the nature of the product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - Damage in materials is a cumulative process, the product of hundreds or thousands of little events (e.g. the tightening and loosening of the belts) or one big event (a crash). So, the product has to be designed to always have a certain damage capacity throughout its intended lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 - Plastics break down and their mechanical properties degrade over time. Specifically they lose their impact resistance and become more brittle. Both of which are bad things for a car seat intended to protect a child in the event of a crash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 - Standards do change slowly, but safety technology changes at a much faster pace. The safety standards in many cases are simply a minimum requirement and not an optimal one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Validating the performance of a product like a car seat requires testing...a lot of testing and testing to a very high standard since the consequences of a defect in the seat can be lethal. Testing is expensive. Making the high standards higher to extend the product life makes the testing even more expensive. The further out you go, it becomes exponentially more expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the plastic materials decline over years, this means the manufacturer has to artificially age these materials and there are limits to how good these procedures actually are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to product lifetimes and guarantees, for the manufacturer, it's a lot like looking into a crystal ball that gets fuzzier and more costly the further into the future you look.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expiration dates are NOT a legal requirement in that they are not legislated, but they are there, in part for legal reasons. The expiration date is a way of communicating to the consumer that the company is only willing to guarantee a standard of performance for a certain period of time due to those factors. It's also a means of limiting the company's liability in the event an older seat is involved in an accident where injury occurs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can bet that these companies have looked long and hard at what the average consumer's usage cycle is for car seats and compared that with what consumers are willing to payfor a car seat when they are developing these products. Could manufacturers make a car seat that they would have an expiration date that's further out in the future? Sure. How much more are you willing to pay for a car seat that can still be used when your infant is a teenager even though it will be outdated (and they won't fit in it anyway)?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: 20yr of product design experience and the answers I got when I researched the same question after buying a car seat for my first child 16yr ago&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2017-06-06T14:33:01.193" LastActivityDate="2017-06-06T14:33:01.193" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15720" PostTypeId="1" AcceptedAnswerId="15726" CreationDate="2017-06-06T13:23:31.433" Score="0" ViewCount="46" Body="&lt;p&gt;We're making a vibration polisher for polishing laser-cut aluminum and steel parts, and we're looking for the abrasive medium to fill it with. We need about 30 liters of it - and the one we found (for jewelry use) was prohibitively expensive in the required amounts. We tried with several gravel kinds and the results were underwhelming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What material, that would be easy to obtain and inexpensive is a common choice for a vibration polisher?&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-10-28T02:50:07.083" Title="Budget abrasive for vibration polishing" Tags="|materials|polishing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15721" PostTypeId="1" AcceptedAnswerId="15722" CreationDate="2017-06-06T12:30:38.417" Score="1" ViewCount="911" Body="&lt;p&gt;I am a non-engineer and I am using a gas power plant data base. Some plants listed are of the type &quot;combustion engine&quot;, some of the type &quot;gas turbine&quot;, some &quot;steam turbine&quot; and some &quot;combined cycle&quot;. I believe to understand that combined cycle is a combination of combustion engine and steam turbine. But what is the difference between combustion engine and gas turbine then? Or does it mean the same?&lt;/p&gt;&#xA;" OwnerUserId="11639" OwnerDisplayName="LenaH" LastActivityDate="2017-06-06T14:08:59.913" Title="Difference between gas combustion engine and gas turbine?" Tags="|power-engineering|gas|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15722" PostTypeId="2" ParentId="15721" CreationDate="2017-06-06T14:08:59.913" Score="1" Body="&lt;p&gt;Combustion engine - an internal combustion engine like a car motor, with pistons moved in cycles. It's inefficient but can be easily scaled down to almost arbitrarily small sizes; usually used as a backup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gas turbine - in power plants, these are very similar in construction to jet engines, where gas is the fuel - a multi-stage turbine compressor, a turbine on exhaust, high RPM; the torque produced is used to run the generator. It's not as efficient as steam or combined cycle, though more than combustion engine - but the power output can be rapidly tuned to needs, providing a response to changing demand (which is typical for the &quot;more inert&quot; types that use steam.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steam turbine - gas heats water in a boiler; overheated steam runs through turbines, then is cooled. This is the same principle as great most other thermal power plants (coal, nuclear, geothermal). It's usually a large installation and may take hours to get up to speed (so no rapid response to demand + waste of energy as demand rapidly vanishes) but it has a very good efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combined cycle - exhaust from gas turbine (the &quot;jet engine&quot;) is used to heat water into steam and run a steam turbine. Better efficiency than both above, and provides the much desired rapid response. Of course cost of construction is similar to sum of costs of construction of the two, and maintenance is more complex, but the operational costs are reduced.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-06T14:08:59.913" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15723" PostTypeId="1" CreationDate="2017-06-06T13:27:00.487" Score="2" ViewCount="108" Body="&lt;p&gt;I am working on fabricating a large scale revolving disco ball for an event in Cincinnati. The scale is 10'diameter and the frame is flat bar steel. the exterior is cardboard, duct tape, and adhesive &quot;mirror stickers.&quot; I am looking to have the ball rotate at 1 RPM. Total weight is 350 lbs. What type of motor should I research to purchase? The motor will be hard mounted at the top of the structure that is to be suspended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Joe&lt;/p&gt;&#xA;" OwnerDisplayName="JoeG" LastEditorUserId="86" LastEditDate="2017-06-06T14:15:47.910" LastActivityDate="2017-10-05T12:44:12.520" Title="Motor needed for large scale disco ball!" Tags="|motors|bearings|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="15725" PostTypeId="2" ParentId="15717" CreationDate="2017-06-06T14:45:03.047" Score="1" Body="&lt;p&gt;You can find the formulas in petroleum industry ,like API, for cylinders = tubulars. This is a routine calculation done in all oil/gas well design. It will be listed under &quot;collapse&quot;. It is in API bulletin 5C3, unfortunately they now charge $ for this.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2017-06-06T15:43:41.640" LastActivityDate="2017-06-06T15:43:41.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15726" PostTypeId="2" ParentId="15720" CreationDate="2017-06-06T15:43:04.443" Score="0" Body="&lt;p&gt;You can use whatever is suitable from a powder to steel balls, sand to shells. This link gives you some info : &lt;a href=&quot;http://www.kramerindustriesonline.com/tumbling-media.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.kramerindustriesonline.com/tumbling-media.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-06T15:43:04.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15727" PostTypeId="1" AcceptedAnswerId="15730" CreationDate="2017-06-06T16:40:31.173" Score="8" ViewCount="7432" Body="&lt;p&gt;When I watch videos of high speed trains I always see explosions of electricity near the top, or arcing. Why does that happen? I know that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Acela_Express&quot; rel=&quot;noreferrer&quot;&gt;Acela&lt;/a&gt; does it a lot but other high speed trains have it, too.&lt;/p&gt;&#xA;" OwnerUserId="9893" LastEditorUserId="11649" LastEditDate="2017-06-09T09:25:39.960" LastActivityDate="2017-06-09T09:25:39.960" Title="Why do electric trains arc at high speeds?" Tags="|mechanical-engineering|electrical-engineering|power-engineering|rail|" AnswerCount="7" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15728" PostTypeId="2" ParentId="15727" CreationDate="2017-06-06T16:44:56.540" Score="8" Body="&lt;p&gt;It is due to the high voltage still causing a connection when the contacts separate due to an irregularity (bump etc) between the contact and the wire.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-06T16:44:56.540" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15729" PostTypeId="2" ParentId="15717" CreationDate="2017-06-06T16:55:42.467" Score="3" Body="&lt;p&gt;The mathematically determined formula for a sphere with a thin wall under external pressure (which is very inaccurate - keep reading!) is simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{2Et^2}{r^2\sqrt{3(1-\nu^2)}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $E$ is the modulus of elasticity, $t$ is the thickness, $\nu$ is the Poisson's Ratio and $r$ is the radius of the sphere (to the midplane).  This is a result of Euler's method of buckling when applied to sphere's.  Ignoring how we derive this formula for now, it was tested in various laboratories before use.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When tested, however, this formula was found to be completely inaccurate.  Spheres would collapse way before this value was ever reached.  It turns out that any imperfections - any flat spots, or any wrinkles would get vastly exacerbated during the collapse and cause premature failure.  There was no rhyme or reason to the methods.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually, instead of the profound mathematical $\frac{2}{\sqrt{3(1-\nu^2)}}$, the empirically determined factor for most metals was determined to be &lt;strong&gt;0.365&lt;/strong&gt;. (Only &lt;strong&gt;32%&lt;/strong&gt; of the maximum possible).  Other materials (or even different manufacturing methods for the same material!) have different empirical values, and various regulatory agencies and scientific bodies have dedicated hundreds of people and millions of testing dollars to uncovering these factors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different shapes, with different formulas also began to creep in, with similar problems - if they were perfectly curved they could reach the theoretical values, but they would cave within about 1/3 of the maximum theoretical value! &#xA; To uncover the proper values for various shapes and materials you'd have to discuss with ASME or ASTM or other engineering organizations to find out what they've discovered.  Mil Handbook 17F has a useful formula for most composites to reach close values, but it requires taking 900 computations and finding the smallest one!&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-06-06T16:55:42.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15730" PostTypeId="2" ParentId="15727" CreationDate="2017-06-06T19:19:55.850" Score="18" Body="&lt;p&gt;Several factors influence this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.mts.com/cs/groups/public/documents/library/mts_2010192.pdf&quot; rel=&quot;noreferrer&quot;&gt;at high speed&lt;/a&gt;, there's a higher chance of the pantograph losing contact with the catenary wire: at higher speed, bumps in the wire cause a more violent excursion that can exceed the capability of the suspension of the pantograph. &lt;/li&gt;&#xA;&lt;li&gt;Low-speed trains can show arcing too.&lt;/li&gt;&#xA;&lt;li&gt;high-speed trains often use high voltage (15 or 25 kV), which is capable of arcing across longer distances than the lower voltage (e.g. 1500 V) used in older trains.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The point where the pantograph of an electric train makes contact with the trolley wire creates one of the most complex and challenging environments for rail component manufacturers and test engineers to understand, let alone predict and improve. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For trains to operate efficiently, the pantograph must maintain constant contact with trolley wires suspended from catenary systems. Yet these wires and their support structures exhibit different vertical stiffnesses along any given section. The catenary system zigzags in 30 to 100 meter intervals to prevent grooving. The force the pantograph applies to the wire must stay&#xA;  within a well-defined range (70N to 120N). If it is too low, loss of contact results in arcing, which not only causes the train to lose power but damages the trolley wire and the contact bar through etching and overheating. If the force is too high, the resulting friction wears down the&#xA;  wire and contact bar prematurely. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Delivering the right amount of force requires variable vertical motion. But when trains move at higher speeds, pantographs lose their ability to react appropriately. Even when the trolley wire is as flat as possible, it is only flat when it hangs undisturbed. When the pantograph lifts the wire, the&#xA;  resulting deformation creates a wave. If there is too much uplift, the pantograph creates a much larger waveform that causes contact problems for the next pantograph coming down the line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/TGV_world_speed_record#Catenary_modifications&quot; rel=&quot;noreferrer&quot;&gt;catenary wire isn't stationary&lt;/a&gt;: it gets moved around by trains and by the wind. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In general, when a pantograph runs underneath the catenary, it sets up a wave-like disturbance which travels down the wire with a speed determined by the tension in the wire and its mass per unit length. When a train approaches this critical speed, the pantograph catches up with the disturbance, resulting in dangerously large vertical displacements of the wire as well as contact interruptions. The top speed of the train is then limited by the critical speed of the catenary. This problem was central to the test runs, since it was desired to test set 325 at speeds well above the critical speed of standard TGV catenary. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11645" LastEditorUserId="11645" LastEditDate="2017-06-07T09:07:32.760" LastActivityDate="2017-06-07T09:07:32.760" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15731" PostTypeId="2" ParentId="15727" CreationDate="2017-06-06T19:24:19.140" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;When I watch videos of high speed trains I always see explosions of&#xA;  electricity near the top, or arcing. Why does that happen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is a gap in the contact, the electrons shooting through the gap turn the air into a plasma and breakdown the air. Because the air is a plasma, it can conduct current, this happens at around 3kV/mm so you know there is some voltage involved. &lt;/p&gt;&#xA;" OwnerUserId="4921" LastActivityDate="2017-06-06T19:24:19.140" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15732" PostTypeId="2" ParentId="15723" CreationDate="2017-06-06T19:38:15.597" Score="0" Body="&lt;p&gt;A worm / gear with the gear driving a shaft supported by a suitable ball race for the bearing may work. May need extra gearing though.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-06T19:38:15.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15733" PostTypeId="1" CreationDate="2017-06-06T23:45:17.953" Score="2" ViewCount="460" Body="&lt;p&gt;I need to design a planet carrier for a planetary gearbox to which I'm gonna be doing some simple vibration experiments. It basically consists of the ring, sun and 3 planets and the ring it's no more than 20cm diameter.&#xA;I've seen many types of designs for planet carriers surfing the web, mainly star type or triangle. I only need the carriers to keep the planets in place but eventually I will need it as the input or the output, so a shaft is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I don't know what kind of bearings I should use on each end of the carrier for the shaft that will go into the planet gears. I'm not a mechanical designer so I don't know about the tolerances either. I was thinking of just measuring the inner diameter of the planets, and design a small shaft to fit very tightly into the gear and the bearing, and have the bearing fit into a bore in the carrier by pressure also, but I don't know if that's also the correct design to go with. I will send everything to be manufactured so I won't have opportunity to mend any mistakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sun = 32 teeth (1.5 module), 48 PD, Ring = 80T 120 PD &amp;amp; Planets = 24T 36 PD&lt;/p&gt;&#xA;" OwnerUserId="4088" LastEditorUserId="4088" LastEditDate="2017-06-07T23:10:06.367" LastActivityDate="2017-06-07T23:10:06.367" Title="Considerations for planet carrier design" Tags="|mechanical-engineering|design|gears|cad|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15734" PostTypeId="2" ParentId="15727" CreationDate="2017-06-07T00:21:01.247" Score="6" Body="&lt;p&gt;As others have posted, a temporary gap between pantograph and overhead conductor is part of the answer, however that's not the complete story.   The other big factor is that the train's motors are an &lt;strong&gt;inductive load&lt;/strong&gt;, which seriously complicates what happens when the circuit is interrupted.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When there is an interruption of a circuit with an inductive load, the &lt;em&gt;current through the load&lt;/em&gt; &lt;strong&gt;cannot go to zero instantly.&lt;/strong&gt;   Instead, current continues to flow through the load, generating an voltage spike at the point of interruption.  (The extra energy to do this is actually come &lt;em&gt;from&lt;/em&gt; the inductive load.)  The voltage rises suddenly until a breakdown (e.g. arc) occurs.   One an arc has formed, the voltage drops, but less voltage is needed to sustain an arc because plasma is more conductive than air at typical temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The currents flowing for a high-speed train will typically be much higher than their low-speed counterparts, so the voltage developed when the circuit is interrupted will be higher.&lt;/p&gt;&#xA;" OwnerUserId="11649" LastEditorUserId="11649" LastEditDate="2017-06-08T22:46:03.290" LastActivityDate="2017-06-08T22:46:03.290" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15736" PostTypeId="1" CreationDate="2017-06-07T02:50:33.657" Score="3" ViewCount="611" Body="&lt;p&gt;This device is about five feet tall.  It is located in a ground-floor alcove in the exterior of a commercial building on a public street in Santa Monica, California.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V3z4j.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V3z4j.jpg&quot; alt=&quot;Red pipes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the chained manual valve handles seems to also be attached to some sort of electric switch.  (The junction box for the cables is visible on the wall behind the pipes in the upper-right quadrant of the picture.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The small black attachment in the lower left quadrant is a water flow or&#xA;volume meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our best guess is that it has something to do with fire suppression.  But what?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it located on the outside of the building?  What are the valve handles for?  Presumably the chains are to prevent tampering by passersby, but why are the handles there at all, if the apparatus is outside the building?  Who is expected to use them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the two seemingly disconnected pipes at the bottom for?  Some sort of drainage?  The two bottom pipes seem to be linked by some sort of feedback apparatus; is this right?  What is it for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more details in your answer, the better.  I am totally mystified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="11653" LastEditorUserId="11653" LastEditDate="2017-06-07T03:07:48.887" LastActivityDate="2017-06-07T05:28:54.097" Title="What is this plumbing apparatus?" Tags="|pumps|valves|fire-sprinkler|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15737" PostTypeId="2" ParentId="15736" CreationDate="2017-06-07T03:05:57.593" Score="4" Body="&lt;p&gt;it has to be water for fire because its red. its on the out side so they can control the water without going into the building. the open pipe is to drain the water for maintenance. it does not have a meter on it so they are not paying for the flow of water.so for fire. all the valves are open. chains are so you cant open or close the valves. the electric system is used to see if valve is tampered with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;also the two valves on top where it splits to 3 pipes they are flow valves. so when it detects the flow of water it sends a alarm to the fire department leting them know the fire systems are going. and there is a fire&lt;/p&gt;&#xA;" OwnerUserId="11651" LastEditorUserId="11651" LastEditDate="2017-06-07T05:28:54.097" LastActivityDate="2017-06-07T05:28:54.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15738" PostTypeId="2" ParentId="7605" CreationDate="2017-06-07T03:30:42.613" Score="1" Body="&lt;p&gt;There are several issues going on that I believe are making the model not very accurate. They all in some way relate to your question about the convection:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Breaking up the flow into 5 discrete chunks is a good start, but probably not a sufficient number. Keep increasing the number of nodes in your model until the answer stops changing. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The next thing to check is the Reynold number of your flow and the length it will take to become fully developed. If your flow is fully developed for most of the length, the heat transfer coefficient will do a good job modeling the problem you are describing with the air not contacting the wall. If the flow is not fully developed (which is my gut guess) then the heat transfer coefficient won't do a good job. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In general, you should consider the high level thermal resistances between the coil and the duct wall. Since air is such a poor conductor and has such a low density, it is a poor heat transfer medium. Since the heat transfer coefficient on the surface of the coil to the air will be similar to the heat transfer coefficient from the air to duct, the extreme difference in dT between the coil and air and the air and wall will result in a very large Q to the air and a very small Q to the duct wall from the air. It is not impossible that the air actually cools the duct wall instead of heating. If you work out the air-to-duct heat transfer coefficient you may find that it is so trivial that it is likely not the source of error in your model.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Without numbers I can't really say for sure, but my guess is that the air in your model is not actually the primary means of heat from the coil to the wall. Instead, the radiation in dominant. Thus, I would look to improve the accuracy of that calculation first if you want to predict wall temperature. For example, what emissivity are you using for the duct wall? How are you calculating the view factor between the coil's interior edge and the wall? Etc etc. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hope this helps and that this is still even remotely relevant 1 year late. &lt;/p&gt;&#xA;" OwnerUserId="2780" LastActivityDate="2017-06-07T03:30:42.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15743" PostTypeId="2" ParentId="15733" CreationDate="2017-06-07T09:31:12.547" Score="2" Body="&lt;p&gt;The carrier can be pretty much any shape - if you don't have the manufacturing capacity, a circle will do just fine. The fancy shapes are to reduce weight (and as result friction and moment of inertia) by reducing the amount of material used, but they are in no way necessary. Only layout of the axes is important: central plus three planets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll need axles and bearings on the carrier for the planets (the usual approach is just friction bearings, e.g. brass cylinders, but if your loads are higher, you may go with small ball bearings) - but you don't need bearings for the central axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your carrier is where you output the torque (ring gear is fixed) you just fix the central axis to it firmly. If it's dual-sided (say, axles with bolts, edges extending over the sides of the ring - self-stabilizing against axial direction forces) - the other side just requires a hole larger than the axis propelling the sun gear. (you can use a bearing there, but it's better to fix the bearing to the chassis than to the carrier. It's also not necessary at all; the planets acting as balls of a ball bearing, the gearbox acting as bearing for both the sun and ring.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use a variant with fixed carrier, moving sun and ring (axial reversal of rotation) then you just fix the carrier to the chassis, no bearings for anything except the planets.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-07T09:31:12.547" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15744" PostTypeId="2" ParentId="15727" CreationDate="2017-06-07T10:30:07.657" Score="2" Body="&lt;h1&gt;It's not about voltage*, it's about current.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;*line voltage&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a high-current circuit is interrupted (especially an inductive one), an arc is formed between the contacts breaking away. The high current then sustains the arc: ohmic heating turns air into plasma, while the plasma conducts the current. It's the basis of arc welding, which uses hundreds of amps at voltage as low as 20V. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=-qH-muG4d0M&quot; rel=&quot;nofollow noreferrer&quot;&gt;Welding at high speed 5000 fps (close-up arc, spatter visible)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even low-voltage (usually 600-800V) trams moving at walking pace do produce arcing and sparks at breakpoints in catenary, while subway does same at the level of the power rail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=gEe0OIq_4UA&quot; rel=&quot;nofollow noreferrer&quot;&gt;Subway Sparks | New York Snowstorm 2017&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the high-current requirement, sparks happen mostly when the train is accelerating (eg from a standstill), or when it draws lots of power to sustain high speed, but they never happen when it's coasting at idle power, despite voltage being the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In low-speed operation, this happens mostly when a break of contact is introduced to the wire-shoe system externally, eg by a physical gap separating different circuits or by contamination from ice, snow or leaves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In high-speed operation, in addition to all the low-speed ones, extra breaks are created by the pantograph jumping on catenary irregularities, just like an off-road truck which gets it's wheel momentarily airborne when it goes too fast on bumps. Some of those irregularities are introduced by the pantograph itself: one can imagine pantograph on a catenary as an upside-down acrobat on a tightrope. Instead of gravity acting on the acrobat downwards, the pantograph pushes the catenary upwards via a spring, so the whole system jumps up and down when passing under suspension points.&lt;/p&gt;&#xA;" OwnerUserId="1931" LastEditorUserId="1931" LastEditDate="2017-06-09T07:55:05.950" LastActivityDate="2017-06-09T07:55:05.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15745" PostTypeId="2" ParentId="15588" CreationDate="2017-06-07T11:17:46.300" Score="2" Body="&lt;p&gt;Actually, no effort is put into balancing it. Every loco obeys the command from the heading unit, eg &amp;quot;power at 70%&amp;quot; and applies it to the wheels. It doesn't really matter which one does most of the pulling. When units are of same model, then the power will be roughly same. But sometimes completely dissimilar locomotives are used together (eg helper engine at the back), so it's bit like communism: &amp;quot;From each according to his ability&amp;quot;. More curious still, most modern locos and MUs have single traction motor per axle, so even with single locomotive your concern is still valid, balancing the power between axles.&lt;/p&gt;&#xA;&lt;p&gt;The worse that could happen is wheelslip. But it's not related only to the difference of power between units, but also (or rather mostly) to applying correct power for a given speed and the friction between a single wheelset and the rail (&lt;a href=&quot;https://en.wikipedia.org/wiki/Adhesion_railway&quot; rel=&quot;nofollow noreferrer&quot;&gt;adhesion&lt;/a&gt;). Eg. the first unit could be on track contaminated with slippery leaves while others still have it clear. And the answer is that each unit (sometimes even each axle) manages slip on it's own.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;When certain wheels start slipping, power to them is reduced, until they stop slipping. That's all.&lt;/strong&gt; (Additionally, sand is also applied to increase grip.) Basically, it's same concept as ABS but applied for acceleration instead of braking.&lt;/p&gt;&#xA;&lt;p&gt;//edit: See &lt;a href=&quot;https://www.youtube.com/watch?v=7kmkGSSaxMU&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; video about duplex steam locomotive T1. It's basically two 2-axle steam engines sharing one boiler, so most of the muti-engine concerns apply here. At about 1:12 there is a clear example of the front engine slipping and overspeeding - they both are at same power, as T1 had one throttle controlling both engines, yet acceleration causes the front to lift slightly and suddenly it turns out that the front set has too much power while the rear one is perfectly fine. This is visible example that the engines should not be at exactly same power setting. (I've used steam example because connecting rods make observing slip easier)&lt;/p&gt;&#xA;&lt;p&gt;//another edit: as the previous video is no longer available, &lt;a href=&quot;https://www.youtube.com/watch?v=UFrXabF-1SI&quot; rel=&quot;nofollow noreferrer&quot;&gt;here's&lt;/a&gt; a video of assorted wheelslips, including diesel multiple-operation. At 0:46 there's another T1 situation, but more relevant to your question are one diesel unit losing all adhesion spectacularly at 6:07, and another diesel unit slipping one bogie/truck out of two at 11:07.&lt;/p&gt;&#xA;" OwnerUserId="1931" LastEditorUserId="1931" LastEditDate="2022-10-19T08:32:01.183" LastActivityDate="2022-10-19T08:32:01.183" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="15746" PostTypeId="2" ParentId="15723" CreationDate="2017-06-07T11:26:31.013" Score="1" Body="&lt;p&gt;There are two things the motor needs to be able to do. Firstly provide enough starting torque to overcome static friction in the bearings and accelerate the ball up to the required speed in a reasonable amount of time. Secondly to be able to provide enough torque to maintain the desired speed against the friction in the bearings for however long you want to run it for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind that not all motors are rated for continuous running, this really comes down to how well cooled they are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally on of the big determining factors will be how good the bearings are and how well balanced the whole thing is. For a target speed of 1rpm you will need a fairly substantial reduction ratio gearbox anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the most convenient solution will be  a DC motor with an integrated planetary gearbox as these are an off the shelf solution to high torque low speed applications. At a very rough guess I would say you are looking at something in the couple of hundred watt range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also want to look as something like the hub and stub axle from a small trailer for the bearing as these are again a cheap and ready made solution. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-06-07T11:26:31.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15747" PostTypeId="2" ParentId="15727" CreationDate="2017-06-07T16:19:33.323" Score="0" Body="&lt;p&gt;Another factor is that overhead line profile changes much more rapidly at high speed. The contact wire isn't at exactly the same distance from the rail all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pantograph is constantly readjusted to apply a constant pressure on the contact wire, but at high speed that doesn't happen fast enough. When the pressure on the contact wire is insufficient, a small bump is all it takes to send the pantograph down a few mm, creating a visible arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for reference, low-voltage trains are capable to create quite visible arcs just as well (the lower voltage is usually compensated by the fact that it's DC), if they go fast enough or the contact wire is in a bad shape.&lt;/p&gt;&#xA;" OwnerUserId="5550" LastEditorUserId="5550" LastEditDate="2017-06-07T16:35:45.993" LastActivityDate="2017-06-07T16:35:45.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15748" PostTypeId="2" ParentId="15727" CreationDate="2017-06-07T16:53:55.603" Score="3" Body="&lt;p&gt;The upforce on a pantograph is 15-40 pounds, 60 pounds at the outside.  (7-18kg, max 30 or so). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trolley (contact) wire is made of solid bronze or copper, typically 4/0 to 400kcmil (107-200mm^2), with a stranded steel messenger (catenary) wire of 3/8-1/2&quot; (10-13mm) diameter.   The messenger wire is supported every 100-200 feet (30-60m) and it supports the contact wire every 6-10' (2-3m).  So the contact wire is free to rise up even a foot (0.3m) as the train passes.  It often has a stabilizer bar to keep it from moving laterally but is free to move vertically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As discussed, any irregularity in the contact wire, or in how it's hung, can cause pantograph and wire to separate for a moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wave action in the wire can also cause a momentary separation.  Sufficient wire or train movement can cause the wire to move out onto the curved &quot;horn&quot; of the pantograph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/15Sn3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/15Sn3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/OJ8GK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OJ8GK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pantograph_(transport)&quot; rel=&quot;nofollow noreferrer&quot;&gt;src&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irregularities in the pantograph's running surface can also cause arcing.  There are typically inset copper or bronze slides; physical damage to a slide or simply a burned spot from arcing may cause the wire to lose contact.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also a pantograph typically has two slides, fore and aft, and the pantograph has either linkage or strong springs to keep it level. If there is any binding or broken linkage or a fatigued or broken spring, it may not be level and may ride on its heel or toe, causing poor contact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arcing of course is caused by current.  Current might remain continuous through the arc (that tendency being proportional to voltage, more likely on the high voltage systems used in high-speed rail) -- however high speed air movement is likely to snuff out the arc, severing power to the train momentarily.  Talk about voltage spikes! &lt;/p&gt;&#xA;" OwnerUserId="11673" LastEditorUserId="11649" LastEditDate="2017-06-08T19:39:24.077" LastActivityDate="2017-06-08T19:39:24.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15749" PostTypeId="1" AcceptedAnswerId="15751" CreationDate="2017-06-07T17:05:38.747" Score="1" ViewCount="47" Body="&lt;p&gt;I am trying to build a reed instrument played by the wind (kind of like these aeolian instruments: &lt;a href=&quot;http://www.zmescience.com/science/physics/aeolian-harps&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eerie musical instruments played by the wind from around the world&lt;/a&gt; but using reeds rather than strings), but given the necessary pressure to make the reed vibrate, I am having difficulties. For example if I place a reed instrument such as a duck call near a fan, it has not enough power to drive it. I tried to place a funnel in between, but still had no luck. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I have two things to consider:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;To find the most suitable type of reed or make one from scratch (something similar to these, but all the online examples are too narrow to capture actual wind: &lt;a href=&quot;http://www.instructables.com/id/Hypotooter-a-mini-musical-instrument&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hypotooter -- a Mini-musical Instrument&lt;/a&gt;) that starts vibrating in the weakest possible wind&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To direct the wind through a funnel-like piping (maybe not a usual funnel shape, but a different shape?), so that the reed is exposed to more power, if this is a good idea at all.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am stuck at this point and don't even know if this is possible at all. I have heard of wind whistles or flutes played by the wind, but only when the wind very strong, however I am looking into a reed instrument because of its different sound quality + the ability for it to be played by lighter winds as well. Is this possible or is there any directions that you could suggest for me to experiment?&lt;/p&gt;&#xA;" OwnerUserId="11674" LastEditorUserId="8055" LastEditDate="2017-06-07T23:16:26.203" LastActivityDate="2017-06-07T23:16:26.203" Title="Reed played by the wind" Tags="|fluid-mechanics|airflow|vibration|acoustics|wind-tunnels|" AnswerCount="1" CommentCount="2" ClosedDate="2017-06-08T18:32:29.937" ContentLicense="CC BY-SA 3.0" />
  <row Id="15751" PostTypeId="2" ParentId="15749" CreationDate="2017-06-07T18:03:52.593" Score="0" Body="&lt;p&gt;Do you &lt;strong&gt;really&lt;/strong&gt; want to make a reed instrument or a reed like instrument?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reeds used in musical instruments generally are very rigid &amp;amp; they need a lot of air pressure on them for the reed to vibrate. A similar effect can be achieved using durable flexible materials such as a tensioned rubber like membrane/diaphragm. A single sheet of membrane can be used, or the membrane can be doubled over to form two vibrating surfaces. One of the advantages of using a membrane is that the force of air required to make a sound would be significantly less than that required for a reed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A kazoo might be an initial place to start investigations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a funnel to concentrate the air pressure is a good idea, but the size of the funnel will be important, as will be the initial strength of the wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do end up with something you like, you might want to consider placing it on a rotating platform, similar to a wind vane so that a change in wind direction can easily be accommodated.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-06-07T18:03:52.593" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15753" PostTypeId="2" ParentId="15702" CreationDate="2017-06-07T20:42:54.040" Score="-1" Body="&lt;p&gt;I would add #5, &lt;strong&gt;dumbing down parental advice so even very stupid parents can get the message clearly without confusion&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classic example is &quot;put your baby in the back seat&quot;.  That advice is applicable to &lt;em&gt;just a few model years of cars from the mid-90s&lt;/em&gt; where they had passenger side airbags, but did not have the seat-occupied detector or the &quot;baby seat locked in&quot; detector which would shut off the airbag in those cases. Nor to cars without passenge air bags.  So, inapplicable to most cars.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Yet this advice was hammered into the skull of every parent&lt;/em&gt;, and made law in many places.  A parent would be stopped and confronted by parents, officials, and police.  Explaining was hopeless. The lower ranks say &quot;don't wanna hear that technobabble&quot;, and the upper ranks would say &lt;strong&gt;&quot;You know that, I know that, but you're also smart enough to know we have to send a consistent, simple message that everyone can understand, and enforce fairly and uniformly.  As a smart one, you have a social responsibility to set a good example.&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That last bit is what it's all about.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, you as an engineer can carefully inspect a car seat, look for the telltale crazing of UV damage or whiting of imapact damage... Evaluate the provenance of friends and family's seats... But the officials are worried that &lt;em&gt;others cannot&lt;/em&gt;, and would recklessly use a seat with obvious damage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'm not so sure that advice is right.  Even a damaged car seat is better than no car seat.  And also better than a family who can't afford a new seat foregoing food or vaccines to get the new seat.  So in that sense, I think the advice is self-serving to both the NHTSA and the manufacturers, with little consideration to other consequences.**  And so I consider #2 to be also valid.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;** such as the epidemic of infants being forgotten in the back seats of cars, dying to the tune of 30-40 a year, a higher rate than babies &lt;em&gt;ever were&lt;/em&gt; hurt by passenger side air bags even when those cars were common. Those cars are mostly off the road, but the forgotten-baby fatalities continue apace.  What a legacy! &lt;/p&gt;&#xA;" OwnerUserId="11673" LastEditorUserId="11673" LastEditDate="2017-06-07T21:09:00.373" LastActivityDate="2017-06-07T21:09:00.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15754" PostTypeId="2" ParentId="7955" CreationDate="2017-06-07T21:32:45.343" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oFs6G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oFs6G.png&quot; alt=&quot;Alpine Maximum Reactions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.sbcmag.info/news/2017/mar/alpine-gives-guidance-reading-component-drawing&quot; rel=&quot;nofollow noreferrer&quot;&gt;SBC Magazine&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently was looking for this same information. Found this thread and then the article. Hope this is helpful in the future!&lt;/p&gt;&#xA;" OwnerUserId="11680" LastActivityDate="2017-06-07T21:32:45.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15759" PostTypeId="1" CreationDate="2017-06-08T10:09:49.167" Score="2" ViewCount="88" Body="&lt;p&gt;What would be the shape having the highest difference of Drag Coefficients regarding opposite sides?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, as per the following table, a hollow semisphere would have respectively Cd=0.38 on one side, and =1.42 on the opposite side. Would there be another shape with greater Cd differences? (with similar Reynolds, around 10^4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aM4ff.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aM4ff.png&quot; alt=&quot;Drag coefficients&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(from Sighard Hoerner's book &quot;Fluid Dynamic Drag&quot;, Chapter 3)&lt;/p&gt;&#xA;" OwnerUserId="8432" LastEditorUserId="8432" LastEditDate="2017-06-08T12:31:05.470" LastActivityDate="2018-06-09T21:01:19.723" Title="Shape with highest difference of Drag Coefficients on opposite sides" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15760" PostTypeId="1" AcceptedAnswerId="15766" CreationDate="2017-06-08T10:30:13.990" Score="1" ViewCount="755" Body="&lt;p&gt;I am using the publication &lt;a href=&quot;https://www.diw.de/documents/publikationen/73/diw_01.c.424566.de/diw_datadoc_2013-068.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Current and Prospective Costs of Electricity Generation until 2050&lt;/a&gt; to source ramping constraints for an economic model of three different gas power plants - CCGT, combustion turbine and steam turbine. As I'm not an engineer I have the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The model is a little simplistic. For that reason, I cannot use&#xA;three different time and cost parameters for hot, warm and cold&#xA;start. For a gas power plant with load factors of between 10 and&#xA;60%, which start would you think is most relevant (i.e. most often&#xA;done): hot (shut down 8h or less), warm (8-50h) or (probably not)&#xA;cold start (more than 50h)? Would you apply something in between hot&#xA;and warm start?&lt;/li&gt;&#xA;&lt;li&gt;Regarding start-up time, there are not many studies for steam&#xA;turbines (p.71 in pdf numbering). Someone advised my to use a rather&#xA;slower start-up time for steam than for combustion. So could I use a&#xA;value similar to CCGT, for example?&lt;/li&gt;&#xA;&lt;li&gt;For the &quot;ramping load gradient limit in %-Pn/min&quot; (p.74 in pdf numbering), I have the same issue with steam turbines as described in 2).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11639" LastEditorUserId="11639" LastEditDate="2017-06-13T09:46:06.783" LastActivityDate="2017-06-13T09:46:06.783" Title="What ramping constraints for gas plant - CCGT, combustion and steam turbine?" Tags="|gas|power-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15761" PostTypeId="1" CreationDate="2017-06-08T11:42:53.493" Score="4" ViewCount="3390" Body="&lt;p&gt;I know the normal process of how bored piles are normally constructed, but I need to know in case we have a water table at 1.5 m under the surface level and the pile must be at a depth of 25 m. How do they prevent the water from filling into the bored pit. Because when they pour concrete they must remove the water from the drilled pit but if the water keep filling the pit and at very deep depth how could they prevent this ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more question I was thinking about is How much does the strength of the concrete get affected in a pile (due to the presence of water and some unwanted soil in it) ?&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2019-07-27T18:16:41.350" Title="Bored Pile construction Procedures When having Water in the ground" Tags="|civil-engineering|geotechnical-engineering|foundations|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15763" PostTypeId="1" CreationDate="2017-06-08T12:15:13.887" Score="1" ViewCount="3260" Body="&lt;p&gt;Hello fellow engineers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend is part-way building an outside storage chest. For whatever reason, the materials and size of storage that he is going for means that the door is heavy to lift and not very practical. I've suggested that he manufacture a counterweight system to aid in opening the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made an attempt to calculate the required weight by modelling using simple moments and balancing forces, but I'm not happy with the result as I'm pretty sure I'm missing things from my model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see this sketch and my 2 calculation sets. (I've modeled this as a simple statics pulley sort of problem.)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rVXAe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rVXAe.jpg&quot; alt=&quot;Sketch&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YCYSB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YCYSB.jpg&quot; alt=&quot;Calc 2&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DRcHI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DRcHI.jpg&quot; alt=&quot;Calc 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I've modeled this as a statics problem, my issue becomes how to account for the force applied by the person lifting the door, which I have assumed would be applied at the eyelet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming my equations are correct, as theta increases, T1 should decrease since M2 is obviously constant. But isn't tension constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'm nearly there but perhaps need a push in the right direction... It's been a while since I've done something like this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="307" LastActivityDate="2017-08-06T14:15:22.820" Title="Help with a simple calculation of the required mass for a counter-weight for a horizontal hatch" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15764" PostTypeId="2" ParentId="15763" CreationDate="2017-06-08T12:53:03.477" Score="-3" Body="&lt;p&gt;This is an incomplete answer, so for that I am sorry. (Limited time to answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that tension is constant, but the vector in the positive y direction is not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps at all!&lt;/p&gt;&#xA;" OwnerUserId="11668" LastActivityDate="2017-06-08T12:53:03.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15765" PostTypeId="1" CreationDate="2017-06-08T13:14:53.540" Score="3" ViewCount="2154" Body="&lt;p&gt;The NSPE stresses the importance of licensing and makes several points about the seal being important.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Only a licensed engineer may prepare, sign and seal, and submit engineering plans and drawings to a public authority for approval, or seal engineering work for public and private clients. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://www.nspe.org/resources/licensure/what-pe&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nspe.org/resources/licensure/what-pe&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Only a licensed engineer, for instance, may prepare, sign, seal and submit engineering plans and drawings to an public authority for approval, or to seal engineering work for public and private clients.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Only PEs can sign and seal engineering drawings;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://www.nspe.org/resources/licensure/why-get-licensed&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nspe.org/resources/licensure/why-get-licensed&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I'm having trouble understanding &lt;em&gt;specifically&lt;/em&gt; when a seal is required and who or what enforces necessary use of the seal. Is there a general approach for this across any given discipline or project?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could imagine these sources of &quot;rules&quot; for using a seal.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Regulation&lt;/li&gt;&#xA;&lt;li&gt;Legislation&lt;/li&gt;&#xA;&lt;li&gt;Insurance requirements&lt;/li&gt;&#xA;&lt;li&gt;Other contractual obligations&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the question is too broad and needs to be narrowed, here's an example...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I worked on an FDA regulated Class III medical device. I read the relevant titles/sections of the CFR and do not recall any mention of using the PE seal. I do not recall any of our hardware design documents or outputs being sealed. I worked on the device control software and I know that none of our software designs, design controls, or outputs were sealed.&lt;/p&gt;&#xA;" OwnerUserId="11690" LastEditorUserId="11690" LastEditDate="2017-06-08T20:08:01.423" LastActivityDate="2021-11-01T22:04:25.410" Title="When is a PE seal required?" Tags="|licensure|seals|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15766" PostTypeId="2" ParentId="15760" CreationDate="2017-06-08T13:15:16.190" Score="0" Body="&lt;p&gt;The primary advantage of gas plants is much faster ramping of output than others - they are energy-inefficient by comparison, but they are meant to handle spikes in demand in the grid. All of them will be of hot start (&amp;lt;8h) as even the steam ones use relatively small boilers that are quick to get up to speed, compared to conventional power plants. As result, they are kept on stand-by or operating at very low output except for when the spike needs to be handled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can definitely use the longer times found with CCGT when describing ST power plants - although &lt;em&gt;even longer&lt;/em&gt; times will be found (gas burners and big boilers are cheaper to build than gas turbines; the CCGT will usually be smaller than ST plant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be more concerned over the inflated numbers for CCGT; they reach ~40-60% of capacity in time typical to GT plants, and then ramp up the output to 100% over the time typical for ST. That's also why you find the wildly varying values for the gradient in the document - not only is there the slow climb of steam part trailing a long way after the gas part, the power output of the gas turbine depend strongly (and non-linearly) on RPM (which again climbs at rate dependent on that power output) - generally, first 30% or so will be rather slow, then the engine picks up and the 50%-80% ramp is really steep. Regardless, 5-10 minutes is an adequate estimate of the time from ignition to full power (not sure what preparations the operator must perform beforehand). And then water in the steam part only starts heating slowly, steam turbine a long way from budging...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any kinds of averages or momentary measurements in CCGT are seriously misguiding due to that awfully non-linear output ramp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, for GT, the 100% output time and climb rate will depend strongly on each other. If you allow the GT to spin up with coils off, all its power directed towards increasing RPM, it will reach peak torque RPM quickly, and provide 100% output in a really short time since. OTOH if you ramp it up while drawing power, the extra load will reduce climb rate of RPM and delay reaching 100% output. So, a choice between 5 minutes of nothing, then 100%, or 12 minutes of output that is maybe 20% over the first 7 minutes, and then climbs to 100% over the remainder of the time.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-08T13:15:16.190" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15767" PostTypeId="2" ParentId="8408" CreationDate="2017-06-08T14:02:07.420" Score="3" Body="&lt;p&gt;First a hydraulic jump is where fluid will transition from supercritical (having a Froude Number greater than one, to subcritical (having a Froude Number less than one). When the Froude number is near one, the jump is very weak and somewhat gradual such that you might not even be able to tell it is &quot;jumping&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out where the jump occurs you need to know when the Froude Number will transition from less than one to more than one. To help with this is an extremely useful equation for modeling open channel flow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dy}{dx}=S\frac{1-\left(\frac{y_n}y\right)^3}{1-\left(\frac{y_c}y\right)^3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; represents the depth of flow in a riverbed, x is the distance along the flow, &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; is the slope of the riverbed (positive being flowing downhill, negative, flowing uphill), &lt;span class=&quot;math-container&quot;&gt;$y_c$&lt;/span&gt; is the depth of flow corresponding to &lt;span class=&quot;math-container&quot;&gt;$Fr=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y_n$&lt;/span&gt; is the &lt;em&gt;normal depth&lt;/em&gt;, the depth of flow corresponding to flow that is balancing the frictional losses with gravitation gains.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$y_c=\sqrt[3]{\frac{q^2}g}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$y_n=\sqrt[3]{\frac{q^2}{C^2S}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; is the flow rate per width of river, &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is Chezy's coefficient (approximated as constant), and &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is acceleration due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A depth greater than &lt;span class=&quot;math-container&quot;&gt;$y_c$&lt;/span&gt; corresponds to a slower velocity, and thus subcritical flow (having a Froude number less than 1). When this is the case, the bottom of the fraction is positive between 0 and 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, when the depth is less than &lt;span class=&quot;math-container&quot;&gt;$y_c$&lt;/span&gt; the flow is supercritical and the bottom of the fraction is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this range therefore, if the top of the fraction is also negative, then the flow will be getting deeper, but if it is positive, the flow will be getting shallower. This corresponds to a depth of flow less than and greater than &lt;span class=&quot;math-container&quot;&gt;$y_n$&lt;/span&gt; respectively. This means that while the flow is supercritical, the depth of flow will always be moving towards &lt;span class=&quot;math-container&quot;&gt;$y_n$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that if you have a steep slope such that &lt;span class=&quot;math-container&quot;&gt;$y_n$&lt;/span&gt; is less than &lt;span class=&quot;math-container&quot;&gt;$y_c$&lt;/span&gt; the depth of flow will move to and then stay at &lt;span class=&quot;math-container&quot;&gt;$y_n$&lt;/span&gt;. If you would like the flow to have a hydraulic jump simply decrease the slope till &lt;span class=&quot;math-container&quot;&gt;$y_n$&lt;/span&gt; is greater than &lt;span class=&quot;math-container&quot;&gt;$y_c$&lt;/span&gt; then the flow will deepen at progressively faster rates until you reach the hydraulic jump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note since &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; is based on your width it's also possible to modify the width rather than slope to cause &lt;span class=&quot;math-container&quot;&gt;$y_n$&lt;/span&gt; to be greater than &lt;span class=&quot;math-container&quot;&gt;$y_c$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wait! After the hydraulic jump the flow will be subcritical, which means the flow downstream can now influence the water level and push the hydraulic jump upstream. How far upstream? Depends on the flow downstream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In subcritical flows it is easiest to work upstream. If you try to work downstream from an initial guess, you'll end up with non-physical flows unless you happen to guess perfectly. This is because in subcritical flows, as you move downstream, the depth moves &lt;em&gt;away&lt;/em&gt; from the normal depth. If you're less than the normal depth, you can quickly end up with a negative depth, and if you're more than the normal depth you can end up with a depth the would overflow your walls quick quickly. However, if you work backwards, going upstream then the depth always moves towards the normal depth, which should always give a valid depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So then the question is where to get your initial condition? Downstream of subcritical flow, and upstream on subscrital flow. The opposite transition of hydraulic jump. This transition is always smooth and thus according to the equation always happens when &lt;span class=&quot;math-container&quot;&gt;$y=y_n=y_c$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$y_n$&lt;/span&gt; decreasing downstream. An extreme example would be the lip of a waterfall or edge of an overflowing dam. For given flow rate, the depth will always be easy to predict at these choke points. As an added bonus the slope at these choke points is largely flow rate independent:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y_c=y_n$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sqrt[3]{\frac{q^2}g}=\sqrt[3]{\frac{q^2}{C^2S}}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac1g=\frac1{C^2S}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$S=\frac{g}{C^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So anytime the slope goes from shallower than &lt;span class=&quot;math-container&quot;&gt;$\frac{g}{C^2}$&lt;/span&gt; to steeper there might be a transition to supercritical flow right there, and anytime the slope goes from steeper than &lt;span class=&quot;math-container&quot;&gt;$\frac{g}{C^2}$&lt;/span&gt; to shallower there might be a hydraulic jump a bit upstream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say might because it is possible that a choke point further downstream is causing the depth to be too high to cause any transitions (you could be at the bottom of a lake for example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the slope is constant and shallower than &lt;span class=&quot;math-container&quot;&gt;$\frac{g}{C^2}$&lt;/span&gt; for a long time (many miles), then odds are, near the top of this section the depth will have converged to &lt;span class=&quot;math-container&quot;&gt;$y_n$&lt;/span&gt;. So by this shortcut you don't have to start all the way back from a choke point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either case, you can model the flow going upstream from the subcritical section and downstream from the supercritical section. Then having modeled the supercritical section you can plot vs position the depth of flow that would result from a hydraulic jump at that location. Any time that plot intersects the subcritical plot is a location where a hydraulic jump could be stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Equations are derived in full in &lt;a href=&quot;https://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-090-introduction-to-fluid-motions-sediment-transport-and-current-generated-sedimentary-structures-fall-2006/course-textbook/ch5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;MIT's open courseware chapter on open channels&lt;/a&gt;. My analysis of those equations was also guided by the analysis presented there.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="1122" LastEditDate="2019-04-30T12:01:47.793" LastActivityDate="2019-04-30T12:01:47.793" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="15769" PostTypeId="1" AcceptedAnswerId="15771" CreationDate="2017-06-08T19:01:03.437" Score="0" ViewCount="175" Body="&lt;p&gt;I ran across the diagram below in the &lt;a href=&quot;https://i.stack.imgur.com/fvlRy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;mindtomachine blog&lt;/a&gt; post describing an homebrew arc welder.   It shows two transformers with inputs in parallel and outputs in series.   Is there some advantage to this arrangement rather than a single transformer with a doubled turns ratio?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it's just what he had at hand, but I wanted to confirm with someone having EE background. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5a829.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5a829.png&quot; alt=&quot;Welder Circuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.  I'm &lt;em&gt;not&lt;/em&gt; planning on building this, just curious.&lt;/p&gt;&#xA;" OwnerUserId="11649" LastEditorUserId="11649" LastEditDate="2017-06-08T19:19:07.560" LastActivityDate="2017-06-09T01:14:24.370" Title="Transformers in parallel/series" Tags="|electrical-engineering|power-electronics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15771" PostTypeId="2" ParentId="15769" CreationDate="2017-06-08T20:44:30.330" Score="1" Body="&lt;p&gt;I would suggest that this is a case where they had used two old high current output transformers with the outputs in series to double the voltage ie just using what was available (second-hand) instead of buying new, a good example of re-purposing.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-08T20:44:30.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15772" PostTypeId="1" CreationDate="2017-06-08T21:22:25.543" Score="1" ViewCount="128" Body="&lt;p&gt;While I was looking for the definition of the center of mass , I found this text :&lt;br&gt;&#xA; &quot;If a single force acts on a    body and the line of action of the force passes through the centre of mass, the body will have linear    acceleration but possess no angular acceleration.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why does the body have an no angular acceleration if the applied force passes the center of mass ?&lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="9742" LastEditDate="2017-06-09T12:28:48.757" LastActivityDate="2017-06-09T12:28:48.757" Title="why does the body have an no angular acceleration if the applied force passes the center of mass?" Tags="|mechanical-engineering|statics|dynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15773" PostTypeId="2" ParentId="15772" CreationDate="2017-06-08T22:42:46.090" Score="2" Body="&lt;p&gt;That's just a really fancy way of saying &quot;if you push on something right in the center, it will move but won't spin&quot;.  Are you familar with pool/billiards?  If you hit the ball right in the exact center, it will move, but it won't spin.  If you hit it a little to the side, or a little above or below center, it will both move and spin.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(of course, the problem with this analogy is that the friction will the table will always cause it to roll as it moves... think of playing billiards in space...)&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-06-08T22:42:46.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15774" PostTypeId="2" ParentId="15769" CreationDate="2017-06-09T01:14:24.370" Score="1" Body="&lt;p&gt;Doubling the turns ratio is not exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, you get double the output voltage but you get different Magnetizing Inductance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, the real reason to do something like that could probably be related to heat dissipation. The more wire layers you put, the hotter the center of the core will be for a given power. You could run thicker wire to offset this, but then you run out of core to wind. This could be specifically true for an arc welder in which voltage is actually reduced and you run very very high currents on the output.&lt;/p&gt;&#xA;" OwnerUserId="11699" LastActivityDate="2017-06-09T01:14:24.370" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15775" PostTypeId="1" AcceptedAnswerId="15776" CreationDate="2017-06-09T02:33:01.530" Score="0" ViewCount="245" Body="&lt;p&gt;2-stroke engines seem to offer a massive advantage over 4-stroke engines.  In the case of gasoline engines, there is a need for forced induction, but in the case of large diesels (which invariably have forced induction anyway) it seems that their advantages outweigh the disadvantages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why is it that all newly developed diesels (not counting large ship engines and research designs such as Achates Power) are 4-stroke?&lt;/p&gt;&#xA;" OwnerUserId="11700" LastActivityDate="2017-06-11T16:10:35.023" Title="Why are 4-stroke diesel engines preferred in new developments?" Tags="|automotive-engineering|diesel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15776" PostTypeId="2" ParentId="15775" CreationDate="2017-06-09T04:53:50.263" Score="1" Body="&lt;p&gt;There have been smaller two stroke diesels produced (the Comma TS3 or the Detroit Diesels - very popular) but to make the TS3 work successfully they need forced induction -a Rootes blower was used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4-stroke tends to be used as their the fuel consumption is generally better while there are more components, another consideration is the ancillary component layout - a 4 stroke may be easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The are questions on other stack exchanges with more / different information : you could start with this one : &lt;a href=&quot;https://mechanics.stackexchange.com/questions/24213/two-stroke-diesel-engine-do-they-use-forced-induction-on-them?rq=1&quot;&gt;https://mechanics.stackexchange.com/questions/24213/two-stroke-diesel-engine-do-they-use-forced-induction-on-them?rq=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-09T04:53:50.263" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15777" PostTypeId="1" CreationDate="2017-06-09T07:01:23.657" Score="1" ViewCount="64" Body="&lt;p&gt;I am looking for a light emitting device, which can produce a narrow and accurate beam, with a small &quot;dot&quot; size, for a significant distance (around the 100-150 meters) similar to that of a laser, but that it would be safer when comes in contact with eyes / other body parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**Edit: My goal is that I want to be able to point 1 device at another, and that the other device will detect it was pointed at (But it should be placed on the body so that's why I want it to be safe), so the beam doesn't actually needs to be visible&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such device exist?&lt;/p&gt;&#xA;" OwnerUserId="11703" LastEditorUserId="11703" LastEditDate="2017-06-09T07:59:13.970" LastActivityDate="2017-06-09T13:03:55.157" Title="Device that emits a safe beam of light - doesn't have to be visible (Like laser, but safer)" Tags="|optics|lasers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15780" PostTypeId="1" CreationDate="2017-06-09T08:18:52.323" Score="-1" ViewCount="83" Body="&lt;p&gt;I hope this question is okay for here. I've been interested in engineering for quite some time, but havent really ever had the motivation to do anything about it. I took a small principles of engineering class last year but it wasn't very serious. We kind of just messed around with 3D printing and computer games. &#xA;I'm not very good at math. It's not so much that I can't do it, but I just forget it and am often too scatterbrained to focus and get it done. I've also missed years of math and so lack a lot of basic knowledge. &#xA;In my class, we were mostly given formulas to follow for simple problems, and I didn't really run in to too much difficulty. But I'm wondering, in the world of engineering, when you're actually out in the field, let's say mechanical/civil particualry, how much math do you use?&#xA;I've always felt sort of held back from getting involved because I worry I'm not &quot;smart enough&quot; &lt;/p&gt;&#xA;" OwnerUserId="11697" LastActivityDate="2017-06-09T08:34:32.480" Title="Does engineering require extensive math?" Tags="|mathematics|" AnswerCount="1" CommentCount="7" ClosedDate="2017-06-09T11:36:15.663" ContentLicense="CC BY-SA 3.0" />
  <row Id="15781" PostTypeId="2" ParentId="15780" CreationDate="2017-06-09T08:34:32.480" Score="2" Body="&lt;p&gt;Yes, most fields of engineering may need integration, diferentiation, laplace transforms and also trig, solid body rotations etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I don't like math but still managed to complete - if you like engineering or are motivated by it then you will be ok.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-09T08:34:32.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15783" PostTypeId="1" CreationDate="2017-06-09T11:36:30.437" Score="0" ViewCount="449" Body="&lt;p&gt;I'm a competent DIYer and have done a little small-scale ferrous and non-ferrous brazing in the past. It was in a professional workshop the equipment was provided suitable on request, so I didn't need to know how to choose the right gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So right now I'm looking at a few jobs around the home and household workshop scale personal work, and thinking that brazing is the appropriate technique. The work I have backlogged includes a copper/brass/bronze sculpture which has a brazed copper &quot;leaf&quot; shaped piece broken off where it was joined, small scale iron/steel/alum (&amp;lt;3 mm flat/5mm rod ferrous), and other minor and broken things from time to time. It's enough to be worth investing in a small brazing torch and other equipment, although I might only use it once or twice a year if that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a huge amount of info out there on fluxes, techniques, and rods, so that's easy. But there is nothing solid that I can find, about torch style and gas choices, is the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've found so far is, temps suitable for different materials (eg above ~ 500C and for copper below about 1000C from memory), temps for different gases (~1500C to 4000C roughly, from memory, sorry if my memory's wrong, the general idea is right I know). I'm also aware of the difference between flame temp and target temp and to expect that the metals and equipment will act as a heatsink. But beyond those basics, I can't find anything to guide me in choosing an appropriate brazing torch type and gas to use. (Solder yes. Flux yes. Protective gas yes. Technique yes. Combustible fuel and torch selection no.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To an extent I can move the torch or adjust the gas flow to change the temp based on the flux and joint appearance, that's easy. But what features or spec items would make a torch suitable for my use, and which gas combinations are most sensible that more experienced brazers would go for? Should I look at heat output or gas flow to know if it's going to be adequate for brazing? I'm going to have very light use (&amp;lt; 1 hour/year?) and for now I'm not expecting heavy duty work, so perhaps some gas + air will be enough and makes economic sense. (Or should that be gas/LPG? No idea.) I might use it more on ferrous steel or filling, once I gain experience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which fuels are likely to be good for me? What makes the difference between a &quot;not hot or consistent enough in actual use&quot; torch and a decent budget-level torch? What sort of gear should my shortlist include (to look up &quot;comparison of...&quot; articles)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all, I don't need more than a basic simple torch but I'd like to get one that can handle a little more than I'm expecting, rather than find after starting a job, that I've bought a lightweight plumbing torch that can't really pump sufficient heat into the joint. That's what I most want to learn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other relevant info might include common &quot;warning signs&quot; to help identify a cheap, poorly built  torch, and maybe also some suggestion how much gas is sensible for me to buy (0.5 L? 10 L? I don't want to run out halfway through my 2nd job and gas lasts for ages on the cupboard shelf).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for helping&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2017-06-09T11:43:35.127" LastActivityDate="2017-06-09T11:59:19.443" Title="Type of brazing torch / gas that would be appropriate?" Tags="|equipment-selection|brazing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15784" PostTypeId="2" ParentId="15783" CreationDate="2017-06-09T11:59:19.443" Score="1" Body="&lt;p&gt;The most versatile brazing setup is oxy/fuel. Here the main hardware is a mixer head with interchangeable nozzle and hoses/regulators for oxygen and fuel. You can use either disposable gas bottles or larger industrial type rental bottles, which is best really depends on how much gas you are likely to use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of setup gives a high flame temperature and (depending on the nozzle shape and size) a wide variety of flame shapes. as such the same basic kit can be used for a variety of tasks from fine soldering to quite heavy brazing on steel etc. The potential high output allows you to work on quite heavy section materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the heat output is directly related to the nozzle size and different nozzle styles can give different flame shapes for different jobs. Also the basic kits come in a range of sizes, micro, standard, heavy etc &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical thing here is not so much the flame temperature as the heat output as you need to be able to apply heat to the joint faster than it is conducted away clearly you need a lot more output for brazing a steel frame than you do for soldering jewellery.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any oxy/fuel setup, propane is generally a good choice of fuel. Unless you want to do fusion welding there is little point in using acetylene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other alternative is to use a plumbers type torch which requires fuel only and uses a venturi burner. These are adequate for small brazing and hard soldering jobs and a bit cheaper to buy but don't have the same versatility to tackle larger jobs as oxy/fuel.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-06-09T11:59:19.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15785" PostTypeId="2" ParentId="15777" CreationDate="2017-06-09T13:03:55.157" Score="3" Body="&lt;p&gt;There are eye-safe lasers, both in the visible and IR.   For your detection system, make sure you understand the power required for your  target (receiver) to register a hit.  Pretty much any commercial diode laser will have a small enough divergence (assuming you include a collimating lens) that point-spread over 150 m is minimal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you also need to consider dwell time on target. If the person is moving, the beam may hit the detector too briefly to produce an alarm.  Check the pulse-width or, equivalently, minimum energy (not power) required by the detector you pick out. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-06-09T13:03:55.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15786" PostTypeId="1" AcceptedAnswerId="15790" CreationDate="2017-06-09T20:54:14.887" Score="2" ViewCount="1382" Body="&lt;p&gt;In order to find the coordinates of the center of an area, we can use this formula: &#xA;$$\begin{align}&#xA; \overline{x} &amp;amp;= \frac{\int x\text{d}A}{A} \\&#xA; \overline{y} &amp;amp;= \frac{\int y\text{d}A}{A}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we derive this formula? Can you help me understand it so it becomes intuitive?&lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="1832" LastEditDate="2017-06-10T13:13:21.923" LastActivityDate="2021-09-19T04:27:03.417" Title="How can we derive the formula of the centroid?" Tags="|statics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15787" PostTypeId="2" ParentId="15786" CreationDate="2017-06-09T21:27:47.857" Score="1" Body="&lt;p&gt;$x dA$ is the differential moment around the y-axis. Xbar is the x coordinate where the integrated differential moments cancel each other out. Obviously the same is true for $y dA$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the area as a very thin plate with uniform weight per area. The centroid is the point where you could hold it on your finger tip without falling down. &lt;/p&gt;&#xA;" OwnerUserId="8707" LastActivityDate="2017-06-09T21:27:47.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15788" PostTypeId="1" AcceptedAnswerId="23699" CreationDate="2017-06-10T09:28:31.310" Score="3" ViewCount="519" Body="&lt;p&gt;As you know in linear theory of elasticity equations for strains are given as (according to picture):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4pGfL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4pGfL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what will be if an element will just rotate (strain equation is on the picture):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0mBo7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0mBo7.png&quot; alt=&quot;rotation of an element&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively i think there should be no strains. The main question is how to deal with it? &lt;/p&gt;&#xA;" OwnerUserId="11713" LastActivityDate="2018-09-11T01:39:47.550" Title="How to deal with rotation of an element in theory of elasticity?" Tags="|mechanical-engineering|structural-engineering|structural-analysis|stresses|elastic-modulus|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15789" PostTypeId="1" CreationDate="2017-06-10T09:48:45.623" Score="0" ViewCount="12494" Body="&lt;p&gt;It's from Apple's latest Earth Day commerical called &lt;strong&gt;“Can a building breathe”&lt;/strong&gt;.&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=tNzCrRmrtvE&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=tNzCrRmrtvE&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The commercial is about Apple's new naturally ventilated building called &quot;Apple Park&quot;. It says &quot;We run chilled water tubing through that concrete mass, and then we use outside air to be our dominant cooling system, and you're using very little energy to get the &lt;strong&gt;tempered water&lt;/strong&gt; to do this. So it's a breathing building.&quot;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JCilQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JCilQ.jpg&quot; alt=&quot;From Apple&amp;#39;s Earth Day commercial &amp;quot;Can a building breathe&amp;quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've searched on the internet for the meaning of &lt;strong&gt;&quot;tempered water&quot;&lt;/strong&gt; and some of them say &lt;strong&gt;&quot;tempered water&quot;&lt;/strong&gt; means &quot;Water in the temperature range from 85°F (29°C) to 110°F (43°C)&quot;. So it's like mild hot water. Can somebody explain what exactly &lt;strong&gt;&quot;tempered water&quot;&lt;/strong&gt; is and how it helps cool the air? &lt;/p&gt;&#xA;" OwnerUserId="11714" LastActivityDate="2021-04-07T04:52:07.720" Title="What does “tempered water” exactly mean and how does it help cool the air?" Tags="|thermodynamics|aerodynamics|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15790" PostTypeId="2" ParentId="15786" CreationDate="2017-06-10T10:54:00.347" Score="5" Body="&lt;p&gt;Think about the centroid that way:&#xA;If you have this curve $y(x)$ (first graphic), where its &quot;mass&quot; is its area $A$.&#xA;The centroid $(x_s,y_s)$ is the place where you would have to put a point mass of equal mass to produce the same static moment about any arbitrary point, thus you could balance this thin plate on your finger if you put your finger right at the centroid.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HULWJ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HULWJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to derive the first moment of area:&#xA;We know that the moment of a point mass wrt. to a point equals the mass multiplied by the distance to said point.&#xA;In the second graphic you have a bar chopped out of the first one.&#xA;The moment of the hatched area wrt to $x_0=0$ equals:&#xA;$$ S_{y'}= m'y = (y'+\Delta y - y)\Delta x \cdot y'= y'\Delta y \Delta x $$&#xA;$$ \Delta y \Delta x =\Delta A \qquad \mbox{(that's the hatched area)} $$&#xA;So the total &lt;em&gt;First Moment of Area&lt;/em&gt; of the bar equals&#xA;$$ S_{y,bar}= y\cdot dA $$&#xA;If you integrate that for the whole $y(x)$ you get the entire &lt;em&gt;FMoA&lt;/em&gt;:&#xA;$$ S_y=\int y\cdot dA $$&#xA;Now the moment generated must equal to that of a point mass at the centroidal coordinate $y_s$, which equals:&#xA;$$ S_{y,centroid}=m\cdot y_s $$&#xA;As stated before the mass is equal to the entire area $m=A=\int dA$&#xA;Thus:&#xA;$$ S_y=\int y\cdot dA = \int dA \cdot y_s$$&#xA;$$ \to y_s=\frac{\int y dA}{\int dA} $$&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2017-06-10T10:54:00.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15791" PostTypeId="2" ParentId="15570" CreationDate="2017-06-10T13:18:32.190" Score="1" Body="&lt;p&gt;Because along z direction, body thickness is so small (ideally zero) that practically it consists of only the two surface boundaries (top and bottom, see figure) . Because no pressure is allowed in these boundaries (otherwise we violate plane stress conditions), equillibrium of internal stresses and external actions cant't be satisfied on them, unless $\sigma_z = 0$. Because no material exists (ideally) between these two boundaries, no $\sigma_z$ is possible &quot;internally&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical applications of plane stress conditions include thin shells, loaded only in their plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X3ebF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X3ebF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9620" LastActivityDate="2017-06-10T13:18:32.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15793" PostTypeId="2" ParentId="14304" CreationDate="2017-06-10T17:13:11.143" Score="1" Body="&lt;p&gt;I had a peugeot 106 electric (battery only, not hybrid) and it was possible to avoid using the brake pedal completely if you read the traffic correctly as it had regenerative braking - the energy went back into the batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if the batteries were already at 100% this won't happen but an absolute max charge is very rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been mentioned the energy available from slowing the vehicle down is what is available to increase the charge less losses. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-10T17:13:11.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15794" PostTypeId="1" CreationDate="2017-06-10T17:32:06.187" Score="3" ViewCount="506" Body="&lt;p&gt;For residential design, looking at IBC 2012 Table 1806.2 it seems a safe soil load to design for is 1500 psf if you want to save time and cost isn't yet an issue.  (assuming you're not building on a swamp)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the problems of swelling and shrinking clay with the weather, for the types of soil listed in the IBC, is there a factor to use to safeguard against settlement to a certain small amount, or is this already 'built in' to the tables?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what the IBC limits its settlement to with the 'presumptive load bearing' table?  &lt;/p&gt;&#xA;" OwnerUserId="11698" LastActivityDate="2017-07-11T20:47:26.780" Title="What is a safe soil load to avoid settlement?" Tags="|structural-engineering|geotechnical-engineering|soil|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15798" PostTypeId="1" AcceptedAnswerId="15800" CreationDate="2017-06-10T21:20:48.730" Score="21" ViewCount="229766" Body="&lt;p&gt;I tried to deliver a refrigerator and the customer saw that the refrigerator was lying down in the truck instead of standing up. He refused to accept it, claiming that the compressor would be damaged by having it on its side.  I tried to explain that that made no sense, but his friend came to me and also said the same thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is transporting a refrigerator on its side a problem? If so, why?&lt;/p&gt;&#xA;" OwnerUserId="10921" LastEditorUserId="255" LastEditDate="2017-06-12T09:30:40.170" LastActivityDate="2020-10-13T11:09:42.687" Title="What's wrong with transporting a refrigerator on its side?" Tags="|refrigeration|" AnswerCount="6" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15799" PostTypeId="2" ParentId="15798" CreationDate="2017-06-10T22:34:05.080" Score="2" Body="&lt;p&gt;The theory is that the fluid can cause an air (well gas) lock that prevents the fluid moving around the system and the lack of fluid damages the compressor, or the fluid can get into the compressor in a liquid state and cause hydraulic lock. Have not tested it myself as have always allowed plenty of time before turning it on - 5 hours or more... posted as an answer on friendly advice.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-10T22:34:05.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15800" PostTypeId="2" ParentId="15798" CreationDate="2017-06-10T22:40:16.993" Score="27" Body="&lt;p&gt;After referring to some good online resources such &lt;a href=&quot;http://removeandreplace.com/2015/10/30/refrigerator-laid-on-side-when-moving-now-will-not-cool-properly/&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt;, I know why we shouldn't transport it laying down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compressor is filled with oil which is critical to its operation.  In the normal upright position gravity keeps the oil in the compressor.  When we lay the refrigerator flat, some of the oil can leave the compressor and go into the cooling lines.  The oil is a thick viscous fluid and can clog the cooling lines thus hampering the refrigerator's ability to cool.  Lack of oil in the compressor can also damage the compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we must lay the fridge down, it is better not to lay it 100% flat, but rather to keep an angle so gravity keeps the oil in the compressor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it has lied down flat for some time, then wait at least 24 hours with the refrigerator in the normal upright position before turning it on.&#xA;This will allow sufficient time for oil that may have gotten into the cooling lines to flow back into the compressor.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-06-10T22:40:16.993" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15801" PostTypeId="2" ParentId="15789" CreationDate="2017-06-11T00:31:21.953" Score="0" Body="&lt;p&gt;The chilled water is used to cool the air as the air passes over the water filled tubes which would be colder than the air, thus reducing the air temperature and increasing the water temperature. This is meant to replace the traditional AC system of buildings. Tempered water likely refers to the water that is being used as the cooling fluid (which would have been refrigerated some how to get to a certain temperature) but it can't be definitively said what temperature it is at. &lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2017-06-11T00:31:21.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15802" PostTypeId="1" CreationDate="2017-06-11T04:17:44.070" Score="1" ViewCount="174" Body="&lt;p&gt;I want to design a tank to hold liquid propane. The proposed tank would have one inlet valve, one output valve, and no sharp edges. I've read that the pressure required to maintain propane in a liquid form at 100 degrees Fahrenheit is 177 psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example analogous to my case; I have a spherical, 40 sq. in. tank in which I want to store liquid propane. If propane must be held at a pressure of 177+ psi to remain in liquid form at 100 degrees Fahrenheit, then the tank must be able to withstand 177 psi * 40 sq. in. = 7080 lbs at all points of the tank's surface. What material properties should I look at to find a material suitable to withstand that force? I want to think tensile strength, but would like an expert's advice.&lt;/p&gt;&#xA;" OwnerUserId="11723" LastActivityDate="2019-05-17T06:23:10.410" Title="What material properties matter when selecting a material to withstand a specific pressure (as a sealed tank)?" Tags="|materials|hydraulics|pneumatic|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15803" PostTypeId="1" AcceptedAnswerId="15804" CreationDate="2017-06-11T08:35:17.263" Score="5" ViewCount="561" Body="&lt;p&gt;I am trying to understand / interpret one recommendation I read on a Caterpillar website for maximizing excavator life:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Never dig over final drive.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;**&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain what this means? What's the &quot;final drive&quot; and what does digging over it mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NSYoS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NSYoS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zKWqN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zKWqN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6277" LastEditorUserId="13416" LastEditDate="2017-11-09T03:33:09.330" LastActivityDate="2022-07-12T00:52:03.533" Title="Understanding a Caterpillar Excavator Recomendation for Increasing Life" Tags="|mechanical-engineering|equipment-selection|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15804" PostTypeId="2" ParentId="15803" CreationDate="2017-06-11T09:08:07.233" Score="6" Body="&lt;p&gt;The final drive is the planetary gear set and hydraulic motor that move the track. Not digging over it means not operating the arm over that area as it stresses the components unnecessarily - so to avoid this it means positioning the machine correctly before starting work or changing its position during work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xyjQh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xyjQh.jpg&quot; alt=&quot;final drive&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="13416" LastEditDate="2017-11-09T19:12:55.400" LastActivityDate="2017-11-09T19:12:55.400" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="15805" PostTypeId="1" CreationDate="2017-06-11T10:50:08.193" Score="4" ViewCount="752" Body="&lt;p&gt;I am from The Netherlands, where noise barriers like these are commonly built along motorways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/9/91/Geluidswal_123.jpg&quot; alt=&quot;noise barrier&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I live in England, where such structures are almost non-existent.  In The Netherlands, living 200 metre from a motorway, I experienced far less noise than I do in England, living roughly 1.5 km from a motorway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much noise does a motorway noise barrier typically block?  I realise the answer will depend on a lot of factors, probably primarily noise barrier design and distance from motorway, and likely others.  The extreme case would be to completely bury the motorway, which should reduce noise to essentially zero.  Ideally, I would like to have a map like the &lt;a href=&quot;http://www.extrium.co.uk/noiseviewer.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;England noiseviewer&lt;/a&gt; where the user can toggle on or off a basic, intermediate, or advanced noise barrier and observe how noise at particular distances gets reduced, although I expect such a map does probably not exist.&lt;/p&gt;&#xA;" OwnerUserId="3599" LastActivityDate="2017-07-11T18:27:18.077" Title="How much noise does a road noise barrier block?" Tags="|highway-engineering|acoustics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15806" PostTypeId="1" CreationDate="2017-06-11T11:03:27.403" Score="0" ViewCount="38" Body="&lt;p&gt;I want to optimise the size and shape of a magnet for a desired pulling force. (Please forgive my limited knowledge of magnetism).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cylindrical magnet fixed to an object will hold it in place by placing a steel plate underneath. There will be a plastic sheet separating them. The poles are the ends of the cylinder magnet, (I know that's not ideal for maximising force); the magnet will be encased by aluminium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I Would like: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I will measure the holding strength of such a cylindrical magnet (known dimensions) directly on the steel plate. Given that, I hope to:&lt;/li&gt;&#xA;&lt;li&gt;calculate the expected holding force of a similar magnet, with the variables being the new cylindrical magnet's dimensions, and the separator's thickness. &lt;/li&gt;&#xA;&lt;li&gt;Accuracy is not crucial. (10-20% perhaps). Ultimately I will use that to optimise factors such as cost. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Factors that can be ignored: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Grade of magnet. (Assume materials and magnetisation are identical). &lt;/li&gt;&#xA;&lt;li&gt;Thickness of steel plate: assume thick enough to simplify calculations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hope this is feasible!&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2017-06-11T11:03:27.403" Title="Calculate the relative pulling force as function of size for neodymium magnets" Tags="|magnets|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15807" PostTypeId="2" ParentId="14304" CreationDate="2017-06-11T14:25:02.610" Score="1" Body="&lt;p&gt;An important energy storage method that is being given very little consideration is gas compression.  The kinetic energy of the vehicle can drive an air pump or compressor that would compress air into a tank specially designed for pressures and heat of compressed air.  Later the energy stored as compressed air can be released through a regulated nozzle to convert to electrical or mechanical energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air pump would be connected mechanically directly to the drive train or the wheels or axle to have the braking effect.  The storage process is not efficient as much heat is generated with the compression.  However, the energy can be stored in the compression tank for a very long time with little loss.  The capacity for energy storage is only limited by the volume of the storage tank and the maximum design pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special consideration has to be made for the lubricating oil used for the air pump/compressor.  It has to be non-combustible other wise it will ignite with the oxygen in the heated compressed air.  This would rule out most hydrocarbon based oils.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastEditorUserId="10921" LastEditDate="2017-06-11T14:36:36.727" LastActivityDate="2017-06-11T14:36:36.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15808" PostTypeId="1" CreationDate="2017-06-11T14:55:31.247" Score="0" ViewCount="240" Body="&lt;p&gt;An important energy storage method that is being given very little consideration is gas compression.  The kinetic energy of the vehicle can drive an air pump or compressor that would compress air into a tank specially designed for pressures and heat of compressed air.  Later the energy stored as compressed air can be released through a regulated nozzle to convert to electrical or mechanical energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air pump would be connected mechanically directly to the drive train or the wheels or axle to have the braking effect.  The storage process is not efficient as much heat is generated with the compression.  However, the energy can be stored in the compression tank for a very long time with little loss.  The capacity for energy storage is only limited by the volume of the storage tank and the maximum design pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Special consideration has to be made for the lubricating oil used for the air pump/compressor.  It has to be non-combustible other wise it will ignite with the oxygen in the heated compressed air.  This would rule out most hydrocarbon based oils.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastEditorUserId="10921" LastEditDate="2017-06-11T15:08:58.103" LastActivityDate="2017-07-13T15:24:48.413" Title="What is a good non-combustible oil to be used in air compression regenerative braking/energy storage system for cars, buses, and trucks?" Tags="|mechanical-engineering|automotive-engineering|chemical-engineering|compressed-air|compressors|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15809" PostTypeId="2" ParentId="15789" CreationDate="2017-06-11T14:57:52.720" Score="0" Body="&lt;p&gt;I think  by&quot;tempered water&quot; , Apple means water at ambient air temperature. So they use power to run fans and pumps for  ambient air/water heat exchangers. Evaporation coolers are usually more efficient but require engineering/maintenance/costs. This is effective when outside temperature is lower than inside temperature ( inside heat caused by lights, computers, people .etc.)  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-06-11T14:57:52.720" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15811" PostTypeId="2" ParentId="15775" CreationDate="2017-06-11T16:10:35.023" Score="1" Body="&lt;h2&gt;It's all about what comes out&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Two-stroke, forced-induction diesels have been a mainstay of heavy industry for quite some time (Detroit Diesel 71 series, EMD 567/645/715, never mind ship engines).  However, they aren't as amenable to precise emissions control as their four-stroke counterparts.  This can be shown by EMD's switch to the four-stroke J (EMD 1010) engine for the SD70ACe-T4 locomotive, as they were unable to get the EMD 715 to meet EPA Tier 4 without eating an unacceptable weight penalty or using SCR (which is a logistical problem for railroads, having to deal with large quantities of yet &lt;em&gt;another&lt;/em&gt; consumable).&lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2017-06-11T16:10:35.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15813" PostTypeId="2" ParentId="15808" CreationDate="2017-06-11T17:00:31.407" Score="0" Body="&lt;p&gt;You might find this link helpful : &lt;a href=&quot;http://www.dailymail.co.uk/sciencetech/article-2568732/Car-runs-air-set-hit-streets-year-Peugeot-claims-new-hybrid-117mpg.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.dailymail.co.uk/sciencetech/article-2568732/Car-runs-air-set-hit-streets-year-Peugeot-claims-new-hybrid-117mpg.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-11T17:00:31.407" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15814" PostTypeId="1" AcceptedAnswerId="18918" CreationDate="2017-06-11T17:23:30.977" Score="2" ViewCount="614" Body="&lt;p&gt;Enthalpy seems to be commonly used to characterize geothermal reservoirs, but how is it calculated? I have an example here where the enthalpy is cited as 800 $kJ kg^{-1}$ for a reservoir with a temperature of 190 °C. In that case the enthalpy was apparently just taken to be the specific heat of water (4200 $J kg^{-1} °C^{-1}$) multiplied with the temperature, something like $H = c\Delta T$ (meaning that the reference temperature must be 0 °C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that the correct way to do it? And if yes, is this really the enthalpy in the thermodynamical sense and why is the reference temperature 0 °C?  &lt;/p&gt;&#xA;" OwnerUserId="11732" LastActivityDate="2018-01-15T20:50:21.857" Title="How do you calculate the Enthalpy of a geothermal resource?" Tags="|thermodynamics|energy|renewable-energy|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15815" PostTypeId="1" CreationDate="2017-06-11T19:15:14.407" Score="1" ViewCount="444" Body="&lt;p&gt;In a statically indeterminate structure, solving the beam using slope deflection method how to cater for the beams with hinges? How would i calculate the FEMS for such beam? Do i have to consider hinge as a Far End Pinned or else? Also If The load is applied directly on the hinge, then how to cater for such a scenario. I have an upcoming Exam and if someone can clear that doubt to me?&lt;/p&gt;&#xA;" OwnerUserId="11733" LastActivityDate="2017-06-11T19:15:14.407" Title="How To Deal With Hinges In Statically Indeterminate Structures?" Tags="|civil-engineering|" AnswerCount="0" CommentCount="2" ClosedDate="2017-06-12T13:17:45.450" ContentLicense="CC BY-SA 3.0" />
  <row Id="15816" PostTypeId="2" ParentId="651" CreationDate="2017-06-12T05:21:14.383" Score="2" Body="&lt;p&gt;I agree that it doesn't make sense for the refrigerator to heat up an air-conditioned space in the summer.  Also, if the temperature is below freezing outside, why not just locate the fridge in an uninsulated and unheated part of the house i.e. garage or balcony?  If the ambient temperature is already cold enough, the refrigerator's thermostat will not turn the unit on.&lt;/p&gt;&#xA;&lt;p&gt;I solved the winter problem by using the fridge in my garage.  The summer problem is more complicated.  Commercial walk-in coolers and freezers do have their condensers outside, typically on the roof.  The next time you stop by a Dunkin' Donuts, wawa, or large restaurant in a strip mall, look up at the roof.  You will see at least 3 condensers, or 2 condensers and a package unit (large unit that houses condenser for air conditioning, and furnace for heating), where the refrigerator, freezer, and air conditioner each have their own condensers.&lt;/p&gt;&#xA;&lt;p&gt;Well, a commercial walk-in fridge or freezer is around 15,000 USD for low end, much more money than the 600 to 1200 USD for a home fridge.  As the other comments and answers have pointed out, it doesn't make economic sense to separate the condenser outdoors for a home fridge.  My approach was just use an insulated flexible duct to vent the exhaust hot air from the fridge condenser outside.  If we put the fridge on a wall that faces outdoors on the other side, the duct would only need to be a foot or two.&lt;/p&gt;&#xA;&lt;p&gt;However, it is difficult to attach a duct to fit the exhaust area of the fridge that is typically at the bottom for the upright fridges.  I mean you can do it as a hack with a lot of duct tape, but it would look funny and unprofessional.  Perhaps, with a lot of requests for this type of adaptability, the manufacturers might consider incorporating a standard for duct outside venting of condenser exhaust.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastEditorUserId="9448" LastEditDate="2022-10-26T12:20:08.170" LastActivityDate="2022-10-26T12:20:08.170" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15817" PostTypeId="1" CreationDate="2017-06-12T07:48:20.280" Score="1" ViewCount="100" Body="&lt;p&gt;When current is injected into a Peltier device/thermo-electric cooler, it heats up, when the current is reversed, it cools down. Why does it takes less time to heat up than to cool down?&lt;/p&gt;&#xA;" OwnerUserId="11743" LastEditorUserId="90" LastEditDate="2017-06-12T09:16:19.753" LastActivityDate="2017-06-12T11:36:29.807" Title="Peltier effect Asymmetrical perforance" Tags="|electrical-engineering|heat-transfer|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="15818" PostTypeId="1" AcceptedAnswerId="15825" CreationDate="2017-06-12T11:03:06.570" Score="1" ViewCount="1418" Body="&lt;p&gt;I want to cast items out of epoxy resin. They  will be tools used to handle ESD sensitive devices, so they need to at dissipate static charge, or perhaps somewhat conduct. Resins used will vary depending on the part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, resin suppliers seem to never give any data on electrical conductivity. I presume it's extremely low. So to have it conduct, am I right in saying I could mix in something like graphite to help conduction? I presume it would reduce strength, or at least change some physical properties, but that's fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've previously experimented mixing graphite powder with non-conductive powders - No liquids or glue, and just compacted by hand in a glass tube. I found that the resistivity could not be controlled easily / effectively by varying the ratio; it seemed it either conducted quite well, or not at all, to the degree where fine tuning ratios would too unreliable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Might that be the case with epoxy resin as well? How might I increase the conductivity of a resin? And how might it be tunable? Principles / formulas? I presume the geometry, size and other physical properties of the powder used is important, as well as, perhaps, the resin?&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2017-06-14T21:23:46.950" Title="How to make epoxy resins dissipate static charge" Tags="|composite-resin|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15819" PostTypeId="2" ParentId="15817" CreationDate="2017-06-12T11:36:29.807" Score="3" Body="&lt;p&gt;Peltier devices are quite inefficient.  A simple first-order approximation for a Peltier is a perfect heat pump in series with a resistor.  The heat pump moves heat symmetrically proportional to the current, but the resistor always makes heat proportional to the square of the current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unavoidable heater in a Peltier device makes it overall better at making heat than cold.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-06-12T11:36:29.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15820" PostTypeId="2" ParentId="651" CreationDate="2017-06-12T14:02:02.897" Score="0" Body="&lt;p&gt;It can make sense to install the condensor outside and this is done in large industrial systems (&lt;a href=&quot;https://www.google.de/search?q=air%20chiller&amp;amp;client=firefox-b-ab&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwigqtHnu7jUAhXMJlAKHYvOBd0Q_AUIDCgD&amp;amp;biw=1308&amp;amp;bih=876&quot; rel=&quot;nofollow noreferrer&quot;&gt;This is what condensors of large AC or fridges look like&lt;/a&gt;, look for those on the roof or behind any installation you'd expect to have significant cooling needs). However, it has drawbacks. Separating condensor from fridge means you need to install piping onsite, this includes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;vents in the high points and low points of the piping, for emptying and filling&lt;/li&gt;&#xA;&lt;li&gt;Piping through walls&lt;/li&gt;&#xA;&lt;li&gt;quality control or even pressure test&lt;/li&gt;&#xA;&lt;li&gt;filling and checking the system after installation, refridgerants are at least potentially hazardous&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A longer piping run would neccessitate a larger compressor - though m gut feeling is not by much (Pressure needed for condensation will likely dwarf losses of a few m piping.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would turn a mass produced consumer product into a piece of bespoke plant design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that a trained technician + 1 assistant can do this for most applications in half a day, so you would a person-day skilled labor to the cost of the fridge. Or in other words: You're now introducing contractors that are (usually) in a hurry to get to the next job into the equation vs the factory tested package system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hattip to secundus who made the point I wanted to make clearer.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2017-06-14T06:52:16.833" LastActivityDate="2017-06-14T06:52:16.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15821" PostTypeId="1" CreationDate="2017-06-12T15:16:57.407" Score="1" ViewCount="1681" Body="&lt;p&gt;Would it be possible to have an electric motor that uses an electromagnet to force a piston up and down to turn a shaft, similarly to an internal combustion engine, but without burning fuel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for this kind of motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there be any advantages of this style of motor compared to other electric motors?&lt;/p&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="368" LastEditDate="2017-06-12T16:30:57.853" LastActivityDate="2019-09-22T02:34:15.257" Title="Can an electric motor use reciprocating pistons like an internal combustion engine does?" Tags="|electrical-engineering|motors|engines|pistons|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15822" PostTypeId="2" ParentId="15821" CreationDate="2017-06-12T15:32:35.867" Score="0" Body="&lt;p&gt;A standard electric motor turns the shaft directly (ok we may or may not need gearing) which keeps the efficiency high. As soon as you add pistons, connecting rods, a crank you reduce the efficiency and add weight. The piston / crank is also used in pumps. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-12T15:32:35.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15825" PostTypeId="2" ParentId="15818" CreationDate="2017-06-12T20:07:50.327" Score="4" Body="&lt;p&gt;If you have to use a resin system like epoxy, it should be no problem to lower the volume resistivity of the finished parts by adding carbon black with a suitable coupling agent to make sure the carbon is evenly dispersed into the epoxy resin. The great loudspeaker designer Harold Beveridge used this epoxy/carbon mix to make the cast electrodes for his electrostatic loudspeakers. You can read how he did this in his US Patent # 3,668,335. To get the desired ESD time constants he mixed roughly 5% by weight of carbon into the epoxy resin to get a volume resistivity of about 10^10 ohms/sq. You can vary the amount to get your desired ESD time constants. Beveridge did not use a coupling agent in his formula but I recommend that you do so , materials science has come a long way since Mr.Beveridge was building his speakers (circa 1973) :-)&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2017-06-12T20:07:50.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15826" PostTypeId="1" CreationDate="2017-06-12T21:24:14.120" Score="1" ViewCount="338" Body="&lt;p&gt;I know that friction helps car to move forward. The engine of the car turns the wheels, and the tyres have to grip the road in order for the car to start moving forward. If there is no friction, tyres cannot grip the road! It is similar to a geared wheel moving on the ground &#xA;&lt;a href=&quot;https://i.stack.imgur.com/K6Qnr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K6Qnr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't imagine how does friction act as a centripetal force when the car is turning on a roundabout? What is happening at the point of contact between the wheel and the road ? &lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="9742" LastEditDate="2017-06-13T22:22:31.153" LastActivityDate="2017-06-13T22:22:31.153" Title="How does friction help a car to turn?" Tags="|mechanical-engineering|automotive-engineering|applied-mechanics|dynamics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15828" PostTypeId="2" ParentId="15826" CreationDate="2017-06-12T22:53:06.950" Score="2" Body="&lt;p&gt;The key phrase here is &lt;a href=&quot;https://en.wikipedia.org/wiki/Slip_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;slip angle&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a very easy concept to visualise but in essence comes down to the fact that when you turn the wheels the sidewalls of the tyre flex a bit relative to the contact patch between the tyre and the road so the whole tyre twists a bit. Of to put it another way the wheel it self wants to turn but the tread of the tyre wants to keep going straight which creates an elastic force in the sidewall. It is this force which pulls the car towards the centre of the turn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easy way to see this in practice is with a bicycle (it is more obvious if you lower the tyre pressure a bit). With the bike stationary lean your weight over the handlebars and turn gently and you should be able to both see the tyre flex and feel the resistance through the handlebars. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-06-12T22:53:06.950" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15831" PostTypeId="1" CreationDate="2017-06-13T05:18:36.820" Score="1" ViewCount="78" Body="&lt;p&gt;Say I want N-ADCs, M-amplifiers, K-filters, and why not, an oscillator too. Is there a platform to combine whatever device i need, and that this platform is implemented on a single chip? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know an FPGA has a pool of logic gates and flip-flips, and so anyone can implement a huge number of functionalities with some coding. But is there something similar in the analog and mixed signal domain of electronics? If so, how do you build your system? is it like a drag and drop sort of thing, where you choose whatever device you need from a library and place it in a layout, and then fab it? Could this be ASIC? can I layout my own ASIC using, for example, Cadence?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please some details!&lt;/p&gt;&#xA;" OwnerUserId="11743" LastEditorUserId="1832" LastEditDate="2017-06-19T15:58:01.900" LastActivityDate="2017-10-17T17:19:24.743" Title="Can different analog and mixed signal devices be combined on a single chip" Tags="|electrical-engineering|circuit-design|microelectronics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15832" PostTypeId="1" AcceptedAnswerId="15839" CreationDate="2017-06-13T05:29:15.587" Score="10" ViewCount="2515" Body="&lt;p&gt;I wonder how can I make hard disk drive more durable? Specifically, I works in a factory environment. And almost every few months, hard disk drive in some of the factory machine are corrupted, and even unrepairable. We already implement several SSD on some of them, but that's not much of help. They're being corrupted too on factory machine with heavy task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So every time it happened, we always restoring using our backup image. And we already attach all of the hard disk drive with shock absorber to reduce the shake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other option or prevention we could use? Perhaps adding any anti magnet material to prevent magnet friction, or something else? And what material we need?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;Factory machine with heavy task I mention above basically machine to create car metal body and car frame mould.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And when I meant the disk is corrupted, it means unreadable. The whole disk. Not the program files or software related. So it won't boot at all.&lt;/p&gt;&#xA;" OwnerUserId="11753" LastEditorUserId="11753" LastEditDate="2017-06-14T07:36:01.037" LastActivityDate="2017-06-14T12:09:10.907" Title="How to make factory machine hard disk drive more durable? How much its estimate life span?" Tags="|mechanical-engineering|machining|computer-hardware|" AnswerCount="7" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15833" PostTypeId="2" ParentId="15832" CreationDate="2017-06-13T05:57:15.120" Score="3" Body="&lt;p&gt;Estimated lifespan is given by the manufacturer usually as &quot;MTBF = 2000hrs&quot; but in &quot;normal conditions&quot; - what you describe is not normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are the SSD drives failing - physical damage or poor connections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One anti shock mounting used in the past was a mercury bath but you probably won't be allowed that !! But you could make an oil bath version...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-06-13T06:09:42.743" LastActivityDate="2017-06-13T06:09:42.743" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="15834" PostTypeId="2" ParentId="15821" CreationDate="2017-06-13T07:27:29.507" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Linear_motor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Linear electric motors&lt;/a&gt; exist, and are used in applications where you want a linear motion. As Solar Mike said, using a linear motor to drive a crank is pointless. &lt;/p&gt;&#xA;" OwnerUserId="11645" LastActivityDate="2017-06-13T07:27:29.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15835" PostTypeId="1" AcceptedAnswerId="15837" CreationDate="2017-06-13T08:14:56.023" Score="0" ViewCount="304" Body="&lt;p&gt;I have read that variable speed hydraulic drives have very low part-load efficiency - due to losses in the internal pumps and motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is part-load?&lt;/p&gt;&#xA;" OwnerUserId="11758" LastActivityDate="2017-06-13T08:47:22.730" Title="Part load definition with respect to hydraulics" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15836" PostTypeId="2" ParentId="15832" CreationDate="2017-06-13T08:24:22.677" Score="2" Body="&lt;p&gt;We are assuming that the problem is caused by shock or vibration. There can be several other causes, such as temperature, humidity, corrosion, chemicals, as pointed out by others. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more approach would be to get the drives to a non-hostile place and extend the connection with cables. This may need your computers to run from external drives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you cannot take the external disks far away, you can still place them in a cushion (sponge like material) to avoid shock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you still cannot escape the vibration or shock, investigate changing your computers with tough ones. This will probably cost your company some dollars, but it's probably better than production stopping.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-06-13T08:24:22.677" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15837" PostTypeId="2" ParentId="15835" CreationDate="2017-06-13T08:47:22.730" Score="0" Body="&lt;p&gt;Part-load is when the full power possible of the drive / motor is not used or needed. The efficiency is lower due to the losses becoming more significant compared to the lower power produced as some of the losses may be independent of power produced.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-13T08:47:22.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15838" PostTypeId="2" ParentId="15831" CreationDate="2017-06-13T08:55:13.323" Score="1" Body="&lt;p&gt;It is perfectly possible to build your own mixed circuit chip, but you will indeed need to learn quite a few things if you want to do it by yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I would like to say that what you will do is not a drag and drop but a  &lt;a href=&quot;https://en.wikipedia.org/wiki/Place_and_route&quot; rel=&quot;nofollow noreferrer&quot;&gt;place and route&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to do it by yourself, here are the steps to follow:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Determine the technology you want to use, if you don’t know what to choose try with a CMOS 250/130nm technology (cheapest), CMOS-biCMOS  250/130nm (more flexible also compatible with large range of frequencies) or with 45/65/130nm CMOS (less flexible, with good performances and a lot of design blocks available)&lt;/li&gt;&#xA;&lt;li&gt;Get in contact with the foundries that provides the chosen technology, and they will provide you a design platform (most likely Cadence, but few foundries would also propose platforms compatible with some cheaper – and less complete - design software)&lt;/li&gt;&#xA;&lt;li&gt;Learn how to use the design software and the company platforms (Usually companies propose tutorials) – it will NOT take just a day to be comfortable! – &lt;/li&gt;&#xA;&lt;li&gt;Use the blocks provided on the platform, and the best of you knowledge to bring a properly connected semi-custom design. (Full custom not recommended without years of experience) Send it to the foundry, and get your chip, packed as you asked.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Otherwise you could also just contact a company with circuit design services and explain your project, they are very qualified, and this will avoid you a jump into the vast world of circuit design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; As noted in the comments by  Tonys, there is too the possibility to use programmable system on chips. The limitations will be that the electrical performances and number of devices will be quite limited.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="10684" LastEditDate="2017-06-13T14:28:37.190" LastActivityDate="2017-06-13T14:28:37.190" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15839" PostTypeId="2" ParentId="15832" CreationDate="2017-06-13T10:06:19.857" Score="14" Body="&lt;p&gt;Quite likely, the SSD killer is electrical. We can't entirely rule out mechanical vibration, but SSDs are pretty robust mechanically. A simple rubber mounting would increase the resiliency even further. Also make certain that both power and data cables have enough slack. Vibration might cause them to come loose, when under tension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to address the electrical reliability, we have to consider two factors. Firstly, the heavy machinery may draw large currents from the power supply. This could cause voltage drops, which in turn can negatively affect the SSD. This is easily solved by an online UPS. Essentially, this type of UPS powers the computer from a battery, while the mains power is used to charge the battery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more unusual problem could be electro-magnetic radiation. High-power machinery will have large currents running, often at 50 or 60 Hz. Unintentionally, this will cause cables to act as antennas. The big cables in the machine act as senders, and the cables to the SSD can act as receivers. The solution here is to have a proper Faraday cage, ideally grounded. That's why normal PC cases are made of metal; they work as Faraday cages, keeping EM radiation out in frequency bands between 50 Hz and several Ghz.&lt;/p&gt;&#xA;" OwnerUserId="669" LastEditorUserId="669" LastEditDate="2017-06-14T12:09:10.907" LastActivityDate="2017-06-14T12:09:10.907" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15840" PostTypeId="2" ParentId="15832" CreationDate="2017-06-13T10:35:17.170" Score="11" Body="&lt;p&gt;First off, electrical and magnetic problems are not as bad as problems of vibrations and air contamination. Moisture in air plus dust or chemicals can corrode or short paths quite easily, and in our installations they are the primary reasons of failures if the devices aren't protected properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best option is just keeping all that's not necessary on the production floor off-site. Keep minimalist embedded controllers by the machines, keep the PCs in a neat office communicating with the controllers over LAN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's not possible, you need sealed cases. Possibly with heat transfer elements, if needed; airtight boxes that keep most of moisture out, some silica inside to absorb the rest of moisture - neither dry dust nor clean air moisture alone are a big problem, but combined they quickly lead to oxidation of contacts, other corrosion-related problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience, EM disturbances are rarely powerful enough to cause any lasting damage. They may knock a device out, forcing a reboot, but a well-built device will recover from that. Power surges are a different matter; without a good surge protection you may see random damage of parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, vibration. The vibration conducted by the floor is easily reduced to negligible levels through a sponge mat or similar. The vibration of a machine, in case the device is directly attached to the machine... there's little that can be done about it. There are dampening systems, but they are only efficient against certain vibration scales... really, just move that control box 2 meters away. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, temperature range must be &quot;within acceptable levels&quot;. You WILL see corruption on overheating devices, and moisture will condense on too cold ones. This is rarely a concern on a production floor, where too many machines depend on it, but as you seal the disk (intentionally, or unintentionally, e.g. through dust) you'll see overheating.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-13T10:35:17.170" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15841" PostTypeId="1" CreationDate="2017-06-13T10:43:33.900" Score="1" ViewCount="49" Body="&lt;p&gt;I am researching on mechanical design of remote compliance devices (RCC)†. In the RCC, manufactured by ATI, they use shear pads for lateral compliance. They mention that shear pads are elastomers bonded with metal shims. I couldn't find how it is being made. Are there layers of elastomers between wedging pieces (shims) or something else? How is adhesion being achieved?&lt;a href=&quot;https://i.stack.imgur.com/ceQlH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ceQlH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11761" LastActivityDate="2017-06-13T10:43:33.900" Title="How is a shear pad of remote compliance device formed?" Tags="|robotics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15842" PostTypeId="2" ParentId="15832" CreationDate="2017-06-13T11:18:43.443" Score="2" Body="&lt;p&gt;As others said, SSDs are resistant against vibrations - there are no moving parts, unlike magnetic hard disk drives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these technologies are, however, vulnerable to electromagnetic fields, as others stated as well. Providing protection against that may help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should, however, also not discount other factors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A quick note: this list definitely isn't complete. Just look at scope of the other answers - from voltage levels, to heat, to software - there is a LOT that can cause these issues. Unless you're confident with computing, you might want to look at hiring someone to figure this out for you, because they might be looking at different factors on-site that you didn't think of. That said, here are a few factors that you should also consider.)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There might be a problem with the cable, and such errors can be very subtle in that they just sporadically show up. Test the &quot;defective&quot; hard disks in a normal environment on a different PC, with different cables - to make sure they actually have physical damage. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It can be your memory as well. Unless you're using ECC memory, this can be difficult to identify. If your bits flip in memory, and that just happens to be where your program, the operating system or its drivers reside, then all bets are off. It might do nothing, it might crash, or it might just write garbage all over your disk.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It might not be a hardware issue at all. A software bug can also corrupt data. Having an exotic driver stack can make your system more prone to corrupting data.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Depending what exactly is the cause (you need to determine that first!) we can recommend possible solutions. There are plenty of solutions - from isolation, to RAID, to checksumming file systems such as ZFS - but you need to determine the cause first.&lt;/p&gt;&#xA;" OwnerUserId="11762" LastActivityDate="2017-06-13T11:18:43.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15843" PostTypeId="2" ParentId="15832" CreationDate="2017-06-13T11:26:23.217" Score="2" Body="&lt;p&gt;In addition to the other answers: in the environment you mention, it's possible there is metallic dust in the air. When that gets into the computer, you can get electrical shorts. &#xA;A sealed case (or ventilation with high-quality air filtering) can help if that's the case. &lt;/p&gt;&#xA;" OwnerUserId="11645" LastActivityDate="2017-06-13T11:26:23.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15844" PostTypeId="2" ParentId="15821" CreationDate="2017-06-13T12:28:28.933" Score="1" Body="&lt;p&gt;Every time you convert energy from one form to another, the conversion is not 100%, there is some loss. For example: electrical energy to mechanical energy (motor), chemical energy to mechanical energy (internal combustion engine), linear mechanical energy to rotary mechanical energy (piston/connecting rod/crank). Many systems will have multiple different conversion steps, and the losses can really add up. Therefore, you are always better off doing as few conversions as possible. If you want linear mechanical energy, start with that in the first place. If you want rotary mechanical energy, start with that in the first place. Only do conversion if you absolutely have to.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-06-13T12:28:28.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15845" PostTypeId="2" ParentId="15821" CreationDate="2017-06-13T14:15:13.600" Score="0" Body="&lt;p&gt;Typical electromagnetic actuators create this kind of motion when you want linear movement in few (2-3) large, discrete steps, e.g. a latch mechanism (for door etc) or other back-and-forth movement, more or less rapid (electric bell, membrane pumps, relays) usually employing a spring for one position, electromagnet for one or two more - you'll rarely see ones for more discrete steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be possible to employ these like pistons of a combustion engine, but the efficiency would be inferior - they don't provide nearly as much force/torque (for the same mass and wattage) as classic (rotary) electric motors.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-13T14:15:13.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15846" PostTypeId="1" AcceptedAnswerId="15852" CreationDate="2017-06-13T14:20:37.233" Score="0" ViewCount="757" Body="&lt;p&gt;I was checking the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mobile_country_code#India_-_IN&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of MCC (Mobile Country Codes) on Wikipedia&lt;/a&gt; and found that India has two MCC: 404 and 405. 404 contains only GSM/UMTS/LTE operator bands and 405 contains both GSM/UMTS/LTE and CDMA operator bands. The problem is that I do not understand the reason for having 2 MCC for the same country. We also have mobile network code (MNC) to differentiate between various operators, circles and the bands.&lt;/p&gt;&#xA;" OwnerUserId="11767" LastEditorUserId="368" LastEditDate="2017-06-14T15:47:44.097" LastActivityDate="2017-06-14T15:47:44.097" Title="Why does India have more than 1 MCC (Mobile Country Code)?" Tags="|telecommunication|wireless-communication|" AnswerCount="2" CommentCount="2" ClosedDate="2017-06-15T16:45:47.113" ContentLicense="CC BY-SA 3.0" />
  <row Id="15848" PostTypeId="1" CreationDate="2017-06-13T15:33:22.020" Score="4" ViewCount="109" Body="&lt;p&gt;I saw a few minutes of a documentary that mentioned the domed roof of the largest mosque in Istanbul is supported by the strength of sets of three arches assembled so they share a common pillar. I'd like to learn more about that building technique but I'm not sure what the name is for that kind of structure. Also, can four stone arches share a central column or would the shared column likely collapse?&lt;/p&gt;&#xA;" OwnerUserId="11771" LastEditorUserId="30651" LastEditDate="2023-01-02T16:50:31.747" LastActivityDate="2024-01-27T20:06:51.007" Title="Is there a name for a structure/building technique comprising three arches that share a central column? (Answer Found - Triple Arch System)" Tags="|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="15850" PostTypeId="2" ParentId="15846" CreationDate="2017-06-13T16:09:20.803" Score="0" Body="&lt;p&gt;Two codes to provide sufficient bandwidth springs to mind.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-13T16:09:20.803" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15852" PostTypeId="2" ParentId="15846" CreationDate="2017-06-13T17:39:41.077" Score="1" Body="&lt;p&gt;The Mobile Network Code (MNC) is either a two digit (&quot;European&quot; standard) or three digit (&quot;North American&quot; standard) identifier, with in the Mobile Country Code (MCC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/International_mobile_subscriber_identity&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/International_mobile_subscriber_identity&lt;/a&gt;, the 3GPP specification 23.003 recommends that the &quot;length of the MNC is uniform within a given MCC area&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the OP's link, presumably India was allocated MCC = 404 and started using two-digit MNC numbers, but then came close to running out of numbers. The additional MCC = 405 has blocks of three-digit numbers as well as two-digit, so they may be unlikely to ever need another MCC allocation.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-06-13T17:39:41.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15854" PostTypeId="2" ParentId="15789" CreationDate="2017-06-14T00:11:23.823" Score="3" Body="&lt;p&gt;I would argue that &amp;quot;tempered water&amp;quot; was the unfortunate result of a marketing department not doing their homework.  Typically in plumbing design, I come across &amp;quot;tempered water&amp;quot; on emergency fixtures (emergency eyewash or emergency showers) or for things like the mixing valves on showers. OSHA has requirements for tempered water delivered to emergency fixtures, which would appear to be what you were turning up when you looked it up.&lt;/p&gt;&#xA;&lt;p&gt;Tempered water must be exactly at a temperature where in an emergency a person could use it immediately. For example, if a person was on fire, or had spilled acid on themselves. In the middle of winter a person would head to the safety shower if there was no fire extinguisher/acid spill kit close by. In winter, depending on location, this water would be near freezing. Enough to give anyone hypothermia and put them into shock (which has happened). Thus, tempered water is always around 22 &lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;C to 25 &lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;C (72-77 &lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;F) to prevent people from either getting cold too fast or burning themselves with scalding hot water. This is far different from what Apple is using it for, but that is technically the scientific term for tempered water.&lt;/p&gt;&#xA;&lt;p&gt;Furthermore, they even say right in the ad that they are using &amp;quot;chilled&amp;quot; water.  From my experience, they are referring to two different concepts in the same sentence.  On a related note, they may not be spending money on refrigerant systems, but you still have to cool the water somehow.  It may be a net gain for their energy calculation by using the water over a refrigerant based system, but if they are not also using some sort of energy recovery system and are just blowing outside air through the system, they have probably negated any gains they made.&lt;/p&gt;&#xA;" OwnerUserId="9770" LastEditorUserId="90" LastEditDate="2021-04-07T04:52:07.720" LastActivityDate="2021-04-07T04:52:07.720" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="15855" PostTypeId="2" ParentId="15832" CreationDate="2017-06-14T01:35:19.960" Score="3" Body="&lt;p&gt;Really a comment but too long:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've dealt with PCs on the factory floor (woodworking), they proved quite resilient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our initial setup which was basically trouble free:  We mounted the PCs inside a cabinet, the front was clear plastic slats (think what you sometimes see on a walk-in refrigerator freezer).  The original intent was to maintain a slight positive pressure with clean air but this was never done and proved to not be needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, after that plenty of machines got installed with less care.  The usual &quot;failure&quot; mode was thermal shutdown, take the cover off and blow it out, it would work fine, although these did prove more problematic as the dust did some damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problems, however, came from their wires.  We specced shielded cables but management went cheap on us, the building was wired with ordinary network wire and later modifications were often made by electricians rather than computer guys.  This caused a &lt;strong&gt;lot&lt;/strong&gt; of interference and was probably responsible for the high failure rate of the network cards.  (Really, now, a Cat-5 tossed over a 480V, 400?A main power bus???  Or even more extreme, a Y connection in a cat 5--which actually worked, albeit with network error problems!)  Don't put a computer on the same circuit as a heavy motor.  Don't run any computer wires parallel to heavy power wires even if they're on separate circuits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial machines were all diskless (not an option these days) and even after that everything of importance was stored on the network so if a machine did act up it could be swapped out very quickly--it took longer to carry the new machine to the station than to get it up and running in place of the problematic one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top lesson--don't let the electricians be anything but carefully-watched assistants when wiring things.&lt;/p&gt;&#xA;" OwnerUserId="589" LastActivityDate="2017-06-14T01:35:19.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15860" PostTypeId="1" CreationDate="2017-06-14T14:29:21.053" Score="4" ViewCount="93" Body="&lt;p&gt;I have a non-linear system shown:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x}_1 = x_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x}_2 = -(1-\vert x_1\vert)x_2 - x_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I determine the region in the state plane where the equilibrium state at the origin is asymptotically stable?&lt;/p&gt;&#xA;" OwnerUserId="11787" LastEditorUserId="6695" LastEditDate="2017-06-14T19:06:29.993" LastActivityDate="2017-06-29T15:53:00.010" Title="How to determine the region in a state plane where the equilibrium state is asymptotically stable" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15863" PostTypeId="1" CreationDate="2017-06-14T19:51:44.957" Score="0" ViewCount="41" Body="&lt;p&gt;High-frequency currents induce skin-effect resistance that is used on some applications to weld metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you have two steel rectangles of dimension $h_0, w, l$ and $h_1, w, l$ placed together along the $h$ axis. I'm wondering if placing a strong enough current through appropriate probes in the transverse faces of the rectangle, will generate enough heat to weld the touching faces uniformly, with no sticks or rods involved&lt;/p&gt;&#xA;" OwnerUserId="11789" LastEditorUserId="90" LastEditDate="2017-06-15T16:08:56.087" LastActivityDate="2017-06-15T16:08:56.087" Title="Steel surface welding with high-frequency current" Tags="|steel|electromagnetism|welding|current|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15865" PostTypeId="2" ParentId="15863" CreationDate="2017-06-14T21:05:00.540" Score="1" Body="&lt;p&gt;Look up ERW ( electric resistance welding ) , this is commonly used to weld steel pipe and tubes ( not necessarily round). Spot welding and projection welds have similar aspects.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-06-14T21:05:00.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15867" PostTypeId="2" ParentId="15818" CreationDate="2017-06-14T21:23:46.950" Score="1" Body="&lt;p&gt;For non conductive samples in an electron microscope , silver paint was used to conduct away any charge. I would expect copper powder or small fibers may be available also. Caution , I tried mixing fine zinc powder into an epoxy  and it immediately reacted with the liquid epoxy.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-06-14T21:23:46.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15868" PostTypeId="1" CreationDate="2017-06-14T22:49:40.577" Score="1" ViewCount="110" Body="&lt;p&gt;I have an 80 ft metal tower, that is cemented into a dirt ground, that houses two antennas to send and receive Wifi. The tower's power source is not a home. It has a dedicated line, that runs directly to it's own power meter. The tower is about 300 feet from a concrete home. The internet company says, that since the tower is buried in dirt, it is &quot;already naturally grounded&quot;. I have other people who are saying it is extremely dangerous to not have it properly grounded with a copper rod. What are the potential dangers if lightening were to strike the tower?&lt;/p&gt;&#xA;" OwnerUserId="11794" LastEditorUserId="9" LastEditDate="2017-06-19T18:13:05.670" LastActivityDate="2017-06-19T19:47:39.427" Title="Ungrounded Tower - Lightning Damage" Tags="|mechanical-engineering|electrical-engineering|structural-engineering|" AnswerCount="0" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15869" PostTypeId="1" CreationDate="2017-06-15T02:01:30.473" Score="0" ViewCount="914" Body="&lt;p&gt;I'm trying to build an hourglass, but I don't know what all the specs are that determine exactly how long it takes a given volume of what material to finish within a certain time. Are there any hourglass equations anywhere? &lt;/p&gt;&#xA;" OwnerUserId="11795" LastActivityDate="2017-06-15T13:50:03.243" Title="How do you calculate the measurements for an hourglass?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2017-06-15T22:39:41.953" ContentLicense="CC BY-SA 3.0" />
  <row Id="15871" PostTypeId="1" CreationDate="2017-06-15T04:31:47.947" Score="3" ViewCount="551" Body="&lt;p&gt;I created a structural model of a laced strut using plate elements. To study second order effects of bow imperfection, I applied a joint displacement of 20 mm in a SAP 2000 v18 model. When the analysis was completed, I double checked the results and found a large joint reaction in the range of 300,000 kN at the joint with applied displacement. Elsewhere, joint reactions do not exceed 1,000 kN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model ran with no warnings or errors. I expected the sum of all forces to be zero for equilibrium, but that seems not to be the case. The global displacement looks fine with the laced strut bending to the specified displacement. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Should all joint reactions sum to zero, if nodal displacement is applied?&lt;/li&gt;&#xA;&lt;li&gt;Should the joint reaction at the node with specified displacement be disregarded?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11443" LastEditorUserId="368" LastEditDate="2017-10-13T15:35:34.153" LastActivityDate="2017-10-13T15:35:34.153" Title="Should sum of all joint reactions be zero when nodal displacement is applied?" Tags="|civil-engineering|structural-analysis|modeling|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15872" PostTypeId="1" CreationDate="2017-06-15T05:47:32.080" Score="0" ViewCount="42" Body="&lt;p&gt;I forgot what all of it is called but there's this specific way you draw the different components of a device or machine, kind of like blue prints but on plain write paper. You break the device down into larger pieces and draw each piece orthographically from like 3 different perspectives with labels and uncertainty in measurement requirements and all that, very standard. Anyone got a clue? &lt;/p&gt;&#xA;" OwnerUserId="11795" LastActivityDate="2017-06-15T07:39:26.973" Title="What is the formal term/process for writing down a design?" Tags="|drafting|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15873" PostTypeId="2" ParentId="15872" CreationDate="2017-06-15T07:11:19.080" Score="1" Body="&lt;p&gt;First angle projection, third angle projection and isometric projection spring to mind - all types of drawing layouts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The name &quot;Blue print&quot; comes from the copying of the final original for production or use - the type of copying that was available back in the day. Now that full colour copiers exist the term still hangs around.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-06-15T07:39:10.573" LastActivityDate="2017-06-15T07:39:10.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15874" PostTypeId="2" ParentId="15872" CreationDate="2017-06-15T07:11:27.567" Score="3" Body="&lt;p&gt;The process of creating an &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Engineering_drawing&quot; rel=&quot;nofollow noreferrer&quot;&gt;engineering drawing&lt;/a&gt;&lt;/em&gt; is called either &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Technical_drawing&quot; rel=&quot;nofollow noreferrer&quot;&gt;technical drawing&lt;/a&gt;&lt;/em&gt; or &lt;em&gt;drafting&lt;/em&gt;. How to draw is covered under the standard &lt;em&gt;ISO 128&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The separate drawings from different directions are called &lt;em&gt;views&lt;/em&gt;, how to use views is covered in exacting detail in standard &lt;em&gt;ISO 128-30&lt;/em&gt; and &lt;em&gt;128-34&lt;/em&gt;.  number of views and choice of views is governed by desire to use of as few views as possible while describing object without ambiguity, with following principles in mind:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;avoiding the use of hidden lines,&lt;/li&gt;&#xA;&lt;li&gt;avoiding repeating same thing.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As such you do not need three views, when one is sufficient. This commonly happens in rotationally symmetrical objects as well as when cutting sheets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: A blueprint is just a residue of how the technical drawings were copied, originals were most likely black on white or black on translucent (to make the pencil drawing or previous original used as guide easy to see). The technical drawing standards does not specify color.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-06-15T07:39:26.973" LastActivityDate="2017-06-15T07:39:26.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15875" PostTypeId="1" CreationDate="2017-06-15T07:27:49.360" Score="-2" ViewCount="60" Body="&lt;p&gt;so my question is, if a swivel hook used in duffel bags has spring inside for compression, then would it help to reduce the weight on our shoulders when we are running around airport with bag hung on our shoulders either sideways or crossed. See attached image for clarification.&lt;a href=&quot;https://i.stack.imgur.com/7S3dI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7S3dI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11797" LastActivityDate="2018-06-14T20:08:42.913" Title="Will including spring inside the swivel hooks used in duffel/messanger bags help?" Tags="|mechanical-engineering|springs|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15876" PostTypeId="2" ParentId="15875" CreationDate="2017-06-15T07:41:28.283" Score="1" Body="&lt;p&gt;No, it won't change the weight or mass of the bag, it may help absorb the shock of it landing on your shoulder as you move. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-15T07:41:28.283" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15877" PostTypeId="1" CreationDate="2017-06-15T08:17:44.767" Score="3" ViewCount="256" Body="&lt;p&gt;Instead of creating multiple bridge models with different boundary conditions for differential settlement, and combining the forces output manually in Excel, is it possible to specify different boundary conditions within the same model, using different load cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using both MIDAS and SAP v18 software and I would like to know if there is any way to do this in any bridge structural modelling software. Or is there any software that doesn't require you to add a restrain to a node for which you are applying a movement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This limitation on the software introduces significant additional effort on the part of the Engineer. Has anyone thought of some way to overcome this issue when dealing with differential settlement or other soil movements?&lt;/p&gt;&#xA;" OwnerUserId="11443" LastActivityDate="2019-09-30T22:01:05.150" Title="Is it possible to apply different boundary conditions on the same model for different load cases?" Tags="|civil-engineering|structural-analysis|modeling|bridges|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15878" PostTypeId="2" ParentId="15871" CreationDate="2017-06-15T08:19:31.833" Score="1" Body="&lt;p&gt;The sum of all joint reactions (resolved in any direction) should be equal and opposite to the sum of all applied forces (resolved to the same direction). An applied displacement is &lt;em&gt;not&lt;/em&gt; an applied force. Therefore, assuming no other loads are present in the loadcase with applied displacement, the joint reactions should sum to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reaction at the node with applied displacement is expected. Something needs to hold that node in its displaced position; the force will do that. But as the sum of all reactions should be zero, other supports should have reactions to balance the reaction at the displaced node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should not disregard the joint reactions from your model. For some reason or another your model is not behaving as you expect. You need to determine &lt;em&gt;what&lt;/em&gt; is causing that behaviour, so that you can either realise that it is the correct behaviour, or fix it.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2017-06-15T08:39:37.000" LastActivityDate="2017-06-15T08:39:37.000" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15879" PostTypeId="2" ParentId="15869" CreationDate="2017-06-15T13:50:03.243" Score="1" Body="&lt;p&gt;I can pretty much guarantee that the only way to calibrate an hourglass is to build it,  fill it with your choice of particles (preferably non-hygroscopic!) and time it.  Add or remove material until you get the timing desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Granted, a mass-production line only needs to do this once and then reproduce the glass figure to tight tolerance, but you aren't going that way. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-06-15T13:50:03.243" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15880" PostTypeId="1" CreationDate="2017-06-15T16:48:38.237" Score="2" ViewCount="942" Body="&lt;p&gt;So suppose I have 4 files.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Part 1&lt;/li&gt;&#xA;&lt;li&gt;Part 2&lt;/li&gt;&#xA;&lt;li&gt;Sub-assembly&lt;/li&gt;&#xA;&lt;li&gt;Assembly&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now in sub-assembly, part1 and part 2 are &quot;connected&quot; to each other and there is an articulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything works well until I place my sub-assembly in my main assembly. When I do that, part1 and part 2 can't &quot;move&quot; anymore. How do I solve that problem?? &lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2017-06-15T16:48:38.237" Title="Inventor parts that move in sub-assembly won't move in main assembly" Tags="|cad|autodesk-inventor|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15884" PostTypeId="1" AcceptedAnswerId="15894" CreationDate="2017-06-16T06:49:23.337" Score="6" ViewCount="3285" Body="&lt;p&gt;I bought sunglasses in China that claims to have UV400 protection. I'd like to test this as my eyes' health is on the line :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to check the maximum UV amount these glasses can protect me against?&lt;/p&gt;&#xA;" OwnerUserId="11807" LastEditorUserId="8055" LastEditDate="2017-06-17T08:12:29.567" LastActivityDate="2020-04-06T16:09:05.703" Title="Test UV protection on sunglasses" Tags="|product-testing|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15885" PostTypeId="1" CreationDate="2017-06-16T07:58:02.583" Score="-1" ViewCount="51940" Body="&lt;p&gt;I have to draw the front,top,left and right views of a given solid objects without scaling them i.e freehand. I am much beginner in this drawings. So I am getting no idea of how the given solid object is break into top,left,right and front view. So kindly please explain all the basics of how to break the figure into parts of the engineering drawing with few examples (without scaling). &lt;/p&gt;&#xA;" OwnerUserId="11809" LastActivityDate="2017-06-19T02:07:42.083" Title="Top View,Front View,Left view and right view help" Tags="|mechanical-engineering|civil-engineering|drawings|technical-drawing|" AnswerCount="3" CommentCount="0" ClosedDate="2017-06-19T18:11:53.793" ContentLicense="CC BY-SA 3.0" />
  <row Id="15886" PostTypeId="2" ParentId="15885" CreationDate="2017-06-16T08:22:30.257" Score="0" Body="&lt;p&gt;Look at the object in front of you, then the face you see is the front view and if you moved your position to the left side, again what you see is the left view etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are two systems : do you draw what you see after (behind) the object or between you and the object but this won't change what you need at the moment.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-16T08:22:30.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15887" PostTypeId="2" ParentId="15885" CreationDate="2017-06-16T08:37:28.063" Score="3" Body="&lt;p&gt;When you have an object to draw, you must follow some simple rules to explain your drawing for anyone to understand it easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 example explanatory pictures below.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Example 1&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lbEOp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lbEOp.jpg&quot; alt=&quot;Example 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Example 2&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oaUgn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oaUgn.jpg&quot; alt=&quot;Example 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope these helps!&lt;/p&gt;&#xA;" OwnerUserId="7468" LastActivityDate="2017-06-16T08:37:28.063" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15889" PostTypeId="2" ParentId="15547" CreationDate="2017-06-16T09:11:12.517" Score="-1" Body="&lt;p&gt;Castellated beam. Cut the web with torch, and re-weld web at higher flange distance.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2017-06-16T09:11:12.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15890" PostTypeId="1" CreationDate="2017-06-16T10:16:01.597" Score="3" ViewCount="846" Body="&lt;p&gt;What is the difference between capacity use factor (CUF) of solar power plant and plant load factor (PLF) of coal based power plant? Can we compare both since they seem analogous?&lt;/p&gt;&#xA;" OwnerUserId="3793" LastEditorUserId="163" LastEditDate="2017-09-14T12:29:42.020" LastActivityDate="2017-09-14T12:29:42.020" Title="CUF v/s PLF of power plants" Tags="|electrical-engineering|power|renewable-energy|solar-energy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15893" PostTypeId="1" CreationDate="2017-06-16T13:13:54.033" Score="-4" ViewCount="547" Body="&lt;p&gt;Let's assume I want to build, for example, a coffee machine with a special twist from scratch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I would need plans, mechanical evaluations, CAO, then machining or 3D printing. I also guess I would need to interface my mechanical parts with some electronics to control flow or some more specific gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: what is the general procedure about this ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which knowledge do I need to learn/know ?&lt;/p&gt;&#xA;" OwnerUserId="6866" LastActivityDate="2017-06-16T13:39:14.353" Title="Making a machine from scratch" Tags="|mechanical-engineering|pumps|" AnswerCount="1" CommentCount="6" ClosedDate="2017-06-16T14:37:10.207" ContentLicense="CC BY-SA 3.0" />
  <row Id="15894" PostTypeId="2" ParentId="15884" CreationDate="2017-06-16T13:35:24.260" Score="3" Body="&lt;p&gt;Your question is a little naive: &quot;Maximum UV protection&quot; doesn't mean much.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you want to know is the spectral  attenuation values in the UVA and UVB wavelength bands.   The glasses reduce the fraction of UV light that passes through them, so the stronger the source (e.g. noon Sun vs. sunset) the more photons per second will be transmitted.  Since biological damage is cumulative (and nonlinear to boot), both the power level and the time of exposure matter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, back to actual sunglasses.  Quoting from &lt;a href=&quot;http://www.allaboutvision.com/askdoc/sunglasses.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;allaboutvision.com&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Q: How can I find inexpensive sunglasses that are guaranteed to block&#xA;  100 percent UV? Many are made in China and say 100 percent UV&#xA;  protection. But how do I really know? — M.D., North Carolina&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A: The amount of UV protection that sunglasses provide cannot be&#xA;  determined by the color of the lenses, the country of origin or the&#xA;  price of the eyewear. The determining factor is the lens material and&#xA;  any additional UV-blocking coatings that may be applied. Uncoated&#xA;  plastic sunglass lenses block about 88 percent UV; polycarbonate&#xA;  lenses block 100 percent UV.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Many cheap sunglasses use a material called triacetate. This material&#xA;  absorbs only about 40 percent of the UV rays. Lenses may also be made&#xA;  of other acrylic materials, which will vary in how much UV protection&#xA;  they provide.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;All sunglasses with polarized sunglasses block 100 percent UV,&#xA;  regardless of the lens material or price of the eyewear (though&#xA;  polarized sunglasses tend to be more expensive than sunglasses without&#xA;  this extra glare-blocking feature).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Personally, I far prefer polarized lenses in any case because of their glare-removal capability.  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-06-16T13:35:24.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15895" PostTypeId="2" ParentId="15893" CreationDate="2017-06-16T13:39:14.353" Score="1" Body="&lt;p&gt;Studying engineering or a course on practical design would be a good start.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;general procedure&quot; is &quot;design-build-redesign&quot; (or just design-build if you intend for this to be a 1-off).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no real good answer to this, the design process is complicated and not standard, you can follow many different methods to design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The knowledge you need to learn and know will be whatever you don't know when you try and design and build this.  There are numerous areas where you will likely have to learn this; but this depends &lt;em&gt;entirely&lt;/em&gt; on what your current level of understanding is.  It could be as easy as reading some chapters from a few textbooks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's more likely that you could spend years looking into the design if you wanted to, optimizing and iterating as you learn more.  You could make your design very precise, or have everything loosely controlled for less consistent results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt; What you would need to know are all the things that are currently preventing you from building this.  There is no easy answer.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-06-16T13:39:14.353" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15896" PostTypeId="1" CreationDate="2017-06-16T13:45:09.393" Score="3" ViewCount="8939" Body="&lt;p&gt;The thin steel float used in my car (which was built in the UK in the early 60s, which I'm assuming is why it's not originally plastic) has rusted to a point where it's leaking in fuel and therefore failing to operate as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm planning to replace it with a plastic float, but obviously want to make sure I'm not using something that could dissolve in the petroleum. I notice that 5 litre plastic canisters are HDPE so I intend to use that, but I'm intrigued as to just how much it matters. Plenty of discussion around the net suggests that most plastic bottles are likely to dissolve, most of which in my experience are PET. But this chart suggests PET is as good as HDPE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.plasticsintl.com/plastics_chemical_resistence_chart.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.plasticsintl.com/plastics_chemical_resistence_chart.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That chart doesn't have data on PP or PE interestingly (though I assume HDPE has the same properties as PE).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question, respecting that storing petroleum in plastic containers not explicitly rated for the task is a bad idea, which plastics will actually dissolve? If it's a grey area is there a minimum thickness?&lt;/p&gt;&#xA;" OwnerUserId="11812" LastActivityDate="2017-08-20T22:06:57.547" Title="Which plastics dissolve in petroleum?" Tags="|plastic|petroleum-engineering|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15898" PostTypeId="2" ParentId="15821" CreationDate="2017-06-16T16:19:35.800" Score="0" Body="&lt;p&gt;It would be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it exists, its uncommon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any application it might serve, it would have to compete with solenoids, where a spring accomplishes the return stroke. In my youth I worked for a company that used a solenoid and ratchet to drive the sprocket wheel in a paper tape reader. It had respectable speed, but ear protection (should have been) recommended.&lt;/p&gt;&#xA;" OwnerUserId="11766" LastActivityDate="2017-06-16T16:19:35.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15899" PostTypeId="2" ParentId="15884" CreationDate="2017-06-16T16:22:05.900" Score="2" Body="&lt;p&gt;The only real way to test this is to get a stable &lt;a href=&quot;https://www.thorlabs.com/thorproduct.cfm?partnumber=LED370E&amp;amp;gclid=Cj0KEQjwg47KBRDk7LSu4LTD8eEBEiQAO4O6r7LjCo6E6_jD4NRnDr2iOQY67TIPrqfprhxED81pSA4aAi8D8P8HAQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;UV source&lt;/a&gt; and a &lt;a href=&quot;https://www.google.com/shopping/product/7687335357554755840?q=UV%20detector%20diode&amp;amp;biw=1500&amp;amp;bih=1184&amp;amp;safe=high&amp;amp;sa=X&amp;amp;ved=0ahUKEwi53Nbo4cLUAhVryoMKHYKcBBgQ8wIItgIwAg&quot; rel=&quot;nofollow noreferrer&quot;&gt;diode that can detect that UV&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measure the relative voltages from the diode with and without the sunglasses, and then you will know how much UV the sunglasses deflect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is pretty much the only way to test it yourself! Without these tools, you cannot test the sunglasses: what is transparent to UV is not always transparent in other parts of the spectrum. You cannot see UV rays, so merely looking at the glasses will not be enough.&lt;/p&gt;&#xA;" OwnerUserId="11631" LastActivityDate="2017-06-16T16:22:05.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15906" PostTypeId="1" CreationDate="2017-06-17T08:15:54.850" Score="1" ViewCount="435" Body="&lt;p&gt;I'm trying to calculate the axial force that is generated in a vertical threaded axis (arrow in picture) when a force of 100N is applied to a handle perpendicular to the axis (circled in picture). &lt;a href=&quot;https://i.stack.imgur.com/o1D1y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o1D1y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found various websites and calculators but I'm not sure which one is correct as they give me different solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque generated in the axis is 14.35Nm and the thread is an M30x2. How can I calculate the correct axial force?&lt;/p&gt;&#xA;" OwnerUserId="11825" LastEditorUserId="90" LastEditDate="2017-06-17T11:41:01.303" LastActivityDate="2017-06-18T06:51:03.437" Title="Axial force in threaded axis" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15907" PostTypeId="1" AcceptedAnswerId="15908" CreationDate="2017-06-17T09:02:58.393" Score="1" ViewCount="32" Body="&lt;p&gt;when designing a pond it is customary to produce a chart for the local council, specifying the height of the water for the volume stored. What is the name of this chart?&lt;/p&gt;&#xA;" OwnerUserId="11826" LastActivityDate="2017-06-17T10:24:54.790" Title="Chart that describes how full the pond is" Tags="|civil-engineering|water-resources|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15908" PostTypeId="2" ParentId="15907" CreationDate="2017-06-17T10:24:54.790" Score="1" Body="&lt;p&gt;Storage curve or water supply curve are both possible depending on the planned use.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-17T10:24:54.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15909" PostTypeId="1" AcceptedAnswerId="16132" CreationDate="2017-06-17T15:48:08.220" Score="1" ViewCount="4729" Body="&lt;p&gt;I need to convert 7.62cm/3in linear motion from an actuator to 150-180° (150 is minimum) rotary motion. I am aware of the existence of gears, but would probe here to see if there is some simpler method. I could use rods mounted in some way. It has to be compact and rugged, however. I'd also be interested in formulas for this type of conversion. Thanks! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &#xA;I found this contraption &lt;a href=&quot;https://i.stack.imgur.com/le6em.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/le6em.gif&quot; alt=&quot;Mystery device&quot;&gt;&lt;/a&gt; for microlinear precision, whilst googling for scotch yoke. Is this a scotch yoke? If not, what is it? I can probably awfulhack one of these with a file, some aluminium rods and strips. But it would be prettier if I bought one.&lt;/p&gt;&#xA;" OwnerUserId="11578" LastEditorUserId="11578" LastEditDate="2017-06-17T23:23:03.940" LastActivityDate="2017-07-02T12:20:49.103" Title="How to convert 7.62cm linear motion to 150-180° rotary motion" Tags="|mechanical-engineering|actuator|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="15910" PostTypeId="2" ParentId="15906" CreationDate="2017-06-17T17:35:37.093" Score="3" Body="&lt;p&gt;The truth is that you can't just calculate it easily, mostly because a good portion of the axial force depends upon how much of the torque goes into overcoming the friction in the joint. The total friction is dependent on the coefficient of friction between the threads and between the parts of the fastener(s) that contact the surface of whatever is being bolted together. Those coefficients in turn are dependent on things like surface finish of the mating surfaces, lubrication and thread tolerances all of which vary, even within &quot;identical&quot; joints.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2017-06-18T06:51:03.437" LastActivityDate="2017-06-18T06:51:03.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15911" PostTypeId="1" AcceptedAnswerId="15914" CreationDate="2017-06-17T17:39:25.960" Score="0" ViewCount="433" Body="&lt;p&gt;I have this skateboard truck for my off-road skateboard. I want to mig weld a piece of aluminum to it to mount a motor. How would I test if it is something I can weld to with a MIG welder? I think it is aluminum, how can I be sure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;if magnets don't stick to it, it is aluminum. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if I test for continuity with my multimeter and I don't have continuity, it is probably anodized. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;if it were polished it would be very shiny and probably reflect light. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How do tell if it is in fact annodized, and if so, what do I need in order to weld through it? Do I sand it down? I was going to use a flux core mig welder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/17MF1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kOh3G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xfwaQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reflection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also here is the inside where they have an existing weld. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/NtGsC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NtGsC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11626" LastEditorUserId="11626" LastEditDate="2017-06-17T23:23:50.830" LastActivityDate="2017-06-17T23:23:50.830" Title="Can I mig weld to my skateboard truck? How can I test if it's weldable?" Tags="|metallurgy|aluminum|welding|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15912" PostTypeId="1" CreationDate="2017-06-17T21:18:01.710" Score="1" ViewCount="257" Body="&lt;p&gt;When Tesla introduced the dual motor model s it had better range and performance (so better in everyway except price). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a quadruple motor car (one per wheel) have even better range and performance?&#xA;Or&#xA;Was the increase in range and performance not due to the number of motors but due to the ability to remove the rear drive shaft etc?&lt;/p&gt;&#xA;" OwnerUserId="10361" LastActivityDate="2017-09-16T12:47:19.163" Title="Advantages of quadruple motor electric car" Tags="|mechanical-engineering|power-electronics|electric-vehicles|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15913" PostTypeId="2" ParentId="15912" CreationDate="2017-06-17T21:45:05.257" Score="1" Body="&lt;p&gt;It's 4 * 4 ... And also the reduced friction of the drivetrain  - the more direct the drive the better and also may have had a weight effect, but given the weight of the batteries that may not really be noticed.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-17T21:45:05.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15914" PostTypeId="2" ParentId="15911" CreationDate="2017-06-17T22:56:50.707" Score="2" Body="&lt;p&gt;That does look a lot like aluminium. All grades of aluminium are non-magnetic as  are some stainless steels.  The best test is weight as aluminium is significantly less dense than steel but has lower tensile strength so judging by the bulk aluminium seems likely. a final test is that if you grind aluminium with an abrasive wheel it will produce virtually no sparks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've not heard of flux core wire for aluminium, stick rods for aluminium alloys do exist but tend not to give great results in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally TIG is the best option for aluminium alloy welding but a decent MIG machine with pure argon shielding gas and the appropriate filler wire can produce good results. The difficulty with MIG is that aluminium wire is significantly softer than steel so tends to have more problems with wire feed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue is that not all grades of aluminium have good weldability t&lt;a href=&quot;http://www.lincolnelectric.com/en-gb/support/welding-solutions/Pages/aluminum-faqs-detail.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;his article&lt;/a&gt; gives good overview. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue, which applies to welding any finished part is that the welding process inevitably puts a significant amount of heat into it which can affect its mechanical properties on a material level  an also cause dimensional distortion, which can be mitigated but never entirely eliminated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is that it may not be impossible to do but is likely to be considerably more difficult than just welding a bracket to  bit of mild steel and modifying finished parts by welding is always a bit risky when you don't know the specific of their material and manufacturing process.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-06-17T22:56:50.707" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15915" PostTypeId="2" ParentId="15909" CreationDate="2017-06-18T13:10:05.223" Score="2" Body="&lt;p&gt;There are commonly used methods used in backhoes and cherry picker trucks to move the arms about 180 degrees with a single hydraulic piston. If you search for photos of backhoes or power line maintenance trucks you may see a type of linkage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a photo &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rPAuD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rPAuD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This video is crude but you can definitely see the 3 moving parts and how they interact. The red line is the hydraulic cylinder. The white line is a link that connects the 2 green arms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=tb4jiscwhvw&quot; rel=&quot;nofollow noreferrer&quot;&gt;backhoe linkage&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this second video the link proportions and pivot points are way off. But the essential pieces are there and the concept is the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=09jZsBDHalE&amp;amp;list=PL_AXzsnrjstigELkDUtQINEmLcngU53FD&quot; rel=&quot;nofollow noreferrer&quot;&gt;Converting Linear to Rotary motion&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transfer function won't be linear though.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastEditorUserId="8055" LastEditDate="2017-06-19T16:35:12.767" LastActivityDate="2017-06-19T16:35:12.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15917" PostTypeId="2" ParentId="15885" CreationDate="2017-06-19T02:07:42.083" Score="1" Body="&lt;p&gt;Speaking in Engineering Drawing terms, there are two methods to generate &lt;a href=&quot;https://en.wikipedia.org/wiki/Multiview_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;projections of an object&lt;/a&gt;. First angle projections and Third-angle projections.These projections are developed based on assuming how the object is conceptually viewed in a quadrant system. &lt;a href=&quot;https://i.stack.imgur.com/SAcTt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SAcTt.jpg&quot; alt=&quot;Quadrants&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In First Angle Projection we place our object in the First Quadrant&#xA;  (see above figure). This means that the Vertical Plane is behind the&#xA;  object and the Horizontal Plane is underneath the object.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In Third Angle Projection the Object is placed in the Third Quadrant.&#xA;  This means that the Vertical Plane is in front of the object and the&#xA;  Horizontal Plane is above the object.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;These changes in the position of the views are the only difference&#xA;  between projection methods.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;http://www3.ul.ie/~rynnet/orthographic_projection_fyp/webpages/third_angle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SjWGA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SjWGA.jpg&quot; alt=&quot;Projections&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, visualize an object being viewed from these different planes. The  Frontal plane gives you the side view. Horizontal plane gives you the top or bottom views (based on the angle of projection). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And visualization is the key to becoming good in CAD drawings, 3D modelling etc. It takes a bit of practice and imagination; but it's easy to get the hang of :)  &lt;/p&gt;&#xA;" OwnerUserId="8778" LastActivityDate="2017-06-19T02:07:42.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15918" PostTypeId="1" CreationDate="2017-06-19T04:07:54.677" Score="0" ViewCount="466" Body="&lt;p&gt;What is the correct type of material, 304L or 316L stainless steel tubing, for transport argon gas? Is it required to be an electro-polished finished?&lt;/p&gt;&#xA;" OwnerUserId="11837" LastEditorUserId="90" LastEditDate="2017-06-19T08:56:45.830" LastActivityDate="2017-06-20T22:29:32.177" Title="What is the correct type of stainless steel for transporting argon gas?" Tags="|materials|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15919" PostTypeId="1" CreationDate="2017-06-19T04:46:21.293" Score="0" ViewCount="574" Body="&lt;p&gt;The error is &quot;&lt;em&gt;There is at least 1 small equation solver pivot term (e.g, at the UY degree of freedom of node 2144). Please check for an insufficiently constrained model&lt;/em&gt;.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This error occurs while applying forces on a bogie frame structure in ANSYS mechanical APDL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I correct this error? because the charged data is correct&#xA;&lt;a href=&quot;https://i.stack.imgur.com/d8JP9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d8JP9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11838" LastEditorUserId="90" LastEditDate="2017-06-19T09:01:40.397" LastActivityDate="2017-06-19T12:50:42.113" Title="Stress Analysis on a Bogie Frame Structure using ANSYS" Tags="|ansys-apdl|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15920" PostTypeId="2" ParentId="15919" CreationDate="2017-06-19T05:32:39.683" Score="0" Body="&lt;p&gt;The error is telling you that node 2144 has a degree of freedom error at UY.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-19T05:32:39.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15921" PostTypeId="1" CreationDate="2017-06-19T11:08:30.773" Score="2" ViewCount="218" Body="&lt;p&gt;The vertical bending capacity of a twin UB laced strut can be calculated by $M_c = p_y \cdot 2Z_x$. Where $Z_x$ is the section modulus of a single UB section about its major axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For lateral bending capacity, can we calculate this by using $I_c = 2I_y + 2As^2/4$, where $s$ is the separation distance between centroid of the two UBs, and $Z_c = I_c/e$, where $e$ is the distance from centre of laced strut to extreme fibre?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the built-up member have a lower bending capacity due to the two beams being laced together at fixed intervals instead of being joined continuously together via welding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, can we design laced strut for significant lateral bending instead of mainly axial compression?&lt;/p&gt;&#xA;" OwnerUserId="11443" LastEditorUserId="1832" LastEditDate="2017-06-19T11:36:36.247" LastActivityDate="2017-06-20T14:04:21.243" Title="How to calculate the lateral bending capacity of a laced strut?" Tags="|structural-engineering|civil-engineering|design|structural-analysis|steel|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15922" PostTypeId="2" ParentId="15921" CreationDate="2017-06-19T11:41:39.957" Score="1" Body="&lt;p&gt;Yes, so long as $s$ is the distance between the centroid of one section and the centroid of the combined section, and $e$ is the distance from the centroid of the combined section to the extreme lateral fiber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the comparison of laced vs. continuously-welded sections, you'll need to check the relevant code in your region. That being said, the codes I'm aware of have conditions which, if met, allow the beams to be considered as working perfectly in unison, in which case there is no loss of strength.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-06-20T14:04:21.243" LastActivityDate="2017-06-20T14:04:21.243" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15923" PostTypeId="2" ParentId="15919" CreationDate="2017-06-19T12:50:42.113" Score="1" Body="&lt;p&gt;Usually, this type of error doesn't mean there is something &quot;wrong&quot; with node 2144. The finite element solver only discovers the error when it gets to the &lt;em&gt;last&lt;/em&gt; node and degree of freedom that it processes, which just happens to be node 2144.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most likely, there is something wrong with they way the model is constrained, and it can move as a rigid body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason for the problem might be that you created the mesh in several parts, and the parts are not properly connected together, so that one part can move freely relative to another part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't find the error by looking at the input, do a vibration analysis with the same constraints. You will probably find that the first mode has a frequency of 0 Hz, and the mode shape will show what the unwanted rigid body motion looks like. Then figure out how to restrain the model to fix the problem.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-06-19T12:50:42.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15924" PostTypeId="2" ParentId="15912" CreationDate="2017-06-19T15:04:25.913" Score="1" Body="&lt;p&gt;The total battery energy capacity is what defines range.  The dual-motor system is provided for improved traction and general control, not range.  There is no &quot;drive shaft&quot; in any electric car.  The single-motor Teslas have the motor over the rear wheels, not at the opposite end of the vehicle the way (obsolete dinosaur absurd design with transmission gear) ICE cars do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of a 4-motor config is that you get rid of differential gears on each axle.  The disadvantages include the need for a more advanced 4-way torque balancing controller (essentially software differential), increasing the probability of motor failure by a factor of sqrt(2),  and probably more total weight (two small motors  weigh more than one bigger motor).  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-06-19T15:04:25.913" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15925" PostTypeId="1" CreationDate="2017-06-19T16:37:39.190" Score="5" ViewCount="167" Body="&lt;p&gt;I'm dealing with a somewhat intractable problem. We're running dSPACE 7.4 (RTI 1006) on MATLAB 2011a, which I understand uses the dSPACE DS1006 C Compiler v. 2.2. When I try to compile one of our models, I wind up with a linker error claiming that fscanf is undefined (it's being invoked in a custom function). I have verified that stdio.h is where it should be in the system hierarchy and that it contains a definition for fscanf. Following are the steps I took to try to resolve the problem:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;linker initially complains about undefined fscanf()&lt;/li&gt;&#xA;&lt;li&gt;extra definition inserted in model-specific file where I know it will be compiled causes linker to complain about inconsistent definitions for fscanf()&lt;/li&gt;&#xA;&lt;li&gt;commenting out fscanf() definition in stdio.h then allows successful compilation/linking&lt;/li&gt;&#xA;&lt;li&gt;inserting stub definition into stdio.h and commenting out original (and stub in model-specific file) results in 'multiple definitions of fscanf' errors (this is the weirdest part of the whole story)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The stub definition (I am assured by a higher-up that this shouldn't break our model but I haven't yet had a chance to verify this) is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;int fscanf(int a, char *cp, void *b, ...){return 0;}&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I'm reading it correctly it's a version of fscanf that won't actually do what fscanf is supposed to do. In summary, I have to hack my compiler's library to get this model to build, and I'm not entirely confident that it will run on our hardware at present. I also don't want to tell everyone in the office to modify their dSPACE installation and break fscanf unless I know that it works and it's the only way forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone give me some insight on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: It does indeed run on our hardware. But I don't like breaking fscanf(), so I'd still like to figure this out.&lt;/p&gt;&#xA;" OwnerUserId="11842" LastEditorUserId="7083" LastEditDate="2017-06-29T11:52:27.340" LastActivityDate="2024-03-25T16:01:50.203" Title="dSPACE compiler linker error -- fscanf undefined" Tags="|automotive-engineering|software|matlab|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15927" PostTypeId="1" CreationDate="2017-06-19T21:12:47.190" Score="1" ViewCount="43" Body="&lt;p&gt;For the purpose of this situation. Lets say there are 4 small electromagnets within a MRI. All of them activate at the same time. The entire setup is comprised up of non ferrous metals. The metals within the core of each solenoid are only magnetized when the current is applied to the electromagnet. The MRI is on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen to the non ferrous metal when it is suddenly magnetized? Would it immediately succumb to the power of the MRI? Could the current be turned off quickly enough for the MRI to not immediately pull the non ferrous metal towards it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my understanding, the non ferrous metal would instantly lose its magnetic properties once current is no longer affecting the electromagnet so the MRI would have no affect on the metal. &lt;/p&gt;&#xA;" OwnerUserId="11846" LastEditorUserId="9" LastEditDate="2017-06-21T17:03:59.527" LastActivityDate="2017-06-21T17:03:59.527" Title="How would an electromagnet function within an MRI?" Tags="|electrical-engineering|electromagnetism|biomedical-engineering|medical-devices|" AnswerCount="0" CommentCount="1" ClosedDate="2017-06-28T16:53:52.997" ContentLicense="CC BY-SA 3.0" />
  <row Id="15928" PostTypeId="2" ParentId="15593" CreationDate="2017-06-19T22:19:49.220" Score="1" Body="&lt;p&gt;I'm not sure what you're trying to figure out, but if you want to know how well your insulation is doing, I suggest that you add a thermostat and measure the total amount of power supplied to your heating element to maintain a constant temperature inside the box.&lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-06-19T22:19:49.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15929" PostTypeId="2" ParentId="651" CreationDate="2017-06-20T00:14:09.473" Score="0" Body="&lt;p&gt;Modern materials would make separation an easy thing to do. Homes could be built with ducting to accommodate such a system. Right now a length of tubing is already used for the water supply for an ice maker. The connections can be designed to fit together without venting problems requiring re/filling the system. Many refrigerators sit and accumulate dust for years gradually loosing efficiency and causing degradation of components. Separating the parts may allow for easier cleaning. Refrigerators generate heat and moisture which can attract insects. They are noisy and do add heat to a room. My guess is that the suggestion will become something people will want and pay extra for as electricity savings becomes more apparent when dependent upon alternative power sources for the home. I would also like to see designs that retain the refrigerated air instead of dumping it when a door is opened. There surely is a better way.&lt;/p&gt;&#xA;" OwnerUserId="11848" LastEditorUserId="11848" LastEditDate="2017-06-20T00:25:21.663" LastActivityDate="2017-06-20T00:25:21.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15931" PostTypeId="1" CreationDate="2017-06-19T04:38:20.737" Score="2" ViewCount="2508" Body="&lt;p&gt;I would like to make a beach cart that supports 200lbs and uses this design &lt;a href=&quot;https://i.stack.imgur.com/7sXKi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7sXKi.jpg&quot; alt=&quot;Beach Cart&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type and dimension wheels and tires should I buy? I have a budget of $100.00 for the tires and wheels. I cannot afford expensive sand tires like Wheelez. Will All Terrain Tires (AT) sufficiently work in sand with the expected load?&lt;/p&gt;&#xA;" OwnerUserId="11943" OwnerDisplayName="Orthopaedic RN" LastEditorUserId="9" LastEditDate="2018-10-28T18:09:29.770" LastActivityDate="2019-06-15T07:02:06.080" Title="What type of tires and wheels do I need to make a PVC beach cart?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="15932" PostTypeId="2" ParentId="15925" CreationDate="2017-06-20T02:46:20.617" Score="0" Body="&lt;p&gt;This presumably worked before.  So the fundamental first step is to regress any and all of the changes you made (you are using source control, right?) until you get it back to a known good state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, you need to examine the change you made.  C supports methods having identical names so long as they have different method signatures.  I.e. the input parameters must be different between the two methods.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C also supports letting you define those two identically named methods in two different areas.  So you can use the standard &lt;code&gt;fscanf&lt;/code&gt; method as well as this custom implementation that you've run into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition you mentioned:&lt;br&gt;&#xA;&lt;code&gt;int fscanf(int a, char *cp, void *b, ...){return 0;}&lt;/code&gt;&#xA;is likely where you're having problems.  And you likely have an additional definition of &lt;code&gt;fscanf&lt;/code&gt; beyond the standard lib definition.  And the two &quot;home grown&quot; &lt;code&gt;fscanf&lt;/code&gt; methods likely have conflicting method signatures.  The &lt;code&gt;...&lt;/code&gt; as the fourth parameter indicates that the method can take a variable number of inputs.  My best guess is that there's another &lt;code&gt;fscanf&lt;/code&gt; definition with 4 inputs defined and / or followed by the variable input indicator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short - hunt down &lt;strong&gt;all&lt;/strong&gt; of the definitions you have in your application for &lt;code&gt;fscanf&lt;/code&gt;.  That's where you'll find the conflicting definition and the source of your linker error.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2017-06-20T02:46:20.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15933" PostTypeId="1" CreationDate="2017-06-20T06:43:39.907" Score="2" ViewCount="411" Body="&lt;p&gt;I have this setup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Pully with a mass attached. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Worm gear drive to lower the mass with a constant speed. (Section A-A)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yTy5q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yTy5q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we look at the cross-section A-A, we can see the worm gear drive: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;M1&lt;/strong&gt; is the torque applied by the mass on the pully and hence on the gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;M2&lt;/strong&gt; is the torque applyed by the worm gear. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/DMz0Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DMz0Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;strong&gt;M1 &gt; M2&lt;/strong&gt; is it still possible to lower the mass at constant speed ? ...or in other words, is it possible to make the gear (above the worm gear) turn at a constant rotational speed ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;...or will it just block the system due to friction ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will rephrase my question: Let's suppose it is a autolocking worm gear, how can I calculate how much M2 needs to be for the mass to lower at a constant speed ?&lt;/p&gt;&#xA;" OwnerUserId="9039" LastEditorUserId="9039" LastEditDate="2017-06-20T12:13:15.583" LastActivityDate="2017-06-20T12:13:15.583" Title="Worm Drive Calculation" Tags="|mechanical-engineering|gears|applied-mechanics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15936" PostTypeId="1" CreationDate="2017-06-20T07:26:37.450" Score="2" ViewCount="5412" Body="&lt;p&gt;Is there a standard chart in Eurocode for the maximum and minimum bending moment in uniformly loaded beam for indeterminate structures where we can use directly? Many textbooks give different coefficient values for the moment for continuous beam, some wl2/10, some 0.046wl2. I know that hogging and sagging moment are different. Is there an authoritative reference for this?&lt;/p&gt;&#xA;" OwnerUserId="11443" LastEditorUserId="11443" LastEditDate="2017-06-21T02:34:59.230" LastActivityDate="2017-08-20T15:56:22.807" Title="What is the bending moment formula for continuously supported beam of equal span?" Tags="|civil-engineering|structural-analysis|beam|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15938" PostTypeId="2" ParentId="15936" CreationDate="2017-06-20T08:55:04.637" Score="0" Body="&lt;p&gt;I would suggest Timoshenko, Strength of Materials is an authoritative reference (used in the UK ...) : This is found on the www, and also, copies can be found in second-hand bookshops but even then they are not cheap...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link removed to avoid &quot;link rot&quot;...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-06-20T14:13:27.103" LastActivityDate="2017-06-20T14:13:27.103" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15941" PostTypeId="1" CreationDate="2017-06-20T12:20:28.917" Score="2" ViewCount="337" Body="&lt;p&gt;I'm watching construction of a new building across from my office. The old building has been demolished and the demo team has completed taking everything away and a new company has come in to start working. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new building will have a five floor underground parking lot and 19 floors above it. The site used to be an industrial area and sits on what used to be marsh. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the question: The excavators are digging holes maybe 6 to 10 feet deep (hard to tell from my vantage point), supporting the hole with vertical walls and filling them with concrete then covering them with dirt. Several of these are along the perimeter but at least one is in the middle of the construction site. What are these for? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I could add that might help is to say that they appear to have surveyed for underground utilities and appear to have marked telco and sewage pipes judging by the orange and green markings. At at least one spot they indicated 14'. I'm not sure if it's at all related to the holes though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/TC5CWDm.jpg&quot; alt=&quot;Construction&quot;&gt;&#xA;&lt;img src=&quot;https://i.imgur.com/HOVqO1U.jpg&quot; alt=&quot;Construction&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11855" LastEditorUserId="1832" LastEditDate="2017-06-20T19:33:29.707" LastActivityDate="2017-06-20T19:33:29.707" Title="Why do we dig holes and fill them in with concrete?" Tags="|concrete|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15942" PostTypeId="1" AcceptedAnswerId="15950" CreationDate="2017-06-20T13:13:04.043" Score="1" ViewCount="897" Body="&lt;p&gt;As a complete novice, I have been reading about outside engineering projects such as decking and patios.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the information I read tells me that to lay structural joists as a subframe / sleeper joists, the joist should be positioned with the thin side down, the wider side running perpendicular to the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my understanding, this makes for a stronger frame which can take more weight. My question is does it &lt;em&gt;have&lt;/em&gt; to be this way? For example, if the goal was to create a lower deck, is it structurally sound for the joists to be laid &quot;wide side&quot; down?&lt;/p&gt;&#xA;" OwnerUserId="11856" LastActivityDate="2017-06-20T18:21:30.897" Title="Does a joist have to lay &quot;thin side&quot; down?" Tags="|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15943" PostTypeId="2" ParentId="15942" CreationDate="2017-06-20T13:25:24.800" Score="0" Body="&lt;p&gt;The joist is positioned &quot;thin side down&quot; as that means more material is absorbing the force (mass ) applied to it and that is where the strength comes from. The thickness is necessary to stop the joist buckling or folding with the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before we had the detailed stress and structures calculations, joists were simply trees cut that looked big enough and left round or just flattened a bit. Now, we minimise the amount of material or make more efficient use by using material, the joist, with the strength focused where it needs to be.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-20T13:25:24.800" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15944" PostTypeId="1" AcceptedAnswerId="15945" CreationDate="2017-06-20T13:35:29.547" Score="0" ViewCount="175" Body="&lt;p&gt;I am about to present my work on vibrations to a NVH specialist at work and he is supposed to help me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated PSD's, now my question (subjective, I know) is how should I plot the data? I've been plotting $PSD \big[\frac{g^2}{Hz}\big]$ against frequency $\big[Hz\big]$, both in linear axes. I've been told the frequency axis should be in logarithmic scale, which makes sense, what about the PSD axis?&lt;/p&gt;&#xA;" OwnerUserId="11858" LastActivityDate="2017-06-20T14:18:45.743" Title="Best format to plot vibration data?" Tags="|mechanical-engineering|vibration|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15945" PostTypeId="2" ParentId="15944" CreationDate="2017-06-20T14:18:45.743" Score="1" Body="&lt;p&gt;Typically, a PSD for vibration will have a log scale for both axes. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-06-20T14:18:45.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15946" PostTypeId="1" CreationDate="2017-06-20T14:58:35.697" Score="2" ViewCount="483" Body="&lt;p&gt;I'm an electrical engineer, so please bear with me :). I came across this situation and I tried to find some equations to solve it, but all I could find was some Finite Element Theory, and I was wondering, if there is an easier way to solve the following situation:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fbPNh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fbPNh.png&quot; alt=&quot;Shells&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thin shell is supported in the middle and a force is applied on the four corners (by a screw or a bolt for instance), or (I believe it is equivalent) a thin shell is supported in one corner and a force acts on the other corner. I am interested in finding the resulting bending (assuming a linear material) or finding the necessary strength of the material to avoid breaking. Can someone point me to a textbook with a (perhaps simplified) solution, or is it necessary to use a Finite Element software?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="11859" LastActivityDate="2017-06-20T23:05:35.887" Title="How to calculate bending of shell" Tags="|mechanical-engineering|structural-engineering|structural-analysis|structures|applied-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15950" PostTypeId="2" ParentId="15942" CreationDate="2017-06-20T18:21:30.897" Score="4" Body="&lt;p&gt;Beams have a property we call moment of inertia ($I$) which controls how strong they are to resist bending. For rectangular beams, the equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \frac{bh^3}{12}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $b$ is the cross-section's base and $h$, its height. This means that the height of the cross-section is far more important than the base. Indeed, if you need a 2x6&quot; joist &quot;thin-side-down&quot;, then you'd need a 2x54&quot; (yeah, that's not a typo) joist &quot;wide-side-down&quot; to withstand the same load. The two sections below have the same strength against bending:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zc8Za.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zc8Za.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And obviously, this doesn't even take into consideration that joists are usually spaced some 16&quot; apart, which is impossible with the necessary wide-side-down section (since it's 54&quot; wide).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-06-20T18:21:30.897" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15951" PostTypeId="2" ParentId="15946" CreationDate="2017-06-20T18:57:10.163" Score="2" Body="&lt;p&gt;You'd do well with minimum potential energy here.  It's a wild ride, hang on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential energy associated with a set of stresses and strains in a flat panel of thickness t is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{2}\int\int_A\int^{t}_{0}(\sigma_x\epsilon_x + \sigma_y\epsilon_y + \sigma_{xz}2\epsilon_{xz}+\sigma_{yz}2\epsilon_{yz}+\sigma_{xy}2\epsilon_{xy}dzdA - \int\int_Ap(x,y)w(x,y)dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting to displacements and defining the following functions $w(x,y)$ as the displacement of the beam, $u_0(x,y)$ as the x-ward displacement of the shell, midplane, $v_0(x,y)$ as the y-yard displacement of the shell, and assigning some constants based off the modulus, E, and the poisson's ratio, $\nu$&#xA;$$A = \frac{Et}{1-\nu^2} $$&#xA;$$G = \frac{Et}{2(1+\nu)} $$ &#xA;$$D = \frac{Et^3}{1-\nu^2} $$&#xA;$$H = \frac{Et^3}{2(1+\nu)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the potential energy is simply&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \frac{1}{2}\iint_A\left[ A \left((\frac{\partial u_0}{\partial x})^2 + 2\nu\frac{\partial u_0}{\partial x}\frac{\partial v_0}{\partial y} + (\frac{\partial v_0}{\partial y})^2 \right) + D\left((\frac{\partial^2 w}{\partial x^2})^2 + 2\nu\frac{\partial^2 w}{\partial x^2}\frac{\partial^2 w}{\partial y^2} + \frac{\partial^2 w}{\partial y^2}^2 \right) + G\left((\frac{\partial u_0}{\partial y})^2 + 2\nu\frac{\partial u_0}{\partial y}\frac{\partial v_0}{\partial x} + (\frac{\partial v_0}{\partial x})^2 \right) + H\left((\frac{\partial^2 w}{\partial y^2})^2 + 2\nu\frac{\partial^2 w}{\partial y^2}\frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial x^2}^2 \right)\right]dydx - Fw(\frac{L}{2},\frac{W}{2}) - Fw(\frac{-L}{2},\frac{-W}{2}) - Fw(\frac{-L}{2},\frac{W}{2}) - Fw(\frac{L}{2},\frac{-W}{2})+ 4Fw(0,0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the procedure is to guess!  Trying a bunch of different polynomials, one that is bound to fit with the boundary conditions (or you relax a boundary condition) to find one that fits.  My suggestion is polynomials, but trigonometric functions have been used before as well.  The goal is to find a simple function that satisfies the conditions, then perform variance with the arbitrary coefficients to find the best coefficients to minimize the potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a step, I assume the following, which satisfy many of the boundary conditions: $w(x,y)=Kx^2y^2$, $u_0(x,y)=B_1x+B_2y$, $v_0(x,y)=C_1x+C_2y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in above, we achieve a function for the potential function.  Focusing only on the w(x,y) function, we find &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V = \frac{K^2(D+H)}{1440}(9L^5W +9LW^5+10\nu L^3W^3)-FKLW + other_stuff$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating V with respect to the arbitrary constant K, we find that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K = \frac{720F}{(D+H)(9L^4+9W^4+10\nu L^2W^2)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar work with the other functions will find that $B_1 = 0, B_2 = 0, C_1 = 0, C_2 = 0$.  So, we know the deflection of the panel.  From there, it's simple enough to follow appropriate formulas to derive the stresses.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2017-06-20T23:05:35.887" LastActivityDate="2017-06-20T23:05:35.887" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15952" PostTypeId="1" AcceptedAnswerId="15961" CreationDate="2017-06-20T18:58:10.213" Score="2" ViewCount="1326" Body="&lt;p&gt;Let's start with arbitrary transfer function as an example, such as:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;obj = tf([1 2],[1 0 2 5]);&#xA;nyquist(obj);&#xA;grid on;&#xA;axis([-2 2 -2 2]); axis equal&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As far as I do understand what is gain (expressed in dB), I am unable to figure out, nor find any information about circles that appear after we use &lt;code&gt;grid&lt;/code&gt;. For some reason there is vertical line at &lt;code&gt;Re=-0.5&lt;/code&gt; associated with &lt;code&gt;0 dB&lt;/code&gt; value. On the left and right of this line we see non coincident circles in the shape of peacock's eye with decreasing or increasing values. Circles converge to points &lt;code&gt;[-1,0]&lt;/code&gt; and &lt;code&gt;[0,0]&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that system's curve starts at point &lt;code&gt;[-0.4,0]&lt;/code&gt; which corresponds to &lt;code&gt;-7.9588 dB&lt;/code&gt; gain, crosses &lt;code&gt;-10 dB&lt;/code&gt; circle twice, where on both occurrences system's gain doesn't match &lt;code&gt;-10 dB&lt;/code&gt; value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, just from the fact that that we have infinite line denoted with &lt;code&gt;0 dB&lt;/code&gt; value, one can infer that it has absolutely nothing to do with actual gain, which on Nyquist plot should be circular around point &lt;code&gt;[0,0]&lt;/code&gt;. In this case my question is simple: what does those dB values  and circles mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5SWqB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5SWqB.png&quot; alt=&quot;Nyquist plot with grid on&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6464" LastActivityDate="2017-06-21T12:09:58.433" Title="Nyquist plot - what is the meaning of circles with dB values on complex plane" Tags="|control-theory|matlab|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15954" PostTypeId="1" AcceptedAnswerId="15955" CreationDate="2017-06-20T19:49:48.417" Score="2" ViewCount="1733" Body="&lt;p&gt;Our house air conditioner went out last Saturday afternoon on a +100 F day. Was able to diagnose the problem itself as a problem with the starting circuit of the big compressor fan of the outside compressor unit. If I gave the fan a strong push with a stick to start it rotating, then the fan motor would start up and the air conditioning would work fine. An AC technician later determined that the compressor fan needed a new start-up capacitor, and now the AC works fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: Why are large 2-ton house air conditioners designed so that their compressor fans require start-up capacitors to turn on? The diameter of the compressor fan on our house AC appears to be roughly the same as that of an inexpensive box fan that I can get at Walmart, and the ac motors on those simple box fans don't require a start-up capacitor to turn on, do they? You just plug them in and they smoothly start revving up to speed all by themselves. So why do house air conditioners use compressor fans that need these trouble prone start-up capacitors, and why can't a simple self-starting ac motor like that of a box fan be used instead for the compressor fan?&lt;/p&gt;&#xA;" OwnerUserId="11865" LastActivityDate="2017-06-21T08:25:56.733" Title="Why does the Compressor Fan for an Air Conditioner need a Starter Capacitor?" Tags="|electrical-engineering|motors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15955" PostTypeId="2" ParentId="15954" CreationDate="2017-06-20T20:28:37.523" Score="3" Body="&lt;p&gt;The simple answer is that for a box fan the load on the motor is proportional to its speed so its starting torque is very low. In the case of a compressor the motor is working against significant resistance just to turn over. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple single phase AC motors in high starting torque applications are often configured so that they are wired differently for starting to provide more starting torque, in this case the capacitor is essentially part of an automatic  switching circuit which alters the motor wiring from staring to running configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways to achieve this  such as star/delta configuration or by driving the motor by a digitally controlled inverter. But the capacitor method is a fairly cheap and simple way to drive high powered single phase AC motors. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-06-20T20:28:37.523" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15956" PostTypeId="2" ParentId="15918" CreationDate="2017-06-20T22:29:32.177" Score="1" Body="&lt;p&gt;The mechanical and manufacturing properties of 304 and 316 are not greatly different and both have good weld-ability (for stainless). The main difference is that 316 has a bit better corrosion resistance and is genreally rated as performing better in more corrosive environments (salt spray etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that industrial argon cylinders are generally non-stainless so unless there are other considerations either should be fine in principal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as noted in comments there are usually strict regulations associated with the design and manufacture of pressure vessels and anything used with high pressures may well need to be certified. This is greatly simplified for piping is you specify a grade which is already certified for its intended use along with the appropriate fittings and connectors. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-06-20T22:29:32.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15957" PostTypeId="1" AcceptedAnswerId="15962" CreationDate="2017-06-21T00:41:23.523" Score="-1" ViewCount="122" Body="&lt;p&gt;Concrete/brick homes in tropical areas are hot due to direct sunlight. Please suggest external siding materials or design to give heat shielding. Please give reasons (and if possible pros &amp;amp; cons).&#xA;Some ideas are using wood siding, because wood thermal capacity is low so that any heat will blown away by wind (but interstitial space behind the wood may infested with animals). Or using living plant leaves, since they give shade and heat on leaves blown away by wind (but may have problems with mould growth, or plant maintenance).&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2017-06-21T07:18:12.203" Title="House / home external skin for heat shield materials" Tags="|materials|heat-transfer|architecture|" AnswerCount="1" CommentCount="1" ClosedDate="2017-06-22T02:44:59.280" ContentLicense="CC BY-SA 3.0" />
  <row Id="15958" PostTypeId="2" ParentId="15896" CreationDate="2017-06-21T02:25:47.763" Score="0" Body="&lt;p&gt;The carburetor float on my'69 MGB sank. It happened abruptly while I was driving. I diagnosed it before taking anything apart when I saw a stream of gas on the road following me, while the engine died at the same time. When I pulled the translucent plastic float I saw that it was more than half full of fuel. I tried squeezing it in various ways to empty it without any luck until I poked it with a needle.  After that it stayed afloat for several months. When it sank again I bought a new, identical float which lasted as long as the car. I hated buying parts from the dealer because the clerk always bid farewell by saying, &quot;See you tomorrow.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, I wouldn't spend time searching for a better designed replacement. You should carefully open up your corroded float, then fill it with something like styrofoam before gluing it back together. &lt;/p&gt;&#xA;" OwnerUserId="11766" LastActivityDate="2017-06-21T02:25:47.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15959" PostTypeId="2" ParentId="13539" CreationDate="2017-06-21T04:51:08.803" Score="3" Body="&lt;p&gt;My calculation got the same answer as you.  Here's what I did.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned that the arm has some weight so I added that force to your diagram.  Also, I labeled the center of the elbow joint as point 'E'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4mLss.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4mLss.png&quot; alt=&quot;http://i.imgur.com/T9KOncw.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The pully is mounted on a frictionless pivot (friction will reduce the amount of force required)&lt;/li&gt;&#xA;&lt;li&gt;The arm is perfectly rigid (if the arm bends, the geometry changes so the forces will change)&lt;/li&gt;&#xA;&lt;li&gt;The system is at rest (rotating motion means acclerations which need forces)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Because nothing is in motion, we can solve for the force using statics (&lt;a href=&quot;https://en.wikipedia.org/wiki/Statics&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Statics&lt;/a&gt;).  Summing of the moments (torques) acting on the blue arm around point E lets us solve for the force with just one equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma M_E=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F\dfrac{D}{2}-ca-WL=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F=\dfrac{2}{D}(ca+WL)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see from the equation that the resultant force is sensitive to the pully diameter.  As the diameter tends towards zero, the force tends towards infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like you already calculated above, the force from the weight is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$W=(225kg)(9.81\frac{m}{s^2})=2,207N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we set the arm weight to zero $(a=0)$ then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F=\dfrac{2}{0.3m}(2207N)(6m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F=88290N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overhead cranes usually require a high factor of safety in their design.  I found an OSHA reference which says use 5:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OSHA 1926.753(c)(1)(i)(C) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.osha.gov/pls/oshaweb/owadisp.show_document?p_table=standards&amp;amp;p_id=10789&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.osha.gov/pls/oshaweb/owadisp.show_document?p_table=standards&amp;amp;p_id=10789&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2408" LastActivityDate="2017-06-21T04:51:08.803" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15960" PostTypeId="1" CreationDate="2017-06-21T05:14:22.100" Score="1" ViewCount="1338" Body="&lt;p&gt;Given a plant with a transfer function of K/s, how can I tune a PID to control it? I found articles explaining how to tune an integrator with time delay, but not for just a pure integrator system.&lt;/p&gt;&#xA;" OwnerUserId="11579" LastEditorUserId="7083" LastEditDate="2017-06-29T11:52:14.870" LastActivityDate="2017-08-28T12:54:37.460" Title="How to tune a PID for a pure integrator system?" Tags="|control-engineering|control-theory|pid-control|transfer-function|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15961" PostTypeId="2" ParentId="15952" CreationDate="2017-06-21T07:03:07.637" Score="4" Body="&lt;p&gt;See &lt;a href=&quot;https://www.mathworks.com/help/control/ref/nyquist.html&quot; rel=&quot;noreferrer&quot;&gt;&lt;code&gt;doc nyquist&lt;/code&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The nyquist function has support for M-circles, which are the contours of &#xA;  the constant closed-loop magnitude. M-circles are defined as the locus of &#xA;  complex numbers where&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$$T(j\omega) = \left|\dfrac{G(j\omega)}{(1+G(j\omega))}\right|$$&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;is a constant value. In this equation, $\omega$ is the frequency in &#xA;  radians/TimeUnit, where TimeUnit is the system time units, and $G$ is the&#xA;  collection of complex numbers that satisfy the constant magnitude &#xA;  requirement.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="11869" LastEditorUserId="1832" LastEditDate="2017-06-21T12:09:58.433" LastActivityDate="2017-06-21T12:09:58.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15962" PostTypeId="2" ParentId="15957" CreationDate="2017-06-21T07:18:12.203" Score="0" Body="&lt;p&gt;Use an over-hanging roof which will then shade the walls. Over-hanging roofs are seen in many countries, but not only for sun but also for snow (eg Switzerland) to stop or reduce the chance of snow piling up against the wooden chalet walls &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This type of design is called &quot;passive solar design&quot; and there are many links / books that can be found.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-21T07:18:12.203" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15963" PostTypeId="1" CreationDate="2017-06-21T07:58:24.390" Score="1" ViewCount="457" Body="&lt;p&gt;I want to do some analysis of an axially loaded simply supported beam considering P-delta effect. I set up a non-linear load case and apply axial load to it. Subsequent analysis of live load and dead load of the beam is done linearly using stiffness matrix from the non-linear case. However, the non-linear analysis results in the beam hogging upwards, instead of sagging downwards, which mitigate effects of live load and dead load, resulting in a reduced bending moment on the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if above analysis is done correctly, or is there a way to force the beam to sag downwards in the non-linear case? Why does it cause beam hogging rather than sagging given that the beam is doubly symmetric with axial load applied to its centroid? I am using SAP 2000 v18 software for above analysis.&lt;/p&gt;&#xA;" OwnerUserId="11443" LastActivityDate="2017-06-21T13:37:05.557" Title="How to analyse a beam considering P-delta effect?" Tags="|civil-engineering|structural-analysis|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15964" PostTypeId="2" ParentId="15954" CreationDate="2017-06-21T08:25:56.733" Score="1" Body="&lt;p&gt;An &lt;a href=&quot;https://en.wikipedia.org/wiki/Induction_motor&quot; rel=&quot;nofollow noreferrer&quot;&gt;induction motor&lt;/a&gt; works by trying to sync up the rotating squirrel cage to the sine wave of the mains voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a single phase supply there is no difference between going forwards or backwards so you need some way of setting the direction on startup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For low torque applications a &lt;a href=&quot;https://en.wikipedia.org/wiki/Shaded-pole_motor&quot; rel=&quot;nofollow noreferrer&quot;&gt;shaded pole&lt;/a&gt; is enough. That is a small loop of copper on the core placed in such a way that the magnetic field gives a push in the right direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For higher torques they instead have a secondary winding perpendicular to the main winding. A capacitor is used to create a &lt;a href=&quot;https://en.wikipedia.org/wiki/LC_circuit&quot; rel=&quot;nofollow noreferrer&quot;&gt;LC circuit&lt;/a&gt; to run the secondary winding slightly out of phase with the main winding. Once the motor is up to speed a centrifugal switch may be used to disconnect the secondary winding and starter capacitor which are often not built for continuous use.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-06-21T08:25:56.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15965" PostTypeId="2" ParentId="15963" CreationDate="2017-06-21T08:44:35.327" Score="1" Body="&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt; - &lt;em&gt;Your current results are incorrect and unsafe to use. Carry out the analysis again, including &lt;strong&gt;all&lt;/strong&gt; load in the P-delta analysis, not just the axial load.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you apply axial load &quot;P&quot;, alone, to a perfectly straight beam, then there is no &quot;delta&quot; for the P-delta effect to occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various methods for creating an initial &quot;delta&quot;; I have never used SAP 2000 so have no idea whether there is a built-in method, or whether you have to do it manually to some extent. In software I have used, I would carry out an eigenvalue analysis, pick the first mode-shape, and use it with a code-specified magnitude for imperfections as the initial shape for my P-delta analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue with this is selecting an appropriate mode shape. If you pick a hogging mode shape, then P-delta will tend to increase the hogging; if you pick a sagging mode shape then it will increase sagging. Numerically, for a perfectly straight beam, the software could easily choose either hogging or sagging, as they are both as numerically valid as each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We now come to your case. An important thing to learn about non-linear analysis is that you cannot carry out linear superposition - it isn't valid. You need to have all loads in the non-linear case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back through the previous methodology, if you carry out an eigenvalue analysis with axial load and with transverse load causing sagging, then the mode shape &lt;em&gt;will&lt;/em&gt; be a sagging one (caused by the transverse load). In the P-delta analysis, again including all load, the transverse load will increase the delta still further, giving an even greater P-delta effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your current results showing the axial load reducing the bending moment are incorrect and unsafe to use.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-06-21T08:44:35.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15967" PostTypeId="2" ParentId="15963" CreationDate="2017-06-21T13:37:05.557" Score="0" Body="&lt;p&gt;In a nonlinear analysis, the reason the beam displaced &quot;the wrong way&quot; is probably because you didn't define &lt;em&gt;all&lt;/em&gt; the loads acting on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the beam was horizontal and you included its self-weight as well as the axial load, it would be very strange if it deflected upwards rather than downwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a civil engineer, but as a general principle, trying to use the stiffness from a nonlinear analysis as the basis for further &quot;pseudo-linear&quot; analyses seems theoretically &quot;just plain wrong&quot; unless it is justified by some civil engineering codes, or assumptions about the relative sizes of the loads, that I don't know about.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-06-21T13:37:05.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15968" PostTypeId="2" ParentId="15936" CreationDate="2017-06-21T14:24:17.413" Score="1" Body="&lt;p&gt;A continuous beam with two spans with uniformly distributed load looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jijcJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jijcJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is symmetric, and we know by inspection that the rotation of the beam at the central support will be null. So let's make our lives a bit easier and only look at the right half of the beam, considering the central support as fixed. Obviously, this is only possible because the beam and its loading is symmetric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gbJMP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gbJMP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This beam is still statically indeterminate, however. But let's get rid of the rightmost support for now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hSkgk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hSkgk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an isostatic structure we can trivially solve. However, the deflection at the extremity will obviously be non-null, which isn't correct for our true condition. Let's calculate what that deflection would be, using the fundamental beam equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\text{d}^2}{\text{d}x^2}\left(EI\frac{\text{d}^2w}{\text{d}x^2}\right) = q$$&#xA;which tells us that the fourth derivative of the deflection is equal to the applied load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, integrating the applied load we get the shear force:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;Q = \int\limits_0^L q\text{d}x = qx + C_1 \\&#xA;Q(0) = C_1 = -qL \\&#xA;\therefore Q = qx - qL&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating the shear force we get the bending moment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;M = \int\limits_0^L (qx - qL)\text{d}x = \frac{qx^2}{2} - qLx + C_2 \\&#xA;M(0) = C_2 = \frac{qL^2}{2} \\&#xA;\therefore M = \frac{qx^2}{2} - qLx + \frac{qL^2}{2}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating the bending moment we get the curvature times the stiffness:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;EI\theta = \int\limits_0^L \left(\frac{qx^2}{2} - qLx + \frac{qL^2}{2}\right)\text{d}x = \frac{qx^3}{6} - \frac{qLx^2}{2} + \frac{qL^2x}{2} + C_3 \\&#xA;\theta(0) = C_3 = 0 \\&#xA;\therefore EI\theta = \frac{qx^3}{6} - \frac{qLx^2}{2} + \frac{qL^2x}{2}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating the curvature we get the deflection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;EI\delta = \int\limits_0^L \left(\frac{qx^3}{6} - \frac{qLx^2}{2} + \frac{qL^2x}{2}\right)\text{d}x = \frac{qx^4}{24} - \frac{qLx^3}{6} + \frac{qL^2x^2}{4} + C_4 \\&#xA;\delta(0) = C_4 = 0 \\&#xA;\therefore EI\delta = \frac{qx^4}{24} - \frac{qLx^3}{6} + \frac{qL^2x^2}{4}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, at the end of the cantilever the deflection is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;EI\delta(L) &amp;amp;= \frac{qL^4}{24} - \frac{qL^4}{6} + \frac{qL^4}{4} \\&#xA;\delta(L) &amp;amp;= \frac{qL^4}{8EI}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in order to regain compatibility with our true condition, we need to bring that deflection down to zero. That's done by the vertical reaction force of the rightmost support. So let's replace that support with a concentrated vertical force and calculate the deflection caused by that force:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;Q = \int\limits_0^L 0\text{d}x = C_1 \\&#xA;Q(0) = C_1 = -P \\&#xA;\therefore Q = -P \\&#xA;M = \int\limits_0^L -P\text{d}x = -Px + C_2 \\&#xA;M(0) = C_2 = PL \\&#xA;\therefore M = -Px + PL \\&#xA;\theta = \int\limits_0^L (-Px + PL)\text{d}x = \frac{-Px^2}{2} + PLx + C_3 \\&#xA;\theta(0) = C_3 = 0 \\&#xA;\therefore EI\theta = \frac{-Px^2}{2} + PLx \\&#xA;\delta = \int\limits_0^L \left(\frac{-Px^2}{2} + PLx\right)\text{d}x = \frac{-Px^3}{6} + \frac{PLx^2}{2} + C_4 \\&#xA;\delta(0) = C_4 = 0 \\&#xA;\therefore EI\delta = \frac{-Px^3}{6} + \frac{PLx^2}{2} \\&#xA;\delta(L) = \frac{PL^3}{3EI}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, we can calculate the necessary force to counteract the deflection at the rightmost extremity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;\frac{PL^3}{3EI} = -\frac{qL^4}{8EI} \\&#xA;\therefore P = -\frac{3qL}{8}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, looking at the component equations for the bending moment from the uniform load and the reaction force, we can add them together to get the final bending moment equation for the span:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;M &amp;amp;= \frac{qx^2}{2} - qLx + \frac{qL^2}{2} - Px + PL \\&#xA;M &amp;amp;= \frac{qx^2}{2} - qLx + \frac{qL^2}{2} - \left(\frac{-3qL}{8}\right)x + \left(\frac{-3qL}{8}\right)L \\&#xA;M &amp;amp;= \frac{qx^2}{2} - \left(qL - \frac{3qL}{8}\right)x - \frac{3qL^2}{8} + \frac{qL^2}{2} \\&#xA;M &amp;amp;= q\left(\frac{x^2}{2} - \frac{5Lx}{8} + \frac{L^2}{8}\right)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum moment is found at the central support and is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M(0) = \frac{qL^2}{8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum moment is trickier, since it isn't necessarily at midspan. First we need to calculate the moment's derivative zero to find its position:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;M' &amp;amp;= q\left(x - \frac{5L}{8}\right) = 0 \\&#xA;\therefore x &amp;amp;= \frac{5L}{8} \\&#xA;\therefore M_{max} &amp;amp;= M\left(\frac{5L}{8}\right) \\&#xA;&amp;amp;= q\left(\frac{\left(\dfrac{5L}{8}\right)^2}{2} - \frac{5L\left(\dfrac{5L}{8}\right)}{8} + \frac{L^2}{8}\right) \\&#xA;&amp;amp;= -\frac{9qL^2}{128} \approx -\frac{qL^2}{14.2}\\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That approximation is a classic numerical shortcut for the maximum moment on a fixed-and-pinned beam. And obviously all my results assume that $q &amp;gt; 0$ points upwards. So all the loads in the beams at the start of my answer have $q = -1\text{ kN/m}$. If you prefer to think of such loads as positive, just flip the sign in my results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then, to confirm our results, here is a beam with $L=5\text{ m}$ (per span) and $q=128\text{ kN/m}$. The equations give us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;M_{min} &amp;amp;= \frac{qL^2}{8} = -400\text{ kNm}\\&#xA;M_{max} &amp;amp;= -\frac{9qL^2}{128} = 225\text{ kNm}\\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UZ7Mc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UZ7Mc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-06-21T14:53:11.107" LastActivityDate="2017-06-21T14:53:11.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15972" PostTypeId="2" ParentId="15896" CreationDate="2017-06-21T19:16:42.257" Score="1" Body="&lt;p&gt;There is a wide world of plastics, and a wider world of corrosive fluids to pit them against.  Then within each plastic are different resins, brands, each with unique manufacturing methods.  Some plastic resins have actually been formulated to be more corrosion resistant against one type of chemical and less against another while still being chemically identical.  Most companies just lock the vast data of corrosion testing in giant matrices that become to vast for casual comparison, and you immediately jump to your favorite chemical and just run the row of plastics.  &lt;a href=&quot;http://www.hipco.com/ChemGuide.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.hipco.com/ChemGuide.cfm&lt;/a&gt; is usually a good source for this, then.  There are others, and official books, but the bottom line is - investigate thoroughly before making plastics decisions.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2017-06-21T20:59:06.840" LastActivityDate="2017-06-21T20:59:06.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15973" PostTypeId="1" AcceptedAnswerId="15974" CreationDate="2017-06-21T22:15:13.990" Score="4" ViewCount="1185" Body="&lt;p&gt;How does the trigger of a typical modern multi speed drill (maybe Ryobi for example) work? Is it just a potentiometer? Or is it more involved?&lt;/p&gt;&#xA;" OwnerUserId="11626" LastActivityDate="2017-06-22T00:48:27.500" Title="Is the trigger of a drill a potentiometer?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15974" PostTypeId="2" ParentId="15973" CreationDate="2017-06-22T00:48:27.500" Score="6" Body="&lt;p&gt;I opened a Ryobi type of cordless drill. The control appeared to be a potentiometer connected to a PWM circuit module which varied the duty cycle of the power applied to the motor.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-06-22T00:48:27.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15975" PostTypeId="1" CreationDate="2017-06-22T02:55:39.483" Score="1" ViewCount="135" Body="&lt;p&gt;I  am trying to make a homemade projector using my phone. Screen size 5.5 inches, focal length of lens is 20 cm, diameter of lens is 6 cm. The problems I am facing are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the edges are distorted;&lt;/li&gt;&#xA;&lt;li&gt;the brightness is too low.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The farther I move my phone away from the lens, the lesser the distortions​ but this also results in less light passing through the lens, so the brightness decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using stops to remove spherical aberration is an option but that will reduce the brightness of image. I want to preserve the little light that my phone can emit. How can I minimize these problems?&lt;/p&gt;&#xA;" OwnerUserId="11880" LastEditorUserId="368" LastEditDate="2017-06-22T16:01:17.807" LastActivityDate="2017-06-22T16:01:17.807" Title="How to get a brighter, sharper image from a projector lens" Tags="|optics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15976" PostTypeId="1" AcceptedAnswerId="15979" CreationDate="2017-06-22T06:30:45.217" Score="0" ViewCount="605" Body="&lt;p&gt;Recently I visited a sericulture facility in India and I saw that people there 'degummed' (&lt;a href=&quot;https://www.youtube.com/watch?v=h13fctiTylI&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=h13fctiTylI&lt;/a&gt;) silk in a batch process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a Chemical Engineer, I thought of this as an unsteady state unit operation and &#xA;I have decided to work on making this process a continuous or steady-state operation as a summer project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far this is my schematic &#xA;&lt;a href=&quot;https://i.stack.imgur.com/ndkJM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ndkJM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The silk thread will be pulled over a large drum and will pass over this drum a number of times in a helical fashion. This drum is immersed in a bath of boiling water with added chemicals so that the silk is degummed. The thread will come out of the other end another end of the cylinder and will be wound on another bobbin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have already determined what sort of chemicals and mass transfer I am gonna get and what should be my temperature of the bath. I am not able to work out on the residence time any part of the silk thread should spend in the bath ? for this I need to determine the RPM of the rotating drum? how do I calculate that? standard degumming time is about 60 minutes. So if I say that the length of the drum is L and its diameter is D and say it is rotating at an angular velocity of w. How do I develop a set of equations to determine my rpm so that I can change my drum dimension if the residence time is too long or too short.&lt;/p&gt;&#xA;" OwnerDisplayName="user11882" LastActivityDate="2017-06-22T14:46:17.390" Title="I need to determine the RPM of a rotating cylinder over which a silk thread is wound and is moving along in the fashion of a screw." Tags="|mechanical-engineering|chemical-engineering|applied-mechanics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="15977" PostTypeId="1" AcceptedAnswerId="15980" CreationDate="2017-06-22T12:33:27.103" Score="4" ViewCount="1419" Body="&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=RIXSzw2ksiE&quot; rel=&quot;noreferrer&quot;&gt;Soap bubble foam made with helium&lt;/a&gt; floats up, but due to extreme fragility hardly counts as &quot;material&quot;. There are many solid foam materials though - PUR foam, or styrofoam to name the most common. They typically use carbon dioxide for inflation though (usually produced from precursors of the foam, as a desirable side effect of their reaction). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it shouldn't be too difficult to make solid foam filled with helium (or hydrogen) in proportions assuring positive buoyancy in air, and I can imagine desirability of it, at least as a filler in applications where mass costs a premium (transport, aviation) even if its structural properties were to be too poor for any other purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is such material produced? Is it used anywhere? Or if not, why?&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-25T00:55:28.077" Title="Is there a lighter-than-air foam material?" Tags="|materials|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15979" PostTypeId="2" ParentId="15976" CreationDate="2017-06-22T13:18:48.993" Score="0" Body="&lt;p&gt;Length $l$ of silk on the immersed part of the drum $m$ divided by feed rate $f$  is the time the thread is immersed &lt;em&gt;T&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l / f = T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feed rate is same as linear velocity of the surface of the drum (assuming negligible silk thickness comparing to drum radius). $f = v = \omega r$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of thread on the drum ($l'$) is the drum circumference ($C$) times number of loops ($n$); number of loops is the usable length of the drum ($L$) divided by pitch of the helix ($p$; which must be no smaller than the silk thickness, preferably at least 2x as large). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l' = nC$; $n = L/p$ ; $C = 2 \pi r$ ; $l' = 2\pi r L/p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the whole drum is immersed, $ l = l'$ so, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ {{2\pi r L/p} \over {\omega r}} = T$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, &#xA;$$ \omega = {{ 2 \pi L } \over {pT}} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the drum of length L and helix pitch p rotating at $\omega$ holds the thread immersed for time T. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Interestingly, makes the radius non-issue). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your drum is partially immersed, you'll need to use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Circular_segment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Circular Segment&lt;/a&gt; length to determine percentage of the circumference that remains submerged, and adjust the time accordingly.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-22T13:18:48.993" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="15980" PostTypeId="2" ParentId="15977" CreationDate="2017-06-22T13:47:43.557" Score="1" Body="&lt;p&gt;No matter how good you seal it, when you inflate a balloon with Helium it will stay up for a while, but after a few days, it will lose its pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enough to realise that in solids, there is also a phenomenon of &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass_diffusivity&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass diffusivity&lt;/a&gt;, and therefore your foam will not retain the gas. This phenomenon is also called &lt;a href=&quot;https://en.wikipedia.org/wiki/Permeation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Permeation&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diffusivity in solids is very complex, and 'mostly' cannot be described with an equation as simple as with &lt;a href=&quot;https://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fick's laws&lt;/a&gt;, and, in many cases, is not even isotropic. But there is still a condition that needs to be met to diffuse: the particle/atom you consider can place itself within the crystallography/pattern of your material. unfortunately for Helium, it is too small and will diffuse through all reasonable materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diffusivity in solids is unfortunately making us unable to isolate a gaz but it &#xA;is also positively used in a lot of fields. &#xA;The most investigated is probably in microelectronics to make local implants of ions in semiconductors and therefore change locally its electrical properties.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="10684" LastEditDate="2017-06-22T14:21:50.643" LastActivityDate="2017-06-22T14:21:50.643" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="15983" PostTypeId="1" CreationDate="2017-06-22T16:03:28.900" Score="0" ViewCount="36" Body="&lt;p&gt;I recently got dental braces and have been using foam earplugs to protect the inside of my lips from the braces. I place them between the lips and the braces and they work quiet well. However I am concerned whether harmful chemicals could potentially leach from the foam. Please advise.&lt;/p&gt;&#xA;" OwnerUserId="11890" LastActivityDate="2017-07-23T12:51:28.133" Title="Health Hazard from leaching?" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15985" PostTypeId="2" ParentId="15860" CreationDate="2017-06-22T16:48:36.507" Score="-1" Body="&lt;p&gt;I suggest you graph $x_1$ and $x_2$ for half unit intervals between -3 and 3 for both variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then work out the derivative of $x_1$ and $x_2$, and mark these on the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then examine the area where the marks go towards the fixed point (ie the origin). Which area around the fixed point end up going towards the f.p.?&lt;/p&gt;&#xA;" OwnerUserId="11867" LastActivityDate="2017-06-22T16:48:36.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15986" PostTypeId="2" ParentId="15977" CreationDate="2017-06-22T17:12:09.717" Score="-2" Body="&lt;p&gt;There is a material type named &lt;em&gt;aerogel&lt;/em&gt; and you can find some information on this &lt;em&gt;quora&lt;/em&gt; topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.quora.com/Is-it-possible-to-create-a-lighter-than-air-aerogel&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-06-22T17:12:09.717" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15988" PostTypeId="1" CreationDate="2017-06-22T18:21:41.917" Score="1" ViewCount="2086" Body="&lt;p&gt;What is the proper and recommended weld preparation for true tangential nozzle on pressure vessel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can something about this thematic be found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SHiTT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SHiTT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="5855" LastEditDate="2017-06-23T15:26:27.940" LastActivityDate="2017-09-21T19:50:33.430" Title="Tangential nozzle on vessel - weld preparation" Tags="|welds|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15989" PostTypeId="1" CreationDate="2017-06-22T20:20:35.023" Score="2" ViewCount="193" Body="&lt;p&gt;I have a 4700 gallon tank and am looking to create a table of the water content by mass in the tank given relative humidity %. The tank is held at 70 degC and 18 psig, and the only thing inside the tank is normal air at varying humidity %.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this has something to do with partial pressures and steam tables.&lt;/p&gt;&#xA;" OwnerUserId="11893" LastEditorUserId="2015" LastEditDate="2017-10-21T17:01:31.640" LastActivityDate="2017-10-21T17:01:31.640" Title="Calculating water content of a tank given humidity?" Tags="|mechanical-engineering|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="15991" PostTypeId="1" AcceptedAnswerId="16001" CreationDate="2017-06-22T21:30:26.747" Score="0" ViewCount="107" Body="&lt;p&gt;When creating a detention pond I have only came across examples where the pond is created by digging. Is it possible that a pond is created by creating retaining walls / embankments? and if it is can you please provide examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note! this is a detention pond not decorative pond.&lt;/p&gt;&#xA;" OwnerUserId="11826" LastEditorUserId="1592" LastEditDate="2020-02-01T17:52:37.833" LastActivityDate="2020-02-01T17:52:37.833" Title="Are detention ponds always dug?" Tags="|civil-engineering|water-resources|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="15993" PostTypeId="2" ParentId="15991" CreationDate="2017-06-23T01:34:23.193" Score="0" Body="&lt;p&gt;Yes, you can build a detention pond in fill slope. It isn't recommended but can be hard to avoid in some cases. For example if the pond must fit at the edge of a site where there is a slope away from the development. In that case you should have as much of the pond in cut area as you can. Also closely consider soil drainage properties and consider a liner of some kind and anti-seep collars to prevent failure due to infiltration in the embankment. Local regulations may also have a say in this design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.elibrary.dep.state.pa.us/dsweb/Get/Document-68005/6.6.3%20BMP%20Dry%20Extended%20Detention%20Basin.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this design guide&lt;/a&gt; for help.&lt;/p&gt;&#xA;" OwnerUserId="40" LastEditorUserId="1832" LastEditDate="2017-06-23T01:59:53.443" LastActivityDate="2017-06-23T01:59:53.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15994" PostTypeId="1" CreationDate="2017-06-23T01:59:47.570" Score="1" ViewCount="68" Body="&lt;p&gt;I want to perform a RULA based analysis on a set of participants and in order to calculate the RULA score I would need the upper arm(shoulder) angle. My question is would the upper arm angle be based of the trunk or the vertical axis? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the person had a bent back and if he was not aligned with the vertical axis, then the upper arm angle would be calculated with respect to the back or the vertical axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the RULA assessment sheet: &lt;a href=&quot;https://i0.wp.com/www.ergo-plus.com/wp-content/uploads/RULA-2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i0.wp.com/www.ergo-plus.com/wp-content/uploads/RULA-2.jpg&lt;/a&gt; the vertical axis and the back are both aligned with each other. But what if they were not? With what other line, ie, the back or the vertical axis, we would obtain the upper arm angle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have so far tried both the methods and both yield decent results but I dont really know which one would be the one RULA actually suggests to use!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You!&lt;/p&gt;&#xA;" OwnerUserId="11897" LastEditorUserId="1832" LastEditDate="2017-06-23T02:11:31.803" LastActivityDate="2018-01-19T09:15:00.033" Title="RULA upper arm angle" Tags="|biomedical-engineering|biomechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15995" PostTypeId="2" ParentId="15994" CreationDate="2017-06-23T05:51:33.713" Score="1" Body="&lt;p&gt;Why not simply record back angle wrt vertical axis in your data then state your assumption that you base your analysis on. Then, the results can be analysed both ways.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-23T05:51:33.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="15996" PostTypeId="2" ParentId="15988" CreationDate="2017-06-23T08:02:23.750" Score="1" Body="&lt;p&gt;The welding codes and standards will apply relevant to your country / location / insurance specifications no matter what angle the nozzle is welded at.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-23T08:02:23.750" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="15997" PostTypeId="1" CreationDate="2017-06-23T08:46:54.383" Score="0" ViewCount="1616" Body="&lt;p&gt;I want to know how to calculate the speed of an electric skateboard &lt;a href=&quot;https://i.stack.imgur.com/jJiID.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jJiID.jpg&quot; alt=&quot;Mount&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If i have that situation(from the picture below) i will need an &lt;strong&gt;acceleration=0.16m/s^2&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a force &lt;strong&gt;F~=15N&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So i will need a power output &lt;strong&gt;P&lt;em&gt;o&lt;/em&gt;=150W&lt;/strong&gt; to move that skateboard with a &lt;strong&gt;velocity=10m/s&lt;/strong&gt; ???&#xA;&lt;a href=&quot;https://i.stack.imgur.com/o3y4V.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o3y4V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Can i calculate the &lt;strong&gt;torque that i need&lt;/strong&gt; with this &lt;strong&gt;T=r*F&lt;/strong&gt; ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;strong&gt;r=radious of the wheel&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;strong&gt;F i think that's the force that i need&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if yes,i will need a &lt;strong&gt;torque of 0,036m*15&lt;/strong&gt;=&lt;strong&gt;0.54Nm&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,in that case,if i have a &lt;strong&gt;motor&lt;/strong&gt; that has a &lt;strong&gt;torque=0.54Nm&lt;/strong&gt; with a &lt;strong&gt;power output=150W&lt;/strong&gt; will move the skate with a speed of 10m/s? With what i'm wrong??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can calculate the torque of the motor with this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;T = (I * V * E *60) / (rpm * 2pi)&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in my case if i have a motor &lt;strong&gt;270kv&lt;/strong&gt; 1665W with Continuous current: 80A with 60%efficiency with a Lipo Battery 22.2V 10.4Ah 30C, I will get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Power Input&lt;/strong&gt;   =22.2*10.4  =&lt;strong&gt;230.88W&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Power Output&lt;/strong&gt;  =230.88*0.6=&lt;strong&gt;138.52W&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RPM&lt;/strong&gt;=270*22.2=&lt;strong&gt;5994RPM&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Torque&lt;/strong&gt;=(10Ah*22.2V*60%ef*60)/(5994RPM*2*3.14)=7.992/37.642,32=&lt;strong&gt;0.21Nm&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what speed the board will have??&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;With what i'm wrong?Please explain&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How the &lt;strong&gt;diameter of the wheel&lt;/strong&gt; affect the speed and acceleration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if i have 72mm wheels,so 0.22m circumference with 5994 rpm will have 0.22*5994=1.318m/min? so 21.96m/s?? if i'm not wrong,&lt;strong&gt;how the mass affect that speed&lt;/strong&gt;? and the not enought torque?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How the diameter of the pulley A and B and the number of teeths and the distance between the teeths affect the torque and speed?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How can i calculate the &lt;strong&gt;output torque and speed&lt;/strong&gt; of the &lt;strong&gt;pulley A&lt;/strong&gt; ??&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With what i'm wrong,how i can calculate what i need please be patient i never take physics/engineering class i just searched on internet &lt;strong&gt;don't be annoyed with my calculus&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update: if i have torque=r&lt;em&gt;m&lt;/em&gt;(v/t) i will get v=(time&lt;em&gt;torque)/(r&lt;/em&gt;m),in my case v=(60*0.21)/(0.036*90)=3.88m/s?? It's that right? If i get more torque with the pulleys from the motor to the wheel i can get something like 7m/s?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Ok Update2: i have 2 gears ,one with 27T other with 14T ,i founded that the ratio is 27/14=1.92 so the torque is 0.21*1.92=0.4Nm? and the speed will be 5994/1.92=3121RPM?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes,i will have v=(time&lt;em&gt;torque)/(r&lt;/em&gt;m) &lt;strong&gt;V= 7.4m/s?&lt;/strong&gt; i can't understand what i'm doing wrong..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe if i think that i have enought torque to move the board,and if the wheel rotate with &lt;strong&gt;3121rpm&lt;/strong&gt; and 72mm diameter with &lt;strong&gt;0.22m&lt;/strong&gt; circumference,it will make &lt;strong&gt;0.22m each rpm&lt;/strong&gt; so 3121*0.22=686.62m/min=&lt;strong&gt;11.44m/s&lt;/strong&gt;. What is the true ??#&lt;strong&gt;Please help !&lt;/strong&gt;#&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Update 3:&#xA;if i will use a 36T pulley with a 14T pulley on the motor with 138.52W power output,5994RPM,0.21Nm torque,70mm wheels &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and i need a torque=0.54Nm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i will have 36/14=2.57 ratio ,so 0.21*2.57=&lt;strong&gt;0.54Nm torque&lt;/strong&gt; and 5994/2.57=&lt;strong&gt;2332RPM&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so each min i will make &lt;strong&gt;0.22m&lt;/strong&gt;(circumference of the wheel)&lt;strong&gt;*2332rpm&lt;/strong&gt;=513m/min=&lt;strong&gt;8.55m/s??&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11899" LastEditorUserId="11899" LastEditDate="2017-06-23T08:56:23.597" LastActivityDate="2018-07-27T11:37:22.453" Title="Torque and Speed of a DC Motor with Pulleys and Wheels" Tags="|motors|gears|torque|battery|pulleys|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="15998" PostTypeId="2" ParentId="15989" CreationDate="2017-06-23T09:08:37.657" Score="1" Body="&lt;p&gt;Partial pressures: &#xA;&lt;img src=&quot;https://wikimedia.org/api/rest_v1/media/math/render/svg/5b4fc737d5a978313a6e220bbc391b2c7d053071&quot; alt=&quot;From wikipedia&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;partial volume equals total volume times partial pressure over total pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 70 degrees, saturation vapour pressure is 4.5psi (from tables).&#xA;Absolute pressure of tank is 18 + 14.7psi = 32.7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partial volume comes out as 646.8 US Gallons, 2.45m^3.&#xA;Density of water vapour is 0.2 kg/m^3 (from tables) at 70C, which comes out to a maximum of 490g of water in the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As relative humidity is just actual over saturation density, your going to have a straight line graph&#xA;&lt;a href=&quot;http://www.engineeringtoolbox.com/density-air-d_680.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/density-air-d_680.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11900" LastActivityDate="2017-06-23T09:08:37.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16000" PostTypeId="2" ParentId="15983" CreationDate="2017-06-23T11:59:48.590" Score="1" Body="&lt;p&gt;Here is what you should do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) DON&quot;T PANIC  (HHGTTG)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Read the label on the bag the earplugs came in.  Does it say &quot;warning: do not eat&quot;? (no).&lt;br&gt;&#xA;3) Consider: your ear is a warm, moist place. Why do you  think there would be no leaching there, even if less than in your mouth?&lt;br&gt;&#xA;4) Why don't you get dental wax like everyone else who uses braces does? I'm sure your orthodontist can provide you some.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, finally, you could look up the MSDS for the foam material online, which should allay all fears forever.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-06-23T11:59:48.590" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16001" PostTypeId="2" ParentId="15991" CreationDate="2017-06-23T12:54:28.280" Score="0" Body="&lt;p&gt;It is possible to construct a dam without digging a hole to form the dam, but earth must be brought in from elsewhere to create bund walls that will enclose the dam. In the past, the inside of the dam used to be lined with clay, to provide a leak proof seal. These days a &lt;a href=&quot;http://www.fabricsolutions.com.au/turkey-nest-dam/&quot; rel=&quot;nofollow noreferrer&quot;&gt;polymer lining&lt;/a&gt; is placed in the smaller turkey's nest dams before filling it with water. Such &lt;a href=&quot;http://soilandhealth.org/wp-content/uploads/01aglibrary/010126yeomansII/010126ch20.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;dams&lt;/a&gt; are usually constructed on flat terrain and tend to be filled via a pipe or hose, generally attached to a pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From page 11 of &lt;em&gt;&lt;a href=&quot;http://irrigationep.weebly.com/uploads/2/2/3/5/22358350/design_of_earthen_tanks.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Design of Earthen Tanks&lt;/a&gt;&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The  turkey's  nest  tank  is  similar  to  the  ring  tank.  But  the  material  for  the  bund &#xA;  construction is brought from the outer side of the bund. The level inside the tank is same as that &#xA;  out side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MVq2L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MVq2L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d9boW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d9boW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/263tW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/263tW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-06-23T12:54:28.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16003" PostTypeId="2" ParentId="15988" CreationDate="2017-06-23T15:42:21.033" Score="1" Body="&lt;p&gt;All code considerations aside, at high pressures the stress distributions in this nozzle configuration demand significant reinforcement and superior welding. The vessel begins to act less like a solid shell and more like a loosely held together curved beam. It is not uncommon for the reinforcement pad to extend out so far that the nozzle + pad take up half of the circumference of the shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anytime a nozzle places this much of a gouge in the shell wall, it should have higher than normal QA/QC reviews and extensive engineering review to confirm compliance with codes and satisfactory factors of safety.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-06-23T15:42:21.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16004" PostTypeId="2" ParentId="7718" CreationDate="2017-06-23T17:10:17.257" Score="4" Body="&lt;p&gt;It really depends on the application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat Pipes are sort of a middle-ground between heat sink and liquid cooling technology.  They take advantage of phase-changing a working gas/liquid to transfer thermal energy from a higher density of thermal energy (greater temperature) to a lower density.  At any rate, no matter the method, you still are ultimately going to be transferring thermal energy to an ambient medium (namely air).  Think about cars and motorcycles.  Motorcycles are typically passively air-cooled and have engines with heat sinks built into the cylinders.  There is little reason to employ a liquid cooling system here because they have plenty of exposure to air due to their smaller, open designs.  Cars on the other hand, virtually always have a liquid cooling system with radiators mounted at the front of the vehicle, evacuating thermal energy from the closely packed engine cylinders to the more open and exposed front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because air (at 1 atmosphere) is such a poor conductor of thermal energy, how effectively you can get rid of heat is going to ultimately boil down to how much surface area is exposed to air, how much air you can move across the exposed surfaces, and the ambient temperature of the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When to use a Heat Sink:&lt;/strong&gt;&#xA;Most straight-forward option.  Use a heat sink, add a fan and you're all set.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Simplest, generally most cost-effective Solution&lt;/li&gt;&#xA;&lt;li&gt;Smallest Possible Footprint&lt;/li&gt;&#xA;&lt;li&gt;High thermal capacity and tolerance&lt;/li&gt;&#xA;&lt;li&gt;The greater the difference in temperatures (object to be cooled and ambient temperature), the more effective the heat sink becomes&lt;/li&gt;&#xA;&lt;li&gt;Air flow is either available or can be readily controlled&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When to use Heat Pipes:&lt;/strong&gt;&#xA;A little more planning may be involved, and space may get to be a concern, but all in all, heat pipes are a very effective means of cooling components within a specified range of temperatures.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Minimal Hassle and Maintenance&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Reduced Weight/Portability (for same capacity heat sink or liquid cooling)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Component to be cooled has relatively low surface area by which to transfer heat (such as a computer cpu)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Vertical Mounting Available: Many heat pipes MUST be mounted so that gravity can properly transfer the liquid after it condenses at the top of the unit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Specific temperature range (to work properly, working medium must phase change).  If component and ambient temperatures are too low, the heat pipes will only function as a mediocre heat sink.  If component and/or ambient temperatures are too high, the gas won't be able to condense and thermal handling will be reduced.&lt;/li&gt;&#xA;&lt;li&gt;Use when space isn't an issue or heat pipes can be custom designed to fit --- for example, the Acer Predator 21 X uses heat pipes produced specifically to fit inside of the laptop.  Stock heat pipe solutions otherwise tend to be somewhat bulky to accommodate the working medium (liquid and gas) and transfer heat through a radiator.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;When to use Liquid Cooling:&lt;/strong&gt;&#xA;If you need to cool a component which is tightly packed amongst other components and otherwise has little access to air flow, liquid cooling is the way to go, so long as you have space elsewhere for a radiator and fan setup.  Essentially, you can use this method to transfer heat from a space-restricted location to a more open location.  Also, depending on the working liquid, liquid cooling will function optimally within a greater range of temperatures than heat pipes, but is also prone to more serious failures if temperatures become to low or high.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When Space is a serious issue --- a heat exchanger can take up very little space and plumbing can be flexible&lt;/li&gt;&#xA;&lt;li&gt;When Air flow is a serious issue --- components to be cooled do not have ready access to air&lt;/li&gt;&#xA;&lt;li&gt;Heat needs to be transferred directly away from components without building up&lt;/li&gt;&#xA;&lt;li&gt;Excellent thermal capacity and heat exchange rate&lt;/li&gt;&#xA;&lt;li&gt;When properly used, can maintain temperature closest to ambient&lt;/li&gt;&#xA;&lt;li&gt;May be combined with other cooling options to lower temperature below ambient&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11911" LastActivityDate="2017-06-23T17:10:17.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16012" PostTypeId="2" ParentId="15977" CreationDate="2017-06-25T00:04:51.333" Score="-1" Body="&lt;p&gt;Here is a link to an aerogel &lt;a href=&quot;https://www.extremetech.com/extreme/153063-graphene-aerogel-is-seven-times-lighter-than-air-can-balance-on-a-blade-of-grass&quot; rel=&quot;nofollow noreferrer&quot;&gt;7 times lighter than air&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are mainly in experimental phase. To make them into practice, they should be cheaper as the equivalent system filled with Helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are mainly porous materials derived from a &lt;a href=&quot;https://en.wikipedia.org/wiki/Aerogel&quot; rel=&quot;nofollow noreferrer&quot;&gt;gel&lt;/a&gt;, from which the liquid component has been removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is like this (it is a different aerogel as in the first link):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uE7kcm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uE7kcm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditDate="2017-06-25T00:55:28.077" LastActivityDate="2017-06-25T00:55:28.077" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16014" PostTypeId="1" CreationDate="2017-06-25T01:28:26.267" Score="-2" ViewCount="135" Body="&lt;p&gt;According to the WTC report there were 3000 gallons of jet fuel per tower. So at energy density of 50 MJ/kg and 3 kg/gal that's 50 GJ of energy. The WTC report also says there was 500 tons of steel and concrete per floor with specific heat over 0.5 kJ per kg Kelvin. That means the fire would reach less than 250 Celsius even if it all burned on one floor.&lt;/p&gt;&#xA;" OwnerUserId="11922" LastEditorUserId="90" LastEditDate="2017-06-25T08:46:20.597" LastActivityDate="2017-06-26T16:19:37.013" Title="How do I refute this WTC argument?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="4" ClosedDate="2017-06-26T09:25:44.440" ContentLicense="CC BY-SA 3.0" />
  <row Id="16016" PostTypeId="1" AcceptedAnswerId="18076" CreationDate="2017-06-25T03:12:18.473" Score="4" ViewCount="3167" Body="&lt;p&gt;I am wondering if there is a formula (or how to derive one) for calculating the minimum radius a rod (with a given radius/diameter) can be bent around while staying in the elastic range.  Obviously, this will depend on the modulus of elasticity for the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am searching for a general formula for this rather than a specific solution for a particular material.  With a general formula, I can evaluate trade offs using different materials (with different moduli), different rod radii, and different radii of bending.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stated more simply, if I wanted to wind a rod of radius X made of a material with modulus of elasticity M around a spool, how would I calculate the minimum radius S of the spool?&lt;/p&gt;&#xA;" OwnerUserId="11925" LastEditorUserId="11925" LastEditDate="2017-06-25T03:36:22.737" LastActivityDate="2017-11-18T21:48:33.013" Title="How to Calculate the Minimum Radius a Rod Can Can Be Bent Staying in the Elastic Range?" Tags="|applied-mechanics|elastic-modulus|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16018" PostTypeId="1" CreationDate="2017-06-25T07:48:00.980" Score="1" ViewCount="270" Body="&lt;p&gt;Say you have a state vector, 1, with states [position, velocity] and you have corresponding A1 and B1 2X2 matrices. Say also that you the same state vector,2, but in a different order [velocity, position] with corresponding A2 and B2 2X2 matrices. Are any calculations, say for a SS controller, you make yield different results or just the same results but in a different order?&lt;/p&gt;&#xA;" OwnerUserId="11930" LastActivityDate="2017-06-29T15:14:22.453" Title="Does the order of the states in a state vector matter?" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16019" PostTypeId="1" CreationDate="2017-06-25T08:10:38.987" Score="0" ViewCount="49" Body="&lt;p&gt;What is preferred material for plate rolling (roll bending) when talking about stainless steel - hot rolled stainless steel plate or cold rolled stainless steel plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final product would be for example shell of pressure vessel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would prefer to use hot rolled stainless steel plate ( for example 6 mm thick), which will ensure that less deforming of material occurs when rolling.&lt;/p&gt;&#xA;" OwnerUserId="5855" LastEditorUserId="5855" LastEditDate="2017-06-25T12:38:57.890" LastActivityDate="2017-06-25T12:38:57.890" Title="Plate rolling - Hot or Cold rolled plate stainless steel" Tags="|steel|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16021" PostTypeId="1" CreationDate="2017-06-25T11:34:57.940" Score="1" ViewCount="1915" Body="&lt;p&gt;I made an electric skateboard (it works great, goes about 25mph) and the motor takes 4 screws to mount it. Motor is mounted to a 1/4 piece of aluminum. The screws go right into the motor and don't come out the other side so bolts are not possible. Over time that crazy vibration (from the road and pebbles and little potholes, from the motor itself being pretty powerful) being makes the screws come loose. Usually happens in less than an hour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried the strongest loctite but nothing seems to hold the screws in place. The loctite lasts a little longer but still they come loose. I have a Ryobi drill that I tightened it with. Maybe I am using the loctite wrong. I am very new to building stuff that is meant to be used in a more extreme way. In a way where lots of vibration is happening and things are moving very fast. And where you life may depend on it. Plus if loctite is so permanent than how would I be able to fix it if things went wrong? And if I use a less permanent loctite then would it even stay for this type of vibration? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am well aware that building something this crazy is a good way to get hurt, but I really enjoy learning. I would love to learn how to make something that lasts for years and doesn't come loose. Something that is &quot;skookum&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you tighten screws to aluminum that will vibrate a lot and get them to not rattle loose over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor is &lt;a href=&quot;https://www.goldenmotor.com/frame-bldcmotor.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. See BLDC-108.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram of the motor. It does not have the sliding brackets as pictured. It came without those, for some reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ARVwd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ARVwd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11626" LastEditorUserId="11626" LastEditDate="2017-06-25T15:01:51.003" LastActivityDate="2021-09-09T23:10:07.070" Title="How to make screws stay tight under high vibration?" Tags="|mechanical-engineering|structural-engineering|bolting|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16022" PostTypeId="1" CreationDate="2017-06-25T11:58:45.797" Score="3" ViewCount="144" Body="&lt;p&gt;This question contains chemistry, but the main focus is DOE design--hence posting on the engineering stack exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is a chemical reaction:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A + B -&gt; Product&lt;/li&gt;&#xA;&lt;li&gt;Product + B -&gt; Impurity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The goals are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;to minimize the impurity.&lt;/li&gt;&#xA;&lt;li&gt;to minimize leftover A. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Reacting the mixture for a long time will get rid of A but increase the Impurity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing B will decrease the Impurity but increase leftover A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To optimize the reaction, I'm planning a DOE with temperature, reagent amounts, and reaction time as variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used software to generate a central composite design. However, the reaction time inputs for each experiment aren't physically reasonable. Look at some of the experiments:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;+-------------+---------------+-------------------+&#xA;| Temperature | Reaction time | Amount of B       |&#xA;+-------------+---------------+-------------------+&#xA;| 20 C        | 4 hrs         | 1 mole equivalent |&#xA;+-------------+---------------+-------------------+&#xA;| 50 C        | 4 hrs         | 1 mole equivalent |&#xA;+-------------+---------------+-------------------+&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At 20C, I think the reaction will proceed slowly enough that it won't be done after 4 hours. However, at 50C, I think the reaction will be done in less than an hour. The reaction output at 4 hours will be useless--all the B will have been long reacted (i.e. the maximum amount of impurity for that temperature/B input will have formed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I handle the fact that reaction time inputs cannot be assigned randomly?&lt;/p&gt;&#xA;" OwnerUserId="11933" LastEditorUserId="8055" LastEditDate="2017-06-26T22:01:52.463" LastActivityDate="2018-06-09T21:06:21.727" Title="DOE to optimize a chemical reaction with impurity forming in series" Tags="|chemical-engineering|statistics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16023" PostTypeId="2" ParentId="16021" CreationDate="2017-06-25T15:02:24.670" Score="7" Body="&lt;p&gt;There are several ways you might try to fix this, for example.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Don't use screws. Put some &lt;em&gt;studs&lt;/em&gt; into the screw holes in the motor and fix them permanently (like the studs that hold the wheels onto a car). Then you can attach the studs to the mounting plate (or the skateboard frame) with nuts and shakeproof washers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ignore the screw fixings, and run long bolts along the 8 slots in the motor heatsink. You will need to make another plate for the back of the to motor, attached to the skateboard somehow. (The OP added a comment saying the motor does not have the slots shown in the picture, but I'll leave this idea in the answer because it might be useful to somebody else).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Design something so the motor can &quot;vibrate&quot; along with the wheel, and absorb the vibration somewhere else - for example bolt the mounting plate to the board with spring-loaded bolts. (That is similar to option 1, but not using studs).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-06-25T15:07:47.253" LastActivityDate="2017-06-25T15:07:47.253" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16026" PostTypeId="2" ParentId="16014" CreationDate="2017-06-25T21:45:39.393" Score="5" Body="&lt;p&gt;Easy.  You go to the house of the person making the claim with a gallon of gasoline.  You pour the gasoline on the garage floor.  You inform the claimant that the garage floor contains approximately 10,000 kg of concrete, and the gallon of gasoline contains approximately 100 MJ of energy, and therefore the temperature will only go up by about 20 degrees C when the gasoline burns.  You hand the claimant a match, and tell him/her to strike it once you're a safe distance away.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2017-06-25T21:45:39.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16027" PostTypeId="1" CreationDate="2017-06-26T01:39:38.543" Score="0" ViewCount="212" Body="&lt;p&gt;In context of machine design, what is mean fatigue strength? (as I am thinking I am mixing it up with midrange strength)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my class we deal a lot with steels and this is a low strength steel, but I can't find those three words &quot;mean fatigue strength&quot; put together followed by a definition and/or formula...&lt;/p&gt;&#xA;" OwnerUserId="11941" LastActivityDate="2017-06-26T21:09:08.707" Title="mean fatigue strength" Tags="|mechanical-engineering|materials|design|strength|fatigue|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16028" PostTypeId="1" CreationDate="2017-06-26T03:27:02.890" Score="0" ViewCount="286" Body="&lt;p&gt;I am building a cart from (&lt;a href=&quot;https://engineering.stackexchange.com/questions/15931/what-type-of-tires-and-wheels-do-i-need-to-make-a-pvc-beach-cart&quot;&gt;What Type of Tires and Wheels Do I Need to Make a PVC Beach Cart?&lt;/a&gt;). I have a wheel whose central diameter is 3 inches and the PVC for the axle is 1 inch in diameter. I am planning on cutting a wooden circle and attaching it to the wheel and cutting a 1.25 inch hole through the center of the wood so the PVC pipe can pass through the wood. Is this a good idea or are there better alternatives? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture of the wheel and tire &lt;a href=&quot;https://i.stack.imgur.com/D7OI7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D7OI7.jpg&quot; alt=&quot;Wheel and Tire&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11943" LastEditorUserId="2015" LastEditDate="2017-10-24T18:03:07.987" LastActivityDate="2017-10-24T18:03:07.987" Title="How To Mount a Large Diameter Tire To A Smaller Diameter Axle?" Tags="|mechanical-engineering|wheels|" AnswerCount="1" CommentCount="3" ClosedDate="2017-11-06T18:07:39.650" ContentLicense="CC BY-SA 3.0" />
  <row Id="16029" PostTypeId="2" ParentId="2265" CreationDate="2017-06-26T05:44:24.797" Score="0" Body="&lt;p&gt;Had a similar problem with my electrolux microwave turntable rotating when door was opened and stopped when closed. Also while rotating you could force it into the opposite direction. After checking the 3 safety microswitches which were found ok. Have noticed that mains polarity, live and neutral swapped influences this. The mains sockets i have are European so the plug can be inserted any way around, not like US or UK type. This has supprised me that some kitchen equipment can be polarity sensitive.&lt;/p&gt;&#xA;" OwnerUserId="11945" LastActivityDate="2017-06-26T05:44:24.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16030" PostTypeId="2" ParentId="16014" CreationDate="2017-06-26T05:47:11.020" Score="5" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;This assumes even a paper clip worth of steel would still suffice to support the construction. This is blatantly false - only a moderate dent in the structural integrity is enough to make the building collapse.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This assumes heat was distributed evenly, fuel heating all parts of the floor at the same rate. No localized zones of much higher temperature, or zones barely affected. Blatantly false, it takes some extremely smart engineering to build furnaces that distribute heat evenly. A building fire is definitely nothing of that kind.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Steel doesn't need to melt to lose structural durability - while 250 °C is not enough, even ~800 °C is quite sufficient to compromise the construction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;These were offices. While the construction was steel and concrete, there was a plenty of paper, plastic, cloth, and other flammable materials. Jet fuel wasn't the only source of heat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="203" LastEditorUserId="368" LastEditDate="2017-06-26T16:19:37.013" LastActivityDate="2017-06-26T16:19:37.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16031" PostTypeId="2" ParentId="16018" CreationDate="2017-06-26T05:51:51.757" Score="-1" Body="&lt;p&gt;No it does not matter! Matrix rows and columns are interchangeable and all it does is change the order of items in them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can even transpose the items and all you have achieved is the reversal of your calculations.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2017-06-26T05:51:51.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16032" PostTypeId="2" ParentId="16028" CreationDate="2017-06-26T06:06:24.383" Score="1" Body="&lt;p&gt;What may be significantly easier is to make an adapter plate that matches the hole pattern on the wheel hub and a pvc pipe flange. The other really easy thing to do at this point is using a frame from a new four wheeled wheelbarrow. The wheelbarrow is belt from those environment, has roughly 1 ton capacity engineered in and it saves you a ton of time which I would assume is worth way more than the wheelbarrow. I hope that this helps.&lt;/p&gt;&#xA;" OwnerUserId="11944" LastActivityDate="2017-06-26T06:06:24.383" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16033" PostTypeId="2" ParentId="16021" CreationDate="2017-06-26T06:24:03.807" Score="0" Body="&lt;p&gt;There are two options to improve the vibrations and keep that motor attached. The first would be to isolate the motor with dampeners, but that doesn't eliminate the vibration from the road. The other option that is used extensive in aviation is to use fasteners that can be safety wried. Here is a &lt;a href=&quot;https://youtu.be/XktwpWPFvHs&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; to explain safety wiring. This company has some aircraft grade hardware including some screws that are drilled for safety wiring. The company is &lt;a href=&quot;https://www.aircraftspruce.com/catalog/hapages/NAS1351H.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aircraft Spruce&lt;/a&gt;. You can also get safety wire from that company. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="11944" LastActivityDate="2017-06-26T06:24:03.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16035" PostTypeId="2" ParentId="16027" CreationDate="2017-06-26T10:58:57.293" Score="1" Body="&lt;p&gt;Fatigue is failure cause by cycling loads which do not exceed the yield stress of the material, typically over tens of thousands of cycles. Not to be confused with work hardening which occurs over relatively few cycles at stresses exceeding the yield stress of the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fatigue strength is usually defined by the stress at which a material will exhibit no fatigue however many load cycles it is exposed to (aka &lt;em&gt;endurance limit&lt;/em&gt;)  or the load at which a sample will fail after N cycles. Also fatigue properties are a statistical distribution rather than an absolute value hence the qualification with 'mean', although this is often omitted as it is implicit in the definition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases low alloy steels have a fatigue strength which is close enough  to their yield stress that once factors of safety have been taken into account it becomes a non-issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fatigue limts can also be affected by the surface condition of the sample eg surface roughness, notches etc. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-06-26T10:58:57.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16038" PostTypeId="1" CreationDate="2017-06-26T16:50:04.550" Score="1" ViewCount="337" Body="&lt;p&gt;I would like to construct a seconds pendulum. The length of a 1/2 Hz pendulum is 994 mm. My question, is that length the distance to the center of mass? And how does the density of the arm effect the period? e.g.; will the length a steel arm and weight be different than an aluminum of wood arm and weight? &lt;/p&gt;&#xA;" OwnerUserId="7220" LastActivityDate="2017-06-26T18:25:18.000" Title="Length of a seconds pendulum" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16039" PostTypeId="2" ParentId="16038" CreationDate="2017-06-26T18:02:41.207" Score="1" Body="&lt;p&gt;The formula which the &lt;a href=&quot;https://en.wikipedia.org/wiki/Seconds_pendulum&quot; rel=&quot;nofollow noreferrer&quot;&gt;seconds pendulum&lt;/a&gt; is derived is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pendulum#Period_of_oscillation&quot; rel=&quot;nofollow noreferrer&quot;&gt;simple gravity pendulum&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple gravity pendulum assumes a few things when you derive the solution to the differential equation.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The arm is massless and does not stretch or compress.&lt;/li&gt;&#xA;&lt;li&gt;There is no air resistance/friction opposing the motion.&lt;/li&gt;&#xA;&lt;li&gt;It is a point mass&lt;/li&gt;&#xA;&lt;li&gt;The angle of movement is small&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(there are some other assumptions; but those are the key ones for what you are asking).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is modeled as a single point of mass on an arm with no weight.  This means that if the mass of the arm can't be ignored; it will affect the equation.  In this case, we have a very long arm that ideally will support the pendulum without stretching or compressing, so it will need to be somewhat strong; and is therefore likely to have noticeable weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a real sized pendulum bob on a weightless arm, you would need to have $994 \  mm$ from the pivot point to the centre of the pendulum bob; because it is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Center_of_mass&quot; rel=&quot;nofollow noreferrer&quot;&gt;centre of mass&lt;/a&gt; of the pendulum where the simple gravity pendulum equation is derived.  When the arm weight is important, it will raise your centre of mass closer to the arm from the middle of the bob.  It also may create substantial differences due to the moment of inertia; as we are no longer a point mass; but instead a rotating rigid body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have to consider the weight and geometry of your arm + bob to determine the centre of mass.  The tricky part will be designing the two in a way so that the centre of mass is $994 \ m $ from your pivot.  It may take trial and error.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastEditorUserId="7736" LastEditDate="2017-06-26T18:25:18.000" LastActivityDate="2017-06-26T18:25:18.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16040" PostTypeId="1" CreationDate="2017-06-26T19:38:10.240" Score="1" ViewCount="278" Body="&lt;p&gt;Can you offer a solution to my question/problem?&#xA;I want to extend the the length of the tongue on my trailer, for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Allow far more clearance for turning when backing the trailer, and allow backing to be to less sensitive to turning inputs in general. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;But primarily to reduce the existing tongue weight of the trailer, which is excessive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Given that I know (physically measured with gauge) the current tongue weight of my trailer, is there a way I can calculate the percentage (%) reduction of the tongue weight per 12&quot; (1 ft.) increase in the length of the tongue?&lt;/p&gt;&#xA;" OwnerUserId="11953" LastEditorUserId="1832" LastEditDate="2017-06-26T20:19:38.413" LastActivityDate="2017-07-27T08:07:44.577" Title="Trailer tongue weight" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16041" PostTypeId="1" AcceptedAnswerId="16071" CreationDate="2017-06-26T20:42:10.937" Score="0" ViewCount="139" Body="&lt;p&gt;I have an object that is launched from the side of a plane (it is a flare). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: what does the flare do while it is proximate to the plane?  Will it hit the tail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After accounting for the motion of the aircraft, I have the motion of the flare decently understood and plotted, according to wind resistance without accounting for the propeller wash. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Then it occurs that the flare will be in the prop wash of the aircraft while it is in motion near the plane. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Understanding that the prop wash will be turbulent, and assuming that I find a way to &lt;a href=&quot;https://www.physicsforums.com/threads/how-fast-is-air-blown-from-a-propeller.606908/&quot; rel=&quot;nofollow noreferrer&quot;&gt;estimate the wind speed of the prop wash&lt;/a&gt;, what will be the effect on the force of wind resistance due to the prop wash?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e., how will the turbulence of the prop wash mitigate the increased wind resistance due to the airspeed of the prop wash? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To expand on this, and the flexibility of the answer, the parameters are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The velocity of the prop wash is an estimate of the upper bound of the &lt;em&gt;average&lt;/em&gt; velocity of the prop wash&lt;/li&gt;&#xA;&lt;li&gt;The purpose of this is to verify that an object coming out of the side of the plane will not hit the tail&lt;/li&gt;&#xA;&lt;li&gt;It is necessary simply that I establish an upper bound on what the wind resistance conditions are going to be, as I already have a code written for the position estimate, relative to the aircraft geometry, speed, altitude, etc.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My current solution is as follows (not ready to post as an answer):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Get an upper bound on power output of the engine (~1900 kilowatts for the &lt;a href=&quot;https://en.wikipedia.org/wiki/General_Electric_T700&quot; rel=&quot;nofollow noreferrer&quot;&gt;turboprop in question&lt;/a&gt; here)&lt;/li&gt;&#xA;&lt;li&gt;Multiply the watts by the reciprocal of the forward velocity of the aircraft and reciprocal of area affected by the propeller to get a force per sq meter &lt;/li&gt;&#xA;&lt;li&gt;Solve for $v_w$ in the equation $F = \rho A_{prop} v_o (v_w - v_o)$, which is a correction to the estimate in response 3 of the linked forum thread. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11577" LastEditorUserId="11577" LastEditDate="2017-06-28T14:50:34.287" LastActivityDate="2017-09-26T19:11:39.797" Title="Wind resistance inside turbulent flow (&quot;prop wash&quot;)" Tags="|aerospace-engineering|aerodynamics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16042" PostTypeId="2" ParentId="16027" CreationDate="2017-06-26T21:09:08.707" Score="0" Body="&lt;p&gt;Steel has an endurance limit, other metals do not. It is about half of the tensile strength.So steels cyclically stressed in a range between zero and one half of their tensile strength, will have very long life ( I hate to say infinite).  Otherwise , fatigue strength for other metals and steel stressed above half of tensile is a specific number of cycles at a specific stress level. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-06-26T21:09:08.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16043" PostTypeId="1" CreationDate="2017-06-27T00:10:37.990" Score="0" ViewCount="759" Body="&lt;p&gt;I’m working on a prototype and I’m having trouble picking out a motor for that job. I need it to be able to move a rope that is about 2-4 mm thick outward (like a winch in reverse almost) and I also need it to be quick. Quick being about 0.025 mm per second. How would I determine this exactly? Unfortunately, math has never been my strong suit and while I have been looking over datasheets I’m not understanding them too well. What I’m asking is how do you determine if a motor has enough torque AND enough speed for an application. &lt;/p&gt;&#xA;" OwnerUserId="6887" LastEditorUserId="9770" LastEditDate="2017-06-29T11:30:55.753" LastActivityDate="2017-06-29T11:30:55.753" Title="How to calculate if a motor is fast AND strong (torque) enough?" Tags="|motors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16044" PostTypeId="2" ParentId="417" CreationDate="2017-06-27T01:40:21.927" Score="1" Body="&lt;p&gt;Depending on the Mach numbers and Reynolds numbers, the size of the seeding rakes have different effects on the flow over the test object. Here is a &lt;a href=&quot;http://www.smokemachines.com/wind-tunnel.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;commercial manufacturer&lt;/a&gt; of smoke machines for wind tunnel testing. Check out their Q&amp;amp;A section for some great information about different types of seeding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another &lt;a href=&quot;http://scitekconsultants.co.uk/products/wind-tunnel-seeders/&quot; rel=&quot;nofollow noreferrer&quot;&gt;consultant group&lt;/a&gt; that talk about issues when developing seeding systems. &lt;/p&gt;&#xA;" OwnerUserId="11944" LastActivityDate="2017-06-27T01:40:21.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16045" PostTypeId="2" ParentId="16040" CreationDate="2017-06-27T03:03:23.120" Score="1" Body="&lt;p&gt;the calculation is actually very simple, if you multiply the tongue weight by the length of the trailer from wheel center to tongue (assuming single axle), that quantity is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so say you want a 25% reduction, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       (1-.25) * tongue weight * new length = &#xA;           tongue weight * current length&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or    &lt;code&gt;new len = 1.33* old len&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note you do not want the tongue weight less than about 5% of the total weight as it may become unstable on the highway. (ok for a trailer you never haul fast though)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;btw for those who aren't familiar, it is much easier to back a longer trailer.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2017-06-27T03:03:23.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16046" PostTypeId="2" ParentId="16043" CreationDate="2017-06-27T03:07:09.327" Score="1" Body="&lt;p&gt;There are a few things that are important, including the characteristics of the load. Is it a constant load? Does it change slowly? Quickly? Make sure to take this into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get some mileage out of the standard formula for mechanical power, which is $P=\tau\omega$. In other words, for a constant power motor, you can have high torque &amp;amp; low velocity, or low torque &amp;amp; high velocity. So, you can start by calculating the torque required to drive your load. Then, the angular velocity to move the rope at your desired rate of 250 microns/sec. It's straightforward to compute power from that, but don't forget to add a safety factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've got the power requirement, you can consult the datasheets to see which motors can deliver that amount of power. I would also recommend using motor sizing tools provided by vendors. For example, this one from &lt;a href=&quot;http://www.portescap.com/resources/motor-selection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Portescap&lt;/a&gt;, or this one from &lt;a href=&quot;http://www.kollmorgen.com/en-us/service-and-support/technical/motioneering-online/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kollmorgen&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11924" LastActivityDate="2017-06-27T03:07:09.327" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16047" PostTypeId="1" CreationDate="2017-06-27T04:18:20.863" Score="5" ViewCount="219" Body="&lt;p&gt;Nanocrystalline diamonds can be synthesized by direct conversion of graphite or carbon black mixed with low amount of boron (0.1%-10%) at 8-10 GPa, according to US patent No.20120255236.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If some scientists decided to make very large monolithic diamonds (tens of cm size) to make for example large diamond anvil cells with a centimeter sample size, would it be practical to use large pressure vessels with very thick steel walls to reach such very high hydrostatic pressures for such a large volume?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to calculate the required material thickness to withstand such high pressures?&lt;/p&gt;&#xA;" OwnerUserId="11954" LastEditorUserId="10902" LastEditDate="2017-06-27T07:44:24.077" LastActivityDate="2017-07-28T15:32:38.073" Title="Is it possible to make very large diamonds using a pressure vessel at several GigaPascal pressure?" Tags="|mechanical-engineering|structural-engineering|materials|stresses|industrial-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16048" PostTypeId="2" ParentId="16040" CreationDate="2017-06-27T07:52:16.433" Score="0" Body="&lt;p&gt;The best way to control the tongue weight is to have the axle(s) central under the load bed. Then making sure you have the 5% or whatever the amount can be achieved by controlling the load or moving the axle back past the centreline during construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then a long tongue (drawbar) helps reversing if the turning circle of the towing vehicle is smaller than the turning circle of the trailer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very short trailers (trailer axle to hitch shorter than wheelbase of towing vehicle) are difficult to reverse : long ones are easy (trailer axle to towing hitch longer than wheelbase of towing vehicle) - of course this does depend on the skill of the driver :)&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-27T07:52:16.433" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16051" PostTypeId="1" CreationDate="2017-06-27T10:22:47.733" Score="0" ViewCount="80" Body="&lt;p&gt;This is a solved example in my text book&quot;Engineering mechanics dynamics. R.C.Hibbeler&quot; : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 3 kg disk is attached to the end of a cord.  The other end of the cord is attached to a ball and socket joint located at the center of the platform.  If the platform rotates rapidly,  and the disk is placed on it and released from rest,  determine the time it takes for the disk to reach a spead great enough to break the cord.  The maximum tension the cord can sustain is 100 N..  And the coefficient of the kinetic friction between the disk and the platform is 0.1 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;when the free body diagram of  the particle was drawn,  the friction force was considered in the tangential direction,,  but it wasn't considered in the normal direction(tension was only considered) why?  I was expecting that there is kinetic friction in the normal direction,  since the disk is moving relative to the platform.&lt;br&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6cSxq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9742" LastActivityDate="2017-06-27T11:14:02.210" Title="Why is friction force is not considered as centripetal force?" Tags="|mechanical-engineering|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16052" PostTypeId="2" ParentId="16051" CreationDate="2017-06-27T11:14:02.210" Score="1" Body="&lt;p&gt;Because friction resists either a centrifugal force or a centripetal force - ie friction opposes any motion.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-27T11:14:02.210" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16056" PostTypeId="2" ParentId="3174" CreationDate="2017-06-27T16:41:56.833" Score="-2" Body="&lt;p&gt;For anyone who doesn't want to over-think every variable, the long story short is to use HOT air, regardless of any other circumstance.  You want to get above the dew point, and dry it out.  Warm/Hot air will do this the best.&lt;/p&gt;&#xA;" OwnerUserId="11959" LastActivityDate="2017-06-27T16:41:56.833" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16057" PostTypeId="1" CreationDate="2017-06-27T17:20:47.977" Score="3" ViewCount="867" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VavO8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VavO8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking at linear compression dampers and finding them most commonly &lt;a href=&quot;https://www.mcmaster.com/#adjustable-dampers/=188lcer&quot; rel=&quot;nofollow noreferrer&quot;&gt;specified only with a &quot;Max Force&quot; – e.g., 270 lbs&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since the damping mechanism is provided by hydraulic or pneumatic resistance, the damping force must be a function of the force applied.  &lt;a href=&quot;https://www.asraymond.com/compression-dampers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a supplier&lt;/a&gt; that specifies the damping rate as the resulting &quot;inches/second of motion given a 40lb load.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As best I can tell, neither of these figures completely describes the damping characteristics of a hydraulic compression damper.  Nor do they describe the same characteristic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there some standard model or specification of such dampers that can characterize the damping resistance or speed that will result from &lt;em&gt;any&lt;/em&gt; force (within some design range)?&lt;/strong&gt;  (And do either of these sample specifications determine the model?)&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2017-06-28T19:19:09.553" Title="Specification of linear speed dampers" Tags="|modeling|springs|specifications|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16060" PostTypeId="1" CreationDate="2017-06-27T22:28:39.447" Score="0" ViewCount="53" Body="&lt;p&gt;Consumer-level electrical portable cords come in a variety of types (and gauge, but that's not the subject here) and each must have its rationale, considering the cost constraint (performance/price sweet spot) and environmental use cases (exposure to water/oil/others liquids, abrasion, visibility, temperature...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a hard time finding out a good reference (&lt;a href=&quot;https://en.wikipedia.org/wiki/Portable_cord&quot; rel=&quot;nofollow noreferrer&quot;&gt;the most appropriate wikipedia page&lt;/a&gt; is poor on pointers) on how these standard cord types came to be, and how a product designer would choose a particular cable type over another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;SVT/SJT cables are used for computer power cords;&lt;/li&gt;&#xA;&lt;li&gt;HPN cables are used in North America on kettles and toasters;&lt;/li&gt;&#xA;&lt;li&gt;I've seen electric irons with cotton-braided cables, I'd expect them to have  constraints as kettle cables, plus mobility;&lt;/li&gt;&#xA;&lt;li&gt;lawn mower cords are typically SJTW with flashy colors (not green...);&lt;/li&gt;&#xA;&lt;li&gt;extreme temperature car block heater cables are STOW, it seems that rubber starts to go hard at -40°C&lt;/li&gt;&#xA;&lt;li&gt;household vacuum cleaner cords are often SV (and often at the gauge limit but are pretty long and tend to be used for a few minutes only...) while garage vacuums are using SJTW (also with bigger gauge in my experience);&lt;/li&gt;&#xA;&lt;li&gt;SJOOW seem to be used in construction&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11963" LastEditorUserId="368" LastEditDate="2017-12-14T18:14:45.000" LastActivityDate="2017-12-14T18:14:45.000" Title="Choosing the best type of portable cord for a consumer electronics product" Tags="|materials|power-engineering|product-engineering|electrical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16062" PostTypeId="2" ParentId="16060" CreationDate="2017-06-28T06:06:21.587" Score="2" Body="&lt;p&gt;The choice of cable is down to the environment that the cable is working in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Designers will be looking at mechanical abuse - by accident or not, temperature, chemical hazards etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, taking your some of your examples:&#xA;Computer power -assuming you mean desktop little movement ambient temperature low chemical hazard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kettles / toasters - temperature risk, kitchen environment so liquid etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric irons the cotton braid was used against temperature risk due to possibility of cable coming into contact with the hot surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lawn mowers cords - visibility is an extra one here many people run over the cable with a mower (how to design for stupidity...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no one design document that I am aware of, the usual process is the designer has to evaluate and mitigate the risks inherent in the situation and also comply with any laws / regulations / codes which is the minimum - also fire departments may have applicable codes and so may insurance companies.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-06-28T06:06:21.587" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16063" PostTypeId="2" ParentId="16047" CreationDate="2017-06-28T10:43:15.587" Score="1" Body="&lt;p&gt;Such a pressure vessel is not technically achievable with current technology. See &lt;a href=&quot;https://engineering.stackexchange.com/questions/2551/gigapascal-pressures-in-a-small-package&quot;&gt;this question&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two main methods of making synthetic diamonds. HPHT (High pressure, high temperature), which uses something like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Diamond_anvil_cell&quot; rel=&quot;nofollow noreferrer&quot;&gt;Diamond Anvil Cell&lt;/a&gt; to achieve pressures in the Gigapascal range and temperatures over 1400 $^o$C. The largest diamonds produced this way seem to be in the 1-2 cm range but it is difficult to get reliable information on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other main method is CVD (Chemical Vapor Deposition) which doesn't require high pressures. A research team at the Carnegie Institute of Washington has developed a method to produce larger diamonds using CVD. See &lt;a href=&quot;https://www.newscientist.com/article/dn16036-artificial-diamonds-now-available-in-extra-large/&quot; rel=&quot;nofollow noreferrer&quot;&gt;article in New Scientist&lt;/a&gt;. Apparently diamonds in the inch range are possible using their method. They also believe that larger sizes will be achievable in the future.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2017-06-28T15:31:48.963" LastActivityDate="2017-06-28T15:31:48.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16064" PostTypeId="2" ParentId="16022" CreationDate="2017-06-28T10:52:22.160" Score="0" Body="&lt;p&gt;You could specify a bit more your knowledge on the system. Which is the limiting reagent? Do you expect a first-order or second order reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my understanding that A is in small excess and that you have a certain knowledge of the reaction rates (as you can forecast that the reaction will be completed at 50 degC by not at 20 degC)&#xA;You can define dimensionless time in a similar way to the Damköhler numbers such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \tau = k\,t$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $k$ is your expected rate constant and $t$ is time. This expression is valid for first-order reactions, but you can define similar one for arbitrary reaction rate expressions. By this approach, your are somehow combining temperature and time into a single variable and the variables for your DOE would be the dimensionless time $\tau$ and the reagent amounts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, you can use &quot;expected conversion of B&quot; as an input variable instead of $\tau$. &lt;/p&gt;&#xA;" OwnerUserId="3991" LastActivityDate="2017-06-28T10:52:22.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16065" PostTypeId="2" ParentId="15960" CreationDate="2017-06-28T12:41:55.860" Score="2" Body="&lt;h2&gt;Ideal Water Tank&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The closed-loop water level control system looks like Figure.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/W3Jk9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W3Jk9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure. Closed-loop water level control system&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 3 can be used as mentioned in comment above :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T(s) = 1 / ( A * s ) where Flow = Area * ( dHeight / dTime )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If all parameters set (&lt;strong&gt;positively&lt;/strong&gt;), this system will be &lt;strong&gt;stable&lt;/strong&gt; also. Changing controller parameters will change the response of system but not the stability. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;MATLAB Simulink can be also used in the design process.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Selecting Controller Parameters&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;One can choose different controller parameters for K1, K2*s, K3/s terms. This will only change the general response behaviour of system but this will not change the stability of system, so it will remain stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple ways to select them;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Inspect the Previous Applications&lt;/li&gt;&#xA;&lt;li&gt;Inspect the Articles about the subject if unique&lt;/li&gt;&#xA;&lt;li&gt;Design&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the design process, one must consider the requirements of the system. Also in some preferred programs like MATLAB can be useful with Simulink in the design process. There are number of common terms in transient response characteristics and which are;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Delay time&lt;/strong&gt; (td) is the time required to reach at 50% of its final value by a time response signal during its first cycle of oscillation.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Rise time&lt;/strong&gt; (tr) is the time required to reach at final value by a under damped time response signal during its first cycle of oscillation. If the signal is over damped, then rise time is counted as the time required by the response to rise from 10% to 90% of its final value.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Peak time&lt;/strong&gt; (tp) is simply the time required by response to reach its first peak i.e. the peak of first cycle of oscillation, or first overshoot.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Maximum overshoot&lt;/strong&gt; (Mp) is straight way difference between the magnitude of the highest peak of time response and magnitude of its steady state. Maximum overshoot is expressed in term of percentage of steady-state value of the response. As the first peak of response is normally maximum in magnitude, maximum overshoot is simply normalized difference between first peak and steady-state value of a response.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The controller parameters so the characteristics of the system must satisfy the system requirements. There are more formulas about each of them depending on the systems characteristic equation's order etc. While changing parameters, it will be useful to simulate the response to visualize the problem.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Desing section can be extended more. Tutorials for: &lt;a href=&quot;http://ctms.engin.umich.edu/CTMS/index.php?example=Introduction&amp;amp;section=SystemAnalysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;System Analysis&lt;/a&gt; and &lt;a href=&quot;http://ctms.engin.umich.edu/CTMS/index.php?example=Introduction&amp;amp;section=ControlPID&quot; rel=&quot;nofollow noreferrer&quot;&gt;PID-Design&lt;/a&gt;. There are detailed lectures in the site too. I can update the answer if there is some missing points.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7468" LastActivityDate="2017-06-28T12:41:55.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16066" PostTypeId="2" ParentId="16057" CreationDate="2017-06-28T13:21:11.910" Score="1" Body="&lt;p&gt;I think:  These ratings are showing the max speed allowed by the damper. Certainly for a specified load, there's a startup acceleration, and further the manufacturers are assuming the load is applied nowhere near an impulse/delta function.  Every system has a resonant frequency, meaning that if the load were applied in a very short pulse (or even a short rise time) the damper would transmit the shock rather than absorb or respond with resistive force.&lt;br&gt;&#xA;The intended applications don't involve impulse forces, which is why (I'm guessing) the PSD is not of interest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While automotive shock absorbers are not exactly the same as these dampers, they can be illustrative.  They're designed to limit the amount of energy transferred from the wheels to the chassis, and  more important,to spread impulse input energy over a much longer output period, thus smoothing the ride.  But if you go over a bump at just the right speed, the shock absorber will 'resonate' and collapse without transmitting any energy, releasing the energy back to the tire/road after the bump.  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-06-28T13:21:11.910" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16067" PostTypeId="1" CreationDate="2017-06-28T14:23:17.463" Score="0" ViewCount="281" Body="&lt;p&gt;I always wonder in directional flow which fan would send colder.air and cool better. I am biased toward table fan as it throws fresh air while ceiling fan throws stale hot air accumaled at top. Please tell which is better according to aerodynamic?&lt;/p&gt;&#xA;" OwnerUserId="10977" LastActivityDate="2017-06-28T20:50:01.793" Title="Ceiling or table fan better" Tags="|aerodynamics|experimental-physics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16068" PostTypeId="2" ParentId="16057" CreationDate="2017-06-28T14:34:38.613" Score="1" Body="&lt;p&gt;The OP's second link, gives damping rates albeit in a most confusing way. If you work it out, for example, the first damper has 1.4 in/sec as the damping and references 40 lbf which means that the damper applies 40lbf when a velocity of 1.4 in/sec is applied. In this case:  $$C_{example} = \frac{40(lbf)}{1.4 \frac{in}{sec}} = 343(\frac{lbf\cdot sec}{ft})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which you multiply by the velocity to get the force:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{damping}(lbf) = V (\frac{ft}{s})*C(\frac{lbf\cdot sec}{ft})$$&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="77" LastEditDate="2017-06-28T19:19:09.553" LastActivityDate="2017-06-28T19:19:09.553" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16070" PostTypeId="2" ParentId="14450" CreationDate="2017-06-28T16:00:23.947" Score="0" Body="&lt;p&gt;First i will give you the solution, then i will highlight some important concepts related to steam cycles and steam tables. So, here goes the answer. Point 3 will be in  compressed liquid region. From compressed water data, we get an approximate value of h&lt;sub&gt;3&lt;/sub&gt; as 160.05 MPa, approximate because pressure,temperature data here is for 15 MPa and 35&lt;sup&gt;o&lt;/sup&gt;C , instead of 15.9 MPa and 38&lt;sup&gt;o&lt;/sup&gt;C given in the problem.You can though through interpolation obtain the exact value.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bJchW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bJchW.jpg&quot; alt=&quot;Compressed water data&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you should know, that when you have information about any two intensive variables such as pressure, temperature or pressure, specific entropy etc, then you can find out all the other information of water such as temperature, pressure, specific enthalpy, specific internal energy, specific volume etc from steam table(assuming the system can be approximated as simple compressible system), which is the case for most problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can find h&lt;sub&gt;3&lt;/sub&gt; by another way, but that would be long and unnecessary. For e.x h&lt;sub&gt;3&lt;/sub&gt; will be equal to the enthalpy at state 1 plus the enthalpy increase of water due to pump work. Again for state 1, we have two variables given, therefore we can find its value from steam table data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, regarding your question on T-S diagram. The T-S diagram is complementary, but not necessary for the solution. As i said before, since you have two variables given at point 3, therefore that is sufficient for you. However, you need to know in which table to find this information i.e compressed water, or saturated water or super-heated steam data. From common sense you can say that the water will either be in the saturated or compressed water state. However, since for saturated region temperature and pressure are dependent properties, therefore you would have needed only one of the two to solve the problem. Because both temperature and pressure have been given to you in the problem, therefore it is likely that water is not in the saturated state. &lt;/p&gt;&#xA;" OwnerUserId="9537" LastActivityDate="2017-06-28T16:00:23.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16071" PostTypeId="2" ParentId="16041" CreationDate="2017-06-28T17:33:34.620" Score="0" Body="&lt;p&gt;Starting with thrust, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nautilus.fis.uc.pt/personal/pvalberto/aulas/cef_mestrado/Air.Resistance.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://nautilus.fis.uc.pt/personal/pvalberto/aulas/cef_mestrado/Air.Resistance.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T_{upper} = P_{shaft} * \frac{1}{v_o}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming propeller efficiency is 100% (worst case, in the context of this question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the relation of thrust to propeller wash velocity with respect to the turbo prop in question is described by the following equation (ignoring exhaust): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T_{upper} = Force = \dot{m}(v_w - v_0)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.grc.nasa.gov/WWW/K-12/airplane/turbprp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.grc.nasa.gov/WWW/K-12/airplane/turbprp.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{m}$ is only as large as the air which the airplane can reach.  Assuming that the velocity remains constant (for a worst case analysis) at whichever thrust level is chosen (presumably, max thrust): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{m} = v_o \rho_{air} A_{prop} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, solving for $v_w$, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;v_w = \frac{F}{\dot{m}} + v_o &#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="11577" LastActivityDate="2017-06-28T17:33:34.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16072" PostTypeId="1" CreationDate="2017-06-28T17:01:48.143" Score="0" ViewCount="322" Body="&lt;p&gt;When pulling a cable to the sides at large angles it has a tendency to come off a caster wheel pulley, how could you prevent it from sliding off the pulley?&lt;/p&gt;&#xA;" OwnerUserId="11969" OwnerDisplayName="Arthur Mamou-Mani" LastEditorUserId="7083" LastEditDate="2017-06-29T11:51:26.197" LastActivityDate="2017-09-28T13:39:18.657" Title="How to prevent a cable going off a caster wheel pulley?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16074" PostTypeId="1" AcceptedAnswerId="16079" CreationDate="2017-06-28T18:51:36.607" Score="4" ViewCount="8445" Body="&lt;p&gt;I am specifying pipes for a water system with 500 ppm chloride in solution.  Ordinarily we would specify 316L stainless steel to limit the corrosion, but I was asked if we could instead specify carbon steel pipe with a corrosion allowance.  So instead of the typical schedule 40 pipe, could I realistically use schedule 80 or 120 etc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My gut reaction is that you will never really even have a &quot;protective&quot; oxide layer with water continually flowing, so you will just continually corrode the pipes out.  All that will be gained with thicker pipe is that it will last longer before failure, but you will have increased corrosion products to remove from the flow stream over time.&#xA;I could not really find any good resources talking about chloride effects on carbon steel pipe since the general consensus is that stainless should be used for this application.&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2019-03-22T12:34:51.113" Title="Chloride corrosion of carbon steel pipe" Tags="|corrosion|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16075" PostTypeId="2" ParentId="16067" CreationDate="2017-06-28T20:50:01.793" Score="4" Body="&lt;p&gt;There are far too many factors involved to definitively state that a table fan cools more than a ceiling fan or vice versa.&#xA;For starters, there are different styles of both.  It also depends highly on what you consider &quot;cools better&quot;.  For example, if you were to come inside and are sweaty, and are looking to cool down quickly, standing in front of a small fan will largely work better because the air volume being moved can be concentrated directly on you.  It will, therefore, provide a more efficient means of removing latent heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the confusion here is actually regarding what a fan does.  In general, a fan is a fan, so a ceiling fan and table fan that are designed to move the same amount of air will perform equally well.  Neither one of them is cooling the air.  While you are correct that air near the ceiling will possibly be warmer and &quot;more stale&quot; this is not due to the performance of the fan.  If you have a supply outlet for an HVAC system right above a ceiling fan, for instance, you will now be blowing cold air down from that outlet (I'm not saying that is the correct placement of an air outlet, just using an example).  Open air fans like you are talking about are really designed for mixing air more than anything.  All of the heat sources in a closed room will seek to equalize when the air is mixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the last part of your question regarding aerodynamics, I'm not really sure what you mean.  There are different blade styles available for both fan types, but I'm not sure what it is you're actually trying to figure out here since you have also tagged the question with &quot;experimental physics&quot;.&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2017-06-28T20:50:01.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16076" PostTypeId="1" AcceptedAnswerId="16077" CreationDate="2017-06-28T22:13:56.503" Score="0" ViewCount="754" Body="&lt;p&gt;I'm planning to build a machine which at some point tows a heavy weight on a rope in the air. Once the weight is heigh enough up in the air I want to keep it there for about an hour or so, before I want to put it down again in a controlled manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, I can brake a DC motor, but I think this takes energy so I would like to avoid doing that for as long as an hour. I guess there should be other options, such as putting a lock of some sort on the rope, but I have no clue how something like that would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone give me some tips on what a usual/best way of blocking a rope for a long time is?&lt;/p&gt;&#xA;" OwnerUserId="11975" OwnerDisplayName="kramer65" LastEditorUserId="11975" LastEditDate="2017-06-30T14:10:00.557" LastActivityDate="2017-08-01T01:10:31.633" Title="How to stop DC-motor with heavy weight from moving?" Tags="|electrical-engineering|motors|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16077" PostTypeId="2" ParentId="16076" CreationDate="2017-06-28T22:29:42.533" Score="3" Body="&lt;p&gt;You'll need a separate braking mechanism on the motor shaft, or somewhere in the gearing between it and the rope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get brakes that are normally engaged, and release on the application of power, which you would do at the same time that you energize the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also use something like a worm drive between the motor and the rope, which is normally self-locking &amp;mdash; the motor can easily move the rope, but the rope cannot back-drive the motor.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2017-06-28T22:29:42.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16078" PostTypeId="2" ParentId="16076" CreationDate="2017-06-28T23:00:37.587" Score="2" Body="&lt;p&gt;The normal solution to this is called an &lt;strong&gt;Electric Winch&lt;/strong&gt; or gear reduction motor with wired remote control and mounting bracket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any requirement there mut be specs or &quot;requirements&quot; which usualy stem from ; budget, time frame, skill (make or buy) then power, acceleration , top speed, mass, which are all related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus every problem is easier to solved when you start by defining needs, or specs, without which, it becomes a &quot;witch hunt&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason a Winch is preferred is because it has the mechanical advantages of transforming power by trading off speed for force and then this ratio with friction in the gears also provides a natural brake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are low cost and must be driven by a car battery with a trickle charger or equiv.  A $100 unit weights 5kg and can lift up a weight of 500kg or more (750?1000?)  I got mine on sale at half price.  an you shop for DC gear motors  DPDT 100A contactor switch  and make your own fixtures but its better to buy one.   You can imagine it runs 100x slower on RPM due to gear ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Life all about tradeoffs and specs for Engineers. What's yours?&lt;/p&gt;&#xA;" OwnerUserId="8847" LastEditorUserId="8847" LastEditDate="2017-06-28T23:05:58.137" LastActivityDate="2017-06-28T23:05:58.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16079" PostTypeId="2" ParentId="16074" CreationDate="2017-06-29T00:03:19.317" Score="3" Body="&lt;p&gt;Corrosion can usually, but not always, be modeled as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rate_equation#Zero_order_reactions&quot; rel=&quot;nofollow noreferrer&quot;&gt;0-th order reaction&lt;/a&gt; which means linear with time.  &lt;a href=&quot;http://hghouston.com/resources/technical-faq/corrosion/calulating-a-corrosion-allowance-for-asme-b313-pi&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASME B31 piping code&lt;/a&gt; treads lightly into this, mainly discussing factors to be used when specifying a corrosion allowance, and no more.  However, we really just need a rate of reaction and some basic principals, and we can derive whatever we want!  &lt;a href=&quot;https://chemistry.stackexchange.com/questions/22043/does-concentration-of-salt-increase-or-decrease-rate-of-rusting&quot; title=&quot;Chemistry Stack Exchange&quot;&gt;Chemistry Stack Exchange&lt;/a&gt; and a &lt;a href=&quot;http://cssf.usc.edu/History/2009/Projects/S0516.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;California Science Fair Project&lt;/a&gt; are used here.  We basically see that the reaction for rusting is: $4Fe +3O_2 +6H_2O \rightarrow 4Fe(OH)_3$, with an overabundance of iron and water in a pipe, the limiting factor is oxygen, and how conductive the solution is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While oxygen is the limiting factor, concentration of dissolved oxygen can heavily impact corrosion because it would be cubed in the reaction rate equation, if not for the fact that the entire inside surface of the pipe is covered in iron, which typically dominates the reactions.  The only reason the reaction doesn't occur faster is because there is such a small concentration of dissolved oxygen, and the multiple reaction steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The science test showed that about &lt;em&gt;0.001&quot;&lt;/em&gt; of steel was lost &lt;em&gt;every year&lt;/em&gt; for salt water solutions (1 mpy).  Running your own tests with the water you will be treating would be the only way to know for certain.  But, as a basis for design, it can be seen that for a &lt;strong&gt;50 year lifetime&lt;/strong&gt; the pipe would need to be $50 * 0.001{in} = 0.05{in}$ thicker for a corrosion allowance.  Such a large corrosion allowance is typically practical, but in the same article water with process contaminated was found to be 10 times more aggressive. Hence why 316 SST is typically used until the veracity of the attack can be determined.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-05-15T21:56:10.887" LastActivityDate="2018-05-15T21:56:10.887" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="16080" PostTypeId="2" ParentId="16074" CreationDate="2017-06-29T02:19:42.067" Score="0" Body="&lt;p&gt;Oxygen and pH will generally be the most important corrosion factors in aqueous  corrosion. If this is a potable water as described ,there is no reason to consider anything but carbon or galvanized steel. If the Cl is present as HCl you have a problem that 316 won't solve ; If you you have temperatures above 150F ,you could have stress corrosion cracking of 316 ( although unlikely below 200 F ). &#xA;If you have experience with similar conditions , that would be the most useful. Otherwise figure on 5 mpy ( 0.005&quot; ) as a worst case. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-06-29T02:19:42.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16081" PostTypeId="1" CreationDate="2017-06-29T02:20:19.560" Score="0" ViewCount="178" Body="&lt;p&gt;Which thermodynamic property does the PdV for a free expansion define if work is zero? Considering there is a finite change in PdV.&lt;/p&gt;&#xA;" OwnerUserId="11974" LastActivityDate="2018-07-18T16:13:49.063" Title="Chemical engineering thermodynamics" Tags="|thermodynamics|chemical-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16082" PostTypeId="2" ParentId="16081" CreationDate="2017-06-29T04:03:44.070" Score="1" Body="&lt;p&gt;I don't think Pdv &lt;strong&gt;on its own&lt;/strong&gt; represents any meaningful quantity in free expansion, as you rightly pointed out that, work done in free expansion is zero, since there is no external pressure to work against. Thus pressure of the gas remains same*(see edit), even though the volume changes. Now, consider the set-up, where you have a adiabatic container, having a removable partition, with one side containing ideal gas(for simplicity) and the other side being vacuum. If we now, remove the partition, and allow the gas to fill the container and come into equilibrium, then at this stage entropy has increased. This increase in entropy is given by, &#xA;Integral of Pdv/T, between the limits Vi &amp;amp; Vf. Here Temperature T, will also remain constant, since the container is adiabatic and W=0.&lt;a href=&quot;https://goo.gl/photos/cWeM7d3FNDWBGvPDA&quot; rel=&quot;nofollow noreferrer&quot;&gt;See the link for mathematical form&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increase in entropy takes place because of the &lt;strong&gt;finite pressure difference&lt;/strong&gt; that exists between the two partitions, which is an &lt;strong&gt;irreversibility&lt;/strong&gt; and hence leads to entropy change. &#xA;* Edit: The pressure doesn't remain constant in free expansion, it changes. Consider the same  set up i have described above, since temperature is constant and amount of substance remains unchanged, therefore according to boyle's law, as volume increases pressure decreases.&lt;/p&gt;&#xA;" OwnerUserId="9537" LastEditorUserId="9537" LastEditDate="2018-07-18T16:13:49.063" LastActivityDate="2018-07-18T16:13:49.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16083" PostTypeId="2" ParentId="15587" CreationDate="2017-06-29T08:06:02.717" Score="1" Body="&lt;p&gt;For any shape, if you know the two moments, $I_{xx}$ and $I_{yy}$, there is a fascinating amount of study to arrive at the principal moments.  You'd need to know $I_{xy}$ as well, which is harder to find tables on.  Fortunately, an angle simply can be broken into two rectangles.  Rectangles by definition have 0 for their product of inertia.  So, the product of intertia for the angle is simply the area of the rectangles times the distance between the centroid of each rectangle in both x and y directions:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9P8PR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9P8PR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{xy} = A_1d_{x_1}d_{y_1} + A_2d_{x_2}d_{y_2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that dimensions are, unlike we're used to, relative vectors - as such for the first rectangle, $d_{y_1}$ is positive, but $d_{x_1}$ is negative.  For rectangle 2, the reverse is true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this, then the principal moments are simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{I_{xx}+I_{yy} \pm \sqrt{I_{xx}^2+I_{yy}^2+4I_{xy}^2-2I_{xx}I_{yy}}}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle can be found as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = \frac{1}{2}\arctan{\frac{2I_{xy}}{I_{yy}-I_{xx}}}$$&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-06-29T08:06:02.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16084" PostTypeId="1" AcceptedAnswerId="16088" CreationDate="2017-06-29T12:10:48.270" Score="1" ViewCount="270" Body="&lt;p&gt;This question emerges from Earth Science SE since recently there was asked &lt;a href=&quot;https://earthscience.stackexchange.com/questions/10695/why-are-co2-sensors-so-expensive-when-co-sensors-arent&quot;&gt;this&lt;/a&gt;. There so far no one had the engineering/ measurement device expertise to answer this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that these are fundamentally different in the case that the purposes of CO2 analyzer is to accurately monitor the levels of CO2 in the atmosphere probably in making a difference between a few parts per million (ppm) in air (as there is currently around 400ppm in the atmosphere and it is a well mixed gas in the atmosphere). On the other hand CO sensors probably need to be able to tell when the concentration of CO exceeds some predetermined threshold where it is safe for humans to breathe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to hear the answer on why and what causes the large price difference. My guesses are: complexity of manufacturing of CO2 analyzer, parts or the large scale production of CO sensors. It is probably the sum of many but from an expert on this or an enlightened opinion I would like to hear from this more accurately.&lt;/p&gt;&#xA;" OwnerUserId="11979" LastActivityDate="2017-06-29T15:06:54.403" Title="Why are CO2 analyzers much more expensive that CO detectors?" Tags="|measurements|product-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16087" PostTypeId="1" AcceptedAnswerId="16105" CreationDate="2017-06-29T13:15:15.710" Score="1" ViewCount="91" Body="&lt;p&gt;When dealing with powders, metal or otherwise, it is often stated weather the powder grains are spherical or flakes. Is it the grinding process that determines the shape of the powder, or is the shape obtained through some other means?&lt;/p&gt;&#xA;" OwnerUserId="4726" LastActivityDate="2017-06-30T16:21:43.770" Title="Particle shape of powders" Tags="|materials|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16088" PostTypeId="2" ParentId="16084" CreationDate="2017-06-29T15:06:54.403" Score="2" Body="&lt;p&gt;Part of the answer is that CO is chemically reactive, while CO2 is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Cheap and simple&quot; CO sensors either mimic the reaction of CO with the oxygen-carrying chemicals in the bloodstream and detect the color change during the reaction, or they contain a simple fuel cell which combusts the CO to CO2 and measures the electrical current produced. The fuel cell method is now the most commonly used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CO2 sensors measure the absorption of infra-red light by the CO2 (i.e. they use the same physics which causes the greenhouse effect). This is an accurate method, but not cheap technology. There are cheaper sensors available which use chemical reactions, but the calibration and accuracy is not very stable and they have a relatively short life (e.g. 2 years).&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-06-29T15:06:54.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16089" PostTypeId="2" ParentId="16087" CreationDate="2017-06-29T15:09:37.187" Score="0" Body="&lt;p&gt;The manufacturing process determines the shape. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-06-29T15:09:37.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16090" PostTypeId="2" ParentId="16018" CreationDate="2017-06-29T15:14:22.453" Score="3" Body="&lt;p&gt;A linear state space model is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\mathbf{x}}=\mathbf{A}\mathbf{x}+\mathbf{B}\mathbf{u},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which $\mathbf{x}=[x_1,x_2,\ldots,x_n]^T$. The reason why we write it in matrix notation is the compactness. Actually, this system is equivalent to the following system of scalar equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x}_1 = a_{11}x_1+a_{12}x_2+\ldots+a_{1n}x_n+b_{11}u_1+b_{12}u_2+\ldots+b_{1m}u_m$$&#xA;$$\dot{x}_2 = a_{21}x_1+a_{22}x_2+\ldots+a_{2n}x_n+b_{21}u_1+b_{22}u_2+\ldots+b_{2m}u_m$$&#xA;$$\vdots$$&#xA;$$\dot{x}_n = a_{n1}x_1+a_{n2}x_2+\ldots+a_{nn}x_n+b_{n1}u_1+b_{n2}u_2+\ldots+b_{nm}u_m.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that you can rearrange the order of the equations and get a new matrix representation. Note, that you can only change the order of the whole equation, so you cannot change the order in $\mathbf{A}$ without changing the order in $\mathbf{x}$, $\mathbf{B}$ and $\mathbf{u}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, there are no standards how one should assign the state variables, as it has no effect on the mathematics as long as the equations are correctly represented by the system. It is essential that you note your order of the state variables in your work / MATLAB / Python / Maple file. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also try to choose the order of the equations in such a way, that the matrix $\mathbf{A}$ has a nice structure. What this actually means is difficult to describe, but a good starting point is to first write down the simple equations and then follow up with the longer equations. Another philosophy which is used when you are transforming scalar equations into state space representation. Then you would first start with the state space representation of the first equation, then do the same for the second equations, and so forth. The advantage is that you will then be able to quickly identify to which scalar equation your lines in the state space represenation belongs. &lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-06-29T15:14:22.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16091" PostTypeId="2" ParentId="15860" CreationDate="2017-06-29T15:40:14.383" Score="4" Body="&lt;p&gt;As you might already know your system is nonlinear, which means that it is not trivial at all. See below for the plot of the system. The linearization around $(0,0)$ gives you the information that the eigenvalues have both negative real parts, which guarantees asymptotic stability in the sense of Lyapunov for the origin. From the plot, one can conjuncture a half-stable limit cycle. Which would imply that trivial equilibrium solution is stable inside the half-stable limit cycle. I doubt that you will be able to get the exact equation for the limit cylce. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to estimate the region of asymptotic stability, you could try to come up with a Lyapunov function, which is also not trivial. Alternative methods would be to use the index theory / Bendixson–Dulac theorem / Poincaré–Bendixson theorem (for the existence of the limit cycle), Method of harmonic balance / Poincaré-Lindstedt method / Multiple Scales to find an approximate equation to the limit cycle. Multiple Scales also allows you to investigate the stability of the limt cycle.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fXHeP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fXHeP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-06-29T15:53:00.010" LastActivityDate="2017-06-29T15:53:00.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16092" PostTypeId="1" CreationDate="2017-06-29T15:42:30.047" Score="2" ViewCount="82" Body="&lt;p&gt;I am designing a part that requires multiple air channels with an intricate structure inside. I would like to know if there is any state of the art injection molding technique or other technology that will enable that solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amount of air channels: 14.&#xA;Volume of the design: 40x40x40mm.&#xA;Distribution: curved and some channels are connected with others, others not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can not show the design, I signed an NDA agreement.&lt;/p&gt;&#xA;" OwnerUserId="1262" LastActivityDate="2017-06-29T20:11:26.387" Title="How to manufacture parts with complex air channels inside?" Tags="|manufacturing-engineering|air|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16093" PostTypeId="2" ParentId="16092" CreationDate="2017-06-29T16:08:35.773" Score="0" Body="&lt;p&gt;If you use 3d printing you can get any internal passages you want. But that could be expensive if you need to make a lot of them. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-06-29T16:08:35.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16096" PostTypeId="2" ParentId="16092" CreationDate="2017-06-29T17:42:54.110" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lost-foam_casting&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lost Foam Casting&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can produce many smaller structures&lt;/li&gt;&#xA;&lt;li&gt;Can be scaled up for reasonable costs&lt;/li&gt;&#xA;&lt;li&gt;Relatively cheap (compared to many other castings types and 3D printing).&lt;/li&gt;&#xA;&lt;li&gt;Can produce items with channels, which would normally require cores to be inserted.&lt;/li&gt;&#xA;&lt;li&gt;Foam is very easy to form and shape&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11631" LastActivityDate="2017-06-29T17:42:54.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16097" PostTypeId="1" AcceptedAnswerId="16098" CreationDate="2017-06-29T17:59:42.383" Score="1" ViewCount="1491" Body="&lt;p&gt;In my lab we have dressing tools for sharpening grinding wheels. The tools are diamonds mounted on a steel shank and the shank fits into a custom holder with a set screw in our machine (pictured).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ud22v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ud22v.jpg&quot; alt=&quot;Dressing tool holder with set screw&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hole in the tool holder is rather large so that it can fit a variety of tool shank diameters. We have a tool with a shank diameter much smaller than the hole diameter, so it has this metal ring wrapped around it to increase its diameter so that it fits tightly in the holder hole. The ring has a gap on one side so that it can be tightened on the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QMwC4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QMwC4.jpg&quot; alt=&quot;Metal part to increase the diameter of the shaft, bottom view&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iqIzL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iqIzL.jpg&quot; alt=&quot;Metal part to increase the diameter of the shaft, side view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;What do you call this piece of metal used to increase the diameter of a small shaft?&lt;/strong&gt; My first thought was to call it a &quot;split collar&quot; or a &quot;split ring&quot; but that doesn't yield anything similar on a Google search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I am wondering is that we have other dressing tools that are too small to fit in this holder and I would like to buy more of these metal pieces so that we don't need to keep re-using the one we have. But I can't find it if I don't know what it's called.&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2017-06-30T13:11:41.653" Title="Part to &quot;increase&quot; the diameter of a small shaft, what is is called?" Tags="|machining|manufacturing-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16098" PostTypeId="2" ParentId="16097" CreationDate="2017-06-29T19:01:42.497" Score="2" Body="&lt;p&gt;In this case, as it is split allow it to be clamped on to fit I would call it a &lt;a href=&quot;https://en.wikipedia.org/wiki/Collet&quot; rel=&quot;nofollow noreferrer&quot;&gt;collet&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are fairly widely used for tool and work holding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any machine shop should be able to make some up for you to the required dimensions pretty easily. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-06-29T19:01:42.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16099" PostTypeId="1" CreationDate="2017-06-29T19:20:13.387" Score="2" ViewCount="474" Body="&lt;p&gt;I have a set of shafts that need to be rotated at a high frequency (~1000 RPM) but not at a constant speed. That is to say, &lt;strong&gt;the angular velocity of the shaft needs to be a periodic function of the shaft's angular position.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This bizarre predicament has arisen in the context of operating an unsteady, transonic wind tunnel. We have a set of vanes whose angular positions determine the Mach number of the flow through the tunnel's test section. For the research we hope to conduct, the test section Mach number will have to oscillate in a specific pattern as a function of time. Technically, there are several ways we could modify the tunnel's design to accomplish this. We are looking into all of them, but in theory, having a motor whose angular velocity is precisely controlled as a function of its angular position would be the most ideal solution to our problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm no expert in electrical engineering. I understand that some motors, like servomotors, are made to rotate at any desired speed within their operating range, and that their speed can be precisely controlled by a closed-loop feedback system. Does this mean that I might be able to program a servomotor to rotate in such a way that its angular velocity is constantly changing as it rotates, even at average rotation speeds of up to 1000 RPM?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, what type of equipment might I need to do this? Is this what a Variable-Frequency Drive (VDF) is for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a servomotor is not appropriate for this task, what type of motor might I need?&lt;/p&gt;&#xA;" OwnerUserId="11802" LastActivityDate="2017-06-29T19:20:13.387" Title="Angular Velocity of motor shaft as function of Angular Position" Tags="|electrical-engineering|control-engineering|motors|servo|wind-tunnels|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16100" PostTypeId="2" ParentId="16092" CreationDate="2017-06-29T20:11:26.387" Score="3" Body="&lt;p&gt;If you are opposed to 3D printing due to the requirement for mass production, then you may be able to work with an injection molding company to design a product that at least has the curved channels done.  Post molding, you can go back out and drill out the straight runs as needed.  If you need to connect two channels, you can cross drill and plug the holes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Designing complex, custom molds is an iterative process that needs to be worked with an injection molding shop.  If you go into that conversation with them and have the expectation that you can get 80-90% of the channels and do the rest after the fact, you are still ahead of the curve compared to waiting on 3D printed models.&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2017-06-29T20:11:26.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16101" PostTypeId="1" CreationDate="2017-06-29T20:59:07.910" Score="0" ViewCount="567" Body="&lt;p&gt;I'm wondering how to interpret artifacts on spectrograms like &lt;a href=&quot;https://i.stack.imgur.com/Z1TVG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;the following&lt;img src=&quot;https://i.stack.imgur.com/Z1TVG.jpg&quot; alt=&quot;attached spectrogram&quot;&gt;&lt;/a&gt; which shows the recording of a linear frequency sweep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The recording is done with a cheap microphone and cheap sound card, and playback was done using an off-the-shelf system, which could have its own issues.&lt;/li&gt;&#xA;&lt;li&gt;There are obviously harmonics which can be seen at the beginning of the chirp, but I can't remember where these can come from, then also those &quot;X&quot; patterns that are catching my attention.&lt;/li&gt;&#xA;&lt;li&gt;There are some fixed-frequency perturbations, some constant, some pulsed and periodic&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm looking for hints, ideas to perform root cause analysis, pointers to common and less common phenomena causing unwanted artifacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Collected hints so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Playback issue: speaker saturation or sound clipping, which can cause harmonics, and can be witnessed using another recording device.&lt;/li&gt;&#xA;&lt;li&gt;Playback issue: clock desynchronization can cause waveform skips which cause very short pulses (of one sample).&lt;/li&gt;&#xA;&lt;li&gt;Short pulses (not on the attached spectrogram, they're short vertical lines) can be caused by power line transient perturbations.&lt;/li&gt;&#xA;&lt;li&gt;Missing anti-aliasing filter can cause capture of aliases of non-audible signals.&lt;/li&gt;&#xA;&lt;li&gt;Out-of-band noise caused by sigma-delta ADCs (&lt;a href=&quot;http://www.ti.com/lit/an/slaa313/slaa313.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref&lt;/a&gt;) which can also be filtered by low-pass filter.&lt;/li&gt;&#xA;&lt;li&gt;Audion noise caused by capacitors in power circuits, eg. &lt;a href=&quot;https://electronics.stackexchange.com/questions/178818/why-does-my-lcd-screen-emit-sound-when-displaying-black-and-white-lines&quot;&gt;in LCD panel amplifiers&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11963" LastEditorUserId="11963" LastEditDate="2017-07-14T21:18:21.130" LastActivityDate="2018-03-12T02:17:33.847" Title="Audio frequency sweep spectrogram artifacts... hints?" Tags="|consumer-electronics|acoustics|audio-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16102" PostTypeId="2" ParentId="16101" CreationDate="2017-06-30T00:01:05.847" Score="1" Body="&lt;p&gt;The &quot;X patterns&quot; look like anti-aliasing. If you project the first downward-sloping line of the X back upwards, it will probably intersect one of the &quot;harmonics&quot; lines at the Nyquist frequency corresponding to your sampling rate. For example If you are sampling at 48k samples/sec, your FFT will show a signal at $(24+x)$ kHz as if its frequency was $(24-x)$ kHz. On the frequency response plot, this looks the same as if the horizontal line at 24 kHz acts like a mirror, reflecting anything that should have been plotted above it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to get rid of them is to use an &lt;em&gt;analog&lt;/em&gt; filter &lt;em&gt;before&lt;/em&gt; you digitize the signal. But if you are using &quot;consumer level&quot; equipment, just ignore them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to tell guess if the &quot;harmonics&quot; are something you actually measured, or an artifact of your digital signal processing. If they are low amplitude, and you don't have access to any better analysis tools, I would be inclined to just ignore them. I'm assuming you don't have any reliable calibration curves for your sound source, microphone, ADC, etc - a few numbers and/or a little graph pulled out of the marketing material don't count as &quot;reliable&quot; IMO. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheap sound cards often are a long way from having linear response in either amplitude or phase at high frequencies - if most people use them for listening to low-bit-rate MP3 files, there's no meaningful high frequency audio content anyway, only random noise!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant-level noise lines at 7.5, 9.5 and about 18.5 KHz are presumably coming from somewhere &lt;em&gt;after&lt;/em&gt; the microphone input. Again, if you can't find out exactly where they come from (for instance by cutting the signal path to see when they go away) you don't have much alternative but to ignore them. Don't forget that those three frequencies shown on the plot might also be aliases of higher frequency signals!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; Note that a chirp signal is never a &quot;perfect&quot; sine wave. If you look at one cycle of the signal, it is distorted slightly because the frequency is changing. The amount of distortion depends on the &lt;em&gt;ratio&lt;/em&gt; of the sweep frequency to the chirp frequency. This might explain why you see &quot;harmonics&quot; at the start (low frequencies) of the chirp but then they disappear. Try changing the sweep frequency and see if the amplitude of the &quot;harmonics&quot; changes.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-06-30T01:15:52.927" LastActivityDate="2017-06-30T01:15:52.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16103" PostTypeId="1" AcceptedAnswerId="16114" CreationDate="2017-06-30T06:03:35.980" Score="2" ViewCount="23053" Body="&lt;p&gt;I'm struggling to calculate with any confidence the allowable shear force for a metric bolt holding two aluminium pieces together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only need an approximate figure, since I will allow a very large margin of safety. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the pieces being joined are sufficiently strong. Assume the bolt is threaded throughout its length. It is acceptable for the thread to be damaged, but no permanent deformation to the bolt core. Assume a single bolt holds the two pieces such that the pieces &lt;em&gt;can&lt;/em&gt; freely rotate, but that they won't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two pieces being joined are are 5mm and 3mm thick. The bolt is M8 with 1mm thread. Assume the bolt is made of a weaker steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the value of the shear force for the above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, the most progress I have made is by using &lt;a href=&quot;http://www.amesweb.info/Screws/Metric_Bolt_Grades_Strength.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;this online calculator&lt;/a&gt;, but I think it's for tension strength. Nevertheless, the &quot;proof load&quot; comes out at 8.82kN for for &quot;property grade&quot; 4.6. I'm guessing the area strength is of similar magnitude.&lt;/p&gt;&#xA;" OwnerUserId="64" LastActivityDate="2017-06-30T14:39:46.587" Title="Maximum shear force of metric steel bolt" Tags="|bolting|shear|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16104" PostTypeId="2" ParentId="8405" CreationDate="2017-06-30T06:39:33.977" Score="1" Body="&lt;p&gt;What you've got there is a maximum boiling azeotrope mixture. It occurs when the negative deviations are very large, and the total pressure curve, in this case, passes through a minimum, giving rise to a maximum in the temperature (i.e. boiling point). Azeotropic mixtures cannot be easily separated by ordinary distillation methods. For example, in the case of ethanol-water, one cannot recover more than 89.4 mole% ethanol using ordinary distillation, as the mixture becomes azeotropic at this point. Other separation techniques (such as azeotropic distillation) must be used. Often the equipment and set-up used are unique for each mixture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest not using the McCabe-Thiele method at all it would yield poor results. Your best bet would be using Ponchon-Saviart or Lewis-Sorell method, they are bit tedious but will definitely yield better results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically though, one would not use a distillation directly, a precursor step would be to use extractive distillation followed by rectification.  Such a problem usually demands a binodal graph. &lt;/p&gt;&#xA;" OwnerDisplayName="user11882" LastActivityDate="2017-06-30T06:39:33.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16105" PostTypeId="2" ParentId="16087" CreationDate="2017-06-30T07:21:27.213" Score="4" Body="&lt;p&gt;It depends partly on the properties of the material and partly on the manufacturing process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;High quality metal powders used for sintering and powder metallurgy for example are often produced by gas atomisation. Here molten metal is sprayed through a fine jet in an inert atmosphere and the droplets formed are allowed to solidify. This process naturally produces spherical beads of reasonably uniform and controllable size. Variations on this process may use a centrifuge or various gas and liquid jets or an electric arc to disperse the metal. You also get a similar effect as a side effect of some arc welding processes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fumed silica, which is widely used as a desiccant, thickening agent and filler for, paints, resins, adhesives etc. is produced by pyrolysis of quartz sand in a flame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brittle materials such as hard metal alloys and many minerals can be pulverised in a ball mill, here the particle size and shape will depend on the mill setup and the fracture characteristics of the material in question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Powders can also be produced by various chemical and electrolytic precipitation processes. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-06-30T16:21:43.770" LastActivityDate="2017-06-30T16:21:43.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16106" PostTypeId="2" ParentId="16103" CreationDate="2017-06-30T07:35:57.863" Score="2" Body="&lt;p&gt;For metric bolts the tensile strength can be calculated from the grade of the bolt expressed as A.B (eg 8.8). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first number is the ultimate tensile stress in 100s of Mega Pascals and the number after the point is the yield stress in tenths of the UTS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example an 8.8 bolt has a UTS of 800 MPa and the yield stress is 8/10 of that (80%). Multiply by the cross sectional area to get the corresponding tensile loads. If you stick to those units you get an answer in Newtons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically for ductile metals shear strength will be about 50% of tensile strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the joint is able to slip you also need to consider the stresses on the hole, which in practice may be the limiting factor, especially if yo are using a steel bolt to join aluminium.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-06-30T07:35:57.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16107" PostTypeId="1" CreationDate="2017-06-30T11:16:26.087" Score="2" ViewCount="535" Body="&lt;p&gt;I have to visualize flow over some designs in my wind tunnel. However, my advisor has stated that previous students have wrecked the machine when they have used oils and I think Dry Ice might interfere with the results owing to its different density and the fact that it will probably sink to the bottom of the tunnel (it is the horizontal kind). &lt;/p&gt;&#xA;" OwnerUserId="11987" LastActivityDate="2017-09-29T00:49:29.407" Title="How can I generate smoke (or mist) that leaves no residue?" Tags="|aerodynamics|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16108" PostTypeId="2" ParentId="16107" CreationDate="2017-06-30T11:24:54.853" Score="1" Body="&lt;p&gt;In any normal wind tunnel, the wind speed will completely nullify the effects of thermal/density convection - unless you're testing some ultra-light apparatus that would work with airspeeds of order of 1m/s, just use dry ice - at ~5m/s airspeed the effects of density change and sinking rate will already be below threshold of other errors, and normally wind tunnel tests should involve airspeeds much higher.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-30T11:24:54.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16109" PostTypeId="2" ParentId="15601" CreationDate="2017-06-30T11:35:39.687" Score="1" Body="&lt;p&gt;If you pump into the same pipe, then yes, you will need to worry about these. You'd need check valves or such, and the performance of the pair would be lower than just sum of the two performances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the easy solution is to draw two separate pipes to the balance tank - make the tank be the point where the two flows meet. That way one flow won't interfere with the other, until the tank overflows. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-30T11:35:39.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16110" PostTypeId="2" ParentId="16072" CreationDate="2017-06-30T12:05:47.597" Score="1" Body="&lt;p&gt;Simple, with guards like these:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/j4E99.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j4E99.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vs8tg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vs8tg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-06-30T12:05:47.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16111" PostTypeId="2" ParentId="16097" CreationDate="2017-06-30T13:11:41.653" Score="1" Body="&lt;p&gt;This could be called a &lt;a href=&quot;http://www.thefreedictionary.com/Bush+(mechanical)&quot; rel=&quot;nofollow noreferrer&quot;&gt;bush&lt;/a&gt; but as no rotary motion it's probably more correct to call it a sleeve.&lt;/p&gt;&#xA;" OwnerUserId="10415" LastActivityDate="2017-06-30T13:11:41.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16113" PostTypeId="1" CreationDate="2017-06-30T14:21:09.473" Score="0" ViewCount="69" Body="&lt;p&gt;Consider a negative feedback loop with input $r(t)$ and output $x(t)$. The difference&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ e(t) = r(t) - x(t) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the input to a controller $C(s)$ and the output is $u(t)$. Then, $u(t)$ is the input to $G(s)$ and the output is $x(t)$. The transfer function for this system is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H(s) = \frac{G(s)C(s)}{1 + G(s)C(s)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that $H(s)$ has the behavior of a forced-damped system. Given a $G(s)$, $C(s)$ is chosen so that $H(s)$ will reflect a forced-damped system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $C(s)$ is always chosen to give the intended behavior for $H(s)$, is there any reason that one form of $G(s)$ may be better than another? For example, consider $G(s)$ &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G_{LP}(s) = \frac{1}{s}$&lt;/li&gt;&#xA;&lt;li&gt;$G_{HP}(s) = \frac{1}{M s^2}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;where $M$ is a constant. The corresponding $C(s)$ are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;linear low-pass filter: $C_{LP}(s) = \frac{K}{p + s}$&lt;/li&gt;&#xA;&lt;li&gt;linear high-pass filter: $C_{HP}(s) = \frac{Ks}{p + s}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;where $K$ and $p$ are constants. The quantities  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$G_{LP}(s)C_{LP}(s) = \frac{K}{s(p + s)}$&lt;/li&gt;&#xA;&lt;li&gt;$G_{HP}(s)C_{HP}(s) = \frac{K}{Ms(p + s)}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;only differ by a constant $M$, but $K$ can always be tuned for each choice of $C(s)$ so that the appearance of this extra constant $M$ does not matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reiterate, would one choice of $G(s)$ be preferable from a physics/engineering perspective if the behavior of the overall system remains the same? &lt;/p&gt;&#xA;" OwnerUserId="11991" LastActivityDate="2018-02-23T19:38:25.713" Title="Choice of transfer function in a system where the controller adapts to give the intended behavior" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16114" PostTypeId="2" ParentId="16103" CreationDate="2017-06-30T14:39:46.587" Score="2" Body="&lt;p&gt;The &lt;em&gt;Guide to Design Criteria for Bolted and Riveted Joints&lt;/em&gt; 1974 by Fisher and Struik, you can use the approximation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{Ult-Shear}\sim 0.62\cdot S_{Ult}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and from the Distortion Energy Theory: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S_{YieldShear}\sim 0.58\cdot S_{Yield}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear force to yield the bolt is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = S_{YieldShear}\cdot A_{bolt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $$A_{bolt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the cross sectional area of the bolt in the shear plane&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the shear plane is across the threaded portion of the bolt, use the minor diameter for an external M8x1 thread, 6.596mm, to calculate the area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're interested in the force required to break the bolt, simply substitute the Ultimate Shear Strength for Yield Shear Strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all likelihood though, the aluminum will tear long before you get to the point of shearing the bolt.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-06-30T14:39:46.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16115" PostTypeId="2" ParentId="12543" CreationDate="2017-06-30T16:09:39.983" Score="2" Body="&lt;p&gt;As &lt;a href=&quot;https://engineering.stackexchange.com/questions/12543/anomalous-oxygen-in-aluminium-sample-sem-eds#comment22223_12543&quot;&gt;@starrise mentioned&lt;/a&gt;, it is likely either a software error (secondary peak) or your sample could be aluminum reinforced with aluminum oxide. With aluminum there will always be some oxygen (even in vacuum coating we see some) but not that high from what I have seen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is if your accelerating voltage is too low, it might not go to the depth you intend it. Are you within the 15-20&amp;nbsp;kV range?&lt;/p&gt;&#xA;" OwnerUserId="11994" LastEditorUserId="1832" LastEditDate="2017-06-30T20:29:11.093" LastActivityDate="2017-06-30T20:29:11.093" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16116" PostTypeId="1" AcceptedAnswerId="16119" CreationDate="2017-06-30T17:17:31.307" Score="1" ViewCount="1673" Body="&lt;p&gt;I'm going through some vibration theory and I need to represent the equations in state space for vibration control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main forced general equation of motion is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{M{\ddot q}+{\Omega_c}G{\dot q}+{[K_b+K_m-\Omega_c^2K_\Omega]q}=T(t)+F(t)}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$K_b$: bearing stiffness symmetrical matrix&lt;/li&gt;&#xA;&lt;li&gt;$K_m$: mess stiffness matrix&lt;/li&gt;&#xA;&lt;li&gt;$K_\Omega$: diagonal centripetal stiffness matrix&lt;/li&gt;&#xA;&lt;li&gt;$T$: applied external torque&lt;/li&gt;&#xA;&lt;li&gt;$F$: static transmission error excitation&lt;/li&gt;&#xA;&lt;li&gt;$q$: the vector with the traslation and rotational coordinates for the sun,ring and carrier&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If I'm not mistaken, for this equation the input variables would be the forces $F(t)$ and $T(t)$ plus the $q$ terms, which would form the $B$ matrix and $u$ input vector in state space form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the free vibrations I have the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{M{\ddot q}+{[K_b+K_m]q}=0}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the input variables in this case just be equal to zero?  and the output variables just $\mathbf{{\ddot q}} $ ? I'm also trying to code this into Matlab and Maple, which can directly get me the state space form of really long equations by just inputting the equation, input variables and output variables.&lt;/p&gt;&#xA;" OwnerUserId="4088" LastEditorUserId="1832" LastEditDate="2017-06-30T20:30:53.970" LastActivityDate="2017-06-30T22:51:54.753" Title="How do I represent equations of motion of a dynamic system in State Space?" Tags="|mechanical-engineering|control-engineering|dynamics|vibration|eigenvalue-analysis|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16117" PostTypeId="2" ParentId="8110" CreationDate="2017-06-30T18:54:37.593" Score="1" Body="&lt;p&gt;If you can use metalization to create your film, you could create a mask in the required pattern and just coat on top of it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I assume you only have a laser at your disposal. Your heat affected zone will depend on the speed at which you remove material. Higher power will tend to drill through faster and affect a smaller area but it may create more warping as you create a steeper heat gradient. There are two things I can think of trying that worked in other applications:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;use a backing that will spread the heat (aluminum/copper), it may reduce the amount of stress in the material&lt;/li&gt;&#xA;&lt;li&gt;try using different settings (power, travel speed) to reduce the stress in the material. Document your trials and see what works best. (Sorry I can't help further on what settings to use, my application/material was a bit different)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11994" LastEditorUserId="1832" LastEditDate="2017-06-30T20:30:09.473" LastActivityDate="2017-06-30T20:30:09.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16118" PostTypeId="1" CreationDate="2017-06-30T22:13:24.417" Score="6" ViewCount="730" Body="&lt;p&gt;Near my apartment is a building site with a tower crane; great fun to watch of course. This crane seems to have a lift / elevator / vertical-people-transporter (arrowed in the picture).&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hAhNa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hAhNa.jpg&quot; alt=&quot;lift on tower crane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can see the lift is only used four times a day (morning-lunch(2x)-afternoon) which is no doubt very nice for the operator but this can hardly be cost-effective. I have seen the operator climb between the operator's cabin and the lift but never any other movement.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably there is some other reason for the lift, so what is it? Also, why doesn't the lift go higher? Currently the operator has to climb about 20m, I see no reason why the lift couldn't go 10m higher (the next floor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS Tagging was tricky and I don't have the rep to make a &quot;crane&quot; tag. Advice?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In response to comments that it might be a climbing crane, added hi-res photos with lift &lt;a href=&quot;https://i.stack.imgur.com/8dGDF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;down&lt;/a&gt;. There's also a photo with lift up on i.stack.imgur.com/yT6bn.jpg; I don't have the rep to add a link to it.&lt;/p&gt;&#xA;" OwnerUserId="2737" LastEditorUserId="2737" LastEditDate="2017-07-12T21:36:56.753" LastActivityDate="2023-12-19T21:06:00.700" Title="Tower crane with lift (elevator)" Tags="|construction-management|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16119" PostTypeId="2" ParentId="16116" CreationDate="2017-06-30T22:38:30.377" Score="3" Body="&lt;p&gt;Karlo mentioned to use the state space of $[q,\dot{q}]^T$ this allows you to write the system as a first order differential equation of state space. If $X$ were the state space, you could write $\dot{X} = f(X)$. As far as the work you need to do, you write that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;X = \begin{bmatrix} q\\ \dot{q}\end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the derivative of $\frac{dq}{dt}=\dot{q}$ which is part of X and $\frac{d\dot{q}}{dt}=\ddot{q}$ which can be solved for algebraically in the ODE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{dX}{dt} = \begin{bmatrix} \dot{q}\\ KM^{-1}q \end{bmatrix}&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the undamped system. If you intend on using a numerical solver in matlab (maybe &lt;code&gt;ode45&lt;/code&gt;) it would look something like this assuming all your system parameters are defined as matrices with similar names and &lt;code&gt;IC&lt;/code&gt; is the initial condition and &lt;code&gt;tmin&lt;/code&gt; and &lt;code&gt;tmax&lt;/code&gt; are time bounds you're interested in for a solution.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;dxdt = @(x) [x(2); M\K*x(1)];&#xA;[q,t] = ode45(dxdt, [tmin, tmax], IC);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;ode45&lt;/code&gt; specifically expects this form: where the first parameter is a vector that is the derivative of a coordinate/state space vector. Notice we would have the same code if we instead used something that wasn't second order by notation, but the condition that $\dot{s}=t$ makes it second order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{dX}{dt} = \frac{d}{dt}\begin{bmatrix} s \\ t \end{bmatrix} = \begin{bmatrix} t \\ KM^{-1}s \end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this technique, we can write any system that is a Nth order differential equation as a first order equation of an Nth order state space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: After seeing your most recent comment, your state space could be written as:&#xA;$$&#xA;X = \begin{bmatrix} x\\y\\u \\ \dot{x}\\\dot{y}\\\dot{u}\end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I misunderstood what you meant by using matlab and maple. Your inputs are known\controlled values, so &lt;strong&gt;$F(t)$&lt;/strong&gt; and &lt;strong&gt;$T(t)$&lt;/strong&gt; along with the system parameters, like &lt;strong&gt;$K$&lt;/strong&gt;, outputs are system behavior, like $x,y,u$ and their derivatives, what you don't know upon the formulation of the equation, but you don't need maple to construct the equation for you, it's just the ones you find from the algebra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, I somehow didn't notice your controller forces &lt;strong&gt;$T(t)$&lt;/strong&gt; and &lt;strong&gt;$F(t)$&lt;/strong&gt; and didn't write them into the first parameter of the solver, but those can be written in as well, as they are. It would just be that $\ddot{q}=M^{-1}(Kq-T(t)-F(t))$.&lt;/p&gt;&#xA;" OwnerUserId="10020" LastEditorUserId="10020" LastEditDate="2017-06-30T22:51:54.753" LastActivityDate="2017-06-30T22:51:54.753" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16121" PostTypeId="1" CreationDate="2017-07-01T13:31:21.737" Score="0" ViewCount="116" Body="&lt;p&gt;I'm trying to use relay as high current motor driver. &lt;a href=&quot;https://i.stack.imgur.com/IC8EF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IC8EF.jpg&quot; alt=&quot;2 SPDT relay module&quot;&gt;&lt;/a&gt; I'm not sure how to connect flyback diodes. This is my current sketch &lt;a href=&quot;https://i.stack.imgur.com/vfUfx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vfUfx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I've read a lot of on DC motors, but I couldn't find anywhere a schematic for 2 SPDT relay motor driver with flyback diodes protection. Because of SPDT relays I can't short circuit the PSU (S1 + S2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be glad with any advice/guidance.&lt;/p&gt;&#xA;" OwnerUserId="11998" LastActivityDate="2017-07-01T20:50:55.480" Title="2 SPDT relays as motor driver, flyback diodes" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16122" PostTypeId="1" AcceptedAnswerId="16129" CreationDate="2017-07-01T17:03:27.483" Score="2" ViewCount="5049" Body="&lt;p&gt;Can a Prince Rupert's drop be made of metal instead of glass?&lt;/p&gt;&#xA;" OwnerUserId="12002" LastEditorUserId="12002" LastEditDate="2017-07-01T17:38:05.773" LastActivityDate="2021-03-08T19:46:16.490" Title="Is a metal Prince Rupert's drop possible?" Tags="|materials|pressure|stresses|metals|cooling|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16123" PostTypeId="2" ParentId="16121" CreationDate="2017-07-01T20:50:55.480" Score="1" Body="&lt;p&gt;The configuration you have in your sketch looks like a good solution. Another approach is to use 2 zener diodes rated for just over V+, connect them in series in opposing directions, across the motor terminals.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-07-01T20:50:55.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16125" PostTypeId="1" CreationDate="2017-07-01T22:48:02.020" Score="2" ViewCount="39" Body="&lt;p&gt;There exist devices (e.g. Bosch PTD 1) which measure the temperature and humidity/moisture of the room and remotely (without contact) of a wall they are pointed at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that you can measure temperature remotely using basically an infrared camera and Stefan–Boltzmann law (and making assumptions about the emissivity). I also understand that you can measure moisture in a wall by placing electrodes on it and measuring the conductance or dielectric constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering how moisture is measured by these devices &lt;em&gt;remotely&lt;/em&gt; when they are typically about one metre from the wall? The device in question includes a laser, but this might just be an aid to show the area being measured.&lt;/p&gt;&#xA;" OwnerUserId="12005" LastActivityDate="2017-07-01T22:48:02.020" Title="Measuring moisture in walls remotely" Tags="|measurements|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16126" PostTypeId="1" CreationDate="2017-07-02T03:00:36.193" Score="1" ViewCount="1621" Body="&lt;p&gt;I want to create a custom field output in ABAQUS. For the purpose of proof on concept, I want to display the maximum shear stress as calculated from the Mohrs circle, as discussed &lt;a href=&quot;http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/calc_principal_stress.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for a 2D shell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have my code below for reference:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from abaqusConstants import *&#xA;from odbAccess import *&#xA;from math import *&#xA;from copy import deepcopy&#xA;from caeModules import *&#xA;from driverUtils import executeOnCaeStartup&#xA;&#xA;# ******************************************************************************&#xA;#Items in this box require student input when changing files&#xA;&#xA;#must input the file path here.&#xA;odbPath = &quot;/home/MohrsTest.odb&quot;&#xA;&#xA;odb = session.openOdb(name=odbPath, readOnly=FALSE)&#xA;odb = session.odbs[odbPath]&#xA;#this will display the instance names. Please choose one to input in line 14.&#xA;print odb.rootAssembly.instances.keys()&#xA;grout_instance = odb.rootAssembly.instances['SQUARE-1']&#xA;# ******************************************************************************&#xA;keys = odb.steps.keys()&#xA;for key in keys:&#xA;    step = odb.steps[key]&#xA;    for frame in step.frames:&#xA;        print frame.description&#xA;        Stress = frame.fieldOutputs['S']&#xA;        #try modifying scalar fields rather than creating new var element by element.&#xA;        S11=Stress.getScalarField(componentLabel=&quot;S11&quot;)&#xA;        S22=Stress.getScalarField(componentLabel=&quot;S22&quot;)&#xA;        S12=Stress.getScalarField(componentLabel=&quot;S12&quot;)&#xA;        TauMax=((S11+S22)*0.5+sqrt(power((S11-S22)/2, 2)+power(S12, 2)))-((S11+S22)*0.5-sqrt(power((S11-S22)/2, 2)+power(S12, 2)))/2&#xA;        ThetaP=(atan2(2 * S12, (S11 - S22))/2) * 180/pi&#xA;        frame.FieldOutput(name='Tau Max', description='Max Tau from Mohrs circle',field=TauMax)&#xA;        frame.FieldOutput(name='Theta P', description='Thetap as measured ccw from 0 degree',field=Thetap)&#xA;odb.save()&#xA;odb.close()&#xA;#  must re - open the output database to see the new custom field output&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Abaqus throws an error immediately upon trying to calculate TauMax because of &quot;Type Error: A float is required.&quot; However, I tried using &quot;Tools-&gt;Field Outputs-&gt; Create from Fields&quot; and then created a field output in the cae for the individual frame. &#xA;If I look at the replay file for this action, I can see the following code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s1f1_S = session.odbs['/home/MohrsTest.odb'].steps['Step-1'].frames[1].fieldOutputs['S']&#xA;tmpField = (((s1f1_S.getScalarField(componentLabel=&quot;S11&quot;)+\&#xA;    s1f1_S.getScalarField(componentLabel=&quot;S22&quot;))*0.5+sqrt(power((&#xA;    s1f1_S.getScalarField(componentLabel=&quot;S11&quot;)-s1f1_S.getScalarField(&#xA;    componentLabel=&quot;S22&quot;))/2, 2)+power(s1f1_S.getScalarField(&#xA;    componentLabel=&quot;S12&quot;), 2)))-((s1f1_S.getScalarField(componentLabel=&quot;S11&quot;)+\&#xA;    s1f1_S.getScalarField(componentLabel=&quot;S22&quot;))*0.5-sqrt(power((&#xA;    s1f1_S.getScalarField(componentLabel=&quot;S11&quot;)-s1f1_S.getScalarField(&#xA;    componentLabel=&quot;S22&quot;))/2, 2)+power(s1f1_S.getScalarField(&#xA;    componentLabel=&quot;S12&quot;), 2))))/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thus, clearly math operations on a FieldObject must be possible. Why does my code not permit this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am happy to provide all .odb and .cae files for reference and verification. &lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2018-02-02T09:38:18.350" Title="Creating a custom field output in ABAQUS" Tags="|mechanical-engineering|finite-element-method|abaqus|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16127" PostTypeId="1" CreationDate="2017-07-02T05:44:59.493" Score="0" ViewCount="70" Body="&lt;p&gt;Imagine I'm breathing through a wide straw/tube in my mouth. At the far end of the straw is a resistance mechanism. I want it to be that as I exhale there is no resistance, but as soon as I start to inhale I'm met with strong resistance so that little or no air can pass. But over the course of a few seconds, I want that resistance to drop to zero (either to drop smoothly or to &quot;pop&quot; to little or no resistance after a couple seconds). Furthermore, I want the timing of that resistance fall off or &quot;pop&quot; to be independent of how hard I am inhaling -- either a strong or a light inhale will produce the same result: strong resistance dropping (or popping) to little or no resistance over a couple seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to think of a way to do this purely mechanically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three mechanical timing mechanisms come to mind:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the balance wheel of a mechanical watch (which causes ticking at a constant rate regardless of how tightly the watch's mainspring is wound)&lt;/li&gt;&#xA;&lt;li&gt;an hourglass (the concept here is that an inhale of any strength would flip the hourglass, and the number of grains fallen would represent the decreasing air resistance)&lt;/li&gt;&#xA;&lt;li&gt;a hydraulic door-closing mechanism (where the door closes at roughly the same speed regardless of the pressure applied)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of those are just &quot;inspirations&quot; in the sense that they seem like they can be used as the basis for a mechanical timing mechanism, but I don't know how to turn any of them into a concrete design for this purpose. Any suggestions welcome.&lt;/p&gt;&#xA;" OwnerUserId="12007" LastActivityDate="2018-02-27T11:44:17.980" Title="air resistor decreasing resistance over fixed time" Tags="|mechanical-engineering|airflow|timing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16128" PostTypeId="2" ParentId="16127" CreationDate="2017-07-02T07:41:56.070" Score="1" Body="&lt;p&gt;Use a flap valve - allows air out and if you design the &quot;catch&quot; correctly then won't allow air to pass until a certain load or depression is achieved.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-02T07:41:56.070" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16129" PostTypeId="2" ParentId="16122" CreationDate="2017-07-02T10:08:15.113" Score="3" Body="&lt;p&gt;Sort of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is rendered a bit of a moot point as most metals already have the the sort of hardness and impact resistance that heat treating imparts on glass anyway so you are essentially approaching the problem from the opposite direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it is certainly true that heat treating can create residual stresses in susceptible metals in a similar fashion, stainless steel with its high coefficient of thermal expansion and low thermal conductivity is particularly susceptible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also steels with sufficient carbon content can be dfferentially  hardened by quenching so if you quench a thick enough section of high carbon steel you will get a hard surface layer with a softer core. Also the hard section will increase in volume, creating residual stress throughout the whole sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also various processes for surface hardening of steels,  such as case hardening which works by diffusing carbon into the surface of low carbon steels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest real world application is in ball bearings which tend to have hard surfaces and soft cores to combine surface wear resistance with toughness. This isn't &lt;em&gt;exactly&lt;/em&gt; the same as a Prince Rupert's drop as you have a change of micro-structure going on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the complication that molten high carbon steel tends to start to burn in air so making an actual droplet might require an inert atmosphere, equally molten steel is generally a lot less viscous than glass so forming the 'tail' wouldn't work in quite the same way. Having said that you could achieve a similar effect by quenching from solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fully hardened (ie quenched but not tempered) high carbon steels aren't that far away from the brittleness of glass. Indeed it is not unknown for quenched parts to spontaneously crack between quenching and hardening. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-07-02T17:26:41.497" LastActivityDate="2017-07-02T17:26:41.497" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16132" PostTypeId="2" ParentId="15909" CreationDate="2017-07-02T12:20:49.103" Score="2" Body="&lt;p&gt;As suggested a rack and pinion will provide a linear transfer function and avoids complex linkages. A variant on the rack and pinion is to replace the straight rack teeth with a taught cable and the pinion teeth with a pulley. This mechanism is sometimes referred to as a capstan and bowstring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Capstan_and_Bowstring&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Capstan_and_Bowstring&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a guarantee against slippage then the cable can be wrapped multiple times and attached to the pulley at the half-way point.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-07-02T12:20:49.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16133" PostTypeId="1" CreationDate="2017-07-02T17:58:52.633" Score="1" ViewCount="3884" Body="&lt;p&gt;The floor the elevator moves to, is controlled by the lift PLC with all the algorithms developed over the years, and that I understand well. What I am not able to understand is how does the lift stop at the exact landing position at a particular floor, neither too high nor too low, but on the same level as the elevator landing on the floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that with modern Variable Frequency motor drive controllers and sensors this is not an unachievable task, but how did they do this in the olden days of relays? And even in today where the speed of the driving AC motor in many current elevator systems, is controlled through pole-switching? Do they use some special PID algorithm or some variant of that?&lt;/p&gt;&#xA;" OwnerUserId="7057" LastActivityDate="2018-09-25T08:59:40.353" Title="How do elevators stop at the exact landing spot when reaching a particular floor?" Tags="|mechanical-engineering|electrical-engineering|control-engineering|pid-control|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16134" PostTypeId="1" CreationDate="2017-07-02T19:34:21.870" Score="1" ViewCount="96" Body="&lt;p&gt;Enclosed within a consumer electronic, I would like to spin a small balanced load.  The device will be rotated relative to gravity, but no other direct forces will be applied to the load.  A coreless DC motor (e.g. often used for pager vibrators) will be used, but in contrast I would like to minimize noise/vibration.  Here is what a current prototype looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hF7ry.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hF7ry.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I just hang the bare motor by its wire leads and let it spin the balanced load, there is low noise/friction relative to the above assembly.  I believe this is because the above assembly adds rotation resistance via the shaft hole and acts as a 'vibrational conduit' to the motor.  However, because the device will be rotated w.r.t. to gravity, and the load is of moderate mass (2-4g), I think there will need to be some sort of shaft support to prevent radial play.  There are two basic designs I can think of to achieve this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;External shaft hole (as seen in above image)&lt;/li&gt;&#xA;&lt;li&gt;Bi-axial motor (seen below):&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YWSAD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YWSAD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main criteria to decide between the two should be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;low noise/vibration&lt;/li&gt;&#xA;&lt;li&gt;low axial/radial play&lt;/li&gt;&#xA;&lt;li&gt;easy to assemble/difficult to incorrectly assemble&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given these criteria, it seems to me that using a biaxial motor is the way to go.  However, there is one catch: they are not commonly produced, and I can only find one &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B01MRIF4RT&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;.  Furthermore, the only model I could find operates at 24V (vs typical 3.7V), which will require stepping up logic voltages.  So, if using a biaxial motor is indeed preferable for this application, would it be reasonable to expect a manufacturer to make me 'custom' biaxial 3.7V motors of quantity 5-10K?  If such a customization is not feasible for this relatively low order quantity, does it make more sense to develop a shaft support assembly for a uniaxial motor, or work around the non-ideal electrical characteristics of the 24V biaxial motor?&lt;/p&gt;&#xA;" OwnerUserId="12011" LastEditorUserId="12011" LastEditDate="2017-07-03T22:18:54.457" LastActivityDate="2017-07-03T22:18:54.457" Title="Spinning a balanced load with DC motor" Tags="|mechanical-engineering|motors|manufacturing-engineering|friction|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16135" PostTypeId="1" AcceptedAnswerId="16154" CreationDate="2017-07-02T22:15:55.710" Score="1" ViewCount="53" Body="&lt;p&gt;Going through some theory about epicyclic gears I came accross a comment that says that the spin velocity of the gear irrelevant if perfectly balanced, I don't understand why this might be true, is it because it would be constant and thus can be measured without variations?&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2017-07-04T12:35:58.607" Title="Why is the spin velocity of a gear irrelevant if perfectly balanced?" Tags="|mechanical-engineering|gears|dynamics|vibration|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16136" PostTypeId="1" CreationDate="2017-07-03T00:14:57.853" Score="1" ViewCount="10801" Body="&lt;p&gt;I am very new to gears I am fairly confused about planetary gears. I want to know the load on the gears contact point between the sun and the planets and the planets and the ring, assuming the ring not moving. Given I have the solid works file ,the pressure angle, and the gear ratios, I can easily calculate the distances, teeth numbers, etc. I also know the initial driving torque of the sun gear. &#xA;So, my question is, utilizing the solid works file and the driving sun gear torque how do I calculate the load on the contact point for each of the gears? Is there a general equation I can use without knowing the power and the rpm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You&lt;a href=&quot;https://i.stack.imgur.com/CW1Lu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CW1Lu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c99G0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c99G0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12014" LastEditorUserId="12014" LastEditDate="2017-07-03T01:10:43.683" LastActivityDate="2019-08-28T18:33:49.720" Title="Calculating load on planetary gear from driving torque" Tags="|mechanical-engineering|gears|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16138" PostTypeId="2" ParentId="15115" CreationDate="2017-07-03T10:52:46.480" Score="0" Body="&lt;p&gt;Electronic components such as the list you provided will normally not be damaged by any magnetic field, static or dynamic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let still note:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A magnetic field can influence the electrical property of your devices, it is difficult to really define strength, (depends on a lot of parameters) but you could consider some impacts already at 50KG (~field of a fridge magnet, ~100 times earth magnetic field). This means the devices may not behave as you would expect but still will not be damaged and will work again once the field is removed. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If that is a concern, most devices are already protected against charge accumulations (due to antenna effects) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;other electronic devices magnetic based (for instance hard disks) are nowadays not very sensible to static magnetic fields, you can have a look at &lt;a href=&quot;https://superuser.com/questions/568336/what-is-the-tolerable-magnetic-field-in-tesla-for-a-hard-disk&quot;&gt;this question/answer&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;If the field is really too strong you may have some physical damages, but it will not happen at device level but rather on the connections/packaging&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10684" LastActivityDate="2017-07-03T10:52:46.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16139" PostTypeId="2" ParentId="16133" CreationDate="2017-07-03T11:08:58.870" Score="2" Body="&lt;p&gt;Alephzero's comment pretty well covers it.  &quot;exact spot&quot; need to be quantified. For an elevator, a mm or two is sufficient error.  For a phased-array optical telescope,  the error must be closer to 100 nm (fraction of visible light wavelength).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the only rule designers apply to elevator algorithms is that the system must be significantly overdamped.  If it were underdamped, the elevator would overshoot a little and have to &quot;back up/down&quot;.  That would be a VeryBadThing(TM) from the users' perspective.  The overdamped system will take a few extra tenths of a second to come to a halt, which is unnoticeable by the occupants.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-07-03T11:08:58.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16140" PostTypeId="2" ParentId="16133" CreationDate="2017-07-03T11:30:49.217" Score="1" Body="&lt;p&gt;In the olden days of relays, they did it &lt;em&gt;imprecisely&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be a relay along the way that reduced the speed of the lift early enough to slow to a crawl before it reached the desired level (sometimes annoyingly early...) Then another relay would stop the lift just the right distance before the stop, that it would stop at the right position. How was &quot;the right position&quot; found? Eyeball the position, let the lift filled with average load (dummy) stop, measure the error, move the nub that toggles the contact up or down the measured distance. Repeat. Error from different distance traveled under different load was just accepted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For industrial lifts with e.g. tracks for hauling minecarts, sometimes physical stops would be employed, so the lift cage would lean against such a stop and stop at a precise position. In other cases, an operator would manually drive the lift to the precise position. In most common applications though, a centimeter of error was common; with wear, tear, or non-standard loads the error would sometimes be larger.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-07-03T11:30:49.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16141" PostTypeId="1" CreationDate="2017-07-03T13:49:48.533" Score="1" ViewCount="170" Body="&lt;p&gt;I've been working on a flat pattern design to send to the machine shop and have run into a problem making the support piece. The problem I've been hitting is I can't figure out how to flange off of the sides in a triangle shape to meet the top. The best way to explain this is through the attached screenshots.&lt;a href=&quot;https://i.stack.imgur.com/vYOQw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vYOQw.png&quot; alt=&quot;Shot to show what the piece is supposed to look like... I used the face tool which I&amp;#39;m pretty sure is not the right to for the job.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The triange support piece in between the bend is the problem child. I'm basically trying to make that piece but as a bend off of the side of the part. The reason I want to do this is so that the flat pattern will come out properly. If anyone can explain to me how to make this bend correctly or a better way of doing this it would be greatly appreciated. I'm going to attach one more screenshot of my attempt on the other side. &lt;a href=&quot;https://i.stack.imgur.com/Urkqj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Urkqj.jpg&quot; alt=&quot;Attempt two&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first post so if anyone has any pointers on what else to include in the post please let me know. To anyone that replies thank you in advance. &lt;/p&gt;&#xA;" OwnerUserId="12021" LastEditorUserId="12021" LastEditDate="2017-07-03T18:43:44.983" LastActivityDate="2017-07-03T18:43:44.983" Title="Sheet metal working. Making a flange" Tags="|mechanical-engineering|autodesk-inventor|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16143" PostTypeId="2" ParentId="15115" CreationDate="2017-07-03T15:03:09.220" Score="2" Body="&lt;p&gt;&lt;strong&gt;Static field effects on devices&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strong static magnetic fields will have little effect on most electronics, including most of the ones on your list.  We routinely placed basic electronics in 1-2T fields, and never had any real problems.  There are some exceptions though (this is probably not a complete list):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Relays&lt;/strong&gt; work by using a magnet to move a piece of ferrous material to close a switch. Strong fields may open them, close them, or get them stuck in whatever position they are in.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Inductors and transformers&lt;/strong&gt; have magnetically soft cores, which couple magnetic flux much better than air.  They are designed for a certain flux density based on the properties of the coils in them, and beyond that they saturate, becoming less effective.  The field they create adds with the external field, so the external field can cause early saturation. This reduces the inductance of inductors and ferrites, and reduces the efficiency of transformers.  It may also introduce harmonic content in transformers.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hall sensors&lt;/strong&gt; obviously measure magnetic field, and will measure the external field.  Watch out for them in motors and in some sensors.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;LVDT&lt;/strong&gt; position sensors probably won't be remotely accurate in an external field.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Motors&lt;/strong&gt; almost always use magnetic fields internally and may or may not work. They tend to have half decent shielding, so you may be pleasantly surprised.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Magnetic recording media&lt;/strong&gt; such as hard disks, tape, floppy disks... are very suceptable, but also usually well shielded (particularly modern hard disks).  So will usually be OK.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you have any items in the above list, it would probably be wise to check with the manufacturer with your specific field strength &lt;em&gt;and field gradient&lt;/em&gt;.  Or look for alternatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Shielding static fields&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can shield against static fields with magnetically soft materials and careful geometry.  The cheap way is to use iron plates or tubes.  Mu-metal is a more expensive but also much more compact and effective option.  The most effective shielding is a series of concentric tubes aligned along the field direction, with the delicate electronics inside.  In the unlikely event you have a cryogenic system, superconductors also make excellent shields - wrap things in lead foil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dynamic fields&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are very complicated, so I'm not going to give a complete answer here.  Broadly speaking, they are rather worse than static fields of the same strength, and may cause all sorts of problems with any electronics.  If you give a rough guide to field strength and frequencies, I might be able to add more.  A few Hz can be treated as above, hundreds of Hz will be more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dynamic fields can be shielded by a combination of the magnetically soft iron described above and highly conductive metal sheets. The soft iron works well at low frequencies, and the copper sheet at high frequencies.  Top end shielded rooms use alternating layers of mu-metal and copper, with air gaps.&lt;/p&gt;&#xA;" OwnerUserId="12022" LastActivityDate="2017-07-03T15:03:09.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16144" PostTypeId="1" AcceptedAnswerId="16160" CreationDate="2017-07-03T18:10:12.927" Score="1" ViewCount="1200" Body="&lt;p&gt;I want to channel light from an LED indicator through some clear, circular acrylic rod. I want a rough-ish idea of the smallest bend radius through which the acrylic can be bent and maintain TIR. I will be experimenting, and would like an equation to guide the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surrounding medium will be air. I don't know if there are different acrylic grades that need to be factored in, hopefully not because retailers don't seem to give further information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wavelength will vary, but only visible light is of interest. So I'd like an equation to give minimum bend radius, given a particular wavelength and rod diameter.&lt;/p&gt;&#xA;" OwnerUserId="64" LastEditorUserId="7083" LastEditDate="2018-01-28T22:18:47.163" LastActivityDate="2018-01-28T22:18:47.163" Title="Tightest bend radius of circular acrylic rod light pipe for total internal reflection" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16146" PostTypeId="1" CreationDate="2017-07-03T19:33:34.403" Score="-2" ViewCount="64" Body="&lt;p&gt;So, I have a project where I have a bicycle that powers a type of generator. I have everything else done, but I need to pick a generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first problem is that I don't really know which type of generator to use. I am relatively new to electromagnetism, electricity, and engineering in general(I am doing this project by myself, and I am in high school; I do not have a profound background in this type of stuff).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am debating between using a doubly-fed generator(which is AC), or some type of dynamo(which is DC). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to charge a relatively large battery. In this regard, which type of generator would be better in terms of general electrical power and cost-effectiveness?&lt;/p&gt;&#xA;" OwnerUserId="12026" LastEditorUserId="11854" LastEditDate="2017-07-05T12:21:57.433" LastActivityDate="2017-07-05T12:21:57.433" Title="Which Generator is &quot;Better&quot;: DC or AC?" Tags="|battery|generator|ac|" AnswerCount="1" CommentCount="1" ClosedDate="2017-07-05T15:48:20.400" ContentLicense="CC BY-SA 3.0" />
  <row Id="16147" PostTypeId="1" AcceptedAnswerId="16279" CreationDate="2017-07-03T20:19:06.337" Score="1" ViewCount="84" Body="&lt;p&gt;This is the Jacobian of a robot arm (IBB IRB 120) with already specified joint angles (0,0,0,0,0,0). This Jacobian is singular for this configuration. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;J =&#xA;[       0,    17/50,    7/100, 0,      0, 0]&#xA;[ 187/500,        0,        0, 0,      0, 0]&#xA;[       0, -187/500, -187/500, 0, -9/125, 0]&#xA;[       0,        0,        0, 1,      0, 1]&#xA;[       0,        1,        1, 0,      1, 0]&#xA;[       1,        0,        0, 0,      0, 0]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What motions are unachievable with this specific configuration? How to find that out conveniently with MATLAB?&lt;/p&gt;&#xA;" OwnerUserId="10870" LastActivityDate="2017-07-12T20:29:26.363" Title="Unachievable motions because of singular Jacobian" Tags="|robotics|kinematics|matlab|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16148" PostTypeId="1" AcceptedAnswerId="16184" CreationDate="2017-07-03T22:11:49.710" Score="1" ViewCount="466" Body="&lt;p&gt;I've been puzzling over a question / class of questions I'd like to solve via rough estimate before turning to CFD analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have a pressure vessel with one inlet and one outlet. The pressure vessel is at some initial pressure $P_0$. The outlet is at some fixed pressure $P_{out}$. The inlet has some mass flow rate, $\dot{m}$ (assume it's attached to an infinite upstream reservoir and a mass flow controller).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a compressible fluid, a known mass flow rate, and a set volume in the vessel, what mechanism would you use to approximate the final steady state (volume-average, we'll say) pressure in the vessel? Assume that the outlet pressure is such that the flow will choke. Assume no heat transfer to the walls and that the gas is ideal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution I have thought of is to perform a dumb iterative solution:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pick some time step, $dt$&lt;/li&gt;&#xA;&lt;li&gt;Add $\dot{m} * dt$ mass to the tank&lt;/li&gt;&#xA;&lt;li&gt;Compute the new pressure in the tank via the ideal gas law&lt;/li&gt;&#xA;&lt;li&gt;Use choked pipe flow equation to determine $\dot{m}$ at exit (knowing exit area)&lt;/li&gt;&#xA;&lt;li&gt;Subtract $\dot{m}_{exit} * dt$ mass from the tank&lt;/li&gt;&#xA;&lt;li&gt;Repeat until &quot;convergence&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Possibly you could do some Bernoulli-streamline-style analysis, but I'd like to also apply this sort of approximation to a multi-inlet / multi-outlet problem. This feels like some form of the Hardy Cross method might be applicable as well. Interested in anybody's suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphs of what the above technique gets me for a 1 kg/s flow rate into a 5 m^3 volume vessel, initial conditions T = 290K &amp;amp; P = 1 Pa in the vessel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dQU0o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dQU0o.png&quot; alt=&quot;Mass flow rate convergence&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2i4uy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2i4uy.png&quot; alt=&quot;Pressure convergence&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Added a few figures from my &quot;estimation.&quot;&lt;/p&gt;&#xA;" OwnerUserId="3583" LastEditorUserId="3583" LastEditDate="2017-07-05T23:54:06.437" LastActivityDate="2017-07-06T19:50:10.060" Title="Steady state approximation for compressible fluid flow (manifolds and other pressure vessels)" Tags="|fluid-mechanics|cfd|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16151" PostTypeId="2" ParentId="2063" CreationDate="2017-07-04T06:24:42.033" Score="1" Body="&lt;p&gt;The critical thing to be aware of is the suitability of application, and stability for that application. Noting that generally Finite Volume methods result in better accounting for volume hence mass balance errors are minimal.&#xA;Then whether it has shock capturing to track hydraulic jumps if modelling steeper catchments, and if it is capable of direct rainfall without instabilities and without mass balance. For example the ANUGA Hydro model is capable of all of these. HEC-RAS I believe is not capable of rainfall on grid as yet.&lt;/p&gt;&#xA;" OwnerUserId="12031" LastEditorUserId="90" LastEditDate="2017-07-04T08:24:49.220" LastActivityDate="2017-07-04T08:24:49.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16153" PostTypeId="2" ParentId="16135" CreationDate="2017-07-04T08:33:00.230" Score="3" Body="&lt;p&gt;If the gear is not balanced then the center of mass will oscillate as it spins which will induce additional load on the shaft and create vibrations. This effect greatly depends on the speed of the imbalanced gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the gear is perfectly balanced then there will be no oscillations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However at &lt;em&gt;extremely&lt;/em&gt; high speeds you will need to ensure that the tensile strength of the material is enough to keep the gear together. By the time you get to those speeds however you will have to deal with similar issues elsewhere.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-07-04T08:33:00.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16154" PostTypeId="2" ParentId="16135" CreationDate="2017-07-04T12:35:58.607" Score="3" Body="&lt;p&gt;This depends greatly on the context.  Here is a counter-example, a situation in which the spin velocity of the gear is relevant to the dynamics of the overall system, even if that velocity is constant.  That situation is high speed gears on aircraft (e.g. in the engines).  At sufficiently high speed, the gear will exhibit the gyroscopic effect. Then when the plane executes a maneuver angular velocity (e.g. nose pitch up on takeoff), there is a gyroscopic reaction moment on the gear, which change the loading on the bearings, and possibly shift the gear off center a little.  Of course, this is probably a rare example.  Most gears do not get fast enough to have significant gyroscopic moment, and most gearboxes are not rotated like in an aircraft. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-07-04T12:35:58.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16155" PostTypeId="2" ParentId="16146" CreationDate="2017-07-04T14:51:12.800" Score="1" Body="&lt;p&gt;AC generators are &lt;em&gt;simpler&lt;/em&gt;. Their inherent construction property is that the sine wave of AC is directly derived from the rotary motion of the rotor. There is no commutator, which is a part that suffers wear over the course of use of a &lt;em&gt;dynamo&lt;/em&gt; (a DC generator). Even with a dynamo, the current isn't &lt;em&gt;exactly&lt;/em&gt; DC, just an approximation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither is &quot;better&quot; or &quot;worse&quot;, they both have their own applications. Digital electronics definitely prefer low-voltage DC. Heavy machinery definitely prefers AC, three-phase. Batteries naturally produce DC while generators naturally produce AC. Transforming between various voltages is possible with very little loss for AC, in a very lossy but very easy way for low-power DC. Rectification (AC-&gt;DC) is much easier than changing DC into AC, or changing frequency of AC (by any other means than changing RPM of the generator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you should use in your project should primarily depend on what you want to power, and what parts you have at hand. The decision is still yours, but hopefully now you know a little more to make it.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-07-04T14:51:12.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16156" PostTypeId="1" AcceptedAnswerId="16157" CreationDate="2017-07-04T14:59:17.123" Score="0" ViewCount="39" Body="&lt;p&gt;I'm trying to calculate the values of pressure, temp, speed and density of air going through an axial compressor (12-stage) a helicopter. I've read a few bachelors' thesis and papers but wasn't able to find anything particularly helpful for this task. Can anyone offer advice or, ideally, a paper/textbook with relevant information? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="10383" LastActivityDate="2017-07-04T16:58:16.517" Title="Help me find a paper/textbook on axial compressor in turboshaft engine" Tags="|fluid-mechanics|aerospace-engineering|compressors|" AnswerCount="1" CommentCount="2" ClosedDate="2017-07-05T04:05:53.300" ContentLicense="CC BY-SA 3.0" />
  <row Id="16157" PostTypeId="2" ParentId="16156" CreationDate="2017-07-04T16:58:16.517" Score="0" Body="&lt;p&gt;This would be a good start :&#xA;Engineering Thermodynamics, Work and Heat Transfer (4th Edition). / Rogers, GFC; Mayhew, YR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still got mine and use it..&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-04T16:58:16.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16158" PostTypeId="1" CreationDate="2017-07-04T18:16:39.893" Score="0" ViewCount="254" Body="&lt;p&gt;I have a contained and thermally insulated cylinder of radius r. Inside this cylinder, there is water that reaches up to a certain height inside the cylinder (hw). There is a heater underwater that will supply thermal energy to the water at a certain wattage. The rest of the space is filled with air at atmospheric pressure (ha is the height the air occupies so that ha + hw = htotal).&#xA;I start supplying thermal energy at a given wattage to the water. My goal is to keep the water bubbling at a fixed rate for as long as I can control the pressure inside the vessel. I want to know:&#xA;How can I determine at what rate does the pressure increase? How do I determine the increase boiling point of water? How do I relate the input heat in wattage to bubbling rate? (evaporation rate?) I want to keep the water bubbling at a constant rate, how do I determine how the heater wattage should increase to match the increasing boiling point as the pressure builds up? Is there a reference for what a bubbling rate (evaporation rate?) looks like?&lt;/p&gt;&#xA;" OwnerUserId="12030" LastActivityDate="2017-11-04T11:35:20.470" Title="How can I determine pressure inside contained cylinder when heat is added?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16160" PostTypeId="2" ParentId="16144" CreationDate="2017-07-04T21:21:53.533" Score="3" Body="&lt;p&gt;You can figure this out from first principles.  Here's how:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you will need the critical angle for the acrylic material.  This is the steepest angle that light can strike the acrylic/air interface and still be totally reflected, and it follows from the basic laws of refraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;sin(t) = n1/n2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where n1 and n2 are the refractive indices of the air and rod. Then, you have a geometry problem to work out how that fits in with the curve of the rod.  Here's a diagram; the edges of the rod are black, the ray of light is red, the blue and green are constructions to help solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hqn6A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hqn6A.png&quot; alt=&quot;diagram of light reflecting in rod&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rod of diameter d is bent around a curve of radius r.  The light beam is tangent to the inner edge at B, and strikes the outer edge at angle t - so it stays in, but only just.  Angle OBA is between a radius and a tangent, so is a right angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then express t using some trigonometry as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;sin(t) = r/(r+d)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Serendipity! we already know sin(t) in terms of refractive indices, so we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n1/n2 = r/(r+d)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And re-arrange for r:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r = d/(n2/n1 -1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use the refractive index of air (n1=1) and acrylic (n2=1.49) then this simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;r &gt; 2.04d&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there may be slight variations in refractive index for different acrylic grades, and for different wavelengths, so you'll probably want to keep to a bend radius a bit larger than that, but it should give a good starting point.&lt;/p&gt;&#xA;" OwnerUserId="12022" LastActivityDate="2017-07-04T21:21:53.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16161" PostTypeId="1" CreationDate="2017-07-04T22:06:01.193" Score="6" ViewCount="452" Body="&lt;p&gt;My (highly limited) engineering intuition says: Dual railing is good - for balance, for switching of tracks, for lightening load and making sure it is applied in basically the same direction always, and perhaps other reasons I haven't thought of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why ever have monorails? I mean, ok, if you want to fit into some super-narrow crevice then maybe I can see the motivation, but other than that - what are the benefits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I'm interested in non-magnetic monorails, although if you want to provide an answer for maglev's as an interesting side-note that's also fine.&lt;/p&gt;&#xA;" OwnerUserId="12045" LastEditorUserId="12045" LastEditDate="2017-07-05T17:08:17.980" LastActivityDate="2022-01-25T18:37:48.657" Title="Monorails - why even have them?" Tags="|rail|transportation|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16162" PostTypeId="1" AcceptedAnswerId="16166" CreationDate="2017-07-04T23:53:24.837" Score="1" ViewCount="719" Body="&lt;p&gt;I'm an engineering coop at a vibrations consultant. I was given a graph plotting a series of applied frequencies against a floor's acceleration response to them. I was asked to find the natural frequency and damping ratio, but I haven't been able to find the damping ratio. I've taken a course in vibrations, but we were always given more about the scenario (mass of the object, spring constant, etc.). I've asked him to at least give me a starting point, but he's refused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My boss also asked me to find the theory behind how the damping ratio is determined empirically. All the resources I've found online either reiterate what I already learned in my courses, or are so advanced that I can't even begin to understand them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are: how is the damping ratio determined empirically, and what is the general way to solve this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, my vibrations course focused on 1 DOF systems, teaching the basics behind damping ratio, natural frequency, and general vibrations. We did questions about base excitation and rotating imbalances in machinery.&lt;/p&gt;&#xA;" OwnerUserId="12042" LastActivityDate="2017-09-03T11:55:10.457" Title="Determining damping ratio empirically?" Tags="|mechanical-engineering|vibration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16163" PostTypeId="2" ParentId="16144" CreationDate="2017-07-05T07:53:52.003" Score="1" Body="&lt;p&gt;I've realised something which Jack B's answer didn't mention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have a very long, straight rod section, of which one end has light entering, at all angles. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any beams of light which are beyond the critical angle immediately escape the rod.&lt;/li&gt;&#xA;&lt;li&gt;the remaining rays continue bouncing indefinitely.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The moment there is any bend of any kind, it is inevitable that some light will exit. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="64" LastEditorUserId="64" LastEditDate="2017-07-05T17:18:28.113" LastActivityDate="2017-07-05T17:18:28.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16164" PostTypeId="2" ParentId="16161" CreationDate="2017-07-05T08:22:39.910" Score="5" Body="&lt;p&gt;One of the factors is reduction of amount of infrastructure per kilometer (= cost).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Monorail is usually elevated on pylons, which means little disruption to the land below (which, in case of normal railway is immense). A single narrow line will exert less torque on the pylons than a broad dual rail would. Monorail is strictly for passenger transport, so there's no need for broad rails that would distribute the load over a broad area, as with cargo transport. The shape of the train allows better silencing (the train is own sound screen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these factors add up to reduction of mass, width, extra infrastructure elements, land usage, and as final result - cost. The trains are more expensive to build, but with railway that's usually a small fraction of cost - the ground infrastructure is the lion share. Monorail minimizes that.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-07-05T08:22:39.910" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="16165" PostTypeId="1" CreationDate="2017-07-05T10:02:03.740" Score="1" ViewCount="369" Body="&lt;p&gt;I have a three-phase synchronous active rectifier &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sMFig.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sMFig.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to control it with a voltage oriented scheme like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rR7I6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rR7I6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have no idea of how to tune multiple PIs, I'm trying to do it by trial but I need a &quot;systematic&quot; method.&lt;/p&gt;&#xA;" OwnerUserId="11497" LastEditorUserId="8055" LastEditDate="2017-07-06T14:11:40.490" LastActivityDate="2017-07-06T19:09:55.987" Title="PID tuning of a classic voltage controlled three-phase rectifier" Tags="|electrical-engineering|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16166" PostTypeId="2" ParentId="16162" CreationDate="2017-07-05T11:12:04.113" Score="1" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Q_factor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; is somewhat electrically focused, but it should be good enough to get you going.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: determine the frequency at which the response peaks, call it fc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 2: determine the frequencies at which the power is 1/2 of the peak power (i.e. amplitude is sqrt(2)/2 times the peak). there are 2 such frequencies, one higher than the peak and one lower.  call them f2 and f1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;step 3: Q=fc / (f2-f1), and damping ratio zeta = 1/(2*Q)&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-07-05T11:12:04.113" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16167" PostTypeId="1" CreationDate="2017-07-05T13:57:18.783" Score="2" ViewCount="259" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mdG0p.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mdG0p.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm designing a lifting device powered by a motor, i would like to know if the forces exerted by the load on pulley 1 and 2 are equal when the motor star running , or if in pulley 1 is exerted more force because the location of the load at the end of the lifting platform. The lifting platform only moves vertically, restricted by rails. What I want to know if the load will exert a perpendicular force that could create friction on the rails &lt;/p&gt;&#xA;" OwnerUserId="12046" LastActivityDate="2017-11-02T19:27:34.753" Title="Pulley system for lifting device" Tags="|pulleys|" AnswerCount="2" CommentCount="3" ClosedDate="2017-11-09T18:08:06.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="16168" PostTypeId="2" ParentId="16144" CreationDate="2017-07-05T15:38:29.367" Score="1" Body="&lt;p&gt;This is a well-known and multiply-engineered problem, as you might imagine.  The simplest fix, at least conceptually, is to coat the exterior with a metallic layer so that you no longer depend on internal reflection &amp;amp; Snell's Law.  At this point, your don't need the core (glass) any more, just a shiny tube.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for situations (glass,  or cladded optical fibers) where you do depend on total internal reflection,  you want a material with the highest possible index of refraction. Follow the equations in other answers to see why.  The usual problem is that high-index materials tend to have a higher absorption coefficient as well.  One way or the other you lose some of your signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, if you have money,  you might be interested in &lt;a href=&quot;https://en.wikipedia.org/wiki/Photonic-crystal_fiber&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;holey fibers&quot;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-07-05T15:38:29.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16169" PostTypeId="2" ParentId="6212" CreationDate="2017-07-05T16:03:12.427" Score="0" Body="&lt;p&gt;Having studied electronic and electrical engineering, I covered both the power/analogue and the digital part. There were certain power designs where using a digital microcontroller (to generate a square wave of a certain time period and duty ratio) helped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are methods of obtaining oscillators using discrete analogue components (a Wien Bridge oscillator), but incorporating a 555 timer (digital) makes things much simpler. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Digital Electronics are used in systems control, so it still has relevance to eletrical engineering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Digital Logic Design and VLSI technology were still my favourite topics as it shows how the theoretical boolean logic is implemented using hardware a nano-scale.&lt;/p&gt;&#xA;" OwnerUserId="12047" LastEditorUserId="90" LastEditDate="2017-07-06T02:11:50.207" LastActivityDate="2017-07-06T02:11:50.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16170" PostTypeId="2" ParentId="16167" CreationDate="2017-07-05T16:19:55.750" Score="0" Body="&lt;p&gt;Of course the load will exert a moment, to study the behaviour of your system you can simply model it as a constrained beam in pure bending if the elasticity of your platform is not negligible or as a constrained rigid body. Every point of the rail is a constrained point, analyzing the strain diagrams you can compute how much perpendicular force you will have. &lt;/p&gt;&#xA;" OwnerUserId="11497" LastActivityDate="2017-07-05T16:19:55.750" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16171" PostTypeId="2" ParentId="213" CreationDate="2017-07-05T20:09:31.320" Score="2" Body="&lt;p&gt;The only major difference the alcohol choice will make is the &lt;a href=&quot;http://www.indsci.com/docs/manuals/VX500_IP.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ionization energy&lt;/a&gt;.  While the ionization energy for Methanol is 10.85 (&lt;a href=&quot;http://www.goodforgas.com/documents/appnotes/TN2004_PID_gas_table_01_16_09.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;), ethanol is 10.62 eV, while Isopropyl is only 10.10 eV.  The slight difference will allow for more molecules at the lower potential to be ionized, making for slightly longer trails.  However, this shouldn't be a major impact.  Design, flammability, corrosion, partial pressures, and other factors would be a bigger consideration.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-07-05T20:09:31.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16172" PostTypeId="2" ParentId="16148" CreationDate="2017-07-05T22:53:01.240" Score="1" Body="&lt;p&gt;Interesting question, but well worth discussion of the confusion of compressible flow equations.  At steady state, we will always have the mass flow rate be constant, even with choked flow.  The flow rate at the exit has to be the same as the rate going in.  This seems counter intuitive - choked flow states that there is a maximum flow rate which a nozzle can produce, and no more.  However, choked flow states this is the case for variance of the outlet conditions, the flow rate will not increase.  It states nothing about variances on the inlet conditions. Indeed, The main idea behind choked flow is that it depends on a singularity in the differential equation (which is derived on constant mass flow rate assumption):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\partial V}{V} = -\frac{\partial A}{A}\frac{1}{1-M^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Mach number is exactly 1, then the differential change in area must be 0, or the differential change in velocity would be infinite!  This must be at the exit nozzle of your vessel.  However, the scenario proposed is intentionally attempting to feed a vessel additional mass from an infinitely pressurized reservoir without a converging/diverging nozzle.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This situation could be resolved if more information was known about the inlet stream (such as temperature, or pressure).  We couldn't use the reservoir properties as it is unknown how the mass flow meter affects the stream.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-07-05T22:53:01.240" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16173" PostTypeId="2" ParentId="15643" CreationDate="2017-07-05T23:00:55.247" Score="0" Body="&lt;p&gt;The short answer is yes, there is a minimum, but it's device dependent and usually not specified. The GE Transistor Manual (no longer in print, but considered to be holy writ as recently as the 1970s)explains why ordinary leakage currents don't turn on SCRs. Referring to the 2 transistor SCR model (easily found online) it says that beta for transistors is dependent on collector current, and that SCR equivalent transistors have betas less than 1 for expected leakage currents.&lt;/p&gt;&#xA;" OwnerUserId="12054" LastActivityDate="2017-07-05T23:00:55.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16174" PostTypeId="1" CreationDate="2017-07-05T23:20:05.660" Score="0" ViewCount="273" Body="&lt;p&gt;I'm trying to make most of an engine from scratch (cylinder, crank case, piston). I've been reading about pistons and I've learned that many modern pistons are cam ground--slightly elliptical in shape (&lt;a href=&quot;http://courses.washington.edu/engr100/Section_Wei/engine/UofWindsorManual/Piston%20Design.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference article&lt;/a&gt;). As I currently understand it, the reason for this is so that when the piston heats up to the operating temperature, it forms a more precise circular shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only have access to a metal lathe and that got me thinking, is it possible to design pistons so that they can be machined circularly and they expand evenly? It seems that it would save on manufacturing costs; then again, why aren't modern pistons designed so they don't have to be cam ground and that they expand uniformly?&lt;/p&gt;&#xA;" OwnerUserId="7503" LastActivityDate="2017-07-06T02:03:54.627" Title="Why aren't modern pistons circular?" Tags="|mechanical-engineering|automotive-engineering|manufacturing-engineering|engines|pistons|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16175" PostTypeId="2" ParentId="16174" CreationDate="2017-07-06T02:03:54.627" Score="-1" Body="&lt;p&gt;This is my best guess - I could be wrong:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Fig. 6 in the reference article, you can see that the piston rod has an I-shaped cross-section (presumably because the rod itself is subject not only to axial compression but to some bending while the piston head is not directly in line with the axis of the rod). Possibly, the connection between the piston head and the pinned connection with the rod has a similar configuration (although it's not shown in the Figure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that at this connection, a great deal more restraint is provided in one direction compared with the other. I imagine that as the piston head heats up, it is inclined to expand more in the direction in which less restraint is provided by the (relatively) cooler piston rod/pinned connection. In that case, the minor axis of the elliptical piston head would coincide with this direction of least restraint.&lt;/p&gt;&#xA;" OwnerUserId="11424" LastActivityDate="2017-07-06T02:03:54.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16177" PostTypeId="1" CreationDate="2017-07-06T10:29:11.530" Score="2" ViewCount="4715" Body="&lt;p&gt;I am an fire safety engineer trying to build a fire model which I could use to demonstrate basic fire behavior in enclosures. In general, the model is a box (for example 0,3*0,3*0,2 m or larger) with one opening. Inside the box is the burner (gas burner or some material which burns).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the university, we always used models built out of gypsum plaster. It was cheap, easy to connect and it could resist quite high temperatures (around 700 degrees Celsius).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any kind of transparent material which could be used in such an application? If I understand correctly, the highest temperature which thermoplastics could withstand are about 270 degrees Celsius. Quartz glass is too brittle I think, and probably too hard to cut out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a material like this?  Or if there are multiple materials, which one(s) are least expensive?&lt;/p&gt;&#xA;" OwnerUserId="12057" LastEditorDisplayName="user16" LastEditDate="2017-07-06T12:03:33.483" LastActivityDate="2017-07-06T23:27:20.290" Title="Transparent material which could withstand high temperatures" Tags="|materials|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16178" PostTypeId="2" ParentId="16177" CreationDate="2017-07-06T12:49:20.877" Score="2" Body="&lt;p&gt;My first thought was &lt;a href=&quot;https://en.wikipedia.org/wiki/Pyrex&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pyrex&lt;/a&gt;, but it's a brand name for heat resistant glassware. Glassware for laboratories and kitchen is made from Pyrex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the US, heat resist glassware seems to be made from tempered &lt;a href=&quot;https://en.wikipedia.org/wiki/Soda-lime_glass&quot; rel=&quot;nofollow noreferrer&quot;&gt;soda-lime glass&lt;/a&gt;, but outside the US &lt;a href=&quot;https://en.wikipedia.org/wiki/Borosilicate_glass&quot; rel=&quot;nofollow noreferrer&quot;&gt;borosilicate glass&lt;/a&gt; is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how the borosilicate glass is made, glass with different softening points can be made. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The softening point (temperature at which viscosity is approximately 107.6 poise) of type 7740 Pyrex is 820 °C (1,510 °F).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For your purposes, borosilicate glass would be more appropriate.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-07-06T12:49:20.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16180" PostTypeId="1" AcceptedAnswerId="16213" CreationDate="2017-07-06T14:47:10.883" Score="0" ViewCount="55" Body="&lt;p&gt;Are these pieces of equipment independent from each other, or one requires also the other one to be in a system (chiller)?&lt;/p&gt;&#xA;" OwnerDisplayName="user5026" LastActivityDate="2017-07-07T20:02:19.820" Title="Difference between Variable Speed Compressor and Variable Speed Condenser(Fan controlled)?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16181" PostTypeId="1" AcceptedAnswerId="22211" CreationDate="2017-07-06T14:48:14.617" Score="1" ViewCount="2066" Body="&lt;p&gt;Hot humid east TX; Large subdivision pool is warm ,92F. A plan is being make to cool it with evaportive spray . One data point is : dry bulb= 82.0 , wet bulb = 79.7 F. This is likely typical average of 24 hours. The cost of pump and spray may be $ 3000 ( to purchase). How much cooling could be expected ? My thought is at this high humidity it will not be cost effective but I can't put numbers to it.  My bad: by &quot;large subdivision pool &quot;, I mean 10,000 sq ft / 450,000 gal.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditorDisplayName="user4139" LastEditDate="2020-06-03T23:37:55.693" LastActivityDate="2020-06-03T23:37:55.693" Title="Is evaporative cooling for a pool effective?" Tags="|thermodynamics|cooling|evaporation|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="16182" PostTypeId="2" ParentId="16165" CreationDate="2017-07-06T19:09:55.987" Score="1" Body="&lt;p&gt;This question is much too broad to answer properly; the &quot;best&quot; approach depends on system properties, available controller architecture, available control design tools,...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you have a MIMO system (i.e. the inputs/outputs of the system are coupled and the controllers can therefore not be considered as independent) the general approach would be to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Characterize the properties of the system to be controlled. This is usually done by either building a model of the plant to be controlled or, if the plant is available, by measuring the frequency response functions between its inputs and outputs (including cross terms).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Design a controller based on the measured or modeled plant characteristics and a MIMO control design technique of your choice. There are many possible choices here; in case a loop-shaping like approach is preferred sequential loop closing might be an option. In case a more formal approach is needed $H_\infty$ control might be an option.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Apply the designed controller to your actual system (but only after you are certain that stability and performance requirements are met).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A good reference, explaining many of the details related to MIMO control design, is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multivariable Feedback Control: Analysis and Design, 2nd Edition&#xA;Sigurd Skogestad, Ian Postlethwaite&lt;/p&gt;&#xA;" OwnerUserId="11869" LastActivityDate="2017-07-06T19:09:55.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16184" PostTypeId="2" ParentId="16148" CreationDate="2017-07-06T19:50:10.060" Score="1" Body="&lt;p&gt;We model the system simply for small mass flow rates compared to the size of the vessel simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FlS5O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FlS5O.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At point 1, the pressure can be near infinite, or the density can be near infinite, or simply the capacity may be near infinite as far as we can tell.  At this point, one thing is known for certain - no matter what the mass flow regulator needs, this system has enough energy to feel the need.  However the reservoir supplies it, compared to the other points, $h_1 \approx \infty$.  However, since the mass flow rate is fairly small compared to the remainder, each removal of mass will keep the temperature in the reservoir fairly constant.  It's fair to state that $T_1 = T$, since $V_1 \approx 0$.  After all, the tank is so massive compared to the tiny amount that leaves that the tank will warm or cool the fluid compared to the amount that has left so the outside will keep $h_1$ constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the dynamics of the mass flow regulator, we know that $\dot m$ mass leaves the reservoir.  We don't know much about point 2, but it's very clear that $h_2 \lll h_1$.  If this wasn't the case, the reservoir in question wouldn't be able to supply a consistent energy supply and mass supply through the line in question.  The pressure is still completely unknown, but $P_1 \ggg P_2 &amp;gt; P_3 \ggg P_4$, or the fluid wouldn't leave the system at all.  However, across the regulator, the tiny mass moving of the fluid compared to the size of the piping calls into question the &lt;a href=&quot;http://www.thermopedia.com/content/1145/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stanton Number&lt;/a&gt; of the fluid.  At a high Stanton number, it can be shown that a lot of heat moves from the pipe back into the fluid, reestablishing $T_2 = T$, and bringing $h_2$ up, but less than $h_1$ since $P_2 &amp;lt; P_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As established above, the temperature of $T_2$ should be brought back to room temperature, $T$.  While the energy is unknown, we are attempting to find $P_3$.  Since we enter a reservoir, $V_3 \approx 0$, and the fluid stagnates.  The reservoir walls may be warming the fluid inside.  Since the mass flow rate is small compared to the mass of the fluid inside the reservoir during steady state situation, it can be seen the reservoir should warm the fluid until it reaches $T$.  Thus, the stagnation temperature, $T_{0_3} = T$.  With this temperature known, the choked flow equation can be modified to reveal $P_{3_{final}}$, since the fluid, having stagnated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot m = \dot m_{choked} = A_e(P_0 = P_3)\sqrt{\frac{k}{R(T_0 = T_3 = T)}}\left(\frac{2}{k+1}\right)^{\frac{k+1}{2(k-1)}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_{3_{final}} = \frac{\dot m}{A_e\sqrt{\frac{k}{RT}}\left(\frac{2}{k+1}\right)^{\frac{k+1}{2(k-1)}}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, with these assumptions, you've effectively turned the reservoir at point 3 into it's own mass flow regulator, at the price of keeping the pressure and temperature inside of the reservoir 3 constant.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-07-06T19:50:10.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16185" PostTypeId="1" CreationDate="2017-07-06T20:34:43.280" Score="2" ViewCount="710" Body="&lt;p&gt;There are eight symmetry classes in linear elasticity: triclinic, monoclinic, orthotropic, tetragonal, trigonal, transversely isotropic, cubic, and isotropic. I was wondering if there existed a material in each of those symmetry classes which almost satisfies the Cauchy relations: $C_{ijkl} = C_{ikjl}$ for all $i,j,k,l \in \mathbb{R}^3$ where $[C_{ijkl}]$ is the linear elasticity tensor (basically the elasticity tensor is completely symmetric). Any help would be greatly appreciated.  &lt;/p&gt;&#xA;" OwnerUserId="12065" LastActivityDate="2017-07-09T20:05:36.177" Title="List of materials (almost) satisfying the Cauchy relations in each symmetry class in linear elasticity" Tags="|materials|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16187" PostTypeId="2" ParentId="16177" CreationDate="2017-07-06T22:36:57.977" Score="0" Body="&lt;p&gt;Borosilicate glass with an indium tin oxide (ITO) coating? A mixture of indium oxide and tin oxide as ITO is used as a transparent coating for - usually - display applications.  However,  it had a melting point in excess of 1500 degrees C &lt;/p&gt;&#xA;" OwnerUserId="11590" LastActivityDate="2017-07-06T22:36:57.977" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16188" PostTypeId="2" ParentId="16177" CreationDate="2017-07-06T23:27:20.290" Score="0" Body="&lt;p&gt;The best answers on this site are when &lt;a href=&quot;https://what-if.xkcd.com/122/&quot; rel=&quot;nofollow noreferrer&quot;&gt;xkcd&lt;/a&gt; has an answer as well.  In addition to borosilicate glass, &lt;a href=&quot;https://accuglassproducts.com/home.php?cat=567&quot; rel=&quot;nofollow noreferrer&quot;&gt;sapphire windows&lt;/a&gt; are pretty common, and can handle a wide variety of high intensity flames.  Quartz (fused silica) also works well - in fact there are quite a few &lt;a href=&quot;https://www.newport.com/c/parallel-windows?q1=%3Aranking-desc%3Acompatibility%3AALL%3AnpCategory%3Aparallel-windows%3Asuppress%3Afalse&amp;amp;facetName=parallelwindowsdiameter&amp;amp;75.0%20mm=&amp;amp;text=&amp;amp;actualMinPrice=&amp;amp;actualMaxPrice=&amp;amp;ajax=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;window materials&lt;/a&gt; to chose from, including calcium fluoride and magnesium fluoride.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the plain window is impossible unless you can mount it properly.  As you mentioned, quartz or any other brittle material glass is difficult to machine and cut properly.  Fortunately, holders for these windows are already made as &lt;a href=&quot;http://www.johnernst.com/sight_windows_index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;industrial products&lt;/a&gt;, that can be procured at any time.  The cost comes down to viewing diameter.  These wind up being a larger impact on cost then the actual window, so window material selection is usually determined based upon long-term costs, not short term costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The downside to these glass ports is the viewing area isn't very big.  There is enough for you to see through with your eyes, but nobody else.  So they're not very good for demonstration purposes.  But they've been tested against pressure loading and temperature loading and are designed not to shatter.  Just be prepared to purchase a few of them so you can give a good demonstration to several people simultaneously.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-07-06T23:27:20.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16189" PostTypeId="2" ParentId="7209" CreationDate="2017-07-06T23:53:40.117" Score="1" Body="&lt;p&gt;Some processes are reversible, but this one isn't.  When a compressor compresses a fluid, it can do so via increasing the stream velocity (such as a piston or a rotor), or it can do by increasing the pressure/density from direct compression (such as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Peristaltic_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;peristaltic pump&lt;/a&gt;).  The first is reversible - the high stream velocity can push back on the solid object, slowing down, and transferring momentum back into a shaft.  The second is irreversible - the high pressure fluid can't compress the in-compressible solid and get it to move preferentially in one direction.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-07-06T23:53:40.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16190" PostTypeId="2" ParentId="7426" CreationDate="2017-07-07T00:10:20.140" Score="3" Body="&lt;p&gt;DC motors typically have a torque to speed curve that's linear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DeKH3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DeKH3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rated torque is usually kept at the very low end of the stall torque, simply to prevent overheating, and breaking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BrICX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BrICX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, $8 kg_f cm$ isn't especially high of a torque.  If the 1:3.4 ratio was reversed, so it took more turns of the motor to turn the crank, this could lift 3 pounds and maintain the 55 rpm.  However, it's more likely with more current you could overload the motor and guarantee even more - though that point is based upon the stall torque of the motor, which is not specified here.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-07-07T00:10:20.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16191" PostTypeId="2" ParentId="16158" CreationDate="2017-07-07T07:01:19.067" Score="1" Body="&lt;p&gt;Nukiyama had a similar setup in his experiment, which failed to account for the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Leidenfrost_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;leidenfrost effect.&lt;/a&gt; The net result is he fried his heater. This is why boiling with electrical wiring is such a tricky matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are &lt;a href=&quot;http://www.thermopedia.com/content/589/&quot; rel=&quot;nofollow noreferrer&quot;&gt;several equations &lt;/a&gt;that relate the heat transfer to the water. The main question you ask is how an enclosed vessel will react. This is pretty easy to see when you look at a Pv diagram for water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bmJmp.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bmJmp.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air above the container is never just air, but a mix of air and water vapor, or steam. Thus we will only be moving up in pressure as we heat to newer and newer constant temperatures. The temperature throughout the chamber can be roughly approximated as constant in both phases throughout this, if you're heating so fast this assumption isn't valid, then Lidenfrost effects will indicate that most heat is being transferred via radiation anyways and you're probably melting your heater with most materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific volumes of either phase change (as the liquid drops or more mass moves to the vapor phase), but the total specific volume of the entire water must remain the same. In other words, given any temperature as we climb, there is a specific saturation pressure that the vessel remains at, and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_gm_g +v_lm_l +v_am_a = V_{vessel}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation has two unknowns, the mass in the liquid phase and the mass in the gas phase, but the specific volumes of each phase are known at a given temperature, using &lt;a href=&quot;http://www.thermopedia.com/content/1150/&quot; rel=&quot;nofollow noreferrer&quot;&gt;steam tables&lt;/a&gt;. The mass of the air is known from the start, and the specific volume of the air at a given temperature is given via the ideal gas laws $v_a = \frac{RT}{MP}$ However, since $m_g + m_l = m_w$, which is known at the start, the masses of each at that temperature can be known. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding what time you will reach a specific temperature, and therefore pressure, is more complicated, but not terrible. This involves finding the enthalpy of each phase, subtracting the starting enthalpy, and dividing by the wattage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that above the critical point, the liquid acts like a gas and the gas acts like a liquid and you can't find a line anymore. Then the line goes straight up. You may not reach the critical point - if your specific volume isn't identical to the critical point's specific volume than the path simply goes straight up from there until you reach maximum pressure.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2017-07-07T09:54:53.710" LastActivityDate="2017-07-07T09:54:53.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16192" PostTypeId="1" AcceptedAnswerId="16197" CreationDate="2017-07-07T07:31:26.470" Score="0" ViewCount="74" Body="&lt;p&gt;Watching Apollo 13 and rereading Lost Moon, Jim Lovell's exciting recollection of the events on the crippled spaceship, I began to wonder. Why does NASA maintain the gasses at the triple point? A pressure vessel designed to handle very high vacuum (which H2 and O2 triple points are near complete vacuum) and low temperatures is thicker and more dense then a pressure vessel designed to handle modest amounts of pressure. Why wouldn't they store these gasses at higher pressures, at or near atmospheric pressure?&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-07-07T12:58:21.723" Title="Triple point storage in space travel" Tags="|mechanical-engineering|thermodynamics|pressure|aerospace-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16193" PostTypeId="2" ParentId="15763" CreationDate="2017-07-07T08:13:00.407" Score="1" Body="&lt;p&gt;Since you considered your model as static - it's correct until things start to move. You are right, as theta increases, T1 should decrease to keep things balanced (static problem). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tension is constant. But when you apply additional force it's not $M_2g$ anymore and that is why everything is moving. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to choose correct weight $M_2$,consider two following  static situations.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Closed hatch: If you choose $M_2$ in a way that system is balanced when hatch is closed (max theta ), then additional small force applied to open the hatch will lead to move of the system and opening of the hatch. You already done it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Opened hatch: If you choose $M_2$ in a way when hatch is opened (theta is max) and everything is balanced. Then small additional force to close the hatch will lead to movement, because additional moment due to weight of the hatch will close the hatch. (That is the minimum required mass $M_2$, if it is less, then your hatch will be closing all the time if not fixed with some additional device.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would choose the mass of the counterweight taking as equilibrium point somewhere in between these two values. Depending on the exact value it will be easier close/open the hatch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important note, if horizontal size of hatch L is equal to vertical elevation of the pulley, you have two problems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Clashing when hatch is opened, definitely something will be broken  at some point of time.&lt;/li&gt;&#xA;&lt;li&gt;Counterweight will touch the ground before the hatch is fully opened, thus it's impossible to stabilize hatch in fully opened position only by use of the counterweight. (or make sure that it than travel lower than hatch level)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you move pulley higher, than angle theta at the eyelet will not be equal to angle between vertical and rope at pulley (it will be 90 deg- theta).&lt;/p&gt;&#xA;" OwnerUserId="12070" LastEditorUserId="1832" LastEditDate="2017-07-07T13:26:52.747" LastActivityDate="2017-07-07T13:26:52.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16194" PostTypeId="1" CreationDate="2017-07-07T12:13:15.697" Score="2" ViewCount="56" Body="&lt;p&gt;There's an 4 inch exhaust stream that's emitting air at 2000CFM, I want to divide the exhaust in a 'Y' kind of outlet with two pipes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If State 0- Exhaust out through one pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If state 1- Exhaust out through another pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What electrically actuated valves should I look at?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="12071" LastActivityDate="2017-08-06T13:11:56.333" Title="What valve to use to divert incoming air from a pipe to another pipe stream?" Tags="|mechanical-engineering|airflow|valves|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16195" PostTypeId="2" ParentId="16194" CreationDate="2017-07-07T12:25:28.160" Score="1" Body="&lt;p&gt;You'll want a three way valve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6ug9m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6ug9m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They come in a top down (above) and a one plane configuration (below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zJw1E.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zJw1E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top down configuration allows L style and T style, while the one plane is typically only L style:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/suTXa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/suTXa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your application a one plane configuration should suit you nicely, and is typically cheaper.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-07-07T12:25:28.160" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16196" PostTypeId="1" AcceptedAnswerId="16199" CreationDate="2017-07-07T12:47:09.093" Score="1" ViewCount="41" Body="&lt;p&gt;I recall there was a set of &quot;engineering numbers&quot; that were to be used when deciding how big to make something. So that if, for example, you wanted your screw to be 1.36mm you'd instead round up or down to the nearest of these prescribed numbers, perhaps to 1.2 or 1.5. This way everyone's things were more likely to be the same size because instead of randomly selecting sizes that were particular to a specific situation, we'd all be using the same set of pre-selected numbers instead. Does anybody know what this concept was called? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="12072" LastActivityDate="2017-07-07T17:09:59.470" Title="Pre-Defined Numbers to be used in Sizing Mechanical Parts" Tags="|engineering-history|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16197" PostTypeId="2" ParentId="16192" CreationDate="2017-07-07T12:58:21.723" Score="1" Body="&lt;p&gt;If you want a few hundred pages of gory detail on gas management, I found an original research report &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19720016146.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at excerpts from a &lt;a href=&quot;https://books.google.com/books?id=8MHBDAAAQBAJ&amp;amp;pg=PA117&amp;amp;lpg=PA117&amp;amp;dq=Apollo%20mission%20oxygen%20storage%20at%20triple%20point&amp;amp;source=bl&amp;amp;ots=skmrTx4iiz&amp;amp;sig=dPb7-GyxXf7hC1UPlI7cBGQkoY8&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjbkODVmPfUAhXD6YMKHUUeBwUQ6AEIQzAI#v=onepage&amp;amp;q=Apollo%20mission%20oxygen%20storage%20at%20triple%20point&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;book on cryogenics&lt;/a&gt;, one thing that's pointed out is that the triple point provides a stable pressure inside the container even as energy is added or removed, because the energy just changes some of the material from one state to another.  By comparison, a can of LOX needs pressure management because a little heat produces a gas head, and so on. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-07-07T12:58:21.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16198" PostTypeId="1" CreationDate="2017-07-07T13:13:22.273" Score="1" ViewCount="1513" Body="&lt;p&gt;Sandpapers are labeled to show their coarseness according to different standards. &lt;a href=&quot;http://www.doityourself.com/stry/a-complete-guide-to-sandpaper-grit-classification&quot; rel=&quot;nofollow noreferrer&quot;&gt;The CAMI system is used in e.g. the USA, and the FEPA system in Europe&lt;/a&gt;. I am interested in the FEPA system. It labels sandpaper with a number such as 220, 360, 500, 1000 etc. where higher numbers indicate finer coarsenesses. Exactly how does this number tell coarseness? I am interested in the FEPA number definition only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found &lt;a href=&quot;https://sizes.com/tools/sandpaper.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this website&lt;/a&gt;. It states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] by the FEPA standard for macrogrits grade F180, no more than 3% by mass of the grit can have a particle size larger than 90 microns, and at least 94% must be larger than 53 microns. In F220 (a “microgrit”), no more than 3% can be larger than 75 microns, at least 50% must be in the range 50.0 to 56.0, and at least 94% must be larger than 45 microns. See the Standards section below for detailed information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;180&lt;/strong&gt;: Grain sizes mainly within $[53;90]\;\mathrm{\mu m}$&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;220&lt;/strong&gt;: Grain sizes mainly within $[45;75]\;\mathrm{\mu m}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This indicates that the FEPA number (I don't know why they call the numbers &lt;code&gt;F&lt;/code&gt;; they are usually &lt;a href=&quot;http://www.doityourself.com/stry/a-complete-guide-to-sandpaper-grit-classification&quot; rel=&quot;nofollow noreferrer&quot;&gt;denoted&lt;/a&gt; with a &lt;code&gt;P&lt;/code&gt;) indicates a range of &lt;em&gt;grain sizes&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do anyone know more and can confirm or better explain the definition and for example know where to find an overview of more FEPA numbers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many websites, &lt;a href=&quot;http://www.doityourself.com/stry/a-complete-guide-to-sandpaper-grit-classification&quot; rel=&quot;nofollow noreferrer&quot;&gt;such as this&lt;/a&gt;, explain how to understand the standard but not how the number relates to a physical value. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Sandpaper&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; is not of much help as it doesn't dive any deeper into the definition details than other websites I can find. And the &lt;a href=&quot;https://www.fepa-abrasives.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;FEPA website&lt;/a&gt; is of no help either since it seems to require payment for the information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for you help.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Update&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I have been looking into sandpaper types.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;FEPA P220 equals CAMI 220. Labeled grain size: $68\;\mathrm{\mu m}$&lt;/li&gt;&#xA;&lt;li&gt;FEPA P500 equals CAMI 360. Labeled grain size: $30\;\mathrm{\mu m}$&lt;/li&gt;&#xA;&lt;li&gt;FEPA P1000 equals CAMI 500. Labeled grain size: $18\;\mathrm{\mu m}$&lt;/li&gt;&#xA;&lt;li&gt;FEPA P4000 equals CAMI 1200. Labeled grain size: $5\;\mathrm{\mu m}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the number is the number of grains (e.g. 220 grains when the number is &lt;code&gt;P220&lt;/code&gt;) that stacked reach some height (for example 1 inch which is 2.54 cm) lengths would be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The P220 gives ~1.5 cm and the CAMI 220 gives 1.5 cm.&lt;/li&gt;&#xA;&lt;li&gt;The P500 gives 1.5 cm and the CAMI 360 gives 1.08 cm.&lt;/li&gt;&#xA;&lt;li&gt;The P1000 gives 1.8 cm and the CAMI 500 gives 0.9 cm.&lt;/li&gt;&#xA;&lt;li&gt;The P4000 gives 2 cm and the CAMI 1200 gives 0.6 cm.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is no system and none come close to for example one inch (not even the American system). Otherwise there must be gaps and spaces counted in as well that I am not aware of. From these quick calculations, the number does not seem to be a &quot;number of grains per length&quot; value or alike.&lt;/p&gt;&#xA;" OwnerUserId="94" LastEditorUserId="94" LastEditDate="2017-07-08T14:25:49.243" LastActivityDate="2022-12-08T19:54:20.517" Title="How is sandpaper grit size (the FEPA number system) defined?" Tags="|standards|grinding|polishing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16199" PostTypeId="2" ParentId="16196" CreationDate="2017-07-07T13:16:30.647" Score="2" Body="&lt;p&gt;You probably mean &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Preferred_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;preferred numbers&lt;/strong&gt;&lt;/a&gt; or any of the other names they go under. There are actually many such series defined in the standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like you are talking of the 1-2-5  series in your example.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2017-07-07T13:16:30.647" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16201" PostTypeId="2" ParentId="16198" CreationDate="2017-07-07T14:06:51.577" Score="2" Body="&lt;p&gt;In general grit numbers refer to the size of mesh that the particles will pass through. There are various ways of indicating this but it is typically some variation of the number of strands per unit length of a woven mesh. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is generally defined as strands per inch, even if it is part of an otherwise metric system, in which case it will be technical strands per 25.4mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So P36 grit would pass through a square grid with 36 lines per inch ie a hole 0.706 mm square. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is based on the assumption that the resulting grit would also not pass through the next mesh in the sequence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't necessarily how abrasives are actually made or sorted but is the basis of the standard. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-07-07T17:01:14.947" LastActivityDate="2017-07-07T17:01:14.947" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16202" PostTypeId="1" AcceptedAnswerId="16204" CreationDate="2017-07-07T14:38:34.993" Score="0" ViewCount="314" Body="&lt;p&gt;From time to time I come across bicycles with a particular geometry (or rather a lack of). Most of them have no seat and a single tube which simply connects wheels and has the bottom bracket (if that still makes sense) in the middle of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm talking about are bicycles like these:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sBLvt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sBLvt.jpg&quot; alt=&quot;occam cycle&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/C5QS8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C5QS8.jpg&quot; alt=&quot;micro pedalflow&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't the frame split in two by the force put on the &quot;bottom bracket&quot; while pedaling? Is the &quot;bottom bracket&quot; stay strengthened in any way? Does the lack of the seat tube imply there is less load on the frame? I assumed that the &quot;diamond&quot; shape of a bicycle offsets the bending and compressing forces that act on a beam structure.&lt;/p&gt;&#xA;" OwnerDisplayName="user12075" LastActivityDate="2017-07-07T16:48:48.400" Title="How are single tube bicycles prevented from collapsing?" Tags="|mechanical-engineering|structural-engineering|design|structural-analysis|bicycles|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16203" PostTypeId="2" ParentId="16202" CreationDate="2017-07-07T14:48:10.100" Score="1" Body="&lt;p&gt;The beam is simply made to be strong enough to support the live load of a fully grown man pedaling. Modern materials can be made stronger than back when the diamond frame was first designed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wheel base looks to be shorter than in a standard road bike (thanks to the smaller wheels) so the bending forces are reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first picture the beam is pretty wide vertically and the axle for the pedals is put through the middle. That helps resist the bending load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second bike doesn't have any reinforcements against braking at the pedals except the welds that attach the hub. but I expect that an engineer rand the numbers and was comfortable with the results.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-07-07T14:48:10.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16204" PostTypeId="2" ParentId="16202" CreationDate="2017-07-07T14:53:31.607" Score="2" Body="&lt;p&gt;While a single tube like that isn't as structurally efficient as a triangulated frame there is no reason it can't take the required loads as long as it is chunky enough. Note that, especially in the top picture the beam is clearly a lot more substantial in section than the tubes in more standard cycles. Equally you can't see how thick the walls of the tube are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the chances are that any weight saving from having less tubes is offset by the fact that the ones you do have need to be more substantial but here I suspect that the main aim is to make the whole thing as compact as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in most cases the limiting factor in bike frame design is stiffness rather than outright strength and most failures are due to either impact damage, manufacturing defects or long term fatigue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From experience something like 25x50x2mm rectangular mild steel tube will support an adults body weight over this sort of length with no trouble whatsoever. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-07-07T16:48:48.400" LastActivityDate="2017-07-07T16:48:48.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16205" PostTypeId="2" ParentId="3075" CreationDate="2017-07-07T16:42:52.050" Score="3" Body="&lt;p&gt;Other answers have well described a road is a lot more than just the surface layer and the vast majority of the cost is the planning and logistics involved rather than the actual materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current road building materials are generally pretty much as cheap as engineering materials get, indeed many are effectively waste products from other industries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider asphalt/tarmac. It is hard to think of a material better suited to its purpose. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It provides excellent grip in both wet and dry conditions for rubber tyres and maintains this pretty much regardless of wear. &lt;/li&gt;&#xA;&lt;li&gt;It is cheap &lt;/li&gt;&#xA;&lt;li&gt;Easy to lay, you basically just dump it out of the back of a truck, grade it and compact it. Once you have the sub grades a tarmac laying machine can put down a road surface at a moderate walking pace. &lt;/li&gt;&#xA;&lt;li&gt;It's easy to make local repairs and if done properly they will blend seamlessly with the existing surface. &lt;/li&gt;&#xA;&lt;li&gt;The basic repair kit is tar and stone chips, you can even get premixed bags of the stuff that you can just dump in a hole. &lt;/li&gt;&#xA;&lt;li&gt;It can be renovated by planing the surface and relaying the top layer. For minor roads you can just spray them with tar and spread stone chips over the top. &lt;/li&gt;&#xA;&lt;li&gt;You can accommodate any arbitrary carriageway width, curve or camber. &lt;/li&gt;&#xA;&lt;li&gt;It is very abrasion resistant and can self repair to some extent. &lt;/li&gt;&#xA;&lt;li&gt;Bitumen, the binding agent is a byproduct of crude oil refining, it doesn't really have any other use  so in environmental terms it is essentially free as long as there is any market for other crude oil products. It is also highly recyclable.  &lt;/li&gt;&#xA;&lt;li&gt;Aside from earthquakes or bombs asphalt roads can't really fail catastrophically &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-07-07T16:42:52.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16206" PostTypeId="1" CreationDate="2017-07-07T16:49:34.143" Score="0" ViewCount="160" Body="&lt;p&gt;We are trying to design a metal mineral deposit detection system using  low frequency radio waves.  I am not sure on the physics of why the ultra low frequency electromagnetic radiation can penetrate the earth and sea.  We need to be able to scan to a depth of 3 miles from the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia classifies &lt;a href=&quot;https://en.wikipedia.org/wiki/Extremely_low_frequency&quot; rel=&quot;nofollow noreferrer&quot;&gt;ELF&lt;/a&gt; (Extremely Low Frequency) radiation as between 3 Hz and 300 Hz, and &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultra_low_frequency&quot; rel=&quot;nofollow noreferrer&quot;&gt;ULF&lt;/a&gt; (Ultra Low Frequency) from 300 Hz to 3 kHz.  Both ranges can penetrate the earth, but I don't know which would be better for our application.  Which frequency would yield the most precise, accurate and reliable results for this depth?&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-07-17T20:12:42.153" Title="Should we use ELF or ULF for very deep metal detection and which specific frequency?" Tags="|electrical-engineering|metals|electromagnetism|rf-electronics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16207" PostTypeId="2" ParentId="3547" CreationDate="2017-07-07T17:02:43.163" Score="0" Body="&lt;p&gt;On average allow a circle with a diameter of 13m (Radius 6.5m) for a carriage driveway.&lt;/p&gt;&#xA;" OwnerUserId="12077" LastActivityDate="2017-07-07T17:02:43.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16208" PostTypeId="2" ParentId="16196" CreationDate="2017-07-07T17:09:59.470" Score="0" Body="&lt;p&gt;It is also worth adding that preferred numbers are highly contextual as what numbers are actually sensible will depend a lot on manufacturing processes and existing standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example many standard steel sections, even when nominally metric are actually based on inch sizes so you often see fairly odd mm dimensions on rolled steel sections, simply because it is too much trouble to convert to round metric numbers for no real benefit. Especially for things like structural sections and pipe where new stock needs to be compatible with legacy installations. Similarly BSP is still the norm for gas and hydraulic fittings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally with the rise of international trade most things that need to be standardised are so this is much less of an issue that it once was. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This probably should be a comment really but is a bit too long&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-07-07T17:09:59.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16209" PostTypeId="2" ParentId="3547" CreationDate="2017-07-07T17:34:29.383" Score="1" Body="&lt;p&gt;Why not just take the car for a test drive and see if it can make the turn?&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2017-07-07T17:34:29.383" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16210" PostTypeId="1" AcceptedAnswerId="16215" CreationDate="2017-07-07T17:47:43.793" Score="0" ViewCount="128" Body="&lt;p&gt;I'm learning theory of axial compressors for my summer internship and I find some of the equations puzzling, although I do understand the way they were derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation bellow is for temperature increase on rotor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XMuYQ.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XMuYQ.gif&quot; alt=&quot;Temperature change equation in Axial compressor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone elaborate on why isn't the $\Delta T$ proportional to other factors like the number of blades? Is it correct to assume that even one blade is considered to be able to accelerate all the fluid going through the compressor to the tangential velocity considered in this equation, given by $U =\omega$r? This would then mean that the number of blades and their geometry, except the main angles, will only play a role in energy losses and distribution of load on individual blades? &lt;/p&gt;&#xA;" OwnerUserId="10383" LastEditorUserId="10383" LastEditDate="2017-07-07T20:06:26.607" LastActivityDate="2017-07-07T21:06:11.593" Title="Axial compressor - change of temperature not proportional to number of blades and their geometrical properties?" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|aerospace-engineering|compressors|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16211" PostTypeId="2" ParentId="14676" CreationDate="2017-07-07T18:00:01.167" Score="0" Body="&lt;p&gt;Attempting to understand, I finally plugged in a few of the first terms in Sloan's.  Sequence &lt;a href=&quot;https://oeis.org/A033581&quot; rel=&quot;nofollow noreferrer&quot;&gt;A033581&lt;/a&gt; for $6n^2$ is a close match to these terms (within manufacturing precision, the exact numbers could have been put on there to match the actual mark position).  $6.1n^2$, rounded up is even closer.  No constant in front of n^2 fits precisely, but it's a close fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The benefit of this is that it would measure the surface area of a cube whose side was n long.  My educated guess is that this can be used to make an estimate for the maximum possible surface area of the object, using the very faded line in the center bar that corresponds to 0 when the calipers are closed.  The estimate would assume the object was a cube, with a 10% safety factor to cover irregularities.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-07-07T18:00:01.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16212" PostTypeId="1" CreationDate="2017-07-07T18:17:16.763" Score="3" ViewCount="114" Body="&lt;p&gt;For a Sun-Planet mesh in a planetary gear train, there will be a deflection $\delta_{sn}=\delta_{s}-\delta_{n}$ along the pressure line and the position of the planet will be determined by $\psi_{n}$, and by looking at the geometry of the picture the sun-planet mesh angle will be $\psi_{sn}=\psi_{sn}-\alpha_s$, where $\alpha_s$ is the pressure angle of the gears. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the &lt;strong&gt;Planet-Ring mesh&lt;/strong&gt;, I can't find any literature defining the deflection and meshing angles, considering the ring is an internal gear, how would the geometry work here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xqktn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xqktn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ePwAk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ePwAk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2017-07-10T15:40:13.570" Title="What is the deflection of planet-ring Gear Mesh along the pressure line and the position of the Plant gear?" Tags="|mechanical-engineering|gears|dynamics|geometry|stiffness|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16213" PostTypeId="2" ParentId="16180" CreationDate="2017-07-07T20:02:19.820" Score="0" Body="&lt;p&gt;They are 2 distinct pieces of equipment that both utilize variable speed drives. and one is not a requirement for the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compressor is used to adjust the pressure/flow rate of the working fluid in the system, adding a variable speed drive gives highly efficient control over this. This can make the system more efficient at part load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the condenser the variable speed drive controls fans that pass air (not the working fluid) over an atmospheric heat exchanger, effectively allowing you to alter its thermal conductivity. This gives additional control over the temperature at the condenser exit.&lt;/p&gt;&#xA;" OwnerUserId="8551" LastActivityDate="2017-07-07T20:02:19.820" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16215" PostTypeId="2" ParentId="16210" CreationDate="2017-07-07T21:06:11.593" Score="1" Body="&lt;p&gt;It is usually a completely different group of people doing compressor aerodynamic design versus overall thermodynamic cycle design.  The compressor aero group, using a variety of computer simulation codes (ranging from simple spreadsheets all the way up to 3 dimensional unsteady CFD) are going to generate a &lt;a href=&quot;https://en.wikipedia.org/wiki/Compressor_map&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;compressor map&quot;&lt;/a&gt;.  The compressor map gives the pressure ratio, mass flow through the compressor, and efficiency for given rotor speeds (and perhaps other variables).  This allows the thermodynamic cycle guys to treat the entire compressor with just a few simple numbers.  All of the details like the number of blades, the number of vanes, the stagger angles of the blades, etc, etc get boiled down to a simple map.  That way the thermodynamics guy does not have to know or even care about how many blades there are. He doesn't even care how many stages are in the compressor. He just knows that at 10,000 rpm the compressor is giving him 10 lbm/s, 15:1 pressure ratio and is 90% efficient (or whatever), and at 12,000 rpm it's giving 13 lbm/s at 18:1 ratio, etc. etc. The thermodynamic cycle then takes that compressor map, along with similar input from the turbine aerodynamics guys, the combustor aerodynamics guys, the inlet and exhaust aerodynamics guys, and combines it into an overall picture of the engine performance.  There will be iteration between those groups.  For example, the cycle guy may determine that he has more than enough thrust but he needs more efficiency.  So the compressor aero guy then changes number of blades or something, now maybe it is 9.5 lbm/s at 14.5:1 ratio but 93% efficient at 10,000 rpm.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-07-07T21:06:11.593" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16216" PostTypeId="1" CreationDate="2017-07-07T22:24:06.170" Score="0" ViewCount="76" Body="&lt;p&gt;I've got the following idea in mind. I have a water-cooled PC, I wish to minimize fan/pump noise by lowering their speeds, while maximizing the cooling that can be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is the simple solution of setting a target water to air delta, (water temp taken from a reservoir, operating under the assumption that this will be the average water temperature) and allowing for (PID?) control to ramp up speeds until that delta is met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is there a better way? For example I can sample die temperatures of the CPU/GPU, the inlet and outlet temperature of each heat exchanger (to within +/- .2 degrees C), and obtain flow speed (+/- 2%) . Can I do something interesting with these data points in real time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What equations should I be looking at to solve this? Am I over-engineering by attempting to calculate all of this? For example I know from test data that at most the delta across most radiators for this application will be approximately 1-3 degrees C&lt;/p&gt;&#xA;" OwnerUserId="12079" LastEditorUserId="12079" LastEditDate="2017-07-08T03:44:23.577" LastActivityDate="2017-09-06T05:26:13.917" Title="Maximizing radiator cooling while minimizing fan noise" Tags="|mechanical-engineering|heat-transfer|heat-exchanger|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16217" PostTypeId="2" ParentId="16216" CreationDate="2017-07-07T22:55:42.417" Score="1" Body="&lt;p&gt;In practice this is likely to be empirical problem as it;s hard to predict exactly what nose is related to a given output or indeed how annoying a particular type of noise  will be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact that with water cooling you have a lot of flexibility in where you actually place you heat exchange ie you can make the pipes a bit longer with relatively little additional pump load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sounds like an optimization problem so hard to generalise . My intuition is that pump speed will be better for dealing with bulk load while fan speed will be able to respond faster to load spikes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately the best thing is to write some software which will allow you to test and optimise the system alternatively you could manually tweak it with pots and have fail-safe override which turns everything up to full if temps get iffy, which might actually be more fun. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also wouldn't really want to miss the opportunity to have an alarm which said '&lt;strong&gt;core temperature approaching critical&lt;/strong&gt;'. at which point you can mutter '&lt;em&gt;setting coolant flow to 90%&lt;/em&gt;'....'&lt;strong&gt;temperature approaching normal&lt;/strong&gt;'. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-07-07T22:55:42.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16219" PostTypeId="1" CreationDate="2017-07-08T11:29:01.083" Score="6" ViewCount="3045" Body="&lt;p&gt;I'm interested in whether a flat rubber washer or equivalent is good enough to replace an o-ring in a low pressure application (edit: I meant low pressure &lt;em&gt;differential&lt;/em&gt;. My bad.). The reason is that flat washers might be easier to design for if you have a dynamic seal application where one tube has to &quot;air lock&quot; against another tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The application is to mount an air sampling tube in such a way as I can automatically sample air and then, later, automatically heat the tube to 200C and pull the vapours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tube is just a thin walled (0.25mm) steel tube 6mm or so in diameter and 40mm or so long). Here's how I initially imagined the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e5i4o.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e5i4o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So some sort of automatic airlock connector tube could move out, seal against the o-ring and then stop at the green washer. The green washers might be something like mica or teflon, something able to take the 200C while insulating the holder - I want to heat just the tube, not everything connected to it! Here's a cross section:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y59td.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y59td.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some issues with this - firstly I'm not convinced that a metal to metal seal (tube against holder) will air seal properly, even when pushed by the automatic connector. Perhaps another o-ring is necessary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, mica washers aren't flexible (I think) and getting a washer into a groove like I've shown might not be possible. If I replace the washer with some form of flat seal which captures itself against a lip, perhaps this type of form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tRWZn.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tRWZn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I would have to make the connector by machining or other process, perhaps this would make the design easier. But is such a v type seal effective at air sealing? The difference is that in the O-Ring, positioning is all that is necessary but with a flat washer type, you need force too (related link: &lt;a href=&quot;https://engineering.stackexchange.com/questions/2640/what-are-the-common-failure-points-of-waterproof-cases/2646#2646&quot;&gt;What are the common failure points of waterproof cases?&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Clarifications:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I believe I tried to present too much information at once and missed the important points!&lt;/li&gt;&#xA;&lt;li&gt;The pressure differential should not be very high. I'm considering a flow, from outer atmosphere, of a few litres per minute through a sorbent. If the pressure diff was more than 1/2 atm I'd be surprised.&lt;/li&gt;&#xA;&lt;li&gt;The green washers are really to provide heat isolation, not air sealing. If the sorbent tube is 200C I didn't want the entire system to try and be 200C too!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2555" LastEditorUserId="2555" LastEditDate="2017-07-10T10:23:13.210" LastActivityDate="2017-07-10T15:24:39.260" Title="O-Ring vs flat washer for air seal" Tags="|airflow|temperature|seals|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16221" PostTypeId="2" ParentId="9092" CreationDate="2017-07-09T16:02:03.870" Score="1" Body="&lt;p&gt;Due to the dangers involved with storing and transporting compressed or liquid acetylene, you would probably have to make use of an onboard acetylene generator, fueled with calcium carbide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acetylene generators are safe, but generally produce hot gas, which reduces specific power, and has a tendency towards preignition and even backfiring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it can be done, and the barriers against doing so are fairly minor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously you wouldn't be able to mass produce your car, as the automobile industry is, in some ways, extremely conservative, but there is nothing at all to prevent you from building one or two such cars.&lt;/p&gt;&#xA;" OwnerUserId="12089" LastActivityDate="2017-07-09T16:02:03.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16223" PostTypeId="2" ParentId="16185" CreationDate="2017-07-09T19:51:10.177" Score="2" Body="&lt;p&gt;It may be easier to analyze this problem using &lt;a href=&quot;https://en.wikipedia.org/wiki/Voigt_notation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voigt or reduced notation&lt;/a&gt;, in which, e.g., $\mathbb{C}_{1112}=C_{16}$, where $\mathbb{C}$ is the 3×3×3×3 stiffness tensor and $C$ is the reduced 6×6 stiffness matrix. In &lt;em&gt;Physical Properties of Crystals&lt;/em&gt;, for example, Nye presents certain implications involving the elasticity tensors using this reduced notation:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nWBuw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nWBuw.png&quot; alt=&quot;Constraints on 6×6 elasticity matrices&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From physical arguments, we already have $\mathbb{C}_{ijkl}=\mathbb{C}_{jikl}=\mathbb{C}_{ijlk}=\mathbb{C}_{klij}$. You're asking about the additional constraint that $\mathbb{C}_{ijkl}=\mathbb{C}_{ikjl}$. In reduced notation, this can be verified to imply 6 additional relationships:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{12}=C_{66}$$&#xA;$$C_{13}=C_{55}$$&#xA;$$C_{14}=C_{56}$$&#xA;$$C_{23}=C_{44}$$&#xA;$$C_{25}=C_{46}$$&#xA;$$C_{36}=C_{45}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see from the image above that these all hold for any isotropic material (including fully amorphous materials, isotropic crystals, and &lt;em&gt;any&lt;/em&gt; crystalline material with sufficiently small and equioriented grains) if $\frac{1}{2}(C_{11}-C_{12})=C_{66}=C_{12}$, or $C_{11}=3C_{12}$. Since&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C_{11}=\frac{E(1-\nu)}{(1+\nu)(1-2\nu)}$$&#xA;$$C_{12}=\frac{E\nu}{(1+\nu)(1-2\nu)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E$ is the Young's elastic modulus and $\nu$ is the Poisson's ratio, we have $\nu=0.25$, which is allowed. (Stable materials must satisfy $-1&amp;lt;\nu&amp;lt;0.5$.) Ceramics, for example, often exhibit a Poisson's ratio of around 0.25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since materials do exist (under this Poisson's ratio constraint) that obey $\mathbb{C}_{ijkl}=\mathbb{C}_{ikjl}$ under this most symmetric and tightly restricted crystal class of isotropy, I'd argue that it wouldn't be physically prohibited for such materials to exist in any of the other crystal classes (although it may require that some of their $\mathbb{C}_{ijkl}$ indices that need not be identically zero are nevertheless approximately zero).&lt;/p&gt;&#xA;" OwnerUserId="10123" LastEditorUserId="10123" LastEditDate="2017-07-09T20:05:36.177" LastActivityDate="2017-07-09T20:05:36.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16224" PostTypeId="2" ParentId="7209" CreationDate="2017-07-09T20:37:46.567" Score="0" Body="&lt;p&gt;I can't imagine an Archimedes screw ( figure) being effective moving air/ gas. If you want to get work out of high pressure air , A reciprocal machine, ie a &quot;steam&quot; engine would be about the only effective mechanism. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-07-09T20:37:46.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16226" PostTypeId="2" ParentId="12543" CreationDate="2017-07-09T20:54:39.933" Score="1" Body="&lt;p&gt;I think you need to check on the grinding and polishing materials. Normal materials used are silicon carbide, alumina and diamond ( We never used quartz that I know of). Were any of these materials used and/or embedded in the samples ? Can you do an elemental display to determine how they are dispersed ? &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-07-09T20:54:39.933" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16227" PostTypeId="1" AcceptedAnswerId="16235" CreationDate="2017-07-09T21:19:50.677" Score="1" ViewCount="221" Body="&lt;p&gt;I'm interested specific on electricity consumption changes on 2 different situations inside an elevator.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When the elevator is stopped and people enters the elevator, does it change in any way the electricity consumption? Simple logic tells as that the elevator in this state is stopped by brakes, and brakes pressure is not affected by the amount of people being inside the elevator, so there should be no difference on electricity consumption on this case.&lt;/li&gt;&#xA;&lt;li&gt;If elevator is going up/down, and a passenger is walking, jumping inside the elevator does this makes any difference on power consumption of the elevator? or regarding the external motor we can see all the cabinet as a unit and the forces that interact inside the cabinet doesn't change anything regarding the external force that is trying to make all the cabinet move ? If it's true how is called this principle?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12091" LastEditorUserId="1832" LastEditDate="2017-07-10T00:29:32.113" LastActivityDate="2017-07-10T18:31:15.223" Title="Elevator standby + moving passanger , electricity consumption changes" Tags="|energy|electrical|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16228" PostTypeId="2" ParentId="16219" CreationDate="2017-07-09T21:50:32.247" Score="2" Body="&lt;p&gt;This really depends on what sort of vacuum / ultimate pressure you're looking for. For something in the millitorr range, this probably won't work. Maybe if you have a big pump and a small internal volume, you could get down to 50-100 torr. Your metal-metal seal here should be fine, assuming your O-ring groove is sized correctly (and for the right pressure direction). You may want to make your O-ring groove on the deep end of the tolerance range for proper sealing so it does not get pinched when your outer tube moves into place. The face seal between the outer tube and your washer will be a problem, I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like PTFE isn't usually flexible to get a correct seal in most vacuum seal applications. Mica definitely won't be. I would opt here to put a second O-ring where your washer is; I'm not sure the washer, even if it was rubber, would necessarily have the right flex and shape to seal the gap. There are high temp vacuum O-rings like Kalrez (and others that are not as pricey) that should handle 200C without a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest thing might be to simply embed the O-ring in the plate where you would have your washer located. I tend to opt for the conservative method of a proper O-ring because with a minimum amount of design, they do a great job of a very predictable and consistent seal. It will be much harder to guarantee that with a washer, and it's always annoying and costly to send something back to the machine shop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you can move the O-ring out further on the plate so it gets sealed against when your outer tube moves into place (probably the best choice, now that I think about it...), or embed the O-ring in a groove on the tip of the outer metal tube. Unloading the tubes with an O-ring in the plate seems like it wouldn't be too bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Terrible drawing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NYGBz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NYGBz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3583" LastEditorUserId="3583" LastEditDate="2017-07-09T22:10:20.510" LastActivityDate="2017-07-09T22:10:20.510" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16229" PostTypeId="2" ParentId="14087" CreationDate="2017-07-10T05:47:00.070" Score="2" Body="&lt;p&gt;Yes, shear &lt;em&gt;flow&lt;/em&gt; in a cross sectional element will always be in the same direction as the shear &lt;em&gt;force&lt;/em&gt; in that same cross sectional element because the shear flow is essentially the distribution of the shear force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Confusion may arise due to the distinction between the &quot;shear force&quot; in an element and the externally applied load that is producing said shear. The net shear force in the cross section will be in the opposite direction of the applied shear-producing load. (At least, that's the sign convention I use, and that appears to be what is happening in the two examples you provided.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple sketches based on the video example you provided:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BVSCjl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BVSCjl.jpg&quot; alt=&quot;shear flow&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the net horizontal shear force will be zero and the net vertical shear force will be equal and opposite to the applied force. (Congruent with basic statics)&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2017-07-10T05:47:00.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16230" PostTypeId="2" ParentId="16219" CreationDate="2017-07-10T11:09:26.527" Score="1" Body="&lt;p&gt;My push on connector with flat washer rubber is good for 2 bar diff press.&#xA;Actually, o-ring seal inside socket connection maybe easier for assembly &amp;amp; disassembly. Difficult fabricate, yes.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2017-07-10T11:09:26.527" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16232" PostTypeId="1" AcceptedAnswerId="16233" CreationDate="2017-07-10T13:21:52.353" Score="4" ViewCount="5948" Body="&lt;p&gt;I'm trying to calculate how much simply-supported round beams of several materials and diameters deflect under their own weight plus an additional weight in the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the material densities so I can calculate weight of the entire beam, and I know the material Young's Modulus so I can calculate Moment of Inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got the deflection due to an added weight calculated correctly, but I'm stuck on deflection due to the shaft's own weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several places have given the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta C = \frac {5 q L^4} {384 E I}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eg: &lt;a href=&quot;https://en.wikipedia.org/wiki/Deflection_(engineering)#Uniformly-loaded_simple_beams&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Deflection_(engineering)#Uniformly-loaded_simple_beams&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am not sure what &lt;code&gt;q&lt;/code&gt; represents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia describes that as &quot;Uniform load on the beam (force per unit length)&quot; and I see other sites mention it in Newtons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that gravity's acceleration (&lt;code&gt;9.80665 m/sec^2&lt;/code&gt;) times the beam's own weight in kilograms to get Newtons of downward force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;eg: &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton_(unit)#Examples&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Newton_(unit)#Examples&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a 102g beam would exert 1N of downward force, so &lt;code&gt;q&lt;/code&gt; would be &lt;code&gt;1&lt;/code&gt;?&lt;/p&gt;&#xA;" OwnerUserId="12095" LastActivityDate="2017-07-21T01:52:35.813" Title="Round beam deflection under its own weight" Tags="|structural-engineering|civil-engineering|beam|deflection|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16233" PostTypeId="2" ParentId="16232" CreationDate="2017-07-10T13:32:42.867" Score="4" Body="&lt;p&gt;No, as the Wikipedia article states, a uniform load's unit is in &quot;force per unit length&quot; ($F/L$), which means it is a load applied along a given distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you have a 102&amp;nbsp;g beam, it weighs a total of 1&amp;nbsp;N. However, that unit is merely &quot;force&quot; ($F$), so we still need to get it into &quot;per unit length&quot; ($1/L$). So, if your beam is 2&amp;nbsp;m long, then $q = \frac{0.102\cdot9.8}{2} = 0.5\text{ N/m.}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of calculating this would be by multiplying the beam's cross-sectional area (unit is $L^2$) by its specific weight ($F/L^3$), which gives you the beam's linear weight ($L^2 \cdot F/L^3=F/L$).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-07-10T13:32:42.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16234" PostTypeId="2" ParentId="16219" CreationDate="2017-07-10T14:07:10.537" Score="2" Body="&lt;p&gt;A properly designed O-ring seal will be your best option, though for 200C operatiing temperature, you'll need O-rings made from a material than is designed to function at that temp, like Silicone or Fluorocarbon (FKM) rubbers. O-rings are popular because they work well in so many applications and they're cheaper overall than anything that functions as well.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With any O-ring seal, static or dynamic, the key parameter is &quot;O-ring Squeeze&quot; which is defined as the percentage reduction in cross sectional area when compressed from the free state. Since the o-ring acts like a spring, this squeeze is directly related to the sealing force. The greater the squeeze, the greater the sealing force, and the greater resistance to leakage...at least in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with most things, more is not always better, too much squeeze can result in a poor joint since it can cause excess stress in the mating parts, excess friction (in the case of a dynamic seal), and O ring damage since there's a greater risk of the o-ring getting pinched between the two mating surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In designing O-ring seals in stainless steel with good surface finish, I would try to stick to &amp;lt; 15% squeeze for dynamic seals and 15%-30% for static seals using a standard NBR O-ring from Parker, but it really depends on the type of O-Ring, the materials involved and all of the other specifics of the application. In this case, you'll want to look at the properties of whatever material you choose and as with any quasi-dynamic joint like the one you describe, smooth mating surfaces and lubrication are essential. (I recommend Krytox grease, pricy, but a little goes a loooooong way. You just need enough grease to make the O-ring look wet.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot that goes into the design of a good O-ring joint. The best resource I've found is the &lt;a href=&quot;http://promo.parker.com/promotionsite/oring-ehandbook/us/en/ehome&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parker O-ring Handbook&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: I designed and tested engineered mechanical seals for 5yr&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2017-07-10T15:24:39.260" LastActivityDate="2017-07-10T15:24:39.260" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16235" PostTypeId="2" ParentId="16227" CreationDate="2017-07-10T15:19:15.463" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When power fails the elevator must remain safe, which means the brakes will be released by applying power.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The force required to move the cabin will change according to how the passenger is accelerating himself. Similar how a phone can vibrate all on its own by oscillating a mass inside of it. The mean power required to lift the cabin+passenger will remain the same regardless of what he's doing in there.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2017-07-10T18:31:15.223" LastActivityDate="2017-07-10T18:31:15.223" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16236" PostTypeId="2" ParentId="12772" CreationDate="2017-07-10T18:18:07.067" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Pour your wall slowly, .5 meters an hour?, that way you wouldn't have the pour pushing full force. A lot of times folks will specify pour rate for this exact reason. It reduces the pressure and allows you to dial back your formwork needs a bit. put some bar in at the bottom of your pour - and attach ties to it holding down your formwork. Let the bottom set up a little bit (pour slowly to attain consolidation) and your ties will hold down your formwork.&lt;/li&gt;&#xA;&lt;li&gt;Fasten your formwork to the ground if you can.. Pour a mat of concrete or a footer that you can anchor your wall to? Again, pour in phases or stages.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12098" LastEditorUserId="12098" LastEditDate="2017-08-02T17:41:18.277" LastActivityDate="2017-08-02T17:41:18.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16239" PostTypeId="1" AcceptedAnswerId="16242" CreationDate="2017-07-10T20:52:45.583" Score="0" ViewCount="457" Body="&lt;p&gt;I am interested in designing a scale model semi-monocoque wing for a glider. I've done some research on the framework and internal structure. I have found out about the spars, ribs and stringers and I am trying to produce a design with the highest strength to weight ratio. I am planning on making the wing 0.5m long with a chord length of 0.1m along the cross section (so no tapering).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my design so far: &lt;a href=&quot;https://i.stack.imgur.com/a2b9w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a2b9w.jpg&quot; alt=&quot;design plans of glider wing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anything is unclear please let me know. In my first diagram, I have placed 2 spars through the wing about 10% of the chord length from the leading edge and 15% from the trailing edge, leaving 75% of the chord length between the two spars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second diagram is a rough plan of how many ribs would be used and where they would be placed: the solid outline is the wing outline, the thick black lines are the spars and the dashed lines are the ribs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used a triangular configuration, as the triangle is the strongest shape and resists shear, torsion, compression and tension better than any other shape. Then I have placed a rib going down from the top corner of each of the 3 triangles to the bottom of each. Then I have placed ribs (or stringers) between each of the bottom corners of the triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can I improve or optimise that design? Is there a certain number of triangles I should use and how big should they be relative to the wing? Should they be equilateral triangles or is isosceles stronger? Also, how many ribs should I be considering using?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for anyone to figure out exact quantities. If there is some kind of calculation I need to carry out to find the optimal number or triangles and ribs, I would be happy to do that. Unless it is way beyond A-level maths/physics standard then I probably will not be able to do it. But I would like advice on how I can figure it out. If my question is vague or unclear, please inform me, and I will gladly edit it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="7802" LastActivityDate="2017-07-10T23:19:02.490" Title="Advice for designing internal structure of glider wing" Tags="|mechanical-engineering|design|structural-analysis|aerospace-engineering|geometry|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16241" PostTypeId="1" CreationDate="2017-07-10T22:52:08.103" Score="0" ViewCount="131" Body="&lt;p&gt;Steam boilers obviously need to be protected from the catastrophe of running out of water -- you then either get a meltdown if the boiler is fired unwatered for an extended period of time or a boiler explosion if the water returns before then, neither of which are good.  Hence, automatic low water cutoffs are required on steam (and hydronic, but that's beyond the scope of this question) boilers in all modern boiler codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current low water cutoff technology consists of two types of cutoffs: &lt;em&gt;probe&lt;/em&gt; and &lt;em&gt;float&lt;/em&gt;.  Floats have the advantage of not requiring an external electrical power source to work, but can get hung up due to deposits (scale), causing them to stop working.  Probes lack moving mechanical parts, but require external electrical power to work, as they use the conductivity of the water as part of the switching circuit, and can also be fouled by boiler scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are other properties we can measure to determine how much water's in the boiler -- one of them being the static head (pressure) of the boiler contents.  This can be done either as a partial head measurement, measuring the pressure of the water above the minimum waterline with respect to steam pressure and cutting out at 0 differential, or as a total head measurement, measuring the static pressure of all the boiler water with respect to the boiler steam pressure and cutting out at a differential equal to the static head of the water at the minimum waterline.  In both cases, the pressure switch would have a trapped steam line piped to it as well as the connection to the boiler water volume, and it would be wired as is typical for low-water cutoffs, opening the fuel control circuit when it trips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How well would such a cutoff scheme work at protecting a boiler?  Would a pressure-switch based safety be more or less vulnerable to failing hazardously (due to fouling or mechanical malfunction) than the existing float-type or probe-type cutoffs?  Are there other flaws in this scheme I should be aware of?&lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2017-07-11T09:27:13.040" Title="Can a water pressure switch be used as a low-water cutoff means for a steam boiler?" Tags="|mechanical-engineering|safety|combustion|pressure-vessel|boilers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16242" PostTypeId="2" ParentId="16239" CreationDate="2017-07-10T23:19:02.490" Score="1" Body="&lt;p&gt;If you want to design something, the first step is to find out what the design needs to do - i.e. what are the forces which will act on the spar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the wing loading from the weight of the plane plus the maximum &quot;g&quot; force you want it to survive in flight (and in crash landings, while you are learning to fly it!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to think about the torsion forces on the wings - read about the difference in the chord position of the &quot;center of pressure&quot; and the &quot;neutral axis&quot; of the beam created by your wing spars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, you can find the bending and torsional stresses in the spars, and calculate how big they need to be to carry those loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;nice&quot; thing about beams is that the stresses are &lt;em&gt;statically determinate&lt;/em&gt; - i.e. changing the shape of the beam doesn't change the loading pattern (except for the change in weight of the wing, of course). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should get you started answering your question. The lightest design will probably have spars that are tapered along their length - that may be more important than the exact layout of the &quot;web&quot; connecting the spars, for weight saving.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-07-10T23:19:02.490" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16243" PostTypeId="1" CreationDate="2017-07-11T00:53:46.983" Score="1" ViewCount="155" Body="&lt;p&gt;Our company is installing flow meters on an old polyurethane foam machine.&#xA;After some research, ultrasonic flow meters were chosen as they have no moving parts and are easy to clean.  Polyurethane foam is produced by mixing polyol with isocyante (not as bad as it sounds) at a fairly precise ratio.  However, the ultrasonic flow meter will not work with the polyol.  The polyol we are using is the polyether type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the manufacturer any polyol should not be a problem, nor should the viscosity be a problem.  Apparently they have used the same flow meters for honey and glycerin and they work well in those applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would the flow of polyether polyols not be able to be measured in an ultrasonic flow meter?&lt;/p&gt;&#xA;" OwnerUserId="5310" LastActivityDate="2017-07-11T00:53:46.983" Title="Measuring flow of polyol with ultrasonic flow meter" Tags="|fluid-mechanics|ultrasound|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16245" PostTypeId="1" CreationDate="2017-07-11T06:28:41.357" Score="1" ViewCount="46" Body="&lt;p&gt;I have following system of coupled Partial differential equations.&#xA;How can I solve the system by Maple?&#xA;\begin{align} 		m_1\frac{\partial^2 u_1}{\partial t^2}+A_1\frac{\partial ^4u_1(x,t)}{\partial x^4}+k(u_1-u_2)=F_1(t) \delta(x-x_1), 	\end{align}&#xA;\begin{align} 		 m_2\frac{\partial^2 u_2}{\partial t^2}-A_2\frac{\partial ^2u_2(x,t)}{\partial x^2}+k(u_2-u_1)=F_2(t)\delta(x-x_2). 	\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In here,  $A_i, m_i, x_i$ and $k$ are constants where $i=1,2$ and $\delta$ is Dirac Delta function.&#xA;&lt;strong&gt;Boundary conditions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_1(0,t)=\frac{\partial^2 u_1}{\partial x^2}(0,t)=u_1(l,t)=\frac{\partial^2 u_1}{\partial x^2}(l,t)=0$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_2(0,t)=u_2(l,t)=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Initial conditions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_i(x,0)=w_{i0}(x),$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial u_i}{\partial x}(x,0)=y_{i0}(x)$ for $i=1,2.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $F_1(t)$ and $F_2(t)$ are unspecified (ungiven) functions, solutions $u_1,u_2$ which we seek will be depended on  $F_1(t)$ and $F_2(t)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Maple Code:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;PDE1:=m1*diff(u1(x,t),t$2)+A1*diff(u1(x,t),x$4)+k*(u1(x,t)-u2(x,t))=F1(t)*delta(x-x1);&#xA;PDE2:=m2*diff(u2(x,t),t$2)-A2*diff(u2(x,t),x$2)+k*(u2(x,t)-u1(x,t))=F2(t)*delta(x-x2);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="11352" LastEditorUserId="11352" LastEditDate="2017-07-12T12:56:52.037" LastActivityDate="2017-07-12T12:56:52.037" Title="solution of system of coupled partial differential equations" Tags="|mechanical-engineering|structural-engineering|automotive-engineering|" AnswerCount="0" CommentCount="2" ClosedDate="2017-07-24T16:40:16.077" ContentLicense="CC BY-SA 3.0" />
  <row Id="16246" PostTypeId="2" ParentId="16241" CreationDate="2017-07-11T09:27:13.040" Score="2" Body="&lt;p&gt;Hydrostatic pressure of water changes per depth according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Pascal%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pascal's law&lt;/a&gt; which equates to $1 \text{Pa} = 1000 \text{kg}/\text{m}^3 * 9.81 \text{m}/\text{s}^2 * h$ or $101 \text{Pa} /\text{m}$ or $1.46*10^{-2}\text{PSI}/\text{m}$ or $4.45*10^{-3}\text{PSI}$ per foot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is not a lot of margin to measure especially when the water is actively boiling which will add noise to the measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better non-contact water detection scheme would be to measure the difference between the magnetic permittivity of water vs. steam/air. However that requires that the immediate area around the sensor is non-metallic. There is usually not a lot of space inside the boiler to put that exclusion zone in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better way of preventing overheating is to put in a temperature sensor to cut off the power when it exceeds margins.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-07-11T09:27:13.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16247" PostTypeId="1" AcceptedAnswerId="16249" CreationDate="2017-07-11T09:34:48.917" Score="0" ViewCount="1086" Body="&lt;p&gt;I am asking here for your help and advice on how to read holes schematic notations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the attached picture I've marked with a green color part of notation that I don't know how to read / understand. And will be thankful for any advice on how to interpret this notes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the red color I've marked a holes with countersinks. But I'm failing to pick up necessary details from the sketch to understand the size/angle/depth of this countersinks. (I understand that they can be missing, but I tend to think that they might be in the notations that I don't understand)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for considering this question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: the drawing was taken from this youtube video - &lt;a href=&quot;https://www.youtube.com/watch?v=2FDeeIlVAtY&amp;amp;feature=youtu.be&amp;amp;list=PL4mNbJmXLK0dn7wdGYOncwFRAQoFIgXiA&amp;amp;t=3219&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oW9Dx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oW9Dx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12107" LastActivityDate="2017-07-11T15:26:36.280" Title="Read holes schematic drawing notation. How to? Particular case" Tags="|measurements|drilling|drawings|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16248" PostTypeId="1" AcceptedAnswerId="16251" CreationDate="2017-07-11T10:06:17.423" Score="1" ViewCount="11995" Body="&lt;p&gt;Yes, the one where you put lighter close to nozzle of deodorant, light it up and start to spray out the deodorant. I suppose that this is simillar to how gas stoves work, so it shouldn't explode. Even when I stop pushing on spray button, flame gets closer to deodorant and it feels like its your last moment, flame can't get back into the can and cause a fireshow, becasue there's no air in the can. Worst thing that can happen is nozzle meltig down over time and flame getting bigger but not out of control. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when things can go bad? I know that i shouldn't throw deodorant can into flame, but can it overheat just from being used as a flamethrower and boom?&lt;/p&gt;&#xA;" OwnerUserId="12108" LastActivityDate="2017-07-11T11:32:44.120" Title="Will deodorant flamethrower ever explode?" Tags="|fire|" AnswerCount="1" CommentCount="3" ClosedDate="2017-07-11T16:01:54.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="16249" PostTypeId="2" ParentId="16247" CreationDate="2017-07-11T10:21:17.467" Score="0" Body="&lt;p&gt;The upside down T means the hole must be perpendicular to the surface. The stuff following it is likely the tolerance of perpendicularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an annotation for &quot;all holes&quot; underneath the cursor, I presume that is about the countersinks.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-07-11T10:21:17.467" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16250" PostTypeId="1" CreationDate="2017-07-11T10:52:10.670" Score="0" ViewCount="75" Body="&lt;p&gt;I have a 50 HP screw compressor that has just began to load and unload within 10 seconds when previously would take between 40 and 60 seconds. Nothing has been added or removed from the system. Any idea why this is happening?&lt;/p&gt;&#xA;" OwnerUserId="12110" LastActivityDate="2020-07-25T08:06:55.783" Title="Rotary Screw Air Compressor" Tags="|pneumatic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16251" PostTypeId="2" ParentId="16248" CreationDate="2017-07-11T11:32:44.120" Score="3" Body="&lt;p&gt;No, the flame can't go upstream back into the can and explode what is in the can.  This is no different than the flame in your gas stove can't go back up the pipe and explode the whole distribution network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fire is a chemical reaction between fuel and oxygen.  There can't be fire in the nozzle or in the can because there is no oxygen there.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-07-11T11:32:44.120" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16252" PostTypeId="2" ParentId="16250" CreationDate="2017-07-11T11:36:02.180" Score="1" Body="&lt;p&gt;Sounds like whatever spring/latch mechanism that kept slide valve from cycling too &quot;easily&quot; is broken or worn out - it takes much lower pressure differential to move the slide valve either way as it doesn't resist being pushed by pressure nearly as strongly - the hysteresis of operation is way more shallow than it should be; the element that provided mechanical latching in end positions failed.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-07-11T11:36:02.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16253" PostTypeId="1" AcceptedAnswerId="16254" CreationDate="2017-07-11T13:58:17.060" Score="1" ViewCount="78" Body="&lt;p&gt;I was wondering whether the hysteresis (in particular, mechanical motion systems) can be considered under the class of friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From theory, I know that the frictional forces are direction, and velocity dependent. In the same sense, they also possess memory in the special cases of frictions. So, is it correct (without abusing any definitions) to consider the hysteresis observed in motion systems to be under the class of friction. Because the hysteresis in motion systems also characterises the same characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply put, under what conditions one can classify the hysteresis under the frictional nonlinearity, and if not, under which conditions this fails to do so?&lt;/p&gt;&#xA;" OwnerUserId="10681" LastActivityDate="2017-07-11T15:22:22.587" Title="Can the hysteresis phenomenon be considered as a (type of) friction" Tags="|mechanical-engineering|friction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16254" PostTypeId="2" ParentId="16253" CreationDate="2017-07-11T15:22:22.587" Score="1" Body="&lt;p&gt;Hysteresis is not the same as friction, but friction is part of hysteresis, it is normally considered as the play or &quot;free movement&quot; in joints etc that cause a system to take a slightly different route &quot;down or return&quot; compared to the &quot;up or outward&quot; travel.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-11T15:22:22.587" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16255" PostTypeId="2" ParentId="16247" CreationDate="2017-07-11T15:26:36.280" Score="0" Body="&lt;p&gt;Just to add.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green is perpendicularity in respect to datum C. Meaning, the called out feature must be perpendicular to datum feature C within 0.003.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also seeing the condition &quot;45 degrees 0.030 - All holes&quot;. The countersink will be an additional 0.030 at 45 degrees from the thru hole values (since the countersink won't be thru). I'm not entirely sure if the larger hole callouts are pointing to the OD or ID of the countersunk hole but you'll simply add or subtract in accordance to the condition: &quot;45 degrees 0.030 - All holes&quot;. Hope that helped.&lt;/p&gt;&#xA;" OwnerUserId="11821" LastActivityDate="2017-07-11T15:26:36.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16256" PostTypeId="1" CreationDate="2017-07-11T17:41:20.550" Score="0" ViewCount="106" Body="&lt;p&gt;I want to measure the velocity of hot gas coming out from an incinerator. The temperature can easily shoot upto 400-500ºC. I've checked online for hot-wire anemometers, even they claim to work in the range of 0-50 ºC or 0-70ºC. Moreover they are dearly expensive. Can someone suggest a good and reliable solution?&lt;/p&gt;&#xA;" OwnerUserId="12114" LastActivityDate="2017-07-11T17:57:12.253" Title="How to measure velocity of hot gas ( temperature range: 100 - 450 ºC)?" Tags="|hvac|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16257" PostTypeId="2" ParentId="16256" CreationDate="2017-07-11T17:57:12.253" Score="1" Body="&lt;p&gt;Use a pitot tube made of a suitable metal and measure the profile across the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did a lab experiment like this years ago to measure the flow profile before and after an axial flow fan.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-11T17:57:12.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16258" PostTypeId="1" AcceptedAnswerId="16289" CreationDate="2017-07-11T18:25:58.337" Score="1" ViewCount="2027" Body="&lt;p&gt;I have a shaft with splines on it that I want to create a drawing for in Solidworks because I may need to have this specific spline replicated on a part in the future. Does anyone have any documentation on how to identify a spline? Or possibly a table of standard splines used in industry based off of shaft size?&lt;/p&gt;&#xA;" OwnerUserId="12119" LastActivityDate="2020-07-28T22:31:26.510" Title="Shaft Spline Information" Tags="|mechanical-engineering|manufacturing-engineering|computer-aided-design|documentation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16259" PostTypeId="2" ParentId="15805" CreationDate="2017-07-11T18:27:18.077" Score="4" Body="&lt;p&gt;The &lt;a href=&quot;https://www.fhwa.dot.gov/environMent/noise/regulations_and_guidance/polguide/polguide05.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;US Federal Highway Administration&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Effective noise barriers can reduce noise levels by 10 to 15 decibels, cutting the loudness of traffic noise in half. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It goes on to say that a 5 dB reduction is &lt;em&gt;simple&lt;/em&gt; to obtain and a 20 dB reduction is &lt;em&gt;nearly impossible&lt;/em&gt; to obtain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even completely burying a highway does not eliminate the noise. While not a highway, there are many buildings in London where you can feel the tube running underneath them.&lt;/p&gt;&#xA;" OwnerUserId="12118" LastActivityDate="2017-07-11T18:27:18.077" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16260" PostTypeId="2" ParentId="15794" CreationDate="2017-07-11T20:47:26.780" Score="2" Body="&lt;p&gt;The PRESUMPTIVE LOAD-BEARING VALUES table in the IBC is meant to be used with ASD loads (see section 1806.1). As such there are already factors of safety &quot;built in&quot; to those values. The only thing I am aware of that the IBC has to say specifically about shallow foundation settlement limitations is in section 1808 (not even worth quoting). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The values in this table have been inherited from a number of sources, including the Uniform Building Code (UBC) and the BOCA, which have tables with very similar numbers. The commentary for the UBC (1991 edition) indicates that the values are based on:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;a factor of safety against general bearing capacity failure of 3.0, and  &lt;/li&gt;&#xA;&lt;li&gt;a total allowable settlement of no more than 0.5 inches. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Settlement and allowable bearing capacity, although related, are different conversations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The industry standard factor of safety used for allowable bearing capacity for a very long time has been in the range of 2.5 to 3.0 (depending on the type of data that is available), so the table is on the higher end as one would expect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, settlement is often the governing design criteria and the typical allowable settlement limit is considered to be on the order of 1 inch to 2 inches, with a differential settlement limit of about half of that. So again, at 0.5 inches it appears that the values in the table seem to be on the higher end of the safety range as one would expect. &lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2017-07-11T20:47:26.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16261" PostTypeId="2" ParentId="16258" CreationDate="2017-07-11T20:53:41.817" Score="0" Body="&lt;p&gt;Many &quot;spline&quot; designs around in industry - load dependant etc start with square, then hex, splines 6, 8 and upward, also &quot;lemon&quot; shafts - lots of those used in the agricultural branch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to give some more info : diameter inner and outer, no of splines, length, sliding joint, application etc.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-11T20:53:41.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16262" PostTypeId="2" ParentId="14410" CreationDate="2017-07-11T21:58:44.713" Score="1" Body="&lt;p&gt;Percent finer is the amount of material from a sample that is smaller than a certain size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the example table provided, the Sieve No. 12 (150 micron) has a % finer of 18.2%. This means that 18.2% of the material passed through that sieve, and 81.8% (which is 100% - 18.2%, which you can see in the &quot;Cumulative Retained&quot; column) was &quot;caught&quot;, or retained, by that sieve. &lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2017-07-11T21:58:44.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16263" PostTypeId="2" ParentId="14410" CreationDate="2017-07-11T22:39:42.333" Score="0" Body="&lt;p&gt;&quot;% Finer&quot; represents the percentage of material of the total sample (100%) that will pass through the current sieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if you want to express it the other way round:&#xA;100% - &quot;% Finer&quot; = &quot;% Retained cumulatively&quot;&#xA;This will represent the cumulative material retained of all sieves with a larger aperture as well as the current sieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Colloquially one could compare it to a single sieve in a flour mill when separating the proverbial wheat from the chaff. The &quot;% Finer&quot; being the wheat and the the &quot;% Retained cumulatively&quot; being the chaff.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2017-07-11T22:39:42.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16265" PostTypeId="2" ParentId="63" CreationDate="2017-07-12T03:53:25.483" Score="1" Body="&lt;p&gt;There is really no argument against the traffic circle. They use less space (contrary to what some have said here). They reduce accidents, injuries, and fatalities substantially. They improve traffic flow, which increases fuel efficiency. They are more free and cooperative, whereas traffic lights are totalitarian. Lighted intersections should be abolished and replaced with circles or grade separated intersections for rush hour situations that might cause a flow problem (which I think is exaggerated). Lights are an outdated panic reaction that has no place in modern society.&lt;/p&gt;&#xA;" OwnerUserId="12125" LastActivityDate="2017-07-12T03:53:25.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16268" PostTypeId="1" AcceptedAnswerId="16269" CreationDate="2017-07-12T11:52:23.310" Score="2" ViewCount="177" Body="&lt;p&gt;Imagine some thing put in the presence of loud pure white noise, say at a level that might come from a teenager's HI-FI set to annoying. And the thing is like a champagne glass, sheet metal box or other thin walled vessel.  I'm not considering anything like a cannon ball or anvil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can foresee two scenarios:-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The thing vibrates at it's natural self resonant frequency, but only with limited amplitude as the power content of white noise at a specific frequency is relatively small.&lt;/li&gt;&#xA;&lt;li&gt;The thing does nothing whatsoever as all the various frequencies present in the white noise induce vibrations that cancel each other out.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Which is it, or might there be a third possibility?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2017-07-13T02:50:56.720" Title="Can a thing resonate whilst exposed to white noise?" Tags="|dynamics|vibration|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16269" PostTypeId="2" ParentId="16268" CreationDate="2017-07-12T12:15:16.430" Score="5" Body="&lt;p&gt;You won't get much resonance because the phase the thing is being driven with keeps changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resonant thing will act like a notch filter, so you are left with the frequencies near its resonance.  However, for resonant energy to build up, the system has to get pumped for a while.  While a short segment of the filtered white noise could resonate the object, the next segment could just as well be out of phase and reduce the energy stored in the resonance.  In the long run, the total power &lt;i&gt;at a particular phase&lt;/i&gt; is 0, so no resonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White noise is random, so sometimes there can be long enough sequences of the right frequency that happen to be with a predominant phase to get a little resonance going.  However, later sequences will always cancel that in the long run.  The question is then is how long &quot;long&quot; is, and how long it takes for the object to build up enough resonant energy to do something interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it comes down to probability.  It is very very unlikely that &quot;white noise&quot; from something like a boombox next to a wine glass can get enough resonance going for the glass to break.  Put another way, the mean time to breaking is likely a lot longer than you're willing and able to wait.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-07-12T12:15:16.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16270" PostTypeId="1" CreationDate="2017-07-12T13:12:03.133" Score="0" ViewCount="38" Body="&lt;p&gt;This topic is slightly related to &lt;a href=&quot;https://electronics.stackexchange.com/questions/1287/electronics-used-for-cooking&quot;&gt;this one&lt;/a&gt;, with some extra details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to go off-grid and therefore buy a PV kit. I can see that the prices for PV systems with a power of +3kW are very high, so I was thinking how much I can get from a 300-500W system, which is way cheaper. The fridge, the PC, and other households can easily work with such a power if stored in batteries, but the problem is the stove. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current stove has a power between 1 and 2 kW, which means that I would need to have this instant power from the PV or have it stored in some batteries. This instant power is of course not available, but I can get more batteries to store let's say at least 1-2kWh of energy, which also costs a lot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing that I have in mind is to find a stove/oven that has a low power (under 300W). Is the crock pot the only option for this? I would like to use the stove/oven for cooking different dishes, including frying and baking. Any idea if a &quot;standard&quot; 1-2kW oven can be modified to work with lower power? Or does any of you have any other idea how to cook with such a little power? &lt;/p&gt;&#xA;" OwnerUserId="8419" LastActivityDate="2017-07-12T16:36:35.097" Title="Is it possible to cook with low power PV systems (under 300 W)" Tags="|photovoltaics|consumer-electronics|home-improvement|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16271" PostTypeId="1" CreationDate="2017-07-12T13:38:10.013" Score="2" ViewCount="93" Body="&lt;p&gt;What is the best way to measure the helix angle of a helical cut gear? I have a sun gear from a planetary set that I would like to have custom made but I'm not sure the best way to get the helix angle, the rest of the information I can measure by myself but this one is proving to be a bit more difficult. Does anyone have a tried and true method to measure this?&lt;/p&gt;&#xA;" OwnerUserId="12119" LastActivityDate="2017-07-12T13:38:10.013" Title="Measuring Helix Angle" Tags="|gears|measurements|power|manufacturing-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16272" PostTypeId="2" ParentId="44" CreationDate="2017-07-12T13:40:38.653" Score="1" Body="&lt;p&gt;Steam Traps can be classified in to many categories which are Mechanical, Thermodynamic and Thermostatic. Venturi Orifice Steam Traps is also another classification even though in reality it is not a typical steam trap. None of the types cannot be called the best since application and operating conditions decide the best steam trap. For ex: Free Float from TLV are the best when it comes to Energy Efficiency (Steam Loss of 0.03 kg/h as per tests according to ISO 7841 and ASME PTC 39) but cannot be used alone in cu-tracing and applications where accurate temperature control below 100 degree centigrade. Similarly Mechanical traps are not suited for extreme pressure applications (above 150-200 bar) where Thermodynamic traps are the best. Thermodynamic traps is the most versatile and economical model compare to Mechanical steam traps. TD traps can also discharge high loads to its size when compared to TD traps with most manufacturers offering a maximum of 3 tonnes/hour @ 40 - 50 barg differential pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is difficult to answer your question since the only information you have provided is the pressure and the batch operation. Following are the important parameters required to select/troubleshoot a Steam Trap:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Design Pressure and Temperature&lt;/li&gt;&#xA;&lt;li&gt;Operating Pressure and Temperature&lt;/li&gt;&#xA;&lt;li&gt;Condensate Load (kg/h or m3/h)&lt;/li&gt;&#xA;&lt;li&gt;Application (Tracing / Process / Drip) (since you mentioned its batch operation I am guessing it might be a process equipment like Heat Exchanger or Reheater.&lt;/li&gt;&#xA;&lt;li&gt;Back Pressure / Differential Pressure&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Kindly provide this information which will enable me to give you an answer. To give you a simple answer based on the provided details, if your venture orifice steam trap is not the correct one for the load and operating conditions.&lt;/p&gt;&#xA;" OwnerUserId="12128" LastActivityDate="2017-07-12T13:40:38.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16274" PostTypeId="2" ParentId="16270" CreationDate="2017-07-12T16:36:35.097" Score="2" Body="&lt;p&gt;Have you considered a solar oven? The kind that sits in the sun and is heated by the sun's rays with no electricity or other power source. For cooking it'd be more efficient than a PV system. Obviously it wouldn't work at night so it might not meet your needs.&lt;/p&gt;&#xA;" OwnerUserId="12131" LastActivityDate="2017-07-12T16:36:35.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16277" PostTypeId="1" CreationDate="2017-07-12T18:46:11.063" Score="3" ViewCount="1932" Body="&lt;p&gt;Electric motor bikes were sold initially on the idea that there would be lesser maintenance and it would remove the requirements of the gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, a lot of people like having a geared system. Some argue it makes it more efficient during cruise mode or high speeds. I want to understand if this is correct. Are there added benefits to having a gear system on an electric motor bike?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hypothesis states that it is not dependent on the torque generated and has to do more with the load generated at the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: this is not a cycle. I'm talking about replacing IC engine powered two wheelers. No pedals. Not a cycle. No assisting. Purely run by the motor&lt;/p&gt;&#xA;" OwnerUserId="6215" LastEditorUserId="6215" LastEditDate="2017-07-15T07:45:32.693" LastActivityDate="2020-12-09T19:23:12.290" Title="Do gears make an electric motor bike more efficient?" Tags="|mechanical-engineering|electrical-engineering|electric-vehicles|" AnswerCount="6" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16279" PostTypeId="2" ParentId="16147" CreationDate="2017-07-12T20:29:26.363" Score="1" Body="&lt;p&gt;In MATLAB use ''svd'' to find the singular vectors corresponding to zero singular values of the Jacobian. The corresponding right singular vectors describe the input directions (i.e. movement in the joint space) that result in zero movement in the task space; the corresponding left singular vectors describe the output directions (in the task space) in which no movement is possible through any combination of movements in the joint space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Jacobian defines the relation between infinitesimal displacements $dx$ in the task space and infinitesimal displacements $dq$ in the joint space:&#xA;$$dx = \mathbf{J} dq$$&#xA;One way to calculate unachievable directions of (infinitesimal) motion from the Jacobian is to use it's singular value decomposition:&#xA;$$\mathbf{J} = U \Sigma V^T$$&#xA;The left singular vector matrix $U$ and the right singular vector matrix $V$ are orthogonal and span the (infinitesimal / local linearized) task and joint spaces respectively; the diagonal matrix of singular values $\Sigma$ determines the scaling between the individual input and output directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unachievable motions are scaled versions of singular vectors corresponding to zero singular value(s). A right singular vector $V$, corresponding to a zero singular value, implies that this input direction (i.e. movement in the joint space) will result in zero movement in the task space. A left singular value $U$, corresponding to a zero singular value, implies that (due to the zero scaling factor) movement in this output direction is not possible.&lt;/p&gt;&#xA;" OwnerUserId="11869" LastActivityDate="2017-07-12T20:29:26.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16280" PostTypeId="2" ParentId="16268" CreationDate="2017-07-13T02:50:56.720" Score="3" Body="&lt;p&gt;The thing will vibrate at all frequencies. It will vibrate more (be amplified) at frequencies that are close to a natural frequency.  Effectively, the system will act as a band pass filter (NOT a band notch filter)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a very similar experiment in graduate school many times. It wasn't a teenagers hi fi set or a wine class, but it was a random input to a vibratory system. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-07-13T02:50:56.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16281" PostTypeId="2" ParentId="16277" CreationDate="2017-07-13T06:12:45.860" Score="-3" Body="&lt;p&gt;As we know people day by day started preferring automatic transmission instead of manual. in case of electric bike we can use CVT for automatic transmission as transmission is required to separate engine and wheel for eg if we directly connect engine to the wheels then we have to face many problems such as speed control, when engine starts, wheel also start moving respectively. therefore we need a transmission or we can say it gear. but normal gear box couldn't work here and in case of electric bike it consumes more power that makes our bike less efficient so we have to use a medium to separate engine a wheel which consumes less power and while we accelerate then it will engage with engine and wheel simultaneously. so basically CVT are best type of transmission system available in market and try to put your engine near wheels for maximum efficiency and less power consumption. for more details you can check the concepts of Tesla model s sports car.  &lt;/p&gt;&#xA;" OwnerUserId="12135" LastActivityDate="2017-07-13T06:12:45.860" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16282" PostTypeId="1" AcceptedAnswerId="16302" CreationDate="2017-07-13T07:54:45.803" Score="0" ViewCount="646" Body="&lt;p&gt;I am trying to find reliable information and comparisons about the energy production of renewable energy technologies, specifically, how solar and wind alternatives compare to nuclear fusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it generally consider, as I have read, that nuclear fusion is the ultimate renewable energy? Is it really that clean? Would a fusion reactor be safe? Is the infrastructure required cheaper than solar and wind (considering long-term returns)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the answer to all or some of the above questions is affirmative, then why countries around the world invest anything at all in solar and wind? Why aren't the efforts directed to obtaining nuclear fusion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing a reasonable answer to these questions could be the inmediate mitigation of ongoing climate change and shortage of non-renewable resources. But then we should say that solar and wind alternatives are a 'transition' towards nuclear fusion. In this case, the main objective for countries worldwide should be obtaining nuclear fusion. But, in the future, when we harness nuclear fusion, wouldn't solar and wind become obsolete?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I just asked this question in &lt;a href=&quot;https://physics.stackexchange.com/&quot;&gt;Physics&lt;/a&gt; and it wasn't well received by some user because it is about 'politics'. Please, note that I am not interested in a political opinion, but a scientific assessment. The political and economic issues that may be involved are inevitable given the nature of the question. I posted here because now I think it is more appropriate for this site.&lt;/p&gt;&#xA;" OwnerUserId="12137" LastEditorUserId="12137" LastEditDate="2017-07-13T08:15:05.930" LastActivityDate="2017-07-14T20:31:14.273" Title="How is the energy production of solar and wind compared with the prospect of nuclear fusion?" Tags="|energy-efficiency|renewable-energy|nuclear-technology|infrastructure|" AnswerCount="2" CommentCount="6" ClosedDate="2017-08-17T12:20:27.903" ContentLicense="CC BY-SA 3.0" />
  <row Id="16283" PostTypeId="2" ParentId="16282" CreationDate="2017-07-13T09:28:44.850" Score="3" Body="&lt;p&gt;The reality of generating electricity from nuclear fusion on an industrial scale is still a very long ways off.  It is speculative high cost, high risk venture, with a potential high payoff far off into the future.   Test reactors alone are &gt;$100 million. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existing technical problems and challenges with nuclear fusion are very complex.  We're talking about generating the equivalent of the suns core at will in a controlled manner here on Earth vs. harnessing the energy of the actual sun from 93 million miles away.  Only one of these can be put into practice now and for the foreseeable future.  &lt;/p&gt;&#xA;" OwnerUserId="12138" LastActivityDate="2017-07-13T09:28:44.850" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="16284" PostTypeId="1" CreationDate="2017-07-13T11:58:02.100" Score="1" ViewCount="180" Body="&lt;p&gt;I am working on capturing air pollution particulate matter from Diesel Generators. Consider the following scenario. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A 15KVA generator that outlets smoke from the exhaust&lt;/li&gt;&#xA;&lt;li&gt;The smoke goes into our capture setup. &lt;/li&gt;&#xA;&lt;li&gt;After every 30-40 mins of operation we need to electro-mechanically clean the capture setup. The cleaner assembly takes approx. 60 seconds. &lt;/li&gt;&#xA;&lt;li&gt;For that duration we need to re-direct the flow of the exhaust to another outlet with some kind of timing/logic&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;i.e.  The best example is a ‘Y’ configuration attached to the exhaust, where the flue gas will flow only in one exhaust at a time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the better understanding I am enlisting the design parameters as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Flue gas velocity: 15 m/s&lt;/li&gt;&#xA;&lt;li&gt;Diameter of the pipe at inlet: 4 inches &lt;/li&gt;&#xA;&lt;li&gt;Diameter of the pipe at outlet: 4 inches&lt;/li&gt;&#xA;&lt;li&gt;Operational temperature: 450°C - 500°C&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am looking at a lowcost/robust way to solve this problem :) as we're building this prototype on a budget, hence we want to examine all the options/suggestions that might be suitable in achieving our above goal/purpose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I've already looked at the standard diverter valves- but they seem to be an overkill/very expensive(like 2500$)/not designed for our application where temp. is high.&lt;/p&gt;&#xA;" OwnerUserId="12071" LastEditorUserId="1832" LastEditDate="2017-07-13T12:32:01.560" LastActivityDate="2017-07-14T06:22:31.847" Title="Looking for elegant ways to divert airflow" Tags="|mechanical-engineering|fluid-mechanics|airflow|flow-control|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16285" PostTypeId="1" CreationDate="2017-07-13T13:44:45.417" Score="2" ViewCount="440" Body="&lt;p&gt;I see there are two different Wikipedia pages for &lt;a href=&quot;https://en.wikipedia.org/wiki/Strain_wave_gearing&quot; rel=&quot;nofollow noreferrer&quot;&gt;strain wave gearing&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Harmonic_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;Harmonic drive&lt;/a&gt;. To my knowledge they are the same. Harmonic Drive is the trademarked name of strain wave gearing by United Shoe Machinery Corp. Am I doing any mistakes here or what?&lt;/p&gt;&#xA;" OwnerUserId="12127" LastActivityDate="2017-07-13T20:56:04.160" Title="What is the difference between a strain wave gearing and a Harmonic drive?" Tags="|gears|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16287" PostTypeId="2" ParentId="15808" CreationDate="2017-07-13T15:24:48.413" Score="1" Body="&lt;p&gt;When I have looked at explosions in compressed air systems the pressures have been high like 2000 + psi.However , phosphate esters were synthetic oil that was recommended. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-07-13T15:24:48.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16289" PostTypeId="2" ParentId="16258" CreationDate="2017-07-13T16:58:33.417" Score="1" Body="&lt;p&gt;Depending on the type of spline you're defining, there are different characteristics that you'll need to identify the spline to either a machinist or to the customer who will need to design the mating part. Some of them are pretty straightforward and will be needed regardless of the type of spline, such as physical diameters (major and minor), number of teeth, the class of fit and spline shape (involute, flat, etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the most basic splines I work with, we can define it with only a major and minor diameter, the number of teeth, and the distance between two parallel tooth edges, because it's a flat side spline. However, something like an involute spline requires more information, such as a pitch diameter, pressure angel, effective tooth width, and distance between pins of a given size. This is because the complex shape makes it impossible to directly measure the spline geometry in a meaningful way, so we require both design dimensions and measurable dimensions. The spline table below is an example for an involute spline - the dimensions listed fully define the spline and allow us to measure it for quality purposes. I would include a table like this on any spline machining print so that you have the information readily available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jmw8Vm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jmw8Vm.jpg&quot; alt=&quot;Spline Table Example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are using a standard spline, such as a spline defined by ANSI B92.1, you could refer to it on a customer print as just the ANSI standard spline with only the information necessary to locate it in the standards table (should be pressure angle, class of fit, and either number of teeth or pitch diameter.) Then the standard will provide all other information, or the methods to calculate the information necessary to manufacture the spline.&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2017-07-13T16:58:33.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16290" PostTypeId="1" CreationDate="2017-07-13T18:46:22.040" Score="0" ViewCount="149" Body="&lt;p&gt;I'm trying to model the planar dynamics of a mass travelling about an epitrochoidal path due to being rigidly attached to a planet gear that rotates about a sun gear. Assume the gears to be circles that rotate about each other with no slip for reasons of simplicity.&lt;/p&gt;&#xA;&lt;p&gt;Both the sun and the planet can rotate. The input to the system is a torque on the planet carrier, pushing the planet parallel to its path about the sun. The output is the torque of the sun.&lt;/p&gt;&#xA;&lt;p&gt;As i have very little experience with modelling dynamics, i don't know where to start. I was thinking that separating the problem into a fixed and inertial frame may be best, but i'm not sure.&lt;/p&gt;&#xA;&lt;p&gt;How should i go about showing the output torque as a function of the input torque and the system constants (e.g. planet mass, sun diametre, etc.)?&lt;/p&gt;&#xA;&lt;p&gt;I drew a quick diagram of the system as a simple model with circles for reference.&lt;/p&gt;&#xA;&lt;p&gt;Cheers :)&lt;/p&gt;&#xA;&lt;p&gt;EDIT:&#xA;To clarify, i've already solved the kinematics of the problem and have equations governing position, velocity, and acceleration for the mass according to the rotation, rotational velocity, and rotational velocity acceleration of the planet and Sun. What I want to find out is the dynamics. Specifically how torque on the planet carrier will affect torque on the sun.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v3fQZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v3fQZ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11435" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-07-18T12:24:40.630" Title="Dynamics of Mass in Epitrochoidal Motion via Ringless Planetary Gear System" Tags="|mechanical-engineering|gears|dynamics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16291" PostTypeId="1" CreationDate="2017-07-13T19:33:15.827" Score="2" ViewCount="112" Body="&lt;p&gt;I am looking to determine the critical speed for a shaft with universal joints at each end, where the halves are joined by telescoping profile members. I can calculate the critical speed of each member individually, but I'm running into problems determining how to combine these numbers to get a critical speed for the whole assembly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shafts do operate at variable lengths, but I can prescribe the maximum length, so I will know the configuration at the worst-case scenario. The shaft can be divided into 3 sections, one with the inner member, one where the inner and outer member overlap, and one with the outer member. I have lengths and area moments of inertia for each cross-section, so I could calculate individual critical speeds, but I can't assume these simply combine linearly since the deformation of one section will contribute to the next. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried some calculations using the Dunkerley equation -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{\omega^2}=\frac{1}{\omega_1^2}+\frac{1}{\omega_2^2}+\frac{1}{\omega_3^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the numbers I got didn't seem to square with my experience in reality. Is this the correct way to do this, in which case I should look at other factors to address my vibration concerns with these shafts? Or is there another way to combine the three critical speeds to obtain a max speed for the whole shaft? (Note also that I don't need an exact answer necessarily - if there is a straightforward way that will give a decent estimate or the form of an equation, I can use existing data and experience to improve that estimate.)&lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2017-07-13T19:33:15.827" Title="Combining critical speeds for a shaft of variable cross-section" Tags="|mechanical-engineering|dynamics|vibration|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16292" PostTypeId="2" ParentId="16290" CreationDate="2017-07-13T19:46:31.380" Score="1" Body="&lt;p&gt;This may be what you are looking for - graph and source below :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sBzEE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sBzEE.png&quot; alt=&quot;epitrochoidal path &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.researchgate.net/figure/229731352_fig7_Fig-710-A-graph-of-the-epitrochoidal-path-that-the-moving-pins-follow-relative-to-the&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.researchgate.net/figure/229731352_fig7_Fig-710-A-graph-of-the-epitrochoidal-path-that-the-moving-pins-follow-relative-to-the&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : the following have some examples of the equations necessary&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mechdesigner.support/epitrochoid-the-curve.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mechdesigner.support/epitrochoid-the-curve.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mechdesigner.support/maths-fb-velocity-eqtns-2d-cam&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mechdesigner.support/maths-fb-velocity-eqtns-2d-cam&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-07-14T07:46:54.543" LastActivityDate="2017-07-14T07:46:54.543" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16293" PostTypeId="2" ParentId="16285" CreationDate="2017-07-13T20:56:04.160" Score="1" Body="&lt;p&gt;To quote the opening line of your &lt;a href=&quot;https://en.wikipedia.org/wiki/Harmonic_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;linked wikipedia article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;A harmonic gear is a strain wave gear&quot;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The first line is quite explicit in saying that they are the same thing.  The second line says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Harmonic Drive&lt;/em&gt; is trademarked by the Harmonic Drive company. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't really see any reason to assume they are not the same thing.  &quot;Harmonic Drive&quot; is the brand name.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-07-13T20:56:04.160" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16296" PostTypeId="2" ParentId="16284" CreationDate="2017-07-14T06:22:31.847" Score="1" Body="&lt;p&gt;Elegant, robust, and low cost don't often occur at the same time ;-) What you are wanting to do is not trivial, especially reliably at high temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would start with an &quot;automatic butterfly damper&quot;. Most the &lt;a href=&quot;http://www.homedepot.com/p/Suncourt-6-in-Automated-Damper-Normally-Closed-ZC106/202824481&quot; rel=&quot;nofollow noreferrer&quot;&gt;available dampers&lt;/a&gt; will be for hvac equipment and it will take two to achieve your diversion functionality, but they are relatively low cost. You will probably need to cool the actuator or increase its distance from the pipe as it will not be rated for the temperatures you are running it at. Here is &lt;a href=&quot;http://stores.hvacexpress.com/zone-damper-4-normally-closed-120-volt/&quot; rel=&quot;nofollow noreferrer&quot;&gt;another damper&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also look at fabricating a more elegant solution yourself and just source the &lt;a href=&quot;https://www.zoro.com/johnson-controls-electric-actuator-70-in-lb-4-to-122f-m9108-gga-2/i/G5077414/&quot; rel=&quot;nofollow noreferrer&quot;&gt;actuator(s)&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.ebay.com/itm/CAROLINA-CONVEYING-FLAP-TYPE-4-DIVERTER-VALVE-PNEUMATIC-ACTUATED-/131649288104?_trksid=p2385738.m2548.l4275&quot; rel=&quot;nofollow noreferrer&quot;&gt;pneumatic conveying diverter valve&lt;/a&gt; would likely work well too as they usually handle pretty high heat, but sounds like they would probably not be in your price range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the unit can be at the end of the exhaust system and you dont need a perfect seal, you could look at using a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B007SJAHW2&quot; rel=&quot;nofollow noreferrer&quot;&gt;linear actuator&lt;/a&gt; or two to physically move it away from the end of the exhaust pipe.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-07-14T06:22:31.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16300" PostTypeId="1" CreationDate="2017-07-14T15:05:16.120" Score="2" ViewCount="55" Body="&lt;p&gt;I am designing a 1m thick raft slab and the drainage guys have thrown in 100mm diameter cast in pipes all over the raft. See Attached images 1 &amp;amp; 2. Wondering how this will affect the shear in the raft and how i can go about calculating required shear reinforcement for a) the column and b) the pipes. I have read the presence of the pipes can be ignored however this doesn't seem right to me especially when columns are surrounded by pipes. Any feedback or pointers to get me in the right direction would be greatly appreciated. the Raft has an H20-150 Basic Matt throughout T&amp;amp;B.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/VdZ9V.jpg&quot; alt=&quot;Image 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/vN91N.jpg&quot; alt=&quot;Image 2&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="10969" LastActivityDate="2017-07-14T15:05:16.120" Title="Affect of Cast in Pipes in a 1m thick Raft Slab on shear" Tags="|structural-engineering|structural-analysis|foundations|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16301" PostTypeId="1" AcceptedAnswerId="23148" CreationDate="2017-07-14T16:49:05.313" Score="2" ViewCount="174" Body="&lt;p&gt;does anyone have any resources for designing a bulk material handling system for an injection molding outfit? Currently we get material in gaylords and vacuum it to the hopper on each molding machine independently. I would like to start buying pellets by the truckload and stored in a silo to be fed to each machine from a central location. I am unsure of what equipment I would need other than the silo itself to do such a thing. Any leads?&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-08-10T04:42:35.803" Title="Designing bulk material handling system for injection molding" Tags="|mechanical-engineering|manufacturing-engineering|plastic|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16302" PostTypeId="2" ParentId="16282" CreationDate="2017-07-14T19:27:14.403" Score="2" Body="&lt;p&gt;&lt;strong&gt;Fusion is widely extolled by advocates as being safe (but is that collective opinion justified)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would prefer to begin my discussion comparing fusion with renewable energy systems with an examination of safety as in my opinion, safety should always come first when we compare energy alternatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Reasons offered why Fusion should be Safe - &#xA; - Only tiny amounts of fusion fuel are inside a fusion reactor at any one time. Continuous operation of a fusion power plant is maintained by continual refuelling with the fuel mixture (deuterium-tritium or deuterium-deuterium), so the fuel inventory in the plasma chamber at any time is gram quantities or less and sufficient only for about one minute of operation. The ITER tokamak will contain only grams of fusion fuel as it runs. A NIF laser fusion capsule contains only about 0.17 milligrams of Deuterium-Tritium fuel (Basic premise – If the fuel is not inside the reactor, it can’t even theoretically explode inside the reactor or become a factor in an accident)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-No greenhouse gasses produced &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;No reactor meltdowns or decay heat burn downs (Chernboyl, Three Mile Island, Fukushima incidents cannot happen)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No mining of fuel (fuel is extracted from water)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Billions of years of fuel available on earth (versus hundreds for fossil fuels)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1/1000ths of the radioactive waste of fission power. Nearly all materials become activated to some degree by energetic neutron bombardment. Neutron reactions in DT fusion reactors will inevitably create radioisotopes. The principal radioactive materials present in a DT or DD fusion reactor will therefore be tritium and neutron activated structural materials surrounding the reaction volume in the fusion chamber. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Radioactive waste, mostly neutron activated fission chamber structural material remains radioactive ~20 years versus hundreds of thousands of years for Minor Actinides in spent fuel produced by fission reactors. DT fusion reactors would however burn and require storage of significant quantities of Tritium, a moderately radioactive nuclear isotope of hydrogen. Tritium gas is hard to secure and small leaks in reactor tanks and plumbing could result in some loss of tritium to the environment. Tritium decays while emitting weak Beta radiation and under most circumstances is not seriously hazardous to humans. The deuterium-deuterium (DD) fusion process, dispensing with tritium as a fuel and the requirement for Tritium storage. DD fusion does however produce Tritium as part of the DD fusion reaction, but this Tritium does not tend to accumulate but is immediately burned in place inside a DD fusion reactor. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Fuel for energy would no longer be a geopolitical issue (huge benefit for humanity) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No nuclear weapons from waste products (again, huge benefit for humanity)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To have success in fusion and renewable energy innovation -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you have to get the BIG PICTURE right&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ultimate potential of D-D Fusion is very great and eclipses the potential of all other forms of energy production including nuclear fission and renewable energy systems-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fusion Energy longer than the earth has existed or the sun will burn&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complete conversion of deuterium nuclear fuel releases an energy content of 250 x 10^15 joules per metric ton of deuterium. The quantity of deuterium in the world’s oceans is estimated at 4.6 x 10^13 metric tons. Deuterium present in seawater will yield around 5 x 10^11 TW-years of energy. In the year 2016 the entire planet consumed around 17 TW-years of energy, which means that the energy content of the deuterium in seawater would be enough for 29.5 billion years of energy supply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give all 10 billion people expected to live on the planet in 2050 the level of energy prosperity we in the developed world are used to, a continuous average use of power of 6 kilowatts per person as is typical in Europe, we would need to generate 60 terawatts as a planet—the equivalent of 900 million barrels of oil per day. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time since the earth first formed = 4.54 billion years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time until the sun burns out = 5 billion years. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deuterium in the sea is capable of completely powering planet earth at a level of 60 Terawatts for 8.33 billion years &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(longer than the earth has existed or the sun will burn).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;D-D fusion of pure Deuterium separated from sea water is a practical fusion reaction and practical fuel for ICF fusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When talking about near term prospects for fusion, fusion scientists frequently neglect to mention that mankind's first successful terrestrial release of fusion energy (Ivy-Mike experimental test shot of 1952) produced in excess of an exawatt of fusion power from D-D (not D-T) fusion. Use of cheap and ubiquitous Deuterium separated from sea water was the very first fusion reaction to work. Futher, it deserves to be pointed out that Ivy-Mike D-D fusion worked the very first time it was tried, and never failed to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hundreds of Magnetic Confinement Fusion devices that have been built over the course of 60+ years have never once even for milliseconds produced break-even fusion power or fusion with energy gain. ICF fusion of cheap and common Deuterium from sea water worked the first time it was tried (Ivy-Mike 1952) and thereafter never failed to work in hundreds of underground nuclear tests held at NTS and on the Pacific Test Range (Marshall Islands). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large tokamak magnetic fusion designers often tend to discount the deuterium-deuterium (D-D) fusion reaction because it is not possible to achieve the necessary fusion reaction rates and break-even energy with D-D fusion using tokamaks and stellarator devices such as we currently are able build. The situation is different, however, with Inertial Confinement Fusion which typically employs fusion plasmas that are literally 10^11 times more dense than the typical ion plasma densities used in MCF tokamaks or stellarators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power than can be drawn from a plasma at fusion conditions is proportional to the second power of the plasma ion density. Even small improvements in plasma ion density will result in significant improvements in the performance of a fusion experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an impressively large Range of plasma density in current and planned fusion experiments&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bGzJp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bGzJp.png&quot; alt=&quot;Plasma ion density of typical current fusion experiments&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is literally a factor of 10^10 to 10^11 difference in plasma ion density between MCF and ICF fusion experiments. This large difference has implications for commercial fusion technology as the denser the plasma, the more power that can be realized from a fusion burn and the smaller the physical fusion reactor can be while achieving fusion energy with energy gain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The D-D reaction takes place at higher temperatures than can typically be achieved in magnetic confinement tokamaks and typically has a lower fusion cross section leading to lower fusion reaction rates on earth relative to D-T fusion . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;D-D fusion is more complex and proceeds with multiple fusion side reactions, but when all of the side-chains run to completion, D-D fusion actually produces more energy than D-T fusion per starting mass of reactants. D-D fusion produces T and He3, which in a secondary reaction burn with D. This was practically demonstrated in early thermonuclear tests like the 15 Megaton Ivy Mike fission triggered deuterium experimental test in 1952. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;D-D fusion does normally produce Tritium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D-D Fusion of Deuterium fuel produces energy through four reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D + D -&gt; He-3 + n + 3.268 MeV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D + D -&gt; T + p + 4.03 MeV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(side chains)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D + T -&gt; He-4 + n + 17.588 MeV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He-3 + D -&gt; He-4 + p + 18.34 MeV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net effect of these four fusion reactions taken together is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6 D -&gt; 2 He-4 + 2p + 2n + 43.243 MeV&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Fusion and neutrons -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DT and DD fusion release large amounts of their fusion energy in the form of fast neutrons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task of shielding fast neutrons requires&lt;/p&gt;&#xA;&#xA;&lt;p&gt;• Fast Neutrons (E &gt;1 MeV) must first be thermalized with high&lt;/p&gt;&#xA;&#xA;&lt;p&gt;neutron scattering materials (typically materials that have a high percentage of  light atoms like hydrogen)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;• These thermal neutrons can then be absorbed or shielded with material having high cross sections for neutron absorption&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absorption of thermalized neutrons can be done by relatively light materials like polyethylene plastic unlike gamma radiation that require lots of high-z material (lead, concrete, water) to absorb the penetrating gamma. LLNL's Dr. Ralph Moir has pioneered several molten salt fusion reactor concepts (HyLife II and PACER Revisited). A thick liquid falling wall of molten salt loaded with a neutron absorber like 50 micron particles of Boron Carbide as a slurry can absorb damaging fast neutrons. Such a thick falling wall of molten salt can effectively protect fusion chamber structure and largely prevent neutron activation of fusion reactor materials. This permits molten salt ICF fusion reactors to have long 30+ year commercial lifetimes and to at reactor end of life produce little or no neutron activated waste.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Energy Density&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue of producing fusion with energy gain while simultaneously achieving a high enough power density are dominant themes of the near future in selection of which energy technology is best suited to produce competitive  power plants that communities will want to build to produce safe and dependable  power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comparison of fusion and “renewable” energy systems from the standpoint of energy density&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source              Joules per cubic meter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solar Radiation =   0.0000015&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geothermal  =       0.05&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wind at 10 mph =        7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal water =       0.5-50&lt;/p&gt;&#xA;&#xA;&lt;p&gt;deuterium in the form of D2O (heavy water) =     6.88 x 10^10 MJ/cubic meter &lt;/p&gt;&#xA;&#xA;&lt;p&gt;note: about 20% by mass of heavy water or D2O is deuterium&#xA;Deuterium is a gas at standard temperature and pressure but heavy water D2O is a easily and safely stored as a liquid. D-D fusion is in excess of million times more energy dense than any renewable energy system.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Practical fusion will always be 50+ years ago (not 50 years away)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mankind came into possession of a practical way of generating energy from fusion over 50 years ago with the Ivy-Mike nuclear test that produced fusion energy from pure Deuterium via DD fusion.&#xA;Practical fusion will always be 50+ years ago (not 50 years away)&#xA;Today, there are smaller pure fusion devices designed to make clean energy (not blast effects) from pure DD fusion of Deuterium separated from sea water. One such design is called mini-Mike, which produces a small predictable controlled energy yield of 250 GJ.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Note: Deuterium separated from sea water is totally non-radioactive and fusion of this fuel produces only totally non-radioactive nuclear waste&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Inertial Confinement Fusion has already been PROVEN to work and produce fusion ignition and fusion with energy gain&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both LANL and LLNL were successful in producing full ICF fusion ignition during the last years of the cold war era. ICF fusion is experimentally demonstrated and proven to work and this places ICF fusion in a different category than any other form of fusion..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy needed to ignite an inertially confined thermonuclear fusion reaction in liquid (or solid) deuterium-tritium (DT) is not that large; it is on the order of not more than 10 MJ or about the same amount of chemical energy stored in about 1.25 cups of automotive gasoline.&#xA;The problem is that this energy must be compressed in space (focused down to an area less than a 2 mm) and in time (to less than 3 nanoseconds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure Inertial Confinement Fusion that does not use nuclear fission to produce the conditions for fusion is today driver limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is still not experimentally possible to build a laser (or ion particle accelerator) large enough to produce DT fusion ignition. Still, many people, including Congress, in the late 1980s and early 1990s wanted to know for certain if inertial confinement fusion would ultimately work and actually produce net energy from fusion. To answer this question, in the final few years on underground nuclear testing, both LANL and LLNL designed a series of test shots called Halite-Centurion. Halite-Centurion series shots were fusion related add on shots piggy-backed onto shots already on the schedule. These shots were designed to utilize a small portion of the X-rays produced from the primary of an experimental device through a line of sight to a remote fusion experiment housed some distance away in the underground experimental test canister. Lasers and Ion-beam fusion drivers such as were available at that time could not provide the driver energy required to produce fusion ignition - but X-rays from a remotely ignited fission device could provide the driver energy needed (&gt;10 MJ energy delivered into a spot of about 2 mm in a time of less than 3 nanoseconds) .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Halite-Centurion fusion experiments in the Nevada desert worked and reliably and repeatedly and produced full fusion ignition of small sub-gram samples of DT fuel (in small filled spheres). These experiments were once classified but DOE allowed senior scientist Dr. John Lindl to declassify and reveal about half of the fusion related project information. Halite-Centurion experiments are referred to in a fundamental document for the design of the NIF facility: the 91 pages paper of John Lindl entitled “Development of the indirect-drive approach to inertial confinement and the target physics basis for ignition and gain”, published in 1995 in the AIP/Physics of plasmas. Halite-Centurion success was used by LLNL managers to sell the NIF program to Congress in the mid-1990s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is lacking in current ICF fusion experiments to support building of practical ICF fusion power plants is a fusion driver with characteristics close enough to the driver employed by LANL and LLNL in Halite-Centurion tests to permit full fusion ignition with energy gain. A driver capable of delivering greater than 10 MJ into a focused area less than a 2 mm in time of less than 3 nanoseconds. The current NIF driver laser driver is capable of delivering about 2 MJ total energy and is only able to achieve &quot;alpha heating&quot; phase of fusion operation - a stage that immediately precedes fusion ignition. LASNEX and HYDRA computer simulations on which LLNL based the engineering design of the NIF laser were overly optimistic in predicting the required fusion driver characteristics to produce ignition. Failure of the LASNEX and HYDRA simulations resulted in NIF being built too small to achieve the goal of fusion ignition and for NIF to serve as the basis of LIFE ICF fusion power plant technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source documents (with links where possible) -&#xA;NY Times article published at the time of Halite-Centurion field tests - Secret Advance in Nuclear Fusion Spurs a Dispute Among Scientists&#xA;&lt;a href=&quot;http://www.nytimes.com/1988/03/21/us/secret-advance-in-nuclear-fusion-spurs-a-dispute-among-scientists.html?pagewanted=all&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nytimes.com/1988/03/21/us/secret-advance-in-nuclear-fusion-spurs-a-dispute-among-scientists.html?pagewanted=all&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following document contains what John Lindl was permitted to release publicly regarding Halite-Centurion ICF by DOE&#xA;&quot;Development of the Indirect‐drive Approach to Inertial Confinement Fusion and the Target Physics Basis for Ignition and Gain.&quot; John Lindl. Page: 3937. AIP Physics of Plasma. American Institute of Physics, 14 June 1995.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sdpaL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sdpaL.png&quot; alt=&quot;Light ion driver pure fusion power plant concept&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12152" LastEditorUserId="12152" LastEditDate="2017-07-14T20:31:14.273" LastActivityDate="2017-07-14T20:31:14.273" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16304" PostTypeId="1" CreationDate="2017-07-14T23:01:31.827" Score="1" ViewCount="32" Body="&lt;p&gt;I am a recent graduate who is working on a brand new hotel project. I am helping to design the electrical system. I was wondering if anyone who is in the field would be willing to help me by giving me step by step advice on the entire process of designing an electrical system for a commercial building (A hotel in particular)? Any advice and guidance will be appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="12081" LastActivityDate="2017-07-14T23:01:31.827" Title="New electrical engineer seeking advice on electrical design for hotel!" Tags="|design|electrical|" AnswerCount="0" CommentCount="3" ClosedDate="2017-07-14T23:21:39.730" ContentLicense="CC BY-SA 3.0" />
  <row Id="16305" PostTypeId="2" ParentId="16277" CreationDate="2017-07-14T23:20:37.453" Score="1" Body="&lt;p&gt;Consider this Torque Curve from the mathworks (&lt;a href=&quot;https://www.mathworks.com/help/physmod/elec/ref/servomotor.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mathworks.com/help/physmod/elec/ref/servomotor.html&lt;/a&gt;):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GgcWD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GgcWD.png&quot; alt=&quot;motor curve&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://www.mathworks.com/help/physmod/elec/ref/serv_torque_speed_env2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;mathworks.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;It's their idea of available torque for a brush-less AC motor.  At low speed, you can get the max torque, but as speed increases, available torque decreases in a way that looks logarithmic.  Efficiency under this curve will change, but I think the bigger issue will be the loss of torque output at high speeds.  If you think you'll use the motor more at low speeds (launch-assist is great on a bicycle), you're probably alright to have a single high-speed pedaling gear for cruising and getting the bike rolling with the motor.&lt;/p&gt;&#xA;" OwnerUserId="4058" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-02-27T15:57:38.927" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="16306" PostTypeId="1" AcceptedAnswerId="16308" CreationDate="2017-07-15T00:10:44.897" Score="1" ViewCount="5205" Body="&lt;p&gt;Hy everyone, new here, I don't know much about how high pressure water pumping works - ( i'm just a programmer ) but i need to know if we can create water pressures &lt;code&gt;above 100 000 psi&lt;/code&gt; for a crazy &lt;code&gt;water-jet cutting project&lt;/code&gt; i have in mind :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the maximum pressure we can reach inside a single pump / or by connecting pumps in series or parallel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;strong&gt;would like to go to pressures of 500 000 psi or even higher if possible&lt;/strong&gt;. I have no practical intuition of what that actually means :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i know that the higher the pressure the better - my reasoning is - if it takes &lt;code&gt;5 minutes&lt;/code&gt; to cut a big thick block of iron at &lt;code&gt;100 000 psi&lt;/code&gt;- will take &lt;code&gt;1 minute&lt;/code&gt; with &lt;code&gt;5 times that pressure&lt;/code&gt; - ( assume that the other parameters will be scaling as well.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taken &lt;a href=&quot;http://waterjets.org/archive/buyers-guide/general/choosing-a-pump/#crankshaft_vs_intensifier_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At pressures of 60,000 PSI (4,100 bar) and higher, metal fatigue&#xA;  becomes a serious issue with many components. Although pumps that can&#xA;  reach 100,000 PSI have been around for many years, nobody runs them at&#xA;  such pressures because of the extreme maintenance issues involved. For&#xA;  this reason, most manufacturers purposely limit their pumps to below&#xA;  60,000 PSI (4,100 bar).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This sounds serious - but i'm not too concerned about efficiency here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested to &lt;strong&gt;know more about the theoretical limits&lt;/strong&gt; that exist with creating such pressures.. not so much if a pump doing this is available or not for commercial purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i have in mind is suppose to be a very serious industrial machinery - custom build - all it matters is to cut as fast as possible. All the other concerns like cost or &lt;code&gt;metal-fatigue&lt;/code&gt; are secondary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also can you please help me picture how big the dimensions of this &lt;code&gt;pump / system of pumps&lt;/code&gt; will actually be? What volume it needs to take?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much :) Have a nice day!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Form @alephzero and @Solar Mike comments (thank you both) i realized there is an important difference between the pressure and the flow rate. This is an xy question and my question title is bad.. sorry about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is - i need this machine to cut faster - as faster as it can get. I don't think we reached the limits of what's possible with our water jet cutting technologies - and if we did, i want to understand why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XoYeb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XoYeb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what are the parameters we need to change - for this cut to happen 5 times faster?&lt;/strong&gt; Or 10 times faster - or whatever i pick. When we will reach the maximum theoretical and then the practical limit of that this pumps can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It will help if we put pumps in series or parallel?&lt;/strong&gt; &lt;strong&gt;It will help if we increase the pressure?&lt;/strong&gt; &lt;strong&gt;Or the flow rate?&lt;/strong&gt; (but how to increase the flow rate - if the diameter is fixed) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that by increasing the pressure we can make the cut faster. @Solar Mike comment made me realize that i don't need pressure - i need a 5 times higher flow rate - in order to cut 5 times faster. But because the orifice is a certain diameter - i think we can't do it without increasing the pressure.(maybe i'm wrong about this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for now don't worry about how the sand will be mixed with this higher speed jet. Based &lt;a href=&quot;https://youtu.be/YBqgofmdKc8?t=91&quot; rel=&quot;nofollow noreferrer&quot;&gt;on this video&lt;/a&gt; it seems you can't mix sand if the water jet is at higher speed/ orifice is smaller - but i think i can get around that by combining more water jets into a single one inside a diamond chamber, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PdKjR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PdKjR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Not sure if this schematic will mess up the jet shape - but anyway the first problem is to figure out how to get that jet moving faster. Then we worry about how to add sand to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on @Mark answer (thank you) i understand now that is not even possible to build such a pump - because the the maximum pressure we can possibly have is &lt;code&gt;150 000 psi&lt;/code&gt; - if that &lt;code&gt;pump is made out of titanium&lt;/code&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no material stronger than that - except diamond, graphene and such -  which are not an option. (I &lt;a href=&quot;https://www.youtube.com/watch?v=iH_-SIE4zlM&quot; rel=&quot;nofollow noreferrer&quot;&gt;found a company that makes nozzles out of pure diamond&lt;/a&gt;, specially designed for water-jet cutting - but they don't make any pumps :)) - and for sure what @alephzero said about diamond strengths depends on the angle that force acts on it - applies here. )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to keep things simple let's stay with 100 000 psi for now. (such a bummer ..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I still want to cut faster. What can i do?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If cutting faster is related to flow rate - then i need to increase flow rate somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It follows that the only way to do that - is to increase the orifice diameter. (the other way - is to increase the pressure - which is not an option)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If i double the diameter - and increase the power appropriately - that will &lt;code&gt;quadruple the flow rate&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now because the &lt;code&gt;diameter is double&lt;/code&gt; - the &lt;code&gt;water-jet cross section has quadrupled&lt;/code&gt; - the &lt;code&gt;amount of material&lt;/code&gt; i need to cut out &lt;code&gt;has quadrupled also&lt;/code&gt; - the same as with flow rate. (not 100% sure about this - but is my first intuition)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i'm trying to understand now is: &lt;strong&gt;So what will happen to cutting speed by increasing the diameter - is this a zero-sum game ??&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, &lt;code&gt;trying to increase speed&lt;/code&gt; -&gt; i &lt;code&gt;increase the flow rate&lt;/code&gt; -&gt; by &lt;code&gt;increasing the diameter&lt;/code&gt; -&gt; which &lt;code&gt;which increases the jet cross section&lt;/code&gt;-&gt; which then   &lt;code&gt;increases the amount of material i need to cut&lt;/code&gt; -&gt; which &lt;code&gt;lowers the cutting speed all over again&lt;/code&gt;. &lt;strong&gt;So i go nowhere with this?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Or can be  beneficial to increase the diameter?&lt;/strong&gt; My intuition can't confirm anything here- i need someone to help me understand what will happen in the case i increase the diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to get out positive form this situation? Or is a hard theoretical limit like the one with pressure? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I rely don't like increasing the diameter - but seems we are out of options here - that's kind of sad..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks everyone, this question got out of hand here - but promise i will not ask more questions :)&lt;/p&gt;&#xA;" OwnerUserId="12155" LastEditorUserId="12155" LastEditDate="2017-07-15T21:37:51.040" LastActivityDate="2020-12-04T00:02:47.160" Title="What's the maximum water jet psi we can create?" Tags="|fluid-mechanics|pumps|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16308" PostTypeId="2" ParentId="16306" CreationDate="2017-07-15T13:34:35.913" Score="2" Body="&lt;p&gt;I think the theoretical limit is based on materials, as alephzero said.  You have to make it out of something, and the yield strength of the material will have to be greater than the internal pressure.  There is a practical limit as well, because there has to be a margin of safety - at such high pressures the system would fail catastrophically if the pump or piping ruptured.  If you use a factor of 3, which is fairly common, that means you need a material with a yield strength of 1,500,000 psi.  Steels are typically something like 70,000 psi.  Titanium might be 150,000 psi.  I don't know where you go next, but that's what you'll need to look for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you say that cost and metal fatigue are secondary, keep in mind that even if the machine cuts metal 5 times faster than anything else when it's running, it really doesn't gain you anything if it's shut down for maintenance and replacement of major components for four days every week. &lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2017-07-15T13:34:35.913" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16310" PostTypeId="1" AcceptedAnswerId="16312" CreationDate="2017-07-15T17:00:41.263" Score="1" ViewCount="96" Body="&lt;p&gt;I am researching a system that appears to have a hydraulic arm with some kind of &quot;handbrake&quot; as pictured below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;System Description:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Table that can raise and lock into place&lt;/li&gt;&#xA;&lt;li&gt;Locking mechanism triggered by user releasing lever&lt;/li&gt;&#xA;&lt;li&gt;Locking mechanism released by user pulling lever&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Yo3F8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yo3F8.png&quot; alt=&quot;desk&quot;&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm new to hydraulics, and I don't know what kind of component this is called. I've searched hydraulic cylinder with handbrake and pneumatic cylinder, but I'm not getting any good information. Can anyone point me in the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="12163" LastActivityDate="2017-07-16T00:52:13.183" Title="Hydraulic Arm - Handbrake" Tags="|hydraulics|pneumatic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16311" PostTypeId="1" CreationDate="2017-07-15T23:33:24.707" Score="1" ViewCount="213" Body="&lt;p&gt;I know how to get the stiffness matrix of a beam with any type of end releases (hinges and rollers) by applying Guyan reduction (static condensation) to the beam stiffness matrix in local axes. However, by doing this, the rollers sliding direction is parallel to the beam local axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First question: Would I get the rollers parallel to the global axes if I apply the Guyan reduction to the beam stiffness matrix expressed in global (rather than local) axes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question: What if I wish to get a roller in a direction which is not parallel to neither local nor global axes? Is it just a matter of transforming the matrix in two steps? I mean: first step, apply a transformation matrix from local axes to roller axes. Apply Guyan reduction there. Then apply a second transformation matrix from roller axes to global axes, and you have the beam stiffness matrix with the desired end release, ready for assembly because it's in global axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be as easy as that? Is there any book covering this? (I found end releases explained in several books, but only in local axes).&lt;/p&gt;&#xA;" OwnerUserId="12165" LastActivityDate="2023-12-23T12:00:25.040" Title="Beam stiffness matrix with sliding end release in arbitrary angle" Tags="|beam|stiffness|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16312" PostTypeId="2" ParentId="16310" CreationDate="2017-07-16T00:52:13.183" Score="3" Body="&lt;p&gt;It is a gas spring. They are rated in units of pounds and travel in inches. Typically ranges are 10 to 200 pounds, and 6 to 24 inches. Here is one example :&#xA;&lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B002UCAHAE&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/JR-Products-GSNI-5300-60-Gas-Spring/dp/B002UCAHAE/ref=lp_16413861_1_3?s=industrial&amp;amp;ie=UTF8&amp;amp;qid=1500165912&amp;amp;sr=1-3&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particular one being shown in your diagram is a specialized version with a locking valve which keeps the gas spring compressed until the valve button is pressed. They are a less common because they are used in more specialized applications, whereas the plain devices are used anywhere a door, lid, or window might be raised easier with the aid of a spring.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-07-16T00:52:13.183" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16313" PostTypeId="1" CreationDate="2017-07-16T10:10:25.277" Score="1" ViewCount="15" Body="&lt;p&gt;I currently doing research on 3d printing on circuit boards. The technique we are using is to put a conductive material, I.e ceramics/metal in a gel/polymer, to print the suspension and then when is complete to sinter away the gel/polymer. The problem I am having is the suspension needs to be thixotropic. Furthermore it must have a high viscosity i.e 500 cpa under low shear and to quickly drop to low viscosity as the shear increases&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far the closest thing I have been able to find is in this &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0955221909003860&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; but if you look at the graphs none of them have a high enough viscosity and more importantly the drop to low viscosity under increasing shear is too shallow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to use models like the ones in this &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0377025714001463#s0055&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; but I am finding it quite difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any suggestions to find more materials like the one I require? Thanks&lt;/p&gt;&#xA;" OwnerUserId="10480" LastActivityDate="2017-07-16T10:10:25.277" Title="Help finding a thixotropic suspension with specific properties" Tags="|fluid-mechanics|fluid|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16314" PostTypeId="2" ParentId="16306" CreationDate="2017-07-16T15:17:14.647" Score="1" Body="&lt;p&gt;Based on your last edit adding reasoning about the effect of diameter / flow rate etc, I would suggest you should seriously look at cutting with two nozzles at the same time - your analysis based on all the other answers / comments doesn't leave much else.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-16T15:17:14.647" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16321" PostTypeId="2" ParentId="16206" CreationDate="2017-07-17T20:12:42.153" Score="1" Body="&lt;p&gt;ELF will likely be very problematic for you practically; remember that your antenna will have to be on the order of a wavelength to transmit or receive such a signal. Even for a 1/4 wavelength antenna, a 30 Hz antenna is 2500 km long. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ULF has some history of amateur radio use, so at least it is feasible - and there are commercial antennas made by Aaronia AG. Still, it may be difficult to find other commercial products (signal generators, amplifiers, etc). You may want to do a simple estimate of the cost of your proposed system including commercial parts before designing it. As in much of engineering, we must deal with what's possible rather than what is perfect.&lt;/p&gt;&#xA;" OwnerUserId="3583" LastActivityDate="2017-07-17T20:12:42.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16322" PostTypeId="1" AcceptedAnswerId="16346" CreationDate="2017-07-17T19:56:58.680" Score="1" ViewCount="100" Body="&lt;p&gt;This is part of an electric resin-extruding machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the part, and if possible, who makes it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any additional information regarding what the part is, or what the assembly is called, is greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kd05d.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kd05d.jpg&quot; alt=&quot;Connected plate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the back side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s79q7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s79q7.jpg&quot; alt=&quot;Plate back side&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's where it connects:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wt7l0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wt7l0.jpg&quot; alt=&quot;Connect to assembly&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UH29Y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UH29Y.jpg&quot; alt=&quot;Connected example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12183" OwnerDisplayName="techjwalker" LastEditorUserId="12183" LastEditDate="2017-07-18T21:36:47.147" LastActivityDate="2017-07-18T21:36:47.147" Title="Identifying a connected plate on a gear liquid-metering pump" Tags="|mechanical-engineering|gears|pumps|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16323" PostTypeId="2" ParentId="16322" CreationDate="2017-07-17T20:50:13.200" Score="3" Body="&lt;p&gt;I would guess that it may be a torque limiter - similar to shear bolts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the gear with the two slotted cheese or pan head screws freely rotate on its shaft ? Then with the plate fitted it is then locked? If so, then those &quot;teeth&quot; are designed to break first before something else if a blockage etc happens.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-17T20:50:13.200" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16325" PostTypeId="1" CreationDate="2017-07-17T23:01:09.957" Score="0" ViewCount="361" Body="&lt;p&gt;Background: This is for a science project I'm creating where I use an airpump connected to a sealed 5gal pail bucket to create a vacuum. That vacuum is then connected to water source to &quot;suck&quot; liquid into the bucket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem: After I shut the airpump off, liquid continues to flow from the water source to the vacuum due to siphoning. Any lowcost ideas on how to prevent the siphoning from happening when I shut down the airpump? Preferably would like a passive solution, but if it has to be electronic, so be it. TIA!&lt;/p&gt;&#xA;" OwnerUserId="12186" LastActivityDate="2020-08-13T11:03:50.373" Title="How to quickly equalize pressure in a low pressure container?" Tags="|pressure|vacuum|siphon|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16326" PostTypeId="2" ParentId="16325" CreationDate="2017-07-17T23:11:21.453" Score="0" Body="&lt;p&gt;I'm assuming you mean the air pump is connected to the sealed bucket seperately from the tube you use to suck water into the bucket? if so, maybe just glueing a rubber disc infront of the tube sucking the water (bucket end) should do the trick. When there's vacuum inside the bucket, water shouldn't have any trouble pushing the rubber flap open, but when pressure equalises, the  rubber would seat back over the tube, stopping the water flow. Can't say i've tried anything like this, just a theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit- again with the rubber flaps idea, if you drill a hole in the top of the bucket and glue another flap on the top of it, when you turn the pump off just lift the flap and that'll equalise pressure double quick!&lt;/p&gt;&#xA;" OwnerUserId="12187" LastActivityDate="2017-07-17T23:11:21.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16327" PostTypeId="2" ParentId="16136" CreationDate="2017-07-17T23:42:48.680" Score="1" Body="&lt;h2&gt;Calculating forces on teeth&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;While determining the forces acting on the teeth of each of the gears requires using Newton's 2nd Law, the calculation of these forces is surprisingly involved. This is because it is important to find the locations on the teeth where these forces act, and the directions of the forces: it is important to analyse the geometry of gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost always in industry will gears be &lt;em&gt;involute&lt;/em&gt; (with the notable exception of gears in clockmaking), and this indeed appears to be the case of the gears in your images. Involute gears have teeth whose shape are based on the involute curve, as shown in Fig. 1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zCO8T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zCO8T.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fig. 1:&lt;/strong&gt; Close-up of a typical involute gear's tooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The involute curve is displayed in cyan, and defines the shape of the part of the tooth that is designed to make contact with the teeth of the mating gear. It is this involute profile that is responsible for the smooth transmission of power from one gear to the next, and is the location of the force that one gear tooth exerts on another. For gears with sufficiently many teeth, the involute profile of the tooth starts at the base circle of the gear, shown in magenta in Fig. 1, and extends to the tips of the teeth. The involute curve can be expressed as a parametric equation with parameter $t\ge0$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;        \begin{matrix}&#xA;        x(t)=r_b(\cos{t}+t\sin{t}) \\&#xA;        y(t)=r_b(\sin{t}-t\cos{t}), \\&#xA;        \end{matrix}&#xA;\tag{1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $r_b$ is the base circle radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note $r_b=r_p\cos{\phi}=0.5Nm\cos{\phi}$, where $r_p$ is the pitch circle radius, $N$ is the number of teeth on the gear, $m$ is the &lt;em&gt;module&lt;/em&gt; and $\phi$ is the pressure angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Aside on undercutting:&lt;/strong&gt;&#xA;For standard gears with less than 18 teeth, the involute profile might not reach down to the base circle, significantly so as the number of teeth becomes very small. This &lt;em&gt;undercutting&lt;/em&gt; of the involute profile is often a bi-product for some of the manufacturing methods, and is generally undesirable. However, it can also be a necessary evil: for gears with low tooth counts, the lowest part of the involute profile obstructs the path of the tips of the teeth of the mating gear; undercutting provides the clearance necessary for the mating teeth to move without jamming. Fig. 2 shows the effects of undercutting on the involute profile for a gear with only 6 teeth:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9gdYt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9gdYt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fig. 2:&lt;/strong&gt; Close-up of a heavily undercut gear tooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the mathematical involute curve (in cyan) does reach down to base circle (in magenta), the corresponding involute profile on the gear tooth itself does not: some of the involute profile has been lost to undercutting. This results in a reduction of the useful geometry of the gear responsible for smooth power transmission. Instead, the involute profile reaches down to a circle called the &lt;em&gt;undercut diameter&lt;/em&gt; (in yellow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pressure Line:&lt;/strong&gt; The involute profile is important when considering the forces as a special feature arises from the contact of involute profiles of two teeth: as the gears rotate, the point at which teeth contact one another moves in a straight line called the &lt;em&gt;pressure line&lt;/em&gt;, as shown in Fig. 3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zjRw5.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zjRw5.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fig. 3:&lt;/strong&gt; Animation demonstrating how contact points move along the pressure line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Fig. 3, the base circles of the gears are shown in magenta. The pressure line is shown in red, with the points of contact as red dots. Note how the pressure line lies tangent to both base circles, and that the contact points move at a constant speed along this line. The blue circles are the pitch circles of the gears, and the orange circles are the outer diameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle the pressure line makes to the vertical is equal to the pressure angle. In the case shown in Fig. 3, the left gear is the driving gear, the right is the driven gear, and so the pressure line lie clockwise to the vertical. If the right gear was made the driving gear, the pressure line would instead lie anticlockwise. Note how the pressure line lies though the pitch point, i.e. the point at which the two pitch circles touch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most importantly, note how the involute profiles of the teeth at the contact point is normal to the pressure line. Therefore, the force that one tooth exerts on another will lie along this pressure line (hence its name!). Since the tooth forces run in a straight line, the distance between the line of action of the forces and the centre of one gear remains constant, this distance being equal to the base circle radius $r_b$. As a result of this, provided the input torque $T_i$ remains constant, the &lt;strong&gt;sum&lt;/strong&gt; of the tooth forces will be equal to a constant value. For a pair of mating gears will negligible friction, this value should equal $F=T_i/r_b$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this information, it is now possible to calculate the sum of the tooth forces one gear exerts on another: between the sun and planet gears and between the planet and ring gears. First, let's look at the sun-planet tooth forces. Let $T_i$ be the known input torque applied to the sun gear, and $F_{PS}$ be the unknown sum of the tooth forces that one planet gear exerts on the sun gear. Consider the forces acting on the sun gear, as shown in Fig. 4:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GWu5X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GWu5X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fig. 4:&lt;/strong&gt; Schematic showing the forces acting on the sun gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Fig. 4, the pitch circles of the sun and planet gears are shown, with the pressure lines included in red, noting that the pressure lines should go through the points the pitch circles meet, rotated by the pressure angle away from the common tangent of the pitch circles. Note how the sum of the tooth forces act along these lines (the exact location along the pressure line at which these forces act is not important). To determine the sum of the tooth forces, applying Newton's 2nd Law would appear to be a sensible course of action. However, it is often satisfactory to just simply enforce static equilibrium of moments, ignoring dynamic effects, especially if the gears are travelling at relatively low speeds where the inertia forces are much lower than the tooth forces. In this answer, only static equilibrium will be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When considering equilibrium, note how the fact that there are three planets engaging with the sun gear means that the total tooth force that each planet gear exerts on the sun gear (and vice versa) has effectively been split in three. The total sun-planet tooth forces for a three-planet planetary gear set turn out to be as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{PS} = \frac{T_i}{3\times 0.5N_S m \cos{\phi}}\tag{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N_S$ is the number of teeth on the sun gear. Note how the value of the module $m$ should be the same for all the gear in the planetary gear set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The static equilibrium of moments can be considered for one of the planet gears, and upon doing so, it can be seen that the sum of the tooth forces the ring gear exerts on one planet gear is equal to the sum of the sun-planet tooth forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are therefore able to determine the sun-planet and planet-ring &lt;em&gt;sum&lt;/em&gt; tooth forces. Note though that this is the sum: between any two mating gears, it is possible to have more than one point of contact at any one time, as can be seen in Fig. 3. If there are multiple points of contact, it can be assumed that the sum tooth force is split equally between the two contact points. The more contact points, the less load individual teeth will experience. (Note however that some gears may have considerable profile inaccuracies from its manufacturing, to the extent that, even if in theory there may be multiple points of contact, all the force falls onto just one contact point.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the force applied onto a single tooth, it is necessary to determine the number of contact points. Note that for a given pair of meshing gears, this number will fluctuate between two values over times; in Fig. 3, in can be seen that the number of simultaneous contact points varies regularly between 1 and 2. To determine this number, the &lt;em&gt;contact ratio&lt;/em&gt; is calculated: this is defined as the length of &lt;em&gt;line of action&lt;/em&gt; divided by the &lt;em&gt;base pitch&lt;/em&gt;. The line of action is the portion of the pressure line along which contact points exist, and the base pitch is the distance between two neighbouring teeth measured along the base circle, equal to $p_b=\pi m \cos{\phi}$. The contact ratio can be interpreted as the average number of contact points, and so can be rounded down and up to obtain the two integer values of number of contact points. Often the line of action is bounded by the outer diameters of the two gears, and so the contact ratio can be calculated as follows (equation adapted from &lt;em&gt;Dudley's Handbook of Practical Gear Design and Manufacture&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho_c = \frac{\sqrt{r_{o1}^2-r_{b1}^2}+\sqrt{r_{o2}^2-r_{b2}^2}-C\sin{\phi}}{p_b}\tag{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_{o1},r_{o2}$ are the radii to the outer circle of gears 1 and 2 respectively, $r_{b1},r_{b2}$ are the respective base circle radii, $C$ is the distance between the centres of the two gears, and $p_b$ is the base pitch. If one of the gears is an internal gear, as is the case for the ring gear, replace the appropriate radius of the outer circle with the radius of the inner circle of the internal gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula assumes that there is no undercutting. However, in situations where undercut does exist, it is still possible to determine the contact ratio by locating in the CAD model points in space where the contact points start and end (assuming it is possible to manipulate the CAD files as an assembly), and measuring the distance between those points to get the length of the line of action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the contact ratio, the force per tooth can determined.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Calculating contact stresses&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now with the force per tooth determine, it is also possible to determine the contact stresses that develop as a result of the teeth making contact. The contact stresses are the stresses that arise on the surfaces of the gear teeth about the points of contact, and are important to gauge so that surface damage can be avoided. The Hertzian model for contact stresses that arise due to line contacts will be useful for determining such stresses. According to Hertz's model, the maximum value of the contact pressure is given as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{\text{max}}=\sqrt{\frac{FE^*}{\pi w R}}\tag{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F$ is the force per tooth, $E^*$ is the &lt;em&gt;contact modulus&lt;/em&gt;, $w$ is the width of tooth face that contacts the other tooth, and $R$ is the effective radius of curvature of the contacting surfaces. $E^*$ can be calculated from the Young's modulus and Poisson's ratio of the materials that comprise the contacting surfaces of gears 1 and 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{E^*}=\frac{1-\nu_1^2}{E_1}+\frac{1-\nu_2^2}{E_2}\tag{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only variable that requires determining is the effective radius of curvature. This can be approximated from the radii of curvature of the individual contacting surfaces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\tag{6}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concave surfaces (such as those of internal gears) will have a negative radius of curvature. Each of the contacting surfaces are involute in shape, and so has the useful properties that the radius of curvature of a point on the involute curve is equal to distance from that point to the base circle, measured along the pressure line:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DlynH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DlynH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fig. 5:&lt;/strong&gt; How to calculate the radii of curvature, $R_1$ and $R_2$ at a contact point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that since the radii of curvatures vary on both involute profiles as the contact point moves along the line, and so the contact pressures will also vary. Therefore, it may require careful consideration of where the contact points will be. Take should be taken in making sure contact points are spaced apart even by the base pitch $p_b$, and to note the configurations at which the number of contact points change. Occasionally, for the sake of obtaining a reasonably useful value of contact stress quickly, it is possible to just consider a contact point lying at the pitch point where the pitch circles of the gears meet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contact pressures are useful to keep in might as excessive contact pressure can cause pitting of the gear surfaces, wearing, and fatigue.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Calculating bending stresses&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The tooth forces will also cause stresses to develop at the roots of the teeth, these stresses being the bending stresses. The complex geometry of gears make it impossible to use analytical methods to figure out the bending stresses. Instead, a Finite Element Analysis should be performed. While it would seem most proper to use the Hertzian contact pressure to define the forces to be applied onto the gear model, it is more than sufficient to use a concentrated force as the force acting on the gear tooth, the magnitude and direction of which as determined in the force calculations above. Bending stresses are important to observe as excessive bending stresses can cause to fracture and loss of the tooth, resulting in sudden failure that is more difficult to gauge and predict that damage due to contact pressures.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2017-07-17T23:42:48.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16328" PostTypeId="1" CreationDate="2017-07-17T23:53:05.210" Score="3" ViewCount="197" Body="&lt;p&gt;Let us assume that we are North American traffic engineers, and have the (hypothetical, plausible, yet pathological) intersection configuration depicted (not to scale and lacking lane lines, sorry -- I did not have time to build the model to scale or paint it!) below on our hands.  Both intersecting roads are four-lane, arterial roads with an AADT of 10000, a 40' bus or tractor-semitrailer for the design vehicle, and 35mph speed limits, while both railroad tracks are signaled, mainline railway tracks with 20+ trains per day, albeit also speed-limited to 35mph through the curves due to cant and curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nPsYk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nPsYk.png&quot; alt=&quot;very crude diagram of intersection configuration&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The traffic signals (not shown in the diagram) are standard 8 phase signals with pedestrian heads and leading protected left turns controlled by a NEMA TS-2, while the crossings are protected by flashing lights with cantilevers and gates controlled by grade crossing predictors.  The nearest crossing (west approach leg) provides 200' of storage space, and the far crossing (south approach leg) provides 300'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This configuration has been in service for several years after a road-widening project on the north-south road.  However, a near-miss happened recently at this intersection where a car had to dart off onto the shoulder to evade an approaching train as it had queued onto the west approach crossing, so preemption is being considered for this location. But, there's a problem -- it's possible that a train move on the south crossing could preempt the intersection, then another train could show up on the other track, causing a &lt;em&gt;second&lt;/em&gt; request for preemption before the traffic signal has finished servicing the first preemption request.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we handle this situation? I can see that dropping the second preempt request on the floor, which is the default behavior as I understand it of a traffic signal controller during preemption, could cause undesirable results including:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No track clearance green for the second train, leaving a possible queue on the track&lt;/li&gt;&#xA;&lt;li&gt;Queues tailing back into the intersection from the second crossing, possibly causing track clearance to fail for the first track due to the  queue from the second direction obstructing the intersection approach from the first direction direction.&lt;/li&gt;&#xA;&lt;li&gt;Illogical turn movements during the preemption hold if limited service operation is used -- these shunt cars into a queue waiting for the second crossing to clear&lt;/li&gt;&#xA;&lt;li&gt;Driver befuddlement at intersection operations with multiple trains present&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, simply servicing the second pre-emption request doesn't work either, because it could yield a track clearance green that allows movements that queue waiting for the first train, or worse, cut off the first train's track clearance green prematurely, leaving queued cars stranded on the tracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The railroads involved both are more than happy to give you all the advance preemption time you want, and are both providing 12-wire preemption interconnects to you.  Furthermore, there's enough traffic here that everyone wants limited service even during a double preemption, and enough trains here that double preemption would occur every three months or so based on data logged by the GCPs.&lt;/p&gt;&#xA;" OwnerUserId="964" LastEditorUserId="964" LastEditDate="2017-07-18T00:22:08.013" LastActivityDate="2018-01-14T17:23:59.720" Title="How can traffic signal controllers handle multiple conflicting preemption requests of equal priority without dropping one?" Tags="|rail|highway-engineering|traffic-intersections|traffic-light|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16329" PostTypeId="1" CreationDate="2017-07-18T01:37:39.510" Score="2" ViewCount="102" Body="&lt;p&gt;According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Betz&amp;#39;s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page on the Betz Limit&lt;/a&gt;, the outgoing wind velocity behind a wind turbine can be as low as 33% of the incoming wind velocity. However, this figure assumes that the turbine is operating at the theoretical Betz limit. Since most turbines operate below this limit, is there a straightforward formula for calculating a more realistic estimate of the wind velocity behind a turbine?&lt;/p&gt;&#xA;" OwnerUserId="12189" LastActivityDate="2017-07-18T06:52:17.987" Title="Estimating wind velocity behind a wind turbine" Tags="|airflow|turbines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16330" PostTypeId="1" AcceptedAnswerId="16338" CreationDate="2017-07-18T02:57:45.107" Score="0" ViewCount="127" Body="&lt;p&gt;I already know solution, I just want to understand something in part b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vertical cylinder fitted with a piston contains air initially at 120kPa and 100 celcius. The piston is restrained by a set of stops. The piston diameter is 0.1m, its mass is 5kg, and the ambient pressure and temperature are 100kPa and 20 celcius. The cylinder is then cooled as heat is transferred to the surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) At what temperature does the piston fall below the stops?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) How far down (as a fraction of its initial height) has the piston moved when the contents of the cylinder has cooled to the ambient temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a)&#xA;So this is assumed to be a 3-step process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5zWNL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5zWNL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ P_2 = P_{external} + \frac{mg}{A} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume is &lt;em&gt;supposedly&lt;/em&gt; constant from 1 to 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{P_1 V_1}{T_1} = m R = \frac{P_2 V_2}{T_2} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $ \frac{P_1}{T_1} = \frac{P_2}{T_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ T_2 = T_1 \cdot \frac{P_2}{P_1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since 2 -&gt; 3 is &lt;em&gt;supposedly&lt;/em&gt; a constant pressure process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \frac{P_3 V_3}{T_3} = \frac{P_2 V_2}{T_2} $ or $ \frac{V_3}{V_2} = \frac{T_3}{T_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So $ \frac{V_2 - V_3}{V_2} = \frac{\delta h}{h_2} $&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So my question is, if the ideal gas equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ P V = mR \cdot T$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m$, $R$ are constants&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if the temperature changes, then on the left hand side either pressure of volume can change, how is it known that from process 1 to 2, volume stays constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem postulates that heat of system will be cooled down by the surrounding temperature (20 C), nothing's said about neither pressure nor volume. So either of them can change... how does one know that volume stays the same from process going from 1 to 2?&lt;/p&gt;&#xA;" OwnerUserId="12190" LastEditorUserId="11867" LastEditDate="2017-07-19T22:51:35.690" LastActivityDate="2017-07-19T22:51:35.690" Title="Thermodynamic, ideal gas problem" Tags="|thermodynamics|gas|pistons|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16331" PostTypeId="1" AcceptedAnswerId="16332" CreationDate="2017-07-18T04:16:47.473" Score="1" ViewCount="496" Body="&lt;p&gt;I know major oil pipelines do exists either underground or under the seabed. However, these usually move crude/unrefined oil from refinery to refinery.  Yet we don't have refined petroleum or gas directly piped from refinery to the gas station? I mean we have done it with electricity, water, sewage, stormwater and piped LPG gas (for cooking), why do we still see petrol takers plying the roads to top up gas stations that you go to top up your car?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I understand this is easier to achieve in big cities and Urban areas rather than laying pipework for rural areas. But I so not yet have concrete info this is implemented in long term gas stations located in the cities. &lt;/p&gt;&#xA;" OwnerUserId="11378" LastEditorUserId="11378" LastEditDate="2017-07-20T06:48:44.340" LastActivityDate="2017-07-20T06:48:44.340" Title="Oil refinery pipeline direct to gas stations" Tags="|pipelines|petroleum-engineering|infrastructure|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16332" PostTypeId="2" ParentId="16331" CreationDate="2017-07-18T04:57:42.070" Score="1" Body="&lt;p&gt;To begin with, gas/petrol filling stations may or may not be located at one site for a long period of time. Dealing with a disused pipeline for a closed station will be a problem. Continually extending the pipeline network will cause disruptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the main issues, is like water distribution pipelines, pipelines for oil/petroleum products can burst, rust out and leak. Water leaks can be disruptive but are environmentally benign. Leaks from oil/petroleum pipelines can be environmentally catastrophic. No-one wants to have ground or ground water contaminated by oil/petroleum products.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-07-18T04:57:42.070" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16333" PostTypeId="2" ParentId="16329" CreationDate="2017-07-18T06:52:17.987" Score="2" Body="&lt;p&gt;The wind behind an operating turbine is turbulent. So no, there is no straightforward formula for calculating a more realistic estimate of the wind &lt;strong&gt;velocity&lt;/strong&gt; behind a turbine. The existence of such a formula is literally &lt;a href=&quot;http://www.claymath.org/millennium-problems/navier%E2%80%93stokes-equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;a million-dollar question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, calculating the approximate &lt;strong&gt;speed&lt;/strong&gt; of the air isn't too hard: we know the energy in any given mass of air is $\frac{1}{2}mv^2$. So if we know the speed upwind of the turbine, the density of that air, and the amount of electricity generated, we can calculate the energy in the air downwind of the turbine, and thus its speed. Assume 95% efficiency for the turbine mechanics &amp;amp; electrics (for kinetic energy that has been harvested and is being converted to electricity), and you'll be near enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The study of the wakes of wind turbines is a rich and complex one, with &lt;a href=&quot;https://scholar.google.co.uk/scholar?as_ylo=2016&amp;amp;q=%22wind%20turbine%22%20wake&amp;amp;hl=en&amp;amp;as_sdt=1,5&quot; rel=&quot;nofollow noreferrer&quot;&gt;quite a lot of work going on at the moment&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some wakes that nature has kindly visualised for us, at Horns Rev windfarm in Denmark (photo courtesy Vattenfall Wind Power)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b1FqM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b1FqM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2017-07-18T06:52:17.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16335" PostTypeId="2" ParentId="16330" CreationDate="2017-07-18T07:35:10.897" Score="0" Body="&lt;p&gt;The process 1&gt;2 is constrained to a change in temperature as the volume is held constant by the stops as in &quot;The piston is restrained by a set of stops.&quot;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-18T07:35:10.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16337" PostTypeId="2" ParentId="16325" CreationDate="2017-07-18T10:45:19.457" Score="0" Body="&lt;p&gt;You don't show a diagram of the setup, but change the pipe entry point to the tank - put it at the bottom then you should prevent the syphon effect. The depression you create with the pump will still cause the water to get into the tank and, while the depression holds the water will stay in the tank as long as it is sealed...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-18T10:45:19.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16338" PostTypeId="2" ParentId="16330" CreationDate="2017-07-18T10:49:07.823" Score="1" Body="&lt;p&gt;The solution for this problem relies upon the fact that the process is reversible, that is every state undergone by the system is in equilibrium.&#xA;Consider the initial state. Now the pressure inside the cylinder is 120 kPa. If we consider the equilibrium of the piston, 4 forces are acting upon it. They are: gravitational force(weight of piston) acting downward, atmospheric pressure acting downward,reaction force exerted by the stops acting downward andpressure inside cylinder acting upward. By doing the calculations we can see that the net downward force exerted by the atmospheric pressure and the weight of piston is 834 N. But the upward force exerted by the gas in the cylinder is 942 N. Hence a downward force of 108 N is required to be applied by the stops as a reaction force. Since reaction force is a surface force this force cannot be applied unless the piston and the stops remain in contact, ie the volume of the gas inside the cylinder remains constant.&#xA;The need for such a reaction force to maintain equilibrium will exist until the pressure inside the cylinder reduces to about 106.24 kPa . This is the process 1-2. &#xA;After reaching the state point 2 , the force exerted by the stops will be zero. So further reduction in pressure inside cylinder is impossible since it would disrupt the equilibrium. Hence, if temperature is again reduced the volume will decrease, as seen in process 2-3&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2017-07-18T10:49:07.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16339" PostTypeId="2" ParentId="13148" CreationDate="2017-07-18T12:12:31.677" Score="2" Body="&lt;p&gt;Under the menu &lt;em&gt;PlotCtrls &gt; Symbols&lt;/em&gt;, you can find the following window where the option &lt;em&gt;ESYS Element coordinate sys&lt;/em&gt; you are looking for is available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I2GMv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I2GMv.png&quot; alt=&quot;Symbols window&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As previously answered, it's also available via the following command:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;/psymb,esys,1 ! or 0 to remove it.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can look for &lt;code&gt;/help,/psymb&lt;/code&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="12198" LastActivityDate="2017-07-18T12:12:31.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16341" PostTypeId="2" ParentId="16328" CreationDate="2017-07-18T14:19:33.220" Score="1" Body="&lt;p&gt;As you might well imagine,  there is a metric f-ton (the technical term :-)  )  of research covering both queuing theory and prioritization analysis.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct, if unsatisfying answer to your title question is: there is never acase where two items have exactly equal priority, just as in American baseball there's never an exact tie between ball reaching glove and runner's foot touching bag.  If your algorithm produces exact matches, then shame on you, essentially. Take a  couple steps back and decide how to assign a priority earlier in the sequence,even if it involves a random assignment,so that there is never a conflict which leads to undesirable outcomes (like a crash).&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-07-18T14:19:33.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16342" PostTypeId="1" CreationDate="2017-07-18T15:39:10.620" Score="0" ViewCount="343" Body="&lt;p&gt;I'm interested in building a ram pump to pump water from a creek that runs at the bottom of my property to the top where i'm going to make my little homestead. I've seen allot of stuff online on building DIY ram pumps and I had a few questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I have to use a metal feed pipe as people suggest or can I use plastic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand it, the distance the water can be pumped is a function of the force that the water in the feed pipe can exert. The rule of thumb is to make the feed pipe twice the output pipe, but if I had two feed pipes, would that not  double the force, and there fore double the distance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the purpose of the stand pipe, will a stand pipe increase the distance the pup will pump?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some of the resources I've found &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.bae.ncsu.edu/extension/ext-publications/water/use/ebae-161-92-ram-pump-jennings.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.bae.ncsu.edu/extension/ext-publications/water/use/ebae-161-92-ram-pump-jennings.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=4y_WWxWdn5A&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=4y_WWxWdn5A&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as well as numerous other youtube videos, describing the construction but not the theory.&lt;/p&gt;&#xA;" OwnerUserId="539" LastEditorUserId="539" LastEditDate="2017-07-19T21:48:03.777" LastActivityDate="2017-07-19T21:48:03.777" Title="How do I build a water Ram Pump that will pump 400 feet and 30 feet vertically" Tags="|mechanical-engineering|fluid-mechanics|pumps|water-resources|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16343" PostTypeId="1" CreationDate="2017-07-18T15:47:39.390" Score="1" ViewCount="193" Body="&lt;p&gt;I'm sorry if this sounds like a beginner question. I've tried to search around, but I may be looking for the wrong terms. I have basic knowledge of fluid-dynamics, but not enough to properly measure a ventilation system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I've been redirected here from Physics Stack Exchange - I hope this is the right place for my question.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to build a fume extractor. Basically, I will have a fan sucking air into a flexible duct, which I will send out of the window, approximately at the same height as the fan, or possibly slightly higher (no more than a couple of meters). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've learned that if I use a duct that is too small (with regards to cross section) the airflow of the fan will be limited to the point that air may even be pushed back towards the fan. However, I'm afraid that, if I use a duct with a cross section much larger than the section of the fan, inefficiencies may arise and limit the air intake of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm already thinking of a larger fan (140 mm instead of the usual 120 mm), but what should be a good size for the duct? Would it be bad to use a 150mm one? Should the sizes match? Should it be larger? Does it matter how long the duct is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also aware that a rigid duct would perform better than the flexible one, but I'm not really trying to optimize the performance. I'm just trying to discover most of the stupid things I should avoid doing. Thanks for understanding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: More details&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's simple schematics for my project:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C5TYg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C5TYg.png&quot; alt=&quot;Schematics of my project&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fan is secured to a box (made of cardboard, or possibly 3D-printed plastic), and at the other end a duct guides away the extracted air, out of the window or into a carbon filter. In a separate room of the box, cables and electronics don't interfere with the air flow to avoid turbulence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the size of the box matter? Should it just fit the fan and the duct, or could it be any larger? Should it be tapered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the shape of the folds in front of the fan have any importance? Are they really helpful or should I skip them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I may be overthinking this: I see many people who just glue some cardboard to a fan and they're done. But I see their designs sometimes don't actually extract air well, and I would like to make sure I'm not overlooking important details. Again, I'm not trying to squeeze any little bit of performance out of the system. &lt;/p&gt;&#xA;" OwnerUserId="12201" LastEditorUserId="12201" LastEditDate="2017-07-19T08:09:04.577" LastActivityDate="2017-07-19T08:09:04.577" Title="Sizing the air duct that comes after a fan" Tags="|fluid-mechanics|airflow|" AnswerCount="0" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="16344" PostTypeId="1" CreationDate="2017-07-18T19:22:28.733" Score="0" ViewCount="272" Body="&lt;p&gt;Hy everyone :) I don't know much about structural engineering, maybe &lt;code&gt;spiral lining&lt;/code&gt; and &lt;code&gt;cylindrical lining&lt;/code&gt; are not the proper terms - but take a look at the pictures below to understand what i mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need help to form some intuition about which one of this 2 is better for lining a tunnel. I want to know specifically about earthquake resistance  And also other important factors which play a role in choosing between this 2 designs - factors i'm not aware of right now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d4jcb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d4jcb.jpg&quot; alt=&quot;cylindrical tunnel, with a cylindrical lining&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/FzJBp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FzJBp.jpg&quot; alt=&quot;cylindrical tunnel, with a spiral lining&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't find a proper spiral lining image - this just gives you an idea - but let say it was a proper lining with joints and such - like the cylindrical one has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the spiral components will be connected in the most practical way possible - and assuming they have the same wall thickness, same concrete composition, etc..  which one is better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Will they behave the same under stress ? Or the cylindrical one is better ? But why?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also i want to understand how a cylinder compares with this shape in terms or resistance..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iNyVl.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iNyVl.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one of 2 geometries ( cylinder or arch ) is most effective for preventing damage in case of earthquakes, or just in general ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Explanation: - why i'm interested in this?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the drawbacks of spiral lining. I think that from a construction perspective spiral lining is better - because can be constructed continuously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, if your TBM moves fast enough - the cylindrical lining requires stopping of the hole machine - to insert each segment all at once. And then another cycle begins. This takes valuable time. Time that can be used for drilling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For increasing tunneling speed - i think spiral lining has this amazing advantage - will not require stopping.. you build as you go, smooth and continuous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i don't want to trade safety for tunneling speed. I want to understand if &lt;code&gt;spiral lining&lt;/code&gt; is actually viable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe i'm wrong about this stopping, but that's another topic :) Let's keep this discussion  only about geometry for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you, and have a nice day!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer: &#xA;With input from Solar Mike, i will add this disclamer: neither of this images are mine,&#xA;Tunnel image is taken from here: &lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1674775515001195&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sciencedirect.com/science/article/pii/S1674775515001195&lt;/a&gt;&#xA;Spiral image is taken from here: &lt;a href=&quot;http://www.sunshinecoastformatubesupplies.com.au/spiraltube-steel/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.sunshinecoastformatubesupplies.com.au/spiraltube-steel/&lt;/a&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12155" LastEditorUserId="12155" LastEditDate="2017-07-19T01:15:18.187" LastActivityDate="2018-04-25T13:52:44.247" Title="What lining design holds better for a tunnel? Spiral lining or cylindrical lining?" Tags="|structural-engineering|structural-analysis|structures|tunnels|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16345" PostTypeId="1" CreationDate="2017-07-18T19:30:53.163" Score="0" ViewCount="466" Body="&lt;p&gt;If I am pneumatically conveying solids in slug flow, is there a method that works in a general case to determine the pressure drop or saltation velocity? By a general case method, I mean one that is like doing the Bernoulli balance, friction loss term and by extension Reynolds number, etc. for incompressible fluids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be really neat/worthwhile in my mind to know at which point in a piece of piping that the solids would plug due to dipping below the saltation velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some suspicions of mine: Would the potential relationships involved change significantly for vertical or horizontal piping? Would dilute flow make this problem easier if you assume equal spacing of air between particle-to-particle? Surely you would not be able to consider the air and solids as separate pressure drops, right?&lt;/p&gt;&#xA;" OwnerUserId="12203" LastActivityDate="2018-02-11T21:26:41.397" Title="How to do a pressure drop or saltation velocity calculation on pneumatically conveyed solids in slug flow?" Tags="|chemical-engineering|piping|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16346" PostTypeId="2" ParentId="16322" CreationDate="2017-07-18T21:34:20.823" Score="3" Body="&lt;p&gt;After extensive searching, it appears the part is referred to as a &quot;driving flange&quot; by the manufacturer's technical diagram, which I've included below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the pump itself seems to be an SPT gear metering pump made by Slack &amp;amp; Parr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JwyZm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JwyZm.jpg&quot; alt=&quot;Technical Diagram&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1mOX7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1mOX7.jpg&quot; alt=&quot;Technical CAD Drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12183" LastActivityDate="2017-07-18T21:34:20.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16347" PostTypeId="1" CreationDate="2017-07-19T03:47:45.487" Score="1" ViewCount="1197" Body="&lt;p&gt;I have no formal training or professional experience as a mechanical engineer. I would like to design a rotating base that is powered by a motor. I would like to accomplish the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Smooth, quiet operation&lt;/li&gt;&#xA;&lt;li&gt;Self-locking (as could be accomplished with a worm drive)&lt;/li&gt;&#xA;&lt;li&gt;Minimal wobbling of the rotating part of the base&lt;/li&gt;&#xA;&lt;li&gt;Minimal loss of power (forgive me if &quot;power&quot; is the wrong word here) due to friction.&lt;/li&gt;&#xA;&lt;li&gt;The rotating part of the base doesn't fall off if the device is held up-side-down.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of using a worm drive (with a helical worm wheel) and thrust bearings. But I'm confused about how the thrust bearings would be attached such that the thing would be able to handle the axial loads from the worm drive, as well as avoid simply falling apart. Believe it or not, I've been researching this question for weeks and I'm still confused. Maybe I'm not mechanically inclined. Can anyone point me in the right direction?&lt;/p&gt;&#xA;" OwnerUserId="12208" LastActivityDate="2019-12-22T12:03:20.423" Title="How to design a rotating base?" Tags="|bearings|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16348" PostTypeId="1" CreationDate="2017-07-19T10:44:55.597" Score="3" ViewCount="272" Body="&lt;p&gt;I saw this &lt;a href=&quot;https://engineering.stackexchange.com/q/15228/11443&quot;&gt;question&lt;/a&gt; asking about calculating monosymmetry index of an unequal angle. What I need is a formula to calculate $\int v_i{(u_i}^2 + {v_i}^2)\text{d}A$, given the depth $d$, breadth $b$ and thickness $t$ of an angle. $A$ is the area and $v$ and $u$ are the coordinates of each small element $i$ of the section. Is there a simplified method to derive the above, neglecting root radius and toe radius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1lNXB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1lNXB.png&quot; alt=&quot;unequal angle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11443" LastEditorUserId="11443" LastEditDate="2017-07-25T05:51:03.870" LastActivityDate="2018-03-06T16:05:02.563" Title="How to integrate ∫vᵢ(uᵢ² + vᵢ²)dA in calculating monosymmetry index of an angle?" Tags="|civil-engineering|structural-analysis|mathematics|cross-section|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16350" PostTypeId="1" CreationDate="2017-07-19T13:42:07.107" Score="0" ViewCount="32" Body="&lt;p&gt;I am putting a tapered splined sprocket on a tapered splined shaft that requires approximately 50 tons of press to seat the sprocket on the shaft. How many foot pounds of torque will it take on the nut to achieve the desired tons of press for the sprocket? The threaded part is approximately 4 inches in diameter. The splined part starts at approximated 4.25 inches and tapers up to approximately 6 inches.&lt;/p&gt;&#xA;" OwnerUserId="12213" LastActivityDate="2017-07-19T16:37:16.577" Title="convert foot pounds of torque on a nut to sideways tons of press" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2017-07-27T14:43:58.343" ContentLicense="CC BY-SA 3.0" />
  <row Id="16352" PostTypeId="2" ParentId="16350" CreationDate="2017-07-19T16:37:16.577" Score="1" Body="&lt;p&gt;The way this is done in industry is to apply 50 tons of force then tighten the nut(s). ie, for high pressure flanges , each stud is hyraulically pulled to a specific load then the nuts are tightened.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-07-19T16:37:16.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16354" PostTypeId="2" ParentId="12421" CreationDate="2017-07-19T17:10:58.813" Score="0" Body="&lt;p&gt;As has been noted; It costs money to make &quot;octane&quot;. And because there are many processes to make these many octane components , there are many different costs and &quot;stories&quot; . One story; xlyenes are very high octane and key components in  polyester raw material , so very high competing value. Second story; reforming produces many high octane components, but because of the high hydrogen pressure and high total pressure and platinum group catalyst make this an expensive process.  Maybe get a book on how refineries work.   &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-07-19T17:10:58.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16355" PostTypeId="1" CreationDate="2017-07-19T19:33:57.963" Score="1" ViewCount="34" Body="&lt;p&gt;I'm considering using a vacuum cylinder in my project (like &lt;a href=&quot;http://www.fipa.com/en_GB/products/211441-lifting-cylinders-vacuum-operated&quot; rel=&quot;nofollow noreferrer&quot;&gt;these ones&lt;/a&gt;) to lift sheets of cardboard, but being new to mechanical engineering I'm not at all sure what I need to attach to operate one, and the supplier's site definitely offers no details. I think I'll need a vacuum pump, but I don't know what size is necessary (for example, will something like &lt;a href=&quot;http://www.ebay.ca/itm/Micro-Vacuum-Air-Pump-Mini-Air-Pressure-Pump-Medical-Pump-30KPa-0-28L-CA-/142183067500?hash=item211ac56f6c:g:wGYAAOSwXeJYLm1L&quot; rel=&quot;nofollow noreferrer&quot;&gt;this mini pump&lt;/a&gt; do?), and whether I just hook up the pump to the hole on the cylinder, or there are other parts that are required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I hope this is the right SE site for this; it was a toss-up between this one and the Robotics one.)&lt;/p&gt;&#xA;" OwnerUserId="12218" LastActivityDate="2017-07-19T19:33:57.963" Title="Proper way to power/use a vacuum cylinder?" Tags="|vacuum|pneumatic|vacuum-pumps|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16356" PostTypeId="1" AcceptedAnswerId="16370" CreationDate="2017-07-19T20:49:31.327" Score="-2" ViewCount="2522" Body="&lt;p&gt;What are the &lt;strong&gt;properties&lt;/strong&gt;, &lt;strong&gt;purposes&lt;/strong&gt; and &lt;strong&gt;relationships&lt;/strong&gt; between: &lt;strong&gt;Markers&lt;/strong&gt;, &lt;strong&gt;Blocks&lt;/strong&gt; and &lt;strong&gt;Parts&lt;/strong&gt; in MSC Adams?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand how to use them properly.&lt;/p&gt;&#xA;" OwnerUserId="12221" LastEditorUserId="368" LastEditDate="2017-07-20T15:47:06.257" LastActivityDate="2017-07-20T15:47:06.257" Title="How do markers, blocks, and parts work in MSC Adams?" Tags="|simulation|cad|software|" AnswerCount="1" CommentCount="2" ClosedDate="2017-08-17T12:15:11.437" ContentLicense="CC BY-SA 3.0" />
  <row Id="16357" PostTypeId="2" ParentId="16347" CreationDate="2017-07-19T21:39:32.187" Score="0" Body="&lt;p&gt;In the past I have used trailer hubs for moderately heavy duty rotating platforms (around a few 100 kg capacity). This vastly simplifies the problem as you have a complete assembly which you can just weld or bolt onto a base. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also look as a belt drive as these tend to be inherently fairly smooth and quiet and are reasonably simple to set up, you can get motors with built in very low ration gearboxes off the shelf. If you use a hub  assembly you might be able to drive it with a belt straight onto the outside of  the hub itself&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't find a suitable hub you should probably look at taper roller bearings as these will support both thrust and radial loads&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-07-20T18:11:04.743" LastActivityDate="2017-07-20T18:11:04.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16358" PostTypeId="2" ParentId="16331" CreationDate="2017-07-19T22:07:54.133" Score="0" Body="&lt;p&gt;The statement part of you questions is mostly incorrect. There are a great&#xA;number of product pipelines ( At least in the US) . There are a few stations that have pipelines to refineries or bulk plants but that is unusual. Some of the problems have been mentioned ; certainly the biggest problem would be getting the plethora of permits for each line. Another tremendous problem is the ignorant public ; For existing pipelines the most common cause of significant leaks is &quot;third party&quot; damage. Where a contractor for a city, state, etc, puts a hole in an 18&quot; pipe because he didn't see it or all the warning signs. Can you imagine if there were a million 1&quot; diameter pipes running all over? Corrosion caused by interference in cathodic protection is already a problem with pipelines in urban areas. However , there are many more local supply lines than you can imagine, for example I expect about every major airport is supplied by pipe ( jet and av gas). Most interstate truck stops have pipelines,etc.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2017-07-20T00:00:52.070" LastActivityDate="2017-07-20T00:00:52.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16359" PostTypeId="1" AcceptedAnswerId="16410" CreationDate="2017-07-19T22:25:30.210" Score="1" ViewCount="307" Body="&lt;p&gt;I have a recessed shelf which holds three components that each generate heat:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A projector&lt;/li&gt;&#xA;&lt;li&gt;A computer&lt;/li&gt;&#xA;&lt;li&gt;An audio power amp&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The shelf is located at the top of the wall and open at the front but closed on the 5 other sides.  Because of this, the airflow created by the component fans is simply not enough to cool the space and things are beginning to overheat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to install a fan to cool the space but not sure what the best placement and airflow direction would be to cool the space.  One consideration is the direction of the existing projector fans which create a bit of an airflow pattern on their own.  I'd also like to keep the heat inside the house since it's beneficial in the winter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is basically, what is the optimal direction and placement of a single fan in order to exchange the air and cool the space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; To clarify, I intend to cut a hole in one of the existing surfaces and install the fan to move air between rooms, &lt;em&gt;and not&lt;/em&gt; just to blow air around inside the shelf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rkGVF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rkGVF.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12223" LastEditorUserId="12223" LastEditDate="2017-07-20T16:23:45.397" LastActivityDate="2017-07-24T04:33:01.873" Title="Most efficient airflow pattern to cool a recessed shelf" Tags="|airflow|cooling|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16360" PostTypeId="2" ParentId="16325" CreationDate="2017-07-20T00:12:10.607" Score="0" Body="&lt;p&gt;Just add a valve (manual or solenoid valve) at water suction line. When you want to stop syphoning, just close the valve.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2017-07-20T00:12:10.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16361" PostTypeId="2" ParentId="11263" CreationDate="2017-07-20T00:35:00.517" Score="5" Body="&lt;p&gt;Lead is more corrosion resistant to sulfuric acid . And copper is more expensive. That is; copper in the battery would dissolve in less than one year.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2020-06-13T14:25:17.170" LastActivityDate="2020-06-13T14:25:17.170" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16362" PostTypeId="2" ParentId="16347" CreationDate="2017-07-20T01:38:02.290" Score="1" Body="&lt;p&gt;If you are not mechanically inclined, and depending on your budget, you may find it easier to just buy one than to build your own.  Check out &lt;a href=&quot;https://www.mcmaster.com/#turntables/=18kl9om&quot; rel=&quot;nofollow noreferrer&quot;&gt;these offerings&lt;/a&gt; from McMaster Carr. Costs a few hundred, but that might be better than spending weeks trying to figure out how to build it yourself. If that is not the size or speed you want, they also sell non-motorized versions that you could then just add a motor to.  Other companies might sell other sizes, this is just the first company I tried.    &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-07-20T01:38:02.290" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16363" PostTypeId="2" ParentId="16359" CreationDate="2017-07-20T04:55:58.843" Score="0" Body="&lt;p&gt;Considering this as simillar to a PC case, a large fan (200mm or greater) in the lower side would be fine if you can manage to create a cutout. The flow direction can be outwards (i.e: exhaust) and it should work fine to create a good enough airflow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HYYUz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HYYUz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8778" LastActivityDate="2017-07-20T04:55:58.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16364" PostTypeId="2" ParentId="16359" CreationDate="2017-07-20T05:18:20.257" Score="1" Body="&lt;p&gt;You should also consider placing the 3 components so that they don't &quot;fight&quot; each other for air or one blows onto the other - they all need free access to air. One blowing hot air into the intake of another won't help the second one's cooling.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-20T05:18:20.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16365" PostTypeId="2" ParentId="11263" CreationDate="2017-07-20T05:20:56.167" Score="1" Body="&lt;p&gt;Copper also has the tendency to work harden and can crack / split over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, lead can also split if it is tightened too much - there are plenty of people that over tighten them.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-20T05:20:56.167" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16367" PostTypeId="1" AcceptedAnswerId="16368" CreationDate="2017-07-20T10:19:52.037" Score="1" ViewCount="42" Body="&lt;p&gt;From communication prospective why can't higher frequency ranges be reflected from atmospheric layers and what is the effect of parameters (like temperature, pressure, etc) of the atmospheric layer on reflection of waves for sky wave and space wave propagation.&lt;/p&gt;&#xA;" OwnerUserId="7113" LastEditorUserId="7113" LastEditDate="2017-07-21T12:54:51.137" LastActivityDate="2017-07-21T12:54:51.137" Title="Why only the spectrum range of radio waves can be reflected from atmospheric layers while higher ranges (Infrared-Visible Light-Ultraviolet) can't?" Tags="|telecommunication|wireless-communication|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16368" PostTypeId="2" ParentId="16367" CreationDate="2017-07-20T11:30:26.630" Score="3" Body="&lt;p&gt;Different materials reflect, absorb, and scatter different wavelengths of light.  The ionized layer at the edge of earth's atmosphere happens to reflect a certain range of wavelength while transmitting others.  This is due to the physics of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you can't use the atmosphere to reflect shorter wavelengths for the purposes of communication, is because the atmosphere doesn't reflect those wavelengths in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary parameters affecting the atmosphere's reflectivity is the amount of ionization and the depth of the ionized layer.  That largely depends on solar activity.  There is a significant difference in this layer between night and day.  It is also subject to individual solar phenomena, like flares or coronal mass ejections.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-07-20T11:30:26.630" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16370" PostTypeId="2" ParentId="16356" CreationDate="2017-07-20T12:13:57.457" Score="0" Body="&lt;p&gt;Markers are coordinate systems they represent a position and orientation. Parts represents separate rigid bodies and ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each actionable part consists of atleast one marker which represents the center of gravity of a part.  Parts and most other modeling elemenrs act as groups to markers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Markers represent the mathematical unerpinnings of systems, each joint (constraint) has a marker in both parts, forces emanate to and from markers, and measuremenrs are made to markers. For the simulation to work all you need are parts, markers, forces and joints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rest are convenience, you do not need geometry for example. For example geometry it does make easier to generate innertia values, softbody stfnesses and interpretting the result animations.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2017-07-20T12:13:57.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16372" PostTypeId="1" CreationDate="2017-07-20T20:15:58.497" Score="1" ViewCount="985" Body="&lt;p&gt;My friend and I are working on a project requiring for us to propel a syringe forward in a small container with significant force. We are debating whether to use an elastic band wrapped around the syringe (in the form of a small hand slingshot) or a flat piece of spring steel that upon activation will revert into its original, curved shape, pushing out on the syringe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there such a type of spring steel that will be able to do this effectively and with a large force? If so, are there any downsides to using it as opposed to an elastic band? The goal is to make a product viable on the market for long-term, one-shot use.&lt;/p&gt;&#xA;" OwnerUserId="12234" LastActivityDate="2017-07-24T03:51:59.380" Title="Elastic bands vs Spring Steel?" Tags="|materials|steel|elastic-modulus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16373" PostTypeId="1" CreationDate="2017-07-20T20:19:18.043" Score="0" ViewCount="46" Body="&lt;p&gt;As I understand it, there are 5 key factors that determine the performance potential of a refrigerant gas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Vapor density&lt;/li&gt;&#xA;&lt;li&gt;Enthalpy of vaporization&lt;/li&gt;&#xA;&lt;li&gt;Thermal Conductivity&lt;/li&gt;&#xA;&lt;li&gt;Critical temperature&lt;/li&gt;&#xA;&lt;li&gt;Specific heat&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What are the relevant equations or calculations to weigh these factors? In other words, if we consider the following performance metrics:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;amount of energy required to perform a specific amount of cooling&lt;/li&gt;&#xA;&lt;li&gt;speed at which cooling take place&lt;/li&gt;&#xA;&lt;li&gt;pressure required&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How can I relate these 3 metrics to the 5 molecular characteristics above? Note that the answer is complex because, for example, item #1 above is a curve. For example, in other words the amount of energy required may not be constant for different rates of cooling (#2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that in this analysis I am ignoring safety issues, long-term maintenance issues like corrosion and leak tendency, and secondary performance issues, like dielectric strength; I am only considering primary performance.)&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2017-07-22T07:05:27.853" Title="How to compute the performance of a refrigerant?" Tags="|refrigeration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16376" PostTypeId="2" ParentId="16232" CreationDate="2017-07-21T01:52:35.813" Score="0" Body="&lt;p&gt;q is going to be the distributed load along the span of the beam. If you are looking at just the self weight, you need to calculate that weight (based on density and cross sectional area) as a distributed force and apply that as q. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important stuff to understand and learn, but you might be able to use this &lt;a href=&quot;https://skyciv.com/free-beam-calculator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;beam deflection calculator&lt;/a&gt; to check your answer. &lt;/p&gt;&#xA;" OwnerUserId="12236" LastActivityDate="2017-07-21T01:52:35.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16378" PostTypeId="2" ParentId="15583" CreationDate="2017-07-21T04:28:07.813" Score="1" Body="&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; Theoretically this is true, there would not be any horizontal deflection as the fixed supported (left column) would offer complete rigidity and the hinge connection at node 2 would not transfer any bending. There would also be no horizontal force from the top member as the force is completely vertical:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CrlQt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CrlQt.png&quot; alt=&quot;deflection of frame&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Source: &lt;a href=&quot;https://skyciv.com/3d-structural-analysis-software-features/&quot; rel=&quot;nofollow noreferrer&quot;&gt;SkyCiv Frame Analysis Software&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in real life this would not be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt; Would be true in reality. This is because the fixed support is not going to offer complete rigidity about node 2 like it does in the above diagram. As the top member deflects, it will cause some forces to transfer in the x direction, not just the y. For applications in structural engineering you would generally consider this negligible. &lt;/p&gt;&#xA;" OwnerUserId="12236" LastActivityDate="2017-07-21T04:28:07.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16380" PostTypeId="2" ParentId="15559" CreationDate="2017-07-21T08:43:55.337" Score="1" Body="&lt;p&gt;Deriving the expression for $H$ is relatively straightforward, if you understand how to manipulate block diagrams. The discussion below outlines the general approach (but you should probably google for a more academic description) and can be applied to either SISO or MIMO block diagrams. However, for the latter, you need to make sure that you maintain the correct order of multiplication (generally, for matrices, $AB \neq BA$). &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Define the input and output signals of interest. In this case, by comparing both block diagrams:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The output of the block $H$, i.e. the input to the block $\Delta$, equals the signal running between the blocks $K$ and $G$. Refer to this signal as the output $x$.&lt;/li&gt;&#xA;&lt;li&gt;The input to the block $H$, i.e. the output of the block $\Delta$, equals the signal running between $\Delta$ and $W_2$. Refer to this signal as the input $u$.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Derive the relation between input $u$ and output $x$ based on the block diagram. The most straightforward approach is to trace the signal path through the block diagram in reverse (i.e. from $x$ to $u$). This results in: &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$ x = K(r - W_2u - Gx)$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Solving for the output $x$ in terms of the inputs $r$ and $u$:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$ x = (I + GK)^{-1}K\cdot r - (I + GK)^{-1}KW_2 \cdot u$$&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;In order to determine the expression for $H$ neglect/ignore all input signals that are not of interest (i.e. set them equal to zero). As $H$   describes the relation $u$ and $x$, $r$ should be set equal to zero. Resulting in the following expression for $H$:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;$$ H = -(I+GK)^{-1}KW_2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above equation assumes that the block diagram describes a MIMO system. In case the system is SISO (and only in that case) this is equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ H = W_2\frac{-K}{1+GK} $$ &lt;/p&gt;&#xA;" OwnerUserId="11869" LastActivityDate="2017-07-21T08:43:55.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16381" PostTypeId="2" ParentId="15583" CreationDate="2017-07-21T09:55:59.517" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/15584/255&quot;&gt;mg4w's answer&lt;/a&gt; has some great examples of how your modelling assumptions aren't true in reality. But even ignoring all of those there will &lt;em&gt;still&lt;/em&gt; be deflection in the left hand column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may not find any deflection in your analysis, see for example &lt;a href=&quot;https://engineering.stackexchange.com/a/16378/255&quot;&gt;Sam's answer&lt;/a&gt;. &lt;em&gt;But this is because you're doing an over-simplistic analysis&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We often assume that we have first-order linear static behaviour. This is because it is easier/faster and normally gives sufficiently useful results. However, a second-order geometrically non-linear (GNL) analysis is closer to reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you hold a piece of string, with a weight applied in the middle, you will find that the ends of the string are being pulled inwards. This is because string has very low bending stiffness, and instead axial stiffness governs. With the top beam in your example it is likely that bending stiffness is significantly greater than axial stiffness, but there will still be &lt;em&gt;some&lt;/em&gt; component of axial load. GNL analysis will pick this up. The axial force in the beam will apply a horizontal force on to the top of both columns; this horizontal force will cause the columns to deflect inwards.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-07-21T09:55:59.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16385" PostTypeId="1" AcceptedAnswerId="16395" CreationDate="2017-07-21T21:01:02.353" Score="2" ViewCount="2627" Body="&lt;p&gt;This question might look silly. I am not knowing the correct engineering terms to use in this question. I am trying to explain using the words what I know in English. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have been hired some construction workers for a work in our backyard. I have some doubt on reinforcing beam for slabs. I came through following 2 models &lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; Model one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mFk9z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mFk9z.png&quot; alt=&quot;Insert the steel bar for slab through the steel bear created for the beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; Model two&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5C5us.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5C5us.png&quot; alt=&quot;Put the steel bar for slab above the bar made for beam.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my perception, I believe the second one is much better in terms of strength. Because the beam is supporting the weight of slab. But my friend suggested me the first one is the right way. But I am not aware why the first one is better than the second. Could any one please help me out in this case to identify the better one.&lt;/p&gt;&#xA;" OwnerUserId="12248" LastActivityDate="2017-07-22T12:51:50.823" Title="Reinforcing beam and slab with correct steel bar structure" Tags="|structural-engineering|beam|reinforced-concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16386" PostTypeId="1" CreationDate="2017-07-22T01:10:39.997" Score="0" ViewCount="4528" Body="&lt;p&gt;Im specifically referring to motorcycle engines, but for arguments sake a car engine wouldn't matter either&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say 100Nm was measured at the wheel using a rolling dyno, then based on the reduction ratios(say 5) from wheel to sprocket to gearbox to the clutch, would the torque produced at the crankshaft be less? 100/5 = 20Nm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming no losses, the power produced at the engine would be the same at the wheel but since there are multiple gear ratios, the torque is increased whilst the angular speed has reduced.  So if the crankshaft was turning at 10000 rpm... with an overall gear ratio of 5. The wheels are turning at 2000rpm?&lt;/p&gt;&#xA;" OwnerUserId="12250" LastActivityDate="2017-11-28T12:08:46.057" Title="Is torque produced at the crank of an engine lower than at the wheels?" Tags="|torque|power|engines|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16387" PostTypeId="1" CreationDate="2017-07-22T02:26:53.387" Score="2" ViewCount="161" Body="&lt;p&gt;A 100ton H frame machine press can use 1x100 ton ram to exert this amount of force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can 3x33 ton rams do the same job if they are applied to same area as the 100ton ram?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in relation to a H frame does the bed hanging rods/bolts take most of the ram force or do the legs disperse the force as well as support weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My apologies if this is a very basic question or if the basic understanding is wrong&lt;/p&gt;&#xA;" OwnerUserId="9794" LastEditorUserId="61" LastEditDate="2018-04-18T13:53:32.510" LastActivityDate="2021-11-28T21:06:40.777" Title="Equivalent Force: Hydraulic rams in H frame machine press" Tags="|mechanical-engineering|pressure|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16390" PostTypeId="2" ParentId="16387" CreationDate="2017-07-22T06:57:05.697" Score="1" Body="&lt;p&gt;Pressure is force * area so what area are you working with? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How will you keep all 3 rams synchronised? Ie same pressure at he same moment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The support bolts share the total load so if there are 4 each support or resists 1/4 of the load. Same for the frame legs - if there are 2, then each supports half the load.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-22T06:57:05.697" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16391" PostTypeId="2" ParentId="16373" CreationDate="2017-07-22T07:05:27.853" Score="-1" Body="&lt;p&gt;To calculate the performance use this calculator , easy and fast to use , its free &lt;a href=&quot;https://www.irc.wisc.edu/properties/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.irc.wisc.edu/properties/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12252" LastActivityDate="2017-07-22T07:05:27.853" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16392" PostTypeId="2" ParentId="16386" CreationDate="2017-07-22T07:11:05.747" Score="0" Body="&lt;p&gt;assuming equal diameter (wheel and crankshaft) the torque is highest on the engine, with some losses it arrives less to the wheels. But assuming no losses , the torque is mainly affected by the diameter so u will have to convert the torque to a linear value unaffected by the radius of the shaft 1 &lt;strong&gt;rpm&lt;/strong&gt; = 1*2*pi*shaft Radius*60 &lt;strong&gt;m/s&lt;/strong&gt; so now this value is fully transmitted to the wheels assuming no losses then u reconvert it to RPM using the wheels diameter.&lt;/p&gt;&#xA;" OwnerUserId="12252" LastActivityDate="2017-07-22T07:11:05.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16394" PostTypeId="1" CreationDate="2017-07-22T11:59:30.540" Score="0" ViewCount="57" Body="&lt;p&gt;3d printers are still at a relatively basic stage and currently not capable of printing entire complex machines that require multiple materials. Is it possible that they will ever be capable of printing copies of themselves? Or are there essential components which probably will never be possible to print.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If essential parts cannot be produced directly, could machines to make them be produced by 3d printing?&lt;/p&gt;&#xA;" OwnerUserId="12257" LastActivityDate="2017-07-22T22:08:41.757" Title="Is it likely that a 3d printer will ever be able to print a copy of itself?" Tags="|3d-printing|" AnswerCount="1" CommentCount="2" ClosedDate="2017-07-23T20:39:04.713" ContentLicense="CC BY-SA 3.0" />
  <row Id="16395" PostTypeId="2" ParentId="16385" CreationDate="2017-07-22T12:51:50.823" Score="4" Body="&lt;h2&gt;Model 1 is always better, but may need some modifications&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Whenever you have two reinforced concrete elements, you always need to facilitate the transfer of internal stresses between them. This is done by &quot;mixing&quot; their reinforcement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, model 2 will have no steel between the slab steel and the beam steel. This means that the connection between them will be very fragile and weak to horizontal shear, meaning that there's a risk of a horizontal crack like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KMpB1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KMpB1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With Model 1, this isn't a risk because the vertical steel from the beam (&quot;stirrups&quot;) will resist these forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Model 1 also has an advantage in that it makes the beam itself stronger, since it allows the beam to behave as a much taller beam (including the height of the slab). Indeed, it even allows the beam to behave not as a rectangular beam, but as a stronger T-shape beam, using some of the slab to resist some of the beam's internal forces (how much depends on your country's standards and codes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the beam and the slab will be poured simultaneously, then that's it, and you can stop reading this answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if there's a chance that the beam will be built first and then the slab will be poured over it, then that means that the beam's stirrups and negative reinforcement (the longitudinal bars at the top of the beam) will be outside of the concrete when the beam is initially built but the slab hasn't yet been poured. This means that the beam's weight and that of the freshly-poured slab (before it gains enough stiffness to do anything) will have to be resisted by the &quot;short&quot; beam alone (without the added height from the slab), so some additional steel will be necessary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6bPIv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6bPIv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are basically reinforcing the same beam twice: once considering it as the short beam (without the slab), and once as the tall beam (with the slab). To avoid confusion as to how to define how much reinforcement goes into each position, here's a list:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;positive reinforcement (longitudinal steel at the bottom of the beam): calculate the necessary reinforcement for the short beam to resist the beam's and the slab's self-weight, and then calculate the reinforcement for the tall beam to resist any additional loads (dead loads, live loads, etc). Add these two numbers and you have the total necessary positive reinforcement.&lt;/li&gt;&#xA;&lt;li&gt;negative reinforcement (longitudinal steel at the top of the beam) and the shear reinforcement (stirrups) for the short beam: calculate the necessary reinforcement for the short beam to resist the beam's and the slab's self-weight. You may need to just put the minimum reinforcement.&lt;/li&gt;&#xA;&lt;li&gt;negative reinforcement (longitudinal steel at the top of the beam) and the shear reinforcement (stirrups) for the tall beam: calculate the reinforcement for the tall beam to resist any additional loads (dead loads, live loads, etc). You may need to just put the minimum reinforcement.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-07-22T12:51:50.823" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16396" PostTypeId="1" AcceptedAnswerId="16397" CreationDate="2017-07-22T13:44:41.550" Score="2" ViewCount="278" Body="&lt;p&gt;Since an evaporator also a heat exchanger, how can the inlet and outlet temperatures of the HEX accepted as the same, i.e. isothermal? In this case, cold stream (inlet and outlet of the evaporator in the refrigeration cycle) does not incur any change in temperature, which requires an infinite size of the heat exchanger.&lt;/p&gt;&#xA;" OwnerDisplayName="user5026" LastActivityDate="2017-07-22T14:57:36.537" Title="Why is it generally assumed that the evaporator works isothermally?" Tags="|evaporation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16397" PostTypeId="2" ParentId="16396" CreationDate="2017-07-22T14:57:36.537" Score="7" Body="&lt;p&gt;Assuming you mean evaporator as part of a vapour compression cycle, then because the fluid enters as a liquid and leaves as a gas, which means that the temperature can be the same but the internal energy changes due to the latent heat of vapourisation.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-22T14:57:36.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16398" PostTypeId="2" ParentId="16394" CreationDate="2017-07-22T15:20:32.453" Score="1" Body="&lt;p&gt;Unlike a lathe, that can bootstrap itself. A 3D printer needs electronic components for its control routines. Even if we excluded those you would need something in heat resistant in the nozzle which obviously your print can not make. Or in the case of laser sintering mirrors and lenses. While you could use some trickkery around this with a lot of work and cooling apparatus you would still end up with a way inferior machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarily you would have hard time manufacturing the straight supports for the mechanisms as well as the precision screws and ballbearings. Altough a sintering machine could do somewhat good fascimiles. Not to mention the fact that these are usually bigger components than the printer can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yes given a long time and a lot of manual work you could design and print one today altough it would be pretty poor, and bad value for money. would it become better? Well essentially a 3D printer is a CNC machine so it can do most of the tasks required (3D printer is a bad description for a diverse family of devices that arbitrarily exclude some and include others). Bearings and screws would be a problem still...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it become better? Yes. One will eventually see printers that can print the electrical conduits maybe resistors and coils, but integrated circuits will be outside scope for a while yet. But evetually yes you could build one that could replicate itself.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-07-22T22:08:41.757" LastActivityDate="2017-07-22T22:08:41.757" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16399" PostTypeId="2" ParentId="16386" CreationDate="2017-07-22T18:53:23.470" Score="1" Body="&lt;p&gt;Generally, most road vehicles are geared so that the wheel speed is less than engine speed and with top gear usually being around about 1:1. Ratios which produce wheel speed greater than engine speed are sometimes called overdrives. On older cars with relatively few selectable gears these are sometimes fitted as a separate module to the gearbox and selected by a separate lever or switch not unlike the hi/lo ratio selector in some 4wd transmissions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conventionally when we talk about &lt;em&gt;reduction&lt;/em&gt; ratios that refers to a reduction in speed not torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in your example with reduction ration of 5:1 the torque at the wheels would be 500Nm not 20Nm. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-07-22T18:53:23.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16400" PostTypeId="2" ParentId="6486" CreationDate="2017-07-23T01:57:27.090" Score="0" Body="&lt;p&gt;I &lt;em&gt;finally&lt;/em&gt; found the answer to this in a BIA (Brick Industry Association) &lt;a href=&quot;http://mha-net.org/docs/BIA01.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;test protocol&lt;/a&gt; for standardized brick fireplace designs.  To quote what they're saying about UL-127:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In general, after temperature equilibrium has been reached (or after&#xA;  12 hours of fueling), UL-127 temperature limits on combustible&#xA;  surfaces are:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;90&amp;deg;F over ambient room temperature for unexposed&#xA;  (ie, in contact and covered) surfaces and&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;117&amp;deg;F over ambient room temperature for surfaces&#xA;  exposed to ambient room air&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="964" LastActivityDate="2017-07-23T01:57:27.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16401" PostTypeId="1" CreationDate="2017-07-23T07:42:32.720" Score="0" ViewCount="840" Body="&lt;p&gt;I'm quite new in NASTRAN enviroment, and I am wondering how to use failure criteria which I can define when creating my isotropic MAT1 material for 1D elements (like CBEAM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Quick Reference Guide, (MAT1) I am able to define 3 stress limits during the material creation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type of related output should I expect? &#xA;Can I obtain failure indices or security indices? (like PLY composites)&#xA;I've tried to launch a SOL101 but I can't figure out how to obtain these values, unless they are in the colums of F06 file named &lt;code&gt;M.S.-T&lt;/code&gt; and &lt;code&gt;M.S.-C&lt;/code&gt; (I can't find a reference in the guide, though!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other types of SOLutions effectively use these stress parameters? &#xA;I'm actually interested in SOL111.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for any help. &lt;/p&gt;&#xA;" OwnerUserId="12262" LastEditorUserId="12262" LastEditDate="2017-07-23T09:06:45.360" LastActivityDate="2021-12-29T08:59:58.640" Title="How to use failure criteria in isotropic material (NASTRAN)" Tags="|modeling|finite-element-method|failure-analysis|nastran|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16403" PostTypeId="1" CreationDate="2017-07-23T12:11:25.780" Score="1" ViewCount="49" Body="&lt;p&gt;I am newbie to embedded programming. My problem is I can write and read data to ds1307(by saying using I2C debugger). But my received data on LCD showing 0xE1 constantly for every changing second. Is it I2C debugger showing wrong data as reading data? Or my conversion is wrong, if it so, my data should change but it doesn't. I'm stuck here so hard.please comment where did I go wrong. &lt;/p&gt;&#xA;" OwnerUserId="12225" LastEditorUserId="7083" LastEditDate="2018-01-28T22:45:40.153" LastActivityDate="2018-01-28T22:45:40.153" Title="RTC interface with PIC16F877A" Tags="|embedded-systems|microelectronics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16404" PostTypeId="2" ParentId="16401" CreationDate="2017-07-23T13:09:28.603" Score="1" Body="&lt;p&gt;Any NATRAN solution should output the margin-of-safety factors if you output the stresses. Whether they are meaningful for any particular analysis is up to the user to decide, of course. For example, they probably are not meaningful for a vibration analysis since the mode amplitudes are arbitrary, but they would be meaningful for a forced response or transient dynamic analysis where the displacements represent something physical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of the stress output should be described for each element type. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, for a CROD element the margin of safety is defined as (ST/SA) - 1.0 when SA is positive, and -(SC/SA) - 1.0 when SA is negative, where SA is the calculated axial stress in the element, and ST and SC are the allowable stress in tension and compression given on the corresponding MAT1 card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Ref: &lt;a href=&quot;https://docs.plm.automation.siemens.com/data_services/resources/nxnastran/10/help/en_US/tdocExt/pdf/User.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://docs.plm.automation.siemens.com/data_services/resources/nxnastran/10/help/en_US/tdocExt/pdf/User.pdf&lt;/a&gt; section 4.20)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For composite shell elements, the margin of safety is calculated for each layer - Google will find you references to the details.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-07-23T13:09:28.603" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16407" PostTypeId="1" CreationDate="2017-07-23T23:51:21.900" Score="1" ViewCount="357" Body="&lt;p&gt;Imagine a closed pipe like below, filled with a liquid. The pipe is copper.  Then imagine a small point heat source of X watts applied to the outside of the pipe in the position shown.  Heat warms the pipe, then the liquid, and circulation develops.  As there's only heat applied on one side of the loop, the liquid naturally cools in the other half.  There'll be convection and radiation there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DBHfT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DBHfT.png&quot; alt=&quot;water loop&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way I could begin to model this so that I could establish a temperature rise at the heat source?  So for example, X watts input would lead to a Y degree temperature rise in the pipe at the point of application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, I'm looking at a heat source of say 10W, and a maximum rise of 20 degrees.  So the design element here is the size of the loop and the diameter of the pipe.  I would like to nominally target ⌀22mm for the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have access to finite elements. A back of the envelope, on a toilet wall or a fag packet calculation would be appreciated.  Is it even possible or does this have to be determined empirically?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2017-07-23T23:51:21.900" Title="Calculating heat transfer in a cooling loop" Tags="|thermodynamics|heat-transfer|cooling|convection|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16408" PostTypeId="2" ParentId="3858" CreationDate="2017-07-24T03:27:12.933" Score="-1" Body="&lt;p&gt;Please have a look at the details in the book &quot;Elements of information theory&quot; by A. Thomas et al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ergodic capacity is Shannon capacity, i.e. the maximum mutual information over all input probability distributions.&lt;/p&gt;&#xA;" OwnerUserId="12271" LastActivityDate="2017-07-24T03:27:12.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16409" PostTypeId="2" ParentId="16372" CreationDate="2017-07-24T03:51:59.380" Score="1" Body="&lt;p&gt;The down side of rubber/elastic is that it can degrade over time from oxygen, oil, and uv exposure. If you go this route make sure to use silicone rubber with uv inhibitors. Shielding from the environment also helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The down side of spring steel is the increased cost, and additional structure to hold all the pieces in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you considered a coil spring; it seems like the geometry would work out better. You could put the entire assembly in a capped tube. They make all different shapes and sizes, many of which publishing force coefficients. Use &lt;a href=&quot;https://en.wikipedia.org/wiki/Spring_(device)#Hooke.27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hooke's Law&lt;/a&gt; to calculate your force at a displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mcmaster.com/#compression-springs/&quot; rel=&quot;nofollow noreferrer&quot;&gt;McMasterCarr Spring Catalog&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know your application, but do some more searching for &quot;spring syringe&quot;. You may find a solution for sale or at the very least some design ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qnqOX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qnqOX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.indigo-orb.com/clinical-studies/episure-epidural-stanford-clinical-studies.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stanford University image article&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-07-24T03:51:59.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16410" PostTypeId="2" ParentId="16359" CreationDate="2017-07-24T04:33:01.873" Score="2" Body="&lt;p&gt;You have a couple good options. Don't worry about the fans that cool the pc or projector. They will be of little influence since they are far from the cool air that needs exchanged. Even if they were near a boundary, any tiny benefit in complex considerations like this lost when you upgrade hardware. The key is to look for the source of reliably cool air, existing hvac airflows, and to work with natural convection.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Cool in, existing hole.&lt;/strong&gt; Place a fan at the bottom of the box near the already open face and blow inward. The projector throw will probalby keep you from going all the way to the front. Closer to the front will be more effective. This requires no new holes, and its flow is not limited by closed doors or other hvac airflows in your house. Simple and reliable.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hot out, existing hole.&lt;/strong&gt; Place the fan at the top of the box near the already open face and blow hot air out. Same benefits and limitations as number 1. The difference between these only comes into play when you are worried about enclosure positive pressure and filtering.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hallway hole, hot out top or cold in bottom.&lt;/strong&gt; Make sure to cut the hole larger and select a larger grating than just the fan. This will reduce performance with some recirculation, but will ensure your hvac system doesn't fight it and closed doors don't seal it off. The down side is that loud audio will leave your theater room through this hole.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Stairway hole, cold in from bottom.&lt;/strong&gt; I do not recommend air out the bottom, because you are working against natural convection. This could also have the same noise leaking issues.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;On a side note, make sure you purchase a &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B00EON40CS&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;80+&quot; rated powersupply&lt;/a&gt; for your pc. They are 80% efficient or greater and will reduce the amount of heat you have to get rid of. In my experience they are also quite a bit more reliable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When selecting fans get one with a high cfm (cubic feet per minute) rating and a low dB (decibel) noise rating. Something like this &lt;a href=&quot;http://rads.stackoverflow.com/amzn/click/B008G3OLRQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;200mm 700rpm one&lt;/a&gt;. If you use more than one fan put them in parallel with both of them on &quot;blow cold in bottom&quot; or &quot;blow hot air out the top&quot;. This will effectively double your airflow. Putting them in series with one on hot and one on cold will not increase total air exchange by very much.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-07-24T04:33:01.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16411" PostTypeId="1" AcceptedAnswerId="16451" CreationDate="2017-07-24T05:20:44.503" Score="0" ViewCount="653" Body="&lt;p&gt;I am, in actuality, a network engineer and not a trained mechanical engineer. It was kind of a hobby of mine until recently, when my boss gave me the opportunity to try my hand at mechanical engineering through CAD design (Autodesk Inventor) about two years ago and I took to it like a duck to water. I have been working in that subject since then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem, however, is that I am unfamiliar with most of the best-practices of mechanical engineering and it has been a &quot;trial by fire&quot; sort of situation. Currently my main problem has to do with the alignment of a chain-drive system used for balancing a counterweight and lifting mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Details:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Horizontal alignment is great in my CAD design, however when I actually order and build out the products, I find it extremely difficult to get the alignments properly set and I end up spending hours fidgeting with the axle and chain sprocket alignment... I am currently using aluminum extrusions (via MISUMI) for the frame design.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the images below, you will find an example (several components have been redacted due to proprietary design). The images show the main components for operating the lift mechanism. One side houses the sprockets for the Counterweight, the other for the forklift itself. It is controlled by an AC servo motor which attaches to a custom gearbox, and finally to one of the chain axles (the counterweight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**The design must use these aluminum extrusions, however this makes it very difficult to properly align the bearing housings for the axles, often making everything off-center or skewed, Even when using an accurate measure and square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest some improvement points or alternative types of components for mounting the axle/sprockets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side Profile:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/owZZx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/owZZx.png&quot; alt=&quot;Side Profile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perspective Profile:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WkTnE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WkTnE.png&quot; alt=&quot;Perspective Profile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8543" LastEditorUserId="8543" LastEditDate="2017-07-31T02:24:28.007" LastActivityDate="2017-07-31T02:24:28.007" Title="how can I improve my chain-drive lifting system?" Tags="|mechanical-engineering|design|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16413" PostTypeId="1" AcceptedAnswerId="16419" CreationDate="2017-07-24T06:47:23.790" Score="0" ViewCount="1964" Body="&lt;p&gt;I'm designing a water proof case for a power bank to be used in an occasional underwater condition. The input and out put port components already have a gasket or sealing ring. Now to close the case I am planning to mill a metal with all walls and floor and then screw the top along with and O-ring. I haven't used an O-Ring before and wanted to know how to select an O-ring for a rounded rectangle groove (Does the circumference being equal to the grove length , enough). Is there any handbook for the same?&lt;/p&gt;&#xA;" OwnerUserId="12274" LastActivityDate="2017-08-23T12:22:39.553" Title="O-Ring Selection for a round rectangle groove" Tags="|waterproofing|fluid|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16414" PostTypeId="1" AcceptedAnswerId="16417" CreationDate="2017-07-24T07:58:38.007" Score="3" ViewCount="306" Body="&lt;p&gt;I have googled for a while and I can't seem to find the correct term to refer to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you have a tunnel in where trains, cars, or anything goes through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tunnel might elements along its extension, such as cracks, fissures, weak spots, etc. To visualize this in paper, a 3D figure must be &quot;projected&quot; into 2D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for the word to refer to this kind of map.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest I've come to is calling it &quot;A projection map&quot; but is there a more suitable term for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z9ZND.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z9ZND.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In japanese it can be called a &quot;展開図(てんかいず) Tenkaizu&quot;, but a dictionary translation, or a google translation don't fit since that term is also used to describe the diagrams of how to fold paper to build something. (like a cube)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried in the &quot;English&quot; site, but I was referred to this site instead.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://english.stackexchange.com/questions/402469/how-do-you-call-a-map-that-shows-the-characteristics-of-a-train-tunnel-3d-in-p?noredirect=1#comment955561_402469&quot;&gt;What do you call a 2D map that shows the characteristics of a 3D&#xA;train tunnel?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12276" LastEditorUserId="8055" LastEditDate="2017-07-24T15:39:04.810" LastActivityDate="2018-04-25T13:54:56.810" Title="How do you call a map that shows the characteristics of a train tunnel (3D) in paper (2D)?" Tags="|civil-engineering|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16415" PostTypeId="2" ParentId="16414" CreationDate="2017-07-24T10:42:17.770" Score="0" Body="&lt;p&gt;Not quite sure which kind of map you want.  A &quot;projection&quot; is certainly a valid option.&lt;br&gt;&#xA;You might also be thinking of a &quot;plan view&quot; which is simply the X-Y layout of the tunnel, ignoring vertical behaviors.  A plan view is more typically used to look at the layout of a building floor-by-floor, showing object locations (walls, doors, etc).  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-07-24T10:42:17.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16416" PostTypeId="1" CreationDate="2017-07-24T11:04:39.520" Score="1" ViewCount="115" Body="&lt;p&gt;Im trying to build a cold isostatic press with a internal vessel area of 300mm3.&#xA;A cold isostatic press forms green metal components using pressures greater than 90,000 psi.&#xA;Using an oil or water/emulsified oil solution do they use a high pressure pump and connect it to a hydraulic intensifier?&#xA;Any direction is greatly appreciated thank you.&lt;/p&gt;&#xA;" OwnerUserId="9794" LastActivityDate="2017-07-24T13:37:39.780" Title="How do you get a closed vessel to &gt;90,000 psi internal pressure" Tags="|pressure|hydraulics|process-engineering|pressure-vessel|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16417" PostTypeId="2" ParentId="16414" CreationDate="2017-07-24T11:09:20.787" Score="5" Body="&lt;p&gt;An engineering drawing that shows a curved surface &quot;unwrapped&quot; onto a plane is called a &lt;strong&gt;development.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is commonly used in sheet metal working, where making the object follows the reverse process - you cut out the correct shape(s) on a &lt;em&gt;flat&lt;/em&gt; sheet of material and then bend them into the correct 3-D form(s) and join them together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://nptel.ac.in/courses/112103019/36&quot; rel=&quot;noreferrer&quot;&gt;http://nptel.ac.in/courses/112103019/36&lt;/a&gt; for some examples - google will find plenty more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In engineering drawing, a &lt;strong&gt;projection&lt;/strong&gt; is &quot;what you would see if you looked at the object from a long distance away, through a telescope&quot; which is &lt;em&gt;not&lt;/em&gt; what your example shows.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-07-24T11:16:22.313" LastActivityDate="2017-07-24T11:16:22.313" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16418" PostTypeId="2" ParentId="16414" CreationDate="2017-07-24T12:14:40.360" Score="2" Body="&lt;p&gt;The type of drawing you mentioned is common for a number of underground mines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surveyors produce the overall outline of the tunnel &amp;amp; then geologists manually drawn on the geological features and structures. Such drawings are produced on an ongoing basis, as tunnels are extended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, they are referred to as maps of the walls &amp;amp; backs. Backs being the term for the ceiling or roof of a tunnel. Civil engineers may use the term ceiling or roof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term you seek may vary between industries such as civil and mining, but as you state a project map would suffice. If you want to be more specific, you can call it a projection map of the walls and ceiling.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-07-24T12:14:40.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16419" PostTypeId="2" ParentId="16413" CreationDate="2017-07-24T12:14:55.957" Score="1" Body="&lt;p&gt;With any O-ring seal, static or dynamic, the key parameter is &quot;O-ring Squeeze&quot; which is defined as the percentage reduction in cross sectional area when compressed from the free state. Since the o-ring acts like a spring, this squeeze is directly related to the sealing force. The greater the squeeze, the greater the sealing force, and the greater resistance to leakage...at least in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with most things, more is not always better, too much squeeze can result in a poor joint since it can cause excess stress in the mating parts, excess friction (in the case of a dynamic seal), and O ring damage since there's a greater risk of the o-ring getting pinched between the two mating surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In designing O-ring seals in stainless steel with good surface finish, I would try to stick to &amp;lt; 15% squeeze for dynamic seals and 15%-30% for static seals using a standard NBR O-ring from Parker, but it really depends on the type of O-Ring, the materials involved and all of the other specifics of the application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you'll want to use an O-ring that has a circumference close to the mean perimeter of your groove. You can, of course buy custom O-rings, but those can be expensive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a lot that goes into the design of a good O-ring joint. The best resource I've found is the &lt;a href=&quot;http://promo.parker.com/promotionsite/oring-ehandbook/us/en/ehome&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parker O-ring Handbook&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-07-24T12:14:55.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16420" PostTypeId="2" ParentId="16416" CreationDate="2017-07-24T13:37:39.780" Score="3" Body="&lt;p&gt;Contact an engineering firm with experience designing and building such systems and vessels to code. Otherwise, you're just building a bomb.&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-07-24T13:37:39.780" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16421" PostTypeId="1" CreationDate="2017-07-24T16:52:25.307" Score="1" ViewCount="830" Body="&lt;p&gt;While building the foundation of a house, the concrete columns were built 15cm (6 inches) from the planned position, which is reducing the space for a door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position that should have been there is shown below. But, C2 is shifted to left by 6 inches, and the column shown at the bottom of pic is shifted 6 inches to right. This leaves no space for the door to be placed as expected. Need a workaround.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CgxxQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CgxxQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In below pic, the upper column is shifted to left by 6&quot;, and lower 6&quot; to right, leaving just 2'6&quot; for door rather than required 3'6&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0CooK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0CooK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need to shift the column by 6 inches. How to do it? It's built up to the ground level only yet, and beam isn't  built over it.&lt;/p&gt;&#xA;" OwnerUserId="12281" LastActivityDate="2017-07-24T16:52:25.307" Title="Shift Concrete Column by few inches" Tags="|civil-engineering|concrete|foundations|columns|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16426" PostTypeId="1" AcceptedAnswerId="16428" CreationDate="2017-07-24T23:27:28.390" Score="1" ViewCount="350" Body="&lt;p&gt;I have a set of equations of motion describing a planetary gear train of 18 DoF (sun, 3 planets, carrier and ring), they have the general form of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{M{\ddot q}+{\Omega_c}G{\dot q}+{Kq}=F(t)}$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$K$: stiffness symmetrical matrix&lt;/li&gt;&#xA;&lt;li&gt;$M$: Mass symmetrical matrix&lt;/li&gt;&#xA;&lt;li&gt;$G$: Gyroscopic skew symmetrical matrix&lt;/li&gt;&#xA;&lt;li&gt;$\Omega_c$: angular speed&lt;/li&gt;&#xA;&lt;li&gt;$F$: excitation&lt;/li&gt;&#xA;&lt;li&gt;$q$: generalized coordinates&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While trying to solve the eigenvalue problem with Matlab, I noticed there´s a whole bunch of ways to represent the equations in state space form and I don´t understand what the main difference is, or which one I should use, since when calculating I get different results in the eigenvectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the free vibrations I have the equation without damping:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{M{\ddot q}+{Kq}=0}$$&#xA;&lt;strong&gt;1.&lt;/strong&gt;&#xA;and to solve the standard eigenvalue problem of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf {\omega^2M\phi_i-{K}\phi_i=0}$$&#xA;I use the Matlab code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[v,D]=eig(K,M)&#xA;root=sqrt(D)&#xA;natural_frq=root/(2*pi)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;which gives me all real natural frequencies and eigenvector per frequency.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt;&#xA;I also tried calculating them in the form of:&#xA;$$\mathbf{{\dot x}=Ax+Bu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{bmatrix} \mathbf{\dot q} \\ \mathbf{\ddot q} \end{bmatrix} = \begin{bmatrix} \mathbf{0} &amp;amp; \mathbf{I} \\ \mathbf{-M^{-1}{K}} &amp;amp; \mathbf{-M^{-1}{G}} \end{bmatrix} \begin{bmatrix} \mathbf{ q} \\ \mathbf{\dot q} \end{bmatrix}+ \begin{bmatrix} \mathbf{0} \\ \mathbf{-M^{-1}} \end{bmatrix}u$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and by using:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[v,D]=eig(A)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I get similar results but in complex numbers (there´s no need for the square root), so with some extra code I choose only the imaginary parts and select the positive values of the eigenvalues vector, which instead of 18 elements is now 36. &lt;strong&gt;The frequencies I get with this method against the first ones are paractically the same, but the eigenvectors are not&lt;/strong&gt;, they are also complex and the results don´t look correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt;&#xA;Then there's another state space form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \begin{bmatrix} \mathbf{I} &amp;amp; \mathbf{0} \\ \mathbf{0} &amp;amp; \mathbf{M} \end{bmatrix} \begin{bmatrix} \mathbf{\dot q} \\ \mathbf{\ddot q} \end{bmatrix} +  \begin{bmatrix} \mathbf{0} &amp;amp; \mathbf{-I} \\ \mathbf{K} &amp;amp; \mathbf{G} \end{bmatrix}\begin{bmatrix} \mathbf{ q} \\ \mathbf{\dot q} \end{bmatrix} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use the code:&#xA;    [v,D]=eig(-B,A)&#xA;I get similar results as the last method, I'm not sure which form of state space this form is, I would guess it's $$A{\dot x} = -Bu$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main question is that I don't know which is the best solution to my problem and why I get complex frequencies in the second method, if my angular speed $$\Omega_c$$ is zero, all damping terms should also go to zero and thus not affect the system results , right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: why do I get very different values when calculating the eigenvectors through the standard eigenproblem form $$\mathbf {\omega^2M\phi_i-{K}\phi_i=0}$$ and with the state space?&lt;/p&gt;&#xA;" OwnerUserId="4088" LastEditorUserId="4088" LastEditDate="2017-07-27T13:50:37.767" LastActivityDate="2017-07-27T16:43:52.750" Title="What's the difference of using different state-space forms for solving eigenvalue problem?" Tags="|mechanical-engineering|gears|dynamics|frequency-response|eigenvalue-analysis|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16427" PostTypeId="2" ParentId="16426" CreationDate="2017-07-25T00:17:54.663" Score="2" Body="&lt;p&gt;For any sinusoidal motion, the $\mathbf{q}$ really represents the motion $\Re(\mathbf{q}\,e^{i \omega t})$ in the time domain. $\mathbf{q}$ is complex because in general, the different elements of the eigenvector have different phase angles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will be the case for your problem since your first equation seems to contain a gyroscopic term - if the rotation is about the $x$ axis, the Eigenvector components if the $y$ and $z$ directions will be 90 degrees out of phase to represent the &quot;whirling&quot; motion of the structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $\mathbf{\dot q} = i \omega\, \mathbf{q}$, so the vector $\begin{bmatrix} \mathbf{q} &amp;amp; \mathbf{\dot q} \end{bmatrix}^T$ can not have &lt;em&gt;all&lt;/em&gt; its terms real, except for the special case where $\omega = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that the eigenvectors are to some extent arbitrary even for a real eigenproblem. Even if you insist they are mass normalized, you can still multiply them by $-1$. For a complex eigenproblem, you can multiply them by any complex value on the unit circle, of the form $\cos \theta + i \sin \theta$ for an arbitrary $\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to compare the solutions for two complex eigenproblems, you have to fix these arbitrary differences - for example, by &quot;rotating&quot; the vectors around the unit circle so that one element of the eigenvector is real and positive, and &quot;scaling&quot; the vectors so that the real positive value is $+1.0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These arbitrary factors don't matter if you just want to &lt;em&gt;use&lt;/em&gt; the eigensolution to compute something which &lt;em&gt;is&lt;/em&gt; real for physical reasons. When you take a linear combination of the eigenvectors, the scaling factors multiplying each vector are also complex, and they &quot;clean up&quot; the arbitrariness in the eigensolution, in exactly the same way that it doesn't matter if you calculate $\mathbf{+x}$ or $\mathbf{-x}$ for a real eigenvector.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-07-25T00:26:28.140" LastActivityDate="2017-07-25T00:26:28.140" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16428" PostTypeId="2" ParentId="16426" CreationDate="2017-07-25T01:08:30.107" Score="2" Body="&lt;p&gt;Regarding the different state space formulations.  First, M might not be invertible.  In many cases it will be invertible, but sometimes it will not be.  In your specific case, this would correspond to a gear with zero mass. Obviously that never happens in the real world as all gears would have some finite mass. But there are situations, where the modeling choices and assumptions lead to a DOF with zero mass.  In that case, you cannot use the first form of the state space equations.  You must use the second one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if M is invertible, then &lt;em&gt;analytically&lt;/em&gt;, both formulations will give the same answer.  But &lt;em&gt;numerically&lt;/em&gt;, they will invariably give slightly different answers due to floating point roundoff error.  Depending on the solver that you use, one form or the other might end up being more accurate.  Speed differences are also possible.  This depends on the exact structure of your matrices and the details of your solver. If you want more details, pick up a book on numerical methods that describes the inner workings of the various eigenvalue solvers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;based on your latest edit, I'm not that my original answer really answers your question.  These two methods really should give you similar answers.  The fact that they are not implies that you are probably doing something wrong, but I don't know what.  Suggestion: don't start with an 18x18 system.  Start with a 2x2 system, something small enough that you can work out the exact answer by hand on one sheet of paper, or even better look up the answer in a textbook. That should give you insight into where your matlab code is going wrong.  If not, a 2x2 system is small enough that you can post the entire thing here (all the inputs, all the outputs, all the code), and we can probably figure it out.  As is, without seeing your input matrices, or the whole code, and only portions of some of the eigenvectors, I'm not sure if I can be of more help.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="9275" LastEditDate="2017-07-26T11:51:49.277" LastActivityDate="2017-07-26T11:51:49.277" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16430" PostTypeId="1" CreationDate="2017-07-25T06:00:52.483" Score="1" ViewCount="1702" Body="&lt;p&gt;I was reading a steel catalog and saw that it categorizes steel sections separately into pipe section, hot-finished circular hollow section, and cold formed circular hollow section. Given the same design strength, for example S355, is there any fundamental differences when designing a section of say, 610mm depth and 10mm thickness, using steel material from these three categorizations?&lt;/p&gt;&#xA;" OwnerUserId="11443" LastActivityDate="2017-07-25T22:49:59.087" Title="What is the difference between pipe section, hot-finished, and cold formed circular hollow section?" Tags="|civil-engineering|design|steel|cross-section|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16432" PostTypeId="1" CreationDate="2017-07-25T10:48:19.363" Score="0" ViewCount="87" Body="&lt;p&gt;I live somewhere where it's relatively warm the entire year (around 10ºC is the coldest you'll get in some days during a normal winter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am dreaming of a house design that I intend to build in the near to mid-term future. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have dogs and I want to build a room dedicated for them outside, where they'll sleep, have a shade and be protected from the rain and wind, that they can go in and out whenever they want. This room will have it's entrance from the garden only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of this question is: Is there any kind of wall design I can use that will give them a clear entrance but at the same time won't allow (at least most of) the wind rage through?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sketch is a design roughly like this, but I am not sure if this would really stop the wind from going in or not:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QtFC1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QtFC1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking it here because I am sincerely interested in the physics of such design and how it could control the flow of the wind.&lt;/p&gt;&#xA;" OwnerUserId="12289" LastActivityDate="2017-07-25T12:59:21.320" Title="Windproof open doors room design" Tags="|building-physics|wind-tunnels|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16434" PostTypeId="2" ParentId="16432" CreationDate="2017-07-25T12:11:57.877" Score="1" Body="&lt;p&gt;Using your posted design, one approach you can use is to have a longer entrance chamber and to have two doors (dog/pet type swinging doors): one at each end of the entrance chamber. But have the doors heavy enough so they are not affected greatly by the wind, but light enough for your dogs to pass through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two doors will increase the resistance to wind. Also, when one door is open the other door will be closed, minimizing the effects of any wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the chamber should allow for both dogs to be in the chamber so that only one or neither of the doors will be open at any one time. You don't want both doors open simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing you could do is, particularly with a longer entrance chamber is to add a baffle, so the entrance chamber looks a bit like a maze. This will further increase resistance to any wind.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-07-25T12:11:57.877" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16435" PostTypeId="2" ParentId="14756" CreationDate="2017-07-25T12:27:24.990" Score="3" Body="&lt;p&gt;I would start by resolving the loads into a single load/moment system at the center of the foundation - in both the x and y directions. It will then be possible to use the Terzaghi equations to calculate the bearing capacity of the foundation, taking into account a reduced effective breadth and length calculated using the following equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$B'=B-2e_B$$&#xA;$$L'=L-2e_L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will also be necessary to check that the resultant loading lies within the middle third of the foundation.&lt;/p&gt;&#xA;" OwnerUserId="7329" LastActivityDate="2017-07-25T12:27:24.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16437" PostTypeId="2" ParentId="16432" CreationDate="2017-07-25T12:59:21.320" Score="1" Body="&lt;p&gt;By the complexity of your drawing, I assume this isn't meant to be a little dog house, but an actual room-sized structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, an important question is whether or not this room will have windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the room won't have any (open) windows or other means of exchanging air between the inside and outside of the room (other than the single door shown), then you honestly don't need anything. Wind is caused by differential pressure in the atmosphere: air moves from high-pressure to low-pressure areas. If your room has only one door, the wind will cause a quick gust into the room, increasing the air pressure there until it is roughly equal to the external air pressure, meaning there will be almost no wind in the room. The same applies in conditions whether the wind would actually cause suction within the room; air would leave the room until its air pressure equalizes with the low air pressure outside. In this case, you don't really even need a door, you can just leave a &quot;hole in the wall&quot; for the dogs to go through. Obviously, not having any windows would make the room rather dark and, over time, it would probably get really stinky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it will have (open) windows or some other means of transferring air between the inside and the outside of the house (other than the door, of course), then there will be wind and you'll need to handle it. &lt;a href=&quot;https://engineering.stackexchange.com/a/16434/1832&quot;&gt;Fred's answer&lt;/a&gt; has some good suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another suggestion is to study the wind's preferential direction. You'll find that wind tends to move in a given direction. For instance, near the sea (other geographic considerations notwithstanding), wind tends to move perpendicular to the shoreline (moving into the mainland or into the sea). Try to orient the doghouse with the doors and windows perpendicular to this preferential direction, so that the wind doesn't hit them directly. This will greatly reduce the effect of the wind within the doghouse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Iy2R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Iy2R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-07-25T12:59:21.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16438" PostTypeId="1" CreationDate="2017-07-25T14:36:32.330" Score="0" ViewCount="84" Body="&lt;p&gt;I am currently taking a course titled &quot;Electric machinery and drives&quot;. We have started with the very basics, i.e with the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau_{motor} = \tau_{load} + \frac{\partial (J\omega_m)}{\partial t}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $\tau$ is the notation for torque and $J$ is the notation for &quot;polar moment of inertia, $\omega_m$ being the angular velocity of the motor shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what is the source of the load torque $\tau_{load}$?, i.e is there some external agency that tries to rotate the shaft in the opposite direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, please also explain what the equation simplifies to if the load is just simply a massive flywheel&lt;/p&gt;&#xA;" OwnerUserId="11619" LastActivityDate="2017-09-23T15:33:06.263" Title="Source of a load torque" Tags="|motors|torque|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16439" PostTypeId="2" ParentId="16438" CreationDate="2017-07-25T14:44:31.407" Score="1" Body="&lt;p&gt;The load torque is from what you want the motor to power or drive be it a saw or conveyor or drill which, depending on the process, may vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the load can change on a conveyor depending on the amount of material to be moved - keep increasing the material and the load increases.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-25T14:44:31.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16440" PostTypeId="1" CreationDate="2017-07-25T15:38:16.270" Score="1" ViewCount="66" Body="&lt;p&gt;I'm looking into manual pellet presses and the one in mind lists a 12T clamping force and then goes on to detail that the analogue gauge goes from 0- 24,000 lbs. Are these two numbers related? When I try to convert T to lbs I usually get 26,880 lbs for 12T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="12295" LastActivityDate="2017-07-25T21:25:42.690" Title="How to convert ton force to pound force, manual pellet press" Tags="|pressure|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16441" PostTypeId="2" ParentId="16403" CreationDate="2017-07-25T17:50:27.667" Score="1" Body="&lt;p&gt;Ok firstly calm down, everything has an answer. you say you read and write data from real time counter by I2C debugger but can not by microcontroller; MCU stands between LCD and RTC right? and there is alternate read/write path to RTC, either MCU can access it, either I2C debugger is that right? so from all this i assume you have some kind of Development board with RTC,MCU,LCD, and a I2C debugger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;first, always try to think of these  electronic components as tiny micro cities. they all live by their own separate lives and communicate to each other by wires just as buses are used to connect cities to each other. if RTC responds to I2C debugger we can assume that RTC is powered and feels ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but LCD shows frozen values. you are not sure is this because MCU could not communicate to RTC...  or because MCU can communicate with RTC but MCU fails to pass this data to LCD. so there are 2 joints, either MCU-&gt;RTC is dead or MCU-&gt;LCD is in wrong format. firstly try to send couple of things to LCD. something, does not matter what. Hello world or something. if it shows well, then problem is certainly with MCU-&gt;RTC communication. &lt;/p&gt;&#xA;" OwnerUserId="12297" LastActivityDate="2017-07-25T17:50:27.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16442" PostTypeId="2" ParentId="16440" CreationDate="2017-07-25T18:53:37.553" Score="1" Body="&lt;p&gt;Since tons are being related to pounds, you should not start by assuming metric tons.  A ton = 2000 pounds, so 12 tons would be 24,000 pounds.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-07-25T18:53:37.553" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16444" PostTypeId="2" ParentId="16430" CreationDate="2017-07-25T20:18:07.710" Score="1" Body="&lt;p&gt;The geometrical tolerances are usually smaller for cold-formed sections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material properties and also be different. For example the final grain structure of the metal will be different, and cold-formed sections usually have higher residual stresses than hot-formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface finishes like galvanizing can only be done &quot;hot.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This contains quite a lot of relevant material (too much to summarize here): &lt;a href=&quot;http://www.condesa.com/pdf/en/TUBO_ESTRUCTURAL_CASTV3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.condesa.com/pdf/en/TUBO_ESTRUCTURAL_CASTV3.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without more context, one difference between &quot;pipe sections&quot; and &quot;circular hollow sections,&quot; could be their intended use. A pipe would typically be used to contain a fluid, and therefore would be have a specification for maximum internal pressure, and tested against leaks. A &quot;hollow section&quot; is just a structural component. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Pipes&quot; usually have circular sections since that is the most efficient use of material to resist internal pressure, but &quot;hollow sections&quot; can also have other shapes such as elliptical, rectangular, etc. &lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-07-25T22:49:59.087" LastActivityDate="2017-07-25T22:49:59.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16445" PostTypeId="1" AcceptedAnswerId="16455" CreationDate="2017-07-25T20:52:30.507" Score="0" ViewCount="114" Body="&lt;p&gt;I am building a device and I tried to Google an answer to this question but I am an electrical person and lack knowledge in this field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know what the relative humidity inside an enclosure (of my device) will be for a given ambient temperature. If I know:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relative humidity outside the enclosure (95% RH)&#xA;The ambient temperature (40 deg C)&#xA;The air temperature inside the enclosure (60 deg C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the air inside the enclosure is hotter because of the electronics I know the RH inside will be lower but I do not know how to go about quantifying and justifying it on paper.&lt;/p&gt;&#xA;" OwnerUserId="12302" LastEditorUserId="90" LastEditDate="2017-07-26T02:15:15.560" LastActivityDate="2017-07-26T08:08:49.937" Title="Correlating relative humidity outside and inside an enclosure" Tags="|mechanical-engineering|temperature|environmental-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16446" PostTypeId="2" ParentId="16445" CreationDate="2017-07-25T21:22:11.643" Score="0" Body="&lt;p&gt;There is a psychrometric chart (ASHRAE) that will give you the relative humidity, but you will need the Dew Point or condensation temperature as well as the ambient temperature. This you can measure with a Wet and Dry bulb thermometer but you will need to access the thermometer to wet the wet bulb. Charts can be found &lt;a href=&quot;https://www.ashrae.org/resources--publications/bookstore/psychrometrics&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps your easiest solution will be one of the cheap meters available which will give you a reading directly in %rh.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-07-26T05:29:09.860" LastActivityDate="2017-07-26T05:29:09.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16447" PostTypeId="2" ParentId="16440" CreationDate="2017-07-25T21:25:42.690" Score="1" Body="&lt;p&gt;Which tons are you using? Ton-force (short) converts to 24,000 lbs – Solar Mike 5 hours ago&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Posted this as an answer, having had it confirmed by the OP as helping him with his issue, confirmed by his comment : &#xA;Ah, I had been using Ton-force (long)...** face palm ** Thanks a lot Solar Mike! &lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-25T21:25:42.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16448" PostTypeId="1" CreationDate="2017-07-25T22:20:48.447" Score="0" ViewCount="47" Body="&lt;p&gt;Hello I am hoping someone can share their knowledge I encountered a problem on a recent project at a site I visited. They are encountering enormous amounts of waste copper wire which they are sending away to scrap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My background is in the mining sector and I wanted to know what the process is to get copper wire into granulated copper and the wire armor and insulation in separate containers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from research that a grinder will break up the cables. My main confusion is within separation of the copper/metals and insulation. Is this achieved used hydrocyclones, magnetic separation, vibration screens or flotation cells (only because a YouTube video appeared to mention this). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to confirm I am specifically asking how they separate the different materials as in &lt;a href=&quot;https://youtu.be/ZKN5MtrY5Cg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/ZKN5MtrY5Cg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12303" LastEditorUserId="-1" LastEditDate="2017-07-27T11:56:47.207" LastActivityDate="2017-07-27T11:56:47.207" Title="Copper recycling process" Tags="|recycling|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16449" PostTypeId="2" ParentId="16448" CreationDate="2017-07-25T23:09:46.533" Score="1" Body="&lt;p&gt;There are quite a few ways I've seen this happen. The lowest tech way (and least environmentally friendly) is burning the insulation off like they do in 3rd world countries, up to slicing/crushing the insulation off. &lt;a href=&quot;https://www.youtube.com/watch?v=-wUvk-6fGP4&quot; rel=&quot;nofollow noreferrer&quot;&gt;This video&lt;/a&gt; shows both of those last two methods. I was searching for a video I saw a long while back that used this method on a vast scale but couldn't find it. I've no experience of a fully automated process, all of these still involve a fair amount of manual labour.&lt;/p&gt;&#xA;" OwnerUserId="12187" LastEditorDisplayName="user748" LastEditDate="2017-07-25T23:38:51.197" LastActivityDate="2017-07-25T23:38:51.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16451" PostTypeId="2" ParentId="16411" CreationDate="2017-07-26T00:11:16.543" Score="1" Body="&lt;p&gt;A very common method for shaft alignment is by using shims and a &lt;a href=&quot;https://ae01.alicdn.com/kf/HTB1vXQjMpXXXXaKXXXXq6xXFXXXh/2-IN-1-font-b-Dial-b-font-font-b-Indicator-b-font-0-0-8mm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;DTI&lt;/a&gt; (dial test indicator). Having a &lt;strong&gt;solid bed plate&lt;/strong&gt; for the bearing housings is a must-have. Depending on the load, the aluminium extrusions may flex, which makes any fine adjustments to the bearing housings redundant. The alignment process generally involves using a  or better yet a &lt;a href=&quot;https://i.ytimg.com/vi/PNMfJLaoUqo/maxresdefault.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;laser alignment kit&lt;/a&gt;. There are plenty of videos on YouTube that document the process of using a DTI to align shafts with a central coupling, but these can be extrapolated to using it with one shaft across two bearings.&lt;/p&gt;&#xA;" OwnerUserId="12187" LastActivityDate="2017-07-26T00:11:16.543" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16452" PostTypeId="1" AcceptedAnswerId="16454" CreationDate="2017-07-26T02:22:28.040" Score="0" ViewCount="37" Body="&lt;p&gt;Pressure pumps are used to apply pressure to a system. What would happen if someone attempted to fill, for example, a bicycle tire (with a check valve) to 2 ATM of pressure using a pressure pump that could only supply 1.5 ATM of pressure? As the gas cannot go backwards due to the check valve during filling, could the pressure be expected to rise in the tire to 2 ATM?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2017-07-26T06:31:10.473" Title="Compression and check valves" Tags="|pressure|compressors|compressed-air|wheels|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16453" PostTypeId="1" CreationDate="2017-07-26T02:26:36.787" Score="0" ViewCount="165" Body="&lt;p&gt;I can't find what i'm looking for in the engineering stack exchange because it's all way over my head.  need basic help answer the basics to get started. so maybe a link or video that helps me understand how to even ask the right question on motor needed. I'm not going to become an engineer or learn everything i simply want to understand how to solve one type of problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so here's my basic need: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a water pump bought at a store that needs to be driven by a drill. But i want to use a motor from another appliance to run the water pump so i can use my drill for other things and have the motor that runs the water pump just working on that.  I took apart a spice/ coffee grinder and rewired it and hooked it up the water pump and its simply doesn't move from lack of power ( it works fine and spins fine not attached to the water pump).  So how do i calculate the motor i need to run the water pump? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;also i will need this calculation to serve other projects... I'm also going to want to make a yeast stirer and use the motor from an orange juicer to spin a magnet that will affect a stirer stick through a glass container to stir yeast for 24- 48 hours.  Not a lot of power needed but will needed to run all day and night. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so I want to understand just the very basics i need to determine how to choose a motor or how to very simply adjust or reconfigure any  motor set up (i.e. use cogs or some kind of added system of wheels to increase the motors power) to: &#xA;1) be able to run a water pump or spin anything that requires a lot of strength and can run continuously for an hour to two hours. &#xA;2) Secondly, I need to calculate and determine needs to run a motor that spins a light load but continuously for 48-72 hours without burning out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cant find answers on google or stack exchange because when i ask the question i get a ton of irrelevant or very complex answers. i'd like help in getting enough info to keep my projects going without having to study engineering for weeks.  i dont need to understand every single thing but enough to get these jobs done and repeat the process for similar future projects.  Thank you!&lt;/p&gt;&#xA;" OwnerUserId="12304" LastEditorUserId="1832" LastEditDate="2017-07-26T03:08:08.803" LastActivityDate="2021-03-09T05:16:14.577" Title="How to determine motor needed for water pump" Tags="|electrical-engineering|motors|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16454" PostTypeId="2" ParentId="16452" CreationDate="2017-07-26T06:31:10.473" Score="2" Body="&lt;p&gt;No, because the pump is designed to limits and if you exceed its working or maximum pressure then, for example, its seals fail and it just does not pump any more.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-26T06:31:10.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16455" PostTypeId="2" ParentId="16445" CreationDate="2017-07-26T08:08:49.937" Score="1" Body="&lt;p&gt;Using the +10 to +120°C psychrometric chart with sensible heating from 40°C, 95% RH to 60°C gives approximately 35% RH. If you need a more precise answer I would look up the formulas which can be found in ASHRAE Fundamentals or CIBSE Guide C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See image below for how to look this up on the chart:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KCDCu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KCDCu.png&quot; alt=&quot;psychrometric chart with markup&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4210" LastActivityDate="2017-07-26T08:08:49.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16457" PostTypeId="2" ParentId="16448" CreationDate="2017-07-26T10:14:36.730" Score="0" Body="&lt;p&gt;For the wire armour you gave the answer : magnetism as long as the armour is not aluminium based...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-26T10:14:36.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16459" PostTypeId="1" AcceptedAnswerId="16461" CreationDate="2017-07-26T12:35:28.490" Score="0" ViewCount="80" Body="&lt;p&gt;Instead the pump relied on the diesel generators, why didn't it just use the residual heat to generate electricity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.s. The question is very specific to the residual heat generated after inserting the control rods and the cooling pump which failed to cool down the system since it has no electricity when the diesel generators failed.&lt;/p&gt;&#xA;" OwnerUserId="12311" LastEditorUserId="9" LastEditDate="2017-08-04T21:37:44.180" LastActivityDate="2017-08-04T21:37:44.180" Title="Why didn't the Fukushima turbine supply electricity to the residual heat removal system?" Tags="|nuclear-engineering|" AnswerCount="1" CommentCount="0" ClosedDate="2017-07-28T00:34:41.790" ContentLicense="CC BY-SA 3.0" />
  <row Id="16461" PostTypeId="2" ParentId="16459" CreationDate="2017-07-26T13:06:42.743" Score="2" Body="&lt;p&gt;According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fukushima_Daiichi_nuclear_disaster#Events&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia entry&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[The earthquake] produced maximum ground g-forces of 0.56, 0.52, 0.56 (5.50, 5.07, and 5.48 m/s2) at units 2, 3, and 5 respectively. This exceeded the earthquake tolerances of 0.45, 0.45, and 0.46 g (4.38, 4.41, and 4.52 m/s2). The shock values were within the design tolerances at units 1, 4, and 6.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When the earthquake struck, units 1, 2, and 3 were operating, but units 4, 5, and 6 had been shut down for a scheduled inspection. Reactors 1, 2, and 3 immediately shut down automatically; this meant the plant stopped generating electricity and could no longer use its own power. One of the two connections to off-site power for units 1–3 also failed, so 13 on-site emergency diesel generators began providing power.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;:EDIT:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/16459/why-didnt-the-fukushima-turbine-supply-electricity-to-the-residual-heat-removal/16461?noredirect=1#comment29580_16461&quot;&gt;OP asked&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;yes, but isn't &quot;immediately shut down automatically&quot; means inserting Control rods? There is still unstoppable nuclear fission which create the residual heat, I'm just wondering why they don't utilize the residual heat to cool itself&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Well, the &quot;residual heat&quot; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Decay_heat#Power_reactors_in_shutdown&quot; rel=&quot;nofollow noreferrer&quot;&gt;decay heat&lt;/a&gt;) &lt;em&gt;is&lt;/em&gt; an effect that generates heat, but (speaking from military reactor experience, not civilian) one of the goals after a scram or fast insertion is to recover the plant, and that means not letting the primary cool down too much. This is assisted by automatically closing the main steam valves on the steam generators, which isolates the secondary plant and thus also isolates the turbine generators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to recover the plant, you need to establish &lt;em&gt;why&lt;/em&gt; the shutdown occurred, assess any damage, and then start bringing the plant back online. The entire time this is happening though, the secondary plant piping is cooling down. You can get into a scenario where you may need to do an accelerated secondary warmup, which still takes time despite being &quot;accelerated.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address your question directly, I'll make a couple points that are &lt;em&gt;pure speculation on my part.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Biggest point in my mind is that there was an assumption that backup power would be available regardless of current catastrophic conditions. There's no need to cool the plant down too quickly if you have the luxury of time thanks to a bank of diesel generators. &lt;/li&gt;&#xA;&lt;li&gt;Next biggest point - In the heat of the moment, you need to respond to &lt;em&gt;what is happening&lt;/em&gt;. You can't take time to consider the &quot;what ifs&quot; that &lt;em&gt;could&lt;/em&gt; happen down the line; you need to address the current casualty. You can say, &quot;The reactor just tripped offline, what if the generators go down?&quot; but you could also say, &quot;The reactor just tripped offline, what if the main steam valves get stuck shut?&quot; or, &quot;..., what if the cooling water pumps won't restart?&quot; The list of hypothetical questions goes on. It was up to the engineers that designed the power plant to consider the likely scenarios, and they codify the actions you should take in a book of &lt;em&gt;emergency procedures.&lt;/em&gt; If they were doing their jobs correctly, &lt;strong&gt;the people running the plant were performing the actions dictated in a checklist that was written before the plant was commissioned.&lt;/strong&gt; If the checklist said not to use decay heat for backup power (or failed to instruct them to use decay heat for backup power) then that would be why they didn't.&lt;/li&gt;&#xA;&lt;li&gt;Available decay heat may have been insufficient or borderline. Bear in mind that when you make enough power to &quot;just cool the primary,&quot; that implies that you also need to make enough power to run all of the auxiliary machinery for the secondary, too. Depending on the plant, you need to run multiple primary coolant pumps - these are NOT small pumps! Picture at the end with people for scale. But you need a means to return the condensed steam back to the steam generator, so you need condensate pumps running, and booster pumps running, and main feed pumps running, you need to have the lubricating oil system online, so you need to run oil pumps (again, not small for a large commercial turbine), and you need to have cooling water to cool the condenser, so you need to have those pumps online, also. I don't have a number on the percentage of full power that the auxiliary machinery consumes, but it is not insignificant. All this is to say that using decay heat to cool the primary plant down may result in a rapid cooldown that exceeds safety limits. The reactor vessel itself has stainless steel walls that are 8-10+ inches thick (20-25+ cm), and the temperature gradient across the thickness there and in the primary piping could result in stresses that cause &lt;em&gt;permanently bending&lt;/em&gt; (plastic deformation). This is generally the concern for rapid cooldowns; heating up too fast could lead to &lt;a href=&quot;https://en.wikipedia.org/wiki/Fracture#Brittle_fracture&quot; rel=&quot;nofollow noreferrer&quot;&gt;brittle fracture&lt;/a&gt;, where part of the primary plant could suddenly and completely fail. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2017-07-27T13:39:51.787" LastActivityDate="2017-07-27T13:39:51.787" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16463" PostTypeId="1" CreationDate="2017-07-26T14:21:15.957" Score="1" ViewCount="1848" Body="&lt;p&gt;When to assume ideal gas law without incurring too intolerable error. I have the idea that it must be at high temperature and low pressure. Is there any specific values, and are there any other things to look for?&lt;/p&gt;&#xA;" OwnerUserId="12300" LastActivityDate="2017-07-27T13:37:25.073" Title="A rule of thumb when assuming ideal gas" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16464" PostTypeId="1" AcceptedAnswerId="16466" CreationDate="2017-07-26T15:54:11.493" Score="0" ViewCount="295" Body="&lt;p&gt;I know how to use thermal resistance circuits to calculate heat transfer in Cartesian and cylindrical coordinates, but I don't know how to deal with real-world insulation values, and I was not able to find much about this online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have this hot water tank that has an insulation jacket with 2&quot; R-12.5 insulation. For simplicity, let's assume the insulation is the only thing that separates the hot water from cold outside air (no heat transfer from bottom of tank, top of tank, etc. Just through the insulation jacket). Assume water tank thickness aside from the 2&quot; insulation jacket is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it goes hot water -&gt; insulation -&gt; cold air outside. I can look up a free convection coefficient for the tank's hot water and can look up a convection coefficient for air as well. That allows me to use the formula&#xA;$$&#xA;R=1/(h*2*pi*r*heightofinsulationjacket)&#xA;$$&#xA;twice (once for water and once for air).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem is how to deal with the insulation jacket. How do I interpret 2&quot; R-12.5 insulation? How do I factor in the thickness of the insulation? The surface area is taken care of in the heat transfer equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Q''=(Tf-Ti)/R&#xA;$$&#xA;or&#xA;$$&#xA;Q=(Tf-Ti)/R*A&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Updated: How do I use an R-value in cylindrical coordinates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g.&#xA;$$&#xA;R/A=1/(hW*2*pi*r1)+Rins/(whatAreaHere)+1/(hA*2*pi*r2)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am missing the area for my R-value insulation&lt;/p&gt;&#xA;" OwnerUserId="12217" LastEditorUserId="12217" LastEditDate="2017-07-26T22:50:48.877" LastActivityDate="2017-07-26T22:50:48.877" Title="How do you use an insulation jacket's R-value for calculating heat transfer from a hot water tank?" Tags="|heat-transfer|energy-storage|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16465" PostTypeId="2" ParentId="16448" CreationDate="2017-07-26T16:30:41.047" Score="0" Body="&lt;p&gt;You say &quot;sent to scrap&quot;, I will bet a large amount of money that somewhere a scrap dealer/ recycler is separating all the copper . Single source scrap is easy to separate, likely by hand. Likely , then it will be processed ( chopped ,maybe burned, etc) to remove insulation and armor ( no doubt armor is steel so magnetic separation.). Because of it's high value copper base waste is often sorted by hand . An experienced sorter can do an impressive job separating various coppers, brasses, and bronzes into several categories. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-07-26T16:30:41.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16466" PostTypeId="2" ParentId="16464" CreationDate="2017-07-26T17:32:36.383" Score="3" Body="&lt;p&gt;The insulation R-value already accounts for thickness, so you can ignore it. You just have to account for the effect of surface area and temperature difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SI units of R are $\frac{K \cdot m^2}{W}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat transfer increases as the surface area increases, so Q is calculated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_{insulation}=\frac{A (T_f-T_i)}{R}$$ and&#xA;$$Q''{insulation}=\frac{(T_f-T_i)}{R}$$&lt;/p&gt;&#xA;" OwnerUserId="11151" LastActivityDate="2017-07-26T17:32:36.383" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16467" PostTypeId="2" ParentId="203" CreationDate="2017-07-26T17:52:16.917" Score="1" Body="&lt;p&gt;I'd agree that you're looking for an epoxy that stays relatively flexible upon curing, especially if the thermal expansion rates of the epoxy and substrates are different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you haven't already done so, I'd recommend searching for &quot;underfill&quot; epoxy. There are a few Loctite and Masterbond products that fit the bill. They typically flow very well during application, have good structural stability (don't expand/contract), survive solder re-flow temperatures (~250C), and remain flexible upon curing. The cryogenic properties you're looking for might be harder to satisfy.&lt;/p&gt;&#xA;" OwnerUserId="11151" LastActivityDate="2017-07-26T17:52:16.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16469" PostTypeId="2" ParentId="16464" CreationDate="2017-07-26T18:15:21.843" Score="1" Body="&lt;p&gt;@geekly: agreed.  Also, insulation thickness will increase surface area for external convection, as:&#xA;$$ Q_{jacket \to air} = 2\pi (l_{heater})(r_{heater} + t_{jacket})h(T_{jacket} - T_{air})$$&#xA;This effect prompted the critical radius of insulation for cyclinders:&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_insulation&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thermal_insulation&lt;/a&gt;:&#xA;$$ r_{critical} = \frac{k}{h} = \frac{t_{jacket}}{R_{jacket}h}$$&lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2017-07-26T18:15:21.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16470" PostTypeId="2" ParentId="2032" CreationDate="2017-07-26T19:25:20.427" Score="3" Body="&lt;p&gt;I would focus on two things - 1) the difference in heat transfer coefficients between the two materials and 2) the difference in heat capacity of the two materials. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The heat transfer coefficient is dependent on the physical interface between the air and the solid. Surface area of the materials and the amount of airflow would both factor in. As mentioned above, the smaller the particles, the more surface area, but the more restrictive it will be to airflow. There's a happy balance there that you might have to determine experimentally.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The heat capacity of the sink material determines how quickly the material's temperature will respond to a change in ambient temperature. The higher this is, the better the sink will perform. An increase in density and specific heat make for a better heatsink material. This is independent of the size of the rocks or the rate of airflow - bigger heat capacity will always be better.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As for the shape of the curves, I would never expect the rate of temperature change to be linear in this case, because the rate of change will change with difference in temperature. It's an exponential relationship. The granite warming curve looks most like what I'd expect to see for convection cooling/heating in a heat exchanger. The shape of the curve is pretty predictable, and by fitting it to a curve of the form $T = C-A e^{-b x}$ we can predict that the room temperature is about 24ºC. The initial rise in temperature of the glass cooling is especially perplexing.&lt;/p&gt;&#xA;" OwnerUserId="11151" LastEditorUserId="1832" LastEditDate="2017-07-26T20:20:50.897" LastActivityDate="2017-07-26T20:20:50.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16471" PostTypeId="2" ParentId="6537" CreationDate="2017-07-26T19:47:28.587" Score="0" Body="&lt;p&gt;Based on surface texture, it looks like this might be a sand casting. That said, I'm not familiar with the distinction between inside and outside dimensions. We do commonly differentiate between dimensions that cross a parting line and those that don't. The ones that cross a parting line have a greater tolerance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is an inside dimension one that's formed by a core, or is contained completely on one side of the mold? I'm definitely interested for my own reference.&lt;/p&gt;&#xA;" OwnerUserId="11151" LastActivityDate="2017-07-26T19:47:28.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16472" PostTypeId="1" CreationDate="2017-07-26T22:27:28.347" Score="1" ViewCount="18" Body="&lt;p&gt;I'm working on a university project about making an absorber function. It's made out of glass and it's very thin (about 6 in diameter). The packing is Berl Saddle made out of ceramic and it's about 0.5 inches. It has a bed height of about 1.406 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how to determine the best liquid flow for a determined air flow? Since we won't be able to get a supply of NH3-Air mixture in time (nor the time to mix them ourselves),we only need to calculate the best flow rates for the absorber to not flood. We only need to calculate maximum liquid rate so the gas is enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm new to this, and my calculations were for reducing from X% NH3 to Y% NH3. Any help?&lt;/p&gt;&#xA;" OwnerUserId="12317" LastActivityDate="2017-07-26T22:27:28.347" Title="Best flow rates of gas and liquid for a specific absorber" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16475" PostTypeId="1" CreationDate="2017-07-27T09:41:02.327" Score="4" ViewCount="8719" Body="&lt;p&gt;Generally, downlink freq. &gt; uplink freq. in mobile communications. I was able to gather some answers as to why is so. But my seniors (in my company) seem not to be satisfied with the below answers.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;High frequency means higher attenuation. So lets suppose that BTS (as in case of GSM) transmits using low frequency on the downlink. So, that means lower attenuation. Hence, data is received as it should be (without attenuation). So, MS (mobile) has to transmit on uplink using higher freq. in order to avoid interference. But high freq. means higher attenuation. So, in order to overcome the attenuation, MS has to transmit on high power. But due to the constraint on the amount of energy available on the MS battery, this approach is not feasible. Hence, downlink frequency should be greater than uplink frequency.&lt;/li&gt;&#xA;&lt;li&gt;This point is from user perspective. Generally, people download more than upload (however, its converging now). So, we require more bandwidth in DL than in UL and it can be attained only if DL freq. &gt; UL freq. Hence DL freq. &gt; UL freq.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have searched almost everything (however, that's not possible). If you could help me regarding the same, it would be very kind.&lt;/p&gt;&#xA;" OwnerUserId="11767" LastEditorUserId="7083" LastEditDate="2018-12-28T00:04:55.873" LastActivityDate="2024-02-03T22:07:15.983" Title="Why downlink frequency is greater than uplink frequency in mobile communications?" Tags="|telecommunication|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16477" PostTypeId="2" ParentId="16463" CreationDate="2017-07-27T13:37:25.073" Score="1" Body="&lt;p&gt;One way to describe how gasses deviate from ideal gasses is by the compressibility factor. Compressibility is defined as:&#xA;$$Z=\frac{PV}{nRT}$$&#xA;For an ideal gas, Z = 1, which leads to the familiar ideal gas law:&#xA;$$PV=nRT$$&#xA;You can look of various compressibility factors for air at different temperatures and pressures. When the compressibility deviates from 1.0, air is not behaving like an ideal gas. You can find a table of values here:&#xA;&lt;a href=&quot;http://www.enggcyclopedia.com/2011/09/air-compressibility-factor-table/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.enggcyclopedia.com/2011/09/air-compressibility-factor-table/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty may be in determining how much deviation of compressibility equates to error. It's not possible to answer that without knowing more about your application and your definition of intolerable error.&lt;/p&gt;&#xA;" OwnerUserId="11151" LastActivityDate="2017-07-27T13:37:25.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16478" PostTypeId="1" AcceptedAnswerId="16487" CreationDate="2017-07-27T13:50:35.320" Score="4" ViewCount="143" Body="&lt;p&gt;This is a theoretical question.&#xA;It is known that in &quot;classic&quot; vibration analysis natural frequencies can be found by solving an eigenvalue problem, from the undampened vibration equation like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{M}\mathbf{\ddot{u}} + \mathbf{K}\mathbf{u} = \mathbf{0}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbf{u}$ is the displacement vector. &#xA;But, with reference to this situation&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MfooU.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MfooU.png&quot; alt=&quot;Cantilever beam with cable&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started to wonder if this type of analysis is not correct here because $\mathbf{K}$ (stiffness) of the cable is not constant but $\mathbf{u}$-dependant (equal to zero when $\mathbf{u}$  is a displacement of compression). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would like to ask if someone could suggest me another strategy, or show why the eigenvalue analysis could be used also in this case.&#xA;Thank you all.&lt;/p&gt;&#xA;" OwnerUserId="12262" LastEditorUserId="12262" LastEditDate="2017-07-27T13:56:30.270" LastActivityDate="2017-07-28T12:55:09.413" Title="Strategies for cable vibration: eigenvalues are suitable?" Tags="|vibration|eigenvalue-analysis|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16481" PostTypeId="1" AcceptedAnswerId="16488" CreationDate="2017-07-27T15:20:54.547" Score="4" ViewCount="1846" Body="&lt;p&gt;What is meant by a &quot;system curve&quot; that is used to determine the operating point of a pump. I know what a &quot;pump curve&quot; is, but I often hear about a &quot;system curve&quot; being generated so that the pump operating point can be determined by where the pump curve and system curve intersect.&lt;/p&gt;&#xA;" OwnerUserId="10344" LastEditorUserId="2015" LastEditDate="2018-08-16T14:00:29.293" LastActivityDate="2018-08-16T14:00:29.293" Title="System curve for use in determining pump operating point" Tags="|mechanical-engineering|fluid-mechanics|pumps|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16482" PostTypeId="2" ParentId="15007" CreationDate="2017-07-27T15:33:45.777" Score="0" Body="&lt;p&gt;I've always known this to be classed as a 'locating pin' or simply 'location' or even 'mating'. It also depends on the type of fit you are trying to achieve, whether it is an interferance, clearance or transition fit.&lt;/p&gt;&#xA;" OwnerUserId="12187" LastActivityDate="2017-07-27T15:33:45.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16483" PostTypeId="2" ParentId="16453" CreationDate="2017-07-27T15:42:14.030" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So how do i calculate the motor i need to run the water pump? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The ideal hydraulic power required to drive a pump is simply a function of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mass flow rate&lt;/li&gt;&#xA;&lt;li&gt;Fluid density&lt;/li&gt;&#xA;&lt;li&gt;Total head loss of the system.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, the ideal power required is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P_{hydraulic-kW} = (q*\rho*g*h)/(3.6*10^6)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$q = flow rate (\frac {m^3}{hr})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \rho = density(\frac {kg}{m^3})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ g = gravity(9.81 \frac {m}{s^2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ h = Head Loss (m)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's pretty much plug 'n chug to get this far, but this is just the amount of power that the pump has to impart to the fluid in order to achieve the flow rate. Since you're interested in the motor and the pump is not perfect, the efficiency of the pump in converting mechanical power fom the motor into fluid power has to be taken into account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P_{shaft-kW} = P_{hydraulic-kW}(\eta_{pump}) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\eta_{pump}$ is the pump efficiency&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the value for $\eta_{pump}$ of your drill pump is either something you'll have to look up, estimate or measure. (I'd guess it's 0.4-ish, but that's a guess, nothing more) &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-07-27T15:42:14.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16484" PostTypeId="2" ParentId="16481" CreationDate="2017-07-27T16:03:10.780" Score="1" Body="&lt;p&gt;Usually a pressure / flow characteristic curve for a given power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are usually provided by the manufacturer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you do an engineering degree where you get to measure the flow rate, pressure, power, calculate the efficiency etc and then have to plot all the dimensionless groups etc to show what is happening.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-27T16:03:10.780" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16485" PostTypeId="2" ParentId="16426" CreationDate="2017-07-27T16:43:52.750" Score="1" Body="&lt;p&gt;You basically want to solve for the homogeneous solutions of the system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\textbf{M}\, \ddot{\textbf{q}} + \textbf{D}\, \dot{\textbf{q}} + \textbf{K}\, \textbf{q} = \textbf{0}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be shown that each solution will be of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\textbf{q}_i(t) = \textbf{u}_i\,e^{\mu_i\,t},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\textbf{u}_i$ is a vector of the same dimension as $\textbf{q}$ (usually normalized to a length of one) and $\lambda_i$ is a scalar. In general any homogeneous solution can be constructed by taking a linear combination these solution (I will ignore for now that you can also have &lt;a href=&quot;http://mathworld.wolfram.com/JordanCanonicalForm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jordan blocks&lt;/a&gt; of bigger then one by one, which will increase the possible solutions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the solution into the homogeneous equation yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\textbf{M}\, \textbf{u}_i\,\mu_i^2\,\,e^{\mu_i\,t} + \textbf{D}\, \textbf{u}_i\,\mu_i\,\,e^{\mu_i\,t} + \textbf{K}\, \textbf{u}_i\,e^{\mu_i\,t} = \left(\textbf{M}\, \mu_i^2 + \textbf{D}\, \mu_i + \textbf{K}\right) \textbf{u}_i\,e^{\mu_i\,t} = \textbf{0}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $e^{\mu_i\,t}$ is always non-zero (for finite time) it can be ignored since the remaining terms on the left hand side of the equation still has equal to zero in order to satisfy the right hand side. Doing so already starts to make the equation look more like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix#Generalized_eigenvalue_problem&quot; rel=&quot;nofollow noreferrer&quot;&gt;generalized eigenvalue problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\textbf{A}\, \textbf{v} = \lambda\, \textbf{B}\, \textbf{v},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;however $\mu_i$ appears both linearly and quadratically. In the special case when $\textbf{B} = 0$ then the generalized eigenvalue problem can be formulated using $\textbf{A} = \textbf{K}$, $\textbf{B} = \textbf{M}$, $\textbf{v} = \textbf{u}_i$ and $\lambda = -\mu_i^2$. Another special case is when $\textbf{M}$, $\textbf{D}$ and $\textbf{K}$ are all simultaneously diagonalizable by pre and post multiplying them by the same matrices. Namely in that case the differential equation can be written as a system of decoupled second order differential equations, like a single mass-spring-damper system. One way such transformation could be constructed when $\textbf{M}$, $\textbf{D}$ and $\textbf{K}$ are symmetric is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\mathbf{U} = \begin{bmatrix}\mathbf{u}_1 &amp;amp; \mathbf{u}_2 &amp;amp; \cdots &amp;amp; \mathbf{u}_n\end{bmatrix},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbf{u}_i$ are the eigenvectors of the eigenvalue problem with $\textbf{B} = 0$. The matrices with which you pre and post multiply would then be $\mathbf{U}^\top$ and $\mathbf{U}$ respectively. Often if the system is not simultaneously diagonalizable and the damping is small, then assuming it is can be a good approximation (so setting the non-diagonal elements of $\mathbf{U}^\top \mathbf{B}\, \mathbf{U}$ to zero). A specific form for which this is the case is called Rayleigh damping, when $\mathbf{B} = \alpha\,\mathbf{M} + \beta\,\mathbf{K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of all the previous mentioned methods is that the generalized eigenvalue problem you have to solve are of the same dimension as $\mathbf{q}$. However if none of cases mentioned previously apply to your system you would have to solve the generalized eigenvalue problem of twice the dimension of $\mathbf{q}$. There are multiple ways of formulating this. A few of them are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;\mathbf{I} &amp;amp; \mathbf{0} \\ &#xA;\mathbf{0} &amp;amp; \mathbf{M}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\dot{\mathbf{q}} \\ \ddot{\mathbf{q}}&#xA;\end{bmatrix} = &#xA;\begin{bmatrix}&#xA;\mathbf{0} &amp;amp; \mathbf{I} \\ &#xA;-\mathbf{K} &amp;amp; -\mathbf{D}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\mathbf{q} \\ \dot{\mathbf{q}}&#xA;\end{bmatrix},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;\mathbf{I} &amp;amp; \mathbf{0} \\ &#xA;\mathbf{0} &amp;amp; \mathbf{I}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\dot{\mathbf{q}} \\ \ddot{\mathbf{q}}&#xA;\end{bmatrix} = &#xA;\begin{bmatrix}&#xA;\mathbf{0} &amp;amp; \mathbf{I} \\ &#xA;-\mathbf{M}^{-1}\mathbf{K} &amp;amp; -\mathbf{M}^{-1}\mathbf{D}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\mathbf{q} \\ \dot{\mathbf{q}}&#xA;\end{bmatrix},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;\mathbf{D} &amp;amp; \mathbf{M} \\ &#xA;\mathbf{M} &amp;amp; \mathbf{0}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\dot{\mathbf{q}} \\ \ddot{\mathbf{q}}&#xA;\end{bmatrix} = &#xA;\begin{bmatrix}&#xA;-\mathbf{K} &amp;amp; \mathbf{0} \\ &#xA;\mathbf{0} &amp;amp; \mathbf{M}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\mathbf{q} \\ \dot{\mathbf{q}}&#xA;\end{bmatrix}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this extended state space it can easily be seen that the solutions will be of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;\mathbf{q}_i(t) \\ \dot{\mathbf{q}}_i(t)&#xA;\end{bmatrix} = &#xA;\begin{bmatrix}&#xA;\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i&#xA;\end{bmatrix} e^{\mu_i\,t}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting this into the three equations above and factoring out $e^{\mu_i\,t}$ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;\mathbf{I} &amp;amp; \mathbf{0} \\ &#xA;\mathbf{0} &amp;amp; \mathbf{M}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i&#xA;\end{bmatrix} \mu_i = &#xA;\begin{bmatrix}&#xA;\mathbf{0} &amp;amp; \mathbf{I} \\ &#xA;-\mathbf{K} &amp;amp; -\mathbf{D}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i&#xA;\end{bmatrix},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;\mathbf{I} &amp;amp; \mathbf{0} \\ &#xA;\mathbf{0} &amp;amp; \mathbf{I}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i&#xA;\end{bmatrix} \mu_i = &#xA;\begin{bmatrix}&#xA;\mathbf{0} &amp;amp; \mathbf{I} \\ &#xA;-\mathbf{M}^{-1}\mathbf{K} &amp;amp; -\mathbf{M}^{-1}\mathbf{D}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i&#xA;\end{bmatrix},&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;\mathbf{D} &amp;amp; \mathbf{M} \\ &#xA;\mathbf{M} &amp;amp; \mathbf{0}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i&#xA;\end{bmatrix} \mu_i = &#xA;\begin{bmatrix}&#xA;-\mathbf{K} &amp;amp; \mathbf{0} \\ &#xA;\mathbf{0} &amp;amp; \mathbf{M}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\mathbf{u}_i \\ \mu_i\,\mathbf{u}_i&#xA;\end{bmatrix}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are all just generalized eigenvalue problems. I am not an expert on numerically solving these problems, however I have few things that can be noted. Namely the second problem formulation requires a matrix inverse, which might not exist, but does reduce it to a normal eigenvalue problem (not sure if this requires any less computation time in general). The third problem formulation will yield symmetric matrices if $\textbf{M}$, $\textbf{D}$ and $\textbf{K}$ are symmetric as well, which might have some attractive properties when doing numerical calculations. So depending on your system and solver a different formulation might give more accurate results.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2017-07-27T16:43:52.750" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16486" PostTypeId="2" ParentId="16481" CreationDate="2017-07-27T18:32:39.387" Score="0" Body="&lt;p&gt;System curve is simply a plot of the required head against change in system flow rate, as the required head (mainly major and minor losses) increases with the increase in flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, for a pump curve, you can see that the available head by the pump decreases with the increase in flow rate. So, if you put the two curves on a single plot (and if the pump is properly selected), you will have an intersection point between the two curves, and that's your operating point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5MzL7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5MzL7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2017-07-27T18:32:39.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16487" PostTypeId="2" ParentId="16478" CreationDate="2017-07-27T21:14:33.533" Score="4" Body="&lt;p&gt;Mathematically, an eigenvalue analysis assumes the modal displacements are infinitesimally small, so the change in stiffness in the cable caused by the vibration is a second order effect which can be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This very-small-amplitude vibration is a good place to &lt;em&gt;start&lt;/em&gt; understanding the behaviour of the structure - for example, look at how the frequency depends on the angle of the beam to the horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if you are interested in large amplitude vibrations, the change in stiffness during each cycle of oscillation can not be ignored, but the vibrations are not sinusoidal anyway. Really, you can only simulate this by a time-marching transient dynamics analysis, updating the stiffness at each time step. The stiffness will vary along the length of the cable because its self weight will cause the tension to change along its length, and also because it is not straight!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simple situations (for example oscillations of a simple pendulum) you can investigate the large amplitude behaviour analytically, but that assumes you know the deformed shape of the structure (which is trivial for a simple pendulum, because the massless string is always straight so long as it remains in tension). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to formulate the large-amplitude motion of your structure the same way as for a pendulum, if you assume the beam is rigid and the cable is massless and stretches along its length. You then will have only one degree of freedom, measured by the change in angle of the beam (or the cable) from its equilibrium position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even for large amplitude oscillations of a pendulum, the period is an elliptic function of the amplitude which can only be evaluated numerically (e.g. by a power series approximation). For anything more complicated, attempting an analytic solution is unlikely to give much quantitative information - though a phase-plane plot may tell you some interesting things qualitatively.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-07-27T21:23:18.203" LastActivityDate="2017-07-27T21:23:18.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16488" PostTypeId="2" ParentId="16481" CreationDate="2017-07-27T22:38:47.953" Score="3" Body="&lt;p&gt;Imagine if you were pumping through a single pipe.  That's all the pump does, is take in water from a source and pumps it along a very long length of pipe.  Since this is just plain piping, the friction loss isn't difficult to determine, but you need to know the velocity to calculate the &lt;a href=&quot;https://en.wikipedia.org/wiki/Reynolds_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reynolds number&lt;/a&gt;.  Without it, you can't use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Moody_chart&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moody Chart&lt;/a&gt; to calculate a friction factor to find out the loss.  But, with a good estimate of the flow rate, the estimated range of the flow rate could be guessed at, and a friction factor can be guessed.  With a friction factor estimate, but a flow rate unknown, you finally settle on the pressure loss across this long length of pipe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta P =f_D \frac{\rho V^2}{2}\frac{L}{D} = f_D \frac{8\rho Q^2}{\pi^2}\frac{L}{D^5}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where P is pressure, Q is the flow rate, $\rho$ is the density, L is pipe length, D is diameter, and $f_D$ is the friction factor.  Let's take cast iron 100mm pipe, highly turbulent regime, running for 10km.  Plugging all the numbers in this theoretical situation, we come to (hypothetically):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta P = 0.304 Q^2 \frac{kPa}{(\frac{L}{s})^2} $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have a formula to give an idea for pump sizing!  More importantly, we have a curve.  Plugging in any arbitrary value of Q (in L/s) would give you the pressure loss across the pipe in kPa.  This curve is the system curve, and it naturally looks like a parabola.  This is, in general, true for all fluid systems without any type of controls response (acting under natural behavior). You can plot this curve on top of the pump curve, and find out where the system will reach equilibrium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, it isn't hard with such a simple system to get a lot of liquid out.  More complex system have a lot more intricate mechanics, but the general parabola rule still applies.  Thus, most people work through the complex mechanics of their system to simplify it into a single point.  In our case, operating at 25 L/s would mean a pump that has to pump 190 kPa.  This is the single operating point.  Typically, many engineers will slightly increase these values, so they will always find an operating point that's safe.  In this case, going to 30 L/s would mean a 275 kPa.  Thus, the only parabola that goes through the point 30 L/s, 275 kPa and the origin (there is only one parabola that does this) would be the system curve.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-06-07T15:42:29.727" LastActivityDate="2018-06-07T15:42:29.727" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="16489" PostTypeId="1" AcceptedAnswerId="16491" CreationDate="2017-07-27T01:36:15.730" Score="1" ViewCount="146" Body="&lt;p&gt;I am analysing the forces within a certain frame, and for some reason there do not seem to be sufficient equations of equilibrium to solve for all the internal forces. Is there something about this structure than makes it unsolvable without more information?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a frame, ABCDE, shown below. Two equal forces are applied to it (both shown as F). Assume we know F, w, and h (and from the illustration, l = w/2). Let's try to find the forces within member ADC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/la6mJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/la6mJ.png&quot; alt=&quot;Frame illustration&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we can analyse the frame as a whole. It is straightforward to show that Ay = By = F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we should be able to analyse member ADC by itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Db9WX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Db9WX.png&quot; alt=&quot;Member ADC&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we sum forces in the y-direction, we get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ay + Cy - F = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Since we already found that Ay = F when we analysed the whole frame, we know that Cy = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we sum forces in the x-direction, we get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Ax + Dx + Cx = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And if we sum moments around A, we get:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;F*w/4 + Dx*h + Cx*w/2 = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Without going any further, we can see that we have two equations with three unknowns (Ax, Dx, and Cx).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the other members in the frame, we don't get any new information, I don't think... So what's missing? How can this simple frame be an unsolvable physical conundrum?&lt;/p&gt;&#xA;" OwnerUserId="12330" OwnerDisplayName="PProteus" LastEditorUserId="12330" LastEditDate="2017-07-27T23:53:05.990" LastActivityDate="2017-07-28T08:04:50.973" Title="Structural Analysis Conundrum" Tags="|structural-engineering|statics|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16490" PostTypeId="1" CreationDate="2017-07-28T04:52:58.397" Score="2" ViewCount="139" Body="&lt;p&gt;Sky scrapers are often found in dense urban environments. Is there a known method to controllably demolish a sky scraper with minimal collateral damage?&lt;/p&gt;&#xA;" OwnerUserId="602" LastActivityDate="2017-08-19T04:33:21.217" Title="Controlled demolition of a sky scraper?" Tags="|civil-engineering|building-physics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16491" PostTypeId="2" ParentId="16489" CreationDate="2017-07-28T08:04:50.973" Score="2" Body="&lt;p&gt;It's statically indeterminate.  It's far from unique - most real world structures are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trivially / intuitively, the frame 'works' if member DE is omitted (ie, you could take it out and the frame still stands up), so we can say immediately that there's a valid solution with member DE having zero axial force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, we can see that if you put a tension into that member (for example, if it's slightly too short and you stretch it to fit it in place) the frame would still work.  Thus, there's a valid solution with member DE having a non-zero axial force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To analyse it, you need to know more.  The normal assumption would be that the members are all unstressed if no loads are imposed (ie, if the members are weightless and F is removed, there's no member actions anywhere).  Then, if you know the relative stiffness of the members you can work out to what degree the forces distribute.  If DE is very axially compliant and AC and CB are very flexurally stiff, the frame would tend towards acting as if DE wasn't there - DE would have a very low force.  If DE is very stiff, and the sides are relatively flexible, then DE would tend to prop the sides apart and pick up lots of load.&lt;/p&gt;&#xA;" OwnerUserId="902" LastActivityDate="2017-07-28T08:04:50.973" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16493" PostTypeId="1" CreationDate="2017-07-28T10:08:01.683" Score="0" ViewCount="459" Body="&lt;p&gt;Why is -180° taken as a reference or phase margin in Bode plots? If the phase plot (grid impedance and inverter impedance versus frequency) exceeds -180° at the gain crossover frequency, then the system will be unstable in a grid-tied inverter system. So why is -180° taken as a critical point?&lt;/p&gt;&#xA;" OwnerUserId="12339" LastEditorUserId="12330" LastEditDate="2017-07-28T14:54:21.233" LastActivityDate="2017-09-29T16:58:28.247" Title="Bode Plot Stability Criteria" Tags="|control-theory|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16494" PostTypeId="1" CreationDate="2017-07-28T11:03:16.847" Score="-1" ViewCount="63" Body="&lt;p&gt;I work in engineering company that doing pipe installation. Just now our company got inquiry from Japanese company. They ask me to manufacture a pipe for exhaust ducting. The thing is they use Japanese standard (JIS). The size that they want to make is 150A, 250A, and 600A. My question is what is the thickness of these pipe?&lt;/p&gt;&#xA;" OwnerUserId="12341" LastActivityDate="2018-05-24T20:19:15.463" Title="What is the Thickness of JIS Standard Pipe" Tags="|mechanical-engineering|materials|standards|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16495" PostTypeId="2" ParentId="16490" CreationDate="2017-07-28T12:33:44.663" Score="3" Body="&lt;p&gt;It is certainly possible, although far from easy and requires very specialist knowledge. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First the building will be surveyed in great detail to determine how the loads are carried throughout the structure. &lt;/li&gt;&#xA;&lt;li&gt;As necessary internal walls and cladding will be removed to expose load bearing structures. &lt;/li&gt;&#xA;&lt;li&gt;The building may be weakened by partially cutting through beams, columns etc &lt;/li&gt;&#xA;&lt;li&gt;Explosive charges are set on predetermined parts of the building. The intent is to &lt;strong&gt;cut&lt;/strong&gt; through beams and columns in a clean and controlled manner rather than pulverise the whole building. Different explosive may be used for different materials and thermite may be used to cut steel beams. &lt;/li&gt;&#xA;&lt;li&gt;Charges may be wrapped in ballistic textile to prevent shrapnel. &lt;/li&gt;&#xA;&lt;li&gt;The danger area will be evacuated and the charges fired in a precise sequence. The intent being to use the weight of the building to pull it in on itself.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If everything goes right the building should collapse onto its own footprint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the comments it is easy to find videos of controlled demolitions on the internet and in many cases you can see the sequence of charges firing. The explosions themselves are usually pretty small, it's the collapse under its own weight which breaks the structure up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concrete buildings are generally demolished from the bottom. Steel framed buildings may need additional charges through out the height of the structure to sever beams and columns into manageable sections and stop the whole thing toppling over. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases explosive demolition may not be practical or adequately safe. In this case it may be necessary to disassemble the whole building from the top down. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit : A comment has raised a good point about the safety of the dust produced. Clearly any airborne dust is something you want to avoid. Also part of the preparation process should be to remove any material which would pose a potential hazard including things like asbestos, glass/glass fibre or any material heavily contaminated with anything especially nasty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in general it is mostly just masonry dust and isn't a massive cause for concern as long as people are kept out of the affected area untill it has settled. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-07-28T17:22:53.763" LastActivityDate="2017-07-28T17:22:53.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16496" PostTypeId="2" ParentId="16478" CreationDate="2017-07-28T12:55:09.413" Score="1" Body="&lt;p&gt;The equation you showed is an ordinary differential equation. It works for solving natural frequencies of discrete systems  (ie lumped rigid masses connected by massless springs). Here you have a continuous system. You will need to set up two partial differential equations, one for the cable and one for the beam. The resulting PDE can be discretized into a system of ODEs (eg galerkin method, finite element method). Thr details of how to do this are too complicated to go into here. Suggest that you pick up a textbook on vibrations of continuous systems &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-07-28T12:55:09.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16498" PostTypeId="2" ParentId="16494" CreationDate="2017-07-28T15:33:09.900" Score="0" Body="&lt;p&gt;You need the JIS standard. When you have the dimensions you will find they very likely correlate to ASTM / NPS dimensions. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-07-28T15:33:09.900" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16499" PostTypeId="1" AcceptedAnswerId="16502" CreationDate="2017-07-28T15:35:48.117" Score="1" ViewCount="356" Body="&lt;p&gt;How do those mechanisms such as the ones in cars work - when you open a tray or glove-box, and it opens slowly by &quot;easing in&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to build my own compartment with a sliding cover that would slowly open by with the push of a button.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="12344" LastActivityDate="2017-07-28T18:03:58.420" Title="Achieving a &quot;Soft&quot; open mechanism" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ClosedDate="2017-08-17T12:14:53.150" ContentLicense="CC BY-SA 3.0" />
  <row Id="16500" PostTypeId="1" CreationDate="2017-07-28T16:04:49.790" Score="0" ViewCount="2428" Body="&lt;p&gt;I know that a redundant link is a link that doesn't affect the motion of the mechanism even if it is removed . But how could we recognize these links ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was solving a problem abuout finding the mobility of a mechanism. I have noticed that the mechanism is negative , although the configuration of the mechanism appears as if the mechanism is able to move , so this is a contradiction. Is this method used to investigate if there are redundant links ? (By finding the mobility first and checking whether the mechanism is able to move or not ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this step , I think we must recalculate the true mobility , by first removing the redundant links . How could we determine these links ? Is there a certain method ? or I must check all the links , by imagining what will happen to the motion of the mechanism if each link was removed ? &lt;/p&gt;&#xA;" OwnerUserId="9742" LastActivityDate="2017-07-29T21:57:40.233" Title="redundant links in a mechanism" Tags="|mechanical-engineering|design|mechanisms|" AnswerCount="1" CommentCount="1" ClosedDate="2017-08-17T12:18:37.210" ContentLicense="CC BY-SA 3.0" />
  <row Id="16502" PostTypeId="2" ParentId="16499" CreationDate="2017-07-28T16:44:18.223" Score="3" Body="&lt;p&gt;One method is they use a piston/ cylinder arrangement that has a small hole to allow the air in / out to slow down the rate of movement caused by the weight or a spring .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a name : damper or damping piston comes to mind.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-07-28T18:03:58.420" LastActivityDate="2017-07-28T18:03:58.420" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16503" PostTypeId="1" CreationDate="2017-07-28T17:01:15.833" Score="1" ViewCount="127" Body="&lt;p&gt;I want to feed USB data through a glovebox, using binding posts seen in the first picture. My thinking was that, in theory, the USB cord is just made of four wires (one for power, 1 ground, two for data). I tried cutting the cable, and soldering the four wires to banana plugs as seen in the first picture. These plugs can be plugged into the binding posts to make contact with the other side of the post and connect to the usb. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when I plug my cable into my computer, it gives me USB malfunction. Although this may be a stupid question, what is potentially the issue causing this malfunction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than my lack of knowledge about how a USB actually works, two obvious issues include using different banana plugs on each wire, and horrendous soldering skills. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f06oM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f06oM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I realize that this isnt the best picture. When I plug this cord in, I have used electrical tape to isolate the wires from one another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i6hxX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i6hxX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2017-09-26T18:11:39.713" Title="Trouble shooting a DIY USB Feedthrough" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16504" PostTypeId="2" ParentId="16503" CreationDate="2017-07-28T17:20:07.060" Score="2" Body="&lt;p&gt;This is a all-around bad idea.  For one thing, it will mess up the impedance of the cable.  You might be able to get away with that at &quot;full speed&quot; 12 Mbit/s and short distances, but will likely cause trouble with &quot;high speed&quot; USB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You only have 4 wires.  Using only two wire colors is asking for trouble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the bottom pair of red and black wires in the top picture are shorted together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom picture looks like you only have 3 wires connected.  USB can work without the power connected if the device is self-powered, but you have to know that, and you have to know which one the power wire is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not feed the USB cable thru a small hole, instead of feeding large lugs thru 4 separate holes?&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-07-28T17:20:07.060" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16506" PostTypeId="1" CreationDate="2017-07-28T18:06:10.057" Score="0" ViewCount="77" Body="&lt;p&gt;Sort of a general question, but does there exist or might there exist a sort of thermal switching material? A material that could be switched from a thermal insulator (maybe more of a thermal insulator than air) to a thermal conductor by some means like through electricity, light or magnetism? Kind of like a thermal switch? It would be a useful material a device had to remain hot or cool off quickly, without using a fan.&lt;/p&gt;&#xA;" OwnerUserId="6084" LastActivityDate="2017-07-28T18:06:10.057" Title="Heat switch material" Tags="|thermodynamics|thermal-conduction|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16507" PostTypeId="1" CreationDate="2017-07-28T18:52:38.843" Score="0" ViewCount="39" Body="&lt;p&gt;I'm building an ultrasound cleaner. The main chamber is filled with water and it is heated. The chamber is completely sealed so the pressure inside increases with the temperature. An ultrasound emitter will be added to the bottom to increase cleaning effectiveness. Do I have to worry about the temperature/pressure increasing considerably inside the chamber because of the added ultrasound energy? Are the cavitation bubble's effects localized do the bubble itself or should I expect a considerable increase in pressure/temperature inside the chamber? &lt;/p&gt;&#xA;" OwnerUserId="12030" LastActivityDate="2017-08-28T05:53:52.100" Title="Do I need to worry about increase in temperature/pressure in ultrasound cleaner?" Tags="|mechanical-engineering|materials|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16508" PostTypeId="1" CreationDate="2017-07-28T19:35:18.253" Score="-1" ViewCount="44" Body="&lt;p&gt;Pic(1): I tried to make a chart with date axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pic(2): The right chart that I want to draw. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to change the axis properties,  maximum  &amp;amp; minimum values to 1/24 system, but it does not  work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me to find a way to make it.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2yJIH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2yJIH.jpg&quot; alt=&quot;Pictures of the question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user12132" LastEditorDisplayName="user16" LastEditDate="2017-07-29T13:19:53.940" LastActivityDate="2017-07-29T13:19:53.940" Title="How to convert to date axis in excel?" Tags="|civil-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16509" PostTypeId="2" ParentId="16508" CreationDate="2017-07-28T20:11:57.537" Score="1" Body="&lt;p&gt;Have a column (or row) with dates with matching the data values in the next column, then select both sets of data and plot using x-y scatter - can't test as on a smartphone...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-07-28T20:11:57.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16510" PostTypeId="1" CreationDate="2017-07-28T22:28:29.100" Score="1" ViewCount="94" Body="&lt;p&gt;I have a long-duration EMG signal (~4 hours long) with a sampling frequency of 1&amp;nbsp;kHz. If I want to filter the data with a low-pass filter in 10&amp;nbsp;Hz in MATLAB, should I filter the signal with moving window? I am using Butterworth (4th order) IIR filter.&lt;/p&gt;&#xA;" OwnerUserId="12029" LastEditorUserId="1832" LastEditDate="2017-12-10T17:47:01.337" LastActivityDate="2017-12-10T17:47:01.337" Title="Filtering Long-duration Signal" Tags="|electrical-engineering|signal-processing|signal|biomechanics|electronic-filters|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16511" PostTypeId="2" ParentId="16507" CreationDate="2017-07-29T05:49:02.300" Score="1" Body="&lt;p&gt;If the vessel can go above 15psi, you will need a pressure relief device. This is the case for any pressure vessel irrespective of the ultrasonic consideration. A standard water heater pop off will probably work fine provided your vessel is rated for that pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most ultrasonic cleaners run at atmospheric pressure. There is no real reason to run at higher pressure unless you are using a solution other than water and don't want it to evaporate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat addition can be calculated from power input. For example if the unit is 12vdc at 2amps it will add 24watts of heat to the system. Your ultrasonic piezo element will have a maximum design temperature. You will need to keep below this temperature by either cooling the piezo element or the whole vessel.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-07-29T05:49:02.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16513" PostTypeId="5" CreationDate="2017-07-29T06:56:44.057" Score="0" Body="&lt;p&gt;SOLIDWORKS solutions cover all aspects of your product development process with a seamless, integrated workflow—design, verification, sustainable design, communication and data management.&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="7006" LastEditDate="2017-07-29T22:23:53.063" LastActivityDate="2017-07-29T22:23:53.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16514" PostTypeId="4" CreationDate="2017-07-29T06:56:44.057" Score="0" Body="SOLIDWORKS solutions cover all aspects of a product development process with a seamless, integrated workflow—design, verification, sustainable design, communication and data management." OwnerUserId="7006" LastEditorUserId="90" LastEditDate="2017-07-29T08:34:10.487" LastActivityDate="2017-07-29T08:34:10.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16517" PostTypeId="2" ParentId="16508" CreationDate="2017-07-29T12:40:23.567" Score="0" Body="&lt;p&gt;It looks like you're showing only the time, but the first column also contains the date.  Try formatting the first column as a number instead of a date/time.  The integer portion that shows is the date and the decimal portion is the time.  In order to display as you've shown, the AM times shown have to be the next day.  So if the PM times begin with a zero followed by the decimal portion, then the AM times will have to begin with a 1 followed by the decimal portion.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2017-07-29T12:40:23.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16518" PostTypeId="1" CreationDate="2017-07-29T14:43:59.830" Score="0" ViewCount="333" Body="&lt;p&gt;I tried looking online, but all I can find is what pitch ratio is, butI can't find anything related to the question above. Help will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="12356" LastActivityDate="2018-09-24T18:01:29.153" Title="What is the importance of pitch ratio? And why is the value different for different aircrafts?" Tags="|mechanical-engineering|aircraft-design|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16521" PostTypeId="2" ParentId="15603" CreationDate="2017-07-29T20:30:07.760" Score="0" Body="&lt;p&gt;I can't give you a black and white answer to this without seeing it but a fair bet would be using ultrasonic sensors. These systems work by having pairs of ultrasonic transducers placed on the outside of an enclosure at angles, pointing towards eachother in the axis the airflow is travelling. One transducer will 'ping' the other, and the recieving one will record how long that ping took to reach it. It will then ping the first sensor, and that in turn records how long it took to receive. If you have air flow inside the enclosure, one ping will take longer than the other because one sensor is effectively travelling slower than the speed of sound (fighting upstream against the air flow) and the other faster (sailing downstream with the air flow). The time difference is directly propotional to the air speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you have access to the outsides of the enclosure at your desired measurement points this could work, but they do have limits with regards to the diameter of the enclosure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://www.uk.endress.com/en/Field-instruments-overview/Flow-measurement-product-overview/Ultrasonic-flowmeters&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; from E&amp;amp;H has a great video which demonstrates what i've written much more clearly, Their clamp-on sensors have a distance limit of 4m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another indirect way to find dead spots could maybe be to use humitidy sensors along the line (in known good spots if possible) and then place others in suspected dead spots. I'd assume the humidity would be higher in the dead-spots wouldn't you think?&lt;/p&gt;&#xA;" OwnerUserId="12187" LastEditorUserId="12187" LastEditDate="2017-07-29T20:39:45.500" LastActivityDate="2017-07-29T20:39:45.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16522" PostTypeId="2" ParentId="16500" CreationDate="2017-07-29T21:16:01.537" Score="2" Body="&lt;p&gt;The mobility criterion is not perfect. If there is a redundant link then either the mechanism is locked or the link is replicating a different links movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thespecial cases are somewhat self evident because it follows one of the predefined patterns.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;essentially same joint is calculated several times:&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A Slider constraint that is connected from multiple points, you can identify it by two parallel sliders.&lt;/li&gt;&#xA;&lt;li&gt;A revolute constraint that connects in multiple places of the same axis. For example a realworld door hinge is connected in 2 places but has only one joint!&lt;/li&gt;&#xA;&lt;li&gt;A planar constraint constrained to same rigid section on same plane &lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;Its a parallellogram of somekind.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also quite often, for beginners, its not actually redundant but you have miscalculated or misunderstood the notation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway you can find the redunant joint by searching for  sub graphs that are locked apparently locked and replacing them with rigid links if they dont fall under special cases. Usually you find triangles.  Another way is to split and calculate mobility on both sides and infer from that.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-07-29T21:57:40.233" LastActivityDate="2017-07-29T21:57:40.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16523" PostTypeId="1" AcceptedAnswerId="16526" CreationDate="2017-07-29T21:48:16.887" Score="-1" ViewCount="2235" Body="&lt;p&gt;How much would it cost to build an oil rig type platform, that floats like an oil rig, but doesn't have the functions of drilling oil and things like that? Basically what I'm trying to say is, how much would it cost to build a large, floating metal platform to put in the ocean? &lt;/p&gt;&#xA;" OwnerUserId="12359" LastEditorDisplayName="user16" LastEditDate="2017-07-30T13:07:11.830" LastActivityDate="2017-07-30T13:07:11.830" Title="How much would a regular sized oil rig type platform cost to build?" Tags="|drilling|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16526" PostTypeId="2" ParentId="16523" CreationDate="2017-07-30T08:44:57.370" Score="0" Body="&lt;p&gt;Google found an answer in about 5 minutes for me:&#xA;&lt;a href=&quot;http://www.offshore-mag.com/articles/print/volume-72/issue-7/rig-report/reviewing-rig-construction-cost-factors.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.offshore-mag.com/articles/print/volume-72/issue-7/rig-report/reviewing-rig-construction-cost-factors.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Drilling equipment - 10% to 30% of total cost.&lt;/li&gt;&#xA;&lt;li&gt;Non-drilling-related equipment - similar.&lt;/li&gt;&#xA;&lt;li&gt;Total equipment cost - 30% to 60% of total.&lt;/li&gt;&#xA;&lt;li&gt;Cost of cheapest jackup rigs - approx. \$200m&lt;/li&gt;&#xA;&lt;li&gt;Total cost cheapest semi-submersible or drillship rig - approx. \$500m - \$600m&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So you are looking at something like \$100m minimum for a bare platform with no equipment.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-07-30T08:44:57.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16530" PostTypeId="1" CreationDate="2017-07-30T17:35:25.030" Score="0" ViewCount="1169" Body="&lt;p&gt;I am accessing the feasibility of using a gear rack and pinion gear to move a moderately heavy object. I have never used gear track for a larger project so I don't have an intuitive basis for judging the fitness of gear rack for this application. To set aside the factors of velocity and acceleration let me ask the question this way : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What approximate size gear rack and pinion gear would be needed to vertically suspend a 2000 pound weight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, how much weight could be reliably suspended with a 1&quot; wide 6 tooth per inch gear rack?&lt;/p&gt;&#xA;" OwnerUserId="11222" LastEditorUserId="13505" LastEditDate="2017-11-01T15:37:55.050" LastActivityDate="2017-11-01T15:37:55.050" Title="Force limits for gear rack and pinion gear" Tags="|mechanical-engineering|gears|torque|mechanical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16532" PostTypeId="1" CreationDate="2017-07-31T00:06:35.497" Score="0" ViewCount="98" Body="&lt;p&gt;In both the Fukushima and Chernobyl incidents, widespread contamination would have been prevented had both facilities been enclosed within a larger structure. While existing plant designs call for hardened structures for the reactor itself, they often don't contain secondary and support infrastructure (Fukushima) or are insufficient to contain more explosive incidents (Chernobyl).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious if existing plant designs could be augmented with the addition of a &#xA;large, non-hardened, dome structure that covers the entire plant. The dome would be able to contain escaped gasses simply by being hermetically sealed - and a sufficient distance between the dome and any potential sources of explosions means there's enough normal air to buffer and absorb the explosive shockwave, if any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a structure could be built very cheaply - a geodesic hemisphere, for example, constructed of plastic links with inward-opening transparent plastic panels for natural light and ventilation. A fail-safe system would automatically close all panels to prevent the escape of unwanted gasses after an explosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is &lt;em&gt;some&lt;/em&gt; precedent for this, insofar as how's Chernobyl's reactor complex &lt;a href=&quot;https://en.wikipedia.org/wiki/Chernobyl_New_Safe_Confinement&quot; rel=&quot;nofollow noreferrer&quot;&gt;is now enclosed within the New Safe Confinement structure&lt;/a&gt; - except it's an ugly, opaque, and very expensive structure, constructed in a dangerous environment. Presumably my dome proposal could be built cheaply - and the cost/benefit of construction vs clean-up costs (multiplied by actuarial risk) make it a straight-forward decision - and given human penchant for &lt;em&gt;safety theatre&lt;/em&gt; it would help &quot;sell&quot; nuclear power to the masses, but that is beyond the scope of my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my engineering questions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How big would a dome or other enclosing structure have to be in order to enclose all of a plant's at-risk structures?&lt;/li&gt;&#xA;&lt;li&gt;And wold the volume of air inside the dome be sufficient to buffer an explosion to eliminate the need to excessively harden or reinforce (and so, add weight) to the dome structure?&lt;/li&gt;&#xA;&lt;li&gt;Assuming that all of the panels in the dome structure are capable of opening and used for ventilation, would that be sufficient to prevent a greenhouse effect?&lt;/li&gt;&#xA;&lt;li&gt;In your opinion, is this proposal feasible or realistic?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11721" LastActivityDate="2017-08-04T21:53:01.307" Title="Would a large dome structure prevent widespread effects of nuclear plant explosions?" Tags="|civil-engineering|nuclear-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16534" PostTypeId="2" ParentId="16532" CreationDate="2017-07-31T01:29:15.480" Score="1" Body="&lt;p&gt;Chernobyl did not have a containment structure which was Russian practice at the time. The graphite moderator caught fire which was very difficult to control without containment. The heat from the fire carried the radioactive ash material high into the atmosphere causing the extreme area of contamination.  I doubt anything would contain a nuclear explosion.   And completely unrelated: Russia had some nuke power plants that used once through water instead of separate reactor and turbine water sections . This caused a low but steady release of radioactive materials. Again showing Russia had a different perspective.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2017-07-31T01:37:56.853" LastActivityDate="2017-07-31T01:37:56.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16535" PostTypeId="2" ParentId="16076" CreationDate="2017-07-31T02:00:08.603" Score="0" Body="&lt;p&gt;One thing that I did to implement a brake system in one of my projects was to install a 20:1 worm-gear reduction. The worm gear, when fully metal and mounted in the proper housing, can act like a brake when stopped, since the interlocking gear cannot turn the worm, but the worm can turn the gear. For this, you will want the motor attached to the worm and whatever equipment you are moving/rotating/mounting/etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to note, that while this DOES work - it can and will cause strain and wear on your gears if the load is very heavy. I would recommend that you get a larger gear (more expensive) for heavier loads.&lt;/p&gt;&#xA;" OwnerUserId="8543" LastEditorUserId="8543" LastEditDate="2017-08-01T01:10:31.633" LastActivityDate="2017-08-01T01:10:31.633" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16537" PostTypeId="1" CreationDate="2017-07-31T05:43:01.850" Score="1" ViewCount="1666" Body="&lt;p&gt;I am using an MQ3 sensor for alcohol detection but the sensing distance is short. Is there any alternative approach or solution to detect alcohol present in air from a longer distance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the drunken driver exhales ethanol vapours, the sensor placed at steering should detect the alcohol content. &#xA;But MQ3 sensors cannot detect the ethanol vapours at that distance. Is there any solution to this?&lt;/p&gt;&#xA;" OwnerUserId="12367" LastEditorUserId="7083" LastEditDate="2018-06-25T22:48:12.217" LastActivityDate="2018-06-25T22:48:12.217" Title="How to increase the sensitivity range of MQ3 Alcohol sensor?" Tags="|electrical-engineering|sensors|" AnswerCount="1" CommentCount="11" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16542" PostTypeId="1" CreationDate="2017-07-31T08:42:56.150" Score="1" ViewCount="41" Body="&lt;p&gt;I notice that many airports allow cars to loop around the arrivals/departures. Is there a specific term for this design feature in airports?&lt;/p&gt;&#xA;" OwnerUserId="12371" LastActivityDate="2017-07-31T08:42:56.150" Title="Is there a specific term for the car loops at an airport?" Tags="|civil-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16544" PostTypeId="2" ParentId="16518" CreationDate="2017-07-31T12:55:00.650" Score="1" Body="&lt;p&gt;Pitch ratio is important because it is a key parameter in propeller performance. As you have probably read already, it reflects the angle of attack of the blades, and thus their lift. Many propellers can very their pitch in flight to adapt to different requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the entire system: an engine produces torque, the amount of which depends on its rpm.&#xA;The propeller converts said torque into thrust for the aircraft. &#xA;The rpm at which the prop-engine system will settle is the one at which the torque from the engine matches the torque exerted by the airflow on the prop. This can be varied by altering the propeller pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional use of pitch control is to &quot;feather&quot; the propeller, which is to place the blades in a least-resistance configuration when the engine fails, so the now useless propeller will generate the least possible drag.&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2017-07-31T12:55:00.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16545" PostTypeId="2" ParentId="16493" CreationDate="2017-07-31T15:52:41.987" Score="1" Body="&lt;p&gt;As already stated in the comments taking a closer look at the Nyquist stability criterion should provide the necessary explanation and details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: using the Nyquist stability criterion it is possible to determine the stability of a closed loop system based on the open loop transfer function. &#xA;Assuming a system consisting of a plant $P$, a controller $C$ and negative feedback the open loop transfer equals $PC$ while the closed loop transfer equals $(I + PC)^{-1}PC$. The closed loop response blows up (i.e. becomes unstable) as $PC$ approaches $-1$. An open loop transfer $PC$ equal to $-1$ implies a magnitude of $1$ and a phase equal to $-180$ degrees.&lt;/p&gt;&#xA;" OwnerUserId="11869" LastActivityDate="2017-07-31T15:52:41.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16546" PostTypeId="1" CreationDate="2017-07-31T17:01:49.190" Score="0" ViewCount="72" Body="&lt;p&gt;Typical acoustic speakers come mounted in boxes. The shape of the boxes is important because the speaker throws pressure both forwards and backwards, so the back pressure goes into the box and resonates, thus the shape and material of the box has a noticeable effect on the sound quality. A constraint on speaker manufacturers is the size of this box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a home, however, if a speaker is mounted in a wall or other special enclosure, it may be possible to build a large or unconventional sound box for the speaker. Has there been any research done on the design and characteristics of such custom enclosures?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2017-07-31T17:01:49.190" Title="Custom cavity for speaker system?" Tags="|acoustics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16547" PostTypeId="1" CreationDate="2017-07-31T17:54:22.507" Score="1" ViewCount="2447" Body="&lt;p&gt;I asked this question over at &lt;a href=&quot;https://aviation.stackexchange.com/questions/42509/how-to-draw-naca-6-series-airfoils&quot;&gt;&lt;em&gt;Aviation.stackexchange&lt;/em&gt;&lt;/a&gt; but after that I figured it might be better to place it here. Especially since there were really good answers on the &lt;a href=&quot;https://engineering.stackexchange.com/questions/13526/how-to-generate-naca-5-digit-airfoil-coordinates&quot;&gt;NACA 5-digit-Series&lt;/a&gt; airfoil generation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to calculate the profile NACA 64-2A015.&#xA;With the help of &lt;a href=&quot;https://aviation.stackexchange.com/questions/42496/is-there-an-analytical-description-of-the-naca-642-015a-profile&quot;&gt;&lt;em&gt;Aviation.stackexchange&lt;/em&gt;&lt;/a&gt; I learned that the A-Version of the profile was created to ease manufacturing by thickening the trailing edge-section (by a straight contour from 80% chord backwards).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am struggling to plot the profile based on the equations given &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20050092351.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19930090976.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qHTYY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qHTYY.png&quot; alt=&quot;equations&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The implementation in MATLAB looks like this: &#xA;Here is the Matlab code I used:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a = 0.4;&#xA;b = 1.0; % caution for NON-unity entries change the equation for h&#xA;% c = 1; to simplifiy the equation the chord is set to 1&#xA;&#xA;cl = 1;&#xA;&#xA;g = -1/(b-a) * (a^2 * (1/2  * log(a) -1/4) - b^2 * (1/2 * log(b) -1/4)); % g  = -1/(1-a) * (a^2 * (1/2 * log(a) -1/4) + 1/4)&#xA;h = 1/(1-a) * (1/2*(1-a)^2 * log(1-a) -1/4*(1-a)^2)+g; % simplified version for b = 1: h =  1/(b-a) * (1/2*(1-a)^2 * log(1-a) - 1/2 * (1-b)^2 * log(1-b) + 1/4*(1-b)^2 - 1/4*(1-a)^2) + g&#xA;&#xA;x = (0:0.001:1);&#xA;y  = cl/(2*pi*(a+b)) * ( 1/(b-a) .* (1/2 * (a-x).^2 .* log(abs(a-x)) - 1/2 .* (b-x).^2 .* log(abs(b-x)) + 1/4 .* (b-x).^2 - 1/4 .* (a-x).^2) - x.*log(x) + g - h.*x); % y = cl/(2*pi*(a+1)) * ( 1/(1-a) .* (1/2 * (a-x).^2 .* log(abs(a-x)) - 1/2 .* (1-x).^2 .* log(abs(1-x)) + 1/4 .* (1-x).^2 + 1/4 .* (a-x).^2) - x.*log(x) + g - h.*x);&#xA;&#xA;L6j01_x = [0, 0.005, 0.0075, 0.0125, 0.025, 0.05, 0.075, 0.1, 0.15, 0.2, 0.25, 0.3, 0.35, 0.4, 0.45, 0.5, 0.55, 0.6, 0.65, 0.7, 0.75, 0.8, 0.85, 0.9, 0.95, 1];&#xA;L6j01_y = [0, 0.01193, 0.01436, 0.01815, 0.02508, 0.03477, 0.04202, 0.04799, 0.05732, 0.06423, 0.06926, 0.0727, 0.07463, 0.07487, 0.07313, 0.06978, 0.06517, 0.05956, 0.05311, 0.046, 0.03847, 0.03084, 0.02321, 0.01558, 0.00795, 0.00032];&#xA;&#xA;plot(x,y), axis equal, hold on&#xA;plot(L6j01_x, L6j01_y,'ro'), hold off&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But the results do not match the data points from the Nasa-Reports&#xA;&lt;a href=&quot;https://i.stack.imgur.com/i5mAg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i5mAg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What did I miss to implement in the equation or is there another representation of the airfoil (excluding leading and trailing edge radius)?&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2017-07-31T18:12:06.647" LastActivityDate="2021-06-05T12:50:24.997" Title="How to generate a NACA 6-Series Airfoil analytically?" Tags="|aerodynamics|matlab|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16549" PostTypeId="1" AcceptedAnswerId="16582" CreationDate="2017-07-31T20:08:00.603" Score="2" ViewCount="44" Body="&lt;p&gt;What is the advantage of adding alloy elements in a steel in order to increase the tempering temperature while keeping hardness and ductility the same as before?&#xA;If I wasn't clear enough, I want to know what difference makes a higher tempering temperature, if hardness and ductility will be the same.&lt;/p&gt;&#xA;" OwnerUserId="12376" LastEditorUserId="1832" LastEditDate="2017-08-01T02:42:17.963" LastActivityDate="2017-08-02T11:24:42.917" Title="Why adding alloy elements in a steel for increasing tempering temperature while keeping the same hardnes and ductility?" Tags="|materials|heat-treatment|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16550" PostTypeId="2" ParentId="16547" CreationDate="2017-07-31T20:52:05.030" Score="0" Body="&lt;p&gt;There seems to be a syntax error at the first term of y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Cli&lt;/em&gt; is defined as the design coefficient of lift, not the &lt;em&gt;chord length&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be other misinterpretations too. I'd suggest checking all terms once again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, in the report it says that these equations are just a simplification of real coordinates of airfoils. (Just above equation 6). They probably did a curve matching, so even if all the equations are entered correctly, there will still be some error in comparison to a given set of coordinates. &lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-07-31T20:52:05.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16551" PostTypeId="1" AcceptedAnswerId="19455" CreationDate="2017-07-31T22:13:03.393" Score="3" ViewCount="1443" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T1piV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T1piV.png&quot; alt=&quot;Circuit 1, gain of 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0o3Xf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0o3Xf.png&quot; alt=&quot;Circuit 2, gain of ~30&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the first circuit, The ratio between the input voltage Vin and the voltage at the collector and the ratio between the input voltage Vin and the voltage at the emitter are supposed to be around 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the second circuit, the ratio between the input voltage Vs and the voltage at the output Vout is around 30. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of the circuit (please correct me if this is incorrect):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I know the purpose of the resistors in the center region of circuit 2 is to bias the transistor and make it operate at a specific Q-point. &lt;/li&gt;&#xA;&lt;li&gt;The purpose of the 10uF capacitors is to filter any DC component of the input and output signal, as well as to prevent changing the biasing.&lt;/li&gt;&#xA;&lt;li&gt;The capacitor across Re is a bypass capacitor, which is supposed to prevent the AC component from affecting the gain through negative feedback.&lt;/li&gt;&#xA;&lt;li&gt;The purpose of the emitter resistor is to provide negative feedback.&lt;/li&gt;&#xA;&lt;li&gt;The purpose of the collector resistor (and I am not sure about this) might be to limit the current from the source.&lt;/li&gt;&#xA;&lt;li&gt;The purpose of the resistor at the output (I am not 100% sure about this either) might be to work as a load&lt;/li&gt;&#xA;&lt;li&gt;The resistor next to the input signal Vs is used to step down the voltage of the signal, as to use a small signal model in the circuit instead of having to use the large signal model.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Assuming the following: a beta (Ic/Ib) of around 250, R1 of 75k and R2 of 33k for both circuits, and an input voltage Vin of 4 volts peak to peak with a frequency of 1khz for the first circuit, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What controls the voltage gain in circuit 1 and 2 at the collector, and why is the gain in the first circuit only 1?&lt;/p&gt;&#xA;" OwnerDisplayName="user12377" LastEditorUserId="8055" LastEditDate="2018-02-15T18:55:32.863" LastActivityDate="2021-06-29T08:12:47.210" Title="What controls the gain in a BJT amplifier, and why does this circuit have a gain of unity?" Tags="|electrical-engineering|circuit-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16554" PostTypeId="1" AcceptedAnswerId="16565" CreationDate="2017-08-01T02:17:38.923" Score="0" ViewCount="112" Body="&lt;p&gt;I have a project where I need to be able to measure the volume of remaining liquid in a bottle based on weight. The idea is to use a Force Sensing Resistor: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(model FSR402-Short Tail: &lt;a href=&quot;http://www.interlinkelectronics.com/FSR402short.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.interlinkelectronics.com/FSR402short.php&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lGXrj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lGXrj.jpg&quot; alt=&quot;image&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://www.interlinkelectronics.com/assets/images/FSR402_short.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;interlinkelectronics.com&lt;/a&gt;)&lt;/sub&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My issue is that I can't seem to get an accurate consistent reading on the pressure exerted on the device when any kind of shifting (such as replacing / refilling the bottle, moving the device location, etc) would create a huge variance in the data points. It can vary from 20-150 points easily... which is too large of a margin since 150 points could equate to about 500ml of liquid.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The idea was to loosely hang the bottle &amp;amp; connector to the pressure plate (see second image) using two centrally located screws. the pressure plate would have a set of two rubber feet perpendicular to the screw placements to allow the pressure to be focused on the rubber footing with the FSR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This produced two main issues that I did not foresee with the CAD design:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;over-tightening/under-tightening of the screws&lt;/li&gt;&#xA;&lt;li&gt;inaccurate or inconsistent readings of the pressure from very minor shifts or bumps.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What can I do to improve my design, shown below?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Top perspective view of pivot mechanism.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ibAHH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ibAHH.png&quot; alt=&quot;Top perspective view of pivot mechanism.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Side perspective view, translucent with descriptions.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WoAnP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WoAnP.png&quot; alt=&quot;Side perspective view, translucent with descriptions.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8543" LastEditorUserId="6767" LastEditDate="2019-02-26T03:56:10.790" LastActivityDate="2019-02-26T03:56:10.790" Title="How can I *more accurately* measure the weight of a hanging bottle?" Tags="|mechanical-engineering|design|cad|mechanisms|autodesk-inventor|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="16555" PostTypeId="1" CreationDate="2017-08-01T08:56:53.300" Score="2" ViewCount="946" Body="&lt;p&gt;I know that different geometries of piping, orifices etc... have different critical Reynolds values. If the value is exceeded laminar stream will turn to a turbulent form which causes a variety of problems. But why shouldn´t the value be too low?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. for pressure piping up to around 300 bars I assume that I have to be inbetween of 3 to 12 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do too low velocities cause problems like cavitation? What are the reasons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI: To have a practical example. I am asking because I want to design a hydraulic system but I only have a NG25 valve available but NG6 is probably already enough.&lt;/p&gt;&#xA;" OwnerUserId="5007" LastActivityDate="2018-05-01T22:12:05.667" Title="Why to avoid low and high fluid velocities in hydraulic systems" Tags="|mechanical-engineering|hydraulics|valves|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16558" PostTypeId="1" AcceptedAnswerId="16559" CreationDate="2017-08-01T10:25:10.320" Score="3" ViewCount="743" Body="&lt;p&gt;I would like to convert a continuous rotation into two 180° intermittent rotations. A standard Geneva drive layout can not do this, is there any variation of a Geneva drive that can do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ei4Ae.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ei4Ae.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is an example of a 4 slot Geneva drive capable of four 90° intermittent rotations.&lt;/p&gt;&#xA;" OwnerUserId="12382" LastEditorUserId="368" LastEditDate="2017-08-01T15:57:37.670" LastActivityDate="2017-08-01T15:57:37.670" Title="How can I produce two 180-degree intermittent rotations with a Geneva drive?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16559" PostTypeId="2" ParentId="16558" CreationDate="2017-08-01T11:10:40.943" Score="5" Body="&lt;p&gt;It's simpler to add a 1:2 ratio gear ration on the output of the geneva drive to get your 180° rotation per 90° of the intermittent&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-08-01T11:10:40.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16562" PostTypeId="1" CreationDate="2017-08-01T13:51:20.407" Score="5" ViewCount="140" Body="&lt;p&gt;Consider a multibody robotic system with Lagrangian $$ L =\frac{1}{2} \dot{q}^{T}M\dot{q} - V $$ and equations of motion of the form&#xA;$$M(q)\ddot{q}+C(q,\dot{q})\dot{q}+A^{T}(q)\lambda+N(q)=0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $N=\frac{\partial V(q)}{\partial q}$  and $\lambda$ represents the Lagrange multipliers for the constraint forces (MLS pg. 269; pg. 287 in pdf). $A$ is the constraint matrix from the Pfaffian form $A\dot{q}=0$. We are concerned with &lt;strong&gt;holonomic&lt;/strong&gt; constraints, which can be differentiated to obtain the Pfaffian form.&#xA;$ C=\sum_{k=1}^{n} \Gamma_{ijk}  \dot{q_{k}} $ where $\Gamma_{ijk}$ are the Christoffel symbols of the first kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lagrange multipliers can be calculated using the formula (MLS pg. 270, pg. 288 in pdf)&#xA;$$\lambda =-(AM^{-1}A^{T})^{-1}AM^{-1}(C\dot{q}-N)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the holonomic constraint forces $A^{T}\lambda$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I would like to see some examples of this method for calculating the holonomic constraint forces for planar systems.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen this method being applied to calculate the tension in a simple pendulum (MLS pg. 270, pg. 288 in pdf), but I would like to see examples of &lt;strong&gt;planar&lt;/strong&gt; systems with more degrees of freedom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm specifically interested in the case of a 2R manipulator shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fQMXv.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fQMXv.png&quot; alt=&quot;Planar 2R manipulator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations of motion are on pg. 164-165 (pg. 182-183 in the pdf) in MLS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My objectives are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Apply the holonomic constraint $\theta_{1}(t)=0$ and obtain the holonomic constraint force.&lt;/li&gt;&#xA;&lt;li&gt;Show that we get the dynamics of a simple pendulum for the&#xA;constrained system for initial conditions $\theta_{2}(0)=0, \dot{\theta_{2}}(0)=0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have done the calculation for the constraint $\theta_{1}(t)=0$ but I'm not able to show 2. The expression I get for $\lambda$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda=&#xA;\begin{array}{c}&#xA; \frac{g (d \text{m2} \text{r2} (\sin  \text{t2})-\text{m2} \text{r2} (\cos  \text{t2}) ((\cos  \text{t2}) b+d))+b d (\sin  \text{t2}) \dot{\text{t2}} \dot{\text{t2}}}{d} \\&#xA;\end{array}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $(t1, t2)=(\theta_{1},\theta_{2})$ and $a, b, d$ are the geometric and inertial parameters $\alpha, \beta, \delta$ described in MLS (pg. 165; pg. 183 in the pdf).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggestions for other methods to calculate holonomic constraint forces would be helpful too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;References:&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MLS refers to &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Murray, R. M., Li, Z., Sastry, S. S., &amp;amp; Sastry, S. S. (1994). &lt;a href=&quot;http://www.cds.caltech.edu/~murray/books/MLS/pdf/mls94-complete.pdf&quot; rel=&quot;noreferrer&quot;&gt;A&#xA;  mathematical introduction to robotic manipulation&lt;/a&gt;. CRC press.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6266" LastEditorUserId="6266" LastEditDate="2017-08-01T15:27:29.547" LastActivityDate="2023-02-26T03:05:00.630" Title="Holonomic constraint forces using Lagrange multipliers" Tags="|mechanical-engineering|modeling|dynamics|robotics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16563" PostTypeId="1" CreationDate="2017-08-01T15:19:33.370" Score="2" ViewCount="3331" Body="&lt;p&gt;I am designing a printed piece that will deflect air from an exhaust fan on a computer 90 degrees. My question is, what would be most efficient, a circular radius, parabolic radius, or hyperbolic radius in the ducting to change the flow of the air 90 degrees upward. I'm trying to minimize the resistance of the ducting to maximize my exhaust air flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know very little in this domain, so please forgive me for not knowing even the terms to search for.&lt;/p&gt;&#xA;" OwnerUserId="12386" LastActivityDate="2017-09-05T22:11:25.743" Title="Fluid dynamics right angle air duct" Tags="|fluid-mechanics|aerodynamics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16564" PostTypeId="2" ParentId="16563" CreationDate="2017-08-01T15:25:00.107" Score="0" Body="&lt;p&gt;What you are referring to is minor loss&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.slideshare.net/ZazaEureka/8-fm-9-flow-in-pipes-major-loses-co-3-copy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flow in pipes major loses CO3&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;check out slide 32 and 35&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slide #32&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8qkIr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8qkIr.jpg&quot; alt=&quot;Slide 32&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slide #35&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aRy44.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aRy44.jpg&quot; alt=&quot;Slide #35&quot;&gt;&lt;/a&gt;&#xA;you'll see that head loss is reduced in a curved pipe versus a right angle pipe&lt;/p&gt;&#xA;" OwnerUserId="6488" LastEditorUserId="8055" LastEditDate="2017-09-05T22:11:25.743" LastActivityDate="2017-09-05T22:11:25.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16565" PostTypeId="2" ParentId="16554" CreationDate="2017-08-01T16:01:08.873" Score="2" Body="&lt;p&gt;This is not an appropriate sensor for what you are trying to do. This type of flat pressure sensor is sensitive to localised pressures. For example, from a piece of dirt or an uneven surface. You should use a load cell which is designed for your application. For example, something like: &lt;a href=&quot;http://www.omega.co.uk/pptst/LCMKD.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Omega LCMKD-20N&lt;/a&gt; would be more suitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also consider this &lt;a href=&quot;http://www.omega.co.uk/literature/transactions/volume3/weigh.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;brief introduction to designing a weighing system&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3010" LastActivityDate="2017-08-01T16:01:08.873" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16566" PostTypeId="1" CreationDate="2017-08-01T17:28:55.543" Score="0" ViewCount="56" Body="&lt;p&gt;I am currently attempting to solve the following problem. Say that I have an object that can expand and contract, given some external force. What I want to know is how much it has expanded, or contracted, over time. ( the diameter difference ) In addition, while this expansion or contraction is occurring, the object is free to move around. ( Example, object rolling around on the floor while wearing these sensors on itself )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am considering of solving this problem by using two sensors, where one lies at the front of the object and one on its complete opposite ( back of the object ). An example of this can be seen below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jazan.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jazan.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the object expands, here is the size difference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3izQe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3izQe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the diameter of the second object is substantially bigger, and it is something I'd like to record as it increases / decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there some good sensors that I can use to record this change over time? ( Note, sensors that can attached to a Raspberry pi board, or wirelessly connected to, that is ). I'd also like to stay away from sensors that would easily be lost if movement of the object was introduced, or require cords. ( i.e Capacitance belts, conductive yarns, etc. as they can easily give false readings of &quot;contraction&quot; and &quot;expansion&quot; given that the movement is too large. ex: Capacitative sensor gets stuck on the side of the object, and gives enormous diameter readings ) &lt;/p&gt;&#xA;" OwnerUserId="9644" LastEditorUserId="9644" LastEditDate="2017-08-01T21:16:50.520" LastActivityDate="2017-09-01T14:43:01.227" Title="Determining change of object over time ( diameter )" Tags="|sensors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16567" PostTypeId="2" ParentId="16566" CreationDate="2017-08-01T17:51:25.640" Score="1" Body="&lt;p&gt;Sounds like you're looking for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Strain_gauge&quot; rel=&quot;nofollow noreferrer&quot;&gt;strain gauge&lt;/a&gt;.  They are quite common when analyzing stress on materials in specific directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sticks to a surface and can record the change in length in one or more axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you're only concerned about diameter change, you would only need one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Might take some playing to configure it with a Raspberry Pi, but it should be relatively easy to find one for your application.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-08-01T17:51:25.640" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16568" PostTypeId="2" ParentId="16563" CreationDate="2017-08-01T18:45:10.650" Score="3" Body="&lt;p&gt;A curved elbow will be considerably better than square elbow, because it reduces turbulence. The exact mathematical shape of the curve is not really important; usually a circular radius is chosen because it is easier to fabricate. It is important to keep the cross sectional area the same as the air moves through the turn; two circular arcs that share the same center point can accomplish this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To improve efficiency beyond a radius you can also add turning veins. It is probably of less importance on a small pc fan than on a large 100+hp fan, but might be something to look into. These also need to follow the same profile as the elbow and have the same cross sectional area between them. Again this can be accomplished with circular arcs that share the same center point as your duct walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aOe09.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aOe09.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.slideshare.net/narasimhaswamy/noise-control-55355735&quot; rel=&quot;nofollow noreferrer&quot;&gt;Diagram sourced from this link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2017-08-02T05:32:38.197" LastActivityDate="2017-08-02T05:32:38.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16570" PostTypeId="2" ParentId="16547" CreationDate="2017-08-01T20:13:20.693" Score="1" Body="&lt;p&gt;Sorry, I got this completely wrong!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is not the profile contour but the chamber-line. Since a symmetric airfoil does not produce any lift, this ($y_c$) should be a straight line (which is exactly what happens for $c_{l}=0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct equation for the thickness distribution is a conformal mapping of a circle as described in &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19970008124.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nasa TM 4741&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d0PDe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d0PDe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2017-08-01T20:13:20.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16571" PostTypeId="1" CreationDate="2017-08-01T23:27:11.430" Score="2" ViewCount="802" Body="&lt;p&gt;I recently found a loupe with a thread pitch reticle displaying common UTS thread pitches... except that rather than 40, 48, or 56 it has a 50. What would this be used for? It doesn't appear to date from earlier than the 1960s - the housing and dioptre lens are plastic, the reticle and retaining ring are glass/metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/El0Jm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/El0Jm.jpg&quot; alt=&quot;image of thread count reticle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12392" LastActivityDate="2022-02-14T19:41:16.613" Title="In what context is 50 threads per inch common?" Tags="|measurements|threads|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16573" PostTypeId="1" AcceptedAnswerId="16579" CreationDate="2017-08-02T01:56:56.997" Score="2" ViewCount="1643" Body="&lt;p&gt;If we consider a stationary car, a moderate wind hitting it broadside might tilt it a bit, but not move it otherwise. Now, consider the car moving perpendicular to that wind (e.g., driving straight forward). If the car were a rolling rigid body, it would still move straight ahead so long as the wind is not strong enough to overcome the friction between the tires and the road. However, experience says that it can sometimes require a significant steering input into the wind to keep the car moving straight ahead. What about the dynamics and architecture of a vehicle make it tend to steer itself when subjected to a crosswind?&lt;/p&gt;&#xA;" OwnerUserId="12319" LastActivityDate="2017-08-02T06:40:42.083" Title="Why does a vehicle under crosswind need corrective steering?" Tags="|mechanical-engineering|automotive-engineering|applied-mechanics|dynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16575" PostTypeId="1" CreationDate="2017-08-02T04:17:43.117" Score="0" ViewCount="1570" Body="&lt;p&gt;The definition of dry steering in wikipedia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dry steering is the act of turning the steering wheel of a vehicle while the vehicle is stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Dry_steering&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Dry_steering&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately the article has no citation to scientific article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm curious is the &quot;negative effects&quot; of dry steering a thing based on real facts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it now only a myth, unrelevant, since the increase of strength in materials we use in cars?&lt;/p&gt;&#xA;" OwnerUserId="9474" LastEditorUserId="9474" LastEditDate="2017-08-02T05:52:27.440" LastActivityDate="2017-08-02T06:05:24.887" Title="Dry steering in cars" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16577" PostTypeId="2" ParentId="16573" CreationDate="2017-08-02T05:48:39.887" Score="0" Body="&lt;p&gt;When cars are designed, the engineers calculate various things such as centre of gravity, turning circle radius etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of these things is the centre of pressure and the effect it has on the handling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wind is a force acting on the vehicle - same as a steering input or the effect of a side slope or camber on the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wind is hitting the vehicle from directly in front it may not affect the direction so much but it will slow it down so needing more power to keep a constant speed. Have you ever ridden a bicycle with the wind in front of you then behind you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effects of a strong side wind can not only cause a change of direction (even noticed when you overtake a truck as it shields you from the wind ), but it can cause some vehicles to tip over.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-02T05:48:39.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16578" PostTypeId="2" ParentId="16575" CreationDate="2017-08-02T06:05:24.887" Score="1" Body="&lt;p&gt;The effects of dry steering are increased load and wear on components compared to steering with movement even slowly such as when parking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increase in strength and, of course quality has also been offset by the increase in the size of tyres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fitting of power steering (ps) has &quot;hidden&quot; the problem as older cars without ps were harder to steer especially at low speeds and people had to be sufficiently strong to be able to properly control a vehicle. Now....&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-02T06:05:24.887" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16579" PostTypeId="2" ParentId="16573" CreationDate="2017-08-02T06:40:42.083" Score="2" Body="&lt;p&gt;This is primarily caused by &quot;side slip&quot; or &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Slip_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;slip angle&lt;/a&gt;&quot;. This occurs because tires are not rigid, but pneumatic bladders of flexible rubber and steel cord. During normal operation (including crosswind and turns) your tires operate with static friction between the road and surface. Exceptions would be skidding to a stop or &quot;peeling out&quot;. When the car is moving, it is easier to steer because the tire can flex and recover, flex and recover as it travels around and around. If you have ever tried to turn a steering wheel (without power-steering) when a car is stationary, it is quite difficult because you are dragging the rubber across the surface as opposed to the &quot;flex and recover&quot; mode of operation when it is rolling. This is the reason the car doesn't steer itself or move horizontally when it is parked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side slip is caused by any horizontal force on the tire during rotation. The image below shows how the front tires flex in a turn. This is very similar to how the tires would flex during crosswind loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IPOlU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IPOlU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://technicalf1explained.blogspot.com/2012/10/f1-tirespart-2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Formula 1 image source&lt;/a&gt;&#xA;&lt;br&gt;&lt;br&gt;&lt;br&gt;&#xA;Here is another image showing side slip with the surface contact patch.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uAJMj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uAJMj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.mgf.ultimatemg.com/group2/suspension/chassis_and_handling/slip_angle.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Slip angle article and image source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;br&gt;&#xA;A secondary cause is that the crosswind force and &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Caster_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;caster&lt;/a&gt;&quot; in the front wheels causes the vehicle to want to turn away from the crosswind. Castor essentially means the pivot point is placed ahead of the place the wheel contacts the surface. Some castor in automobile design is desirable because it makes the steering self correcting. This is responsible for the feedback force you feel in the steering wheel during a crosswind. The drivers steering wheel torque input is required to oppose this crosswind force or the car will quickly drift off the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be easier to visualize on a shopping cart. While you push a shopping cart, a person pushes it to the left causing a moment on the two front caster wheels rotating them in to turn in the direction the force was applied. After the force leaves the wheels will realign with the single force pushing them forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/197dv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/197dv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-08-02T06:40:42.083" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16580" PostTypeId="2" ParentId="16563" CreationDate="2017-08-02T06:42:22.217" Score="0" Body="&lt;p&gt;Given the low speed and flow rate of the application then a simple angled flat plate will cause the airflow to change direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the other answers are correct but for the application a flat plate will work, but a curved duct would be best. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-02T06:42:22.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16581" PostTypeId="2" ParentId="16571" CreationDate="2017-08-02T06:47:25.890" Score="1" Body="&lt;p&gt;Fine threads (40 to 120 TPI) are used in adjustable mountings in fields such as optics and mirrors as well as other fields with small parts : watch making etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More detail can be found here:&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fine_adjustment_screw&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fine adjustment screw&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorDisplayName="user16" LastEditDate="2017-08-02T11:52:05.787" LastActivityDate="2017-08-02T11:52:05.787" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16582" PostTypeId="2" ParentId="16549" CreationDate="2017-08-02T11:24:42.917" Score="2" Body="&lt;p&gt;Temperature resistance is especially useful for machine tools eg drills, lathe tools ect etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechanised cutting generates a lot of heat and equally cutting tools need to be tempered fairly hard for obvious reasons and it is very easy for something like a drill to get well into the normal tempering range of carbon steel, thus losing its original temper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;High speed steels&lt;/em&gt; for example are widely used for general purpose drill bits etc as they allow higher speeds and/or feed rates to be used while remaining in an acceptable temperature range. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-08-02T11:24:42.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16583" PostTypeId="2" ParentId="16566" CreationDate="2017-08-02T13:07:28.180" Score="0" Body="&lt;p&gt;A strain gauge is the simplest answer. You affix it to the object, and when it deforms, there is a small change in the resistance of the strain gauge which can be read using a circuit such as a wheatstone bridge. If this thing is rolling around, I suppose using computer vision to try to see the deformation is out of the question.&lt;/p&gt;&#xA;" OwnerUserId="12401" LastActivityDate="2017-08-02T13:07:28.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16584" PostTypeId="1" CreationDate="2017-08-02T13:59:25.310" Score="3" ViewCount="85" Body="&lt;p&gt;I have this idea of creating my own curriculum for studying Architectural Engineering. I aim at talking with actual architectural engineers in my surroundings and ask them for advice on the topics I should study and help me with advancing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By looking at the programs that universities offer all over the world I got a good idea of the topics to be covered, but since I'm in charge, I would like to personalize my own program. I'd like to cover both the engineering and design part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're an architectural engineer, what topic would you suggest me to start with? What are the topics that you find especially useful and important in your career as an architect/engineer? What would you advise universities to change in their curriculum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some context: I'm especially interested in building and repair of houses, combining more classical designs with modern technologies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: I am aware that by studying by myself I won't obtain the necessary license to perform as an engineer.&lt;/p&gt;&#xA;" OwnerUserId="12398" LastEditorUserId="12398" LastEditDate="2017-08-02T14:10:56.623" LastActivityDate="2017-08-02T15:02:05.510" Title="Where to start when studying Architectural Engineering by myself?" Tags="|structural-engineering|building-physics|architecture|construction-management|" AnswerCount="1" CommentCount="10" ClosedDate="2017-08-09T18:25:11.070" ContentLicense="CC BY-SA 3.0" />
  <row Id="16585" PostTypeId="2" ParentId="16584" CreationDate="2017-08-02T15:02:05.510" Score="1" Body="&lt;p&gt;I would suggest that you take a look at Architectural Engineering Professional Societies in your local area. Most of these societies have recommended educational curriculum and continues educational opportunities. Attach is a link to &lt;a href=&quot;http://www.asce.org/architectural-engineering/architectural-engineering-institute/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Architectural Engineering Institute&lt;/a&gt; which I believe is part of &lt;a href=&quot;http://www.asce.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;American Society of Civil Engineers&lt;/a&gt;. Another organization that might be of interest to is &lt;a href=&quot;http://cmaanet.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Construction Management Association of America&lt;/a&gt;. Little bit of research in to these professional societies should enable you to create your own curriculum. Also take a look at &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_engineering_societies&quot; rel=&quot;nofollow noreferrer&quot;&gt;List of engineering societies&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope these will help you get started.&lt;/p&gt;&#xA;" OwnerUserId="8055" LastActivityDate="2017-08-02T15:02:05.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16586" PostTypeId="1" CreationDate="2017-08-03T03:39:18.157" Score="0" ViewCount="72" Body="&lt;p&gt;A gyro measures the angular velocity of a drone around 3 axes: yaw, roll and pitch, but what physical change does it make to the drone after measuring this value to give it stable flight? As it is a gyro chip, and not an actual gyroscope, would it still precess? does it change motor speed etc.? Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="12356" LastActivityDate="2017-09-06T18:57:44.907" Title="What action does a 3 axis gyro stabiliser do to stabilise the drone?" Tags="|mechanical-engineering|aircraft-design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16587" PostTypeId="1" CreationDate="2017-08-03T07:27:23.783" Score="0" ViewCount="333" Body="&lt;p&gt;What would be most effective way to transfer heat in the following situation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Specific conditions:&lt;/strong&gt; A vertical cylinder or pipe, 30-100cm long, 5-10 cm in diameter, heavily thermally insulated except at the bottom, placed in a solid medium (except at the top, which is in air) whose vertical &lt;strong&gt;temperature profile varies on an hourly basis&lt;/strong&gt; and will change polarity (delta T=T_top - T_bot will change sign), and has a temperature difference of around 2-10ºC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom of the cylinder is attached to a heat &quot;sink&quot; embedded in the solid surrounding medium. The top has a thin insulation layer between it and the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The goal is to tie the top temperature as close to the bottom temperature as possible&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZLlh6.jpg&quot; alt=&quot;On the right, not entirely to scale, the lagged vertical pipe is shown embedded in a solid medium whose temperature profile (on the left) varies during the day. Basal temperature is maintained at ambient (for that depth) with a heat sink embedded in the solid medium. The goal is to keep the top temperature as close to that of the base as possible by fluxing heat up and down through the pipe, and always compromised by the poorly insulating top lid.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought was to use a solid metal cylinder (likely aluminum not copper for cost reasons) since conductive heat transport is agnostic to the top/bottom temperature polarity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's tempting to use free convection in a pipe using water etc, which is presumably far more effective at driving heat transport. However, if the top is warmer than the bottom, full-cylinder convection won't happen. Forced convection or pumping is out of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the best strategy here? Would a mix of metal conducting rods in a cylinder otherwise full of liquid be even better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: diagram upload as requested. &lt;/p&gt;&#xA;" OwnerUserId="12407" LastEditorUserId="12407" LastEditDate="2017-08-04T00:08:59.373" LastActivityDate="2017-08-04T04:36:25.143" Title="Most effective heat transfer in a vertical pipe with small top/bottom temperature difference?" Tags="|heat-transfer|heat-exchanger|thermal-conduction|convection|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16589" PostTypeId="1" CreationDate="2017-08-03T09:54:17.550" Score="0" ViewCount="444" Body="&lt;p&gt;I'm trying to measure the rotational speed of a turbocharger, but the laser tachometer I have can only measure up to 99,999 RPM and it isn't enough (display starts to output letters) and devices that can measure higher than that are quite expensive. I have seen Arduino based tachometers that use a photodiode coupled with a laser but it seems to require the beam to be broken, something that can't be done with the turbocharger. Is there any other option to measure its rotational speed, even it is less accurate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's for a small project so buying a $200 tachometer isn't justifiable.&lt;/p&gt;&#xA;" OwnerUserId="12409" LastActivityDate="2017-08-03T19:10:21.423" Title="How to measure above 100K RPM?" Tags="|measurements|sensors|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16590" PostTypeId="2" ParentId="16589" CreationDate="2017-08-03T11:27:52.137" Score="1" Body="&lt;p&gt;The people I worked with replaced the nut on the inlet side (as it is cooler than the exhaust side...) with a magnetised one and picked up the changing field with a Hall effect sensor (if memory serves...) to get the rpm.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-03T11:27:52.137" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16591" PostTypeId="2" ParentId="16555" CreationDate="2017-08-03T12:05:58.073" Score="0" Body="&lt;p&gt;Most of these &quot;guidelines&quot; in engineering math is there purely for economic reasons. If the value goes very low it just means that your fluid is creeping along the pipe. Normally in a plant you would not want that, hence the guidelines of this must be above 3000 or for this pipe size you have this velocity. If you can work it out and get an answer then it is possible. You must personally decide what your requirements are.&lt;/p&gt;&#xA;" OwnerUserId="7927" LastActivityDate="2017-08-03T12:05:58.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16592" PostTypeId="2" ParentId="16555" CreationDate="2017-08-03T12:13:26.467" Score="0" Body="&lt;p&gt;In a place where I worked they used as a rule of thumb was up to 7m/s for delivery (the pipes have to be a higher grade to support higher pressures and become expensive...) and less than 2m/s for the return as return was much lower pressure (with larger diameter pipes of a lower grade) - think of the losses and also think of the noise generated by the moving fluid, it is not always the economic reasons that come first but pumping power is also considered.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-03T12:13:26.467" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16593" PostTypeId="1" CreationDate="2017-08-03T12:32:21.707" Score="2" ViewCount="3973" Body="&lt;p&gt;I've been asked to describe the physical significance of a dimensionless group (Biot Number) . I understand the definition of the group but I am unsure what the term &quot;physical significance&quot; means in the context of engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for helping.&lt;/p&gt;&#xA;" OwnerUserId="12385" LastActivityDate="2021-02-19T07:58:59.050" Title="Physical Significance" Tags="|heat-transfer|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16596" PostTypeId="1" AcceptedAnswerId="16598" CreationDate="2017-08-03T12:53:23.490" Score="1" ViewCount="89" Body="&lt;p&gt;I have to calculate a ratio of a planetary gear set. It's an existing gear set and had been already implemented for years ago(works fine). Somehow it does not fit the general formula of planetary gear set which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R = 2*P + S (number of teeth)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On my gear set:&#xA;R = 62&#xA;P = 21&#xA;S = 18&lt;br&gt;&#xA;62 != 21*2 + 18&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how can i calculate the correct ratio?&lt;/p&gt;&#xA;" OwnerUserId="12411" LastActivityDate="2017-08-03T18:24:43.863" Title="Marginal Planetary Gear Set" Tags="|gears|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16598" PostTypeId="2" ParentId="16596" CreationDate="2017-08-03T18:19:08.063" Score="0" Body="&lt;p&gt;While R = 2*P + S is typical, the deviation doesn't stop you from calculating the gear ratio.  The carrier is over sized from a tight fit, but it seems to be functioning.  A 22 gear planet would help, but these are unusual.  So let's just calculate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall gear ratio of a simple planetary gearset can be calculated using the following two equations, representing the sun-planet and planet-ring interactions respectively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S\omega_s + P \omega_p - (S + P)\omega_c = 0 $$&#xA;$$R\omega_r - P \omega_p - (R - P)\omega_c = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\omega_r,\omega_s,\omega_p,\omega_c$ is the angular velocity of the Ring, Sun, Planet and Planet Carrier respectively, and $R,S,P$ are the Number of teeth of the Ring, the Sun gear and each Planet gear respectively. From the above we can deduce that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S\omega_s + R\omega_r = (S + R)\omega_c$$&#xA;$$-\frac{R}{S} = \frac{\omega_s - \omega_c}{\omega_r - \omega_c}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming $\omega_{r} \neq \omega_c$.  If we know which are fixed, we can go further.  Calling $n=\frac{R}{S}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1, Sun fixed: $$\frac{\omega_c}{\omega_r}=\frac{n}{1+n} =\frac{R}{R+S}$$&#xA;Case 2, Carrier fixed: $$\frac{\omega_r}{\omega_s}=-\frac{1}{n}=-\frac{S}{R}$$&#xA;Case 3, Ring Fixed: $$\frac{\omega_c}{\omega_s}=\frac{1}{1+n} =\frac{S}{R+S}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, $n=62/18 = 3.44$, so Case 1 is 31:40, Case 2 is 9:31, and Case 3 is 9:40.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2017-08-03T18:24:43.863" LastActivityDate="2017-08-03T18:24:43.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16599" PostTypeId="2" ParentId="16589" CreationDate="2017-08-03T19:10:21.423" Score="0" Body="&lt;p&gt;100,000 RPM is only about 1.6KHz which isn't a challenging frequency to measure. All you need is a transducer that will produce a &quot;once per revolution&quot; signal. An LED, a photodiode, and a stripe of reflecting paint on the rotor (or black paint if the rotor itself is reflecting) would probably work fine. Any cheap PC sound card and free audio software will sample the signal at 44.1 KHz or even higher, which is more than enough.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-08-03T19:10:21.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16600" PostTypeId="1" CreationDate="2017-08-03T20:26:09.087" Score="1" ViewCount="57" Body="&lt;p&gt;What optical methods can be used to measure the concentration of reacting gases in a packed bed reactor during an exothermic and fast reaction? I need to monitor and measure the local concentration of gases used to produce methane (i.e. methane, carbon monoxide, carbon dioxide, water, and hydrogen) along the length of the reactor bed in order to be able to clarify the kinetics of this reaction.&lt;/p&gt;&#xA;" OwnerUserId="12414" LastEditorUserId="10344" LastEditDate="2017-08-04T18:23:53.470" LastActivityDate="2017-08-07T08:51:02.693" Title="Optical methods for reaction engineering" Tags="|mechanical-engineering|design|chemical-engineering|optics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16601" PostTypeId="2" ParentId="16600" CreationDate="2017-08-03T21:14:48.920" Score="1" Body="&lt;p&gt;Would some form of spectroscopy work? I.e. measuring the absorption of particular wavelengths that the gasses you're concerned about absorb and using the resultant data to infer concentrations? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This company (&lt;a href=&quot;http://www.vertilas.com/content/spectroscopy-gas-sensing&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.vertilas.com/content/spectroscopy-gas-sensing&lt;/a&gt;) seems to have a product available. &lt;/p&gt;&#xA;" OwnerUserId="2768" LastActivityDate="2017-08-03T21:14:48.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16602" PostTypeId="1" CreationDate="2017-08-03T21:50:16.187" Score="0" ViewCount="1283" Body="&lt;p&gt;I am looking for assistance to understand why, with SolidWorks 2017, in the &quot;Flow Simulation wizard - Analysis Type&quot; I have no &quot;Rotation&quot; option?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do I need to do in order to see it?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Rotating Region&quot; is not active, it is a gray icon in &quot;Command Manager&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U9g16.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;wizard - Analysis Type&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YttVN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rotating Region&quot; is not active&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12415" LastEditorUserId="8543" LastEditDate="2017-08-04T14:34:10.630" LastActivityDate="2018-06-30T19:50:57.333" Title="Solidworks Flow Simulation the Rotation option missing" Tags="|finite-element-method|simulation|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16603" PostTypeId="1" AcceptedAnswerId="16609" CreationDate="2017-08-03T23:12:17.583" Score="3" ViewCount="1978" Body="&lt;p&gt;How does a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bicycle_brake#Side-pull_caliper_brakes&quot; rel=&quot;nofollow noreferrer&quot;&gt;single pivot side-pull caliper bicycle brake&lt;/a&gt; work? The brake cable only pulls on one of the arms, so what is causing the force on the other arm? I understand that it is a normal force due to the cable housing, but why is this force developed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll elaborate on my questions a bit. Let $A$ be the point where the cable is attached, $B$ be the point where the housing pushes on the other arm, $P$ be the point of pivot, $C$ be the point where the first arm touches the rim and $D$ be the point where the second arm touches the rim. $APC$ is then the arm pulled on by the cable and $BPD$ is the arm potentially pushed by the housing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the cable pulls at $A$ with some tension force $T$. A normal force $N$ from the wheel must then push on $C$, in order for the torque with respect to $P$ to be zero. We have $N = (|AP|_x/|CP|_yS) $, if $x$ and $y$ are in the horizontal and vertical direction respectively, and $T$ is vertical and $N$ horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force $N$ gives a lateral force on the rim and a torque on the wheel. They could be balanced by a normal force of the same magnitude at the other brakepad. By symmetry, $B$ would then have to be pushed by a force of magnitude $T$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But couldn't they also be balanced at the hub? Does it have to do with the construction of the wheel, i.e. because it's spoked? If the wheel were in the form of a solid cylinder (which would of course be very impractical!), would the situation be different?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ya5kZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ya5kZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;image source &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Bicycle_caliper_brake_highlighted.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="12416" LastEditorUserId="288" LastEditDate="2017-08-04T08:44:03.703" LastActivityDate="2023-05-02T03:01:42.527" Title="How does a single pivot side-pull caliper bicycle brake work?" Tags="|mechanical-engineering|" AnswerCount="7" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16605" PostTypeId="2" ParentId="16603" CreationDate="2017-08-04T00:19:54.933" Score="1" Body="&lt;p&gt;The brake cable doesn't &lt;em&gt;apply a force&lt;/em&gt; to the arms. It &lt;em&gt;reduces the size of the gap&lt;/em&gt; between the arms, and therefore between the two brake pads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cyclist can only to apply a &lt;em&gt;force&lt;/em&gt; to the brake lever because the wheel rim prevents the arms moving together. If the bike wheel is removed, it requires very little force to pull the brake lever as far as it can move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the brake assembly can pivot around the point where it is attached to the bike frame, by symmetry the force applied to brake pad on each side will be equal and opposite. The two arms and the pivot form a symmetrical lever arrangement.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-08-04T00:19:54.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16606" PostTypeId="2" ParentId="16587" CreationDate="2017-08-04T04:36:25.143" Score="1" Body="&lt;p&gt;Some solar water heating systems use a tube with a fluid that evaporates / condenses which moves the heat from end to end as you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are designed to work preferentially in one direction as one end has a bulb or larger diameter to better transfer the heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will post an image when on the laptop, but, for the moment here is a link : &lt;a href=&quot;http://www.apricus.com/html/solar_collector_heat_pipe.htm#.WYP5GYXRaaM&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.apricus.com/html/solar_collector_heat_pipe.htm#.WYP5GYXRaaM&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-04T04:36:25.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16607" PostTypeId="2" ParentId="16602" CreationDate="2017-08-04T04:40:21.303" Score="1" Body="&lt;p&gt;Make sure that your unit system and parameters are properly defined (you'll want &quot;SI&quot;), you have saved your file, and also that you have selected the object(s) before clicking the wizard. If this doesn't work, save your file and restart SolidWorks (it may be a glitch).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a quick search on Youtube shows a video of exactly what you want:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=PwX7bJXSP5Q&quot; rel=&quot;nofollow noreferrer&quot;&gt;Youtube Tutorial: Solidworks Rotation Simulation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8543" LastEditorUserId="8543" LastEditDate="2017-08-04T04:49:09.460" LastActivityDate="2017-08-04T04:49:09.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16608" PostTypeId="1" CreationDate="2017-08-04T06:33:27.313" Score="1" ViewCount="107" Body="&lt;p&gt;Suppose I have a flat plate of arbitrary but connected shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I try to connect an axle perpendicular to the plate it is very difficult to align it with the centre of mass, creating wobble when the axle and plate are rotated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a mechanical device that can be attached to the axle and plate that causes the axle to be self-centering to the centre of mass of the plate?&lt;/p&gt;&#xA;" OwnerUserId="12418" LastActivityDate="2017-08-04T06:33:27.313" Title="Is there a way to construct a self-centering axle?" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16609" PostTypeId="2" ParentId="16603" CreationDate="2017-08-04T08:59:46.980" Score="4" Body="&lt;p&gt;Inside The pivot there is a spring holding the 2 arms apart. This adds a torque to the arm and keeps the cable under tension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In normal state the brake pads are not in contact with the rim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the cyclist starts pulling $A$ and $B$ will start to move together until one of the pads contacts the rim. This will then prevent that arm from moving further and apply a little bit of drag on the rim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further pulling on the cable will still try and move the pads together but not only 1 arm can move because the other is blocked by the rim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once both pads are in contact with the rim any tension that isn't compensated for by the spring will put force on the pads squeezing the rim and applying the breaking force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force applied by both brakepads don't need to be equal. The arms are very rarely perfectly centered and balanced and the brake line will also apply a bit of force putting things further off balance. Any sideways force can indeed be compensated for by the hub. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at a bicycle wheel head-on you will see that where the spokes are attached to the hub is much wider than the single line where they are attached on the rim. This creates a triangle shape that can compensate for the sideways force on the wheel and prevents buckling in turns. Any sideways force on the rim will result in a upwards force on one side of the hub and a downwards force on the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dealing with force diagrams you have to take into account &lt;em&gt;all forces&lt;/em&gt; which can be quite the rabbit hole in complex situations.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-08-04T08:59:46.980" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16610" PostTypeId="2" ParentId="16600" CreationDate="2017-08-04T10:56:14.290" Score="1" Body="&lt;p&gt;You need to tell us much more about which species or functional groups you want to follow, concentration range, sampling rate, temperatures, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infrared can be useful for some molecules (water, CO2, N2O), fluorescence for O2 or nitric oxide...&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Edit after comment:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;From your comment I can say that methane, CO, CO2 and water are all infrared active with specific bands well apart from each other so you could measure all those with an infrared probe that can read (simultaneously or sequentially) different wavenumbers. I would say that the sampling time should not be shorter than 1 minute for each species. I cannot tell you much about the temperature requirements. I think it should be fine but you need to check with a provider&lt;/p&gt;&#xA;" OwnerUserId="3991" LastEditorUserId="3991" LastEditDate="2017-08-07T08:51:02.693" LastActivityDate="2017-08-07T08:51:02.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16612" PostTypeId="2" ParentId="15663" CreationDate="2017-08-04T13:20:56.243" Score="0" Body="&lt;p&gt;This problem has nothing to do with Euler buckling. It's a Mechanics of Materials problem. The extra relationship equation you are looking for will come from the internal stress/strain in the members of the structure. Note that all members have pinned connections which cannot carry moments. You are therefore dealing with a truss problem, where all members will only carry axial forces, stresses, and strains. The equations for stress and strain in axial members will give you the final expression you need.&lt;/p&gt;&#xA;" OwnerUserId="12401" LastActivityDate="2017-08-04T13:20:56.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16613" PostTypeId="1" CreationDate="2017-08-04T14:01:28.003" Score="1" ViewCount="102" Body="&lt;p&gt;Cassette tape players use a read head where the magnetic field from the tape is directed through an induction coil. The induced current in the coil is then amplified and sent to the speakers. Now I'm wondering if magnetic tape technology could also work with a permanent magnet that vibrates from the changes in magnetic field in the tape (for playback) or that is made to vibrate externally to create a changing magnetic field on the tape (for recording).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this is just a theoretical question as I'm not going to use a permanent magnet tape player for anything. I am just wondering if this would be possible with any level of sound quality.&lt;/p&gt;&#xA;" OwnerUserId="12424" LastActivityDate="2018-01-01T19:00:29.340" Title="Could a magnetic audio tape recorder/player be made using a vibrating permanent magnet?" Tags="|magnets|audio-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16615" PostTypeId="2" ParentId="16613" CreationDate="2017-08-04T14:20:59.557" Score="1" Body="&lt;p&gt;No. The magnetic flux density available on a tape is way to small to move a magnet with any significant mass enough to produce audible sound.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2017-08-04T14:20:59.557" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16616" PostTypeId="1" CreationDate="2017-08-04T16:45:30.590" Score="3" ViewCount="918" Body="&lt;p&gt;I recently looked up the recommended insulation value for my area (7) and was surprised how much lower the wall insulation value was shown to be. As you can see it is almost a third of the recommended attic value. I am fairly certain that these numbers have more to do with the limited space in the typical framed wall than the actual theoretical ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the idea ratio (can it be represented as a ratio?), or what is the calculation for the ideal spread of insulation over a house assuming you had as much room as you needed in every wall, floor, and ceiling?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WM0iF.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WM0iF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12425" LastActivityDate="2018-09-29T13:01:35.867" Title="Heat Loss in Typical House" Tags="|thermodynamics|heat-transfer|thermal-conduction|thermal-insulation|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16617" PostTypeId="1" CreationDate="2017-08-04T18:20:00.553" Score="1" ViewCount="367" Body="&lt;p&gt;I'm building a machine to make drinks and I would like to be able to put a straw into the cup. To achieve this, I need to build some kind of mechanism, that will be able to feed me with one straw on demand.&lt;/p&gt;&#xA;&lt;p&gt;What I was thinking about is some V-shaped box where all the straws will be stored, with a hole at the bottom, but there will be a gear blocking the hole, so the straws will not fall out of the box. However, the gear will have a cut in it and will be rotating. So each turn, one straw will fall into this cut. Something like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EkkGg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EkkGg.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Will it work? Or is there any better way of achieving this?&#xA;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12428" LastEditorUserId="12428" LastEditDate="2021-07-22T17:45:21.030" LastActivityDate="2021-07-22T17:45:21.030" Title="Will this straw feeder work?" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="16618" PostTypeId="1" CreationDate="2017-08-04T21:05:14.150" Score="2" ViewCount="852" Body="&lt;p&gt;I am currently designing a planetary gear set, and I have designed the sun gear with a module of 2.25 and 35 teeth. Is there any formulae to calculate the internal gears size and amount of teeth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally I had like to have 3 small planet gears and 1 central gear.&lt;/p&gt;&#xA;" OwnerUserId="10930" LastEditorUserId="9" LastEditDate="2017-08-04T21:38:39.263" LastActivityDate="2017-08-07T15:53:15.487" Title="How to design planetary gears?" Tags="|gears|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16619" PostTypeId="2" ParentId="16532" CreationDate="2017-08-04T21:53:01.307" Score="0" Body="&lt;p&gt;It is feasible and realistic. It was already build many times:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jy59J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jy59J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tNxKv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tNxKv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This structure has likely a positive effect to the public opinion. It could have defended the region against the Chernobyl and Fukushima disasters. And it had been much more cheap as building this after an event.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probable reason, why there is no such sphere around all currently built reactor is some similar as there are no catapults in passanger airplanes. Safety is reached on different ways. The balance between the&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;public opinion&lt;/li&gt;&#xA;&lt;li&gt;costs&lt;/li&gt;&#xA;&lt;li&gt;and safety&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;reasons are simply not for build them. But they could be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: maybe integrating the concrete of the sphere into the reactor building (i.e. building it with thicker walls and with more complex safety mechanisms) could reach the same goal.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2017-08-04T21:53:01.307" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16622" PostTypeId="1" CreationDate="2017-08-05T03:32:03.987" Score="0" ViewCount="408" Body="&lt;p&gt;Someone please explain me Mohr's circle in detail in a simple language&lt;/p&gt;&#xA;" OwnerUserId="12430" LastActivityDate="2017-08-05T04:26:01.130" Title="Mohr's Circle explanation in detail in simple languagr" Tags="|mechanical-engineering|mechanical|" AnswerCount="1" CommentCount="1" ClosedDate="2017-08-09T14:40:06.540" ContentLicense="CC BY-SA 3.0" />
  <row Id="16623" PostTypeId="2" ParentId="16622" CreationDate="2017-08-05T04:26:01.130" Score="2" Body="&lt;p&gt;Mohr's circle is a graphical representation of the state of stress at a point. Once the Mohr's circle for a point is constructed, we can easily determine the normal and shear stress components on any plane passing through that point. &#xA;While using the Mohr's circle the vertical axis represents shear stress and the horizontal axis represents normal stress. If we draw a radius at the same angle of inclination with horizontal axis as that of the plane whose stress components are to be determined, then the vertical projection of the radius gives the shear stress and the horizontal projection gives the normal stress in the scale you have chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For constructing the Mohr's circle for a point, given the stress components are  N&lt;code&gt;x&lt;/code&gt; and N&lt;code&gt;y&lt;/code&gt; being normal stress components and T being the shear stress component in a particular plane inclined at angle Z. First establish a coordinate system with vertical and horizontal axes. Then locate the point on the horizontal axis such that its coordinate is the average of N&lt;code&gt;x&lt;/code&gt; and N&lt;code&gt;y&lt;/code&gt;. Then draw a circle with radius equal to vector sum of 0.5(N&lt;code&gt;x&lt;/code&gt;- N&lt;code&gt;y&lt;/code&gt;) and T. ie the square root of the sum of the squares of these. This completes the Mohr's circle.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2017-08-05T04:26:01.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16625" PostTypeId="1" CreationDate="2017-08-05T16:13:32.187" Score="-2" ViewCount="214" Body="&lt;p&gt;Is next possible pls :&#xA;planetary gear system with&#xA;--  fixed sun gear&#xA;-- 2, 3 or 4 planetary gears ( which is possible ?? )&#xA;so that output speed of ring gear is exactly 2x speed of planetary system ?&#xA;I mean : for 1 revolution of the planetary &quot;holder&quot;, the ring makes exactly 2 revolutions. As the planetary system is the drive, the ring system is driven. The result is in the same direction, if I understand this correct. &lt;/p&gt;&#xA;" OwnerUserId="12436" LastActivityDate="2017-08-06T17:43:42.623" Title="planetary gear(s) : possible or not?" Tags="|gears|" AnswerCount="1" CommentCount="4" ClosedDate="2017-08-17T12:21:46.470" ContentLicense="CC BY-SA 3.0" />
  <row Id="16626" PostTypeId="1" AcceptedAnswerId="17890" CreationDate="2017-08-05T17:05:26.860" Score="3" ViewCount="103" Body="&lt;p&gt;As my first serious mechanical engineering project, I've built a wall-mounted folding table using of a pair of 4-bar mechanisms (WIP, ignore the wall mounts): &lt;a href=&quot;https://i.stack.imgur.com/Dw1kf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dw1kf.jpg&quot; alt=&quot;folding table with 4-bar mechanism&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using fisheye bar ends with spherical bearings, like this: &lt;a href=&quot;https://i.stack.imgur.com/TCPNJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TCPNJ.jpg&quot; alt=&quot;fisheye spherical bearing bar-end&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism behaves more or less the way I'd intended, but the spherical bearings have introduced a lot of lateral motion into the mechanism, which I'd like to minimize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a good way to minimize or eliminate this lateral motion? Should I replace the fisheyes with clevis bar ends?&lt;/p&gt;&#xA;" OwnerUserId="12437" LastActivityDate="2018-01-21T08:53:31.007" Title="Eliminating lateral motion in a pair of 4-bar mechanisms" Tags="|mechanical-engineering|stability|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16627" PostTypeId="1" CreationDate="2017-08-05T17:29:29.720" Score="1" ViewCount="64" Body="&lt;p&gt;I have a dynamic mechanical system of the form:&#xA;$$\mathbf{M{\ddot q}+{\Omega_c}G{\dot q}+{Kq}=F(t)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I need to introduce some uncertainty for the $K$ parameter, which is a symetric matrix of $q$ degrees of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a bit of research of Monte Carlo simulation but I´m a bit confused of how different textbooks approach the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example , my actual stiffness value is of $1e8$, and I wish to generate values uniformly distributed between $K_l = 1e6$ and $K_u = 1e12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that I can do this by applying:&#xA;$$\lambda_i = \mu_{\lambda_i} + \epsilon_i$$&#xA;where $\mu_{\lambda_i}$ is the mean value of $\lambda_i$ and $\epsilon$ is a random parameter, and also &#xA;$$\sigma_{\lambda_i}^2 = E[{(\lambda_i-\mu_{\lambda_i})^2}]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be using Matlab for running the simulation , with the command &lt;code&gt;rand&lt;/code&gt; I can generate a randomly distributed numbers between 0 and 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I find with this is that I don't know know to input the boundaries of the $K$ stiffnes values I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I found another text book that explains how to generate random variables using:&#xA;$$ X_i = A + (B-A)R_i$$&#xA;Where $R_i$ is the random number which can be generated by Matlab with rand command to generate a new random variable $X_i$, with lower and upper limits $A$ and $B$. So in matlab I would do a for loop like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for i=1:num&#xA;    Ki=Kl+(Ku-Kl)*rand;&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My main question would actually be, what is the difference between these two methods and for the first one, would $\epsilon$ just be a random 0-1 number adding to the average designated value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2017-08-05T18:31:23.940" Title="How to implement a Monte Carlo Simulation for studying uncertainty in dynamic stiffness?" Tags="|dynamics|simulation|stiffness|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16628" PostTypeId="2" ParentId="16627" CreationDate="2017-08-05T18:31:23.940" Score="3" Body="&lt;p&gt;My advice would be to forget about the mathematical theory and start thinking about what the engineering means. It's hard to think of any situation where the stiffness of a sensibly designed structure might vary &quot;randomly&quot; by a factor of a million, unless all the stiffness values are so high that the response is very insensitive to the stiffness value - and in that case, why bother doing a simulation at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the stiffness of a dynamic system by a factor of 1 million will change the frequency of the fundamental modes by a factor of about 1 thousand. That isn't &quot;a design&quot; in any sense of the words that I can recognise in real life engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about what &lt;em&gt;causes&lt;/em&gt; your stiffness to change - geometrical tolerances, unknown material properties, poorly controlled nonlinear behaviour (e.g. buckling, plasticity, creep), etc. Then use those &lt;em&gt;physical&lt;/em&gt; parameters to &lt;em&gt;calculate&lt;/em&gt; the stiffness for each simulation point in your Monte Carlo study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this approach will change the different entries in the $q$-degrees of freedom stiffness matrix by &lt;em&gt;different&lt;/em&gt; amounts, and that is probably much more realistic than multiplying the whole matrix by one scalar factor. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-08-05T18:31:23.940" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16630" PostTypeId="1" AcceptedAnswerId="16639" CreationDate="2017-08-05T21:24:35.643" Score="1" ViewCount="102" Body="&lt;p&gt;If I was designing a simple gabled house with a 21 ft vaulted ceiling as the peak of the roof, and the house was about 32 ft x 40 ft with no load bearing walls (other than the 4 exterior walls)...  would I need a beam across the entire peak of the house? If so would it have to be steel cause of the 40 ft span?&lt;/p&gt;&#xA;" OwnerUserId="12438" LastEditorUserId="2246" LastEditDate="2017-08-11T13:27:19.000" LastActivityDate="2017-08-11T13:27:19.000" Title="Structural engineering: Would I need a beam for this?" Tags="|structural-engineering|beam|" AnswerCount="1" CommentCount="3" ClosedDate="2017-08-13T11:07:09.270" ContentLicense="CC BY-SA 3.0" />
  <row Id="16631" PostTypeId="2" ParentId="16617" CreationDate="2017-08-05T21:40:44.487" Score="2" Body="&lt;p&gt;Mechanisms like this can work, although I think in this case you would be better with one roller mounted directly under the hopper. As it is the two cutouts don't line up untill the 3 and 9 o'clock positions so there is nothing to s top both notches trying to grab a straw each and ending up with both trying to occupy the same space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should also be prepared for a lot of fine adjustment in prototyping to get it to work properly. I've worked on a few mechanisms like this and they tend to be a bit of a nightmare to get right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to ask yourself whether you really need an automatic straw dispenser. Presumably at some point a human being will pick up the drink in their hands and is it really a benefit to save them the effort of picking up a straw from a manual dispenser?&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-10-06T11:33:19.603" LastActivityDate="2017-10-06T11:33:19.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16632" PostTypeId="2" ParentId="16625" CreationDate="2017-08-06T02:15:22.800" Score="3" Body="&lt;p&gt;It does not appear possible. When the planetary holder rotates 1 time the ring gear rotates 1 full time plus the number of teeth on the sun gear. So in order for the ring gear to rotate twice as many times the ring gear and sun gear would need to have the same number of teeth. So in that design it would not be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If however the planetary holder is fixed then the ratio between the ring gear and the sun gear is simply the ratio of the diameters. For example a 1&quot; sun gear and a 2&quot; ring gear would provide a 2-to-1 ratio. This would not take advantage of the planetary action, and likely could be achieved without the planetary carrier.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastEditorDisplayName="user16" LastEditDate="2017-08-06T17:43:42.623" LastActivityDate="2017-08-06T17:43:42.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16636" PostTypeId="1" AcceptedAnswerId="16637" CreationDate="2017-08-06T14:18:57.383" Score="0" ViewCount="1334" Body="&lt;p&gt;What is difference between the &lt;strong&gt;tailboard&lt;/strong&gt; and the &lt;strong&gt;bood lid&lt;/strong&gt;? Can you show me where each part is? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for help. &lt;/p&gt;&#xA;" OwnerUserId="12343" LastActivityDate="2017-08-06T14:34:54.253" Title="Difference between the tailboard and boot lid." Tags="|automotive-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16637" PostTypeId="2" ParentId="16636" CreationDate="2017-08-06T14:34:54.253" Score="1" Body="&lt;p&gt;A tailboard would be the name of the opening panel found at the rear on a pick-up type vehicle, while a boot lid is the name for the panel that closes the boot or trunk. An SUV has the rear panel called a door and if the rear window opens it is likely to be called a hatch - also hatch-back.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-06T14:34:54.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16639" PostTypeId="2" ParentId="16630" CreationDate="2017-08-06T16:16:45.423" Score="0" Body="&lt;p&gt;You would not need a steel beam . Although steel would give you the lowest profile. An engineered wood beam would be more typical ; truss type or laminated beam . I guess a wood beam would be about 2 feet high/deep. They will both meet code anywhere but you you will need to know the weight of your roof and determine how you will provide lateral stability. So professional design will be needed.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2017-08-09T02:15:00.403" LastActivityDate="2017-08-09T02:15:00.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16640" PostTypeId="1" CreationDate="2017-08-06T16:50:12.987" Score="1" ViewCount="290" Body="&lt;p&gt;In the calibration procedure of an instrument (e.g. a pressure transducer), when we want to determine the calibration curve, is it a good choice to force the passage of the calibration curve through the origin?&lt;/p&gt;&#xA;" OwnerUserId="12446" LastActivityDate="2017-08-07T17:59:21.310" Title="Calibration curve of an instrument" Tags="|instrumentation|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16641" PostTypeId="1" AcceptedAnswerId="16643" CreationDate="2017-08-06T17:20:14.423" Score="4" ViewCount="1258" Body="&lt;p&gt;I'm trying to measure the actuation force of buttons on the &lt;a href=&quot;https://en.wikipedia.org/wiki/PlayStation_Vita&quot; rel=&quot;nofollow noreferrer&quot;&gt;Playstation Vita&lt;/a&gt;, in preparation of building an external device that presses these buttons automatically. Specifically, I'm trying to measure the force required to push the buttons on the D-Pad on the left, and the action buttons on the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling &quot;how to measure actuation force&quot; will give you pages of results of mechanical keyboard enthusiast forums, and they often mention measuring the actuation force by stacking pennies / whatever coins you have until the key is depressed and the computer recognises the keypress. I can't do that normally here, since the buttons are really small (&amp;lt;1cm for the action buttons) and even the smallest coin my currency has cannot be stacked onto it without toppling over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an attempt to mitigate the problem, I made the following setup. It still uses coins as weights to push down the button, but I rested the coins on the ruler that in turn rests on another object of somewhat equal height/thickness to the game console, in this photo, the remote control seen in the background. I conducted the experiment by continuously adding coins with a game running on the console until it registers a keypress, and then summing up the weights of the coins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/CBRPL.jpg&quot; alt=&quot;experimental setup&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the results are rather inconsistent and questionable. I repeated my experiments several times (by knocking down the coin tower and stacking it up again) as standard practice, and I got values ranging from &lt;b&gt;100g to 180g&lt;/b&gt;, which feels extremely high for buttons on a handheld game console, and that large spread seems to be telling me that my experimental technique is just plain wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some possible causes of inconsistencies from the setup:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The coins are not the same - I'm stacking 5 different coins at once&lt;/li&gt;&#xA;&lt;li&gt;The coins are not stacked in the same order each time I redo the experiment&lt;/li&gt;&#xA;&lt;li&gt;The center of the coin tower is not aligned with the center of the button on the game console - I don't know how to do this without the coin tower making contact with the neighboring buttons or it toppling over&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some steps I have taken to possibly improve accuracy:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The ruler used is a stiff metal ruler and does not bend significantly&lt;/li&gt;&#xA;&lt;li&gt;The ruler is kept horizontal&lt;/li&gt;&#xA;&lt;li&gt;The only points of contact of the ruler are the remote control seen in the background and a single button on the D-Pad.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;TL;DR:&lt;/b&gt; How can I improve the accuracy of my setup, or other better ways to measure actuation force of such small buttons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;b&gt;Update &amp;amp; Final Results&lt;/b&gt; I used the weighing scale method, and the results are 80g for the D-Pad and about 120g for the action buttons on the right. Thanks everyone!&lt;/p&gt;&#xA;" OwnerUserId="12445" LastEditorUserId="12445" LastEditDate="2017-08-14T13:34:19.343" LastActivityDate="2017-08-14T13:34:19.343" Title="How can I accurately measure the actuation force of small buttons (&lt;1cm diameter) using home equipment?" Tags="|mechanical-engineering|measurements|torque|experimental-physics|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16642" PostTypeId="2" ParentId="16626" CreationDate="2017-08-06T19:19:06.347" Score="1" Body="&lt;p&gt;Can you cross brace between the left and right side of the mechanism. If I understand correctly there are two parallel four bar mechanism that in tandem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am afraid that if you place more constraints in the joints, if they are not aligned correctly the mechanism will not work correctly as it will try to bend some of the linkages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proposal&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weld the red bars below to the linkages&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FDACa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FDACa.png&quot; alt=&quot;pic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-08-06T19:19:06.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16643" PostTypeId="2" ParentId="16641" CreationDate="2017-08-07T01:24:16.443" Score="4" Body="&lt;p&gt;If you want a quick and somewhat inaccurate hill-billy engineering answer, I would take a digital kitchen scale and place the PSV ontop of it, tare it, then load up a program to test the buttons (game, menu, etc). with the scale device zero'd out, slowly depress the button until it is read by the program. a full actuation can and will differ from the triggering of the circuit. Wait for the program to read the triggering rather than the button's actuation, and add a few extra grams to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a precise measurement... I have no idea what you have available for building as far as &quot;home equipment&quot; is concerned.&lt;/p&gt;&#xA;" OwnerUserId="8543" LastActivityDate="2017-08-07T01:24:16.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16644" PostTypeId="1" CreationDate="2017-08-07T02:03:01.717" Score="0" ViewCount="172" Body="&lt;p&gt;I'm trying to make an involute spur gear generator but can't seem to get it right. I know that the diameter of the base circle of the involutes is equal to the pitch diameter * cos(pressure angle). But this means that as I increase the number of teeth and the pitch circle gets larger, its scale relative the base circle also increases. And at a certain point, the pitch circle passes the the intersection point of the two involute curves that will make up the sides of a tooth! (My tooth sides are made of an involute in standard position, and one that is mirrored and rotated by an appropriate angle)&lt;/p&gt;&#xA;" OwnerUserId="12450" LastActivityDate="2017-08-08T15:27:48.833" Title="Can the pitch circle of a gear be farther out than the intersection point of the involutes on opposite sides of a tooth?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16645" PostTypeId="2" ParentId="16593" CreationDate="2017-08-07T02:28:05.673" Score="1" Body="&lt;p&gt;Note: My understanding is that the question is regarding the term &quot;physical significance&quot;, not specifically Biot's number's physical significance. Therefore my question will focus on that and will merely indicate how to obtain Biot's number's significance. Actually doing so is left as an exercise to the reader.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A property's &quot;physical significance&quot; means exactly what it seems to mean: what the property describes in the physical world. Basically, &quot;physical significance&quot; is a fancy term for &quot;definition&quot;. Another way of looking at it is imagining you have an object or system and you want to measure some property; what exact physical characteristics do you need to measure to calculate this property?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, what is the physical significance of temperature? Well, temperature is a description of how quickly the molecules of a given object are moving. That is temperature's physical significance. So, if you want to measure temperature, how would you do so? &quot;Use a thermometer&quot; isn't the answer I'm looking for, because it is merely an instrument, and I'm looking for the physical properties it measures. So, if you want to measure temperature, you could freeze the object in time and measure the speed of each of its particles (ignoring quantum uncertainty and the fact that &quot;speed&quot; makes no sense if you freeze time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biot's number is a representation of the temperature gradient to be expected within an object: if Biot's number is large, then the object's thermal conductivity is much smaller than its thermal film coefficient (times a volume/area coefficient).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, again, what physical properties would you need to measure to calculate Biot's number? Thermal conductivity and film coefficients aren't (in my opinion) a satisfactory answer because they are derived properties, much as temperature is. So, keep it going. What physical properties would you need to measure to calculate thermal conductivity, for instance? Once you reach the fundamental physical properties of the object which define Biot's number, you can then define what relationship Biot's number actually describes, at which point you'll have obtained its physical significance.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-08-07T02:28:05.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16648" PostTypeId="2" ParentId="16618" CreationDate="2017-08-07T15:53:15.487" Score="1" Body="&lt;p&gt;For any standard planetary gearset, the following formula relating the numbers of teeth of the sun, annulus (internal) and planet gears applies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N_s + 2N_p = N_a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $N$ is the number of teeth, and the subscripts correspond to sun, planet and annulus gears respectively. This is a purely geometric result, and comes from the fact that the pitch circles of the gears should be tangent to one another, and that the diameters of the pitch circles are directly proportional to the number of teeth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how, with the number of teeth of the sun gear given to you, the number of teeth of the annulus gear can be varied by selecting the number of teeth of each planet. For example, given a sun gear of 35 teeth:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;choosing a planet with 20 teeth would yield an annulus with $35+2\times 20=75$ teeth.&lt;/li&gt;&#xA;&lt;li&gt;choosing a planet with 35 teeth would yield an annulus with $35+2\times 35=105$ teeth.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While there is a freedom to select the number of teeth on the planet and annulus, note that this selection will impact the value of the gear ratio. For example, having the sun/planet/annulus with 35/20/75 teeth will yield a 22:7 (i.e. 3.14:1) gear ratio from the sun gear to the planet carrier while fixing the annulus gear, and 35/35/105 teeth will yield a 4:1 gear ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the diameter of each of the gears' pitch circles, including the annulus gear, is calculated as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d=N\times m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is the module of the gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The annulus gear's outermost diameter is arbitrary, but it should be kept greater than the root diameter, which for standard &lt;em&gt;internal&lt;/em&gt; gears is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d_{root} = (N+2.5)\times m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cPvss.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cPvss.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2017-08-07T15:53:15.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16650" PostTypeId="2" ParentId="16640" CreationDate="2017-08-07T17:10:23.517" Score="0" Body="&lt;p&gt;If you're developing the calibration curve for a sensor (a device like a pressure transducer), since your measurement system will account for the conversion of the signal from an electrical signal in V or mA to Pressure using the sensor's calibration curve, then there's no reason to do so unless the sensor is really giving a strong bias at zero input. If it is, then I would suspect a defective sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have a load cell and I want to develop the calibration curve for it. I will set up my calibrator which applies a precise load in N and reads the electrical output of the transducer in V. I'll do this at 0, the max load I expect and a few points in between. If my sensor has a linear response, I should have a nice straight line graph with an equation like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{Newtons}= m*Volts + b  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $m$ has units of $ \frac{Newtons}{Volt}$ and $b$ has units of $Newtons$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $b$ is large, relative to the $m*Volts$ term, I would question the integrity of the load cell.  &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-08-07T17:10:23.517" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16651" PostTypeId="2" ParentId="16640" CreationDate="2017-08-07T17:53:01.360" Score="1" Body="&lt;p&gt;This depends on the sensor.  In some sensors, an offset it to be expected, and should be included in a calibration curve.  In others, there is a good physics reason that the actual output should be zero when the thing being measured is zero, and you should force it through the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets look at two examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A half-bridge strain sensor with two active elements.  This type of sensor consists of two strain gauges stuck on a metal beam such that when the beam bends one is compressed and one is tensioned. The two are connected in series, between the supply rail and ground (lets say 5V and ground for arguments sake). The zero position will be when neither is stressed, and the output will be close to 2.5V, but if one strain gauge is slightly different from the other due to manufacturing variations, then it will be slightly off.  In this case, you should not force the fit through the origin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A current-sensing shunt resistor.  These are low value resistors designed to allow you to measure the current through them without dissipating too much energy.  The voltage measured is directly proportional to the current thanks to ohms law.  If there is no current flowing through, then there cannot possibly be a voltage generated, and any voltage you do measure is probably a fault in the measurement kit not the shunt.  So in this case, you might well choose to force the calibration curve through the origin.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In general, you would need to look at the sensor's datasheet (if you bought it) or think carefully though how it works (if you made it) to decide whether you expect an offset.  If you tell us what type of pressure transducer it is and how it works, then we might be able to tell you which category it falls into. Mechanical pressure sensors based on a movable membrane will usually (but not always) fall into the first category, where ionisation gauges would usually fall into the second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, after doing the fit, you should check whether the observed offset is what you expect, and that the fit to the data is good.&lt;/p&gt;&#xA;" OwnerUserId="12022" LastEditorUserId="12022" LastEditDate="2017-08-07T17:59:21.310" LastActivityDate="2017-08-07T17:59:21.310" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16652" PostTypeId="1" AcceptedAnswerId="16653" CreationDate="2017-08-07T18:27:29.103" Score="-4" ViewCount="9092" Body="&lt;p&gt;I want to lift a mass of 1kg so how much force is required ?&lt;/p&gt;&#xA;" OwnerUserId="12457" LastEditorUserId="12457" LastEditDate="2021-10-21T15:04:33.053" LastActivityDate="2021-10-21T15:04:33.053" Title="Force Required to Lift a WEIGHT of 1Kg" Tags="|motors|torque|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ClosedDate="2017-08-09T01:16:35.407" ContentLicense="CC BY-SA 4.0" />
  <row Id="16653" PostTypeId="2" ParentId="16652" CreationDate="2017-08-07T19:20:06.107" Score="3" Body="&lt;p&gt;Mass is measured in kg and weight measured in N. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorDisplayName="user16" LastEditDate="2017-08-09T12:51:39.493" LastActivityDate="2017-08-09T12:51:39.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16655" PostTypeId="1" AcceptedAnswerId="16656" CreationDate="2017-08-07T22:53:56.067" Score="-2" ViewCount="205" Body="&lt;p&gt;I want to measure the distance between the back of a human body, and the chest. This is denoted in the image below, where Point A is the back and Point B is the chest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V7GOV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V7GOV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not wish to measure the diameter of the chest. Rather, the direct path from Point A, to Point B ( Through the Ribcage ). Do any &lt;strong&gt;mobile sensors&lt;/strong&gt; exist that can help me capture this data, which can be worn by the applicant at all times? ( During sleep, walking, running, etc.)&lt;/p&gt;&#xA;" OwnerUserId="9644" LastEditorUserId="9644" LastEditDate="2017-09-05T20:13:29.690" LastActivityDate="2017-09-05T20:13:29.690" Title="Accurately measure the distance from chest to back" Tags="|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16656" PostTypeId="2" ParentId="16655" CreationDate="2017-08-08T00:16:30.667" Score="3" Body="&lt;p&gt;The simplest answer would be &quot;a large pair of calipers&quot;. In fact, they make such a tool for exactly this task. See &lt;a href=&quot;http://www.seritex.com/gpm-large-spreading-caliper-rounded-ends-108/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.seritex.com/gpm-large-spreading-caliper-rounded-ends-108/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The GPM Large Spreading Caliper with rounded ends provides accurate&#xA;  measurement of the distance between two points, where one or both&#xA;  points may not be on a flat surface. This Large Spreading Caliper can&#xA;  also be used for the measurement of Anterior-Posterior chest depth.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The ends of the spreading branches are rounded for the patient's&#xA;  comfort, as opposed to the GPM Large Spreading Caliper model 109,&#xA;  which has pointed ends.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The GPM Large Spreading Caliper with rounded ends is manufactured from&#xA;  chrome plated steel. Hinged at the base, the two curved branches with&#xA;  rounded ends are connected by a pivotal ruler, which is engraved in&#xA;  millimeter intervals from 0 to 60 cm. The branches can be immobilized&#xA;  in any position by a screw located on the slider which guides the&#xA;  ruler. The ruler gives the precise distance between the two points&#xA;  over which the measurement has been taken.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Or you could improvise by using something like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fvrmi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fvrmi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: The OP changed the question to include the &quot;mobile and wearable&quot; requirements, after this answer was written.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="8055" LastEditDate="2017-08-08T18:20:42.493" LastActivityDate="2017-08-08T18:20:42.493" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16661" PostTypeId="1" CreationDate="2017-08-08T06:48:11.043" Score="-1" ViewCount="98" Body="&lt;p&gt;Hermetic Compressor is used by split Air Conditioning unit. The compressor compressed freon gas coming from indoor unit. The gas is usually quite cold (about 2 deg.C lower than inside room temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: Is the compressor better inlet with cold or hot freon gas? What are the effects or reasoning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, would it be good that compressor gets cooling from cold freon but this might considered waste of freon work (add compression work because superheat condition)? The compressor cooling should be from outside. The next problem maybe if cold freon could cold the compressor oil.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastEditorUserId="1953" LastEditDate="2018-03-31T17:22:31.490" LastActivityDate="2018-03-31T17:22:31.490" Title="Good or bad (AC Split compressor input with cold freon)" Tags="|hvac|compressors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16662" PostTypeId="1" AcceptedAnswerId="16816" CreationDate="2017-08-08T08:54:05.730" Score="0" ViewCount="40" Body="&lt;p&gt;I have a Problem with a transformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy function is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I(u_1,u_2,P)=\int_R [\frac{\mu}{2}(u_{i,j} + u_{j,i}) u_{i,j}-Pu_{i,i} dx_1 dx_2 - \int_R f_i u_i dx_1dx_2 +\int _S t_iu_idS$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_{i,j}=\frac{du_i}{dx_j}$ and Einstein's summation convention is applied (sum over repeating indices).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The task is to compute the variation $\delta I=0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the result sheet they do following transformation, which I do not understand:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\tag{1}I(u_1,u_2,P)&amp;amp;=\int_R [\frac{\mu}{2}[(\delta u_{i,j} + \delta u_{j,i})u_{i,j}+(u_{i,j} + u_{j,i}) \delta  u_{i,j}] dx_1 dx_2 +[...]\\&#xA;\tag{2}I(u_1,u_2,P)&amp;amp;=\int_R [\mu(u_{i,j} + u_{j,i})\delta u_{i,j}dx_1 dx_2 +[...]&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="12352" LastEditorUserId="1832" LastEditDate="2017-08-08T12:30:54.113" LastActivityDate="2017-08-19T10:39:38.480" Title="Variational Problem of slow steady flow of viscuous fluid in 2D" Tags="|fluid-mechanics|finite-element-method|fluid|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16663" PostTypeId="1" CreationDate="2017-08-08T09:50:04.073" Score="0" ViewCount="753" Body="&lt;p&gt;I have a gas mixture (steam and air) contained in a tank. &#xA;I know the tank pressure ($P$), temperature ($T$) and the massic air fraction ($x_a$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to determine the density of both gases to perform some calculations afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, I would have done the following :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\rho_{steam} = IAPWS97('rho\_PT', P,T)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. asking IAPWS tables to provide me the density of steam at the tank conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems based on available ressources that this is wrong. The correct way do do it would rather be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\rho_{steam} = IAPWS97('rho\_PT', P_{steam},T)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore taking into account the partial steam pressure to compute its density within the tank rather than the total tank pressure &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that ? Intuitively, I would use total tank pressure because it seems to me that both steam and air experiences the total tank pressure and not only their partial pressure and therefore I would use total tank pressure to compute their properties.&lt;/p&gt;&#xA;" OwnerUserId="8261" LastActivityDate="2017-08-08T12:42:43.873" Title="Determining gas properties in a steam and air gas mixture" Tags="|thermodynamics|gas|steam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16664" PostTypeId="2" ParentId="16663" CreationDate="2017-08-08T12:42:43.873" Score="2" Body="&lt;p&gt;Considering that air and steam behave as ideal gases, you can write the following equations:&#xA;$$ P V = nRT  $$&#xA;$$ P_A V = n_A RT$$&#xA;$$ P_W V = n_W RT $$&#xA;where I use subscripts $_A$ and $_W$ for air and steam respectively. Taking $M_A$ and $M_W$ as the molar masses of air and water, the density of each of the gases would be&#xA;$$ P_A V = (m_A/M_A)  RT \rightarrow \underbrace{(m_A/V)}_{\rho_A} =\frac{P_A M_A}{RT}  $$&lt;br&gt;&#xA;$$ P_W V = (m_W/M_W)  RT \rightarrow \underbrace{(m_W/V)}_{\rho_W} =\frac{P_W M_W}{RT}  $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence if you want to find the density of &lt;strong&gt;each&lt;/strong&gt; of the gases, you only care of the partial pressure because it relates with the amount of matter of each of the gases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wanted to find the density of the mixture of gases, you would use the total pressure and an &quot;equivalent molar mass&quot; for the mixture &lt;/p&gt;&#xA;" OwnerUserId="3991" LastActivityDate="2017-08-08T12:42:43.873" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16665" PostTypeId="2" ParentId="16661" CreationDate="2017-08-08T14:19:17.450" Score="0" Body="&lt;p&gt;If the compressor is designed to work with the fluid as a gas then the temperature is part of that design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas entering the compressor is not primarily there for cooling - the primary function is for the cycle to move heat from one place (evaporator) to another (condenser).&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-08T14:19:17.450" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16666" PostTypeId="1" CreationDate="2017-08-08T14:32:12.317" Score="4" ViewCount="10153" Body="&lt;p&gt;I am having problem with the application of Saint Venant principle.I have heard that quite often we use that knowingly or unknowingly.I would be very greatful to anyone who would explain this very explicitly to me.&lt;/p&gt;&#xA;" OwnerUserId="12431" LastActivityDate="2017-08-08T18:08:15.213" Title="What is Saint Venant Principle?" Tags="|mechanical-engineering|structural-engineering|applied-mechanics|mechanical-failure|mechanical|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16667" PostTypeId="2" ParentId="16644" CreationDate="2017-08-08T15:27:48.833" Score="1" Body="&lt;p&gt;That is impossible. The whole point of the pitch diameter is that the thickness of the tooth $X$ equals the gap $Y$ at that radial distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take one profile and rotate it half the tooth separation angle and where it intersects the other profile is where the pitch diameter is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7tHqz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7tHqz.png&quot; alt=&quot;PD&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-08-08T15:27:48.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16668" PostTypeId="2" ParentId="16666" CreationDate="2017-08-08T15:58:40.907" Score="9" Body="&lt;p&gt;The intuitive meaning of Saint Venant's principle is that &quot;if two different sets of applied loads are statically equivalent, then the differences between the two stress patterns they create are only significant close to where the loads are applied&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is used very frequently in engineering modelling. For example if a beam is bolted to a support with several bolts, the details of the loading at the bolt positions can often be replaced by a single force and bending moment, applied to the end of the beam. That simplified model won't tell you the detailed stresses around the circumference of each bolt hole, but it will tell you the stresses and deflections of the complete beam accurately enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to convert the intuitive meaning of the principle into a precise mathematical statement about the solutions to differential equations for different boundary conditions, but engineers usually just assume the principle is correct!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few situations where Saint Venant's principle does &lt;strong&gt;not&lt;/strong&gt; apply. A well-known example is the torsion of a split cylinder, where a &quot;small&quot; change in the loading distribution can change the stresses and displacements along the whole length of the cylinder. Similar exceptions to the principle can occur (sometimes, but not always!) in any curved thin shell structure which is only curved in &lt;em&gt;one&lt;/em&gt; direction measured along the surface, and flat in the perpendicular direction.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-08-08T15:58:40.907" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16670" PostTypeId="2" ParentId="16666" CreationDate="2017-08-08T18:08:15.213" Score="8" Body="&lt;p&gt;To add to &lt;a href=&quot;https://engineering.stackexchange.com/a/16668/1832&quot;&gt;@alephzero's answer&lt;/a&gt;, here's an layman's explanation of Saint-Venant's Principle: far enough away from the load's point of application, equivalent loads can be treated as identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the stress profile of a column of area $A$ with a concentrated load $F$ applied at the top is effectively identical to the stress profile of the same column with a uniform load $q = F/A$ applied over the entirety of its top face. Or, to explain it visually:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D5hbe.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D5hbe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as 99% of the column is concerned, it doesn't matter if the force applied is perfectly concentrated at one point or distributed over its surface. Those loads are equivalent and generate the same stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you study the stress profile near the top of the column, then Saint-Venant can't help you. For instance, what's the stress near the top-right corner of the column under the concentrated load? At first sight, it would seem to be zero (it actually isn't, as will be seen below). Meanwhile, under the uniform load, the stress profile is simple and can be trivially determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see how the stress profile is different near the point where the load is applied, trace the &quot;load paths&quot; in each case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MH30S.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MH30S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one is trivial, so I won't spend any time in it. The first one is a bit more complicated, so let's simplify it into a strut-and-tie model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cdrw7.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cdrw7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that some tension is created perpendicular to the applied force in the process of distributing the force along the entire section. That tension, however, is concentrated very close to the area where the load is applied. Farther away, the beam behaves exactly as if it were under a uniformly distributed load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you're dealing with a concrete column (or some other material with low tensile strength) and you don't handle the tension near the column's head, you might end up with something more like this (think knife and warm butter):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/61HpL.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/61HpL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-08-08T18:08:15.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16671" PostTypeId="1" CreationDate="2017-08-08T18:20:57.077" Score="0" ViewCount="255" Body="&lt;p&gt;Is there formula to calculate the compressive force required to achieve x% of squeeze in an O-ring? I have referred the Parker O-ring and Apple Rubber Handbooks, but they provide just approximations. I am trying to calculate the linear force per length for a 4&amp;nbsp;mm o-ring and I want to achieve 25% squeeze.&lt;/p&gt;&#xA;" OwnerUserId="12465" LastEditorUserId="1832" LastEditDate="2017-08-08T23:06:19.563" LastActivityDate="2023-11-18T15:20:23.037" Title="O-ring compressive force" Tags="|waterproofing|seals|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16672" PostTypeId="2" ParentId="8315" CreationDate="2017-08-08T19:20:16.087" Score="0" Body="&lt;p&gt;Your approach is absolutely acceptable so long as you realize its limits. The reason you are not seeing a method like this in a text book is because it is a mix of numerical methods and conventional empirical methods that is not particularly common. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it is not common to approach a problem this way is because the natural convection heat transfer coefficient (htc) is a bulk property. It is based on the overall geometry and is going to give you a prediction of the heat transfer over the entire surface of the object. Thus the htc is based on the overall system geometry and properties and not the local geometry and properties right around each node. My hope is this image will make that clear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/swula.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/swula.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have a hot plate in some cold fluid. We have broken the hot plate into 6 nodes to estimate the surface temperature, but we can't calculate the htc based on only each nodes temperature because that is not the way natural convection works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most robust way to solve this is to solve the natural convection numerically as well. This is what a commercial CFD package like comsol would do. However, this would require that you model the fluid numerically which as you say above would get very complicated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other approach you could take is to continue using the bulk htc but recalculate it each time you iterate through your internal numerics. Then you would come up with a bulk htc and you could calculate the heat loss from each node via &lt;code&gt;Q=dA*htc*(T_i-T_bulk)&lt;/code&gt;. When you are running your solver you just need to recalculate the htc based on a new average surface temperature each iteration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other compromises you could make here to increase the accuracy of your model, but the overarching theme is to know the limits of each compromise you make. Using a bulk htc is an ok move if your fluid has a high Cp and thus will not have a big dT as it flows over the cube. It is also an ok move if the geometry is a cube or a sphere where the water is not flowing over the part for a long time. This would be a bad approach if your fluid had a low Cp and thus had a big dT as it flowed over your part or if the part had a very long flow path as it does not capture the change in duty caused by the change in dT between the first and last node the water flows past. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, I think the model is simple enough you should just write it and see what the results are. You can then start messing with the temperature you use for the bulk temperature to try to capture the water's temperature better. &lt;/p&gt;&#xA;" OwnerUserId="2780" LastActivityDate="2017-08-08T19:20:16.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16673" PostTypeId="2" ParentId="16671" CreationDate="2017-08-08T19:31:30.237" Score="3" Body="&lt;p&gt;To be honest, if you need more precision than what the handbooks tell you, your best bet is to buy some o-rings and test them under the range of conditions under which you expect them to function. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-08-08T19:31:30.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16674" PostTypeId="1" CreationDate="2017-08-08T19:48:41.077" Score="0" ViewCount="61" Body="&lt;p&gt;I am looking for a weight sensor which has a range from 2000 kg to 30000 kg to measure the weight of passing trucks. &#xA;Accuracy is not very important. Power use is very important, typically 3.3v. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do there exist these kind of sensors?&lt;/p&gt;&#xA;" OwnerUserId="12467" LastEditorUserId="1586" LastEditDate="2017-08-22T05:26:07.430" LastActivityDate="2017-12-20T08:09:54.377" Title="Low power, large weight sensor needed" Tags="|electrical-engineering|sensors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16675" PostTypeId="2" ParentId="16674" CreationDate="2017-08-08T19:53:22.147" Score="1" Body="&lt;p&gt;You can look at this product, which is rated up to 100 T and has a calibration certificate &lt;a href=&quot;http://www.novatechloadcells.co.uk/ds/f317.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;F317&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Made by Novatech : &lt;a href=&quot;http://www.novatechloadcells.co.uk/ds/f317.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.novatechloadcells.co.uk/ds/f317.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-08T19:53:22.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16678" PostTypeId="1" CreationDate="2017-08-09T01:33:28.387" Score="1" ViewCount="1286" Body="&lt;p&gt;I'm trying to calculate piston force pressures for double acting pistons for pneumatic and hydraulic pressures.  I found the following formula for calculating both &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Force} = \frac{\text{Pressure} \times \pi  \times (\phi_\text{Bore}^2 - \phi_\text{Rod}^2)}{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that there should be some kind of conversion factor when going from pneumatic to hydraulic or vice versa.  Is this formula valid for both pneumatic and hydraulic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got the formulas from the following sites:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/hydraulic-force-calculator-d_1369.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/hydraulic-force-calculator-d_1369.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/pneumatic-cylinder-force-d_1273.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/pneumatic-cylinder-force-d_1273.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12471" LastEditorUserId="1832" LastEditDate="2017-08-09T03:54:24.267" LastActivityDate="2017-08-09T14:54:54.173" Title="Calculating piston force for pneumatic vs hydraulic" Tags="|hydraulics|pneumatic|pistons|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16679" PostTypeId="2" ParentId="16678" CreationDate="2017-08-09T03:07:55.483" Score="4" Body="&lt;p&gt;The formula is valid independant of the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force in an hydraulic system is based on pressure and area - a height difference between pistons is usually small compared to the acting pressure from the pump - this is why is is usually ignored. Think of the pressure created by the difference in height of the pistons in tipper truck systems for example compared to the system pressure of 30 Bars or 700 Bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking at the pressure created by a column of fluid of 30m height then that will be affected by the different fluids (p = density * gravity * height).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you come to calculate the motor power necessary then the fluid becomes relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit aded in response to question below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is valid independant of the fluid ie Pressure = Force / Area, so a fluid (air, water, hydraulic oil) at a given pressure acting on the surface of a piston creates a force proportional to the Area.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-08-09T14:54:54.173" LastActivityDate="2017-08-09T14:54:54.173" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16680" PostTypeId="1" AcceptedAnswerId="16687" CreationDate="2017-08-09T06:04:44.057" Score="6" ViewCount="448" Body="&lt;p&gt;Quite often while calculating angular velocity we come across the quantity radians per second. How do I treat radian dimensionally?&lt;/p&gt;&#xA;" OwnerUserId="12431" LastEditorUserId="-1" LastEditDate="2021-04-11T20:43:34.653" LastActivityDate="2021-04-11T20:43:34.653" Title="Is radians a dimensionless quantity?" Tags="|mechanical-engineering|fluid-mechanics|applied-mechanics|mathematics|mechanical-failure|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="16682" PostTypeId="2" ParentId="16680" CreationDate="2017-08-09T09:03:57.853" Score="1" Body="&lt;p&gt;Radians are a &lt;strong&gt;fundamental&lt;/strong&gt; unit of measurement, as are metres, grams, seconds, ampere, kelvin, candela, steradian &amp;amp; mol.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Metre is the fundamental unit for length&lt;/li&gt;&#xA;&lt;li&gt;Gram is the fundamental unit for mass (kilograms in SI units)&lt;/li&gt;&#xA;&lt;li&gt;Second is the fundamental unit for time&lt;/li&gt;&#xA;&lt;li&gt;Ampere is the fundamental unit for electrical current&lt;/li&gt;&#xA;&lt;li&gt;Kelvin is the fundamental unit for temperature&lt;/li&gt;&#xA;&lt;li&gt;Candela is the fundamental unit for luminous intensity&lt;/li&gt;&#xA;&lt;li&gt;Mol is the fundamental unit for the amount of substance&lt;/li&gt;&#xA;&lt;li&gt;Steradian is the fundamental unit for solid angles&lt;/li&gt;&#xA;&lt;li&gt;Radians is the fundamental unit for plane angles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-08-09T09:03:57.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16684" PostTypeId="1" AcceptedAnswerId="16686" CreationDate="2017-08-09T09:51:44.157" Score="0" ViewCount="197" Body="&lt;p&gt;How do I say my weight? In which SI units? I see people are using Kg as a unit of denoting weight from my childhood. Answer this question &quot;What's your weight?&quot;.&lt;/p&gt;&#xA;" OwnerUserId="12457" LastEditorDisplayName="user16" LastEditDate="2017-08-09T12:52:19.127" LastActivityDate="2017-08-09T12:52:19.127" Title="How do I denote someone weight" Tags="|unit|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16685" PostTypeId="2" ParentId="16684" CreationDate="2017-08-09T10:39:09.580" Score="1" Body="&lt;p&gt;As this is an Engineering stack we use the prescribed units, mass is in kg and weight is in Newtons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a common misconception by the masses (great unwashed...) and it is also covered in this question :&#xA;&lt;a href=&quot;https://engineering.stackexchange.com/questions/16652/force-required-to-lift-a-weight-of-1kg&quot;&gt;Force Required to Lift a WEIGHT of 1Kg&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also obvious in various phrases such as &quot;I am gong to boil the kettle&quot; which is assumed to mean boil the water that is inside the kettle and not put the kettle in a big cooking pot and boil it...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-09T10:39:09.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16686" PostTypeId="2" ParentId="16684" CreationDate="2017-08-09T10:40:40.903" Score="5" Body="&lt;p&gt;Weight is a force and is expressed in Newton (N). Mass is expressed in kilogram (kg).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in informal (non-scientific) language, people often express weight in kg, although this is not correct strictly speaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relation between the two is $F=mg$, with $F$ the weight (N), $m$ the mass (kg) and $g$ earth's gravity constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://en.wikipedia.org/wiki/Mass_versus_weight&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3633" LastActivityDate="2017-08-09T10:40:40.903" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16687" PostTypeId="2" ParentId="16680" CreationDate="2017-08-09T12:10:12.643" Score="15" Body="&lt;p&gt;radian is a &lt;em&gt;derived unit&lt;/em&gt;, defined as the ratio of arc length to radius. As the ratio of two lengths it is dimensionless.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2017-08-09T12:10:12.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16688" PostTypeId="1" CreationDate="2017-08-09T14:38:46.110" Score="2" ViewCount="106" Body="&lt;p&gt;I am conducting few trials for checking the placement of a thermocouple on a valve for measuring the temperature of the valve and the material is an X45 alloy. You can see the slot taken on the valve and the thermocouple is inserted in the slot (picture is attached).&lt;a href=&quot;https://i.stack.imgur.com/ozbSb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ozbSb.jpg&quot; alt=&quot;Thermocouple placed on the neck of the valve&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Procedure:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Turned on the muffle furnace with a set point of 300 deg C&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After the furnace temperature is stable at 300 deg C, kept the valve inside the furnace.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Kept the valve inside the furnace and let it attain an equilibrium.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Took the valve outside after set amount of time and measured the temperature using a microcontroller (arduino + MAX6675). IT took about 10 seconds to measure the temperature and I could not run the wires inside the furnace as I did not have access to high temperature withstanding cables for the thermocouple. Also, as the valve is taken out for measuring the temperature,  there will be some heat loss due to atmospheric exposure.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;After 2 hours of this trial, I found the temperature of the material to be around 250 deg C.&lt;/p&gt;&#xA;&lt;p&gt;Due to the huge difference between setpoint and the material temperature, I wanted to raise a question.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; The thermocouple and the furnace are calibrated.&lt;/p&gt;&#xA;&lt;p&gt;My question is, if a material is placed in a furnace with a setpoint temperature, would the temperature of the material be same as the setpoint after a certain amount of time? My guess is there will be some temperature difference but since I am not a mechanical engineer, I want to clarify this in this wonderful forum.&lt;/p&gt;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;&lt;p&gt;Vish&lt;/p&gt;&#xA;" OwnerUserId="12474" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-08-10T15:00:06.253" Title="Temperature of a material in a furnace" Tags="|mechanical-engineering|thermodynamics|heat-transfer|thermal-conduction|thermal-expansion|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16690" PostTypeId="1" CreationDate="2017-08-09T15:38:06.603" Score="0" ViewCount="153" Body="&lt;p&gt;As part of some experimentation on a bolted joint, I have used an instrumented bolt with a number of strain gauges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The testing has produced results of strain with respect to time. By using the Elastic Modulus of the bolt I can easily convert this into mean stress in the bolt. My question is how this relates to clamping force in the joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it as simple as averaging this stress over the loading area (i.e. bolt head and nut clamping surfaces) to attain a clamping load?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="12475" LastActivityDate="2017-08-09T17:59:16.753" Title="Instrumented Bolt - Conversion of measure strain/stress to clamping force" Tags="|bolting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16691" PostTypeId="2" ParentId="16688" CreationDate="2017-08-09T17:21:16.453" Score="1" Body="&lt;p&gt;You can use &lt;a href=&quot;http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton's Law of Cooling&lt;/a&gt; to model both the heating and cooling of your valve. You'd use the data you already collected to establish the model, then use the model to see how long it takes the valve to approach equilibrium in the furnace. You can use the same model to estimate the temperature behavior between the time you removed the part from the furnace and when the Arduino was able to get a fix on the thermocouple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the detailed process for modeling the heating/cooling of the part. I'll be working with the following sample temperature ($T$) vs time ($t$) data. The data represents temperature data taken every minute, starting 5 minutes after the part is removed from the furnace.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;t (min)   T (C)&#xA;5         234&#xA;6         222&#xA;7         211&#xA;8         201&#xA;9         191&#xA;10        182&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Take the natural logarithm of the temperature data, it should be roughly linear with respect to time. Now do a linear regression through $t$ vs. $ln(T)$ and you should get a slope ($\alpha$) and y-intercept of -0.05 and 5.704 respectively. I used Excel to do the regression; LibreOffice Calc, Matlab, Octave, et al. should also do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of the part just as it came out of the furnace ($T_0$) will be approximately equal to $e$ raised to the power of the y-intercept: $e^{5.704}$, or 300 C in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the slope and initial temperature from the linear fit combined with Newton's Law to model the cooling process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;T(t)=T_0\cdot e^{\alpha t}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the same equation to model the heating process in the furnace by flipping the sign of $\alpha$, as long as the heat transfer modes within the furnace are similar to those during cooling, e.g. natural convection and radiation.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastEditorUserId="1560" LastEditDate="2017-08-10T13:31:59.777" LastActivityDate="2017-08-10T13:31:59.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16692" PostTypeId="2" ParentId="16690" CreationDate="2017-08-09T17:59:16.753" Score="1" Body="&lt;p&gt;A &quot;do it yourself&quot; approach to this will probably be inaccurate, because you don't know the friction force between the bolt and the threads, and the stress distribution within the bolt is also unknown (and non-uniform). The results will probably have the same accuracy and repeatability as measuring the torque applied to the bolt head - typically, about $\pm20\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;professional&quot; way to do it is use bolts manufactured with &lt;em&gt;internal&lt;/em&gt; strain gauges located on the neutral axis of the bolt, for example from &lt;a href=&quot;http://www.strainsert.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Strainsert.&lt;/a&gt; These are typically accurate to 1%, and can be individually calibrated if required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://catalog.strainsert.com/Asset/Overview-and-Application-of-Internally-Gaged-Force-Sensing-Fasteners_new.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://catalog.strainsert.com/Asset/Overview-and-Application-of-Internally-Gaged-Force-Sensing-Fasteners_new.pdf&lt;/a&gt; for more details.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-08-09T17:59:16.753" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16695" PostTypeId="2" ParentId="16688" CreationDate="2017-08-09T22:07:01.047" Score="1" Body="&lt;p&gt;The standard technique is to weld a thermocouple to the work. Then the time to temperature depends on the furnace,  size , heat source, fan , etc. A serious furnace will have a survey ( like Mil H 6872 ) of temperature variations available from the manufacturer. If this is a typical small lab muffle , just weld the thermocouple to the valve you already sacrificed. For such a small part you should be looking at much less than one hour.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-08-09T22:07:01.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16696" PostTypeId="2" ParentId="13545" CreationDate="2017-08-09T23:32:53.290" Score="0" Body="&lt;p&gt;There is a chance 250F might be above the Tg of the part if it is cured at room temperature. If this is the case then you will definitely get sagging if you postcure without a ramp at 250F. You might want to try ramping slowly up to 250F so that the temperature of the oven never shoots past the Tg of the material( which is steadily increasing with degree of cure). &lt;/p&gt;&#xA;" OwnerUserId="12479" LastActivityDate="2017-08-09T23:32:53.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16700" PostTypeId="2" ParentId="16680" CreationDate="2017-08-10T09:58:16.103" Score="1" Body="&lt;p&gt;Yes. Radian is a dimensionless quantity. One simple way to prove it is to use the relation between angle, radius and arc length.&#xA;$$ l = \phi \cdot r $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly the angle in radians has to be dimensionless, otherwise this basic equation would make no sense.&lt;/p&gt;&#xA;" OwnerUserId="10383" LastActivityDate="2017-08-10T09:58:16.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16701" PostTypeId="1" CreationDate="2017-08-10T14:03:01.247" Score="7" ViewCount="647" Body="&lt;p&gt;What design standards exist for the design of ballast against uplift forces, and what factor of safety against uplift do these design standards promulgate? I am familiar with a typical FS of roughly 1.5 against sliding forces due to active soil pressure for the design of many types of structures (including cantilevered retaining walls and gravity dams), but I am not aware of any such standard FS against uplift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible- perhaps likely- that the safety factor can vary depending on the nature of the loading. Specifically I have in mind uplift due to wind. But other forms of applied uplift may include: buoyant forces, dynamic/static pressure applied by some other fluid (e.g., water, etc.), and load transference forces (perhaps in a machine where ballast is used to counteract forces applied to some arm/member).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I have done this in the past is to use standard load combination equations in design codes. For example, the ASCE 7 2010 load combinations for design of the limit state considering wind uplift&lt;sup&gt;*&lt;/sup&gt; are as follows&lt;sup&gt;**&lt;/sup&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;ASD: $~~~~~~~~~~~~~~~~0.6D+1.0W_{working~stress/service~state}^{***}$&lt;/li&gt;&#xA;&lt;li&gt;Strength design: $0.9D+1.67W_{working~stress/service~state}^{***}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, the problem with this approach is twofold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, using these load combinations directly leads two different, inconsistent effective factors of safety: &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;$Effective~FS_{ASD~Method} = 1.0/0.6 = 1.67$&lt;/li&gt;&#xA;&lt;li&gt;$Effective~FS_{Strength~Method} = 1.67/0.9 = 1.85$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Secondly, the intention of a design code such as ASCE 7 is to provide guidance for the &lt;em&gt;load side&lt;/em&gt; of the design equation. The &lt;em&gt;resistance side&lt;/em&gt; of the design equation is usually left to the various engineering groups/societies who publish standards for different materials used to resist loads, e.g. the American Concrete Institute (ACI), or the American Institute of Steel Construction (AISC). These groups provide standard factors of safety (ASD design) and resistance factors (LRFD design). However as far as I am aware, there is no American Institute of Ballast. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that for this question I am not considering uplift due to seismic forces, because seismic events occur in the form of a spectral acceleration which &lt;em&gt;results&lt;/em&gt; in a force, and that force gets higher as the mass of the ballast increases. Therefore ballast is not effective on its own in preventing uplift due to seismic acceleration (though in some cases it may be effective in restraining lateral seismic acceleration via friction).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;*&lt;sub&gt; Note that in ASCE 7 2010/IBC 2012 forward, wind loads are factored loads (factored up by 1.6); prior to 2010, wind loads were working stress and the ASD load factor was 1.0 (LRFD load factor of 1.6). &lt;/sub&gt;&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;**&lt;sub&gt;The actual ASCE 7 load combinations utilize the ASCE 7 2010, factored wind uplift load, i.e.:&lt;/sub&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sub&gt;$~~~~$7. ASD: $~~~~~~~~~~~~~~~~0.6D+0.6W_{ASCE~7~2010}$&lt;/sub&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;sub&gt;$~~~~$6. Strength design: $0.9D+1.0W_{ASCE~7~2010}$&lt;/sub&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;***&lt;sub&gt; Due to the above notes, $W_{working~stress/service~state}=0.6W_{ASCE~7~2010}$. &lt;/sub&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2018-04-17T18:04:58.937" LastActivityDate="2023-12-18T20:04:45.593" Title="Industry standard factor of safety for ballast design" Tags="|civil-engineering|safety|engineering-general|" AnswerCount="2" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16702" PostTypeId="2" ParentId="16688" CreationDate="2017-08-10T15:00:06.253" Score="0" Body="&lt;p&gt;Another answer : parts about the size of your valve, smallish ( 3 cubic feet) electric furnace with strong circulation .Parts heated to 2150 F required less than 25 minutes but this included a couple minutes soaking time ( at temperature , ie. essentially no difference in temperature between parts and furnace).&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-08-10T15:00:06.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16704" PostTypeId="1" AcceptedAnswerId="16727" CreationDate="2017-08-10T15:40:38.833" Score="2" ViewCount="151" Body="&lt;p&gt;I have an O-shaped aluminum (7075) profile WxDxH - 40mm x 200mm x 25mm&#xA;The inner channel is 20 mm wide and 10 mm deep - runs through the entire length of the bar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside, I have a 20 mm x 10 mm bar (also Al 7075). As you can see from the attached picture, on the opposite face there's a hole in which a flat screw is used to fasten the two pieces in position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pvpxg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pvpxg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This telescopic mechanism needs to withstand large loads &gt;50 kg (&gt;110 lbs) on both ends and I'm worried that using only a screw would mean that there will be slippage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better way to design a secure clamp for such a telescoping mechanism? I need to be able to do fine positioning and have the confidence that it won't move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help.&lt;/p&gt;&#xA;" OwnerUserId="11266" LastActivityDate="2017-08-17T19:12:32.750" Title="Aluminum sliding mechanism clamping" Tags="|mechanical-engineering|aluminum|mechanisms|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16705" PostTypeId="1" CreationDate="2017-08-10T15:51:52.740" Score="1" ViewCount="3393" Body="&lt;p&gt;I have a box (as shown below). I know the temperature outside of the box (for example you could use 293K) and I would assume the temperature inside the box is initially the same as outside of the box. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 6000 chickens in the box. They produce a total of 15.8 W of heat with 8 W of that being sensible heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the box is sealed and there is no ventilation how would I calculate the change in temperature over time? Would it be a simple case of  $Q=mcΔT$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously in the real world I would not leave 6000 chickens in a box to overheat. How could I calculate the temperature change over time if I ventilated the box at a rate of 4m3/s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: Material would be PVC coated material (like on the side of lorries/trucks). I have an found an R value of 0.16 m²K/W for this type of material and it's thickness would be around 0.75mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/AymGO.jpg&quot; alt=&quot;The box in question&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12487" LastEditorUserId="12487" LastEditDate="2017-08-11T12:56:37.767" LastActivityDate="2018-10-11T19:00:17.623" Title="How do I calculate the temperature change inside a box with a heat source? What if I ventilate the box?" Tags="|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16706" PostTypeId="1" AcceptedAnswerId="16708" CreationDate="2017-08-10T18:06:17.690" Score="0" ViewCount="905" Body="&lt;p&gt;I am creating subwoofer boxes out of MDF or Baltic Birch.  My local makerspace has  a CNC machine.  I am going to make all of the cuts with it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What software is generally used or will work to get a file for a CNC machine?  In the &lt;a href=&quot;http://www.sector67.org/blog/faq/#target-toc-12&quot; rel=&quot;nofollow noreferrer&quot;&gt;makerspace's FAQ&lt;/a&gt; they say DXF or STL but it's under the section regarding Laser Cutting.  So I'm not sure if that's the needed file format.&lt;/p&gt;&#xA;" OwnerUserId="12491" LastActivityDate="2023-10-23T12:23:05.717" Title="Looking for program and file format for CNC Milling" Tags="|modeling|cnc|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16708" PostTypeId="2" ParentId="16706" CreationDate="2017-08-10T18:57:33.677" Score="1" Body="&lt;p&gt;Properly prepared DXF is the safest choice. Corel Draw CDR is another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While most CNC machines use STL, there are subtle differences between dialects, there are idiosyncrasies concerning spindle speed, tool changers, tool profiles, material removal with thick tool, fine finish with fine tool, and so on, and so forth. And editing a wrongly generated STL to fit the machine is usually far more headaches than it's worth. You just come with the properly prepared &quot;high-level&quot; format file like DXF and let the CNC operator convert it to the &lt;em&gt;correct&lt;/em&gt; STL. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does &quot;Properly prepared&quot; mean?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;all elements are closed - closed curves or sets of segments; contours enclosing the shapes. If you need a 0.5mm thick line that's 10cm long, you don't draw a 10cm line and set its thickness to 0.5mm in properties - you draw a rectangle of 0.5x100mm using default style line, or any other style; the styles are all stripped when converting. Same, dashed line is a row of rectangles, not a line with style: dashed. &lt;/li&gt;&#xA;&lt;li&gt;all texts are transformed to curves (you can't be sure the makerspace has all the same fonts).&lt;/li&gt;&#xA;&lt;li&gt;no intersections. That means, merge all partially overlaying elements, don't use invisible elements to mask parts of the visible ones, put (even minuscule) distance between corners that touch - or opposite, extend a little and leave a tiny bridge; no joint at a single point.&lt;/li&gt;&#xA;&lt;li&gt;no overlaying parallel lines. Similarly to above, all &quot;zero width&quot; gaps between elements should be replaced by the elements being fused together. &lt;/li&gt;&#xA;&lt;li&gt;no bitmaps whatsoever. If you absolutely must include a bitmap graphics, submit one in possibly big resolution, to be vectorized. Preferably don't try to vectorize by yourself; programs for vectorization love to make a mess of tiny segments that's about impossible to straighten out and make usable for CNC.&lt;/li&gt;&#xA;&lt;li&gt;watch out for narrow gaps. The tool has a certain thickness; making a notch half millimeter thick is quite tricky and often impossible. The outer edges may be quite fancy and very precise but all gaps/incisions/grooves are limited by the tool thickness.&lt;/li&gt;&#xA;&lt;li&gt;your project should be 2D. If there are different milling depths, draw them as separate layers and annotate with proper depths. If you need slopes or other irregular 3D surfaces, discuss with your CNC operator how to supply them best.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Picture of some of these caveats: &quot;Tak&quot; = &quot;Yes&quot;. &quot;Nie&quot; = &quot;No&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dnwUG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dnwUG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2017-08-10T19:11:13.800" LastActivityDate="2017-08-10T19:11:13.800" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16710" PostTypeId="1" CreationDate="2017-08-10T19:30:42.377" Score="-1" ViewCount="58" Body="&lt;p&gt;I am designing a television stand for 55 inch television that weighs 24 kilograms. Now, i have made a stand and attached it to the back of the television. Now, what should be the weight of the stand to prevent the tv from tipping. And is a straight rectangular design that stands behind the tv can be stable, or do i need to extend it to the front of the tv. And also, another question - When designing any new product how can i be sure that the product will be stable and not tip over. Please explain as you would to a high school student, because i am not good at physics. Please see the attached image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QTnFf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QTnFf.png&quot; alt=&quot;tv stand&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8783" LastEditorUserId="1832" LastEditDate="2017-08-11T03:23:00.240" LastActivityDate="2017-08-11T03:23:00.240" Title="Need help to understand why something tips over and how to prevent it" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2017-08-17T12:13:25.580" ContentLicense="CC BY-SA 3.0" />
  <row Id="16711" PostTypeId="2" ParentId="16710" CreationDate="2017-08-10T19:49:15.927" Score="1" Body="&lt;p&gt;The main discipline that studies this tip over phenomena is &quot;statics&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simple terms, for stability, the center of gravity of an object must lie in between its touch down points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One more thing is, if someone tilts the TV, or if you put it on an inclined place, the weight vector must still be inside the touch down points (base) of the tv. Else, it will  fall on its own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these inclined or dynamic situations, the lower the CG, the better the stability of an object.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-08-10T19:49:15.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16712" PostTypeId="1" AcceptedAnswerId="16714" CreationDate="2017-08-10T19:54:42.457" Score="0" ViewCount="312" Body="&lt;p&gt;I'm trying to determine the amplitude vs the excitation frequency of a multi dof dynamic system. I'm very confused from reading different methods, although I understand there's a very general one, which is not giving me the results that I think should be when plotting the response.&#xA;Considering a system with no damping  $\mathbf{M{\ddot x}+{Kx}=F(t)}$, and by assuming the harmonic excitation of the force is a cosine the system becomes $\mathbf{-MX{cos(\omega t)}+{KX}cos(\omega t)=fcos(\omega t)}$. From here the cosines get cancelled out so&#xA;only $\mathbf{[K-M{\omega^2}]X=f}$ remains. By solving for the $X$ vector in the end I get $X = \frac{F/M}{|\omega_n^2-\omega^2|}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with Matlab I wrote the following code, with parameters I found on an example from brown.edu, in which the explanation is not complete, my system is actually a 24 DoF system, so I need the code a bit automated:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    k1 = 2;&#xA;k2 = 1;&#xA;k3 = 1;&#xA;m1 = 1;&#xA;m2 = 1;&#xA;&#xA;M=[m1 0; ...&#xA;   0 m2];&#xA;K=[k1+k2 -k2; ...&#xA;  -k2 k3+k3];&#xA;F=[1;...&#xA;   1]; &#xA;&#xA;[V,D] = eig(K,M);      &#xA;D     = diag(sqrtm(D)); &#xA;frequencies   = (D);           &#xA;[rows,~] = size(V);&#xA;&#xA;f=linspace(0,0.7,100);&#xA;omega=2*pi*f;&#xA;Mag = M\F; &#xA;Xt = zeros(rows, length(omega));&#xA;    for i = 1:rows&#xA;        for j = 1:length(omega)          &#xA;            Xt(i,j) =  abs((frequencies(i)^2 - omega(j)^2)\Mag(i)) ;&#xA;        end&#xA;    end&#xA;figure&#xA;    for m = 1:length(frequencies)&#xA;        plot(omega, Xt(m,:));&#xA;        set(gca,'YScale','log')&#xA;        hold on&#xA;        axis tight&#xA;    end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where the main part is the  &lt;code&gt;Xt(i,j) = (abs (frequencies(i)^2 - omega(j)^2)\Mag(i) );&lt;/code&gt; , which follows the determined formula for the displacements of each degree of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When plotting what I get is:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/l8RLs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l8RLs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which shows the masses vibrations being at resonance, but from what I understand there should be more peaks and some antiresonance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing in the general math?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT_1:&lt;/strong&gt; I changed the bit of the code to calculate the absolute values of the equation like: $$X(\omega) = \left|\sum_n \frac{F/M}{\omega_n^2 - \omega^2}\right|$$ but I get the same when plotting. I found another solution but I don't understand how its determined or why choose it.&#xA;From ${M{\ddot x_i}+{Kx_i}=f(t)sin\omega t}$ the steady state solution is $x_{iss} = x_{is} sin \omega t + x_{ic} cos \omega t$ , where $i = 1,2$ , then &#xA;$$ \begin{bmatrix} \mathbf{-(\omega)^2M+{K}} &amp;amp; \mathbf{0} \\ \mathbf{0} &amp;amp; \mathbf{-(\omega)^2M+{K}} \end{bmatrix} \begin{bmatrix} \mathbf{x_{ic}} \\ \mathbf{x_{is}} \end{bmatrix} = \begin{bmatrix} \mathbf{0} \\ \mathbf{f} \end{bmatrix} $$&#xA; which is then solved to get ${X_{i}} = (x_{ic} + x_{is} )^{1/2}$ &#xA;First, I don´t understand very well if just $x_{iss}$ is substituted in ${M{\ddot x_i}+{Kx_i}=f(t)sin\omega t}$ for $x_i$. And is this just assuming a sum of $cos$ and $sin$ instead of my first approach of only $cos$? Then, do I need to consider adding the transient response also, which I´m actually not sure how it would be. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT_2:&lt;/strong&gt; I think I figured out what I was doing wrong. I changed the equation from $$X(\omega) = \left|\sum_n \frac{F/M}{\omega_n^2 - \omega^2}\right|$$ to $$ X = [K-M\omega^2]^{-1}F$$ which mathematically I though they were the same??? The code part that changed is as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for j = 1:length(omega)       &#xA;       Xt(:,j)  =  abs((K-M*omega(j)^2)\F);&#xA;    end&#xA;figure&#xA;    for m = 1:length(frequencies)&#xA;        plot(omega,Xt(m,:))&#xA;        hold on&#xA;        axis tight&#xA;    end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And the result plot is:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/49106.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/49106.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my extended MDoF system I got:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oacKP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oacKP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In which I excited all DoF, even when I only excite a single one, it is still very hard to notice what's going on, is there a better way of analysing the response?&lt;/p&gt;&#xA;" OwnerUserId="4088" LastEditorUserId="4088" LastEditDate="2017-08-11T03:05:29.230" LastActivityDate="2017-08-11T03:05:29.230" Title="Undamped forced vibration for a multi dof system" Tags="|dynamics|vibration|frequency-response|eigenvalue-analysis|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16713" PostTypeId="2" ParentId="16680" CreationDate="2017-08-10T21:30:10.633" Score="3" Body="&lt;h2&gt;TL/DR&lt;/h2&gt;&#xA;&lt;p&gt;As the other answers have stated, radians are dimensionless.&lt;/p&gt;&#xA;&lt;h2&gt;Cycles and Turns&lt;/h2&gt;&#xA;&lt;p&gt;This is a question that seems to come up often, and from many different quarters; not only from students and lay people, but I have seen plenty of professionals for whom the concept of radians causes difficulty.&lt;/p&gt;&#xA;&lt;p&gt;One way that seems to help some people's intuition is to put aside the idea of radians entirely, and instead talk about &lt;em&gt;cycles&lt;/em&gt; and &lt;em&gt;turns&lt;/em&gt; of a circle.&lt;/p&gt;&#xA;&lt;h2&gt;As the World Turns&lt;/h2&gt;&#xA;&lt;p&gt;Consider, for example, the rotation of the Earth around the Sun. We can call each time the Earth completes a trip around the sun as a &lt;em&gt;cycle&lt;/em&gt;, or a &lt;em&gt;full turn&lt;/em&gt; about its path. Note that it &lt;strong&gt;doesn't matter&lt;/strong&gt; that the Earth's trip around the Sun is NOTHING like a circle; the &lt;em&gt;trip itself&lt;/em&gt; is the cycle.&lt;/p&gt;&#xA;&lt;p&gt;Using this terminology, calculating the average angular velocity of the Earth is very simple: &lt;span class=&quot;math-container&quot;&gt;$1 \frac{cycle}{year}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The conversion from per year to per seconds is rather trivial:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1 \frac{cycle}{year}\frac{1~year}{365~days}\frac{1~day}{24~hours}\frac{1~hour}{3600~seconds}=1 \frac{cycle}{31.5576E6~seconds}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Furthermore, the unit for &lt;span class=&quot;math-container&quot;&gt;$1 \frac{cycle}{second}$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$hz$&lt;/span&gt;. So:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1 \frac{cycle}{year}=1 \frac{cycle}{31.5576E6~seconds}=316.9~\text{microhz, or μhz}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Again: &lt;span class=&quot;math-container&quot;&gt;$1 hz = 1 \frac{cycle}{second}$&lt;/span&gt;, or we can also say &lt;span class=&quot;math-container&quot;&gt;$\frac{turn}{second}$&lt;/span&gt;. Remember: &lt;strong&gt;a cycle and a turn are the same thing&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Circles as Useful Models&lt;/h2&gt;&#xA;&lt;p&gt;It turns (&amp;lt;-- funny!) out that &lt;em&gt;circles&lt;/em&gt; are a very useful way to &lt;em&gt;model&lt;/em&gt; cycles/turns. Even though the path of the Earth &lt;em&gt;not a circle&lt;/em&gt;, we can use a circle to &lt;em&gt;map&lt;/em&gt; the position of the Earth at any point in time we are interested in against other things we are interested in. We'll use the seasons as an illustration.&lt;/p&gt;&#xA;&lt;p&gt;To make things easier, let's assume that a solar year- 1 cycle- starts at the Winter solstice: December 21st. Therefore, we can have the following model of a year:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v96Dp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v96Dp.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;So now, we can make the following fairly intuitive statements:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Winter occurs at 0 turns, 1 turn, 2 turns, etc etc.&lt;/li&gt;&#xA;&lt;li&gt;Spring occurs at 0.25 turns, 1.25 turns, 2.25 turns, etc etc.&lt;/li&gt;&#xA;&lt;li&gt;Summer occurs at 0.5 turns, 1.5 turns, 2.5 turns, etc etc.&lt;/li&gt;&#xA;&lt;li&gt;Spring occurs at 0.75 turns, 1.75 turns, 2.75 turns, etc etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Because we have chosen a circle as a &lt;em&gt;model&lt;/em&gt; of a solar cycle or turn, we now have some very useful tools available for use &amp;quot;for free&amp;quot;. These tools can help us to analyze our model in a mathematically rigorous way. To put it another way: we chose a circle precisely because we know all kind of cool things about circles that we can use to help us understand a solar cycle better. I'll provide an example at the end.&lt;/p&gt;&#xA;&lt;h2&gt;What Is a Radian, Really?&lt;/h2&gt;&#xA;&lt;p&gt;One of the things we get for free: for circles, we know that the relationship between the &lt;em&gt;radius&lt;/em&gt; and the &lt;em&gt;circumference&lt;/em&gt; is a &lt;em&gt;constant ratio&lt;/em&gt;. In keeping with what we have done so far, we are just going to refer to the &lt;em&gt;circumference&lt;/em&gt; of the circle as &lt;em&gt;one turn&lt;/em&gt;, or &lt;em&gt;one cycle&lt;/em&gt;. That ratio is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{1~\text{turn}}{1~\text{radius}}=2\pi\rightarrow 1~\text{turn}=2\pi~\text{radii}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If we divide both sides of that equation by &lt;span class=&quot;math-container&quot;&gt;$2\pi~\text{radii}$&lt;/span&gt;, we finally arrive at the definition of a radian. Which is the &lt;em&gt;ratio of the circle radius to one full turn of the circle&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1~\text{radian}=\frac{1~\text{turn}}{2\pi~\text{radii}}=\frac{1}{2\pi}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;But recall that for our &lt;em&gt;circle model&lt;/em&gt;, the turn and the circumference are the same thing. Therefore the units for the radii and the turn cancel each other out, and we have a dimensionless radian as expected.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{1~\text{circumference}}{2\pi~\text{radii}}=\frac{X\require{cancel} \cancel{Unit}}{Y\require{cancel} \cancel{Unit}}=\frac{X}{Y}=\frac{1}{2\pi}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h2&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=ZfYBsuBzulE&quot; rel=&quot;nofollow noreferrer&quot;&gt;What Does it All Mean, Basil?&lt;/a&gt;&lt;/h2&gt;&#xA;&lt;p&gt;It turns out, &lt;em&gt;radians&lt;/em&gt; are often a lot more convenient to use than &lt;em&gt;turns&lt;/em&gt; and/or &lt;em&gt;cycles&lt;/em&gt;. Here's a couple examples:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Because we understand how radians and &lt;em&gt;turns&lt;/em&gt; are related, we can &lt;strong&gt;intuit&lt;/strong&gt; much of the rest of geometry without having to memorize formulae. One example is the arc length formula:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$L_{arc}=\theta\cdot radius$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is in radians.&lt;/p&gt;&#xA;&lt;p&gt;Where does this formula come from? It is simply a result of the &lt;em&gt;definition&lt;/em&gt; of a radian. Observe:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1~\text{radian}=\frac{1~\text{turn}}{2\pi~\text{radii}}=\frac{1}{2\pi}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The arc length &lt;span class=&quot;math-container&quot;&gt;$L_{arc}$&lt;/span&gt; is some &lt;em&gt;portion&lt;/em&gt; of a total turn, or &lt;span class=&quot;math-container&quot;&gt;$Decimal\cdot\text{turns}$&lt;/span&gt;. So:&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{equation}&#xA;\begin{split}&#xA;&amp;amp; Decimal~\text{radians}=\frac{Decimal\cdot\text{turns}}{radius} \Rightarrow \\&#xA;\\&#xA;Decimal\cdot\text{turns}= &amp;amp; L_{arc}=Decimal~\text{radians}\cdot radius=\theta\cdot radius&#xA;\end{split}&#xA;\end{equation}&lt;/span&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Another very useful set of tools we get for free by virtue of choosing a circle as our model is the entire field of trigonometry. Trigonometric relationships are almost always expressed using radians, degrees, or sometimes gradians. However, in keeping with the above, you could easily use &lt;em&gt;turns&lt;/em&gt; of the circle instead of radians. For example, the equation for &lt;span class=&quot;math-container&quot;&gt;$\text{sin}$&lt;/span&gt; is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{sin}~\theta=\frac{opposite}{hypotenuse}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;If we make &lt;span class=&quot;math-container&quot;&gt;$\theta=\frac{1}{4}\text{turn}$&lt;/span&gt;, then:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{sin}~1~quarter~turn=\frac{1~radius}{1~radius}=1.0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/28G0C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/28G0C.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;But if you want to use a calculator, the calculator doesn't know about &lt;em&gt;turns&lt;/em&gt;. It knows about radians and degrees. But from above we know that &lt;em&gt;1 turn&lt;/em&gt; is &lt;span class=&quot;math-container&quot;&gt;$2\pi$&lt;/span&gt; radians, so 1 quarter turn is just &lt;span class=&quot;math-container&quot;&gt;$\frac{2\pi}{4}$&lt;/span&gt; radians, or &lt;span class=&quot;math-container&quot;&gt;$\frac{\pi}{2}$&lt;/span&gt; radians.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h2&gt;Back to Earth&lt;/h2&gt;&#xA;&lt;p&gt;We started with the Earth revolving around the Sun. Can we do anything interesting relating the circle model, radians, and the seasons together? Indeed we can.&lt;/p&gt;&#xA;&lt;p&gt;We know that the Earth's rotation is tilted at an axis of about 23.5 degrees from vertical (with respect to the sun), which is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$23.5/1~\text{turn}=23.5/360=0.0653~\text{radians}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We also know that this tilt is at maximum at the solstices; the Earth's Southern hemisphere is pointed toward the Sun on December 21st, and the Northern hemisphere pointed toward the Sun on July 21st. The angle with respect to the Sun is zero at the equinoxes: March 20th and September 22nd.&lt;/p&gt;&#xA;&lt;p&gt;Can we approximate the tilt on some arbitrary date? We can easily do that using our circle model. How about on George Washington's birthday: February 22nd.&lt;/p&gt;&#xA;&lt;p&gt;We can ascertain using geometry that the relative inclination of the axis varies with the cosine of the angle of the Earth's position around the Sun, starting with position 0 at the Winter solstice:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Winter: &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; = 0 turns &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt; 23.5 deg tilt &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow \text{cos}~\theta=1.0$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Spring: &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; = 0.25 turns &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt; 0 deg tilt &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow \text{cos}~\theta=0.0$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Summer: &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; = 0.5 turns &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt; 23.5 deg tilt &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow \text{cos}~\theta=1.0$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Spring: &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; = 0.75 turns &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt; 0 deg tilt &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow \text{cos}~\theta=0.0$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Starting at the Winter solstice, how many turns are we on February 22nd?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(Jan~1-Dec~21)=9~days$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(Feb~1-Jan~1)=31~days$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Feb~22=22~days$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Total=62~days$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;How many turns is that?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$62/365.2425=0.16975~\text{turns}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We'll need to convert to radians to use a calculator. So what is the approximate portion of the 1 yr trip that has been traveled from the starting position measured in radians?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$0.16975~\text{turns}=0.16975\cdot 2\pi~\text{radians}=1.0666~\text{radians}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;As shown above, the cosine varies from 1.0 at position 0 (23.5 deg tilt) to 0.0 at 1 quarter turn (0 deg tilt). So:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Tilt_{Feb~22}≈\text{cos}(1.0666~\text{radians})\cdot 23.5~\text{deg}=11.35~\text{deg}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is approximate because we have used a circle to model the change in tilt over time. However, in reality the Earth's speed around the sun is not constant, so this is just an estimation. Our circle model is only an approximation of reality.&lt;/p&gt;&#xA;" OwnerUserId="88" LastEditorUserId="88" LastEditDate="2021-03-25T13:12:25.603" LastActivityDate="2021-03-25T13:12:25.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16714" PostTypeId="2" ParentId="16712" CreationDate="2017-08-10T22:57:45.350" Score="1" Body="&lt;p&gt;The code in your OP seems to be for a 2 DOF system not a 24 DOF system. You could check that by printing the &quot;frequencies&quot; variable from your eigensolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you are not getting anti-resonances is because you are taking the absolute values in the wrong place. You should be calculating $$X(\omega) = \left|\sum_n \frac{F/M}{\omega_n^2 - \omega^2}\right|$$ not $$X(\omega) = \sum_n \frac{F/M}{\left|\omega_n^2 - \omega^2\right|}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contribution to $X$ from each mode &lt;em&gt;changes its phase&lt;/em&gt; by 180 degrees, and therefore &lt;em&gt;changes sign,&lt;/em&gt; when the frequency passes through the corresponding $\omega_n$. The antiresonances occur when the contributions from different modes cancel out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, you will probably see the antiresonances better on a plot if you use a log scale for $X$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT following comments by the OP:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to decompose $F$ into its modal components and apply them to the corresponding terms in the sum. If $$M\ddot x + K x = F.$$ Pre-multiply by the eigenvectors $\Phi$: $$\Phi^T M\ddot x + \Phi^TK x = \Phi^T F$$&#xA;But you know $\ddot x = -\omega^2 x$, so $$(-\omega^2 \Phi^T M  + \Phi^TK) x = \Phi^T F$$ Also $\Phi^T M \Phi = I$ for mass-normalized eigenvectors, and $\Phi^T K \Phi = \text{diag}\,(\omega_n^2)$, so inserting $\Phi \Phi^{-1}$ into the left hand side gives $$(-\omega^2  + \text{diag}\,(\omega_n^2)) \Phi^{-1}X = \Phi^T F.$$ Decomposing $X$ into its modal components gives $$X = \sum_n \Phi_n\xi_n,$$ so you end up with the decoupled equations $$(-\omega^2  + \omega_n^2) \xi_n = \Phi_n^T F$$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, if $\mathbf{f}$ is the vector applied forces, the $F$ in each term of your sum should be $\Phi_n^T \mathbf{f}$ where $\Phi_n$ is the $n$th column of the eigenvector matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you need to recombine the $\xi_n$ to get $X$, and &lt;em&gt;then&lt;/em&gt; take its absolute value.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-08-11T01:32:32.477" LastActivityDate="2017-08-11T01:32:32.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16716" PostTypeId="1" AcceptedAnswerId="16980" CreationDate="2017-08-11T06:46:34.920" Score="2" ViewCount="573" Body="&lt;p&gt;i'm trying to apply free body diagram to simplified excavator consept. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/vwTuc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vwTuc.jpg&quot; alt=&quot;Excavator Consept&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see, earth drill causes moment on  the arm. If we know the &quot;L&quot;, we can find F = M / L. But let's say ground friction is very very high. The F force cant move the excavator. Then the arm is going to try take excavator down. Because the arm comes from upper not the ground level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem is how can i find the force and its moment arm which causes taking down?&lt;/p&gt;&#xA;" OwnerUserId="12411" LastActivityDate="2017-09-07T00:32:48.823" Title="Effect of Moment on Excavator arm" Tags="|structural-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16717" PostTypeId="2" ParentId="16081" CreationDate="2017-08-11T10:45:21.083" Score="0" Body="&lt;p&gt;During a gas free expansion in an adiabatic chamber, experimentally (according to Joule's experiment) no change in temperature is observed; then, if the gas can be considered ideal, its internal energy $U$ does not change during the transformation.&lt;br&gt;&#xA;Let us now state the First Principle of Thermodynamics in its differential form:&#xA;$$dU=dQ+dW$$&#xA;We can make the following assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;we have just said that $U$ does not change, then $dU=0$;&lt;/li&gt;&#xA;&lt;li&gt;remember that the chamber is adiabatic, then $dQ=0$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Eventually, what we obtain is&#xA;$$dW=0$$&#xA;so no work is performed within the system.&lt;br&gt;&#xA;&lt;em&gt;Q.E.D.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please notice that I have not used the definition $dW=PdV$.&lt;br&gt;&#xA;That is because free expansion is an &lt;strong&gt;irreversible&lt;/strong&gt; transformation (to perform the reverse process, i.e. compression, a non-null work is required) and elementary work cannot be expressed as $PdV$, since this definition is valid only for reversible work.&lt;/p&gt;&#xA;" OwnerUserId="2541" LastEditorUserId="2541" LastEditDate="2017-08-11T10:51:08.963" LastActivityDate="2017-08-11T10:51:08.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16718" PostTypeId="2" ParentId="16510" CreationDate="2017-08-11T11:10:36.020" Score="1" Body="&lt;p&gt;First, I'm assuming that by &quot;moving window&quot;, you mean breaking the signal up into smaller chunks and filtering each chunk separately.  In the signal processing word, &quot;moving window&quot; would often mean an FIR filter (e.g. fir1() in matlab), but because you explicitly said you are using butterworth filter, which is an IIR filter, I assume that is not what you are talking about.  If I am wrong, please edit the question to clarify.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, 4 hours at 1000 S/s would occupy about 100 Mb of ram.  That's big, but not outlandish.  Assuming you have a 64 bit processor and a decent amount of RAM (2 GB or more), there is no reason to break this up into smaller chunks.  I filter vectors of 100 MB in one single pass (i.e. without breaking them up) all the time. Any computer built within the past 3 or 4 years should be able to handle this just fine.  10 years ago the answer would have been different.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-08-11T11:10:36.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16719" PostTypeId="1" AcceptedAnswerId="16925" CreationDate="2017-08-11T11:59:50.277" Score="0" ViewCount="179" Body="&lt;p&gt;I've recently learned about free-flow intersections. In particular, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Diverging_diamond_interchange#Double_Crossover_Merging_Interchange&quot; rel=&quot;nofollow noreferrer&quot;&gt;DCMI&lt;/a&gt;, and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stack_interchange&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stack&lt;/a&gt; interchanges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the disadvantages of using them, compared to the &quot;normal&quot; signaled intersections? do they cost significantly more?&lt;/p&gt;&#xA;" OwnerUserId="12497" LastEditorUserId="12497" LastEditDate="2017-08-11T12:59:27.477" LastActivityDate="2017-08-31T09:48:52.220" Title="What are the disadvantages of free-flow intersections?" Tags="|highway-engineering|traffic-intersections|traffic-light|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16720" PostTypeId="1" AcceptedAnswerId="16987" CreationDate="2017-08-11T13:47:19.647" Score="1" ViewCount="2467" Body="&lt;p&gt;I am using Dynamic Implicit to model a rolling problem. But I am not really sure what to choose as Initial increment. The default suggestion by Abaqus is the step time. &lt;/p&gt;&#xA;" OwnerUserId="3345" LastActivityDate="2018-05-04T03:18:00.803" Title="How to decide initial size increment in Abaqus Dynamic Implicit?" Tags="|finite-element-method|abaqus|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16721" PostTypeId="1" AcceptedAnswerId="16726" CreationDate="2017-08-11T13:47:40.923" Score="1" ViewCount="941" Body="&lt;p&gt;Regarding Solidworks 2014 (SP2)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an assembly I have three identical copies of a single transparent acrylic component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How may I give them different colours?  When I change the colour of the first one to green (material...edit) they all become green even though only the first one is selected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly this change is getting pushed all the way back to the .sldprt file, but I only want this to happen at the assembly:component instance level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to do this using a design table in the .sldprt file but I cannot figure out how to configure the columns.  Also I couldn't persuade the appearance manager to understand what I want as the &quot;copy/paste Appearance&quot; features seem to overwrite the Appearance rather than adding new ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: @SBarry's suggestion worked a treat...&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Iqhzi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iqhzi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastEditorDisplayName="user6335" LastEditDate="2017-08-14T07:48:48.640" LastActivityDate="2017-08-14T07:48:48.640" Title="How to have multiple copies of the same component all with different colours in Solidworks?" Tags="|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16722" PostTypeId="1" AcceptedAnswerId="16724" CreationDate="2017-08-11T15:38:49.983" Score="0" ViewCount="26" Body="&lt;p&gt;I have taken old speakers I found lying around and connected them to a 3.5mm auxiliary connection, however when I plug this into my laptop/phone the sound comes through just really quiet and sometimes only the bass. How can I get the speakers working more efficiently/louder? I'm assuming the Auxiliary connection that I took from old broken earphones is causing the problem, can I get something better?&lt;/p&gt;&#xA;" OwnerUserId="12500" LastActivityDate="2017-08-11T15:53:04.030" Title="Wiring old speakers" Tags="|materials|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16724" PostTypeId="2" ParentId="16722" CreationDate="2017-08-11T15:53:04.030" Score="0" Body="&lt;p&gt;You will need an amplifier of some sort : stereo and after that the choice is huge.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-11T15:53:04.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16726" PostTypeId="2" ParentId="16721" CreationDate="2017-08-11T18:08:33.220" Score="1" Body="&lt;p&gt;Couple of ways you could do it, but this method is pretty easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Go to the DisplayManager (Appearances) tab - 4th tab in the Feature Manager area (left side of the screen).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Search for the appearance that matches your plastic material.  It might say &quot;polycarbonate plastic&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Right click the appearance and click &quot;Copy Appearance&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Go back to the Feature Manager (your tree listing all the parts)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Select the individual components you want to be a specific color from your feature tree by CTRL + clicking the name of each component&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) Right click your selection and click &quot;Paste Appearance&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;7) Go back to the Appearances tab, right click &amp;amp; edit the new appearance to be the color you want.  It should maintain transparency &amp;amp; general material appearance&lt;/p&gt;&#xA;" OwnerUserId="4708" LastActivityDate="2017-08-11T18:08:33.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16727" PostTypeId="2" ParentId="16704" CreationDate="2017-08-11T19:24:40.050" Score="0" Body="&lt;p&gt;If I were building this I would do three things to make the assembly secure, you may think it is over-kill but I will throw this out to you anyway. &#xA;Where you have the threaded hole to receive the locking screw, I would have a flangemount locknut threaded insert instead.&#xA;Instead of a plain screw I would use a set screw with a pointed tip that would dig in to the aluminum to give you more holding power.&#xA;And last but not least , have several of these set screw assemblies spaced along the length of the tube, don't rely on just one screw to hold 100 lbs!&lt;/p&gt;&#xA;" OwnerUserId="628" LastEditorUserId="628" LastEditDate="2017-08-17T19:12:32.750" LastActivityDate="2017-08-17T19:12:32.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16729" PostTypeId="1" CreationDate="2017-08-11T23:03:08.097" Score="2" ViewCount="71" Body="&lt;p&gt;I have inherited a set of templates for creating drawings for solar panel  projects.  Each set is about 12 pages and only 2-4 pages are different between projects.  Several pages are actually just embedded PDF cutsheets of the equipment used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The person that created the template has everything in one drawing - all of the PDF,  several different versions of one line and three line drawings that might be used,  blocks of language,  etc are all scattered about the model space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are layouts for each page that have a view port zoomed into the content for that page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is this a standard practice?  I feel like it's a disorganized mess and am looking for a better way to do it.  My thought was to split things out into individual drawings,  and used sheet sets or a template to reference everything.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is using a combination of sheet sets and a template for new projects the way to go here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I have an individual drawing for each PDF?  Or maybe a cut sheet drawing that contains all of them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking for some guidance on how this is typically done.  I'm very new to autocad and am not sure how a set of drawings is organized.&lt;/p&gt;&#xA;" OwnerUserId="12507" LastEditorDisplayName="user16" LastEditDate="2017-11-14T12:55:58.943" LastActivityDate="2017-11-14T12:55:58.943" Title="Autocad - multipage drawing with all content in the same model space" Tags="|cad|autocad|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16732" PostTypeId="1" CreationDate="2017-08-12T10:18:50.217" Score="0" ViewCount="470" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JekOK.jpg&quot; alt=&quot;Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spec:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aim: Drain a 50 litre tank of oil using a positive displacement pump (see Diagram), so that it drains quicker than gravity alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pump should not be starved at any point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flow rate from the 50 litre tank should be at least twice the flow rate from the 5 litre tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current setup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 5 litre tank is present because it's used to fill the 50 litre tank (in a different configuration) and also to ensure that the pump is never starved. Once the 50 litre tank is fully drained, the pump should be able to draw oil from the 5 litre tank and then the pump is switched off by a minimum level sensor in the 5 litre tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All hoses are 1&quot; inner diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I'm stuck:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my current setup, will the pump draw the same volume from both tanks at the same rate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I do anything to ensure that the 50 litre tank drains at a much higher rate than the 5 litre tank so I don't have to constantly top up the 5 litre tank?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I reduce the diameter of the hose from the 5 litre tank or use a valve to restrict the flow, can I achieve a greater flow from the 50 litre tank compared to the 5 litre tank?&lt;/p&gt;&#xA;" OwnerUserId="12499" LastEditorUserId="12499" LastEditDate="2017-08-13T18:46:19.730" LastActivityDate="2017-09-13T05:53:21.160" Title="How to drain large tank using pump without starving pump?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16733" PostTypeId="1" AcceptedAnswerId="16783" CreationDate="2017-08-12T11:07:53.443" Score="3" ViewCount="1245" Body="&lt;p&gt;I have a pretty basic question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to install a timing pulley on a ball screw shaft (6mm):&#xA;&lt;a href=&quot;http://uk.rs-online.com/web/p/products/2816419/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://uk.rs-online.com/web/p/products/2816419/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The processed end (the pointy side) of the shaft is secured by a support mount, I need to place a timing pulley on the other side. I'm not sure how to go about this. E.g. this pulley: &lt;a href=&quot;http://uk.rs-online.com/web/p/timing-belt-pulleys/0744508/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://uk.rs-online.com/web/p/timing-belt-pulleys/0744508/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would a mounting hub or a pulley with a holding screw work? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="11266" LastActivityDate="2021-02-10T04:11:47.220" Title="Mounting a timing pulley on a ball screw shaft" Tags="|mechanical-engineering|gears|pulleys|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16734" PostTypeId="1" CreationDate="2017-08-12T13:11:44.617" Score="5" ViewCount="1466" Body="&lt;p&gt;My research group recently purchased a 40W CO2 laser cutter. We work in the biosensing field and some of the materials we wish to cut are very thin nano/micro fibers that will be cut into relatively small pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When cutting the material we hope to keep the samples as clean and uncontaminated as possible. So the current grid is not the best option as it will most likely get contaminated by other materials we cut and also the small samples will most likely fall through the grid anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest an alternative solid material that can be used instead of the  grid? Ideally the material should not be affected by laser cut at all or at least not give off any smoke, particles or other contamination. Also the material should not reflect the laser back towards the sample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus far I have tried two options on an external laser cutter before we got our own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st) Ceramic tile as I was told that it would not reflect any of the laser underneath the samples and also not get cut by the laser. BUT after testing it was found that a whole lot of molten ceramic was deposited on the samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd) Stainless steel metal sheet that was orbital sanded to reduce reflection. This was a little better than the ceramic but I think there was some reflection of the laser under the laser that reduced the quality of the cut. I only assume this because there was a little burnt material left on the steel after cutting the material.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking possibly glass above the grid but I have heard that even glass can be etched and so if the sample material retains heat this may cause the glass to etch or shatter? We are yet to setup and switch on our new laser cutter and some experimentation and optimization is needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any advice!!&lt;/p&gt;&#xA;" OwnerUserId="11187" OwnerDisplayName="Nick Law" LastActivityDate="2022-07-15T19:25:25.567" Title="Laser Cutting backing material instead of grid?" Tags="|cutting|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16737" PostTypeId="1" AcceptedAnswerId="16763" CreationDate="2017-08-13T08:01:29.020" Score="0" ViewCount="93" Body="&lt;p&gt;I'm getting more serious about a combat robot design I've been contemplating for several years now (BattleBots, Robot Wars, etc) and find myself overwhelmed by the sheer variegation of ASTM and AISI grades and alloys.  I hope that by detailing the application, those familiar with the nomenclature can steer me in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall design is a full-body spinner - somewhere around 0.75m in diameter, at least 2500 RPM.  (Approximately 205 MPH weapon speed, though this is a minimum target.)  The circumference is lined with five or six solid-mount blades approximating the size and shape of forestry mauls, though slightly thinner - while primarily relying upon inertia and brute force, I would like to retain some degree of sharpness/cutting.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Structural integrity is the key attribute, remaining intact (if deformed to some degree) in spite of high-speed impacts with fixed hazards, opponents, and/or similar spinning weaponry.  Any significant separation would be of serious concern owing to the high-speed rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blades will be mounted to the outside of the body with high-strength bolts, anchoring roughly half the length and 60% the mass of each individual blade.  (I have no intention of ever welding these pieces as even a fractured blade would be swiftly replaced.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Withstanding impacts and remaining affixed with minimal distortion is highly desirable.  Retaining a sharp edge would be nice, but this is mostly wishful thinking - for the sake of other traits, some sacrifice must be made.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What little I understand of this sort of metallurgy has me looking at ASTM-S1, which seems to be recommended for high-speed impacts and sudden shock-loading.  (Also assuming those terms mean what I think they do.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any guidance/recommendations in this department would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="12517" LastActivityDate="2017-08-15T00:55:54.377" Title="Which types of steel would be most appropriate for blunt cutting weaponry on a combat robot?" Tags="|materials|steel|robotics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16738" PostTypeId="1" AcceptedAnswerId="16746" CreationDate="2017-08-13T09:06:22.510" Score="1" ViewCount="67" Body="&lt;p&gt;The wikipedia page for groundworks (earthworks) does mention contour trenching and cut &amp;amp; fill but not a word about bring tunnels. Is a bored tunnel part of groundworks discipline/field? If not what discipline does it belong to.&lt;/p&gt;&#xA;" OwnerUserId="11826" LastActivityDate="2017-08-14T09:10:10.520" Title="Is tunnel boring considered groundworks?" Tags="|civil-engineering|tunnels|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16740" PostTypeId="2" ParentId="16732" CreationDate="2017-08-13T19:01:07.563" Score="0" Body="&lt;p&gt;A suggestion I have is that you put the 5 litre tank, or a second 5 litre tank, in the pipe between the 50 litre tank and the pump like so:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dnWoE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dnWoE.png&quot; alt=&quot;extra tank&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as soon as the 50 litre tank is empty, then the level starts to drop in the 5 litre tank and that cabe sensed to switch off the pump. The 50 litre tank is feeding via the upper pipe and the pump drawing from the lower pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you have the vertical space is a different question....&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-13T19:01:07.563" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16741" PostTypeId="2" ParentId="16737" CreationDate="2017-08-13T21:33:46.947" Score="0" Body="&lt;p&gt;S-1 is not an ASTM designation , likely AISI tool steel designation  ( likely it is included in some ASTM spec for bar, ). First , what do the other guys use, no use re-inventing the wheel ? Second , is cost a concern? Third, is heat-treating a concern? I would stay away from tool steels ; &quot;all &quot; tool steels primarily hold an edge. So grades like S-1, S-2, H-11, H-12 etc, are the toughest among steels that are intended to hold an edge. And you will pay a premium for &quot;tool steel&quot;.  Most of these have about 0.5 % C; C generally lowers toughness. I don't think any contain nickel ; the first alloy you add for toughness. So, I think low alloy steels with medium carbon would give maximum toughness with some hardness. ( AISI/ SAE) 4330 &gt; 8630 &gt; 4130 ; These have 0.3% carbon and are listed in descending Ni content. When quenched and tempered a good one may stop a Charpy hammer ( 240 ft.lb) ( standard toughness test). Also things like 4320 with a carburized case could be used.   &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-08-13T21:33:46.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16742" PostTypeId="2" ParentId="11519" CreationDate="2017-08-13T22:02:58.233" Score="0" Body="&lt;p&gt;Mechanical connection is the first choice. I would have said brazing is too hot for any aluminum alloy , but checking a reference I was surprised.  There is a loop hole; somehow, using aluminum /silicon fillers ( no copper) can still be caused &quot;brazing&quot;. However the steel must be coated with aluminum or copper first. So , mechanical or solder are the main options.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-08-13T22:02:58.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16744" PostTypeId="1" CreationDate="2017-08-14T07:35:41.940" Score="2" ViewCount="1535" Body="&lt;p&gt;While calculating hoop's stress we consider projected area. Infact in bearing stress from pins in a truss something similar is applicable.Can you explain why we are taking something like this instead of the curved area?&lt;/p&gt;&#xA;" OwnerUserId="12431" LastEditorUserId="12597" LastEditDate="2017-08-19T12:44:50.213" LastActivityDate="2017-08-19T12:44:50.213" Title="Projected area in hoop's stress structural mechanics" Tags="|mechanical-engineering|structural-engineering|civil-engineering|structures|applied-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16745" PostTypeId="2" ParentId="16744" CreationDate="2017-08-14T08:32:32.947" Score="0" Body="&lt;p&gt;Because some of the curved surface is not being loaded.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-14T08:32:32.947" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16746" PostTypeId="2" ParentId="16738" CreationDate="2017-08-14T09:10:10.520" Score="1" Body="&lt;p&gt;In my experience &lt;strong&gt;groundworks&lt;/strong&gt; refers to the site activity of shallow work in the ground, such as trenches (for foundations or drainage) and slab foundations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tunnel construction comes under the term &lt;strong&gt;tunnelling&lt;/strong&gt;. Tunnelling is a specialist activity: you wouldn't want a general groundworks contractor to try boring tunnels, and you wouldn't want a tunnelling contractor to dig shallow foundations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the design side they are all part of the discipline of &lt;strong&gt;geotechnical engineering&lt;/strong&gt;. Again, tunnelling is specialised on the design side too, and will be done be specialised geotechnical engineers. Foundation design may be carried out by structural engineers or geotechnical engineers, mainly depending on how complicated it is. Small residential building foundations are designed by rules of thumb based on the soil type; skyscraper foundations take much bigger loads and rules of thumb are not sufficient.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Research&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It's surprisingly difficult to find a definition for &lt;strong&gt;Groundworks&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are plenty for &quot;groundwork&quot; as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the foundation, basis, preliminary or basic works&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;a href=&quot;https://en.oxforddictionaries.com/definition/groundwork&quot; rel=&quot;nofollow noreferrer&quot;&gt;oxford dictionaries&lt;/a&gt;, &lt;a href=&quot;http://www.dictionary.com/browse/groundwork&quot; rel=&quot;nofollow noreferrer&quot;&gt;dictionary.com&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of the usage of &quot;groundwork&quot; seems to be metaphorical, comparing &quot;preliminary work&quot; for any type of project to laying the foundation for constructing a building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Groundworker&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; does have a page for &quot;groundworker&quot; as somebody who:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;clears the site, lays a foundation, installs drainage and other pipework, and may build roads if necessary&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A typical &lt;a href=&quot;http://www.malcolmgroup.co.uk/construction/civil-engineering-groundworks/&quot; rel=&quot;nofollow noreferrer&quot;&gt;website&lt;/a&gt; for a &quot;groundworks&quot; contractor talks about:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Re-profiling the earthworks to required design levels&lt;/li&gt;&#xA;  &lt;li&gt;Drainage and roads&lt;/li&gt;&#xA;  &lt;li&gt;Foundations and buildings&lt;/li&gt;&#xA;  &lt;li&gt;Ground floor slab&lt;/li&gt;&#xA;  &lt;li&gt;SurfacingFinishing touches&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems consistent with the definition for &quot;groundworker&quot; - someone who does shallow depth work in the ground, typically for buildings.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Geotechnical engineering&lt;/strong&gt; is defined as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Geotechnical engineering is the branch of engineering concerned with the analysis, design and construction of foundations, slopes, retaining structures, embankments, tunnels, levees, wharves, landfills and other systems that are made of or are supported by soil or rock.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;a href=&quot;http://www.ejge.com/Definition.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;ejge.com&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-08-14T09:10:10.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16747" PostTypeId="1" CreationDate="2017-08-14T09:36:20.327" Score="2" ViewCount="42" Body="&lt;p&gt;These are the discretized drift diffusion equations as taken from the book &quot;Analysis and Simulation of Semiconductor Devices&quot; by Siegfried Selberherr. The electron continuity equation is:&#xA;$$\begin{align}&#xA;&amp;amp;((\Delta_{j-1}^y+\Delta_j^y)/2\Delta_i^x).D_{i+1/2,j}^n.B((\varphi_{i+1,j}-\varphi_{i,j})/V_T).n_{i+1,j} \\&#xA;+&amp;amp;((\Delta_{j-1}^y+\Delta_j^y)/2\Delta_{i-1}^x).D_{i-1/2,j}^n.B((\varphi_{i-1,j}-\varphi_{i,j})/V_T).n_{i-1,j} \\&#xA;-&amp;amp;[(((\Delta_{j-1}^y+\Delta_j^y)/2\Delta_i^x).D_{i+1/2,j}^n.B((\varphi_{i,j}-\varphi_{i+1,j})/V_T)) \\&#xA;+&amp;amp;(((\Delta_{j-1}^y+\Delta_j^y)/2\Delta_{i-1}).D_{i-1/2,j}^n.B((\varphi_{i,j}-\varphi_{i-1,j})/V_T)) \\&#xA;+&amp;amp;(((\Delta_{i-1}^x+\Delta_i^x)/2\Delta_j^y).D_{i,j+1/2}^n.B((\varphi_{i,j}-\varphi_{i,j+1})/V_T)) \\&#xA;+&amp;amp;(((\Delta_{i-1}^x+\Delta_i^x)/2\Delta_{j-1}^y).D_{i,j-1/2}^n.B((\varphi_{i,j}-\varphi_{i,j-1})/V_T))].n_{i,j} \\&#xA;+&amp;amp;((\Delta_{i-1}^x+\Delta_i^x)/2\Delta_j^y).D_{i,j+1/2}^n.B((\varphi_{i,j+1}-\varphi_{i,j})/V_T).n_{i,j+1} \\&#xA;+&amp;amp;((\Delta_{i-1}^x+\Delta_i^x)/2\Delta_{j-1}^y).D_{i,j-1/2}^n.B((\varphi_{i,j-1}-\varphi_{i,j})/V_T).n_{i,j-1} \\&#xA;=&amp;amp;R_{i,j}^{EFFECTIVE}.((\Delta_{i-1}^x+\Delta_i^x)/2).((\Delta_{j-1}^y+\Delta_j^y)/2)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hole continuity equation is:&#xA;$$\begin{align}&#xA;&amp;amp;((\Delta_{j-1}^y+\Delta_j^y)/2\Delta_i^x).D_{i+1/2,j}^p.B((\varphi_{i,j}-\varphi_{i+1,j})/V_T).p_{i+1,j} \\&#xA;+&amp;amp;((\Delta_{j-1}^y+\Delta_j^y)/2\Delta_{i-1}^x).D_{i-1/2,j}^p.B((\varphi_{i,j}-\varphi_{i-1,j})/V_T).p_{i-1,j} \\&#xA;-&amp;amp;[(((\Delta_{j-1}^y+\Delta_j^y)/2\Delta_i^x).D_{i+1/2,j}^p.B((\varphi_{i+1,j}-\varphi_{i,j})/V_T)) \\&#xA;+&amp;amp;(((\Delta_{j-1}^y+\Delta_j^y)/2\Delta_{i-1}^x).D_{i-1/2,j}^p.B((\varphi_{i-1,j}-\varphi_{i,j})/V_T)) \\&#xA;+&amp;amp;(((\Delta_{i-1}^x+\Delta_i^x)/2\Delta_j^y).D_{i,j+1/2}^p.B((\varphi_{i,j+1}-\varphi_{i,j})/V_T)) \\&#xA;+&amp;amp;(((\Delta_{i-1}^x+\Delta_i^x)/2\Delta_{j-1}^y).D_{i,j-1/2}^p.B((\varphi_{i,j-1}-\varphi_{i,j})/V_T))].p_{i,j} \\&#xA;+&amp;amp;((\Delta_{i-1}^x+\Delta_i^x)/2\Delta_j^y).D_{i,j+1/2}^p.B((\varphi_{i,j}-\varphi_{i,j+1})/V_T).p_{i,j+1} \\&#xA;+&amp;amp;((\Delta_{i-1}^x+\Delta_i^x)/2\Delta_{j-1}^y).D_{i,j-1/2}^p.B((\varphi_{i,j}-\varphi_{i,j-1})/V_T).p_{i,j-1} \\&#xA;=&amp;amp;R_{i,j}^{EFFECTIVE}.((\Delta_{i-1}^x+\Delta_i^x)/2).((\Delta_{j-1}^y+\Delta_j^y)/2&#xA;\end{align}$$&#xA;where $\Delta_i$ and $\Delta_j$ are the mesh spacing for finite difference grid along X and Y directions. $D_n$,$D_p$ are electron and hole diffusion coefficients.$B(x) = x/(e^x-1)$ is Bernoulli's function and $V_T$ is the thermal voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, $$R^{EFFECTIVE} = R^{SRH} + R^{AU} + R^{OPT} + R^{II}$$ where&#xA;$R^{EFFECTIVE}$ is the effective generation/recombination term. $R^{SRH}$, $R^{AU}$, $R^{OPT}$ and $R^{II}$ are SRH, Auger, Radiative and Impact Ionization Recombination terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am applying the coupled Newton's iterative scheme to solve the above equations where I need to differentiate the equations. When differentiating with respect to potential '$\varphi$', I noticed that $\varphi$ appears in three places namely,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;High field Electron/Hole Mobility, $\mu_{n,p}^{high}=\mu_{n,p}^{low}⁄ [1 + (\mu_{n,p}^{LICN}.E_{n,p}/v_{n,p}^{sat})^{\beta_{n,p}}]^{1/\beta_{n,p}}$ and $D_{n,p} = \mu_{n,p}.V_T$;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bernoulli's function, $B(x)$ where 'x' here means difference between potentials at neighboring grid points and&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Impact Ionization $R^{II} = -\alpha_n.(|J_{i,j}^n|/q) - \alpha_p.(|J_{i,j}^p|/q)$, $\alpha_n = a_n.e^{-b_n/|E_{i,j}|}$, $\alpha_p = a_p.e^{-b_p/|E_{i,j}|}$ where $J^{n,p}$ and $E$ represent electron/hole currents and electric field respectively.&#xA;Current terms are made of potential terms as in,&#xA;$J_{i+1/2,j}^{nx}=D_{i+1/2,j}^n.[B(\frac{\varphi_{i,j}-\varphi_{i+1,j}}{V_T}).n_{i,j} - B(\frac{\varphi_{i+1,j}-\varphi_{i,j}}{V_T}).n_{i+1,j}]/\Delta_i^x  \rightarrow  (31)$, $J_{i,j+1/2}^{ny}=D_{i,j+1/2}^n.[B(\frac{\varphi_{i,j}-\varphi_{i,j+1}}{V_T}).n_{i,j} - B(\frac{\varphi_{i,j+1}-\varphi_{i,j}}{V_T}).n_{i,j+1}]/\Delta_j^y  \rightarrow  (32)$, $J_{i+1/2,j}^{px}=D_{i+1/2,j}^p.[B(\frac{\varphi_{i,j}-\varphi_{i+1,j}}{V_T}).p_{i+1,j} - B(\frac{\varphi_{i+1,j}-\varphi_{i,j}}{V_T}).p_{i,j}]/\Delta_i^x  \rightarrow  (33)$, $J_{i,j+1/2}^{py}=D_{i,j+1/2}^p.[B(\frac{\varphi_{i,j}-\varphi_{i+1,j}}{V_T}).p_{i,j+1} - B(\frac{\varphi_{i+1,j}-\varphi_{i,j}}{V_T}).p_{i,j}]/\Delta_j^y  \rightarrow  (34)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, basically I need to differentiate all of these to find the Jacobi with respect to $\varphi$. But, the problem arises when the difference between potentials become zero. This means that the entire electron/hole equation becomes a constant when the potential difference is zero. Hence, Jacobi is also zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would mean that the final Jacobian matrix would have entire columns and rows made out of zeros thus making the matrix singular and it is not possible to solve a singular matrix.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this the correct approach to solving using the Newton's method. What should I do when the potential difference is zero ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12525" LastEditorUserId="1832" LastEditDate="2017-08-15T14:38:02.423" LastActivityDate="2017-08-15T14:38:02.423" Title="How to calculate the Jacobi of electron/hole continuity equations for semiconductors with respect to potential when applying Newton's method?" Tags="|electrical-engineering|simulation|cad|diffusion|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16749" PostTypeId="2" ParentId="475" CreationDate="2017-08-14T12:21:07.213" Score="2" Body="&lt;p&gt;Re-iterating some of above as manufacturer I can advise that Weathering Steel TCBs are generally available in M24 diameter only, with the following lengths: 70, 80, 90, 100, 105, 110, 120, 125, 130, 140, 150, 160, 170, 180, 190, 200, 220, and 240&amp;nbsp;mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CSK &amp;amp; alternative diameters can be manufactured bespoke - dependant on quantity. &#xA;Galv / Zinc Flake  / GK Coatings in WS structures are susceptible to bi-metallic corrosion. &lt;/p&gt;&#xA;" OwnerUserId="12528" LastEditorUserId="1832" LastEditDate="2017-08-14T15:06:11.313" LastActivityDate="2017-08-14T15:06:11.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16750" PostTypeId="2" ParentId="87" CreationDate="2017-08-14T13:52:37.757" Score="0" Body="&lt;p&gt;I do have a commercial interest however article below is relevant to the question asked.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Greenkote treated bolts to be used to repair Leadenhall Building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;London’s Leadenhall Building, or as it is more popularly known, the “Cheesegrater” hit the headlines again earlier this year when, not long after completion, reports of three of the giant studs securing the exoskeleton suddenly snapped and fell, in one case from the 15th floor, before landing on the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the studs used in the construction of the 738 ft. tower are reported as being approximately 72mm diameter. The distinctive wedge-shaped 52 floor tower is 738ft tall and is close to both the Lloyds Building and the Gherkin. It opened in July 2014 and was designed by Rogers Stirk Harbour + Partners&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analysis of the problem revealed the cause as hydrogen embrittlement, which usually occurs when hydrogen atoms diffuse into the steel and weaken the overall molecular structure, making it brittle and likely to snap suddenly. Often blamed as a side effect of galvanising, a process using acids, in this case the studs had not been treated, so the fault was probably down to a manufacturing error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UK specialists Tension Control Bolts, whose products are being used in the assembly of the new safe containment shelter to encase the Chernobyl reactor in the Ukraine, were commissioned to identify a suitable solution and specified a Greenkote treated fasteners to replace the damaged and missing studs. The work is expected to be finished by the end of 2016.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Greenkote coating process offers a number of critical advantages, depending upon the application. Using no acids, hydrogen embrittlement is avoided, whilst the process offers high corrosion resistance, excellent paint adhesion and exceptional levels of coating consistency. When coating threads, neither internal nor external surfaces need to be reworked. A small number of companies globally, including TC Bolts, are licensed to use the coating process.&#xA;Mark Gore, CEO of Greenkote commented, “Greenkote’s thermal diffusion process offers exceptional corrosion protection and critically in this instance, protection against hydrogen embrittlement, as the coating does not rely upon aggressive acids at any stage. We are delighted that our system has been chosen to help repair this iconic new building so helping to ensure its success as a significant London landmark.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For avoidance of ambiguity TCB Ltd did not manufacture or supply the sub-standard steel studs; rather we Greenkoted the replacement studs. &lt;/p&gt;&#xA;" OwnerUserId="12528" LastEditorUserId="1832" LastEditDate="2017-08-14T14:59:06.053" LastActivityDate="2017-08-14T14:59:06.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16751" PostTypeId="1" AcceptedAnswerId="16755" CreationDate="2017-08-14T13:53:30.890" Score="3" ViewCount="105" Body="&lt;p&gt;I want to build drawers for my kitchen. On Youtube I found this Channel, where a guy does woodworking. In a video about making kitchen drawers, he says that the bottom surface of the drawer will be stronger if the joint connecting the bottom to the sides is positioned lower. &lt;a href=&quot;https://youtu.be/wLMQ_6Y29pY?t=5m15s&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/wLMQ_6Y29pY?t=5m15s&lt;/a&gt; (5:15 to 6:03).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, it should make pretty much no difference. Am I wrong? &lt;/p&gt;&#xA;" OwnerUserId="12532" LastActivityDate="2017-08-14T16:21:48.977" Title="Which drawer design can hold more weight?" Tags="|structural-engineering|stresses|wood|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16752" PostTypeId="2" ParentId="6724" CreationDate="2017-08-14T14:10:06.203" Score="0" Body="&lt;p&gt;depends on the application but a TCB Stud may be a possibility - this has a left hand thread on the end opposite the spline and a right hand thread on the end nearest the spline.  The left hand thread is tightened by the installation tool into a left-handed threaded nut and then the right hand threaded nut is tightened against the washer against the steel. &lt;/p&gt;&#xA;" OwnerUserId="12528" LastActivityDate="2017-08-14T14:10:06.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16753" PostTypeId="1" CreationDate="2017-08-14T15:36:12.127" Score="2" ViewCount="191" Body="&lt;p&gt;There is discrepancy between 2 labs regarding what the normal pump back pressure should be given the following dimensions of an LC system flowing at a constant rate.  I have searched for equations to determine this and check if the software read back is correct but I have been unsuccessful at determining an appropriate equation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;System dimensions and set up are as follows: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;1.254m of PEEK tubing going from the pump to the MS inlet with a single 2uM PEEK filter attached in-line (no column).  &lt;/li&gt;&#xA;&lt;li&gt;The inner diameter of the PEEK tubing is 127 um.&lt;/li&gt;&#xA;&lt;li&gt;Flow rate 1mL/min&lt;/li&gt;&#xA;&lt;li&gt;Solvent = 84% Acetonitrile in water with 0.1% formic acid.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In this instance, assume an open ended system in which the end of the tubing is disconnected from the MS inlet and collected into beaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an equation to determine what the pressure should be?&lt;/p&gt;&#xA;" OwnerUserId="12533" LastEditorUserId="1832" LastEditDate="2017-08-14T16:28:38.363" LastActivityDate="2017-08-14T16:28:38.363" Title="What is expected back pressure at a constant flow rate?" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16754" PostTypeId="1" CreationDate="2017-08-14T16:20:40.097" Score="-2" ViewCount="1286" Body="&lt;p&gt;Is the moment of force taken in a clockwise and anticlockwise fashion? Why then do we curl our fingers and check it?Explain the equivalence or similarity between the two.&lt;/p&gt;&#xA;" OwnerUserId="12431" LastEditorUserId="12431" LastEditDate="2017-08-14T16:29:39.860" LastActivityDate="2017-08-15T00:53:33.680" Title="Moment of force or torque" Tags="|mechanical-engineering|mechanical-failure|moments|" AnswerCount="2" CommentCount="2" ClosedDate="2017-08-15T12:16:51.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="16755" PostTypeId="2" ParentId="16751" CreationDate="2017-08-14T16:21:48.977" Score="6" Body="&lt;p&gt;Using a tie-and-strut model makes it easier to explain the difference here. A tie-and-strut model is a simplification of the structure which describes where the structure is under tension, and where it is under compression. The model gets effectively simplified into a truss. Below we have two examples of the axial load on members of two different trusses with identical loading (positive values are tension, negative are compression):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KGh4y.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KGh4y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the values, you'll see that the results are mirror images of each other: same values, but different signs. Importantly, notice that the first truss has a negative (compressive) force of 634.9 between the supports and the upper chord, while the second one has a positive (tensile) force of 634.7 between the supports and the lower chord.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, looking at the video, we basically have two possible layouts, which are converted into the overlaid trusses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a7fLe.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a7fLe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/74YKI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/74YKI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing with the truss results shown above, it is clear that the first diagonal emanating from the support will suffer compression in the first format and tension in the second layout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what will resist the forces represented by that diagonal? The combined plywood layers, of course. However, that force needs to be transmitted from layer to layer. In the case of compression, that transfer is done by contact. However, in tension, that transfer occurs via the glue. If the glue were to fail, then the plywood layers will separate and the structure will progressively collapse:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IJSuY.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IJSuY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, with the other orientation, that force is compressive, so such a failure is impossible (other failure states will certainly happen before the &quot;compressive strut&quot; collapses).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-08-14T16:21:48.977" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16756" PostTypeId="1" CreationDate="2017-08-14T17:02:45.990" Score="1" ViewCount="1033" Body="&lt;p&gt;I'm making a pressurized bottle rocket out of a plastic bottle, vinegar, baking soda and a cork.  When the vinegar &amp;amp; baking soda react in the closed bottle, the pressure builds up but i want to measure how high this pressure is.  Ideally, i'd like to use a small, light weight gauge so that it does not affect the rocket's launch dynamics significantly.   I hope that i can also mount the instrument to the inside of the bottle so that i can measure the pressure entirely from within the bottle. Is there a small, lightweight instrument i can use to measure the pressure build-up from inside the bottle?  &lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2017-09-03T03:46:39.990" Title="How to measure the pressure inside a bottle rocket?" Tags="|fluid-mechanics|measurements|instrumentation|" AnswerCount="4" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16758" PostTypeId="2" ParentId="16756" CreationDate="2017-08-14T18:33:02.623" Score="3" Body="&lt;p&gt;Why not mount the pressure gauge to the cork itself,so that the gauge remains on the ground with the cork?  (i.e. gauge inserted in a hole in the cork or similar setup) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were hoping to measure pressure over time as the rocket rises, you can get a pretty decent estimate as follows.  This method does assume that nearly all of the gas is generated prior to launch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Measure the pressure prior to launch. Also measure the gas (air) volume in the container prior to launch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) record as best you can the time from launch until the fluid is fully ejected.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a more  or less linear expulsion rate, you know the gas volume has gone from its starting value to the full-container value in that time.  PV=nRT and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I agree that this is not a high-precision operation, but it's cheap and simple.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-08-14T18:33:02.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16759" PostTypeId="1" AcceptedAnswerId="16760" CreationDate="2017-08-14T19:40:03.243" Score="1" ViewCount="110" Body="&lt;p&gt;In order to identify &lt;a href=&quot;https://en.wikipedia.org/wiki/Steinhart%E2%80%93Hart_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Steinhart-Hart coefficients&lt;/a&gt; for a thermistor, three known temperatures may be used to find three resistances.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to choose these temperatures to calibrate from?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it more beneficial to use three temperatures within the typical operating temperature of the thermistor, or to take a broad range of temperatures (say 0, 50, and 100°C)?&lt;/p&gt;&#xA;" OwnerUserId="12536" LastActivityDate="2017-08-14T20:12:58.650" Title="Thermistor - Steinhart-Hart Coefficients" Tags="|electrical-engineering|sensors|measurements|temperature|calibration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16760" PostTypeId="2" ParentId="16759" CreationDate="2017-08-14T20:12:58.650" Score="2" Body="&lt;p&gt;It depends to some extent. Normally you'd assume you should look at the datasheet, see what it's operating limits are and then use the centre point for your 3 points. If the thermistor is not entirely linear however, you'll be off at certain temperatures across the range. Personally i'd find out what my expected limits are for the application and calibrate for those limits. I'd even take more than 3 points for that reason.&lt;/p&gt;&#xA;" OwnerUserId="12187" LastActivityDate="2017-08-14T20:12:58.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16761" PostTypeId="2" ParentId="87" CreationDate="2017-08-14T21:01:58.240" Score="1" Body="&lt;p&gt;Get a few books on the subject , eg, Hydrogen Effects in Metals, AIME, 1981 , and ISO 15156 ; if you want more than the misleading answers found on this forum. If you have a specific question , try Eng-Tips forum . The first thing you will find is the topic has a dozen names because it is so widespread. One thing you can eliminate is High Temperature Hydrogen Attack - wherein H dissolves into the steel and reacts with carbides to from methane causing cracking of the steel ; that has nothing to do with the subject topic( see API 941 ).   One small facet of hydrogen cracking seems to be your question : High strength fasteners, hardness above Rockwell C 22 / yield strength above 90,000 psi ,  are subject to failure when exposed to about any corrosion or cathodic protection. Use larger, lower strength fasteners.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-08-14T21:01:58.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16762" PostTypeId="2" ParentId="16754" CreationDate="2017-08-14T22:03:33.720" Score="3" Body="&lt;p&gt;A moment has a direction, which you can imagine when you look at the direction of the force and the axis you're calculating the moment from. Depending on which direction you're looking at it from, however, it can be seen as either clockwise or counter-clockwise. To remedy this confusion, we use the right-hand rule to identify whether a moment is positive or negative around an axis, instead of as clockwise or counter-clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A moment is positive when, curling the fingers of your right hand in the direction of the moment around the axis, your thumb points in the positive direction of the axis. Accordingly, it is negative when your thumb points in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right-hand rule is just a convention; it could very well be the left-hand rule that defines a positive direction. However, as far as I know, the right-hand rule is the leading convention. It is also used for angular velocity and magnetic fields, among other things.&lt;/p&gt;&#xA;" OwnerUserId="12505" LastEditorUserId="12505" LastEditDate="2017-08-14T22:36:00.890" LastActivityDate="2017-08-14T22:36:00.890" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16763" PostTypeId="2" ParentId="16737" CreationDate="2017-08-14T22:52:17.120" Score="0" Body="&lt;p&gt;Answer 2: Because you are not building thousands of the Bots, You need to look at materials that are already heat-treated for strength and toughness. Just thinking of the material and not the form : rear axel( of rear wheel drive ), Torsion bars , leaf and coil springs. Construction equipment ; bulldozer blade ( edge if it is  separate material from the bulk, bucket teeth ( if replaceable ), bulldozer treads, blades/arms of big chipper shredder ( hammer mill ) used to chop tree trunks ( Asplendunk is one name). The chipper is almost what you are making. These are educated guess ; all likely have more carbon than optimum for toughness. But it gives some ideas where to look.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2017-08-15T00:55:54.377" LastActivityDate="2017-08-15T00:55:54.377" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16764" PostTypeId="2" ParentId="16754" CreationDate="2017-08-15T00:53:33.680" Score="-1" Body="&lt;p&gt;The right hand rule dictates what a positive moment is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Km4RN.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Km4RN.gif&quot; alt=&quot;rhl&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a planar force, a positive moment is along the +z axis (out of the plane). Also a positive rotation is also along the +z axis (out of plane). See diagrams below on the left for what positive moments and rotations looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/anjgk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/anjgk.png&quot; alt=&quot;planar&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the right are the exact same moments and rotations, just expanded out by components. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rule for the moment of a force $\pmatrix{F_x &amp;amp; F_y}$ located at $\pmatrix{x&amp;amp;y}$ is $$M = x\,F_y - y\,F_x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for motions the velocity if a point located at $\pmatrix{x&amp;amp;y}$ while being rotated by $\omega$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\pmatrix{v_x &amp;amp; v_y} = \pmatrix{ -y\, \omega &amp;amp; x\, \omega}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are kind of hard to remember, so you kind of have to visualize it to come up with these. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively use the 3D versions, which are &lt;em&gt;much simpler&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{M} = \vec{r} \times \vec{F} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \vec{v} = \vec{\omega} \times \vec{r} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\times$ is the vector cross product, applied as 3D vectors. Here again you have to memorize that position comes before force in the 1st expression, but after rotation in the 2nd one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. In the UK, they  do use the left-hand-rule with a different permutation of vectors (zyx instead of xyz) in order for the result to be the same.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-08-15T00:53:33.680" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16766" PostTypeId="2" ParentId="8792" CreationDate="2017-08-15T09:35:56.803" Score="1" Body="&lt;p&gt;See NASA's computer program &lt;a href=&quot;https://www.grc.nasa.gov/WWW/CEAWeb/ceaHome.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;CEA&lt;/a&gt; (Chemical Equilibrium with Applications). From this site you can download an application to perform calculations. This application contains a text file called &lt;code&gt;trans.inp&lt;/code&gt; with coefficients, $A$ through $D$, for an empirical dynamic viscosity equation:&#xA;$$\ln\eta=A\ln T+\frac{B}{T}+\frac{C}{T^2}+D$$ &#xA;and for thermal conductivity:&#xA;$$\ln\lambda=A\ln T+\frac{B}{T}+\frac{C}{T^2}+D$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $T$ in Kelvin, $\eta$ in $\mu$P (or $10^{-7}$Pa s), and $\lambda$ in $\mu$W/cm K (or $10^{-4}$W/m K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The file contains coefficients for multiple temperature ranges (up to 10,000 K for CO$_2$ and 15,000 K for H$_2$, N$_2$, and O$_2$) and also for binary interaction parameters, e.g., $\eta_{ij}$ for species $i$ and $j$, required for obtaining the viscosity of mixtures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this site one can download NASA report RP-1311 which describes how to obtain viscosity of mixtures (see section 5.2.1) and RP-1311-P2 describing the file format (see TABLE E.1).&lt;/p&gt;&#xA;" OwnerUserId="6096" LastEditorUserId="6096" LastEditDate="2017-08-15T09:41:17.007" LastActivityDate="2017-08-15T09:41:17.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16769" PostTypeId="2" ParentId="16729" CreationDate="2017-08-15T14:21:30.807" Score="1" Body="&lt;p&gt;Yes and no. My experience is that it is something of an outdated method of formatting an autocad file. I believe the vports interfacing of autocad matured quite a bit in the 2000's. &lt;/p&gt;&#xA;" OwnerUserId="8635" LastActivityDate="2017-08-15T14:21:30.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16770" PostTypeId="1" AcceptedAnswerId="16776" CreationDate="2017-08-15T14:32:30.230" Score="0" ViewCount="39" Body="&lt;p&gt;Originally I was going to use accelerometers, but they are to big for the disk I'm trying to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I have a set of planetary gears for which I need to measure the vibrations in the x and y components, mainly. I´ll be performing modal testing with an instrumented hammer and a laser single point vibrometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since my gears are straight teeth spur gears I didn't take into account the &quot;z&quot; coordinate or vibrations of the system for my dynamic model, so if I point the laser perpendicularly to the gear´s plane the vibrations being measured by the laser will be those in z, which I'm not interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that if I had two Laser Doppler Vibrometers I could measure the in-plane motions I'm looking for, but since I only have one I was wondering if there´s a way of decomposing the projected component by pointing the laser at certain angle?&lt;/p&gt;&#xA;" OwnerUserId="4088" OwnerDisplayName="spe4ker" LastActivityDate="2017-08-16T10:47:32.790" Title="Is it possible to measure x &amp; y displacement components with a single point (out-of-plane) vibrometer?" Tags="|optics|vibration|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16775" PostTypeId="1" CreationDate="2017-08-16T06:13:47.890" Score="0" ViewCount="91" Body="&lt;p&gt;In &lt;em&gt;The Physiology of the Joints&lt;/em&gt; by Kapandji, he provides a diagram on page 21 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CGFfw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CGFfw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He does not formally define the concept of &quot;resistance&quot; but just creates this variable $R$ and claims that, if $n$ is the number of curves, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R = n^2 +1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n2q4Q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n2q4Q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He also references an equation called &quot;Delmas Index&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$DI = \frac{\text{length of vertebral column from tip to tip with curves}}{\text{length of vertebral column from tip to tip fully extended}} \times 100$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, he provides no citation for any of this. Upon Googling both of these a fair amount, I have sadly found other physiology textbooks reference &lt;strong&gt;only&lt;/strong&gt; this book. There doesn't seem to be any actual scientific analysis given. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are no references given anywhere in the book. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does an actual quadratic relationship exist between the curves and the resistance of the rod to compression? If so, what is the real equation used in fields like continuum mechanics? &lt;/p&gt;&#xA;" OwnerUserId="648" LastActivityDate="2017-08-16T14:15:44.727" Title="Does the resistance of a curved rod to compression depend on the number of curves?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16776" PostTypeId="2" ParentId="16770" CreationDate="2017-08-16T10:47:32.790" Score="1" Body="&lt;p&gt;There's just no way that you are going to turn a single Z measurement into an X or Y measurement.  With multiple Z measurements, you could calculate pitch and yaw angles, but for a spur gear, that's not going to get you want you want as the gear is going to translate, not bend or rotate.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggestion 1: Glue something to the spur gears that sticks up far enough out of plane that you can get the laser to reflect off of it in the X and Y direction.  e.g. glue a small reflective cube to the disk.  Point the laser at it in the X direction, perform the impact test, and then move the laser to the Y direction, and repeat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suggestion 2: regarding accelerometers... How big of a disk are we talking here?  They do make some very small accelerometers.  e.g. &lt;a href=&quot;https://buy.endevco.com/22-accelerometer.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Endevco Model 22&lt;/a&gt;  is a mere 0.14 grams.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-08-16T10:47:32.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16778" PostTypeId="2" ParentId="16775" CreationDate="2017-08-16T14:15:44.727" Score="1" Body="&lt;p&gt;When a curved solid is subjected to compressive stress, the displacement depends on the curvature (all other things being equal). When you have a section of material with a certain bending stiffness, and the entire section is curved with a radius $R$ and a subtended angle $2\phi$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/loyyl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/loyyl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then an axial load F will give a bending moment $M=Fh$ and change $\phi$ by $\Delta \phi \propto \frac{M}{EI}$. Displacement of the ends of the beam (towards each other) will depend on $\phi$ and $R$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta x = 2R\left(\sin(\phi+\Delta \phi) - \sin\phi)\right)\\&#xA;\approx 2R\cos\phi \Delta \phi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To relate this to the force, we need to know $h=R(1-\cos\phi)$, where I am assuming the initial curvature (at $\phi = \phi_0$, if you like) occurs without needing to apply any stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta x = 2R\cos\phi \frac{F R(1-\cos\phi)}{EI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you can translate this into the equations you quote really depends on the assumptions you make. If you assume the radius of curvature to be constant, then two shorter arcs with the same curvature have much lower value of $h$ and will be stiffer; but if you assume that the displacement $h$ is the same, then the radius of curvature must be much smaller, and the object will be more compliant. If you assume constant total length $L$, you can convert $\frac{L}{R}=\phi$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you can play around with these equations to convince yourself that the &quot;standard rule&quot; requires a lot of assumptions. I am not sure it's worth going much deeper into this - but ask questions in the comments and I can clarify if needed.&lt;/p&gt;&#xA;" OwnerUserId="565" LastActivityDate="2017-08-16T14:15:44.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16779" PostTypeId="1" AcceptedAnswerId="16819" CreationDate="2017-08-16T15:18:20.947" Score="2" ViewCount="16619" Body="&lt;p&gt;Why does a ductile rod in torsion fail at an angle perpendicular to its axis? I know the reason for brittle material failing in pure shear but I don't understand the reason for ductile material.&lt;/p&gt;&#xA;" OwnerUserId="12558" LastEditorUserId="368" LastEditDate="2017-08-16T16:12:10.930" LastActivityDate="2017-08-19T14:37:27.803" Title="Failure in torsion" Tags="|mechanical-engineering|applied-mechanics|mechanical-failure|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16780" PostTypeId="2" ParentId="1883" CreationDate="2017-08-16T17:21:44.087" Score="2" Body="&lt;p&gt;Texas Instruments has extensive material on this Piezo Electric disc for applications of measurements fluid level. You can search their website for &lt;a href=&quot;https://www.steminc.com/PZT/en/piezo-disc&quot; rel=&quot;nofollow noreferrer&quot;&gt;STEMINC&lt;/a&gt; and you will see two discs used on Thickness mode. Both discs have frequency 1MHz and are used with a couple of their controllers. The discs are SMD10T2R111WL and &lt;a href=&quot;https://www.steminc.com/PZT/en/piezo-ceramic-disc-1-mhz&quot; rel=&quot;nofollow noreferrer&quot;&gt;SMD15T21R111WL&lt;/a&gt;. STEMINC also has these same dimensions discs with Radial frequency but the Texas Instruments Controller TDC100 for liquid level measurement will work with the 1MHz version.&lt;/p&gt;&#xA;" OwnerUserId="12560" LastEditorUserId="1832" LastEditDate="2017-08-16T21:23:45.113" LastActivityDate="2017-08-16T21:23:45.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16781" PostTypeId="2" ParentId="16756" CreationDate="2017-08-16T21:38:56.687" Score="2" Body="&lt;p&gt;&lt;strong&gt;Use a Manometer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A manometer can be mounted to your launch platform and stick into your rocket. The manometer can be made of a simple transparent tube affixed to a vertical rod with dimension tick marks along it. Measure the density of your vinegar (you could just use water density and probably be fine to a close approximation), and then use standard pressure formula using density gravity and height of fluid above datum.&lt;/p&gt;&#xA;" OwnerUserId="7619" LastActivityDate="2017-08-16T21:38:56.687" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16782" PostTypeId="1" CreationDate="2017-08-17T00:06:43.223" Score="0" ViewCount="5721" Body="&lt;p&gt;Would it need hot forging or hot pressing before heat treatment to be equal in strength and toughness to commercial AR500 steel plates?&lt;/p&gt;&#xA;" OwnerUserId="12564" LastEditorUserId="9" LastEditDate="2017-08-27T15:36:17.223" LastActivityDate="2020-01-02T01:42:50.687" Title="Is it possible to make AR500 steel at home by just melting its components and applying appropriate heat treatment?" Tags="|mechanical-engineering|structural-engineering|materials|design|steel|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16783" PostTypeId="2" ParentId="16733" CreationDate="2017-08-17T06:07:53.490" Score="1" Body="&lt;p&gt;What you need is to find one that has a configurable machined length for the support end. This way, you will have the screw flats on the fixed end and a machined &quot;stepped&quot; end on the support side for the pulley to attach to, just after the supporter. Just make sure the stepped end on the support side is long enough to contain BOTH the pulley and the support mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4mGho.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4mGho.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8543" LastActivityDate="2017-08-17T06:07:53.490" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16787" PostTypeId="1" CreationDate="2017-08-17T11:06:55.150" Score="1" ViewCount="150" Body="&lt;p&gt;I am trying to design a heat source for use with testing the functionality of an infrared thermometer. I am happy with my circuit, I have a surface heated to a nice, constant 50°C (as constantly monitored by a thermocouple). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface is just an aluminium alloy, which I have sprayed with a matt black paint. Unfortunately, finding the exact emissivity of the surface is proving difficult. I am unsure whether to keep this as my surface, or find something else. The matt black paint is supposed to have an emissivity of 0.93-0.97, yet aluminium alloy has a very low emissivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A table can be found &lt;a href=&quot;http://www-eng.lbl.gov/~dw/projects/DW4229_LHC_detector_analysis/calculations/emissivity2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;HERE&lt;/a&gt; which shows the emissivity of different surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermal conductivity of different materials can be found &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_thermal_conductivities&quot; rel=&quot;nofollow noreferrer&quot;&gt;HERE&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I chose aluminium was because it is easy to get hold of, it is not expensive and has good thermal conductivity. Because of its low emissivity though, I had to paint the surface in order to test the infrared thermometers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a lot of research on the subject and there are so many places that specify using a black surface to get the best emissivity (as in closer to 1.0), yet there are some black surfaces which have a very poor emissivity as seen &lt;a href=&quot;https://www.design1st.com/Design-Resource-Library/engineering_data/ThermalEmissivityValues.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;HERE&lt;/a&gt;. That document also shows some white surfaces as having good emissivity, and white is the furthest from black you can get!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other articles such as this one &lt;a href=&quot;https://www.design1st.com/Design-Resource-Library/engineering_data/ThermalEmissivityValues.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;HERE&lt;/a&gt; that look at the effect of the smoothness of a surface, which I found interesting as some materials with smooth surfaces have a good emissivity, even if not black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: What is more important when it comes to emissivity (getting closer to 1.0), colour or surface finish? Do different colours or surface finishes affect the wavelength of the thermal radiation? If so, why and how do these factors have an effect?&lt;/p&gt;&#xA;" OwnerUserId="12568" OwnerDisplayName="MCG" LastEditorUserId="368" LastEditDate="2017-08-17T16:58:06.307" LastActivityDate="2017-09-17T00:10:53.120" Title="Governing emissivity" Tags="|electrical-engineering|materials|heat-transfer|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="16790" PostTypeId="2" ParentId="16787" CreationDate="2017-08-17T13:24:13.703" Score="2" Body="&lt;p&gt;Since the paint covers the aluminum, the emissivity of Al is irrelevant. The heat is conducted from metal to paint, and about the best you can do is to assume that both are at the same (surface) temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calibrate a thermometer, what's important is not the emissivity value (1 vs 0.99 vs 0.6) but the uniformity across the spectral band of the thermometer.  That's because the temperature is calculated based on the ratio of power at different wavelengths.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, you have to consider not only the paint's spectral emissivity but also whether you need a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lambert%27s_cosine_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lambertian&lt;/a&gt; output distribution.  If you are just testing IR thermometers, you probably don't- just point the thermometer perpendicularto the radiating surface and be consistent in your alignment procedure. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-08-17T13:24:13.703" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16791" PostTypeId="1" CreationDate="2017-08-17T13:55:04.183" Score="1" ViewCount="135" Body="&lt;p&gt;Can anyone give me a few pointers to calculating the time take for an electric shower to reach it's set temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set temperature is actually a power setting and a flow rate setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that with 3500 watts of heating power and a flow rate of 3.5 litres per min, I will get a 14.35 degree rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.5 l/min = 58.3 g/sec&lt;br&gt;&#xA;58.3 * specific heat of water 4.18 = 243.8&lt;br&gt;&#xA;3500w / 243.8 = 14.35 degrees rise  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, is it possible to work out the time taken to reach this temp? I assume I need to know something about the efficiency of the heat exchanger to calculate this as well as the specific heat of the materials between the heater and my measurement point?&lt;/p&gt;&#xA;" OwnerUserId="5419" LastEditorUserId="5419" LastEditDate="2017-08-17T17:16:35.320" LastActivityDate="2017-08-17T20:30:31.907" Title="How to calculate time taken for an electric shower to reach set temperature" Tags="|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16793" PostTypeId="2" ParentId="16782" CreationDate="2017-08-17T15:14:28.950" Score="0" Body="&lt;p&gt;I read the ad ; AR500 bullet proof vest has pretty much nothing to do with steel. Vest design and the aramide fiber are the critical factors. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2017-08-17T19:29:59.020" LastActivityDate="2017-08-17T19:29:59.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16794" PostTypeId="1" CreationDate="2017-08-17T16:12:23.690" Score="1" ViewCount="132" Body="&lt;p&gt;If the waste heat was moved to areas of lower pressure such behind the car-green area-&#xA;&lt;img src=&quot;https://i.stack.imgur.com/Pst1Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Could this reduce the drag from these areas of lower pressure.  From what I understand this hot air would increase pressure in those areas but equally hot air is less dense so would it increase drag?&lt;/p&gt;&#xA;" OwnerUserId="11324" LastActivityDate="2017-09-20T20:22:36.363" Title="Could car waste heat be used to lower drag?" Tags="|thermodynamics|heat-transfer|drag|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16795" PostTypeId="1" CreationDate="2017-08-17T18:03:36.527" Score="1" ViewCount="44" Body="&lt;p&gt;I'm currently studying Aerodynamics, and one thing that I noticed is that the maximum and minimum &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; coordinate of the airfoils (which are necessary to compute the chord) on the transformed plane (let it be &lt;span class=&quot;math-container&quot;&gt;$z(x,y)$&lt;/span&gt;) correspond to the transformed intersection points of the circumference on the original plane  (let it be &lt;span class=&quot;math-container&quot;&gt;$\zeta(\xi,\eta)$&lt;/span&gt;) with the real axis. I don't find any proof of this statement, and so I tried to do it by myself. The problem is that it got too complicated to be solved analytically (too many different powers of trignometric functions). So I'm requesting someone to try demonstrate this too. I think that there's a simpler way to do it.&lt;/p&gt;&#xA;&lt;p&gt;There's my introduction to the problem:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Consider the original circumference defined on the complex &lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt; plane:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ymvqg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ymvqg.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is the circumference radius, and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; the intersection of the circumference with the real positive axis, &lt;span class=&quot;math-container&quot;&gt;$\xi$&lt;/span&gt;. The parameter &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; is the angle between the horizontal line and the line that links &lt;span class=&quot;math-container&quot;&gt;$\zeta_0$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;. The center of the circumference is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\zeta_0=-b\varepsilon+ia\sin(\beta)=-b\varepsilon+i(1+\varepsilon)b\tan(\beta)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, this circumference is defined by:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\zeta=-b\varepsilon+b\frac{1+\varepsilon}{\cos(\beta)}\left(e^{i\theta}+i\sin(\beta)\right),\hspace{15pt}\theta \in [0,2\pi]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Now I need to show that for the Joukowski transform &lt;span class=&quot;math-container&quot;&gt;$z=\zeta+\frac{b^2}{\zeta}$&lt;/span&gt; the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; coordinate (real coordinate on the plane &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;) has&#xA;a maximum on &lt;span class=&quot;math-container&quot;&gt;$\zeta=b$&lt;/span&gt; (intersection of the circunference with the positive real axis) and a minimum on &lt;span class=&quot;math-container&quot;&gt;$\zeta=-b(1+2\varepsilon)$&lt;/span&gt;, (intersection of the circunference with the negative real axis).&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="9747" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-08-17T18:03:36.527" Title="Determination of a Joukowski airfoil chord (demonstration)" Tags="|fluid-mechanics|aerodynamics|airfoils|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16796" PostTypeId="2" ParentId="16701" CreationDate="2017-08-17T18:33:19.290" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It is possible- perhaps likely- that the safety factor can vary depending on the nature of the loading. Specifically I have in mind uplift due to wind. But other forms of applied uplift may include: buoyant forces, dynamic/static pressure applied by some other fluid (e.g., water, etc.), and load transference forces (perhaps in a machine where ballast is used to counteract forces applied to some arm/member).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A machine where ballast can be used to counteract forces is simpler than you think - it happens frequently in tank / pressure vessel design.  The liquid inside the tank is a ballasting force that could counteract a wide variety of loads:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Vacuum pressure inside the vessel causing caving in a flat bottom&lt;/li&gt;&#xA;&lt;li&gt;External flooding for buoyancy causing the bottom to cave in&lt;/li&gt;&#xA;&lt;li&gt;External flooding for buoyancy causing the vessel to float&lt;/li&gt;&#xA;&lt;li&gt;Wind forces from a wind event&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The ultra conservative ASME (specifically ASME RTP-1 - I have not verified for other sections of ASME code) will not allow the liquid inside to be used to counteract any of these forces.  However, they will allow the maximum allowable strain to be doubled during these events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ASTM D3982 will allow the liquid inside to counteract the first loading for tanks with a vacuum loading less than .217 psi - or 6&quot; of liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice however, before a hurricane reaches land everyone fills their storage tanks before evacuating, regardless of how well they were designed.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-08-17T18:33:19.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16797" PostTypeId="2" ParentId="8889" CreationDate="2017-08-17T19:32:19.153" Score="-3" Body="&lt;p&gt;Not all ductile alloys neck.  See, for example, Haynes 25 (L605).&lt;/p&gt;&#xA;" OwnerUserId="12582" LastActivityDate="2017-08-17T19:32:19.153" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16798" PostTypeId="2" ParentId="16791" CreationDate="2017-08-17T20:30:31.907" Score="2" Body="&lt;p&gt;With SI units you can derive the formula for yourself.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;More power -&gt; lower time. $ t \; \alpha \; \frac {1}{P} $.&lt;/li&gt;&#xA;&lt;li&gt;More water (mass) -&gt; longer time. $ t \; \alpha \; {m} $.&lt;/li&gt;&#xA;&lt;li&gt;Higher specific heat capacity -&gt; longer time. $ t \; \alpha \; {SHC} $.&lt;/li&gt;&#xA;&lt;li&gt;Higher temperature -&gt; longer time. $ t \; \alpha \; \Delta T $.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Putting them all together you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t = \frac {\Delta T \cdot m \cdot SHC}{P} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t = \frac {14.35 \cdot \frac {3.5}{60} \cdot 4180}{3500} = 1 \; s $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes sense as there at steady state we must heat the inflow per unit time to the steady state temperature otherwise it won't be steady state!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information we're missing is the capacity of the heating tank. If it held 1 L of water then:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t = \frac {14.35 \cdot 1 \cdot 4180}{3500} = 17 \; s $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then add in the copper of the cylinder, losses to ambient, ...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My sanity check on this is to calculate time to boil 1 L of water from 20°C using a 2.1 kW electric kettle. Since we're in kW mode we'll convert the SHC to k as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ t = \frac {\Delta T \cdot m \cdot SHC}{P} = \frac {80 \cdot 1 \cdot 4.2}{2.1} = 160 \; s $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little under three minutes sounds about right.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2017-08-17T20:30:31.907" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16799" PostTypeId="2" ParentId="16782" CreationDate="2017-08-18T00:44:06.110" Score="1" Body="&lt;p&gt;If you want to make the equivalent of &lt;strong&gt;commercial&lt;/strong&gt; grade AR500, you are going to need &lt;strong&gt;commercial&lt;/strong&gt; grade equipment. This will include a professional grade foundry, press, roller, etc - not to mention the SPACE required to do all this which will cost you in excess of several 100,000$ to purchase. This is not something that DIY equipment will be able to do efficiently. Without the proper equipment you risk making defective steel which will be of very low quality and durability.&lt;/p&gt;&#xA;" OwnerUserId="8543" LastActivityDate="2017-08-18T00:44:06.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16800" PostTypeId="1" AcceptedAnswerId="16802" CreationDate="2017-08-18T09:08:20.903" Score="0" ViewCount="127" Body="&lt;p&gt;I am trying to figure out how much torque this piece of metal that I have can deal with before breaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any tools online that can help me calculate these items?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hardness of this iron is 167-229Ha&#xA;tensile strength is &gt;=58kg/mm2&#xA;elasticity is &gt;= 45%&#xA;density is 0.785&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently the pipe is a hexagaon that's 55mm in diameter, then the ends are cylinders that are 40mm in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is early in the design phase I'd like to be able to adjust torque and the size of the metal pipe so that it wouldn't break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I have 1200RPM 15HP motor that I have reduced down to 85RPM which if I calculated correctly should be around 926 ft-lbs of torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any tools that will help me convert from ft-lbs to lbs-ft3 or calculate these values?&#xA;&lt;a href=&quot;http://www.endmemo.com/sconvert/kg_m3lb_ft3.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;one such online calculator.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12586" LastEditorUserId="12586" LastEditDate="2017-08-18T10:27:36.290" LastActivityDate="2017-08-18T12:22:57.087" Title="ft-lbs to ft-lbs3 or newtons as well as strength of metal" Tags="|motors|gears|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16802" PostTypeId="2" ParentId="16800" CreationDate="2017-08-18T12:22:57.087" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.engineeringtoolbox.com/torsion-shafts-d_947.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page&lt;/a&gt; and &lt;a href=&quot;http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect07.d/IAST.Lect07.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; would be good places to get you started.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to my comment above, you'll probably want to design to the yield strength and not the ultimate tensile strength.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-08-18T12:22:57.087" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16803" PostTypeId="1" AcceptedAnswerId="16805" CreationDate="2017-08-18T17:31:42.530" Score="1" ViewCount="8738" Body="&lt;p&gt;Most cheap pump can achieve, according to the Amazon page, around 5 PSI and I would need something a way higher that can still be controlled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have &lt;a href=&quot;https://www.amazon.ca/4inch-Normally-Closed-Electric-Solenoid/dp/B01M0UEISP/ref=sr_1_3?ie=UTF8&amp;amp;qid=1503076215&amp;amp;sr=8-3&amp;amp;keywords=electric%20air%20valve&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; solenoid valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1R5VP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1R5VP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I use pulse-width modulation (PWM) with a longer time period to control the flow of a pressurized liquid? Will that  damage the valve? What is the accuracy of the valve?&lt;/p&gt;&#xA;" OwnerUserId="10037" LastEditorUserId="1586" LastEditDate="2017-08-22T05:26:04.230" LastActivityDate="2022-09-21T19:19:34.850" Title="Can PWM be used to control the flow of a solenoid valve?" Tags="|electrical-engineering|pressure|pumps|valves|liquid|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16804" PostTypeId="2" ParentId="15086" CreationDate="2017-08-18T18:46:41.063" Score="2" Body="&lt;blockquote&gt;There is a back EMF in a motor, due to current flow in a magnetic field.&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  A motor develops back EMF due to conductors cutting across magnetic field lines.  This has nothing to do with current.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;So, in a generator, is there a reverse rotation (opposed to original rotation) due to a current flow in magnetic field?&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reverse rotation doesn't make sense.  A generator doesn't spin backwards due to it being loaded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is reverse &lt;i&gt;torque&lt;/i&gt; due to current.  The generator makes a voltage proportional to its rotation speed.  The load the generator is connected to then draws current.  You can't separate the generator and the motor inside the unit.  The current that the load draws causes the inherent motor to make torque opposing the direction of rotation of the generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is this torque times the shaft rotation which represents mechanical power being put into the generator.  If a generator could always be freely spun regardless of the electrical power it puts out, then it would violate the laws of physics.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-08-18T18:46:41.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16805" PostTypeId="2" ParentId="16803" CreationDate="2017-08-18T18:55:31.567" Score="3" Body="&lt;p&gt;It is possible to modulate the drive of some types of solenoid fluid valves to ultimately regulate the pressure of the fluid at their output port.  I have designed a product that does excatly that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can't easily use any valve.  You want a valve that is intended for part-way operation between full on and full off.  There are such solenoid valves, but you shouldn't assume they can do this unless this is explicitly stated.  These are often referred to as &lt;i&gt;proportional&lt;/i&gt; valves, althogh they aren't necessarily all that linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using PWM to control a solenoid is quite feasible.  The product I mentioned did exactly that.  You should make the PWM period short enough so that the current thru the valve changes little each cycle.  Only the average current goes to driving the valve.  The high speed AC component only heats it without doing anything useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Valves intended only for on or off operation are not suitable for this.  You can PWM the solenoid, but PWMing the fluid flow causes a lot of mechanical stresses, pulses, and other problems.  On/off valves aren't intended for that kind of operation.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-08-18T18:55:31.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16806" PostTypeId="2" ParentId="16113" CreationDate="2017-08-18T19:04:01.393" Score="1" Body="&lt;p&gt;Yes, the choice of $G(s)$ and $C(s)$ is preferable from a physics/engineering perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C(s)$ stands for controller and $G(s)$ for plant.&#xA;As an engineer you want these transfer functions to match the real physical system including the corresponding parameters.&#xA;Otherwise you would rather write down $H(s) = \frac{K}{s(p+s)+K}=\frac{K}{s^2+ps+K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if your plant is an unconstrained mass $G(s) =\frac{1}{Ms^2}$ is preferred. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your plant is an integrator (like a watertank) $G(s)=\frac{1}{s}$ is preferred.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is known that $H(s)$ has the behavior of a forced-damped system, the notation&#xA;$H(s)=\frac{K\cdot\omega_n^2 }{s^2+ 2 \beta \omega_n s+\omega_n^2 }$ or $H(s)=\frac{K}{(s+\lambda )^2+\omega_d^2}$ would be preferred, since the parameters have a real physical representation.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastEditorUserId="8216" LastEditDate="2018-02-23T19:38:25.713" LastActivityDate="2018-02-23T19:38:25.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16807" PostTypeId="2" ParentId="14376" CreationDate="2017-08-18T20:16:27.140" Score="1" Body="&lt;p&gt;When using a 4/3 directional valve your control options are very limited forward, off or backward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A PID controller will make no sense, since its has an output which is a finite number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only controller you can use is a 3-state controller&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UqYts.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UqYts.png&quot; alt=&quot;3-state controller&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand, create and use a 3-state feedbackcontroller, you should read the section &quot;In engineering&quot; on &lt;a href=&quot;https://en.wikipedia.org/wiki/Hysteresis&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Hysteresis&lt;/a&gt; which explains it for a 2-state controller i.e. on/off controller.&#xA;And for the 3-state feedbackcontroller: &lt;a href=&quot;http://www.jumo.de/en_DE/support/faq/controllers/T2/Q6.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.jumo.de/en_DE/support/faq/controllers/T2/Q6.html&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2017-08-18T20:16:27.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16808" PostTypeId="1" CreationDate="2017-08-18T20:31:53.847" Score="1" ViewCount="506" Body="&lt;p&gt;I am trying to understand what the drawing states for a transmission tower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, the underlined portion. Any help would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BC65X31 + CO45&#xA;CT41X38&#xA;CO45&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S9I15.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S9I15.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12595" LastActivityDate="2017-08-18T20:31:53.847" Title="Tower design engineering drawing" Tags="|structural-engineering|civil-engineering|" AnswerCount="0" CommentCount="1" ClosedDate="2018-10-23T21:37:18.207" ContentLicense="CC BY-SA 3.0" />
  <row Id="16810" PostTypeId="2" ParentId="16744" CreationDate="2017-08-19T00:51:57.593" Score="2" Body="&lt;p&gt;We cut the vessel in half as shown below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the vertical projection considered because the pressure inside the cylindrical vessel due to fluid acts perpendicular to the surface of the cylindrical vessel, such that differential force is perpendicular at any point. Assuming that we take the horizontal component and vertical component of each of the differential forces, all horizontal component of the differential force is actually countered on the other side of the vessel, therefore giving an equilibrium in horizontal direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hoop now only considers vertical component of the force, therefore, whatever configuration of the vessel, we need to take only the vertical component, thus we only need the area of the vessel perpendicular to the force being considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed in a comment, this analysis is similar to bearing pins in truss member in a way because in analysis of pins, we take the diameter of the pin as equal to the diameter of the hole, thus the pressure in the pin is &lt;strong&gt;EQUAL&lt;/strong&gt; in the half of the pin, as the other side is unloaded. The bearing area is then calculated as the projected area because of what is explained above. In reality, this is not what happens because we tend to increase the hole diameter for ease in installation. Pressure vessels on the other hand is analyzed similar to this because fluid will tend to pressurized all directions with the same value. Thus, bearing area projection is similar in these two cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fH9KN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fH9KN.jpg&quot; alt=&quot;Pressure acts on side of the vessel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-08-19T11:34:32.620" LastActivityDate="2017-08-19T11:34:32.620" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="16812" PostTypeId="2" ParentId="15663" CreationDate="2017-08-19T03:35:08.897" Score="1" Body="&lt;p&gt;Creating an FBD in joint A:&lt;a href=&quot;https://i.stack.imgur.com/ccnCU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ccnCU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can derive that member AC is under compression. &lt;strong&gt;Note: you don't need to solve for the reactions since we can answer directly using method of joints&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euler's buckling formula states that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jGwXU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jGwXU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if you check your column there, the critical buckling load is equal to 37.285 kN (Compression). &lt;strong&gt;Note: use K = 1.0 since both ends are pin-connected&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer the first question: &lt;strong&gt;NO, it is not strong enough&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the second question reverses the function mentioned above, given capacity of AC = 37.285 kN, we can get AB = 74.570 kN, therefore, &lt;strong&gt;the point load capacity at joint A is equal to 64.580 kN.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2017-08-19T03:35:08.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16814" PostTypeId="2" ParentId="14428" CreationDate="2017-08-19T04:37:31.843" Score="0" Body="&lt;p&gt;The answer for your first question is because the DC motor is probably not mechanically loaded. When the voltage started to rise up, the motor withdrew a current to rotate it's inertia. When the voltage reaches to rated value, the speed settles and the current will be almost zero as long as there is no mechanical torque applied on DC motor shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding EMF that appears in your model, It is a simulation block to model the interaction between the &lt;strong&gt;electrical&lt;/strong&gt; side of the circuit (voltage, current) and the &lt;strong&gt;mechanical&lt;/strong&gt; side (torque,speed). In real life the you will not see such thing as the DC motor inherently convert the electrical energy to mechanical one. As we just modeled the machine electrical losses by R, a real life parameters such as Back EMF can be included in EMF model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An idea about back EMF in DC motors: it is a voltage that appears in the opposite direction to current flow as a result of the motor’s coils moving relative to a magnetic field. It is this voltage that serves as the principle of operation for a generator. The back EMF is directly related to the speed of the motor, so knowing the value of back EMF allows us to calculate the speed of that motor. &lt;a href=&quot;https://www.precisionmicrodrives.com/application-notes/ab-021-measuring-rpm-from-back-emf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9302" LastActivityDate="2017-08-19T04:37:31.843" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16815" PostTypeId="2" ParentId="16794" CreationDate="2017-08-19T10:27:22.777" Score="1" Body="&lt;p&gt;The thermic energy output of a car is too small to make a difference. Also the technical implemention would be very difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you ever hold your hand near to the exhaust. Its really not that hot compared to a camp fire. But this was in idle so when driving it might be different. On the other hand when driving you also have a lot of air mass to be heated. So when driving the air will be cooler compared to standing in idle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly the technical solution would be very complicated and counterintuitive. Usually car makers want the exhaust gases to have only a very small amount of energy. They try to get energy out of it with a turbocharger for example. The added weight by some kind of exhaust distribution system would counter the effect of your idea in terms of fuel efficiency.&lt;/p&gt;&#xA;" OwnerUserId="12530" LastActivityDate="2017-08-19T10:27:22.777" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16816" PostTypeId="2" ParentId="16662" CreationDate="2017-08-19T10:39:38.480" Score="1" Body="&lt;p&gt;I found an answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As $I= f(u_1,u_2)$ it follows that $\{i,j\}=1,2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\tag{1} u_{ij}\delta u_{ji}+ u_{ji}\delta u_{ij}=&amp;amp; \sum_i \sum_j (u_{ij}\delta u_{ji}+ \delta u_{ij}u_{ji})  \\&#xA;\tag{2} =&amp;amp;u_{11}\delta u_{11}+ u_{12}\delta u_{21}+u_{21}\delta u_{12}+u_{22}\delta u_{22}+ \\ &amp;amp;\delta u_{11}u_{11}+ \delta u_{12} u_{21}+\delta u_{21} u_{12}+\delta u_{22} u_{22} \\&#xA;=&amp;amp;2u_{11}\delta u_{11}+ 2u_{12}\delta u_{21}+2u_{21}\delta u_{12}+2u_{22}\delta u_{22}\\&#xA;=&amp;amp;2u_{ij}\delta u_{ji}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\tag{1}I(u_1,u_2,P)&amp;amp;=\int_R [\frac{\mu}{2}[(\delta u_{i,j} + \delta u_{j,i})u_{i,j}+(u_{i,j} + u_{j,i}) \delta  u_{i,j}] dx_1 dx_2 +[...]&#xA;\\ &amp;amp;= \int_R [\frac{\mu}{2}[\delta u_{i,j}u_{i,j} + \delta u_{j,i}u_{i,j}+u_{i,j}\delta  u_{i,j} + u_{j,i}\delta  u_{i,j}] dx_1 dx_2 +[...]&#xA;\\ &amp;amp;= \int_R [\frac{\mu}{2}[2u_{i,j}\delta  u_{i,j}+  \delta u_{j,i}u_{i,j}+  u_{j,i}\delta  u_{i,j}] dx_1 dx_2 +[...] &#xA;\\ &amp;amp;= \int_R [\frac{\mu}{2}[\delta u_{i,j}u_{i,j} + \delta u_{j,i}u_{i,j}+u_{i,j}\delta  u_{i,j} + u_{j,i}\delta  u_{i,j}] dx_1 dx_2 +[...]&#xA;\\ &amp;amp;= \int_R [\frac{\mu}{2}[2u_{i,j}\delta  u_{i,j}+ 2u_{j,i}\delta  u_{i,j}] dx_1 dx_2 +[...] &#xA;\\ &amp;amp;=\int_R [\mu(u_{i,j} + u_{j,i})\delta u_{i,j}dx_1 dx_2 +[...]&#xA;\end{align}&lt;/p&gt;&#xA;" OwnerUserId="12352" LastActivityDate="2017-08-19T10:39:38.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16817" PostTypeId="2" ParentId="16779" CreationDate="2017-08-19T10:45:05.753" Score="0" Body="&lt;p&gt;The stresses of a rod in shear and in torsion are of the same kind. So they fail of the same reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a conversion factor for materials. It is sqrt(3). So with the same stress induced to a rod under torsion compared to one under tensile load is more likely to break by this factor.&lt;/p&gt;&#xA;" OwnerUserId="12530" LastActivityDate="2017-08-19T10:45:05.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16818" PostTypeId="1" AcceptedAnswerId="16820" CreationDate="2017-08-19T12:22:44.133" Score="1" ViewCount="120" Body="&lt;p&gt;Currently software and maps represent earth in many different ways, the latest is Point Clouds as this is the most common surveying tools currently available. For the needs of visualisation, construction etc. this seems to suffice for now. Not sure this is an engineering question but I tried in the philosophy forum and been advised this is applied science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the most accurate representation of earth in a digital format (or mathematical equation):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3D splines&lt;/li&gt;&#xA;&lt;li&gt;solids and primitives&lt;/li&gt;&#xA;&lt;li&gt;breaklines/contours&lt;/li&gt;&#xA;&lt;li&gt;polygonal mesh, TIN&lt;/li&gt;&#xA;&lt;li&gt;square 3D pixels&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;?&lt;/p&gt;&#xA;" OwnerUserId="12604" LastActivityDate="2017-08-19T18:11:07.023" Title="What is the most accurate geometry to represent earth" Tags="|cad|geometry|surveying|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16819" PostTypeId="2" ParentId="16779" CreationDate="2017-08-19T14:37:27.803" Score="5" Body="&lt;p&gt;As a rule of thumb:&#xA;When brittle materials are subjected to torsion they fail in the plane, where tension is at its highest, i.e. at a 45° angle.&#xA;Ductile materials on the other hand fail in the plane of maximum shear stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at Mohr's Circle for pure shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JGq6F.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JGq6F.png&quot; alt=&quot;Mohr&amp;#39;s Circle for pure shear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Maximum-Shear-Stress Theory&lt;/em&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] yielding of the [ductile] material begins when the absolute maximum shear stress in the material reaches the shear stress that causes the same material to yield when it is subjected &lt;em&gt;only&lt;/em&gt; to axial tension.&lt;br&gt;– &lt;em&gt;R.C. Hibbeler, Mechanics of Materials, p.525&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for ductile materials: $$ \tau_{max}=\frac{\sigma_y}{2} $$&#xA;As you can see in the graphic, $\tau_{max}$ is reached before $\sigma_y$, thus the plane of failure is perpendicular to the primary axis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For brittle materials on the other hand, &lt;em&gt;Maximum-Normal-Stress Theory&lt;/em&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[...] a brittle material will fail when the maximum tensile stress, $\sigma_1$, in the material reaches a value that is equal to the ultimate normal stress the material can sustain […]&lt;br&gt;– &lt;em&gt;R.C. Hibbeler, Mechanics of Materials, p.528&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;for brittle materials: $$ \tau_f=\sigma_f $$&#xA;So, under pure shear it fails in tension at a 45° angle.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2017-08-19T14:37:27.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16820" PostTypeId="2" ParentId="16818" CreationDate="2017-08-19T18:11:07.023" Score="2" Body="&lt;p&gt;You are confusing methods of organizing data with the accuracy of that data or the accuracy of the interpolation method.  The two are orthogonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, you could describe a surface in Z as a function of X and Y as a rectangular grid of points.  At each point, the Z for that X,Y is known explicitly.  For X,Y between points, you do a 2D linear interpolation between the 4 nearest points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you could describe the surface as a set of contour lines, like a topo map.  For any point not on a contour, you interpolate between the two nearest countours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one is more accurate?  Either can be made arbitrarily accurate.  If you put a grid point every mile and compare that to a map with 10 foot countours, the contour map will usually give you far better accuracy.  On the other hand, consider a grid point every meter versus 100 foot contours.  Most of the time the grid will be much more accurate.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-08-19T18:11:07.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16821" PostTypeId="1" CreationDate="2017-08-19T19:08:17.103" Score="1" ViewCount="39" Body="&lt;p&gt;Most phone batteries will have a connector of some type with 4 contacts.  Of course two of those I'd expect to be +- Voltage, but the other two compose some sort of mystery input to the phone (thermistor? signal?  I don't know).  I was hoping to substitute a primary battery instead of the factory one for my project (I have a 3.6v primary battery), but my device always seems to have a problem with it, presumably due to the battery protection circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I might just leave the protection circuit in place and wire power to both input and output- that way whatever input the phone needs to see is still there, but even that approach seems not to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me bypass this system so that I can use a custom battery in this device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cj5Cp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cj5Cp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above is a diagram with some measured resistance/voltage values.  If a value between points isn't shown, you can consider it an open circuit (although I didn't measure battery resistance)&lt;/p&gt;&#xA;" OwnerUserId="7935" LastEditorUserId="1832" LastEditDate="2017-08-21T14:34:30.867" LastActivityDate="2017-08-21T14:34:30.867" Title="I would like to run a smartphone using a primary battery instead of rechargeable for a project. What are the 4 contacts used for on the factory LiPo?" Tags="|electrical-engineering|power-electronics|power-engineering|battery|consumer-electronics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16822" PostTypeId="1" AcceptedAnswerId="16825" CreationDate="2017-08-19T23:59:26.493" Score="1" ViewCount="590" Body="&lt;p&gt;I'm attempting to make the rigid equivelent of a gate tensioner/turnbuckle. I'm doing this for a couple of reasons:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I like the idea of not fussing with cables&lt;/li&gt;&#xA;&lt;li&gt;I'm using this experience to learn about the technology behind the turnbuckle&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I've mocked up something that &lt;em&gt;almost&lt;/em&gt; works. I've connected two hook-bolts to a threaded rod (5/16-18-36&quot;) via connectors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seemed like it would work in my head; I thought I would just be able to tighten one of the connectors to tension the gate. However, in practice, all I accomplish when tightening a connector is tightening one side, but then loosening the other side. The devise comes apart and fails miserably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What kind of mechanism would it take to make this work as expected?&lt;/strong&gt; In summary, I want to shorten the length of the device/rod by tightening a connector.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ab3E9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ab3E9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KHmzh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KHmzh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For bonus points: What is the principle called in Mechanical Engineering theory that would explain the problem where one side tightens, but the other side loosens?&lt;/p&gt;&#xA;" OwnerUserId="10536" LastActivityDate="2017-08-26T14:07:36.613" Title="Rigid tensioner/turnbuckle: One side tightens, but the other side loosens" Tags="|mechanical-engineering|threads|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16823" PostTypeId="1" AcceptedAnswerId="16861" CreationDate="2017-08-20T02:50:46.623" Score="0" ViewCount="125" Body="&lt;p&gt;I did some calculations to find out the moment of inertia and physical qualities of some metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using typical medium carbon steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll use a density of 7.8 g/m^3 until I get better numbers.&#xA;There's a solid circular shaft that's 1000mm long&#xA;The diameter of this shaft is 40mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives me: Polar moment of area 251327 mm^4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next part is the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will be a shaft connected to a 15HP 1200RPM motor but I'll be adding a gearbox to bring the RPM down to about 85RPM this will give me about 1256 newtons/meter of force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a bit lost at this junction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to convert the newtons to MPa or the Polar moment to some other units?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I compare the 1256 newtons to the amount of force this cylinder can withstand 251327 mm^4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would this pipe bend or break under the max torque load?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I relate 1256N.m to 251327mm^4 to see if the pipe would break?&lt;/p&gt;&#xA;" OwnerUserId="12586" LastActivityDate="2017-08-23T12:49:33.770" Title="moment of inertia forces and torque" Tags="|gears|modeling|moments|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16825" PostTypeId="2" ParentId="16822" CreationDate="2017-08-20T07:19:10.193" Score="4" Body="&lt;p&gt;To make this work, one side will need the &quot;normal&quot; righthand thread and the other a lefthand thread or  a rotating connection , but threads on both ends of the nut will not change the tension.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-20T07:19:10.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16826" PostTypeId="2" ParentId="16823" CreationDate="2017-08-20T10:44:44.743" Score="1" Body="&lt;p&gt;If you don't want plastic deformation to occur in the shaft you need to make sure that you don't exceed the yield strength $\sigma_y$ of the material throughout the cross-section of the material.&#xA;To calculate the maximum elastic shear stress in the (solid circular) cross-section you can use the following formula:&#xA;$$ \tau_{max}=\frac{Tr}{J} $$&#xA;$ T\quad…\quad $applied Torsional Moment&lt;br&gt;&#xA;$r \quad…\quad$Radius&lt;br&gt;&#xA;$J=\frac{\pi}{2}r^4 \quad…\quad $Polar Moment of Area&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as discussed in &lt;a href=&quot;https://engineering.stackexchange.com/questions/16779/failure-in-torsion/16819#16819&quot;&gt;this&lt;/a&gt; post, for ductile materials subjected to pure shear stress $t_{y}\approx\frac{\sigma_y}{2}$.&#xA;For medium carbon steel $\sigma_y$ is in the range of $500-700$ $MPa$&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2017-08-20T10:44:44.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16827" PostTypeId="1" CreationDate="2017-08-20T13:17:14.273" Score="1" ViewCount="2623" Body="&lt;p&gt;These are the mechanisms I'm aware of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Crankshaft&lt;/li&gt;&#xA;&lt;li&gt;Scotch yoke&lt;/li&gt;&#xA;&lt;li&gt;cam-shaft&lt;/li&gt;&#xA;&lt;li&gt;wobble drive (a.k.a nutator or Z-crank)&lt;/li&gt;&#xA;&lt;li&gt;swash-plate&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are other mechanisms to convert linear motion to rotation?&lt;/p&gt;&#xA;" OwnerUserId="12127" LastActivityDate="2017-08-20T13:17:14.273" Title="Mechanisms to convert linear motion to rotation" Tags="|mechanical-engineering|mechanisms|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16828" PostTypeId="1" CreationDate="2017-08-20T14:16:39.267" Score="2" ViewCount="97" Body="&lt;p&gt;I'm undertaking a civil hydraulics question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It involves the flow of water in an open channel down a river at $5m^3/s$ until the flow is split in two by a weir acting as a sluice gate allowing flow into another stream. A weir pool is located on the upstream side of the gate. $2m^3/s$ flows into a stream not dictated by said weir and I assume we apply conservation laws so that $3m^3/s$ flows into the stream under the gate. It's stated that the gate is set to a certain height and the weir pool height is a function of the flow rate into the other stream. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial question pertained to finding the height of the gate above the river bed, however, the following question states that subsequent flow measurements were taken and indicate that when the gate is set at the calculated height, the flow into the stream under the weir is actually 10% less than what is calculated (I assume $2.7m^3/s$). It then asks, why could that be the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thoughts were that this could be due to frictional effects from the gate and possibly across the river bed, maybe a change in the channel geometry or possibly a change in slope. Have I covered all the possible options here or is there more that could be affecting the flow rate in this situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers &lt;/p&gt;&#xA;" OwnerUserId="11405" LastActivityDate="2021-04-18T20:59:12.957" Title="Difference in flow measurements in an open channel" Tags="|civil-engineering|hydraulics|open-channel-flow|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16829" PostTypeId="1" AcceptedAnswerId="16833" CreationDate="2017-08-20T15:30:39.723" Score="1" ViewCount="288" Body="&lt;p&gt;Please explain me the difference between $\lim_{x-&amp;gt;0}\frac{\partial E}{\partial x}$ and $\lim_{x-&amp;gt;0}dE/dx$.In physics I encountered something similar while reading about Newton's Law of Fluids.While in F.M. White its done using partial derivatives.I want to know the physical difference instead of the highly mathematical one. Having said that I am well conversed with the first principle of derivatives.&lt;/p&gt;&#xA;" OwnerUserId="12558" LastEditorUserId="12558" LastEditDate="2017-08-20T17:52:23.623" LastActivityDate="2017-08-21T15:04:12.383" Title="Difference between partial derivatives and derivatives in physics" Tags="|mechanical-engineering|electrical-engineering|mathematics|experimental-physics|" AnswerCount="2" CommentCount="3" ClosedDate="2017-08-30T03:32:37.143" ContentLicense="CC BY-SA 3.0" />
  <row Id="16830" PostTypeId="1" AcceptedAnswerId="16839" CreationDate="2017-08-20T18:08:45.530" Score="2" ViewCount="13477" Body="&lt;p&gt;I understand that Hydrostatic pressure and Soil pressure exist concurrently. I'm curious how these separate loads interact with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full disclaimer: I am not an engineer nor do I pretend to be one. I recognize that these are very broad &amp;amp; elementary questions; I am at the early stages of my understanding. Please be assured I am not, and will not, attempt to engineer  actual structures. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does the pore size limit, in any way, the surface area available for&#xA;hydrostatic loading on the back of a retaining wall? I assume it&#xA;does not. It would stand to reason that water under pressure would&#xA;suspend the soil particles and push its way to the point where most&#xA;of the surface area on the back of the wall was under a hydrostatic&#xA;load?&lt;/li&gt;&#xA;&lt;li&gt;Soil particules become suspended in water while a hydrostaic load is&#xA;present. Does this (or any other water related factor) affect the&#xA;overall soil load or does it remain the same?&lt;/li&gt;&#xA;&lt;li&gt;Is the total load on the wall simply $\frac{H^2}{2}(\gamma_s K_a + \gamma_w)$ (Where $\gamma_s$ and $\gamma_w$ are the soil and water densities, respectively)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2812" LastEditorUserId="90" LastEditDate="2022-08-03T00:51:18.457" LastActivityDate="2022-08-03T00:51:18.457" Title="Retaining Wall: Saturated Soil &amp; Hydrostatic Pressure" Tags="|pressure|geotechnical-engineering|retaining-wall|soil-mechanics|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16831" PostTypeId="2" ParentId="16830" CreationDate="2017-08-20T19:58:38.193" Score="0" Body="&lt;p&gt;See this link : &lt;a href=&quot;https://books.google.ch/books?id=C7rny3qA6RMC&amp;amp;pg=PA245&amp;amp;lpg=PA245&amp;amp;dq=geology+soil+soaked+in+water&amp;amp;source=bl&amp;amp;ots=itgaiJ0008&amp;amp;sig=2nPX04ETVBeyGzKs-cH5WXPmyDw&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwj_ms7Ry-bVAhWBvBoKHWWfBaIQ6AEIYjAN#v=onepage&amp;amp;q=geology%20soil%20soaked%20in%20water&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://books.google.ch/books?id=C7rny3qA6RMC&amp;amp;pg=PA245&amp;amp;lpg=PA245&amp;amp;dq=geology+soil+soaked+in+water&amp;amp;source=bl&amp;amp;ots=itgaiJ0008&amp;amp;sig=2nPX04ETVBeyGzKs-cH5WXPmyDw&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwj_ms7Ry-bVAhWBvBoKHWWfBaIQ6AEIYjAN#v=onepage&amp;amp;q=geology%20soil%20soaked%20in%20water&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the subject is large...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZaaJU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZaaJU.png&quot; alt=&quot;soil plasticity&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-20T19:58:38.193" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16832" PostTypeId="2" ParentId="12838" CreationDate="2017-08-20T20:03:20.900" Score="5" Body="&lt;p&gt;This is actually two different questions. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mechanisms to convert linear motion to rotational: There are &lt;a href=&quot;https://engineering.stackexchange.com/questions/16827/mechanisms-to-convert-linear-motion-to-rotation&quot;&gt;different mechanisms around&lt;/a&gt;, the most common of all crankshafts.&lt;/li&gt;&#xA;&lt;li&gt;what is the best mechanism for your problem, coupling the opening of the door to the sliding of the can. I don't think you can use a rack-pinion mechanism unless you have pinion with big radius. IMHO the best mechanism for your need should be a 1DOF mechanism mechanism with one prismatic and one revolute joint. The revolute joint will be on the door and the sliding part in front of the can. Something like this:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PRtk9.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PRtk9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12127" LastEditorUserId="12127" LastEditDate="2017-08-20T20:11:00.317" LastActivityDate="2017-08-20T20:11:00.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16833" PostTypeId="2" ParentId="16829" CreationDate="2017-08-20T20:04:13.057" Score="5" Body="&lt;p&gt;For a detailed explanation search &lt;code&gt;WikiPedia&lt;/code&gt; for &lt;code&gt;derivative&lt;/code&gt;, &lt;code&gt;partial derivative&lt;/code&gt; and &lt;code&gt;total derivative&lt;/code&gt;. For a short, non-mathematical, summary see below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partial derivative of a function of several variables is it's derivative with respect to one of those variables, assuming that all other variables are constant. The total derivative does not make this assumption and includes all indirect dependencies to find the overall dependency with respect to the variable of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, consider $\frac{df}{dx}$, the total derivative of the function $f(x,y)$ with respect to the variable $x$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\operatorname df}{\operatorname dx}=\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{\operatorname dy}{\operatorname dx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which depends on both $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ (i.e. the partial derivatives of $f$ with respect to $x$ and $y$).&lt;/p&gt;&#xA;" OwnerUserId="11869" LastActivityDate="2017-08-20T20:04:13.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16834" PostTypeId="1" CreationDate="2017-08-20T20:26:54.807" Score="2" ViewCount="362" Body="&lt;p&gt;Is there a standard that transportation engineers use for the number of vehicles a road should carry daily  for it to be widened to four, six, eight lanes, etc? Curious because I've noticed that some states seem to be unnecessarily widening their interstates to six lanes in rural areas, and others refusing to do so in places that need it.&lt;/p&gt;&#xA;" OwnerUserId="12615" LastActivityDate="2017-08-21T09:11:01.180" Title="What should a road's traffic volume to justify the number of lanes?" Tags="|civil-engineering|transportation|highway-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16835" PostTypeId="1" AcceptedAnswerId="16844" CreationDate="2017-08-20T22:02:21.190" Score="1" ViewCount="639" Body="&lt;p&gt;I know that a 230/400V 3phase motor requires 230V on each coil. No more and no less. So on a 3*230V grid it would be wired in Delta, on a 3*400V it would wired in Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went to evening school to learn about this and my course on the subject has several diagrams on this, most resembling these 2:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7mXXA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7mXXA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Wo4xv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wo4xv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also put this in practice and wired the motors and tested errors etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But thinking about it now I noticed that we never grounded the center point of these connections. How can a Y-config at 3*400V produce 230V on the coil? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a mathematical way of calculating this? It's confusing since there's 3 voltages that are all averaged at 400V but measured at any given moment they're 3 very different voltages connected to eachother. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have however also seen images online of a delta/Y connections that do ground the centerpoint:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/PSCre.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PSCre.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;And here it totally makes sense that the coil get 230V because we know that L-N voltage is sqrt(3) of L-L voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I'm missing something here. &lt;/p&gt;&#xA;" OwnerUserId="12264" LastActivityDate="2017-08-21T13:33:38.463" Title="How does a Y connection in a 3 phase motor create 230V without a neutral/ground" Tags="|electrical-engineering|motors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16836" PostTypeId="1" CreationDate="2017-08-21T00:13:10.110" Score="3" ViewCount="64" Body="&lt;p&gt;I'm looking for any tips on a simple and cheap means of generating a regular spark for ignition of propane in air, could be electrical or other but needs to be simple and fairly reliable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Criteria are : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;needs to remotely ignite/re-ignite a propane jet through a 6 mm OD copper pipe, will be burning in open air (not an enclosed furnace etc) &lt;/li&gt;&#xA;&lt;li&gt;it is a prototype so doesn't need to be packaged as a consumer product and will be closely monitored so needs not be completely foolproof but equally shouldn't be inherently unsafe. &lt;/li&gt;&#xA;&lt;li&gt;Can have bulky remote elements via wires etc but the working end needs to be reasonably compact. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is the prototype &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/zagkIKq17x0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/zagkIKq17x0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="4357" LastEditorUserId="6767" LastEditDate="2018-05-03T17:30:00.333" LastActivityDate="2018-07-03T11:46:30.360" Title="Advice on simple spark ignition system" Tags="|combustion|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16839" PostTypeId="2" ParentId="16830" CreationDate="2017-08-21T03:07:21.863" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XnDL3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XnDL3.jpg&quot; alt=&quot;Retaining Wall&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look on above photo (on your right hand side).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see how the loads are distributed if the water level is less the the retaining wall height.&#xA;The area of the green surface will be the lateral loads on the wall.&#xA;Hope this will give you a basic understanding about distribution of loads.&lt;/p&gt;&#xA;" OwnerUserId="12620" LastActivityDate="2017-08-21T03:07:21.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16840" PostTypeId="1" AcceptedAnswerId="16868" CreationDate="2017-08-21T03:52:08.810" Score="2" ViewCount="1152" Body="&lt;p&gt;(This is probably a simple question, but I've been staring at it so long I'm not sure I know how to do it any more. I have an Excel spreadsheet with like 3 different ways to calculate this, and they all give different results, so I know I've screwed up somewhere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to build a do-it-yourself 1700-degree-Celsius (though will settle for 3000F) electric muffle furnace because I like electricity and playing with fire. I am trying to get an idea of the cost, which means I need to know how thick the insulation need to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, materials rated for those kinds of temperatures tend to both be expensive and have relatively high thermal conductivity values, so I plan to use enough high-temperature insulation to reduce the chamber temperature to about 1300C/2400F, surrounded by cheaper, lower-conductivity insulation to bring the exterior temperature down to the target. I want the inside of the furnace as large as I can affordably make it, so instead of a rule of thumb I'd like some real equations I can play with the parameters of.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The furnace will be of a front-loading box design.&lt;/li&gt;&#xA;&lt;li&gt;The furnace will run on 120V AC, with a target power dissipation of 1200W.&lt;/li&gt;&#xA;&lt;li&gt;For convection calculations, I am assuming a 10 W/m^2K heat transfer coefficient as this will be free-standing. I also assume room temperature is 21C and I'd like to target a maximum of 43C on the exterior surface.&lt;/li&gt;&#xA;&lt;li&gt;My current calculations use a 12&quot; x 12&quot; x 8&quot; furnace chamber, though I'll probably settle for whatever it turns out I can afford.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am considering the following refractory materials as insulation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Kast-O-Lite 30 LI Insulating Castable: Max 3000F/1650C, max ~0.75 W/mK&lt;/li&gt;&#xA;&lt;li&gt;Alumina Fiberboard: Max 3100F/1700C, max ~0.45 W/mK&lt;/li&gt;&#xA;&lt;li&gt;NC-28 Insulating Fire Brick: Max 2800F/1540C, max ~0.44 W/mK&lt;/li&gt;&#xA;&lt;li&gt;NC-26 Insulating Fire Brick: Max 2600F/1430C, max ~0.32 W/mK&lt;/li&gt;&#xA;&lt;li&gt;NC-23 Insulating Fire Brick: Max 2300F/1260C, max ~0.27 W/mK&lt;/li&gt;&#xA;&lt;li&gt;8# Ceramic Fiber Wool: Max 2000F/1100C, max ~0.3 W/mK&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;(I'm thinking about using Satanite, ITC-100HT, and ITC-296A as coating materials, but I can't find data about how much that would improve the conductivity ratings.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also pretty new to this, so any advice at all is appreciated. (Hell, maybe I'm overthinking this, the ratings mean nothing, and I should just put &quot;a bunch&quot; of insulation in an see what happens.) I plan to use MoSi2 elements and a nitrogen atmosphere, and will be sintering various powdered metals, including steel. I'll be slapping a PID controller on it for controlled ramp-soak phases.&lt;/p&gt;&#xA;" OwnerUserId="12619" LastActivityDate="2017-08-24T06:19:53.100" Title="How thick do I need to make the insulation in my homemade furnace?" Tags="|thermodynamics|thermal-conduction|thermal-insulation|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16841" PostTypeId="2" ParentId="16835" CreationDate="2017-08-21T05:26:52.270" Score="0" Body="&lt;p&gt;Simply put a balanced load does not need, but may have, a neutral connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the effective resistance of any supplied phase is its resistance plus the parallel resistance of the other two as the current is shared. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-21T05:26:52.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16842" PostTypeId="2" ParentId="16834" CreationDate="2017-08-21T08:25:55.130" Score="3" Body="&lt;p&gt;&quot;Vehicles per day&quot; is a pretty bad metric to use. Instead you should look at rush hour and see how many vehicles are trying to get through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every lane can carry 1800 vehicles each hour on a freeway. In residential areas that is reduced to 1500. With traffic lights and priority based intersections it depends on how the lights are regulated and how much crossing traffic there is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that: the point where you are stuck in a jam is not the point where you should add a lane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead you should try and find the &lt;strong&gt;actual bottleneck&lt;/strong&gt; and improve that. This bottleneck is where you can start going the speed limit again. This may be an on-ramp where the traffic added saturates the lanes that are there, or an off-ramp where &lt;em&gt;everyone&lt;/em&gt; wants to get off and that is saturating the interchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However just improving that bottleneck will add traffic downstream and probably create a new jam.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2017-08-21T09:11:01.180" LastActivityDate="2017-08-21T09:11:01.180" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16844" PostTypeId="2" ParentId="16835" CreationDate="2017-08-21T13:33:38.463" Score="3" Body="&lt;p&gt;Your understanding so far is correct. If you draw your phasor diagrams to scale you can solve this from basic trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8oA9M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8oA9M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Phasor diagram for delta and star connections&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should, when you have finished, see that the phase to phase voltages are $ \sqrt 3 $ times the phase to star voltages.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But thinking about it now I noticed that we never grounded the center point of these connections. How can a Y-config at 3*400V produce 230V on the coil?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now realise that without a neutral connection the star / wye point will only be in the centre if all three phases are balanced. If one phase is more highly loaded then the star point will be pulled towards that phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that we have a neutral star point at all is that the currents into this point all sum to zero at every instant. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a mathematical way of calculating this? It's confusing since there's 3 voltages that are all averaged at 400V but measured at any given moment they're 3 very different voltages connected to each other.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tend to think graphically and my illustration below may help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wSkKT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wSkKT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 2. (1) The red phase current is the geometric sum of the black and blue. (2) When black and blue are equal and opposite the red must be zero. (3) When black falls to zero the other two must sum to zero.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that helps.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2017-08-21T13:33:38.463" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16846" PostTypeId="2" ParentId="16829" CreationDate="2017-08-21T15:04:12.383" Score="0" Body="&lt;p&gt;Consider a surface &lt;code&gt;S&lt;/code&gt; in 3-space and its gradient &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ {\hat{i} }\frac{\partial{S}}{\partial{x}} + {\hat{j} }\frac{\partial{S}}{\partial{y}} $  . Then each partial represents the slope along the given axis, as though you'd taken a 2-dimensional slice of the surface and looked at the slopeof the line that resulted.&lt;br&gt;&#xA;If you then find the direction (vector in x and y) which maximizes the gradient, you'll find the path of steepest descent. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-08-21T15:04:12.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16847" PostTypeId="1" CreationDate="2017-08-22T05:25:52.677" Score="2" ViewCount="98" Body="&lt;p&gt;There seems to be a very strong trend of the number of wheel lugs on cars and trucks increasing over time. For instance, many small cars in the 80s used 4 lugs, but have since switched to 5 (e.g., Honda Civic, Volkswagen Golf).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More stark examples can be seen in trucks. Consider the M35 series of trucks used by the US Army. The standard model weighs nearly 13,000 lbs empty, and is capable of carrying a load of around 5000 lbs. Its wheels use 6 lugs. (See top photo)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latest version of the Ford F-150, a truck weighing about 4500 lbs with payloads in the 1500-3000 lb range, now has 6 lugs. (See center photo)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, a vehicle more comparable in weight and cargo specs to the M35 is the Ford F-550. It has a Gross Vehicle Weight Rating of 17,500-19,500 lb range. It uses 10 lugs per wheel. (See bottom photo)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WXPAE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WXPAE.jpg&quot; alt=&quot;Photo collage of an M35 variant, a 2017 F-150, and a 2017 F-550.&quot;&gt;&lt;/a&gt;&#xA;(credits: Wikipedia &amp;amp; ford.com)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why the lug inflation? &lt;a href=&quot;https://engineering.stackexchange.com/a/14610&quot;&gt;This answer&lt;/a&gt; states &quot;To some extent, the number of bolt studs depends on the anticipated torque on the wheel.&quot; While it's true that engine output and braking power have increased over the years, have they really increased enough to merit a 66% increase in the number of lugs in this example?&lt;/p&gt;&#xA;" OwnerUserId="12629" LastActivityDate="2017-09-21T09:50:28.930" Title="Why is the wheel lug count on vehicles increasing over time?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16848" PostTypeId="2" ParentId="16847" CreationDate="2017-08-22T09:32:16.293" Score="3" Body="&lt;p&gt;As the diameter and width have increased so has the number of fixing points. Also, as many manufacturers use the same wheels across models then having the same fixings reduces component numbers...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the type of wheel and its material have changed, older wheels tended to be thicker and could therefore use fewer studs.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-22T09:32:16.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16849" PostTypeId="1" CreationDate="2017-08-22T10:31:49.213" Score="0" ViewCount="2461" Body="&lt;p&gt;I am hoping for some help with calculating the size of pipe for a given length.&#xA;The system design is a 10000L supply fills a 5 tonne water tank that then feeds a variable speed pump. The pump sends the water around a 150M loop/ring of pipe then returns to the 5 tonne water tank. The return pipe will be approximately 600mm from the inside base of the tank, so it will be in the water creating a closed loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to move the water at a velocity of above 4.5 m/s to decrease biofilm on the pipe surface. along the 150M loop there will be 4 take off points, 1 of which requires 2000 l/h output, the other 3 require 500 l/h output. So if they were all open at the same time a total of 3500 l/h is required. However in the future the loop/ring will be increased to double its size to 300M with similar outputs, so a total of 7000 l/h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the pipe size for:&#xA;A - 3500 l/h @ 150M lenght (with bends)&#xA;B - 7000 l/h @ 300M length (with bends)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also once A is calculated, would increasing the pump speed and pressure increase the flow rate to achieve B?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the math will be a bit beyond me, so please can you go gentle with me.&lt;/p&gt;&#xA;" OwnerUserId="12633" LastEditorUserId="8055" LastEditDate="2017-08-22T21:21:08.947" LastActivityDate="2017-11-21T20:26:16.313" Title="Pipe diameter, Flow Rate, Pressure &amp; Velocity for water" Tags="|pressure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16851" PostTypeId="2" ParentId="16849" CreationDate="2017-08-22T13:56:09.537" Score="1" Body="&lt;p&gt;This is a complicated question - it may be useful to add a diagram if you can, since each output and input to the pipe system will influence your question's answer, and by reading we may miss one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution to problems like yours (where we know our requirements about the inputs and outputs of a pipe system) are often solved using the Hardy Cross method: &lt;a href=&quot;https://en.wikipedia.org/wiki/Hardy_Cross_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Hardy_Cross_method&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We set each pipe input and output with their fixed conditions and solve for the flow inside the &quot;network&quot; of pipes using the Darcy-Weisbach equation and some initial guesses. Then we vary the non-fixed conditions until the solution converges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the Darcy-Weisbach equation for a simple one-input one-output pipe system but you must use something more like Hardy-Cross for a complex system as the one you've suggested. The Darcy-Weisbach equation essentially relates flow speed ($v$) to pressure loss ($\delta_{P}$) and characteristic length of the pipe ($L$, $D$):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{\delta_{P}}{L} = F_d \rho \left(\frac{v^2}{2 D}\right)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\delta_P$ - pressure loss [$ML^{-1}T^2$]&lt;/li&gt;&#xA;&lt;li&gt;$L$ - unit length of pipe [$L$]&lt;/li&gt;&#xA;&lt;li&gt;$F_d$ - friction factor [$1$]&lt;/li&gt;&#xA;&lt;li&gt;$\rho$ - density of fluid [$ML^{-3}$]&lt;/li&gt;&#xA;&lt;li&gt;$v$ - speed of fluid [$LT^2$]&lt;/li&gt;&#xA;&lt;li&gt;$D$ - diameter of pipe [$L$]&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So you can see from this that since you know your required flow rates, you will perhaps need to vary your upstream pressure (by changing your pump) or your pipe sizes / friction factors to achieve the velocities you want. And you'll definitely need to know your upstream pressure at your pump or reservoir! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you will provide an initial guess of your pipe size and friction factors and vary them until you get a network that has continuity of flow (i.e. mass conservation). For complex pipe shapes (such as bends, junctions, etc) you may need to look into using the resistance coefficient method (also known as the $k$-value method) to find your pressure loss. There are very good sets of experimental fits for loss factors in Robert Blevin's &quot;Applied Fluid Dynamics Handbook&quot; which I highly suggest you pick up if you commonly encounter these kinds of problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Description of the $k$-value method can be found here: &lt;a href=&quot;https://neutrium.net/fluid_flow/pressure-loss-from-fittings-excess-head-k-method/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://neutrium.net/fluid_flow/pressure-loss-from-fittings-excess-head-k-method/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good reference would be the Crane TP-410 publication on pipe flow. The math isn't super hard for pipe flow, but doing complicated pipe networks can be tricky.&lt;/p&gt;&#xA;" OwnerUserId="3583" LastEditorUserId="3583" LastEditDate="2017-08-23T18:26:52.623" LastActivityDate="2017-08-23T18:26:52.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16852" PostTypeId="1" AcceptedAnswerId="16853" CreationDate="2017-08-22T14:49:31.697" Score="1" ViewCount="3229" Body="&lt;p&gt;I need to sketch some diagrams involving basic symbols like springs and dampers, specifying angles, perpendicularity etc... something like the images below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d5Up7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d5Up7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q1AkE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q1AkE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously my first option was word, then power point, the visio, in which the latter was a bit more productive, but I haven't found any other software that will let me do the diagrams more productively and have an elegant professional look. &lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2017-08-23T11:20:57.473" Title="What software is used for drawing dynamic &amp; mechanical diagrams?" Tags="|mechanical-engineering|design|dynamics|diagram|technical-drawing|" AnswerCount="1" CommentCount="2" ClosedDate="2017-08-23T11:51:49.110" ContentLicense="CC BY-SA 3.0" />
  <row Id="16853" PostTypeId="2" ParentId="16852" CreationDate="2017-08-22T16:37:18.300" Score="4" Body="&lt;p&gt;This is more of a long comment than an answer as i can not advice any specific software. First, if you intend to do anything professional in print or web productions than yous shouldn't be looking in direction of Microsoft for anything. Much abused does not mean any good.&lt;/p&gt;&#xA;&lt;p&gt;First tier would be to use direct vector drawing apps. In this category you have:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Illustrator&lt;/li&gt;&#xA;&lt;li&gt;Autocad&lt;/li&gt;&#xA;&lt;li&gt;Inkscape&lt;/li&gt;&#xA;&lt;li&gt;Corel draw etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;They are more than capable of doing these.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kLmTR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kLmTR.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: 12 minutes of playing with Illustrator, and that's about all I have time for.&lt;/p&gt;&#xA;&lt;p&gt;You can also use pacakges designed to do programmatic input like Tex and TiKz for example. Also &lt;strong&gt;Mathematica&lt;/strong&gt; works quite well for both images, and you can animate in it.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-08-23T08:07:17.777" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16854" PostTypeId="1" AcceptedAnswerId="16856" CreationDate="2017-08-22T18:22:00.203" Score="0" ViewCount="917" Body="&lt;p&gt;I&quot;m working on a design and having difficulty calculating possible misalignment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a shaft (diameter D, tolerance t) that is running through two bushings. The two bushings are separated by a distance L. The two bushings are of width W1 and W2 (measured along the axis parallel to the shaft), with internal diameter of D1 and D2 with tolerances of T1 and T2. (see image below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DVCJC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DVCJC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about calculating the potential misalignment of the shaft? When I started this I though it was going to be a quick 5 minute challenge but the more I work on it the more difficult it's seeming.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-11-21T11:18:01.123" Title="Bushing and Shaft Alignment" Tags="|mechanical-engineering|tolerance|" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="16855" PostTypeId="1" CreationDate="2017-08-22T22:16:47.570" Score="3" ViewCount="60" Body="&lt;p&gt;I have a setup with 4 cylinders placed next to each other in following configuration. The cylinders are made of .2mm stainless steel. I would ideally want to rotate them along their axis using a single actuator/motor. What would be the best mechanism to do so? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to keep the cylinders as close to each other, due to space constraints. Any references would also help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iGhuH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iGhuH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12071" LastEditorUserId="12071" LastEditDate="2017-08-23T06:47:03.887" LastActivityDate="2017-08-23T06:47:03.887" Title="Rotating four cylinders with one motor/actuator" Tags="|motors|mechanisms|bearings|kinematics|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16856" PostTypeId="2" ParentId="16854" CreationDate="2017-08-23T08:03:15.880" Score="0" Body="&lt;p&gt;I would like to point out that your CAD application can most likely do this for you. Just make a parametric drawing and sweep the values you need. This may be more practical as you go forward since the cad application can effortlessly make changes based on geometric constraints and you don't need to calculate this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E1pM9.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E1pM9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: Parametric sweep of one value. No harder than typing 3 values. Note the drawing s intentionally exaggerated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the sweep by script is not much harder. This opens up statistical analysis by montecarlo simulation. Doing it this way reduces chance of error made during the mathematical analysis. Of course, this is both a good thing and a bad thing. Analytical analysis can give insights into global minima, but as the complexity of your analysis grows this might not be feasible so a local minima may suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yeah you can calculate this by hand too, the problem is that one you start adding the things you now consider trivial this may no longer be a trivial extension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N19CA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N19CA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 2&lt;/strong&gt;: Vector expression&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can express this as a equation of vector expressions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;  \vec a + \vec b + \vec c = 0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\vec a$ is known, magnitude of $\vec b$ is known and direction of $\vec c$ is linked to vector  $\vec b$. So vector  $\vec b$ is $D \cdot \{sin(\theta), cos(\theta)\}$ therefore  $\vec c$ is $x \cdot\{-cos(\theta), sin(\theta)\}$. Since you have 2 directions and 2 unknowns this is solvable. But i wont take this further.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-08-23T09:27:16.050" LastActivityDate="2017-08-23T09:27:16.050" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16857" PostTypeId="1" AcceptedAnswerId="16858" CreationDate="2017-08-23T10:49:22.653" Score="1" ViewCount="3014" Body="&lt;p&gt;I am trying to figure out am I doing this calculation correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the maximum shear stress for a solid cylindrical pipe I need two formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment of inertia = pi/2*r^4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure what this formula is called but its:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;max = T*r/ Moment of inertia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to calculate these numbers for a solid shaft that is 20mm in diameter with a torque of 1255nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1255*0.02/((pi/2)*0.02^4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pascals = 99869726.7902&#xA;MegaPascals = 99.87&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material that I have has a yield strength of 490 MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the calculation saying that even at 20mm radius the stress will be below the 490MPa that would cause this pipe fail elastically?&lt;/p&gt;&#xA;" OwnerUserId="12586" LastEditorUserId="615" LastEditDate="2017-08-23T13:53:38.870" LastActivityDate="2020-08-05T16:31:17.237" Title="Maximum torsion shear stress in cylinder" Tags="|mechanical-engineering|applied-mechanics|modeling|moments|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16858" PostTypeId="2" ParentId="16857" CreationDate="2017-08-23T11:06:29.657" Score="2" Body="&lt;p&gt;I have not repeated your maths, but the formulae are shown in the image. If you have the numbers correct the you have a shaft that will support nearly 5 times the load you are applying... &#xA;&lt;a href=&quot;https://i.stack.imgur.com/zgYg4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zgYg4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source : &lt;a href=&quot;http://www.engineersedge.com/material_science/shear_stress_in_shafts_13117.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineersedge.com/material_science/shear_stress_in_shafts_13117.htm&lt;/a&gt;&#xA;where you can enter your values and have them confirmed.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-23T11:06:29.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16861" PostTypeId="2" ParentId="16823" CreationDate="2017-08-23T12:49:33.770" Score="1" Body="&lt;p&gt;Andrew has provided some good information with regard the formula for shear stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I disagree with is the standard yield strength ($\sigma_y$) values for a medium carbon steel.  I think Andrew has forwarded UTS values.  For this I think you should be working in the $275MPa$ for low strength, to $450MPa$ for high strength range, with most structural steel being $355MPa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add apply this to your query, you have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$T = 1256 N m$ (torsional moment)&lt;/li&gt;&#xA;&lt;li&gt;$r = 0.02 m$ (radius)&lt;/li&gt;&#xA;&lt;li&gt;$J = 25.133 cm^4$ (polar moment of area)&lt;/li&gt;&#xA;&lt;li&gt;$\tau_{allow} = \frac{\sigma_y}{2} $&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, if the calculation results in $\tau_{max} &amp;gt; \tau_{allow}$ then you know you are in trouble regarding any deformations / damage to the shaft.&lt;/p&gt;&#xA;" OwnerUserId="8212" LastActivityDate="2017-08-23T12:49:33.770" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16862" PostTypeId="1" CreationDate="2017-08-23T18:21:05.687" Score="1" ViewCount="232" Body="&lt;p&gt;I'm trying to conduct a thermo-mechanical analysis in ansys workbench, I've been able to obtain some results using the transient thermal and static structural modules, however, I would like to perform a more accurate simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until this moment I just have reach a linear solution for the static structural module, as shown in the next images,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v1DUc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v1DUc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LD0w4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LD0w4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I get the solution from the thermal module I import it to the static structural module and solve, but there are some elements that cross each other, for example, inside the leaf of the door there are some stiffeners that cross the frontal door plate. Also the leaf interferes with the frame of the door in various locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I fix this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to use the static structural module with the large deflection set &quot;on&quot;, but I'm new to this kind of analysis and I'm not been able to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can send the files if it's needed.&lt;/p&gt;&#xA;" OwnerUserId="12649" LastEditorUserId="1832" LastEditDate="2017-08-23T18:27:17.493" LastActivityDate="2017-08-23T18:27:17.493" Title="Nonlinear simulation in ANSYS workbench" Tags="|applied-mechanics|simulation|ansys-workbench|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16863" PostTypeId="1" CreationDate="2017-08-23T19:16:30.017" Score="2" ViewCount="7642" Body="&lt;p&gt;When I look at the yield of one material then you find more than one yield for different thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why the yield changes with changing the thickness. is there an equation to prove this ? &lt;/p&gt;&#xA;" OwnerUserId="12650" LastEditorUserId="16486" LastEditDate="2018-11-23T08:20:12.277" LastActivityDate="2020-09-12T00:53:17.420" Title="Why does the yield strength of steel go up if I reduce the thickness of the specimen?" Tags="|mechanical-engineering|materials|yield-point|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="16864" PostTypeId="1" CreationDate="2017-08-23T20:32:29.907" Score="3" ViewCount="1725" Body="&lt;p&gt;I'm designing a product to accept a 5-gallon water jug. This is the light blue, transparent plastic kind for potable water that you can buy, swap, or fill at grocery stores. I can take measurements of jugs I own, but I'd like to ensure the product will accept any jug conforming to the standard. Does such a standard exist, particularly for the mating surface of the cooler?&lt;/p&gt;&#xA;" OwnerUserId="12651" LastActivityDate="2017-08-24T05:05:51.177" Title="Industry standard dimensions for a 5-gallon water cooler?" Tags="|mechanical-engineering|design|product-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16866" PostTypeId="2" ParentId="2964" CreationDate="2017-08-24T02:27:12.890" Score="0" Body="&lt;p&gt;The starting torque is the  torque required by the starting motor to rotate from idle position during the engine cranking.This torque must be high so that the starting motor should rotate.&lt;/p&gt;&#xA;" OwnerUserId="12656" LastActivityDate="2017-08-24T02:27:12.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16867" PostTypeId="2" ParentId="16864" CreationDate="2017-08-24T04:57:11.590" Score="1" Body="&lt;p&gt;I did some searching and did not find a standard. Majority of these bottles are used inverted on a water cooler and seal via the container holding vacuum (like an inverted dog watering dish) so the dimensions are not critical (and likely not held to a tight standard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I own two different standards of thread on type 5 gallon jugs that I use in a &lt;a href=&quot;https://primowater.com/dispensers-list/bl-cooler-pewter-hotcold-817206010890/&quot; rel=&quot;nofollow noreferrer&quot;&gt;bottom loaded pump style water cooler&lt;/a&gt;. Since a hose pumps the water out, the interface is again not critical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would recommend is contacting some large manufacturers and ask them for drawings or critical dimensions. You wont get a response for all of them, but some will be interested in a product being compatible with their bottles. They may also have information on any standard if it exists. If push comes to shove, you may have to ask for a quote on 50,000 bottles annually and require a detailed schematic with the quote. Tricks you learn ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.oberk.com/containers/plastic-bottle/1002680-5gallon-clear-aqualite-water-bottle-with-handle-round-pvc-with-55mm-neck&quot; rel=&quot;nofollow noreferrer&quot;&gt;oberk.com&lt;/a&gt;&#xA;&lt;br&gt;&lt;a href=&quot;http://www.napcor.com/PET/manufacturers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;napcor.com&lt;/a&gt; &#xA;&lt;br&gt;&lt;a href=&quot;http://www.parker-plastics.com/5-gal-water/&quot; rel=&quot;nofollow noreferrer&quot;&gt;parker-plastics.com&lt;/a&gt; &#xA;&lt;br&gt;&lt;a href=&quot;http://sterlingbottles.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;sterlingbottles.com&lt;/a&gt; &#xA;&lt;br&gt;&lt;a href=&quot;https://www.alibaba.com/showroom/5-gallon-water-bottle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;alibaba.com&lt;/a&gt; &#xA;&lt;br&gt;&lt;a href=&quot;http://www.canamplastics.com/productpages/5gallon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;canamplastics.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2017-08-24T05:05:51.177" LastActivityDate="2017-08-24T05:05:51.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16868" PostTypeId="2" ParentId="16840" CreationDate="2017-08-24T06:19:53.100" Score="2" Body="&lt;p&gt;First off, the design problem is not trivial and it looks like you have done your homework. On high temperature, small vessels like this you need to consider the 3 dimensional transfer of heat (the wall area is increasing as you move out). Or you can be conservative and much more easily calculate it with all the insulation acting at the outside dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets say you want a 6&quot;x6&quot;x6&quot; firebox with 6 inches (0.152m) of insulation surrounding. Your external surface area of the 18&quot;x18&quot;x18&quot; box would be 18*18*6walls = 5184in^2 or 3.4m^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a 1700-20 =1680C differential and 0.4 W/mK&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_conductivity#Equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thermal Conductivity Equation&lt;/a&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1680K * (3.4m^2 / 0.152m) * 0.4 W/mK = 15kW worst case&#xA;(Notice how units cancel)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check ourselves, this &lt;a href=&quot;http://sentrotech.wpengine.com/wp-content/uploads/2015/03/1700C-666-Data-Sheet.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;commercially available 6&quot;x6&quot;x6&quot; unit&lt;/a&gt; uses 4kW. So our numbers were probably a bit too conservative. Calculating it at the 3/4 thickness (14x14x14) gives us 0.76m^2 and 3.4kW. This is likely pretty close. The manufacturer will make sure their amount of heating is some amount greater than the heat loss at maximum temperature so the max temperature can be reached in a reasonable amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radiation losses will also become significant at 1700C. A stainless or aluminum reflective shield should be used for your outer layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stacking different rated materials is possible, by running a duplicate calculation (furnace in a furnace). Make sure to have &lt;strong&gt;lots&lt;/strong&gt; of safety factor on the maximum material temperature, as those conductivity ratings are not always very close. They will often provide you &quot;at least&quot; the value they are publishing. More insulation is better for most customers, but if you are counting on that external oven to only get to a certain temperature; too high of thermal resistance will result in that material over heating and failing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, a good test of what is possible in a DIY type environment is to look at what is &lt;a href=&quot;http://www.mtixtl.com/1700CCompactMuffleFurnace4.7x4.7x4.71.7L-KSL-1700X-S.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;commercially available&lt;/a&gt;. Look at the watts vs insulation vs firebox size of what vendors are selling. Then remember they do this for a living so your diy design (while much cheaper for you) will at best require 50% more power, 50% more insulation, or achieve say 200F less temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds like a fun and challenging project! Targeting a &lt;a href=&quot;https://www.youtube.com/watch?v=M5oKvHktmNU&quot; rel=&quot;nofollow noreferrer&quot;&gt;1300C furnance&lt;/a&gt; might be a better starting point if you have not built any high temperature devices before. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-08-24T06:19:53.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16869" PostTypeId="1" CreationDate="2017-08-24T11:39:12.823" Score="-2" ViewCount="52" Body="&lt;p&gt;Petrol powered cars are sometimes described as having an &quot;Internal Combustion Engine&quot; (ICE). This describes the chemical processes happening inside the engine: mainly the combustion of petrol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could electric cars be described with the word &quot;Internal Ionisation Engine&quot; (IIE). This would describe the chemical processes happening inside the batteries: oxidation/reduction and movements of ions.&lt;/p&gt;&#xA;" OwnerUserId="12657" LastActivityDate="2017-08-24T11:55:15.207" Title="Could Internal Ionisation Engine be a description of electric cars?" Tags="|automotive-engineering|energy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16870" PostTypeId="2" ParentId="16869" CreationDate="2017-08-24T11:55:15.207" Score="4" Body="&lt;p&gt;It's a very poor analogy, especially in presence of &lt;a href=&quot;https://en.wikipedia.org/wiki/Ion_thruster&quot; rel=&quot;nofollow noreferrer&quot;&gt;ion engines&lt;/a&gt; used for space travel, where the propellant is ionized inside the engine, then accelerated outwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides, the electric cars are based on standard electric motors - all the ion transmission happens in batteries, which are not a part of the engine. The engine is based upon rotating (or switching) induced magnetic field. Nothing &quot;ion&quot; to it.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-08-24T11:55:15.207" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16872" PostTypeId="2" ParentId="16438" CreationDate="2017-08-24T15:19:09.233" Score="1" Body="&lt;p&gt;I suspect that moment of inertia terms is that of the rotating parts of &lt;em&gt;the motor itself&lt;/em&gt;, note that it disappears when angular velocity is constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't make much sense to model the external load on the motor as a moment of inertia as many common types of load are essentially friction of one sort of another. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-08-24T15:19:09.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16873" PostTypeId="1" AcceptedAnswerId="16889" CreationDate="2017-08-24T15:30:59.660" Score="2" ViewCount="1363" Body="&lt;p&gt;I have a shape that I'd like to calculate the polar moment of inertia for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tKlv5m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tKlv5m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the diameter is 600mm; the hexagon in the middle sides are 70mm each; the holes around the hexagon are 130mm from the center of the disk and they are 16mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\pi D^4}{32} - \frac{5\sqrt3}{8}s^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I remove the 5 holes from the calculation? Is that even something that's done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT]&#xA;The disk is 30mm thickness.&#xA;The mass of the disk is  66.5Kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the small holes is total 0.3Kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hexagon mass is 3Kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;weight without holes is 66Kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mass with the circular cutouts is 66.5-0.3 = 66.2kg&#xA;mass with the hexagon cutout is 66.5-3 = 63.5Kg&#xA;mass with all the cutouts is 63.3Kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which percent change in weight would help calculate the polar moment of inertia for this shape?&lt;/p&gt;&#xA;" OwnerUserId="12586" LastEditorUserId="12586" LastEditDate="2017-08-25T10:40:05.550" LastActivityDate="2017-08-26T12:09:41.917" Title="polar moment of inertia for compound shape" Tags="|mechanical-engineering|applied-mechanics|modeling|moments|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16875" PostTypeId="1" AcceptedAnswerId="16881" CreationDate="2017-08-24T17:36:37.983" Score="0" ViewCount="541" Body="&lt;p&gt;I am working with a normalized ball-screw drive where the nut has holes for screws to screw onto the moving part. The catalogue says these holes have diameter 13.5 mm. What is the appropriate screw size (metric) to use on such a part?&lt;/p&gt;&#xA;" OwnerUserId="8590" LastEditorDisplayName="user16" LastEditDate="2017-08-24T23:11:43.960" LastActivityDate="2017-08-25T02:16:42.573" Title="Hole diameter vs Screw Diameter" Tags="|design|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16876" PostTypeId="1" CreationDate="2017-08-24T18:07:55.357" Score="3" ViewCount="3254" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;After passing 9,000 metres (30,000 ft), one of the outer Plexiglas window panes cracked, shaking the entire vessel.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Project_Nekton&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Wikipedia)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why did the first layer of plexiglass Window panes crack on the Bathyscaphe Trieste while descending down to the Mariana Trench? Weren't they tested under such amounts of pressure? As far as I know, plexiglass can hold pressure of up to 20,000 PSI before it can actually crack.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Related cross-post: &lt;a href=&quot;https://physics.stackexchange.com/questions/353497/why-did-the-plexiglass-on-the-bathyscaphe-trieste-crack&quot;&gt;https://physics.stackexchange.com/questions/353497/why-did-the-plexiglass-on-the-bathyscaphe-trieste-crack&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12660" LastEditorUserId="12661" LastEditDate="2017-08-24T19:46:28.300" LastActivityDate="2017-08-24T19:46:28.300" Title="Why did the Plexiglass on the Bathyscaphe Trieste crack?" Tags="|pressure|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16877" PostTypeId="1" CreationDate="2017-08-24T19:07:17.100" Score="3" ViewCount="698" Body="&lt;p&gt;I had been told that a silt soil, when professionally compacted (from all the laboratory essays to the compaction with a vibratory roller), ends hard as a rock, and when hit with a metal bar it makes a metallic noise.&lt;/p&gt;&#xA;&lt;p&gt;I find that hard to believe. Is it true? What is the common reality?&lt;/p&gt;&#xA;&lt;p&gt;I mean the question from the viewpoint of a earth pavement construction, or as a sub base for another type of pavement.&lt;/p&gt;&#xA;&lt;p&gt;I have interest in the answer from the viewpoint of a fast checking of the quality of the compacted soil (before any essay to check the level of compaction)&lt;/p&gt;&#xA;" OwnerUserId="12663" LastEditorUserId="90" LastEditDate="2022-08-03T00:31:05.820" LastActivityDate="2023-08-01T16:15:29.770" Title="Is it possible for a silt soil to be compacted hard as a rock?" Tags="|civil-engineering|geotechnical-engineering|soil|soil-mechanics|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="16878" PostTypeId="5" CreationDate="2017-08-24T19:18:07.717" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-08-24T19:18:07.717" LastActivityDate="2017-08-24T19:18:07.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16879" PostTypeId="4" CreationDate="2017-08-24T19:18:07.717" Score="0" Body="Questions related to the use of soils in engineering, as a material, or as object of research.&#xD;&#xA;&#xD;&#xA;Some common uses in Civil engineering are as surface for water flow, for foundations, as part of pavement structure, as seepage medium, or as origin of loads over other structures.&#xD;&#xA;&#xD;&#xA;Other uses in agronomic engineering, or biology, are as medium for storage of water, cultivation, as an erosionable agent, as a souce of receiver of contaminants." OwnerUserId="12663" LastEditorUserId="12663" LastEditDate="2017-08-25T02:29:57.820" LastActivityDate="2017-08-25T02:29:57.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16880" PostTypeId="2" ParentId="16876" CreationDate="2017-08-24T19:21:21.620" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;http://www.plexiglas.com/export/sites/plexiglas/.content/medias/downloads/sheet-docs/plexiglas-general-information-and-physical-properties.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this datasheet&lt;/a&gt; (Table 12, page 10), modern Plexiglass has the following properties:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Tensile strength (0.25&quot; specimen-0.2&quot;/min)&#xA;--------------------------+---------+------------&#xA;Maximum psi               | 10,500  | 10,200&#xA;Rupture psi               | 10,500  | 10,200&#xA;Elongation, maximum %     | 4.9     | 4.5&#xA;Elongation, rupture %     | 4.9     | 4.5&#xA;Modulus of elasticity psi | 450,000 | 450,000&#xA;Poisson’s ratio           | 0.35    | —&#xA;--------------------------+---------+------------&#xA;Flexural strength (span depth ratio 16, 0.1&quot;/min)&#xA;--------------------------+---------+------------&#xA;Maximum psi               | 16,000  | 15,000&#xA;Rupture psi               | 16,000  | 15,000&#xA;Deflection, maximum in    | 0.6     | 0.5&#xA;Deflection, rupture in    | 0.6     | 0.5&#xA;Modulus of elasticity psi | 450,000 | 450,000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The datasheet also has the following statement (page 5):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Design stresses&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Plexiglas sheet has good tensile and flexural strength properties. However, stress considerably below the values shown in Table 12 (page 10) will produce light surface cracks known as crazing.To avoid stress-crazing, design limits for continuously imposed loads should not exceed 1,500 psi for Plexiglas G sheet and 750 psi for Plexiglas MC sheet. Stresses of greater magnitude but short duration will not generally cause stress-crazing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Whether the four or so hours it took to reach 9000 meters is enough to cause stress-crazing, I can't say, but maybe that was it. Given that the cracking was apparently violent enough to shake the craft, I doubt it, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a square Plexiglass sheet with a span/depth ratio of 16 will fail at 16000&amp;nbsp;psi. What isn't clear to me, however, is what precisely is meant by that. Is that the ultimate applied load or the maximum internal stress it can withstand? I'd guess it's the applied load, but it isn't clear (to me, at least) from the datasheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we (or at least I) have no idea what the span/depth ratio was in the Trieste, nor do we know if perhaps Plexiglass was weaker back then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how it wasn't tested for such loads, it almost certainly was. One possible explanation would be a manufacturing or testing failure, where the conditions in the tests were different in some slight but significant way, or where the tests actually unknowingly weakened the bathyscaphe.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-08-24T19:21:21.620" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16881" PostTypeId="2" ParentId="16875" CreationDate="2017-08-25T02:16:42.573" Score="2" Body="&lt;p&gt;I would suggest you use the largest diameter bolt you can find that fits through the hole, and that you can be sure you'll be able to install properly. By &quot;install properly&quot; I mean that you leave enough clearance between the bolt and 13.5mm hole to account for misplacement of the holes in whatever you're mounting the nut to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my work, when I design a clearance hole for a bolt, I usually go 0.5mm above the nominal size of a bolt, and we feel comfortable going down to 0.2mm above nominal if we need a more snug feet. A 13.5mm hole for a 12mm bolt is a lot of slop, so I would suggest a small washer if you go this route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have imperial sizing available, a 1/2&quot; bolt will have a nominal diameter of 12.7mm, so that would probably actually be your best bet. 13mm bolts do exist, but they're not common. &lt;/p&gt;&#xA;" OwnerUserId="41" LastActivityDate="2017-08-25T02:16:42.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16882" PostTypeId="1" CreationDate="2017-08-25T02:20:42.497" Score="5" ViewCount="70" Body="&lt;p&gt;How does one consider the moment capacity of a combined pipe plus H-section if the infill material is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sand, friction angle 30 degree&lt;/li&gt;&#xA;&lt;li&gt;Grout, C20&lt;/li&gt;&#xA;&lt;li&gt;Concrete, C45&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h0lLR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h0lLR.jpg&quot; alt=&quot;Pipe with H-section&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eurocode has a section on composite design, which requires use of shear connectors. Where shear connectors are not feasible due to lack of space, can bearing of infill material be relied on for moment transfer? For example, what would be the moment capacities of a 600mm outer diameter pipe section with 10mm thick wall containing a 300x300mm H-section with 25mm thick flange and 16mm thick web, both grade S355 steel, based on each of the three infill materials?&lt;/p&gt;&#xA;" OwnerUserId="11443" LastEditorUserId="33" LastEditDate="2018-08-24T18:56:31.360" LastActivityDate="2018-09-23T19:00:28.440" Title="How does infill material affects the capacity of a tubular section with H-section inside?" Tags="|structural-engineering|civil-engineering|steel|building-design|concrete|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16883" PostTypeId="1" CreationDate="2017-08-25T02:49:26.770" Score="-1" ViewCount="57" Body="&lt;p&gt;while representing capacitors (1/jwc) and inductors (jwl).&#xA;Here generally, j- complex number. Then what is the use for representing with j of these components?&#xA;what is the link to use it?&lt;/p&gt;&#xA;" OwnerUserId="12592" LastActivityDate="2017-08-27T13:13:51.287" Title="Why complex functions are used to represent some passive components in an electric circuits?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16885" PostTypeId="1" CreationDate="2017-08-25T13:44:08.310" Score="4" ViewCount="1038" Body="&lt;p&gt;Are there any applications in which a Nyloc nut is preferred to a proper cap nut?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ko0Gz.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ko0Gz.png&quot; alt=&quot;Nyloc nuts&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/1GSJs.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1GSJs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As best I can tell: a Nyloc nut is just a specialized cap nut that will fail under lower torque and temperatures than a metal cap nut.&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2017-08-25T17:48:18.863" Title="When are Nyloc nuts indicated instead of cap nuts?" Tags="|fasteners|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16886" PostTypeId="2" ParentId="16885" CreationDate="2017-08-25T13:55:41.737" Score="5" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Nyloc_nut&quot; rel=&quot;noreferrer&quot;&gt;Nyloc nut&lt;/a&gt; is a form of &lt;a href=&quot;https://en.wikipedia.org/wiki/Locknut&quot; rel=&quot;noreferrer&quot;&gt;locknut&lt;/a&gt;.  The purpose is to prevent unintended loosening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Acorn_nut&quot; rel=&quot;noreferrer&quot;&gt;cap nut&lt;/a&gt; is not necessarily a locknut.  You can get cap nuts that have some form of locknut on the threads; but in general this is not a standard feature of all cap nuts.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-08-25T13:55:41.737" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16887" PostTypeId="2" ParentId="16885" CreationDate="2017-08-25T15:40:07.773" Score="7" Body="&lt;p&gt;Not all cap nuts are self locking, some are, although they can be a bit weird looking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nylocs are generally the first choice for general use as they provide good resistance to loosening due to vibration or flexing of the joint and will stay in position on a thread even when not under tension (unlike spring washers). Unlike locking adhesives they can be removed and adjusted without  losing their locking properties and can, to some extent be reused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are certainly many alternative types of lock-nut which may be preferred for specific applications for example structural applications may use washers which bite into the joint, there are a whole variety of thread sealing adhesives, spring and star washers have applications and you can also get nuts with asymmetric threads which exert a radial load on the thread. all of these have their various advantages but nylocs are pretty ubiquitous for general use whre you just want something chap simple and reliable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So typically they would be used for general engineering assemblies which aren't economical to analyse in exhaustive detail and need to be made to a price. Obviously for high temperature applications nylocs aren't suitable but that is usually reasonably obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cap nuts are generally used to prevent having an exposed thread protruding on the nut side of the joint this generally looks neater, reduces the risk of snagging and can help prevent the nut from seizing due to corrosion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The major downside of cap nuts is that the bolt or stud needs to be within a few mm of the exact length required and there are a number of applications where they cannot be used eg if you need a nut some way along the length of a stud or need a reasonable range of adjustment. Also if you get the bolt length wrong you can end up with an apparently tight nut which isn't exerting the expected compression on the joint. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-08-25T17:48:18.863" LastActivityDate="2017-08-25T17:48:18.863" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16888" PostTypeId="2" ParentId="16822" CreationDate="2017-08-25T18:26:02.830" Score="0" Body="&lt;p&gt;I managed to find a turnbuckle that had thread that fit standard threaded rod. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Turnbuckles seem to often have non-standard thread. Maybe the thread is tighter for the heavy loads they sustain (relative to the length of thread)?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used a turnbuckle and replaced one of the hooks with a threaded rod (bent in a hook shape). All parts have 3/8&quot; thread. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to do the job well enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yHZ31.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yHZ31.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VH480.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VH480.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/jW3tW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jW3tW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10536" LastEditorUserId="10536" LastEditDate="2017-08-26T14:07:36.613" LastActivityDate="2017-08-26T14:07:36.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16889" PostTypeId="2" ParentId="16873" CreationDate="2017-08-25T21:05:19.093" Score="4" Body="&lt;p&gt;Without looking up all the formula's for you, the approach to this problem is rather simple.  Find the polar moment of a solid cylinder, and subtract off the &#xA;polar moment of the holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the  off center holes you need to use the parallel axis therem, so you will have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_\text{assembly} = I_\text{solid cylinder} - I_\text{hexagon} - 6 (I_\text{small cylinder} + m r^2 )$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the expressions you need are all on &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_second_moments_of_area&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2017-08-26T12:09:41.917" LastActivityDate="2017-08-26T12:09:41.917" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16890" PostTypeId="1" CreationDate="2017-08-26T00:05:09.807" Score="1" ViewCount="87" Body="&lt;p&gt;How can I calculate field strength of a Neodymium Ring magnet at an angle?&#xA;I have a Neodymium Ring magnet, and I rotate the magnet around an axis perpendicular to the poles. (Not continuous rotation). For a given distance from the surface, and or for a given angle away from the center of the field, how can I calculate the field strength? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at some methods to calculate field strength, but they refer to electro-magnets, and current. Where this is a rare earth magnet, I am unsure how to perform such calculations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The application is a device that exerts a magnetic field into a device, where either a weak, or non-existent field is not already present. &lt;/p&gt;&#xA;" OwnerUserId="4282" LastActivityDate="2017-08-26T00:05:09.807" Title="How can I calculate field strength of a Neodymium Ring magnet at an angle?" Tags="|magnets|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16891" PostTypeId="2" ParentId="15013" CreationDate="2017-08-26T06:37:25.817" Score="0" Body="&lt;p&gt;I don't think the maximum value on span 2 is correct, it should be 8.44 and not 18.47 &#xA;Check it again. Result are correct below based on 3 moments method&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oTN1M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oTN1M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12620" LastEditorUserId="12620" LastEditDate="2017-08-26T07:34:58.927" LastActivityDate="2017-08-26T07:34:58.927" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16893" PostTypeId="1" CreationDate="2017-08-27T02:49:14.787" Score="3" ViewCount="78" Body="&lt;p&gt;Suppose there's a pool  which is supported by 4 beams (Perimeter beams)and 4 columns. When designing the beam should we consider the moment generated  due to the horizontal water pressure? (I'm using BS 8110 for designing)&lt;/p&gt;&#xA;" OwnerUserId="12105" LastEditorUserId="90" LastEditDate="2017-08-27T08:39:15.510" LastActivityDate="2021-03-01T22:44:25.750" Title="Is there a need to consider the moment generated by horizontal water pressure in a pool?" Tags="|structural-engineering|beam|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16894" PostTypeId="2" ParentId="16893" CreationDate="2017-08-27T03:01:50.783" Score="2" Body="&lt;p&gt;The answer depends on the design. The pool walls are going to resist the pressure and in return they are going to &quot;bulge out&quot; a bit. If there is enough clearance between the beams and the walls then the beams are only going to take any incidental side loading of the structure. That is if the column supports are designed correctly and do not induce any moments in the beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The devil is in the details, and there aren't much of them included in the question to effectively answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS. Do you have access to any FEA software?&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-08-27T03:01:50.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16895" PostTypeId="2" ParentId="16883" CreationDate="2017-08-27T13:13:51.287" Score="3" Body="&lt;p&gt;In engineering, math is less a thing that you do, and more of a language to express ideas.  The language is written to be able to express complicated idea very concisely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of complex functions in electrical engineering is precisely for this reason: it is a way to express a complicated idea concisely. It is entirely possible to derive a theory of electric components that does not use any complex functions.  In that case, everything is a combination of sines and cosines.  e.g. $V=A\cos(\omega t) + B\sin(\omega t)$ The problem with this notation is that it will get really cumbersome to describe how capacitors and inductors work.  E.g. you need to say things like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;if the input current $i$ is a sine wave at frequency $\omega$, then the output voltage $V$ is a cosine wave at frequency $\omega$ with amplitude $V=i \omega L$&quot;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While this is a correct description, it is long. Because inductors and capacitors occur all the time in nearly every circuit, we don't want to have to write that entire sentence every single time we have an inductor. If we did, an EE textbook would be ten thousand pages long. The complex notation allows us to be more compact. If all agree on the notation that we can represent combinations of sines and cosines by $e^{i\theta}=\cos\theta + i \sin\theta$, and then you say that an inductor has complex impedance $j \omega L$, and then &quot;$V=i(j\omega L)$&quot;, means that exact same thing as my sentence above, but it's so much shorter.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-08-27T13:13:51.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16896" PostTypeId="1" AcceptedAnswerId="16898" CreationDate="2017-08-27T20:54:57.747" Score="0" ViewCount="125" Body="&lt;p&gt;&lt;strong&gt;&quot;Ponder the path of your feet, and let all your ways be established.&quot; Proverbs 4:26&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over the last two years I have had a strong will to become an inventor within the field of &#xA;computer-technology/robotics,a field where I have many visions and concepts for projects, &#xA;and have been persistent and consistent in my ways, so that I may reach a point that I &#xA;can assume the title of engineer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently working full-time(lot's of over-time as a laborer), yet with all my free time I try to mold myself&#xA;into the man that can execute these projects I visualized. I'm currently trying to design and develop my first project, and I wasted a lot of time picking different books&#xA;up, not knowing where to start my research and discipline. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I have started with programming? Electronics? Mechanics? Eventually I realized that I had a strong &#xA;apptitude for mathematics, and decided to order a used college algebra book from online, and I have been making&#xA;progress ever since(would be done with the txtbook if it wasn't for the amount of hours I worked every week.)&#xA;    Electronics was a path that I always felt was on this path that I am becoming aware of, and in different periods&#xA;of my life I would find myself picking up Eectronics books and for reasons beyond my understanding I would never get far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet now that I have developed a solid foundation of mathematics, I am beginning to understand and comprehend Electronics &#xA;theory, and I am confident that I can now begin to progress to tinkering with components. &#xA;    I have also decided that maybe it would be a good thing not to try to discipline myself so much, so I &#xA;have enrolled into a computer science degree program, and my long term educational goal Is to become a computer engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all this being stated, what is the path? We all have a journey, we all have a point in space-time that we &#xA;have visualized, but how do you arrive at that point? How do I arrive at the point where I have designed and developed &#xA;my first robotics project?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm seeking feedback from the community so that I may arrive at this point as efficiently as possible. Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**P.S.: Excuse me for my lack of formatting, this is only my second post on Stack-Exchange, and it is a learning curve for me.&lt;/p&gt;&#xA;" OwnerUserId="12689" LastEditorUserId="683" LastEditDate="2017-08-27T21:14:45.873" LastActivityDate="2017-08-27T21:48:05.033" Title="I Say To myself:&quot;I Am An Engineer!&quot;...How Do I Arrive At That Point?" Tags="|robotics|computer-engineering|" AnswerCount="1" CommentCount="15" ClosedDate="2017-08-28T00:43:49.570" ContentLicense="CC BY-SA 3.0" />
  <row Id="16897" PostTypeId="1" CreationDate="2017-08-27T21:28:02.187" Score="2" ViewCount="63" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CicJ9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CicJ9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm only a HS physics student, so please bear with my lack of jargon. My question goes like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the scenario of two coaxial rotors spinning independently in two different directions, each with the same amount of torque, are the thrusts produced by each rotor the same as if the two rotors were side-by-side?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ua18o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ua18o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;My first guess was that they had to be different because of things like air displacement and pressure from the above rotor being applied to the below rotor, but it seems everything I find treats each rotor's thrust like it is completely unaffected by the one above or below it. Could someone shed some light on this? In what way is thrust from the above or below rotor affected by its companion? Thanks a lot!&lt;/p&gt;&#xA;" OwnerUserId="12687" LastActivityDate="2017-08-27T21:28:02.187" Title="A Question of Coaxial Rotors and Lift" Tags="|mechanical-engineering|torque|aerodynamics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16898" PostTypeId="2" ParentId="16896" CreationDate="2017-08-27T21:36:20.130" Score="0" Body="&lt;p&gt;Starting from a fairly low level academic background, I think the best way to &quot;develop your first robotics project&quot; is to focus on the practical side, not the theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get a robotics kit like Lego Mindstorms, build some of the tutorial projects, then follow your own ideas. Mindstorms is not just a &quot;kid's toy&quot; - MIT were involved in designing it, and it is used at university level for robotics lab work. It has the advantage of providing ready-made &lt;em&gt;subsystems&lt;/em&gt; that do relevant tasks which you can interconnect to make your complete robot, instead of having to design everything yourself from scratch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you &lt;em&gt;can&lt;/em&gt; design your own subsystems from scratch and use them with the rest of the Mindstorms kit, if you need to or want to learn how to do that. But in the real world, engineers don't design stuff from scratch just &quot;because they know how to&quot;, if they can buy it ready-made at a reasonable price.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't discourage anyone from starting a computer science degree program - so long as you understand that computer science is fundamentally a &lt;em&gt;theoretical&lt;/em&gt; subject, not a &lt;em&gt;practical&lt;/em&gt; one. Realistically, your math background might not be strong enough to start a CS course right now. You might do better to switch to computer engineering, or mechatronics - they are more focused on &quot;practical engineering&quot; and actually &quot;making stuff&quot; rather than the theory of &quot;why stuff works&quot; - though of course CE and mechatronics will also cover the &lt;em&gt;essential&lt;/em&gt; parts of the theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, &quot;making stuff&quot; in the previous paragraph includes &quot;making software&quot; as well as &quot;making hardware&quot;. You probably don't need to know how to design and write your own compiler for a new computer language (which you would learn how to do in a CS course) or design a new type of CPU chip to run your newly invented language efficiently - but you do need to know how to &lt;em&gt;use&lt;/em&gt; a &quot;standard&quot; programming language and like C, C++ or Java and &quot;standard&quot; embedded computing hardware like an Arduino, a Raspberry Pi, or a smaller microcontroller system to interface with the hardware of your robot. A CE course is likely to be more relevant for that than CS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies for &quot;learning electronics&quot;. Get some (fairly cheap and simple) kits, build them, then modify them. When (note, &quot;when&quot; not &quot;if&quot;!) your modifications don't work, you can always ask why on Electronics SE! The theory will make more sense if it you can relate it to a practical context. Theory on its own is all very well, but the &lt;em&gt;real&lt;/em&gt; learning starts when  you try to understand what just caused some of your circuit components to &quot;release their magic smoke&quot; and die!&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-08-27T21:48:05.033" LastActivityDate="2017-08-27T21:48:05.033" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16899" PostTypeId="2" ParentId="5591" CreationDate="2017-08-28T15:15:36.040" Score="0" Body="&lt;p&gt;I am researching on this topic actually. Hydraulic jump is not an easy matter as it seems to be. the location of the jump also depends on the thickness after the jump. if you increase the height after the jump(for example, put a bump). the location of the jump will move upstream. it depends on a lot of parameters, such as the upstream speed, the downstream configuraion, surface tension. etc. with all being said, the true reason behind this phenomenon is still unknown.&lt;/p&gt;&#xA;" OwnerUserId="12698" LastActivityDate="2017-08-28T15:15:36.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16902" PostTypeId="1" CreationDate="2017-08-29T01:53:47.420" Score="1" ViewCount="91" Body="&lt;p&gt;I am always fascinated to watch construction on field as it  helps me to build my knowledge deeper. The answer to a question what I am looking upto is HOW is the brick size decided ? Like in India we have standarized size of 19×9×9 cm But actually how only this size became standard? &lt;/p&gt;&#xA;" OwnerUserId="12703" LastActivityDate="2017-08-29T01:53:47.420" Title="How is brick size decided?" Tags="|masonry|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16903" PostTypeId="2" ParentId="16734" CreationDate="2017-08-29T06:00:32.060" Score="2" Body="&lt;p&gt;Steel with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bluing_(steel)&quot; rel=&quot;nofollow noreferrer&quot;&gt;black passified surface&lt;/a&gt; (as mentioned by Jason in the comments) would probably work very well. This surface is &lt;a href=&quot;http://ferrocell.us/references/Magnetite%20(Fe3O4)_%20Properties%20Synthesis%20and%20Applications.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magnetite&lt;/a&gt; and has a melting point of 1590°C. The carbon steel has a high thermal mass and reasonable thermal conductivity to pull the heat away from the focal point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, copper plate &lt;a href=&quot;https://en.wikipedia.org/wiki/Black_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;black oxide surface treated&lt;/a&gt; to have a layer of &lt;a href=&quot;https://en.wikipedia.org/wiki/Copper(II)_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;cupric oxide&lt;/a&gt; would also work well. Cupric oxide can withstand 1,326°C and the very high thermal conductivity of copper will pull the heat away from the focal point. Either of these solutions will likely require buying a metal plate and applying the surface treatment yourself, as they are not widely available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is the focal length of your laser. A shorter focal length will permit the kerf of the beam to be smaller. However for normal cutting a shorter focal length also reduces the depth of cut, so manufactures seek to strike a balance between these. Since your depth of cut is likely very small, you will want to source the shortest focal length you can and closely test the focus each time before proceeding with a cut.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/evE98.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/evE98.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also important to note that cheap lenses will suffer from &lt;a href=&quot;https://en.wikipedia.org/wiki/Spherical_aberration&quot; rel=&quot;nofollow noreferrer&quot;&gt;spherical aberration&lt;/a&gt; because a sphere is cheaper to manufacture. An &lt;a href=&quot;https://en.wikipedia.org/wiki/Aspheric_lens&quot; rel=&quot;nofollow noreferrer&quot;&gt;aspheric lens&lt;/a&gt; will cost more money, but will result in a smaller focal point. &lt;a href=&quot;https://www.thorlabs.com/thorproduct.cfm?partnumber=AL72525-E&quot; rel=&quot;nofollow noreferrer&quot;&gt;1 inch focal length lens from ThorLabs&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hfgKP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hfgKP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to experiment with different laser powers and different cutting gasses to reduce contamination. Selection of the material in the product you are cutting is also important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an absorption graph for uncoated steel and copper at the 10.6um wavelength. I unfortunately did not find one for the coating materials. If the material you are using is fixed, and the other items above have not addressed the issue, you may want to look a different wavelength cutting laser.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wE4fL.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wE4fL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2017-08-29T06:13:05.473" LastActivityDate="2017-08-29T06:13:05.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16904" PostTypeId="1" CreationDate="2017-08-29T08:08:20.617" Score="0" ViewCount="647" Body="&lt;p&gt;I have 6 inch borewell of 1200 feet (domestic use). With camera test, we could see water storage is at 960 feet. i.e approx storage of 200 feet. i.e around 1000 liters. my question is, can we put motor? will it work? What happens if we pump and water is just enough for pipe above? after we stop motor, will it discharge water back or will it hold in pipe and next pumping will push out?&lt;/p&gt;&#xA;" OwnerUserId="12705" LastEditorUserId="1586" LastEditDate="2017-11-01T22:15:02.953" LastActivityDate="2017-11-01T22:15:02.953" Title="How borewell pump works? does water stays in pipe after stop?" Tags="|motors|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16905" PostTypeId="2" ParentId="16904" CreationDate="2017-08-29T12:11:43.203" Score="0" Body="&lt;p&gt;You should be able to find submersible motor / pumps for this and most, if not all, will have a one way valve to stop the backflow, this means delivery is quick when the motor is turned on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just need to source one that will work at the pressure created by the depth.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-08-29T12:11:43.203" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16906" PostTypeId="2" ParentId="16904" CreationDate="2017-08-29T12:37:52.393" Score="0" Body="&lt;p&gt;Place a retaining valve and the water will stay in the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XhyLA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XhyLA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8464" LastActivityDate="2017-08-29T12:37:52.393" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16908" PostTypeId="2" ParentId="13244" CreationDate="2017-08-29T14:19:10.153" Score="1" Body="&lt;p&gt;I have worked with specifying optical materials for use as windows in visible and IR applications. Optical material selection will be determined by two constraints:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The portion of the EM spectrum you need to observe. You specified LWIR, but in my limited experience only certain wavelength bands are observed. You will find that many materials used for IR imaging have uneven transparency ratings as a function of light wavelength. For IR applications ZN-S or Cleartran is the go to material for optical windows.&lt;/li&gt;&#xA;&lt;li&gt;Manufacturing costs. Optical materials must be grown and formed into what is called a blank. Then initially machined into your desired shape, and then polished. Depending on the size of your application, and how tight your optical tolerances are, the polishing process can take months. Other things to consider are homogeneity of index of refraction in the material, it can vary slightly in the material, number of occlusions allowed per volume (defects, bubbles, and other stuff that blocks transmission), and then finally coatings.  If your lens is exposed to the elements, you will need one coating for abrasion (usually the one you mentioned), coatings for anti reflection may be required, and so on.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To sum up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your application will drive your selection to one or two materials and then the manufacturing process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Material performance is one metric, manufacturing cost is your second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best bet for a real answer is to contact one or three optical material vendors. The more information you can provide to them about what your use case is, the better their answer will be.&lt;/p&gt;&#xA;" OwnerUserId="12709" LastEditorUserId="1586" LastEditDate="2017-11-01T22:15:27.550" LastActivityDate="2017-11-01T22:15:27.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16909" PostTypeId="1" AcceptedAnswerId="16912" CreationDate="2017-08-29T15:23:45.607" Score="0" ViewCount="92" Body="&lt;p&gt;I have a 20 cm long glass tube of outer diameter &lt;code&gt;6.4 mm&lt;/code&gt; and inner diameter &lt;code&gt;4.3 mm&lt;/code&gt;. Air at &lt;code&gt;Ti = 20 °C&lt;/code&gt; is pushed through the glass tube at &lt;code&gt;2 litres per minute&lt;/code&gt;. The glass tube has a wire wound resistance heater on it which we can assume is perfectly thermally insulated from the ambient air.&#xA;My aim is to find out the power needed to heat the air flowing through the tube to &lt;code&gt;Te=650 °C&lt;/code&gt; at the tube exit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After doing the calculation using &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5Cdot%7BQ%7D=%5Cdot%7Bm%7Dc_p(T_e-T_i)&amp;space;=%5Crho%5Cdot%7Bv%7Dc_p(T_e-T_i)%5C%5C&quot; title=&quot;\dot{Q}=\dot{m}c_p(T_e-T_i) =\rho\dot{v}c_p(T_e-T_i)\\&quot; /&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Where &lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5Cdot%7BQ%7D&quot; title=&quot;\dot{Q}&quot; /&gt; is the power supplied by the heater (&lt;code&gt;W&lt;/code&gt;),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5Cdot%7Bm%7D&quot; title=&quot;\dot{m}&quot; /&gt; is the mass flow rate (&lt;code&gt;kg/s&lt;/code&gt;),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5Crho&quot; title=&quot;\rho&quot; /&gt; is the density of the air &lt;code&gt;kg/m^3&lt;/code&gt; at the bulk temperature   &lt;img src=&quot;https://latex.codecogs.com/gif.latex?T_b=(T_i&amp;plus;T_e)/2&quot; title=&quot;T_b=(T_i+T_e)/2&quot; /&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get the power required as being &lt;code&gt;12.5 W&lt;/code&gt;. This seems to me to be too low. Unfortunately I do not know where I am going wrong. Does anybody have any idea where I am making a mistake in my calculation?&lt;/p&gt;&#xA;" OwnerUserId="12713" LastActivityDate="2017-08-29T20:11:48.303" Title="Problem finding temperature gain of air through heated glass tube" Tags="|thermodynamics|heat-transfer|heating-systems|glass|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16910" PostTypeId="2" ParentId="15603" CreationDate="2017-08-29T16:20:48.020" Score="2" Body="&lt;p&gt;I picked out several things of interest from your question.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;You seem to be interested in measuring the airflow near the walls of the tunnel&lt;/li&gt;&#xA;&lt;li&gt;You seem to be interested in finding areas of (relatively) low flow, but don't seem to be interested in the specific flow rate or velocity.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From these two items, I offer that you would be better served by a flow visualization technique, rather than a free stream velocity measurement technique.  For your application, I would recommend the use of strings, as is commonly done in low speed wind tunnels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These have several advantages.  They are VERY low cost, show differences fairly easily, and can be placed anywhere on the walls of the tunnel and later removed.  Unfortunately, they require visual monitoring (e.g. you have to see them or be able to take video) and they provide limited quantifiable data.  However, they are a very simple and efficient way to visualize the flow field in the tunnel, and thereby determine where the areas of low flow may be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See NASA's informative site &lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/tunvsmoke.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for more information.&lt;/p&gt;&#xA;" OwnerUserId="5865" LastActivityDate="2017-08-29T16:20:48.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16912" PostTypeId="2" ParentId="16909" CreationDate="2017-08-29T20:11:48.303" Score="0" Body="&lt;p&gt;If I perform a back of the envelope calculation (using just the density of air at standard conditions, 1.225 kg/m3):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Converting to SI units of mass flow rate:&#xA;&#xA;2 SLM of air -&amp;gt; 0.04 grams / sec = 0.00004 kg / sec&#xA;Cp = 1005 J / kg K&#xA;Te = 923.15 K&#xA;Ti = 293.15 K&#xA;&#xA;P = 0.00004 * 1005 * (923.15-293.15) = 25.326 W (or J / s)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Your density is probably different by a factor of two or so: @ 350C, air density = 0.566 kg / m3. Everything looks in order to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, the required power may seem low, but look at how little mass is flowing through the section that is being heated every second. 2 liters a minute sounds like a lot until you break it down into grams per second. Electrothermal heating is also a fairly lossless process, so if you perform this experiment, you will likely not need to increase this number by too much to account for inefficiencies.&lt;/p&gt;&#xA;" OwnerUserId="3583" LastActivityDate="2017-08-29T20:11:48.303" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16913" PostTypeId="1" CreationDate="2017-08-29T22:33:07.280" Score="2" ViewCount="395" Body="&lt;p&gt;I am struggling to solve this problem and my available time conflicts with my professors available time. Could any of you offer me some help, perhaps describe the method I would use to solve such a problem in a way that I can learn from it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you. (Problem follows):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XwuEY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XwuEY.jpg&quot; alt=&quot;Voltage Divider&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circuit designer wants to change the values of R_1 and R_2 so that the design output voltage v_o = 5 V is achieved when the load resistance is R_L = 200 kΩ rather than at no-load. The actual output voltage must not drop below 4.5 V when R_L = 100 kΩ . What is the smallest resistor value that can be used for R_1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(R_1 and R_2 are unknown quantities.)&lt;/p&gt;&#xA;" OwnerUserId="12717" LastEditorUserId="8055" LastEditDate="2017-08-31T08:21:58.700" LastActivityDate="2017-08-31T08:21:58.700" Title="Designing a Voltage Divider" Tags="|electrical-engineering|circuits|circuit-design|" AnswerCount="1" CommentCount="1" ClosedDate="2017-09-02T22:37:02.540" ContentLicense="CC BY-SA 3.0" />
  <row Id="16914" PostTypeId="2" ParentId="16913" CreationDate="2017-08-29T23:50:43.450" Score="4" Body="&lt;p&gt;R2 and RL are in parallel so the voltage across them will be the same so can be treated as a single resistor when thinking about the voltage across them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Req(lower) = (1/R2 + 1/100kohms)^-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Req(upper) = (1/R2 + 1/200kohms)^-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vs/(R1 + Req) = I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this gets your started, you're looking to produce a system of 2 equations that you can use to solve for R1 and R2&lt;/p&gt;&#xA;" OwnerUserId="12691" LastActivityDate="2017-08-29T23:50:43.450" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16915" PostTypeId="1" AcceptedAnswerId="17080" CreationDate="2017-08-30T04:05:19.460" Score="2" ViewCount="4002" Body="&lt;p&gt;I am working on a small project where I, from a relatively confined space in a well that has a diameter of 30 cm, will need to transport the water upwards about 2 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking at ram pumps but cant really see a way to utilise that as i dont have a &quot;higher&quot; water source. I have also been looking at electric pumps on Alibaba and it seems that i'll land at around 2W, which seems like a lot for an iot device - i really dont want to have to recharge the device every week, esp. since there is little energy to harvest nearby.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The requirements are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ability to pump water upwards 2 meters&lt;/li&gt;&#xA;&lt;li&gt;Flow of around 12 liters per hour would be enough&lt;/li&gt;&#xA;&lt;li&gt;Ideally 12V or lower&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are the best options? :) I know that product recommendations are not allowed, but what techniques should I look at overall? Is there a resource somewhere that compares the efficiency of different pumping methods?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT] RE comments: In terms of effeciency, I mean as low wattage as possible. I wont have easy access to recharging (solar and wind are out, but potentially the energy of the water flowing down into the well could produce a minor charge for the battery). If we can get to a point where the battery only needs to be changed once a month, that would be perfect.&lt;/p&gt;&#xA;" OwnerUserId="12719" LastEditorUserId="12719" LastEditDate="2017-08-30T13:18:16.313" LastActivityDate="2017-09-12T06:13:52.283" Title="Most efficient/low power consumption way to pump water" Tags="|pumps|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="16919" PostTypeId="1" CreationDate="2017-08-30T10:47:11.057" Score="1" ViewCount="543" Body="&lt;p&gt;I am doing a project at the moment which involves improving the packaging operation at work. Right now I am looking at ways to improve packaging selection based on engineering principles, rather than guess work, which is what's being used currently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We make many shapes and sizes of bottles and a major problem that we have is pallets which collapse because of poor packaging selection. The main problem is shown in the below image - the bottles in my (crude) sketch show a cross section of 1 stack of bottles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MQ6SZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MQ6SZ.png&quot; alt=&quot;Bottle stack sketch.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the 'void' in the base of the bottle, which we call the 'push' is often large enough to allow the neck of the lower bottle to move a considerable distance in to it. This causes the corrugated board to be pinched between the 2 bottles and in effect is a 4-point beam loading situation, with the orientation of the forces flipped 180 degrees (relative to their orientation in the testing standard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to know how I can relate the values from a 4-point bending stiffness test (BS ISO 5628:2012) and use these values to select an appropriate corrugated cardboard grade, based on the expected 4-point loading situation for each bottle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my limited knowledge of beams, I think I would need to:-&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Define a maximum load based on the expected mass of bottles above the&#xA;point at which the maximum load is expected (i.e. on the neck of the&#xA;lowest bottle in the stack.) &lt;/li&gt;&#xA;&lt;li&gt;define a maximum allowable deflection to&#xA;work back and find a stiffness that would correspond to this&#xA;deflection. &lt;/li&gt;&#xA;&lt;li&gt;input the lengths and positions of the forces based on&#xA;the bottle geometry.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some sticking points that I haven't been able to address:-&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is my fundamental assumptions of this being a 4-point bending&#xA;situation correct? We make round and square bottles which vastly complicate the loading situation. Obviously in my model I am only considering a cross section to relate this to a 4-point loading test.&lt;/li&gt;&#xA;&lt;li&gt;There are many bottles on each layer of a pallet. In my approach I&#xA;have opted to simplify the model by only considering the mass&#xA;directly above a single bottle. I am not sure, for example, how&#xA;significant the contribution of the frictional forces from the&#xA;surrounding bottles is.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; We do not used cardboard boxes. We used either pads or trays. In either case, the loading is always normal to the flat face of the cardboard. Edge compression never occurs with our packaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd really appreciate any feedback. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="307" LastEditorUserId="307" LastEditDate="2017-08-30T11:39:23.690" LastActivityDate="2017-08-30T11:39:23.690" Title="How can I use the bending stiffness of corrugated cardboard to determine maximum stacking heights of pallets loaded with glass bottles?" Tags="|structural-analysis|beam|" AnswerCount="0" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="16920" PostTypeId="2" ParentId="16915" CreationDate="2017-08-30T19:36:31.583" Score="1" Body="&lt;p&gt;Does the battery need to also fit down into your well area?  If not, you should then just be able to get a large battery (i.e. car battery or marine deep-cycle) that may have capacity for a month at a time.  Then at your once-per-month changeout switch the battery out with one that's been trickle charged over the month in another location that does have solar/wind/hydroelectric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to look at is whether you could use smaller pumps with less capacity/power draw in cascade where each one wouldn't need to pump up as high, reducing their power requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just make sure to fuse it appropriately as car batteries can provide A LOT of current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But overall, you have a certain amount of work that has to be done (raising a volume of water up a given distance).  There is a set minimum amount of power required to do your work, no matter how close to perfect efficiency there is in your lift mechanism.&lt;/p&gt;&#xA;" OwnerUserId="11065" LastActivityDate="2017-08-30T19:36:31.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16922" PostTypeId="1" CreationDate="2017-08-31T06:52:34.730" Score="4" ViewCount="484" Body="&lt;p&gt;As is probably known a traffic light may be turned off in which case normal rules applies to the intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the traffic light is active then red &amp;amp; red+yellow means that drivers are required to stop, while green means that the drivers are allowed to drive. Yellow also means that you're allowed to drive, but cautions the driver that it will/may turn red on which the driver are required to stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by what sequence do one start/stop the traffic lights? Stopping seem quite obvious, just make sure that all are showing red (or only those that will have the right of way have green) and then just turn off the light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how to turn the lights on again. I think that just reversing the sequence may not be fair - if the light all the suddenly turns red you're required to stop (but may or may not have the time to stop your vehicle). On the other hand entering with green light might be dangerous.&lt;/p&gt;&#xA;" OwnerUserId="12730" LastActivityDate="2017-08-31T12:13:29.547" Title="What is the starting/stopping sequence of a traffic light" Tags="|traffic-intersections|traffic-light|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16924" PostTypeId="2" ParentId="16922" CreationDate="2017-08-31T09:36:18.943" Score="1" Body="&lt;p&gt;There is another state the lights can be in: blinking yellow. Which means that the intersection is not being controlled and normal priority rules are in force. Going from off to all blinking does not change how you should behave on the intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going from blinking yellow to solid yellow (longer than the normal 3 seconds) to red gives drivers enough warning that the light will turn red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a major and minor road on the intersection where if the lights are off the minor yields to major, you can make the minor road turn red and the major turn green. Possibly showing the minor road an extended solid yellow first.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-08-31T09:36:18.943" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16925" PostTypeId="2" ParentId="16719" CreationDate="2017-08-31T09:48:52.220" Score="0" Body="&lt;p&gt;Free flow intersections take up a lot more real estate due to the wider curves needed for the higher speeds. Every time traffic crosses the roads need to be put onto a different grade. This means bridges and ramps to lift the traffic to that grade and back down again. Ramps (and the levies they are built on) also take up space and moves interaction with the main roads away from the actual intersection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If traffic flow is not heavy then a simple priority regulated diamond interchange with just single lane traffic all around is more than enough. With a little forethought adding a third lane on the bridge to allow for left-turning traffic to not block other traffic will help flow for not that much extra cost.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-08-31T09:48:52.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16926" PostTypeId="2" ParentId="16922" CreationDate="2017-08-31T12:13:29.547" Score="3" Body="&lt;p&gt;What country? This is different between different countries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can give the sequence in Poland, which is standard in most of EU.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;180 seconds of Blinking Yellow. (Groups &lt;sup&gt;[**]&lt;/sup&gt; with Yellow blink yellow, the rest remains off. If the controller was already in this mode, past duration is counted towards the 180 seconds.)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;All Yellow. Whichever groups don't have the yellow signal, display red.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;All Red. Duration of the shortest Intergreen&lt;sup&gt;[**]&lt;/sup&gt; time, no less than 5 seconds.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;(optionally) - Startup program. A short custom sequence that opens most prioritized lanes (through normal sequence, like 1s of Red+Yellow, followed by Green).&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;Standard traffic program. Normally side lanes are opened first, main direction later.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Ending just requires minimum intergreen of red signal after stopping all groups normally, before transition to YB - the controller can transition to dark (off) 180 seconds later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;[*] Group - a set of signal lights (often just one signal light) that display the same signal. May be a set of lanes that allow transition to the same exit, or a signal light hanging above the lane, repeated by another on the side of the lane - or two signal lights on two sides of the same zebra crossing.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;[**] Intergreen time - time between stopping one group and activating another, necessary for vehicles/pedestrians to vacate the crossing without risk of collision with traffic from the newly opened lane. The intergreen times are assembled into a matrix, each starting group to each ending, with non-colliding entries empty, the remaining containing time that needs to count down from the end of the ending group) before the starting group is allowed to start.  &lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a standard stop and start sequence of a controller; diagram states of each group. The numbers above the diagram are the second of given phase, and phase number of given program. The names on the left are the group names. The program is made for development purposes, so it doesn't retain the 180 seconds restriction on yellow blinker (just to save the time of the developer); actual traffic program would require a 180 second gap between the start and end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2tBlh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2tBlh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-08-31T12:13:29.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16928" PostTypeId="1" AcceptedAnswerId="16930" CreationDate="2017-08-31T18:52:22.497" Score="1" ViewCount="55" Body="&lt;p&gt;I see some example of DLX assembly instructions such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;LW Ri, 0 (Rz)&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I can't find out why &lt;code&gt;Rz&lt;/code&gt; has brackets around it.&lt;/p&gt;&#xA;" OwnerUserId="12737" LastEditorUserId="7083" LastEditDate="2018-01-28T23:19:32.707" LastActivityDate="2018-01-28T23:19:32.707" Title="DLX assembly synthax" Tags="|electrical-engineering|embedded-systems|computer-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16929" PostTypeId="2" ParentId="16915" CreationDate="2017-08-31T18:56:34.040" Score="2" Body="&lt;p&gt;Consider a &lt;a href=&quot;http://wgbis.ces.iisc.ernet.in/energy/water/paper/drinkingwater/wellsconstruction/fig156.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;bucket-and-chain pump&lt;/a&gt;, as depicted from &lt;a href=&quot;http://wgbis.ces.iisc.ernet.in/energy/water/paper/drinkingwater/wellsconstruction/appendix7.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantages are that operation can be intermittent, priming isn't a problem, and the pump will work with a low battery, as long as the motor turns.&lt;/p&gt;&#xA;" OwnerUserId="10706" LastActivityDate="2017-08-31T18:56:34.040" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16930" PostTypeId="2" ParentId="16928" CreationDate="2017-08-31T20:27:20.493" Score="1" Body="&lt;p&gt;Basically, &quot;Rn&quot; means &quot;the contents of register n&quot; and &quot;(Rn)&quot; means the &quot;data in the memory location whose address is in register Rn&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For memory access, you can also add a constant to (Rn), so 4(R8) means &quot;add 4 to the address in register 8 (but don't change the data that is in R8!), and use the data in that memory location&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.cs.utexas.edu/~pingali/CS378/2015sp/lectures/DLX.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The DLX Instruction Set Architecture&lt;/a&gt; for examples (e.g. on Page 8.)&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="8055" LastEditDate="2017-12-05T19:41:26.613" LastActivityDate="2017-12-05T19:41:26.613" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16934" PostTypeId="1" CreationDate="2017-09-01T02:06:07.280" Score="1" ViewCount="61" Body="&lt;p&gt;&quot;Calculators stackexchange&quot; doesn't exist, so I guessed this would be the right place to ask. Can a calculator solve $-0.2 \cdot e^{-0.06-4.2i}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or any $e$ to a complex for that matter. I've been solving this with Wolfram Alpha or Matlab when in a project, but I'd like yo know if is there anyway to get it done in a calculator.&lt;/p&gt;&#xA;" OwnerUserId="12740" LastEditorUserId="9" LastEditDate="2017-09-02T22:32:48.497" LastActivityDate="2017-09-02T22:32:48.497" Title="Can a casio fx series do this operation?" Tags="|telecommunication|radio|" AnswerCount="0" CommentCount="21" ContentLicense="CC BY-SA 3.0" />
  <row Id="16937" PostTypeId="1" CreationDate="2017-09-01T04:32:55.867" Score="2" ViewCount="546" Body="&lt;p&gt;In ASCE 7-10, with regards to strength design,for fluid load, it mentions that where fluid loads are present, they shall be included with the same load factor as dead load D in combination 1 through 5 and 7. How about for load combination 6?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Load combination 6 is as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.9D + 1.0W,&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which I expected to be&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0.9D + 0.9F + 1.0W&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However due to the statement in ASCE 7-10, I do not know what factor to assume for fluid load as fluid load will be present in  my situation. The same applies to Allowable stress design load combination albeit with different combination.&lt;/p&gt;&#xA;" OwnerUserId="1693" LastActivityDate="2017-09-16T01:18:23.840" Title="ASCE 7-10, Fluid load factor for basic combinations for strength and allowable stress design" Tags="|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16938" PostTypeId="1" AcceptedAnswerId="16941" CreationDate="2017-09-01T09:05:56.590" Score="1" ViewCount="1070" Body="&lt;p&gt;We know that moment of inertia is commonly known as second moment of area. Which is $\int_{}{} y^2 dA$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there another moment of inertia called as mass moment of inertia given by $\int_{}{} r^2 dM$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also we know that radius of gyration is by definition,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_{mass} =\sqrt{ \frac{ \int_{}{} r^2dM  }{\int_{}{}dM} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$r_{area} = \sqrt{ \frac{ \int_{}{} r^2dA  }{\int_{}{}dA} }$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is whether this $r_{mass}$ and $r_{area}$ are same for any arbitrary object? Of course the axes about which integral is evaluated are the same. &lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2017-09-01T15:12:50.783" Title="Is radius of gyration same for both types of moment of inertia?" Tags="|mathematics|moments|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16940" PostTypeId="1" CreationDate="2017-09-01T10:38:36.293" Score="1" ViewCount="56" Body="&lt;p&gt;I'm building a portable emission analyser which measures emissions from a tailpipe, to aid in troubleshooting and tuning an engine. For this, I need a $\text{CO}_{2}$ , a $\text{CO}$, and an $\text{O}_{2}$ sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While these sensors are widely available, I am in search for the type of sensor that has the lowest response time, while being able to measure in the range of 5-10% (gas dilution is not an option for my application). I already found a $\text{CO}_{2}$  and an $\text{O}_{2}$ sensor that are able to do so, but I can't find a $\text{CO}$ sensor with these capabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those all seem to have a response-time of around 30 seconds, and most are only able to measure up to around 2000ppm. I need a sensor as fast as 1 second or faster and able to measure at least 5% $\text{CO}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, Is there a $\text{CO}$ sensor on the market that is able to measure in the range of 5%, while responding within 1 second?&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="1586" LastEditDate="2018-10-08T19:28:21.533" LastActivityDate="2018-10-08T19:28:21.533" Title="Is there a 5% range CO sensor with a response rate higher than 1Hz?" Tags="|electrical-engineering|measurements|sensors|gas|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="16941" PostTypeId="2" ParentId="16938" CreationDate="2017-09-01T15:07:29.300" Score="2" Body="&lt;p&gt;Area moment of inertia and mass moment of inertia are two different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A) Mass moment of inertia (or moment of inertia):&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the resistance offered by a &lt;em&gt;solid body&lt;/em&gt; when subjected to rotation (or application of torque). The mass moment of inertia is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I = \int r^2 dM $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of gyration is the distance at which the mass can be considered to be concentrated (with respect to the axis of rotation). For two wheels with the same overall mass but different distribution of the mass, the one with mass concentrated near the axis of rotation has less moment of inertia. This is the 'rotational' equivalent of the 'mass' in translational motion, thus sometime also known as the 'angular mass'.  Moment of inertia has units $kg m^2$. &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_moments_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a list&lt;/a&gt; of moments of inertia some regular geometric shapes. Check out &lt;a href=&quot;https://www.youtube.com/watch?v=7PRZ5IIT5Tg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; by the YouTube channel 'smarter every day' for a physical sense of the moment of inertia.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B) Area moment of inertia (or second moment of area):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area moment of inertia is a property of a $2$D section or surface of a body. For a solid body, the area moment of inertia can be defined only for the cross sections. The second moment of area is not to be confused with mass moment of inertia. The area moment of inertia is used to characterize the resistance offered by a body (such as a beam) for deflection under load. Its units are $m^4$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_second_moments_of_area&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a list&lt;/a&gt; of area moments of inertia of a few regular $2$D shapes. Note that, for a circular shaft, &lt;a href=&quot;https://en.wikipedia.org/wiki/Polar_moment_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;the polar moment of inertia&lt;/a&gt; is more relevant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming to the question, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So my question is whether this $r_{mass}$ and $r_{area}$ are same for&#xA;  any arbitrary object? Of course the axes about which integral is&#xA;  evaluated are the same.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Provided that a solid body has uniform cross sectional area and uniform second moment of area (both about the axis of rotation), $r_{mass}$ and $r_{area}$ will be same if the material is isotropic (i.e. uniform density). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an object with uniform thickness $h$ unit and constant cross sectional area $dA$ and constant second moment of area $I_{area}$ along the thickness, the total mass will be &#xA;$$ M = \int dM =  \rho h \int dA $$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In such as case, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_{mass} = \sqrt{\frac{\int r^2 dM}{\int dM}} = \sqrt{\frac{\rho h \int r^2 dA}{\rho h \int dA}}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $\rho$ is the constant density of the material. This is same as $r_{area}$. However, it is to be noted that the mass moment of inertia was found out for the entire body whereas the area moment of inertia is only valid for a surface of the body (cross section) thus it needs to be same along the axis, same holds true for area of the cross-section itself.&lt;/p&gt;&#xA;" OwnerUserId="223" LastEditorUserId="223" LastEditDate="2017-09-01T15:12:50.783" LastActivityDate="2017-09-01T15:12:50.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16942" PostTypeId="1" CreationDate="2017-09-01T19:14:33.763" Score="2" ViewCount="259" Body="&lt;p&gt;Apologies for the probably daft question, I am thinking of buying a small KURODA ball screw (180 mm overall length, 8mm diameter, 2mm lead). I have a photo from the seller and I noticed a wider area towards one end of the screw. I suspect it is for mounting a pulley (it seems to have a screw flat area). Can you please confirm if my understanding is correct?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zBbOJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zBbOJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11266" LastActivityDate="2018-04-03T05:20:20.570" Title="Ballscrew screw flat area" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16944" PostTypeId="1" CreationDate="2017-09-02T08:52:35.323" Score="-2" ViewCount="61" Body="&lt;p&gt;Consider this example.&#xA;Take a reverse osmosis filter,and the outlet is an hose long 8km with distilled water inside. Let the apparatus sink deep in the ocean for 8Km. Now you have a long vertical hose with pure water inside,and the reverse osmosis filter at the bottom of ocean. Pure water exert 800Kg of pressure from inside,while in the outside we have 1,035Kg per 8000m ,828Kg of pressure. In a perfect world the salty water should naturally be separated from salt and emerge through the hose. Few months ago i did some math to understand if it is possible do &quot;that&quot;,separating through containers the water so that at the base there is only a situation of 1bar-28bar and not 800bar-828bar. I put inside the compressibility of water too that at this pressure is important,and i computated a solid possibility to distillate water with only the different weight of two columns of water. I come up with an idea that with two, 350m of columns of water, we are able to produce energy...problem is that would cost 600€ per Kw/h due the expensive filters. I red somewhere that the sand is able too to filter salty water...in that case the system would be remarkable&lt;/p&gt;&#xA;" OwnerUserId="12413" LastEditorUserId="12413" LastEditDate="2017-09-04T00:03:36.753" LastActivityDate="2017-09-04T00:03:36.753" Title="Free Distillation" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16945" PostTypeId="1" AcceptedAnswerId="16949" CreationDate="2017-09-02T09:19:52.353" Score="0" ViewCount="90" Body="&lt;p&gt;There are Tait-Bryan angular changes $(\dot{\phi}, \dot{\theta}, \dot{\psi})$, which describe the angular change rates of the body-frame within the inertial frame (world frame).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then there are angular change rates $(p,q,r)$ of the aircraft within its own body frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert these angular change rates between each other, a rotation matrix is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\begin{bmatrix} p \\ q \\ r \end{bmatrix} = \begin{bmatrix} 1 &amp;amp; 0 &amp;amp; -sin(\theta) \\ 0 &amp;amp; cos(\phi) &amp;amp; sin(\phi)cos(\theta) \\ 0 &amp;amp; -sin(\phi) &amp;amp; cos(\phi)cos(\theta) \end{bmatrix} \cdot \begin{bmatrix} \dot{\phi} \\ \dot{\theta} \\ \dot{\psi} \end{bmatrix}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is that? Why aren't they exactly the same?&lt;/p&gt;&#xA;" OwnerUserId="10870" LastActivityDate="2017-09-02T15:40:48.523" Title="Rotation of aircraft's body-frame" Tags="|aerospace-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16949" PostTypeId="2" ParentId="16945" CreationDate="2017-09-02T15:40:48.523" Score="1" Body="&lt;p&gt;When plane is flying north, roll axis goes from north to south. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the plane is flying west the roll axis goes from west to east.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no they are not the same.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-09-02T15:40:48.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16950" PostTypeId="2" ParentId="16944" CreationDate="2017-09-02T16:07:45.953" Score="2" Body="&lt;p&gt;It's not clear what you are really asking, but it seems it is something about getting &quot;free&quot; reverse osmosis by using the pressure sufficiently deep in the ocean instead of having to spend energy in pumping to create that pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, you can use pressure from ocean depth to overcome the osmotic pressure.  However, then the de-salinated water must be at surface pressure.  That means you spend the energy pumping the clean water to the surface instead of pushing the salty water against the membrane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no free lunch.  You're going to pay for the flow volume times the pressure across the membrane one way or another.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-09-02T16:07:45.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16951" PostTypeId="1" AcceptedAnswerId="16978" CreationDate="2017-09-02T17:12:18.110" Score="1" ViewCount="81" Body="&lt;p&gt;According to James Cameron, the submarine barely had any space for him to move around etc., and as he approached the bottom of the Mariana Trench, the submarine shrieked down by a few inches (due to pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why was the challenger deep designed to be just big enough for James Cameron to fit in?&lt;/p&gt;&#xA;" OwnerUserId="12660" LastActivityDate="2017-09-05T05:37:22.267" Title="Why was the challenger deep so small?" Tags="|pressure|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16952" PostTypeId="1" CreationDate="2017-09-02T19:47:36.627" Score="2" ViewCount="361" Body="&lt;p&gt;Thinking experiment about a flying device having the fewest moving parts, whats the slowest cruise speed achievable using a ramjet, no matter the scale of this device. Working altitude is troposphere. What would section cut view of this engine look like?&lt;/p&gt;&#xA;" OwnerDisplayName="user12750" LastActivityDate="2018-11-07T01:16:48.940" Title="What's the slowest airflow speed a ramjet could work?" Tags="|thermodynamics|aerodynamics|engines|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16957" PostTypeId="1" CreationDate="2017-09-03T02:05:49.510" Score="-2" ViewCount="610" Body="&lt;p&gt;The question goes as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exercise machine consists of a lightweight cart which is mounted on small rollers so that it is freeto move along the inclined ramp.Two cables are attached to the cart-one for each hand.if the hands are together so that the cables are parallel and if each cable lies essentially in a vertical plane,the question is to deter the force $P$ which each hand must exert on its cable on order to maintain an equilibrium position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PSWjL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried writing the equation for equilibrium as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$2P=mg\sin(\theta)$$ from the FBD of cart and man together.However this is not yielding correct result.Any ideas?Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastActivityDate="2017-11-02T13:11:00.000" Title="Equilibrium condition in a cart on a inclined plane" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2018-08-09T12:29:52.523" ContentLicense="CC BY-SA 3.0" />
  <row Id="16958" PostTypeId="2" ParentId="16756" CreationDate="2017-09-03T03:00:45.533" Score="0" Body="&lt;p&gt;Discaimer - I haven't actually done this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put an ordinary balloon in the bottle with the mouth folded over the mouth of the bottle. Add the vinegar and baking soda to the empty balloon and cork. The size of the balloon (reduction of original air volume) should give you a reasonable estimate.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2017-09-03T03:00:45.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16959" PostTypeId="2" ParentId="16756" CreationDate="2017-09-03T03:46:39.990" Score="0" Body="&lt;p&gt;Firs of all, I would measure pressure only during [lots of] ground tests.  What's the actual need for making these measurements in-flight?  Does it justify the difficulties of miniaturizing your instrumentation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would use a pressure sensor that can work with liquid (such as this &lt;a href=&quot;https://www.digikey.com/product-detail/en/honeywell-sensing-and-productivity-solutions/MLH200PSL01A/480-5152-ND&quot; rel=&quot;nofollow noreferrer&quot;&gt;Honeywell MLH series&lt;/a&gt;).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would use a digital storage oscilloscope for data acquisition.  It will show you the time history for the pressure.  Most of today's oscilloscopes can export data to file, which allows you to process it later. &lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2017-09-03T03:46:39.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16961" PostTypeId="1" CreationDate="2017-09-03T08:35:17.587" Score="2" ViewCount="40" Body="&lt;p&gt;When producing groundwork designs especially in Bentley software, I noticed people like to produce them with contours and vertical lines or even tadpole-shaped lines instead of simply doing triangulation. Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rOuA0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rOuA0.jpg&quot; alt=&quot;pond with vertical lines&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the name of the vertical lines?&lt;/p&gt;&#xA;" OwnerUserId="11826" LastEditorUserId="33" LastEditDate="2018-08-24T23:25:42.160" LastActivityDate="2018-08-24T23:25:42.160" Title="What is the name of the vertical lines in groundwork plans" Tags="|design|cad|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="16963" PostTypeId="2" ParentId="16957" CreationDate="2017-09-03T13:05:28.157" Score="1" Body="&lt;p&gt;We are not here to do your homework for you, so we will only point you in the right direction so that &lt;i&gt;you&lt;/i&gt; can solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by finding the force required in the direction of the rope to hold the cart stationary.  That's the force the sum of the two ropes must apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the tension of a rope is the same everywhere.  The force it exerts on one end is the same as it exerts on the other.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-09-03T13:05:28.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16964" PostTypeId="2" ParentId="14708" CreationDate="2017-09-03T13:11:22.407" Score="1" Body="&lt;p&gt;What you have have posted I believe is Bill of Materials that includes Assembly level representation. Below is the complete BOM which I believe is the one you have partially referenced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yuzXM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lWNFC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you were to create a BOM from the attached spreadsheet I would remove following line items and to create the BOM.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Everyroad, GPS Car Navigation Unit&lt;/li&gt;&#xA;&lt;li&gt;Everyroad, Front Bezel Assembly&lt;/li&gt;&#xA;&lt;li&gt;Everyroad, Rear Assembly&lt;/li&gt;&#xA;&lt;li&gt;Everyroad, PCBA, model 300&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of the above higher level assembly part numbers that require components like capacitors, screws, labels etc to create higher level assemblies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bottom up or top down approach can be followed to create assembly drawings. Below is an example for bottom up assembly structure pseudo steering system.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;| BOM Level | 0 | 1 | 2 | 3 | Part Number | Part Name                                | Revision | Quantity |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   | * |   |   | 20-0001     | Steering System                          |     A    |     1    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   | * |   | 30-0002     | Steering Wheel Assmbly                   |     B    |     1    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-7911     | Spring                                   |     D    |     1    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-7984     | Bearing                                  |     D    |     1    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-7940     | Support                                  |     A    |     1    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-7952     | Screws                                   |     C    |    10    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-7912     | Screws                                   |     A    |     3    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-7918     | Hub                                      |     B    |     1    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   | * |   | 30-0005     | Steering Column Assembly                 |     B    |     1    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-6002     | Flange Nut                               |     B    |     2    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-6005     | Column Holder                            |     A    |     1    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-6010     | Flange Bolt                              |     C    |     2    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-6010     | Steering Joint Gromment                  |     A    |     1    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   | * |   | 30-0010     | Rack and Pinion Assembly                 |     C    |     1    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-5001     | Front Suspension Steering Ball Dust Seal |     B    |     2    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-5010     | Steering Gear                            |     A    |     1    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-5020     | Front Wheel Spindle Connecting End       |     C    |     2    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   | * |   | 30-0020     | Wheel Assembly                           |     C    |     2    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-8012     | Tire                                     |     C    |     2    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-8013     | Bolt                                     |     A    |    10    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   |   | * | 40-8015     | Cap                                      |     C    |     2    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   | * |   | 40-7952     | Screws                                   |     C    |    20    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;|           |   |   | * |   | 40-7912     | Screws                                   |     A    |     5    |&#xA;+-----------+---+---+---+---+-------------+------------------------------------------+----------+----------+&#xA;&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Assembly drawing of a Steering Assembly&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3L8Ws.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FqRtz.jpg&quot; alt=&quot;Steering Assembly&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.arenasolutions.com/resources/articles/bill-of-materials-example/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bill of Materials (BOM) Examples &amp;amp; Samples&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/126/how-to-reduce-engineering-bill-of-material-cost-for-electronics-products&quot;&gt;How to reduce Engineering Bill of Material Cost for electronics products?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://Text%20Tables%20Generator&quot; rel=&quot;nofollow noreferrer&quot;&gt;Text Tables Generator&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2017-09-05T02:13:46.227" LastActivityDate="2017-09-05T02:13:46.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16965" PostTypeId="1" AcceptedAnswerId="16967" CreationDate="2017-09-03T13:35:55.863" Score="0" ViewCount="26" Body="&lt;p&gt;In the context of physical asset management (BIM etc) what is a Design Element? I have come across the term several time but never in an official document - can anyone explain the term and provide a definition in case it is an actual professional term&lt;/p&gt;&#xA;" OwnerUserId="11826" LastActivityDate="2017-09-03T16:28:04.460" Title="What is Design Element in the context of asset management" Tags="|cad|construction-management|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16967" PostTypeId="2" ParentId="16965" CreationDate="2017-09-03T16:28:04.460" Score="1" Body="&lt;p&gt;Wikipedia explains the term &quot;Design Element&quot; quite well. I don't expect there is a formal &quot;official definition&quot; of the term Design Element, but the relevance to BIM (Building Information Modelling) seems fairly obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Visual_design_elements_and_principles&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Visual_design_elements_and_principles&lt;/a&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Graphic_design#Experiential_graphic_design&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Graphic_design#Experiential_graphic_design&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-09-03T16:28:04.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16968" PostTypeId="2" ParentId="16586" CreationDate="2017-09-03T21:02:58.323" Score="1" Body="&lt;p&gt;The IMU (Inertial Measurement Unit) is a component within the aircraft which generates a signal that is proportional to the accelerations experienced on all 9 DOF (degrees of freedom). Modern IMUs are extremely sensitive to acceleration and can be calibrated to measure minute angular changes, very rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IMU signal can be analog or digital and it is processed by a flight controller which is programmed to manage the flight behaviour of the aircraft. It does this by adjusting the control surfaces (ailerons, rudder,...) via a servo, or the propeller speed/pitch in the case of a rotor wing via an ESC or electronic speed controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small &quot;drones&quot; (correctly called UAV or UAS) which use 3 or more horizontal propellers but no wing, only use the relative speed of the motors to manoeuvre the aircraft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the motor speed affects the propeller thrust and so each pair or motors can be adjusted to create differential torque to rotate the aircraft and unbalanced lift to change the aircraft attitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a hovering aircraft is tilted, it will move sideways and therefore the IMU can sense the angle which is then used by the FC to adjust the motors to maintain that angle until the pilot moves the controls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IMU simply generates signals - the flight controller processes them - the ESC is the throttle that controls the motor.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="7021" LastEditDate="2017-09-06T18:57:44.907" LastActivityDate="2017-09-06T18:57:44.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16970" PostTypeId="1" CreationDate="2017-09-04T13:02:46.460" Score="1" ViewCount="74" Body="&lt;p&gt;Being a student in the field of Mechanical Engineering, I am very new to Machine Learning.I did a couple of basic online courses of Machine Learning (including the one by Andrew Ng on Coursera). So, I am a bit curious about using Machine Learning in this field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know is where can I find monitoring data in large quantities related to the use of any mechanical component  so that I can use it to train and predict results(just for practicing so as to know how everything works)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also any other tips/tutorials/guidance on how to start working on the combination of these two fields would be helpful.   &lt;/p&gt;&#xA;" OwnerUserId="8340" LastActivityDate="2018-04-02T16:59:28.893" Title="Historical data for using supervised learning algorithm in the field of Mechanical Engineering" Tags="|mechanical-engineering|design|mechanical-failure|data|failure-analysis|" AnswerCount="1" CommentCount="1" ClosedDate="2018-04-03T11:51:09.230" ContentLicense="CC BY-SA 3.0" />
  <row Id="16972" PostTypeId="2" ParentId="16970" CreationDate="2017-09-04T13:30:23.743" Score="0" Body="&lt;p&gt;I don't know about the data, but currently there is an intensive research on the application of machine learning on solar radiation prediction. Maybe these papers might be useful to you if you can access them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S0959652615004096&quot; rel=&quot;nofollow noreferrer&quot;&gt;The artificial neural network for solar radiation prediction and designing solar systems: a systematic literature review&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1364032115003470&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar radiation forecasting with multiple parameters neural networks&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S1364032115008679&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comparison of regression and artificial neural network models for estimation of global solar radiations&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2017-09-04T13:30:23.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16974" PostTypeId="2" ParentId="16942" CreationDate="2017-09-04T16:32:18.240" Score="1" Body="&lt;p&gt;It could be used for a sprocket/gear/pulley but that is not its purpose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The driven end of the screw is usually attached directly to a motor shaft by a flexible collar which allows some minor misalignment. The part marked &quot;End support mount&quot; has a threaded portion for the nut to lock the shaft into the end bearing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The undriven end is allowed to &quot;float&quot; in its bearing to allow for expansion, usually secured with a circlip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part with the flat is then on the inside of the bearing and inside the machine. It is not a required part of the mechanism and appears unique to that manufacturer.  &lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-09-04T16:32:18.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16976" PostTypeId="1" CreationDate="2017-09-05T02:33:21.107" Score="3" ViewCount="2127" Body="&lt;p&gt;I need to make an nonlinear observer of a system and need to know how the charactaristics of a tire are to get the final conclusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really need some good articles about how the &lt;strong&gt;effective radius&lt;/strong&gt; of a tire will vary depending on velocity and pressure. &#xA;I would appreciate any articles about &lt;strong&gt;rolling resistance&lt;/strong&gt; too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate if someone could give me a hint.&lt;/p&gt;&#xA;" OwnerUserId="8884" LastEditorUserId="8632" LastEditDate="2018-03-05T12:11:44.703" LastActivityDate="2021-01-19T02:07:00.617" Title="Effective radius of a tire and rolling resistance based on information on velocity and tire pressure" Tags="|automotive-engineering|dynamics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16977" PostTypeId="1" CreationDate="2017-09-05T02:38:46.310" Score="1" ViewCount="123" Body="&lt;p&gt;I'm not sure if 'engineering' is the right place to post this question. I'm studying how color filters for a liquid crystal display are fabricated. I found a table on the web which compares pros and cons of each pattern (Please take a look at the attached image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LIaWx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LIaWx.jpg&quot; alt=&quot;color filter patterns&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why are the mosaic and delta patterns difficult to fabricate?&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;Why is the driving circuit of the mosaic pattern complex?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12770" LastEditorUserId="1832" LastEditDate="2017-09-05T23:37:46.637" LastActivityDate="2017-09-05T23:37:46.637" Title="Question about color filters for a liquid crystal display" Tags="|electronic-filters|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="16978" PostTypeId="2" ParentId="16951" CreationDate="2017-09-05T05:37:22.267" Score="2" Body="&lt;p&gt;The design of the submarine to be used in such purpose depends on the following (arranged from least important to most):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Cost - The amount used to build the submarine. With respect to Solar Mike's and Fred's comments, the larger the submarine, the more money you will use to fund the project.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Buoyancy - The larger the submarine, the harder it gets to make it remain submerged. Remember that buoyant force is equal to the weight of the fluid displaced by the object, therefore, having bigger submarine requires bigger engines, thus requires more fuel. Fuel efficiency is very important as depth is about 11 kms below MSL.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Design strength - The walls of the submarine shall be designed to resist 107,910 kPa of pressure (gauge) because the pressure is directly proportional to the depth. Stress = PD / 4t. That means, as the diameter becomes bigger, the thicker the vessel should get to resist pressure. If the submarine becomes too thick, this goes back to item no. 1 (cost efficiency) and no. 2 (submarine now becomes very heavy and will need a lot of power to propel upwards)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2017-09-05T05:37:22.267" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="16979" PostTypeId="2" ParentId="16937" CreationDate="2017-09-05T06:14:06.490" Score="2" Body="&lt;p&gt;The combination of loads shown in the code implies that loads &lt;strong&gt;shall be combined to test all possible structure behavior under different conditions&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Load combinations 1 to 5 incorporates the &lt;em&gt;fluid load&lt;/em&gt;, while load 6 does not. It does not mean that fluid is not present in loading, instead directs the users to consider instances wherein fluid load &lt;strong&gt;will not be present&lt;/strong&gt; e.g. during low tides, or low water table level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take note that sometimes &lt;em&gt;having fewer loads considered leads to larger stresses&lt;/em&gt;. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example of this case is in the analysis of slab in buildings. We used checkered load assignment (I'm not sure how they call it) wherein we assign floor loads alternately in &lt;em&gt;checkered&lt;/em&gt; pattern to simulate behavior of slab when it is not loaded while the others are. This leads to different compression and tension stresses in beams and columns. &lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12936" LastEditDate="2017-09-16T01:18:23.840" LastActivityDate="2017-09-16T01:18:23.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16980" PostTypeId="2" ParentId="16716" CreationDate="2017-09-05T07:28:59.360" Score="1" Body="&lt;p&gt;I will try to answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem, from what I can understand, goes something like this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How are you going to determine the maximum weight the excavator arm can support without tipping over?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your understanding is quite wrong there. It is not as simple as M = F*d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vRITn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vRITn.jpg&quot; alt=&quot;See image for the representation of points and forces considered.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the excavator arm will create a vertical force downward, say F. The arm is hydraulic-system controlled therefore you can safely assume that the arm is free from rotation and can be considered rigid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your drawing, the F is located in the point where the arm is connected to the body. This is quite wrong because that point has nothing to do with the overall tipping action. You should consider the &lt;strong&gt;pivot point&lt;/strong&gt; wherein your excavator would tip. In that case, it is the front end of the steel track of the excavator (in case that the arm is directly in parallel to the tracks). It is labelled as point &lt;strong&gt;O&lt;/strong&gt; in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up moment at point O yields the following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;M &amp;amp;= FL &amp;amp;&amp;amp; \text{(overturning moment)}\tag1 \\&#xA;M &amp;amp;= F_1L_1 &amp;amp;&amp;amp; \text{(righting moment)}\tag2&#xA;\end{align}$$&#xA;where $F$ is the load carried in the arm; $F_1$ is the load (weight) of the counterweights, including the engine, the body, etc; $L$ is the distance between $F$ and $O$, $L_1$ is the distance between $F_1$ and $O$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equating EQ-1 to EQ-2 yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;FL &amp;amp;= F_1 L_1 \\&#xA;F &amp;amp;= F_1 L_1 / L&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the arm capacity is dependent on the weight of the body, the center of gravity of the excavator and its location, and the distance of the excavator arm from the body.&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-07T00:32:48.823" LastActivityDate="2017-09-07T00:32:48.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16981" PostTypeId="1" CreationDate="2017-09-05T08:23:50.947" Score="4" ViewCount="265" Body="&lt;p&gt;Let's take a 2009 Camry as an example. It runs on a 2.4l, 2360cc to be exact, engine with 4 cylinders. This means that the capacity of all the 4 cylinders combined would be around 2.4 litres in volume. I have noticed with Japanese cars after around 3.0L they add two more cylinders and make the engine a V6. Why can't you have a 3.0L I4? Or a 2.4L V6? How does it change the engine? I'm assuming that for a 3.0L I4 the cylinders would have to be larger to hold the extra volume. Would that mean that it would take them longer to burn the fuel inside and lead to slower acceleration maybe? Finally does it matter if the car is a 4WD or a 2WD?&lt;/p&gt;&#xA;" OwnerUserId="12771" LastActivityDate="2018-03-04T12:42:51.170" Title="Engine capacity and number of cylinders" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16982" PostTypeId="2" ParentId="16981" CreationDate="2017-09-05T08:58:25.943" Score="0" Body="&lt;p&gt;You can have a 1.6 Litre V6 if you want - currently used in formula one...&#xA;As to how it changes the engine the inherent balancing of inline 4, v4, V6's, inline 6, straight 8 or v8 does make a difference to how smooth the engine runs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to 2wd or 4wd it does not affect the engine as long as it has sufficient power.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-09-05T08:58:25.943" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="16983" PostTypeId="1" CreationDate="2017-09-05T11:00:15.660" Score="0" ViewCount="1505" Body="&lt;p&gt;Many consumer grade pressure washers specify a minimum pressure for the supplied water, which is sometimes well above the low end of household water pressure.  Our water pressure is often low and I'm looking for a way to verify without a pressure meter that it is adequate for this purpose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data from the utility company isn't useful.  The pressure is very affected by the usage of other houses in the area, plus there are endless effects on the pressure between the water meter and the hose bib.  However, I can use the water at off-peak times and test it at the hose at the time I want to use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulas I've found online tend to deal with what happens inside the pipe or between one point and another.  They require all kinds of parameters that aren't really relevant to this problem.  I'm only interested in what comes out the end of the hose.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that for pressure washer nozzles, the area of the hole relates the flow rate to the pressure, and one factor can be calculated from the other two.  A garden hose is at a very different scale; characteristics that influence flow at high pressure on a tiny scale aren't likely to extrapolate well.  Nevertheless, the same physical laws should apply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end of the hose is a 3/4&quot; hole and I can measure the GPM coming out.  Can the PSI of the system driving the water out of the hose be calculated from just those two factors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked at the standard model of a water tank with a hole in it, which would seem to represent this.  However, the results were nowhere near the same ballpark as household water pressure.&lt;/p&gt;&#xA;" OwnerUserId="12772" LastEditorUserId="12772" LastEditDate="2018-02-23T02:18:01.247" LastActivityDate="2024-03-23T20:07:05.990" Title="Estimate PSI from pipe diameter and GPM" Tags="|fluid-mechanics|pressure|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="16985" PostTypeId="1" CreationDate="2017-09-05T13:05:01.987" Score="2" ViewCount="1638" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lMpQu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lMpQu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this is the truss I have to solve, more precisely I have to find forces in all of the members of the truss. That is why I used method of joints. I started from the joint A and realised without calculating that members attached to it (Fab and Fal) are the zero force members. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dRfSY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dRfSY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;PICTURE ON THE LEFT: After that I continued the process with joint B and calculated forces in members Fbl and Fbc. If I have made a mistake here, I will appreciate  you also point that out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PICTURE ON THE RIGHT: The only joint with 2 unknown members that has left after calculating members of joint B is joint L. But here is the problem. Since I can only use 2 equations - sum of the forces in the X and Y direction (in method of joints), I don't know how to find Flk and Flc, since those are not neither vertical nor horizontal forces. I know I should split those into their components  (vertical and horizontal), but since they are both unknown members (so are the components), I don't see way to calculate any of them.&lt;/p&gt;&#xA;" OwnerUserId="12776" LastEditorUserId="12778" LastEditDate="2017-09-05T14:00:18.943" LastActivityDate="2017-09-06T05:18:45.630" Title="Solving truss using method of joints" Tags="|mechanical-engineering|civil-engineering|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16987" PostTypeId="2" ParentId="16720" CreationDate="2017-09-05T15:47:30.583" Score="2" Body="&lt;p&gt;It really depends on two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What do you hope to get out of the analysis?&lt;/li&gt;&#xA;&lt;li&gt;How much discretization do you need to integrate the equations?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the first item, if you need 100 snapshots of stresses during the rolling process, then try the step time divided by 100 as your initial and maximum time increment. If all you need is the final deformed state of the structure, then you might be able to get away with the step time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Abaqus might reduce your initial increment if it can't integrate the equations. This gets us into the second item. Many of the models in Abaqus (e.g., the material response) require a certain level of time discretization to converge to an acceptable solution. Abaqus will (try to) reduce the time increment to the maximum value that will allow it to integrate the equations. Once you know what this maximum time increment is, best practice is to perform a convergence study by reducing the time increment and checking that your quantities of interest do not change very much. That is, you need to confirm that the time increment selected by Abaqus is accurate enough for your purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, Abaqus may not be able to converge to an acceptable solution using the step time as the initial time increment. In that case, I usually reduce the initial time increment by a factor of ten until Abaqus converges to a solution. The convergence study is still needed after that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have experience with the material, the loading, the contact algorithms, etc..., you'll be in a better position to pick the initial time increment. However, the step time isn't a bad first guess if you know absolutely nothing at all about the system's response. If nothing else, the analysis will break early.&lt;/p&gt;&#xA;" OwnerUserId="12781" LastActivityDate="2017-09-05T15:47:30.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16989" PostTypeId="1" CreationDate="2017-09-05T19:00:38.620" Score="3" ViewCount="462" Body="&lt;p&gt;For pedagogic reasons I am looking for applications in engineering for the simple spring harmonic oscillator, i.e. something like this (&quot;harmonic&quot; implies in particular &quot;without damping&quot;): &lt;a href=&quot;https://i.stack.imgur.com/xe1zg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xe1zg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle to the horizontal my be between 0 and 90 degree and one spring may be missing.&lt;/p&gt;&#xA;" OwnerUserId="12782" LastActivityDate="2018-06-04T23:30:13.593" Title="Applications of simple spring harmonic oscillator" Tags="|mechanical-engineering|applied-mechanics|education|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="16990" PostTypeId="1" CreationDate="2017-09-05T19:02:44.857" Score="2" ViewCount="1640" Body="&lt;p&gt;I am studying this problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The rigid bar ABC shown in Fig. P-212 is hinged at A and supported by a steel rod at B.&#xA;  Determine the largest load P that can be applied at C if the stress in the steel rod is&#xA;  limited to 30 ksi and the vertical movement of end C must not exceed 0.10 in. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rP8sJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rP8sJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solution is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hGYJU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hGYJU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTIONS:&#xA;1. Why is the elongation distribution in the length of the beam a triangle?&#xA;2. Why is the elongation $\delta_{st}$ computed as the elongation in the cable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody please explain the solution?&lt;/p&gt;&#xA;" OwnerUserId="12783" LastEditorUserId="1832" LastEditDate="2017-09-06T14:21:29.707" LastActivityDate="2020-04-24T02:16:27.253" Title="Computing strain in beams" Tags="|structural-engineering|civil-engineering|stresses|deformation|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="16991" PostTypeId="2" ParentId="16990" CreationDate="2017-09-05T19:42:33.517" Score="1" Body="&lt;p&gt;1: The problem describes the beam as rigid, this means it will not deform and will always remain straight. Therefore when loaded in point C it only rotates about A. This is why you get a triangle when drawing the deformed shape together with the original position and the deformation at the points of interest in the same figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: There is no reason they calculate the deformation of the cable first. However, there are two limits states in this problem. One is at failure of the rod at 30 kip and the other is when the rod has deformed so that point C has moved 0.1 in. You have to check them both. &lt;/p&gt;&#xA;" OwnerUserId="11242" LastActivityDate="2017-09-05T19:42:33.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16993" PostTypeId="1" AcceptedAnswerId="16996" CreationDate="2017-09-06T01:01:18.950" Score="2" ViewCount="146" Body="&lt;p&gt;Does anyone know what this mechanism for linear movement is called? I have a different model of something like this, and the black chain appears rigid, so I assume it is the mechanism moving the laser head. I could be wrong though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JvXR2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JvXR2.jpg&quot; alt=&quot;The mechanism in question.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12748" LastEditorUserId="12748" LastEditDate="2017-09-06T01:14:08.080" LastActivityDate="2017-09-06T09:33:42.253" Title="What is this linear movement mechanism called?" Tags="|mechanisms|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="16994" PostTypeId="2" ParentId="16990" CreationDate="2017-09-06T04:55:33.270" Score="2" Body="&lt;p&gt;I will try to explain the solution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;QUESTIONS: 1. Why is the elongation distribution in the length of the beam a triangle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &quot;real-life&quot; visualization of the displacement of the system is supposed to be circular. &lt;a href=&quot;https://i.stack.imgur.com/PtaZs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PtaZs.png&quot; alt=&quot;See image for reference of terms.&quot;&gt;&lt;/a&gt; In the picture, the red lines indicates the exaggerated actual movement of points in the system. As depicted, the end of beam, &lt;em&gt;C'&lt;/em&gt;, displaces by $x_2$ and $y_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, the relationship between the original position and the deflected position of the &lt;strong&gt;rigid&lt;/strong&gt; beam lies in the concept of trigonometry. In civil engineering, we design structures to avoid excessive deflections and angular rotations, such that the angle of depression on the beam, denoted by &lt;em&gt;B&lt;/em&gt; in the figure, is very small, e.g. B &amp;lt; 5 degrees. Therefore, the expected _vertical displacement, $y_2$ of the system is &lt;strong&gt;relatively small&lt;/strong&gt; compared to the length of the beam itself. Also, it is important to note that since $y_2$ is small, $x_2$ will also be extremely small, which is close to a negligible value. &#xA;To provide an easier analysis, we therefore then &lt;em&gt;assume&lt;/em&gt; that the deflected beam projects a triangular profile because &lt;em&gt;C'&lt;/em&gt; is almost vertically collinear with &lt;em&gt;C&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: $X_2 = Y_2\tan(B)$&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 2. Why is the elongation $_{st}$ computed as the elongation in the cable&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The explanation goes similar with question 1, but this does not include arcs. The simplest I can get is that: &lt;strong&gt;For small angles, say angle &lt;em&gt;A&lt;/em&gt;, the hypotenuse of a triangle is approximately equal to one of its leg.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the position of the end of the cable after displacement does not matter anymore as it is very close to zero.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The above explanations makes engineering simpler. Take note that using that concept, you can easily get the displacement of point &lt;em&gt;C&lt;/em&gt; just by ratio and proportion.&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-16T05:53:23.797" LastActivityDate="2017-09-16T05:53:23.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16995" PostTypeId="2" ParentId="16985" CreationDate="2017-09-06T05:09:03.467" Score="3" Body="&lt;p&gt;I cannot view your image clearly, but your problem is solvable:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me try to explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since FBL is known (FBL = 45 tension), the remaining members are FLK and FLC.&#xA;You need to find the inclination of FLK. Solve for angles A and B &lt;a href=&quot;https://i.stack.imgur.com/EaXZJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EaXZJ.png&quot; alt=&quot;in this image,&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up forces along Y gives equation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;FLK (sin A) - FLC (sin B) - FBL = 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Summing up forces along X gives equation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;FLK (cos A) + FLC (cos B) = 0&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then solve the two equations using either: Method of elimination, Method of substitution or by calculator. Please refer to your algebra notes for this :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me give you the first step of Method of Substitution:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;FLK  = - FLC (cos B) / (cos A) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;substitute in the first equation to solve FLC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-06T05:18:45.630" LastActivityDate="2017-09-06T05:18:45.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16996" PostTypeId="2" ParentId="16993" CreationDate="2017-09-06T09:33:42.253" Score="4" Body="&lt;p&gt;The head probably moves by rack and pinion or a hidden timing belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cable management is done by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Cable_carrier&quot; rel=&quot;nofollow noreferrer&quot;&gt;cable carrier&lt;/a&gt;. It works by having each joint only have a limited rotation so that it can support itself when held horizontally.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-09-06T09:33:42.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="16998" PostTypeId="1" CreationDate="2017-09-06T12:47:50.070" Score="1" ViewCount="648" Body="&lt;p&gt;I am setting up a test-bed for an experiment. Here's what I have. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A circular car exhaust like this&#xA;&lt;a href=&quot;https://t1.uc.ltmcdn.com/en/images/5/9/3/img_what_does_white_smoke_from_the_car_exhaust_mean_395_orig.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The gas comes out of the exhaust at a velocity, x.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;We need to lower the gas velocity to X/3 &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Conditions&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The 'reducer' should exert very less backpressure on the flow&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What possible directions should I look into, if you know any pre-filters, baffles, geometries that can help me reduce the flow rate in minimum size, that'd be awesome.&lt;/p&gt;&#xA;" OwnerUserId="12071" LastActivityDate="2017-09-06T16:15:56.920" Title="Reducing air flow speed without much backpressure" Tags="|airflow|aerodynamics|flow-control|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="16999" PostTypeId="2" ParentId="16998" CreationDate="2017-09-06T16:15:56.920" Score="1" Body="&lt;p&gt;Try using a diffusor. But to give you an equation which helps your problem, we need more information about your air flow. Most important is: Air temperature, velocity, heat capacity ratio of your gas (for air approximately around 1.4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This helps to calculate the Mach-number. If it is lower than 0.3 you can just use the Bernoulli equation to get your result. Otherwise the easiest way is to look up gas dynamic tables.&lt;/p&gt;&#xA;" OwnerUserId="12796" LastActivityDate="2017-09-06T16:15:56.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17000" PostTypeId="1" AcceptedAnswerId="17023" CreationDate="2017-09-06T17:12:13.427" Score="1" ViewCount="128" Body="&lt;p&gt;I'm currently writing a simulation environment in python/c for heat networks and got to the point where I'm implementing the controls for the environment.&lt;br&gt;&#xA;The PID controller is already implemented and should be working, but now I stumbled upon a question where I need some advice which way is the best or most common way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I've got a heat generator with the power $P=40\,kW$, through which water with a heat capacity $c_p \approx 4000\,\frac{J}{kgK}$ is flowing. The inlet flow temperature is $T_{in} = 50\,^°C$, current outlet flow temperature at timestep $n$ is $T^{n}_{out} = 70\,^°C$ and the massflow is $\dot m^{n} = 0.5\,\frac{kg}{s}$. The desired outlet flow temperature $T^{*}_{out} = 90\,^°C$.&lt;br&gt;&#xA;For a constant $c_p$ I need to tell the pump to reduce the massflow to $\dot m^{n+1} = 0.25\,\frac{kg}{s}$ to get $T^{n+1}_{out} = 90\,^°C$ (assuming infinitely long timesteps to get a steady state condition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the example fast and easy, I assume to only have a P-controller with a proportional factor of $K_p = 1.2$ and a &lt;em&gt;pump control algorithm&lt;/em&gt;, which is an equation to correlate massflow and temperature, but has no P/I/D-parts!&#xA;The &lt;em&gt;pump control algorithm&lt;/em&gt; equation is:&#xA;$$\dot m = \frac{P}{c_p \left(T_{out} - T_{in}\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've got two possibilites which values to pass as &lt;em&gt;setpoint&lt;/em&gt;-values to the P(ID)-controller:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I can pass the desired outlet temperature $T^{*}_{out} = 90\,^°C$ as setpoint-value and the current outlet temperature $T^{n}_{out} = 70\,^°C$ as process-value to the controller.&lt;br&gt;&#xA;This means that the control variable output of the controller will be $C = \left(90\,^°C - 70\,^°C\right)K_p = 24\,^°C$.&lt;br&gt;&#xA;This will be fed into my &lt;em&gt;pump-control-algorithm&lt;/em&gt; equation with $T_{out} = T^{n}_{out} + C$ which then yields $\dot m^{n+1} = \approx 0.2272 \frac{kg}{s}$&#xA;and will then be set as the massflow.&lt;/li&gt;&#xA;&lt;li&gt;Or I can pass the desired outlet temperature $T^{*}_{out} = 90\,^°C$ to the &lt;em&gt;pump-control algorithm&lt;/em&gt; to get the desired massflow &lt;strong&gt;and then pass the desired massflow as setpoint variable&lt;/strong&gt; to the P(ID)-controller.&lt;br&gt;&#xA;Thus the desired massflow will be according to the &lt;em&gt;pump control algorithm&lt;/em&gt; equation: $\dot m^{*} = 0.25\,\frac{kg}{s}$. This will be passed to the P(ID)-controller as setpoint-value and the current massflow $\dot m^{n}_{out} = 0.5\,\frac{kg}{s}$ as process-value, which yieds the control variable: $C = \left(0.25\,\frac{kg}{s} - 0.5\,\frac{kg}{s}\right)K_p = -0.3\,\frac{kg}{s}$.&lt;br&gt;&#xA;So the massflow set by the pump will be: $\dot m^{n+1} = \dot m^{n} + C = 0.2\,\frac{kg}{s}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Considering both possibilities, the second path with passing the massflow to the controller seems alot more aggressive. Since I have little to no experience with controlling water flows with a pump, I don't know which is the common way to go. Or is there anything I forgot to consider?&lt;br&gt;&#xA;What also could be interesting is the region/nation in which it is made in one or the other way and for which applications. My application would be on the one hand a district heating network and on the other hand potable hot water stations, both situated in germany.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help in advance!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;br&gt;&#xA;As several people keep telling me, that a system like this is unlikely to be built: It is common and very much likely in central europe, where energy efficiency is rated quite highly!&lt;br&gt;&#xA;In fact systems like this &lt;strong&gt;are being built&lt;/strong&gt; often! (I definetely know this, as I am working at operation optimization for systems like this. Thus I'm working with alot of systems like this.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this question is &lt;strong&gt;not&lt;/strong&gt; about if systems like this are likely to exist. It is if the control flow usually first goes to the controller and then to some algorithm which calculates a value to set for an actuator, &lt;strong&gt;or vice versa&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that usually a voltage or PWM is passed to the actuator and &lt;em&gt;not&lt;/em&gt; a massflow which has to be set. This is the only point where I agree that it is unlikely it is going to be done in reality. But this is how it works best for my simulation environment and it &lt;strong&gt;does not affect&lt;/strong&gt; my question which is about the order of the control flow!&lt;br&gt;&#xA;So... Help is &lt;em&gt;really&lt;/em&gt; appreciated, but I don't want to discuss if this system exists or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for help in advance!&lt;/p&gt;&#xA;" OwnerUserId="12796" LastEditorUserId="12796" LastEditDate="2017-09-07T12:56:10.870" LastActivityDate="2017-09-08T11:26:39.333" Title="Layout of a controller for controlling a water pump" Tags="|control-engineering|pid-control|optimal-control|flow-control|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17006" PostTypeId="1" AcceptedAnswerId="17008" CreationDate="2017-09-07T08:05:55.277" Score="1" ViewCount="270" Body="&lt;p&gt;Let's say I'm designing a table surface which has to be stiff, rigid, but at the same time, not too heavy. It is rectangular in shape and ratio of length:width is quite large. It is only supported on both ends of its length. If I use a 50 mm slab of gray cast iron for the surface of the table, during maximum loading the table will bend and the Z-axis displacement at the center would be 0.04 mm. This table will be used as a machining table, and the maximum tolerance of the workpiece will be 0.005 in (0.127 mm), so I think the 0.04 mm deflection of the table is already good because it will give room for tolerance from other component of the machine, such as cutting tools, spindle, etc.&#xA;Problem is, with 50 mm steel slab, the maximum weight criterion is exceeded. The whole table is designed to be able to tilt on A-axis (it's a trunnion table), driven by a motor. The heavier the table is, the bigger and exponentially more expensive the motor would be. I have to reduce the weight of this structure, but I have to keep the stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if instead a simple flat 50 mm slab I make it like 20 mm slab with extra stiffener profile on its underside? Like honeycomb pattern, or any other stiffener configuration. which is possible because it's a cast iron anyway, just have to create the mould.. by doing that I could reduce the weight by almost 40kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: Is it theoretically possible? 50 mm slab gives 0.04 mm bending displacement, now I want to make it like 20 mm slab plus some stiffener to reduce weight, but I would like to keep same performance with 0.04 mm maximum displacement. The maximum load is 1200 N, concentrated in small area in the middle of the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: previously I wrote the requirement as 0.005 mm deflection, but it is actually 0.005 in (0.127 mm). However, this +- 0.127 mm figure is basically the machining tolerance of the workpiece that will be made on top of this table, so I guess the tolerance of table itself should be much lower than that, I don't know the exact number yet but it just won't go as low as 0.005 mm that's probably too extreme. I'm still in the early stage of the design so a lot of design requirements is still open to revision..&lt;/p&gt;&#xA;" OwnerUserId="12801" LastEditorUserId="12801" LastEditDate="2017-09-08T01:01:01.330" LastActivityDate="2020-04-21T18:51:52.390" Title="Increasing the stiffnes of a structure by adding stiffener, but at the same time reducing main surface thickness" Tags="|mechanical-engineering|structural-analysis|machining|deformation|stiffness|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17007" PostTypeId="2" ParentId="17006" CreationDate="2017-09-07T08:16:31.137" Score="1" Body="&lt;p&gt;Wooden tables have been doing that for a long time using an apron. In fact a steel clad wooden surface may be enough. But if you seal the top with a slab of steel you will also want to seal the underside to avoid moisture related warping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add a T-profile beam on the underside connecting the supports and your table will be much stiffer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If necessary you can also add a cross shape or some beams connecting the other two beams to add some support in the center.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-09-07T08:16:31.137" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17008" PostTypeId="2" ParentId="17006" CreationDate="2017-09-07T11:01:02.503" Score="0" Body="&lt;p&gt;The formula for central displacement shows that the displacement is inversely proportional to both the elastic modulus, E, and the second moment of area, I.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to increase either E or I, eight times. Increasing E is not allowed because it is an intrinsic property of the material that you are using. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I is proportional to the cube of height, so doubling height would directly solve your problem. However, the increase in mass would not be allowed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you can do is make a hollow rectangular section and thus increase the height of the cross section as much as possible. This will take some calculations ( by hand or by using a spreadsheet program). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems quite doable.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastEditorUserId="-1" LastEditDate="2020-04-21T18:51:52.390" LastActivityDate="2020-04-21T18:51:52.390" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="17009" PostTypeId="1" CreationDate="2017-09-07T11:30:04.193" Score="0" ViewCount="564" Body="&lt;p&gt;What method should i use in order to remove moisture from air of a container completely? Should i flush it with N2 or argon (its air-tight) or is there a device that i could use to remove moisture from the container? The container is prone to condensation due to extremely low temperatures. I need that water out :D&lt;/p&gt;&#xA;" OwnerUserId="12805" LastActivityDate="2017-09-07T12:10:31.440" Title="Drying a air-tight container" Tags="|compressed-air|drying|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17011" PostTypeId="2" ParentId="17009" CreationDate="2017-09-07T12:10:31.440" Score="1" Body="&lt;p&gt;The de-facto standard way of getting all the water out of a cyrostat before cooling it is to pump it out with a vacuum pump, then back-fill with dry gas from a cylinder.  Nitrogen is cheap and easy, helium is preferred if the cryostat will be cooled below -190C (but I guess &quot;extremely low temperatures&quot; probably isn't that low).  Depending on the achievable pressure with the vacuum pump and how strict the requirement is, you might pump and flush more than once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the requirement is not as strict, it may be enough to blow out the air with dry gas without pumping down to a vacuum first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get rid of a small amount of water after purging, or take up moisture seeping in through a leak, you can use silica desiccant.  It's dirt cheap, but has limited capacity.  When it's full, it can be dried out in an oven and used again.  It's the same stuff that you get in little &quot;do not eat&quot; bags with clothes and goods, but you can buy it in big containers too.&lt;/p&gt;&#xA;" OwnerUserId="12022" LastActivityDate="2017-09-07T12:10:31.440" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17013" PostTypeId="2" ParentId="16989" CreationDate="2017-09-07T16:39:17.193" Score="1" Body="&lt;p&gt;While nothing in the real world truly has zero damping, a good example, though it doesn't use a coil spring would be a tuned mass vibration absorber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.google.com/patents/US5637938&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Tuned Dynamic Vibration Absorber&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's basically a mass attached to a motor, or another vibrating object by a stiff rod. By varying the amount of mass and the stiffness of the rod, the absorber can be &quot;tuned&quot; to have a natural frequency matching the dominant frequency of the motor's vibration. (For most electric motors driven with 60Hz AC, this will be 120Hz)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the motor runs, the 120Hz vibration of the motor will drive the absorber at resonance instead of driving the attached structure which would create noise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While these absorbers are similar in concept to the &quot;Tuned Mass Dampers&quot; seen in large buildings, a tuned mass absorber does not require damping to function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a &lt;a href=&quot;http://www.acs.psu.edu/drussell/Demos/absorber/DynamicAbsorber.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;nice explanation from Penn State&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2061" LastEditorUserId="2061" LastEditDate="2017-09-07T16:44:17.453" LastActivityDate="2017-09-07T16:44:17.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17014" PostTypeId="1" CreationDate="2017-09-07T22:55:28.340" Score="5" ViewCount="393" Body="&lt;p&gt;We have a problem with mounting a PCB at a different angle than the mold direction angle. I have graphic below to help explain. If we were to build standoffs at the same angle as the mold direction but we need that PCB to be 90 degrees from the ground plane. What is the best method for mounting this? Any help would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4pCra.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4pCra.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12820" OwnerDisplayName="Tim T" LastEditorUserId="2819" LastEditDate="2017-09-12T05:34:07.187" LastActivityDate="2020-05-07T13:29:20.920" Title="Mounting a pcb at a different angle than mold direction" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17015" PostTypeId="2" ParentId="17014" CreationDate="2017-09-07T23:51:23.537" Score="1" Body="&lt;p&gt;Does it have to be mounted directly to the housing, or could you use a mount-adjustment plate between the housing and the PCB? it seems to me like the mounting adjustment plate would make your life much easier as you wouldn't need to worry about the screw-hole direction of the PCB.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;suggested solution 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/62IrE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/62IrE.png&quot; alt=&quot;suggested solution 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image, of course, is not to size. you could essentially get away with a 5-6mm gap between the mount and the PCB. Be creative ;)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;suggested solution 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another suggestion would be, depending on the diameter of the housing you could essentially do a straight drill into the housing, if you don't want to deal with the adjustment mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/44Uju.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/44Uju.png&quot; alt=&quot;suggested solution 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;suggested solution 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third option i just thought of would be to use the same idea as your original image, with mounting screws that are 1-2 mm smaller diameter than the through holes on the PCB, and support it with a rubber spacer/washer combination.&lt;/p&gt;&#xA;" OwnerUserId="8543" LastEditorUserId="8543" LastEditDate="2017-09-08T00:05:33.377" LastActivityDate="2017-09-08T00:05:33.377" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17016" PostTypeId="1" CreationDate="2017-09-08T03:38:53.760" Score="0" ViewCount="431" Body="&lt;p&gt;Currently, I'm studying about Ultrasound wave. My aim is set up a system that ultrasound wave can go through ( not reflect and not absorb ultrasound wave). Besides that, this material can contain liquid. &#xA;Can you give me what materials can adapt this experiment?&#xA;Best regards.&lt;/p&gt;&#xA;" OwnerUserId="12813" LastActivityDate="2017-09-08T08:41:06.200" Title="What materials that can contain water and ultrasound wave can go through" Tags="|materials|ultrasound|" AnswerCount="2" CommentCount="2" ClosedDate="2017-09-10T12:37:49.103" ContentLicense="CC BY-SA 3.0" />
  <row Id="17017" PostTypeId="2" ParentId="17016" CreationDate="2017-09-08T05:36:42.210" Score="1" Body="&lt;p&gt;Since ultrasound is often used for medical purposes on humans and animals, I would think that ballistic gelatin could be used for this experiment. It supposedly has the same density as muscle tissue, and is comprised primarily of water, much the same as muscle tissue. The best part is, you can set objects (fake skeletal structure or other foreign objects) into the mold before pouring in the mixture!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to an instructable on how to make regulation test grade ballistics gel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.instructables.com/id/Ballistics-Gel-Step-by-Step-Regulation-Testing-Den/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ballistics &quot;Ordnance&quot; Gel - Step by Step&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8543" LastActivityDate="2017-09-08T05:36:42.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17018" PostTypeId="1" CreationDate="2017-09-08T05:49:49.550" Score="2" ViewCount="553" Body="&lt;p&gt;In order to calculate the elastic critical moment of a laced double I-section, I need the torsional and warping constants. Is there a general formula to obtain this? BS 5950 only has formulas for single I-section. I tried Googling but it seems such information is not readily available.&lt;/p&gt;&#xA;" OwnerUserId="11443" LastActivityDate="2017-09-08T05:49:49.550" Title="General formula for calculating torsional and warping constants of built-up section" Tags="|steel|building-design|eurocodes|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17019" PostTypeId="2" ParentId="17016" CreationDate="2017-09-08T08:41:06.200" Score="1" Body="&lt;p&gt;Any common lab materials will do if all you need is for the ultrasound waves to pass through them, like glass or aluminum plate. More importantly is that the material must not porous and have a good surface condition. Also the angle of incident should be perpendicular to the material surface.&#xA;Look at the table on the bottom of this page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.kayelaby.npl.co.uk/general_physics/2_4/2_4_1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.kayelaby.npl.co.uk/general_physics/2_4/2_4_1.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real value should varies based on surface treatment etc, but it's good for comparison of which one is better than the others. I would suggest aluminum or glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really need a material with as low attenuation as possible for your project, however, you might want to use single crystal metals. The grain boundaries in polycrystalline materials causes scattering in sound wave and increase attenuation, single crystal material has no grain boundaries, it will perform better.&#xA;Casting a large size single crystal metal alloys is possible , they do so with nickel alloy for turbine blades since long ago. However I'm sure it will be very expensive.&lt;/p&gt;&#xA;" OwnerUserId="12801" LastActivityDate="2017-09-08T08:41:06.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17020" PostTypeId="1" AcceptedAnswerId="17026" CreationDate="2017-09-08T09:40:03.983" Score="0" ViewCount="278" Body="&lt;p&gt;Isoparametric triangular finite elements can be formulated using &quot;triangular coordinates&quot; $(\zeta_1, \zeta_2, \zeta_3)$, also sometimes called area/areal/barycentric coords, or using &quot;Cartesian coordinates&quot; $(r,s)$ defined on a master element. These two systems can be related to each other without difficulty, and seem to be functionally equivalent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that several texts and journal papers use triangular coords  $(\zeta_i)$ exclusively, particularly those that are more recent and/or mathematically oriented. On the other hand, I have older textbooks and use certain commercial FEA packages that consistently present element formulations with $(r,s)$ coordinates. I have a feeling that this might be due to historical reasons, but perhaps there is more to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided that the element geometry is restricted (straight sides, and any side nodes are uniformly spaced), I understand that triangular coords may be useful because they lend themselves to analytical integration, and thus avoid potentially costly numerical integration procedures. However, when these conditions are not met, numerical integration must be used. In that case, is there any benefit from using triangular coordinates, conceptual or computational?&lt;/p&gt;&#xA;" OwnerUserId="11602" LastEditorUserId="11602" LastEditDate="2017-09-08T10:25:05.253" LastActivityDate="2017-09-08T19:27:17.837" Title="Do triangular/area coordinates offer computational advantages for numerically integrated triangles?" Tags="|finite-element-method|mathematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17021" PostTypeId="2" ParentId="17014" CreationDate="2017-09-08T09:51:05.180" Score="0" Body="&lt;p&gt;How are the screws to be fitted into the standoffs? Moulded-in inserts, or self-tappers? Is a drilling process after the moulding acceptable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a drilling process is acceptable, I'd elongate the standoffs in the vertical direction - or simply make the upper side vertical (increasing the draw angle), so that drilling normal to the PCB doesn't break out through the top of the standoff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're using moulded-in inserts, again, I think you can elongate the standoffs to accomodate them normal to the PCB (and skewed wrt the mould)&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2017-09-08T09:51:05.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17022" PostTypeId="2" ParentId="17014" CreationDate="2017-09-08T11:23:32.073" Score="0" Body="&lt;p&gt;One possibility is to use fairly long screws and place a short length of eg soft silicone tube either side, this should take up the difference caused by the screws not being perpendicular to the board and also give some vibration isolation as a bonus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar solution is to use angled spacers/washers (eg short length of plastic tube cut at an appropriate mitre angle, This might be more fiddly to assemble though. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-09-08T11:23:32.073" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17023" PostTypeId="2" ParentId="17000" CreationDate="2017-09-08T11:26:39.333" Score="1" Body="&lt;p&gt;Control problems are usually easier if you specify&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The measured variables, I assume $T_{in}$ and $T_{out}$&lt;/li&gt;&#xA;&lt;li&gt;The manipulated variables, I assume the pump speed according to your description&lt;/li&gt;&#xA;&lt;li&gt;The objectives of the controller. Do you need to be exactly at 90 degC or would and outlet with a certain offset be acceptable? Or is the temperature very important and you would vary the pump outlet a lot if needed in order to be exactly at 90 degC?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Your approaches are almost feedforward: you rely on a basic model of how water gets heated in order to control your system. Although that can be a sensible option, it can fail if there are heat losses, changes in $T_{in}$ that you are not measuring, etc. For such a system good old feedback should work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can accept a certain offset, the easiest by far is to use just a proportional controller using $T_{out}$ as a controlled variable (with setpoint 90 degC or slightly higher) and the pump speed (or voltage) as the manipulated variable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to be exactly at 90 degC, then you would require integral action. The tricky part is that you can make your controller unstable if you specify a very tight integral action or have very long oscillations if the integral action is too slow. However, for a system like yours, it should not be a problem. &lt;/p&gt;&#xA;" OwnerUserId="3991" LastActivityDate="2017-09-08T11:26:39.333" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17024" PostTypeId="1" AcceptedAnswerId="17025" CreationDate="2017-09-08T16:15:59.243" Score="2" ViewCount="3519" Body="&lt;p&gt;What are the tolerance specs on the major diameter of imperial bolts? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, a 1/4-20 imperial bolt has a major diameter of 0.25in, but what is the tolerance on that dimension. I (and everyone I've asked) have always taken that value to be exact, but that's obviously incorrect. &lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-09-08T17:43:25.760" Title="Tolerance on bolt major diameters" Tags="|standards|tolerance|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17025" PostTypeId="2" ParentId="17024" CreationDate="2017-09-08T17:43:25.760" Score="2" Body="&lt;p&gt;Here is a link to a screw thread chart.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineersedge.com/screw_threads_chart.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineersedge.com/screw_threads_chart.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note there are different tolerance classes, and also note the max is never over the nominal size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the example 1/4-20&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;class 2: 0.2408 &amp;lt; d &amp;lt; 0.2489&#xA;class 3: 0.2419 &amp;lt; d &amp;lt; 0.25&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2017-09-08T17:43:25.760" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17026" PostTypeId="2" ParentId="17020" CreationDate="2017-09-08T19:27:17.837" Score="0" Body="&lt;p&gt;Triangular coordinate systems (or tetragonal systems, in 3D) have the advantage of giving all the vertices of the element equal status. That becomes clear in numerical integration procedures, when you look at the positions and weights of the integration points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three coordinates are not independent - they should sum to 1 at every point in the element. But usually, preserving the symmetry in the formulas is more useful than losing the independence of the coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Early papers often picked one vertex (arbitrarily) as &quot;special&quot;, and mapped that vertex to the origin of a local system and the adjacent sides along the X and Y axes. For an arbitrary shaped triangular element, this will give &lt;em&gt;different results&lt;/em&gt; depending which vertex you put at the origin, which can be confusing when trying to understand why two identical-looking models don't give the same answers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the era of automatic mesh generation and refinement, triangular meshes often had a systematic bias anyway (e.g. start with a rectangular mesh and draw all the diagonals pointing in the same direction) so treating one vertex of the triangle as &quot;special&quot; might have been a &lt;em&gt;good&lt;/em&gt; thing to do, in conjunction with that type of mesh. But that's all history now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you &lt;em&gt;could&lt;/em&gt; re-map a &quot;symmetrical&quot; numerical integration procedure onto an &quot;old-school&quot; local coordinate system, if you really wanted to - but it's easier not to bother doing that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hardly anyone uses analytic integration for &quot;real world&quot; problems any more - even if the geometry is suitable for it, properties like Youngs Modulus often vary (e.g. with temperature) within a single element. In any case, you can create numerical integration rules that give &lt;em&gt;exact&lt;/em&gt; results for the special cases that can be integrated analytically, so why bother to use two different integration methods when one method will handle both situations?&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-09-08T19:27:17.837" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17027" PostTypeId="1" CreationDate="2017-09-08T20:02:15.087" Score="1" ViewCount="3200" Body="&lt;p&gt;Let's say I have a Sphere made out of Acrylic, with a diameter &lt;code&gt;d&lt;/code&gt; meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I want to submerge it down in the ocean. The depth is such that it reaches up to &lt;code&gt;p&lt;/code&gt; PSI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How thick does the Sphere have to be, in order to withstand pressure of up to &lt;code&gt;p&lt;/code&gt; PSI?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I used the variables &lt;code&gt;d&lt;/code&gt; and &lt;code&gt;p&lt;/code&gt; so I can be given a formula that I can learn/understand instead of a raw answer.)&lt;/p&gt;&#xA;" OwnerUserId="12660" LastActivityDate="2023-06-25T14:01:25.670" Title="Material thickness to withstand Pressure" Tags="|mechanical-engineering|pressure|waterproofing|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17028" PostTypeId="1" CreationDate="2017-09-08T20:32:58.390" Score="3" ViewCount="715" Body="&lt;p&gt;I am confused about the difference between NPT, outer diameter of a pipe, and nominal pipe size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have a 1/8&quot; NPT pipe. Does this mean the outer diameter of my pipe is 1/8&quot;? Is the nominal pipe size equal to the outer diameter of the pipe?&lt;/p&gt;&#xA;" OwnerUserId="12821" LastActivityDate="2017-09-09T13:43:25.383" Title="What is the difference between NPT and pipe diameter?" Tags="|piping|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17029" PostTypeId="2" ParentId="17028" CreationDate="2017-09-08T21:11:19.157" Score="0" Body="&lt;p&gt;The NPS designation is a name that indicates a size but it does not define that size.  Like &quot;two by four&quot; is the name of a certain size of lumber, it not the actual size of the lumber. There is a particular size of pipe that is defined by the name AND schedule ; such as 2 inch, schedule 80.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-09-08T21:11:19.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17030" PostTypeId="2" ParentId="17027" CreationDate="2017-09-08T21:12:41.070" Score="2" Body="&lt;p&gt;take note if you are talking about dropping to the deepest parts of the ocean the pressure can be as high as 15,000 psi.  the Yield strength of acrylic is maybe 10,000 psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your wall will be very thick! You can not use most of the thin shell based formulae you might look up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can take an extreme view and imagine a large block of material with a tiny void. The wall hoop stress at the void wall is approximately 3/2*pressure, or over 20,000 psi. Which is to say your cavity is liable to collapse no matter how thick you make it.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2017-09-08T21:12:41.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17032" PostTypeId="2" ParentId="17027" CreationDate="2017-09-09T00:28:10.930" Score="1" Body="&lt;p&gt;The basic formula used in analyzing &lt;em&gt;thin-walled spherical pressure vessel&lt;/em&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;stress = PD / 4t&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where P is the pressure inside or outside, D is the inner diameter, and t is the wall thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;t = PD / [(allowable stress of the material) * 4]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Note that this formula is only applicable if D/t &gt;= 10&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that allowable stress in acrylic is 10,000 psi @ &lt;strong&gt;room temperature&lt;/strong&gt;. This changes when you submerge this in water, so might want to consider this too.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;The unit of measurement in the formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For English system: stresses should be in &lt;em&gt;pounds per sq.inch (psi)&lt;/em&gt; while dimensions should be in &lt;em&gt;inches (in.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For SI system: stresses should be in &lt;em&gt;megaPascal (MPa)&lt;/em&gt;, and dimensions should be in &lt;em&gt;millimeters (mm)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-09T00:42:04.210" LastActivityDate="2017-09-09T00:42:04.210" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="17033" PostTypeId="1" AcceptedAnswerId="17038" CreationDate="2017-09-09T03:39:16.520" Score="1" ViewCount="72" Body="&lt;p&gt;If a 4x4 (true 4x4 not AWD) has 3 wheels slipping, does the one tire with traction receive all the torque available from the transmission? Is there typically a reduction in power transmission in 4x4 systems to the wheels that are not normally powered (i.e. front wheels on a pickup)?&lt;/p&gt;&#xA;" OwnerUserId="1174" LastActivityDate="2017-09-09T12:57:59.327" Title="Equal power transmission in 4x4 vehicles?" Tags="|transmission|car|traction|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17034" PostTypeId="2" ParentId="16081" CreationDate="2017-09-09T06:47:10.527" Score="1" Body="&lt;p&gt;For a closed system (no mass exchange, only heat exchange is possible), the chagne in the internal energy is the obtained by subtracting the expansion work from heat flow in to the system. So, if there is no free expansion ,i.e. no work is done on the system or by the system. The change in internal energy is equal to the heat supplied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;dQ = dU&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a open system (where there is mass and heat exchange), enthalpy is the term that corresponds to free expansion. Enthalpy is given by,&#xA;H = U + (P.V)&lt;/p&gt;&#xA;" OwnerUserId="12825" LastActivityDate="2017-09-09T06:47:10.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17035" PostTypeId="1" CreationDate="2017-09-09T07:43:33.623" Score="1" ViewCount="33" Body="&lt;p&gt;I'm control engineering student.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As seen on the video link below, how the controller controls the tip vibration and the link movement at the same time? Please explain using the block diagram of the system, since i am quite confused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know about simple feedback controllers. For instance if i have a simple dc motor position control. How it works is if the &lt;em&gt;now position&lt;/em&gt; of the motor is less than the setpoint, the controller add a little bit power to the motor so the motor turns. Otherwise, if the &lt;em&gt;now position&lt;/em&gt; is more than the setpoint, controller trying to reduce power or inverse the polarity of the power given to match by the setpoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the flexible manipulator i have no idea about how the control system works. Since the controller must regulate the motor speed and direction of rotation to somehow reduce the tip vibration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=VF673MGugjY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Single Link Flexible Manipulator Controller&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12826" LastEditorUserId="12826" LastEditDate="2017-09-09T07:54:54.630" LastActivityDate="2017-09-09T07:54:54.630" Title="How Single Link Flexible Manipulator Controller controls the vibration of the link tip?" Tags="|control-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17037" PostTypeId="2" ParentId="17033" CreationDate="2017-09-09T12:38:25.897" Score="1" Body="&lt;p&gt;Only if the differential is limited slip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no limited slip diff then the wheels with no traction will just spin wildly and no torque is transferred to the wheel with traction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some off-road vehicles let you lock the diffs so that torque can be delivered to all wheels regardless of traction.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-09-09T12:38:25.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17038" PostTypeId="2" ParentId="17033" CreationDate="2017-09-09T12:57:59.327" Score="4" Body="&lt;p&gt;It depends on what type of differential it has. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nature of an open differential (as found on normal road cars) is that the spinning wheel(s) will get &lt;em&gt;all&lt;/em&gt; of the torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Proper off-road vehicles will usually have a diff-lock selector which works on some of all of the differentials which means that both driveshafts the diff is connected to get the same torque, regardless of slip. This should only be used at low speeds on slippery surfaces as the locked differential won't be doing its normal job anymore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More sophisticated vehicles may have limited slip differentials which are a compromise between an locked and open differential and in some cases allow adjustment of the degree of slip either by manual selection of computer control. Depending on the degree of slip selected the differential will always send some fraction of torque to both wheels regardless of their relative speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Limited slip differentials are particular useful on vehicles which are intended to drive at speed on loose surfaces eg in rallying and for permanent AWD cars as they can also be used to adjust turn-in characteristics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is traction control which monitors the relative speed of the driven wheels and applies the brakes to lock the spinning wheel which transfers torque to the opposite wheel. The are also manual versions of this known as 'fiddle brakes' where the driver has separate brake levers for each side of the vehicle (or front/rear) this system can also be used for 'skid steer' a bit like the way that most tracked vehicles are steered. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-09-09T12:57:59.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17039" PostTypeId="2" ParentId="17028" CreationDate="2017-09-09T13:43:25.383" Score="2" Body="&lt;p&gt;Pipe comes in different wall thicknesses, designated schedule 40, Sch. 80, Sch. 10, etc.  Higher numbers are thicker walls.  The outer diameter is the same for all schedules, so the inner diameter varies.  The nominal size is close to - but not exactly - the actual inner diameter for schedule 40 pipe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a chart &lt;a href=&quot;http://www.engineeringtoolbox.com/ansi-steel-pipes-d_305.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;summarizing the dimensions&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2017-09-09T13:43:25.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17040" PostTypeId="1" AcceptedAnswerId="17078" CreationDate="2017-09-09T22:54:46.257" Score="1" ViewCount="33" Body="&lt;p&gt;Has anyone designed a pond that extends beyond its surface perimeter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example cross section:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ywPlj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ywPlj.png&quot; alt=&quot;sub surface example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;brown - existing surface (or brought to grade)&#xA;black - designed pond&#xA;blue - capacity&lt;/p&gt;&#xA;" OwnerUserId="11826" LastEditorUserId="2819" LastEditDate="2017-09-12T04:43:53.583" LastActivityDate="2017-09-12T05:03:17.173" Title="Is it possible to create a sub-surface pond?" Tags="|water-resources|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17041" PostTypeId="1" AcceptedAnswerId="17043" CreationDate="2017-09-10T00:11:34.397" Score="0" ViewCount="4853" Body="&lt;p&gt;QUESTION(see image attached):  &lt;a href=&quot;https://i.stack.imgur.com/S14iQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S14iQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really need help here can someone help me show how we get the solution as shown in the image file attached above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;okay since I was asked by one of the users commenting asking me to show what I had done; please see attached image file below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fFKwB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fFKwB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12830" LastEditorUserId="12830" LastEditDate="2017-09-10T20:59:36.887" LastActivityDate="2017-09-10T20:59:36.887" Title="Determine the reactions at A and C, knowing that α = 30º" Tags="|mechanical-engineering|statics|moments|" AnswerCount="1" CommentCount="2" ClosedDate="2017-09-20T12:00:08.047" ContentLicense="CC BY-SA 3.0" />
  <row Id="17043" PostTypeId="2" ParentId="17041" CreationDate="2017-09-10T02:44:26.657" Score="1" Body="&lt;p&gt;First you have to notice that the contact at A is a roll support, so the reaction force there is perpendicular to the surface. Therefore Fa has 60 degree against horizontal axis. When you draw free body diagram, decompose the Fa reaction force into its horizontal and vertical components:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fax = Fa Cos60&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fay = Fa Sin60&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you solve the balance equations of X-axis, Y-axis, and Moments. I would start from the  sum of Moments at C equal zero. You will get the value of Fa.&#xA;Then with the balance equations of X-axis and Y-axis with Fa already known you will get the value of Fcx and Fcy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get a slightly different result, probably due to rounding:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fa = 365.24 N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fcx = 782.62 N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fcy = 316.3 N&lt;/p&gt;&#xA;" OwnerUserId="12801" LastActivityDate="2017-09-10T02:44:26.657" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17044" PostTypeId="1" CreationDate="2017-09-10T06:50:30.477" Score="4" ViewCount="616" Body="&lt;p&gt;I'm 17 years old and learning first-year college math. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What college level math do I need for mechanical linkage design?&lt;/li&gt;&#xA;&lt;li&gt;Can I learn designing and creating mechanical linkage without math but with my geometric intuition? or there's resource I can upgrade these skill?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It will be very good if you introduce some source textbooks which treat mechanical linkage design.&lt;/p&gt;&#xA;" OwnerUserId="12834" LastEditorUserId="8055" LastEditDate="2017-09-11T18:54:59.363" LastActivityDate="2023-11-01T15:30:39.857" Title="what college level teaches mechanical linkage design?" Tags="|mechanical-engineering|mechanisms|linkage|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17045" PostTypeId="1" AcceptedAnswerId="17046" CreationDate="2017-09-10T07:49:23.540" Score="7" ViewCount="3564" Body="&lt;p&gt;Let's assume I want to build a subway system. I dig two subway stations and now I start digging a tunnel from both of these endpoints simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I ensure that the two tunnels dug from the different endpoints meet in the middle? I don't believe I can use GPS underground. What methods do professionals use to ensure that the two tunnels meet? Are there some positioning technologies that work underground?&lt;/p&gt;&#xA;" OwnerUserId="12836" LastActivityDate="2021-06-05T07:26:12.480" Title="How are tunnels dug from two endpoints joined?" Tags="|civil-engineering|tunnels|positions-measurement|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17046" PostTypeId="2" ParentId="17045" CreationDate="2017-09-10T08:45:22.130" Score="9" Body="&lt;p&gt;The alignment of a tunnel constructed from two ends and meeting somewhere is the middle is achieved via &lt;a href=&quot;http://www.aboutcivil.org/tunnel-survey.html&quot; rel=&quot;noreferrer&quot;&gt;old fashioned surveying methods&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tunnel breaks through to the surface, surveying stations are accurately established on the surface of the Earth at each end of the tunnel. These days GPS surveying techniques would most likely be used for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the surface surveying stations have been established, surveying stations are established underground, as the tunnel progresses from each end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to the use of lasers and electronic distance measuring devices, &lt;a href=&quot;https://en.wikipedia.org/wiki/Theodolite&quot; rel=&quot;noreferrer&quot;&gt;theodolites&lt;/a&gt; &amp;amp; tape measures were used to take surveying measurements so that co-ordinates could be assigned to each underground surveying station. These days, &lt;a href=&quot;https://en.wikipedia.org/wiki/Total_station&quot; rel=&quot;noreferrer&quot;&gt;total stations&lt;/a&gt; tend to be used to undertake the measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ensure the alignment of the tunnel is true, &lt;a href=&quot;https://en.wikipedia.org/wiki/Gyrotheodolite&quot; rel=&quot;noreferrer&quot;&gt;gyrotheodolites&lt;/a&gt; will be used periodically to check the position of the underground surveying stations. These use a gyroscope to determine the orientation of true north.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tunnel will not break through to the surface, vertical shafts must be first constructed by which people and equipment can be lowered to access the shaft. Unlike tunnels that break through to the surface, where people can walk from the surface into the underground tunnel, surveying lines and stations cannot be taken directly underground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To establish the first surveying station underground two plumb bobs need to be lowered down, at least one shaft. From the surface the co-ordinates of each plumb bob line can be determined. This gives a surveying plane that can be transferred underground. From this, the first underground surveying station will be established and the remainder of the underground surveying technique will be the same a for a tunnel that breaks through to the surface.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-09-10T08:45:22.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17048" PostTypeId="1" CreationDate="2017-09-10T11:11:06.453" Score="1" ViewCount="13" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Szujg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Szujg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the shear force vertical when the flow is horizontal? Also, why is the shear force at the top surface of the control volume negative and the bottom is positive when they are both in the same direction?&lt;/p&gt;&#xA;" OwnerUserId="8607" LastActivityDate="2017-09-10T11:11:06.453" Title="For a flow driven by a pump between 2 parallel, stationary plates, why is the shear force vertical and is there positive and negative shear force?" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17049" PostTypeId="1" CreationDate="2017-09-10T12:46:34.303" Score="3" ViewCount="168" Body="&lt;p&gt;For an experiment I need to use an electric motor, unfortunately I'm not an expert of electric machines. The electric motor I need has to comply with some requests:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It must be brushless.&lt;/li&gt;&#xA;&lt;li&gt;It has always to turn at low speed, from 0 to max 6 rad/s.&lt;/li&gt;&#xA;&lt;li&gt;Its width can't exceed 2cm and its diameter can't exceed 11cm.&lt;/li&gt;&#xA;&lt;li&gt;It is energized by a battery.&lt;/li&gt;&#xA;&lt;li&gt;It has to develop a quite intense torque.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So I'd like to know which type of electric motor meets these requests (if any). &lt;/p&gt;&#xA;" OwnerUserId="12837" LastEditorUserId="1586" LastEditDate="2017-11-01T17:35:54.770" LastActivityDate="2017-11-01T17:35:54.770" Title="Which type of electric motor I need?" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="17050" PostTypeId="2" ParentId="17049" CreationDate="2017-09-10T13:28:54.257" Score="1" Body="&lt;p&gt;Here are some source for you investing with regard to Brushless motors. Take look at the Hobby Robotic Motor Vendor. For the most part they a good source to visualize possible solution. Also it might be best to narrow down a target supply voltage and possible battery options. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;High Volume Vendors&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://orientalmotor.com/brushless-dc-motors/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brushless DC Motor Speed Control Systems&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.moonsindustries.com/Moons-Category/Brushless-DC-Motors/c/A05&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brushless DC Motors&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.nmbtc.com/brushless-dc-motors/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MinebeaMitsumi Group BRUSHLESS DC MOTORS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.deltaww.com/Products/CategoryListT1.aspx?CID=030501&amp;amp;PID=ALL&amp;amp;hl=en-US&quot; rel=&quot;nofollow noreferrer&quot;&gt;Delta Group Brushless DC Motors (H/A Motors)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.hansen-motor.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hansen Precision Electric Motors&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.czleili.com/english/products_class_87.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Leili Micro-motor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.zwgearbox.com/StandardPlanetaryGea.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;ZHAOWEI - zwgearbox - Standard Planetary Gearbox&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://jvlusa.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;JVL INTEGRATED SERVO MOTORS AND STEPPER MOTORS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hobby Robotic Motor Vendors&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.adafruit.com/category/229&quot; rel=&quot;nofollow noreferrer&quot;&gt;ROBOTICS &amp;amp; CNC / MOTORS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.pololu.com/category/22/motors-and-gearboxes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pololu Mechanical Components Motors and Gearboxes&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sparkfun.com/categories/178?page=all&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sparkfun Motors and Drivers&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hope these resource will help you get started.&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2017-10-28T14:04:46.330" LastActivityDate="2017-10-28T14:04:46.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17051" PostTypeId="1" AcceptedAnswerId="17059" CreationDate="2017-09-10T14:47:09.877" Score="1" ViewCount="2029" Body="&lt;p&gt;The other day I was sawing some wood on my mitre saw. I have a single light bulb in that room and it was on. The moment I press the button for the saw to turn on, the lights dim a little and go back to full brightness while the saw is running. The saw was connected to the same room. Why does this happen only when the saw is started? The saw has supposedly slightly  less than 2 kW power and uses a universal electric motor.&lt;/p&gt;&#xA;" OwnerUserId="12532" LastEditorUserId="2819" LastEditDate="2017-09-12T05:38:49.203" LastActivityDate="2017-09-12T05:38:49.203" Title="Why do the lights dim when I turn on a mitre saw?" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17052" PostTypeId="1" AcceptedAnswerId="17076" CreationDate="2017-09-10T16:21:50.790" Score="0" ViewCount="2151" Body="&lt;p&gt;Most if not all of the off-the-shelf epoxies I've used have generated a fair amount of heat while they are hardening. This seems to indicate to me that these are exothermic reaction, i.e. they generate heat from the reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The common practice at my last company was to apply heat (heatgun or hairdryer) to epoxied joints to get them to cure faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand the chemistry though, adding heat to an exothermic reaction should slow the reaction and effectively take longer for the joint to cure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was my company's 'standard practice' wrong to be adding heat to epoxy joints and extending the curing time, or is epoxy some type of anomalous material that cures faster when adding heat to the exothermic reaction. OR, is epoxy actually an endothermic reaction and the heat perceived during curing is something else? &lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-09-12T04:24:02.353" Title="Epoxy curing, heat or no heat?" Tags="|epoxy|chemistry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17053" PostTypeId="1" AcceptedAnswerId="17077" CreationDate="2017-09-10T17:20:25.727" Score="0" ViewCount="85" Body="&lt;p&gt;I'm doing a project on Theo Jansen's Mechanism, where I have to modify the mechanism. What is the most efficient way to design these mechanisms? Is there any good software where I can just draw the links and input crank and test out the mechanism? I have used GeoGebra, but it doesn't simulate the mechanism physically, and while using Creo I found that the pin joints often get stuck - like they suddenly fold up and do not move. Also, creating the links one by one takes quite a lot of time.&lt;/p&gt;&#xA;" OwnerUserId="12842" LastEditorUserId="2819" LastEditDate="2017-09-12T04:36:17.620" LastActivityDate="2017-09-12T04:38:36.323" Title="How to quickly design and simulate crank based mechanisms?" Tags="|mechanical-engineering|mechanisms|linkage|" AnswerCount="1" CommentCount="0" ClosedDate="2017-09-20T11:58:25.723" ContentLicense="CC BY-SA 3.0" />
  <row Id="17054" PostTypeId="1" CreationDate="2017-09-10T17:30:30.370" Score="0" ViewCount="62" Body="&lt;p&gt;I've been working on a design that requires a plastic snap feature. I've assumed up to this point that the material will be injection molded ABS, with material properties I pulled off of matweb. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After several dozen iterations on the design, I've pushed the geometry of the snap as far as possible give the geometric conditions of the feature and I'm right on the edge of acceptable (Factor of safety of ~1.1). If my loading conditions end up being too conservative or any features of my part come out slightly out of spec I could easily have parts that fail. I'd be much more comfortable with an FOS up above 1.8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally I'd like to lower the Youngs modulus of my material and I'm wondering if ABS can have its material properties 'toyed' with by using a different formulation of the plastic. Also, how consistent are material properties of injection parts? &lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-09-25T05:30:00.753" Title="Material Properties of ABS" Tags="|materials|design|abs|injection|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17055" PostTypeId="1" CreationDate="2017-09-10T19:10:41.793" Score="1" ViewCount="49" Body="&lt;p&gt;It is said when a planet gear revolves around a fixed sun wheel, the number of revolution it makes might be more than what appears to be logical. For example, if the planet and sun are fitted with 20 and 40 teeth respectively, the planet will rotate three times as it travels around the circumference of the sun wheel. Why three not two? Is there an explanation or formulae for this?&lt;/p&gt;&#xA;" OwnerUserId="12843" LastActivityDate="2017-09-11T11:09:19.597" Title="Planetary gearing" Tags="|mechanical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17056" PostTypeId="1" CreationDate="2017-09-10T21:51:15.577" Score="0" ViewCount="676" Body="&lt;p&gt;QUESTION (see attached image file): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o6fb2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o6fb2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got stuck when trying to find tension for part A; see my working below.&#xA;and help me solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kn4ie.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kn4ie.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12830" LastEditorUserId="12597" LastEditDate="2017-09-11T02:42:27.297" LastActivityDate="2017-09-16T01:25:08.947" Title="Finding the force in a cable of a rigid system" Tags="|mechanical-engineering|statics|moments|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17057" PostTypeId="1" CreationDate="2017-09-10T23:13:28.513" Score="2" ViewCount="17" Body="&lt;p&gt;Let's say I am running a FDTD simulation of a wave-equation to determine a transfer function of an LTI system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;H(f) = \dfrac{Y(f)}{X(f)}\&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $Y(f)$ and $X(f)$ are Fourier transforms of signals measured at some point in the FDTD grid for a grid with an &quot;obstacle&quot; inserted and an empty grid, respectively. Assume that it's an acoustic wave simulation and that the &quot;obstacle&quot; is modelling a fully reflective object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: does the type of an excitation signal matter in that kind of problem? For example, how broadband the signal is? And if does matter, then - how? My intuition tells me since in the end we're computing a relation of two signals, any propagation issues of the input signal will be the same in both simulations and therefore cancel each other in the above equation. The only issue that I can think of is the SNR.&lt;/p&gt;&#xA;" OwnerUserId="12845" LastActivityDate="2017-09-10T23:13:28.513" Title="Does excitation matter in a time-domain simulation/computation of a transfer function of an LTI system?" Tags="|simulation|acoustics|numerical-methods|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17058" PostTypeId="2" ParentId="17056" CreationDate="2017-09-11T00:31:59.710" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tZgH5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tZgH5.png&quot; alt=&quot;See image for reference.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is a trigonometric problem. The trick is for you to solve the value of A, B, C, and x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve for x, you'll need the &lt;em&gt;cosine law&lt;/em&gt; such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angle C is very obvious to be equal to 70 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x^2 = 10^2 + 12^2 - 2(10)(12)cos(70)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x   = 12.7246 \ cm\ = 12.72 cm.$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;now, solve for the value of angle (A+B). Angle A is equal to 20 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$cos (A+B) = (12^2 - 10^2 - 12.72^2) / (-2*10*12.72)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore, $$(A+B) = 62.416\  degrees$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wiNVh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wiNVh.png&quot; alt=&quot;Rotated image.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, rotating the FBD by 20 degrees clockwise will give the image above. Summing up moments along the hinge (C) will give:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T sin(62.416) * 10 = 200 sin (70) * 15$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$$T = 318.06 \ N \ ----&amp;gt;answer$$&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12936" LastEditDate="2017-09-16T01:25:08.947" LastActivityDate="2017-09-16T01:25:08.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17059" PostTypeId="2" ParentId="17051" CreationDate="2017-09-11T01:51:25.067" Score="6" Body="&lt;p&gt;The wires in your home have resistance. Current going through those wires causes a voltage drop. The startup (inrush) current of electric motors can be several times their normal operating current. It's impossible to know exactly what the startup current is for your motor without knowing specifics supplied by its manufacturer. The high startup current occurs because the wires in the armature windings appear almost as a short until they set up a magnetic field (become energized). The momentary high current doesn't last long enough to cause the breaker to trip or fuse to blow.  There's a pretty good explanation &lt;a href=&quot;http://sciencing.com/calculate-motor-inrush-current-7791777.html&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;. &lt;br&gt;So...the momentary high current causes a like momentarily high voltage drop which in turn causes the light to dim.&lt;/p&gt;&#xA;" OwnerUserId="12535" LastActivityDate="2017-09-11T01:51:25.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17061" PostTypeId="1" CreationDate="2017-09-11T07:48:34.113" Score="1" ViewCount="59" Body="&lt;p&gt;I have simulated a 2D model in Abaqus which indicates 2 main part, first a sector that represent as wheel and, second a rectangular that represent rail. the wheel and rail are in contact and wheel should rotate and move from left to right. there is a vertical small crack in middle of the rail to studying the result of contact on fracture mechanics. my model has been completed, and in the 1/10 of total time good result has been achieved but after that the stress distribution is completely wrong and although the contact seems to exist, the stress is almost zero. Any guide, comment or advice would be appreciated. thank you&lt;/p&gt;&#xA;" OwnerUserId="12849" LastActivityDate="2017-09-11T13:58:36.217" Title="Mechanical Engineering- Modeling contact in Abaqus" Tags="|simulation|rail|numerical-methods|abaqus|mechanical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17062" PostTypeId="1" AcceptedAnswerId="17068" CreationDate="2017-09-11T09:21:51.083" Score="-2" ViewCount="42" Body="&lt;p&gt;What is the relationship between nozzle size and sound produced? If there are references on it,kindly put it the comment space.&lt;/p&gt;&#xA;" OwnerUserId="12321" LastActivityDate="2017-09-11T13:52:09.053" Title="relationship between nozzle radius and sound produced" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17063" PostTypeId="1" CreationDate="2017-09-11T09:37:18.267" Score="2" ViewCount="177" Body="&lt;p&gt;I'm a little confused by the differences between LQR control and numerical optimal control (such as direct multiple shooting, direct collocation and pseudospectral optimal control). It seems that LQR uses linear system dynamics to minimize a quadratic cost function and can find a solution rapidly, so it's suitable for on-board control. Whereas numerical optimal control uses full non-linear system dynamics to minimize a cost function (not necessarily quadratic), but requires a fairly good initial guess to achieve convergence to the desired solution, and can take a relatively long time to find a solution, so it isn't generally suitable for on-board control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having had some experience implementing numerical optimal control algorithms, I'm hoping to try and implement an LQR controller in similar simulated problems to see how each differs. If, for example, we consider the problem of minimizing the fuel used when vertically landing the first-stage of a rocket back on Earth, using 3-DOF dynamics in a Cartesian coordinate system, we get the following output for the system's states, $s = (x,y,v_{x},v_{y},m)^{T}$, and controls, $u=(u_{x},u_{y})^{T}$, versus time (the weird jagged control inputs that occur for most of the control profiles are due to the rocket's engines being turned off (coasting) and so control input is arbitrary since it has no effect on the rocket's trajectory).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wKDWz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wKDWz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/viHsE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/viHsE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GLf1x.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GLf1x.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lSheC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lSheC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DODGK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DODGK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iELje.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iELje.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6wzWm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6wzWm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kitRF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kitRF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our required final state is zero x- and y-velocity $(v_{x}(t_{f}),v_{y}(t_{f}))=(0,0)$, and, if we consider our landing site to be at the origin of our coordinate system, a final x- and y-position of $(x(t_{f}),y(t_{f})) = (0,0)$. As can be seen, given our initial conditions, numerical optimal control produces a trajectory made up of a series of &quot;waypoints&quot; every few seconds, giving the control profile required to achieve the desired final state (a two-point boundary value problem), and simultaneously tries to minimize the fuel used to achieve such a trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, does an LQR controller act in a similar way, where I only have to give it my system's initial conditions, and it'll work out the series of control input values to use in order to meet the desired final conditions whilst minimizing fuel use? It seems almost too good to be true! I feel like I'm missing something fundamental to my understanding of their differences.&lt;/p&gt;&#xA;" OwnerUserId="9755" LastEditorUserId="9755" LastEditDate="2017-09-11T15:42:35.070" LastActivityDate="2017-09-11T15:42:35.070" Title="LQR vs. Numerical Optimal Control" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17064" PostTypeId="2" ParentId="17055" CreationDate="2017-09-11T11:09:19.597" Score="1" Body="&lt;p&gt;It's one more than you think due to the gear going around.  So 2 due to the teeth ratio, +1 for going around.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-09-11T11:09:19.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17065" PostTypeId="1" AcceptedAnswerId="17072" CreationDate="2017-09-11T11:24:39.927" Score="7" ViewCount="416" Body="&lt;p&gt;Below is a stack of reinforced concrete slabs with hollows (camera is positioned along the hollows' axis). Such slabs are used for floors - you build walls of the next floor, then lay such slabs such that they rest onto walls with the end shown on the picture and the opposite end. The dark round dots are steel rebars covered in rust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QqSej.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QqSej.jpg&quot; alt=&quot;stack of slabs with hollows&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen slabs with round hollows and with oval hollows but this shape is new to me. It's something like a rounded hexagon with an extra cavity on the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the name of this shape and why is it used for hollows in a reinforced concrete slab?&lt;/p&gt;&#xA;" OwnerUserId="374" LastActivityDate="2022-01-25T23:25:04.367" Title="What's the name and purpose of this specific shape of hollows in a reinforced concrete slab?" Tags="|structural-engineering|civil-engineering|concrete|reinforced-concrete|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17066" PostTypeId="2" ParentId="17052" CreationDate="2017-09-11T11:46:40.987" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As I understand the chemistry though, adding heat to an exothermic reaction should slow the reaction and effectively take longer for the joint to cure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Higher temperature means higher reaction speed, due to higher molecular kinetic energy. What changes for exothermic vs. endothermic reactions is their equilibrium point, not their speed. In the case of epoxy polymerization, you have an irreversible reaction, so the point is moot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Epoxy can self-degrade if the heat release is not acounted for&lt;/strong&gt;; but this heat is, by itself, not enough to even fully polimerize the adhesive, except in very thick parts where heat cannot be conducted away fast enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the aerospace industry, for example, epoxy is cured at high temperatures and pressures, with a careful temperature ramp to avoid local overheats from the exothermic release in thick sections.&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2017-09-11T11:46:40.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17067" PostTypeId="2" ParentId="17065" CreationDate="2017-09-11T12:16:27.953" Score="1" Body="&lt;p&gt;The solid bits between the holes are probably more informative. They remind me of the shape of steel rails use for railroads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solid bits are buttressed at their bases, with the buttresses encroaching into the holes, resulting in the holes having an unusual shape. The tops of the holes look like they're arched. Notice also, that the reinforcing steel rods have been placed in the middle of the buttressed walls between the holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all suggests such slabs would be used in high stress situations and the holes are present to reduce the mass of the slabs.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-09-11T12:16:27.953" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17068" PostTypeId="2" ParentId="17062" CreationDate="2017-09-11T13:52:09.053" Score="1" Body="&lt;p&gt;Imho the nozzle diameter should only affect the impedance, but not the frequency. The frequency is affected by the length of the nozzle. Everything else is really easy to look up using some google skills and keywords like &quot;acoustic resonance&quot;.&lt;/p&gt;&#xA;" OwnerUserId="12796" LastActivityDate="2017-09-11T13:52:09.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17069" PostTypeId="2" ParentId="17061" CreationDate="2017-09-11T13:58:36.217" Score="1" Body="&lt;p&gt;There can be thousands of causes for your error. You need to give many more informations to be able to make a guess.&#xA;Like: discretization, solving methods for the DEQs, grid topology, boundary conditions etc.&lt;/p&gt;&#xA;" OwnerUserId="12796" LastActivityDate="2017-09-11T13:58:36.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17070" PostTypeId="2" ParentId="17056" CreationDate="2017-09-11T17:07:18.470" Score="1" Body="&lt;p&gt;It looks like you're grasped the basic concept of the question -- using trigonometry to solve for the inclination of the cable, and then summing moments about point C to find the tension in the cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the trig, angle alpha is 47.6025 degrees (as shown in sketch below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G6k8xl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G6k8xl.jpg&quot; alt=&quot;Geometry&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, summing moments about point C, the tension T is calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SU9ZIl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SU9ZIl.jpg&quot; alt=&quot;Summation of Moments&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2017-09-11T17:07:18.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17072" PostTypeId="2" ParentId="17065" CreationDate="2017-09-11T21:54:30.713" Score="7" Body="&lt;p&gt;It looks like a typical precast concrete &quot;hollow-core&quot; or &quot;voided&quot; slab. A quick Google image search reveals the variety of void shapes employed by manufacturers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XAXZfl.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XAXZfl.jpg&quot; alt=&quot;Voided Slab Examples&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image on the left (from Oldcastle Precast, incidentally) nicely shows the variation of void shape with increasing slab thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in the other answers, we care about the concrete profile more than the void profile. The voids are simply introduced to minimize concrete area where it is not structurally necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, a rectangular void would minimize concrete. However, the corners are better off rounded to reduce the stress concentrations (and therefore minimize cracking). Hence, the oblong voids. In thinner slabs, the voids may indeed approach a circular shape. One possible explanation for the irregular void shape in your posted question, is that the irregular void can set proper concrete cover for the rebar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jd6rRl.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jd6rRl.jpg&quot; alt=&quot;Thoughts on Void Shape&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2017-09-11T21:54:30.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17074" PostTypeId="2" ParentId="17044" CreationDate="2017-09-12T00:09:28.347" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;To answer your title, It's not really a matter of what college&#xA;level, but rather what college. For example, a general studies&#xA;college (one with a broad range of majors) may not offer such a&#xA;class, whereas a technical college that focuses specifically on&#xA;mechanical engineering will offer a variety of these courses at&#xA;multiple levels.&lt;/li&gt;&#xA;&lt;li&gt;To answer your first question - a mixture of simple algebra,&#xA;geometry, and trigonometry are the mathematics required for understanding and designing the more advanced linkages. However, knowledge of physics is also necessary.&lt;/li&gt;&#xA;&lt;li&gt;To answer your last second question - While it is possible to do&#xA;this, you will never fully understand the concepts behind it&#xA;mathematically, and when you do pursue a career in this subject -&#xA;you may find it very difficult to have a timely workflow without&#xA;knowing the fundamental basics or the formulas behind it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8543" LastActivityDate="2017-09-12T00:09:28.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17075" PostTypeId="2" ParentId="17044" CreationDate="2017-09-12T04:19:37.790" Score="2" Body="&lt;p&gt;There are some purely geometric linkage design methods, such as two and three position synthesis (which are easy enough to teach even a 8 year old). But there are methods that are not so geometrically intuitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real problem with linkage design is that the mathemathics behind kinematics require a very good foundation of trigonometry and/or imaginary numbers, you know the things that nearly everybody cited as never needing for anything. It is also useful to know linear algebra for coordinate transforms and inverse kinematics calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even so trigonometric methods only lend themselves to 2D mechanisms. If you want to go further especially into dynamics, then you need multivariable calclulus and larangian mechanics. But beyond basics this gets really nasty, as you have many many singularties. And you end up doing a lot of numerical simulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its one of those things that is deceptively simple at first. Good understanding of statics and equation solving is a plus.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-09-12T05:01:51.520" LastActivityDate="2017-09-12T05:01:51.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17076" PostTypeId="2" ParentId="17052" CreationDate="2017-09-12T04:24:02.353" Score="0" Body="&lt;p&gt;Per the &lt;a href=&quot;https://en.wikipedia.org/wiki/Arrhenius_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arrhenius Equation Wiki&lt;/a&gt;; &quot;the reaction rate doubles for every 10 degree Celsius increase in temperature.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that temperature increases the reaction speed is independent of endothermic or exothermic. The reaction rate increases because the motion of particles is greater when the temperature is higher, which increases the likeliness (and speed) that two reactant molecules will meet one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used a microwave (very briefly) to get a head start on fiberglass epoxy in cold weather, but heating your work space is a much better idea. Since it is exothermic like you mentioned, it is easy for it to have a run-away reaction where the heat it generates further speeds it up. Too fast of cure time will result in a much weaker material; too slow (like days in the cold) and it is possible for the hardener wick into a surface more than the resin and the epoxy will never fully harden.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-09-12T04:24:02.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17077" PostTypeId="2" ParentId="17053" CreationDate="2017-09-12T04:38:36.323" Score="0" Body="&lt;p&gt;3D CAD software has a hard time with lots of pin connections, because without a range limitation the solution is indeterminate. I have not used Creo, but if you can limit the range of motion with interference or other, the model may work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.design-simulation.com/wm2d/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Working Model 2D&lt;/a&gt; is nice for designing mechanisms. I used it as a student; not sure how much it costs commercially.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-09-12T04:38:36.323" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17078" PostTypeId="2" ParentId="17040" CreationDate="2017-09-12T05:03:17.173" Score="3" Body="&lt;p&gt;I would describe this negative incline portion of the pond as a &quot;shelf&quot;. They are common and need to be self-supporting with a material that has compression and tensile strength; concrete and rebar, rock, fiberglass, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of examples of this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;covered irrigation canals&lt;/li&gt;&#xA;&lt;li&gt;dam water turbine intakes&lt;/li&gt;&#xA;&lt;li&gt;water diversion intakes&lt;/li&gt;&#xA;&lt;li&gt;zoo habitats&lt;/li&gt;&#xA;&lt;li&gt;natural shelf under waterfalls&lt;/li&gt;&#xA;&lt;li&gt;naturally in water caves&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-09-12T05:03:17.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17079" PostTypeId="2" ParentId="17014" CreationDate="2017-09-12T05:30:32.233" Score="1" Body="&lt;p&gt;A countersinking (usually wood) screw would allow you to mount the board with angled studs and keep the head from scraping the surface or applying an eccentric load. Remember to oversize the hole to account for the angle.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/eCjIB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eCjIB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of attachment mechanisms beyond screws:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hot Glue (fast and reliable with proper molded guides)&lt;/li&gt;&#xA;&lt;li&gt;Snap fit (if mold limits permit)&lt;/li&gt;&#xA;&lt;li&gt;Capture between two molded pieces (it usually takes to pieces to make an enclosure anyways)&lt;/li&gt;&#xA;&lt;li&gt;Slide over plastic studs and quickly heat the studs with an iron to retain the board&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-09-12T05:30:32.233" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17080" PostTypeId="2" ParentId="16915" CreationDate="2017-09-12T06:13:52.283" Score="2" Body="&lt;p&gt;12 liters per hour and 100% efficiency is:&lt;br&gt;&#xA;12l/hr * 1kg/l * 1hr/3600s * 2m * 9.81m/s^2 = &lt;strong&gt;0.0654 Watt&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best case efficiency on a pump this small without a military budget will probably be 25%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To operate for one day on a 12v battery you would need:&lt;br&gt;&#xA;(0.0654w/12v)/0.25 * 24hr = &lt;strong&gt;0.5AmpHr/day&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the details of your setup, but it would probably be more efficient (for the cost) to pump one liter up to a gravity holding chamber (perhaps with a level switch) 12 times a day, rather than run a very tiny pump continuously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very small &lt;a href=&quot;https://www.aliexpress.com/item/1PCS-used-DC-6-V-submersible-pump-sinking-pump/32690622459.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;centrifugal pump&lt;/a&gt; or &lt;a href=&quot;https://www.aliexpress.com/item/2PCS-Micro-Vacuum-Pump-Medical-Mini-Negative-Pressure-Pump-DC-Pressure-Small-Air-Pump-DC3V-6/32791745569.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;diaphragm pump&lt;/a&gt; would probably be your best bet for the low head application. A &lt;a href=&quot;https://www.aliexpress.com/item/24V-Brushless-DC-Mini-Water-Pump-Submersible-40H-24110-11M-450LPH-with-DC-Plug-Low-Noise/32606474788.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;brushless centrifugal pump&lt;/a&gt; will generally have higher efficiency if you can find one small enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is any available light (even indirect), solar may be applicable for this low demand application.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-09-12T06:13:52.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17083" PostTypeId="1" AcceptedAnswerId="17524" CreationDate="2017-09-12T13:01:12.643" Score="1" ViewCount="138" Body="&lt;p&gt;The CNC spindle is supposed to operate its single shaft in two different modes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;high RPM (30,000-80,000 RPM), variable, possibly quite high torque, a straightforward PID to maintain the RPM as torque varies (&quot;work mode&quot;, RPM control).&lt;/li&gt;&#xA;&lt;li&gt;very low RPM (exact value doesn't matter), nearly no torque (just friction of bearings), rotation of shaft to precise angle. (&quot;optical scan of the tool, from all sides&quot;, angle control). &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The first is achievable through a good AC or DC motor. The latter would be achievable through a stepper motor with microstepping. I can cut off any of the motors (leave it idle) and drive the other to achieve either of the modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't think typical stepper motors are capable of surviving 80,000 RPM applied externally - and I'd prefer to avoid a mechanical clutch decoupling the stepper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there steppers that can just survive these kinds of speeds? Or other neat solutions that don't overly increase the mechanical complexity?&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-10-14T05:23:44.853" Title="Stepper on high-RPM shaft" Tags="|motors|stepper-motor|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17085" PostTypeId="2" ParentId="17044" CreationDate="2017-09-12T19:08:13.217" Score="4" Body="&lt;p&gt;It all depends on how complicated the linkage is. For many simple linkages all you need to know is &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/ttrig.html#c1&quot; rel=&quot;nofollow noreferrer&quot;&gt;trigonometry&lt;/a&gt; (sin, cos, tan) and &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/torcon.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;torque/moment equilibrium&lt;/a&gt; (Force x Distance = Force x Distance). Once you can do this I would say you are in the 95th percentile of linkage design ability. Most non mechanical engineers do not have the ability to calculate or visualize the force transmission of a simple linkage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get better and design more complex systems you will need physics 111 and 112 (calculus based). Calculus 1, 2, and 3 are also necessary, mostly 3 because it involves &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;vectors&lt;/a&gt;. Statics and Dynamics (200 level engineering courses) tie everything together and reinforce the academic rigor to reliably calculate mechanisms. Machine design (300 or 400 level engineering courses) teach how to add in dynamics, strength of the linkage, inertia, how to simply complex gear systems, optimization, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of good &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0321625927&quot; rel=&quot;nofollow noreferrer&quot;&gt;physics&lt;/a&gt;, &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0134160681&quot; rel=&quot;nofollow noreferrer&quot;&gt;statics&lt;/a&gt; and &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0133915387&quot; rel=&quot;nofollow noreferrer&quot;&gt;dynamics&lt;/a&gt; books. Probably have a look through a local college bookstore and see what they have (i usually just got the isbn and bought mine used on half.com to save $). Or visit with a professor; they may be able to give you an old copy or point you in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have a good trig and physics foundation I would recommend &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0073398209&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shigley's Mechanical Engineering Design&lt;/a&gt; and the &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/1591264146&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mechanical Engineering Reference Manual&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-09-12T19:08:13.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17086" PostTypeId="1" CreationDate="2017-09-12T19:56:04.913" Score="1" ViewCount="91" Body="&lt;p&gt;I would like to work on a science project which I will measure people how much pressure I have to put on a pressure point to start to hurt somebody. Is there a device which I can measure the pressure that I apply onto the person's skin?&#xA;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="12867" LastEditorUserId="2819" LastEditDate="2017-09-13T04:34:05.587" LastActivityDate="2017-09-13T04:34:05.587" Title="How to measure force against skin?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17088" PostTypeId="2" ParentId="17086" CreationDate="2017-09-13T04:33:08.170" Score="1" Body="&lt;p&gt;Use a &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B00R1J9DZ6&quot; rel=&quot;nofollow noreferrer&quot;&gt;load cell&lt;/a&gt; or a &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B004164SRA&quot; rel=&quot;nofollow noreferrer&quot;&gt;digital scale&lt;/a&gt; to measure force. A lower accuracy &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B00B887CLS&quot; rel=&quot;nofollow noreferrer&quot;&gt;force-sensing resistor&lt;/a&gt; may be sufficient. If you simplify your test to vertical, gravitation force on a known mass may suffice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As user8055 pointed out, you will need to make sure the area and shape of the end of the protrusion is constant in your testing.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-09-13T04:33:08.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17089" PostTypeId="2" ParentId="17083" CreationDate="2017-09-13T05:11:04.977" Score="1" Body="&lt;p&gt;A &lt;a href=&quot;https://www.vxb.com/One-Way-Bearings-s/231.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;one way clutch bearing&lt;/a&gt; on the stepper motor might give you the clutching action you want without much added mechanical complexity (note that this would require some experimentation). The high rpm motor would be attached to the tool, and would spin in the free wheeling direction. The stepper motor would be attached to the one way clutch bearing and index in the opposite (clutch locked) direction. This obviously limits stepper motor control to one direction. It also depends on the internal friction of the high rpm motor to keep from free wheeling after positioning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stepper motors are rated at lower rpm because they have less and less torque the faster they go; so they generally won't self power up to high speeds. However, it is also very possible that the bearings in the stepper would survive the 30,000 RPM. I am not sure about 80k though; the rotar will likely handle the load (not 100% sure though) and it should be reasonably balanced. You could possibly try tearing one down and pressing in 80k rpm rated bearings (sounds like a project in itself though). Cooling would be another possible road block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible option would be for the high rpm motor to be on the head and a high torque stepper motor (and large bearing) on the bed. Definitely some design compromises and software programming that direction, but something to consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then there is the less attractive, but simple compromise of having two separate heads ;-)&lt;br&gt;&#xA;Good Luck!&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-09-13T05:11:04.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17090" PostTypeId="2" ParentId="16732" CreationDate="2017-09-13T05:53:21.160" Score="1" Body="&lt;p&gt;I think using the 5 liter tank complicates the control; I would avoid using it for the draining process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend hooking the pump directly to the 50 liter tank. Place the pump on the ground so it gets as much suction head as possible (to reduce the risk of cavitation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that a gear pump should not run dry, but a few seconds will not harm it (especially with oil). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method would be to add a restriction or decrease the size of the discharge hose of that pump so that it is pumping against a pressure when at full flow (5psi or 35kPa should do). Then use a pressure switch to shut the pump down if that pressure is not reached (indicating pumping air). It will take some tuning and you will have to have a start switch that bypasses the trip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method would be to turn off the pump when the pump amps go low for a second or two; indicating the pump is pumping air instead of fluid. You could use a current transducer to detect the two states and shut down the pump with a micro controller. &lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-09-13T05:53:21.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17091" PostTypeId="1" CreationDate="2017-09-13T07:37:52.870" Score="1" ViewCount="2473" Body="&lt;p&gt;I am planning to deploy two Ethereum mining rig in my store room. Both rigs will have a total of 14 RX 570 GPU's and rest of usual suspects. The GPUs and PSU will be running at full usage while other items nearly at idle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any ventilation in store at the moment so I will be installing a few 120mm fans (maybe 6 or 9) as intake on one side (in the wall) and exhaust on the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone give me an estimate (very rough one) of how much CFM I should intake/exhause in order to sufficiently cool these 2 rig?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Apologies for the delay but my query was to install fans in the walls on the store room and not on either side of case. The case will have its own fans but I wanted to know how much CFM air to bring in the store room from one wall to exhaust it at another.&lt;/p&gt;&#xA;" OwnerUserId="12873" LastEditorUserId="12873" LastEditDate="2017-09-13T17:32:15.043" LastActivityDate="2017-09-13T17:32:15.043" Title="CFM needed to ventilate a Ethereum mining rig?" Tags="|thermodynamics|heat-transfer|cooling|" AnswerCount="1" CommentCount="4" ClosedDate="2017-09-20T11:54:48.803" ContentLicense="CC BY-SA 3.0" />
  <row Id="17092" PostTypeId="1" CreationDate="2017-09-13T09:30:23.097" Score="1" ViewCount="214" Body="&lt;p&gt;I have one floor mall with 10 shops (4 meter height), I should supply water from the main drain (3 inch) connected with pressure reducer (1 1/2 inch). I don't have the flow rate of at the 1 1/2 inch, but i know that the water can reach the roof.&#xA;I am planning to install 3 plastic water tanks with volume of (3000 litre each).I was thinking to connect those three tanks with valve.&#xA;The owner insist to have for each shop different centrifugal pump with 1 1/2 inch pipe. Is't acceptable and workable to provide 10 small pumps (divided into 3 groups, 4 pumps connected to the first WT, 3 PUMPS TO THE SECOND AND 3 FOR THE LAST WT).&#xA;please advise if that feasible and what should be the best solution in this case.&#xA;thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="12874" LastEditorUserId="12804" LastEditDate="2017-09-13T17:09:06.060" LastActivityDate="2017-09-14T01:32:36.207" Title="Water supply system" Tags="|mechanical-engineering|pumps|water-resources|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17093" PostTypeId="2" ParentId="17091" CreationDate="2017-09-13T09:51:21.800" Score="1" Body="&lt;p&gt;Instead of following Solar Mike's answer of simply adding one fan per side, I'd rather try to make it more compact by thinking about an intelligent layout, which incorporates directing the cooling flow to the GPU surface instead of just blowing parallely to its board.&#xA;An example is shown in the attached image. This could reduce the needed amount of fans to $\left(n-1\right)/2$, with $n$ being the amount of GPUs.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wnpaT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wnpaT.png&quot; alt=&quot;GPU rig with 3 GPUs and two different kinds of flow diverters&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flow diverters between the GPUs are of two different shapes. The left one, the bow shaped ones, are more optimized. The start at an angle parallel to the flow and then divert the flow to the GPU surface. This is good for having a low pressure loss and thus a high mass flow, but difficult to manufacture.&lt;br&gt;&#xA;On the right side is a simple flow diverter. This could be just some metal sheets, which are connected by some simple metal bar at the top and bottem side (shown with dots) which also is with where they could be mounted to the case. Easy to manufacture and quite ok flow diverting.&lt;br&gt;&#xA;Without any diverter you'll need an ALOT higher mass thoughput to get some turbulence with gets the air to the GPU surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:&lt;br&gt;&#xA;If the GPUs already have their own fans, forget about the diverters and just make sure that fresh air is only coming from one side. That means to have the output side of the GPU fans facing away from the case fan. Perhaps even make some kind of a closed case at the backside, top and bottom of the GPUs to avoid pushing fresh air out at these sides!&lt;/p&gt;&#xA;" OwnerUserId="12796" LastEditorUserId="12796" LastEditDate="2017-09-13T10:02:07.600" LastActivityDate="2017-09-13T10:02:07.600" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17094" PostTypeId="1" CreationDate="2017-09-13T13:57:02.443" Score="1" ViewCount="204" Body="&lt;p&gt;I see on internet, that aircraft propellers most effective blades angle only 15 degrees on every blade. Why propellers not use 45 degrees angle on propeller's blade that give more thrust on the aircraft rather than 15 degrees?&lt;/p&gt;&#xA;" OwnerUserId="12876" LastActivityDate="2018-10-21T23:24:27.903" Title="Propeller design" Tags="|pressure|aerospace-engineering|aerodynamics|aircraft-design|thrust|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17096" PostTypeId="1" AcceptedAnswerId="17098" CreationDate="2017-09-13T15:23:15.637" Score="2" ViewCount="805" Body="&lt;p&gt;Historically the longest lasting building materials seem to be stone. But I was wondering what material modern technology can create that would last longer under various conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The various things the material would need to be tolerant to would include extreme heat, cold, water, wind, plants, animals, sunlight, earthquakes, as well as a determined effort by people to deface, damage or destroy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had considered steel blocks with a coating of an alloy or metal that would not react to air or water, or maybe an advanced form of composite glass or plastic such as bullet and bomb resistant glass that would be multiple feet thick and reinforced with metal as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was planning to ask this question in an architecture stack, but all I could find was home improvement and I thought engineering may have more people versed about materials science than do it yourselfers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if a modern team was going to build a pyramid and wanted it to last much longer than those of Egypt what materials would likely be used?&lt;/p&gt;&#xA;" OwnerUserId="12877" LastActivityDate="2023-03-27T05:14:23.133" Title="What is the longest lasting modern building material?" Tags="|materials|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17097" PostTypeId="2" ParentId="17094" CreationDate="2017-09-13T15:24:11.460" Score="2" Body="&lt;p&gt;Just think of the propeller as a mean to thrust air in one direction. The higher the angle of incidence, the more the propeller is pushing the air (up to about 45°, then the force perpendicular to the rotational direction starts to take over).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the fact that air is inertial mass. So accelerating it means building up pressure behind the propeller. Now when the angle of incidence becomes too big, so it thrusts too much air, the pressure &quot;behind&quot; the propeller becomes to high. The energy needed for the air to flow against that pressure is extracted from the air, thus the flow starts to separate from the propeller blade and the blade begins to stall. That means that it is losing most or even all of its aerodynamic properties and no or only little air is being thrusted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This critical angle of incidence depends on many factors, like the rotational speed, density of air, propeller profile, relative axial speed of the propeller to the air flow, etc....&lt;/p&gt;&#xA;" OwnerUserId="12796" LastEditorUserId="77" LastEditDate="2018-10-21T23:24:27.903" LastActivityDate="2018-10-21T23:24:27.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17098" PostTypeId="2" ParentId="17096" CreationDate="2017-09-13T15:32:28.640" Score="2" Body="&lt;p&gt;I guess some kind of steel alloys, most probably featuring a high amount of titanium, will be the best, perhaps with some ceramic coating to weaken the effect of corrosion (even if it is already small for most titanium alloys).&#xA;But ceramic coatings may be too brittle to survive people trying to deface the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same counts for composite glass and other ceramics. Nearly all of them are extremely brittle and will thus deteriorate under the influence of impacting forces. Bomb resistant glass for example also shatters when an explosion hits it, but a plastic foil in between the glass layers keeps the outer structure intact. But this does not apply for constant effort to shatter it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So imho only corrosive resistant metal alloys will be lasting for a long time.&lt;/p&gt;&#xA;" OwnerUserId="12796" LastActivityDate="2017-09-13T15:32:28.640" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17099" PostTypeId="1" CreationDate="2017-09-13T15:46:00.287" Score="1" ViewCount="939" Body="&lt;p&gt;I understand plane strain/stress relations are typically utilized for 2D materials. I was wondering if we consider a 2D material acting in a 2D space, would we have a stiffness matrix such as&#xA;\begin{equation*}&#xA;\left[&#xA;\begin{array}{ccc}&#xA;C_{1111} &amp;amp; C_{1122} &amp;amp; C_{1112} \\&#xA;&amp;amp; C_{2222} &amp;amp; C_{2212} \\&#xA;&amp;amp; &amp;amp; C_{1212}&#xA;\end{array}&#xA;\right]&#xA;\end{equation*}&#xA;rather than just plane strain/stress relations imposed on the classical stiffness tensor in 3D. Are there any situations where such a formulation would be appropriate in the real world? Thanks for the insight!&lt;/p&gt;&#xA;" OwnerUserId="12881" LastActivityDate="2019-01-18T04:38:50.200" Title="Stiffness Tensor in Two Dimensions" Tags="|elastic-modulus|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17100" PostTypeId="1" CreationDate="2017-09-13T16:06:38.393" Score="1" ViewCount="40" Body="&lt;p&gt;Does having multiple angular contact bearings together increase the total dynamic loading of the system linearly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: Angular contact bearing with rated dynamic loading at 50kN, thus 3 of them setup together will result in a dynamic loading of 150kN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See picture for reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5YSfY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5YSfY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9186" LastActivityDate="2017-09-13T16:06:38.393" Title="Angular Contact Bearing Dynamic Loading" Tags="|mechanical-engineering|bearings|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17101" PostTypeId="1" CreationDate="2017-09-13T16:32:55.373" Score="0" ViewCount="101" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/uTCgA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The banana clips attach to a multi-meter out of view, the power supply wires are correctly in place, and the resistors are correctly in place. Would putting the red left-hand banana clip and its respective wire into the H15 tie node place the multi-meter in a correct--i.e. series and not parallel--position to measure total current out?&lt;/p&gt;&#xA;" OwnerUserId="12882" LastEditorUserId="8055" LastEditDate="2017-09-14T16:21:41.323" LastActivityDate="2017-09-14T16:21:41.323" Title="Measuring Current In/Out of Resistors" Tags="|electrical-engineering|circuits|" AnswerCount="1" CommentCount="4" ClosedDate="2017-09-20T11:46:11.210" ContentLicense="CC BY-SA 3.0" />
  <row Id="17102" PostTypeId="1" CreationDate="2017-09-13T17:52:58.883" Score="1" ViewCount="221" Body="&lt;p&gt;I drew a little picture:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bJWsX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bJWsX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to use you imagination to view these pictures as profile of bolt threads. In derivation on torque required to raise load, the angle of friction (ro) is calculated like in the left picture. Then it says that due to the angle of the 'tooth', friction-angle is adjusted, like in the right picture. &#xA;The problem is, a force F2 should exist, but they just ignore it. Why? &#xA;I opend Karl-Heinz Deckers famous Maschinenelemente book, where it shows how to get the formula required to raise load, and also at one point it just ignores the F2 force (at least it acknowledges its existence in one of the earlier steps of the derivation, but then just ignores it).&#xA;Can someone explain this? &lt;/p&gt;&#xA;" OwnerUserId="8851" LastActivityDate="2017-09-13T21:59:30.760" Title="Derivation on torque required to raise load, isn't there a force disapearing?" Tags="|mechanical-engineering|bolting|fasteners|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17103" PostTypeId="1" CreationDate="2017-09-13T18:48:23.710" Score="1" ViewCount="38" Body="&lt;p&gt;Consider a system of the form&#xA;\begin{equation}&#xA;\Sigma\colon\ \dot{x}=f(x),&#xA;\end{equation}&#xA;where $x\in \mathbb{R}^n$ and $f\colon \mathbb{R}^n\to \mathbb{R}^n$ is a sufficiently smooth function vanishing at the origin. Assume that $\Sigma$ has a &lt;em&gt;globally exponentially stable&lt;/em&gt; equilibrium at $x=0$. Now apply a global coordinate transformation&#xA;\begin{equation}&#xA;z=T(x),&#xA;\end{equation}&#xA;where $T\colon \mathbb{R}^n\to \mathbb{R}^n$, $T(0)=0$ is a global $\mathcal{C}^1$ diffeomorphism. Accordingly, we obtain the transformed system&#xA;\begin{equation}&#xA;\Sigma^\prime \colon\ \dot{z}=g(z):=\frac{\partial T(x)}{\partial x}f(x)\bigg|_{x=T^{-1}(z)}.&#xA;\end{equation}&#xA;&lt;strong&gt;Question:&lt;/strong&gt; Is $\Sigma^\prime$ globally exponentially stable at $z=0$?&lt;/p&gt;&#xA;" OwnerUserId="5930" LastActivityDate="2017-09-13T18:48:23.710" Title="Exponential stability under a global coordinate transformation" Tags="|control-engineering|control-theory|stability|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17104" PostTypeId="1" AcceptedAnswerId="17109" CreationDate="2017-09-13T19:12:36.927" Score="1" ViewCount="109" Body="&lt;p&gt;I have a trajectory which is a function of time that returns tuples of position, velocity, acceleration and jerk for each moment. So at every control loop iteration I would like the motion of my actuator to match these values. However it is not clear how to build a control system which will be able to follow position, velocity, acceleration and jerk commands altogether. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual question is that of control system input. In the vast majority of cases input is just a force which in context of PWM is just a number. Therefore natural question arises: should I anyhow map my 4 trajectory parameters to this one input of control system(if yes then how?) or is there any other way?&lt;/p&gt;&#xA;" OwnerUserId="10483" LastActivityDate="2017-09-18T15:47:03.537" Title="How to execute trajectory?" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17106" PostTypeId="1" CreationDate="2017-09-13T20:46:44.270" Score="2" ViewCount="282" Body="&lt;p&gt;IFC files have an IFCMATERIAL property which names a material and is referenced by other components in the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IFC file doesn't point to a materials library, doesn't indicate a system variable that points to a library and uses very general names such as &quot;concrete&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are these relations managed across multiple interfaces and systems? For example, I have an IFC exported from Revit, where I to import this into xBIM on a system without Revit, how would xBIM know what materials to bind to the objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12494" LastEditorUserId="3353" LastEditDate="2018-06-14T09:43:04.693" LastActivityDate="2018-12-18T08:07:52.000" Title="How are IFC file material relationships maintained across platforms" Tags="|cad|building-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17107" PostTypeId="2" ParentId="17102" CreationDate="2017-09-13T21:51:09.487" Score="1" Body="&lt;p&gt;From the diagram of the tooth profile in the upper right corner I see FN, FN' and F2. FN coincides with the lifting force so it is relevant to the solution. FN' results in friction which will affect the required torque. But F2 does not affect either. F2 might be viewed as a component of FN', but it doesn't uniquely affect the solution in a way not already accounted for by FN'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If F2 resulted from, for example, the bolt being too large for the hole, then it, F2, would be a unique force not derived from FN or FN', and would need to be factored in. But that is not the case here.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastEditorUserId="11222" LastEditDate="2017-09-13T21:59:30.760" LastActivityDate="2017-09-13T21:59:30.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17109" PostTypeId="2" ParentId="17104" CreationDate="2017-09-14T00:14:51.430" Score="4" Body="&lt;p&gt;You have to choose a controller that best fits the system you are trying to control. You have to take into consideration the variables you are trying to control when deciding on the controller. Although the trajectory generator outputs four different values you don't have to use all of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Judging from your question I assume you're trying to control the motion of something, maybe the movement of a robot arm or something similar, from position A to B. This has to happen in a smooth manner, which is why you have a trajectory. If this doesn't describe your system you will at least get a notion of how you can &quot;merge&quot; the trajectory outputs together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple controller that is easy to implement is the PID controller. It takes two of the trajectory outputs into account (position and velocity). Its controller law is expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(t) = K_p \cdot e(t) + K_i \cdot \int e(t) + K_d \cdot \dot{e}(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $u(t)$ is the input to the actuator, if you only have one actuator. The unit of $u(t)$ doesn't matter as you have to scale the PID parts anyway (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$e(t)$ is the error, and in this particular case defined as $e(t) = p_d - p$ (difference between desired position and actual measured position). Desired position is what you get from the trajectory. That makes $\dot{e} = v_d - v$. You get the desired velocity from the trajectory. You have to measure the velocity $v$. If you can't measure it you have to estimate it with a state observer. $\int e(t)$ is the accumulated error over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_p$, $K_i$ and $K_d$ are scaling factors (or gains). These are usually constants and have to be chosen by you using dimensional analysis and tuning on the real system. You choose the gains so that $u(t)$ looks reasonable with respect to magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can expand the PID controller to include acceleration and jerk by adding additional parts to it. But as mentioned before, even though the trajectory outputs four parameters doesn't mean you have to use all of them, it just makes the motion smoother if implemented correctly. You could for instance choose to only use a P controller&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$u(t) = K_p \cdot e(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not going to write all the theory behind this controller here. You can read more on &lt;a href=&quot;https://en.wikipedia.org/wiki/PID_controller&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Wikipedia&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2017-09-14T00:14:51.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17110" PostTypeId="1" CreationDate="2017-09-14T00:26:16.073" Score="2" ViewCount="512" Body="&lt;p&gt;I have looked for a way to make a cyclorotor mechanism which allows the blades to rock/rotate (in the picture below, its rocking I believe) to no avail. I am not able to find the required length of the green link or that green circle (refer picture). Any idea on how to find out the required length? also, how to determine the eccentricity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture drawn using paint was the idea I was working on. Here the blade is supposed to spin by 180 degrees when the whole system completes one rotation with intermediate positions as shown in the figure. I am not able to determine the length of the remaining links to connect the system. I was thinking on the lines of 4 (or is it 5?) bar mechanisms as in two of the links have been given in each quadrant but cant find any logical way to find the remaining links.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;please help ! &lt;a href=&quot;https://i.stack.imgur.com/azPL7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/azPL7.png&quot; alt=&quot;my idea&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tRRP5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tRRP5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12885" LastEditorUserId="9" LastEditDate="2020-04-12T23:30:45.210" LastActivityDate="2023-12-24T05:08:10.980" Title="How to make a cyclorotor?" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="17111" PostTypeId="2" ParentId="17092" CreationDate="2017-09-14T01:32:36.207" Score="1" Body="&lt;p&gt;In the beginning you mention supplying water from the main drain, but I'm assuming you mean meter. &#xA;From there, you go to a &quot;pressure reducer&quot;.  You may have a pressure reducer, but given that you are cutting the pipe diameter in half, can you please clarify if it is actually a pressure reducing valve or if you just mean a pipe reducer (will make the pipe sizer smaller).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote an entire answer before I re-read your question and realized you were not asking about 3 pumps vs 10, but, simply, if 10 pumps would work.  First of all, if this is for a commercial application, I think that you should have a plumber or engineer probably designing the system since there is more than just throwing some tanks on a roof and hooking up pipe involved here.  There are some different isolation valves and other items--like strainer(s)-- that should probably be installed.  With that being said, I think having 10 small pumps in this application sounds like an overcomplication.  If you are connecting the 3 tanks together (you say valve, but I think you mean manifold), then you really only need one pipe outlet from the collective tanks.  Then you can run the pipe throughout the mall to the various shops.  Depending on the distance to the furthest run, you will probably need pressure reducing valves for the shops (particularly those close to the pump).  You could also put two pumps by the tanks in parallel to provide a level of redundancy if one pump fails, as well as additional capacity if there is a high demand on occasion.  1 1/2 inch pipe suggests there may be either reasonable demand or the owner was just throwing a number out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully that answer is enough for you to get started or at least enough to figure out what other questions to ask.  If this is a real project and not just a design exercise, I highly encourage you run it by a licensed professional, and, as Solar Mike said, make sure you heed the local codes and regulations being enforced.  Water systems won't explode when poorly designed (which is a possibility with natural gas), but you can cause severe property damage if there are failures or leaks, and you can make people sick if you have things like large sections that are never used which will sit with stagnant water in the pipes until someone eventually turns on a faucet.&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2017-09-14T01:32:36.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17112" PostTypeId="2" ParentId="17101" CreationDate="2017-09-14T01:49:35.357" Score="1" Body="&lt;p&gt;No. Placing the red lead into H15 would not allow you to measure the current, because the black lead of the multimeter would still be in I15.  H15 and I15 are connected together (at least on a standard breadboard), so the two leads of the multimeter are essentially connected together. That won't measure anything. As far as I understand the picture, you would have to do the following (in this order)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) remove the black power supply lead from J 15&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) remove the black multi-meter lead from I 15&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) move the red multi-meter lead to H15&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) connect the black multi-meter lead to the power supply lead.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;DISCLAIMER: these instructions are only as good as my understanding of the circuit.  Without a circuit diagram, I may have misunderstood. If you aren't 100% sure, ask for clarification first. In current mode, the impedance of the multimeter is very low, so if you connect in the wrong place you may overload the meter. Make sure you are using a fused multi-meter, and put it on a setting appropriate for the current you expect to measure.  Many multi-meters have different spots to plug in the leads, one for high current and one for low current.  Some cheaper meters may have unfused inputs.  Only use those if you are very sure about what you are doing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this doesn't make sense, then post a circuit diagram and we can elaborate on the answer.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-09-14T01:49:35.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17115" PostTypeId="1" CreationDate="2017-09-14T03:16:56.790" Score="1" ViewCount="145" Body="&lt;p&gt;I'm attempting to create a suspended shelf. I need to determine the horizontal force at point A to determine whether or not the fasteners will fail. All members are rigid and the supports at A and B are fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am very rusty on statics and your help would be greatly appreciated! I would attach the work I've attempted, but quite frankly, it's embarrassing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rftLX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rftLX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time.&lt;/p&gt;&#xA;" OwnerUserId="12890" LastActivityDate="2017-09-16T01:17:00.647" Title="Suspended Shelf Problem" Tags="|mechanical-engineering|structural-engineering|statics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17116" PostTypeId="2" ParentId="17115" CreationDate="2017-09-14T03:35:35.427" Score="3" Body="&lt;p&gt;The horizontal force in A is solved by summing up moment at B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$300 (14) = Ax (14)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore, $$Ax = 300lbs.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If your construction is to be bolted in the wall, the support condition shall be assumed to be pinned connection. It is safe to assume that point B will be hinged and point A to be horizontal reaction only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VzZIj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VzZIj.png&quot; alt=&quot;See image below&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or the point A to be hinged and B is horizontal reaction only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FC7sL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FC7sL.png&quot; alt=&quot;See image for another possible support condition&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, the value of horizontal reaction at A will still be the same. :)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Other supplemental info based on OP's comment.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A bolted connection is GENERALLY a pinned connection. Examples of this are in BRIDGE TRUSS connections. This is because rigid connections in dynamic systems are catastrophic as it is prone to extreme fatigue due to constant deformation and back to its original shape..&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The vertical reaction can be checked (if you are checking/designing for/vs the shearing of bolts) by distributing the vertical force &lt;strong&gt;equally&lt;/strong&gt; into Ay and By. This is not static problem anymore but more on the design method. Therefore, by designing the shearing strength of bolts, you can assume that &lt;strong&gt;BOTH point A and B carries the same amount of shear caused by 300lb load&lt;/strong&gt;, that is equal to 150lb per joint. This is contrary to the assumption in static check that only one of the two points carry the load.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you want to get the exact force in each point, you might want to use indeterminate methods in solving for the reactions such as force method, virtual work methods, etc. These usually use displacement as 4th parameter as there will be four unknowns in the system. But taking your question as a simple static problem (because this is a shelf), I suggested the simplest approach to your problem. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/Y6o3V.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y6o3V.png&quot; alt=&quot;This is the **real** indeterminate FBD of your system.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Still, by summing up moments at B, the value of &lt;em&gt;Ax&lt;/em&gt; remains the same.&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12936" LastEditDate="2017-09-16T01:17:00.647" LastActivityDate="2017-09-16T01:17:00.647" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17118" PostTypeId="1" AcceptedAnswerId="17119" CreationDate="2017-09-14T09:51:31.930" Score="17" ViewCount="10390" Body="&lt;p&gt;I have to model the behavior of an electric car. For this, I use &lt;a href=&quot;https://books.google.fr/books?id=6aWSaJUZuOEC&amp;amp;pg=SA7-PA47&amp;amp;lpg=SA7-PA47&amp;amp;dq=equation%20electric%20vehicle&amp;amp;source=bl&amp;amp;ots=oHIOhDcJaq&amp;amp;sig=o9-kql0qtUwJb--clA0VRwFspVA&amp;amp;hl=fr&amp;amp;sa=X&amp;amp;ved=0ahUKEwjKhffDsqTWAhVH1RQKHdAvATUQ6AEIPTAB&quot; rel=&quot;noreferrer&quot;&gt;these equations&lt;/a&gt; and I can observe by &quot;playing&quot; with parameters that, when going downhill at constant speed, the car has a negative consumption (i.e. recovers energy). I was wondering if this is realistic?&lt;/p&gt;&#xA;" OwnerUserId="12895" LastEditorUserId="6790" LastEditDate="2017-09-15T12:24:40.110" LastActivityDate="2017-09-16T03:49:01.690" Title="Does an electric vehicle going downhill recover energy?" Tags="|mechanical-engineering|electric-vehicles|regenerative-braking|" AnswerCount="5" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17119" PostTypeId="2" ParentId="17118" CreationDate="2017-09-14T10:34:12.623" Score="32" Body="&lt;p&gt;It depends on how steep the hill is.  On a slight hill, the energy added by gravity is still not enough to overcome rolling friction and air resistance, so the car still needs power to maintain speed.  On a steeper hill, the two may balance out, so no power is used, and no power is generated.  On a hill that's steep enough to require braking to control the speed, the car recovers energy.  It's called regenerative braking.  If the car is going too fast, applying the brakes turns the motor into a generator and charges the battery.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2017-09-14T10:34:12.623" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="17121" PostTypeId="2" ParentId="17118" CreationDate="2017-09-14T11:37:11.093" Score="6" Body="&lt;p&gt;Yes, it can certainly happen.  To recharge the battery when going down hill, the free roll speed needs to be greater than the speed you are trying to go.  That means going down the same hill may result in a different outcome depending on speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's say you put the car in neutral and coast down a particular hill.  For most of the hill, the car settles at 45 MPH.  If you were to drive down the hill at 35 MPH, then the system should be charging the battery a bit.  However, if you drive down at 55 MPH, then additional power is needed than just the gravity assist of the hill, power is taken from the battery to run the motor or motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a Honda Civic Hybrid, and charging the battery when going down a hill is not unusual at all.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-09-14T11:37:11.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17122" PostTypeId="2" ParentId="15890" CreationDate="2017-09-14T12:06:26.457" Score="2" Body="&lt;p&gt;Plant load factor and capacity (use) factor are two names for the same thing: the ratio of mean power to rated nameplate capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Capacity factors between very similar or identical technologies that are serving a similar role on the grid, can be comparable: e.g coal and gas CCGT plants both working as baseload, or both working as mid-merit, or both working as peakers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, capacity factors between very different technologies, or by plants serving very different roles, are not comparable. So no, coal and PV capacity factors are not comparable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wind capacity factors in the short term are mostly driven by weather, but over the long term are mostly driven by the farm design, &lt;a href=&quot;http://energynumbers.info/capacity-factor-of-wind&quot; rel=&quot;nofollow noreferrer&quot;&gt;as I explained elsewhere in more detail, previously&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, all other relevant things being equal, a higher capacity factor is better than a lower one. However, all other relevant things are almost never equal, and it's rarely as simple as higher being better. Indeed, there are some fairly common circumstances where higher can be worse. If a wind farm was achieving 80% over the medium to long term, that almost certainly means that the windfarm designer has made a very expensive mistake - they should have designed for 20-60%, depending on the particular local weather and economics.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2017-09-14T12:06:26.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17124" PostTypeId="2" ParentId="17118" CreationDate="2017-09-14T16:16:34.047" Score="0" Body="&lt;p&gt;Absolutely.  That &quot;recovering energy&quot; will kill you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the freight trains of yore.  In the day, helper districts involved extra steam engines waiting at the bottom of a mountain to help trains up.  Today, it's more important to &lt;a href=&quot;http://trn.trains.com/railroads/abcs-of-railroading/2006/05/dynamic-braking&quot; rel=&quot;nofollow noreferrer&quot;&gt;help trains &lt;em&gt;down&lt;/em&gt; the hill&lt;/a&gt; with their regenerative (dynamic) brakes.  Some railroads require additional units be added &lt;em&gt;for&lt;/em&gt; dynamic braking, even for a train that could climb the hill! Best practice is to plan to use mainly or only dynamic brake, which saves money too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steam trains would awkwardly drag brakes all the way down the mountain, filling the valley with gray smoke. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Automotive brakes (gas or electric) don't have enough gray smoke, so they &lt;em&gt;need&lt;/em&gt; engine braking... except electric cars &lt;em&gt;can't&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's been a huge problem, especially for the early/homebrew DC conversions.  Like the freight trains, they could go up hills they couldn't go down.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even a regenerating car has an issue, if the owner tops up the battery at Summit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is to say, when an electric car &lt;em&gt;loses the ability to regen&lt;/em&gt; because its batteries are full, its only fallback is friction brakes - and &lt;em&gt;as discussed&lt;/em&gt;, those can't sustain a downgrade either.  They're playing with locomotive-style &quot;make heat&quot; dynamic braking, but I don't know how far along it is.  They need it. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;By &quot;don't have enough gray smoke&quot; I mean automotive brakes are not designed nor rated to descend a continuous mountain grade. &lt;/p&gt;&#xA;" OwnerUserId="11673" LastEditorUserId="11673" LastEditDate="2017-09-16T03:49:01.690" LastActivityDate="2017-09-16T03:49:01.690" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17126" PostTypeId="2" ParentId="15912" CreationDate="2017-09-14T17:27:48.377" Score="1" Body="&lt;p&gt;Tesla initially built the Model S &lt;a href=&quot;https://en.wikipedia.org/wiki/Tesla_Model_S#Powertrain&quot; rel=&quot;nofollow noreferrer&quot;&gt;with one motor driving the rear wheels&lt;/a&gt;. The D model has a second motor driving the front wheels. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Was the increase in range and performance ... due to the ability to remove the rear drive shaft etc?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. There was no rear drive shaft in the RWD model, the motor was installed behind the rear wheels, against the differential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The range of the D model is about ~10 miles more than the RWD model with the same battery. I haven't found a statement that says why range has increased. One possibility is there's more scope for regeneration with the AWD model: on the RWD car, regeneration is limited by traction, most of the brake force is supplied by the front wheels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The increased performance is thanks to the second motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going from 2 to 4 motors allows you to eliminate the differentials, at a cost of having to drive each motor at a different speed, making the electronics more complex: not just the control circuitry, but you need to have two power amplifiers instead of just one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether it'll improve performance depends on which motors you choose: if you replace one 100 kW motor with two 100 kW motors you get more performance but a shorter range, for instance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important factor is the power/weight ratio of the motors. I suspect the 2x 50 kW motors would weigh a bit more than a single 100 kW motor, as you have to duplicate some items that don't directly contribute to output power, like the commutator and the motor shaft. The 2 casings will weigh more, too. Whether the extra weight will be offset by eliminating the differential remains to be seen. &lt;/p&gt;&#xA;" OwnerUserId="11645" LastEditorUserId="11645" LastEditDate="2017-09-16T12:47:19.163" LastActivityDate="2017-09-16T12:47:19.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17127" PostTypeId="1" CreationDate="2017-09-14T17:30:23.250" Score="1" ViewCount="112" Body="&lt;p&gt;&lt;strong&gt;EDITED THIS POST FOR THE PEOPLE WHO PROVIDED ME WITH GREAT FEEDBACK&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been building a few simple project circuits, mainly to analyze electronic theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I'm having right now is that I'm using a cheap multi-meter from amazon that is giving me wild and inaccurate readings on DC current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The case at hand is SIMPLE-CIRCUIT-A: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;9V Battery(multimeter is only reading about 7.5 from across the terminals)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2x 200-ohm resistors in series, totaling for 400ohms&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1 small piece of #22 wire&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1 breadboard&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FUzR8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FUzR8.jpg&quot; alt=&quot;Circuit image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I listen to the mutimeters readings on the 9V battery as being only about 7.5V, then i should be pulling about 18.75mA at any point in the series circuit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[![Circuit schematic and math][2]][2]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is what I'm actually getting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4rJzK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4rJzK.jpg&quot; alt=&quot;Inaccurate DC reading&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you could say, &quot;Maybe the battery really is applying 9V across the circuit.&quot;, that conclusion would give us a DC of 22.5mA, yet im getting at best a reading of 24.5mA, that's a 9% error margin!!! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could there be an issue with my multimeter? On top of that, I can never get stable readings from the leads, is that common in all multimeters? Usually when I apply the leads to a component to read DC, the numbers will bounce all over the place.&lt;/p&gt;&#xA;" OwnerUserId="12689" LastEditorUserId="12689" LastEditDate="2017-09-14T20:45:23.963" LastActivityDate="2017-09-14T20:45:23.963" Title="Not getting accurate readings from my multimeter" Tags="|instrumentation|" AnswerCount="0" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="17128" PostTypeId="2" ParentId="17118" CreationDate="2017-09-14T19:03:34.277" Score="4" Body="&lt;p&gt;&lt;a href=&quot;http://www.abb.com/cawp/seitp202/a130bf42c59cc7d8c1257dd20046d0d6.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;ABB train regenerative braking.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some Swiss trains give back energy to the grid upon descending the mountain pass that they previously climbed. The article claims as much of 70% is recovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I rode the &lt;a href=&quot;https://www.jungfrau.ch/shop/en/tickets/bergbahn?___from_store=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jungfraujoch&lt;/a&gt; and the personnel mentioned for every 3 trips down the mountain enough electricity was recovered for a &quot;free&quot; trip up.&lt;/p&gt;&#xA;" OwnerUserId="12908" LastActivityDate="2017-09-14T19:03:34.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17129" PostTypeId="2" ParentId="17118" CreationDate="2017-09-14T22:29:21.520" Score="4" Body="&lt;p&gt;The basic equations governing an electric car are reversible, so a simple analysis would say that the battery is getting charged. In the real world, there are more complications. In the case of an electric bike, there is a ratchet that disengages the engine from the wheel when the wheel is going faster from the engine (just like how on a regular bike -- where &quot;regular&quot; doesn't include track bikes -- if you stop pedaling, the wheel keeps moving even though the chain isn't). In an electric car, the motor can be disengaged from wheels either mechanically or electrically. Electric motors generally can't handle as much power when being driven backwards as when going forward, so electric cars are built to limit how much power can be transferred from the wheels to the battery. Absent those modifications, any time the wheels are going faster than the motor, there's going to be electricity flowing &quot;backwards&quot;. Whether that energy will be charging the battery or frying the motor is a different matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you include heat, all cars &quot;recover&quot; energy in some sense when going downhill, either through increased kinetic energy, or heating of brake pads.&lt;/p&gt;&#xA;" OwnerUserId="12911" LastActivityDate="2017-09-14T22:29:21.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17131" PostTypeId="1" AcceptedAnswerId="17148" CreationDate="2017-09-15T08:45:35.660" Score="2" ViewCount="1125" Body="&lt;p&gt;I have been trying to determine the response coefficients for seismic analysis of Non building structures in &lt;strong&gt;ASCE 7-10&lt;/strong&gt; and found that the value of &lt;strong&gt;&lt;em&gt;R, the response coefficient&lt;/em&gt;&lt;/strong&gt;, is larger in case steel connections are detailed using &lt;strong&gt;AISC-341&lt;/strong&gt; than &lt;strong&gt;AISC-360&lt;/strong&gt;, this would imply a larger design load in case of AISC 341 than AISC 360. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;AISC 341 -&amp;gt; Large seismic design load&#xA;AISC 360 -&amp;gt; Smaller seismic design load&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Does that imply AISC 341 design results in a more conservative steel connection than AISC 360 or is there a difference in steel design philosophy?&lt;/p&gt;&#xA;" OwnerUserId="1693" LastEditorUserId="1832" LastEditDate="2017-09-15T09:09:41.013" LastActivityDate="2024-01-21T23:17:43.543" Title="AISC 360 vs AISC 341" Tags="|civil-engineering|steel|aisc|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17132" PostTypeId="1" CreationDate="2017-09-15T12:12:09.297" Score="2" ViewCount="82" Body="&lt;p&gt;So a product of me being a bored mechanical engineering student and my love of camping has prompted me to attempt to make my own Seebeck generator. It will be powered by a simple wood fire in a can and slightly fan forced so i am expecting 400+ degrees (celsius). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what would be best to conduct the heat from the fire (ideally from stock i already have), a solid copper round bar (8mm), a thicker aluminium or copper heatsink or copper tubing (6.5mm OD 3.7mm ID). the copper tubing will either be left with void in the center or i had the idea of filling it with lead/tin solder as the fire temperature will most likely be above their melting point and hopefully transfer the heat well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is more a thought experiment into whether the solder option will work or it i have missed something crucial like it expanding and exploding on me. i have also heard that fin style heat sinks, like that found on PC's, are terrible conductors of heat as the cross section of their fins are too small to carry large amounts of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any input is great and i'll probably end up testing them all anyway because there is no better way to learn than by doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drew up some quick plans of what I am imagining&lt;a href=&quot;https://i.stack.imgur.com/aFTe6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aFTe6.jpg&quot; alt=&quot;Seebeck generator&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="12919" LastEditorDisplayName="user16" LastEditDate="2017-09-20T11:57:28.140" LastActivityDate="2017-09-20T11:57:28.140" Title="Thermal Conductivity" Tags="|heat-transfer|cooling|" AnswerCount="0" CommentCount="9" ClosedDate="2017-09-20T11:57:34.450" ContentLicense="CC BY-SA 3.0" />
  <row Id="17133" PostTypeId="1" CreationDate="2017-09-15T19:07:43.667" Score="1" ViewCount="111" Body="&lt;p&gt;I have calculated the resistances encountered by the vehicle. Now to find the required power i used formula&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                    Required power= Total resistance force * Max. velocity&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I came out with around 50 kw.....but this is the power required by vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to calculated the motor power?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think 50 kw powered motor is required to propel a vehicle with gross weight around 700 kg&lt;/p&gt;&#xA;" OwnerUserId="12927" LastEditorUserId="8055" LastEditDate="2017-09-16T03:28:32.070" LastActivityDate="2017-09-16T03:28:32.070" Title="How the sizing of electric motor is done for a electric vehicle?" Tags="|electrical-engineering|torque|power|electric-vehicles|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17134" PostTypeId="1" AcceptedAnswerId="17137" CreationDate="2017-09-15T19:15:35.050" Score="2" ViewCount="227" Body="&lt;p&gt;The book &quot;Instrument Engineers' Handbook vol. 1 - Liptak&quot; said that for the first-order system forced by a step or an impulse, the time constant is the time required to complete 63.2% of the total rise or decay; at any instant during the process, the time constant is the quotient of the instantaneous rate of change divided into the change still to be completed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I express the last sentence mathematically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="7006" LastActivityDate="2017-09-16T01:40:33.827" Title="Time constant - formula" Tags="|control-theory|education|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17135" PostTypeId="1" CreationDate="2017-09-15T19:35:54.147" Score="1" ViewCount="61" Body="&lt;p&gt;For Example, the pressure of evaporator can be changed, also the pressure of condenser and many more parameters. What is that parameter which remain constant while others keep changing ?&lt;/p&gt;&#xA;" OwnerUserId="12931" LastActivityDate="2017-09-18T09:27:02.403" Title="What will remain constant during the circulation of working substance in a refrigerating machine?" Tags="|thermodynamics|heat-transfer|compressors|refrigeration|" AnswerCount="1" CommentCount="5" ClosedDate="2017-09-20T11:53:38.763" ContentLicense="CC BY-SA 3.0" />
  <row Id="17136" PostTypeId="2" ParentId="17133" CreationDate="2017-09-15T19:41:10.700" Score="1" Body="&lt;p&gt;The vehicle power that you have found is the ideal power. It does not include any resistance forces that have not been taken into account. For example, air drag, heating (=friction) due to parts interacting, mechanical losses (things that bend / change shape during motion), etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To estimate the required motor power, the idea is to analyze the whole path from the motor shaft until where the main force occurs (tyre to ground traction). By combining (=multiply) the efficiency losses at each chain of force transfer, the general efficiency of the drive mechanism can be estimated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum power is needed when maximum force is required. Mainly the maximum speed for vehicles that aerodynamics dominate and maximum climb angle for vehicles like trucks.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-09-15T19:41:10.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17137" PostTypeId="2" ParentId="17134" CreationDate="2017-09-15T19:55:03.357" Score="4" Body="&lt;p&gt;Does the book not give you the mathematics?  The underlying expression is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$parameter = 1-e^{ -t/ \tau }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so you see at $t = \tau$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$parameter = 1 - e^{-1} = 0.63$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the second sentence says divide &quot;change to be completed&quot; which is $1 -parameter$ , or  $e^{-t/\tau}$, by the first derivative:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{d parameter}{dt} = \frac{e^{-t/\tau}}{\tau}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{e^{-t/\tau}}{\frac{e^{-t/\tau}}{\tau}} = \tau$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pretty simple..?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will say the wording of that sentence is pretty awkward. I would reverse it as &quot;the time constant is the quotient of the change still to be completed &lt;em&gt;divided by&lt;/em&gt; the instantaneous rate of change.&quot;&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="78" LastEditDate="2017-09-16T01:40:33.827" LastActivityDate="2017-09-16T01:40:33.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17139" PostTypeId="1" CreationDate="2017-09-15T22:45:52.860" Score="1" ViewCount="41" Body="&lt;p&gt;I have been pouring through countless articles about Euler's angles, rotation matrices, Rodriguez Rotation Formula, Quaternions, trying to understand how to generate an algorithm for this problem and have as of yet been unsuccessful.  I am trying to figure out, given a pitch and roll along a particular heading, the manner in which one could determine the co-planar  pitch and roll along the magnetic North heading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cdqfv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cdqfv.png&quot; alt=&quot;Reference Image Showing concept&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above image, I have the pitch and roll values at the currentHeading, and am trying to calculate the pitch and roll values of the blue line in the same plane but in the magnetic North direction.  Any help would be highly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="12934" LastEditorUserId="1832" LastEditDate="2017-09-16T01:48:44.340" LastActivityDate="2017-09-16T01:48:44.340" Title="Given Pitch and Roll in one Heading, convert to co-planar pitch and roll for North" Tags="|control-theory|aerospace-engineering|positions-measurement|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17140" PostTypeId="1" AcceptedAnswerId="17141" CreationDate="2017-09-16T00:31:36.987" Score="3" ViewCount="1474" Body="&lt;p&gt;I am currently into an estimating job for a small construction company. In my previous experiences, in calculating the labor cost for a certain job, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Labor cost = material cost * 0.45&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did this to simplify the computation. Are there any other specific labor factor for different job classification for the following class:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;concreting &lt;/li&gt;&#xA;&lt;li&gt;masonry&lt;/li&gt;&#xA;&lt;li&gt;carpentry&lt;/li&gt;&#xA;&lt;li&gt;painting&lt;/li&gt;&#xA;&lt;li&gt;steel&lt;/li&gt;&#xA;&lt;li&gt;rebars, etc. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;that I can adopt? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This will not involve equipment in the process.&lt;/p&gt;&#xA;" OwnerUserId="12936" LastEditorUserId="12597" LastEditDate="2017-09-16T00:35:24.923" LastActivityDate="2017-09-17T14:24:00.377" Title="Relationship between Material cost and Labor cost" Tags="|civil-engineering|construction-management|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17141" PostTypeId="2" ParentId="17140" CreationDate="2017-09-16T00:38:39.243" Score="2" Body="&lt;p&gt;In the Philippines, there are 2 types of labor rates that we consider.&#xA;We have provincial and non-provincial rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For non-provincial rates, we can go as far as 45% material cost, while on provincial rates, we only go up to 35% of the material cost. This is due to our present labor rules and regulations (which i hope should be changed soon).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the above percentage varies according to the difficulty of work, e.g. painting and carpentry requires a lot of skills than concreting that's why I sometimes go as much as 60% for these type of jobs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-16T00:49:36.697" LastActivityDate="2017-09-16T00:49:36.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17142" PostTypeId="5" CreationDate="2017-09-16T01:00:12.550" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-09-16T01:00:12.550" LastActivityDate="2017-09-16T01:00:12.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17143" PostTypeId="4" CreationDate="2017-09-16T01:00:12.550" Score="0" Body="This tag should be used for questions regarding work, material and equipment schedules, cash flow and cost analysis in projects, critical path methods, and other subjects regarding the flow of work, time, and money in construction. This is similar to project management." OwnerUserId="12597" LastEditorUserId="1832" LastEditDate="2017-09-16T01:50:28.833" LastActivityDate="2017-09-16T01:50:28.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17144" PostTypeId="1" AcceptedAnswerId="17145" CreationDate="2017-09-16T01:32:53.053" Score="1" ViewCount="1199" Body="&lt;p&gt;Most of resources I've been reading don't explain the scientific reason for the existence of shear force. Most resources just teaches how to solve problems and about properties. Nothing goes deep (&lt;strong&gt;not even a little bit deep&lt;/strong&gt;, sorry if I'm being unfair, for me it was just &lt;strong&gt;drawing diagrams and solving them&lt;/strong&gt;).&lt;/p&gt;&#xA;" OwnerUserId="12937" LastEditorUserId="90" LastEditDate="2017-09-16T03:36:01.407" LastActivityDate="2017-09-16T03:36:01.407" Title="What causes shear force?" Tags="|structural-engineering|civil-engineering|strength|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17145" PostTypeId="2" ParentId="17144" CreationDate="2017-09-16T01:41:13.823" Score="2" Body="&lt;p&gt;This is from &lt;em&gt;Wikipedia&lt;/em&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_force&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Shear_force&lt;/a&gt; :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Shearing forces are unaligned forces pushing one part of a body in one direction, and another part of the body in the opposite direction. When the forces are aligned into each other, they are called compression forces.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In engineering terms, it is the force that forces the body to be &lt;strong&gt;sliced&lt;/strong&gt; and to be &lt;strong&gt;split&lt;/strong&gt; with regards to the direction of the force. It is like running a boiled potato into your hands, the skin of that potato will be removed. Same goes with the materials. All materials are assumed to have infinite &lt;em&gt;layers&lt;/em&gt; which can be removed like the potato skin when a force parallel to that layer is applied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See images below for the failure caused by shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i7L3W.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i7L3W.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O0MUp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O0MUp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rdWyV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rdWyV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As an answer to OP's comment, take a look at the pictures:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that a block is loaded by two forces acting in the opposite directions, the red layer will be the plane that will carry the shear force.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2l2GA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2l2GA.png&quot; alt=&quot;Original Block&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we don't see is that the block is composed of infinite number of differential planes, see picture below, that also counteracts the applied force.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vP33T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vP33T.png&quot; alt=&quot;Next Block&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we stay by the rule of statics, the force will then be distributed over the other layers giving the &lt;strong&gt;same amount of force&lt;/strong&gt; for each layer. See picture below. &lt;a href=&quot;https://i.stack.imgur.com/b24Sn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b24Sn.png&quot; alt=&quot;Block 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, shear force &lt;em&gt;can&lt;/em&gt; make the any of the differential plain fail.&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-16T02:04:30.743" LastActivityDate="2017-09-16T02:04:30.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17146" PostTypeId="5" CreationDate="2017-09-16T03:36:54.147" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-09-16T03:36:54.147" LastActivityDate="2017-09-16T03:36:54.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17147" PostTypeId="4" CreationDate="2017-09-16T03:36:54.147" Score="0" Body="This is to be used for topics regarding the strength of materials, the analysis of internal forces in a body and all the governing principles which considers deformation of bodies." OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-16T16:24:33.753" LastActivityDate="2017-09-16T16:24:33.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17148" PostTypeId="2" ParentId="17131" CreationDate="2017-09-16T04:58:30.837" Score="2" Body="&lt;p&gt;The R-factor reduces seismic design force from the linear-elastic demand. Thus, the &lt;em&gt;larger&lt;/em&gt; the R-factor, the &lt;em&gt;lower&lt;/em&gt; the seismic design force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, we're approximating inelastic seismic design by accounting for the ductility of the structural system. Connections designed per AISC-360 (which does not specifically address seismic design) can be expected to be less ductile than those designed per AISC-341.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AISC-341 = more ductile = higher R = lower seismic design force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AISC-360 = less ductile = lower R = higher seismic design force&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2017-09-16T04:58:30.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17149" PostTypeId="1" CreationDate="2017-09-16T07:14:54.747" Score="0" ViewCount="150" Body="&lt;p&gt;Is the maximum tensile strength a property of a material ? If it is so then suppose I am given the maximum tensile strength of two flat spiral springs to be $30MN/m^2$ and I have to find the dimensions (length per unit thickness $l/t$) of the spring I need to have to make some spring controlled system in an indicating instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to divide the maximum tensile strength by two, multiply it by two or leave it as it is ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I can calculate the dimensions of the spring by&#xA;$$f_{max} = \frac{Et\theta}{2l}$$&#xA;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E$ is the modulus of elasticity (given to me)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta $ is the maximum angle of deflection $\pi/2$&lt;/p&gt;&#xA;" OwnerUserId="12939" LastActivityDate="2017-09-17T17:06:38.407" Title="Maximum Tensile strength confusion" Tags="|measurements|instrumentation|" AnswerCount="1" CommentCount="2" ClosedDate="2017-09-23T00:38:29.167" ContentLicense="CC BY-SA 3.0" />
  <row Id="17151" PostTypeId="1" CreationDate="2017-09-16T11:16:48.510" Score="1" ViewCount="26" Body="&lt;p&gt;Im currently writing a model for a guidance law named &lt;strong&gt;Maintaining Guidance law&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The law looks as following: $$ a_{M} = [K_1,K_2]*[\dot{\lambda_T} ,\dot{\lambda_D}]'$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since im using a linear model I use the following calculations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \ddot{y} = \ddot{y_T}-\ddot{y_M}=a_T-a_M$$&#xA;$$ \dot{y} = \dot{y_T}-\dot{y_M}$$&#xA;$$ y = y_T-y_M$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where: $$\dot{\lambda} = \frac{y+\dot{y}*t_{go}}{V_{c}*t_{go}^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And K1 and K2 are vectors calculated before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engagement looks as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rxcb1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rxcb1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to transform it into X-Y axis however I got stuck in few things I couldnt figure out as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In this model, $\dot{\lambda}$ depands of $t_{go}$ which becomes small close to the end and makes $\dot{\lambda}$ huge ---&gt; $a_M$ is huge ---&gt; $V_M$ is not constant as taken in linearized models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The same problems occurs when I use $v_{c}$. In all the linear model $V_c=V_{M}+V_T$ where the velocities are constant. How to use it then in my model since its not constant/almost constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any tips of how to transform it to X-Y graph but still keep it linear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="12940" LastActivityDate="2017-09-16T11:16:48.510" Title="Linear Guidance law into X-Y axis" Tags="|modeling|matlab|acceleration|linear-motion|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17152" PostTypeId="1" AcceptedAnswerId="17153" CreationDate="2017-09-16T13:44:42.530" Score="2" ViewCount="253" Body="&lt;p&gt;I'm studying vibrations in a gearhead, for which I have a general dynamic model of 12 degrees of freedom (only translational coordinates) and I have calculated its natural frequencies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to try and correlate my model with a physical system. I did some impact testing with a hammer and measured the response with one accelerometer placed on the horizontal direction of the gearheads case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My theory on mechanical vibrations and modal testing is quite basic and I don't know how to proceed and interpret the results of the readings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried simulating the forced response in order to correlate the model I'm using to the measurements. For example, in one of the experiments, I hit the gearhead horizontally and measured the response on the opposite side of the gearhead with the accelerometer, in the same axis. In the forced response simulation, I applied a unit force on what would be that equivalent degree of freedom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the magnitude response of the test, I have a few peaks ranging from　2khz to 24khz (its also the width of my window res). But in my force response simulation I get resonance peaks much higher, from　9khz up to 80khz. The natural frequencies range from 1khz to 16khz. &#xA;I have a few questions regarding what I'm trying to do:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Should the natural frequencies I calculated all show in the forced response simulation within a frequency range of up to 20khz, or only those close to the degree of freedom of the applied force in the force vector? In my results so far, none of the peaks in the forced response are close to one of the natural frequencies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How can I correlate the test measurements with the forced response simulation? none of the peaks in the magnitude response of the tests seem to be close to the simulation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From the modal test results, I measured some of the peaks and about 3 of them are very close to some of the natural frequencies, does this mean that this frequency were excited in this specific test? or can it be just coincidence? as I explained before, in the forced response simulation I couldn't find any relation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I should also note that my model is undamped, and my equipment was quite limited (only one sensor and small hammer(22khz)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read some theory on varius text books on vibrations and modal testing, they all show how to interpret a general FRF of course, but I can't find practical examples relating a theoretical model and modal testing like I'm doing, not explained in detail at least, most just skip to results, so the general methodology is what I'm lacking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="4088" LastActivityDate="2017-09-16T14:56:32.417" Title="Modal Testing and Frequency Response Interpretation" Tags="|mechanical-engineering|dynamics|vibration|modal-analysis|frequency-response|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17153" PostTypeId="2" ParentId="17152" CreationDate="2017-09-16T14:49:27.850" Score="2" Body="&lt;p&gt;You can't make much use of just one FRF, especially if none of the peak responses line up between the model and the test. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix the accelerometer at the position of one of your model DOFs and then impact the model at all 12 DOF positions (in the correct directions, of course). Move the accelerometer to the other DOF positions in turn, and repeat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be measuring the hammer force as well as the accelerometer response, so you can make the FRFs constisent with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to do some analog filtering of the measured signals &lt;strong&gt;before&lt;/strong&gt; you digitize them, otherwise the results will be a mess to interpret because of artefacts caused by aliasing. For example if you digitize at 22k samples/sec but don't do any filtering, you can't tell the difference between resonances at 11+1 = 12kHz and 11-1 = 10kHz in the digitized data and the FRFs - or even worse, there is no difference between resonances at 21kHz and 1kHz! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will then have 144 FRFs in total. You can cross-correlate them to see which peaks are &quot;real&quot; and which are just random artefacts. More important, you can also find the &lt;em&gt;mode shape&lt;/em&gt; corresponding to each peak, and try to identify that in the modes from your model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need some software to do this, of course. Commercial modal analysis software isn't cheap, but there are some free packages - for example &lt;a href=&quot;http://hosting.umons.ac.be/html/mecara/EasyMod/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hosting.umons.ac.be/html/mecara/EasyMod/index.html&lt;/a&gt; (which also needs Matlab). That is not a personal recommendation for EasyMod - I've never used it - but modal analysis and testing seems to be a speciality in Belgian universities so it &quot;comes from a good home&quot; IMO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good book to read is Ewins, D.J. “Modal Testing; theory, practice and application”. You can probably find the first edition free on the internet - that will tell you more than you ever wanted to know about the subject. Note the words &lt;em&gt;theory, practice and application&lt;/em&gt; in the title. It's a good book because it describes all three aspects. Just having a mathematical understanding of FRFs (which is about as much as you will learn in a typical B-Eng degree) is only a part of the total picture!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, it is not at all unusual for a first attempt at modelling to give modes that are all &quot;way off&quot; from anything you measured - but unless you can identify some corresponding &lt;em&gt;pairs&lt;/em&gt; of modes in the test and model data, you won't get very far trying to improve the situation except by pure good luck.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-09-16T14:56:32.417" LastActivityDate="2017-09-16T14:56:32.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17154" PostTypeId="1" CreationDate="2017-09-16T21:01:02.863" Score="0" ViewCount="137" Body="&lt;p&gt;If a facility buys natural gas at 8 dollars/Mcf to provide heating from an 85% efficient natural gas boiler, what is the equivalent cost of electricity ($/kWh) for a 95% efficient electric heater?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how to relate these.  What equations can I use?  Where do I begin?&lt;/p&gt;&#xA;" OwnerUserId="12945" LastActivityDate="2017-09-17T02:37:32.470" Title="Natural gas boiler Vs. Electric Heater efficiency" Tags="|gas|energy-efficiency|heating-systems|electrical|cost-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17155" PostTypeId="2" ParentId="16787" CreationDate="2017-09-17T00:10:53.120" Score="2" Body="&lt;p&gt;An article on &lt;a href=&quot;https://www.omega.com/prodinfo/infraredthermometer.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Omega&lt;/a&gt;'s website may help.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Five Ways to Determine Emissivity&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;There are five ways to determine the emissivity of the material, to ensure accurate temperature measurements:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Heat a sample of the material to a known temperature, using a precise sensor, and measure the temperature using the IR instrument. Then adjust the emissivity value to force the indicator to display the correct temperature.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;For relatively low temperatures (up to 500°F), a piece of masking tape, with an emissivity of 0.95, can be measured. Then adjust the emissivity value to force the indicator to display the correct temperature of the material.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;For high temperature measurements, a hole (depth of which is at least 6 times the diameter) can be drilled into the object.&lt;/li&gt;&#xA;&lt;li&gt;This hole acts as a blackbody with emissivity of 1.0. Measure the temperature in the hole, then adjust the emissivity to force the indicator to display the correct temperature of the material.&lt;/li&gt;&#xA;&lt;li&gt;If the material, or a portion of it, can be coated, a dull black paint will have an emissivity of approx. 1.0. Measure the temperature of the paint, then adjust the emissivity to force the indicator to display the correct temperature.&#xA;Standardized emissivity values for most materials are available (see pages 114-115). These can be entered into the instrument to estimate the material's emissivity value.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;Masking tape is commonly available, self adhesive, removable and low cost.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-09-17T00:10:53.120" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17156" PostTypeId="2" ParentId="17154" CreationDate="2017-09-17T02:37:32.470" Score="2" Body="&lt;p&gt;I answered a &lt;a href=&quot;https://sustainability.stackexchange.com/questions/6112/is-heating-with-candles-economically-and-or-ecologically-worthwhile/6113#6113&quot;&gt;similar question&lt;/a&gt; on SE Sustainable Living recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before anyone (including yourself) can answer your question you need to supply some more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing you need to find out &amp;amp; provide is, what is the &lt;strong&gt;heating value of the gas&lt;/strong&gt;. In metric units this would be the mega joules (MJ) of heat that one cubic metre of gas produces. In US customary units it will be the amount of BTU (or btu, being British thermal unit) per cubic foot of gas, or the BTU per Mcf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy companies charge by the MJ or BTU value of the gas they supply because  different types of gas have different energy values. Even though natural gas is methane each reservoir of gas has a different composition of natural gas, i.e. 98 percent methane and 2 percent other gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing you need to find out is &lt;strong&gt;how much heat the electric heater will produce&lt;/strong&gt; per kilo watt hour (kWh).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compare between gas and electricity you will have to convert either the gas heating value from BTU/Mcf to kWh/Mcf (if your using metric units J/m3 to kWh/m3). Or, convert electrical BTU/kWh to BTU/Mcf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other things you will need to know are the costs for gas and electricity. The cost of electricity will be in dollars per kWh and the cost of gas will probably be dollars per BTU, but it also might be dollars per Mcf. Which another reason why you need the heating (or energy) value of the gas that could be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having this information you then do a cost comparison between the two.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-09-17T02:37:32.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17157" PostTypeId="2" ParentId="17149" CreationDate="2017-09-17T06:11:15.620" Score="1" Body="&lt;p&gt;You have a few questions there:&lt;/p&gt;&#xA;&lt;h1&gt;Is the maximum tensile strength a property of a material ?&lt;/h1&gt;&#xA;&lt;p&gt;Yes. There is such a thing as maximum tensile strength, or &lt;a href=&quot;https://en.wikipedia.org/wiki/Ultimate_tensile_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;ultimate tensile strength&lt;/a&gt;. This is a property of the material, and therefore the shape does not matter. If any section of the spring experiences a &lt;a href=&quot;https://en.wikipedia.org/wiki/Stress_(mechanics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;stress&lt;/a&gt; equal to or above the ultimate tensile strength of the material, it will break!&lt;/p&gt;&#xA;&lt;p&gt;Geometry and loading conditions can cause different sections of a part to experience more or less stress, but the ultimate tensile strength determines if it will break or not. (This makes sense when you think of two identically shaped springs which are of different material: why does the plastic one break before a metal one, even if they are loaded the same?)&lt;/p&gt;&#xA;&lt;p&gt;It is also worth knowing that there is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Yield_(engineering)&quot; rel=&quot;nofollow noreferrer&quot;&gt;yield strength&lt;/a&gt;, which is the stress limit between permanent deformation and elastic deformation. It is a generally good idea that you should stay below the yield strength of a material to keep parts the same size. Working (well) below a materials yield strength is good design.&lt;/p&gt;&#xA;&lt;h1&gt;Do I need to divide the maximum tensile strength by two, multiply it by two, or leave it as it is?&lt;/h1&gt;&#xA;&lt;p&gt;Leave it. No dividing, multiplying, or anything to that number.&lt;/p&gt;&#xA;&lt;p&gt;For a spring, a quick approximation would be to figure out the maximum force the spring will support (including its own weight), and divide that by its cross sectional area. If that force/area goes above the ultimate tensile strength, it will break for sure. (If that force/area goes above the yield strength, the spring will be permanently deformed, but not always broken.)&lt;/p&gt;&#xA;&lt;p&gt;Of course, you may introduce a &lt;a href=&quot;https://en.wikipedia.org/wiki/Factor_of_safety&quot; rel=&quot;nofollow noreferrer&quot;&gt;factor of safety&lt;/a&gt; which may alter the numbers you are using. Maybe you want the spring to handle 2, 3, or more times the loading you actually think it will experience. In this case, I would multiply the maximum expected stress by this safety factor, which would allow you to easily compare materials you wish to use.&lt;/p&gt;&#xA;" OwnerUserId="11631" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-09-17T17:06:38.407" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17164" PostTypeId="1" CreationDate="2017-09-18T03:19:27.660" Score="1" ViewCount="638" Body="&lt;p&gt;I think this is an electricity question:  When I start my microwave, it starts normally the motor revs down drastically, then back up to normal, then stops.  ..a total of about 10-15 seconds.  Thinking the microwave was shot, I bought a new one.  Same results.  What could be the cause?&lt;/p&gt;&#xA;" OwnerUserId="12956" LastActivityDate="2017-09-18T18:14:57.313" Title="AC power fluctuation microwave oven" Tags="|electrical-engineering|electrical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17166" PostTypeId="2" ParentId="17135" CreationDate="2017-09-18T09:19:49.310" Score="1" Body="&lt;p&gt;Depends how far you want to go with the modifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can change the working parameters of the pump, changing the flow rate. While you can't meaningfully reduce the volume of the coolant in liquid phase through compressing it, through changing the flow rates you may delay the condensation - make it occur farther down the radiator - and so the volume of the liquid phase will change. You can introduce physical changes by modifying the machine: add or remove coolant, replace coolant with a different substance that has different phase change diagrams and heat properties, you may change the tubing modifying pressures and volumes, change the evaporator, replace the coolant with one that only works through compression/expansion without phase change at all, or even change the entire principle of operation - toss the whole vapor-compression refrigeration aggregate and paste a Peltier cell with a fan to the back of the cool box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The, only thing you can't change is that you input work in order to transfer heat, or it will cease to be a refrigerating machine. &lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2017-09-18T09:27:02.403" LastActivityDate="2017-09-18T09:27:02.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17167" PostTypeId="1" AcceptedAnswerId="17168" CreationDate="2017-09-18T09:23:21.940" Score="0" ViewCount="91" Body="&lt;p&gt;I read that trains in pratice instead of acceleration to a certain speed level and holding the speed, they accelerate and then cost, accelerate again, und keep this pattern. This was shown by experiments. What is the explanation why this way of driving is more energy-efficient?&lt;/p&gt;&#xA;" OwnerUserId="12959" LastActivityDate="2017-09-18T11:26:22.213" Title="What is the explanation why Pulse and Glide is more energy-efficient for trains" Tags="|energy-efficiency|" AnswerCount="1" CommentCount="2" ClosedDate="2017-09-20T15:07:26.313" ContentLicense="CC BY-SA 3.0" />
  <row Id="17168" PostTypeId="2" ParentId="17167" CreationDate="2017-09-18T11:26:22.213" Score="0" Body="&lt;p&gt;Just keeping a train at a fixed speed over level terrain takes a small fraction of the engine's power output.  As a result, the engine runs very inefficiently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In pulse and glide, the short bursts from the engine during the pulse phase are much higher power.  These allow the engine to run at higher efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly the same overall average power is produced by the engine either way, but with pulse and glide, that power is produced at a more efficient operating point of the engine.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-09-18T11:26:22.213" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17169" PostTypeId="2" ParentId="17104" CreationDate="2017-09-18T15:47:03.537" Score="1" Body="&lt;p&gt;If you want to track a trajectory and its derivatives is equivalent to just tracking that trajectory. However depending on your controller and reference signal you might get a steady state error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to know how to choose an appropriate controller we have to look at how a potential reference could look like. In the worst case the highest derivative of the reference we want to track is a constant, so of the form $t^n$. The Laplace transform of this is $n!\,s^{-1-n}$. So in order for the steady state error to be zero the &lt;a href=&quot;https://en.wikipedia.org/wiki/Sensitivity_(control_systems)&quot; rel=&quot;nofollow noreferrer&quot;&gt;sensitivity transfer function&lt;/a&gt; should have bigger positive slope at low frequencies than the negative slope of the Laplace transform of the reference. This implies that in general you want your openloop (plant times controller transfer function) to have at least $n+1$ integrators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I should also mention that you can also get a huge performance improvement by using feed forward.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2017-09-18T15:47:03.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17170" PostTypeId="1" AcceptedAnswerId="17171" CreationDate="2017-09-18T17:18:54.027" Score="2" ViewCount="116" Body="&lt;p&gt;How can I create a gear system which keeps, say, 60 rpm stable on output wheel, when I don't care about force? I am using a hand-turned crank as an input wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know a flywheel could be used, but it is weak to small accelerations and slowdowns over time.&lt;/p&gt;&#xA;" OwnerUserId="12964" LastActivityDate="2017-09-18T18:07:26.940" Title="Keeping rpm stable with input of floating rpm" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17171" PostTypeId="2" ParentId="17170" CreationDate="2017-09-18T18:05:09.603" Score="2" Body="&lt;p&gt;Since this is fully analog, if you don't want to delve into extremely complex devices like a continuously variable transmission with ratio governed by deviation of output speed, your best bet would be to ballpark the input torque range and provide a strongly non-linear speed governor creating extra friction on output exceeding rated value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Centrifugal_governor&quot; rel=&quot;nofollow noreferrer&quot;&gt;centrifugal governor&lt;/a&gt; that upon exceeding preset speed activates a brake on the input shaft, with friction climbing rapidly as output speed is exceeded - say, the weights of the governor move jaws of a brake, engaging them as friction reaches the limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat simpler mechanism, frequently used in music boxes, uses viscous friction of air; a little fan driven by a set of gears increasing its speed relative to input, responds with torque of air resistance proportional to square of rotary speed. This is less accurate, but simpler and suffer less wear over time. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-09-18T18:05:09.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17172" PostTypeId="2" ParentId="17170" CreationDate="2017-09-18T18:07:26.940" Score="1" Body="&lt;p&gt;The keyword you are looking for is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Governor_(device)&quot; rel=&quot;nofollow noreferrer&quot;&gt;governor&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most simple type is a parted circle that are hinged to be able to move outward into a ring by centrifugal force and the friction counteracts the increased speed.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-09-18T18:07:26.940" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17173" PostTypeId="2" ParentId="17164" CreationDate="2017-09-18T18:14:57.313" Score="1" Body="&lt;p&gt;What you are observing is the klystron (or other magnetron) tube (the microwave generator) being cycled on and off.  Most home microwave ovens cycle the generator on and off to allow the heated regions of the food to disperse the heat more evenly throughout the entire mass.    In fact, on those models where you select &quot;%heating&quot; the duty cycle changes, not the actual output power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next time,  before tossing your microwave, observe the results: does it boil a cup of water as quickly as it did when new?  Basically, unless it doesn't heat the water at all, it's still working.  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-09-18T18:14:57.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17174" PostTypeId="1" CreationDate="2017-09-18T23:17:33.097" Score="1" ViewCount="78" Body="&lt;p&gt;Does anybody know what that &quot;f&quot; looking sign is next to the 5, left-hand bottom corner of the LED labeling?: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It Indicates the diameter of the LED package.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Summary of Conclusion:&lt;/strong&gt; Don't seek precision unless you have the means to obtain it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;...I'm a young electronics hobbyist, who has the passion to become a robotics engineer at later stages of my maturity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lately I have been tinkering with my first circuit designs, simple stuff really. My first circuits were &quot;plug-and-play&quot;, based off of rough, highly in-precise calculations, but it would get things lit like the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pyERB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pyERB.jpg&quot; alt=&quot;Simple circuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, I decided to revise my scketch, adding in LED's and using a more scientific method to make more precise calculations about the circuits functionality...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story, but I got my circuit working. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12689" LastEditorUserId="12689" LastEditDate="2017-09-26T11:26:11.840" LastActivityDate="2017-09-26T11:26:11.840" Title="Confusing symbol in led labeling: How to Obtaining precision with hobby electronics?" Tags="|power-electronics|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17175" PostTypeId="1" CreationDate="2017-09-19T07:11:01.680" Score="1" ViewCount="132" Body="&lt;p&gt;Suppose, the wheels of a car were to be replaced with solid concrete wheels, which do not deform,flex, etc. Would the car not be able to accelerate, take corners, etc ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not understand why is it necessary for tires to develop slip for them to be able to move the car. &lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2017-09-19T07:11:01.680" Title="Why is longitudinal slip necessary for car to be able to move?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17176" PostTypeId="2" ParentId="17174" CreationDate="2017-09-19T11:45:52.130" Score="1" Body="&lt;p&gt;Note the circle thru the middle of the symbol.  That symbol is telling you that the following number is the diameter.  Somewhere else it probably says that millimeters is the default unit of length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, these LEDs are 5 mm in diameter.  That size is also known as &quot;T1 &amp;frac34;&quot;.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-09-19T11:45:52.130" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17177" PostTypeId="1" CreationDate="2017-09-19T11:47:18.713" Score="1" ViewCount="63" Body="&lt;p&gt;I am wondering the following, and didn't find anything online or here in the forum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) How is the capacity or load factor of a gas power plant calculated: (GROSS electricity generated)/(maximum capacity x time period) or (NET electricity generated)/(maximum capacity x time period)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What range is common for a gas power plant capacity factor (gas turbine, steam turbine, CCGT) when the plant is operated at a profit?&lt;/p&gt;&#xA;" OwnerUserId="11639" LastActivityDate="2017-09-19T11:47:18.713" Title="How to calculate and common capacity factors for gas power plants" Tags="|electrical-engineering|power-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17178" PostTypeId="1" AcceptedAnswerId="17179" CreationDate="2017-09-19T12:09:05.050" Score="2" ViewCount="8933" Body="&lt;p&gt;A door of volume l x b x h and weight w is hanging vertically on a horizontal hinge. How can I determine the force/torque to open or close this door?&lt;/p&gt;&#xA;" OwnerUserId="12970" LastEditorUserId="12970" LastEditDate="2017-09-20T13:10:55.053" LastActivityDate="2019-03-13T03:09:27.590" Title="Force required to open a door of volume lxbxh and weight w" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17179" PostTypeId="2" ParentId="17178" CreationDate="2017-09-19T12:44:23.957" Score="2" Body="&lt;p&gt;Assuming that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the door all one material and the weight is evenly distributed&lt;/li&gt;&#xA;&lt;li&gt;the door must be opened all the way to horizontal&lt;/li&gt;&#xA;&lt;li&gt;you open it by pushing on the edge furthest from the hinge&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;then the force required is half the weight.  In other words if the mass is 10kg, the weight $w = mg = 10kg*9.8ms^{-2} = 98$N, and 49N will be required to open it.  If you want a formula for $F$ then it's probably:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = mg/2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't push on the edge, then more force is needed.  If you can't push directly upwards, then more force is needed.  If there is any friction in the system, slightly more force is needed. If the door doesn't need to go horizontal, slightly less force is needed. If the door has heavy bits at the outside edge, more force is needed. All in all, it would be wise to design the system with a good amount of extra force.&lt;/p&gt;&#xA;" OwnerUserId="12022" LastActivityDate="2017-09-19T12:44:23.957" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17180" PostTypeId="1" CreationDate="2017-09-19T12:46:42.847" Score="1" ViewCount="45" Body="&lt;p&gt;I would like to simulate the following scenario with Simulink/Matlab(/Stateflow): A random number generator is running. Every 10 seconds the numbers that were generated during the last 10 seconds are added up. If the sum is greater than 50 the system is put into an unsafe state (simply a variable 'state' that initially is zero and then gets changed to one). Otherwise (sum &amp;lt;= 50), the state remains unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is how to implement the 'parallelism' of keeping the random number generation going while analysing the generated data in the background. Any hints of how to do this are greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I would like the simulation to somehow look:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nEGhv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nEGhv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12971" LastEditorUserId="12971" LastEditDate="2017-09-19T13:18:57.417" LastActivityDate="2017-09-19T13:18:57.417" Title="Simulink analysis of data parallel to generation" Tags="|matlab|simulink|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17182" PostTypeId="1" AcceptedAnswerId="17199" CreationDate="2017-09-19T16:43:32.853" Score="3" ViewCount="1705" Body="&lt;p&gt;I was solving problems from the Meriam and Kraige Engineering Mechanics Statics vol-1 and I couldn't get the correct way through this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is to find out the force in member $DK$ of the loaded overhead sign truss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/j5ckD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried eliminating member NM,FG,GT,HI.i tried using the method of section but any section passing through DK is of more than 3 members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest how to approach this problem.Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastActivityDate="2017-09-21T03:22:43.573" Title="Force in a member of the loaded overhead sign truss" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17183" PostTypeId="1" CreationDate="2017-09-19T18:05:35.480" Score="1" ViewCount="2770" Body="&lt;p&gt;What is the difference between maximum and absolute shear stress in multi axial stress system? &lt;/p&gt;&#xA;" OwnerUserId="12974" LastActivityDate="2017-09-22T02:05:01.693" Title="Difference between maximum shear stress and absolute shear stress" Tags="|applied-mechanics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17185" PostTypeId="1" CreationDate="2017-09-20T04:58:16.720" Score="3" ViewCount="244" Body="&lt;p&gt;Is it possible to have a drive and driven gear on the same shaft?  Rookie question, I know.  I've been doing a lot of googling and sometimes I think I understand that they can, but reading the definitions, I don't understand how that's possible.&lt;/p&gt;&#xA;" OwnerUserId="12977" LastActivityDate="2017-09-20T06:38:20.630" Title="drive/driven gears" Tags="|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17186" PostTypeId="2" ParentId="17185" CreationDate="2017-09-20T05:17:07.040" Score="3" Body="&lt;p&gt;Yes, if one gear is keyed to the shaft and the other on a bearing or sleeve - has been used in aftermarket overdrive units by Fairey for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : also old belt-driven equipment had fixed / sliding pulleys which could be locked together or unlocked to provide a &quot;neutral&quot; as most were driven from a common shaft in machine shops.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-09-20T06:38:20.630" LastActivityDate="2017-09-20T06:38:20.630" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17187" PostTypeId="1" AcceptedAnswerId="17188" CreationDate="2017-09-20T07:10:05.767" Score="1" ViewCount="222" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NjES8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NjES8.png&quot; alt=&quot;The different support reactions at B and C&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that pins prevent horizontal and vertical translation and that rollers prevent only the vertical one.&#xA;Therefore, what I do not understand here is that while the support at B reacts with two forces C support has only one, even though both look like pins to me.Where am I going wrong here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I could not find anything on the site about whether I can upload images this way. I hope there are no copyright issues related to my post because&#xA; I do not own the images. If there is, please notify me and I will immediately take them.)&lt;/p&gt;&#xA;" OwnerUserId="12979" LastActivityDate="2021-11-19T05:06:54.470" Title="Isn't a pin support supposed to prevent translation both horizontally and vertically?" Tags="|statics|connections|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17188" PostTypeId="2" ParentId="17187" CreationDate="2017-09-20T07:23:49.423" Score="6" Body="&lt;p&gt;It's simple: while B is a pinned support, C is a roller. You can see that in the figure, where B is drawn clearly bolted to the &quot;ground&quot; while C isn't. In fact, the contact between C and the &quot;wall&quot; seems to be a wheel of some sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rollers don't have to &quot;prevent only the vertical [translation]&quot;. They have to prevent translation in one direction while allowing it in another. The most common case of a roller is resting on a horizontal surface, therefore preventing vertical motion but allowing horizontal motion. In this case, however, the roller rests on a vertical surface, therefore horizontal movement is restricted but vertical movement is free.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="78" LastEditDate="2017-09-20T17:33:16.887" LastActivityDate="2017-09-20T17:33:16.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17190" PostTypeId="1" CreationDate="2017-09-20T10:40:00.167" Score="1" ViewCount="951" Body="&lt;p&gt;The formula for newtons to pascals is below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1\text{ Pa} = 1\text{ N/m}^2 = 9.81\text{ kg/m/s}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to solve this problem but the units are a bit confusing. I want to get the shear stress (tau) based on these input variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General stress = Force/Area&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force is 1255Nm the area is length 20mm and height 10mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$tau = 1.255\ x\ 10^6\ Nmm\ /\ 200\ mm^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with straight division, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1,255,000/200 = 6275$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I am unsure about the units;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I mathematically devise the units?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit 2]&#xA;I think I've narrowed down the question and would like to see if I am on the right track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula: Tau=Tr/J &#xA;since I know the Maximum torque, the polar moment of inertia for a solid shaft, I should use the Modulus of rigidity from the material properties sheet for Tau then solve for the radius to get the diameter of the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="12586" LastEditorUserId="12586" LastEditDate="2017-09-22T14:37:50.093" LastActivityDate="2017-09-22T14:37:50.093" Title="Calculate shear stress unit conversions" Tags="|measurements|" AnswerCount="1" CommentCount="2" ClosedDate="2017-09-30T12:28:33.223" ContentLicense="CC BY-SA 3.0" />
  <row Id="17191" PostTypeId="2" ParentId="17190" CreationDate="2017-09-20T11:15:08.147" Score="2" Body="&lt;p&gt;You might be mixing up units here. &#xA;You say the applied force is 1255Nm: Is this is a torque application? If so, you might want to look at the Torsion formula, Tau=Tr/J&#xA;(&lt;a href=&quot;https://en.wikipedia.org/wiki/Torsion_(mechanics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Torsion_(mechanics)&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the units of force are (as they should be) Newtons, then the shear stress comes out to be in N/mm^2, as expected.&lt;/p&gt;&#xA;" OwnerUserId="12981" LastActivityDate="2017-09-20T11:15:08.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17193" PostTypeId="1" AcceptedAnswerId="17195" CreationDate="2017-09-20T13:30:03.870" Score="1" ViewCount="94" Body="&lt;p&gt;I've been working on implementing a new company standard for our file naming convention (super fun), and it has been mentioned to me that there is a &quot;standard&quot; that says that you shouldn't use spaces in file and folder names when saving documents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done some looking and I haven't found a published standard that says one way or the other whether this is acceptable. Does anyone know of a standard relating to this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Released Drawings&quot; as opposed to &quot;ReleasedDrawings&quot; as a folder name&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;123 - Rev 0 - Important part&quot; as opposed to &quot;123-Rev0-Importantpart&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping to find a a specific standard, not just a rule of thumb, or best practice.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-09-21T23:12:42.920" Title="Filename standards and spacing" Tags="|standards|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="17194" PostTypeId="2" ParentId="17193" CreationDate="2017-09-20T15:19:21.503" Score="1" Body="&lt;p&gt;AFAIK, the main reason for the space convention was for url and network oriented systems, as some systems had problems when encoding/decoding spaces as parts of urls (they also look messy in a browser etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various organization have their own standards such as &lt;a href=&quot;http://sitem.herts.ac.uk/secreg/upr/pdf/IM11-Apx5-Standard%20Naming%20Conventions%20for%20electronic%20Files,%20Folders%20and%20Records-v03.0.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this organization's standard&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12804" LastEditorDisplayName="user16" LastEditDate="2017-09-21T23:12:42.920" LastActivityDate="2017-09-21T23:12:42.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17195" PostTypeId="2" ParentId="17193" CreationDate="2017-09-20T16:29:58.050" Score="4" Body="&lt;p&gt;If you are likely to share your files with other organizations that use different operating systems, you need to research what works best on all the main options - i.e. Windows, MacOS, and the various flavors of Unix. The &quot;If it works on Windows 10, that's OK because it's the only system we use&quot; is probably digging a hole to fall into in future!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spaces in file names generally cause pointless aggravation, when a system takes only the first &quot;word&quot; of the name as the complete name, and first complains that the file doesn't exist, and then complains that it can't understand what follows it. You can work round that by putting the whole name in quotes, but why make people do that? If you don't like ReleasedDrawings, use Released-Drawings or Released_Drawings instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other invalid characters - the most obvious being / which is the separator between &lt;em&gt;parts&lt;/em&gt; of the file name (e.g. folders and directories) and the equivalent \ on Windows. Others are quote marks, &quot;&gt;&quot;, &quot;&amp;lt;&quot;, &quot;?&quot;, &quot;*&quot;, etc. These characters have special meanings on command-line instructions in most operating systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some operating systems have case sensitive file names (ReleasedDrawings and Releaseddrawings are different files!) but others do not. Windows is a compromise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some systems have &quot;reserved&quot; file names - for example COM and NUL on Windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be a limit to the total length of a file name (including all the folder names that precede it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Microsoft reference (obviously Windows-specific) is &lt;a href=&quot;https://support.office.com/en-us/article/Invalid-file-names-and-file-types-in-OneDrive-OneDrive-for-Business-and-SharePoint-64883a5d-228e-48f5-b3d2-eb39e07630fa&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://support.office.com/en-us/article/Invalid-file-names-and-file-types-in-OneDrive-OneDrive-for-Business-and-SharePoint-64883a5d-228e-48f5-b3d2-eb39e07630fa&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could do worse than stick with the POSIX specification for &quot;fully portable filenames&quot; which permits only characters A-Z, a-z, 0-9, &quot;_&quot;, &quot;-&quot;, and &quot;.&quot; (&quot;-&quot; must not be the &lt;em&gt;first&lt;/em&gt; character of the name) and a maximum length of 14 characters, or ISO 9660 (used on CDs and similar devices) which is case-insensitive and only allows A-Z, 0-9, &quot;_&quot;, and &quot;.&quot; with a maximum length of about 180 characters.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-09-20T16:29:58.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17196" PostTypeId="1" AcceptedAnswerId="18245" CreationDate="2017-09-20T18:34:43.420" Score="2" ViewCount="536" Body="&lt;p&gt;I would like to build a clock where the face rotates. Imagine a standard clock, but instead of hands pointing to the hour, there would be a rotating face plate with a small window, as it were. The face plate would obscure everything except the current hour, which would be visible through the window. If it makes any difference, I need it to be a 24-hour, not a 12-hour, clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have considered buying a 24-hour high-torque clock movement (because I'm not interested in building the actual gears, etc.), but I get the impression that it would be hard to attach anything to it other than standardized clock hands. I could try to glue the face plate onto the hour hand, but I don't know if that would hold effectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you. (Also, if this isn't the right forum, please send me in the right direction.)&lt;/p&gt;&#xA;" OwnerUserId="12983" LastActivityDate="2017-11-29T01:42:31.493" Title="building a clock with a rotating face" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17197" PostTypeId="2" ParentId="16794" CreationDate="2017-09-20T20:22:36.363" Score="1" Body="&lt;p&gt;I see how you're thinking: the car is a &quot;piston&quot; and if you heat the air behind this &quot;piston&quot;, it will push it forward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now think how inefficient this &quot;engine&quot; is - you are adding fractions of a degree to the air (working gas) which makes for horrible thermodynamic efficiency. On top of that, you have no &quot;piston walls&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1415" LastActivityDate="2017-09-20T20:22:36.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17198" PostTypeId="2" ParentId="17183" CreationDate="2017-09-20T23:12:27.207" Score="1" Body="&lt;p&gt;I think the term is absolute maximum shear stress, meaning which one is the largest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three planes of shear: one is the largest (the maximum of the maxima), the other two are not.&lt;/p&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2017-09-20T23:12:27.207" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17199" PostTypeId="2" ParentId="17182" CreationDate="2017-09-21T02:10:38.810" Score="2" Body="&lt;p&gt;This is a K-Truss. I will try to provide a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrections: There should be a cyan member in the truss, otherwise, it is unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8moUG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8moUG.png&quot; alt=&quot;First Cut&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown above, the system will be cut by curved line (as highlighted by the red line). This is a special case to the method of sections wherein the rule allows cutting of more than 3 members provided that all member except one are collinear at a point. Separating the left side of the system gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S8c24.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S8c24.png&quot; alt=&quot;Section 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactions have been solved by static equations. In the above image, we only need the horizontal unknowns $LK$ and $CD$, thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up moments at C yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$15\ (8)=LK\ (5)$$&#xA;$$LK\ =\ 24\ kN\ (Tension)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will leave the sign as is so as to simplify the analysis of sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next is summing up moments at L yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$15\ (8)\ -\ CD\ (5) = 0$$&#xA;$$CD\ =\ -24\ kN$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we do not need to solve the values of member BP and MP as they are not needed in the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we cut another section from the system as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qaL9T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qaL9T.png&quot; alt=&quot;Section cut 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and separating the left section:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cd0TC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cd0TC.png&quot; alt=&quot;Section 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that members $CD$ and $LK$ has been solved as denoted in yellow color. Two equations and two unknowns will be needed, thus, summing up forces along x-axis and y-axis yields the following equations respectively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$QD\ [4/sqrt(22.25)]\ +\ QK\ [4/sqrt(22.25)]\ +\ CD\ +\ LK\ =\ 0$$&#xA;$$QD\ [2.5/sqrt(22.25_]\ -\ QK\ [2.5/sqrt(22.25)]\ -\ 5\ +\ 15\ =\ 0$$&#xA;$$QD\ =\ -9.434\ kN$$&#xA;$$QK\ =\ +9.434\ kN$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that we cannot still use method of joints at joint $D$, therefore, we will continue to solve for more members of the system as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2eDG6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2eDG6.png&quot; alt=&quot;Section 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Separating the right sections:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bNBX3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bNBX3.png&quot; alt=&quot;Section 4&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up moments at J yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$20\ (8)\ +\ ED\ (5)\ =\ 0$$&#xA;$$ED\ = \ -32\ kN$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can use method of joints at joint $D$ because there are only two unknowns left as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jb2IV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jb2IV.png&quot; alt=&quot;joint D&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up forces along x-axis yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-DC\ -\ DQ\ [4/sqrt(22.25)]\ +\ DE\ +\ DR\ [(4/sqrt(22.25)]\ = \ 0$$&#xA;$$DR\ = \ 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up forces in the y-axis yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-DQ\ [2.5/sqrt(22.25)]\ +\ -DK\ =\ 0$$&#xA;$$DK\ =\ 5\ kN$$&lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2017-09-21T02:10:38.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17200" PostTypeId="2" ParentId="17182" CreationDate="2017-09-21T03:22:43.573" Score="2" Body="&lt;p&gt;At first glance it’s a fairly daunting problem, but it’s actually easily solved using K-truss analysis.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Solve for the external reactions by summing the moments about Point U, and summing vertical forces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use K-truss analysis to determine the net vertical force in diagonals QD/QJ and RD/RK. This is done by making vertical section cuts and then summing the vertical forces. We need only consider the vertical component of the force in the diagonals.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Distribute that equivalent vertical force we calculated in step 2 between the diagonals. (Basically we look at another free body diagram and consider vertical equilibrium again.) Because the diagonals are at the same angle, the equivalent vertical force distributes equally.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use method of joints at Joint K to solve for the force in element DK. (5kN in tension)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GXe3Jl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GXe3Jl.jpg&quot; alt=&quot;Solution Method&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2017-09-21T03:22:43.573" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17201" PostTypeId="1" CreationDate="2017-09-21T03:55:33.473" Score="1" ViewCount="11434" Body="&lt;p&gt;If I know the torque and rpm that the shaft need to operate at, can I calculate the diameter of the shaft?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A motor operating at 60 rpm applying 1,000 Nm of force on a shaft that has a modulus of rigidity of 50 GPa and is 10 cm in length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I calculate the diameter of the shaft so that it would not fail? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tensile strength is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Tensile Strength, Ultimate  686 MPa &#xA;Tensile Strength, Yield     490 MPa &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This will be a horizontal shaft, the length isn't final yet but I am estimating that it will be 2,000 mm long and supported at 200 mm from each end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There shouldn't be any unforseen shocks being applied to the shaft. It will be transmitting the power from a 15 HP motor that's connected via a coupling on one end.&lt;/p&gt;&#xA;" OwnerUserId="12586" LastEditorUserId="368" LastEditDate="2017-09-21T15:24:41.180" LastActivityDate="2019-09-13T11:31:29.700" Title="How to calculate the diameter of a transmission shaft from RPM &amp; Torque" Tags="|mechanical-engineering|materials|elastic-modulus|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17202" PostTypeId="2" ParentId="17201" CreationDate="2017-09-21T07:09:23.783" Score="3" Body="&lt;p&gt;The elastic modulus would give you how much it would twist in use, you would need the tensile strength:&#xA;taken from engineering toolbox &lt;a href=&quot;http://www.engineeringtoolbox.com/torsion-shafts-d_947.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineeringtoolbox.com/torsion-shafts-d_947.html&lt;/a&gt; :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;τ = T r / J                              (1)&#xA;&#xA;where&#xA;&#xA;τ = shear stress (Pa, psi)&#xA;&#xA;T = twisting moment (Nm, in lb)&#xA;&#xA;r = distance from center to stressed surface in the given position (m, in)&#xA;&#xA;J = Polar Moment of Inertia of Area (m4, in4)&#xA;&#xA;Diameter of a Solid Shaft&#xA;Diameter of a solid shaft can calculated by the formula&#xA;&#xA;D = 1.72 (Tmax / τmax)^1/3                            (4)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;this should answer your question with regard to the minimum theoretical value, but of course in real work applications you would need to consider safety factors and deflection (which you can find calculations for on the aforementioned link).&lt;/p&gt;&#xA;" OwnerUserId="12804" LastEditorUserId="12804" LastEditDate="2017-09-21T10:03:47.617" LastActivityDate="2017-09-21T10:03:47.617" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17204" PostTypeId="2" ParentId="17201" CreationDate="2017-09-21T07:15:34.193" Score="1" Body="&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T\ =\ 1000\ Nm^2$, $G\ =\ 50 GPa$, $L\ =\ 10\ cm\ =\ 100\ mm$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque has the wrong units, please double check. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question, a simple google check gives &lt;a href=&quot;https://en.wikipedia.org/wiki/Torsion_(mechanics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link about torsion mechanics.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2017-09-21T07:15:34.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17205" PostTypeId="5" CreationDate="2017-09-21T09:21:14.563" Score="0" Body="&lt;p&gt;Matlab is a high-level language and interactive programming environment for numerical computation and visualization developed by MathWorks. It can be programmed to numerically solve many engineering problems using its core functions and/or the available specialized toolboxes.&lt;/p&gt;&#xA;" OwnerUserId="12990" LastEditorUserId="1832" LastEditDate="2017-09-21T22:49:51.130" LastActivityDate="2017-09-21T22:49:51.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17206" PostTypeId="4" CreationDate="2017-09-21T09:21:14.563" Score="0" Body="Use of Matlab to solve engineering problems." OwnerUserId="12990" LastEditorUserId="12990" LastEditDate="2017-09-21T22:50:10.207" LastActivityDate="2017-09-21T22:50:10.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17207" PostTypeId="1" AcceptedAnswerId="17210" CreationDate="2017-09-21T13:40:07.983" Score="1" ViewCount="107" Body="&lt;p&gt;A colleague of mine sent me this image of some racking and I've been trying to figure out the design intent of some of the features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image one (key holes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y4rfh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y4rfh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the keyholes on this racking the one side of the lobe is curved while the other side is straight, sloping away from where the skids are going to be placed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image 2 (top view paint sketch of the racking profile)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5qnkQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5qnkQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image is a quick and dirty sketch of the shape of the racking profile,, sort of a W shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I looked at these images I was thinking that the two shapes were made to work together, the straight side of the keyhole being made to slightly pull the shelves of the racking apart when placed in them. I.e. not fully resting on the bottom of the keyholes under their own weight, and then the 'W' shape of the racking uprights being used to act as a bit of a spring/dampener feature. My thinking/reasoning for this is so that when a forklift drops/lowers a skid onto the shelves, the racking uprights will have room to flex slightly and absorb some of the impact (forklifts not being particularly precision instruments). This way the racking pins don't get impact loaded every time a skid gets loaded, and also the flexing of the uprights would allow for a greater tolerance in the levelness of the shelving. (Shelving/racking safety requirements being remarkably strict)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is,  does my reasoning sound accurate, or is there another reason for these somewhat 'complicated*' shapes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;note- by complicated I don't mean intricate, these are very simple shapes to make, but they appear to have more design intent because they go beyond simple holes and extrusion processes. I come from industry where all geometry requires justification and calculations not just &quot;because it looks cool&quot;.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-09-22T00:34:50.570" Title="Racking design question" Tags="|mechanical-engineering|structural-engineering|design|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17208" PostTypeId="1" CreationDate="2017-09-21T15:54:10.603" Score="1" ViewCount="36" Body="&lt;p&gt;I have to determine the cooling time of a pipe from its initial temperature $\vartheta_{\text{wall,}1}=500 \text{ °C}$ to the temperature $\vartheta_{\text{wall,}2}=150 \text{ °C}$. The cooling happens due to an air flow through the pipe with the temperature $\vartheta_{\text{inlet}} = 30{\text{ °C}}$ and due to contact to the air in the environment $\vartheta_{0} = 20{\text{ °C}}$ at the outer surface of the pipe. Is there a quick way to estimate it? I thought about using Newton's Law of cooling but the problem is the heat transfer coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data for inlet:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Inlet temperature: $\vartheta_{\text{inlet}} = 30{\text{ °C}}$&lt;/li&gt;&#xA;&lt;li&gt;$p_{\text{inlet}} = 4 \text{ bar}$&lt;/li&gt;&#xA;&lt;li&gt;$u_{\text{in}} = 5 \text{ m/s}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Geometric/matrerial data of pipe:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Initial wall temperature (assumed to be constant) $\vartheta_{\text{Wall},1}=500 \text{ °C}$&lt;/li&gt;&#xA;&lt;li&gt;Outer diameter $D_\text{o}=0.0210 \text{ m}$&lt;/li&gt;&#xA;&lt;li&gt;Wall thikness $t = 0.0005 \text{ m}$&lt;/li&gt;&#xA;&lt;li&gt;Length $L = 0.2000 \text{ m}$&lt;/li&gt;&#xA;&lt;li&gt;Material not sure about it lets assume is some steel material&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-09-21T15:54:10.603" Title="How to estimate cooling time of pipe due to inner flow (instationary heat transfer)" Tags="|airflow|cooling|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17209" PostTypeId="2" ParentId="15988" CreationDate="2017-09-21T19:50:33.430" Score="1" Body="&lt;p&gt;With a few assumptions : Dimensions indicate pressure below where serious codes like ( USA) ASME section 8, div, 1 would apply. Although &quot;lethal&quot; materials would change that. You could still look at any available code to see what those requirements are. In particular weld reinforcement/ thickness. No code = no NDE , but assuming steel, It wouldn't be too expensive to do some magnetic particle testing ( MT). But , for example, if it will contain gasoline in a crowded area, you should get serious.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-09-21T19:50:33.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17210" PostTypeId="2" ParentId="17207" CreationDate="2017-09-22T00:23:33.663" Score="3" Body="&lt;p&gt;The main reason for the shape (W-shape) of one side of the rack is to improve structural strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4EIHu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4EIHu.png&quot; alt=&quot;See picture as reference.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As depicted in picture $a$, the system is to resist forces that will tend to bend the vertical pole in its local x-axis, see picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DVKiF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DVKiF.png&quot; alt=&quot;Deflected shape of the pole.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the moment of inertia of the profile shall be designed to be the maximum considering also the part of contact between the rack and the pole. This will be proven when you compare the moment of inertia of the two parts. Also, if you fold a piece of paper or cardboard into V-shape, it is harder to bend then when it is flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if the profile is to be constructed like in picture $a$, the side in which the racks are connected will easily bulge out the frame. See image below. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/sqK87.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sqK87.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, this design gives room for the deflection to occur without compromising both the strength and the aesthetics of the racking system. See also &lt;a href=&quot;https://qph.ec.quoracdn.net/main-qimg-19ea8f8ed5515c7f13406fa7c347efce&quot; rel=&quot;nofollow noreferrer&quot;&gt;this image.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-22T00:34:50.570" LastActivityDate="2017-09-22T00:34:50.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17211" PostTypeId="2" ParentId="17183" CreationDate="2017-09-22T02:05:01.693" Score="1" Body="&lt;p&gt;given a tensor stress state you can find the shear on any arbitrary plane and direction. Maximum shear stress says we have found the plane with the maximum value.  The maximum shear stress is always positive. (or zero for hydrostatic case)  Note this is an invarient, you can report maximum shear without specifying a coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;absolute shear stress is probably the absolute value of the shear stress on a specified orientation which may or may not be the special orientation that gives the maximum.  I say probably because the question hasn't given enough context, but I expect you are seeing something like &lt;code&gt;abs(sxy)&lt;/code&gt; in a given coordinate system&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2017-09-22T02:05:01.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17212" PostTypeId="1" AcceptedAnswerId="17215" CreationDate="2017-09-22T13:19:19.083" Score="2" ViewCount="38" Body="&lt;p&gt;Is it possible to simulate a layer of dust deposited on top of a PV cell in a controlled manner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One potential method I had was to put a given weight of a particulate/powder (chalk powder, talcum powder, etc.) into water, and pour the water on top of the solar panel, letting it dry and thus leaving the powder on top of the panel. Would this be viable, and if so, what would be the best powder to use?&lt;/p&gt;&#xA;" OwnerUserId="13005" LastActivityDate="2017-09-22T17:04:29.007" Title="Simulate Dust Layer on PV Cell?" Tags="|simulation|experimental-physics|photovoltaics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17213" PostTypeId="1" CreationDate="2017-09-22T13:20:36.257" Score="1" ViewCount="39" Body="&lt;p&gt;I need to carry dc (direct current) electrical power from a solar array to point of use.  I know that an undesirable capacitive load will develop if I just run the power and neutral wires parallel to each other for a long distance.  For this reason, the transmission wire pair would need to be twisted around each other to introduce an inductive load to offset the capacitive load.  This is a form of impedance balancing with the goal of net 0 reactive load, where reactance encompasses both inductance and capacitance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there is a formula to calculate the wire twisting rate in twists per meter needed to cause enough inductance to zero out capacitance but I have had trouble finding it.  I think it is based on the total length of wire, size of wire, voltage, and width of insulation among other variables.  Any help is much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-09-22T13:20:36.257" Title="What is the twisting rate to reduce capacitance load on DC power transmission line?" Tags="|electrical-engineering|power|power-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17215" PostTypeId="2" ParentId="17212" CreationDate="2017-09-22T14:15:29.330" Score="1" Body="&lt;p&gt;A PV cell's system efficiency will vary exactly linearly with the amount of light blocked by the dust.  For the particle sizes and wavelengths involved, refraction and multiple scattering effects are minimal.  What this leaves you with is a way to &quot;map&quot; dust collection to the equivalent uniform, neutral density filter as SolarMike implied in his comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend setting up two cells, one as a clean reference, and the other with arbitrary dust levels.  Make a few measurements and generate a graph of ,say, power loss vs.  total volume of powder deposited.   You may want to repeat a few times to guarantee more or less uniform distribution of powder over the active area.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-09-22T14:15:29.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17216" PostTypeId="1" AcceptedAnswerId="17220" CreationDate="2017-09-22T14:51:22.127" Score="2" ViewCount="286" Body="&lt;p&gt;I would like to plug a 25A welder on a 10A house plug. I understand that it will not accept full power or the circuit breaker would switch.&#xA;However, my question is if I may run under some lower power setting: for instance setting 70A out of 200A on the welder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a very simplistic point of view:&#xA;A welding machine with a maximum of 200A for 25A@220V mean, the welder at 60A would require about 10A@220V &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a welding machine is not such a simplistic situation and uses several variants of waves, AC or DC, several frequencies, etc. which make all more complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&#xA;May I plug this 25A Welder on a 10A-protected house plug?&lt;/p&gt;&#xA;" OwnerUserId="11129" LastEditorUserId="2819" LastEditDate="2017-09-22T19:19:59.300" LastActivityDate="2017-09-22T19:19:59.300" Title="Plugging a 25A device (Welder) on 10A house plug" Tags="|electrical-engineering|welding|welds|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17218" PostTypeId="2" ParentId="17212" CreationDate="2017-09-22T17:04:29.007" Score="1" Body="&lt;p&gt;More to the &quot;even-distribution&quot; question, I'd recommend your idea of water distribution.  I'd use spray nozzles - they're designed for relatively even distribution.  To that end, I'd recommend &lt;a href=&quot;http://www.bete.com/products/is.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;IS&lt;/a&gt; nozzles (lower budget) or &lt;a href=&quot;http://www.bete.com/products/nc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;NC&lt;/a&gt; nozzles (higher budget), with a distribution manifold.  These will provide an easily repeatable test method.  Simply dial down to the minimum levels for consistent spray (1 psi for IS, 3.5 psi for NC).  Then adjust the concentration of dust in your spray solution.  Use a pressure regulator to ensure consistent pressure across the spray system, and a timer to ensure the exact amount of solution sprays across the panels.  Finally, as a test to the system, weigh the dried off powder at the end of the experiment on each panel to verify how much dust was deposited on each panel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the powder, if you have access to a very low humidity area I'd recommend powdered milk or creamer.  Talcum powder is also used readily.  But that's mostly a personal opinion.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-09-22T17:04:29.007" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17219" PostTypeId="1" CreationDate="2017-09-22T18:41:57.790" Score="1" ViewCount="40" Body="&lt;p&gt;Here's one for any of you electronics experts. On the 2nd story of my house, I have a TV antenna. It reaches a height of about 5 ft above the house, and the mounting bracket is attached to the back wall of the restroom. The downlead cable runs into the attic, through a preamp and on to the tv's throughout the house. Now...here is the puzzler...anytime someone goes up to the restroom and opens the door, the picture on all of the tv's will start tiling badly...before that, it was crystal clear....once the person comes out, opening and closing the door again, it will become clear. Now, why in the world would opening and closing the door of the restroom affect the signal from an antenna mounted behind the restroom and at least 5 foot higher than it's roof? I thought the family was nuts when they first told of this...I have tested and retested, it makes no sense at all, but I can't deny that it does happen.Anyone have a theory?&lt;/p&gt;&#xA;" OwnerUserId="13008" LastActivityDate="2017-11-26T19:20:44.107" Title="Antenna reception problem" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="4" ClosedDate="2017-11-27T03:01:24.147" ContentLicense="CC BY-SA 3.0" />
  <row Id="17220" PostTypeId="2" ParentId="17216" CreationDate="2017-09-22T19:17:49.530" Score="2" Body="&lt;p&gt;Yes, you can plug a higher rated load into a lower rated breaker if the load is monitored to minimize trips. If you are using a different plug, make sure that the other plug is removed entirely there are no exposed wires or contacts. I assume you are in europe and the standard wall voltage is indeed 220V (If someone in the US is reading this in the future, a standard wall outlet is 110V and will not work for a 220V device).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The breaker is sized to protect the wiring, not your device. It will allow greater than 10A for a small period of time (say 15A for 30 seconds) for start up loads. If you exceed 10A for a length of time the circuit breaker will trip to protect the wiring. This will not harm your device, it is just like unplugging it or the power going off.  You can increase the load and trip the breaker a couple times to test how high you can run your welder, but like agenpt mentioned in the comments, you should not run it in a way where you are tripping the breaker frequently. Breakers are only rated for so many trips (say 100) before they risk not operating correctly and need to be replaced.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-09-22T19:17:49.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17221" PostTypeId="1" CreationDate="2017-09-22T21:05:33.930" Score="1" ViewCount="18" Body="&lt;p&gt;I'm looking for a high torque stepper motor which has a mount wholes on both sides (rear and front). A double shafted model would be great. &lt;/p&gt;&#xA;" OwnerUserId="12539" LastActivityDate="2017-09-22T21:05:33.930" Title="Are there high torque stepper motors with rear and front mount availabe on the market" Tags="|stepper-motor|" AnswerCount="0" CommentCount="2" ClosedDate="2017-09-25T12:10:18.027" ContentLicense="CC BY-SA 3.0" />
  <row Id="17222" PostTypeId="2" ParentId="15064" CreationDate="2017-09-22T23:33:15.760" Score="1" Body="&lt;p&gt;Lets say the worker (or pair of workers, front and back) that used this press could apply 100lbf or 445N of force. And lets say that force was applied at about 1 meter away from the center of the screw. The screw has a large force contact area and would be well coated in viscous lard for lubrication. The lubrication would be in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lubrication#The_regimes_of_lubrication&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;fluid film&quot; regime&lt;/a&gt; in which the fluid supports all the load; the same regime a modern grease lubricated lead screw works in. Modern lead screws are in the &lt;a href=&quot;https://www.sdp-si.com/D810/PDFS/Ball%20And%20Acme%20Lead%20Screw%20Technical%20Info.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;30 to 50% efficiency&lt;/a&gt; range. Page 9 of &lt;a href=&quot;http://web.mit.edu/2.75/fundamentals/FUNdaMENTALs%20Book%20pdf/FUNdaMENTALs%20Topic%206.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; calculated 25% efficiency with a friction of 0.2 and an angle of 30 degrees. Lets say 10% efficient to be conservative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Screw_(simple_machine)#Actual_mechanical_advantage_and_efficiency&quot; rel=&quot;nofollow noreferrer&quot;&gt;Force * Distance = 2*Pi* efficiency * torque&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force = (2* 3.14159 * 0.10 * (445N * 1m)) / 0.0584m&lt;br&gt;&#xA;&lt;strong&gt;Force = 4788N or 1076lbf&lt;/strong&gt; (per screw)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area it presses is important too. Just guessing based on your photo I'm going to estimate 25cm x 25cm. This gives a pressure of &lt;strong&gt;77kPa or 11psi&lt;/strong&gt;. This must have been sufficient pressure to economically press the paper; otherwise the medieval engineer/artisan would have used a smaller pressing area to get a higher pressing force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More pressure would result in more water removed, but it is not linear. The increase from 0 to 5 psi will remove more water than the increase from 5psi to 10psi, and so on. So even if the press generated more or less pressure, its not super critical from a historical process point of view. The operator was probably most concerned with the paper holding shape after it was pressed. The raw materials in the paper, pressing aids (gypsum), upstream processes, etc will also have a big factor on water retention/removal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressed moisture would have likely still been 60 to 75% (still lots of water to remove). Using the press would have saved a lot of time and fuel for the a secondary drying process. Sun or fire would have been necessary to get the moisture level down to a non-molding level (say 5 to 10%). (My experience in this area is from beet pulp pressing and drying, but those numbers should be close.)&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-09-22T23:33:15.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17224" PostTypeId="1" AcceptedAnswerId="17225" CreationDate="2017-09-23T06:06:40.240" Score="3" ViewCount="433" Body="&lt;p&gt;I am solving problems from the book: Meriam Kraige Engineering Mechanics statics vol 1. The problem is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sliding door rolls on the two small lower wheels $A$ and $B$. Under normal conditions, the upper wheels don't touch their horizontal guide. The question is to find out the force $P$ required to slide the door at a steady speed if wheel $A$ becomes &quot;frozen&quot; and doesn't turn in its bearing. The coefficient of kinetic friction between a frozen wheel and the supporting surface is $0.3$ and the center of mass of the 64 kg door is at its geometric center. We have to neglect the small diameter of the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VmuAN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated the normal reactions at the two wheels to be equal to $32g$ each. I then applied $P=0.3N$ which gave me the incorrect value of $P$ .I couldn't get where I am wrong.Any help shall be highly appreciated.Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastEditorUserId="12597" LastEditDate="2017-09-23T11:54:08.347" LastActivityDate="2017-09-23T11:54:08.347" Title="Force P required to slide the door if one of the wheel becomes frozen" Tags="|civil-engineering|statics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17225" PostTypeId="2" ParentId="17224" CreationDate="2017-09-23T07:19:53.727" Score="1" Body="&lt;p&gt;See picture below for reference. The thing is, the forces are now applied at a body. This is quite different when you are analyzing a particle because the position of each force are significant enough to change the effects of other forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rYv7I.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rYv7I.png&quot; alt=&quot;Reference drawing.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The inertia is neglected because the velocity is to be constant (steady speed)&lt;/li&gt;&#xA;&lt;li&gt;The friction force, $F_f$, depends linearly on the normal force, $R_a$. the relationship is given by: $F_f=\mu\ N$&lt;/li&gt;&#xA;&lt;li&gt;Since the point of contact is at point $A$, the friction force acts at point $A$ as well.&lt;/li&gt;&#xA;&lt;li&gt;The friction in the rotating wheel, $B$ is neglected.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So to sum up, you &lt;strong&gt;need&lt;/strong&gt; to consider the reduction of reaction force at A due to turning moment of force $P$. Such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up forces at $B$, gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P\ (1)\ -\ 64\ (0.35)\ +\ R_a\ (0.7)\ =\ 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and summing up forces along x-axis gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P\ =\ F_f\ =\ \mu_k\ R_a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P\ = 0.3\ (R_a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the two equations simultaneously gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P\ = \ 6.72\ kg$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_a\ = \ 22.4\ kg$$&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-23T07:25:07.947" LastActivityDate="2017-09-23T07:25:07.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17228" PostTypeId="1" AcceptedAnswerId="17230" CreationDate="2017-09-23T10:32:42.323" Score="3" ViewCount="895" Body="&lt;p&gt;Two automobiles both of which have the mass center located midway between the front and rear axles,are identical except that one is front-wheel-drive and the other is rear-wheel-drive.The cars are driven at constant speed at ramps of various inclinations.The question is to judge which car could climb the ramp of higher inclination angle from a theoretical point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YCPa6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea was that the front wheel drive would be more effective if more weight acts on the front wheel.But in this case the center of gravity is midway between the rear and front wheel axles .I couldn't see how I can approach this problem.Any ideas?Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastActivityDate="2017-09-23T12:30:16.173" Title="Car which can climb the ramp of higher inclination" Tags="|statics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17229" PostTypeId="1" CreationDate="2017-09-23T11:56:55.550" Score="1" ViewCount="17" Body="&lt;p&gt;What is homogeneous temperature in thermodynamics? Can homogeneous temperature assumption derive no heat transfer? If yes, why? &lt;/p&gt;&#xA;" OwnerUserId="11335" LastActivityDate="2017-09-23T11:56:55.550" Title="Can homogeneous temperature assumption derive no heat transfer? If yes, why?" Tags="|thermodynamics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17230" PostTypeId="2" ParentId="17228" CreationDate="2017-09-23T12:12:57.477" Score="5" Body="&lt;p&gt;The rear-driven car is able to climb higher inclination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rational:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Where is the engine and the weight has for sure an impact, but it is not related with rear or front driven car: You have rear-driven cars with engine in the front and reverse. Additionally, you specify that the center of gravity is in the same place (center), so the place of the engine has no impact.&lt;/li&gt;&#xA;&lt;li&gt;However, if you consider the forces acting on the car, the gravity will lower the weight component on the front and increase the weight component to the rear. This increase the capacity of a rear-driven car to climb. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note: Consider however, that rear-driven cars will have lower direction control while climbing, and for this reason, they might have big issues in climbing icy roads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2: Driving a front-wheel-driven backward allows to climb higher inclinations than forward. But probably not something to try on the highway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 3: The ideal situation for climbing strong inclinations is the rear-wheel-driven car with rear direction (you do not have the problem of control). You may find several construction machines designed in this way, but it is not practical (difficult to drive) at high speed, and thus not used for cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MJzfn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MJzfn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11129" LastEditorUserId="11129" LastEditDate="2017-09-23T12:30:16.173" LastActivityDate="2017-09-23T12:30:16.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17231" PostTypeId="1" CreationDate="2017-09-23T12:50:41.050" Score="1" ViewCount="453" Body="&lt;p&gt;Aerosol and gaseous emissions are generated in toilets. Drawing exhaust air from the ceiling ensures effective dissemination of airborne contaminants. Why is the exhaust air not drawn from the toilet instead?&lt;/p&gt;&#xA;" OwnerUserId="13015" LastActivityDate="2017-10-13T07:00:20.723" Title="Why are toilets not equipped with exhaust air vents?" Tags="|civil-engineering|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17232" PostTypeId="1" AcceptedAnswerId="17250" CreationDate="2017-09-23T13:05:10.520" Score="3" ViewCount="748" Body="&lt;p&gt;This is a problem from the book Engineering Mechanics Statics vol-1 by Meriam Kraige&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rectangular steel yoke is used to prevent slippage between the two boards under tensile loads $P$ .If the coefficients of static friction between the yoke and the board surfaces and between the boards are all $0.30$ the question is to determine the maximum value of $h$ for which there is no slipping and to find out the corresponding normal force $N$ between the two boards for $P=800N$ if impending motion occurs at all surfaces.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/X076V.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I tried&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first part if $\theta$ denotes the angle between the top part of yoke and horizontal then $\cos \theta = \frac{200}{210}=\frac{150}{h}$ which gives $h=157.5 mm$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part I tried to draw the FBD of the top block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9NVfo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From FBD $$P=f_1+f_2+f_3+f_4$$ and $$mg=N_3-N_1$$ Also by taking moments about COM I got $P=f_3-f_1$.However I am not able to proceed after this.Any help shall be highly appreciated.Thanks&lt;/p&gt;&#xA;" OwnerUserId="12580" LastActivityDate="2017-09-25T00:40:36.607" Title="Normal force between two boards" Tags="|statics|friction|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17233" PostTypeId="1" CreationDate="2017-09-23T13:21:09.373" Score="2" ViewCount="264" Body="&lt;p&gt;I am currently looking for materials, which return into their relaxed form/ position after intense bending/ stretching. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please let me know what is the name of this mechanical property, so that I could search further on this topic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know any examples for such materials? Perhaps some polymers?&lt;/p&gt;&#xA;" OwnerUserId="13016" LastActivityDate="2017-09-26T16:59:03.347" Title="Stress-resistant, elastic materials" Tags="|stresses|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17234" PostTypeId="2" ParentId="17231" CreationDate="2017-09-23T13:39:29.087" Score="2" Body="&lt;p&gt;Because the design of the bowl (assuming a &quot;standard&quot; toilet with flush) is such that it forms an airlock (the U bend) to stop the smell from the drain entering the bathroom.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-09-23T13:39:29.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17236" PostTypeId="2" ParentId="17233" CreationDate="2017-09-23T16:57:15.243" Score="4" Body="&lt;p&gt;It depends exactly what application you are looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of stress a material will tolerate is defined by its tensile strength and the amount it will deform under a given stress is defined by its Young's Modulus (low Young's Modulus is flexible high is stiff) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want something which will stretch by a large percentage of its original dimensions then rubbers (generally classified as elastomers) tend to be good. Typical applications include things like engine mountings and flexible covers for articulated joints. Rubbers also tend to be self-damping and so are good for vibration isolation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of maximum stress resistance then spring steels are widely used, these tend to be medium carbon, low alloy steels which are hardened and tempered. These materials have very good tensile strength and toughness. Although steel as a material has high stiffness, you can control the degree of flexibility of a spring by its shape (eg coil springs) or by mechanical linkages as in the case of torsion springs. Overall steel tends to be the preferred for high performance springs although you do see polymer and composite leaf springs eg in commercial vehicles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option for some applications is polymer textiles and rope, for example the dynamic ropes used in climbing here the way that yarns are formed and woven also contribute to their flexibility. These tend to be used for impact absorption and are effective at moderating high impact loads without failing although they may be damaged by loads close to their ultimate capacity. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-09-23T16:57:15.243" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17237" PostTypeId="1" CreationDate="2017-09-23T19:43:44.310" Score="1" ViewCount="37" Body="&lt;p&gt;Is there a tutorial on how to do non-linear FEA using Siemens Advanced Simulation NX (not remap) with Nitinol, preferably using sol 601. An example would be very helpful, I searched online but most just have Ansys or abacus examples. &lt;/p&gt;&#xA;" OwnerUserId="11930" LastActivityDate="2017-09-23T19:43:44.310" Title="Nitinol FEA using Advanced Simulation NX" Tags="|finite-element-method|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17239" PostTypeId="2" ParentId="17231" CreationDate="2017-09-23T22:32:38.677" Score="1" Body="&lt;p&gt;They do sometimes. Look up long drop toilet. They're used sometimes at Western campsites. The vent is usually painted black. It heats up with solar radiation and causes a strong convection current that really sucks on the cheeks. They're surprising ly effective.&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2017-09-23T22:32:38.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17241" PostTypeId="1" AcceptedAnswerId="17293" CreationDate="2017-09-24T05:22:31.613" Score="4" ViewCount="292" Body="&lt;p&gt;This is a problem from the book Engineering Mechanics Statics vol-1 by Meriam Kraige.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elements of a unidirectional mechanical clutch are shown in the figure.A torque $M$ applied to the outer ring is transmitted to the output shaft through frictional interaction between the outer ring and the balls and between the balls and the inner driven member.If the direction of rotation of the outer ring is reversed,the wedging action of the balls is absent,and no torque can be transmitted to the output shaft.For given values of $r,r_o$ and coefficient of friction $\mu$ which is applicable for both pairs of interacting surfaces,the question is to find the minimum dimension $b$ of inner member which will permit the transmission of torque without slipping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EfxH8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't get how to approach this problem.i tried drawing FBD of the balls but could get nothing out of it.Any help shall be highly appreciated.Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastActivityDate="2017-09-27T01:12:13.147" Title="Dimension of the unidirectional mechanical clutch" Tags="|statics|friction|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17242" PostTypeId="1" AcceptedAnswerId="17245" CreationDate="2017-09-24T07:29:27.510" Score="0" ViewCount="3028" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7xNfU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7xNfU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain why the support at C does not have a vertical reaction force while the support at E has?&#xA;Doesn't the support at C have the same kind of pin connection as the support at E?&lt;/p&gt;&#xA;" OwnerUserId="12979" LastActivityDate="2017-09-24T12:44:36.813" Title="Why are the support reactions of the two pin-connected members different?" Tags="|statics|beam|connections|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17245" PostTypeId="2" ParentId="17242" CreationDate="2017-09-24T12:44:36.813" Score="3" Body="&lt;p&gt;The rod BC is a &lt;strong&gt;two force member&lt;/strong&gt;.  Therefore, the forces on either end must be along the axis of the member.  i.e. since the rod is horizontal, the forces on either end can only be horizontal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beam AGDE is not a two force member.  Therefore, it can have forces on the ends in both directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the reason that the reactions are different does not have to do with the pinned boundary, but one of the members being two force and the other not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this doesn't make sense, I would suggest you look up &quot;two force member&quot; in your textbook.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-09-24T12:44:36.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17247" PostTypeId="1" CreationDate="2017-09-24T18:59:20.663" Score="3" ViewCount="227" Body="&lt;p&gt;I'm working on a school project that requires us to build a compound machine with a mechanical advantage of 9. It should be able to lift a 10oz soup can three or more inches by using the weight of fifteen pennies. My machine consists of an inclined plane with a mechanical advantage of 3.16 (base is 12 in, height is 4 in), a travelling pulley system with a mechanical advantage of 2, and a wheel and axle system with a mechanical advantage ~3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my machine:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CLWKh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CLWKh.jpg&quot; alt=&quot;Compound Machine Pic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total mechanical advantage should be 19, but when I put the weights on the system, it turns out that I need 45 pennies to lift the can, which means my mechanical advantage is 3. I checked for sources of friction, but I couldn't find any of note. I could add a second pulley to increase the MA, but I'd prefer to fix the problem instead of implementing a patch-up solution. Any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="13032" LastActivityDate="2017-09-25T01:05:44.757" Title="Surprisingly Low Efficiency of Compound Machine [Project Help]" Tags="|mechanical-engineering|wheels|pulleys|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17248" PostTypeId="2" ParentId="17247" CreationDate="2017-09-24T19:51:24.923" Score="0" Body="&lt;p&gt;It's not massively surprising that you have a lot of losses in something like this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compound pulleys always generate a lot of friction as you tend to have a lot of points of contact and the bearings are often a bit crude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing I would suggest is to make sure that all of the bearings area well lubricated, this should help a &lt;em&gt;lot&lt;/em&gt;. Maybe try lithium grease or a PTFE lubricant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also suspect that a major source of friction is where the cord unwinds on the axle, substituting something like filament nylon fishing line should help a lot with that. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-09-24T19:51:24.923" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17250" PostTypeId="2" ParentId="17232" CreationDate="2017-09-25T00:40:36.607" Score="2" Body="&lt;p&gt;I will try to answer it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, draw the forces acting upon it. (I considered only the top plank/board as it is symmetric with the bottom board). Refer to the figure below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tlbKs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tlbKs.png&quot; alt=&quot;Reference drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Separating the upper plank and analyzing the forces upon it, you will get the figure below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q13Hn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q13Hn.png&quot; alt=&quot;Figure 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $F_{\text f1}=P_y (0.3)$ and $F_{\text f2}=N(0.3)$, therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up forces in the x-axis gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0.3(P_y)+0.3(N)=800$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and Summing up forces in the y-axis gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$N-P_y=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving both, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_y=N=1,333.33\ N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving also for $F_{\text f1}$ gives $F_{\text f1}=400\ N.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for the angle of inclination of the yoke gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = arctan (400/1333.33)=16.70\ degrees$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore, $h = 150/\cos(16.70)=156.60\ mm$&lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2017-09-25T00:40:36.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17251" PostTypeId="2" ParentId="17247" CreationDate="2017-09-25T01:05:44.757" Score="2" Body="&lt;p&gt;I have a few suggestions when it comes to managing forces:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Try putting the angle of the ramp and the angle of the cord in the same value. This will reduce the force required to pull it up as the system will utilize the wheels. as i can see, the front wheels are up, therefore, the cables tend to lose force.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CK9OY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CK9OY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Try putting the center of gravity of the object &quot;in-line&quot; with the cord so that no moment occurs. You can try putting the can down to lower its Center of Gravity.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mpbOm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mpbOm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2017-09-25T01:05:44.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17252" PostTypeId="1" AcceptedAnswerId="21965" CreationDate="2017-09-25T02:09:50.273" Score="2" ViewCount="1489" Body="&lt;p&gt;I have this question assigned as part of my homework, and I am having trouble figuring out the solution. Below is the question and my attempts at solving it. I need to have the homework ready tomorrow morning, so I would really appreciate any help or direction in finding the solution.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that a composite solid consists of alternating materials $A$ and $B$, with layer thickness $L_A$ and $L_B$, respectively. Both materials are isotropic, with thermal conductivities $k_A$ and $k_B$. It is desired to predict the average, steady-state heat flux in a sample that contains many layers. The flux in the composite is written as &#xA;$$\boldsymbol{ Q=-k\cdot G}$$&#xA;where $\boldsymbol{Q}$ and $\boldsymbol{G}$ are the average heat flux and temperature gradient, respectively, and $\boldsymbol{k}$ is the effective thermal conductivity. With coordinate axis of material interfaces aligned with normal along the $y$-axis, the effective conductivity will be of the form &#xA;$$\boldsymbol{k}=\left[\begin{array}{ccc} k_{xx} &amp;amp; 0 &amp;amp;0\\0&amp;amp;k_{yy}&amp;amp;0\\0&amp;amp;0&amp;amp;k_{zz}\end{array}\right].$$&#xA;(a) Evaluate $k_{xx}$, $k_{yy}$, and $k_{zz}$ in terms of $k_A$, $k_B$, and $\phi$, where $\phi=L_A/(L_A+L_B)$ is the volume fraction of material $A$ ($L_A$ is the length of material $A$, and $L_B$ is the length of material $B$). (&lt;em&gt;Hint&lt;/em&gt;: The steady heat flux $q$ through a slab of thickness $L$ and thermal conductivity $k$ is $q=k\Delta T/L$, where $\Delta T$ is the temperature difference between surfaces.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Show that, in general, the vectors $\boldsymbol{Q}$ and $\boldsymbol{-G}$ will not be parallel to one another. That is, the directions of the heat flux and temperature gradient will tend to differ.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Work:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can consider the orientation of axes such that the positive $x$-axis comes out of the page towards us, the positive $y$-axis goes to the right of the page, and the positive $z$-axis goes to the top of the page. We can also consider a segment of the composite solid with material $A$ on the left and material $B$ on the right along the $y$-axis, i.e., material $A$ is from $0$ to $L_A$ on the $y$-axis, and material $B$ is situated from $L_A$ to $L_B$ along the $y$-axis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vector equation $\boldsymbol{Q=-k\cdot G}$ can be broken down into 3 equations (using the definition $\boldsymbol{G}=\nabla T$):&#xA;$$q_x=-k_{xx}\frac{\partial T}{\partial x}\\q_y=-k_{yy}\frac{\partial T}{\partial y}\\q_z=-k_{zz}\frac{\partial T}{\partial z}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding $k_{xx}$ and $k_{zz}$, since the difference in material is only along the $y$-axis, I would think that $k_{xx}$ and $k_{zz}$ would not depend on $\phi$, $L_A$, or $L_B$, rather only on the thermal conductivity of the particular material $A$ or $B$, depending on the location along the $y$-axis. However, if the materials are assumed to be infinite along the $x$-axis and $z$-axis (this may be a logical inference, since the height and width of the composite solid are not given), then the flux along the $x$ and $z$ directions should be $0$. &#xA;If this assumption is correct, then $\frac{\partial T}{\partial x}=\frac{\partial T}{\partial z}=0$, so this tells me nothing about $k_{xx}$ or $k_{zz}$. Is this assumption correct? Did I miss something regarding the evaluation of $k_{xx}$ and $k_{zz}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out $k_{yy}$, we can construct equations for q for each material along the $y$-axis:&#xA;$$q_A=k_A\Delta T_A /L_A\\q_B=k_B\Delta T_B/L_B$$&#xA;Then, using the given hint, we can consider temperature to the left of material $A$ to be $T_0$, the temperature at the interface between materials $A$ and $B$ to be $T_1$, and the temperature to the right of material $B$ to be $T_2$. Accordingly, $\Delta T_A=T_0-T_1$ and $\Delta T_B=T_1-T_2$, so the equations become the following:&#xA;$$q_A=k_A\left(T_0-T_1\right)/L_A\\q_B=k_B\left(T_1-T_2\right)/L_B$$&#xA;The Equations can be rearranged:&#xA;$$T_0-T_1=q_A L_A/k_A\\T_1-T_2=q_B L_B/k_B$$&#xA;Then the two equations can be added together to form $\Delta T=T_0-T_2$:&#xA;$$\Delta T=\frac{q_A L_A k_B +q_B L_B k_A}{k_A+k_B}$$ &#xA;From here, however, I am not sure how I would proceed; $\phi$ does not explicitly appear, and $q_A$ and $q_B$ do appear but the question implies they should not appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate approach would be to add the two equations together:&#xA;$$q_A+q_B=\frac{k_A L_B \left(T_0-T_1\right)+k_B L_A \left(T_1-T_2\right)}{L_A+L_B}$$&#xA;From the definition of $\phi$, we also know that $1-\phi=L_B/(L_A+L_B)$. Using this fact, the above equation can be rearranged as follows:&#xA;$$q_A+q_B=k_A(T_0-T_1)+\phi\left[k_A(T_1-T_0)+k_B(T_1-T_2)\right]$$&#xA;However, here again I do not know how to proceed; I still have $T_0$, $T_1$, and $T_2$ separate, and this equation is only useful if we assume $q_y=q_A+q_B$. Is this last assumption correct? Is there a simplification of the equation that I missed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding (b), I presume that once I find $\boldsymbol{Q}$ and $\boldsymbol{-G}$, I can show $\boldsymbol{Q\cdot -G}\neq1$, but I do not yet have $\boldsymbol{Q}$ or $\boldsymbol{-G}$ (or even $\nabla T$, from which I can find $\boldsymbol{-G}$) to figure this out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help or direction on any part of this problem would be much appreciated! As I mentioned, I need to have the homework ready tomorrow morning, so I am kind of in a rush. Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="13036" LastActivityDate="2018-05-30T17:45:49.110" Title="Thermal Conductivity of a Layered Composite" Tags="|fluid-mechanics|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17254" PostTypeId="2" ParentId="17231" CreationDate="2017-09-25T04:41:23.277" Score="7" Body="&lt;ol&gt;&#xA;&lt;li&gt;it would complicate toilet designs and make toilet bowls more expensive to manufacture&lt;/li&gt;&#xA;&lt;li&gt;it would require an electrician to install, as well as a plumber, greatly increasing the cost of installation&lt;/li&gt;&#xA;&lt;li&gt;the exhaust vent would suck in splashed urine and faeces-laden bowl water, turning the outlet pipe and the exhaust fan at the end into a stinking health hazard&lt;/li&gt;&#xA;&lt;li&gt;sitting on a vented bowl in winter would result in cold air rushing past your genitals&lt;/li&gt;&#xA;&lt;li&gt;you are conflating emissions and contaminants — bad (but otherwise harmless) smells rise up from toilet bowls, but the nasty bugs are only transmitted by splashing or human activity, so a ceiling exhaust does not help spread contaminants in any meaningful way because it does not generate strong air currents near the bowl&lt;/li&gt;&#xA;&lt;li&gt;exhaust vents in toilets have nothing to do with sanitation — they exist merely to maintain air quality (i.e. prevent the room from smelling bad)&lt;/li&gt;&#xA;&lt;li&gt;given that most replacement air will enter the room underneath the door, a vented toilet bowl will simply draw air along the floor — this will make the extraction of warm, smelly air that has risen towards the ceiling difficult, allowing smells to linger and requiring the exhaust fan to run significantly longer to accomplish its task of clearing the air (with a corresponding increase in electricity consumption and failure rate)&lt;/li&gt;&#xA;&lt;li&gt;if the toilet becomes blocked for whatever reason, the water level in the bowl could rise to the level of the vent and then get sucked directly into the exhaust pipe and down that pipe towards the electric fan at the other end — a scenario that won't end well regardless of what happens next&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Having said all of the above, a number of &lt;em&gt;waterless/composting toilets&lt;/em&gt; are vented and do work well — so the idea does have merit in certain circumstances.&lt;/p&gt;&#xA;" OwnerUserId="12883" LastEditorUserId="12883" LastEditDate="2017-10-13T07:00:20.723" LastActivityDate="2017-10-13T07:00:20.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17255" PostTypeId="2" ParentId="17054" CreationDate="2017-09-25T05:30:00.753" Score="2" Body="&lt;p&gt;Yes, the proportions of acrylonitrile, butadiene, and styrene can vary within some ranges. However unless you will be able to polymerize it yourself you will have to use a commercially available polymer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would look as product information sheets for bulk polymer pellets sold for injection molding. From those you will be able to see the real available range of properties like youngs modulus, tensile strength, shore hardness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of mechanical anisotropy throughout the part, some could exist, but it will be difficult to determine the magnitude of such effects. These effects primarily arise from the differential cooling rates of different areas of the part due to part geometry (which can induce internal stresses). To mitigate these effects its helpful to pay attention in design to uniformity in part thickness so that cooling rates are kept similar. &lt;/p&gt;&#xA;" OwnerUserId="13039" LastActivityDate="2017-09-25T05:30:00.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17257" PostTypeId="1" AcceptedAnswerId="17265" CreationDate="2017-09-25T07:46:57.117" Score="2" ViewCount="551" Body="&lt;p&gt;I am currently rating two motorcycles to decide which one to buy. The criteria which is most important to me is the power-to-weight ratio. Now, I want to calculate the power of the motorcycle by looking at its displacement. The engine displacement is the amount of air-fuel mix that an engine can burn at a time. Looking at google, I cannot find a concrete formula to convert the engine displacement (cubic centimeter) to power (horsepower). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I am looking at two models:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Suzuki Raider J 115 Fi (fuel injection)&lt;/li&gt;&#xA;&lt;li&gt;Honda XRS 125 (Carburetor)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know that the Honda model will provide more power but I want to look at the other model's fuel efficiency. I want to convert to see whether the power difference is not that significant. For the sake of comparison/formula, assume that you feed both with the same amount of fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please advise?&lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2017-09-25T12:53:14.520" Title="Engineering formula to relate engine displacement and power" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17258" PostTypeId="1" CreationDate="2017-09-25T08:30:16.453" Score="1" ViewCount="3346" Body="&lt;p&gt;QUESTION: &lt;a href=&quot;https://i.stack.imgur.com/VXrwr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VXrwr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My working: I had divided the body into 5 common shapes which are shown here: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/zU4Yc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zU4Yc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now that I'm solving for x bar, y bar, and z bar I find it difficult in finding the x, y and z directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see my table of working here: &lt;a href=&quot;https://i.stack.imgur.com/JgNgs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JgNgs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12830" LastActivityDate="2017-09-25T11:52:15.980" Title="For the machine element shown, locate the X, Y and Z coordinate of the center of gravity" Tags="|mechanical-engineering|composite|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17259" PostTypeId="2" ParentId="17257" CreationDate="2017-09-25T08:31:25.503" Score="3" Body="&lt;p&gt;Well, first I would research the manufacturer's specs for each engine and then you can divide the power by the mass which will give you a decent comparison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, another approach, based on more technical theory (used when they designed the engine)is shown here : &lt;a href=&quot;https://www.quora.com/In-a-car-engine-is-compression-ratio-related-to-horsepower&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.quora.com/In-a-car-engine-is-compression-ratio-related-to-horsepower&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IGlX2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IGlX2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An even more detailed explanation is available here, this is the approach taken in courses such as Power Plant Analysis : &lt;a href=&quot;http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node25.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node25.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-09-25T08:31:25.503" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17261" PostTypeId="2" ParentId="17258" CreationDate="2017-09-25T11:43:43.640" Score="4" Body="&lt;p&gt;Try this one. (Note: I don't know how to do tables in this site but please tabulate the data on your own.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Areas are:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$0.75(4)(7)$&lt;/li&gt;&#xA;&lt;li&gt;$\frac{\pi(2^2)}{2}(0.75)$&lt;/li&gt;&#xA;&lt;li&gt;$-\pi(1.25^2)(0.75)$&lt;/li&gt;&#xA;&lt;li&gt;$1(4)(2)$&lt;/li&gt;&#xA;&lt;li&gt;$-\frac{\pi(1.25^2)}{2}(1)$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: Negative areas as &lt;strong&gt;holes&lt;/strong&gt; that need to be subtracted from the system.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;X-Centroids&lt;/strong&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$2$&lt;/li&gt;&#xA;&lt;li&gt;$2$&lt;/li&gt;&#xA;&lt;li&gt;$2$&lt;/li&gt;&#xA;&lt;li&gt;$0.5$&lt;/li&gt;&#xA;&lt;li&gt;$0.5$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Y-Centroids&lt;/strong&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$-0.75/2$&lt;/li&gt;&#xA;&lt;li&gt;$-0.75/2$&lt;/li&gt;&#xA;&lt;li&gt;$-0.75/2$&lt;/li&gt;&#xA;&lt;li&gt;$2/2=1$&lt;/li&gt;&#xA;&lt;li&gt;$2-\frac{4(1.25)}{3(\pi)}$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Z-centroids&lt;/strong&gt; are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$7/2$&lt;/li&gt;&#xA;&lt;li&gt;$7+\frac{4(1.25)}{3(\pi)}$&lt;/li&gt;&#xA;&lt;li&gt;$7$&lt;/li&gt;&#xA;&lt;li&gt;$2$&lt;/li&gt;&#xA;&lt;li&gt;$2$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can then apply &lt;strong&gt;Varignon's Theorem&lt;/strong&gt; to get the centroid.&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12936" LastEditDate="2017-09-25T11:52:15.980" LastActivityDate="2017-09-25T11:52:15.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17262" PostTypeId="1" CreationDate="2017-09-25T11:52:14.177" Score="0" ViewCount="3841" Body="&lt;p&gt;QUESTION: &lt;a href=&quot;https://i.stack.imgur.com/opMqs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/opMqs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need help, can someone help me show how we get the solution as stated in the question; as i am trying my very best to solve.&lt;/p&gt;&#xA;" OwnerUserId="12830" LastActivityDate="2017-09-25T12:12:02.983" Title="Locate the centroid of the plane area shown" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17263" PostTypeId="2" ParentId="17262" CreationDate="2017-09-25T12:03:32.943" Score="3" Body="&lt;p&gt;Remember that you can break a object into pieces.  The centroid of the whole object is the weighted centroids of the pieces.  Since this is a planar object, the centroids of the pieces are weighted according to the area of those pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your object is regular enough so that it can be broken into pieces for which centroids can be easily found, like rectangles and triangles.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2017-09-25T12:12:02.983" LastActivityDate="2017-09-25T12:12:02.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17265" PostTypeId="2" ParentId="17257" CreationDate="2017-09-25T12:53:14.520" Score="4" Body="&lt;p&gt;There's no simple relationship. When manufacturers quote a figure for power it's usually a peak value, which doesn't mean much without knowing the engine speed (rpm) at which this is developed. Power is time based, so an engine capable of spinning twice as fast should be able to burn twice as much air/fuel in the same time for a given displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can find power and torque curves, these will tell you more than a single value. You'll get different power and torque curves depending on how the displacement is achieved (broader piston or longer stroke).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power to weight ratio is a factor, but if you're looking at low speed acceleration you'll probably find that torque is more relevant.&lt;/p&gt;&#xA;" OwnerUserId="13043" LastActivityDate="2017-09-25T12:53:14.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17269" PostTypeId="1" AcceptedAnswerId="17609" CreationDate="2017-09-25T20:26:50.753" Score="1" ViewCount="128" Body="&lt;p&gt;The problem I have at hand is the following: I have a container C with a given volume V. I need to fill it with water, fast, up to a given pressure P. I have a number of high pressure hydropneumatic pumps to choose from, but they are expensive and take a while to get here. I don't even have such pump readily at hand to test the equations. The flow rate for these pumps is easily calculated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Q = Q_{max} - K_{b} P&#xA;$$&#xA;$K_{b}$ is a pump parameter per datasheet. Obviously pressure $P$ won't go higher than $Q_{max}/K_{b}$, the pump stops working. Now I start simplifying things: I consider no thermal factors, no sudden pipe diameter change, no turbulence, no back flow, constant ambient pressure, always the same fluid; basically no hydrodynamics. And the longest shot, I consider the pressure per volume rate of the container constant:&#xA;$$&#xA;\frac{P}{V} = K_{t}&#xA;$$&#xA;I treat $K_{t}$ as some sort of elasticity constant which in reality comprises the actual container elasticity,water compressibility, container's volume, etc. It's the only empiric data I have: how many liters of water I stuffed in the container to get the target pressure. Taking differentials, and considering t=0 the moment the container is full with P = 0, I get:&#xA;$$&#xA;P(t) = \frac{Q_{max}}{K_{b}}(1-e^{-K_{t}K_{b}t})&#xA;$$&#xA;Which makes sense for a single pump. But I can't find a way to adjust it to multiple pumps in parallel with different parameters. I need an equation to later iterate and get the best combination of pumps possible. Trying to get an &quot;equivalent&quot; pump, the flow rate is the sum of flow rates as in: $Q(t) = Q_{1}(t) + Q_{2}(t) ...$, easily resolving to an equivalent pump having parameters $[\sum{Q_{{max}_i}},\sum{K_{{b}_{i}}}] $, but that does not make sense with the equation above. Keep in mind this is for pressures up to 2000 Bar.&lt;/p&gt;&#xA;" OwnerUserId="13050" LastEditorUserId="1832" LastEditDate="2017-09-25T22:46:16.953" LastActivityDate="2017-10-19T19:34:10.890" Title="High pressure water pumps filling a container. How long does it take to obtain P pressure?" Tags="|fluid-mechanics|pumps|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17271" PostTypeId="1" AcceptedAnswerId="17280" CreationDate="2017-09-26T00:19:42.780" Score="1" ViewCount="150" Body="&lt;p&gt;This is a simple question to follow up the question from the thread below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/17257/engineering-formula-to-relate-engine-displacement-and-power&quot;&gt;Engineering formula to relate engine displacement and power&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motorcycle manufacturers tend to show the displacement value in mass production motorcycles incorporated in their names such as Honda Wave 125i, Suzuki Raider 150, Kawasaki Z1000, etc. Most of these are below 200cc displacement. (Exceptions are Yamaha YZF-R1M, Kawasaki Ninja H2, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: If power is not determined by the displacement alone based on answers and comments on the thread above, does this imply that the displacement is the most important factor in determining the engine's overall performance?&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorDisplayName="user16" LastEditDate="2017-09-26T01:50:30.537" LastActivityDate="2019-06-03T06:21:45.783" Title="Is displacement the most important factor in determining a motorcycle engine's overall performance?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17273" PostTypeId="1" AcceptedAnswerId="17275" CreationDate="2017-09-26T03:51:22.323" Score="0" ViewCount="160" Body="&lt;p&gt;I often seen printers &lt;a href=&quot;http://www.office.xerox.com/multifunction-printer/color-multifunction/xerox-color-c60-c70/enus.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this&lt;/a&gt; in offices and schools. Considering how we can have printers small enough to fit on a desk, why are these printers taking up half a room? What are they capable of doing, and why do they need to be that large?&lt;/p&gt;&#xA;" OwnerUserId="13055" LastActivityDate="2017-09-26T11:24:33.170" Title="Why are corporate printers so large?" Tags="|mechanical-engineering|computer-hardware|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17274" PostTypeId="2" ParentId="17273" CreationDate="2017-09-26T05:09:50.403" Score="2" Body="&lt;p&gt;These environment requires a printer which are supposed to include photocopying, fax, scan and a risograph among other functions. &lt;a href=&quot;https://en.wikipedia.org/wiki/Risograph&quot; rel=&quot;nofollow noreferrer&quot;&gt;Risograph&lt;/a&gt; are highly needed in offices and schools for high-speed duplication of printed materials such as questionnaires, examination papers, forms, etc. The risograph can replicate a copy into 500 pages in about 2 minutes so this saves a lot of time and effort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See its specifications below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/okSQm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/okSQm.png&quot; alt=&quot;Multi-functional Printer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-26T11:07:23.953" LastActivityDate="2017-09-26T11:07:23.953" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17275" PostTypeId="2" ParentId="17273" CreationDate="2017-09-26T11:24:33.170" Score="2" Body="&lt;p&gt;One requirement is serviceability and time between refills (of either ink or paper. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When printing 15 copies of a 100 page report you don't want to have to stand next to it and refill the paper and ink 10 times before it's done. Instead you want to push the button and then pick up the stack for binding later after you had your coffee.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having components be serviceable means that you need to be able to access them or remove them. This means adding space for an arm to reach in (and possibly a head) or adding bulk for the interface of the module and the rest of the machine (or both).&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-09-26T11:24:33.170" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17277" PostTypeId="1" CreationDate="2017-09-26T11:44:55.477" Score="1" ViewCount="125" Body="&lt;p&gt;I have been building my first simple circuits with just resistors, LED's wire, and alkaline batteries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I would like to add capacitors to my components box. Read a couple of articles online on the matter and it seems pretty deep, so to get the next step in the learning process going, I wanted to use the same technique I did with resistors and LED's, which is to plug a capacitor or two up and make some examinations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but first, I wanted to straighten a vague point in my notes on the concept of capacitors: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between the plates of a capacitor is the dielectric, is the built up charge between the two plates eventually suppose to traverse through the dielectric?&lt;/p&gt;&#xA;" OwnerUserId="12689" LastActivityDate="2017-09-29T17:54:11.230" Title="Is current suppose to pass through a capacitors dielectric? What happens if it does?" Tags="|electrical-engineering|power-electronics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17278" PostTypeId="1" CreationDate="2017-09-26T11:55:34.497" Score="0" ViewCount="11300" Body="&lt;p&gt;how to solve internal hinge problem with loading given on hinge&#xA;&lt;a href=&quot;https://i.stack.imgur.com/62uRD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/62uRD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13035" LastActivityDate="2017-09-27T07:23:20.827" Title="how to solve internal hinge problem with loading given on hinge" Tags="|structural-engineering|civil-engineering|structures|statics|beam|" AnswerCount="2" CommentCount="1" ClosedDate="2017-09-27T18:59:41.583" ContentLicense="CC BY-SA 3.0" />
  <row Id="17280" PostTypeId="2" ParentId="17271" CreationDate="2017-09-26T13:05:28.180" Score="2" Body="&lt;p&gt;Me again. It really depends what you mean by performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you increase displacement, you're going to be looking at an increase in mass (bigger cylinders, larger crank and case). Silly example - you could throw a locomotive engine at a bike and it wouldn't get you anywhere useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power will be proportional to displacement per second, so the engine speed is going to be a factor as much as bore squared x stroke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends what you want to do. Green lanes over the moors? Good bit of low down torque. Drag strip? Sheer power. A track or road with lots of corners? Nice progressive delivery.&lt;/p&gt;&#xA;" OwnerUserId="13043" LastActivityDate="2017-09-26T13:05:28.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17281" PostTypeId="2" ParentId="17278" CreationDate="2017-09-26T13:25:07.930" Score="1" Body="&lt;p&gt;A hinge just means that the moment in the beam at that point is 0. Each beam cantilevers from the pin support to the hinge point, but we know that the deflection of each cantilever must be the same at the hinge point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the particular example you have shown, you can therefore use symmetry to split the structure into two at the hinge point, with 10kN in each direction.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2017-09-26T13:25:07.930" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17282" PostTypeId="2" ParentId="17277" CreationDate="2017-09-26T13:41:14.210" Score="2" Body="&lt;p&gt;Here is a good link on &lt;a href=&quot;https://learn.sparkfun.com/tutorials/capacitors&quot; rel=&quot;nofollow noreferrer&quot;&gt;Capacitors&lt;/a&gt;. Capacitor is a energy storage device. Capacitor can act like a battery. Stored energy is discharged to a device like an LED. Take a look at this capacitor battery LED resistor animation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bC2Ez.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bC2Ez.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8055" LastActivityDate="2017-09-26T13:41:14.210" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17284" PostTypeId="2" ParentId="17233" CreationDate="2017-09-26T16:59:03.347" Score="0" Body="&lt;p&gt;The word you are looking for is hyperelasticity. Material models from this field of study are useful for rubberlike materials that are expected to not exhibit plastic deformation.&lt;/p&gt;&#xA;" OwnerUserId="12401" LastActivityDate="2017-09-26T16:59:03.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17287" PostTypeId="2" ParentId="13994" CreationDate="2017-09-26T18:09:02.003" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A chain saw repair person said that chain saws have a carb which needs to be used or it will become unusable.  He said a chain saw needs to be run at least twice per year.  It's a &quot;diaphragm&quot; carb -- which can operate in any position because it has no float. He said a homeowner should have an electric chain saw rather than a 2-cycle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I can't personally attest to the effectiveness of this but an engineer I used to work would prepare his four cycle engines for winter storage by running them with Coleman fuel, AKA white gas, until they went dry.  His explanation was that the naphtha based fuel left very little reside.  As would be expected the engines ran particularly badly with the goofy fuel but once they ran out, what was left would evaporate cleanly.  I have never tried this and would not recommend it unless you are prepared to take the engine apart to recover from the experiment.  Observe, I said four cycle.  If you try this with two cycle equipment, say goodbye first.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2017-09-26T18:09:02.003" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17288" PostTypeId="2" ParentId="16706" CreationDate="2017-09-26T19:19:03.213" Score="0" Body="&lt;p&gt;You'll want to find out what formats the CAM software which they usually use for this machine accepts.&lt;/p&gt;&#xA;&lt;p&gt;For hobbyist machines, SVG is a frequent option. Initially popularized by PartKam:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://partkam.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://partkam.com/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;it had direct support for CAM in the gcodetools plug-in for Inkscape (until that was deprecated), and many computer aided manufacturing (CAM) programs will import SVGs.&lt;/p&gt;&#xA;" OwnerUserId="2380" LastEditorUserId="2380" LastEditDate="2023-10-23T12:23:05.717" LastActivityDate="2023-10-23T12:23:05.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17289" PostTypeId="1" AcceptedAnswerId="17307" CreationDate="2017-09-26T19:23:55.173" Score="1" ViewCount="1114" Body="&lt;p&gt;I'm new to mechanical engineering, gears, and motors. Coming from a software/hardware background, I am trying to understand how to leverage gear reduction to maximize the torque output of a stepper motor for an equatorial mount. The speed can be very slow so I would like to focus on maximizing torque for a heavy payload.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By my understanding, the output torque is the product of the input torque and the gear ratio. Is this the same for the &quot;holding torque&quot; of a gear system? Is the output holding torque the product of the input holding torque and the gear ratio?&lt;/p&gt;&#xA;" OwnerUserId="13061" LastActivityDate="2017-09-27T17:47:31.703" Title="How does gear reduction affect a stepper motor's holding torque?" Tags="|mechanical-engineering|gears|torque|stepper-motor|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17291" PostTypeId="1" CreationDate="2017-09-26T19:59:04.243" Score="0" ViewCount="192" Body="&lt;p&gt;I was wondering if there are any window mounted split air conditioners in US market or anywhere else - One in which compressor and radiator hang out of the window and only evaporator is inside the room. I was able to pull up many patents from as early as 1938 with same idea but don't think a commercial product exist. Any thoughts why it could be so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some patents and more mentioned in their reference section:&lt;br&gt;&#xA;&lt;a href=&quot;https://www.google.com/patents/US6568201&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/patents/US6568201&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://www.google.com/patents/US2268451&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/patents/US2268451&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13062" LastEditorUserId="13062" LastEditDate="2021-08-23T14:39:08.560" LastActivityDate="2021-08-23T14:39:08.560" Title="Window mounted split air conditioner" Tags="|mechanical-engineering|hvac|cooling|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="17292" PostTypeId="1" CreationDate="2017-09-26T21:03:49.840" Score="4" ViewCount="1032" Body="&lt;p&gt;Looking at the capacity of train networks, systems like CBTC help increase capacity, but the spacing of trains is ultimately limited by poor braking performance.  In my understanding, safe distances are calculated by assuming a worst case situation ahead, such as a derailment or collision.  The train behind must have adequate stopping distance plus a safety margin behind the train in front.  Since trains have such poor braking performance, this required spacing is very large, especially for heavy and/or higher speed trains.  Why aren't trains, especially in, for example, very dense subway systems, equipped with some auxiliary braking system, such as additional rubber shoes that could contact the track, or even maybe extreme things like plows off to the side of the track, that could stop the train very quickly in a rare emergency, such as the train ahead derailing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--edit--&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me add a bit more detail about the thinking behind this question: Commuter trains in my area can brake at about 0.05g, so they stop in about 1/2 mile.  Once you multiply out all of the various safety factors, they wind up requiring 10 minute spacing.  By comparison, a bus driving down the highway maintains a following distance of less than 10s.  Since the spacing for trains is based on a worst case scenario of an instantaneous derailment ahead, and braking distance is proportional to the square of deceleration, even a 40% increase in emergency braking deceleration is good for nearly a 2x increase in capacity on a track. 0.1g is unlikely to cause injury to passengers, and this is only applicable in the worst case catastrophic derailment ahead.  Since this scenario almost never happens, even a single use brake that completely destroys the train and damages the track is acceptable as long as the passengers aren't injured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example in my area is BART.  BART's bottleneck is in the transbay tube, with one track in each direction.  There's been talk about building a second tube, perhaps at a roughly 10 billion dollar cost.  This would be necessary because BART says it can't maintain a headway of less than 2.5 minutes safely.  If a bus can maintain a 10s headway safely, the only thing keeping the capacity of BART from being 15x better is braking performance in an emergency.&lt;/p&gt;&#xA;" OwnerUserId="13064" LastEditorUserId="13064" LastEditDate="2017-10-02T08:43:25.643" LastActivityDate="2019-01-27T00:56:32.663" Title="Why do trains not have a high-friction emergency braking system?" Tags="|rail|transportation|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17293" PostTypeId="2" ParentId="17241" CreationDate="2017-09-27T01:12:13.147" Score="2" Body="&lt;p&gt;Let me give you a headstart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T=F*d$, therefore, $F_{f1}=(T/2)/r$. The torque is divided by two because there are two balls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, you can compute $N_1$. Sum up moments at the center of the ball, you will get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{f1}=F_{f2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you have $F_{f2}$, you will get $N_2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the output shaft now, you will get the position of the force as shown in the right figure. You will need to sum up moment at the center of the output shaft. The resulting moment needs to be equal to $T/2$ as we need to transmit the torque to the output shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will require hard mathematics but I think this will do it. I know there is a simpler solution so you we will have to find out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sKp5y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sKp5y.png&quot; alt=&quot;See image below.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2017-09-27T01:12:13.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17294" PostTypeId="2" ParentId="17278" CreationDate="2017-09-27T07:17:49.477" Score="1" Body="&lt;p&gt;Here is the conjugate beam method to start off your analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k6tHT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k6tHT.png&quot; alt=&quot;Conjugate beam method&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have 3 unknowns. $M, R, and\ V$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up moment at the conjugate internal hinge (going towards the free end) gives the first equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up moment at the conjugate pin/hinge support (going towards the free end) gives the second equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summing up forces along y-axis gives a reaction at the conjugate hinge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, using the reaction from the conjugate hinge and summing up moment at the free end (involves the whole figure) gives the third equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three equations, three unknowns, gives the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If the system is non-symmetric, the number of unknowns becomes 6. Therefore, you need to do the summing up of moments to the other side of the beam to get &lt;em&gt;another three&lt;/em&gt; equation. :) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the figure that is described in the other answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zlXyT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zlXyT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-27T07:23:20.827" LastActivityDate="2017-09-27T07:23:20.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17295" PostTypeId="2" ParentId="17292" CreationDate="2017-09-27T07:28:22.167" Score="0" Body="&lt;p&gt;It's a boring answer, but the main reason is cost - in addition to the equipment itself there would need to be a maintenance schedule. There's also the question of additional mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of the quality/speed/cost triangle (&lt;a href=&quot;https://en.wikipedia.org/wiki/Project_management_triangle&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Project_management_triangle&lt;/a&gt;). Additional emergency brakes are a high quality suggestion, but will increase time and cost for an item that in normal use will not be seen as necessary - it's usually only after a problem occurs that public expectations and legislation catch up with quality/safety recommendations.&lt;/p&gt;&#xA;" OwnerUserId="13043" LastActivityDate="2017-09-27T07:28:22.167" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17296" PostTypeId="1" CreationDate="2017-09-27T10:07:32.133" Score="1" ViewCount="2399" Body="&lt;p&gt;What are the advantages and disadvantages of using circular section (&quot;Scotch&quot;) keys, keyseats and keyways?  I am not referring to Woodruff style, I am only interested in the type where the key itself if a full cylindrical piece and the shear force is diametric.  For instance...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RURkp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RURkp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wGmfK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wGmfK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Woodruff style and straight cut style seem overwhelmingly more common than circular section types.  What is the reason for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that it would be very convenient to be able to simply drill a keyway/keyseat into a parts before turning the ID/OD to it's final diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition I would think that a circular feature such as this would eliminate stress concentrations that would be present on broached and milled square section cuts.  Key joint cutting is in any case a precision operation so I doubt there's any major loss in tolerance quality over other types of keys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have any particular application in mind for this, I'm merely curious.  However I would like to know some situations where circular keys are advantageous or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Further thoughts: Wouldn't a circular key actually experience more of a compression force than a shear force, especially under high torque applications?  Given the many non-parallel shear forces all converging at the centre of the key, this would manifest as compression would it not?&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastEditorDisplayName="user6335" LastEditDate="2017-09-27T10:41:21.987" LastActivityDate="2018-03-26T04:19:58.407" Title="Circular keyways - advantages / disadvantages?" Tags="|machining|metallurgy|manufacturing-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17297" PostTypeId="1" AcceptedAnswerId="17298" CreationDate="2017-09-27T11:16:20.450" Score="3" ViewCount="1490" Body="&lt;p&gt;If we have an object on a frictionless inclined bench, why does the minimum force applied to keep it steady have to be parallel to the incline (or perpendicular to the reaction force)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eBEbX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eBEbX.jpg&quot; alt=&quot;Object on an inclined bench&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, why does the angle θ in the image have to be 0 in order for the force to be minimum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider we know the weight of the object and the reaction of the incline.&lt;/p&gt;&#xA;" OwnerUserId="12979" LastEditorUserId="330" LastEditDate="2017-09-27T20:23:47.680" LastActivityDate="2017-09-30T00:19:58.733" Title="Why does the minimum force have to be parallel to the incline?" Tags="|statics|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17298" PostTypeId="2" ParentId="17297" CreationDate="2017-09-27T12:02:57.297" Score="9" Body="&lt;p&gt;The minimum force is in the direction the object would move without that force.  Clearly the object would move along the inclined plain, down and right in your diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, its the component of the force along the direction of travel that matters.  More mathematically, it's the dot product of the force with the acceleration unit vector, with negative values contributing to cancelling the acceleration, and positive values enhancing it.  The dot product is -1 when the force is exactly opposite the acceleration direction.  A sideways force would do nothing, and has a dot product of 0.  A force pushing down the incline has a dot product of 1, and just like the math says, would make the object go down the incline faster.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-09-27T12:02:57.297" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17299" PostTypeId="1" CreationDate="2017-09-27T12:20:10.580" Score="0" ViewCount="246" Body="&lt;p&gt;I am designing a process heat exchanger to cool a gas effluent stream consisting of 70% Nitrogen 23% Carbon Dioxide and 7% Water (all mole percent). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stream is going to enter at 250 degC and 1.51325 bar (i.e. just above atmospheric pressure) and leave at 50 degC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stream is being cooled by an aqueous stream which has been specified as having thermal properties of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this,&#xA;I have calculated that the partial pressure of the water in the mixture initially is approximately 0.106 bar. I used the formula Py=p to find this. P being the total pressure of the system, y being the mole fraction of component in the gas mixture and p being the partial pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this formulation correct for the partial pressure and will partial pressure remain constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it correct to calculate the vapour pressure and compare this to partial pressure to check if the water in the gas mixture will condense?&lt;/p&gt;&#xA;" OwnerUserId="12385" LastEditorUserId="9262" LastEditDate="2018-04-25T19:31:14.487" LastActivityDate="2018-04-25T19:31:14.487" Title="Heat Exchanger Design- Cooling a gas mixture" Tags="|pressure|heat-transfer|heat-exchanger|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17300" PostTypeId="2" ParentId="17297" CreationDate="2017-09-27T12:36:57.500" Score="2" Body="&lt;p&gt;The force you're applying would have an effect that counteracts the body's tendency to move. That effect will be calculated by [applied force] $\cos \theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $\theta$ is not zero, $\cos{\theta}$ would be less than 1, so you would have to apply a larger force to produce the same result. For example, 10 degrees for $\theta$ would equate to a force parallel to the plane of 0.9848 times the applied force. You would need a larger applied force (approximately 1.015 times) to produce the same effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When $\theta = 0$, $\cos{\theta} = 1$.&lt;/p&gt;&#xA;" OwnerUserId="13043" LastEditorUserId="1832" LastEditDate="2017-09-28T07:25:27.903" LastActivityDate="2017-09-28T07:25:27.903" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17301" PostTypeId="2" ParentId="17299" CreationDate="2017-09-27T13:08:12.423" Score="1" Body="&lt;p&gt;&lt;strong&gt;TL;DR&lt;/strong&gt; You are correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this formulation correct for the partial pressure   &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, your Py=p is correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...and will partial pressure remain constant?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Strictly speaking no, because total pressure will not remain constant. Use ideal gas law or one of the real gas laws to find the new pressure of each species at the new temperature. From there find the new total pressure, also compare partial pressure of steam to vapor pressure at 50°C to see if you have condensation.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-09-27T13:08:12.423" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17303" PostTypeId="2" ParentId="17292" CreationDate="2017-09-27T15:57:56.520" Score="5" Body="&lt;p&gt;Another contribution to the issue, if you want to stop a train of any length in a hurry you would need to uniformly brake every car in a controlled way.  If you excessively brake the forward part you risk &lt;em&gt;causing&lt;/em&gt; a derailment.  Excessive braking at the rear also risks derailment if on any sort of curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would seem to preclude any &quot;simple minded&quot; emergency plow/shoe type braking. So now you are left with fitting every car with a fairly sophisticated powered brake and control system.  Feasible for passenger, but cost prohibitive for freight.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2017-09-27T15:57:56.520" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17305" PostTypeId="1" CreationDate="2017-09-27T16:15:00.800" Score="0" ViewCount="1720" Body="&lt;p&gt;I am going to try and ask this question again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this formula to calculate shear stress in a solid shaft:&#xA;τ = T r / J    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to solve for the radius; I should use the modulus of rigidity from the material sheet in the formula for τ ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;example, knowing&#xA;T = 1000Nm&#xA;J = pi * R^2 / 2&#xA;τ = modulus of rigidity from material sheet&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to design for a solid shaft with a known material and known torque then I can use the formula: τ = T r / J&lt;br&gt;&#xA;replacing &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT]&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Density     2.6989 g/cc 0.097504 lb/in³ &#xA;Mechanical Properties   Metric  English Comments&#xA;Hardness, Vickers   15  15  Annealed&#xA;Modulus of Elasticity   68.0 GPa    9860 ksi    &#xA;Poissons Ratio  0.36    0.36    calculated&#xA;Shear Modulus &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I added the above material property. Let's say that I am making a solid shaft that will be under torsional stress, the shaft will be spinning but at a low RPM so there's no shocks to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep things simple forget about about factor of safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;τ = modulus of rigidity from the material property sheet, is that correct? If it's not correct; how would I calculate the diameter of the shaft based on the info that I have which are;&#xA;maximum torque&#xA;the polar moment of inertia&#xA;the mechanical property sheet&lt;/p&gt;&#xA;" OwnerUserId="12586" LastEditorUserId="12586" LastEditDate="2017-09-27T16:59:05.353" LastActivityDate="2017-09-27T16:59:05.353" Title="Tau max value in torsion calculation" Tags="|mechanical-engineering|materials|elastic-modulus|" AnswerCount="1" CommentCount="0" ClosedDate="2017-09-30T21:47:29.400" ContentLicense="CC BY-SA 3.0" />
  <row Id="17306" PostTypeId="2" ParentId="17305" CreationDate="2017-09-27T16:43:36.157" Score="1" Body="&lt;p&gt;No, that is not correct. The modulus of rigidity is not related to the allowable shear stress in the shaft, so you would not set τ equal to the modulus of rigidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need is the allowable shear stress. That depends on the material, what type of loading you are using (such as a static load or a repeating load), and what factor of safety is required.&lt;/p&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2017-09-27T16:43:36.157" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17307" PostTypeId="2" ParentId="17289" CreationDate="2017-09-27T17:47:31.703" Score="4" Body="&lt;p&gt;Yes, the holding torque goes up proportionally with the gear ratio. Plus the friction of the gear train, and any other seals, bearings along the gear train that add friction.  &lt;/p&gt;&#xA;" OwnerUserId="13069" LastActivityDate="2017-09-27T17:47:31.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17308" PostTypeId="2" ParentId="17219" CreationDate="2017-09-27T17:58:11.640" Score="0" Body="&lt;p&gt;What is the person doing in the bathroom? Are they communicating with aliens via a similar or harmonic frequency? HeHe... Honestly, the door is probably a red herring as mentioned, unless it is metal. Add a foil or metallic shield between. Perhaps ground the antenna post to a nearby pipe. Remove the pre-amp from the line and see what happens. The interference is likely RF, so any slight change will minimize or eliminate the problem. Is it possible the shield of the cable is nicked, or grounded on a metal edge? Is the cable shielded on one or both ends between the antenna or pre-amp? But eliminate the pre-amp as a source first. &lt;/p&gt;&#xA;" OwnerUserId="13069" LastActivityDate="2017-09-27T17:58:11.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17309" PostTypeId="2" ParentId="17196" CreationDate="2017-09-27T18:05:41.090" Score="0" Body="&lt;p&gt;Mount a very balanced, VERY flat set of disks (one for hour, one for minute) to the hour and minute shafts. The lighter and thinner the disk is, the better. Clock hands don't weigh very much. You could prototype using some aluminum foil.  &lt;/p&gt;&#xA;" OwnerUserId="13069" LastActivityDate="2017-09-27T18:05:41.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17312" PostTypeId="1" CreationDate="2017-09-27T21:54:50.713" Score="1" ViewCount="103" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/23EN8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/23EN8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I need help with this problem.&#xA;Suppose we have a cantilever with width $-c \leq y \leq c$ and length $x=l$ fixed at $x=0$. We know that it is loaded by uniform shear along the lower edge with magnitude $S$ and the upper edge and the end $x=l$ are free from load. I know how to find the stresses using the airy stress function, but could you please tell me how to find $\sigma_x$ and $\tau_{xy}$ using the elementary tension and bending formula so that I can compare?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The depth of the beam perpendicular to the x axis is 1.&lt;/p&gt;&#xA;" OwnerUserId="13071" LastEditorUserId="110" LastEditDate="2017-09-29T10:34:28.947" LastActivityDate="2017-09-29T10:34:28.947" Title="Calculating stress from elementary tension and bending formulas" Tags="|stresses|shear|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17313" PostTypeId="1" CreationDate="2017-09-27T22:24:10.030" Score="1" ViewCount="124" Body="&lt;p&gt;I have an embarrassing question that I just can't seem to figure out. I have a fracture problem that is infinitely easier to solve in cylindrical coordinates. The boundary conditions, however, are expressed in cartesian coordinates Namely, at a particular surface:$$  \sigma_{xz}=0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just cannot figure out how to represent this in cylindrical coordinates. We were given a similar example in class notes where the boundary condition is: $$\sigma_{yz}=0$$and, we were just told that, in cylindrical coordinates, this is equivalent to $$\mu \frac{1}{r}  \frac{\partial u_3}{\partial \theta}|_{\theta = \pi}=0 $$ where $\mu$ is the shear modulus of the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone point me in the right direction to obtain a similar statement for $\sigma_{xz}$, or, similarly, show me how the statement for $\sigma_{yz}$ was obtained? I get that $$\sigma_{xz}=\frac{\partial u_3}{\partial x_1} + \frac{\partial u_1}{\partial x_3}$$. There's a step in the change of variables that I'm missing.&lt;/p&gt;&#xA;" OwnerUserId="13072" LastEditorUserId="13072" LastEditDate="2017-09-27T23:56:20.437" LastActivityDate="2017-09-27T23:56:20.437" Title="Shear stress derivative in polar/cylindrical coordinates" Tags="|structural-engineering|materials|stresses|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17314" PostTypeId="1" CreationDate="2017-09-27T23:03:34.303" Score="0" ViewCount="81" Body="&lt;p&gt;We are working on a mounted screen design that will have microphones at the bottom of the screen housing. The screen housing will be sitting at a 10.5 degree angle. So we are worried about water sitting in the microphone hole and not giving us good microphone quality. I have included an image below showing the design. I am stumped and cant think of another way to fix the water not being able to drain out of the hole because of the angle. Any suggestions or help would be greatly appreciated. Thanks&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bc55m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bc55m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12820" LastActivityDate="2017-09-30T08:01:53.323" Title="Microphone angled waterproofing" Tags="|fluid-mechanics|waterproofing|audio-engineering|sound-isolation|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17315" PostTypeId="2" ParentId="17277" CreationDate="2017-09-28T02:25:29.290" Score="0" Body="&lt;p&gt;First I want to address some fundamental capacitor properties for you to consider:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Certain capacitors have specific polarity markings that you need to take into account and insert them correctly into your circuit. Not doing so can cause damage to your components. If you don't see any indication of polarity, you should be okay with either direction you place it. &lt;/li&gt;&#xA;&lt;li&gt;As briefly mentioned in another answer, capacitors are &quot;energy storage devices&quot;, specifically charge. Current is simply the flow of charge carriers, so a capacitor will prevent this flow while it is charging and when it discharges, the current will flow.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, think about the circuit implemented in user8055's answer. A switching circuit needs to be thought of as 2 pseudo-independent circuits&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First: When the switch is initially creating the smaller circuit of the DC source and the capacitor, charge builds up in the capacitor. As long as the switch is in the first position and not connected to the resistor/LED branch, this charge will not go anywhere and current will not flow, as the capacitor is fully charged due to the source constantly supplying voltage to it, thus keeping it in an equilibrium state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this first state, think of the capacitor as a fully charged battery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second: As soon as the switch flips to the other branch, the capacitor is no longer being supplied with power, and it begins to discharge. This discharge creates a movement of charge carriers (current) that passes through the resistor and LED, lighting it up. Make sure you notice the DC source isn't actually powering anything at this time, as it isn't part of a complete circuit. The only source of energy is the charged capacitor. Eventually, the capacitor will fully discharge, and the current will stop flowing, thus the LED turning off, due to not having any power source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to consider is how a capacitor will work when the circuit is driven by an AC source, instead of DC. Depending on the configuration of the resistor and capacitor in the circuit, this is how filters are created to restrict specific frequency ranges. &lt;/p&gt;&#xA;" OwnerUserId="13073" LastEditorUserId="8055" LastEditDate="2017-09-29T17:54:11.230" LastActivityDate="2017-09-29T17:54:11.230" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17316" PostTypeId="1" AcceptedAnswerId="17341" CreationDate="2017-09-28T04:04:06.247" Score="1" ViewCount="115" Body="&lt;p&gt;What are the Principle directions? Are they unit normals of three planes perpendicular to each of the principle stresses?&lt;/p&gt;&#xA;" OwnerUserId="13075" LastActivityDate="2017-11-04T21:48:40.137" Title="Principle Directions of a Stress Tensor" Tags="|materials|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17320" PostTypeId="2" ParentId="17314" CreationDate="2017-09-28T12:04:18.613" Score="1" Body="&lt;p&gt;Why have the hole perpendicular to the front surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angle the hole so that it is slightly uphill when the screen is at the angle as shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, could you fit a membrane on the surface of the microphone and mount it closer to the front face with a larger hole for the entry rhs but smaller hole on the face lhs.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-09-28T12:04:18.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17322" PostTypeId="2" ParentId="17314" CreationDate="2017-09-28T12:29:32.743" Score="2" Body="&lt;p&gt;You could make a water drain trap. Drill a vertical hole that meets the microphone hole somewhere along its length, with a drain hole sloping down for the water to escape. You could then make a step change in diameter in the mike hole (for example between the pink and blue parts in our drawing) to stop the water getting past the drain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you appear to have several parts assembled together, you should be able to figure out a way to make something like that - but there isn't enough information on the single view in your drawing to make a useful guess at how to do it!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-09-28T12:29:32.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17323" PostTypeId="1" AcceptedAnswerId="17426" CreationDate="2017-09-28T14:09:23.037" Score="1" ViewCount="203" Body="&lt;p&gt;I am working on split type AC compressor. I actually want to know, if i changed the frequency of the compressor to run on different power level, will it affect the life of the compressor. My actual work to run the Compressor with different power level. I will use VFD ( variable frequency drive) to change the frequency of compressor and it will take different power with respect to the frequency. can you tell me that all rotary compressor is designed for a specific frequency. ?&lt;/p&gt;&#xA;" OwnerUserId="13080" LastEditorUserId="2819" LastEditDate="2018-01-05T04:55:51.090" LastActivityDate="2018-03-06T05:45:01.163" Title="can we run rotary compressor of Air conditioner with different frequency level?" Tags="|compressors|compressed-air|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17325" PostTypeId="1" CreationDate="2017-09-28T14:37:43.233" Score="1" ViewCount="913" Body="&lt;p&gt;I'm building a plane, and one of the solutions I've thought of for connecting the wing spars to the fuselage involves connecting carbon fiber parts.  As mentioned in &lt;a href=&quot;https://aviation.stackexchange.com/a/17029/65&quot;&gt;this answer&lt;/a&gt;, I found this explanation for how to calculate the scarf ratio:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The scarf ratio is determined by the ratio of fiber tensile strength&#xA;  and the shear strength of the resin. If your fiber takes 3000 N/mm²,&#xA;  the resin transmits 20 N/mm² and the fiber content is 60%, the ratio&#xA;  is 1:90, so a 1 mm thick composite layer would need to be joined over&#xA;  a length of 90 mm.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have tried to find more detailed explanations, but have not yet succeeded.  I am interested in knowing where this information comes from.  I have at least two specific questions:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;does the answer depend on how the composite layer was prepared?&lt;br&gt;&#xA;For example, does the answer change if the composite layer is a pultruded carbon-fiber strip? The fiber ratio is similar to what is claimed above as ideal, but the curing temperature is higher.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is proposed above that the resin transmits 20N/mm^2 - this square mm,  which is a surface, how is it oriented?&lt;br&gt;&#xA;If this was 20 N/mm^2 per mm of overlap of two composite layers, it would seem to make more sense. That's what seems to be being calculated. So is it a typo, or is there something more fundamental that I'm missing?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think what I'm missing is that I'm thinking of lap-joints, also mentioned.&#xA;I think that the orientation of the 20 N/mm^2 surface in the above discussion is parallel to the scarf surface, the surface along which the composite layers are joined.  In this case, is the 20 N/mm^2 a tensional strength, or shear strength?  looks like it's a combination of both?  it would be a shear strength in the case of a lap joint, a tensional strength in the case of a but joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar question about lap joints is &lt;a href=&quot;https://engineering.stackexchange.com/questions/17332/how-to-calculate-lap-overlap-for-composites&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13081" LastEditorUserId="13081" LastEditDate="2017-09-28T21:10:17.787" LastActivityDate="2017-09-28T21:10:17.787" Title="How to calculate scarf ratio or lap overlap for composites?" Tags="|composite|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17326" PostTypeId="2" ParentId="17325" CreationDate="2017-09-28T15:46:18.890" Score="1" Body="&lt;p&gt;Formulas from &quot;The Behavior of Structures Composed of Composite Materials&quot;.  Consider this source material for a much deeper understanding of composites.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the answer depend on how the composite layer was prepared?&#xA;  For example, does the answer change if the composite layer is a pultruded carbon-fiber strip? Looks like the fiber ratio is the same as what is claimed above as ideal, but the curing temperature is higher.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, and no.  What you're looking for is the stress present in the mid plane.  With higher fiber content and different preparation, you can have lower or higher stresses in the layer.  As such, if you have a 1mm thick layer of composite, and you are pulling with 1,000 N, then for a 1mm wide section of the plane, you will have approximately 1,000 N/mm^2. The amount of &quot;pull&quot; that goes through your shell will be based upon the loads you are putting it through, which can depend on how the layer was prepared.  This is discussed more in the text below.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is proposed above that the resin transmits 20N/mm^2 - this square mm, which is a surface, how is it oriented? Is there something more fundamental that I'm missing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A significant question.  We are orienting ourselves so the loads are being transmitted in the x-direction, the joint is in the y-direction, and the thickness of the beam is in the z-direction.  Image shown for clarification.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2NpC8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2NpC8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then look at a 1-mm wide &quot;slice&quot; of the joint - taken in the y-direction, and review.  This slice may transmit 3,000 N through the 1-mm wide slice of the joint, and is 1mm thick.  If so, we define the adhesive tensile stress / shear stress as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \sigma_{adh} = E_{adh} \frac{\delta_{z1} - \delta_{z2}}{t} $$&#xA;$$ \tau_{adh} = G_{adh} \frac{\delta_{x1} - \delta_{x2}}{t} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\sigma_{adh}$ is the tensile stress developed in the adhesive&lt;/li&gt;&#xA;&lt;li&gt;$\tau_{adh}$ is the shear stress developed in the adhesive&lt;/li&gt;&#xA;&lt;li&gt;$E_{adh}$ is the young's modulus of the adhesive&lt;/li&gt;&#xA;&lt;li&gt;$G_{adh}$ is the shear modulus of the adhesive&lt;/li&gt;&#xA;&lt;li&gt;$\delta_{z1}, \delta_{z2}$ are the displacements in z of the pieces, respectively&lt;/li&gt;&#xA;&lt;li&gt;$\delta_{x1}, \delta_{x2}$ are the displacements in x of the pieces, respectively&lt;/li&gt;&#xA;&lt;li&gt;t is the thickness of adhesive perpendicular to the angle of the joint. (Approximately in z).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The adhesive may be only able to take 20 N of force of shear stress in a lab specimen.  This documented tensile test is then 20 N/mm^2. Per Vinson:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\tau_{adh}}{\sigma_{fiber}} = -\frac{h}{l}$$&#xA;$$\frac{\sigma_{adh}}{\sigma_{fiber}} = \frac{h^2}{l^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The joint is then analyzed carefully for full analysis.  In this example, since the beam is only being pulled, not bent, the shear stress is to be reviewed.  Since the thickness of the consists ENTIRELY of adhesive and does not contain any fibers, there is no fiber content to be considered, we simply utilize the shear strength of the adhesive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, the answer before is incorrect.  To ensure the adhesive has only 20 N/mm^2 of stress (the maximum the laboratory says it can handle), we apply the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{20\frac{N}{mm^2}}{3000\frac{N}{mm^2}} = -\frac{1}{150} = \frac{h}{l}$$&#xA;$$l/h = 150$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, most joints have minimal tensile load, and typically have the load in bending.  This will cause significant z-direction displacements, which are highlighted into the two formulas.  Most joint designs lie between 7-9 degrees, depending on the bending loading, tensile loading, and displacement offsets expected due to loading at the joint.  Fiberglass is typically done with a 4:1 - 6:1 length ratio, while carbon-epoxy will reach 12:1 ratios.  90:1 ratios are unheard of in practice,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lap Joints, as Solar Mike Mentioned, are a completely separate topic with significantly more complicated formulas (Also detailed in Vinson's Text).&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2017-09-28T20:12:48.437" LastActivityDate="2017-09-28T20:12:48.437" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17327" PostTypeId="1" CreationDate="2017-09-28T18:26:32.977" Score="0" ViewCount="46" Body="&lt;p&gt;I am looking for the term used to describe a situation I've encountered. I'll use a standard hydraulic cartridge that you'd use to keep a door shut as an example, except this one can be locked out with a rotary valve. How would you refer to the valve being pushed open by pressure or load? If under pressure, the valve is unable to close until pressure is relieved. Is there a term for this? I thought it may have been &quot;Jetting&quot;, but that doesn't exactly describe this situation.&lt;/p&gt;&#xA;" OwnerUserId="13086" LastEditorUserId="1832" LastEditDate="2017-09-28T21:06:54.910" LastActivityDate="2017-10-29T10:12:47.483" Title="Fluid-dynamics question: proper term" Tags="|fluid-mechanics|terminology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17328" PostTypeId="1" AcceptedAnswerId="21068" CreationDate="2017-09-28T18:28:09.293" Score="-1" ViewCount="81" Body="&lt;p&gt;Steel prefabricated buildings cost less, require less time and provide a reasonable quality compared to concrete buildings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are common for warehouses, offices and other types of buildings. Also, there are prefabricated steel houses for residential use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have noticed that steel buildings/apartments are rare/non-existent. Is there a reason for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically asking regarding the usage of prefabricated steel in apartment and/or multistory buildings for regular use. &lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2018-03-27T15:11:58.510" Title="Why aren't Steel Prefab Buildings common for residential use?" Tags="|steel|building-design|" AnswerCount="1" CommentCount="7" FavoriteCount="0" ClosedDate="2018-03-27T15:13:42.487" ContentLicense="CC BY-SA 3.0" />
  <row Id="17329" PostTypeId="1" CreationDate="2017-09-28T19:14:41.273" Score="0" ViewCount="34" Body="&lt;p&gt;When were airfoil cross-sections first used in turbine blades? When did this become standard practice? Did people understand the connection to aircraft, or did heavier-than-air aircraft exist at all yet?&lt;/p&gt;&#xA;" OwnerUserId="13087" LastActivityDate="2017-09-28T19:57:54.703" Title="When were airfoil cross-sections first used in turbine blades?" Tags="|turbines|engineering-history|turbomachinery|airfoils|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17330" PostTypeId="2" ParentId="17329" CreationDate="2017-09-28T19:46:54.050" Score="3" Body="&lt;p&gt;Good description in the link, but curved surfaces were in use as sails around 800 years ago and Bernoulli  came up with his formula , Ludwig Prandtl was testing sections in 1017 in his large wind tunnel, the first naca section was tested in 1935 ....  &lt;a href=&quot;http://www.flyingmag.com/technicalities/technicalities-short-history-airfoils&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.flyingmag.com/technicalities/technicalities-short-history-airfoils&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this link to Sir Frank Whittle is also very illuminating as a design error had the shape of the blades incorrect during testing around December 1937...  &lt;a href=&quot;https://books.google.ch/books?id=a_YTYwP4aS4C&amp;amp;pg=PA120&amp;amp;lpg=PA120&amp;amp;dq=turbines+whittle&amp;amp;source=bl&amp;amp;ots=aEH_MZYiEQ&amp;amp;sig=SVPqca_ONdeuOQ3rAXVZHRDm2o4&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwi4guii1MjWAhUEcRQKHQHFAncQ6AEInAEwFg#v=onepage&amp;amp;q=turbines%20whittle&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://books.google.ch/books?id=a_YTYwP4aS4C&amp;amp;pg=PA120&amp;amp;lpg=PA120&amp;amp;dq=turbines+whittle&amp;amp;source=bl&amp;amp;ots=aEH_MZYiEQ&amp;amp;sig=SVPqca_ONdeuOQ3rAXVZHRDm2o4&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwi4guii1MjWAhUEcRQKHQHFAncQ6AEInAEwFg#v=onepage&amp;amp;q=turbines%20whittle&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is a &quot;poem&quot; about the work at the time - source the Frank Whittle link above :&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oghtl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oghtl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-09-28T19:57:54.703" LastActivityDate="2017-09-28T19:57:54.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17331" PostTypeId="2" ParentId="17196" CreationDate="2017-09-28T20:47:32.220" Score="1" Body="&lt;p&gt;Could it be over engineered to ridiculous levels? Perhaps costly, but there would be major advantages.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Faceplate of any size. Mounted directly onto a &lt;a href=&quot;http://uk.rs-online.com/web/p/stepper-motors/5350401/&quot; rel=&quot;nofollow noreferrer&quot;&gt;stepper motor&lt;/a&gt; so&#xA;easily adequate torque for all conceivable plate sizes. 400 steps per&#xA;revolution.  &lt;/li&gt;&#xA;&lt;li&gt;Add a belt drive and reducing wheels for even more&#xA;resolution. &lt;/li&gt;&#xA;&lt;li&gt;Controlled by an &lt;a href=&quot;https://www.arduino.cc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arduino&lt;/a&gt; and motor controller. &#xA;Mains powered so no batteries. &lt;/li&gt;&#xA;&lt;li&gt;Add a GPS receiver for infinitely&#xA;accurate time keeping.&lt;/li&gt;&#xA;&lt;li&gt;Can probably download all the necessary code as someone's already done each bit of this.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Too silly?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2017-09-28T20:47:32.220" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17333" PostTypeId="1" CreationDate="2017-09-28T21:57:55.887" Score="1" ViewCount="512" Body="&lt;p&gt;QUESTION: Air at 35°C enters a 10 m long section of a rectangular duct of cross section 25 cm by 15 cm made of&#xA;commercial steel at an average velocity of 10 m/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Take density of air and dynamic viscosity as 1.145 kg/m^3 and 1.895 x 10^-5 kg/m.s respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine:&#xA;(a) Reynolds number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see my working here below,&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nXxH1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nXxH1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MLAaL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MLAaL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/qRlHQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qRlHQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i take the cross section 25cm * 15 cm as my area. I need help finding the diameter of pipe and please do confirm my workings cause i feel like its somewhat off since from my calculation above my Reynolds number is quite big.&lt;/p&gt;&#xA;" OwnerUserId="12830" LastEditorUserId="12830" LastEditDate="2017-09-28T22:20:43.377" LastActivityDate="2019-03-27T11:07:37.937" Title="Determining Reynolds number" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17335" PostTypeId="2" ParentId="16107" CreationDate="2017-09-29T00:44:03.690" Score="0" Body="&lt;p&gt;I would recommend using an ultrasonic fogger/mister in a containor with distilled water. You would have to pump some air into the sealed containor with a &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B002JLJC0W&quot; rel=&quot;nofollow noreferrer&quot;&gt;diaphragm pump&lt;/a&gt; and then have a line running from the other side of that containor to the location in the wind tunnel where you need it. The shorter the distance the better, but the ultrasonic mister should sufficiently atomize it to travel without much loss. Once in the wind tunnel it will disappear quickly as the mist evaporates. I would recommend just increasing the quantity of mist to go a further distance. It wont create a mess, since you can control how far it travels before the mist evaporates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B00PAJXVJS&quot; rel=&quot;nofollow noreferrer&quot;&gt;small unit&lt;/a&gt;&#xA;Here is a &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B01MRL81BR&quot; rel=&quot;nofollow noreferrer&quot;&gt;larger unit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to go the dry ice route you could do it the same way, as the &quot;smoke&quot; you see is just condensed water vapor from the dry ice temperature. The CO2 will proprell the mist so you won't need the diaphragm pump. It will just cost more labor/money if your test is very long or frequent. It will also be much more difficult to regulate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I agree with SF. that the density difference will not likely affect your results. The amount of visualizing mist is very small compaired to the primary wind tunnel stream.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2017-09-29T00:49:29.407" LastActivityDate="2017-09-29T00:49:29.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17336" PostTypeId="1" CreationDate="2017-09-28T13:19:57.527" Score="1" ViewCount="24" Body="&lt;p&gt;I am working on split type AC compressor. I actually want to know, if i changed the frequency of the compressor to run on different power level, will it affect the life of the compressor. My actual work to run the Compressor with different power level. I will use VDF ( variable frequency drive) to change the frequency of compressor and it will take different power with respect to the frequency. &#xA;can you tell me that all rotary compressor is designed for a specific frequency. ? &lt;/p&gt;&#xA;" OwnerUserId="13080" OwnerDisplayName="fame" LastActivityDate="2017-09-29T03:42:30.793" Title="can we run rotary compressor of Air conditioner with different frequency level ?" Tags="|cooling|" AnswerCount="0" CommentCount="2" ClosedDate="2017-09-30T13:33:04.740" ContentLicense="CC BY-SA 3.0" />
  <row Id="17339" PostTypeId="2" ParentId="17327" CreationDate="2017-09-29T09:25:13.910" Score="1" Body="&lt;p&gt;pressure relief valve - limits or controls the pressure to the set limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least based on your description...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-09-29T09:25:13.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17341" PostTypeId="2" ParentId="17316" CreationDate="2017-09-29T09:41:39.010" Score="1" Body="&lt;p&gt;I think that &quot;Principle directions&quot; can be used both in terms of stress (the direction of principal stresses) and in terms of inertia (the 3 direction where the inertia matrix is diagonal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mainly use &quot;principal direction&quot; in terms of inertia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that you can ganeralize: principal directions are the unit vector where something 3D distribution (inertia, stresses...maybe other things) can be represented as a diagonal matrix. &lt;/p&gt;&#xA;" OwnerUserId="12262" LastEditorUserId="12262" LastEditDate="2017-09-30T15:52:30.797" LastActivityDate="2017-09-30T15:52:30.797" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17342" PostTypeId="1" CreationDate="2017-09-29T19:10:34.683" Score="2" ViewCount="717" Body="&lt;p&gt;I am working on designing a set of gears for 3D printing. I have not done any gear design before - how should I improve the design to prevent gear slipping?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p0Znt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p0Znt.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yY4d5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yY4d5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13097" LastActivityDate="2017-10-01T18:24:47.257" Title="Gear Design - Teeth Fit" Tags="|gears|solidworks|" AnswerCount="2" CommentCount="3" ClosedDate="2017-10-09T12:06:29.553" ContentLicense="CC BY-SA 3.0" />
  <row Id="17343" PostTypeId="2" ParentId="17342" CreationDate="2017-09-29T20:20:18.487" Score="3" Body="&lt;p&gt;Teeth on gears generally will not slip if your pitch and tooth profiles match properly, since the teeth would have to break off to slip. Minimizing backlash can be done by tight tolerance control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the images you have it appears as though you've designed the teeth more freeform and by eye, rather than calculating the profiles (I could be wrong though). If you're working through CAD to completely design the gears, you can write a macro that will generate profiles based on any tooth profile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general process for this is create your drive gear however you like. Then create an idler gear with no tooth profile at all overlapping the drive gear. Subtract the interferance between the two parts from the idler, then rotate each based on your prescribed gear ratio. (how much you rotate each will depend on how accurate you need you profile to be). Repeat until you've completed one full rotation of whichever gear you're trying to make.&#xA;- basic youtube video of how it works here &lt;a href=&quot;https://www.youtube.com/watch?v=9MycFQKnZkA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=9MycFQKnZkA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can go the pure math method of designing the gear tooth profile completely through mathematics. Which is much more complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Easier still, use off the shelf gears!&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-09-29T20:20:18.487" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17344" PostTypeId="2" ParentId="17297" CreationDate="2017-09-29T21:15:40.267" Score="1" Body="&lt;p&gt;Since I posed the question in comment, here is how to treat the friction case, and of course at the end it reduces to the expected friction free result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the applied force has components $f_x$ parallel (up) the plane and a normal component $f_y$ downward into the plane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal force between block and plane is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_n = W\cos(t) + f_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force $f_x$ required to hold against sliding, assuming static coulomb friction with coefficient $u$ is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_x = W\sin(t) - u f_n = W\sin(t) - u (W\cos(t) + f_y)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude of the applied force is then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_{mag} = \sqrt{f_x^2+f_y^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;without showing all the messy steps, setting $d f_{mag} / d f_y == 0$ we can solve for $f_y$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_y = u W\sin(t)\cdot\dfrac{1 - u /\tan(t)}{1 + u^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_x = W\sin(t) ( 1 - u\cdot\dfrac{1/\tan(t) + u}{1 + u^2})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the friction free case ($u=0$) this results in the expected $f_x=W\sin(t),f_y=0$ (ie parallel to the plane) but with friction angling the force toward the plane reduces the required force magnitude.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2017-09-30T00:19:58.733" LastActivityDate="2017-09-30T00:19:58.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17345" PostTypeId="1" CreationDate="2017-09-29T21:33:33.150" Score="1" ViewCount="197" Body="&lt;p&gt;Since my last post on Engineering-Beta, I have composed a few simple simple circuits involving capacitors(super-caps specifically), resistors and LED's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following image is of a branch from a circuit I designed,which highlights the discharge of a capacitor through a resistor and LED. I will be using this as a reference to all my questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ygLP0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ygLP0.jpg&quot; alt=&quot;Circuit C1.3, Branch-B&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment That I allow current to flow from this circuit, Voltage will begin to drop, and current would begin to drop also. I know that the addition of resistance into the circuit can slow the loss of charge at the capacitor...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote a bunch of stuff regarding all this, but it seems trivial now. Basically, I'm dealing with a rate of change that is not constant,  and I'm trying to find precision in my circuit, I would like to be able to calculate the time duration of the discharge but, I believe that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Calculus is concerned with things that do not change at a constant rate&quot;&#xA;-&lt;em&gt;TheMathPage.com&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum everything up, all I know right now is intermediate algebra, and from all my google searches about this topic, they keep talking about derivatives and calculus. So, how can I figure out the rate of discharge of a capacitor? Or have I answered my own question already?&lt;/p&gt;&#xA;" OwnerUserId="12689" LastEditorUserId="8055" LastEditDate="2017-10-02T14:28:24.687" LastActivityDate="2017-11-02T02:09:59.320" Title="Can I Calculate The Discharge of a Capacitor If All I know Is Algebra" Tags="|electrical-engineering|power-electronics|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17346" PostTypeId="2" ParentId="14974" CreationDate="2017-09-29T22:30:00.497" Score="2" Body="&lt;p&gt;The Los Angeles Tall Buildings Structural Design Council (LATBSDC) 2017 guidelines suggest using the following (Table 3, LATBSDC 2017):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Service-Level Evaluation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Axial : $0.8E_cA_g$&lt;/li&gt;&#xA;&lt;li&gt;Flexural : $0.8E_cI_g$&lt;/li&gt;&#xA;&lt;li&gt;Shear : $0.4E_cA_g$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;MCE-Level Evaluation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Axial : $0.5E_cA_g$&lt;/li&gt;&#xA;&lt;li&gt;Flexural : $0.5E_cI_g$&lt;/li&gt;&#xA;&lt;li&gt;Shear : $0.2E_cA_g$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For code-level evaluation you could use some value between those two, as you professor recommended. However, it is common practice to use the full section for PT slabs (i.e., modifier = 1.0).&lt;/p&gt;&#xA;" OwnerUserId="13098" LastActivityDate="2017-09-29T22:30:00.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17347" PostTypeId="2" ParentId="17314" CreationDate="2017-09-30T08:01:53.323" Score="0" Body="&lt;p&gt;You didn't specify any dimensions but assuming the waveguide is small enough to act as a capillary, a membrane to prevent water entering it is your only option.&lt;/p&gt;&#xA;" OwnerUserId="13101" LastActivityDate="2017-09-30T08:01:53.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17349" PostTypeId="2" ParentId="13466" CreationDate="2017-09-30T08:30:33.553" Score="2" Body="&lt;p&gt;The far end formula, $\frac{3EI}{L}$, applies if the beam is discontinuous in that point. This means that the point of reference is the &lt;strong&gt;end of the beam where the support is a hinge/pin&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: $K_{AB}$. Note the $A$ in the $AB$. Point $A$ is the reference point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pin support in picture two is not discontinuous in point B, therefore, the distribution factor is still $\frac{4EI}{L}$.&lt;/p&gt;&#xA;" OwnerUserId="12597" LastEditorUserId="12597" LastEditDate="2017-09-30T10:14:53.197" LastActivityDate="2017-09-30T10:14:53.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17350" PostTypeId="1" AcceptedAnswerId="17358" CreationDate="2017-09-30T09:58:25.223" Score="2" ViewCount="107" Body="&lt;p&gt;This is an above view of a system in which tin cans arrive via conveyor (1) to a &amp;quot;feeding star&amp;quot; (&amp;quot;&lt;strong&gt;подающая звезда&lt;/strong&gt;&amp;quot; in Russian; No. 5) and are placed firmly in slot 4. (&lt;a href=&quot;http://www.freepm.ru/Models/105263.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source in Russian&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OG0oo.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OG0oo.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;What does one call this &amp;quot;feeding star&amp;quot; in English?&lt;/p&gt;&#xA;&lt;p&gt;I'm actually translating a text about a pharmaceutical filling-and-capping line, in which there is a &amp;quot;feeding star&amp;quot; and a &amp;quot;stoppering star&amp;quot;. My text comes with no drawings, but I googled and found out that these &amp;quot;stars&amp;quot; are basically star-like cogwheels that take something from a moving conveyor system and position it precisely for further use. In my text, this &amp;quot;feeding star&amp;quot; probably passes drug-filled vials to the &amp;quot;stoppering star&amp;quot; which is part of the vacuum-operated automatic stoppering unit.&lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="349" LastEditDate="2021-01-11T14:13:30.587" LastActivityDate="2021-01-11T14:13:30.587" Title="What do I call this &quot;cogwheel&quot; that feeds tin cans to a specific position?" Tags="|terminology|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17351" PostTypeId="2" ParentId="17350" CreationDate="2017-09-30T11:56:32.740" Score="2" Body="&lt;p&gt;Sprocket is one possibility, a feed or delivery spider are other possibilities.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="90" LastEditDate="2017-09-30T15:21:28.423" LastActivityDate="2017-09-30T15:21:28.423" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17352" PostTypeId="2" ParentId="17345" CreationDate="2017-09-30T12:01:33.103" Score="1" Body="&lt;p&gt;Correct you need to find the discharge rate of the capacitor. Take a look at &lt;a href=&quot;http://www.mouser.com/pdfdocs/ELNACalcuDYNACAPDischargeTime_Jul2011.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Calculation of Discharge Time&lt;/a&gt;. The equation constant current discharge equation 1. For you equation total capacitance is is 2000uf. Your V1 is 15mA(400)+2V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ltxDh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ltxDh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some good references on this topic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.schoolphysics.co.uk/age16-19/Electricity%20and%20magnetism/Electrostatics/text/Capacitor_charge_and_discharge_mathematics/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mathematical treatment of charging and discharging a capacitor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.electronics-tutorials.ws/rc/rc_1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Home / RC Networks / RC Charging Circuit&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/4951/how-do-i-calculate-how-fast-a-capacitor-will-discharge&quot;&gt;How do I calculate how fast a capacitor will discharge?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.digikey.com/-/media/Images/Supplier%20Centers/T/TT%20Electronics%20Welwyn/tt-electronics-capacitor-discharge-calculator.xls?la=en&amp;amp;ts=452828e7-5a3d-47a6-af87-1bece327a585&quot; rel=&quot;nofollow noreferrer&quot;&gt;Capacitor Discharge Calculator - Digikey&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/4951/how-do-i-calculate-how-fast-a-capacitor-will-discharge&quot;&gt;How do I calculate how fast a capacitor will discharge?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2017-10-03T01:49:21.930" LastActivityDate="2017-10-03T01:49:21.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17353" PostTypeId="1" CreationDate="2017-09-30T15:52:15.940" Score="1" ViewCount="73" Body="&lt;p&gt;I'm working on a deltawing model airplane where I want to implement a PID controller to stabilize the pitch, roll and yaw (which are only controlled by two elevons). However, I don't know too much about the subject. I only working once on a position control PID for a robot, but an aircraft is a bit more complicated. I figured out that I'd start with one rotation first, the pitch since it has the most disturbances during flight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured out that I should first look at the system. The input would be a set point controlled by an external RC controller, which goes through the controller and plant and result into a rotation of the servo motors controlling the elevons. The elevons then give a pitch moment that should lead to the correct set point pitch angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do I go about designing a stabilizing PID controller for a system like this? Thanks for the help!&lt;/p&gt;&#xA;" OwnerUserId="13109" LastActivityDate="2017-09-30T15:52:15.940" Title="PID control for UAV" Tags="|mechanical-engineering|control-engineering|pid-control|aircraft-design|stability|" AnswerCount="0" CommentCount="1" ClosedDate="2017-10-12T12:28:30.373" ContentLicense="CC BY-SA 3.0" />
  <row Id="17354" PostTypeId="1" AcceptedAnswerId="17355" CreationDate="2017-09-30T16:11:52.777" Score="3" ViewCount="132" Body="&lt;p&gt;I read that when you want to double the max speed of e.g. a boat you would have to increase the power eightfold! I could understand that you have to square it (= fourfold) because you are working against &quot;a wall&quot; of water which makes it an area (= power of two)... but why to the power of three?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please give a physical/mathematical explanation but also an intuition on why this is the case. &lt;/p&gt;&#xA;" OwnerUserId="13108" LastActivityDate="2017-10-03T01:38:38.593" Title="Intuitive explanation for why doubling max speed is proportional to eightfold power" Tags="|power|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17355" PostTypeId="2" ParentId="17354" CreationDate="2017-09-30T18:10:31.977" Score="4" Body="&lt;p&gt;Think of it as the volume of water you're displacing per second - the area multiplied by the distance moved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So intuitive I hope I can get away without an equation, but there's some good stuff in the &quot;Power&quot; paragraph here :&lt;a href=&quot;https://en.wikipedia.org/wiki/Drag_%28physics%29#Drag_at_high_velocity&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Drag_%28physics%29#Drag_at_high_velocity&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13043" LastActivityDate="2017-09-30T18:10:31.977" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17356" PostTypeId="2" ParentId="17350" CreationDate="2017-09-30T21:14:27.207" Score="0" Body="&lt;p&gt;The &quot;feeding&quot; and &quot;stoppering&quot; seem to be just literal Russian-to-English translations of &quot;filling&quot; and &quot;capping,&quot; which would be good technical terms in English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would call the &quot;star&quot; an &quot;escapement&quot; by analogy with mechanical clocks etc - it forces the cans to move one at a time in a controlled manner, in the same way that at clockwork escapement allows a gear wheel to move one tooth at a time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=6nhNu-AlAfo&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=6nhNu-AlAfo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Escapement&quot; seems to be used in this context in several US patents for can filling devices that operate on one can at a time, though I couldn't find one that used the same &quot;star&quot; type of device.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-09-30T21:14:27.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17358" PostTypeId="2" ParentId="17350" CreationDate="2017-10-01T01:54:08.953" Score="5" Body="&lt;p&gt;Looks like it may be called &lt;strong&gt;starwheel feeder&lt;/strong&gt; or &lt;strong&gt;starwheel infeed&lt;/strong&gt; - here's an excerpt from &lt;em&gt;&lt;a href=&quot;https://goo.gl/A2FRG3&quot; rel=&quot;noreferrer&quot;&gt;Design of Automatic Machinery&lt;/a&gt;:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6mlAx.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6mlAx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="349" LastActivityDate="2017-10-01T01:54:08.953" CommentCount="0" CommunityOwnedDate="2017-10-03T03:19:19.290" ContentLicense="CC BY-SA 3.0" />
  <row Id="17360" PostTypeId="1" AcceptedAnswerId="17361" CreationDate="2017-10-01T06:44:40.173" Score="3" ViewCount="278" Body="&lt;p&gt;I was browsing for Gibraltar strait tunnel where due to greater depth it is not being build.&#xA;Question popped up as to why tunnel building is difficult at lets say 1000m. we have coal mines which go deeper than that, than why not tunnel?&lt;/p&gt;&#xA;" OwnerUserId="10977" LastActivityDate="2017-10-01T08:14:19.823" Title="Tunnel at depth problem" Tags="|civil-engineering|tunnels|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17361" PostTypeId="2" ParentId="17360" CreationDate="2017-10-01T08:14:19.823" Score="8" Body="&lt;p&gt;&lt;strong&gt;Tunnel Access&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin, to construct and use a tunnel, it needs to be accessed. Access will be via either a vertical or sub-vertical shaft or a ramp that wheeled vehicles can use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shafts may be practical for mines, but for a traffic tunnel access via a shaft is impractical because of the rate of transfer of vehicles from one level to the other. A small number of vehicles will be able to fit into a cage in a shaft &amp;amp; then the cage will be lowered or raised, transporting the vehicles. This is OK for mines but not for a large number of vehicles carrying trade goods or cars with families.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the only practical way of accessing such a tunnel is via a ramp. Irrespective of whether the ramp is straight or spiraled, its overall length will be dictated by the gradient of the ramp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a uniform gradient of 1 in 10 (10 percent), for a tunnel 1000 m (1 km) deep, the horizontal length of the ramp will be 10 000 m and its slope length will be 10 050 m (10.05 km). Such a ramp will need to be constructed at both the entry and exit of the tunnel. Thus, the overall length of the tunnel will need to be an additional 20.1 km. This would add significant cost to the construction of the tunnel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For vehicular traffic, will one tunnel, divided in half length-ways be enough, or will two tunnels be required: one allowing for traffic to flow in one direction and the other tunnel for traffic to flow in the other direction. If two tunnels are needed, then double the cost of construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tunnel Ventilation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Irrespective of the type of vehicles that use the tunnel, electrically powered vehicles or combustion engine powered vehicles, ventilation will need to be provided to the tunnel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ventilation can be provided via ventilation shafts or another tunnel dedicated to ventilation and servicing the vehicular transportation tunnel. If the tunnel passes under water, such as the Straits of Gibraltar, then ventilation shafts cannot be used along the route of the tunnel - a dedicated ventilation tunnel must be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is the cost of powering the ventilation fans and how the energy will be sourced to do so - solar, wind, gas, nuclear, coal power stations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ground Control&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue that will need to be considered is that at 1000 m depth ground stresses may be large which will dictate the form of ground control measures and the subsequent cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tunnel Rock Temperature&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, at 1000 m depth the rock temperature will be high. Depending on location, temperatures of 40 C to 60 C are not unusual. To cater for this, will the vehicular transportation tunnel need to be cooled via an air cooling system. This will cost a lot of money to install and operate. Again, the issue of a power source for such a system will need to be addressed.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-10-01T08:14:19.823" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17365" PostTypeId="2" ParentId="17342" CreationDate="2017-10-01T18:24:47.257" Score="2" Body="&lt;p&gt;In gear design, the most used tooth profile is the involute profile. I am not sure what do you mean by avoiding slipping(Velocity of sliding) because for two gears to move, there will definitely be slip as they have to maintain the contact while they transmit motion(except at the pitch point).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are using a 3D printer, please use the cycloidal profile as a tooth profile which rolls without slipping. The reason people don't use it is the difficulty in mass manufacturing.(Although they are used in clocks and watches)  &lt;/p&gt;&#xA;" OwnerUserId="556" LastActivityDate="2017-10-01T18:24:47.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17366" PostTypeId="2" ParentId="17333" CreationDate="2017-10-01T19:06:17.847" Score="1" Body="&lt;p&gt;It's correct. Even if you take the area to be half of the current area and reduce the velocity of air by a factor of 10, even then you will get turbulent flow.I took these numbers just to put things in perspective. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1.145*1*0.09375/(1.89*10^-5) ~ 5679&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="556" LastActivityDate="2017-10-01T19:06:17.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17368" PostTypeId="1" CreationDate="2017-10-02T04:59:26.493" Score="-1" ViewCount="1679" Body="&lt;p&gt;I have a question concerning if there is any automobile company that has mass &#xA;produced ceramic engines? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not then why is that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried searching but couldn't come up with any answers.&lt;/p&gt;&#xA;" OwnerUserId="13122" LastEditorUserId="12804" LastEditDate="2017-10-03T08:31:46.920" LastActivityDate="2017-10-08T06:34:18.513" Title="Use of ceramics in internal combustion engines - why and why not?" Tags="|ceramics|" AnswerCount="4" CommentCount="0" ClosedDate="2017-10-09T12:07:55.893" ContentLicense="CC BY-SA 3.0" />
  <row Id="17369" PostTypeId="1" CreationDate="2017-10-02T06:27:26.333" Score="-1" ViewCount="556" Body="&lt;p&gt;This parameter defines the structural behavior, defining shear or flexure behavior of the structure. &#xA;How do I use this parameter to define the stiffness matrix of the structure to perform dynamic analyses?                  I am performing RSA manually on a stick model, coding in Matlab. This parameter has no units.(dimensionless, its a ratio). RCC model. This is referred to as ALPHA in &quot;APPROXIMATE SEISMIC LATERAL DEFORMATION DEMANDS IN MULTISTORY BUILDINGS By Eduardo Miranda&quot;&lt;/p&gt;&#xA;" OwnerUserId="13123" LastEditorUserId="13123" LastEditDate="2017-10-02T07:26:12.087" LastActivityDate="2021-02-08T10:00:45.120" Title="Beam-to-Column-Stiffness ratio" Tags="|structural-engineering|structural-analysis|dynamics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17370" PostTypeId="2" ParentId="17368" CreationDate="2017-10-02T06:51:19.800" Score="2" Body="&lt;p&gt;A search gives many links, such as Ford using zirconia ceramics and aircraft manufacturers using ceramics to increase the turbine inlet temperature : images and sources below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source : &lt;a href=&quot;http://ceramicrotaryengines.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ceramicrotaryengines.com&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KIfts.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KIfts.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source : &lt;a href=&quot;http://www.substech.com/dokuwiki/doku.php?id=ceramics_in_combustion_engines&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.substech.com/dokuwiki/doku.php?id=ceramics_in_combustion_engines&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rmZ13.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rmZ13.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-02T06:51:19.800" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17371" PostTypeId="2" ParentId="17368" CreationDate="2017-10-02T10:32:42.067" Score="2" Body="&lt;p&gt;I don't know of anyone who's done this. Ceramic components certainly, but not an entire engine. A significant issue is tensile strength - while compressive strength is excellent with ceramic materials, tensile strength is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ceramic liner for a piston - where it would be surrounded by a metal with high tensile strength and would be operating exclusively in compression - would be a good idea. A ceramic block to contain repeated compression and expansion would be more susceptible to fracture.&lt;/p&gt;&#xA;" OwnerUserId="13043" LastActivityDate="2017-10-02T10:32:42.067" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17372" PostTypeId="2" ParentId="17368" CreationDate="2017-10-02T11:13:27.253" Score="2" Body="&lt;p&gt;From my research, no-one is mass producing complete internal combustion engines. Trials of a reciprocating ceramic engine components were conducted by the &lt;a href=&quot;http://www.ultrahardmaterials.co.uk/engine.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ford Motor Company&lt;/a&gt;, in the USA.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;During the 1980s, in a development program at Ford Motor Company, zirconia-based ceramic components (e.g. a short ceramic cylinder liner) have been successfully tested in reciprocating internal combustion engines for over 500 hours of operation before failure. It was believed that the failure of zirconia ceramic components in internal combustion engines was due to the use of Yttria (Y-TZP) and Magnesia (Mg-TZP) zirconias. These zirconia ceramics are prone to hydrothermal degradation due partly to the moisture present in the fuels (see W.Bunk and H.Hausner &quot;Ceramic Materials and Components for Engines&quot; Proceedings of the Second International Symposium; 14-17 April 1986, Lubeck-Travemunde, FRG). It is believed that these programs were abandoned mainly because of the hydrothermal degradation of the zirconias they used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-10-02T11:13:27.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17373" PostTypeId="1" CreationDate="2017-10-02T15:31:07.823" Score="1" ViewCount="65" Body="&lt;p&gt;I am planning to supply drinking water only (not bath or sanitation or washing)  from a tank on a 10-storey building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that if a gravity feed is used, then the pressure will be different on each floor. For example, the estimated PSI will be only 7 on the top floor and 46 on the bottom. Ideally we want to have a PSI of 40 PSI on every floor. How can this be accomplished?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2017-10-02T16:12:47.730" Title="Managing water supply pressure to a building" Tags="|hydraulics|water-resources|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17374" PostTypeId="2" ParentId="17373" CreationDate="2017-10-02T16:12:47.730" Score="4" Body="&lt;p&gt;The best method is to pressurize your tank on the upper floor, then regulate the flow as it goes down.  This is because pressurizing will require active components (i.e. pumps, and other devices that you do not want to run continuously, or they will cost money while not in use).  Once the tank has 40 psi of pressure, a water pressure regulator at each floor, such as shown below, should be used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CqlXK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CqlXK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of these at each floor can trim the pressure properly to the correct pressure.  &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-10-02T16:12:47.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17375" PostTypeId="1" CreationDate="2017-10-02T22:09:14.443" Score="0" ViewCount="66" Body="&lt;p&gt;If I make a sandwich of wood, thin plastic, carbon-fiber tow (cut to about a foot long) impregnated with epoxy to the right ratio, thin plastic, and wood, and clamp it together and leave it, will I end up with a structure similar to a pultruded carbon-fiber rod?  Will it have anywhere near the compressional strength of a product like these &lt;a href=&quot;http://www.cstsales.com/rod_comp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1]&lt;/a&gt;?  ..I'm not the first to consider this.. &lt;a href=&quot;https://www.rcgroups.com/forums/showthread.php?281722-Homemade-carbon-fiber-rods&quot; rel=&quot;nofollow noreferrer&quot;&gt;[2]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13081" LastEditorUserId="1832" LastEditDate="2017-10-02T22:39:28.877" LastActivityDate="2017-10-03T18:38:26.987" Title="home-made pultruded components?" Tags="|carbon-fiber|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17377" PostTypeId="2" ParentId="17354" CreationDate="2017-10-03T01:38:38.593" Score="4" Body="&lt;p&gt;It's a pretty rough approximation, and only holds where skin friction is the dominant source of drag throughout the speed range considered. Skin friction increases about as speed ^2. So for double speed, the resistance force is quadrupled. Plus the rate is doubled. Power is the rate of doing work, so 4 X force and 2 X rate equals 8 X power. This doesn't work if wave making is significant, and it isn't very good in the planing regime where surface area is changing with speed.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2017-10-03T01:38:38.593" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17378" PostTypeId="1" CreationDate="2017-10-03T07:33:42.590" Score="0" ViewCount="111" Body="&lt;p&gt;Given that momentum is the product of speed and weight (Velocity x Weight) would rapid deceleration or acceleration in weighless space, of Space X's BFR have an impact on its passengers. I ask because if it is to travel at 27000km/h most of its journey would be spent accelerating and decelerating very much limited by earthlike g forces as airliners do, therfore minimizing on the reduction of travel time. Would most of the braking and accelerating have to take place in space at zero gravity?&lt;/p&gt;&#xA;" OwnerUserId="12808" LastActivityDate="2017-10-03T07:58:50.700" Title="Rapid acceleration and deceleration of Space X's BFR" Tags="|acceleration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17379" PostTypeId="2" ParentId="17378" CreationDate="2017-10-03T07:58:50.700" Score="1" Body="&lt;p&gt;The factor here is mass, not weight - weight is the force produced by mass and the effect of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same kinds of forces are produced by acceleration. In &quot;weightless space&quot; (zero gravity) a spacecraft accelerating at 1g (9.81 m/s2) will impose the same force on its occupants as earth's gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a high speed spacecraft, a lot of the time will be spent accelerating and decelerating to keep the force on the occupants within tolerable limits.&lt;/p&gt;&#xA;" OwnerUserId="13043" LastActivityDate="2017-10-03T07:58:50.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17381" PostTypeId="2" ParentId="17375" CreationDate="2017-10-03T11:27:25.983" Score="0" Body="&lt;p&gt;&lt;strong&gt;No, and it would be hard to guess how far off it would be&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main difference is that pultruded rods have a controlled fibre/matrix ratio while an ambient-cured Epoxy will be off, even if you mix properly, as you cannot assure a homogeneous wetting of the tow with hand lay-up techniques and no debulking/vacuum pressing step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, it can work for your application, just don't expect it to be comparable to pultruded rods.&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2017-10-03T11:27:25.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17382" PostTypeId="2" ParentId="17375" CreationDate="2017-10-03T14:07:22.197" Score="1" Body="&lt;h1&gt;No - Pretension can throw off the compression modulus&lt;/h1&gt;&#xA;&lt;p&gt;While the properties can be calculated using formulas (depending upon the levels of reliability and degree of design, as shown in &lt;a href=&quot;https://engineering.stackexchange.com/questions/4182/how-do-i-calculate-an-estimate-for-the-properties-of-a-composite-material&quot;&gt;other answers&lt;/a&gt; on this site), the net result can still be off due to pre-tension in the fibers.&lt;/p&gt;&#xA;&lt;h3&gt;Basis:&lt;/h3&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Composites handle tension well, but are terrible in compression.  This is primarily due to fiber buckling under compression:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qviIW.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qviIW.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;When pultruded components are cured, they are cured while under tension.  &lt;a href=&quot;https://i.stack.imgur.com/itOmr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/itOmr.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The pre-tensioning leaves a residual tensile load in the fibers.  This tensile load can (not always) allow the fibers to absorb more of the compression load before they begin buckling.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The only way to be sure if this has any impact on your parts is to fabricate them using tension and without pre-tension, and verify the compression modulus and strength of each part.  Only after careful testing could you verify if the parts were affected by pre-tension or not.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-10-03T14:07:22.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17383" PostTypeId="1" CreationDate="2017-10-03T15:24:20.683" Score="0" ViewCount="1214" Body="&lt;p&gt;For a project that I am a part of, we are in need of a brake pad material for use against an aluminum surface without causing any damage to it. So far through my research, the only safe alternative that I have found is 100% graphitized carbon brake pads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type of material composition should I be looking at which provides a reasonable coefficient of friction against aluminum while also limiting the wear to the pads and not the aluminum surface? Where would I find the physical properties of such a material?&lt;/p&gt;&#xA;" OwnerUserId="13135" LastActivityDate="2017-10-08T06:29:30.960" Title="Brake pad materials for use against an aluminium surface" Tags="|materials|aluminum|friction|" AnswerCount="4" CommentCount="5" ClosedDate="2017-10-09T12:10:55.700" ContentLicense="CC BY-SA 3.0" />
  <row Id="17384" PostTypeId="2" ParentId="17383" CreationDate="2017-10-03T15:44:30.340" Score="2" Body="&lt;p&gt;Just remember that &quot;soft&quot; materials wear &quot;hard&quot; ones : brake pads on cars wear the steel discs... So, the aluminium will wear as it is friction based.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : What about a water dynamometer? Ie basically pumping water and loading the pump to increase the resistance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One brake that won't wear is a non-friction one : an eddy brake based on electrical current : used on buses , coaches, heavy goods vehicles may be worth considering...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to &lt;a href=&quot;http://www.telmausa.com/products/operating-principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Telma&lt;/a&gt; one of the manufacturers, explaining its function.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-10-03T18:41:52.590" LastActivityDate="2017-10-03T18:41:52.590" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17385" PostTypeId="2" ParentId="17383" CreationDate="2017-10-03T18:17:27.830" Score="0" Body="&lt;p&gt;If your aluminum surface is like a traditional rotor or drum, and significant braking force is required; aluminum does not have good elevated temperature strength and may fatigue/crack.   &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-10-03T18:17:27.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17386" PostTypeId="2" ParentId="17375" CreationDate="2017-10-03T18:38:26.987" Score="1" Body="&lt;p&gt;Many pultruded FRP sucker rods are used in oil well pumping. The only pretension is that load used to pull them through the manufacturing process. String design keeps them in tension ( steel rods in the bottom of the string, FRP at the top). I believe this is because the compression properties are relatively poor. For compressive strength , I doubt that you want to emulate a pultruded rod. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-10-03T18:38:26.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17387" PostTypeId="1" AcceptedAnswerId="17388" CreationDate="2017-10-03T18:58:23.867" Score="1" ViewCount="356" Body="&lt;p&gt;Suppose we have the following truss-like structure, with a load of 1N on the middle joint and two reactions on the bottom joints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JOVqB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JOVqB.png&quot; alt=&quot;Truss&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I try to analyze this using a Cremona diagram, I run into the problem that I do not know how to handle a load on a joint inside the structure. Is it possible to analyze this with a Cremoda diagram, and if so, how does the naming of the different faces and panels work? Because we can only depict forces on joints between faces, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there even a solution?&lt;/p&gt;&#xA;" OwnerUserId="13136" LastActivityDate="2017-10-03T21:32:29.443" Title="Internal load in cremona diagram" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17388" PostTypeId="2" ParentId="17387" CreationDate="2017-10-03T21:27:04.027" Score="2" Body="&lt;p&gt;The vertical reactions at A and B are clearly 0.5N each, by symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are reactions at A and B, those points must be connected to something external. If the connections are fixed points, there is no force in member AB because the length of AB can't change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpreting the diagram literally, there are no horizontal components of the reaction at points A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing AB makes the truss a statically determinate structure and you can solve for the other member forces in the usual way - for example, start at joint A and find the forces in AC and AD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't accept the argument that there is no force in AB, there is no unique answer to the problem, because the structure is statically indeterminate.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-10-03T21:32:29.443" LastActivityDate="2017-10-03T21:32:29.443" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17390" PostTypeId="1" CreationDate="2017-10-03T23:50:55.503" Score="-2" ViewCount="76" Body="&lt;p&gt;I posted this over on physics, they suggested I cross-post it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been looking for a solid square (non-pourus) that can expand to 12x it's size and back again. This needs to grow and shrink on only the x and y axis, I prefer for the z to stay flat on one/both sides but it really doesn't matter whether it changes or not. The expansion and retraction needs to be done smoothly or in small increments and equally (not x first then y or vice versa). This process needs to be done more than once. I was thinking an elastic material but some materials stay permanently stretched once crossing a certain point, staying too long, or show marks of stretching when restored (like a balloon after someone pulls it). Most of the heat-based ideas I've found either don't change to the size requirements or are one-time solutions.&lt;/p&gt;&#xA;" OwnerUserId="13139" LastActivityDate="2017-10-04T10:23:02.727" Title="Is there something that can expand to 12 times it's size on the x &amp; y axis (not concerned about the z), and can still return to it's original shape?" Tags="|design|mechanical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17392" PostTypeId="2" ParentId="17390" CreationDate="2017-10-04T10:23:02.727" Score="0" Body="&lt;p&gt;Lots of synthetic materials like lycra or rubber have even better expansion capabilities. Stretch a lycra sleeve over a pantograph/scissor mechanism frame including diagonal supports that prevent buckling/distortion. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-10-04T10:23:02.727" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17401" PostTypeId="2" ParentId="17383" CreationDate="2017-10-05T10:36:52.020" Score="1" Body="&lt;p&gt;Depending on the loads rubber might work, often used for bicycle brakes on alloy wheel rims. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option might be to use rubber rollers or wheels which contact the rail and fit them with a braking system much like road wheels on a car. The advantage of this is that as long as you don't lock them you are braking vial rolling contact rather than sliding friction directly onto the rail itself which should cause much less wear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously it is also important that the contact surface is kept reasonably clean as any grit that gets between the rail and the braking surface will accelerate wear. One way to achieve this is semim-rigid 'wipers' such as you see on a lathe saddle to keep out swarf. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-10-05T10:36:52.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17403" PostTypeId="1" AcceptedAnswerId="17409" CreationDate="2017-10-05T15:41:13.167" Score="2" ViewCount="206" Body="&lt;p&gt;I need to cut a 7mm solid Aluminium plug (Non-hollow and non-threaded) to replace a part in my car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple drill adapter tool to cut a solid metal cylinder (Plug) from an Aluminium mass ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Hole saw cutters are not in small sizes like 7mm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commercially these plugs are called &quot;Solid Aluminium Rivets&quot;. But 7mm size is not available commercially (Available maximum size is 1/4&quot; or 6.4mm). Actually head part of the rivet is not necessary for the function. It is needed for remove the plug for rebuild. But practically, that head got damaged when I try to remove it by a plug remover. So I had to drill-out the plug. So there is no photo to show. So, since the head is useless, I just need to make a plug to close the Idle port of my old carburetor.&lt;/p&gt;&#xA;" OwnerUserId="13158" LastEditorUserId="13158" LastEditDate="2017-10-06T15:25:37.527" LastActivityDate="2017-10-06T16:02:07.540" Title="Is there a drill adaptor or attachment tool for cut a metal solid cylinder" Tags="|metals|drilling|cutting|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17406" PostTypeId="1" CreationDate="2017-10-05T17:54:42.677" Score="0" ViewCount="969" Body="&lt;p&gt;I am a bit confused about choosing reference signal in order to control state space models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read(however without deep math explanation) that one has to scale his reference signal for to make system to be able to track input signal. Also there is a function which gives a scaling coefficient &lt;code&gt;Nbar&lt;/code&gt;, so I have to multiply my reference signal by it:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = size(A,1);&#xA;Z = [zeros([1,s]) 1];&#xA;N = inv([A,B;C,D])*Z';&#xA;Nx = N(1:s);&#xA;Nu = N(1+s);&#xA;Nbar=Nu + K*Nx;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In my particular case there is the model of a pendulum on a one dimensional cart with arbitrary chosen poles. My state variables are&#xA;${\vec{\textbf{x}}} = [x\ \dot{x}\ \theta\ \dot{\theta}]^{T}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is matlab code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;clear all;&#xA;&#xA;M = 1;&#xA;m = 1;&#xA;l = 1.5;&#xA;g = 9.8;&#xA;I = m*l^2;&#xA;b = 0.05;&#xA;&#xA;denom = M*(m*l^2) + I*(m + m);&#xA;&#xA;a22 = -(b*I + b*m*l^2)/denom;&#xA;a23 = (g*(l^2)*(m^2))/denom;&#xA;a42 = -(b*l*m)/denom;&#xA;a43 = (g*(M+m)*l*m)/denom;&#xA;&#xA;b21 = (I + m*l^2) / denom;&#xA;b41 = l*m / denom;&#xA;&#xA;A = [0 1 0 0; 0 a22 a23 0; 0 0 0 1; 0 a42 a43 0];&#xA;B = [0; b21; 0; b41];&#xA;C = [1 0 0 0];&#xA;D = 0;&#xA;&#xA;% Check for controllability&#xA;co = ctrb(A, B);&#xA;fprintf(&quot;%f\r\n&quot;, rank(co));&#xA;&#xA;% POLES&#xA;P = [-1.5 -0.9 -2.5 -3.5];&#xA;&#xA;% Placing poles&#xA;K = place(A, B, P);&#xA;&#xA;% Reference signal rescaling&#xA;sys = ss(A, B, C, D);&#xA;N = rscale(sys, K);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;My simulink model:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UGZQp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UGZQp.png&quot; alt=&quot;Inverted pendulum on a cart. Simulink&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And position plot:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yWYkH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yWYkH.png&quot; alt=&quot;Inverted pendulum cart position plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything works like a charm. However I totally miss the intuition behind that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine standard PID control, in this case if I would like to control position I would set the error to ´current_position - desired_position´ and apply standard PID formula to that error. From the above it is intuitively clear why position is controlled and if I want to control, for example, velocity I will just set error to velocity divergence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in a state space I have some coefficient which arises after calculating dot product between my state and gain matrix. And for some reason after subtracting it from a scaled version of u I get controlled position. Why is a big secret for me. And how would one control any other state variable in this case, like velocity for example? &lt;/p&gt;&#xA;" OwnerUserId="10483" LastEditorUserId="10483" LastEditDate="2017-10-05T20:20:01.573" LastActivityDate="2017-10-13T18:09:19.003" Title="How to choose reference signal in a state space model?" Tags="|control-engineering|control-theory|matlab|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17407" PostTypeId="1" AcceptedAnswerId="17411" CreationDate="2017-10-05T17:57:39.807" Score="1" ViewCount="393" Body="&lt;p&gt;I was reading a text talking about Joule's experiment(mechanical equivalent of heat)&lt;br&gt;&#xA;The mechanical equivalent of heat was a concept that had an important part in the development and acceptance of the conservation of energy and the establishment of the science of thermodynamics.. The apparatus used in the experiment is shown below: &#xA;&lt;img src=&quot;https://i.stack.imgur.com/O1Tcq.gif&quot; alt=&quot;enter image description here&quot;&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have noticed that there is a spring balance in that apparatus. Why is there a spring balance in the mechanical equivalent of heat apparatus?  What will happen if it is removed? &lt;/p&gt;&#xA;" OwnerUserId="9742" LastActivityDate="2017-10-05T23:40:06.970" Title="Why is there a spring balance in the mechanical equivalent of heat apparatus" Tags="|mechanical-engineering|thermodynamics|heat-transfer|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17408" PostTypeId="2" ParentId="17407" CreationDate="2017-10-05T19:33:09.493" Score="1" Body="&lt;p&gt;To load the mechanism and be able to record and repeat that loading,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But more interesting information is found in the following link, about measuring heat through boring cannons , stirring water and using electricity ie current flow in electrically conducting materials :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.aps.org/publications/apsnews/201506/physicshistory.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;James Prescott Joule&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-10-05T20:03:35.333" LastActivityDate="2017-10-05T20:03:35.333" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17409" PostTypeId="2" ParentId="17403" CreationDate="2017-10-05T20:24:47.957" Score="3" Body="&lt;p&gt;There are cutters that produce circular, slightly tapered plugs, typically 4-12mm diameter. &lt;a href=&quot;https://i.stack.imgur.com/ojDvA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tapered plug cutter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tEwZo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tEwZo.jpg&quot; alt=&quot;Tapered plug cutter diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Used in a rigid drill press or milling machine with slow feeds and plenty of lubricant these can cut plugs from soft metals as well as wood and plastic.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-10-05T20:24:47.957" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17410" PostTypeId="1" AcceptedAnswerId="17412" CreationDate="2017-10-05T21:46:49.833" Score="1" ViewCount="40" Body="&lt;p&gt;I am doing a project looking into the advantages of radio/microwave heating of oil reservoirs. I've seen research indicating that RF/Microwave heating can be used for environments such as shallow, tight, high permeability, fractured, etc zones. What are the disadvantages of applying techniques such as SAGD in these zones?&lt;/p&gt;&#xA;" OwnerUserId="13161" LastActivityDate="2017-10-06T01:57:53.827" Title="What types of oil reservoirs are applicable to radio/microwave heating and advantageous to SAGD?" Tags="|geotechnical-engineering|rf-electronics|petroleum-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17411" PostTypeId="2" ParentId="17407" CreationDate="2017-10-05T23:40:06.970" Score="1" Body="&lt;p&gt;The spring balance is used to load a &lt;a href=&quot;https://en.wikipedia.org/wiki/Prony_brake&quot; rel=&quot;nofollow noreferrer&quot;&gt;Prony brake&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2017-10-05T23:40:06.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17412" PostTypeId="2" ParentId="17410" CreationDate="2017-10-06T01:57:53.827" Score="0" Body="&lt;p&gt;Heating oil makes it less viscous which makes it easier to flow and thus easier to pump. RF/microwave heating is an easy and relatively safe way of reducing the viscosity of the oil a  reservoir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the greater the permeability of a rock mass the more easily a fluid will flow through it. If a rock mass has large pore spaces, such as sandstone, the more permeable it will be. For rock which has small/tight pore spaces the permeability can be increased by fracturing the rock mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SAGD (steam assisted gravity drainage) is another way to reduce the viscosity of the oil. The effectiveness of SAGD would rely on how precise the horizontal steam holes were drilled relative to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rock mass have varying properties and there may be softer and harder portions within a rock mass. This can affect the path of the drill bit and steel when the holes are drilled. Consequently, the holes may not always be exactly where they are required for efficient SAGD recovery of the oil. Some parts of the reservoir may be cooler or hotter than needed. Applying additional heat via RF/microwave heating will increase the likelihood of more of the oil being heated and having its viscosity reduced and thus potentially increasing the recovery of oil from the reservoir.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-10-06T01:57:53.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17413" PostTypeId="1" CreationDate="2017-10-06T02:59:31.243" Score="2" ViewCount="1079" Body="&lt;p&gt;I have an object (shown in green) that I want to pull out distance X and then I want another object (shown in red) to move 2X. What would be the best way to achieve this?  I thought that gear racks connected to each and then linked by a gear train might work, but I can't quite figure out the best way to lay that out (or if there is a simpler way to do this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CLrYu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CLrYu.png&quot; alt=&quot;Objects moving&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3777" LastActivityDate="2017-10-07T06:56:47.593" Title="Use two gear racks to multiply linear motion" Tags="|gears|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17415" PostTypeId="2" ParentId="17413" CreationDate="2017-10-06T06:44:57.000" Score="1" Body="&lt;p&gt;On the toothed gear track have gears (cogs) where one is twice the diameter of the other : with the same rotational input the larger one will move twice the distance...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-06T06:44:57.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17416" PostTypeId="2" ParentId="17403" CreationDate="2017-10-06T11:42:38.317" Score="2" Body="&lt;p&gt;It depends a but what you mean by M7 size. Specifying M7 rather than 7 mm diameter implies that the hole is threaded. If this is the case you are better off with a bit of M7 threaded rod cut to length. The best material will depend on the application aluminium male threads are rare but do exist. In most cases stainless should be ok. If you do use a metal other than aluminium galavnic corrosion may be an issue but using thread locking or anti-seize compound should mitigate that pretty well unless it is constantly exposed to water or coolant (in which case a standard thread may not provide adequate sealing anyway). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If  you do mean a plain 7mm diameter hole then you can just buy 7mm diameter rod and cut it to length, although getting it to stay in the hole may not be trivial and you might be better off tapping the hole anyway. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit : based on the additional information in the question I would be tempted to tap the hole with an M8 thread and use a grub screw with some thread locking compound as the plug. If sealing is an issue you could also use a button head set screw with a soft washer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually if you set a 6.4 mm solid rivet properly it the shank should expand to fill a 7mm hole but it sounds like this will be difficult in this case as I assume you have limited access to the back side of the hole. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-10-06T16:02:07.540" LastActivityDate="2017-10-06T16:02:07.540" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17417" PostTypeId="2" ParentId="63" CreationDate="2017-10-06T11:48:05.273" Score="2" Body="&lt;p&gt;Something that I didn't see mentioned here is a HUGE saving in wasted fuel, burnt by cars and truck sitting at a traditional intersection, waiting for a light to change. To me, this is the best feature. Roundabouts are planet friendly period!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another spinoff to this is the reduced frustration to drivers who are sitting, waiting and waiting and waiting...&lt;/p&gt;&#xA;" OwnerUserId="13164" LastActivityDate="2017-10-06T11:48:05.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17418" PostTypeId="2" ParentId="17413" CreationDate="2017-10-06T12:56:50.753" Score="1" Body="&lt;p&gt;Make one rack static connect a rack parallel to that connected to object X2 and add a gear sandwiched between them with the axle connected to object X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That way as object X moves the other object X2 will move twice as far.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-10-06T12:56:50.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17419" PostTypeId="2" ParentId="11493" CreationDate="2017-10-06T13:27:33.427" Score="-2" Body="&lt;p&gt;i think series combition of radiatiors would be good because water has cooled two times in one round.&lt;/p&gt;&#xA;" OwnerUserId="13167" LastActivityDate="2017-10-06T13:27:33.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17421" PostTypeId="1" AcceptedAnswerId="17468" CreationDate="2017-10-06T15:10:48.693" Score="3" ViewCount="302" Body="&lt;p&gt;I'm currently trying to extend the shock-absorbing capabilities of a welded steel structure that is shown in the following image. It is a structure that supports a slat conveyor. A load is to fall approximately in the middle half of the structure. The base of the structure rests in a concrete foundation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to expand the shock-absorbing capabilities because I'm worried about the structure failing due to pure impact forces. I'm not as worried by vibrations as by the maximum force exerted on it due to the impact. The load impact is difficult to quantify, even for a worst case scenario, but it is safe to say that the structure has resisted an impact force that is 25% lower than the one that is to fall now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HuQwD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HuQwD.png&quot; alt=&quot;Welded steel structure&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would prefer not to change all the beam sizes, since the slat conveyor that rests on top of it is already assembled, but I could add some supports if needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;em&gt;Design of Welded Structures&lt;/em&gt;, however, making the beam span between supports shorter raises the impact stresses suffered by the structure, even if it lowers the static stresses (once the load is in place).Is there anything I can do to minimize these impact stresses?&lt;/p&gt;&#xA;" OwnerUserId="12823" LastEditorUserId="12823" LastEditDate="2017-10-09T15:03:28.010" LastActivityDate="2017-10-09T20:27:34.937" Title="How can I absorb shock on an existing steel structure?" Tags="|structural-engineering|civil-engineering|structures|dynamics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17423" PostTypeId="2" ParentId="17413" CreationDate="2017-10-06T15:46:45.063" Score="3" Body="&lt;p&gt;An alternative to gears is to use a parallelogram mechanism, much like a Pantograph. All you need to do is have 2 parallelograms on top of each other the other with 2 times as long edges while sharing the 2 of the edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; if you allready have an vertical slider then you can simplify the mechanism to 2 cranks essentially eliminating 3 of 6 parts. Though the gear has less parts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fqpdd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fqpdd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: A conceptual sketch of a parallelogram methanism&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have quite a leeway on designing a longer vs wider parallelogram. Even vary the relative speed profile if you so wish. Even though &lt;em&gt;image 1&lt;/em&gt; shows the front most item moving more you can shift the link down wherever you want. Or you can hinge the thing upside down or search a inversion alternate config of this mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do similar things with a &lt;em&gt;Peaucellier–Lipkin linkage&lt;/em&gt; or a Chebyshev's Lambda Mechanism if you wish. But tehy are more involved.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-10-07T06:56:47.593" LastActivityDate="2017-10-07T06:56:47.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17426" PostTypeId="2" ParentId="17323" CreationDate="2017-10-07T01:27:38.860" Score="1" Body="&lt;p&gt;The answer is, we don't know what compressor you have so we can't answer definitively. MANY compressors are indeed designed to operate at a very narrowly specific speed because the lubrication of it is dependent on that speed. Also, are you working with a compressor that would use a 3 phase motor? Because for the most part VFDs only work on 3 phase motors. There are a few that work on SOME types of single phase motors, but the type of motors that will work are not the types that are typically used on compressors.&lt;/p&gt;&#xA;" OwnerUserId="13176" LastActivityDate="2017-10-07T01:27:38.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17428" PostTypeId="1" CreationDate="2017-10-07T05:26:40.493" Score="0" ViewCount="317" Body="&lt;p&gt;QUESTION:&lt;a href=&quot;https://i.stack.imgur.com/gWLOe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gWLOe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I needed help here, I tried solving but still can't prove or arrive at the conclusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;my working: &lt;a href=&quot;https://i.stack.imgur.com/iwajK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iwajK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12830" LastActivityDate="2017-10-07T08:05:02.217" Title="hydraulic diameter" Tags="|fluid-mechanics|dynamics|hydraulics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17429" PostTypeId="1" CreationDate="2017-10-07T05:50:05.387" Score="0" ViewCount="726" Body="&lt;p&gt;This&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gB04y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gB04y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure what to do. At first I ignored what is said about using an integrator and drew the block diagram of the system, which looked a little complicated. So now, using just integrators I am stuck. If we try just taking the first system &lt;/p&gt;&#xA;&#xA;&lt;p&gt;x'1 = x1 + 4u&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if we want to us an integrator, where y' = u, how can we ever draw this while satisfying the previous equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Actually I think I got your hint. Is my system correct ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gB04y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gB04y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13179" LastEditorUserId="13179" LastEditDate="2017-10-08T20:59:00.493" LastActivityDate="2017-10-08T20:59:00.493" Title="Block Diagram using Integrator (Linear Systems Theory - Hespanha)" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17430" PostTypeId="2" ParentId="17428" CreationDate="2017-10-07T08:05:02.217" Score="3" Body="&lt;p&gt;You have the correct answer, you just haven't simplified the terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Root 3 divided by 3 is 1 over root 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ie. x^(1/2)  / x = x^(1/2).x(-1) = x^(-1/2)&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-10-07T08:05:02.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17431" PostTypeId="2" ParentId="17429" CreationDate="2017-10-07T09:11:26.057" Score="1" Body="&lt;p&gt;You can use the following block diagram. Note that your $\boldsymbol{D}=\boldsymbol{0}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gXKxX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gXKxX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to have all the states you could draw it like the following picture. You can neglect all gains with $1$, if gain blocks are not allowed (for whatever crazy reason) then you can replace them by an equivalent number of signals to the summation block. eg. gain of 3 for $x_2$ will be split into three separate lines all connected to the summation block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i3Pnf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i3Pnf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-10-08T16:23:40.273" LastActivityDate="2017-10-08T16:23:40.273" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17432" PostTypeId="1" CreationDate="2017-10-07T12:19:46.220" Score="-2" ViewCount="70" Body="&lt;p&gt;What is meant by &amp;quot;depth of member&amp;quot; and what is the formula of it?&#xA;Can I use it in moment of inertia formula?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V41KN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V41KN.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;How to use the given information to answer the question?&#xA;and Thank you.&lt;/p&gt;&#xA;" OwnerDisplayName="user12132" LastEditorUserId="-1" LastEditorDisplayName="user16" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-10-08T00:44:14.903" Title="HELP Please! Depth of member" Tags="|structural-engineering|civil-engineering|structural-analysis|structures|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17434" PostTypeId="2" ParentId="16989" CreationDate="2017-10-07T18:06:44.693" Score="0" Body="&lt;p&gt;You could compare &lt;a href=&quot;https://en.wikipedia.org/wiki/Spar_buoy&quot; rel=&quot;nofollow noreferrer&quot;&gt;spar buoys&lt;/a&gt; to &lt;a href=&quot;https://en.wikipedia.org/wiki/Weather_buoy#/media/File:Mooredbuoytypes.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;discus buoys&lt;/a&gt; in terms of spring constant of equivalent models, and introduce the idea of driver response as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pedagogically speaking, oscillating systems are part of the engineering environment. We need to recognize, accommodate, and sometimes mitigate their effects more so than design and construct them. Harmonic oscillation is an analysis tool, not a design tool.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2017-10-07T18:06:44.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17438" PostTypeId="1" CreationDate="2017-10-07T21:13:12.880" Score="1" ViewCount="34" Body="&lt;p&gt;I have customized a fan for our centrifuge. How may I design it? Can you recommend me some software / resources ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the the specifications: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type : Radial&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Capacity : 5000 Nm3/h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Head : 100 millibar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed : 2925 RPM&lt;/p&gt;&#xA;" OwnerUserId="13185" LastActivityDate="2017-10-07T21:13:12.880" Title="How may I design a centrifugal fan?" Tags="|design|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17441" PostTypeId="2" ParentId="17432" CreationDate="2017-10-08T00:44:14.903" Score="3" Body="&lt;p&gt;The depth of a member is merely equal to its total height (or whichever total dimension is parallel to the loading direction).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-10-08T00:44:14.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17444" PostTypeId="1" CreationDate="2017-10-08T03:57:57.420" Score="0" ViewCount="92" Body="&lt;p&gt;It really has no informative signs on it just signs warning not climb or careful with your ball. It is in a Japanese park. I thought it is a water mound with its lots of pipes coming out it for aquifer to water the park. &lt;a href=&quot;https://i.stack.imgur.com/popi2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/popi2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GzAb6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GzAb6.jpg&quot; alt=&quot;Tightly stonework like a Japanese castle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U04e5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U04e5.jpg&quot; alt=&quot;Some barbed wire fenced off plants&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="401" LastActivityDate="2017-11-25T04:10:32.033" Title="What is this structure in a park" Tags="|civil-engineering|hydrology|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17445" PostTypeId="1" CreationDate="2017-10-08T06:20:07.450" Score="0" ViewCount="12040" Body="&lt;p&gt;As most westerners know well, ceramic toilets are designed to be sat on, even by those who are quite overweight. However, since it's customary in Asian countries to squat when using the toilet, some visitors to western countries sometimes squat on the bowl of the toilet when using it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When sitting, a proportion of the weight of the user is supported by the legs, but when squatting the user's entire weight is borne by the bowl. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, and at least one report of a squatting traveller breaking a toilet bowl (!), leads me to wonder about the safety of this practice. I haven't been able to find any reliable scientific discussion of this online. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much more force is borne by the toilet when standing/squatting on it? How strong is a typical ceramic toilet bowl? Is it safe to squat on a toilet?&lt;/p&gt;&#xA;" OwnerUserId="13188" LastActivityDate="2021-12-31T10:20:28.787" Title="Is it safe to stand on a toilet bowl?" Tags="|safety|strength|ceramics|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17446" PostTypeId="2" ParentId="17383" CreationDate="2017-10-08T06:29:30.960" Score="0" Body="&lt;p&gt;a vehicle going 500 MPH will contain a huge quantity of kinetic energy which the brakes will need to dissipate. This will be difficult to do without affecting the 6061 aluminum either by heating it to the point where its alloy characteristics are lost or by rubbing material off the aluminum component. for example, consider a large airliner. at full gross weight for takeoff, a panic stop from the decision (takeoff or abort) speed (about 150-200 MPH) down to zero will leave the brake discs on the main landing gear red-hot, blow out the tires, and set them on fire. it is unreasonable to require a stop from 500 MPH to not damage the aluminum part.&lt;/p&gt;&#xA;" OwnerUserId="13189" LastActivityDate="2017-10-08T06:29:30.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17447" PostTypeId="2" ParentId="17368" CreationDate="2017-10-08T06:34:18.513" Score="-1" Body="&lt;p&gt;an engine block, for example, must resist both compressive loads and tensile loads. Ceramics are at a disadvantage compared to metals in this regard because they exhibit brittle failure in tension. &lt;/p&gt;&#xA;" OwnerUserId="13190" LastActivityDate="2017-10-08T06:34:18.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17448" PostTypeId="1" CreationDate="2017-10-08T07:08:55.813" Score="1" ViewCount="745" Body="&lt;p&gt;I have been trying to create a machine to climb coconuts as the existing machines are very slow and strenuous to use and frankly, most climbers prefer to climb it manually than use the existing machine, which all more or less use the same concept of the image given below &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Search Google for more).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanted to automate that process to make it easier and faster. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/nkHcu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nkHcu.jpg&quot; alt=&quot;Coconut Climber&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been experimenting with machines and designs like using circular drums and powering them, using raised platforms attached to vehicles but none of them have properly materialized. So now after many futile attempts, I turn to the engineering community to help me and suggest new ideas on their own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to this and I don't know if I can ask such a question. &#xA;But still I just wanted to try. Give me any ideas if you can. If you feel that this is not appropriate, please comment below and I will correct it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You&lt;/p&gt;&#xA;" OwnerUserId="13191" LastActivityDate="2017-10-09T10:16:21.743" Title="Coconut Climbing Machine Ideas" Tags="|mechanical-engineering|robotics|mechanical-failure|" AnswerCount="1" CommentCount="16" ClosedDate="2017-10-09T12:05:11.523" ContentLicense="CC BY-SA 3.0" />
  <row Id="17449" PostTypeId="2" ParentId="17445" CreationDate="2017-10-08T08:17:38.620" Score="1" Body="&lt;p&gt;Safe for the Toilet Bowl : Yes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Safe for you : If you can balance yourself properly ,Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have climbed on the toilet bowl lots of time and nothing has happened&lt;/p&gt;&#xA;" OwnerUserId="13191" LastActivityDate="2017-10-08T08:17:38.620" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17450" PostTypeId="2" ParentId="17445" CreationDate="2017-10-08T09:51:36.167" Score="2" Body="&lt;p&gt;Toilet bowls aren't all the same so it impossible to say. Some may take it some may not. There are several factor at play.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When climbing on the bowl you are putting load onto one leg and have some impulse going on for lifting the other leg. Depending on where you step this increased load can be stressful for the bowl. While the bowl may generally be able to handle this depending on where you step and how much you weight this may cause the bowl to break (unexpectedly).&lt;/p&gt;&#xA;&lt;p&gt;You can lessen this impact by putting some weight on your hand. Alternatively stepping of some sort of ladder/stool lessens the chance of a accident considerably but that would probably eliminate the need to stand on the bowl in the first place.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The lid on the bowl may buckle under the load and give way. So while the bowl itself may be fine you may not be very good for you.&lt;/p&gt;&#xA;&lt;p&gt;Now lids differ quite much. If you have a hard thick than probably this is not a problem, but for softer chap plastic lids this may in fact be a problem.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The problem is that what is good for the bowl and what is good for the lid are not the same. So the lid stresses as little as possible if you stand on the rim the bowl.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is often a entirely unnecessary risk to take, so you shouldn't take it. Many jurisdictions simply do not allow people to work on platforms not designed to be climbed on for just this reason. Falling from even a slightly elevated platform is a risk factor for serious injury.&lt;/p&gt;&#xA;&lt;p&gt;Unless you really regularly save some significant time by using the bowl as a platform (thus maximizing risk) i wouldn't consider this risk even having any positive time saving features. It is more like bad planning instead.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;So would i stand on the bowl? I would avoid it. Have I done so, yes in the past i have, but i was much younger and lighter then. Like i said not worth a risk taking.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-10-08T09:57:03.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17451" PostTypeId="2" ParentId="17445" CreationDate="2017-10-08T11:06:31.173" Score="3" Body="&lt;p&gt;There is another safety issue - the way the bowl is attached to the structure of the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the UK, the only structural attachment is by two (large) brass screws passing through cast holes in the bowl and screwed into the wooden floor. Since the top of the bowl usually overhangs the base, applying a large off-centered load while climbing on and off the bowl could break the screws or pull them out of the wood, since they were never intended to resist a large &lt;em&gt;upwards&lt;/em&gt; force, but only to stop the bowl moving &lt;em&gt;sideways&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-10-08T11:06:31.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17452" PostTypeId="1" AcceptedAnswerId="17453" CreationDate="2017-10-08T16:59:11.033" Score="0" ViewCount="176" Body="&lt;p&gt;Viscosity is resistance to the motion of fluid layers sliding over one another. Density is a measure of forces of attraction between atoms or molecules of same species. Is there any relation between the two? &lt;/p&gt;&#xA;" OwnerUserId="12974" LastActivityDate="2017-10-08T17:51:47.670" Title="Relation between density and viscosity" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17453" PostTypeId="2" ParentId="17452" CreationDate="2017-10-08T17:51:47.670" Score="1" Body="&lt;p&gt;No, mercury is very dense but pours easily&#xA;Honey is lighter but is very viscous - (changes with temperature though)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density and viscosity are two different characteristics of a fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Viscosity has the units Poise for dynamic viscosity and Stokes for kinematic viscosity.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-08T17:51:47.670" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17454" PostTypeId="1" AcceptedAnswerId="17492" CreationDate="2017-10-08T18:18:39.300" Score="0" ViewCount="94" Body="&lt;p&gt;I have a homework which should solve by me. My problem is questions are really simple or should I think outside of the box? Like, bode diagram, nyquist or etc.? And, are my answers correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Question-1&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$G(s) = K\dfrac{As+1}{Bs+1}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For which values $K, A$ and $B$ is the system always stable? Should I look directly to the pole of the system?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$Bs+1=0$&lt;/li&gt;&#xA;&lt;li&gt;$s=-1/B \implies  \text{So, must } B&amp;gt;0$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is it enough? Or, anything else? What about K, A?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Question-2&lt;/h3&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$$G(s) = K\dfrac{As+1}{(Bs+1)(Cs+1)}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For which values $K, A, B$ and $C$ is the system always stable? Should I look directly to the pole of the system or anything else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Bs+1=0 \wedge Cs+1=0$$&#xA;$$s=-1/B \wedge s=-1/C$$&#xA;$$\implies  B&amp;gt;0 \wedge C&amp;gt;0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it enough? Or, anything else? What about $K$ and $A$?&lt;/p&gt;&#xA;" OwnerUserId="13198" LastEditorUserId="7627" LastEditDate="2017-10-10T14:49:21.160" LastActivityDate="2017-10-11T11:37:57.453" Title="How to solve system's general stability from transfer function?" Tags="|control-engineering|transfer-function|stability|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17455" PostTypeId="1" AcceptedAnswerId="17463" CreationDate="2017-10-08T19:09:01.987" Score="1" ViewCount="397" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3FPvG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3FPvG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m following Anderson’s Fundamentals Of Aerodynamics. However, the proof for the above theorem is not provided. Every time I try it, things get really lengthy and cumbersome. Is there some nice and short method to getting around to prove the time rate of change of circulation over a chose closed loop 0 in an inviscid, incompressible flow ? &lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2018-02-06T14:18:03.513" Title="Proving Kelvin’s Circulation theorem" Tags="|aerodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17457" PostTypeId="1" AcceptedAnswerId="24476" CreationDate="2017-10-08T21:34:32.057" Score="1" ViewCount="753" Body="&lt;p&gt;If the force in member OA is 0.25F and inclined at $\alpha$ degree from the horizontal. Calculate the angle $\beta$ to keep the system in equilibrium and calculate the angle $\omega$. $\alpha$ = 20°. F = 100 N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zONCH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zONCH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By alternate interior angles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e8Hra.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e8Hra.jpg&quot; alt=&quot;poorly drawn alterior angles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\beta - \alpha$ is the angle between OB line and horizontal axis x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum of forces in X direction should be equal to 0:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum F_{x} = -0.25F \cdot \cos(\alpha) -F_{OB} \cdot \cos(\beta - \alpha) + 100N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum F_{y} = 0.25F \cdot \sin(\alpha) - F_{OB} \cdot \sin(\beta - \alpha)= 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is this correct? I'm not sure what $F_{OB}$, it's not given. So I have two equations, and two unknowns: $F_{OB}$ and $\beta$. I tried to solve via &lt;a href=&quot;http://www.wolframalpha.com/widgets/view.jsp?id=ae438682ce61743f90d4693c497621b7&quot; rel=&quot;nofollow noreferrer&quot;&gt;wolframalpha&lt;/a&gt; but it didn't find any solutions. What do?&lt;/p&gt;&#xA;" OwnerUserId="12190" LastEditorUserId="1832" LastEditDate="2017-10-08T22:18:23.047" LastActivityDate="2019-06-27T18:07:04.047" Title="Equilibrium of forces in 2d plane, find angle alpha and w" Tags="|civil-engineering|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17459" PostTypeId="1" CreationDate="2017-10-09T09:22:00.467" Score="4" ViewCount="1639" Body="&lt;p&gt;I am currently collecting accelerometer and attitude data using smartphone sensors. Data is collected at 100Hz over 2.5 seconds. Of which there are 250 accelerometer points and 250 points representing the attitude of the device. I am looking to convert the attitude/accelerometer data to gyroscope Data. For example, 1 of the 250 segments look like;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Accelerometer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;AccX = 0.08458628 &lt;/li&gt;&#xA;&lt;li&gt;AccY = -0.0586758144&lt;/li&gt;&#xA;&lt;li&gt;AccZ = 0.273877382&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attitude&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pitch = 0.111681767&lt;/li&gt;&#xA;&lt;li&gt;Roll = -0.8779668&lt;/li&gt;&#xA;&lt;li&gt;Yaw = 1.91596484&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the corrected gyroscope data from this data? &lt;/p&gt;&#xA;" OwnerUserId="12410" LastEditorUserId="12410" LastEditDate="2017-10-09T10:09:16.577" LastActivityDate="2018-06-26T04:47:45.457" Title="Calculating gyro data from accelerometer, pitch, roll and yaw" Tags="|mechanical-engineering|sensors|data|acceleration|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17460" PostTypeId="1" CreationDate="2017-10-09T09:56:07.157" Score="0" ViewCount="151" Body="&lt;p&gt;Starting from a piece of Mild Steel that is then Pack Carburized, what's the maximum Carbon content that the surface can soak in? Given that we keep it around less than 1000°C and enough time, is it limited to 6.67%, the Carbon content of Cementite?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be helpful also if you can describe the depth of penetration of the Pack Carburization, given enough time. Something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Casedepth ~ $K \sqrt{time}$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where K is the Diffusivity constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, that's wrong as I think there's a limit to the depth considering the alloy, exact temperature and dimension.&lt;/p&gt;&#xA;" OwnerUserId="1915" LastEditorUserId="1915" LastEditDate="2017-10-09T10:02:35.527" LastActivityDate="2019-08-31T04:06:55.850" Title="Maximum Carbon content on the surface of Mild Steel that is Pack Carburized" Tags="|materials|steel|alloys|heat-treatment|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17462" PostTypeId="2" ParentId="17448" CreationDate="2017-10-09T10:16:21.743" Score="1" Body="&lt;p&gt;How about a &quot;&lt;strong&gt;Cherry Picker&lt;/strong&gt;&quot; type machine?  They are readily available for hire or purchase in most countries and come in all shapes and sizes...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Apparently I am not allowed to post pictures of cherry picker machines here  :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small machine with outrigger legs and a small diesel motor could easily (and safely) allow access to the treetops and might even be faster than manual climbing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Placing the machine equidistantly between two or more adjacent trees might make it feasible to harvest from multiple trees without moving the machine (the hydraulic arm can pivot 360 degrees around the fixed base).&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastActivityDate="2017-10-09T10:16:21.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17463" PostTypeId="2" ParentId="17455" CreationDate="2017-10-09T12:23:38.837" Score="1" Body="&lt;p&gt;The circulation is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma = \oint_{C(t)}\boldsymbol{u}d\boldsymbol{r}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, take the material derivative of this expression (remember to use the product rule because our contour is time-dependent) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{D \Gamma}{Dt} =\oint_{C(t)}\dfrac{D \boldsymbol{u}}{Dt}d\boldsymbol{r}+\oint_{C(t)}\boldsymbol{u}d\left[\dfrac{D\boldsymbol{r}}{Dt}\right].$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we assume that the flow in inviscid and all volume forces are conservative (have a potential $\Psi$). Such that we can use Euler's equation for the first integral &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{D \Gamma}{Dt} =\oint_{C(t)}\left[-\dfrac{\nabla  p}{\rho}-\nabla \Psi \right]d\boldsymbol{r}+\oint_{C(t)}\boldsymbol{u}d\boldsymbol{u}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term on the right-hand side is integrable and has an anti-derivative of $\boldsymbol{u}^2$. As we are integrating on a circle (starting point = endpoint) it must be zero. If we assume that $\rho(p)$, which is called barotropic, then we can conclude that the closed contour integral on the left-side is also zero. Note, that the closed contour integral for conservative fields $-\nabla \Psi$ is zero, also called path independence. Hence, we conclude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{D \Gamma}{Dt} =0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that for inviscid, barotropic flow's and conservative volume forces the circulation $\Gamma$ is conserved.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-10-09T12:28:58.353" LastActivityDate="2017-10-09T12:28:58.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17464" PostTypeId="2" ParentId="17460" CreationDate="2017-10-09T16:35:07.130" Score="1" Body="&lt;p&gt;I don't know about a computer game, but in the real world max C in the skin is more like 1.5 % with time like 24 hours. Carbide formers like Cr, Mo , will increase that slightly.Also there are trade secrets in the &quot;pack&quot; , alkali metals and earths can facilitate carburization and affect C content. High surface C has few applications so most commercial carburization is done in gases and limits surface carbon to numbers like 0.8 %. Carburization in gas is directly proportional to square root of time . But something else in a pack ( with long times) because some reactants are consumed and not replaced in the pack.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-10-09T16:35:07.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17468" PostTypeId="2" ParentId="17421" CreationDate="2017-10-09T20:27:34.937" Score="4" Body="&lt;p&gt;Large displacement shock absorbers used to dampen the effects of seismic events may be suitable for this application.  These could be used to isolate the part of the structure that is grounded and the part that experiences the impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seismic dampers sit between the ground slab and the building and can effectively reduce the shock loadings in all directions.  Hydraulic dampers are often used to attenuate forces in a specific direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dampers are available in many sizes and formats, from simple steel springs to sophisticated viscous piston systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3NfDL.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3NfDL.jpg&quot; alt=&quot;Steel spring used in a building footing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aChhs.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aChhs.jpg&quot; alt=&quot;Hydraulic dampers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-10-09T20:27:34.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17470" PostTypeId="1" CreationDate="2017-10-09T23:37:25.530" Score="1" ViewCount="754" Body="&lt;p&gt;Page No. 318, John Anderson’s “Fundamentals Of Aerodynamics”.&lt;a href=&quot;https://i.stack.imgur.com/gUW3J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gUW3J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author here states from Kelvin’s circulation theorem that the initial circulation around a closed loop including the airfoil when the flow is stationary, which is 0, must be equal to the final value of circulation in that closed loop once the flow has reached a steady state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay. So far, so good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then, he argues that the starting vortex’s negative circulation is the reason for the formation of the positive circulation around the airfoil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the above can only be true when the airfoil, once steady-state flow has been reached, is still a part of the closed loop formed by those initial fluid elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, here’s the thing. Those initial fluid elements have by now, crossed the airfoil. There’s no obligation that the airfoil must at all times remain enclosed in that loop, made by those very same fluid elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, then Kelvin’s circulation theorem simply cannot be applied in the way the author presents it for the airfoil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, then the starting vortex should  not be the reason for the development of circulation around the airfoil, or should it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some clarity needed. &lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2017-10-10T07:47:26.987" Title="An Argument from Kelvin’s Circulation Theorem" Tags="|aerodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17471" PostTypeId="1" CreationDate="2017-10-10T01:22:58.253" Score="0" ViewCount="104" Body="&lt;p&gt;I have a question for all experienced in manufacturing of steel products;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible, and or workable to resistance spot weld dissimilar metal thicknesses together? for example, 1.5-2mm thick steel to 3mm thick steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And please elaborate on the limits... such as the minimum thickness to the maximum thickness that is able to be welded reliably and quickly in a manufacturing environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help :).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thomas&lt;/p&gt;&#xA;" OwnerUserId="13218" LastActivityDate="2017-10-10T14:54:44.700" Title="Resistance spot welding multiple metal thinknesses" Tags="|steel|manufacturing-engineering|welding|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17474" PostTypeId="2" ParentId="17470" CreationDate="2017-10-10T07:37:34.573" Score="1" Body="&lt;p&gt;Normally you would apply Kelvin's circulation theorem on the whole system such that everything that is important is enclosed by the contour for all times $t$ (note, that the author also expanded the contour form the first to the second picture). The system is consisting of the wing and surrounding air has a circulation of zero at the beginning. As soon as we observe a relative motion between a wing and surrounding air we can observe that there is a starting vortex generated. As we assume that Kelvin's circulation theorem is valid (conditions: inviscid + barotropic flow and volume forces are conservative) we know that a vortex with opposing orientation has to be created to balance the starting vortex circulation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that circulation is defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Gamma = \oint_{C(t)}\boldsymbol{u}d\boldsymbol{r}.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your system boundaries must always include both vortices. The mathematical explanation is that a vortex is nothing but a mathematical singularity. The velocity $\boldsymbol{u}_{\text{V}}$ of such a vortex is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\boldsymbol{u}_{\text{V}}=\dfrac{\Gamma_0}{2\pi r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is easy to see that $\boldsymbol{u}_{\text{V}}$ blows up for $r \to 0$.&#xA;As soon as this singularity is at the system boundary it will cross the contour $C(t)$, hence the circulation is not well defined from this point onwards because it is the line integral over this contour. If you start from the point in which the contour only encloses one of the vortices then you can again apply Kelvin's circulation theorem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{D\Gamma}{Dt} = 0 \implies \Gamma = \text{const.}$$&#xA;but now the initial circulation is equal to $\Gamma = \text{const.}$ and will stay like this as long as the vortex (=singularity) will not cross the boundary.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-10-10T07:47:26.987" LastActivityDate="2017-10-10T07:47:26.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17475" PostTypeId="2" ParentId="17454" CreationDate="2017-10-10T08:08:19.813" Score="0" Body="&lt;p&gt;You did answer both questions correctly. You could also consider solving the both questions by using the Hurwitz criterion. The first one is directly solvable with the Hurwitz criterion the second one is a little bit involved :).&lt;/p&gt;&#xA;" OwnerUserId="7627" LastActivityDate="2017-10-10T08:08:19.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17480" PostTypeId="2" ParentId="17471" CreationDate="2017-10-10T13:00:53.510" Score="1" Body="&lt;p&gt;Yes, but it may be necessary to pre-heat the thicker part - this depends on the relative thickness differences as well. You should set the current setting for the thinner of the two parts - otherwise excessive damage to that part may result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More detail here : &lt;a href=&quot;https://www.longevity-inc.com/resources/articles/welding_thick_to_thin_parts&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.longevity-inc.com/resources/articles/welding_thick_to_thin_parts&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-10T13:00:53.510" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17481" PostTypeId="1" CreationDate="2017-10-10T13:29:14.437" Score="0" ViewCount="110" Body="&lt;p&gt;I'm working on an FEA model and my mesher uses a &quot;Marching Voxel method&quot; I'm wondering what the relationship is between mesh size and computation time is? I.e. if I half the mesh size (1mm to 0.5mm) is the computation time for the tighter resolution doubled? quadrupled? eight times longer?&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-10-10T13:29:14.437" Title="Marching Voxel FEA Mesh" Tags="|finite-element-method|meshing|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17482" PostTypeId="2" ParentId="17471" CreationDate="2017-10-10T14:54:44.700" Score="1" Body="&lt;p&gt;I never heard of preheating for spot welding. For greater metal thickness , the technique of &quot;Projection&quot; welding is used. A type of spot weld where one component has a small raised area where the weld will occur. I think it is unnecessary for the thicknesses you have.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-10-10T14:54:44.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17483" PostTypeId="1" AcceptedAnswerId="17672" CreationDate="2017-10-10T17:07:24.540" Score="0" ViewCount="126" Body="&lt;p&gt;Just to leave some space here I link to an introduction of linear transformation of state. How is it useful? Can nayone provide an example where it is indeed useful and has to be done?&#xA;&lt;a href=&quot;http://www.staff.ul.ie/burkem/Teaching/st-tr.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.staff.ul.ie/burkem/Teaching/st-tr.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8884" LastActivityDate="2017-10-30T13:39:46.033" Title="Why do we use linear transformation of states in linear control technique" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17486" PostTypeId="1" AcceptedAnswerId="17490" CreationDate="2017-10-11T01:43:43.130" Score="0" ViewCount="1006" Body="&lt;p&gt;Can a Plasma Cutter use for melt metals like Iron, Brass, Stainless Steel ect, ?. Particularly for molding purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If can't, is there any other electric item that doesn't use gases (Acetylene, Butane, Oxygen) for melt metals ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to heat a mass of metal up to the melting point and pour that melted metal to a Plaster Parris impression cast. I want to find a electric solution to melt the metal instead of gas.&lt;/p&gt;&#xA;" OwnerUserId="13158" LastEditorUserId="13158" LastEditDate="2017-10-11T08:13:23.940" LastActivityDate="2017-10-11T10:49:19.010" Title="Can a Plasma Cutter use for melt metals?" Tags="|metals|molding|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17487" PostTypeId="1" CreationDate="2017-10-11T02:32:57.127" Score="2" ViewCount="174" Body="&lt;p&gt;Its function was to expel the air that came inside the cylinder along with the steam to prevent the engine to become &quot;wind logged&quot;. Since the valve was non-return, how did Newcomen stopped steam from leaving after pushing out the air?&lt;/p&gt;&#xA;" OwnerUserId="13235" LastActivityDate="2017-10-11T09:15:18.307" Title="How did the snifting valve in Newcomen's steam engine work?" Tags="|mechanical-engineering|steam|mechanical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17488" PostTypeId="2" ParentId="17487" CreationDate="2017-10-11T09:15:18.307" Score="2" Body="&lt;p&gt;The snifter valve allowed air and some steam to escape - this did not differentiate between air and steam, just opened briefly when steam was entering.&#xA;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Newcomen_atmospheric_engine&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Newcomen_atmospheric_engine&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opening on early engines was manual - but as they increased in speed this had to become automatic which led to the &quot;Potter Cord&quot; and later a beam with cams and tappets to control the valves.&#xA;See &lt;a href=&quot;https://www.gracesguide.co.uk/Newcomen_Engine&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.gracesguide.co.uk/Newcomen_Engine&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-11T09:15:18.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17489" PostTypeId="1" AcceptedAnswerId="17498" CreationDate="2017-10-11T09:52:48.413" Score="0" ViewCount="184" Body="&lt;p&gt;In my application the liquid would be tapped from a fast-varying flowrate, and I want to measure the static pressure change during this by measuring the air pressure above it. Will this work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dVFa0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dVFa0.jpg&quot; alt=&quot;If the image doesn&amp;#39;t work, all it shows is just a fixed volume of gas connected to the top of a liquid filled tube OK&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11394" LastEditorUserId="7083" LastEditDate="2017-10-14T23:16:57.467" LastActivityDate="2017-10-14T23:16:57.467" Title="Is the pressure of the gas identical to the static pressure of the liquid in the diagram?" Tags="|mechanical-engineering|pressure|sensors|compressible-flow|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17490" PostTypeId="2" ParentId="17486" CreationDate="2017-10-11T10:49:19.010" Score="3" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plasma cutters use a jet of air along with the arc to cut they pretty thoroughly oxidise the metal, plus the jet would just spray molten metal everywhere if you tried to contain it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of electrical only heating an induction furnace is efficient and scalable, for low melting point metals like pewter and possible some brasses a straightforward restive heating kiln should work but will be less efficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also plaster isn't suitable for casting ferrous metals, the temperatures involved are way too high. Sand or ceramic shell moulds are required for iron and steel. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-10-11T10:49:19.010" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17492" PostTypeId="2" ParentId="17454" CreationDate="2017-10-11T11:37:57.453" Score="3" Body="&lt;p&gt;I would like to extent the already given answer by MrYouMath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So question 1 is pretty straight forward and you already got it right. If there's no right half plane (RHP) pole then it doesn't matter what gain you chose. Even for $A = B$, $G(s) = K$ yields a finite response. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Question 2 have a look at the Routh Hurwitz Array&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array} {|r|r|}&#xA;\hline&#xA;s^2 &amp;amp; B \cdot C &amp;amp; 1 \\&#xA;\hline&#xA;s^1 &amp;amp; B+C &amp;amp; 0\\&#xA;\hline&#xA;s^0 &amp;amp; 1 &amp;amp; \\&#xA;\hline&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the system to be stable there must not be any sign changes in the first column, hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$BC &amp;gt; 0 \quad \land \quad B+C &amp;gt; 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $BC &amp;gt; 0$ we derive that B and C must have the same sign. $B+C &amp;gt; 0$ yields that the sign has to be positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see neither $A$ nor $K$ are involved in that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to explore other methods like root locus, bode, ... Keep in mind that you have variables ($A$,$B$,$C$) in there. I know that you can see the gain margins for root loci in Python, Matlab, etc. but I think that's it. I don't think (but I stand to be corrected) that you can derive the values for $A$,$B$,$C$ that way. I think with Bode plots this may work, however as you've seen it's much easier to solve with Hurwitz or by just looking at the poles.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2017-10-11T11:37:57.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17494" PostTypeId="1" AcceptedAnswerId="17495" CreationDate="2017-10-11T14:09:25.460" Score="2" ViewCount="59" Body="&lt;p&gt;Suspension often requires different ranges of behavior. Small amplitude and high sensitivity for vibrations and small sharp bumps but also high amplitude and low sensitivity for larger smoother road deformation.&#xA;So my question is whether or not it's possible to use two suspension system on top of the other.&#xA;I was thinking of a small fast shock (strong spring with weak damper) for small amplitude/high frequency and then a long slow shock (weaker spring with a stronger damper) for high amplitude/low frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main concern is of course interactions between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any example of this setup?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="13240" LastActivityDate="2017-10-11T17:30:51.137" Title="Is it possible to &quot;add&quot; suspension systems?" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17495" PostTypeId="2" ParentId="17494" CreationDate="2017-10-11T15:00:21.333" Score="1" Body="&lt;p&gt;There are manufacturers of progressive springs - soft for the first part of the movement then harder for the rest of the travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also shock absorber manufacturers that make various types of shock - depends on the applications you want and what is available &quot;off the shelf&quot; ie fairly standard or you can go to them with your requirements and they will make them to your specifications (for road use within limits, off road ....), but they won't be cheap..!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: also check out Citroen for hydraulic suspension (variable) and the older mini for hydrolastic... and also landrover / Range rover for adaptive settings...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-10-11T17:30:51.137" LastActivityDate="2017-10-11T17:30:51.137" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17496" PostTypeId="2" ParentId="17494" CreationDate="2017-10-11T15:15:40.127" Score="1" Body="&lt;p&gt;The short answer is 'yes,' since each absorber has its own frequency response curve.&lt;br&gt;&#xA;Now, the ideal setup is an active absorber with, say, piezo-driven adjusters which respond to the initial impact profile.  There is a prototype out there somewhere of an active sensor which is designed to recognize road bumps and essentially turn off all damping, so that the wheel assembly goes up, over, and down the bump without transmitting any vertical force to the car body.  More gentle (less delta-functionish) bumps are treated differently, to keep the overall ride as smooth as possible. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-10-11T15:15:40.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17497" PostTypeId="2" ParentId="17494" CreationDate="2017-10-11T16:21:33.850" Score="0" Body="&lt;p&gt;My answer will probably sound simplistic, but examples of this are more common than people think - the spring/damper arrangement of most vehicle suspensions, coupled with a pneumatic tyre, does exactly what you're describing (tyre = low amplitude high frequency, spring/damper = high amplitude, low frequency).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I think joojaa was thinking along similar lines, and any wheel with a damped elastic response will have this effect.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are much smarter ways of doing it, as Mike and Carl have indicated.&lt;/p&gt;&#xA;" OwnerUserId="13043" LastActivityDate="2017-10-11T16:21:33.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17498" PostTypeId="2" ParentId="17489" CreationDate="2017-10-11T18:13:15.943" Score="0" Body="&lt;p&gt;Yes, the pressure will be identical (with some qualifications). If you think about it, the gas will compress until the force on the boundary between the gas and liquid are equal (i.e., the pressure). Even with a compressible liquid, this should also be the case. You will want to have a reasonable initial pressure for your gas, as well - it may not work as well if your gas volume shrinks to a very small size in response to a very high fluid pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will get some amount of gas dissolving into the liquid, according to Henry's law. I'm not sure whether that will be an issue with the fluid you are working with. The gas will evaporate out of the liquid / dissolve in response to the pressure increasing and decreasing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How quickly the gas responds to the increase or decrease in pressure will be another question. It will depend on the gas you are using (based on whether it is monatomic, diatomic, molecular, etc - based on its capacity to store and release energy in various modes). It will also depend on the degree of fluctuation of pressure, as well as the temperature of the gas and the size of the volume pictured here. If either the gas or fluid are supercritical, or if the initial gas pressure is at a low vacuum, things will get weird, but as long as you stay within the regimes where ideal gas assumptions are relevant you should be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to note is that a big tap combined with a small chamber should reach equilibrium pretty quickly. You may want to be careful about getting into very high pressures though if you are designing the seals / shell for this pressure vessel. Using inert gases is obviously preferable. Additionally, if you're using a pressure transducer, you may want to fit it with a snubber to prevent damage from fluid hammer.&lt;/p&gt;&#xA;" OwnerUserId="3583" LastEditorUserId="3583" LastEditDate="2017-10-11T18:21:52.733" LastActivityDate="2017-10-11T18:21:52.733" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17499" PostTypeId="1" CreationDate="2017-10-11T21:03:38.020" Score="0" ViewCount="190" Body="&lt;p&gt;I am an engineering student and recently came upon some articles that provide numerical models that relate electromagnetic fields and temperature. I want to test out these numerical models but am unsure how to begin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was suggested to use either MatLab or Python. Are there any resources that will teach me how to use either to solve numerical models that include differential equations and system of equations?&lt;/p&gt;&#xA;" OwnerUserId="13161" LastEditorDisplayName="user16" LastEditDate="2017-10-11T23:36:45.877" LastActivityDate="2017-10-13T07:12:40.480" Title="What program to implement mathematical models?" Tags="|modeling|mathematics|" AnswerCount="3" CommentCount="3" ClosedDate="2017-10-16T17:43:47.363" ContentLicense="CC BY-SA 3.0" />
  <row Id="17500" PostTypeId="2" ParentId="17499" CreationDate="2017-10-11T21:26:16.870" Score="3" Body="&lt;p&gt;&quot;Numerical Recipes&quot; is probably what you need the most.  It contains tons of useful scientific/numerical computation codes. It's a book, but I believe it was also available online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from that, you probably need some more theoretical background, and some programming experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Linear Algebra&quot; lectures would help a lot with solving sets of equations. Ocw.mit.edu has open courses available, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fortran, C/C++, Python and Matlab would all help you model and solve the equations for you. And there are several good textbooks and open source materials on the Internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matlab takes the most investment (unless it's already licensed at your institution). It has many solvers already built in, and it handles matrices extremely well. Plus it has built in visualization capability. (Though it takes just a few lines of code in the other languages to come up with some graphs, or you could write your data to a file and use another application for plotting). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd rather learn how to do the computations yourself, Fortran and C/C++ would be the best teaching way to go.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-10-11T21:26:16.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17504" PostTypeId="1" CreationDate="2017-10-12T02:30:33.987" Score="-1" ViewCount="328" Body="&lt;p&gt;I'm wondering if molten iron was hurled into outer space if it might create a shape similar to a Prince Rupert's drop. Like, say, if a star had finally exhausted it's fuel and exploded. &lt;/p&gt;&#xA;" OwnerUserId="13244" LastActivityDate="2017-10-12T17:29:04.523" Title="Iron Prince Rupert's drop" Tags="|materials|thermodynamics|structures|" AnswerCount="2" CommentCount="5" ClosedDate="2017-10-13T02:15:03.473" ContentLicense="CC BY-SA 3.0" />
  <row Id="17506" PostTypeId="2" ParentId="17504" CreationDate="2017-10-12T06:45:33.720" Score="4" Body="&lt;p&gt;Unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Space is a vacuum.  Vacuums are very poor thermal conductors (consider Thermos flasks), so the iron ball would cool rather slowly which is precisely the opposite of what you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dropping it into something like a very large vat of liquid nitrogen (with forced circulation), might yield something structurally similar to the glass drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However metal is a great deal more flexible than glass when solid, so anything that would normally cause a glass PRD to shatter spectacularly would probably only cause the metal version to snap in half in a rather disappointing manner.&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastActivityDate="2017-10-12T06:45:33.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17507" PostTypeId="2" ParentId="17499" CreationDate="2017-10-12T08:11:37.227" Score="2" Body="&lt;p&gt;I can wholeheartedly recommend a Python and the Numpy &amp;amp; Scipy libraries, there are plenty of good sources available to learn python and the Frameworks, for free and on the internet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Python is a really good language to learn and it enforces good coding practices by design, as well as being easy to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have access to Matlab and simulink, it may already have the models you require built in, but it costs a significant amount of money unless you have a licence at your institution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is very opinion based, and may therefore be closed.&lt;/p&gt;&#xA;" OwnerUserId="12804" LastEditorUserId="12804" LastEditDate="2017-10-13T07:12:40.480" LastActivityDate="2017-10-13T07:12:40.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17508" PostTypeId="1" CreationDate="2017-10-12T09:24:26.510" Score="2" ViewCount="305" Body="&lt;p&gt;I am conducting research to determine vibration limits for cultural heritage objects (e.g. art, small sculpture, ceramics). The vibration source can be, for example, transport, construction in or next to a museum, or rock concerts in or next to a museum. As a former materials scientist and engineer in the aircraft industry working on component lifing, I am fully aware that vibrations are a form of cyclic load, and that such damage is cumulative. The limits have to be based on some form of an S-N (Wöhler) diagram for fatigue life. S is normally given in stress amplitude, but one could also use displacement/strain, which can be converted to stress. However, in the vibration monitoring world, vibrations are given either in velocity or acceleration, and sometimes power density spectra. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For buildings, vibration criteria are generally given in units of velocity, but in transportation, they are often given in g, and sometimes PDS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been using velocity as a measure for S in the S-N diagrams, since that is what building engineers use, and it is the closest I can get to deformation (one derivative). If I give a vibration limit in velocity (and duration) and the frequency range in which it is valid, then I am indirectly giving a deformation, which is what one sees in building codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the transport industry, acceleration (g) appears to be used more often. I assume that this has to do with the fact that the cause of damage is more likely to be shock than continuous vibrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question: which unit, velocity or acceleration, correlates better to damage due to vibrations, and why? &lt;/p&gt;&#xA;" OwnerUserId="13246" LastActivityDate="2018-05-27T22:30:42.240" Title="Units for vibration measurements" Tags="|vibration|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17509" PostTypeId="2" ParentId="17508" CreationDate="2017-10-12T12:09:35.707" Score="1" Body="&lt;p&gt;Acceleration, since that translates directly to the force on the object.  Constant velocity is irrelevant.  Imagine some object on a jet plane traveling fast but smoothly.  The high velocity itself doesn't cause problems.  It's the sudden change in velocity that does, and that's exactly what acceleration is.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-10-12T12:09:35.707" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17510" PostTypeId="2" ParentId="17499" CreationDate="2017-10-12T12:18:03.803" Score="0" Body="&lt;p&gt;I would look into the subject of &quot;Numerical Methods&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are textbooks that explain numerical methods for many different scenarios; from solving basic systems, to numerical solutions to partial differential equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May also be worth checking your program to see if you will have any courses on this topic.  It's a pretty powerful tool, and I'd actually be pretty surprised if your program didn't have some sort of numerical methods course, or at least a lot of coverage in other courses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The textbook I had (can't remember the author or exact title) was old (1970s/80s, always talked about FORTRAN); but the methods given were generic and could be applied to any language quite easily.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastActivityDate="2017-10-12T12:18:03.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17511" PostTypeId="1" CreationDate="2017-10-12T12:24:02.567" Score="0" ViewCount="209" Body="&lt;p&gt;I have the following parameters in my project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://uupload.ir/files/gpl9_11111.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;enter link description here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/01b2f.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/01b2f.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I submitted a paper. Now, one of the Reviewer's comments is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Validation is not enough and convincing. The CFD results should compared with static and dynamic torque and efficiency performance with LES simulations available in the literature and or experimental measurements.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the exact meaning of &quot;static and dynamic torque&quot;?&#xA;I am grateful that give me useful links and papers in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My used turbulence model in my project is K-W SST(RANS simulation). Also, there are not any LES simulation available in this my field. My work is a 3D turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2017-10-12T12:31:42.450" LastActivityDate="2017-10-12T12:31:42.450" Title="What is the exact meaning of &quot;static and dynamic torque&quot;?" Tags="|turbines|cfd|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17512" PostTypeId="2" ParentId="17508" CreationDate="2017-10-12T13:15:13.750" Score="0" Body="&lt;p&gt;In automotive, we typically develop our durability vibration test profiles based on real world acceleration data using equivalent damage. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-10-12T13:15:13.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17513" PostTypeId="2" ParentId="17504" CreationDate="2017-10-12T17:29:04.523" Score="0" Body="&lt;p&gt;There are a few reasons why this probably wouldn't work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly in order to be hardenable iron needs a fairly specific set of additional alloying elements ie it needs to be steel. For maximum hardness and hardenability you are looking at around 0.7 - 1.2 % carbon much less and it won't get very hard at all, much more and you are getting more towards cast iron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also plain carbon steel needs pretty rapid cooling to harden, generally in the order of a few second from critical temperature (around 800c)  to below the transition temperature (around 200C)  and this generally requires a quenching medium, typically oil or water depending on the mass and section thickness involved. There are steels which will harden in a stream of air but these tend to have a fairly high chromium content eg &lt;a href=&quot;http://www.steelexpress.co.uk/toolsteel/A2-Steel-properties.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;A2 tool steel&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that while space is cold heat transfer tends to be slow because radiation is the only mechanism available and I suspect that it won;t cool fast enough to harden. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you could get a workable alloy from stellar reactions I don't really know. Meteoric iron can certainly contain nickel but I've never heard of naturally occurring steel as such. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-10-12T17:29:04.523" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17518" PostTypeId="1" CreationDate="2017-10-13T06:08:43.340" Score="1" ViewCount="37" Body="&lt;p&gt;Can I use Fiber microscope to see micron sized particles?&#xA;That's mean, can Fiber microscope be used as a typical optical microscope?&#xA;&lt;a href=&quot;https://goo.gl/JWhMqe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Generic Handheld 400x Fiber Optical Microscope Inspection&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j5O73.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j5O73.jpg&quot; alt=&quot;Fiber microscope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13256" LastActivityDate="2017-10-13T06:08:43.340" Title="Use Fiber microscope to view micron sized particles" Tags="|optics|fiber-optics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17519" PostTypeId="1" CreationDate="2017-10-13T17:09:00.247" Score="2" ViewCount="212" Body="&lt;p&gt;What is the name for a protrusion of material that's built to be flexible in order to act as a latching mechanism? These show up in all kinds of products, but one of the most recognizable is in a snap-fit buckle (below). Is there a name for this component?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KKarE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KKarE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13259" LastActivityDate="2017-10-13T17:18:04.883" Title="Terminology: Flexible protrusion for a latch" Tags="|mechanical-engineering|terminology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17520" PostTypeId="2" ParentId="17519" CreationDate="2017-10-13T17:18:04.883" Score="2" Body="&lt;p&gt;&quot;Snap-fit&quot; uses the same idea generally for any assembled part using elastic regime of the material. &lt;a href=&quot;https://en.wikipedia.org/wiki/Snap-fit&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Snap-fit&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13261" LastActivityDate="2017-10-13T17:18:04.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17522" PostTypeId="2" ParentId="17406" CreationDate="2017-10-13T18:09:19.003" Score="2" Body="&lt;p&gt;So assuming that we are dealing with LTI systems of the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\dot{x} &amp;amp;= A\,x + B\,u \\&#xA;y &amp;amp;= C\,x + D\,u&#xA;\end{align} \tag{1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $x\in\mathbb{R}^{n_x}$, $u\in\mathbb{R}^{n_u}$, $y\in\mathbb{R}^{n_y}$ and which is stabilizable and detectable, however I will assume that the full state is known so no need for constructing an actual observer. The goal is to find a control law for $u$ such that as time goes to infinity the steady state of the output, $y_{ss}$, is equal to a known constant reference, denoted by $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally when you use state feedback of the form $u = -K\,x$ such that $A-B\,K$ is Hurwitz, then the only thing that this will do is force the full state $x$ to zero and in turn the the output $y$ as well. So this would not solve the problem. In order to solve the problem stated above one also needs to find what the steady state of the full state, $x_{ss}$, and input, $u_{ss}$, would need to be such that $y_{ss} = r$. Since the model is detectable $\dot{x}$ should be zero when $y = r$. For now I will assume that $x_{ss}$ and $u_{ss}$ can be expressed as a linear combination of $y_{ss}$ and thus $r$, so $x_{ss} = M_x\,r$ and $u_{ss} = M_u\,r$. When we start to solve the problem you will see that such matrices can exists which solve the problem. Namely if we plug this into equation $(1)$ we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;0_{n_x\times n_x} &amp;amp;= (A\,M_x + B\,M_u)\,r \\&#xA;r &amp;amp;= (C\,M_x + D\,M_u)\,r.&#xA;\end{align} \tag{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this should be solvable for every $r$. So factoring out $r$ then this can be rewritten as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;A &amp;amp; B \\ C &amp;amp; D&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;M_x \\ M_u&#xA;\end{bmatrix} = &#xA;\begin{bmatrix}&#xA;0_{n_x\times n_x} \\ I_{n_y\times n_y}&#xA;\end{bmatrix} \tag{3}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be solved with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;M_x \\ M_u&#xA;\end{bmatrix} = &#xA;\begin{bmatrix}&#xA;A &amp;amp; B \\ C &amp;amp; D&#xA;\end{bmatrix}^{-1}&#xA;\begin{bmatrix}&#xA;0_{n_x\times n_x} \\ I_{n_y\times n_y}&#xA;\end{bmatrix}. \tag{4}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a solution if the matrix that needs to be inverted is full rank and if $n_y \leq n_u$ (when $n_y &amp;lt; n_u$ you could need to use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Generalized_inverse#Construction&quot; rel=&quot;nofollow noreferrer&quot;&gt;right inverse&lt;/a&gt;). If $n_y &amp;gt; n_u$ then it might still be solvable, but not for every $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming equation $(4)$ can be solved then by defining a new state variable $\hat{x} = x - x_{ss}$ and new control input $\hat{u} = u - u_{ss}$, then if a control law exist that brings $\hat{x}$ to zero, then $x$ will go to $x_{ss}$ as well by definition. The time derivative of $\hat{x}$ can be expressed as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{\hat{x}} = \dot{x} - \dot{x}_{ss} = A\,x + B\,u - \underbrace{(A\,x_{ss} + B\,u_{ss})}_{0} = A\,(x - x_{ss}) + B\,(u - u_{ss}) = A\,\hat{x} + B\,\hat{u}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated before using the control law $\hat{u} = -K\,\hat{x}$ such that $A-B\,K$ is Hurwitz, should force $\hat{x}$ to zero. By combining the definitions for $\hat{x}$, $\hat{u}$ and the control law for $\hat{u}$ one can solve for $u$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;u = -K\,x + K\,x_{ss} + u_{ss}. \tag{5}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting in the expressions for $x_{ss}$ and $u_{ss}$ as a function of $r$ gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;u = -K\,x + \underbrace{(K\,M_x + M_u)}_N\,r. \tag{6}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using equation $(4)$ then the gain multiplied by $r$ can also be expressed as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;N = \begin{bmatrix}&#xA;K &amp;amp; I_{n_u \times n_u}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;A &amp;amp; B \\ C &amp;amp; D&#xA;\end{bmatrix}^{-1}&#xA;\begin{bmatrix}&#xA;0_{n_x\times n_x} \\ I_{n_y\times n_y}&#xA;\end{bmatrix}. \tag{7}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this gain $N$ will change by choosing a different $K$. However it can be noted that if you design a $K$ and corresponding $N$ based on a state space model whose dynamics not exactly matches the system you are using the control law on, due to some error during the identification of this system, then the steady state error might not go to zero as time goes to infinity. A solution for this might be to extend the state space by the integral of the error $y-r$, such as the &lt;a href=&quot;https://nl.mathworks.com/help/control/ref/lqi.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;LQI controller&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2017-10-13T18:09:19.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17523" PostTypeId="1" AcceptedAnswerId="17525" CreationDate="2017-10-14T05:09:48.997" Score="2" ViewCount="525" Body="&lt;p&gt;I know that Formula One and some sports cars have KERS systems to recover kinetic energy, but is there an expense/practicality barrier to installing them in consumer vehicles? As the technology clearly serves efficiency purposes beyond just providing extra horsepower, and as KERS can capture energy electrically or mechanically, it seems that it would be a useful step towards increased consumer vehicle efficiency.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I recognize that there's a significant cost associated with thermoelectric capture, and that &lt;a href=&quot;https://engineering.stackexchange.com/questions/389/thermoelectric-technology-to-harvest-energy-from-internal-combustions-engines&quot;&gt;this suggested duplicate&lt;/a&gt; raises the concern of how it's impractical to slow heat dissipation. However, if we limit the energy capture to brakes (typically cooled by natural convection and therefore not subject to any air/liquid cooling), could we see a more practical energy capture solution than trying to capture kinetic energy straight from an internal combustion engine? &lt;/p&gt;&#xA;" OwnerUserId="13266" LastActivityDate="2017-11-13T12:57:14.987" Title="Why don't most consumer vehicles use kinetic energy recovery systems?" Tags="|mechanical-engineering|thermodynamics|regenerative-braking|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17524" PostTypeId="2" ParentId="17083" CreationDate="2017-10-14T05:23:44.853" Score="1" Body="&lt;p&gt;Couple the stepper motor to the high speed spindle motor only indirectly by means of magnets rotating around the spindle motor shaft which, at low speeds, will align themselves with specially designed ferrous spots fixed to the high speed spindle shaft and sync with it. The magnets should create a detent force strong enough to move the spindle at slow speeds, but would break free at higher speeds, particularly if the stepper was held in place electrically.&lt;/p&gt;&#xA;" OwnerUserId="11222" LastActivityDate="2017-10-14T05:23:44.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17525" PostTypeId="2" ParentId="17523" CreationDate="2017-10-14T09:00:39.310" Score="2" Body="&lt;p&gt;It is used - check out buses / coaches but size / weight / cost tend to preclude it for cars - also the centrifugal forces can cause catastrophic failure so the enclosing drum adds weight as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a quick search, an interesting paper shows that it could be applied to cars at a cost of approx $2000 - for a possible fuel saving of 35% - given how the bean-counters make adjustments to cars in the cents range the chances of it becoming mainstream are limited. Unless, of course, the Government legislate it in - chances of that with all the lobbyists....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pk4Fw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pk4Fw.png&quot; alt=&quot;Abstract form source below&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.ch/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=8&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0ahUKEwi8tYfj3u_WAhWE7RQKHRr3A2oQFghaMAc&amp;amp;url=http%3A%2F%2Fwww.mdpi.com%2F1996-1073%2F8%2F10%2F10636%2Fpdf&amp;amp;usg=AOvVaw3rFfUNMeDuvzwcLZSogR28&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.ch/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=8&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=0ahUKEwi8tYfj3u_WAhWE7RQKHRr3A2oQFghaMAc&amp;amp;url=http%3A%2F%2Fwww.mdpi.com%2F1996-1073%2F8%2F10%2F10636%2Fpdf&amp;amp;usg=AOvVaw3rFfUNMeDuvzwcLZSogR28&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-14T09:00:39.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17533" PostTypeId="1" CreationDate="2017-10-15T12:42:01.050" Score="1" ViewCount="573" Body="&lt;p&gt;I have a five meter long, 2.40 meter wide, 200 kg portable catamaran I've built at home. To transport, I bought a semi-finished trailer, just the running chassis; and built the upper structure, namely the supports for the catamaran.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally they build trailers for a typical mono-hull boats where most of the weight is on the rear side, so they build the chassis such that the wheels and the axle are closer to the end rather than being near the middle of the length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my catamaran is uniform in weight along its length and most of its weight is on the front part (before the axle). Having the axle in the middle seemed more reasonable, so I changed its place:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LSj5c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LSj5c.png&quot; alt=&quot;The Car - Trailer System&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I measured the amount of weight exerted on the coupling, it was as low as 30 kgs. Not a problem, but when I tested launching the boat (on the ground), with just about half a meter sliding the catamaran backwards, the trailer's bow took off into the air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It needed about 10 kgs to come down to the ground. If I stayed on the trailer during the launch it would not be a problem, but this caused me to think of another possibility - possible instability during driving uphill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LdwuN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LdwuN.png&quot; alt=&quot;Unstable conditions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to formulate the problem in terms of forces and moments. There are two critical pivot points: The trailer wheels (axle) which I believe initiates the unstable action, based on the net moment on the axle. And the coupling which may cause car wheels to take off if the moment is non zero on it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;What maximum incline uphill would cause this instability?  Can an equation be provided with a variable axis position so I can optimize the stability using this equation?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13276" LastEditorUserId="2015" LastEditDate="2017-10-24T18:28:23.060" LastActivityDate="2017-10-28T15:28:37.700" Title="Maximum slope of the road for a car pulling a trailer" Tags="|mechanical-engineering|structural-analysis|statics|dynamics|stability|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17536" PostTypeId="1" CreationDate="2017-10-15T17:55:37.207" Score="0" ViewCount="39" Body="&lt;p&gt;First a bit of background.  I'm building a movie theater in my basement, but the walls are owens finishing system (insuation with fabric), so mounting anything to the walls is impossible.  I want to hang 3 movie poster frames on the back wall.  I had an idea yesterday, but before I start cutting I need to determine if it will work.  Hopefully this is an easy engineering calc, just a bit beyond my expertise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am installing kitchen cabinets on the back wall, with full height cabinets on the left and right side leaving 8' in the middle for counter height cabinets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to attach an aluminum bar to the tall cabinets (spaning 8') to attach the 3 poster frames.  Each one weighs 13 lbs, so 50lbs would be the max load.  I would like the 6 holes in the bar to be slotted if possible to make adjustment horizontally possible.  I would need to know the width and thickness of the bar needed so no sag is experienced, but it is as invisible as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help&lt;/p&gt;&#xA;" OwnerUserId="13280" LastActivityDate="2017-10-16T18:20:36.083" Title="Theater Build Help" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17538" PostTypeId="1" AcceptedAnswerId="17550" CreationDate="2017-10-16T02:21:09.867" Score="0" ViewCount="200" Body="&lt;p&gt;There is a national level competition &lt;strong&gt;Ecokart&lt;/strong&gt; (Electric kart design &amp;amp; manufacturing) has been held every year in Delhi, India. there are some rules for making the kart which are specified in the rule book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ecokart.in/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ECO KART SERIES&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="13278" LastEditorUserId="8055" LastEditDate="2017-10-24T01:09:06.847" LastActivityDate="2017-10-24T01:09:06.847" Title="What are the calculations required to design transmission system for an Electric Kart ( i.e Go Kart)?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="2" CommentCount="2" ClosedDate="2017-10-16T20:39:27.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="17540" PostTypeId="1" CreationDate="2017-10-16T06:51:34.690" Score="2" ViewCount="307" Body="&lt;p&gt;I am trying to modify a micro servo for continuous rotation, but I am not sure how to center the servo's potentiometer without having to use a servo tester. The micro servo that I am using is  &quot;SG90 Micro Servo Motor 9G&quot;.  &lt;/p&gt;&#xA;" OwnerDisplayName="user13284" LastActivityDate="2018-10-13T04:00:08.913" Title="How to center the potentiometer of a 9g micro servo without using a servo tester?" Tags="|servo|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17542" PostTypeId="1" AcceptedAnswerId="39615" CreationDate="2017-10-16T11:04:25.543" Score="3" ViewCount="3327" Body="&lt;p&gt;I'm working on a design where I need a ball bearing with an inner diameter of 8mm. The ball bearings will support small 3D printed wheels, looking kind of like a skateboard wheels. In order to fasten these wheels, I'm planning to use 8mm screws. The screws that I have found are threaded all the way, and I'm wondering if it is possible/recommended to use threaded screws to support ball bearings? Ideally I would use screws that are not threaded all the way, so the ball bearing can sit on the part that is not threaded. However, I couldn't find such screws. So what if I would use threaded screws? Would it work at all? Is there some workaround I can do using washers/extra bolts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the design, plus a new design I came up with...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.dropbox.com/s/c86cxw1wyc9zi7v/IMG_3628.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.dropbox.com/s/c86cxw1wyc9zi7v/IMG_3628.JPG&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have learned so far... There is such a thing as a shaft collar, and that's what I should be using in this case. Here is one that is 3D printed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.thingiverse.com/thing:400921&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.thingiverse.com/thing:400921&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And on there there is also an example of how it is used together with ball bearings. That's what I should be inspired by...&lt;/p&gt;&#xA;" OwnerUserId="13290" LastEditorUserId="255" LastEditDate="2018-02-16T14:30:48.493" LastActivityDate="2021-01-09T02:10:14.363" Title="Can I use a threaded bolt to support a ball bearing?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17543" PostTypeId="1" AcceptedAnswerId="17548" CreationDate="2017-10-16T11:49:28.763" Score="3" ViewCount="8303" Body="&lt;p&gt;I am trying to ascertain the maximum line pull for an old winch I have in my workshop and I am arriving at values that are considerably higher than what is in the original manufacturer's manual. I am not sure if I am doing this right and want to validate my calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The winch is rotated hydraulically with a motor. The motor shaft is connected to a wet brake which in turn is connected to a planetary gear hub. This hub is then connected directly to the drum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some specs:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the motor builds up 3000 PSI and has 38 LPM of flow it produces a torque of 295Nm with 316RPM. (or 30.08163KGm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wet brake disengages and produces negligible friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio in the gear hub is +53.58:1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter of the drum on the winch is 0.81m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter of the cable used to pull is 0.0202m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, to find linear force for a winch I must use this formula: Torque = Force X Arm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where Arm = radius of drum + radius of cable radius drum = 0.81m/2 = 0.405m radium cable = 0.0202m/2 = 0.0101m Arm = 0.405m + 0.0101m = 0.4151m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where Torque = Torque of motor X ratio of gear hub Torque = 30.08163KGm X 53.58:1 = 1611.7737354KGm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore Force = Torque / Arm Force = 1611.7737354KGm / 0.4151m = 3882.856505420381KG&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct to therefore say that the maximum line pull before the winch stalls is 3882.85KG? Therefore We can say it can lift anything bellow this weight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8uVG3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8uVG3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13291" LastEditorUserId="1832" LastEditDate="2017-10-16T12:02:52.643" LastActivityDate="2018-08-04T03:39:31.310" Title="Computing winch maximum line pull (torque vs lift capacity)" Tags="|mechanical-engineering|torque|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17545" PostTypeId="2" ParentId="17538" CreationDate="2017-10-16T15:55:17.477" Score="1" Body="&lt;p&gt;To begin, you need to ask yourself the requirements.  Here's a short list to begin:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Manual or Automatic?  If automatic, how will you automate it?&lt;/li&gt;&#xA;&lt;li&gt;Top speed?&lt;/li&gt;&#xA;&lt;li&gt;Maximum incline at top speed?&lt;/li&gt;&#xA;&lt;li&gt;Maximum weight&lt;/li&gt;&#xA;&lt;li&gt;Maximum dimensions&lt;/li&gt;&#xA;&lt;li&gt;Motor characteristics.  Most DC motors have a maximum torque and maximum speed, and follow a roughly triangular load curve.  These details ensure you can hit all the load points&lt;/li&gt;&#xA;&lt;li&gt;Availability.  If you're building this yourself, the world is your oyster.  If you're buying it, you're settling on some off-the shelf options.  What do you have available from suppliers?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-10-16T15:55:17.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17548" PostTypeId="2" ParentId="17543" CreationDate="2017-10-16T17:24:06.680" Score="0" Body="&lt;p&gt;The main constraint would be the cable braking strength.  Per Engineering Toolbox, this size cable has a maximum safe load of 42.3 kN.  Your device produces 38 kN at stall point, so the cable is under it's safe load.  This implies that the device will indeed handle this level of load without a problem.  The system may have derated the pump as approaching the stall point of most motors, including hydraulic motors, can wear excessively and cause early breaking, so it is best to stay within nameplate parameters for day-to-day operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, due to angles, we could have additional loading beyond the weight.  See this handy diagram and table from Engineering Toolbox:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l3iRr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l3iRr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;α   β   Increased Force&#xA;0   90  1.00&#xA;5   85  1.00&#xA;10  80  1.02&#xA;15  75  1.04&#xA;20  70  1.07&#xA;25  65  1.10&#xA;30  60  1.16&#xA;35  55  1.22&#xA;40  50  1.31&#xA;45  45  1.41&#xA;50  40  1.56&#xA;55  35  1.74&#xA;60  30  2.00&#xA;65  25  2.37&#xA;70  20  2.92&#xA;75  15  3.86&#xA;80  10  5.76&#xA;85  5   11.5&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-10-16T17:24:06.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17549" PostTypeId="2" ParentId="17536" CreationDate="2017-10-16T18:20:36.083" Score="1" Body="&lt;p&gt;The optimum shape to avoid bending would probably be a L-profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For calculating the thickness required given the three loads, some calculation is required. These are basic (essential) static calculations, so there are lots of online sources available:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.aps.anl.gov/APS_Engineering_Support_Division/Mechanical_Operations_and_Maintenance/Calculators/ElasticBeam2.html&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Beam Calculator&quot;&gt;Click here for a beam calculator.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation needs the sectional moment of inertia of the profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The L-profile has a moment of inertia as given in the following web page: &lt;a href=&quot;https://www.engineeringtoolbox.com/area-moment-inertia-d_521.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;click here to visit the page&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll probably do these calculations either online or by using a spreadsheet calculator, or by hand (/calculator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the calculations follow the forward analysis path. This means, you calculate the deflection given a load case and beam design. In order to find the optimum shape, that gives a small enough deflection, you will need to do several calculations. &lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-10-16T18:20:36.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17550" PostTypeId="2" ParentId="17538" CreationDate="2017-10-16T19:58:58.387" Score="1" Body="&lt;p&gt;There is no simple calculation which will give you a single correct answer as gear ratios are essentially an optimisation problem. A higher final drive ratio will give a higher top speed (up to a point at least) while a lower ratio will give you more torque (although usable torque is also limited by the traction of the drive wheels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall approach for designing something like this is to start with fairly simple calculations based on reasonable approximations to work out rough orders of magnitude and where the limits are and then progressively refine your numbers as you get closer to what a sensible design should look like. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also helpful to look at existing real world examples to get a sense of what is likely to be possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter of the wheels and tyres will affect the final drive ratio too so you may want to look at what wheels you are using first. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Th basic concept of &lt;a href=&quot;https://en.wikipedia.org/wiki/Gear_train&quot; rel=&quot;nofollow noreferrer&quot;&gt;gear ratios&lt;/a&gt; is simple enough and similarly for every turn of the road wheels you move forwards by the circumference of the wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;eg. if you have a wheel with circumference of 0.6m turning @ 600 rpm (ie 10 revolutions per second the road speed is 0.6m * 10/s = 6m/s or about 13.4mph &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-10-16T19:58:58.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17552" PostTypeId="1" CreationDate="2017-10-16T20:26:07.287" Score="4" ViewCount="498" Body="&lt;p&gt;My classmates are making mousetrap cars to go 10 feet, and whoever goes the fastest wins. How a mousetrap car works is by attaching a lever arm to the snapper (the part of the mousetrap that shoots forward). The lever arm has a string attached to the end of it, and it is wrapped several times around the axle of a car. When you let go of the lever arm, the snapper goes 180 degrees, spinning the axle and making the car go forward. Here is a link for more information: &lt;a href=&quot;https://www.wikihow.com/Build-a-Mousetrap-Car&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.wikihow.com/Build-a-Mousetrap-Car&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an idea for a car, but want to make sure it's sound before building it. I plan to have six wheels. The back wheels will be adopted for distance, so it will have a longer lever arm and a thinner axle. The middle wheels will be adopted for speed, so it will have a thicker axle and a shorter lever arm. There will be two strings attached to the lever arm. One will be at the end, and will be wrapped around the distance axle, while one will be near the middle and attached to the speed axle. My reasoning is that the middle wheels will provide a lot of speed, and at the same time make it easier for it to start moving. The back will make sure it keeps going once the speed boost wears off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know if doing this will work, or if it would be better to just have two pairs of wheels that are moderate thickness and moderate distance from the mousetrap. This is my concept sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nvn6I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nvn6Im.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13032" LastEditorUserId="368" LastEditDate="2017-10-17T16:39:54.380" LastActivityDate="2019-07-16T12:01:40.180" Title="Six-Wheeled Mousetrap Car [Concept]" Tags="|mechanical-engineering|car|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17555" PostTypeId="1" AcceptedAnswerId="17558" CreationDate="2017-10-17T06:11:58.843" Score="1" ViewCount="674" Body="&lt;p&gt;The aircraft carrier &lt;a href=&quot;https://en.wikipedia.org/wiki/USS_Nimitz&quot; rel=&quot;nofollow noreferrer&quot;&gt;USS Nimitz&lt;/a&gt; cost \$4.5bn. The cruise ship &lt;a href=&quot;https://en.wikipedia.org/wiki/Allure_of_the_Seas&quot; rel=&quot;nofollow noreferrer&quot;&gt;Allure of the Seas&lt;/a&gt; cost \$1.2bn. Why is the difference so huge?&lt;/p&gt;&#xA;" OwnerDisplayName="user11311" LastEditorUserId="6790" LastEditDate="2020-05-02T19:35:50.560" LastActivityDate="2020-05-02T19:35:50.560" Title="Why are aircraft carriers so expensive to build?" Tags="|ships|naval-engineering|" AnswerCount="3" CommentCount="2" ClosedDate="2017-10-21T03:00:21.300" ContentLicense="CC BY-SA 4.0" />
  <row Id="17556" PostTypeId="2" ParentId="17555" CreationDate="2017-10-17T07:03:13.020" Score="1" Body="&lt;p&gt;Because the technology in the aircraft carrier is so advanced : and that is basically most of the systems offence, defence, living, propulsion etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other only has living and propulsion... at a high standard...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-17T07:03:13.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17557" PostTypeId="2" ParentId="17555" CreationDate="2017-10-17T07:15:10.667" Score="1" Body="&lt;p&gt;Most navy ships require very high redundancy of all critical system (Propulsion computer, navigation etc), the material specs are also usually very tight, meaning more costs, for example the paint used on some of the new carriers alone was a multi million dollar project to withstand the F35 Jets. &lt;/p&gt;&#xA;" OwnerUserId="12804" LastActivityDate="2017-10-17T07:15:10.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17558" PostTypeId="2" ParentId="17555" CreationDate="2017-10-17T08:01:23.820" Score="3" Body="&lt;p&gt;The USS Nimitz is nuclear powered, whereas the Allure of the Seas is not. Nuclear reactors are not cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the USS Nimitz, like all aircraft carriers has a catapult to help launch aircraft off the decks. The Allure of the Seas does not have a catapult. Catapults are expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The USS Nimitz has kevlar armor over vital spaces, but civil cruise ships don't&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add to this the radar systems and armaments that aircraft carriers must have but are not required for civil cruise ships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aircraft carriers are build to handle severe weather where some cruise ships may not be. They must also have the propulsion systems and structural integrity to travel long distances at high speed. Civil cruise ship don't.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="9656" LastEditDate="2017-10-17T14:53:42.723" LastActivityDate="2017-10-17T14:53:42.723" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17559" PostTypeId="1" CreationDate="2017-10-17T08:09:37.947" Score="1" ViewCount="38" Body="&lt;p&gt;I have been in touch with suppliers of actuators for subsea valves. They have mentioned their solutions are available for no matter the valve size, as long as the interface is API 17D class 2, 4 or 7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During my search, I could barely find any mention of these classes, even in the API 17D itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are the API classes for torque tools?&lt;/strong&gt; Do they depend on size? Is there a rule of thumb to &quot;convert&quot; class to valve size?&lt;/p&gt;&#xA;" OwnerUserId="7994" LastActivityDate="2017-10-17T08:09:37.947" Title="What is API class in subsea torque tools?" Tags="|mechanical-engineering|petroleum-engineering|offshore|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17561" PostTypeId="2" ParentId="15425" CreationDate="2017-10-17T09:09:40.287" Score="1" Body="&lt;p&gt;In answer to the &quot;&lt;strong&gt;I want to simplify the process&lt;/strong&gt;&quot; part of the Q...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reversing the cut of the worm would allow the final gear to turn in the correct direction without reversing the input shaft direction and would save one spur gear at least.  Perhaps the second spur is needed, perhaps not, the Question isn't clear on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wvKE2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvKE2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastActivityDate="2017-10-17T09:09:40.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17562" PostTypeId="1" AcceptedAnswerId="17569" CreationDate="2017-10-17T10:44:14.957" Score="-1" ViewCount="391" Body="&lt;p&gt;I am looking for something like this to make my robot ideas as prototype. But I don't know it's name! Also if there are better options I will happy to know.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BpW14.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BpW14.jpg&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;&lt;sub&gt;(source: &lt;a href=&quot;https://cdn.instructables.com/FQC/7ENO/IW5JB2UM/FQC7ENOIW5JB2UM.MEDIUM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;instructables.com&lt;/a&gt;)&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://cdn.instructables.com/F3J/BE2K/IW5JB58G/F3JBE2KIW5JB58G.MEDIUM.jpg&quot; alt=&quot;&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="13302" LastEditorUserId="6767" LastEditDate="2020-12-21T20:14:04.703" LastActivityDate="2020-12-21T20:14:04.703" Title="I am looking for some basic mechanical parts to build structures(prototypes)?" Tags="|metals|robotics|building-design|prototyping|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="17564" PostTypeId="1" CreationDate="2017-10-17T13:03:35.767" Score="1" ViewCount="1448" Body="&lt;p&gt;If a jet starts in extreme cold, i.e cold starts, then the jet cannot power the heat exchangers to heat the fuel. Then how do they heat the fuel to start the jets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously you can't use batteries/inductive heating to heat the fuel. Most batteries don't work at the cold start temperatures (-50c or so). Even if you want to use exothermic chemicals to heat fuel, you need stored power to open valves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then how do they do it? Do they just insulate the tank and hope for the best? even if you insulate tank, you need power to pump fuel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;say its a kerosene +oil based jet. or jet fuel A.&lt;/p&gt;&#xA;" OwnerUserId="13304" LastEditorUserId="13304" LastEditDate="2017-10-17T13:21:15.047" LastActivityDate="2017-10-17T16:13:42.760" Title="How do jets/airplane cold start?" Tags="|mechanical-engineering|thermodynamics|aerospace-engineering|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="17566" PostTypeId="2" ParentId="17564" CreationDate="2017-10-17T13:10:20.240" Score="0" Body="&lt;p&gt;The can also be plugged into an external power source in need be, but their primary way of starting in extreme power situations is the backup power module, a gasoline generator stored in the rear of the plane.&lt;/p&gt;&#xA;" OwnerUserId="13306" LastActivityDate="2017-10-17T13:10:20.240" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17567" PostTypeId="1" CreationDate="2017-10-17T13:41:58.510" Score="0" ViewCount="661" Body="&lt;p&gt;I have been researching how to make backlit panels for flight simulators, like this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tqCyL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tqCyL.jpg&quot; alt=&quot;Backlit Boeing 737 panel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across this video:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=vf7tQNYqliw&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=vf7tQNYqliw&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where they use cast transparent acrylic and then paint it with &quot;laser color&quot;, however I cannot find anything on the internet for laser color, laser paint, laser resistant paint whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Came across a post in aviation stackexchange, where they also refer to &quot;white opaque special laser colour&quot; (search for &quot;How are backlit aircraft panels (for simulators) made?&quot;, I cannot post more than 1 link now).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried myself using painted opal acrylic with black and then grey color, but when the laser hits, it burns the paint and damages the acrylic, which gives it smoked appearance when backlighting is off, as opposed to white.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't tried painting white first, because 1. The acrylic is white opal in the first place, and 2. I'm afraid I will have the same result when the laser hits the paint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody provide more insight about the technology and the actual products used? What is that laser color they talk about?&lt;/p&gt;&#xA;" OwnerUserId="13308" LastActivityDate="2018-05-20T00:45:39.000" Title="What paint to use for laser engraved (aircraft and not only) panels with backlighting" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17568" PostTypeId="2" ParentId="17562" CreationDate="2017-10-17T14:25:29.517" Score="1" Body="&lt;p&gt;Try 80-20 structural tubing. It is available is various sectional sizes, often used for robotics and machine framing.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-10-17T14:25:29.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17569" PostTypeId="2" ParentId="17562" CreationDate="2017-10-17T15:09:02.830" Score="1" Body="&lt;p&gt;What you're looking for is described as &quot;Perforated&quot;, So a quick google for &quot;Perforated Square Tubing Steel&quot; produced: &lt;a href=&quot;https://www.discountsteel.com/items/Galvanized_Steel_Perforated_Square_Tube.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.discountsteel.com/items/Galvanized_Steel_Perforated_Square_Tube.cfm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;80-20&quot; is a brand name, and the product that @Donald Gibson was referring to can more generally be found called &quot;T Slot Extrusion&quot;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2017-10-18T21:01:05.160" LastActivityDate="2017-10-18T21:01:05.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17570" PostTypeId="1" AcceptedAnswerId="17576" CreationDate="2017-10-17T15:37:50.817" Score="0" ViewCount="315" Body="&lt;p&gt;Is it only possible to write a User Defined Function for Ansys Fluent in C? All of the tutorials I could access online do so. I was wondering if this could be done using Java, or any other language for that matter?&lt;/p&gt;&#xA;" OwnerUserId="9398" LastActivityDate="2019-11-07T11:32:00.383" Title="Writing a User Defined Function for Ansys Fluent" Tags="|fluid-mechanics|ansys|fluent|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17571" PostTypeId="1" CreationDate="2017-10-17T16:12:58.553" Score="0" ViewCount="5524" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gvoyP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gvoyP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to Solidworks, my part has several holes (created using the Hole Wizard) which needs Centerlines drawn to connect them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However when I click on the Centerline tool in the Drawing, then click on one hole followed by another hole, I get the error message&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Unable to create centerline from the selected entities&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What should be the correct way to create the centerlines?&lt;/p&gt;&#xA;" OwnerUserId="13312" LastActivityDate="2023-03-28T15:47:03.773" Title="Cannot Create Centerline between 2 Holes in Solidworks Drawing" Tags="|solidworks|drafting|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17572" PostTypeId="2" ParentId="17564" CreationDate="2017-10-17T16:13:42.760" Score="4" Body="&lt;p&gt;The basic method of starting jet engines designed for this sort of temperature range is to windmill the engine by blowing air into it. All the power for fuel pumping etc comes from mechanical drives from the main engine rotors, not from batteries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of electrical power needed to operate the engine igniter plugs is small, and there are battery chemistries that will work well enough (even if the maximum current is reduced) down to -40C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You only need to start one engine that way - then you have plenty of hot air available to start the others. Cold starting is irrelevant unless the &lt;em&gt;aircraft&lt;/em&gt; also has anti-icing systems that will allow it to fly at such temperatures, of course - de-icing the aircraft on the ground before takeoff is only the first step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the ground temperature is -40C, you obviously need to provide some heated accommodation for the crew and/or passengers, and that is enough to keep ground starting equipment warm as well. So far as the engine is concerned, a sudden transient temperature rise from say 0C to 1000C when it fires up is not much different from -40C to 1000C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can prevent jelling of fuel (and just as important, oil) with suitable additives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that at passenger jet cruising altitudes, the air temperature is typically -50C in any case, and may go as low as -80C. Aircraft designed for surveillance, or search and rescue, may need special in-flight systems to work in those temperatures while loitering over the mission target, especially if some engines are shut down to save fuel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the old UK Nimrod SAR planes (a modified version of the 1950s Comet passenger jet - now retired from service) might have two of the four engines shut down for 16 hours at altitude. The oil systems were modified so the oil flow path was through both engines on the same wing in sequence - the &quot;shut down&quot; engine continues to windmill, and therefore needs proper lubrication for the entire flight time.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-10-17T16:13:42.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17573" PostTypeId="2" ParentId="17540" CreationDate="2017-10-17T16:39:38.090" Score="1" Body="&lt;p&gt;A hobby r/c servo is centered when the pulse width is (nominally) 1.5mS, or in the case of a continuous rotation servo, when the servo is not moving. i.e. the servo motor is perfectly idle with no current draw. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply connect the servo to a receiver/MCU with an analog input and adjust until the servo stops moving. No need for a servo tester.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="7021" LastEditDate="2017-10-17T19:55:06.670" LastActivityDate="2017-10-17T19:55:06.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17575" PostTypeId="1" AcceptedAnswerId="17588" CreationDate="2017-10-17T19:30:38.700" Score="10" ViewCount="281" Body="&lt;p&gt;I hope this is the right place to ask this question; if not I'd really appreciate some tips as to where else I can go for answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pC24m.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pC24m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symbol above (apologies for the blurry quality) is on a Chinese/Taiwanese-manufactured piece of testing equipment. It's located on the manufacturer's sticker. I can find absolutely no trace of it elsewhere on the Internet (I've absolutely, positively searched).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody shed some light as to what this symbol might mean? E.g. an electrical safety certification, a manufacturing standard...?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="13317" LastActivityDate="2017-10-19T09:26:50.583" Title="What's this mystery symbol on a manufactured machine?" Tags="|mechanical-engineering|electrical-engineering|manufacturing-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17576" PostTypeId="2" ParentId="17570" CreationDate="2017-10-17T21:38:55.293" Score="0" Body="&lt;p&gt;If you go to Ansys themselves the answer is clear :&#xA;Source is : &lt;a href=&quot;http://www.ansys.com/services/training-center/fluids/ansys-fluent-using-user-defined-functions&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ansys.com/services/training-center/fluids/ansys-fluent-using-user-defined-functions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CJVNB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CJVNB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-10-17T22:22:15.273" LastActivityDate="2017-10-17T22:22:15.273" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17577" PostTypeId="1" CreationDate="2017-10-17T21:49:04.083" Score="0" ViewCount="64" Body="&lt;p&gt;I'm currently trying to figure out whether or not we can flame-harden a steel wheel which was supplied with a lower hardness than expected. I'm worried, however, that flame-hardening can either deform the wheel or change its size somehow, especially given that a flame-hardening job has never been done in the company before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wheels are about 13&quot; in diameter and about 6&quot; in thick. They have a few metal details and inner holes, which I'm not exceedingly worried about since other wheels on the same machine will have different designs. The wheel will roll on a plate, although I do not know the hardness of the plate at the moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, I would like to flame harden the wheels from 160 Br to 400 Br, to bring it up to the design specifications, although I would be content taking it up to 300 Br, which is the hardness of some previous wheels we have tested from the same machine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What considerations should I take before attempting to set up the flame-hardening? I have quite a few wheels to spare in case the job goes wrong, but I would also like to know if there is some quantitative analysis that should be done before starting.&lt;/p&gt;&#xA;" OwnerUserId="12823" LastEditorUserId="12804" LastEditDate="2017-10-18T10:57:01.717" LastActivityDate="2017-10-18T13:26:43.993" Title="What considerations should I take when flame-hardening a wheel?" Tags="|mechanical-engineering|manufacturing-engineering|metallurgy|metals|heat-treatment|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17578" PostTypeId="2" ParentId="17542" CreationDate="2017-10-18T02:10:48.030" Score="1" Body="&lt;p&gt;If the bearing has a machined raceway then it would be fine from a bearing operation standpoint. Caution is needed long term because the tips of the threads might wear out inducing slop in the connection. The bearings might walk out or their inner raceway might start spinning causing wear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally if the bearing is designed for specific radial clearance after it has been installed via a press fit, then when placed loosely over threads the clearance will be higher and the ball loads will be more concentrated. The life of the bearing will decrease as the clearance increases. This might not be as important in this application. Try to find the loading capacity of the bearing and you can calculate the life of the bearing from the ratio&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ (\mbox{life in cycles}) =10^6 \left( \frac{\rm capacity}{\rm load} \right)^3 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes fully loaded balls (with zero radial clearance).&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-10-18T02:10:48.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17580" PostTypeId="1" CreationDate="2017-10-18T03:17:02.163" Score="0" ViewCount="624" Body="&lt;p&gt;In my exam, I was given 2 options to design a process. Assuming a saturated liquid coming out from the separator, A) should the stream pass through a valve to reduce pressure, then to heat exchanger to heat it up, or B) should the stream be heat up first, then passes the valve to reduce pressure? Assuming the final temp and pressure in both cases are the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the justification to that selected option?&lt;/p&gt;&#xA;" OwnerUserId="13323" LastActivityDate="2018-02-11T21:10:56.003" Title="Valve first,then heat exchanger OR heat exchanger first then valve" Tags="|chemical-engineering|process-engineering|valves|heat-exchanger|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17582" PostTypeId="1" CreationDate="2017-10-18T10:01:48.900" Score="1" ViewCount="55" Body="&lt;p&gt;I'm interested in the development of electric small airplanes with 4 or 6 seats, that could really happen in the following decades. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the main problems related to this project would be the specific energy of the batteries used on these airplanes. Batteries should have a good energy storage to make the range of the aircraft reasonable, and they also need to be light, to make the lift possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw in this &lt;a href=&quot;http://pubs.rsc.org/en/content/articlepdf/2011/ee/c0ee00777c&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; that the specific energy of the Li-ion batteries could reach 300 Wh/kg until 2030, but it's still a little bit low for a small aircraft. I did some more research, and I found that new different type of batteries are being developed, like lithium-air batteries, that in practice can get a specific energy of 1700 Wh/kg (just read &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Lithium%E2%80%93air_battery&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;).  I found also that glass batteries could be a thing too (read &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Lithium%E2%80%93air_battery&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;), which could have specific energy several times greater that the recent Li-ion batteries and could work at low temperatures (which is good for high altitude flight). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've little knowledge about chemistry of batteries, and so I'm requesting someone that has good formation in this area to share an opinion about the possibility of existence of batteries like lithium air or glass batteries at 2030 for example, and which value of specific energy could be a good prevision around that year.&lt;/p&gt;&#xA;" OwnerUserId="9747" LastEditorUserId="368" LastEditDate="2017-10-18T14:58:05.173" LastActivityDate="2017-10-18T14:58:05.173" Title="Batteries for electric airplanes" Tags="|aircraft-design|battery|energy-storage|electric-vehicles|chemistry|" AnswerCount="0" CommentCount="3" ClosedDate="2017-10-18T14:58:27.680" ContentLicense="CC BY-SA 3.0" />
  <row Id="17583" PostTypeId="2" ParentId="17577" CreationDate="2017-10-18T13:26:43.993" Score="1" Body="&lt;p&gt;With 13&quot; diameter and 6&quot; thickness they are VERY &quot;stocky&quot; to flame harden. It would require very high heat flux to heat the surface before the whole wheel is hot. Like 20 oxy-acetylene tips on a 6&quot; wide burner. The goal is to heat the surface to 1600F while the bulk is no hotter than 500F so that you get a conduction quench. Induction heating would work - not for amateurs. Alternative would be carburizing to some depth and quenching; presumably the low carbon core would not harden. I have assumed this is a low carbon steel like AISI 1020.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-10-18T13:26:43.993" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17585" PostTypeId="1" CreationDate="2017-10-18T15:52:33.447" Score="1" ViewCount="83" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U3AKP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U3AKP.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Notice that x and y are proportionally equal and perpendicular to your line of sight, but z is not. It's not &lt;a href=&quot;https://www.google.com/search?tbm=isch&amp;amp;sa=1&amp;amp;q=30+90+30+isometric&amp;amp;oq=30+90+30+isometric&amp;amp;gs_l=psy-ab.3...6851.10894.0.11955.9.9.0.0.0.0.677.1055.3-1j0j1.2.0....0...1.1.64.psy-ab..7.2.1055...0i7i30k1j0i8i7i30k1j0i8i13i30k1j0i8i30k1.0.uGMqTquNgo4&quot; rel=&quot;nofollow noreferrer&quot;&gt;30-90-30 isometric&lt;/a&gt;. Also z should actually be pointing right at you rather than down and to the left if this was true 3D rendering. Hence this is a style. Does it have a name?&lt;/p&gt;&#xA;&lt;p&gt;Note: z is free to scale any way you like so this style can be made into a kind of isometric. That is x, y and z can all have the same scale. But regardless of that does this have a name that wont be confused with 30-90-30 isometric?&lt;/p&gt;&#xA;" OwnerUserId="8415" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-10-18T16:25:55.990" Title="What is this drawing style that faces the x-y plane but stylisticly pulls the z axis down and to the left?" Tags="|drafting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17587" PostTypeId="2" ParentId="17585" CreationDate="2017-10-18T16:19:03.663" Score="3" Body="&lt;p&gt;This appears to be a &lt;strong&gt;Cavalier projection&lt;/strong&gt; since the oblique lines are at the same scale as the xy plane. The name comes from military fortifications - a &quot;cavalier&quot; was the French term for an artificial mound behind a defensive wall, which allows the enemy to be seen over the wall. As the name might suggest, a low &quot;&lt;em&gt;cavalier&lt;/em&gt;&quot; provided a relatively risk-free way for a &lt;em&gt;cavalry&lt;/em&gt; unit to see over the defensive wall. The projection was intended to show such a view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equal scaling of all three axes makes this easy to draw, but the result is not necessarily very &quot;realistic&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a similar &lt;em&gt;Cabinet projection&lt;/em&gt; which was commonly used in illustrations of furniture, etc. In a cabinet projection the oblique lines are shortened in a &quot;simple&quot; proportion like 2:1 or 3:2, for easy drawing and greater realism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Oblique_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Oblique_projection&lt;/a&gt;, which has some illustrations of both projections.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-10-18T16:25:55.990" LastActivityDate="2017-10-18T16:25:55.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17588" PostTypeId="2" ParentId="17575" CreationDate="2017-10-18T21:58:16.933" Score="5" Body="&lt;p&gt;My best guess after a couple minutes of Googling is that this is a &lt;code&gt;CMC&lt;/code&gt; certification which stands for &lt;code&gt;China Metrology Certification&lt;/code&gt;.  If you search for that however, the first result points to &lt;code&gt;China Metrology Accreditation&lt;/code&gt; which has a logo nearly identical to that one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That first result is at &lt;a href=&quot;http://hqts-qai.com/accreditations-certifications/&quot; rel=&quot;noreferrer&quot;&gt;this link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logo you will find looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/coXD3.jpg&quot; alt=&quot;image CMA&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea what these are for.  The corresponding text from that one site includes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;China Metrology Accreditation (CMA) is a mandatory accreditation system designed to assess the metrology capabilities of testing or calibration laboratories that will test products being sold into the Chinese market.  This accreditation allows HQTS-QAI to issue test reports and certificates bearing the CMA mark, which verifies that the tested products conform to China’s mandatory quality and safety standards.  Test reports bearing the CMA mark are accepted as legal documents in China.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6681" LastActivityDate="2017-10-18T21:58:16.933" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17589" PostTypeId="1" CreationDate="2017-10-18T22:27:59.513" Score="1" ViewCount="576" Body="&lt;p&gt;I have a vacuum chamber that I would like to backfill with 15 mTorr of argon.  The problem with accomplishing this is that most pressure regulators can't do pressures that low.  Are there regulators that accurately regulate mTorr level pressures, or is there some other approach to backfilling a vacuum system at low pressure?&lt;/p&gt;&#xA;" OwnerUserId="5103" LastActivityDate="2017-10-19T21:08:53.800" Title="How to backfill vacuum chamber with low pressure gas" Tags="|pressure|vacuum|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17590" PostTypeId="2" ParentId="17562" CreationDate="2017-10-18T22:33:52.717" Score="1" Body="&lt;p&gt;In addition to 80/20 you could use &quot;unistrut&quot; or &quot;channel strut&quot; framing which may be more readily available at a big box store. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gh9NF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gh9NF.jpg&quot; alt=&quot;unistrut&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-10-18T22:33:52.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17596" PostTypeId="1" CreationDate="2017-10-19T05:51:40.213" Score="0" ViewCount="131" Body="&lt;p&gt;On my belt sander, there is a small cast aluminum drive pulley/wheel about 1.25&quot; diameter that slips over a 5/16&quot; shaft. Two flats on the shaft mate with two flats inside the bore.  The internal flats in the bore have sheared off, so now the wheel just spins on the shaft instead of driving it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manufacturer no longer makes the cast aluminum wheel, and no parts stores via google have any old stock left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two ideas for fixing it 1) mill keyways (too expensive) or 2) drill a hole for a 'shear pin' (spring pin, roll pin; which is my plan at the moment).  I'm open to other suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the pin, what is the optimum diameter and what metal?  If it is small, it'll likely shear under normal use. If it is large, then the shaft will be weakened too much.&lt;/p&gt;&#xA;" OwnerUserId="13338" LastEditorUserId="13338" LastEditDate="2017-10-19T05:58:55.530" LastActivityDate="2020-07-25T14:03:45.380" Title="How to fix sheared flats on a pulley?" Tags="|pulleys|shear|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17597" PostTypeId="2" ParentId="17596" CreationDate="2017-10-19T08:32:41.040" Score="1" Body="&lt;p&gt;One other possibility is to mount the motor / pulley in a drill stand and drill a &quot;keyway&quot; axially between the shaft and pulley then put a pin in that hole - 2mm dia but 10mm long which would work well...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have to do the first pass with a similar metal to the shaft : a nut and then drill again with the pulley as the drill may wonder off as the pulley is soft in comparison...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-19T08:32:41.040" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17598" PostTypeId="1" AcceptedAnswerId="17599" CreationDate="2017-10-19T08:50:35.357" Score="1" ViewCount="54" Body="&lt;p&gt;I have a small device emitting up to ~90 Watts thermal power over a surface of 2x2 square centimeters. What is the best way (both commercially and DiY) for keeping the temperature of the device constant, maybe +- 1 degree at room temperature?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a thermoelectric element I could use, which has a surface area of 18x10 square centimeters for the cold plate. It is capable of removing up to ~75 Watts thermal power at 0 degrees temperature difference. How efficient will it be when I put the electric device on top of it? How much will the size difference effect the efficiency of the TEC? I know that it is rated for lower thermal powers, but it can be used as intermediary solution while keeping the device at lower powers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The device itself is a solid state laser diode, which means that even small changes in the temperature (i.e. degrees) will shift the wavelength, something I do not want (I prefer having it at ~20 degrees). The diode can emit up to 90 Watts, depending on how much output power I request, but the upper limit is 90 Watts.&lt;/p&gt;&#xA;" OwnerUserId="2844" LastEditorUserId="2844" LastEditDate="2017-10-19T09:21:01.670" LastActivityDate="2017-10-19T09:38:39.290" Title="Cooling a small high-power device" Tags="|cooling|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17599" PostTypeId="2" ParentId="17598" CreationDate="2017-10-19T09:13:56.590" Score="2" Body="&lt;p&gt;A CPU cooler is the first thing that springs to mind DIY - Heatsinks coupled with forced convection are a tried-and-tested method of heat removal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided the heat output from the diode is constant, and the room doesn't change, then this should remain at a constant temperature, albeit above the ambient temperature of the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the application you describe, it seems that some form of thermostatic control would be required to keep the diode temperature constant as you vary the output power. DIY, This means hooking up a thermocouple to an arduino or similar, to control the power delivered to the thermoelectric element that you described previously. Adding a Heat-Sink/Fan to the hot side of the element will improve efficiency, and give you more control. Varying the fan speed will have an effect on the cold-side temperature, and this can be used in addition to simply varying the voltage into the element.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2017-10-19T09:38:39.290" LastActivityDate="2017-10-19T09:38:39.290" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17600" PostTypeId="2" ParentId="17575" CreationDate="2017-10-19T09:26:50.583" Score="0" Body="&lt;p&gt;This appears to be related to GB/T 50430-2007, the Chinese &quot;Code for Qualtity management of Engingeering Construction Enterprises&quot;. Unfortunately I can't find the full text translated into english, but found reference to the CMC mark &lt;a href=&quot;https://books.google.co.uk/books?id=nPumDgAAQBAJ&amp;amp;lpg=PA63&amp;amp;ots=d_alubKM3b&amp;amp;dq=CMC%20certification%20test%20equipment%20china&amp;amp;pg=PA63#v=onepage&amp;amp;q=CMC%20marking&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;in section 11.5.1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-10-19T09:26:50.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17601" PostTypeId="1" AcceptedAnswerId="17632" CreationDate="2017-10-19T10:34:07.320" Score="2" ViewCount="115" Body="&lt;p&gt;I am Comp-sci student doing a software engineering paper about designing a program, planning the software like you were actually planning a real-life software project, short of writing actual pseudo-code or code. For my project, I picked doing an app that can detect a physical man in the middle attack. Like monitoring for a tap on the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It got me thinking, can you monitor for someone catching wireless communication? Like the latest WEP2 hack. If someone were monitoring your wireless traffic passively to capture your secure handshake, could you detect them snooping wirelessly without them transmitting packets?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My knowledge of EE stops at MOSFETs and timing of electrical currents through logic gates and metastable currents propagating through circuits because the timing of logic gate was off. My knowledge of EE certainly falls short of radio broadcasting.  So I am not even sure if such a feat is possible, I suspect it isn't but it's out of my ken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect if you were close enough a receiver would leave behind some kind of detectable EM footprint.&lt;/p&gt;&#xA;" OwnerUserId="13340" LastEditorUserId="368" LastEditDate="2017-10-20T18:20:10.750" LastActivityDate="2017-10-21T05:38:04.037" Title="Can you detect the operation of a wireless radio (or other wavelength) receiver?" Tags="|electrical-engineering|sensors|wireless-communication|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17604" PostTypeId="2" ParentId="2333" CreationDate="2017-10-19T15:18:12.820" Score="1" Body="&lt;p&gt;looks like (hex mc) made by Hexcel. it is a (quasi isotropic) laminate pre-preg of random carbon fiber tape. &lt;a href=&quot;http://www.hexcel.com/Products/HexMC-Materials-for-Industries&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.hexcel.com/Products/HexMC-Materials-for-Industries&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="13344" LastActivityDate="2017-10-19T15:18:12.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17605" PostTypeId="2" ParentId="17601" CreationDate="2017-10-19T15:25:48.173" Score="0" Body="&lt;p&gt;Yes, kind of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could create a sensitive/specialized metal detector, which would induce/detect a current in any piece of metal (including an antenna)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are antennas specifically designed to be detected (rfid technology); you broadcast a particular frequency, and they resonate and respond.  However the ratio of produced to response power is like 10^6 or more&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically, no.  Think about this- the sun is emitting EM energy, just like a wifi router.  Does the sun get dimmer or brighter if more people look at it?&lt;/p&gt;&#xA;" OwnerUserId="6488" LastActivityDate="2017-10-19T15:25:48.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17607" PostTypeId="1" CreationDate="2017-10-19T16:53:07.907" Score="0" ViewCount="1376" Body="&lt;p&gt;I originally thought that a higher Young's modulus would increase susceptibility, but now I'm thinking it might be independent. &lt;/p&gt;&#xA;" OwnerUserId="13347" LastActivityDate="2017-10-25T20:06:42.240" Title="For a bridge/beam, how will the Young's modulus affect the susceptibility of it buckling?" Tags="|mechanical-engineering|statics|" AnswerCount="1" CommentCount="3" ClosedDate="2017-10-28T15:25:11.527" ContentLicense="CC BY-SA 3.0" />
  <row Id="17608" PostTypeId="2" ParentId="17571" CreationDate="2017-10-19T18:12:42.310" Score="1" Body="&lt;p&gt;You should be able to sketch a line using the line tool (under the Sketch tab) &lt;a href=&quot;https://i.stack.imgur.com/3DK3k.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3DK3k.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;, and then simply select &quot;For Construction&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the sketch tab, you should also be able to access the Centerline tool that will skip this step for you.  Click the drop-down menu on the Line tool, and select Centerline &lt;a href=&quot;https://i.stack.imgur.com/j4wkq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j4wkq.png&quot; alt=&quot;Centerline Sketch Tool&quot;&gt;&lt;/a&gt; .  Alternatively, select Tools &gt; Sketch Entitities &gt; Centerline&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GfjWE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GfjWE.png&quot; alt=&quot;Centerline&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;centerline&quot; tool you tried using (under the Annotation tab) essentially will draw a line midway between two other lines.  Notice that the icon for the tool appears to be a side profile of a shaft.  &lt;a href=&quot;https://i.stack.imgur.com/Uhmgb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uhmgb.png&quot; alt=&quot;Centerline Icon&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4708" LastEditorUserId="4708" LastEditDate="2017-10-19T18:18:13.407" LastActivityDate="2017-10-19T18:18:13.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17609" PostTypeId="2" ParentId="17269" CreationDate="2017-10-19T19:34:10.890" Score="0" Body="&lt;p&gt;Well, I did a numerical approach on this using a backward euler form, and it turns out the equation above is actually correct. My mistake was that I was not considering anywhere in the formulation that indeed the low pressure pumps shut off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The optimization algorithm must take this into consideration and recalculate the equivalent pump ruling out the lowest pressure pump as soon as that pressure is reached (or at least considered so, at about 70% of the pressure).&lt;/p&gt;&#xA;" OwnerUserId="13050" LastActivityDate="2017-10-19T19:34:10.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17611" PostTypeId="2" ParentId="17589" CreationDate="2017-10-19T21:08:53.800" Score="3" Body="&lt;p&gt;The simplest way to do this would be to design your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A commercially available spring, with an OD of 9mm, a length of 19mm, and a wire dia. of 1mm has a rate of 3.15 N/mm, with 11.5 mm compression length.  Putting the lowest possible regulator in front of your homemade regulator, set a large diaphragm across a 1m diameter expansion chamber.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, place this commercially available spring on both sides of the diaphragm.  At 15 mTorr, the springs should deflect a controllable 0.5 mm, which could be used to regulate the poppet valve.  The key is a very large expansion chamber and a very small spring with a highly precise spring rate.  To prevent the spring from moving, the expansion chamber will need to be evacuated, and the small springs need to be placed carefully to ensure the edges do not bind, but they do not collapse or buckle.  This shouldn't be a problem, as the springs are well designed, but it's something to keep in mind when designing the expansion chamber.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-10-19T21:08:53.800" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17617" PostTypeId="1" CreationDate="2017-10-20T05:37:33.200" Score="0" ViewCount="38" Body="&lt;p&gt;Is there any pump available which can suck the water just like solenoid pump ( which means i can just put the input pipe in water and it will suck the water and get it out unlike DC motor pump which have to be submurged in water, which is not feasible for my project) but the good pressure is also needed.&#xA;I am thinking about piston pump but i am not being able to understand if it's going to work with my project or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my project,i have a jug of water from which i want to get the water and send it to other but I can't submerge the motor in water all the time.&lt;/p&gt;&#xA;" OwnerUserId="9965" LastActivityDate="2017-10-20T11:32:05.643" Title="Pump like solenoid but more powerful?" Tags="|mechanical-engineering|electrical-engineering|biomedical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17619" PostTypeId="1" CreationDate="2017-10-20T07:45:59.910" Score="1" ViewCount="69" Body="&lt;p&gt;There are several machines that have (rotating) cutting blades. For example the kitchen aid Vorwerk Thermomix can cut pieces of Food in hundreds of pieces within a few seconds. But why it is so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the blade hits something, there must act some Forces that cut it into pieces. How to compute the Forces that act on a blade to the Thing that the blade is cutting? Can I also compute stress values during the cutting process? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any literature in Engineering mechanics that unravels the mystery of cutting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I wonder why knives are classified with Rockwell hardness. I would be happy about a Basic introduction into this topic. &lt;/p&gt;&#xA;" OwnerUserId="8941" LastActivityDate="2017-10-20T07:45:59.910" Title="Engineering mechanics of cutting blades" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17620" PostTypeId="2" ParentId="17617" CreationDate="2017-10-20T11:32:05.643" Score="2" Body="&lt;p&gt;It sounds like you need a Diaphraghm pump. These are self-priming, like solenoid pumps, and come in a wide range of sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far above the water is the highest point in your system compared to the jug? And what volume flow rate do you require?&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-10-20T11:32:05.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17623" PostTypeId="1" CreationDate="2017-10-20T15:10:50.157" Score="3" ViewCount="47" Body="&lt;p&gt;I am currently developing an interstellar mission for a school project. I am thinking whether to design the propulsion system (Bussard scoop) or the main hull and structure of the spaceship first. What do aerospace engineers begin with usually?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am somewhat more inclined towards choosing propulsion first because most of the ship's mass will consist of the engine anyway and the rest can be &quot;built around&quot; it, whereas the engine will be optimized upon completion of the rest of the ship.&lt;/p&gt;&#xA;" OwnerUserId="11717" LastActivityDate="2017-10-20T17:25:15.657" Title="Development of an interstellar mission: should I design the propulsion system or the main hull of the ship first?" Tags="|aerospace-engineering|propulsion|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17624" PostTypeId="2" ParentId="17623" CreationDate="2017-10-20T17:25:15.657" Score="1" Body="&lt;p&gt;The order of problem solving for most designs comes down to a fairly simple order: Specify Requirements, Evaluate, Test, Refine, Optimize, Detail.  (Repeating the last five steps in iteration).  So, we need requirements.  Unfortunately, requirements are tied somewhat into the design, so we've got even more iteration to do.  Here's the order I'd recommend:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Kinematics&lt;/strong&gt;  There's a reason you walk before you run.  Solve your planned distance, time, velocity, acceleration, delta V, turn radius, etc.  Consider your gravitational well you'll need to pull out of and the brakes you'll need to put yourself on once you get to your new one.  This way you have an idea of your loads.  Anything you can figure out that doesn't require knowing a particle's mass, such as velocity, time, and acceleration, are open here.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Material Science&lt;/strong&gt; Materials are a very real requirement, and this is where I'd start with for everything.  This is for both the hull and the engine, as well as for the life support systems and everything else. Somehow we'll need to find materials that can hold everything up.  We don't need to get into too much detail here, but we need to get a mass estimate and a basic size estimate to finalize the loads.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Final Mass Estimate&lt;/strong&gt; We then need to place and figure out the main components mass.  Depending upon our constraints, we need to know the mass for passengers, life support, experiments, electrical, engines, etc.  Finalizing this mass estimate will allow for sizing of the main components.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Engine Sizing&lt;/strong&gt; Once we finalize loading, we're in a better world to size the engine.  We begin here using our existing materials science requirements and finalize a high specific impulse engine to meet out preliminary mass estimate.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Main Component&lt;/strong&gt; The hull, whose main purpose is to keep everything from falling apart during maneuvers, can't be finalized until we size the remainder of our components.  Design them up based upon the duration of the trip (if that had to be changed during the engine sizing).  If that changes the mass estimate, go back and redo the engine sizing.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Hull Design&lt;/strong&gt; Now with the final requirements in place, we need to test the hull.  If it falls within our mass estimate, now we can move to optimization.  Otherwise, back to recheck all of the assumptions that were based on the mass estimate.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Optimize&lt;/strong&gt; All along the way, and after you finish, optimize as much as you can.  With more and more details in place, and with mass and other specifications in place, you can finalize a lot of details and find ways to optimize your design.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-10-20T17:25:15.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17625" PostTypeId="1" CreationDate="2017-10-20T18:18:41.970" Score="-1" ViewCount="107" Body="&lt;p&gt;I was reading an article on them, and I thought should these become the main weapons of the ship then a single EMP could disarm them.&lt;br&gt;&#xA;Then that brought up the questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are EMP s still just sci-fi?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the military have solid countermeasures against EMP usage/EMP made irrelevant against us ships?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And has serious damage ever been caused by one?&lt;/p&gt;&#xA;" OwnerUserId="13357" LastEditorUserId="9" LastEditDate="2017-12-11T19:35:12.337" LastActivityDate="2021-02-03T02:01:40.343" Title="How dangerous are these new rail guns on the U.S. Navy ships?" Tags="|electrical-engineering|structural-analysis|power-electronics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17627" PostTypeId="2" ParentId="8505" CreationDate="2017-10-20T19:19:45.497" Score="2" Body="&lt;p&gt;The reason is due to slight instability of the fans, and bad fan stability selection.  To begin with, all backwards inclined fans have, to some extent, an unstable range.  All fans are designed to operate at an RPM using an RPM chart.  This includes the stable and unstable ranges of the fan:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ax5cF.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ax5cF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So long as the fan is in the &quot;downward slope&quot;, this is considered acceptable.  This has been discussed as &lt;strong&gt;&lt;em&gt;fan surge&lt;/em&gt;&lt;/strong&gt; with &lt;a href=&quot;https://engineering.stackexchange.com/questions/14604/how-to-design-a-vacuum-climbing-robot/14633#14633&quot;&gt;other answers&lt;/a&gt; on this site.  One of the big notices with fan surge is that it stops at the local maximum of the fan, and goes no farther.  Many times, a single fan is considered to operate at this point and this is fine for operation.  However, operating multiple fans in parallel can cause problems if they are in the normally stable region, but above the maximum pressure at no flow point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xki3Y.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xki3Y.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because the double fans can oscillate as you describe, with one fan operating at the left point of the curve, and one fan operating at the right point of the curve, then rapidly switching back and forth.  Only by ensuring that the static pressure across the fans is less than the no-flow pressure, can you ensure completely stable operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more research, see below some additional sources.  &lt;a href=&quot;https://www.ksb.com/centrifugal-pump-lexicon/parallel-operation/191648/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Math Source&lt;/a&gt; &lt;a href=&quot;http://www.greenheck.com/library/articles/42&quot; rel=&quot;nofollow noreferrer&quot;&gt;Graphical Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-10-20T19:19:45.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17628" PostTypeId="2" ParentId="8360" CreationDate="2017-10-20T19:25:31.243" Score="2" Body="&lt;p&gt;Have you tried a gauge guard?  It's a simple diaphragm that allows pressure to transmit, without the liquid crossing the barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bxubz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bxubz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-10-20T19:25:31.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17632" PostTypeId="2" ParentId="17601" CreationDate="2017-10-21T05:38:04.037" Score="1" Body="&lt;p&gt;many radio receivers contain a circuit called an IF oscillator. the output of this oscillator is added to the radio frequency signal that the radio is receiving to downconvert the incoming high frequency to a lower frequency that is easier to process. Old radios used IF oscillators that were powerful enough to actually transmit signals on their IF frequency which other receivers could then  pick up and home in on. during the second world war, this technique was used to find enemy ships at night, until they learned how to shield their receivers so they would not inadvertently transmit on their IF frequency and thereby reveal their location. I do not know how well-shielded radios of recent manufacture are against &quot;spurious transmission&quot;, which is what this effect is called.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-21T05:38:04.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17633" PostTypeId="2" ParentId="17596" CreationDate="2017-10-21T05:42:02.003" Score="0" Body="&lt;p&gt;you can also try mixing some 5 minute epoxy with fine metal dust as a filler/thickener, to yield a heavy paste, and then apply enough of it to the shaft to fill the gaps between it and the bore. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-21T05:42:02.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17636" PostTypeId="1" CreationDate="2017-10-21T11:07:29.480" Score="0" ViewCount="379" Body="&lt;p&gt;I want to calculate the density of CO2 using a cubic equation of state in order to plot the P-v diagram. From theory I know that I get one real root when the substance is in single-phase region and two when it is in two-phase region. In the latter case the smaller volume is associated to liquid while the other to vapor. I'm using Matlab to solve the cubic eos, incrementing the pressure and the temperature at fixed intervals. How can I merge the two solutions to represent the actual Pv plot?&lt;/p&gt;&#xA;" OwnerUserId="13368" LastActivityDate="2024-03-21T20:03:20.377" Title="PV diagram from cubic equation of state" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17637" PostTypeId="1" AcceptedAnswerId="17638" CreationDate="2017-10-21T13:39:37.783" Score="4" ViewCount="312" Body="&lt;p&gt;I was wondering about the ethics of an interviewer asking to see your design notebooks in an interview.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First off, I'm not sure if it's even legal for someone to ask that, since many of the designs I have are covered by (Non disclosure Agreements) NDA'S  with various companies which the interviewer is not a part of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, I'm not sure if the request to review my notes is even ethical?  Your design notebook is basically a screen shot of EVERYTHING you've worked on good, bad, right and wrong asking to see it is almost like asking to see your work email account from another company, in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guidance on this would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastEditorUserId="90" LastEditDate="2017-10-21T15:16:09.803" LastActivityDate="2017-10-22T00:51:35.857" Title="Ethics in an interview" Tags="|ethics|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17638" PostTypeId="2" ParentId="17637" CreationDate="2017-10-21T14:21:00.990" Score="5" Body="&lt;p&gt;Only offer work you wish to show and have the right to show.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If work is subject to an nda, then they should not ask and if they do, they should respect that you are under an nda...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-21T14:21:00.990" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17641" PostTypeId="2" ParentId="17637" CreationDate="2017-10-21T19:36:56.540" Score="3" Body="&lt;p&gt;Although already answered, for such an interview question the best response would be to show them some portfolio containing a few sample design sketches, instead of NDA protected materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they really wish to see a notebook related to a previous company, I would fear about their ethical basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been to several interviews as the interviewer. Occasionally, I've asked candidates to draw some sketches while explaining their previous experience or a technical topic. I would tell them especially not to explain anything proprietary. My motivation was to see how they explain their ideas through drawing and words. This practice both verifies the depth of their experience and helps understand the communication style and strength of the candidate.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-10-21T19:36:56.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17642" PostTypeId="1" AcceptedAnswerId="17730" CreationDate="2017-10-21T20:15:00.317" Score="1" ViewCount="112" Body="&lt;p&gt;Can polycarbonate be used to store a cryogenic liquid for a short duration, let's say 15-20 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard that PVC will be really brittle at low temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to do is a rocket and I need a tank to store my LOX. I need it to be something like PVC or polycarbonate because they are light and cheap enough for my purposes. Also, they are often sold for tubing, the shape I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other suggestions of material will be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10037" LastActivityDate="2017-10-26T06:56:49.853" Title="Polycarbonate as a short-term cryo-tank?" Tags="|materials|cryogenics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17643" PostTypeId="1" AcceptedAnswerId="17649" CreationDate="2017-10-21T21:53:54.827" Score="0" ViewCount="65" Body="&lt;p&gt;Question: &lt;a href=&quot;https://i.stack.imgur.com/BSWNf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BSWNf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what i thought; see below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mJ5Iz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mJ5Iz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we do not know density if i where to use that formula on the image file .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and i a m wondering why i was given the Reynolds number .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;need help.&lt;/p&gt;&#xA;" OwnerUserId="12830" LastActivityDate="2017-10-22T07:01:43.883" Title="Coefficient of Drag (Cd) Question" Tags="|fluid-mechanics|airflow|drag|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17644" PostTypeId="1" CreationDate="2017-10-21T22:16:04.900" Score="3" ViewCount="51" Body="&lt;p&gt;Does a test body subjected to a strain that causes plastic deformation still have elastic recovery?&lt;/p&gt;&#xA;" OwnerUserId="11623" LastEditorUserId="1832" LastEditDate="2017-10-21T23:10:13.060" LastActivityDate="2017-10-25T14:02:40.580" Title="Behavior of materials on application of tension" Tags="|mechanical-engineering|materials|mechanical-failure|deformation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17645" PostTypeId="2" ParentId="17644" CreationDate="2017-10-21T23:16:42.653" Score="3" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a perfectly elastoplastic material (where the stress-strain relationship is perfectly linear until the yield point and is then constant and equal to the yield point), the material will elastically recover all of its elastic deformation, but the plastic deformation will remain. This is the blue line in the figure below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In (not perfectly) elastoplastic materials, the same general rule still applies, with recovery following the elastic deformation slope. This is the brownish line in the figure below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VYgHq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VYgHq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;&lt;a href=&quot;https://materials.ucsb.edu/~matclass/101/pdffiles/Lecture_9.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material has become &quot;strain-hardened&quot;, which means it has a new yield point (at the point of release) but retains the same modulus of elasticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In non-elastoplastic materials (perfectly plastic, visco-elastic, visco-plastic, etc), things get messier, but I don't think the OP wants to go into that sort of detail.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-10-25T14:02:40.580" LastActivityDate="2017-10-25T14:02:40.580" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17646" PostTypeId="2" ParentId="17637" CreationDate="2017-10-22T00:51:35.857" Score="2" Body="&lt;p&gt;The simplest way to handle is if the work is propitiatory to your previous employer then indicated up front. Also remember looks like you are interviewing for a new position, so take the opportunity to ask if there anything that is of particular interest to the interviewer that you use alternative methods to answer the interviews questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, sometime interviewer ask this questions to validate how you protect propitiatory information. So if you offer up propitiatory information from previous employers, chances are you will also offer up their propitiatory information too.     &lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2017-10-22T00:51:35.857" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17647" PostTypeId="1" CreationDate="2017-10-22T01:38:21.663" Score="3" ViewCount="62" Body="&lt;p&gt;I am trying to model a machine that uses eyelets to support a piece of fabric, and this design uses a chain to hold the eyelets in a loop; the links are pictured below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iInDN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iInDN.png&quot; alt=&quot;Eyelet Chain Link&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this machine runs more and less of these links will be needed, and so I'm trying to design a support to keep the chain under tension. The ideas i have are segments of a circle to act as a hand to pull outward from outside, or a loop that is made to push outwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circle segments could be good but there would need to be more of these segments when you increase the number of links. This means either the operator or machine would have to know when to add these supports and be able to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loop could be good as well, but it would need to able to range in size from ~1cm to 20cm in circumference. This would be difficult to make durably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other way that machines have supported adjusting chains?&lt;/p&gt;&#xA;" OwnerUserId="13374" LastActivityDate="2017-10-22T06:58:22.043" Title="Supporting a Chain While Its Size is Varied" Tags="|mechanical-engineering|mechanisms|systems-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17648" PostTypeId="2" ParentId="17647" CreationDate="2017-10-22T03:37:41.393" Score="1" Body="&lt;p&gt;Consider having a &quot;loop&quot; of chain on an idler pulley that can move to add / remove slack and sprung or weighted (gravity is free) as necessary : this could be on the return side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;This image gives an idea : just assume the lhs is vertical, those, or only one, wheels can be pulled by a spring or even simpler just have weights hanging on them. The total can be adjusted for the amoun t of chain or the effect of different fabrics...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rhs shows a little slack - that before adjustment gets longer; where you add extra chain to provide the loop - need an extra gear for the return though...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g2Bs0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g2Bs0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit for idea 2 : sorry for poor image - just a hack with copy/paste...&#xA;&lt;a href=&quot;https://i.stack.imgur.com/EplPs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EplPs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-10-22T06:58:22.043" LastActivityDate="2017-10-22T06:58:22.043" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17649" PostTypeId="2" ParentId="17643" CreationDate="2017-10-22T07:01:43.883" Score="1" Body="&lt;p&gt;So the formula you need is Density = P /(R*T) and the value of R you have been given is for air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have re-worked your answer then you should calculate the Reynolds number and then you can check your result here : &#xA;&lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/dragsphere.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Drag on a sphere by NASA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-22T07:01:43.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17651" PostTypeId="1" AcceptedAnswerId="17653" CreationDate="2017-10-22T12:43:33.683" Score="1" ViewCount="1991" Body="&lt;p&gt;I have to determine the centre of mass for a 3D object (bounding box: approximately $0.30 \text{ m} \times 0.20 \text{ m} \times 0.10 \text { m}$) that is not regular and does not have a uniform density. I know how to determine the centre of mass for a flat object. I thought that I could try to use a similar approach for the 3D object, but the problem that I have is that I cannot determine the intersection of the planes as it is somewhere inside the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I thought I could use a balance for three directions and determine the centre of mass by finding the point of balance for all three &quot;main directions&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the last method a good method? What is the best way to determine&#xA;  the centre of mass by measurement?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-10-22T14:15:44.450" LastActivityDate="2018-05-09T14:43:09.877" Title="Experimental method to determine the centre of mass for 3D object" Tags="|measurements|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17652" PostTypeId="2" ParentId="17651" CreationDate="2017-10-22T13:26:38.887" Score="2" Body="&lt;p&gt;Two options, you can pick the more suitable / practical one.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;option 1: hanging&lt;/strong&gt; : you can hang the object and draw a gravitational (vertical) line or many of them. Then you hang it from another point and again draw the line(s). They coincide at the CG point. You may have to do some analysis to find metric values for the CG location. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;option 2: tilt and weigh&lt;/strong&gt; : measuring it at a certain pitch angle gives you the CG_x and CG_y, if you locate three scales on a perfect floor and place your object pressing three scales. If you measure it after tilting it (like 30 degrees) you can also find out the CG_z value.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Your method is similar to &lt;strong&gt;option 2&lt;/strong&gt;. But I couldn't be sure whether one measurement with the suggested setup is sufficient.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-10-22T13:26:38.887" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17653" PostTypeId="2" ParentId="17651" CreationDate="2017-10-22T16:20:58.810" Score="3" Body="&lt;p&gt;I regularly determine the 3d centroid of UAVs (drones) in this way:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;a) Find a container with flat, orthogonal faces (a box) that contains&#xA;the object tightly &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Pad the interior with tissue or styrofoam peanuts so that the&#xA;object does not move &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) Use a wooden dowel as a fulcrum and balance the container  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) Mark the location of the dowel vs. the box face &lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) Repeat for the other sides &lt;/p&gt;&#xA;&#xA;&lt;p&gt;f) the intersection of the planes marked on the container coincides&#xA;with the mass centroid.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The mass of the container + peanuts should be small compared to the object to be measured.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-10-22T16:20:58.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17654" PostTypeId="1" CreationDate="2017-10-22T22:07:45.100" Score="0" ViewCount="141" Body="&lt;p&gt;Question: &lt;a href=&quot;https://i.stack.imgur.com/mn1BH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mn1BH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See what I tried below and verify if I'm wrong. I need help here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YUs8r.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YUs8r.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12830" LastActivityDate="2017-10-22T23:47:21.307" Title="Rotation of Fluid in a Cylinder" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17655" PostTypeId="2" ParentId="17654" CreationDate="2017-10-22T23:47:21.307" Score="2" Body="&lt;p&gt;You just need to get the height of the paraboloid (air) in the cylinder. Knowing that the volume of the paraboloid is half of the cylinder of the same height,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H_{paraboloid}=2*H_{cylinder}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $H_{cylinder}$ is the height of air when it is not rotating, $0.2\ m$, therefore $$H_{paraboloid}=0.4m=y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;substituting the value in the equation, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y=\frac{\omega^2x^2}{2g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x=0.1\ m$, (radius of paraboloid) and $g=9.81\ m/s^2$, you will get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega=28.014\ rad/s = 267.51 rev/min$$&lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2017-10-22T23:47:21.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17656" PostTypeId="2" ParentId="17651" CreationDate="2017-10-23T02:38:46.537" Score="1" Body="&lt;p&gt;Another method is to pendulum the thing and calculate the pendulum arm from the swing period that corresponds to the CG. Note, there are two options here, a single pendulum where the object also rotates with the swing, or multiple suspension points of the same length where the object stays flat as it swings. As in the other methods, to get the exact cg, you need to do this at least 3 times in noncoplanar orientations and calculate the intersection. This is actually an approved method for computing the VCG of smallish boats. &lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2017-10-23T02:38:46.537" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17658" PostTypeId="1" CreationDate="2017-10-23T09:49:40.973" Score="1" ViewCount="786" Body="&lt;p&gt;Abstract:The detergency property in oil comes from a chemical additive called detergent which has a property of preventing the deposition of carbon deposits and wash it away.&#xA;There is also this property of lube oil being Dispersant where an additive is added to divide the larger size deposits into tiny particles to be carried in a colloidal suspension evenly throughout the bulk of oil, which can be removed while filtration of the oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:Can this property of detergency and dispersancy measured so as to ascertain how much or how easily it can react and dislodge unwanted particles in an engine? &lt;/p&gt;&#xA;" OwnerUserId="8750" LastActivityDate="2017-10-23T09:49:40.973" Title="Detergency and dispersancy" Tags="|fluid-mechanics|measurements|engines|marine-engineering|quality-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17661" PostTypeId="1" CreationDate="2017-10-23T10:30:29.007" Score="0" ViewCount="90" Body="&lt;p&gt;I have always used the term &quot;invert level&quot; as the &quot;floor level&quot; of pressure and gravity pipelines alike. A long section for pressure pipe from another engineer came across my desk where his &quot;invert level&quot; indicates the bottom of the trench  as the level below the pipeline (includes bedding)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have gone through various handbooks at the office and did an online search, I can not find a distinction between pressure vs gravity pipes, but the reference often refer to gravity pipelines or channels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the term &quot;invert level&quot; applicable as the &quot;floor&quot; of a pressure pipe or the bottom of the trench?&lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2017-10-23T10:30:29.007" Title="Is the term &quot;invert level&quot; applicable as the &quot;floor&quot; of a pressure pipe or the bottom of the trench?" Tags="|pipelines|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17662" PostTypeId="1" AcceptedAnswerId="17671" CreationDate="2017-10-23T11:33:16.233" Score="2" ViewCount="48" Body="&lt;p&gt;I'm designing a probe to provide a stable and continuous stream of exhaust gas sample to an analyser. The probe has to be inserted into the exhaust for about 20–30 cm to keep outside air from entering the probe. The static pressure there is about 1000 Pa above atmosphere. The dynamic pressure is probably way higher. The material of the probe is brass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that the fluctations in the environment inside the exhaust may have significant influence on my sampling. The analyser precisely dilutes the exhaust gas to match the sensor's specs, and environmental changes may alter the dilution ratio, resulting in incorrect readings. The probe has to offer a sample at a pressure as constant as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure inside the exhaust will force exhaust gas into the probe, and with changing engine load, the temperature, static, and dynamic pressure will change, forcing a different amount of exhaust gas into the probe. I want to have that pressure as stable as possible, so that I can correct for it in my calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have studied the pitot tube and my design is now as illustrated below:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pDv4n.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pDv4n.png&quot; alt=&quot;probeShape&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to eliminate the influence of dynamic pressure, but I'm not sure if this is the best way to create a sampling of constant pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;strong&gt;tl;dr&lt;/strong&gt;, consider image above, is there a better way to offer samples of constant pressure, in an environment that has fluctuating static and dynamic pressure, and temperature?&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="163" LastEditDate="2017-12-27T07:12:27.160" LastActivityDate="2017-12-27T07:12:27.160" Title="Can I improve this exhaust gas probe to be immune to pressure changes?" Tags="|pressure|airflow|gas|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17663" PostTypeId="1" AcceptedAnswerId="17664" CreationDate="2017-10-23T12:02:11.097" Score="5" ViewCount="370" Body="&lt;p&gt;I have a question about the &lt;em&gt;significant figures&lt;/em&gt; topic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&#xA;Find the volume of a cylinder having a diameter measured as 2.3 ft and length measured as 8.25 ft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution:&lt;/strong&gt;&#xA;V = π(D^2 /4)h = π((2.3)^2)(8.25)/4 = 34.276739 cu ft = 34 cu ft&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;br&gt;&#xA;In my workbook is explained that the final answer is rounded to two significant figures because the quantity 2.3 has two significant figures.&lt;br/&gt;&#xA;Why will I have to round to two significant figures if the quantity 8.25 have three significant figures? &lt;/p&gt;&#xA;" OwnerUserId="13389" LastEditorUserId="63" LastEditDate="2017-10-23T17:08:24.497" LastActivityDate="2017-10-25T04:16:57.317" Title="Significant figures" Tags="|civil-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17664" PostTypeId="2" ParentId="17663" CreationDate="2017-10-23T12:10:40.030" Score="4" Body="&lt;p&gt;You can only assume the accuracy of the least precise component of the input. so if you have elements with only 2 SF then you can not assert a higher level of precision.&lt;/p&gt;&#xA;" OwnerUserId="12804" LastActivityDate="2017-10-23T12:10:40.030" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17665" PostTypeId="2" ParentId="17662" CreationDate="2017-10-23T12:50:18.393" Score="0" Body="&lt;p&gt;One solution to your problem I saw was to take the gasses through a water trap, at least that controlled the temperature for them. Did it add extra moisture ? I don't have the answer for that, but it did give them a smooth gas flux into the analyser...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-23T12:50:18.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17666" PostTypeId="1" CreationDate="2017-10-23T13:19:51.910" Score="3" ViewCount="140" Body="&lt;p&gt;I am a university student and I need some help in a project. For the past few years now, many accidents have occurred whereby in overcrowded buses, passengers are accidentally propelled from the back emergency door. Or, sometimes because the rear door is not properly closed.The driver are unaware when this incident occurs as he has no control on the rear door. And after a deep analysis, I found that accidents like these have occurred in US and in European countries as well.&#xA;Hence, I wanted to find a solution which can be implemented in any buses which will prevent these type of accidents in the future. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;An automated system which will prevent the rear door from opening while the bus is in motion and will automatically unlock when the bus is at rest.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some links describing accidents that have occurred:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://abc7ny.com/news/boy-opens-exit-door-on-moving-school-bus-falls-out/358411/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boy opens exit door on moving school bus, falls out&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.dailymail.co.uk/news/article-3716265/Family-calls-tighter-safety-rules-son-nine-falls-60mph-coach-saved-passing-surgeon-nurse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Family calls for tighter safety rules after son, nine, falls out of 60mph coach and is only saved by a passing surgeon and nurse&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Actually my problem is that I can't find a device which can tell me if the bus is in motion or not.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some initial tests with an accelerometer. &lt;strong&gt;But when the bus moves at constant speed, its acceleration is zero and hence the device assumes the bus has stopped.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can use an RPM sensor but I need a solution which can be easily plugged-in in any buses. That is, the system can be implemented by anyone... something which can easily be mass produced. &lt;/p&gt;&#xA;" OwnerUserId="13390" LastEditorUserId="368" LastEditDate="2017-10-26T15:57:02.590" LastActivityDate="2017-10-26T20:06:15.093" Title="How to detect whether a bus is in motion or not" Tags="|mechanical-engineering|electrical-engineering|control-engineering|measurements|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17670" PostTypeId="2" ParentId="17666" CreationDate="2017-10-23T19:01:37.660" Score="1" Body="&lt;p&gt;I would personally use a GPS to detect movement of the bus. GPSes already outputs speed, which means you don't have to calculate it yourself. If the speed is non-zero then the bus is moving. This will also detect movement with locked wheels, e.g. sliding on slippery surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all GPSes sends output in the form of NMEA telegrams (which is ASCII) over serial line (usually RS-422). It can also be plain ASCII over Ethernet. This is easy to interface to with an ordinary computer. Most modern computers don't have COM ports, but USB-to-COM adapters are cheap and pretty much plug-and-play. It shouldn't be a problem to interface to microcontrollers either. Modern cars (and buses for that matter) already utilize GPS speed, but I don't know if it's easy to get information from that particular GPS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed telegram is called &lt;a href=&quot;http://aprs.gids.nl/nmea/#vtg&quot; rel=&quot;nofollow noreferrer&quot;&gt;VTG&lt;/a&gt;. It looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;$GPVTG,360.0,T,348.7,M,000.0,N,000.0,K*43&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last number before the K is ground speed. The K indicates that the unit is knots, but it will vary between GPSes and how they are configured. A GPS will typically output this telegram once per second, but the frequency can usually be increased to 10 Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The read speed will be somewhat noisy, so a deadband around zero might be needed when considering whether the bus is moving or not. A lowpass filter can also be used, but this introduces phase lag, so I'd go for deadband instead. Here is some data I gathered from a ship on DP (not moving). A bus doesn't roll and pitch like a ship, so the data would not be as noisy in your case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dTIUr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dTIUr.png&quot; alt=&quot;movement of vessel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this particular graph I would consider anything below 0.1 m/s as not-moving and anything above as motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disadvantages:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this solution has some disadvantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is no GPS coverage in tunnels.&lt;/li&gt;&#xA;&lt;li&gt;Very tall building can cause shadowing.&lt;/li&gt;&#xA;&lt;li&gt;Doesn't work when &lt;a href=&quot;http://www.telegraph.co.uk/news/2016/10/21/is-kremlin-cyber-warfare-behind-moscow-gps-quirk-sending-uber-ca/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Putin&lt;/a&gt; is in town.&lt;/li&gt;&#xA;&lt;li&gt;A plain old speedometer connected directly to wheel will always be more reliable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditDate="2017-10-26T20:06:15.093" LastActivityDate="2017-10-26T20:06:15.093" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17671" PostTypeId="2" ParentId="17662" CreationDate="2017-10-23T19:56:01.697" Score="1" Body="&lt;p&gt;It sounds like you are unable to affect the analyser end of things. My first suggestion would be to adjust the analyser software to undersample the probe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this does not sound possible and without knowing more about your design constraints, I would suggest adding an accumulator. With only 1K pa over Atmos. you should be able to calculate the volume of a small accumulator that could smooth pressure spikes and allow the gas to cool slightly for a better read. I've seen designs in applications similar to what you are describing which have a two port accumulator (more of a settling chamber) where one is intake from the probe front end and the second contains an orifice calculated to provide a set flow for a given pressure range to the end sensor.&lt;/p&gt;&#xA;" OwnerUserId="13394" LastActivityDate="2017-10-23T19:56:01.697" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17672" PostTypeId="2" ParentId="17483" CreationDate="2017-10-23T21:53:07.257" Score="2" Body="&lt;p&gt;The examples in which it is useful are already given in your PDF file. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first reason is modal decomposition (use $x=Vz$, in which $V$ is the matrix containing the eigenvectors of the matrix $A$). Assuming we have n distinct eigenvalues, then modal decomposition helps us to turn a difficult system into a diagonal system, in which we can solve each line of the ODE by a simple integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second application is the controllable canonical form. This system also has a very simple structure. Additionally, it is used to simplify the pole placement method when doing it by hand, without using the Ackermann formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third application is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kalman_decomposition&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kalman decomposition&lt;/a&gt; of a state space system into controllable &amp;amp; observable, controllable &amp;amp; not observable, not controllable &amp;amp; observable and not controllable &amp;amp; not observable subsystems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you can conclude that a linear transformation might be used to transform the system into a form which is easier to handle hand the original system or it gives you additional information that you couldn't simply see without the transformation.&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2017-10-30T13:39:46.033" LastActivityDate="2017-10-30T13:39:46.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17675" PostTypeId="2" ParentId="8059" CreationDate="2017-10-24T05:23:42.747" Score="2" Body="&lt;p&gt;what I would do instead is omit the capacitors and immerse the load resistor R1 in a small, well-insulated container of water with a thermocouple in it. regardless of the nature of the waveform being generated by the motor while it is being used as a brake, the temperature rise of the water will furnish an accurate accounting of the energy being dumped by the motor, as long as the resistance of the motor is small compared to the load resistance.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-24T05:23:42.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17676" PostTypeId="1" CreationDate="2017-10-24T05:37:30.307" Score="2" ViewCount="48" Body="&lt;p&gt;I'm designing a drill powered vehicle that needs to be able to drive uphill and downhill. &lt;a href=&quot;https://i.stack.imgur.com/P6siG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;The drive train roughly looks like this&lt;/a&gt;. The drill locks up when it's not running so I need something that will allow the shaft to freely spin when the power is off. I looked into centrifugal clutches but my drill doesn't produce high enough rpm for that to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything else that I could purchase or manufacture that would make this possible?&lt;/p&gt;&#xA;" OwnerUserId="11194" LastActivityDate="2017-10-24T06:21:03.077" Title="How can I power a driveshaft while also allowing it to freely spin when not under power?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17677" PostTypeId="2" ParentId="13082" CreationDate="2017-10-24T05:41:20.307" Score="2" Body="&lt;p&gt;It appears the spline-like grooves pressed into the shaft's surface are upsets, which are there to derive an interference or press fit between the shaft and the gear. This is intended to fasten the gear onto the shaft so hard that it will not come loose, but if too much torque is applied to the gear by the worm drive, it will shear loose from the shaft and spin. Glue would not help since the strength of the press fit is derived from the shear strength of the stuff the gear is made out of, which is probably be greater than that of any glue you might try. If this is a common problem, then the assembly method of the gear on the shaft probably needs to be revisited.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="90" LastEditDate="2021-04-17T06:06:19.657" LastActivityDate="2021-04-17T06:06:19.657" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="17678" PostTypeId="2" ParentId="12051" CreationDate="2017-10-24T05:47:22.600" Score="1" Body="&lt;p&gt;your graphs indicate that the volumetric efficiency of the machine is a function of the temperature of the working fluid for high values of the pressure ratio. This is not too surprising; but it is likely that the pump manufacturer furnishes an efficiency curve for only one temperature, hence only one line on the chart. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-24T05:47:22.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17679" PostTypeId="2" ParentId="10486" CreationDate="2017-10-24T05:59:40.863" Score="0" Body="&lt;p&gt;this system is constituitively linear but contains one time-varying coefficient, so it is technically nonlinear. even if your modeling software does not permit time-varying coefficients, you can do a piecewise model in which you start the run, halt it, change the system mass, and restart it. this will get you close, and by shortening the run time at a given value of mass you will approach the behavior of the &quot;real&quot; system. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-24T05:59:40.863" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17680" PostTypeId="2" ParentId="17676" CreationDate="2017-10-24T06:21:03.077" Score="4" Body="&lt;p&gt;yes, it is called a freewheel and most bicycle rear hubs have them, so you can coast without having to pedal the bike. in the mechanical engineering world, this is called an overrunning clutch.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-24T06:21:03.077" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17683" PostTypeId="2" ParentId="14016" CreationDate="2017-10-24T12:04:40.003" Score="0" Body="&lt;p&gt;Intel HEX format is well documented out there.  Basically, it contains a bunch of bytes at specific addresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surely there are also many utilities out there that read HEX files and do various things with the information in them.  For example, here is the output of my HEX_DUMP utility run on the snippet you provided:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;000001C0-000001DF (32): 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00&#xA;  00 00 00 00 00 00 00 00 00 00 00 00 00&#xA;000001E0-000001FF (32): 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00&#xA;  00 00 00 00 00 00 00 00 00 00 00 00 00&#xA;00000200-0000021F (32): 54 71 07 80 80 9A 07 80 80 9A 07 80 80 9A 07 80 8C 53 06&#xA;  80 00 00 11 A0 14 00 11 A0 00 00 00 00&#xA;00000220-0000023F (32): 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00&#xA;  00 00 00 00 00 00 00 00 00 00 00 00 00&#xA;00000240-0000025F (32): 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00&#xA;  00 00 00 00 00 00 00 00 00 00 00 00 00&#xA;00000260-0000027F (32): 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00&#xA;  00 00 00 00 00 00 00 00 00 00 00 00 00&#xA;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This output is showing you the address range resulting from each line, the number of data bytes on that line, then the actual data bytes in HEX.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What that data means and what you do with it is, of course, up to you.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-10-24T12:04:40.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17684" PostTypeId="1" AcceptedAnswerId="17703" CreationDate="2017-10-24T13:04:39.607" Score="4" ViewCount="10449" Body="&lt;p&gt;I’m pretty sure it’s something to do with the fitting, possibly specifying to use the minimum circumscribed or maximum inscribed circle to tolerance the diameter but I can’t remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tolerance would be in the form&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;⌀123 ± 0.01 Ⓔ&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In case the Unicode doesn’t work correctly that’s&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[Diameter Symbol] [value] [plus minus symbol] [value] [E in a circle]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13405" LastActivityDate="2017-10-25T09:34:27.497" Title="What does an E in a circle after a diameter mean?" Tags="|drawings|technical-drawing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17686" PostTypeId="2" ParentId="11799" CreationDate="2017-10-24T15:05:38.580" Score="-2" Body="&lt;p&gt;The k number is the rate constant given in pounds per inch, which up to a multiplying factor should be the k in Hooke's law.&lt;/p&gt;&#xA;" OwnerUserId="13406" LastActivityDate="2017-10-24T15:05:38.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17687" PostTypeId="2" ParentId="17684" CreationDate="2017-10-24T15:15:09.097" Score="2" Body="&lt;p&gt;The Ⓔ is one of the options in the &quot;feature control frame&quot; as defined in the ISO 8015 standard, or ASME Y14.5M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It stands for &quot;envelope requirement.&quot; The basic idea is that the tolerance of the feature should not exceed the &lt;em&gt;perfect&lt;/em&gt; geometrical shape at the maximum material size, which is identified by the symbol Ⓜ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.globalspec.com/reference/43764/203279/chapter-10-envelope-requirement&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.globalspec.com/reference/43764/203279/chapter-10-envelope-requirement&lt;/a&gt; for an example. The standard documents themselves give the definitive explanation, of course.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-10-24T15:15:09.097" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17688" PostTypeId="1" CreationDate="2017-10-24T15:25:01.643" Score="0" ViewCount="253" Body="&lt;p&gt;I'm looking into plane strain and I'm a little confused about the concept. If we assume the material is much longer in the $z$-direction, the typical assumptions I see are&#xA;\begin{equation*}&#xA;|\varepsilon_{12}|, |\varepsilon_{13}|, |\varepsilon_{33}| \ll 1&#xA;\end{equation*}&#xA;where $\boldsymbol{\varepsilon}$ is the strain tensor. Even if we suppose those terms are zero, it seems to yield the following stresses are nonzero:&#xA;\begin{equation*}&#xA;\begin{split}&#xA;\sigma_{33} =&amp;amp; C_{1133} \varepsilon_{11} + C_{2233} \varepsilon_{22} + C_{3312} \varepsilon_{12} \\&#xA;\sigma_{23} =&amp;amp; C_{1123} \varepsilon_{11} + C_{2223} \varepsilon_{22} + C_{2312} \varepsilon_{12} \\&#xA;\sigma_{13} =&amp;amp; C_{1113} \varepsilon_{11} + C_{2213} \varepsilon_{22} + C_{1312} \varepsilon_{12} \\&#xA;\end{split}&#xA;\end{equation*}&#xA;In practice are the $\sigma_{i3}$ typically ignored or assumed to be small? This formulation seems to imply we have $\sigma_{33}$ is not necessarily small even in the case of isotropy, for instance $C_{1133}$ or $C_{2233}$ $\gg 1$.  As we already have three (hopefully independent) equations for $\sigma_{11},\sigma_{12},\sigma_{22}$ in terms of $\varepsilon_{11}, \varepsilon_{12},\varepsilon_{22}$ we could I suppose rewrite $\sigma_{33}, \sigma_{23},\sigma_{13}$ in terms of $\sigma_{11},\sigma_{12},\sigma_{22}$, but I wonder if I am perhaps missing a core assumption of plane strain. For instance we assume $\frac{\partial u_1}{\partial z} \approx -\frac{\partial u_3}{\partial x}$ since $|\varepsilon_{13}| \ll 1$, but would it be wrong to strengthen this and assume $\frac{\partial u_1}{\partial z} = \frac{\partial u_3}{\partial x} = 0$, where $u$ is the displacement?   &lt;/p&gt;&#xA;" OwnerUserId="13407" LastActivityDate="2018-01-22T18:08:04.417" Title="Plane Strain in Anisotropic Materials" Tags="|elastic-modulus|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17689" PostTypeId="2" ParentId="17663" CreationDate="2017-10-24T15:52:58.950" Score="3" Body="&lt;p&gt;Selter's answer follows the book's approach.  A proper analysis, however, should go a little deeper.&lt;br&gt;&#xA;First, we stick with the convention that the uncertainty in any provided value is +/- half the least sig digit, i.e. 2.3 means 2.3 +/- 0.05 .   Next, we notice that the diameter enters into the formula squared and the height  linearly.   The standard approach is to look at the partial derivatives and do variational analysis.  Following the mechanisms presented, e.g., in &lt;a href=&quot;https://smile.amazon.com/Reduction-Error-Analysis-Physical-Sciences/dp/9339221206/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bevington&lt;/a&gt;  Chapter 3, we will get (&quot;Var&quot; is variance, or sigma^2) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$   \frac{Var(volume)}{Vol^2} = \frac{\pi}{4} *( 2*\frac{Var(diameter)}{Diameter^2} + \frac{Var(height) }{height^2} ) $&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-10-24T15:52:58.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17690" PostTypeId="2" ParentId="17688" CreationDate="2017-10-24T16:33:42.923" Score="1" Body="&lt;p&gt;There are two cases typically considered for stress on a 2d material, plane stress and plane strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In plane stress, the stress is considered negligible in the 3 direction.  This is the case for a material that has no forces in these directions.  Note the strain can change - the material can get thinner or thicker, but ultimately no new stress is being applied in the 3 direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In plane strain, the strain is considered negligible in the 3 direction.  This is the case for a material that is constrained to not get thicker or thinner in this direction.  Note that the stress can change - there is some force holding the material to be a constant thickness, to ensure the strain is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many pressure loading cases are combinations of plane strain and plane stress.  An analysis of both scenarios is taken, and a few FEAs of the 3D scenario can lead to an empirical equation.  For example, one of my analysis of a knuckle on a flat top head for FRP composites (with a 1 1/2&quot; radius knuckle) shows that the highest stress (at the knuckle portion), regardless of diameter, was 0.7 x the plain strain stress and 0.3 x the plain stress stress.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2017-10-24T16:40:52.643" LastActivityDate="2017-10-24T16:40:52.643" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17691" PostTypeId="2" ParentId="11394" CreationDate="2017-10-24T16:47:10.413" Score="3" Body="&lt;p&gt;Non-integer gear ratio reduces gear noise and improves service life. Micromo has done a nice job of explainining why [&lt;a href=&quot;https://www.micromo.com/media/wysiwyg/Technical-library/Gearheads/25_gearhead_construction_use.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;]. In short: Assume a gear tooth on gear A is faulty. Let's say the gear ratio is 3:1 (See picture 1). Then, this faulty tooth engages with three locations on the larger gear (gear B) at every rotation of the gear. Imagine this in the long run and it will cause wear only on those three locations. Now imagine the gear ratio was 3.444:1 (See picture 2). then the engaging locations on the larger gear (gear B) keeps shifting at every rotation (see the picture where the numbers on the wheel shows engagement at that number of rotation of gear A), hence wear is distributed among all gear B teeth and each tooth takes less wear in the long run thus more service life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rm7cr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rm7cr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dChTq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dChTq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13408" LastActivityDate="2017-10-24T16:47:10.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17692" PostTypeId="1" CreationDate="2017-10-24T19:05:58.777" Score="1" ViewCount="232" Body="&lt;p&gt;I was wondering how blenders used in the kitchen for the purpose of creating smoothies, soups, etc. operate, particularly, how the rotation of their blade is achieved, i.e. what sort of motor is used for this purpose. I am additionally interested in finding additional Bibliography in the bespoked subject, as well as the industrial methods used to generate each component of blenders (meaning not only the mechanical parts). If anyone has ideas on the Bibliography as well, please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="13411" LastActivityDate="2024-03-17T01:04:05.473" Title="Operation of Fruit Blender-Mixer" Tags="|mechanical-engineering|electrical-engineering|design|manufacturing-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17694" PostTypeId="1" CreationDate="2017-10-24T22:02:40.517" Score="5" ViewCount="62" Body="&lt;p&gt;Here on Earth, the atmosphere is ideal for jet engines which combust a fuel using the native oxygen as an oxidizer. On another planet, might it make sense to carry oxidizer in the fuel tanks and scoop the fuel from the air (assuming the planet contained some suitable Fuel in its atmosphere)? This is just the reverse of what jets here on Earth do. Would there be some reason for this to not work?&lt;/p&gt;&#xA;" OwnerUserId="13412" LastActivityDate="2017-10-24T22:02:40.517" Title="Could a Fuel Breathing Engine Rather Than an Oxidizer Breathing Engine Work?" Tags="|aerospace-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17699" PostTypeId="2" ParentId="11799" CreationDate="2017-10-25T05:50:26.337" Score="1" Body="&lt;p&gt;With McMaster, the part number usually is more related to where it is located in the warehouse or page number (or supplier's catalog item) then anything else.  They may list it as &quot;Aisle 96, Column 37, Shelf K, Bin 25&quot; as an example.  Other manufacturers and distributors use other part numbers.  It has very little, if anything to do with the part's performance typically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, sometimes the part number does correspond to the performance, and it does help.  Spears Plastics will list their schedule 40 pipe with a 4 in front, and an 8 in front of schedule 80 fittings.  They put a C in every piece for CPVC fittings instead of PVC.  And they have a -XXX code at the end, 010 for 1&quot;, 020 for 2&quot;, etc. &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-10-25T05:50:26.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17700" PostTypeId="1" CreationDate="2017-10-25T06:14:48.860" Score="0" ViewCount="852" Body="&lt;p&gt;I'm trying to create a flowchart of the operational process for a widget that I'm designing. What would be a good way to indicate a &quot;background&quot; process that occurs but can be affected by other steps in the flowchart? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I want to indicate that a vehicle will drive continuously and scan for obstacles continuously. If an obstacle is detected, the vehicle will then stop driving, change direction, and then continue driving and scanning again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I most effectively indicate this in a flowchart?&lt;/p&gt;&#xA;" OwnerUserId="13417" LastEditorDisplayName="user16" LastEditDate="2018-04-24T00:27:08.560" LastActivityDate="2018-11-20T05:01:30.290" Title="Indicate continuous background process in flowchart" Tags="|design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17701" PostTypeId="1" CreationDate="2017-10-25T06:23:24.767" Score="0" ViewCount="158" Body="&lt;p&gt;Say you have an LED light that emits green light. Are there sensors out there that could accurately determine the colour information of the green light? Such as the relative values of red, blue and green light being emitted from the LED.&lt;/p&gt;&#xA;" OwnerUserId="13418" LastEditorUserId="8055" LastEditDate="2017-10-25T17:00:50.540" LastActivityDate="2021-06-10T15:16:49.267" Title="Is there a way to determine the colour information of light emitted from an LED?" Tags="|electrical-engineering|sensors|lighting|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17702" PostTypeId="2" ParentId="17701" CreationDate="2017-10-25T08:28:05.850" Score="1" Body="&lt;p&gt;You want to do a light spectrum analysis on the led. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this with a prism that will spread out the light into colors and then find where the peaks and valleys are in the spectrum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That or get a &lt;a href=&quot;https://en.wikipedia.org/wiki/Optical_spectrometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Optical spectrometer&lt;/a&gt; which will give better data than just a colored barcode.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-10-25T08:28:05.850" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17703" PostTypeId="2" ParentId="17684" CreationDate="2017-10-25T09:34:27.497" Score="3" Body="&lt;p&gt;Like promised, I am now looking at &lt;em&gt;EN ISO 14405-1:2016&lt;/em&gt; where this E symbol (Ⓔ) is defined. More accurately is defined in section &quot;3.8 envelope requirement&quot;. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The upper value of a external feature defines a virtual cylinder that the entire feature has to fit inside. Likewise a internal features minimum value defines a cylinder that has to fit inside the hole. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Each slice of needs to be smaller or bigger (for internal features) than the other tolerance value measure as a two point size (imagine measuring with a caliper)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note: Nowehere does it say the cylinder is aligned to the centerline&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h4NmH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h4NmH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: Cylinder needs to be inside the envelope and not point measured smaller than the minimum size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See relevant standard for more robust explanation.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2017-10-25T09:34:27.497" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17705" PostTypeId="1" CreationDate="2017-10-25T13:56:28.387" Score="0" ViewCount="120" Body="&lt;p&gt;I need to mount a heavy panel against a wall.  The panel will be first mounted on two legs which go from the top of the panel to the floor.  These legs will transfer most of the panel's weight to the floor, and will also be bolted to the wall to prevent the panel falling forward.  Since there are only two legs (behind the panel), there will be some force exerted on the bolts and thus on the wall as the panel tries to fall forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate this horizontal force exerted on the wall?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a drawing:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8LHVh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8LHVh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1558" LastEditorUserId="1558" LastEditDate="2017-10-25T21:54:34.813" LastActivityDate="2017-10-25T22:23:42.777" Title="Calculating horizontal force exerted on a wall by a panel" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17708" PostTypeId="2" ParentId="17607" CreationDate="2017-10-25T15:20:51.737" Score="1" Body="&lt;p&gt;A column is typically modeled using the Euler's formula.  The idea is that as a column is compressed, the column will displace via compressing, or it can maintain its original length by deflecting into an arc.  This outward deflection is known as buckling. As a curved arc, the length between the two endpoints may have a shorter distance, as the shortest path between two points is a straight line.  However, this deflection will cause the overall length of the beam to remain the same.  A column made with material of a higher modulus of elasticity will have less deformation under a given stress, so overall it should handle more load before buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can verify this concept by looking at the Euler's formula for buckling:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F = \frac{n\pi EI}{L^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where E is the Young's modulus, I is the moment of Inertia, and L is the effective length (n is a constant based on the end conditions).  Reviewing the force required, we can see that two given beams of the same cross section and the same length, but with different modulus of elasticity, the stiffer beam can handle more force before buckling is onset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus a beam with a higher modulus of elasticity will have a higher factor of safety against buckling than an identically designed beam with a lower modulus of elasticity.  An engineer would therefore state that a stiffer beam is less susceptible to buckling.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2017-10-25T20:06:42.240" LastActivityDate="2017-10-25T20:06:42.240" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17709" PostTypeId="1" CreationDate="2017-10-25T16:25:03.920" Score="1" ViewCount="955" Body="&lt;p&gt;Suppose you have an LED that emits a spectrum of colours such that the resultant colour emitted is white. Will changing the temperature if the LED itself change the composition of the light's colour spectrum, hence changing its resultant colour?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My immediate answer to this is no but after a bit of research, I came accross this graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CgbmT.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CgbmT.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.lrc.rpi.edu/programs/nlpip/lightingAnswers/led/heat.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;website&lt;/a&gt; states that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Figure 9 shows the light output of several LEDs as a function of junction temperature. The temperature dependence is much less for InGaN LEDs (e.g., blue, green, white) than for AlGaInP LEDs (e.g., red and yellow).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As the graph clearly shows, the relative light outputs of the red, white and blue light changes at different rates as temperatures increase. This shows how the intensity of the light overall changes with the temperature of the LED system. &lt;strong&gt;However, does this mean that the colour composition of the emitted light changes with temperature as well?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13418" LastEditorUserId="8055" LastEditDate="2017-10-27T08:12:42.873" LastActivityDate="2021-06-10T09:09:58.950" Title="Does temperature affect the colour of the light emitted by an LED?" Tags="|electrical-engineering|lighting|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17712" PostTypeId="2" ParentId="17709" CreationDate="2017-10-25T19:07:39.883" Score="1" Body="&lt;p&gt;There is no such single LED.  White is produced either with a mixing of color LEDs or by embedding an LED, typically UV,  inside a &quot;bulb&quot; with a phosphor coating such that the coating absorbs the photons and re-emits a wide spectrum of light.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, if you are mixing tricolor LEDs and their power outputs change at different gains with temperature, then obviously the color mix will change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A given LED's bandgap will change slightly with temperature as well, but this is probably not significant in the situations you are interested in. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-10-25T19:07:39.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17713" PostTypeId="1" AcceptedAnswerId="17741" CreationDate="2017-10-25T19:09:14.373" Score="1" ViewCount="76" Body="&lt;h1&gt;Some background for context only&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I am in the process of designing a pressure system where my pressurized fluid needs to be kept incredibly clean, with typical contaminants at the ppt or below level. The fluid in question is more compressible than a typical hydraulic fluid, so I need to build in an expansion volume for when I cycle to lower pressure.  This expansion volume is going to look very much like a hydraulic accumulator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common hydraulic accumulators are pneumatic, and the gas charge is separated from the fluid by a bladder, diaphragm, piston, or metal bellows. All of these are problems for my cleanliness requirements, as the moving parts will tend to shed metal shavings, etc.  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The question&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;My question is, why is the separation of the two volumes necessary? In my case, this is for a fixed piece of equipment, so I can guarantee that the accumulator will always be vertical.  Can I just have a nitrogen-filled cylinder with the fluid connected at the bottom and rely on the fluid to act as my piston?  If this does work, why isn't it more common? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to address some of the comments:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not asking for advice on how to design a clean accumulator; I only want to know the fairly general question of whether it is possible to not separate the charge gas from the fluid.  &lt;/p&gt;&#xA;" OwnerUserId="13424" LastEditorUserId="13424" LastEditDate="2017-10-25T19:41:11.750" LastActivityDate="2017-10-26T06:24:35.067" Title="possibility for hydraulic accumulator with no moving parts?" Tags="|design|hydraulics|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="17714" PostTypeId="1" CreationDate="2017-10-25T19:09:07.600" Score="4" ViewCount="2109" Body="&lt;p&gt;&lt;strong&gt;Should Heavy Trucks drive more slowly on weak (bitumen) roads to reduce potholes in the road surface?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am discussing with a feasibility engineer what causes more potholes.  He says it is simply the 'Weight' on the road (ie of the vehicle touching the road), and I argue that it is the acceleration AND weight combined...(Perhaps the Work to be a little more exact... Something that relates to acceleration/direction, mass and time).  My thoughts are, that a truck travelling fast will bounce and produce more short term weight, and also accelerating back and forth will put more pressure on the road surface - more than the truck weighs in fact, all combining to create potholes or enlarge existing potholes.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(What 'he' means by Weight Only - is that they can predict how a road will wear down simply by a factor of the Weight of the trucks that will drive on it.  They do not consider the speed (limit), and I also have no real idea of how accurate they are but it seems roads wear out a little faster than they think)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain how I am wrong?  Or why the feasibility engineer is correct?  (Or please let me know if this question is not in the correct forum, for which I apologise in advance)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB:  We have forestry trucks running on our small road, which 'seems' to be tearing it up - and they drive fast!  Faster than the 30 km/h recommended limit (legal limit 50...People, dogs, kids, deer, moose, bikes around).  So I would like to argue that the forestry trucks should drive super slowly to preserve our precious roads, which are ostensibly for local traffic only.  Not that I mind the trucks on the roads every now and then, I just want them to reduce impact.  The roads are probably only 'made' for local car traffic, and the odd tractor.&lt;/p&gt;&#xA;" OwnerUserId="13428" OwnerDisplayName="Grantly" LastEditorUserId="13428" LastEditDate="2017-10-25T21:33:47.627" LastActivityDate="2024-02-20T21:00:33.650" Title="Does the speed of Heavy Vehicles make more potholes" Tags="|civil-engineering|highway-engineering|car|" AnswerCount="7" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17715" PostTypeId="2" ParentId="17714" CreationDate="2017-10-25T19:56:40.567" Score="3" Body="&lt;p&gt;If you have a 20 ton truck sitting stationary on the road, it's exerting 20 tons of force on the roadway.  If that truck is traveling at 100km/hr, it is still exerting 20 tons of force on the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure this is why your friend says that weight is the only factor that matters.  The truck stopping or starting isn't even a major concern because that happens infrequently and randomly on the road.  Intersections and the roads around them might need to be stronger because of the constant braking and acceleration, but that's about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that's only for straight roads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On bumps and curves, speed is going to matter because speed + weight = momentum.  The inflection point of any bumps or elevation changes in the road will have more force on them as speed increases, and curves in the road will have more lateral (or perpendicular if the road is banked) forces as speed increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a road develops a slight depression, faster vehicles will hit it harder which could make it grow faster than slower traffic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, trying to add variable speeds to the calculations might be too complex or error prone for the engineers.  Using a static weight might be the best they can do accurately.&lt;/p&gt;&#xA;" OwnerUserId="2859" OwnerDisplayName="JPhi1618" LastEditorUserId="2859" LastEditDate="2017-10-25T20:48:37.103" LastActivityDate="2017-10-25T20:48:37.103" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17716" PostTypeId="1" CreationDate="2017-10-25T21:12:44.887" Score="2" ViewCount="98" Body="&lt;p&gt;I am used to strength being measured in units of stress, such as pounds per square inch, or Pascals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming what the author is trying to say is, &quot;this filament will break when holding 18 kg,&quot; is that correct?&lt;/p&gt;&#xA;" OwnerUserId="13429" LastActivityDate="2017-10-25T21:37:31.517" Title="&quot;The monofilament wire has a tensile strength of 18 kg&quot; What does this mean?" Tags="|mechanical-engineering|materials|strength|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17717" PostTypeId="2" ParentId="17716" CreationDate="2017-10-25T21:37:31.517" Score="2" Body="&lt;p&gt;Material strength is generally expressed in the form of stress but for a component using a load is perfectly logical. Equally while using Newtons would be more precise quoting an ultimate load as a mass is often more practically useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is less clear is whether this is a 'safe working load' (ie you can hang a mass of 18kg off it and it won't break) or if it is the ultimate load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Safe working loads generally include a factor of safety to account for dynamic loading and quoting the load in kg rather than N sort of implies this but assuming 18kgf as the breaking load would be more conservative in the absence of any other information. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-10-25T21:37:31.517" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17718" PostTypeId="2" ParentId="17700" CreationDate="2017-10-25T21:37:37.090" Score="1" Body="&lt;p&gt;I use sequence function chart for PLC code design and general industrial process flow diagrams. I find that separating out parallel loops simplifies the structure and legibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X8XT7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X8XT7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Independent process flows.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For PLC relay ladder logic I will number each step and can then make conditional branch by referencing a step in another flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a free online charting utility at &lt;a href=&quot;https://www.draw.io&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.draw.io&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2017-10-25T21:37:37.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17719" PostTypeId="2" ParentId="17705" CreationDate="2017-10-25T22:23:42.777" Score="2" Body="&lt;p&gt;As an order of magnitude approximation, do a moment balance. Assume the weight is in the center of the box, so 6 inch from the wall.  Suppose the wall force is at a single point half way up (54 inches from the floor).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; force = 150lb * 6 / 54 ~= 17 lb.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In reality that force will be distributed among (I imagine) 6 or 8 fasteners. Size them for 20lb &lt;em&gt;each&lt;/em&gt; should be no problem and give plenty of margin.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2017-10-25T22:23:42.777" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17720" PostTypeId="1" CreationDate="2017-10-25T23:09:27.917" Score="0" ViewCount="23" Body="&lt;p&gt;What would be the best method of accelerating a continuous flow of hydrogen plasma to 300 000 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have thought about a rail gun approach but don't think this would be suitable for these speeds or a continuous flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plasma can be flowing in anything.&lt;/p&gt;&#xA;" OwnerUserId="10361" LastEditorUserId="11324" LastEditDate="2017-10-26T17:21:11.937" LastActivityDate="2017-10-26T17:21:11.937" Title="methods of propulsion of hydrogen plasma" Tags="|acceleration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17721" PostTypeId="1" CreationDate="2017-10-26T00:21:09.103" Score="2" ViewCount="1102" Body="&lt;p&gt;The conventional wisdom consistently holds that &quot;oil (or grease) attracts dirt&quot;, giving rise to the need for seals and dry-lubricants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I've always considered that to be loose thinking, the truth being closer to &quot;oil &lt;em&gt;retains&lt;/em&gt; dirt (dust, grit)&quot; and that there are no electrostatic, magnetic, or gravitational forces at work actually drawing loose particles into an oil film.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my intuition correct or does an oil or grease film actually exert some attractive force on airborne particles?&lt;/p&gt;&#xA;" OwnerUserId="3937" LastActivityDate="2017-10-26T06:00:35.843" Title="Does oil actually &quot;attract&quot; dirt?" Tags="|fluid-mechanics|materials|friction|bearings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17723" PostTypeId="1" CreationDate="2017-10-26T01:43:20.280" Score="1" ViewCount="105" Body="&lt;p&gt;I'm afraid I'm completely out of my depth here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to use a tungsten filament to radiatively heat something near to it (in vacuum).  My issue is I cannot figure out the easiest and/or cheapest way to control the power to the filament.  The answer I keep circling around is something involving an SCR and maybe a transformer to step down the line voltage.  But some things I've read (Chromalox's product materials, mostly) indicate that I don't need a step-down transformer, or if I did I just need a phase-angle fired SCR.  Other places I've looked (product literature from other people, such as Omega) indicate that an SCR, even phase-angle fired, can't drive an inductive load reliably. I've also considered putting the transformer first, but most SCRs I've seen require line voltage input (either for auxiliary electronics or for their own operation). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another point of confusion is that I am not sure how, if at all, I am controlling the current in the filament (high temperature resistance=0.1 ohms ~= 15x low temperature resistance).  The SCR just turns the output on and off, after all, to produce some sort of average power, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have experience with this sort of system or goal?  The filament in question is designed to be driven at 7-ish volts and up to 62 amps, although I assume I could always drive it with higher voltage.  I would like to plug into a 120V single phase wall socket.  I am not wedded to the SCR, or really any element of the design, as long as it can be computer-controlled eventually (and get real hot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2017-10-26T05:54:15.357" Title="Control power to a tungsten filament?" Tags="|power-electronics|power|heating-systems|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17724" PostTypeId="2" ParentId="17714" CreationDate="2017-10-26T01:44:27.307" Score="1" Body="&lt;p&gt;Speed makes it worse; big vehicles cause the great majority of road damage. The damage cause by cars is mostly limited to'polishing&quot; ; the pavement is smoothed which reduces the friction. Many roads a scarified , those small grooves about 1/4 &quot; deep to offset the polishing damage. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-10-26T01:44:27.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17725" PostTypeId="2" ParentId="17709" CreationDate="2017-10-26T01:52:51.870" Score="1" Body="&lt;p&gt;To add to Carl Witthoft's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In diode-based lasers (similar concept to LEDs) temperature control is important for maintaining a precise, stable wavelength.  This is related to the band edge moving slightly with temperature.  However, as Carl Witthoft said, this is a small effect on the scale of &quot;colors&quot; -- my recollection is a degree Celsius corresponds to sub-nm changes in wavelength for a blue diode laser (405 nm). &lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2017-10-26T01:52:51.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17726" PostTypeId="2" ParentId="203" CreationDate="2017-10-26T02:04:18.977" Score="3" Body="&lt;p&gt;A few things, hopefully helpful:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Differential thermal contraction is almost certainly your enemy.  For most engineering materials, the vast majority of thermal contraction occurs between 300 and 77 K, the two temperatures you are working at.  Your PCB is almost certainly shrinking much more than the thing attached to it and cracking your epoxy (normal resin epoxies are notorious for cracking in cryogenic environments).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I work with cryogenics for my 9-5 and we use &quot;GE varnish&quot; for just about everything.  Also called IM7031 varnish.  Thins with ethanol/toluene mixture, and can be baked dry.  Tends not to crack in cryogenic environments.  It will hold pretty well without a cure, too.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another, more permanent option is Stycast, which comes in different flavors for different thermal properties.  If you want a LESS permanent option, Apiezon N or H grease works well.  H grease is thicker (maybe necessary if you have a big sample that weighs ~1g rather than ~10 mg).  Both undergo a glass transition at low temperature and hold on tight while giving electrical isolation and thermal contact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you are concerned about intermittent electrical contact, cigarette paper can be wetted by just about all of the &quot;goos&quot; I mentioned, and will make sure there is no accidental contact.  Put a layer between your two samples.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A good general purpose reference on cryogenic techniques is Jack Ekin's book, Experimental Methods for Low Temperature Measurements.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2017-10-26T02:04:18.977" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17727" PostTypeId="2" ParentId="16734" CreationDate="2017-10-26T02:10:02.503" Score="0" Body="&lt;p&gt;If money is no object, an alumina plate could work well, maybe 1/4&quot; thick.  It is a fairly nonreactive material and we use it in crucibles up to 1500C.&lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2017-10-26T02:10:02.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17728" PostTypeId="2" ParentId="15161" CreationDate="2017-10-26T02:15:03.403" Score="4" Body="&lt;p&gt;The short answer is, anyone can make graphene, but knowing you did or doing anything with it requires a bit more equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I work with people who work with graphene.  This is not the only way to make graphene, but it is how they do it.  They use Scotch tape to pull apart graphite crystals repeatedly.  Then the Scotch tape with lots of little flakes gets attached to a Si wafer with an oxide layer, and is carefully peeled away.  Some flakes are left behind.  Usually the wafer already has had &quot;align markers&quot; for an SEM patterned on it using lithography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is using an optical microscope to identify suitably thin flakes that may be monolayers.  The nearby align markers are noted.  Then, further cleanroom processing allows them to deposit electrical contacts and do electrical measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that answers your question!&lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2017-10-26T02:15:03.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17729" PostTypeId="2" ParentId="17666" CreationDate="2017-10-26T02:20:28.170" Score="0" Body="&lt;p&gt;Every bus I've been on bounces up and down while it moves.  I think an accelerometer (or three) would work fine.  Might take a bit of calibration, but it would be a simple solution for detecting the motion of the bus.&lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2017-10-26T02:20:28.170" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17730" PostTypeId="2" ParentId="17642" CreationDate="2017-10-26T02:23:38.767" Score="1" Body="&lt;p&gt;Short answer is no, as other commenters have suggested.  The best solution is a small Dewar flask, which is stainless steel.  Depending on how much you need, a commercial Thermos could do in a pinch.  (Edit: I saw you want to use this for a tank on your rocket.  If you are using it for a tank, there is not much in the cheap+light phase space.  Cheap would be stainless and light would be titanium.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As some unsolicited advice, liquid oxygen is very dangerous and if you are asking this question I would encourage you to read about the hazards associated with it a bit more.  In particular, many things that you don't think are combustible will find a way to burn when they are impregnated or in contact with liquid oxygen.  (Also, there are the obvious cryogenic hazards as well.)&lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2017-10-26T02:23:38.767" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17731" PostTypeId="2" ParentId="17625" CreationDate="2017-10-26T02:30:07.867" Score="0" Body="&lt;p&gt;Almost every weapons system on a ship would be susceptible to an &quot;EMP&quot; because almost every weapon is controlled by an electronic or computer-based targeting system.  However, to answer your question, the sort of large area EMP you are imagining (e.g. as in the movie &quot;The Matrix&quot;) is not something that &quot;exists&quot; in any meaningful sense outside of nuclear detonations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that was helpful.&lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2017-10-26T02:30:07.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17733" PostTypeId="2" ParentId="17444" CreationDate="2017-10-26T02:40:52.643" Score="1" Body="&lt;p&gt;As Jem Eripol points out, it is to prevent erosion / collapse of the earth above it.  Likely the land was cleared and leveled to produce the park, and to prevent the hillside above it from filling in the space, the retaining wall was added.  There is also the element of the designer thinking it &quot;looked cool&quot; -- that is a function in and of itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helped.&lt;/p&gt;&#xA;" OwnerUserId="2283" LastActivityDate="2017-10-26T02:40:52.643" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17735" PostTypeId="1" CreationDate="2017-10-26T03:18:12.433" Score="1" ViewCount="1651" Body="&lt;p&gt;I need to find the torque that will rotate a stainless steel drum that is filled with small steel balls. The total volume of this drum is 210 cubic feet, and there will be 38 cubic feet of material inside at any given point in time. It is a horizontal cylinder that will rotate through its central axis. What is the torque required to start turning the drum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass of Drum: 13,500 lb&lt;br&gt;&#xA;Mass of Material inside: 13,700 lb&lt;br&gt;&#xA;Diameter of Drum: 38&quot;&lt;br&gt;&#xA;Length of Drum: 297&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drum to Rotate at 10 rpm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friction in the bearings and the drive is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I have to consider that the material will move up the side of the drum as it rotates, but I'm not sure how to go about doing this.&lt;/p&gt;&#xA;" OwnerUserId="13431" LastEditorUserId="13431" LastEditDate="2017-10-26T11:33:46.387" LastActivityDate="2017-10-26T11:33:46.387" Title="Torque Required to rotate a horizontal drum/cylinder with material inside" Tags="|mechanical-engineering|design|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17736" PostTypeId="2" ParentId="17723" CreationDate="2017-10-26T03:47:39.843" Score="2" Body="&lt;p&gt;We have a test rig that works something like this, to heat up small objects (typically cubes of material with sides a few mm long) to about 1000C fairly quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat source is a &quot;tube&quot; of ordinary (and fairly cheap) tungsten-halogen car headlight bulbs surrounding the test piece, rated 12V 55W each, with home made parabolic reflectors covered in cooking foil behind them. The bulbs have their own internal reflectors to provide a directional beam, as required for a car headlight. The foil reflectors are just to mop up any stray light and heat output - mainly from the radiation that missed the test piece and needs reflecting back from the opposite side of the &quot;tube.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So long as the power supply doesn't exceed 12V DC the bulbs will survive OK. So far we have only fried one - and that was when somebody knocked the thing of the bench onto the floor while it was running!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We used a PID controller that &quot;just happened to be lying around&quot; to control the voltage of a high current DC power supply, with a thermocouple providing the feedback loop into the controller. We never bothered to tune the PID much - the light output is continually flickering &quot;randomly&quot; at a few Hz, but the thermal inertia of the rig and the test piece keeps their temperatures pretty stable, and the bulbs don't seem to mind that level of abuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So don't make your design more complicated than it needs to be - throw something together, try it, and fix any problems you hit. We never really had to do any hard thinking about the &quot;fix the problems&quot; step - it worked &quot;well enough&quot; almost from the first test run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A phase-angle supply switching at 50 or 60 Hz might give the bulbs a harder time than this &quot;very noisy DC voltage&quot; approach - but we haven't tried it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: everything in this rig this running in air, not in a vacuum, so there is a lot of heat lost through convection. Even so, four bulbs (220W) can get it up to temperature in 10 or 20 minutes.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2017-10-26T03:53:09.803" LastActivityDate="2017-10-26T03:53:09.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17737" PostTypeId="2" ParentId="17723" CreationDate="2017-10-26T05:54:15.357" Score="1" Body="&lt;p&gt;I recommend a variac (variable ac transformer) rated at 500 watts, This will downconvert the wall outlet voltage 120VAC to anything in the range of 0 to 120. Alternatively, to get easier control, you could put a 10 to 1 stepdown transformer after the variac; as the variac voltage is turned up or down over the range of 0 to 120VAC, the output of the stepdown tranformer will vary from 0 to 12VAC. Note that the stepdown transformer has to be rated at 500 watts too. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-26T05:54:15.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17738" PostTypeId="2" ParentId="17721" CreationDate="2017-10-26T06:00:35.843" Score="2" Body="&lt;p&gt;you are right, a surface that is wet with oil will capture dirt that falls on it (and not let go) but the oil itself does not attract dirt. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-26T06:00:35.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17739" PostTypeId="2" ParentId="17735" CreationDate="2017-10-26T06:13:34.530" Score="1" Body="&lt;p&gt;start with the following approximation: at the moment the drum starts turning, assume the balls and whatnot resting in the bottom of the drum do not move around but are for that moment stuck together so they will rotate as a single mass around the axis of the rolling drum. This distribution of mass will have a well-defined moment of inertia you can calculate, which we call (I). Next, determine how fast you wish the drum to accelerate, this number is (alpha). the starting torque is then torque (tau) = (I) x (alpha). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;alternatively, you can go to the website of someone who manufactures ball mills (which is what you described) and see what they recommend for the torque and RPM of the motor drive of a ball mill of a certain shape and mass.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-26T06:13:34.530" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17740" PostTypeId="2" ParentId="17720" CreationDate="2017-10-26T06:19:10.117" Score="-2" Body="&lt;p&gt;that is an extremely high speed. Is the plasma flowing in a pipe or vacuum system, or is it in some sort of containment vessel, which also has to be accelerated along with the plasma?&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-26T06:19:10.117" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17741" PostTypeId="2" ParentId="17713" CreationDate="2017-10-26T06:24:35.067" Score="1" Body="&lt;p&gt;gases are soluble in oils. this means that as your oil enters the gas-filled accumulator and the pressure in it builds up, gas will probably dissolve into the oil. Then, when you release the pressure on the system, the gas will boil out of solution and cause the oil to foam up vigorously. the foamy oil then enters your plumbing and creates big problems, especially if it is ingested by a pump.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-26T06:24:35.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17742" PostTypeId="2" ParentId="17714" CreationDate="2017-10-26T06:31:36.183" Score="1" Body="&lt;p&gt;I think JPhi is correct. if you have a slight unconformity in the road surface- say a small depression- then the leading surface of a fast-moving tire will strike the edge of the depression with a significant impact force, which is entirely different from the response of the road to a purely static or even a suddenly-applied load. In this way, fast-moving vehicles will &quot;hammer&quot; the unconformity in a way that a slow-moving vehicle will not. On this basis, then, fast-moving vehicles should do more damage to an &quot;imperfect&quot; road surface than slow-moving vehicles would. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-26T06:31:36.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17743" PostTypeId="2" ParentId="17692" CreationDate="2017-10-26T06:40:33.357" Score="0" Body="&lt;p&gt;for low speed blenders like food processors, typical motors are induction type, as used in electric fans or table saws. for high speed blenders for crushing ice for drinks and so forth, brush-type AC motors are used, as in old-style (corded, 120VAC) electric drills. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-26T06:40:33.357" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17744" PostTypeId="2" ParentId="17666" CreationDate="2017-10-26T06:48:49.417" Score="-1" Body="&lt;p&gt;you could pick up a signal from a safety switch in the door jamb which detects whether the door is open or closed. assuming the bus has an automatic transmission, when it is stopped to pick up passengers, the driver's foot is on the brake and hence the brake lights are energized. You then design a simple circuit which sounds an alarm buzzer if the brake light is off (that is, if the driver has pulled his foot off the brake in order to start moving)  AND the door is not closed. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-26T06:48:49.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17745" PostTypeId="2" ParentId="17642" CreationDate="2017-10-26T06:56:49.853" Score="0" Body="&lt;p&gt;in designing the flask to hold your LOX, please note that the LOX will slowly warm up as it sits in the flask, and will then pressurize the flask.  In addition to preventing heat loss and not being brittle at low temperatures, the flask will also have to be strong enough to withstand pressure buildup inside it without exploding. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-26T06:56:49.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17746" PostTypeId="1" CreationDate="2017-10-26T07:01:17.187" Score="2" ViewCount="923" Body="&lt;p&gt;As far as I know, moments/torques occur when forces applied are not in line. However, I often see moments/torques being applied at a single point in free body diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can moment/torque be applied at a single point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Illustration Below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikibooks/en/4/45/Beam_moment_Solid_Mechanics.png&quot; alt=&quot;Diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caption: The above image shows an ideal moment acting at the center of a beam. An ideal moment is one which is not associated with a force.&#xA;&lt;a href=&quot;https://en.wikibooks.org/wiki/Strength_of_Materials/Loading_of_Beams&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link for the page&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13433" LastEditorUserId="2015" LastEditDate="2017-10-26T16:41:09.303" LastActivityDate="2017-10-27T01:45:06.923" Title="Moment and Torque at a Single Point" Tags="|mechanical-engineering|structural-engineering|beam|moments|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17748" PostTypeId="1" AcceptedAnswerId="18613" CreationDate="2017-10-26T09:54:18.667" Score="3" ViewCount="343" Body="&lt;p&gt;I have come across a problem, which has troubled me for some time now.  What needs to be done is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mass on a rod 0.6m (mass less) has a mass of 1 kg attached at the end of it. The rod needs to be rotated 60°, within t=120 sec (see image). What I would like to do is size a rotational spring (located at the pivot point) and a damping system, such that it that will damp the spring force. Thus the rotation happens within the specified amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have written the generic differential equation of the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J\theta'' + C\theta' + K\theta = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(typical differential equation of damped spring system)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for a critical damped system, and for $t=0$, $\theta=0$ I have the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta(t)=A t \exp(-bt)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is a constant, and $b$ is the damping coefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how can I continue, such that I can size the damping coefficient and the spring constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if I can continue from here, how should I then proceed to size my system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note $g=0$, no gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CYLbM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CYLbM.png&quot; alt=&quot;problem decription, PLEASE NOTE g=0, no gravity&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13437" LastEditorUserId="2015" LastEditDate="2017-10-26T22:59:30.553" LastActivityDate="2018-08-26T01:01:07.697" Title="Simple mass spring damping system sizing K, and C factors" Tags="|mechanical-engineering|vibration|linear-motion|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17751" PostTypeId="1" CreationDate="2017-10-26T14:17:42.253" Score="0" ViewCount="46" Body="&lt;p&gt;I am an engineer currently designing a system in which the leakage of a hose is tested. The system consists of the following components:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;An air data system used to generate a pressure inside the hose.&lt;/li&gt;&#xA;&lt;li&gt;An automated pressure measurement device used to measure the pressure in the hose.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The goal of the test is for the air data system to exert a pressure on the hose and make sure that the pressure doesn't drop while the system maintains that pressure. The air data system will maintain the pressure and the measurement device will periodically read the pressure over a length of time. After some time has passed there is a limit to which the pressure should have changed. If that limit is exceeded the test will be considered a failure otherwise considered a pass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure measurement device is not perfect and leaks a small amount of pressure on its own. Is there a way to account for how much pressure is leaked by the pressure measurement device? I'd like to do this in order to get a slightly more accurate value for pressure in the hose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it helps, here is some other information I know about the system:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The air data system is +/- 2% accurate.&lt;/li&gt;&#xA;&lt;li&gt;The air data system can be detached from the pressure measurement device and measure its own pressure it's putting out.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is there a way where maybe I can measure the pressure of the air data system in isolation, and then with the pressure measurement device attached... in order to maybe obtain some sort of mathematical/physical relationship or system of equations? I am thinking that something along those lines might work.&lt;/p&gt;&#xA;" OwnerUserId="5892" LastActivityDate="2017-10-26T14:30:23.383" Title="How can I determine pressure leakage in a pneumatic system?" Tags="|pressure|measurements|product-testing|instrumentation|air|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17752" PostTypeId="2" ParentId="17751" CreationDate="2017-10-26T14:30:23.383" Score="2" Body="&lt;p&gt;Fit a dummy test hose with no leaks - a solid tube for example, then pressurise the system to a known value perhaps the maximum used +10%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then take readings every 10 seconds / 5 minutes (whatever is relevant for your real tests) for a defined amount of time ie 10* the length of a normal test. Repeat 10 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Average the results for the &quot;loss per unit time&quot; of the standard system parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Test a real hose and see if the results deviate form the known leak curve... If they do you have a leak and its severity...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-26T14:30:23.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17753" PostTypeId="1" CreationDate="2017-10-26T14:54:19.673" Score="1" ViewCount="19" Body="&lt;p&gt;I have only been in the HVAC business for 3 years and I have already encountered 2 cases of leaking air conditioning evaporators.  In both cases the evaporators were located above the central furnace in an up-flow forced air system.  My diagnosis is that while the evaporators may have been designed to withstand the hot air temperatures leaving the furnace heat exchanger and flowing through the evaporator, the repeated expansions and contractions of the metallic evaporator coils due to these extreme temperature fluctuations shortened their lifespan.  In other words, if the AC evaporator did not have to contend with the heat from the furnace it would last much longer and less prone to develop leaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for a better design where both furnace and AC evaporator have to share the same forced air system, wouldn't it be smarter to place the evaporator and its casement beneath the furnace but above the forced air fan?&lt;/p&gt;&#xA;" OwnerUserId="10921" LastActivityDate="2017-10-26T14:54:19.673" Title="What would be the disadvantage if any to placing evaporator casement beneath furnace for central forced air HVAC sustem?" Tags="|hvac|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17754" PostTypeId="2" ParentId="17666" CreationDate="2017-10-26T16:31:11.680" Score="0" Body="&lt;p&gt;I would recommend you take inputs from multiple sensors for a 'foolproof' solution. Since this is a safety device, redundancy is important as well. Here are a couple of inputs that you could use.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Vehicle speed reported by the ECU, you could read this through the vehicle communication bus, I'm not sure if commerical busses have a standard OBD port(like in cars), but there should be a standard electrical communication bus interface that you could tap into to get onboard data.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Combined data from IMU and GPS modules, similar to the ones used in drones, can help you get the velocity as well. You could find these modules for cheap on eBay/Alibaba. There will be some effort involved in writing firmware for a microcontroller to fuse data from the sensors to  calculate Velocity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13262" LastActivityDate="2017-10-26T16:31:11.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17755" PostTypeId="2" ParentId="17746" CreationDate="2017-10-26T16:34:12.160" Score="1" Body="&lt;p&gt;As you say, torques are caused by forces not in line. In this case you have a &quot;couple,&quot; which is caused by TWO equal and opposite forces not in line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendation is that you draw an EQUIVALENT problem with an upward force and downward force of equal magnitude acting the same distance away from the point of application of the couple. Notice then that it doesn't actually matter which point we take the moment about, you get the same couple about every point. Thus, a couple, or a &quot;pure moment&quot; is a FREE vector. You can place it anywhere on the body and it has the same effect on the deformation or the motion.&lt;/p&gt;&#xA;" OwnerUserId="13429" LastActivityDate="2017-10-26T16:34:12.160" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17756" PostTypeId="2" ParentId="17746" CreationDate="2017-10-26T17:13:03.413" Score="0" Body="&lt;p&gt;Isolated moments, as shown here are more of a mathematical construct than a realistic construct.  That said, in three dimensions this is completely possible.  A mounted pneumatic drill turning a bolt at the midpoint of the beam (with minimal loading into the beam) would cause this concentrated moment.  The motor applies a torque to the bolt.  This model works well to simplify the complex drill forces, and accurately describes the stresses in the beam when the distance between the flats on a bolt is small compared to the the length of the beam.  This substitution from a complex motor to a simple concentrated moment is known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Saint-Venant%27s_Principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saint-Venant's Principle&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A shorter beam would result in more complex boundary conditions and the need to consider higher order elastic effects.  This is the primary difference between a continuum mechanics researcher, who would model these effects with partial differential equations, and an engineer, who would multiply the Saint-Venant stresses with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Stress_concentration#Concentration_factor_calculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;stress-concentration factor&lt;/a&gt; and design accordingly.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-10-26T17:13:03.413" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17757" PostTypeId="1" AcceptedAnswerId="17765" CreationDate="2017-10-26T20:45:11.510" Score="3" ViewCount="290" Body="&lt;p&gt;It is easy to found stepper motors with torque over 2NM for a few coins, however, only expensive industrial servomotors achieve this torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Stepper motor: 10usd for 1.3nm with 2.8V at 3.5A rpm 0.6kg&lt;/li&gt;&#xA;&lt;li&gt;Servo AC motor: 278usd for 2.39nm with 220V~ at 3A 3000rpm 6kg&lt;/li&gt;&#xA;&lt;li&gt;Servo DC motor: 67usd for about 0.4nm 24V at 7.8A 3000rpm 1Kg&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Torque is not power, and closed loop has many benefits, but I am still confused:I believe many industrial purpose require quite high torque (CNCs, robotic, etc.) and it would be preferable (to my understanding) a lower rpm with higher torque. Additionally, it seem to me that any purpose for a servomotor require more than a few 0.4nm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this torque that different? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the benefit / purpose of those non-hybrid servo-motors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="11129" LastActivityDate="2017-10-27T14:21:27.300" Title="Why servomotor has such low torque compared with stepper motors?" Tags="|motors|stepper-motor|servo|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17759" PostTypeId="2" ParentId="17748" CreationDate="2017-10-26T21:20:50.307" Score="0" Body="&lt;p&gt;Correct me if I'm wrong, but it looks like you have your values for $\theta$ and $t$ as 60° and 120s respectively. I'm not sure what $A$ is, it might be a design choice for you to make. For your critically damped system, you would only need to know the damping coefficient $b$. You would just rearrange the equation until it became&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$b = \frac{\ln(\theta/A)}{\ln(t) t}$$&lt;/p&gt;&#xA;" OwnerUserId="13442" LastEditorUserId="2015" LastEditDate="2017-10-26T22:58:19.013" LastActivityDate="2017-10-26T22:58:19.013" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17760" PostTypeId="1" CreationDate="2017-10-26T21:33:50.800" Score="1" ViewCount="2978" Body="&lt;p&gt;Is there a preferred solution to the problem of driving parallel rotary shafts from a single motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have six shafts that I want to drive from one motor. The shafts are all:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;rotating the same direction&lt;/li&gt;&#xA;&lt;li&gt;parallel&lt;/li&gt;&#xA;&lt;li&gt;planar (although this is not a hard constraint)&lt;/li&gt;&#xA;&lt;li&gt;low speed and moderate torque (~100W/0.1HP)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If I were to wrap one timing belt on pulleys around each shaft, most of the shafts would have very low contact angles and would likely slip. Running six timing belts, one from each shaft to the motor, is ungainly and expensive. Gearboxes would be difficult to align, have low efficiency (40-80% from my research), and require frequent maintenance (lubrication).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems like it must be a solved problem, but none of the solutions I can find seem very elegant. Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2023-05-29T15:12:04.943" Title="Driving many parallel shafts from one motor" Tags="|mechanical-engineering|power-transmission|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17762" PostTypeId="2" ParentId="17746" CreationDate="2017-10-27T00:59:38.270" Score="1" Body="&lt;p&gt;There's a reason why such moments are called &quot;ideal&quot;. That's because they're idealized, not real. They are a simplification of reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way you could get an ideal moment in a beam is by welding two rigid vertical bars at the midpoint, one above and one below the bar. And then apply opposing forces at the extremities of the bars. As far as the beam is concerned, it will suffer an ideal bending moment, without any associated forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sae1K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sae1K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For something a bit more reasonable, imagine the following structure. Beam C is the only one with any loading (ignore self-weight). It is supported by beams B, which are themselves supported by beams A, which are supported by pinned supports at their extremities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4Oigd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Oigd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fixity at the connection of beams B and C means that beam C will suffer a (small but non-zero) bending moment at the connection, which gets transferred to beam B as a concentrated torsional moment (along with the shear support reactions). Likewise, the connection between beams A and B will have the same bending moment in B transferred as torsion to A. However, the torsion in B also gets transferred to A as a concentrated bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, these concentrated bending-moment/torsion transformations can be easily described in terms of internal force couples as well.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-10-27T00:59:38.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17763" PostTypeId="2" ParentId="17746" CreationDate="2017-10-27T01:45:06.923" Score="0" Body="&lt;p&gt;An applied pure torque does not have a point associated with it and it is said to shared with the &lt;em&gt;entire&lt;/em&gt; body, as opposed to forces that act through a point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is entirely equivalent to the statement that rotational motion is shared with the &lt;em&gt;entire&lt;/em&gt; body, as opposed to linear velocity which is only defined at a single point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is almost impossible to produce a pure torque in real life, with net force zero. You can transmit forces without a torque using a bearing, but you cannot transmit a torque without any forces. At least not mechanically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torques represent a force at a distance, and a pure torque is mathematically equivalent to a zero force at infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to rigid body motion, where linear velocity is a representation of a rotation at a distance, and a pure translation is zero rotation about a infinite point.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-10-27T01:45:06.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17765" PostTypeId="2" ParentId="17757" CreationDate="2017-10-27T05:29:32.707" Score="3" Body="&lt;p&gt;Operating principle is different. In essence if you buy just a motor it will be just that. You can now add gearing that will reduce speed and increase torque. Since torque is directly proportional to gear ratio and mechanical advantage;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;MA = \frac{v_{input}}{v_{output}}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You get  a hundred times more torque if you gear 3000 rpm down to 30 rpm. So now that DC motor you listed would be 40Nm and that AC servo a 239Nm monster. So indeed those are way more powerful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further there may be gearing inside your stepper motor, its quite common. You can get geared servos. It is just more uncommon as designing the drive train needed is kind of the point of using normal motors. As the drive train is the part where you optimize things for your application. That or manufacture/design the motor yourself.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-10-27T14:21:27.300" LastActivityDate="2017-10-27T14:21:27.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17766" PostTypeId="1" CreationDate="2017-10-27T08:15:41.150" Score="2" ViewCount="3149" Body="&lt;p&gt;The picture below represents a refrigeration unit &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/43aGh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/43aGh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to calculate the power produced by the motor, two quantities must be measured &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Angular speed of the motor. This parameter can be observed from the tachometer, which indicates the rotational speed of the the compressor. The belt pulley ratio is used to find the rotational speed of the motor. &lt;/li&gt;&#xA;&lt;li&gt;The torque produced by the motor. In my lab's manual, it is written that the spring balance reading times the troque arm is equal to the torque produced by the motor. Can you help me understand how does this spring balance reading indicates the torque produced by the motor? (I know that the the torque is equal to Fxd, but I can't understand how does the spring balance reading represent the force produced by the motor) &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also, it is written in the manual&#xA;&quot;Assume typical compressor friction force =5N&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this value represent the friction between the belt and the motor? If not, then What does this value represent? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I think that since there is friction, then power dissipation will occur due to friction and thus Power delivered to the compressor = Power produced by the motor minus power dissipated by friction&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, if there is something wrong, correct me &lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="8055" LastEditDate="2017-10-30T14:53:15.493" LastActivityDate="2017-11-02T20:50:56.127" Title="How to determine the compressor power in a refrigeration system?" Tags="|mechanical-engineering|thermodynamics|motors|dynamics|power|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17767" PostTypeId="2" ParentId="17760" CreationDate="2017-10-27T08:38:53.277" Score="0" Body="&lt;p&gt;Connect a belt from each shaft to the next. Space taken up is twice the width needed for a single pair connection. There may be a bit of slack introduced that way which can be reduced by doing a binary split so each shaft powers 2 others which will need up to $\log n$ belts on the first shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you add a single smaller gear between each shaft that they couple onto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use tension wheels to push the single timing belt into the gears to increase contact pressure and area.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-10-27T08:38:53.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17769" PostTypeId="1" CreationDate="2017-10-27T11:25:58.047" Score="1" ViewCount="100" Body="&lt;p&gt;So I have this problem, every Sunday we have about 200 kids come over to our house in town, and they make a LOT of noise. We also have mildly nice neighbors that do not like noise Sunday afternoon when they are trying to watch a movie on their front porch. They live directly across the street, with only two vine-covered fences on either side of the road. They are getting somewhat fed up with all the noise, (imagine 100/200 kids yelling at the top of their lungs) and my dad had the idea that we build a brick wall to stop the sound. So my question is, how high would this have to be to be efficient, and is it worth the cost? (a 7ft wall would cost about $200-500) Or would it not work at all? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I previously posted this question as an acoustics physics question, it was closed for being off-topic, they suggested engineering as better, so I posted it here. &lt;/p&gt;&#xA;" OwnerUserId="13444" LastEditorUserId="2015" LastEditDate="2017-10-27T15:30:28.113" LastActivityDate="2020-02-12T04:41:21.487" Title="How much does a wall stop sound?" Tags="|mechanical-engineering|measurements|acoustics|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="17771" PostTypeId="1" AcceptedAnswerId="18751" CreationDate="2017-10-27T15:55:00.083" Score="5" ViewCount="92" Body="&lt;p&gt;In the event of an explosion through a high pressure gas pipe, is there a way to vent the pressure wave, in order to protect downstream objects? The mechanism must be faster than sound; pressure relief valves and even rupture discs are actually too slow and will not respond to such a &quot;fast&quot; pressure wave.&lt;/p&gt;&#xA;" OwnerUserId="13446" LastEditorUserId="2015" LastEditDate="2017-10-27T16:39:51.023" LastActivityDate="2018-01-06T03:08:11.650" Title="Is there a pressure relief mechanism that operates faster than sound in a material?" Tags="|mechanical-engineering|fluid-mechanics|pressure|mechanisms|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17772" PostTypeId="1" CreationDate="2017-10-27T17:04:35.227" Score="3" ViewCount="108" Body="&lt;p&gt;I'm wondering, if you put an air compressor pipe into a micro turbine, lets say &lt;a href=&quot;https://www.banggood.com/Mini-Engine-High-Speed-Turbine-Model-p-1058723.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;something like that&lt;/a&gt; attached to a motor, does air that get out of the turbine have the same energy as when it gets in ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because i get that the compressor might have to push harder, but the air itself still has to get out of the pipe, basically i don't get why you can't put like 10 of those turbines attached together and somehow get free energy, i know that it's impossible but i don't get why..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to calculate the energy of that thing?&lt;/p&gt;&#xA;" OwnerUserId="13448" LastEditorUserId="163" LastEditDate="2017-12-27T07:10:32.680" LastActivityDate="2018-01-26T08:15:16.827" Title="Compressed air energy storage?" Tags="|turbines|energy|compressed-air|" AnswerCount="2" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17773" PostTypeId="2" ParentId="17760" CreationDate="2017-10-27T17:23:28.510" Score="1" Body="&lt;p&gt;Consider a single large gear connected to the motor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller spur gears can be arranged around the central gear to drive as many shafts as the geometry allows.  Each shaft will rotate in the same direction, but opposite to the central gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of the size of the central gear vs. the outer gears will determine the speed and relative torque of each driven shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Designing the system to account for torque, inertia, shock loading etc is a much more complex problem.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-10-27T17:23:28.510" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17774" PostTypeId="1" CreationDate="2017-10-27T18:15:21.167" Score="1" ViewCount="234" Body="&lt;p&gt;I have some doubt on fluid mechanics that need some confirmation. I am on the section that discusses about compressible flow and the effect of variation of flow cross section area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It stated that that an isentropic flow would obey conservation of mass and what I wonder about is that any chance that an isentropic flow would disobey conversation of mass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What made me wondering this was I was considering the mass flow rate of isentropic flow in converging-diverging nozzle. If my understanding towards isentropic flow is not wrong, the mass flow rate of an isentropic flow would always be constant no matter of what position it is in the nozzle.&lt;/p&gt;&#xA;" OwnerUserId="13447" LastEditorUserId="13447" LastEditDate="2017-10-27T18:21:53.760" LastActivityDate="2017-12-27T01:27:59.523" Title="Isentropic Flow and Conservation of Mass" Tags="|fluid-mechanics|compressible-flow|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17775" PostTypeId="1" CreationDate="2017-10-27T19:04:15.007" Score="1" ViewCount="463" Body="&lt;p&gt;The image below shows a common technique for working on valves in the petroleum industry. A secondary valve 2 is closed and allows servicing of Valve 3.&lt;/p&gt;&#xA;&lt;p&gt;My question is about the blue area. What is the pressure in this area after the valve has been closed? Does it stay at 10,000 PSI? I opened Valve 3 after some time and some liquid spewed out at a safe velocity. I was told by a technician that the pressure trapped in the blue area is safe, and that it will diminish even further if you allow it to sit for more time. I though that pressure could be contained for a reasonable amount of time in such a setting. I thought that we rely on the ability to contain pressure in storage tanks, for example. I thought that storage of something pressurized would be possible in a valve setting like this. The technician said it is not possible and that he knows first hand from experience. Is this technician wrong? Where does the pressure go?&lt;/p&gt;&#xA;&lt;p&gt;I have reasoned that it is related to energy, and I have come up with a similar thought experiment to explain this phenomenon. I have a pipe with closed ends and many springs pushed together in series contained within the pipe. I also have a device similar to the blade-like one a magician uses when he does the magic trick to cut off people's limbs in a box. I use a device like this and slide it into the pipe and I isolate only the last spring in the pipe. When I open the end of the pipe, only one spring comes out with limited energy. If I had opened the pipe without using the blade-like isolation device, then many springs would have sprung out with incredible energy. Is this a sound comparison and understanding of the concept?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/apTd8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/apTd8.png&quot; alt=&quot;Valve Picture&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D4jpI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D4jpI.png&quot; alt=&quot;Spring Picture&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13451" LastEditorUserId="9448" LastEditDate="2022-10-04T18:03:03.607" LastActivityDate="2022-10-04T18:03:03.607" Title="When a valve under pressure is closed and creates a separate cavity, what happens to the pressure in the separate cavity? Will it leak out over time?" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="17777" PostTypeId="2" ParentId="17775" CreationDate="2017-10-27T20:09:45.407" Score="1" Body="&lt;p&gt;the pressure in the piping system which existed when the valves were closed will remain in the isolated segment of the pipe, which then behaves like a pressure vessel. Note that since the wall of the pipe is stiff and the fluid inside it nearly incompressible, the amount of energy stored in the pressurized &quot;vessel&quot; is small and if you crack open a vent valve connected to it, only a small amount of liquid will jet out of it before the pressure falls to zero. Note also that since no valve is perfect, it would be natural to notice that the pressure in the &quot;vessel&quot; would decay with time. Finally, note that the situation is different if the pipe contained air, which would compress readily under pressure and store a dangerously significant amount of energy in the vessel. In this case, you would need to be very careful to safely bleed off the pressure before servicing the valve.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-27T20:09:45.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17778" PostTypeId="2" ParentId="17766" CreationDate="2017-10-27T20:15:55.953" Score="2" Body="&lt;p&gt;The motor frame isn't bolted to anything. The pillow block bearings locate the entire motor. The torque arm is the only thing stopping the motor frame from spinning instead of the drive pulley. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what the 5N refers to. If I had to guess, It would say it's the piston rings on the cylinder wall.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2017-10-27T20:15:55.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17779" PostTypeId="2" ParentId="17771" CreationDate="2017-10-27T20:19:23.110" Score="1" Body="&lt;p&gt;I will propose a possible answer here. If you have an optical sensor in the pipe which detects the flash of the explosion, it may be possible to use that flash signal to open a pressure dump valve or a series of such valves prior to the arrival of the pressure wave at the downstream location you wish to protect. this solution uses the idea that the electric signal can be transmitted to the valve faster than the sonic velocity inside the pipe but it also requires a dump valve which can be opened very quickly. Ratchet freak may be right; dissipating the energy of the explosion may be the better answer, especially since it does not require instrumentation and fast-acting valves. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-27T20:19:23.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17780" PostTypeId="1" AcceptedAnswerId="17781" CreationDate="2017-10-27T20:48:52.903" Score="-1" ViewCount="45" Body="&lt;p&gt;When a reservoir is used to store water for electricity, it's pipes are usually laid in several segments on the ground to reach its source, such as a lake. It seems that if the pipes went directly to the lake by being suspended, less piping could be used and less energy lost due to friction would occur. Why are systems not built like this?&lt;/p&gt;&#xA;" OwnerUserId="13453" LastActivityDate="2017-10-27T20:59:46.240" Title="Why do pumped storage hydroelectric facilities use bent pipes?" Tags="|water-resources|energy|renewable-energy|energy-storage|" AnswerCount="1" CommentCount="1" ClosedDate="2017-10-30T14:55:04.610" ContentLicense="CC BY-SA 3.0" />
  <row Id="17781" PostTypeId="2" ParentId="17780" CreationDate="2017-10-27T20:59:46.240" Score="1" Body="&lt;p&gt;Suspended pipes are heavy and need support - any change of direction also creates a force - and there is the aerial impediment to consider : Birds, parachutes planes etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also some &quot;pipes&quot; are cut directly into the rock saving on material costs and consider that some of these pipes have a water flow rate of 45m3/s so the inertial forces are massive.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-27T20:59:46.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17782" PostTypeId="2" ParentId="17774" CreationDate="2017-10-28T00:59:11.457" Score="1" Body="&lt;p&gt;As far as I understand, especially assuming steady conditions, your mass flow rate is held constant. The thing that is normally changing is the mass flow rate per unit area, or mass flux. Knowing that maximum mass flux occurs under stagnation property conditions, you can actually use simple relationships like Fleigner's formula to solve for your mass flux and if you know your cross-sectional area then your mass flow rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as far as your question, mass flow rate wouldn't change through the nozzle, just pressure and density properties that affect fluid velocity. &lt;/p&gt;&#xA;" OwnerUserId="13454" LastActivityDate="2017-10-28T00:59:11.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17783" PostTypeId="2" ParentId="11419" CreationDate="2017-10-28T02:28:13.723" Score="1" Body="&lt;p&gt;there are non-silicone flexible foam products commonly and cheaply available; these are used as weatherstripping for door and window frames in houses (sold in hardware stores) and for sealing gaps between the sides of a pickup truck bed and a camper shell (sold in auto supply stores).  of the foam sealing materials available, the highest quality ones are used to fabricate replacement door gaskets for cars, but these require application of an adhesive to stick them together since they do not have stickytape on one side. Both home and auto products are usually sold in narrow rolls and have stickytape on one side. They come in different roll widths and foam thicknesses. Please note however that because they are foam, they consist mostly of air and hence are not very thermally conductive. all of them are good electrical insulators. Another possibility is foam-core doublesticky mounting tape from 3M. This is thinner than the gasketing material described above and the adhesive used is extremely strong, and on both sides of the tape, but the product is a lot more expensive than door gasket. It is available in roll form at industrial supply houses or as precut squares (as for sticking things like towel racks to walls) in hardware stores. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-28T02:28:13.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17784" PostTypeId="2" ParentId="15720" CreationDate="2017-10-28T02:50:07.083" Score="1" Body="&lt;p&gt;try something called sharp plaster sand. this is sold in building supply stores and is basically a high grade of beach sand with very little dirt or silt in it, and is sold in bags. it is cheap. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-28T02:50:07.083" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17785" PostTypeId="2" ParentId="17772" CreationDate="2017-10-28T03:06:55.627" Score="1" Body="&lt;p&gt;When you compress air it gives off heat per the ideal gas law. When the air re-expands it doesn't get that heat back. There has been some research in using large salt caverns to store compressed air and using water or molten salt do store the thermal energy and return it to the air as it expands in the turbine. Even then, there is still loss because heat transfer requires a differential. This results in a lower expansion temperature and a lower energy extracted. &lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-10-28T03:06:55.627" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17786" PostTypeId="1" CreationDate="2017-10-28T03:13:20.120" Score="0" ViewCount="156" Body="&lt;p&gt;As I understand from theory that two integrators are required to provide zero velocity lag in a type 2 servo system.  My question is how does 2 integerators help to nullify velocity errors. Less of maths with intuitive  explanation will be useful &lt;/p&gt;&#xA;" OwnerUserId="13455" LastActivityDate="2018-03-27T11:34:24.740" Title="Type 2 servo velocity tracking" Tags="|servo|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17787" PostTypeId="2" ParentId="7699" CreationDate="2017-10-28T03:14:47.407" Score="0" Body="&lt;p&gt;the least sound-conductive materials are dense (so most of the incoming sound wave reflects off without being transmitted) and have lots of internal friction (to dissipate the transmitted wave). a good compromise for engineering purposes is a product called quiet steel, which is two thin sheets of steel with a layer of rubbery material laminated in the middle. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-28T03:14:47.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17788" PostTypeId="2" ParentId="7289" CreationDate="2017-10-28T03:25:58.313" Score="1" Body="&lt;p&gt;per pound, steel is far cheaper than plastic and far stronger. to make a plastic structure as strong as a steel one would require a very large amount of plastic which would cost far more than if steel were used. Corrosion of steel can be managed by coatings and alloying composition. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-10-28T03:25:58.313" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17789" PostTypeId="2" ParentId="17786" CreationDate="2017-10-28T08:42:15.560" Score="1" Body="&lt;p&gt;Integrators is just a fancy way of saying you want to check your results. A type 0 system simply does not check the result at all. A type 1 system is reading the position directly from the shaft for example. So if there would be any unknown load etc in the system then you could compensate for this position error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whatabout a type 2 system? Well since you can measure the position you can deduce what the position should be If you were moving at speed x. This system is supprise supprise an integrator, integrating velocity gives position. So a type 2 system just calculates where the position ought to be if velocity is what you think it should be versus where it is. It can then compensate for this error.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2017-10-28T08:42:15.560" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17791" PostTypeId="1" AcceptedAnswerId="17848" CreationDate="2017-10-28T10:41:43.673" Score="-1" ViewCount="243" Body="&lt;p&gt;I would like to know what industrial engineers use for planning and developing a process:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the most widely &lt;code&gt;business analysis&lt;/code&gt; modeling language used (Think of it as the stage&#xA;consisting of 'How the product is intended to be?')? &lt;strong&gt;(Before the process)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;What is the type of &lt;code&gt;software&lt;/code&gt; and what are the &lt;code&gt;methods&lt;/code&gt; used for &lt;code&gt;managing&lt;/code&gt; the process? &lt;strong&gt;(Within the process)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13157" LastEditorUserId="13157" LastEditDate="2017-11-08T16:14:11.303" LastActivityDate="2017-11-11T18:31:12.473" Title="How does an industrial engineer develop a process?" Tags="|process-engineering|industrial-engineering|project-management|" AnswerCount="1" CommentCount="49" ClosedDate="2017-11-01T11:55:42.190" ContentLicense="CC BY-SA 3.0" />
  <row Id="17792" PostTypeId="2" ParentId="16904" CreationDate="2017-10-28T15:04:06.117" Score="1" Body="&lt;p&gt;You cannot possibly use any pump with suction at that depth. You must have a multistage submersible pump. This would be a small ESP in the oil industry where liquids are pumped from as deep as about 15,000 ft with over 100 pump stages. A good domestic water well company will have some applicable pumps and check valves. You will need some serious electric power and equipment to install and recover the pump for service. You may need  &quot;pump off&quot; control to shut off the pump when the water level drops too low and other things . It sounds like you need an actual engineer to set it up.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-10-28T15:04:06.117" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17793" PostTypeId="2" ParentId="15603" CreationDate="2017-10-28T18:56:24.697" Score="0" Body="&lt;p&gt;You can measure evaporation directly: have smal glass vials with a (known) amont of water and a small opening on top, place these at various places and document evaporation over a cycle. If the tunnel is a dryer, you will have very moist air at some points and so no ecvaporation of water - then you should find something else, maybe an organic solvent with a not too high vapor pressure. In this case consider health and safety aspects - I imagine you use a total of 100ml or so at most, but think it through anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do thermal measurements: measure  the current needed to maintain a sensor at a certain $\Delta T$ above air temp.The more energy needed, the more flow. Thisis available as propersensor for 2000 € or so (&lt;a href=&quot;http://www.bindergroup.info/binder/en/products/flow/gasflowmeasure.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.bindergroup.info/binder/en/products/flow/gasflowmeasure.php&lt;/a&gt;) but since you only need a qualitative statement you can maybe improvise something yourself.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2017-10-28T18:56:24.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17794" PostTypeId="1" AcceptedAnswerId="17795" CreationDate="2017-10-28T19:52:13.167" Score="0" ViewCount="144" Body="&lt;p&gt;I was asked to solve the following problem on my test: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tank is filled up to height H with a liquid having specific gravity 1.6. The rest of the tank is filled with air. A pressure gauge attached on top of the tank measures the pressure to be 11.5 m of water and another pressure gauge attached at the bottom of the tank reads 16.8 m of water. Now, what is the value of H? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know where to start. I have solved U tube manometer problems but never faced something like this one. I don't have a single clue. Even the total height of the tank is also not given. Is it really possible to solve the problem with only single data (the SG of the liquid) ? What do you think?&lt;/p&gt;&#xA;" OwnerUserId="9540" LastActivityDate="2017-10-28T21:03:11.643" Title="Pressure measurement problem" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17795" PostTypeId="2" ParentId="17794" CreationDate="2017-10-28T21:03:11.643" Score="1" Body="&lt;p&gt;I'll only outline the steps : &#xA;1) SG to density : Sg of 1.6 gives density as 1600kg/m3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) pressure difference is given from the two gauges 5.3m of WATER...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) calculate the pressure created by 5.3m of water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) re-calculate for Height as Pressure / (density * gravity) but this time use 1600...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neat that you have to spot that the gauges are in metres of water....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed gravity as 9.81 m/s2,&#xA;and ignored gauge pressure as it's the difference...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-28T21:03:11.643" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17796" PostTypeId="1" CreationDate="2017-10-29T01:19:10.313" Score="1" ViewCount="34" Body="&lt;p&gt;I have a situation where I have two water supply lines and both need to be the same pressure. One of the lines is normally always a little higher than the other, but by different amounts depending on circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a special kind of valve that will automatically restrict the high line so that it always matches the pressure of the low line?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2017-10-29T01:19:10.313" Title="Device to balance water pressure" Tags="|hydraulics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17800" PostTypeId="1" CreationDate="2017-10-29T08:34:40.980" Score="1" ViewCount="666" Body="&lt;p&gt;I am trying to write a program on Aspen Hysys. Hysys has a script manager to write scripts using Visual Basics for Applications syntax. Having no background in VBA, I'd like some help to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Import the current value of a controller's output signal (OP).&lt;/li&gt;&#xA;&lt;li&gt;Compare it with it's previous value (in a dynamic simulation).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="13461" LastEditorUserId="8055" LastEditDate="2017-10-30T14:53:46.240" LastActivityDate="2017-10-30T14:53:46.240" Title="How to use VBA on Aspen Hysys?" Tags="|control-engineering|chemical-engineering|dynamics|process-engineering|signal-processing|" AnswerCount="0" CommentCount="4" ClosedDate="2017-11-09T22:58:38.247" ContentLicense="CC BY-SA 3.0" />
  <row Id="17801" PostTypeId="1" CreationDate="2017-10-29T08:46:24.820" Score="0" ViewCount="494" Body="&lt;p&gt;In one of my project, I need deformation gradient or displacement gradient for some elements in my model that made by ABAQUS software and I can't find it directly, any suggestion or help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="12849" LastEditorUserId="8055" LastEditDate="2017-10-30T14:53:29.540" LastActivityDate="2018-05-07T22:49:10.423" Title="How to derive deformation gradient or displacement gradient?" Tags="|mechanical-engineering|finite-element-method|numerical-methods|abaqus|mechanical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17802" PostTypeId="1" AcceptedAnswerId="17863" CreationDate="2017-10-29T08:55:25.683" Score="2" ViewCount="250" Body="&lt;p&gt;While focusing on Quality and strength of welded joints produced but both methods..&lt;/p&gt;&#xA;" OwnerUserId="13462" LastEditorUserId="1586" LastEditDate="2017-11-01T22:15:47.067" LastActivityDate="2017-11-23T19:24:00.400" Title="How does Electric arc welding compared to gas welding?" Tags="|mechanical-engineering|welding|welds|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17803" PostTypeId="1" CreationDate="2017-10-29T11:16:31.057" Score="2" ViewCount="503" Body="&lt;p&gt;You have a gas of the following compositions : 40% N2, 30% CO2, 30%,CH4 in a tank . What is the Average Molecular Weight of the gas if your compositions is based on molar ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that the AMR with basis set at $100$kg-Mol = $29.2$kg/kg-mol&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Im still learning , I tried to give a basis of $100$kg ... but I don't get the same AMR. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this the case ? &lt;/p&gt;&#xA;" OwnerUserId="11561" LastActivityDate="2018-10-13T22:26:11.203" Title="Chemical engineering calculations - basis" Tags="|chemical-engineering|chemistry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17812" PostTypeId="1" CreationDate="2017-10-30T08:51:18.507" Score="1" ViewCount="43" Body="&lt;p&gt;I was passing column in waiting area of Pearson airport when my sweater cought some kind of device on the wall. Device was without cover and looks like old thermostat. When my sweater caught it, it fell off and started to blow air (I think). Maybe someone knows what is this device and why it blows compressed air? Yes, I informed airport employee.&lt;/p&gt;&#xA;" OwnerUserId="9518" LastActivityDate="2017-10-30T11:31:15.223" Title="What is this &quot;johnson controls&quot; device on the wall in airport?" Tags="|compressed-air|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17813" PostTypeId="1" CreationDate="2017-10-30T09:39:28.153" Score="1" ViewCount="50" Body="&lt;p&gt;I am new to Ansys Fluent and am trying to model heat transfer through a system of fins. The analysis is in 3D. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have set the Energy Equation as ON. The problem is that when I set heat loss by convection in the fins, a heat transfer coefficient is demanded by the software. However I am looking to evaluate the overall heat transfer coefficient of the fin system in the first place. Shouldn't the model be completely solved simply by knowing the material properties of the fin and the fluid and the velocity and temperature of fluid at the inlet?&lt;/p&gt;&#xA;" OwnerUserId="9398" LastActivityDate="2017-10-30T09:39:28.153" Title="Modelling Convection in Fluent" Tags="|ansys|fluent|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17814" PostTypeId="2" ParentId="17812" CreationDate="2017-10-30T11:31:15.223" Score="2" Body="&lt;p&gt;It might have been a old thermostat.  In ancient times, thermostats sometimes worked pneumatically.  When the temperature got too low, the thermostat would open a valve that released the pressure on some pressurized air.  This would then directly actuate the valve on the radiator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There was a small bleed system so that when the system was closed, it would pressurize over a minute or so.  The valve in the thermostat was significantly larger, so that when that valve was open, the pressure would be near ambient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of system was common in large buildings where a boiler would be running most of the time, since some zone somewhere always wanted heat.  The hot water or steam was always available, so the thermostat and valve combinations controlled whether it would flow thru the radiator for that zone only.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-10-30T11:31:15.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17816" PostTypeId="1" CreationDate="2017-10-30T15:31:40.497" Score="1" ViewCount="778" Body="&lt;p&gt;I have this stepper motor: &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B071LTXQGW&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stepper Motor, Nema 17 Stepper Motor 4-lead 1.8 Deg 40N.cm Holding Torque 1.7A 42 Motor with Motor Mounting Bracket and Screws for 3D Printer/CNC&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know how can I calculate how much weight it motor can lift in this situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3210x.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3210x.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13478" LastEditorUserId="8055" LastEditDate="2017-11-02T18:06:55.777" LastActivityDate="2017-11-02T18:06:55.777" Title="How can I calculate how much weight can lift my stepper motor?" Tags="|mechanical-engineering|electrical-engineering|structural-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17820" PostTypeId="1" CreationDate="2017-10-30T16:59:09.953" Score="4" ViewCount="1435" Body="&lt;p&gt;I am working on a project. A water level indicator using ULN2003. Now I want to integrate a 7 segment display to indicate the water level. How?&lt;/p&gt;&#xA;" OwnerUserId="13481" LastEditorUserId="7083" LastEditDate="2018-01-28T23:19:28.387" LastActivityDate="2018-01-28T23:19:28.387" Title="How to integrate 7 segment display with ULN2003?" Tags="|electrical-engineering|control-engineering|communication|" AnswerCount="1" CommentCount="1" ClosedDate="2017-11-06T18:09:27.710" ContentLicense="CC BY-SA 3.0" />
  <row Id="17822" PostTypeId="1" CreationDate="2017-10-30T18:22:20.257" Score="1" ViewCount="385" Body="&lt;p&gt;I've been working on a schematic and its been going beyond my comprehension; so i thought i should consult better structural engineers here. (I'm a fresh Civil Engr Graduate though)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i have this scheme shown below, where the task is to design a truss carrying a conveyor mechanism. i'm able to calculate trusses of any type with correct loading parameters, but this scheme has been eluding me for hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z4Ih1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z4Ih1.png&quot; alt=&quot;Elevation of scheme&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so i simplified it to this for proper analysis&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8Jhsj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Jhsj.png&quot; alt=&quot;Elevation of scheme(Simplified)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is that, i know trusses cannot carry load on their members as done in the scheme where 7.3 KN is applied. This is where the problem is, am i missing something? this is my first conveyor truss analysis, i really hope this QA site will shed a lot more light on this matter. Thanks&lt;/p&gt;&#xA;" OwnerUserId="13484" LastActivityDate="2018-03-16T10:21:00.747" Title="Conveyor Truss Analysis Problem with loading" Tags="|structural-engineering|civil-engineering|structures|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17823" PostTypeId="1" AcceptedAnswerId="17826" CreationDate="2017-10-30T20:31:01.867" Score="4" ViewCount="323" Body="&lt;p&gt;I bought recently 10 meters of Kanthal A1 AWG18 wire, which is a type of heating element (FeCrAl alloy). I intend to make a coil out of it, but it's no use doing that right now, I need more of it for my project (electric furnace), have ordered more of it online and am waiting a couple weeks till it arrives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this, if I use the heating element a few times in a different project (assume here that the wattage is the maximum possible), in a way that I won't twist or wind it in any way, will this affect the physical properties of the wire in ways that might make it harder for me to turn it into a coil later on, for my initial project? (I'll be winding the coil manually, it will be between 5 and 10 mm of diameter)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask that because heating elements tend to get more brittle with time, but I don't know how long it might take for such changes to happen.&lt;/p&gt;&#xA;" OwnerUserId="13485" LastEditorUserId="2819" LastEditDate="2017-10-31T02:39:40.993" LastActivityDate="2017-10-31T16:10:54.243" Title="Winding a wire heating element after a few uses" Tags="|materials|thermodynamics|heat-transfer|heating-systems|thermal-conduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17824" PostTypeId="2" ParentId="17820" CreationDate="2017-10-30T22:37:31.067" Score="3" Body="&lt;p&gt;I am assuming you have a micro controller driving the ULN2003. How about the following sketch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/44NU6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/44NU6.png&quot; alt=&quot;connecting 7 segment display to ULN2003&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This show how to connect the segments and provide power to the display. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following chart has some guidelines on how to drive the 7 segment display.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fbdt0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fbdt0.jpg&quot; alt=&quot;7 segment display drive signals&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These should get you started.&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2017-10-30T22:37:31.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17826" PostTypeId="2" ParentId="17823" CreationDate="2017-10-31T02:36:24.000" Score="2" Body="&lt;p&gt;I have not used &lt;a href=&quot;https://en.wikipedia.org/wiki/Kanthal_(alloy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kanthal&lt;/a&gt;, but I imagine it has similar properties to &lt;a href=&quot;https://en.wikipedia.org/wiki/Nichrome&quot; rel=&quot;nofollow noreferrer&quot;&gt;nicrome&lt;/a&gt;. Basically when you heat the wire up and cool it quickly, it &lt;a href=&quot;https://en.wikipedia.org/wiki/Quenching&quot; rel=&quot;nofollow noreferrer&quot;&gt;quenches&lt;/a&gt; the wire. To reverse this process (or avoid it all together) you need to bring the metal to a specific temperature (without looking it up, red hot is plenty), hold it there for some amount of time, then slowly cool the wire by decreasing current to it over time. This is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Annealing_(metallurgy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;annealing&lt;/a&gt;. The amount of time depends on the metal, but you can repeat the process again and again until you get the right amount of time. The annealing time may be something unrealistic for diy operation, and may not give you exactly the same properties you started with. But knowing this relationship and cooling the wire down slower will help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kanthal is rated for high temperature operation, but if you approach or exceed the rated temperature of the alloy, oxidation may occur which will progressively damage the material. In this case the damage would not be reversible. I'm not a chemist or metallurgist, but heating the material in the presence of carbon and other materials can result in chemical reactions that will modify the alloy (usually deteriorating its favorable properties).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color of the wire during operation can give you an idea of its operating temperature, see the wiki &lt;a href=&quot;https://en.wikipedia.org/wiki/Incandescence&quot; rel=&quot;nofollow noreferrer&quot;&gt;incandescence&lt;/a&gt; chart below:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dA8Hu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dA8Hu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-10-31T02:36:24.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17827" PostTypeId="1" AcceptedAnswerId="17830" CreationDate="2017-10-31T06:29:26.127" Score="1" ViewCount="26" Body="&lt;p&gt;Is there a purely mechanical device that takes in clockwise or counter-clockwise rotation 'indiscriminately' and outputs in both cases clockwise rotation?&lt;/p&gt;&#xA;" OwnerUserId="13488" LastActivityDate="2017-10-31T09:56:38.090" Title="Automatic translation of either-sided rotation to fixed directions?" Tags="|mechanical-engineering|gears|transmission|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17830" PostTypeId="2" ParentId="17827" CreationDate="2017-10-31T09:56:38.090" Score="0" Body="&lt;p&gt;How about something like this, uses two one-way clutches :&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZznI1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZznI1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Output is clockwise from the front...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-10-31T09:56:38.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17831" PostTypeId="1" CreationDate="2017-10-31T10:44:56.510" Score="0" ViewCount="103" Body="&lt;p&gt;Take the &lt;a href=&quot;https://www.gepower.com/gas/gas-turbines/lms100&quot; rel=&quot;nofollow noreferrer&quot;&gt;LMS100&lt;/a&gt; gas turbine for example. Why does the efficiency change when generating power in 50 or 60hz waveform? &lt;/p&gt;&#xA;" OwnerUserId="13492" LastActivityDate="2018-04-01T17:51:52.057" Title="Thermal efficiency affected by hertz (Gas turbine)" Tags="|electrical-engineering|turbines|energy-efficiency|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17833" PostTypeId="2" ParentId="17831" CreationDate="2017-10-31T12:19:52.900" Score="0" Body="&lt;p&gt;In gas turbines, peak efficiency is reached at a given design speed. When a gas turbine is used to drive an alternator, the reduction gearing is designed according to the best design speed of the turbine and the frequency and number of poles of the alternator, In the case of two poles and 50Hz, the alternator speed is 3000 rpm, but if the frequency is 60 Hz, the speed is 3600 rpm. If the same reduction gearing is used, the working speed of the turbine will vary by 20%, thus, when designing the reduction gearing, a choice of frequency must be made.&lt;/p&gt;&#xA;" OwnerUserId="10324" LastEditorUserId="10324" LastEditDate="2017-10-31T16:21:06.533" LastActivityDate="2017-10-31T16:21:06.533" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17836" PostTypeId="1" AcceptedAnswerId="17845" CreationDate="2017-10-31T16:08:03.483" Score="1" ViewCount="57" Body="&lt;p&gt;I'm trying to design a device that can stretch a tissue sample of set length, right now about 2 mm, to desired stretch ratios. This will be controlled using a interface where the user can input their desired ratio, and the controller will tell a 6 in stroke WindyNation linear actuator to move to a position based on the input ratio. The controller also needs to communicate with a force sensor and display the force that is being exerted on the sample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this type of problem, is it better to use an Arduino, Raspberry Pi, or both to power the controller? I've found that the Pi won't be very good due to issues with timing. Will this be a serious issue?&lt;/p&gt;&#xA;" OwnerUserId="13053" LastActivityDate="2017-10-31T20:50:43.890" Title="Designing of a tissue stretching device using a linear actuator" Tags="|design|biomedical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17837" PostTypeId="2" ParentId="17823" CreationDate="2017-10-31T16:10:54.243" Score="1" Body="&lt;p&gt;Kanthal and nichrome/ chromel are rather different; iron base vs nickel base. Neither has a &quot;martensite&quot; transformation so can not be hardened by fast cooling. Little change would occur on use , possibly grain growth at high temperatures could cause limited embrittlement. You could maximize ductility by annealing: Anneal by heating to red or orange heat and quench into water.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-10-31T16:10:54.243" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17838" PostTypeId="1" CreationDate="2017-10-31T16:59:04.530" Score="1" ViewCount="165" Body="&lt;p&gt;If one has a specified allowed X length and Y height, into which a fillet can be placed to reduce stress concentrations, is there an equation which can be used to determine the optimal shape of the fillet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked into Baud Fillets and Thum-Bautz fillets, but I find that although these fillets work fantastically for a desired y height OR a desired x height, these don't work well when one has constraints on both of these dimensions. Furthermore, it seems scaling the fillet shape along the axis does not work well, in many cases ending up worse than a circular fillet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is relevant, this is for fillets that are to be placed on the ID or OD of steps in tubing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know one option is various shape optimization algorithms, but I was hoping that there was some kind of out of the box solution that I could use to hopefully avoid this.&lt;/p&gt;&#xA;" OwnerUserId="13498" LastActivityDate="2017-10-31T19:05:27.803" Title="Most efficient fillet design given an envelope (for stress)" Tags="|mechanical-engineering|design|stresses|finite-element-method|strength|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17839" PostTypeId="1" CreationDate="2017-10-31T17:42:02.703" Score="0" ViewCount="5665" Body="&lt;p&gt;Help me to find out the hydrostatic force acting on an inclined immersed surface that is actually the diagonal of a 4mx3m rectangular tank filled with water. Given that the tank is 2m deep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know, F = Specific weight x Area x Depth of the center of gravity of the immersed surface from the free water surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;the problem is that I have no idea how to calculate that depth of the center of gravity of the immersed surface&lt;/strong&gt; from the free water surface. Any help will be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="9540" LastEditorUserId="14498" LastEditDate="2018-03-02T15:25:57.253" LastActivityDate="2018-03-02T23:07:30.180" Title="Hydrostatic force on inclined immersed surface" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17840" PostTypeId="2" ParentId="8434" CreationDate="2017-10-31T17:58:30.277" Score="1" Body="&lt;p&gt;ode solvers are all different. I've recently compared several fixed-time-step and variable-time-step ode solvers, with the same initial conditions, with the same model. Summary: They all give different solutions, sometimes fundamentally different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matlab/Simulink documentation gives a rough outline of picking the most useful (not the BEST) solver for the application under consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mathworks.com/help/matlab/math/choose-an-ode-solver.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Choose an ODE solver&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mathworks.com/matlabcentral/answers/92280-what-are-the-general-guidelines-for-choosing-ode-solvers-for-simulating-a-model-using-simulink&quot; rel=&quot;nofollow noreferrer&quot;&gt;Matlab Answers page&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mathworks.com/help/simulink/ug/types-of-solvers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Decision tree to pick an ODE solver&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mathworks.com/help/matlab/ordinary-differential-equations.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;about ordinary differential equations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;start with a variable time-step solver (ode45, for example)&lt;/li&gt;&#xA;&lt;li&gt;if solver performance is not as desired (too long to compute a timestep), start with the simplest method (ode1 Euler) and iteratively try higher order solutions until you arrive at an acceptable accuracy.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you do not have design constraints like using a fixed-time-step ode solver, going with the highest one that you can afford (ode45?) would be the best idea.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-10-31T17:58:30.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17841" PostTypeId="1" CreationDate="2017-10-31T18:15:52.007" Score="1" ViewCount="278" Body="&lt;p&gt;The picture below shows a refrigeration unit &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pVLa3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pVLa3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was reading the operation manual of this unit. It is written that the pressure of the condenser is controlled by the water flow rate. Reducing water flow rate will increase the pressure of the condenser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the pressure of the evaporator is controlled by the refrigeration load or cooling capacity (Q dot L). Increasing Q dot L is acheived by adjusting the voltage of the electric heater. The pressure of the evaporator is increased by increasing Q dot L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you explain to me how could reducing water flow rate cause the condenser pressure to increase ? And how could increasing the cooling capacity cause the pressure of the evaporator to increase ?&lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="9742" LastEditDate="2017-11-01T08:37:06.647" LastActivityDate="2017-11-01T08:37:06.647" Title="Adjusting the pressure of the condenser and the evaporator" Tags="|mechanical-engineering|thermodynamics|pressure|compressors|refrigeration|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17842" PostTypeId="2" ParentId="15235" CreationDate="2017-10-31T18:20:20.457" Score="1" Body="&lt;p&gt;Comparing two systems (or system to reality) can be quite complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;System identification methods (test inputs) can be utilized to simulate and compare dynamic responses in the time and frequency domains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, defining some response evaluation criteria (overshoot to a step response, modes and frequencies of systems) and running test cases using both models and comparing the results could be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dr. Jategoankar from DLR in Germany has books and some materials online regarding System Identification. He's a well known authority in the field.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-10-31T18:20:20.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17843" PostTypeId="2" ParentId="17838" CreationDate="2017-10-31T19:05:27.803" Score="1" Body="&lt;p&gt;I'm not an expert in this area, but what you might do is constrain just x, see what the resulting FEA stress is, and compare that to just constraining y. That would at least get you the best out of those formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a good &lt;a href=&quot;https://medsi.lbl.gov/SysIncludes/retrieve.php?url=https://medsi.lbl.gov/files/page_139/Session_3_High_Precision_Position_Mechanisms__Vibration_Stability/04-25_Investigation_of_optimised_notch_shapes_for_flexural_hinges_S.Zelenika_S.Henein_L.Myklebust.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;fillet research paper&lt;/a&gt; I skimmed. Looks like the grodzinski filet shape might be something to look into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.researchgate.net/figure/264601337_fig3_Fig-3-Geometrical-construction-of-Grodzinski-profile&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt; (might be a good paper but you have to buy it)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CyUb9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CyUb9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-10-31T19:05:27.803" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="17844" PostTypeId="1" AcceptedAnswerId="17849" CreationDate="2017-10-31T20:23:24.980" Score="4" ViewCount="263" Body="&lt;p&gt;Can someone explain the how/why Nylons brittleness is to some degree a function of it's moisture content. I don't understand HOW the water content changes the elongation at break of the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water doesn't bond or change the chemistry of the nylon when absorbed, my understanding is that it just fills some of the voids within the nylon. so then how is this changing the slope of the stress/strain curve?&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-11-01T03:23:21.973" Title="Nylon brittleness as function of moisture content" Tags="|mechanical-engineering|materials|chemical-engineering|chemistry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17845" PostTypeId="2" ParentId="17836" CreationDate="2017-10-31T20:50:43.890" Score="0" Body="&lt;p&gt;All depends on the sampling rate you require:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RaspberryPi will run an OS, which has multi-tasking scheduler, drivers/IO interruptions, etc. (mostly like any computer). In general computing, this is great, but for sampling a sensor, it make difficult to get a precise frequency: you will have an average of -+25us jitter and up to nearly +500us in some cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, Arduino is a MCU, it has no complex operating system. It cannot run a desktop application, but will have no issue in sambling precisely a sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if it is fine to sample at 1000Hz or slower, then either systems are fine. The RaspberryPi would give more processing power and many tools for showing in screen, sending by network or anything else with the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is required to get high sampling rate (e.g. 50KHz), then the Arduino will do a better job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, you could sample with the Arduino, send the data to a RaspberryPi which could do complex stuff with the data.&lt;/p&gt;&#xA;" OwnerUserId="11129" LastActivityDate="2017-10-31T20:50:43.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17847" PostTypeId="2" ParentId="17844" CreationDate="2017-10-31T22:22:57.697" Score="0" Body="&lt;p&gt;Water acts as a plasticizer, and plasticizers decrease the attractive forces between molecular chains to make the material more flexible.&lt;/p&gt;&#xA;" OwnerUserId="13121" LastActivityDate="2017-10-31T22:22:57.697" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17848" PostTypeId="2" ParentId="17791" CreationDate="2017-11-01T03:00:18.527" Score="1" Body="&lt;p&gt;SysML is a good choice when it comes to modeling use cases and understanding systematic approach (business modeling), in a generic industrial context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SQL is used to mine business data. R is great for analysis and simulation.  VBA has good portability when coupled with Excel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PDM software is good for setting up a document control process. Checking files into a database, monitoring revisions, and controlling the approval process, of BOM's (bill of materials) (Product Lifecycle Management (PLM)).&lt;/p&gt;&#xA;" OwnerUserId="13416" LastEditorUserId="13157" LastEditDate="2017-11-11T18:31:12.473" LastActivityDate="2017-11-11T18:31:12.473" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17849" PostTypeId="2" ParentId="17844" CreationDate="2017-11-01T03:23:21.973" Score="5" Body="&lt;p&gt;Nylons are a class of polymers formed from di-acid and di-amine monomers. The amine groups bond with the acid groups to form amide groups, end to end. The resulting amide groups have a positive charge at the amine hydrogen, and negative charge at the acid double-bond oxygen. The result is a polymer with periodic polar moments across the chain. The polar moments align between neighboring chains, creating a strong van der Waals force between them, in the form of hydrogen bonding. This is part of why nylon fibers are so strong: they resist chain sliding due to the strong inter-chain bonding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is also a polar molecule and also exhibits hydrogen bonding, giving it some interesting properties compared with similar molecules. Because of its polar nature, water readily infiltrates nylon materials and is attracted to the polar amide groups of nylons. Water creates space between chains, increasing the ability for chains to move around one another more easily, improving ductility. As Chris Johns noted, it acts as a plasticizer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that nylon and other amide polymers are somewhat special in that they are hydrophilic and oleophobic, unlike other common engineering polymers.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2017-11-01T03:23:21.973" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17850" PostTypeId="2" ParentId="17766" CreationDate="2017-11-01T03:33:02.653" Score="2" Body="&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;length(arm) is known (dynomometer reaction arm)&lt;/li&gt;&#xA;&lt;li&gt;Current to motor (I) is known&lt;/li&gt;&#xA;&lt;li&gt;Voltage to motor (V) is known&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Measurements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Angular momentum of motor (rpm)&lt;/li&gt;&#xA;&lt;li&gt;Force applied to load cell of dynomometer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Computations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Torque = sum(F) X length(arm)&#xA;--&gt; T = (F(load cell) - F(spring)) x length(arm)&#xA;--&gt; T = (F(load cell) - 5N) x length(arm)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;P(mech) = T x ω(rad/s)    where, ω(rad/s) = ω(rpm) x (2*pi)/60  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Compressor Power&lt;/strong&gt; = P(mech) = (F(load cell) - 5N) x length(arm) x ω(rpm) x (2*pi) / 60&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;P(elec) = I * V&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;P(elec) = P(mech) + P(loss)&lt;/li&gt;&#xA;&lt;li&gt;P(loss) = P(elec) - P(mech)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore: P(loss) = (I * V) - ((F(load cell) - 5N) x length(arm) x ω(rpm) x (2*pi)/60)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Balance the work/power of the system, while accounting for losses due to inefficiencies (frictional, electrical and thermodynamic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out this link for more info on how a dynomometer works.  &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Dynamometer&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13416" LastEditorUserId="13416" LastEditDate="2017-11-02T20:03:42.873" LastActivityDate="2017-11-02T20:03:42.873" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17851" PostTypeId="1" CreationDate="2017-11-01T04:09:15.813" Score="4" ViewCount="189" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o0Ows.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o0Ows.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi.. The question above is one of the tutorial question from fluid mechanics textbook. In the textbook solution for this question, the density of the methane gas stream at downstream was computed with the ideal gas law equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8gIpL.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8gIpL.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't understand is it possible for us to compute the density of the methane stream at downstream with the equation on below provided we know the density of methane gas and mach number at downstream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EEIzZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EEIzZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From datasheet, the density of methane gas = 0.678 kg/m^3 and Mach number at downstream, Ma= 0.4537&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer from the textbook solution was 2.858 kg/m^3&lt;/p&gt;&#xA;" OwnerUserId="13447" LastActivityDate="2018-12-23T17:01:55.423" Title="Methane Gas Density in Adiabatic Compressible Flow" Tags="|fluid-mechanics|fluid|compressible-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17853" PostTypeId="2" ParentId="16530" CreationDate="2017-11-01T08:37:40.883" Score="3" Body="&lt;p&gt;The life time of gears (which a rack and pinion mechanism basically is) is limited by teeth failure, wear, and pitting. So the answer to your question not only depends on the force acting on the rack, but also the required lifetime, lubrication, geometry of the gear and rack, and material combination. Meaning that while your rack might hold this force for an hour, it might fail after some weeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several books and guidelines out there about the dimension process of gear pairs. Just use them since they apply to you problem as well. The difference between a gear pair and a rack an pinion mechanism is that one of your gears has an infinite pitch diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an estimate, I would recommend you to search for rack and pinion with your favorite search engine and look in supplier's catalogs. Some of them report maximum force.&lt;/p&gt;&#xA;" OwnerUserId="13505" LastActivityDate="2017-11-01T08:37:40.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17855" PostTypeId="1" CreationDate="2017-11-01T11:26:22.120" Score="5" ViewCount="282" Body="&lt;p&gt;I have a product where we transmit quite a lot of shear force through a small bolted join. On our current design the contact area is about 10mm², the bolt is M2, and the force is 45N, and it works quite well. But now we want more, the requirements are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We want to go up to 300N, in either direction along one axis.&lt;/li&gt;&#xA;&lt;li&gt;No sliding between the parts, even at the micron level.  Elastic deformation of a few microns is OK though.&lt;/li&gt;&#xA;&lt;li&gt;Both parts are grade 4 (pure) titanium.&lt;/li&gt;&#xA;&lt;li&gt;The top of the lower part must be flat, nothing sticking up when the second part is removed. Shallow holes are OK.&lt;/li&gt;&#xA;&lt;li&gt;Ideally, the position of the top part in the direction of the force should be adjustable. Preferably continuously, but failing that then in about 0.5mm steps.&lt;/li&gt;&#xA;&lt;li&gt;The join can be made and un-made about 100-1000 times over the life of the product. By a competent and careful operator.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As is, at 300N the joint as is slides around all over the place.  I could go up to a M2.5 machine screw, or maybe M3 at a pinch, but some back-of-the-envelope scribbling suggests that won't be enough. I could specify a removable epoxy, but can't find one strong enough. So I thought about mechanically keying the two parts so that they cannot slide. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could cut both surfaces with a zig-zag profile, like a screw thread.  In fact I guess it might be possible for a machine shop to do that with an M3 thread cutting mill (if I pick the right profile) or the corner of an end mill (for a 45 degree profile).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've not seen such a thing before, at least not at this scale.  Does it sound like a good plan? What should I watch out for?  Is there an easier option I have missed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a rough sketch of what I mean:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cUtBQ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cUtBQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Left: as we have it now, for 45N. Right: What I mean by zig-zag profile.  I've sketched it as if it were an M3 that's been &quot;unrolled&quot; onto the yellow part.&lt;/p&gt;&#xA;" OwnerUserId="12022" LastEditorUserId="12022" LastEditDate="2017-11-01T15:43:36.327" LastActivityDate="2018-08-25T17:00:31.337" Title="Cutting flattened out &quot;threads&quot; to prevent small-scale slipping" Tags="|mechanical-engineering|threads|" AnswerCount="3" CommentCount="11" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17856" PostTypeId="2" ParentId="17839" CreationDate="2017-11-01T12:15:08.720" Score="2" Body="&lt;p&gt;First of all a drawing would really help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xhAdI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xhAdI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possibility #1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make a sketch and add the forces to the inclined area. You will either get a triangle or a trapezoidal, depending on the exact setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can look up the center of gravity or calculate it yourself (&lt;a href=&quot;https://en.wikipedia.org/wiki/Parallel_axis_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parallel axis theorem&lt;/a&gt;). Then you need to calculate the pressure at that point, let's call it $p_\text{res}$ for resulting Force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = p_\text{res} \cdot A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possibility #2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just integrate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\vec{F} = \int p \vec{n} dA$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\vec{n}$ being the normal vector. I'd prefer this way if you don't really need to know where the resulting force is attacking.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2017-11-01T12:15:08.720" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17858" PostTypeId="1" AcceptedAnswerId="17860" CreationDate="2017-11-01T14:29:21.613" Score="1" ViewCount="1011" Body="&lt;p&gt;There is ice build in my Samsung RF266 refrigerator. I have posted the following questions in &lt;a href=&quot;http://diy.stackexchange.com&quot;&gt;Home Improvement Stack Exchange&lt;/a&gt; hoping to find a solution. The questions are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://diy.stackexchange.com/questions/125403/samsung-rf266-refrigerator-not-cooling-fan-making-noise&quot;&gt;Samsung RF266 refrigerator not cooling - Fan making noise&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://diy.stackexchange.com/questions/126094/ice-buildup-inside-samsung-rf266-refrigerator&quot;&gt;Ice buildup inside Samsung RF266 refrigerator&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Bel&#xA;below is an image of the ice build up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F4xx8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F4xx8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the purpose of the fan is to circulated the cold air and coils are what generates the cold air. I don't believe the ice build up is normal.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; From an engineering standpoint, how is this suppose to work? What would cause the ice build up? What design feature are present in a refrigerator to prevent ice buildup? &lt;/p&gt;&#xA;" OwnerUserId="8055" LastActivityDate="2017-11-02T20:10:50.297" Title="How to prevent ice build up in a refrigerator?" Tags="|refrigeration|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17860" PostTypeId="2" ParentId="17858" CreationDate="2017-11-01T15:18:47.013" Score="1" Body="&lt;p&gt;Excessive ice build up can occur when the refrigeration system operates continuously for prolonged periods - hours. This can make the coils very cold and ice to form on them. By operating contiguously, the cooling process never stops and ice just accumulates. The system needs to stop so the ice can melt and the resultant water evaporate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be due to a poorly designed refrigeration system or cold air continuously leaking from inside the refrigerator, so the inside of the refrigerator never reaches a stable cold temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ice build up like this can be exacerbated by humid conditions - such as living in a tropical climate, a period of heavy rain, trying to flash freeze/cool large amounts of very hot food.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-11-01T15:18:47.013" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17861" PostTypeId="1" AcceptedAnswerId="17875" CreationDate="2017-11-01T18:07:46.737" Score="3" ViewCount="135" Body="&lt;p&gt;I am trying to configure a force sensor (a piezoresistive force sensor) in a device to measure small forces (less than 0.5 mN). This is difficult due to noise of the system. One of the ways I have considered to address this problem is to mechanically amplify the force applied to the force sensor with a gearbox. The small force applied would be transferred to the gearbox with a gear on a drive bar, and the amplified force from the increased torque would be transferred to the sensor with another gear and drive bar (see schematic). However, I have no experience with gearboxes and would like some opinion on feasibility or if this system even makes sense. I am particularly worried that static friction forces in the gears would cause inaccuracies, especially because gearboxes are designed for a completely different purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/biLdg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/biLdg.jpg&quot; alt=&quot;Schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13531" LastActivityDate="2017-11-02T13:48:43.933" Title="Can a hobby gearbox be used to mechanically amplify small forces for better force sensor resolution?" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17862" PostTypeId="2" ParentId="14091" CreationDate="2017-11-01T19:48:34.900" Score="0" Body="&lt;p&gt;1.5mm would be the &lt;strong&gt;expected&lt;/strong&gt; value for M2 hex. This can, however, vary between manufacturers - there may be legitimate reasons for deviating from standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing would of course be to ask the manufacturer what driver size they recommend. The second best thing is to get a measurement. 1.5mm can be measured using calipers, perhaps aided by a magnifying glass.&lt;/p&gt;&#xA;" OwnerUserId="12173" LastActivityDate="2017-11-01T19:48:34.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17863" PostTypeId="2" ParentId="17802" CreationDate="2017-11-01T22:32:45.870" Score="2" Body="&lt;p&gt;Both processes require the base material temperature to reach welding temperature.  Arc welding does this using an electric arc, and the process is quite fast.  Also, the deposition of the filler material is fast.  This is important because the heat affected zone is smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat affected zone is the region when a phase change in the parent material takes place due to the energy being introduced to the material.  For steel the crystalline structure changes from face centered cubic to body centered cubic as the molten metal freezes.  Therefore, the material shrinks and induces stress in the parts being welded.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.lincolnelectric.com/en-us/support/welding-how-to/Pages/weld-distortion-detail.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weld Distortion Info on Arc Welding&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.twi-global.com/technical-knowledge/job-knowledge/distortion-corrective-techniques-037/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Weld Distortion Info on Oxyacetylene Gas Welding&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Gas welding is a slower process, and the torch flame is on the work piece longer to reach the material’s welding temperature.  The parent material conduct more heat away from the torch, and the heat affected zone is larger.  There is more distortion to the base metals with gas welding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both processes have the pros and cons.  Typically, gas welding is reserved for thin, low stress parts.  Arc welding is good for structural welding, and higher stress location.  Thin materials can be arc welded, but with increased difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="13416" LastEditorUserId="13416" LastEditDate="2017-11-02T00:53:30.760" LastActivityDate="2017-11-02T00:53:30.760" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17864" PostTypeId="2" ParentId="16555" CreationDate="2017-11-01T22:39:22.577" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Low velocities can be problematic because they are difficult to meter (control) precisely.&lt;/li&gt;&#xA;&lt;li&gt;High velocities can cause erosion in valves, and generate more turbulence due to the high momentum in geometry changes.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13416" LastActivityDate="2017-11-01T22:39:22.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17867" PostTypeId="1" AcceptedAnswerId="17886" CreationDate="2017-11-02T02:22:04.797" Score="1" ViewCount="442" Body="&lt;p&gt;I'm building a product prototype for a client. We are trying to figure out the best way to build a hollow helix capable of transporting liquid at home. Something similar to this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ftFPo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ftFPo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far our best idea is to use standard tubing and try to use rings mounted on the side of a support cylinder which the helix loops through but this seems almost impossible to pull off at a small scale, and wouldn't look too good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have also considered heating sand and using that to bend acrylic tubing, but this certainly still doesn't feel ideal. &lt;/p&gt;&#xA;" OwnerUserId="12523" LastEditorUserId="90" LastEditDate="2017-11-03T10:59:23.367" LastActivityDate="2022-05-06T17:57:48.373" Title="How to build a transparent double hollow helix at home" Tags="|plastic|molding|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17869" PostTypeId="2" ParentId="17861" CreationDate="2017-11-02T05:27:26.003" Score="3" Body="&lt;p&gt;You may actually be interested in using a lever instead of a gearbox, as its a simpler construction and has less friction. But produces same torque increase, as a gear is essentially a folded up lever. If you want to be really fancy you couple this with two pins but that is not nessesary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just be sure to calibrate the new system somehow after you have built it. On the other hand you can just invest in a strain gauge and connect it to a shaft with known dimensions, as this may be more reliable than relying on the accuracy of your rotating joint to transfer force form one place to another&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2017-11-02T05:27:26.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17870" PostTypeId="2" ParentId="17867" CreationDate="2017-11-02T06:05:30.703" Score="1" Body="&lt;p&gt;Wrap the tube around a shaft then release it (it will grow slightly). You can find many youtube videos describing this process using a drill or lathe to do this for metal springs. The process for a tube is/is nearly identical. Your tube material just has to be suitably deformable to hold the shape like this.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2017-11-02T06:05:30.703" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17871" PostTypeId="2" ParentId="17855" CreationDate="2017-11-02T08:46:06.903" Score="1" Body="&lt;p&gt;A &quot;Chamfer End Mill&quot; tool can be used to cut a V groove into a surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p7xkn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p7xkn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure to specify a tool that is capable of cutting material at its point as well as along its radius.  (Although arguably, the groove doesn't need a sharp point, trapezoidal grooves would work just as well).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose your &quot;feeds and speeds&quot; carefully.  Titanium isn't good at dissipating heat, so you want to make sure you don't overheat the tool or the workpiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you use a carbide (or better) chamfer end mill, in a very rigid machine with low RPM and feed rate, and plenty of flood cooling.  Consult with your machinist and do some experimental cuts on scrap stock first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you're looking for very small feature sizes, you might have good results using an engraving bit, although this would need to be carefully tested to avoid breaking the delicate tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to get good surface finish which will ensure good contact between the metal parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Alternatively&lt;/strong&gt;, you could simply mill a pocket into the yellow part for the blue part to drop down into.  That would only require milling to be done on the yellow part and you wouldn't need the grooves at all.&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastEditorDisplayName="user6335" LastEditDate="2017-11-02T09:16:09.700" LastActivityDate="2017-11-02T09:16:09.700" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17872" PostTypeId="1" AcceptedAnswerId="17874" CreationDate="2017-11-02T10:47:51.403" Score="2" ViewCount="77" Body="&lt;p&gt;I'm solving this problem in mechanics, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two cylinders A and B, weighing 100 lb and 200 lb respectively, are connected by a rigid rod curved parallel to the smooth cylindrical surface shown in Fig. P-329. Determine the angles α and β that define the position of equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ST0bl.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ST0bl.gif&quot; alt=&quot;Fig. P-329&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already understood that the relation between $\beta$ and $\alpha$ is  $\cos \alpha = 2 \cos \beta$ by doing summation of moment with respect to O and equating to 0. What I don't get is how come the angle from this picture becomes 90 degrees? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zPpws.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zPpws.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10099" LastActivityDate="2017-11-02T12:05:30.797" Title="Included angle in a semicircle" Tags="|statics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17873" PostTypeId="2" ParentId="17867" CreationDate="2017-11-02T11:20:03.207" Score="0" Body="&lt;p&gt;Not sure of the diameter you want, but the scaffolding wheels can be adjusted for height and the &quot;screw&quot; is quite coarse - that may give you a &quot;nice&quot; profile to work with... heat and gloves spring to mind...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-02T11:20:03.207" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17874" PostTypeId="2" ParentId="17872" CreationDate="2017-11-02T12:05:30.797" Score="3" Body="&lt;p&gt;It's given by the question. If you pay attention to the original image, you'll see that the internal angle ($180−\alpha−\beta$) has the classic right-angle symbol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/07NUKs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/07NUKs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;sup&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/6/6c/Right_angle.svg/1200px-Right_angle.svg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-11-02T12:05:30.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17875" PostTypeId="2" ParentId="17861" CreationDate="2017-11-02T13:48:43.933" Score="5" Body="&lt;p&gt;Any gearbox or lever linkage is going to have backlash to go with your other concerns. By the time you get to a system using this kind of mechanical amplifier to get the sensitivity you need along with the low noise, accuracy and precision you want, you're into making a what's essentially a precision watch... Can it be done? Sure. It's going to be time consuming and expensive though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be faster, cheaper and more efficient to simply purchase a &lt;a href=&quot;https://www.novatechloadcells.co.uk/ds/f329.htm&quot; rel=&quot;noreferrer&quot;&gt;load cell designed to measure micro Newton level Forces&lt;/a&gt;. If you're a student, doing a project on a student budget with no corporate sponsor, you should call the company anyway, they might help you anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really need to build the measurement device yourself, I suggest looking into a &lt;a href=&quot;https://www.britannica.com/technology/torsion-balance&quot; rel=&quot;noreferrer&quot;&gt;Torsion Balance&lt;/a&gt;. They're typically used to measure things like the gravitational attraction force between two objects, so the concept is in line with your measurements. You could also build a strain gage based sensor, perhaps some kind of canteliever beam.   &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2017-11-02T13:48:43.933" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17876" PostTypeId="2" ParentId="16555" CreationDate="2017-11-02T19:01:58.617" Score="1" Body="&lt;p&gt;There are lots of reliable and precisely controlled hydraulic systems that have very low or zero velocities.&#xA;Some examples include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Slow Hydraulic Positioners (rate controlled band saw arm)&lt;/li&gt;&#xA;&lt;li&gt;Hydraulic Vises&lt;/li&gt;&#xA;&lt;li&gt;Hydraulic Belt Tensoioners&lt;/li&gt;&#xA;&lt;li&gt;Truck Scales&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There is no base reason to imply a minimum velocity on a hydraulic system. Systems can be designed to work with zero flow. In some circumstances however it does make sense to have a minimum velocity:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When line sizing for power transfer to a hydraulic motor, you wouldn't want to design for too low of velocity or lines become unnecessarily large, costing more money. This also increases the systems total volume requiring more oil to fill the system.&lt;/li&gt;&#xA;&lt;li&gt;In equipment for cold weather operation, we would drill small holes in the hydraulic cylinder heads to provide a small amount of flow to an intermittent or zero flow device (like a vise cylinder) in order to keep the hydraulic fluid and cylinder warm for responsive operation.&lt;/li&gt;&#xA;&lt;li&gt;Metering at very low velocities can be difficult with low quality proportional valves. High quality valves with positional encoder or pressure feedback can servo control at as low of flow-rate as you would like.&lt;/li&gt;&#xA;&lt;li&gt;Some low cost centrifugal hydraulic power units do not have pressure bypass or electric bypass valves. These units require some amount of continuous flow to keep from burning up.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2017-11-02T19:07:44.627" LastActivityDate="2017-11-02T19:07:44.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17877" PostTypeId="2" ParentId="16167" CreationDate="2017-11-02T19:27:34.753" Score="0" Body="&lt;p&gt;The drawing is lacking detail on how to the load and platform will not rotate.  Despite this, you can assume accelerations are zero; therefore static. The sum of the forces, and the sum of the moments in the system must equal zero.  Then a safety factor of should be computed per ASME B30.&lt;/p&gt;&#xA;" OwnerUserId="13416" LastActivityDate="2017-11-02T19:27:34.753" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17878" PostTypeId="2" ParentId="17766" CreationDate="2017-11-02T19:34:59.697" Score="1" Body="&lt;p&gt;OP has messaged me that his problem is understanding how the spring force setup measures torque. Let's see if this furnishes the answer he wants: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine an electric motor which is connected to a load. Assume that it is performing work on the load, which means that &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the motor is exerting a force on the load &lt;/li&gt;&#xA;&lt;li&gt;the load is moving&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Assuming a constant force, the net work done by the motor on the load is then w = (force) x (distance) and the power is  p = (force x distance)/time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since velocity = distance/time, we can rearrange the power equation to read p = (force) x (velocity) and since we are talking about rotary motion, the force involved is a twisting action and the velocity is a rotational speed. Thus the power equation is translated to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;shaft horsepower = (torque) x RPM, taking care to keep our units consistent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we are ready to look at the motor and the apparatus connected to it. First we note that the electric motor is applying a twisting force (torque) to the load which means the load is applying an equal and opposite twisting force to the motor. this means the motor casing is applying that same twisting force to the foundation it is bolted to, which means those bolts are transferring the torque reaction. In turn, this means that if we can determine the stresses in those mounting bolts, we can solve for the torque being produced by the motor, multiply that torque by the RPM at which the motor is running, and then know the power being produced by the motor and absorbed by the compressor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram above shows the motor is mounted on a pivot so that anything that applies a torque to the motor will cause the entire motor to rotate about that pivot. To prevent this from happening, a spring force gauge is also connected to the motor mount through a lever. the spring force gauge is simply a spring which deflects a known distance under the influence of a force and hence lets you measure that force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is the solution: the spring force gauge can be thought of as a mounting bolt which lets us measure the stress in it which is exerted upon it by the motor. the force that the spring gauge senses is being applied to the motor via a lever arm as specified in the drawing, and since torque = force x lever arm, the spring force gauge reading can be converted to the torque being applied to the load by the motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if this answers your question. Please note that this arrangement of a force-gauge-and-lever-arm plus an RPM gauge is known as a dynamometer and is the standard tool for measuring the horsepower being transferred by rotating shafts. You can find more examples of dynamometers on the web.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="8055" LastEditDate="2017-11-02T20:50:56.127" LastActivityDate="2017-11-02T20:50:56.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17879" PostTypeId="2" ParentId="17858" CreationDate="2017-11-02T20:10:50.297" Score="0" Body="&lt;p&gt;the presence of severe ice buildup in a refrigeration system strongly suggests a door gasket leak.  Check to see if the gasket is in good shape and also if the door is closing properly against the body of the refrigerator. In this connection, note that most refrigerators have reversible door hinges which let the owner remount the door so it opens left-handed instead of right-handed. If the hinge attachment screws are loose or if the top and bottom hinges are skewed, then the door won't seal properly when closed. A simple way to check the gasket seal integrity is to place a lit flashlight inside the refrigerator and carefully observe the entire gasket seal for light escaping- indicating a bad seal. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-11-02T20:10:50.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17881" PostTypeId="1" CreationDate="2017-11-02T21:52:50.657" Score="1" ViewCount="75" Body="&lt;p&gt;I was reading some articles about acoustic levitation and I was wondering whether it is possible to lift a 100 kips load off the ground for about 0.5-1 inches with that. If it is, how should I calculate how much energy I need to deliver to a ultrasound speaker and effectively concentrate with a lens that can be used to produce such lift on an object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not very familiar with this subject, so forgive me if I am asking some pretty dum question here. Thanks!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit - Somebody apparently downvoted the question without saying why, I don't exactly know why. But I thought I should put some additional info to help people understand what I am referring to. Showing some &quot;effort&quot;.... It is such a new field that there are not a lot of articles on how you calculate something like this and what is realistic. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://phys.org/news/2015-01-acoustic-levitation-simple.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Acoustic levitation made simple&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://phys.org/news/2016-08-acoustic-levitation-large-sphere.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Researchers demonstrate acoustic levitation of a large sphere&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13526" LastEditorUserId="8055" LastEditDate="2017-11-06T18:06:39.900" LastActivityDate="2017-11-06T18:06:39.900" Title="Calculating the force produced by acoustic levitation" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="5" ClosedDate="2017-12-18T01:45:41.173" ContentLicense="CC BY-SA 3.0" />
  <row Id="17882" PostTypeId="1" CreationDate="2017-11-03T01:58:56.160" Score="2" ViewCount="1063" Body="&lt;p&gt;I've just started to learn Gate, Globe, Ball, and Butterfly valves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking about &quot;dirty services&quot; now. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which types of valves are most suitable for dirty services, and what characteristics must the valves have? &lt;/p&gt;&#xA;" OwnerUserId="11561" LastEditorUserId="19930" LastEditDate="2019-04-30T16:12:12.517" LastActivityDate="2019-04-30T16:12:12.517" Title="Best types of valves for dirty services" Tags="|valves|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="17885" PostTypeId="1" CreationDate="2017-11-03T10:37:43.360" Score="2" ViewCount="32" Body="&lt;p&gt;Is it possible, under current technology, to have an electronic filter in the microwave range (100 MHz to 300 GHz) with very low bandwidth , say 100 to 1000 Hz?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes,  then what is the basic technology behind it? Any example would be great. Smallest bandwidth filters I am aware is that of GSM mobiles at 200kHz. &lt;/p&gt;&#xA;" OwnerUserId="13533" LastEditorUserId="8055" LastEditDate="2017-11-03T17:56:11.820" LastActivityDate="2017-11-03T17:56:11.820" Title="Low bandwidth electronic filter" Tags="|electrical-engineering|rf-electronics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17886" PostTypeId="2" ParentId="17867" CreationDate="2017-11-03T11:26:49.857" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Buy, or make two metal springs of desired helix diameter and pitch; wire diameter considerably narrower than the desired helix.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Coat the springs in paraffin, immersing in molten paraffin, possibly a couple times, until coated spring diameter is as desired.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Twist the two springs together to create the double helix.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Place the helix in a tube (either openable length-wise cast, or something easily destructible, say, thin plastic or glass. Fix it in place so that it won't move&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pour clear acrylic resin to fill the tube; leave it to cure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Remove the item from the tube cast. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat gently, to melt the paraffin, but not to damage the resin. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Screw the springs out of the canals. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wash the surplus paraffin out with warm ethanol.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For actual production process (after prototyping) consider extruding pure paraffin or other easy to melt material into the desired form dual helix, without wire core (use extruder similar to one used to make pasta). &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-11-03T11:26:49.857" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17887" PostTypeId="1" CreationDate="2017-11-03T12:40:02.840" Score="0" ViewCount="284" Body="&lt;p&gt;What is the smallest capacitor in size, which has been engineered and how much is it's capacitance? I presume it must be something etched on chip. &lt;/p&gt;&#xA;" OwnerUserId="13534" LastEditorUserId="13534" LastEditDate="2017-11-03T14:03:17.960" LastActivityDate="2021-07-20T14:08:00.707" Title="Smallest capacitor in size" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17888" PostTypeId="1" AcceptedAnswerId="17909" CreationDate="2017-11-03T14:34:05.923" Score="1" ViewCount="325" Body="&lt;p&gt;Can you please suggest what kind (type) of hinge / pins I should use to make the mechanism in the attached file. I was considering to use the pins shown in the second image. But I am not sure if it will work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7XlU2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7XlU2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v5zWym.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v5zWym.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13536" LastEditorUserId="1832" LastEditDate="2017-11-03T18:00:58.410" LastActivityDate="2017-11-05T04:58:23.157" Title="Type of hinge I should use" Tags="|mechanical-engineering|applied-mechanics|mechanisms|" AnswerCount="1" CommentCount="2" ClosedDate="2017-11-09T18:05:26.307" ContentLicense="CC BY-SA 3.0" />
  <row Id="17889" PostTypeId="1" CreationDate="2017-11-03T14:34:23.037" Score="1" ViewCount="245" Body="&lt;p&gt;What is most effective way to apply linear actuator to move a lift within a small sized space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. for example, for 14&quot;, which is total 820mm, to move a 330 mm lift in 740mm height box (width and depth are not limited)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yZ2aW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yZ2aW.jpg&quot; alt=&quot;linear actuator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13535" LastActivityDate="2017-11-03T15:59:53.740" Title="Using linear actuator in the limited space" Tags="|actuator|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17890" PostTypeId="2" ParentId="16626" CreationDate="2017-11-03T14:40:18.643" Score="0" Body="&lt;p&gt;For reference, I'm answering my own question with the solution I used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I replaced half of the fish-eye bearings with clevis bearings (on the wall end of each of the bars), which has eliminated most of the lateral motion while still allowing enough flexibility to actually mount the thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is still some lateral motion but, because the table has one side flush with the wall in both the fully-open and fully-closed configurations, it doesn't matter in practise.&lt;/p&gt;&#xA;" OwnerUserId="12437" LastEditorUserId="12437" LastEditDate="2018-01-21T08:53:31.007" LastActivityDate="2018-01-21T08:53:31.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17891" PostTypeId="2" ParentId="17887" CreationDate="2017-11-03T15:13:30.253" Score="1" Body="&lt;p&gt;You can create caps with values down to 10fF or less (that's femto-farads) with size about 1 micron squared, but whether they are useful depends on the amount of stray capacitance in the rest of the chip wiring.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2017-11-03T15:13:30.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17893" PostTypeId="2" ParentId="17889" CreationDate="2017-11-03T15:59:53.740" Score="1" Body="&lt;p&gt;Multiply the distance moved using a lever, but the input force needed increases...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-03T15:59:53.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17896" PostTypeId="1" CreationDate="2017-11-03T18:51:44.353" Score="3" ViewCount="231" Body="&lt;p&gt;We are planning to start a basic course on Computer Aided Designs (2D and 2D) for high school kids. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea  is to help them shape their innovative ideas into something tangible.   We shall finally help them develop prototype using 3D printers or mass produce later on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would like to know what all open source  options are available for 2d and 3d designs. Most importantly we are looking for tools which are simple and easy enough so that high school students  can use it (we will be limiting to simple designs).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions ? &lt;/p&gt;&#xA;" OwnerUserId="13540" LastActivityDate="2017-11-04T20:05:53.377" Title="Simple and easy Open Source CAD tools for basic 2D and 3D designs" Tags="|cad|autocad|3d-printing|computer-aided-design|" AnswerCount="1" CommentCount="3" ClosedDate="2017-11-16T19:11:57.400" ContentLicense="CC BY-SA 3.0" />
  <row Id="17899" PostTypeId="1" AcceptedAnswerId="17928" CreationDate="2017-11-04T04:16:16.027" Score="0" ViewCount="26" Body="&lt;p&gt;I'm solving this problem, &lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aLweG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aLweG.jpg&quot; alt=&quot;Problem&quot;&gt;&lt;/a&gt;&#xA;&lt;br&gt;&#xA;I used the formula $Q_p = mc_p\Delta T$. Since it is a mixture, I did this:&#xA;$Q_{Total} = 0.3 × 0.867 × (50-150) + 0.7 × 1.043 × (50-150 ) $ &lt;br&gt;&#xA;$ = -99.02 kJ/kg $&lt;br&gt;&#xA;Is my solution and final answer correct? If not, where am I doing wrong? Thankss. &lt;/p&gt;&#xA;" OwnerUserId="10099" LastActivityDate="2017-11-06T21:37:53.540" Title="Heat released by gas mixtures that undergo constant pressure cooling" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17902" PostTypeId="2" ParentId="17896" CreationDate="2017-11-04T20:05:53.377" Score="3" Body="&lt;p&gt;It isn't open source, but for students &lt;a href=&quot;https://www.onshape.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;Onshape&lt;/a&gt; is free. Onshape runs in the browser so it will work on pretty much any current computer regardless of operating system. It is real 3D solid modeling with good 2D drafting. I've played with it some and it works remarkably well and as CAD programs go is easy to learn. Because it is cloud based, it greatly fosters collaboration which could be great for student teams. In addition, Onshape provides a lot of &lt;a href=&quot;https://learn.onshape.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;learning resources&lt;/a&gt; including &lt;a href=&quot;https://www.onshape.com/video&quot; rel=&quot;nofollow noreferrer&quot;&gt;videos&lt;/a&gt; so teaching it should be enhanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w2Oqg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w2Oqg.png&quot; alt=&quot;Onshape Screen Shot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9506" LastActivityDate="2017-11-04T20:05:53.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17903" PostTypeId="1" CreationDate="2017-11-04T20:08:35.727" Score="0" ViewCount="114" Body="&lt;p&gt;is it possible that if i have a cabin which is insulated from outside but is built with some sort of material like copper and there is a heating element of sort which is providing 100C constantly but because room is totally insulated and shut can the temperature goes upto 1000C or not ?&#xA;if it is what kind of material can be used other can copper ?&lt;/p&gt;&#xA;" OwnerUserId="9965" LastActivityDate="2017-11-05T19:26:25.113" Title="is it possible to achieve 1000C temperature in a cabin?" Tags="|chemical-engineering|heating-systems|chemistry|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="17905" PostTypeId="2" ParentId="16976" CreationDate="2017-11-04T22:16:03.460" Score="1" Body="&lt;p&gt;Measure the effective circumference to calculate the diameter.  To do this mark the tire, and the ground with paint or a soapstone.  Move the vehicle until the mark on the tire contacts the ground again, and mark the ground.  Then measure the distance between the marks on the ground.  This distance is the effective circumference.  C = 2 * pi * r.  Solve for r.  r = C / (2 * pi).&lt;/p&gt;&#xA;" OwnerUserId="13416" LastActivityDate="2017-11-04T22:16:03.460" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17907" PostTypeId="1" AcceptedAnswerId="17931" CreationDate="2017-11-05T02:01:35.780" Score="3" ViewCount="5136" Body="&lt;p&gt;I'm used to calculate the minimum radius required for bending operations on both aluminum and steel sheets, given the following parameters: &lt;a href=&quot;https://i.stack.imgur.com/yxEBc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/yxEBc.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem statement:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For a sheet metal stock with  inch thickness, determine the minimum&#xA;  tool radius for both the steel and aluminum alloys that will not tear&#xA;  the material. Assume the sheet of material is in pure bending (i.e.,&#xA;  no additional tension is applied during forming).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have already learned about the following formulas:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;R = 1/k&lt;/li&gt;&#xA;&lt;li&gt;k(y) = e(y) / (h/2)&lt;/li&gt;&#xA;&lt;li&gt;e(y) = k.y&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given that R = 1/k, the next step is to find the maximum curvature k for each sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I don't know how to find the maximum curvature, and I'm still stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="13157" LastEditorUserId="421" LastEditDate="2017-11-07T05:17:09.403" LastActivityDate="2017-11-07T05:17:09.403" Title="Sheet Metal Bending: stuck on calculating the minimum radius required" Tags="|mechanical-engineering|manufacturing-engineering|metals|moments|yield-point|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17908" PostTypeId="2" ParentId="17903" CreationDate="2017-11-05T02:41:02.893" Score="3" Body="&lt;p&gt;If the heating element is constantly at 100°C, there is no way it can cause the room can get warmer than 100°C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, perpetual motion machines would be possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you meant the heating element is providing a constant 100 Watts, then it is possible for the room to get to 1000°C with enough (meaning a lot) of insulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your copper building might weaken to the point of collapse at 1000°C because copper melts at 1083°C.&lt;/p&gt;&#xA;" OwnerUserId="13553" LastActivityDate="2017-11-05T02:41:02.893" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17909" PostTypeId="2" ParentId="17888" CreationDate="2017-11-05T04:58:23.157" Score="2" Body="&lt;p&gt;What is shown in the picture is what people usually refer to as a pin connection. There are multiple ways you can design a pin connection as suggested in the comments above and there is no one right design for it. Disregarding practical considerations, such as materials, load cycles and etc, and in pure design engineering terms, you will need to find the shear strength of the items you have selected for your pin, and make sure it is reasonably stronger than what you have calculated for the shear load on the pin based on the configuration you have. The shear load on the pin can be determined from the normal force at the pin connection, and ut can be something less if you have double shear or triple shear mechanisms in your pin connection in question.&lt;/p&gt;&#xA;" OwnerUserId="13526" LastActivityDate="2017-11-05T04:58:23.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17910" PostTypeId="1" CreationDate="2017-11-05T09:57:46.690" Score="-1" ViewCount="720" Body="&lt;p&gt;In my research over the internet everyone is telling that brass is not safe but still there are solenoid valves, joints, and piping made of brass used in everything like coffee machines, water dispensers etc.&lt;/p&gt;&#xA;" OwnerUserId="9965" LastEditorUserId="13505" LastEditDate="2017-11-07T02:03:07.387" LastActivityDate="2017-11-07T02:03:07.387" Title="Are brass and copper utensils safe for humans to eat from?" Tags="|mechanical-engineering|chemical-engineering|corrosion|chemistry|" AnswerCount="1" CommentCount="13" ClosedDate="2017-11-06T18:41:55.870" ContentLicense="CC BY-SA 3.0" />
  <row Id="17912" PostTypeId="2" ParentId="17910" CreationDate="2017-11-05T12:19:57.110" Score="2" Body="&lt;p&gt;It always depends on the concentration and where you place the copper within the human body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, the most interesting example are implants which are coated with copper and implanted within the human body. The idea behind this is, that copper is toxic at very low concentrations for bacteria, and the toxic concentration for human cells is slightly higher. If you hit the window between these two concentrations, you can prevent bacteria from settling down on your implant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the human body needs copper. The recommendations about the daily intake differ. Some studies in the US have shown that the general copper intake in the USA is too high, others showed that it's too low (&lt;a href=&quot;https://en.wikipedia.org/wiki/Copper_in_health&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some more information are provided by WHO (&lt;a href=&quot;http://www.who.int/water_sanitation_health/dwq/chemicals/copper.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;WHO - copper&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During designing a product, it gets more complex since the amount of released copper is influenced by complex corrosion processes which highly depend on the medium, time, etc. I think e.g. copper pans are coated most often (&lt;a href=&quot;http://www.nytimes.com/health/guides/nutrition/cooking-utensils-and-nutrition/overview.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Copper pans&lt;/a&gt;) to prevent copper from entering your body.&lt;/p&gt;&#xA;" OwnerUserId="13505" LastActivityDate="2017-11-05T12:19:57.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17915" PostTypeId="1" CreationDate="2017-11-05T19:11:44.483" Score="0" ViewCount="399" Body="&lt;p&gt;I am in the process of designing an off road vehicle and require a form of articulation between the front and back half and cannot find relevant information on the types of joints available.&#xA;I'm looking for movement similar to the vehicle in the video posted&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=WT73uOhKSAY&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=WT73uOhKSAY&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10468" LastActivityDate="2024-04-04T07:07:10.893" Title="Types of articulation" Tags="|mechanical-engineering|mechanisms|joining|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17916" PostTypeId="2" ParentId="17903" CreationDate="2017-11-05T19:26:25.113" Score="2" Body="&lt;p&gt;To answer the question the way it's asked&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;heating element of sort which is providing 100°C constantly but because room is totally insulated and shut can the temperature goes upto 1000°C or not ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, this is not possible.  A 100°C heater can't heat its surroundings above 100°C.  Heat will not pass from cold part of the system to hot part of the system.  Heat passes only from a hotter to a colder system.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Laws_of_thermodynamics#First_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;First law of thermodynamics.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2017-11-05T19:26:25.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17917" PostTypeId="1" AcceptedAnswerId="17938" CreationDate="2017-11-05T21:41:05.130" Score="0" ViewCount="38" Body="&lt;p&gt;I'm trying to build a step stool that has a rotating chair back that will rotate from top position (e.g. chair) into bottom position (e.g. step stool). I do not want a free rotation of the chair back so I want to avoid it slamming on the floor. I've been looking for a bearing that acts as a mechanical stepper to click every 15 degrees with enough force requirement to stop the chair back from falling to the ground without force being applied (other than gravity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been searching google for a while now and cannot figure out what this part would be called.&lt;/p&gt;&#xA;" OwnerUserId="13559" LastActivityDate="2017-11-07T11:30:48.977" Title="What is the hardware called that can step a lever N degrees upon manual force?" Tags="|mechanical-engineering|bearings|bolting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17918" PostTypeId="1" CreationDate="2017-11-05T23:23:49.510" Score="1" ViewCount="59" Body="&lt;p&gt;When searching for this online I did not seem to be able to find any examples of this, which might imply that this idea does not work (or I did not use the right search terms). I have seen the use of two counter rotating fans, but this would probably still impart rotational energy to the flow after the second fan. I think a jet engine is an example of what I mean, but I at least have not seen other applied use cases.&#xA;Intuitively I would think that adding stationary blades after the driven rotating ones could improve performance (if the total weight is not an issue). Namely those stationary blades can convert some of the rotational energy of the fluid, imparted by the rotating fan blades, into (linear) kinetic energy of the fluid. So increase the mean flow rate and/or head pressure. Also converting and thus reducing the rotational energy should also reduce friction between the fluid and the wall of the duct. Since a rotating fluid flowing through a duct has a higher velocity near the wall then a none rotating fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would (some of) these gains be counteracted by a higher torque required to keep the rotating blades spinning at the same speed? Or might the gains at typical speeds be too low?&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2017-11-06T16:02:22.317" Title="Can adding stationary bades after an axial ducted fan increase efficiency?" Tags="|fluid-mechanics|airfoils|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17919" PostTypeId="1" CreationDate="2017-11-05T23:23:56.883" Score="2" ViewCount="642" Body="&lt;p&gt;So far, I have learned of three quantities that are related to the failure of a beam (axial and longitudinal loads). The first illustrates the stress under which balsa wood will undergo plastic buckling:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PdVbQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PdVbQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where m is a constant (estimate =2 for balsa, sigma ys is yield strength, and rho s is 1500 kg/m^3. rho is obviously density. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is a graph of compressive failure strength of balsa wood at varying densities:&lt;a href=&quot;https://i.stack.imgur.com/qkALh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qkALh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%27s_critical_load&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler's Critical Load&lt;/a&gt; equation: &quot;The critical load is the maximum load which a column can bear while staying straight.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uP3t2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uP3t2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, E = elastic modulus, I = area moment of inertia, K is a constant varying from 1-2 (based on whether the member is pinned or not on ends, estimate =1), L is the length, and Pcr is the critical load in Newtons (I believe.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two are from &lt;a href=&quot;http://www.its.caltech.edu/~mvural/Balsa_stat_linked.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; research paper, and the last is from Wikipedia. My question is, if there is a length at which the load will not bend the beam, then why is there a plastic buckling stress, which only depends on area? Shouldn't the beam not buckle under any load as long as the length is small enough? I believe there is something I am missing with the fundamental definitions of these. If stress is F/A, then the first equation could be changed to F = A *... which does not include length, as Euler's equation does. And how does failure, as shown in the graph, play into this?&lt;/p&gt;&#xA;" OwnerUserId="9176" LastEditorUserId="421" LastEditDate="2017-11-06T17:56:38.967" LastActivityDate="2017-11-06T17:56:38.967" Title="What's the difference between critical load and yield stress, and failure in general?" Tags="|civil-engineering|materials|stresses|wood|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17920" PostTypeId="2" ParentId="17919" CreationDate="2017-11-06T05:30:36.087" Score="3" Body="&lt;p&gt;Reading the paper, plastic buckling in this context means the collapse of micro-scale cells forming the structure of the wood, causing local buckling of the cell walls. The net effect of this is a global failure which presumably looks like a crushed band in the sample. In a sense, each individual cell wall is undergoing a constrained version of elastic buckling. The overall sample permanently deforms as a result, hence &lt;em&gt;plastic&lt;/em&gt; buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall effect of plastic buckling is in contrast to elastic buckling, which is a global bending phenomenon caused by an eccentric loading instability in sufficiently slender columns and plates loaded in compression along their long axes. If the centers of pressure of the two sample-fixture interfaces don't align perfectly along the compression direction, there is a slight eccentricity of the loading. The eccentricity leads to the formation of a moment couple, subsequent bending, and ultimately elastic buckling. Note that there is no way such perfect alignment can occur, and so elastic buckling should always be considered in slender objects loaded in compression along their length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the significance of the graph, it appears to be showing that density and yield strength of balsa are linear over the given range of densities, regardless of other common considerations or test methods. Note that the material can undergo special failure modes not available to monolithic materials due to its porous microstructure. Balsa has a hollow cell microstructure, and individual cells can collapse in one of two ways. The first is by plastic buckling, where cell walls buckle, causing axial cell collapse. The second is by kink-band formation, or a coordinated cell collapse where individual cells shear, appearing to tip over. Because these failure mechanisms don't require much physical space to occur around the sample, unlike with elastic buckling, they allow balsa to fail regardless of confinement. As a result, constraint has no effect on the density-yield relationship. To support this, from the conclusion section of the paper,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Independence of strength from confinement is attributed to the highly&#xA;  porous cellular microstructure of balsa.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The linearity of the relationship is explained by the fact that different mechanisms come into play at different densities. Despite that they may individually have nonlinear density-yield relationships, as in the first equation you posted, at some density another mechanism takes over and becomes energetically favorable. On either side of that density, different mechanisms will be observed at failure. It appears that, over the range of samples observed, the mechanisms coincidentally line up. At different densities, different failure mechanisms are seen because they are more energetically favorable. To support this, again from the conclusions section,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The post-mortem SEM examinations reveal that failure mode transition&#xA;  from elastic/plastic buckling to kink band formation occurs as the&#xA;  density increases.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;To sum up, the samples failing by plastic buckling aren't buckling like a beam does, but rather there is a coordinated buckling of a microstructural feature resulting in permanent deformation of the sample. Because balsa has a hollow cell microstructure, some failure modes can occur even when the sample is mechanically confined. Also, different failure modes occur at different densities in such a way that density happens to form an overall linear relationship with yield-strength.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2017-11-06T17:56:25.447" LastActivityDate="2017-11-06T17:56:25.447" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="17921" PostTypeId="2" ParentId="17915" CreationDate="2017-11-06T08:40:01.477" Score="0" Body="&lt;p&gt;One category of vehicle that does transmit power from the drive unit to the other unit is the agricultural tractor : have a look at the different shaft sections that have been used square, lemon (oval type) etc with the use of Hardy-Spicer or constant velocity joints for articulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as the Landrover as I mentioned above, I think there was a large truck with driven trailer in the Paris-Dakar but can't remember the manufacturer..&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-06T08:40:01.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17922" PostTypeId="1" CreationDate="2017-11-06T12:25:40.100" Score="1" ViewCount="41" Body="&lt;p&gt;Frequency shift keying uses at least two different frequencies. How much should be the difference between the two frequencies so that the receiver reads the signal properly reliably? Is there any theoretical law or empirical norm?&lt;/p&gt;&#xA;" OwnerUserId="13569" LastEditorUserId="8055" LastEditDate="2017-11-06T16:53:23.537" LastActivityDate="2017-11-06T16:53:23.537" Title="Gap between the two frequencies in FSK" Tags="|electrical-engineering|rf-electronics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17923" PostTypeId="5" CreationDate="2017-11-06T15:32:01.427" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2017-11-06T15:32:01.427" LastActivityDate="2017-11-06T15:32:01.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17924" PostTypeId="4" CreationDate="2017-11-06T15:32:01.427" Score="0" Body="This is a special tag that is designed to be used only by the system for questions that have had all of their other tags removed. Do not add this tag to existing questions, it is reserved for use by the system to identify posts with no valid tags." OwnerUserId="6767" LastEditorUserId="6767" LastEditDate="2017-11-06T18:07:00.827" LastActivityDate="2017-11-06T18:07:00.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17925" PostTypeId="2" ParentId="17918" CreationDate="2017-11-06T16:02:22.317" Score="1" Body="&lt;p&gt;Yes, it depends on if it is cost effective; for a consumer window fan = No. For a consumer vacuum cleaner =Yes. Of the jet engines I have seen , every stage of blades ( rotating) has a stage of vanes ( stationary).... Except the first stage/fan.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-11-06T16:02:22.317" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17928" PostTypeId="2" ParentId="17899" CreationDate="2017-11-06T20:08:20.863" Score="0" Body="&lt;p&gt;You're assuming that both gases have the same mass, which isn't true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{mass} = \text{density}\cdot\text{volume}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V_{\text{CO}_2}=0.3 V_\text{total}$ and $V_{\text{N}_2}=0.7 V_\text{total}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the calculated mass to find $Q_p$ unit is in $\text{KJ}/\text{m}^3$.&lt;/p&gt;&#xA;" OwnerUserId="13577" LastEditorUserId="1832" LastEditDate="2017-11-06T21:37:53.540" LastActivityDate="2017-11-06T21:37:53.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17931" PostTypeId="2" ParentId="17907" CreationDate="2017-11-06T22:26:09.103" Score="3" Body="&lt;p&gt;I hope the following helps.  This is from a book I used in college. Manufacturing Engineering and Technology, 5th Ed. by Kalpakjian and Schmid&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R1aHB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R1aHB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7ApdV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7ApdV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iUmbu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iUmbu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6T4Tq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6T4Tq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HFfS1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HFfS1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nHkmW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nHkmW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h1&gt;Minimum Bend Radius&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;Where **r** is the tensile reduction of area.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QQghx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QQghx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;" OwnerUserId="13416" LastEditorUserId="13416" LastEditDate="2017-11-07T01:05:03.707" LastActivityDate="2017-11-07T01:05:03.707" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="17936" PostTypeId="1" CreationDate="2017-11-07T10:10:51.820" Score="2" ViewCount="98" Body="&lt;p&gt;I have a question about design of concrete shells with three-layer method described in p.7.3.2.2 fib Model Code for concrete structures.&#xA;The main idea of method is that we divide shell on three layers. The outer layers provide resistance to the in-plane effects of both the bending and the in-plane axial loading, while core provides a shear transfer between the outer layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The action effects of applied loads are expressed as eight components, three moments per unit width($m_x,m_y,m_{xy}$), three axial forces per unit width ($n_x,n_y,n_{xy}$) and two shear forces per unit width ($v_x,v_y$) in directions parallel to the orthogonal reinforcement.&#xA;(there is positive directions of forces on picture)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Mv6he.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mv6he.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we use some formula to find forces in the outer layers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{1}&#xA;n_{x,inf,sup} &amp;amp;= \frac{n_x}{2} &amp;amp;&amp;amp;\pm \frac{m_x}{z} &amp;amp;&amp;amp;+ \frac{v_x^2}{2v_0}\cot(\theta) \tag{7.3-1}\\&#xA;n_{y,inf,sup} &amp;amp;= \frac{n_y}{2} &amp;amp;&amp;amp;\pm \frac{m_y}{z} &amp;amp;&amp;amp;+ \frac{v_y^2}{2v_0}\cot(\theta) \tag{7.3-2}\\&#xA;n_{xy,inf,sup} &amp;amp;= \frac{n_{xy}}{2} &amp;amp;&amp;amp;\pm \frac{m_{xy}}{z} &amp;amp;&amp;amp;+ \frac{v_x v_y}{2v_0}\cot(\theta) \tag{7.3-3}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&#xA;The subscript notation &lt;em&gt;inf&lt;/em&gt; and &lt;em&gt;sup&lt;/em&gt; refer to the interior and superior faces on the element. The interior face is the tensile face for an element in positive bending.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_0=\sqrt{v_x^2+v_y^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta$ is the inclination of compression stresses in the core layer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have the picture for the equations:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/g53PW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g53PW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is my question(s):&#xA;Why is plus sign in equations (7.3-1 - 7.3-3) in front of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;\frac{v_x^2}{2v_0}\cot(\theta) \\&#xA;\frac{v_y^2}{2v_0}\cot(\theta) \\&#xA;\frac{v_x v_y}{2v_0}\cot(\theta)&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why $v_0\cot(\theta)/2$ direction in &lt;em&gt;sup&lt;/em&gt; part of shell is the same as &lt;em&gt;inf&lt;/em&gt; part?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand how we got this part in equation.&#xA;Does anybody have any information about the three-layer model? Can you explain me this part? &#xA;Do you know any articles explained this?&lt;/p&gt;&#xA;" OwnerUserId="13199" LastEditorUserId="1832" LastEditDate="2017-11-07T11:47:34.920" LastActivityDate="2017-11-07T11:47:34.920" Title="How to consider shear forces in CEB/fib 2010 three-layer model for concrete slab design?" Tags="|concrete|shear|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17937" PostTypeId="1" AcceptedAnswerId="17940" CreationDate="2017-11-07T11:04:06.037" Score="2" ViewCount="63" Body="&lt;p&gt;I want to provide a gradual increase in force provided to a piston (perhaps 0-100N over the space of a couple of seconds maximum), up till the point it begins moving, and then drive it to a varying speed after that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of using a linear actuator and applying a variable voltage to control the speed and force, but apparently using such motors in stall like this is generally a bad idea due to burning out the coils. However stepper motors and solenoid actuators are both designed for holding loads in place while applying a certain torque - would a stepper motor be able to do this without damage? Or could a solenoid actuator provide the gradual force increase in stall, and then the rest of the movement be accomplished using a seperate motor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance &lt;/p&gt;&#xA;" OwnerUserId="11394" LastActivityDate="2017-11-07T12:46:35.263" Title="How to vary the force output of a motor while stalled?" Tags="|motors|pressure|torque|stepper-motor|linear-motors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17938" PostTypeId="2" ParentId="17917" CreationDate="2017-11-07T11:30:48.977" Score="1" Body="&lt;p&gt;It sounbds like you are describing a &lt;a href=&quot;https://en.wikipedia.org/wiki/Detent&quot; rel=&quot;nofollow noreferrer&quot;&gt;detent&lt;/a&gt; mechanism &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the design this can be a positive lock like pin engaging with a hole which needs to be manually disengaged with each step or a 'soft' latch for example a ball bearing engaging with a conical pit which can be overcome just by applying a bit more force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can buy the sprung balls as a complete assembly which could then be mated to a disk with a shallow hole every 15 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also a pawl and ratchet can be modified so that the teeth work like cams against a sprung pawl rather than providing a positive stop. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-11-07T11:30:48.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17939" PostTypeId="1" AcceptedAnswerId="17948" CreationDate="2017-11-07T12:17:33.103" Score="0" ViewCount="134" Body="&lt;p&gt;I want to model a chemical process. I was searching in the literature to find similar works. In some of papers they say that &quot;we proposed an unstructured model&quot;. What does it mean? What is the difference between a structured model and an unstructured model?&lt;/p&gt;&#xA;" OwnerUserId="6566" LastActivityDate="2017-11-08T09:44:51.670" Title="What is an unstructured mathematical model?" Tags="|chemical-engineering|modeling|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17940" PostTypeId="2" ParentId="17937" CreationDate="2017-11-07T12:46:35.263" Score="3" Body="&lt;p&gt;There is nothing inherently wrong with running a motor in stall.  The difficulty is that the power dissipation &lt;i&gt;may&lt;/i&gt; be higher than the motor can handle long term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every motor can dissipate &lt;i&gt;some&lt;/i&gt; amount of power safely indefinitely.  In stall, the power being dissipated as heat is the full voltage times current.  As long as that doesn't exceed what the motor can handle, you can run the motor as long as you want in stall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque (or force for a linear motor) is proportional to current, and the power dissipation proportional to the square of the current.  Both these are independent of the motor speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to run a motor safely at any speed is to regulate the current thru it.  As long as that does not exceed the maximum current the motor can handle continuously, you can apply it continuously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, you want to control torque.  That works out well with the above.  Find the current to torque constant, and regulate the current to get the desired torque.  As long as you limit this current to what the motor is rated for continuously, there is no special issue about the motor being stalled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible problem for you is the &quot;lumpy&quot; nature of the torque over a magnetic cycle.  This is also called &quot;cogging&quot;.  Stepper motors usually have high cogging, so may not be appropriate for what you want.  A brushless DC motor being properly controlled can have quite low torque ripple.  That is probably a better choice for you, since you are trying to control torque.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-11-07T12:46:35.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17941" PostTypeId="1" CreationDate="2017-11-07T16:52:36.410" Score="1" ViewCount="127" Body="&lt;p&gt;I need help to figure out the torque needed by a motor to lift an arm with fixed weight at the end like in the picture....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the green circle is the fixed motor and the arm is welded to the motor shaft and the movement is only 90 degrees goes from vertical to horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arm length is 55cm, arm weight is 4kg and the fixed attached weight is 13 kg &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanna know what is the torque (n-m) required to rotate the arm from a vertical to horizontal position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2xIZ7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2xIZ7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13587" LastEditorUserId="8055" LastEditDate="2017-11-07T20:35:46.680" LastActivityDate="2017-11-10T07:58:46.317" Title="help in calculation of motor torque" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17942" PostTypeId="1" CreationDate="2017-11-07T19:07:57.987" Score="0" ViewCount="51" Body="&lt;p&gt;I have undertaken a project where I need to design a brake system for a miniature locomotive. To test the effective of the brakes I have decided to create a rig where the brake system is attached to a flywheel which is spun to a certain angular speed in order to simulate the momentum of the locomotive as if it was on a track. I don't have a track available which is why I am doing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the locomotive is 800KG and assumed to be travelling at a const velocity of 4.2 m/s therefore will possess a linear momentum of 3,360 kg.m/s and a linear kinetic energy of 7065J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to simulate this momentum in the form of a flywheel, in order to determine its dimensions I need to know how much angular momentum the flywheel needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure how I can correctly mathematically/physically be able to &quot;convert&quot; linear momentum into angular momentum. I know that these two quantities are independent which is why I am asking this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I have ended up so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy in an object has two components: linear K.E. (El) and rotation K.E.(Er) Et = El + Er&lt;/p&gt;&#xA;&#xA;&lt;p&gt;El = 0.5*m*(v^2) , Er = 0.5*I*w^2, Et = total energy where, 'v' is linear velocity, 'w' is angular speed and 'I' is moment if inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I assume the locomotive only has linear component: 7065J and I assume that the flywheel only has rotation component - can I assume it to have 7065J of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I question myself whether a linear joule is the same as rotational joule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any comment is helpful - i can work on from a hint&lt;/p&gt;&#xA;" OwnerUserId="13590" LastActivityDate="2017-11-21T13:30:08.357" Title="Simulating linear momentum/KE through angular momentum/KE" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17944" PostTypeId="1" AcceptedAnswerId="17945" CreationDate="2017-11-07T19:10:58.120" Score="1" ViewCount="37" Body="&lt;p&gt;I am working on a project which involves applying electrical load across a metal contact filled with lubricants (bearing applications). However, I am missing one of the blocks as attached and I cannot figure it out what material this is. Looking at the posts here, I am presuming this must be some copper alloy such as brass, however, I know that the rotating disc that this block sits on is of brass material and that can raise issues such as wear due to the presence of similar materials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions what material I can use for this application to not damage the discs while providing good conductivity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g9roUl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g9roUs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ENbOsl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ENbOss.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gEhXjl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gEhXjs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13588" LastEditorUserId="13588" LastEditDate="2017-11-08T12:20:43.227" LastActivityDate="2017-11-08T12:20:43.227" Title="Selection of copper alloy for electrical and wear applications" Tags="|electrical-engineering|materials|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17945" PostTypeId="2" ParentId="17944" CreationDate="2017-11-07T19:49:40.320" Score="0" Body="&lt;p&gt;It looks like phosphor bronze to me which makes sense as it has good bearing properties and reasonable electrical conductivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could also potentially be sintered (ie porous), oil filled bronze bushings are fairly common. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-11-07T19:49:40.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17948" PostTypeId="2" ParentId="17939" CreationDate="2017-11-08T09:44:51.670" Score="1" Body="&lt;p&gt;A model for a biological reaction, bioreactor, etc. is called unstructured when the biomass components are not diferentiated, i.e. the cells with their internal metabolites are all lumped into a single variable (commonly expressed as X). We normally use Monod kinetics to model growth in unstructured models. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, a structured model may model as different variables the cell count, their content in i.e. pyruvate, fats, ATP, etc. Using this information, they can also account for different growth phases for the cell&lt;/p&gt;&#xA;" OwnerUserId="3991" LastActivityDate="2017-11-08T09:44:51.670" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17950" PostTypeId="1" AcceptedAnswerId="17951" CreationDate="2017-11-08T14:47:04.497" Score="1" ViewCount="227" Body="&lt;p&gt;I want to lift something in the air with a stepper motor. If the object is 1kg and I want to keep it in the air for 12 hours. Is it a good idea to keep the power on the motor all that time? Will it draw excessive amount of power? Ideally I would like to cut power to the motor when the movement is done, but then, will the motor be able to hold its position?&lt;/p&gt;&#xA;" OwnerUserId="13604" LastEditorUserId="8055" LastEditDate="2017-11-09T01:29:03.493" LastActivityDate="2017-11-09T01:29:03.493" Title="How much power does a stepper motor draw when holding weight?" Tags="|electrical-engineering|stepper-motor|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17951" PostTypeId="2" ParentId="17950" CreationDate="2017-11-08T15:52:02.567" Score="2" Body="&lt;p&gt;Depends on how much weight and the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some stepper motors have detent torque. This is the torque they will hold without consuming any power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also have a holding torque for which you need to supply power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most fool-proof solution is to use a external brake.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2017-11-08T15:52:02.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17952" PostTypeId="1" AcceptedAnswerId="17964" CreationDate="2017-11-08T16:03:00.023" Score="0" ViewCount="44" Body="&lt;p&gt;I'm making sort of a &lt;a href=&quot;https://www.google.com/search?tbm=isch&amp;amp;q=desk%20cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;desk cycle&lt;/a&gt;. What are the ways to make silent circular motion with resistance ? (I don't need to adjust resistance and resistance should be just light.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I didn't need resistance, I could use a bearing. Or maybe there are regular bearings that have some specific resistance ?&#xA;I see there are also magnetic bearings but I had really difficulties to find their prices and those ones that I found costs in hundreds. Are there some cheap ones, ideally ~ 15 or less bucks ?&#xA;Or do you have an idea how to achieve silent circular motion with resistance ? I was thinking of something like winding/unwinding - continual never ending loop against some weight but that would require changing of the motion from wind to unwind to make it never ending and honestly it seems it isn't very simple to design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks&lt;/p&gt;&#xA;" OwnerUserId="13607" LastActivityDate="2017-11-09T15:22:40.210" Title="Silent circular motion with resistance" Tags="|design|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="17953" PostTypeId="1" CreationDate="2017-11-08T17:41:18.097" Score="2" ViewCount="65" Body="&lt;p&gt;So we are doing this project on conducting an experiment to investigate the Statical Equilibrium of Plane Frameworks. Why do we need to conduct such experiments when we can easily calculate by hand and by pin-joint analysis.&lt;/p&gt;&#xA;" OwnerUserId="13610" LastActivityDate="2017-11-23T23:06:46.740" Title="Why conduct experiment for load distribution within three triangulated structural frameworks when you can calculate theoretically" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17954" PostTypeId="1" AcceptedAnswerId="17960" CreationDate="2017-11-08T18:16:45.140" Score="-1" ViewCount="67" Body="&lt;p&gt;I have several (at least six) piles of mineral adding up to 10 000 tons. I want to mix them in one homogeneous big pile. How can I do it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="13611" LastEditorUserId="9" LastEditDate="2017-11-10T06:30:53.770" LastActivityDate="2017-11-10T06:30:53.770" Title="How to obtain an homogeneous mixture of 10 000 tons of minerals, each one in a different pile?" Tags="|manufacturing-engineering|" AnswerCount="1" CommentCount="6" ClosedDate="2017-11-10T22:22:27.320" ContentLicense="CC BY-SA 3.0" />
  <row Id="17955" PostTypeId="2" ParentId="17953" CreationDate="2017-11-08T19:47:56.980" Score="2" Body="&lt;p&gt;To compare the theoretical and practical results, perhaps understand the differences and evaluate the sources of errors, maybe due to measurement for example.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-08T19:47:56.980" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17956" PostTypeId="2" ParentId="17801" CreationDate="2017-11-08T19:51:35.230" Score="3" Body="&lt;p&gt;From a theoretical standpoint, the displacement gradient is equivalent to strain (assuming a structural problem). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Numerically, you can obtain the derivate of a quantity through multiplication with the derivative of the shape functions, which is often referred to as the B matrix:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;du = B u&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="13602" LastActivityDate="2017-11-08T19:51:35.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17957" PostTypeId="2" ParentId="17953" CreationDate="2017-11-08T21:40:12.753" Score="7" Body="&lt;p&gt;There are several general reasons for conducting practical  tests in addition to theoretical analysis. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;to verify that the theoretical approach and associated assumptions are valid. Pretty much all loading models involve  some simplifications and assumptions. &lt;/li&gt;&#xA;&lt;li&gt;for quality assurance :  practical tests can pick up manufacturing/material defects,  design errors and mis-communications  between the design and manufacture stages&lt;/li&gt;&#xA;&lt;li&gt;regulatory approval : slapping a load on a structure is a lot easier to implement than verifying calculations especially in non-standard cases. &lt;/li&gt;&#xA;&lt;li&gt;practical testing verifies the whole product which can actually be less trouble than documenting every single stage of design and production. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You don't say if this is an academic project, if so it may simple be a case that the course wants to teach/evaluate both theoretical and practical skills. A huge aspect of real world engineering is having a sense of how theoretical models relate to reality and understanding that all models contain assumptions which may not always be valid. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-11-23T23:06:46.740" LastActivityDate="2017-11-23T23:06:46.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17958" PostTypeId="1" CreationDate="2017-11-09T00:29:35.570" Score="0" ViewCount="429" Body="&lt;p&gt;I have a conceptual design that I've been evaluating which can be summarized as a simple pendulum extended to 3D (i.e. swing not limited to x-z plane). If I want to actually build the conceptual design, is it possible to have a pivot point that does not permit rotation about the pendulum leg axis, but does allow for it to swing &quot;freely&quot; in 3D space?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easier way to probably ask is, if the pendulum is hanging statically, can I constraint the motion such there is no rotation about the leg-axis (statically coherent with z-axis) and later, dynamically, allow for it to swing &quot;freely&quot; in 3D space?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should emphasis, I'm looking for a &quot;product&quot; idea that permits such motion. I thought a universal joint may work, but I think that in order to swing &quot;freely&quot; the joint must be able to rotate about the z-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, the more and more I think about it, I'm wondering if it's theoretically possible? Does this give rise the the &quot;elastic pendulum&quot; idea, where a spring must be involved?&lt;/p&gt;&#xA;" OwnerUserId="13360" LastActivityDate="2018-02-08T11:27:40.860" Title="Pivot joint of pendulum with no yaw" Tags="|mechanical-engineering|structural-engineering|structural-analysis|applied-mechanics|dynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17960" PostTypeId="2" ParentId="17954" CreationDate="2017-11-09T01:51:15.147" Score="5" Body="&lt;p&gt;This happens all the time in the minerals industry where people need to blend material from different stockpiles to produce a supposedly uniform product for a processing plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the size of each stockpile either a front end loader or a bucket wheel excavator are used to take some material from one pile and deposit it elsewhere. You can get fancy and have a number of loader feeding hoppers for conveyor belt and have a number of conveyor belt form one pile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what is being optimized: grade of a metal, minimization of contaminant minerals/metals, hardness or softness of the material being fed to crushers/grinders the mix from each stockpile will be varied - such as one bucket from from Pile 1, three from Pile 2 &amp;amp; one from Pile 3. The blend mix will be up to you to decide.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-11-09T01:51:15.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17961" PostTypeId="1" AcceptedAnswerId="17962" CreationDate="2017-11-09T06:00:29.347" Score="3" ViewCount="258" Body="&lt;p&gt;I had a typical &lt;a href=&quot;https://previews.123rf.com/images/mybaitshop/mybaitshop1005/mybaitshop100500092/6989210-A-pink-bendy-straw-on-a-white-background--Stock-Photo-bent.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;bending drinking straw&lt;/a&gt; and when I twisted the outside tip, I expected the straw to rotate in an L-shape, so that the other end of the straw would be rotating sideways. Instead, I found that the straw remained perfectly in place, and the other end twisted in place regardless of the bent angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was just curious: Do they ever use this particular mechanism in engineering, rather than, say, gears?&lt;/p&gt;&#xA;" OwnerUserId="13617" LastEditorUserId="13617" LastEditDate="2017-11-09T21:06:11.600" LastActivityDate="2017-11-10T14:20:05.923" Title="Is a mechanism similar to a bendy-straw ever used to change the direction of a rotation in engineering?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17962" PostTypeId="2" ParentId="17961" CreationDate="2017-11-09T07:00:23.657" Score="3" Body="&lt;p&gt;Yes, on some ratchet &amp;amp; socket sets there is a flexible drive : two hardened steel springs, one wound one way one the other. But it won't change the direction unless it points 180 degrees the other way. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-09T07:00:23.657" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17964" PostTypeId="2" ParentId="17952" CreationDate="2017-11-09T08:04:10.433" Score="1" Body="&lt;p&gt;You could try agitating a viscous fluid.  Connect your pedal crankshaft to a vertical shaft (via a pair of quiet bevel gears) which has a paddle on the end of it that is surrounded by some thick gloopy fluid like motor oil (or something preferably water soluble).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The viscosity is proportional to the resistance.  You could vary the resistance (while pedaling) by moving the paddle up and down so that it descends into the fluid to different depths.  Maximum resistance will be when the paddle is fully submerged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will be pretty much silent if you choose a good linkage between the shafts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a paint stirrer, the principle would be the same.&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastEditorDisplayName="user6335" LastEditDate="2017-11-09T08:54:37.577" LastActivityDate="2017-11-09T08:54:37.577" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17966" PostTypeId="1" CreationDate="2017-11-09T09:23:37.830" Score="-1" ViewCount="320" Body="&lt;p&gt;I want to know what is the meaning of dσ/du in this solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iQ0mV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iQ0mV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OtPXs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OtPXs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13620" LastEditorUserId="8055" LastEditDate="2017-11-11T22:30:16.133" LastActivityDate="2022-12-27T04:02:25.287" Title="mechanics of materials Mohr" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17967" PostTypeId="2" ParentId="17941" CreationDate="2017-11-09T12:25:42.787" Score="3" Body="&lt;p&gt;As a quick approach you can calculate the torque needed to hold the arm + weight in a horizontal position&#xA;$$&#xA;M=M_{weight}+M_{arm}&#xA;$$&#xA;$$&#xA;M=m_{weight} \cdot g \cdot l+\frac{1}{2} \cdot m_{arm} \cdot g \cdot l&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your torque (after the gear) must be higher than that to be able to rotate in that position.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;M=13\ kg \cdot 9.81\ m/s^2 \cdot 0.55\ m+0.5 \cdot 4\ kg \cdot 9.81\ m/s^2 \cdot 0.55\ m&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="8282" LastEditorUserId="78" LastEditDate="2017-11-10T07:58:46.317" LastActivityDate="2017-11-10T07:58:46.317" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17968" PostTypeId="2" ParentId="17952" CreationDate="2017-11-09T15:22:40.210" Score="0" Body="&lt;p&gt;you might turn a generator and either dump the power into a resistor or use it to power something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course silent is a bit of a subjective term.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2017-11-09T15:22:40.210" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17969" PostTypeId="2" ParentId="17958" CreationDate="2017-11-09T17:10:57.493" Score="0" Body="&lt;p&gt;Universal and very low friction 360degree joints can be made by linking two spherical magnets.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the contact force is limited by the size of the magnet this type of connection satisfies the requirement for free low friction 3D motion. The contact force is approximately 5Kg for a pair of 2cm rare earth magnetic spheres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These joints are sometimes used on hex manipulators such as articulating 3D printer heads used in delta style printers. Additional mechanical parts are obviously required. &lt;a href=&quot;https://i.stack.imgur.com/bw1jd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bw1jd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-11-09T17:10:57.493" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="17971" PostTypeId="1" CreationDate="2017-11-09T20:56:46.493" Score="1" ViewCount="53" Body="&lt;p&gt;My group and I are currently working on a biomedical engineering project for our class. Our design idea is a wearable metal detector that uses metallic tracers in the blood to determine the concentration of glucose in the blood. There are multiple aspects of this design, but we wanted to reduce this project down to a proof of concept. We want to make a metal detector that is REALLY sensitive that can detect objects that we can make as small as possible (think iron fillings) that could simulate the tracer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We plan to use an Arduino to read the changes in coil inductance. The main problem that we are running into is that, being biomed majors, we do not really know much about the electronics and circuits that will make this concept work. We found a few circuits for metal detectors, but we really don't understand what's going on in them. Here's a link to one that I found &lt;a href=&quot;http://3.bp.blogspot.com/-iu_ERGb6vzQ/Ue2DyIWZYFI/AAAAAAAAAd8/yl2Wk05B37A/s1600/IMG_20130722_202500.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://3.bp.blogspot.com/-iu_ERGb6vzQ/Ue2DyIWZYFI/AAAAAAAAAd8/yl2Wk05B37A/s1600/IMG_20130722_202500.jpg&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main questions so far:&#xA;Can someone explain whats going on in this circuit? What would the input and output waveforms look like?&#xA;Can we use a simpler circuit and still get around the same sensitivity?&#xA;If we want a more sensitive detector would a small coil with a ton of turns be a safe option?&#xA;We'd like to use two coils concentric setup (one transmitter on receiver) would that add to sensitivity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="13626" LastActivityDate="2017-11-09T20:56:46.493" Title="Metal detectors for medical applications" Tags="|electrical-engineering|biomedical-engineering|" AnswerCount="0" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17972" PostTypeId="1" AcceptedAnswerId="17991" CreationDate="2017-11-10T05:20:13.943" Score="6" ViewCount="1926" Body="&lt;p&gt;I was at a well-known electric car company today and they claimed &quot;instant torque.&quot; The quote was something like the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Driving this car is very different from a gas-powered vehicle. For example, it has instant torque when you press the accelerator. Unlike a gas vehicle where there is a slight delay.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is, where does this &quot;instant torque&quot; come from? Is the explanation as simple as the fact that electrons/electricity travel/s at the speed of light. But it takes time for an internal combustion engine to convert chemical potential energy into kinetic energy via thousands of tiny explosions? Or is there something else that could illuminate this explanation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;a href=&quot;https://mechanics.stackexchange.com/questions/25944/why-do-electric-motors-produce-peak-torque-instantly-vs-gasoline-engines/25946&quot;&gt;I just found this related question on a separate SE.&lt;/a&gt; Can anyone add more of a &quot;fundamental physics explanation&quot; to the set of answers?&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2017-11-10T08:55:05.840" LastActivityDate="2018-11-10T07:21:55.187" Title="Why do electric cars claim &quot;instant torque?&quot;" Tags="|automotive-engineering|electrical|car|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17973" PostTypeId="2" ParentId="17972" CreationDate="2017-11-10T06:38:59.460" Score="3" Body="&lt;p&gt;No, it's a characteristic that electric motors produce most, if not all, of their torque from zero rpm - which is why the electric cars are so good at getting off the line and how electric motors don't always need gearing to start heavy machinery - just lots of energy...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-10T06:38:59.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17974" PostTypeId="2" ParentId="17972" CreationDate="2017-11-10T10:29:01.873" Score="2" Body="&lt;p&gt;The reason is inherent losses of the engine. Internal combustion engines consume a lot of torque they produce for own operation - they need to compress the fuel-air mix, they need to eject the combustion products, inject fuel-air mix, run the cooling system, run the fuel pump, run the alternator for providing electricity for spark plugs, overcome mechanical friction of the gearbox and so on. This all takes from the energy/torque produced through combustion and decompression, and the energy intake/output characteristics are quite non-linear; at low RPM there's really little usable surplus output torque and increasing it requires increasing RPM - the growth of surplus soon outpaces growth of demand, and the engine reaches peak power (...then the losses begin catching up again, and the engine reaches peak RPM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also connected with the combustion process: through pressing the accelerator you modify the fuel to air ratio in the mix, making it more energetic, but you can only go so far extracting certain amount of energy from a single combustion (single piston cycle) - further increasing of amount of fuel will just lead to ejecting unburnt fuel or flooding the spark plug, instead of increasing the amount of energy produced. Instead, you're better off increasing the number of combustion events per second - the frequency of piston cycles - the RPM of the engine. That way instead of minuscule amount of very fuel rich mix producing modest amount of energy, you have a much large amount of moderately fuel-rich mix producing a lot of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In electric motor there's both very little of such 'self-maintenance' losses and the amount of output torque doesn't depend on some kind of cycles, like frequency of combustion events - energy input can be ramped all the way up, no problems like flooding the spark plugs with too much of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this is the torque-RPM relation. The resulting time-torque relation is simple: as a combustion engine starts, its RPM are low, low torque output. If you want to increase it, you need to &quot;spin it up&quot;, increase RPM - and that takes torque, which you don't have in abundance yet - so the process takes time. In electric engines, you get full output immediately.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2017-11-10T22:59:27.060" LastActivityDate="2017-11-10T22:59:27.060" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="17975" PostTypeId="2" ParentId="17958" CreationDate="2017-11-10T10:41:09.720" Score="1" Body="&lt;p&gt;About the simplest approach is to use two single-axis joints (hinges) in line, 90 degrees to each other. That way one provides X-axis swinging, the other Y-axis, and nothing allows rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to maintain tip orientation (&quot;parallel to the floor&quot;), you'll need frame constructions, a single parallelogram shape using four hinges in the corners maintains the bottom's orientation relative to the floor. A rectangular prism based on two walls of such parallellograms, swinging on four to eight (4 is geometrically sufficient but 8 provide better rigidity) hinges perpendicular to the original 8 (2x4), will provide two-axis swinging retaining orientation of the bottom part relative to the floor.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-11-10T10:41:09.720" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17976" PostTypeId="2" ParentId="17972" CreationDate="2017-11-10T12:38:10.277" Score="8" Body="&lt;p&gt;ICEs (internal combustion engines) take some time to ramp up their output torque.  Various mechanical systems have to react before more mixture is injected into cylinders, and that then makes more pressure upon combustion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a throttle-body engine as example.  You push on the gas pedal, which opens the throttle valve more.  That causes more air to rush into the intake manifold, which drags more gas with it due to the venturi in the carburetor.  This lower pressure mixture gets sucked into a piston on the next intake stroke.  Then there has to be a compression stroke before the mixture is ignited, and you finally get more torque as a result.  Other things have to react to possibly change the timing, fuel/air ratio, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite the above, the real perceived lag is due to the torque curve of the engine.  Even though the engine is putting out more torque within a revolution or so, the engine's output is limited at low speed.  Now matter how optimal the mixture being fed into the cylinders, it can't produce significantly more torque until it gets to a higher speed.  That takes a while.  Or, the transmission system has to shift to allow the engine a more optimum operating point to produce the additional output power.  Either way, it takes time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For electric motors, on the other hand, torque is proportional to current.  This is independent of speed, so works just as well from a standing stop to cruising at highway speed.  The current is controlled by producing pulses thru transistor that can do this 100s of 1000s of times per second.  There is some lag to build up more current due to the inductance of the motor windings, but this lag is on the order of microseconds, milliseconds at most, and well below the human perception range.  Remember that people have a hard time detecting a lag below about 50 ms.  Having the motor be able to produce torque faster than humans can perceive the lag pretty much happens without any particular design effort to make it so.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-11-10T12:38:10.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17977" PostTypeId="1" CreationDate="2017-11-10T12:55:55.603" Score="-1" ViewCount="64" Body="&lt;p&gt;Most of the RF filters have a bandwidth 1 MHz or more. Mobiles have a 200 kHz bandwidth filter. What is the narrowest bandwidth possible for a RF filter?&lt;/p&gt;&#xA;" OwnerUserId="13635" LastEditorUserId="90" LastEditDate="2017-11-10T15:01:12.273" LastActivityDate="2018-01-25T03:55:27.540" Title="Narrowest bandwidth RF filter" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2018-01-26T09:56:18.187" ContentLicense="CC BY-SA 3.0" />
  <row Id="17978" PostTypeId="1" CreationDate="2017-11-10T14:02:38.003" Score="0" ViewCount="187" Body="&lt;p&gt;I have 2 pipe lines, both water . that combine to heat up water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming at steady state , &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat gained by cold water = heat lost by hot water &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say , according to formula , I calculated the combined temperature to be 62.94 degrees C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However , the temperate transmitter tells me that it's 63.06 degrees C, slightly higher than 62.94.. why is that the case ?&lt;/p&gt;&#xA;" OwnerUserId="11561" LastActivityDate="2018-10-20T13:24:17.187" Title="Heat balance in a pipe flow network" Tags="|heat-transfer|chemical-engineering|heating-systems|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17979" PostTypeId="2" ParentId="17961" CreationDate="2017-11-10T14:20:05.923" Score="1" Body="&lt;p&gt;There are a few types of joint which fall broadly into this category as well as the spring type arrangement used in flexible drives for pendant drills etc there is a joint called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Rag_joint&quot; rel=&quot;nofollow noreferrer&quot;&gt;rag joint&lt;/a&gt; which uses reinforced rubber of leather to transmit rotation through an angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally any sort of flexible tubing like fuel hose can be an effective way to improvise a flexible shaft coupling for prototyping etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While things like paired universal joints and constant velocity joints are now more common for power applications an elastic joint has some advantages in that it is simple, easy to miniaturise and can help to isolate vibration and shock loading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally CV joints are often protected by convoluted rubber gaiters (boots) which work on the same principal, although these are there to provide an external seal to contain grease and keep out contamination rather than transmit power. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-11-10T14:20:05.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17980" PostTypeId="1" AcceptedAnswerId="18017" CreationDate="2017-11-10T14:29:06.513" Score="2" ViewCount="146" Body="&lt;p&gt;I thought I understood what a proportionality was, but it was recently called into question. I want to make sure I fully understand how the term is generally used in the engineering world. The basic explanation of proportionality is that two variables are related by a constant multiplier such that a change in one of them results in a change in the other, scaled by this constant of proportionality. So we could say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A(x_1, x_2, ... x_n) \propto B(x_1, x_2, ... x_n)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A(x_1, x_2, ... x_n) = cB(x_1, x_2, ... x_n)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ and $B$ are the proportional variables and $c$ is the constant of proportionality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question deals specifically with what restrictions are placed on this constant. In my previous description, $A$ and $B$ were functions of a number of variables, they are necessarily the same variables. Is this the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, in engineering, few things are every truly a constant, so our constant of proportionality could be a function of a number of parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A(x_1, x_2, ... x_n) = c(y_1, y_2, ... y_n)B(x_1, x_2, ... x_n)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is my understanding that when discussing proportionality it must be assumed that all of those $y_n$ parameters must be held constant within your system. This is where my understanding starts to come into question. It was suggested that $c$ need not be independent of $A$ or $B$ and this is still a proportionality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = c(B)B&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I had always assumed that this is a prime example of a situation where $A$ is not proportional to $B$. But upon thinking about it further, I came up with 3 limited domain cases given the following relationship:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = (e^B - 1)B&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;case 1: $B &amp;lt; -10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;case 2: $B &amp;gt; 10$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;case 3: $-\infty &amp;lt; B &amp;lt; \infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that for case 1, we could perhaps say that $A \propto B$ given the small change in the coefficient. But for cases 2 and 3, would it ever be reasonable to consider $A \propto B$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify my questions: Is anything about my understanding as outlined above wrong? Are there any situations where it is correct to say $A \propto B$ and that the constant of proportionality is a strong function of $B$? Does this answer change if we can consider $B$, and therefore $A$ to be nearly a constant? Is that a useful situation to consider?&lt;/p&gt;&#xA;" OwnerUserId="13636" LastActivityDate="2017-11-13T13:51:55.130" Title="What exactly is a proportionality?" Tags="|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17981" PostTypeId="1" CreationDate="2017-11-10T16:15:40.127" Score="2" ViewCount="35" Body="&lt;p&gt;I've represented with Matlab a system with three degrees of freedom. I'm trying to plot a pole-zero map from my Frequency Response Matrix &lt;code&gt;H(w)&lt;/code&gt; in the case of Damped-Forced : &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;w=0:0.01:30;&#xA;for ii=1:length(w)&#xA;    FRF=inv(-w(ii)^2*M_gen+1i*w(ii)*C_gen_approx+K_gen);&#xA;    FRF11(ii)=FRF(1,1); FRF12(ii)=FRF(1,2); FRF13(ii)=FRF(1,3); &#xA;    FRF21(ii)=FRF(2,1); FRF22(ii)=FRF(2,2); FRF23(ii)=FRF(2,3);&#xA;    FRF31(ii)=FRF(3,1); FRF32(ii)=FRF(3,2); FRF33(ii)=FRF(3,3);&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Does anybody know if it is possible do this directly with some possible solution or do I have to get values of my FRF end, then generate pole-zero maps?&lt;/p&gt;&#xA;" OwnerUserId="13637" LastEditorUserId="13637" LastEditDate="2017-11-10T16:29:02.147" LastActivityDate="2017-11-10T16:29:02.147" Title="Plot zero-pole from frequency response matrix" Tags="|mechanical-engineering|dynamics|vibration|modal-analysis|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17983" PostTypeId="1" AcceptedAnswerId="17986" CreationDate="2017-11-10T19:07:23.243" Score="0" ViewCount="44" Body="&lt;p&gt;I'm working on a design that uses fairly large plastic components (ABS, ~400mm longest dimensions). I'm looking at adding a steel component that encapsulates this plastic part. And, the steel part needs to be tight against the ABS part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the thermal properties of ABS, the plastic part will change size fairly drastically relative to the Steel part. Since the steel part needs to tightly encapsulate the ABS part, I've been considering cooling the ABS part to around -10 to -20 Celsius before attaching the steel which would be at around +20 Celsius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, with ABS cooled to that temperature would riveting the steel part to it still be an option, or should I expect fracturing from the rivet? I've tried looking into the hardness, brittleness, and Younges modulus of ABS as a function of temperature but I can't find any good data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rivet would be a 7/32 sized rivet through a 3mm thick piece of ABS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2017-11-10T21:27:17.173" Title="Temperature vs Material Properties - ABS" Tags="|mechanical-failure|thermal-expansion|abs|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17985" PostTypeId="1" CreationDate="2017-11-10T20:15:13.970" Score="1" ViewCount="737" Body="&lt;p&gt;I'm making an model of a DC motor, and following the guide for &lt;a href=&quot;http://support.ctc-control.com/customer/elearning/younkin/motorParameters.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;measuring motor parameters&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having trouble measuring the motor's inertia, as the guide for this section is vague. This is a graph of my RPM over time (in milliseconds):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eDmp0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eDmp0.jpg&quot; alt=&quot;RPM vs Milliseconds&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the graph is noisy, but its the best I can capture at the moment.&#xA;What I can't figure out is, to find the acceleration, what section of this graph is the linear region the guide refers to? It states &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The acceleration is therefore&#xA;  the change in velocity for the linear part of the exponential curve divided by the time&#xA;  elapsed for the detected rate of change in velocity&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Another document, states this when saying how to measure motor inertia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The rotor inertia can be estimated from measurements of the mechanical time constant, kt&#xA;  ,&#xA;  and R. Alternatively, a constant current could be applied to the motor and the motor’s position differentiated&#xA;  twice to get acceleration α. Then the inertia can be estimated using the relation J = T /α = ktI/α. Note&#xA;  that constant current/torque/acceleration is not possible for long periods of time.&#xA;  (from &lt;a href=&quot;http://hades.mech.northwestern.edu/images/6/61/Asst7.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mystery Motor Data Sheet&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This implies there is another way to measure inertia using kt and R (which I know both of), but doesn't seem to state it. I would love to know what it is.&lt;/p&gt;&#xA;" OwnerUserId="13640" LastEditorUserId="8055" LastEditDate="2017-11-11T19:59:25.793" LastActivityDate="2017-11-11T19:59:25.793" Title="Determining brushed DC motor inertia for modeling a motor" Tags="|motors|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17986" PostTypeId="2" ParentId="17983" CreationDate="2017-11-10T21:27:17.173" Score="0" Body="&lt;p&gt;Cooling ABS to -20 C should be within its operating conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why are you concerned that your encapsulation does not sit tight? What is the temperature range in which this part is used? If its above room temperature, that ABS part will expand more and most likely „press-fit“ into the encapsulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have a drawing of that ABS part and encapsulation?&lt;/p&gt;&#xA;" OwnerUserId="8282" LastActivityDate="2017-11-10T21:27:17.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="17991" PostTypeId="2" ParentId="17972" CreationDate="2017-11-11T16:44:13.293" Score="5" Body="&lt;p&gt;I think the &lt;em&gt;instant torque&lt;/em&gt; claim mostly applies to &quot;off the line&quot; acceleration. That is from a standstill and electric motor has 100% of its available torque available at 0 rpm (mostly). The tradeoff is the an electric motor is &lt;em&gt;always&lt;/em&gt; going to see a drop in torque with speed. Generally when you compare wheel torque availability as a function of speed between electric and a gas engine (with 5 forward gears) you have the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9RXq6.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9RXq6.png&quot; alt=&quot;EC&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;strong&gt;Figure 1&lt;/strong&gt;. The defining characteristic of an electric car is that peak torque is available at 0 rpm, in contrast to a gas engine which peak torque occurs at speed.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also because of gearing, even though the torque and power values are much higher compared to gas engines, electric cars are limited to lower speeds because at some point the available torque goes to zero (even with ignoring air drag), compared to gas cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think secondary is the fact that having only one gear torque is delivered to the wheels quicker with electric, compared to a gas car which takes much longer. Remember gasoline engines are practically an air pump (with fuel added later to match the air). The more air that can be pushed through the more efficient the engine is (and the more torque it makes). This means that it takes time to spool up the engine to the right rpm (usually requiring a downshift in gears) and for the inertia of the engine and the inertia of the air at the intake to be overcome. You might not realise it, but at higher rpm the air going through the intake ports approaches supersonic speeds around bends and curves. It takes a lot of energy to get air to those speeds and that energy is taken from the energy available for acceleration.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-11-11T16:44:13.293" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="17992" PostTypeId="1" AcceptedAnswerId="18048" CreationDate="2017-11-11T18:49:14.297" Score="9" ViewCount="11720" Body="&lt;p&gt;We have a two year-old, two-story town house with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;fiberglass insulation in the walls;&lt;/li&gt;&#xA;&lt;li&gt;a Tyvek vapor barrier under foam board on the exterior walls;&lt;/li&gt;&#xA;&lt;li&gt;blown cellulose insulation in the attic;&lt;/li&gt;&#xA;&lt;li&gt;stucco outside;&lt;/li&gt;&#xA;&lt;li&gt;and a refrigeration cycle air conditioning.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We live in Phoenix, AZ (desert climate) and our indoor humidity levels are consistently 20% to 35% higher than outside. During the summer we consistently have humidity levels in excess of 80% (not 70%), even with the damper fully open. It has been so humid inside that the tile floors had sweat on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have no indoor plants or water features. Our dryer vent is operational, and our exhaust fans from the bathrooms are vented externally through the roof. There are no unusual ceiling vaults and we do have ceiling fans to stir the air. Ours is an end unit, so we have one shared wall with a neighboring town house, which has no vapor barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have had the builder heavily involved for warranty work to determine the root cause of the humidity sources. We have tried:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Slowing the AC blower motor down to allow the water in air to condense;&lt;/li&gt;&#xA;&lt;li&gt;Looking for water leaks with an infrared camera (FLIR);&lt;/li&gt;&#xA;&lt;li&gt;Looking for blockages in the vents in the eaves;&lt;/li&gt;&#xA;&lt;li&gt;Calibrating the thermostat (Ecobee);&lt;/li&gt;&#xA;&lt;li&gt;Balancing the air flow by adjusting the vents inside with flow meter hood;&lt;/li&gt;&#xA;&lt;li&gt;Adjusting the air conditioner refrigerant cycle pressures;&lt;/li&gt;&#xA;&lt;li&gt;Fully opening the damper to let in dry air;&#xA;&#xA;&lt;h2&gt;- Increasing the exhaust fan to run for 30 min in a 60 min cycle.&lt;/h2&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The problem hasn't gone away and we're not sure what to do next. Is there anything else we can try before we resort to installing a whole house dehumidifier, or adding additional attic ventilation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are your thoughts on the RH trend over time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/trhP4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/trhP4.jpg&quot; alt=&quot;Indoor Temp and Humidity over time&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sRZAG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sRZAG.jpg&quot; alt=&quot;Aug 1, 2018 indoor temp and RH&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I did a rough calculation that the home needs to express ~350 pints of water per day (ASHRAE) psychometric chart corrected for altitude, but the HVAC company measured the AC system is operating properly and measured the AC will remove 77 pints of water per day.  I read that whole home dehumidifiers typically remove ~100 pints per day.  What are your recommendations on how to best ventilate the attic while controlling cost?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://github.com/derkooh/HumidityData&quot; rel=&quot;nofollow noreferrer&quot;&gt;Historical Indoor RH and Temp. Including limited outdoor data&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13416" LastEditorUserId="13416" LastEditDate="2018-08-11T18:19:54.923" LastActivityDate="2018-08-11T18:19:54.923" Title="Humidity in home is much higher than outside" Tags="|thermodynamics|hvac|building-design|environmental-engineering|" AnswerCount="4" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="17994" PostTypeId="1" CreationDate="2017-11-12T00:06:55.763" Score="1" ViewCount="386" Body="&lt;p&gt;When the absolute pressure dips below the vapour pressure, cavitation occurs. Why would dynamic pressure be taken into account in this case?&lt;/p&gt;&#xA;" OwnerUserId="13433" LastActivityDate="2019-06-20T03:15:43.460" Title="Why does net positive suction head (NSPH) include velocity head?" Tags="|pumps|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="17995" PostTypeId="1" CreationDate="2017-11-12T00:18:00.243" Score="2" ViewCount="343" Body="&lt;p&gt;I am currently working on a prototype wire winch. I encountered some difficulties when trying to join the shaft and the cable drum.&#xA;Indeed the joint must be a welding, so I presume no keys are needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However since I'm not used to welding, I'm just wondering what type of welding would be the best for a Shaft/Drum connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also suppose that welding type would be a &quot;Tee weld&quot; or am I wrong?&lt;/p&gt;&#xA;" OwnerUserId="13645" LastEditorDisplayName="user16" LastEditDate="2017-11-12T15:32:32.210" LastActivityDate="2017-11-12T15:32:32.210" Title="Types of welding (Shaft)" Tags="|mechanical-engineering|welding|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="17999" PostTypeId="2" ParentId="17995" CreationDate="2017-11-12T07:20:27.310" Score="1" Body="&lt;p&gt;It is possible to consider a female spline in the drum and a splined shaft : this would make for easy replacement in case of failure and may help the assembly process.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-12T07:20:27.310" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18001" PostTypeId="2" ParentId="17994" CreationDate="2017-11-12T11:55:15.183" Score="2" Body="&lt;p&gt;Because dynamic pressure is contributing to the total pressure and, when checking for the possibility of cavitation all terms must be included or at least accounted for.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-12T11:55:15.183" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18002" PostTypeId="2" ParentId="17995" CreationDate="2017-11-12T12:05:27.807" Score="1" Body="&lt;p&gt;There are several issues to consider here :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it is pretty unusual to weld assemblies directly to the output shaft of a motor as is significantly complicates assembly and maintenance. &lt;/li&gt;&#xA;&lt;li&gt;Shafts are often plated and/or high strength steels which may require special preparation and treatment for welding. &lt;/li&gt;&#xA;&lt;li&gt;In some cases you want to isolate the motor from vibrations, shock loading and bending from the load, especially in the case of something like a winch. &lt;/li&gt;&#xA;&lt;li&gt;Welding almost inevitably introduces some distortion to a joint so you generally want to avoid welding parts where alignment is critical unless you have some other means of adjusting afterwards or are prepared to construct a very rigid jig. &lt;/li&gt;&#xA;&lt;li&gt;There is quite a lot of potential for damaging the motor by heat, stray current or sparks/spatter. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also in many cases it if preferable for the load on the winch to be carried by its own set of bearings as the motor bearings may not be designed for those sorts of radial and bending forces. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-11-12T12:05:27.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18003" PostTypeId="1" CreationDate="2017-11-12T16:34:25.460" Score="2" ViewCount="71" Body="&lt;p&gt;I am currently working on a problem involving volatile organic compounds (VOC) source monitoring on some oil and gas transportation environment. I think using infrared (IR) cameras could be a straightforward approach. The core idea is that temperature gradients will indicate the points where there are emission sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently comparing IR camera specifications. Some cameras can be literally portable as a smartphone gadget (e.g. Flir ONE) but I'm quite lost on which properties I should look for (gases' and cameras'), in order to find the most practical and cost effective solution. I've thought to start by finding out:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Which are the most frequently-occurring VOC in these environments;&lt;/li&gt;&#xA;&lt;li&gt;What are these compounds' most interesting properties (maybe emissivity, absorptivity, transmittance, reflectivity?) Note: Please correct me if I'm wandering off here, as it's been a while since I read any Heat Transfer literature;&lt;/li&gt;&#xA;&lt;li&gt;And how broad can the cameras be in terms of energy spectrum registering.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am looking for guidance from anyone with practical experience using this method, especially with a gadget like the Flir ONE, or any method of inexpensive, portable VOC monitoring.&lt;/p&gt;&#xA;" OwnerUserId="13651" LastEditorUserId="368" LastEditDate="2017-11-13T16:26:34.123" LastActivityDate="2017-11-13T16:26:34.123" Title="How to select a smartphone IR camera for VOC monitoring" Tags="|fluid-mechanics|heat-transfer|environmental-engineering|air-quality|" AnswerCount="0" CommentCount="1" ClosedDate="2017-11-13T16:32:19.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="18006" PostTypeId="2" ParentId="17714" CreationDate="2017-11-12T17:43:34.260" Score="0" Body="&lt;p&gt;It seems some official documentation does indeed mention 'speed' as a determining factor in the amount of road damage, especially with respect to heavy vehicles...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will post the link, as the document is very long, but they even mention a milestone speed of 40 kmh where speeds over this are known to cause more damage by heavy vehicles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.nvfnorden.org/lisalib/getfile.aspx?itemid=261&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nvfnorden.org/lisalib/getfile.aspx?itemid=261&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13428" LastActivityDate="2017-11-12T17:43:34.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18007" PostTypeId="1" AcceptedAnswerId="18008" CreationDate="2017-11-12T19:21:30.203" Score="4" ViewCount="116" Body="&lt;p&gt;In the making of firework explosives, the explosive mixtures are handled with great care to avoid static electricity induced detonation of the explosives. Is it a good idea that they store the explosives in a plastic container - isn't there a risk of static buildup that could detonate the explosives?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=3mlrjrLUpyg&quot; rel=&quot;nofollow noreferrer&quot;&gt;How It's Made Fireworks&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="8055" LastEditDate="2017-11-15T00:44:33.593" LastActivityDate="2017-11-15T00:44:33.593" Title="Quantifying Static Electricity Hazard with High Explosives" Tags="|electrical-engineering|materials|design|electromagnetism|electrical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18008" PostTypeId="2" ParentId="18007" CreationDate="2017-11-12T23:40:56.000" Score="3" Body="&lt;p&gt;Static electricity is a major issue when handling or dealing with explosives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When blast holes are loaded with ANFO (ammonium nitrate fuel oil) as the main explosive, and a specific density of ANFO is required in the holes, the ANFO is blown into the holes via compressed air and a tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For short holes drilled with hand held machines, the tube is made of aluminum because of static electricity concerns. For large holes drilled with larger machines, the tube is made of plastic. To ensure that static electricity will not be an issue low static plastic tubing is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally black plastic tubing is used. To differentiate low static tubing for conventional tubing, low static tubing has a red or orange stripe that runs the full length of tubing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Providing the plastic containers used to store high explosives are made of low static plastic, static electricity will not be an issue. However, the plastic containers need to be identifiable as being made from low static plastic.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-11-12T23:40:56.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18009" PostTypeId="2" ParentId="17714" CreationDate="2017-11-13T00:25:56.423" Score="2" Body="&lt;p&gt;To answer this question, you need to consider the failure modes you are protecting against. To understand the failure modes, you need to have the full details related to the materials of the road, thicknesses of the respective materials used to pave the road, and etc. A bitumen road running heavy trucks will most likely fail by excessive rutting in my opinion, in which the main factor of consideration would be related to the weight of the traffic running on the road, and less so about how fast the trucks are traveling. Therefore, without knowing the full details of the scenario, your feasibility engineer is correct in a sense that the weight is probably the more important factor in this case when the road is new, and speed may/may not be neglected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when will speed make a difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed generally makes a difference when you have localized irregularities on the road surface, such as a pothole. A presence of such an irregularity will deteriorate much quicker in the presence of higher speed axle loads - this is what people generally refer to as &quot;impact loading&quot; or &quot;dynamic impact load factor&quot; if you are working on doing some calculations. Dynamic impact load generally increases exponentially with speed (not necessarily exponentially in mathematical terms, but you get the idea).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I personally work in the rail industry but the underlying principal is the same. I am not familiar with the numbers for trucks, but for HAL axle loads in north America (higher axle loads than EU), the amount of impact that is generated by a wheel impact from a rail gap can be up to 400% +/- of the normal load for a train wheel traveling at 70 mph. I would imagine that trucks would experience a lesser increase due to the attenuation produced by the rubber tyres. But impact loads are certainly a significant force in accelerating deterioration of a road once potholes or localized failures on the surface has started. &lt;/p&gt;&#xA;" OwnerUserId="13526" LastEditorUserId="13526" LastEditDate="2017-11-13T05:04:53.007" LastActivityDate="2017-11-13T05:04:53.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18010" PostTypeId="1" CreationDate="2017-11-13T02:23:32.027" Score="0" ViewCount="115" Body="&lt;p&gt;Is it possible to determine plastic hinge location of a frame in SkyCiv? Is it also possible determine moment -curvature relationship curve in SkyCiv? If yes, how?&lt;/p&gt;&#xA;" OwnerUserId="13659" LastEditorUserId="1832" LastEditDate="2017-11-13T02:31:17.880" LastActivityDate="2018-01-12T22:08:43.880" Title="Plastic Hinge by Skyciv" Tags="|structural-engineering|moments|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18013" PostTypeId="1" CreationDate="2017-11-13T08:58:33.317" Score="2" ViewCount="115" Body="&lt;p&gt;I have a treaded wheel which is 16'' diameter and 4'' across (trailer wheel).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want it to spin constantly at 2750 RPM, what amount of torque do I need to apply to combat friction and air resistance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find this out for one of my projects and I am only knowledgeable of ideal world physics. Is there an equation for measuring non-conservative forces on a spinning wheel (like F= b*v^2)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice would be appreciated. Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Thanks for all the insights. I wanted to get an estimate because I have to choose the proper motor to use in this situation. Much cheaper to know which motor/wheel to go for now than to realize the motor/wheel combination is too weak.&#xA;In case I have been misunderstood, I purely want the force to keep the wheel spinning. The wheel isn't moving an object body. So, would any drag (v^2) still apply?&lt;/p&gt;&#xA;" OwnerUserId="13664" LastEditorUserId="13664" LastEditDate="2017-11-13T15:19:22.617" LastActivityDate="2017-11-13T15:19:22.617" Title="Torque needed to keep a wheel spinning at high RPM" Tags="|friction|wheels|" AnswerCount="0" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18014" PostTypeId="1" CreationDate="2017-11-13T09:03:54.927" Score="0" ViewCount="34" Body="&lt;p&gt;One of the worldbuilding questions was asking about the efficiency&#xA;of a possible siege and I wondered if there was something like a scale&#xA;in former times (when they did not have machines) how difficult a ground&#xA;is for using a pickaxe (Something like 1 = compacted earth and 6 = granite).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Does one/several qualitative scale(s) exist (comparable like the Mohs scale for hardness) which judges the difficulty of the ground for a pickaxe?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If not, are there some qualitative tables how difficult a specific ground is for using a pickaxe?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1137" LastActivityDate="2017-11-22T07:50:19.027" Title="Have there been/is there a scale about the difficulty of a ground for using a pickaxe?" Tags="|materials|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18015" PostTypeId="1" AcceptedAnswerId="18079" CreationDate="2017-11-13T10:49:37.763" Score="3" ViewCount="2680" Body="&lt;p&gt;Need some help designing a bolt pattern for a pressure tight enclosure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do i determine the force applied at a distance from a bolt location so that  I meet the minimum force to compress the gasket along the entire length of the enclosure but while using the minimum number of bolts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if anyone know how to model this on solidworks that would be very useful as well!&lt;/p&gt;&#xA;" OwnerUserId="13665" LastEditorUserId="1832" LastEditDate="2017-11-13T11:44:02.400" LastActivityDate="2017-11-20T12:50:06.177" Title="Determine force distribution from a bolt pattern" Tags="|mechanical-engineering|civil-engineering|solidworks|bolting|pressure-vessel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18016" PostTypeId="2" ParentId="17523" CreationDate="2017-11-13T12:57:14.987" Score="1" Body="&lt;p&gt;Actually, a lot of consumer cars &lt;i&gt;do&lt;/i&gt; use kinetic energy recovery.  Every hybrid and electric car that I know of does this, and there are significant numbers of hybrids and electric cars out there now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since these cars already have a electric motor connected to the wheels, it only takes a little addition to the control electrics to run the motor as a generator when you want to slow down.  The cost of the additional electronics is minimal compared to the cost of the motor and the electrical system as a whole, so this is a no-brainer in most cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same system is used to charge the battery when going down a hill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Honda Civic hybrid works this way, for example.  Stepping on the brake lightly while moving first runs the motor in reverse.  I can see the battery charge/discharge indicator going to charge.  Stepping harder on the brake causes more charging, and therefore more drag.  Eventually the friction brakes are also engaged.  This is all done quite smoothly and seamlessly.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-11-13T12:57:14.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18017" PostTypeId="2" ParentId="17980" CreationDate="2017-11-13T13:51:55.130" Score="3" Body="&lt;p&gt;Proportionality is a little more complex than that. In general, it only means they are correlated, and the correlation is monotonic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally two variables are proportional, if $$A = f(B)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where f is a &lt;em&gt;&lt;strong&gt;monotonic&lt;/strong&gt;&lt;/em&gt; function. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The basic variant is &quot;directly proportional&quot; - a linear relation:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = cB + d&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with constant (or independently parametric) $c$ and $d$; $c &amp;gt; 0$. Usually, when you talk about proportionality without other qualifiers, it's what is meant. $d$ is called &lt;em&gt;offset&lt;/em&gt; or &lt;em&gt;bias&lt;/em&gt;, and you may talk about &lt;em&gt;biased proportionality&lt;/em&gt; when it's non-zero.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Then there's the very common &quot;inversely proportional.&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A = {c \over B} + d&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$A = cB + d$ for negative $c$ is &quot;linearly proportional, with negative coefficient&quot; - sometimes mislabeled as &quot;inversely proportional&quot;. General $A = cB + d$ for non-zero $c$ with sign not given is just &quot;linearly proportional&quot;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then there are others:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;quadratically proportional&quot;, $A = cB^2 (+dB +e)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;exponentially proportional&quot;, $A = e^{cB} (+ d)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;logarithmically proportional&quot;, $A = c \ log B (+ d)$&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In other cases, you'll see &quot;proportional to &lt;em&gt;f&lt;/em&gt; of B&quot;, e.g. the name of $A = cB^4$ is not called &quot;tesseractically proportional&quot;, just &quot;proportional to fourth power of B&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But note all these functions are strongly monotonic. There's no such thing as &quot;sinusoidally proportional&quot;. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-11-13T13:51:55.130" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18019" PostTypeId="2" ParentId="16729" CreationDate="2017-11-13T19:12:39.353" Score="1" Body="&lt;p&gt;I have had a single drafter that did this, the rest did not.  I found this method worked better in the long run for complex projects, but was poor on simple projects.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-11-13T19:12:39.353" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18020" PostTypeId="2" ParentId="18010" CreationDate="2017-11-13T20:46:25.193" Score="1" Body="&lt;p&gt;What you are after is a physically (material) non-linear analysis of a frame. From the SkyCiv &lt;a href=&quot;https://skyciv.com/documentation/structural-3d/solving-the-structure/non-linear-analysis/&quot; rel=&quot;nofollow noreferrer&quot;&gt;documentation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;The SkyCiv Non-Linear solver only takes into account geometric non-linearities.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can use the linear elastic moment distribution to determine the point at which the first plastic hinge will occur (the point with the largest bending moment). However, without a physically non-linear analysis, you will not be able to model what happens after the first plastic hinge forms in one analysis run.&lt;/p&gt;&#xA;" OwnerUserId="13602" LastActivityDate="2017-11-13T20:46:25.193" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18023" PostTypeId="2" ParentId="16729" CreationDate="2017-11-14T09:06:04.410" Score="1" Body="&lt;p&gt;AutoCAD has made great advances in recent version when it comes to dynamic blocks. If you have a lot of line work which is repetitive and just needs some scaling, stretching or rotation dynamic blocks are very helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can even create drop-down boxes to select various options you created...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One should use blocks as much as possible&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You only use what you need, the rest remains in the blocks list (no  more orphaned information. Once the drawing is complete, your can purge the unused blocks to reduce size.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can move blocks around, rotate them etc. no battle with items on hidden layers which don't move or rotate along&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You edit the block only once, and everywhere you have used it, the changes apply (caution for dynamic blocks though)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2017-11-14T09:06:04.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18024" PostTypeId="1" CreationDate="2017-11-14T12:01:18.353" Score="2" ViewCount="428" Body="&lt;p&gt;I'm not a mechanical engineer and am not used to looking at mechanical drawings. I'm trying to understand what these tolerance features mean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iOnj0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iOnj0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WCu2e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WCu2e.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;There is a positional tolerance feature, which references datum A. But, what exactly is it specifying? That the middle point of the flat must be within 0.1 mm of the rotational axis? Or the thickness of the disc? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bX3t4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bX3t4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13673" LastEditorUserId="3617" LastEditDate="2020-07-09T08:52:41.060" LastActivityDate="2020-07-09T08:52:41.060" Title="Explain tolerances in technical drawing" Tags="|mechanical|technical-drawing|tolerance|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="18025" PostTypeId="1" CreationDate="2017-11-14T14:41:59.350" Score="1" ViewCount="453" Body="&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt;&#xA;I'm going to build a homemade wind turbine. It will be all fancy and rotational so that it's always pointing to the wind, so it will be mounted on a frame etc.&#xA;This will be mounted on a pole that fits vertically through the roof of a barn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the place I'm going to build it for quite often experiences high wind speeds that would probably cause the wind turbine to break or get damaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I would like the turbine to be retractable/extendable. My current best idea is to have a gear system and a handcrank, with some sort of worm-gear-esque arrangement at the pole end that will use the rotational force of the gear system to raise or lower the turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vhbKC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vhbKC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How best can I translate the energy from the gear system into raising/lowering force? &lt;/li&gt;&#xA;&lt;li&gt;Is there any way to lock the system so that the pole doesn't simply fall when I stop cranking, and can be raised/lowered safely by handcrank?&lt;/li&gt;&#xA;&lt;li&gt;Is this idea feasible to make by hand for an amateur with no background in engineering?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13679" LastActivityDate="2017-11-14T14:41:59.350" Title="How to raise a pole with a handcrank and have it lock in place?" Tags="|gears|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18026" PostTypeId="2" ParentId="18024" CreationDate="2017-11-14T15:58:43.897" Score="3" Body="&lt;p&gt;I believe this is a related mating envelope related to datum A for the three plates, per ASME Y14.5-2009.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pTRle.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pTRle.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13416" LastEditorUserId="13416" LastEditDate="2017-11-20T14:11:32.593" LastActivityDate="2017-11-20T14:11:32.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18027" PostTypeId="1" AcceptedAnswerId="18033" CreationDate="2017-11-14T16:36:59.887" Score="0" ViewCount="73" Body="&lt;p&gt;For Linear actuators, a common mechanism is to use ballscrews with ballnuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ballscrew is supported at the two sides by a set of bearings which permit the rotation of the screw while keeping the axial reaction force to the ballnut movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking for those ballscrew supports, I found there are several categories:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Floating (identified by &lt;strong&gt;F&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Fixed (Identified by &lt;strong&gt;K&lt;/strong&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An other letter indicate several types/shapes of supports: &lt;strong&gt;E, B, F,&lt;/strong&gt; etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those pieces are usually sold in a set of one fixed and one floating support:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;e.g. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;BK + BF &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;EK + EF&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;FK + FF&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not find understandable information about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What types of ballscrew/leadscrew supports exits?&#xA;What is the difference between them?&lt;/p&gt;&#xA;" OwnerUserId="11129" LastActivityDate="2017-11-15T17:35:53.813" Title="What type of supports exists for ballscrews or leadscews?" Tags="|mechanical-engineering|terminology|linear-motion|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18028" PostTypeId="2" ParentId="18024" CreationDate="2017-11-14T16:44:24.460" Score="3" Body="&lt;p&gt;It says the flats shall be within 0.100 units of radial dimension positioned with reference to 'A'.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. if you were to make the rim which is a perfect circle, then mill the flats at the three positions, what you should see is that the flats are exactly the same sectional shape and size if the machinist was perfectly precise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some minor deviation from that precision (+/- 0.1 units) is permitted which limits the variability of size and position of the cut. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? Perhaps a precision tool is needed or the mating piece has a similar tolerance, to ensure a good fit. &lt;/p&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="7021" LastEditDate="2017-11-14T19:28:22.217" LastActivityDate="2017-11-14T19:28:22.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18029" PostTypeId="1" CreationDate="2017-11-14T17:24:23.153" Score="1" ViewCount="51" Body="&lt;p&gt;Most of the RF receivers have a clock for synchronization. Are there RF receivers which do not need clock? If yes,  how do they work?&lt;/p&gt;&#xA;" OwnerUserId="13682" LastEditorUserId="8055" LastEditDate="2017-11-15T16:18:47.747" LastActivityDate="2017-11-15T16:18:47.747" Title="RF receivers without clock" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18030" PostTypeId="2" ParentId="17992" CreationDate="2017-11-15T03:38:57.823" Score="-2" Body="&lt;p&gt;Your relative humidity (RH) of 70% doesn't seem that high. If you really want to decrease it, then you need to increase your ventilation flow, since your living and outside air both contribute to a high humidity. Your hot showers, washing, cooking, breathing, sweating, adds a substantial amount of water vapor to the inside atmosphere (some estimates put this at 20 lb per day for a typical household, which means you would reach 100% RH after a few days if your house was a sealed plastic bag). At best, all of this air with high humidity is replaced via ventilation with outside air, or condensed out when passing through the HVAC cooling coils . Now, that outside air at, say, an average temperature of 95 °F and 25% RH will have a RH of around 55% at an inside temperature 75 °F. So at best case, your air will have a 55% RH. Now adjust for the fact that your ventilation and heat pump/AC unit throughput is normal and not very high, and your 70% RH is to be expected. &lt;/p&gt;&#xA;" OwnerUserId="13388" LastEditorUserId="368" LastEditDate="2017-11-15T16:40:10.463" LastActivityDate="2017-11-15T16:40:10.463" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18031" PostTypeId="1" CreationDate="2017-11-15T05:06:17.717" Score="0" ViewCount="704" Body="&lt;p&gt;I have an electric motor which is intended for a bicycle/scooter and has a 11-teeth chain sprocket. &#xA;I plan to use it to spin a trailer wheel (which have 4 holes for bolts about 3'' from the center). In these cases, do I have to design a mount to attach the wheel to the motor or is it possible to buy a mount that would fit my needs?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice would be appreciated. Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wheel: &lt;a href=&quot;http://recstuff.com/trailer-tires-wheels/tire-wheel-combos/8-tire-and-rim-packages/4-80x8-loadstar-trailer-tire-lrc-on-4-bolt-galvanized-wheel/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://recstuff.com/trailer-tires-wheels/tire-wheel-combos/8-tire-and-rim-packages/4-80x8-loadstar-trailer-tire-lrc-on-4-bolt-galvanized-wheel/&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VSUEA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VSUEA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13664" LastActivityDate="2018-03-15T07:21:36.573" Title="How do I attach an electric motor to a trailer wheel?" Tags="|motors|wheels|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18032" PostTypeId="2" ParentId="18031" CreationDate="2017-11-15T05:13:09.667" Score="1" Body="&lt;p&gt;Since a scooter has a chain drive and the rear wheel is approximately the same size as some of the small trailer wheels, then you should mount the trailer wheel on an axle with bearings and a sprocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose the sprocket diameter to give the required speed range as you should know the max rpm of the motor , the dia of the trailer wheel and already the 11 tooth drive...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-15T05:13:09.667" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18033" PostTypeId="2" ParentId="18027" CreationDate="2017-11-15T06:16:39.293" Score="1" Body="&lt;p&gt;As you see, you use one fixed and one floating bearing to compensate for stress due to thermal expansion of the rod, manufacturing or assembly tolerances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Floating bearing can’t take axial loads, where fixed bearings take both axial and radial loads.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I guess the letters e, b, f specify the mounting type, for example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bMYBt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bMYBt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YocWt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YocWt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to be sure you have to provide a datasheet or link&lt;/p&gt;&#xA;" OwnerUserId="8282" LastEditorUserId="8282" LastEditDate="2017-11-15T08:26:59.397" LastActivityDate="2017-11-15T08:26:59.397" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18034" PostTypeId="1" CreationDate="2017-11-15T10:31:31.347" Score="1" ViewCount="100" Body="&lt;p&gt;&quot;... a multiview projection is a technique of illustration by which a standardized series of orthographic two-dimensional pictures is constructed to represent the form of a three-dimensional object. Up to six pictures of an object are produced (called primary views), ...&quot; (Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Multiview_projection&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia, Multiview projection&lt;/a&gt;). While the resulting six views of orthographic projection are the same, a difference in arrangement of these views around the viewing box, lead to two different projection schemes: first-angle and third-angle projections, with the chance to mix things up (to say the least) when drawings are transferred from one convention to another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To mitigate this issue, the projection angle is denoted by a projection symbol on engineering drawings. This international symbol shows the projection (top and front view) of a truncated cone, that defines and identifies either first-angle or third-angle projection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that national and supra-national standardization tries to minimize confusion by the two concflicting systems. From &lt;a href=&quot;https://en.wikipedia.org/wiki/Multiview_projection#Multiviews&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia, Multiview projection&lt;/a&gt; we learn that first-angle projection is standard throughout Europe and Asia (excluding Japan). Whereas third-angle is standard in the USA, Japan, Canada, and Australia. The UK or at least some British companies, so the article claims, switched from first-angle to third-angle during World War II.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which are the default multiview projection schemes for countries in Central and South America as well as Africa? Preferably backed by national regulations or standards, if applicable.&lt;/p&gt;&#xA;" OwnerUserId="4800" LastActivityDate="2019-12-07T23:02:00.253" Title="Default multiview projection schemes" Tags="|drafting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18036" PostTypeId="2" ParentId="18029" CreationDate="2017-11-15T12:39:00.330" Score="1" Body="&lt;p&gt;RF receivers generally need to receive signal over a known and narrow frequency range.  As such, they need some mechanism to accomplish this frequency selection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some methods of frequency selection utilize a signal at a known frequency, which you could possibly call a &quot;clock&quot;, although I think that's stretching the term a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a basic AM radio receiver can be made selective by utilizing the narrow passband of a LC filter (inductor and capacitor working together).  This is exactly how early AM radios worked.  Then the &lt;i&gt;super-heterodyne&lt;/i&gt; method came along, which does use a specific synthesized frequency to allow for a more selective overall filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean by &quot;clock&quot; that the receiver needs to know absolute time, then no, the statement &lt;i&gt;&quot;most RF receivers have a clock for synchronization&quot;&lt;/i&gt; is just nonsense.  Some higher level systems may need to know the time for other reasons, but simply receiving and demodulating the RF is not one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could conceive of higher level transmission protocols that might, for example, use frequency-hopping on a known schedule.  In that case, the receiver might use a clock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;GPS decoding ultimately requires synchronization, but the modulation scheme is designed so that this synchronization can be derived from watching the RF signals for long enough.  Eventually finding position requires knowing the time very precisely, but that's at a higher level above the RF receivers.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-11-15T12:39:00.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18037" PostTypeId="1" AcceptedAnswerId="18050" CreationDate="2017-11-15T13:56:22.277" Score="2" ViewCount="869" Body="&lt;p&gt;I am interested in comparing the different materials that can be used to make castings of cast irons, and therefore was wondering which material is commonly used/which materials can be used to make permanent casting molds (to be used specifically for cast iron). Thank you.&lt;/p&gt;&#xA;" OwnerUserId="13691" LastActivityDate="2017-11-18T11:21:43.617" Title="Which materials can be used to make permanent casting molds for cast iron?" Tags="|materials|metallurgy|metals|casting|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18041" PostTypeId="2" ParentId="18027" CreationDate="2017-11-15T17:35:53.813" Score="1" Body="&lt;p&gt;BK, BF refer to the mount format - in this case a rectangular face mounted block with one (BF) or two (BK) bearings and mount holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st Letter == mount format;&#xA;2nd letter == bearing set ;&#xA;numeric == ID of the bearings in mm;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick internet search will identify all of these bearing types and specifications.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cQh6U.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cQh6U.png&quot; alt=&quot;F series face mounted&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HvJvC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HvJvC.png&quot; alt=&quot;E series base mount&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HEZeH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HEZeH.jpg&quot; alt=&quot;B series heavy duty&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2017-11-15T17:35:53.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18042" PostTypeId="2" ParentId="17992" CreationDate="2017-11-15T17:45:49.967" Score="2" Body="&lt;p&gt;While your AC is running, what is it's typical duty cycle?  In other words, how long does it run during each cycle (i.e. 5 minutes on, 15 minutes off) to keep things cooled to your set point?  If it doesn't run very long at all to keep things cool each time it runs, you may have a vastly over sized AC unit that isn't running long enough to actually remove the moisture in the air before the temperature gets to the set point.  An ideal AC (or heating system) should be the most efficient and do the best job when it runs close to 100% of the time in the worst-case design scenario.  Bigger isn't necessarily better when it comes to AC and furnaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this (short cycling) could be the problem, you should have your AC contractor run a load calculation on your town house to verify that your AC unit is sized properly for your specific unit.  Depending on the quality of the contractor doing the install, they may have just slapped in whatever unit they may have on hand, or just put in a big one because you're in a hot environment and figured that they could charge you (or the builder) for a larger unit than necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing to check is the location of your thermostat.  Make sure it's not causing short cycling of your AC unit by being right in the blast of the vents.  Ideally you'd want it to be away from any direct air movement from the vents so it's more accurately measuring the overall room temperature and not just the air right around the vents.&lt;/p&gt;&#xA;" OwnerUserId="11065" LastActivityDate="2017-11-15T17:45:49.967" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18043" PostTypeId="2" ParentId="18037" CreationDate="2017-11-15T17:52:51.477" Score="1" Body="&lt;p&gt;Solid graphite is used. Simple shapes that are cut into the graphite. like wheels.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-11-15T17:52:51.477" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18044" PostTypeId="2" ParentId="17822" CreationDate="2017-11-15T20:19:38.650" Score="0" Body="&lt;p&gt;Several things you may want to know:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Ussualy, those schematics are only orientative. Maybe you can move members. If you really need it you can put a profile under the load. You have to draw a quick schema and send it to the PM for aproval.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A truss can only carry loads on nodes but, you can design a chord as a continuous beam. This chord may carry loads wherever you want. The truss will be designed as a typical truss but your chord will need complementary design. You will need to check both axial and lateral loads (of course combined).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Be careful designing that chord. Remember to take into account the bendings when designing buckling and lateral buckling. Your analysis program will do it wrong.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="5083" LastActivityDate="2017-11-15T20:19:38.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18045" PostTypeId="1" CreationDate="2017-11-15T23:48:09.693" Score="4" ViewCount="114" Body="&lt;p&gt;I have a grid that I plan to discretize in 1D. It extends from 0 to L using N nodes, and I need to refine it at each end. I plan to have something that looks as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hpri9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hpri9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried the to specify these two parameters:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;growth rate (r)&lt;/li&gt;&#xA;&lt;li&gt;initial node spacing (dx1)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then the x points are calculated in a series such as the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x(i) = dx1 * r ^ x(i-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This works fine if I know N, and specify r and dx1. But ideally, I would need r and dx1 to be functions of N to properly scale the grid when N increases. I.e. with N = 10, r should be about 1.7 but when N = 100, r should be about 1.1 to have the same &quot;shape&quot;. Similar problem with dx1. Anyone know how to make those two parameters functions of N that will scale with it?&lt;/p&gt;&#xA;" OwnerUserId="13698" LastEditorUserId="13698" LastEditDate="2017-11-16T02:50:58.530" LastActivityDate="2018-07-15T20:42:44.890" Title="What is a good method (equation) for locally refining a 1D grid at each end?" Tags="|finite-element-method|cfd|meshing|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18046" PostTypeId="1" CreationDate="2017-11-15T23:54:43.633" Score="2" ViewCount="78" Body="&lt;p&gt;I have to project a mechanism that transform a rotation of $30 \pi/\text s$ from a engine in a rotation of $(\pi/4)/\text s$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the gears should be 120x bigger than the other, but this is too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gears should have a ray $&amp;lt; 30\text{ cm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I create this mechanism?&lt;/p&gt;&#xA;" OwnerUserId="13700" LastEditorUserId="1832" LastEditDate="2017-11-29T22:58:20.520" LastActivityDate="2017-11-29T22:58:20.520" Title="Gears and rotation" Tags="|gears|mechanisms|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18047" PostTypeId="2" ParentId="18045" CreationDate="2017-11-16T04:22:20.787" Score="0" Body="&lt;p&gt;Since 1D would likely not be too computationally intensive even for very large sets, you could just sum all the lengths for a given N, then divide all the distances in that mesh array by that value. It would not be computationally efficient like an explicit function, but will do exactly what you need without too much experimentation. If you need the computational efficiency and dont get any derrivations here, you may want to repost in mathamatics.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-11-16T04:22:20.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18048" PostTypeId="2" ParentId="17992" CreationDate="2017-11-16T05:13:20.883" Score="5" Body="&lt;p&gt;Sounds like you have covered your bases. The only thing I can recommend is gathering data and considering some scenerios to evaluate that data against. I am not an HVAC guy but can give you the 1000ft engineering perspective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any of these trends would be helpful in debuggig the situation:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Relative humidity and temperature at a discharge of the HVAC system trended over time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Condensate discharge from the air conditioner into a bucket, gallons per day.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Relative humidity and temperature in each room trended over time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Surrounding water table level&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Condensation locations and times of day&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Possible scenerios in order of likelyness:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A portion of the condensate being removed from the air is not being drained off and is being revaporated. Either due to installation issue, mfg defect, or possibly a very frequent cycle time.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To small of airflow is being cooled then mixed with the main air stream. When a small stream of air is cooled very cold, most all the water is removed from this stream. However, no water is removed from the larger main stream it mixes with. This can result in the mixed stream being 100% humidity (think fog from dry ice). To mimimize humidity, fresh air from outside should be introduced upstream of the air conditioner before being distributed to the rest of the house.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Moisture wiking through concrete. May be exacerbated by irrigation or a high water table.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Unnoticed moisture producing item. A fridge with bad defrost settings and an icemaker. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-11-16T05:13:20.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18050" PostTypeId="2" ParentId="18037" CreationDate="2017-11-16T12:24:56.243" Score="3" Body="&lt;p&gt;You can cast iron into cast iron moulds as long as the moulds have plenty of mass compared to the quantity of metal you are pouring into them. Small artisan foundries sometimes use cast iron moulds to make small ingots as a 'signature piece' to use up small quantities of surplus metal. Often these are open face moulds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously steel works as well but steel moulds tend to be significantly more difficult to make. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically in industry multiples of the same casting are made in green sand moulds using a pattern which is pressed into the mould which means that although the specific mould is broken up after each casting the sand can be reused and it is fairly straightforward to mass produce moulds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With metal moulds you need to be quite careful about draft angles and avoiding undercuts. Shrinkage stresses can also be a problem with more complex and hollow  shapes which can end up too constrained by the mould. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also need to allow additional venting compared to sand moulds as unlike sand gasses can't escape into the walls of the mould &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-11-18T11:21:43.617" LastActivityDate="2017-11-18T11:21:43.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18051" PostTypeId="1" AcceptedAnswerId="18052" CreationDate="2017-11-16T12:43:15.897" Score="0" ViewCount="84" Body="&lt;p&gt;This comes from the composites store website:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;These high performance carbon shapes are produced by packing a high volume of carbon fiber into epoxy resin under tension resulting in an unmatched tensile and compression stength.&lt;/p&gt;&#xA;&lt;p&gt;These shapes are twice the stiffness of Aluminum.&lt;/p&gt;&#xA;&lt;p&gt;Compared to a wet lay-up of Carbon Tow the shapes are:&lt;/p&gt;&#xA;&lt;p&gt;350% Stronger in Tension&lt;/p&gt;&#xA;&lt;p&gt;550% Stronger in Compression&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.cstsales.com/graphlite_data.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quote source&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;It doesn't explain why.  Carbon Tow are long continuous strands of carbon fiber.  So somehow putting it under tension first, then fixing it in position with the epoxy makes it stronger?  How?&lt;/p&gt;&#xA;" OwnerUserId="13081" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-11-16T13:32:35.177" Title="Why is graphlite so strong in tension?" Tags="|composite|carbon-fiber|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18052" PostTypeId="2" ParentId="18051" CreationDate="2017-11-16T13:16:59.977" Score="1" Body="&lt;p&gt;Its not about the tension in the final product. [They are claiming] tension during fabrication buys them a much higher fiber loading. See 67 vol%.. a typical &quot;wet layup&quot; is maybe half that.   Having the fibers very very straight probably contributes somewhat as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also the 3+GPa claimed strength is for some &quot;micro&quot; composite. If you look elsewhere on that site you will see practical size rods of &quot;graphlite&quot; are more like 2GPa.  &lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2017-11-16T13:32:35.177" LastActivityDate="2017-11-16T13:32:35.177" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18053" PostTypeId="2" ParentId="18046" CreationDate="2017-11-16T15:48:49.330" Score="1" Body="&lt;p&gt;In practice probably the best solution is to find an off the shelf gearbox with the reduction ratio that you require for example  : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ondrivesus.com/gearboxes-speed-reducers&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ondrivesus.com/gearboxes-speed-reducers&lt;/a&gt; &lt;strong&gt;(not a recommendation)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you have specialist knowledge designing and manufacturing a high ratio gearbox from scratch is unlikely to be very cost effective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That way you just need to find one which meets your requirements for reduction ration and torque capacity. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-11-16T15:48:49.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18055" PostTypeId="1" CreationDate="2017-11-17T01:19:02.183" Score="-1" ViewCount="801" Body="&lt;p&gt;One way is to go to the configurations manager, right click, and select add configuration.  This is super slow if you have many to add.  There must be a faster way, similar to how you create new configurations with a design table.&lt;/p&gt;&#xA;" OwnerUserId="13081" LastActivityDate="2017-11-17T01:52:23.833" Title="How do you create multiple configurations of an assembly quickly in Solidworks?" Tags="|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18056" PostTypeId="2" ParentId="18055" CreationDate="2017-11-17T01:22:46.210" Score="-1" Body="&lt;p&gt;Switch back from the configuration manager to the tab that shows all the parts that are in the assembly.  Right click any of them, select configure component.  The window that pops up will provide an excell-like interface with an option to add another configuration.  Click on it, give it a name, hit enter.  Immediately type the next name.  Keep going until they are all entered.  Voila.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=yI6YEAt6kq8&quot; rel=&quot;nofollow noreferrer&quot;&gt;An even better answer!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13081" LastEditorUserId="13081" LastEditDate="2017-11-17T01:52:23.833" LastActivityDate="2017-11-17T01:52:23.833" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18058" PostTypeId="1" CreationDate="2017-11-17T02:50:17.550" Score="1" ViewCount="32" Body="&lt;p&gt;For an arbitrary articulated structure subject to non-affine (non homogeneous) deformations, the components of the strain tensor are defined locally. Is there a way to calculate a numerical strain tensor that represents the whole volume? Can a Poisson's ratio be determined for such situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first guess is that the volume average of the fields should do it, but I'm not sure if this is the right approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll appreciate any help.&lt;/p&gt;&#xA;" OwnerUserId="13712" LastEditorUserId="90" LastEditDate="2017-11-17T03:16:12.327" LastActivityDate="2017-11-17T03:16:12.327" Title="How to calculate a global strain from a field disitribution" Tags="|structural-engineering|structures|deformation|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18059" PostTypeId="1" CreationDate="2017-11-17T05:17:54.980" Score="1" ViewCount="31" Body="&lt;p&gt;I was trying to simulate the simplified model of RDC (AD2S90) provided in link &lt;a href=&quot;https://i.stack.imgur.com/Z6bsX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z6bsX.jpg&quot; alt=&quot;tracking servo model&quot;&gt;&lt;/a&gt;. The issue is to design proper compensator for this model to track constant velocity with zero velocity lag. Not very clear how to proceed. As suggested in the data sheet (page No.9) I tried to add a compensator (100/s * (s+100)/(s+1000)) between demodulator and the VCO. But from simulation(step response), it appears to oscillate. Where am I going wrong? Almost all the data sheets that I came across recommend the compensator with the above mentioned transfer function, but in simulation, it does not work. How to proceed? &lt;/p&gt;&#xA;" OwnerUserId="13455" LastActivityDate="2017-11-17T05:17:54.980" Title="modeling tracking servo" Tags="|control-theory|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18061" PostTypeId="1" AcceptedAnswerId="18064" CreationDate="2017-11-17T08:46:22.640" Score="3" ViewCount="72" Body="&lt;p&gt;I have a temperature sensor and I want to normalize the sensor values. The measured room has a pretty constant temperature but the sensor shows slightly different values e.g. &lt;code&gt;30.20, 30.00, 30.30, 30.00&lt;/code&gt;. For display reasons, I want to have it less variation. Also sometimes it deviates unreasonable e.g. &lt;code&gt;30.00, 30.00, 10.00, 30.00, 30.00&lt;/code&gt;. But it can change from one moment to another also in a reasonable way e.g. opening a window: &lt;code&gt;30.00, 30.00, 30.00, 30.00, 25.00, 15.00,10.00&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not have target curve otherwise I thought about a PID.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I sum the last values and build an average. What can I do to&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;normalize the output values&lt;/li&gt;&#xA;&lt;li&gt;eliminate runaway values &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13714" LastEditorUserId="8055" LastEditDate="2017-11-17T15:24:59.027" LastActivityDate="2017-11-17T15:24:59.027" Title="Control Variations in a Measurement (Temperature Sensor)" Tags="|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18064" PostTypeId="2" ParentId="18061" CreationDate="2017-11-17T10:12:14.547" Score="1" Body="&lt;p&gt;I simple &lt;a href=&quot;https://en.wikipedia.org/wiki/Low-pass_filter&quot; rel=&quot;nofollow noreferrer&quot;&gt;lowpass&lt;/a&gt; filter will probably solve the problem. It is easy to implement and you can adjust how much you want to filter with the time constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of a low-pass filter in actual code. It's technically Python but reads like pseudo-code. It takes a list of temperature values as input and outputs the filtered values as a list.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Input:  input_series, time_constant&#xA;Output: output_series&#xA;&#xA;&#xA;output_series = []&#xA;output_series.append(input_series[0])&#xA;&#xA;for current_unfiltered_value in input_series:&#xA;    previous_filtered_value = output_series[-1]&#xA;&#xA;    B = 1.0 / time_constant&#xA;    A = 1.0 - B&#xA;&#xA;    new_filtered_value = (A * previous_filtered_value&#xA;                          + B * current_unfiltered_value)&#xA;&#xA;    output_series.append(new_filtered_value)&#xA;&#xA;return output_series&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The downside of using lowpass filters is that it introduces phase lag. The filtered value will lag slightly behind the measured signal, depending on the time constant. A highly filtered signal will lag more than a less filtered signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other filters you can use for better filtering result. For example the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kalman_filter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kalman filter&lt;/a&gt; or the &lt;a href=&quot;https://en.wikipedia.org/wiki/Alpha_beta_filter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alpha-beta filter&lt;/a&gt;. They are both model-based, which means you have to make a model for your temperature changes for it to work. There is an excellent &lt;a href=&quot;https://github.com/rlabbe/Kalman-and-Bayesian-Filters-in-Python&quot; rel=&quot;nofollow noreferrer&quot;&gt;notebook&lt;/a&gt; on Github that introduces these filters if you are unfamiliar with them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would personally go with a simple lowpass filter in the beginning. If you don't control a process based on the filtered values, but want to filter the values to make them prettier in a user interface, then that's usually more than good enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For eliminating runaway values I would calculate the &lt;a href=&quot;https://en.wikipedia.org/wiki/Variance&quot; rel=&quot;nofollow noreferrer&quot;&gt;variance&lt;/a&gt; of the measured signal at every timestep (you could, for example, calculate the variance for the last 50 samples). If the calculated variance passes a certain threshold then it can be assumed to be a spike, which can then be rejected.&lt;/p&gt;&#xA;" OwnerUserId="78" LastEditorUserId="78" LastEditDate="2017-11-17T10:45:19.790" LastActivityDate="2017-11-17T10:45:19.790" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18065" PostTypeId="2" ParentId="1925" CreationDate="2017-11-17T13:24:46.183" Score="2" Body="&lt;p&gt;If you find two peaks in the E&quot; chart it usually indicates that this material has two Tg's which can occur if the material a blend or a block co-polymer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For rubbery material the E can rise slightly in the rubbery region with respect to  Temp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did you ever figure this out? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;see&#xA;&lt;a href=&quot;https://www.doitpoms.ac.uk/tlplib/glass-transition/printall.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.doitpoms.ac.uk/tlplib/glass-transition/printall.php&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13716" LastEditorUserId="13716" LastEditDate="2017-11-17T13:36:12.810" LastActivityDate="2017-11-17T13:36:12.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18066" PostTypeId="2" ParentId="11562" CreationDate="2017-11-17T13:25:43.023" Score="0" Body="&lt;p&gt;I'm unable to post a comment as I don't yet have 50 reputation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks to me like you've gone about this problem in a sensible and logical manner. I have a few questions, however.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can you confirm if the pressure that your friend uses is really &quot;4 bar&quot; (absoute), or 4 barg (gauge, or above atmostpheric)? You have calculated for 4 barg.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Your suspicion seems to stem from a mismatch between your human strength and the numbers. This is an exceptionally good way to 'sanity check' your answers! How does the rate at which you are able to push the clay through 'by hand' compare to the 4 bar(g) compressed air system? If the air can push it through at double the rate, then it will require approximately double the force. Does this clear up the discrepancy?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-11-17T13:25:43.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18067" PostTypeId="2" ParentId="18034" CreationDate="2017-11-17T13:53:02.690" Score="1" Body="&lt;p&gt;In my experience, Africa is broadly split into First Angle for former European Colonies, and Third Angle for anywhere where American development is taking place. As such, finding a standard for each country would be exceptionally time consuming! Do you have any specific countries in mind that you're dealing with?&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-11-17T13:53:02.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18068" PostTypeId="1" AcceptedAnswerId="18073" CreationDate="2017-11-17T15:09:46.160" Score="2" ViewCount="868" Body="&lt;p&gt;I am currently trying to design a simple truss (pin-jointed) structure able to withstand a 1.2 kN load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I carried all the analysis and found the forces, stresses and strains in the members. My question now is how good and accurate are those forces I calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I found that in a particular member the force was 10000N (tensile) and the proof stress of the alloy I'm using = 100 MPa. Theoretically, using any bar with area &gt; 100mm^2 should be fine, right? However, in the analysis I made several assumptions (massless bars, no firction, loaded through centroid...) and the calculated Force might not be exactly the real-life one. As a rule of thumb, what margins should I take to account for these assumptions? Should I systematically allow for 10% error for example? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, how safe is it (with regards to failure of the member) to go close to the proof stress? Should I there again allow for a margin for error?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your answers!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best wishes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB1: I do not have access to a Finte Elements Sotware to run it computationally and check!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB2: The structure is statically determinate.&lt;/p&gt;&#xA;" OwnerUserId="13717" LastActivityDate="2017-11-18T14:41:29.857" Title="Percentage error in calculated Forces during simple structural analysis for a truss pin-jointed strucures" Tags="|structural-engineering|civil-engineering|structures|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18069" PostTypeId="2" ParentId="5950" CreationDate="2017-11-18T10:13:14.530" Score="1" Body="&lt;p&gt;Adding to these above ideas&amp;amp;techniques, &#xA;In flexible pipes, along with the flexibility we can have filaments(flagellum ) on the walls. These could dissipate the turbulence kenetic energy for the same $Re$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we have to pay interms of pressure loss.&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2017-11-18T10:13:14.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18071" PostTypeId="1" CreationDate="2017-11-18T11:34:14.533" Score="1" ViewCount="46" Body="&lt;p&gt;I am interested how to check a nominally pined joint. Especially how to check its rotational capacity and whether i can define it as nominally pinned in structural fem model. I understand that there are criteria for nominally p. Joints but also there is a criterium that this joint must have enough of rotational capacity to accept rotation of the member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I give an example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say i designed a nominally pinned joint according to eurocode criteria. Ran a calculation and got design forces. (shear force and maybe axial force). Joint satisfied all code checks but now i want to if this joint has rotational capacity to accept members end rotation (also calculated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that there certain amount of estimation and engineering experience. But i would like to know what would do an experienced engineer when designing a joint that has stiffness somewhere between pinned and semirigid. And dont want to define it as semirigid in structural calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="13724" LastActivityDate="2017-11-18T11:34:14.533" Title="Check of a nominaly pinned steel joint" Tags="|structural-engineering|civil-engineering|structural-analysis|structures|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18072" PostTypeId="1" AcceptedAnswerId="18184" CreationDate="2017-11-18T13:25:54.333" Score="0" ViewCount="54" Body="&lt;p&gt;I had the question myself it those high voltage transmission lines, disturb other electrical devices? Let's say I have an high voltage next to a mobile basestation, does it effekt the electrical components in the base station or the wireless communication of the antennas ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way I could calculate that?&lt;/p&gt;&#xA;" OwnerUserId="11980" LastActivityDate="2017-11-26T02:29:59.993" Title="Do high votlage transmission lines disturb other electrical devices?" Tags="|electrical-engineering|electrical-grid|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18073" PostTypeId="2" ParentId="18068" CreationDate="2017-11-18T14:41:29.857" Score="1" Body="&lt;p&gt;This is a question which cannot be answered conclusively, because the issues you are raising are matters of risk and probability. There is no such thing as a calculation which &lt;em&gt;&lt;strong&gt;guarantees&lt;/strong&gt;&lt;/em&gt; that the final structure will be secure. All you can do is state that the adopted calculations imply in an acceptable (and very, very low) level of risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After all, as you mention, there are a multitude of random variables with behaviors out of your control:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Might the load be greater than what you actually expect?&lt;/li&gt;&#xA;&lt;li&gt;Might the material be faulty, with a strength below what you expect?&lt;/li&gt;&#xA;&lt;li&gt;Will the structure behave exactly like the structural model, or will imperfections modify the internal stresses?&lt;/li&gt;&#xA;&lt;li&gt;And countless other potential risks...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, what is that acceptable level of risk? Who has the authority to determine that? Well, the answer is &quot;not you.&quot; Instead, every country/state/whatever has a set of codes which determine the adequate methods and considerations to be taken into account when performing your calculations in order to reach the level of risk which the country/state/whatever believes is acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dealing with structures, there tend to be two general methodologies: allowable strength design (ASD) and load and resistance factor design (LRFD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In ASD, the required strength to resist a given load is calculated and compared to the &quot;allowable strength&quot; of the material, which is a fraction (50%, for example) of the material's nominal strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In LRFD, the expected loads are increased by a given factor and then the required strength for this increased load is compared to the allowable strength of the material, which is again a fraction of the nominal strength (but higher than in ASD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in Brazil, bridges have their dead loads increased by 35% and their live loads by 50% (and the dynamic factor). Meanwhile, concrete bridges have their strength reduced by ~30%, while steel bridges are reduced by only ~13%. This is an example of the LRFD method. (Brazil doesn't really use ASD nowadays).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are dead loads increased by 35% while live loads are only increased by 50%, though?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, it's because you can be pretty reasonably sure of what the final weight of your bridge will be. After all, you are in control: you selected the dimensions and the material. Something will need to go pretty seriously wrong in either of those variables (dimensions or concrete density, for example) for your bridge to be significantly off your expected weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you have no control over the live loads that drive over your bridge. Bridges are usually calculated with a codified live load, which describes a heavy truck or train. But, who knows what trucks will weigh in 20 years? And codes usually allow you to only consider one (or whatever) trucks driving by your bridge at a given moment. Sure, the odds of having two of these massive trucks driving by at once are astronomical, but what if you have one of them followed by a bunch of other, lighter trucks? Basically, you have no control over what happens on top of your bridge, so you need to put a much higher safety factor on live loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same concept applies to why concrete strength is discounted so severely, compared to steel. Concrete is often done &lt;em&gt;in-situ&lt;/em&gt; (on site), where quality control is more problematic. Even when it's mixed in a controlled environment and then trucked over to the site, there are a multitude of factors which govern concrete strength: was it mixed properly, are the aggregates adequate for the site's environment, was it poured correctly, was it hydrated adequately post-pour, what will its strength profile over time be, etc, etc, etc. Meanwhile, steel is manufactured in a factory, put in a truck, positioned on the site, and then welded/screwed in. There are basically no random variables as to the structural element's strength once it leaves the factory, which means that quality control at the factory door is essential and sufficient to be very sure of the material's actual strength. So the smaller safety factor is basically to cover the potential for screwups at the factory and for potential loss of strength over time due to the elements. The only other variable to be considered for steel is the linkages (welding/screwing), but guess what, there are codes for that as well.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-11-18T14:41:29.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18074" PostTypeId="1" AcceptedAnswerId="18075" CreationDate="2017-11-18T20:05:32.907" Score="4" ViewCount="7697" Body="&lt;p&gt;I would like to learn to weld (stick welding) for my own purposes (mostly for small repairs and building simple metal constructions). So far I'm trying to educate myself by reading and watching instructional videos, but so far I haven't found an answer to one question which is bothering me (I know it may sound stupid):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that welding equipment is fully functional (free of any defects and everything is set up properly), is there a risk of electric shock if a welder inadvertently touch part of construction which he is welding?&lt;/p&gt;&#xA;" OwnerUserId="13727" LastEditorUserId="330" LastEditDate="2017-11-19T02:00:45.367" LastActivityDate="2019-02-03T23:53:31.430" Title="Welding and risk of electric shock" Tags="|electrical-engineering|welding|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18075" PostTypeId="2" ParentId="18074" CreationDate="2017-11-18T20:17:27.947" Score="3" Body="&lt;p&gt;No. Current will always take the path of least resistance. The path of least resistance is through the ground clamp that you secure to the workpiece, or another conductor that the workpiece is touching. If this ground clamp becomes disconnected, you can't weld (or start to), because the circuit would be broken. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're welding in wet conditions or providing a path to ground (i.e. holding the ground clamp and touching the electrode) then there is certainly a risk of electrocution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should always make efforts to provide the safest possible path to ground though, do not clamp to oxidised/painted metal and do not weld in wet conditions. The main risks involved in welding are more concerned with burns, corneal inflamamtion and retinal burning from the UV generated by the arc. &lt;/p&gt;&#xA;" OwnerUserId="13673" LastEditorUserId="13673" LastEditDate="2017-11-18T20:22:59.797" LastActivityDate="2017-11-18T20:22:59.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18076" PostTypeId="2" ParentId="16016" CreationDate="2017-11-18T20:46:17.237" Score="7" Body="&lt;p&gt;If we're staying within the realms of beam theory, we can go with this approach which is valid for any material that exhibits linear-elastic behaviour before yield:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curvature of a beam is related to the applied moment and it's flexural stiffness:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\kappa = \frac{M}{EI}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, for a cicular rod, $I = \frac{\pi d^4}{64}$. The curvature is also equal to the inverse of the bent radius, giving the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;R = \frac{1}{\kappa} = \frac{EI}{M}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rod of circular cross-section, the stress at the extreme fibre is related to the applied bending moment and the elastic section modulus: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\sigma = \frac{M}{Z}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where for a circular rod, $Z = \frac{\pi d^3}{32}$. This gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\therefore M = \frac{\sigma \pi d^3}{32}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if a given material has a certain yield strength, $\sigma = f_y$, the minimum radius can be determined through the substitution of the above equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;R_{min} = E \, \frac{\pi d^4}{64} \frac{32}{f_y \pi d^3}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\boxed{\therefore R_{min} = \frac{E d}{2 f_y}}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $E$ is the elastic modulus of the material, $d$ is the diameter of the rod and $f_y$ is the yield strength of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer to &lt;a href=&quot;http://homepage.tudelft.nl/p3r3s/b17_exam_2017_july.pdf&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; solution to question 6a on page 8 for a similar solution for a rectangular section. Note that the end result is the same for both a rectangle and circle as both have a distance of $D/2$ from the neutral axis to the extreme fibre.&lt;/p&gt;&#xA;" OwnerUserId="13602" LastEditorUserId="13602" LastEditDate="2017-11-18T21:48:33.013" LastActivityDate="2017-11-18T21:48:33.013" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18077" PostTypeId="2" ParentId="18074" CreationDate="2017-11-18T21:10:21.870" Score="-2" Body="&lt;p&gt;To get shocked, the voltage applied to your skin (by touching an electrode) must be large enough to establish a current in your skin from the point of contact to the ground connection (usually your feet in contact with the ground).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the voltages used for welding (just a few volts) this doesn't happen: at a potential of a couple of volts, your skin is a good enough insulator to prevent a current from flowing which is big enough for you to sense as a shock. Similarly, if you grabbed the two terminals of a 12-volt battery with your hands, the 12 volts isn't enough to force a current into your skin, up one arm, across your chest, and down the other arm. 120 volts will do the trick however!&lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="330" LastEditDate="2017-11-19T00:06:00.447" LastActivityDate="2017-11-19T00:06:00.447" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18078" PostTypeId="2" ParentId="18074" CreationDate="2017-11-18T22:22:36.237" Score="0" Body="&lt;p&gt;Unlike TIG and MIG, stick welding does carry some risk of electric shock for the following reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is no trigger switch on a stick welder, rather there is always a voltage between the electrode and the ground clamp.&lt;/li&gt;&#xA;&lt;li&gt;You need to touch the (consumable) electrode of a stick welder in order to replace it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Therefore, you can give yourself an electric shock if you are replacing the electrode without turning off the welder, and you make skin contact with the grounded workpiece, or anything electrically connected to it. Perhaps the most likely scenario would be if the workpiece was on a metal workbench, you were wearing thin clothing or pants and a shirt rather than coveralls, and your hip brushed against the workbench while you were replacing the electrode. This risk can easily be avoided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that stick welder voltages are lower than mains voltage (perhaps reaching only 80-100V), and are usually DC (unless welding aluminium), so the risk of death or injury is much lower than a mains voltage electric shock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Touching the workpiece does not present a risk of electric shock unless you are also touching the electrode, and vice-versa. Like all forms of welding though, you can burn yourself due to the object still being hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, welding involves far greater risks than electrocution, such as eye damage from UV light (avoidable by using a welding hood), exposure to hazardous fumes (avoidable through extraction and/or respirator use), and burns due to UV exposure or direct heat (avoidable though use of gloves and coveralls), and the potential for starting fires (avoidable by removing all flammable materials from the vicinity).&lt;/p&gt;&#xA;" OwnerUserId="11438" LastEditorUserId="11438" LastEditDate="2017-11-18T22:36:53.510" LastActivityDate="2017-11-18T22:36:53.510" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18079" PostTypeId="2" ParentId="18015" CreationDate="2017-11-19T01:59:42.520" Score="1" Body="&lt;p&gt;I remember learning about this in design class. Each bolt has two components to it's shear force. One is equal to a fraction of the applied load directly, and the other is the shear force needed to create an equipollent moment at the center of the bolt pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, with $n$ bolts, and $\mathbf{r}$ the location of the load from the bolt pattern center, the equipollent moment is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M =\| \mathbf{r} \times \mathbf{F} \| = r_x F_y - r_y F_x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the position of the &lt;em&gt;i&lt;/em&gt;-th bolt is $\mathbf{b}_i = R_i \mathbf{e}_i$ where $R_i$ is the radial distance and $\mathbf{e}_i$ the radial direction vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider also the perpendicular direction $\mathbf{n}_i$. This points tangentially from each bolt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total shear force for the &lt;em&gt;i&lt;/em&gt;-th bolt is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{S}_i = \frac{1}{n} \left( \mathbf{F} + \frac{M}{R_i} \mathbf{n}_i \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can show that this satisfies the equilibrium equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{cases} \sum \limits_{i=1}^n \mathbf{S}_i = \mathbf{F} \\ \sum \limits_{i=1}^n \| \mathbf{b}_i \times \mathbf{S}_i \| = M \end{cases} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assumes that the connecting parts are compliant enough to give an equal load distribution to the bolts and that the bolt holes are a loose fit and don't push against the bolts when not loaded.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-11-19T01:59:42.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18080" PostTypeId="1" CreationDate="2017-11-19T08:52:44.460" Score="1" ViewCount="54" Body="&lt;p&gt;Both have a linear relationship, is there any merit in plotting one over the other?&lt;/p&gt;&#xA;" OwnerUserId="13732" LastEditorUserId="1832" LastEditDate="2017-11-19T14:36:20.443" LastActivityDate="2017-11-19T14:36:20.443" Title="Column Buckling: Which is more useful to plot; Buckling Load vs Slender Ratio, or, Stress vs Slender Ratio" Tags="|mechanical-engineering|structural-engineering|materials|structures|mechanical-failure|" AnswerCount="0" CommentCount="1" ClosedDate="2017-11-21T23:20:37.197" ContentLicense="CC BY-SA 3.0" />
  <row Id="18082" PostTypeId="2" ParentId="18074" CreationDate="2017-11-19T20:00:06.347" Score="3" Body="&lt;p&gt;As long as the equipment isn't faulty and you don't do anything really silly you are very unlikely to  get electrocuted from stick welding simply because the voltages involved are not large enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed in most welding processes it is fairly normal to be touching the work at some point during welding, albeit you are usually wearing gloves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it  isn't impossible to get a shock from stick welding and of course you need to be aware of general electrical safety it's not an immediate hazard that you need to actively avoid, in the way that burning yourself on the hot metal is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you absolutely can get a shock is TIG welding with a machine with high frequency start. If the work isn't well earthed and you're touching it you will get an unpleasant shock as the high frequency element superimposed on the current is much better at being conducted through the human body than the main DC or low frequency AC current. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-11-23T19:00:45.807" LastActivityDate="2017-11-23T19:00:45.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18084" PostTypeId="1" AcceptedAnswerId="18089" CreationDate="2017-11-20T00:27:55.087" Score="1" ViewCount="337" Body="&lt;p&gt;Given that hot air rises, it's easy to understand how attic insulation can help keep a building warm; say, in the winter, for example. But when winter changes to summer, does the attic insulation help keep the building cool? Or even worse, does attic/roof insulation make the building hotter by resisting the flow of heat up and out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the below diagram, my idea is that the R-factor of the attic insulation will resist the flow of heat up and out of the building by impeding flow from zone 1 to zone 2 in the below diagram.&lt;/p&gt;&#xA;&#xA;Figure  1. Premise: R-factor resists heat flow from zone 1 to zone 2 thus trapping warm air inside the house in the summertime?&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kYlgf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kYlgf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would we quantify the overall effect on house temperature based on this and other competing effects? Both theoretically and empirically?&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2017-11-20T07:42:58.523" LastActivityDate="2017-11-20T07:42:58.523" Title="Does attic insulation help cool a building?" Tags="|thermodynamics|building-design|building-physics|thermal-insulation|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18085" PostTypeId="2" ParentId="15798" CreationDate="2017-11-20T00:34:12.427" Score="2" Body="&lt;p&gt;I have been shipping refrigerators on their sides since 1976; that adds up to thousands of times. I tell the customer that it is the biggest wives' tale ever.  I tell the customer to allow two hours for the oil to return back to the compressor and there will be no problems. I lay them down to where the suction line on the compressor is facing down. That way when the compressor starts up the oil in the suction line is pulled right back into the compressor. The oil does not go through the piston valve as it enters the &quot;pot&quot; section. The whole compressor is under the low side of the refrigeration cycle. Laying it down on the discharge side can discharge any oil into the condenser that may clog the cap tube that feeds the evaporator. THIS WILL PASS the compressor will push it through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just love people who speculate with their answers with no refrigeration degree and no practical application.&lt;/p&gt;&#xA;" OwnerUserId="13740" LastEditorUserId="1832" LastEditDate="2017-11-20T00:57:14.663" LastActivityDate="2017-11-20T00:57:14.663" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18086" PostTypeId="2" ParentId="18084" CreationDate="2017-11-20T02:54:31.907" Score="2" Body="&lt;p&gt;in the summer, the outside of the house- especially its roof- is hotter than the inside of the house. In this case, attic insulation does not physically &quot;cool&quot; the house- it just impedes heat transfer from the attic to the inside of the house. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-11-20T02:54:31.907" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18087" PostTypeId="1" AcceptedAnswerId="18155" CreationDate="2017-11-20T03:49:20.673" Score="0" ViewCount="71" Body="&lt;p&gt;I am trying to understand how to turn a basic click button into something that behaves like a pressure sensitive button, meaning the harder you press it the greater the value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are PWM modules that you can use to get different current but I am thinking of something like a button.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could I do this with a micro-controller, a click button and some software or are there special buttons that do the job and send signals in a range based on how hard/ fast it's pressed?&lt;/p&gt;&#xA;" OwnerUserId="12586" LastActivityDate="2017-11-24T13:38:41.163" Title="Applying PWM to get analog button" Tags="|electrical-engineering|pwm|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18088" PostTypeId="2" ParentId="18087" CreationDate="2017-11-20T06:03:02.837" Score="1" Body="&lt;p&gt;A standard on/off push button will not read analog.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To accurately read analog force you need a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Load_cell?wprov=sfla1&quot; rel=&quot;nofollow noreferrer&quot;&gt;load cell&lt;/a&gt;. This is what is used in a bathroom or letter scale for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other (generally less acurate and lower cost) technologies that use the spring constant of a material in combination with a distance sensing method. This &lt;a href=&quot;https://www.sparkfun.com/products/9375&quot; rel=&quot;nofollow noreferrer&quot;&gt;resistive force sensor&lt;/a&gt; uses the spring constant of its internal polymer and the resistance vs thickness relationship to opperate. It will not provide nearly as linear of response as a load cell but would be simple to implement with a pwm analog sensing circuit like you were origionally intending.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-11-20T06:03:02.837" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18089" PostTypeId="2" ParentId="18084" CreationDate="2017-11-20T06:12:37.757" Score="2" Body="&lt;p&gt;Attics are much hotter than the outside air when the sun is shining on them (say 120F in an attic). When the air conditioning system is running the internal air temperature is less than the outside (lets say 72F in the house and 85F outside). The ceiling temerature might be warmer than the floor, but both will be much cooler than the attic (say 73F for the ceiling and 71F for the floor). Heat will transfer from the hotter attic (120F) into the house (73F ceiling temp), so ceiling insulation is very important for keeping a house cool.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-11-20T06:12:37.757" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18090" PostTypeId="1" CreationDate="2017-11-20T07:42:14.587" Score="0" ViewCount="47" Body="&lt;p&gt;Is there any way to determine either empirically or by convention a rule-of-thumb when one should spend money on insulating windows rather than insulating the attic of a house or building to keep it warm in winter and cool in summer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, how do we determine where to place the next piece of insulation to maximize the increase in total building system insulation effective R-value?&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2017-11-20T08:38:12.927" LastActivityDate="2017-11-22T07:19:49.177" Title="When does window insulation have more impact than attic?" Tags="|thermodynamics|building-physics|thermal-insulation|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18091" PostTypeId="1" CreationDate="2017-11-20T10:06:47.453" Score="0" ViewCount="6790" Body="&lt;p&gt;I recently installed &lt;strong&gt;SIMATIC STEP 7-Micro/WIN V4.0.8.06&lt;/strong&gt; but cancelled the installation after some steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the program appears in the control panel and when I try to uninstall it, the error:- &quot;&lt;em&gt;Did not find Shared Comm client list in registry at:.....&lt;/em&gt;&quot; appears and the unistallation process terminates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried to uninstall the registry entries by opening  &quot;&lt;em&gt;HKEY_LOCAL_MACHINE &gt; SOFTWARE&lt;/em&gt;&quot; directory.&#xA;Still it won't uninstall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help!!&lt;/p&gt;&#xA;" OwnerUserId="13741" LastActivityDate="2017-12-20T16:13:02.233" Title="How to uninstall SIMATIC STEP 7-Micro/WIN V4.0.8.06?" Tags="|software|pneumatic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18092" PostTypeId="2" ParentId="18015" CreationDate="2017-11-20T12:50:06.177" Score="0" Body="&lt;p&gt;I found these documents relating to my question. However i suspect they're all for circular gasket designs with the bolt pattern being circular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EngineersEdge:&#xA;&lt;a href=&quot;https://www.engineersedge.com/general_engineering/gasket_cover_compression.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineersedge.com/general_engineering/gasket_cover_compression.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Journal Paper:&#xA;International Journal of Pressure Vessels and Piping 120-121 (2014) Comparative study of bolt spacing formulas used in bolted joint designs&#xA;Abdel-Hakim Bouzid&lt;/p&gt;&#xA;" OwnerUserId="13665" LastActivityDate="2017-11-20T12:50:06.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18093" PostTypeId="2" ParentId="18091" CreationDate="2017-11-20T15:40:50.053" Score="1" Body="&lt;p&gt;Your best bet, if Control Panel/Add or Remove programs isn't working, is to navigate to your STEP 7 Micro/WIN directory in the Windows Explorer and delete it (default directory: &lt;code&gt;C:\Program Files\Siemens\STEP 7 MicroWIN V4.0&lt;/code&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To remove the registry items associated with STEP 7:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to:&#xA;&lt;code&gt;HKEY_LOCAL_MACHINE \ Software \ Microsoft \ Windows \ CurrentVersion \ Uninstall.&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Click the  &lt;code&gt;Uninstall&lt;/code&gt; folder and go to &lt;code&gt;Edit &amp;gt; Find&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Enter: &lt;code&gt;STEP 7 Micro/WIN&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check whether the &lt;code&gt;DisplayName&lt;/code&gt; code matches one of the &quot;MicroWin Release&quot; versions i.e. &lt;code&gt;V4.0.8.06&lt;/code&gt;. Delete the corresponding folder... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer to &lt;a href=&quot;https://support.industry.siemens.com/cs/document/14618173/how-do-you-uninstall-step-7-micro-win-v4-0-after-an-aborted-installation-?dti=0&amp;amp;lc=en-RU&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; document for help. &lt;/p&gt;&#xA;" OwnerUserId="13673" LastActivityDate="2017-11-20T15:40:50.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18094" PostTypeId="1" CreationDate="2017-11-20T19:08:45.580" Score="0" ViewCount="114" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q3HFF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q3HFF.jpg&quot; alt=&quot;Find the hibrid (h) parameters for the circuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help me, please. I im studying circuit analysis and i couldn't solve this problem. &lt;/p&gt;&#xA;" OwnerUserId="13747" LastEditorUserId="8055" LastEditDate="2017-11-21T00:55:50.127" LastActivityDate="2017-11-21T00:55:50.127" Title="Help with two-port network parameters problem!" Tags="|electrical-engineering|circuits|" AnswerCount="1" CommentCount="2" ClosedDate="2017-11-21T01:56:02.263" ContentLicense="CC BY-SA 3.0" />
  <row Id="18095" PostTypeId="2" ParentId="12861" CreationDate="2017-11-20T19:55:46.813" Score="1" Body="&lt;p&gt;There are some works by professor Wiciak, where vibrations or noise are controlled by piezoelements. &#xA;&lt;a href=&quot;http://www.sciencedirect.com/science/article/pii/S2212017316305503&quot; rel=&quot;nofollow noreferrer&quot;&gt;Development of noise reduction panel using piezoelectric material&lt;/a&gt;&#xA;Piezomaterials can be used for active vibration control. &#xA;Maybe you could apply this principle to this problem. How about bimetal? Can you use heat? Or cooling of predeformed shape?&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2017-11-21T17:53:51.683" LastActivityDate="2017-11-21T17:53:51.683" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18098" PostTypeId="2" ParentId="18094" CreationDate="2017-11-20T22:59:09.340" Score="1" Body="&lt;p&gt;First, try to understand what the h-parameter equations are asking of you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_{11} = \frac{V_1}{I_1}\bigg\rvert _{V2=0}$$ I.e. short-circuit $V_2$, such that no potential can be developed accross it. Then, find the corresponding voltage and current. &lt;/p&gt;&#xA;" OwnerUserId="13673" LastActivityDate="2017-11-20T22:59:09.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18099" PostTypeId="2" ParentId="12861" CreationDate="2017-11-21T08:53:06.047" Score="0" Body="&lt;p&gt;You will want to choose a thickness of plate to resist the water flow conditions, then select an acuator with enough force to deflect the plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your stroke length will be set by how much deflection you need. Your actuator type will be selected based on the required accuracy and budget. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pizeo would work great since it is the highest accuracy and waterproof(probably only very clean water though). I probably wouldnt mess with shape memory alloys as that is likely to introduce more complexity into a simple design requirement. A gear motor linear actuator would work well too if the accuracy is sufficent and it can be kept out of the water. Radio controlled servo actuators come in a wide range of torques and accuracies would also likely work. Possibly hydraulics if the location is to hazardous for electric actuators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as attaching the actuator; the easiest would be in tension like you show. Piano wire could be used to allow the actuator to be located in a more convenient or dryer location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actuator could also be in compression from the end if necessary. You just need to make sure the load is pinned and loaded eccentrically so it deflects the same way every time.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2017-11-21T09:02:51.600" LastActivityDate="2017-11-21T09:02:51.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18102" PostTypeId="1" CreationDate="2017-11-21T11:23:52.007" Score="1" ViewCount="841" Body="&lt;p&gt;I work in a lab and i was given the mission of identifing the reynolds in a square pipe, in different flow rates&#xA;I have the dimentions of the rectangle - 5cm X 6cm, and I have the flow rate. the fluid is air (so I also have the charecteristics of the fluid)&#xA;what is the characteristic length here? and can i assume the speed of the air is the same in all the cross section? (we are talking about q between 30-100 m^3 per hour)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you.&lt;/p&gt;&#xA;" OwnerUserId="13755" LastActivityDate="2017-11-21T13:02:34.657" Title="How to calculate Reynolds in rectangular pipe" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18103" PostTypeId="2" ParentId="12861" CreationDate="2017-11-21T11:26:01.893" Score="0" Body="&lt;p&gt;Presumably you will also want a reasonably well defined curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create a smooth, repeatable and adjustable curve you will want to stay withinthe elastic limit of the the material you are using which will depend on the material itself and its thickness. Steel or a composite sheet would seem like reasonable candidates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also puts a limit on the stiffness of the surface as the elastic deformation will be limited by the  thickness of the plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The profile you end up with can be calculated by beam deflection theory and will depend on where you apply loads, in what direction and how the ends of the plate are supported.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the simplest solution is to support one end and apply a load at the other (much like your example mechanism). Although there are lots of potential permutations of load and support arrangements. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-11-21T11:26:01.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18104" PostTypeId="2" ParentId="17942" CreationDate="2017-11-21T11:51:05.630" Score="0" Body="&lt;p&gt;To answer your question &quot;whether a linear joule is the same as rotational joule&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are both forms of energy that are considered equal when using the conservation of energy methods - which I believe is what you're trying to do.&lt;/p&gt;&#xA;" OwnerUserId="8212" LastActivityDate="2017-11-21T11:51:05.630" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18106" PostTypeId="2" ParentId="18102" CreationDate="2017-11-21T13:02:34.657" Score="3" Body="&lt;p&gt;So, using the formula Hd = (4 * A) / wet perimeter then with 5 and 6 given the Hd is 5.455.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference see here :&#xA;&lt;a href=&quot;https://www.engineeringtoolbox.com/hydraulic-equivalent-diameter-d_458.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydraulic Diameter&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-21T13:02:34.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18107" PostTypeId="1" CreationDate="2017-11-21T13:11:05.100" Score="1" ViewCount="6005" Body="&lt;p&gt;How to change the material properties, i.e. Young's modulus and Poisson's ratio, of a set of elements which is defined by a named selection in Ansys workbench using a command snippet of APDL code?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The named selection is denoted &lt;code&gt;steelElements&lt;/code&gt;.&#xA;An appropriate material, named &lt;code&gt;Structural Steel&lt;/code&gt; is defined in an &quot;engineering data&quot; system.&lt;/p&gt;&#xA;" OwnerUserId="13738" LastEditorUserId="13738" LastEditDate="2017-11-21T15:55:54.737" LastActivityDate="2018-06-19T20:37:12.803" Title="Change material properties of elements in named selection in Ansys workbench using APDL commands" Tags="|ansys-workbench|ansys-apdl|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18108" PostTypeId="2" ParentId="17942" CreationDate="2017-11-21T13:30:08.357" Score="1" Body="&lt;p&gt;Your fundamental assumptions are correct, but I would question the practicality of the system that you are suggesting for testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with a heavy flywheel at a large diameter, you would need to spin it to a very high rotational speed in order to store the 7065J of energy that you are trying to mimic. Unfortunately brake-force is not independent of the surface speeds, and other important effects such as overheating/longevity of the system may not be accurately modeled in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend dimensional analysis to allow you to model the system at a smaller scale (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Buckingham_%CF%80_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Buckingham_%CF%80_theorem&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Similitude_(model)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Similitude_(model)&lt;/a&gt;) &lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-11-21T13:30:08.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18110" PostTypeId="2" ParentId="18107" CreationDate="2017-11-21T14:49:27.240" Score="1" Body="&lt;h1&gt;Strategy:&lt;/h1&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Identify maximum number of material properties&lt;/li&gt;&#xA;&lt;li&gt;Create material&lt;/li&gt;&#xA;&lt;li&gt;Assign material to selected elements&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;h1&gt;Code:&lt;/h1&gt;&#xA;&lt;pre&gt;&lt;code&gt;    !   Commands inserted into this file will be executed just prior to the ANSYS SOLVE command.&#xA;    !   These commands may supersede command settings set by Workbench.&#xA;&#xA;    !   Active UNIT system in Workbench when this object was created:  Metric (mm, t, N, s, mV, mA)&#xA;    !   NOTE:  Any data that requires units (such as mass) is assumed to be in the consistent solver unit system.&#xA;    !                See Solving Units in the help system for more information.&#xA;    FINISH&#xA;    /PREP7&#xA;    *DEL,mySelectioName,,nopr &#xA;    *DIM, mySelectioName, string, 248, 1&#xA;    !################### Start User Input&#xA;    myEModul = 210000&#xA;    myPoissionration = 0.3&#xA;    mySelectioName(1, 1) = 'steelElements'&#xA;    !################### End User Input&#xA;    ALLSEL&#xA;    !Get maximal material number assigned and increment&#xA;    *GET,maxMaterialNumber,ELEM,0,MATM&#xA;    maxMaterialNumber = maxMaterialNumber + 1&#xA;    !Create new material properties&#xA;    MP, EX, maxMaterialNumber, myEModul&#xA;    MP, NUXY, maxMaterialNumber, myPoissionration&#xA;    !Select elements and change material property number of those materials&#xA;    CMSEL,s,mySelectioName(1, 1)&#xA;    MPCHG, maxMaterialNumber, all&#xA;    !Select all entities and leave code snippet&#xA;    ALLSEL&#xA;    FINISH&#xA;    /SOLU&#xA;&#xA;    ! Doublecheck in Classic:&#xA;    !   /pnum,mat,1 !0=off, 1=on&#xA;    !   /number,1   !0=on, 1=off&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="13738" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-11-21T14:49:27.240" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18111" PostTypeId="2" ParentId="12869" CreationDate="2017-11-21T15:51:55.400" Score="1" Body="&lt;p&gt;&lt;strong&gt;Answer to: How to synchronize the motor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a precise synchronization of the two motors (electronically), a closed loop is required:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The rotation is captured by a sensor&lt;/li&gt;&#xA;&lt;li&gt;The sensor input a microcontroller&lt;/li&gt;&#xA;&lt;li&gt;The MCU drive the motor&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The typical mechanism consists of a few permanent magnets on the motor rotor, which field is detected by a hall-effect sensor.&lt;br&gt;&#xA;An MCU with a Analogic-to-digital converter would sample this input at high frequency (at very least twice the maximum RPS of the motors).&lt;br&gt;&#xA;The MCU drive the motors, this depends on the type of motors: for a simple DC engine, the electric circuit is pulsed (PWM) to achieve the correct speed without reducing power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer to: What seem the problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rail contact against the wheels naturally synchronize their rotation, trying to synchronize front and rear wheels electronically would just be useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any of the two motors overheat compare to the other, it is probably a malfunction of some component (motor or driver circuit). I suggest swapping the motors and see if it's the motor or the driver. In any case, replace the part if needed.&lt;br&gt;&#xA;In most of the cases, a little difference is acceptable.&lt;/p&gt;&#xA;" OwnerUserId="11129" LastActivityDate="2017-11-21T15:51:55.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18112" PostTypeId="1" AcceptedAnswerId="18122" CreationDate="2017-11-21T18:06:46.203" Score="0" ViewCount="2710" Body="&lt;p&gt;This is a more technical rather than professional question. I have a hyper-threaded quad core CPU. While building models in SpaceClaim I noticed that it only uses one logical core. A single core will be on a very high usage percentage while the others are nearly idle. How can I set SpaceClaim to use more cores and make better benefit of the CPU?  &lt;/p&gt;&#xA;" OwnerUserId="13759" LastActivityDate="2017-11-22T06:37:05.070" Title="How to use more CPU cores in SpaceClaim?" Tags="|cad|ansys|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18115" PostTypeId="2" ParentId="8794" CreationDate="2017-11-21T18:53:58.237" Score="0" Body="&lt;p&gt;If you use Mathematica or similar program you can create finite difference solution for bending of a plate. I have done similar calculation based on Timoshenko's work and rather accurate if you use reasonable number of elements. You can vary thickness and dimension and test your formulas, also you can upgrade code to include large deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tested my code with FEM but is too large to paste it here. It is based on this article: &lt;a href=&quot;http://www.teamsociety.eu/team_conferences/team_2010/default.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sertic&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2017-11-21T19:37:13.977" LastActivityDate="2017-11-21T19:37:13.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18116" PostTypeId="2" ParentId="17966" CreationDate="2017-11-21T20:52:11.457" Score="1" Body="&lt;p&gt;For same shearing stress if absolute value of &lt;span class=&quot;math-container&quot;&gt;$\sigma_x$&lt;/span&gt;-&lt;span class=&quot;math-container&quot;&gt;$\sigma_y$&lt;/span&gt; is larger, points on Mohr circle are wider apart and &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is larger thus maximum shear stress is larger. Variable &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; is introduced to represent this connection and derivative was tried to obtain solution but it failed and solution was obtained using geometric properties of Mohr's circle. You should try to draw circle for better understanding.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="20245" LastEditDate="2021-04-05T17:03:07.957" LastActivityDate="2021-04-05T17:03:07.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18117" PostTypeId="1" AcceptedAnswerId="18121" CreationDate="2017-11-21T21:26:32.150" Score="1" ViewCount="61" Body="&lt;p&gt;I'm trying to design a hover car track. As in, you have a 'track/circuit' that have objects being slowly moved along by the force of the air from beneath and sides. The propulsion to move the objects forwards comes from the air from the sides that are angled in the direction of the track. Picture &lt;a href=&quot;https://en.wikipedia.org/wiki/Scalextric&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scalextric&lt;/a&gt; cars combined with air hockey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These objects that are like carriers that hold things and are under 750 g (in total) and regarding contact with track is a circle that is 135 mm in diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Track length isn't known at the moment but I think 3–5 m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I effectively 'power' and 'transport' the air supply? I've looked into centrifugal fans in series and parallel and their designs and outputs but I'm stuck at calculating what the requirements would be to do this and if it can be done effectively at low cost, like something I could do in a medium sized garage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The track will split and merge too. And there would be multiple carriers on the track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carriage is cylindrical in shape, weight = 750g, diameter =135mm, height 110mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;track is flat with no change in elevation. width of track would be big enough to fit a single carriage approx 150mm&#xA;The track would be like a square trench for the carriage to fit in. so the walls would be 110mm too. &#xA;The walls would be a manifold that has jets angled in the direction of the track, say 60 degrees away from the normal of the wall and parallel with the floor of the track. &lt;/p&gt;&#xA;" OwnerUserId="13762" LastEditorUserId="13762" LastEditDate="2017-12-15T08:06:38.530" LastActivityDate="2017-12-15T08:06:38.530" Title="How to power and transport air to a small hover car track" Tags="|mechanical-engineering|airflow|compressed-air|" AnswerCount="1" CommentCount="6" ClosedDate="2017-12-14T18:32:05.420" ContentLicense="CC BY-SA 3.0" />
  <row Id="18118" PostTypeId="1" CreationDate="2017-11-19T21:41:40.577" Score="1" ViewCount="1876" Body="&lt;p&gt;The shape of the sand grains from our planet's deserts is too round to be used in concrete (&lt;a href=&quot;https://www.economist.com/blogs/economist-explains/2017/04/economist-explains-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;http://www.bbc.com/capital/story/20160502-even-desert-city-dubai-imports-its-sand-this-is-why&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;https://www.newyorker.com/magazine/2017/05/29/the-world-is-running-out-of-sand&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;). This is why we are drawing the last sand resources from natural beaches for making concrete for the ever-increasing human demands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it not make sense to crush the round desert sand grains in a mill to produce the sand required for concrete? If we break each sphere into two fragments, the resulting grains are sharp-edged and asymmetric, with &quot;grip&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a matter of cost?&lt;/p&gt;&#xA;" OwnerDisplayName="Raabe" LastEditorUserId="255" LastEditDate="2017-11-22T10:26:04.580" LastActivityDate="2020-10-28T17:43:05.373" Title="Why is desert sand not processed to make it suitable for concrete?" Tags="|materials|concrete|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18119" PostTypeId="2" ParentId="18118" CreationDate="2017-11-20T11:29:53.633" Score="1" Body="&lt;p&gt;Yes, cost seems to be the limiting factor.  Specifically costs to haul the sand where it is needed.  Where I live there are huge piles of sand being removed from the local river within a few miles of the concrete plants.  In the New Yorker article, reference #3 in your question, one of the engineers they interviewed said after about 60 miles it becomes too costly to transport the sand for construction activities.  If cost isn't a limiting factor, and if the milled sand could be sorted to an appropriate size for the &lt;a href=&quot;http://www.wisegeek.com/what-is-the-difference-between-concrete-and-cement.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;cement&lt;/a&gt;, other sources further from where the sand is needed could be utilized.&lt;/p&gt;&#xA;" OwnerDisplayName="Kurt" LastActivityDate="2017-11-20T21:21:52.637" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18120" PostTypeId="1" AcceptedAnswerId="18124" CreationDate="2017-11-22T03:00:44.637" Score="-1" ViewCount="111" Body="&lt;p&gt;I am working on a project for an Environmental Engineering class. Our concept is a system that collects rain water for use in toilet flushing, i.e. requires minimal filtration. I have no experience in engineering statics, so I am curious if it is possible, with calculations to back it up, to attach a 50 gallon drum for water collection to the side of a house for purposes of gravity feeding the water into a toilet on a lower level of the house. And if not, could this be accomplished at low cost with the rain barrel instead at ground level, collecting runoff from the rain gutters on the roof, using a water pump to displace the water to the bathroom? Any help is appreciated thank you :D &#xA;Additional details: we are students at the University of Washington, so our system is initially proposed for rainy cities such as Seattle where collecting large volumes of rainwater is fairly easy, and the availability of the rainwater isn't of much concern. This solution intends to solve problems off runoff, and to lower monthly water utility cost, as well as to save on water usage.&lt;/p&gt;&#xA;" OwnerUserId="13764" LastActivityDate="2017-11-22T09:47:02.317" Title="Collecting Rainwater off a Domestic Roof for Flushing Toilets" Tags="|mechanical-engineering|structural-engineering|civil-engineering|mechanical|" AnswerCount="2" CommentCount="12" ClosedDate="2017-12-14T18:16:26.250" ContentLicense="CC BY-SA 3.0" />
  <row Id="18121" PostTypeId="2" ParentId="18117" CreationDate="2017-11-22T06:12:48.543" Score="-1" Body="&lt;p&gt;Have a look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hovertrain&quot; rel=&quot;nofollow noreferrer&quot;&gt;hovertrain&lt;/a&gt; wiki.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, &lt;a href=&quot;https://en.wikipedia.org/wiki/Pneumatic_tube&quot; rel=&quot;nofollow noreferrer&quot;&gt;pneumatic tube transport&lt;/a&gt; is often used at bank drive up windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small demo could be constructed low cost in a garage. Do some research on those links above and if you have specific questions regarding fan selection or design; ask them at that time.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-11-22T06:12:48.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18122" PostTypeId="2" ParentId="18112" CreationDate="2017-11-22T06:37:05.070" Score="0" Body="&lt;p&gt;Software must be programmed specifically to use multiple cores. There is nothing restricting SpaceClaim or any other CAD software from being programmed this way, but it is currently more complex/expensive to program and there is no immediate incentive for companies to work this direction. Their software is already very expensive, so their customers don't mind buying a faster computer every year. Customers won't demand multi threading support until the price of software drops below the price of new computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an (unfortunately old 2007) list I copied from &lt;a href=&quot;http://www.tomshardware.co.uk/forum/243332-28-programs-cores-quad-core-chip&quot; rel=&quot;nofollow noreferrer&quot;&gt;toms hardware&lt;/a&gt;. It helps paint the picture of how few programs are multi threaded. The main thing a multiple core processor buys you is the ability to run different programs at 100% simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;3D Studio MAX using Mental Ray Renderer (&gt;99 % of 4 cores)&lt;/li&gt;&#xA;  &lt;li&gt;Adobe Premiere Elements v3.0.2 (52 - 85 % of 4 cores depending on source type, filters, etc.)&lt;/li&gt;&#xA;  &lt;li&gt;AutoGK v2.40 (30 - 53 % of 4 cores depending on source type, filters, etc.)&lt;/li&gt;&#xA;  &lt;li&gt;Cinema 4d Rendering (&gt;99 % of 4 cores)&lt;/li&gt;&#xA;  &lt;li&gt;Dr. DivX v2.0.0 (47 - 65 % of 4 cores depending on source type, filters, etc.)&lt;/li&gt;&#xA;  &lt;li&gt;DVDShrink v3.2 (~90 % of 4 cores)&lt;/li&gt;&#xA;  &lt;li&gt;Lightwave 3D (&gt;99 % of 4 cores)&lt;/li&gt;&#xA;  &lt;li&gt;Nero Suite 7.x (&gt;90 % of 4 cores when encoding)&lt;/li&gt;&#xA;  &lt;li&gt;Noise Ninja v2.13 (~80 % of 4 cores when doing the noise reduction on an image)&lt;/li&gt;&#xA;  &lt;li&gt;Sony Vegas 7.0e (83 - 100 % of 4 cores depending on source type, filters, etc.)&lt;/li&gt;&#xA;  &lt;li&gt;TMPG XPress v4.2.3.193 (65 - 100 % of 4 cores depending on source type, filters, etc.)&lt;/li&gt;&#xA;  &lt;li&gt;Winrar v3.70   (~85 - 90 % of 4 cores on benchmark; ~75% in practice)&lt;/li&gt;&#xA;  &lt;li&gt;x264 v0.55.663 (&gt;99 % of 4 cores when doing the 2nd pass of a 2 pass encode)&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-11-22T06:37:05.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18123" PostTypeId="2" ParentId="18090" CreationDate="2017-11-22T07:19:49.177" Score="1" Body="&lt;p&gt;A project makes economic sense when the return from the project exceeds the rate you can borrow money at. Usually we want a fairly large return before we concern ourselves with the project however. Lets say we want a 3 year break even or 33%APR. All aspects of that project should target that rate of return to be optimized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate this you first determine the return. In this case it is the amount of heat saved by upgrading to new windows (kW * 24hr * 365 days), multiplied by the cost of heat ($/kWhr). Then take this amout divided by the cost of the window project. If it is more than your target APR, consider increasing to the next quality level of windows. If it is less than your target APR, try decreasing the quality of the new windows or abandon that aspect of the project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then run the same calculation for all the other items in your project: attic insulation, wall insulation, HVAC COP, etc.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-11-22T07:19:49.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18124" PostTypeId="2" ParentId="18120" CreationDate="2017-11-22T07:28:18.060" Score="1" Body="&lt;p&gt;At 8 pounds a gallon, a 50 gallon drum when full will weigh 400 pounds. This load cannot safely be fastened to the side of a house which was not designed to carry it. You will require a separate pedestal to support the weight of the drum, and it will need to be anchored in place so it will not tip over in an earthquake. If, on the other hand, you position the drum at ground level, you do not have to worry about the effect the drum's weight would have on your house and it does not have to be earthquake-proof. A simple pump can then be used to move the water where you want it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that in tropical locations that get a lot of rain, it is common to design the roof of a house to support a rainwater catchment tank that can hold hundreds of gallons of water for toilet flushing. I think your idea is worth pursuing but  ideally you want the house to be designed from scratch to accomodate your design.  Good luck to you. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-11-22T07:28:18.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18126" PostTypeId="2" ParentId="18014" CreationDate="2017-11-22T07:50:19.027" Score="1" Body="&lt;p&gt;The best metric for this would be a volume or mass flowrate: m^3/hr, ft^3/day, kg/hr, lbm/day, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any of these metrics would be easily measurable, and smooth out the variables of localized rock conditions and noise from distractions, weather, etc. You make a rough quantification of this when you are digging a hole for example to determine if you should switch to a different tool, change conditions, or give up. Mining has certainly used these metrics for as long as measurement has been available.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-11-22T07:50:19.027" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18128" PostTypeId="1" CreationDate="2017-11-22T09:17:19.177" Score="0" ViewCount="41" Body="&lt;p&gt;I would like to know more about how to connect pieces of gypsum plaster effectively, without space between them. I have a few pieces of gypsum plaster (around 10 mm thickness, 20 cm width, 30 cm height) and would to connect them and form a simple box. I was thinking about some kind of steel clamps which could effectively connect the pieces, but I am not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in connecting gypsum plaster board pieces without using screws e.g. without destroying the material in any way. It is about a model, which could be disassembled and reassembled again. That is why I first though about some steel clamps or some steel frames in which the parts could be inserted and removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas or where to look?&lt;/p&gt;&#xA;" OwnerUserId="12057" LastEditorUserId="255" LastEditDate="2017-11-22T10:34:45.693" LastActivityDate="2017-11-22T13:37:07.533" Title="Connecting gypsum plaster pieces without space between them" Tags="|materials|connections|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18131" PostTypeId="2" ParentId="18120" CreationDate="2017-11-22T09:47:02.317" Score="1" Body="&lt;p&gt;You should evaluate the source : ie amount of rainfall over the year, when it is available, how many days it rains etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then evaluate the use over the year : flushes per day etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can work out the storage volume necessary based on the number of days between rainfall filling the tank and the use per day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the volume of tank required then you can work out the weight and start designing a support or decide where it can be located (my parents house had the water tank halfway up the hill to provide pressure with no pumps necessary...).&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-22T09:47:02.317" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18133" PostTypeId="2" ParentId="18128" CreationDate="2017-11-22T13:37:07.533" Score="1" Body="&lt;p&gt;Use some steel channel and have the pieces slide in and out, but you will see that this type of board will suffer damage...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steel or aluminium channel is available at many stores...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-22T13:37:07.533" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18134" PostTypeId="2" ParentId="11638" CreationDate="2017-11-22T14:59:15.360" Score="0" Body="&lt;p&gt;That is a big cable. I am making an estimation here, but let's follow along with my cable analysis:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;1000 Kcmil&lt;/code&gt; copper cable has 4 layers over the king strand with &lt;code&gt;61&lt;/code&gt; individual strands&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;2000 Kcmil&lt;/code&gt; copper cable with 5 layers of the king strand has $n=$&lt;code&gt;91&lt;/code&gt; individual strands&lt;/li&gt;&#xA;&lt;li&gt;&lt;code&gt;2000 Kcmil&lt;/code&gt; with &lt;code&gt;91&lt;/code&gt; strands =&gt; each strand $d=$&lt;code&gt;0.148 in^2&lt;/code&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use $d =\sqrt{ \frac{{\rm Kcmil}}{1000\,n}}$ where $n$ is the strand count.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Overall size of the cable is $D=$&lt;code&gt;1.927 in&lt;/code&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use $D = d (1+2 ( {\rm layers}+1))$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Total strand area is $A=$&lt;code&gt;1.571 in^2&lt;/code&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use $A = n \frac{\pi d^2}{4}$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The strand strength factor for aluminum (not pure &lt;code&gt;-O&lt;/code&gt; spec) is $\rm sf=$&lt;code&gt;0.965&lt;/code&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use ${\rm sf} = 1.1277 - 1.096 \, d$ (taken from a table in the Alcoa conductor manual and curve-fitted with a line.)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;I don't have the copper equivalent of the strength factor, but I am assuming it is the same ( approx. similar strength ratio of strand surface to center).&lt;/li&gt;&#xA;&lt;li&gt;The ultimate strength of copper is $\sigma_U=$&lt;code&gt;62100 psi&lt;/code&gt; for fully hard copper.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use solid copper tables and back-calculate from one strand of &lt;code&gt;0.162 in&lt;/code&gt; having rated strength &lt;code&gt;1280 lbs&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The rated strength of &lt;code&gt;2000 Kcmil&lt;/code&gt; copper cable is &lt;strong&gt;&lt;code&gt;94100 lbs&lt;/code&gt;&lt;/strong&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use $T = ({\rm sf})×A ×\sigma_U$&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Reference: &lt;a href=&quot;http://www.southwire.com/ProductCatalog/XTEInterfaceServlet?contentKey=prodcatsheet46&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.southwire.com/ProductCatalog/XTEInterfaceServlet?contentKey=prodcatsheet46&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-11-22T14:59:15.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18135" PostTypeId="2" ParentId="12582" CreationDate="2017-11-22T17:58:03.593" Score="1" Body="&lt;p&gt;It's important to note that &quot;Rule 1&quot; is considered to apply automatically when referencing ASME Y14.5, but not necessarily so if using the ISO version. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, the concept of Rule one extends the size tolerance to be inclusive of the FORM tolerance.  In other words, if the pin was as it Maximum Material Boundary (MMB) - it's largest size - it would have to be perfectly straight and be able to fit down a perfectly sized &quot;thru-hole&quot; of that same diameter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it departs from it's largest size, it can start to bend, waist, taper, barrel etc. WITHIN it's limits of size. HOWEVER, its local size at any point along its length can never be smaller than the Least Material Boundary (LMB) - smallest diameter size - tolerance.  Imagine always being able to pull a perfect sphere of LMB diameter along the axis of the part and never having it be larger or &quot;protrude through&quot; the part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this concept understood, we can now consider how your question regarding Cylindricity comes into play.  Cylindricity essentially controls the outside &quot;shape&quot; and &quot;form&quot; of the part similarly to how Rule 1 does, except that it is used to FURTHER REFINE that tolerance.  Say, for example, you had a pin with a rather large tolerance on its diameter. You might be willing to accept the pin at any size, but you need it to be relatively straight and round.  In this case, Rule 1 - although it also controls the &quot;roundness&quot; and &quot;straightness&quot; inherently, might actually allow TOO much variation in the part.  To fix that, you'd apply an additional Cylindricity tolerance.  Now that diameter of the part can fluctuate within it's limit of size, but would have to be consistent within the limits of its Cylindricity tolerance along it's length.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the limits of size - in this case 33g6 - are often sufficient enough to control the form of the object provided the requirements of Rule 1 are also implied.  Where even tighter control is needed, you might add an additional Cylindricity tolerance.&lt;/p&gt;&#xA;" OwnerUserId="4897" LastEditorUserId="1832" LastEditDate="2017-11-22T22:01:36.683" LastActivityDate="2017-11-22T22:01:36.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18136" PostTypeId="1" CreationDate="2017-11-22T19:01:52.870" Score="3" ViewCount="1323" Body="&lt;p&gt;I'm looking to get some collective agreement on a question that I have come across a few times now.  When referring to features that are technically independent, what is the general consensus on dimensioning them?  Take, for example, the part below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rC2gS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rC2gS.jpg&quot; alt=&quot;Dimension Example 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 8.5&quot; and 7&quot; diameters are technically independent.  And when being inspected, would typically be reported back as individual diameters. A metrologist would provide me with measurements of the two different  8.5&quot; diameters and the four different 7&quot; diameters. These measurements would be calculated off of each of the individual arcs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineering drawing practice (as far as I've always learned) is to call out the number of places where that feature exists.  However, often it makes much more sense to have the feature reported back based on multiple surfaces.  For example, the 8.5&quot; diameter would be much more practical (and likely more inclusive of design intent) if it were based on the combination of the top and bottom (as shown) surfaces - essentially a single diameter.  Similarly, the 7&quot; diameters would be based on two surfaces on each side of the part.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I cannot find any guidance in any of the ASME Y14 standards that clearly indicate how to dimension this.  Is there a recognized difference between saying &quot;2X&quot; vs &quot;2 SURFACES&quot; as shown here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y4kzW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y4kzW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This scenario comes up quite often especially when considering features like crush ribs, or splines, and as I am often reviewing engineering drawings, I've been looking for a specific reference to point to when explaining the differences to other engineers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was taught that stating &quot;2 SURFACES&quot; informs manufacturing/metrology that the feature is based on the combination of both (or multiple) surfaces, where 2X implies that they are unique. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the ASME Y14.5-2009 standard introduces a concept called &quot;Continuous Feature&quot; which MIGHT be a solution to this situation, but the examples and definition are still somewhat unclear and a little ambiguous in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have different thoughts?&lt;/p&gt;&#xA;" OwnerUserId="4897" LastEditorUserId="4897" LastEditDate="2018-02-15T19:43:46.370" LastActivityDate="2018-02-15T19:43:46.370" Title="GD&amp;T for Multiple Surfaces as Single Feature" Tags="|mechanical-engineering|drafting|standards|technical-drawing|drawings|" AnswerCount="0" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18140" PostTypeId="1" AcceptedAnswerId="18141" CreationDate="2017-11-23T00:12:12.960" Score="1" ViewCount="121" Body="&lt;p&gt;So recently I went to Supercomputing 17 conference and picked this from a booth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know as to what this device is? What's its name? Where to find its documentations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;a href=&quot;https://i.stack.imgur.com/t4pJp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t4pJp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13776" LastEditorUserId="7083" LastEditDate="2018-01-28T22:04:18.077" LastActivityDate="2018-07-31T22:56:12.657" Title="What is this device?" Tags="|electrical-engineering|sensors|embedded-systems|iot|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18141" PostTypeId="2" ParentId="18140" CreationDate="2017-11-23T00:23:29.803" Score="1" Body="&lt;p&gt;There should be writing on the top (metal) portion and/or some silk screening on the bottom. Either an ESP32 or (more likely) an ESP8266. That should give you enough info to get started...&lt;/p&gt;&#xA;" OwnerUserId="9682" LastActivityDate="2017-11-23T00:23:29.803" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18144" PostTypeId="1" CreationDate="2017-11-23T01:30:43.527" Score="0" ViewCount="3120" Body="&lt;p&gt;I am trying to simulate contact between a flat rigid plate and a sphere. I am applying a load on the flat rigid plate and the sphere is flexible. How can I avoid this flat from penetrating into the sphere?&lt;/p&gt;&#xA;" OwnerUserId="13779" LastEditorUserId="6790" LastEditDate="2020-05-02T20:41:10.453" LastActivityDate="2024-02-13T00:08:25.983" Title="How to avoid contact penetration in Ansys?" Tags="|finite-element-method|ansys|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18145" PostTypeId="2" ParentId="10623" CreationDate="2017-11-23T06:47:23.320" Score="0" Body="&lt;p&gt;It is very important to mention in your case(given in figure), It not only increasing, but growing like a $parabolic$ profile(dotted line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding to paul, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine boundary layer growth as a momentum diffusion process. (air loosing momentum but not the mass in the boundary layers.) momentum diffusion is attributed to the viscosity of the fluid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand this process, let us take the momentum balance for your problem,&#xA;It looks like 1D, So,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ u \frac{\partial u}{\partial x}  = - \frac{1}{\rho} \frac{\partial  p}{\partial x} + \nu\frac{\partial^2 u}{\partial^2 x}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can assume $\nu\frac{\partial^2 u}{\partial^2 y}$ is smaller (imagine you are rubbing your hands one against another, one of them is retarted against another in the rubbing direction.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;also your geometry dont have any curvature. So $- \frac{1}{\rho} \frac{\partial  p}{\partial x} = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now your momentum balance eqn will become,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ u \frac{\partial u}{\partial x}  = \nu\frac{\partial^2 u}{\partial^2 x}$ it is the balance beween convection of your mass by $u$ and diffusion. !!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is mathematically interesting equation and restrict yourself for the region $\delta$- boundary layer thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get good agreement between experiment and this theory, when you assume  $\delta = \sqrt{\frac{\nu x}{ U }}$. (this is called similarity solution in boundary layer theory, credited to prandtl). &#xA;So your profile grows as $\delta \sim x^{\frac{1}{2}}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming to second question, now you have curvature on the blades. So, $- \frac{1}{\rho} \frac{\partial  p}{\partial x} \neq 0$. &#xA;If you know the velocity profile as a function of $x$ around your turbine blade(one can get through potential flow), then, using bernoulli's equation, convert pressure term  interms of velocity and function of $x$. (bernoulli's equation appalicable above the edge of the boudray layer $\sim$ inviscid region)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once velocity polynomial relation is substituted in the momentum balance equation for pressure, work out the similarity analysis (though complicated)to get your boundary layer profile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my knowledge, &#xA;Tha boundary layer profile over the $isolated$ turbine blade grows very slowely upto 1/4 th of the chord (favourable pressure), and grows faster after(adverse pressure) that if the boundary layer is atatched till the trailing edge.&lt;br&gt;&#xA;(adverse pressure region is pron to separation, where boundary layer theory breaks). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a cascade turbine you have to find the effect of adjacent blades. still you can proceed with the potential flow theory to get velocity in terms of x co-ordinates. &lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2017-11-23T06:47:23.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18146" PostTypeId="1" CreationDate="2017-11-23T08:32:40.793" Score="1" ViewCount="36" Body="&lt;p&gt;Which material can be used to protect a Si layer from Oxidation by a sub nanometer layer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eY65W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eY65W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9039" LastEditorUserId="90" LastEditDate="2017-11-23T11:11:08.073" LastActivityDate="2017-11-23T11:11:08.073" Title="How to avoid oxidation of Si using a sub nanometer layer" Tags="|materials|manufacturing-engineering|surface-preparation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18147" PostTypeId="1" CreationDate="2017-11-23T09:00:46.587" Score="4" ViewCount="220" Body="&lt;p&gt;I’m trying to understand how anti-squat works in motorcycles and how the geometry plays a major role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/741yF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/741yF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don’t understand, however, the reason why the force from the tires that is transmitted to the chassis of the bike via the swing arm is always along the axis/centre line of the swing arm and not parallel to the ground.  &lt;/p&gt;&#xA;" OwnerUserId="10075" LastEditorUserId="10075" LastEditDate="2017-11-24T08:06:39.177" LastActivityDate="2021-08-05T22:06:15.593" Title="Anti-Squat Force Transmission" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18148" PostTypeId="2" ParentId="18146" CreationDate="2017-11-23T09:23:36.460" Score="1" Body="&lt;p&gt;1nm layer is not much: let me remind you that this is roughly 10 atoms thickness...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the good news is that we can easily create very thin films with a process of &lt;a href=&quot;https://en.wikipedia.org/wiki/Atomic_layer_deposition&quot; rel=&quot;nofollow noreferrer&quot;&gt;atomic layer deposition&lt;/a&gt; (ALD) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The isolating layer could be the classically grown $SiO_2$ (usually grown 0.5-2nm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also use $TiO_2$ or $TiO_2/SiN_x$ stack which are also great for surface passivation but usually grown thicker 4-15nm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to add that even if it will protect from a pure $O_2$ oxidation, at such dimensions the diffusivity of smaller particles is quite strong, the layer may also break from ions, and the layer will not be uniformly 1nm if the surface roughness is too important (will only work on properly polished Silicon wafer and not on the edges).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore in general we do not passivate (protection of aggressive elements) with such thin layer.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastEditorUserId="10684" LastEditDate="2017-11-23T10:16:39.367" LastActivityDate="2017-11-23T10:16:39.367" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18149" PostTypeId="1" AcceptedAnswerId="18152" CreationDate="2017-11-23T09:38:19.090" Score="2" ViewCount="3408" Body="&lt;p&gt;Does anybody know what the thread size is on a bicycle pump connector, such as this one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2dh0w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2dh0w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;: I'm not talking about the female end that goes on the valve stem on the bike (that's a Schrader valve thread), I'm talking about the male end that connects with the pump. It looks almost like an M5, but doesn't quite fit in an M5 female connector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have spent quite a bit of time googling the question, but have drawn a blank so far.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2023-12-11T20:00:07.733" Title="Bicycle pump connector thread size" Tags="|mechanical-engineering|threads|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18150" PostTypeId="2" ParentId="18147" CreationDate="2017-11-23T10:07:47.153" Score="1" Body="&lt;p&gt;Because you have bearings at each end of the swing arm which can only take over radial or axial forces and no torques in z-direction. Therefore, you can only push or pull the swing arm and the force transmitted has to be in the same direction (in order to not create torque). The remaining forces have to be taken over by the dampers.&lt;/p&gt;&#xA;" OwnerUserId="13505" LastActivityDate="2017-11-23T10:07:47.153" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18152" PostTypeId="2" ParentId="18149" CreationDate="2017-11-23T14:52:04.870" Score="2" Body="&lt;p&gt;I'm pretty sure it's the same thread as a Presta valve thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is defined by ISO 4570 designation &quot;5v2&quot;, and is similar to &quot;12-24 UNC&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit to add ISO reference)&lt;/p&gt;&#xA;" OwnerUserId="8212" LastEditorUserId="8212" LastEditDate="2017-11-23T14:59:54.217" LastActivityDate="2017-11-23T14:59:54.217" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18153" PostTypeId="1" CreationDate="2017-11-23T15:09:20.743" Score="1" ViewCount="166" Body="&lt;p&gt;I am looking for some material which is resistant to higher temperatures (around 1000 degrees Celsius). So far, I found this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Gypsum plaster board&lt;/li&gt;&#xA;&lt;li&gt;Perlite board&lt;/li&gt;&#xA;&lt;li&gt;Calcium silicate board&lt;/li&gt;&#xA;&lt;li&gt;Sodium silicate board.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Which of this is mostly resistant to damage? Or is there any high temperature resistant material which has good mechanical properties (especially resistant to damage done on edges of boards)?&lt;/p&gt;&#xA;" OwnerUserId="12057" LastEditorUserId="9" LastEditDate="2018-01-24T20:50:18.770" LastActivityDate="2018-07-26T13:37:13.533" Title="Material resistant to higher temperatures and damage" Tags="|materials|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18155" PostTypeId="2" ParentId="18087" CreationDate="2017-11-23T19:02:29.980" Score="1" Body="&lt;p&gt;I would go for a so called series elastic element (SEE). Just make a small guide and slider, attach a compression spring to one end of the guide and to the slider and measure the position of the slider by using a linear potentiometer (super easy to read out with an Arduino, you basically just need GND, 5V and an analog input). If you push the slider, the spring generates force and you can read it out by getting the position of the lin pot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with force gauges most often is that they move only for some micrometers. If you implement a SEE you will combine force and position, making it much easier to use your button (better control for us stupid humans). Furthermore, you have a linear relationship between force and position $F = - k \cdot (x_o - x)$, as well as between position and signal provide by the lin pot. --&gt; Easy calibration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck :)&lt;/p&gt;&#xA;" OwnerUserId="13505" LastActivityDate="2017-11-23T19:02:29.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18156" PostTypeId="2" ParentId="17802" CreationDate="2017-11-23T19:24:00.400" Score="1" Body="&lt;p&gt;It depends a bit on which arc process you compare it to. Broadly speaking there is no fundamental reason why one should be any better or worse than the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the biggest limitation of gas welding is shielding. A correctly adjusted gas flame will protect the joint from oxidation but is certainly not entirely inert which can be a problem with more active metals (especially aluminium) and alloys which are particularly susceptible to contamination. Certainly for welding stainless steel and aluminium MIG and TIG are almost universally preferred. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also gas welding is now much less common in commercial and industrial applications so you are more likely to find up to date standards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to an electric arc (especially TIG) a gas flame is cooler and broader and so the heat affected zone tends to be larger and you end up putting more heat into the work overall. This is more of an issue for some materials than others   &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-11-23T19:24:00.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18157" PostTypeId="1" AcceptedAnswerId="18158" CreationDate="2017-11-23T19:39:41.560" Score="0" ViewCount="33" Body="&lt;p&gt;So I've noticed that computer science has actually been developed so that it has become more human and less technological in the sense that development has raised the abstraction level up to a &quot;more human&quot; level. A modern computer scientist is able to think about computing in very &quot;humanly&quot; ways, rather than as merely machine-like ways. That's because programming languages have become to be able to express concepts more naturally, rather than as some sort of memory address manipulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a similar kind of phenomenon observable in some other fields of engineering?&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2017-11-23T21:39:15.883" Title="Can you perceive similar kind of &quot;humanizing&quot; of technology in other fields of engineering than computer science?" Tags="|computer-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18158" PostTypeId="2" ParentId="18157" CreationDate="2017-11-23T21:39:15.883" Score="3" Body="&lt;p&gt;An important factor is that computing inherently has a numbers of layers of abstraction and function between the end user and the actual functionality. To some extent this is a from of automation in that regularly used actions are packaged into a single simple set of commands. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact that more computing power allows more visually orientated interfaces which, can to a degree hide what is going on not to far under the surface. Equally most people experience computing via software, which to a large extent represents picking form a menu of pre-defined options created by somebody else, even when it doesn't necessarily feel like it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are analogies to this in other forms of engineering. A trivial example is standardisation of something like screw threads. It is rare that anyone would now bother to design a thread from scratch, most of the time you are picking form a list of available options and the same goes for things like bearings, gears and motors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also ways in which the same advances you are talking about in computing have had a direct effect on mechanical and manufacturing engineering. An obvious example is the proliferation of CAD/CAM software. You can now design a part on a computer, send the file to a manufacturer and have it delivered shortly afterwards and (at least in theory) it could be more or less untouched by human hands in between. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process cuts out what would otherwise be a huge amount of labour in terms of producing accurate, dimensioned drawings again it is more a case of automating and streamlining a previously existing process than anything entirely radical. But it does strip away one layer of specialist skill/knowledge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where it does become more radical is when you st art to integrate physical modelling with automated drafting so you can test and optimise parts and assemblies in a way which isn't really possible with physical prototypes, especially when dealing with cost-prohibitive manufacturing processes.     &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-11-23T21:39:15.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18160" PostTypeId="2" ParentId="18087" CreationDate="2017-11-24T13:30:30.130" Score="0" Body="&lt;p&gt;&quot;&lt;strong&gt;Quantum-tunnelling composite&lt;/strong&gt;&quot; is a rubber material impregnated with metallic particles.  Its resistance varies as a function of mechanical pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, put a piece of QTC membrane between a pair of metal plates and squeeze it, the resistance of the QTC will decrease with an increase in pressure.  This allows more or less current to flow from one contact to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stuff works remarkably well, I've used it as the lower half of a resistor voltage divider and you can get very smooth and reproducible results over a wide range of pressure.  You can even use it to form a crude but effective speed controller for (very small) motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case you'd use the ADC of your microcontroller to measure the voltage and convert that into an integer value that quantifies the pressure on the button.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Quantum_tunnelling_composite&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Quantum_tunnelling_composite&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's widely available and not very expensive if you only need a couple of square inches of it.&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastEditorDisplayName="user6335" LastEditDate="2017-11-24T13:38:41.163" LastActivityDate="2017-11-24T13:38:41.163" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18161" PostTypeId="1" AcceptedAnswerId="18162" CreationDate="2017-11-24T16:15:54.433" Score="1" ViewCount="63" Body="&lt;p&gt;I want to repair a dent in a piece of cast iron with brass. I want to fill the site with brass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason, I want to do it by myself. I can buy an instrument for this work in a limited budget(within $200)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dont like to buy a gas welder to be used at my home since the storage is a risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other option for this work?&lt;/p&gt;&#xA;" OwnerUserId="13158" LastActivityDate="2017-11-25T16:31:55.123" Title="Is there any option other than a gas welder" Tags="|gas|welding|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18162" PostTypeId="2" ParentId="18161" CreationDate="2017-11-24T16:43:43.290" Score="2" Body="&lt;p&gt;Brazing is generally a good way to repair cast iron but it tends not to be very easy to fill large gaps. You certainly can braze with a sufficiently powerful venturi burner using propane only but if you want to build up metal you will need the more precise flame that you get from oxy/gas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oxy/propane will certainly work and propane is much less dangerous to store than acetylene. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stick welding can be successful for repairing cast iron, generally nickel electrodes are used, brass electrodes don't exist but you can get bronze ones which should work. However you will probably still need to pre-heat the work to reduce the risk of cracking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another alternative, which is quite attractive if you don't need high temperature resistance is to use a metal filled resin. You can make this up yourself from eg epoxy casting resin and fine brass powder. This has the advantage that you don't have the thermal stresses which are what cause the problems in welding or brazing cast iron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also techniques which involve cutting keyed slots into cast iron and peening in cold metal slugs (usually nickel) but this is more for crack repair and may not be suitable for this application. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-11-25T16:31:55.123" LastActivityDate="2017-11-25T16:31:55.123" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18163" PostTypeId="1" AcceptedAnswerId="18164" CreationDate="2017-11-24T17:18:41.950" Score="0" ViewCount="385" Body="&lt;p&gt;There is a type of screw/ bolt that is flat on top so that it's almost flush when fully seated, but it is &lt;em&gt;not&lt;/em&gt; counter-sunk, and it usually takes a hex driver (allen key). I see it used on Ikea furniture frequently, and the only pic I could find of it was from an Ikea manual:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6ir15.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6ir15.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that it is not counter-sunk so from the side it looks like a 'T'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what the correct mech eng jargon term for this screw would be? E.g. if I'm telling someone to switch from a pan-head to this type of screw? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="13793" LastActivityDate="2020-06-17T04:36:09.150" Title="Correct term for non-countersunk flat hex screw/ bolt?" Tags="|mechanical-engineering|bolting|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18164" PostTypeId="2" ParentId="18163" CreationDate="2017-11-24T18:31:16.057" Score="3" Body="&lt;p&gt;I would call that a wafer head machine screw. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-11-24T18:31:16.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18165" PostTypeId="1" CreationDate="2017-11-24T18:41:10.897" Score="1" ViewCount="162" Body="&lt;p&gt;A hydropower plant uses a river to generate electrical energy. It has a drop of $107m$ and the flow rate of water is $275$ $m^3$/$s$. The flow rate of the water is the same in and out of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was asked to determine whether the system is open or closed. &#xA;Ans: open&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to simplify the general energy balance for the system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q + W = ΔH + ΔE_k + ΔE_p$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is where I am a bit uncertain as to how to simplify it, I think that only $ΔH$ can be removed from the equation since the enthalpy won't change. Is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so if $Q + W =ΔE_k + ΔE_p$ how can I use that to calculate the maximum amount of electrical energy that can be generated? &lt;/p&gt;&#xA;" OwnerUserId="13796" LastActivityDate="2018-02-23T09:44:44.023" Title="Energy balance for a system" Tags="|chemical-engineering|chemistry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18166" PostTypeId="1" CreationDate="2017-11-24T21:14:15.057" Score="2" ViewCount="385" Body="&lt;p&gt;I understand that double-helical gears are used over regular helical gears to balance axial forces in a shaft but, depending on the direction, it seems that these balanced forces can be either compressive or tensile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick Google search shows that many suppliers recommend the compressive configuration over the tensile one when installing these gears, because it is more &quot;stable.&quot; However, most of the websites I have looked at have similar wording when saying this, and I have not yet found any source that is credible (academic textbooks, well-know suppliers, etc.). This makes me suspect that this is either not a very important consideration, or it is not true. From what I understand, if the forces are compressive, they will correct any axial misalignment, but I would think that tensile configuration would also correct the misalignment. In other words, I can't really picture the &quot;unstable&quot; scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, I do not understand how both the pinion and the gear could be both in a &quot;stable&quot; configuration, since any compressive forces in the gear would have to cause a tensile force in the pinion. If one direction is truly &quot;stable,&quot; which of the two (pinion or gear) should be stabilized? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anybody has any credible online sources or intuitive explanations on this stability problem, that would be a great help &lt;/p&gt;&#xA;" OwnerUserId="12823" LastActivityDate="2017-11-26T08:55:04.490" Title="Why do herringbone/double-helical gears have a stable and unstable configuration?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18167" PostTypeId="2" ParentId="18153" CreationDate="2017-11-24T21:24:20.760" Score="0" Body="&lt;p&gt;1000 C is pretty hot. You need steel making and treating materials. One of the few solids that is not heavy is magnesium oxide.It won't tolerate much mechanical abuse. Otherwise the only options are various types of fire brick and castables which I think of as pretty dense .  Of the things you mentioned only cal-sil has a chance .  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-11-24T21:24:20.760" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18168" PostTypeId="2" ParentId="18165" CreationDate="2017-11-24T21:46:06.123" Score="0" Body="&lt;p&gt;What's wrong with P = QgH * density ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P = Power, Q is flow rate, g = gravity, density of the fluid water and H is height...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And most water power engineers take gravity as 10 for a first stab - the error gets &quot;lost&quot; in the efficiency later...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-24T21:46:06.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18169" PostTypeId="1" AcceptedAnswerId="18170" CreationDate="2017-11-25T00:33:15.857" Score="1" ViewCount="899" Body="&lt;p&gt;How is &quot;16.7 million colors&quot; in a display calculated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it a software or a hardware thing (the 16.7 million colors)?&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2017-11-25T00:50:31.880" Title="How is &quot;16.7 million colors&quot; in a display calculated?" Tags="|computer-hardware|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18170" PostTypeId="2" ParentId="18169" CreationDate="2017-11-25T00:50:31.880" Score="3" Body="&lt;p&gt;Each color red, green, blue that makes a pixel has a range of 0 to 255. So each of the colors has 256 possible settings. 256*256*256=16.77 million possible colors.&lt;/p&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2017-11-25T00:50:31.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18173" PostTypeId="1" CreationDate="2017-11-25T06:09:32.580" Score="1" ViewCount="1519" Body="&lt;p&gt;While Surveying, the North is always considered the reference direction, for the purposes of mapping. I have specifically the following issues: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The magnetic compass, has a least count of 30 minutes, and it deeply affected by local disturbances. &lt;/li&gt;&#xA;&lt;li&gt;The pole star, which denotes the astronomical north, is not visible during the day time, while it is the best time to conduct the surveying process. &lt;/li&gt;&#xA;&lt;li&gt;The sun, definitely cannot be taken as the reference object, as it is not fixed during the day, and also suffers from the sun moving from the north to south and backwards during various parts of the year. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Therefore, I would like to know, as to why the North was chosen as the reference direction for Surveying and Mapping. &lt;/p&gt;&#xA;" OwnerUserId="13802" LastActivityDate="2018-01-03T21:29:17.697" Title="Why is North considered the reference direction in Surveying?" Tags="|civil-engineering|surveying|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18174" PostTypeId="2" ParentId="18173" CreationDate="2017-11-25T12:03:50.730" Score="2" Body="&lt;p&gt;Firstly True North is a well defined point which can be determined fairly easily by the pole star and indirectly (but still fairly easily) by a compass. Even if there was no pole star or compass  this is still a fairly logical way to define direction on a rotating globe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also relates to the most sensible way to draw a grid on a globe. Having parallel lines of latitude and converging lines of longitude is as sensible as anything else, allowing for a simple global coordinate system which can easily be translated to a flat local map. If you define true north as the axis of rotation then from that simple definition you can work out the precise point as accurately as you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To put it another way if you draw an arrow on your map which points north it is then very easy for anyone else to determine exactly what direction that is supposed to be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also once you have established the direction of true north on a particular site you don't need to make every measurement from it directly. You can set up a physical marker or calibrate a magnetic compass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you absolutely &lt;strong&gt;can&lt;/strong&gt; use the sun as a reference point as long as you know the accurate local time relative to a reference time (noon at Greenwich being the traditional reference) in fact with the right instruments this can tell you absolute position as well as the direction of north. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to why you use north rather than south, west or east, it doesn't makes any fundamental difference as they are all still defined relative to each other and historically north is the most convenient direction to find by the pole star, a magnetic compass or by observing the track of shadows. Plus if you have a universal convention that the top of a map is north it makes it easier for everyone to use and prevents confusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I'm not sure what else you would use, you can pick any arbitrary direction you want but you still have to define it in terms of something and the same limits of accuracy of direction finding equipment still apply.   &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-11-25T15:32:08.960" LastActivityDate="2017-11-25T15:32:08.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18175" PostTypeId="1" AcceptedAnswerId="18183" CreationDate="2017-11-25T12:47:54.067" Score="0" ViewCount="133" Body="&lt;p&gt;$$\propto  \dfrac {\text {Cross-Sectional Area}}{\text {Volume}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen this equation on my textbook (physics). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us imagine that there are two pen with $7$ and $3$ lenght. Which is harder to break? According to the equation, $3$ is harder to break. Can you explain it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kindest Regards!&lt;/p&gt;&#xA;" OwnerUserId="13576" LastEditorUserId="1832" LastEditDate="2017-11-25T14:59:08.457" LastActivityDate="2017-11-25T20:31:39.443" Title="What does this equation mean?" Tags="|building-physics|geometry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18178" PostTypeId="2" ParentId="18175" CreationDate="2017-11-25T15:24:06.857" Score="4" Body="&lt;p&gt;The $\propto$ symbol represents proportionality. So, whatever is on the left hand side is directly proportional to cross-sectional area and inversely proportional to volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as to the actual relationship you are asking about, that's impossible to confidently answer given how broad your question is. That being said, that relationship is correct for two of the most common failure modes: bending and compression (the third being tension, which isn't a function of element length).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For elements under transverse loads, bending moments are generated which are a function of the element's length (and of many other things). So longer elements suffer greater bending moments under the same load, and can therefore withstand lower loads than (longitudinally) shorter elements. This can be easily displayed with a toothpick: get one and snap it in half; get one of the halfs and repeat; repeat to satisfaction; you'll notice it gets harder the shorter the segment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elements under compressive loads frequently collapse not to due excessive compressive stress, but due to buckling. An element's buckling load is inversely proportional to its length, so (longitudinally) shorter elements under compression can withstand larger loads than longer ones.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-11-25T15:24:06.857" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18179" PostTypeId="2" ParentId="18173" CreationDate="2017-11-25T16:01:04.327" Score="1" Body="&lt;p&gt;The Chinese invented the compass &amp;amp; ancient &lt;a href=&quot;http://www.learnchinesehistory.com/history-chinese-compass/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chinese compasses&lt;/a&gt; pointed south, not north. Because of this, &lt;a href=&quot;http://america.aljazeera.com/opinions/2014/2/maps-cartographycolonialismnortheurocentricglobe.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Chinese used to draw their maps with south at the top&lt;/a&gt;. Arab map makers did the same, possibly because the Chinese did it that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having North at the top became the dominant situation by the beginning of the 16th Century, when Europeans were in a world exploration phase. Map makers of the era, took note of what Ptolemy did in the second century AD/CE. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the knowledge of the ancient Greeks, Ptolemy, who lived in Alexandria, in Egypt, realized &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;he was in the northern half of a very large globe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Previously though, there was a tradition among some Europeans to put East at the top of maps.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-11-25T16:01:04.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18180" PostTypeId="1" CreationDate="2017-11-25T18:39:17.810" Score="0" ViewCount="46" Body="&lt;p&gt;Let's say a cargo ship uses a large amount of fuel while on a voyage. Could you direct the exhaust i.e. Carbon dioxide into a chamber utilising a check valve so the emissions don't backflow into the engine? Allow the gas to travel through a hose towards the front of the ship and release through a gas control valve which regulated that the gas is at a specific pressure. Before releasing it so that it blocked the wind and cleared the ship essentially eliminating or reducing frontal drag?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this method also be utilised on planes and other vehicles? the gas wouldn't weigh much due to heat from just being emitted from an engine so I can't figure out why this isn't used in formula one cars and so forth?&lt;/p&gt;&#xA;" OwnerUserId="7287" LastActivityDate="2017-11-25T19:09:29.067" Title="Is it possible to use a vehicles own carbon emissions to shield it from wind drag?" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|applied-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18181" PostTypeId="2" ParentId="18180" CreationDate="2017-11-25T19:09:29.067" Score="2" Body="&lt;p&gt;No, due to Newton's Third Law of Motion, which basically states that a force applied generates an equal and opposite reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, the exhaust gas could be released at the front of the car and cancel (or at least mitigate) the air drag, but in so doing, the exhaust itself would push the car in the opposite direction. So instead of air drag, you have an equivalent force from the exhaust gasses being propelled out of the car's nose.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-11-25T19:09:29.067" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18183" PostTypeId="2" ParentId="18175" CreationDate="2017-11-25T20:31:39.443" Score="2" Body="&lt;p&gt;I suppose your statement is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $$\text{Cross-Sectional Area}\times \text{Length} = \text{Volume}$$&#xA;then,&#xA;$$x\propto \frac{\text{Cross-Sectional Area}}{\text{Cross-Sectional Area}\times \text{Length} }$$&#xA;$$\therefore x\propto \text{Length}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if $x$ is a measure of how hard something is to break, then, yes, something of three units length will be comparably harder to break than something of seven units length.&lt;/p&gt;&#xA;" OwnerUserId="13673" LastActivityDate="2017-11-25T20:31:39.443" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18184" PostTypeId="2" ParentId="18072" CreationDate="2017-11-26T00:53:11.723" Score="1" Body="&lt;p&gt;Generally, no. Transmission lines can act like antennas, but they operate at around 50 Hz. The wavelength of a 50 Hz EM-wave is $6\times 10^6$ m. In order to effectively couple that energy, you'd have to have a receive antenna of $3\times 10^6 $ m length. That's not feasible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only this, but of course the gain of the antenna scales with frequency and effective area. For low frequencies, even antennas of large effective area tend to be bad radiators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if it COULD feasibly couple with basestation electronics, it just wouldn't happen. A base station's carrier frequency is more likely to interract with its own electronics than a power line's. And, because of this, basestation electronics is very well shielded from most frequencies of radiation, especially RF/HF radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get calculations for all of this stuff in basic antenna design books i.e. Modern Antenna Design by Thomas Milligan.&lt;/p&gt;&#xA;" OwnerUserId="13673" LastEditorUserId="13673" LastEditDate="2017-11-26T02:29:59.993" LastActivityDate="2017-11-26T02:29:59.993" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18186" PostTypeId="2" ParentId="17977" CreationDate="2017-11-26T02:26:25.580" Score="1" Body="&lt;p&gt;It &lt;strong&gt;&lt;em&gt;REALLY&lt;/em&gt;&lt;/strong&gt; isn't true to say &quot;most RF filters have a bandwidth of 1 MHz or higher&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, there is no limit. You could design a multi-order bandpass filter with very high selectivity, such that it 'rejects' all other frequencies. In actually fact, it will have a gradual 'roll off' either side of the designed passband frequency, which will attenuate any frequencies which aren't in the passbad. Frequencies below which the gain is -3 dB, are said to be 'cut-off'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this is written about in frequency design textbooks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one such filter topology exists, Bessel (4th order):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wPD96.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wPD96.png&quot; alt=&quot;filter topology&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That filter has a frequency response that looks like: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Lgvj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Lgvj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that this is a fairly selective filter. The passband region is only 190 Hz wide... Filters which are highly selective are said to have a very high Q factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=\frac{f_c}{\Delta f}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $\Delta f$ is the bandwidth (190 Hz) and $f_c$ is the cut-off frequency, 10 kHz.  &lt;/p&gt;&#xA;" OwnerUserId="13673" LastActivityDate="2017-11-26T02:26:25.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18187" PostTypeId="1" AcceptedAnswerId="18208" CreationDate="2017-11-26T05:40:27.770" Score="1" ViewCount="226" Body="&lt;p&gt;I have to work out the air mass flow coming out each vent on a server and have proposed to use a hot wire anemometer to measure the air velocity at each vent by placing it slightly inside one outlets of the vent (it is a grid pattern), then multiply that by the vented area and air density to find mass flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concern is that I am instructing technicians to do it on site, and I am worried that the measurements may vary significantly depending on where on the vent they are taken from i.e. if they take it from the corner or side of the vent it may be different from the centre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first time trying to measure something like this and was wondering how others in the industry do it. Is there a better way of doing this?&lt;/p&gt;&#xA;" OwnerUserId="13810" LastActivityDate="2018-03-30T18:14:09.667" Title="How to accurately measure mass flow through server vents using an anemometer" Tags="|fluid-mechanics|airflow|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18188" PostTypeId="1" CreationDate="2017-11-26T07:53:07.760" Score="1" ViewCount="576" Body="&lt;p&gt;What are cascaded and parallel PID loop and what are the differences between them. I searched on google and came across few websites but still i don't understand the the two types of control loops ?&lt;/p&gt;&#xA;" OwnerUserId="13811" LastActivityDate="2017-12-12T11:30:49.787" Title="Cascaded PID loop and parallel PID loop" Tags="|electrical-engineering|control-engineering|control-theory|robotics|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18189" PostTypeId="2" ParentId="18166" CreationDate="2017-11-26T08:55:04.490" Score="1" Body="&lt;p&gt;I imagine this may have something to do with compressive forces being favorable to tension forces in respect to fatigue failure. Pulling apart bonds vs smashing them together. &lt;/p&gt;&#xA;" OwnerUserId="11821" LastActivityDate="2017-11-26T08:55:04.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18190" PostTypeId="1" AcceptedAnswerId="18192" CreationDate="2017-11-26T11:17:39.917" Score="4" ViewCount="3889" Body="&lt;p&gt;Most of the real life structures are statically indeterminate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the benefit of designing a statically indeterminate structure instead of a statically determinate one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't mean only beams and trusses.Per example, the shaft of a merchant vessel is considered as a typical statically indeterminate structure.&lt;/p&gt;&#xA;" OwnerUserId="5825" LastEditorUserId="5825" LastEditDate="2017-11-27T05:59:37.703" LastActivityDate="2017-12-29T18:07:31.207" Title="What is the advantage of statically indeterminate structures?" Tags="|mechanical-engineering|structural-engineering|structural-analysis|structures|statics|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18192" PostTypeId="2" ParentId="18190" CreationDate="2017-11-26T15:51:09.400" Score="11" Body="&lt;p&gt;For example, take a look at the following static systems.&#xA;Assume they have the same length and the same (constant) cross-section. Thus an equal allowed bending moment $M_u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/S8qjr.png&quot; width=&quot;300&quot; &gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first system is statically determinate, as it is supported only by simple supports. The maximum moment developing within the beam is $M=\frac{QL}{4}$, thus the load under which the beam fails is $$Q_u=4\frac{M_u}{L}$$&#xA;This coincides with the formation of a plastic hinge at point B, which leaves the systems statically under-determinate, a mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second system, the beam is supported by two clamped supports, which both reduce the maximum moment at the point where the load is applied. If you determine the static forces within the beam you will find that, neglecting residual stresses, the maximum moment $M=\frac{QL}{8}$. Thus,&#xA;$$ Q_u=8\frac{M_u}{L} $$&#xA;To turn into a mechanism, three plastic hinges have to be formed, which requires more work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the second system theoretically can bear twice the load of the first system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statically indeterminate systems in general are more stable and more rigid, but harder to calculate. Another important factor are the above-mentioned residual stresses, resulting amongst others from thermal expansion or the manufacturing process (e.g. inhomogeneous cooling of steel).&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2017-11-26T15:51:09.400" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18194" PostTypeId="1" CreationDate="2017-11-26T18:25:48.943" Score="0" ViewCount="290" Body="&lt;p&gt;In Ansys workbench is it possible to import parameters with part geometry? Eg. if I create part or assembly in Creo or SolidWorks and define parameters for this geometry, can I export parameters with geometry and import and use those parameters in Ansys workbench (structural analysis)? Can you describe process? &#xA;If you answer question please also write which version of ansys and what program can be used in such way.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2017-11-27T20:22:05.797" LastActivityDate="2017-11-27T20:22:05.797" Title="Can Ansys import parameters" Tags="|mechanical-engineering|finite-element-method|ansys|ansys-workbench|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18195" PostTypeId="1" CreationDate="2017-11-26T19:00:11.893" Score="4" ViewCount="183" Body="&lt;p&gt;Can someone please tell you what to start with to construct a Kids Bike Trailer as 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qXymE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qXymE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should select material and geometry. I think I should start sketching with forces (children's weight) and type of load (whether it is tensile, compressive or torsional), then select the appropriate material for it. Am I on the right track?&lt;/p&gt;&#xA;" OwnerUserId="11393" LastEditorUserId="1832" LastEditDate="2017-12-15T02:01:41.523" LastActivityDate="2017-12-19T01:49:11.400" Title="3D construction" Tags="|mechanical-engineering|structural-analysis|structures|statics|industrial-engineering|" AnswerCount="1" CommentCount="0" ClosedDate="2017-12-15T21:21:51.273" ContentLicense="CC BY-SA 3.0" />
  <row Id="18196" PostTypeId="2" ParentId="18187" CreationDate="2017-11-26T19:02:49.773" Score="0" Body="&lt;p&gt;Remote site measurements can be problematic, especially if the technicians have not already done a similar measurement. It sounds like this is a measurement that has not been done previously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the measurement will be sufficient for the purpose, here are some tips for remote-site instructions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;draw a technical sketch, showing where you want the sensor to be placed&lt;/li&gt;&#xA;&lt;li&gt;Submit a procedure text, instructing tools and items to use (power supply, voltage measurement, etc) and cabling parameters (type of cable, length limitation, etc)&lt;/li&gt;&#xA;&lt;li&gt;do the measurement yourself at your location, to see if more info is needed&lt;/li&gt;&#xA;&lt;li&gt;Take and use a photograph to show how to validate the location of the placed sensor (sensor edge to vent edge)&lt;/li&gt;&#xA;&lt;li&gt;make a phone call to check if everything is understood as intended, and to convince the technician that this setup is important for your project, etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Extra note for the experiment:&lt;/em&gt; this seems to be a rather rough method to measure the mass flow. And I am assuming that the measurement will not result in critical decisions (like design change, server replacement, etc). And I guess the output temperature of the vent air would be an important parameter, too. &lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2017-11-26T19:02:49.773" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18197" PostTypeId="1" AcceptedAnswerId="18199" CreationDate="2017-11-26T19:49:26.350" Score="2" ViewCount="414" Body="&lt;p&gt;I am getting confused with the maths in the max distortion strain energy part of the question. Plus there is a misprint as the formula for distortion should be a^2 +b^2-2ab=(yield stress)^2, here they have missed the 2 in 2ab. Or maybe I am wrong&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please solve it correctly showing the working(maths). I just need a step by step working as I am making some error solving it myself. And the one solved here is missing steps.&#xA;*F.S=1 that's why not mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;img src=&quot;https://i.stack.imgur.com/D4soc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13816" LastEditorUserId="13816" LastEditDate="2017-11-26T20:59:58.847" LastActivityDate="2017-11-29T22:29:21.527" Title="Machine design problem" Tags="|mechanical-engineering|design|mathematics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18198" PostTypeId="2" ParentId="18195" CreationDate="2017-11-26T21:01:33.470" Score="3" Body="&lt;p&gt;There are several pages about building kids trailer: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.industrial-newsroom.com/news-detail/t/new-eu-standard-en-159182011-for-bicycle-trailers/&quot; rel=&quot;nofollow noreferrer&quot;&gt;New EU Standard EN 15918:2011 for Bicycle Trailers&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Bicycle_trailer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bicycle trailer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is mentioned EU standard:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A voluntary European standard (EN 15918) for two-wheel bicycle trailers was published in September 2011.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This standard applies to trailers carrying cargo and/or up to 2 passengers, and enforces a maximum gross weight of 60 kg. There are many safety-related requirements, including restraints, structural integrity, stability, small parts, sharp edges, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should check if your country has national regulations because they supercede EU regulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My toughts about design process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make 3d model using software like Solidworks and also run structural anlysis, check stresses and deformations. Maybe use parametric model and later optimize construction when you see stress profile. Also check available materials, aluminium alloys are used often for bikes, light and not as expensive as carbon fiber, shaping and connecting parts may be problem with sophisticated materials. &#xA;Stability should be also your main concern and I would advise getting this EU standard.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="8055" LastEditDate="2017-12-19T01:49:11.400" LastActivityDate="2017-12-19T01:49:11.400" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18199" PostTypeId="2" ParentId="18197" CreationDate="2017-11-26T21:14:03.437" Score="2" Body="&lt;p&gt;Maximum distortion strain energy formula from your example is reduced &lt;a href=&quot;https://www.google.hr/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;url=https://nnf.mit.edu/sites/default/files/documents/sr-2009-2b.pdf&amp;amp;ved=0ahUKEwi_q6-wkt3XAhXLLFAKHZ1GDRkQFgiuATAT&amp;amp;usg=AOvVaw0Bt-cRVhhDf2ZPbKHfkIGt&quot; rel=&quot;nofollow noreferrer&quot;&gt;von Mises&lt;/a&gt; formula,&#xA; so the calculation should be correct. It does not comply the binomial theorem because of specific physical properties of material and stress tensor components relations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very easy ask google question. Though always check wikipedia's answer and try finding some lecture notes or other literature if possible peer reviewed.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2017-11-29T22:29:21.527" LastActivityDate="2017-11-29T22:29:21.527" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18200" PostTypeId="1" CreationDate="2017-11-27T01:55:30.650" Score="0" ViewCount="38" Body="&lt;p&gt;&lt;strong&gt;problem 1-5, James M. Gere's mechanics of materials text book&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the given example : &#xA;&lt;a href=&quot;https://i.stack.imgur.com/IO3IO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IO3IO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a, b and c he used the full 12k value of P to get the bearing and shear values. my logic was to analyse the forces when getting the shear on the pin G. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;my instructure said we were using the max values of P, but that wasn't a satisfactory answer for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;also : &#xA;in a, the value for the bearing area is chosen to be 2dt where d is the diamter as shown. why isnt the bearing area as shown here? (hopefully that's a clear enough aiding diagram )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YgHm6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YgHm6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13823" LastEditorUserId="13823" LastEditDate="2017-11-27T02:02:40.763" LastActivityDate="2017-12-27T20:36:39.187" Title="when is an inclined force analysed?" Tags="|materials|bearings|shear|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18201" PostTypeId="1" CreationDate="2017-11-27T02:12:10.613" Score="1" ViewCount="144" Body="&lt;p&gt;Thank you for taking time to assist me.  I have a rectangular surface that i want to hang from it's 4 corner points.  The corners on two sides could be joined together clipping on to a centre support beam (like in my excel drawing - sorry only software I have available) and then the support beam is hanging from a centre swivel hook which hangs from the ceiling.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how would one achieving horizontal balance if a weight was placed off centre on the horizontal surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't use a vertical counter balance so have tried looking at linear counter balance.  I have found some examples online but have no idea how this would be implemented here.  I have no engineering knowledge so any assistance would be greatly appreciated.  If you could even just tell me what I should be searching for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;img src=&quot;https://i.stack.imgur.com/YfXYC.png&quot; alt=&quot;![Excel drawing&quot;&gt;]&lt;a href=&quot;https://i.stack.imgur.com/YfXYC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13822" LastActivityDate="2018-02-27T14:44:21.650" Title="Balancing a horizontol flat surface when weight applied to one side" Tags="|mechanical-engineering|design|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18202" PostTypeId="1" AcceptedAnswerId="18211" CreationDate="2017-11-27T03:05:02.320" Score="2" ViewCount="89" Body="&lt;p&gt;I am implementing a second order filter.&#xA;It simulates hydraulic servo on an airplane. Its about 35-40 years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Wlurl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wlurl.png&quot; alt=&quot;Second order filter to simulate servo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What confuses me is the notation. Like &quot;2(.7)(52)S&quot;, what does it mean? If its just products, why then not write the resulting number?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried with doing 2704/(S^2+72.8S+2704), but it seem a bit slow.&lt;/p&gt;&#xA;" OwnerUserId="7236" LastActivityDate="2017-11-27T10:02:08.987" Title="What filter is this?" Tags="|control-engineering|aerospace-engineering|simulation|servo|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18203" PostTypeId="1" CreationDate="2017-11-27T06:06:15.407" Score="3" ViewCount="130" Body="&lt;p&gt;What's the best way to precisely measure intricate mechanical parts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, notice the tiny differences between the pitch in these two threads, and the tiny variations in external width of the cylinder:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iA3ru.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iA3ru.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I've been using a micrometer and a vernier caliper. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/hxqti.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hxqti.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/g2ITo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g2ITo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The micrometer is capable of more precise measurements, but it's no good at getting into tight spots. The jaws of the verniers have a sharp, bevelled edge that can fit into some tight spots, but it's less accurate, and hard to know if the jaws are making firm contact to the desired surface, or catching an edge or another surface (like the outer diameter of a thread, for instance).&lt;/p&gt;&#xA;" OwnerUserId="2829" LastActivityDate="2017-12-02T01:19:33.853" Title="Making precise, intricate measurements" Tags="|mechanical-engineering|materials|measurements|threads|tools|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18204" PostTypeId="2" ParentId="8577" CreationDate="2017-11-27T06:09:50.800" Score="0" Body="&lt;p&gt;In a isolated blade, the flow separates in the upper side due to adverse pressure gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the case of turbines, It is well guided flow between the blades. &#xA; So the effect of adjacent blades come into picture. for the same mass flow rate but different passagae dimensions, it alters the pressure on the blades such that the flow don't seperate. but there are cases in which flow seperates for the same solidity (no of blades or thickness of blades) with less mass flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can also refer the below plots (from my own potential flow code.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AOTgV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AOTgV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the above is coefficient of pressure plot for isolated airfoil. &#xA;When we put the same aerofoil as a cascade of aerofoils(NACA  0012), We can see how the pressure distribution changes for different solidity. the outer most curve is for very less no of blades (~isolated) and the inner most profile is for high solidity. So we can't think it as a isolated blade anymore. we have to consider effect of other blades too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4qckV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4qckV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we can't say there is a suction pressure on the upper side of cascade airfoil as compared to isolated. Hope this helps for your first two quastions.&lt;/p&gt;&#xA;" OwnerUserId="13721" LastEditorUserId="13721" LastEditDate="2017-11-27T06:15:00.900" LastActivityDate="2017-11-27T06:15:00.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18205" PostTypeId="1" AcceptedAnswerId="18210" CreationDate="2017-11-27T06:11:20.867" Score="1" ViewCount="158" Body="&lt;p&gt;I'd noticed that for the small motors, the ones used in toy cars, there are rpm values and voltage values written on the specifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have a motor at a range of (12V - 24V) and rpm of (4600 - 9300) respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does this mean that I need to provide 12V to the toy car so that the wheels will rotate at 4600 rpm and 24V to rotate the wheels at 9300 rpm?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like to know does the other way work? I mean if I were to use the motor as a generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does this mean that if I were to provide 4600 rpm, I can generate 12V of voltage?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any response is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="10633" LastActivityDate="2017-11-27T09:52:01.950" Title="Motor for Toy Cars" Tags="|mechanical-engineering|electrical-engineering|motors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18206" PostTypeId="2" ParentId="18203" CreationDate="2017-11-27T06:29:20.490" Score="2" Body="&lt;p&gt;Usually threads are done by some standard. You can use calipers to measure and than find closest in standard, if your product is from eu, its easy you need eu standards like this &#xA;ISO 68-1:1998 or maybe &lt;a href=&quot;https://en.m.wikipedia.org/wiki/British_Standard_Pipe&quot; rel=&quot;nofollow noreferrer&quot;&gt;BS pipe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer would be to measure as accurate as possible but also use literature. Mechanical engineering handbooks like &lt;a href=&quot;http://www.wikiwand.com/en/Bojan_Kraut&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kraut&lt;/a&gt; are great to have at hand to check each machine component. If your measurements are for quality control and you do them all the time in same range, thread etalons  would be great. You can make them using some precise technology and give them to measuring institution or some accredited lab that have 3d optical measurements and if they are accurate use them instead of calipers for quality control. Optical 3d measuring device is also great piece that is used when high accuracy is needed.&#xA;Also some special measuring devices are available &#xA;&lt;a href=&quot;http://di.com/micrometers-c11/screw&quot; rel=&quot;nofollow noreferrer&quot;&gt;Micrometer&lt;/a&gt; or &lt;a href=&quot;https://www.google.hr/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;url=http://grafker.hu/wp-content/uploads/2013/09/C_2010_EN_XB_Wires.pdf&amp;amp;ved=0ahUKEwiSiIOQkd7XAhXIJOwKHbxhA4sQFgiJATAT&amp;amp;usg=AOvVaw3TeaCt17LR4N0nb6QnnxmZ&quot; rel=&quot;nofollow noreferrer&quot;&gt;wires&lt;/a&gt;.&#xA;Depending on your industry needs you should upgrade your measuring equipment.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2017-11-27T07:53:09.600" LastActivityDate="2017-11-27T07:53:09.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18207" PostTypeId="1" AcceptedAnswerId="18219" CreationDate="2017-11-27T08:01:21.203" Score="3" ViewCount="686" Body="&lt;p&gt;Diesel engines are considered &quot;steady torque&quot;, which means that torque is same no matter the revolutions whereas torque depends on revolutions for an engine that runs on Otto cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for this?&lt;/p&gt;&#xA;" OwnerUserId="5825" LastActivityDate="2017-11-28T05:17:34.980" Title="Why diesel engines are considered steady torque engines?" Tags="|mechanical-engineering|automotive-engineering|torque|diesel|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18208" PostTypeId="2" ParentId="18187" CreationDate="2017-11-27T08:34:26.247" Score="0" Body="&lt;p&gt;The Hot-wire anemometer useful for accurate turbulence measurements. Before using it, one has to calibrate the hot-wire with a fine Pitot tube for the range of velocity interested in. In which case you might require local laboratory/workplace temperature compensation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, unless you need a turbulence-intensity/spatial or temporal correlations, you can go for a well designed pitot tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the measurements have to be carried out in the entire outlet region with reasonable spatial resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can calculate the mass flow rate using the below formula (assuming data taken in a 2D-plane )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot m =   \bar{\rho(T)}\int\int u(x,y) dx dy = \bar{\rho(T)} \sum_{i=1}^{n_x}\sum_{j=1}^{n_y} u(i,j) \Delta x_{i,j} \Delta y_{i,j}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, $\bar{\rho(T)}  $ is density of air compensated for temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps you. &lt;/p&gt;&#xA;" OwnerUserId="13721" LastEditorUserId="13721" LastEditDate="2017-11-28T04:10:38.453" LastActivityDate="2017-11-28T04:10:38.453" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18209" PostTypeId="2" ParentId="18207" CreationDate="2017-11-27T08:40:24.007" Score="3" Body="&lt;p&gt;Compared to Otto cycle engines, diesel engines have a smaller rpm range and, as they tend to have more mass (due to the higher compression ratio etc), tend to develop the torque earlier in the rev band and it tends to be a flatter curve through the range.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-27T08:40:24.007" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18210" PostTypeId="2" ParentId="18205" CreationDate="2017-11-27T09:44:38.980" Score="4" Body="&lt;p&gt;It means that if you were to apply 12V to the motor terminals, and the motor was not attached to the car in any way (i.e. no resistance to the shaft turning at all), then it would spin at approximately 4600 ±tolerance (and likewise ≈ 9300rpm @ 24v). As soon as you add in any resistance, then the speed will drop as the torque required to keep the motor moving increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QuZng.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QuZng.gif&quot; alt=&quot;Typical PMDC Motor Performance&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, it's likely that there's some sort of gear-train between the motor output shaft and the wheels, which will reduce the speed of the wheels compared to the numbers printed on the motor even further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You won't get 12V back for 4600rpm input, due to the efficiency of the motor. Note also that peak efficiency is not measured at 'free speed'. Neglecting that for an example, however, and assuming the motor is 80% efficient in both directions, this means 4600rpm corresponds to 9.6V of ideal input, and converting back, 4600rpm could only create ~7.7V.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2017-11-27T09:52:01.950" LastActivityDate="2017-11-27T09:52:01.950" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18211" PostTypeId="2" ParentId="18202" CreationDate="2017-11-27T10:02:08.987" Score="8" Body="&lt;p&gt;It might be written that way because the &lt;a href=&quot;http://www.kves.uniza.sk/kvesnew/dokumenty/DREP/Filters/2nd%20order%20transfer%20function.pdf&quot; rel=&quot;noreferrer&quot;&gt;transfer function&lt;/a&gt; of a second order system is written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;h(s) = \frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\zeta$ is the relative damping ratio and $\omega_n$ is the natural frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in your system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\zeta &amp;amp;= 0.7 \\&#xA;\omega_n &amp;amp;= 52&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's easy to see what the parameters are when it's written with $\zeta$ and $\omega_n$ factored out.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2017-11-27T10:02:08.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18213" PostTypeId="2" ParentId="18200" CreationDate="2017-11-27T20:20:06.383" Score="1" Body="&lt;p&gt;Force is normal to plane in which cross section area is d*t on both sides thus 2dt. It is full force because steel strut caries only axial force and full force is transmitted to the pin and force is normal to this cross section plane. This is similar to calculation of stress in cylindrical vessel where the area used to calculate stress is the cross section of the vessel.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2017-11-27T20:31:46.947" LastActivityDate="2017-11-27T20:31:46.947" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18214" PostTypeId="1" CreationDate="2017-11-27T21:03:34.390" Score="1" ViewCount="71" Body="&lt;p&gt;First of all I am just a starter in Matlab, so the program is not optimized. For now I just want to get it running. This is a 3D heat problem with boundary conditions. Fixed temperature from the north, insulated from the bottom, and we have convection on the 4 surrounding sides with the same h. when I run it it does not give me the right convergences. I have the analytical and the solidworks approximations, so I know the results I get in Matlab are wrong. Could someone advise me please. Your help is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;==============================&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%===Solution of the problem of 2D in the bookmarks:Case2a=====&#xA;clc;&#xA;close all;&#xA;clear all;&#xA;&#xA;DX=0.01; % step size&#xA;DY=0.01;&#xA;DZ=0.01;&#xA;Lx= 0.6; %length along x-axis in cmn&#xA;Ly=0.4;   %length along y-axis&#xA;Lz=0.3;   %length along z-axis&#xA;Nx=Lx/DX+1;    %Number of nodes along x-axis&#xA;Nx=floor(Nx)+1; % I shouldn't add 1. I should get it automatically &#xA;Ny=(Ly/DY+1);    %Number of nodes along y-axis&#xA;Ny=floor(Ny);&#xA;Nz=(Lz/DZ+1);&#xA;Nz=floor(Nz)+1;&#xA;&#xA;K=2;&#xA;h=500;&#xA;T_inf=20;&#xA;&#xA;X=0:DX:Lx;&#xA;Y=0:DY:Ly;&#xA;Z=0:DZ:Lz;&#xA;&#xA;T=zeros(Nx,Ny,Nz);                                                            &#xA;&#xA;% I added imaginary nodes outside then it gave me numbers&#xA;&#xA;tol=1;&#xA;s=0; %s=2000 could be set as the maximum number of allowed iteration for &#xA;example&#xA;err=1; &#xA;T_old=1;&#xA;&#xA;while err &amp;gt;=tol %&amp;amp;&amp;amp; s&amp;lt;2001&#xA;    s=s+1;&#xA;&#xA;%------------boundary conditions------------------------------------------&#xA;&#xA;%====North temperature (Boundary)&#xA;T(:,Ny,:)=90;  &#xA;%===South Boundary &quot;insulation&quot; ==============&#xA;for k=1:Nz&#xA;    for i=2:Nx-1&#xA;   T(i,1,k)=0.25*[2*T(i,2,k)+T(i-1,1,k)+T(i+1,1,k)];&#xA;  %T(i-1,1,k)=4* T(i,1,k)-2*T(i,2,k)-T(i+1,1,k);&#xA;    end&#xA;end&#xA;%================Back Boundary &quot;Convection&quot; ================&#xA;for j=2:Ny-1 %changed&#xA;    for i=2:Nx-1&#xA;  T(i,j,1)=0.5*K/(h*DX+2*K)*[2*T(i,j,2)+T(i-1,j,1)+T(i+1,j,1)+2*h*DX*T_inf/K];&#xA;    end&#xA;end&#xA;%================Front Boundary &quot;Convection&quot; ================&#xA;for j=2:Ny-1&#xA;    for i=2:Nx-1&#xA;  T(i,j,Nz)=0.5*K/(h*DX+2*K)*[2*T(i,j,Nz-1)+T(i-1,j,Nz-1)+T(i+1,j,Nz-1)+2*h*DX*T_inf/K];&#xA;    end&#xA;end&#xA;&#xA;%================West Boundary &quot;Convection&quot; ================&#xA;for j=2:Ny-1&#xA;for k=2:Nz-1&#xA;  T(1,j,k)=0.5*K/(h*DX+2*K)*[2*T(2,j,k)+T(1,j,k-1)+T(1,j,k+1)+2*h*DX*T_inf/K];&#xA;    end&#xA;end&#xA;%================East Boundary &quot;Convection&quot; ================&#xA;for j=2:Ny-1&#xA;    for k=2:Nz-1&#xA;  T(Nx,j,k)=0.5*K/(h*DX+2*K)*[2*T(Nx-1,j,k)+T(Nx,j,k-1)+T(Nx,j,k+1)+2*h*DX*T_inf/K];&#xA;    end&#xA;end&#xA;&#xA;&#xA;&#xA;% The interior nodes&#xA;for k = 2:Nz-1 % I think it needs to go loop by loop(pairs of loops)&#xA;    for j = 2:Ny-1&#xA;        for i=2:Nx-1&#xA;    T(i,j,k)=0.25*(T(i-1,j)+T(i+1,j)+T(i,j-1)+T(i,j+1));&#xA;        end&#xA;    end&#xA;end&#xA;err=max(max(abs(T-T_old)));&#xA;T_old=T;&#xA;end&#xA;&#xA;T=rot90(T)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="13830" LastActivityDate="2017-11-27T21:03:34.390" Title="Temperature distribution" Tags="|matlab|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18215" PostTypeId="1" CreationDate="2017-11-27T21:53:48.710" Score="3" ViewCount="575" Body="&lt;p&gt;I'm researching iris mechanisms with the goal of building a functioning, weather-sealed door. I am learning much as I go along, but am faced with a challenge in nomenclature (I am not an engineer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using Google to search for non-overlapping iris mechanism designs that have the smallest possible distance from the outer edge of the opening to the outer edge of the 'housing' (the word that naturally comes to me to describe that part of the device) - but I am stumped about how to narrow down my search results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the terms &quot;small housing&quot;, &quot;narrow housing&quot;, &quot;smallest perimeter&quot; etc are not proving to be useful when searching. No meaningful changes in the search results (and there are a gazillion results - I didn't know this mechanism was so popular in the DIY home-cnc world). &quot;Enclosure&quot; and &quot;concealment&quot; do not seem to impact the results much either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone advise me on &lt;strong&gt;&lt;em&gt;what terms I should use that will enable me to narrow my results&lt;/em&gt;&lt;/strong&gt;? &amp;lt;--- THAT is the question being asked.&lt;/p&gt;&#xA;" OwnerUserId="7381" LastEditorUserId="7381" LastEditDate="2017-12-01T23:11:21.827" LastActivityDate="2024-03-31T02:01:40.220" Title="Iris mechanism nomenclature - need help to enable better search results" Tags="|mechanical-engineering|nomenclature|research|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18216" PostTypeId="1" CreationDate="2017-11-27T23:17:06.777" Score="3" ViewCount="294" Body="&lt;p&gt;Although it's different in each state, most highway bridge rails used to contain either poles or elevated concrete slabs. Very few were just solid concrete. Then, in the mid to late 1970s (some places a little earlier), solid concrete walls resembling Jersey barriers took over in practically every state as the predominant bridge rails. Why? Was there some study that came out around this time that determined these to be safer? &lt;/p&gt;&#xA;" OwnerUserId="12615" LastActivityDate="2018-01-08T13:12:15.043" Title="Why did Jersey barrier bridge rails become the dominant type of highway bridge rail?" Tags="|civil-engineering|transportation|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18217" PostTypeId="1" CreationDate="2017-11-28T00:57:02.203" Score="0" ViewCount="151" Body="&lt;p&gt;I'm working on a project which utilizes an iPad to change colors of individual lights in array of ~200 RGB LEDs. I have little knowledge of large-scale LED configurations and could use some direction. All lights will be on, it's only a matter of programmability to  assign a color to a certain LED in the array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a controller that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can handle connections for iPad use over WiFi (Raspberry Pi-esque... Python is ideal)&lt;/li&gt;&#xA;&lt;li&gt;Facilitates color-changing of individual LEDs in the array.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A design question based on the above requirements: I assume there are no LED strips where diodes on the strip may be individually controlled; am I wrong in assuming this? If not, what is the optimal wiring option?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a look at the Pixelpusher by Heroic Robotics, but that board seems overkill for what I'm trying to accomplish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm well-versed in using the Raspberry Pi for the situation of remotely-controlling a few LEDs, but I don't think it has enough power to handle the 200 diodes I need. If I'm missing some configuration details that would make this possible, please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="13831" LastEditorUserId="1832" LastEditDate="2017-11-28T10:03:21.033" LastActivityDate="2018-01-27T12:17:10.477" Title="LED controller to control individual RGB diodes remotely?" Tags="|electrical-engineering|lighting|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18218" PostTypeId="2" ParentId="18207" CreationDate="2017-11-28T01:44:34.963" Score="2" Body="&lt;p&gt;The higher torque comes from a relatively longer stroke, which is needed to develop the higher compression for a diesel. The longer stroke gives a longer &quot;arm&quot; for the piston to push on the crankshaft. Or , think of the longer crankshaft throws; distance from crankshaft center-line to the throw center-line. The affect is more apparent when the throws are in the 3 O'clock and 9 O'clock positions. Diesels have an inherently narrower power range than a gasoline engine, that may give the appearance of a more steady torque output. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-11-28T01:44:34.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18219" PostTypeId="2" ParentId="18207" CreationDate="2017-11-28T05:17:34.980" Score="2" Body="&lt;p&gt;The diesel engines are working on diesel cycle. The heat addition(ingnition) is taking place during the constant pressure process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After added energy is spent by the expansion of piston.&#xA;The amount of expansion is fixed in terms of stroke length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for given fixed pressure and fixed expansion we ought to get constant work output. That's why the torque is relatively fixed in diesel engines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of otto cycle/petrol/gasoline engines the heat is added during constant volume process. So you can rise the pressure as much you can for the same expansion by adding more heat. Hence you can get quick pickup/high torque in petrol/gasoline engines by adding more fuel while starting the engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2017-11-28T05:17:34.980" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18220" PostTypeId="2" ParentId="17851" CreationDate="2017-11-28T09:10:14.403" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wJWje.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wJWje.png&quot; alt=&quot;static thermodynamic varibales before and after shock&quot;&gt;&lt;/a&gt;The text book answer is correct. &#xA;I think you missed to calculate the density of the downstream of the shock.&#xA;the equation you entered is valid only in the region 1 or in region 2. means the isentropic relations are not valid across the shock. Across the shock you have to use normal shock relations. Then you would get the right answer&lt;/p&gt;&#xA;" OwnerUserId="13721" LastEditorUserId="13721" LastEditDate="2017-11-28T10:03:51.500" LastActivityDate="2017-11-28T10:03:51.500" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18222" PostTypeId="1" CreationDate="2017-11-28T09:48:26.410" Score="2" ViewCount="43" Body="&lt;p&gt;&lt;a href=&quot;https://mathworks.com/help/mpc/ug/terminal-weights-and-constraints.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Terminal Penalty&lt;/a&gt; is used to ensure stability of Model Predictive Control (MPC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the initial cost function is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J(u)=\Sigma_{i=1}^{\infty}[x(k+i)^TQx(k+i)+u(k+i−1)^TRu(k+i−1)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is modified to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J(u)=\Sigma_{i=1}^{p−1}[x(k+i)^TQx(k+i)+u(k+i−1)^TRu(k+i−1)]+x(k+p)^TQ_px(k+p)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terminal penalty matrix $Q_p$ is calculated from solving a Riccati equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_p=A^TQ_pA−A^TQ_pB(B^TQ_pB+R)^{−1}B^TQ_pA+Q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering what if a matrix is used which has higher magnitude members? Does it still guarantee the stability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if $\beta&amp;gt;1$, can I replace $Q_p$ with $\beta Q_p$ yet ensuring stability?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or if $(Q'_p-Q_p)$ is positive definite, can I replace $Q_p$ with $Q'_p$ yet ensuring stability?&lt;/p&gt;&#xA;" OwnerUserId="5794" LastActivityDate="2018-03-03T20:34:38.650" Title="Can I replace terminal weight with an even stricter matrix yet ensuring stability?" Tags="|control-theory|stability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18223" PostTypeId="2" ParentId="18217" CreationDate="2017-11-28T10:49:13.120" Score="1" Body="&lt;p&gt;I use leds with the chip ws2812 aka neopixel and are very easy to control with Arduino. With Raspberry Pi is a bit coplex due to timing requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With neopixel every led may be individually controlled with only one pin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can find the led strip: &lt;a href=&quot;https://www.adafruit.com/product/1376&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.adafruit.com/product/1376&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a Christmas decoration with neopixel and Arduino Yun controled form arduino embedded web page opened from my iPad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The code: &lt;a href=&quot;https://github.com/jecrespo/NeoPixel&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/jecrespo/NeoPixel&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13758" LastActivityDate="2017-11-28T10:49:13.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18224" PostTypeId="2" ParentId="16386" CreationDate="2017-11-28T11:15:27.653" Score="0" Body="&lt;p&gt;I would propose another approach, let's calculate the speed at which a vehicle should go while the engine is at $1000RPM$ so that torque at the wheel is equal  to the torque at the engine. This depends only on the diameter of the wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical car tire have a radius of is $25cm$. The circumference is around $PI \over 2$ meters. So when the engine is at $1000RPM$, in order for the wheels to rotate at the same speed and thus having the same torque, the car speed will have to be $1570m/min$ or $94Km/h$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since usually a motorcycle have bigger wheels, and also low torque, fast engines, it seems highly unlikely than torque at the wheel will ever be lower than that of the engine, even with the taller gear. &lt;/p&gt;&#xA;" OwnerUserId="13837" LastEditorUserId="13837" LastEditDate="2017-11-28T12:08:46.057" LastActivityDate="2017-11-28T12:08:46.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18226" PostTypeId="2" ParentId="15997" CreationDate="2017-11-28T12:52:22.327" Score="2" Body="&lt;p&gt;Maximum speed is affected by air drag, rolling resistance and power of the motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force pushing the skate is simple to calculate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_a = k_m{ P \over v }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $P$ is the power output of the motor (including transmission losses) at the current skate speed $v$. With the right transmission you will achieve peak power at maximum speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air drag force can be approximated to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_d = k_dv^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $k_d$ is the drag constant which you need to know, this depends on your body area, your shape, your height, density of the air and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there is the wheels friction force which is highly variable and depends on many things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_f = f(v,m,typeOfTerrain) ??$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friction depends on your weight, the current speed, type and diameter of wheels, type of terrain, etc... It is a very complex phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate max speed you will have to write the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_a = f_d + f_f$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And solve for v. When the pushing force of the engine is equal to the losses you are at maximum speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is pretty much impossible without having a pretty good friction model for your skate wheels and also a given &quot;standard&quot; surface.&#xA;Imagine trying to move with your skate on sand, you will be much slower than on a perfectly flat concrete surface...&lt;/p&gt;&#xA;" OwnerUserId="13837" LastActivityDate="2017-11-28T12:52:22.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18227" PostTypeId="1" CreationDate="2017-11-28T15:02:27.943" Score="0" ViewCount="641" Body="&lt;p&gt;I've got a damascus steel knife, which I accidentaly broke the wooden handle. I teared it apart and all I was left was a bare blade. I've created a new wooden handle, but while making it, I've also brushed the blade a bit, while fine sanding the wooden handles while attached to the blade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any option I could re-etch the blade again, and if so - what acid should I use (I've heard for Ferric III Chloride (Fe3Cl)) but I don't know the concentration neither the time or temperature of etching or the whole procedure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how do I etch a damascus steel knife?&lt;/p&gt;&#xA;" OwnerUserId="5797" LastActivityDate="2018-02-26T23:41:32.830" Title="Damascus steel etching" Tags="|steel|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18229" PostTypeId="2" ParentId="18201" CreationDate="2017-11-28T16:49:05.060" Score="1" Body="&lt;p&gt;Instead of developing a complex mechanism to allow the surface to balance, you could support it by a rigid connection. Any deviation from horizontal will be opposed by the bending moment in the support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use two bearings at either end of a tube (much like the headset on a pushbike) in order to allow your support to rotate, even when a bending moment is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine solidly fixing a bike frame to the ceiling, and then joining your Red Crossbar rigidly to the forks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0qyXH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0qyXH.jpg&quot; alt=&quot;Rough Diagram of Rigid Ceiling Support Allowing Rotation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the rough diagram above, the red head-tube is firmly fixed to the ceiling. The black forks are rigidly joined to the black crossbar, and the blue surface hangs from the ends of the crossbar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This provides a single (and circular) fixing point to the ceiling (as you correctly stated is a requirement, but allows the system to remain rigid with respect to the off-centre mass on the blue surface.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2017-11-29T12:59:54.340" LastActivityDate="2017-11-29T12:59:54.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18230" PostTypeId="1" CreationDate="2017-11-28T18:21:58.467" Score="2" ViewCount="1344" Body="&lt;p&gt;So i have a little diagram to visualize:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ThCSM.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets say i have a 20 mm plate connected with two M8 bolts, and i apply a 30kN force in the direction shown in picture. Question is - would the force be equally distributed between bolts - i.e., each bolt will have a 15kN of shear stress? Or each bolt will still take full stress of 30kN?&lt;/p&gt;&#xA;" OwnerUserId="13844" LastEditorUserId="1832" LastEditDate="2017-11-28T22:14:23.887" LastActivityDate="2017-11-28T22:25:55.160" Title="Difference in shear stress - one bolt vs two" Tags="|structural-analysis|statics|shear|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18231" PostTypeId="2" ParentId="18230" CreationDate="2017-11-28T18:26:05.667" Score="0" Body="&lt;p&gt;Are the holes / bolts a machined fit? If not then one is most likely taking more stress than the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when engineers do this type of analysis they take the load and divide it equally between the bolts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar result is obtained when you have one bolt and three plates ie the middle plate is sandwiched...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-28T18:26:05.667" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18232" PostTypeId="1" AcceptedAnswerId="18239" CreationDate="2017-11-28T18:43:28.757" Score="0" ViewCount="625" Body="&lt;p&gt;We define the durability with the equation as seen below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x\propto \frac{\text{Cross-Sectional Area}}{\text{Cross-Sectional Area}\times \text{Length} }$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\therefore x\propto \text{Length}^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation was only for matters which has same density. Let us imagine that there is a tree with $3\text h$ and an iron pole with $3\text h$. Now we see that our equation doesn't work in this condition. The iron pole will be more durable than the tree even if they have same lenght. Then, how do we define the durability as per density? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to our equation, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text {Tree} = \frac {1}{3h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text {Iron Pole} = \frac {1}{3h}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like their durabilities are same. However, Iron pole has greater density than tree. By the way, Iron pole is more durable. How do you explain this? or Is there any equation that we can use when there are two matter which doesn't have same density?&lt;/p&gt;&#xA;" OwnerUserId="13576" LastEditorUserId="1832" LastEditDate="2017-11-29T11:56:14.660" LastActivityDate="2017-11-29T11:56:14.660" Title="How do we define the durability as per density?" Tags="|building-physics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18233" PostTypeId="1" AcceptedAnswerId="18250" CreationDate="2017-11-28T19:26:10.770" Score="0" ViewCount="243" Body="&lt;p&gt;I am having a problem with transferring the heat flux boundary conditions into a temperature to be able to put it into a matrix. I understand that &#xA;deltat = deltax*q''/k but I do not know how to code it so that I can loop it into the matrix in MATLAB. I have not had heat transfer and it is a steady state problem, so it should be relatively simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could you index code to be able to loop through since deltat = T(i,j) - T(I+1,J)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/seZCR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/seZCR.png&quot; alt=&quot;code&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/E3KgP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E3KgP.png&quot; alt=&quot;Schematic for Problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any help!&lt;/p&gt;&#xA;" OwnerUserId="13487" LastActivityDate="2017-11-29T12:28:37.130" Title="Laplace Equation -- Heated Plate -- Heat Flux Boundary Condition" Tags="|heat-transfer|mathematics|matlab|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18234" PostTypeId="2" ParentId="18230" CreationDate="2017-11-28T20:15:01.613" Score="3" Body="&lt;p&gt;Assuming a material like cold rolled steel: The steel will yield and deform slightly at small points of very high stress , equalizing the load/stress. So the bolts will have very similar loads. This plastic deformation and cold work is what makes steel so useful or &quot;forgiving&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-11-28T20:15:01.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18235" PostTypeId="2" ParentId="18227" CreationDate="2017-11-28T20:29:34.870" Score="0" Body="&lt;p&gt;After removing scratches with silicon carbide paper ( 320 &gt; 400 &gt; 600 , etc) try Nital etch. It is the standard for etching for steel metallographic examination . As I remember there is 2 and 5 % nitric acid in methanol. Cotton swab , room temperature , a few seconds , rinse, dry. Ferric chloride is aggressive , I guess it could be used.   &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-11-28T20:29:34.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18236" PostTypeId="1" CreationDate="2017-11-28T21:52:48.447" Score="1" ViewCount="30" Body="&lt;p&gt;An elderly friend of mind whose eyesight is fading  recently asked me if there was a way to illuminate her new induction stove top so that the &quot;burner rings&quot; would be easier to see, to her it just looks like a big black non descript surface. I was thinking maybe a blacklight would illuminate the little square marks embedded in the glass that define the pan rings or locations. I suppose I could just buy one and try it but to save time and money I thought I would ask the experts here to see if anyone had any experience with black light being used to make these kinds of markings pop out visually.&lt;/p&gt;&#xA;" OwnerUserId="13848" LastActivityDate="2017-11-28T21:52:48.447" Title="Illumination properties of black lights" Tags="|photonics|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18237" PostTypeId="2" ParentId="18227" CreationDate="2017-11-28T22:12:07.517" Score="1" Body="&lt;p&gt;From Damascus type steels you only need quite a gentle etch normally a you are really just exposing the crystal structure and selectively oxidising the metal rather than aiming to remove much material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ferric chloride should work fine, the concentrations isn't critical, you are aiming for a fairly dilute solution, orange fanta sort of colour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;White vinegar should also work and is less of a problem if you have already permanently attached the handle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may need to give the blade a polish to get rid of any scratches and existing oxides form the surface and you definitely need to make sure it is thoroughly clean &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-11-28T22:12:07.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18238" PostTypeId="2" ParentId="18230" CreationDate="2017-11-28T22:25:55.160" Score="3" Body="&lt;p&gt;As others have mentioned, the load will be distributed between the bolts. How they will be distributed is more complicated, but it is usually assumed to be an equal division between the bolts (in this case), as &lt;a href=&quot;https://engineering.stackexchange.com/a/18231/1832&quot;&gt;SolarMike mentioned&lt;/a&gt;. This assumption is made easier by the cold work &lt;a href=&quot;https://engineering.stackexchange.com/a/18234/1832&quot;&gt;explained by blacksmith37&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an analytical standpoint, all that can be inferred is that the sum of the forces acting on the bolts must be equal to the applied force of 30&amp;nbsp;kN. This equality must be satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why, don't look at the bolts, but at the plate. It suffers the effects of three different forces: the applied 30&amp;nbsp;kN force at the bottom (let's call it $P$), and the reactions of each of the two bolts (let's call them $R_A$ and $R_B$). If the sum of all of these forces (let's call it $F$) were not zero&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = P + R_A + R_B \neq 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then that would imply that there is a net force applied to the plate. And as we know from &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton's Second Law of Motion&lt;/a&gt;, force equals mass times acceleration. If there's a non-null net force applied to the plate, then that means that the plate will need to be accelerating!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if the applied load is too large and the bolts snap, then the applied load will no longer be balanced, there will indeed be a net force applied to the plate, and the plate will indeed accelerate in the direction of the force.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-11-28T22:25:55.160" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18239" PostTypeId="2" ParentId="18232" CreationDate="2017-11-28T22:47:10.377" Score="2" Body="&lt;p&gt;This is clearly building on &lt;a href=&quot;https://engineering.stackexchange.com/q/18175/1832&quot;&gt;this question&lt;/a&gt;, and specifically its accepted answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to toot my own horn, but I recommend you also read my answer to that question. More specifically, the first paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's important to notice here is that the equation given above isn't describing an equation, but a relationship of &lt;em&gt;&lt;strong&gt;proportionality&lt;/strong&gt;&lt;/em&gt;. When you say that &quot;durability&quot; (an unclear term) $x \propto 1/L$, you are stating that durability is inversely proportional to length. As I stated in my answer to your other question, that is correct for most common failure states.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this statement makes no attempt at being a complete description of the variables that determine an element's &quot;durability&quot;. It merely states that one of the variables that goes into &quot;durability&quot; is the inverse of the element's length. There may be (and indeed are) other variables that also influence &quot;durability&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abandoning that useless term and taking the example of a column under uniaxial load, we can calculate that the buckling load is &lt;strong&gt;&lt;em&gt;equal&lt;/em&gt;&lt;/strong&gt; to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_E = \frac{\pi^2 EI}{(KL)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't bother describing what the variables are because that's immaterial to this question. What's important to note here is that this is an &lt;strong&gt;&lt;em&gt;equation&lt;/em&gt;&lt;/strong&gt;, as described by the use of the equality symbol ($=$) instead of the proportionality symbol ($\propto$). An equation attempts to be a complete description of the dependent variable (in this case $P_E$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So long as you accept the assumptions that go into deriving this equation, then $P_E$ is exactly and exclusively equal to $\dfrac{\pi^2 EI}{(KL)^2}$. That is a true, undeniable fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, here are a few other true, undeniable facts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$P_E \propto E$&lt;/li&gt;&#xA;&lt;li&gt;$P_E \propto I$&lt;/li&gt;&#xA;&lt;li&gt;$P_E \propto 1/K^2$&lt;/li&gt;&#xA;&lt;li&gt;$P_E \propto 1/L^2$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That is, $P_E$ is directly proportional to $E$. If you double $E$, you double $P_E$. There are other variables that are relevant to actually determine the value of $P_E$, sure, but there's no denying that a greater $E$ implies a greater $P_E$. The same applies to all of the other statements of proportionality above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's what's wrong with your case. You are looking at a statement of proportionality describing that &quot;durability&quot; is proportional to the inverse of the element's length and assuming this states that there are no other relevant variables. That is not true. All that statement is saying is that an increase in length reduces &quot;durability&quot;. It makes no assertions as to the existence (or lack thereof) of other variables which may also influence the &quot;durability&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-11-28T22:47:10.377" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18240" PostTypeId="1" CreationDate="2017-11-28T22:52:58.577" Score="3" ViewCount="237" Body="&lt;p&gt;I would like to begin a case study in structural design using bamboo, which has currently become available to me. The design instance I would like to structurally copy is the &lt;a href=&quot;https://www.roguefitness.com/rogue-rm-6-bolt-together-monster-rack-2-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rogue RM-6 &quot;Monster Rack&quot;&lt;/a&gt; and a simple bench. Knowing that bamboo is used in scaffolding there must be details on structural reinforcement when binding stems together to make a structure. I would like to apply these kinds of principles to this rack design making necessary modifications where compromise is necessary. I do not know a good source to learn how make such structural binding techniques, so I thought to ask here if anyone here knows of one.&lt;/p&gt;&#xA;" OwnerDisplayName="user13851" LastEditorDisplayName="user13851" LastEditDate="2017-11-28T22:58:59.700" LastActivityDate="2018-04-17T11:55:50.650" Title="How to structurally bind bamboo stems for loads?" Tags="|structural-engineering|civil-engineering|structures|wood|reinforcement|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="18241" PostTypeId="1" CreationDate="2017-11-28T23:36:30.140" Score="0" ViewCount="37" Body="&lt;p&gt;I'm throwing a party with themed drinks, and one is a shot topped with 151, and set on fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll need to serve about 15 at a time, and don't own that many shot glasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any disposable drinking glass that can be safely held at ~130 f (the only actual number I found online, but probably not too precise) for less than a minute?  What plastic number (the recycling number) should I look for?&lt;/p&gt;&#xA;" OwnerUserId="6488" LastActivityDate="2017-11-29T13:26:34.093" Title="consumer plastics that can take a bit of heat" Tags="|materials|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18243" PostTypeId="2" ParentId="18203" CreationDate="2017-11-28T23:48:23.783" Score="4" Body="&lt;p&gt;A relatively inexpensive and quite useful device would be a &quot;pocket optical comparator&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k7tZQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k7tZQ.jpg&quot; alt=&quot;comparator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G7J89.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G7J89.jpg&quot; alt=&quot;comparator device&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They come with Several etched glass plates with features of size that you optically/visually compare to your part. I'm happy to explain more if need be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, you can also use a two-part rapid set silicone mold material to take molded impressions of your features and then slice up the molds and use them with the comparator. This can be cumbersome in a production environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=fowler+pocket+optical+comparator&amp;amp;safe=off&amp;amp;client=safari&amp;amp;hl=en-us&amp;amp;prmd=sivn&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwjHq8nZuOLXAhUE4SYKHUwmCqUQ_AUIEygC&amp;amp;biw=375&amp;amp;bih=559&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/search?q=fowler+pocket+optical+comparator&amp;amp;safe=off&amp;amp;client=safari&amp;amp;hl=en-us&amp;amp;prmd=sivn&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwjHq8nZuOLXAhUE4SYKHUwmCqUQ_AUIEygC&amp;amp;biw=375&amp;amp;bih=559&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, depending of the size of the features you're measuring, you could get a set of Gage Pins and then you can use them to measure a feature width. For example, the width of a groove. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2017-12-01T02:54:40.770" LastActivityDate="2017-12-01T02:54:40.770" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18244" PostTypeId="1" AcceptedAnswerId="18249" CreationDate="2017-11-29T01:29:45.420" Score="1" ViewCount="175" Body="&lt;p&gt;*Sorry everyone for the very wordy question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to write up a method to accurately measure the airflow through the inlet and various outlets on one of our servers in order to measure the heat output from the server partitions/cards. The boss wants to know the rate of heat production from each card, the current airflow provided to each card, and what the airflow actually should be to keep the cards running at a certain temperature. The server is partitioned into three sections and the outlet vents look similar to these:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uWPmc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uWPmc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have purchased a TSI 9565 velocity meter that comes with a hot wire anemometer and a small pitot tube that measures down to 0.15m/s. I plan on using the pitot tube attachment to measure the air velocity, but my concern is that the velocity will vary depending on where on the vent I take it from, so I won't just be able to average it. The problem server is on site and the technician has limited time to access the server, which means he will only be able to take one measurement at each vent. We do have an exact copy of the server setup here in the main office, but the rack is quite different which means the airflow going into each server will be different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was thinking that because the server setups are the same, maybe the ratios between the velocities for each vent would also be the same. I don't have any justification for this, but if it is roughly true, I could record each individual velocity reading from each vent perforation on the server here in the office, where I have plenty of time to do it. Then I could specific a certain spot on each vent for the technician on site to take a single velocity reading from and find the factor of difference between that server and the office server. From there I could multiply all the velocity values for that vent that I recorded on the office server. It's a bit hard to explain so I've included a quick example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9yr3r.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9yr3r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is my assumption that the servers would have the same relative velocity profiles valid? If not, what would be an accurate way to measure the airflow through each vent?&lt;/p&gt;&#xA;" OwnerUserId="13810" LastActivityDate="2017-11-29T13:29:19.000" Title="How to measure airflow going through individual server vents" Tags="|fluid-mechanics|thermodynamics|airflow|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18245" PostTypeId="2" ParentId="17196" CreationDate="2017-11-29T01:42:31.493" Score="1" Body="&lt;p&gt;With the disks light enough, mounting (gluing) them to the clock hands (trimmed short enough not to obscure the disks) shouldn't pose much of a problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Print the disks on transparency (transparent foil for overhead projectors; printable with inkjet or laser, be sure to pick the type that matches the printer you have), obscuring (printing over) everything that's not supposed to show, leaving the windows unprinted (transparent) plus a ring on the bottom disk to let the top disk's window &quot;see through&quot; all the way down to the fixed face plate with the digits (you can print that one too). Carefully cut the transparencies to shape, detach the clock hands (they are usually push-on, plain friction / tight fit mount), cut short not to obscure the windows and paste them to the reverse side of the transparencies, then mount on the clock. With each disk weighing about a gram, and being a centered weight, they won't affect performance of the clock in the least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the windows to be immobile and the number dials to turn instead, you need to print three disks: bottom with hour numbers, middle with minutes and a transparent ring to reveal the hours beneath, and topmost which would be larger than the two, affixed to the clock case, with the windows.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2017-11-29T01:42:31.493" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18247" PostTypeId="1" CreationDate="2017-11-29T07:28:53.443" Score="0" ViewCount="47" Body="&lt;p&gt;I'm not a mechanical engineer so please be patient with my descriptions and question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on a design for a device that needs to measure out different amounts of powders by volume (not weight!), for example ground spices. Without a mechanical engineering background I am basically winging it. But coming from a software engineering background, I imagine that there must be established patterns for this type of thing in mechanical engineering as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are some go-to patterns and designs for measuring out powders in mechanical engineering design?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't have to be incredibly precise. Not as precise as, for example, a gun powder measuring device. These need to be rough measures, though precision wouldn't be a problem either.&lt;/p&gt;&#xA;" OwnerUserId="13860" LastEditorUserId="13860" LastEditDate="2017-11-29T07:42:20.540" LastActivityDate="2017-11-29T21:23:05.910" Title="Established Patterns and Designs for Measuring Powder by Volume" Tags="|mechanical-engineering|measurements|flow-control|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18248" PostTypeId="2" ParentId="8339" CreationDate="2017-11-29T10:17:12.577" Score="2" Body="&lt;p&gt;I'm thinking of metal drawer tracks or slides, like those for kitchen drawers. Lightweight, easy to instal, you could create sort of cage so that your construction moves along the tracks, low friction, always horizontal, never stuck.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastActivityDate="2017-11-29T10:17:12.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18249" PostTypeId="2" ParentId="18244" CreationDate="2017-11-29T12:16:31.080" Score="0" Body="&lt;p&gt;&quot;Is my assumption that the servers would have the same relative velocity profiles valid?&quot;&#xA;Assuming your server's output air temperature is $343~K$ max. And from the experimental data, max velocity is about ~$3~m/s$ also assuming the individual port of  ~$5~mm$ in size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the field server, the Reynold's number ($Re$) is approximately ~$10^4$ . and in office (same temperature &amp;amp; size but the velocity is 1.5 m/s) ($Re$) &amp;lt; 10000. both are well within in the laminar region (it is only the exit condition). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the velocity distribution is caused by same geometry(in field &amp;amp; office) and Reynold's no indicates that the regime of flow are ~same (laminar regime), assuming linearity between two of your measurements might work.  (for velocity profiles &amp;amp; mass flow rate only).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linking from the previous question, you are intended to find mass flow rate &amp;amp; heat thrown out from the server.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you come to heat transfer calculations, you have to measure the temperature profile for both the servers in the same way you did for velocity profiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature profiles not necessarily similar because, total heat transfer is depends on local heat transfer coefficients (mainly convection by air) which are non linear function of local Nusselt number &amp;amp; local Re number. &#xA;So it may not bring the similar temperature profile at the outlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the flows with similar velocity profiles may not have similar temperature profiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps. &lt;/p&gt;&#xA;" OwnerUserId="13721" LastEditorUserId="13721" LastEditDate="2017-11-29T13:29:19.000" LastActivityDate="2017-11-29T13:29:19.000" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18250" PostTypeId="2" ParentId="18233" CreationDate="2017-11-29T12:28:37.130" Score="2" Body="&lt;p&gt;You haven't specified which method you are using, so i am going to assume Finite Volume.&#xA;In that case you have a staggered grid like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u1GIe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u1GIe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertical lines are the faces of you cells and the circles are the centers of you cells. Your left boundary is located at the face at $x=0$ and your right boundary is located at the face at $x=L$. The domain of size $L$ is divided into $N$ cells. To simplify the boundary treatment, I have included two ghost / virtual nodes $i=0$ and $i=N+1$. I assume a uniform gridspacing $\Delta x=L/N.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how do you impose a flux condition at $x=0$ in terms of local temperatures? Well from a Taylor expansion around $x=0$ we find:&#xA;$$&#xA;T_{i=0} = T_{x=0} - \left.\frac{dT}{dx}\right|_{x=0}\left(\frac{\Delta x}{2}\right) + \frac{1}{2}\left.\frac{d^2T}{dx^2}\right|_{x=0}\left(\frac{\Delta x}{2}\right)^2 + O\left(\Delta x^3\right)&#xA;$$&#xA;and:&#xA;$$&#xA;T_{i=1} = T_{x=0} + \left.\frac{dT}{dx}\right|_{x=0}\left(\frac{\Delta x}{2}\right) + \frac{1}{2}\left.\frac{d^2T}{dx^2}\right|_{x=0}\left(\frac{\Delta x}{2}\right)^2 + O\left(\Delta x^3\right)&#xA;$$&#xA;Subtracting:&#xA;$$T_{i=1} - T_{i=0} = \left.\frac{dT}{dx}\right|_{x=0}\Delta x + O\left(\Delta x^3\right)$$&#xA;If $\Delta x \ll 1$ we can neglect the $O\left(\Delta x^3\right)$ terms and we find the flux at $x=0$ in terms of the local temperatures at $i=0$ and $i=1$:&#xA;$$\left.q&quot;\right|_{x=0}=-\lambda\left.\frac{dT}{dx}\right|_{x=0} = -\lambda\frac{T_{i=1} - T_{i=0}}{\Delta x}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to impose the flux at $x=0$ you need to impose the local temperature of the ghost node at $i=0$ according to above equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar analysis can be done for the boundary at $x=L$, I'll leave that for OP.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastActivityDate="2017-11-29T12:28:37.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18251" PostTypeId="2" ParentId="18241" CreationDate="2017-11-29T13:26:34.093" Score="3" Body="&lt;p&gt;There are plastics out there which can deal with high temperatures, but they certainly won't present a more cost-effective solution than glass in this instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kU7wC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kU7wC.jpg&quot; alt=&quot;(c) Copyright Granta Design Ltd, Cambridge, England www.grantadesign.com&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-11-29T13:26:34.093" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18252" PostTypeId="1" CreationDate="2017-11-29T15:22:43.227" Score="1" ViewCount="153" Body="&lt;p&gt;I'm interested in locating a database containing the elastic constants (the tensor matrix) for a variety of materials (preferably for a variety of symmetry classes). I don't have a particular material in mind, so just some general list would be very helpful. If someone knows of a good resource I'd appreciate it. Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="13867" LastActivityDate="2017-11-29T15:22:43.227" Title="Database of elastic constants" Tags="|elastic-modulus|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18255" PostTypeId="1" CreationDate="2017-11-29T16:31:22.543" Score="25" ViewCount="8445" Body="&lt;p&gt;I was wondering, from an engineering point of view, for what reasons can too much accuracy in modeling be detrimental?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a scientific point of view it seems like it's almost always a benefit, apart from when you need to take less computation time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from an engineering point of view, besides time (or computing power) why should you avoid that?&lt;/p&gt;&#xA;" OwnerUserId="13871" LastEditorUserId="9" LastEditDate="2017-12-04T05:51:02.773" LastActivityDate="2017-12-04T05:51:02.773" Title="From an engineering point of view what are the downsides of a too accurate model?" Tags="|modeling|systems-engineering|" AnswerCount="11" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18256" PostTypeId="1" CreationDate="2017-11-29T16:46:58.333" Score="4" ViewCount="219" Body="&lt;p&gt;I am trying to model laminar, forced convection heat exchange for water in a round pipe and rectangular duct. In both cases, a portion of tube / duct is maintained at a constant surface temperature while the rest of the tube / duct is insulated. I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a reference that gives the Nusselt number for fully-developed, laminar flow in a round tube when a portion of the tube is insulated? (What I'm looking for would be similar to the results presented here: &lt;a href=&quot;http://www.dtic.mil/dtic/tr/fulltext/u2/736260.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.dtic.mil/dtic/tr/fulltext/u2/736260.pdf&lt;/a&gt; in Table 14 pg. 151, but for round pipes instead of rectangular ducts)  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there a reference that gives the Nusselt number for either Graetz flow, or simultaneously-developing flow in a tube / duct when one or more of the sides are insulated? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="4406" LastActivityDate="2018-05-25T14:22:12.113" Title="Nusselt number for partially adiabatic tubes / rectangular ducts" Tags="|heat-transfer|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18257" PostTypeId="2" ParentId="18255" CreationDate="2017-11-29T17:01:45.623" Score="0" Body="&lt;p&gt;Cost : the cost of time or the cost of computing power and the cost of accuracy - if other variables have a tolerance of 5% for example why compute results to 1%...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-11-29T17:01:45.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18258" PostTypeId="2" ParentId="18255" CreationDate="2017-11-29T17:07:43.270" Score="7" Body="&lt;p&gt;An extremely accurate model may require a prohibitive amount of input data. It might be possible to generate an excellent model of weather systems, for example, by taking as input the position and velocity of every gas molecule in the atmosphere. In practice, such a model would not be useful, since there's no realistic way to generate the proper input. A less accurate model that only requires limited input data would be preferable in this case.&lt;/p&gt;&#xA;" OwnerUserId="1549" LastActivityDate="2017-11-29T17:07:43.270" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18259" PostTypeId="2" ParentId="18255" CreationDate="2017-11-29T17:13:12.707" Score="27" Body="&lt;p&gt;The most obvious downside is cost, all engineering projects have a finite budget and spending more money than you need to is clearly a bad thing not to mention wasting time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can also be more subtle issues. Things like FE analysis are always approximations and sometimes adding unnecessary detail can introduce artefacts and make it more difficult to troubleshoot a model. For example you can get discontinuities which result in stress raisers &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the consideration that even if you have the computing power to comfortably handle a large chunk of data suppliers and customers may not and in many cases transferring big files is still a bit of a bottleneck. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly if you have more parameters than you need you are potentially creating extra work down the line in managing and debugging files. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again even if you have abundant time and resources now it may well be that someone further down the line needs to use that model without the same luxury, especially if it ends up being part of a product that youe are selling to customers. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-12-01T11:16:11.533" LastActivityDate="2017-12-01T11:16:11.533" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18260" PostTypeId="1" AcceptedAnswerId="18429" CreationDate="2017-11-29T17:29:45.073" Score="1" ViewCount="350" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/94VLZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/94VLZ.jpg&quot; alt=&quot;Venturi Tube Problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to calculate the volume flow. I've tried setting up Bernoulli and using the hydrostatic pressure and the conservation of mass, as one does to solve a horizontal tube, but I keep hitting up against needing to know the heights z1 and z2, which are not given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't figure out a way to remove it from my equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density is constant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is in equilibrium.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No turbulence.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No friction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressures are constant across the diameters.  &lt;/p&gt;&#xA;" OwnerUserId="9065" LastActivityDate="2017-12-12T11:29:03.100" Title="Volume flow in a non-horizontal Venturi tube" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18261" PostTypeId="2" ParentId="18255" CreationDate="2017-11-29T17:33:59.133" Score="0" Body="&lt;p&gt;In the previous answers input and cost  were mentioned. If you want accuracy eg. in optimization of production parameters you probably need more measurements and first you need to analyze how much you can reduce costs vs how many work hours thus increased expence is for increasing freaquency of measurements or the cost of automized system which will replace manual data collecting.&#xA;Second example if you get very accurate resaults in which you invested time and other resources to obtain, do you have adequate equipment for quality control, industrial measurements etc. or even technology. If your results are in vain than time spent obtainig them is misspent.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastActivityDate="2017-11-29T17:33:59.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18262" PostTypeId="2" ParentId="18255" CreationDate="2017-11-29T17:42:42.293" Score="14" Body="&lt;p&gt;There are a few reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a purely pragmatic perspective, it's due to time constraints. The requisite time to solve a model increases &lt;em&gt;far, far faster&lt;/em&gt; than the level of precision, and whichever level is adopted is subjective, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also affected by the fact that excessive accuracy is mostly useless. After all, your model might be 99.999% accurate for the given input values, but the real world is imprecise. Steel's modulus of elasticity has a tolerance of $\pm5$-$15\%$, for example. So why bother with a super accurate model if one of your key inputs can be off by 10%? (it goes without saying that the margins of error for other materials such as concrete or soil and for other variables such as loading are significantly higher).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to this, there is no point in being too precise. But indeed, it may be beneficial to not even try to be too precise. The reasons for this are mostly psychological, however. Mainly, you don't want your model to be too precise, and you don't want to output your results with seven decimal places, because you don't want to evoke a false sense of confidence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The human brain is hardwired to think that 1.2393532697 is a more accurate value than 1.2. But that's actually not the case. Due to all the real-world uncertainties your model cannot possibly take into consideration (especially given current hardware limitations), 1.2 is almost certainly just as valid a result as 1.2393532697. So don't ilude yourself or whoever sees your model. Just output 1.2, which transparently indicates that you don't really know what's going on after that second digit.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2017-11-29T17:42:42.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18263" PostTypeId="2" ParentId="18255" CreationDate="2017-11-29T19:04:53.550" Score="1" Body="&lt;p&gt;In reality there is the data we have, and there is the data we &lt;strong&gt;don't&lt;/strong&gt; have. Almost always, the amount of data we don't have is much much more than we could ever hope to gather for practical or economical reasons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By trying to fit data obnoxiously well to the few samples we have therefore will risk making our model do really bad estimates into areas where we honestly have no clue (due to lack of data). Then our model will give us a false sense of security.&lt;/p&gt;&#xA;" OwnerUserId="13875" LastActivityDate="2017-11-29T19:04:53.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18264" PostTypeId="2" ParentId="18247" CreationDate="2017-11-29T20:55:29.677" Score="3" Body="&lt;p&gt;I suppose you need some kind of &lt;a href=&quot;http://www.lambda-instruments.com/?pages=powder_dosing_description&quot; rel=&quot;nofollow noreferrer&quot;&gt;powder doser&lt;/a&gt; or &lt;a href=&quot;http://www.tbma.com/en/products/dosing-feeders/volumetric-dosing-feeders/&quot; rel=&quot;nofollow noreferrer&quot;&gt;volumetric feeder&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might work on measuring time. Very easy principle, you need valve (maybe solenoid valve or for some sort of flap door) and timer switch. Mark volumes on your container, measure time needed for filling each volume for each spice you need and program your valve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use Arduino to operate your system, cheap and accesible. If you have programming experience you might be able to program it very fast, maybe even find some already made  solutions online. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different spices will have different times because they have different density or bit (particle) size, so keep that in mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may encounter all sort of problems like spices tending to stuck in pipe etc. You might need to use vibrating sieve or sifter, to prevent clogging, also control humidity, maybe use dehumidifiers or smth. But probbably some of the problems are solved during rest of the production process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can buy doser or pay patent rights to build it if this &lt;a href=&quot;https://www.google.dk/patents/US20130126041&quot; rel=&quot;nofollow noreferrer&quot;&gt;one&lt;/a&gt; is sufficient for your needs.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2017-11-29T21:23:05.910" LastActivityDate="2017-11-29T21:23:05.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18265" PostTypeId="1" CreationDate="2017-11-29T21:18:17.450" Score="2" ViewCount="907" Body="&lt;p&gt;I am trying to calculate the amount of energy required to heat 12,300 kg of water from 70 °C at 25 MPa to 120 °C at 40 MPa. I have thought of using &lt;strong&gt;Q = mcT&lt;/strong&gt; but I know that the specific heat capacity of water varies slightly with pressure. How can I calculate the energy required since &lt;strong&gt;c&lt;/strong&gt; varies?&lt;/p&gt;&#xA;" OwnerUserId="13161" LastEditorUserId="368" LastEditDate="2017-12-04T16:29:23.337" LastActivityDate="2017-12-04T17:33:39.000" Title="How to calculate the required energy to heat water at high pressure?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18266" PostTypeId="2" ParentId="18215" CreationDate="2017-11-29T22:24:47.517" Score="0" Body="&lt;p&gt;I have found iris diaphragm &lt;a href=&quot;https://www.google.com/patents/US5576792&quot; rel=&quot;nofollow noreferrer&quot;&gt;patent&lt;/a&gt; and there are other patents cited, you can find shape of the blade in one of them. Modifying shape of the blade you can get full closure. I do not think that size is relevant. Though waterseal might mean hydrostatic pressure thus increased moment and stresses so size might be important. There are also questions of intelectual property. Because your design would be based on someone's patent in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second source that might be relevant was &lt;a href=&quot;https://www.google.hr/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;url=http://www3.imperial.ac.uk/pls/portallive/docs/1/375907.PDF&amp;amp;ved=0ahUKEwjikZzn6-TXAhUGnRQKHaxCAh4QFgiAATAV&amp;amp;usg=AOvVaw3vm0s86JvdqQPUepIEKaeF&quot; rel=&quot;nofollow noreferrer&quot;&gt;sliding iris mechanism&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2017-12-02T07:55:32.280" LastActivityDate="2017-12-02T07:55:32.280" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18267" PostTypeId="2" ParentId="18255" CreationDate="2017-11-30T00:59:48.687" Score="6" Body="&lt;p&gt;&quot;Too accurate&quot; is not monotonic.  It can actually create an illusion of fidelity which makes you think it's worth pumping more money into the simulation.  This becomes very important when you're presenting data from mixed-fidelity models, where some parts are very detailed and other parts are very coarse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A real life example I had involved sampling altitudes over terrain.  The team had decided to sample the terrain in 1024 chunks to maximize fidelity.  Our customer wanted a ReallyGood(tm) answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I was bothered by the runtime hits this particular algorithm caused, and I wanted to understand how much fidelity I was actually paying for.  I hadn't seen any terrain data, so I asked them how they loaded it.  The answer was &quot;oh, we don't have terrain.  It's just flat.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it sounded like I had an awesome high-fidelity model which sampled 1024 points.  What I actually had was a low-fidelity model which did no better than sampling 1 point 1024 times, but ran a whole ton slower, and masqueraded as a higher-fidelity model!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the real engineering world, leaders don't always have the opportunity to learn the entire architecture of a model.  In fact, I'd say they &lt;em&gt;never&lt;/em&gt; have the time.  Our leadership was making decisions off the assumption that we had an awesome 1024 point model.  Nobody was at fault, it's just what happens when you tune the fidelity up too high on one part of the model, and have low fidelity on the other.  Its the nature of the beast with mixed-fidelity.&lt;/p&gt;&#xA;" OwnerUserId="678" LastActivityDate="2017-11-30T00:59:48.687" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18268" PostTypeId="2" ParentId="10582" CreationDate="2017-11-30T02:24:07.180" Score="1" Body="&lt;p&gt;This is well after your post but may answer some of your questions.&lt;br&gt;&#xA;I was an engineer in Papua New Guinea durng the late 1970's early 1980s when Mabey began supplying Compact Bailey (now Compact 100), and we had a fair bit of older bridging to the &quot;standard&quot; design as well.  At the time we used the data from the Bailey Uniflote book, but even then we had issues, as tables of available live load were not transparent on true capacity - they assumed a lowish dynamic load factor and were calcualted using working stresses.&#xA;We tendered the supply (with Acrow and Mabey competing - well before the Chinese and Indians started supplying) based on the standard capacities.  I don't recollect how much extra the Compact gave, if any, but they were only purchased in competitive tender against the standard Acrow product so we took no account of any extra capacity even if they had it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, these days we still find ourselves occasionally load rating them and we still use the same approach - that is we use the data for the standard Bailey via the live load capacity tables, adjusted for dynamic factors and anything varying the dead loads.  I believe, despite all the alloy and section changes, that the panels were designed to meet proof loading requirements, so in the end it still comes back to the table ratings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have only seen one compact among about a half dozen we have encountered - you can tell them easily by the absence of the transom clamp - the compact has a &quot;permanent&quot; bolted end.  We treated it just the same.&lt;/p&gt;&#xA;" OwnerUserId="13882" LastActivityDate="2017-11-30T02:24:07.180" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18269" PostTypeId="2" ParentId="18255" CreationDate="2017-11-30T04:19:29.550" Score="39" Body="&lt;p&gt;Beware of &lt;a href=&quot;https://en.wikipedia.org/wiki/Overfitting&quot; rel=&quot;noreferrer&quot;&gt;overfitting&lt;/a&gt;. A more accurate model of gathered data from a system may not be a better predictor of future behavior of a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9raxh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9raxh.png&quot; alt=&quot;Overfitting Example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above image shows two models of some data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linear line is somewhat accurate on the training data (the points on the graph), and (one would expect) it will be somewhat accurate on the testing data (where the points are likely to be for x &amp;lt; 5 and x &gt; -5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By contrast, the polynomial is 100% accurate for the training data, but (unless you have any reason to believe the 9th degree polynomial is reasonable for some physical reason), you would assume this will be an extremely poor predictor for x &gt; 5 and x &amp;lt; -5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linear model is 'less accurate', based on any comparison of errors with the data we have gathered. But it is more generalisable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, Engineers have to worry less about their model, and more about what people will do with the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I tell you that we're going on a walk on a hot day and it's expected to last 426 minutes. You are likely to bring less water than if I tell you the walk will last 7 hours, and even less than if I say the walk will last 4-8 hours. This is because you are responding to my implied level of confidence in my forecast, rather than the mid point of my stated times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you give people an accurate model, people will reduce their margin of error. This leads to bigger risks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the walk on a hot day example, if I know the walk will take 4-8 hours in 95% of cases, with some uncertainty around navigation and walking speed. Perfectly knowing our walking speed will decrease the uncertainty of the 4-8 figure, but it won't significantly effect the 'chance of us taking so long that water becomes an issue', because that is driven almost entirely by the uncertain navigation, not the uncertain walking speed.&lt;/p&gt;&#xA;" OwnerUserId="13887" LastEditorUserId="-1" LastEditDate="2017-12-01T19:13:25.843" LastActivityDate="2017-12-01T19:13:25.843" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18270" PostTypeId="2" ParentId="18255" CreationDate="2017-11-30T08:46:16.267" Score="0" Body="&lt;p&gt;Would hou need a satellite image at centimeter-resolution in order to identify forests by color? Sure not. I would be detrimental, as you'd have to decide about any non-green 10 square centimeters patch. The same for modelling: the detail resolution should fit the resolution of your target features. If not, you'll loose time downsizing.&lt;/p&gt;&#xA;" OwnerUserId="13889" LastActivityDate="2017-11-30T08:46:16.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18273" PostTypeId="2" ParentId="18255" CreationDate="2017-11-30T16:53:21.043" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So from an engineering point of view, besides time (or computing power) why should you avoid that&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Coming from a mechanical engineering perspective the biggest reason is you only commit to the additional effort if it produces significantly different results.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the level of accuracy in your model is orders of magnitude higher than the level of accuracy your would be able to deliver in execution of your design you are wasting your effort.  If the level of accuracy described in your model is higher than what is required that has impact for the client. You are wasting money.  For example if you are specifying higher precision than the design actually requires (eg +/- .00001mm in the length of a vent pipe) you are wasting your clients money because a 350mm vent to atmosphere does about the same job as a 350.0005mm vent to atmosphere but the latter is significantly more expensive to produce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In university we all learned to model the physical world using Newtonian physics even though it is well established that post-Newtonian physics present a more accurate model of physical behavior.  In spite of this I know of no mechanical engineering program that by default eschews Newtonian models as too inaccurate.  If we use the more accurate model and come up with an answer that is 0.1% closer to the theoretical truth that will not impact our final design in the vast majority of cases.  If our yield stress is 0.1% different that gives us an insignificant difference in our required cross section which leads us to choosing the exact same size of I-beam based on either method.  In this circumstance the costs of additional effort deliver no additional benefit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are situations where precision is required to produce a workable design, for example modeling of some satellites required relativistic physics.  In these circumstances we need to find a model that delivers the level of accuracy required and we need to design to the model.  If we need to calculate dimensions to +/- 0.0001% it's totally wasted effort if our part dimensions are +/- 0.1%.  In real world applications part dimensions of the latter degree of accuracy are much more common than the former.&lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditDate="2017-11-30T17:04:53.873" LastActivityDate="2017-11-30T17:04:53.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18274" PostTypeId="1" CreationDate="2017-11-30T18:16:18.970" Score="5" ViewCount="1366" Body="&lt;p&gt;I'm confused on how to differentiate mixture control from the main metering system, and why we even need one at all. Let's say we had a large resovior, and a needle type mixture control system. The differential pressure would push/pull fuel from the discharge nozzle at our set mixture ratio? Then where does the need for the main metering system come in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Float-type carburetor &#xA;&lt;a href=&quot;https://i.stack.imgur.com/xdR3Q.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xdR3Q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated, thank you!&lt;/p&gt;&#xA;" OwnerUserId="9066" LastEditorUserId="9066" LastEditDate="2017-11-30T18:24:59.290" LastActivityDate="2018-08-28T03:01:38.213" Title="Whats the purpose of a main metering system in a carburetor?" Tags="|automotive-engineering|fuel-economy|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18275" PostTypeId="2" ParentId="18274" CreationDate="2017-11-30T19:35:14.527" Score="0" Body="&lt;p&gt;as shown, the main metering system is the portion of the carb which is active through most of its operating range (i.e., when not idling, for which you have an idling system, and when not accelerating, for which you have the accelerator pump system). So the point of your question is not clear, because without the main metering system, you wouldn't have a functioning carb of this type. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-11-30T19:35:14.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18276" PostTypeId="1" CreationDate="2017-11-30T19:39:51.917" Score="1" ViewCount="1783" Body="&lt;p&gt;I took apart a cheap domestic vacuum cleaner because I am making a small vacuum-forming system, and I only need the motor/fan assembly, so I repackaged it to save space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the vacuum cleaner is a valve, which I gather is called a bypass valve and is present on most/all vacuum cleaners.  In this case it's just a circular plate held against a gasket by a spring:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bC9Xb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bC9Xb.jpg&quot; alt=&quot;bypass valve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on my measurements, this valve will open and let air through if the pressure difference across the opening is greater than about 10 or 11 kPa.  It sits inside the machine where the user can't get to it, between the fan intake and the outside atmosphere.  In my hacked mini-vacuum, the gasket is on the outside of the box, where I can block it if I want to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the point of this valve?  For my purposes, the more vacuum I can get, the better.  But I don't want to block the valve if it would permanently damage / set fire to the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(it's a 700W 240V AC motor, and for vacuum forming I will be running it for 1 or 2 minutes at a time).&lt;/p&gt;&#xA;" OwnerUserId="4815" LastActivityDate="2017-11-30T20:54:31.293" Title="What is the purpose of a vacuum cleaner's bypass valve?" Tags="|motors|airflow|turbomachinery|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18277" PostTypeId="2" ParentId="18276" CreationDate="2017-11-30T19:56:28.743" Score="3" Body="&lt;p&gt;It’s purpose is to allow sufficient air to cool the motor when the filter / dirt bag is blocked - otherwise it will overheat and could catch fire.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-11-30T20:54:31.293" LastActivityDate="2017-11-30T20:54:31.293" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18278" PostTypeId="1" AcceptedAnswerId="18304" CreationDate="2017-11-30T22:17:58.583" Score="-1" ViewCount="676" Body="&lt;p&gt;in stress concentrations, say in a hole made in a beam loaded axially, we need the value of K ( the stress concentration value) which equals max stress\ avg stress for the given body. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the avg stress is given by the apllied force over the net area, but im not sure what's the max stress that satisfies this relation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;sorry if the question is more of a defenition. &lt;/p&gt;&#xA;" OwnerUserId="13823" LastActivityDate="2017-12-03T04:49:42.070" Title="what's the max stress value ?" Tags="|mechanical-engineering|materials|stresses|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18279" PostTypeId="1" CreationDate="2017-11-30T22:26:32.073" Score="2" ViewCount="23" Body="&lt;p&gt;My lab and I are looking to conduct tensile test experiments on polymer samples with integrated planes of bubbles or voids (see image). The goal of this is to see how well we can manipulate crack propagation during tensile testing with the induced bubbles by varying the angle this plane makes with the predicted crack plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Preparing the currently tested polymer (Ecoflex) involves mixing two liquid components together, pouring the mixture into a petri dish, putting the dish in a vacuum chamber to remove bubbles, and setting the dish out on a level surface to cure for several hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our current ideas on creating this integrated bubble plane:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;manually inputting bubbles in a curing solution on a petri dish using a syringe&lt;/li&gt;&#xA;&lt;li&gt;creating a layer of polymer which we let cure without placing in a vacuum chamber. Once cured, pour another layer of uncured liquid polymer on one side of the cured (non-vacuumed) polymer, place the combined polymer in the vacuum chamber to remove bubbles from the uncured part, and allow to cure. Repeat process for the other side of the non-vacuumed part&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Samples are cut from the cured polymers to achieve tensile specimens with desired orientations of the bubble plane and the initial crack is created by cutting the specimen with a razor blade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are still experimenting with these processes and are looking for other ideas as well. Any input on how to create this bubble/void plane would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TqoLD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TqoLD.png&quot; alt=&quot;Sample Concept&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13919" LastActivityDate="2017-11-30T23:07:10.620" Title="Creating Planes of Bubbles in Polymer Samples" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18281" PostTypeId="2" ParentId="18279" CreationDate="2017-11-30T23:07:10.620" Score="1" Body="&lt;p&gt;You might try to vacuum the entire sample and then, once cured to a (to be determined) stage,  superheat the plane with laser or high frequency ultrasound until bubbles form. &lt;/p&gt;&#xA;" OwnerUserId="13922" LastActivityDate="2017-11-30T23:07:10.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18282" PostTypeId="2" ParentId="18274" CreationDate="2017-11-30T23:20:00.563" Score="1" Body="&lt;p&gt;IF I understand your question, you're asking why we need a main metering system with any moving parts. It sounds like your assumption is that, as the throttle is opened, the increased airflow should pull a commensurately larger volume of fuel through the jet to keep the mixture correct. In a rough sense, that's true, but... That passage is a limiting factor, and as fuel demand increases, it becomes less efficient and the mixture suffers. Because there is such a wide throttle (airflow) range, it is more efficient to have a graduated, or stepped increase on the fuel inlet side. Find a diagram of an old motorcycle carb for a simple image of this. Most of them use a tapered needle in a main jet which is opened wider via the accelerator cable. &lt;/p&gt;&#xA;" OwnerUserId="13922" LastActivityDate="2017-11-30T23:20:00.563" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18284" PostTypeId="1" CreationDate="2017-12-01T07:02:21.130" Score="0" ViewCount="105" Body="&lt;p&gt;In a post apocalyptic setting, rebuilding humanity's technologies would be a pretty prevalent idea. I am in the process of becoming a mechanical engineer. Something that I have noticed is most classes don't teach how to make persion tools from scratch. How would someone, given unlimited raw resources but no references(rulers,tools,ext) be able to build something like a persicion lathe or a vernier scale?&lt;/p&gt;&#xA;" OwnerUserId="3976" LastActivityDate="2017-12-01T11:50:34.213" Title="How would a person be able to rebuild modern machining?" Tags="|measurements|lathe|" AnswerCount="1" CommentCount="8" ClosedDate="2017-12-01T13:12:55.330" ContentLicense="CC BY-SA 3.0" />
  <row Id="18287" PostTypeId="2" ParentId="18278" CreationDate="2017-12-01T09:57:41.270" Score="0" Body="&lt;p&gt;When talking about stress concentration, the goal is to calculate max stress, or you can measure it &lt;a href=&quot;https://pdfs.semanticscholar.org/5731/30327d30986da31fd30e1cb48635242351b8.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;optically&lt;/a&gt;  by measuring displacement like &lt;a href=&quot;http://www.gom.com/metrology-systems/aramis.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aramis&lt;/a&gt; and calculating stresses.&lt;br&gt;&#xA;In the &lt;a href=&quot;http://www.ewp.rpi.edu/hartford/~ernesto/Su2012/EP/MaterialsforStudents/Aiello/Roark-Ch06.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;literature&lt;/a&gt; about Mechanics of solids i.e. stress concentration factors eg.&lt;a href=&quot;http://www.amesweb.info/StressConcentrationFactor/TransverseCircularHoleInRoundBar.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roark&lt;/a&gt; you can see all relevant formulas and definition. &#xA;Best is to use your course literature because different authors and different lecturers often use different notation. &lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2017-12-01T10:55:36.767" LastActivityDate="2017-12-01T10:55:36.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18288" PostTypeId="2" ParentId="18284" CreationDate="2017-12-01T11:50:34.213" Score="2" Body="&lt;p&gt;In many cases it is possible to make measurements to a much greater degree of precision than the parts required by the instrument doing the measurement. Just think of things like plumb lines, spirit levels, dividers etc and once you can get straight lines and circular arcs you can do pretty accurate geometric constructions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also techniques like lapping and honing naturally tend to create geometrically precise surfaces on flats and bores respectively and don't require precision tools. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is white metal bearings where a low melt allow (eg &lt;a href=&quot;https://en.wikipedia.org/wiki/Babbitt_(alloy)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Babbitt metal&lt;/a&gt;) is cast into a bearing surface and then scraped by hand to fit a shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly when making parts the first priority is that they fit relative to each other, this doesn't necessarily require very accurate absolute measurements. One fundamental technique is to offer up tow parts with one of them coated in marking blue or lamp black and use that to see where any high or low spots are then carefully scrape or file them down and repeat. Indeed similar techniques are still relevant today for things like refurbishing machine tools. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a good book on this subject called &lt;em&gt;&lt;a href=&quot;https://www.amazon.co.uk/Art-Blacksmithing-Alex-W-Bealer/dp/0785803955&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Art of Blacksmithing&lt;/em&gt; by Alan W Bealer.&lt;/a&gt; This includes a lot of techniques used by fitters on steam ships which due to the need to make improvised repairs with limited resources was one of the last trades to use traditional hand-fitting techniques. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often apprentices would make many of their own tools as part of their training.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this approach is that any one set of parts only fits with itself. To get the interchangeability of parts needed for proper mass manufacture you need a standardised system of measurement and tolerances. For obvious reasons screw threads were one of the first things to be standardised. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in summary geometric tolerances and fit come first than absolute measurements. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-12-01T11:50:34.213" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18290" PostTypeId="2" ParentId="207" CreationDate="2017-12-01T12:31:31.600" Score="2" Body="&lt;p&gt;To give you a practical example: my dad was a structural engineer working for a large national company; his specialty was to take the drawings for constructions (mainly building facades), which usually were reasonable &quot;OK&quot;, and calculate specific things like size of screws/bolts, spacing, necessary dimension of struts and so on. They worked on very large structures, like airports, opera buildings, skyscrapers. A little change in calculation (say, screws which are a bit smaller, or a bit fewer) can mean hundreds of thousands of € saved. Too small, and bad things happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In his last decade before his pension, he mainly used GWBasic (!) with little self-written programs for his work. This means, he directly worked the methods he knew and had used long before the advent of computers in his field into GWBasic programs. You could call this some kind of trivial numerical simulation, but in fact it was just a glorified pocket calculator (actually he had done the same on pocket calculaters with programmable magnetic strips, before).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of his working days, professional Finite Element software started to appear, and he used those for very complicated projects from time to time. It was never about actually coming up with new results, but always to find out whether a certain approach was feasible. I.e., in his line of work, it's all about loads on steel bars and such; and the manual calculations are, for obvious reasons, mostly reduced to linear cases (and then with 100-200% security margins added to that). Finite Elements opens up whole new worlds for architecturaly interesting buildings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the Finite Elements, he could get &lt;em&gt;much&lt;/em&gt; closer to real necessities (or so people believe), but obviously now it is hard (or, for people like him) outright impossible to verify the results. And believe me, &quot;risk&quot; is a very prominent thing in that respect; if the facade of a big building in a city comes down, people die and engineers end up in jail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; Engineers use numerical simulations similarly to physicians/scientists, to verify assumptions, or iteratively find sweet spots and such. But it is very much required that they need to know what, in general, to expect. It's the same as in science, where an experiment for which you did not reason about the expected results beforehand, is just junk.&lt;/p&gt;&#xA;" OwnerUserId="13945" LastActivityDate="2017-12-01T12:31:31.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18291" PostTypeId="2" ParentId="207" CreationDate="2017-12-01T13:28:50.850" Score="1" Body="&lt;p&gt;There isn't much left to say but that knowing result before runing simulation isn't knowing exact numerical value but to have certain expectations regarding solution based on understanding the physics of the problem. Usually engineers set problem and choose general method and when we finally formulate problem as set of equations and boundaries we seek help from mathematicians to help us solve it in most effective way. Usually engineers are those who define equations, mathematicians solve them. &#xA;If you have no understanding of bending than, though you can solve biharmonic equation, your solution will Probably not be set of right deflections.&#xA;When mathematician learnes to use tools for solwing pde he can solve most pde problems but eg. mechanical engineer though will understand basics of solwer will not try to use it to solve radar imaging or electric flow.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2017-12-01T13:42:18.210" LastActivityDate="2017-12-01T13:42:18.210" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18292" PostTypeId="2" ParentId="17839" CreationDate="2017-12-01T16:38:26.603" Score="1" Body="&lt;p&gt;For the general case of an inclined plane, given the angle $\alpha$, the heights $h_1$ and $h_2$, the water density $\rho$, and the (constant) width $w$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure at a certain depth $h$ is equal to $p=\rho\cdot g\cdot h$, so the force on the incline is the integral of the green area.&#xA;For a trapeze:&#xA;$$ F=\int p\cdot dx dw = \frac{h_1+h_2}{2}xw \rho $$&#xA;where $x$ is the length of the base of the trapze, in this case $x=\frac{h_2}{\sin\alpha}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/XcwSN.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the $x_c$-centroid line of a trapeze you can use the following formula:&#xA;$$ x_c=\frac{x}{3}\cdot\frac{h_2+2h_1}{h_1+h_2} $$&#xA;The height above base of the centroid $h_c=x_c\cdot \sin\alpha$&lt;/p&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="8632" LastEditDate="2018-03-02T18:25:06.063" LastActivityDate="2018-03-02T18:25:06.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18293" PostTypeId="1" AcceptedAnswerId="18294" CreationDate="2017-12-01T17:14:50.487" Score="0" ViewCount="87" Body="&lt;p&gt;I was told I should ask here.  Please note I am not an engineer and I apologize if this isn't an appropriate place, as I suspect this is for engineers to exchange ideas (though I found no &quot;rules&quot; at all).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With no expectation of a high degree of precision, what should I do to mestimate (measure/estimate) the force required to completely close the lid of a large box?  It is hinged and latches, much like the hood of a car, and has gas charged struts holding it open.  It doesn't take much to get it down - I think the struts only slightly offset gravity - but from there it takes considerable force to fully close/engage the latch mechanism (significant friction plus a strong spring).  It is too difficult for the local seniors to get closed, but it is important for theft prevention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to put a small motor inside that grabs the latch and pulls it closed that last inch.  I realize I'll have a lot to do to design the &quot;grab&quot; and also to consider things like pinch protection... but for now I just want to see if a suitable motor exists.  What I don't know how to do is calculate the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I simply stand on a scale and take the delta when it latches, along with the distance the latch traveled?  Intuitively - for my non-engineering brain anyway - I can't really see how the distance matters (if the force is enough to move it an inch isn't it enough to move it a mile?), but the definition for the term &quot;foot pounds&quot; suggests that shorter distance requires less force.  So let's say my scale tells me I weigh 200 lbs and when I press on the hood at the point it latches it says I weigh 100 lbs, and it traveled an inch... does that mean I need a motor rated for 8.33 foot pounds?   Or am I completely off base?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your indulgence.&lt;/p&gt;&#xA;" OwnerUserId="8960" LastActivityDate="2017-12-01T18:50:25.977" Title="Practical low-tech real world method for measuring how much force to close a lid" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18294" PostTypeId="2" ParentId="18293" CreationDate="2017-12-01T18:13:28.530" Score="2" Body="&lt;p&gt;I would suggest that a good way to estimate this sort of thing is by using sand bags (or similar). Just pile them on the lid untill it closes and then weight them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that it sounds like your real issue is a bit more nuanced and the real problem is getting the latch closed and there might be better ways to approach this than just adding a powered actuator. To me it sounds like a less stiff latch is an all round better solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to change the relationship between the rate of the gas struts and the weight of the lid. As you will be aware a car trunk usually closes itself with relatively little effort so it may be that using different struts (or changing their mounting position) or adding weight to the lid or just lubricating the hinges will solve the problem. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In response to comments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;100lb seems like a lot to close a latch, but if it is designed to have someone lean their weight on it then maybe. The difficulty here though is that when you lean on something to close it, especially in this sort of situation is that dynamic forces and momentum come into play and the 'effort'required to operate a 'sticky'mechanism can be hard to quantify. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial instinct in approaching this problem would be to just add a bit more weight to the lid and see if that helps. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-12-01T18:50:25.977" LastActivityDate="2017-12-01T18:50:25.977" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18295" PostTypeId="1" AcceptedAnswerId="18331" CreationDate="2017-12-01T18:42:44.697" Score="0" ViewCount="190" Body="&lt;p&gt;For a low-carbon steel part that will be put into a long lifetime use (10 years plus) what are recommended load cases to consider creep? The part I'm specifically working with will have a relatively low cycle count (20k load cycles over ten years) at about 10% of it's yield strength Loads are compression and tension loads with no load being it's 'normal' position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a percentage of yield strength that calculating creep worth considering? Are there material properties that are more critical? Is cycle count more critical? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, how do you go about calculating creep? I vaguely remember doing this in university but can't find any simple questions.&lt;/p&gt;&#xA;" OwnerUserId="10590" LastEditorUserId="1832" LastEditDate="2017-12-01T19:00:22.997" LastActivityDate="2017-12-04T18:07:34.707" Title="Calculating Creep" Tags="|structural-engineering|strength|creep|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18296" PostTypeId="1" AcceptedAnswerId="18298" CreationDate="2017-12-01T20:29:54.020" Score="0" ViewCount="1879" Body="&lt;h2&gt;Question in short&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The following drawing is a ball-screw with both ends mechanized and a ball-nut on it.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the meaning of the indicated notations?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3v730.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3v730.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Context&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What is clear to me are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Indications of sizes/distances&lt;/li&gt;&#xA;&lt;li&gt;Angles&lt;/li&gt;&#xA;&lt;li&gt;Parameters of precision/error&lt;/li&gt;&#xA;&lt;li&gt;Diameters&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It seem that the little letter (e.g. &lt;code&gt;2&lt;/code&gt;) are related with the values in the box, but I can't understand the meaning. The &lt;code&gt;1&lt;/code&gt; is completely obscure to me.&lt;/p&gt;&#xA;" OwnerUserId="11129" LastActivityDate="2017-12-01T21:20:45.713" Title="Understanding some drawing notations" Tags="|technical-drawing|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18297" PostTypeId="2" ParentId="18296" CreationDate="2017-12-01T20:50:25.670" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;This is an indication of surface finish.&lt;/li&gt;&#xA;&lt;li&gt;This is an indication of a datum&lt;/li&gt;&#xA;&lt;li&gt;This is an indication of concentricity&lt;/li&gt;&#xA;&lt;li&gt;This is an indication of runout&lt;/li&gt;&#xA;&lt;li&gt;This is an indication of perpendicularness&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Almost all of these relate to GD&amp;amp;T, which I would recommend you read up on. The fact that you don't know these shows a general lack of knowledge. I would recommend skimming something like ASME y-14.5, or taking a class on drafting. &lt;/p&gt;&#xA;" OwnerUserId="13498" LastEditorUserId="11129" LastEditDate="2017-12-01T21:20:45.713" LastActivityDate="2017-12-01T21:20:45.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18298" PostTypeId="2" ParentId="18296" CreationDate="2017-12-01T21:05:16.423" Score="1" Body="&lt;p&gt;What you are asking is mainly in reference to the GD&amp;amp;T feature control frames on the drawings. Your drawing looks like it's metric, so it may refer to an ISO standard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your drawing references a specification, you should research that as this answer would essentially need to teach you GD&amp;amp;T and how to interpret it. Maybe start here:&#xA;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Geometric_dimensioning_and_tolerancing&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Geometric_dimensioning_and_tolerancing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are several resources online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Machined Surface Finish requirement(material removal required) as indicated by the integrated horizontal line below the value. &lt;/li&gt;&#xA;&lt;li&gt;It looks like a datum callout&lt;/li&gt;&#xA;&lt;li&gt;Concentricity or Coaxiality(depending on spec and interpretation)(not commonly specified) &lt;/li&gt;&#xA;&lt;li&gt;Circular Runout(typically used over concentricity)&lt;/li&gt;&#xA;&lt;li&gt;Perpendicularity&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What GD&amp;amp;T does is to clearly define what features are important and how they are to be inspected. They are used to control position and/or form of the part. In your case, that's a rotating shaft so typically the cylinders need to be concentric so they are balanced and aligned for assembly. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-12-01T21:05:16.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18300" PostTypeId="2" ParentId="18203" CreationDate="2017-12-02T01:19:33.853" Score="0" Body="&lt;p&gt;In the oil industry where it is critical to get the many hundreds of different threads correct; they use working thread gauges  ( precision hardened steel ) that are traceable to master gauges , which are traceable to grand master gauges. The carbide tools used to cut these threads are generally verified on optical comparators. A machine shop would not encounter these type threads, but that is how they are measured. The answer to measuring any threads is usually a set of gauges, as has been recommended .  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-12-02T01:19:33.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18303" PostTypeId="1" AcceptedAnswerId="18311" CreationDate="2017-12-02T10:40:26.650" Score="1" ViewCount="122" Body="&lt;p&gt;I’m trying to understand the effects of  a bike’s trail on its steering stability. There, however, is a concept of slippage velocity of the tires during steering of which I cannot make heads or tails of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Book- Motorcycle Dynamics&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Author- Vittore Cossalter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page no. 13-14&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0afjY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0afjY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1st Question: Why should the final resultant velocity of the front tire be in the same initial direction of motion ? Shouldn’t it be in the plane or rotation of the front tire, now that a steering input has been given?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd Question: What is this slippage velocity ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve tried Google as well. But there isn’t any information on this topic in simple words. &lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2017-12-03T02:58:25.717" Title="What is slippage velocity?" Tags="|mechanical-engineering|automotive-engineering|dynamics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18304" PostTypeId="2" ParentId="18278" CreationDate="2017-12-02T14:45:58.800" Score="1" Body="&lt;p&gt;In your example (a hole in a bar loaded axially), the hole stretches. This causes the stress around the hole to vary. The locations where the hole changes curvature the most is where the stress is a maximum. From this, the stress concentration can be calculated: k=max stress/average stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image from a finite element analysis (FEA) showing the elongation and stress around the hole. Red is high stress, dark blue is low stress. Note that the minimum stress is lower than the average stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YYLTE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YYLTE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The hole in the original bar is circular, and the above image greatly exaggerates the displacement. Also, you need to check the reference for the stress concentration factor K to see if the average stress = force/w/t (the total area) or average stress = force/(w-d)/t (the net area), where w is the width of the bar, d is the diameter of the hole, and t is the thickness of the bar. &lt;/p&gt;&#xA;" OwnerUserId="12984" LastEditorUserId="12984" LastEditDate="2017-12-03T04:49:42.070" LastActivityDate="2017-12-03T04:49:42.070" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18305" PostTypeId="2" ParentId="18278" CreationDate="2017-12-02T14:58:57.580" Score="1" Body="&lt;p&gt;In manual calculations you will pick or calculate a value of K from a table or graph based on the geometry or calculate from a formula. Depending on the material there may also be additional factors to take into account relating to surface roughness etc. There are also various online &lt;a href=&quot;http://www.amesweb.info/StressConcentrationFactor/CentralCircularHoleInFiniteWidthPlate.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;calculators&lt;/a&gt; which can work it out for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially K is the ratio between average stress and max stress so you need to know at least two of the three values to calculate the third. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in practice the average stress is easily calculable, multiply this by the appropriate K value and you get the max stress (assuming that the assumptions used to calculate K are valid). &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-12-02T14:58:57.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18306" PostTypeId="1" CreationDate="2017-12-02T16:01:11.747" Score="1" ViewCount="664" Body="&lt;p&gt;How can I calculate the volume (cubic meter per second) of water flowing out a tank? (or the velocity in meter per second)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A horizontal pipe with a diameter of 1 meter is mounted (within the tank) 10 meters below the surface (that is 10 meter for the center line of the pipe).&#xA;What's the formula to calculate the water volume flowing out the tank?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the volume of the tank important?&#xA;Let's say one tank with a diameter of 5 meter and another with a diameter of 20 meter, the volume of the water flowing out at the beginning is the same I guess? Off course for the first tank volume will decrease quicker than for the second tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a difference if the pipe is mounted within or outside the tank?&lt;/p&gt;&#xA;" OwnerUserId="13985" LastEditorDisplayName="user16" LastEditDate="2017-12-03T18:32:45.967" LastActivityDate="2017-12-03T18:32:45.967" Title="How to calculate the volume of water flowing out of a tank?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18307" PostTypeId="2" ParentId="18306" CreationDate="2017-12-02T16:47:27.430" Score="1" Body="&lt;p&gt;The pressure driving the flow is down to the height of water above the pipe - the diameter will have no effect : pressure = density * gravity * height, you should be able to take it from here - this does seem like a homework question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing to consider is how the pipe connects to the tank : sharp edge joint, rounded edge, pipe extends into tank...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;If the pipe is small relative to the height of the tank then the following where A is the area:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;V = Cd A (2 g H)^1/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If the pipe is large then use the following where b is the width of the outlet in m :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;V = 2/3 Cd b (2 g)^1/2 (H^3/2 - h^3/2) &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and H is the height from the liquid surface to the top of the pipe and h is the height from the liquid surface to the bottom of the pipe.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-12-03T17:01:16.797" LastActivityDate="2017-12-03T17:01:16.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18309" PostTypeId="1" AcceptedAnswerId="18328" CreationDate="2017-12-02T20:15:23.167" Score="1" ViewCount="60" Body="&lt;p&gt;I would like a lock for a garage door (roll up style). I would like the lock to be something I can control electronically, and also I would like it to retain it's state when the power is removed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power requirement seems to rule out any type of electromagnetic lock, and I considered a solenoid, but I'm not sure if I can get one that stays in the current position when the power is removed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there are two types that will work - one with a latch on a solenoid (&lt;a href=&quot;https://www.adafruit.com/product/1512&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.adafruit.com/product/1512&lt;/a&gt;) or possibly a small linear actuator that could be extended through a plate to lock the door. Are there any other options I'm overlooking? &lt;/p&gt;&#xA;" OwnerUserId="13797" LastActivityDate="2017-12-04T13:59:27.967" Title="Type of electronic lock for a garage door?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18310" PostTypeId="2" ParentId="18309" CreationDate="2017-12-02T21:58:18.647" Score="1" Body="&lt;p&gt;You could consider using a motor/servo set-up. These would only use power when they're actually opening/closing the lock. For a small lock there's plenty of servo controlled units (from looking mostly for central locking on cars &lt;a href=&quot;https://www.ebay.co.uk/i/112350866434?chn=ps&amp;amp;adgroupid=47842945425&amp;amp;rlsatarget=pla-380312364851&amp;amp;abcId=1129946&amp;amp;adtype=pla&amp;amp;merchantid=110247205&amp;amp;poi=&amp;amp;googleloc=1007080&amp;amp;device=c&amp;amp;campaignid=974959912&amp;amp;crdt=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this&lt;/a&gt;. They need a controller to drive the servo though. &#xA;That said, a linear actuator would also work, but would need to know when it  was fully closed/opened. Whether that be by limit/reed switches or the like.&lt;/p&gt;&#xA;" OwnerUserId="12187" LastActivityDate="2017-12-02T21:58:18.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18311" PostTypeId="2" ParentId="18303" CreationDate="2017-12-03T02:58:25.717" Score="2" Body="&lt;p&gt;When a wheel is rolling in a straight line, the velocity of the wheel at the point of contact with ground is zero. That is, the backwards spinning speed of the wheel cancels the forward speed of the bike. In the frame of the wheel, everything is moving in the x-direction. Here $\bar{v}$ is the speed of the wheel at the contact point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar{v}_{straight,x} = v_{bike} - \omega r = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar{v}_{straight,z} = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the perspective of the wheel, the relative speed of the bike is different after the wheel is turned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar{v}_{turned,x} = v_{bike}\cos\theta - \omega r = -(1-\cos\theta)v_{bike} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar{v}_{turned,z} = -v_{bike}\sin\theta $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms $\bar{v}_{turned}$ terms define the sliding velocity of the wheel.&lt;/p&gt;&#xA;" OwnerUserId="13995" LastActivityDate="2017-12-03T02:58:25.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18313" PostTypeId="2" ParentId="18255" CreationDate="2017-12-03T07:11:11.357" Score="0" Body="&lt;p&gt;Most of the real answers are excluded by your artificial constraint that computing power and calculation times are not to be considered. A model that takes hours or days to evaluate does not allow rapid design iterations and will slow things down on a human scale, increasing cost and possibly leading to inferior results. Cleverly simplifying models without losing too much accuracy can be a very useful approach, then the brute-force model can be used to validate the final iteration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible overly complex models may mask fundamental errors in the model, or that the work required to gather information to practically use the model to the maximum will outweigh any potential benefit. For example if you need to know the characteristics of a material to a greater degree of accuracy than the supplier can control them, you can either accept the error bands or go and test each batch of material to tweak the model. &lt;/p&gt;&#xA;" OwnerUserId="1276" LastActivityDate="2017-12-03T07:11:11.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18315" PostTypeId="1" AcceptedAnswerId="18335" CreationDate="2017-12-03T14:05:27.950" Score="0" ViewCount="1113" Body="&lt;p&gt;I'm looking for the best orientation (North-East-South-West) of a greenhouse (simplifiable as a rectangular solid).&lt;br&gt;&#xA;My question is more specifically about the &lt;strong&gt;sun position effects&lt;/strong&gt; (over the times of the day and over the days of the year).&lt;br&gt;&#xA;With &lt;em&gt;sun effect&lt;/em&gt; I mean both &lt;strong&gt;solar exposure&lt;/strong&gt; for plants and &lt;strong&gt;temperature increase&lt;/strong&gt; within the greenhouse. Anyway I think both aspects go hand-in-hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, how can I calculate the position of the greenhouse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;PS&lt;/strong&gt;: if can help to better &lt;em&gt;focus&lt;/em&gt; my question, the question could also be re-arranged e.g. for house's rooms: which is the coldest/hottest room in my house? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; solar exposure's issue was solved. Now it remains how to deal with temperature gain. In short: with maximum solar exposure I will get automatically the higher temperature inside the greenhouse?&lt;/p&gt;&#xA;" OwnerUserId="2157" LastEditorUserId="2559" LastEditDate="2017-12-23T13:38:20.187" LastActivityDate="2017-12-24T18:02:13.020" Title="Place a greenhouse: how to orient it for the best effect?" Tags="|civil-engineering|building-design|building-physics|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18317" PostTypeId="2" ParentId="18222" CreationDate="2017-12-03T19:12:54.133" Score="1" Body="&lt;p&gt;This approach is just assuming a roll-out policy, so using a control policy after and including time step $k+p$ which is known to stabilize the system. For this roll-out policy the input and output constraints are not taken into consideration. So in order to guarantee stability $p$ has to be chosen to be sufficiently large such that these constraints will not be violated given some initial condition $x(k)$ within some compact set and roll-out policy. Using a more aggressive roll-out policy would usually mean bigger inputs, so you might need to increase $p$ to still ensure that the constraints will not be violated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now onto answering you real question. You are allowed to choose any $Q_p$ such that you can come up with a control policy which stabilizes the system, whose cost adds up to $x(k+p)^\top Q_p\,x(k+p)$. I am assuming you are dealing with a linear time invariant system, in which case a constant state feedback could stabilize the system. One example of such a controller would be the &lt;a href=&quot;https://en.wikipedia.org/wiki/Linear%E2%80%93quadratic_regulator#Infinite-horizon.2C_discrete-time_LQR&quot; rel=&quot;nofollow noreferrer&quot;&gt;linear quadratic regulator&lt;/a&gt; (LQR). This requires solving the discrete algebraic Riccati equation (DARE)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Q_p = A^\top Q_p A − (A^\top Q_p B + N) (B^\top Q_p B + R)^{−1} (B^\top Q_p A + N^\top) + Q&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;however $N$ is often chosen to be zero, so this would yields the same equation as in your question. The corresponding roll-out policy would then be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;u = -(R + B^\top Q_p B)^{-1} (B^\top Q_p A + N^\top)\,x.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally for LQR you start of with a given $(A,B)$, which is stabilizable, and then choose $R$, $Q$ and $N$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;Q &amp;amp; N \\ N^\top &amp;amp; R&#xA;\end{bmatrix} \succ 0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a &lt;a href=&quot;https://nl.mathworks.com/help/control/ref/dare.html#f0-192314&quot; rel=&quot;nofollow noreferrer&quot;&gt;positive definite matrix&lt;/a&gt;. But if you start with $Q_p$ this turns the problem into finding $R$, $Q$ and $N$ which satisfy both the previous condition and the DARE. If you assume $N=0$ then this comes down to finding a positive definite $R$ such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;Q = Q_p - A^\top Q_p A + A^\top Q_p B (B^\top Q_p B + R)^{−1} B^\top Q_p A&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is positive definite as well. But this assumes that you can choose any positive definite matrix for $Q_p$. If you limit yourself to multiplying a previously found $Q_p$ by a positive scalar, then it can be shown that scaling both $Q_p$, $Q$, $R$ and $N$ by $\beta &amp;gt; 0$ still satisfies the DARE and this will also ensure that matrix constructed from $R$, $Q$ and $N$ is still positive definite. Since $(\beta\,M)^{-1} = M^{-1} / \beta$, so $\beta$ can be factored out from the DARE. And multiplying a positive definite matrix by a positive scalar will also return a positive definite matrix. It can be noted this actually gives the same roll-out policy, because the cost of every $x$ compared to every $u$ has not changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So using $\beta\,Q_p$, with $\beta &amp;gt; 1$ will still ensure stability, since there exists a stabilizing roll-out policy associated with that terminal cost. Since this roll-out policy is actually identical to the previous roll-out policy, therefore the constraints on the system should not be violated if they also weren't previously. Since the terminal cost is higher this should actually reduce the magnitude of the state and thus the input after $p$ time steps. So with this terminal cost you might actually be able to lower the value of $p$.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2017-12-03T19:12:54.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18318" PostTypeId="1" CreationDate="2017-12-03T20:19:59.980" Score="1" ViewCount="33" Body="&lt;p&gt;I'm going over pipelining and stalls and forwards with MIPS architecture, and there's a couple things I'm uncertain about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If two adjacent instructions, such as the first two of each set bellow, both use the same register, but the first isn't actually modifying that register, is there still a stall? Like if the add and first &lt;code&gt;addi&lt;/code&gt; of the second set each modified &lt;code&gt;$t0&lt;/code&gt;, then yea, you'd want a stall because you'd want the updated &lt;code&gt;$t0&lt;/code&gt; value to be used in the next instruction. But here the result of the next instruction wouldn't actually be affected since &lt;code&gt;$t0&lt;/code&gt; doesn't change. So is there no stall in this case? Or is there a stall just because the register is in use despite whether or not it's being written to or only read? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in the first set is the first &lt;code&gt;addi&lt;/code&gt; between the add and second &lt;code&gt;addi&lt;/code&gt; enough of a gap not to need a stall over &lt;code&gt;$t1&lt;/code&gt;'s use? Or would you stall need forwarding to help there?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;add $t1, $t0, $t0&#xA;addi $t2, $t0, 1&#xA;addi $t3, $t1, 1&#xA;&#xA;addi $t1, $t0, 1&#xA;addi $t2, $t0, 2&#xA;addi $t3, $t0, 3&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="8733" LastEditorUserId="9" LastEditDate="2017-12-05T15:59:37.620" LastActivityDate="2017-12-05T15:59:37.620" Title="MIPS Assembly stalls and forwarding when a register's value isn't modified in previous step" Tags="|embedded-systems|computer-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18320" PostTypeId="1" CreationDate="2017-12-04T00:32:25.373" Score="1" ViewCount="37" Body="&lt;p&gt;I've got an 1/8&quot; thick steel rail with 1.5&quot; diameter and 15' in length sitting on two 5/32&quot; thick steel rails with 2.5&quot; diameter and 100&quot; in length 14' apart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VCjuq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VCjuq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to add a diagonal support for a rope that is supported with &lt;strong&gt;3/32&quot; thick wires&lt;/strong&gt;. I started into the initial design, and stopped here because I don't feel I'm heading in the right direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tomqB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tomqB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can I structurally strengthen a support to a rail with wires to allow 164&quot; of vertical clearance from the rail to the extended end of said support? Giving a total of ~264&quot; of ground-to-end rope length.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it's not clear, the rope will hang down to the ground from the end of the extending support rail. Also, if it's of any relevance, the three rails in the first image are schematics from the actual existing physical structure that is already welded together and cemented into the ground (18&quot; deep).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I am now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/urROF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/urROF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagining, here, that the two lower triangular shaped supports are rails to be welded and shaped to the curvature of the rails they attach to. The area below and between the main rails are used for gymnastic rings, some hanging ab straps, and a lowered hanging pull up bar. I mention this because the design, as it stands, seems to clutter up the &quot;work area,&quot; if you will.&lt;/p&gt;&#xA;" OwnerUserId="14026" LastEditorUserId="14026" LastEditDate="2017-12-04T01:09:58.190" LastActivityDate="2017-12-04T01:09:58.190" Title="How to support height extender on a rail?" Tags="|structural-engineering|materials|design|structures|building-design|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18321" PostTypeId="2" ParentId="4178" CreationDate="2017-12-04T03:39:37.737" Score="1" Body="&lt;p&gt;I would have drilled 2 holes in a bracket and then welded it onto the flat plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reduce friction and absorb high frequency vibrations, I would have then used cheap nylon bushings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o9RtI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o9RtI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a bolt with a smaller diameter than that of the holes of the swing-arm. Now the space that would be left hollow in between the bolt and the holes should be filled with the nylon bushings mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6hCqP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6hCqP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2017-12-04T03:39:37.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18322" PostTypeId="1" CreationDate="2017-12-04T04:04:05.723" Score="0" ViewCount="116" Body="&lt;p&gt;The motor in a washing machine oscillates at set frequencies as compared to a motor that spins very fast in a fixed direction in a blender/mixer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lH581.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lH581.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don’t understand however is : Don’t they both have the same job to perform -To mix a liquid (usually water) with solid stuff ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/81pp9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/81pp9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more friction the motors can create between the solid materials( food or clothes ), the more rubbing action there would be, hence a faster operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why don’t the motors of a mixer spin like that of a washing machine ? Or are there those types of mixers as well ? If so, which one does the job faster ?&lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2017-12-04T13:55:56.553" Title="Washing Machines vs. Mixers" Tags="|design|motors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18323" PostTypeId="2" ParentId="18322" CreationDate="2017-12-04T05:02:16.963" Score="2" Body="&lt;p&gt;The motor in a washing machine does indeed spin the basket containing the clothes, but it does not oscillate. Usually, it rotates in one direction and turns on and off intermittently, which action is intended to mix the soapy water with the clothes. Any other oscillation present in a washer is produced by its transmission and not the motor directly. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="78" LastEditDate="2017-12-04T13:55:56.553" LastActivityDate="2017-12-04T13:55:56.553" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18324" PostTypeId="1" AcceptedAnswerId="18353" CreationDate="2017-12-04T05:08:05.777" Score="3" ViewCount="2040" Body="&lt;p&gt;I have been assigned to work out how much air flow is needed through each partition of our servers in order to keep our processors at (or below) 60C (333K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have worked out the current mass flow rate going through each partition, and the heat flow rate, and I know the temperature of each processor from the system logs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I don't know how to calculate the required mass flow rate to bring down the temperature of the CPU's to a certain value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, let's assume on one partition I have found:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;air mass flow rate = 0.01 kg/s&#xA;heat flow rate = 150 J/s&#xA;temperature of inlet air = 298K&#xA;temperature of outlet air = 308K&#xA;temperature of CPU = 353K&#xA;Goal CPU temperature = 333K. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So I need to reduce the CPU temperature by approximately 6% (1 - 333/353).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean I simply need to increase the airflow by 6%? I originally figured this could be true because a decrease in temperature of the CPU would be a linear decrease in the heat flow. So according to the heat capacity formula (q=mc$\Delta$T), where q is the heat flow, m is the mass flow, c is the heat capacity at constant pressure, and $\Delta$T is the change in temperature:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1.06 * 150 J/s = 1.06 * 0.01kg/s * 1000J/kgK * (308K - 298K)&#xA;Therefore the mass flow rate needs to be 0.0106m/s.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The problem with this is I am assuming the outlet temperature remains the same, which I don't expect would happen. And the mass flow increases by such a marginal amount. I just don't feel like an increase in mass flow by 6% would decrease the CPU temperature by 20C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please show me how I should be doing this? (Please note the goal is simply to find how much I need to increase the airflow)&lt;/p&gt;&#xA;" OwnerUserId="13810" LastEditorUserId="13810" LastEditDate="2017-12-04T05:51:06.027" LastActivityDate="2017-12-05T21:32:02.550" Title="How to calculate required air mass flow to keep CPU below certain temperature?" Tags="|fluid-mechanics|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18325" PostTypeId="2" ParentId="18322" CreationDate="2017-12-04T05:09:25.420" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gn8BR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gn8BR.jpg&quot; alt=&quot;Centrifugal force&quot;&gt;&lt;/a&gt; acts on the drum of the washing machine because of the uneven distribution of the mass caused by the clotes. There is also forces act on the drum in conventional belt driven systems which also causes incresing oscilations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jNjCp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jNjCp.png&quot; alt=&quot;Belt Drive vs Direct Drive&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1599" LastActivityDate="2017-12-04T05:09:25.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18326" PostTypeId="1" CreationDate="2017-12-04T08:55:16.327" Score="3" ViewCount="204" Body="&lt;p&gt;How would you calculate the eigenfrequency of such a cantilever?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9F3bY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9F3bY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HbPwF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HbPwF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9039" LastActivityDate="2018-07-03T14:58:34.157" Title="Eigenfreuency of beam" Tags="|mechanical-engineering|frequency-response|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18327" PostTypeId="2" ParentId="16981" CreationDate="2017-12-04T11:56:39.663" Score="1" Body="&lt;p&gt;To supplement the other answers, there's definitely a little bit of marketing sprinkled in there. In your case, a V6 &quot;seems&quot; like &lt;em&gt;more&lt;/em&gt; engine than a I4. Therefore, a greater perceived value by the consumer. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-12-04T11:56:39.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18328" PostTypeId="2" ParentId="18309" CreationDate="2017-12-04T13:59:27.967" Score="2" Body="&lt;p&gt;You can get latching solenoids which fire back and forth given a current pulse. They stay in position when the current is removed.&lt;/p&gt;&#xA;" OwnerUserId="1992" LastActivityDate="2017-12-04T13:59:27.967" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18329" PostTypeId="1" CreationDate="2017-12-04T15:46:08.633" Score="2" ViewCount="164" Body="&lt;p&gt;Why is it that utilizing clockwork motors never came into fruition as power sources for things like vehicles and bolt shooters, in place of compressed air and strings?&lt;/p&gt;&#xA;" OwnerUserId="14036" LastActivityDate="2017-12-04T18:10:41.740" Title="Clockwork Motor feasibility?" Tags="|motors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18330" PostTypeId="2" ParentId="18265" CreationDate="2017-12-04T17:33:39.000" Score="1" Body="&lt;p&gt;The energy required to make this change is the difference in internal energy between the two states&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ ∆Q=m((c_{v_2}×T_2 )-(c_{v_1}×T_1 )) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ m=12300kg $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ c_{v_1}=4074\;Jkg^{-1}K^{-1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ c_{v_2}=4029\;Jkg^{-1}K^{-1} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ T_1=343.2\;K $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ T_2=393.2\;K $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ ∆Q=12300((4029×393.2)-(4070×343.2))=2305MJ $&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-12-04T17:33:39.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18331" PostTypeId="2" ParentId="18295" CreationDate="2017-12-04T18:07:34.707" Score="1" Body="&lt;p&gt;creep is caused when applied stresses in a metal part can be relieved by movement of metal atoms via diffusion within the crystal lattice. diffusive transport only becomes important when the service temperature exceeds about 1/3 to 1/2 the melt temperature (rough rule of thumb, using degrees K)- meaning steel doesn't creep at room temperature but lead/tin solder does. the materials science text by Van Vlack covers this topic in more detail. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-12-04T18:07:34.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18332" PostTypeId="2" ParentId="18329" CreationDate="2017-12-04T18:10:41.740" Score="4" Body="&lt;p&gt;The fundamental issue is energy density. A spring simply doesn't store that much energy per unit mass and is ultimately limited by the yield stress of the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where springs are good is in providing large forces over a fairly short range of movement and time so they are good for things which require a fast response.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There area also issues with managing the power delivery of springs as they naturally provide a force proportional to their extension. There are ways around this which have been developed for watches etc but they add a lot of complexity and are difficult to implement for power systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Springs are ok for things like watches because a watch is (in a sense)  a fair attempt at a perpetual motion machine and work well with motive power which provides a lot of torque but don't need much actual energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compressed air is only a bit better than springs in terms of energy density. It is just about feasible for vehicles in special circumstances eg compressed air locomotives have been used for situations like mines where there is an explosion risk. The advantage of compressed air for workshop tools etc is that it can be produced centrally (with a big compressor) and distributed, also air  tools tend to be lightweight, compact and reliable compared to electric ones, especially for high speed, low torque applications like die grinders and polishers. pneumatic are also good for very high speed, compact actuators (F1 cars have pneumatic valves for example). &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-12-04T18:10:41.740" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18334" PostTypeId="1" CreationDate="2017-12-04T21:20:57.167" Score="1" ViewCount="62" Body="&lt;p&gt;I am doing some labor work on a rig and am confused about something. I heard from my superiors that we are injecting steam at 180°C with the wellbore pressure being 2&amp;nbsp;MPa. If I look at these conditions on a steam table, shouldn't the steam be compressed water?&lt;/p&gt;&#xA;" OwnerUserId="13161" LastEditorUserId="1832" LastEditDate="2017-12-04T23:40:51.353" LastActivityDate="2018-03-22T19:56:34.827" Title="Shouldn't steam be compressed water in SAGD?" Tags="|thermodynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18335" PostTypeId="2" ParentId="18315" CreationDate="2017-12-04T22:49:01.420" Score="2" Body="&lt;p&gt;In general: if you are in the northern hemisphere, the sun will pass through the southern part of the sky. Therefore, a greenhouse which is built onto the side of an existing structure will be placed against its south-facing wall so it is not in the shadow of that structure, as it would be if it were affixed to its north-facing wall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and as pointed out by others in the comment section, for maximum solar gain you want the largest area of the structure facing south as well. So if the greenhouse is rectangular, you want its long axis pointing east-west. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-12-04T22:49:01.420" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18336" PostTypeId="1" CreationDate="2017-12-04T23:55:29.163" Score="2" ViewCount="129" Body="&lt;p&gt;I have a physics professor for whom I'd like to build a machine. Featured in a problem he is notorious for assigning students, originally from Kleppner and Kolenkow's &quot;Mechanics,&quot; the machine causes all three blocks to accelerate, including the large block M1. A picture is featured below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem features frictionless surfaces, massless ropes, and massless frictionless pulleys. I'd like to build the machine for the professor as a thank you gift, but I ask if anybody has suggestions for how to get relatively close to the frictionless and massless constraints above with objects I can realistically build with, e.g. teflon and steel for low-friction surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any advice and suggestions!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IIpYO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IIpYO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14043" LastActivityDate="2017-12-07T10:56:32.770" Title="Building a machine from an ideal physics problem" Tags="|mechanical-engineering|materials|friction|building-physics|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18337" PostTypeId="2" ParentId="18326" CreationDate="2017-12-05T00:53:19.527" Score="1" Body="&lt;p&gt;Are you looking for a single number for one particular beam, or a closed form solution for any possible combination of dimensions?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the former, the easiest will be ANSYS (or Abaqus or NASTRAN or any other commercial FEA software).  Model the two different beams as shell elements.  With shell elements you can use a single layer for each sub-beam.  With brick elements, you'll need at least 7 layers for each sub-beam, which means a longer computation time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the later, you'll probably want to pick up a graduate level textbook on the vibration of continuous systems.  You'll need to set up and solve partial differential equations. It is probably easiest to set up two differential equations that are linked by a boundary condition that the motion at the interface is the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Material 0 would have to be modeled as a plate (&lt;a href=&quot;https://en.wikipedia.org/wiki/Bending_of_plates&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Bending_of_plates&lt;/a&gt;). If $w1&amp;lt;&amp;lt;w0$,then Material 1 could potentially be modeled as an Euler-Bernoulli beam (&lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&lt;/a&gt;), which would be a little easier.  But otherwise they'd both have to be plates.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the eigenfunctions in closed form would be pretty hard. Instead, if $h1&amp;lt;h0$ and $w1&amp;lt;w0$, you could probably use the Raleigh-Ritz method (&lt;a href=&quot;https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Ritz_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Rayleigh%E2%80%93Ritz_method&lt;/a&gt;) using the eigenfunctions of the Material 0 plate by itself as the basis functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit to address your comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link you gave shows the math for a &lt;em&gt;beam&lt;/em&gt;.  A defining characteristic of a beam is that it is a one dimensional problem.  i.e. the deflection at the left of the beam is the same as at the right of the beam.  the only coordinate that matters is the length along the beam (x in the link's notation).  What you are asking about in this question is one step more complicated.  In your case, the deflection at the left side can and will be different than the right side, because of the extra material making the left side stiffer.  So you have a two dimensional problem. This is called a &lt;em&gt;plate&lt;/em&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="9275" LastEditDate="2017-12-05T12:08:14.930" LastActivityDate="2017-12-05T12:08:14.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18338" PostTypeId="1" CreationDate="2017-12-05T00:57:09.887" Score="0" ViewCount="58" Body="&lt;p&gt;I have been thinking of a design for a standard bicycle monorail. I don't have the resources to prototype it right now so I would like to bounce the design off the community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically thinking of a half cylinder track that is larger diameter than a bicycle wheel. For a prototype I was considering ripping 2 inch pvc in half for the straight sections. In application people could ride a standard department store quality bike on to the track at an enterance and off at an exit without any modifications or reconfigurations to their bicycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My biggest concern is balance stability on the track (even on a straight section not considering corners). On a normal path the operator corrects from a disturbance by gyroscopic precession (and resultant turning) of the front wheel. The track will limit this but will provide an elevation change toward the outside of the track that may assist. Thoughts? Any prior art you know of? Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't want to debate the usefulness of the idea, but just so everyone is in the know, there are three potential benefits:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lower cost for longer distance bicycle path infastructure.&lt;/li&gt;&#xA;&lt;li&gt;Lower rolling resistance than a normal flat path.&lt;/li&gt;&#xA;&lt;li&gt;Lower cost for snow removal and maintenance.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2017-12-08T13:56:55.223" Title="Standard bicycle monorail design considerations" Tags="|mechanical-engineering|applied-mechanics|transportation|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18339" PostTypeId="2" ParentId="18336" CreationDate="2017-12-05T01:42:11.477" Score="0" Body="&lt;p&gt;The massless ropes won't be a problem. Just use steel cable. Close enough. Get good ball bearings for the pulley and that should also be close enough. The problem will be the interface between m1 and m2. Even Teflon is probably too much friction. Probably the easiest is to just put wheels on m2 so that it rolls on m1 instead of sliding. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-12-05T01:42:11.477" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18340" PostTypeId="2" ParentId="18338" CreationDate="2017-12-05T02:28:25.800" Score="2" Body="&lt;p&gt;it would be nearly impossible to stay balanced on a bike with the steering constrained like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit/&#xA;Not sure the really warrants a dissertation on bicycle stability, but briefly when riding you are in a state of meta-stable equilibrium. If your center of gravity moves slightly off center you would tend to tip over. The corrective action, which makes the activity possible, is that you counter steer. This is so natural that you don't even think about it, but if you put your wheels in a groove so no steering is possible you will learn in a hurry.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2017-12-08T13:56:55.223" LastActivityDate="2017-12-08T13:56:55.223" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18341" PostTypeId="1" AcceptedAnswerId="19459" CreationDate="2017-12-05T02:42:54.080" Score="0" ViewCount="99" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EmGun.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EmGun.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I am having some difficulty determining the constants $a, b, k$ in the open circuit voltage transfer function $T(s) = v_{0}/v_{i}(s)$, which is given by $T(s) = \dfrac{k [s + a]}{[s + b]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the constants must be in terms of the elements $\{C_{1}, C_{2}, gm1, gm2, GL[=1/RL]\}$ positive. I also know that the $gm$ can be negative when switching the input terminal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="14046" LastActivityDate="2018-02-16T18:41:05.343" Title="Small Signal OTA Circuit" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18344" PostTypeId="1" CreationDate="2017-12-02T04:33:34.810" Score="0" ViewCount="83" Body="&lt;p&gt;I understand this is a multifaceted question and I apologize . All of it is relevant to be understood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On Earth Fly wheel storage devices are used as an alternative to batteries. Even with magnetic bearings these devises expand and contract making the clearance between parts an issue. Between the Earth and Moon temperature can rang from −270 °C and −454 °F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maintaining the bearings in them and the weight make them less efficient then batteries. Transporting them while they are spinning is also an issue. Moving parts that touch wear out. Bearing must be maintained and has it own set of issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I propose this as an alternative: This type of magnetic levitation below can be used in place of the magnetic bearings at 0 gravity and possibly on Earth or moon. Are flywheels used in space and in this way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lower drawn model below on the left will work on Earth but the gravity and air slow it down. No air on moon even better in space. The flywheel can be much heavier and be contained/entrapped using 2 magnetic bases and the magnetic core in the center of the flywheel.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using timers to apply Gauss to a curved edge of the magnetic core it could be spun faster similar to spinning a basket ball on your finger with out knocking it off the axis, but also like a brushless motor. This would also be easier to transport. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there still be any friction other than when electricity is collected or Gauss is applied while spinning? The picture below is my version of a &lt;strong&gt;bearingless&lt;/strong&gt; version of a flywheel storage devise . The design is basic, but is it plausible?  When up scaled? Why not? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am hoping someone could improve the design. The disks would be 2 parts to a space station. The flywheel would be large and heavy to the point of simulating gravity if you where standing inside the drum also heavy enough where power drawn shouldn't cause a noticeable shift in gravity.  It doesn't have to spin fast if it is heavy. Since the main part do not touch they will last indefinitely. Constructing the magnets can be done in parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LXTMV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LXTMV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L0RWD.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L0RWD.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This link demonstrates well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://media.giphy.com/media/OffasXBv2RmOQ/source.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://media.giphy.com/media/OffasXBv2RmOQ/source.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z1ADS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z1ADS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13974" OwnerDisplayName="Procreator" LastEditorUserId="13974" LastEditDate="2017-12-09T05:28:59.460" LastActivityDate="2017-12-09T05:28:59.460" Title="Magnetic Entrapped Storage at 0 gravity" Tags="|design|dynamics|energy|magnets|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18345" PostTypeId="2" ParentId="18344" CreationDate="2017-12-02T09:15:06.403" Score="7" Body="&lt;p&gt;Flywheels are used in space, but not for storage of energy. They are used for attitude adjustment, as the spinning of a large mass has significant implications in an object which is free to move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike on Earth, where a flywheel or gyro can be spun up against the force holding it in place (bolted to foundations), and change in spin speed will cause the craft it is attached to to rotate the other direction. This is true whether or not you are accelerating or decelerating the flywheel, so when putting energy in, or trying to recover energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary - very useful for rotating a satellite or craft, not so useful for storing energy (in fact if a gyro has been spun up to its maximum and further attitude adjustments are required in that same direction, there is a whole process for bleeding off that spin in order to have a useful adjustment range once more)&lt;/p&gt;&#xA;" OwnerUserId="381" OwnerDisplayName="Rory Alsop" LastActivityDate="2017-12-02T09:37:23.037" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18346" PostTypeId="2" ParentId="18344" CreationDate="2017-12-05T15:21:40.300" Score="1" Body="&lt;p&gt;If you build this device on the scale you suggest (a space station with 2 toruses as the 2 flywheels), the variations in rotation speed lead to variations in gravity in the toruses. Variable gravity makes things difficult for humans: for example, extending your arm to place a cup of coffee on a table. The weight of arm+cup informs the amount of muscle tension needed in the arm, lots of practice allows you to do this movement without thinking about it. When the gravity changes, your calibration is off and you end up banging into the edge of the table instead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also won't be able to get away with a design with no bearings: you have two pressurized spaces rotating in opposite directions that have to be connected to each other with a reasonably airtight seal. Plus numerous connections for power, data, water, gases etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speculation: having two masses rotate in opposite directions is a good start, but I'm not convinced it's enough to counteract the torque you get when trying to change the station's attitude. The two masses are each at a distance to the center of mass of the station, so their moments don't cancel each other out. &lt;/p&gt;&#xA;" OwnerUserId="11645" LastActivityDate="2017-12-05T15:21:40.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18347" PostTypeId="2" ParentId="18336" CreationDate="2017-12-05T15:36:24.397" Score="1" Body="&lt;p&gt;I think I would be tempted to put both M2 and M3 on rails with wheels if they are inset into the respective blocks you should be able to keep a pretty small gap between the respective blocks which should preserve the visual effect, minimise friction and also has the advantage that you are constraining the movement to a single plane and you could also included stops and buffers to make it all easy to reset. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you could even use sets of three wheels like roller-coaster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach would probably also make it easier to build as you could fabricate the blocks and use off the shelf precision ground steel stock for the rails and not have to worry about machining super flat surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;maybe use small Vee pulleys for wheels running on round silver steel rails. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;M1 could just go on fairly big wheels, again if they are inside a hollow block the exterior faces could sit just above the surface it sits on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would probably use something like dyneema cord for the string as it has a very high strength to weight ratio and is more flexible than steel cable or wire and so behaves well with small pulley systems and can be knotted. It's used for things like fishing line and climbing accessory cord and is available in a wide variety fo diameters and colours fairly cheaply &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-12-05T15:36:24.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18348" PostTypeId="2" ParentId="18334" CreationDate="2017-12-05T15:49:22.050" Score="1" Body="&lt;p&gt;You are correct, that water at 180°C and 2 MPa will be a liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the steam is not being injected at the &quot;wellbore pressure&quot;, or its temperature rises during the injection process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YJh0a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YJh0a.png&quot; alt=&quot;Water Phase Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2017-12-05T16:14:07.790" LastActivityDate="2017-12-05T16:14:07.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18349" PostTypeId="1" AcceptedAnswerId="18350" CreationDate="2017-12-05T17:28:35.537" Score="0" ViewCount="312" Body="&lt;p&gt;Why don't we use fillets when making buildings and houses (i.e why do we leave the edges sharp?)? Does stress concentration factor not matter in making structures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am talking about the building like this--&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RnQrR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RnQrR.jpg&quot; alt=&quot;Building with sharp edges&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this edge is not &lt;code&gt;highly stress concentrated&lt;/code&gt;(only the atmospheric pressure is acted on these edges, which is easily bearable by this building.) but assume the time of any natural disaster, at that time these type of edges would be more likely to fail.And making the edges round doesn't affect the cost much.I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOW ANOTHER EXAMPLE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ADZfW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ADZfW.jpg&quot; alt=&quot;Stiars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also stress concentrated.As we climb the stairs there will definitely some forces will act.&lt;/p&gt;&#xA;" OwnerUserId="14063" LastEditorUserId="14063" LastEditDate="2017-12-07T08:08:37.377" LastActivityDate="2017-12-07T08:08:37.377" Title="Why we don't use fillets while making buildings?" Tags="|mechanical-engineering|civil-engineering|structures|" AnswerCount="2" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18350" PostTypeId="2" ParentId="18349" CreationDate="2017-12-05T19:47:54.860" Score="3" Body="&lt;p&gt;The issue is simply cost. The loads in structures such as benches (which you describe in comments under the question) are very low and such structures often have huge safety margins. These margins are not intentional for such trivial structures, but are a consequence of working with reinforced concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reinforced concrete requires some concrete cover to protect the steel reinforcement from oxidation. For structures under small loads such as benches, just this concrete cover around very light steel reinforcement is already far more than sufficient to withstand the loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, stress concentrations are no issue: worst case, a bit of concrete around the edges will be chipped off, but the structure will still be more than capable of withstanding the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So fillets become an added construction cost: it's cheaper to build a simple rectangular cross-section than a filleted one. Comfort is not a primary concern.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2017-12-06T13:18:09.210" LastActivityDate="2017-12-06T13:18:09.210" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18351" PostTypeId="1" AcceptedAnswerId="51560" CreationDate="2017-12-05T20:35:01.053" Score="2" ViewCount="4648" Body="&lt;p&gt;I am dealing with a setup similar to the image below.  The horizontal plate is bolted to a horizontal concrete surface (Bridge Abutment). And a horizontal force is being applied to the the vertical.  My question is how do you determine how thick the stiffener plate needs to be.  I am looking to understand the general process and then I will turn around and try to apply my code (CHBDC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/arjvr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/arjvr.png&quot; alt=&quot;Steel stiffener&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts are you need to check to see if it will buckle.  Treat it like a column and performa k*L/r check.  k=2 for fixed fixed connection due to welds.  not sure about section properties as they vary.  Also not how to convert a horizontal point load in the middle of the plate into axial load? in the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was looking to stay out of FEA. Wondering if there was some basic first principles formula or something along those lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am actually trying to size the green peices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o3V5x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o3V5x.png&quot; alt=&quot;double stiffeners&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6993" LastActivityDate="2022-06-30T23:58:58.480" Title="How to calculate the required thickness of a steel stiffener plate in a corbel/bracket" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18352" PostTypeId="2" ParentId="18349" CreationDate="2017-12-05T20:45:24.653" Score="1" Body="&lt;p&gt;In general architecture there is a cost pressure towards standardisation and it is often cheaper to use standard parts which are easy to assemble, specify and transport. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the fact that in architecture the crucial loads are often supporting the span of the roof so the big loads tend to be compression and thrust on the walls with pinned joints where fillets don't really add much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally in steel framed buildings it is often cheaper to over-specify the sections used rather than optimise them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that there are certainly plenty  of example of architecture where material limited (eg stone) have produced extrmely elegant structurally optimised designs : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kOulb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kOulb.jpg&quot; alt=&quot;Vaulted Ceiling&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="13311" LastEditDate="2017-12-06T11:49:26.017" LastActivityDate="2017-12-06T11:49:26.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18353" PostTypeId="2" ParentId="18324" CreationDate="2017-12-05T21:32:02.550" Score="1" Body="&lt;p&gt;For anyone interested in how to solve this, I have reached out to my thermo professor who has come back with a very helpful way of working this out using the formula:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$rate.of.cooling [W] = rate.of.heat.production = h A (T_{surface} - T_{air})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, $h$ is the convection coefficient, $A$ is the surface area of the processor, $T_{surface}$ is the surface temperature of the processor and $T_{air}$ is the average temperature of the inlet air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of cooling will remain the same no matter the airflow. The convection coefficient is proportional to the velocity of the air, so any changes to it's value will require an equivalent change in velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial $T_{surface}$ value is $353K$, so the temperature difference that is driving the heat exchange is $353K - 298K = 50K$. The goal $T_{surface}$ value is $333K$, which has a temperature difference of only $30K$ driving the cooling process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the rate of cooling remains the same at $150J/s$, the other side of the equation must also remain the same. So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$hA(50) = x\cdot h A (30)$$&#xA;$$x = 50/30 = 1.7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the convection coefficient is increased by a factor of 1.7, and velocity is proportional to the convection coefficient, the velocity must be increased a factor of 1.7 in order to maintain the same rate of cooling, but decrease the surface temperature of the processor to $333K$. This is assuming the vented area remains the same.&lt;/p&gt;&#xA;" OwnerUserId="13810" LastActivityDate="2017-12-05T21:32:02.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18354" PostTypeId="1" CreationDate="2017-12-05T22:10:41.833" Score="2" ViewCount="41" Body="&lt;p&gt;I am trying to discretise the PDE: $$\phi \frac{\partial c}{\partial t}+\frac{\partial j}{\partial x}=0$$ where $c$ is a function of $x$ and $t$, and $j=qc-D(q)\frac{\partial c}{\partial x}$, $q$ is the Darcy flux (function of $x$ and $t$) and $D(q)$ is the coefficient of diffusion/dispersion, and $\phi$ is a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;over the interval $(x_{i-1/2}, x_{i+1/2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Progress&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align*} &amp;amp;\phi \frac{\partial c}{\partial t}+\frac{\partial j}{\partial x}=0&#xA;\\ &amp;amp;\Rightarrow \frac{\partial c}{\partial t}+\frac{\partial}{\partial x}\left(\frac{qc}{\phi}-D(q)\frac{\partial c}{\partial x}\right)=0&#xA;\\&amp;amp;\Rightarrow \frac{\partial}{\partial t}\int_{x_{i-1/2}}^{x_{i+1/2}}c \ dx \ +\left|\frac{qc}{\phi}-D(q)\frac{\partial c}{\partial x}\right|_{x_{i-1/2}}^{x_{i+1/2}}=0&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do I do to simplify the expression in terms of the values of the $c_i, q_i$ terms. (I am trying to get matlab to run this iteratively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conventions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_i$ is the midpoint of the $i$th interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\delta x$ is the length of the interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$f_i=\frac1{\delta x}\int_{x_{i-1/2}}^{x_{i+1/2}}f \ dx$$&lt;/p&gt;&#xA;" OwnerUserId="14067" LastActivityDate="2017-12-05T22:10:41.833" Title="Discretise this transport equation" Tags="|matlab|numerical-methods|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18355" PostTypeId="1" AcceptedAnswerId="18360" CreationDate="2017-12-06T01:56:23.627" Score="0" ViewCount="41" Body="&lt;p&gt;I have been wondering about how a positively charged plasma might be used to generate electrical power. So, I was wondering if, similar to a moving negative charge, a moving (or oscillating) positive charge would generate a voltage across a coil.&lt;/p&gt;&#xA;" OwnerUserId="14072" LastActivityDate="2017-12-06T17:27:31.880" Title="Does a moving positive charge generate a magnetic field?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18356" PostTypeId="1" CreationDate="2017-12-06T03:48:32.047" Score="0" ViewCount="114" Body="&lt;p&gt;I am designing an Actual working extendable welding station for my college project. The idea is to lift the unit up/down with the help of 2 lead screws and 2 DC gear-motors, one on each side. The total weight needed to lift will be around 500 lbs. I want to select suitable dc gear-motor for that. Lead-screw will be 3/4&quot; diameter of 0.2&quot; pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the torque calculations done to select a dc motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done this much so far: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = m \cdot a = 227 \mathrm{kg} \cdot 9.8 \mathrm{m}/\mathrm{s}^2 = 2224.6 \mathrm{N}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, we are using 2 motors to drive this, so i just divide the force/2 = 1112.3 N.&#xA;I need the table to lift 15&quot;. So, the lead screws can be around 20&quot; long.&#xA;Considering 20&quot; long lead screw:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = F \cdot r = 1112.3 \mathrm{N} \cdot 0.508 \mathrm{m} = 565 \mathrm{Nm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, that is the absolute minimum torque just to keep the system in steady state right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I need assistance in finding the real torque required and how to assume the minimum acceleration needed to find the real torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how to calculate Power for the motor.&#xA;Based on this calculations, I will be buying the motors to actually install it in my design so I need the calculations to be accurate! Thanks to all in advance.&lt;/p&gt;&#xA;" OwnerUserId="14075" LastEditorUserId="13505" LastEditDate="2017-12-06T14:47:37.007" LastActivityDate="2018-01-05T15:41:16.283" Title="Selection of correct DC gearmotor with given load" Tags="|mechanical-engineering|electrical-engineering|motors|torque|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18357" PostTypeId="1" AcceptedAnswerId="18363" CreationDate="2017-12-06T03:54:01.437" Score="0" ViewCount="71" Body="&lt;p&gt;In order to conduct a study, I need to place a rectangular plate freely placed (1 cm by 2cm) inside water and control it from outside (better wireless). I need to give oscillatory motion with specific frequency/amplitude. &#xA;At present I used LEGO Next servo motors and I could not achieve it successfully due to mechanical instability and methods to control freq and amp. Also the gear assemblies used affected the results.&#xA;May I ask people here to suggest a way to produce vibration motion of this plate with minimal interfering mechanical set up?&#xA;I can choose any kind of plate (plastic or metal) I can employ magnetic or electric methods to produce vibration.&#xA;any direction way forward will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="7200" LastActivityDate="2017-12-08T16:12:32.447" Title="producing vibration motion of a plate (1cmx2cm) placed in water" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18359" PostTypeId="2" ParentId="18356" CreationDate="2017-12-06T14:17:25.640" Score="1" Body="&lt;p&gt;No, it does not work that way ;) 565 Nm is much more than you need for this. The error is in the second part of your calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between force which a lead screw can work against and the torque which has to be applied to the screw is depending on some variables. Assuming that you have steel on steel ($\mu$ = 0.25), using your diameter ($d$ = 19 mm), pitch (which is the same as the lead for normal screws, $l$ = 5.08 mm) and this formula (&lt;a href=&quot;https://en.wikipedia.org/wiki/Leadscrew&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiki&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = \frac{F \cdot d}{2} \cdot \frac{l + \pi \cdot \mu \cdot d}{\pi \cdot d-\mu \cdot l}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives you a required torque of 3.6 Nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;However&lt;/em&gt;, consider that this does not include acceleration. After correcting your $F$ by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ F = \frac{m \cdot (g + a)}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $g = 9.81 \frac{\mathrm{m}}{\mathrm{s}^2}$ and $a$ the required acceleration ($1 \frac{\mathrm{m}}{\mathrm{s}^2}$ should be more than sufficient). I would furthermore add a security factor of 2 before selecting the motor. Keep in mind that you have to include all moving parts into your $m$, meaning that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m = m_{table} + m_{everything~positioned~on~the~table} + m_{lead~screw}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correct your values and recalculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A stepper motor might be more suitable for your application than a geared DC motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="13505" LastEditorUserId="13505" LastEditDate="2017-12-06T14:25:47.617" LastActivityDate="2017-12-06T14:25:47.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18360" PostTypeId="2" ParentId="18355" CreationDate="2017-12-06T17:27:31.880" Score="1" Body="&lt;blockquote&gt;Does a moving positive charge generate a magnetic field?&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.  The real question is why would you think it wouldn't.  Maxwell's equations are clear and have stood the test of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The polarity of the charge only effects the polarity of the resulting magnetic field.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-12-06T17:27:31.880" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18361" PostTypeId="1" AcceptedAnswerId="19099" CreationDate="2017-12-06T18:34:48.007" Score="1" ViewCount="935" Body="&lt;p&gt;western United States.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Problem / Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The bad power in my house is somehow producing EMI in my audio equipment. Will a simple Faraday cage help? (Using an &lt;a href=&quot;https://www.homedepot.com/p/Andersen-20-11-16-in-x-55-13-32-in-Stone-Aluminum-Casement-Insect-Screen-C5-1344020/100655642&quot; rel=&quot;nofollow noreferrer&quot;&gt;aluminum metal screen&lt;/a&gt;) &#xA;If not that, then what would be the best, most practical, cost-effective solution to blocking this interference.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As is usual on S.E., if you think I'm wrong about something, please tell me why.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm not inexperienced with engineering / physics, although not a professional. Technical explanations are welcome but not required.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Story&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I recently moved into a new rental house. It has old wiring (pre-60s) which is mostly un-grounded and quite degraded. The landlord offered to ground a few of the outlets, but not replace all the wiring. For now, I'm going to have to live with the wiring as-is.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why I think this is EMI&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Using an amp with a cord as an antenna, the noise behaves exactly like radio wave reception in a TV or a radio. It changes depending on angle, and dissipating somewhat when I block it with my body.&lt;/li&gt;&#xA;&lt;li&gt;I'm using a new power conditioner with ground noise eliminating circutry.&lt;/li&gt;&#xA;&lt;li&gt;There was no noise at all, even without the power conditioner, at my last place.&lt;/li&gt;&#xA;&lt;li&gt;Several other non-audio electrical appliances around the house hum, and sound quite a bit like the noise I hear in the amps.&lt;/li&gt;&#xA;&lt;li&gt;The produced audio frequencies from the noise are all above 500hz or so. It's my understanding that ground loops are usually low-frequency.&lt;/li&gt;&#xA;&lt;li&gt;Using other outlets, grounded or not, all have the same issue.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Things I already tried that didn't work&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Plugging in to the grounded outlets around the house using an extension cord.&lt;/li&gt;&#xA;&lt;li&gt;Turning off as many things as possible to find the source of the noise.&lt;/li&gt;&#xA;&lt;li&gt;Using different circuit breakers, turning off as many as possible.&lt;/li&gt;&#xA;&lt;li&gt;Using different combination of audio equipment to reduce noise. Using shielded, balanced audio cables greatly reduces the hum. However, that's not possible with everything I own.&lt;/li&gt;&#xA;&lt;li&gt;Standing in such a way as to reduce the noise. This works somewhat, but is untenable. (this lead me to the Faraday cage idea).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14088" LastEditorUserId="14088" LastEditDate="2017-12-06T19:26:10.717" LastActivityDate="2018-01-25T20:21:27.610" Title="Faraday cage to reduce EMI in audio amplifier" Tags="|electrical-engineering|rf-electronics|power-transmission|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18362" PostTypeId="1" CreationDate="2017-12-06T20:21:53.697" Score="0" ViewCount="123" Body="&lt;p&gt;I work frequently with materials sensitive to oxygen and moisture, within a glove box. However, there are frequently times I want to move these samples from the glove box to a scanning electron microscope (SEM) or other device without exposing them to air. I can easily use some small module (eg tupperware) which can be sealed manually in the glove box to allow me to safety transfer my samples to another glove box. But, if I want to see it in a SEM or another device, I must have another way to transfer it without exposure. These samples can be very very small, 1 inch in size or less.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could easily build a small box or module for transferring my samples, and seal that box by hand in the glove box. However, I need a mechanism I can use to unseal the module once it has been placed inside of another device, where it is inaccessible by hand. If I could find a small valve that could be opened automatically, that would be suitable for example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest such a mechanism for breaking a sealed environment automatically?&lt;/p&gt;&#xA;" OwnerUserId="3419" LastEditorUserId="90" LastEditDate="2017-12-07T08:38:33.550" LastActivityDate="2024-04-06T09:04:50.617" Title="Automated valve for testing module" Tags="|measurements|valves|instrumentation|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18363" PostTypeId="2" ParentId="18357" CreationDate="2017-12-06T22:23:41.757" Score="1" Body="&lt;p&gt;get a loudspeaker and glue a tiny push rod to the center of the cone and to the plate. Then send an AC signal at the frequency of choice into the speaker. This will work for frequencies below the fundamental free-air resonance of the cone, for a 12&quot; speaker this is usually about 1 to 2 kHz. for frequencies higher than that (up to 5kHz) you can use the driver unit off of a high-power horn type tweeter. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-12-06T22:23:41.757" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18364" PostTypeId="2" ParentId="18336" CreationDate="2017-12-07T06:04:29.273" Score="1" Body="&lt;p&gt;Some issues are considered in other answers, I'll just add that you can cut part of the surface of M1 which is in contact with M2 and M3 and build in small cylinders so that you get rolling in stead of sliding, friction should be very small. Something similar to &lt;a href=&quot;https://www.lewcoinc.com/conveyors/products/gravity-roller-conveyor/1418.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;roller conveyor&lt;/a&gt; like in the link just be careful to align them with the surface to hide them, should be simple enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also maybe think about creating exchangeable parts so your professor can show students what happens when friction is or isn't acting.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2017-12-07T10:56:32.770" LastActivityDate="2017-12-07T10:56:32.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18365" PostTypeId="1" AcceptedAnswerId="18403" CreationDate="2017-12-07T08:05:56.607" Score="3" ViewCount="500" Body="&lt;p&gt;On &lt;a href=&quot;https://en.wikipedia.org/wiki/Water_jet_cutter&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Water jet cutter&quot;&gt;Wikipedia&lt;/a&gt; it states that the main benefit of a water jet cutter is that there is no heat-affected zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering about the physics behind this. I have found experiments supporting this, but I cannot produce a theoretical argument. Am I wrong in assuming that by the time it would take to heat up the material, a new influx of cold water keeps the temperature down?&lt;/p&gt;&#xA;" OwnerUserId="10979" LastEditorUserId="13311" LastEditDate="2017-12-07T17:51:41.497" LastActivityDate="2017-12-10T18:30:58.597" Title="Why is there no heat-affected zone with water jet cutting?" Tags="|thermodynamics|heat-transfer|cutting|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18366" PostTypeId="1" CreationDate="2017-12-07T17:14:55.900" Score="1" ViewCount="12" Body="&lt;p&gt;I would like to perform parameters estimation for a differential equation from experiment. If the value of these parameters are obtained from several experiments (repetitions) where for each experiment different final cost function is obtained, how can I calculated the final parameters values from these experiments? &#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="14099" LastActivityDate="2017-12-07T17:14:55.900" Title="How to combine ODE parameters values from several estimations?" Tags="|experimental-physics|transfer-function|cost-estimation|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18367" PostTypeId="1" CreationDate="2017-12-07T17:39:36.423" Score="1" ViewCount="405" Body="&lt;p&gt;What is the NORMAL STRESS in a thin walled pressure vessel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of only 2 types of stresses in thin walled pressure vessels: Longitudinal &amp;amp; Tangential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is any one of these 2 is considered as the NORMAL STRESS? If yes, which one and why?&lt;/p&gt;&#xA;" OwnerUserId="9540" LastActivityDate="2018-08-28T18:41:58.307" Title="Thin walled pressure vessel" Tags="|mechanical-engineering|applied-mechanics|pressure-vessel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18368" PostTypeId="2" ParentId="18365" CreationDate="2017-12-07T18:10:03.133" Score="3" Body="&lt;p&gt;I can't speak to the mathematics of the heat generated during water jet cutting, but the following points may clarify the situation for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat affected zone is generally with metals where the cutting process will generate enough heat to change the mechanical properties of the metal, essentially locally heat treating the material at and nearby the cutting area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Waterjet does not get near hot enough to create a heat affected zone. Remember, It's water at ambient pressure so it would become steam if it were any hotter than boiling temperature. That would make it an ineffective medium to carry the abrasive to blast through the metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, let's assume that the machine's water reservoir/supply is large enough where its temperature is stable. The pumping/pressurizing will create heat and the abrasive cutting action will create heat, but there is a &quot;constant&quot; supply of water at a stable temperature being applied to the material which would regulate the materials temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, a waterjet part could still become warm to the touch after being cut.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-12-07T18:10:03.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18369" PostTypeId="1" CreationDate="2017-12-07T19:21:46.067" Score="1" ViewCount="98" Body="&lt;p&gt;I am trying to work out the compressive stress in a material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a strut diameter  of 30mm. Length is 2.5m. Load 0f 30kN.&#xA;Modulus of elasticity for the material is 160GPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find the amount it compresses by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have came up with the answer: 1.130973355x10^10m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using Stress/Strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stuck on the 160GPa. I have wrote it in the equation as: 160x10^9.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not sure if this is correct. And it should just be 160.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And not sure if I have calculated it correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help or give advice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="14100" LastActivityDate="2018-01-07T15:53:41.053" Title="The compressive stress in a material" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18370" PostTypeId="2" ParentId="18367" CreationDate="2017-12-07T20:14:06.230" Score="5" Body="&lt;p&gt;Both Axial stress, (parallel to the axis of cylindrical symmetry) and Hoop Stress (in the tangential direction) are &quot;Normal Stresses&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are &quot;Normal&quot;, because, if you consider an infinitesimal unit volume, the stresses can be considered to be entirely axial in the X and Y directions respectively. There is no stress in the Z direction because of the 'thin wall' assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other types of stress are Bending, and Shear, neither of which can be induced by the pressure of the gas, in a symmetrical vessel.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-12-07T20:14:06.230" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18371" PostTypeId="1" AcceptedAnswerId="18373" CreationDate="2017-12-07T22:43:49.893" Score="2" ViewCount="112" Body="&lt;p&gt;Here is how this artist rendered &lt;a href=&quot;https://en.wikipedia.org/wiki/Isambard_Kingdom_Brunel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brunel's&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Royal_Albert_Bridge&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saltash Bridge&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/axaDo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/axaDo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is how it appears today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0P9km.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0P9km.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did the appearance change, or was some artistic licence taken in the painting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;Did this artist take some licence with the Royal Albert Bridge - or has its appearance changed?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="14103" LastActivityDate="2023-02-07T21:46:42.680" Title="Did this artist take some licence with the Royal Albert Bridge - or has its appearance changed?" Tags="|bridges|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18372" PostTypeId="1" AcceptedAnswerId="18374" CreationDate="2017-12-07T23:19:05.033" Score="1" ViewCount="47078" Body="&lt;p&gt;Most air conditioners have three modes: Auto, Sun, Snowflake&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference it terms how it works?&lt;/p&gt;&#xA;" OwnerUserId="14104" LastActivityDate="2020-11-28T19:01:52.307" Title="What's the difference between aircon modes Auto, Sun, Snowflake?" Tags="|airflow|temperature|electrical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18373" PostTypeId="2" ParentId="18371" CreationDate="2017-12-08T00:09:27.670" Score="3" Body="&lt;p&gt;Probably artistic interpretation. #i see the differences you are alluding  to but they don't look like things you could reasonably alter in a structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The artist's rendering looks like an engraving or etching and as such the final image would have gone through several stages probably starting with a sketch in the field in pencil or watercolours before being worked up into a more detailed composition and then being adapted by the engraver. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each stage there would have been some degree of interpretation to produce a pleasing composition as well as practical limitations imposed by the processes used. It is certainly not uncommon for landscape artists to idealise scenes a bit and engravers would need to interpret the image to convert it into a printable grey-scale format.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-02-06T18:16:57.617" LastActivityDate="2018-02-06T18:16:57.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18374" PostTypeId="2" ParentId="18372" CreationDate="2017-12-08T02:24:14.620" Score="4" Body="&lt;p&gt;&lt;strong&gt;The sun is heating mode&lt;/strong&gt;. When the room temperature reaches the set temperature, the air conditioner stops operating until the temperature falls below the set temperature and the starts operating again. When in heating mode, the air conditioner does not cool. This setting is used during cold weather periods, such as in winter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The snowflake is cooling mode&lt;/strong&gt;. When the room temperature reaches the set temperature, the air conditioner stops operating until the temperature rises above the set temperature and starts operating again. When in cooling mode, the air conditioner does not heat the room. This setting is used in hot weather periods, such as summer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Auto is automatic mode&lt;/strong&gt; where the air conditioner can either heat or cool as required. It tries to achieve the set temperature by switching from heat mode to cooling mode automatically. If you don't want to change between heat and cool modes when the seasons change, just set the air conditioner to automatic mode.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2017-12-08T10:58:53.260" LastActivityDate="2017-12-08T10:58:53.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18375" PostTypeId="1" CreationDate="2017-12-08T08:01:43.987" Score="1" ViewCount="31" Body="&lt;p&gt;I have a tiny plastic disc of dia 1cm submerged inside water in a container in the halfway.&#xA;I have to make the disc to oscillate sinusoidal with a frequency(1Hz to 10Hz) and amplitude (1 cm to 5 cm). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got an &lt;a href=&quot;https://engineering.stackexchange.com/questions/18357/producing-vibration-motion-of-a-plate-1cmx2cm-placed-in-water&quot;&gt;answer&lt;/a&gt; for similar question based on a mechanical method using a loud speaker. But this does not serve the purpose as a vibrating connector creates ripples and affect my observations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I produce such oscillation and control externally without any wires?&#xA;I made a failed attempt using Lego gears owing to mechanical instability and interference with other moving parts.&#xA;I can use magnetic electric or any other means and I can replace plastic with any other material that can respond to external field variation. Therefore looking for a non mechanical method&lt;/p&gt;&#xA;" OwnerUserId="7200" LastEditorUserId="7200" LastEditDate="2017-12-09T06:07:54.463" LastActivityDate="2017-12-09T06:07:54.463" Title="vibrate a floating microbody" Tags="|fluid-mechanics|applied-mechanics|" AnswerCount="0" CommentCount="0" ClosedDate="2017-12-09T05:02:27.013" ContentLicense="CC BY-SA 3.0" />
  <row Id="18376" PostTypeId="2" ParentId="18369" CreationDate="2017-12-08T09:11:34.913" Score="1" Body="&lt;p&gt;For a one-dimensional linear-elastic case you can use Hooke's Law:&#xA;$$ \varepsilon=\frac{\Delta L}{L_0} \qquad (1) $$&#xA;where $L_0$ … initial length of the member, $\Delta L$ … change in length&#xA;(The convention that I use is, that tensile stresses cause positive strains and compressive stresses cause negative strains.)&#xA;Assuming a uniform distribution of stress:&#xA;$$ \sigma=\frac{F}{A} \qquad (2) $$&#xA;where $\sigma$ … normal stress, $F$ … Force acting on member, $A$ … cross-sectional area of member&#xA;Now, Young's Modulus is defined as the slope of the stress-strain diagram in the linear-elastic region:&#xA;$$ \sigma=E\cdot \varepsilon \qquad (3) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By combining equations (2) and (3) you get the resulting strain:&#xA;$$ \varepsilon =\frac{F}{A\cdot E} \qquad (4)$$&#xA;and by inserting equation (1) and solving for $\Delta L$ you get:&#xA;$$ \Delta L= \frac{F\cdot L_0}{A\cdot E} \qquad (5) $$&lt;/p&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="13602" LastEditDate="2017-12-08T15:52:45.653" LastActivityDate="2017-12-08T15:52:45.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18377" PostTypeId="1" CreationDate="2017-12-08T11:55:07.883" Score="1" ViewCount="72" Body="&lt;p&gt;I actually don't know on which SE to ask this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there are quite often too low bridges/tunnels, or too high trucks or trailers, why isn't there 100m ahead of those bridges, some light frames (like football goal cage) the size and height of the entrance, which isn't dramatic to destroy, allowing driver to know they won't go through, and giving them time to brake before causing damage to both tunnel/bridge, truck,... ? Or are there already?&lt;/p&gt;&#xA;" OwnerDisplayName="user12750" LastActivityDate="2018-10-22T21:19:25.707" Title="Light frame at tunnels and bridges" Tags="|bridges|architecture|tunnels|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18378" PostTypeId="1" CreationDate="2017-12-08T14:31:17.613" Score="1" ViewCount="143" Body="&lt;p&gt;I am trying to figure out if there is a decrease in electricity consumption of a cooling unit if the ambient temperature decreases from 35°C to 20°C. More detailed, a company is planning to install an HVAC system in a production hall with huge machines. Currently, each machine is cooled by a cooling unit attached to the machine. The question, will the electricity consumption of the cooling units decrease if the temperature in the production hall decreases?&lt;/p&gt;&#xA;" OwnerUserId="14108" LastActivityDate="2017-12-09T01:15:20.407" Title="Temperature dependent electricity consumption of HVAC System" Tags="|hvac|cooling|mechanical|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18379" PostTypeId="2" ParentId="18378" CreationDate="2017-12-08T15:44:30.370" Score="1" Body="&lt;p&gt;A machine consumes energy, part of which end-up as heat in the machine structure itself and in the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cooling unit dissipate heat from a place and &quot;push&quot; this heat somewhere else. In our case, the cooling unit dissipate from the machine and push this heat to the room. The difference of temperature between this other place (the room) and the machine determinate the efficiency of the cooling system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the OP question, the cooling unit seem &lt;code&gt;active&lt;/code&gt;, meaning it consume energy to actively dissipate heat. For this purpose, the energy consumed by the cooling unit will end-up in the room as well as the dissipated heat of the machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any decent cooling unit has a target temperature, and would adjust energy consumption for  cooling just enough to get the target temperature (thermostat). If that is the case, lowering the room temperature will increase efficiency of the cooling unit, which will reduce it consumption accordingly. &lt;strong&gt;(Yes to your answer)&lt;/strong&gt;&lt;br&gt;&#xA;If the unit is not able to adjust it own consumption, it will consume the same energy to cool the machine to a lower temperature due to the higher efficiency. (No to your answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, another alternative would be to conduct the machine heat or the cooling unit heat to the exterior(*) this would reduce the heat generated inside the room and by this way, reduce the room temperature, lower the cooling unit consumption, etc.. without requiring the HVAC. If the outer temperature is low enough, improving ventilation of the room (with exterior) might already have a huge impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) This is supposing you have a lower temperature outside.&lt;/p&gt;&#xA;" OwnerUserId="11129" LastActivityDate="2017-12-08T15:44:30.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18380" PostTypeId="1" CreationDate="2017-12-08T15:58:13.673" Score="3" ViewCount="170" Body="&lt;p&gt;I am interested in developing an AutoCAD plugin and am trying to understand the relationships between several different types of AutoCAD plugin files:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Managed DLLs that ship with AutoCAD plugins&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://fileinfo.com/extension/arx&quot; rel=&quot;nofollow noreferrer&quot;&gt;ARX files&lt;/a&gt; that ship with AutoCAD plugins&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://fileinfo.com/extension/cuix&quot; rel=&quot;nofollow noreferrer&quot;&gt;CUIX files&lt;/a&gt; that ship with AutoCAD plugins&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From what I can tell these three files are all inter-related and work together to initialize and load a plugin, I'm just not seeing the &lt;em&gt;forest through the trees&lt;/em&gt; as to what information each file type contains, which files loads or depends on the others, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;sounds&lt;/em&gt; like a plugin's core functionality is supposed to go inside the ARX file, and written in C++ and compiled for the native/target platform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also &lt;em&gt;sounds&lt;/em&gt; like, at startup, AutoCAD loads its plugin DLLs, which are all managed .NET libraries (typically written in C#), and in turn they invoke their native/compiled respective ARX files. As to where the CUIX file comes into play, I still have no idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can anyone help explain the relationships &amp;amp; dependencies of these files, and what types of code/logic/data goes inside of them?&lt;/strong&gt; Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="14110" LastEditorUserId="14110" LastEditDate="2017-12-08T16:15:40.887" LastActivityDate="2024-04-02T04:08:00.940" Title="Understanding AutoCAD plugin files by Example" Tags="|autocad|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18381" PostTypeId="2" ParentId="18357" CreationDate="2017-12-08T16:12:32.447" Score="0" Body="&lt;p&gt;Professor Wiciak has done some similar researches using &lt;a href=&quot;http://acoustics.ippt.gov.pl/index.php/aa/article/view/1419&quot; rel=&quot;nofollow noreferrer&quot;&gt;piezoelements&lt;/a&gt; to induce vibrations.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastActivityDate="2017-12-08T16:12:32.447" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18382" PostTypeId="2" ParentId="18377" CreationDate="2017-12-08T16:24:33.920" Score="3" Body="&lt;p&gt;Such measures do exist, but they don't always prevent careless drivers from ignoring the measures and crashing into low bridges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One bridge that I know of has been hit at least 100 times and has been hit many times since a &lt;a href=&quot;http://www.abc.net.au/news/2016-05-27/montague-street-bridge-gets-new-safety-measures/7450628&quot; rel=&quot;nofollow noreferrer&quot;&gt;safety gantry was installed in 2016&lt;/a&gt; to prevent such crashes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a picture of the safety gantry that was installed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LPDsi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LPDsi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The the bottom of theblack and yellow poles hanging from the gantry swing is the same height above the road as the clearance height of the bridge. When a vehicle that is too high for the bridge passes under the gantry, the poles make a loud noise as the vehicle drives under the gantry, alerting the driver there is a low height hazard ahead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bridge is so notorious for being hit, a &lt;a href=&quot;http://howmanydayssincemontaguestreetbridgehasbeenhit.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;website&lt;/a&gt; has been created which shows pictures of vehicles that have crashed into it and there is a count of days since the bridge was last hit.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-12-08T16:24:33.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18385" PostTypeId="2" ParentId="18362" CreationDate="2017-12-08T17:55:35.163" Score="0" Body="&lt;p&gt;Interesting problem. The box is under vacuum, so thankfully that works in your favour when you're transporting it in atmosphere as the air will push down on any seals, helping to keep it closed. This means you don't need to go overboard with any gaskets/compounds/o-rings etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the box is safely back inside a vacuum (the SEM) The pressure on any seals due to air will be negated so whatever is driving the lid open/closed will not have to be large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't think of any off the shelf ideas so here's one potential way. My solidwork skills leave A LOT to be desired so don't judge me too harshly! I've just put in what i thought was needed to get the point across. There's many different mechanisms to do this same thing, just converting rotational to linear action (scotch yoke/ leadscrew etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the box actually choose when to open/close is another issue. You can implement a timer (with an arduino perhaps) to give you enough time to transport it and let the SEM pump down, or even add an external pressure sensor to the box to only open when the external pressure of the box is acceptable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One major thing to be wary of with any home-made system though is potential contamination of the vacuum system. If any capacitors or even the battery outgas (according to EEVBLOGs forum lithium battery's can survive due to the lack of liquid in the cell) then you'll be at fault.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eUARK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eUARK.png&quot; alt=&quot;Lid Open&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/Is9Fn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Is9Fn.png&quot; alt=&quot;Lid closed&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your SEM has external handles to manipulate the object under scan this problem gets much easier! &lt;/p&gt;&#xA;" OwnerUserId="12187" LastActivityDate="2017-12-08T17:55:35.163" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18387" PostTypeId="1" CreationDate="2017-12-08T22:42:33.520" Score="2" ViewCount="90" Body="&lt;p&gt;A typical nonlinear FE problem can be given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\mathbf{R} = \mathbf{K}\rho + \mathbf{F}_{NL}(\rho),&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where R is the vector of applied forces, K is the material or linear stiffness matrix, $F_{NL}$ is the nonlinear force function and $\rho$ is the vector of displacements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tangent stiffness matrix is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;K_T = K + \dfrac{\partial F_{NL}}{\partial \rho}(\rho)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an incremental procedure is applied, $\dfrac{\partial F_{NL}}{\partial \rho}$ will need to be updated at load application step, since it is a function of the deformed configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does the material stiffness matrix $K$ not need to be updated? $K$ is formed by rotating local stiffness matrices and assembling them into a global frame of reference. Surely when the geometry becomes deformed $K$ will need updating also, even if it is the linear stiffness matrix?&lt;/p&gt;&#xA;" OwnerUserId="10539" LastActivityDate="2018-05-08T01:49:12.537" Title="In an incremental nonlinear FE procedure does the material stiffness matrix need to be updated?" Tags="|mechanical-engineering|finite-element-method|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18388" PostTypeId="2" ParentId="18387" CreationDate="2017-12-08T23:48:47.383" Score="1" Body="&lt;p&gt;Depends on what you mean by &quot;nonlinear &quot;.  Elastic plastic material constitutive model? Large deflections? Large rotations? Follower forces?  Element failure?  Contact elements? In some situations you will update everything, in other situations you will have a linear matrix which does not get updated and another one that does. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-12-08T23:48:47.383" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18389" PostTypeId="2" ParentId="18378" CreationDate="2017-12-09T01:15:20.407" Score="0" Body="&lt;p&gt;Assuming that the cooling unit attached to the machine is an &lt;strong&gt;air-cooled direct expansion unit&lt;/strong&gt;, the cooling coefficient of performance of the unit (COP), which dictates how much energy the unit will use, is mainly dependent on the dry bulb temperature of the ambient air surrounding the condensing unit of the cooling system and the actual cooling output the system has to provide. We can assume that the cooling output required by the machine will stay the same (even if it will actually probably decrease due to the fact the the temperature of the hall will be reduced, which will reduce further the energy use of the machine cooling system) so the COP will mostly change because of the reduced ambient temperature. The cooling COP of refrigeration cycle increases with the a decrease in temperature of the air surrounding the condensing unit (here the hall but usually outdoors) so the cooling system of the machine will use less energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the cooling energy use of the machine cooling system will be reduced but this will be offset by the energy use of the cooling system installed to condition the hall. By how much? We would need a lot more information.&lt;/p&gt;&#xA;" OwnerUserId="8047" LastActivityDate="2017-12-09T01:15:20.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18390" PostTypeId="1" CreationDate="2017-12-09T07:16:28.817" Score="0" ViewCount="64" Body="&lt;p&gt;Modulate a laser's wavelength? What are the inner workings of a laser, and is it possible to change the wavelength mechanically? Are there any lasers that can change wavelength?&lt;/p&gt;&#xA;" OwnerUserId="14115" LastEditorUserId="14115" LastEditDate="2017-12-13T07:02:51.897" LastActivityDate="2017-12-13T07:02:51.897" Title="FM transmission over laser" Tags="|lasers|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18391" PostTypeId="2" ParentId="18390" CreationDate="2017-12-09T14:40:40.953" Score="2" Body="&lt;p&gt;No, it is not possible to modulate the wavelength of a laser. The particular wavelength of light is determined by the amount of energy released when an excited electron drops to a lower orbit, which in turn is defined by the specific glass/crystal/gas used in the laser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you could design some sort of mechanism to swap in/out different emission media to make step-changes in wavelength, you would not be able to reproduce a meaningful signal.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-12-09T14:40:40.953" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18392" PostTypeId="1" AcceptedAnswerId="18458" CreationDate="2017-12-09T15:46:05.237" Score="5" ViewCount="639" Body="&lt;p&gt;I am researching about ways to continuously determine black spots (size 50+ micron) in our product stream. This stream consists of imperfect spherical polymer beads (size 3 mm) of different materials (ABS, SEBS, etc.) and colors (opaque and transparent). I want to be able to count them as well as characterize them (size). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking into a couple of options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Visual observation using computer vision&lt;/li&gt;&#xA;&lt;li&gt;NIR spectroscopy:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3PsWp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3PsWp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other options do I have and which one is most effective (preferably also least expensive)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;In our production plant we produce at 500 kg/hr and the pellets are transported along a conveyor as a monolayer of around 50 cm width at around 5 cm/s. However i can imagine that is simply too complex to begin with and i am not opposed to having a secondary stream with much lower massflow and then upscale in a later stage or just have labscale setup to check samples taken of the productstream. I am ok with an approximate count as long as it is reproducible. I would also be ok with initially only detecting 100+ micron black spots.&lt;/p&gt;&#xA;" OwnerUserId="3100" LastEditorUserId="3100" LastEditDate="2017-12-13T12:46:23.063" LastActivityDate="2017-12-13T23:59:38.100" Title="How can I detect black spots?" Tags="|materials|measurements|sensors|optics|spectrometry|" AnswerCount="1" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="18394" PostTypeId="2" ParentId="2300" CreationDate="2017-12-09T22:48:00.023" Score="1" Body="&lt;p&gt;lbm and lbf are not the same -- they are only of the same value in one situation, when dealing with gravity at sea level...  examine a situation without gravity, the force produced by a jet of water.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;density of water: 62.4 lbm/ft&lt;sup&gt;3&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;area of nozzle: 0.06 ft&lt;sup&gt;2&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;velocity: 10 ft/s&lt;/li&gt;&#xA;&lt;li&gt;volume flow = area * vel = 0.6 ft&lt;sup&gt;3&lt;/sup&gt;/s&lt;/li&gt;&#xA;&lt;li&gt;F = dwater * volume flow * vel = 374.4 lbm ft/s&lt;sup&gt;2&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;to convert to lbf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F =  374.4 lbm ft/s&lt;sup&gt;2&lt;/sup&gt; divide by 32.2 lbm-ft/lbf-s&lt;sup&gt;2&lt;/sup&gt; = 11.63 lbf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is just counter intuitive to think of quantity of lbm as greater than the quantity lbf, you expect them to be same as they are often interchanged, pound can be used for mass or force --  that it has to be divided by 32.2 lbm-ft/lbf-s&lt;sup&gt;2&lt;/sup&gt; not just 32.2 and not gravity. In the SI system&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;density of water 1000 kg/m&lt;sup&gt;3&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;area of nozzle 0.005574 m&lt;sup&gt;2&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;velocity 3.048 m/s&lt;/li&gt;&#xA;&lt;li&gt;volume flow = area * velocity = .01699 m&lt;sup&gt;3&lt;/sup&gt;/s&lt;/li&gt;&#xA;&lt;li&gt;F = dwater * volume flow * velocity  = 51.78 kg m/s&lt;sup&gt;2&lt;/sup&gt; which is a newton so 51.78 N&lt;/li&gt;&#xA;&lt;li&gt;1 lbf = 32.2 ft/s&lt;sup&gt;2&lt;/sup&gt; lbm&lt;/li&gt;&#xA;&lt;li&gt;1 lbm = .03106 s&lt;sup&gt;2&lt;/sup&gt;/ft lbf  -- just bizarre -- in that you have to add units to the conversion&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;which leads to the question -- what are lbs???  if not lbf  and lbm is nothing more then a mathematical manipulation which creates a lot of confusion, but SI system has a similar problem.  When you weigh sometime you are measuring a force, yet in SI we record this force in terms of mass (kg). Why we can not create a system that make sense is beyond me. The confusion comes from the English system, we should not ask what is your weight, but what is your mass.  Instead of weighing 170&amp;nbsp;lbs, I would respond saying that I have a mass of 5,474 lbm ft/s&lt;sup&gt;2&lt;/sup&gt; (170*32.2) -- time to diet I think. Of course this is ridiculous. The confusion comes from an overgeneralization, i.e. 12 inches in a foot, therefore 32.2 lbm in a lbf in not true.  lbm (mass) must be accelerated before the gravitational constant (gc) can be applied. If I want to find my mass, I would take my weight 170 lbs divide out the local gravitational pull, lets say 30ft/s2 = 5.667 lbf /(ft/s2) and then multiply it by the gc (gravitational constant) 32.2 lbm-ft/(lbf-s2) to get 182.5 lbm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I think the guy who came up with the pound mass (lbm) was dyslexic.  What I think he really wanted to do was state that;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 lbm * 32.2 ft/s2 = 32.2 lbf  that would have been perfect, a lbf = lbm ft/s2, but some idiotic reason he decided that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 lbm * 32.2 ft/s2 should = 1 lbf at sea level on earth, so to make the units work you have to either divide the left side or multiply the right side by gc i.e. 32.2 lbm-ft / lbf-s2.  This means that lbm is not really a mass unit, but a mass gravitational constant unit (which is ridiculous) so when you multiply lbm by an acceleration you have to divide out the gravitational constant before you can obtain a force.  Other than by mistake why would anyone come up with such a unit????  and why do we precise in keeping such a unit???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how much easier it would be that water has a density of 2 lbm/ft3, so that 2 lbm/ft3 * 32.2 ft/s2 = 64.4 lbf/ft2  instead of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;62.4 lbm/ft3 * 32.2 ft/s2 / (32.2 lbm-ft / lbs-s2) = 62.4 lbf/ft2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the logic fails me ... please, someone enlighten me......&lt;/p&gt;&#xA;" OwnerUserId="14121" LastEditorUserId="14121" LastEditDate="2017-12-11T21:40:00.817" LastActivityDate="2017-12-11T21:40:00.817" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18395" PostTypeId="1" CreationDate="2017-12-10T07:14:19.427" Score="3" ViewCount="3239" Body="&lt;p&gt;There's a EU &lt;a href=&quot;https://www.cen.eu/work/products/ENs/Pages/default.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;EN standard&lt;/a&gt; 1822 for air filters  that has seen several iterations. The pre-2009 versions of EN 1822 (which seems to have come out in 1998) were pretty confusing as they allowed poorer filters to be labeled as H or HEPA; e.g. a &lt;a href=&quot;https://www.kaercher.com/int/accessory/hepa-12-filter-64148050.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;HEPA 12&quot; filter&lt;/a&gt; would filter 99.9% of particulates, but this is strictly inferior the US &lt;a href=&quot;https://en.wikipedia.org/wiki/HEPA&quot; rel=&quot;nofollow noreferrer&quot;&gt;HEPA&lt;/a&gt; which filters 99.97% of particles of 0.3 um (assuming Wikipedia is correct about this). The 2009 revision of EN 1822 somewhat fixed this as anything below &lt;a href=&quot;https://www.camfil.com/FileArchive/Brochures/Standards/EN_1822%20eng.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;99.95% is now called &quot;EPA&quot;&lt;/a&gt; instead, so the &quot;HEPA 12&quot; from EN 1822:1998 is now &quot;EPA 12&quot; under EN 1822:2009.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm still not sure if the two standard are really comparable though. I don't know what the particulate size is for EN 1822, nor am I sure what is the difference between &quot;integral value&quot; and &quot;local value&quot; for the emissions under EN 1822. The latter in particular is somewhat worrisome because it carries a footnote that allows manufacturers to change this by agreement with the purchaser, which I suspect some may turn into a sort of click-wrap agreement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is HEPA 13 under EN 1822:2009 (integral value ≥ 99.95%, local value ≥ 99.75%) approximately the same as US HEPA or does one need to move to H14 for that (integral value ≥ 99.995, local value ≥ 99,975)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe there's something else that I'm missing because I've found &lt;a href=&quot;http://www.filtac.com/wp-content/uploads/EpahepaUlpa_E_2-09.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;a comparison&lt;/a&gt; that puts the Mil Std 2&lt;strong&gt;9&lt;/strong&gt;2 numbers below those from EN 1822, so I guess there's a difference in testing methodology that accounts for this. Does someone know the details to clarify?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RZ94i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RZ94i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This latter table surely refers to pre-2009 versions of EN 1822, as no year is stated anywhere in the document and has &quot;H12&quot; and so forth in it. A similar version but using the newer EN terminology &lt;a href=&quot;http://www.venfilter.com/normativa/en-1822-normative-absolute-filters-epa-hepa-and-ulpa&quot; rel=&quot;nofollow noreferrer&quot;&gt;can be found&lt;/a&gt; on the website of a different manufacturer, but oddly there's a E11 but H12 in there. These tables don't inspire much confidence because I'm not sure 292 is even the US Mil. Std. number for HEPA filters, (the 292 standard is about rockets). &lt;a href=&quot;http://golighthouse.nl/sites/default/files/Lighthouse%20Filtertest%20seminar%202014%20presentatie%20EN1822%20FT&amp;amp;C_0.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Different documents&lt;/a&gt; state it as Mil Std 282, which is indeed about fitlers for protective cloting... but otherwise the equivalence stated is about the same [H]EPA 12 being the same as the US HEPA of 99.97%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p1Qr7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p1Qr7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct though, and if it is what in the testing method explains the difference?&lt;/p&gt;&#xA;" OwnerUserId="480" LastActivityDate="2022-12-14T23:06:05.730" Title="EU EN standard 1822 vs US HEPA filters" Tags="|standards|air-quality|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18396" PostTypeId="2" ParentId="18395" CreationDate="2017-12-10T07:44:10.047" Score="1" Body="&lt;p&gt;(Partial answer) it does look like the efficiency depends convexly on particle size, and has a minimum at MPPS (most penetrating particle size);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that could explain the difference in standards. There's one &lt;a href=&quot;https://www.osti.gov/scitech/biblio/6241348/&quot; rel=&quot;nofollow noreferrer&quot;&gt;1982 LLNL tech report&lt;/a&gt; that says that for design considerations, the particle size should be lower at around 0.21 um, at least for a class of (single-fiber) filters&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In this paper the published results of several experiments are compared with a modern filtration theory that predicts single-fiber efficiency and the particle size of maximum penetration. For high-efficiency particulate air (HEPA) filters used under design conditions this size is calculated to be 0.21 ..mu..m diam. This is in good agreement with the experimental data. The penetration at 0.21 ..mu..m is calculated to be seven times greater than at the 0.3 ..mu..m used for testing HEPA filters.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If we take 0.0003 ( = 1 - 0.9997) and multiply it by 7 as suggested in this last report, we get 0.0021, corresponding to an efficiency of 99.79% at MMPS, which would put in in the EN 1822:2009 class E 12 (and/or in ISO class 25E) of 99.95% (integral) efficiency, but not in the class above it (ISO 30E being 99.9%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will accept another answer if someone provides a better (or more in-depth) explanation.&lt;/p&gt;&#xA;" OwnerUserId="480" LastActivityDate="2017-12-10T07:44:10.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18398" PostTypeId="2" ParentId="16510" CreationDate="2017-12-10T12:55:24.893" Score="0" Body="&lt;p&gt;The short answer is &lt;strong&gt;you should not use&lt;/strong&gt; a &lt;em&gt;moving window&lt;/em&gt; because the &lt;em&gt;moving window&lt;/em&gt; itself creates a &lt;a href=&quot;https://en.wikipedia.org/wiki/Moving_average&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;low-pass-ish&lt;/em&gt;&lt;/a&gt; filtering effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as @DanielKiracofe already stated you could slice the signal up into smaller segments and filter those. Of course you need to be careful that the your window size fits your low-pass filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in order to circumvent the problem of stitching the slices together I would use a short segment (duration: 10s) of the signal to test your code and and than run &lt;strong&gt;&lt;a href=&quot;https://de.mathworks.com/help/signal/ref/filtfilt.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;filtfilt&lt;/a&gt;&lt;/strong&gt; on the whole signal to reduce phase-shifting.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2017-12-10T12:55:24.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18401" PostTypeId="1" AcceptedAnswerId="18402" CreationDate="2017-12-10T17:32:02.407" Score="1" ViewCount="1800" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Calculate the radius of iridium atom, given that Ir has an FFC crystal&#xA;  structure, a density of $22.4\text{g/cm}^3$. and atomic weight of&#xA;  $192.2\text{g/mol}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Everyone whom I asked this question is giving different answers. By the way, I was going to ask something about this. Why are they finding that the atomic radius of iridium atom is $1.34\text{pm}, 1.35\text{pm}, \text{1.36pm}$? According to textbook solution, it seems $1.36\text{pm}$. However, they said that all answers correct. Thereupon, I wanted to check their answer from wikipedia, that gives anohter answer too. What would you say? Are all answers correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kindest Regards!&lt;/p&gt;&#xA;" OwnerUserId="13058" LastActivityDate="2017-12-10T19:30:52.220" Title="Calculate the radius of iridium atom, given that Ir has an FFC crystal structure, a density of 22.4 g/cm3 and atomic weight of 192.2g/mol" Tags="|materials|chemical-engineering|chemistry|" AnswerCount="1" CommentCount="4" ClosedDate="2017-12-11T02:58:47.113" ContentLicense="CC BY-SA 3.0" />
  <row Id="18402" PostTypeId="2" ParentId="18401" CreationDate="2017-12-10T18:11:00.920" Score="1" Body="&lt;p&gt;The radius $r$ can be calculated purely from geometry. Consider the FCC lattice. Its lattice parameter $a$ is related to radius through the arrangement of atoms. What is the close-packed direction in the FCC lattice? Along the face diagonal, because two corner atoms touch the face-centered atom. The length of the face diagonal is $\sqrt{a}$ by the pythagorean theorem. That length is also four $r$. You can draw the lattice and the atoms of a face to convince yourself this is true. So $4r=\sqrt{a}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now it remains to calculate the lattice parameter. How many atoms per unit cell? Again you can draw the lattice and atoms of the unit cell to convince yourself there are 4 atoms per unit cell (remember corners count for 1/8 atoms and faces 1/2). How many unit cells per mole? Well there are $6.022\times 10^{23}$ atoms per mol, divided by atoms per unit cell gives unit cells per mole. How big is the unit cell? Of course $a^3$ cm cubed per unit cell. Multiply by unit cells per mole to get cm cubed per mole. Now multiply by density (grams per cm cubed) to get grams per mole. Since those are units of atomic weight we can set what we have equal to the value given. Solve for $a$, then substitute into the earlier expression to obtain $r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The probable reason there are different values is that different methods may have been used to calculate the atomic radius. What we just did is one method, and accurate enough for bulk calculation purposes in an engineering context. Another method is to use Density Functional Theory to model the electrostatic force given lattice parameters and electron orbitals and other properties. It is also possible that the values are for Ir present in various alloys or compounds, or at different temperatures. Atomic radii vary for the same atom when bonded to others, and for different temperatures due to thermal expansion.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2017-12-10T19:30:52.220" LastActivityDate="2017-12-10T19:30:52.220" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18403" PostTypeId="2" ParentId="18365" CreationDate="2017-12-10T18:30:58.597" Score="0" Body="&lt;p&gt;The item to be cut is completely submerged in a large water filled tank. Some as large as 10x10 feet, or more. What little heat that is created by the abrasion, and the pump, is immediately carried away from the kerf. I've had slots cut in Kydex® plastic with no apparent heat increase in the material. &lt;/p&gt;&#xA;" OwnerUserId="11068" LastActivityDate="2017-12-10T18:30:58.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18406" PostTypeId="1" CreationDate="2017-12-11T03:34:24.570" Score="2" ViewCount="63" Body="&lt;p&gt;I am trying to design a gearbox of sorts that would climb a rope from turning a gear (motor, handcrank, etc). The largest issue is that, while both the top and the bottom of the rope will be attached to something, I cannot predict how taut the rope will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about just wrapping the rope around a large worm gear and hoping that the weight of the device itself would generate enough friction to stop the rope from slipping off, but it doesn't seem like the best solution. I also thought about using a few helical gears radially around the rope, turning them with an internal gear ring that wraps around them, but this seems like an overly engineered solution that might not even work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any and all input is greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Lauren&lt;/p&gt;&#xA;" OwnerUserId="14135" LastActivityDate="2018-04-10T09:02:34.847" Title="Efficient Gearing to Climb a Rope Vertically" Tags="|mechanical-engineering|gears|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18408" PostTypeId="1" AcceptedAnswerId="21237" CreationDate="2017-12-11T09:57:12.473" Score="-1" ViewCount="493" Body="&lt;p&gt;There are cars with petrol engines having 1,0 l or even less engine displacement but I have never seen a car with diesel engine having less than 1,5 l. Are there technical reasons that diesel engine cannot have less than about 1,5 l? &lt;/p&gt;&#xA;" OwnerUserId="14137" LastEditorUserId="368" LastEditDate="2017-12-11T15:58:04.220" LastActivityDate="2018-04-08T04:02:21.147" Title="Why do diesel engines tend to have larger engine displacements?" Tags="|diesel|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18410" PostTypeId="1" CreationDate="2017-12-11T10:43:19.620" Score="1" ViewCount="114" Body="&lt;p&gt;What kind of positioning holding Hinges do Laptops &amp;amp; Notebook Computers use?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a set of Mech/ Engineering terms for hinges similar to these and are there a category of them that I can study/ look into?  &lt;/p&gt;&#xA;" OwnerUserId="6649" LastActivityDate="2017-12-11T17:35:56.853" Title="What kind of positioning holding Hinges do Laptops &amp; Notebook Computers use?" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18411" PostTypeId="1" AcceptedAnswerId="18412" CreationDate="2017-12-11T11:29:42.863" Score="1" ViewCount="944" Body="&lt;p&gt;The biggest diameter of my pressure sensor input port has a size of 1.93 mm. The silicon hose which I want to connect with the sensor's port has an inner diameter size of 2.0 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My honeywell pressure sensor: &lt;a href=&quot;https://i.stack.imgur.com/qPYAa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qPYAa.jpg&quot; alt=&quot;Honeywell pressure sensor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My silicon hose: &lt;a href=&quot;https://i.stack.imgur.com/aPcwr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aPcwr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can imagine - it does not fit. The port is to small or the inner diameter of the silicon hose to large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to bring both together? Are there adapters availabe? What is the name therefore? What is a good way in engineering to fix such an issue?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help!&lt;/p&gt;&#xA;" OwnerUserId="14138" LastActivityDate="2017-12-11T13:57:28.663" Title="How can I connect a silicon hose with 2mm inner diameter to a smaller 1,9mm port" Tags="|mechanical-engineering|pressure|sensors|pneumatic|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18412" PostTypeId="2" ParentId="18411" CreationDate="2017-12-11T11:50:58.130" Score="1" Body="&lt;p&gt;The fittings on your pressure sensor look like barb fittings. If that's the case, a 1.6mm ID silicone tubing will easily stretch over a 1.93mm barb fitting. 1.6mm (or 1/16&quot;) tubing is a fairly common size and you should be able to get it from standard laboratory suppliers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll then need an adapter to go from your 2mm ID to 1.6mm ID silicone tubing. Again, 2.4mm (3/32&quot;) is a fairly common size and a 2mm ID silicone tubing will easily stretch over a 2.4mm barb fitting. So something like AD-6005 from &lt;a href=&quot;http://www.nordsonmedical.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nordson Medical&lt;/a&gt; or equivalent should do it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/St0S5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/St0S5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2017-12-11T13:57:28.663" LastActivityDate="2017-12-11T13:57:28.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18413" PostTypeId="2" ParentId="18408" CreationDate="2017-12-11T14:35:01.983" Score="3" Body="&lt;p&gt;Diesel engines use compression to ignite fuel. Compression ignition requires higher compression ratios than Otto cycle engines. This results in parts that have to tolerate higher stresses, which means heavier parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to shrink a diesel engine down, you'll have small cylinders and everything else will be heavy (even the engine block). It's just not an efficient setup—lots of work goes into moving those heavy parts!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the main benefit of diesel is higher torque at low RPMs. A small engine is probably in a small vehicle. You don't need lots of torque to get a small vehicle going.&lt;/p&gt;&#xA;" OwnerUserId="6488" LastEditorUserId="368" LastEditDate="2017-12-11T15:59:15.657" LastActivityDate="2017-12-11T15:59:15.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18414" PostTypeId="2" ParentId="18410" CreationDate="2017-12-11T17:35:56.853" Score="3" Body="&lt;p&gt;They are commonly called Friction Hinges or torque hinges. They come in many configurations. They work by placing layer of a material between each side of the hinge along the pin axis. Also there's a design with a circumferential &quot;damper&quot; between the pin and the housing which will create the resistance to hold the hinge in position by friction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find that some use several layers to create a sandwich construction of the friction hinge. Found in compact designs. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-12-11T17:35:56.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18415" PostTypeId="1" AcceptedAnswerId="18416" CreationDate="2017-12-11T18:20:54.707" Score="0" ViewCount="1372" Body="&lt;p&gt;Define &quot;production tooling&quot;?&#xA;What are its advantages?&#xA;And why production tooling is essential for manufacturing?&lt;/p&gt;&#xA;" OwnerUserId="13632" LastEditorUserId="13632" LastEditDate="2017-12-12T16:52:03.030" LastActivityDate="2017-12-12T16:52:03.030" Title="What is production tooling?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ClosedDate="2017-12-12T18:47:02.770" ContentLicense="CC BY-SA 3.0" />
  <row Id="18416" PostTypeId="2" ParentId="18415" CreationDate="2017-12-11T18:28:46.633" Score="0" Body="&lt;p&gt;If you are to make one part or only samples, you would have prototype tooling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are to mass produce a part(high volume), then you have production tooling. Depending on your budget and quantity to be mass produced, it will drive the design of the production tooling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The considerations for production tooling are as follows but not limited to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Quantity to be made before service is required&lt;/li&gt;&#xA;&lt;li&gt;Setup&lt;/li&gt;&#xA;&lt;li&gt;Serviceability&lt;/li&gt;&#xA;&lt;li&gt;Mistake proofing/human factors&lt;/li&gt;&#xA;&lt;li&gt;Safety&lt;/li&gt;&#xA;&lt;li&gt;List item&lt;/li&gt;&#xA;&lt;li&gt;Quality&lt;/li&gt;&#xA;&lt;li&gt;Process Monitoring/inspection system&lt;/li&gt;&#xA;&lt;li&gt;Speed of production&lt;/li&gt;&#xA;&lt;li&gt;Budget&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Your question is vague, so I don't know what you are to manufacture, but apply the above points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Production Tooling is the tools used to mass&#xA;Produce a product. Production tooling considers the above points when being designed and manufactured. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2017-12-11T18:42:50.240" LastActivityDate="2017-12-11T18:42:50.240" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18418" PostTypeId="1" CreationDate="2017-12-11T19:15:47.570" Score="1" ViewCount="24" Body="&lt;p&gt;I need to produce pulses of light that turn on and off for equal intervals at 30 GHz. I know there are lasers that are capable of this frequency, so I need to know what materials they use to produce pulses of light at that frequency within those lasers. How could I search for them?&lt;/p&gt;&#xA;" OwnerUserId="14143" LastEditorUserId="90" LastEditDate="2017-12-12T00:56:32.730" LastActivityDate="2017-12-12T00:56:32.730" Title="Where could I find, or what is a material that could be used with 30 GHz pulsed DC current to produce 30 GHz pulses of light?" Tags="|optics|lasers|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18419" PostTypeId="1" CreationDate="2017-12-11T19:53:59.473" Score="2" ViewCount="17" Body="&lt;p&gt;Differential elevation at time of closure pours is considered a short&#xA;term concern. What are the recommended remediation techniques available to address the concern during phase construction of bridges.&lt;/p&gt;&#xA;" OwnerUserId="14142" LastActivityDate="2017-12-11T19:53:59.473" Title="Closure Pours having Differenial Elevation in Bridge Construction" Tags="|structural-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18420" PostTypeId="2" ParentId="2343" CreationDate="2017-12-11T20:27:51.393" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1I0VP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1I0VP.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have had the same exercise in school, maybe it helps, you just know $B= 1/2 AC$, and you know C is in horizontal line from 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you  want the acceleration diagram let me know, but it is just the same.&lt;/p&gt;&#xA;" OwnerUserId="14145" LastEditorUserId="1832" LastEditDate="2017-12-11T21:17:55.913" LastActivityDate="2017-12-11T21:17:55.913" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18423" PostTypeId="1" CreationDate="2017-12-12T03:01:05.360" Score="1" ViewCount="70" Body="&lt;p&gt;Is there a liquid or compounds that can change from gas then to liquid and back in a vacuum or with air? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture below the hour glass looking devise liquid comes in contact with light or heat it vaporizes passes the neck turning a electric turbine. At the absence of heat or light it would drain back into the lower half. Water is not the only liquid that could be used. Can this work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a vacuum water cannot be a liquid but under enough pressure like on the lowest picture, the weight would be measured allow enough pressure for water to be in a liquid state when it is out of the sun. A balance between a super cooled liquid, gas state and heat/or light.&#xA;Black dye could be added to increase performance.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2j5rk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2j5rk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Uz6ir.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uz6ir.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13974" LastEditorUserId="13974" LastEditDate="2017-12-12T17:04:55.200" LastActivityDate="2018-01-11T17:58:14.133" Title="Is there a generator that uses night and day this way?" Tags="|fluid-mechanics|thermodynamics|pressure|generator|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18424" PostTypeId="2" ParentId="18423" CreationDate="2017-12-12T04:26:56.993" Score="2" Body="&lt;p&gt;what you are describing here is basically a reversed heat pump that uses a temperature difference to run a heat engine in which the working fluid is so chosen as to undergo a phase change during the engine's power extraction cycle, and the high temperature source is a solar collector of some sort and the low temperature sink is night-time ambient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Systems like this are indeed practical and extensively documented- especially in the case of using an expansion turbine to extract the shaft work, but the efficiency limiter is the temperature at which the heat is added to the working fluid. to beat this limit requires the use of a concentrating collector to get the high temperature portion of the cycle as hot as possible, but this step adds significant cost because then the collector has to be designed to actively track the movement of the sun across the sky. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-12-12T04:26:56.993" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18425" PostTypeId="2" ParentId="18423" CreationDate="2017-12-12T06:15:23.000" Score="1" Body="&lt;p&gt;At the very basic level, this is how &lt;strong&gt;steam turbines&lt;/strong&gt; or an &lt;strong&gt;organic rankin cycle&lt;/strong&gt; in solar thermal power plants work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.barber-nichols.com/sites/default/files/wysiwyg/images/rankine_cycle_thermodynamic_diagram_large.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.barber-nichols.com/sites/default/files/wysiwyg/images/rankine_cycle_thermodynamic_diagram_large.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However you would be limited/constrained by the volume of the lower container. By increasing its volume it takes longer to be heated. A smaller volume would heat faster but would not produce so much energy per day (cycle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn’t the working fluid drain back into the lower container from itself (using gravity)? Is it really necessary to have a weight to help?&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2017-12-12T06:15:23.000" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18426" PostTypeId="1" CreationDate="2017-12-12T08:35:46.517" Score="0" ViewCount="1834" Body="&lt;p&gt;I'm a chemical engineer student and I have just started to learn about mass balance without reaction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my problem - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/852Ah.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/852Ah.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find Out the mass composition and mass flowrate of stream Y (the stream into the divider).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How am I suppose to do a mass balance around process 1 (involving the divider, before process 2) ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the problem , I don't know any of the compositions of the recycle 1 , I only know the flow rate . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only tells me that the flowrate into process 1 is $50000+ 1500 $ . With this alone, how do I find mass composition and mass flowrate of stream Y &lt;/p&gt;&#xA;" OwnerUserId="13646" LastActivityDate="2023-07-09T23:13:03.310" Title="Mass balance of a recycle stream" Tags="|chemical-engineering|chemistry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18427" PostTypeId="1" CreationDate="2017-12-12T08:52:21.613" Score="2" ViewCount="205" Body="&lt;p&gt;I have an underdamped control system for a DC motor. It is only a mathematical model so I can't visualise the motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can change the value of the &quot;k&quot; or compensator. I have noticed that changing this value changes the frequency of oscillation in the step response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering what the effect of increased oscillation in a control system would be on a DC motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14150" LastEditorUserId="14150" LastEditDate="2017-12-12T08:58:07.947" LastActivityDate="2017-12-12T11:50:03.013" Title="Oscillations in motor control systems" Tags="|control-engineering|motors|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18429" PostTypeId="2" ParentId="18260" CreationDate="2017-12-12T11:29:03.100" Score="2" Body="&lt;p&gt;If I understand correctly, the $\Delta p$ is the net pressure drop shown by the gauge. The pressure indicator will always show the 'net effect', i.e. it doesn't understand the physical reasons contributing towards the reading. Thus, it &lt;em&gt;includes&lt;/em&gt; the hydrostatic pressure drop as well as dynamic pressure drop. Assuming there is the pressure drop along the direction of flow, and assuming that the pressure gauge shows absolute value of the pressure drop (i.e. magnitude), we can write the Bernoulli's equation (which we can safely do since the flow is irrotational, incompressible and inviscid) along the streamline as,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;p_1 + \frac{1}{2} \rho v_1^2 + \rho g z_1 = p_2 + \frac{1}{2} \rho v_2^2 + \rho g z_2&#xA;\end{equation} &#xA;i.e.&#xA;\begin{equation}&#xA;\frac{1}{2} \rho \left( v_1^2 - v_2^2 \right) = \underbrace{\overbrace{\left(p_2 - p_1 \right)}^{-ve} + \overbrace{\rho g \left( z_2 - z_1 \right)}^{+ve}}_{-\Delta p}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see that, tilting of the Venturi exerts back-pressure at section $1$. i.e. it resists the flow.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the continuity equation, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}\label{cont}&#xA; v_1 = \frac{A_2}{A_1} v_2&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting above equation in Bernoulli's equation, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;\frac{1}{2} \rho \left( \left(\frac{A_2}{A_1}\right)^2 -  1\right)v_2^2 = {-\Delta p}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation}&#xA;v_2 = \sqrt{ \left(\frac{-2 \Delta p}{\rho}\right) \left( \frac{A_1^2}{A_2^2 - A_1^2}\right) } = 0.1326 \frac{m}{s}&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and hence, $v_1 = 0.0398 \frac{m}{s}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required flow rate is then; $v_1 A_1 = v_2 A_2 = 0.004 \frac{m^3}{s}$. &lt;/p&gt;&#xA;" OwnerUserId="223" LastActivityDate="2017-12-12T11:29:03.100" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18430" PostTypeId="2" ParentId="18188" CreationDate="2017-12-12T11:30:49.787" Score="1" Body="&lt;p&gt;In cascade, the output of one controller will be the input to the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In parallel the controllers are stacked and the outputs from these controllers will combine to give 1 output. &lt;/p&gt;&#xA;" OwnerUserId="14150" LastActivityDate="2017-12-12T11:30:49.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18431" PostTypeId="1" CreationDate="2017-12-12T11:40:23.970" Score="4" ViewCount="296" Body="&lt;p&gt;I have recently finished Matriculation and currently trying to build up certain skills I think will be useful. I have applied for university and one of the courses I am looking at is mechatronics. I was wondering if anyone had information on what programming languages to learn and where I could learn them (hopefully in a form that best fits a busy schedule and allows for me to pause and resume learning at my choice such as a video series on YouTube or something).&lt;/p&gt;&#xA;" OwnerUserId="14153" LastEditorUserId="7083" LastEditDate="2018-08-14T00:10:22.013" LastActivityDate="2018-08-14T00:10:22.013" Title="Where to learn programming for mechatronics and overall?" Tags="|automotive-engineering|" AnswerCount="3" CommentCount="3" ClosedDate="2018-01-12T12:33:45.960" ContentLicense="CC BY-SA 4.0" />
  <row Id="18432" PostTypeId="2" ParentId="18427" CreationDate="2017-12-12T11:50:03.013" Score="2" Body="&lt;p&gt;An under-damped control system with a DC motor, particularly if there is backlash in the system, can cause the direction of the motor to switch very rapidly, causing large inductive loads, and potential damage to your circuitry.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-12-12T11:50:03.013" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18433" PostTypeId="1" CreationDate="2017-12-12T14:41:34.913" Score="0" ViewCount="272" Body="&lt;p&gt;I have a base with 4 wheels, the rear axle is a motor which drives the base forward or backwards, this contains a diff, so there's no problem there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The front axle is a steering axle like that in a car, except where the steering column would be extends vertically, and not at a diagonal in a car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In spare parts, I have an arm which is capable of pushing and pulling (depending on which way the DC current goes), and this is affixed to a plate at the rear of the wheel base, and points at 90 degree angle from the plate, towards the steering column. The arm is affixed 20cm to the left of the steering column. I want to use this arm to turn the steering axle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I fit another (fixed, non extending) arm onto the steering axle so it can be pushed and pulled, what could I do to connect the two arms so that the extending arm can turn the steering axle? Would the non extending arm also need to be extendable (like a piston)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a very poorly drawn diagram that I've rushed a bit. The yellow section is the extending arm, the blue bar is the bit for steering. The black bar that extends from the blue bar is the steering column (with the proposed fixed arm pointing left)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zS42U.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zS42U.png&quot; alt=&quot;Supporting Image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="14155" LastEditorUserId="14155" LastEditDate="2017-12-12T14:58:20.550" LastActivityDate="2023-12-12T12:05:05.687" Title="Push / Pull Steering on an axle" Tags="|mechanical-engineering|linear-motion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18434" PostTypeId="2" ParentId="18431" CreationDate="2017-12-12T14:55:46.063" Score="0" Body="&lt;p&gt;In my experience programming is something you can teach yourself from books etc very well, certainly well enough to get a good grounding in preparation for a course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I taught myself C++ from a book designed for self learning and I found the learning curve to be not too difficult. I'm certainly not a particularly skilled programmer but it was enough to put together basic control software. I'm not sure how useful videos would be as it's not really a very visual skill. A lot of it is looking up specific ways to do specific tasks and getting familiar with the rules and  terminology of your chosen language. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the comments it is also sensible to look at control systems and feedback which involves some mathematics which is not especially hard but not immediately intuitive either at first and it can take a while to get a grasp of the relationship  between the theory and practical applications. Here even the most basic practical experimentation is hugely useful as being able to see the relationship between the functions and variables and the actual physical response is invaluable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also worth looking at the nuts and bolts of how software and hardware communicate and interface with each other ie how you turn the output from a programme to a control signal. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2017-12-12T14:55:46.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18435" PostTypeId="2" ParentId="18433" CreationDate="2017-12-12T15:47:42.893" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YAXbO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YAXbO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;A 90 degree crank would do it : like an L : the yellow bar pushes or pulls the horizontal prt of the L and the other end of the L moves the black bar left or right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit added pic, something like :&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-12-12T17:33:05.947" LastActivityDate="2017-12-12T17:33:05.947" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18439" PostTypeId="1" CreationDate="2017-12-12T19:16:34.130" Score="1" ViewCount="32" Body="&lt;p&gt;I'm working on an AI Home Temperature controller for my computer science project, and I would like to know how much Ni-chrome wire (in terms of length and diameter) is used in an typical home electric heater? &lt;/p&gt;&#xA;" OwnerUserId="14158" LastActivityDate="2017-12-12T19:16:34.130" Title="How much Ni-chrome wire is used in your average home electric heater?" Tags="|heat-transfer|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18440" PostTypeId="2" ParentId="2300" CreationDate="2017-12-12T20:55:29.497" Score="6" Body="&lt;p&gt;I wrote this paper in response to a declaration made by Dynamics Professor that &amp;quot;there is no difference between a lbm and lbf.&amp;quot; The discussions from the students that followed exposed a huge concept error that seems to stem from the misuse of the above statement. It has some comedic relief, so it makes it more bearable ;) Enjoy!&lt;/p&gt;&#xA;&lt;h2&gt;The lbm-lbf Relationship: Why it Matters&lt;/h2&gt;&#xA;&lt;p&gt;&lt;em&gt;by Kevin McConnell&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Is there really a difference between a pound-mass and a pound-force? Many people might even ask, “What the hell is a pound-mass?” Well, you can point the finger at your sixth-grade physics teacher (or anyone else who may have misled you) for the confusion that surrounds this simple question. But don’t worry, it’s never too late to learn something new (and something undeniably important).&lt;/p&gt;&#xA;&lt;p&gt;Here’s something to mull over: let’s say that you step on a scale and it reads “150.” The readout of the scale may even provide you with units of “lbs.” Well, a scale measures the amount of force that an object exerts so we can assume that the units then are lbf (pound-force). And your physics teacher told you that there is no difference between a pound-mass and a pound-force so that must mean that your body is composed of 150 pounds of mass as well, right? What your physics teacher DIDN’T tell you are the hidden assumptions that must be true for that relationship to exist. There is something so fundamentally wrong with the statement, “pounds-mass and pounds-force are the same thing!”&lt;/p&gt;&#xA;&lt;p&gt;First off, pounds-mass is a unit of mass, and pounds-force is a unit of force (wait… WHAT?!). Newton’s second law of motion tells us that net force is equated by the product of mass and acceleration. So, we can see that a relationship exists between mass and force, but we would NEVER say, “mass and force are the same thing!”&lt;/p&gt;&#xA;&lt;p&gt;Let’s say that I took the same scale from above on a trip to mars; what would the scale read there? Would you be surprised if the scale readout as “57 lbs?” Or what if I brought the scale to Jupiter and it told me I weighed “380 lbs?” Is the scale correct? Absolutely! As we learned previously, the scale is measuring the amount of force that you exert due to gravity (acceleration). And we know that gravity on these planets differs because of a difference in their size and mass.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;KEY CONCEPT&#xA;Note that your mass DOES NOT change from planet to planet; only the amount of force exerted by your mass.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;So why do we keep hearing that there is no difference between pounds-mass and pounds force? Because English units were created such that 1 lbm exerts 1 lbf here on Earth! And without further ado, here is the relationship that makes it happen:&lt;/p&gt;&#xA;&lt;p&gt;1 lbf = 32.174 lbm ft/s^2&lt;/p&gt;&#xA;&lt;p&gt;So, the statement that people are trying to say should sound something more like “on earth, pounds-mass subject to gravity IS pounds-force!” To further illustrate this point, lets use newtons second law to calculate the force exerted by a 1 lbm object here on earth:&lt;/p&gt;&#xA;&lt;p&gt;Force=mass x acceleration&lt;/p&gt;&#xA;&lt;p&gt;let acceleration=g=32.174 ft/s^2    (this is Earth' s gravitational constant)&lt;/p&gt;&#xA;&lt;p&gt;F=m x g= 1 lbm x (32.174 ft/s^2) = 32.174  (lbm ft)/s^2&lt;/p&gt;&#xA;&lt;p&gt;But we can’t really conceptualize the units lbm-ft /s2, so we use the relationship from above to convert it to pound-force (lbf):&lt;/p&gt;&#xA;&lt;p&gt;F=  32.174 lbm-ft/s^2 x (1 lbf / 32.174 lbm ft/s^2) = 1 lbf&lt;/p&gt;&#xA;&lt;p&gt;We have just proved that 1 lbm exerts 1 lbf here on Earth! If this is new to you, you should drink a beer tonight to celebrate a breakthrough in your understanding! Let’s go one step further to demonstrate why the scale would read differently on Mars and Jupiter&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;‘NOTHER KEY CONCEPT&#xA;The relationship (eq. 1) from above DOES NOT change if you’re on a different planet just because the gravity changes; this wouldn’t make sense and you’ll see why&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;Force = mass x acceleration&lt;/p&gt;&#xA;&lt;p&gt;let acceleration = g = 12.176 ft/s^2    (this is the gravitational constant on Mars)&lt;/p&gt;&#xA;&lt;p&gt;let mass = m = 150 lbm&lt;/p&gt;&#xA;&lt;p&gt;F = m x g = 150 lbm x 12.176 ft/s^2 = 1826.4  (lbm ft)/s^2&lt;/p&gt;&#xA;&lt;p&gt;Once again, lets convert this quantity from lbm-ft /s2, to something we know (lbf) by using the relationship illustrated above:&lt;/p&gt;&#xA;&lt;p&gt;F=(1826.4 lbm ft/s^2) x (1 lbf / 32.174 lbm ft/s^2) = 56.8 lbf&lt;/p&gt;&#xA;&lt;p&gt;Even though I imagine that you now have a firm grasp on this concept, let’s try it out on Jupiter to really send it the point home:&lt;/p&gt;&#xA;&lt;p&gt;Force = mass x acceleration&lt;/p&gt;&#xA;&lt;p&gt;let acceleration = g = 81.336 ft/s^2    (this is the gravitational constant on Jupiter)&lt;/p&gt;&#xA;&lt;p&gt;let mass=m=150 lbm&lt;/p&gt;&#xA;&lt;p&gt;F = m x g = 150 lbm) x 81.336 ft/s^2  x (1 lbf / 32.174 lbm ft/s^2 )=379.2 lbf&lt;/p&gt;&#xA;&lt;p&gt;Now you’ve seen it and you can say that you understand it! So, let’s highlight the crucial points to everything that we just went over:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;pounds-mass (lbm) and pounds-force (lbf) are NOT the same&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;an object’s mass is constant from place to place (i.e. from Earth to&#xA;Mars) but the force that it exerts IS different&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The following relationship is key to understand the link between lbm&#xA;and lbf:&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;1 lbf=32.174 lbm ft/s^2&lt;/p&gt;&#xA;&lt;p&gt;Arm yourself with this knowledge so that you can fight the good fight: the next time you hear someone say that pound-mass and pound-force are the same thing, you can confidently say they are not!&lt;/p&gt;&#xA;" OwnerUserId="14160" LastEditorUserId="-1" LastEditDate="2022-02-02T17:18:14.110" LastActivityDate="2022-02-02T17:18:14.110" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18441" PostTypeId="1" AcceptedAnswerId="18442" CreationDate="2017-12-12T21:23:23.573" Score="4" ViewCount="254" Body="&lt;p&gt;I have seen this kind of mark on various cast metal parts. This is an image of the base casting of a hard disk drive made by HGST.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UzKGi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UzKGi.jpg&quot; alt=&quot;Markings on an HGST HDD base casting&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are these marks for?&lt;/li&gt;&#xA;&lt;li&gt;What are they called?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14161" LastActivityDate="2017-12-12T22:03:59.497" Title="Purpose and name of fine pitch marks on a cast metal part" Tags="|manufacturing-engineering|metals|casting|molding|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18442" PostTypeId="2" ParentId="18441" CreationDate="2017-12-12T22:03:59.497" Score="6" Body="&lt;p&gt;these are part-tracking marks. they indicate which cavity in the mold the part was cast from, in which mold set, in what molding machine, in what year, month, day and shift. this way, a part that fails its finished goods inspection or failed in the field can be tracked backwards through the factory to its source, and the reason for the failure fixed.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-12-12T22:03:59.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18444" PostTypeId="1" AcceptedAnswerId="32631" CreationDate="2017-12-13T00:48:03.733" Score="2" ViewCount="134" Body="&lt;p&gt;I'm relatively new to hydraulics, and I'm trying to understand this schematic. I've got the majority of the system figured out, but I cant for the life of me understand what the symbols at positions 18/20/23 represent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unit actively controls a constant pressure in the two cylinders attached using an electronic pressure switch. Its the control unit of a hydraulic tensioning system in a ski lift. My feeling is that 18/20 may have something to do with electrical pressure cells?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WNch1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WNch1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="14163" LastEditorUserId="14163" LastEditDate="2017-12-13T04:12:16.270" LastActivityDate="2022-08-17T10:20:17.813" Title="Need help identifying component on hydraulic schematic" Tags="|mechanical-engineering|hydraulics|instrumentation|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18445" PostTypeId="1" AcceptedAnswerId="18447" CreationDate="2017-12-13T00:53:03.557" Score="5" ViewCount="3193" Body="&lt;p&gt;This question got into my mind when I am cleaning my motorcycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If spring shock-absorbers like the one below absorbs shock from road imperfections, reducing strain in the motorcycle body as your drive, then why does car manufacturers use rigid bumper for cars instead of semi-rigid or flexible bumpers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't it good for the car that its bumper dissipates some energy in collision so that the time of impact increases, therefore reducing impact force? What are the pro's and con's of this idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/21i8h.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/21i8h.jpg&quot; alt=&quot;shock absorber&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2017-12-19T05:00:57.193" Title="Rigid bumpers vs shock absorber bumpers" Tags="|mechanisms|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18446" PostTypeId="2" ParentId="18445" CreationDate="2017-12-13T01:11:35.943" Score="3" Body="&lt;p&gt;it is a good idea. 70 years ago, bumpers were built out away from the bodywork by about 5&quot; to 6&quot; to take hits without damaging the car, but styling changed and by 60 years ago, bumpers became part of the bodywork. 40 years ago, shock-absorbing bumpers were required by law in the USA, but they made the cars look ugly so people didn't like them. the law was revoked and now the bumpers do not have to sustain shocks without damage.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-12-13T01:11:35.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18447" PostTypeId="2" ParentId="18445" CreationDate="2017-12-13T01:24:04.120" Score="8" Body="&lt;p&gt;To answer your later questions first:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Isn't it good for the car that its bumper dissipates some energy in&#xA;collision so that the time of impact increases, therefore reducing&#xA;impact force?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Yes, it is good, for the reasons you describe. Your misconception, however, seems to be that a &amp;quot;rigid&amp;quot; bumper would not do these things. The bumper, and indeed much of the car bodywork constitutes a &amp;quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Crumple_zone&quot; rel=&quot;noreferrer&quot;&gt;Crumple Zone&lt;/a&gt;&amp;quot;, which is designed to exhibit exactly the behaviour you describe, by increasing the impact time, and reducing the consequent peak force.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;To answer your first question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If spring shock-absorbers like the one below absorbs shock from road&#xA;imperfections, reducing strain in the motorcycle body as your drive,&#xA;then why does car manufacturers use rigid bumper for cars instead of&#xA;semi-rigid or flexible bumpers?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The key difference between using a spring to reduce the impact force vs using a crumple zone, is that the spring &lt;em&gt;returns that force&lt;/em&gt;. In a big crash, you really don't want to be bouncing backwards. Bumpers are optimised for safety in an emergency, rather than day-to-day driving into things, as it were.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2017-12-13T01:24:04.120" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18448" PostTypeId="1" AcceptedAnswerId="18451" CreationDate="2017-12-13T11:21:08.667" Score="0" ViewCount="32" Body="&lt;p&gt;I really know that the real calulation of the range of UMTS, GSM or LTE antennas is depending on a lot of issues, reflection, and other stuff which makes it really complex. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need only a rough model which I can calculate the range of them, so I thought I use the free-space path loss formula and look for the &quot;r&quot; in the formula. Is there anything wrong with my idea? Despite the fact that of course in really you'll almost never have just a free space but rather buildings,hills and other stuff but lets say, for my model I say there is nothing that disturbs my signal.....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I can use the formula, my question would be so the only difference of those systems would be the frequency and the transmitted power or received power of the mobile phones, is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Free-space_path_loss&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Free-space_path_loss&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11980" LastActivityDate="2017-12-13T16:17:00.950" Title="Calculating the link budget of antenna and receiver with free space-path-loss formula" Tags="|electrical-engineering|radio|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18450" PostTypeId="1" AcceptedAnswerId="18460" CreationDate="2017-12-13T15:56:32.410" Score="2" ViewCount="1602" Body="&lt;p&gt;I'm in a bit of a pickle with one of the quality techns that works for us (the company) over a profile of surface tolerance. I initially submitted something resembling figure 1 (top capture) and quality came back to me saying figure 2 (bottom capture) would be better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never seen ''auto-referencing'' of datum and I don't see the point because it doesn't add anything to the GD&amp;amp;T. I've never seen anything like this in any book, reference or drawing I stumbled upon over my (short) carreer. Are there any arguments to which practice would be best? I'm open to any source saying one way should be the preferred one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4skyG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4skyG.png&quot; alt=&quot;What I&amp;#39;m proposing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zZtt3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zZtt3.png&quot; alt=&quot;What quality guy suggests&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14168" LastEditorUserId="14168" LastEditDate="2020-07-10T11:19:48.900" LastActivityDate="2020-07-10T11:19:48.900" Title="Drawing best practice and datum self-referencing" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18451" PostTypeId="2" ParentId="18448" CreationDate="2017-12-13T16:17:00.950" Score="0" Body="&lt;p&gt;The usual  $ r^{-4} $ round-trip link loss is based on the antenna being a point radiator, or at least so far from the target as to appear almost like a point radiator.  The full formula has all sorts of correction factors (aka &quot;fudge factors&quot;) for the actual beam profile, and targets are represented as a standard shape of equivalent total reflectance, e.g. a sphere or a flat plate mounted perpendicular to the propagation axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That should suffice for a rough approximation of the link loss.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2017-12-13T16:17:00.950" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18452" PostTypeId="1" CreationDate="2017-12-13T17:00:05.247" Score="1" ViewCount="424" Body="&lt;p&gt;$$d = \dfrac{V_1^2-V_2^2}{2g\left(\dfrac{a}{g}\pm G\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if the car travel on the road of a sag curve or crest curve, there is two gradients of the curve. (G1, G2). Which gradient value should be used in the formula ? The one with larger gradient?&lt;/p&gt;&#xA;" OwnerUserId="14169" LastEditorUserId="1832" LastEditDate="2017-12-13T19:57:27.850" LastActivityDate="2018-06-13T20:43:04.317" Title="How to calculate stopping distance with grade?" Tags="|civil-engineering|highway-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18453" PostTypeId="2" ParentId="18452" CreationDate="2017-12-13T17:58:35.740" Score="1" Body="&lt;p&gt;Recall that tires have a no slip condition when they interface with the road and the car is rolling.  One gradient will cause the car to rock about an axis aligned with the direction of travel (passenger - driver) and the other gradient will cause the car to dive upon stopping (front - back).  The gradient that is transverse to the direction of travel (i.e. the gradient that causes rocking not diving) will not be a factor unless the dynamic forces are large enough to exceed the breaking friction of the tire/road interface.  Depending on how sophisticated your model is, this is also a function of speed and ambient temperature as the breaking friction coefficient is going to depend on the tire temperature (and road temperature).  Assuming normal road conditions and normal speeds (not race track calculations) the only gradient that would contribute to stopping distance in a meaningful way is the gradient that that is aligned with the direction of travel.  &lt;/p&gt;&#xA;" OwnerUserId="14091" LastActivityDate="2017-12-13T17:58:35.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18454" PostTypeId="1" CreationDate="2017-12-13T18:44:52.137" Score="8" ViewCount="831" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KJZIF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJZIF.png&quot; alt=&quot;Case study, the oscillation of the spring and the source of coulomb damping&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is my first time posting anything, but I could not figure out the derivation of the amplitude of the spring for the 4 first oscillations. I am not acquainted with the effect and calculations of damping since I am only in high school. This is an assignment I have, I think I am in over my head, but I have invested too much time in this to change the topic. I figure that the only deviation to the classic coulomb damping where the source of energy dissipation is due to the weight sliding on top of a surface, and the force of friction is constant and does not change, is that the friction coefficient changes as more coils enter the paper cylinder which is compressing the spring. The values that I have is the frictional force of when the entire spring is within the paper cylinder, the mass of the weight, the initial elongation of the spring (x naught) and the spring constant. &lt;a href=&quot;https://i.stack.imgur.com/XSRZy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XSRZy.png&quot; alt=&quot;classical coulomb damping case&quot; /&gt;&lt;/a&gt; . If anyone can help me with my assignment I would be eternally thankful! If not, I would gladly take any help with deriving the amplitude of the classic coulomb damping case as seen above as a backup. Moreover, If there is any rule that I am breaking with this post, let me know. This is as previously stated, the first time that I post anything, and it is due to pure desperation as I am way in over my head! Thanks in advance!&lt;/p&gt;&#xA;&lt;p&gt;I got 15 experimental values for the amplitudes at half revolution increments for 4 revolutions, took the averages and inserted them into a graph. The graph can be seen here: &lt;a href=&quot;https://gyazo.com/3e2d1c18b05a31d1eca88c411bdbb2cb&quot; rel=&quot;noreferrer&quot;&gt;https://gyazo.com/3e2d1c18b05a31d1eca88c411bdbb2cb&lt;/a&gt; .&lt;/p&gt;&#xA;&lt;p&gt;To solve the equation for x(t) in the equation ma = -kz +F_r, where you would get Asin(wt) + Bcos(wt) + F_r/k, you merely need to equate it to the initial displacement to calculate B, and equate it to the derivative of displacement i.e. initial velocity to get A. A is zero if there is no initial velocity, but the question still stands, how do you get -sgn(v) so that it can be used as the value of friction over spring constant which would give the damping term. I suppose it could be gotten by taking -v/magnitude of v, but how would you calculate the magnitude of v? Since solving the equation without it would merely result in a regular cosine function.&lt;/p&gt;&#xA;" OwnerUserId="14170" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2024-02-14T06:04:20.477" Title="Deriving the amplitude of a coulomb (dry friction) damped spring" Tags="|mechanical-engineering|friction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18455" PostTypeId="2" ParentId="18454" CreationDate="2017-12-13T20:16:32.680" Score="0" Body="&lt;p&gt;As this is for school, I don't want to just give you the answer.  I will do my best to give you some direction without completely giving it away. Don't feel bad that you are feeling desperate, these are the problems where you learn the most.  Do not give up and stay persistent.  Your ability to understand has nothing to do with being in high school but with your dedication to learning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spring displacement is purely a function of force and displacement.  in this case you should not just consider the static forces acting on the spring but also the d'lambert force (dynamic) force.  Recall that F=Ma.  do your free body diagram but also consider the dynamic force in your FBD (which opposes changes in direction) for your derivation, start with the equation of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x(t) = a0+a(t)^2 + v0+v(t)*t + x0.  where a0, v0 and x0 are the values of acceleration, velocity and position at time=0 which you know.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your dynamic load will be mass X the acceleration term in the equation of motion.  you can solve for the acceleration component by rearranging the equation above.  to get the force, multiply both side by the mass.  you can then plug and chug in excel to get the answer(0&#xA;&#xA;&lt;p&gt;the effect of friction will depend on velocity.  you can google to learn more about damping coefficients and its time dependent nature.&lt;/p&gt;&#xA;" OwnerUserId="14091" LastActivityDate="2017-12-13T20:16:32.680" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18458" PostTypeId="2" ParentId="18392" CreationDate="2017-12-13T23:59:38.100" Score="4" Body="&lt;p&gt;The simplest method will be a multi-step approach.  The first step will be a defective product detection, that gathers general data about the product stream and quarantines defective products.  On a parallel path, a small batch of samples will be also sequestered from the main line to be analyzed continuously for detectable defects.  A laboratory analysis could then confirm size and count using digital tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recognizing that these are, for the most part, polarized plastics, they should have a fairly consistent low to modest &lt;a href=&quot;https://en.wikipedia.org/wiki/Relative_permittivity&quot; rel=&quot;nofollow noreferrer&quot;&gt;relative permittivity&lt;/a&gt;, double to triple the value of air.  Regardless of opacity and/or material, a product line with high amount of defects will be unpolymerized or coked plastics, consisting primarily of carbon, which has a detectable higher relative permittivity, nearly five times the value of even the most polar plastics.  A static (ultra low frequency) dielectric detector can detect these defects, and quarantine the offending batch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x60N1.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x60N1.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the relative voltage across the plates changes, the increase in stored voltage can be detected as a defect for sequestering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a parallel path, a small portion can run through a dielectric spectrometer.  These have had some uses before, (See &lt;a href=&quot;https://www.osapublishing.org/as/abstract.cfm?uri=as-71-3-456&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for detection of different polymers based upon response spectra, and &lt;a href=&quot;http://ieeexplore.ieee.org/document/4772907/?reload=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; where it was used to detect areas of low polymerization in LDPE.)  It works the same way as the dielectric detector, but uses an alternating current.  This can get a full spectrum analysis for comparison between polymers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Wj7W7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wj7W7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this detection method is slower, it will need to be able to analyze samples one at a time.  However, this will ensure the main detector is not fluctuating substantially due to contaminants in the air or other problems.  It will also review a history of defective compounds.  As the library of spectrum is built, this could replace entirely the need to utilize laboratory detection, except for the cases of control chart defects (see below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some advantages of this is that it can detect defects inside the beads that would not be visible via normal spectroscopy alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Final laboratory counts to measure black spots could then be utilized to verify the automated machines are working within parameters.  The 7 tools of quality, particularly the control chart, will be useful as a countermeasure to ensure these detectors are working properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9BfLY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9BfLY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2017-12-13T23:59:38.100" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18459" PostTypeId="2" ParentId="2777" CreationDate="2017-12-14T01:04:43.720" Score="1" Body="&lt;p&gt;Coupling two motors used to be a problem. They would balance the two by monitoring current draw between them and balance. Coupling two motors is not longer a problem anymore with the advent of digital drives, which can pair each up with one another. This solves the old coupling of two shafts with a jack shaft.&lt;/p&gt;&#xA;" OwnerUserId="14175" LastEditorUserId="1832" LastEditDate="2017-12-14T08:48:12.363" LastActivityDate="2017-12-14T08:48:12.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18460" PostTypeId="2" ParentId="18450" CreationDate="2017-12-14T04:47:26.407" Score="2" Body="&lt;p&gt;Don't do it. It's probably not legal in any GD&amp;amp;T spec. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't reference a feature back to itself(think: chicken or egg?). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first version is more appropriate, but not ideal. How do you plan to inspect the position of the first hole(datum b)? I see what you're trying to accomplish(intent), but it should be clear from an inspection stand point. I think you need to remove the tolerance of position for that datum B hole and give datum C twice the tolerance...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Btw, do your holes sizes have a tolerance or fall back on a general note?&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-12-14T04:47:26.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18461" PostTypeId="1" CreationDate="2017-12-14T08:11:31.430" Score="1" ViewCount="60" Body="&lt;p&gt;Vehicles and Factories produce air-pollution like Nitrogen Dioxide(NO&lt;sub&gt;2&lt;/sub&gt;) and Sulfur Dioxide(SO&lt;sub&gt;2&lt;/sub&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plants use Fertilizer with those same compounds (NO&lt;sub&gt;2&lt;/sub&gt; &amp;amp; SO&lt;sub&gt;2&lt;/sub&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by what method can air-pollution be captured, and recycled as fertilizer ? &lt;/p&gt;&#xA;" OwnerUserId="14179" LastActivityDate="2017-12-16T19:06:02.053" Title="Is there any method to recycle air-pollution into fertilizer?" Tags="|mechanical-engineering|control-engineering|manufacturing-engineering|process-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18463" PostTypeId="2" ParentId="18461" CreationDate="2017-12-14T11:16:52.277" Score="1" Body="&lt;p&gt;One of the problems with such an idea is the concentration of such polluting gases, within the exhaust gas stream of individual producers is relatively low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar problem exists with coal fired power stations and schemes to capture the carbon dioxide produced so that it can be sequestrated underground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, the polluting gasses must be removed from the exhaust stream, concentrated and stored so that can be disposed of, or utilized in another way - such as making fertilizer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equipment to do so is large and expensive, which would deem many industrial plants uneconomic, particularly if it had to be retrofitted to existing plants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fitting such a system to motor vehicles, would increase the size of the vehicles and their weight and it would reduce fuel efficiency. It would be better to abandon the use of fossil fuelled vehicles and use another souce of energy, such as electricity.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2017-12-14T11:16:52.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18465" PostTypeId="1" CreationDate="2017-12-14T15:25:47.453" Score="1" ViewCount="899" Body="&lt;p&gt;Let's say I buy a barbecue grill, is there a material testing kit that I can buy that'll allow me to find out what metal the grill is made from and if it were galvanized or cured in a particular substance?&lt;/p&gt;&#xA;" OwnerUserId="7287" LastActivityDate="2017-12-15T12:55:12.437" Title="How to find out what a material is made of specifically?" Tags="|materials|manufacturing-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18466" PostTypeId="2" ParentId="18465" CreationDate="2017-12-14T15:55:16.103" Score="2" Body="&lt;p&gt;The test kit is a magnet and horse- sense. If magnetic = steel (most likely) , shiny and magnetic= chrome stainless. Non-magnetic and shiny = austenitic stainless aka 18-8 . Non-magnetic and silver grey but not shiny and feels cold and relatively light = aluminum ( likely ). Magnetic and silver grey = galvanized steel , a dumb material choice for a grill.  How do you &quot;cure&quot; a metal ? &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2017-12-14T17:04:01.843" LastActivityDate="2017-12-14T17:04:01.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18467" PostTypeId="2" ParentId="18465" CreationDate="2017-12-14T16:15:28.420" Score="1" Body="&lt;p&gt;There is no &quot;material testing kit that [you] can buy&quot;, but there are recognised processes/sequences for determining both &lt;a href=&quot;http://www.weldinghandbook.com/types-of-metals/metal-identification/&quot; rel=&quot;nofollow noreferrer&quot;&gt;metals&lt;/a&gt;, and &lt;a href=&quot;https://www.doitpoms.ac.uk/tlplib/artefact/images/poster1changed.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;plastics&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.weldinghandbook.com/types-of-metals/metal-identification/&quot; rel=&quot;nofollow noreferrer&quot;&gt;weldinghandbook.com&lt;/a&gt; has an excelent article covering this, which I will not repeat in full!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary - check first if your material is magnetic, and then follow one of these three processes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i1.wp.com/www.weldinghandbook.com/wp-content/uploads/2017/01/metal-identification-tests-1-1.png?w=600&quot; rel=&quot;nofollow noreferrer&quot;&gt;Non Magnetic&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i1.wp.com/www.weldinghandbook.com/wp-content/uploads/2017/01/metal-identification-tests-2.png?resize=289%2C249&quot; rel=&quot;nofollow noreferrer&quot;&gt;Slightly Magnetic&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i2.wp.com/www.weldinghandbook.com/wp-content/uploads/2017/01/metal-identification-tests-3-1.png?w=600&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magnetic&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-12-14T16:15:28.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18468" PostTypeId="1" CreationDate="2017-12-14T19:05:58.590" Score="1" ViewCount="225" Body="&lt;p&gt;I have to join metal to metal (Aluminium or SS piece onto a vertical SS wall) that could take a shear load of say, 600gm with an area of contact of around 2cm^2. The environment is above boiling surface of water. So, the adhesive should be resistant to 100 deg C temperature, and moisture - with water droplets. Check the image for setup. &lt;strong&gt;What are the best adhesives for this purpose?&lt;/strong&gt; I have checked Araldite, but specs say its max temp is around 65 deg C. Any suggestions? Also desirable is if the adhesive is semi-permanent: If I need to remove it and make the joining at any other location, it shouldn't be terribly difficult.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ngotP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ngotP.jpg&quot; alt=&quot;Setup&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13790" LastActivityDate="2024-01-24T08:49:11.667" Title="What is the best heat and moisture resistant adhesive?" Tags="|mechanical-engineering|materials|design|adhesive|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18469" PostTypeId="1" AcceptedAnswerId="18532" CreationDate="2017-12-14T21:39:56.063" Score="2" ViewCount="191" Body="&lt;p&gt;The water held underground in a geothermal system is under high temperature and pressure. Are pumps used to raise the very hot underground water to the surface or does the water rise under its own pressure? One source says that the water is pumped up: &lt;a href=&quot;https://energy.gov/eere/geothermal/electricity-generation&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://energy.gov/eere/geothermal/electricity-generation&lt;/a&gt; while another says that it rises on its own: &lt;a href=&quot;https://www.youtube.com/watch?v=kjpp2MQffnw&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=kjpp2MQffnw&lt;/a&gt; I am not sure if I am just misunderstanding the wording. Does anyone know which is correct?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2017-12-21T09:03:25.997" Title="Are pumps used in geothermal systems" Tags="|pressure|heat-transfer|pumps|geotechnical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18470" PostTypeId="1" CreationDate="2017-12-15T01:48:52.553" Score="1" ViewCount="219" Body="&lt;p&gt;I am doing a gerotor pump application design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the velocity of the suction port should be 1 ~ 2 m/s to avoid cavitation. Since I have enough space for suction port, it is not a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet, the discharge port's diameter must be 16 pi because of the space and sealing problem. I read that discharge velocity should be 3 m/s, but I found the discharge velocity goes up to 4.5m/s when I calculated it with diameter 16.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen if the discharge velocity is over 3 m/s?&lt;/p&gt;&#xA;" OwnerUserId="14182" LastActivityDate="2017-12-15T01:48:52.553" Title="What happens, when the pump discharge velocity is faster than 3 m/s?" Tags="|design|pumps|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18471" PostTypeId="1" CreationDate="2017-12-15T03:01:23.017" Score="1" ViewCount="7865" Body="&lt;p&gt;A home windows installation rep gave me a quote today and conducted the following &lt;em&gt;temperature differential&lt;/em&gt; demonstration.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Alice pointed a heat gun (with a surface temperature of &lt;strong&gt;165F&lt;/strong&gt;) at the &lt;em&gt;interior&lt;/em&gt; surface of a window.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bob measured the &lt;em&gt;exterior&lt;/em&gt; surface of the window (with a no-contact laser thermometer) to be &lt;strong&gt;45F&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is there any way to calculate R-value (or any other standard thermal insulation reference) given the temperature differentials resulting from the above setup and procedure? Can the setup or procedure be improved in order to get an R-value or any other standard and useful benchmark or metric? If so, how?&lt;/p&gt;&#xA;&#xA;Fig. 1. Test Setup&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sy0Vj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sy0Vj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Per one of the answers, let's make the simplifying assumption the temperature of the window's interior surface is equal to the temperature of the heat gun.&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2017-12-16T20:02:38.117" LastActivityDate="2023-03-12T22:03:30.723" Title="Calculating R-value from temperature differential" Tags="|heat-transfer|thermal-conduction|thermal-insulation|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18473" PostTypeId="2" ParentId="18465" CreationDate="2017-12-15T12:55:12.437" Score="2" Body="&lt;p&gt;You can buy a device that tells you what elements are in a lump of something.  This is gross overkill for a barbecue grill.  As others have said, a magnet and a little common sense goes a long way for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the device I'm talking about is a X-ray spectrometer.  These are available in a handheld &quot;gun&quot; configuration, as one example.  You point the device at your unknown lump of stuff and pull the trigger.  It blasts the lump with x-rays with controlled wavelengths, and looks at the resulting emissions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, bombarding with high enough energy photons makes pretty much everything act like a phosphor.  The high energy photons cause electrons to move to higher states.  The photons emitted by falling back to the lower states is distinctive of the particular element.  Collect enough of these re-emission photons to get a nice spectrum, and you can tell what elements are in the lump of stuff to surprising accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One company that makes such devices is Thermo Fisher.  Uses include assaying of minerals, looking for lead in paint on old houses, and lead and other bad things in toys being imported.  These devices have tables of common alloys built in to their firmware.  They not only tell you directly that your lump is stainless steel, but what type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While such a thing would definitively answer your question, it would be gross overkill since its cost is many times that of a new barbecue grill.  If you are deciding which area is best to mine for gold though, the cost is likely well worth it.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-12-15T12:55:12.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18474" PostTypeId="1" AcceptedAnswerId="18475" CreationDate="2017-12-15T13:07:23.373" Score="0" ViewCount="42" Body="&lt;p&gt;I'm trying to derive the first order response of a sensor with step input. However, I'm stuck at step just before conducting the Laplace transform. In the following image, how is the left side equal to the right side? I know the right side equation is necessary to obtain to perform Laplace transform. &lt;a href=&quot;https://i.stack.imgur.com/faNZG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/faNZG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13336" LastActivityDate="2017-12-15T15:01:00.227" Title="Deriving the First order response of a sensor" Tags="|mechanical-engineering|sensors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18475" PostTypeId="2" ParentId="18474" CreationDate="2017-12-15T15:01:00.227" Score="0" Body="&lt;p&gt;It's simple maths. Starting from the RHS to get to the LHS:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;AK \left[ \frac{1}{s} - \frac{1}{s - \frac{-1}{\tau}} \right]&#xA;= AK \left[ \frac{1}{s} - \frac{1}{s + \frac{1}{\tau}} \right]&#xA;= AK \left[ \frac{1}{s} - \frac{\tau}{\tau s +1} \right]&#xA;= AK \left[ \frac{\tau s + 1 - \tau s}{s(\tau s +1)} \right]&#xA;= \frac{AK}{s(\tau s + 1)}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2017-12-15T15:01:00.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18476" PostTypeId="2" ParentId="18452" CreationDate="2017-12-15T18:15:29.047" Score="1" Body="&lt;p&gt;There are 4 grades on each Vertical curve.&#xA;G1 up and down &#xA;G2 up and down&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most critical grade is the bigger of the two while going downwards (-ve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think about it, trying to stop a car going down hill is harder than a car going uphill at the same speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ex:&#xA;if a curve goes up 2.71% then goes down 3.50% &#xA;then the value of the G term in the denominator is&#xA;+ (- 3.50/100) &lt;/p&gt;&#xA;" OwnerUserId="14192" LastActivityDate="2017-12-15T18:15:29.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18477" PostTypeId="2" ParentId="18471" CreationDate="2017-12-15T18:21:25.383" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/7917/3820&quot;&gt;Borrowing from this answer (and its associated question)&lt;/a&gt;: You must factor in the time you are taking the measurement as well as the starting temperature of the window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature change over time for different materials looks like the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nh1zL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nh1zL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Material 3 has the highest R-value and 1 has the lowest. R-value is proportional to the inverse of thermal conductivity, $k$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R \propto \frac{1}{k}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relevant heat-transfer equation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{Q} = k A \frac{\Delta T}{l}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$l$ is thickness&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$ is surface area&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this follows that besides other factors the rate of heatflow is proportional to the thermal conductivity, $k$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{Q} \propto k$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R \propto ln(\frac{\Delta t}{\Delta T})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\Delta t$ is the length of time the heat lamp was on the window prior to the temperature measurement reading.&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2017-12-15T20:30:59.973" LastActivityDate="2017-12-15T20:30:59.973" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18478" PostTypeId="1" CreationDate="2017-12-15T18:24:48.663" Score="1" ViewCount="29" Body="&lt;p&gt;Which properties of flexible pavement layers are reduced (discounted) in a stage construction design with an overlayed layer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thickness of stage 1 Layers?&#xA;Structural Number ?&#xA;Modulus of Resilience ?&#xA;layer coefficient ?&#xA;Other properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, do all layers get effected by the reduction?&lt;/p&gt;&#xA;" OwnerUserId="14193" LastActivityDate="2017-12-15T18:24:48.663" Title="Pavement stage construction; reduction of Structual Properties" Tags="|civil-engineering|design|highway-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18479" PostTypeId="1" CreationDate="2017-12-15T21:49:53.110" Score="3" ViewCount="2715" Body="&lt;p&gt;This question bothers me for quite a while and is about bicycle frame construction, specifically a BMX frame and even more specifically it's front triangle.&#xA;It is a heavy duty frame by definition, and most common point of frame failure in this area, is lowest point around area D - and may happen even if type1 gusset is present.&#xA;It's usually a crack along weld (which, if unnoticed, may result in tube C snapping off upon next considerable impact and is highly unwanted).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6JS8N.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6JS8N.png&quot; alt=&quot;gusset comparison&quot;&gt;&lt;/a&gt;&#xA;I am looking into building a frame with type2 gusset, which appears more reliable overall, but unfortunately lack knowledge to confirm this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tend to think additional heat stress from second weld during type1 gusset install does not do good and useful weld surface does not increase a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With type2 gusset, however, useful weld surface increases considerably, also holding all three tubes together, which should relieve stress on area D as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, few reputable people suggest type2 construction would still need type1 gusset between tubes B and C, but cannot back up their argument with real data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize - by what amount does type1 gusset does actually improve situation here, and would it be any use along with type2 gusset?&#xA;I would rather increase tube B wall thickness around weld area rather than add type1 gusset, if this proves an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional info:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frame tube material is 4130 chromoly steel, TIG welded.&#xA;Tubes A and B are usually around 35~38mm outer diameter and have 1.25mm thick wall.&#xA;Tube C wall is usually 2-3mm thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frame experiences repeated cycles of 'sharp' dynamic loads (i.e. handling rider mass upon landing from height).&lt;/p&gt;&#xA;" OwnerUserId="14196" LastActivityDate="2017-12-16T08:07:42.067" Title="Actual amount of reinforcement gained by adding gusset to welded steel tube construction" Tags="|steel|mechanical-failure|welding|heat-treatment|bicycles|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18480" PostTypeId="2" ParentId="18471" CreationDate="2017-12-15T22:42:06.090" Score="1" Body="&lt;p&gt;There are a lot of things wrong with the demonstration, including:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The temperature of the heat gun isn’t relevant - to measure the R value you need to look at the temperature difference across the window, not from the heat gun to the far side of the window.  Kind of like calculating the heat loss through your walls based on the outside temperature and the temperature in your fireplace.&lt;/li&gt;&#xA;&lt;li&gt;A lot of the heat loss through windows is through the frame rather than through the glass.  Window energy ratings typically include an overall value, and a glazing-only value.&lt;/li&gt;&#xA;&lt;li&gt;Heat loss has to be measured under steady-state conditions using a test setup that measures temperatures and heat transfer.  &lt;/li&gt;&#xA;&lt;li&gt;The relevant equation is Q = U A DT where Q is heat flow, U is the overall heat transfer coefficient (1/R), A is the area, and DT is the temperature difference.  To measure R or U, you have to know both of the temperatures and the heat flow.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the end though, you really don’t need to do all this.  The windows should have a label that provides a U value which you can compare to other windows, or use to calculate the R value.  &lt;a href=&quot;http://www.nfrc.org/energy-performance-label/&quot; rel=&quot;nofollow noreferrer&quot;&gt;See here for more information&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2017-12-15T22:42:06.090" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18481" PostTypeId="2" ParentId="18479" CreationDate="2017-12-16T08:07:42.067" Score="2" Body="&lt;p&gt;This answer is from the riding/cycling side, not an engineering view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think of the forces involved on a head tube, then the &quot;static&quot; force ( while it is standing still) is upward through the head tube.  The top tube is under compression and the downtube is under tension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When riding, its roughly the same, but there are sideways forces from steering which are approximately balanced by the rider's leaning.  Plus minor oscillations in he force magnitude due to pedalling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all well and good for a nana bike ridden gently, but BMXs have additional forces.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Standing to pedal puts an opposite force through the handlebars (pulling on the bars for more power)   A BMX rider in a sprint start could be pushing 1000 to 1500 watts through the frame for 10 seconds.&lt;/li&gt;&#xA;&lt;li&gt;Landings - a big increase in the static forces.&lt;/li&gt;&#xA;&lt;li&gt;Potholes and lumps - the wheel will push upward and backward if the bike is ridden into a pothole or a berm/curb/speed-bump etc&lt;/li&gt;&#xA;&lt;li&gt;Crashes - everything happens in a crash, tension compression rotation and vibration.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another relevant stress situation is a frontal impact.   Here's a road bike that has bashed into something with its front wheel (IE, straight line impact with a non-moving object.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lsDRd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lsDRd.jpg&quot; alt=&quot;From http://www.ellisbriggscycles.co.uk/buying-a-secondhand-steel-frame/&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the subtle top tube bend, and the more significant downtube bend.  With time, both sites will become a ring of rust.  The fork and front wheel may have been a write-off, or they mat have survived largely unscathed - both are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why?  The welding process on steel will harden the area of the weld, and increase its brittleness.   So the front section of top/downtube is less flexible, and its more likely to bend at the first soft bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;gusset&lt;/strong&gt; helps stiffen and protect that area of the bike's frame, and the leverage that can be exerted on the first softer section is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't want a bike to be all stiff, because then it would be uncomforable to ride - one of the plusses of a steel bike is that it has give, or flex, to take the buzz out of the road.  An aluminium frame doesn't do that as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pictured road bike also has downtube shifter bosses brazed or welded into place, which probably provided a gathering point for stresses and encouraged the larger bend under load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also notice this frame is &lt;strong&gt;lugged&lt;/strong&gt; which is the ornate couplers around the tube joints.  These do a similar job to the gusset.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As a rider - the last thing you want to happen is have your bike become a spontaneous unicycle.  If any major part of it breaks, you're probably going to hit the road.  This is to be avoided at all costs, even if it means a small weight penalty.  Teeth don't grow back.&lt;/p&gt;&#xA;" OwnerUserId="12158" LastActivityDate="2017-12-16T08:07:42.067" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18486" PostTypeId="2" ParentId="18461" CreationDate="2017-12-16T19:06:02.053" Score="0" Body="&lt;p&gt;Just found a patent for this: &lt;a href=&quot;https://www.google.com/patents/US20070154374&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/patents/US20070154374&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;Method for removing sulfur dioxide and other acid gases, mercury, and nitrogen oxides from a gas stream with the optional production of ammonia based fertilizers&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IId04.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IId04.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14179" LastActivityDate="2017-12-16T19:06:02.053" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18487" PostTypeId="1" CreationDate="2017-12-16T20:29:01.740" Score="1" ViewCount="406" Body="&lt;p&gt;i want to print a worm gear for a robot. This robot will work every day for at least 10 hours. The gears size is about the half of a sheet paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched the internet for example in industrial use. I can not find much about the life span of 3d printed gears. The only thing is that abs is good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are 3d printed parts applicable for industrial use? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can i except for the life span?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerDisplayName="user14157" LastActivityDate="2017-12-17T19:18:06.740" Title="Are 3d printed gears (ABS) suitable for long term use?" Tags="|design|gears|robotics|3d-printing|quality-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18489" PostTypeId="1" AcceptedAnswerId="18501" CreationDate="2017-12-17T00:52:42.527" Score="0" ViewCount="2349" Body="&lt;p&gt;Symmetrical truss is given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7DVfu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7DVfu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have issue with the sign convention for joint H.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See below the solution for joint I. As you can see, when calculating equilibrium in y (vertical) direction, forces that act upwards are written with &quot;+&quot; sign, and forces acting downwards with &quot;-&quot; negative sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tly7C.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet for joint H below, forces acting downwards are written as positive, which is why $F_{HJ}$ turns out to be a negative force (compression force). But it's not consistent with the solution for previous joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8glFL.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you go about finding force in member HJ?&lt;/p&gt;&#xA;" OwnerUserId="12190" LastEditorUserId="1832" LastEditDate="2017-12-17T01:28:10.950" LastActivityDate="2017-12-18T10:37:37.550" Title="Structural Analyses of Trusses, method of joints" Tags="|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18490" PostTypeId="1" AcceptedAnswerId="18497" CreationDate="2017-12-17T12:20:40.443" Score="1" ViewCount="163" Body="&lt;p&gt;A component of a control loop is approximated by the following relation between input $x$, and output, $y$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = 5x^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During normal operation, the value of input to this component ranges between $1$ and $1.5$. In&#xA;stability analysis of the overall control loop, what is the gain of this component?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I'm not sure how to solve this problem. I don't know how to create a transfer function by performing a Laplace transform because this isn't a differential equation and there is only one component. Would really appreciate it if someone could point me in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="14204" LastActivityDate="2017-12-18T00:23:17.170" Title="PID Gain Calculation" Tags="|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18491" PostTypeId="1" CreationDate="2017-12-17T12:54:09.540" Score="0" ViewCount="95" Body="&lt;p&gt;What techniques were used to create the maps of a world, country or region when satellite or similar technology did not exists few decades or centuries ago?&lt;/p&gt;&#xA;" OwnerUserId="5484" LastEditorUserId="90" LastEditDate="2022-08-03T00:54:59.813" LastActivityDate="2022-08-03T00:54:59.813" Title="How maps of globe, country or region were created few decades/centuries ago when modern technology like today did not exists?" Tags="|surveying|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18493" PostTypeId="1" CreationDate="2017-12-17T14:27:09.723" Score="5" ViewCount="170" Body="&lt;p&gt;This will sound like a super silly question, but:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it called &quot;Model Predictive Control&quot;?&#xA;I mean : the other types of control also compute the control input based on prediction of what is going to happen based on the &quot;Model&quot;, no?&lt;/p&gt;&#xA;" OwnerUserId="14207" LastEditorUserId="14288" LastEditDate="2022-07-29T10:45:04.150" LastActivityDate="2022-07-29T10:45:04.150" Title="Why is it called &quot;Model predictive control&quot;?" Tags="|control-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18494" PostTypeId="2" ParentId="18491" CreationDate="2017-12-17T15:38:05.250" Score="3" Body="&lt;p&gt;You don't need satellite images to make accurate maps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accurate &lt;a href=&quot;https://en.wikipedia.org/wiki/Surveying&quot; rel=&quot;nofollow noreferrer&quot;&gt;surveying techniques&lt;/a&gt; have been around for a very long time and are essentially based on fairy simple technology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic principals of surveying require the ability to measure direction, elevation and distance relative to some fixed reference point and some basic geometry can allow you to make indirect measurements, especially of heights. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly you can make up a simple sighting device with a spirit level or plumb line and have a second person with a target to sight on of known height to plot lies of equal height ie contour lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Land based topographical mapping really came into its own with the advent of artillery as accurate maps became a military necessity in fact in the UK the map making agency still called the Ordnance Survey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically you choose a convenient local reference point and accurately determine its absolute position (ie latitude and longitude) by reference to the sun at a known time of day (or noon if you don't have an accurate chronometer) or the pole star. Then you can measure relative angles to other features using eg a theodolite and you can fix the location of any point by trigonometry without having to do a lot of actual distance measuring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is this combination of local measurements and the ability to fix reference points to a global coordinate system which is key  to accurate large scale mapping. Often on very old maps distances are reasonably accurate but overall shapes of landmasses and coastlines are a bit distorted. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-12-30T16:27:05.987" LastActivityDate="2017-12-30T16:27:05.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18495" PostTypeId="1" CreationDate="2017-12-17T17:38:25.343" Score="0" ViewCount="639" Body="&lt;p&gt;Ion thruster has very low thrust due to its low mass flow rate. However is it possible to increase the mass flow rate? Together with its high exhaust velocity, won't the ion thruster be able to reach that of chemical rocket's thrust considering how the exhaust velocity of chemical rocket is much lower?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it not possible to increase the mass flow rate due to the limited amount of electricity that the rocket can carry? I know there are other things to consider like the screen which will affect its efficiency(prevent all the ions from leaving) but I don't know how much it will affect the maximum mass that can flow.&lt;/p&gt;&#xA;" OwnerUserId="14208" LastActivityDate="2017-12-18T01:31:07.723" Title="Increase thrust of ion thruster?" Tags="|mechanical-engineering|propulsion|rocketry|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18496" PostTypeId="2" ParentId="18487" CreationDate="2017-12-17T19:18:06.740" Score="1" Body="&lt;p&gt;Ask these questions:&#xA;What is the industrial application High/lo speed? Hi torque? Is it a continuous load? Shock loading? Direction changes?  Actual gear teeth amount? What gear tooth size standard?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are needed to provide an actual estimate to gear life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What 3D printing process?? FDM? SLS? Etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it feasible? Possibly, with a SLS printed gear set, relatively low speed and low torque with a continuous load.. also, assuming you can get the center holes in the right position and get your gear center-to-center properly aligned.  Use some standard gear life calculations and maybe de-rate it to 50%. 3D printed materials behave differently than conventional &quot;equivalent&quot; material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Off the cuff, for an industrial application, I would find a readily available molded plastic gear set rather than print a set. You need something reliable and readily available.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-12-17T19:18:06.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18497" PostTypeId="2" ParentId="18490" CreationDate="2017-12-18T00:23:17.170" Score="1" Body="&lt;p&gt;The input and the output of the component need to have a linear relation (such as y = 5 * x), so you have to linearize that ecuation in order to aproximate it to a linear ecuation that you could use to obtain the gain of the component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To linearize :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If x = a;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = (f(a) + f’(a) * (x - a))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you take a = 1 (one of the values you gave)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y = 5 + 10 * (x - 1)&#xA;y = 10 * x - 5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could use a gain of 10 and add a sumator with a minus for substracting 5. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps you.&lt;/p&gt;&#xA;" OwnerUserId="14214" LastActivityDate="2017-12-18T00:23:17.170" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18498" PostTypeId="2" ParentId="18495" CreationDate="2017-12-18T01:31:07.723" Score="1" Body="&lt;p&gt;Only by increasing the area of the emitter. For a given species, you can't increase the density of the mass flow beyond a certain limit. And that limit decreases if you lower the emitting velocity. What you would need to do gain thrust at constant power, or even constant specific power, would be kick more reaction mass at a lower velocity. If the energy per unit reaction mass was 1/4, you could get 4 times the mass to 1/2 the velocity, which changes the density in the ion jet by a factor of 8. Unfortunately, these are all charged the same, and you have a bit of an issue with &lt;a href=&quot;https://en.wikipedia.org/wiki/Space_charge&quot; rel=&quot;nofollow noreferrer&quot;&gt;space charge&lt;/a&gt; density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wiki on ion drives actually does a decent job with the physics, but you do need a bit of electromagnetics in your toolkit.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2017-12-18T01:31:07.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18499" PostTypeId="2" ParentId="18468" CreationDate="2017-12-18T03:30:29.487" Score="1" Body="&lt;p&gt;You want to look at high temp RTV silicone adhesives. They can handle temps way above 100c. Typically found and used as make-a-gasket material on engines and commonly found at automotive stores. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'd like to see you use SS for the bracket to keep your thermal expansion co-efficient the same as the base part..I know it's not a lot of temperature change but it may help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, I second the fasteners method per the comments because you will have a captured connection between the two parts.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-12-18T03:30:29.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18500" PostTypeId="1" AcceptedAnswerId="18504" CreationDate="2017-12-18T09:54:01.530" Score="3" ViewCount="77" Body="&lt;p&gt;I am searching for information on how finite element code (abaqus /ansys) could be used to model the contact stress behaviour between 2 bearing surfaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key point is that there is an additional hard coating n both surfaces so as to reduce wear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would one model this additional hard layer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this hard layer be represented by an additional layer of elements with a higher elastic modulus?&lt;/p&gt;&#xA;" OwnerUserId="14216" LastEditorUserId="1832" LastEditDate="2017-12-23T17:14:07.647" LastActivityDate="2017-12-23T17:14:07.647" Title="Contact stress of bearing surfaces" Tags="|mechanical-engineering|structural-engineering|materials|gears|structural-analysis|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18501" PostTypeId="2" ParentId="18489" CreationDate="2017-12-18T10:37:37.550" Score="2" Body="&lt;p&gt;The answer presented is not inconsistent, but it is badly laid out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have not visibly defined &quot;upwards as positive&quot; (they should have...). The equation that's confusing you is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum F_y = 0;  F_{HJ}\cos(45) + 100\sin(75) + 273.2\cos(45) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this instance, every single force on the Free Body Diagram in the 'y' direction is pointing downwards, so, to follow the &quot;upwards is potitive&quot; convention, we could write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ -F_{HJ}\cos(45) - 100\sin(75) - 273.2\cos(45) = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which, I'm sure you'll agree, is equivalent to the originally presented formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When answering questions like this, you should have always define the positive directions by an annotation with the FBD, and show any extra steps (such as multiplying an equation through by '-1'), to prevent confusion for casual readers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note further, that a 'trailing minus' at the start of an equation can get lost, and is often the root of a sign-error; what they did made sense, but they should have made it clearer.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-12-18T10:37:37.550" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18503" PostTypeId="2" ParentId="14428" CreationDate="2017-12-18T21:02:27.657" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;The components in your model represent internal equivalant circuit of a motor such as internal resistor and internal inductor characteristics. Using &lt;em&gt;mobility anology&lt;/em&gt;, you can replace inertia with a capacitor. So circuit becomes a RLC network. In steady state, the capacitor will be loaded with electrical charges and voltage drop on the capacitor will approach to the input voltage. The voltage drop on the resistor will be very low, and thus there will be little current passing through it. Thinking mechanically, the inertia requires energy to change its angular velocity in transient. But after reaching steady state, it wants to keep its velocity. The only thing against this is the damping. So, as long as the circuit supply enough power to compansate the damping, the velocity will be constant.&lt;/li&gt;&#xA;&lt;li&gt;Yes, your interpretation is correct. You modelica component explicitly state that with equations &lt;em&gt;kw=v&lt;/em&gt; and &lt;em&gt;tau=-ki&lt;/em&gt;. As an alternative, different constants such as $k_v$ and $k_i$ can be used. Because of the conservation of energy, you must pay attention to input-ouput energies. In your model $k=1$, so it is ideal energy converter. Input power to EMF component, $P_{in}=v_{emf}\cdot i$ and output power of the EMF component $P_{out}=w\cdot\tau$ is equal (on the EMF component). The heat lost of the motor is represented by resistor.&lt;/li&gt;&#xA;&lt;li&gt;In reality, it depends on how much detail you want to experiment (simulate/analyse). You may model the windings, permanant magnets, magnetic fields, current, relative positions of magnets and windings and resulting force etc (in the case of PMDC). But if you will not analyse DC motor deeply, and just use it as a component in your system, an approximation is good enough. Usually, variable efficiency is modeled, so using k as not a constant but a function of w or something.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1599" LastEditorUserId="1599" LastEditDate="2017-12-19T05:27:27.567" LastActivityDate="2017-12-19T05:27:27.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18504" PostTypeId="2" ParentId="18500" CreationDate="2017-12-19T00:29:02.247" Score="4" Body="&lt;p&gt;I would suggest a two step process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Do a high resolution simulation (with four layers) of a small region to estimate the effective contact stiffness of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Do a two surface simulation with the contact stiffness from step 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing a high resolution simulation of the whole system will be computationally expensive, and not suggested unless you have access to parallel computing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is that the thin layer will have to be discretized using at least three elements through the thickness; leading to millions (if not billions) of elements for the entire assembly.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2017-12-19T00:29:02.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18505" PostTypeId="1" CreationDate="2017-12-19T03:08:33.957" Score="0" ViewCount="175" Body="&lt;p&gt;I am learning engineering drawing and confused on this.&lt;a href=&quot;https://i.stack.imgur.com/e5LdN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e5LdN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14227" LastActivityDate="2018-01-18T10:04:13.927" Title="In the image below,only 4 quadrants are mentioned but my intuition says there should be 8, please clarify me?" Tags="|mechanical-engineering|technical-drawing|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18506" PostTypeId="1" CreationDate="2017-12-19T04:05:42.070" Score="1" ViewCount="226" Body="&lt;p&gt;I am trying to create a simple R2R DAC, maybe 8-16 bits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to be able to that type of DAC inline to a headphone line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the an 8 bit R2R DAC will need 8 pins, how exactly do I wire up a TRS connector to that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I have to put an analog to digital converter inline first?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.5mm TRS analog audio player output -&gt;&#xA;[create analog to digital output] -&gt;&#xA;[apply the digital signal to my 8bit DAC] -&gt;&#xA;output to 3.5mm TRS analog&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WK38c.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WK38c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12586" LastActivityDate="2017-12-19T05:24:27.540" Title="Adding an R2R Digital to analog converter inline with TRS connector" Tags="|circuits|circuit-design|audio-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18507" PostTypeId="2" ParentId="18445" CreationDate="2017-12-19T05:00:57.193" Score="0" Body="&lt;p&gt;The difference is the amount of crash energy dissipated and limited play course of a spring loaded dumper.&#xA;The energy absorbed by a spring is the area under force line: &#xA;1/2 k.x^2.m. - m is the bumper mass.&#xA; which is very small compared to inertia of the car. And at the end of it's course it delivers the huge leftovers of impact energy to the rest of the car while now adding it's own mass to attacking car's inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea behind integrated crash structural design is a highly energy absorbing gradual and ductile crumpling of the entire structure surrounding the occupants as a scrifice zone while reducing the negative acceleration to tolerable levels. No loose part is going to be of any help.&#xA;If you look at the corrogated pressed patterns on the body frame of the new cars you see they have been designed to absorb much energy in a crash while crumpling away from the cab of the car. You will have the entire front of the car's mass and length to help dampen the impact.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2017-12-19T05:00:57.193" CommentCount="0" CommunityOwnedDate="2017-12-19T05:00:57.193" ContentLicense="CC BY-SA 3.0" />
  <row Id="18508" PostTypeId="2" ParentId="18506" CreationDate="2017-12-19T05:24:27.540" Score="1" Body="&lt;p&gt;Okay, I will go ahead and answer my own question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I'll have to deal with an analog signal anyways, coming from the audio output of a mobile device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll need an ADC [analog to digital converter]&#xA;then I'll match that up with a DAC [digital to analog converter]&#xA;and send that out a TRS plug.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This thing will need it's own power supply for it all but I think it might be a fun project to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.allaboutcircuits.com/textbook/digital/chpt-13/digital-analog-conversion/&quot; rel=&quot;nofollow noreferrer&quot;&gt;enter link description here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12586" LastActivityDate="2017-12-19T05:24:27.540" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18509" PostTypeId="2" ParentId="18505" CreationDate="2017-12-19T06:53:51.283" Score="2" Body="&lt;p&gt;Tha author is describing First Angle Projection : accept it as it is - trying to change it will only lead to confusion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will soon be shown Third Angle Projection - again you will need to accept how it is done as both are used in industry - a symbol on the drawing will tell you what convention has been used. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-12-19T09:48:35.520" LastActivityDate="2017-12-19T09:48:35.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18510" PostTypeId="1" AcceptedAnswerId="18543" CreationDate="2017-12-19T10:00:46.237" Score="4" ViewCount="177" Body="&lt;p&gt;I want to calculate the deformation of the following structure:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ynvss.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ynvss.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have numbered the vertical beam 1 and the horzontal beam 2. The hight of beam 2 is given by:&#xA;$h(x)=h\sqrt{1-x/L_2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I thought I could solve this with Castigliano's method. Where beam 1 undergoes an axial load $F$ and a constant bending moment of $Fa$. Beam 2 undergoes a bending moment of $0$ on the left and $Fa$ on the right. The energy is given by: $U=\int_0^L\frac{F(x)^2}{2AE}dx+\int_0^L\frac{M(x)^2}{2EI}dx$ &#xA;So for beam 1 and 2 the energy would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ U_1 = \int_0^{L_1}\frac{F^2}{2bhE}dx+\int_0^{L_1}\frac{F^2a^2}{2E\frac{1}{12}bh^3}dx = \frac{F^2L_1}{2bhE} + \frac{F^2a^2L_1}{2E\frac{1}{12}bh^3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_2 = \int_0^a\frac{F^2x^2}{2E\frac{1}{12}b(h\sqrt{1-x/L_2})^3}dx = &#xA;\frac{4F^2L_2^2}{Ebh^3}\left(\frac{\sqrt{1-a/L_2}(a^2+4aL_2-8L_2^2)}{a-L_2}-8L_2\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $b$ is the thickness of the part&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I applied Castigliano's method, which gives the deformation of the shape in the same direction as $F$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta=\frac{\partial U}{\partial F}= \frac{F}{E}\left(\frac{L_1}{bh} + \frac{a^2L_1}{\frac{1}{12}bh^3} + \frac{8L_2^2}{bh^3}\left(\frac{\sqrt{1-a/L_2}(a^2+4aL_2-8L_2^2)}{a-L_2}-8L_2\right)\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this equation gives me wrong results. For instance, with $F=1500, E=113.76*10^9, b=0.005, h=0.007, L_1=0.05, L_2=0.04$ and $a=0.032$&#xA;I get a deformation of 9.8 mm, where as a FEA simulation (Nastran In-Cad) gives a deformation of 6.2 mm (in the same direction as the force).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried this without the parabolic property and this does generate correct results. That equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{F}{E}\left(\frac{L_1}{A_1}+\frac{L_2^2L_1}{I_1}+\frac{L_2^3}{3I_3}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think there something wrong with my parabolic equation, but I can't figure out what it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: did some trial and error, and if I fill in $h(h/2)$ for $h$ in the last term I get pretty accurate results. To clarify, the height used (in this situation) in the last therm ($\frac{8L_2^2}{bh^3}$) is the hight of beam 2 where beam 1 ends, on the left. But I still have know idea why this is and if this works for every situation&lt;/p&gt;&#xA;" OwnerUserId="14151" LastEditorUserId="14151" LastEditDate="2017-12-19T14:41:39.077" LastActivityDate="2018-01-02T13:24:51.143" Title="Deformation of two perpendicular beams with parabolic properties" Tags="|mechanical-engineering|beam|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18512" PostTypeId="1" CreationDate="2017-12-19T11:57:29.557" Score="0" ViewCount="627" Body="&lt;p&gt;I need help with the following factor of safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the infomation to apply to the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tensile Strength = 460MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear Stress = 280MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bolt Diameter = 8mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angle = 50 Degrees&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to determine the tensile stress and shear stress in the bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the equation for Shear stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{Ultimate Stress}{Allowable Stress}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far i have this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;201.061mm^2 being Area of the 8mm bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{9kN}{201.061mm^2} = 0.04476\frac{kN}{mm^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x1000 = 44.76\frac{N}{mm^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$= 44.76MPa$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{280MPa}{44.76MPa} = 6.255MPa$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Shear Stress = 6.255MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And i think i work it backwards to work out tensile strength?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{Ultimate Load}{Allowable Load}$$&#xA;The problem i have is how to apply the 50 degree angle. I have read to multiply the equation by cos or sin. But i am really stuck on this part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone help me please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attached diagram. Sorry for not attaching this before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a is 50 degrees. And is what i am stuck on. I do not know how to apply the degrees to the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kdgAa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kdgAa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14101" LastEditorUserId="14101" LastEditDate="2017-12-20T08:47:31.593" LastActivityDate="2017-12-20T21:03:50.973" Title="Factor of safety with an angle" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18513" PostTypeId="1" CreationDate="2017-12-19T13:35:49.497" Score="0" ViewCount="169" Body="&lt;p&gt;I came across this snippet in a vendor manual for gaskets for a flanged joint for vacuum duty (tongue &amp;amp; groove). What exactly is a compression stop? Is it evident in the sketch below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.flexitallic-gmbh.eu/wp-content/uploads/2015/02/Gasket_Design_Criteria.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.flexitallic-gmbh.eu/wp-content/uploads/2015/02/Gasket_Design_Criteria.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZijkY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZijkY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gQgeC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gQgeC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6277" LastActivityDate="2017-12-19T16:22:56.733" Title="What is a &quot;compression stop&quot; for a tongue &amp; groove joint?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18514" PostTypeId="2" ParentId="18513" CreationDate="2017-12-19T16:22:56.733" Score="1" Body="&lt;p&gt;A compression stop is a method of preventing the gap between the two pieces being zero ie providing a minimum space for the gasket. This could be achieved by the two “teeth” or annular ring being slightly longer than the cutout or groove they go in.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-12-19T16:22:56.733" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18517" PostTypeId="1" AcceptedAnswerId="18519" CreationDate="2017-12-20T05:57:19.473" Score="0" ViewCount="63" Body="&lt;p&gt;At my workplace I have been assigned to create an accurate 3D model of our Dell PowerEdge R820 servers so we can insert our custom cards and run a CFD and thermal analysis on them. The end goal is to have an exact understanding off all the airflow and heat transfer within the servers and server racks to help in designing future systems and fix overheating issues we are having with some of our current ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that the insides of these servers are quite complex and it is going to take an era for me to reverse engineer all the parts with a pair of verniers and find out how much heat is produced and where.So I am wondering if anyone else has come across something like this before and how they created an accurate CFD of their system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would 3D scanning be an option? I am concerned that 3D scanning will be too accurate, in the sense that what should be a straight edge might be turned into 200 different vertices etc. And I am not sure of how much cleanup it would require. I plan to do the CFD in Solidworks which requires solid modelling - it may have issues with the surfacing of the scans. Would like to hear from anyone with experience regarding this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, does anyone have experience with getting 3D CAD models from companies like Dell? I work at a large company and we buy a lot of their products for our projects but I wouldn't have a clue what department to call to see what needs to be organised to get CAD files. Would welcome any input here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, for the thermal modelling, my plan of attack is to use some IR cameras to see where the heat is being produced. But I don't know how to work out HOW much heat is being produced at these spots so I can model the heat production from these processors in the CFD. What do people normally do for this?&lt;/p&gt;&#xA;" OwnerUserId="13810" LastActivityDate="2017-12-20T08:33:52.440" Title="How to model the inside of a server for CFD and thermal studies?" Tags="|cad|cfd|solidworks|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18518" PostTypeId="2" ParentId="18517" CreationDate="2017-12-20T07:20:55.347" Score="0" Body="&lt;p&gt;For your IR camera, you need an image of a known object at a known temperature in the same environment as a temperature reference , then when you have the IR image you can cross-reference the colours to approximate the temperatures. Check out some building thermal images for the idea.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-12-20T07:20:55.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18519" PostTypeId="2" ParentId="18517" CreationDate="2017-12-20T08:22:51.857" Score="1" Body="&lt;p&gt;Your options are a bit limited. Either you get the model from Dell, or you model it yourself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting the model is a long shot, but if you are a really big customer that might happen. However this is not necceserily a godsend, since:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The model may not be accurate enough for your need. They may have ripped some data out to protect their IP&lt;/li&gt;&#xA;&lt;li&gt;It may be too accurate in places that are of no interest for you. And mostly the data is just in your way.&lt;/li&gt;&#xA;&lt;li&gt;The model may be unsuitable for your task. So its geometry but your cfd needs the material info and heat generation info.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That said modeling a suitably absteacted version that simulates faster than imported models is not necceserily  a huge job. Its just very scary, as it is just raw work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would scanning help? In my experience not really, though it might speed up data gathering, as you now have a digital model to check against. So you do not need to go and measure that one dimension you realized you need. Thats a good thing if you can not have the object on the table next to you. However, scanning is another can of worms as a work process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concentrate on abstracting things as much as you can. Do you really need every detail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note thermal camera does not directly neasure what you want it measures how much things radiate at a specific band which varies by material and its surface quality. worse some materials like the silicon in your chips are transparent at typical infrared camera ranges. You still need to know how much the materials transfer energy to air. That said the thermal camera is a good validator target for your model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a lot of work no matter how you look at it.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2017-12-20T08:33:52.440" LastActivityDate="2017-12-20T08:33:52.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18520" PostTypeId="1" CreationDate="2017-12-20T08:56:26.917" Score="1" ViewCount="140" Body="&lt;p&gt;If we look at the Bill of Materials of BlueROV 2 they have written the thrusters separately as clockwise and anticlockwise. Why not use Reverse config. ESC instead of anticlockwise propellers?&lt;/p&gt;&#xA;" OwnerUserId="14181" LastActivityDate="2018-02-18T14:57:10.770" Title="Why does the BlueROV2 have three clcokwise and three anti-clockwise propellers?" Tags="|design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18521" PostTypeId="2" ParentId="18520" CreationDate="2017-12-20T13:19:38.460" Score="1" Body="&lt;p&gt;As explained on &lt;a href=&quot;https://www.bluerobotics.com/store/thrusters/t200-thruster/&quot; rel=&quot;nofollow noreferrer&quot;&gt;their website&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The T200 comes with &lt;strong&gt;clockwise and counter-clockwise propellers&lt;/strong&gt; to counter torque.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And when you look at the body of the vessel, you can see six thrusters distributed around it.  Given that it's intended for work in confined spaces, I think that it makes sense that you would want the propulsion to be balanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.bluerobotics.com/wp-content/uploads/2016/06/iso-bottom-300x300.png&quot; alt=&quot;base&quot;&gt; &#xA;&lt;img src=&quot;https://www.bluerobotics.com/wp-content/uploads/2016/06/top-down-top-300x300.png&quot; alt=&quot;top&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2017-12-20T13:19:38.460" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18522" PostTypeId="1" CreationDate="2017-12-20T17:22:41.007" Score="0" ViewCount="63" Body="&lt;p&gt;I'm looking into developing a system that supplies wasted heat of the transformer of my aquarium LED light (which is always quite) to the aquarium water in order to save resources. The heat which is currently emitted into the room ought to be transported from the transformer into the aquarium water using heat exchange and the water flow in tubes which are already available because of the filtration setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's gonna be a hobby project, but I was planning to design a 3D printable enclosing for the transformer and the transmission of heat into the circulating water for different sized transformers in order to increase reusability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know from PC hardware cooling systems that water cooling is generally more efficient than air stream cooling (the LED transformer for this project works without moving air). However, I'd like to check what to consider during design of the heat transmission in order to ensure that the water cooling always is more efficient than the static air cooling so that I and potential other users don't risk damage on their transformers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following details reduce reusability, but I guess it's fine to develop the prototype localized at first: In the hottest summer (Berlin, Germany) the aquarium remains below 23 degree whereas I'd like to have it at 25 constantly. Thus, I assume that the water can always take all the heat of the transformer.&lt;/p&gt;&#xA;" OwnerUserId="9780" OwnerDisplayName="Karl Richter" LastEditorUserId="9780" LastEditDate="2017-12-20T20:11:52.360" LastActivityDate="2017-12-20T20:11:52.360" Title="Is circular water cooling reliably more efficient than static air cooling?" Tags="|thermodynamics|electrical-engineering|cooling|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18523" PostTypeId="2" ParentId="18522" CreationDate="2017-12-20T18:30:03.660" Score="1" Body="&lt;p&gt;Karl. &#xA;All performance characteristics are dependant on heat exchanger design parameters, which are  used to carry heat away from the process. In your case returning heated water to the aquarium is not an option,  making a passive aircooled setup the obvious choice for a system of no complexity and minimal build challenges. &lt;/p&gt;&#xA;" OwnerUserId="14245" LastActivityDate="2017-12-20T18:30:03.660" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18524" PostTypeId="2" ParentId="18522" CreationDate="2017-12-20T19:21:46.547" Score="1" Body="&lt;p&gt;the easiest and safest way to cool the transformer and heat the water would be to get some heatsink compound (thermally-conductive grease), grease up the back of the transformer housing, and simply press it firmly against the aquarium glass near the bottom of the aquarium- and hold it there with some duct tape. the heat transfer through the glass will not be optimal, but it might well be good enough to serve your purposes. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-12-20T19:21:46.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18525" PostTypeId="2" ParentId="7403" CreationDate="2017-12-20T20:46:01.127" Score="1" Body="&lt;p&gt;It will depend on the type of glue you will use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Super glue (cyanoacrylate adhesives) requires very thin bond line and their final strength weakness as the bond line thickness increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2K Adhesive is quite the opposite as they are a gap filling material and if the bond line is smaller than a certain value (depending on the one you use) the final strength decreases and off-course it has a maximum bond line thickness    &lt;/p&gt;&#xA;" OwnerUserId="14247" LastActivityDate="2017-12-20T20:46:01.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18526" PostTypeId="2" ParentId="18512" CreationDate="2017-12-20T21:03:50.973" Score="1" Body="&lt;p&gt;You can solve the problem at various levels of complexity.  But the first thing to do is to consider the simplest possible situation where all you have is the inclined plane and forces at an angle to the plane.  From &lt;a href=&quot;http://www.themathpage.com/abooki/propI-27-28.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euclid&lt;/a&gt;, and your figure, that angle is $\theta$ = 50 degrees.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the bolt is perpendicular to the inclined plane.  Then the force can be resolved into components that are normal and tangential to the plane.  The two components are $F_n = F \sin\theta$ = 6900 N and $F_t = F \cos\theta$ = 5800 N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the 8 mm bolt is $ A = \pi d^2/4$ = 5e-5 m$^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal and tangential stresses are $\sigma_n = F_n/A$ = 137 MPa and $\sigma_t = F_t/A$ = 115 MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factors of safety are $f_n = \sigma^{\text{allow}}_n/\sigma_n$ = 460/137 = 3.3 and $f_t = \sigma^{\text{allow}}_t/\sigma_t$ = 280/115 = 2.4.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2017-12-20T21:03:50.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18528" PostTypeId="1" CreationDate="2017-12-20T23:39:39.967" Score="1" ViewCount="396" Body="&lt;p&gt;I am working on a robotic arm, I am going to use servo motors, those are going to be controlled from the PC with command (it is a chess-game arm) so the command will be some thing like move piece from A1 to E7 for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the angle that I am going to pass to the servos for each specific command?&lt;/p&gt;&#xA;" OwnerUserId="14250" LastEditorUserId="90" LastEditDate="2017-12-21T00:28:21.280" LastActivityDate="2018-01-21T05:35:49.247" Title="Angles Measurement to control (move) a robotic arm" Tags="|electrical-engineering|control-engineering|robotics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18532" PostTypeId="2" ParentId="18469" CreationDate="2017-12-21T09:03:25.997" Score="4" Body="&lt;p&gt;They are both correct. There are actually a number of different types of geothermal power plant with the most common being, dry steam, flash steam, and binary cycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dry steam plants require pumps to draw from underground resources of steam. This is piped directly from underground wells to the power plant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flash steam plants (most common) use geothermal reservoirs that have naturally high temperatures (180 Celsius +). The hot water from the reservoir flows up through a well under its own pressure. As it reaches the surface the pressure decreases and the water begins to boil and turn in to steam. Any water left over from this process is then pumped back in to the reservoir. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Binary cycle power plants are more complex but are capable of operating at much lower temperatures (100-180 Celsius). These use hot water pumped from an underground reservoir to heat an organic fluid with a low boiling point. This organic fluid is vaporised in a heat exchanger which turns a turbine. The leftover water is &quot;injected&quot; back in to the reservoir. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to read up more on the specific types of geothermal plant, this website has more information: &lt;a href=&quot;https://www.conserve-energy-future.com/geothermalpowerplanttypes.php&quot; rel=&quot;noreferrer&quot;&gt;https://www.conserve-energy-future.com/geothermalpowerplanttypes.php&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="14150" LastActivityDate="2017-12-21T09:03:25.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18534" PostTypeId="1" CreationDate="2017-12-21T15:55:35.623" Score="1" ViewCount="54" Body="&lt;p&gt;I am a mechanical engineer who is interested in learning different processes and equipment mainly used in process industry namely, textile, chemical, sugar mill, and other food process industry.&#xA;Is there a one place to go solution available or if someone can help me and guide me to resources available online that can help me in understanding how these industries work in term of processes and equipment. &lt;/p&gt;&#xA;" OwnerUserId="7777" LastActivityDate="2017-12-21T15:55:35.623" Title="Industrial machinery and equipment" Tags="|mechanical-engineering|manufacturing-engineering|process-engineering|petroleum-engineering|" AnswerCount="0" CommentCount="3" ClosedDate="2017-12-24T05:16:32.763" ContentLicense="CC BY-SA 3.0" />
  <row Id="18536" PostTypeId="1" AcceptedAnswerId="28566" CreationDate="2017-12-21T16:12:23.943" Score="4" ViewCount="3205" Body="&lt;p&gt;Are there some general guidelines for approximating the plastic modulus of steel under tension so that a bilinear stress strain curve can be defined for Loading beyond the elastic region?&lt;/p&gt;&#xA;" OwnerUserId="14216" LastEditorUserId="1832" LastEditDate="2017-12-23T17:13:23.980" LastActivityDate="2019-05-28T15:50:20.097" Title="Plastic modulus" Tags="|structural-engineering|materials|structural-analysis|applied-mechanics|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18538" PostTypeId="1" CreationDate="2017-12-21T17:32:35.877" Score="1" ViewCount="42" Body="&lt;p&gt;I am currently attending a radio technology course. We continuously work on radio sets that are sometimes used as manpacks by the Army. In order to learn how to operate them we work in close proximity to the sets and their antennas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The antennas are generally whip type or long rods. Such as these :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6UMNw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6UMNw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/37YL4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/37YL4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radio sets generally output around 25-50W in the 30-90MHz frequency range. Can that possibly cause damage to my iPhone?&lt;/p&gt;&#xA;" OwnerUserId="14258" LastActivityDate="2017-12-21T17:32:35.877" Title="Can an antenna damage an iPhone?" Tags="|rf-electronics|electromagnetism|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18539" PostTypeId="2" ParentId="18528" CreationDate="2017-12-21T23:10:49.410" Score="1" Body="&lt;p&gt;&lt;strong&gt;Angle measurement&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to measure the angle (that is what I read in your question), I would go for rotary potentiometers. They are super cheap, lightweight, and easy to read out with a simple microcontroller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Control&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control of the servos should work with a PID loop to which you feed your setpoint (desired angle) and the measured angle. By adapting the three constants P, I, and D you can define the &quot;behavior of your movements&quot;. To control the motors I would recommend you to use a microcontroller instead of the computer. Let the computer calculate the angles for the robotic arm, pass it to your microcontroller, and the &#xA;latter should take care of measuring the angles and moving your servos. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Angle calculation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your question is how you calculate the angles of the robot's arm as a function of the end-effector's position, we need to know how many axis you are using, which length the robotic limbs have etc. However, I would have a look at coordinate transformations. Tell us more about the robot's geometry and I will try to help you :).&lt;/p&gt;&#xA;" OwnerUserId="13505" LastActivityDate="2017-12-21T23:10:49.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18541" PostTypeId="2" ParentId="18528" CreationDate="2017-12-22T05:07:10.260" Score="2" Body="&lt;p&gt;The act of calculating the angles of a robot from the position of the end eflector is called inverse kinematics. Incidenttally the reverse, calculating end efflector position from angles is called forward kinematics. Kinematics being the study of motion of interconnected bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are many ways to attack kinematics problems, and the exact solution depends on your particular geometry and hardware in the control loop. You can do either iterative or closed form solutions for the IK. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally i would look into using something IKFast for closed form solutions for your chain (but that depends on the chain really). Although seems to me based on your design and control unit description that you probably dont have much motion planning in mind. You are more planning to use more pose to pose quick moves which may not work too well. &lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2017-12-22T05:07:10.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18542" PostTypeId="1" AcceptedAnswerId="18544" CreationDate="2017-12-22T08:52:01.853" Score="-1" ViewCount="53" Body="&lt;p&gt;The core challenge of building a tank is deciding where to focus its armor and how much speed / cost you're willing to trade for protection, so this got me thinking, with modern detection technology, like laser, could we potentially place a single very thick plate of armor on an axle and once a projectile is detected, rotate it (around the outside of the tank) in front of the projectile in time to stop it (assuming it isn't a close quarter engagement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, tanks could stay much lighter and much cheaper and still have all the benefits of extremely tough armor, and more importantly, have that same protection on all sides of the tank rather than just the front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is: Is that physically possible? Is it physically possible with known modern technology to detect an incoming projectile, then rotate a wheel with enough torque and speed to intercept the projectile's path before it passes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factors I know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fastest projectile should be traveling at 2000 m/s. &lt;/li&gt;&#xA;&lt;li&gt;Radius to be traveled around the sides of an average tank is 15 meters including the gun barrel.&lt;/li&gt;&#xA;&lt;li&gt;A 0.5 x 0.5 meter steel plate with a thickness of 120 mm will weigh 246 kg.&lt;/li&gt;&#xA;&lt;li&gt;Processing / detection time isn't relevant. The time between scanning the incoming projectile (twice, to calculate angle and speed) and sending the correct information to the armor acceleration mechanism is easily less than 1ms, during which the projectile has only traveled 2 meters since detection.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I don't know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Assuming we use the best viable method known, how fast can that 246kg plate be accelerated to the correct location? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Would it be better if the plate was kept in constant motion and simply adjusted once the projectile is detected so it doesn't need to be accelerated from a zero energy state.? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other factors to consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The armor doesn't &lt;em&gt;have&lt;/em&gt; to be stopped right away. It can be accelerated to exactly the correct speed to be in front of the projectile at the calculated time and place of collision, then freely continue swinging around the tank and slow down gradually. &lt;/li&gt;&#xA;&lt;li&gt;There might need to be 2 equally sized plates on opposite ends of the tank to counter each other so the tank can remain stable - so potentially twice the weight. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13617" LastEditorUserId="13617" LastEditDate="2017-12-22T13:12:57.970" LastActivityDate="2017-12-22T13:16:46.787" Title="Would it be physically possible for rotating tank armor to move in front of an incoming projectile fast enough to intercept?" Tags="|sensors|" AnswerCount="1" CommentCount="6" ClosedDate="2017-12-22T14:01:25.810" ContentLicense="CC BY-SA 3.0" />
  <row Id="18543" PostTypeId="2" ParentId="18510" CreationDate="2017-12-22T10:34:28.657" Score="0" Body="&lt;p&gt;Thanks to &lt;a href=&quot;https://engineering.stackexchange.com/questions/18510/deformation-of-two-perpendicular-beams-with-parabolic-properties/18543#comment34150_18510&quot;&gt;Biswajit&lt;/a&gt; for his comment, which pointed out my mistake. My coordinates where wrong. The correct equation for $U_2$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U_2 = \int_0^a\frac{F^2(x-a)^2}{2E\frac{1}{12}b(h\sqrt{1-x/L_2})^3}  dx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the correct equation for the deformation is:&#xA;$$ \delta= \frac{F}{E}\left(\frac{L_1}{bh} + \frac{a^2L_1}{\frac{1}{12}bh^3} + \frac{8L_2}{bh^3}\left(-3a^2+12aL_2+8L_2 \sqrt{1-\frac{a}{L_2}}(L_2-a)-8L_2^2\right)\right) $$&lt;/p&gt;&#xA;" OwnerUserId="14151" LastEditorUserId="14151" LastEditDate="2018-01-02T13:24:51.143" LastActivityDate="2018-01-02T13:24:51.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18544" PostTypeId="2" ParentId="18542" CreationDate="2017-12-22T13:06:18.737" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming we use the best viable method known, how fast can that 246kg&#xA;  plate be accelerated?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not just a plate to accelerate, it's a &lt;a href=&quot;https://en.wikipedia.org/wiki/Moment_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;moment of inertia&lt;/a&gt; to fight and it depends on the position of the plate regarding the center of rotation, its geometry, thus depends on the design of the tank, then it depends on the acceleration curve of the motor and the initial position of the plate, etc.…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is no engineer answer to your question, too many guesses and unsupported assumptions need to be made on too many variables without access to (classified) accurate data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;&lt;strong&gt;How far a laser can detect an incoming projectile accurately&lt;/strong&gt;&quot; : &quot;accurately&quot; needs to be defined, what is your maximum acceptable error of position ? Then, is it by clear weather or in the dust/storm/etc. ?&lt;/li&gt;&#xA;&lt;li&gt;&quot;&lt;strong&gt;What's the fastest scan rate that can be achieved by the laser detection system? We need to read 2 locations to determine the projectile's attack vector.&lt;/strong&gt;&quot; : you certainly need to use satellites here, since the tank might be blinded by mountains etc. Thus, you need to take into account the latency of the transmission, the risk of data loss, etc. And you need at least 3 readings to achieve proper accuracy in a short amount of time. After that, you need to achieve a third-order accuracy reading on the missile acceleration in order to get a first-order accuracy on the position of the impact (since it's  a double integral). As you see, it's not just sensing and measuring, it's &lt;a href=&quot;https://en.wikipedia.org/wiki/Numerical_methods_for_ordinary_differential_equations&quot; rel=&quot;nofollow noreferrer&quot;&gt;differential equations&lt;/a&gt; to solve in real-time, thus the numeric overhead is significant.&lt;/li&gt;&#xA;&lt;li&gt;&quot;&lt;strong&gt;The armor doesn't have to be stopped right away. It can be accelerated to exactly the correct speed to be in front of the projectile at the calculated time and place of collision, then freely continue swinging around the tank and slow down gradually.&lt;/strong&gt;&quot; : actually, using a moving armor is quite clever and might make good use of the conservation of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_momentum&quot; rel=&quot;nofollow noreferrer&quot;&gt;angular momentum&lt;/a&gt;. However, the very aspect of the automation control is a challenge here. Assuming you have the perfect motor (actuator) to do so, ensuring a proper control to achieve precision on position, speed and acceleration based on an acceleration and position input needs at least a 3rd-order accurate controler with a very fast numeric solver. Since your oversized motor won't probably have a linear response, you will have to correct that as well. So it's a lot of math to process with a lot of measures to gather and a lot of errors to handle in a short amount of time, and by the time you get the parameters to input in your motor controler, you might just be dead already.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Yes, physically, this is just basic newtonian mechanics and tennis playing with missiles, and it is possible in theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is no definitive answer to the feasibility aspect until someone does it, and this is more sci-fi than engineering.&lt;/p&gt;&#xA;" OwnerUserId="11617" LastEditorUserId="11617" LastEditDate="2017-12-22T13:16:46.787" LastActivityDate="2017-12-22T13:16:46.787" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18545" PostTypeId="1" CreationDate="2017-12-22T13:13:24.910" Score="6" ViewCount="796" Body="&lt;p&gt;I'm trying to design a kind of water valve with inexpensive materials as a first prototype. &#xA;The water flow from the PVC pipe (1) reach the body of the valve and pass through an aluminum grid (3) to the water tank. When the water level goes up pushes the float closing the water intake at point (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CLSax.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CLSax.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the buoyancy force needed to stop the water flow? And, what will be the mass of the float?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s back to the basics; here I present the problem and some math that I been doing, I would like your opinion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Connected to the PVC pipe (2) I have a garden hose with a water flow pressure of, let's say... 49&amp;nbsp;kPa (I need to check this with a manometer), and I attached a 25&amp;nbsp;mm diameter and 0.5&amp;nbsp;m long PVC pipe. Let’s pretend that the float seals the other side of the PVC pipe, so I need to calculate the force generate the water flow pressure against the float.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please take in consideration that I'm not a fluid mechanic expert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I open the garden hose, the PVC pipe starts to fill, so based on this situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zo9Fc.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zo9Fc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_1+\dfrac{\rho gh_1+v^2\rho}{2}=P_2+\rho gh_2+\dfrac{v^2\rho}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I took the height of $P_1$ as the reference, $h=0$, and the diameter of the PVC pipe and the garden hose pipe are the same (25&amp;nbsp;mm), the water flow velocity at those points are equals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_1=P_2+\rho gh_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the garden hose pressure is 49&amp;nbsp;kPa:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;49000\text{ kg}/\text{ms}^2=P_2+9.8\text{ m}/\text{s}^2\cdot 1000\text{ kg/m}^3\cdot 0.5\text{ m} \\&#xA;P_2=53900\text{ kg/ms}^2 \\&#xA;P_2=53.9\text{ kPa}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, assuming this math is correct…now I have to calculate the force against the bottom of the PVC pipe at point 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P=F/A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to simplify this example, I took the diameter of the PVC pipe as the contact area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}A=\pi r^2=\pi\cdot0.025^2=0.002\text{ m^2} \\&#xA;F=107.8\text{ N}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pressure of the water flow generates a force of 107.8&amp;nbsp;N, I need an opposite force with a higher value to counteract it. Is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to find a material (mass; area) that generate enough buoyancy force to stop the water flow through the valve and seal the water intake, and when the water level goes down, the float valve will let pass the water flow to continue to fill the water tank.&lt;/p&gt;&#xA;" OwnerUserId="14265" LastEditorUserId="1832" LastEditDate="2017-12-22T14:10:13.083" LastActivityDate="2023-02-20T06:07:35.913" Title="Buoyancy force problem" Tags="|mechanical-engineering|fluid-mechanics|flow-control|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18548" PostTypeId="2" ParentId="14652" CreationDate="2017-12-22T17:19:20.550" Score="4" Body="&lt;p&gt;It doesn’t really make much sense to talk about the &lt;em&gt;torque&lt;/em&gt; of a rack for power transmission, since racks do not rotate but instead translate. Instead, it makes more sense to talk about the &lt;em&gt;force&lt;/em&gt; of the rack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Also, I’d be careful about defining a gear ratio between a rack and pinion, since the ratio involves two different kinds of “speed”: translational and rotational. Unlike gear ratios between typical gears, the gear ratio as you define it between a rack and pinion ($\frac{v_R}{\omega_P}$) is not a non-dimensional number: it instead has units of length (metres, inches). Note how doubling the size of each of the rack and pinion also doubles this gear ratio, whereas such a change of gear ratio doesn’t occur for regular gears. By kinematical arguments, the gear ratio is equal to pitch radius of the pinion: $\frac{v_R}{\omega_P}=r_P$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, a power balance argument can be used to show how the mechanical advantage depends on the gear ratio. Let’s consider a pinion driving a rack: (excuse the rough drawing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cFYNQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cFYNQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming 100% efficiency, the power into the pinion is equal to the power out of the rack. That is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_P \omega_P = F_R v_R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $T_P$ is the pinion torque, $\omega_P$ the pinion angular speed, $F_R$ the rack force and $v_R$ the rack translational speed. Note how rearranging provides a relationship between mechanical advantage and gear ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{v_R}{\omega_P} = \frac{T_P}{F_R} = r_P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note how it is possible to derive a corresponding expression for cases with efficiency below 100% by modifying the power balance argument: $T_P \omega_P\eta = F_R v_R$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how this all fits in with a drive train connected to the pinion, let’s suppose the drivetrain starts with gear $i$ (which has torque $T_i$ and angular speed $\omega_i$) and ends with gear $o$ ($T_o$, $\omega_o$). Gear $o$ is connected via gearshaft to the pinion (of the rack and pinion set) ($T_o=T_P$, $\omega_o=\omega_P$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the drivetrain has gear ratio of $G=\frac{\omega_P}{\omega_i}=\frac{T_i}{T_P}$. Notice that if we multiply the gear ratios of the drivetrain and rack-pinion set together we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$r_P G = \frac{v_R}{\omega_i} = \frac{T_i}{F_R}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, rearranging, the force on the rack is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_R = T_i r_P G$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with gear $i$ as the input to the overall system, and the rack as the output, you can see how the gear ratio $G$ of the geartrain does indeed contribute to the mechanical advantage of the pinion-rack gearset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second part of your question, I’m afraid I can’t help much without seeing any sort of diagram of the configuration of the gearing.&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2017-12-22T17:36:55.050" LastActivityDate="2017-12-22T17:36:55.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18549" PostTypeId="1" CreationDate="2017-12-22T18:00:58.843" Score="6" ViewCount="2794" Body="&lt;p&gt;You're probably thinking, &quot;Well..  Why not? It's gotta be &lt;em&gt;something..&lt;/em&gt;&quot; Right?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rvtrp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rvtrp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But 55° is kind of an odd angle if you think about it.. and it does seem like a pretty specific choice  considering a perfect equilateral/equiangular triangle consists of 60° angles. Wouldn't 60° be the default/simple/obvious choice for a triangular profile? Wouldn't it have been easier to manufacture? It seems like 60° would at least be the most logical starting point (like starting at zero) during the design phase. Or any other common, suitable, trigonometric angle; i.e. 30°, 45°, even 90°.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L544M.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L544M.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; Also, &lt;a href=&quot;https://en.wikipedia.org/wiki/Joseph_Whitworth#Accuracy_and_standardisation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Whitworth&lt;/a&gt; strikes me as a pretty meticulous, character; detail oriented, with a careful consideration for all aspects of his designs. Plus, when it comes to mechanical engineering, he has a history of favouring experimental &amp;amp; &lt;a href=&quot;https://en.wikipedia.org/wiki/Polygonal_rifling&quot; rel=&quot;nofollow noreferrer&quot;&gt;unorthodox geometry.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2829" LastEditorUserId="2829" LastEditDate="2017-12-23T09:17:52.073" LastActivityDate="2022-01-17T16:42:52.177" Title="Why are Whitworth threads angled at 55°?" Tags="|mechanical-engineering|design|geometry|threads|fasteners|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18550" PostTypeId="1" CreationDate="2017-12-22T20:35:42.213" Score="1" ViewCount="883" Body="&lt;p&gt;I am trying to figure out what stepper motor is best for my Arduino project.  I want to lift 5kg in steps, doesn't have to be fast, its more important to lift the weight.  I have done some calculations and come to the conclusion that I need a 3NM motor (rounding up for additional margin etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5kg x 9.8ms2 = 49N&#xA;49N x 0.05 = 2.45Nm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5kg x 0.1ms2 = 0.5N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;49.5N = upwards force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;49.5N x 0.05m = 2.475Nm&#xA;49.5N x 0.5m/s = 24.75 Nm/s = 24.75W (0.1m/s would also work bringing down the power requirement)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've concluded that I can use &lt;a href=&quot;https://www.amazon.co.uk/Nema-23-Stepper-Step-Motor-23HS45-4204S/dp/B075MDT6VJ/ref=sr_1_6?ie=UTF8&amp;amp;qid=1513962528&amp;amp;sr=8-6&amp;amp;keywords=nema%2023&quot; rel=&quot;nofollow noreferrer&quot;&gt;this motor&lt;/a&gt; or &lt;a href=&quot;https://www.amazon.co.uk/STEPPERONLINE-Stepper-4-wires-Plasma-Router/dp/B06W2KJ52G/ref=sr_1_2?ie=UTF8&amp;amp;qid=1513963229&amp;amp;sr=8-2&amp;amp;keywords=nema%2023&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;.  Looks like I need a driver too so this &lt;a href=&quot;https://www.ebay.com/itm/Nema23-3Nm-1-8-degree-Stepper-Motor-2ph-4-wire-4-2A-L112mm-with-MB450A-Driver/112664212024?hash=item1a3b4f7238:g:7rAAAOSwyGZaJgiS&quot; rel=&quot;nofollow noreferrer&quot;&gt;motor and driver&lt;/a&gt; looks like it would do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are my conclusions correct?  If so is it simply a case of connecting the motor to the driver and the driver to the Arduino using a motor shield such as &lt;a href=&quot;https://www.amazon.co.uk/dp/B00F9JFE26/ref=pe_385721_37038051_TE_3p_dp_1&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; (for multiple motors).  Looks like I can't connect the motor to the shield directly because of power requirements.  I assume then when its all connected I can use &lt;a href=&quot;http://www.hobbytronics.co.uk/tutorials-code/arduino-tutorials/stepper-motor-sn754410&quot; rel=&quot;nofollow noreferrer&quot;&gt;this code&lt;/a&gt; to run it.  I think I used the stepper motor in that code example before (almost four years ago so I can't quite remember).  Also how exactly is the power supplied?  To connect two motors I'm thinking it would need 4 PSUs, one for the Arduino, one for the motor shield and one for each of the motor drivers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice to help me better understand this would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="14269" LastActivityDate="2017-12-22T20:35:42.213" Title="selecting correct stepper motor to lift 5kg for Arduino project - newbie questions" Tags="|electrical-engineering|stepper-motor|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18551" PostTypeId="1" AcceptedAnswerId="18552" CreationDate="2017-12-22T23:09:52.120" Score="3" ViewCount="151" Body="&lt;p&gt;The part is a clevis and was insterted in a steering rack . Part was in Tension, torsion and bending. It failed at the end of the threaded bolt part. Is it a common problem for bolted joints ? How can it be resolved ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first a bend was observed in the threaded part in the direction in which it is kept in the picture before failure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XduOE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XduOE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/10OcP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/10OcP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/mbb8E.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mbb8E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8UPOs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8UPOs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14271" LastEditorUserId="14271" LastEditDate="2017-12-23T15:39:18.043" LastActivityDate="2017-12-23T16:18:18.993" Title="What type of failure is this?" Tags="|materials|failure-analysis|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18552" PostTypeId="2" ParentId="18551" CreationDate="2017-12-23T01:32:55.820" Score="2" Body="&lt;p&gt;this is a classic fatigue fracture, caused by a cyclic/reversed bending load superimposed on a static tension load. The cyclic load is caused by the reversed bending; the neutral axis shows clearly as a straight line running diagonally across the center of the part. the fact that it is a straight line demonstrates that the torsion component was not the dominant load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the initiator of the fracture was the sharp tip of the cut thread in the part and the reversed cyclic loads caused the fatigue fracture crack to develop on both sides of the bending neutral axis. Finally, note that the cross-section of the final failure caused by overload (the area defined by the shear failure zone along the diagonal line as defined above) is small compared to the unfractured cross-section of the part; this demonstrates that the loads originally imposed on the part were small compared to the tensile strength of the part itself as designed. The conclusion therefore is that this part was not designed for fatigue service; it should have had the threads pressed into it with a set of roller die rather than having them cut on a lathe or with a threading die. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-12-23T01:32:55.820" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18553" PostTypeId="2" ParentId="18551" CreationDate="2017-12-23T16:18:18.993" Score="0" Body="&lt;p&gt;I agree , reversed bending fatigue, no torsion, likely sharp thread roots, maybe a loose nut. But the surface looks granular, not smooth as typical of fatigue. That suggests very low strength steel. So, radius the thread roots and be sure the nut is properly torqued and check if the hardness meets spec.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-12-23T16:18:18.993" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18556" PostTypeId="1" AcceptedAnswerId="32801" CreationDate="2017-12-23T21:44:17.103" Score="2" ViewCount="885" Body="&lt;p&gt;This is a clevis made for linking the tie-rods to the steering rack. This part is under dynamic loading and is getting loosened. We tried using jam nuts but it did not work well. Due to loosening the thread came out of the rack and the part fractured maybe due to bending fatigue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NXnQs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NXnQs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iZ8Sy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iZ8Sy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cdFGY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cdFGY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14271" LastActivityDate="2020-01-24T01:00:23.877" Title="How to prevent screw from loosening?" Tags="|mechanical-engineering|mechanical-failure|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18558" PostTypeId="1" CreationDate="2017-12-24T12:09:58.297" Score="1" ViewCount="168" Body="&lt;p&gt;Beams 1 to 7 carry the load of the roof (for example 200N/m2)&#xA;The maximum bend of these beams is 5mm&#xA;How do I calculate the forces working on these beams?&lt;img src=&quot;https://i.stack.imgur.com/nhMXL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14283" LastActivityDate="2017-12-28T00:11:09.660" Title="Calculation of beams in flat roof structure" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="1" CommentCount="4" ClosedDate="2018-01-15T13:16:55.853" ContentLicense="CC BY-SA 3.0" />
  <row Id="18561" PostTypeId="2" ParentId="18315" CreationDate="2017-12-24T18:02:13.020" Score="0" Body="&lt;p&gt;The angle will also depend on your objective. I use my green house to winter over various plants so my angle is set to maximize winter light , normal to the sun in mid November/January. If you are growing bedding plants from seed ,you probably want maximum sun in March / April ( northern hemisphere). I found trying to make a variable roof angle was too complicated for a home greenhouse. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-12-24T18:02:13.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18562" PostTypeId="2" ParentId="17701" CreationDate="2017-12-24T18:10:45.987" Score="2" Body="&lt;p&gt;What you are asking for is called a &lt;em&gt;spectrometer&lt;/em&gt;.  There are two main ways spectrometers work: with a prism or a diffraction grating.  Either way, different wavelenght of light end up getting spread out spacially.  You then detect the amplitude at various points along this spread-out spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another crude way is with multiple filters, then detecting the light amplitude out of each filter.  That's usually only good if you are looking for a particular wavelength, not trying to find the spectrum in general.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-12-24T18:10:45.987" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18564" PostTypeId="1" CreationDate="2017-12-25T01:15:27.713" Score="0" ViewCount="176" Body="&lt;p&gt;I'm getting a shipping container delivered to my property that weighs about 5000 pounds. Once it's delivered I will need to re-position it. I'm thinking of trying to use a number of cylindrical logs, together with some mechanical advantage to move it. My car has a towing capacity of 1500 pounds, so I'm contemplating getting a 6:1 block and tackle, effectively allowing my car to pull 10,000 pounds. I'm a little iffy on a few points though. What is the actual force that is required to pull a 5000 pound container over rolling logs on grass ( slope will also factor in) I have no idea what the coefficient of friction is here. Does the rope that feeds the pulley have to be able to withstand the full force or just 1/6 of the force? Assuming I need an anchor point, does the anchor point need to hold the full force of the pull? (any suggestions on how I can create an anchor point in a grass field .... was thinking maybe driving a steal pipe down a few feet at an angle). What else am I missing?&lt;/p&gt;&#xA;" OwnerUserId="539" LastActivityDate="2017-12-25T03:34:15.070" Title="How can i move a 5000 lb object with 1500 pound towing capacity?" Tags="|mechanisms|pulleys|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18565" PostTypeId="2" ParentId="18493" CreationDate="2017-12-25T03:01:57.373" Score="2" Body="&lt;p&gt;This refers to Model and Predictive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Model&lt;/strong&gt;: This control type highly depends on the model. In a PID controller, maybe you do not know the model clearly, but PID works fine. In MPC, wrong model leads to loss of control. Your controller highly depends on the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Predictive&lt;/strong&gt;: This controller does not depend on the history of the states (unlike PID). It only depends on the current states and what it predicts about the future. It estimates (predicts) the future and takes the best control action to optimize the future in a finite horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note&lt;/em&gt;: If you have the system model, current state and series of current and future control inputs, then you can exactly tell what will be the open-loop future outputs.&lt;/p&gt;&#xA;" OwnerUserId="14288" LastEditorUserId="14288" LastEditDate="2018-02-25T20:28:42.610" LastActivityDate="2018-02-25T20:28:42.610" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18566" PostTypeId="2" ParentId="18564" CreationDate="2017-12-25T03:34:15.070" Score="1" Body="&lt;p&gt;Your Cars towing capacity is considering towing on a highway where you have to be able to accelerate to traffic speed and decelerate safely. Considering you are only moving this off road you should be able to safely pull a load as heavy as you have traction for.If you have a tow hitch already on your vehicle the weakest point is likely to be the chain or cable you use. You will need to exercise extreme caution that should that fail there are no people in the danger zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a block and tackle you distribute the load over the number of lines carrying the load. Easy way to look at it is if there are 3 ropes between the pulleys they share 1/3 the load each but if you have a single rope going from the pulley to the load or the pulling vehicle it will of course have the entire load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How far are you needing to move this container? If it is more than a short distance you might want to consider paying someone to move it for you.&lt;/p&gt;&#xA;" OwnerUserId="10516" LastActivityDate="2017-12-25T03:34:15.070" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18567" PostTypeId="1" CreationDate="2017-12-25T09:27:44.833" Score="2" ViewCount="477" Body="&lt;p&gt;I am replacing my dead diesel engine in my sailboat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to replace it with a new diesel and an electric motor in parallel hybrid.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hybrid_vehicle_drivetrain&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hybrid vehicle drivetrain&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This would allow me to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Run the propeller using the diesel engine only.&lt;/li&gt;&#xA;&lt;li&gt;Run the propeller using the electric motor.&lt;/li&gt;&#xA;&lt;li&gt;Generate power by running the diesel and motor as a generator with a disengaged propeller.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In order to to do this I need a clutch so that I can disengage the propeller, please see the attached drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tbX2q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tbX2q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;It would look kinda like this.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Bxgdv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bxgdv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clutch requirements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bi-directional - must be able to go forward and reverse.&lt;/li&gt;&#xA;&lt;li&gt;Ability to engage/disengage remotely, electrically or mechanically.&lt;/li&gt;&#xA;&lt;li&gt;Stay engaged with springs. A power would need to be applied to disengage the clutch. &lt;/li&gt;&#xA;&lt;li&gt;Torque &gt;= 600Nm. &lt;/li&gt;&#xA;&lt;li&gt;Does &lt;strong&gt;not&lt;/strong&gt; need to be able to engage/disengage with differential shaft speeds, although it wouldn't hurt.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;9.5488 x Power (W) / Speed (RPM) = Torque (N.m) =&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;9.5488 x 50 000 (W) / 800 (RPM) = &lt;strong&gt;596 Nm&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I won't get full 50 kW from the Diesel engine at 800 rpm shaft speed but I am using it as a worst case scenario, especially since the electric motor can generate a lot of torque during acceleration and develop much more than 20 kW for short periods of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is, do you have any suggestions regarding what clutch I could use?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have spent days now searching the internet and I am simple out of my league.&#xA;Please help with links to products, ebay or anything else that would help me find a clutch that would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would love to be able to use something that is readily available such as the clutch that is between the engine and the gearbox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate your help, thank you.&lt;/p&gt;&#xA;" OwnerUserId="14291" LastEditorUserId="7083" LastEditDate="2018-03-28T02:16:53.953" LastActivityDate="2018-07-26T13:09:13.170" Title="Find clutch for parallel hybrid sailboat" Tags="|mechanical-engineering|automotive-engineering|power-transmission|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18568" PostTypeId="1" AcceptedAnswerId="18569" CreationDate="2017-12-25T10:29:43.173" Score="-2" ViewCount="64" Body="&lt;p&gt;i am designing a new car and planning to use an existing engine, my field is aerospace engineering so i have a good knowledge in machine design and thermodynamics, at first where can i find the technical specs of an engine other than disassembling and figuring out needed information? secondly, what are the good software that are suitable for individual?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: looks like some people don't understand what is written above, the question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where do i obtain specifications, data, blueprints, whatever of an engine, for example GM's LS3 or Honda K20?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;know one of the users below told me to purchase a workshop manual, i will appreciate any additional info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="14292" LastEditorUserId="14292" LastEditDate="2017-12-26T09:34:42.333" LastActivityDate="2017-12-26T09:34:42.333" Title="how can i improve a current engine by redesigning internals such as pistons , ...etc?" Tags="|design|automotive-engineering|car|" AnswerCount="1" CommentCount="4" ClosedDate="2017-12-26T02:34:09.107" ContentLicense="CC BY-SA 3.0" />
  <row Id="18569" PostTypeId="2" ParentId="18568" CreationDate="2017-12-25T11:04:51.260" Score="0" Body="&lt;p&gt;Design a stroker kit - first pick an engine where there is room to change the distance the piston travels - most engines are now designed with interference ie the piston and the valves occupy the same space but at different times.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-12-25T11:04:51.260" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18570" PostTypeId="2" ParentId="18567" CreationDate="2017-12-25T13:02:11.460" Score="0" Body="&lt;p&gt;Consider a gearbox such that either the engine can be connected or the electric motor and possibly both at the same time, but the main advantage is that while one is connected the other is free.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-12-25T13:02:11.460" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18571" PostTypeId="1" CreationDate="2017-12-25T13:07:58.430" Score="0" ViewCount="272" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vUQfK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vUQfK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, motion of the ball on the beam depends on the beam angle alpha, shouldn't this system have a single degree of freedom?&lt;/p&gt;&#xA;" OwnerUserId="14293" LastActivityDate="2017-12-27T19:59:57.243" Title="How to determine the degrees of freedom for ball and beam system such as shown in pic below" Tags="|dynamics|modeling|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18573" PostTypeId="1" CreationDate="2017-12-25T15:22:35.903" Score="2" ViewCount="96" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RfYQm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RfYQm.png&quot; alt=&quot;heat flow&quot;&gt;&lt;/a&gt;I have a machine that has power output of 2750 watt, which is blasted to a insulator foam with thermal resistance of 2.1 m2 kelvin/watt where the insulator is 5 cm thick and it drives the heat into ventilation and out of the roof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to make sure that the insulator can withstand the heat blast from the machine. So if I know that 2750 watt can output 9383 btu/hr where 1 watt outputs 3.412 btu/hr, how can I be sure that the insulator can withstand the heat without it getting spread out? regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited additional info: exhaust temp is 40 degree Celsius, room temp is 25 degree Celsius, and outside temp is 32 degree Celsius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flammability limit: &#xA;lower limit for hydrocarbon gas (part of the foam) 1.8% by volume&#xA;upper limit for hydrocarbon gas: 8.5 % by volume &#xA;flash point of hydrocarbon gas: -84 degree Celsius &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume of the exhaust = 50x 60 x 200 cm cube = 600,000cm cube&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4GBMu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4GBMu.png&quot; alt=&quot;machine exhaust from different angle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14294" LastEditorUserId="14294" LastEditDate="2018-01-02T12:17:34.093" LastActivityDate="2018-07-31T16:00:48.380" Title="Insulator thermal resistance" Tags="|thermodynamics|thermal-conduction|thermal-insulation|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18574" PostTypeId="1" CreationDate="2017-12-25T16:49:31.367" Score="8" ViewCount="1850" Body="&lt;p&gt;What are the factors limiting the height of a &quot;tower&quot; made of typical fired brick and mortar?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;tower&quot;, I mean a structure whos point is to be tall, and can support a human standing on or near its top.  I'm interested in the self-supporting structural aspects, not limitations of construction techniques, or environmental factors like wind loading, possiblity of earthquakes, etc.  I want to understand what limits such a structure mostly from its own weight and properties of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is to maximize height for any given land area the thing sits on.  I'm trying to leave reasonable latitude for land area to understand the constraints, but let's not get silly like wrapping around the earth or end up with a star made of diamond, as a similar questions mentioned in a comment devolved to.  If a maximum is required to avoid silly answers, let's say about the area of a typical Manhattan city block, with reasonable latitude on shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand this material can handle high compressive loads, practically no tension load, and has a high density as building materials go.  My gut feel (I'm a electrical engineer, not so fluent with structural engineering) is that the optimum (least material for highest result) shape would be a circular crossection, probably with exponentially decreasing diameter with height.  How well does that scale due to ground loading?  What if this is built on bedrock?  Wouldn't that be about the same properties, if not better, than the brick and mortar?  Would that allow scaling some shape aribtrarily, or do the proportions of the shape change with size?&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2017-12-25T17:46:04.487" LastActivityDate="2021-01-14T15:39:34.633" Title="Height limits of brick and mortar tower?" Tags="|structural-engineering|statics|" AnswerCount="5" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18576" PostTypeId="2" ParentId="18573" CreationDate="2017-12-25T17:52:00.847" Score="1" Body="&lt;p&gt;You need to investigate the flammability properties of the insulating materials : wood can be a good insulator if it is dry, but it burns well etc.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-12-25T17:52:00.847" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18578" PostTypeId="2" ParentId="18574" CreationDate="2017-12-25T20:02:19.710" Score="3" Body="&lt;p&gt;since the structure is loaded only by its own weight, the limiter will be the maximum compressive strength of the material, beyond which the bottom floor gets crushed. as pointed out by others, this can be at least partly mitigated by tapering the sidewalls. in any case, knowing the compressive strength, the density, and the dimensions of the tower, its height at the point of compressive failure of the first floor can be calculated. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-12-25T20:02:19.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18579" PostTypeId="2" ParentId="18574" CreationDate="2017-12-25T22:23:33.793" Score="2" Body="&lt;p&gt;There is a brick building in Chicago that is claimed to be the tallest that brick can build ( without crushing the bottom). It is about 200 ft. But it will have the additional weight of floors and other interior components, so I guess it is not quite an answer. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-12-25T22:23:33.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18580" PostTypeId="1" CreationDate="2017-12-26T07:13:14.773" Score="0" ViewCount="149" Body="&lt;p&gt;I'm working on a disc brake balancing machine, as shown below this is a belt pulley drive and for alignment purposes I want motor to be able to move &lt;em&gt;vertically as well as inwards outwards&lt;/em&gt; (to tighten the belt). Is there any clamp/adjuster design to achieve this?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Qk796.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qk796.png&quot; alt=&quot;Disk brake balancing machine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14301" LastActivityDate="2018-01-04T23:16:05.197" Title="How to make a clamp for this motor to move it in two directions?" Tags="|mechanical-engineering|design|applied-mechanics|mechanisms|mechanical|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18583" PostTypeId="1" AcceptedAnswerId="18590" CreationDate="2017-12-26T11:31:58.617" Score="1" ViewCount="427" Body="&lt;p&gt;I have a system that has the following transfer function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$G(s) = \frac{1}{s^2(s^2+4+1)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be seen it is a 4th order system.&#xA;This is the bode plot of the system:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HWMlb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HWMlb.png&quot; alt=&quot;bode plot of plant&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to use a lead compensator in order to generate sufficient phase margin. If I use a tame PDD controller I am limiting my crossover frequency around the position of the green arrow, as the double derivative action can theoretically only provide 180 phase lead. However if I use a tame PDDD action I can place my crossover frequency in the vicinity of the orange arrow. This also means that I can make my system infinitely fast, if there is no limit on control action...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is it realistic to use a triple tame derivative action? I can't find a similar controller anywhere on the internet!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="14305" LastActivityDate="2017-12-27T07:37:23.633" Title="What are the disadvantages of PDDD controller" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18584" PostTypeId="2" ParentId="18256" CreationDate="2017-12-26T12:55:13.653" Score="-1" Body="&lt;p&gt;What is the need in this case of obtaining the Nusselt number? Before the advent of powerful computers able to solve numerical solutions to the problem you describe, these adimensional numbers were used to help in obtaining analytical solutions to such problems. That's why the paper you cite was published more than 45 years ago. Now, these adimensional numbers are used to simplify problems by scaling analysis so you can quickly get an order of magnitude estimate of the solution. In your case, with the problem highly specified (shape of duct, only a portion of it insulated), to avoid being anachronistic, wouldn't it be better to just solve your problem using a package like Comsol?     &lt;/p&gt;&#xA;" OwnerUserId="13388" LastActivityDate="2017-12-26T12:55:13.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18585" PostTypeId="1" AcceptedAnswerId="18587" CreationDate="2017-12-26T15:29:48.383" Score="2" ViewCount="7889" Body="&lt;p&gt;I know little about internal combustion engines and I feel a little confused about brake and shaft powers. I read a book that only talks about shaft power, and other that only talks about brake power, without making a proper definition of the physical meaning of these quantities. I'm now thinking that they both mean the same thing: &lt;strong&gt;useful power on the crankshaft (the power produced by the engine cylinders less the losses and the power that is used in accessory components)&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in &lt;em&gt;Mechanics and thermodynamics of propulsion (1992) (Hill Peterson)&lt;/em&gt;, pag. 155, the the brake specific fuel consumption uses shaft power in its definition, and not the brake power (as the name suggests it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate some clarification, or some book suggestions, since the ones that I read are too vague.  &lt;/p&gt;&#xA;" OwnerUserId="9747" LastEditorUserId="9747" LastEditDate="2017-12-26T19:28:49.020" LastActivityDate="2020-02-08T05:58:27.770" Title="What is the difference between shaft power and brake power?" Tags="|mechanical-engineering|aerospace-engineering|power|propulsion|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18586" PostTypeId="2" ParentId="18585" CreationDate="2017-12-26T19:26:59.617" Score="3" Body="&lt;p&gt;You mean &lt;strong&gt;BRAKE&lt;/strong&gt; horsepower. It is the same as shaft horsepower. It is called brake horsepower (abbreviated BHP) because it is measured at the output shaft of the engine while it is running on a test stand with a mechanical device called a brake. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other definitions of the power output of an engine which rely not on measurements but on calculations based on things like piston area, which is why shaft or BHP are named this specific way. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="90" LastEditDate="2017-12-27T00:33:16.213" LastActivityDate="2017-12-27T00:33:16.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18587" PostTypeId="2" ParentId="18585" CreationDate="2017-12-26T23:03:27.200" Score="6" Body="&lt;p&gt;The short answer is these terms are defined by standards organizations based on specific test methods. These methods specify the engine condition, environmental factors, fuel chemistry, inlet and exhaust systems configurations, and the correction factors applied to the measurements. The specific tests and standards used vary by country and by industry. So neither term has a specific meaning unless you call out the document that defines the terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, though, shaft horsepower measures power after the transmission, brake horsepower measures before the transmission. On a cruise ship, you might have many gensets feeding a couple of high voltage buses that power the prop shaft motors. You want to talk about shaft power here, not brake power. Similarly, turboshaft jet engines have builtin gearsets to drive, say, a helicopter rotor. Shaft power is the relevant term here as well. Shaft power usually makes more sense when comparing different types of power plants that use different transmission arrangements.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2017-12-26T23:08:54.470" LastActivityDate="2017-12-26T23:08:54.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18588" PostTypeId="2" ParentId="18580" CreationDate="2017-12-27T00:44:08.247" Score="1" Body="&lt;p&gt;Use a dovetail guide for the vertical movement and place the motor on it using elevating screws for the horizontal movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Miizu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Miizu.jpg&quot; alt=&quot;Dovetail guide&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you place the bottom plate of the contraption above (picture &lt;a href=&quot;https://www.norelem.de/xs_db/BILD_DB/2/www/750/21030.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;) on the box and use the threaded holes on the top plate to mount the motor, you can move the motor up or down. Use screws in the holes at the side of the upper plate to adjust friction and lock both in one position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kgyvv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kgyvv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, use something like the screws above (picture &lt;a href=&quot;https://img0.fastenal.com/productimages/3430206.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;) to fix the motor to the upper plate. &lt;/p&gt;&#xA;" OwnerUserId="14306" LastActivityDate="2017-12-27T00:44:08.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18589" PostTypeId="2" ParentId="17772" CreationDate="2017-12-27T07:07:26.393" Score="0" Body="&lt;p&gt;No, the energy coming out of the turbine does not have the same energy as when it goes in. The compressed air does work on the turbine by decompressing: this means that the pressure downstream from the turbine is less than the pressure upstream from it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason you don't get more energy when you put them in series, is that you've still only got the same energy being input - the energy in the compressed air. You can't get more energy out than you put in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism that prevents the energy out being proportional to the number of turbines in series, is as follows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've got just one turbine, then the upstream pressure is the pressure from the compressor or compressed storage, and the downstream pressure is about equal to ambient air pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you've got two turbines in series, then the upstream pressure on the first turbine is still equal to the compressor pressure; and the downstream pressure from the last turbine is still about equal to ambient air pressure. And that means that the air pressure between the two turbines is somewhere between the two, which means that the air has done less work on the first turbine, and less work on the second turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all, the two turbines combined had the potential to do no more work than one turbine on its own, because the pressure gradient from start to end is the same: except now, that work is spread across two turbines instead of one, so you've got twice the losses, and hence in reality less work is done by two turbines than one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this applies because the two turbines have same kind of work applied to them by the same kind of energy. Combined-cycle gas turbines get more work out of two turbines in series than one, by burning fuel to create an expansionary force that drives the first turbines; the heat in that exhaust air is then used to heat water, creating steam, which then drives a second turbine. This means that the second turbine does not have to compromise the pressure downstream from the first turbine.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2017-12-27T07:07:26.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18590" PostTypeId="2" ParentId="18583" CreationDate="2017-12-27T07:37:23.633" Score="1" Body="&lt;p&gt;In the synthetic example you provide there is no real disadvantage to the triple derivative. In a practical application the story is different. Generally, the triple derivative will amplify (high-frequent) measurement noise and additional low-pass filtering is needed to guarantee performance. The phase lag associated with the low-pass filter will limit the achievable bandwidth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is generally (in any controller) a good idea to include a low-pass filter as this will limit control action at higher frequencies. This is desirable as generally measurement noise will increase at higher frequencies (which limits the effectiveness of the feedback loop) and it is generally necessary to limit forces generated by the actuators (e.g. to limit heat dissipation, to limit the maximum generated force or because the available actuator bandwidth is limited).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatives to including a triple derivative with additional low-pass filters in the controller include lead-lag filters or band-limited differentiators.&lt;/p&gt;&#xA;" OwnerUserId="11869" LastActivityDate="2017-12-27T07:37:23.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18591" PostTypeId="1" AcceptedAnswerId="18595" CreationDate="2017-12-27T08:05:18.800" Score="-1" ViewCount="4865" Body="&lt;p&gt;&lt;h1&gt;Background&lt;/h1&gt;&lt;br&gt;&#xA;I am currently preparing for Entrance Examination for Engineering  college , somehow I noticed that I don't even know which branch I want to take and what is the kind of Education I am opting for .I always wanted to make cool stuffs not like Robots in particular but like something worth usage like that, but I just know there are few branches for but which is the one which can lead me into getting the required knowledge needed to make devices maybe like TV or a phone things dealing with semiconductors , Transistors ?&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;h1&gt;Main&lt;/h1&gt;&lt;br&gt;&#xA;So what is the branch that can enable me to build gadgets like modern day use somehow maybe I want to relate with the creation of gadgets that can acts as an interface b/w user and the digital world &#xA;Thats all what I want to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;h2&gt;Note: &lt;/h2&gt;I couldn't find the required tag for the need so maybe it wont be good enough to rely on that.&lt;/p&gt;&#xA;" OwnerUserId="14313" LastEditorUserId="14313" LastEditDate="2017-12-28T08:16:46.297" LastActivityDate="2020-02-04T18:05:20.177" Title="Which is the branch of engineering that deals with making gadgets?" Tags="|building-design|circuits|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18592" PostTypeId="2" ParentId="18591" CreationDate="2017-12-27T08:30:31.393" Score="1" Body="&lt;p&gt;I think you would be interested in &lt;a href=&quot;https://en.wikipedia.org/wiki/Electrical_engineering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electrical Engineering&lt;/a&gt;, more specifically in &lt;a href=&quot;https://en.wikipedia.org/wiki/Electronic_engineering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electronics Engineering&lt;/a&gt;.&lt;br&gt;&#xA;Do browse through these links.&lt;/p&gt;&#xA;" OwnerUserId="3345" LastActivityDate="2017-12-27T08:30:31.393" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18593" PostTypeId="2" ParentId="18591" CreationDate="2017-12-27T09:41:18.703" Score="0" Body="&lt;p&gt;My degree in engineering crossed various fields as it was called Combined Engineering and you had a modular or “pick and mix” style so you may want to look at the degrees on offer carefully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our department had courses from digital signal processing, thermofluids, power plant analysis, stress and structures and many in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact I don’t remember anyone doing an identical set of subjects amongst my colleagues.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2017-12-27T16:47:36.360" LastActivityDate="2017-12-27T16:47:36.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18595" PostTypeId="2" ParentId="18591" CreationDate="2017-12-27T15:26:40.550" Score="0" Body="&lt;p&gt;This question is leading. It leads people to see that you want to be a some sort of an electrical engineer. However, none of the answers actually answers you not knowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making gadgets is not a engineering branch. Its a interdiciplinary branch combining many different fields. At the end of the day wether your a industrial designer or an engineer matters little. You get a suitable foundation studying mostly any of the engineering branches. Be it EE, Mechatronics, electrical engineering, mechanical engineering etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What matters after this is doing the work. If you want to make innovative designs then by neccesity you will need to work with what you want to do on your own time, and sadly atleast partially with your own money.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2017-12-27T15:26:40.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18596" PostTypeId="2" ParentId="17775" CreationDate="2017-12-27T16:58:24.053" Score="1" Body="&lt;p&gt;As noted &quot;no valve is perfect&quot; so any trapped fluid will leak someplace eventually. Common practice is to use &quot;block and bleed&quot; valves with double block valves. Wherein a valve like 12 inch nominal size will incorporate a 1 inch valve that will bleed off to an appropriate location. This would apply to process valves. For well head valves, likely the best sealing valves, block and bleed is not normally used as it is difficult enough to seal ,eg,  10,000 psi without bleed valves as a possible leak source. Apparently it is assumed that anyone operating wellhead valves knows what they are doing.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2017-12-27T16:58:24.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18597" PostTypeId="2" ParentId="18571" CreationDate="2017-12-27T19:49:46.983" Score="1" Body="&lt;p&gt;Try to think of degrees of freedom as objects that can move independently of each other. They may be coupled like in your system such as alpha and theta so you can describe one of them with the other... however the ball is not. I would suggest &quot;the art of modeling in science and engineering with mathematica&quot; &#xA;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/1584884606&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/Modeling-Science-Engineering-Mathematica-Second/dp/1584884606&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you'd like a good read.&lt;/p&gt;&#xA;" OwnerUserId="14318" LastEditorUserId="9" LastEditDate="2017-12-27T19:59:57.243" LastActivityDate="2017-12-27T19:59:57.243" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18598" PostTypeId="2" ParentId="18558" CreationDate="2017-12-28T00:11:09.660" Score="0" Body="&lt;p&gt;The best way to do this is to model it in a program like STAAD or Tekla. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is most important is to understand your load - earthquake, wind, snow, self-load from the roof.&#xA;You also need to define where the columns are in the above diagram after you define that you can start to solve the problem. &lt;/p&gt;&#xA;" OwnerUserId="14319" LastActivityDate="2017-12-28T00:11:09.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18599" PostTypeId="1" CreationDate="2017-12-28T00:13:00.717" Score="1" ViewCount="159" Body="&lt;p&gt;There are many complex mathematical shapes like the cycloid, cissoid, umbilic torus (&lt;a href=&quot;http://fab.cba.mit.edu/classes/863.15/section.CBA/people/VanWyk/images/UmbilicTorus.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;picture&lt;/a&gt;) or the shape of the &lt;a href=&quot;https://www.archdaily.com/774683/ad-classics-azadi-tower-hossein-amanat&quot; rel=&quot;nofollow noreferrer&quot;&gt;Azadi tower&lt;/a&gt;. Sure one can design complex objects like this in CAD software today but how does one construct such objects accurately in the real world. What techniques do builders use to take an architect's zany design and accurately construct the prescribed shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know modern techniques to acccomplish this and also techniques used by &lt;em&gt;pre&lt;/em&gt; 20th century engineers. &lt;/p&gt;&#xA;" OwnerDisplayName="smilingbuddha" LastEditorUserId="7083" LastEditDate="2018-01-28T22:18:27.673" LastActivityDate="2018-01-28T22:18:27.673" Title="Making complicated curves / surfaces in the real world accurately" Tags="|cad|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18600" PostTypeId="1" CreationDate="2017-12-28T06:44:52.590" Score="1" ViewCount="2419" Body="&lt;p&gt;I was not sure on which site to ask this question, but I guess Engineering might do well from the aspect of matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering, what is the shelf life of a HEPA filter, can I just buy a few of them for my air cleaner and let them sit in my drawer?&lt;/p&gt;&#xA;" OwnerUserId="14322" LastActivityDate="2018-02-26T16:34:57.023" Title="What is the shelf life of HEPA filters?" Tags="|air-quality|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18601" PostTypeId="1" CreationDate="2017-12-28T08:07:40.523" Score="1" ViewCount="62" Body="&lt;p&gt;Is it acceptable for fire-rated gypsum board to be stored outdoor (covered parking area) in a humid city?&#xA;Is it acceptable to be damaged at edges and sides?&#xA;Does it need spacers when stored flat on each others?&lt;img src=&quot;https://i.stack.imgur.com/t3t67.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14323" LastActivityDate="2018-02-03T15:37:29.833" Title="Gypsum board damages(fire rated)" Tags="|civil-engineering|construction-management|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18603" PostTypeId="2" ParentId="10486" CreationDate="2017-12-28T10:51:27.097" Score="-1" Body="&lt;p&gt;It's a nonstationary non-linear system. It can be easily linearized with a large adequate answer around it's working point.&lt;/p&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2017-12-28T10:51:27.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18606" PostTypeId="2" ParentId="18599" CreationDate="2017-12-28T13:41:26.847" Score="2" Body="&lt;p&gt;Most CAD software will allow you to produce arbitrary curves using control points eg &lt;a href=&quot;https://en.wikipedia.org/wiki/Non-uniform_rational_B-spline&quot; rel=&quot;nofollow noreferrer&quot;&gt;NURBS&lt;/a&gt; and indeed this is analogous to the more traditional method of using a flexible strip of wood or steel held in place with pegs under tension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important point is to consider the manufacturing process and how you translate an arbitrary curve on a drawing or model into a real part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest type of curve to deal with is one made up of circular arcs and conic sections. These can be rolled fairy easily on &lt;a href=&quot;https://www.baileighindustrial.co.uk/hydraulic-roll-bender-r-h150-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;roll benders&lt;/a&gt;, slip rollers or mandrel benders (for tight radius curves in hollow sections). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more complex compound curves modern multi-axis CNC processes haev made a huge difference, indeed you can see this is the progression of mass produced car bodies since the 1960s. Having said that the traditional process of hand sculpting patterns in clay is still  not entirely obsolete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also a lot of artisan techiques for making complex forms in sheet metal. Hand built car bodies are generally constructed over a plywood '&lt;a href=&quot;https://scontent-lht6-1.xx.fbcdn.net/v/t1.0-9/26166392_1547191905328533_7469270906038984097_n.jpg?oh=8938cebeb0dc1ef802972e26fafdb23f&amp;amp;oe=5AB316CB&quot; rel=&quot;nofollow noreferrer&quot;&gt;buck&lt;/a&gt;' &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7VEx2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7VEx2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the background of that picture you can also see an '&lt;a href=&quot;https://en.wikipedia.org/wiki/English_wheel&quot; rel=&quot;nofollow noreferrer&quot;&gt;English wheel&lt;/a&gt;' which is essentially a pair of adjustable rollers used to shape sheet metal as well as various specialist hammers and dies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;these sheet metal forming processes require a lot of skill from the operator both in terms of the physical process and having a good 'eye' for interpreting a drawing or model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also interesting to look at Gothic cathedrals which mostly consist of intersecting circular arcs which can be generated using essentially quite simple geometry with dividers and a square (which is still the symbol of freemasonry).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important construction is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Catenary&quot; rel=&quot;nofollow noreferrer&quot;&gt;catenary curve&lt;/a&gt; , often seen in bridges, essentially the shape formed by a chain hanging under its own weight. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In many cases part of the design process is simplifying an arbitrary curve to elements which can be made by reasonably straightforward geometric constructions &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-01-05T19:56:53.000" LastActivityDate="2018-01-05T19:56:53.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18607" PostTypeId="2" ParentId="18591" CreationDate="2017-12-28T14:05:10.467" Score="1" Body="&lt;p&gt;Consumer electronics are generally multi-disciplinary projects encompassing product design, UI design, electronic engineering and software engineering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design of automated machines with mechanical actuators and electronic control (robots if  you like) tends to be classified as &lt;em&gt;mechatronics&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it really depends which particular aspect you are interested in. Mechanical Engineering will give you a fairly broad base but may also include elements of thermodynamics and fluid dynamics which are less relevant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Broadly speaking it sounds like you should be looking at courses which cover &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Engineering/mechanical design (especially CAD) &lt;/li&gt;&#xA;&lt;li&gt;Product design &lt;/li&gt;&#xA;&lt;li&gt;manufacturing &lt;/li&gt;&#xA;&lt;li&gt;Electrical and electronic engineering &lt;/li&gt;&#xA;&lt;li&gt;Software and computer science &lt;/li&gt;&#xA;&lt;li&gt;Control systems engineering &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As well as the basic engineering staples like statics, kinematics, dynamics and materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often engineering courses will have some sort of specialism and overall ethos. I would suggest that you look for course which have a practical, project based focus. You might find pure electrical or electronic engineering a bit narrow. And software design is a whole different thing.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like something with a product design focus would suit you. Of the standard courses maybe automotive engineering or a broad mechanical or general engineering course. In fact some of the biomedical engineering courses may be a good fit as these tend to focus on product development. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to undergraduate level the specific content probably doesn't matter that much as long as the course content  is interesting and engaging, but it sounds like you would prefer something which take a practical, project based approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically an engineering degree indicates a reasonable level of understanding of the subject but isn't really a vocational qualification and professional engineers will generally need some additional level of industrial experience or higher level academic research based qualification to be useful. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit in response to comment by OP. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not formal education is the best path is somewhat subjective. Formal qualifications are certainly not essential but you will certainly need an alternative strategy for getting the sort of technical knowledge which definitely is essential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly there are also significant advantages in terms or a career and employment in having recognised qualifications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest unless you have a very clear idea of exactly what alternative route  you have in mind then a structured course is probably a sensible option. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also at this stage you aren't really committing yourself to a specific field and the base skills you will learn are eminently transferable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you are unsure it is also a good idea  to look in more detail into course syllabuses and get a better idea of the precise content. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2017-12-30T01:26:25.510" LastActivityDate="2017-12-30T01:26:25.510" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18609" PostTypeId="2" ParentId="18600" CreationDate="2017-12-28T15:20:39.293" Score="1" Body="&lt;p&gt;Air filters are made of fibrous material (paper typically) that is basically inert. There is no reason that storing it in a cool, dry place would cause it to expire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The HEPA designation has more to deal with how the air flows through it rather than the material that it is made from&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2017-12-28T15:20:39.293" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18610" PostTypeId="2" ParentId="18599" CreationDate="2017-12-28T15:53:48.567" Score="0" Body="&lt;p&gt;This is a very broad question and here is my broad answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding &lt;em&gt;old&lt;/em&gt; methods(pre-CAD and CNC) you can look at how Gaudi designed and built many of his buildings. In particular he used Catenary curves. Also, there are many craftsman to create the features by hand. This is well documented. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;New&lt;/em&gt; buildings leverage the power of the CAD/CNC systems aid in the complex fabrication. If everything is unique, so be it, and CAD/CNC/hard work is done to build the job. Depending on the materials, will drive the fabrication methods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the planning and detailing phase and/or quoting phase, typically some repetition is identified or created. The surfaces are rationalized into modules of repeating parts to simplify the entire system. If possible, splines are tried to be converted into sets of radial/tangential curves. Software and software automation is used to help with this rationalization. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will drive down cost, decrease production time, decrease paperwork, simplify site work, decrease unique part count, decrease model size, etc, etc. In these projects Time is Money. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, this type of architecture requires highly skilled expert contractors to build this type of work. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-12-28T15:53:48.567" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18611" PostTypeId="1" CreationDate="2017-12-28T17:09:00.677" Score="3" ViewCount="302" Body="&lt;p&gt;I have two spur gears, each of a diametrical pitch of 16. One gear has 36 teeth and the other 72. I need to know what the distance should be center line to center line of the gear set. This would be the distance between the two shafts that the gears would rotate on. &lt;/p&gt;&#xA;" OwnerUserId="14327" LastActivityDate="2018-03-01T11:54:50.910" Title="Center line to centerline of pivots for two 16 diametrical pitch gears. One gear 72 tooth other 36 tooth" Tags="|gears|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18613" PostTypeId="2" ParentId="17748" CreationDate="2017-12-28T21:43:28.870" Score="0" Body="&lt;p&gt;The equation should be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ J \, \ddot{\theta} + K \, (\theta-\theta_T) + C \, \dot{\theta} =0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\theta_T$ is the target angle and the rest position for the spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things simplify wit the following substitution &#xA;$$ \begin{cases} &#xA;  K  = J \Omega^2 \\ &#xA;  C  = 2 \zeta J \Omega &#xA;\end{cases} $$&#xA;where $\Omega$ is a parameter the relates to the stiffness, and $\zeta$ a parameter that relates to the damping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general solution of the equation $\ddot{\theta} + \Omega^2 (\theta-\theta_T) + 2 \zeta \Omega \dot{\theta} =0$ is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta = \theta_T + C_1 \exp\left( -\Omega t \left(\sqrt{(\zeta^2-1)}+\zeta \right) \right)+ C_2 \exp\left( -\Omega t \left(\sqrt{(\zeta^2-1)}-\zeta \right) \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the coefficients $C_1$ and $C_2$ are found based on the boundary conditions. In this case, the rod is at rest when $t=0$ and thus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{cases} &#xA;  C_1  =\frac{ \theta_T}{2} \left( \frac{\zeta}{\sqrt{(\zeta^2-1)}}-1 \right) \\&#xA; C_2  =-\frac{ \theta_T}{2} \left( \frac{\zeta}{\sqrt{(\zeta^2-1)}}+1 \right)&#xA;\end{cases} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fastest response is when $\zeta \rightarrow 1$ since that minimizes the value of the first exponent. With optimum damping the solution becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \theta = \theta_T \left( 1 - {\rm e}^{-\Omega t}(1+\Omega t) \right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $C = 2 J \Omega$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as you stated, you cannot find the correct $\Omega$ to reach the target angle at the specified time because $\Omega$ cannot be isolated from the solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it can be isolated from the general solution when $\zeta&amp;gt;1$. From the two terms, the first one approaches zero much faster than the second one. We can find the coefficient $\varphi = \Omega t$ which makes the second exponent near zero by a value $\epsilon$ (in degrees).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left. \frac{ \theta_T}{2} \left( \frac{\zeta}{\sqrt{(\zeta^2-1)}}+1 \right) \exp\left( -\varphi \left(\sqrt{(\zeta^2-1)}-\zeta \right) \right) = \epsilon \right\} $$ $$ \varphi = \frac{ \ln\left( 2 \frac{\epsilon}{\theta_T} ( \zeta \sqrt{(\zeta^2-1)}-\zeta^2+1)\right)}{\sqrt{(\zeta^2-1)}-\zeta} = \Omega\, t_T $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $t_T$ is the target time. A good compromize occurs when $\varphi = \frac{1}{\sqrt{(\zeta^2-1)}}$ which makes $\frac{{\rm d}\varphi}{{\rm d}\zeta}=0$ and hence minimizes $\varphi = \Omega \, t_T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This produces the solution&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{cases} &#xA;  K = \frac{J}{t_T^2 \gamma (\gamma+2)} \\&#xA;  \zeta  = \frac{ \sqrt{(J+K t_T^2)}}{t\,\sqrt{K} } \\&#xA;  C = \frac{2}{t_T} \sqrt{J (J+K t_T^2)} &#xA;\end{cases} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma \ll 1$ is a small positive value (defined as $\zeta = 1+\gamma$).&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2017-12-28T21:43:28.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18614" PostTypeId="1" CreationDate="2017-12-28T21:52:01.687" Score="0" ViewCount="35" Body="&lt;p&gt;In full-hybrid cars like the Toyota Prius, usually both internal combustion engine (ICE) and electric motor produces mechanical power to the wheels at cruise. But some of the mechanical power produced by the internal combustion engine is converted to electric power through a generator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I would like to know typical values for the fraction of the total power generated by the ICE, which is transmitted to the generator at cruise.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9747" LastActivityDate="2018-01-28T10:25:18.800" Title="What is the fraction of power transmitted to the generator in full-hybrid cars?" Tags="|motors|energy-efficiency|propulsion|generator|car|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18618" PostTypeId="1" AcceptedAnswerId="18622" CreationDate="2017-12-29T03:27:29.883" Score="2" ViewCount="13528" Body="&lt;p&gt;In control theory, are manipulated variables and control variables the same concept or is there a subtle difference between them?&lt;/p&gt;&#xA;" OwnerUserId="5794" LastEditorUserId="5794" LastEditDate="2017-12-29T13:27:46.647" LastActivityDate="2018-01-01T20:04:36.197" Title="The differences between manipulated variables and control variables" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18619" PostTypeId="1" AcceptedAnswerId="18672" CreationDate="2017-12-29T04:50:55.460" Score="0" ViewCount="134" Body="&lt;h2&gt;The Problem&lt;/h2&gt;&#xA;&lt;p&gt;I'm proposing a 2D problem in the x-y plane. Say there is a free-floating disk of radius R and mass Md (no velocity), at which a projectile is shot. The projectile (velocity Vp, mass Mp) has a velocity which is parallel to the x axis. If taking the disk's center for the origin, the projectile contacts the disk at a distance Y above the x axis, and it sticks (no energy lost).&lt;/p&gt;&#xA;&lt;h2&gt;The Question&lt;/h2&gt;&#xA;&lt;p&gt;I'd like to know what the ending velocity and angular velocity of the disk/projectile system is. Using conservation of energy, I end up with two variables, and I can't figure out what else to use to create a system (I'm very rusty on conservation of momentum, but I figure that's where the solution lies?).&lt;/p&gt;&#xA;&lt;h3&gt;Disclaimer&lt;/h3&gt;&#xA;&lt;p&gt;I'm not a student, if you're worried about doing my homework. Back in those days, I wouldn't have blinked at this, but now... software engineering has done me wrong XD&lt;/p&gt;&#xA;" OwnerUserId="14321" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-01-02T18:13:08.693" Title="Momentum of a free-floating body being hit by a projectile" Tags="|mechanical-engineering|energy|" AnswerCount="2" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="18620" PostTypeId="2" ParentId="18614" CreationDate="2017-12-29T10:16:36.933" Score="1" Body="&lt;p&gt;It will vary between 0 and maximum depending on the demands of the drive system - ie the engine output is shared whatever the wheels need and the remainder to the generator to charge the batteries.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2017-12-29T10:16:36.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18621" PostTypeId="2" ParentId="659" CreationDate="2017-12-29T11:05:11.883" Score="1" Body="&lt;p&gt;Drill small hole in the aluminum, big enuf to put in the TC junction head, then pound the edge of the hole with a center punch, that'll deform the hole to hold the TC in the hole mechanically.  Works great in high temperature applications.&lt;/p&gt;&#xA;" OwnerUserId="14336" LastActivityDate="2017-12-29T11:05:11.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18622" PostTypeId="2" ParentId="18618" CreationDate="2017-12-29T12:04:55.013" Score="2" Body="&lt;p&gt;Manipulated variables and control variables are the same concept. They are the variables that are used to control the process, e.g. the output of the controller. I'm not sure why there are two different names for the same thing, but I guess it's because the industry and academia often develop different lingo because they have different goals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variable that is subject to control is called the &lt;em&gt;controlled&lt;/em&gt; variable by some people. It shouldn't be confused with the &lt;em&gt;control&lt;/em&gt; variable. I prefer to use the name process variable (PV) instead of controlled variable.&lt;/p&gt;&#xA;" OwnerUserId="78" LastActivityDate="2017-12-29T12:04:55.013" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18624" PostTypeId="1" CreationDate="2017-12-29T17:06:25.200" Score="1" ViewCount="59" Body="&lt;p&gt;So I'm designing an AUV which will have wheels filled with air/helium(*) to ascent to the Antarctica ice shelves from underwater. My goal is retrieve data, images and information about the underwater world of Antarctica.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what kind of tires and what radius would be ideal to use on this AUV? Note that the wheels will be rolling on ice underwater, so I need a material that will increase the friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*) I have also taken in consideration that the force of the AUV towards the surface (which will be blocked by thick flat ice). For that I was thinking that using helium in the tires and/or other parts(**) of the AUV would increase the buoyant force which would mean greater friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(**) Maybe I could add a separate sphere(***) somewhere filled with helium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(***) Sphere because I can get the most amount of volume possible for the least surface area needed.&lt;/p&gt;&#xA;" OwnerUserId="12819" LastActivityDate="2017-12-30T15:37:17.443" Title="Tires on ice underwater (AUV Design)" Tags="|robotics|water-resources|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18626" PostTypeId="1" AcceptedAnswerId="18632" CreationDate="2017-12-29T18:40:51.140" Score="2" ViewCount="270" Body="&lt;p&gt;For a single 4x4 wood column loaded with say 300lbs of weight is there any horizontal thrust being generated in any direction? Columns don't seem to be referenced in this manual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.engineeringwiki.org/wiki/Main_Page&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineering Wiki - Civil Engineering&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="14332" LastEditorUserId="8055" LastEditDate="2018-01-03T21:47:03.547" LastActivityDate="2018-01-04T00:40:38.403" Title="Does a column produce horizontal thrust?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18627" PostTypeId="1" CreationDate="2017-12-29T18:57:00.763" Score="2" ViewCount="720" Body="&lt;p&gt;Usually in cores and shear walls, one extends vertical bars above the construction joint a distance at least equal to the lap splice length to allow for splices away from the construction joint. But is there a problem if the splice does occur at the construction joint (such that part of the splice is below and part is above the joint)?&lt;/p&gt;&#xA;" OwnerUserId="14337" LastEditorUserId="1832" LastEditDate="2018-01-04T00:34:11.620" LastActivityDate="2018-07-14T19:28:25.743" Title="construction joint in shear wall passing through vertical bar splices" Tags="|structural-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18629" PostTypeId="1" CreationDate="2017-12-29T20:50:53.647" Score="2" ViewCount="55" Body="&lt;p&gt;Absorbent cotton wool is a good bio-compatible example, but it has low availability in the US and is not reusable.&lt;/p&gt;&#xA;" OwnerUserId="14341" LastActivityDate="2018-01-29T01:39:45.787" Title="What materials absorb water and release it steadily when compressed?" Tags="|fluid-mechanics|materials|design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18630" PostTypeId="2" ParentId="18599" CreationDate="2017-12-29T21:42:42.990" Score="0" Body="&lt;p&gt;I was involved in the early stages of design of Azadi tower, my partners were building the study models for the project. The architect, Hossein Amanat had hired a famous structural engineering firm Ov Arup. They changed and modified the curves on the arcs and legs many times, working in close coordination and relying heavily on feed back from model making team.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The models where made of acrylic resin hardened into cylinders of roughly 12 inch diameter and where cut to rough dimensions by a dremel and then sanded down to precision by an assortment of custum build files or custum made tracks to slide the model by the band saw paper. But the finish shaping was done by hand by one of our friends, Amir Tabrizian, who was really knowledgeable of the geometry of ancient Iranian arcs and curves which were used in the Azadi design, and at the same time had perfect ambidexterity to hold and file or trim the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I heard his input and feed back as to using generating lines and curves was very useful for the study of the geometry of project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later they decided to use the white stone used to surface the building as the actual formwork for structural concrete, so they created detailed models and shop drawings for each individual piece of stone clading including instructions on how to cut them to form the intricate surfaces.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2017-12-29T22:20:12.480" LastActivityDate="2017-12-29T22:20:12.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18631" PostTypeId="2" ParentId="18629" CreationDate="2017-12-30T00:48:48.977" Score="2" Body="&lt;p&gt;reticulated (i.e., open-celled) foams of polyurethane are excellent for this purpose, and commercially available in bulk. their nominal pore sizes can be specified as well. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2017-12-30T00:48:48.977" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18632" PostTypeId="2" ParentId="18626" CreationDate="2017-12-30T02:00:41.537" Score="4" Body="&lt;p&gt;Columns can and do create horizontal reaction if that's what you mean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some examples are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A column loaded by a vertical force not passing through its axial center.&lt;/li&gt;&#xA;&lt;li&gt;A column rigidly or semi-rigidly connected to a beam.&lt;/li&gt;&#xA;&lt;li&gt;A column adjacent to a non-structural member such as a partition wall, incidental stacked material such as in a storage.&lt;/li&gt;&#xA;&lt;li&gt;A column subject to lateral forces of a shear strap or lateral force of shear wall.&lt;/li&gt;&#xA;&lt;li&gt;A column not loaded symmetrically.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2018-01-04T00:40:38.403" LastActivityDate="2018-01-04T00:40:38.403" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18633" PostTypeId="2" ParentId="3326" CreationDate="2017-12-30T02:26:44.653" Score="2" Body="&lt;p&gt;W = P/L. The load is assumed to be uniform along the top rol length. Another L is the spacing of bottom rollers, called the bending span. This L only before the bending starts. Say at the top 12 o'clock at each bottom roller. Once it does, the L becomes shorter as the support of the plate moves to 1 o'clock at the left and 11 o'clock at the right roller. The shorter the span L the more power is needed.&lt;/p&gt;&#xA;" OwnerUserId="14343" LastActivityDate="2017-12-30T02:26:44.653" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18634" PostTypeId="1" CreationDate="2017-12-30T03:06:20.607" Score="5" ViewCount="101" Body="&lt;p&gt;Let me state I was once an Engineering student but am now a professional software engineer working on a side project that involves something similar to the question below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a box hanging from a beam and the box weighs 200 pounds. The beam is parallel to the ground and it can safely support the weight of the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is does the ability of the beam to safely support the weight of the box change as the box moves from one end of the beam to the other. Does the speed at which the box moves affect the ability of the beam to support the weight? (for example, 1 mph vs. 300 mph). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically if the beam supports the weight of the box when it's not moving, is it safe to assume it will continue to support the weight when it's zooming from one end of the beam to the other?&lt;/p&gt;&#xA;" OwnerUserId="14344" LastActivityDate="2019-01-07T02:39:24.857" Title="Beam support weight in motion" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18635" PostTypeId="2" ParentId="18634" CreationDate="2017-12-30T08:24:05.610" Score="2" Body="&lt;p&gt;At &lt;em&gt;low&lt;/em&gt; speed and acceleration it's probably ok. &#xA;At &lt;em&gt;high speed&lt;/em&gt;, please consider the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to make a couple assumptions and will speak to concepts because we only have a detail about the suspended mass:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The hanging mass is hanging significantly far below the beam&lt;/li&gt;&#xA;&lt;li&gt;The hanging mass is rigidly connected to beam with 1 degree of freedom. (Not suspended by a cable or something which would make your mass a pendulum(think wrecking ball)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Moving a mass at significant speed is going to have momentum. When you start and stop the momentum you have an acceleration and deceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Momentum = mass  * velocity (for reference)&#xA;Force = mass * acceleration&#xA;Acceleration = (delta)speed/time&#xA;Torque or Moment = force*distance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you significantly accelerate or decelerate your mass you have a high force. Multiply that force by the distance it's suspended from from the beam/track and now you have induced a moment load into the beam. This alone may be an issue for your beam. It may not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, depending on the motion of your mass, for example if it moves only between two fixed locations, you might see fatigue at the two locations due to the cyclic or repeated stress at those same locations. If the mass can start and stop in any location along the beam it will have different implications. Also, you would need to calculate the combined load in the worst location(furthest point away from any supports.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last thoughts, the design of your &quot;trolley&quot; or skate which lets the mass slide along the beam could better distribute the load applied along the beam making it more capable to sustain the loads applied as described above. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2017-12-30T08:24:05.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18636" PostTypeId="2" ParentId="18634" CreationDate="2017-12-30T08:26:00.843" Score="0" Body="&lt;p&gt;If the beam is initially supporting your box hanging at its center where it creates the most moment, usually moving the load to the sides of the beam back and forth should be ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the case with most common beams made up of extruded prismatic matterials such as wood or metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you need to make sure you have to check your moment and shear diagrams. Shear forcce in a typical point load case decreases from a maximum on one side going to zero under the load to the other maximum on the other side, under what is called influence line shear. The maximum shear is equal to your load when it is applied right over the support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as moving the load fast back and force that is a whole new ballpark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you move your load across the length of the beam you are vertically inducing a perturbation to the beam and hence the beam and your weight may start to vibrate vertically and therefore increase momentary live load force fluctuations beyond the static design of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the frequency of this back and forth vibration is close to natural frequency of this system then you are dealing with harmonic resonance, which could increase the amplitude of vibration in this beam and lead to catastrophic failure.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="16486" LastEditDate="2019-01-07T02:39:24.857" LastActivityDate="2019-01-07T02:39:24.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18638" PostTypeId="2" ParentId="18624" CreationDate="2017-12-30T15:30:31.520" Score="1" Body="&lt;ol&gt;&#xA;&#xA;&lt;li&gt;Land-based vehicles routinely use spikes for good traction on ice.  This seems like a obvious thing to at least investigate.  Look up &lt;i&gt;studded snow tires&lt;/i&gt;.&#xA;&#xA;&lt;li&gt;I don't get the point of helium.  It is very slightly more boyant than air underwater.  The extra handling and expense relative to air just doesn't sound worth it.&#xA;&#xA;&lt;li&gt;I'm skeptical that the underside of the ice is smooth enough for long enough to make such a vehicle worthwhile.  Rolling wheels are not good at dealing with obstructions or gaps larger than their diameters.&#xA;&#xA;&lt;li&gt;Look into treads.  Those are used on the land-based vehicles that need to traverse and grip rough terrain.  Your bouyance would have to come from elsewhere, but a tank of air is cheap and simple.&#xA;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="624" LastActivityDate="2017-12-30T15:30:31.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18639" PostTypeId="2" ParentId="18624" CreationDate="2017-12-30T15:37:17.443" Score="0" Body="&lt;p&gt;Something like &lt;a href=&quot;https://www.youtube.com/watch?v=RbDe5dEu07I&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; would allow buoyant force, traction on ice (think ice skating) and hydrodynamic propulsion (screw in water) all in one design.&lt;/p&gt;&#xA;" OwnerDisplayName="user12750" LastActivityDate="2017-12-30T15:37:17.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18640" PostTypeId="1" AcceptedAnswerId="18655" CreationDate="2017-12-30T16:00:25.773" Score="-1" ViewCount="4288" Body="&lt;p&gt;I'm having a trouble drawing this isometric shape.&#xA;I hope to find the drawing for this shape. ( would be so grateful if it's drawn in that isometric cube where each square of the cube is one unit of the dimensions )&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nP2qa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nP2qa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;My try according to my understanding to the answer of @kamran&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lTQX0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lTQX0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct ?&lt;/p&gt;&#xA;" OwnerUserId="14349" LastEditorUserId="14349" LastEditDate="2017-12-31T07:24:58.137" LastActivityDate="2017-12-31T12:02:52.363" Title="Drawing isometric from two orthographic views" Tags="|drawings|" AnswerCount="2" CommentCount="13" ClosedDate="2018-01-02T18:16:45.357" ContentLicense="CC BY-SA 3.0" />
  <row Id="18641" PostTypeId="1" AcceptedAnswerId="18676" CreationDate="2017-12-30T16:13:48.690" Score="4" ViewCount="209" Body="&lt;p&gt;There is a common problem with fruit trees in &lt;strong&gt;temperate climates&lt;/strong&gt; where the trees &lt;strong&gt;flower before the last frost&lt;/strong&gt;. Generally speaking, if the tree has flowered, and the temperature drops below freezing, the flowers die and do not produce fruit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is certainly not a new problem. It's my understanding that the problem affects almost the entire fruit farming industry in temperate climates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard that fruit farmers have &lt;a href=&quot;http://www.omafra.gov.on.ca/english/crops/facts/85-116.htm#freeze&quot; rel=&quot;nofollow noreferrer&quot;&gt;large-scale commercial techniques&lt;/a&gt; that they use to mitigate the issue, such as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Covering&lt;/li&gt;&#xA;&lt;li&gt;Fog or smoke&lt;/li&gt;&#xA;&lt;li&gt;Wind Machines&lt;/li&gt;&#xA;&lt;li&gt;Sprinkling&lt;/li&gt;&#xA;&lt;li&gt;Heating&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While these methods might be practical for commercial farmers, &lt;strong&gt;not all of them are practical for small-scale urban farmers like me&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One technique that us gardeners often wonder about is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Instead of actively protecting the open flowers, would it be possible to trick fruit trees into not flowering as early?&lt;/strong&gt; In other words, keep the ground frozen longer than it would be naturally, to delay flowering (I'm making an assumption that trees don't flower when the ground is frozen). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While such a technique would likely be impractical for large-scale farmers, it might be resonably practical for small-scale gardeners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I imagine that there are more factors at play than just ground temperature. For example, the amount of sun, angle of the sun, and the temperature of the air might also be factors when it comes to the flowering timing.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that prolonged frozen ground conditions could significantly delay the tree from flowering (maybe by a couple of weeks) which would make a huge difference when it comes to avoiding late frosts from killing flowers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can I prolong frozen ground conditions to trick fruit trees into flowering later?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Idea:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An idea would be to insulate the ground around the tree with snow. The theory being that the insulation would prevent heat from the sun from warming up the ground where the roots are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WfMSI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WfMSI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, an obvious problem with snow is that it melts, and not always when you want it to. From past experience, I'm guessing that this pile of snow will mostly melt before it is helpful to me (before late spring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm dealing specifically with plants in the prunus genus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Prunus tomentosa&lt;/li&gt;&#xA;&lt;li&gt;Prunus besseyi x prunus salicina hybrids&lt;/li&gt;&#xA;&lt;li&gt;Prunus avium x prunus salicina hybrids&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10536" LastEditorUserId="10536" LastEditDate="2018-01-09T14:19:03.730" LastActivityDate="2021-12-14T05:25:55.323" Title="Prolong frozen ground conditions (to trick fruit trees into flowering later)" Tags="|temperature|environmental-engineering|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18642" PostTypeId="2" ParentId="18640" CreationDate="2017-12-30T18:47:45.743" Score="1" Body="&lt;p&gt;I can help you imagine its shape and leave the actual drawing to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shape was a basic 13x9x9 cube. They first cut from the bottom face a channel of 5x3, then they chamfered (cut to an angle) one side, making it to a slope!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then they cut a trapezoid prism from top right and left side of the object but not all the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;May be if you consider the side view as front view, and front view as side view it can help!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2017-12-30T18:47:45.743" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18643" PostTypeId="1" AcceptedAnswerId="18645" CreationDate="2017-12-30T19:29:44.530" Score="1" ViewCount="713" Body="&lt;p&gt;Basically, I'm thinking about a system, that filtrates, and makes a lake's water drinkable, drives water into my house's faucets without electricity, and all these should be able to work when it is -20 Celsius outside. The water heating is not a problem, it is going to happen with a good old wood consumer stove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I even thought about adding a microsystem, that makes energy, like this: &lt;a href=&quot;https://youtu.be/rPdpnunr1k0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/rPdpnunr1k0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ram pump I would like to build is something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/CG0laNqJWY0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/CG0laNqJWY0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/nFZYD05I29s&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/nFZYD05I29s&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I draw a basic thing I don't even dare to call a plan:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pWcFd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pWcFd.png&quot; alt=&quot;Water moving&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lake water goes down, into a simple filtration system and makes the mentioned hydro energy tool work, then flows (underground) into the other filtrating tool (under the house), that makes drinkable water. The next stop is a huge container or a few smaller, then the ram pump makes the water go up about 3 meters into a cold water container, and from that to a heatable storage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;The ram pump makes &quot;wastewater&quot; but I could make that into a smaller pond or something.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to know more about this topic so I appreciate every idea to make this notion better and make the pump, the hydropower tool and the whole system more efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: The electricity made with the turbine is to store it in a battery or more batteries, so I could use it to make light or charge a laptop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I have is to make the system work in -20 Celsius and to use a turbine to make electricity &lt;em&gt;efficiently.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14351" LastEditorUserId="14351" LastEditDate="2017-12-31T00:42:31.333" LastActivityDate="2017-12-31T01:18:46.360" Title="How to make water go into my house, from a lake without energy?" Tags="|mechanical-engineering|pumps|heating-systems|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18644" PostTypeId="2" ParentId="18643" CreationDate="2017-12-30T23:20:38.597" Score="0" Body="&lt;p&gt;1 meter of head is not a lot. You will need a huge flow rate to make this work. Let's just take the &quot;heats a part of the water (for showering) &quot; part. Typical tankless hot water heaters are about 10 to 30 kW. Let's just pick 10 kW. For a 1 liter per second flow at a 1 meter drop, youll get 9.8 watts (see &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Hydropower&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Hydropower&lt;/a&gt;) assuming a 100% efficient turbine, so you'll need 1000 liters per second flow rate. At that rate you'd drain an Olympic size swimming pool in about 40 minutes. You'll either need a very big lake or take very short showers. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2017-12-30T23:20:38.597" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18645" PostTypeId="2" ParentId="18643" CreationDate="2017-12-31T00:27:16.310" Score="1" Body="&lt;p&gt;My grandfather's house had a system like that. According to my dad, it used a tiny pump in a stream to supply the house. A plate about the size of a playing card flopped up and down in the tiny creek and drove a piston pump the size of my pinky. It fed a hose to a tank on the roof about 100 yards away. The tubing was maybe 1/4 inch ID. Assuming about 1/4 cubic inch per stroke and 60 strokes per min, that's 15 cubic inches/min or 3.9 gal per hour. Sounds reasonable. This was more than a century ago, though. You could carry the whole thing in your pocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a vastly overcomplicated version called &lt;em&gt;The Plata Pump&lt;/em&gt;: &lt;a href=&quot;http://www.fao.org/docrep/010/ah810e/AH810E214.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.fao.org/docrep/010/ah810e/AH810E214.gif&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/brDvf.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/brDvf.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="14351" LastEditDate="2017-12-31T01:18:46.360" LastActivityDate="2017-12-31T01:18:46.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18647" PostTypeId="1" AcceptedAnswerId="18734" CreationDate="2017-12-31T02:20:50.517" Score="2" ViewCount="3217" Body="&lt;p&gt;Which part of the LED is affected when it is exposed to liquid nitrogen? I have seen experiments showing LED's brightness increasing and color shifting, when the LED is submerged in liquid nitrogen. I have also seen one where the color went DOWN in frequency, how can you explain this inconsistancy?&lt;/p&gt;&#xA;" OwnerUserId="14115" LastEditorUserId="8055" LastEditDate="2018-01-08T19:08:03.070" LastActivityDate="2018-01-08T19:08:03.070" Title="LED color change in extreme cold" Tags="|electrical-engineering|lighting|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18648" PostTypeId="2" ParentId="18619" CreationDate="2017-12-31T03:34:02.810" Score="0" Body="&lt;p&gt;Suggested approach and a bit of reasoning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are multiple ways to view the final situation. You can look at it at the instant of contact and realize that the final velocity of the projectile is Vd + Wd * y. This works, but isn't fun. A better way is to think of the final &lt;em&gt;average&lt;/em&gt; velocity of the projectile is just Vd, but it is also stuck to a spinning object off center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the initial masses, inertial moments, and offset y to calculate the final inertial moment (I), neutral axis, and the offset of the projectile from the neutral axis. Call the offset r.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there are two energy terms, translational and rotational. Total mass * V^2 + I * W^2 = the original projectile energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we know the ratio of V to W, so we can substitute and get rid of either one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The collision resulted in some impulse (a force integrated over time). The same impulse caused the rotation rate and the translation rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impulse = M * V&lt;br&gt;&#xA;Impulse * r = I * W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so W = M * V * r / I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;substitute and solve for V, then W.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2017-12-31T03:34:02.810" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18649" PostTypeId="1" CreationDate="2017-12-31T03:52:38.740" Score="0" ViewCount="2226" Body="&lt;p&gt;Given two identical samples of heat treatable steel in both composition and geometry, if one of the pieces is annealed or hardened would there be a measurable difference in their thermal expansion coefficient?&lt;/p&gt;&#xA;" OwnerUserId="14353" LastActivityDate="2018-01-01T18:27:01.683" Title="Does heat treating steel change its Thermal Expansion Coefficient?" Tags="|mechanical-engineering|structural-engineering|materials|steel|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18650" PostTypeId="1" CreationDate="2017-12-31T04:34:12.607" Score="1" ViewCount="49" Body="&lt;p&gt;First post here, forgive me any faux-pas. In regards to compressible flow and accelerating a mass flow, such as in a converging nozzle, I know that you can't accelerate a mass past the speed of sound (Mach 1, that is, the velocity at which an infinitesimally small mechanical wave would propagate itself through the same medium, i.e. the mass of air itself). So, you need a diverging section to allow you to continue to de-pressurize the mass and continue to convert its enthalpy into kinetic energy (velocity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know &lt;em&gt;how&lt;/em&gt; this works, via the following equation (and its analogues):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://latex.codecogs.com/gif.latex?%5Cfrac%7BdA%7D%7BA%7D&amp;space;=&amp;space;-%5Cfrac%7BdV%7D%7BV%7D(1-M%5E2)&quot; title=&quot;\frac{dA}{A} = -\frac{dV}{V}(1-M^2)&quot; /&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I understand choked flow and all sorts of analogies for how much mass flow could be possible (traffic jams, etc). My question is more of a &lt;em&gt;why&lt;/em&gt;. Why is the speed of &lt;em&gt;sound&lt;/em&gt; the max velocity of some given mass at the throat? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first, that was my question, then, I recognized that sound is just a wave of pressure difference - so then, I asked, what is the relationship between pressure and mass? Well, pressure is force over area, and force is mass accelerated - so is that the key to the relationship? But pressure has no mass - so then do I have to treat the mass flow as a wave (like in wave-particle duality for matter)? I think by this point I've gone way too far and missed something simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why the &quot;speed of sound&quot;? Why pressure waves, instead of electromagnetic waves, or some arbitrary constant with respect to the medium? What does the velocity of pressure have to do with the velocity of matter in this context?&lt;/p&gt;&#xA;" OwnerUserId="14350" LastActivityDate="2017-12-31T04:34:12.607" Title="What is the theoretical relationship between sound and mass with respect to the sound barrier as a max velocity?" Tags="|mechanical-engineering|fluid-mechanics|aerospace-engineering|aerodynamics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18651" PostTypeId="1" AcceptedAnswerId="18789" CreationDate="2017-12-31T07:28:20.850" Score="0" ViewCount="335" Body="&lt;p&gt;All I know is sand blasting is used for cleaning and etching purposes and abrasive jet machining (AJM) is used for making cavities and slots. But I think AJM can well also be used for cleaning and etching since it involves spraying abrasive particles just like sand blasting. If thats true isnt sand blasting a type of AJM process? How are they different? Are they different in machining procedure, or the equipment used?&lt;/p&gt;&#xA;" OwnerUserId="14355" LastActivityDate="2018-01-08T21:51:26.707" Title="How is sand blasting different from abrasive jet machining" Tags="|mechanical-engineering|machining|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18652" PostTypeId="2" ParentId="18649" CreationDate="2017-12-31T09:46:55.130" Score="1" Body="&lt;p&gt;In my experience, tables/data listing the thermal expansion coefficients of different materials do not make any differentiation between hardened/annealed etc. states of any given material. The degree of accuracy to which the TEC can be measured is limited by the tolerances to which each material is made up of, and this variability will likely be larger than any change from heat treatment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One related thing to note, is that when you heat treat one of your samples, it will likely change size as a result of the treatment. See &lt;a href=&quot;https://vacaero.com/information-resources/the-heat-treat-doctor/1316-dimensional-changes-after-heat-treatment.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://vacaero.com/information-resources/the-heat-treat-doctor/1316-dimensional-changes-after-heat-treatment.html&lt;/a&gt; for more details&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2017-12-31T09:46:55.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18654" PostTypeId="2" ParentId="18611" CreationDate="2017-12-31T10:51:21.687" Score="3" Body="&lt;p&gt;The pitch diameter $d$ of any spur gear can be calculated from the number of teeth $N$ and the diametral pitch $D_P$ (or &lt;em&gt;module&lt;/em&gt; $m$ in metric units):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d = \frac{N}{D_P} \left( = Nm\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the (nominal) centreline distance $C$, note how the centreline distance is the sum of the pitch radii of each gear in mesh:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$C = \frac{d_1+d_2}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$d_1$ and $d_2$ are the pitch diameters for gears 1 and 2 respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, $d_1=36/16=2.25\text{in}$ and $d_2=72/16=4.5\text{in}$. Therefore, $C=\frac{2.25+4.5}{2}=3.375\text{in}.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.375in is the nominal value of centreline distance. (Note, however, that (provided the gears are &lt;em&gt;involute&lt;/em&gt; gears, which is usually a fair assumption) it is possible to opt for slightly higher (and maybe lower) values of centreline distance than the nominal value without significantly impacting the smoothness of operation of the gears, provided such a change in centreline distance doesn’t cause the teeth to stop contacting or start jamming with one another.)&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2017-12-31T10:51:21.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18655" PostTypeId="2" ParentId="18640" CreationDate="2017-12-31T12:02:52.363" Score="3" Body="&lt;p&gt;Close!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AfVtt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AfVtt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. The error.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there is no hidden line on the cut-away face then that edge must be in-line with the front edge. Project the bottom corner straight back as shown.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2017-12-31T12:02:52.363" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18658" PostTypeId="2" ParentId="2349" CreationDate="2017-12-31T18:57:32.190" Score="3" Body="&lt;p&gt;I know it's an old thread, but I can't resist taking a stab at it.  Especially when there's so much bad or unrelated info in the other answers.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Thermal efficiency and fuel mileage ARE related - there are other factors to consider though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fuel energy density is a factor.  For gasoline engines it's about 44.5 HP-hours (since mpg occurs over some time and torque is instantaneous, I'll use HP, if it bothers you just multiply by 5252 and divide by RPM to translate.  Despite all the argument about which measures engine output &quot;better&quot;, they are directly related algebraically) and if you use 20 HP and burn 1 gallon of fuel you get 20/44.5=44.9% NET thermal efficiency; if you could do this at 60 MPH you'd get 60/1=60 MPG (MPH/GPH).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;One example given above used 10% efficiency at 34 MPG.  I'll assume 60 MPH to keep the math simple; 10%x44.5=4.45 HP-hours per gallon.  Achieving 34 MPG at 60 MPH requires: 60/34=1.765 gallons per hour of gasoline (MPH/MPG=GPH). If you extract 4.45 HP from 1.765 gallons of gasoline then you use a total of 7.85 HP. You cannot get 340 MPG from a 100% efficient gasoline engine powered car in the real world (even if you allow for the acknowledged impossibility of a 100% efficient engine) because you can't get a car to travel at 60 MPH using less than 8 HP!  There are vehicles that get even greater fuel mileage, but there not street worthy cars and the engines certainly run much higher than 10%.  Bottom line, the numbers ARE related and you can't just pair up random numbers or you end up with ridiculous claims.   &lt;/li&gt;&#xA;&lt;li&gt;At low highway speeds (again, I like to use 60 MPH to keep the calculations simple), most modern engines operate around 15% to 18% T.E. (and there is some variability) and most modern sedans require somewhere around 15 to 20 BHP at the flywheel. This means MOST mid-size sedans get between 20 and 32 MPG (T.E. x 44.5 x gal. = HP-hours: 15/(.18x44.5)=&lt;strong&gt;1.87 gal&lt;/strong&gt;. and 20/(.15x44.5)=&lt;strong&gt;3.00 gal.&lt;/strong&gt;; 60/1.87=&lt;strong&gt;32.0 MPG&lt;/strong&gt;  and 60/3.00= &lt;strong&gt;20.0 MPG&lt;/strong&gt;).&lt;/li&gt;&#xA;&lt;li&gt;The reason manufacturers are downsizing engines is that smaller engines tend to operate at greater efficiency AT THE SAME POWER OUTPUT.  Cars that get better fuel mileage than those calculated above &lt;strong&gt;usually&lt;/strong&gt; have smaller than 2.0 Liter engines.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14358" LastActivityDate="2017-12-31T18:57:32.190" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18659" PostTypeId="1" CreationDate="2017-12-31T23:06:20.707" Score="1" ViewCount="385" Body="&lt;h1&gt;Given This Diagram, I am to:&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5VXbY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5VXbY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine Principal Stresses&lt;br&gt;&#xA; (My Ans: 92000 and -8500kPa confirmed with Answer Key)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine Max Shear&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My Ans: 50500 kPa confirmed with Answer Key)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Find the Angle between the plane of the maximum (positive) shear stress and the original plane of stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(My Ans 28deg Answer Key &lt;strong&gt;-61.8deg&lt;/strong&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My Mohr's Circle has C=42,000 and R=50,477 It is my understanding that my original point would be at (84,000 , 28,000) and the max shear stress is at the highest Y value of the circle (42,000,50,477)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such I calculate: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;cos^-1(42/50.5) = 33 deg &lt;/li&gt;&#xA;&lt;li&gt;90deg-33.7deg=56.2deg&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;56.2deg/2 = 28.1 deg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where am I going wrong?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14359" LastActivityDate="2018-04-05T02:14:18.770" Title="Maximum Normal Stress From Mohr's Circle" Tags="|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18660" PostTypeId="2" ParentId="18649" CreationDate="2018-01-01T05:36:01.727" Score="0" Body="&lt;p&gt;I don't see how they can have the same value.  Hardening or annealing affects the crystal structure of material and in the case of steel, how much of the carbon (and other hardening elements) gets absorbed into the iron. Both of these will affect how they react to changes in temperature. Is the coefficient of thermal expansion higher or lower?  That part, I can not tell you. I have searched and searched on the web looking for examples where the hardening is listed with the coefficient of thermal expansion and I came up empty. I expect that the variability due to differences in steel chemistry has more effect than that of hardening.  (Keep in mind that, say, 1040 steel can have between 0.370 and 0.440 % carbon and between 0.60 and 0.90 % manganese.)&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2018-01-01T05:36:01.727" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18664" PostTypeId="2" ParentId="2695" CreationDate="2018-01-01T18:08:00.703" Score="1" Body="&lt;p&gt;Or you could just buy one. They are used in oil industry to measure liquid level in wells. They are necessary to turn the pump jacks off /on. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-01-01T18:08:00.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18665" PostTypeId="2" ParentId="18649" CreationDate="2018-01-01T18:27:01.683" Score="0" Body="&lt;p&gt;In a career as a metallurgist , I never saw a distinction noted for different thermal expansion rates. However in a hardened condition steel will be expanded slightly. And for some high alloys that may be ferritic or austenitic ( depending on heat-treatment) at ambient temperatures, there will be a coefficient difference.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-01-01T18:27:01.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18666" PostTypeId="2" ParentId="16613" CreationDate="2018-01-01T19:00:29.340" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_force_microscope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magnetic force microscopy&lt;/a&gt; should have sufficient resolution.  The principle of operation here is that an external force causes a very small (few micrometers) magnetic probe to vibrate, and then the magnetic field from the sample (e.g. the tape) will cause tiny changes to the amplitude or phase of the vibration. Certainly not practical for a cassette player, and maybe not exactly what you were after, but it should work.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-01-01T19:00:29.340" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18667" PostTypeId="2" ParentId="18651" CreationDate="2018-01-01T19:04:19.177" Score="1" Body="&lt;p&gt;I think  AJM is water jet blasting with added sand  ; If so, the pressure is 100 to 1000 times higher than routine sand blasting.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-01-01T19:04:19.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18668" PostTypeId="2" ParentId="18618" CreationDate="2018-01-01T20:04:36.197" Score="0" Body="&lt;p&gt;Manipulated variable and controlled variable are not necessarily the same thing. For example, take a water tank level control system where make up water flow into the tank is the manipulated variable and tank level is the controlled variable. There are many other examples besides this one.&lt;/p&gt;&#xA;" OwnerUserId="5393" LastActivityDate="2018-01-01T20:04:36.197" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18672" PostTypeId="2" ParentId="18619" CreationDate="2018-01-02T17:25:01.037" Score="1" Body="&lt;p&gt;When it comes to solving any collision problem, it is always important to apply &lt;em&gt;conservation of linear momentum&lt;/em&gt; and &lt;em&gt;conservation of angular momentum&lt;/em&gt;. This gives us the first two equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conservation of linear momentum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two rigid bodies 1 and 2, the conservation of linear momentum states the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_1 \mathbf{u}_1+ m_2 \mathbf{u}_2 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_i$ is the mass of body $i$, $\mathbf{u}_i$ is the velocity vector of body $i$ just before the collision, and $\mathbf{v}_i$ just after the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let body 1 be the projectile, and body 2 be the disc. Since the disc is initially stationary, we can simplify this to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_1 \mathbf{u}_1 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first equation we’ll need to solve the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conservation of angular momentum&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For angular momentum about an arbitrary point $P$ to be conserved, the following mess of an equation should be (briefly) observed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\mathbf{r}_1-\mathbf{r}_P\right)\times\left(m_1 \mathbf{u}_1\right)+I_1\omega_1\mathbf{k}+ \left(\mathbf{r}_2-\mathbf{r}_P\right)\times\left(m_2 \mathbf{u}_2\right)+ I_2\omega_2\mathbf{k} = \left(\mathbf{r}_1-\mathbf{r}_P\right)\times\left(m_1 \mathbf{v}_1\right)+I_1\Omega_1\mathbf{k}+ \left(\mathbf{r}_2-\mathbf{r}_P\right)\times\left(m_2 \mathbf{v}_2\right)+ I_2\Omega_2\mathbf{k} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\mathbf{r}_i$ is the position vector of the centre of mass of body $i$ , $\mathbf{r}_P$ is the position vector of point $P$, $I_i$ is the moment of inertia about the axis of rotation for body $i$, $\omega_i\mathbf{k}$ is the angular velocity vector of body $i$ just before the collision, and $\Omega_i\mathbf{k}$ is the angular velocity vector of body $i$ just after the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By setting $\mathbf{r}_P=\mathbf{r}_2$, noting that body 2 is initially stationary, and noting that body 1 is a particle and thus has no moment of inertia, the equation simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{r}\times\left(m_1 \mathbf{u}_1\right)= \mathbf{r}\times\left(m_1 \mathbf{v}_1\right)+ I_2\Omega_2\mathbf{k}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2=-\sqrt{R^2-Y^2}\mathbf{i} + Y\mathbf{j}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that all the terms are parallel to the (out-of-plane) $z$-axis, i.e. parallel to $\mathbf{k}$. Therefore we can use the dot product to multiply all of the terms by $\mathbf{k}$ without loss of information from the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_1\left(\mathbf{r}\times\mathbf{u}_1\right)\cdot\mathbf{k}= m_1\left(\mathbf{r}\times\mathbf{v}_1\right)\cdot\mathbf{k} + I_2 \Omega_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By noting that:&#xA; $$\left(\mathbf{a}\times\mathbf{b}\right)\cdot\mathbf{c}= \left(\mathbf{b}\times\mathbf{c}\right)\cdot\mathbf{a} = \left(\mathbf{c}\times\mathbf{a}\right)\cdot\mathbf{b} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can rewrite the equation as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_1\mathbf{r^*}\cdot\mathbf{u}_1= m_1\mathbf{r^*}\cdot\mathbf{v}_1 + I_2 \Omega_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\mathbf{r^*}=\mathbf{k}\times\mathbf{r}=-Y\mathbf{i}-\sqrt{R^2-Y^2}\mathbf{j}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this form, we now have the second of the equations we need to solve the collision problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The third and final equation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need one more equation to solve this problem. This final equation will arise from the fact that the two bodies stick to one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;((In your question, you specify that there is no energy loss as a result of the collision. However, it is not actually possible for the total kinetic energy of the system to be conserved if both bodies stick* after the collision. If the collision were perfectly elastic (no energy loss), both bodies must rebound from each other after colliding. For this reason, it is important that conservation of kinetic energy is not applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*If the bond that sticks one body to the other is not perfectly rigid (i.e. it is possible to pull apart the body apart as if there was a spring joining them), then it would be theoretically possible for mechanical energy to be conserved. However, this would result in complex oscillating behaviour between the two bodies, much like an undamped mass-spring system.))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the two bodies are to stick post-collision, it important that velocities of the bodies at the point of contact are the same. This results in the following kinematic condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{k}\times\mathbf{r}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be simplified to give the final equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{r^*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In summary&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three equations have been derived:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_1 \mathbf{u}_1 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_1\mathbf{r^*}\cdot\mathbf{u}_1= m_1\mathbf{r^*}\cdot\mathbf{v}_1 + I_2 \Omega_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{r^*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three unknowns: &#xA;$$\mathbf{v}_1 \quad \mathbf{v}_2 \quad \Omega_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the velocity of the projectile, the velocity of the centre of the disc, and the angular speed of the disc, respectively, immediately after the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using vector algebra, it is possible to solve these equation for the three unknowns. Then it may be necessary to perform some substitutions, including:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_1=M_P$$&#xA;$$m_2=M_D$$&#xA;$$I_2 = \frac{1}{2}M_D R^2$$&#xA;$$\mathbf{u}_1=V_P \mathbf{i}$$&lt;/p&gt;&#xA;" OwnerUserId="276" LastEditorUserId="276" LastEditDate="2018-01-02T18:13:08.693" LastActivityDate="2018-01-02T18:13:08.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18675" PostTypeId="1" CreationDate="2018-01-02T19:54:29.223" Score="1" ViewCount="498" Body="&lt;p&gt;I want to be sure that the way I like to think about the displacement method of analysis is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that a beam as given by the figure below is to be solved. &lt;a href=&quot;https://i.stack.imgur.com/3KjvK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3KjvK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under this conditions, the only unknown is the rotation of joint $B$, $\alpha_B$.&#xA;As I see it, the steps taken to compute $\alpha_B$ are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) apply a moment $M_B^F$ at point $B$ of the beam such that no rotation of $B$ takes place. Such a moment is clockwise. Due to this, node $B$ is under the action of an equal and opposite moment of anticlockwise direction and, therefore, according to the standard conventions, of positive value $PL/8$, and, as such, not in equilibrium;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) for node B to be in equilibrium, as it effectively is in the original structure, apply at beam $B$ a moment whose value will depend on $\alpha$; Node $B$ will be subject to an equal and opposite moment $M(\alpha_B)$ such that $$M(\alpha_B)+ PL/8=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a correct interpretation of this method of analysis?&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2018-01-02T22:37:11.440" LastActivityDate="2018-04-04T22:14:18.140" Title="An interpretation of the slope-deflection method (or displacement method)" Tags="|structural-analysis|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18676" PostTypeId="2" ParentId="18641" CreationDate="2018-01-02T20:07:21.020" Score="3" Body="&lt;p&gt;Delaying flowering would definitely be a better solution than attempting to save the flowers after the fact. Its definitely going to take some experimentation and hypothesis testing to determine what works for that tree variety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done a bit of plant research for an indoor aeroponic system that I operate with strawberries. Putting them in the refrigerator allows me to trick them into a winter cycle. I have read that most plants also have a process that measures the length of darkness that tells them when to grow and when to flower. Not sure if this occurs in the tree stems/buds or not; but regardless it is very hard to control the ambient lighting of a tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as your technique for keeping roots cold; I think piling snow is a great idea. You can then cover the snow with straw, sawdust, cardboard, blankets, or other to help insulate it and make it last longer. You could also use a low flow rate sprinkler or mist nozzle at the base of the tree on a cold night to make some additional ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the ice will not keep the ground &quot;frozen&quot; because soil has salts that make it freeze at a temperature below that of clean water. However it will certainly keep the soil colder for longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, it would be good to test it on one tree while leaving a the same variety of tree alone as a control. This way you will be able to roughly see how many days you were able to delay flowering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps something additional to test is cooling the whole tree. A large amount of uninsulated ice in the surrounding area will create a temporary micro climate that will keep the whole tree cooler for longer. Shading the tree from direct sunlight, and blocking from wind would increase this effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good Luck! Please share a link to your results, I'd be interested to see how it turned out.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-01-02T20:07:21.020" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18677" PostTypeId="1" AcceptedAnswerId="18757" CreationDate="2018-01-02T22:02:56.327" Score="3" ViewCount="86" Body="&lt;p&gt;I have a heat pump which moves heat from outside to inside. I also have a solar heating array that collects radiant heat from the sun. Given that my goal is to heat the inside (with minimal energy consumption from the heat pump), my question is: &lt;strong&gt;Is it better to heat the air inside with the solar array or to heat the heat pump's evaporator outside?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Breaking this down, it's a question of whether it's better to use these sources in parallel or in series. In parallel, the solar array heats the inside directly and the heat pump heats the inside directly. In series, the solar array heats the evaporator and the heat pump heats the inside directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other relevant details, the evaporator exchanges heat by convection and is limited to a minimum temperature of 0°C. This is because any colder and you risk freezing the convector, thus stopping convection. The environment can get as low as 3°C at night, meaning heat is exchanged across a minimum of a 3°C difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions for this, the heat pump can move 5kW of heat across a 12.8°C difference (8.3-21.1°C), consuming 2kW. The indoor heat loss is (let's say) 5kW for a 15°C temperature difference (5-20°C). The solar array acts as a heat reservoir with 10kWh stored at 40°C (5°C to 40°C). Finally, let's assume the heating time is 10 hours.&lt;/p&gt;&#xA;" OwnerUserId="11676" LastEditorUserId="11676" LastEditDate="2018-01-02T22:27:48.443" LastActivityDate="2018-01-07T01:00:33.990" Title="Is it better to heat a heat pump evaporator or the load directly?" Tags="|heat-transfer|heating-systems|solar-energy|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18678" PostTypeId="1" CreationDate="2018-01-03T14:58:13.963" Score="1" ViewCount="18" Body="&lt;p&gt;I routinely work on building designs that incorporate rooftop upblast fans over large industrial spaces such as this one:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2RO3U.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2RO3U.jpg&quot; alt=&quot;Axial Upblast Fan&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When selecting these fans, as with every other fan, you need to determine the external static pressure that the fan will encounter.  I have heard opinions ranging from &quot;negligible&quot; to 0.15&quot; of static pressure up to 1&quot; without any technical basis to back the responses up.  Generally, they are exhausting a space at a slightly higher rate than the space is being supplied air in order to negatively pressurize the area relative to adjacent spaces, with an intent of maintaining 0.1&quot; of static pressure.&lt;br&gt;&#xA;What guidelines/calculations are available to allow the external static pressure for this fan configuration to be calculated?&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2018-01-03T14:58:13.963" Title="External Static Pressure on Open Gallery Fan" Tags="|hvac|equipment-selection|specifications|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18680" PostTypeId="1" AcceptedAnswerId="18688" CreationDate="2018-01-03T17:56:33.140" Score="5" ViewCount="10889" Body="&lt;p&gt;I've been watching drag races of semi trucks and found that, especially during the initial phase of acceleration, they all bend to their right-hand side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand, that the huge torque in rear wheels would lift the whole front of the truck, but why does it slant to one side? &#xA;Can rotational inertia of the drive shaft be strong enough to cause such a phenomenon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iSlRR.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iSlRR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14380" LastActivityDate="2021-03-03T10:13:46.847" Title="Why do drag race semi trucks slant to one side?" Tags="|mechanical-engineering|automotive-engineering|torque|traction|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18681" PostTypeId="2" ParentId="18680" CreationDate="2018-01-03T18:45:21.550" Score="1" Body="&lt;p&gt;I think your assessment is correct. The rear wheels effectively allow the truck to 'pop a wheelie', i.e. lifting the centre of mass by a certain amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frame of these trucks is flexible enough to allow some torsion, and it wouldn't take much to bias the rotation to one direction. I think you're seeing the RHS of the cab (driver's perspective) being pushed down towards the road, rather than the LHS being pulled up towards the sky. The reaction forces to the drive shaft, the weight of the driver (and associated steering components) on the RHS of the cab, etc. could all encourage the cab to tilt that way. &lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-01-03T18:45:21.550" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18682" PostTypeId="1" CreationDate="2018-01-03T19:49:50.980" Score="1" ViewCount="131" Body="&lt;p&gt;I have a construction where water falls on/through a perforated plate via a waterfall (see image below). The plate needs to evenly distibute the water over the underlying structure. My question is how to calculate the hole sizes (or total hole area) in the plate, so the water distributes evenly and does not overflow? Flow rate coming from the waterfall is 30m3/h, some dimensions are in the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully someone can help me, thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O9EwH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O9EwH.png&quot; alt=&quot;image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14382" LastActivityDate="2018-01-03T19:49:50.980" Title="Calculate aperture size in plate where water falls on, at a certain water flow" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18683" PostTypeId="1" AcceptedAnswerId="18758" CreationDate="2018-01-03T20:07:30.460" Score="4" ViewCount="189" Body="&lt;p&gt;I work with left ventricular assist devices in patients with heart failure. These devices operate by maintaining a fixed speed of a centrifugal or axial rotor. The native heart continues to contract creating pressure differences between the pump inflow (heart) and outflow (aorta).&lt;/p&gt;&#xA;&lt;p&gt;The device and the heart are in parallel by my reckoning. Volume A enters the heart. Volume B exits the aortic valve. Volume C exits the pump. Flow in the body is A= B+C. If the pump is off, flow in the body is A=B. If the aortic valve is sewn shut, the flow in the body is A=C.  It is considered dogma that when flow through the device increases, device power consumption increases. The heart and the pump share the pulsatility of the heart and are subject to the same afterload in most situations.&lt;/p&gt;&#xA;&lt;p&gt;I want to understand why power increases when pressure across the pump decreases. I understand that there will be more mass-flow through the pump with a lower delta-P across the pump. What I don't understand is why this mass-flow slows the pump, ostensibly requiring more power to overcome apparent resistance. It seems more logical that increased flow decreases energy of rotor rotation as the reduction in pressure difference would seem to have the action of imparting energy to the rotor.&lt;/p&gt;&#xA;&lt;p&gt;Why is more power required to turn the rotor with a lower delta P across the pump and what principles/equations explain this effect?&lt;/p&gt;&#xA;" OwnerUserId="14383" LastEditorUserId="14383" LastEditDate="2020-12-28T18:44:20.390" LastActivityDate="2020-12-28T18:44:20.390" Title="Power in a continuous flow pump" Tags="|pumps|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="18684" PostTypeId="2" ParentId="18680" CreationDate="2018-01-03T20:16:30.047" Score="3" Body="&lt;p&gt;Inertia plays a small part, but because the mass of the engine and drive train is low compared to the mass of the vehicle, it is a small torque compared to the torque being transmitted through the drive shaft. You can see the drive train of a normal semi below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tGqpp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tGqpp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this normal semi you can also see the black steel frame running from the engine mounts to the rear wheels. This solid frame provides the equal and opposite reaction torque with very little deflection or rotation. You may see 1&quot; of suspension deflection or so when a normal semi accelerates from parked at a stop light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green truck in the photo is likely modified for aesthetic/theater appeal. It has a modified frame to permit that extreme rotation. This modification would actually reduce acceleration performance of the dragster (since energy is wasted in the process of lifting), but the real goal of the actors (i mean drivers ;-) is to entertain the crowd and viewers. &lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-01-03T20:16:30.047" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18685" PostTypeId="1" CreationDate="2018-01-03T20:46:12.013" Score="-2" ViewCount="60" Body="&lt;p&gt;Can anyone please suggest me a material which can produce heat with DC current.&#xA;I want it to be low of cost.&#xA;I should be able to turn it into a sheet.&#xA;And if there no such material. Please tell me how i make it. Or what can i use to do so.&lt;/p&gt;&#xA;" OwnerUserId="14384" LastActivityDate="2018-01-04T01:46:25.820" Title="Suggest a material which can produce heat with DC current" Tags="|materials|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18686" PostTypeId="2" ParentId="18173" CreationDate="2018-01-03T21:22:49.667" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Northern_Hemisphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Northern Hemisphere is home to approximately 6.57 billion people which is around 90% of the earth's total human population of 7.3 billion people.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From ancient times, it is logical that navigation over land and later across the sea would be conducted based upon the so-called North Star and magnetic north.  A northerly reference point would therefore become a conventional orientation for the larger population of the planet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that it is more a matter of culture and demographics rather than purely functional or technical origins.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastEditorUserId="3266" LastEditDate="2018-01-03T21:29:17.697" LastActivityDate="2018-01-03T21:29:17.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18688" PostTypeId="2" ParentId="18680" CreationDate="2018-01-03T22:27:54.603" Score="5" Body="&lt;p&gt;as the engine produces torque which is applied to the driveshaft, the engine block produces a counter-torque which must be blocked by bolting the block to the frame so the engine itself does not simply spin around in the sense opposite to the rotation of its crankshaft. for a fully-loaded truck starting from a standstill with a large diesel engine in it, that countertorque is huge enough to twist the frame of the truck and hoist one of its wheels off the ground, as illustrated in the OP's photo.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-03T22:27:54.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18689" PostTypeId="2" ParentId="18641" CreationDate="2018-01-03T22:36:06.317" Score="1" Body="&lt;p&gt;One thing that comes to my mind are ice rink coolers. You could design similar pipe system that would keep surrounding ground cool. Few problems that could occur:&#xA;- pipes are filled with antifreeze and leaks could cause ecological disaster and forbiden substances would penetrate fruit so you should find ecofriendly antifreeze&#xA;- you need strong cooler unit which will have high energy consumption so using insulation like styrofoam or mineral wool would be advisable&#xA;-together with building adjustable shutters or sun blockers, price of such system would be high and financial study should be done&#xA;-using energy from sun to power such system would also increase initial investment maybe other alternative energy sources would be better&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2018-01-03T23:08:47.190" LastActivityDate="2018-01-03T23:08:47.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18690" PostTypeId="2" ParentId="18683" CreationDate="2018-01-03T22:36:41.430" Score="1" Body="&lt;p&gt;If one wishes for the flow rate of blood through the pump to increase, the power output of the pump must increase and the consequence of this is that the output pressure of the pump increases.  the pressure differential is not decreased if the heart is assisting the mechanical pump by increasing the inlet pressure applied to the pump; note that the heart and the pump are in series which means that the pressures developed by the heart and the pump are additive.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-03T22:36:41.430" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18691" PostTypeId="2" ParentId="18675" CreationDate="2018-01-03T22:53:07.027" Score="1" Body="&lt;p&gt;You should get same final result as the table you gave in comment, this kind of tables are formulas obtained by solving indeterminate beam. &#xA;There is procedure for solving statically indeterminate beams as you can see in the following link  &lt;a href=&quot;https://www.google.hr/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;url=https://www.civil.iitb.ac.in/~minamdar/ce317/Lectures/Slope-Deflection.pdf&amp;amp;ved=0ahUKEwi8-dbF8rzYAhVOJ1AKHTnvD38QFgi6ATAc&amp;amp;usg=AOvVaw0lFZI0628bKVYKvzE2KhB0&quot; rel=&quot;nofollow noreferrer&quot;&gt;slope deflection&lt;/a&gt;. Those solutions should guide you to solve your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind basics for real beam from theory of conjugate beam&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lrDIN.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lrDIN.gif&quot; alt=&quot;conjugate beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2018-01-04T20:49:28.117" LastActivityDate="2018-01-04T20:49:28.117" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18692" PostTypeId="2" ParentId="17992" CreationDate="2018-01-03T23:30:52.307" Score="0" Body="&lt;p&gt;If I understood correctly you have one year old house. Do your neighbours who don't have your problem also have one year old houses? Once I have heard of people having similar problems with humidity due to moving in before construction materials had time to dry so humidity was stored in concrete floors and it cused problems. This link could help explain better &lt;a href=&quot;https://blog.yourpacesetter.com/blog/understanding-humidity-in-your-new-home&quot; rel=&quot;nofollow noreferrer&quot;&gt;humidity in new house&lt;/a&gt;. If this is your case maybe it could be solved using industrial dehumidifiers for some period of time, maybe there are companies who deal with such requests and can give you better answers.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastActivityDate="2018-01-03T23:30:52.307" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18693" PostTypeId="2" ParentId="18685" CreationDate="2018-01-03T23:32:04.457" Score="0" Body="&lt;p&gt;Depends on how much heat you want, what voltage you want to use, and what current.  But pretty much any metal will produce heat when you flow current through it.  Copper, aluminum, iron, steel, take your pick.  Take apart an old hair dryer or a space heater or an electric blanket and see how they do it.  Those are typically designed with AC current, but they could just as easily be re-designed to use DC.  You just have to get the voltage and current right.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-01-03T23:32:04.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18694" PostTypeId="1" CreationDate="2018-01-04T00:58:55.257" Score="1" ViewCount="55" Body="&lt;p&gt;I can’t seem to find an answer that doesn’t involve more than a ton and a half of concrete. The manual for the wind turbine reports that the load on the turbine in a 110&amp;nbsp;mph wind is only 155&amp;nbsp;lbs. When I add the weight of the pole above ground that’s just over 200&amp;nbsp;lbs. Shouldn’t 3x or 4x that amount of weight in ballast be sufficient? The county won’t let me set up my alternative power system until I get a solid answer to this question.&lt;/p&gt;&#xA;" OwnerUserId="14388" LastEditorUserId="1832" LastEditDate="2018-01-04T01:02:14.097" LastActivityDate="2018-01-04T13:27:42.427" Title="How much ballast is required in 110 mph winds to stabilize: a 60 lb., 20’ pole; 3' below ground/17’ above ground, with a 24lb. wind turbine atop it?" Tags="|structures|stresses|torque|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18695" PostTypeId="1" AcceptedAnswerId="18698" CreationDate="2018-01-04T01:16:02.583" Score="0" ViewCount="1523" Body="&lt;p&gt;Consider this triangle with its centroid at $C$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GW2LT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GW2LT.png&quot; alt=&quot;Text-book triangle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then this is how I believe we can calculate the &lt;em&gt;Second Moment of Area&lt;/em&gt; along the $x_{C}$ and $y_{C}$ axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;I_{xc} &amp;amp;= \frac{bh^3}{36} \\&#xA;I_{yc} &amp;amp;= \frac{hb(b^2-st)}{36}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far so good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then we have this triangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bTXV1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bTXV1.png&quot; alt=&quot;Slanted triangle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there any similar way of calculating the second moment of area for a slanted triangle, like the one above?&lt;/li&gt;&#xA;&lt;li&gt;Will even the same equations work?&lt;/li&gt;&#xA;&lt;li&gt;How is $s$ and $t$ derived in that case? Can they be negative?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Please assume we can &lt;strong&gt;NOT&lt;/strong&gt; rebase the triangle — we must use $LR$ as the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.engineeringtoolbox.com/area-moment-inertia-d_521.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source for equations&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14387" LastEditorUserId="1832" LastEditDate="2018-01-04T02:07:51.163" LastActivityDate="2018-01-05T19:31:47.177" Title="Second Moment of Area for slanted triangle?" Tags="|statics|moments|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18696" PostTypeId="2" ParentId="18685" CreationDate="2018-01-04T01:46:25.820" Score="0" Body="&lt;p&gt;sheet heaters have been on the market for years. the cheapest ones use a carbon black-rich paste painted onto a sheet of fireproof paper-like material, onto which is sealed another sheet of the same to protect it. strips of copper tape running along 2 opposite sides of the sheet are used to feed electricity into the carbon black, which gets hot when current flows through it. These heaters are used in foot massage units, yogurt makers, food warmers and waterbed heaters. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-04T01:46:25.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18697" PostTypeId="2" ParentId="18694" CreationDate="2018-01-04T02:02:13.010" Score="1" Body="&lt;p&gt;Using very conservative assumptions, it makes sense that you'll need massive ballast weights to stabilize your pole. This is because the ballast is extremely inefficient for this task. I won't go into the math itself, but here's a conceptual demonstration of what's going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ap4Vx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ap4Vx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine your pole is a vertical lever, with the fulcrum at the soil level. At the top of the pole is a horizontal force, and at the bottom is a vertical force. When the pole is precisely vertical, the vertical force at the bottom won't do anything, but the horizontal force at the top will want to topple the pole, so it will rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the structure has rotated a bit, the vertical force will start to work against the pole's rotation, restraining it. However, it is very inefficient in doing so. Without going into the math, the following equilibrium must be reached:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{ballast}\ell_{underground}\theta = F_{wind}\ell_{aboveground}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\theta$ is the pole's rotation in radians at stability. Therefore, the weight of the ballast must be equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_{ballast} = \dfrac{F_{wind}\ell_{aboveground}}{\ell_{underground}\theta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows us that the greater the aboveground/underground ratio, the heavier the ballast. Also, the less the structure is allowed to rotate to achieve stability, the heavier the ballast as well. Since structures aren't allowed to deflect far, $\theta$ can be assumed very small, so the ballast must be very, very large. Tons seems absurd, but is actually reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, however, uses the very conservative assumption of ignoring any work done by the ground to restrain the pole's motion. Assume for a moment that your pole is embedded in solid diamond (or titanium, or whatever). Will any ballast be required? Of course not. The wind will try to rotate your pole, but that'll make the underground section push against the diamond, which will push right back and impede any such rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same applies to any type of soil. When something pushes it (like the pole trying to push the soil out of its way so it can rotate), the soil develops what's known as &quot;passive pressure&quot; and pushes back. The soil will do this until the pressure crosses its ultimate strength, at which point it collapses and loses any resistive capacity. Using the diagram from before, this becomes equivalent to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XDHDa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XDHDa.png&quot; alt=&quot;![enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This distributed force is far more efficient and can dramatically reduce the necessary ballast. However, it is a function of the type of soil. If your pole is embedded into solid rock, you probably won't need any ballast. If your pole is embedded into a marsh, this force will probably be insignificant and you'll need your massive ballast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine how much the soil can help you, you'll need to hire a qualified geotechnical engineer, who'll need to extract samples from the exact spot (and surrounding area) where you intend to plant your pole. This will need to be combined with a reasonable structural model, according to applicable codes, to determine the appropriate size of the ballast. All of this should be done by qualified engineers and this should not be considered a DIY project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, another possibility would be to adopt a different structural model, such as support cables around your pole, which could be significantly more effective than simple ballast. To judge the feasibility of this, however, we'd need a lot more information regarding your &quot;alternative power system&quot;, and it would probably be out-of-scope for this question anyway.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-01-04T02:02:13.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18698" PostTypeId="2" ParentId="18695" CreationDate="2018-01-04T02:04:13.330" Score="2" Body="&lt;p&gt;Oscar, I'm not sure what your math background is, but there are a lot of different ways to get there. The most direct is to just reason from the properties of affine transforms, specifically &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_mapping&quot; rel=&quot;nofollow noreferrer&quot;&gt;shear mapping&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During a horizontal shear, as in from triangle 1 to triangle 2, the distances moved by different points are proportional to their y coordinate. This tells us, in effect, that if your Iyc equation is a general solution for acute triangles, it must also be a general solution for obtuse triangles, since they are generated from the same transform. The only difference is the magnitude of the transform. So you just have to be consistent with how you measure &lt;em&gt;s&lt;/em&gt; and &lt;em&gt;t&lt;/em&gt;. &lt;em&gt;s&lt;/em&gt; starts at L and goes to T, and &lt;em&gt;t&lt;/em&gt; starts at T and goes to R, measuring horizontally. This is easy to verify by comparing two triangles that are almost right angles (tiny t value), one of each type.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-01-04T02:04:13.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18699" PostTypeId="1" CreationDate="2018-01-04T02:26:32.663" Score="2" ViewCount="502" Body="&lt;p&gt;I want to make a wheel that behaves pretty much like a mouse wheel does — when you rotate it with your finger and release, it rotates back or forward by itself to match a nearest discrete position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the common ways to do so?&lt;/p&gt;&#xA;" OwnerUserId="14389" LastActivityDate="2018-04-05T21:17:40.057" Title="What mechanism can make a wheel turn in discrete steps?" Tags="|mechanical-engineering|wheels|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18700" PostTypeId="2" ParentId="18699" CreationDate="2018-01-04T03:05:11.840" Score="0" Body="&lt;p&gt;a stepper motor. they have discrete detents, furnish propulsion, are cheaply available in a variety of sizes, and easy to drive using arduino, raspberry pi, beaglebone, etc. and lots of user information regarding them exists in the robotics community.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-04T03:05:11.840" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18702" PostTypeId="1" CreationDate="2018-01-04T05:07:28.370" Score="2" ViewCount="196" Body="&lt;p&gt;I recently was at the Naval Air Museum in Pensacola Florida.  There are lots of models of aircraft carriers, including the latest class (Gerald Ford class as I recall).  This latest class has electromagnetic rails instead of steam catapults to launch the jets.  However it still has arresting cables. So I got to thinking that with the sophisticated servo systems we have today, it seems possible to have a servo system on rails on the landing strip that follows the jet as it lands and automatically grabs the tail hook and has a controlled deceleration. On the Science channel show &quot;Outrageous Acts of Science&quot;  there was an amateur inventor who created a servo system to follow his dart throws to give him a bullseye every time. A servo system would improve the chances of a successful landing. I read astronaut Scott Kelly's autobiography and he said one time as a Navy pilot in bad weather at night he had to try dozens of times to land because his tail hook would not grab a cable. Also, the servo approach would eliminate the scrapping of steel cables.  From what I have read, steel cables get replaced fairly frequently due to wear.  Am I missing a fundamental flaw for this idea?&lt;/p&gt;&#xA;" OwnerUserId="5922" LastActivityDate="2018-01-05T19:41:10.437" Title="Aircraft carrier arresting cables" Tags="|mechanical-engineering|control-engineering|aerospace-engineering|mechanisms|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18704" PostTypeId="1" CreationDate="2018-01-04T06:59:11.720" Score="-1" ViewCount="162" Body="&lt;p&gt;I have been searching for a material that can heat up when we supply dc current.&#xA;Many give me a suggestion for use nichrome wires. Or carbon rich paste paint(conductive paint).&#xA;I want a sheet of such material(nichrome)&#xA;But i couldn't find sheets online or in the stores. So first of all i want to know how can i make A4 size sheet from those wires.? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second of all is. If i use carbon rich paste paint and apply it to a normal sheet of paper. Will i be able to obtain heat from it by applying dc current.? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to apply dc current. I can't apply AC current. &lt;/p&gt;&#xA;" OwnerUserId="14384" LastEditorUserId="14401" LastEditDate="2018-01-04T23:46:26.023" LastActivityDate="2018-01-05T00:59:33.580" Title="How can i obtain heat from conductive paint using dc current" Tags="|electrical-engineering|materials|chemical-engineering|manufacturing-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18705" PostTypeId="2" ParentId="18699" CreationDate="2018-01-04T07:46:39.100" Score="-1" Body="&lt;p&gt;Genova or Geneva mechanism is stepper mechanism, there are plenty of websites explaining Geneva mechanism or drive, like this wikipedia article &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Geneva_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geneva drive&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/9/9b/Geneva_mechanism_6spoke_animation.gif/240px-Geneva_mechanism_6spoke_animation.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/9/9b/Geneva_mechanism_6spoke_animation.gif/240px-Geneva_mechanism_6spoke_animation.gif&quot; alt=&quot;Geneva Drive Animation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13311" LastEditDate="2018-03-05T11:33:55.673" LastActivityDate="2018-03-05T11:33:55.673" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18707" PostTypeId="2" ParentId="18647" CreationDate="2018-01-04T11:38:00.317" Score="2" Body="&lt;p&gt;This is a rather in-depth question but broadly speaking - The part of the LED which is affected by the liquid nitrogen is called the p-n junction. The two halves of the junction are a set distance apart, this is the &quot;band-gap&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heating and cooling of the LED causes the crystal lattice of the p-n junction materials to either expand or retract. This causes the band-gap to increase and decrease. In this specific case the liquid nitrogen causes the crystal lattice to become more closely packed. This increases the size of the band-gap. The result of which is the decrease in wavelength of the light being emitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An LED exposed to liquid nitrogen would last only a few seconds before failing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please refer to &lt;a href=&quot;http://solarcellcentral.com/junction_page.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;p-n junctions and band gaps&lt;/a&gt; for additional information.  &lt;/p&gt;&#xA;" OwnerUserId="14150" LastEditorUserId="8055" LastEditDate="2018-01-08T19:07:59.260" LastActivityDate="2018-01-08T19:07:59.260" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18708" PostTypeId="2" ParentId="18704" CreationDate="2018-01-04T11:57:18.593" Score="1" Body="&lt;p&gt;For your first question, If you google &quot;nichrome sheet&quot;, the very first hit is &lt;a href=&quot;http://www.goodfellow.com/E/Nickel-Chromium-Sheet.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.goodfellow.com/E/Nickel-Chromium-Sheet.html&lt;/a&gt;, which will sell you nichrome sheets in several different thicknesses and sizes. None of the options is quite A4 size, but it's at least close.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your second question, it is very unclear what you are asking.  How much heat do you want (watts or BTU/hr, etc)?  What temperature do you want the surface of the sheet to get to?  Do you have limitations on the power supply or can you use literally any DC voltage and current? Do you care if the paper catches on fire? As I said in the answer to your original question, basically anything in the world will heat up if you pass enough current through it, so the literal answer to your question is &quot;yes&quot;, but I suspect that this is not a useful answer.  If you can describe your application in more detail, we can give you a more helpful/useful answer.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-01-04T11:57:18.593" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18710" PostTypeId="1" CreationDate="2018-01-04T11:56:35.613" Score="2" ViewCount="5496" Body="&lt;p&gt;I saw this picture today:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4LmOB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4LmOB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They give an ratio of 5/352. I think this is not possible with a simple gear. Maybe the upper sides ratio is the cause for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which formula do I need to calculate the ratio for first half and which one for the upper half?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use it for a stepper motor (nema 17) as a gear box to increase the torque. &lt;/p&gt;&#xA;" OwnerDisplayName="user14157" LastEditorUserId="8055" LastEditDate="2018-01-04T16:13:20.270" LastActivityDate="2018-07-03T20:19:11.070" Title="How to calculate the gear ratio of planetary gear?" Tags="|stepper-motor|torque|mechanical|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18711" PostTypeId="2" ParentId="18704" CreationDate="2018-01-04T13:12:26.133" Score="3" Body="&lt;p&gt;If you end up with something that conducts current, it will heat up with applied current.  One way or another, you have what is called a &lt;i&gt;resistor&lt;/i&gt;.  A resistor dissipates power proportional to the square of the applied voltage or current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick is to find a material that ends up causing a convenient resistance.  A metal sheet might be so low in resistance that you end up dissipating more power in the connections and the circuit that makes the current than in the sheet itself.  Conversely, if you end up with too high a resistance, it would take too much voltage to get a meaningful amount of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing you should do is create a spec of how much power the whole sheet should dissipate.  That will give you a rough range of total end to end resistance you want.  The power dissipated by a resistor is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp; &amp;nbsp; W = A&lt;sup&gt;2&lt;/sup&gt; &amp;Omega; = V&lt;sup&gt;2&lt;/sup&gt;/&amp;Omega;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where W is the power in Watts, A the current in Amperes, V the EMF in Volts, and &amp;Omega; the resistance in Ohms.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-01-04T13:12:26.133" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18712" PostTypeId="2" ParentId="18694" CreationDate="2018-01-04T13:27:42.427" Score="0" Body="&lt;p&gt;Sounds about right for an unstayed structure. You can reduce the ballast somewhat if you stay your mast (probably 2/3 the way up, design to keep the blades well clear of the stays under worst case sag/vibration/bending. But now you need three or four ballast masses to hold the stays into the ground.&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2018-01-04T13:27:42.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18713" PostTypeId="2" ParentId="18702" CreationDate="2018-01-04T14:04:35.260" Score="1" Body="&lt;p&gt;As SolarMike points out,  there's a ton (pun intended) of energy the arresting cable has to dissipate, along with the springs it's attached to.  Remember that the pilot hits full throttle the moment he touches the deck, so that if he doesn't hook in,  he can get airborne again -- instead of going for a fatal swim.  Your servo would have to be incredibly quick to respond in both horizontal and vertical position to guarantee catching the tailhook. Fast, agile and strong don't really go together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tailhook itself, BTW, is mounted to the strongest part of the aircraft frame -- if you thought of a mixed approach such as having the landing gear double as a cluster hook,  forget it.  They'd snap right off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose, for the sake of argument, you wanted to go with a SciFi approach of using a mega-electromagnet.  It might be fun to calculate the field strength needed to stop the jet, not to mention the hell that would play with all the electronics in a kilometer range!  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-01-04T14:04:35.260" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18714" PostTypeId="1" CreationDate="2018-01-04T14:14:07.307" Score="3" ViewCount="7774" Body="&lt;p&gt;I was wondering while parking my car, what is the axis of rotation/pivot of a turning car? Does it differ when it is turning frontwards of backwards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it different for a 4x4?&lt;/p&gt;&#xA;" OwnerUserId="14395" LastActivityDate="2018-10-14T21:18:43.947" Title="What is the axis of rotation of a turning car?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18715" PostTypeId="1" CreationDate="2018-01-04T14:48:41.790" Score="-1" ViewCount="2503" Body="&lt;p&gt;I need help with the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pile-driver has a hammer mass of 1000 kg which is dropped onto a pile from a height of 5 meters. The pile is driven 0.5 meters into the ground. The pile has a mass of 500 kg. Calculate the retarding force exerted by the ground on the combined mass of the pile and hammer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qXNkW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qXNkW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14101" LastEditorUserId="163" LastEditDate="2018-01-12T13:44:21.710" LastActivityDate="2018-01-12T13:44:21.710" Title="Retarding Force" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="8" ClosedDate="2018-01-12T22:08:52.870" ContentLicense="CC BY-SA 3.0" />
  <row Id="18716" PostTypeId="2" ParentId="18601" CreationDate="2018-01-04T14:48:54.573" Score="1" Body="&lt;p&gt;Officially Gypsum panels should be stored inside where they are not exposed to inclement weather, the Gypsum Association states &quot;Gypsum panel products shall not be stored in areas of excessive humidity&quot;, unfortunately they do not specify what constitutes excessive humidity. However, exposure of Gypsum to high moisture levels can lead to water stains, discolouration, mold, paper de lamination, and sag.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The damage shown in the picture looks severe enough that it would need to be repaired in order for the panels to maintain the required fire-rated separation between areas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no need to put spacers in between the boards providing they are stacked on a flat surface on top of each other. There is a limit on the height of stacks which is 17 feet (5m).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gypsum Association Handling Procedures: &lt;a href=&quot;https://www.pabcogypsum.com/sites/default/files/GA-801-07.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.pabcogypsum.com/sites/default/files/GA-801-07.pdf&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gypsum Association - Repair of Fire-Rate Gypsum Panels: &lt;a href=&quot;https://www.gypsum.org/wp-content/uploads/2015/06/GA-225-15.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.gypsum.org/wp-content/uploads/2015/06/GA-225-15.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14150" LastActivityDate="2018-01-04T14:48:54.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18717" PostTypeId="2" ParentId="18714" CreationDate="2018-01-04T15:18:41.890" Score="0" Body="&lt;p&gt;Take the axles of all the wheels and extend them. Where they all intersect is the pivot point of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a front-wheel-steering car that will be somewhere in line with the rear wheels.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2018-01-04T15:18:41.890" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18718" PostTypeId="2" ParentId="18710" CreationDate="2018-01-04T16:30:23.173" Score="0" Body="&lt;p&gt;There are simple steps to solve this.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Count the number of sun gear teeth and the number of ring gear teeth if the planetary gear is set in underdrive with the carrier as the output. Add the two numbers together.&lt;/li&gt;&#xA;&lt;li&gt;Divide the resulting number by the number of teeth on the driving&#xA;member for the gear in underdrive when the carrier is the output.&#xA;The total is the gear ratio for that gear setting.&lt;/li&gt;&#xA;&lt;li&gt;Divide the number of teeth on the driving member by the sum of the&#xA;sun gear and ring gear teeth on the planetary gear for a system set in overdrive with the carrier as the input. The resulting number is the gear ratio for that transmission setting.&lt;/li&gt;&#xA;&lt;li&gt;Count the number of teeth on the driven gear and the number of teeth&#xA;on the driving gear if the planetary gear is set in underdrive with the carrier held.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Divide the number of driven gear teeth by the number of driving gear&#xA;teeth to obtain the gear ratio for this system setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read More : &lt;a href=&quot;https://woodgears.ca/gear/planetary.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://woodgears.ca/gear/planetary.html&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14063" LastEditorUserId="14063" LastEditDate="2018-01-04T16:38:18.530" LastActivityDate="2018-01-04T16:38:18.530" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18719" PostTypeId="1" CreationDate="2018-01-04T16:42:35.220" Score="2" ViewCount="126" Body="&lt;p&gt;Recently I have been trying to find out more about the process that is used to coat glasses lenses with gold. I asked about it in another forum last night and I was directed here. I initially thought that I could use the same process that NASA used to coat astronaut's helmet visors in gold, but now that is too expensive of a process. I want a thin transparent layer of gold on my glasses (lenses) basically. I have access to jewelry equipment.&lt;/p&gt;&#xA;" OwnerUserId="14399" LastActivityDate="2018-03-12T03:17:33.910" Title="Coating gold on to sunglasses lenses?" Tags="|metallurgy|metals|metal-folding|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18720" PostTypeId="2" ParentId="18702" CreationDate="2018-01-04T16:43:45.423" Score="1" Body="&lt;p&gt;I don't see any reason the same kind of motor used for launching the aircraft could not be used to decelerate it. The energy would have to be dumped somewhere, but I suspect that would not be a big problem in a ship. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the primary advantages touted for electromagnetic launching is that it will increase the &lt;strong&gt;lifespan&lt;/strong&gt; of the airframe due to reduced stress. It's probably less of a problem on landing because the aircraft generally come back lighter than when they left, having burned fuel and dumped their payloads of death and destruction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take a fully loaded 27000 kg &lt;a href=&quot;https://en.wikipedia.org/wiki/Lockheed_Martin_F-35_Lightning_II&quot; rel=&quot;nofollow noreferrer&quot;&gt;F-35B&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Shenyang_J-15&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shenyang J-15&lt;/a&gt; or landing at (say) 150 knots (~75m/s), the kinetic energy is about 75MJ. If you allow for heating aluminum by 70°C you'd need 1.2MT of aluminum (and you'd need to cool it down enough before the next landing..). Doesn't sound exceptionally difficult. &lt;/p&gt;&#xA;" OwnerUserId="1276" LastActivityDate="2018-01-04T16:43:45.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18721" PostTypeId="1" AcceptedAnswerId="18727" CreationDate="2018-01-04T16:52:38.737" Score="8" ViewCount="819" Body="&lt;p&gt;My office is afforded a prime view of a construction project where the workers are currently covering the leveled ground in a wire mesh, raking gravel across the entire mesh, and then covering that gravel layer in another layer of dirt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the function of this, engineering-wise? I know gravel in a foundation typically is used to help with water runoff, but I can't guess why the mesh is being rolled out underneath, and this is going to become the foundation of a 3-story health pavilion, not some single-family house or storage shed (the entire footprint here with gravel and mesh will have a building on top of it, not a parking deck or a parking lot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this simply a base layer of some kind, upon which concrete and rebar will be placed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oxKX8.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oxKX8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am in the United States, in Georgia.&lt;/p&gt;&#xA;" OwnerUserId="14397" LastActivityDate="2018-01-04T18:34:09.787" Title="Why is this construction site pouring gravel over a wire mesh, and then covering it with dirt?" Tags="|structural-engineering|geotechnical-engineering|foundations|soil|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18722" PostTypeId="2" ParentId="16830" CreationDate="2018-01-04T16:56:42.897" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Does the pore size limit, in any way, the surface area available for&#xA;  hydrostatic loading on the back of a retaining wall? ...&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It does not limit the surface contact area. The hydrostatic pressure and the soil load act on the wall concurrently. However, as you noted, the soil load is reduced because the effective weight of the soil is reduced due to buoyant forces. See below answer for more about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the pore-size does do is potentially control other aspects of design- namely, whether the design should consider drained or undrained conditions. A small pore size should consider undrained conditions. Undrained conditions may not need to be considered for granular soils with proper draining, depending on the assumed location of the water table.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Soil particules become suspended in water while a hydrostaic load is&#xA;  present. Does this (or any other water related factor) affect the&#xA;  overall soil load or does it remain the same?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The presence of water impacts the soil load, as does movement of the wall away from the wall (active soil pressure) or toward the soil (passive soil pressure), or nonmovement of the wall (at-rest soil pressure).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;Is the total load on the wall simply $\frac{H^2}{2}(\gamma_s K_a + \gamma_w)$ (Where $\gamma_s$ and $\gamma_w$ are the soil and water&#xA;  densities, respectively)?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the effective unit soil weight is multiplied by the lateral pressure coefficient ($K_0$ which denotes at-rest, $K_a$, or $K_p$, depending on whether the wall is moving and what direction it is moving), it can be combined with the unit weight of water to determine an equivalent fluid pressure (EFP). The effective soil unit weight $\gamma'$ is the saturated weight $\gamma_{sat}$ minus the unit weight of water, $\gamma_W$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\gamma_{EFP}=K\gamma'+\gamma_W&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and your calculation becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1}{2}H^2\gamma_{EFP}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The EFP has often been used in ASD design. However as things have moved towards LRFD design, it is less often used because the water and horizonal soil forces can have different load factors, making the EFP approach impractical for calculations. The loads are instead kept separate.&lt;/p&gt;&#xA;" OwnerUserId="88" LastActivityDate="2018-01-04T16:56:42.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18723" PostTypeId="2" ParentId="18714" CreationDate="2018-01-04T17:01:59.587" Score="1" Body="&lt;p&gt;When turning/banking the steering mechanism linkage design e.g., Ackerman, is such that axes of rotated front wheels intersect on fixed rear wheel axis line at P, so around P the entire vehicle instantaneously rotates in the plane of the ground. Operates same way for 4X4 or 2X4 driven wheels.&lt;/p&gt;&#xA;" OwnerUserId="1270" LastActivityDate="2018-01-04T17:01:59.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18724" PostTypeId="2" ParentId="18659" CreationDate="2018-01-04T17:59:34.517" Score="1" Body="&lt;p&gt;This sould probably be written as comment, as I will give only partial answer. Your angle is wrong, you are in wrong quadrant, either you just forgotten to take in consideration sign (positive or negative stress) or you mixed stress axis orientation. It would help if you provide a sketch of Mohrs circle. One starting point should be ($\sigma_x$, $\tau_{xy}$), other ($\sigma_y$, -$\tau_{xy}$). From this pic you gave I suppose your $\sigma_y$ equals 0. &#xA;&lt;a href=&quot;https://www.google.hr/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;url=http://users.ox.ac.uk/~kneabz/Stress6_ht08.pdf&amp;amp;ved=0ahUKEwjhhYPg777YAhXCh7QKHXlIDXsQFghMMAo&amp;amp;usg=AOvVaw2kXOKoyPQrAYdxbNGafu3-&quot; rel=&quot;nofollow noreferrer&quot;&gt;This link&lt;/a&gt; provides procedure and examples of constructing Mohr's circle.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="14306" LastEditDate="2018-01-05T01:38:09.630" LastActivityDate="2018-01-05T01:38:09.630" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18725" PostTypeId="2" ParentId="18695" CreationDate="2018-01-04T18:27:31.060" Score="1" Body="&lt;p&gt;The short answer is that the same equations are applicable in both cases for $I_{xc}$. For $I_{yc}$, the equation you have is applicable only with a sign change if you assume that $t$ is positive in both cases. However it is possible to have an equation in $b$ and $s$ that is valid in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long math-based answer. I am going to call the vertex of right angled vertex 0. In both cases, the final triangle &lt;em&gt;TLR&lt;/em&gt; is obtained from triangles &lt;em&gt;TOL&lt;/em&gt; and &lt;em&gt;TOR&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area, centroid, and second moment of areas about the centroid for each triangle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{cc}&#xA;  &amp;amp; area &amp;amp; centroid &amp;amp; moment \\&#xA; TOL &amp;amp; \frac{h s}{2} &amp;amp;\left\{\frac{2 s}{3},\frac{h}{3}\right\} &amp;amp; \left\{\frac{h^3 s}{36},\frac{h s^3}{36}\right\} \\&#xA; TOR   &amp;amp; \frac{h t}{2} &amp;amp; \left\{\frac{2 s+b}{3},\frac{h}{3}\right\} &amp;amp; \left\{\frac{h^3 t}{36},\frac{h t^3}{36}\right\} \\&#xA;\end{array}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will now use the parallel axis theorem to compute the moment about the centroid of TLR which is at $\left\{\frac{b+s}{3},\frac{h}{3}\right\}$. (The formula is $I=I_c+Ad^2$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new moments of &lt;em&gt;TOL&lt;/em&gt; are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left\{\frac{h^3 s}{36},\frac{h s^3}{36}\right\}+\frac{1}{2} h s&#xA;   \left\{\left(\frac{h}{3}-\frac{h}{3}\right)^2,\left(\frac{b+s}{3}-\frac{2&#xA;   s}{3}\right)^2\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which simplifies to &#xA;$$ \left\{I_{XTOL}, I_{YTOL}\right\} = \left\{\frac{h^3 s}{36},\frac{1}{36} h s \left(2 b^2-4 b s+3 s^2\right)\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly the new moments of &lt;em&gt;TOR&lt;/em&gt; are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\left\{\frac{h^3 t}{36},\frac{h&#xA;   t^3}{36}\right\}+\frac{1}{2} h t&#xA;   \left\{\left(\frac{h}{3}-\frac{h}{3}\right)^2,\left(\frac{1}{3} (b+2&#xA;   s)-\frac{b+s}{3}\right)^2\right\}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that simplifies to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left\{I_{XTOR}, I_{YTOR}\right\} = \left\{\frac{h^3 t}{36},\frac{1}{36} h t \left(2 s^2+t^2\right)\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$I_{xc}$ for TLR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1 ($t+s=b$):&#xA;$$I_{xc} = I_{XTOL}+I_{XTOR}= \frac{h^3 s}{36}+\frac{h^3 t}{36}= \frac{h^3 b}{36}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2 ($s-t=b$):&#xA;$$I_{xc} = I_{XTOL}-I_{XTOR}= \frac{h^3 s}{36}-\frac{h^3 t}{36}= \frac{h^3 b}{36}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;$I_{yc}$ for TLR&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1 ($t=b-s$):&#xA;$$I_{yc} = I_{YTOL}+I_{YTOR}= \frac{1}{36} b h \left(b^2-b s+s^2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2 ($t=s-b$):&#xA;$$I_{yc} = I_{YTOL}-I_{YTOR}= \frac{1}{36} b h \left(b^2-b s+s^2\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $I_{xc}$ you can use $\frac{h^3 b}{36}$ for both cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $I_{yc}$ you can use $ \frac{1}{36} b h \left(b^2-b s+s^2\right)$ for both cases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use the expression you have for case 1, then for case 2 you need to use $\frac{1}{36} b h \left(b^2+st\right)$ if you assume $t$ is positive, or you can use the same expression for both cases but must have the proper sign for $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used Mathematica for the rather tedious computations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JTBfo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JTBfo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2018-01-04T18:27:31.060" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18726" PostTypeId="1" CreationDate="2018-01-04T18:30:04.637" Score="6" ViewCount="690" Body="&lt;p&gt;I kind of have an intuition or at least imagination about how fillet|chamfer of inner edges|vertexes reduce stress (concentration).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But does that apply to outer edges|vertexes aswell? Maybe to a lesser degree or not at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So should one use outer fillets|chamfers for &lt;strong&gt;this&lt;/strong&gt; reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Lets take a thin hollow brick-shaped object (case or box) as an example.)&lt;/p&gt;&#xA;" OwnerUserId="14400" LastActivityDate="2018-06-28T19:07:28.997" Title="Do outer fillets|chamfers also reduce stress as inner ones do?" Tags="|mechanical-engineering|design|applied-mechanics|stresses|geometry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18727" PostTypeId="2" ParentId="18721" CreationDate="2018-01-04T18:34:09.787" Score="4" Body="&lt;p&gt;You are correct, this is the subbase that the foundations will be built on top of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gravel can be used as you say for water run off but it also allows for a level base on which to lay the concrete. In areas where the ground becomes highly saturated it can protect the concrete foundation from erosion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wire mesh is usually used for load-bearing concrete, simply to give it some extra strength and prevent cracking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dirt going on top of the subbase is the base material. This is used to help the workers get the correct grade and to keep the foundation flat. It also supports the workers and equipment during the concrete pouring.&lt;/p&gt;&#xA;" OwnerUserId="14150" LastActivityDate="2018-01-04T18:34:09.787" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18728" PostTypeId="2" ParentId="18591" CreationDate="2018-01-04T20:16:31.487" Score="0" Body="&lt;p&gt;To manufacture something like an elctronic device means you should have an interdisciplinary knowldege,You will need sensors which requires you to know instrumentation,If the sensor is electronic in nature you should know how to interface it thus you need knowledge of embedded systems,for embedded systems to work you need to program the chip,To program a chip efficently with proper resource managment then you should understand the arhitecture of chip which involves embedded and vlsi the list will go on so what I meant to say is you should first decide whether you should get a degree in either a hardware oriented or software oriented field then go on to specialize usually people with passion in this field take up course in Electrical and elcectronics if more hardware oriented or electronics and communication then take a masters in one of the fields in computer science.&lt;/p&gt;&#xA;" OwnerUserId="14401" LastActivityDate="2018-01-04T20:16:31.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18729" PostTypeId="2" ParentId="18726" CreationDate="2018-01-04T20:28:55.703" Score="12" Body="&lt;p&gt;Sorry for not taking your &quot;Thin Hollow Brick-Shaped Object&quot; as an example - I felt a standard notched test specimen would illustrate the principle more clearly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be seen that adding external radii to the test sample has no effect on the stress. In this example, cutting even more material away from the corners would make no difference to the failure of the sample, since that would always occur at the point of highest stress, i.e. the root of the notch.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/T4Wuym.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T4Wuym.png&quot; alt=&quot;Von Mises Stress in Notch Sample with/without External Radii&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited to add: in a 90-degree bend specimen, the removal of material on the external edge will actually cause an increased stress on any corresponding internal edges, since there is a reduced cross sectional area over which the stress can be spread. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/QrBuQl.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QrBuQl.png&quot; alt=&quot;Von Mises Stress in 90deg Bend Sample with/without External Radii&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are real-world examples where ommitting an external radius might cause a component to fail early, but these do not fall within the theoretical scope of your question. For example, it is desirable in injection moulded components to keep the wall sections as constant as possible so that the plastic can cool evenly. If your 'hollow brick' example had internal radii, but none externally, then the plastic would be locally thicker, cool more slowly, and this could introduce voids or residual stresses that would not otherwise have been present.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-01-04T23:23:54.747" LastActivityDate="2018-01-04T23:23:54.747" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18730" PostTypeId="2" ParentId="18580" CreationDate="2018-01-04T23:16:05.197" Score="0" Body="&lt;p&gt;For vertical movement you could use a round guide and a tight fit rubber ring inside the holder to enable movement when you want to move it but for friction to keep it in place when you want to keep it still.&#xA;For horozontal movement you could use scissor mechanism with screw which could control movement and act as a brake.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastActivityDate="2018-01-04T23:16:05.197" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18731" PostTypeId="1" CreationDate="2018-01-05T00:28:26.857" Score="0" ViewCount="133" Body="&lt;p&gt;I am looking for a mechanism that would allow me to support a fixture attached to a 3-axis motorized stage. I have attached a picture of my current design. I trying to design/incorporate some sort of hardware that will support the green fixture to travel in the z-axis as the grey stage moves during its analysis procedure. I originally tried a lab top scissor lift but that requires me to manually adjust the stage. I was thinking of a gas spring platform but wasn't sure how to incorporate it properly. Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://image.ibb.co/dKY1uw/Moving_Stage.jpg&quot; alt=&quot;fixture example&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14235" LastEditorUserId="1832" LastEditDate="2018-01-05T21:06:35.820" LastActivityDate="2023-12-26T19:03:30.383" Title="Z-axis auto adjusting fixture device" Tags="|design|mechanical|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18733" PostTypeId="2" ParentId="18714" CreationDate="2018-01-05T02:39:29.627" Score="2" Body="&lt;p&gt;Turning center of the cars with fixed rear wheels is on a line extending from rear axel inward to the point interesting with the two front wheel axel lines. Providing two tracks, one inside for rear wheel and one outside for front wheel. Only in some rough terrain vehicle which have both the rear and front wheels attached to steering system this does not apply!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This creates a larger radius of turning for the front wheels, usually in the order of 3-4 feet tracking at a position on a tangent to the rear wheel turn circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normal range of rdius of turn for midsize seddans is 16-20 feet inside dimension and 34-40 feet outside dimension!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back up same pattern applies! Unless you make the car skid by going too fast or doing stunt turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This discrepancy in radius is critical when designing turning ramps and driveways! one hase to allow for gradual increase in the width of the lane to accomodate this extra radius asthe car travels into the turn! All graphic standard enging/architectural handbooks have digrams for different residential and commercial settings!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our driveway is both turning and sloping up. Many times in hot summer days I see the intricate dark track of our cars turning on the driveway. They start from two parallel lines, then separate into four track lines with the ones arcing outsid left by front wheels then they merge back and become two lines after the ramp ends in front of the garage space.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-05T02:39:29.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18734" PostTypeId="2" ParentId="18647" CreationDate="2018-01-05T06:28:59.733" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Light-emitting_diode#Colors_and_materials&quot; rel=&quot;nofollow noreferrer&quot;&gt;LED&lt;/a&gt; is a semiconductor that was designed for a certain voltage at a certain temperature. This certain voltage is called &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Band_gap#List_of_band_gaps&quot; rel=&quot;nofollow noreferrer&quot;&gt;band gap&lt;/a&gt;&quot; and is set by properly &lt;a href=&quot;http://www.chemistry.wustl.edu/~edudev/LabTutorials/PeriodicProperties/MetalBonding/MetalBonding.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;doping&lt;/a&gt; it during manufacturing. The &quot;band gap&quot; is the voltage (think pressure) required for the electrons to hop the &quot;gap&quot;. The larger the band gap, the higher the voltage to cross, the higher the energy of light emitted, and the higher the frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HdiGc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HdiGc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a semiconductor is cooled its resistance increases as well its band gap. This larger band gap requires a larger corresponding voltage for electrons to hop. This increases the frequency (blue shift) of the light emitted, like the orange to green shift from liquid nitrogen demonstrated in this &lt;a href=&quot;https://www.youtube.com/watch?v=4qgPhNHJPB8&quot; rel=&quot;nofollow noreferrer&quot;&gt;youtube video&lt;/a&gt;. Notice that the measured voltage drop across the LED also increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, solar photo-voltaic cells are also semiconductors and increase output voltage as the temperature gets colder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the voltage to frequency relationship doesn't always make sense though. I am not in the LED manufacturing industry, but other things are going on physically inside the LED when the opposite occurs. For example, this &lt;a href=&quot;https://youtu.be/hxBJ9ToVWmA?t=34&quot; rel=&quot;nofollow noreferrer&quot;&gt;youtube video&lt;/a&gt; shows a yellow LED change to green and green to yellow when submersed in liquid nitrogen. Yellow to green makes sense as we previously discussed. Green to yellow doesn't though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the doping chemical in that specific green led becomes more conductive at low temperatures and actually reduces the band gap. I am not a material scientist, but something similar to that has to occur, because the band gap voltage sets the emitted light frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also see color changes by increasing the voltage on an LED beyond its design voltage as shown in this &lt;a href=&quot;https://www.youtube.com/watch?v=vCY_HS_RCqc&quot; rel=&quot;nofollow noreferrer&quot;&gt;youtube video&lt;/a&gt;. The life of the LED is significantly reduced, but its a simple experiment. This color shift is most likely due to over heating the semiconductor inside its plastic housing. At higher temperatures, the band gap decreases and the corresponding frequency decreases (red shift).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a good &lt;a href=&quot;https://www.digikey.com/en/articles/techzone/2013/may/thermal-effects-on-white-led-chromaticity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Digikey article on chromatic shift in LEDs due to temperature&lt;/a&gt;. Mid way down the page it better explains the physical nature of the band gap:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the temperature of the LED rises, the atoms in the lattice vibrate more, which slightly increases the lattice constant. This in turn decreases the band gap which increases the wavelength of the emitted photon.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-01-05T06:28:59.733" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18736" PostTypeId="1" CreationDate="2018-01-05T08:57:23.673" Score="2" ViewCount="712" Body="&lt;p&gt;[![enter image description here][1]][1]Is it possible for a hypoid bevel gear like the one pictured below to have a 1:1 ratio? If 1:1 isn’t possible, any idea what the smallest ratio could be in order to output the least amount of gear reduction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/fVLuK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gqmou.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="7869" LastEditorUserId="7869" LastEditDate="2018-01-06T05:40:12.087" LastActivityDate="2018-01-06T05:40:12.087" Title="Can a hypoid bevel gear set have a 1:1 ratio?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18737" PostTypeId="1" CreationDate="2018-01-05T10:02:11.990" Score="5" ViewCount="364" Body="&lt;p&gt;I've gone through a control theory course, and after looking back at what we covered and how the subject was taught to us, I still have quite a few questions (which I unfortunately did not ask in class). As such, I thought I'd ask a few of them here, as separate posts, in order to clear up a few points that I don't understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we linearize our dynamic equations/transfer functions when doing classical and state-space control? Is this only to allow us to be able to analyse stability and do initial specification design (rise time, overshoot etc.), where we know that the real nonlinear system won't actually act exactly in the above manner? I would have thought that since we'll generally be dealing with nonlinear systems in the real-world, it would be better to test our control implementation on a model that more closely resembles our real system.&lt;/p&gt;&#xA;" OwnerUserId="9755" LastActivityDate="2018-05-09T07:31:02.467" Title="Linearization of dynamic equations/transfer function in control theory" Tags="|control-engineering|control-theory|dynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18738" PostTypeId="2" ParentId="18736" CreationDate="2018-01-05T11:08:07.743" Score="1" Body="&lt;p&gt;As far as I know, there isn't any mathematical limitation that would rule out using Hypoid gears for a 1:1 transmission (known as Mitre /Miter gears). You won't find any commercially, however, since they are less mechanically efficient than spiral gears (due to sliding along the teeth). Hypoid gears are favoured in some situations for their ability to achieve a high reduction in a small space through behaving much like a worm-gear (see the high-ratio hypoid set below). For a 1:1 ratio, there is no benefit in selecting a Hypoid form, and so Spiral or Straight is the industry preference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hDuibl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hDuibl.jpg&quot; alt=&quot;Spiral Miter Gears&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spiral Miter Gears&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wLcYXl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wLcYXl.jpg&quot; alt=&quot;High Reduction Hypoid Gears&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High Reduction Hypoid Gears&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-01-05T11:08:07.743" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18739" PostTypeId="2" ParentId="18737" CreationDate="2018-01-05T12:40:24.373" Score="1" Body="&lt;p&gt;Majority of systems are nonlinear. However, many of them can be linearized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why do we linearize them? Nonlinear systems are too complicated to analyze. Sometimes impossible. Calculating them also requires a huge computation power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the solution? We linearize the system. We approximate. But a complicated problem is replaced with a much simpler one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stability is a part of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linearization helps performing many of calculations offline. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In optimal control, Linear Quadratic Regulator &lt;a href=&quot;https://ocw.mit.edu/courses/mechanical-engineering/2-154-maneuvering-and-control-of-surface-and-underwater-vehicles-13-49-fall-2004/lecture-notes/lec19.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1]&lt;/a&gt; is solved offline thanks to the linearization.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In my own field, Model Predictive Control (MPC) &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0957417417306000&quot; rel=&quot;nofollow noreferrer&quot;&gt;[2]&lt;/a&gt; predicts the future system dynamics to optimize the control input. If you set e.g. prediction horizon $N_p=100$ and control horizon $N_c=3$, it means that your system should calculate 100 samples ahead during its optimization. But, as the system is linear, a part of calculations is performed offline.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can achieve 1% higher performance by calculation of nonlinear system. But as far as a system works properly and it is reliable, the industry is not interested in the headache of complexities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complicated system is also hard no identify and unreliable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, there are a lot of noises and disturbances which are much higher than your improvement so it does not often worth caring unless really required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in advance linearization, have a look at &lt;a href=&quot;https://au.mathworks.com/help/mpc/ug/gain-scheduling-mpc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gain-Scheduled MPC&lt;/a&gt; and also &lt;a href=&quot;http://au.mathworks.com/help/mpc/adaptive-mpc-design.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adaptive MPC Design&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14288" LastEditorUserId="14288" LastEditDate="2018-05-09T07:31:02.467" LastActivityDate="2018-05-09T07:31:02.467" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="18740" PostTypeId="1" CreationDate="2018-01-05T13:12:44.680" Score="4" ViewCount="122" Body="&lt;p&gt;Can weathering Steel be left unpainted? Are there any environments when the properties of weathering steel will prevent the steel from performing as designed, i.e. surface rust forming to encapsulate the remaining steel core.&lt;/p&gt;&#xA;" OwnerUserId="14414" LastEditorUserId="33" LastEditDate="2018-04-06T02:31:43.107" LastActivityDate="2018-04-06T02:31:43.107" Title="Weathering steel?" Tags="|structural-engineering|materials|steel|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18742" PostTypeId="1" CreationDate="2018-01-05T14:45:50.273" Score="3" ViewCount="395" Body="&lt;p&gt;Consider the wheel of a bicycle . &lt;a href=&quot;https://i.stack.imgur.com/DKrph.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DKrph.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now from what I understand, all the spokes that are above the upper half of the wheel happen to be in tension as the hub of the wheel hangs on these. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don’t understand, however, is the loading (tensile or none) experienced by the spokes that are in the lower half of the wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is all the weight of the bicycle  supported by the rim of the wheel, as that is where the Normal Force acts at vertically upwards ? Are rims then designed to undertake compressive loading ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if so, then why ? Why not instead use tubes or rods instead of spokes which can handle both tensile and compressive loading ?&lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2018-01-05T17:44:09.537" Title="Does the Rim of a wheel handle all the compressive force?" Tags="|mechanical-engineering|design|automotive-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18743" PostTypeId="2" ParentId="18742" CreationDate="2018-01-05T16:03:26.967" Score="5" Body="&lt;p&gt;I remember several years ago watching a documentary about building Wembley Stadium, that had a good segment on wooden vs wire wheels: &lt;a href=&quot;https://youtu.be/aqOzcKdvFDI?t=2m32s&quot; rel=&quot;noreferrer&quot;&gt;https://youtu.be/aqOzcKdvFDI?t=2m32s&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise: The spokes are pretensioned, to the point that under load, or even impact, the lower spokes never go slack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further to this, the spokes pulling in on the rim mean that the rim itself is in pulled in all the way around, which makes it significantly stronger. If you squash a circle you will end up with an elipse shape. The tensioned spokes resist this by pulling back on the part of the rim that is now further away from the axle than usual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fCcbum.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fCcbum.jpg&quot; alt=&quot;Circle vs elipse with equal circumference&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-01-05T16:03:26.967" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18744" PostTypeId="2" ParentId="18742" CreationDate="2018-01-05T17:34:03.113" Score="1" Body="&lt;p&gt;Let's assume we are looking at a wheel of a bicycle turning clockwise and we refer to positions on this wheel by the clock hours: 12,1,2,3...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper spokes gradually get loaded in tension as the wheel turns and the leftmost spoke passes the 9  position and moves to 10. It gets loaded by a factor of sin(a), a being the angle clockwise from horizontal or 9 position. Gradually this tension increases till at vertical, or 12 it is maximum. So the spokes above horizon contribute to supporting the load proportional to their angle. As mentioned in other answers there is also residual load of pretention in spokes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's assume that you hit a bump or a small object on your way. The rim flexes and wants to deflect inside into the circle. This deflection will create bending moment in the rim just like a horizontal beam and will be resisted by the well known equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;m= sigma.I/c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with sigma being allowable stress of rim, I it's bending  second moment of area and c half of thickness of rim, approximately.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-01-05T17:44:09.537" LastActivityDate="2018-01-05T17:44:09.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18746" PostTypeId="2" ParentId="18695" CreationDate="2018-01-05T19:15:59.273" Score="0" Body="&lt;p&gt;Slanted triangle can be obtained by cutting off piece of right triangle, and you can calculate moment ot inertia or second moment of area by subtracting smaller right triangle from the larger one using Steiner's rule. You can express all dimensions considering your base and edges using trigonometric functions.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastEditorUserId="13748" LastEditDate="2018-01-05T19:31:47.177" LastActivityDate="2018-01-05T19:31:47.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18747" PostTypeId="2" ParentId="18702" CreationDate="2018-01-05T19:41:10.437" Score="0" Body="&lt;p&gt;I think some automated means of grabbing the tailhook and using the existing cable system or new arresting system could be a viable idea. Dissipating the energy isn't that tricky, and no saying it has to be dissipated by a motor. Remember that the aircraft isnt always perfectly lined up so the system would need to account for a fairly large amount of miss alignment. Also the ship attitude and winds are not constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main issue I see is risk vs reward. How many aircraft have been lost because of the current system. How many aircraft could be lost if this system has glitches during the first year. How many aircraft will be lost with the new system in the long term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The military isn't concerned with cost or waste, only reliability. Its easier for people to see the reliability in simple springs and cables than in servos and more complicated systems. If you could come up with a system that could leave the existing system in place, that is a much easier pitch. Some sort of computer controlled secondary hooking system, that requires pilots to still practice their skills on the old system, but provides a backup guarantee to get them landed during a combat situation.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-01-05T19:41:10.437" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18748" PostTypeId="2" ParentId="18740" CreationDate="2018-01-05T20:09:32.920" Score="1" Body="&lt;p&gt;Anywhere within about 5 miles of a sea coast , they should be coated to resist corrosion . Otherwise they are pretty good. They do slowly rust and turn reddish brown , I have heard of them being painted because important people don't like the color , but otherwise it is generally not necessary to paint them . They have tiny amounts of P , Cu , Cr , and a few other elements depending on the manufacturer.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-01-05T20:09:32.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18751" PostTypeId="2" ParentId="17771" CreationDate="2018-01-06T02:39:25.100" Score="4" Body="&lt;p&gt;They are called blast valves. You need a delay loop in the pipe so the trigger can get ahead of the shock wave.  Our interest was in closing off ventilation air conduits before the nuclear weapon's overpressure could penetrate the site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the pipe loops over the Launch Control Equipment Building? Those are the delay loops for ventilation air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l3MFI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l3MFI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RVSPk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RVSPk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.dtic.mil/get-tr-doc/pdf?AD=AD0636321&quot; rel=&quot;nofollow noreferrer&quot;&gt;BLAST-CLOSURE VALVES&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-01-06T03:08:11.650" LastActivityDate="2018-01-06T03:08:11.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18755" PostTypeId="1" CreationDate="2018-01-06T09:48:19.110" Score="1" ViewCount="4400" Body="&lt;p&gt;Wonder if anyone would know what is the average thermal emissivity value of air relative to blackbody at near room temperatures? I am looking for that magical number between 0 and 1. Looked everywhere but couldn't find the value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to enumerate thermal losses when heating air inside an envelope made of Polyethylene with very high transmissivity and hence low emissivity. I am thinking that actual radiative thermal losses of air itself might play an important role with regard to the temperature of the heated air. I am completely neglecting convective and conductive heat loss at this stage. (I understand that these are actually major contributors to the thermal balance.)&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2018-01-06T09:48:19.110" Title="What is the gray-body emissivity value of air" Tags="|heat-transfer|heating-systems|thermal-conduction|thermal-insulation|radiation|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18756" PostTypeId="2" ParentId="18683" CreationDate="2018-01-06T23:30:43.397" Score="2" Body="&lt;p&gt;I believe the aspect that you are missing with your intuition, is efficiency. For a given impeller diameter, there will be a specific flow rate through the pump at which it is able to operate at maximum efficiency. Have a look at the standard graph below. You can see how despite the Head (delta P for you) dropping off at higher flow rates, and the Power keeps rising to compensate for the falling efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yMf7Tl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yMf7Tl.jpg&quot; alt=&quot;Generic Pump Efficiency Curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-01-06T23:30:43.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18757" PostTypeId="2" ParentId="18677" CreationDate="2018-01-07T01:00:33.990" Score="1" Body="&lt;p&gt;Assuming you don't have solar panels on your heating array, it would be more efficient to heat the load directly as you're only moving the heat once, so you wouldn't lose as much in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you have solar panels, these work more efficiently at lower temperatures.  Depending on your climate, you might be able to move more heat from the solar array into the heat pump evaporator than you would be able to move indoors.  This could mean you are able to cool the panels a little more, making them generate slightly more power.  (You'd have to check this though, if you're getting down to 3°C you probably don't need any panel cooling!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't say where the heat pump evaporator is located, but in some places it is located underground, below the frost line.  If this is the case, then pumping heat from the solar array to the underground evaporator will also allow some level of heat storage by heating up the ground, which will then be accessible to the heat pump overnight, improving efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will likely also be able to get away with a simpler system if you move heat from the solar array to the heat pump evaporator, as you won't have to worry about maintaining a comfortable temperature and can just pump everything.  The heat pump won't complain if it gets too warm, unlike the indoor occupants.&lt;/p&gt;&#xA;" OwnerUserId="14429" LastActivityDate="2018-01-07T01:00:33.990" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18758" PostTypeId="2" ParentId="18683" CreationDate="2018-01-07T01:14:13.110" Score="1" Body="&lt;p&gt;This is entirely a design choice. If you imagine a positive displacement pump such as a piston pump running at constant speed, then the work done is basically a function of pressure difference, since the flow rate is constant. The greater the pressure, the more work required. The opposite extreme would be a pump like the squirrel cage fan in your house's heating system. If you apply enough back pressure, the flow goes to zero. Inside the scroll, the air is just spinning around with the fan blades. There is no torque and no work required to just keep the system running, doing nothing. As you begin to reduce back pressure, the fan begins to move air. Now there is a torque on the blades because the incoming air mass has to be accelerated to scroll perimeter speed. The loaded motor will slow down a bit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Axial pumps can be designed to be very stiff, having a big power rise in the face of backpressure rise, or soft, having a small power rise in the face of increasing back pressure. But if the flow really gets choked down, the power drops in all cases. These devices would be set up to top out below normal diastolic pressure, so they would be choked and just free spinning during the ventricular contraction. What the power response looks like as flow increases from stall is at the designers discretion.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-01-07T03:06:01.943" LastActivityDate="2018-01-07T03:06:01.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18759" PostTypeId="1" AcceptedAnswerId="18797" CreationDate="2018-01-07T12:55:36.193" Score="4" ViewCount="468" Body="&lt;p&gt;I am currently reading “Fundamentals Of Vehicle Dynamics” by Thomas D. Gillespie. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PsY1r.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PsY1r.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author mentions how the asymmetric pressure distribution across the length of the contact patch of a tire while rolling results in Rolling resistance. This, then inevitably creates a moment about the axis of the wheel which resists its movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author states, however, that this asymmetric pressure distribution only occurs when a tire is rolling. There isn’t any asymmetry when it’s in a dead-stop position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know now is the reason for this asymmetric pressure distribution. What causes it? How does Rolling generate asymmetry?&lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2018-01-09T11:39:59.187" Title="What causes Rolling Resistance?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18760" PostTypeId="1" CreationDate="2018-01-07T13:26:45.603" Score="2" ViewCount="64" Body="&lt;p&gt;I would like to know what methods are usually used to strengthen &lt;strong&gt;existing&lt;/strong&gt; masonry vaults and domes, especially the ones called Catalan Vaults, and how to design them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="413" LastEditorUserId="413" LastEditDate="2018-01-07T19:44:48.697" LastActivityDate="2018-04-09T05:06:57.457" Title="How to strengthen existing masonry vaults and domes?" Tags="|strength|" AnswerCount="2" CommentCount="4" ClosedDate="2018-04-14T03:04:19.543" ContentLicense="CC BY-SA 3.0" />
  <row Id="18762" PostTypeId="1" CreationDate="2018-01-07T14:25:14.957" Score="2" ViewCount="4371" Body="&lt;p&gt;I am currently reading “Fundamentals Of Vehicle Dynamics” by Thomas D. Gillespie. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HgTFF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HgTFF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author states how tires generate the required frictional forces for movement by 2 methods :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Hysteresis &#xA;2)Adhesion &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s a no-brainier in understanding how adhesion can help generate friction. What I don’t understand however is how Hysteresis can help produce the same ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the mechanism using which tires generate forces by stretching and un-stretching ? How does this work?&lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2021-06-10T16:17:50.990" Title="How does Hysteresis in tires helps create friction?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18763" PostTypeId="1" AcceptedAnswerId="18799" CreationDate="2018-01-07T15:02:01.907" Score="0" ViewCount="3189" Body="&lt;p&gt;I have been having a debate with a coworker  regarding heat exchanger fluid temperatures and while I thought I firmly grasped the principles, I was not arguing my point very well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking at an air to water coil for an air handler. I was selecting the coil based on a leaving air temp of ~60F (it's an industrial space with a lot of ventilation). I am using condensing boilers so I sized for 180 leaving water temperature with 140 return on the coldest day which gives a 40 degree temperature delta. &#xA;On the air side, I am bringing in 5 degree air (EAT) at a rate of 16000 CFM on one of the air handlers. My coworker wants to lower the temperature to 140 supply and 100 degree return, his reason being that it would still be a 40 degree temperature delta on the water side so the thermal output would be the same. I believe the hotter supply temperature will result in more rapid heat transfer but I didn't really have any good argument against his logic. He was simply relating the two rules of thumb equations for air and water: [Btu/hr]=1.08*CFM*deltaT and [Btu/hr]=500*gpm*deltaT to prove his point. &#xA;My argument was that this does not account for the heat transfer taking place from one side of the coil to the other, but I was not able to adequately illustrate my side to the point that he would agree. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, am I even right about operating the system at 180 degrees being a more efficient solution, and, if so, how do I fit the heat exchanger heat transfer into the system of equations?&lt;/p&gt;&#xA;" OwnerUserId="9770" LastEditorUserId="9770" LastEditDate="2018-01-08T20:21:07.320" LastActivityDate="2019-06-06T14:43:08.280" Title="Heat Exchanger water to air temperatures" Tags="|thermodynamics|heat-transfer|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18764" PostTypeId="1" AcceptedAnswerId="18856" CreationDate="2018-01-07T17:02:44.080" Score="2" ViewCount="246" Body="&lt;p&gt;Most large cities these days have some sort of metropolitan commuter train system. What are the reasons for the commuter trains, why couldn't we instead of having train tracks why not clear a large distance of straight road and have a vehicle, such as a large bus riding on it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the speed of trains (in my city they go at an average of around 70 km/h) surely the speed alone is not a factor since it would probably be feasible to have some sort of large vehicle achieve this speed. Trains also require large infrastructure such as tracks and power lines. On the other hand, such large vehicles would probably be safer on tracks instead of roads, and a train on tracks would probably be easier to automatize than a vehicle with wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of a lot of reasons for and against trains over wheeled vehicles, but what are the real reasons why commuter trains are so popular a choice for cities?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastEditorUserId="90" LastEditDate="2018-01-08T00:49:25.393" LastActivityDate="2018-01-11T20:41:23.173" Title="Advantages of a commuter train instead of a vehicle travelling in a straight road" Tags="|transportation|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18765" PostTypeId="1" CreationDate="2018-01-07T17:20:10.967" Score="2" ViewCount="1186" Body="&lt;p&gt;I am shopping for a spring on McMaster and there is a quantity listed as &quot;rate constant&quot;.  Is this the same thing as spring constant?  If so, what are the units of this value? (lbs/in, lbs/ft, other)?&lt;/p&gt;&#xA;" OwnerUserId="14431" LastActivityDate="2018-01-07T20:08:32.683" Title="Are spring constant and rate constant the same thing?" Tags="|springs|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18766" PostTypeId="2" ParentId="18765" CreationDate="2018-01-07T19:24:56.400" Score="3" Body="&lt;p&gt;Spring rate can be specified either for a whole spring or per unit length which is often more convenient if you are buying spring stock to cut to length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant of a whole spring is force per unit length of extension the rate constant will be F/L/L so force. This isn't particularly physically meaningful but is convenient if you are buying spring stock in bulk lengths. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So say you buy spring stock with a rate constant of 106 lb then if you make a 6&quot; spring form that then the spring constant for that spring is 106lb/6&quot; ie 17.66lb/inch ie it exerts a return force of 17.7 pounds for every inch it is compressed. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-01-07T20:08:32.683" LastActivityDate="2018-01-07T20:08:32.683" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18767" PostTypeId="2" ParentId="18760" CreationDate="2018-01-07T19:36:52.853" Score="0" Body="&lt;p&gt;The design and construction of Vaults are mostly done by the experienced masons for these reasons!&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Inconsistency in the specifications of bricks, sand, lime, gypsum and even the water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The adhesive strength of the mortar they use and the time it takes for initial setting and final hardening.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The permeability of the bricks and weight change when they are wet.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The arc of the vault has the optimal camber to meet all these conditions and hold itself during the time the mason lays them and cure into a strong structure. And its pattern of joints is meant to provide maximum ease during construction by dividing the wide spans into smaller ones and also later offer more strength to resist gravity and lateral loads.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Many masons by experience know what pattern and which mixture to use depending on the climate and location of the job-site. That's is the reason for finding typical patterns and graceful architectural themes spreading through a geographical area. and then changing gradually from one city to another one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask some of the masons currently working on the same area and compare there methods to other parts you can glean information that you need.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-07T19:36:52.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18768" PostTypeId="2" ParentId="18762" CreationDate="2018-01-07T21:43:32.943" Score="2" Body="&lt;p&gt;I'm guessing that you are confusing the terms &lt;em&gt;friction coupling&lt;/em&gt; with &lt;em&gt;traction&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I pull a trailer over a flat, level surface, most of the resistance is due to the hysteresis in the contact patch. The adhesion prevents the tire from deforming and then rebounding and recovering it's energy. Pulling a trailer across wet ice is easier than pulling it across sanded concrete. This is the coupling between deformation forces and the adhesion which fights against, say, lateral stretching and return of the squashed tire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, just realize that adhesion and deformation interact with each other and can't usefully be separated in the real world. They are strongly coupled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The coupling also detracts from traction since some of the available adhesion is now expended on opposed forces caused by the deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a convenient PDF - see page# 3. &lt;a href=&quot;http://www.mchenrysoftware.com/forum/UMRollingResistance.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rolling Resistance&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-01-07T21:43:32.943" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18769" PostTypeId="1" AcceptedAnswerId="18781" CreationDate="2018-01-07T22:23:57.260" Score="2" ViewCount="216" Body="&lt;p&gt;I'm building a machine that operates over 20 orders of magnitude with logarithmic process characteristics. I have tried regular PID and also all sorts of variations of PID and did not get good results. What other alternatives to PID I have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start with 21% oxygen in a closed vessel (at high temperature) and pump it out ion by ion down to log10(pO2)=-20 of original amount. I have sensitive enough sensor, pump and power supply for it. This part all works great if I set pump currents by hand or babysit the process. To go up in concentration I just lower the pump current and leakage takes care of adding oxygen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as you see, in context of trying to automate the control with PID, what works well at one region is completely unsuitable at another region. It does not help that the full loop of measurements takes about 10 seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have limited the current used to 500mA, and at worst conditions the current needed to cope with the leakage is 700µA, so the range for the current is not so huge. Nor is the range for sensor voltage, which is about 100mV to -1V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe I could control pump voltage instead of pump current. Current results directly to change in amount of oxygen (which is logarithmic). The resistance of the pump depends on temperature, and most of all the oxygen gradient, and would yield much more linear and automatically regulated response.&lt;/p&gt;&#xA;" OwnerUserId="14435" LastEditorUserId="14435" LastEditDate="2018-02-07T14:29:20.863" LastActivityDate="2018-02-07T14:29:20.863" Title="Tricks for controlling process with logaritmich characteristics over 20 orders of magnitude" Tags="|pid-control|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18770" PostTypeId="2" ParentId="12823" CreationDate="2018-01-07T22:38:54.503" Score="1" Body="&lt;p&gt;The proof is, as should be expected, correct. But it can be made clearer if broken down in additional steps.&lt;/p&gt;&#xA;&lt;p&gt;Consider, again, a small particle of the deformable body. The total work done by the virtual forces, &lt;span class=&quot;math-container&quot;&gt;$\text{d}W_t,$&lt;/span&gt;  system is given by the equality &lt;span class=&quot;math-container&quot;&gt;$$\text{d}W_t=\text{d}W_r+\text{d}W_d,$$&lt;/span&gt;where &lt;span class=&quot;math-container&quot;&gt;$\text{d}W_r$&lt;/span&gt; is the work done due to the displacement of the particle as a rigid body, and &lt;span class=&quot;math-container&quot;&gt;$\text{d}W_d$&lt;/span&gt; is the work done due to the displacement of the points on the boundaries of the particle as they deform. As said before, &lt;span class=&quot;math-container&quot;&gt;$\text{d}W_r=0$&lt;/span&gt;, and so &lt;span class=&quot;math-container&quot;&gt;$$\text{d}W_t=\text{d}W_d.$$&lt;/span&gt;&#xA;Integration of the last equation over the entire volume gives &lt;span class=&quot;math-container&quot;&gt;$$W_t=W_d.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;To evaluate &lt;span class=&quot;math-container&quot;&gt;$W_t$&lt;/span&gt; let's consider &lt;em&gt;one&lt;/em&gt; of the internal boundaries of the particle in question (one should imagine the particle as having a cubic or square shape). Call it boundary &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;. The total work done at such boundary by the internal virtual stresses, &lt;span class=&quot;math-container&quot;&gt;$\text{d}W_t^B,$&lt;/span&gt; is given by the equality  &lt;span class=&quot;math-container&quot;&gt;$$\text{d}W_t^B=\text{d}W_r^B+\text{d}W_d^B,$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\text{d}W_r^B$&lt;/span&gt; is the work done by the internal virtual stresses due to the displacement of the points of boundary &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; as the particle undergoes a displacement as a rigid body, and &lt;span class=&quot;math-container&quot;&gt;$\text{d}W_d^B$&lt;/span&gt; is the work done by the internal virtual stresses as boundary &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; deforms. It should be noted that &lt;span class=&quot;math-container&quot;&gt;$\text{d}W_r^B$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\text{d}W_d^B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\text{d}W_r^B+\text{d}W_d^B$&lt;/span&gt; are generally not equal to zero.&lt;/p&gt;&#xA;&lt;p&gt;Now, for every internal boundary of a particle there is an adjoining particle that has a common boundary to the particle initially considered, and therefore these adjacent boundaries are displaced exactly the same amount. Since the forces acting on the two adjacent internal boundaries are numerically equal but opposite in direction, the total virtual work done on the pair of adjoining internal boundaries is zero. Hence, since all internal boundaries occur in pairs, there is no net virtual work done by the forces on all the internal boundaries. Hence, since all internal boundaries occur in pairs, there is no net virtual work done by the forces on all the internal boundaries. &lt;span class=&quot;math-container&quot;&gt;$W_t$&lt;/span&gt; therefore consists only of the work done by the external forces on the external boundaries.&lt;/p&gt;&#xA;&lt;p&gt;So, what should be noted in the proof is that it is the &lt;em&gt;&lt;strong&gt;total&lt;/strong&gt; work done at boundary &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; at the original particle plus the &lt;strong&gt;total&lt;/strong&gt; work done at boundary &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; of the adjoining particle that adds up to zero.&lt;/em&gt; If we were to consider the sum &lt;span class=&quot;math-container&quot;&gt;$\text{d}W_d^B + \text{{d}W_d^B}'$&lt;/span&gt; (where &lt;span class=&quot;math-container&quot;&gt;$\text{{d}W_d^B}'$&lt;/span&gt; is the work done by the virtual stresses on the adjoining particle as its boundary &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; deforms), then, for the reasons pointed by Brian Moths in his answer, such sum would not be zero.&lt;/p&gt;&#xA;" OwnerUserId="4421" LastEditorUserId="4421" LastEditDate="2022-04-22T22:41:21.273" LastActivityDate="2022-04-22T22:41:21.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18772" PostTypeId="1" CreationDate="2018-01-08T04:11:04.893" Score="-1" ViewCount="67" Body="&lt;p&gt;Ok so. When you look in videos(or in real life) with the railroad signals, when a train passes over it, it takes around 3 to 5 seconds for it to actually change to the correct aspect. What is up with that?&lt;/p&gt;&#xA;" OwnerUserId="9893" LastActivityDate="2018-01-10T11:47:31.530" Title="Why is there a delay with railroad signals?" Tags="|rail|" AnswerCount="1" CommentCount="3" ClosedDate="2018-01-10T13:26:12.693" ContentLicense="CC BY-SA 3.0" />
  <row Id="18773" PostTypeId="1" AcceptedAnswerId="18791" CreationDate="2018-01-08T04:56:52.193" Score="3" ViewCount="294" Body="&lt;p&gt;I've come across this phenomenon several times and never found any real reason for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I consider a cylindrical object like a rod, I need to calculate several things. One of those is the stress tensor in cylindrical coordinates represented by $r$, $\theta$, and $z$ directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in many books and in my lecture notes there's always this one line that says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{\partial\bullet}{\partial\theta}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that when compressing a steel rod (assuming there is little to no friction at the contact points) the steel rod should get compressed in the $z$-direction and expand in the radial direction. &#xA;&lt;strong&gt;Why is there no stress in the circumferential direction?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've looked in many places and I hope that I didn't accidentally miss the other SE question with respect to this specific topic.&lt;/p&gt;&#xA;" OwnerUserId="14437" LastEditorUserId="12754" LastEditDate="2018-01-10T01:38:54.413" LastActivityDate="2018-01-10T03:22:52.030" Title="Why is the partial derivative of stress in the circumferential direction equal to zero in axial symmetric bodies?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18774" PostTypeId="1" CreationDate="2018-01-08T06:17:46.080" Score="2" ViewCount="71" Body="&lt;p&gt;From my understanding, the more surface area the flow passes, the more drag happens due to skin friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this is the new Ford GT, which has a converging cockpit and separated side body panels. I did some research and the answer I found is related to decreasing the frontal area of the vehicle, but when I compare it with the increase of the surface area it's almost double the area of similar cars.&lt;img src=&quot;https://m.ford.com/campaignlibs/content/dam/ford_com/en_us/gtreveal/Recut_120215/Overlay_1250PX/GT_TertreRouge_P02_19390_v03_final151112-tablet.jpg&quot; alt=&quot;the new ford gt convergent cockpit and separated taillight panels&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image source: ford.com&lt;/p&gt;&#xA;" OwnerUserId="14292" LastEditorUserId="90" LastEditDate="2018-01-08T07:35:22.547" LastActivityDate="2018-02-07T08:21:19.897" Title="Why do some cars have flow channels in their bodies to decrease drag?" Tags="|automotive-engineering|aerodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18775" PostTypeId="2" ParentId="18774" CreationDate="2018-01-08T07:07:20.300" Score="2" Body="&lt;p&gt;I think  those tunnels are there to reduce the amount of flow separation drag occurring at the blunted end of the body, while still maintaining the &quot;fastback&quot; roof line of the original car. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-08T07:07:20.300" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18776" PostTypeId="2" ParentId="2403" CreationDate="2018-01-08T07:12:57.203" Score="1" Body="&lt;p&gt;buy a propane lighter refiller cartridge, these are about 1&quot; in diameter and 5&quot; long. make the cartridge part of the fuselage. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-08T07:12:57.203" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18779" PostTypeId="2" ParentId="18216" CreationDate="2018-01-08T11:14:17.540" Score="6" Body="&lt;p&gt;I've done some research on New Jersey barriers a few years ago and came across &lt;a href=&quot;https://www.fhwa.dot.gov/publications/publicroads/00marapr/concrete.cfm&quot; rel=&quot;noreferrer&quot;&gt;this article&lt;/a&gt; by the FHWA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic operating principal of the barrier is to dissipate some energy and redirect and reorient the crash to be parallel to the barrier by lifting the front tyres to reduce friction between the tyres and asphalt in order to assist the redirection. Many configurations and tests were done and many improvements were made over the years, as also explaned in the article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mentalfloss.com/article/51603/why-are-road-partitions-called-jersey-barriers&quot; rel=&quot;noreferrer&quot;&gt;This article&lt;/a&gt; from Mental Floss is a bit less technical and explains the history a bit more colloquially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for bridges specifically, the same safety principles will apply. Many states have their own bridge design manuals which will specify the barrier based on the bridge location and use case and many of the higher safety barriers are designed similar to the jersey barriers with some decorative twists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found New Jersey's &lt;a href=&quot;http://www.state.nj.us/transportation/eng/documents/BSDM/&quot; rel=&quot;noreferrer&quot;&gt;bridge design manual&lt;/a&gt; and section 23 sheds some light on the barrier selection criteria, but many states' manuals are similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oldest standard I could find pertaining to barrier selection based on traffic use case is &lt;a href=&quot;http://onlinepubs.trb.org/Onlinepubs/nchrp/nchrp_rpt_230.pdf&quot; rel=&quot;noreferrer&quot;&gt;NCHRP Report 230&lt;/a&gt; from 1981. This was subsequently replaced by NCHRP 350 which seems to be in the process of being replaced by MASH (Manual for Assessing Safety Hardware); if not already.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Older guidelines as early as 1962 are mentioned &lt;a href=&quot;http://www.roadsystems.com/mash-and-nchrp-350/&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; but I couldn't find them to see what they specify and how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some (modern) bridge rail guidelines from &lt;a href=&quot;http://www.virginiadot.org/business/resources/bridge/Manuals/Part2/Chapter25.pdf&quot; rel=&quot;noreferrer&quot;&gt;Virginia&lt;/a&gt; and &lt;a href=&quot;http://www.state.nj.us/transportation/eng/structeval/pdf/bridgerail.pdf&quot; rel=&quot;noreferrer&quot;&gt;New Jersey&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this sheds some light on the topic.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastEditorUserId="14440" LastEditDate="2018-01-08T13:12:15.043" LastActivityDate="2018-01-08T13:12:15.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18780" PostTypeId="1" AcceptedAnswerId="18795" CreationDate="2018-01-08T13:25:25.440" Score="4" ViewCount="72" Body="&lt;p&gt;I am used to the metric system where everything is simple and consistent. I can do quick conversions between metric and imperial in my head, but especially when it comes to volumetric measures, I cannot be sure if British or US Customary units are used. Here I refer to units such as gallons, pints, etc. not cubic-linear units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions arising from this confusion is how to know which unit is being used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In technical publications, will the author or publisher note the applicable units somewhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For everyday use, is there a convention that says U.S. gallons or otherwise must be specified, or is it up to the whims of the manufacturer? If not, how can I reduce my chances of making an error if I cannot ask or measure?&lt;/p&gt;&#xA;" OwnerUserId="14440" LastActivityDate="2018-01-09T10:32:54.890" Title="Is there a way to distinguish between U.S. Customary and British Imperial volumetric units?" Tags="|measurements|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18781" PostTypeId="2" ParentId="18769" CreationDate="2018-01-08T13:40:42.237" Score="0" Body="&lt;p&gt;An alternative to PID is Model Predictive Control (MPC) &lt;a href=&quot;https://en.wikipedia.org/wiki/Model_predictive_control&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1]&lt;/a&gt;, &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0957417417306000&quot; rel=&quot;nofollow noreferrer&quot;&gt;[2]&lt;/a&gt;, &lt;a href=&quot;http://www.uni-obuda.hu/users/szakala/SMC%202016%20pendrive/1139_smc2016.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;[3]&lt;/a&gt;, &lt;a href=&quot;https://web.stanford.edu/class/ee364b/lectures/mpc_slides.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;[4]&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your system is highly nonlinear, I would recommend using Nonlinear Model Predictive Control (NMPC) &lt;a href=&quot;http://control.ee.ethz.ch/~apnoco/Lectures2015/07-Nonlinear%20Model%20Predictive%20Control.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;[5]&lt;/a&gt;, &lt;a href=&quot;http://www.springer.com/gp/book/9780857295002&quot; rel=&quot;nofollow noreferrer&quot;&gt;[6]&lt;/a&gt;, &lt;a href=&quot;https://pdfs.semanticscholar.org/4238/9407bc489b9d42a3a3d9a00cc45e44dc79ea.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;[7]&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Model predictive controller, predicts the future dynamic of the system and takes action before hand. This controller performs an online optimization at each sampling time and handles the system constraints explicitly. &lt;/p&gt;&#xA;" OwnerUserId="14288" LastActivityDate="2018-01-08T13:40:42.237" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18782" PostTypeId="2" ParentId="18780" CreationDate="2018-01-08T13:49:05.327" Score="2" Body="&lt;p&gt;It should be clear which is being used from the environment :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are in the UK on a farm in Kent for example and you are told &quot;20 gallons&quot; then that will be 20 imperial gallons, but if you are in the USA on a farm in Texas and they say 20 gallons then that will be 20 US gallons (which by the way are smaller).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are being sent a product from abroad then usually the units are specified clearly ie US gallons, Imperial gallons (Imp. Gallons).&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-01-08T13:49:05.327" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18784" PostTypeId="1" CreationDate="2018-01-08T16:06:14.463" Score="1" ViewCount="68" Body="&lt;p&gt;I have read that the longer the whistle the lower the pitch and the shorter the whistle the higher the pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are other factors. Pressure, size of the inlet, size of the outlet, diameter of the whistle... what changes occur to the whistle when these variables are changed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, is there a way to change the threshold of when the whistle starts making noise? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: I want a whistle that only makes noise between 1 and 2 pounds per sqft of pressure. Outside that range it is quiet. Is it possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source of the 20 mph airflow is after the whistle and increases in pressure over time, 30+ days, so when the pressure goes beyond a specific point, I'd like a whistle so I can hear it. What variables need to be considered when designing a whistle that can do this? And how would you do this?&lt;/p&gt;&#xA;" OwnerUserId="14448" LastEditorUserId="9" LastEditDate="2018-01-09T02:01:58.027" LastActivityDate="2018-01-09T02:01:58.027" Title="What changes can be made to a whistle to change these variables?" Tags="|fluid-mechanics|airflow|" AnswerCount="0" CommentCount="11" ClosedDate="2018-01-10T15:24:59.943" ContentLicense="CC BY-SA 3.0" />
  <row Id="18785" PostTypeId="2" ParentId="18773" CreationDate="2018-01-08T16:29:13.980" Score="1" Body="&lt;p&gt;This assumption always requires justification, and when this justification is not given by your lecturer, you should ask for it if it is not clear to you. For example when investigating bending, derivatives in $\theta$ direction will not generally be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common way to justify this assumption is to assume it in the first place, calculate a solution, and then check if the solution satisfies the full set of differential equations you had in the first place (without any simplifications). If it does, then you have found a valid solution. If it does not, then the assumption is nonsense. Sometimes the final check is not performed, probably because the authors think this is trivial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific case of a uniaxially compressed rod, especially when neglecting friction at the contact surfaces, is one of the cases where this works.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2018-01-08T16:29:13.980" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18786" PostTypeId="2" ParentId="18759" CreationDate="2018-01-08T17:28:43.450" Score="2" Body="&lt;p&gt;When the tire rolls forward, there is gradual loading of the weight of the car on the tire that causes it to deflect and set in to a small flattening motion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This creates bending on the bottom face threads and the walls buckling. This constant folding of the tire creates heat and consumes energy and resist the rolling forward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is one of the reasons that having fully pressured tires saves gas.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-01-08T18:45:44.660" LastActivityDate="2018-01-08T18:45:44.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18787" PostTypeId="2" ParentId="18759" CreationDate="2018-01-08T18:01:43.917" Score="3" Body="&lt;p&gt;At the leading edge of a tire, I would presume that as the tire tip has a velocity with a component heading in the negative z direction, the resulting change to a &quot;deformed patch&quot; will result in it having an acceleration to stop this velocity, which would require a force.  This force would take the form of additional stress on this leading edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would have a reverse scenario on the trailing edge, where the tire tip will gain velocity in the positive z direction, and the resulting force will reduce the stress in the tire on the trailing edge.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-01-08T18:01:43.917" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18788" PostTypeId="2" ParentId="18762" CreationDate="2018-01-08T19:48:25.603" Score="3" Body="&lt;p&gt;When you compress or stretch a material, the work done is partly converted into elastic energy, which causes the body to return (approximately) to it's initial length, once you remove the load. On the other hand, some of the energy is dissipated into heat. This is the primary cause for hysteresis.&#xA;(Tyre-)rubber, due to it's visco-elastic properties is strongly affected by hysteresis, shown by the following stress-strain-diagram (force-extension-diagram, to be precise)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/c/c6/Elastic_Hysteresis.svg/930px-Elastic_Hysteresis.svg.png&quot; width=&quot;400&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt; &lt;em&gt;Fig 1 Source: &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/c/c6/Elastic_Hysteresis.svg/930px-Elastic_Hysteresis.svg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://upload.wikimedia.org/wikipedia/commons/thumb/c/c6/Elastic_Hysteresis.svg/930px-Elastic_Hysteresis.svg.png&lt;/a&gt;&lt;/em&gt; &lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means if you apply a load to rubber and release it afterwards, you will measure a different force for the same deformation during load and release.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you now consider a tyre on a flat surface, the contact surface between the two materials forms a plane, meaning that in the front of the contact surface (in the direction in which the tyre is rolling) the rubber is in a loading phase (blue curve). In the back of the contact surface the rubber is unloading (red curve). This results in a non-symmetrical stress-distribution, as shown in the following figure. (Albeit for both of the bodies being cylinders, but the principle is the same)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/8/8c/Pressure_distribution_for_viscoelastic_rolling_cylinders.png&quot; width=&quot;400&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt; &lt;em&gt;Fig 1 Source: &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/8/8c/Pressure_distribution_for_viscoelastic_rolling_cylinders.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://upload.wikimedia.org/wikipedia/commons/8/8c/Pressure_distribution_for_viscoelastic_rolling_cylinders.png&lt;/a&gt;&lt;/em&gt; &lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This results in a counterclockwise moment, with respect to the axis of rotation (clockwise), in other word: a retarding moment, or rolling resistance.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2018-01-08T19:48:25.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18789" PostTypeId="2" ParentId="18651" CreationDate="2018-01-08T21:51:26.707" Score="1" Body="&lt;p&gt;Sand blasting is not a type of AJM because you would never use it to machine anything. Sand blasting would never be used to cut/mahine an intricate shape in a brittle material, the jet does not have enough velocity and has too big a diameter which means it would never be accurate enough for a machining task. It's mainly used just to clean say rust off a casting. And AJM would never be used to clean a thread in a casting because it could potentially damage it.&lt;/p&gt;&#xA;" OwnerUserId="14451" LastActivityDate="2018-01-08T21:51:26.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18790" PostTypeId="2" ParentId="18549" CreationDate="2018-01-08T22:03:34.333" Score="1" Body="&lt;p&gt;The angle is derived from the amount of force it would take to pull in apart. IE: If the angle was 70% is would take less force to pull it apart.&lt;/p&gt;&#xA;" OwnerUserId="14451" LastActivityDate="2018-01-08T22:03:34.333" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18791" PostTypeId="2" ParentId="18773" CreationDate="2018-01-08T22:22:46.867" Score="1" Body="&lt;p&gt;1) axisymmetry requires symmetry of the &lt;em&gt;entire problem&lt;/em&gt;. The body and the loading and the material properties.  For the bending case the boundary condition is not axi-symmetric and so you can not use an axisymmetry assumption to simplify the governing equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) zero circumferential stress &lt;em&gt;does not follow&lt;/em&gt; from axisymmetry.  It happens to be so for this simple problem. You have zero transverse stress in a square bar in axial load too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a trivial example consider a cylinder under uniform pressure P over the entire surface. All normal stress components are equal to P, so clearly sigma_theta is not zero.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2018-01-10T03:22:52.030" LastActivityDate="2018-01-10T03:22:52.030" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18793" PostTypeId="1" CreationDate="2018-01-09T08:41:17.027" Score="1" ViewCount="30" Body="&lt;p&gt;I need to calculate the dose rate leaking from sealed Co-60 sources.&lt;/p&gt;&#xA;&lt;p&gt;Let's say the activity of a source is 7 TBq. The source is tube shaped, 2 mm radius and 10 mm in height.&lt;/p&gt;&#xA;&lt;p&gt;There is 5 cm of tungsten and 3 cm lead in the radiation's way.&lt;/p&gt;&#xA;&lt;p&gt;The measured dose rate according the NCRP report 102: &amp;quot;the maximum and the average leakage kerma rates through the housing shall not exceed 100 uGy/h&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How could I calculate if the 5 cm tungsten and 3 cm lead will be enough to limit the leakage dose rate to the 100 uGy/h range?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;We have built a Monte Carlo model of the above scenario and ran the simulation.&lt;/p&gt;&#xA;&lt;p&gt;There are a number of particles with a given energy arriving to the other side of the shielding. We save the state of these particles in a phase-space. So in this phase space we have the type of the particle: e-, gamma, position, direction, energy.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Calculating dose from the particle energies&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Dose is Joule/kg, so we need a volume with some material in which the energy of the particles will be absorbed. So we put a 1 cm^3 lead cube next to the area where the energy is the most dense in the phase space. Restart all the particles. Now we look at how much energy was absorbed in the lead cube and divide by the weight of the cube. The result is somewhat sensitive to the size of the cube because the shielding is not homogeneous.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Calculate dose rate.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Activity is 7 TBq which is 7x10^12 decays/s for Co-60 let's say that that is 7x10^12 gammas of 1.1732 MeV plus 7x10^12 gammas of 1.3325 MeV. So modelling 14x10^12 initial events would equal to 1 sec of real time. Let's say we model 1x10^9 events so now we can say have how much of real time is simulated and thus we have the dose rate.&lt;/p&gt;&#xA;&lt;p&gt;This is what I would like to verify.&lt;/p&gt;&#xA;" OwnerUserId="14454" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-01-09T11:29:56.903" Title="Dose rate through shielding from Co-60 to verify Monte Carlo result?" Tags="|nuclear-engineering|radiation|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18794" PostTypeId="2" ParentId="6866" CreationDate="2018-01-09T08:43:03.613" Score="0" Body="&lt;p&gt;The angles of those lines are determined by yield line theory. They are the axes of bending, should the slab decide to fail, which will require the least internal work to yield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://theconstructor.org/structural-engg/yield-line-theory/6839/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a very simple, watered down explanation, but if you Google it, you'll find some white papers (or whatever colour you decide to print on :P) with more technical explanations and all the math.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastActivityDate="2018-01-09T08:43:03.613" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18795" PostTypeId="2" ParentId="18780" CreationDate="2018-01-09T10:32:54.890" Score="1" Body="&lt;p&gt;According to a quick scan of &lt;a href=&quot;https://en.wikipedia.org/wiki/Imperial_units#Current_use&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, no country actually officially uses imperial units any more. There are plenty of &lt;em&gt;casual&lt;/em&gt; uses of imperial (for example, I live in the UK and measure my height in feet and inches, my weight in stone and pounds, and my milk and beer in pints) but no &lt;em&gt;technical&lt;/em&gt; uses that I know of.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In technical publications, will the author or publisher note the applicable units somewhere?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In technical publications from the US, US Customary units will be used. In technical publications from the UK, metric units will be used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this should work for the vast majority of cases. i.e. pints, gallons etc. are US ones. The main caveat is the publication date: the further back in time you go, the more likely it is that imperial units could have been used (see the Wikipedia article for further detail).&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-01-09T10:32:54.890" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18797" PostTypeId="2" ParentId="18759" CreationDate="2018-01-09T11:39:59.187" Score="4" Body="&lt;p&gt;I came across &lt;a href=&quot;http://the-contact-patch.com/book/road/c2020-the-contact-patch&quot; rel=&quot;nofollow noreferrer&quot;&gt;a site&lt;/a&gt; that, among others, tries to explain exactly the phenomena in your question. It goes quite into depth about the different components of the tyre and how they interact with each other and the effects it has on the contact patch and rolling resistance. I also especially enjoyed the part speaking of the standing wave.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as a post script:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really like @Mark's &lt;a href=&quot;https://engineering.stackexchange.com/a/18787/14440&quot;&gt;answer&lt;/a&gt; and I've drawn a sketch showing how I understand it:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DEtWF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DEtWF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the moment when an element of the tyre touches the ground it has to stop to a standstill (relative to the ground). In order to have this happen, a force in the opposite direction needs to be applied to decelerate it. fx isn't shown in the diagram but that's the required forward force to decelerate the element, also causing rolling resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite obviously applies to the trailing edge, needing an upward force to accelerate it again.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastActivityDate="2018-01-09T11:39:59.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18799" PostTypeId="2" ParentId="18763" CreationDate="2018-01-09T14:03:05.843" Score="1" Body="&lt;p&gt;A proper analysis requires a log mean delta T calculation, and an understanding of the configuration of the heat exchanger (i.e parallel flow, cross flow, etc.).  But at a very high level it comes down to a variation of the basic convective heat flow equation Q = U&lt;em&gt;A&lt;/em&gt;DT where the DT is the temperature difference between the water and the air.  Assuming the heat transfer coefficient and area are the same, then a higher DT will provide a higher heat transfer rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations you cited apply to the required water flow to provide the heat, and the heat increase in the air.  They don't have anything to say about how much heat will actually be transferred.  Using those equations alone, you could bring in water at 73 degrees and send it out at 33 degrees, still getting the 40 degree drop, but obviously it won't work, because they don't account for the heat transfer rate from water to air.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2018-01-09T14:03:05.843" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18800" PostTypeId="2" ParentId="18715" CreationDate="2018-01-09T14:28:36.350" Score="1" Body="&lt;p&gt;Taking upwards as positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_1=1000kg,\;h_1=5m $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_2=500kg,\;h_2=0m $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m_3=m_1+m_2=1500kg,\;h_3=-0.5m $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_0 = Initial\;State$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_1 = At\;Impact$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$t_2 = End\;State$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since $h_2=0$, the energy of the system at $t_0$, $E_0$, is entirely due to the Gravitational Potential Energy of $m_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_0=m_1*g*h_1=1000*9.81*5=49.05kJ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $t_1$, neglecting air resistance, the GPE of $m_1$ has been converted entirely into Kinetic Energy, and as such&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_1=E_0=49.05kJ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At $t_2$, nothing is moving, and so once again all energy is in the form of GPE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_2=m_3*g*h_3=1500*9.81*-0.5=-7.36kJ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy change between $t_1$ and $t_2$ is therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta E=E_2-E_1=-7.36-49.05=-56.41kJ$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This energy change is brought about by the retarding force acting on $m_3$. Assuming a constant force over the time that $m_3$ is decelerating, and noting that the Work Done took place over a distance of $h_3-h_2=-0.5m$,  the Retarding Force can be calculated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\frac{-56.41}{-0.5}=112.8kN$$&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-01-09T23:18:02.527" LastActivityDate="2018-01-09T23:18:02.527" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18801" PostTypeId="1" CreationDate="2018-01-09T14:39:09.597" Score="0" ViewCount="4624" Body="&lt;p&gt;I am attempting to mount a device by screwing into ~2&quot; of rubber (not sure what durometer or material, but it is decently firm, maybe 70 durometer). However, I am not sure what type of anchor to use (corkscrew, wall-anchor type, ground anchor, etc). Shown below are some examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dp4aO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dp4aO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the advantages/disadvantages of each for soft materials? I assume since the material is closest to a cork, a corkscrew would likely be best (assuming there is sound reasoning behind the design of this ageless device). The worst case scenario is if the fastener tears out, and pulls out a chunk of the material along with it. I would think a helix would minimize the risk of damage, however the lack of surface area might not provide the most shear strength/holding force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a guide, or published paper, discussing the tear-out strength of these fasteners in rubber/elastomers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As per the comments, here is some additional information: The rack (with measurement equipment on it), is ~100 lbs. Unfortunately, I do not have access to the inside plate, or other side of the rubber layer. This is also a temporary installation (hence screws), so I cannot bond to it (epoxy), which was my initial thought.&lt;/p&gt;&#xA;" OwnerUserId="1933" LastEditorUserId="1933" LastEditDate="2018-01-09T17:40:23.213" LastActivityDate="2018-01-10T19:42:09.503" Title="Screwing into Rubber - Screw Selection" Tags="|materials|fasteners|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18802" PostTypeId="2" ParentId="18549" CreationDate="2018-01-09T16:34:50.413" Score="3" Body="&lt;p&gt;In 1841 Sir Joseph Whitworth produced a paper on a universal system of screw threads. He then collected a variety of screws and proposed a universal thread using their average pitch and depth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result was the 'Whitworth thread' with the depth and pitch of constant proportion, giving the 'V' thread a mean angle of 55 degrees and the number of threads per inch was specified for various diameters. The thread was first introduced in his own workshop and was in universal use by 1858. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(ref: The Whitworth Society whitworthsociety.org/history.php?page=2 ) &lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2018-01-09T16:34:50.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18804" PostTypeId="1" CreationDate="2018-01-09T19:59:09.443" Score="0" ViewCount="863" Body="&lt;p&gt;I am doing a project on controlling wheels of a wheelchair.&#xA;The diameter of the wheel is about 25-30 inches, the weight which the wheelchair is supposed to carry is upto 80 Kgs and it should run at a speed of 0.5-0.7 metres per second.&#xA;I thought of using a 24V dc motor, what speed and Power must the dc motor provide to move a wheelchair at above specifications.&lt;/p&gt;&#xA;" OwnerUserId="14463" LastActivityDate="2019-02-04T02:22:55.733" Title="Dc motors for controlling wheels of a wheelchair" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18805" PostTypeId="2" ParentId="18801" CreationDate="2018-01-09T20:02:54.677" Score="1" Body="&lt;p&gt;In tire rubber (~&lt;a href=&quot;http://mykin.com/rubber-hardness-chart&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shore 60A durometer&lt;/a&gt;), &lt;a href=&quot;https://www.homedepot.com/p/Grip-Rite-8-x-2-in-Phillips-Bugle-Head-Coarse-Thread-Sharp-Point-Polymer-Coated-Exterior-Screws-1-lb-Pack-PTN2S1/100197689&quot; rel=&quot;nofollow noreferrer&quot;&gt;standard wood screws&lt;/a&gt; would work well. They are available, easy to install/remove, and will leave a smaller mark after removed. 2&quot; is a pretty generous depth. While the pull out force will obviously not be as great as wood, it will be substantial. I would estimate over 100lbf tension on a single 2&quot; wood screw; in shear it would easily be double that. Tension keeps the equipment cabinet from tipping away from the wall, and shear keeps it from sliding down the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A corkscrew or expansion shield is more likely to pull out a chunk of material under excess load. While a wood screw would not hold as much load per fastener, it will most likely fail in just the area around the threads and strip out a hole a little smaller than its maximum diameter.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-01-09T20:02:54.677" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18806" PostTypeId="1" AcceptedAnswerId="18812" CreationDate="2018-01-09T21:39:39.337" Score="17" ViewCount="13567" Body="&lt;p&gt;From an engineering perspective, what limits the maximum speed you can reach with a regular car? I understand that some of the faster cars are for safety reasons limited to not run faster than say 250 km/h, but that's not my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of several reasons, but not sure which of these is relevant:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the limit set by some part (which?) breaking if I increase the rpm, as suggested by red marks on rpm meters?&lt;/li&gt;&#xA;&lt;li&gt;Or is it rather that you cannot get in fuel fast enough to keep increasing the rpm?&lt;/li&gt;&#xA;&lt;li&gt;Or is it that friction/drag increases as you speed up and the engine cannot overcome this as it can only generate a maximum amount of force/torque? If yes, what does this amount of torque/force depend on?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12005" LastActivityDate="2019-09-22T23:24:27.983" Title="What limits the speed of a car?" Tags="|mechanical-engineering|car|" AnswerCount="6" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="18807" PostTypeId="1" CreationDate="2018-01-09T21:41:10.923" Score="1" ViewCount="84" Body="&lt;p&gt;I apologize for the complete beginner question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm an Undergraduate Research Assistant trying to design a shelving 8'tall  by 10' Long, 4' wide shelving unit to support about 200 lbs. I'm designing this to go over 4' 8' table, so the folks in the lab can reach up and grab any of our optical equipment easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, is how to get started? I've had luck finding components on McMaster-Carr for Struts(&lt;a href=&quot;https://www.mcmaster.com/#struts/=1b21hmv&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mcmaster.com/#struts/=1b21hmv&lt;/a&gt;) and braces, but I'm completely new to building things in general, and I'm trying to avoid any stupid mistakes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone point me to any resources or tips to get started? Any help would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example of the thing I'm trying to build&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Uysjx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uysjx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14465" LastEditorUserId="8055" LastEditDate="2018-01-11T15:56:29.463" LastActivityDate="2018-01-11T15:56:29.463" Title="How to design a shelving unit out of McMaster Carr Struts?" Tags="|mechanical-engineering|structures|statics|beam|" AnswerCount="0" CommentCount="3" FavoriteCount="0" ClosedDate="2018-01-21T20:31:48.620" ContentLicense="CC BY-SA 3.0" />
  <row Id="18808" PostTypeId="2" ParentId="18806" CreationDate="2018-01-09T21:53:48.407" Score="11" Body="&lt;p&gt;So, what limits the speed is a combination of two things : the power from the engine with the gearing and the rolling and air resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to approx 40mph the rolling resistance is the largest resistance, but above that speed air resistance is the dominant factor and increases the faster you go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engine power is fixed (ok tuning etc) but the gearing is also important - an engine of 200bhp can either power an agricultural tractor or a neat sports car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the total resistance is equal to the power available at the wheels then you don’t go any faster.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-01-09T21:53:48.407" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18809" PostTypeId="2" ParentId="18804" CreationDate="2018-01-09T23:06:42.793" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What speed and Power must the dc motor provide to move a wheelchair at above specifications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Wheel RPM = chair_speed / Wheel-diameter x 60&lt;br&gt;&#xA;Motor speed = wheel speed/Gear_ratio.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor power = hard_without_more_data.&lt;br&gt;&#xA;Factors include: Friction, gear train efficiency, motor efficincy, carpet-lino-concrete- ..., slope or flat, ...   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at existing chairs.&lt;br&gt;&#xA;2 x 500 Watts seems good in many cases. &#xA;Lower or even much lower may be OK in your case.&lt;br&gt;&#xA;In a demonstration version on a smooth hard floor with a good motor and gearbox and smooth hard wheels I'd expect that 100 Watts would probably be acceptable.&lt;br&gt;&#xA;At your target speed losses from windage are minimal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adjust (increase) any of the following for losses due to motor efficiency (say perhaps 80-90%) and gearbox efficiency (90-95%?)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;Windage&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power due to air resistance can be determined from the drag equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drag = 0.5 x Air density x Area x Coefficient-of-drag x velocity-squared&lt;br&gt;&#xA;Power = Drag x velocity.&lt;br&gt;&#xA;Whence:&lt;br&gt;&#xA;Power (Watts) = 0.5 x 1.2 kg/m^3 x A x 1 x V^3.&lt;br&gt;&#xA;A in m^2&lt;br&gt;&#xA;V in m/s&lt;br&gt;&#xA;Cd = 1 = flat plate drag - about right for a wheelchair&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 1 m/s and 1 square metre area  Power = 0.6 x 1 x 1 x 1^3 = 0.6 Watt!&lt;br&gt;&#xA;At 2 m/s (7 kph) Power 5 Watts.&lt;br&gt;&#xA;At 5 m/s (18 kph) Power = 75 Watts  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ie in a &quot;still air&quot; environment or where minor air movement occurs power loss to windage is liable to be under 10 Watts. In external gusty wind&quot; environments substantial power peaks may occur. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Climbing:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the wheelchair travels on an ascending (non level) surface then extra power will be required to provide the potential energy thus achieved by the gain in height.&lt;br&gt;&#xA;Energy required to move m kg upwards against gravity g a height h&lt;br&gt;&#xA;E = mhg&lt;br&gt;&#xA;g id about 9.8 m/s/s which can be approximated to g = 10 for this sort of working. In SI units m kg, h m, v m/s, g m/s/s:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy = 10.m.h Watt,seconds.&lt;br&gt;&#xA;To achieve this in time t seconds&lt;br&gt;&#xA;Power = E/t = 10.m.h/t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power required = 10.m x the height increase per second.&lt;br&gt;&#xA;For a wheelchair travelling a V m/s up an incline of d degrees,&lt;br&gt;&#xA;Height increase per second = v.tan(d)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power due to slope = 10.m.V x tan(d)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For eg a 5 degree slope at 1 m/s and a 100 kg all up load&lt;br&gt;&#xA;Watts = 10.m.v.tan(d) = 10.100.1.tan(5) ~= 90 Watts&lt;br&gt;&#xA; ie power required to ascend a slope is non trivial above a few degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 1 in X slope&lt;br&gt;&#xA;x = 1/(tan(d))&lt;br&gt;&#xA;So a 5 degree slope = 1 in 1/(tan(5)) = 1 in 11.4&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2018-01-12T12:10:05.530" LastActivityDate="2018-01-12T12:10:05.530" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18810" PostTypeId="1" CreationDate="2018-01-09T23:38:05.730" Score="2" ViewCount="125" Body="&lt;p&gt;I'm going back to college next year to pursue a career in mechanical engineering. I was wondering, along with the theoretical and CAD material you learn, is it customary that ME students learn how to use lathes, mills and other shop equipment, or is that solely taught to technicians? &lt;/p&gt;&#xA;" OwnerUserId="12609" LastEditorDisplayName="user16" LastEditDate="2018-01-10T02:56:07.040" LastActivityDate="2018-01-10T20:30:07.897" Title="Do mechanical engineering students learn to use shop equipment?" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18811" PostTypeId="2" ParentId="18810" CreationDate="2018-01-09T23:49:06.673" Score="1" Body="&lt;p&gt;That depends on the Uni and the course and who they do or don’t work with in industry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My course had 4 weeks of worshop courses : welding etc but that was rare as Unis go, I got an exemption as I was apprentice-trained...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of my friends in other Unis were jealous of the extent of our equipment : wind tunnels (subsonic and supersonic) single cylinder engines petrol and diesel, a jet engine, water pumps axial, kaplan and pelton turbines, water flow channels for Reynolds numbers etc, tribology equipment, also some state of the art stress/strain test gear that was also doing research for industry... Electrical, electronic and digital stuff etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, check out what is specified for the course...&#xA;Just re-read your question : most colleges do have a larger practical element , in my apprenticeship many things were covered that were not covered at work, but work covered some things in more detail.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-01-09T23:49:06.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18812" PostTypeId="2" ParentId="18806" CreationDate="2018-01-10T01:57:59.097" Score="27" Body="&lt;p&gt;There are a few simple reasons why the speed of a vehicle (road conditions notwithstanding) may be limited:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Gearing&lt;/strong&gt; -- Production vehicles with conventional transmissions have a limited number of gears. For most modern cars, this is usually 5 or 6, whereas older vehicles may have as few as 2 or 3. If the gear ratio of the highest gear is too low (&quot;lower&quot; gears are expressed as larger numerical ratios), it's entirely possible that the engine will redline before air resistance becomes a factor at all. This ties into your first point about the red zone on the tachometer. If you have reached redline, which is the maximum rotational speed of your engine, but don't have a higher gear to shift into, then you cannot go any faster without damaging your engine.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Drag&lt;/strong&gt; -- Like any physical object, cars are subject to air resistance and other sources of drag (rolling resistance, etc). If the drag on the car exceeds the amount of power that the engine is capable of producing to the wheels, then your speed is once again limited.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Speed Limiters&lt;/strong&gt; -- It's worth mentioning that production vehicles are almost always speed limited in the ECU (Engine Control Unit) for safety or legal reasons. If the ECU detects the wheels are spinning at a fast enough speed, it will cut power to the engine to prevent the vehicle from going faster. It's possible to circumvent this protection with an aftermarket ECU or with a modification to the stock one. As an example, some model years of the Honda Civic are speed-limited to around 120 mph (190 km/h).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Tire Ratings&lt;/strong&gt; -- All tires have a certain speed rating, which is likely much less than the actual top speed of your vehicle. The speed rating is a single letter and is part of the tire code (see &lt;a href=&quot;https://www.tirerack.com/tires/tiretech/techpage.jsp?techid=35&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt; for more information). For example, temporary spare tires may be limited to only 80 mph (130 km/h) before they are in imminent danger of suffering a blowout.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Stablity/Aerodynamics&lt;/strong&gt; -- This is less of an engineering perspective than a practical one, but for a &quot;regular&quot; car there will be a certain point that the suspension and other components are not sufficient to keep the car driving straight down the road in a safe fashion. Anecdotal evidence to support this point comes in the form of a story about an old '70s full-size American car that, when taken up to triple-digit speeds, did not have the aerodynamic capabilities to keep the front wheels on the ground. In short, the lift from the air moving under the car lifted the front wheels off the ground into a terrifying high-speed &quot;wheelie&quot;. Though this process did not technically cause the car to slow down, it certainly made it difficult to control at that speed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14467" LastEditorUserId="14497" LastEditDate="2018-01-11T13:41:57.957" LastActivityDate="2018-01-11T13:41:57.957" CommentCount="19" ContentLicense="CC BY-SA 3.0" />
  <row Id="18813" PostTypeId="2" ParentId="18810" CreationDate="2018-01-10T05:31:17.137" Score="1" Body="&lt;p&gt;I agree with Solar Mike. It completely depends on the university.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would take my university as an example. During my time as an undergraduate, there are two types of internships I have to do, &lt;strong&gt;basic internship&lt;/strong&gt; and &lt;strong&gt;advanced internship&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic one was rather broad and primarily focussed on presenting me the vast fields of mechanical engineering (the practical part, like turning objects on a lathe or repairing broken or defect parts), the advanced one was rather focussed on topics that you would encounter as an engineer (construction, like CAD drawings, research, and analysis, project planning, testing and so on).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following list shows the preferred fields that a ME student of my university should work in (GP [Grundpraktikum] = basic internship, FP [Fachpraktikum] = advanced internship). Since I'm in a German university, the list will be in German, but the main point of this post was already mentioned in the above text.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;GP1: Spanende Fertigungsverfahren&lt;/li&gt;&#xA;&lt;li&gt;GP2: Umformende Fertigungsverfahren&lt;/li&gt;&#xA;&lt;li&gt;GP3: Urformende Fertigungsverfahren &lt;/li&gt;&#xA;&lt;li&gt;GP4: Thermische Füge- u. Trennverfahren &lt;/li&gt;&#xA;&lt;li&gt;GP5: Instandhaltung, Wartung, Reparatur &lt;/li&gt;&#xA;&lt;li&gt;GP6: Montage, Demontage&lt;/li&gt;&#xA;&lt;li&gt;FP1: Wärmebehandlung&lt;/li&gt;&#xA;&lt;li&gt;FP2: Werkzeug-, Prüfstands- und Vorrichtungsbau &lt;/li&gt;&#xA;&lt;li&gt;FP3: Oberflächentechnik&lt;/li&gt;&#xA;&lt;li&gt;FP4: Messen, Prüfen, Qualitätskontrolle&lt;/li&gt;&#xA;&lt;li&gt;FP5: Fertigungssteuerung&lt;/li&gt;&#xA;&lt;li&gt;FP6: Fachrichtungsbezogene praktische Tätigkeit nach Absprache mit dem Praktikumsamt&lt;/li&gt;&#xA;&lt;li&gt;FP11: Konstruktion, Entwicklung, Forschung &lt;/li&gt;&#xA;&lt;li&gt;FP12: Erprobung, Prüfstandsversuche&lt;/li&gt;&#xA;&lt;li&gt;FP13: Arbeitsvorbereitung&lt;/li&gt;&#xA;&lt;li&gt;FP14: Projektierung, Produktprogrammplanung &lt;/li&gt;&#xA;&lt;li&gt;FP15: Produktions- und Fabrikplanung&lt;/li&gt;&#xA;&lt;li&gt;FP16: Ingenieurdienstleistungen&lt;/li&gt;&#xA;&lt;li&gt;FP17: Fachrichtungsbezogene praktische Tätigkeit nach Absprache mit dem Praktikumsamt&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another way to get in contact with it is to work in an institute. I, for example, work in an institute of my university. We focus on forming technologies and we (&lt;strong&gt;student assistants&lt;/strong&gt;) do the testing and the analysis for the engineers that work on their Ph.D.'s or other work-related subjects. But we also have a lot of external institutes (like the Fraunhofer Institute) nearby, so some of my friends work there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if your university of choice doesn't offer this kind of work or experience, you could always check and see, if there are companies that offer &quot;&lt;strong&gt;student jobs&lt;/strong&gt;&quot;, as it is called. It is a part-time job (sometimes even less) and there are many benefits (something to add to your CV, and this is a huge plus!!, getting real working experience, learning new programs and things you never learned in university, getting more or less flexible hours, especially in your exams period and so on.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I'm already off track by now, I would also go over some of your assumptions. Not all university offer the same treatment. For example, my university didn't offer as much CAD as I'd hoped for. We learned a little bit about AutoCAD in my 4th semester and that was it. I had to take special classes to learn a little bit more about CATIA V5 and CAD in general. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize everything, when you plan to study ME and you have the choice, where to go to, I would recommend that you check out their homepages and see for yourself what they offer (sometimes even calling and talking to them, where you can directly ask questions about some parts that are unclear). And you should inform yourself about the surroundings of the university. What companies are there? Which of those of student jobs? What kind of field do I want to work in? Do I want a sole university experience or do I want to work in an external company (benefit here: they might hire you after your graduation)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope, you have a better picture of what you should expect and what you should consider when going to study ME (although I just covered the practical part. You should definitely think about your field of interest and if your university has the right specializations if you already know what you want to do after that).&lt;/p&gt;&#xA;" OwnerUserId="14437" LastActivityDate="2018-01-10T05:31:17.137" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18815" PostTypeId="1" CreationDate="2018-01-10T08:26:30.363" Score="1" ViewCount="707" Body="&lt;p&gt;&lt;br/&gt;&#xA;I have a few questions relating to a multiple layered spinner that uses ball bearings to reduce friction, kind of similar to a fidget spinner. The image below is an example of the multilayered bearing spinner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/geMaa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/geMaa.png&quot; alt=&quot;Bearing Spinner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are as follows:&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does having multiple layer of bearing rings decrease the overall friction on the outermost ring and allow it to spin for a longer period of time?&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;Does increasing the number of ball bearings in each ring decrease friction?&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;Are there any formulas or relation that would allow me to calculate the overall run time of the outermost ring for a given tangential force? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14472" LastEditorUserId="8055" LastEditDate="2018-01-11T15:56:45.393" LastActivityDate="2019-02-06T07:00:36.483" Title="Multilayered Bearing Spinner" Tags="|friction|bearings|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18816" PostTypeId="2" ParentId="18815" CreationDate="2018-01-10T08:34:33.627" Score="1" Body="&lt;p&gt;The more you increase the number of balls and tracks increases the friction, what does change is the speed differential between the any track and its predecessor / successor - but guaranteeing that the middle tracks will share the changes of speed....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a suggestion for a long life spinner : use a spinner that is supported in a magnetic field then it is basically only air friction and, if you put it in a vacuum not even that.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-01-07T07:00:36.067" LastActivityDate="2019-01-07T07:00:36.067" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="18817" PostTypeId="2" ParentId="18806" CreationDate="2018-01-10T08:41:37.920" Score="9" Body="&lt;p&gt;Physics stops you.  Ever ride a bicycle?  25 mph is easy, 30 is hard, 40 requres special gear or a very pumped up body, and 50 is nigh impossible.  Why does the effort get so steep for such small speed increases? Air. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Aerodynamic drag&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Rolling resistance is the dominant factor when your car is going slow - that's why they're so hard to push.  But at higher speeds, the ruling factor is aerodynamic drag.  That's because rolling resistance is fairly linear.   Aero drag is &lt;em&gt;not&lt;/em&gt;, and is at least a second-order factor -- it comes on like a freight train at higher speeds.  In fact I've heard coal trains that can do 40 mph loaded, can only do about 35 empty. That's because each empty coal car is a big scoop, catching the wind.  And rolling resistance is close to nil for a train. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Available power&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Obviously, that aero drag is balanced against the power the engines can apply against it.   They could get that empty coal train up to 70 mph if they put four times the locomotives on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're talking about cars, so it's a matter of how much the batteries can supply and the motor-controller can push without taking damage.   Or since you're assuming fuel-powered cars, it's about &lt;strong&gt;how much &lt;em&gt;air&lt;/em&gt; the engine can move through&lt;/strong&gt;.  Adding the correct amount of fuel is easy.  Airflow is decided by tuning - intake/exhaust ducting/porting/resonance, valve size and numbers, camming, stuff like that.  (Also the engine redline (max RPM), but if your ducting/resonance is tuned so raising redline will make a difference, that engine is going to run quite poorly on the street.)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gearing can also be a factor.  I had a car which struggled at its top speed.  Fourth gear was too tall for it to be able to accelerate.  Third gear put it past its horsepower peak, so the faster you went, the &lt;em&gt;less&lt;/em&gt; horsepower you had.  The gearing was all wrong for speed, but it was excellent for street-driving fun, refinement and MPG, which is what I paid for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical non-sporty street car like a Ford Flex is likely to hit physics limits of aerodynamic drag before it hits a point where the computer would rev-limit or MPH-limit.  &lt;/p&gt;&#xA;" OwnerUserId="11673" LastEditorUserId="11673" LastEditDate="2018-01-10T16:35:00.330" LastActivityDate="2018-01-10T16:35:00.330" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18818" PostTypeId="2" ParentId="18806" CreationDate="2018-01-10T09:44:40.777" Score="2" Body="&lt;p&gt;For modern high performance cars, built after around the year 2000 up until today (2018), the answer is simple: &lt;strong&gt;tire grip&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engine performance and management has already reached a level that can outperform how much grip tires have on the road. What happens when you try to exert more energy than your tires can handle is wheel slip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people would be familiar with wheel slip from standing starts (like in drag racing) and when cars are moving slowly (when people deliberately burn rubber). This type of slip was eliminated by letting a CPU control the brakes (traction control).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next comes the wheel slip we're concerned with. Wheel slip while moving at top speed. The traditional fix for this was to modify the car's aerodynamics and add down force to basically push the car to the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have reached the stage where adding even more down force would increase drag and reduce performance. But our top performing engines can still generate wheel slip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in parallel to engine development, advances in chemistry and tire design have also contributed to increasing a car's top speed by upping the speed the tire is able to handle before losing grip. Now that we have reached a point where our best engines can consistently fail tires the current limiting factor is tire grip.&lt;/p&gt;&#xA;" OwnerUserId="403" LastActivityDate="2018-01-10T09:44:40.777" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18819" PostTypeId="1" CreationDate="2018-01-10T10:07:23.360" Score="1" ViewCount="126" Body="&lt;p&gt;I need to construct a 3D wooden bicycle trailer for Kids. The weight of bicycle trailers must also be light!&#xA;The Frame should be made of wood.&#xA;what criteria should I use to construct a Frame,Children's weights?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8ODvN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8ODvN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11393" LastEditorUserId="11393" LastEditDate="2018-01-11T01:49:01.270" LastActivityDate="2018-01-11T01:49:01.270" Title="Which frame is optimal for children's bicycle trailers? tubular or rectangular frame and why?" Tags="|mechanical-engineering|structural-engineering|structural-analysis|wood|mechanical|" AnswerCount="2" CommentCount="8" ClosedDate="2018-01-10T20:37:49.950" ContentLicense="CC BY-SA 3.0" />
  <row Id="18820" PostTypeId="2" ParentId="18819" CreationDate="2018-01-10T10:40:01.447" Score="1" Body="&lt;p&gt;From a structural viewpoint, please let me enumerate the pros and cons of this idea - pipe vs rectangular hollow steel section (HSS).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of strength, circular profiles have the strongest moment of inertia (the one that resists bending) among structural profiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ex. For an HSS of 2'x3'x1.0mm THK, the moment of inertia is  $201,744.50 \  mm^4$.&#xA;For an equivalent section, we take a pipe of 79.59mm diameter, 1mm thick. The moment of inertia is approx. $366,967.62 \ mm^4$. That's roughly 45% more strength than HSS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of workability, HSS is a lot more easier to work than pipes. Imagine connecting two pipes into a junction. One has to cut the end part of the other pipe in order to fit it into the body of the other pipe. And cutting requires more precision in it. See image below: &lt;a href=&quot;https://i.stack.imgur.com/gHB6c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gHB6c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing is that pipes cost more (with reference to our local suppliers) per kilogram of weight than HSS. This is maybe due to the difficulty in making circular profiles. &lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2018-01-10T10:40:01.447" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18821" PostTypeId="2" ParentId="18772" CreationDate="2018-01-10T11:47:31.530" Score="1" Body="&lt;p&gt;Its for safety. Electric trains especially can create lots of freak currents on the train tracks that may or may not cause false detections, so you want to be sure that the train detection has been triggered for several seconds continuously instead of by just a single short pulse.&lt;/p&gt;&#xA;" OwnerUserId="14479" LastActivityDate="2018-01-10T11:47:31.530" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18824" PostTypeId="2" ParentId="18819" CreationDate="2018-01-10T14:48:12.977" Score="3" Body="&lt;p&gt;I assume that you are talking about a steel frame with a wooden body mounted on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In general&lt;/em&gt; round tube is more efficient, but if the loading is primarily bending in one direction rectangular section can be better. So for something like a space frame chassis or roll cage where the whole structure can be subject to quite complex loading round tube is typically more weight efficient but a ladder or A-frame chassis tends to act more like a simple beam. So rectangular section may actually be better in terms of structural efficiency as well. Certainly almost all ladder frame chassis are made from RHS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rectangular section is also a lot more practical to work with as it just needs to be mitred rather than fish-mouthed for welded joins and you can also make bolted joints with flanges. Similarly having flat surfaces tends to make it a lot easier to attach brackets and drill holes, a major consideration for a trailer frame. It also has the advantage that it can be tapered fairly easily by cutting out elongated triangles and bending and welding it back together. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-01-10T17:52:15.213" LastActivityDate="2018-01-10T17:52:15.213" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18825" PostTypeId="1" CreationDate="2018-01-10T17:13:20.413" Score="-1" ViewCount="91" Body="&lt;p&gt;so, soda is under pressure and has gas dissolved in it.  But, when you open it, the gas is still dissolved in it.  But, if we wait a few hours, the gas has escaped into the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What factors determine the rate at which gas escapes the soda-gas solution?&lt;/p&gt;&#xA;" OwnerUserId="6488" LastActivityDate="2018-01-10T18:18:47.253" Title="why doesn't soda go flat immediately after opening?" Tags="|fluid-mechanics|gas|" AnswerCount="2" CommentCount="5" ClosedDate="2018-01-13T13:52:58.217" ContentLicense="CC BY-SA 3.0" />
  <row Id="18827" PostTypeId="1" CreationDate="2018-01-10T17:21:25.623" Score="1" ViewCount="143" Body="&lt;p&gt;I'm working on a project where we are trying to prototype a whistle that works at the lowest possible air pressure and still produce an audible sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions on whistle type? and any additional factors or insight that might help, would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume, Open air 72 degree Fahrenheit environment using very consistent airflow of regular 72 degree Fahrenheit air.&lt;/p&gt;&#xA;" OwnerUserId="14448" LastEditorUserId="12754" LastEditDate="2018-01-14T21:29:39.350" LastActivityDate="2018-01-14T21:29:39.350" Title="Which Whistle Type works with the lowest PSI?" Tags="|fluid-mechanics|pressure|acoustics|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="18828" PostTypeId="2" ParentId="18825" CreationDate="2018-01-10T17:52:46.893" Score="0" Body="&lt;p&gt;The solubility of carbon dioxide in soda is related to both the temperature and pressure. A cold, pressurized drink can hold more CO2 per ml of liquid. When you open a soda, the gas immediately starts to come out of solution. Bubbles form and rise to the surface - this is the familiar sight of a fizzy drink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you drink half of a large bottle of soda, and put the lid tightly back on, it will still go partially 'flat', since there is a large volume of compressible air above the liquid. Some CO2 comes out of solution until the pressure in the bottle rises, and an equilibrium is found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep the remainder your drink fizzy if you only want to drink half of the large bottle, simply decant the part you wish to save for later into a smaller bottle, leaving less air inside.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-01-10T17:52:46.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18829" PostTypeId="2" ParentId="18825" CreationDate="2018-01-10T18:18:47.253" Score="2" Body="&lt;p&gt;dissolved CO2 in water remains in solution even after removal of pressure because of the absence of nucleation sites at which the gas can come out of solution. without a population of those seed sites, significant amounts of gas will remain in solution for hours after the removal of pressure. Dropping a Mento mint into a bottle of soda provides tens of thousands of active nucleation sites and will &quot;catalyze&quot; the exsolvation process, causing a spectacular geyser to shoot out of the top of the bottle. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-10T18:18:47.253" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18830" PostTypeId="2" ParentId="18806" CreationDate="2018-01-10T19:25:08.943" Score="0" Body="&lt;p&gt;You didn't specify a gear-driven (direct-drive) car, for one thing.  Even for direct-drive vehicles, the limiting speed is set ultimately by designing aerodynamics to keep the car from taking off and flying.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last I checked, land-speed records were set with extreme-design cars, jet or rocket-powered, with incredibly specialized wheels and brakes (standard discs would just fly apart at the high RPMs involved) among other parts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sum, as new materials are developed with ever-greater strength-to-mass ratios, land speed records are likely to continue to be broken.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-01-10T19:25:08.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18831" PostTypeId="2" ParentId="18801" CreationDate="2018-01-10T19:36:37.120" Score="0" Body="&lt;p&gt;I recommend using a bracket as opposed to a single screw. This way you can separate the pull out stress from shear and avoid bending/ tilting of fastener which exacerbates the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bracket in the shape of letter H 1/8&quot; thick, approximately 3&quot; tall and 1.5&quot; wide which has  two pre-drilled holes on top and bottom flanges and two holes in the vertical web, ready to receive 1/8&quot; by 2&quot; long decking screws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I venture to guess each bracket would support at least 200lbs weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can start calculating the allowable loads by using the NDS American Wood Council equations by adjusting them to your rubber shear values!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-01-10T19:42:09.503" LastActivityDate="2018-01-10T19:42:09.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18832" PostTypeId="2" ParentId="15204" CreationDate="2018-01-10T20:02:01.433" Score="2" Body="&lt;p&gt;The problem with your a approach is, that this is a statically indeterminate system. Basically, you could for example turn &lt;em&gt;one&lt;/em&gt; fixed bearing into a simple support and the structure would still be statically stable.&#xA;Remove another reaction force and the system turns into a mechanism. &#xA;Thus the degree of static indeterminacy is $n=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a consequence, the reaction forces depend on the materials elastic properties and cannot be determined simply by solving the forces and moment equations.&#xA;My approach was using the force method, assuming the elastic modulus $E$ and the area moment of inertia $I$ to be constant throughout the entire beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v3O36.png&quot; width=&quot;400&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this more general case of beam-lengths $a$ and $b$ I got the following solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A=F\frac{b^3}{a^3+b^3} \qquad B=F\frac{a^3}{a^3+b^3} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus for the case $a=L$ and $b=2L$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A=\frac{8}{9}F \qquad B=\frac{1}{9}F $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;PS: It may look like the solution is not depending on the elastic properties of the beam, but this is due to the assumption of a homogeneous beam, allowing me to cancel out $EI$ during my calculations.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;force-method:&lt;/em&gt;&lt;/strong&gt; &lt;em&gt;(see &lt;a href=&quot;https://i.stack.imgur.com/snfU2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;attachment&lt;/a&gt;)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle behind the force method is, that you remove e.g. the moment restraint at point $B$ (replace it with a simple support, so it becomes determinate) and then calculate the moment distribution of this determinate system.&#xA; Then, by implementing a unitary moment $X_1=1$, you calculate the required moment $X_1=M_B$ to satisfy the conditions of a fixed restraint: no rotation at $B$ $\quad \to\delta_1=\delta_{10}+X_1\cdot\delta_{11}=0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;note: $\delta_1$ is the rotation at $B$, which must be zero, $\delta_{10}$ is the rotation caused by $F$ in the reduced system (marked as GS), $X_1\cdot\delta_{11}$ is the rotation caused by the moment $X_1$ (marked as ÜS), thus when you solve for $X_1$ you get $M_B$&lt;/p&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="8632" LastEditDate="2018-01-12T12:51:44.947" LastActivityDate="2018-01-12T12:51:44.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18833" PostTypeId="2" ParentId="18810" CreationDate="2018-01-10T20:30:07.897" Score="0" Body="&lt;p&gt;It depends on the course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the UK most undergraduate engineering degrees will have content which varies a lot. However, even in the most practically focused courses there just isn't time to get any sort of serious grounding in hands-on manufacturing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My degree ended up being split between Bristol and Birmingham. At Bristol they one of the first things they got us to do was to make a load cell from scratch and fairly early on we had to make, in groups, a machine for sorting various 'casino' chips in steel, aluminium and opaque and transparent plastic and we had to demonstrate it in front of the whole year, they also got us to design an aircraft gearbox on actual paper. Birmingham was much more theory focused and there was much more CAD/CAM and lab work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But an engineering degree is definitely not a particularly good way to learn how to make stuff in a practical sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what its worth I now make metal sculpture and I learned more about making tings doing that than I did in my engineering degree BUT the theoretical background is actually pretty useful as it teaches you how to analyse practical problems. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-01-10T20:30:07.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18834" PostTypeId="1" AcceptedAnswerId="21316" CreationDate="2018-01-10T20:32:45.410" Score="2" ViewCount="2284" Body="&lt;p&gt;Assume you can use any whistle type, is there a way to choose the pressure point at which a whistle makes noise? For example I have a whistle that starts making noise at 2 psi but I would like to lower that to 1 psi. Would I change the area of the inlet, outlet? what feature would you change to accomplish this?&lt;/p&gt;&#xA;" OwnerUserId="14448" LastEditorUserId="15506" LastEditDate="2018-04-14T11:57:04.543" LastActivityDate="2018-04-14T11:57:22.717" Title="Whistle Design - Optimizing sound initiation pressure and volume" Tags="|fluid-mechanics|pressure|airflow|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18835" PostTypeId="2" ParentId="18827" CreationDate="2018-01-10T21:47:07.393" Score="1" Body="&lt;p&gt;&lt;a href=&quot;http://www.flutopedia.com/refs/Goss_2013_IntraoralPressureInEthnicWindInstruments.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Native american flutes&lt;/a&gt; from: Intraoral Pressure in Ethnic Wind Instruments Clinton F. Goss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This study measured intraoral pressure when playing six classes of ethnic wind instruments (N = 149): Native American flutes (n = 71) and smaller samples of ethnic duct flutes, reed instruments, reedpipes, overtone whistles, and overtone flutes.&lt;/p&gt;&#xA;" OwnerUserId="6488" LastEditorUserId="9" LastEditDate="2018-01-11T12:55:15.990" LastActivityDate="2018-01-11T12:55:15.990" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18836" PostTypeId="2" ParentId="18719" CreationDate="2018-01-11T01:55:34.920" Score="1" Body="&lt;p&gt;You will have to use vapor deposition under high vacuum. This requires a piston vacuum pump to get a rough vacuum, then a turbomolecular or diffusion pump to get to high vacuum. Then heat gold in this vacuum till it melts. Gold vapor will deposit a thin layer on all surfaces of the container including your glasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/L41ej3vHQT8?t=880&quot; rel=&quot;nofollow noreferrer&quot;&gt;glasslinger demonstrates vapor deposition with gold on youtube&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While possible, it is not easy and it will probably cost less to get it done professionally by an optical coatings company than the gold you will waste in your attempts.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-01-11T01:55:34.920" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18837" PostTypeId="1" CreationDate="2018-01-11T03:35:16.110" Score="2" ViewCount="80" Body="&lt;p&gt;I am using a 1HP pump to move water from a ground tank to overhead tanks that are at 30 feet height. The pipe is 1 inch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor is currently exposed and open and I would like to know if I can move the motor to second floor at 20 feet height, where it will be protected from elements? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the pump will be drawing water from tank that will be about 25 feet below and delivering water that is about 10 feet above.&lt;/p&gt;&#xA;" OwnerUserId="14495" LastActivityDate="2018-09-03T18:32:04.860" Title="Water pump location" Tags="|motors|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18838" PostTypeId="2" ParentId="18837" CreationDate="2018-01-11T07:22:27.580" Score="2" Body="&lt;p&gt;The maximum water suction height is 33.9 ft ie: 10.3 m in perfect conditions.&lt;br&gt;&#xA;Considering losses and pump efficiency you &quot;could&quot; get water to the pump level but I don't think you'd be able to pump it up 10 more feet.&lt;/p&gt;&#xA;" OwnerUserId="14496" LastActivityDate="2018-01-11T07:22:27.580" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18840" PostTypeId="1" AcceptedAnswerId="18843" CreationDate="2018-01-11T09:34:56.407" Score="8" ViewCount="10297" Body="&lt;p&gt;I have little/no experience with pipe networks and have been searching for a methodology to calculate the discharge from multiple outlets of an oil pipe network with a single inlet. As far as i can tell I have too many unknowns to use something like the Hardy Cross method (and often i don't have loops, just multiple branches and outlets as shown below), and Hardy-Cross seems to be used to find flows in pipes with known inflows and outflows, which I don't have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example may be something like this: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Inlet flow rate is known &lt;/li&gt;&#xA;&lt;li&gt;Fluid Properties are known&lt;/li&gt;&#xA;&lt;li&gt;Geometry of all pipes/bends/outlets are known&lt;/li&gt;&#xA;&lt;li&gt;The height above a datum of all pipes and outlets are known&lt;/li&gt;&#xA;&lt;li&gt;Flow is each pipe branch is UNKNOWN&lt;/li&gt;&#xA;&lt;li&gt;Discharge from each outlet is UNKNOWN (ext. pressure can be assumed ambient)&lt;/li&gt;&#xA;&lt;li&gt;I have target outflows, and could modify pipe length/diameters to obtain them&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;See my very crude diagram: &lt;a href=&quot;https://i.stack.imgur.com/Z5Bog.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z5Bog.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can anybody suggest an analytical method I can use to calculate the outflows and/or optimise the pipe lengths to obtain a set of target outflows?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A worked example of a similar problem would also be appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The person who tried to solve this problem before me just divided the inflow up based on the area of the outlets, without considering losses, bends or even the lengths of the pipes anywhere in the network. I have been searching for a while with limited resources for a worked example or solution method for this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will probably write a code to solve these problems if I find an appropriate methodology. (I'm aware commercial software &lt;a href=&quot;https://www.cadreanalytic.com/cadreflo.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;does exist&lt;/a&gt; that could calculate this, but I don't have access). &lt;/p&gt;&#xA;" OwnerUserId="14498" LastEditorUserId="2015" LastEditDate="2018-06-08T01:44:03.677" LastActivityDate="2020-09-03T06:25:25.970" Title="How Can I Determine Multiple Discharges from a Pipe Network? (Knowing the Inlet Conditions)" Tags="|mechanical-engineering|fluid-mechanics|civil-engineering|pipelines|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18841" PostTypeId="2" ParentId="18837" CreationDate="2018-01-11T09:46:57.963" Score="1" Body="&lt;p&gt;You need to consider &lt;em&gt;net positive suction head&lt;/em&gt;, or NPSH. Wikipedia article &lt;a href=&quot;https://en.wikipedia.org/wiki/Net_positive_suction_head&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. This is typically influenced by pump height, water temperature (vapour pressure) and friction losses. The higher you lift you pump w.r.t. your fluid source, the less NPSH you'll have available. Increasing fluid temperature and lengthening the suction pipe will have a negative impact on NPSH available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for pumps to work properly, you require a minimum amount of NPSH, which you can get from your pump's datasheet. This is typically influenced by the pump type, operating pressure and flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have less NPSH available than required, cavitation will happen which will decrease pumping efficiency and damage the impeller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing a very rough, quick calc puts your NPSH available at around 2m. I don't have much experience with typical pump NPSH required curves but I don't recall typical values much lower than 2.5m, so I think your pump will have a hard time pumping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.enggcyclopedia.com/2011/07/sample-problem-pump-npsh-calculation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a link to a sample calculation which is a bit simpler than the Wikipedia examples. Pay close attention to consider the lowest liquid level you'll be required to pump in order to be conservative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the type of pump, you'll have to prime it, i.e. you have to fill the suction pipe with water every time before you start the pump because the pump cannot form a vacuum with air to draw the water up, so it will run dry and burn out the pump. &lt;/p&gt;&#xA;" OwnerUserId="14440" LastEditorUserId="14440" LastEditDate="2018-01-11T15:33:04.680" LastActivityDate="2018-01-11T15:33:04.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18842" PostTypeId="1" CreationDate="2018-01-11T11:08:42.863" Score="5" ViewCount="4335" Body="&lt;p&gt;isn't the bending moment supposed to be zero on roller supports at beams?&#xA;then why isn't it zero here in this picture at x=2?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZteVC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZteVC.png&quot; alt=&quot;Bending moment at x=2 is not zero!&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14501" LastActivityDate="2022-12-06T09:23:22.227" Title="is bending moment on roller supports at beams zero?" Tags="|mechanical-engineering|structural-engineering|statics|moments|diagram|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18843" PostTypeId="2" ParentId="18840" CreationDate="2018-01-11T11:32:27.470" Score="5" Body="&lt;p&gt;The flow of current in a circuit is broadly analogous to fluid down a pipe, with the resistance of wires/resistors combining according to standard laws. If you can break your pipe network down into small segments of known &quot;resistance&quot;, then you can use current laws, and widely available circuit simulators to get a good approximation of the outlet flow (or at least, much better than your predecessor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have taken your diagram, and approximated the lengths of each discrete segment, and given them a relative resistance value. Clearly those segments with corners in might have a higher resistance than a straight of equivalent length - It's up to you to determine the correct values here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fLoIZm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fLoIZm.jpg&quot; alt=&quot;Annotated diagram - lengths approximated to resistances&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.falstad.com/circuit/circuitjs.html?cct=$%201%200.000005%2010.20027730826997%2050%205%2043%0Ar%20240%2080%20240%20144%200%2010%0Ar%20240%20144%20176%20144%200%2010%0Ar%2080%20144%2080%20368%200%2020%0Ar%20176%20144%20176%20224%200%2010%0Ar%20176%20224%20176%20368%200%2010%0Aw%20128%20224%20176%20224%200%0Ar%20128%20224%20128%20368%200%203%0Ar%20240%20144%20336%20144%200%2020%0Aw%20336%20144%20336%20224%200%0Aw%20336%20224%20272%20224%200%0Ar%20272%20224%20272%20368%200%2020%0Ar%20336%20224%20336%20368%200%2010%0Ar%20336%20224%20432%20224%200%2020%0Ar%20432%20288%20432%20368%200%205%0Aw%20432%20288%20400%20288%200%0Ar%20400%20288%20400%20368%200%203%0Ar%20432%20224%20496%20224%200%205%0Ar%20496%20224%20496%20368%200%203%0Aw%20496%20224%20544%20224%200%0Ar%20544%20224%20544%20368%200%206%0Aw%20176%20144%2080%20144%200%0Ar%20432%20224%20432%20288%200%2015%0Ag%2080%20368%2080%20384%200%0Ag%20128%20368%20128%20384%200%0Ag%20176%20368%20176%20384%200%0Ag%20272%20368%20272%20384%200%0Ag%20336%20368%20336%20384%200%0Ag%20400%20368%20400%20384%200%0Ag%20432%20368%20432%20384%200%0Ag%20496%20368%20496%20384%200%0Ag%20544%20368%20544%20384%200%0AR%20240%2080%20240%2048%200%200%2040%205%200%200%200.5%0A&quot; rel=&quot;noreferrer&quot;&gt;You can make an equivalent circuit using http://www.falstad.com/circuit/&lt;/a&gt;, Screenshot below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DjqPrm.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DjqPrm.png&quot; alt=&quot;screenshot of falstad link in answer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once your resistances are set, you can adjust the input voltage until you get a nice 'round' current value at the input (e.g. 100mA), and then see the percentage flow at each outlet by reading the current at those points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also add a &quot;Current Source&quot; at any outlets where there is a known flow rate, e.g. provided by some sort of limiting device, and all the other outlets will calculate accordingly.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-01-11T11:32:27.470" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18844" PostTypeId="1" CreationDate="2018-01-11T11:37:38.787" Score="6" ViewCount="1109" Body="&lt;p&gt;If I stood on a sailing ship and had a very strong pair of lungs (possibly motorized) then if I blew the sail would I be able to move the ship forwards?&lt;/p&gt;&#xA;" OwnerUserId="14504" LastActivityDate="2018-07-31T22:03:36.497" Title="How would you move a sail boat with no wind or paddles" Tags="|building-physics|" AnswerCount="7" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18845" PostTypeId="2" ParentId="18842" CreationDate="2018-01-11T12:32:24.127" Score="7" Body="&lt;p&gt;no. The &lt;strong&gt;change&lt;/strong&gt; in moment is zero, as you can see on your plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you can see if you imagine sectioning the beam slightly to the right of the support and constructing a free body diagram, the moment there is not zero. the support does not provide any concentrated moment, so the moment does not change there and so the moment at the support can not be zero.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2018-01-11T15:59:27.470" LastActivityDate="2018-01-11T15:59:27.470" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18846" PostTypeId="2" ParentId="18844" CreationDate="2018-01-11T13:03:22.687" Score="-3" Body="&lt;p&gt;If you would be in the water or standing ashore (and would have VERY powerful lungs) then yes.&#xA;But like Solar Mike said: &quot;for every action there is an equal and opposite reaction&quot;. So the ship won't move if you would be standing on the ship itself.&lt;/p&gt;&#xA;" OwnerUserId="14479" LastEditorUserId="14479" LastEditDate="2018-01-11T14:24:24.950" LastActivityDate="2018-01-11T14:24:24.950" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18847" PostTypeId="2" ParentId="18840" CreationDate="2018-01-11T14:34:11.997" Score="5" Body="&lt;p&gt;You have 9 unknown values you are seeking (the flow at each of 9 outlets), and one that you need to find in the process (the total pressure drop)  From the given information, you can assemble equations that you can solve using linear algebra methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of the flows at the outlets will equal the inlet flow: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_t = Q_a + Q_b + Q_c + Q_d + Q_e + Q_f + Q_g + Q_h + Q_j$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also set up equations based on the fact that the pressure drop will be equal along any path from a given node to an outlet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure drop in any segment can be found using many methods.  In this case, the Darcy-Weisbach equation would probably be most useful:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_f = f\cdot\dfrac{L}{D}\cdot\dfrac{v^2}{2g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$h_f$ = head loss (m)&lt;/li&gt;&#xA;&lt;li&gt;$f$ = friction factor &lt;/li&gt;&#xA;&lt;li&gt;$L$ = length of pipe work (m) &lt;/li&gt;&#xA;&lt;li&gt;$d$ = inner diameter of pipe work (m) &lt;/li&gt;&#xA;&lt;li&gt;$v$ = velocity of fluid (m/s) &lt;/li&gt;&#xA;&lt;li&gt;$g$ = acceleration due to gravity (m/s²) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Equivalent length information for fittings can be found here: &lt;a href=&quot;https://www.engineeringtoolbox.com/resistance-equivalent-length-d_192.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineeringtoolbox.com/resistance-equivalent-length-d_192.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An excel template for using this method can be found here:&#xA;&lt;a href=&quot;https://www.engineeringtoolbox.com/equivalent-pipe-length-method-d_804.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineeringtoolbox.com/equivalent-pipe-length-method-d_804.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your example is much more complex than the example given at that page (9 paths vs. 2), but the spreadsheet is scaleable.&lt;/p&gt;&#xA;" OwnerUserId="7763" LastEditorUserId="7763" LastEditDate="2018-01-13T14:33:09.287" LastActivityDate="2018-01-13T14:33:09.287" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18848" PostTypeId="2" ParentId="18837" CreationDate="2018-01-11T15:12:41.303" Score="0" Body="&lt;p&gt;Water (any fluid) has a maximum suction height.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is the the water boils.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even at 70 F water will evaporate and has a vapor pressure of 0.0231 ATM.&lt;br&gt;&#xA;Water will boil at 70 F and 0.0231 ATM.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure of a liquid column is density * height&lt;br&gt;&#xA;the density or water is 62.428  lb/ft3&lt;br&gt;&#xA;1 atm is 14.69595 lbs / sq inch = 2116.2168 lbs/ft2&lt;br&gt;&#xA;can just use zero for vapor pressure of water in the equation&lt;br&gt;&#xA;62.428 * height = 2116.2168&lt;br&gt;&#xA;&lt;strong&gt;height = 33.9 feet&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pump cannot produce a perfect suction. The realistic suction height is about 25 ft.  So you are right on the edge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A suction pump typically has a different design.  It is more efficient to push a liquid than pull (suck) it.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-01-11T15:12:41.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18849" PostTypeId="2" ParentId="18844" CreationDate="2018-01-11T15:19:16.110" Score="1" Body="&lt;p&gt;Blowing you push the boat backwards.  If the sail caught all your wind then it would net to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would be better off just drop the sail.  Face forward to suck and face rear to blow.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or leave the sail up and rock the boat side to side.  On a small boat you can move slow but definitely move.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-01-11T15:49:14.113" LastActivityDate="2018-01-11T15:49:14.113" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18851" PostTypeId="2" ParentId="18844" CreationDate="2018-01-11T15:34:51.237" Score="4" Body="&lt;p&gt;The answer, despite what several responders think, is unequivocally &lt;strong&gt;YES&lt;/strong&gt; . However, you will get more efficient propulsion just by blowing directly backwards. All a sail does is deflect wind, thus changing the direction of momentum (with the help of your keel or centerboard).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this works is that when you blow, you are expelling air more or less purely in the stern direction, and equal/opposite reaction rules apply.  However, when you &lt;strong&gt;inhale&lt;/strong&gt;, you draw air in from more or less all directions, and there is next to no effect on the boat's momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone who continues to doubt this is referred to swamp air-propellor-powered boats, as in agentp's comment.  &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Airboat&quot; rel=&quot;nofollow noreferrer&quot;&gt;airboat&#xA;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-01-11T15:34:51.237" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18852" PostTypeId="2" ParentId="18431" CreationDate="2018-01-11T17:29:51.487" Score="2" Body="&lt;p&gt;Octave(free and open source) and Matlab(commercial and paid) are languages that are helpful in designing and analyzing control systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They both have scripting and modeling capabilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there are several tutorials, sample scripts, and free lecture notes for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C++, Java or python could also be useful. But for engineering purposes, the simplicity of Matlab (or Octave) is a great advantage, compared to the learning curve of languages, i.e. visualization methods, finding and installing toolboxes etc for the programming languages.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2018-01-11T17:29:51.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18853" PostTypeId="2" ParentId="18764" CreationDate="2018-01-11T17:53:30.453" Score="0" Body="&lt;p&gt;The primary advantage is a sense of prestige and permanence that encourages more ridership and more development than buses. Technically heavy rail can achieve larger passenger volumes with minimal headway compared to buses (which can achieve smaller headways but are more limited in size), but commuter trains cannot achieve the same headways as heavy rail with multiple doors and level boarding, and very few heavy rail systems run enough trains to minimize those headways. So the technical advantages are more of a theoretical advantage that is not often put to the test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the US particularly, rail is seen as OK for the middle class and upper middle class commuter, while these same commuters tend to ignore bus service--even when the bus might provide faster travel times. This rather intangible preference for rail tends toward higher ridership of rail systems over bus transit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Development is also an important consideration. Developers can see a train station and expect a good return on investment by building transit oriented development around a train station--but they are wary of doing the same for a bus. A sign in the ground, a bench, maybe a shelter, can be moved around much easier than the train station and would change the appeal of a development in the typical 20 year lifespan. Bus rapid transit stations are rarely seen in the US so developers are not confident how to develop for them, either.&lt;/p&gt;&#xA;" OwnerUserId="2606" LastActivityDate="2018-01-11T17:53:30.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18854" PostTypeId="2" ParentId="18842" CreationDate="2018-01-11T19:06:38.100" Score="1" Body="&lt;p&gt;Moment in gerber joint or hinge equals zero, but moment in roller support is not necessarily zero. Moment is double integral of beam load. Reaction force in roller support causes sudden change of moment curve tangent slope. &lt;/p&gt;&#xA;" OwnerUserId="13748" LastActivityDate="2018-01-11T19:06:38.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18855" PostTypeId="2" ParentId="18842" CreationDate="2018-01-11T19:08:40.703" Score="2" Body="&lt;p&gt;Bending moment is the area under the shear diagram which is definitely increasing by a slope of 5kn/m as it gets closer to support in a straight line, so it is maximum on the support. And this moment is balanced by the reaction of roller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you remove the continuity in beam over the roller and allow a joint there thus removing the moment there, the cantilever part will immediately lose its equlibrium and rotate clockwise down and will turn into a pendelum swinging back and forth!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-11T19:08:40.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18856" PostTypeId="2" ParentId="18764" CreationDate="2018-01-11T20:41:23.173" Score="1" Body="&lt;p&gt;I don't know if its a driver, but an advantage to rail is that you can run parallel lines very close together and in close proximity to structures and such.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you'd want buses approaching each other at 100+mph on a paved road that would fit in the space of an existing rail corridor.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2018-01-11T20:41:23.173" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18857" PostTypeId="1" CreationDate="2018-01-11T20:51:11.660" Score="-1" ViewCount="90" Body="&lt;p&gt;I figured I would ask here before going out and buying an anemometer. for a 20&quot; by 25&quot; HVAC Return (or just state the size W x H that your data is based off). Assume a 3 ton system. what is the ball park expected air pressure, air speed, or CFM. with out a filter, with a new cheap filter, and if I have no filter but cover up half the opening so it becomes a 12.5&quot; by 20&quot; opening?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="14448" LastActivityDate="2018-01-12T16:35:56.257" Title="what is the air speed of an HVAC central air system return at the filter?" Tags="|fluid-mechanics|airflow|hvac|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18858" PostTypeId="1" AcceptedAnswerId="18860" CreationDate="2018-01-11T21:08:51.743" Score="1" ViewCount="59" Body="&lt;p&gt;These items: &lt;a href=&quot;https://www.parktool.com/product/dial-indicator-gauge-set-for-ts-2-2-and-ts-2-truing-stands-ts-2di&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.parktool.com/product/dial-indicator-gauge-set-for-ts-2-2-and-ts-2-truing-stands-ts-2di&lt;/a&gt; are mechanical gauges with roller tips that are intended to measure deviations from true in bicycle rims.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this kind of instrument called in general?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, what are good keywords to look for devices like this the output of which can be read electronically?&lt;/p&gt;&#xA;" OwnerUserId="14511" LastEditorUserId="12754" LastEditDate="2018-01-12T01:42:08.523" LastActivityDate="2018-01-12T06:59:22.470" Title="What are the electronic readout versions of Dial Indicators called?" Tags="|measurements|tools|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18860" PostTypeId="2" ParentId="18858" CreationDate="2018-01-11T21:26:40.860" Score="2" Body="&lt;p&gt;In machine shops, the purely mechanical types are called dial indicators. For ones that can be read electronically, the classic ones are called LVDT's or Linear Variable Displacement Transducers. There are many types of these in use today as what are called proximity sensors which measure gap widths or clearances without having to touch the object being measured. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="12754" LastEditDate="2018-01-12T06:59:22.470" LastActivityDate="2018-01-12T06:59:22.470" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18861" PostTypeId="2" ParentId="18431" CreationDate="2018-01-11T21:55:37.283" Score="1" Body="&lt;p&gt;I've taken the NI Labview Core 1,2 classes. They were very helpful. Labview also has a student version that is affordable, I believe I paid $60 USD for the license. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arduino is another very great learning tool.&lt;/p&gt;&#xA;" OwnerUserId="14512" LastActivityDate="2018-01-11T21:55:37.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18863" PostTypeId="2" ParentId="18844" CreationDate="2018-01-12T09:10:55.577" Score="1" Body="&lt;p&gt;Bit of lateral (ha!) thinking - set the sail to an angle, stand to one side of it and blow across the beam of the boat - perpendicular to the axis and intended direction of motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That way, the reaction of the person blowing is perpendicular to the desired direction of travel and is counteracted by the keel. Forward motion is provided by the forward component of the reaction of the sail deflecting a crosswind, with the keel counteracting other component forces.&lt;/p&gt;&#xA;" OwnerUserId="13043" LastActivityDate="2018-01-12T09:10:55.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18864" PostTypeId="2" ParentId="18844" CreationDate="2018-01-12T09:23:51.037" Score="0" Body="&lt;p&gt;From my interpretation of the OP the airboats fan is round the wrong way. It's blowing the air out the back which moves the boat forward. You would need to turn the fan around 180 degrees and attach a sail in the middle of the boat for it to be an accurate representation of the OP. Or make a frame and attach a sail behind the air boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you did turn the fan around and attach a sail in the middle from a basic force point of view the air would hit the fan and be transferred down the mast, back through the hull and back up through the fans frame which would cancel itself out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when the sails introduced the force dynamics are changed as pointed out by Phil. And if the sail does create twice the reaction force then the boat would move forward.    &lt;/p&gt;&#xA;" OwnerUserId="14451" LastEditorUserId="14451" LastEditDate="2018-01-12T09:35:35.657" LastActivityDate="2018-01-12T09:35:35.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18866" PostTypeId="1" CreationDate="2018-01-12T11:41:08.617" Score="7" ViewCount="8891" Body="&lt;p&gt;I was reading the book Mechanics of Materials by Beer and Johnston.The author points out in torsion chapter that ductile materials generally fail in shear.And brittle materials materials are weaker in tension than in shear.From this they concluded that when subjected to torsion a circular shaft made of ductile material breaks along a plane perpendicular to it's longitudinal axis and the brittle material break along surfaces forming angle 45° with the longitudinal axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't get how they come to the fact that ductile material fail in shear and brittle material fail in tension.Any ideas?Thanks.&lt;img src=&quot;https://i.stack.imgur.com/tbDEW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="12580" LastActivityDate="2018-07-23T23:20:00.110" Title="Why ductile materials fail in shear when subjected to torsion?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18867" PostTypeId="1" CreationDate="2018-01-12T15:57:47.893" Score="2" ViewCount="46" Body="&lt;p&gt;Using a varistor is a common practice  in alternator excitation system. It is used with rotating rectifier diode.&#xA;Is it used for transient overvoltage protection?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6aM9f.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6aM9f.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it protecting diode or rotor winding.&#xA;How this overvoltage may occur? &lt;/p&gt;&#xA;" OwnerUserId="14514" LastActivityDate="2018-01-12T22:13:31.783" Title="Transient overvoltage protection by varistor" Tags="|electrical-engineering|power-electronics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18868" PostTypeId="2" ParentId="18857" CreationDate="2018-01-12T16:35:56.257" Score="0" Body="&lt;p&gt;&quot;400 is the standard CFM per ton, but you can go from 350-450 and still be in a normal range.&quot; &lt;a href=&quot;http://hvac-talk.com/vbb/showthread.php?121424-CFM-per-ton-question&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quote Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for a dirty filter and blocking half the filter... &quot;At higher static pressures the motor is using more energy to maintain airflow. Higher pressures mean higher operating speeds. &quot;&#xA;&lt;a href=&quot;http://hvac-talk.com/vbb/showthread.php?403522-Does-a-blower-motor-with-a-dirty-filter-pull-more-or-less-than-normal-amps&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quote Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HVAC Systems vary but variable speed system will speed up to compensate for the restricted airflow to keep a constant airflow. so by blocking half the in-take the air pressure/suction will increase to keep the CFM the same.&lt;/p&gt;&#xA;" OwnerUserId="14448" LastActivityDate="2018-01-12T16:35:56.257" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18869" PostTypeId="1" AcceptedAnswerId="18917" CreationDate="2018-01-12T18:06:59.283" Score="1" ViewCount="81" Body="&lt;p&gt;I have following assembly:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lv1cl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lv1cl.png&quot; alt=&quot;Problematic assembly&quot;&gt;&lt;/a&gt;&#xA;and I have following &lt;code&gt;Dowel holes&lt;/code&gt; misaligned:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KOr3H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KOr3H.png&quot; alt=&quot;Problematic dowel holes&quot;&gt;&lt;/a&gt;&#xA;Now, both sets of holes (for both parts) were created using &lt;code&gt;Linear Sketch Pattern&lt;/code&gt; tool and now I want to mate these parts &quot;through&quot; these dowel holes sets. The first option is to recheck &lt;code&gt;Linear Sketch Pattern&lt;/code&gt; parameters for both sets, however is it possible to automatic creation of holes at exact position from first part?&lt;/p&gt;&#xA;" OwnerUserId="6290" LastActivityDate="2018-01-15T18:19:22.477" Title="Solidworks 2017 Dowel holes misaligned" Tags="|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18871" PostTypeId="2" ParentId="18834" CreationDate="2018-01-12T20:10:25.033" Score="1" Body="&lt;p&gt;A whistle works based on flow. For a given whistle design, a certain flow is required to generate the pressure oscillations which generate the sound. You can think of it like suddenly filling the resonator chamber with air, then draining it with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Venturi_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;venturi effect&lt;/a&gt;, over and over. Below a certain flow rate, the flow is laminar and just exhausts without oscillation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of the chamber will change the freqency of the sound because it takes more time to fill and drain a larger volume. This will in turn change transmission and energy characteristics of the emitted sound. However, it will have a very small effect on minimum flow required to produce sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to have a whistle trigger off of pressure you will need to configure your system to have your pressure trigger point deliver the minimum airflow to the whistle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go from a low pressure to a higher pressure trigger point on the whistle, you can restricting the airflow with a throttling valve, orifice, or a small pipe. Alternatively, you could divert part of the flow to a different path or even a second whistle in parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To go from a high pressure to a lower pressure trigger on the whistle, you will have to decrease the air passage area such that the operational pressure drop across the whistle is high enough for the lower pressure to deliver enough flow.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="15506" LastEditDate="2018-04-14T11:57:22.717" LastActivityDate="2018-04-14T11:57:22.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18874" PostTypeId="1" CreationDate="2018-01-12T22:21:20.897" Score="2" ViewCount="13035" Body="&lt;p&gt;We've got drawings for a prototype that we're trying to mimic by altering an original mold of the product.  For us to mold it would be quite expensive to just test it to see if it works or not.  We're trying to see if there's way to cut the Food Grade Silicone in a uniform way that would leave the edges smooth? This would allows us to test our drawing in the real world and get feedback. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We haven't had any success so far.  I'm not a materials guy so this is new to me.  Looking for any advice.  &lt;/p&gt;&#xA;" OwnerUserId="14518" LastEditorUserId="19930" LastEditDate="2019-08-20T01:36:11.960" LastActivityDate="2022-10-24T00:17:22.733" Title="How to cut Silicone and leave smooth edges?" Tags="|mechanical-engineering|materials|product-testing|cutting|prototyping|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="18876" PostTypeId="1" CreationDate="2018-01-13T05:12:41.073" Score="1" ViewCount="745" Body="&lt;p&gt;I am reading Mechanics of Materials by Beer and Johnston.The author points out if the torque in the shaft is removed the resulting reduction of stress and strain at a point considered will take place along a straight line .As was the case for normal stresses the shearing stress will keep on decreasing until it has reached a value equal to its maximum value at C minus twice the yield strength of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/IzR3E.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone throw some insight into it?I am not aware regarding normal stresses as well.Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastActivityDate="2018-01-15T14:27:44.010" Title="Residual stress in circular shaft subjected to torsion" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18877" PostTypeId="1" AcceptedAnswerId="18881" CreationDate="2018-01-13T06:47:25.473" Score="0" ViewCount="39" Body="&lt;p&gt;I have a wheel which has a radius of 0.1m that needs to spin at 10000RPM. The outer surface of this wheel is pressed up against another free spinning wheel like a football or tennis ball launcher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am feeding metal powder through these two wheels, which I need to travel at 104.72m/s (basically the wheel needs to spin at 10000RPM while the powder is feed through). The powder is fed at a rate of 0.013 kg/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I could work out the force on the wheel then I could work out the torque and hence power, however that is where I am not sure on what to do. The powder is being fed from almost standstill to a wheel traveling at over 100 m/s. It would seem the acceleration is almost instantaneous, but I know that's not the case. How would I work something like this out? Would I need the frictional coefficients of the wheels and powder?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Assume this is operating in a vacuum and there is no air resistance)&lt;/p&gt;&#xA;" OwnerUserId="13810" LastEditorUserId="13810" LastEditDate="2018-01-13T23:49:26.330" LastActivityDate="2018-01-14T11:39:06.657" Title="Power needed to accelerate metal powder through two spinning wheels" Tags="|dynamics|acceleration|tribology|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18878" PostTypeId="1" CreationDate="2018-01-13T08:39:57.713" Score="0" ViewCount="1244" Body="&lt;p&gt;Suppose there is a vehicle with difference size of wheels in front and in rear. Specifically bicycle, There would be 2 cases that front wheel larger than rear wheel and vice versa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what is advantage and disadvantage between each setting. From normal setting to larger front and to smaller front. While rear wheel would be a driven wheel and front wheel would be a control. If control wheel smaller or larger which would become harder? Or which setting would be less efficient?&lt;/p&gt;&#xA;" OwnerUserId="1606" LastEditorUserId="7461" LastEditDate="2018-01-13T16:40:23.560" LastActivityDate="2018-01-13T17:43:56.240" Title="What is an advantage/disadvantage of difference wheels size in front and rear of a bicycle?" Tags="|wheels|bicycles|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18880" PostTypeId="2" ParentId="18844" CreationDate="2018-01-13T11:14:25.650" Score="1" Body="&lt;p&gt;Well After some googling I found this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=uKXMTzMQWjo&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=uKXMTzMQWjo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does indeed work possibly because the sail deflects the wind backward meaning that you go forward with conservation of momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it wouldn't work but hadn't figured that the wind would bounce off the sail&lt;/p&gt;&#xA;" OwnerUserId="14504" LastActivityDate="2018-01-13T11:14:25.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18881" PostTypeId="2" ParentId="18877" CreationDate="2018-01-13T13:48:00.273" Score="2" Body="&lt;p&gt;You actually don't need to know the force on the wheel to work on the power.  Just look at the kinetic energy of the powder.  Kinetic energy = $(1/2)mv^2$.  For 0.013 kg of powder, the kinetic energy is $(1/2)(0.013)(104.72)^2=71$ Joules.  You need to provide that much Joules every second, so the power is just 71 Joules/second = 71 Watts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this assumes that the bearings in your wheels are frictionless.  They are not.  It will take some amount of power just to spin the wheels with no powder. This will depend on how good your bearings are.  That's something that you could easily measure if you have the wheels already.  If not, probably just round up to something like 100W and that should be more than enough.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="9275" LastEditDate="2018-01-14T11:39:06.657" LastActivityDate="2018-01-14T11:39:06.657" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18882" PostTypeId="2" ParentId="18878" CreationDate="2018-01-13T14:51:54.740" Score="1" Body="&lt;p&gt;In the case of a bicycle a larger diameter wheel rolls up and over stuff like a root easier.  Since the rear is power you can power over the root.  You see some 29' front and 26' rear but the inconvenience of two size of tube and tire is a major drawback.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-01-13T14:51:54.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18883" PostTypeId="2" ParentId="18878" CreationDate="2018-01-13T17:43:56.240" Score="0" Body="&lt;p&gt;There are several that come to mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drag is due to rolling resistance at the contact patch, bearing resistance, and wind drag. These all operate with different scaling laws, both with respect to speed, and size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inertia is very important to the feel of a bike. Getting the caster and inertia right for the bike has a huge effect on steering. Wheel size affects weight, and therefore inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wheel size affects the gearing needed. This is the big one for conventional chain drive bikes. If you want reasonable speeds, you need a decent sized rear wheel because of the constraints on single stage chain drive gear ratios. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Propulsion and braking. These often aren't equally distributed, and you need to provide acceptable tire wear. You can do it with width or diameter. But in the case of a farm tractor, width is constrained by the space between rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a kid, I owned the only bike in town with a 24&quot; rear and 20&quot; front tire (thanks dad), and I got made fun of over it. I was not aware of any offsetting benefit at the time. Although the practice I got from being chased probably helped me to become a competitive racer later on.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-01-13T17:43:56.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18884" PostTypeId="1" AcceptedAnswerId="18887" CreationDate="2018-01-13T19:12:30.520" Score="1" ViewCount="723" Body="&lt;p&gt;I've seen the terms &quot;SCADA boxes&quot;, &quot;SCADA out-of-the-box&quot;, &quot;SCADA turnkey systems&quot; used in various articles but without a definition. I'd like to know whether the terms are synonymous and if not what is the &quot;generally accepted&quot; concept of a &quot;SCADA box&quot;?&lt;/p&gt;&#xA;" OwnerUserId="14523" LastActivityDate="2018-01-13T23:22:33.793" Title="What are SCADA Boxes?" Tags="|control-engineering|industrial-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18885" PostTypeId="1" CreationDate="2018-01-13T21:03:49.107" Score="0" ViewCount="2608" Body="&lt;p&gt;I was under the impression that the entropy reduction in a system only works with either mass or heat transfer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I was researching intercooling of compressors I stumbled across a peculiar sketch (Figure 8-46 from Çengel: Introduction to Thermodynamics and Heat Transfer). I cross-checked with my thermodynamics textbook (Figure 9-43 from Çengel: Thermodynamics - An Engineering Approach). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fq8Ur.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fq8Ur.png&quot; alt=&quot;cengel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entropy gets smaller throughout the complete compression process. While I do understand that it gets smaller when cooled I do not understand why the entropy reduces prior and after the intercooling.&#xA;Figure 9-43 shows the compression as isentropic (which is often done to simplify the analysis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing? The only idea would be a heat-loss during the compression. I then realised that a bit later in the book a similar graph shown in Fig. 7-34:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sLPky.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sLPky.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for a reduction in entropy other that heat-loss in these cases?&#xA;Is it maybe a didactic choice?&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="12754" LastEditDate="2018-01-14T21:28:59.203" LastActivityDate="2018-01-14T21:28:59.203" Title="What is the reason for a reduction in entropy in a compressor" Tags="|thermodynamics|compressors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18886" PostTypeId="1" CreationDate="2018-01-13T23:13:53.793" Score="2" ViewCount="3592" Body="&lt;p&gt;I'm trying to find the required Newton or Nm of torque to move 6000&amp;nbsp;kg up a grade at 100&amp;nbsp;km/h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to work out the closest suited engine platform to do the job while been efficient, durable,cost effective and most of all required torque.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mass: 6000&amp;nbsp;kg&lt;/li&gt;&#xA;&lt;li&gt;Speed: 27.77&amp;nbsp;m/s = 100&amp;nbsp;km/h&lt;/li&gt;&#xA;&lt;li&gt;Grade: 7° = 12%&lt;/li&gt;&#xA;&lt;li&gt;Wheel size: 830&amp;nbsp;mm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14525" LastEditorUserId="1832" LastEditDate="2018-01-14T19:56:47.933" LastActivityDate="2018-01-15T22:34:11.623" Title="Required torque to move 6000kg at 27.77m/s" Tags="|torque|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="18887" PostTypeId="2" ParentId="18884" CreationDate="2018-01-13T23:22:33.793" Score="2" Body="&lt;p&gt;The plant I used to work at had a Supervisory Control And Data Acquisition (SCADA).  They had all kinds of transducers measuring everything in sight and we could sit in the control room and check on things all over the plant.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;em&gt;warstory&lt;/em&gt;)&#xA;One night (when I should have been working) I was poking around and noticed the temperature of the Feedwater pump A bearing was high and just then, the call came in from the field that the Feedwater pump B was burning up and we needed to take it down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I showed the high temp on the A pump to the operator on duty and he checked with the field operator who confirmed it was Feedwater pump B so he cut Feedwater B.  The operator in the field immediately called back with &quot;A, the A pump, not the B pump.&quot;  The good news was we were able to bring B back up without too much delay and the plant had no difficulty.&#xA;(&lt;em&gt;end warstory&lt;/em&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, I have no recollection of being able to actually control anything with the SCADA system, such as turn on equipment, it was all for data logging.  How did that differ from the Plant Monitoring System (PMS)?  Again, I don't recall except to say that it allow monitoring all kinds of things that were not available to the PMS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The SCADA system had these Remote Terminal Units (RTUs) out in the field that would interface with the transducers and send the data back to the system that was used for display.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am suspicious of anybody that says they have a turn-key SCADA system.  I expect they will sell you their off-the-shelf display software and the RTUs (or whatever their modern equivalents are) and a then charge you for the (possibly huge amount of) time spent on configuration.  I would rather sign up to buy an &quot;off-the-shelf&quot; SCADA than try to cook something up in-house but it is hardly turn-key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;But that is not your question.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I can find on the web, it appears the SCADA-boxes you refer to are the modern equivalent of RTUs.  They receive the data from the transducers, package it up and send it back to the main data logging and display system via some sort of network.  The out-of-the-box and turnkey systems are the same thing.  You buy display and data logging software from the vendor and either play hell configuring it yourself or pay them big bucks to do the configuration for you.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2018-01-13T23:22:33.793" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18888" PostTypeId="1" CreationDate="2018-01-14T00:52:09.017" Score="1" ViewCount="555" Body="&lt;p&gt;I would like to find the full (12 by 12) Timoshenko beam element stiffness matrix for a variable cross section with mass center axis and elastic center does not coincide. I have the full stiffness matrix for a constant cross section with mass center axis and shear center axis does not coincide as shown &lt;a href=&quot;https://www.researchgate.net/post/Beam_element_full_stiffness_matrix_for_a_variable_cross_section_with_non_coincidence_of_CG_and_shear_center&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Now I need to find the stiffness matrix for a tapered beam such as the one below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O1mbx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O1mbx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14526" LastEditorUserId="1832" LastEditDate="2018-05-29T19:11:28.850" LastActivityDate="2018-05-29T19:11:28.850" Title="Beam element stiffness matrix for variable area cross section with mass and shear center not coinciding" Tags="|structural-engineering|structural-analysis|structures|aerospace-engineering|finite-element-method|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="18889" PostTypeId="1" CreationDate="2018-01-14T01:34:44.453" Score="0" ViewCount="24" Body="&lt;p&gt;I'v been preparing a seminar related to vibrations_damping systems when it come to the two topics: the active controller systems and the passive ones, could anyone please tell the simplest definitions and differences between them both, thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="14527" LastActivityDate="2018-01-14T19:10:55.233" Title="active vibrations_damping systems" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18890" PostTypeId="1" CreationDate="2018-01-14T07:20:20.940" Score="0" ViewCount="150" Body="&lt;p&gt;Lets pretend I designed a plastic Lunchbox, like the ones pictured below in solid-works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SQcDP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SQcDP.jpg&quot; alt=&quot;this&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What processes or options are involved in obtaining a physical sample of my design?&lt;/p&gt;&#xA;" OwnerUserId="14529" LastActivityDate="2020-11-14T06:03:55.407" Title="How would I get my hands on something I designed in Solid works?" Tags="|mechanical-engineering|design|cad|plastic|solidworks|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18892" PostTypeId="1" CreationDate="2018-01-14T09:24:28.837" Score="0" ViewCount="63" Body="&lt;p&gt;I would like to evaluate failure in steel using FEM, especially in geometries with stress concentrations (notches etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possibility is to calculate the stress distribution using a purely linear equation of state (elastic = Hooke's law) and evaluate the von Mises stress, using appropriate stress concentration factors (plastische Stützzahlen in German). The drawback is that you need to know (or approximate) the appropriate stress concentration factor(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is to calculate the stress distribution using a nonlinear equation of state. For steel I read that the bilinear elasto-plastic model is realistic enough for most cases. However, I'm unclear what to evaluate as failure criterion. I noticed that when I change the load to a certain model by a factor of two, the maximum von Mises stress does not change significantly (once it is in the plastic region). I suppose this is due to the values of the two moduli (elastic modulus and tangent modulus) being at least one order of magnitude apart (210 GPa vs. 1-10 GPa). My assumption is that I can use the maximum strain or maximum plastic strain as failure criterion, however I was unable to find a reference supporting this assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that I am interested in failure from a single loading event only. I understand that the fatigue failure mechanisms (when load cycles are applied) are quite different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is: What can I use as failure criterion, and why? Please support your answer with a (link to a) book chapter or publication - thanks.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2018-01-14T09:24:28.837" Title="Plastic failure evaluation of steel using FEM" Tags="|mechanical-engineering|finite-element-method|mechanical-failure|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18894" PostTypeId="2" ParentId="18890" CreationDate="2018-01-14T11:55:12.350" Score="1" Body="&lt;p&gt;Without additional information regarding your design, the answer is difficult and has to cover multiple aspects. As noted in the comments, injection molding is expensive to set up, especially to produce one unit. You have not indicated that your design requires this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your design lends itself to 3D printing, there are online 3D printing services such as Shapeways and others which will take your design and return to you a model created in the material of your choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even a 3D printed design such as the lunch box, which would have fragile walls due to lack of thickness, could be created if you did not require structural strength. The same model could be used as the master of a home-made mold which would then allow for resin casting at relatively low cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the design does not lend itself to 3D printing, there are fabrication services such as emachineshop.com but they would require a different source reference, likely in DXF format. SolidWorks can generate such documents, to facilitate the construction/fabrication of your work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all cases, you would have each component created separately and assembled after creation. An exception to this is SLS 3D printing, in which the nylon powder used to create the part also serves as the support and separator material. Moving parts can be created in one print, fully assembled. SLS 3D printing is very expensive compared to other forms of additive manufacturing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the reverse side of the coin is machining. Your design could be machined on a CNC milling center. This process would require a block of material slightly larger than your design and all portions of the block that are not part of the design are carved away by high speed cutters of various shapes. Because this process requires a large block of raw material and removes so much, it is also quite expensive. Solidworks is capable of creating the files necessary for producing an item in this manner.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-01-14T11:55:12.350" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18895" PostTypeId="1" AcceptedAnswerId="18922" CreationDate="2018-01-14T12:24:06.590" Score="0" ViewCount="87" Body="&lt;p&gt;By finite element type I specifically mean element geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of Beam/Strut, Quad, Tri, Tet &amp;amp; Hex but are there other types, eg a hexagonal prism, rhombihedral etc.. ?&lt;/p&gt;&#xA;" OwnerUserId="14531" LastActivityDate="2018-01-16T00:03:28.123" Title="Are there other finite element types besides the usual?" Tags="|finite-element-method|meshing|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18896" PostTypeId="1" CreationDate="2018-01-14T13:35:08.020" Score="2" ViewCount="342" Body="&lt;p&gt;&lt;strong&gt;Goal&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am creating a project where I will need to move small beans around in turbulent air flow. I will have a heat source to which I do not want the heat to leave easily (I want the beans to be allowed to heat up). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially tried to purchase a fan that simply had high CFM (about 400 cfm or so) but that seemed to be a poor choice as it just sucks all of my heat away and does not move my beans very effectively.  This fan was simply a 120mm computer case fan. &lt;strong&gt;I believe my problem was that it pushes air with high velocity but with little momentum behind it.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My thought process&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I feel that this &lt;a href=&quot;https://pdhonline.com/courses/m213/m213content.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; gives a pretty great definition of the different types of fans and motors. I am not product searching but rather just trying to pick the appropriate blower / fan type that would best help my application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that I want &lt;strong&gt;low CFM&lt;/strong&gt; and &lt;strong&gt;high static pressure&lt;/strong&gt; if I am understanding correctly? Based on this I think that the squirrel cage type blowers may be good enough. Please do correct me if someone has more experience with this though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Follow up&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once a type is established, what is the best method to actually size up the right sized blower?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yMukp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yMukp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14534" LastEditorUserId="14534" LastEditDate="2018-01-14T19:46:52.673" LastActivityDate="2018-02-15T22:58:51.020" Title="Which fan / blower type should I chose?" Tags="|airflow|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="18897" PostTypeId="2" ParentId="18885" CreationDate="2018-01-14T14:40:59.737" Score="2" Body="&lt;p&gt;\begin{equation}&#xA;{\displaystyle \Delta S=nC_{v}\ln {\frac {T}{T_{0}}}+nR\ln {\frac {V}{V_{0}}}.}&#xA;\end{equation}&#xA;Snipped from &lt;a href=&quot;https://en.wikipedia.org/wiki/Entropy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Entropy: wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node40.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;5.4 Entropy Changes in an Ideal Gas&lt;/a&gt;, which I recommend you read starting at Section 5. It's taken from some of MIT's online resources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Chemomechanics says in his coments, the necessary condition is that there is enough heat loss from the fluid during compression such that the volume term  decreases by more than the temperature term increases. &lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-01-14T15:51:17.440" LastActivityDate="2018-01-14T15:51:17.440" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18898" PostTypeId="2" ParentId="18889" CreationDate="2018-01-14T19:10:55.233" Score="1" Body="&lt;p&gt;a passive controller system contains only passive components: springs, masses, and friction devices. it has no motors, sensors, actuators nor an external power source. an active controller system has all of these. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-14T19:10:55.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18899" PostTypeId="1" CreationDate="2018-01-14T20:13:31.903" Score="0" ViewCount="43" Body="&lt;p&gt;I'm currently thinking about creating a small piano-like (but of course nowhere as big or complicated) music box which can be digitally controlled with a microcontroller. Therefore I need to know which kind of electronic component has the ability of creating a rapid movement throwing up a kind of &quot;hammer&quot; against a string. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be able to have a maximum speed of 0.2 m/s with a maximum force of ~45N. Minimal speed should be approximately 0.05 m/s with a force of only 5N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a component which fits these requirements? (Or do you have any other suggestions how I could solve this?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I initially wanted to post this on the electronics-forum, but I assume the one for engineering fits my question better, if I'm incorrect, I apologize]&lt;/p&gt;&#xA;" OwnerUserId="14540" LastEditorUserId="2819" LastEditDate="2018-01-15T19:33:13.740" LastActivityDate="2018-01-15T19:33:13.740" Title="Type of electric actuator for a music box?" Tags="|mechanical-engineering|electrical-engineering|mechanisms|equipment-selection|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18900" PostTypeId="1" CreationDate="2018-01-14T21:19:54.650" Score="3" ViewCount="377" Body="&lt;p&gt;During my time revising for my control engineering module, i've come across this question  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The automatic cruise speed control for a vehicle is described by the following equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigl( 1\bigl)$ $m \frac{dv_0(t)}{dt} + Fv_0(t)=u(t)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigl( 2\bigl)$ $u(t)=K[v_0(t)-v_r(t)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where m is the mass of the car, 1000$kg$, F is the frictional constant 50 $Nm^{-1}s$ and $u(t)$ is the throttle actuator force and K is the proportional gain control for the actuator.&#xA;Also $V_0(t)$ is the actual speed while $V_r(t)$ is the desired speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two parts to this question, and im a bit confused on both of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;part 1 says to draw a block diagram of the speed control system. In the control engineering module we generally draw the block diagrams as related to s, so i thought i'd get the laplace transform of the function (to get the transfer function) and draw it from the general model of a control system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This coincided with part 2, which wants me to show that the transfer function in terms of K is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{V_0(s)}{V_r(s)} = \frac{K}{1000s+50+K}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went about &lt;a href=&quot;https://i.stack.imgur.com/nMD2u.jpg&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;working it out&quot;&gt;working it out&lt;/a&gt; and eventually ended up with this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{V_0(s)}{V_r(s)} = \frac{-K}{1000s+50-K}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me figure out where i went wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, i attempted making a block diagram in the t domain for this system, however since there are two equations with equal outputs (not summed) i was completely stumped on how i would display that as a model that is accurate, without making assumptions like equating K with the system gain of the other equation&lt;/p&gt;&#xA;" OwnerUserId="14544" LastActivityDate="2018-01-15T10:40:36.267" Title="Determining the transfer function of cruise control given characteristic equations" Tags="|electrical-engineering|control-engineering|control-theory|" AnswerCount="3" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18901" PostTypeId="1" CreationDate="2018-01-14T21:36:33.790" Score="1" ViewCount="163" Body="&lt;p&gt;I need a heat sink for my Peltier cooling fan but also need to saves as much vertical space as possible. I've been exploring various Heat Sink options but many of them extrude outwards making compact variants hard to find. I've found compact options such as Flared Fin Heat Sinks: &lt;a href=&quot;https://i.stack.imgur.com/cg1kg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cg1kg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it would really be optimal if the fins just went to one side for my application as need the heat dispersal to be concentrated in one direction. My question is could something like this be at all effective for my peltier TEC: &lt;a href=&quot;https://i.stack.imgur.com/ENvkJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ENvkJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14542" LastActivityDate="2018-01-15T19:49:23.503" Title="Would if be possible to have an efficient, diagonal heat sink to save vertical space?" Tags="|mechanical-engineering|heat-transfer|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18902" PostTypeId="2" ParentId="18900" CreationDate="2018-01-14T22:31:22.920" Score="3" Body="&lt;p&gt;If you are using $(2)$ as your definition for $u(t)$ then your expression is correct. However in control it is often the convention to use negative feedback. In this case equation two would become&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;u(t) = K [v_r(t) - v_0(t)] \tag{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is equivalent to negating your value of $K$. This would yield the same as the given result.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-01-14T22:31:22.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18903" PostTypeId="2" ParentId="18899" CreationDate="2018-01-15T00:37:35.740" Score="2" Body="&lt;p&gt;you can use a solenoid mechanism. these are commonly used to actuate door locks in cars with electric locks, meaning that 1) they run on 12VDC (convenient to work with) and 2) a junk yard would be a good place to search for an example you could test. bear in mind that there are two basic types of electric door lock actuators; one uses a solenoid assembly and the other uses a DC gearmotor. Ask the junkyard guys where to find a solenoid type. Bring your tools!&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-15T00:37:35.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18904" PostTypeId="1" AcceptedAnswerId="18915" CreationDate="2018-01-15T05:32:06.287" Score="1" ViewCount="1484" Body="&lt;p&gt;In 802.11n 2.4Ghz, the spectrum is varied from 2400Mhz to 2500Mhz and consists of 13 (or 14) channels, each channel having 20Mhz band width. &#xA;When I search google for best channel, all websites say 1,6 and 11 are the best channels since they won't overlap. But from my study, i found the following.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;channel 1 : 3 [2, 3, 4]&#xA;channel 2 : 4 [1, 3, 4, 5]&#xA;channel 3 : 5 [1, 2, 4, 5, 6]&#xA;channel 4 : 6 [1, 2, 3, 5, 6, 7]&#xA;channel 5 : 6 [2, 3, 4, 6, 7, 8]&#xA;channel 6 : 6 [3, 4, 5, 7, 8, 9]&#xA;channel 7 : 6 [4, 5, 6, 8, 9, 10]&#xA;channel 8 : 6 [5, 6, 7, 9, 10, 11]&#xA;channel 9 : 6 [6, 7, 8, 10, 11, 12]&#xA;channel 10 : 6 [7, 8, 9, 11, 12, 13]&#xA;channel 11 : 5 [8, 9, 10, 12, 13]&#xA;channel 12 : 4 [9, 10, 11, 13]&#xA;channel 13 : 3 [10, 11, 12]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Channel 1 overlaps with 3 channels, Channel 6 overlaps with 6 channels and finally channel 11 overlaps with 5 channels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from my study, I conclude that channel 1,2,3,11,12,13 have less overlap interference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct?&lt;/p&gt;&#xA;" OwnerUserId="14547" LastEditorUserId="14498" LastEditDate="2018-01-15T12:09:00.487" LastActivityDate="2018-01-15T17:07:58.630" Title="Why channel 1, 6 and 11 are best for Wifi communication 802.11n?" Tags="|rf-electronics|wifi|waves|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18905" PostTypeId="1" CreationDate="2018-01-15T07:24:44.640" Score="3" ViewCount="384" Body="&lt;p&gt;I'm performing an experiment based on slag corrosion behavior of the refractory material. I need to store molten metal at $\approx$ 1600-1700 K for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I'm working on slag corrosion and penetration behaviour of refractory (MgO-MgAl2O4). I'm going to use immersion method, so in this method &quot;Cylindrical or square pillar shaped samples are held in the corrosive liquid(slag) in a furnace. Immersion time, temperature and atmosphere can be varied&quot; So, for storing slag+molten metal, I need to use a crucible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest what would be the best choice of material for crucible.&#xA;What is a crucible material?&lt;/p&gt;&#xA;" OwnerUserId="14549" LastEditorDisplayName="user16" LastEditDate="2018-01-16T00:12:30.153" LastActivityDate="2018-10-13T06:00:04.950" Title="What material should I use to store molten metal?" Tags="|metallurgy|corrosion|ceramics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18906" PostTypeId="2" ParentId="18904" CreationDate="2018-01-15T08:45:04.127" Score="1" Body="&lt;p&gt;All channels good but:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;National standards can control some parameters (gain, numbers of channels, gain in channel ...). Gain of Channel 1 can be less in some cases.&lt;/li&gt;&#xA;&lt;li&gt;I have 19 visible ssids in my workplace so channel is best if it only my and have no neighbors.&lt;/li&gt;&#xA;&lt;li&gt;often if no activity wifi routers choose channel 6, and it`ll be the most loaded.&lt;/li&gt;&#xA;&lt;li&gt;often all routers automatically select channels and can change it next time. So you can plan channels load only if you can set channels manually for all wifi routers.&lt;/li&gt;&#xA;&lt;li&gt;you can set come nonstandard frequency (if your router can) but client devices must understand it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lkrgk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lkrgk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use 'auto' channel or use wifi-analyzer before setting number of channel manually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EO1VY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EO1VY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14551" LastEditorUserId="8055" LastEditDate="2018-01-15T15:03:34.600" LastActivityDate="2018-01-15T15:03:34.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18907" PostTypeId="2" ParentId="18900" CreationDate="2018-01-15T10:40:36.267" Score="1" Body="&lt;p&gt;If it was me, and bearing in mind what the other answer said about negative feedback, I would do something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/69Vov.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/69Vov.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change the gain blocks from &lt;code&gt;1&lt;/code&gt; to the correct parameters. &lt;code&gt;vr&lt;/code&gt; is obviously your input/reference speed profile. I don't think there's any need to compute the transfer function to do a block diagram.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2018-01-15T10:40:36.267" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18908" PostTypeId="2" ParentId="18886" CreationDate="2018-01-15T11:21:48.483" Score="1" Body="&lt;p&gt;It seems you are looking for power, not torque. Torque (together with RPM) is just one of the factors for calculating power and can be amplified or reduced by using gears. It is not possible to just give a needed torque figure for an engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all there are too many variables missing anyway.&#xA;For all i know we could be talking about a moving vehicle or a stationary setup like a cable elevator, which could lead to a different solution.&lt;/p&gt;&#xA;" OwnerUserId="14479" LastEditorUserId="14479" LastEditDate="2018-01-15T11:27:31.600" LastActivityDate="2018-01-15T11:27:31.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18909" PostTypeId="1" CreationDate="2018-01-15T11:35:06.067" Score="0" ViewCount="1181" Body="&lt;p&gt;While I’m studying about a fired heater (furnace) in chemical engineering , I learnt that the process fluids enters from the convection section and leaves at the radiant section of the furnace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this the case ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this linked to the temperature of the hot flue gas rising up? The temperature of the hot flue gas is lower in the convection section. So how does it link to why the process fluid is entering from the convection section and not radiant section ? Thanks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7ZEoO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7ZEoO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture shows the process fluid entering from the convection section and leaves at the radiant section. So to clarify, my question is asking why does it enter from the convection section and not the radiant section. ? &lt;/p&gt;&#xA;" OwnerUserId="13646" LastEditorUserId="13646" LastEditDate="2018-01-15T11:44:18.410" LastActivityDate="2018-02-14T15:23:40.480" Title="Process fluid flow in a fired heater" Tags="|heat-transfer|chemical-engineering|heating-systems|heat-exchanger|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18910" PostTypeId="1" AcceptedAnswerId="18936" CreationDate="2018-01-15T14:26:47.413" Score="1" ViewCount="291" Body="&lt;p&gt;I own this vaporizer: &lt;a href=&quot;https://www.dynavap.com/dynashop/vaporizers/omni/omnivap/xl-titanium&quot; rel=&quot;nofollow noreferrer&quot;&gt;(product link)&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7LQmC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7LQmC.png&quot; alt=&quot;Dynavap vaporizer&quot;&gt;&lt;/a&gt;&#xA;It has no electrical parts. To use it, you take off the cap at the left, put some &quot;herbs&quot; in it to smoke (vaporize), and put the cap back on. Then you heat the cap with a torch lighter until it says click. The click indicates that the cap (made of stainless steel) is at the right temperature. Then you can inhale from the right part and hot air with vaporized herbs will enter your body. It will click again when it has cooled down. &lt;a href=&quot;https://youtu.be/i86qSoFolVs&quot; rel=&quot;nofollow noreferrer&quot;&gt;(video)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this click work? Can I make this myself with a CNC machine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The click system is at the end of the cap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R6alB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R6alB.jpg&quot; alt=&quot;Cap&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14554" LastActivityDate="2018-01-16T18:28:58.870" Title="How does this mechanical heat click work?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18911" PostTypeId="2" ParentId="18876" CreationDate="2018-01-15T14:27:44.010" Score="1" Body="&lt;p&gt;When loaded above the yield point ($Y$ in your sketch), ductile material will start to deform plastically. When you stop at some point before the material fails and reduce the load again, the material will relax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Citing the &lt;a href=&quot;https://en.wikipedia.org/wiki/Plasticity_(physics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia article on plasticity&lt;/a&gt;: &quot;Plasticity in a crystal of pure metal is primarily caused by two modes of deformation in the crystal lattice, slip and twinning. Slip is a shear deformation which moves the atoms through many interatomic distances relative to their initial positions. Twinning is the plastic deformation which takes place along two planes due to a set of forces applied to a given metal piece.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the relaxation will be elastic. However, since the internal structure has been changed somewhat during plastic deformation, residual stresses will remain inside your specimen. There also is some lasting deformation. When loaded into the opposite direction, the yield limit is reduced. The length of the elastic line is the same for the pre-deformed specimen as for a new specimen (which is $2\tau_Y$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The underlying principle is the same for torsion and for pure tension. The major difference is that when twisted, only an outer layer of the material will undergo plastic deformation when loaded above the yield point, while the tension probe will undergo plastic deformation in the whole cross section. This is because strain and stress increase linearly with radius in a torsion probe (see &lt;a href=&quot;http://www.ae.msstate.edu/tupas/SA2/chA6.2_text.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for an example), but are independant on the location in a tension probe.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2018-01-15T14:27:44.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18912" PostTypeId="2" ParentId="18909" CreationDate="2018-01-15T14:55:16.813" Score="2" Body="&lt;p&gt;It is just getting as much heat as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat flows from hot to cold.  The convection section is a much lower temperature. The convection section is basically waste heat in flue gas. The cold process fluid  is (pre) heated to recover some of the waste heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exit process fluid may be hotter than the flue gas.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any heat exchanger you should save the hottest heat for last.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-02-14T15:23:40.480" LastActivityDate="2018-02-14T15:23:40.480" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18913" PostTypeId="1" CreationDate="2018-01-15T15:36:11.163" Score="3" ViewCount="328" Body="&lt;p&gt;I'm trying to model the effects of solar radiation on a (thick) concrete wall. My assumptions are followings:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The wall behaves as an half continuum with parameters $k = 0.8 W/Km$, $\rho=2500 kg/m^3$, $c_p=880J/kg K$&lt;/li&gt;&#xA;&lt;li&gt;Solar radiation is simulated as an heat flux on the wall surface $\dot{q}(0,t) = 650 W/m^2$&lt;/li&gt;&#xA;&lt;li&gt;There is an exchange of heat between wall and surrounding air,  modeled using an heat transfer coefficient $h=15W/m^2K$ and considering air temperature as constant with $\vartheta_\text{air}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now I should solve the heat equation &#xA;$$ u_{,t} -\alpha u_{,xx}=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under following IC and BC:&#xA;$$\begin{cases} u(x,0) = \vartheta_0 \\ 650+ h(\vartheta_\text{air}-u(0,t))= -k u_{,x}(0,t) \end{cases}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to solve this PDE using Mathematica, but gave me a nonsensical result. Taking $h=0$ however leads to good results. So I'm quite sure the error lies in the consideration of the flux between wall and air (boundary condition), but don't see it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here is the code I used in mathematica.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;k=0.8; c=880;  rho=1950; a=k/(c*rho); h=15; theta_air=0;&#xA;&#xA;heqn=D[u[x,t],t] == a*D[u[x,t],{x,2}];&#xA;ic1=u[x,0] == 0;&#xA;bc1 = h*(theta_air-u[0,t])+650 == -k*Derivative[1,0][u][0,t]&#xA;theta_0 = 20;&#xA;&#xA;sol=DSolveValue[{heqn,ic1,bc1},u[x,t],{x,t}][[1,1,1]]&#xA;&#xA;f[x_,t_]=sol;&#xA;theta_0+f[0.,7200]&#xA;theta_0+f[0.1,7200]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which gives:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;15.3649+0.i&#xA;2.6153+0.i&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Again, this PDE gives the correct result assuming $h=0$ (no heat exchange between wall and air). This result was checked with the solution of an exercise found on a book.&lt;/p&gt;&#xA;" OwnerUserId="12220" LastEditorUserId="12220" LastEditDate="2018-02-15T09:04:32.587" LastActivityDate="2018-02-15T09:04:32.587" Title="Heat equation for solar radiation on a wall" Tags="|civil-engineering|thermodynamics|heat-transfer|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="18914" PostTypeId="2" ParentId="18627" CreationDate="2018-01-15T16:27:07.870" Score="2" Body="&lt;p&gt;There's no structural issue to any code that I know of. In the final state for design strength calculations construction joints are ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it may make construction more difficult. Ask the contractor (if he's already been appointed for the project).&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-01-15T16:27:07.870" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18915" PostTypeId="2" ParentId="18904" CreationDate="2018-01-15T17:07:58.630" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Q: &quot;Why channel 1, 6 and 11 are best for WiFi ... ?&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The are three major sources of interference: Co-Channel, Adjacent-Channel, and non-WiFi.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://community.arubanetworks.com/t5/Technology-Blog/An-Intro-to-Co-Channel-Interference/ba-p/273248&quot; rel=&quot;nofollow noreferrer&quot;&gt;Co-Channel interference&lt;/a&gt; is simply where other WiFi equipment is using the same channel.  and you need to share timeYour equipment waits until the channel is clear before trying it's luck, if there's enough competition or a fault in the equipment you get a collision; meaning you must wait and try again.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When setting up &lt;strong&gt;your own&lt;/strong&gt; equipment on &lt;strong&gt;different channels&lt;/strong&gt; and seek to not interfere &lt;strong&gt;with yourself&lt;/strong&gt; you would choose channels 1, 6, 11, and 14 (if it's available in your area).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gtMWR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gtMWR.png&quot; alt=&quot;2.4 GHz channel widths&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avoiding interference from other people's WiFi involves using a WiFi APP on your cellphone (or test equipment) to determine which channels are least congested. If you enable &lt;a href=&quot;https://en.wikipedia.org/wiki/Channel_bonding#Wi-Fi&quot; rel=&quot;nofollow noreferrer&quot;&gt;channel bonding&lt;/a&gt; you might double your transmission speed but you're more likely to double your collisions if anyone else is nearby.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Adjacent channel interference occurs when a nearby channel is used by either you or your neighbor. If you use channel 1 and everyone else uses channel 6 or higher then your in luck.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Interference can also occur from non-WiFi sources such as cordless phones or microwaves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lfhar.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lfhar.png&quot; alt=&quot;Sources of interference&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people only use a single WiFi channel in their home. In that case you're not going to want to stick with 1,6,11 but instead survey the channels and attempt to find one that is least used (all of the time).&lt;/p&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2018-01-15T17:07:58.630" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18916" PostTypeId="1" CreationDate="2018-01-15T17:30:09.723" Score="1" ViewCount="72" Body="&lt;p&gt;Before to present my question, I have to say that I'm not a hydraulic expert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm working on a water valve design and I found this model which work with a diaphragm and a spring. I want to consider all the variables acting in this mechanism in order to understand how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how I think it works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water flow pushes the diaphragm to let the water flow pass through the valve, the dynamic pressure in the pipe is about 14 PSI. When the water tank starts to fill, buoyancy force acts against the float closing a little hole which acts like a water leak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cxalA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cxalA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water inside the diaphragm chamber generates a force that pushes the diaphragm up, sealing the water intake and cutting the water flow through the water tank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O0EBp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O0EBp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the water valve cut the water flow the static pressure in the pipe reach 40 PSI.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this, I have some questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the function of the spring? &lt;/li&gt;&#xA;&lt;li&gt;The sealing force generates by the diaphragm is higher than the force according to the 40 PSI of static pressure, or they are equal?&lt;/li&gt;&#xA;&lt;li&gt;The pressure inside the diaphragm chamber rises when the water leak is sealed?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14265" LastEditorUserId="8055" LastEditDate="2018-01-15T17:57:32.480" LastActivityDate="2018-01-15T17:57:32.480" Title="Water Valve Functionality" Tags="|fluid-mechanics|hydraulics|compressible-flow|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18917" PostTypeId="2" ParentId="18869" CreationDate="2018-01-15T18:19:22.477" Score="1" Body="&lt;p&gt;Perhaps I'm not fully understanding you but it seems like what you are trying to do is create the dowels and the dowel holes in the same action; this is fairly straightforward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumable you used &lt;strong&gt;Extruded Cut&lt;/strong&gt; to make these holes, what you want to do is replace that feature with an &lt;strong&gt;Extruded Boss/Base&lt;/strong&gt; using the same sketch and be sure to uncheck the &lt;strong&gt;Merge Entities&lt;/strong&gt; box so that the extrusion is a separate body from your main part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to &lt;strong&gt;Select all bodies&lt;/strong&gt; in the scene and create an &lt;strong&gt;Intersect&lt;/strong&gt; feature. Once again make sure that &lt;strong&gt;Merge Entities&lt;/strong&gt; is turned &lt;strong&gt;off&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final step is to use &lt;strong&gt;Insert&gt;Feature&gt;Split&lt;/strong&gt; to export each of the bodies in the scene as individual parts so that you can combine them in assembly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this was the answer you were looking for&lt;/p&gt;&#xA;" OwnerDisplayName="user10618" LastActivityDate="2018-01-15T18:19:22.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18918" PostTypeId="2" ParentId="15814" CreationDate="2018-01-15T19:00:06.417" Score="1" Body="&lt;p&gt;&lt;em&gt;Nitpicky&lt;/em&gt; Answer: &lt;strong&gt;you can't&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;More Detailed Answer&lt;/em&gt;: While &lt;a href=&quot;https://en.wikipedia.org/wiki/Enthalpy&quot; rel=&quot;nofollow noreferrer&quot;&gt;enthalpy&lt;/a&gt; is commonly treated as an absolute quantity it is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermodynamic_potential&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermodynamic potential&lt;/a&gt;. This is why textbooks might show the definition as &lt;a href=&quot;https://itp.uni-frankfurt.de/~valenti/WS13-14/all_1314_chap5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;\1\&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${\displaystyle \mathrm{d}H=\mathrm{d}U+\mathrm{d}\left(pV\right)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means (essentially) you need a reference-point and all values for &quot;enthalpy&quot; are the differences in &lt;em&gt;thermodynamic potential&lt;/em&gt; with respect to this reference point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's illustrate this with a thought experiment. Imagine a reservoir (artificial lake behind a dam) and let's pretend it contains 1000l. The question how much energy can be produced by a pelton-turbine can only be answered when you add information about the height (In this case its even called potential energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to quantify the thermodynamic potential of a geothermal resource it is necessary to define a meaningful reference point and this reference point needs to be disclosed otherwise the numbers are not meaningful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\1\ &lt;a href=&quot;https://itp.uni-frankfurt.de/~valenti/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://itp.uni-frankfurt.de/~valenti/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2018-01-15T20:50:21.857" LastActivityDate="2018-01-15T20:50:21.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18919" PostTypeId="2" ParentId="18901" CreationDate="2018-01-15T19:49:23.503" Score="2" Body="&lt;p&gt;There are a lot of factors that go into &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_sink#Fin_arrangements&quot; rel=&quot;nofollow noreferrer&quot;&gt;heat sink design&lt;/a&gt;.  Usually the optimization process of reducing the distance between the heat surface and the air results in fins that are perpendicular to the surface.  A slanted fin like you have shown would work just fine, it would just be a little less efficient per mass/volume of heat sink.  You would also have to have it custom made, as you will likely not find one that is mass produced in that configuration. If it is forced convection on this heat sink, a few extra cfm of airflow will make a bigger difference than heat sink design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would stick to standard available heat sink designs if possible. If you have a lot of space constraints, have a look at using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_pipe&quot; rel=&quot;nofollow noreferrer&quot;&gt;heat pipe&lt;/a&gt; design like most modern laptops.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-01-15T19:49:23.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18920" PostTypeId="2" ParentId="18886" CreationDate="2018-01-15T22:34:11.623" Score="3" Body="&lt;p&gt;To overcome hill resistance, you need &lt;strong&gt;2977 Nm or 2195 ft*lbs&lt;/strong&gt; of torque at the wheels.&lt;br&gt;&#xA;Assumed that the gravitional acceleration in your situation is $9.81m/s^2$, as it approximtely is everywhere on Earth, your weight is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_g=m*g=6000kg*9.81=58860N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hill resistance which drags you backwards is  :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_h=sin(a)*F_g=sin(7deg)*58860=7173N$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep any given speed, the force must be the same at the outer side of your wheel. So any speed that is achieved, is maintained. (theoretically, no friction included)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumed you meant that the diameter(not radius) of your wheels is 830mm,&#xA;this translates in a torque &lt;strong&gt;at the wheels&lt;/strong&gt; of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_w=F*r=7173*(0.830/2)=2977Nm$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To drive your vehicle to the speed of 27.78m/s up a hill of 7 degrees, you need &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P=F*v=7173 * 27.78 = 199.257 kW$ (or roughly 270hp) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;of power, when there wouldn't be any friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there will be friction, so you'll need more power and torque. This isn't calculable though, since there isn't enough information supplied in the question. At a normal asphalt road with an expectable loaded truck of 6 tonnes, i'd guess that you need roughly 50% more power, and therefore 50% more torque at the same speed. You also need a margin to accelerate to that speed in not-too-much time. &lt;strong&gt;NB:&lt;/strong&gt; this is just a wild guess, don't rely on it for any important decisions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget that the transmission in a vehicle will lower the torque demand at the motor, at the price of the demand for a higher angular speed of the engine, relative to that of the wheels.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-15T22:34:11.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18921" PostTypeId="2" ParentId="18905" CreationDate="2018-01-16T00:00:39.287" Score="1" Body="&lt;p&gt;I would suggest a silicon carbide crucible.  The melting point is like 3000k which should serve your purposes.  Also as I recall, it is inert to most molten metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say you are testing slag with molten metal.  What is the metal?  Do you know what the slag is composed of?&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2018-01-16T00:00:39.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18922" PostTypeId="2" ParentId="18895" CreationDate="2018-01-16T00:03:28.123" Score="3" Body="&lt;p&gt;Certain geometries can benefit from polyhedral elements or elements with edge degrees of freedom.  I can think of three main developments in that direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Voronoi cell finite elements, e.g., &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/0045794994904359&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sciencedirect.com/science/article/pii/0045794994904359&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Isogeometric elements, e.g., &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0045782504005171&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sciencedirect.com/science/article/pii/S0045782504005171&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Finite elements based on exterior calculus, e.g., &lt;a href=&quot;https://www.cambridge.org/core/journals/acta-numerica/article/finite-element-exterior-calculus-homological-techniques-and-applications/1A2AEB067BCA561D9ED6D674026539B9&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cambridge.org/core/journals/acta-numerica/article/finite-element-exterior-calculus-homological-techniques-and-applications/1A2AEB067BCA561D9ED6D674026539B9&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These elements are still being actively designed, but attention today is mostly on isogeometric and exterior calculus-based elements; primarily based of meshing issues and the need to solve multiphysics problems and to avoid element locking.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-01-16T00:03:28.123" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18923" PostTypeId="1" CreationDate="2018-01-16T02:18:10.313" Score="1" ViewCount="84" Body="&lt;p&gt;As a side kick I came up w/ a wet scrubber / air purifier design using trapped charged water droplets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking about building a proof of concept but would like the community's feedback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schematic attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Existing scrubbers mix pretreat gas stream w/ liquid droplets and then collect the dirty droplets via a cyclone, mist condenser, or charged plates (for charged droplets).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea is to use an ion trap to hold a cloud of charged scrubbing droplets stationary against the gas influx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;scrubber cloud&quot; essentially acts as a air filter but w/ much less pressure drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time we're conserving scrubbing liquid usage and dramatically reducing reactor volume compared to a wet scrubber.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ion trap can simply be 2 mesh screens separated by voltage diff on order of few kV&lt;/li&gt;&#xA;&lt;li&gt;Droplets hundreds of microns&lt;/li&gt;&#xA;&lt;li&gt;Flow velocity a few m/s&lt;/li&gt;&#xA;&lt;li&gt;Charge per droplet a few uC&lt;/li&gt;&#xA;&lt;li&gt;Back of env calc shows electric field force can counter air drag thereby trapping the droplets.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Over time droplets get dirty, coalesce, and may leave the trap. Then they can be captured via gravity or a demister downstream&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are your thoughts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qOLhy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qOLhy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14536" LastEditorUserId="9" LastEditDate="2018-01-16T05:47:09.173" LastActivityDate="2018-01-16T05:47:09.173" Title="Design review needed for novel ion trap air scrubber / purifier concept" Tags="|mechanical-engineering|electrical-engineering|civil-engineering|chemical-engineering|environmental-engineering|" AnswerCount="0" CommentCount="1" ClosedDate="2018-01-24T13:10:24.110" ContentLicense="CC BY-SA 3.0" />
  <row Id="18924" PostTypeId="1" CreationDate="2018-01-16T04:47:16.630" Score="1" ViewCount="242" Body="&lt;p&gt;Given the reactions of studs from the loads above them, how much does the bottom plate distribute the loads over the surface the bottom plate lays on?&lt;/p&gt;&#xA;" OwnerUserId="14562" LastEditorUserId="14562" LastEditDate="2018-01-16T22:41:46.753" LastActivityDate="2018-01-16T22:41:46.753" Title="How are point loads from studs onto a bottom plate distributed along the bottom plate?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18925" PostTypeId="2" ParentId="18924" CreationDate="2018-01-16T06:46:44.147" Score="2" Body="&lt;p&gt;Talking about lumber studs and 2x double bottom plate, under normal load applications and type 5 construction, the base plate as prescribe by applicable code, such as type 5 sheets handed out in the building departments, are adequate and no calculation as for the distribution of loads is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an even distribution is assumed!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if there is concentrated load on a 3x or larger size post, it has to be calculated and dealt with, with additional possible load factors and penalties for discontinued loading and other special cases.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-16T06:46:44.147" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18927" PostTypeId="2" ParentId="10767" CreationDate="2018-01-16T12:41:25.033" Score="0" Body="&lt;p&gt;In South Africa we use the &lt;a href=&quot;https://www.csir.co.za/sites/default/files/Documents/Red_BookVol2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Guidelines for Human Settlement Planning and Design&lt;/a&gt;, or what we just call the (Civil) Red Book. This is a really good resource for smaller projects or quick feasibility estimates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chapter 9 in Volume 2 is about water supply and and it states on page 24: &quot;Velocities in pipes should be approximately 0,6 m/s and should not exceed 1,2 m/s.&quot; This is in line with the 1 m/s you've picked up from reports.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastActivityDate="2018-01-16T12:41:25.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18929" PostTypeId="1" AcceptedAnswerId="18931" CreationDate="2018-01-16T13:47:01.747" Score="3" ViewCount="412" Body="&lt;p&gt;I am a newbie at this!! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an experiment, I have a diesel generator that releases hot flue gases at 150-160 Degree Celcius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I want the output gas to be at less than 75 degrees for a duration of ~ 1 hour without 'circulating' water. I'd like to do this with static water tank/bath&lt;/strong&gt; (because 1 hr is usually the maximum time of operation at once for a generator.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-The location I am in is where ambient temperature is ~ 18-30 degree celcius.&#xA;-Exhaust diameter is 4 inches&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to be able to design a kind of heat exchanger: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Can I just fill a tub of water, and divide the output exhaust into 4/5 tubes, take them through a tank of water? (like MUX-DEXMUX)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What's the amount of water that'd be required for the setup to be able to give that much heat reduction, for ~ an hour? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What sort of geometry should I build to not have much back-pressure on the gas flow.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What'll likely be the best material to build these pipes?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What's the best direction/references/designs to look into for design such, water bath/non-circulating type heat exchangers? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My experiment till now. &lt;a href=&quot;https://i.stack.imgur.com/f1Hoy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f1Hoy.jpg&quot; alt=&quot;Like this i took a spiral tubing, through a bucket of water and it almost worked. However this isn&amp;#39;t a optimized design as it&amp;#39;d cause backpressure due to spirals&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12071" LastActivityDate="2018-02-01T18:46:01.743" Title="Rapidly cool a flue gas without water circulation" Tags="|mechanical-engineering|heat-transfer|automotive-engineering|hvac|thermal-conduction|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18931" PostTypeId="2" ParentId="18929" CreationDate="2018-01-16T16:31:12.033" Score="1" Body="&lt;p&gt;Your question is a bit open-ended, so here are a couple recommendations.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if possible,  replace that hose with some ductwork that has finned exterior so as to dump as much heat as possible into the water, as well as directly to air.  Your comment about multiple exhaust tubes touches on this, as you'd be increasing the surface area - to - volume ratio.  You might run into backpressure problems at the interface, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, try to get a water container that is as long &amp;amp; wide as possible, both to maximize the length of tubing in the bath and to maximize the area at the water-air interface.  Again, if possible use a container with finstock on the outside.   This is because your overall heat problem is one of transferring heat not only to the water but also out of the water.  Heat transfer depends directly on the delta-temperature, so if you can keep the water cooler, you'll be able to extract more heat from the gas tube.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all that said, first calculate the amount of energy (heat) in the exhaust gas.  You can look up the equations relating temperature, specific heat of the air, and total flow (cubic ft/minute, e.g.).  Then you will know how much energy must be removed per minute to reduce the exit temperature to the desired level.   The next step is more difficult: calculating the heat transfer rate for the tube--water interface.  I doubt you can get a good estimate for this without running experiments.   Once that's known (as a function of water temperature), along with the known specific heat of water, you can get an estimate of the amount of water required to dump the total heat load. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-01-16T16:31:12.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18932" PostTypeId="2" ParentId="18929" CreationDate="2018-01-16T16:42:23.650" Score="3" Body="&lt;p&gt;Your approach is not an unusual one, engine brakes to measure horsepower usually make use of such a method, but they use much bigger water barrels. To answer your questions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&amp;quot;Can I just fill a tub of water, and divide the output exhaust into 4/5 tubes, take them through a tank of water?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;A: Yes you can, but the heat exchange between the water and the exhaust gas is not known, so you don't know the diameter or lenght of pipe needed for the desired temperature drop. The only thing known is that the cooling gets worse when the difference in temperature between water/exhaust gas gets lower. That happens when the water heats up after you have tested for a while. Using more tubes gives better cooling.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What's the amount of water that'd be required for the setup to be able to give that much heat reduction, for ~ an hour?&lt;/p&gt;&#xA;&lt;p&gt;A: That heavily depends on the efficiency of the pipes regarding heat exchange. More pipes is better, more surface is better(bigger or folded pipes), thinner wall thickness is better, materials with lower heat resistance is better(like copper or else aluminum).&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What sort of geometry should I build to not have much back-pressure on the gas flow.&lt;/p&gt;&#xA;&lt;p&gt;A: I'd suggest a cooling spiral within your barrel, that will give the least resistance to the gas flow. Sharp turns in your piping make things worse.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What'll likely be the best material to build these pipes?&lt;/p&gt;&#xA;&lt;p&gt;A: Copper is an excellent heat exchanger and still afforable. Aluminum is also good at it, and even less expensive. You can buy those crumpled up, thin walled aluminum pipes for aircon units and such. I'd recommend these with a 50mm/2in diameter:&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://cdn.shopify.com/s/files/1/1592/3963/products/Aluminum_Flex_400_Entry_Shots_Large_00f12a61-97c4-4f15-ad45-9c45e6348e00_1024x1024.jpg?v=1495485384&quot; alt=&quot;alu hose&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;You just want to create as much surface as possible, as thin as possible to reduce the total heat resistance. Crumpled up pipes create more surface, adding more pipes adds more surface. The more crumpled up, the better. It's just like air filters. The crumpliness of the pipes does add a tiny bit of resistance to gas flow, but you won't notice it.&lt;/p&gt;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;What's the best direction/references/designs to look into for design such, water bath/non-circulating type heat exchangers?&lt;/p&gt;&#xA;&lt;p&gt;A: I wouldn't know, but I think you've now got enough information to proceed.&#xA;Maybe take a look at instructables.com for some inspiration if you get stuck.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The needed amount of water will depend on the airflow from the exhaust.&#xA;Assuming a 2 litre four stroke engine revving at WOT at 4000rpm and 90% volumetric efficiency, you'd get an &lt;strong&gt;inlet&lt;/strong&gt; volume airflow of:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Volume flow = 2*(1/2)*(4000/60)*0.9 = 0.06 m3/s$&lt;/span&gt; or about 127CFM.&lt;/p&gt;&#xA;&lt;p&gt;The ambient temperature you give is 23 degrees. Air at that temperature has a density of about 1.185kg/m3 at 1 bar. Assuming an absolute manifoldpressure of 1 bar at WOT, that would translate in a mass airflow of:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Massflow=1.185*0.06=0.0711kg/s$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Air has a specific heat of 287 J/kgK, so every degree temperature drop per kilogram gives you 287 Joules of heat energy. We're not dealing with dry air, but with exhaust gas which also containt about 1/15 part fuel which has different properties. So let's take 300J/kgK for specific heat to compensate that.&lt;/p&gt;&#xA;&lt;p&gt;A temperature drop of 90 degrees C (150-60) over our mass flow thus yields a cooling power of:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P=dT*c*m=90*300*0.0711=1920W$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;That's nearly exactly the power that the average watercooker has. You can imagine how fast the water will be heated up. We can calculate that. Water has a specific heat of about 4200J/kgK. So let's say your barrel contains 20Litre(thus ~20kg) of water, then our 1920Watts of power will heat up the water&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$dt/s=1920/(4200*20)=0.023 K/sec$&lt;/span&gt; or about 1 degree every 44 seconds.&lt;/p&gt;&#xA;&lt;p&gt;The surrounding air also cools down the barrel in turn, but let's ignore that for the sake of simplicity here.&lt;/p&gt;&#xA;&lt;p&gt;That means that our 20L of water will boil in 56 minutes, given that the cooling will be constant. But it won't, it will cool less and less the warmer it gets.&#xA;And if you want an exhaust gas temperature of 60 degrees, you obviously have to maintain water temperature (a lot) lower than that. Let's say it can't rise above 45 degrees. That means that the water can rise 22 degrees above ambient, which gives us 16 minutes of cooling before we reach a water temperature of 45 degrees.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;NB:&lt;/strong&gt; this is a calculation that takes a &lt;strong&gt;lot of assumptions&lt;/strong&gt;, and your mileage will vary wildly if the operating conditions of the engine are different from what I assumed. Fill in your own engine properties to get an idea of your situation.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;My advice:&lt;/strong&gt; create some margin, i'd recommend to find an old 200L oil barrel, and use 5 50mm aluminium hoses of about 3 metre, as shown in the picture. Create a little spacing between the hoses for water to flow between, to maximise cooling efficiency. Make a spiral of the hoses in the barrel to keep air restriction at a minimum. Observe the water temperature in the barrel to get an idea how fast it will rise to a level that it won't cool anymore. Preferably make the air flow upwards in the spiral. Thermo syphon working will keep the water circulating in the barrel for maximum cooling efficiency. You can vary the hose length to maintain your 60 degrees air output as the water temperature rises.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-01-16T18:07:36.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18933" PostTypeId="1" CreationDate="2018-01-16T17:40:15.673" Score="1" ViewCount="331" Body="&lt;p&gt;I’m a chemical engineering student and I’m learning about heat exchangers,I came across this question and had a doubt , &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cold water is heated from 20 degree C to 85 degree C while hot process stream is cooled from 130 to 55. Overall heat transfer coefficient is $348 W/m^2 . DegreesC$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface area of each tube is 0.092$m^2$ , how many tubes would be required to construct this heat exchanger ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rate of heat transfer = 165800W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated LMTD = 39.79 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I calculated the area , which is about 11.97$m^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Number of tubes = $11.97/0.092 = 130.4$ tubes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I need 130.4 tubes . Do I round up to 131 or round down to 130? and why ?&lt;/p&gt;&#xA;" OwnerUserId="13646" LastActivityDate="2018-01-16T18:11:34.177" Title="Deciding on number of tubes to construct a heat exchanger" Tags="|heat-exchanger|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18934" PostTypeId="2" ParentId="18933" CreationDate="2018-01-16T18:11:34.177" Score="1" Body="&lt;p&gt;In this situation I suggest you round up - as rounding down means that the number of tubes is not sufficient.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-01-16T18:11:34.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18935" PostTypeId="1" CreationDate="2018-01-16T18:22:12.343" Score="3" ViewCount="74" Body="&lt;p&gt;I would like to find n by n matrices P and Q that minimize&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;J = norm(P) + w*norm(Q), where w is a given weight,&#xA;subject to&#xA;P&gt;=0, Q&gt;=0, and f(P,Q)=0, where f(P,Q)=0 is a given function of P and R.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to solve this problem using the lmi solver of MATLAB, but have no idea how to deal with the equality constraint: f(P,Q)=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f(P,Q)=0 is equivalent to f(P,Q)&gt;=0 and f(P,Q)&amp;lt;=0, but MATLAB only solves strictly feasible constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the above problem contains not strictly feasible constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any good solution to handle this kind of problem using MATLAB?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="14568" LastActivityDate="2018-01-16T18:22:12.343" Title="How to solve LMIs with equality constraints using MATLAB?" Tags="|matlab|optimal-control|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18936" PostTypeId="2" ParentId="18910" CreationDate="2018-01-16T18:28:58.870" Score="1" Body="&lt;p&gt;Flashers in your car that makes your blinkers blink used to work on the same principle. If you lightly bend a thin metal sheet, it will resist more to bending in the perpendicular direction. You can demonstrate it by holding a sheet of paper with one hand. It won't hang down if you push your thumb to curve the paper somewhat while holding it. But lightly push the other end of the paper down with your other hand, and eventually it will flip over and hang down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your vaporizer will probably work with that principle, but with a small metal sheet. And it is probably fitted in a location that compresses it, just like you pushed the other end of the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That resistance to flipping over decreases as the temperature of the metal increases. So when heated, at some point, it will become weak enough to snap; creating the snapping sound.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-16T18:28:58.870" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18938" PostTypeId="2" ParentId="18896" CreationDate="2018-01-16T20:58:24.127" Score="1" Body="&lt;p&gt;I assume you want to keep the beans suspended in the air, like they do with persons at skydive simulators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to do that, you do need a high speed airflow (or CFM). The (static) pressure that the fan can maintain doesn't matter here. The high speed airflow creates dynamic pressure that acts on the beans and keeps them suspended, which is what your looking for if I understand correctly. At terminal velocity, the force that acts on the object due to drag, is equal to the force that pulls it down, which we call gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following formula for the terminal velocity of an object:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v=\sqrt(2mg/\rho*C_d*A)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $v$ is the velocity on an object, relative to the medium/stream in metres per second.&lt;br&gt;&#xA;$m$ is the mass of the object(beans) in kilogram.&lt;br&gt;&#xA;$g$ is the gravitational acceleration, which is about $9.81m/s^2$ everywhere on Earth.&lt;br&gt;&#xA;$\rho$ is the density of the medium, which is 1.2kg per cubic metre for air at standard conditions.&lt;br&gt;&#xA;$C_d$ is the drag coefficient of the object. This is a factor usually between near 0 and 1. It's defined by the shape and skin texture of the object. A sphere has a $C_d$ of about 0.9, so i'd expect a bean to have a $C_d$ of about 0.85. It depends which orientation the bean has relative to the airflow.&lt;br&gt;&#xA;$A$ is the frontal area of the object. This also depends on the orientation of the bean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we can't change the properties of the beans, nor can we significantly change the density of the air, all we're left with is the velocity as tunable parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find the properties of a bean in my physics book, but let's suppose it's as dense as average wood: about 700kg/m3, and that the bean has the shape of a sphere with a diametre of 8mm. (about 1/3rd of an inch) Then the frontal area of the bean will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A=1/4*\pi*D^2=1/4*\pi*8^2=50.3mm^2$ or 5.03e-5 m2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of the spherical bean will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V=1/6*\pi*D^3=268mm^3$ or 2.68e-7 m3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which at a density of 700kg/m3, gives us a mass of the bean of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$m=\rho*V=700*2.68e-7=0.188grams$ or 1.88e-4 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we fill in these values in the first formula, it gives us a velocity of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sqrt((2*1.88e-4*9.81)/(1.2*0.85*5.03e-5))=8.5m/s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the needed velocity of the air to keep them suspended. If you multiply it by the surface of your fan, you get the needed airflow. This is assumed the surface of the airduct above the fan is the same surface as the fan. Your 120mm fan has a surface of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A=1/4*\pi*D^2=1/4*\pi*0.12^2=0.0113m^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means that the volume air flow must be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Volumeflow=v*A=8.5*0.0113=0.096m^3/s$ or 96L/s or &lt;strong&gt;203CFM&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd choose a fan that has a higher CFM rating(&lt;strong&gt;300CFM&lt;/strong&gt; or so), to create a margin to make up for any errors made by assumptions in these calculations. You can always easily throttle down the fan. And also, your mesh, the heater, the beans, and any other construction will restrict the airflow, which makes it impossible for the fan to provide advertised CFM rating. That rating is probably based on an unrestricted fan. That's why you need to aim higher for CFM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to warm up the beans, i'd recommend to create a construction that makes the fan recirculate the air. So create your airduct like a donut or a square donut. Not only does it allow you to gradually heat up the air and the beans without losing your heat, it also keeps the load of the fan at a minimum.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-16T20:58:24.127" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="18939" PostTypeId="1" AcceptedAnswerId="18940" CreationDate="2018-01-16T22:16:14.180" Score="4" ViewCount="1304" Body="&lt;p&gt;I've learned that a high viscosity means that the fluid is more thick. &lt;a href=&quot;https://en.wikipedia.org/wiki/Viscosity#Dynamic_(shear)_viscosity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dynamic Viscosity&lt;/a&gt; thus represents the &lt;em&gt;resistance&lt;/em&gt; to shear forces. Presumably, higher resistances are represented by higher numbers in the &lt;a href=&quot;http://www.viscopedia.com/viscosity-tables/substances/iso-viscosity-classification/&quot; rel=&quot;nofollow noreferrer&quot;&gt;relevant ISO spec&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus I'm puzzled about the unit of &lt;a href=&quot;https://en.wikipedia.org/wiki/Viscosity#Kinematic_viscosity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kinematic Viscosity&lt;/a&gt; (which is basically Dynamic Viscosity modulated by a measure of the density of the fluid). The unit is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;m^2/s&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But how can this unit be understood intuitively? The higher the viscosity, the lesser the flow-rate and thus less m^2 fluid should flow per second in whatever measurement setup they are using.&#xA;Looking at the unit for kinematic viscosity, it seems that a higher value means that the fluid has a &lt;em&gt;lower&lt;/em&gt; viscosity (i.e. is less thich), because, well, more square meters of fluid seems to flow per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the unit of kinematic viscosity be understood intuitively, when seen in relation to the definition of viscosity?&lt;/p&gt;&#xA;" OwnerUserId="14572" LastEditorUserId="9275" LastEditDate="2018-01-17T00:26:02.730" LastActivityDate="2023-10-26T08:12:26.557" Title="Is there an intuitive explanation for the Kinematic Viscosity unit of mm^2/s?" Tags="|machining|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18940" PostTypeId="2" ParentId="18939" CreationDate="2018-01-17T00:25:17.493" Score="9" Body="&lt;p&gt;Dynamic viscosity represents the resistance of fluid to shear forces as you said.  This is what lay people think of when they think viscosity.  High dynamic viscosity = more resistance to flow.  E.g. honey has much higher dynamic viscosity than water, or cold motor oil has higher dynamic viscosity than warm motor oil.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinematic viscosity is something totally different. Don't try to think of it as resistance of fluid to flowing, because that's not what it is. Thinking of it as so many $m^2$ of fluid flowing per second is not what it is about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To understand what it is, you have to understand that there are many different types of fluid flow, and they behave very differently.  One of the fundamental distinctions has to do with the ratio of inertia forces to viscous forces, which is called the Reynold's number.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Reynolds_number&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; has a pretty good page on it.  Flows with high reynolds number will behave completely differently than flows with low reynolds number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Reynold's number is defined as $Re = \rho s L / \mu$, where $\rho$ is density, $s$ is velocity, L is a length, and $\mu$ is dynamic viscosity. Of these four quantities, 2 are intrinsic to a particular fluid (density and dynamic viscosity) and 2 are more to do with the situation (the length and the velocity). The reynold's number is super important and it comes up all the time in fluid mechanics.  Because it comes up so often, and because both density and viscosity are intrinsic to the type of fluid, we simplify the Reynold's number equation to just $Re = s L / \nu$, where $\nu$ is kinematic viscosity.  Now it's 3 equations, exactly 1 of which has to do with the type of fluid, and 2 that have to do with the particular situation.  This is just for convenience.  We could have just kept the original two quantities.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the units, the other two quantities in the reynold's number equation are length and velocity.  Length * velocity will give you m^2/s.  Note that is the same units as kinematic viscosity.  So when you calculate $Re = s L / \nu$, you come out with a pure number that has no units.  Unitless numbers occur very frequently in fluid mechanics.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-01-17T00:25:17.493" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18941" PostTypeId="2" ParentId="18929" CreationDate="2018-01-17T00:38:05.423" Score="1" Body="&lt;p&gt;The higher heating value, HHV, of diesel is taken at 19,300 Btu/lbm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sanity check on the fuel consumption - 3l/h. For small diesels, brake specific fuel consumption of about 1/3 lb/hp-hr is the norm. So about 3 * 3 * 1.8lb/l is 16.2hp or 12kw. So we're in the ballpark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For non-turbo diesels, the heat budget is roughly 30% work out, 35% lost to liquid coolant loop, and 35% lost through exhaust. Turbos are a bit trickier. For a small turbo genset, I'm going to go with 40% out the manifold. If I looked up the boost at 12kw, we could estimate exhaust temps more accurately, buts lets guess 900F at the turbo exit. Essentially, we are subtracting the aftercooler heat out of the exhaust system figure. So I end up at about 32% of the fuel's HHV in the exhaust downstream of the turbo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;19,300btu/lb * 1.8lb/l * 3l *.32 = 33,350Btu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Btu raises 1 lb of water 1 degree F. You aren't taking 100% of the heat out of the exhaust, either. Some water vapor will remain, and the exhaust temp is a little above ambient, giving the estimate a small cushion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow for a 20 degree C rise in water temp, you have 25 degrees C remaining for your 75C outlet. That isn't very much, but worth a look. That works out to about 926 lb of water, or 111 gallons US. As a practical starting place, 4 55 gallon steel oil drums each with a pipe loop ought to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better starting point might be a non-turbo powerhead with a closed-loop, water-jacketed exhaust manifold to handle some of the load. &lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-01-17T00:54:18.430" LastActivityDate="2018-01-17T00:54:18.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18943" PostTypeId="2" ParentId="18939" CreationDate="2018-01-17T07:03:16.890" Score="1" Body="&lt;p&gt;Suppose you want to compare the performance of a lifting surface like an airfoil when operating in two different fluids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lift equals some f1(geometry) * density * velocity squared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the major drag components equals some  f2(geometry) * dynamic viscosity * velocity squared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to have a convenient lift to drag ratio expression, we take the drag formula and multiply it by density over density and factor (density * velocity squared). Now we have L/D = f1/f2 * density/viscosity. This is very convenient when density is constant as in liquids or incompressible flows. For instance, it tells us that a foil in water can operate at a higher L/D than one in air at the same lift force, because the kinematic viscosity of water is about 18 times less than that of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's all there is to it. It's handy when you want to cancel density out of other terms in the equation; it has a simpler set of units to keep track of; and it facilitates more compact notations.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-01-17T07:03:16.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18945" PostTypeId="1" CreationDate="2018-01-17T07:39:17.573" Score="4" ViewCount="170" Body="&lt;p&gt;If I have a state-space model, so that matrices &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; are known, how can I design the right input &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt;, so that &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; is a desired signal, say, a sine wave with constant amplitude?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{aligned} \dot{x} &amp;amp;= A x + B u \\ y &amp;amp;= C x + D u \end{aligned}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="14577" LastEditorUserId="6790" LastEditDate="2020-05-04T13:36:02.253" LastActivityDate="2020-05-04T13:36:02.253" Title="How to design the right input to obtain a desired output for a linear system?" Tags="|control-theory|aerospace-engineering|matlab|systems-engineering|linear-control|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="18947" PostTypeId="1" CreationDate="2018-01-17T10:29:02.267" Score="1" ViewCount="943" Body="&lt;p&gt;I've got a simple question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What material should hammer's head be made of? We're talking about medium sized ball-peen hammer, used for riveting brass, copper, aluminum and steel rivets. I have acess to lathe and miling machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about using material CK-45 (AISI 1045; Mat. number 1.1191), is it proper to use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dTK5s.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dTK5s.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for all the help!&lt;/p&gt;&#xA;" OwnerUserId="5797" LastActivityDate="2018-01-17T15:42:13.000" Title="Proper metal for turning/milling a ball-peen hammer" Tags="|materials|metallurgy|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18948" PostTypeId="1" AcceptedAnswerId="18952" CreationDate="2018-01-17T15:04:57.377" Score="3" ViewCount="3131" Body="&lt;p&gt;I have some doubts about a quick-release mechanism that I'm working on and I'd like to run it by you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have hollow tube in which there's a spring-loaded rod. Part of it is made wider and there's a metal round pin that restrains the entire mechanism as shown below: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/xsGhI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xsGhI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spring rate is 1.23 N/mm its free length is 135mm but it is compressed to 35 mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using a servo quick release mechanism similar to this: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/OSL4Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OSL4Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is driven by a servo with a stall torque of 9.8kg/cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, would the lateral force the rod is exerting on the pin be a problem for the motor? E.g. the friction between the pin and the hole of cylinder being so great that the motor won't be able to move? Is there a way to mitigate this? &lt;/p&gt;&#xA;" OwnerUserId="14581" LastActivityDate="2018-01-17T19:12:48.313" Title="Quick release for spring-loaded rod" Tags="|mechanical-engineering|springs|servo|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18949" PostTypeId="2" ParentId="18947" CreationDate="2018-01-17T15:42:13.000" Score="2" Body="&lt;p&gt;Carbon steel bar stock you suggest would be fine for a modest size , moderate duty hammer as you describe. As noted,the heat-treatment is more important. Quench and temper, you don't want the face any harder than about H Rc 30. A better choice would be a Ni containing steel for toughness; eg. 4330 /4340. Also, for extreme duty, a forging would be better as bar stock has the risk of longitudinal imperfections ; inclusions of silicates, etc.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-01-17T15:42:13.000" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18950" PostTypeId="2" ParentId="8324" CreationDate="2018-01-17T15:55:48.257" Score="0" Body="&lt;p&gt;Poor method to get a known clamping force; frictions are large unknowns. In the real world (when clamping force is important) ,a hydraulic tensioner pulls the stud/bolt and then the nut is tightened. For ordinary applications like car wheel lugs or headbolts , the manufacturer has the experience to know the torque levels to apply. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-01-17T15:55:48.257" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18952" PostTypeId="2" ParentId="18948" CreationDate="2018-01-17T19:12:48.313" Score="4" Body="&lt;p&gt;The force which is needed to pull the pin out, will depend on the friction of the pin and the neighbouring material, and the surface at which the friction works. Let's assume a friction factor $\mu$ of 1, though it's probably less for plastics. That means that theoretically, just as much force is needed to remove the pin, as the force put on the pin by the spring, which is 123N(or 12.5kg) in your case, assuming the springrate is constant. If the servo has an arm of 2 cm, then the maximum force the servo will be putting on the pin, is half it's stall torque, so 4.9kg. That wouldn't be enough to remove the pin. However, if the friction factor, or the arms length is lower, it may be enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'd rather be concerned that the pin doesn't get damaged after a few times, as your servo pulls it out of the tube under force. That used to be the one lifelimiting factor in the BB guns I had as a kid. Those used the same construction as you show here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd advice you to use a solenoid to pull the pin out of it's hole, it does so with enough force, and much faster than a servo, greatly reducing the stress on the pin. You can easily find a solenoid strong enough to pull the pin out.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-17T19:12:48.313" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18953" PostTypeId="1" CreationDate="2018-01-17T21:15:57.520" Score="1" ViewCount="703" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y5Kzs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y5Kzs.png&quot; alt=&quot;Pumping Curve and Pump Details&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have just received a recommendation to use a pump with a &quot;Rated Flow&quot; of 1cu.m/h and a &quot;Rated Head&quot; of 91m. I am realizing I am actually not familiar with these two terms (Rated Flow and Head). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My total dynamic head is a little more than the &quot;Rated Head&quot; (It's 110m), and I intend to pump at a flow rate a little less that the &quot;Rated Flow&quot; (I intend to pump at 0.5cu.m). Both these values--though not very efficient--fall well-within the constraints of the pumping curve. Should I be worried by the &quot;Rated&quot; labels? What exactly do they signify?&lt;/p&gt;&#xA;" OwnerUserId="11147" LastActivityDate="2018-03-20T22:39:56.617" Title="Can my total dynamic head (TDH) be more than the &quot;Rated Head&quot; on the pump?" Tags="|mechanical-engineering|fluid-mechanics|pumps|water-resources|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18954" PostTypeId="1" CreationDate="2018-01-18T02:14:12.077" Score="1" ViewCount="36" Body="&lt;p&gt;My system comprises of: a light source, a set of lenses, a projected image of the light source.&#xA;I want the projected image to be both further away from the lenses and smaller than the source. Which is in contradiction with the optic law that :&#xA;source size / source to lens distance= image size/ image to lens distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know optical tools that could help me ? &lt;/p&gt;&#xA;" OwnerUserId="5926" LastActivityDate="2018-01-18T02:14:12.077" Title="What optics system can create an image both smaller and further away from the system than the source object?" Tags="|optics|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="18956" PostTypeId="1" CreationDate="2018-01-14T21:55:46.930" Score="0" ViewCount="29" Body="&lt;p&gt;During my time revising for my control engineering module, i've come across this question  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The automatic cruise speed control for a vehicle is described by the following equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigl( 1\bigl)$ $m \frac{dv_0(t)}{dt} + Fv_0(t)=u(t)$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\bigl( 2\bigl)$ $u(t)=K[v_0(t)-v_r(t)]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where m is the mass of the car, 1000$kg$, F is the frictional constant 50 $Nm^{-1}s$ and $u(t)$ is the throttle actuator force and K is the proportional gain control for the actuator.&#xA;Also $V_0(t)$ is the actual speed while $V_r(t)$ is the desired speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two parts to this question, and im a bit confused on both of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;part 1 says to draw a block diagram of the speed control system. In the control engineering module we generally draw the block diagrams as related to s, so i thought i'd get the laplace transform of the function (to get the transfer function) and draw it from the general model of a control system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This coincided with part 2, which wants me to show that the transfer function in terms of K is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{V_0(s)}{V_r(s)} = \frac{K}{1000s+50+K}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went about &lt;a href=&quot;https://imgur.com/a/Zlc8p&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;working it out&quot;&gt;working it out&lt;/a&gt; and eventually ended up with this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{V_0(s)}{V_r(s)} = \frac{-K}{1000s+50-K}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me figure out where i went wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, i attempted making a block diagram in the t domain for this system, however since there are two equations with equal outputs (not summed) i was completely stumped on how i would display that as a model that is accurate, without making assumptions like equating K with the system gain of the other equation&lt;/p&gt;&#xA;" OwnerDisplayName="mudabir sadeed" LastActivityDate="2018-01-18T06:40:03.567" Title="Determining the transfer function of a system given the time-domain characteristic equations" Tags="|control-theory|" AnswerCount="0" CommentCount="1" FavoriteCount="0" ClosedDate="2018-01-18T13:10:51.117" ContentLicense="CC BY-SA 3.0" />
  <row Id="18957" PostTypeId="2" ParentId="18900" CreationDate="2018-01-15T18:41:28.023" Score="1" Body="&lt;p&gt;If you take the Laplace transform of the system you obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$msV_0(s)+FV_0(s)=U(s)$$&#xA;$$U(s)=K\left[V_0(s)-V_\text{r}(s) \right]$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plug the second equation into the first equation and solve for $V_0(s)/V_\text{r}(s)$: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{V_0(s)}{V_\text{r}(s)}=\dfrac{-K}{ms+(F-K)},$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your answer seems to be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the block diagram (in time domain) is given by in the following picture.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZFAyX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZFAyX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7627" OwnerDisplayName="MrYouMath" LastActivityDate="2018-01-15T18:41:28.023" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18958" PostTypeId="1" CreationDate="2018-01-18T09:39:27.793" Score="3" ViewCount="106" Body="&lt;p&gt;The three Mohr's circles depict all the possible normal stress / shear stress combinations when material is taken and rotated along one of it's axes of principal stress. Nevertheless, it is clear that rotation around just one principal axis does not yield all the possible configurations. For any point on the Mohr's circle, one of the axes is aligned with one of the principal axes by definition - however, you can rotate a block in such a way so that no axes are aligned with the principal axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, aligned the axes $x$, $y$, $z$ with the principal stresses $\sigma_1$, $\sigma_2$, $\sigma_3$.  If one rotates clockwise around $z$ by $30$ degrees giving the axes normal to the sides of the block $x'$, $y'$ and $z$, and &lt;em&gt;then&lt;/em&gt; rotates about $x'$ by say 30 degrees to $x'$, $y''$, $z'$ none of the axes are aligned with the principal stresses anymore. Such a combination of normal and shear stresses &lt;strong&gt;does&lt;/strong&gt; exist in the structure however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is then: where would such a point fall on the coordinate axes of normal stress vs shear? Would it fall between Mohrs' circles? How do we know such a combination of rotations would not yeild a shear higher than that of the largest Mohr's circle?&lt;/p&gt;&#xA;" OwnerUserId="14592" LastActivityDate="2018-01-18T09:39:27.793" Title="Normal stress - shear stress pairs not lying on Mohrs' Circles" Tags="|structures|stresses|shear|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18961" PostTypeId="1" AcceptedAnswerId="19020" CreationDate="2018-01-18T15:08:19.843" Score="0" ViewCount="93" Body="&lt;p&gt;I have been playing OpenRails Train SImulator recently and I thought i made my own train(which i did), then the physics part came up. It required me to fill in a lot of parameters, like braking, sanders, and of course, engine. Now the engine part is confusing me, it requires me to enter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass(KG), Max Power(kW), Max Force(kN), Continuous Force(kN), and Max Velocity(mph).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the Mass, MaxForce and MaxContinousForce is easy to find, but what about the rest? In order for me to know more about what i am doing, i would like to know how the speed in the game is calculated using all these values? Because i thought the speed of the train was only calculated using the MaxForce and the Mass value, but all these values seem to have some connection as if one of those listed values is changed, speed is different. So how would I calculate the speed of a train using these parameters(What is/are the formula('s)? And how are the horsepower values and force values related to each other in terms of final speed?(As again, i though kW didn't matter in speed)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;So mainly what i want to know is:&#xA;Is the calculation of speed in general affected by horsepower/kW(does it have anything to do with speed/force at all?)&lt;/p&gt;&#xA;" OwnerDisplayName="user14594" LastEditorDisplayName="user14594" LastEditDate="2018-01-18T19:15:37.917" LastActivityDate="2018-01-22T02:02:09.663" Title="How is train speed calculated in OpenRails?" Tags="|mechanical-engineering|acceleration|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18962" PostTypeId="2" ParentId="18945" CreationDate="2018-01-18T19:56:32.790" Score="1" Body="&lt;p&gt;You can design an asymptotic output tracker based on feedback linearization if the residual dynamics are stable. The theory for this can be found in the book 'Nonlinear Control Systems' by Isidori [&lt;a href=&quot;http://www.springer.com/us/book/9783540199168&quot; rel=&quot;nofollow noreferrer&quot;&gt;Springer&lt;/a&gt;]. You can find examples worked out using Mathematica &lt;a href=&quot;http://reference.wolfram.com/language/ref/AsymptoticOutputTracker.html#1230668630&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://www.wolfram.com/mathematica/new-in-10/nonlinear-control-systems/asymptotic-output-tracking-with-estimator.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way is to develop a LQR tracking controller. See, for example, Chapter 4 of the book 'Optimal Control' by Anderson and Moore [&lt;a href=&quot;http://users.cecs.anu.edu.au/~john/papers/BOOK/B03.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;available online&lt;/a&gt;].&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2018-01-18T19:56:32.790" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18964" PostTypeId="1" CreationDate="2018-01-19T01:51:27.200" Score="4" ViewCount="97" Body="&lt;p&gt;It's well-known that a control system with integral control can track a constant reference signal (command) with zero steady-state error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is needed to generate a controller that can track a ramping reference signal (i.e. one that changes at a constant rate) with zero steady-state error?&lt;/p&gt;&#xA;" OwnerUserId="508" LastActivityDate="2018-01-19T04:36:50.047" Title="What's a control system architecture that can track a ramping reference signal?" Tags="|control-engineering|pid-control|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18965" PostTypeId="1" AcceptedAnswerId="18990" CreationDate="2018-01-19T02:14:15.010" Score="1" ViewCount="103" Body="&lt;p&gt;I am not able to understand the relation between ride frequency and wheel deflection of a car, as per given in Race car vehicle dynamics book by milliken. Is it something to relate with harmonic motion of a forced oscillator? I still don't get why there is a 188 in the numerator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GzjM0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GzjM0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14271" LastActivityDate="2018-01-24T14:40:46.680" Title="Relation between ride frequency and wheel deflection of a car" Tags="|mechanical-engineering|design|dynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18966" PostTypeId="1" AcceptedAnswerId="18970" CreationDate="2018-01-19T02:47:58.097" Score="1" ViewCount="65" Body="&lt;p&gt;(Please move this question to the appropriate exchange site if it is not this one; I originally posted this &lt;a href=&quot;https://travel.stackexchange.com/questions/108460/how-do-locks-for-shoe-lockers-in-sentos-work&quot;&gt;here&lt;/a&gt;, and a comment suggested that I move it here.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Danganronpa it appears that the lock at 9:27 in this &lt;a href=&quot;https://www.youtube.com/watch?v=qOVU8UZRrHc&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; has a weird shape, and I cannot figure out how it might work.  Using Google, it appears that these are locks for shoe lockers in &lt;a href=&quot;https://en.wikipedia.org/wiki/Sent%C5%8D&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sentōs&lt;/a&gt;, but I haven't been able to find a picture showing how they work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comment on the original place I put this question pointed me to this &lt;a href=&quot;https://bramanswanderings.com/2014/04/03/japanese-locker/&quot; rel=&quot;nofollow noreferrer&quot;&gt;website&lt;/a&gt;, which says &quot;The wooden key was interesting as it had grooves in it. If the pattern of the grooves match the pins in the lock then the key will go all the way down and trip the catch on the lock.&quot;  However, this seems a little suspicious because then it looks like you could open any lock by using a small stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some more googling I found this &lt;a href=&quot;https://www.ebay.com/i/323002263864?rt=nc&quot; rel=&quot;nofollow noreferrer&quot;&gt;auction&lt;/a&gt; which has pictures of inside of the lock (and also pictures of the key, which are different from those in the game; I am now guessing that the key shape in the game probably does not exist, but only because I cannot come up with a lock mechanism for it).  It does not explain how it works, though, and I am still interested in this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do they work?  More specifically,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the picture of the lock in the video realistic?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is the description from that &lt;a href=&quot;https://bramanswanderings.com/2014/04/03/japanese-locker/&quot; rel=&quot;nofollow noreferrer&quot;&gt;website&lt;/a&gt; accurate, and if not then how does it work?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14601" LastActivityDate="2018-01-19T10:22:52.390" Title="How do locks for shoe lockers in Sentōs work?" Tags="|design|mechanisms|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18967" PostTypeId="2" ParentId="18964" CreationDate="2018-01-19T04:22:49.260" Score="3" Body="&lt;p&gt;Consider a controller $C(s)$ for plant $G(s)$ and feedback $H(s)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed loop is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{Y(s)}{U(s)}=\frac{CG}{1+CGH}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ramp is $U(s)=\frac1{s^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the error is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=Y(s)-\frac{1}{s^2}=(\frac{CG}{1+CGH}-1)\frac1{s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the final value theorem,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lim_{t\to \infty} e(t)=\lim_{s\to 0} sE(s)=\lim_{s\to 0} (\frac{CG-1-CGH}{1+CGH} \times \frac1{s})=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a known case, you can find out how many $s$ in the denominator of $C$ will reduce the steady state error to zero.&lt;/p&gt;&#xA;" OwnerUserId="14288" LastEditorUserId="14288" LastEditDate="2018-01-19T04:36:50.047" LastActivityDate="2018-01-19T04:36:50.047" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18970" PostTypeId="2" ParentId="18966" CreationDate="2018-01-19T10:22:52.390" Score="2" Body="&lt;p&gt;The key in the game is not a realistic representation of keys used in traditional Japanese shoe lockers - it has lateral groves besides normal lengthwise groves used in these locks. They are usually low-security locks that act on principle similar to warded locks (only fixed obstructions blocking a mismatching key; a key that is significantly slimmer will just work) - and similarly, easy to bypass. Possibly there exist more secure variants; I haven't encountered any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lock like in the video would be fairly easy to implement as a variant of a &lt;a href=&quot;https://www.youtube.com/watch?v=ehA8WUX4xaY&quot; rel=&quot;nofollow noreferrer&quot;&gt;dimple lock&lt;/a&gt; - a lock where the key has holes on the flat surface and engages the tumblers with its sides instead of edges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of drill holes, groves can be used to engage the tumblers at various locations to various depth; to avoid necessity of turning, and thanks to the size of the key, the tumblers could be considerably bigger and act directly as the latches on the door. Still, this sort of key, made of wood, would sooner or later get dented by the tumblers, failing eventually. So, a lock like this is possible, is not very practical, and it's highly dubious if it's used in reality.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-01-19T10:22:52.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18971" PostTypeId="1" CreationDate="2018-01-19T13:31:37.417" Score="1" ViewCount="2969" Body="&lt;p&gt;I am trying to find the buckling load of following column:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZYMfx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZYMfx.jpg&quot; alt=&quot;column with buckling load&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started from the system of equations $P = (K + K_g(N))U$, where $K$ is the stiffness matrix of a single 2D frame element and $K_g$ is the geometric stiffness matrix of the same element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary conditions give zero displacements and rotation in the node at the bottom of the column. So I tried to solve the equations for the 2nd part of the column, where the only unknowns are the displacements and the rotation in node 2, with $N = P$. But this gives me the same result as for a standard cantilever beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example on how to calculate the buckling load for a cantilever beam:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4z5mO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4z5mO.jpg&quot; alt=&quot;cantilever beam example&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/31CBV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/31CBV.jpg&quot; alt=&quot;cantilever beam example&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Spez8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Spez8.jpg&quot; alt=&quot;cantilever beam example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried solving the complete system of equations with unknown displacements in node 1 and 2, but without success. And since I need to be able to calculate it by hand, this doesn't seem to be the most efficient way to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what I'm doing worng. Could someone help me to start in the right  way?&lt;/p&gt;&#xA;" OwnerDisplayName="user14605" LastEditorDisplayName="user14605" LastEditDate="2018-01-20T20:19:19.247" LastActivityDate="2024-03-13T06:01:40.237" Title="Calculate the Euler buckling load of a column using the finite element method" Tags="|structural-engineering|civil-engineering|finite-element-method|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18972" PostTypeId="1" CreationDate="2018-01-19T16:06:36.577" Score="1" ViewCount="108" Body="&lt;p&gt;I'd like to write something in Python to automate and maintain a specific dosing formula for a couple of our processes. We need a dosing pump that either has a USB port that can interact with a computer or a microcontroller. Do any manufacturers make pumps like this? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14606" LastActivityDate="2018-06-14T22:08:43.430" Title="Metering pump that is either computer compatible or microcontroller compatible?" Tags="|mechanical-engineering|chemical-engineering|industrial-engineering|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="18973" PostTypeId="1" AcceptedAnswerId="18974" CreationDate="2018-01-19T16:38:36.030" Score="-1" ViewCount="58" Body="&lt;p&gt;There is a large drilling crane or something, which is connected by a giant hose to a skip full of water (central skip, blue). It looks like a jacuuzi but I suspect they are using it for something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g3NBm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g3NBm.jpg&quot; alt=&quot;construction_site_jacuuzi&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps it is not a jacuuzi? What is it for if not?&lt;/p&gt;&#xA;" OwnerUserId="14607" LastEditorUserId="14607" LastEditDate="2018-01-19T16:47:38.793" LastActivityDate="2018-01-19T16:57:55.020" Title="What is this construction site jacuuzi for" Tags="|civil-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18974" PostTypeId="2" ParentId="18973" CreationDate="2018-01-19T16:51:01.500" Score="5" Body="&lt;p&gt;It is likely to be  tanks for &lt;a href=&quot;https://en.wikipedia.org/wiki/Drilling_fluid&quot; rel=&quot;noreferrer&quot;&gt;drilling fluid&lt;/a&gt; (aka 'mud'). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a slurry of various substances, such as bentonite clay suspended in oil or water which serves various functions. In particular it provides cooling and lubrication for the drill head and helps to clear debris created by drilling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When drilling into pressurised oil and gas wells it also provide a hydrostatic head which resists the well pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically it will be pumped down the centre of the drill pipe and returns up through the gap between the drill pipe and the bore. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drilling fluid is usually made up on site as needed by mixing dry ingredients with oil or water plus any additional additives. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-01-19T16:51:01.500" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18977" PostTypeId="1" CreationDate="2018-01-19T20:18:38.713" Score="2" ViewCount="320" Body="&lt;p&gt;I recently took a quiz from POWER magazine that asked the following question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In a typical cooling system for lubricant oil in a gas turbine, which configuration is preferred to prevent leaks and contamination?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no clarification about what we are preventing leaks or contamination to or the type/size of the gas turbine, but their &quot;correct&quot; answer was to put cooling water on the tube side and oil lubricant on the shell side. They did not provide any further explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't see why this would be the preference. There are more consequences of the water contaminating the oil, so you should put the higher pressure oil through the tube so if there is a leak oil leaks into the water. Plus, if the oil goes on the shell side and is higher pressure, the exchanger will cost more because there is more material that needs to be able to withstand a higher pressure. Lastly, I would assume oil will be more prone to contaminants than water and more fouling will occur on the oil side. Thus, you ought to put the oil in the tube for maintenance purposes as well. Am I missing something here or did they select the wrong answer?&lt;/p&gt;&#xA;" OwnerUserId="14613" LastActivityDate="2018-01-19T20:18:38.713" Title="Choosing sides of a shell and tube heat exchanger for oil cooling" Tags="|cooling|heat-exchanger|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18978" PostTypeId="2" ParentId="18953" CreationDate="2018-01-19T20:36:45.473" Score="1" Body="&lt;p&gt;In this case, &quot;Rated&quot; refers to the design point of the pump. Just look at your pump curve, there is no way the pump's shutoff head is 91 m at a flow of 1 m^3/h. That is right in the middle of the curve. &quot;Rated&quot; is referring to the design criteria used to select the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to find more information for this pump to really address your question. The curve that you attached does not give a maximum head or minimum continuous stable flow. There is also not a line for maximum flow, although you can just assume that you need to stay away from the end of the curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are really going to run this pump at 110 m, you may want to select a different pump. You will be running this pump way up on its curve. Plus, you are going to get a really poor efficiency doing that (which also is not provided).&lt;/p&gt;&#xA;" OwnerUserId="14613" LastActivityDate="2018-01-19T20:36:45.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18979" PostTypeId="2" ParentId="14659" CreationDate="2018-01-19T21:14:53.737" Score="2" Body="&lt;p&gt;An &lt;a href=&quot;https://en.wikipedia.org/wiki/Isentropic_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;adiabatic and reversible&lt;/a&gt; process (i.e. isentropic) for ideal gases can be described as follows&lt;sup&gt;1&lt;/sup&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p \cdot V^\gamma = \mathrm{const.}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\gamma$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_capacity_ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;heat capacity ratio&lt;/a&gt;. Other processes can be described by changing the exponent. The equation for a &lt;a href=&quot;https://en.wikipedia.org/wiki/Polytropic_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;polytropic process&lt;/a&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p \cdot V^n= \mathrm{const.}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $n=1$ this equation describes a &lt;a href=&quot;https://en.wikipedia.org/wiki/Isothermal_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;isothermal&lt;/a&gt; process and for values greater $\gamma$ this equation can be used to describe irreversible processes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[1]: &lt;a href=&quot;https://en.wikipedia.org/wiki/Isentropic_process#Table_of_isentropic_relations_for_an_ideal_gas&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia: Isentropic process&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2018-01-19T21:14:53.737" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18980" PostTypeId="1" AcceptedAnswerId="18983" CreationDate="2018-01-19T22:25:24.197" Score="3" ViewCount="202" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0PicO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0PicO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I know if a rod is hinged at its two ends and if there is no load act upon it along its length then the forces exist only along its length but i don't see how  this characteristic comes into play ?  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-01-20T08:26:40.073" LastActivityDate="2018-01-20T12:31:54.120" Title="Why in the construction of power transmission towers , the rods are hinged oblique?" Tags="|civil-engineering|beam|torque|power-transmission|" AnswerCount="3" CommentCount="4" ClosedDate="2018-01-21T20:31:28.593" ContentLicense="CC BY-SA 3.0" />
  <row Id="18981" PostTypeId="1" AcceptedAnswerId="19001" CreationDate="2018-01-20T05:01:34.453" Score="0" ViewCount="130" Body="&lt;p&gt;I have a room of computer rack that with 2750 watt power needs, which according to &lt;a href=&quot;https://www.enviromon.net/advanced-server-btu-calculator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;, I need 9000 BTU/h AC. I'm deciding if I should put a vent in the room or not. When there's a ventilation, it will look like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1zJIV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1zJIV.png&quot; alt=&quot;Scenario 1&quot;&gt;&lt;/a&gt;&#xA;Note that there are two wind turbine ventilators on the roof. Also the attic is cool because of the good air circulation.&#xA;Question for this scenario:&#xA;1. Will the AC and the grill vent vanquish the humidity from the air outside, which is warm and humid?&#xA;2. Will dust enter the room via vent despite the fact that the airflow is upward to the wind turbine ventilator?&#xA;3. Is 9000 BTU an optimum number? Is it too much or too little? I'm thinking most of the heat is drawn to the vent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second scenario will look like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BAVMe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BAVMe.png&quot; alt=&quot;Second scenario&quot;&gt;&lt;/a&gt;&#xA;In this scenario, instead of a ventilation with a chimney, there is a ducting that caries heat, which escapes to the attic and goes to the wind turbine vents. Question:&#xA;Will this give faster heat reduction whereas there is no way humid air will come to the room?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards&lt;/p&gt;&#xA;" OwnerUserId="14294" LastActivityDate="2018-01-21T12:23:47.810" Title="Room with computer racks, with or without ventilation" Tags="|thermodynamics|heat-transfer|airflow|heat-treatment|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18982" PostTypeId="2" ParentId="18980" CreationDate="2018-01-20T05:37:41.053" Score="1" Body="&lt;p&gt;It could be just to meet a section of a code, or facilitate construction.&#xA;Or it could be part of the design that is meant to work with configuration and tension of power cables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically in long truss sections they use L shaped metal profiles and a gusset plate and a set of bolts, but use of hinges is not prohibited.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-20T05:37:41.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18983" PostTypeId="2" ParentId="18980" CreationDate="2018-01-20T09:53:52.823" Score="3" Body="&lt;p&gt;I think this question is about additional cross bracing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a rectangular frame in a vertical plane the ability to carry compressive loads is excellent, but a comparatively small lateral force can deform the rectangle into a parallelogram which will collapse under the same loading. Since the cross bracing will operate to restrict primarily tensile and (to a lesser extent) compressive forces, these struts can be hinged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there's a possibility of the main cross bracing struts bending, compressive forces could cause buckling. Additional cross bracing can be applied to the main cross bracing to keep the strut in the line of the force it is designed to restrict. These will also act in tension and compression (mainly tension), and can be hinged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases the idea of the hinge is to reduce the effect of lateral forces that might bend the strut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power transmission towers are pure structure, and have several levels of cross bracing - as can be seen in the picture. Though each level of bracing adds mass, it adds less mass (and less resistance to air flow) than using a thicker beam or a reinforcing plate would to achieve the same rigidity.&lt;/p&gt;&#xA;" OwnerUserId="13043" LastActivityDate="2018-01-20T09:53:52.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18984" PostTypeId="2" ParentId="18980" CreationDate="2018-01-20T12:31:54.120" Score="2" Body="&lt;p&gt;If you imagine a square or rectangular shape made from rods joined with points at the corners.  If those pins act as hinges then the whole shape can easily deform, effectively acting as a 4 bar link. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if the shape is a triangle then even if the corners are hinged then it can't be deformed without stretching one of the bars. Also, whichever way  you try to deform it one of the bars will always be in tension. So the  stiffness of the structure doesn't depend on the rigidity of the joints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the fundamental reason why breaking up a structure into triangular sections tends to be an efficient way to make a stiff, lightweight structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach also tends to be fairly straightforward to design and analyse. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-01-20T12:31:54.120" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18986" PostTypeId="1" CreationDate="2018-01-20T18:00:36.420" Score="3" ViewCount="26" Body="&lt;p&gt;I have created a simple xcos simulation for quarter wheel using the equations given at the bottom of the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the total torque on the wheel will be the braking torque and the friction torque. My question was right till the end the friction torque is really high( I am calculating mu from Burckhardt tire model). Even as the velocity goes to zero, the friction force gets higher (mu*mg). Is this realistic? Will a tire right up to the point it actually stops experience such high torque?&#xA;Thanks!&#xA;&lt;a href=&quot;https://i.stack.imgur.com/x8Dxh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x8Dxh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14622" LastActivityDate="2018-01-20T18:00:36.420" Title="Forces on a wheel during braking (Quarter Vehicle Model)" Tags="|mechanical-engineering|abs|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18987" PostTypeId="2" ParentId="18939" CreationDate="2018-01-20T19:11:44.993" Score="1" Body="&lt;p&gt;Let's start from the definition of dynamic viscosity, which is the proportionality constant in Newton's linear relationship between shear and velocity gradient: shear = (dynamic) viscosity x velocity gradient.  Thus, dynamic viscosity has dimensions of stress per velocity gradient.  Physically, it is a measure of how much stress a fluid experiences for a certain velocity gradient.  Kinematic viscosity is defined as the dynamic viscosity divided by fluid density.  Thus, kinematic viscosity is a measure of the stress a fluid experiences for a certain velocity gradient per unit of its density.&lt;/p&gt;&#xA;" OwnerUserId="14623" LastEditorUserId="14623" LastEditDate="2018-04-19T02:53:14.920" LastActivityDate="2018-04-19T02:53:14.920" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="18988" PostTypeId="2" ParentId="18981" CreationDate="2018-01-20T20:26:00.107" Score="0" Body="&lt;p&gt;Don't vent the cool dry air to the atmosphere.  That is just wasting energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't have a single cool air input with a seprate distribution duct.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a cool air source with 1 or more ducts into the room (like your house).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most computer rooms use the sub-floor to distribute the cool dry air.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-01-20T20:26:00.107" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18990" PostTypeId="2" ParentId="18965" CreationDate="2018-01-21T00:50:46.747" Score="1" Body="&lt;p&gt;This is a log-log plot in which $log(\omega)$ is plotted against $log(x)$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the frequency for $x=5$ is $\frac{188}{\sqrt{5}}=84.0762$ and the frequency for $x=10$ is $\frac{188}{\sqrt{10}}=59.4508$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the log scale, the tick marks show the original values, but the points $\{5,84.0762\}$ and $\{10,59.4508\}$ are actually plotted at $\{\log (5),\log(84.0762)\}=\{1.60944,4.43172\}$ and $\{log(10),\log(59.4508)\}=\{2.30259,4.08515\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9ZYvd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ZYvd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastEditorUserId="1627" LastEditDate="2018-01-24T14:40:46.680" LastActivityDate="2018-01-24T14:40:46.680" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18991" PostTypeId="1" AcceptedAnswerId="25915" CreationDate="2018-01-21T01:36:48.313" Score="1" ViewCount="1177" Body="&lt;p&gt;I am designing a  mock-up aircraft throttle, like one of these:&lt;a href=&quot;https://i.stack.imgur.com/CMNbG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CMNbG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an airplane, it it slides forward and backward to control the thrust of the engines. The problem is, I am not sure how to go about adding more friction to the sliding motion. If I were to just design a sliding handle, If you let go of it, it wouldn't be held in place, and it would slide back down. I need something sort of &quot;reverse bearing&quot; that will add friction to the rotation of the handle, so that when you let go of the handle, it wont slide around-- it will just stay in the same position until it is moved again.&lt;/p&gt;&#xA;" OwnerUserId="14625" LastEditorUserId="14625" LastEditDate="2019-02-10T02:51:37.180" LastActivityDate="2019-02-10T20:27:35.817" Title="Mechanical device that creates friction when rotating?" Tags="|mechanical-engineering|friction|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="18992" PostTypeId="1" AcceptedAnswerId="18997" CreationDate="2018-01-21T02:31:21.780" Score="1" ViewCount="4942" Body="&lt;p&gt;I hear a lot about open loop and closed loop Model Predictive Control (MPC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference between an open loop MPC and a closed loop MPC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any block diagram demonstrating the difference is also appreciated.&lt;/p&gt;&#xA;" OwnerUserId="14627" LastActivityDate="2023-02-24T10:41:10.373" Title="Open loop versus closed loop Model Predictive Control" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18993" PostTypeId="1" AcceptedAnswerId="19000" CreationDate="2018-01-21T04:30:08.513" Score="0" ViewCount="74" Body="&lt;p&gt;I am working on a model city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have utility poles A, B and C (brown) that go around a sharp corner. Looking from above, how do I connect the poles with power lines (gray, solid)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the crossarms of utility pole B rotated correctly with respect to the direction of the lines? Do the crossarms of pole B the correct length, or do they need to be longer than normal? Are there any other problems I should keep in mind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lBiUb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lBiUb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14628" LastEditorUserId="14628" LastEditDate="2018-01-21T04:54:04.140" LastActivityDate="2018-01-21T12:20:59.143" Title="Connecting power lines around corners" Tags="|civil-engineering|power|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="18994" PostTypeId="1" AcceptedAnswerId="19003" CreationDate="2018-01-21T05:11:17.203" Score="0" ViewCount="31" Body="&lt;p&gt;This is a plot of Yaw moment vs Lateral acceleration of my vehicle. I know that it can give me the analysis of trimmed conditions during cornering, yet unable to properly quantify it. What all can I achieve out of this plot. Thanks&lt;a href=&quot;https://i.stack.imgur.com/1HjbI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1HjbI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14271" LastActivityDate="2018-01-21T12:53:07.963" Title="Interpretation of the testing data of my vehicle" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="18997" PostTypeId="2" ParentId="18992" CreationDate="2018-01-21T10:58:59.527" Score="3" Body="&lt;p&gt;In short: it is both. Model Predictive Control is a repeated open-loop control in a feedback fashion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The explanation comes not from the general concept of open-loop and closed-loop, but from how MPC works. We solve an optimization problem with a certain prediction horizon N, i.e. we &lt;em&gt;predict&lt;/em&gt; the next N states until a certain point in the future (that's why it's called predictive). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives us an optimal input sequence. &#xA;We then only take the first element of this input sequence and apply it to the system. Until here, we use an open-loop control scheme, because we compute an input more or less far into the future without knowing all states, we only assume them based on our model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then wait one time-step for the systems reaction, measure the state (there comes the closed-loop) and start all over. All this happens on-line, during runtime of the system (there are subforms of MPC that get computed before, but computing online is the general concept of MPC). By &lt;strong&gt;repeatedly&lt;/strong&gt; solving this optimization problem and applying the beginning of the result, we establish a feedback. &lt;/p&gt;&#xA;" OwnerUserId="14629" LastEditorUserId="14629" LastEditDate="2018-01-21T17:06:31.950" LastActivityDate="2018-01-21T17:06:31.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="18998" PostTypeId="1" AcceptedAnswerId="19005" CreationDate="2018-01-21T11:52:47.427" Score="1" ViewCount="87" Body="&lt;p&gt;I am working on a problem at the moment and would appreciate some of your expertise regarding this issue. If you see the image below,  I am trying to calculate heat transfer from a fluid entering a vessel/tank on the grey cuboid within. To begin with, I am focussing on side C.&lt;/p&gt;&#xA;&lt;p&gt;I would like to understand how we would work out the Velocity of the fluid at point C when water is pumped in at A (intake), and, when discharging from point (B), i.e A flow, B return.&lt;/p&gt;&#xA;&lt;p&gt;The overall problem I am trying to solve is how long it would take to heat the cuboid from convection heat transfer (forced + also natural). At the moment I am calculating (Re) and need to determine velocity.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;It is not a home work question, it is for general research. There is no actual question I’m working from.&lt;/p&gt;&#xA;&lt;p&gt;I would like to understand how to work out velocity at C when there is a flow into A and out of B. This will allow me to work out Re at C.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ReR7Q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ReR7Q.jpg&quot; alt=&quot;Example&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14630" LastEditorUserId="-1" LastEditorDisplayName="user16" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-01-21T19:03:49.050" Title="Convection Heat Transfer" Tags="|mechanical-engineering|heat-transfer|convection|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="18999" PostTypeId="2" ParentId="18991" CreationDate="2018-01-21T11:55:28.753" Score="1" Body="&lt;p&gt;I assume your looking for that smooth, expensive feel, just like with volume buttons on higher-segment audio installations. Their resistance and damping allow you to easily make very precise adjustments, since the knob gradually gives in when you try to rotate it. I assume plane controls also give you this feeling, since they also need to be controlled precisely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way used to achieve this effect is using a spring loaded arm or friction plate, against another friction plate, all submerged in high-viscosity grease in a sealed container. The friction plate gives the resistance to rotating, the grease gives damping, to prevent the knob from abruptly rotating once enough force is applied. The grease also makes for lubrication, elemininating the abrupt effect. You could use other methods of damping, but this one's very easy to make yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For optimal effect, create a friction plate not too small; a larger plate will decrease any abrupt reaction. Use a heavy grease, like faucet grease or something. Thicker grease will give more damping, and will increase that smooth, expensive feel. (only to a certain point, but it's hard to find a grease to go beyond that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, you could attach a rubber band or spring to gently bring back the lever to its neutral position. Obviously, you don't want that for your throttle, but other controls like rudder pedals might benefit from this behaviour. Last, make a rubber bump stop at the end for the lever to hit, so the lever will silently and smoothly stop at the end of its reach. Otherwise it will hit e.g. the end of the plastic cover, causing an unpleasant sound and feel.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-21T11:55:28.753" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19000" PostTypeId="2" ParentId="18993" CreationDate="2018-01-21T12:20:59.143" Score="2" Body="&lt;p&gt;You can easily see that the wires don't allign(parallel) with the dotted line which would be the line they should follow to maintain the same mutual distance. To calculate the widht the mast should have, you can take the width of the first mast, and divide that by $cos(\alpha)$ where $\alpha$ is &lt;em&gt;half&lt;/em&gt; the angle the middle mast makes in degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality(at least in my country), the masts in sharp corners like those of yours, are lower and stronger than normal masts. Normal masts directing wires in straight lines mostly only get loaded perpendicular to the ground. Masts in corners, colloquially called 'little soldiers' in my country, are build thicker and shorter since they need to take on lateral forces much more. They are not necessarily as wide or wider than normal masts. I believe they are also build at an angle relative to the ground, further increasing resistance to lateral forces.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-21T12:20:59.143" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19001" PostTypeId="2" ParentId="18981" CreationDate="2018-01-21T12:23:47.810" Score="1" Body="&lt;p&gt;Additionally, switch rooms and IT rooms tend to just recirculate the room volume with some small DX cooling. It is not viewed as an occupied space, hence no need for extract for heat rejection, you would need some provision for smoke extract though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also another design consideration, exposing an IT space to external environment is not a good idea if you are likely to get high/very low RH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thats my 5 cents.&lt;/p&gt;&#xA;" OwnerUserId="14630" LastActivityDate="2018-01-21T12:23:47.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19003" PostTypeId="2" ParentId="18994" CreationDate="2018-01-21T12:53:07.963" Score="1" Body="&lt;p&gt;What are you trying to find in the data?&lt;br&gt;&#xA;Under normal circumstances, ie. no drifing or such the following can be assumed; at the same vehicle speed, lateral acceleration is inherent to yaw speed, and the yaw speed is inherent to the corner radius. So yaw acceleration, or yaw moment as you can also define it, is created when steering into a corner. Theoretically, it is 0 the moment you keep the steering wheel at any steady position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is theoretic, and based on formulas and basis physics. In automotive world, the highly dynamic reality ofter differs vastly from the theory we've learned. Explaining the V shape would be very hard to explain with only formulas and theoretic blahblah. The below is what my intuition tells me happens in reality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't suddely steer, and keep the steering wheel in position like a step response, but you gradually steer in, creating yaw moment during steering since the corner radius decreases. Yaw moment is higher in slow sharp corners, since you turn the steering wheel fast. Those sharp corners are made at a &lt;em&gt;relatively&lt;/em&gt; high speed, making for high lateral acceleration, explaining the upper right and left corners of your V. At the highway, you would steer relatively gently and slower, since situations are more dangerous at higher speeds, and your tyres aren't able to pull the same stunts at higher speeds without breaking out.&#xA;This explains why the upper corners meet in a mutual point, creating a V shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the upper left corner is higher likely indicates that your turns in one direction(probably right turns) are made faster than the other direction. Probably caused by the fact that you can oversee corners better, and that you have the right of way, and have to slow down less.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-21T12:53:07.963" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19004" PostTypeId="1" AcceptedAnswerId="19026" CreationDate="2018-01-21T12:56:51.393" Score="5" ViewCount="14292" Body="&lt;p&gt;I have tried searching for this, but I am unable to find a good answer.  There is all sorts of stats about the longest bridge in the world, the highest, the longest span, but I think one of the most difficult things is the depth of the pier.&lt;/p&gt;&#xA;&lt;p&gt;From my internet search I found a San Francisco article that says the deepest pier in the world is on the &lt;a href=&quot;https://www.extranomical.com/HTML_PAGES/OAKLAND_BAY_BRIDGE.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bay Bridge at 242 feet&lt;/a&gt; (~74m).  This was built in the 1930's, so does this feat of engineering still remain at the deepest, or has it been superseded?&lt;/p&gt;&#xA;&lt;p&gt;I initially wondered about this because of Boris Johnson refloating the idea of the idea of a road bridge between England and France.  At 26 miles, that would not be a world breaking engineering challenge, but it seemed to me the depth of the piers.  The Strait of Dover deepest point is 68m, so this does seem like a feasible project from a engineering point of view, but I think the question remains whether it is an efficient use of resource.  Particularly when the Channel Tunnel utilisation is only running at 48% or so.&lt;/p&gt;&#xA;" OwnerUserId="289" LastEditorUserId="289" LastEditDate="2023-11-05T17:47:48.363" LastActivityDate="2023-11-05T17:47:48.363" Title="What is the deepest bridge pier in the world?" Tags="|civil-engineering|bridges|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="19005" PostTypeId="2" ParentId="18998" CreationDate="2018-01-21T13:04:18.717" Score="1" Body="&lt;p&gt;Assuming incompressibility you can work with volume-flow to calculate (a first guess) the velocity, $u$ with the following equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably you have some volume or mass $\dot{m}$ flow entring at (A).&#xA;This mass-flow is entering the box through a opeing with a certain cross-section $A_\mathrm{A}$. The relation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_\mathrm{A} = \frac{\dot{m}}{\rho\, A_\mathrm{A}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same relation should also be valid for the exit B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_\mathrm{B} = \frac{\dot{m}}{\rho\, A_\mathrm{B}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the velocity around the cube we will now assume a constant velocity $u_\mathrm{C}$ which is of course not really the case but it is a start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the equation has the same structure you just need to put in the correct area ($A_\mathrm{C}$) which is the area of the box minus the area of the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xrq99.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xrq99.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_\mathrm{C} = \frac{\dot{m}}{\rho\, (A_\mathrm{box} - A_\mathrm{cube})}$&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2018-01-21T13:12:41.987" LastActivityDate="2018-01-21T13:12:41.987" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19006" PostTypeId="1" CreationDate="2018-01-21T14:40:36.513" Score="2" ViewCount="681" Body="&lt;p&gt;In case of incompressible flow, divergence of velocity field is zero so we can say that the velocity field is solenoidal. However, given that a flow field is solenoidal, is it necessarily incompressible? &lt;/p&gt;&#xA;" OwnerUserId="12974" LastActivityDate="2018-01-22T06:19:36.393" Title="Incompressible flow field and Solenoidal flow field" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19007" PostTypeId="1" CreationDate="2018-01-21T15:20:31.253" Score="1" ViewCount="2018" Body="&lt;p&gt;I’m a chemical engineer learning about fired heaters. I’m learning about draft and I picked this out of a textbook- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example - Drafts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Just above the burners = $-0.3\text{in H}_2\text{O}$&lt;/li&gt;&#xA;&lt;li&gt;At the bridgewall = $-0.05\text{in H}_2\text{O}$&lt;/li&gt;&#xA;&lt;li&gt;At the breeching = $-0.6\text{in H}_2\text{O}$&lt;/li&gt;&#xA;&lt;li&gt;Above the stack damper = close to zero.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Why is the draft at the breeching more negative than the draft just above the burned and at the bridgewall  ? Isn’t the draft when moving up the furnace be less negative ? &lt;/p&gt;&#xA;" OwnerUserId="13646" LastEditorUserId="1832" LastEditDate="2018-01-21T17:15:17.017" LastActivityDate="2018-07-30T18:11:59.383" Title="Draft of a fired heater" Tags="|pressure|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19009" PostTypeId="1" AcceptedAnswerId="19025" CreationDate="2018-01-21T17:18:38.250" Score="2" ViewCount="4238" Body="&lt;p&gt;I've been trying to find actual figures for the braking power of cars (in order to determine things like stopping time). As I understand it this comes in two parts: the maximum force that the brakes can apply on the tires, and the maximum force the tires can apply on the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd love to find figures for particular cars (or brake systems or tires) but I would be fine with any figures for actual cars under reasonable conditions. It's easy to find math problems of this sort online but I don't expect their figures to have much in common with the real world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could point me in the right direction (car data sheets, third party testing, government regulations, etc.) I'd be very much obliged.&lt;/p&gt;&#xA;" OwnerUserId="5362" LastActivityDate="2018-01-22T10:24:06.050" Title="How much braking force do cars generate?" Tags="|mechanical-engineering|automotive-engineering|energy|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="19010" PostTypeId="1" AcceptedAnswerId="19023" CreationDate="2018-01-21T17:26:12.227" Score="6" ViewCount="205" Body="&lt;p&gt;&lt;a href=&quot;https://www.moneyweb.co.za/news-fast-news/drought-hit-cape-town-at-point-of-no-return-tightens-water-targets/&quot; rel=&quot;noreferrer&quot;&gt;Cape Town, South Africa is experiencing an extremely serious drought&lt;/a&gt;, with estimates putting April 21, 2018 as the day that the city will be forced to shut off all municipal water supplies and begin water rationing, except to essential services such as hospitals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible solution is to delay this date by turning the supply off and on for scheduled periods of time over the next few months. However, experts have said that this is not a solution that can be implemented, because the pipes aren't supposed to deal with this variation of pressure and leaks and other infrastructure damage will occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way that a city could implement a water rationing system where water is shut on and off periodically, while minimizing leaks and damage to infrastructure?&lt;/p&gt;&#xA;" OwnerUserId="14632" LastActivityDate="2018-01-31T12:06:57.140" Title="Is there a way to reduce leaks and infrastructure damage when turning water on and off at the scale of city?" Tags="|civil-engineering|water-resources|pipelines|piping|infrastructure|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19011" PostTypeId="1" CreationDate="2018-01-21T17:35:39.190" Score="1" ViewCount="2821" Body="&lt;p&gt;Domestic ovens doesn't have any form of thermal insulation AFIK, and even have vents to circulate fresh air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IT83J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IT83J.jpg&quot; alt=&quot;domestic electric oven&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be more energy efficient to add thermal insulation to prevent heat escaping from oven. As a second benefit, it will be reach temp faster, reducing cooking times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why doesn't anybody does this?&lt;/p&gt;&#xA;" OwnerDisplayName="Look Alterno" LastEditorUserId="7083" LastEditDate="2018-01-28T22:35:19.447" LastActivityDate="2019-07-26T06:30:03.267" Title="Why do domestic electric oven do not use thermal insulation?" Tags="|electrical-engineering|" AnswerCount="5" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="19012" PostTypeId="2" ParentId="19011" CreationDate="2018-01-21T18:27:09.027" Score="0" Body="&lt;p&gt;Your photo shows a microwave oven. These heat the food directly using, as the name suggests, microwaves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you boil a cup of water in a microwave oven and then feel the walls of the inside of the oven you will find that they are not hot. Therefore there is no point in insulating them.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;[comment] I thinks it's a standard 'toaster oven' beloved by students and efficiency dwellers.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've never used on but if the elements are on top it suggests that the food is cooked by direct infrared radiation rather than by heating the walls and air as in a baking oven. Maybe a student can answer.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2018-01-21T22:07:21.977" LastActivityDate="2018-01-21T22:07:21.977" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19013" PostTypeId="2" ParentId="19011" CreationDate="2018-01-21T18:46:53.053" Score="1" Body="&lt;p&gt;That’s a very cheap oven - you can see the door is just a piece of glass (presumably heat resistant) with no seals to keep heat in.  Compare that with a decent quality oven.  Everything about this says “cheap”.  As is usually the case, better, more efficient designs usually cost more to make.  But with that said, I don’t know that anyone can really say it doesn’t have insulation without taking it apart and looking.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2018-01-21T18:46:53.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19014" PostTypeId="1" CreationDate="2018-01-21T19:21:32.473" Score="1" ViewCount="143" Body="&lt;p&gt;I was watching &lt;a href=&quot;https://www.youtube.com/watch?v=WAwDvbIfkos&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Feynman video&lt;/a&gt; and (as usual) his explanations make a lot of sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I just took a train, and am not so sure anymore. Looking at the train wheels they did not appear cone shaped to me. Also, if the train was running on cone shaped wheels I would expect that there is very little contact area between the wheel and the rails. However the rails looked completely shiny on the top suggesting that the wheels are in contact with all of the rail (i.e. flat wheels).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are modern trains using a different method to deal with this problem or are the cones almost flat anyway (what angle)?&lt;/p&gt;&#xA;" OwnerUserId="12005" LastActivityDate="2018-01-21T19:21:32.473" Title="Cone shaped train wheels" Tags="|rail|wheels|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19015" PostTypeId="2" ParentId="19009" CreationDate="2018-01-21T19:39:20.567" Score="1" Body="&lt;p&gt;braking systems in cars can be rated in terms of braking horsepower capacity, which is the net frictional force exerted on the brake disc or drum times the rotational speed of the disc or drum at the point of contact of that force. When described this way, car brakes can dissipate hundreds of horsepower in a hard stop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know if there are any mandated braking standards based on braking horsepower ratings, but as a practical matter the brakes in a car can be considered adequate if they are capable of locking up the wheels at any forward speed the car can achieve (in the absence of any ABS system) using the driver's muscle forces. This would guarantee that the braking system can exert the optimum braking force, defined as that generated just before rolling friction gives way to sliding friction. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-21T19:39:20.567" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19016" PostTypeId="2" ParentId="19011" CreationDate="2018-01-21T19:46:55.417" Score="1" Body="&lt;p&gt;It is a very cheap toaster oven, and does what it's built to do: heat things up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many construction workers and field technicians use it. It is cheap so that after receiving the abuse of harsh job sites and rarely if ever getting cleaned, it is rather easy to throw away and get a new one.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-01-22T14:11:18.980" LastActivityDate="2018-01-22T14:11:18.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19017" PostTypeId="2" ParentId="19011" CreationDate="2018-01-21T20:41:35.260" Score="1" Body="&lt;p&gt;It almost certainly has fiberglass insulation between the metal walls, and the vents on the &lt;strong&gt;outside&lt;/strong&gt; that you see are there to keep the electrics relatively cool. &lt;/p&gt;&#xA;" OwnerUserId="1276" LastActivityDate="2018-01-21T20:41:35.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19018" PostTypeId="2" ParentId="19006" CreationDate="2018-01-21T20:44:07.240" Score="1" Body="&lt;p&gt;&lt;strong&gt;TD;DR: No&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Compressibility&quot; rel=&quot;nofollow noreferrer&quot;&gt;Compressibility&lt;/a&gt; is a characteristic of the fluid (or maybe also the way you describe a problem).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial \rho}{\partial p} \neq 0 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the a pressure variation will also cause a variation in density.&lt;br&gt;&#xA;In most cases water and air at low velocities can be treated &lt;strong&gt;incompressible&lt;/strong&gt; because the fraction above is very close to Zero. This means that the fluid’s density does not change (much) when pressure is applied. Hence, in order to see compressibility effects one needs a pressure variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Solenoidal_vector_field&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solenoidality&lt;/a&gt; is a characteristic of a vector-field.&#xA;It is a phenomenological description of the (e.g. velocity or pressure) flow-field. But this only means that a particular flow field is divergence free. It does not (necessarily) state anything about the physical characteristics of the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means a compressible fluid can be solenoidal if the boundary conditions and forces create one.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2018-01-22T06:19:36.393" LastActivityDate="2018-01-22T06:19:36.393" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19019" PostTypeId="1" CreationDate="2018-01-21T21:01:15.543" Score="5" ViewCount="120" Body="&lt;h2&gt;Background&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The normal space elevator (that reaches from the Earth's surface to a counterweight beyond geostationary) needs a material that's strength/weight ratio is better than any material we have available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've another idea, that I assume isn't possible either, but I'd like to know what makes it impossible?...&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The idea&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;What if, rather than starting the tether from the Earth's surface, it starts from 2000km above the Earth? This will &lt;em&gt;massively&lt;/em&gt; reduce the necessary strength/weight ratio of the material. Not only is the part closest to Earth experiencing the greatest gravitational force (per unit mass) but there's a sort of vicious circle: To add a metre to the length of the tether we need to strengthen the whole tether (above this new segment), which means making it heavier, which means making it stronger, and heavier, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found, using the numbers for Zylon, it works.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&quot;How's it stay up?&quot;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This elevator, unlike a real space elevator is in 'proper orbit', or at least it is, on average. The part that hangs towards the Earth will be suborbital (indeed will be going quite slowly relative to low-earth orbit), while the part that extends beyond geostationary will be at superorbital speed. It is in orbit.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&quot;But how do we get to the start of the tether if it’s 2000km up?&quot;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Going up into space is easy, getting into orbit is the expensive bit. A rockoon might be a neat way to get to the 2000km mark with a very modest rocket (the rocket equation means that we can use a very small rocket to get to 2000km, compared to the rocket required to get to 2000km &lt;em&gt;orbit&lt;/em&gt;) - about 3kg of propellant per kg of payload, I think.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&quot;But won’t you pull the whole thing down as you climb it?&quot;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Yes. To correct for this, ion-engines will be arranged on the tether for station-keeping. Some of the payload can be used for refuelling these (they have a specific impulse 10 times better than a rocket launcher, and the rocket equation helps here, so hopefully we’ll need far less fuel!)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Numbers&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I wrote a python script with a few assumptions (that it's 'suspended' from something geostationary, so it has a length of 27415km), it ends 2000km above the Earth's surface. It has a tensile strength of 5.8GPa and density of 1.54 g/cm^3 (Zylon). It can hold a 400kg payload at the bottom of the tether. I found it needed to have a total weight of 2100 tonnes and a maximum diameter of 10.4mm.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Questions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9dNBP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9dNBP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;caption: &lt;em&gt;simulated space elevator - plotted every two minutes. Not sufficiently high to be geostationary, and I suspect either the initial conditions or modelling errors mean that it failed to remain orbital!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this work? I've &lt;a href=&quot;http://www.michaeltsmith.org.uk/?p=165&quot; rel=&quot;nofollow noreferrer&quot;&gt;modelled a simple 7-segment tether&lt;/a&gt;, but I need to tweak the initial conditions a little to get it into a good orbit. I worry that something weird happens with orbital mechanics that means this isn't going to be a stable object? Can a tether stay 'vertical' without too much station keeping?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it does work, it's neat, as it'll take less than 100 flights of the falcon heavy to get it up there, and after than geostationary orbit will be cheap :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Code for the &lt;a href=&quot;https://github.com/lionfish0/experimentation/blob/master/Modelling%20Space%20Tether.ipynb&quot; rel=&quot;nofollow noreferrer&quot;&gt;model&lt;/a&gt; and &lt;a href=&quot;https://github.com/lionfish0/experimentation/blob/master/Space%20Tether.ipynb&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass calculations&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Edit #1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To look at @kamran's point that &quot;The system is...doomed to...pull the higher eleveation parts into lower/ heavier orbits and crash eventually.&quot; It does seem it's inherently overly sensitive to initial conditions, if it is slightly too low, then the gravitational force on it scales much quicker than it would if the object were a point mass. I wonder if this can be mitigated by shortening the tether: The centre of such a contraction would be the centre-of-mass, thus it would 'pull up' the centre-of-gravity out of the gravity well a little. Maybe this would be the mechanism to control this effect? But it does feel like it's not in a stable state and will require active control. See 'fall-to-earth' scenario below!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IPo53.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IPo53.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14634" LastEditorUserId="14634" LastEditDate="2018-01-22T10:04:02.360" LastActivityDate="2018-01-22T19:47:41.607" Title="Would a partial space elevator work?" Tags="|aerospace-engineering|dynamics|stresses|modeling|strength|" AnswerCount="1" CommentCount="9" ClosedDate="2018-01-23T14:19:47.887" ContentLicense="CC BY-SA 3.0" />
  <row Id="19020" PostTypeId="2" ParentId="18961" CreationDate="2018-01-22T02:02:09.663" Score="1" Body="&lt;p&gt;Let's put aside this specific game, and just say we want to calculate the speed of some generic vehicle.  I think that will answer what you are really trying to get at.  Our vehicle is going to obey Newtons\'s second law $F=ma$, force = mass times acceleration.  So, let's add up all of the forces that act on the train.  The main ones will be the driving force from the engine, and then rolling resistance and air drag as pointed out by @SolarMike.  Let's ignore rolling resistance to simplify the problem, and just look at the others.  So &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{engine} + F_{drag} = ma$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we know from &lt;a href=&quot;https://en.wikipedia.org/wiki/Drag_(physics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;, that $F_{drag} \propto v^2$, where $v$ is the velocity.  To simplify the discussion, I'll lump all of the constants into just one, so $F_{drag}=-Cv^2$, where the minus sign is indicating that the drag force is in the opposite direction of the applied engine force. So at this point we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{engine} - Cv^2 = ma$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the formula for the engine?  Well, it's complicated and will depend on your specific engine. If you have a transmission, it will depend on what gear you are in.  But two important points: it is dependent on speed, and above a certain speed it will start to decrease with speed.  i.e. as you go faster and faster, the engine is producing less torque. You can find some real engine curves &lt;a href=&quot;http://web.itu.edu.tr/~sorusbay/ME422/LN11.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;,  but for this discussion, let's oversimplify it a lot and just assume a linear relationship $F_{engine}=A-Bv$. So now we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A-Bv - Cv^2 = ma$.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, imagine your train starts at v=0.  The left hand side is positive, so acceleration is positive and your train speeds up.  Now it has some velocity. The left hand side is still positive, but less than before. The engine is producing less force, and the air drag is acting against us. So the train continues to accelerate, but not as much. Finally, at some speed, the left hand side will be zero.  The air drag will have increased to the point where the force produced by the engine exactly equals the air drag.  At this point, acceleration is zero.  The train has reached its top speed.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-01-22T02:02:09.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19021" PostTypeId="2" ParentId="19019" CreationDate="2018-01-22T04:10:44.637" Score="1" Body="&lt;p&gt;I can see the image: A long thin cable arced in a sharp slant backwards, strained by the huge forces of orbital velocity trying to tear it apart and cut it to free orbiting segments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what is there to balance a possible whiplash effect which could possibly run up and down the length of cable and wind the whole system into a violant dance like a wounded snake?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the balancing force keeping the end at approximately 2000 kilometers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about tidal effects? This is a classical case of effects of moon's gravity working with different intensity on higher and lower parts of the cable twice daily!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is inherently imbalances and doomed to coil down and pull the higher eleveation parts into lower/ heavier orbits and crash eventually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes it is coceivebale to have counter balance boosters and rockets to try to maintain the geometry of the system, but most likely whith anacceptable expenses and logistics!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-01-22T19:47:41.607" LastActivityDate="2018-01-22T19:47:41.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19022" PostTypeId="2" ParentId="19010" CreationDate="2018-01-22T08:36:16.927" Score="2" Body="&lt;p&gt;The problem with older and under-maintained (this is South Africa after all) systems is that the pipes get brittle, rusted and joints fragile and the network won't be able to handle the transient forces of constantly opening and closing valves, even if it's done slowly. The networks are in general not designed to withstand such large constant cyclic loading in any way as the water supply is &quot;usually&quot; constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another solution might be to close the valves partway to slow down the flow to a trickle so people consume the water more slowly. Other advantages are that leaks in the system and faucets will lose much less water and might possibly stop due to the pressure reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problems with this, however, are that some people might not have water at all due to head losses in the network and cavitation might happen at the half closed valves causing damage to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just an update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a radio interview this morning, 24 Jan, the Western Cape's premier noted that they are actually drastically reducing the water pressure of suburbs that refuse to reduce their water consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit based on Johnathan's comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are regulations on the minimum and maximum water pressure to serve the system. The pressure could still be regulated to stay within these limits, but as a temporary solution to extend water availability, having water pressure lower than the norm is a bit better than having people walk or drive to collect their minimum daily ration of 25l of water; which is what will happen if nothing changes by 21 April 2018.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastEditorUserId="14440" LastEditDate="2018-01-24T05:27:46.970" LastActivityDate="2018-01-24T05:27:46.970" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19023" PostTypeId="2" ParentId="19010" CreationDate="2018-01-22T10:04:12.900" Score="2" Body="&lt;p&gt;It is difficult to provide a useful answer to this question without additional information.  For example, what is the system pressurization - operating psi/bars?  Having installed ductile iron water mains, and having pressure tested new installations to destruction (resulting from hairline cracks in new pipes), I can confirm that the effects can potentially be devastating to infrastructure - especially roads and possibly other foundations/buried structures.  In addition to the obvious negatives associated with traffic disruption and reinstatement, there will also be substantial water losses should this happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MTMo7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MTMo7.jpg&quot; alt=&quot;Road collapse due to pipe failure.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where HDPE pipe is used, one of the predictors of service life is the cyclic nature of the pressurization.  Cyclic pressurization may not lead to catastrophic failure, as it could for ductile iron, but it will certainly shorten the lifespan of the network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the nature of the network design in terms of control points?  Can the network even be controlled?  The valves, tees and joint connections will likely be the most fragile part of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonetheless, in my view the best chance of implementing such a programme successfully would be to partially shut down the network on a rotational schedule in such a way that essentially the entire network would have a chance to stay pressurized - or at least ensure that the portion of the network that is exposed to risk would be minimized.  This would require endpoint valves however.  It could be possible, depending upon the nature of the network, to install these at strategic locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether such an approach is economic, or would provide any benefits at all in terms of a reduction of water consumption, is another question.  My guess is it may not, but then again a rationing programme may not be enforceable either.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastActivityDate="2018-01-22T10:04:12.900" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19024" PostTypeId="1" CreationDate="2018-01-22T10:07:51.457" Score="2" ViewCount="593" Body="&lt;p&gt;In the not too distant future I will be looking to simulate lubricant oil flows in a gear box, with rotating gears located in a bath of oil. I will be looking at the distributions at various angles of incline. Generically, a model may look something like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KKDqz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KKDqz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does anybody have experience with Smoothed Particle Hydrodynamics (SPH) CFD packages that can provide advice on its limitations?&lt;/strong&gt; I am less interested in absolute values of density/velocity etc., primarily I need to see the distribution of the fluid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to this branch of CFD and all I can find is advertising guff from the people who make the software and are trying to sell it. As I understand it, pre-processing time will be considerably lower than LES CFD since SPH does not require a mesh, where as an LES simulation would require a complex mesh with multiple rotating frames. Since we aren't looking at discrete volumes with SPH, instead at particles, the convection term is done away with making computation easier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The companies selling SPH seem to be targeting exactly my application, I'm guessing that's for a good reason...&lt;/p&gt;&#xA;" OwnerUserId="14498" LastEditorUserId="14498" LastEditDate="2018-01-24T10:02:12.763" LastActivityDate="2019-07-31T20:09:09.647" Title="Smoothed Particle Hydrodynamics (SPH) Vs. Large Eddy Simulation (LES) for Gearboxes" Tags="|fluid-mechanics|automotive-engineering|gears|cfd|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19025" PostTypeId="2" ParentId="19009" CreationDate="2018-01-22T10:24:06.050" Score="1" Body="&lt;p&gt;To elaborate on my comment; obviously generated braking force is defined by how hard you push the braking pedal. The required force to lock the wheels (given that ABS is disabled) depends on how much traction the tyre has. Let's say our tyre has a constant friction coefficient of 1. That's a simplified situation where the tyre has as much traction force, as there is force on the tyre by the vehicle's weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows you to decelerate(or accelerate) with a maximum of 1G or 9.81m/s2, before your tyres start slipping. That means that you'd get to a stop in 2.83s when travelling 100km/h. You can see from this figure, that most supersports cars are limited in their 0-100 time by the tyres. They use as wide tyres as they can on the driven wheels, to maximise traction. Heating up the tyres and lowering their pressure, and a slick surface on dry hot asphalt further increases their traction. But it doesn't allow you to accelerate to 2G or something.(except for crazy top fuel dragsters) But that's why i'm rather sceptical about Tesla's promised 0-100 figures for the announced roadster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, a friction coefficient is not close to constant and way more complex. In reality, your tyres are &lt;em&gt;always slipping,&lt;/em&gt; which is the main reason they wear out. It gets worse in corners, acceleration, braking etc. but it's always present in a certain amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, traction sharply decreases once the wheel starts slipping by a certain amount. I believe that was usually 20% or something. You only have kinetic friction at that point, which is always lower than static friction. That's the reason ABS works, it constantly regains static friction by releasing the brakes briefly. I hope this gives you an better understanding of how and why tyres work.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-22T10:24:06.050" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19026" PostTypeId="2" ParentId="19004" CreationDate="2018-01-22T11:52:09.983" Score="4" Body="&lt;p&gt;It's not a bridge, but if you wish to consider the depth of marine structures generally, a good example is the Troll A platform in the North Sea.  From &lt;a href=&quot;https://en.wikipedia.org/wiki/Troll_A_platform&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The Troll A platform has an overall height of 472 metres (1,549 ft), weighs 683,600 tons (1.2 million tons with ballast) and has the distinction of being the tallest and heaviest structure ever moved by mankind. The platform stands on the sea floor 303 metres (994 feet) below the surface of the sea ...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8p7wJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8p7wJ.jpg&quot; alt=&quot;Troll A Platform Norway&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a starting point for bridges you have a &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_highest_bridges&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of the highest bridges&lt;/a&gt; with the highest bridge being the Duge Bridge in China with a deck height of 565m:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6CQlm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6CQlm.jpg&quot; alt=&quot;Duge Bridge China&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for marine crossings, there is a bridge (that apparently exceeds San Francisco Oakland Bay Bridge you have referenced) that has received instruction to proceed during 2010 but has since been put on hold.  The &lt;strong&gt;Bridge of the Horns&lt;/strong&gt; is a Red Sea crossing at the Island of Perim (300m sea depth) from Yemen to Djibouti that is described by &lt;a href=&quot;http://www.cowi.com/menu/newsandmedia/news/newsarchive/newbridgewilllinkthemiddleeastwithafrica&quot; rel=&quot;nofollow noreferrer&quot;&gt;COWI DK&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The length and navigational clearance required west of Perim will likely mean at least 10km of multi-span suspension bridge, with the remainder being approach bridges. The concrete pylons will have to be supported up to depths of 300m and at the same time tower 400m over the water's surface to support the ultra-long span suspension bridge. Each pylon will be 700m. It will be enormous.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ADWv9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ADWv9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for information, Professor TY Lin has proposed a Gibraltar crossing that would have a height of 910m. I could not find specific information on depth, but one would assume that it would exceed 300m (the sea depth there ranges between 300 and 900m). A tunnel was originally considered but has been dismissed as it would be 900m deep, the deepest tunnel in the world is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Eiksund_Tunnel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eiksund Tunnel&lt;/a&gt; at 287m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming back to the underlying purpose of your question; ie Boris Johnson's proposal, I have pointed out these extreme data points to illustrate that the project is well within the bounds of current engineering capability.  In my view the most significant challenge of the project would be determining the location and magnitude of the main span required so as not to interfere with shipping vessels as this would likely determine the size of piers and depth of foundations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that these depths are referenced to the sea bed, they do not include depths that might be required for foundations.  Checking the &lt;a href=&quot;https://maps.ngdc.noaa.gov/viewers/bathymetry/&quot; rel=&quot;nofollow noreferrer&quot;&gt;bathymetric survey data&lt;/a&gt; at the San Francisco Oakland Bay Bridge crossing does not support a sea bed depth of 74m (looks to be in the range of 20 to 30m), therefore the depth you have quoted (74m) may include foundations; for comparison purposes the bathymetric data at the Strait of Dover yields a sea bed depth of about 50m - as you have indicated an extreme of 68m to the sea bed.)&lt;/p&gt;&#xA;" OwnerUserId="3266" LastEditorUserId="3266" LastEditDate="2018-01-22T13:56:14.230" LastActivityDate="2018-01-22T13:56:14.230" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19027" PostTypeId="1" CreationDate="2018-01-22T12:15:08.950" Score="2" ViewCount="315" Body="&lt;p&gt;Few days ago someone told me that the amount of bolt's preload in a single lap joint can influence the natural frequency of the beam. So to test that out I modeled a 3D single lap joint in ABAQUS and defined bolt load for it. I realized that the difference between the natural frequencies related to preload of 1&amp;nbsp;N.m and 13&amp;nbsp;N.m is in the order of 0.01. Did I model it wrongly or the amount of preload can’t change the natural frequencies?&#xA;the screenshot of abaqus model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4ILEe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4ILEe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14379" LastEditorUserId="1832" LastEditDate="2018-01-22T12:28:36.937" LastActivityDate="2018-10-18T10:55:43.413" Title="How the amount of bolt's preload in a single lap joint can change the natural frequencies?" Tags="|bolting|frequency-response|abaqus|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19029" PostTypeId="1" AcceptedAnswerId="21074" CreationDate="2018-01-22T16:05:30.047" Score="1" ViewCount="8378" Body="&lt;p&gt;I am currently working on my dissertation project using abaqus, my model is an arch bridge consisting of a shell and solid elements, there are concrete and steel materials also, I have checked all of the constraints, and I found that there are some constraints giving the error messages 'time increment required is less than specified' and 'too many attempts made for this increment', specifically, those constraints happened in shell-solid tie constraints. the other constraints between solid to solid were fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody help me resolve these problems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SqnMu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SqnMu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/AZqpV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AZqpV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14641" LastEditorUserId="1832" LastEditDate="2018-01-22T16:52:13.187" LastActivityDate="2021-07-21T04:54:38.067" Title="How to Solve this error message on Abaqus 'time increment required is less than specified' and 'too many attempts made for this increment'?" Tags="|structural-engineering|finite-element-method|abaqus|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19030" PostTypeId="1" CreationDate="2018-01-22T16:46:54.153" Score="2" ViewCount="38" Body="&lt;p&gt;In a perfect world, there should be a constant assigned to the angle difference between voltage and current , no inductance.   I suppose the angle is unique to the amount of inductance in each circuit. Even a straight wire has a certain amount of inductance. The frequency of the AC voltage would determine the reactance of each component in the circuit. Both resistance and reactance impede current flow . In a circuit with a steady frequency, would the capacitive reactance and inductive reactance be combined with the resistance and multiplied by the average voltage to give the actual wattage ? The reststance, reactance would be determined using the rules for determining resistance for a circuit ?&lt;/p&gt;&#xA;" OwnerUserId="14642" LastActivityDate="2018-01-22T16:46:54.153" Title="power calculation" Tags="|electrical|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19031" PostTypeId="2" ParentId="19029" CreationDate="2018-01-22T18:47:40.793" Score="0" Body="&lt;p&gt;The first thing you can try is to go into the step manager and change the minimum increment size and maximum number of increments.   However as this appears to be on the very first increment there is likely something wrong with the model such as inconsistent boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the &lt;code&gt;.msg&lt;/code&gt; file is usually enlightening.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2018-01-22T18:47:40.793" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19032" PostTypeId="1" AcceptedAnswerId="19044" CreationDate="2018-01-22T19:36:41.477" Score="1" ViewCount="59" Body="&lt;p&gt;I've been trying to find the value of thermal resistivity for some linear disc resistors. I know that Thermal Resistivity is the reciprocal of Thermal Conductivity but the thermal conductivity on the data sheet is given in W/cm^2.C/cm     which is the units for the heat transfer coefficient per cm not conductivity if the &quot;period&quot; is actually a multiplier and not just a period? Is this some convention I am unaware of on data sheets like this? I do have the thickness of the resistor disc in use so I think I can get the actual conductivity by multiplying the value given by the thickness squared? &lt;/p&gt;&#xA;" OwnerUserId="14647" LastActivityDate="2018-01-23T21:45:32.867" Title="Thermal Conductivity? Units On a Data Sheet" Tags="|materials|thermodynamics|thermal-conduction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19033" PostTypeId="1" AcceptedAnswerId="19040" CreationDate="2018-01-22T21:20:57.873" Score="0" ViewCount="720" Body="&lt;p&gt;I know how to solve it by moment line methode but i want to use singularity function here and i'm very confused. My strategy consist of writing the correspond singularity function for each element but i have absolutely no clue how can i write the right singularity function for the inverse ramp distributed load at the right side? &lt;a href=&quot;https://i.stack.imgur.com/QmeIH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QmeIH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-01-24T11:27:18.980" Title="How to deal with inverse distributed load ramp starts and ends somewhere along the beam?" Tags="|mechanical-engineering|statics|beam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19034" PostTypeId="2" ParentId="19032" CreationDate="2018-01-23T01:00:37.170" Score="0" Body="&lt;p&gt;I suspect what is meant is &lt;code&gt;W/(cm^2*C)/cm&lt;/code&gt; which reduces to &lt;code&gt;W/(cm C)&lt;/code&gt;&#xA;(I readily found examples of this by google..) Its not unusual for industry data sheets to play fast and loose with technical correctness. (Who needs those pesky parenthesis)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hvrint.com/LinearDisc.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.hvrint.com/LinearDisc.pdf&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HN7EV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HN7EV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;4 W/(m k)&lt;/code&gt; (more common unit) is a pretty plausible order of magnitude number for the material they describe.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2018-01-23T01:07:32.557" LastActivityDate="2018-01-23T01:07:32.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19035" PostTypeId="1" CreationDate="2018-01-23T02:04:02.217" Score="6" ViewCount="78336" Body="&lt;p&gt;Zippo lighters are notorious for needing regular refilling. The fuel evaporates from the cotton regularly.&#xA;Some people put a bit of rubber around the seal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What atmospheric conditions keep fuel from evaporating? Temperature, humidity, etc.&#xA;Should I keep the lighter near my body  (for heat) or in an outside pocket?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I keep it topped off, or will fuel evaporate faster that way?&lt;/p&gt;&#xA;" OwnerUserId="6488" LastEditorUserId="615" LastEditDate="2018-01-23T19:17:01.400" LastActivityDate="2021-03-07T23:13:59.783" Title="How can I keep a Zippo lighter from losing fluid to evaporation or leaks?" Tags="|fluid-mechanics|thermodynamics|fire|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19037" PostTypeId="1" AcceptedAnswerId="19042" CreationDate="2018-01-23T10:06:42.480" Score="2" ViewCount="186" Body="&lt;p&gt;I've Googled the hell out of this and it doesn't seem to be, i.e.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It's structured like a standard stud (not reinforced top)&lt;/li&gt;&#xA;&lt;li&gt;It doesn't run along a floor joist&lt;/li&gt;&#xA;&lt;li&gt;The uprights are pretty badly cut and not snug&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But it looks to me like the remains of an attic support structure (it an attic conversion/extension into a dormer-like structure) and to my untrained eye, it seems like a lot of roof (80&quot; from nearest edge to back wall) to support itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was planning on building a stud wall on the outside edge and running a support beam along the length to be safe but is there any way I can be sure that the support isn't even needed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AitHD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AitHD.jpg&quot; alt=&quot;Wall I want rid of&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8529" LastEditorUserId="8529" LastEditDate="2018-01-24T12:54:36.007" LastActivityDate="2018-01-24T12:54:36.007" Title="Is there any way I can be sure a wall is NOT load bearing?" Tags="|building-design|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19038" PostTypeId="2" ParentId="19037" CreationDate="2018-01-23T10:31:05.693" Score="1" Body="&lt;p&gt;Is that stud mounted onto the plasterboard? If so, it is not supporting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the plasterboard buts up to the stud then it is supporting, but I don’t think it is - it looks to be added to make a « cupboard » or just hide stuff.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-01-23T10:31:05.693" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19039" PostTypeId="1" CreationDate="2018-01-23T10:47:41.767" Score="2" ViewCount="1517" Body="&lt;p&gt;I have access to lake water.  My land is situated in the back of this lake. In summer, water in the lake shrinks to a distance of about 2000 to 2350 ft away from my land. My land is 10 feet above the water surface level. So I want to draw this water through a siphon. Can this work, if I dug up 10 ft depth near the land as the distance is 2000 to 2500 ft? The depth of water from the surface level of the water is around 15 ft.&lt;/p&gt;&#xA;&lt;h2&gt;ETA:&lt;/h2&gt;&#xA;&lt;p&gt;&amp;quot;350 lpm is sufficient for my 15.5 Acres of lands.&amp;quot;  &amp;quot;I would like to use 2.5-inch pipe [ID or OD not specified by the OP]&amp;quot;&lt;/p&gt;&#xA;" OwnerUserId="14654" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-01-24T04:37:49.223" Title="Can Siphoned water travel a distance of 2500 ft with a height difference of 10 ft?" Tags="|fluid-mechanics|fluid|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19040" PostTypeId="2" ParentId="19033" CreationDate="2018-01-23T10:47:58.987" Score="0" Body="&lt;p&gt;Are you familiar with the &lt;em&gt;Dirac delta function&lt;/em&gt;, &lt;em&gt;Heaviside step function&lt;/em&gt; and the &lt;em&gt;Ramp function&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could use Laplace Transforms together with those three handy functions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dirac Delta function&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta(x)= \left\{\begin{array}{l}+\infty\qquad&amp;amp;x=0\\0&amp;amp;x\neq0\end{array}\right. $$&#xA;The &lt;em&gt;Dirac Delta function&lt;/em&gt; is equal to infinity, at $x=0$, and $0$ everywhere else. Respectively $\delta(x-a)=+\infty$ if $x=a$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the &lt;em&gt;Euler-Bernoulli beam theory&lt;/em&gt;, where &#xA;$$ EI\frac{d^4w}{dx^4}=q(x) $$ &#xA;a static interpretation of the Dirac function is, that a point load $Q$ at point a represents a distributed load with infinite density, thus can be written as $q(x)=Q\cdot \delta(a-x)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Heaviside step function&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The step function is the integral of the &lt;em&gt;Dirac Delta function&lt;/em&gt;&#xA;$$ H(x)=\int_{-\infty}^x\delta(s)ds $$&#xA;or&#xA;$$ H(x)= \left\{\begin{array}{l}1\qquad&amp;amp;x\geq0\\0&amp;amp;x&amp;lt;0\end{array}\right. $$&#xA;Thus, a distributen load, with magniuted $q$, between $x=a$ and $x=b$ can be written as $q(x)=q\left[ H(x-a)-H(x-b) \right]$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ramp function&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ramp function again is the integral of the Heaviside function.&#xA;$$ R(x)= \left\{\begin{array}{l}x\qquad&amp;amp;x\geq0\\0&amp;amp;x&amp;lt;0\end{array}\right. $$&#xA;Thus, a distributied load , starting at $x=a$, with a slope $m$ can be written as $ q(x)=m\cdot R(x-a) $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following distributed load, for example:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hRJ7X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hRJ7X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This load can be written as:&#xA;$$ q(x)=\frac{q}{b-a}\left[R(x-a)-R(x-b)\right]-\frac{q}{d-c}\left[R(x-c)-R(x-d)\right]  $$&#xA;&lt;em&gt;please see &lt;a href=&quot;https://i.stack.imgur.com/H0xuR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;attached file&lt;/a&gt; for the derivation&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further explanation of the Dirac Delta function&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take, for example the point force $Q$. That force can be approximated by a distributed load $Q=q\cdot dx$, therefore $q=\frac{Q}{dx}.$ Now, as a point load acts on an interval $dx\to0$, we can say $\lim_{dx\to0} q=+\infty$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also represented, if you look at a shear force diagram. We know, that $V'(X)=q(x)$, and we know that a point load $Q$ (external force or support) at point $a$ causes a &quot;jump&quot; in the shear force diagram, according to its magnitude, this can be represented by the Heaviside function. $V=Q\cdot H(x-a)$, now &#xA;$$V'(x)=q(x)=(Q\cdot H(x-a))'=Q\cdot H'(x-a)$$&#xA;As explained above, the derivative of the &lt;em&gt;Heaviside function&lt;/em&gt; is the &lt;em&gt;Dirac delta function&lt;/em&gt;, therefore&#xA;$$ q(x)=Q\cdot \delta(x-a) $$&lt;/p&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="8632" LastEditDate="2018-01-24T11:27:18.980" LastActivityDate="2018-01-24T11:27:18.980" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19041" PostTypeId="2" ParentId="19039" CreationDate="2018-01-23T11:19:42.697" Score="8" Body="&lt;p&gt;As long as you can (1) set the free water surface of your destination at a lower level than the free water surface of your source, (2) use a completely sealed, air tight siphon, and (3) initially prime the system, it should work.  Your source being your datum, for example, setting the depth of your destination to the same level (in this case 10 ft) will not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must set the destination lower than your datum (or source), by digging a pit deeper than 10 feet, and also digging a pit depth that will be deeper than the lowest seasonal water level (including that due to your consumption).  This could be, in your case, for example only, a pit that is 12 to 15 feet in depth, but you would have to determine the exact value for yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition you should consider the water pressure that you require at the destination.  The more pressure you require, the deeper your pit, relative to your water source.  The difference in free water surface (from the source to the destination, in feet, for example) represents the water &lt;strong&gt;head&lt;/strong&gt; and this can be converted to pressure by using such tools as are found at &lt;a href=&quot;https://www.engineeringtoolbox.com/pressure-head-water-d_1354.html&quot; rel=&quot;noreferrer&quot;&gt;this website&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastEditorUserId="3266" LastEditDate="2018-01-23T12:16:43.110" LastActivityDate="2018-01-23T12:16:43.110" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19042" PostTypeId="2" ParentId="19037" CreationDate="2018-01-23T11:45:13.207" Score="5" Body="&lt;p&gt;That wall has not been designed to be load bearing by a structural engineer. As you say, it only has a single header (top beam) rather than a double, and the verticals aren't cut accurately enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What worries me... is that you say this was an attic conversion. Attic conversions are sometimes done by people who don't understand what they're doing and don't follow structural regulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as long as you're happy that the conversion was done properly, you're completely safe to take out that stud wall.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;to my untrained eye, it seems like a lot of roof (80&quot; from nearest edge to back wall) to support itself.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Roof tiles sit on purlins (small horizontal timbers) that sit on rafters (the big diagonal members that basically define the shape of the roof). In a normal roof you have a truss, which supports the rafters. However, it is completely possible to design rafters such that there is just a triangle (left inclined rafter, right inclined rafter, and a ceiling/floor joist joining them). It's just that you'll need bigger rafters, and better joints, then if you have a truss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why I worry about attic conversions: if someone just cuts out the inner truss without redesigning the rafters, you can get understrength roofs. And a non-load bearing wall might be mitigating that, until it's removed.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-01-23T11:45:13.207" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19043" PostTypeId="2" ParentId="19035" CreationDate="2018-01-23T12:45:28.103" Score="4" Body="&lt;p&gt;Higher temperature definitely increases evaporation rate. Keeping it away from body heat and not letting it heat up a lot from flame will keep the fuel from evaporating too fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than that, concentration gradient - a seal will reduce the evaporation around the container; won't stop evaporation through the wick. Also, it may make opening it for refilling more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Topping it off will increase the concentration (and its gradient) and so the initial evaporation will be faster - but the total time until &quot;dried up&quot; will be longer. (meaning, if you fill it with 2ml of fuel at once, it will last longer than if you filled it with 1ml, but shorter if you filled it with 1ml &lt;em&gt;twice&lt;/em&gt;, the other refill after the first one evaporates). Considering minuscule amounts and low price of fuel there's really no reason to skimp on it (other than risk of spills if you overfill).&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-01-23T12:45:28.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19044" PostTypeId="2" ParentId="19032" CreationDate="2018-01-23T14:10:18.620" Score="0" Body="&lt;p&gt;They are missing parentheses, as noted by agentp, but it should be said that &lt;code&gt;W/(cm^2*C)/cm&lt;/code&gt; is actually a relatively useful way of describing Thermal conductivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's re-format that as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Conductivity =\frac{Power\ Transfer\ Rate}{ \left(\frac{Surface\ Area\ *\ Temperature\ Difference}{Thickness} \right)}=\frac{W}{ \left(\frac{{cm}^2 *\Delta °C}{cm} \right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this is useful, is that it incorporates the values that you may already know about your system, in the units that you are likely to use them. Namely,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rate of Power Transfer through the material - ($W$)&lt;/li&gt;&#xA;&lt;li&gt;Surface area over which Power Transfer is occuring - ($cm^2$)&lt;/li&gt;&#xA;&lt;li&gt;Thickness through which Power Transfer is passing - ($cm$)&lt;/li&gt;&#xA;&lt;li&gt;Temperature difference between hot and cold sides of material - ($°C$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Setting the units out as they have done means that you can intuitively input the values you have, combined with the published conductivity values, to find the Power Transfer Rate that you are most likely looking for&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Power\ Transfer\ Rate = Conductivity*\left(\frac{Surface\ Area\ *\ Temperature\ Difference}{Thickness} \right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that, for a fixed conductivity, doubling the surface area (dividing by a larger number) will mean that you will experience double the power transfer.&#xA;Likewise, doubling the thickness will halve the power transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is slightly more difficult to see when using the SI units &lt;code&gt;W/(m.K)&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-01-23T21:45:32.867" LastActivityDate="2018-01-23T21:45:32.867" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19045" PostTypeId="1" CreationDate="2018-01-23T15:45:43.187" Score="2" ViewCount="374" Body="&lt;p&gt;Let’s say I have a circular rod that has an outside lip with a larger diameter and certain height. The rod moves axially, and the intent of the lip is to mechanically stop the rod at a maximum axial displacement. How would I calculate the maximum static axial force I could apply to the rod before the lip shears off of the rod? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0lfy6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0lfy6.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a pretty good understanding of how to apply this problem in a simple cantilevered beam with a load orthogonal to the beam surface that provides a shear stress, however I can’t seem to wrap my head around this kind of problem, especially regarding second moment of area with a cross section that is radial instead of on a constant plane. As far as I can tell, the process is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Draw a FBD to find resultant forces on the shaft.&lt;/li&gt;&#xA;&lt;li&gt;Determine the resultant normal and shear stresses due to axial, shear, torsion, and bending forces.&lt;/li&gt;&#xA;&lt;li&gt;Use Mohr’s circle to determine sigma max, sigma min, and tau max.&lt;/li&gt;&#xA;&lt;li&gt;Apply one of the failure criteria for brittle/ductile materials.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can anyone provide some guidance or direction?&lt;/p&gt;&#xA;" OwnerUserId="14658" LastActivityDate="2018-01-24T00:43:30.020" Title="Failure due to shear stress on rod lip?" Tags="|statics|stresses|shear|failure-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19046" PostTypeId="2" ParentId="19035" CreationDate="2018-01-23T18:22:51.473" Score="4" Body="&lt;p&gt;The big issue is that Zippos aren't that well sealed, especially the lid which allows fuels to evaporate through the wick and it's this wick to air interface which is the most critical thing as the whole function of the wick is to provide a big surface area for fuel to evaporate and thus burn.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the fill level probably doesn't make a big difference as the wick will suck up fuel by capillary action as long as there is fuel to suck. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature will make a difference as it will have a direct effect on the vapour pressure of the fuel humidity not so much as the air won't get saturated with fuel vapour in the same way as with moisture so they aren't competing as such. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;also it is not easy to improve the sealing on a Zippo, at least not around the lid where is really matters as the interface between the lid and the body is fairly thin and there is no easy way to keep the lid firmly in place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare to something like &lt;a href=&quot;https://www.heinnie.com/maratac-peanut-lighter&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; which has a thread and O-ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From and engineering perspective the classic Zippo design is simple and convenient but the price you pay is wasting a  bit of fuel through evaporation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the practical  best solution is to carry a small screw-top container with extra fuel. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-01-23T18:22:51.473" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19047" PostTypeId="2" ParentId="19039" CreationDate="2018-01-23T19:14:59.773" Score="2" Body="&lt;p&gt;I'm going to suggest that use of a siphon-based system is just plain unworkable here.  What you need is a pipe permanently installed in or below the lake bottom, with a good filter at the inlet end, and an electric pump.  Personally I'd go with a submersible pump at the inlet end (and thus a proper power cable run alongside the pipe) so that priming the system is never required.  Just imagine trying to refill your 2500-foot input pipe if your shore-based pump loses its seal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you haven't made it clear what your pressure requirements are, and you also haven't mentioned your latitude, I'm leaving questions of the pump's pressure head and the system's survivability in winter for later. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-01-23T19:14:59.773" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19048" PostTypeId="2" ParentId="19045" CreationDate="2018-01-23T20:02:37.883" Score="1" Body="&lt;p&gt;This is very similar to the punching shear in column caps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would calculate the shear at a distance d/2 away from the rod, with d being the thickness of flange. Hwever if your guid rail or slidin track has prticular constrains you need to consider the impact of that too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would apply ultimate stress method and just divid the factored load by the surface of the failure cylinder which is the diameter of the rod plus d/2 of flange.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-01-24T00:43:30.020" LastActivityDate="2018-01-24T00:43:30.020" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19049" PostTypeId="1" AcceptedAnswerId="19086" CreationDate="2018-01-23T20:13:07.880" Score="5" ViewCount="304" Body="&lt;p&gt;I would like to independently check a vendor's silo design for proper explosion venting. Worst case, it is essentially cylinder filled with a stoichiometric mixture atomized sugar dust and atmospheric oxygen initially at atmospheric pressure. The roof of the cylinder is designed to detach from the structure with, lets assume, zero force. The roof also has some amount of mass per square inch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lots of different ways to go about it, but initially to skip transient calcs I was thinking I could calculate the worst case by assuming 100% of oxygen reacts instantly. Put that energy into the ideal gas law and calculate the final pressure if unvented. Not sure if this will truly be the max pressure or if the transient shock wave pressure would exceed this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum pressure would be the psi required to lift and accelerate the roof up. The vented roof pressure loading should be pretty easy, again not considering transients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on how to better roughly calculate wall pressures without cfd? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-01-25T08:00:30.607" Title="How to calculate the pressure loading of a vented silo explosion?" Tags="|structural-engineering|fluid-mechanics|combustion|pressure-vessel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19050" PostTypeId="2" ParentId="19039" CreationDate="2018-01-23T20:21:33.787" Score="3" Body="&lt;p&gt;As mensioned above it works. You would need to make sure the level of discharge at your end is lower than the lowest seasonal level of the lake and also make sure it never gets submrged at your end to interrupt the flow. Both ends of the siphon need to be terminating at vertical drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would use shut off valves at both ends for temporary priming and maintenace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend the inlet to source lake to terminate into a 90 degree elbo vertically submerged under the lowest seasonal level of the lake witha one way chack valve for stoping loss of prime in case the lake water recedes more.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-23T20:21:33.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19051" PostTypeId="2" ParentId="17580" CreationDate="2018-01-23T20:37:46.007" Score="1" Body="&lt;p&gt;Lower pressure heat exchanger is going to be easier to build, but will also need a larger surface area because the heat transfer coefficient is higher for compressed fluids. I suggest trying to ballpark your heat transfer coefficient for both pressures (and velocities) and seeing how that affects your other parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5302" LastActivityDate="2018-01-23T20:37:46.007" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19052" PostTypeId="1" AcceptedAnswerId="19098" CreationDate="2018-01-23T21:29:25.167" Score="3" ViewCount="137" Body="&lt;p&gt;I have read about the application of sewage sludge incinerators&lt;sup&gt;1&lt;/sup&gt; but they do not have a net-positive power-generation.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/jjGGe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jjGGe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Is there a way to generate net-power from sewage sludge?&#xA;What are the processes to do so?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;1: German: &lt;a href=&quot;http://www.berlin-klimaschutz.de/de/projekte/energieerzeugung-durch-klaerschlammverbrennung&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.berlin-klimaschutz.de/...&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2018-01-25T20:04:27.350" Title="Is it Possible to Generate Power From Sewage Sludge?" Tags="|power|waste-water-treatment|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="19053" PostTypeId="2" ParentId="18992" CreationDate="2018-01-23T21:36:08.290" Score="5" Body="&lt;p&gt;I think the other answer is not complete.  Model predictive control (or 'receding-horizon control' is a technique in which a predictive system model is used to evaluate a sequence of future control inputs; of all such control input sequences, an optimization algorithm chooses the best one.  Typically, the &lt;em&gt;first&lt;/em&gt; input of the sequence is implemented.  After a certain amount of time, the process is repeated to find a new control input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the predictive model is deterministic, this is straightforward.  Now, when the predictive model is &lt;em&gt;uncertain&lt;/em&gt; (e.g., stochastic or adversarial), one often differentiates between &lt;em&gt;open-loop MPC&lt;/em&gt; and &lt;em&gt;closed-loop MPC&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Open-loop MPC&lt;/strong&gt; considers only a single, fixed sequence of future control inputs; this sequence must yield good performance under all possible realizations of the uncertainty.  In other words, when constructing our initial plan, we ignore the fact that we're going to re-plan (with more information) soon.  We plan as though we were committed to our input sequence (even though we aren't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Closed-loop MPC&lt;/strong&gt; considers the &lt;em&gt;effect of feedback&lt;/em&gt; (or &lt;em&gt;recourse&lt;/em&gt;). This is the fact that, before the controller makes future decisions, it will have more information available than it does now.  For this reason, the controller must optimizes over control &lt;em&gt;policies&lt;/em&gt;, not just inputs.  These problems are quite often intractable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll add one more:  &lt;strong&gt;Certainty-equivalent MPC&lt;/strong&gt; replaces the uncertain predictive model with a deterministic one (usually by replacing all random quantities with an average or nominal value).  At this point, there's no uncertainty, and we simply precede by optimizing over input sequences, as in open-loop MPC.  This is often simply referred to as &lt;em&gt;MPC&lt;/em&gt; (without a qualifier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out &lt;a href=&quot;https://web.stanford.edu/class/ee364b/lectures/stoch_mpc_slides.pdf&quot; rel=&quot;noreferrer&quot;&gt;Boyd's MPC description here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14664" LastActivityDate="2018-01-23T21:36:08.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19056" PostTypeId="2" ParentId="19037" CreationDate="2018-01-24T01:51:18.500" Score="2" Body="&lt;p&gt;I have seen many of these DIY jobs by home owners that has been defying any sense of logic. Not only this could be a part of support of the roof by cutting the diagonal 2x4s that supported the rafters. It may not be even sitting on a solid member of framing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would try to get a read of the framing both of roof and floor of attic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An experienced Carpenter could help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a small probe that can send it's camera through a 1/4 inch hole and has a few LED lights to light up dark places. This probe connects to your droid phone and you can record what you see. I bought it for $28 from Amazon, it is really a handy tool.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-24T01:51:18.500" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19057" PostTypeId="2" ParentId="19024" CreationDate="2018-01-24T08:49:45.013" Score="1" Body="&lt;p&gt;I don't use SPH for fluid dynamics personally, but I've seen some presentations where it has been used for high strain rate material behavior which shares some effects (and equations) with standard fluid dynamics. SPH is good for large deformations and material separation processes. The &lt;a href=&quot;https://de.wikipedia.org/wiki/Smoothed_Particle_Hydrodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;German wikipedia article&lt;/a&gt; has a list of advantages and disadvantages of SPH which seem quite reasonable to me. I've tried to translate them to English; please edit this answer if the translation is not good enough. Note that I cannot really comment on the individual items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Advantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;SPH is a Lagrangian method with mass always conserved&lt;/li&gt;&#xA;&lt;li&gt;very robust code, i.e. finds results under most circumstances&lt;/li&gt;&#xA;&lt;li&gt;comparatively simple implementation, including testing of kernels&lt;/li&gt;&#xA;&lt;li&gt;when using Gaussian normal distribution as kernel, allows interpretation of theoretical results quite easily&lt;/li&gt;&#xA;&lt;li&gt;in modern codes, the calculation effort grows with $N \, \log N$, with $N$ being the number of particles&lt;/li&gt;&#xA;&lt;li&gt;SPH shows good global results for a small number of particles&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Disadvantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;often, the code is too robust; i.e. using incorrect models, physically meaningless results can be calculated&lt;/li&gt;&#xA;&lt;li&gt;error estimation is often problematic and can only be performed in comparison with other methods&lt;/li&gt;&#xA;&lt;li&gt;SPH is highly dispersive&lt;/li&gt;&#xA;&lt;li&gt;For good results, large particle numbers are required. Therefore it is usually not less computationally expensive than other methods&lt;/li&gt;&#xA;&lt;li&gt;dealing with discontinuities is not easy, since structures on scales smaller than the smoothing length will be smoothed&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4122" LastEditorUserId="14498" LastEditDate="2018-01-24T08:58:29.593" LastActivityDate="2018-01-24T08:58:29.593" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19058" PostTypeId="2" ParentId="18165" CreationDate="2018-01-24T09:22:23.457" Score="1" Body="&lt;p&gt;You can eliminate Q and E&lt;sub&gt;k&lt;/sub&gt; as well. This is because the heat transfer between the water and ambient is negligible in hydro-electricity(for steam power plants Q might be significant) . Also, at the the reservoir/dam where water is stored, the water usually has negligent speed, thus low value of K.E. The potential energy of water is the significant term, with it being equal to mgh or Qρgh in the rate form; where Q is discharge or volume flow rate of water. This potential energy(p.e) is responsible for the work done W by the turbine runner. The electrical work generated would be some factor times the p.e, because of losses such as penstock losses, losses incurred in shaft, runner etc.&lt;/p&gt;&#xA;" OwnerUserId="9537" LastActivityDate="2018-01-24T09:22:23.457" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19059" PostTypeId="1" CreationDate="2018-01-24T10:18:17.317" Score="1" ViewCount="29" Body="&lt;p&gt;I need a good estimate of the air demand for a system to fluidizde a powder. We a have silo with a 60° cone and a rotary cell feeder attached to the opening in the cone. The powder (fine powdered activated coal) shall be fluidized by blowing some compresses air into it. Perrys chemical engineering handbook gives some tables and nomogram for full fluzidization (air velocity of 0.15 - 6 m/s over free area of silo). However, I really only need fluidization near the walls of the cone, so I should be able to get away with less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silo diameter is 3-4m, mass flow rate up to 60kg/h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I estimate the air need for fluidization?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-01-24T10:18:17.317" Title="Air demand for powder fluidization system" Tags="|pneumatic|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19060" PostTypeId="1" CreationDate="2018-01-24T11:59:42.177" Score="3" ViewCount="1988" Body="&lt;p&gt;I have a warehouse ( 36m x 14.4m x 10m ) that I want to keep it under 30 degrees Celsius but everyday the temperature will goes above that. It has two turbine ventilator that I don't know the specification and for the ventilation it only relies to two rolling doors ( 3x2m ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a thought that it is caused by bad airflow in the warehouse so&#xA;I have an idea to add 3pcs of 8ft HVLS Fan and adding 5 more turbine ventilators. But will it works because I will have to do a big investments to that fans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in Jakarta, Indonesia. And the ambient temperature ranges from 28 - 35 degree C in the afternoon.&lt;/p&gt;&#xA;" OwnerUserId="14671" LastEditorUserId="14671" LastEditDate="2018-01-25T05:11:43.090" LastActivityDate="2018-01-25T05:11:43.090" Title="How to cool down a warehouse?" Tags="|airflow|hvac|cooling|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19061" PostTypeId="2" ParentId="19060" CreationDate="2018-01-24T12:26:16.420" Score="0" Body="&lt;p&gt;One possibility, if you only have the fans and no hvac is to consider a « puits canadienne ».&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is a pipe or pipes in parallel buried underground about 2m or 6 feet down, where the ground temperature is lower than air ambient - depends on the season of course and country... (in Switzerland the air ambient can be -10 deg C and underground +5 deg C , used this for the air ventilation on the house).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are sites which will help with sizing the pipes and issues of moisture getting trapped in the pipes must be addressed : legionnaires disease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I find a site I will add it ...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-01-24T12:26:16.420" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19062" PostTypeId="2" ParentId="19060" CreationDate="2018-01-24T12:34:13.187" Score="3" Body="&lt;blockquote&gt;keep it under 30 degrees Celcius but everyday the temperature will goes above that&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No amount of ventilation when it's hot out is going to fix that.  In fact, more ventilation will make that worse.  Ventilation will only cool the warehouse when the outside temperature is less than the inside temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way ventilation can help, if it's managed properly, is to ventilate aggressively when it is cooler out than in, then NOT ventilate when the outside temperature goes above the inside temperature.  Whether that is feasible depends on a number of factors you haven't told us.  It may not be acceptable to decrease the daytime ventilation sufficiently due to fresh air requirements, doors must be open, etc.  Also, with little insulation, the warehouse will heat up when its hot outside anyway.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-01-24T12:34:13.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19064" PostTypeId="2" ParentId="19027" CreationDate="2018-01-24T15:50:06.563" Score="1" Body="&lt;p&gt;I'm unsure about the magnitude of the change, but providing tension by preloading an object definitely alters the natural frequency. It is commonly adopted to avoid resonance from external disruptions. I have seen the idea adopted with shafts, but in a single lap joint like your example I can't imagine the affect is very big, as you observed. When undertaking modal analysis of a large assembly, such as an engine case, the tension of the bolts at the mounting locations can be an important consideration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of two beams bolted together, I have not seen examples of the affect of preload. But common sense suggests the impact would be much less significant than preloading an entire shaft, for example. Each beam in your model has it's own natural frequency, as does the bolt; the bolt's natural frequency will probably have increased non-negligibly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you preloaded a single beam (i.e. applying forces at each end, think guitar string) you should see a more pronounced effect. &lt;/p&gt;&#xA;" OwnerUserId="14498" LastEditorUserId="14498" LastEditDate="2018-02-06T11:36:20.697" LastActivityDate="2018-02-06T11:36:20.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19065" PostTypeId="1" CreationDate="2018-01-24T16:21:05.357" Score="1" ViewCount="139" Body="&lt;p&gt;I would like to build a heavy duty cargo bike. Heavy = approx 200Kg of load on the front rack. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want 2 front wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, there are 2 options that I know of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;the front wheels are fixed to the front rack/ load and the whole part is moving when I want to turn. This uses a single pivot. This seems appropriate, but I'm worried about the single pivot and the tube size (regarding the tube between the bottom bracket and the pivot)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The rack is fixed with the chassis, everything including the wheels can move without moving the rack/load. This includes the whole load is not always in the same plan, and it worries me because i'm not sure it's a good idea to move 200kg, it is a lot of inertia&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What's your opinion?&lt;/p&gt;&#xA;" OwnerUserId="14673" LastEditorUserId="14498" LastEditDate="2018-01-26T16:10:59.523" LastActivityDate="2018-01-26T16:10:59.523" Title="heavy duty cargo bike - which direction to choose" Tags="|bicycles|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19066" PostTypeId="1" CreationDate="2018-01-24T16:33:58.157" Score="1" ViewCount="1956" Body="&lt;p&gt;basically I am trying to build an oven that we are going to use to cure prepreg carbon fiber. I found a company that can make it but they are requesting the termal load in order to continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the information I already have:&#xA;Dimensions of the oven: 23.7in x 59in x 59in (height, width, depth)&#xA;Desired temperatures: max 284F, min 158F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found the formula to calculate this, but honestly, I don't know how to begin:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;q = -k A dt/dx&#xA;q = heat transfer rate, Btu/hr&#xA;k = thermal conductivity, Btu/(hr-ft-F)&#xA;A = area normal to heat flow, ft^2&#xA;dt/dx = temperature gradient, F/ft&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="14674" LastActivityDate="2018-01-24T20:19:29.847" Title="How to calculate thermal load?" Tags="|heat-transfer|heating-systems|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19067" PostTypeId="2" ParentId="19066" CreationDate="2018-01-24T16:54:55.723" Score="1" Body="&lt;p&gt;I'd tell them what you're aiming to do e.g. the required temperature range, heat up time, is it open or closed, is there material moving through it continually or is it a batch process, what is the ambient temperature, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They should do the thermal load calculations and guarantee the performance. If they can't, you should probably look for another supplier.  Otherwise, if it doesn't work they're going to tell you that you gave them the wrong thermal load.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2018-01-24T16:54:55.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19068" PostTypeId="1" CreationDate="2018-01-24T17:16:22.043" Score="2" ViewCount="185" Body="&lt;p&gt;I am trying to calculate prior to testing the expected back pressure resulting from injecting gaseous oxygen through an injector orifice (choked) into a rocket combustion chamber before ignition. I have tried using the Borda-Carnot equation with the assumption that the chamber mean flow is sufficiently low and that the pressure differential between the chamber and atmosphere is sufficient to drive the same mass flow through the nozzle that was injected into the chamber. Calculating the delta P under these assumptions yields an unreasonably small delta P across the nozzle. Is there a hand-calc method available that would provide some insight to this? I know that there are highly localized dynamics in the gas flow that are likely going to cause significant deviation from the analytical values, but I am mostly looking for a first-order estimate.&lt;/p&gt;&#xA;" OwnerUserId="14676" LastActivityDate="2020-03-06T09:53:42.143" Title="Combustion Chamber Back Pressure due to gas injection" Tags="|fluid-mechanics|compressible-flow|rocketry|injection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19069" PostTypeId="2" ParentId="19065" CreationDate="2018-01-24T17:17:37.757" Score="0" Body="&lt;p&gt;You will probably be much better off not having the load directly attached to the steering as this will put a lot of load on the steering linkage and probably make the steering very heavy and it will be heavy enough as it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to consider steering geometry. As soon as you have a pair or steered wheels you need to look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Ackermann_steering_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ackerman steering&lt;/a&gt; which steers the inner wheel more than the outer one. Without this you are forcing one or both of the steered tyres to slip which is wasting a lot of energy and will make the steering heavier still. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a single pivot you are essentially attempting to drag the wheels across the direction of travel without any mechanical advantage. Indeed what is more likely to happen in practice is that you end up with rear wheel steering &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-01-24T17:17:37.757" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19070" PostTypeId="2" ParentId="12340" CreationDate="2018-01-24T17:31:50.520" Score="1" Body="&lt;p&gt;One of the assumptions necessary for your pipe-loss formula you used is that the flow is incompressible (M&amp;lt;0.3). You could conceivably break up the problem into smaller segments where the compressible fluid density doesn't vary significantly and approximate those as incompressible individually. The fluid velocity will be directly related to you gasses density and will necessarily be limited to M&amp;lt;1 within the capillary. You'll want to start with the density of oxygen at p1 and calculate the loss going from p1 to a point sufficiently close to p1 where the density does not change significantly, and then recalculate the new pressure, density. You can continue to do this with iterated mass flow until you get convergence with the total delta P.&lt;/p&gt;&#xA;" OwnerUserId="14676" LastActivityDate="2018-01-24T17:31:50.520" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19071" PostTypeId="2" ParentId="19066" CreationDate="2018-01-24T20:19:29.847" Score="0" Body="&lt;p&gt;The equation for heat transfer is not of much value here.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You literally need the heat to cure the prepreg carbon fiber.  At 284F how much heat is the absorbed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need an engine to go 60 mph.  A motorcycle would need a different hp than a truck.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-01-24T20:19:29.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19072" PostTypeId="1" CreationDate="2018-01-24T20:45:50.690" Score="3" ViewCount="58" Body="&lt;p&gt;Let’s say I have an irregularly shaped bearing with a cross section that looks like the angled surface of a conical frustum, shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oy1Q0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oy1Q0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a bending moment is applied which causes the inside diameter to ‘cave outwards’, how would I perform a static loading failure analysis to see at what bending moment the shape would cave outwards? How would I determine the second moment of area and distance from centroid values required to apply the bending stress formula sigma = My/I? Or is there a manufacturing method in the form of some sheet metal bending formula that would be more helpful in looking at failure mode in a cylindrical cross section instead of a planar one.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The end goal is to have the curved surface completely vertical, as if it was the surface area of a rod. If there is an easier method of calculating the forces required, let me know!&lt;/p&gt;&#xA;" OwnerUserId="14658" LastEditorUserId="14658" LastEditDate="2018-01-24T21:11:34.313" LastActivityDate="2018-01-24T22:12:17.483" Title="How to apply static failure criteria for bending stress in an angled cylindrical surface?" Tags="|statics|stresses|mathematics|mechanical-failure|failure-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19073" PostTypeId="2" ParentId="19072" CreationDate="2018-01-24T22:12:17.483" Score="1" Body="&lt;p&gt;This is a typical case of stresses on shells and plates and obviously lends itself to analyzing using polar coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your element is a thin shell and is loaded in a way that only membrane stresses are involved there are some formulas in references such as Roark's formulas for stress and strain. Otherwise you may need to use FEM and define your loading.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-24T22:12:17.483" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19074" PostTypeId="2" ParentId="18731" CreationDate="2018-01-24T22:34:42.913" Score="0" Body="&lt;p&gt;A normal spring will not work, because as you move it, the force changes.  Beyond a certain point, the force will get so high that your positioning stage has to fight the spring, i.e. the spring will be making the load on the stage higher, not lower.  What you want instead is called a &quot;constant force spring&quot;. As the name says, it will exert a constant force over a fairly wide range of motion.  For example&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.leespring.com/learn-about-constant-force-springs&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.leespring.com/learn-about-constant-force-springs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.vulcanspring.com/conforce-constant-force-springs&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.vulcanspring.com/conforce-constant-force-springs&lt;/a&gt;  &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.centuryspring.com/constant-force&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.centuryspring.com/constant-force&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and many other manufacturers as well.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="9262" LastEditDate="2019-12-17T11:37:38.543" LastActivityDate="2019-12-17T11:37:38.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19075" PostTypeId="2" ParentId="19065" CreationDate="2018-01-24T23:04:42.670" Score="1" Body="&lt;p&gt;You shouldn't put the load on the front, &lt;strong&gt;above&lt;/strong&gt; the front wheel(s). That's the worst possible location for multiple reasons.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It'll get you smacked on the road:&lt;br&gt;&#xA;200kg located above the front wheel sure will be heavy enough to pull you over for a nice face-plant on the road upon braking. The inertia will 'pull you over' when you brake. The higher you place it, the worse it gets. You get a bonus at that when leaning in on a corner, worsening it even further.&lt;/li&gt;&#xA;&lt;li&gt;Steering will be heavy:&lt;br&gt;&#xA;The rotational inertia means a large mass can be hard to rotate. In other words. It'll be much heavier to turn your steerer with a heavy load attached to it.&lt;/li&gt;&#xA;&lt;li&gt;The front tyre(s) already gets the most abuse upon breaking, putting a load on it worsens that.&lt;/li&gt;&#xA;&lt;li&gt;Steering behaviour such as turning-radius, or minimum speed required to make a turn may get worse.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'd make a three-wheeler with two wheels at the rear, and a load space between them, just in front of the rear axle, close to the ground. This will give you the a good result because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Your steering can remain as simple and lightweight as that of a normal bike. You won't have to take into account Ackermann. The steering like on old horse carriages would also eliminate the need for that, but it would create an awful ride experience.&lt;/li&gt;&#xA;&lt;li&gt;Mass is located close to ground, keeping its negative effect upon steering and 'leaning in' to a minimum.&lt;/li&gt;&#xA;&lt;li&gt;Mass is located at the rear, preventing it from causing unexpected effects upon braking and steering, which are hard to counter yourself.&lt;/li&gt;&#xA;&lt;li&gt;You can easily create more space there for cargo than on a rack on your steerer.&lt;/li&gt;&#xA;&lt;li&gt;You can access the cargo easily with eg. a handtruck. You'll save your back/spine and energy.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In short, I recommend to design it like this, preferable i'd be locating the mass closer to the ground though: &lt;a href=&quot;https://i.stack.imgur.com/Rk4Mx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rk4Mx.jpg&quot; alt=&quot;bikeGood&quot;&gt;&lt;/a&gt;&#xA;Rather than this: &lt;a href=&quot;https://i.stack.imgur.com/hMG6v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hMG6v.jpg&quot; alt=&quot;bikeBad&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="13311" LastEditDate="2018-01-26T00:29:59.387" LastActivityDate="2018-01-26T00:29:59.387" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19076" PostTypeId="2" ParentId="19052" CreationDate="2018-01-24T23:40:15.770" Score="1" Body="&lt;p&gt;In order to make use of it as a fuel, you could consider these options:&lt;br&gt;&#xA;- Processing the stuff with micro-organisms like bacteria that can create methane from it, or other usable fuels.&lt;br&gt;&#xA;- Drying it and using it directly as a combustible fuel. The dryer you can get it, the more efficient your combustion will be.(less energy will be wasted to vapourising the water it's containing)&lt;br&gt;&#xA;- You can gassify it after it's been dryed, like they do with wood. This gets you a more versatile/dense/clean end product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all this will waste a lot of other usable components within the sludge.&#xA;Granted that the sludge isn't contaminated, I suppose it would be best to use it as fertiliser. Almost all of the components will be put to use then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also extract usable compounds that are present abundantly enough, like ammonia. You can sell that to fuel manufacturers to make ad blue from. That way, our pee/poo will contribute to a cleaner environment.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-24T23:40:15.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19077" PostTypeId="2" ParentId="1910" CreationDate="2018-01-25T00:36:41.937" Score="2" Body="&lt;p&gt;Turbo compounding is used currently in truck engines by Detroit diesel and Volvo, but they are suited to long haul applications where the engine is at a constant revs/load, the power turbine it is said extracts up to 50 horsepower to put back into the crankshaft but the trade off as explained earlier is increased exhaust backpressure meaning the pistons have to push harder to get the exhaust gas out, so in transient (engine up and down the rev range) operation the power turbine costs energy, in a steady state it can reduce fuel consumption 2-5% &lt;/p&gt;&#xA;" OwnerUserId="14681" LastActivityDate="2018-01-25T00:36:41.937" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19078" PostTypeId="2" ParentId="19049" CreationDate="2018-01-25T00:43:33.373" Score="2" Body="&lt;p&gt;The pressure in the silo is determined by the temperature rise, caused by the combustion. The end temperature after combustion can be calculated like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T_{end}= \dfrac{E}{(m_{air}+m_{fuel})C_V}+T_{begin}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $E$ is the added energy added per space(eg. m&lt;sup&gt;3&lt;/sup&gt;), $m$ is the mass of the fuel and air in kg present in that space, and $C_V$ is the specific heat of the mixture at constant volume (take 724&amp;nbsp;J/kgK for air), and $T_{begin}$ is temperature of the mixture before igniting (room temperature, 293&amp;nbsp;K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume we have a temperature rise of 980 degrees, which makes the end temperature 1000°C or 1273&amp;nbsp;K. Then when begin pressure is 1&amp;nbsp;atmosphere, the end pressure will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$p_{end}=p_{begin}\cdot\dfrac{T_{end}}{T_{begin}}=1013.25\cdot\dfrac{1273}{293} =4402\text{ hPa/mbar}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's 4.34 times as high; 4.5&amp;nbsp;bar or about 64&amp;nbsp;psi. Obviously there's a lot of assumptions and guesses made here, and it won't fully combust instantaneously,  but I would be surprised if the figures differ a lot. I'd bet it is indeed about 4 to 5 bar. It's obviously enough to blow a roof away.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="1832" LastEditDate="2018-01-25T01:03:00.557" LastActivityDate="2018-01-25T01:03:00.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19081" PostTypeId="1" CreationDate="2018-01-25T03:11:50.380" Score="2" ViewCount="247" Body="&lt;p&gt;Here's what I'm proposing - a small turbine directly coupled to a permanent magnet that's spinning at, say, 150k RPM. The magnet is enclosed one or several wire loops parallel to the axis of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pquGa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pquGa.png&quot; alt=&quot;illustration&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generated EMF is proportional to the magnetic flux and its rate of change. Without an iron core, the flux would be a lot smaller, but it would be in part compensated by the much faster rate of change. There would also be no core losses as well as no eddy current losses. Is something like this practical? Why is it said that driving generators from turbines without any gear reduction is inefficient?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are iron cores added to generators to increase their specific power or their efficiency? Is it both?&lt;/p&gt;&#xA;" OwnerUserId="5302" LastEditorUserId="5302" LastEditDate="2018-01-25T05:41:53.757" LastActivityDate="2018-01-25T05:41:53.757" Title="Generating practical power without an iron core, but with high RPM" Tags="|power-electronics|turbines|magnets|electromagnetism|generator|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19082" PostTypeId="2" ParentId="19081" CreationDate="2018-01-25T03:51:06.947" Score="3" Body="&lt;p&gt;the iron structure(s) inside a generator serve the purpose of minimizing flux leakage, which maximizes the utilization of the magnetic field in generating electricity. One of the ways in which they accomplish this is by furnishing what is called a return path for the lines of force, so as to &quot;close&quot; the magnetic circuit inside the device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For things rotating at 150,000 RPM, mechanical stresses due to rotation become the design limiter which means that if the armature had a core made of, say, laminations of silicon steel, it would fly apart and perform significant damage on its surroundings in the process if rotated at that speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know why it is asserted that direct drive is inefficient, except in the sense that the generator mass (due to including an iron core in the armature) limits the maximum speed of the machine to speeds much lower than those at which a turbine is most efficient.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-25T03:51:06.947" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19083" PostTypeId="2" ParentId="18567" CreationDate="2018-01-25T03:53:15.983" Score="0" Body="&lt;p&gt;electric clutches, as used for example to engage and disengage air conditioner compressors in cars, are abundantly available in a variety of sizes and power ratings. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-25T03:53:15.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19085" PostTypeId="2" ParentId="2303" CreationDate="2018-01-25T05:18:49.990" Score="1" Body="&lt;p&gt;Well, the concept of g force measurement goes by measuring the net acceleration considering all effects. That is assume an object in free fall it will have a 0g acceleration, meaning Force= M(g-a), where a is pseudo acceleration=g. Hence F = M*A = M(g-g), A=The acceleration in g. Therefore, A=0g, which is freefall acceleration.&lt;/p&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.m.wikipedia.org/wiki/G-force&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Gravitation acting alone does not produce a g-force, even though g-forces are expressed in multiples of the acceleration of a standard gravity. Thus, the standard gravitational acceleration at the Earth's surface produces g-force only indirectly, as a result of resistance to it by mechanical forces. These mechanical forces actually produce the g-force acceleration on a mass. For example, the 1 g force on an object sitting on the Earth's surface is caused by mechanical force exerted in the upward direction by the ground, keeping the object from going into free-fall. The upward contact-force from the ground ensures that an object at rest on the Earth's surface is accelerating relative to the free-fall condition. (Free fall is the path that the object would follow when falling freely toward the Earth's center). Stress inside the object is ensured from the fact that the ground contact forces are transmitted only from the point of contact with the ground.&lt;/p&gt;&#xA;&lt;p&gt;Objects allowed to free-fall in an inertial trajectory under the influence of gravitation only, feel no g-force acceleration, a condition known as zero-g (which means zero g-force). This is demonstrated by the &amp;quot;zero-g&amp;quot; conditions inside a freely falling elevator falling toward the Earth's center (in vacuum), or (to good approximation) conditions inside a spacecraft in Earth orbit. These are examples of coordinate acceleration (a change in velocity) without a sensation of weight. The experience of no g-force (zero-g), however it is produced, is synonymous with weightlessness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14685" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-01-25T11:20:59.497" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19086" PostTypeId="2" ParentId="19049" CreationDate="2018-01-25T08:00:30.607" Score="1" Body="&lt;p&gt;You have two pressures which are important here. One pressure (the one you are considering and that @Bart has calculated) is caused by the chemical reaction, i.e. instead of sugar dust and air you have some new gases (carbon oxides and water and probably nitrogen oxides) at elevated temperature. This is called the constant volume equilibrium pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally there is a flame front travelling through your explosive mixture that causes a jump in pressure (and density and temperature). This is the transient explosion pressure, and this pressure is much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to estimate the explosion pressure, for a first order guess you can calculate the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chapman%E2%80%93Jouguet_condition&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chapman-Jouguet pressure (CJ pressure)&lt;/a&gt;. There is a &lt;a href=&quot;https://www.youtube.com/watch?v=2_9q0f6Lnvk&quot; rel=&quot;nofollow noreferrer&quot;&gt;good youtube video&lt;/a&gt; that explains how to do that. Note that this is a 55 minutes lecture video.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if you do not want to do the calculation by hand, you can use &lt;a href=&quot;https://www.grc.nasa.gov/WWW/CEAWeb/ceaguiDownload-win.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA's CEA software (Chemical Equilibrium with Applications)&lt;/a&gt;. This will probably take a similar amount of time to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, you just use a number from &lt;a href=&quot;http://www.dustexplosion.info/dust%20explosions%20-%20the%20basics.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; (8.5 bar) or &lt;a href=&quot;https://www.fauske.com/blog/bid/316347/kst-and-pmax-tests-for-combustible-dust-who-or-what-are-they&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; (8.5 bar) or &lt;a href=&quot;https://www.crystalsugar.com/media/277431/Powdered-Sugar.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; (9 bar) website, which all seem reasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that you will have higher pressures in edges and especially corners due to concentration of shock waves. For this to calculate you need special CFD software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, 8.5 bar are enough to blow away the roof.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2018-01-25T08:00:30.607" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19087" PostTypeId="2" ParentId="1910" CreationDate="2018-01-25T08:01:11.900" Score="2" Body="&lt;p&gt;I limit my comments to aircraft: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;turbocompounding, as used in the R3350, was a way of extracting the kinetic energy of the exhaust gases with a turbine and putting it back on the crankshaft. Exhaust gas backpressure buildup was not a design limiter for cruise at 25,000 feet. Today, the exhaust gas energy recovery turbine is used instead to drive the supercharger, which on the 3350 was driven off the crankshaft, making turbosupercharging less complex than turbocompounding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is moot for aircraft engines bigger than 600SHP or so, where turboprop engines have taken over. they are far more reliable in the 1500-2000 SHP range than the R3350 or the R4360 as well as having better power-to-weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, turbocompounding was a bridge technology sitting between piston engines and turboprop technology. Once engines like the Allison 501 came along, 18 and 28 cylinder piston engines, with or without turbocompounding, were finished.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-25T08:01:11.900" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19088" PostTypeId="1" AcceptedAnswerId="19095" CreationDate="2018-01-25T08:53:03.320" Score="0" ViewCount="448" Body="&lt;p&gt;Need help with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of a sea defense consists of a section of a concrete wall 4 meters high, and 6 meters wide.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3M6jM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3M6jM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Calculate the resultant thrust on the section when the sea reaches a height of 3 metres relative to the     base of the section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Calculate the overturning moment experienced by the section about point A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my answer. Have i done this correctly? I am not sure if i am on the right track with this or not. Especially the overturning moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Thrust&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Density of water = 1000kg/m^3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Width = 6m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Height of fluid = 3m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f = pg \frac{h} {2} A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ft = 1000 x 9.81 x 1.5 x (6x3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ft = 264,870 N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Overturning Moment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = Ft \frac{h} {3}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = 264.870 x \frac{3} {3} = 264.870 Nm$&lt;/p&gt;&#xA;" OwnerUserId="14101" LastEditorUserId="6790" LastEditDate="2020-05-04T11:28:55.493" LastActivityDate="2020-05-04T11:28:55.493" Title="Hydrostatic Pressure" Tags="|mechanical-engineering|hydrostatics|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19089" PostTypeId="1" AcceptedAnswerId="19094" CreationDate="2018-01-25T09:24:17.987" Score="0" ViewCount="600" Body="&lt;p&gt;I am currently working on abaqus modeling, I am trying to find a quick way to get the strain and stress values shown in the output file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do I need to put in the .inp file? Does anyone know how to resolve this?&lt;/p&gt;&#xA;" OwnerUserId="14641" LastEditorUserId="10902" LastEditDate="2018-01-25T10:01:26.690" LastActivityDate="2018-01-25T20:52:10.000" Title="How to obtain strain and stress values by editing abaqus input file?" Tags="|structural-engineering|finite-element-method|abaqus|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19092" PostTypeId="1" CreationDate="2018-01-25T13:56:37.917" Score="3" ViewCount="646" Body="&lt;p&gt;I want to convert the pressured air exiting a valve to heat. I know that a vortex tube can be used to do this, but I am wondering if I need to make modifications to the vortex tube to make it more efficient or if there is just another, more efficient, way to generate heat.&lt;/p&gt;&#xA;" OwnerUserId="14691" LastEditorUserId="14691" LastEditDate="2018-01-25T18:19:07.557" LastActivityDate="2019-04-18T12:34:34.190" Title="What is the best method to generate heat from a compressed air canister?" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|cfd|compressed-air|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19093" PostTypeId="2" ParentId="19065" CreationDate="2018-01-25T15:48:30.993" Score="0" Body="&lt;p&gt;2nd is a bad design in my opinion          &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The load is higher has it must be above the wheels (less stable)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Load is transferred via the head set&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When you turn the stance is not as wide so easy to tip&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I suggest you draw up the 2 designs to get more input.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-01-25T15:56:10.720" LastActivityDate="2018-01-25T15:56:10.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19094" PostTypeId="2" ParentId="19089" CreationDate="2018-01-25T16:29:07.030" Score="2" Body="&lt;p&gt;just guessing you might be seeking ascii human readable reporting, you can put in the inp file, (after your first &lt;code&gt;*STEP&lt;/code&gt; line.)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;   *EL PRINT,ELSET=setname&#xA;   E&#xA;   S&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;this will dump the data into your &quot;.dat&quot; file.  This is primarily useful if you want data for a small element set (like one particular element).  If you need the whole model data do yourself a favor and learn to directly read the output database (odb) file.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="1832" LastEditDate="2018-01-25T20:52:10.000" LastActivityDate="2018-01-25T20:52:10.000" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19095" PostTypeId="2" ParentId="19088" CreationDate="2018-01-25T17:59:03.360" Score="0" Body="&lt;p&gt;Usually with this kind of problems what I would do is to analyse what effect is caused by what source. In your specific problem, there is water under gravitational load. The gravitational field imposes a &lt;a href=&quot;https://en.wikipedia.org/wiki/Vertical_pressure_variation&quot; rel=&quot;nofollow noreferrer&quot;&gt;vertical pressure distribution&lt;/a&gt; inside the water. This pressure acts on the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the first answer I would give is the distribution of pressure $p$ as a function of the gravity of earth $g$, the density of water $\rho$ and the vertical coordinate $z$ (I would choose $z=0$ at sea level and positive values of $z$ going downward, but you can choose a different coordinate system). This gives you a function $p(z)$. This is not directly asked for, but is required to answer the two questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second step, since the &lt;a href=&quot;https://en.wikipedia.org/wiki/Influence_line#Distributed_loads&quot; rel=&quot;nofollow noreferrer&quot;&gt;load is distributed&lt;/a&gt;, I would integrate this distributed load (force per unit length) along the height:&#xA;$$&#xA;F=\int_0^h w \cdot p(z) \, \mathrm{d}z&#xA;$$&#xA;with width $w=6\,\mathrm{m}$ and height $h=3\,\mathrm{m}$. The distributed load is the product of pressure and wall width, because the pressure does not vary with width. If it did, I would have to integrate not only along the height axis $z$, but also along the width axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, to calculate the momentum, I would integrate the local &lt;a href=&quot;https://en.wikipedia.org/wiki/Torque&quot; rel=&quot;nofollow noreferrer&quot;&gt;torque&lt;/a&gt;, with is the product of the distributed load $w \cdot p(z)$ and the lever arm length $(h-z)$:&#xA;$$&#xA;M=\int_0^h w \cdot p(z) \cdot (h-z) \, \mathrm{d}z&#xA;\quad\mbox{.}&#xA;$$&#xA;Now you can calculate everything and check if my results are the same as yours. :-)&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2018-01-25T17:59:03.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19096" PostTypeId="1" AcceptedAnswerId="19227" CreationDate="2018-01-25T18:17:14.813" Score="0" ViewCount="26" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/20Gcp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/20Gcp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently making a corroded model in Abaqus, but not all locations are suffering from corrosion. As shown in the picture there are some parts (the blue parts) that do suffer from corrosion, so since the picture is only part of the overall bridge structure, I am wondering whether there exists many ways to make finer mesh size in corroded locations only without creating a partition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please comment if you know how to resolve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You&lt;/p&gt;&#xA;" OwnerUserId="14641" LastEditorUserId="14498" LastEditDate="2018-01-26T08:51:14.890" LastActivityDate="2018-02-02T05:52:45.663" Title="is there any other ways to refine mesh size in particular place without making a partition?" Tags="|corrosion|abaqus|meshing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19098" PostTypeId="2" ParentId="19052" CreationDate="2018-01-25T20:04:27.350" Score="0" Body="&lt;p&gt;In order to generate net-positive power from sewage-sludge the chemical energy contained in the the sludge needs to be big enough in order to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Dry the sludge to an acceptable level&lt;/li&gt;&#xA;&lt;li&gt;Overcome all process inefficiencies&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In a study from Dieter O. Reimann&lt;sup&gt;1&lt;/sup&gt; the necessary energy content to maintain incineration of different wast types was calculated.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0eP96.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0eP96.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Reimann concludes:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To achieve self-propagating incineration, a minimum calorific value in the sewage sludge app. 2,500 kJ/kg above the energy required for drying should therefore be aimed at.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An overview of the properties of sewage-sludge shows that even in dried form sewage-sludge is under the minimum value of 2,500 kJ/kg.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Gpsyl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gpsyl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A energy-content of 20MJ/kg for &lt;em&gt;dried&lt;/em&gt; sewage-sludge is reported by Gerhardt et al.&lt;sup&gt;2&lt;/sup&gt;: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/JyH6k.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JyH6k.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means there exist the possibility to produce electrical-power using sewage-sludge if the water-content is low enough. Which could be achieved by a solar-thermic-process&lt;sup&gt;3&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://i.stack.imgur.com/TRvEe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TRvEe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;[1]: &lt;a href=&quot;http://ec.europa.eu/environment/archives/waste/sludge/pdf/workshoppart4.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Problems about sewage sludge incineration&lt;/a&gt;, founded by the &lt;a href=&quot;http://ec.europa.eu/environment/index_en.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;European Union&lt;/a&gt;&lt;br&gt;&#xA;[2]: &lt;a href=&quot;https://web.anl.gov/PCS/acsfuel/preprint%20archive/Files/43_1_DALLAS_03-98_0197.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fuel Characteristics Of Sewage Sludge...&lt;/a&gt;&lt;br&gt;&#xA;[3]: &lt;a href=&quot;https://www.isah.uni-hannover.de/36.html?&amp;amp;tx_tkforschungsberichte_pi1%5BshowUid%5D=545&amp;amp;tx_tkforschungsberichte_pi1%5Bbackpid%5D=62&amp;amp;L=1&amp;amp;cHash=65d685bfd16ebb33ee7ceb807b8529e1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Universität Hannover - ISAH&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2018-01-25T20:04:27.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19099" PostTypeId="2" ParentId="18361" CreationDate="2018-01-25T20:21:27.610" Score="0" Body="&lt;h2&gt;After having the power company fix my mains connection, most of the noise dissapeared&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I finally got around to calling my power company, and had them come out and test my mains connection. They told me that the grounding on it was bad. After a repair team fixed it, I found about 75% of the noise was gone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still more noise than past places, (again, I suspect the old wiring) but at least now the noise is at a manageable level.&lt;/p&gt;&#xA;" OwnerUserId="14088" LastActivityDate="2018-01-25T20:21:27.610" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19100" PostTypeId="1" CreationDate="2018-01-25T21:00:19.850" Score="2" ViewCount="14022" Body="&lt;p&gt;What is the difference between the formulas $\tau = VQ/It$ and $\tau = V/A$ for finding shear stress due to transverse loading? I understand that the use of first moment of area, is the only reason for the more complex formula to account for shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a square cross section the maximum shear stress used for failure analysis is simply 4V/3A, which relates to tau = V/A except it has a multiplication factor of 1.33. What is the point of this safety factor, since the rectangle is such a simple shape, and which formula should I use? The factory of safety ends up being very different due to the extra 1/3. &lt;/p&gt;&#xA;" OwnerUserId="14658" LastEditorUserId="8632" LastEditDate="2018-01-27T15:13:46.413" LastActivityDate="2018-01-27T15:13:46.413" Title="What is the difference between tau = VQ/It and tau = V/A?" Tags="|mechanical-engineering|stresses|mechanical-failure|shear|failure-analysis|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19101" PostTypeId="2" ParentId="19100" CreationDate="2018-01-26T00:25:09.357" Score="-1" Body="&lt;p&gt;The first formula takes the stress of the fiber parallel to its axis or perpendicular to the plane of consideration. &#xA;The second takes the stress perpendicular to the axis or in some cases, parallel to the plane of consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See image below. &#xA;The red line in top drawing represents the fibers being stressed out in shear, VQ/Ib.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red line in bottom drawing represents the fibers being stressed out in shear, V/A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DGMd2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DGMd2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12597" LastActivityDate="2018-01-26T00:25:09.357" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19102" PostTypeId="2" ParentId="19092" CreationDate="2018-01-26T04:47:15.360" Score="2" Body="&lt;p&gt;Any engine that can run on compressed air will do this. There are quite a few, but as an example imagine a cylinder that has a bit of compressed air in it. The pressure will push on the piston doing work until it has equalized with the atmospheric pressure. Then a valve opens and perhaps a flywheel or another piston on a crankshaft pushes the piston back and vents the air out to the atmosphere, a new bit of compressed air is admitted and the cycle repeats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the absence of a load, the engine will convert all of the pressure energy into heat from friction in the cylinder or bearings as well as air resistance.&lt;/p&gt;&#xA;" OwnerUserId="5302" LastActivityDate="2018-01-26T04:47:15.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19103" PostTypeId="1" CreationDate="2018-01-26T08:46:17.887" Score="2" ViewCount="244" Body="&lt;p&gt;The system that I'm working on has a rotating cylinder (cleaning head). It needs to rotate vertically between two arms, much like a cleaning brush in a car wash. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I have not considered how the shafts of the cleaning head will be connected to the motor exactly but it would either be directly connected to a coupler or with a right-angled gearbox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have added some illustrations just in case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E97G9m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E97G9m.jpg&quot; alt=&quot;rough sketch of system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifications of cleaning head:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weight : 4~5kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Length : 1.5 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rpm    : 100 RPM MAX&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done a bit of research and I know that the shafts need to be shouldered but how do I make one end of the shaft fixed? Tapered? Have a screw on the end of the shaft and lock the bearing with a washer and a lock nut?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm thinking of having the top arm be where the shaft is &quot;fixed&quot; and the bottom be &quot;floating&quot; and connected to the motor. Is this a good decision?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, what bearings and bearing housings should I use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deep groove ball bearings seems to be able to handle the load but since there will be axial and radial loading(the cleaning head will need to tilt about 15 degrees at all times), so roller bearings maybe more suitable. I'm not certain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are split plummer (pillow) block housings and flanged housings, which is better for my use?&lt;/p&gt;&#xA;" OwnerUserId="14667" LastEditorUserId="13311" LastEditDate="2018-01-26T23:40:50.720" LastActivityDate="2018-07-12T07:56:40.780" Title="Bearing and Bearing Housing for Said Design" Tags="|mechanical-engineering|design|mechanisms|bearings|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19104" PostTypeId="2" ParentId="19092" CreationDate="2018-01-26T13:07:45.147" Score="3" Body="&lt;p&gt;One possibility would be to use the air to drive a turbine (probably a radial turbine as they are used in turbochargers). The turbine could be connected to a generator and the generator to a resistor. You will most likely need to cool the bearing of the turbine and the generator as well, so you will generate a lot of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mLoe2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mLoe2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that your pressurised air (@4500PSI=320bar) is at room-temperature, 20°C the extraction of the potential pressure energy will result in pretty low temperature at the exhaust of your process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used &lt;a href=&quot;http://www.ethermo.us/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ethermo.us&lt;/a&gt; to calculate the thermodynamic states. For a ideal (efficiency=1.0) process the expansion from 320 bar and 20°C to 1 bar (ambient pressure) would result in the condensation of air, which makes everything a lot more complicated, so I expanded only to 12 bar. But even at 12 bar the exhaust air will have -160°C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I would appreciate if someone could cross-check those numbers and calculations!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: Mass of compressed air in the container&lt;br&gt;&#xA;Volume: &lt;span class=&quot;math-container&quot;&gt;$V = 1l$&lt;/span&gt;,&lt;br&gt;&#xA;density: &lt;span class=&quot;math-container&quot;&gt;$\rho_\mathrm{@310bar, 20°C} = 330,33 \frac{kg}{m^3}$&lt;/span&gt;&#xA;Mass of air in the container, &lt;span class=&quot;math-container&quot;&gt;$m_\mathrm{air}=V \cdot \rho = 0,33kg$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Specific enthalpy-difference for isentropic expansion. Both states have the same specific entropy (&lt;span class=&quot;math-container&quot;&gt;$5\frac{kJ}{kgK}$&lt;/span&gt;)&lt;br&gt;&#xA;Initial state: &lt;span class=&quot;math-container&quot;&gt;$h_\mathrm{@310bar, 20°C} = 248,39 \frac{kJ}{kg}$&lt;/span&gt;&lt;br&gt;&#xA;Final state: &lt;span class=&quot;math-container&quot;&gt;$h_\mathrm{@12bar, -160°C} = 91,86 \frac{kJ}{kg}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: Available energy&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\Delta H = \left(h_\mathrm{@310bar, 20°C} - h_\mathrm{@12bar, -160°C} \right) \cdot m_\mathrm{air} = 51,71 kJ $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is unlikely that a process will extract &lt;span class=&quot;math-container&quot;&gt;$52 kJ$&lt;/span&gt; so it is likely to be less. I am afraid (given no errors) that this amount of energy will not justify the technical challenges.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Remark:&lt;br&gt;&#xA;The working mechanism of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Vortex_tube&quot; rel=&quot;nofollow noreferrer&quot;&gt;vortex-tube&lt;/a&gt; is analytically described by Polihronov and Straatman&lt;sup&gt;1&lt;/sup&gt;. The author states the following with respect to heating&lt;sup&gt;2&lt;/sup&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In conclusion: if one needs to convert the energy of pressurized gas into &lt;strong&gt;heat&lt;/strong&gt; by using the vortex tube effect, the best approach is: remove the hot exit of the tube.(...)and drive a heat pump with it to produce heat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;[1]: Polihronov, Jeliazko G.; Straatman, Anthony G. (2012). &quot;Thermodynamics of angular propulsion in fluids&quot;. Physical Review Letters.&lt;br&gt;&#xA;[2]: Polihronov, &lt;a href=&quot;https://www.quora.com/Heating-using-a-vortex-tube&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heating using a vortex tube&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="6767" LastEditDate="2019-04-18T12:34:34.190" LastActivityDate="2019-04-18T12:34:34.190" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="19105" PostTypeId="2" ParentId="19092" CreationDate="2018-01-26T13:21:33.303" Score="1" Body="&lt;p&gt;Since the expanding air will try to cool it's immediate surroundings, you need isolate the expanded air supply from the area to be heated. As long as you have a heat sink, you can use an air motor to run a heat pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you can capture heat during the filling of the cylinder too.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-01-26T13:21:33.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19106" PostTypeId="2" ParentId="18567" CreationDate="2018-01-26T13:46:18.900" Score="0" Body="&lt;p&gt;It sounds like you actually need to disengage the drive at two points. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;To disengage the engine from the drive altogether so you can run with just the motor without the engine acting as a brake&lt;/li&gt;&#xA;&lt;li&gt;To disconnect the propeller so you can use the engine/motor as a genset. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As you have shown a gearbox connected to the engine in your diagram a combination of a clutch and neutral gear as found in car engines would meet case 1. this really depends on what clutch/flywheel/gearbox sets are available for your engine. If you're using a road vehicle engine this should be pretty easy to find. Marine engines don't necessarily have clutches and gearboxes as the propeller is only loosely coupled to the load and inherently acts a bit like a  torque converter in an automatic gearbox. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For case (2) you may find that a dog clutch is the simplest and most reliable solution and will probably also be easier to find as a stand-alone component. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="7083" LastEditDate="2018-03-28T02:18:08.827" LastActivityDate="2018-03-28T02:18:08.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19110" PostTypeId="2" ParentId="19100" CreationDate="2018-01-26T19:29:19.620" Score="0" Body="&lt;p&gt;The first one is the shear stress in a beam with rectangular cross section and because of parabolic distribution of shear on such cross sections it's factored by 1.5 to get the average stress on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second one is the shear flow at a certain height of a beam cross section, usually considered for composite beams to analyse the shear connection and shear stress at that level. Then one can calculate the shear studs or nails or any mechanical attachment to connect for example the web of a composite beam to its flange!  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-26T19:29:19.620" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19111" PostTypeId="1" CreationDate="2018-01-26T20:10:00.330" Score="0" ViewCount="106" Body="&lt;p&gt;I've been working with a team of 'designers' that are highly concerned over a potential failure mode of ABS plastic. Their theory is that under a static load well under the yield point of the material that a plastic ABS part will eventually stretch and fail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific example is a handle that I've calculated to hold over 190 lbs before yielding is being loaded to 30 lbs and held there for seven plus days. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only type of failure I can think of that this could be would be creep, but since we're working at constant room temperature, constant load, evenly distributed load over a wide surface (3 square inches) and so far below the yield point I can't imagine that this test would produce any useful results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone think of anything this test would show? Is there a failure mode that I'm not thinking of (specific to ABS or not)? Or is this more likely a test in futility?&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2018-01-26T20:10:00.330" Title="Long term deflection" Tags="|mechanical-failure|product-testing|fatigue|abs|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19113" PostTypeId="1" AcceptedAnswerId="19115" CreationDate="2018-01-26T21:33:17.913" Score="3" ViewCount="536" Body="&lt;p&gt;I understand bevel gears are used in situations were a 90 degree change is required, but I can't visually understand why. Why can't we just use standard spur gears at 90 degrees, what do we actually benefit from using a bevel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially I'm asking how the engagement between gear teeth react if we just used spur gears for a 90 degree change, rather than bevel gears.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/724Qq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/724Qq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9066" LastEditorUserId="9066" LastEditDate="2018-01-26T22:37:39.553" LastActivityDate="2018-02-02T00:07:05.427" Title="Why do we use bevel gears?" Tags="|gears|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19114" PostTypeId="1" AcceptedAnswerId="19137" CreationDate="2018-01-26T21:44:03.120" Score="6" ViewCount="347" Body="&lt;p&gt;I'm reading this article on afterburners:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.airspacemag.com/flight-today/how-things-work-afterburners-18481403/&quot; rel=&quot;noreferrer&quot;&gt;https://www.airspacemag.com/flight-today/how-things-work-afterburners-18481403/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says in the article:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A typical jet engine uses only about half the oxygen it ingests, leaving a large amount of potential energy. The afterburner, which is a long extension at the back of the engine, combines much of the remaining oxygen with jet fuel, squirted into the high-speed exhaust stream from the engine’s turbine, and ignites the mixture.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've searched many places but I've never really understood why the engine only uses about half of the oxygen it ingests. Instead, the afterburner is needed to burn more oxygen. But why is not possible to design jet engines that burn more oxygen instead of adding another separate component, the afterburner? Why do jet engines only burn about a half of the oxygen ingested?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2018-01-28T22:18:13.603" Title="Why afterburners are needed instead of burning more fuel inside the turbine?" Tags="|combustion|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19115" PostTypeId="2" ParentId="19113" CreationDate="2018-01-26T22:44:28.973" Score="5" Body="&lt;p&gt;It is beveled to maintain continuous and stable flow of power with the least amount of metal on metal skidding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is designed to optimize the flow of lubricant and maintain the contact surface of the gears within acceptable range of stress and temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also to reduce impact and undue vibration to the housing and bearings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also reduces the hiccups in drivetraine and jerking of suspension.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-26T22:44:28.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19116" PostTypeId="1" CreationDate="2018-01-26T23:09:43.417" Score="3" ViewCount="267" Body="&lt;p&gt;I was trying to understand this stress curve in the notch effect. $\sigma_y$ I can understand but $\sigma_x$ and $\sigma_t$ not. Could you please explain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/7b26C.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="11393" LastEditorUserId="1832" LastEditDate="2018-01-27T01:09:50.013" LastActivityDate="2018-01-27T15:15:03.150" Title="stress curve in the notch effect?" Tags="|mechanical-engineering|structural-engineering|materials|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19118" PostTypeId="2" ParentId="19100" CreationDate="2018-01-27T01:29:58.550" Score="-1" Body="&lt;p&gt;There is a difference between Tau=VQ/It and Tau=V/A. The first is the shear stress the corresponded V is perpendicular on the biggest dimension of the beam (the axis of the beam) while the second tau is the tensile or compressive stress, in this case V is along  the axis of the beam the stress : if V is positive, it's a tensile stress otherwise it's a compressive stress. &lt;/p&gt;&#xA;" OwnerUserId="10043" LastActivityDate="2018-01-27T01:29:58.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19119" PostTypeId="2" ParentId="19116" CreationDate="2018-01-27T02:55:47.247" Score="2" Body="&lt;p&gt;$\sigma_x$ is tension and the reason for its curve is it is higher where there is more material to affect stretching the rod to go back to fill in the notch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma_\tau$ is shear and it increases toward center in a hyperbolic curve because of additive stresses of vertical and horizontal tensions: $\sigma_y + \sigma_x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you draw the Mohr's stress circle over a few points in that region you see it! &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2018-01-27T15:15:03.150" LastActivityDate="2018-01-27T15:15:03.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19121" PostTypeId="1" CreationDate="2018-01-27T06:10:47.770" Score="3" ViewCount="2868" Body="&lt;p&gt;I've been tasked to design a steam coil that will be immersed into a tank in order to maintain the fluid temperature above 25C (but not to exceed 40C), and I need to calculate its surface area &amp;amp; length. The fluid is a highly viscous oil additive and it's not agitated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach was the calculate the total energy needed to heat the coil, using $Q = m.c_p\Delta T$. where m is the mass of fluid in the tank, $c_p$ is the specific heat capacity of the fluid, and $\Delta T$ is the difference between the initial temperature of fluid (25C) and what it should be maintained at (~35-40C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I equated this energy to $UA\Delta T_m$, the energy delivered by the steam, where $U$ is the overall heat coefficient for steam, $A$ is the outer surface area of the coil and $\Delta T_m$ is the log mean temperature difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming this approach is correct, but I'm doubting my calculations because &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It doesn't match with &lt;a href=&quot;http://www.titanmf.com/immersion-coil-surface-area-calculator-for-heating-with-steam/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Immersion Coil Surface Area Calculator For Heating with Steam&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;It seems quite large&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Apart from the temperatures I've mentioned above, these are the only parameters I've been provided with from the company:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;inner diameter of the tank = $3.35\,m$&lt;/li&gt;&#xA;&lt;li&gt;outer diameter of the tank = $3.6\,m$&lt;/li&gt;&#xA;&lt;li&gt;height of tank = $6.5\,m$&lt;/li&gt;&#xA;&lt;li&gt;density of the fluid = $850\,kg/m^3$&lt;/li&gt;&#xA;&lt;li&gt;viscosity of the fluid = $30, 000\,Cst$&lt;/li&gt;&#xA;&lt;li&gt;pressure at which steam enters the pipe = $150\,psi$&lt;/li&gt;&#xA;&lt;li&gt;Heat time of $2\,hours\,(7200s)$&lt;/li&gt;&#xA;&lt;li&gt;pipe diameter of $2'' (0.0508\,m)$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So here's the calculation I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M_{fluid} = V * \rho = \frac{PI * 3.35^2 * 6.5}{4} * 850 = 49000\,kg$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the $C_p$ to be $1800\,J/kg.K$,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = 49000 * 1800 * (40 - 25) = 1.323\,GJ$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I estimated $\Delta Tm$ to be $86$ using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{(ΔT_1 - ΔT_2)}{ln(ΔT_1/ΔT_2)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $T_1$ being the inlet steam - the final temperature of the fluid = 145 C.&lt;br&gt;&#xA;With $T_2$ being the outlet 'steam' - initial temperature of the fluid = 45 C (I estimated the outlet 'steam' to be around 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the $U factor$ to be 100 (&lt;a href=&quot;https://www.engineeringtoolbox.com/heat-transfer-coefficients-coils-d_178.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;estimated from this site&lt;/a&gt;),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A = 1.323 GJ / (100 * 86 * 7200) = $21m^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the corresponding length would be $\pi DL = A$ and $L = 131m$. Clearly something (or a few things) are wrong here since I don't think the tank can be that big so help would be appreciated on where I went wrong.&lt;/p&gt;&#xA;" OwnerUserId="14707" LastEditorUserId="282" LastEditDate="2018-01-27T18:51:01.403" LastActivityDate="2018-01-27T18:51:01.403" Title="estimating the surface area of a steam coil immersed in a tank" Tags="|thermodynamics|heat-transfer|heat-exchanger|steam|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19122" PostTypeId="2" ParentId="19121" CreationDate="2018-01-27T06:19:30.560" Score="4" Body="&lt;p&gt;One thing to consider is that pipes used for heating like this are normally « finned » to increase the surface area - fins can be small discs fitted to the pipe or even wire loops - both increase the surface area massively and reduce the length required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This article explains the finned pipe although they are collecting heat: &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/030626199190038Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;The performance of a coiled finned-tube heat-exchanger submerged in a hot-water store: The effect of the exchanger's orientation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="282" LastEditDate="2018-01-27T16:16:04.153" LastActivityDate="2018-01-27T16:16:04.153" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19124" PostTypeId="2" ParentId="19116" CreationDate="2018-01-27T13:57:24.860" Score="3" Body="&lt;p&gt;In these cases, it is always worth thinking about how the stress &quot;flows&quot; through the element. In this case, that's quite simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jtOiEm.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jtOiEm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stress near the center of the element is basically unaffected by the notching. The stress near the edges, however, is diverted by the notches. From the image it is already quite clear that there's quite literally a stress concentration near the notches, which explains why $\sigma_y$ peaks there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we simplify this diagram into a strut-and-tie model (ignoring the stress along the core, which simply goes straight through the middle), we end up with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S4LKzm.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S4LKzm.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember the convention that arrows pointing away from the node represent tension and those pointing into the node are in compression. We can therefore see that the main verticals (and the diagonals representing the stress &quot;detours&quot;) in red are entirely in tension, as would be expected. However, the detours create transversal imbalances near the notch (in blue). At the center of the notch, both (red) diagonals point &quot;out&quot; of the section: their vertical components cancel out, however their horizontal components add up to a force pointing &quot;out&quot;. This force must therefore be cancelled out by another force (blue) on the other side of the node pointing &quot;in&quot;. Since this is an arrow pointing away from the node, it is in tension. This is $\sigma_x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the outskirts of the notch (above and below it), where the stress lines start to be deflected, there is another horizontal force imbalance, however in this case the deflected tension points &quot;in&quot;, so the balancing force must point &quot;out&quot;. In this case, that force points into the node, so that area is under compression.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-01-27T13:57:24.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19125" PostTypeId="2" ParentId="19100" CreationDate="2018-01-27T14:51:21.513" Score="3" Body="&lt;p&gt;As mentioned, the formula $\tau_{avg}=\frac{V}{A}$ is called &lt;em&gt;average shear stress&lt;/em&gt;. In some calculations it might suffice to calculate $\tau_{avg}$ for a widely used cross-section (let's say a standardized beam,e.g. HEA or IPE in Europe), and apply a safety coefficient to come up for shear flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other formula $\tau=\frac{V\cdot Q}{I\cdot t}$ (beam shear formula) takes into account, that variations of shear stress throughout the cross-section occur. Here, $\tau_{avg}$ might be used to simplify the formula of the true shear stress, that is, as a multiple of $\tau_{avg}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take for example a rectangular cross-section:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rXnUF.png&quot; width=&quot;300&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear stress in function of $z$ is:&#xA;$$ \tau(z)=\frac{6Vz(h-z)}{wh^3}=\frac{6z(h-z)}{h^2}\cdot \tau_{avg} $$&#xA;which for $z=\frac{h}{2}$ (the center point) yields:&#xA;$$ \tau(z=\frac{h}{2})=\frac{3}{2}\frac{V}{wh}=\frac{3}{2}\cdot\tau_{avg} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, it might sometimes be a bit easier to give the formula of the shear stress distribution as a multiple of the average shear stress $\tau_{avg}$&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2018-01-27T14:51:21.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19127" PostTypeId="1" AcceptedAnswerId="19136" CreationDate="2018-01-27T19:01:16.817" Score="0" ViewCount="246" Body="&lt;p&gt;I was given this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A current of 1,400 Amps flows in a copper bus bar.  The bus bar is 0.6 cm wide by 9 cm long.  You can assume the current flows only on the surface of the bus bar (so you can ignore the “depth” of the copper bar).  Find the average electron velocity in the copper bus bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper has a free electron density of 1029 electrons/cubic meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charge C = 1.6022e-19 Coulombs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current = i = charge times velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = amounts of charge per volume&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However because Im asked to ignore the third dimension of the copper bar &#xA; Im not sure if Im solving this problem correctly&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;i &amp;amp;= \frac{dq}{dt} \ldots \frac{n(volume)q}{dt}=\frac{n{A(V_dt)q}}{dt} \\&#xA;i &amp;amp;= \frac{n*A*V_dq}{dt} \\&#xA;V_d&amp;amp;=1.618E^{-8}um/s&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="14713" LastEditorUserId="1832" LastEditDate="2018-01-27T22:45:47.110" LastActivityDate="2018-01-28T16:49:09.527" Title="Find average electron velocity on copper bus bar(um/s)" Tags="|electrical-engineering|circuits|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19128" PostTypeId="2" ParentId="11299" CreationDate="2018-01-27T19:55:29.607" Score="0" Body="&lt;p&gt;There are some typos on Page 13 which may add to the confusion.  In figure 8.9, the right figure, the k graph is labelled ki, and the ki graph is labelled k.&#xA;In the text, the stability margins have been misassigned: 0.67 should go with 0.18, 0.60 with 0.30, 0.55 with 0.36, 0.48 with 0.58.&lt;/p&gt;&#xA;" OwnerUserId="14715" LastActivityDate="2018-01-27T19:55:29.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19129" PostTypeId="1" AcceptedAnswerId="22542" CreationDate="2018-01-27T23:05:59.587" Score="1" ViewCount="941" Body="&lt;p&gt;I've been trying to get this sweep to work for a while now. It works with a solid circle. It doesn't work with an annulus. I can provide more info if necessary. I made sure the order of the features matched what I've seen in other threads about this issue but that doesn't seem to help. Every time I try to build it, I get the error &quot;The intermediate profile # 2 could not be solved.&quot;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uea3z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uea3z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14716" LastActivityDate="2018-07-01T22:56:20.497" Title="Sweep with guide curve in SolidWorks" Tags="|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19130" PostTypeId="2" ParentId="19114" CreationDate="2018-01-27T23:25:13.653" Score="3" Body="&lt;p&gt;After burner primary task is to add extra thrust to a jet engine by heating the jet stream after the turbine via injecting and burning fuel in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason they just don't add more fuel in the jet engine before turbine even though it would be more economical is the very high temperatures of the engine just before the turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fuel injection into the jet engine is restricted by the temperatures not available oxygen consumption.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-27T23:25:13.653" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19131" PostTypeId="1" AcceptedAnswerId="19132" CreationDate="2018-01-28T00:56:34.527" Score="3" ViewCount="54" Body="&lt;p&gt;I have this pickup truck that I love dearly. I often change the aerodynamic configuration of the truck when I use my camper shell, or roof top tent, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a reasonably easy way to create a mesh model of my truck from images so that I can use ANSYS to model the aerodynamic effects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14717" LastActivityDate="2018-01-28T01:45:20.820" Title="Scanning and modeling my personal vehicle for aerodynamic simulation" Tags="|automotive-engineering|aerodynamics|mechanical|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19132" PostTypeId="2" ParentId="19131" CreationDate="2018-01-28T01:45:20.820" Score="4" Body="&lt;p&gt;What you seek is called 3D photogrammetry and is a relatively exciting segment of 3D modeling. You can use those terms to search with your favorite search engine and be overwhelmed with options. I found a link to a summary performed a bit more than a year ago, which might be more useful to you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://pfalkingham.wordpress.com/2016/09/14/trying-all-the-free-photogrammetry/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://pfalkingham.wordpress.com/2016/09/14/trying-all-the-free-photogrammetry/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The author of the summary took photos of a couple different models to present reasonable challenges to the software available and posted his results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Tm4y0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tm4y0.jpg&quot; alt=&quot;3d model&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image above came from the linked site. You'd be learning to use image processing software to clean up the results in order to accomplish your goal, but it's time well spent, considering the image file you'd have in return.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-01-28T01:45:20.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19134" PostTypeId="1" AcceptedAnswerId="19135" CreationDate="2018-01-28T15:06:10.623" Score="0" ViewCount="113" Body="&lt;p&gt;I am a chemical engineering student learning about material balance, so the problem below will not be a ‘real’ one , but just an exercise for me to understand better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This system below is not a reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: can I say that the overall material balance for this system is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F1 + F2 + F3 = P1 + P2 + P3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vICgQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vICgQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this doubt because Feed 3 comes in after Product 1 and Product 2 have been removed. Does this matter at all? &lt;/p&gt;&#xA;" OwnerUserId="13646" LastEditorUserId="1832" LastEditDate="2018-01-28T21:00:11.090" LastActivityDate="2018-01-28T21:00:11.090" Title="Concept on material balance" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19135" PostTypeId="2" ParentId="19134" CreationDate="2018-01-28T16:08:13.290" Score="0" Body="&lt;p&gt;I'm assuming R is a recycled material that gets reprocessed from product stream P1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the fullness of time and the complete reprocessing of R so that at the end of production, no R is present then you can say that F1 + F2 + F3 = P1 + P2 + P3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, processes and processing plants rarely operate that way. Usually there will be an amount of R within the system. To account for this F1 + F2 + F3 = P1 + P2 + P3 + R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With material balance, conservation of mass is paramount - what goes in must come out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your concerns about F3 being added after P1 and P2 have been removed, F3 is added to the residual from the P1 and P2 process to produce P3. Your concern would be valid if there was no residual feeding into the second process to which F3 is added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to consider the system as whole when it comes to material balance. There are three inputs/feeds and three products, with some of the feed recirculating.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-01-28T16:08:13.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19136" PostTypeId="2" ParentId="19127" CreationDate="2018-01-28T16:49:09.527" Score="0" Body="&lt;p&gt;I would argue that your free electron density isn't correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless I have misunderstood your question, the equation you need is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_e = \frac{I}{nAq}$$&#xA;$$v_e = \frac{1400}{1029\times 5.4\times10^{-6} \times 1.6\times10^{-19}}$$&#xA;$$v_e = 1.572\times10^{25} \text{ m/s}$$&#xA;$$v_e = 1.572\times10^{31} \text{ $\mu$m/s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n$ should be about $8.5\times10^{28}$, though, which makes the actual answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_e = \frac{1400}{8.5\times10^{28}\times 5.4\times10^{-6} \times 1.6\times10^{-19}}$$&#xA;$$v_e = 0.01906 \text{ m/s}$$&#xA;$$v_e = 1.906\times10^{-8} \text{ $\mu$m/s}$$&lt;/p&gt;&#xA;" OwnerUserId="13673" LastActivityDate="2018-01-28T16:49:09.527" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19137" PostTypeId="2" ParentId="19114" CreationDate="2018-01-28T17:55:00.780" Score="0" Body="&lt;p&gt;There are some parts to your question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Part 1: The highest temperature is achieved with a air-fuel-ratio below 1.&lt;/li&gt;&#xA;&lt;li&gt;Part 2: The temperature limit of available materials cooling processes is even lower.&lt;/li&gt;&#xA;&lt;li&gt;Part 3: Given an upper temperature limit the thermodynamic-cycle of a jet-engine can only deliver more power by re-heating the exhaust-gases.&lt;/li&gt;&#xA;&lt;li&gt;Part 4: Every design-process is a trade-off between multiple factors. For the after-burner two factors are efficiency and complexity of the engine.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 1: Combustion Reactions&lt;/strong&gt;&lt;br&gt;&#xA;Depending on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Air%E2%80%93fuel_ratio&quot; rel=&quot;nofollow noreferrer&quot;&gt;air-fuel-ratio&lt;/a&gt; the combustion-temperature will vary.&#xA;The following figure shows the relation. Observe that the maximum power (think temperature) is not achieved for a air-fuel-ratio of 1 (, i.e. &lt;a href=&quot;https://www.engineeringtoolbox.com/stoichiometric-combustion-d_399.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;stoichiometric combustion&lt;/a&gt;, $\lambda = 1$). This is due to dissociation of the combustion products.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tebUR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tebUR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 2: Temperature Limit of Materials&lt;/strong&gt;&lt;br&gt;&#xA;Due to the high mechanical loads on the rotating turbine parts (aero- and centrifugal-loads) the maximum allowable temperature is below the theoretical maximum limit. The following figure shows the how the development of new materials did allow for higher temperatures in the jet-engine&lt;sup&gt;1&lt;/sup&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/P94ji.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P94ji.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 3: Thermodynamic Cycle&lt;/strong&gt;&lt;br&gt;&#xA;The basic (idealised) thermodynamic cycle of a jet-engine is called &lt;a href=&quot;https://en.wikipedia.org/wiki/Brayton_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Brayton-Cycle&lt;/a&gt;. The following figure shows this cycle.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JCL33.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JCL33.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Enthalpy&quot; rel=&quot;nofollow noreferrer&quot;&gt;enthalpy&lt;/a&gt;-difference between state 5 and 6 is what is converted into thrust. Given the temperature limits the only way to increase the power output of the thermodynamic cycle (vertical distance between state 5 and 6) is to re-heat the turbine-exhaust.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lwAz6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lwAz6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Observe that the temperature limit for stationary parts is higher, this is also due to the lower operating hours of the after-burner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part 4: Engineering considerations, trade-offs&lt;/strong&gt;&lt;br&gt;&#xA;Usually the additional thrust produced by the after-burner is not needed throughout the whole mission. Therefore it is acceptable to sacrifice efficiency for mechanical simplicity. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/bYVoF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bYVoF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The shown &lt;a href=&quot;https://de.wikipedia.org/wiki/Eurojet_EJ200&quot; rel=&quot;nofollow noreferrer&quot;&gt;EJ-200&lt;/a&gt; also show the engine stations. It is easily seen the difference in mechanical complexity between stations 4-5 and stations 5-6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But why is not possible to design jet engines that burn more oxygen instead of adding another separate component, the afterburner? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is theoretical possible to design an engine which would burn all oxygen. However, until now this was not necessary.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why do jet engines only burn about a half of the oxygen ingested?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Because fuel is burned for maximum heat not to deplete oxygen. The maximum temperature is limited by material and dissociation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;[1]: &lt;a href=&quot;http://www.virginia.edu/ms/research/wadley/high-temp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;engineering.virginia.edu&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2018-01-28T22:18:13.603" LastActivityDate="2018-01-28T22:18:13.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19139" PostTypeId="1" AcceptedAnswerId="19140" CreationDate="2018-01-29T01:22:41.297" Score="1" ViewCount="6217" Body="&lt;p&gt;I have this data sheet for a stepper motor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b1bu0.png&quot; alt=&quot;stepper motor data sheet&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pitch diameter of the pinion attached to it is listed as &quot;Ø6&quot;. What unit is this? Looking at the &lt;a href=&quot;https://en.wikipedia.org/wiki/https://en.wikipedia.org/wiki/%C3%98&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia page for Ø&lt;/a&gt; it seems like this is just a nonce symbol indicating that the measurement in question is a diameter, but then the question still remains, what unit is it? Does it mean 6 millimeters? Is there some convention about default units in data sheets I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="14725" LastActivityDate="2018-01-29T08:19:49.320" Title="Meaning of O-slash in data sheet about gear diameter?" Tags="|mechanical-engineering|motors|gears|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19140" PostTypeId="2" ParentId="19139" CreationDate="2018-01-29T01:29:28.247" Score="7" Body="&lt;p&gt;In the lower right corner of the document/drawing, you can see four (or five) lines of text upward: UNIT: mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another piece of useful information in the data panel is the scale, shown there as 3:1. You would be able to measure the components of the drawing and divide by 3 to determine real dimensions, but only if the drawing is printed to 1:1 and not &quot;fit to page&quot; as if sometimes done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your other research result is correct.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-01-29T01:29:28.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19141" PostTypeId="2" ParentId="19139" CreationDate="2018-01-29T03:22:52.883" Score="0" Body="&lt;p&gt;Yes this drawing is all in metric. Therefore,  &quot;Ø6&quot; = 6mm diameter &lt;/p&gt;&#xA;" OwnerUserId="14726" LastActivityDate="2018-01-29T03:22:52.883" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19143" PostTypeId="2" ParentId="19139" CreationDate="2018-01-29T07:14:28.637" Score="1" Body="&lt;p&gt;Engineering drawings come in two flavours metric and north american (mainly US only). Metric drawings are using mm as a unit, even in architecture where a mm is a slightly small unit. North american drawings are using inches as their base unit. If they would use anything else then the unit would read next in the dimension (this is incredibly rare). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the sheet states that the unit is mm in the title block. The title block applies to the whole drawing. Other much weaker hints are the A4 size paper (ISO size indicates mm), projection rotation rule (image is drawn in first angle, US almost exclusively use third) and scale info. Although you really shouldn't measure distances on drawings since the numbers are what counts and there can be reproduction errors and/or dimensions not in scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possible hints would be invoking any ISO standards vs ASME standards which here are not present (for example a general tolerance). Also simply assuming mm* would work for the bulk of drawings made for any other purpose than US consumption. Besides in this case inhes would sound incredibly big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* Although, inches are much more common in electronics than mechanical drawings, so best not guess anything.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-01-29T08:19:49.320" LastActivityDate="2018-01-29T08:19:49.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19144" PostTypeId="1" AcceptedAnswerId="19145" CreationDate="2018-01-29T14:46:57.397" Score="-3" ViewCount="70" Body="&lt;p&gt;I'd like a simple explanation of the function of &lt;strong&gt;by pass&lt;a href=&quot;https://i.stack.imgur.com/nFU5c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nFU5c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/strong&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is an image of the project am working on and if it can help clarify.&lt;/p&gt;&#xA;" OwnerUserId="14732" LastEditorUserId="14732" LastEditDate="2018-01-30T09:37:35.400" LastActivityDate="2018-01-30T09:37:35.400" Title="What is the use of a &quot; by pass &quot;" Tags="|mechanical-engineering|pipelines|petroleum-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2018-01-29T22:15:06.850" ContentLicense="CC BY-SA 3.0" />
  <row Id="19145" PostTypeId="2" ParentId="19144" CreationDate="2018-01-29T14:56:38.427" Score="1" Body="&lt;p&gt;“By-pass” generally means to take an alternative path or route around something, so could be a village or a valve or a flow meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes meters have a by-pass arrangement with 3 valves so that they can be replaced without stopping the flow.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-01-29T14:56:38.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19146" PostTypeId="2" ParentId="18140" CreationDate="2018-01-29T15:21:37.840" Score="1" Body="&lt;p&gt;It looks like the devkit associated with the NodeMCU community.&#xA;If you turn it around with the metal portion facing away from you it should say something like &quot;devkit v1.x&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a developer board with an ESP8266 built into it. You can actually see the ESP8266 board sitting on top of the other board in the photo!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ESP8266 specifies only the microcontroller (under the metallic cover) whereas the ESP-12N specifies the board on which the microcontroller is included (the board piggy backing on the bigger board). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function of the underlying board is to simply provide you with a USB interface (UART - USB) and a voltage stabilization for convenience and standardization of the I/Os.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this wikipedia article for more information: &lt;a href=&quot;https://en.wikipedia.org/wiki/NodeMCU&quot; rel=&quot;nofollow noreferrer&quot;&gt;NodeMCU&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14680" LastEditorUserId="7083" LastEditDate="2018-07-31T22:56:12.657" LastActivityDate="2018-07-31T22:56:12.657" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19147" PostTypeId="1" AcceptedAnswerId="19148" CreationDate="2018-01-29T15:48:13.467" Score="1" ViewCount="9831" Body="&lt;p&gt;I have done some research about the subject but I can't find the difference between Temperature Element (&lt;strong&gt;TE&lt;/strong&gt;) and  Temperature Indicator (&lt;strong&gt;TI&lt;/strong&gt;) &#xA;&lt;a href=&quot;https://i.stack.imgur.com/k0zhA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k0zhA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image of the P&amp;amp;ID of the project am working on if it can help clarify my question.&lt;/p&gt;&#xA;" OwnerUserId="14732" LastEditorUserId="368" LastEditDate="2018-01-30T17:01:15.150" LastActivityDate="2022-07-21T11:04:29.823" Title="What is the difference between TE and TI?" Tags="|mechanical-engineering|electrical-engineering|instrumentation|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19148" PostTypeId="2" ParentId="19147" CreationDate="2018-01-29T16:10:15.460" Score="1" Body="&lt;p&gt;The element does the measuring and the indicator shows the value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think of a simple thermometer, the bulb at the bottom measures the temperature and the glass tube indicates the value...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-01-29T16:10:15.460" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19149" PostTypeId="1" CreationDate="2018-01-29T16:50:39.283" Score="3" ViewCount="385" Body="&lt;p&gt;I want to see how different combination/position of fan on airplane affect stability of airplane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need simulator to test one of the idea which involves specific position for propeller on the plane. Before moving on to 3d print and test the model I need to see how aerodynamic affect my flying machine and if any kind of stabilization required at least in normal weather condition with uniform airflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using the Ansys but it only shows the statistics with respect to lift, air flow directions, it does not actually simulate object flying. I need to see if idea even work, is there any good simulator for that? Or someone can give me reference in Ansys software that can allow me to achieve my purpose.&lt;/p&gt;&#xA;" OwnerUserId="8900" LastActivityDate="2023-12-01T17:02:05.220" Title="Physics motion simulator for flying object" Tags="|simulation|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19150" PostTypeId="1" AcceptedAnswerId="19177" CreationDate="2018-01-29T18:46:39.440" Score="1" ViewCount="523" Body="&lt;p&gt;I am trying to find the torque exerted on the shaft of a capstan and bow drive. In particular, the system I'm looking at has two wire rope pulleys, and two idler pulleys, and is used to drive a carriage as seen Below:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/B2ndK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B2ndK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the drive pulley consists of two separate pulleys tensioned against one another, with the end of two wire ropes terminated in the pulleys. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naively, I would assume that the force F is equal to the Torque T * the radius r of the drive pulley. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any additional effects due to a multiple number of wraps on the pulley (one side reels out, while the other side reels in, keeping the total angle of wraps equal throughout the motion)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the capstan effect play a large part with this mechanism? From what I understand the capstan effect should cause less tension at the wire rope stops embedded in the drive pulley, but not effect F. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14734" LastActivityDate="2018-01-30T22:36:05.033" Title="Torque Required to Drive Capstan and Bow" Tags="|torque|pulleys|power-transmission|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19151" PostTypeId="2" ParentId="19150" CreationDate="2018-01-29T19:35:21.433" Score="0" Body="&lt;p&gt;The force is actually F= Torque/ radius of drive pulleys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F=T, tension in cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ignore the friction, the force is not changed by your configuration of pulleys.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-29T19:35:21.433" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19152" PostTypeId="1" CreationDate="2018-01-29T20:08:56.500" Score="2" ViewCount="389" Body="&lt;p&gt;I want to use a winch to raise and lower a load, but now I'm unsure, if the winch can handle the lowering? Normally, winches are only used to pull a load, right? When I now reverse the motor direction, the winch unrolls the rope and gets pulled in addition by the load. Where goes now the energy? A big part should go into the gearbox i think. The motor cannot act as a generator, since it is powered when unrolling the winch, because of the stand still brake. So I think the load will accelerate until the friction in the gearbox is high enough to balance the system. Am I right with this assumption?&lt;/p&gt;&#xA;" OwnerUserId="8784" LastActivityDate="2018-01-30T11:00:39.020" Title="Behaviour of a winch when unrolling" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19153" PostTypeId="2" ParentId="19152" CreationDate="2018-01-29T20:36:34.673" Score="3" Body="&lt;p&gt;There are many types of winches with different mechanism to control the lifting and lowering of the load. A winch has to be able to Lift the load to a desired height and hold it there indefinitely and lower it safely with control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some winches are hydraulic, some use chain as the cable and the chain engages the moving wheel for safety. Most of light duty winches have friction controls for braking and lowering. These controls clutch onto the rope or cable and lower it in a controlled way and have release settings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some winches have levers that work like a Car jack and lift or lower the load in steps.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="90" LastEditDate="2018-01-30T11:00:39.020" LastActivityDate="2018-01-30T11:00:39.020" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19155" PostTypeId="2" ParentId="19152" CreationDate="2018-01-30T02:39:28.000" Score="2" Body="&lt;p&gt;inexpensive, commercially-available 12VDC winches (as for mounting on front bumpers of 4x4 trucks) contain worm gear drives which exhibit a characteristic called SELF-LOCKING. this prevents the load from causing the electric motor to rotate backwards, unwind the cable spool, and lower the load when the motor is not energized. if you reverse the drive voltage on one of these, it will unspool and lower the load UNTIL you shut the motor off, at which point the winch will stop and the load will remain suspended at that point. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-01-30T02:39:28.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19156" PostTypeId="1" CreationDate="2018-01-30T04:24:30.970" Score="2" ViewCount="1179" Body="&lt;p&gt;I am building a small robot having pneumatic system. I am using 12v car tyre inflator as my compressor.But i didn't find any 3ltr to 6ltr standardrd air tank. Can i convert my old 3ltr lpg tank into air tank&lt;img src=&quot;https://i.stack.imgur.com/xM7Bc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14739" LastActivityDate="2018-03-01T17:07:58.327" Title="Lpg cylinder to compressed air tank" Tags="|gas|compressed-gases|pneumatic|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19157" PostTypeId="1" AcceptedAnswerId="19245" CreationDate="2018-01-30T05:24:00.817" Score="4" ViewCount="1755" Body="&lt;p&gt;Why the closed loop dominant poles of the root locus can show the response of the system? Wont it neglect the effect of the closed loop zeros?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I read on the books, root locus method deal with the closed loop poles. It sketch the locus of the close-loop poles under an increase of one open loop gain(K) and if the root of that characteristic equation falls on the RHP. It means the close loop pole fall into RHP and make system unstable. But the exercises and examples also treat Root locus as a method of designing compensation or gain to fulfill the system requirement like damping factor and setting time. Isn't the root locus method for stability only?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the system response should include the closeloop zero. Did I miss something? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, can Nyquist plot also used to show the system response and to design the system? As I thought its some how same as Root locus, with frequency being the varying K(gain).&lt;/p&gt;&#xA;" OwnerUserId="14740" LastActivityDate="2021-05-17T01:53:58.413" Title="Control theory: Closed loop zeros, Root locus and its dynamic response" Tags="|control-engineering|control-theory|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19158" PostTypeId="1" CreationDate="2018-01-30T06:09:01.427" Score="1" ViewCount="104" Body="&lt;p&gt;While comparing magnetic bearings to others ceramic bearings, a rough first point maybe to compare would be the two different bearings' coefficients of friction. This is just a thumbnail comparison apart from load bearing, and rpm, and any other metric that one might have with the application in concern. Is there a rough estimate for passive magnetic bearings, for an example to compare with a ceramic bearing might have 0.2, steel bearing 0.8, or this bearing called an Autonomous Decentralized Bearing&quot; : 0.0027, any numbers for a passive magnetic bearing, or for reference any numbers for a similar active magnetic bearing? &lt;/p&gt;&#xA;" OwnerUserId="14742" LastEditorUserId="14742" LastEditDate="2018-01-30T06:14:50.623" LastActivityDate="2018-01-30T06:14:50.623" Title="Coefficient of Friction, Passive Magnetic Bearing" Tags="|bearings|electromagnetism|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19159" PostTypeId="1" CreationDate="2018-01-30T06:35:16.100" Score="12" ViewCount="6415" Body="&lt;p&gt;So I understand that buoyancy happens because the fluid exerts larger amounts of pressure from beneath an object compared to above, like in this picture&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SCfkm.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SCfkm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, what if the object is forced to the bottom of the container, so that no fluid is beneath it. The logic is, if there is no fluid beneath, there is nothing to push it up. So, would that object still experience buoyancy? If yes, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: interesting to see some answers disagreeing with each other. One thing to note - according to my textbook buoyancy is a force that happens because of hydrostatic pressure - it has nothing to do with density of the objects. So those who say then can will experience buoyancy because it is less dense are wrong I guess, that is not buoyancy.&lt;/p&gt;&#xA;" OwnerUserId="8851" LastEditorUserId="7736" LastEditDate="2018-02-06T13:03:21.130" LastActivityDate="2018-02-06T13:03:21.130" Title="Would an object at the bottom of the sea still experience buoyancy?" Tags="|fluid-mechanics|pressure|fluid|" AnswerCount="4" CommentCount="6" ClosedDate="2018-02-06T16:55:37.953" ContentLicense="CC BY-SA 3.0" />
  <row Id="19160" PostTypeId="1" CreationDate="2018-01-30T06:47:21.717" Score="3" ViewCount="1123" Body="&lt;p&gt;Sorry if this is the wrong site for the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I power this household lamp off the cigarette lighter in my car? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ikea.com/us/en/catalog/products/40315673/&quot; rel=&quot;nofollow noreferrer&quot;&gt;JANSJÖ LED wall/clamp spotlight, white&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cqvdJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cqvdJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It comes with an AC adapter:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Intertek  &lt;/li&gt;&#xA;&lt;li&gt;4005088 &lt;/li&gt;&#xA;&lt;li&gt;kmv-070-030-na-2 &lt;/li&gt;&#xA;&lt;li&gt;Input 100-120V ~ 50/60Hz 0.09A&lt;/li&gt;&#xA;&lt;li&gt;Output 7V = 0.43A 3W&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this picture the bottom has the standard 2 AC wall outlet prongs. They are difficult to see, but are there. The top has the two female sockets for the detachable cord that runs to the LED lamp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yrZ1L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yrZ1L.jpg&quot; alt=&quot;https://imgur.com/a/2dPkG&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the output DC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose I could just plug it into an inverter, but was thinking there might be a DC adapter to go from the vehicle's 12-15V to the required 7V for the LED.&lt;/p&gt;&#xA;" OwnerUserId="14743" LastEditorUserId="8055" LastEditDate="2018-02-02T16:44:32.880" LastActivityDate="2018-02-02T16:44:32.880" Title="Run this LED household lamp in a car?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19162" PostTypeId="2" ParentId="19159" CreationDate="2018-01-30T06:58:39.677" Score="2" Body="&lt;p&gt;Yes it will - the space the object occupies is lighter than the fluid around it so it wants to rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same as pushing a ball to the bottom of the bath - does it stay there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: for those who say the shape of the ball makes all the difference:&#xA;try it with a plastic hollow cube (filled with air) so the cube can lay flat on the surface...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-01-30T12:05:49.080" LastActivityDate="2018-01-30T12:05:49.080" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19163" PostTypeId="2" ParentId="19156" CreationDate="2018-01-30T07:03:45.550" Score="0" Body="&lt;p&gt;Have you checked the tanks used for gases for blowing up party balloons - they are smaller : about 30cm long and 15cm dia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They may be a better option as they are probably lighter than the one you are looking at - weight must be a consideration for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have kept one for something I am planning that is why I have the size...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibilty is to check out the Nitrous tanks used in tuning cars ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, even, some of the compressed air tanks used for vehicle brakes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or possibly the smaller divers air tanks...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-01-30T07:14:02.763" LastActivityDate="2018-01-30T07:14:02.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19164" PostTypeId="1" CreationDate="2018-01-30T08:07:24.730" Score="3" ViewCount="113" Body="&lt;p&gt;I'm simulating the flow in a capillary with a &lt;strong&gt;sudden expansion&lt;/strong&gt;. The small capillary has a diamater of &lt;strong&gt;250 µm&lt;/strong&gt;. The big capillar has a diamater of &lt;strong&gt;1000 µm&lt;/strong&gt;. My flow rate ist $ 1\frac{ml}{min} $. Using standard properties of water I have a Reynolds number of 85 in my small pipe. For my mesh I choose one similar to this case: &lt;a href=&quot;https://openfoamwiki.net/index.php/Sig_Turbulence_/_Dellenback_Abrupt_Expansion&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://openfoamwiki.net/index.php/Sig_Turbulence_/_Dellenback_Abrupt_Expansion&lt;/a&gt;. I am using a low-Re mesh. I simulated it laminar first, but it seems unphysical to me, because I couldn't see any vortex, but because of the high expansion ratio one would expect a vortex after the sudden diameter change. Now I'm using k-omega-SST and getting similar results (no vortex at all). My simulation is converging well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I have to use a special turbulence model for these situation? Or am I wrong in expecting a vortex in this situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RRBdu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RRBdu.png&quot; alt=&quot;Flow Field&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/36pRm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/36pRm.png&quot; alt=&quot;Nut&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gxtpq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gxtpq.png&quot; alt=&quot;U_Centerline&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QnO2o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QnO2o.png&quot; alt=&quot;Geometry&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards,&#xA;Gesetzt&lt;/p&gt;&#xA;" OwnerUserId="14744" LastEditorUserId="14744" LastEditDate="2018-02-03T12:23:08.197" LastActivityDate="2018-02-14T13:58:24.883" Title="Turbulence model for low reynolds problem" Tags="|cfd|turbulence|meshing|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19166" PostTypeId="2" ParentId="19160" CreationDate="2018-01-30T09:55:53.487" Score="3" Body="&lt;p&gt;Yes, the output will be DC. Because LED's are (by name and nature) Diodes, they only allow current to pass through in one direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further to this, the picture you supplied shows the symbol for DC current:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/km3zr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/km3zr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are widely available DC-DC 'Buck Converters' available, which can accept a variable input voltage, and give a constant output voltage at a different level. The adaptors for charging your phone (5V) are a perfect example of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;USB car chargers are available from 'pound shops' ('dollar stores'), so if you are on a very tight budget, and you can deal with your LED being slightly less bright, you could use one of these (and solder a male USB connector onto your lamp)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need that extra brightness, then simply buy the 'correct' converter, e.g. &lt;a href=&quot;https://www.amazon.co.uk/DEOK-Synchronous-Rectification-Conversion-Efficiency/dp/B00HY3FWEK&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.co.uk/DEOK-Synchronous-Rectification-Conversion-Efficiency/dp/B00HY3FWEK&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would still get a USB car charger, however, because it will have a ready-made reliable plug-and-play connection to your 12V socket. Just take the insides out and solder your new Buck Converter to the appropriate terminals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-01-30T16:37:26.647" LastActivityDate="2018-01-30T16:37:26.647" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19167" PostTypeId="2" ParentId="19159" CreationDate="2018-01-30T11:39:20.480" Score="2" Body="&lt;p&gt;&lt;strong&gt;This question is a theoretical/academic edge-case.&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A body in the water will experience two forces:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pressure&lt;/strong&gt; acting on all surfaces in contact with water&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Gravity&lt;/strong&gt; acting on the mass of the body&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The articel on &lt;a href=&quot;https://en.wikipedia.org/wiki/Buoyancy&quot; rel=&quot;nofollow noreferrer&quot;&gt;buoyancy&lt;/a&gt; over at Wikipedia explains very good how the following equations are set-up. This article also gives the defintion of buoyancy as:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In physics, buoyancy or upthrust, is an &lt;strong&gt;upward force&lt;/strong&gt; exerted by a fluid that &lt;strong&gt;opposes the weight&lt;/strong&gt; of an immersed object.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(The reader has to decide whether a body on the ground is still immersed.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The buoyancy force, $F_\mathrm{B}$, can be calculated by integrating the stress (here: pressure), $\sigma$ across the whole surface, $A$,  of the body:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_\mathrm{B} = \oint \sigma\, \mathrm{d}A $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an immersed body on can use &lt;a href=&quot;https://en.wikipedia.org/wiki/Gauss_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gauss theorem&lt;/a&gt;. This means one can replace the area-integral with a volume-integral. However, in this edge-case the aera-integral of the body is not &quot;closed&quot;. As the can sits on the ground there is no water(pressure) at the bottom side of the can (see also the explanation over at Physics.SE &lt;a href=&quot;https://physics.stackexchange.com/questions/59866/how-does-the-buoyant-force-on-a-cube-at-the-bottom-of-a-tank-of-water-manifest-i&quot;&gt;1&lt;/a&gt;, &lt;a href=&quot;https://physics.stackexchange.com/questions/73515/&quot;&gt;2&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means for the edge-case, that the body has contact with the ground it is not possible to use the equation based on the volume-integral: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_\mathrm{B} = \rho \cdot V_\mathrm{displaced} \cdot g$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to compute the buoyancy force is to integrate the pressure-vectors on the surface of the body.&lt;br&gt;&#xA;This means for a perfect flat ground and a perfect can the aera-integral becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_\mathrm{B} = -p_\mathrm{at-top-of-can} \cdot A_\mathrm{top}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net-force (buoyancy and gravitational force) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_\mathrm{net} = -p_\mathrm{at-top-of-can} \cdot A_\mathrm{top} - m_\mathrm{can} \cdot g$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Whether $F_\mathrm{B}$ in this case should be called buoyancy needs to be discussed.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very similar effect are &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Thermal&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermals&lt;/a&gt;. When sun light wars the air on the ground its density reduces as with your object under water you have no upward (pressure-)force because there is nothing beneath the war air bubble with a higher density. You need a disturbance if this stable system, which brings some higher density fluid underneath the low density area in order to get buoyancy. The following figure &lt;a href=&quot;http://www.wetterklima.de/flug/soaring/Thermalaufbau.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt; illustrates these steps. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/HTTeN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HTTeN.png&quot; alt=&quot;picture of thermals&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2018-02-02T13:06:45.910" LastActivityDate="2018-02-02T13:06:45.910" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19168" PostTypeId="1" CreationDate="2018-01-30T12:01:27.593" Score="1" ViewCount="912" Body="&lt;p&gt;I have been trying to model a complex pipe network recently and have come across something I can't find any information on. In my model, I have been assigning equivalent lengths of pipe to various fittings, but I commonly encounter fittings such as this: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Inlet to T: 10mm diameter pipe&#xA;t-through: 10mm diameter pipe&#xA;t-branch: 1mm diameter pipe &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s6GPp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s6GPp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially I have a T junction where the branch is also a sudden contraction, but the dominant flow direction is perpendicular to the branch. &lt;strong&gt;Does anybody know a method of calculating equivalent length of pipe in a case like this?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think a regular vena contracta, as described by normal theory for sudden contractions, would occur in the 1mm branch, since it is perpendicular to the main flow and not coaxial with the main 10mm pipe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have various instances where this occurs, and the amount of the contraction varies (e.g. the branch pipes can be 1,2 or 5mm) but the normal method of equivalent pipe length for a t-junction does not account for the changing area. I need something that will account for it. &lt;/p&gt;&#xA;" OwnerUserId="14498" LastEditorUserId="14498" LastEditDate="2018-01-30T12:16:48.720" LastActivityDate="2018-01-30T12:16:48.720" Title="Flow in a T-Junction with pipes of different diameters" Tags="|mechanical-engineering|fluid-mechanics|pipelines|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19169" PostTypeId="1" AcceptedAnswerId="19182" CreationDate="2018-01-30T12:02:18.263" Score="4" ViewCount="85" Body="&lt;p&gt;A contractor delivered three components, all are supposed to be in 1.4571. Part A is actually done in &lt;a href=&quot;http://www.m-woite.de/en/materials/14404.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;1.4404 / (X2CrNiMo17-12-2)&lt;/a&gt; (according to pretty conclusive documentation). Part B is allegedly done in &lt;a href=&quot;http://www.m-woite.de/en/materials/14571.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;1.4571 / (X6CrNiMoTi17-12-2)&lt;/a&gt;, but the material looks like part A (shiny, silvery). Part C meanwhile is also allegedly done in 1.4571, but looks different (silvery gray). &quot;Allegedly&quot; means that for the documentation for parts B and C is far less conclusive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, from someone who has encountered both steels: Would I expect them to look the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, to rephrase my question: Part A and C look the same but, according to the documentation provided, should be different materials - does this tell me something is amiss?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2018-01-30T13:15:44.203" LastActivityDate="2018-01-31T02:36:28.557" Title="Do the steels 1.4404 and 1.4571 look the same?" Tags="|mechanical-engineering|materials|steel|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19170" PostTypeId="1" CreationDate="2018-01-30T15:48:48.160" Score="2" ViewCount="302" Body="&lt;p&gt;I am a first year engineering student and I am currently struggling with the dynamics prt of my course. It’s not that I don’t understand the theory behind it’s just that every problem I tackle seems to be different from the last one and brings with it different tricks and methods that need to be used. Any advice on how to get good at it?&lt;/p&gt;&#xA;" OwnerUserId="14746" LastActivityDate="2018-01-30T22:20:51.943" Title="How to get good at dynamics" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2018-01-31T00:37:09.773" ContentLicense="CC BY-SA 3.0" />
  <row Id="19171" PostTypeId="2" ParentId="19159" CreationDate="2018-01-30T16:02:37.500" Score="2" Body="&lt;p&gt;I'm not sure why SolarMike deleted his response.  The only thing holding the can to the ground (&quot;proud&quot; in naval terms) is the vacuum force, i.e. the same pressure that keeps you from lifting the can off a table if there's a perfect seal to the table.&lt;br&gt;&#xA; So long as the can's density is less than that of the surrounding fluid it'll experience a buoyant force.  Don't confuse an existing &lt;strong&gt;force&lt;/strong&gt; with the &lt;strong&gt;net force&lt;/strong&gt; .  Once there is a channel to allow water to flow under the can, the delta pressure of the water with depth will cause the can to rise to the surface.  (It's a matter of the &lt;strong&gt;pressure&lt;/strong&gt; vs. depth, not the density).  As shown on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Buoyancy&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt;, the pressure at the bottom of the can (water pressure) is greater than that at the top of the can, thus forcing the can to rise.  This pressure differential exists even when the can is proud; it's just the &lt;em&gt;lack&lt;/em&gt; of pressure that would result if a vacuum were to form there that holds the can to the ground.   So, in sum, the can always sees a buoyant force.  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-01-30T16:02:37.500" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="19172" PostTypeId="2" ParentId="19159" CreationDate="2018-01-30T16:06:23.317" Score="3" Body="&lt;p&gt;Yes your can will still have buoyancy when it is submerged to the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of the submersion depth, any object will lose weight equal to the weight of water it has displaced, even when held at the bottom. You confuse the hydrostatic pressure with buoyancy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrostatic pressure will increase with depth, to a point that it may even crush the can in. But the buoyancy exerted by water on the can stays more or less the same, because water is almost incompressible, so its density is more or less the same at shallow and deep water. Hence the displaced water will weight the same on the bottom and buoyancy it causes would be the same. &lt;/p&gt;&#xA;&#xA;&#xA;" OwnerUserId="4816" LastEditorDisplayName="user6335" LastEditDate="2018-01-31T00:36:32.913" LastActivityDate="2018-01-31T00:36:32.913" CommentCount="15" ContentLicense="CC BY-SA 3.0" />
  <row Id="19173" PostTypeId="2" ParentId="19156" CreationDate="2018-01-30T16:13:32.597" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;https://gashosesandregulators.com/propaneregulatorfacts.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;one of many resources&lt;/a&gt; regarding pressure in propane tanks, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a tank with 20 pounds of gas at 70°F has a pressure of 145 psi, at 90°F would have 180 psi, at 105°F would have 235 psi, and at 130°F would have 315 psi.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Those numbers convert to 21°C / 10 bar, 32.2°C / 12.4 bar, 40.5°C / 16.2 bar and 54.4° / 21.7 bar respectively. Quite substantial numbers for &quot;ordinary&quot; use as an air storage tank. My commercially produced tank of approximately the same size has a red-zone on the gauge at the 150 psi level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One aspect of using a propane tank is the inherent danger of the fuel within. Exhaust the tank through ordinary use or similarly safe manner. Remove the valve and fill the tank with water. This will cause any remaining fuel vapors to be displaced, rendering the tank safe(r).&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-01-30T16:13:32.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19174" PostTypeId="1" AcceptedAnswerId="19186" CreationDate="2018-01-30T16:24:56.690" Score="2" ViewCount="3560" Body="&lt;p&gt;I am looking for a way to double check readings I am getting through a mass flow meter that appear to be reading lower than expected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once flow starts in the system, the pressure regulator at the supply tank drops in pressure readout. This drop should be due to the dynamic pressure of the flow of gas through the regulator so I was thinking it may be possible to correlate the mass flow meter's readings with the calculated mass flow resulting in the drop in pressure at the regulator. I started by using the ideal gas law (low enough pressures) to calculate the density of the gas at the static pressure  read by the regulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho=\frac{PM}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I took the pressure difference read from the tank regulator between the static no-flow to flowing as the dynamic pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P_d=\frac{1}{2}\rho v^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above equation I calculated the velocity of the fluid flow, and then just used the cross sectional area to calculate the mass flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{m}=\rho A_c v$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When calculating the density I am getting an unreasonable value for fluid velocity so I was wondering if this procedure makes sense?&lt;/p&gt;&#xA;" OwnerUserId="14676" LastEditorUserId="14676" LastEditDate="2018-01-30T18:43:19.267" LastActivityDate="2018-02-02T21:52:39.487" Title="Using pressure drop to correlate mass flow" Tags="|fluid-mechanics|compressible-flow|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19175" PostTypeId="2" ParentId="15608" CreationDate="2018-01-30T17:48:35.063" Score="0" Body="&lt;p&gt;No real difference between a shim and a spacer.  A shim is used to adjust the space/clearance between faying surfaces (where two pieces connect). Generally as a means of getting away from the cost of tight tolerances.  Sometimes as a wear part that gets replaced to hold a particular clearance (e.g. the selectable shims that were used to adjust the clearance in my overhead cam on my old motorcycle).  A washer is used to distribute the load from the head of a bolt or screw (or to keep your faucet from leaking).  Also may perform a locking function.  Nothing to keep you from using a washer, or stack of washers, as a shim - to adjust a clearance.  I'd label the part by function.  If you're using it to adjust a clearance call it a spacer.  If you're using it under the head of a screw or bolt, call it a washer.  In my opinion the function of the part in its application is more useful and relevant than an arbitrary name.&lt;/p&gt;&#xA;" OwnerUserId="14749" LastActivityDate="2018-01-30T17:48:35.063" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19176" PostTypeId="2" ParentId="19170" CreationDate="2018-01-30T22:20:51.943" Score="0" Body="&lt;p&gt;I'm in my final year as automotive engineering student, and I have had the same problem, mostly because of the bedrock level teaching quality of my teachers, and because I found my books' explaining vague, abstract, and needlessy complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I resorted to WikiPedia and forums/sites for theory and formula's(still needlessly complicated sometimes), but I benefited most by watching videos. For me, seeing a visual example of the theory in a video, helps me extremely well to understand it. Maybe the same holds true for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So i'd recommend following a YouTube channel that explains the theory that you have to know. There are many of them in various languages. You have a lot of teachers uploading courses, and some of them can explain very well. There are also channels that visualise the theory, and I recommend that to get a good grasp of the concept. Theory and practice are inherent to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be a (good) engineer, you need to know both the theory, and how it works in practice. If you miss one of them, learning will be harder and boring, and you won't understand what you're doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you start really understanding the theory by visualising it, you'll likely get motivated to learn more about it, and it will also make further learning much easier. Just my $0.02 as a fellow student.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-30T22:20:51.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19177" PostTypeId="2" ParentId="19150" CreationDate="2018-01-30T22:36:05.033" Score="0" Body="&lt;p&gt;In theory, the force is just the torque, divided by the radius.&#xA;In practice, wrapping the rope a few times around the pulley will rob a little torque from it, but it may even be unnoticeable. At the same time, friction between rope and pulley will increase, thus allowing you to transfer more force to the rope before it starts slipping over the pulley. It also helps to tension the rope to reduce slipping(much like a derailleur on mountainbikes), but again it'll cost you a little torque. So only consider it if the rope keeps slipping.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-30T22:36:05.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19178" PostTypeId="2" ParentId="19149" CreationDate="2018-01-30T22:46:36.440" Score="1" Body="&lt;p&gt;I recommend researching a few different plane designs in terms of propeller(not fan) layout, and you'll most likely have your answer. AFAIK, a prop at the back is more efficient, a prop at the front makes the plane more stable, since airflow is directed over the wings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have to consider engine location and weight distribution. It's extremely important for aerial vehicles. I wouldn't advice you to reinvent the wheel, learning from existing plane models is most likely giving you more chance of success. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, i'm satisfied with the flow simulation in SolidWorks. Depending on your situation, you may be able to obtain a student license for it.&#xA;I also recommend asking at the Aviation section of this forum. They'll most likely have treated questions like yours before.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-01-30T22:46:36.440" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19179" PostTypeId="1" CreationDate="2018-01-31T00:20:09.197" Score="1" ViewCount="41" Body="&lt;p&gt;I have this variable voltage battery eliminator and the &quot;switch&quot; to hold the wire in place is broken. Is there any specific name for this &quot;switch&quot; and can I buy it somewhere to replace it or can you suggest some other way to overcome this please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8qpds.jpg&quot; alt=&quot;Back of the battery eliminator&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/aosC8.jpg&quot; alt=&quot;The broken part along with it&amp;#39;s spring&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14751" LastEditorUserId="110" LastEditDate="2018-01-31T02:46:54.490" LastActivityDate="2018-01-31T07:40:17.993" Title="Switch at the back of a battery eliminator" Tags="|mechanical-engineering|mechanisms|mechanical-failure|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19180" PostTypeId="1" AcceptedAnswerId="19191" CreationDate="2018-01-31T00:37:42.570" Score="2" ViewCount="1559" Body="&lt;p&gt;In order to build a lighting system for a microscope (with a digital camera) I want to make a very small diameter laser beam (parallel, rather than convergent). If possible around 1 um diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I build such a laser using a common diode laser? &lt;/p&gt;&#xA;" OwnerUserId="5926" LastEditorUserId="5926" LastEditDate="2018-02-01T06:39:28.767" LastActivityDate="2018-02-01T06:39:28.767" Title="How can I build a micrometer diameter laser beam from a common laser diode?" Tags="|optics|lasers|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19181" PostTypeId="2" ParentId="19179" CreationDate="2018-01-31T01:49:29.597" Score="5" Body="&lt;p&gt;It's difficult to be certain, but I suspect what you have is commonly known as a spring speaker clip. Using those terms in your search engine to find one that has listing for maximum and voltage. My search, done in the US, resulted in primarily US sources, &lt;a href=&quot;http://www.techtoolsupply.com/ProductDetails.asp?ProductCode=IC107BSSBK&quot; rel=&quot;noreferrer&quot;&gt;one of which&lt;/a&gt; lists a set that can handle 300 volts and 15 amps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n1utb.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n1utb.jpg&quot; alt=&quot;speaker spring clip&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-01-31T01:49:29.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19182" PostTypeId="2" ParentId="19169" CreationDate="2018-01-31T02:36:28.557" Score="3" Body="&lt;p&gt;Not familiar with Russian specs but these look like ( in USA names) 316 L and 316 ; If so , you could not tell them apart by appearance. Also , if they have similar composition tolerances to USA specs you may not be able to separate them with chemical analyses ; 316 may have identical composition to 316 L. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-01-31T02:36:28.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19184" PostTypeId="2" ParentId="19179" CreationDate="2018-01-31T07:40:17.993" Score="-1" Body="&lt;p&gt;If you don’t need it to stay original, then you could use screw- block connectors aka chocolate block connectors of a sufficient size for the wires necessary &amp;amp; current necessary.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-01-31T07:40:17.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19185" PostTypeId="1" AcceptedAnswerId="19192" CreationDate="2018-01-31T08:31:33.413" Score="3" ViewCount="202" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VbOaj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to have a star dot above each bar chart. I don't know how to present that index. Please anyone help me to resolve this problem,&lt;/p&gt;&#xA;" OwnerUserId="14641" LastEditorUserId="110" LastEditDate="2018-01-31T11:48:04.450" LastActivityDate="2018-01-31T15:41:39.520" Title="How do I include statistical significant index in bar chart in excel?" Tags="|statistics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19186" PostTypeId="2" ParentId="19174" CreationDate="2018-01-31T09:58:42.247" Score="1" Body="&lt;p&gt;It is not entirely clear from the question, where the instrument readings take place. The following equations might need to be adapted in order to reflect the sensor positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic equation for the mass-flow $\dot{m}$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot{m} = A \cdot \rho \cdot v$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$A$, $\rho$, and $v$ need to be evaluated or measured at the same location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Ideal_gas_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;ideal gas law&lt;/a&gt; can be used here but be aware the static values for temperature $T$ and pressure $p$ have to be used here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho = \frac{p_\mathrm{s}}{RT_\mathrm{s}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Gas_constant#Specific_gas_constant&quot; rel=&quot;nofollow noreferrer&quot;&gt;specific gas constant&lt;/a&gt; for air is something around $287.058\frac{J}{kgK}$. It changes with temperature but this does not seem to be the problem right now.&lt;br&gt;&#xA;Usually the problem is now to compute $T_\mathrm{s}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following equations use an isentropic process to compute first the Mach-Number and then the static Temperature. Step 1 is to solve the equation for $M$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{p_\mathrm{s}}{p_\mathrm{t}} = \left( 1 + \frac{\gamma - 1}{2} M^2\right)^{\frac{\gamma}{\gamma-1}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $\gamma$ is the &lt;em&gt;isentropic coefficient&lt;/em&gt; which is around $1.4$ for air depending on the humidity. $p_\mathrm{t}$ and $p_\mathrm{s}$ are the static and total pressure. Their relation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_\mathrm{t} = p_\mathrm{s} + p_\mathrm{d}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_\mathrm{d}$ is often calculated with $\frac{1}{2}\rho u^2$ this simplification creates an error in the density $\rho$ of 5% for a Mach-Number $M=0.3$. So it is possible with the result of step 1 to check if this was the source of the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the calculation of the Mach-Number the isentropic relation of total and static temperature can be used to calculate the static temperature in step 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{T_\mathrm{s}}{T_\mathrm{t}} = \left( 1 + \frac{\gamma - 1}{2} M^2\right)^{-1}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the total temperature $T_\mathrm{t}$ is the temperature of the gas in the pressure supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final step 3 is the calculation of the velocity $v$. This will be done using the equation for the Mach-Number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$M = \frac{u}{a}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the only unknown is $a$ (the speed of sound) which can be computed using the following relation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a = \sqrt{\gamma R T_\mathrm{s}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing these calculations by hand is laborious but it's easy to set up a script or sheet to do these calculations. There are three constants:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Through flow area, $A$&lt;/li&gt;&#xA;&lt;li&gt;Specific gas constant, $R$&lt;/li&gt;&#xA;&lt;li&gt;Isentropic coefficient, $\gamma$ (sometimes its $\kappa$)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And there are three measurement values:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Total pressure, $p_\mathrm{t}$&lt;/li&gt;&#xA;&lt;li&gt;Static pressure, $p_\mathrm{s}$&lt;/li&gt;&#xA;&lt;li&gt;Total temperature, $T_\mathrm{t}$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2018-01-31T09:58:42.247" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19188" PostTypeId="2" ParentId="19180" CreationDate="2018-01-31T11:41:57.370" Score="2" Body="&lt;p&gt;If you have one convex and one concave lens, you can effectively make the laser beam smaller and more concentrated while still having a parallel ray. I believe it's called a galilean lens. It would take some effort to place them accurately that you have a width of only 1um though. If it's concentrated enough (which it &lt;strong&gt;certainly&lt;/strong&gt; is if converged to 1um) it can easily &lt;em&gt;burn your skin&lt;/em&gt; and much more easily &lt;em&gt;your eyes&lt;/em&gt;. Your eyes don't repair themselves when damaged more than the slightest bit. Take care when working with lasers.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="110" LastEditDate="2018-01-31T13:26:50.763" LastActivityDate="2018-01-31T13:26:50.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19189" PostTypeId="1" CreationDate="2018-01-31T13:30:07.320" Score="1" ViewCount="44" Body="&lt;p&gt;I am trying to specify a motor, but I don't know where to start. I need it to be able to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lift a weight of $200kg$,&#xA;through a height of $2m$,&#xA;Using a lead-screw with the following dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$OD = 26mm\\ Pitch = 5mm\\ Core =21mm\\ Mean\ Diameter=23.5mm$$ &lt;/p&gt;&#xA;" OwnerUserId="14757" LastEditorUserId="13311" LastEditDate="2018-01-31T15:09:55.720" LastActivityDate="2018-02-01T00:50:51.890" Title="Can you suggest a motor that would be suitable for these parameters?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="1" ClosedDate="2018-02-01T14:08:01.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="19191" PostTypeId="2" ParentId="19180" CreationDate="2018-01-31T15:04:09.260" Score="7" Body="&lt;p&gt;&lt;strong&gt;Why this is hard&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One micron is on the border of what is theoretically possible. You cannot create a parallel ray at this scale. One micron is almost as small as the wavelength of the light you are trying to focus, so the usual rules of optics don't apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lasers normally produce what is known as a Gaussian beam, and they have a distinctive shape.  You can focus them down to a small dot, but only if you are willing to let the beam get a lot wider before and after the spot. Looking side on at the beam, you'll get something like this:&lt;a href=&quot;https://i.stack.imgur.com/CSrT4.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CSrT4.png&quot; alt=&quot;A side-on drawing of a gaussian beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;(image from &lt;a href=&quot;https://en.wikipedia.org/wiki/Gaussian_beam&quot; rel=&quot;noreferrer&quot;&gt;wikipedia&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point where the beam is at it's narrowest is known as the &quot;waist&quot;. At that point the beam has a diameter of $W_0$.  At a distance of $Z_R$ from the waist, in either direction, the beam has grown to twice the area, or $\sqrt2$ times the diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation which links $W_0$ and $Z_R$ is:&#xA;$$Z_R = \frac{\pi W_0^2}{\lambda} $$&#xA;Where $\lambda$ is the wavelength of the laser you are using.  If we assume a 1000nm cheapish laser diode, and a 1um waist, then $Z_R$ is 3.1um.  If you go to an expensive blue laser, you can get $Z_R$ up to 7.7um.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How to do it&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to bring the laser beam into a convergent lens, and place the sample at the focus. The convergent lens must be very high quality, and it must be held and oriented with micron accuracy. Realistically, you probably won't be able to obtain a good enough single lens, so you will need an assembly of lenses. A microscope objective would probably work, but you won't be able to get a second one close enough to the sample.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect the best option would be to buy a microscope with this feature built in. That also has the added advantage that it will have all the safeguards necessary to make sure the beam never makes it to the eyepieces.&lt;/p&gt;&#xA;" OwnerUserId="12022" LastActivityDate="2018-01-31T15:04:09.260" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19192" PostTypeId="2" ParentId="19185" CreationDate="2018-01-31T15:41:39.520" Score="3" Body="&lt;p&gt;Quite seriously:   &lt;em&gt;Friend Don't Let Friends Use Excel&lt;/em&gt;.   Learn to use Matlab/Octave, or python,  or R,  or just plain Gnuplot.&lt;br&gt;&#xA;No matter what tool you use, since &quot;significance&quot; is a separate variable, you need to convert your collection of &quot;*&quot; , &quot;**&quot;  etc to numeric values and plot the desired symbols at the matching category locations. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-01-31T15:41:39.520" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19193" PostTypeId="1" CreationDate="2018-01-31T15:44:22.047" Score="2" ViewCount="122" Body="&lt;p&gt;I want to build my own 4 point bending tester. First, I need to see what parts are available for sale. Can anyone tell me if the circled L-shaped part has a name I can use to search with, and if so what? Alternately, any suggestions as to parts I could use in a similar fashion would be useful. &lt;a href=&quot;https://i.stack.imgur.com/zGVF2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zGVF2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2018-02-01T23:49:05.970" Title="How to describe or call this part?" Tags="|mechanical-engineering|measurements|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19194" PostTypeId="2" ParentId="19180" CreationDate="2018-01-31T15:49:40.290" Score="1" Body="&lt;p&gt;If you want a small illumination area on the object plane,  start by focussing your source onto a 1-micron pinhole, then relay that pinhole (a single biconvex lens will suffice) to the object plane.  Be aware that:&lt;br&gt;&#xA;1) rather little light will make it through the pinhole&lt;br&gt;&#xA;2) due to diffraction effects, you will get a center spot and surrounding rings of illumination.  As Jack B pointed out in his answer (for a parallel beam), you can't do better than this. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-01-31T15:49:40.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19195" PostTypeId="2" ParentId="17831" CreationDate="2018-01-31T16:07:22.373" Score="1" Body="&lt;p&gt;There are multiple effects how net-frequency impacts the GT-efficiency. But it also depends on the GT-architecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following effects also impact the efficiency even if the 50 and 60Hz version would get their unique design in order to operate them at their design-point (at peak performance)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aero-derivatives for 50 and 60Hz usually use the same „gas-generator“ but differ in the power-turbine. The power-turbine is not on the same shaft/rotor as the gas-generator. This means the 60Hz-version will spin faster. Without going into the aerodynamic details the efficiency of a turbine is higher with a higher speed. Besides this you will also see that the 60Hz version has a higher power-output as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Industrial GTs (for higher power-output) are designed so that generator and GT are on the same rotor/shaft. In order to use as many designs in a 50 and 60Hz version the manufacturers usually ‘scale’ the GTs which means the Mach-Number is kept constant. This means that a 50Hz version of a given Model will have a higher power-output and will also have a slightly higher efficiency. This is because bigger engines have relatively smaller tolerances compared to smaller (60Hz) engines.&lt;/p&gt;&#xA;" OwnerUserId="1339" LastActivityDate="2018-01-31T16:07:22.373" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19196" PostTypeId="1" CreationDate="2018-01-31T19:46:18.577" Score="2" ViewCount="1002" Body="&lt;p&gt;Imagine a house/building, seen from outside, at sunrise. As the the sun moves in the horizon, a) the shadow of the house is displaced/deformed, and b) some faces of the house have more/less sunlight in the afternoon than in the morning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to predict those changes along the day and along the year. Besides the external 3D shape of the house, this only depends on space &amp;amp; time. Space: the latitude, longitude and orientation of the house/building. Time: the hour, day, month of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the exterior measurements of the house. At this point I treat the house as a solid object (ie I don't care about the light inside the house; I also don't care about cases where light which passes thorough the building eg two opposite windows).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only need to predict two interrelated things (along the day/year) for planning and visualization purposes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The shadow of the house&lt;/li&gt;&#xA;&lt;li&gt;The faces of the house which will be receiving light&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With (1) I can predict how the adjacent buildings/parks/spaces would be affected by sunlight decrease and with (2) I can plan to put more/less and/or smaller/larger windows in different places of each wall of the house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that many years ago this was possible with a plugin for Google Sketchup, but for different reasons I would like not to use this option but a free application (if FOSS, even better). Given the recent surge of 3D modeling/design programs I thought maybe there are other programs which are capable of doing this.&lt;/p&gt;&#xA;" OwnerUserId="14759" LastEditorUserId="8055" LastEditDate="2018-02-01T01:13:18.593" LastActivityDate="2018-02-01T05:53:09.263" Title="How to predict shadows of a house (3D modeling + sunlight)?" Tags="|building-design|lighting|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19197" PostTypeId="2" ParentId="19196" CreationDate="2018-01-31T21:52:53.830" Score="1" Body="&lt;p&gt;Autodesk 3D Studio used to do this when I was rendering with it around years 2000 to 2005. It is not free though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you could define one or many light sources and one or several cameras and then animate your model around a path and it would calculate the shadows and reflections and even you could define opacity, translucency, surface mesh any building material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no doubt that it is more capable now and has many more features! But they may call it a different name now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the tracing of your earth orbit and daytime and sun angles there are many formulas that can be easily manipulated to be able to use them as input for the light source path, or you can get the stepwise coordinates in 'MS Excel' and plot them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How ever for a task as simple as what you need you can approximate you angles by 4 intervals a day and four seasons!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the design can not be sensitive to any level of change more than that! &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-01-31T21:52:53.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19199" PostTypeId="1" AcceptedAnswerId="19200" CreationDate="2018-01-31T22:50:57.180" Score="3" ViewCount="92" Body="&lt;p&gt;If trying to achieve a 1:4 gear ratio, will a planetary gear system help reduce the loss of torque? Or is it essentially no different than the loss of torque when using a large drive gear to rotate a small driven gear?&lt;/p&gt;&#xA;" OwnerUserId="7869" LastActivityDate="2018-02-01T00:39:08.297" Title="Gear acceleration and maintaining torque" Tags="|mechanical-engineering|gears|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19200" PostTypeId="2" ParentId="19199" CreationDate="2018-02-01T00:34:04.857" Score="5" Body="&lt;p&gt;Apart from frictional losses, which should be fairly low for a 1:4 ratio, the torque/speed ratio is inescapable ie if you increase the angular speed by a factor of 4 the torque must be reduced by a factor of 4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power is torque x angular velocity so if this was not the case you would get more power out than you put in, which as we know is not possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planetary gears can be used to produce quite high ratios in a compact package and also have the ability to change ratios by locking different combinations of input/output/carrier eg in automatic transmissions but their fundamental principal of operation is the same as any other type of gear or indeed any type of mechanical advantage. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-02-01T00:39:08.297" LastActivityDate="2018-02-01T00:39:08.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19202" PostTypeId="2" ParentId="19193" CreationDate="2018-02-01T03:26:14.217" Score="0" Body="&lt;p&gt;These parts are purpose built and you would be very unlikely to find these other than as replacement parts for an item like this one (and likely very expensive for what they are).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were going to build such an instrument, you would want access to a machine shop (or great patience with hand tools) and the metalworking skills to go with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, most of these parts look pretty simple, at least the ones I can make out from the picture, and most of them could be fashioned from readily available bar stock and assembled with standard hardware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might consider adapting the design to use commonly-available bar stock, like aluminum extrusions, or perhaps fabricating some of the components from individual pieces of bar stock and fastening them together with screws (like a separate base and upright composing the circled part). Then the parts could be perhaps be fashioned using only a saw and a file (and patience :).&lt;/p&gt;&#xA;" OwnerUserId="3937" LastActivityDate="2018-02-01T03:26:14.217" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19203" PostTypeId="2" ParentId="19196" CreationDate="2018-02-01T05:35:56.193" Score="0" Body="&lt;p&gt;All raytracers can do these, any system supporting depthmaps can do these. Most systems 3D can even do it analytically for each timestep. Many 3D applications have a dedicated sky simulation pakage so no need to even calculate the sun path jus give coordinates and time. They will even gladly simulate the contribution of ambient sky if you want (even heat loss/input).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So wether you use 3DS max, Maya, Sketchup, Rhino or blender matters little. Hell even most CAD packages meant for mechanical engineering can do this without any problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you want free use blender, using something like &lt;a href=&quot;https://wiki.blender.org/index.php/Extensions:2.6/Py/Scripts/3D_interaction/Sun_Position&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;. Though to be honest even your browser is capable of doing this.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-02-01T05:53:09.263" LastActivityDate="2018-02-01T05:53:09.263" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19204" PostTypeId="1" CreationDate="2018-02-01T08:14:07.027" Score="1" ViewCount="305" Body="&lt;p&gt;I have a question that maybe will be very easy for you. &#xA;I am working on a district heating model using Dymola, and now I have to incorporate a boiler into my model, such as &lt;a href=&quot;https://simulationresearch.lbl.gov/modelica/releases/latest/help/Buildings_Fluid_Types.html#Buildings.Fluid.Types.EfficiencyCurves&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was asked to present my results in the form of &quot;energy curve consumption of the boiler&quot; but I do not know how to get it, I have been reading about, but the concept isn't fully in my mind, can anyone help a little here?&lt;/p&gt;&#xA;" OwnerUserId="14565" LastEditorUserId="1832" LastEditDate="2018-02-01T09:35:32.403" LastActivityDate="2018-09-06T01:14:44.897" Title="Energy consumption of a boiler" Tags="|thermodynamics|modeling|energy-efficiency|boilers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19205" PostTypeId="1" CreationDate="2018-02-01T11:55:19.593" Score="6" ViewCount="229" Body="&lt;p&gt;I would like to have a 3d simulation of a helicopter landing on a ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about using matlab for this, as a far as I know you can connect models from Simulink to virtual worlds using blocks, which is the main feature of the Simulink 3D Animation toolkit (&lt;a href=&quot;https://es.mathworks.com/products/3d-animation/features.html&quot; rel=&quot;noreferrer&quot;&gt;https://es.mathworks.com/products/3d-animation/features.html&lt;/a&gt;). You can even edit 3D objects, or you can import it from CAD packages. My main question is the following, where do I define the dynamics equations of the helicopter? I haven't really worked with CAD software neither animation, so I don't know how to do this. I'm not even sure if this is the right way to do it but it should look like this.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BEFb1.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BEFb1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model of the helicopter is simple. The controller is feed (SENSORS) by the position and velocities of the center of mass of the helicopter. The controller outputs (ACTUATORS) are the main thrust, the tail thrust and the tilt angles. These four are the inputs to the helicopter. The reference is the vector position 0 which is the landing point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I said my main problem is the animation. If it wasn't for this, I could just create a model of the helicopter in simulink blocks and another model of the controller in simulink, connect them together, play it and see the plots of the desired tracking variables. But I want to do a 3D animation of the tracking of the reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attach a photo of the helicopter dynamics to make myself clear. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/QJrIx.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QJrIx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14768" LastActivityDate="2023-12-24T03:05:25.590" Title="Need some advice, control system animation. Appreciate some help!" Tags="|mechanical-engineering|electrical-engineering|control-engineering|aerospace-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19206" PostTypeId="1" AcceptedAnswerId="19209" CreationDate="2018-02-01T13:41:04.173" Score="4" ViewCount="12972" Body="&lt;p&gt;In theory of bending of beams, often these terms are used. There are no clear definition present. Most of them contradicting. Which of the property(bending moment and shear force) is zero in which case? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition of &lt;a href=&quot;https://en.wikipedia.org/wiki/Contraflexure&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;contraflexure&lt;/code&gt;&lt;/a&gt; is defined as the point of zero bending moment. Then why have another term, i.e &lt;code&gt;inflection point&lt;/code&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;NOTE:The wikipedia link has no references.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14771" LastEditorUserId="90" LastEditDate="2019-01-31T13:27:39.613" LastActivityDate="2019-02-01T19:04:02.593" Title="Are the points of inflection and contraflexure the same?" Tags="|mechanical-engineering|structural-engineering|applied-mechanics|beam|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="19208" PostTypeId="1" CreationDate="2018-02-01T14:16:48.493" Score="2" ViewCount="9321" Body="&lt;p&gt;When looking at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lithium-ion_battery&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; for Lithium-ion batteries, one can see that there are some technical details such as Specific Energy and Specific Power, but I don't know how to connect them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one can see this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Specific Energy: 100–265 Wh/kg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Specific Power: 250 - 340 W/kg&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to the theory, power equals energy divided by time; i.e. 1 W = 1 Wh/t. So can guess that t is the discharge time. Li-ion batteries usually have a discharge rate of 1 C, which means that the battery would be discharged in around one hour. Looking at those numbers, it doesn't look like my logic is correct, therefore I would like to know if this is the case or if those numbers don't really match? &lt;/p&gt;&#xA;" OwnerUserId="8419" LastActivityDate="2018-02-01T19:05:48.253" Title="Specific energy vs Specific power in batteries" Tags="|power|energy|battery|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19209" PostTypeId="2" ParentId="19206" CreationDate="2018-02-01T14:31:50.913" Score="5" Body="&lt;p&gt;A point of inflection is defined as the point where a function changes from convex to concave or vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a function &lt;span class=&quot;math-container&quot;&gt;$f(x)$&lt;/span&gt;, this is frequently mathematically defined as the point where &lt;span class=&quot;math-container&quot;&gt;$f''(x) = 0$&lt;/span&gt;, since &lt;span class=&quot;math-container&quot;&gt;$f''(x)$&lt;/span&gt; describes &lt;span class=&quot;math-container&quot;&gt;$f(x)$&lt;/span&gt;'s curvature&lt;sup&gt;[1]&lt;/sup&gt;. Being more pedantic about it, this is a necessary but not sufficient condition: a function might have no curvature &lt;span class=&quot;math-container&quot;&gt;$\left(f''(x) = 0\right)$&lt;/span&gt; but not actually change between convex and concave (this is strictly called an undulation point). That being said, in common parlance any point which satisfies &lt;span class=&quot;math-container&quot;&gt;$f''(x) = 0$&lt;/span&gt; is considered an inflection point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above has nothing to do with bending moment, it is a characteristic of any mathematical function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contraflexure, on the other hand, is as you've said: the point where a beam suffers no bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, here's the bending moment of a cantilever beam with a linear load which generates a point of contraflexure and a point of inflection (notice that to the left of the point, the curve has negative curvature, and to the right of the point, it has positive curvature). The point of contraflexure is easily identifiable on a diagram (it's where the bending moment diagram equals zero). The point of inflection is harder to identify precisely; you can quickly tell that there is a point of inflection and that is it &quot;around here&quot;, but you can't easily point to its exact location (without doing the math, I can tell it's somewhere in the circled region).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cSm2i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cSm2i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when speaking of beams, it is possible to state that the point of contraflexure is the same as the point of inflection. &lt;strong&gt;&lt;em&gt;However&lt;/em&gt;&lt;/strong&gt;, one will actually be talking about two different things at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As previously stated, the point of contraflexure is that point where bending moment is equal to zero. However, according to the fundamental beam equation, we know that bending moment is the second derivative of deflection (times the stiffness). So, since the point of inflection is that point where the second derivative of a function is equal to zero&lt;sup&gt;[2]&lt;/sup&gt;, the point of contraflexure is equal to the &lt;strong&gt;&lt;em&gt;deflection's&lt;/em&gt;&lt;/strong&gt; inflection point.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;[1]&lt;/sup&gt; As mentioned by &lt;a href=&quot;https://engineering.stackexchange.com/a/25745/1832&quot;&gt;@Mohan's answer&lt;/a&gt;, there is a possible exception to this rule: continuous functions &lt;span class=&quot;math-container&quot;&gt;$f(x)$&lt;/span&gt; with continuous &lt;span class=&quot;math-container&quot;&gt;$f'(x)$&lt;/span&gt; but discontinuous &lt;span class=&quot;math-container&quot;&gt;$f''(x)$&lt;/span&gt;. If &lt;span class=&quot;math-container&quot;&gt;$f''(x)$&lt;/span&gt; is discontinuous at &lt;span class=&quot;math-container&quot;&gt;$x=a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\text{sign}\left(\lim\limits_{x\rightarrow a^+}f''(x)\right) \ne \text{sign}\left(\lim\limits_{x\rightarrow a^-}f''(x)\right)$&lt;/span&gt; (in English: the sign of &lt;span class=&quot;math-container&quot;&gt;$f''(x)$&lt;/span&gt; to the left of the discontinuity is different than to its right), then &lt;span class=&quot;math-container&quot;&gt;$x=a$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$f(x)$&lt;/span&gt;'s inflection point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This only works in cases where &lt;span class=&quot;math-container&quot;&gt;$f'(x)$&lt;/span&gt; is continuous. This is because for a point to serve as an inflection point for &lt;span class=&quot;math-container&quot;&gt;$f(x)$&lt;/span&gt;, there must be a single valid tangent at that point. So if the derivative is continuous, that condition is met even if &lt;span class=&quot;math-container&quot;&gt;$f''(x)$&lt;/span&gt; (the curvature) is discontinuous).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;[2]&lt;/sup&gt; The exception described above in [1] applies to beams. If a beam suffers a concentrated bending moment, that leads to a discontinuous bending (or curvature) diagram, but the deflection's derivative (tangent angle diagram) will still be continuous. So if that concentrated bending moment leads to a change in the bending diagram's sign at that point, then that will be an inflection point in the beam's deflection.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-02-01T19:04:02.593" LastActivityDate="2019-02-01T19:04:02.593" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19210" PostTypeId="2" ParentId="19164" CreationDate="2018-02-01T17:16:56.937" Score="0" Body="&lt;p&gt;It is likely, in my opinion, that you cannot clearly observe any vortex due to the RANS nature of your simulation. You are seeing a time averaged view of the flow across a single time step. I like this &lt;a href=&quot;http://www.cradle-cfd.com/tec/column04/008.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;description&lt;/a&gt; of RANS in layman's terms: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ALF2Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ALF2Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Let us think about a concept of a time-averaged field calculated by RANS. Imagine a lot of people crossing a diagonal crosswalk in central city at a busy commute time. It may be hard to follow the movement of each person if you look down from the top of a building, but you can tell that most people are walking from the subway station towards the sidewalk on the opposite side. This general view is the “time-averaged field”. That is to say, in RANS, you would not be conscious of detailed movements or routes of each individual; instead, you would calculate how many and in which direction people cross in a fixed amount of time. Nevertheless, time-averaged fields are affected by turbulence as the nature of flow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are seeing a general view of your flow field (influenced by turbulence and therefore by your choice of turbulence model), if you want to be sure of capturing the detailed movements, such as a vortex, you may wish to try URANS (although be careful to chose a total simulation time that is long enough to capture everything - URANS still uses time averaging in a given time step) or if you have a very nice PC (and lots of time) LES would do a nice job - maybe overkill for this problem though. &lt;/p&gt;&#xA;" OwnerUserId="14498" LastActivityDate="2018-02-01T17:16:56.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19211" PostTypeId="1" CreationDate="2018-02-01T18:18:42.830" Score="1" ViewCount="95" Body="&lt;p&gt;Suppose that you have a bar of length 2 meters. This bar is along the x-axis. At x=0 there is a pin support. Also, at x=1 meter there is a pin support. A forceof magnitude 50Newtons is applied along the x-axis or along the axis of the bar at x=2 meters. What are the reactions at the supports? &lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="9742" LastEditDate="2018-02-01T18:54:21.893" LastActivityDate="2018-02-01T18:54:21.893" Title="What are the reactions at the supports?" Tags="|mechanical-engineering|civil-engineering|statics|" AnswerCount="0" CommentCount="11" ClosedDate="2018-02-03T12:42:11.447" ContentLicense="CC BY-SA 3.0" />
  <row Id="19212" PostTypeId="1" CreationDate="2018-02-01T18:24:52.540" Score="1" ViewCount="23" Body="&lt;p&gt;From my understanding, the Euler Bernoulli Beam theory is mostly used for small angle displacements, and an implication of this assumption is that the length of the beam is subject to distortion and stretching, meaning that when the beam is bent, its length is greater than when relaxed, due to the fact that the end of the rod always remains in line with the end of the rod at equilibrium. The theory thus fails for larger displacements when the change in length of the beam actually affects the result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does Timoshenko beam theory suffer from this issue as well? Do beams in Timoshenko theory have consistent length? Are there simple ways of conserving the length of the beam in these theories? Or is the only way to use Geometrically Exact Beam theory and do finite element methods?&lt;/p&gt;&#xA;" OwnerUserId="14773" LastActivityDate="2018-02-01T18:24:52.540" Title="Consistent length in beam theories" Tags="|stresses|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19213" PostTypeId="2" ParentId="18929" CreationDate="2018-02-01T18:46:01.743" Score="1" Body="&lt;p&gt;alright! After all the comments and what was available this is what we ended up designing&lt;a href=&quot;https://i.stack.imgur.com/Do1fh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Do1fh.jpg&quot; alt=&quot;DIY heat exchanger&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inlet flue gas at 140 degrees was 65-75 degrees at outlet after an hour of continuous usage, with outdoor temperature as 25 degrees.&lt;/p&gt;&#xA;" OwnerUserId="12071" LastActivityDate="2018-02-01T18:46:01.743" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19214" PostTypeId="1" AcceptedAnswerId="19402" CreationDate="2018-02-01T18:50:43.967" Score="1" ViewCount="210" Body="&lt;p&gt;I have an extractive distillation column set up in AspenPlus. Methylcyclohexane (MCH) is expected to come out from the top (in the distillate stream) and the bottoms stream is mostly Toluene/Phenol, with a bit of MCH. The feeds are MCH+Toluene (at stage 8 out of 25) and Phenol (at stage 18 out of 25).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature - 100 C. Pressure - 1 atm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I use the N.R.T.L (non-random two-liquid) model for the simulation, I get weird results. However, when I use UNIFAC, I get what I expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought NRTL was supposed to take into account deviations from ideality (caused by the phenol group?), so I don't know what's throwing it off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="14774" LastEditorUserId="14774" LastEditDate="2018-02-09T18:00:31.357" LastActivityDate="2018-02-13T16:20:06.430" Title="Failure of N.R.T.L method to model Phenol / Methylcyclohexane / Toluene system in AspenPlus. Why?" Tags="|thermodynamics|chemical-engineering|modeling|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19215" PostTypeId="2" ParentId="19204" CreationDate="2018-02-01T18:51:48.903" Score="2" Body="&lt;p&gt;Boiler efficiency curves are typically available from the boiler manufacturer, and will typically define the efficiency (ratio of output energy to input energy) as a function of flue gas temperature, return water temperature, loading (as percentage of full load), fuel type or some other operating parameter or combination of parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good primer for different variables affecting boiler efficiency can be found &lt;a href=&quot;http://www.cleaverbrooks.com/reference-center/insights/boiler-efficiency-guide.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;here- the Boiler Efficiency Guide from Cleaver-Brooks, a leading boiler manufacturer&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="7763" LastActivityDate="2018-02-01T18:51:48.903" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19217" PostTypeId="2" ParentId="19208" CreationDate="2018-02-01T19:05:48.253" Score="4" Body="&lt;p&gt;Batteries can't deliver unlimited current. Internal resistance, and limitations of the chemical reaction within the battery make that it can only deliver so much current. And the more current is drawn, the bigger the voltage drop will be because of above mentioned reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific power (W/kg) and capacity, specific charge or specific energy(Wh/kg or J/kg) are pretty much totally unrelated. You can compare it with humans. Let's take two athletes. Usain Bolt is able to run faster than a lightning bolt for 100 metres, but Phidippides is able to run 42km in one run. Both persons are about the same weight/size. The first says something about specific power, the latter about capacity. They can't compete with each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Battery A may be able to store only enough charge to power a lightbulb for 1 minute, while still being able to deliver 100 Ampere if needed.&#xA;Battery B may be able to store enough energy to power the exact same bulb for an hour, while only being able to deliver 1 Ampere if needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some battery types like LiPo's can deliver much current, and without having too much voltage drop. Li-ion batteries perform somewhat less, but can store a slightly bigger charge at the same size of a LiPo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific power means how much power a (one-celled)battery can deliver at a certain weight. So a Li-ion cell of 1 kg is on average able to deliver 300W. It doesn't matter how long you will need to consume power before recharging, you just can't get more power out of that cell. It's intrinsic capabilities limit that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More expensive batteries may deliver more, but it's an average number. So in this analogy, if you need 600W, you just need 2kg of li-ion. Totally unrelated to that is the charge they can hold. It's just that there is a correlation most of the times, but not necessarily.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-02-01T19:05:48.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19218" PostTypeId="1" CreationDate="2018-02-01T21:18:36.403" Score="2" ViewCount="6276" Body="&lt;p&gt;I am designing something with a sandwich construction that needs to squeeze the internal layer uniformly. The lower plate is to be made from cast iron and the upper plate needs to be made from copper. The squeezing will be done with four threaded rods, nuts and some spring washers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rt1SE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rt1SE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The squeezing force I need to apply is 30720N, so I guess 7680N per nut (or is it not so simple?). And what I really don't know is, how do I determine a thickness of material required, given its properties and my design details? My requirements are &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Plates will not deform like playdough between 20˚-150˚C&lt;/li&gt;&#xA;&lt;li&gt;Plates will not buckle like a spring very much, so that the sandwich pressure is not applied unevenly.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;kamran's answer describes the red line below as the lever distance in his model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JahEy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JahEy.png&quot; alt=&quot;Kamran points out I can model my fulcrum distance as this length&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried my best to read up on Young's Modulus and Shear Modulus, etc., but I do not grok how to correctly apply these to real life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Can someone kindly point me to a formula or a workflow so that I can calculate the answer myself?&lt;/p&gt;&#xA;" OwnerUserId="14775" LastEditorUserId="14775" LastEditDate="2018-02-03T08:53:54.690" LastActivityDate="2018-02-03T16:51:39.767" Title="How can I determine the required thickness of these plate materials?" Tags="|structural-engineering|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19219" PostTypeId="2" ParentId="19218" CreationDate="2018-02-01T23:26:53.033" Score="2" Body="&lt;p&gt;From the information you've given it is difficult to be too specific, but looking at the loads and design you might get some ideas by considering elastomeric bridge bearing plates or sole plate design standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of useful references may be found for &lt;a href=&quot;http://www.bgstructuralengineering.com/BGSCM13/BGSCM008/Design/BGSCM0080604.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;bearing plates&lt;/a&gt; and &lt;a href=&quot;http://www.eng-tips.com/viewthread.cfm?qid=108837&quot; rel=&quot;nofollow noreferrer&quot;&gt;sole plates&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also another post that considers stiffness versus thickness of steel plate &lt;a href=&quot;https://engineering.stackexchange.com/questions/68/how-does-width-and-thickness-affect-the-stiffness-of-steel-plate&quot;&gt;here&lt;/a&gt;.  Essentially it comes down to stiffness, which is a function of Moment of Inertia and Modulus of Elasticity.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastActivityDate="2018-02-01T23:26:53.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19220" PostTypeId="2" ParentId="19218" CreationDate="2018-02-01T23:28:31.233" Score="3" Body="&lt;p&gt;First off you want to make sure if you should divide the required squeezing force by four. Assuming that is correct we can proceed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of symmetry you can assume the flange overhang will have to resist a bending moment that would want to bend it down like a cantilever slab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are actually many steel column base-plates that are loaded very similar to this and by checking a column base-plate design manual you'll find an identical configuration and use the equations and recommendation there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative as a reasonable simplification is to assume the plate's bending moments, once on x-x direction and then on y-y direction, so we disregard the stress concentration at the corner and assume the flange is under a concentrated load by one bolt acting at the distance measured perpendicular to the edge of the plate from center of the bolt to inside edge (0.707 x your call out distance), let's call it $l$, and let's call your bolt tension $p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your moment is $ M= pl$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your slab bending moment $I=bh^3/12$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $b$ is your plate width and $h$ is its thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your stress $\sigma  =My/I= Mh/2I$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you can test if your plate is strong enough to support the bolt load if you have its allowable stress. The only catch is you apply only one bolt tension per edge.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="14775" LastEditDate="2018-02-02T02:34:48.967" LastActivityDate="2018-02-02T02:34:48.967" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19221" PostTypeId="2" ParentId="19193" CreationDate="2018-02-01T23:49:05.970" Score="1" Body="&lt;p&gt;The part you've circled is typically referred to as an &quot;anvil&quot;, or sometimes a &quot;loading strut&quot; Two are usually secured at the bottom, and either one or two anvils are used on top (depending on a 3 or 4 point bend test).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would certainly take a look at products made by Instron:&#xA;&lt;a href=&quot;http://www.instron.us/en-us/products/testing-accessories/grips/flexure-fixtures&quot; rel=&quot;nofollow noreferrer&quot;&gt;Instron General Purpose Flexure Fixtures&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the Image you show in your post is actually from their website:&#xA;&lt;a href=&quot;http://www.instron.us/en-us/products/testing-accessories/grips/flexure-fixtures/3-4-point-flexure-fixture-options/2810-403&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deflectometer Plunger&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this answers your question. &lt;/p&gt;&#xA;" OwnerUserId="4897" LastActivityDate="2018-02-01T23:49:05.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19222" PostTypeId="2" ParentId="19113" CreationDate="2018-02-02T00:07:05.427" Score="0" Body="&lt;p&gt;One of the key design characteristics of a gear is its tooth profile. This is typically based on what is called an Involute Curve, and the shape of the tooth is designed to maintain consistent force through the tooth at its pitch line during rotation and to minimize backlash within the series. The interweaving of the teeth themselves is referred to as the gears &quot;meshing&quot; and the curvature of the teeth allow them to clear one another as the gear rotates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Wikipedia illustration give a good visualization of how the teeth actually &quot;mesh&quot;: &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/c/c2/Involute_wheel.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://upload.wikimedia.org/wikipedia/commons/c/c2/Involute_wheel.gif&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also take a look at &lt;a href=&quot;http://Geargenerator.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://Geargenerator.com&lt;/a&gt; which give a great animation of gear tooth interaction and allows you to vary the tooth parameters to see how the shape of the teeth interact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, reconsidering your original question, you can easily see that simply putting two flat spur gears at a right angle to one another would defeat the entire concept of proper gear &quot;mesh&quot; and in fact the teeth would bind because they wouldn't &quot;fit&quot; into one another.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A set of bevel gears solves this problem by placing the teeth on an angle so that they face one another and mesh properly!&lt;/p&gt;&#xA;" OwnerUserId="4897" LastActivityDate="2018-02-02T00:07:05.427" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19223" PostTypeId="1" CreationDate="2018-02-02T01:22:14.263" Score="1" ViewCount="24" Body="&lt;p&gt;What is the required: an overcurrent protection for a 3phase, 150kVA, 277/480V-120/208V transformer. With an expected load of 20a mps (3 phase 208v).&lt;/p&gt;&#xA;" OwnerUserId="14776" LastEditorUserId="9" LastEditDate="2018-02-02T02:31:21.623" LastActivityDate="2018-02-02T02:31:21.623" Title="How do I determine overcurrent protection for a transformer?" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="2" ClosedDate="2018-02-02T14:47:14.933" ContentLicense="CC BY-SA 3.0" />
  <row Id="19225" PostTypeId="1" CreationDate="2018-02-02T04:38:55.950" Score="2" ViewCount="65" Body="&lt;p&gt;I am checking to see if this is done correctly, the slot is a little confusing for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;B is a pin in a slot, solve for the spring constant (k) so the 50 lb force applied does not move pin B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving for the moment about A I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50lbs*2&quot; - sin45(N)*3&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N = 47.1 lbs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then solving for forces in the Y direction I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;k*.125&quot; - sin45(N) ==&gt; k = 33.2lbs/(.125&quot;) = 266 lbs/in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4Yoq1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Yoq1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14778" LastActivityDate="2018-02-02T04:38:55.950" Title="Pin, Slot, and Spring Equilibrium" Tags="|moments|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19227" PostTypeId="2" ParentId="19096" CreationDate="2018-02-02T05:52:45.663" Score="0" Body="&lt;p&gt;I have solved this problem&#xA;I used partition and adjust the mesh seed size &lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-02-02T05:52:45.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19228" PostTypeId="1" AcceptedAnswerId="19269" CreationDate="2018-02-02T08:56:23.410" Score="4" ViewCount="5273" Body="&lt;p&gt;I asked a &lt;a href=&quot;https://engineering.stackexchange.com/questions/19168/flow-in-a-t-junction-with-pipes-of-different-diameters&quot;&gt;previous question&lt;/a&gt; about pipe t-junctions with pipes of varying diameters, and it got me wondering... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How are losses handled and calculated in pipes with holes drilled directly into them?&lt;/strong&gt; Flow runs perpendicular to the hole and most of the theory I can see for flow through a pipe exit assumes the pipe and orifice are coaxial. Or for an orifice plate directly in the pipe flow, neither of which seem appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing some theory must exist since common objects like lawn sprinklers contain these pipes with many holes drilled into them. I have been thinking about a pipe with one hole drilled in it of small diameter. Essentially we have a free jet exiting perpendicular to the main pipe flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://zavax.files.wordpress.com/2015/09/vlcsnap-2015-10-05-14h19m29s85.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://zavax.files.wordpress.com/2015/09/vlcsnap-2015-10-05-14h19m29s85.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14498" LastActivityDate="2018-02-06T11:31:43.667" Title="Losses through holes drilled in a pipe" Tags="|mechanical-engineering|fluid-mechanics|civil-engineering|pipelines|fluid|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19230" PostTypeId="1" AcceptedAnswerId="21911" CreationDate="2018-02-02T10:47:11.907" Score="6" ViewCount="361" Body="&lt;p&gt;I would like to compare the shape of the frequency spectrum of a plastic and aluminium component. Now if we assume that the plastic and aluminium component only differ in the material, while the mass and stiffness stay the same (I know, large assumption), how would generally differ the shape of the frequency spectrum from each other? Obviously, the damping plays a major role here and would be much higher in case of plastic, therefore the response amplitude would be lower, the resonance zone wider and a shift (which is many times neglected) towards a lower frequency would be made... &lt;strong&gt;Could I in a very simplified version, just for the sake of understanding, draw it like this?&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QbU0S.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QbU0S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Another question would be the temperature dependence of the elastic modulus. If the temperature rises, the modulus is reduced which also affects the eigenfrequency. In case of aluminium not so much, but in case of plastics, the difference can be enormous. My understanding is that in case of elevated temperatures, the damping increases; in case of plastic much more than in case of aluminium. &lt;strong&gt;Could I visualize this as:&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Ap5XC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ap5XC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any constructive answers are much appreciated!&#xA;Kind regards,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luka&lt;/p&gt;&#xA;" OwnerUserId="7234" LastEditorUserId="7234" LastEditDate="2018-02-02T11:25:57.463" LastActivityDate="2018-05-25T20:57:15.850" Title="Elastic (Young's) Modulus relation to the Eigenfrequency" Tags="|mechanical-engineering|structural-engineering|materials|frequency-response|eigenvalue-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19232" PostTypeId="1" CreationDate="2018-02-02T15:34:59.233" Score="1" ViewCount="137" Body="&lt;p&gt;Are there analytical expressions (functions of geometry and applied pressure) for the stress in a hollow prolate spheroid, like there are expressions for the stress in a hollow sphere/cylinder?&lt;/p&gt;&#xA;" OwnerUserId="14783" LastActivityDate="2018-02-02T15:34:59.233" Title="Equations for stress in a hollow prolate spheroid under external pressure" Tags="|stresses|pressure-vessel|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19233" PostTypeId="1" AcceptedAnswerId="19234" CreationDate="2018-02-02T17:40:26.280" Score="-2" ViewCount="1952" Body="&lt;p&gt;What is the meaning of this statement &quot;In a reciprocating engine the flow of intake charge takes place through the intake valve opening which is varying during the induction operation. &lt;strong&gt;Also&lt;/strong&gt;, &lt;strong&gt;the&lt;/strong&gt; &lt;strong&gt;maximum&lt;/strong&gt; &lt;strong&gt;gas&lt;/strong&gt; &lt;strong&gt;velocity&lt;/strong&gt; &lt;strong&gt;through&lt;/strong&gt; &lt;strong&gt;this&lt;/strong&gt; &lt;strong&gt;area&lt;/strong&gt; &lt;strong&gt;is&lt;/strong&gt;  &lt;strong&gt;limited&lt;/strong&gt; &lt;strong&gt;by&lt;/strong&gt; &lt;strong&gt;the&lt;/strong&gt; &lt;strong&gt;local&lt;/strong&gt; &lt;strong&gt;Sonic&lt;/strong&gt; &lt;strong&gt;velocity&lt;/strong&gt;&quot;&lt;/p&gt;&#xA;" OwnerUserId="14786" LastEditorUserId="14786" LastEditDate="2018-02-02T18:52:04.027" LastActivityDate="2018-02-02T19:39:42.277" Title="What is Inlet-valve Mach index?" Tags="|mechanical-engineering|engines|car|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19234" PostTypeId="2" ParentId="19233" CreationDate="2018-02-02T19:39:42.277" Score="-1" Body="&lt;p&gt;If the local velocity of the inlet charge through that area hits mach 1, the flow is &quot;choked&quot; by the presence of a normal shock wave. I have never heard of flow rates high enough to induce choked flow in an intake runner tube, however; this seems unlikely. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-02-02T19:39:42.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19236" PostTypeId="2" ParentId="19174" CreationDate="2018-02-02T21:52:39.487" Score="0" Body="&lt;p&gt;If I am not mistaken, you are looking at the drop in pressure right at a pressure reducing regulator when flow initiates, and wanting to infer mass flow from the value of this drop.  If this is true, I would like to add that there are many reasons why a pressure regulator would &quot;droop&quot; other than the velocity head effect.  Regulators have springs that compress when the fill valve opens, and there are several friction losses that together comprise the flow performance curve.  Many designs have &quot;lock up&quot; that presents a significant offset between the static and flowing conditions.  Have I interpreted your question correctly?&lt;/p&gt;&#xA;" OwnerUserId="14790" LastActivityDate="2018-02-02T21:52:39.487" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19237" PostTypeId="1" CreationDate="2018-02-02T21:54:59.873" Score="2" ViewCount="56" Body="&lt;p&gt;For the purpose of detecting a defect or failure in a machine, it is often practical to physically listen to any abnormal noises that a machine might be making, and even use a simple stopwatch to determine the frequency of these noises. This practice often gives clues as to what part of a machine is having problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of detecting failure in a large plant, filled with many machines that might not have many sensors, one can also often isolate a machine with a problem by listening to abnormal ambient noises, and tracing them back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any sort of practice in which a microphone is placed at a certain location to detect the &quot;ambient sound signature,&quot; and then is used as a sort of sensor for problems in plants? I can imagine a baseline being made with this microphone and then used as a reference for abnormal ambient noises, then paired with a monitoring camera to detect whether there is any explanation for non-regular noises.&lt;/p&gt;&#xA;" OwnerUserId="12823" LastEditorUserId="12754" LastEditDate="2018-02-03T00:55:26.883" LastActivityDate="2018-02-12T22:17:07.920" Title="Are there general-purpose acoustic FFT sensors for plant diagnostics?" Tags="|sensors|acoustics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19238" PostTypeId="1" AcceptedAnswerId="19333" CreationDate="2018-02-03T07:47:07.043" Score="0" ViewCount="563" Body="&lt;p&gt;I have made bridge model in Abaqus,&#xA;I also have done experimental study and I have the load-displacement data from my experiment,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am now going to compare the results from Abaqus and the results from experiment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this step, I want to compare the load-displacement curves between experiment and analysis in Abaqus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how to get this information from abaqus to excel, please help me resolve this problem, I really appreciate it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-02-09T04:16:18.923" Title="How to get all forces and displacements values in abaqus and export it to excel?" Tags="|finite-element-method|abaqus|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19239" PostTypeId="1" CreationDate="2018-02-03T10:09:25.977" Score="1" ViewCount="117" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LsSs3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LsSs3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MIomC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MIomC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hsJNh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hsJNh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/46aUT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/46aUT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tryl9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tryl9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9eAg8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9eAg8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I am not very good at understanding the ANN theory, but please if anyone can explain me about what does (a) as input of the activation function mean?&#xA;and how to get this value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="14641" LastEditorUserId="14641" LastEditDate="2018-02-03T12:12:38.027" LastActivityDate="2018-02-04T23:31:48.177" Title="What does (a) the input of the activation function mean?" Tags="|statistics|" AnswerCount="1" CommentCount="17" ContentLicense="CC BY-SA 3.0" />
  <row Id="19240" PostTypeId="1" CreationDate="2018-02-03T13:23:00.980" Score="2" ViewCount="37" Body="&lt;p&gt;An animal's blood pressure and/or body movements can be used to generate a few mW of electrical energy. Can an RF transmitter which radiates 1 mW be detected by receivers (without any parabolic reflectors) on satellites? The receiver has to only detect the signal and power level, no data transfer is required. This could be used to locate the animal.&lt;/p&gt;&#xA;" OwnerUserId="14794" LastEditorUserId="1832" LastEditDate="2018-02-03T15:20:14.410" LastActivityDate="2018-02-03T15:20:14.410" Title="Detection of weak signal by satellites" Tags="|electrical-engineering|power-electronics|rf-electronics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19241" PostTypeId="1" CreationDate="2018-02-03T13:50:48.850" Score="2" ViewCount="82" Body="&lt;p&gt;A piezo material like pzt produces voltage upon mechanical stress application. But how much power it can generate or how much current it can generate as a function of piezo characteristics and force applied? &lt;/p&gt;&#xA;" OwnerUserId="14794" LastActivityDate="2018-04-05T03:14:18.260" Title="Current generated by piezoelectric material, say pzt" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19242" PostTypeId="1" CreationDate="2018-02-03T15:03:22.580" Score="-2" ViewCount="16" Body="&lt;p&gt;One can estimate energy intensity for line of sight radio communication using the inverse square law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming perfect reflection from a flat surface, is there any formula, or is there a way to estimate energy intensity for radio waves that reflect once from a surface before reaching a destination?&lt;/p&gt;&#xA;" OwnerUserId="14795" LastEditorUserId="90" LastEditDate="2018-02-06T01:11:01.253" LastActivityDate="2018-02-06T01:11:01.253" Title="Energy intensity in non line of sight communication" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19243" PostTypeId="1" AcceptedAnswerId="19280" CreationDate="2018-02-03T16:28:53.570" Score="-1" ViewCount="327" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I4vHx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I4vHx.jpg&quot; alt=&quot;This is the question &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to solve this question I arrived at this diagram but taking the moments about the center of the beam I arrived at x=L/2. Can someone tell me what I need to do to get the correct answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xuNpn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xuNpn.jpg&quot; alt=&quot;Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zw6mX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zw6mX.jpg&quot; alt=&quot;This is my answer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14797" LastEditorUserId="14797" LastEditDate="2018-02-06T17:21:23.967" LastActivityDate="2018-02-08T01:00:16.233" Title="Beams question help" Tags="|mechanical-engineering|statics|" AnswerCount="1" CommentCount="3" ClosedDate="2018-02-07T15:45:57.950" ContentLicense="CC BY-SA 3.0" />
  <row Id="19244" PostTypeId="2" ParentId="19218" CreationDate="2018-02-03T16:51:39.767" Score="1" Body="&lt;p&gt;Okay, you're not going to like this answer, but you need to space the individual devices in the array to accommodate intermediate fasteners. The assembly as you drew it would cost a couple hundred dollars in materials and fabrication, and would be very heavy. So leave 3 6mm gutters between devices perpendicular to the wire gutters, drill a 6mm hole between each unit, drop in a 4x4mm nylon shoulder washer, and install a 4M machine screw. This gives you 12 intermediate fasteners. Then run 16 more around the perimeter. These last could be 3M in a 3x4mm shoulder washer if you wanted to get fussy. This gets the thickness of copper down to about 6mm for good results. Going with fancier spring isolation washers of at least 10mm dia can get you down to 5mm copper. A happier solution would be to find a Peltier form factor that has fastener notches so you can close-pack them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that you have a thermal gradient across the copper makes it want to distort. So you don't really want too thick of a copper plate, the thermal distortion forces would soon overwhelm the clamping forces.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-02-03T16:51:39.767" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19245" PostTypeId="2" ParentId="19157" CreationDate="2018-02-04T00:03:38.537" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But the exercises and examples also treat Root locus as a method of designing compensation or gain to fulfill the system requirement like damping factor and setting time. Isn't the root locus method for stability only?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What determines stability? zeros or poles?&#xA;Poles on RHP always cause instability. That is what we would like to avoid. In many cases, as long as the system is stable, engineers do not care. They just like a system work either good or bad. But the system requirements is also important. But it is meaningless if the system is not stable. Therefore, stability is the first requirement. The system quality is a bonus.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why the closed loop dominant poles of the root locus can show the response of the system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe what I am saying is not exactly what you mean by this question. However,&#xA;here, I emphasize on the term 'dominant poles'. When the absolute value of a pole $||p_i||$ is small, it means that the effect of the pole in this fraction is high compared with $s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{1}{(s-p_i)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, $p_i$ is called dominant. But, if the absolute value of the pole is very big, it means that the fraction is not very sensitive to $s$ and hence $s$ can be neglected. Therefore, nondominated poles can be approximated by a gain.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wont it neglect the effect of the closed loop zeros?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Zeros are important as well. Zeros on RHP (nonminimum phase) are not desirable but they do not cause instability directly. But, since they attract the poles by the increase of the controller gain, they attract the poles to RHP if a high gain is used. Nonminimum phase systems are bad but not necessarily instable.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;can Nyquist plot also used to show the system response and to design the system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I personally do not use them because root locus does the job for me. In practice, It has not have any application for me but I am not sure about the others.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But the exercises and examples also treat Root locus as a method of designing compensation or gain to fulfill the system requirement like damping factor and setting time. Isn't the root locus method for stability only?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;See page 2 of this &lt;a href=&quot;https://ocw.mit.edu/courses/mechanical-engineering/2-004-systems-modeling-and-control-ii-fall-2007/lecture-notes/lecture20.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;slide&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This paragraph is out of the scope of your question but worth mentioning. You mentioned RHP. For systems with fractional order such as &#xA;$$\frac{(s^{0.6}+0.4)(s^{0.6}+0.7)}{(s^{0.6}+1.3)(s^{0.6}+0.9)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stability criteria could be more relaxed. See Figure (6-a) at page 13 of &lt;a href=&quot;https://arxiv.org/pdf/0811.4102.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; article.&lt;/p&gt;&#xA;" OwnerUserId="14288" LastActivityDate="2018-02-04T00:03:38.537" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19246" PostTypeId="2" ParentId="19241" CreationDate="2018-02-04T01:23:25.647" Score="1" Body="&lt;p&gt;Piezo devices produce hundreds of millivolts or more upon flexure, but the current they produce is far smaller. Think of it this way: the typical output impedance of a piezo transducer is of order ~megaohms. Even if it produces a volt of output, this means the current is of order ~1/million, and the power generated is correspondingly tiny. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-02-04T01:23:25.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19248" PostTypeId="1" AcceptedAnswerId="19249" CreationDate="2018-02-04T08:33:58.437" Score="2" ViewCount="1557" Body="&lt;p&gt;I have a vapor system that requires a 50 psi feed of DI water. Without getting fancy, is there a simple commercial off-the-shelf part I can buy, fill with water and hand-pressurize? Or something with an NPT inlet &amp;amp; outlet that I could pressurize with nitrogen or some other inert gas? At maximum, I think I'd need a 10 gallon reservoir - something that can sit on a lab bench.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative is to slap a small pump in the reservoir, but that seems like overkill. &lt;/p&gt;&#xA;" OwnerUserId="3583" LastEditorUserId="3583" LastEditDate="2018-02-04T21:29:32.917" LastActivityDate="2018-02-04T21:29:32.917" Title="Simplest Way to Make A Pressurized Water Reservoir" Tags="|pumps|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19249" PostTypeId="2" ParentId="19248" CreationDate="2018-02-04T11:33:25.837" Score="4" Body="&lt;p&gt;A well storage pressure tank may serve your purpose. There are reverse osmosis tanks that are designed to handle up to 100 psi (per Amazon item post) and ordinary well tanks with internal bladders that provide at least your specified 50 psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bladder tanks will be larger than the water capacity, as half the volume contains a rubber bladder. Air pressure fills the bladder, pressurizing the water as the well pump fills the water volume portion. When the pump reaches cut-off pressure, the bladder is compressed and maintains that pressure as the water is drawn off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without a bladder tank, as your tank empties, the pressure would drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you will be hand-pressurizing the tank, you are the cut-off switch and can adjust via air-pressure gauge on the air pump or fill nozzle of the bladder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick search discovered bladder well storage tanks as small as 2 gallon, 4 gallon, and 14 gallon which makes it more likely one will find a 10 gallon capacity.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-02-04T11:33:25.837" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19250" PostTypeId="2" ParentId="18737" CreationDate="2018-02-04T14:41:21.360" Score="3" Body="&lt;p&gt;In addition to @arash’s answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Designing directly with a nonlinear system is cumbersome, not only computationally expensive.&lt;/li&gt;&#xA;&lt;li&gt;By linearization one can cover the whole nonlinear space, by generating as many linearization points as required. &#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Basically, a design is optimized and proposed for those linear models, and then it is validated against the more realistic non-linear model.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This process gives sufficient confidence for the whole design space, while allowing engineers to break up the problem to several linear systems, and corresponding automation. &lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2018-02-04T14:41:21.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19251" PostTypeId="1" AcceptedAnswerId="19260" CreationDate="2018-02-04T19:19:59.343" Score="10" ViewCount="4231" Body="&lt;p&gt;I'm quite curious about the 3-ring release system used in parachutes, as described in this Wikipedia article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/3-ring_release_system&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/3-ring_release_system&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder why is this system so popular? I've been told it's nearly ubiquitous in the industry, but even after Googling and asking some skydivers I haven't been able to find out what makes it better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, the Wikipedia article mentions that each of the loops multiply the mechanical advantage of the small cord loop that is secured by the yellow wire. Could somebody describe in detail how this multiplication of mechanical advantage takes place, and why is the number of ring chosen to be 3 instead of say, 5?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/2/25/3Ring_release_animation.gif/220px-3Ring_release_animation.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/2/25/3Ring_release_animation.gif/220px-3Ring_release_animation.gif&quot; alt=&quot;3-ring release system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13372" LastEditorUserId="13311" LastEditDate="2018-02-05T11:16:02.873" LastActivityDate="2018-02-07T19:39:16.297" Title="Mechanical advantage of the 3-ring release system" Tags="|mechanical|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19252" PostTypeId="2" ParentId="19251" CreationDate="2018-02-04T20:13:26.287" Score="6" Body="&lt;p&gt;I can't address all the questions you've posed, but at least one of the answers comes directly from the inventor, Bill Booth, as noted in the wiki. He holds the patent for the device (held?) and I had the opportunity to fly with him a couple of decades ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the following is word of mouth. Perhaps there is documentation to back it up, but I heard it from Bill and that's good enough for me. I did see the patent documentation in his office, but that was before digital cameras and camera phones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was explained to me that prior to the 3-ring release, the mechanism most commonly used was called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Cut-away&quot; rel=&quot;noreferrer&quot;&gt;Capewell&lt;/a&gt; release. As he tells it, the mechanism is prone to jamming and is particularly difficult to release when under load, as would be the case when plummeting to earth with a fouled parachute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;His genius in developing the 3-ring release resulted in a system that would release completely when under high loads, would release completely when under light loads and be mechanically simple. It's also easy to manufacture without requiring special machining or unusual skills.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current configuration, again as described by Bill Booth, is that there is a ten to one reduction of force for each ring. If one ignores the friction of the cable at the ring lock loop, the total reduction is one thousand to one over the three rings. That means you can hang one thousand pounds from each release (two per canopy) and the force required to break the locking cable free, perpendicular from the attachment point, is one pound. I don't know newtons so well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding in the laterally oriented securing mechanism of the cable, the force required by the parachutist is even less. If there is one pound of force pulling against the cable at the loop that secures it, how much force is required to slide that cable out and free up the assembly? I don't know that answer, but it's bound to be very low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've experimented with this concept for a payload release mechanism for a multi-rotor radio controlled helicopter. Using the math of ordinary levers, I had a slightly less than 5000 to 1 reduction in force, using rigid linkage rather than loops to retain the levers. The components were necessarily light, and would tolerate up to 2 pounds payload, perhaps more, but the copter was too sluggish at that point. The linkage components weighed 20 grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eJE0o.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eJE0o.png&quot; alt=&quot;lever classes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image above comes from a &lt;a href=&quot;https://en.wikipedia.org/wiki/Lever&quot; rel=&quot;noreferrer&quot;&gt;wiki entry regarding levers&lt;/a&gt;. The 3-ring release is a class 2 lever, pivoted at one end. Referencing the animation in the wiki, the first lever load is the large ring from which the 'chutist is suspended. It applies force to the second ring upwards (relative to the animation) of the pivot of the second ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second ring is latched by the third ring, applying force to the third ring in the same manner. The mechanical advantage from which the lever's math comes into play is due to the spacing of the pivot relative to the next ring's location.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2018-02-04T20:23:13.930" LastActivityDate="2018-02-04T20:23:13.930" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19253" PostTypeId="1" CreationDate="2018-02-04T20:25:21.593" Score="7" ViewCount="495" Body="&lt;p&gt;I am a bachelor's level student tasked with modeling the vehicle dynamics of an RC Car that will be made to drive autonomously. I lack prior experience or particularly specific education with vehicle dynamics so I have been studying the field a bit, particularly Vehicle Dynamics and Control by Rajesh Rajamani. By advice from a supervisor I will not be using the Pacejka Magic Formula. I think I have a good enough idea of the problem and how to approach it but I've hit a wall quite badly regarding the parameter 'Cornering stiffness'. As I understand this is a property of the tires, and something that I would find on data sheets on tires. However in my case the tires are considerably more simple, they're just solid rubber rings around the wheels. I have tried to read up on Cornering Stiffness to see if I can determine it myself somehow but I haven't really found anything I've understood. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how should I go about determining the cornering stiffness? Otherwise, considering the shape of these tires are there any assumptions I can make that would simplify the model?   &lt;/p&gt;&#xA;" OwnerDisplayName="user14808" LastActivityDate="2024-01-12T00:09:26.100" Title="Questions regarding Cornering Stiffness for RC Car" Tags="|mechanical-engineering|control-engineering|automotive-engineering|dynamics|modeling|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19255" PostTypeId="2" ParentId="19239" CreationDate="2018-02-04T22:50:50.977" Score="3" Body="&lt;p&gt;Just a quick summary of my concluding thoughts on this step in your research.  I understand that you have been advised to use the &lt;strong&gt;Sigmoidal Function&lt;/strong&gt;, so this discussion will try to explain it's potential application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I don't think that an &lt;strong&gt;Artificial Neural Network (ANN)&lt;/strong&gt; would apply to your problem.  Typically ANN would consider a range of factors on some outcome.  When this process is considered, a lot of data is required along with specialized software.  This is because training of the ANN is necessary to establish that the model works.  Sometimes ANN is used to determine which factors are important in a non-linear, multivariate investigation.  Let me provide an example that may illustrate its use.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Application of ANN&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Suppose we are experiencing a routine cracking problem of some concrete elements of a structure.  We may collect data, on cracked elements and uncracked elements, to include:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Temperature of concrete.&lt;/li&gt;&#xA;&lt;li&gt;Temperature of air.&lt;/li&gt;&#xA;&lt;li&gt;Time of delivery.&lt;/li&gt;&#xA;&lt;li&gt;Time of casting.&lt;/li&gt;&#xA;&lt;li&gt;Wind speed.&lt;/li&gt;&#xA;&lt;li&gt;Solar radiation.&lt;/li&gt;&#xA;&lt;li&gt;Cement content.&lt;/li&gt;&#xA;&lt;li&gt;Water content.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;And potentially lots of other parameters (there could be dozens, even hundreds).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in theory we could send perhaps thousands of replicated data sets (measurements) through a neural network to see which factors may or may not be predictive of cracking of concrete elements.  That is how ANN is applied and it requires highly specialized software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a part of this process, we may also try different &lt;strong&gt;weighting factors&lt;/strong&gt;, as $w$, on the input data to try to improve predictive ability.  So, for example we could assemble a vector of weighting factors that can be used to adjust each of the input values.  This is a trial and error process that may or may not result in a better outcome.  We also may choose an &lt;strong&gt;activation function&lt;/strong&gt; for the hidden nodes that receive the weighted input values. In your example the activation function is a sigmoid; there are many others.  The activation function in simple terms will change the behaviour of your ANN in the sense that it changes the response to information.  Also important is how the nodes are connected; a good software package will go a long way to doing this for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I hope that this better explains what ANN is and how it may work.  I do not think that it would apply to your project because I have experience of this technique being applied to large, international concrete durability research using thousands of data sets and the ANN method alone was enough for granting a PhD to a student from a major UK engineering school.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Sigmoid Function&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now putting ANN aside, we look at your &lt;strong&gt;sigmoidal function&lt;/strong&gt; from a different perspective.  From &lt;a href=&quot;https://en.wikipedia.org/wiki/Sigmoid_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A wide variety of sigmoid functions have been used as the activation function of artificial neurons, including the logistic and hyperbolic tangent functions. Sigmoid curves are also common in statistics as cumulative distribution functions (which go from 0 to 1), such as the integrals of the logistic distribution, the normal distribution, and Student's t probability density functions.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance I recognized that this function could be derived from the family of &lt;a href=&quot;https://en.wikipedia.org/wiki/Normal_distribution&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gaussian&lt;/a&gt; distributions, and indeed it is.  The Gaussian (Normal) distribution is the most commonly used statistical function in existence, and is often used as the basis of error computation and regression analysis.  A Gaussian probability density function, or &lt;strong&gt;PDF&lt;/strong&gt;, looks like the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tRt7q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tRt7q.png&quot; alt=&quot;Gaussian PDF&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must think of a PDF as a special class of functions belonging to the class of &lt;strong&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Generalized_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;Generalized&lt;/a&gt;&lt;/strong&gt; (sometimes called &lt;strong&gt;Generic&lt;/strong&gt;) functions. These are described as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In mathematics, generalized functions, or distributions, are objects extending the notion of functions. There is more than one recognized theory. Generalized functions are especially useful in making discontinuous functions more like smooth functions, and describing discrete physical phenomena such as point charges. They are applied extensively, especially in physics and engineering.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can consider any probability density function to simply represent a &lt;strong&gt;state of knowledge&lt;/strong&gt;.  For example, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Uniform_distribution_(continuous)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Uniform PDF&lt;/a&gt; illustrated below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Hetc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Hetc.png&quot; alt=&quot;Uniform PDF&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;represents a maximum uncertainty in that, no matter what $z$ value we are considering within the bounds of $a$ and $b$, we will have the same probability of outcome.  This is different from the Gaussian above because there is a much lower probability associated with the extreme values than the average value.  With the Gaussian, as the standard deviation, or $\sigma$, gets smaller, we can consider that our state of knowledge has improved.  Complete certainty, for example, would be where $\sigma \rightarrow 0$ (indeed this condition represents the Dirac Delta Function that is widely used in physics and engineering, especially Finite Element Analysis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beauty of using a &lt;strong&gt;PDF&lt;/strong&gt; is that it transforms (or some may use the term normalizes) the data to a relative real range of 0 to 1 based upon our state of predictive knowledge.  Thus we may convert and consider any problem in the relative, rather than absolute.  This permits multiple data sets to be compared from different experiments and also allows prediction of behaviour, rather than magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Sigmoid Function&lt;/strong&gt;, just like a Gaussian PDF, is indeed one of these creatures.  A Sigmoid function plot, with &lt;strong&gt;a&lt;/strong&gt; replaced by $\sum{wx} + bias$, is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HcAvd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HcAvd.png&quot; alt=&quot;Sigmoid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outcome is a value between 0 and 1.  The shape of the curve is very near to what is termed a Gaussian Cumulative Distribution Function or &lt;strong&gt;CDF&lt;/strong&gt;.  An excellent discussion of how this is derived may be found at &lt;a href=&quot;https://towardsdatascience.com/multi-layer-neural-networks-with-sigmoid-function-deep-learning-for-rookies-2-bf464f09eb7f&quot; rel=&quot;nofollow noreferrer&quot;&gt;TowardsDataScience.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Possible Application&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The application of the Sigmoid to data carries with it the assumption that your data would follow a Gaussian form.  That is to say that extreme values are far less probable than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your specific data, thickness or effective thickness, I would not expect that weighting factors $w$ would be necessary nor would there be a need for $bias$ but these decisions are up to you.  You have a matrix of measurements that are to be transformed to the range of 0 to 1 using the Sigmoidal, or Gaussian CDF, function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also Excel best practice guides describing how the Sigmoid Function may be used in regression analysis on experimental point data.  This can be reviewed in a document that can be downloaded from &lt;a href=&quot;http://excelcurvefitting.com/pdfs/curve%20fitting%20best%20practice%20part%203.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ExcelCurveFitting.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you apply it depends upon the specific needs of your work.  For example, you could transform each value and run the full simulation, then transform collections of values within 50 by 50 cm grids to a single value and run it again, comparing the outcome of each.  From this you may consider the need for weighting factors or bias values to improve the regression.  It could be that this becomes a trial and error process that would eventually lead you to a deeper insight into the nature of your data and the problem you are attempting to solve.&lt;/p&gt;&#xA;" OwnerUserId="3266" LastEditorUserId="3266" LastEditDate="2018-02-04T23:31:48.177" LastActivityDate="2018-02-04T23:31:48.177" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19256" PostTypeId="1" CreationDate="2018-02-05T11:05:08.020" Score="1" ViewCount="61" Body="&lt;p&gt;Good morning,&#xA;I need to source 30 or 50 bearings such as the picture. This is a double effect roller bearing and the mid washer is drilled as you can see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shaft diameter: 17mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Height: 22.26mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed is not an issue&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not standard and very hard to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, I was thinking to put 2 AXK1730 needle bearing + 2*2 AS1730 washers and make another mid washer with the 2 holes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think about this?&lt;a href=&quot;https://i.stack.imgur.com/DVyCW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DVyCW.jpg&quot; alt=&quot;double effect roller bearing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14673" LastActivityDate="2018-02-05T11:05:08.020" Title="double effect roller bearing or two needle roller bearing" Tags="|bearings|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19258" PostTypeId="1" AcceptedAnswerId="19282" CreationDate="2018-02-05T12:41:03.837" Score="2" ViewCount="75" Body="&lt;p&gt;I tried the question but I didn’t get the correct answer, can anyone tell me what mistake I made?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s8EpC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s8EpC.jpg&quot; alt=&quot;Question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DKaXC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DKaXC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;I &amp;amp;= \int y^2\text{d}A \\&#xA;&amp;amp;= 25^2\cdot10^{-6}\cdot200\cdot50\cdot10^{-6} \\&#xA;&amp;amp;= 25^2\cdot10^{-8} \\&#xA;\sigma &amp;amp;= \dfrac{My}{I} \\&#xA;&amp;amp;= \dfrac{10^3\cdot25\cdot10^{-3}}{25^2\cdot10^{-8}} \\&#xA;&amp;amp;= 4\cdot10^6&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="14797" LastEditorUserId="1832" LastEditDate="2018-02-05T14:14:04.007" LastActivityDate="2018-02-07T17:48:18.213" Title="Beam question to find the maximum stress" Tags="|mechanical-engineering|statics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19260" PostTypeId="2" ParentId="19251" CreationDate="2018-02-05T16:50:13.930" Score="3" Body="&lt;p&gt;As noted by fred_dot_u, each ring acts as a &quot;Class 2 Lever&quot;. Imagine you are trying to lift the orange beam shown below from a position close to the pivot (distance y). The force somebody else would need to apply at the far end (distance x) would be smaller, by a factor of X/Y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LH6uAm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LH6uAm.png&quot; alt=&quot;simple lever&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine now than instead of applying your force directly upwards, you instead chose to apply it at a shallow angle. You can see that the perpendicular distance to the pivot is reduced, despite the force being applied at the same point on the beam. If your applied force (bodyweight in the case of the parachute) remains constant, then the reaction required to stop the beam moving is now even lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3kEBXm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3kEBXm.png&quot; alt=&quot;inclined force lever&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the Three-Ring-Release works, is by linking two of these systems together. It's difficult to show on a 'classical diagram', due to the way the fabric wraps around to form the fulcrum in each lever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the image below illustrates how the reaction force required on the small fabric loop (bottom left arrow) is many times smaller than the force applied by the weight of the human (far right arrow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is critical to minimize this reaction force, since if it were too large, then the friction force against the 'pull cord' (yellow in your original .gif) would be too big, and it would become difficult to remove the cord and release the parachute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2XEKp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2XEKp.png&quot; alt=&quot;Three Ring Release Levers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this answers your question, and explains how the mechanical advantage comes about. As for &quot;why 3, and not 5&quot;? This comes down simply to the fact that three is &lt;em&gt;enough&lt;/em&gt;. 5 would simply increase cost and complexity to manufacture the mechanism, as well as adding more points of failure, and 2 would not reduce the force sufficiently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B. There is no &quot;pulley action&quot; going on - the mechanical advantage is merely to reduce this frictional force and ensure that the yellow cord can be removed easily, regardless of how heavy the human is.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-02-05T16:50:13.930" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19261" PostTypeId="1" CreationDate="2018-02-05T17:19:49.667" Score="-1" ViewCount="74" Body="&lt;p&gt;I am designing a device that can climb up a tube. To do so, I am planning to  use a motor that will utilize spinwheels with high friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I work out how powerful my motor needs to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the device is 1kg.&lt;/p&gt;&#xA;" OwnerUserId="14746" LastEditorUserId="9" LastEditDate="2018-02-23T03:18:00.843" LastActivityDate="2018-02-23T03:18:00.843" Title="How to work out how powerful a motor needs to be to climb up a tube?" Tags="|mechanical-engineering|design|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19263" PostTypeId="1" AcceptedAnswerId="19271" CreationDate="2018-02-05T17:39:18.730" Score="2" ViewCount="91" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hZ0Sk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hZ0Sk.jpg&quot; alt=&quot;Question&quot;&gt;&lt;/a&gt;&#xA;I have got 19.6 as the answer for (a) but the book tells 14 is the answer. Can someone tell what mistake I have done or is there in the book.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Atefp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Atefp.jpg&quot; alt=&quot;This is my solution &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14797" LastActivityDate="2018-02-06T15:11:03.330" Title="Torque and angular acceleration" Tags="|mechanical-engineering|dynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19264" PostTypeId="2" ParentId="19242" CreationDate="2018-02-05T18:26:58.560" Score="0" Body="&lt;p&gt;A &quot;flat perfectly reflective surface&quot; is known as a mirror, and all that does is change the direction of wavefront propagation.  It has no effect whatsoever on the wavefront profile (for you nit-pickers, I'm assuming a bulk reflector, not a multilayer dielectric with polarization sensitivity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if your actual question involves the interaction between a beam which has both LOS to the target and a reflected section, then you have to deal with interference effects.  That's a more complicated situation.  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-02-05T18:26:58.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19266" PostTypeId="1" CreationDate="2018-02-06T05:57:37.073" Score="2" ViewCount="87" Body="&lt;p&gt;Moore's law states that no of transistors on a given area doubles every 18 months. Electronic components are typically made by photolithography. So,  Moores law basically limits the resolution of photolithography. Why resolution of photolithography evolved gradually and not abruptly? &lt;/p&gt;&#xA;" OwnerUserId="14822" LastActivityDate="2018-06-12T14:51:59.257" Title="Why resolution of photolithography evolved gradually?" Tags="|microelectronics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19267" PostTypeId="2" ParentId="19266" CreationDate="2018-02-06T08:11:50.807" Score="2" Body="&lt;p&gt;There's a adage: you have to crawl before you can walk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may as well ask why supersonic aircraft weren't developed first, before people started experimenting with gliders, the Wright brothers developing the first powered aircraft, followed by propeller powered biplanes, then monoplanes, then jet aircraft and finally supersonic jet aircraft. All this over the course of some five to six decades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every technology has its limits, then someone develops a better way and improvements are made and the technology advances. The same applies to photolithographic techniques that are used to make electronic chips. The technology used to develop the first chips couldn't make the chips of today &amp;amp; the engineers of the day didn't know how to develop techniques to make the chips that are made now.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-02-06T08:11:50.807" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19268" PostTypeId="2" ParentId="19258" CreationDate="2018-02-06T09:05:54.470" Score="2" Body="&lt;p&gt;For a rectangular section, you don't need to integrate; &lt;a href=&quot;https://en.wikipedia.org/wiki/Second_moment_of_area#Rectangle_with_centroid_at_the_origin&quot; rel=&quot;nofollow noreferrer&quot;&gt;it's a solved solution&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;I &amp;amp;= \dfrac{bh^3}{12}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the actual error in your integration - I don't know; I'm afraid I left that behind when I graduated 10 years ago. In the real world (as a civil/structural engineer) I've only used simple sections with solved solutions, or used software to calculate it for me.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-02-06T09:05:54.470" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19269" PostTypeId="2" ParentId="19228" CreationDate="2018-02-06T11:31:43.667" Score="3" Body="&lt;p&gt;So, I attempted to solve a simplified problem using Bernoulli, it is an approximation though, since it doesn't account for the change in flow direction and subsequent turbulence caused after the hole. I rearranged equations for volumetric flow rate and head loss. &lt;strong&gt;Please let me know if you spot any errors.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AYuO7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AYuO7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; the last two lines rearrange from an earlier form of the equation to find pressure drop, and &lt;strong&gt;&lt;em&gt;SG&lt;/em&gt;&lt;/strong&gt; refers to specific gravity of the fluid. &lt;/p&gt;&#xA;" OwnerUserId="14498" LastActivityDate="2018-02-06T11:31:43.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19270" PostTypeId="2" ParentId="15877" CreationDate="2018-02-06T12:32:19.320" Score="0" Body="&lt;p&gt;Yes, this is possible in &lt;a href=&quot;http://www.lusas.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;LUSAS&lt;/a&gt; from version 15 onwards (the current version is version 16).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This software has the concept of &quot;multiple analyses&quot; where each analysis can have different attributes, e.g. cross-section properties, material, support conditions. You can combine and/or envelope results across different analyses.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Disclaimer - I am a LUSAS employee. This answer is posted personally and not in my capacity as a LUSAS employee, even though I'm posting it from my work laptop.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-02-06T12:32:19.320" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19271" PostTypeId="2" ParentId="19263" CreationDate="2018-02-06T15:11:03.330" Score="1" Body="&lt;p&gt;You omitted the inertia of the mass of $0.5 kg$. The total inertia of the system is the inertia of the disk plus the inertia of the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I = \frac{M R^2}{2} + m R^2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total torque about the axle is $T= m g R$. Then use $I \alpha =T$ to get the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \left(\frac{M R^2}{2} + m R^2\right)\alpha=m g R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha=\frac{2 g m}{R (2 m+M)}=\frac{(2) (9.8) (0.5)}{(0.2)((2)(0.5)+2.5)}=\frac{9.8}{0.7}=14$$  &lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2018-02-06T15:11:03.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19273" PostTypeId="2" ParentId="19266" CreationDate="2018-02-06T18:32:19.057" Score="2" Body="&lt;p&gt;The component density on chips made by photo lithography is directly dependant on the wavelength of the light source used and as shorter wavelengths equate to higher photon energies you need to develop practical sources for ever increasing frequencies and the masks and dyes to go with them, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally you need more and more precise and clean manufacturing plant. Once you get to UV lithography a stray hair or skin cell is like hitting your machines with a big hammer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Part of this is the investment required to manufacture and develop the tooling required for example you need to be able to create very hard vacuums and incredible precise mirrors. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-02-06T18:32:19.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19277" PostTypeId="1" CreationDate="2018-02-06T22:48:52.030" Score="3" ViewCount="177" Body="&lt;p&gt;As the question states, do vibrations cause any influences on inductive sensors? For example, do vibrations on rail, from a locomotive, affect signalling equipment?&lt;/p&gt;&#xA;" OwnerUserId="14833" LastEditorUserId="90" LastEditDate="2018-02-07T07:40:42.760" LastActivityDate="2018-10-05T16:00:19.377" Title="Do vibrations affect inductive sensors?" Tags="|electrical-engineering|vibration|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19278" PostTypeId="2" ParentId="19277" CreationDate="2018-02-06T23:03:10.090" Score="1" Body="&lt;p&gt;it depends on whether or not there are any ferrous metal parts nearby and whether the vibration of the sensor is out of phase with the vibration of the metal part. If so, then the sensor will pick up a vibration signal. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-02-06T23:03:10.090" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19279" PostTypeId="1" AcceptedAnswerId="19281" CreationDate="2018-02-06T23:17:41.103" Score="4" ViewCount="659" Body="&lt;p&gt;From what I understand most types of steel are strong both in tension and compression, yet they are still usually significantly stronger in tension. Suppose a common I-beam supported on both ends and with a load on top of it; the bottom flange experiences tension, while the top one compression, yet since steel is somewhat stronger in tension, why would the bottom flange be as thick as the top one? Is it not done because it's easier to manufacture symmetrical ones or is there some other reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to use assymetrical I-beam to save weight and steel in that case or is my thinking completely off? &lt;/p&gt;&#xA;" OwnerUserId="14834" LastActivityDate="2018-02-07T03:16:37.780" Title="Why are steel I-beams mostly symmetrical?" Tags="|structural-engineering|steel|beam|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19280" PostTypeId="2" ParentId="19243" CreationDate="2018-02-06T23:57:05.387" Score="1" Body="&lt;p&gt;&lt;strong&gt;Second Attempt:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading your comments, I approached the problem using a different process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1:&lt;/strong&gt;&#xA;Find the y-forces using  $\Sigma$$F_y = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(2\cdot Tsin(\theta))$ - $(W\cdot L)$ $=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2Tsin(\theta)$ = $WL$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2:&lt;/strong&gt; Treat the beam as a collection of beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to break the beam into three parts.  The outside edges can be thought of as cantilever beams that are attached by the supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3:&lt;/strong&gt; Making the first cut at the midpoint to the left support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the formula for bending moment of a simple beam under a uniformly distributed load, we can determine that $M_{mid} = \frac{WS^2}{8}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 4:&lt;/strong&gt; Making the second cut at the left sling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the formula for bending moment of a cantilever beam under a uniformly distributed load, we can determine that $M_{cantilever} = \frac{Wx^2}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 5:&lt;/strong&gt; Setting equations equal and solving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From left end, to get the sum of moments to equal zero, $M_{cantilever} = \frac{M_{mid}}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{WS^2}{16} =\frac{Wx^2}{2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$WS^2=8Wx^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S^2=8x^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$S = 2\sqrt{2}x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Final:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the center span $S = 2\sqrt{2}x$.  Because $S = L-2x$, you can substitute to find the equation $L-2x = 2\sqrt{2}x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L = 2\sqrt{2}x + 2x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$L \approx 4.8284x$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x \approx \frac{1}{4.8284L} \approx 0.2071L$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;_&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $x$ is the distance from a support on the beam to the left end of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $S$ is the span of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $L$ is the length of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, the beam of length $L$ is divided into three sections; the left cantilever of length $x$, the center span $S$ of length $L-2x$, and the right cantilever of length $x$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $Tsin(\theta)$ is y-component of the tension force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $W$ is the weight of the beam.&lt;/p&gt;&#xA;" OwnerUserId="14835" LastEditorUserId="14835" LastEditDate="2018-02-08T01:00:16.233" LastActivityDate="2018-02-08T01:00:16.233" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19281" PostTypeId="2" ParentId="19279" CreationDate="2018-02-07T02:53:59.433" Score="5" Body="&lt;p&gt;Academically, I believe you are correct in that you could potentially optimize an I-beam to have a higher failure-load to mass ratio. You would have to play with the &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_second_moments_of_area&quot; rel=&quot;noreferrer&quot;&gt;area moment of inertia&lt;/a&gt; equations to check for sure. &lt;a href=&quot;https://en.wikipedia.org/wiki/T-beam&quot; rel=&quot;noreferrer&quot;&gt;T-beams&lt;/a&gt; would be the extreme of what you are proposing. They are less common, but composite steel/concrete T-beams are used extensively in parking garages, with additional steel in the lower tensile section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to note that many structures are design-limited by maximum deflection rather than structural failure. Additional area on the tensile member will reduce this defection. It would be interesting to see how deflections would compare between a symmetrical and the hypothetical failure-load optimized beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Symmetrical is superior for torsional loads or compound loads where the flange loading may be variable. Symmetrical is also superior for columns because it reduces eccentricity and the potential for buckling for a given mass/length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being able to use a beam in multiple situations is important because structural steel is a commodity. The price of a particular style of I-Beam is inversely proportional to how much it is used in industry in your region. This is often a bigger factor than the cost of the additional steel in the beam or cost of the additional mass in your design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally if your structure requires lots of trapezoidal shapes, you want to use symmetrical material so you waste less material.&#xA;Right-side-up &gt; Upside-down &gt; Right-side-up &gt; Upside-down...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for every symmetrical beam, it saves the engineers an orientation specification in the drawings, and saves the fabricators a potential error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a good &lt;a href=&quot;http://publications.lib.chalmers.se/records/fulltext/182546/182546.pdf&quot; rel=&quot;noreferrer&quot;&gt;I-beam thesis&lt;/a&gt; to research further.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2018-02-07T03:16:37.780" LastActivityDate="2018-02-07T03:16:37.780" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19282" PostTypeId="2" ParentId="19258" CreationDate="2018-02-07T03:31:20.080" Score="4" Body="&lt;p&gt;While I agree with &lt;a href=&quot;https://engineering.stackexchange.com/a/19268/1832&quot;&gt;AndyT's philosophy&lt;/a&gt; of simply using the solved equations, it is important to learn the origin of these solutions, so here we go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mistake you made was that you didn't perform the integration at all. You simply got the equation for the moment of inertia, ignored the integration sign, set $y$ as half the section's height, and replaced $\text{d}A$ with the section's area. That's not integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;I &amp;amp;= \int_A y^2\text{d}A \\&#xA;&amp;amp;= b\int_{-\frac{h}{2}}^{\frac{h}{2}} y^2\text{d}y \\&#xA;&amp;amp;= b \left.\dfrac{y^3}{3}\right|_{-\frac{h}{2}}^{\frac{h}{2}} \\&#xA;&amp;amp;= \dfrac{bh^3}{3}\left(\dfrac{1}{8}+\dfrac{1}{8}\right) \\&#xA;&amp;amp;= \dfrac{bh^3}{12}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the correct moment of inertia, calculating the stress is straightforward.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2018-02-07T17:48:18.213" LastActivityDate="2018-02-07T17:48:18.213" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19283" PostTypeId="2" ParentId="19157" CreationDate="2018-02-07T07:08:01.443" Score="0" Body="&lt;p&gt;&lt;em&gt;Why the closed loop dominant poles of the root locus can show the response of the system? Wont it neglect the effect of the closed loop zeros?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes , that is indeed considered as a weakness of rootlocus - its lack of capability to predict closed loop zero's effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ref: &lt;a href=&quot;http://lpsa.swarthmore.edu/Root_Locus/RLocusExamples.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://lpsa.swarthmore.edu/Root_Locus/RLocusExamples.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quoting from the above reference&#xA;&quot;While the poles of a system (the roots of the denominator polynomial) are very important in determining the behavior of a system, the zeros of the system (the roots of the numerator polynomial) can also be important.  After performing a root-locus design, it is critical to go back and test the closed loop system to ensure that it behaves as expected&quot;&lt;/p&gt;&#xA;" OwnerUserId="14840" LastActivityDate="2018-02-07T07:08:01.443" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19284" PostTypeId="1" CreationDate="2018-02-07T08:42:04.667" Score="2" ViewCount="98" Body="&lt;p&gt;For photography there are tilt-shift lenses.&#xA;For video projectors there is lens shift lenses.&#xA;With video projector lens shift I get perfect, undistorted video projected up to 100% above, below, left or right from center of the lens.&#xA;Same with the shifting function of tilt-shift camera lenses, the offset changes but photo is undistorted from that, so several photos can be taken and stitched together in photoshop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never had chance to look inside either tilt shift camera lenses or video projection lenses with lens shift feature. Both are too expensive for me to attempt to disassemble. Nor have I had luck finding info on someone else doing it or explaining what's inside.&#xA;My only guess has been that either all some lens elements are physically shifted along the DMD or LCD but I haven't been able to understand how that won't cause distortion as center of lens changes for other lenses and DMD/LCD.&#xA;The only place I've found any info on this has been a youtube video for a DIY video projector where someone used a (fresnel) lens tilting, not shifting.&#xA;&lt;a href=&quot;https://youtu.be/Rp8HIf5eUpo?t=4m45s&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/Rp8HIf5eUpo?t=4m45s&lt;/a&gt;&#xA;The projector in that video is the only one I've seen where someone attempts to use fresnel lens tilting to do keystone correction optically rather than in software, and nothing is done for shifting.&lt;/p&gt;&#xA;" OwnerUserId="14841" LastActivityDate="2018-02-07T08:42:04.667" Title="how do tilt-shift DSLR lenses and video projector lens shift lenses work?" Tags="|mechanical-engineering|optics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19285" PostTypeId="1" CreationDate="2018-02-07T10:21:31.810" Score="3" ViewCount="126" Body="&lt;p&gt;What is the difference between installing a humidifier before a heater and after a heater?&lt;/p&gt;&#xA;" OwnerUserId="13130" LastActivityDate="2018-02-09T11:12:37.603" Title="A humidifier before a heater and after a heater" Tags="|hvac|heating-systems|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19287" PostTypeId="2" ParentId="19285" CreationDate="2018-02-07T12:13:06.183" Score="4" Body="&lt;p&gt;warm air can hold more moisture than cold air so the amount of water you can suspend in the air will change depending on where you install the humidifier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mounting the &lt;em&gt;humidifier&lt;/em&gt; before the heater will give you a final relative humidity below 100% after the heater. This can be important if you want to avoid dew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mounting the &lt;em&gt;heater&lt;/em&gt; before  the humidifier will allow you a higher absolute humidity after the setup.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2018-02-09T11:12:37.603" LastActivityDate="2018-02-09T11:12:37.603" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19288" PostTypeId="1" AcceptedAnswerId="19290" CreationDate="2018-02-07T14:13:03.557" Score="5" ViewCount="436" Body="&lt;p&gt;I am trying to solve this but I am stuck; I have watched loads of YouTube videos but still don't understand how to complete it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A mass of $m=0.12 kg$ of air has an initial temperature of $T_1=500°C$ and pressure of $p_1=0.8 MPa$. If the air is expanded according to the law $pV^{1.2} = c$ to a final volume of $90\ litres$, determine&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i) its initial volume, $V_1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii) its final pressure, $p_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iii) its final temperature. $T_2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For air, take $R_{specific} = 287 Jkg^{-1} K^{-1}.$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I have got these equations which i believe i need to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$pV = nRT$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n = \frac{p  V}{R  T}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these the correct equations to use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also think the fixed amount of gas is the constant?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking at Boyle's law. And Charles' law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="14101" LastEditorUserId="13311" LastEditDate="2018-02-07T17:46:35.333" LastActivityDate="2018-02-08T17:32:13.897" Title="Applications of Thermodynamics" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19290" PostTypeId="2" ParentId="19288" CreationDate="2018-02-07T15:13:00.423" Score="5" Body="&lt;p&gt;First step, is to find the value of '$n$'. The molar mass of air is $29 g/mol$, meaning that in $0.12kg$, you have $4.144mol$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember to convert all your units into standard SI values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$p_1=800,000Pa$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$n=4.144mol$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$R=8.31441JK^{-1} mol^{-1}$ - N.B. You have been suplied with the R in terms of kg, rather than mol. This &lt;em&gt;can&lt;/em&gt; make the maths easier, but is often a source of confusion. I recommend always using the Universal gas constant, and calculating the number of mol that you have, rather than using Specific Constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T_1=773.15K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging $pV=nRT$ to give $V=\frac{nRT}{p}$ allows you to calculate the initial volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i)  $V_1=\frac{4.144*8.31441*773.15}{800,000}=0.03339m^3$ You can verify this &lt;a href=&quot;http://www.wolframalpha.com/input/?i=air%200.12kg%20500degC%200.8MPa%20volume%20in%20m%5E3&quot; rel=&quot;nofollow noreferrer&quot;&gt;via Wolfram|Alpha&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the fact that $pV^{1.2}=c$ means that you can state that $p_1V_1^{1.2}=p_2V_2^{1.2}$, which can be rearranged to give $p_2=\frac{p_1V_1^{1.2}}{V_2^{1.2}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii)  $p_2=\frac{800,000*0.03339^{1.2}}{0.09^{1.2}}=243410Pa=0.243MPa$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, rearranging $pV=nRT$ to give $T=\frac{pV}{nR}$ allows us to calculate the final temperature:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iii) $T_2=\frac{243,410*0.09}{4.144*8.31441}=636.4K=363.3\unicode{x2103}$. This can also be &lt;a href=&quot;http://www.wolframalpha.com/input/?i=air%200.12kg%20363.3degC%200.243MPa%20volume%20in%20m%5E3&quot; rel=&quot;nofollow noreferrer&quot;&gt;verified in the same way as before&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-02-07T16:47:36.793" LastActivityDate="2018-02-07T16:47:36.793" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19291" PostTypeId="2" ParentId="19285" CreationDate="2018-02-07T15:57:47.543" Score="1" Body="&lt;p&gt;There's a secondary consideration: temperature.  It takes a lot of energy to convert liquid water to gaseous water.  If your proposed humidifier is not generating steam (gaseous water), then wherever the humidifier is placed, the air will be cooled as the water or water mist evaporates.  If before the heater, and the heater actively manages the output air temperature, then you won't see an effect.  If after the heater, the temperature of the air sent to the house proper will be lower, and it'll take longer to reach the design temperature. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-02-07T15:57:47.543" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19292" PostTypeId="1" AcceptedAnswerId="19299" CreationDate="2018-02-07T16:06:19.180" Score="4" ViewCount="631" Body="&lt;p&gt;I'm curious about which scenarios one would use a force function when using a PLC programming interface. Examples pertaining to process engineering, manufacturing, or industrial settings would be ideal.&lt;/p&gt;&#xA;" OwnerUserId="14846" LastActivityDate="2018-02-08T01:54:56.713" Title="Real world applications where one would use a force function in PLC programming?" Tags="|manufacturing-engineering|process-engineering|instrumentation|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19293" PostTypeId="1" AcceptedAnswerId="19296" CreationDate="2018-02-07T16:24:03.087" Score="2" ViewCount="729" Body="&lt;p&gt;This may be an obvious question, but here goes anyhow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have two pulleys (different diameters) with infinitely strong bearings and shafts, does increasing the belt tension have any effect on the torque required to turn the system? I.e. Is a super tight belt is harder to turn than a weakly tensioned belt (assuming no belt slip in either case)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the cost and wear implications of doing this as increasing the tension will obviously wear out your belts and bearings significantly faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done a force balance, and it seems as though, no, everything balances out. If you increase the tension on a belt you're increasing the force on both sides and the torque on the pulley input seems to be independent of the belt tension. When the system is moving at a constant speed under a constant load, belt tension doesn't seem to factor in (again ignoring the loads on the bearings and shaft)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However from a practical point of view it seems to make intuitive sense that if you crank up the belt tension it'll be harder to turn them. Is my intuition wrong, or is there something I've over looked?&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2018-02-08T07:52:28.790" Title="Belt tension to rotational force" Tags="|design|pulleys|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19294" PostTypeId="2" ParentId="18760" CreationDate="2018-02-07T16:51:23.450" Score="-2" Body="&lt;p&gt;Your question is too broad, we might come up weigh an answer for a specific scenario but would need to know the specific weaknesses and design for that instance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only general answer I can give is too add strength to the weak areas. This can be determined by modeling the system in a computer and using finite element analysis. A civil engineer has tools and the experience to examine and design strengthening measures. &lt;/p&gt;&#xA;" OwnerUserId="10516" LastActivityDate="2018-02-07T16:51:23.450" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19296" PostTypeId="2" ParentId="19293" CreationDate="2018-02-07T17:25:31.977" Score="3" Body="&lt;p&gt;Your intuition is correct in &quot;real life&quot;, but incorrect within the specific parameters laid out in your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &quot;real life&quot; (where your intuition rightly stems from), the extremely tight belt will put a high radial load on the bearings, and this will introduce friction into the system meaning it takes more torque to turn the pulleys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your imaginary perfect system, the extra tension will simply be balanced by higher reaction forces at the bearings, with no effect on the torque/friction, since the bearings are perfect.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-02-07T17:25:31.977" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19297" PostTypeId="2" ParentId="19293" CreationDate="2018-02-07T18:37:14.793" Score="0" Body="&lt;p&gt;The answer depends on whether the belt is flat/toothed or smooth/vee-profile. In the case of the vee belt, increasing the belt tension increases the work it takes to pull the belt out of the pulley groove as the pulley rotates. This increases the amount of work it takes to rotate the two-pulley system and heats up the belt. This effect does not occur if the belt engages the pulley surface with teeth. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-02-07T18:37:14.793" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19298" PostTypeId="2" ParentId="19251" CreationDate="2018-02-07T19:39:16.297" Score="1" Body="&lt;p&gt;Before the &quot;3 ring circus,&quot; as we jumpers called it, were the Capewell variations.  Here are some pics:&lt;a href=&quot;https://www.flickr.com/photos/43867826@N07/sets/72157622676844920/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.flickr.com/photos/43867826@N07/sets/72157622676844920/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;shot and a half&quot; required moving a cover to access the release mechanism.  &quot;One shots&quot; released by just moving the cover. &lt;/p&gt;&#xA;" OwnerUserId="7768" LastActivityDate="2018-02-07T19:39:16.297" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19299" PostTypeId="2" ParentId="19292" CreationDate="2018-02-08T01:54:56.713" Score="4" Body="&lt;p&gt;A &quot;force&quot; in a PLC is the act of forcing a memory bit on or off from the PLC programming software. It is used exclusively for testing. Forcing a bit on or off over-rides all other PLC logic. You can think of it like both setting it and making it read only. &quot;set&quot; a bit is also an option with most PLCs and is also useful for testing. It differs from &quot;force&quot; in that it changes the value only once and does not over-ride any logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if I wanted to test if a light has failed, I can &quot;force&quot; its output bit in the PLC software to determine if it has failed or if it is an issue with the logic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to test a if some latching logic works, I would &quot;set&quot; the trigger and see if it responds how I expected.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-02-08T01:54:56.713" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19300" PostTypeId="1" CreationDate="2018-02-08T02:54:26.390" Score="2" ViewCount="292" Body="&lt;p&gt;Why can I not tow 50,000 lbs (40,000 cargo and 10,000 Trailer bed) using multiple pickup truck each individually capable of towing 30,000 lbs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7DiEw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7DiEw.png&quot; alt=&quot;Trailer Weight Combinations&quot;&gt;&lt;/a&gt;&#xA;I was told that the towing capacity does not &quot;Add up&quot; where if one SUV can tow 30,000 lbs. Two SUVs can only Tow 30,000 lbs (maybe a little more).&#xA;Is that true?&lt;/p&gt;&#xA;" OwnerUserId="14691" LastEditorUserId="14691" LastEditDate="2018-03-07T15:13:47.020" LastActivityDate="2018-03-07T15:13:47.020" Title="Vehicle Train Towing" Tags="|mechanical-engineering|structural-engineering|automotive-engineering|mechanical|" AnswerCount="1" CommentCount="3" ClosedDate="2018-02-08T19:50:57.380" ContentLicense="CC BY-SA 3.0" />
  <row Id="19301" PostTypeId="1" CreationDate="2018-02-08T03:12:23.993" Score="2" ViewCount="181" Body="&lt;p&gt;I have a machine/robot that has a manual joystick to control forward and reverse. The joystick is connected to a mechanical linkage system that goes forward/backward, which than controls a hydraulic pump. I want to be able to &quot;hack&quot; the joystick to control the forward/backward motion through software. I have a few design criteria that I am trying to work around, hoping to get some input/help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) I need a failsafe. If I cut the power while the machine is moving and then turn it back on, I would like the machine to start in neutral, not where (forward/backward) it last was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Its not a binary forward/backward so some voltage control would be nice  to make it smoother.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Able to control manually and through software. I would like to send X voltage to it through software (arduino + hardware) but not lose functionality if someone wants to take physical control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice on what kind of set up I would need to do this?&lt;/p&gt;&#xA;" OwnerUserId="14849" LastActivityDate="2018-09-06T15:02:59.697" Title="Voltage controlled linear actuator with spring back" Tags="|mechanical-engineering|linear-motors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19302" PostTypeId="2" ParentId="19301" CreationDate="2018-02-08T06:38:00.717" Score="1" Body="&lt;p&gt;I suggest:&#xA;joystick-&gt;arduino -&gt;motor driver for arduino-&gt; micro peristaltic pump&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;1) I need a failsafe.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;no prob as long as joystick returns to center position( joysticks usually have springs for that)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;2) Its not a binary forward/backward so some voltage control&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;arduino has pwm output, if you put a capacitor you get analog output, but maybe that's not even required because the driver will convert that for you.&#xA;Same for forward reverse, if the driver doesnt do it, you'll need a H-bridge to switch polarity on the pump&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;3) Able to control manually and through software.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;both manual and software control will go through the arduino, so it's just a few lines of code to define the priority&lt;/p&gt;&#xA;" OwnerUserId="5926" LastEditorUserId="5926" LastEditDate="2018-02-08T10:52:17.963" LastActivityDate="2018-02-08T10:52:17.963" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19303" PostTypeId="1" CreationDate="2018-02-08T07:07:54.067" Score="1" ViewCount="57" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CQYk4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CQYk4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/evGQh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/evGQh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ttt1y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ttt1y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UaG48.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UaG48.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Grgua.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Grgua.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already had a data for corroded thickness distribution (in matrix form) for corroded surface model in Abaqus with mesh size 5 cm&#xA;I want to create thickness distribution with bigger mesh size (e.g 50 cm) but I dont want to use average thickness to obtain the smaller matrix size,&#xA;I want to use effective thickness by utilizing the equations in the picture.&#xA;But till today I still have no idea how to use those equations.&#xA;Does anyone have any idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you &lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-02-08T07:07:54.067" Title="How to find all the parameters in weight vector in Levenberg-Marquardt Algorithm?" Tags="|abaqus|statistics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19304" PostTypeId="2" ParentId="19293" CreationDate="2018-02-08T07:52:28.790" Score="0" Body="&lt;p&gt;From reading your answers (provided as comments) to the answers given, the “best solution” to increasing the load capacity is to change to multi-belt pulleys: 2 and up to 8 are used in some situations - belts can provide a “softness “ compared to gears or a shaft.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-02-08T07:52:28.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19305" PostTypeId="2" ParentId="19300" CreationDate="2018-02-08T09:51:46.290" Score="3" Body="&lt;p&gt;In your example the pickup trucks are not load bearing because the weight of the trailer is resting on the trailer wheels. As such the drive wheels of the pickup trucks are likely to skid. The goose necks have no additional function since they do not carry any weight from the trailer, using a chain or towstrap between the trucks would have the same effect.&lt;/p&gt;&#xA;" OwnerUserId="14479" LastActivityDate="2018-02-08T09:51:46.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19306" PostTypeId="1" CreationDate="2018-02-08T10:07:52.630" Score="6" ViewCount="117" Body="&lt;p&gt;Given futuristic materials, could a structure ever be strong enough to contain a vacuum, yet light enough to provide buoyancy under standard atmospheric pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously &quot;futuristic materials&quot; is highly ambiguous because we don't know what exists in the future... but what properties would it take? Are there theoretical materials that provide the right properties that are at least compliant with the laws of chemistry?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, if you created a woven carbon fibre bag and then somehow impregnated it with seamless industrial diamond, would that be sufficient?&lt;/p&gt;&#xA;" OwnerUserId="7205" LastActivityDate="2018-02-08T13:16:23.710" Title="Could a vacuum tank ever be light enough to be boyant in air?" Tags="|vacuum|strength|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19307" PostTypeId="1" AcceptedAnswerId="19322" CreationDate="2018-02-08T11:34:52.863" Score="3" ViewCount="3027" Body="&lt;p&gt;When I am drawing things (Assembly) in SolidWorks, on some parts which I work on, I can select the edge of a circle and then measure the dimension to a line, like shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cg60g.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cg60g.png&quot; alt=&quot;Measuring Edge of Circle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem is that on some others I simply can't. It won't allow me to select the Edge of a cirlce when I hold down the SHIFT key. If I &quot;insist&quot; and drag the measurement to an edge, it jumps to the other end, like shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DbeHY.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DbeHY.gif&quot; alt=&quot;Opposite Edge selected&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I'm doing everything right: Doubleclick Sketch to edit it, select a line, then hold SHIFT and then try selecting the edge of a circle. I already searched on the internet around, but no one had really a solution for it. The closest I came was &lt;a href=&quot;https://forum.solidworks.com/thread/19072&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does it work on some projects and on some others it don't? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'm using SolidWork 2016 Premium x64 (SP 0.0)&lt;/p&gt;&#xA;" OwnerUserId="14853" LastEditorUserId="14853" LastEditDate="2018-02-08T11:37:07.643" LastActivityDate="2022-04-11T10:00:03.010" Title="Selecting edge of a circle in Solidworks" Tags="|measurements|solidworks|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19308" PostTypeId="1" AcceptedAnswerId="19311" CreationDate="2018-02-08T11:39:28.003" Score="2" ViewCount="5705" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7bcPB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7bcPB.png&quot; alt=&quot;internal hinge&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a beam with 1 internal hinge just like the picture. We know the internal hinge can not transfer moments between AB and BD. We know the 30 kN force makes some moments on BD &amp;amp; AB. If B (internal hinge) can not transfer moments between the two elements, then why does the 30&amp;nbsp;kN (which is applied on BD) cause moments in AB?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if i'm clear enough but the 30&amp;nbsp;kN force makes some moments on AB while we know that no moments can transfer between AB &amp;amp; BD.&lt;/p&gt;&#xA;" OwnerDisplayName="user14854" LastEditorUserId="1832" LastEditDate="2018-02-08T12:15:07.223" LastActivityDate="2020-10-04T15:04:12.943" Title="Transfer of moments in beams with internal hinges" Tags="|structural-analysis|structures|beam|moments|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19309" PostTypeId="2" ParentId="19306" CreationDate="2018-02-08T11:51:21.157" Score="-3" Body="&lt;p&gt;It is possible with materials that is &quot;lightweight&quot; and &quot;high compressive stress&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or create a wireframe globe that covered with thin rubber on the outside and then we can vacuum it.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2018-02-08T11:51:21.157" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19310" PostTypeId="2" ParentId="19261" CreationDate="2018-02-08T11:55:31.617" Score="3" Body="&lt;p&gt;I agree with agentp. Even a micro motor is enough power but will take long time to climb. The motor torque shall be designed to overcome load torque, off course.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2018-02-08T11:55:31.617" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19311" PostTypeId="2" ParentId="19308" CreationDate="2018-02-08T12:13:35.857" Score="3" Body="&lt;p&gt;Imagine if AB didn't exist, so all we have is BD with the pinned support at D and the force at C. In this case, BD is hypostatic and becomes a mechanism, rotating around D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, we know that doesn't happen when AB is there. This tells us that the hinge is supporting BD, apparently generating an upwards vertical force at B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Newton's Third Law tells us that every action causes an equal and opposite reaction. So if BD &quot;feels&quot; an upwards vertical force at B, then AB will &quot;feel&quot; an equal downwards vertical force at B as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;downwards force&quot; at B (as seen by AB) will then cause a bending moment in AB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more visual demonstration, notice that this structure can be replaced by two individual beams AB and BD. BD is a simply-supported beam, and AB is a cantilever with a concentrated load equal to the reaction found in BD's support at B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ubU21.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ubU21.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Diagram obtained with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis program.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-02-08T12:13:35.857" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19312" PostTypeId="1" CreationDate="2018-02-08T12:27:52.737" Score="1" ViewCount="44" Body="&lt;p&gt;My aim was to damp the vibration of a stepper motor ( Nema 8 ) using only spring systems. I have taken modern control course and I wanted to apply it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I first analyzed my open loop system and found it to be highly unstable. I then used Ackermans formula to find State feedback $K$ after checking it to be controllable and observable, and I found $K$. I plugged my $K$ in Matlab and the solution becomes perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;THE PROBLEM:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is this magical number (matrix) $K$ I need to insert in my system to make this work? What is the value of $K$ in real life? What does $K$ represent?&lt;/p&gt;&#xA;" OwnerUserId="14856" LastEditorUserId="1832" LastEditDate="2018-02-08T14:45:10.327" LastActivityDate="2018-02-08T14:45:10.327" Title="Applying modern control systems to real mechanical engineering problems" Tags="|vibration|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19313" PostTypeId="2" ParentId="19306" CreationDate="2018-02-08T13:16:23.710" Score="5" Body="&lt;p&gt;Start out by doing some basic calculations.  Air is mostly N&lt;sub&gt;2&lt;/sub&gt;, with a atomic mass of 28, and O&lt;sub&gt;2&lt;/sub&gt;, with a atomic mass of 32.  There is more nitrogen than oxygen, so let's say air has a average atomic mass of 29.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 0 &amp;deg;C and 1 atm, one mole of ideal gas occupies 22.4 liters.  Let's say 20 &amp;deg;C is more realistic, so we'll scale that by the absolute temperature ratio (22.4 l)(293 &amp;deg;K)/(273 &amp;deg;K) = 24.0 l.  So the density of what we will call &quot;air&quot; is (29 g)/(24 l) = 1.2 kg/m&lt;sup&gt;3&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for every cubic meter you can enclose, you get to use 1.2 kg of material.  That by itself is not a big deal.  The gotcha is that the surface has to somehow be able to withstand about 15 pounds per square inch, or about 11.4 tons per square meter, or 101 kN/m&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you want to make a sphere 10 m in diameter.  That has a volume of about 524 m&lt;sup&gt;3&lt;/sup&gt;, so you can use up to about 630 kg of material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be pretty straight forward to calculate the thickness of various materials required to make a 10 m diameter sphere than can withstand 1 atm of external pressure.  None of the materials we have today come close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that's a really naïve way to make a vacuum tank.  A suitable internal lattice structure will save well more than its own weight in weight of the skin material.  The real engineering issues are with the clever design of the internal supports.  The optimum arrangement of internal supports depends on the relative strengths and densities of the available materials, so you can't just create one design then work backwards to see what the material properties need to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To really answer the question, you'd have to pick some material properties, iterate to find the optimum structure design, see where you're at, adjust the material properties, adjust the structure to the new tradeoffs, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that scale matters.  A sphere with twice the diameter gives you eight times the mass budget, but only four times the surface area.  There is obviously some win in going larger.  On the other hand, larger diameters give you less advantage from the arch effect due to the curvature, and the interior supporting structures don't scale the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, I expect that this is solved with a multi-dimensional iterative design process.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-02-08T13:16:23.710" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19314" PostTypeId="1" CreationDate="2018-02-08T13:20:01.183" Score="0" ViewCount="818" Body="&lt;p&gt;I am trying to find easier way to get stress and strain values in a specific element in Abaqus.&#xA;Does anyone know how to do it? Is it possible to do with Query?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2023-12-10T18:05:35.613" Title="How to query stress and strain values in a specific element in Abaqus?" Tags="|abaqus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19315" PostTypeId="1" CreationDate="2018-02-08T13:34:05.303" Score="1" ViewCount="143" Body="&lt;p&gt;I am trying to enter frequency dependent viscoelastic properties for an elastomer into abaqus.  I found where to enter the properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit Material-&gt;Mechanical-&gt;Elasticity-&gt;Viscoelastic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Domain: Frequency&#xA;Frequency: Tabular&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The test data that I have to enter is G* (complex) in which the real component is the shear storage modulus and the imaginary component is the loss modulus.  Using this and the Poisson ratio I can compute the complex bulk modulus, K*.  However abaqus is looking for the following test data as a function of frequency:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Omega g* real&#xA;Omega g* imag&#xA;Omega k* real&#xA;Omega k* imag&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these the values for shear and bulk modulus or are they the Fourier transform of these quantities?&lt;/p&gt;&#xA;" OwnerUserId="14857" LastActivityDate="2018-02-08T13:34:05.303" Title="Entering frequency dependent elastomer properties in abaqus" Tags="|abaqus|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19316" PostTypeId="1" AcceptedAnswerId="19325" CreationDate="2018-02-08T15:31:11.583" Score="3" ViewCount="4274" Body="&lt;p&gt;How can I calculate thrust in Newton from mass flowrate and velocity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please  note,I have mathematical skills of 7 year old,the most complex task I can do is multiply and divide,I dont understand any equations.&lt;/p&gt;&#xA;" OwnerUserId="10080" LastActivityDate="2018-02-08T19:55:21.750" Title="How to calculate thrust from mass flowrate and velocity" Tags="|airflow|thrust|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19318" PostTypeId="1" AcceptedAnswerId="19320" CreationDate="2018-02-08T16:58:55.533" Score="2" ViewCount="460" Body="&lt;p&gt;Been brushing up on my control systems knowledge and I keep seeing that a P controller WILL NOT have 0 error and that doesn't make sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a tank problem, the setpoint is 50 cm, you open the drain valve and the level goes down to 49. Controller output goes up and you start putting water into the tank. Controller output does not go to 0 until the level is back to 50. So while this may take a while if your Kp is low, it will get there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that steady-state error is not measured at infinite time but the idea that a P controller will not reach the SP is odd to me. &lt;/p&gt;&#xA;" OwnerUserId="14861" LastEditorUserId="78" LastEditDate="2018-02-09T01:22:39.427" LastActivityDate="2023-09-30T09:38:14.340" Title="Why does a proportional controller have to have a steady state (offset) error?" Tags="|mechanical-engineering|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19319" PostTypeId="2" ParentId="19288" CreationDate="2018-02-08T17:32:13.897" Score="1" Body="&lt;p&gt;You can also solve the problem without knowing the average molar mass air, which is why R(specific) is given. &#xA;Multiplying R(specific) with mass will give you the number of moles multiplied by the universal gas constant.&#xA; i.e mR(specific)=nR&#xA;Proceed on the usual lines to get the answer after this.&lt;/p&gt;&#xA;" OwnerUserId="14751" LastActivityDate="2018-02-08T17:32:13.897" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19320" PostTypeId="2" ParentId="19318" CreationDate="2018-02-08T17:32:23.017" Score="4" Body="&lt;p&gt;If you open your drain valve, and then close it again, then you're correct, the tank will refill back up to 50, leaving you with no offset. If, however, the valve is left open, then the water may not come into the tank fast enough to compensate for the water that's leaving. This means that the level will continue to fall, up to the point where the proportional gain means that the rate of water in = water out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a P controller, the rate of water in will only be non-zero when there &lt;em&gt;is&lt;/em&gt; an error, therefore, if the outlet valve is open, then &lt;em&gt;there must be an error&lt;/em&gt;, or no water would come in at all. Integral action corrects for this offset.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-02-08T17:32:23.017" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19321" PostTypeId="1" AcceptedAnswerId="19324" CreationDate="2018-02-08T17:39:28.363" Score="0" ViewCount="27" Body="&lt;p&gt;I want to mount a small engine on a shaft of a large engine and rotate them both in the same direction, will the RPMs be a sum of the their RPMs or will the resulting number be a multiplication?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2018-02-08T18:28:21.020" Title="Does angular velocity of a body rotating on top of another rotating body add up or multiply?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19322" PostTypeId="2" ParentId="19307" CreationDate="2018-02-08T18:02:23.973" Score="1" Body="&lt;p&gt;You can achieve this simply and quickly by adding a point to the circumference of your circle. I like this method because it's easy to see exactly where the dimension is going to, and it's robust in the case where the line become non-horizontal etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5PuQG.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5PuQG.gif&quot; alt=&quot;point on circle demo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can achieve this more 'officially' by using the 'minimum' arc condition in the 'leader' tab of the dimension properties. I don't like this method, because relying on manually altering dimension properties doesn't sit well in my personal workflow. You may prefer it, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6yXIB.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6yXIB.gif&quot; alt=&quot;dimension options demo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more pertinent point, however, is that it's not good practice to dimension a part like this, since that's not how it will be manufactured. If possible, the dimensions in your sketches should match with those that would be desirable on the 2D manufacturing drawing.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-02-08T18:42:00.417" LastActivityDate="2018-02-08T18:42:00.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19323" PostTypeId="1" CreationDate="2018-02-08T18:26:35.863" Score="1" ViewCount="2175" Body="&lt;p&gt;How can I solve this problem using FEM (longhand calculation)? I want to find deflections at each load point. This is not a homework problem. This is related to a project that I and my brother is doing for the conveyor. The distributed load is conveyor and point loads are some features that we are trying to add to it. I tried to solve this by Integrating moment function but if I try to add/remove point load from this, I have to do that integration part again and in some cases, the distributed by leg supports instead of any beam. &#xA;I was thinking this problem can be generalized through FEA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) I want to find deflection at each node point (where the loads are) and maximum deflection &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I tried to solve this by double integration method but that method will only be specific to this beam that is why I thought of the FEM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) In this example, conveyor is resting on the beam that is the distributed load and when conveyor is supported by leg supports, there is no need for calculation as they are standard&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) I do not have FEA software package.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E=300\times10^6\text{ psi}=2068\ GPa\quad,\quad I=500\text{ in}^4=2.08\times10^{-4}\text{ m}^4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kj7hb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kj7hb.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14863" LastEditorUserId="14863" LastEditDate="2018-02-09T19:32:59.593" LastActivityDate="2018-02-09T19:32:59.593" Title="Simply Supported beam with multiple point loads and distributed load" Tags="|mechanical-engineering|structural-analysis|beam|finite-element-method|" AnswerCount="0" CommentCount="7" ClosedDate="2018-02-09T16:50:21.877" ContentLicense="CC BY-SA 3.0" />
  <row Id="19324" PostTypeId="2" ParentId="19321" CreationDate="2018-02-08T18:28:21.020" Score="2" Body="&lt;p&gt;It's definitely not multiplication! Imagine you had a motor spinning in your hand at 1000rpm, and you decided to turn it over once every 30 seconds (2rpm). Would the output shaft be spinning at 2000rpm, or 1002rpm?&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-02-08T18:28:21.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19325" PostTypeId="2" ParentId="19316" CreationDate="2018-02-08T19:55:21.750" Score="1" Body="&lt;p&gt;Well the best learning happens when there is a need, and the internet can answer almost any question. Easiest way to understand physics is to use &lt;a href=&quot;https://en.wikipedia.org/wiki/SI_derived_unit&quot; rel=&quot;nofollow noreferrer&quot;&gt;SI derived units&lt;/a&gt; and always carry your units through your equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;velocity = 5 m/s&#xA;&lt;br&gt;mass flowrate = ?kg/s&#xA;&lt;br&gt;volumetric flowrate = 5m^3/s&#xA;&lt;br&gt;Thrust: ? Newtons = kg*m/s^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked google what the &lt;a href=&quot;https://www.google.com/search?q=mass%20of%20cubic%20meter%20of%20air&amp;amp;oq=mass%20of%20cubic%20meter%20of%20air&quot; rel=&quot;nofollow noreferrer&quot;&gt;mass of a cubic meter of air&lt;/a&gt; was:&#xA;&lt;br&gt;density = 1.293 kg/m^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density lets us convert your volumetric flow into mass flow:&#xA;&lt;br&gt;(looks like googles calculator carries units now which is cool)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=5m%5E3%2Fs%20*%201.293%20kg%2Fm%5E3%20%3D&amp;amp;oq=5m%5E3%2Fs%20*%201.293%20kg%2Fm%5E3%20%3D&quot; rel=&quot;nofollow noreferrer&quot;&gt;5m^3/s * 1.293 kg/m^3&lt;/a&gt; = 6.465 kg/s&#xA;&lt;br&gt;mass flow = 6.465 kg/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we just multiply mass flow and velocity and confirm our units work out:&#xA;&lt;br&gt;&lt;a href=&quot;https://www.google.com/search?ei=eql8WvnjOpC8jwPgv4egCQ&amp;amp;q=6.465%20kg%2Fs%20*%205%20m%2Fs%20%3D&amp;amp;oq=6.465%20kg%2Fs%20*%205%20m%2Fs%20%3D&quot; rel=&quot;nofollow noreferrer&quot;&gt;6.465 kg/s * 5 m/s&lt;/a&gt; = 32.325N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the thrust is 32.325N or &lt;a href=&quot;https://www.google.com/search?ei=O6p8WtW7GaW4jAPQ9JzADQ&amp;amp;q=32.325%20Newton%20in%20pound%20force&amp;amp;oq=32.325%20Newton%20in%20pound%20force&quot; rel=&quot;nofollow noreferrer&quot;&gt;7.27 pounds force&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-02-08T19:55:21.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19326" PostTypeId="1" AcceptedAnswerId="19328" CreationDate="2018-02-08T20:10:32.683" Score="3" ViewCount="6546" Body="&lt;p&gt;I recently learned the basics of automotive clutches, more specifically friction clutches. Here is a simplified version of their operation explained to me (the car starts at rest): The engine and the transmission are connected via friction plates. When the clutch is pressed down, the plates are separated. When to clutch is released slowly, the plates press harder together and the friction between them causes the car, via transmission, to gain speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what confuses me is this: The car weights many tonnes, and the engine spin very fast. When the plates are coming into contact with each other, they are slipping until the car begins to move. Isn't this causing a huge amount of heat to be produced? And most of all, how are the friction plates manufactured so that they don't &quot;smoothen out&quot; very quickly, reducing the friction coefficient between them? I understand that they &lt;strong&gt;do&lt;/strong&gt; wear out and need to be replaced sometimes, but considering the high torque and speed of the engine, you would think they'd wear extremely quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, isn't the force needed to press the plates together enormous? Considering the large mass of the car, how kind of mechanism is used inside to car the press the plates together? And how can a person then, using only one foot, be able to again separate these plates? I assume the clutch pedal is connected to some kind of hydraulic system to amplify the force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've Googled a lot and watched several Youtube videos on the matter but they all simply show that friction holds the plates together and pressing the clutch releases them.&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2022-10-26T12:20:00.753" Title="Friction clutch and the force required" Tags="|mechanical-engineering|car|transmission|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19327" PostTypeId="2" ParentId="19326" CreationDate="2018-02-08T20:50:41.960" Score="3" Body="&lt;p&gt;Clutches do indeed carry quite lot of force but when you pull away from stationary the correct technique is to use fairly low revs and blend the clutch in quickly but smoothly. In this case the clutch is only slipping for a second or so and the torque involved is fairly small. If you are more aggressive or slip the clutch for long periods you can indeed wear it out very quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friction surface is also typically quite large and or large diameter which provides a mechanical advantage, reducing the torque seen by the friction plates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When changing gears the synchromesh in the gearbox matches the engine speed to the road speed pretty closely so the forces involved are much less. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clutch also bears on the flywheel which makes a pretty effective heat sink. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friction plate material itself is usually a composite of some sort, not dissimilar to brake pads and may include ceramics, glass fibre, various resins and synthetic fibres such as Kevlar. the material is developed such that it retains its friction properties as it wears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the force required, yes the normal force between the two plates is pretty big this is achieved by a combination of hydraulics and mechanical leverage in the linkages. Note that the travel in a clutch pedal is usually quite long while the actual distance moved by the plates may just be a few mm, again it is a very similar situation to the braking system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the clutch is engaged it is held together by springs, pressing the clutch pedal pushed the plates apart against those springs. Also the reason that it is a pedal at all is that extending your leg can provide quite a lot of force, trying to depress the clutch with your hand is more difficult than you expect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some cars, especially older ones with big engines (eg older Jaguars and Land Rovers) do have somewhat heavy clutches which can become difficult in situations like stop-start traffic where you frequently need to fully depress and hold the clutch. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-02-10T22:08:09.803" LastActivityDate="2018-02-10T22:08:09.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19328" PostTypeId="2" ParentId="19326" CreationDate="2018-02-08T22:27:30.643" Score="1" Body="&lt;p&gt;To answer your questions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Yes, a lot of heat gets produced. Most of it will sink into the flywheel, but the clutch does get very hot. That's why it's good to keep the revs low when feathering the clutch. Don't pull away revving more than 1200rpm. Better to keep it at 1000, also with a petrol car. That may require some practice, though. It's not hard at all to burn the clutch and flywheel. Your flywheel then will look like this:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WM8oH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WM8oH.jpg&quot; alt=&quot;burned flywheel&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;If you'd run your fingers across it, you'd feel a crater-like surface. Chips of&#xA;metal just melted and sprung off the flywheel, that's how hot it can become.&#xA;You'd get this when trying to pull away with a caravan on a steep hill for too&#xA;long, or when your clutch is incorrectly tuned (issue on older cars).&#xA;The glue/resin in the clutch can melt and cause a glazed surface on the clutch, reducing its friction.&lt;/p&gt;&#xA;&lt;p&gt;It's not that hard to get a car moving. Try to push a car when it's in neutral, one person is adequate to get it moving. And the clutch has it much easier, being helped by a transmission and a final gearing; after the clutch, the transmission with a gearing of about 3 to 4 (in 1st) and a differential also with a gearing of about 3 to 4 is used to amplify torque. So the torque of the engine is amplified roughly 12 times.&lt;/p&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;They achieve smooth operation by carefully designing the lever ratios in the pedal and clutch arm, and by using diaphragm springs. Older clutches used conventional springs, newer clutches since the 60s use diaphragm springs. Those are springs with a progressive-like behaviour. With older coil spring clutches, you feel a constant force when depressing the clutch pedal. It kind of tires your leg in traffic jams. With modern ones, you feel a 'wobble' in the pedal. At the end of its travel, the spring is pushed with much greater force. It makes it possible to use minimal travel and relatively weak leg strength to control massive amounts of force. I won't elaborate about that here, but you could ask about it in a separate question.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CKSBz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CKSBz.jpg&quot; alt=&quot;diaphragm vs convo&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;It's not a problem if the clutch has a smooth surface, better even. It won't wear down as quickly. If it's flat, that doesn't mean it doesn't create friction. It gets pushed hard against the flywheel and pressure plate, creating enough friction for operation. The material of a clutch is designed to withstand wear by friction, and it gets used relatively little. The brakes are used far more and are loaded much more, and thus they wear out much quicker.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, the force is indeed great. Consider a car that uses a clutch with a mean radius of 12cm, and the width of the padding on the clutch is 3cm. That gives us a surface of about 0.023m2. Assume the clutch has to transmit 100Nm of torque. That means the force on the clutch padding will be about 833N.&lt;/p&gt;&#xA;&lt;p&gt;If our clutch has a friction coefficient of 0.8, we need to push on the clutch with a force of &lt;span class=&quot;math-container&quot;&gt;$F=833/0.8=1042N$&lt;/span&gt; That's more than 100kg. In reality, the clutch has to be able to withstand much more. We're lucky to have levers and hydraulics.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To clean up your main confusion: The pedal travel is relatively long with roughly 25cm, considering the pressure plate only backs away a few mm. The force of your leg is indeed amplified by a lever and hydraulics, the same way you're able to jack up the car with your arms. If your foot travels 25cm, and the clutch only travels 2.5mm, that means the travel is reduced a factor of 100. It also means the force is amplified by a factor of 100. 1kg of pressure by your leg translates in 100kg force on the clutch.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="9448" LastEditDate="2022-10-26T12:20:00.753" LastActivityDate="2022-10-26T12:20:00.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19330" PostTypeId="1" CreationDate="2018-02-09T03:56:37.057" Score="3" ViewCount="3307" Body="&lt;p&gt;Significant amount of heat is lost from an engine by various means, so to recoup the energy, is it possible to increase efficiency via heating of diesel fuel prior to injection using waste engine heat in passenger cars similiar to marine diesels? Are there any recent journals or studies in this matter?&lt;/p&gt;&#xA;" OwnerUserId="14866" LastActivityDate="2019-08-14T12:00:14.620" Title="Preheating Diesel before injection?" Tags="|mechanical-engineering|combustion|diesel|marine-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19331" PostTypeId="1" AcceptedAnswerId="19334" CreationDate="2018-02-09T04:12:14.733" Score="3" ViewCount="50" Body="&lt;p&gt;Currently I am doing a comparison results between analytical results and measurement results, I compare stress and strain values in 32 specified locations, unfortunately I found that there are some locations does not give a good agreement.&#xA;and my professor want me to think about the reason, instead of modifying the model,&#xA;does anyone come up with an idea about a good reason for this problem?&#xA;Thank you&#xA;Cindy&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-02-09T06:03:55.950" Title="What is a good reason if the data from analysis and measurement does not give a good agreement?" Tags="|structural-engineering|civil-engineering|data|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19332" PostTypeId="2" ParentId="19314" CreationDate="2018-02-09T04:13:30.070" Score="0" Body="&lt;p&gt;I found the solution, I used query tool -&gt; probe values &lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-02-09T04:13:30.070" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19333" PostTypeId="2" ParentId="19238" CreationDate="2018-02-09T04:16:18.923" Score="0" Body="&lt;p&gt;I found the answer, I input the separated input files to solve the maximum line number limit in dat file&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-02-09T04:16:18.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19334" PostTypeId="2" ParentId="19331" CreationDate="2018-02-09T06:03:55.950" Score="5" Body="&lt;p&gt;Two possibilities come to mind:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One, the assumptions / theory / model is not describing / modelling the situation correctly,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two, the source data is incorrect / has an error at those locations - perhaps the sensors are not mounted the same or the mounting points have been damaged / moved or the calibration for each sensor has drifted.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-02-09T06:03:55.950" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19335" PostTypeId="2" ParentId="19330" CreationDate="2018-02-09T06:12:42.500" Score="2" Body="&lt;p&gt;Consider how much energy is needed to heat the fuel that is injected each time compared to the energy that can be recovered from the turbo-charger for example. One thing to note is that many of the diesel pumps (style DPA / inline high pressure) will compensate for changes in the density of the fuel according to the temperature, the modern common rail injection systems will compensate for this temperature as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did heat the diesel fuel on one of my cars using the radiator from the automatic version of the car and running the fuel through the oil cooler part (blocking the airflow through that bit), result : helped the running of the engine during the cold months but no appreciable change in fuel consumption was noted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: noticed that you tagged marine-engineering, so large marine diesels use a very different fuel and that is heated before injection so that the viscosity is reduced as the fuel very viscous when cold. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-02-09T09:39:40.100" LastActivityDate="2018-02-09T09:39:40.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19336" PostTypeId="1" AcceptedAnswerId="19337" CreationDate="2018-02-09T08:16:36.603" Score="2" ViewCount="250" Body="&lt;p&gt;In that graph, I want to show the two values and the error percentage from the comparison between two values. I am using microsoft excel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can anyone give me any idea which best graph for this case?&#xA;Thank you&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-02-09T12:23:53.900" Title="What is the best graph type to show a comparison value between two cases?" Tags="|statistics|data|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19337" PostTypeId="2" ParentId="19336" CreationDate="2018-02-09T11:28:00.943" Score="4" Body="&lt;p&gt;Based on the discussion in comments, you probably want to use a bar chart combined with a line for the error. Have a look at &lt;a href=&quot;https://support.office.com/en-us/article/create-a-combination-chart-cf2c969a-b11e-4a62-bb8f-da021177452f&quot; rel=&quot;nofollow noreferrer&quot;&gt;Create a combination chart&lt;/a&gt; for more details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, and using dummy data, it would probably look something like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Data in Excel&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OCYoN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OCYoN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Resulting combination graph&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RoyJX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RoyJX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2018-02-09T12:23:53.900" LastActivityDate="2018-02-09T12:23:53.900" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19340" PostTypeId="1" AcceptedAnswerId="19341" CreationDate="2018-02-09T13:21:52.213" Score="2" ViewCount="73" Body="&lt;p&gt;I am learning about gears, hobs and things of this sort. The main book is the machinist handbook which does a pretty good job of explaining things but things are in imperial scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This causes me a lot of confusion, especially when dealing with gear cutting blades where imperial uses DP and metrics uses Modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any recommended books as thorough as the machinist handbook about gears but one that uses metric system instead?&lt;/p&gt;&#xA;" OwnerUserId="12586" LastActivityDate="2018-02-09T14:22:33.037" Title="Gear cutting books in metric system" Tags="|gears|machining|cutting|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19341" PostTypeId="2" ParentId="19340" CreationDate="2018-02-09T14:22:33.037" Score="2" Body="&lt;p&gt;There is an English translation of &quot;Tabellenbuch Metall&quot;, which is the german equivalent. This uses metric DIN standards. &lt;a href=&quot;https://www.amazon.co.uk/Mechanical-Trades-Handbook-Ulrich-Fischer/dp/3808519134&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.co.uk/Mechanical-Trades-Handbook-Ulrich-Fischer/dp/3808519134&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-02-09T14:22:33.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19342" PostTypeId="1" CreationDate="2018-02-09T15:17:20.887" Score="2" ViewCount="75" Body="&lt;p&gt;I'm creating a simulation in abaqus of a endmill removing material from and aluminium plate. The simulation is almost working, but only the surface level element are being removed from the plate. The interior elements are not interacting with the end mill. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on how I can solve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phil&lt;/p&gt;&#xA;" OwnerUserId="14875" LastActivityDate="2018-02-09T15:17:20.887" Title="Abaqus Milling Simulation" Tags="|simulation|abaqus|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19343" PostTypeId="1" CreationDate="2018-02-09T18:31:08.827" Score="0" ViewCount="20" Body="&lt;p&gt;I'm learning AutoCAD and decided to make a floorplan as my first project. I know how to insert dimensions in AutoCAD but I don't know where to actually put them. E.g. in this image do I put the distance between the outeredge of the wall and the door or the inner edge of the wall and the door? Can I put the written dimensions on the inside of the building instead of the outside? These questions might not be very clear but basically what I'm asking is: is there a step by step guide for putting dimensions in a floor plan. I've looked at some images of floorplans and they're really not helping me much&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HCxa5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HCxa5.jpg&quot; alt=&quot;there is z-ish shaped area (it goes right -&amp;gt; up 5&amp;#39; -&amp;gt; right again) with a door that swings outward into the corner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14879" LastActivityDate="2018-02-10T05:15:51.470" Title="Is there a step by step guide to adding dimensions to a floorplan?" Tags="|architecture|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19345" PostTypeId="1" CreationDate="2018-02-09T20:16:40.297" Score="1" ViewCount="351" Body="&lt;p&gt;I am a math professor teaching an Intro ODEs course, and most of my students are going into engineering. As such, I like to give them lots of examples of how our material is applied. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are coming up on the Laplace Transform and I'm looking for a relatively simple example from Control Theory. I was thinking of perhaps examining a PID control for the cruise control in a car or for a thermostat (these appear to be the classic examples) but I'm having trouble seeing why we really need the Laplace Transform, or transfer functions, at all: we already know how to solve linear ODEs with constant coefficients without Laplace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am hoping someone can show me why we really need (or are at least greatly helped by) the Laplace Transform and transfer functions in these examples, or point me toward a simple example where we do. &lt;/p&gt;&#xA;" OwnerUserId="14880" LastActivityDate="2018-02-13T18:23:50.700" Title="Simple examples to illustrate the utility of the Laplace Transform" Tags="|control-engineering|control-theory|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19346" PostTypeId="2" ParentId="19345" CreationDate="2018-02-10T01:28:27.613" Score="4" Body="&lt;p&gt;Consider $s$ as a derivative operator.&#xA;Therefore, &#xA;$$\frac{Y}{X}=\frac{s+2}{s^2+0.5s+3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;looks like&#xA;$$\frac{Y}{X}=\frac{\frac{d}{dt}.+2}{\frac{d^2}{dt}.+0.5\frac{d}{dt}.+3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\frac{d^2}{dt}.+0.5\frac{d}{dt}.+3) y=(\frac{d}{dt}.+2)x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(\frac{d^2}{dt}y+0.5\frac{d}{dt}y+3y) =(\frac{d}{dt}x+2 x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For linear systems, we more need derivatives and convolutions than time-variance and multiplication (except for scaling). Therefore, it worths moving to Laplace domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$s$ represents frequency as well.&#xA;The beautiful examples are the initial and final value theorems which show how time and frequency are, in the limit, the inverse of each other&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(0)=\lim_{s\to \infty} sX(s)$$&#xA;$$x(\infty)=\lim_{s\to 0} sX(s)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another useful application of Laplace transform is for the design of high-pass and low-pass filters. They are much more comfortable to work in $s$ domain than messy derivatives.&lt;/p&gt;&#xA;" OwnerUserId="14288" LastEditorUserId="11869" LastEditDate="2018-02-12T21:58:38.097" LastActivityDate="2018-02-12T21:58:38.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19347" PostTypeId="2" ParentId="19343" CreationDate="2018-02-10T05:15:51.470" Score="1" Body="&lt;p&gt;From my drafting classes civil drawings were the most difficult because the rules were less strict. I preferred machine drawings where everything had to be precise. As for some general rules, I strongly recommend consistency, if you give wall / room dimensions using inside distances, keep with that. Wall thickness can be listed in a note with a leader to specific walls or like this &quot;Interior walls 4&quot; thick, exterior 6&quot; thick&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall dimensions of the whole house are external but room, door, window and most locating distances are inside unless they make more sense to be outside. Like if you want to ensure the front door is located centered on the front of the house then giving a distance from an interior wall of a room that's not centered will not ensure the door is where you want it but you may want to include those internal dimensions even though they are redundant. &lt;/p&gt;&#xA;" OwnerUserId="10516" LastActivityDate="2018-02-10T05:15:51.470" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19348" PostTypeId="2" ParentId="19330" CreationDate="2018-02-10T05:26:59.223" Score="0" Body="&lt;p&gt;Heating the diesel will have a stronger effect on how it burns in the cylinder than how much energy it adds to the system by recovering waste heat. Diesels depend upon compression to heat and ignite the fuel adding heated fuel would cause it to vaporize sooner and predetonation would be possible. &lt;/p&gt;&#xA;" OwnerUserId="10516" LastActivityDate="2018-02-10T05:26:59.223" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19350" PostTypeId="1" CreationDate="2018-02-10T16:05:25.833" Score="0" ViewCount="116" Body="&lt;p&gt;I have to design a model of a domestic boiler. &#xA;I represent the boiler as a heat exchanger (HEX1), as shown in the figure. &#xA;In many articles I have found that it is schematized with this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ M C \frac{\text{dT}_{10}}{\text{dt}}=\text{c m$\_$dot} \left(\text{Ti}-T_{10}\right)-Q $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $M C$ is the thermal capacity of the boiler, and $Q$ is the heat product by the burner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My &lt;strong&gt;question&lt;/strong&gt; is: &lt;em&gt;why there is $\frac{\text{dT}_{10}}{dt}$ at the left hand side of the equation? Why not $\frac{d\frac{\text{T}_{10} + \text{T}_{i}}{2}}{dt}$ or $ \frac{\text{dT}_{i}}{dt}$?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1ZDK7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ZDK7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;:&#xA;Connor thank you for the answer. However I don't understand:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And finally, the rate at which heat is stored within the control&#xA;  volume is proportional to the rate of change of temperature of the&#xA;  exchanger: $$Q_{stored} = M C \frac{dT_{10}}{dt}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why the temperature is $T_{10}$?&#xA;I have found that a possible answer is ralated to the so colled &quot;well-mixed conditions&quot;, infact in a &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S1359431106001815&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;the time derivative of the total energy in the mth block is&#xA;  calculated on the basis of the well-mixed condition&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is there a better explanation of the &lt;strong&gt;well-mixed condition&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="11940" LastEditorUserId="11940" LastEditDate="2018-02-13T10:27:32.293" LastActivityDate="2018-05-14T12:03:33.877" Title="Model of a domestic boiler" Tags="|fluid-mechanics|thermodynamics|heating-systems|boilers|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19351" PostTypeId="2" ParentId="19345" CreationDate="2018-02-10T23:04:30.800" Score="3" Body="&lt;p&gt;What's the response (to an arbitrary load) of a viscoelastic assembly consisting of a damper (with damping constant &lt;span class=&quot;math-container&quot;&gt;$\eta_1$&lt;/span&gt;) connected in series with a spring (with spring constant &lt;span class=&quot;math-container&quot;&gt;$k_2$&lt;/span&gt;) and another damper (&lt;span class=&quot;math-container&quot;&gt;$\eta_2$&lt;/span&gt;) that are connected in parallel?&lt;/p&gt;&#xA;&lt;p&gt;In other words, the assembly looks like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t1HyK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t1HyK.jpg&quot; alt=&quot;viscoelastic assembly&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;and we're pulling the right side; the left side is fixed. We wish to know the displacement of the right side.&lt;/p&gt;&#xA;&lt;p&gt;(The individual displacement &lt;span class=&quot;math-container&quot;&gt;$u_i(t)$&lt;/span&gt; of a single spring exposed to a load &lt;span class=&quot;math-container&quot;&gt;$F_i(t)$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$u_i(t)=\frac{F_i(t)}{k_i}$&lt;/span&gt;; the individual displacement rate &lt;span class=&quot;math-container&quot;&gt;$\dot u_i(t)$&lt;/span&gt; of a single damper exposed to the same load is &lt;span class=&quot;math-container&quot;&gt;$\dot u_i(t)=\frac{F_i(t)}{\eta_i}$&lt;/span&gt;.)&lt;/p&gt;&#xA;&lt;p&gt;This type of problem arises all the time in the context of automotive engineering, metallurgy, polymer synthesis, and biomechanics, among other fields. But you'll have a tough time writing the response of this and much more complex assemblies if you keep the time derivatives around.&lt;/p&gt;&#xA;&lt;p&gt;Instead, let's take the Laplace transforms for each component and assume zero displacement at &lt;span class=&quot;math-container&quot;&gt;$t=0$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$F_i(s)=k_iu_i(s)$&lt;/span&gt; for a spring and &lt;span class=&quot;math-container&quot;&gt;$F_i(s)=s \eta_i u_i(s)$&lt;/span&gt; for a damper. Now, realizing that displacements add when lumped components are connected in series, whereas forces add when the components are connected in parallel, we find that the total displacement is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u(s)=\frac{F(s)}{s\eta_1}+\frac{F(s)}{k_2+s\eta_2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Thus, the transfer function of the assembly is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{u(s)}{F(s)}=\frac{1}{s\eta_1}+\frac{1}{k_2+s\eta_2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;What happens if I apply a unit step load for 1 second and then let go? We'd write this load as &lt;span class=&quot;math-container&quot;&gt;$F(t)=u(t)-u(t-1)$&lt;/span&gt;, corresponding to &lt;span class=&quot;math-container&quot;&gt;$F(s)=\frac{1}{s}-\frac{\exp(-s)}{s}$&lt;/span&gt; (we can look up these Laplace transforms in a table, for example, or use a symbolic tool such as Wolfram Alpha). The resulting response is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u(s)=\frac{1-e^{-s}}{s^2\eta_1}+\frac{1-e^{-s}}{s(k_2+s\eta_2)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The corresponding time response (letting all system variables equal 1 for simplicity) is &lt;span class=&quot;math-container&quot;&gt;$$u(t, \eta_1=\eta_2=k_2=1)=e^{-t}\left[(e-te^t)u(t-1)+e^t(t+1)-1\right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;or, graphically (with time in seconds on the x-axis and displacement on the y-axis),&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/weEvT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/weEvT.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10123" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-02-13T18:23:50.700" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19352" PostTypeId="1" CreationDate="2018-02-11T01:30:07.130" Score="0" ViewCount="59" Body="&lt;p&gt;Looked at this gravel screen and its a different design wondering how the shaker box moves sideways directly connected to the small gas engine,&#xA;How or what mechanical connection would be involved?&#xA;&lt;a href=&quot;https://youtu.be/BG8YXpPKhLU?t=36&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/BG8YXpPKhLU?t=36&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14892" LastActivityDate="2018-02-11T03:14:27.080" Title="What mechanical design will shake a gravel screen box back and forth not up and down" Tags="|mechanical-engineering|mechanical|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19353" PostTypeId="2" ParentId="19350" CreationDate="2018-02-11T01:58:41.383" Score="1" Body="&lt;p&gt;The short answer is: it appears that the boiler is treated as a &quot;lumped&quot; mass; that is, that the temperature throughout the boiler is a uniform $T_{10}$, and that the working fluid exits the heat exchanger at the same temperature as the temperature of the heat exchanger itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first law of thermodynamics essentially states:&#xA;$$Q_{stored} = Q_{in} - Q_{out}$$&#xA;where $Q_{stored}$ is the heat stored within the control volume, $Q_{in}$ is the heat entering the control volume, $Q_{out}$ is the heat exiting the control volume. I assume a positive sign convention for each of these terms; that is, a positive value of $Q_{stored}$ indicates that the energy stored within the control volume is &lt;strong&gt;increasing&lt;/strong&gt;, a positive value of $Q_{in}$ indicates energy transfer from &lt;strong&gt;outside&lt;/strong&gt; of the control volume &lt;strong&gt;into&lt;/strong&gt; the control volume, and a positive value of $Q_{out}$ indicates energy transfer from &lt;strong&gt;inside&lt;/strong&gt; the control volume &lt;strong&gt;out of&lt;/strong&gt; the control volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that the rate at which heat enters the heat exchanger equals the rate at which the burner supplies heat to the exhanger plus the rate at which heat is transferred into the exchanger by the incoming working fluid:&#xA;$$Q_{in} = Q + \dot{m}c T_i$$&#xA;Furthermore, the rate at which heat leaves the control volume is equal to the rate at which heat is transferred by the exiting working fluid:&#xA;$$Q_{out} = \dot{m}c T_{10}$$&#xA;And finally, the rate at which heat is stored within the control volume is proportional to the rate of change of temperature of the exchanger:&#xA;$$Q_{stored} = M C \frac{dT_{10}}{dt}$$&#xA;Putting it all together, we have:&#xA;$$M C \frac{dT_{10}}{dt} = Q + \dot{m}c T_i - \dot{m}c T_{10}$$&#xA;$$M C \frac{dT_{10}}{dt} = \dot{m}c (T_i - T_{10}) + Q$$&#xA;I grew up in the &quot;HIP to WIN&quot; (&lt;strong&gt;H&lt;/strong&gt;eat &lt;strong&gt;I&lt;/strong&gt;n &lt;strong&gt;P&lt;/strong&gt;ositive, &lt;strong&gt;W&lt;/strong&gt;ork &lt;strong&gt;I&lt;/strong&gt;n &lt;strong&gt;N&lt;/strong&gt;egative) school of thought, and in my answer I have taken $Q$ from the perspective of the heat exchanger; thus $Q$ is positive when there is a net transfer &lt;em&gt;into&lt;/em&gt; the heat exchanger. However, the paper you read may have defined the $Q$ value from the perspective of the burner, in which case a net transfer into the heat exchanger (i.e. net transfer &lt;em&gt;out of&lt;/em&gt; the burner) would give a negative $Q$. That would explain the negative sign ($-Q$) in the equation you referenced.&lt;/p&gt;&#xA;" OwnerUserId="8518" LastActivityDate="2018-02-11T01:58:41.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19354" PostTypeId="2" ParentId="19352" CreationDate="2018-02-11T02:00:44.773" Score="1" Body="&lt;p&gt;You can determine the plane of rotation of the engine by noting the pull starter on the front of the engine. This puts the axis of rotation parallel to the grill slots of the sieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engine must have an eccentric plate of some sort, akin to the linkage of an old time steam engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L6mS3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L6mS3.jpg&quot; alt=&quot;steam locomotive linkage&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be a bearing on the eccentric plate where it connects to the link rod. Visible in the video is the other end of the link rod where it attaches to the shaker box. There appears to be a rubber mount in that attach point, but also might have a bearing inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the shaker box is mounted on chain, which allows for the side-to-side movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the power and speed displayed in the video, very little translation is needed to accomplish the objective.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-02-11T02:00:44.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19355" PostTypeId="2" ParentId="19345" CreationDate="2018-02-11T02:36:53.840" Score="0" Body="&lt;p&gt;For your approach you need to know the exact ODE of the system. However when designing a controller for a system that can be approximated as LTI you do not necessary need a fully parameterized and identified ODE model. You can either use some PID tuning algorithm or, maybe more related to your question, measure the frequency response function (FRF) of the system. You could fit a transfer function onto this FRF, but the FRF itself is already enough information to design a controller using loopshaping.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-02-11T02:36:53.840" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19356" PostTypeId="2" ParentId="19352" CreationDate="2018-02-11T03:14:27.080" Score="0" Body="&lt;p&gt;A watt's linkage has very good linear properties and is very simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Watt%27s_linkage&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Watt%27s_linkage&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-02-11T03:14:27.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19357" PostTypeId="1" CreationDate="2018-02-11T10:09:48.883" Score="0" ViewCount="85" Body="&lt;p&gt;Looking for a way to shift a 30mm lens in two axis perpendicular to another lens. The lens is for shifting a video projection up, down, left, right. Basically what you will find in tilt-shift camera lenses or video projectors with motorized lens-shift feature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The travel distance is very short, 1/4th of the 30mm lens diameter, but precision has to be at least 1024 steps across that 30mm/4 length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would love any ideas how to implement this. Can only think of a gantry based system but can’t find small enough motors. Speed is also a concern, needs to go from one edge to the next in 1/3 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Haven’t been able to teardown tilt-shift camera lenses or motorized video projectors to see how it is implemented there.&lt;/p&gt;&#xA;" OwnerUserId="14841" LastActivityDate="2019-01-08T05:48:15.773" Title="precision positional control needed for shifting position of lenses" Tags="|mechanical-engineering|motors|stepper-motor|servo|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19358" PostTypeId="2" ParentId="19357" CreationDate="2018-02-11T10:54:20.210" Score="0" Body="&lt;p&gt;Well, a lever controlled by a fine thread which can be rotated by a stepper motor. Something like this but you will have to deal with the rotation arc etc:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mVtnG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mVtnG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of the distances &quot;pivot : connection&quot; and &quot;pivot : thread drive&quot;&#xA;will give you the 4:1 bit you are after.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-02-11T10:54:20.210" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19359" PostTypeId="1" CreationDate="2018-02-11T11:14:36.823" Score="0" ViewCount="85" Body="&lt;p&gt;does anyone have any idea what standard shaft size / adapter this is?&#xA;the motor is NEMA 56C and the socket size it self is 27.8 mm in diameter and 32 overall with the key.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OwmAx.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PkzqN.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14896" LastEditorUserId="1832" LastEditDate="2018-02-11T12:10:12.057" LastActivityDate="2018-08-11T06:01:25.023" Title="Shaft Size design / replacement" Tags="|mechanical-engineering|design|applied-mechanics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19360" PostTypeId="2" ParentId="19330" CreationDate="2018-02-11T12:38:54.330" Score="1" Body="&lt;p&gt;Everything that gets the mixture in the cylinder hotter makes for a more efficient combustion. That is because the power of the engine depends purely on the pressure exerted on the piston, and the effective pressure is caused by temperature rise, and nothing else. So yes, injecting hot diesel would make for a more efficient combustion, in theory. But it would be insignificant. You can heat it up to just below the boiling point, but you'd still need the energy to evaporate the diesel, which is far more than needed to heat it up. Installations for this purpose will cost more than the profit they'd give you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an alternate solution though; heating up the air. Once the air is trapped inside e.g. a cylinder, you could heat it up, and it would gain pressure. That pressure will add up to any pressure caused by combustion. You would get a more efficient combustion. The problem is that you're dealing with a continuous air flow, which is hardly trappable. If you heat it up without trapping it, it will just expand and flow out of the cylinder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd need a pressure tank in which you can trap bulk charges of air, you can then heat up that trapped air, and feed it to the engine once it's hot.&#xA;But also those installations would only perhaps be cost effective in large static installations. In cars, downsizing and turbochargers are the way to go for realistic and cost-effective gains in efficiency. That's why you see this more and more today.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For marine applications, heavy oils are preheated to reduce their viscosity, which only then enables them to be used for injection. The heaviest oils are so thick you can nearly walk on them when they're cold.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-02-11T12:38:54.330" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19361" PostTypeId="1" CreationDate="2018-02-11T15:13:01.427" Score="0" ViewCount="44" Body="&lt;p&gt;I am trying to find information about the details of mechanical properties of tin-base alloys, such as ~8% antimony Brittanica. But probably since tin is use primarily in decorative applications or as solder, I have failed to find detailed material properties like you would easily find for more common engineering materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I am interested in fatigue limits as function of the number of cycles, and the effect of casting temperature and cooling rates on material properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I have been able to find is that Brittanica has around a 100MPa yield strength. Does anyone have a more extensive reference?&lt;/p&gt;&#xA;" OwnerUserId="14899" LastActivityDate="2018-02-11T22:59:56.637" Title="Mechanical properties of tin base alloys" Tags="|materials|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19363" PostTypeId="2" ParentId="19214" CreationDate="2018-02-11T20:44:04.797" Score="1" Body="&lt;p&gt;One more question, where did you get the binary interaction parameters? Have they been process or experimentally validated? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding if NRTL from a high level is that it is good for modelling of non ideal or polar vle if the interaction parameters are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could mean validating the interaction parametrs in the range in which they are used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please reply with what you find out!&lt;/p&gt;&#xA;" OwnerUserId="14901" LastEditorUserId="14901" LastEditDate="2018-02-11T20:49:08.560" LastActivityDate="2018-02-11T20:49:08.560" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19364" PostTypeId="2" ParentId="17580" CreationDate="2018-02-11T21:10:56.003" Score="3" Body="&lt;p&gt;I think the key information is that the liquid is saturated. This means any reduction in pressure will lead to partial vaporization and two phase flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For heat tranfer best overall heat transfer coefficient is always direct to liquid rather than vapor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You still will have vaporization once the saturated liquid once it enters the heater, but the effect will be less. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For same heat duty the required size in heat transfer area of a vapor heater will be much larger than a liquid heater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would put the pressure reducing valve after the heater.&lt;/p&gt;&#xA;" OwnerUserId="14901" LastActivityDate="2018-02-11T21:10:56.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19367" PostTypeId="2" ParentId="16345" CreationDate="2018-02-11T21:26:41.397" Score="1" Body="&lt;p&gt;Unfortunately solids handling has a reputation for being more art than science because it is much harder to predict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reference book on this topic which is detailed and where you may find more information on predictive models is Pneumatic Conveying by Klinzing , Rizk, et al, although Im sure there are lots of other places to look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the amount of information required to develop an accurate model is more based on empirical testing or predictive models based on your materials properities. Solids vary so much more than fluids , with particle size distributions,  surface area of particles,  shape of partiles and other properties which are not required for fluid flow models. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generically conveying can be broken into “typical velocities” for dense phase or dilute phase conveying, but for design you typically either need to base on an existing system, work with a partner to do empirical testind on your material, or design with a sufficient level of conservatism.&#xA;Rules of Thumb for Chemical Engineers by Stephen Hall has a section on chapter on conveying that I believe is helpful. Chemical Engineering  Progress also has some articlesnthat may be helpful and if you have an AIChE membership you can access for free.&lt;/p&gt;&#xA;" OwnerUserId="14901" LastActivityDate="2018-02-11T21:26:41.397" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19371" PostTypeId="2" ParentId="19361" CreationDate="2018-02-11T22:43:58.033" Score="1" Body="&lt;p&gt;Physical properties are not generally available because tin is not used in stressed applications. Accordingly, likely no one has measured fatigue properties, etc. Tin is used for bearings , certain solders and coating steel containers. It is also very expensive. I see Britannica ( pewter) also has 3 % Cu and was used for eating utensils about 100 years ago. However , for what it may be worth; American Society for Metals , 1948 Handbook has a few pages on properties and uses of tin up to WW 2. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-02-11T22:59:56.637" LastActivityDate="2018-02-11T22:59:56.637" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19372" PostTypeId="1" CreationDate="2018-02-12T02:14:42.323" Score="0" ViewCount="79" Body="&lt;p&gt;My friends and I are trying to hypothesize how computers use energy. Please settle the debate. Do computers, or any other electrical device of the sort, e.g. 3d printers, use different levels of power from an outlet when completing different tasks? For example, if I keep my charger in the outlet constantly, using it only for, say, emailing, does it use precisely the same amount of power if I were to keep my charger in the outlet constantly, while farming bitcoin, or playing a graphically intensive video game? Can the computer ask the outlet for more power (not more than the outlet is capable of, but in relation to how much it usually uses in menial tasks); or does the computer get the exact same amount of power, keeping it at full throttle, insofar as it is connected to the outlet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: analogous: for example, assuming my charger stays in the outlet at a constant rate (never being unplugged), will my electric bill increase or stay the same depending on whether or not I email, or if I play graphically intensive video games? &lt;/p&gt;&#xA;" OwnerUserId="14905" LastEditorUserId="14905" LastEditDate="2018-02-12T02:27:18.627" LastActivityDate="2018-02-13T09:24:34.580" Title="Computer engineers -- do computers use varying amounts of power?" Tags="|computer-engineering|" AnswerCount="2" CommentCount="4" ClosedDate="2018-02-13T20:13:21.763" ContentLicense="CC BY-SA 3.0" />
  <row Id="19373" PostTypeId="2" ParentId="19359" CreationDate="2018-02-12T02:25:20.320" Score="1" Body="&lt;p&gt;From Baldor motor chart, 5/8&quot; diameter shaft with 3/16&quot; keyway.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-02-12T02:25:20.320" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19374" PostTypeId="2" ParentId="19237" CreationDate="2018-02-12T02:42:51.323" Score="1" Body="&lt;p&gt;The closest thing I can think of is Diagnostic Listening Devices such as STEELMAN 97220 Mini EngineEAR or the multichannel system that are regularly used in auto diagnostics, can also be used in plant machinery.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-02-12T02:42:51.323" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19375" PostTypeId="2" ParentId="19103" CreationDate="2018-02-12T03:06:15.910" Score="1" Body="&lt;p&gt;Since the 5 kg axial weight is small I would suggest pillow block bearings, probably base mount. If you go to McMaster, search pillow block bearings, bearing pictures and ratings will be shown. I think cleaning fluids is involved in this application, so sealed bearings, preferably stainless, and minimize water going into the area of bearing. Motor on top so it doesn't get wet.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-02-12T03:06:15.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19376" PostTypeId="1" AcceptedAnswerId="19592" CreationDate="2018-02-12T04:31:33.320" Score="0" ViewCount="225" Body="&lt;p&gt;I own a warehouse that needs to push 25,000 CFM.  I currently have an old warehouse exhaust fan that doesn't push anywhere close to this.  The diameter of the fan blade is about 36 inches, but the frame in which the fan sits is 41 x 41 inches.  I've done a bit of looking and can't seem to find a fan that is 41 inches or less in diameter that pushes 25,000 or more CFM.  So, I really have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does a fan exist that would push 25,000 CFM which that could fit into the 41 x 41 opening (probably means close to a 36 in diameter fan blade); and&lt;/li&gt;&#xA;&lt;li&gt;Is there a mathematical function that relates RPMs of a fan, the number of blades, and size of the blades to CFM (or a similar function such that when you plug in the size and speed of the fan, you can find the CFM)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="14907" LastActivityDate="2018-02-24T02:43:14.050" Title="Do 36 inch diameter fans exist that push 25,000 CFM?" Tags="|airflow|hvac|cooling|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19377" PostTypeId="2" ParentId="15664" CreationDate="2018-02-12T04:38:14.327" Score="-1" Body="&lt;p&gt;It depends on valve design and the application needs.Can ben matal or teflon lined,like &lt;a href=&quot;http://www.ceramic-valves.com/96-sight-glass-ball-check-valve.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;PFA lined ball check valve&lt;/a&gt; for corrsive applications.&lt;/p&gt;&#xA;" OwnerUserId="14908" LastActivityDate="2018-02-12T04:38:14.327" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19378" PostTypeId="2" ParentId="18972" CreationDate="2018-02-12T04:49:25.360" Score="0" Body="&lt;p&gt;A few pumps might take a 4-20 mA signal, and you might be able to find a microcontroller which would generate 4-20mA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest to look at dosing pumps that are used for boiler feed chemicals. &lt;/p&gt;&#xA;" OwnerUserId="14901" LastActivityDate="2018-02-12T04:49:25.360" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19379" PostTypeId="2" ParentId="19372" CreationDate="2018-02-12T04:56:24.603" Score="3" Body="&lt;p&gt;Yes, a computer uses varying amount of power. Yes a 3D printer uses a varying amount of power. In fact all electrical devices do vary based on load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both motors and processors use more energy when there is more load. This is pretty easy to see in a computer or graphics card when you run something heavy, then eventually the fans kick on maximum to cool the device. While if you idle it does not happen. More than that a idle computer actually shuts down parts of itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes mining bitcoin uses much more power than sending email * or playing graphic intensive games**. Bitcoin mining essentially maxes out everything for 24/7.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* even if you are a bulk e-mailer, because your network connection is limited&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** you can only stay awake for so long&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-02-12T14:42:30.323" LastActivityDate="2018-02-12T14:42:30.323" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19380" PostTypeId="1" CreationDate="2018-02-12T05:51:37.043" Score="5" ViewCount="72" Body="&lt;p&gt;Is there an analytical solution for the Navier Stokes equation for flow through a flattened tube ( flat plate in the center and semicircular lobes at the end).&lt;a href=&quot;https://i.stack.imgur.com/HNwD2.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HNwD2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14909" LastActivityDate="2018-02-12T05:51:37.043" Title="Flow through flattened tube" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19381" PostTypeId="1" AcceptedAnswerId="19383" CreationDate="2018-02-12T06:43:26.063" Score="4" ViewCount="128" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mwCGL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mwCGL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that Absolute pressure is zero-referenced to a perfect vacuum and Gauge pressure is zero-referenced to ambient air pressure. So, Absolute pressure is equal to Gauge pressure plus atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the Part B of this problem, confused me on what the absolute pressure under piston B is, which comes from the fact that the system is open on both piston A and B. In this case, is the (Weight of block)/(Area of piston), Gage or Absolute pressure? &lt;/p&gt;&#xA;" OwnerDisplayName="user13284" LastActivityDate="2018-02-13T09:20:35.603" Title="What are the differences between Gauge and Absolute pressure?" Tags="|thermodynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19383" PostTypeId="2" ParentId="19381" CreationDate="2018-02-12T09:41:28.920" Score="4" Body="&lt;p&gt;Your explanation of the differences between Gauge and Absolute is correct. As you have noted, the difference between them is due to the pressure exerted &lt;em&gt;by the atmosphere&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure exerted by the blocks, therefore, is a &quot;Gauge&quot; pressure, and must be added to the atmospheric pressure (dependent on altitude, etc) to find the Absolute Pressure in the fluid. It is important to note that The pressure exerted by the blocks is independent of atmospheric pressure - it would exist in a vacuum. &lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-02-12T09:41:28.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19387" PostTypeId="1" CreationDate="2018-02-12T17:23:10.603" Score="1" ViewCount="189" Body="&lt;p&gt;I have to tune a state space model for certain parameters in Matlab. &#xA;The transfer function has to be &#xA;&lt;span class=&quot;math-container&quot;&gt;$$T(s)=\frac{k_q(1+0.162s)}{s^2+8.23s+67.7}$$&lt;/span&gt;&#xA;Using the place command, I have been able to place the poles at &lt;span class=&quot;math-container&quot;&gt;$$p_{1,2}=-4.115\pm 7.12i$$&lt;/span&gt; with a full state feedback gain of &lt;span class=&quot;math-container&quot;&gt;$-280$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$-28$&lt;/span&gt;.&#xA;However, I now have to place the zero using a prefilter such that the zero is at &lt;span class=&quot;math-container&quot;&gt;$s=-0.6653$&lt;/span&gt; and the system tracks an input signal properly. &#xA;I fail to see how a prefilter would affect the enumerator such that the location of the zero can be altered. How can I place the zero? &lt;/p&gt;&#xA;" OwnerUserId="14916" LastEditorUserId="18238" LastEditDate="2020-05-10T02:34:39.130" LastActivityDate="2020-05-10T02:34:39.130" Title="Zero placement in transfer function" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19388" PostTypeId="2" ParentId="19381" CreationDate="2018-02-12T22:14:48.843" Score="2" Body="&lt;p&gt;It doesn't seem that the test creator relies on your ability to differentiate between absolute and relative (or gauge) pressure. It's not mentioned anywhere, other than in B where he wants you to give the absolute pressure, which is impossible to answer, since no ambient pressure is given. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not really possible to give an answer at all since the described situation is not static. If the mass is increased to $150\ kg$, it will launch piston 1 rapidly in the air. Furthermore, in a static situation, the pressure below B is/must be the same as below piston A or anywhere else in the fluid. The whole question gives me the idea that the creator of the test doesn't really know what he's talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think he just wants to hear what the relative pressure in the fluid would be, if the mass on piston B is $150\ kg$, and piston A is frozen. Then you would come to &#xA;$3316\ kgf/m^2$ or $32.1\ kPa$ relative, assuming a gravitational acceleration of $9.81\ m/s^2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute pressure doesn't matter anyway, since only relative pressure will be able to do any work. You can mention that the ambient pressure should be given to be able to give an absolute pressure, but in my experience, smartassery is rarely rewarded in school. Although the teacher is just plain wrong here and he should accept his fault...&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="13311" LastEditDate="2018-02-13T09:20:35.603" LastActivityDate="2018-02-13T09:20:35.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19389" PostTypeId="2" ParentId="19237" CreationDate="2018-02-12T22:17:07.920" Score="1" Body="&lt;p&gt;What's the end goal? Typically systems that do what you're talking about are critical, expensive equipment like large pumps and turbines that take time to shutdown and repair or where something like the failure of a bearing could destroy the entire machine. These systems use accelerometers attached near the critical components to sense the operating component's &quot;signature&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most factory equipment should be fine on a maintenance schedule though. You know, the &quot;Replace these bearings and greases these joints every 5000 hr&quot; kind of stuff.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, I have seen production QC checks done on cast gear blanks using a microphone. The blanks were on a conveyor which ran through a small enclosure to reduce the influence of outside noise, there was a pneumatic &quot;hammer&quot; which would strike the gear and the microphone would pick up the ringing of the blank and compare it to the baseline. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure you could try doing something similar with a machine in a factory, but you would probably have to enclose it to keep other machinery from triggering a false alarm or masking a real issue. Creating a baseline acoustic profile for all of the different moving parts could be a complete mess. &lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2018-02-12T22:17:07.920" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19390" PostTypeId="1" CreationDate="2018-02-12T22:34:24.063" Score="2" ViewCount="60" Body="&lt;p&gt;I have a simple, practical engineering question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to design a hot water circulation system with a constant flow rate of 1.3 L/min. I'm having trouble finding any pumps that will do the job at the temperature (80-100C) for a reasonable cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, I'm trying to &lt;a href=&quot;http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0162888&quot; rel=&quot;nofollow noreferrer&quot;&gt;Development of a Consistent and Reproducible Porcine Scald Burn Model&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Where to search for for reliable, inexpensive, high-temperature water pumps? I tried McMaster Carr, but they didn't have anything that fit the bill.&lt;/li&gt;&#xA;&lt;li&gt;Another way to adjust the flow rate? I thought a nozzle might work, but I'm worrying about too much back pressure damaging the pump, or worse, that a restricted area wouldn't even reduce the flow (RTT)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14465" LastEditorUserId="8055" LastEditDate="2018-03-15T15:53:33.373" LastActivityDate="2019-04-21T22:08:15.017" Title="Designing for a specific flowrate" Tags="|fluid-mechanics|pumps|heating-systems|fluid|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19391" PostTypeId="2" ParentId="19390" CreationDate="2018-02-13T02:38:10.050" Score="4" Body="&lt;p&gt;Many commercial hot water recirculation pumps will do that. Just grabbed one at random - Grundfoss Alpha1 15 40 130.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't fetch the page after I ran their calculator. I used temp of 99C and head of 2m. 100C doesn't work in their calculator. Click link then click &quot;view in product center&quot;, and fill in form. &lt;a href=&quot;https://www.grundfos.com/products/find-product/alpha1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ALPHA1 circulator pump - heating, cooling, hot water&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have to edit settings for 60hz.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="6767" LastEditDate="2019-04-21T22:08:15.017" LastActivityDate="2019-04-21T22:08:15.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19392" PostTypeId="1" CreationDate="2018-02-13T07:50:26.357" Score="1" ViewCount="132" Body="&lt;p&gt;&lt;em&gt;Clarification: This question is all about the gas supply to the water heater. It has nothing to do with the water supply.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My hot water heater specifies 5.5-10.5 W.C. (water column inches) of pressure. I measured the pressure to be 7 W.C. Within spec. The problem is, this pressure reading measures the same (7wc) whether the main gas valve is (a) wide-open  (b) half-open or (c) fully closed.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this makes sense. Because let's assume the system pressure at the gas source is 7wc (plus a nominal pressure drop). If there is any flow at all through the pipes (and no gas is exiting through the pipe walls) then we would expect the pressure (at steady-state) at any point inside the pipe to equal the system pressure. If it were any less, then gas would flow from the source of high pressure to the point of lower pressure at the measuring point until the two pressures reach equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why did the engineers specify pressure? It seems like they should, at the very least, specify a mass flow rate of gas. For example, if any obstruction were inside the pipe and impeding flow or the valve were not completely open, this would result in a malfunctioning heating system. But that problem would &lt;em&gt;not&lt;/em&gt; necessarily cause a low pressure sensor reading. But it &lt;em&gt;would&lt;/em&gt; create an unacceptably low mass flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what gives? Is this just an oversight? Or is there some reason  I can't see why &lt;em&gt;pressure&lt;/em&gt; and not &lt;em&gt;mass flow&lt;/em&gt; is specified?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jo9Rw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jo9Rw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cp41S.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cp41S.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; My valve is old and it leaks. So, when it's fully shut off, its mass flow rate through the chamber is, let's say, 1% of that when it's &quot;wide open.&quot;&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2018-02-16T08:26:24.073" LastActivityDate="2018-02-16T08:26:24.073" Title="Why is gas pressure specified instead of mass flow?" Tags="|pressure|gas|heating-systems|pressure-vessel|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19393" PostTypeId="1" AcceptedAnswerId="19395" CreationDate="2018-02-13T08:32:16.463" Score="2" ViewCount="86" Body="&lt;p&gt;Does structural metal tubing (e.g., either as &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Hollow_structural_section&quot; rel=&quot;nofollow noreferrer&quot;&gt;HSS&lt;/a&gt;, or in &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Space_frame&quot; rel=&quot;nofollow noreferrer&quot;&gt;space frames&lt;/a&gt;, or even motorcycle/bicycle frames) ever come with any small metal rods or columns &lt;em&gt;inside&lt;/em&gt; of the tubing itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it might add some weight (and/or manufacturing complexity), could it also significantly improve the rigidity and load-bearing ability of the tubing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Edit: a more &quot;biological&quot; visualization of this could be similar to that of a bird bone's hollow structure (see &lt;a href=&quot;https://qph.fs.quoracdn.net/main-qimg-e3a99fcd31cbfd69c3cbf2192210df5d&quot; rel=&quot;nofollow noreferrer&quot;&gt;this image&lt;/a&gt;), but with possibly far fewer crossbeams (to save on weight), perhaps just one or two small &quot;spokes&quot; or crossbeams every now and then?&lt;/p&gt;&#xA;" OwnerUserId="14919" LastActivityDate="2018-02-24T02:05:36.257" Title="Support columns inside metal tubing" Tags="|structural-engineering|structural-analysis|beam|metals|columns|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19394" PostTypeId="1" AcceptedAnswerId="19398" CreationDate="2018-02-13T08:48:40.017" Score="2" ViewCount="773" Body="&lt;p&gt;-- &lt;em&gt;This issue might be due to a bug in SolidWorks 2014 SP2&lt;/em&gt; --&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to route a water hose from one location to another without passing through a keepout area (the red box shown).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am simply sweeping a circle (red sketch) along a Projected Curve defined by two orthogonal sketches, one on TOP (green curve) and one on FRONT (blue curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EjB08.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EjB08.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very nearly working, apart from that I cannot seem to persuade Solidworks to complete the Projected Curve.  It always seems to stop short of the final location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each of the two control curves begin and end with straight line segments that are tangent to the spline that governs the path.  I need these to get a nice, straight controlled entry/exit from the flat surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control points begin and end at the same locations in space (projected onto their respective planes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong, that is preventing this curve from being completed?&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastEditorDisplayName="user6335" LastEditDate="2018-02-14T10:50:21.617" LastActivityDate="2018-02-14T10:50:21.617" Title="Solidworks - Projected Curve is incomplete" Tags="|solidworks|technical-drawing|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19395" PostTypeId="2" ParentId="19393" CreationDate="2018-02-13T09:15:08.437" Score="1" Body="&lt;p&gt;Structural metal tubing is normally used to take bending, axial and shear loads, i.e. to globally transfer load from one place to another. It also has to resist local loads, i.e. if there is a concentrated force or impact force in a particular location, the member has to not buckle locally. In other words: bending shear and axial and taken by (more or less) the whole cross-section; local loads are resisted by parts of the cross section, and they are then transferred into the whole cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bending and axial loads travel down the outside walls of a hollow member. Internal bracing won't make any difference to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear force and local loads travel across the cross-section. Internal bracing could help with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience as a structural engineer, the critical factors governing design are normally bending and axial force. Sometimes shear is a controlling factor in cross-section selection. I can't think of a single instance when local loading has been of any importance to a structural hollow section. (NB I work in bridges. Local impact may be of more importance in other disciplines.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, when you're using &quot;standard&quot; structural hollow sections, I find it unlikely that internal bracing would be beneficial. And the cost of adding it would make it uneconomically viable.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;There is, however, one aspect in my experience where internal bracing is required. That's when fabricating a &lt;a href=&quot;https://www.google.co.uk/search?biw=768&amp;amp;bih=740&amp;amp;tbm=isch&amp;amp;sa=1&amp;amp;ei=TqyCWtGBCIqVgAa7qb-IDg&amp;amp;q=bridge%20steel%20box%20girder&amp;amp;oq=bridge%20steel%20box%20girder&amp;amp;gs_l=psy-ab.3..0i8i30k1.3755.4562.0.4865.6.6.0.0.0.0.75.315.6.6.0....0...1c.1.64.psy-ab..0.6.314...0i13k1j0i8i7i30k1j0i8i13i30k1.0.DZagy0oOQQY&quot; rel=&quot;nofollow noreferrer&quot;&gt;steel box girder&lt;/a&gt;. These use relatively thin plates spaced a long distance apart, and in order to avoid buckling they need to be braced, internally, at regular distances.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-02-13T09:15:08.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19396" PostTypeId="2" ParentId="19372" CreationDate="2018-02-13T09:24:34.580" Score="0" Body="&lt;p&gt;Imagine that you have a pressurized water tube. If you bore a bigger hole in it, you will get a bigger spatter. So is it going also with the wires. The wire is the tube, and the pressurized water is the electrons in the wire. In the case of wire, it is not so trivial, but there are cheap chinese devices doing it. The computer don't need to say to the connector, what power it wants. The connector and the power supply of the computer work together on a way, that the computer gets always what it wants. More details can you read &lt;a href=&quot;https://en.wikipedia.org/wiki/Switched-mode_power_supply&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2018-02-13T09:24:34.580" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19398" PostTypeId="2" ParentId="19394" CreationDate="2018-02-13T10:13:56.370" Score="2" Body="&lt;p&gt;I'd recommend using a 3D Sketch Spline for creating your hose - this will more accurately model the natural shape that the hose would assume given the boundary conditions of perpendicular to the grey transparent block at the in/out positions. Let me know if you have any questions regarding this method:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://media.giphy.com/media/3ohs4vhSqZtKssDF1m/giphy.gif&quot; alt=&quot;Gif showing how to use 3D Sketch Spline for Hose Creation[1]&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-02-13T11:11:35.270" LastActivityDate="2018-02-13T11:11:35.270" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19400" PostTypeId="1" CreationDate="2018-02-13T15:10:30.363" Score="0" ViewCount="399" Body="&lt;p&gt;I have a pipe network with laminar flow (peak output from the pump gives a Reynolds number around 1100). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some test data for the network and the results are really odd to me; I cannot explain the flow distribution using standard theory for K values or assigning equivalent lengths to pipe fittings. The results look like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WhUJt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WhUJt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Branches A and B are pipes of the same diameter (0.5cm), and are fed, via a T-junction by a pipe with double their diameter (1cm). The branches are the same length, the only difference between them is one has 2 holes in it (of 2mm diameter at the tip, and 1mm diameter at slightly downstream of midspan), the other only has 1 hole (of 1mm diameter, at the tip). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This minor geometrical difference causes 19 times more flow to go down branch B, according to my test results. But the theory I have for equivalent lengths of pipe suggests this shouldn't happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect it's because the data only applies to turbulent flows. I also am doubting the accuracy of the test data (I did not conduct the test) given that large discrepancy between the two branches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does any body know a source of equivalent lengths of pipe/k factors that are valid for slow moving (laminar) flows? (below 2000 Re)&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="14498" LastActivityDate="2018-05-14T21:08:46.563" Title="Equivalent Lengths of Pipe for Laminar Flows" Tags="|fluid-mechanics|civil-engineering|pipelines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19401" PostTypeId="1" AcceptedAnswerId="19426" CreationDate="2018-02-13T15:46:51.520" Score="1" ViewCount="33" Body="&lt;p&gt;I am a beginner in control system for hydraulic servo systems. The reason why I haven't started yet to do simulations of hydraulic servo system is because I have no information about the components I want to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For exampe. To do a simulation of a hydraulic system, you need to know the dynamical characteristics of the components you are using. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at this &lt;a href=&quot;https://home.uia.no/geirh/PDF/C39.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;document&lt;/a&gt;, you sure know what I'm talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's a assume that you are building a system like that and you want to simulate it in MATLAB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, OK. First you need to know what components you want to have. That's easy, just follow the picture. Then when you have found all the components you need, you also need to know the dynamics such as gain factors, damping, leackage etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem that all suppliers I have visit, don't show their specification inside the data sheet of the products they are selling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example, if you want to do a transfer function of a &lt;a href=&quot;https://docs-emea.rs-online.com/webdocs/12c4/0900766b812c44d2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydraulic pump&lt;/a&gt; and you go to Bosch Rexroth to buy a hydraulic pump. All they going to give you are static equations e.g formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that led me to the question: If I want to have dynamical information of hydraulic pump, hydraulic cylinders, electrical motors. What supplier should I visit then? &lt;/p&gt;&#xA;" OwnerUserId="14923" LastActivityDate="2018-02-14T19:51:41.747" Title="Suppliers who write dynamical specification inside their data sheet?" Tags="|control-engineering|control-theory|hydraulics|simulation|servo|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19402" PostTypeId="2" ParentId="19214" CreationDate="2018-02-13T16:20:06.430" Score="3" Body="&lt;p&gt;At moderate temperatures and pressure NRTL should be a good option provided that the binary interaction parameters are correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to check:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Whether the binary interaction parameters for your three compounds are in Aspen Plus database. If not, try to find them in literature or, use UNIFAC to determine only the unknown parameters (most process simulators allow you to do that).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the three binary interaction parameters are in the Aspen Plus database, check whether they represent the experimental behaviour of your system. In particular, try to reproduce binary Pxy and/or Txy diagrams. Otherwise, try to reproduce any experimental bubble or dew point information you might have.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course, if you don't have good binary interaction parameters, UNIFAC being totally predictive is a good choice. As you know, UNIFAC divides a molecule into groups which contribute differently to the Gibbs free energy. The compounds you mention can be decomposed into fairly standard groups (you can check the list &lt;a href=&quot;http://www.ddbst.com/published-parameters-unifac.html#ListOfMainGroups&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) and UNIFAC results should be reliable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, I have to say that, though uncommon, it may happen that process simulators contain errors in interaction parameters (NRTL, Wilson, UNIQUAC) or similar that lead to poor predictive capabilities. I did work in a project (mostly confidential) where a widespread process simulator completely failed to reproduce the VLE behavior of something similar to 2-butanol in water/methanol solution because the interaction parameters were wrong. We detected it comparing with experimental data of Txy and we regressed and corrected the parameters.&lt;/p&gt;&#xA;" OwnerUserId="3991" LastActivityDate="2018-02-13T16:20:06.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19403" PostTypeId="1" AcceptedAnswerId="19467" CreationDate="2018-02-13T16:26:52.227" Score="0" ViewCount="157" Body="&lt;p&gt;I need to design some custom involute gears for a project that I am working on and this is my first dive into gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I worked out how to calculate the size of gears by the required numbers of teeth and module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gear dimensions:&#xA;Module = 1mm&#xA;Teeth  = 36&#xA;Clearance = 0.1&#xA;Pitch Diameter = 72mm&#xA;Outside Diameter = 74mm&#xA;Whole Depth = 2.1mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see it's a tiny gear and I'd like to be able to make the blade to cut this gear myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a standard pressure angle of 20 degrees, how would I calculate the angle of the cutting surface to make this gear profile?&lt;/p&gt;&#xA;" OwnerUserId="12586" LastActivityDate="2018-03-19T03:34:25.947" Title="Calculating the angles for gear cutters" Tags="|gears|machining|cutting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19405" PostTypeId="1" CreationDate="2018-02-13T18:53:59.113" Score="1" ViewCount="48" Body="&lt;p&gt;Suppose a closed tank does not have any pressure relief system and is set to drain from the bottom. It is said that because the headspace pressure drops as the liquid level goes down, the tank will not properly drain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in real situations, I would guess air bubbles start to form in the bottom hole and rise to equilibrate pressure and allow the liquid to drain. I think this ends up depending on the rate of formation and motion of the bubbles which are affected by the geometry and fluid and so on. Is there a systematic way of looking at this? How do you work it out on paper?&lt;/p&gt;&#xA;" OwnerUserId="14924" LastActivityDate="2018-02-16T01:14:04.707" Title="What is the criteria for bubble entrapment in a closed headspace tank set to drain?" Tags="|fluid-mechanics|pressure|pressure-vessel|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19406" PostTypeId="2" ParentId="19400" CreationDate="2018-02-13T19:22:52.343" Score="1" Body="&lt;p&gt;Under laminar flow, you can use Hagen-Poiseuille model of flow, namely $$f = \frac{64}{Re}$$ The Darcy friction factor relates to the head loss coefficient $K$ by $$ K = f \frac{L}{d}$$&lt;/p&gt;&#xA;" OwnerUserId="14924" LastActivityDate="2018-02-13T19:22:52.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19407" PostTypeId="1" CreationDate="2018-02-13T20:24:03.050" Score="3" ViewCount="174" Body="&lt;p&gt;I know in a power distribution system, the power is generated (from a dam or turbine or something) and consumed immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the power consumed is variable. So how does the generation system take that into account. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example what happens to the hydro dam turbines (or wind turbines etc) when I turn my light off or unplug my laptop charger?&lt;/p&gt;&#xA;" OwnerUserId="14925" LastActivityDate="2018-02-15T18:14:58.927" Title="How does power generation (hydro, wind etc...) account for variable power consumption?" Tags="|power-engineering|power-transmission|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19408" PostTypeId="2" ParentId="19407" CreationDate="2018-02-13T21:30:09.297" Score="-1" Body="&lt;p&gt;It generally does not. After the investment is made in a dam generation site (or nuclear) it is kept running. That is where the distribution system is essential, the power is shipped across the county to where it can be used. Wind and solar are different, they may shut down when you need them, so are generally backed up by gas turbines (they can start up and shut down quickly). Distribution also is very important for wind and solar: but, that is why wind and solar require such large taxpayer subsidies; They basically need a matching gas turbine system to back them up. Coal units can be adjusted, but slowly.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="1832" LastEditDate="2018-02-13T21:48:34.940" LastActivityDate="2018-02-13T21:48:34.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19409" PostTypeId="2" ParentId="19393" CreationDate="2018-02-13T21:36:19.227" Score="1" Body="&lt;p&gt;No, you make the tubular strong enough to take the loads. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-02-13T21:36:19.227" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19410" PostTypeId="2" ParentId="19407" CreationDate="2018-02-13T22:09:22.757" Score="3" Body="&lt;p&gt;This is more of an addition to the &lt;a href=&quot;https://engineering.stackexchange.com/a/19408/1832&quot;&gt;previous answer by blacksmith37&lt;/a&gt;.  I think it might be more appropriate as a comment, or perhaps an edit to the previous answer.  With no ability to comment at this time, I'll leave the next step to people smarter than myself.  I believe this is a fairly straightforward answer:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How they predict the amount to generate&lt;/li&gt;&#xA;&lt;li&gt;How they handle excess generation &lt;/li&gt;&#xA;&lt;li&gt;How they handle shortage of generation&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The extent of my knowledge is coming from personal research and experience working in a coal power plant in the Midwest US. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Energy consumption is regulated by independent service operators (ISO).  Essentially, each day they (ISO) take a bid from all the sources willing to sell their electricity.  ISO's are very, very good at predicting the amount of power needed by the public by considering the weather, time of day, and other factors.  The power plants, turbines, panels, etc. make enough electricity to meet this demand each day.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The typical variability from the expected power consumption is small.  If there is &lt;em&gt;excess&lt;/em&gt; power generated, you can look at &lt;a href=&quot;https://electronics.stackexchange.com/questions/117437/what-happens-to-excess-energy-fed-into-the-power-grid&quot;&gt;this question&lt;/a&gt; on the electrical engineering stack exchange.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is not enough power being generated, &quot;peak&quot; plants that can ramp up quickly are used, as described in blacksmith57's answer.  While coal power plants ramp up slower, they don't usually run max capacity.  What this means is that several coal plants can all run at 80% capacity (arbitrary number) and each plant can slowly boost their output during &quot;peak&quot; time, to at least reduce the demand.  Faster reacting gas plants are much better operating as &quot;peakers&quot;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The relationship between the ISO's and the power plants is highly regulated following &lt;a href=&quot;https://en.wikipedia.org/wiki/California_electricity_crisis&quot; rel=&quot;nofollow noreferrer&quot;&gt;market manipulation at the turn of the century&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="14835" LastEditorUserId="1832" LastEditDate="2018-02-14T00:38:38.600" LastActivityDate="2018-02-14T00:38:38.600" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19411" PostTypeId="1" CreationDate="2018-02-13T22:51:09.973" Score="0" ViewCount="147" Body="&lt;p&gt;I want to build a 4-point bending apparatus for operating within a corrosive environment, using EXCO type corrosion solution as seen in ASTM Standard G46. Thus, in this highly corrosive environment, I need a very stiff material to act as the testing apparatus. Obviously, metals would be preferred. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest a metal that would be immune to corrosion, and/or an easy method to coat a more generic metal, say tool steel, such that it is immune?&lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2023-04-25T10:03:15.360" Title="Selection of corrosion resistant material for mechanical testing apparatus" Tags="|mechanical-engineering|corrosion|instrumentation|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19412" PostTypeId="2" ParentId="19407" CreationDate="2018-02-13T23:02:34.330" Score="1" Body="&lt;p&gt;gas and steam turbines, nuclear power plants, and hydro plants can be throttled to match their power output with the demand on an hour-by-hour basis. when demand is high, they are ramped up to full rated power; when it is low, they are trimmed back to lower power levels or the resulting excess generating capacity is shunted over to another portion of the power grid where the power demand is high at that moment.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-02-13T23:02:34.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19413" PostTypeId="1" CreationDate="2018-02-14T00:18:37.507" Score="1" ViewCount="79" Body="&lt;p&gt;I am looking for a cheap source of low power single transverse mode laser diode in order to focus them to a very small point. I thought I would get that from a DVD drive but the pattern I see at the focus point isn't a single point at all (2 or more points). I find that strange. Could you confirm that the DVD diodes aren't single transverse mode? (suggestions and reference of suitable diode are welcomed)&lt;/p&gt;&#xA;" OwnerUserId="5926" LastEditorUserId="7083" LastEditDate="2018-03-11T01:04:18.543" LastActivityDate="2018-03-11T01:04:18.543" Title="Are DVD laser diodes single transverse mode?" Tags="|optics|lasers|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19416" PostTypeId="2" ParentId="19392" CreationDate="2018-02-14T09:09:48.483" Score="1" Body="&lt;p&gt;Before I address the question:&#xA;&quot;My valve is old and it leaks. So, when it's fully shut off, its mass flow rate through the chamber is, let's say, 1% of that when it's &quot;wide open.&quot; GET THIS FIXED IMMEDEATLY. leaky gas installations are no joke. Just because nothings happened yet does not mean that you are safe. Sorry for only noticing this now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To your question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The burner is, in effect, a throttle. Mass flow through the burner is then a function of the pressure at the burner. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, if any obstruction were inside the pipe and impeding flow or the valve were not completely open, this would result in a malfunctioning heating system. But that problem would not necessarily show up on a pressure sensor reading.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not in the static case (valves closed), but in the static case your flow meter would sho zero too. When there's flow, a too low supply would correspond to a low pressure upstream of the burner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;agentp helpfully points out:  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the input BTUH is effecively a mass flow. The heating value of natural gas is about 1000BTU/scf..so you need 360scf/&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You suggest &quot;* A good spec should specify what can be best measured.*&quot; - correct, and that's pressure. An analog barometer is 50 € or less, an accurate gas flow meter goes for upward 1k€. Measuring gas flows is not trivial, either you need to accurately measure flow velocity and absolute pressure &amp;amp; temperature or you measure mass flow directly via thermal dispersion, which is more complicated than it sounds.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-02-14T09:09:48.483" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19417" PostTypeId="2" ParentId="19413" CreationDate="2018-02-14T09:22:16.277" Score="1" Body="&lt;p&gt;Diode lasers consist of many tiny emitters that are stacked together for power scaling, which has negative effects on beam quality. For the same reason, the laser beam also behaves different in both axes in respect to divergence and beam quality. That might be the reason why you see multiple focal points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reaching an actual single transverse mode for a laser is usually done with a special optical resonator (pun with my username not intended lol), containing elements to increase losses for higher modes. There is no room for that in a DVD drive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to wikipedia, you can also achieve single transverse mode by using a special structure in the semiconductor, for example a DBR (distributed bragg reflector).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the laser diodes in a DVD player are cheap mass produced diodes, so I don't expect such sophisticated structures to be used in them when they are not needed. &lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2018-02-14T09:22:16.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19418" PostTypeId="1" CreationDate="2018-02-14T10:04:20.820" Score="5" ViewCount="353" Body="&lt;p&gt;Do you happen to know relevant references in the literature that model the mechanical vibrations of a solid object (preferably antennas) by means of complex representations as a function of time?, i.e. can we say that the radial displacement can be modeled as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ r(t) = \sum_{i=1}^{N}{a_i \exp\left(-j b t \cos(\phi_i)\right)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $a_i, b,$ and $\phi_i$ are constant independent of time $t$ and $j$ is the complex number such that $j^2 = -1$.&lt;/p&gt;&#xA;" OwnerUserId="14930" LastActivityDate="2024-03-24T18:05:55.713" Title="Mechanical vibration of objects" Tags="|mechanical-engineering|vibration|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19419" PostTypeId="2" ParentId="19164" CreationDate="2018-02-14T13:58:24.883" Score="1" Body="&lt;p&gt;I solved my problem. I refined the mesh. Part of the problem was the high aspect ratio of my cells, because my tube is very thin in comparison to the length. I have solved the problem laminar and then take this as a starting value for my turbulence model. My results are physical now and the flow is fully developed some time after the expansion. Thanks to everyone who has contributed. &lt;/p&gt;&#xA;" OwnerUserId="14744" LastActivityDate="2018-02-14T13:58:24.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19420" PostTypeId="2" ParentId="18426" CreationDate="2018-02-14T14:09:46.043" Score="1" Body="&lt;p&gt;In generel for N species you have N mass balances. So in your case 2. Also mass is conserced ( other than moles when it comes to chemical reaction). So you definitely know that the mass flow rate of Y is 1500. Also have in mind that the divider doesn't change the composition of your species and you also know the composition of product 1 ( have in mind that all species together sum up to 1). I think from this point on it's not so hard to find out what Y is.&lt;/p&gt;&#xA;" OwnerUserId="14744" LastActivityDate="2018-02-14T14:09:46.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19421" PostTypeId="1" CreationDate="2018-02-14T14:35:57.397" Score="5" ViewCount="1623" Body="&lt;p&gt;I'm currently learning about how a CPU works on the hardware level and something that keeps coming up is memory organization. From what I've learned (and please, do correct me on anything that is not right), you can have a memory organized in the following two ways:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Byte organized memory: Each address on the address bus points to a memory location where a byte (8 bit) is stored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Word organized memory: Each address on the address bus points to a memory location where a word (multiple of 8 bit) is stored. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, a byte organized memory can access any memory location that is a byte boundary while a word organized memory can only access memory locations that are a word boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now here is where I have some trouble. I am studying for a midterm comming up and I am looking at past assignments and the following question comes up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a set of memory modules with 20 bit address and 8 bit data interface. We need to build a byte organized main memory of 4 MB for a 16-bit data architecture CPU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I know that we need to build a main memory using the memory modules given. To start off, we need to find the capacity of each memory module, whicch is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2^m * n, where m is the address bus in bits and n is the data bus in bits. This gives a capacity of 2^20*8 = 1MBytes for each memory module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to figure out how many memory modules are needed. This is fairly easy and is found as follows: 4MB / 1MB = 4 modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the part where I don't seem to understand, the memory is supposed to be byte organized, but the data bus out of the main memory is 16 bits. How can I design a byte organized memory if each access yields a word? Doesn't byte organized mean that given an address, it would be possible to access that specific byte in the memory? How is that possible when the memory modules must be designed in a way that yields a 16 bit data bus? Here is the solution for the question. Please explain to me in details why the main memory was designed this way, and how it is possible to access each bytes individually. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answer:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VfsWh.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VfsWh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerDisplayName="user12377" LastActivityDate="2018-02-18T17:50:25.700" Title="Byte organized and Word organized memory using memory modules" Tags="|electrical-engineering|computer-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19422" PostTypeId="1" AcceptedAnswerId="22005" CreationDate="2018-02-14T14:53:12.283" Score="4" ViewCount="171" Body="&lt;p&gt;I have laminar flow in a tube. Consider the tube to be 0.2 m long and with an average velocity of 0.05 m/s. The analytical expression for my transfer function is: $E(t)= \frac{\tau^2}{2*t^3}$ for $t$&gt;=$\frac{\tau }{2}$ and $E(t)=0$ for $t &amp;lt; \frac {\tau }{2}$.  $\tau$ is the mean residence time. In this case: $\tau$=0.2m/0.05m/s=4 s. I want to convolute this with an exponential equation: $E_2=(1-exp(\frac{-t}{2.55}))$. This equation descripes the magnetization of a particle in a static magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get the average magnetization of the particles at the outlet of the tube. So I thought I would do the following: $E_{out}(t)=E(t)*E2(t)$. I would then take the value at t=4s. I have the results from CFD which gives the following: &lt;a href=&quot;https://i.stack.imgur.com/TOxPu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;CFD result&lt;/a&gt;, the red line is the magnetic field and the green line is $E_{out}(t)$. I'm interested in the value at x=0,2m which is 0.7942.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My results of the convolution are totally different:&lt;a href=&quot;https://i.stack.imgur.com/7bCdc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Convolution result&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? I'm a little bit confused. Has anyone an idea of how to approach this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best regards,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gesetzt&lt;/p&gt;&#xA;" OwnerUserId="14744" LastEditorUserId="14744" LastEditDate="2018-02-20T15:28:12.987" LastActivityDate="2018-06-02T01:55:43.387" Title="Convolution with RTD laminar flow" Tags="|chemical-engineering|cfd|matlab|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19424" PostTypeId="2" ParentId="19407" CreationDate="2018-02-14T17:36:03.723" Score="1" Body="&lt;p&gt;If the system is not producing enough power voltage will drop so they know to produce more power.  If voltage is too high then producers drop power.  Distribution can bleed off power if required to not over power the consumer but the need for that is rare. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the US there is a grid west of the Mississippi and a grid and east of the Mississippi. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 3 type of sources:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Have almost no control.  For example wind.&lt;/li&gt;&#xA;&lt;li&gt;Have control at a variable cost.  For example coal or natural gas. &#xA;Nuclear also has variable cost but it does not vary much.  Nuclear is typically just run at design capacity.&lt;/li&gt;&#xA;&lt;li&gt;Hydro is unique in you have a finite source.  There is only so much&#xA;water behind the dam. It does not cost more to let out more water&#xA;but when you let out all the water you cannot produce more.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;First you fire up the cheap power (wind and nuclear).  Then fill in the variable power with coal and gas.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a state like WA that has a lot of hydro they have the luxury of selling off power at a premium during the summer when demand is high in the south. In the winter the reverse is true and WA can typically buy power for a discount from the south when demand is low in the south. A damn like Grand Coulee will not vary output by demand on a hourly or daily basis. They will just have like monthly flow rates based on water level.  Since hydro is cheap power it is consumed first.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-02-15T18:14:58.927" LastActivityDate="2018-02-15T18:14:58.927" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19425" PostTypeId="1" CreationDate="2018-02-14T18:28:57.403" Score="2" ViewCount="107" Body="&lt;p&gt;I've read about the idea of putting solar panels in space, or the already used electrodynamic tethers which gain energy as they move through Earth's magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question I have is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you get the generated energy back to Earth?&lt;/p&gt;&#xA;" OwnerUserId="13174" LastEditorUserId="9" LastEditDate="2018-02-14T21:46:39.503" LastActivityDate="2018-02-17T01:50:48.607" Title="How do generators in space transmit/ deposit energy back to Earth?" Tags="|heat-transfer|energy|energy-storage|generator|" AnswerCount="1" CommentCount="10" ClosedDate="2018-02-28T20:30:50.350" ContentLicense="CC BY-SA 3.0" />
  <row Id="19426" PostTypeId="2" ParentId="19401" CreationDate="2018-02-14T19:51:41.747" Score="1" Body="&lt;p&gt;The dynamics you’re looking for are largely governed by how you connect the different components in your design. Although the hydraulic pump you select is important, the equations provided in the supplier catalog should (at least) be sufficient for an initial design analysis (i.e. make an assumption with respect to pump’s efficiency by using the provided charts or google for typical/sane values). Depending on the required detail level of the model (which is determined by the design specifications) additional details might or might not have to be included in the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to your last question; in my experience the level of detail you’re looking for is generally not available in supplier catalogs. Mainly because it is not needed for sizing a pump, motor, etc. However, if you know what additional information you need, try giving the supplier a call. They might be willing/able to help ...&lt;/p&gt;&#xA;" OwnerUserId="11869" LastActivityDate="2018-02-14T19:51:41.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19427" PostTypeId="2" ParentId="19387" CreationDate="2018-02-14T20:14:45.173" Score="1" Body="&lt;p&gt;Remember that zeros are not global system properties but depend on the specifics of the selected/defined inputs and outputs. One possible way of modifying the zeros in your model is to modify the system inputs or outputs by adding filters, transformations, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this specific case, try adding a prefilter with the desired zero and optionally additional poles to cancel any existing zeros. Probably not the best approach, and not preferable in a real application, but it should work in this theoretical example.&lt;/p&gt;&#xA;" OwnerUserId="11869" LastActivityDate="2018-02-14T20:14:45.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19428" PostTypeId="2" ParentId="19421" CreationDate="2018-02-14T20:34:43.803" Score="3" Body="&lt;p&gt;If you notice, you need 20 bits to access the 1Mb byte modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x405c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x405c.png&quot; alt=&quot;Byte module&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But your processor is word aligned.  So 2 1Mb byte modules form a 2Mb word module.  The key to understanding how it works is the Misalignment at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R1pSA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R1pSA.png&quot; alt=&quot;Misalignment of Word addressed modules&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need 20 bits to access the specific byte aligned memory, but each location is a word or two bytes.  So you leave A0 (Address bit 0) out of the memory decoding.  Which is shown at the bottom of the drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A20 to A1 (bits 1 .. 20) select your 1Mb of word-aligned memory locations.  The misalignment is compensated for in the processor, as in high byte D15-D8 and low byte D7-D0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add a second bank and A21 selects upper or lower banks.  Which is misleading from the diagram.  Both banks are accessed and A21 multiplexs between which bank is accessed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other option would be A21 selects between the the two banks by activating chip select.  Appropriate bank puts data directly on data bus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same approach applies to 32, 64 or 128 memory buses.  Leave out the low address bits which are made redundant.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit to clarify OP comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by your reasoning, you need to be physically able to access each byte.  Thus replicating what is already in the processor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 64-bit processor with 8Gbytes of memory fetches 8 bytes/access, so 1G 64-bit locations.  Should we expand decoding so each byte can be accessed individually?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about booleans?  Do we access via bits?  64 or 128 bit floats.  You have the physical structure of the hardware and data structures, which must optimize throughput through processor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you create a structure with a bool, byte, 16 bit integer, byte and a 64 bit float.  A compiler for the 16 bit processor may allocate it as float (4 ints), int (2 separate bytes combined), int, and int (bool - 15 bits wasted).  So 7 ints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you turn the structure into an array, the compiler may allocate each member as 8 ints, so the inherent alignment structure for floats and ints is not disrupted. Another int wasted.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2018-02-18T17:50:25.700" LastActivityDate="2018-02-18T17:50:25.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19429" PostTypeId="1" AcceptedAnswerId="19436" CreationDate="2018-02-15T00:42:26.087" Score="5" ViewCount="5707" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j18gP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j18gP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Source: &lt;a href=&quot;https://www.motherearthnews.com/organic-gardening/low-tunnel-construction-mini-hoop-house&quot; rel=&quot;nofollow noreferrer&quot;&gt;Low Tunnel Construction: How to Build a Mini Hoop House&lt;/a&gt; &lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a small greenhouse that I want to keep as &lt;strong&gt;cold&lt;/strong&gt; as possible in the spring. I plan to cover the greenhouse with a &lt;strong&gt;polyethylene tarp&lt;/strong&gt; to shield it's contents from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure what colour of tarp would be the most effective at shielding heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that light coloured tarps would allow for solar gain. And I wonder if a black tarp might also retain warmth by absorbing heat from the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there specific colours of material that are known to let in the least amount of heat?&lt;/p&gt;&#xA;" OwnerUserId="10536" LastEditorUserId="7083" LastEditDate="2018-03-10T23:21:10.173" LastActivityDate="2021-08-04T13:26:57.100" Title="What colour of tarp will let in the least amount of heat?" Tags="|thermodynamics|heat-transfer|environmental-engineering|thermal-insulation|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19431" PostTypeId="2" ParentId="19429" CreationDate="2018-02-15T02:01:01.933" Score="3" Body="&lt;p&gt;I would suggest a reflective mylar film such as those used for survival blankets. &#xA;That would be the most effective single layer solution. Adding additional layers of plastic would further insulate the ground from the warmer air in the tunnel.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2018-02-15T02:01:01.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19432" PostTypeId="2" ParentId="19429" CreationDate="2018-02-15T02:33:24.893" Score="0" Body="&lt;p&gt;@donald Gibson is correct, i will add this note: a black tarp will exclude heat because it absorbs it. but after absorbing it, its temperature goes up and it begins re-radiating the heat which includes off the shaded side of the tarp- which means that any tarp that excludes the transmission of infrared (heat) radiation by absorbing it will inevitably fail to block heat because that absorption inevitably gives rise to radiation. the incident radiation must be reflected away, as pointed out by @Donald gibson. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13363" LastEditDate="2018-02-15T05:49:48.867" LastActivityDate="2018-02-15T05:49:48.867" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19434" PostTypeId="2" ParentId="6015" CreationDate="2018-02-15T06:07:45.807" Score="1" Body="&lt;p&gt;The concept is not valid. We agree the heavy portion of the tyre moves the suspension to rotate on a larger radius than the lighter sections. Conversly the lighter section opposite the heavy section will be pulled inwards to rotate on a smaller radius. Now we want the beads to congregate on the light portion BUT centrifugal force effects the beads too. As you stated &quot; ... gravity still acts on the beads but is small relative to centrifugal forces.&quot; As long as the centrifugal force is greater than gravity the beads will gather at the point furtherest from the center of rotation, ie. the heavy portion. Thus not helping at all.&lt;/p&gt;&#xA;" OwnerUserId="14944" LastActivityDate="2018-02-15T06:07:45.807" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19435" PostTypeId="1" CreationDate="2018-02-15T08:02:38.580" Score="1" ViewCount="67" Body="&lt;p&gt;I have a structure which I need to know some of its mechanical properties, using the standard values is good enough so looking into a mechanical properties chart of various materials would be ok.&#xA;The problem is that I don't know what is the material the structure is made and I cannot disassembly it and neither do a tensile test or a crystallography analysis &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I identify the material of the structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JmgO2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/NBTnn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14946" LastEditorUserId="14946" LastEditDate="2018-02-16T16:51:27.947" LastActivityDate="2018-02-16T16:51:27.947" Title="How to identify a material without doing a destructive test or crystallography analyses" Tags="|materials|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19436" PostTypeId="2" ParentId="19429" CreationDate="2018-02-15T10:29:55.953" Score="4" Body="&lt;p&gt;As a practical and cost effective solution, I'd advise using aluminium foil, or tin foil as it's also called. Please realise that it's rather fragile, so it's best to glue it to a plastic foil or something to keep the wind from ripping it apart. It doesn't really matter if cracks get in the foil. Use the most shiny, reflective side of the foil as the outside. This maximises reflection of heat in the form of radiation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make your roof wider, so it covers more ground surrounding your crops. If the sun is able to heat up the ground right next to your crops, the heat in the ground will travel to the ground under your crops, and ending up also heating the crops themself. You don't want that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't get or keep it any cooler than the surrounding air, so make sure it's ventilated, to keep temperatures as close as possible to outside air. That's the lowest you'll be able to go. Therefore, don't hermetically insulate your crops like you did in the picture, but enable it to ventilate. No insulation can keep all the heat outside. It will let through more or less heat. Any heat that is let through by the roof is removed by ventilating. If you cover them like in your picture, any heat let through by the roof is instead kept under it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd advise to set up your roof like this. You'll also keep the ground surrounding the crops as cool as possible this way, minimising any heat conducted to the crops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NqQsk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NqQsk.png&quot; alt=&quot;roof&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="368" LastEditDate="2018-02-15T16:24:33.880" LastActivityDate="2018-02-15T16:24:33.880" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19437" PostTypeId="2" ParentId="19429" CreationDate="2018-02-15T10:54:39.817" Score="1" Body="&lt;p&gt;White and silver are the most reflective (not absorb heat from the sun). There is some debate about which is more reflective. A white tarp is probably going to be cheaper. &lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-02-15T10:54:39.817" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19438" PostTypeId="2" ParentId="19435" CreationDate="2018-02-15T13:39:40.570" Score="1" Body="&lt;p&gt;Surely you know &lt;i&gt;something&lt;/i&gt; about the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's metal or otherwise made of &quot;heavy&quot; elements, one way is a X-ray spectrometer.  These come in hand-held gun-style devices.  You basically point it at what you want to know the composition of, pull the trigger, and it tells you within a few seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen these things in action, and they are surprisingly sensitive.  These devices generally come with a built-in library of common materials and alloys.  I've seen one distinguish between several different types of stainless steel, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another common use for these devices is in assaying minerals.  You literally point it at a rock and it tells you what minerals are in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet another common use is to check for lead in paint coating, or in toys imported from locations with questionable practices.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-02-15T13:39:40.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19439" PostTypeId="1" AcceptedAnswerId="19444" CreationDate="2018-02-15T16:21:18.343" Score="0" ViewCount="1244" Body="&lt;p&gt;I drew a quarter of my element and i want to generate the whole part by using 'pattern mirror'. The actual element is totally symmetrical (as you can see in the first image sketched by yellow curves) but in my final results (the second image) that ain't so and I have absolutely no clue why! &#xA;&lt;a href=&quot;https://i.stack.imgur.com/MRorS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MRorS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xDbjS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xDbjS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorUserId="10902" LastEditDate="2018-02-15T16:24:47.867" LastActivityDate="2018-02-21T13:41:52.453" Title="I need some help with 'pattern mirror' tool in Solidworks" Tags="|mechanical-engineering|solidworks|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19440" PostTypeId="1" AcceptedAnswerId="19441" CreationDate="2018-02-15T16:56:51.317" Score="1" ViewCount="66" Body="&lt;p&gt;When reconditioning of a pneumatic cylinder, which fit between piston and cylinder is required? and how much tolerance between mating parts is required?. Will surface finish inside of cylinder play a major role?.&lt;/p&gt;&#xA;" OwnerUserId="14951" LastActivityDate="2018-02-17T02:51:52.633" Title="Cylinder Reconditioning" Tags="|pneumatic|" AnswerCount="2" CommentCount="0" ClosedDate="2018-02-28T20:30:25.713" ContentLicense="CC BY-SA 3.0" />
  <row Id="19441" PostTypeId="2" ParentId="19440" CreationDate="2018-02-15T17:53:05.853" Score="1" Body="&lt;p&gt;It depends on the application. I think it's better described as play. Tolerance says something about the requirements regarding precision of the production, play says something about the space between piston and cylinder. I'll assume you are referring to the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's a fast moving piston, like in a compressor, the play can be relatively much. The piston moves so fast, air doesn't get much chance or time to escape. Relatively, fast moving pistons create a lot of heat, so play is required to prevent them from getting too hot. Leaving more play will effectively give less resistance, giving less heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's a slow moving piston, like a hydraulic jack, a tight fit is required. If there's too much play, air (or oil in the case of a jack) will be able to escape, and you won't be efficient in moving or compressing the air. &#xA;Since these kind of pistons don't move fast, more resistance is allowed since there won't be any significant heat generated anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface finish plays a big role in this situation, a rough surface is much harder to seal, and will also give more resistance. Generally speaking, the flatter the better. A practical example: try to mount the suction mount of your navigation or phone to a window. It will mount just fine since the glass is perfectly flat. If you try to mount it to a wall, it won't even hold; the wall is too rough to give a proper seal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the times, a piston will have one or more piston rings which are relied upon for a proper sealing. Both metal and rubber piston rings are used in resp. fast and slow moving pistons. This method is more efficient in sealing, and makes maintenance cheaper and easier; replacing a ring is cheaper and easier than replacing a whole piston.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-02-15T17:53:05.853" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19442" PostTypeId="1" CreationDate="2018-02-15T18:07:20.907" Score="2" ViewCount="139" Body="&lt;p&gt;I have been involved with a number of metalwork design projects lately and I'm increasingly finding myself asking for quotations for small prototype runs (often one-off jobs) to verify the validity of a design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kind of items I'm talking about are small items like electronic product enclosures, mounting plates for screwing PCBs onto etc.  Items that would probably fit into a shoebox and are made of machined or folded plastic, aluminium, mild steel.  Not heavy engineering, so no welding etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; How would a large custom metalwork manufacturing company (1000+ employees let's say) interpret my design files and what steps would they be likely to take in order to estimate costs for a small run of prototype units?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I give them a 3D STEP file (output from Solidworks for instance), how would they evaluate the complexity of the design and convert that to a list of costs such as tooling charges, material costs and setup-time costs etc?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What I think might happen:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;They load the STEP file, find the smallest stock material size that fits the design - this gives the materials cost.&lt;/li&gt;&#xA;&lt;li&gt;An engineer analyses the features in the design and does the CAM toolpath generation and other operations at this level.  This becomes NRE tooling costs and production line job-change costs.&lt;/li&gt;&#xA;&lt;li&gt;Shipping costs are estimated based on the final object's size, shape and weight.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But this sounds like it would take a long time to organise and would be expensive in terms of man-hours, which they would want to avoid since I haven't given them any money yet.  Quite often the quote comes back to me within a couple of days, which suggests that they might simply be &quot;eyeballing&quot; the project, guessing the price and adding a wavy-hand percentage to cover the guesswork.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this kind of thing work in this industry?  I have asked one of our smaller manufacturers but they seemed reluctant to answer this sort of question as if it was some kind of closely guarded secret.  I want to stay on their good side so I stopped asking :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to hear details from those with experience in custom machining, sheet metal fabrication (stamping, folding etc) and other related activities.&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastEditorUserId="110" LastEditDate="2018-02-18T12:59:48.243" LastActivityDate="2019-04-28T03:51:57.253" Title="Understanding metalwork manufacture -- How are Purchase Quotations prepared?" Tags="|machining|prototyping|cost-engineering|metal-folding|cost-estimation|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19444" PostTypeId="2" ParentId="19439" CreationDate="2018-02-15T22:12:12.367" Score="0" Body="&lt;p&gt;Change your settings to be &quot;&lt;strong&gt;Pattern Bodies&lt;/strong&gt;&quot;, instead of trying to pattern the features. Selecting the 'geometry pattern' option &lt;em&gt;may&lt;/em&gt; also help, but it's not as robust as patterning the bodies, as then you can be 100% certain that your part is symmetrical. You would use feature/face patterns if you wanted to e.g. have a circular pattern of holes in an otherwise asymmetric part.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-02-15T22:12:12.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19445" PostTypeId="1" AcceptedAnswerId="19452" CreationDate="2018-02-15T22:32:43.210" Score="2" ViewCount="2216" Body="&lt;p&gt;I was always curious why lead was chosen as the default for metal acid batteries.  This article describes lead-acid battery operation and there are plenty of resources like this on how lead acid batteries work, but I could not find any resources that explain &quot;why lead?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about lead makes it the standard for metal-acid batteries?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/electric/leadacid.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lead Acid Battery on HyperPhysics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10921" LastEditorUserId="2819" LastEditDate="2018-02-16T06:47:36.170" LastActivityDate="2018-02-16T10:58:25.060" Title="Why lead was chosen as the metal for lead-acid batteries?" Tags="|electrical-engineering|chemical-engineering|battery|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19446" PostTypeId="2" ParentId="18896" CreationDate="2018-02-15T22:58:51.020" Score="0" Body="&lt;p&gt;To add onto Bart's Excellent Answer (above), we have to step out of theoretical and into applications.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Redo the math with air density at the higher temperature! - This changes ${\rho}$ to only 0.55 gm/m^3 at 300C.  Terminal velocity shifts to &lt;strong&gt;12.5 m/s&lt;/strong&gt; Redo the flow rate with new terminal velocity! &lt;strong&gt;You need 300 CFM&lt;/strong&gt; Add onto the safety factor - since beans won't suspend if they go on the non-flat face, we get rid of the safety factor and see we need &lt;strong&gt;450 CFM&lt;/strong&gt;, or 420 gm/hr. 420 gm/hr of cool air at 1.2 gm/m^3 is only &lt;strong&gt;206 CFM.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Now we develop the flow profile.  Through the heater, the losses can be expressed as some sort of mesh.  There are a few ways known to express this, but this needs to be reviewed.  At the end, the fan needs to be able to counter the static pressure loss of flowing 450 CFM through the heater.&lt;/li&gt;&#xA;&lt;li&gt;Similar through the mesh holding up the beans.  450 CFM of air at 0.55 gm/m^3 flows through the mesh, which would be experienced as a pressure loss.&lt;/li&gt;&#xA;&lt;li&gt;The beans themselves will float on velocity pressure alone, not static pressure.  As a result, the final static pressure to push through the mesh and heater can be calculated.&lt;/li&gt;&#xA;&lt;li&gt;The fan operates on a curve.  This has been discussed elsewhere, but the fan will need to hit the static pressure loss for the heater as well as the mesh, while simultaneously providing 450 CFM.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-02-15T22:58:51.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19447" PostTypeId="1" AcceptedAnswerId="19449" CreationDate="2018-02-16T01:01:10.570" Score="2" ViewCount="137" Body="&lt;p&gt;I have this pneumatic cylinder that I wish to connect to a wooden door in order to make it kinda like a Star wars door.  I'm wondering what the best way to connect the cylinder to the door would be.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cylinder has a threaded rod at the end. Looks like the one attached.  I was wondering if there are components like a bracket that can be screwed onto the door and has a threaded loop to which the Rod could be screwed to?&lt;img src=&quot;https://i.stack.imgur.com/nIaxI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XX4L5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="14955" LastEditorUserId="9" LastEditDate="2018-02-16T01:40:31.860" LastActivityDate="2018-02-16T01:40:31.860" Title="How to connect a pneumatic cylinder to a door?" Tags="|pneumatic|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19448" PostTypeId="2" ParentId="19405" CreationDate="2018-02-16T01:14:04.707" Score="1" Body="&lt;p&gt;Good question. I suspect this will be very difficult to predict an accurate answer. Partly it could depend on the discharge line size or line geometry as you point out. Consider the example of emptying a 1 liter coca cola bottle up side down. It does come out, but slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to design the tank to be self draining I would provide some way to break the vapor lock or vacuum in the headspace of the tank if possible.&lt;/p&gt;&#xA;" OwnerUserId="14901" LastActivityDate="2018-02-16T01:14:04.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19449" PostTypeId="2" ParentId="19447" CreationDate="2018-02-16T01:21:42.283" Score="3" Body="&lt;p&gt;Based on the threads at the end of your sample cylinder, you would be seeking a female threaded rod end:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RJLQA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RJLQA.jpg&quot; alt=&quot;female threaded rod end&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a multitude of variations of this product, almost always including the words &quot;rod end&quot; with different modifiers. You could have a forked threaded rod end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I1dCS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I1dCS.jpg&quot; alt=&quot;forked threaded rod end&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would not have to be female threads, of course. Depending on the work load, you may not have to have the bearing insert of the first image, although it allows for longer life and easier alignment in use.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-02-16T01:21:42.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19450" PostTypeId="2" ParentId="19445" CreationDate="2018-02-16T02:09:47.563" Score="5" Body="&lt;p&gt;each electrochemical reaction involving a lead atom in a lead-acid cell releases two electrons into the external circuit, which means it has a relatively good extractable power-to battery mass ratio. in addition, the charge/discharge process retains reversibility over a relatively large number of cycles, giving the cell a long usable lifetime. the materials needed to manufacture the cell (lead and sulfuric acid) are not expensive, and finally, the manufacturing processes are very well-understood after over 100 years of experience. All of these factors contribute to making lead-acid batteries the default technology where battery mass is not a design limiter but cost is. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-02-16T02:09:47.563" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19452" PostTypeId="2" ParentId="19445" CreationDate="2018-02-16T06:45:20.693" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Lead%E2%80%93acid_battery&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lead acid batteries&lt;/a&gt; has been around a long time and is easy to manufacture. They are rechargeable, recyclable, and reasonably safe. &lt;a href=&quot;https://en.wikipedia.org/wiki/VRLA_battery&quot; rel=&quot;nofollow noreferrer&quot;&gt;AGM or Absorbent Glass Mat&lt;/a&gt; lead acid has the added benefit of being sealed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason they are so common is because of the high &lt;a href=&quot;https://en.wikipedia.org/wiki/Comparison_of_commercial_battery_types&quot; rel=&quot;nofollow noreferrer&quot;&gt;watt-hour/$ ratio&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lead acid 6.77–17.41&lt;/li&gt;&#xA;&lt;li&gt;Alkaline  0.48&lt;/li&gt;&#xA;&lt;li&gt;Lithium   2.75&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They are the battery of choice for &lt;a href=&quot;https://en.wikipedia.org/wiki/Lead%E2%80%93acid_battery#Applications&quot; rel=&quot;nofollow noreferrer&quot;&gt;car starters&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Uninterruptible_power_supply&quot; rel=&quot;nofollow noreferrer&quot;&gt;UPS (uninterruptible power supplies&lt;/a&gt;), commerical/industrial battery backup systems, off-grid power systems, electric forklifts, emergency lighting, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per watt-hours in use they are definitely the most common. Per unit, it would most likely be alkaline, followed by lithium and lithium ion.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="90" LastEditDate="2018-02-16T10:58:25.060" LastActivityDate="2018-02-16T10:58:25.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19453" PostTypeId="2" ParentId="19425" CreationDate="2018-02-16T07:16:10.150" Score="-4" Body="&lt;p&gt;Generators in space are mainly used as nuclear power source in space where solar power is not a viable power source. The different space generators in use today are radioisotope thermoelectric generator and radioisotope piezo electric generator. This generator  converts energy stored in a radioactive material directly into motion to generate electricity by the repeated deformation of material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the transmission of energy is considered, it can be achieved by perpetual motion which is most possible in space the on earth. In space the weightless space helps to achieve perpetual motion which oppose the law of conservation of energy. Hence the energy can be tried to transmit back to earth from generators in space through perpetual motion.perpetual motion is motion of bodies that continue indefinitely.the laws of thermodynamiccs apply even at very grand scales.example,motions and rotations of celestial bodies such as planets may appear perpetual,but are subjected to many processes that dissipate kinetic energy,such as solar wind,interstellar medium resistance,gravitational radiation and thermal radiation,so they will not keep moving forever.there is a scientific consensus that perpetual motion in an isolated system voilates either first law of thermodynamics,or second law of thermodynamics or both.Depending on this criteria perpetual motion machines are clasified as&#xA;perpetual motion machine of first kind which produces work without the input of energy,it thus voilates the first law of thermodynamics.&#xA;pertual motion machine of second kind is a machine which spontaneously converts thermal energy into mechanical energy.when the thermal energy is equivalent to workdone it does not voilate first law of thermodynamics but voilates second law.&#xA;perpetual motion machine of third kind is defined as one that eleminates friction and other dessipative forces completely.it is impossible to  make such a machine as the dessipative forces cannot be completely eleminated.but it can be reduced to the limit in space.&#xA;hence using one of the criteria with the generators in space we can try to transmit energy back to earth.&lt;/p&gt;&#xA;" OwnerUserId="14958" LastEditorUserId="14958" LastEditDate="2018-02-17T01:50:48.607" LastActivityDate="2018-02-17T01:50:48.607" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19455" PostTypeId="2" ParentId="16551" CreationDate="2018-02-16T14:37:19.943" Score="1" Body="&lt;p&gt;In this circuit:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NPfnP.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NPfnP.gif&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The gain from Vin to Vout2 is about 1, and to Vout1 about -1.&lt;/p&gt;&#xA;&lt;p&gt;That the two gains have the same magnitude with opposite sign can be deduced from the fact that at first approximation, the emitter and collector currents are equal.  Since the emitter and collector resistors are the same, the voltage changes across them due to the same current will also be the same.&lt;/p&gt;&#xA;&lt;p&gt;Actually the emitter and collector currents aren't exactly equal.  The emitter current is the collector current plus the base current.  However, due to the gain of the transistor, the base current is a small fraction of the collector current.  For example, if the gain of the transistor is 100, then the emitter current relative to the collector current is 101 to 100.  That can be approximated to 1:1 for many practical purposes.&lt;/p&gt;&#xA;&lt;p&gt;So now the question comes down to why the magnitude of either gain is 1.  This can be answered by looking at the emitter voltage relative to the base voltage.  At first approximation, the B-E voltage is fixed, usually around 700 mV for silicon BJTs in their active useful range.  Since gain is about how much &lt;i&gt;changes&lt;/i&gt; are amplified, the fixed B-E offset doesn't matter.  AC-wise, the emitter voltage is the same as the base voltage.&lt;/p&gt;&#xA;&lt;p&gt;Note that the input capacitor is proof that this circuit is not intended to work at DC.  That capacitor specifically blocks DC, and only passes frequencies above some rolloff.  Basically, the capacitor forms a high pass filter against the impedance of R1, R2, and the apparent impedance looking into the base.&lt;/p&gt;&#xA;&lt;p&gt;This circuit is often used with an serial R-C network between the two outputs. The middle point between the R-C added become then a phase shifting network (0 to 180 degrees theoritical)&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="-1" LastEditDate="2021-06-29T08:12:47.210" LastActivityDate="2021-06-29T08:12:47.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19456" PostTypeId="2" ParentId="15401" CreationDate="2018-02-16T15:58:00.160" Score="0" Body="&lt;p&gt;By your description, it happens that you got data from your vehicle running and build a model based on it. If you did it, you got the closed loop model, as you mentioned it has a controller (see the image bellow). But what you need is the plant (process) model. To get it, you need to turn off the vehicle embedded controller and take the samples to build your model.&lt;a href=&quot;https://i.stack.imgur.com/hkI6q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hkI6q.png&quot; alt=&quot;Closed loop system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Image source: &lt;a href=&quot;https://hobbybotics.com/projects/hobbybotics-reflow-controller-v8-03/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hobbybotics Reflow Controller V8.03&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="14962" LastEditorUserId="7083" LastEditDate="2018-03-18T20:17:57.817" LastActivityDate="2018-03-18T20:17:57.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19457" PostTypeId="2" ParentId="14143" CreationDate="2018-02-16T16:12:49.730" Score="0" Body="&lt;p&gt;I'll answer part of your question.&#xA;Pade approximation is used to represent delays in rational form.&#xA;By doing this, you transform exp(-sT) in something like a transfer function, which can be used naturally in state space representation.&lt;/p&gt;&#xA;" OwnerUserId="14962" LastActivityDate="2018-02-16T16:12:49.730" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19458" PostTypeId="1" AcceptedAnswerId="19461" CreationDate="2018-02-16T16:49:15.080" Score="0" ViewCount="64" Body="&lt;p&gt;I recently outfitted a crew to build scaffold in preparation for an upcoming turnaround at our plant.  Among the tools requested were canvas buckets for manually hauling material and tools up and down scaffolds.  In sourcing these I went through a reliable well established industrial supplier with a know brand.  I ordered bags rated for 100 lbs which above the maximum load for manual hoisting permissible on our site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several days after arrival on site one of these bags failed with a 27 lb load in it.  Nobody was hurt but it could have easily turned out differently.  We immediately pulled all bags from use.  In investigating it was discovered that when used outside in winter conditions that the plastic parts of the bag bottoms become brittle.  No temperature restrictions were offered in any documentation related to these bags and these tools were being used for the purpose they were designed for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel awful that a decision I made put people at risk.  Is there some aspect of due diligence in tool selection that a reasonable engineer would have been expected to take that I had missed?  &lt;/p&gt;&#xA;" OwnerUserId="869" LastEditorUserId="869" LastEditDate="2018-02-16T17:03:11.917" LastActivityDate="2018-02-16T20:13:37.893" Title="Potentially fatal near miss: Due diligence in tool selection" Tags="|tools|specifications|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19459" PostTypeId="2" ParentId="18341" CreationDate="2018-02-16T18:41:05.343" Score="0" Body="&lt;p&gt;The op amp in the left of the circuit don't do anything;&#xA;The op amp in the right is connected like a comparator, with positive feedback. That means it's output will be VCC or GND depending on Vi, not intermediate values.&lt;/p&gt;&#xA;" OwnerUserId="14962" LastActivityDate="2018-02-16T18:41:05.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19460" PostTypeId="1" CreationDate="2018-02-16T19:59:05.697" Score="1" ViewCount="65" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZxxLB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZxxLB.png&quot; alt=&quot;fractionationg column&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the book 'Chemical Engineering' by Coulson and Richardson (Volume 2, 5th edition), in the section 'Fractionating Column')(page-560), it's said that,'The feed stream is introduced on&#xA;some intermediate tray where the liquid has approximately the same composition as the&#xA;feed.' This gives no definite position of the feed discharge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our professor explained that the feed is discharged somewhere below the sieve plate, which is a plate situated in the middle of the column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is there a particular reason for the feed stream being discharged here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the feed stream be discharged anywhere in the stripping section above the liquid?'&lt;/p&gt;&#xA;" OwnerUserId="14871" LastActivityDate="2018-02-16T19:59:05.697" Title="position of feed stream discharge in the fractionating column" Tags="|chemical-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19461" PostTypeId="2" ParentId="19458" CreationDate="2018-02-16T20:13:37.893" Score="4" Body="&lt;p&gt;As a busy engineer, I would say you did your due diligence. There are always more variables than we are given time to consider. The higher the risk of failure/injury, the more time we spend on it, but there are limits. I often try to have another engineer or manager look at my work so there is another set of eyes on it. Also, taking time to test new tools/equipment is a good when possible. In hindsight this would have addressed the issue. Sometimes you can do &quot;field testing&quot; by just testing the smallest portion of the system at a time. This could be like first giving out one bag to a worker that you know will give your prompt and reliable feedback before your give out all of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our company just started requiring a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hazard_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;hazard analysis&lt;/a&gt; for all capital projects and major process changes. A small change like bags may have flown under the radar, but a procedure like that might be something worth implementing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other side of things, remember that the workers are not helpless. They consciously or subconsciously know to test their tools before they use them. They should also be wearing proper PPE like hardhats and steel toe boots to reduce risk in general. It also helps to let these guys know that they have a say in what tools they use. If they feel something is unsafe they need to make everyone aware. We recently implemented a &lt;a href=&quot;http://docs.healthandsafetyhub.co.uk/Vennsys/Media_library/vennsys-media-take-5-meter-reader-booklet.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;take 5 safety program&lt;/a&gt; that is designed to assist with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to this, please let the manufacturer know (not just the vendor). You will save more people than just the people at your facility. If they are of any credibility they will refund or replace your bags too.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-02-16T20:13:37.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19462" PostTypeId="1" CreationDate="2018-02-16T21:28:23.427" Score="3" ViewCount="54" Body="&lt;p&gt;Firstly, I am sorry that the title of this question isn't helpful. I have bought these motors for a project:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.banggood.com/5V-7mm-2-Phase-4-Wire-Stepper-Motor-p-1154480.html?cur_warehouse=CN&quot; rel=&quot;nofollow noreferrer&quot;&gt;5V 7mm 2 Phase 4 Wire Stepper Motor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I assumed that the 1.7mm diameter threaded shaft would take an M1.7 nut, but it does not. Standard thread gauges haven't given me a good fit and I am thinking that this thread is probably some industry standard. The motors are designed to adjust the lenses in digital cameras, so does anyone know were I might find standards relating to this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternativley, is there some list of common threads that I can just try all the 1.7mm diameter nuts to find a good fit? I have time and money but don't know where to look.&lt;/p&gt;&#xA;" OwnerUserId="14966" LastEditorUserId="7083" LastEditDate="2018-02-19T22:26:09.647" LastActivityDate="2018-02-19T22:26:09.647" Title="What thread could this camera motor be?" Tags="|motors|measurements|stepper-motor|standards|threads|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19463" PostTypeId="1" CreationDate="2018-02-16T22:47:39.077" Score="3" ViewCount="109" Body="&lt;p&gt;I have a question that I had asked in the stack physics section and they told me that I should ask my question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am basically starting to work on an RC submarine, and I want to use a 200ml syringe as a ballast tank. My plan was to pull the syringe handle back with a gear system, that's being rotated by a small DC motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how can I calculate, how much force(Newtons) does the motor need to pull the handle back and suck water into that syringe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After finding how much force it needs to pull, how can I calculate the minimal electrical power it needs to use in order for it to operate?&lt;/p&gt;&#xA;" OwnerUserId="14967" LastEditorUserId="2819" LastEditDate="2018-02-19T12:03:48.760" LastActivityDate="2018-02-21T15:34:45.863" Title="Air pressure under water" Tags="|fluid-mechanics|pressure|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19464" PostTypeId="2" ParentId="19442" CreationDate="2018-02-17T02:42:15.797" Score="0" Body="&lt;p&gt;First, I'm not sure if any large custom metalwork manufacturing company of 1000+ employees exist. Most local metalworking companies in the Chicago area are about a dozen employees or less. A laser, press brake, and a few mills and lathes can produce many millions $ of metal parts in an year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seen several 20+ years experienced machinist estimate pricing. They look at the print for about 10 seconds and say a price. I'm pretty sure they look at the material, overall dimensions, and machining needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sheet metal cutting is automated. When the dxf or cad file is uploaded, they have software that checks part weight and amount of laser cutting needed. Press brake work is probably # of bends, tooling change, size of part. Haven't been involved in stamping as it is for large quantity production runs.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-02-17T02:42:15.797" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19465" PostTypeId="2" ParentId="19440" CreationDate="2018-02-17T02:51:52.633" Score="0" Body="&lt;p&gt;General comment, this is a direct answer to the question. Most pneumatic cylinder, especially bores of 2&quot; or less are non repairable. Actually I have never seen a pneumatic cylinder repaired, but I think the procedure is to get a new seal kit. I suggest contacting the manufacture of the cylinder.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-02-17T02:51:52.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19466" PostTypeId="2" ParentId="19418" CreationDate="2018-02-17T03:06:15.553" Score="0" Body="&lt;p&gt;No, I didn't check but I don't think these type of equations would be in a mechanical vibrations textbook mainly because a forcing function or initial displacement is not present, nor is the distance from the base of the antenna. In addition to mechanical vibrations equations, Finite element analysis can provide radial displacement. Textbooks links are below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/013436130X&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/Mechanical-Vibrations-6th-Singiresu-Rao/dp/013436130X/ref=sr_1_1?ie=UTF8&amp;amp;qid=1519437095&amp;amp;sr=8-1&amp;amp;keywords=mechanical+vibrations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0070340412&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/Schaums-Outline-Mechanical-Vibrations-Graham/dp/0070340412/ref=sr_1_6?ie=UTF8&amp;amp;qid=1519437095&amp;amp;sr=8-6&amp;amp;keywords=mechanical+vibrations&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4311" LastEditorUserId="4311" LastEditDate="2018-02-24T01:53:31.430" LastActivityDate="2018-02-24T01:53:31.430" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19467" PostTypeId="2" ParentId="19403" CreationDate="2018-02-17T03:24:51.577" Score="1" Body="&lt;p&gt;This is more of comments than an answer.&#xA;module = pitch circle diameter /  Number of Teeth. A gear with 1mm module and 36T would have a 36mm pitch diameter, not 72&quot;&#xA;Also, companies such as &lt;a href=&quot;https://us.misumi-ec.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://us.misumi-ec.com&lt;/a&gt; sells 1mm module 36T gears with selectable bores sizes.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-02-17T03:24:51.577" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19468" PostTypeId="2" ParentId="19411" CreationDate="2018-02-17T03:36:37.327" Score="0" Body="&lt;p&gt;I suggesting contacting a coating company. In the Chicago area a coater is VP Plating, &lt;a href=&quot;http://www.vpfinish.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.vpfinish.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if there any pivot joints you will need bearings. You may want to contact Igus bearings or Pacific Bearing. They various chemical resistant plastics that may hold up in this environment.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-02-17T03:36:37.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19470" PostTypeId="2" ParentId="19462" CreationDate="2018-02-17T06:59:23.303" Score="1" Body="&lt;p&gt;There are some &quot;exotic&quot; thread sizes used for small instrument work. Note that for a specific diameter there can be more than one pitch. It is also possible that the thread is not metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.engineersedge.com/hardware/metric-external-thread-sizes1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Metric fine pitch threads&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rndW8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rndW8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2018-02-17T06:59:23.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19472" PostTypeId="1" AcceptedAnswerId="21838" CreationDate="2018-02-17T16:45:37.907" Score="0" ViewCount="122" Body="&lt;p&gt;In the textbook : &lt;em&gt;Electric Power Principles&lt;/em&gt;, in Chapter 1 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;on the working of a pv cell : &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The source current Is is the result of absorption of photons in&#xA;  sunlight that cause separation of valence electrons from their atoms.&#xA;  The resulting hole/electron pairs fall across the high field gradients&#xA;  present at the diode junction.  Because any voltage resulting from&#xA;  this current tends to forward bias the actual junction, the voltage&#xA;  available is limited&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't understand the train of thought leading to a limited voltage, and I'd appreaciate a better explaination or a reading recomendation. &lt;/p&gt;&#xA;" OwnerUserId="13823" LastEditorUserId="90" LastEditDate="2018-02-18T11:00:10.660" LastActivityDate="2018-05-22T18:10:07.387" Title="Working mechanism of photovoltic cells" Tags="|electrical-engineering|power-electronics|power-engineering|photovoltaics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19473" PostTypeId="1" CreationDate="2018-02-17T17:18:18.853" Score="3" ViewCount="215" Body="&lt;p&gt;I am trying to understand why normal shocks are caused in Converging Diverging Nozzles. My textbook explains that since the exit pressure must match the back pressure, the normal shock is the only mechanism that can make this happen, which I understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, why does the exit pressure have to match the back pressure? What if it did not? Then simply the jet of fluid exiting the converging diverging nozzle would expand or contract under the pressure driving force induced with the back pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any clarifications welcome.&lt;/p&gt;&#xA;" OwnerUserId="9515" LastActivityDate="2022-01-03T13:06:15.463" Title="Why does exit pressure match back pressure in Converging Diverging Nozzles?" Tags="|fluid-mechanics|compressible-flow|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19474" PostTypeId="1" CreationDate="2018-02-17T22:54:21.050" Score="-4" ViewCount="116" Body="&lt;p&gt;I plan to buy a mining rig, with 750 watt PSU. It will be turned on for 24/7. I had like to calculate, how much kw/h will it spend per month?&lt;/p&gt;&#xA;" OwnerUserId="14979" LastEditorUserId="9" LastEditDate="2018-02-17T23:22:01.140" LastActivityDate="2018-02-17T23:22:01.140" Title="Power consumption of mining rig?" Tags="|electrical-engineering|power-electronics|electrical-grid|" AnswerCount="1" CommentCount="2" ClosedDate="2018-02-19T06:07:58.677" ContentLicense="CC BY-SA 3.0" />
  <row Id="19475" PostTypeId="2" ParentId="19474" CreationDate="2018-02-17T23:01:08.010" Score="2" Body="&lt;p&gt;Since I lack the reputation to write comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electricity cost of mining is way higher than the earnings. Small scale mining is not profitable anymore for a long time already, mining farms have 20000 graphics cards or something in that range.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond that: a 750 Watt PSU will (at its limit) consume 0.75kWh per hour and thus 126 kWh per week, roughly 500 kWh per month. How much the mining rig actually draws from the socket depends on the components you use. &lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2018-02-17T23:01:08.010" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19476" PostTypeId="2" ParentId="19442" CreationDate="2018-02-18T12:57:09.820" Score="-1" Body="&lt;p&gt;There is no one size fits all answer. There are many factors that go into during the quotation process. As requester it important to for you know strength and weakness of your supplier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following are few of my experiences, though they are electrical engineer based I believe there are similarities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use external prototyping houses design and manufacture of electrical/electronic/firmware based systems. I have &lt;strong&gt;N&lt;/strong&gt;on-&lt;strong&gt;D&lt;/strong&gt;isclosure &lt;strong&gt;A&lt;/strong&gt;greement's (NDA) with all vendors. Therefore I provide as much information as possible to the vendor. There vendors usually references similar project that they have worked in the past. For example if they have worked on MSP430 microcontroller based project in the past with similar components then they have a references with regard to time they spent on the project. They use this a baseline estimate cost to execute the project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I give same vendor a project with alternate microcontroller such as a LPC5400 from NXP I usually get a quote that is fairly high and usually takes longer the generate a quotation. If the project is cost sensitive (usually they are) I find a vendor that has experience with NXP product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the if the features requested are generic, then chances are you will get a competitive response fairly quickly. For example the if the PCB board is square and a standard size. For most of the standard features the NRE costs are standard. But if the PCB has features that are custom such as circular or alternate shapes it might take longer and chances a quotation will be higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/126/how-to-reduce-engineering-bill-of-material-cost-for-electronics-products?rq=1&quot;&gt;How to reduce Engineering Bill of Material Cost for electronics products?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-02-18T13:04:14.130" LastActivityDate="2018-02-18T13:04:14.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19477" PostTypeId="2" ParentId="19411" CreationDate="2018-02-19T03:41:29.460" Score="0" Body="&lt;p&gt;I see G -46 only covers evaluation of pitting; not a test media : Pitting is not the objective of 4 point bent beam tests , eg. Shell Sc test.  The standard materials for racks in general are Hastelloy C or titanium. But you need to know what the test environments will be. I have seen titanium racks burst into flame when immersed into a test solution without proper evaluation. Contact NACE in Houston , I expect they will give you lists of vendors of corrosion test equipment. ASTM may also be some help ( they used to be on Race St in Philadelphia but I believe they moved.). Your employer has apparently not provided adequate training.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-02-19T12:23:41.363" LastActivityDate="2018-02-19T12:23:41.363" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19479" PostTypeId="1" AcceptedAnswerId="19480" CreationDate="2018-02-19T06:59:13.030" Score="0" ViewCount="29" Body="&lt;p&gt;there&#xA;I installed SINO 942 tracker in my bike, it works well, but then my bike's self start has stopped working (it works when I ride bike for around an hour), so&#xA;I think the tracker is eating up battery when the bike is parked for long hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am already thinking about installing a manual switch, or a rechargable&#xA;12 V battery (so it wont need to connect to main battery), but all of these&#xA;are manual and regularly to-be-interacted-with solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest best solution I can use to rectify this issue.&lt;/p&gt;&#xA;" OwnerUserId="14988" LastActivityDate="2018-02-19T07:15:38.860" Title="Efficient solution for Bike Tracker" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2018-03-01T13:13:53.863" ContentLicense="CC BY-SA 3.0" />
  <row Id="19480" PostTypeId="2" ParentId="19479" CreationDate="2018-02-19T07:15:38.860" Score="1" Body="&lt;p&gt;You can go with your second battery, but use a split-charge system so whenever the bike is running the second battery gets charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of a split charge system / relay is &lt;a href=&quot;https://www.ebay.co.uk/itm/160477-0-727-10-EQUIV-12V-100A-100-AMP-HEAVY-DUTY-SPLIT-CHARGE-RELAY-CAMPER-VAN/201605769301?hash=item2ef0a3e855:g:0f4AAOSwgsxaaEY8&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-02-19T07:15:38.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19482" PostTypeId="1" AcceptedAnswerId="19483" CreationDate="2018-02-19T10:59:39.357" Score="1" ViewCount="125" Body="&lt;p&gt;&lt;strong&gt;Brief&lt;/strong&gt; I am actually of 17 not even in college but I have dream for making my project/dream/aspiration/or So to come true so I am thinking of making a device that performs a function say like Thermometer measures Temperature, now working and doing the function is not a concern of now (a Question for me), the thing is if I got the data of Reading of final Temp (assume it has performed its temperature calculation and given output [Yep its digital]) so now it gave me some value x .&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt; I want to get this value x into my server or website (maybe I am unable to express it but I want to get it to me through Internet) and store that value for data of  &lt;strong&gt;Person P&lt;/strong&gt; so , I am not sure how I would achieve that. Assume all calculation and functions are done and are independent of device now I just need to know how to store that value for use?  Would it require something like connecting it to a wifi card or similar and making an OS (Till now this has came across my mind )  &lt;/p&gt;&#xA;" OwnerUserId="14313" LastEditorUserId="110" LastEditDate="2018-02-19T11:11:36.120" LastActivityDate="2018-02-19T12:08:36.060" Title="How do I get data from my Hardware to server or so?" Tags="|electrical-engineering|telecommunication|" AnswerCount="2" CommentCount="3" ClosedDate="2018-02-25T11:25:04.580" ContentLicense="CC BY-SA 3.0" />
  <row Id="19483" PostTypeId="2" ParentId="19482" CreationDate="2018-02-19T11:37:06.293" Score="0" Body="&lt;p&gt;There are many ways to achieve your goal. Assume your goal is to publish a temperate data to a server or website. You will need at minimum microcontroller, temperature sensor and wireless module. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One good place to start is &lt;a href=&quot;https://www.arduino.cc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Arduino community&lt;/a&gt;. You could purchase a Arduino kit with a temperature sensor and ESP8266 wifi module. There are sufficient resources online to publish data to server or website. Another good place to start is &lt;a href=&quot;https://www.adafruit.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adafruit&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also suggest you take a look at  &lt;a href=&quot;https://www.adafruit.com/product/2680&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adafruit.io Internet of Things ESP8266 Wifi starter kit&lt;/a&gt;. They use &lt;a href=&quot;https://io.adafruit.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;adafruit.io&lt;/a&gt; IOT cloud to publish data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I have not use these resources in a longtime. Therefore I am not update on the latest trends. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other alternate resources online. Let us know how this works out for you. Good luck. &lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2018-02-19T11:37:06.293" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19484" PostTypeId="2" ParentId="19463" CreationDate="2018-02-19T11:57:52.707" Score="2" Body="&lt;p&gt;It depends on the air pressure inside your submarine and the maximum depth you want to operate at. Friction of the syringe will also have to be accounted for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say you want to dive 100meters max and the air pressure inside the unit is atmospheric (which cancels out the air pressure above the water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then calculate the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrostatics#Hydrostatic_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;hydrostatic pressure&lt;/a&gt;:&lt;br&gt;&#xA;&lt;a href=&quot;https://www.google.com/search?q=1000kg%2Fm%5E3+*10m+*+9.81m%2Fs%5E2&amp;amp;oq=1000kg%2Fm%5E3+*10m+*+9.81m%2Fs%5E2&amp;amp;aqs=chrome..69i57j6j69i58.1100j0j9&amp;amp;sourceid=chrome&amp;amp;ie=UTF-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;1000kg/m^3 *10m * 9.81m/s^2 &lt;/a&gt;= 98kPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force is dependent on the diameter of syringe. Looks like the &lt;a href=&quot;http://www.harvardapparatus.com/media/harvard/pdf/Syringe%20Selection%20Guide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;diameter of a 200ml syringe&lt;/a&gt; is 44.75mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?ei=S7eKWtCKDuDEjAP8hrfACQ&amp;amp;q=3.14159*+%28+44.75mm%2F2+%29%5E2&amp;amp;oq=3.14159*+%28+44.75mm%2F2+%29%5E2&amp;amp;gs_l=psy-ab.3...1640.1640.0.1868.1.1.0.0.0.0.107.107.0j1.1.0....0...1.1.64.psy-ab..0.0.0....0.9Yuy2mcneQ0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Area = 3.14159* ( 44.75mm/2 )^2 &lt;/a&gt;= 0.00157 m^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force = &lt;a href=&quot;https://www.google.com/search?q=98kPa+*+0.00157+m%5E2&amp;amp;oq=98kPa+*+0.00157+m%5E2&amp;amp;aqs=chrome..69i57j6.1097j0j4&amp;amp;sourceid=chrome&amp;amp;ie=UTF-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;98kPa * 0.00157 m^2&lt;/a&gt; = 154N  (~35 pounds force)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this method considers the volume of the air in the submarine large compared to the volume in the syringe. The required force will be reduced by the increased air pressure in the submarine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The static friction from the seal will also need to be measured empirically and added to the total force required. To estimate it, this &lt;a href=&quot;http://www.specialtypharma.com/Main/Back-Issues/SYRINGE-STUDY-Gerresheimer-Scrutinizing-20000-Syri-982.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; gives an overview of forces required for a range of syringes; recently-moved up to 3 year old. Based on their graph it looks like about 5 Newtons for a recently-moved (6 or 8mm diamete?) syringe. A 200ml syringe will be more than this and will need tested empirically with a scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add the friction value to the total force required and multiply by a 1.5 safety factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then just buy a linear actuator with the right force and stroke. Then just use the manufacturer specifications on power consumption as it already includes the gear/motor efficiency calculations. This &lt;a href=&quot;https://www.firgelliauto.com/products/premium-linear-actuators&quot; rel=&quot;nofollow noreferrer&quot;&gt;FA-35-12-(X)-P&lt;/a&gt; linear actuator has a maximum draw of 5A at 12v or 60watts.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2018-02-19T18:57:19.660" LastActivityDate="2018-02-19T18:57:19.660" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19485" PostTypeId="2" ParentId="19482" CreationDate="2018-02-19T12:08:36.060" Score="0" Body="&lt;p&gt;Just as Mahendra says, in terms of hardware you'll need your sensors connected to something such as an arduino or raspberry pi and have an internet connection. If using a pi you have both wired/wireless options ready out of the box. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you planning on hosting your own server or buying server space? Either way linux would be an easy way to go. Watch &lt;a href=&quot;https://youtu.be/vzojwG7OB7c&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; to set up your own server, it shows you how to install Apache, PHP and MySQL. If buying some webspace, you'll need to buy a domain name from a registrar such as GoDaddy  or 123reg and then buy the space from somewhere such as hostinguk.net, anyone that hosts linux with PHP and sql. Once you've done both, link your domain with the webspace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have everything, you can program your hardware to read the sensor input and send it to your websites database. There are countless guides on all of this online, this answer should point you in the right direction atleast.&lt;/p&gt;&#xA;" OwnerUserId="12187" LastActivityDate="2018-02-19T12:08:36.060" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19487" PostTypeId="1" CreationDate="2018-02-19T14:17:56.020" Score="3" ViewCount="79" Body="&lt;p&gt;I'm building a tracker for a high altitude balloon but some of the components are out of stock and may not be in before our launch date. I'm good with an iron but have limited engineering knowledge. Here is the list of components for reference: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.mouser.com/ProductDetail/810-CGA4J1X7S1C106KC&quot; rel=&quot;nofollow noreferrer&quot;&gt;Multilayer Ceramic Capacitors MLCC - SMD/SMT TDK&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.mouser.com/ProductDetail/667-ERJ-P06J101V&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thick Film Resistors - SMD Panasonic&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.mouser.com/ProductDetail/863-MBR120VLSFT3G&quot; rel=&quot;nofollow noreferrer&quot;&gt;Schottky Diodes &amp;amp; Rectifiers - ON Semiconductor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.mouser.com/ProductDetail/81-GRM21BR61C226ME4L&quot; rel=&quot;nofollow noreferrer&quot;&gt;Multilayer Ceramic Capacitors MLCC - SMD/SMT Murata Electronics&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm wondering how people go about finding suitable replacements for out of stock parts. I realize that there may be many factors in finding replacements and that this is too broad a question for a quick answer but I just wanted to start somewhere. &lt;/p&gt;&#xA;" OwnerUserId="6631" LastEditorUserId="7083" LastEditDate="2018-02-20T03:23:07.310" LastActivityDate="2020-11-22T14:00:20.060" Title="Replacement components?" Tags="|electrical-engineering|circuits|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19491" PostTypeId="1" CreationDate="2018-02-19T16:44:18.450" Score="0" ViewCount="229" Body="&lt;p&gt;In Callister it says, &quot;two modes of failure are possible ductile and brittle.&quot; Why not malleable?Furthermore why is uniaxial tensile test chosen for predicting failure?&lt;/p&gt;&#xA;" OwnerUserId="12558" LastEditorUserId="12558" LastEditDate="2018-02-19T16:49:53.257" LastActivityDate="2018-04-21T13:59:23.570" Title="Ductile,Brittle and Malleable" Tags="|mechanical-failure|failure-analysis|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19492" PostTypeId="2" ParentId="19491" CreationDate="2018-02-19T17:08:24.113" Score="0" Body="&lt;p&gt;Malleable is the same as ductile. The uniaxial tensile test provides a uniform method to measure the mechanical properties of materials. It only predicts failure of uniaxial test bars. There are a variety of techniques / methods to predict failures or structures ; Such as L E F M , linear elastic fracture mechanics.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-02-19T17:08:24.113" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19493" PostTypeId="2" ParentId="19491" CreationDate="2018-02-19T17:48:37.397" Score="1" Body="&lt;p&gt;Ductile (&quot;able to be drawn out into a thin wire&quot;) makes sense as the failure method, because the test is stretching the sample. Malleable (&quot;able to be hammered or pressed into shape without breaking or cracking&quot;) doesn't make sense here, since the sample is not under compression! You could describe a three-point bend failure as Malleable, however. When @blacksmith37 states that &quot;Malleable is the same as ductile&quot;, this is what he's referring to. They are somewhat analogous, where &quot;Ductile&quot; is in tension, and &quot;Malleable&quot; in compression. It's important to note, however, that they are &lt;em&gt;not&lt;/em&gt; the same physical property. For example, Zinc and Lead both exhibit much higher malleability than they do ductility.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-02-20T13:47:23.817" LastActivityDate="2018-02-20T13:47:23.817" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19494" PostTypeId="1" AcceptedAnswerId="19506" CreationDate="2018-02-19T22:52:08.987" Score="3" ViewCount="84" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YwsLy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YwsLy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Prunus&quot; rel=&quot;nofollow noreferrer&quot;&gt;prunus&lt;/a&gt;&lt;/em&gt; pits that I &lt;a href=&quot;https://en.wikipedia.org/wiki/Stratification_(seeds)&quot; rel=&quot;nofollow noreferrer&quot;&gt;stratify&lt;/a&gt; in a refrigerator during winter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, the seeds germinate in springtime, which is ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, often some of the seeds &lt;strong&gt;germinate prematurely&lt;/strong&gt;. When this happens, it's necessary to grow the seeds/plants indoors during winter until they can be planted outdoors (after the &lt;a href=&quot;https://en.wikipedia.org/wiki/Growing_season&quot; rel=&quot;nofollow noreferrer&quot;&gt;last frost date&lt;/a&gt;). &lt;strong&gt;This is not ideal.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than let the premature pits grow,  an alternative would be to slow germination down (or even prevent it) by keeping them at &lt;strong&gt;just above freezing&lt;/strong&gt;. It's my understanding that the colder plants are, the slower they grow (up until freezing, of course--the plants are too young to go dormant for sub-zero temperatures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question becomes, &quot;Where in my house is the temperature at just above freezing?&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, refrigerators are at a temperature of around 6°C. From experience, I've found that 6°C is too high; the plants grow and eventually die due to lack of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I store the seeds at 1°C in a household setting?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rEgU3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rEgU3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10536" LastEditorUserId="10536" LastEditDate="2018-02-20T12:25:12.497" LastActivityDate="2018-02-20T12:25:12.497" Title="Store seeds at 1°C" Tags="|temperature|environmental-engineering|refrigeration|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19495" PostTypeId="2" ParentId="19487" CreationDate="2018-02-19T23:01:15.237" Score="1" Body="&lt;p&gt;You are correct. There are many factor that need to be considered before you can find replacement components. Lets take for example the first component on the list the multilayer ceramic capacitor (MLCC) from TDK. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HOSBP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HOSBP.jpg&quot; alt=&quot;Multilayer Ceramic Capacitor - TDK&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the parameters capacitance, tolerance and foot print as circled in the above image. You are most likely limited to 0805 foot print capacitor. Depending on the circumstances you might be or might not be able to find a replacement other than a 0805. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the function of the electrical circuit you might be or might not be limited to a 10uF capacitor. Similarly depending on the electrical circuit sensitivity you might be or might not limited to a tolerance of 10%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary it is good idea to design with using generic electrical components. Another option is to secure the parts prior to fabricating the PCB (I am assuming the PCB is already fabricated. &lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2018-02-19T23:01:15.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19496" PostTypeId="2" ParentId="15674" CreationDate="2018-02-20T00:11:53.943" Score="0" Body="&lt;p&gt;Well, I have used regular acrylic tubing for the gaseous state. However, if any spark contact may occur, use spark resistors, a bubbler, ect. If it is in liquid state I would assume that more would be required. My project this year used hydrogen as a fuel source produced through electrolysis that was run through solar energy. My hydrogen output was in a gaseous state for efficiency. &lt;/p&gt;&#xA;" OwnerUserId="14999" LastActivityDate="2018-02-20T00:11:53.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19498" PostTypeId="1" CreationDate="2018-02-20T00:20:01.310" Score="1" ViewCount="667" Body="&lt;p&gt;I'm a newbie of electricity and I would like to know if a piece of tool steel connected to both positive and ground pole of a welder tend to heat up red hot.&#xA;If yes, at wich amperage?&#xA;There are some risks involved such as overheating the welder or a huge waste of electricity due to the hight resistence of the metal.&#xA;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="15000" LastActivityDate="2018-02-20T22:51:52.387" Title="Heating up Tool Steel with a welder" Tags="|electrical-engineering|steel|welding|tools|heat-treatment|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19499" PostTypeId="2" ParentId="19498" CreationDate="2018-02-20T01:10:33.223" Score="1" Body="&lt;p&gt;This is a layman's view, but based on some engineering principles of which I am aware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tool steel as you describe would become a direct short. As such, the resistance isn't necessarily going to be high enough to make it red hot. There may be other components in the circuit (clamps, cable connections, cables, etc.) that would have higher resistance than the piece of steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A stick welder performs as it does because the hot lead (welding rod) and ground (work piece) are brought together only long enough to create an arc. This arc creates high temperatures to melt the rod into the work piece, as well as to heat the work piece to the correct temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you desire to heat a piece of tool steel for tempering purposes, perhaps, consider to use the electricity available in the form of &lt;a href=&quot;https://www.instructables.com/id/DIY-Induction-Heater/&quot; rel=&quot;nofollow noreferrer&quot;&gt;inductive heating&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linked article describes the method of creating a DIY inductive heating device with a variety of electronic components. I would not suggest that this is the precise answer for you, but to provide you with leads in a direction more effective than shorting out a welding rig. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another similar DIY tutorial can be found on the &lt;a href=&quot;https://www.instructables.com/id/12KW-Induction-Heater/&quot; rel=&quot;nofollow noreferrer&quot;&gt;same site&lt;/a&gt;. This one reads as a 12kw device, likely sufficient to perform the tasks you require.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not be surprised to learn that these suggestions are more complex than you would desire, considering the complexity of the latter project:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9UqhS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9UqhS.jpg&quot; alt=&quot;inductive heater&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-02-20T01:10:33.223" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19500" PostTypeId="2" ParentId="19498" CreationDate="2018-02-20T01:43:29.680" Score="0" Body="&lt;p&gt;Depending on the size and amperage it will get hot . It has been done on steel bars at the mill to heat-treat them , but with more amperage than a welder can supply. Why tool steel ?&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-02-20T01:43:29.680" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19502" PostTypeId="1" CreationDate="2018-02-20T04:51:44.333" Score="1" ViewCount="633" Body="&lt;p&gt;I want to compare the same model with solid shape and shell shape,&#xA;since my model is big, I am wondering if there is a way to convert the solid part into a shell in Abaqus easily without creating a new model from the beginning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me resolve this problem&#xA;Thank you very much..&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-02-20T04:51:44.333" Title="How to convert solid part to shell part in Abaqus?" Tags="|finite-element-method|abaqus|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19504" PostTypeId="2" ParentId="19491" CreationDate="2018-02-20T05:50:57.260" Score="0" Body="&lt;p&gt;Ductile: The capacity of materials to allow large deformations or large extension without failure is termed as ductility. The materials with high ductility are termed as ductile materials. Normally if the elongation exceeds 15% the material is ductile.&#xA;Brittle: A brittle material is one which exhibit a relatively small extensions or deformations to fracture and failure occur suddenly without any warning. If the elongation is less than 5% the material is considered to be brittle.&#xA;Malleable:It is a special case of ductility where it can be rolled into thin sheets but it is not necessary to be so strong. &lt;/p&gt;&#xA;" OwnerUserId="15007" LastActivityDate="2018-02-20T05:50:57.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19505" PostTypeId="2" ParentId="15054" CreationDate="2018-02-20T06:00:44.583" Score="1" Body="&lt;p&gt;The cheapest option for you to consider would be to use Esp8266 modules. They technically cost 1 USD. You can import a bulk (lot) order from aliexpress to make it cheaper. For you command centre you can use a NodeMCU (Esp8266 V12E) costing about 10-12 USD (800 INR). &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Program the NodeMCU as a server and the smaller Esp8266s as clients. It is pretty simple and you can choose from many libraries on 'github'. Their GPIOs receive inputs from each sensor and transmit the data over Wi-Fi as there are all connected to a network hosted by your router/modem to your command centre. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The controllers need a 3V input which is not common, so you need to use level controllers (costs ~2 dollars) step down standard voltages. Once you have received all the data on your server you can process it as desired and program it to give necessary feedback to your sensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cost (excluding senors) for approx 5 sensors = 20 USD.&lt;br/&gt;&#xA;Refer the links below for more information.&lt;br/&gt;&#xA;&lt;a href=&quot;http://www.nodemcu.com/index_en.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.nodemcu.com/index_en.html&lt;/a&gt;&lt;br/&gt;&#xA;&lt;a href=&quot;http://esp8266.net/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://esp8266.net/&lt;/a&gt;&lt;br/&gt;&#xA;&lt;a href=&quot;https://iot-playground.com/blog/2-uncategorised/42-esp8266-wifi-pir-motion-sensor-arduino-ide&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://iot-playground.com/blog/2-uncategorised/42-esp8266-wifi-pir-motion-sensor-arduino-ide&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15006" LastEditorUserId="15006" LastEditDate="2018-02-20T06:09:48.827" LastActivityDate="2018-02-20T06:09:48.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19506" PostTypeId="2" ParentId="19494" CreationDate="2018-02-20T07:15:03.320" Score="3" Body="&lt;p&gt;The temperature control in a fridge is pretty crude. The evaporator is in the freezer and the compressor is controlled with an electric thermostat. The fridge portion however relies on convection from the freezer. The temperature control from the fridge is sometimes nothing more than a slider that establishes how quickly the convected air moves. There are likely better methods employed in more expensive units so you will have to look at your specific fridge. In any case, the temperature can fluctuate quite because the designers were not concerned with a little temperature fluctuation harming food.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/5Tdip.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Tdip.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if your freezer has an auto defrost cycle this will also increase the temperature for &lt;a href=&quot;https://en.wikipedia.org/wiki/Auto-defrost#Mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;15 minutes every day&lt;/a&gt; (or whatever the timer settings are).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before we go changing things, it might be helpful to get a &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B01HH7YD2Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;wifi temperature logger&lt;/a&gt; or something similar so you can see how much the temperature fluctuates and how much modifications improve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming this fridge is 100% dedicated to this project; I would try adjusting the fridge temperature to target your 1°C with its current crude control. Fill the rest of the fridge with gallon jugs of 3/4 full of tap water(room for freezing) to help stabilize the temperature. If the freezer is not loaded, also fill it with 3/4 filled gallon jugs of water. This increased thermal mass will help stabilize the temperature. It also helps ensure the temperature does not dip much below 0°C if controls are not ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that doesn't fully address it, figure out how the &lt;a href=&quot;http://www.appliance411.com/faq/howdefrostworks.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;auto defrost cycle&lt;/a&gt; works on your specific refrigerator/freezer and disable it. This &lt;a href=&quot;https://www.repairclinic.com/Shop-For-Parts/a4/Refrigerator-Parts&quot; rel=&quot;nofollow noreferrer&quot;&gt;parts site&lt;/a&gt; has pretty good schematics of most models. Of course, now you will have to remember to do it manually before and after or your freezer/fridge will run all the time and waste power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that doesn't get the temperature stable enough you can replace the passive convection temperature control system with an &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B004YTSB7C&quot; rel=&quot;nofollow noreferrer&quot;&gt;ac fan&lt;/a&gt; and &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B01KNXETWS&quot; rel=&quot;nofollow noreferrer&quot;&gt;pid controller&lt;/a&gt;. The fan turns on when it needs more cold air from the freezer and turns off when it is cold enough. Definitely more work, but it will be able to hold the temperature much closer than with other methods.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-02-20T07:15:03.320" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19508" PostTypeId="1" AcceptedAnswerId="19517" CreationDate="2018-02-20T09:32:41.430" Score="2" ViewCount="786" Body="&lt;p&gt;Shear stress is introduced in statics as $\frac{P}{A}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the maximum shear stress with neutral axis is different.  It is $\frac{3P}{2bh}$ (Where $b$ is width and $h$ is height)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference between these concepts?&lt;/p&gt;&#xA;" OwnerUserId="15001" LastEditorUserId="2015" LastEditDate="2018-02-20T18:38:45.590" LastActivityDate="2018-02-20T18:54:53.140" Title="What is the difference &quot;maximum&quot; shear stress and shear stress" Tags="|mechanical-engineering|structural-engineering|stresses|experimental-physics|shear|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19509" PostTypeId="1" AcceptedAnswerId="19524" CreationDate="2018-02-20T09:45:31.730" Score="2" ViewCount="1146" Body="&lt;p&gt;I was looking into using &lt;a href=&quot;https://www.bosch-motorsport.com/content/downloads/Raceparts/en-GB/51865867208058251.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lambda Sensor LSU 4.9&lt;/a&gt;  to know the lambda value. From the datasheet, its looks though this sensors cannot be used without a special IC such as Bosch lambdatronic LT4. Can we get any information out of the sensor by connecting only 12v to the heater+ and heater-. What would be the minimum circuit to get some information from this sensors?&lt;/p&gt;&#xA;" OwnerUserId="15011" LastEditorUserId="110" LastEditDate="2021-02-08T12:46:11.910" LastActivityDate="2021-02-08T12:46:11.910" Title="Can lamda sensor 4.9 be used without the special LSU-IC?" Tags="|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19510" PostTypeId="2" ParentId="19472" CreationDate="2018-02-20T11:01:34.167" Score="0" Body="&lt;p&gt;A photovoltaic device is a device that generate voltage when exposed to light. Example of photovoltaic device is a solar cell.The photovoltaic effect was&#xA;discovered by Alexander-Edmond Becquerel in 1839, in a junction formed&#xA;between an electrode (platinum) and an electrolyte (silver chloride). It uses n-type semiconductor substrate.  A trivalent impurity is diffused on its top surface such that a thin p-type layer is formed. Metallic contacts are formed on both the layers for external circuit connections. Top surface is coated with antireflecting material to increase absorption of light energy. Finally it is covered with thin protecting glass. While junction is forming, depletion layer and potential barrier builds up because of crossing of majority carriers across the junction. When photon of energy greater than band gap(&gt;Eg) are incident on the cell, the photons are absorbed and electron-hole pairs are generated. The barrier potential segregates them and pushes electrons to n-side and holes in p-side. This displacement of charges sets up a p.d(photo voltages) across two regions, with p-side as +ve and n-side as -ve electrode. If external load Rl is connected, current flows through it and we get electrical energy. Typically a solar cell can generate photovoltage from 0.5V to 1.2V. Solar cells are connected in series and parallel to increase voltage and current capacity. Such combination of solar cells is called photovoltaic array.Photovoltaic cells are used in infrared detectors, light intensity calculators, solar fans, solar heater and solar panels etc.A solar panel consists of hundreds or thousands of photovoltaic cells mounted together in the form of a huge cell module.&lt;/p&gt;&#xA;" OwnerUserId="15007" LastEditorUserId="15007" LastEditDate="2018-02-20T11:57:03.810" LastActivityDate="2018-02-20T11:57:03.810" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19513" PostTypeId="1" AcceptedAnswerId="19521" CreationDate="2018-02-20T12:57:22.833" Score="3" ViewCount="81" Body="&lt;p&gt;So I have laser cutter. And I am thinking about buying a control unit (&lt;a href=&quot;http://www.topwisdom.com.cn/en/product.asp?dlm=11&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;, to be exact). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my concern here is the following: If the input DXF file contains splines, they will be converted to lines (G01) and arcs (G02 or G03), since these are all the entities G-Code has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since have absolutely no knowledge about motors or controlling them, the question is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does the control unit from the link do the same? Does it convert DXF to G-Code?&lt;/li&gt;&#xA;&lt;li&gt;If yes, is there a control unit that would quide my laser cutter over a spline from the DXF and NOT on some series of lines and arcs?&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.topwisdom.com.cn/en/product.asp?dlm=11&quot; rel=&quot;nofollow noreferrer&quot;&gt;This webpage&lt;/a&gt; says that only Z axis is controlled with PID. Why? How are other axis controlled. Does this mean I could in general choose between PID and G-Code?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8667" LastActivityDate="2018-02-20T20:36:50.067" Title="Are all motors controlled with G-Code?" Tags="|control-engineering|motors|pid-control|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19514" PostTypeId="2" ParentId="10211" CreationDate="2018-02-20T15:22:25.000" Score="-2" Body="&lt;p&gt;Pulse Width Modulation (PWM) is a method of manipulating the DC current's duty cycle or switching it ON and OFF at certain repeated intervals to control the current. For example, if you consider a time period of 4 ms, the pulse could be in ON sate for 2 ms and in the OFF state for 2 ms for each repetition. This is a 50% duty cycle. Let us assume the speed of the DC motor or wheel is x or x m/s in this case. If the ON pulse lasts 3 ms and the OFF pulse lasts 1 ms, the speed of the wheel or DC  motor would be greater than x or x m/s as it is getting more overall current in the same period of time. Feeding a constant current is not advisable to drive a DC motor. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the major parameter PWM affects is the speed of rotation of the  DC motor. The torque on the other hand depends on the DC motor itself. This should not be a problem if the motor is powerful to produce the required torque. In case if it is not, PWM impacts the torque as follows.&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The DC motor stops and starts at regular intervals(Not visible to the naked eye at high frequencies). That means it has to overcome the initial friction that many times. Due to this, it gives you an illusion that your torque is being reduced. In fact, the torque is constant but the energy is being consumed to overcome the friction.&lt;br/&gt;&lt;/li&gt;&#xA;&lt;li&gt;The torque appears to be less for lesser durations the pulse in the ON state during a repetition and vice versa.&lt;br/&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To compensate for the reduction in speed, increase the width of the pulse or use a better motor. Increasing the pulse width gives the motor more current. This also depends on the dead weight the motor is carrying. Perform a few trials to determine the pulse length required to compensate for the speed of the motor as the relationship is not linear. &lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other option to consider is to decrease the frequency and keep the integral pulse width/second constant to compensate for the energy drain. &lt;/p&gt;&#xA;" OwnerUserId="15006" LastActivityDate="2018-02-20T15:22:25.000" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19515" PostTypeId="1" CreationDate="2018-02-20T15:32:45.293" Score="1" ViewCount="75" Body="&lt;p&gt;I am using techniques explained in the book of Skogestag, Multivariable feedback control, for feedback control of a MIMO plant. In this book they explain the phenomena of static inverse plant decoupling in which the interaction of the off-diagonal terms are eliminated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a triangular MIMO plant in which one of the off-diagonal components is zero. Is it still usefull to apply static inverse plant decoupling in this case?&lt;/p&gt;&#xA;" OwnerUserId="15015" LastActivityDate="2018-04-22T19:09:31.757" Title="MIMO control: static inverse plant decoupling" Tags="|mechanical-engineering|control-engineering|frequency-response|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19516" PostTypeId="1" AcceptedAnswerId="19520" CreationDate="2018-02-20T16:21:19.840" Score="0" ViewCount="87" Body="&lt;p&gt;I'm translating a Russian text about a drug product manufacturing process. The text mentions briefly the vial stoppering machine, in which rubber stoppers are first fed into the hopper and then directed towards vials, and a &lt;strong&gt;&lt;a href=&quot;https://www.multitran.ru/c/m.exe?l1=2&amp;amp;l2=1&amp;amp;s=%EF%F3%E0%ED%F1%EE%ED&quot; rel=&quot;nofollow noreferrer&quot;&gt;пуансон&lt;/a&gt;&lt;/strong&gt; drives the stopper into the vial. Multitran &lt;a href=&quot;https://www.multitran.ru/c/m.exe?l1=2&amp;amp;l2=1&amp;amp;s=%EF%F3%E0%ED%F1%EE%ED&quot; rel=&quot;nofollow noreferrer&quot;&gt;provides multiple options&lt;/a&gt; for how to translate this &quot;пуансон&quot;. It's basically some shaft-like member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I googled and found a &lt;a href=&quot;http://www.findpatent.ru/patent/231/2319659.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;description of some stoppering machine in Russian&lt;/a&gt;, and it has a diagram, with &quot;пуансон&quot; under number 60:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/75iq5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/75iq5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The text there says that the axle (59) rotates, making the cam (58) rotate and push this.. plunger\piston\punch member (?) towards the vial and drive the rubber stopper into the vial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unsure which word to use for this part. &lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="349" LastEditDate="2018-02-20T16:33:08.543" LastActivityDate="2018-02-20T20:34:33.547" Title="What do I call this shaft that drives the rubber stopper into the glass vial in a vial stoppering machine?" Tags="|terminology|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19517" PostTypeId="2" ParentId="19508" CreationDate="2018-02-20T16:26:47.597" Score="4" Body="&lt;p&gt;The difference is in the assumptions.  The first &quot;shear stress&quot; is simply assuming the stress is uniform over the cross section.  As a result, we see a shearing force $P$ shearing a rod of uniform cross section $A$, and due to the assumption or uniform shear stress across the cross section, we have the resulting $\frac{P}{A}$ shear stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we abandon this assumption of uniform shear stress, and we use the assumptions for beam shear, we find &quot;maximum shear stress&quot;.  Here, we find the shear stress develops parabolically across the cross section, and develops a maximum sheer stress at the center line of the cross section, for symmetric cross sections.  For a rectangular cross section, this maximum sheer stress happens to be $\frac{3P}{2bh}$, but for other cross sections the value differs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the value is $$\frac{PQ}{Ib}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $b$ is the thickness of the cross section at the cut, $Q$ is the first moment of the area above the cut referenced to the centroid of the area: $$\int^{h_t-c}_{h_c-c} x dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $c$ is the height of centroid from the bottom of the cross section $$\frac{\int^{h_t}_{0} x dA}{\int^{h_t}_{0} dA}$$, $h_t$ is the height of the top of the area from the bottom of the cross section, $h_c$ is the height of the cut from the bottom of the cross section. $I$ is the second moment of the area above the cut referenced to the centroid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int^{h_t-c}_{h_c-c}{x^2}dA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note different uses of the reference point other than the bottom of the cross section will result in easier formulas for Q and I).  A &lt;a href=&quot;https://engineering.stackexchange.com/questions/19100/what-is-the-difference-between-tau-vq-it-and-tau-v-a?rq=1&quot;&gt;related post&lt;/a&gt; may be able to help you with more visuals.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-02-20T18:54:53.140" LastActivityDate="2018-02-20T18:54:53.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19519" PostTypeId="1" CreationDate="2018-02-20T19:15:38.653" Score="1" ViewCount="74" Body="&lt;p&gt;My body weight and that of the skateboard is $M = 90\text{ kg}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to move 10 meters in 4 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;x(t) &amp;amp;= (at^2)/2 \\&#xA;x(4) &amp;amp;= (a\cdot4^2)/2 = 10 \\&#xA; &amp;amp;= a\cdot16/2 = a\cdot8 = 10 \\&#xA;\therefore a &amp;amp;= 10/8 = 1.25\text{ m/s}^2&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wheels of the skateboard have a radius $r = 0.027\text{ m}$. I want a maximum velocity of $v = 30\text{ km/h} = 8.333\text{ m/s}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \dfrac{V}{r} = \dfrac{8.333}{0.027} = 308\text{ rad/s} = 2941\text{ rpm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F = ma = 1.25\cdot90 = 112.5\text{ N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque needed $C = Fr = 112.5\cdot0.027 = 3.03\text{ Nm}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the power $P = C\omega = 933\text{ W}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not really sure if it is correct because I only need 3.03&amp;nbsp;Nm when I start moving, same for 2941&amp;nbsp;rpm is just the maximum velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$C\omega = P$ but I'm not sure that it is the torque at start multiplied by $\omega_{max}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am right I need then to buy a DC Motor with 933&amp;nbsp;W and at least 3.03&amp;nbsp;Nm and at least 2941&amp;nbsp;rpm?&lt;/p&gt;&#xA;" OwnerUserId="14830" LastEditorUserId="1832" LastEditDate="2018-02-20T21:00:54.300" LastActivityDate="2018-02-20T21:00:54.300" Title="Choose a DC Motor for a skateboard" Tags="|mechanical-engineering|motors|torque|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19520" PostTypeId="2" ParentId="19516" CreationDate="2018-02-20T20:34:33.547" Score="2" Body="&lt;p&gt;A camshaft is rotating a cam that is part of the shaft. It seems the cam acts on a roller, fixed on a push-rod, and so pushes down the push-rod in the guider.&#xA;I think the push-rod pushes the rubber down with a punch cap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think push-rod is the word you're searching for if you're talking about part 60.&#xA;Plunger may be a synonym, but I think of a kind of pump when hearing that term.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-02-20T20:34:33.547" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19521" PostTypeId="2" ParentId="19513" CreationDate="2018-02-20T20:36:50.067" Score="3" Body="&lt;p&gt;Not really an answer, but I can't comment, so:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this mean I could in general choose between PID and G-Code?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PID and G-code are not two interchangable approaches to control a CNC-machine, they serve different purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G-code is basically the programming language to tell the CNC-machine where to drive with the axes, when to do stuff like turn on the laser, turn on cooling water, turn on process gas, whatever your machine can do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PID is a (simple) closed-loop control algorithm, in case of the control unit you linked, it ensures that the z-axis is at the position it is supposed to be, even under the influence of some disturbance. &lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2018-02-20T20:36:50.067" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19522" PostTypeId="2" ParentId="10211" CreationDate="2018-02-20T20:54:01.473" Score="2" Body="&lt;p&gt;Remember: current is proportional to torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be linear like you expect if it was like:&lt;br&gt;&#xA;PWM pulse high: Instant full current, and thus full torque.&lt;br&gt;&#xA;PWM pulse low:  Instant zero current and thus zero torque.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But just like with objects, you can't instantly speed them up, or slow them down to 0. You'll very clearly feel that when running into a wall.. :p Inertia prevents that. Just like inductance prevents instant changes in current flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in reality it's like:&#xA;PWM pulse high: Current begins to flow, until fully developed.&#xA;Rise-time depends on inductance. With the current rise, torque rises until a steady-state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PWM pulse low: Current starts reducing, and dissipates in eg. a freewheel diode.&#xA;That way, the motor acts as an alternator, creating 'negative' torque, or rather said, torque in reverse direction. This is the culprit here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without the alternator working, 50% PWM would get rather close to 50% torque if inductance is not too high, and the PWM frequency is chosen right. But the 50% time that PWM is off, you get reversed torque, and that's what causes your observation about 50% PWM not giving 50% torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More advanced motor controllers (partially)solve this issue, but if you just drive a DC motor with eg. a MOSFET of BJT, you'll have this issue. And remember to take care of freewheeling the motor, or you'll destroy the FET. You likely can't rely on the substrate diode for that, as it'll probably be underdimensioned.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-02-20T20:54:01.473" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19523" PostTypeId="1" CreationDate="2018-02-20T21:43:27.233" Score="2" ViewCount="403" Body="&lt;p&gt;I need to specify a stepper motor to drive a system of gears that isn't yet produced. I have almost no information about the system except that it will involve a stack of planetary gear sets with a common 3/8&quot; shaft, with a total weight of about 100lbs (mixed metal and plastic). It will need to rotate very slowly, about 1 rpm, and occasionally reverse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be able to get hold of one planetary gear set soon. What would anyone recommend to get a handle on the needed motor? I figured I'd wrap a wire around the shaft and pull it with a fish scale to see how many ounces it takes to move it, then multiply that by the number of sets. Given that value, how could I convert that to a motor spec? I'm assuming that most of the requirement will be in overcoming the initial inertia &amp;amp; friction.&lt;/p&gt;&#xA;" OwnerUserId="14939" LastActivityDate="2018-02-21T13:31:42.550" Title="Determining required motor torque" Tags="|torque|stepper-motor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19524" PostTypeId="2" ParentId="19509" CreationDate="2018-02-20T22:32:37.087" Score="1" Body="&lt;p&gt;No you can't. The LSU4.9 is a wideband sensor, and so differs from the old school narrow-band zirconia sensors that just produced a voltage. The wideband incorporate an oxygen pump along with a zirconia element to make it a wideband. That pump must be controlled for proper working, as well as the temperature which must be kept within limits. You can't just put 12V on the heater, it will destroy the sensor. The heater has to be throttled. And you can't just only use the zirconia element within the sensor to use it as a narrowband.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Driving the 4.9 requires a quite sophisticated construction. The LT4 uses the CJ125 IC, and even without the LT4, you will need that IC. The CJ125 IC is specially designed and made by Bosch to allow users to use the LSU sensors. And it is quite adequate for that purpose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some manufacturers succeeded in making their own circuitry without the CJ125, some with more success than others. But unless you're capable of producing your own IC's, you'll have to use the CJ125. And even then correctly using it is certainly not easy. A lot of peripheral hardware is needed to use the CJ125. At that point, you could best just use the LT4 or a similar tool. Innovate and AEM are manufacturers that also offer tools to read out the 4.9. Both use the CJ125 IIRC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;14Point7 offers a cheap PCB(~$40) that closed-loop heats and reads the sensor, and gives you the lambdavalue and temperature in either an analog voltage or via I2C communication. I have used it and it works fine. I think it's the closest you'll come to successfully reading the 4.9 yourself.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-02-20T22:32:37.087" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19525" PostTypeId="2" ParentId="19498" CreationDate="2018-02-20T22:51:52.387" Score="0" Body="&lt;p&gt;No. A welder isn't made for that. It gets a thin wire of steel hot enough to locally melt and stick to a piece of metal, it can't just heat a chunk of iron red hot. That takes enormous sustained amount of heat. If the piece has any serious mass, it will take much more heat than the welder can provide. You can heat it up. but the transformer would be molten before the steel starts to glow. I'd rather use a torch if you want to get it glowing, that's the proper tool here. Even if you have a powerful enough welder, your electrical connection will probably not be sufficient to deliver so much power. You'd just pop the fuses in your home. A torch however, can easily provide enough power/heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do you want to get it to glow? Do you need to rejoin it with another piece of steel? If you explain your plans, we might be able to help you better.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-02-20T22:51:52.387" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19527" PostTypeId="2" ParentId="19463" CreationDate="2018-02-20T23:02:19.127" Score="0" Body="&lt;p&gt;If energy consumption is important, a springloaded syringe may be effective. You only need to let go of it to submerge. That costs little power, so more is left for driving the boat. Refilling it with air again is an issue, but that could be done with a CO2 charge. You now, those used by airsoft and paintball pistols and what not. Again, you'd only have to open a valve to let in some CO2 and push out the water. That way, you can submerge and surface as often as you want without having to interact with the boat. (as long as the CO2 charge is not depleted of course..)&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-02-20T23:02:19.127" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19530" PostTypeId="1" CreationDate="2018-02-21T08:16:56.653" Score="0" ViewCount="55" Body="&lt;p&gt;I've seen it in the drawing,the expansion joint holds the nozzle.so why the expansion joint looks like this picture.what is the necessity of shaping the tube like this.in this picture the nozzle passes through the expansion joint has a diameter 57 mm.&lt;a href=&quot;https://i.stack.imgur.com/x3VqI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x3VqI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15005" LastActivityDate="2018-02-21T09:12:55.303" Title="what is expansion joint(compensating tube) in pressure vessels?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19531" PostTypeId="2" ParentId="19463" CreationDate="2018-02-21T08:20:02.173" Score="0" Body="&lt;p&gt;Just get a water meter, there are analog ones with an additional impulse output or a 4-20mA signal - where i live, 50-100€ a piece. A water meter without signal output might set you back 30€.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other solutions are more complicated for a worse result:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;pump flow rate is pressure dependant, during a pumping cycle the water level in your well will sink and level in your tank will rise, flow will become lower.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;counting the pumping cycles as per ercnutschs answer would work better, with the caveats in his answer. You'd need to set up to simple sensors and homebrew a system to interpret the signals&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since this is drinking water, any solution must be sure to no compromise water quality. Drinking water meters are made with this in mind.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In summary, I think an off-the shelf water meter is your best bet.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-02-21T08:20:02.173" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19532" PostTypeId="2" ParentId="19530" CreationDate="2018-02-21T09:12:55.303" Score="1" Body="&lt;p&gt;The expansion joint is designed and fitted to compensate or protect the equipment from changes in dimension due to changes in temperature and / or pressure.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-02-21T09:12:55.303" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19533" PostTypeId="2" ParentId="14329" CreationDate="2018-02-21T10:05:24.733" Score="0" Body="&lt;p&gt;right click on mesh edit and generate. once it turns into green tick then right click again and there will be an option to delete&lt;/p&gt;&#xA;" OwnerUserId="15023" LastActivityDate="2018-02-21T10:05:24.733" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19534" PostTypeId="1" AcceptedAnswerId="19536" CreationDate="2018-02-21T10:46:49.140" Score="0" ViewCount="314" Body="&lt;p&gt;I have been struggling to find which node and constraint suffering from 'nodes are used more than once as a slave node' in abaqus error message.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me resolve this problem&#xA;Thank you &lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-02-21T12:44:22.763" Title="How to know which constraints suffer from 'nodes are used more than once as slave node' in abaqus?" Tags="|mechanical-engineering|structural-engineering|finite-element-method|abaqus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19536" PostTypeId="2" ParentId="19534" CreationDate="2018-02-21T12:44:22.763" Score="1" Body="&lt;p&gt;in many cases where this type error is detected during job submission a node set is created to identify the offending nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;open the odb in the visualization module, go into display group manager, look for node sets with names like &quot;ERROR....&quot; or &quot;WARNING...&quot;.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2018-02-21T12:44:22.763" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19537" PostTypeId="2" ParentId="19523" CreationDate="2018-02-21T13:31:42.550" Score="1" Body="&lt;p&gt;It is the same as everywhere else. You do not need to calculate torque based on the design of the reducer/motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you need to figure out the ratio that specific gear vs motor will spin at. Once you do that, calculate the power of the pull, are you pulling with 5lbs or 10lbs of force? once you figure out the pull/push factors, you can translate these values into usable equations. since pull/push is power distribution it will always be able to be converted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;once you calculate the ratio and pull factors, you can then translate these values into torque. These calculators are available all over the net.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(you can use a food scale weight hook to determine the amount of pull in lbs tied to the end of the string)&lt;/p&gt;&#xA;" OwnerUserId="15027" LastActivityDate="2018-02-21T13:31:42.550" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19538" PostTypeId="2" ParentId="19439" CreationDate="2018-02-21T13:41:52.453" Score="0" Body="&lt;p&gt;When you have created a 3D part then you don't have to go to sketch circular pattern for an easy approach. Instead, you can directly go to feature toolbar there you will find and a ‘Linear Pattern’ below which there will be an arrow for a drop-down menu from there select 'Circular Pattern'. Or you can go to &lt;strong&gt;Insert&gt;Pattern/Mirror&gt; Circular Pattern&lt;/strong&gt; and then you can perform the same step by selecting features you want for the circular pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are doing it from ‘Sketch’ then you have to go to ‘Circular Sketch Pattern’ from the drop down of Linear Sketch Pattern. Then select sketches of features, not the features and then you can create a circular sketch pattern. And thereafter extruding the circular sketch pattern will give you complete part.&lt;/p&gt;&#xA;" OwnerUserId="15028" LastActivityDate="2018-02-21T13:41:52.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19539" PostTypeId="2" ParentId="7921" CreationDate="2018-02-21T15:09:21.337" Score="0" Body="&lt;p&gt;Simulating a process is a big task which includes various preprocessing steps. We try to model the real world yet are not able to do it 100%.  So sometimes it’s hard to find errors in your model. Please check for the following errors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cad Modeling Error&lt;/li&gt;&#xA;&lt;li&gt;Material Properties Error&lt;/li&gt;&#xA;&lt;li&gt;Discretization Errors &lt;/li&gt;&#xA;&lt;li&gt;Other small errors&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Check for the modeling error. There may be a problem in the geometry of parts and assembly of your motor. You should check the assembly mates and their constraints. Sometimes constraints are not given correctly which give wrong results while simulating the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check your material properties which you have given to your motor parts. Improper or incomplete material properties will result in different behavior when you will simulate the process.  For example, the limit of Poisson’s ratio in isotropic materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your model may have Discretization errors. If you want to run the proper simulation with highly accurate results you can use the ALTAIR Hypermesh for meshing as there may be a possibility that your element size may be too large or too small. After meshing you can use the solver like ALTAIR RADIOSS or LS-DYNA. You can ANSYS workbench to simulate also. These software has better FEM solving capability and provides better meshing options than CAD software. You can use the mesh refinement method. Coarse meshing always gives the false results.&#xA;Apart from these, you should check for Hourglass effect, element type, contact options, time step, mesh type, damping controls, and erosion controls. Check for proper boundary conditions, if you forget to set boundary conditions you may get wrong results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don’t find any error then try to use different CAE software for simulation. Sometimes CAD software are not able to properly simulate the process. If the error still persists you should consult with a CAD / CAE expert and show him your model. And keep in mind these simulations are approximations.&lt;/p&gt;&#xA;" OwnerUserId="15028" LastActivityDate="2018-02-21T15:09:21.337" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19540" PostTypeId="2" ParentId="19463" CreationDate="2018-02-21T15:34:45.863" Score="0" Body="&lt;p&gt;Your best bet is to calculate this experimentally. The friction of the syringe will likely completely dwarf the pressure requirements - unless you want your model to go deeper than a couple meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a dynamometer (or a spring weight) to determine force needed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;to move the piston with free air access&lt;/li&gt;&#xA;&lt;li&gt;to double the volume with air access restricted (say, going from 5 to 10ml) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This way you'll get the force of friction and force of friction + 0.5 bar pressure differential. Subtract the first readout for pressure force alone. You're getting 1 bar per 10 meters of depth, so the force from pressure will be 0.2 x (your result) per meter of depth. (and your RC transmitter is unlikely to reach deeper than 2-3 meters deep anyway)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the force, the motor torque rating, your mechanical transmission between the torque and linear force (screw actuator?) and your motor power rating at max load, you'll know the power requirements - but you really should go with a pretty high safety factor, like 2x, because varying conditions like friction depending on temperature, sticky moisture, voltage drop, friction of the transmission etc will be large enough to break this in common operation easily.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-02-21T15:34:45.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19542" PostTypeId="1" CreationDate="2018-02-21T15:55:45.257" Score="1" ViewCount="59" Body="&lt;p&gt;I am planning to design an autonomous bot that takes power from a chemical reaction.i was thinking if CO2 produced via some chemical reaction can drive small wheels so as to generate electricity which can be used to move the bot forward.Which chemical reaction for the production of CO2 would be best for that? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastActivityDate="2018-02-21T16:25:21.993" Title="Can CO2 produce energy to drive a bot" Tags="|chemical-engineering|robotics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19544" PostTypeId="2" ParentId="19542" CreationDate="2018-02-21T16:25:21.993" Score="3" Body="&lt;p&gt;Potentially yes. There are a lot of homemade toys which use a reaction between vinegar and bicarbonate of soda to make simple CO2 jet engines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you want to use this sort of reaction to generate more controllable mechanical power for actuators or motors you are giving yourself a lot of problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fighting robots often use compressed CO2 to power pneumatic actuators for weapons and flippers  but this is usually stored on-board as pressurised gas rather than generating it on demand (you can failry easily obtain scrap CO2 fire extinguishers for next to nothing) . And while it is a very simple way to get a high speed, high energy actuator it doesn't have very good power density and for anything other than a fairly simple piston actuator the disadvantages tend to outweigh the benefits. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-02-21T16:25:21.993" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19546" PostTypeId="2" ParentId="19515" CreationDate="2018-02-21T18:35:33.983" Score="1" Body="&lt;p&gt;If a MIMO plant is upper or lower triangular then one can't create an interaction loop which might make the system unstable. So using stabilizing SISO feedback control for each diagonal term should stabilize the system. Therefore from a stability standpoint decoupling would not be necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, if reference is not the equilibrium point then some of the outputs will also be affected by the other references as well. So complete decoupling might improve the tracking performance. But feed forward could be used for this as well. &lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-02-21T18:35:33.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19547" PostTypeId="1" AcceptedAnswerId="21291" CreationDate="2018-02-21T20:19:01.007" Score="6" ViewCount="825" Body="&lt;p&gt;I came up with this concept for a partial rotation, lubricant free, low friction bearing for a robotics application. Two complementary sets of tensioned tendon-like cables (or tapes) serve both to constrain the degrees of freedom of the joint as well as the rolling surface. Here's a render of a model I made to illustrate the idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AGUpe.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AGUpe.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/B9zLU.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B9zLU.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure that I'm &lt;em&gt;far&lt;/em&gt; from the first person to have had this idea, but having no clue about its proper name nor about how to describe it, I'm unable to find anything on the net. I guess it's best described as a kind of &lt;a href=&quot;https://en.wikipedia.org/wiki/Flexure&quot; rel=&quot;noreferrer&quot;&gt;flexure&lt;/a&gt;, but what exactly? Has this mechanism found any real world applications? &lt;/p&gt;&#xA;" OwnerUserId="5641" LastEditorUserId="5641" LastEditDate="2018-02-21T20:27:52.273" LastActivityDate="2018-04-11T14:17:01.573" Title="What is this kind of bearing called" Tags="|mechanical-engineering|mechanisms|bearings|flexures|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19549" PostTypeId="1" CreationDate="2018-02-21T20:35:22.390" Score="0" ViewCount="617" Body="&lt;p&gt;I am new to abaqus and for my final year project I'm meant to simulate a force on a cylinder (testing displacement of articular cartilage) ranging from 1-15&amp;nbsp;Hz and between 15-36 Newtons but I don't know how to do it. I have all the necessary material parameters I just don't know how to apply the loading. Any help, guides or tutorials would be appreciated. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="15033" LastEditorUserId="1832" LastEditDate="2018-02-21T22:14:32.767" LastActivityDate="2018-03-26T21:31:32.103" Title="Cyclic loading on ABAQUS CAE" Tags="|mechanical-engineering|civil-engineering|finite-element-method|abaqus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19550" PostTypeId="2" ParentId="6576" CreationDate="2018-02-21T20:41:56.813" Score="1" Body="&lt;p&gt;The difference between the cut in and cut out of a temperature or a pressure control is the 'differential'. Simple controls have set differentials.&#xA;I would use 'hysteresis' where there is a loss of movement in a mechanical gear and/or lever train when the train reverses direction. This is sometimes termed the 'lash'.&lt;/p&gt;&#xA;" OwnerUserId="15032" LastActivityDate="2018-02-21T20:41:56.813" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19552" PostTypeId="1" AcceptedAnswerId="19554" CreationDate="2018-02-22T04:28:57.117" Score="0" ViewCount="1176" Body="&lt;p&gt;This is a practice question that I am trying. I am a student learning about material balance-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ajacZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ajacZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find the mass composition of stream D. &#xA;The one written in pencil is the values that I have found in advanced- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made a material balance around separator 2 ignoring the recycle stream,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mZv3h.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mZv3h.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I done up a component X balance- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1250 X + (120 \times 0.3) = (379.08 \times 0.374) + (90.92 \times 0.222) $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solve for x $= 0.10076$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly,I done up a component Y balance-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$125 Y + (120 \times 0.7) = (379.08\times0.626) + (90.92\times 0.778)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solve for Y $= 0.1792$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is wrong as the composition of X and Y total does not add up to 1. Where did I go wrong or what must I do ?&lt;/p&gt;&#xA;" OwnerUserId="15037" LastEditorUserId="15037" LastEditDate="2018-02-22T04:35:27.183" LastActivityDate="2018-02-22T05:26:41.050" Title="Material balance without chemical reactions" Tags="|materials|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19554" PostTypeId="2" ParentId="19552" CreationDate="2018-02-22T05:15:38.707" Score="1" Body="&lt;p&gt;How do you get D = 1250?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should it be A + B - C ie 450 + 130 - 230 = 350 based on separator 1 only, ie what leaves separator 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then when you include separator 2 the +900 will give 1250 arriving at separator 2.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-02-22T05:26:41.050" LastActivityDate="2018-02-22T05:26:41.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19555" PostTypeId="1" CreationDate="2018-02-22T05:41:44.667" Score="-1" ViewCount="73" Body="&lt;p&gt;I was working on the pressure vessels and it was difficult for me to draw the parts of the &lt;strong&gt;bottom plate&lt;/strong&gt;. Why are the bottom plate designed to overlap the plates each other? The picture is below &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JCAN7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JCAN7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;     &lt;/p&gt;&#xA;" OwnerUserId="15005" LastEditorUserId="7083" LastEditDate="2018-03-11T01:03:27.120" LastActivityDate="2018-03-11T01:03:27.120" Title="How can I draw each parts of the pressure vessel bottom plate?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19556" PostTypeId="2" ParentId="19555" CreationDate="2018-02-22T05:57:15.623" Score="2" Body="&lt;p&gt;They overlap to avoid butt joints.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-02-22T05:57:15.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19557" PostTypeId="1" CreationDate="2018-02-22T09:55:41.547" Score="0" ViewCount="689" Body="&lt;p&gt;I have a $100\ \text{psi}, 300\ \text{cfm}$ air compressor that we use for running power tools, I need a way to inject mortar deep into a small opening and was wondering if anyone had experience with setups like this.  From doing some research on the topic it sounds like historically it's pretty much what they did before the specialized concrete pumps we have now were around.  I'm not worried about air mixing with the mortar but it would be interesting to know how much force this would have on the person at the operating end of the hose.  I can't see it being more than a fire hose as long as we stick to a relatively small hose (the air input is $1&quot;$ and I was planning on $1/2&quot;$ for the output hose)  All fittings and hoses are rated for $150\ \text{psi}$ and I was planning on buying a $200\ \text{psi}$ rated pressure tank.  Am I missing anything here?   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; Just in case someone wanted to see my math here, I used the the following numbers &#xA;density of (flowable grout) $= 128.7 \ \text{lbs/ft}^3 = 2061.6 \ \text{kg/m}^3$ &#xA;Operating pressure $= 100\ \text{psi} = 689.476\ \text{kN/m}^2$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;using $P =\rho*g*h$, we get $689476 = 2.0616*9.85*h$, where $h=\frac{689476}{2061.6*9.85} = 34\ \text{m}$ of pressure $= 111\ \text{ft}$ of available head (minus what's lost from friction with the hose/pipe/fittings) &#xA;Should still be plenty enough pressure for what I need though also pressure at the nozzle would be $\frac{100}{0.7853} = 127.3\ \text{lbs}$, probably gonna bump that hose up to $3/4&quot;$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YI7f2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YI7f2.jpg&quot; alt=&quot;Rough drawing of setup&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit&lt;/em&gt; - Drawing explanation&#xA;As addressed in the comments I was just going to use ball valves at the grout intake air inlet and grout outlet, the pressure tanks I was looking at were 200 PSI 5 gallon water tanks that have like 8 ports already built into them.  The mortar works out to about 4.7 gallons of flowable mortar/40# bag so it should work out for the smaller jobs we'd be using it for. &lt;/p&gt;&#xA;" OwnerUserId="15040" LastEditorUserId="13311" LastEditDate="2018-02-22T22:48:30.630" LastActivityDate="2024-03-24T10:01:40.180" Title="using air pressure as a liquid pump" Tags="|fluid-mechanics|pressure|pressure-vessel|foundations|masonry|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19558" PostTypeId="2" ParentId="19473" CreationDate="2018-02-22T12:17:05.897" Score="1" Body="&lt;p&gt;In order to answer the question, one more question needs to be answered. The exit pressure is present at the downstream of the shock, how does this information is transferred to upstream of the shock? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normal shocks occur for a supersonic flow. The Euler equations are hyperbolic for supersonic flow but there will be boundary layer at the wall edges. The flow inside the boundary layer is subsonic, making the equations elliptic. Elliptic equations require two boundary conditions in two dimensions one of which is that the pressure at the exit matches with the back pressure. &#xA;But as dealing with viscous flows is hard, we use inviscid flow theory and say that the exit pressure matches with the back pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it is the boundary layer, that's helping to communicate this information to the upstream.&lt;/p&gt;&#xA;" OwnerUserId="15041" LastActivityDate="2018-02-22T12:17:05.897" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19560" PostTypeId="1" AcceptedAnswerId="19561" CreationDate="2018-02-22T14:36:52.747" Score="1" ViewCount="567" Body="&lt;p&gt;Why is hardening and brittle nature so related?What is work hardening or strain hardening as far as true stress strain diagram is concerned?&lt;/p&gt;&#xA;" OwnerUserId="12558" LastActivityDate="2018-02-22T22:57:37.607" Title="Hardening and Brittleness" Tags="|mechanical-engineering|materials|mechanical-failure|failure-analysis|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19561" PostTypeId="2" ParentId="19560" CreationDate="2018-02-22T15:35:01.093" Score="3" Body="&lt;p&gt;Brittleness describes the inability of a material to absorb a considerable amount of plastic deformation. Thus, the material fails without a lot of deformation.&#xA;Hardening, on the other hand is the process of increasing a material's yield strength $\sigma_y$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle used for work/strain hardening has to do with residual strain.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/RJejY.png&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you stretch e.g. steel beyond its elastic limit, i.e. beyond $\epsilon_{el}$, or $\sigma_y$, it starts to yield, any further work done on the material will result in plastic deformation and is therefore irreversible. Only the work done up to the elastic limit is reversible.&#xA;For example, if you stretch the material in the $\sigma$-$\epsilon$-diagram (~steel) to the point $\epsilon_{pl}$ and then remove the load, a residual strain $\epsilon_r$ will remain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you reset your testing apparatus and measure the material's stress-strain diagram again, it will now follow the green line, as the material has undergone strain-hardening. It has now an elastic limit $\epsilon_{el,2}\approx\epsilon_{pl}-\epsilon_r$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elastic modulus $E$ remains more or less the same. ($E$ is rather affected by the alloy components). Ultimately, its yield strength $\sigma_y$ has increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;green&quot; material will now undergo less deformation $\epsilon$ up to its failure than the initial material (blue), thus it can be considered more brittle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if you consider &lt;em&gt;one&lt;/em&gt; material, you can make it harder, but that will also leave it more brittle.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="8632" LastEditDate="2018-02-22T22:57:37.607" LastActivityDate="2018-02-22T22:57:37.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19562" PostTypeId="1" CreationDate="2018-02-22T15:56:33.580" Score="1" ViewCount="98" Body="&lt;p&gt;From standard pipe-loss equations we know that features in piping systems, such as wall surface roughness, or valves along the pipe system, cause major and minor pressure losses respectively. However, my intuition of how such features actually cause pressure losses seems to be incorrect, as I would have thought that surface roughness and valves would actually increase pressure along the pipe system as they tend to cause more restrictions in the flow. Could someone explain on an intuitive level why such features actually cause the opposite to take place?&lt;/p&gt;&#xA;" OwnerUserId="9755" LastActivityDate="2018-02-22T16:08:46.190" Title="Intuition behind pressure losses in pipes" Tags="|pressure|pipelines|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19563" PostTypeId="2" ParentId="19562" CreationDate="2018-02-22T16:08:46.190" Score="2" Body="&lt;p&gt;They do restrict the flow, which would normally cause the pressure to increase &lt;strong&gt;upstream&lt;/strong&gt; of the &quot;roughness&quot; of whatever obstacle you are considering. In practice however, the source of the flow is operating at a certain pressure (e.g. pump), so the pressure upstream remains unchanged, but the pressure &lt;strong&gt;downstream&lt;/strong&gt; decreases. The net effect is a pressure loss along the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, any irregularity in the inside surface of the pipe will provide resistance to flow, and cause a pressure drop along the pipe. An electrical analogy would be a resistor causing a voltage drop across the resistor (electrical current being the electrical equivalent to flow).&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2018-02-22T16:08:46.190" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19564" PostTypeId="1" AcceptedAnswerId="19565" CreationDate="2018-02-22T16:10:51.800" Score="4" ViewCount="11814" Body="&lt;p&gt;I have purchased these copper heat-pipes; however, I am wondering how to bend them without kinking them or cutting them open due to pressure on weak points/bends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G4ACs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G4ACs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are flat copper heat pipes, so I can't use a pipe bender; however, &lt;strong&gt;these are sold specifically to be bent, according to descriptions&lt;/strong&gt;, so I was wondering if anyone here has any experience or tips for bending these flat copper heat pipes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be a little more specific, I want to make a 90 degree bend in the middle of the copper heat pipe, without making kinks (which degrade performance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Furthermore,&lt;/strong&gt; would wrapping copper foil around the heat pipes allow for greater thermal dissipation? Apparently some computer modders have seen success; however, I thought it would be appropriate to ask opinions here as I will try to wrap copper foil set in place by thermal paste to the heat pipes.&lt;/p&gt;&#xA;" OwnerDisplayName="user15044" LastEditorUserId="9" LastEditDate="2018-02-23T02:10:22.233" LastActivityDate="2018-02-23T02:10:22.233" Title="How to bend copper heat pipes?" Tags="|mechanical-engineering|metallurgy|industrial-engineering|home-improvement|" AnswerCount="1" CommentCount="20" ContentLicense="CC BY-SA 3.0" />
  <row Id="19565" PostTypeId="2" ParentId="19564" CreationDate="2018-02-22T18:36:23.717" Score="2" Body="&lt;p&gt;Copper is notoriouos for hardening and breaking under bending or hammering it, or working it any other way. To keep it soft and flexible, you have to anneal it often while working on it. It also bends easier and breaks less soon when you bend it while it's hot. You can heat it with the torch, but you're right about exploding it, getting it too hot may get the pressure too high, and the pipe may pop. Especially since it's filled with some kind of chemical, that may be dangerous. So at least wear goggles and gloves, but rather a full face mask and clothes that cover your arms. Since the pipe is so small, and probably filled with little fluid and mostly gas, i think it will rather 'pop' than explode, but still it can be dangerous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To bend it you can cut three disks from eg. aluminium or any metal. The middle one will have the radius of the inner radius of the pipe bending, the two outer disks will have that radius, plus the width of the pipe. They will keep it from twisting while you bend it. To minimise the risk of breaking the pipe, keep the radius of your bend as large as possible. Heat the pipe and start making a small bend of a few degrees, reheat it again and bend it a few degrees further, repeat until bent enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're right about your foil thing. Heat pipes are the most effective thing in &lt;strong&gt;transporting heat, not dissipating&lt;/strong&gt;. They just relocate the heat from the source to somewhere else. You can't really get rid of any heat with just heat pipes. That's what you need a heatsink for. A heatsink is meant to dissipate heat. So it has to have a large surface, preferably close to the source. Heat pipes are used in laptops since it's not possible to put the heatsink onto the processor. Thus, a heat pipe is used to transport the heat to the side of the laptop, where the heatsink is located. Thermal paste is used for a proper heat conduction between heatsink and heatpipe or source. Keep the paste layer as thin as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copper is the best material for conducting heat, but aluminium is often used since it's cheaper. I'd rather buy a good heatsink than try and craft it yourself. It probably won't be as effective. When searching for a heatsink, look for the right C/W rating. It basically means how hot it will get when dissipating a certain amount of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5C/W means it will get 5C hotter for every Watt of power it has to dissipate.&#xA;When you know the temperature you allow, and the power you have to dissipate, you know how much C/W you need for a heatsink. So lower is better. You can put a fan on the heatsink to make it dramatically more effective. C/W rating is mostly for still air.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-02-22T18:36:23.717" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19566" PostTypeId="2" ParentId="17459" CreationDate="2018-02-22T18:52:02.833" Score="1" Body="&lt;p&gt;Accelerometers are capable of measuring the acceleration they experience relative to free-fall. Accelerometers are used to measure the upwards acceleration that counters gravity when at rest, its a hoax that it measures the acceleration due to gravity. This acceleration is measured as 1&amp;nbsp;g (g&amp;nbsp;=&amp;nbsp;9.8&amp;nbsp;m/s&lt;sup&gt;2&lt;/sup&gt;) on the z-axis, when both pitch and roll angles are zero, but when the sensor is tilted either the x-axis or the y-axis experiences a component of the upward acceleration, whose magnitude depends on the tilt angle. Now, moving to gyroscope, analysing the gyroscope reading means to determine the roll, pitch and yaw axis of a device with respect to its initial position. Now, coming to your question, in your case as we have obtained the acceleration readings from the accelerometer, we can apply the below mentioned formula to obtain gyroscope readings from accelerometer readings.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;(Here, $\phi$ = roll and $\theta$ = pitch)&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan \phi_{xyz} = \left(\frac{G_{py}}{G_{pz}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan \phi_{xyz} = \left(\frac{-G_{px}}{G_{py}\sin \phi + G_{pz} \cos \phi}\right) = \frac{-G_{px}}{\sqrt{G^2_{py}+G^2_{px}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are curious enough to know more about these formulas and the source from which I obtained these formulas you can open the link below and have a look at the document.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://cache.freescale.com/files/sensors/doc/app_note/AN3461.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tilt Sensing Using a Three-Axis Accelerometer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, Pitch equation is defined always to be positive, so the range of the equation is [-90, 90] and roll equation provides [-180, 180] range. When the pitch angle is 90º, the roll axis is directly aligned with the axis of gravity, thus we cannot measure the roll angle in this case. Also, the roll equation is undefined when both Gx and Gz are equal to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this cleared your doubt. If any further doubts persists, feel free to write.&lt;/p&gt;&#xA;" OwnerUserId="15046" LastEditorUserId="7083" LastEditDate="2018-03-28T02:18:21.233" LastActivityDate="2018-03-28T02:18:21.233" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19568" PostTypeId="1" CreationDate="2018-02-22T21:22:03.920" Score="8" ViewCount="485" Body="&lt;p&gt;I am trying to model this system but am having difficulties in setting up the equations of motion. This is a simplified model of a motorcycle suspension system, where P is the ground in the image below. In addition, the variable $u$ is the input to the system, which is the displacement from the ground (like a motorcycle going over a bump). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sesNf.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sesNf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider mass $m_2$ first. Keeping $m_1$ fixed and displacing $m_2$ a bit yields the equation of motion for $m_2$ as &#xA;$$&#xA;m_2 \ddot{y} = -k_2 (y-x) -b_2 (\dot{y} - \dot{x}) &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And similarly, for $m_1$, keeping $m_2$ and $u$ fixed and displacing $m_1$ a little bit, the equation of motion comes out to be&#xA;$$&#xA;m_1 \ddot{x} = -k_1 (x-u) -b_1(\dot{x} - \dot{u}) - k_2(x-y) - b_2(\dot{x}-\dot{y})&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two term on the RHS come from the coupling of $x$ with $u$ while the second and third come from the coupling of $m_1$ and $m_2$, or equivalently from Newton's third law. However, the textbook solution only has the first two terms, and not the interaction between $m_1$ and $m_2$ for the EOM of mass 1. I am extremely confused as to why this is the case. Is it because they are vertical and not horizontal? How would the EOM change if the scenario was all horizontal?&lt;/p&gt;&#xA;" OwnerUserId="15049" LastActivityDate="2022-11-18T21:06:40.840" Title="System Dynamics Modeling" Tags="|dynamics|transfer-function|systems-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19569" PostTypeId="2" ParentId="16983" CreationDate="2018-02-22T23:33:09.903" Score="1" Body="&lt;p&gt;Pascal's law $\text{Pressure} = \rho gh$, $\rho=$ density of the fluid $g=$ gravitational force and $h=$ head pressure.  To find head pressure you could use a clear tube and measure how high the water goes before it equalizes and stops flowing.  Or just point it straight up, hold it against a wall and measure the highest point it makes the wall wet. releasing the fluid from the closed system of the hose isn't a very accurate way to do it but would probably get you close.  Use $\rho = 1000\text{ kg/m}^3$, $g=9.8\text{ N/kg}$ (same as m/s&lt;sup&gt;2&lt;/sup&gt;) and $h = m$ to get kg/m&lt;sup&gt;2&lt;/sup&gt; pressure.  Generally macro level physics is 99% unit manipulation so getting good at understanding how they go together and cancel out will make memorization of the formulas a whole lot easier. the problem is usually with the gravitational constant you know 9.8&amp;nbsp;m/s&lt;sup&gt;2&lt;/sup&gt; but getting that into a figure you can conceptualize is difficult.  That's why I always hated moment of inertia... &lt;/p&gt;&#xA;" OwnerUserId="15040" LastEditorUserId="1832" LastEditDate="2018-02-23T01:01:57.503" LastActivityDate="2018-02-23T01:01:57.503" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19571" PostTypeId="2" ParentId="9121" CreationDate="2018-02-23T05:24:01.933" Score="0" Body="&lt;p&gt;The table will topple when the vector sum of all the forces including the weight of table legs and wings fall outside the polygon created by the footprint of legs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to find the weight of your frame and its CG ( which appears to be at a height of 60% total table height). Then you measure the weights of each wing and find its CG at two positions, tilted down and deployed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately you will have a mass M of the table and one set of x,y,z for its coordinates of CG for each configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you apply your test load P at any point on the table surface outside the polygon of support traced by its legs. If the resultant sum of {[P, multiplied by its distance from the CG of the table in x-direction and M, multiplied by the table's X coordinated of CG] divided by M + P} is bigger than X of CG, the table will topple under that combination of force and position of force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you write a small routine on a spreadsheet and repeat this algorithm with bigger off-center for the position of the load you will obtain a contour line around the legs polygon which delineates what are acceptable safe loads and at what positions. Basically, it would look like the topology of a hillside with a flat pad on top ( the polygon traced by legs) and sloping down from there.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-02-23T05:24:01.933" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19572" PostTypeId="1" CreationDate="2018-02-23T06:11:07.663" Score="3" ViewCount="855" Body="&lt;p&gt;Referring to page 123 of the code in EC2, BS EN 1992-1-1:2004, note 1 above Table 7.3N indicates that maximum bar spacing for crack control is derived based on $k_1$ parameter being 0.8 and $k_4$ parameter being 1.0. According to crack width calculation in clause 7.3.4(3), $k_1$ is 0.8 and $k_4$ is 0.425. Under stress limitation clause 7.2, $k_1$ is 0.6 and $k_4$ is 1.0. So, what do $k_1$ and $k_4$ refer to in Table 7.3N?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know precisely what the parameters mean because I want to verify my crack width calculation based on clause 7.3.4 to see if they match with what is presented on the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kq1oT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kq1oT.png&quot; alt=&quot;crack control&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11443" LastEditorUserId="1832" LastEditDate="2018-02-23T09:53:35.463" LastActivityDate="2018-02-23T19:21:18.863" Title="What do the parameters k1 and k4 adopted in crack control tables in EC2 refer to?" Tags="|reinforced-concrete|eurocodes|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19573" PostTypeId="2" ParentId="15007" CreationDate="2018-02-23T12:04:47.467" Score="0" Body="&lt;p&gt;What you seem to be describing is commonly called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Key_(engineering)&quot; rel=&quot;nofollow noreferrer&quot;&gt;keyway&lt;/a&gt; (and the male portion is called the key). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On rotating shafts with significant torque transmission you can use a spline. A keyway isn't always like your illustration, but it is common in tooling/machining to ensure alignment. &lt;/p&gt;&#xA;" OwnerUserId="14498" LastActivityDate="2018-02-23T12:04:47.467" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19574" PostTypeId="2" ParentId="19557" CreationDate="2018-02-23T14:10:48.307" Score="0" Body="&lt;p&gt;I would look at an auger extruder system like I've sketched out. They're good for extruding abrasive pastes like ceramics, mortar or grout. When the motor turns, the mortar would flow, when it stops flow stops.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vbNfS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vbNfS.jpg&quot; alt=&quot;Crappy sketch of auger extruder&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2061" LastActivityDate="2018-02-23T14:10:48.307" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19575" PostTypeId="1" AcceptedAnswerId="19577" CreationDate="2018-02-23T14:23:09.017" Score="0" ViewCount="133" Body="&lt;p&gt;Specifically, a 100W CO2 laser. Are there any considerations other than the maximum power for a fiber?&lt;/p&gt;&#xA;" OwnerUserId="15059" LastActivityDate="2018-02-23T16:28:53.390" Title="Can I use an optical fiber to direct a laser cutter?" Tags="|optics|lasers|fiber-optics|" AnswerCount="1" CommentCount="2" ClosedDate="2018-03-01T13:13:05.047" ContentLicense="CC BY-SA 3.0" />
  <row Id="19576" PostTypeId="1" AcceptedAnswerId="19590" CreationDate="2018-02-23T14:41:23.753" Score="0" ViewCount="171" Body="&lt;p&gt;I'm currently designing scissors, and I have to describe the tolerance and fits of the parts. What kind of tolerance must be used for the pin holding the two blades together, though allowing for rotations? (i.e. clearance, transition, or interference).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your insights!&lt;/p&gt;&#xA;" OwnerUserId="13717" LastActivityDate="2018-02-24T02:19:05.403" Title="Tolerance of pin in Scissors" Tags="|manufacturing-engineering|tolerance|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19577" PostTypeId="2" ParentId="19575" CreationDate="2018-02-23T14:49:04.753" Score="3" Body="&lt;p&gt;CO2 lasers cannot be coupled into standard silica-based fibres, because these fibres absorb light longer than 2.1um. You could use hollow silica fibres, but they are quite recently developed, so might be expensive. Otherwise I‘d just go with the standard approach of free propagation over mirrors OR use a solid state laser which can be coupled into standard fibres easily&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addition: For any laser, whether CO2 or solid state, the laser needs a minimum beam quality (minimum meaning the $M^2$ should be sufficiently small) so that you can guide it with a given fibre. The numerical aperture NA and the core diameter $d_k$ of the fibre are the important values, giving you the following condition for the maximum beam parameter product your laser beam is allowed to have if it should be coupled into a given fibre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$NA \cdot \frac{d_k}{2} \geq \frac{d_f \cdot \Theta}{4}$&lt;/p&gt;&#xA;" OwnerUserId="14629" LastEditorUserId="14629" LastEditDate="2018-02-23T16:28:53.390" LastActivityDate="2018-02-23T16:28:53.390" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19578" PostTypeId="1" AcceptedAnswerId="19579" CreationDate="2018-02-23T15:51:43.537" Score="0" ViewCount="1990" Body="&lt;p&gt;i'm trying to connect a D-cut shaft to a plate for transmission of torque&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GEBES.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GEBES.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the motor that i am using with a 'diameter' of 4.5+/-.1mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is whether I can use 5mm circular universal mounting hub like this to do so? &lt;a href=&quot;https://i.stack.imgur.com/lcMzJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lcMzJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, i have access to a a laser printer. Would you recommend making a mounting hub similar to the one shown above but with a D-Cut hole? It has a precision of .2mm, so along with the uncertainty in the shaft of 0.1mm, I guess I would have to make the 'hole' 4.8mm. Would this potential difference in size mean that the torque may not be transmitted efficiently?&lt;/p&gt;&#xA;" OwnerUserId="15061" LastActivityDate="2018-02-23T16:01:06.140" Title="How do I connect a plate to a D-cut shaft?" Tags="|motors|torque|power-transmission|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19579" PostTypeId="2" ParentId="19578" CreationDate="2018-02-23T16:01:06.140" Score="1" Body="&lt;p&gt;In answer to your questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes, you can (and should) use a standard 5mm hub, such as the one shown in your question. The flat on the motor shaft is only there to provide somewhere for a grub screw to bite in (better than having a perfectly rounded shaft).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;No, I would stick with a standard hub with a circular hole, for the reasons stated above. I don't know what the tolerance is on your hub, but it should be a tight fit around the shaft (5mm +0 -0.012). Ignore the 4.5mm dimension, that's just the flat for the grub screw, it's irrelevant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="334" LastActivityDate="2018-02-23T16:01:06.140" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19581" PostTypeId="1" AcceptedAnswerId="19582" CreationDate="2018-02-23T17:44:47.567" Score="2" ViewCount="267" Body="&lt;p&gt;Apparently I have a stupid question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got curious and crunched some numbers to determine approximately how much air an engine consumes in total to burn a full tank of gas. Something went horribly wrong and I feel stupid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So theoretically, I can get this number by multiplying the total amount of fuel in the tank, let's say 15 gallons (56.78 liters), by the air part of the air-fuel ratio, which I read is 15:1. So the engine would consume about 225 gallons (851.7 liters) of air per tank of gas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that number makes no sense if you take into account the other way to calculate this: engine size. Let's say we have a four cylinder, two liter engine. Since it's four stroke, the air it will consume should be equivalent to its rpm. So at 1,000 rpm it will consume 1,000 liters of air per minute. Right? That's a very long way from a total of 850 liters for the whole tank of gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what did I do wrong?&lt;/p&gt;&#xA;" OwnerUserId="15051" LastActivityDate="2018-02-23T18:02:29.853" Title="How much air will an engine use per tank of gas?" Tags="|mechanical-engineering|automotive-engineering|fuel-economy|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19582" PostTypeId="2" ParentId="19581" CreationDate="2018-02-23T18:02:29.853" Score="4" Body="&lt;p&gt;You probably want to look at stochiometric ratios by mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this is a difficult problem to solve from first principals as for an IC engine air is essentially free and monitoring its consumption in bulk is not a major priority. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect at the mistake you have made is confusing mass ratios with volume ratios. As air is a gas and as such has very low density this make a very big difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that 15:1 air:fuel ratio by mass then we will need 15kg of air for every kg of fuel burned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air has a density of about 1.2 kg per cubic metre so that's 18 cubic metes of air per kg of fuel (at atmospheric pressure) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative approach is to consider the swept volume per cylinder and compare with rpm and fuel economy &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-02-23T18:02:29.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19583" PostTypeId="2" ParentId="19572" CreationDate="2018-02-23T19:21:18.863" Score="1" Body="&lt;p&gt;They refer to the factors defined in section 7.3.4 on calculation of crack widths (and whatever corresponding national annex that applies to your case). Section 7.3.3 is intended to be an easy to use summary of section 7.3.4 so that is where you should look for any additional information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally I never use section 7.3.3 because the situations where all those assumptions are actually correct are rare, and making corrections can be more work than just using the equations in section 7.3.4. But that's just my opinion.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2018-02-23T19:21:18.863" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19584" PostTypeId="1" CreationDate="2018-02-23T19:23:56.730" Score="0" ViewCount="162" Body="&lt;p&gt;I'm writing a script that converts from a customer's preferred unit of &lt;span class=&quot;math-container&quot;&gt;$\frac{RPM}{s}$&lt;/span&gt; and a device's input/output of &lt;span class=&quot;math-container&quot;&gt;$\frac{s}{100Hz}$&lt;/span&gt; My Logic is as follows:&lt;/p&gt;&#xA;&lt;h1&gt;Converting &lt;span class=&quot;math-container&quot;&gt;$\frac{RPM}{s}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\frac{s}{100Hz}$&lt;/span&gt;:&lt;/h1&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{RPM}{s}=\frac{Revolutions}{(s)(min)}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{Revolutions}{(s)(min)}\right)\left(\frac{1 \;min}{60 \;s}\right)=\frac{Revolutions}{s^2}=\frac{Hz}{s}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{Hz}{s}\right)^{-1}=\frac{s}{Hz}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\left(1\frac{s}{Hz}\right)\left(\frac{100\;[1Hz]}{1\;[100Hz]}\right)=100\frac{s}{100Hz}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Answer:&lt;/h1&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{RPM}{s}\right)^{-1}\left(\frac{60 \;s}{1 \;min}\right)\left(\frac{100\;[1Hz]}{1\;[100Hz]}\right)=\frac{s}{100Hz}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Therefore, I understand that my inverse function would also be:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{\;s}{100Hz}\right)^{-1}\left(\frac{100\;[1Hz]}{1\;[100Hz]}\right)\left(\frac{60 \;s}{1 \;min}\right)=\frac{RPM}{s}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;To me this &lt;em&gt;&lt;strong&gt;FEELS&lt;/strong&gt;&lt;/em&gt; wrong as, per these equations&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1\frac{RPM}{s} = 6000\frac{s}{100Hz}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;AND&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1\frac{s}{100Hz} = 6000\frac{RPM}{s}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Meaning I just need one equation to covert betweeen the two units. Am I missing something or do I really need to only do half the work?&lt;/p&gt;&#xA;" OwnerUserId="14359" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-02-27T09:29:59.267" Title="Convert angular acceleration from (S/100Hz) to (RPM/s)" Tags="|measurements|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19585" PostTypeId="2" ParentId="19584" CreationDate="2018-02-23T22:55:36.330" Score="3" Body="&lt;p&gt;$\frac{s}{100Hz}$ has basic unit dimensions of $[time]^2$, and as such is &lt;em&gt;not&lt;/em&gt; a unit of angular acceleration. You cannot convert directly between these two quantities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we can convert from $rpm/s$ into $Hz/s$. Using the relationship that $$60rpm=2\pi\ rad/s=1Hz$$ we can determine that $1rpm=\frac{1}{60}Hz$, and thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$1\frac{rpm}{s}=\frac{1}{6000} \ \frac{100Hz}{s}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, to convert to &lt;em&gt;from&lt;/em&gt; your customer's units to the input, you would use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[input\ to\ device\ \frac{s}{100Hz}\right]=\frac{1}{6000\left[customer\ number\ in\ \frac{rpm}{s}\right]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to convert &lt;em&gt;to&lt;/em&gt; your customers units from the output, you would use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left[customer \ number\ out\ \frac{rpm}{s}\right]=\frac{1}{\frac{1}{6000}\left[output\ from\ device\ \frac{s}{100Hz}\right]}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: see comments below for details - Removed rogue $\pi$'s.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-02-27T09:29:59.267" LastActivityDate="2018-02-27T09:29:59.267" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19587" PostTypeId="2" ParentId="19393" CreationDate="2018-02-24T02:05:36.257" Score="0" Body="&lt;p&gt;I ride dirt bikes as a hobby and the frame of the bike is a mix of tubes, patch stiffener plates welded over it, reinforcing sleeves and what not. It is designed to be light and tough and take a lot of punishment!&#xA;Here is the frame of Suzuki RMz450, a very agile and powerful dirt bike:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xI1ht.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xI1ht.jpg&quot; alt=&quot;RMZ450 2012 Frame.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in design of tubular trusses I have never had to be concerned with local buckling or concentrated stress mid length of member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because they are designed mostly to have a joint at every possible load or support point! Basically they undergo only tension or compression stresses.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-02-24T02:05:36.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19588" PostTypeId="2" ParentId="17296" CreationDate="2018-02-24T02:10:19.187" Score="2" Body="&lt;p&gt;The only time I have seem them used is on overload clutches where a spring allows the dowel to pop out of the inner member when a torque is exceeded.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-02-24T02:10:19.187" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19589" PostTypeId="2" ParentId="15549" CreationDate="2018-02-24T02:13:49.973" Score="0" Body="&lt;p&gt;Find a vendor with a Water jet cutter.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-02-24T02:13:49.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19590" PostTypeId="2" ParentId="19576" CreationDate="2018-02-24T02:19:05.403" Score="1" Body="&lt;p&gt;Just looked at my kitchen scissors; clearance approximately of +.001&quot;/+.003&quot;.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-02-24T02:19:05.403" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19592" PostTypeId="2" ParentId="19376" CreationDate="2018-02-24T02:43:14.050" Score="1" Body="&lt;p&gt;I did some quick math. Call the opening 3 ft sq or 9 ft^2. Dividing 25,000 ft^3 by 9 ft^2 will give you the third dimension of 2,777 ft, which would need to travel in a minute. This is about 31.5 mph wind speed. I did a quick search of measured wind speed for fans, they were approximately 4 to 8 mph. Probably not for 25,000 CFM.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-02-24T02:43:14.050" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19593" PostTypeId="2" ParentId="15549" CreationDate="2018-02-24T10:57:01.960" Score="1" Body="&lt;p&gt;Approach a local sheet metal fabricator.  They will be able to arrange a custom die or punch tool that suits their machinery.  It is quite possible that they already have a die tool that meets your star requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For such a small part and high quantities (10,000s - millions?) they would charge very good rates as the tooling setup is easy, and can be fully automated.  The waste material can be easily managed too if using appropriate stock material (probably a large roll of half-inch wide thin brass that is fed into the stamping machine like a tape).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically they will use &lt;a href=&quot;https://en.wikipedia.org/wiki/Vibratory_finishing&quot; rel=&quot;nofollow noreferrer&quot;&gt;vibratory deburring&lt;/a&gt; to remove sharp edges from the parts, this is a simple process that is entirely automated.  They would probably pour your freshly stamped parts into a large drum deburrer and set it going for an hour or two, each part would come out burnished and smooth.&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastEditorDisplayName="user6335" LastEditDate="2018-02-24T11:31:23.350" LastActivityDate="2018-02-24T11:31:23.350" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19594" PostTypeId="2" ParentId="11352" CreationDate="2018-02-24T17:20:43.557" Score="1" Body="&lt;p&gt;There is no theoretical reason for this not to work, but there are many practical considerations, such as the fact that BLDC's typically produce 3-phase power, which would require extra components to convert for use as a charging source, and there may be issues with back-driving any internal gearbox that may be present. E-bikes regularly include regenerative braking functionality, so you could look into this. Without more details on your exact setup, we can't be of any further help.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-02-24T17:20:43.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19595" PostTypeId="1" CreationDate="2018-02-24T18:49:21.813" Score="2" ViewCount="3536" Body="&lt;p&gt;In ABAQUS, there are multiple ways to specif your part is in plane stress/strain. &#xA;1. When assigning the section, one can specify that it is a generalized plane strain section.&#xA;2. When assigning the section, one can specify that the section is homogeneous with a plane stress/strain thickness of 1. &#xA;3. When meshing the part, elements within the same section can be assigned an element type of generalized plane strain, plane strain, or plane stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have sorted through ABAQUS documentation but have yet to arrive at a satisfactory answer, regarding the effective differences between above options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that with option 1, there is no thickness specified. Does this mean that abaqus assumes the simulation such that it is infinitely thick in the Z, as is consistent with plane strain? Versus in option 2, the thickness is 1? Meanwhile, how would results be affected if only the elements were plane stress/strain, with a regular solid and homogeneous section?&lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2018-03-26T23:32:51.407" Title="ABAQUS Analysis: Difference between assigning Section as Plane stress/strain vs. Element type?" Tags="|finite-element-method|abaqus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19597" PostTypeId="2" ParentId="19549" CreationDate="2018-02-24T21:11:24.290" Score="1" Body="&lt;p&gt;In general, how you use ABAQUS depends on what you are looking to observe. Are you hoping to simply observe stress or deflection as a function of time? Are you trying to simulate damage or failure? These considerations matter at a basic level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a simple level, there are several steps in creating an ABAQUS file from CAE.&#xA;1. Creating the geometry of a part.&#xA;2. Creating a material.&#xA;3. Assigning a section to your part, thereby specifying the material.&#xA;4. Meshing the part.&#xA;5. Assigning loading, boundary, constraints, and interactions depending on the scenario. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should easily be able to accomplish all of these things with any amount of googling. There are many tutorials out there. &lt;a href=&quot;https://www.youtube.com/watch?v=7keHd1KeGjQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a tutorial showing a cyclic loading on a cylinder, for example. You should be able to follow steps of these tutorials. &lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2018-02-24T21:11:24.290" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19598" PostTypeId="2" ParentId="19595" CreationDate="2018-02-24T22:48:51.677" Score="1" Body="&lt;p&gt;plane stress, plane strain, and &lt;em&gt;generalized&lt;/em&gt; plane strain are three different things.  In each case you need to specify the appropriate element type ( plane stress is default ). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generalized plane strain introduces an additional issue. You must create and assign a reference point to the section. It is the reference point assignment that causes generalized plane strain to be handled separately at the section level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit aside, for &lt;em&gt;classical&lt;/em&gt; generalized plane strain you must pin the rotational degrees of freedom of the reference point. Something that is unfortunately not clear or well documented.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;// other questions&#xA;section thickness really only affect the scaling of edge force boundary conditions ( or reaction forces output ).  If in doubt leave as 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assigning any 2d element type to a 3d geometry will simply throw an error.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastEditorUserId="7461" LastEditDate="2018-02-24T22:55:29.473" LastActivityDate="2018-02-24T22:55:29.473" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19601" PostTypeId="1" AcceptedAnswerId="19607" CreationDate="2018-02-25T01:18:27.017" Score="0" ViewCount="4796" Body="&lt;p&gt;I am trying to CAD up a drawing for a plate for this motor to mount to. I figured it would be easy since they provided an mechanical drawing, but I have never seen one like this before. Can you please help me read this? It is from China and I don't even have any idea what units it is in. I have read these before but I am completely stuck on this one. I am trying to figure out the spacing of the mounting holes on the front of the motor. Here is the drawing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xmxhz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xmxhz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and here is a picture of the actual motor if that helps at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x4Glp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x4Glp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I can tell the mounting holes are on the bottom (or left if you rotate it). Any help is GREATLY appreciated. I can't continue my project until I get this figured out.&lt;/p&gt;&#xA;" OwnerUserId="15075" LastEditorUserId="110" LastEditDate="2018-02-25T02:42:10.317" LastActivityDate="2018-02-25T14:53:03.303" Title="Please Help me Read this Engineering Drawing" Tags="|mechanical-engineering|motors|cad|technical-drawing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19602" PostTypeId="2" ParentId="19601" CreationDate="2018-02-25T01:38:50.380" Score="3" Body="&lt;p&gt;Bottom or left based on what datum?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It matters not, really. The end of the shaft visible in the photo is your mounting area. The drawing shows two sets of mounting bolt options. The inner circle formed by the holes in the first ring appear to be (surprisingly) SAE threads, 5/16&quot; x 18 threads per inch, 10 mm deep. The second ring of bolt holes are showing as 3/8&quot; x 16 threads per inch, 19 mm deep. A nice combination of SAE and metric (grin).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other dimensions appear to be in millimeters, including the diameter of the raised center section of the motor plate at that end. Your mounting plate, assuming a flat plate mount, would require a center hole of 114.3 mm diameter to clear the boss of 3.2 mm thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that the 5/16&quot; holes are in recesses of the boss, one either would not use those holes or would provide for 3.2 mm washers/spacers to fit in the recess. The bolt holes on the outer circle are quite hefty and should take a substantial load transfer, removing the requirement of using all eight holes for mounting.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-02-25T01:38:50.380" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19603" PostTypeId="1" CreationDate="2018-02-25T02:08:19.317" Score="1" ViewCount="46" Body="&lt;p&gt;I was trying to model experimentally the interdiffusion of two miscible liquids. Is there an optical method to find out the extent of diffusion; the addition of fluorescing agent into any of the two fluids would work, but will it not influence the diffusion process by changing the chemistry of the fluid itself?&lt;/p&gt;&#xA;" OwnerUserId="7163" LastEditorDisplayName="user4139" LastEditDate="2018-04-22T11:56:46.983" LastActivityDate="2018-04-22T11:56:46.983" Title="Diffusion of miscible fluids" Tags="|fluid-mechanics|diffusion|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19607" PostTypeId="2" ParentId="19601" CreationDate="2018-02-25T06:18:22.463" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;I am trying to CAD up a drawing for a plate for this motor to mount to.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I read the drawing and TinEye.com searched the photo. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the &quot;Star&quot; bracket you'll need a 114.3 mm recess, 3.2 mm deep to allow for the raised inner section, with a greater than 22.23 mm through hole to allow for the shaft. The bolt hole circles are 106.4 mm and 149.2 mm diameter. Use 5/16&quot; x 18 TPI bolts and 3/8&quot; x 16 TPI bolts; the &quot;Star&quot; bracket uses all 8 bolts, the &quot;L&quot; bracket configuration uses only the 3/8&quot; bolts. Of course you can combine designs and come up with your own.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Metric Conversion (rounded to nearest imperial measurement):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;114.3 millimetres = 4 1/2 inches&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;3.2 millimetres = 1/8 inch&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;22.23 millimetres = 7/8 inch&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;106.4 millimeters = 4 3/16 inches&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;149.2 millimeters = 5 7/8 inches&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here are example mounting plates for that motor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www.electricmotorsport.com/nema-143t-c-face-1-2-inch-aluminum-l-bracket-motor-mount.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.electricmotorsport.com/nema-143t-c-face-1-2-inch-aluminum-l-bracket-motor-mount.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KndTl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KndTl.jpg&quot; alt=&quot;Motor Mount - L Bracket&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://m.ebay.com/itm/Manta-10-hp-electric-motor-MOUNT-Etek-Briggs-Motenergy-Permanent-Magnet-STAR-NC-/391984038365?nav=SEARCH&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://m.ebay.com/itm/Manta-10-hp-electric-motor-MOUNT-Etek-Briggs-Motenergy-Permanent-Magnet-STAR-NC-/391984038365?nav=SEARCH&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0MO2k.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0MO2k.jpg&quot; alt=&quot;Motor Mount - Star&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UxtKs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UxtKs.jpg&quot; alt=&quot;Motor Mount - Star mounted on motor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are considering purchasing one of those brackets you can probably find the &quot;L&quot; bracket for a lower price since it looks easy to make and is priced fairly expensive at the place given in that URL. The &quot;Star&quot; bracket is more complicated and CNC machined, for a custom item it's price is more inline (and you may find it difficult to get someone to make you one for less).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which of the two, or another design entirely, really depends upon what you intend to attach it to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those photos ought to give you a hint about how you want to draw it up and the appropriate dimensions to use. The place you bought it from probably has brackets, a drawing of the bracket might help you more than one of the motor.&lt;/p&gt;&#xA;" OwnerUserId="14556" LastEditorUserId="14556" LastEditDate="2018-02-25T14:53:03.303" LastActivityDate="2018-02-25T14:53:03.303" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19608" PostTypeId="1" CreationDate="2018-02-25T06:29:51.203" Score="-1" ViewCount="1256" Body="&lt;p&gt;I am currently having an error message 'Unknown keyword 'section:steel1'. The keyword may be misspelled, obsolete, or invalid'&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://i.stack.imgur.com/2HDeo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2HDeo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;error message: Unknown keyword 'section:steel8e3'. The keyword may be misspelled, obsolete,&#xA;   or invalid'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have run the model it was successfully completed, but once I edited to inp.file (the coordinate part only I edited) , and I submitted the modified inp.file , I received the error message mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how to resolve this problem?&#xA;Thank you in advance&lt;/p&gt;&#xA;" OwnerUserId="14641" LastEditorUserId="14641" LastEditDate="2018-02-25T09:16:18.077" LastActivityDate="2019-12-17T19:03:10.410" Title="How to resolve this error message 'Unknown keyword 'section:steel1'. The keyword may be misspelled, obsolete, or invalid'" Tags="|abaqus|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19610" PostTypeId="2" ParentId="19608" CreationDate="2018-02-25T09:23:01.503" Score="0" Body="&lt;p&gt;So, why does this definition look like :&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CrX36.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CrX36.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this one seems to be missing bits :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uw5N0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uw5N0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are these meant to be one definition and has an extra return? see : &#xA;&lt;a href=&quot;https://i.stack.imgur.com/10tvp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/10tvp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-02-25T09:23:01.503" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19614" PostTypeId="1" AcceptedAnswerId="19617" CreationDate="2018-02-25T18:11:19.630" Score="0" ViewCount="37" Body="&lt;p&gt;Actually, I've asked this question in &quot;English Language &amp;amp; Usage&quot; but someone told me ask it here!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does it mean by &quot;loops and troughs&quot; in:&#xA;&quot;Source has acquired a pattern of loops and troughs.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some context including this phrase: &lt;a href=&quot;https://books.google.com/books?id=a6LzCAAAQBAJ&amp;amp;pg=PA482&amp;amp;lpg=PA482&amp;amp;dq=%22loops+and+troughs%22&amp;amp;source=bl&amp;amp;ots=0M9eK7xYHc&amp;amp;sig=afrE7rVYxX8Zxda7_L7NDGYWzYA&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjhk4Cqz8HZAhUKZ1AKHSFnBsEQ6AEIKzAB#v=onepage&amp;amp;q=%22loops%20and%20troughs%22&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://books.google.com/books?id=a6LzCAAAQBAJ&amp;amp;pg=PA482&amp;amp;lpg=PA482&amp;amp;dq=%22loops+and+troughs%22&amp;amp;source=bl&amp;amp;ots=0M9eK7xYHc&amp;amp;sig=afrE7rVYxX8Zxda7_L7NDGYWzYA&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjhk4Cqz8HZAhUKZ1AKHSFnBsEQ6AEIKzAB#v=onepage&amp;amp;q=%22loops%20and%20troughs%22&amp;amp;f=false&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15082" LastActivityDate="2018-02-26T01:58:20.097" Title="What does it mean by &quot;loops and troughs&quot;?" Tags="|metallurgy|chemistry|" AnswerCount="1" CommentCount="1" ClosedDate="2018-03-10T23:24:22.560" ContentLicense="CC BY-SA 3.0" />
  <row Id="19615" PostTypeId="2" ParentId="19603" CreationDate="2018-02-25T18:28:37.340" Score="1" Body="&lt;p&gt;if the refractive indices of the two fluids are different, you may be able to use a diffractometer to measure the extent of the diffusion. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-02-25T18:28:37.340" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19616" PostTypeId="1" CreationDate="2018-02-25T22:29:00.600" Score="5" ViewCount="396" Body="&lt;p&gt;I need to build some type of scale where one can place four 600mL beakers on it and measure out its weight down to a milligram level of accuracy. I'm doing this so I can dose a very small amount of liquid (2 microliters) into the beakers at a time. The problem is that load cells and essentially all resistance-based force sensors are only accurate down to ~.03 - .05% of their full-scale load (at best). I've looked into piezoelectric sensor solutions but it seems that they are ill-suited for detecting static loads.&lt;/p&gt;&#xA;&lt;p&gt;I'm completely open to having 4 separate scales, but the problem of having an accurate enough load cell will still exist.&lt;/p&gt;&#xA;&lt;p&gt;Is there any way to detect changes in weight to this level of accuracy? Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="15083" LastEditorUserId="9448" LastEditDate="2022-10-04T18:02:56.000" LastActivityDate="2022-10-04T18:02:56.000" Title="Weight Sensing Problem: Need to detect 10kg load down to milligram accuracy" Tags="|mechanical-engineering|electrical-engineering|measurements|sensors|microfluidics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="19617" PostTypeId="2" ParentId="19614" CreationDate="2018-02-26T01:58:20.097" Score="0" Body="&lt;p&gt;Have you ever looked at finger prints? Many finger prints have patterns of loops &amp;amp; troughs - though for finger prints the troughs may be called arches. It's just a description of a pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get a rubber band and change its shape any way you like and it forms all forms of loops. Break the rubber band to create a rubber string and change its shape and you will get different forms of troughs and arches.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-02-26T01:58:20.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19618" PostTypeId="2" ParentId="19616" CreationDate="2018-02-26T03:53:55.877" Score="0" Body="&lt;p&gt;Based on the information that you provided I would recommend looking at a Wheatstone Bridge Circuit type of arrangement. It is my experience that using similar type if circuitry can help increase accuracy (need conformation) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iHxOK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iHxOK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters I would suggest taking a look at single point bridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qATd4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qATd4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your other electronic circuits I would suggest looking at instrumentation or difference amplifier circuit together with ADC which will feed a micro controller similar to the circuit block below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1X8ZJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1X8ZJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sensing element a pressure sensor base weight sensing scheme or stand strain gauge system might still be you best option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aI5lO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aI5lO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at &lt;a href=&quot;https://www.omega.com/subsection/load-cells-force-sensors-torque-all.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Load Cells, Force Sensors and Torque Transducers&lt;/a&gt;. There is good chance you will find a good solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also suggest that you take a look at &lt;a href=&quot;https://www.sensorprod.com/index.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tactile Pressure Sensor Technologies from Sensor Products Inc&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some good references for you to get started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lit/ml/slyp163/slyp163.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bridge Measurement Systems&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.efunda.com/designstandards/sensors/methods/wheatstone_bridge.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Basic Wheatstone Bridge Circuit&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://physics.stackexchange.com/questions/10538/wheatstone-bridge&quot;&gt;Wheatstone Bridge&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.omega.com/literature/transactions/volume3/strain.html&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Omega Strain Guage&quot;&gt;Omega - The Strain Gage&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.omega.com/literature/transactions/volume3/strain.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Strain Gage - Measuring Circuits&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-02-26T13:35:07.230" LastActivityDate="2018-02-26T13:35:07.230" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19619" PostTypeId="1" CreationDate="2018-02-26T08:52:07.330" Score="3" ViewCount="13919" Body="&lt;p&gt;In EC2 for calculating crack width and deflection, does the effective depth &lt;em&gt;d&lt;/em&gt; refers to the depth from compression face to centroid of the tension reinforcement or is that calculated to the centroid of outermost tension reinforcement? It seems from output file that software OASYS Adsec 8.4 uses the latter definition. So I need to confirm which is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/olO08.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/olO08.jpg&quot; alt=&quot;effective depth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15086" LastActivityDate="2018-02-26T11:06:56.023" Title="What is the effective depth of a beam with two layers of tension reinforcement?" Tags="|structures|building-design|reinforced-concrete|eurocodes|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19620" PostTypeId="2" ParentId="19616" CreationDate="2018-02-26T09:30:50.140" Score="2" Body="&lt;p&gt;Actually, 600mL of liquid (water?) is about 600g, not 10kg, so it makes selecting an off-the-shelf balance a lot easier. Likewise, 2ul is about 2mg so a 1mg accuracy should be good enough. Based on this something like &lt;a href=&quot;https://www.sartorius.com/sartorius/en/EUR/Applications/Laboratory/Weighing/Laboratory-Balances/Cubis-MSA/p/M_Cubis_MSA&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Sartorius balance&lt;/a&gt; should do it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i3jz0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i3jz0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt;: although I have nothing to do with balances, I do work for a company which is part of the Sartorius group.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2018-02-26T09:30:50.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19622" PostTypeId="1" CreationDate="2018-02-26T10:29:27.093" Score="2" ViewCount="333" Body="&lt;p&gt;In the old British standard we use a time dependent modulus of elasticity of concrete to calculate crack width, which is usually about half of the 28-day &lt;em&gt;Ec&lt;/em&gt; for long term modulus. In Eurocode EN 1992-1-1:2004, from clause 3.1.3, it seems that &lt;em&gt;Ecm&lt;/em&gt; is defined as the 28-day modulus of elasticity of concrete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this definition correct, or should the formula be interpreted as &lt;em&gt;Ecm(t)&lt;/em&gt; instead?&lt;/p&gt;&#xA;" OwnerUserId="15086" LastActivityDate="2018-02-27T16:48:51.370" Title="What is the definition of Ecm in equation 7.10 of EC2?" Tags="|building-design|reinforced-concrete|elastic-modulus|eurocodes|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19623" PostTypeId="2" ParentId="19619" CreationDate="2018-02-26T11:06:56.023" Score="1" Body="&lt;p&gt;Looking at section &quot;7.3.3 - Control of cracking without direct calculation&quot;, we have equations (7.6) and (7.7) which define the maximum bar diameter for crack control:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\phi_s &amp;amp;= \phi_s^*\cdot\dfrac{f_{ct,eff}}{2.9}\cdot\dfrac{k_c h_{cr}}{2(h-d)} &amp;amp;&amp;amp;\text{Bending (part of section in compression)}\tag{7.6} \\&#xA;\phi_s &amp;amp;= \phi_s^*\cdot\dfrac{f_{ct,eff}}{2.9}\cdot\dfrac{k_c h_{cr}}{(h-d)}&amp;amp;&amp;amp;\text{Tension (entire section under tension)}\tag{7.7}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables are all defined immediately after the equations are declared. For us, the only variable that matters is&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$d$ is the effective depth to the centroid of the outer layer of reinforcement&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if this is what OASYS is calculating, then it is certainly correct. Nowhere else in EC2 is $d$ given that definition (it's usually the depth to the centroid of the entire reinforcement). That is not to say, however, that the use of this definition to calculate &quot;crack width and deflection&quot; (as per your question) is incorrect, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section &quot;7.3.4 - Calculation of crack widths&quot; has equations (7.8) to (7.10):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;w_k &amp;amp;= s_{r,max}\left(\epsilon_{sm}-\epsilon_{cm}\right) \tag{7.8}\\&#xA;\epsilon_{sm}-\epsilon_{cm} &amp;amp;= \dfrac{\sigma_s-k_t\dfrac{f_{ct,eff}}{\rho_{p,eff}}\left(1+\alpha_e\rho_{p,eff}\right)}{E_s} \geq 0.6\dfrac{\sigma_s}{E_s} \tag{7.9} \\&#xA;\rho_{p,eff} &amp;amp;= \dfrac{A_s + \xi_1^2 A_p}{A_{c,eff}} \tag{7.10}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$A_{c,eff}$ is the effective tension area. $A_{c,eff}$ is the area of concrete surrounding the tension reinforcement of depth, $h_{c,ef}$, where $h_{c,ef}$ is the lesser of $2.5(h-d)$, $(h-x)/3$ or $h/2$ (see Figure 7.1)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That possible value $2.5(h-d)$ does not define which $d$ should be used, however. Looking at Figure 7.1(a) (shown below), it is not entirely clear which $d$ is being described: if it is the standard $d$, then the centroid would be closer to the lower layer; if it is the &quot;outer layer&quot; $d$, then it would be equally spaced between both layers. If you squint, it looks like the shown centroid is indeed a bit closer to the lower layer, but you never know if that's just an imprecision in how the figure was drawn. The fact that it is labelled as &quot;level of steel centroid&quot; isn't very helpful either: is it the centroid of all the steel or just the steel relevant to cracking, which is what's being discussed in this section? Gun to my head, I'd choose the former (all steel), but I can't guarantee that's correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Oc8p9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oc8p9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What bothers me most is that the theory behind sections 7.3.3 and 7.3.4 is almost certainly the same, so I find it odd that $(h-d)$ would have different definitions in these related topics. It's also been a long time since I've looked at cracking theory, so I can't even look at this from that perspective.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-02-26T11:06:56.023" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19624" PostTypeId="1" CreationDate="2018-02-26T16:10:08.280" Score="1" ViewCount="78" Body="&lt;p&gt;I'm looking for 1D software packages that can simulate flows of oil (lubricant) in pipes of a small diameter (typically between 1mm to 12mm in diameter) and have found around 10+ possible candidates, but pretty much all of them seem to be designed for water flows, and much larger pipes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can anybody recommend a software package&lt;/strong&gt; in which I can build a model in a GUI and input the pump and pipe characteristics in order to find the outflow at multiple outlets? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Full CFD seems like overkill, &lt;strong&gt;I'm keen to hear from any body that has experience with simulating lubricant flows, particularly in compact networks.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I couldn't find similar questions (re: software recommendations) elsewhere on the engineering SE, please let me know in the comments if this isn't good practice, and I will remove the question.&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="14498" LastEditorUserId="14498" LastEditDate="2018-02-26T17:08:14.643" LastActivityDate="2018-12-20T18:41:23.577" Title="Software options for simulating oil flow in small diameter pipes" Tags="|fluid-mechanics|simulation|pipelines|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ClosedDate="2018-12-20T15:32:51.190" ContentLicense="CC BY-SA 3.0" />
  <row Id="19625" PostTypeId="2" ParentId="19624" CreationDate="2018-02-26T16:18:36.973" Score="1" Body="&lt;p&gt;As far as I understand any CFD package worth its name can model fluid flow - you just have to specify the fluid characteristics for that fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for size, that should not be an issue - some packages advertise that they have modelled the flow of blood through the heart and wind through a city around the high-rise buildings...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for a package that is purely GUI - that I am not sure. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-02-26T16:18:36.973" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19626" PostTypeId="1" AcceptedAnswerId="19627" CreationDate="2018-02-26T18:03:48.060" Score="2" ViewCount="195" Body="&lt;p&gt;I'm building a LED ceiling lamp. The idea is to mix different tones of white to get desired tone.&lt;br&gt;&#xA;It will initially have 120W LEDs and maybe more later&lt;br&gt;&#xA;It will probably never run at full power but I want to be able to.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It consists of two parts. Upper part where psu and electronics are, and lower part where LEDs are.&lt;br&gt;&#xA;The parts are separated by sheet metal.&lt;br&gt;&#xA;The dimensions are 1020x620x150&lt;br&gt;&#xA;The upper parts height is 50mm&lt;br&gt;&#xA;The upper part is sealed from top (by ceiling) and from bottom (by metal sheet)&lt;br&gt;&#xA;The upper part will have holes in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to make the lamp as shallow as possible and for that reason I've come up with this idea for airflow generation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use large fans to lower noise generation.&lt;/li&gt;&#xA;&lt;li&gt;Turn them on their side.&lt;/li&gt;&#xA;&lt;li&gt;To force air to move in one direction block top on one side of fans, and bottom on other side of fans (images &lt;a href=&quot;https://i.stack.imgur.com/KWQ0g.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;, &lt;a href=&quot;https://i.stack.imgur.com/qVmuo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;4&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Would this kind of configuration work or will I have to use small fans at high rpm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Sbsy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Sbsy.jpg&quot; alt=&quot;3D overview of ceiling lamp&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KWQ0g.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KWQ0g.jpg&quot; alt=&quot;3D close up of fans&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XiDMB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XiDMB.jpg&quot; alt=&quot;2D schema of ceiling lamp&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/qVmuo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qVmuo.jpg&quot; alt=&quot;2D schema of fans&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15089" LastEditorUserId="15089" LastEditDate="2018-02-26T18:35:06.867" LastActivityDate="2018-02-26T18:35:06.867" Title="Cooling with fans in narrow space" Tags="|airflow|cooling|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19627" PostTypeId="2" ParentId="19626" CreationDate="2018-02-26T18:23:45.880" Score="3" Body="&lt;p&gt;No, this will not work well (as I read your plans). You should configure your fans (of any size) to draw air from inside the chassis and expel it outside the chassis. This will require vents on the perimeter of the chassis to allow air in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No 'blocking' is necessary except that the outflow face of the fan should be sealed around the outflow vent perforations to ensure all air entering the fan is drawn from the chassis.&lt;/p&gt;&#xA;" OwnerUserId="15090" LastEditorUserId="15090" LastEditDate="2018-02-26T18:29:25.700" LastActivityDate="2018-02-26T18:29:25.700" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19628" PostTypeId="1" CreationDate="2018-02-26T21:50:22.993" Score="3" ViewCount="1448" Body="&lt;p&gt;The explanations already submitted (ex: &quot;How to calculate deflection of a simple beam with a load in the center&quot;) didn't help me much. I'm building a single beam gantry crane out of Doug Fir and need to know how to calculate the needed dimensions for the beam itself; not just the answer. I see all kinds of wood beam calculators on line but they all ask about the beam spacing of the floor joists, or the rafters or the roof trusses which is irrelevant in my case. I must not be using any of these calculators correctly because I can't get these to show a single-beam calculation. Anyone know how to calculate this long-hand or is there a table available somewhere? I know it can be done because I've lifted many a diesel truck engine block with a quality straight-grained 4 x 10 x 8 foot long Doug Fir beam using block and tackle with nary a problem. The application for what I'd like to calculate and eventually build &quot;Norm Abrams&quot; style is 3000 lbs with perhaps a 10% safety factor. Again I just want to know HOW to calculate the mid-span single-beam lifting capacity based on Wood species, cross dimension, and beam-span. I gotta believe this is much easier than I'm trying to make it. Your thoughts?&lt;/p&gt;&#xA;" OwnerUserId="15092" LastEditorUserId="9" LastEditDate="2018-02-26T22:10:45.887" LastActivityDate="2018-02-27T19:06:14.050" Title="How to calculate the mid-span lifting capability of a single-beam wooden gantry crane?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19629" PostTypeId="1" CreationDate="2018-02-27T00:36:56.847" Score="2" ViewCount="184" Body="&lt;p&gt;For a variable speed centrifugal blower, can a rough measure of mass flow be obtained from only motor speed and power when the outlet restriction is variable?&lt;/p&gt;&#xA;" OwnerUserId="15096" LastEditorUserId="1832" LastEditDate="2018-02-27T01:06:31.583" LastActivityDate="2018-04-29T08:55:45.660" Title="Determination of blower mass flow from limited information" Tags="|fluid-mechanics|compressible-flow|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19630" PostTypeId="1" CreationDate="2018-02-27T00:50:13.807" Score="3" ViewCount="152" Body="&lt;p&gt;I want to build a pod. I am thinking to put a linear actuator on its lid. but my problem is, since linear actuators have this self locking mechanism, if it will be brown out, I cant open or close the lead manually. How can I deal with this scenario? &lt;/p&gt;&#xA;" OwnerUserId="15097" LastEditorUserId="1832" LastEditDate="2018-02-27T01:08:42.573" LastActivityDate="2018-02-27T08:22:15.347" Title="Electric linear actuator" Tags="|mechanical-engineering|actuator|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19631" PostTypeId="2" ParentId="19630" CreationDate="2018-02-27T01:29:50.513" Score="2" Body="&lt;p&gt;Linear actuators commonly have mounting brackets that permit rotation about an axis with the appropriate bracket. In your application, the rotation would be required, as the lid angle changes as it opens. Both ends of the actuator have to rotate about the fastener, usually a pin or bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C1FCz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C1FCz.jpg&quot; alt=&quot;Linear actuator heavy duty mount&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replace either end with a quick release pin, which allows you to disengage the actuator for manual operation. You may want to consider pin designs with securing capability as shown in the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9x3ZV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9x3ZV.jpg&quot; alt=&quot;quick release pin&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be aware that this results in the entire weight of the hatch/lid bearing on the human in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the additional comment, one could add this support mechanism to the lid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OUpLU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OUpLU.jpg&quot; alt=&quot;Cabinet Door Stay with braking&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's called a cabinet door stay with braking. There are various models and designs and it's a simple enough design to be hand-made specific to your purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had envisioned a design in which the support was a simple rod with notches. One end is hinged to allow for the lid travel, while the other end resides in a hinged block with a hole to fit the rod. A pin inserted perpendicular to the rod would engage the notches in the rod, but the cabinet door stay is much simpler and safer!&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2018-02-27T08:22:15.347" LastActivityDate="2018-02-27T08:22:15.347" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19632" PostTypeId="2" ParentId="19628" CreationDate="2018-02-27T02:03:11.173" Score="1" Body="&lt;p&gt;The most basic way is to simplify the effects of dynamic loading caused by the start and stop of the crane or the winch , they could introduce a factor of 2.8 for load magnification, is to basically multiply you load by three and calculate as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M = \dfrac{P\ell}{4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $P$ is your factored load, $\ell$ is span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $f(b) = \dfrac{mc}{I}$, where the second moment of inertia is $I= \dfrac{bh^3}{12}$ for rectangular cross sections and $c=\dfrac{h}{2}$. Depending on the species of timber $f(b)$ ranges between 900 and 1250&amp;nbsp;psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally you need to check for allowable deflection as has been mentioned in other answers, and shear assuming the load is just very near to one support (worst case scenario). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find out many sizes of timber satisfying your calculations, the optimal choice is the fat beam rather slender one $h/b &amp;lt; 2$. And all rails and pulleys need to have damped stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lifting and lowering of loads has dynamic unplanned impacts like off-center torque due to the initial dislodging the load off its cart, so the beam has to resist torsional stress. Imagine the load going back and forth in pendulum-like motion in all directions. Best cross sections for torsion are tubular or at least not slender members, more near square cross sections.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-02-27T19:06:14.050" LastActivityDate="2018-02-27T19:06:14.050" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19633" PostTypeId="1" CreationDate="2018-02-27T02:39:35.037" Score="5" ViewCount="1483" Body="&lt;p&gt;I'm building a rotating lamp for a small DIY project to learn about motors and timing belts/pulleys. As I am learning, finding parts that are small enough, fit a stepper motor shaft and reasonable priced is difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find a way to turn a lazy Susan that has a slip ring through the center within a space of about 6-8 inches. The slip ring is so that I can get electricity on the top with it still being able to freely turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lSeWp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lSeWp.jpg&quot; alt=&quot;Timing belt Mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Issues:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the usual increments of timing belts?  On Robotshop I found 8 in and 25in and that is it.  To me this seems like a small range.  Is this normal and have people found better supplies to offer a wide range of timing belts....in something like 1 in increments.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How much slop is in the timing belt?  When attaching two timing pulleys do you usually have to add idlers or stretch out one of the wheels to get a good fit?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How close can timing pulleys be before there is not enough threads to move.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15098" LastEditorUserId="1832" LastEditDate="2018-02-27T02:53:56.973" LastActivityDate="2018-11-29T20:24:40.330" Title="How to mount timing pulleys/belts for small DIY project" Tags="|gears|stepper-motor|pulleys|timing|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19634" PostTypeId="1" CreationDate="2018-02-27T05:50:34.730" Score="2" ViewCount="151" Body="&lt;p&gt;In EC2 clause 7.3.4(3), cover &lt;em&gt;c&lt;/em&gt; is defined as the cover to longitudinal reinforcement. Say for a beam with &lt;em&gt;cnom&lt;/em&gt; of 40mm and outer links of 16mm, the cover to longitudinal reinforcement will be 56mm. However, from clause 8.2.2 of PD 6687-2:2008, it states that the cover &lt;em&gt;c&lt;/em&gt; should be taken as &lt;em&gt;cnom&lt;/em&gt;. My understanding for &lt;em&gt;cnom&lt;/em&gt; is that it is the nominal cover to the outermost bar, which is the outer link if it is present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since crack width depends very much on the concrete cover, why does PD 6687-2 propose a different definition for the cover? Should the calculation for crack width include outer link diameter? OASYS Adsec seems to include link diameter in its calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7qWaL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7qWaL.png&quot; alt=&quot;PD 6687-2:2008&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15086" LastActivityDate="2018-02-27T05:50:34.730" Title="Clarification of cover definition in Eqn 7.11 of EC2 for calculating crack width" Tags="|building-design|reinforced-concrete|eurocodes|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19635" PostTypeId="1" CreationDate="2018-02-27T06:46:39.810" Score="5" ViewCount="566" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In a fully developed turbulent flow of an in-compressible fluid inside a pipe of radius $R$, velocity at the center is $U_m$. If we define $U^*=\sqrt{\tau_0/\rho}$, where $\tau_0$ is the wall shear stress and $\rho$ is the density, then find the velocity distribution as a function of $y=R-r$ distance from the wall.&#xA;  Consider the $l=k \frac{du/dy}{d^2u/dy^2}$ as Von Karman mixing length.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, if we write $\tau \approx \tau_0=-\overline{\rho u' v'}= \rho l^2 (du/dy)^2$, then we get $$(U^*)^2=k^2 \left(\frac{du/dy}{d^2u/dy^2}\right)^2 (du/dy)^2$$ and $$U^*=k\frac{(du/dy)^2}{d^2u/dy^2}$$ Now let $p=u'$ to get $p'/p^2=k/U^*$. Integrating twice gives $$-1/p=\frac{k}{U^*} y+C_1$$and $$u=-\frac{U^*}{k} \ln \left( \frac{k}{U^*} y +C_1\right)+C_2. $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one of the conditions for finding $C_1$ and $C_2$ is $u(y=R)=U_m$. What will be the other condition? This is the problem I encountered solving a similar problem:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a pipe with diameter $0.8 \ m$ water is flowing (turbulent) and velocity at $y=0.2 \ m$ is $2 \ m/s$. If the relation $u/U^*= C_1 \ln(y/R)+ C_2$ is true, then find $C_1$, $C_2$, and wall shear $\tau_0$ (notation is as same as above).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Should we relate this to viscous sub-layer somehow?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I have posted this in the physics site also. If I get my answer in one SE site I'll delete the question in the other one.&lt;/p&gt;&#xA;" OwnerUserId="6566" LastEditorUserId="6566" LastEditDate="2018-02-27T17:38:23.303" LastActivityDate="2018-06-22T02:28:28.093" Title="Von Karman mixing length $l=k \frac{du/dy}{d^2u/dy^2}$" Tags="|fluid-mechanics|chemical-engineering|turbulence|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19636" PostTypeId="1" CreationDate="2018-02-27T06:53:15.837" Score="2" ViewCount="161" Body="&lt;p&gt;I researched the properties of acrylic, or PMMA, and found different results on its mechanical properties. Tensile strength is a material property, but is it affected by shape or size? I’m working on a project that involves the use of PMMA optical fibers. I want to know how much tension different diameters of fiber can handle. Would diameter affect the tensile strength? If so, I would like to know how to calculate the tensile strength of fibers of several diameters ranging from 0.75mm to 2mm. I don’t think the fibers will be subject to much force to break them, but it would be helpful to know what they can handle. Thanks in advance for any helpful answers.&lt;/p&gt;&#xA;" OwnerUserId="15101" LastActivityDate="2018-02-28T20:40:29.137" Title="How to Determine Tensile Strength of PMMA Optical Fiber?" Tags="|mechanical-engineering|materials|fiber-optics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19637" PostTypeId="1" AcceptedAnswerId="19641" CreationDate="2018-02-27T07:33:31.790" Score="1" ViewCount="84" Body="&lt;p&gt;How might the &quot;shell&quot; in this picture be manufactured? What process could be used to achieve this geometry? It is made from plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yCpZdm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yCpZdm.png&quot; alt=&quot;render of component in question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15097" LastEditorUserId="13311" LastEditDate="2018-02-27T10:21:13.177" LastActivityDate="2018-02-27T17:36:09.130" Title="What manufacturing process could be used for this part?" Tags="|plastic|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19639" PostTypeId="2" ParentId="19637" CreationDate="2018-02-27T13:58:52.183" Score="1" Body="&lt;p&gt;I suspect that given the size, it's probably a polyurethane &lt;a href=&quot;https://en.wikipedia.org/wiki/Reaction_injection_molding&quot; rel=&quot;nofollow noreferrer&quot;&gt;RIM moulding&lt;/a&gt;. I have used &lt;a href=&quot;https://www.midaspattern.co.uk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Midas Pattern&lt;/a&gt; for such shells in the past. Here is an example from their website:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WdFo6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WdFo6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2018-02-27T13:58:52.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19640" PostTypeId="2" ParentId="19622" CreationDate="2018-02-27T15:01:49.890" Score="1" Body="&lt;p&gt;There are two major effects that make the elasticity modulus of concrete vary with time: hardening and creep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a Eurocode context $E_{cm}(t)$ refers to the hardening of concrete until it reaches its 28-day strength. This is section 3.1.3(3). If you need to calculate a crack width at an early stage, you should probably use $E_{cm}(t)$ instead of $E_{cm}$, but this has nothing to do with the long term modulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Creep or long term effects is not included in $E_{cm}$ (refer for example section 3.1.4(3)). For equation this is accounted for by the factor $k_t$ and by the fact that you should account for creep in the calculation of the stresses in the cross section. This might be the confusing part: Even though equation (7.9) uses an elasticity modulus equal to $E_{cm}$, the steel stress in the same equation should be calculated using an elasticity modulus which is not equal to $E_{cm}$ but corrected for creep in accordance with section 3.1.4(3). Unless there are no long term loads, of course, in which case it gets a little simpler.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastEditorUserId="1832" LastEditDate="2018-02-27T16:48:51.370" LastActivityDate="2018-02-27T16:48:51.370" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19641" PostTypeId="2" ParentId="19637" CreationDate="2018-02-27T17:36:09.130" Score="2" Body="&lt;p&gt;It might also have been made by thermoforming, in which a flat sheet of plastic is heated until it softens and then pressed against a replica of the finished part and allowed to cool and harden. If the process of conforming to the replica is aided by applying a vacuum in the space between the sheet and the replica, the process is called vacuum forming. this is popular for making large objects with thin walls especially if the finished part must be inexpensive. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-02-27T17:36:09.130" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19642" PostTypeId="1" AcceptedAnswerId="19730" CreationDate="2018-02-27T17:44:34.170" Score="2" ViewCount="1093" Body="&lt;p&gt;I'm using an engineering software based on &lt;strong&gt;Autodesk AutoCAD 2018 OEM&lt;/strong&gt; and I need to assign &lt;strong&gt;keyboard shortcuts to NumPad keys&lt;/strong&gt; directly, without any modifier (SHIFT, CTRL, ALT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On older version of Autocad this was possible, but now even though it's possible to assign NumPad keys directly, they do not work. According to &lt;a href=&quot;https://knowledge.autodesk.com/support/autocad-lt/learn-explore/caas/CloudHelp/cloudhelp/2018/ENU/AutoCAD-LT/files/GUID-9970E91B-A4F5-405C-99C1-46AD804A8858-htm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; on the AutoCAD LT knowledge base: &lt;em&gt;&quot;Valid modifier and key combinations include Number Pad (NUMPADn) keys containing no modifiers&quot;&lt;/em&gt; and I hope the same is true for the &quot;standard&quot; AutoCAD (i.e. no &quot;LT&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact I'm able to assign a shortcut to - for example - &quot;Scale&quot; function to NUMPAD(8), as in the screenshot below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AMTXt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AMTXt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I'm not able to use it because if I press the NumPad 8 with &lt;strong&gt;Num Lock ON&lt;/strong&gt; the AutoCAD console reads an 8:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qt4cA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qt4cA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and if I keep the &lt;strong&gt;Num Lock OFF&lt;/strong&gt; then the console picks the previous command (as it's the same that pushing the &lt;kbd&gt;&amp;uarr;&lt;/kbd&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if this feature is documented and the &lt;strong&gt;cui&lt;/strong&gt; AutoCAD command allows me to choose that keys, how are they suppose to work? How can I assign a key to speed up my work without using a modifier?&lt;/p&gt;&#xA;" OwnerUserId="15102" LastActivityDate="2018-03-05T17:45:03.727" Title="NumPad shortcut keys in AutoCAD 2018" Tags="|autocad|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19644" PostTypeId="1" CreationDate="2018-02-27T21:48:57.093" Score="3" ViewCount="596" Body="&lt;p&gt;I need to produce a very slow rotational speed for a disk -- 0.25rpm, or 1 rotation in every 4 minutes.  The disk is nominally 12&quot; in diameter.  It will be driven by a DC gearmotor, through a worm and gear with a 1:20 drive ratio.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VOdyt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VOdyt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slowest gearmotor that I can find (within my budget and space parameters) is the Servo City 12rpm motor here: &lt;a href=&quot;https://www.servocity.com/motors-actuators/gear-motors/heavy-duty-gear-motors/premium-planetary-gear-motors&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.servocity.com/motors-actuators/gear-motors/heavy-duty-gear-motors/premium-planetary-gear-motors&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But even at that slowest speed, and with the 1:20 worm gear, that still only gets me down to 0.6rpm, more than twice as &quot;fast&quot; as I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The options that I can think of would be an additional pair of gears being driven by the worm gear, or a planetary gear reduction unit between the gearmotor and the worm gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few more parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All of the components I'm already using for the mechanism are based upon .25&quot; dia. driveshafts (most are from Servo City)&lt;/li&gt;&#xA;&lt;li&gt;I need to go with off-the-shelf components; nothing fabricated (modified is OK)&lt;/li&gt;&#xA;&lt;li&gt;It must be long-wearing and precise (no plastic or cheap alloys)&lt;/li&gt;&#xA;&lt;li&gt;There's limited space between the worm gear and the disk&lt;/li&gt;&#xA;&lt;li&gt;My budget for whatever solution to reduce the speed is =/&amp;lt; $50 &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I had been considering some of the gearboxes from AndyMark ( &lt;a href=&quot;https://www.andymark.com/Gearboxes-s/55.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.andymark.com/Gearboxes-s/55.htm&lt;/a&gt; ) but frankly, I'm put off by the fact that they don't even list the shaft diameters for their gearboxes; most of them seem to be proprietary sizes intended for use only within specific systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="15103" LastActivityDate="2018-03-07T18:33:35.860" Title="reducing rotational speed from gearmotor to below 1rpm" Tags="|gears|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19645" PostTypeId="2" ParentId="19644" CreationDate="2018-02-27T22:03:26.287" Score="3" Body="&lt;p&gt;For precise control of speed, if that is required, consider using a motor controller that uses PWM (pulse width modulation) to control the energy available to the motor. With a matched motor &amp;amp; controller very slow rotational speeds are possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a motor controller in this way, the physical gearing ratio and arrangement is less important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also consider a simple belt &amp;amp; pulley system for your project given the size and mass involved.  Very high ratios can be achieved with simple and inexpensive parts.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2018-02-27T22:03:26.287" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19647" PostTypeId="1" CreationDate="2018-02-27T23:43:06.810" Score="3" ViewCount="2323" Body="&lt;p&gt;I am working on a science fair project. My question is if electricity moves better through a thin wire or a thicker wire. I need to collect data and make charts, and the thing I want to measure is the brightness. The only problem is that I'm not sure how to scientifically measure the brightness of a light bulb. Probably not important to the question but it'll be connected to a single D battery.&lt;/p&gt;&#xA;" OwnerUserId="15105" LastEditorUserId="1832" LastEditDate="2018-02-28T01:31:58.023" LastActivityDate="2018-03-01T02:21:44.110" Title="How do I measure the brightness of a light bulb?" Tags="|lighting|" AnswerCount="4" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19649" PostTypeId="2" ParentId="19624" CreationDate="2018-02-28T00:43:03.943" Score="1" Body="&lt;p&gt;Pipe-flo is a good commercial tool for what you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://eng-software.com/products/pipe-flo/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://eng-software.com/products/pipe-flo/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14901" LastActivityDate="2018-02-28T00:43:03.943" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19650" PostTypeId="1" CreationDate="2018-02-28T01:06:14.587" Score="4" ViewCount="388" Body="&lt;p&gt;I would like to calculate the force exerted on the wall of an underground reservoir. The volume of rocks surrounding the reservoir creates a force equal to the product of the density of the lithosphere, the gravitational acceleration and the height above the reservoir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to understand is what are the forces created by the liquid inside of the reservoir. How can I calculate these forces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8hZTh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8hZTh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3351" LastEditorUserId="1832" LastEditDate="2018-02-28T01:28:09.403" LastActivityDate="2023-12-31T03:08:31.057" Title="How to calculate the force that a liquid creates on the wall of an underground reservoir" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19651" PostTypeId="2" ParentId="19647" CreationDate="2018-02-28T01:31:56.850" Score="2" Body="&lt;p&gt;You will need to use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Light_meter&quot; rel=&quot;nofollow noreferrer&quot;&gt;light meter&lt;/a&gt; to measure &lt;a href=&quot;https://en.wikipedia.org/wiki/Luminous_intensity&quot; rel=&quot;nofollow noreferrer&quot;&gt;luminous intensity&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because light is a form of radiation emitted by a source, it follows the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/vision/isql.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;inverse square law&lt;/a&gt;, where luminous intensity decreases with distance from the source of light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To ensure consistent and valid results for your experiment you will need to ensure the light meter sensor is placed at exactly the same distance from the light source for every reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, to ensure your results are not affected by other sources of light it would be best to place the light bulb &amp;amp; light meter inside a light proof box, which has black internal walls.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2018-02-28T01:44:33.043" LastActivityDate="2018-02-28T01:44:33.043" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19654" PostTypeId="2" ParentId="19629" CreationDate="2018-02-28T08:10:13.830" Score="1" Body="&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; not really. You determine power by your choice of motor size, or if you use an FI, rpm. Motor power is propoertional to the third power of rpm (All else equal). So you really don't have value pairs of power and speed to work with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider thie field of PV curves:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qNsG7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qNsG7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solid curves correspond to a specific speed each, the dashed to an (approximate!) power. See how almost paralell they are and how wide the overlap is. Note that I could be wrong, it's just that with half an hour of staring at affinity laws &amp;amp; PV curves I didn't see a way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't get a flowmeter for your problem, you should be be able to get a simple barometer that you install downflow of the the blower. The intersection of your blower curve for the specific speed you measure with the pressure should give you an idea of Volume flow. Depending on your needs and the pressure levels involved, the barometer can be as simple as a translucent U-hose with water in it.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-02-28T08:10:13.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19655" PostTypeId="2" ParentId="19624" CreationDate="2018-02-28T13:20:03.997" Score="1" Body="&lt;p&gt;I've used Daxesoft's &lt;a href=&quot;http://www.pipeflow.co.uk/public/control.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pipeflow Software&lt;/a&gt; before - it's capable, but a little pricey at ~£1000 ($1380 US) per licence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It comes with a fairly comprehensive database of pipes, pipe fittings and fluids, and allows you to customize and add your own. It should be more than adequate for your needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're on a budget and your piping system isn't massively complex, you could write something in Python or Excel. Check out Crane Fluids Techincal Paper 410 for a concise, practical approach to solving pressure drop problems. It was a seminal paper in it's time.&lt;/p&gt;&#xA;" OwnerUserId="2551" LastActivityDate="2018-02-28T13:20:03.997" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19656" PostTypeId="1" CreationDate="2018-02-28T15:22:30.747" Score="-1" ViewCount="42" Body="&lt;p&gt;Today I was using oven in my office with a plate of meal&#xA;the plate was made of stain less steel, my colleague rushed and asked me&#xA;to take it out otherwise the oven will be short circuit&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So please inform me if S.L.Steel or iron pots are OK for the oven or&#xA;are they really harmful ?&lt;/p&gt;&#xA;" OwnerUserId="14988" LastEditorUserId="10902" LastEditDate="2018-03-01T08:59:25.463" LastActivityDate="2018-03-01T08:59:25.463" Title="Using Stainless steel or iron pots in oven" Tags="|circuits|" AnswerCount="1" CommentCount="1" ClosedDate="2018-03-01T09:40:50.177" ContentLicense="CC BY-SA 3.0" />
  <row Id="19657" PostTypeId="2" ParentId="19656" CreationDate="2018-02-28T15:53:13.403" Score="3" Body="&lt;p&gt;If it was a microwave oven then &lt;strong&gt;all metals&lt;/strong&gt; can &lt;a href=&quot;https://www.gizmodo.com.au/2014/05/what-actually-happens-when-you-put-metal-in-a-microwave/&quot; rel=&quot;nofollow noreferrer&quot;&gt;damage the oven&lt;/a&gt; if they are placed in the oven &amp;amp; the oven is operating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the oven is is thermal oven that is heated by an electric element, gas or wood, then there will no issues with using steel or stainless steel items in such ovens,&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-02-28T15:53:13.403" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19660" PostTypeId="2" ParentId="19647" CreationDate="2018-02-28T17:15:09.383" Score="0" Body="&lt;p&gt;Your question was &quot;whether electricity moves better through a thin wire or a thicker wire&quot;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power is lost due to resistance. Copper wire has a very low resistance and for your experiment a thicker wire and a thinner wire will perform the same. In other words the difference would be so small that it would be very difficult to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wires made from other materials, such as Nichrome have high resistance and if Nichrome was used for your experiment, you would be able to measure the difference, as seen by the brightness of the bulb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring brightness can be achieved in many ways, but generally some sensor that is sensitive to light energy will be required. A web search for &quot;light sensors&quot; will uncover lots of information.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2018-02-28T17:15:09.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19661" PostTypeId="2" ParentId="19647" CreationDate="2018-02-28T18:41:27.220" Score="2" Body="&lt;p&gt;While an incandescent bulb can be a good visual indicator of the current and/or voltage in a circuit measuring its brightness and then converting that to numerical data is a very roundabout and error prone method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a more fundamental problem that you are assuming a relationship between light level and wire thickness which you haven't proved ie your experiment has too many unknown variables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would be much better of using a voltmeter or ammeter to measure either the current through the circuit or the voltage drop along a length of wire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other alternative is just to use comparative data eg just record whether bulb A or bulb B looks brighter for each experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only possible reason for using a bulb is to keep the experiment simple using a light meter to indirectly measure current rather defeats that object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to investigate the relationship between current, voltage and light intensity then that is a perfect valid (and arguably more interesting) experiment but you need to frame your terms to reflect that. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-02-28T18:41:27.220" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19662" PostTypeId="2" ParentId="19635" CreationDate="2018-02-28T18:51:52.567" Score="2" Body="&lt;p&gt;I am not entirely sure I follow your derivations, specifically I'm not sure where you get the equation for the Von Karman mixing length, and why the equivalence of tau and $\tau_0$ give you the Prandtl turbulence model that you write afterwards (apologies since I can't get the equations to format properly, but I am referring to the equations you write immediately after stating $\tau = \tau_0$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, in the context of the last question, you typically don't need to solve any differential equation once they have told you that the logarithmic overlap law desrcibes the flow, which is: $u/U^*=C_1 \ln(y/R)+C_2$. In Fluid Mechanics by White on page 364 he says as much &quot;For turbulent pipe flow, we need not solve a differential equation, but instead proceed with the logarithmic law...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to whether you need to relate the viscous sub-layer, I believe the answer is no. Once the question has told you the flow is determined by the logarithmic relationship, then any viscous sub-layer equation will not be valid for that specific flow. In this case, you can assume the Von Karman constant $k=0.41$ (for pipe flow) and $C_1 = 1/k$ and $C_2 = 5.0$ (also for pipe flow). Admittedly this has just been pulled from tables, so maybe this is not what you're looking for. Nonetheless that's how we were asked to solve these problems. &lt;strong&gt;After using these values,&lt;/strong&gt; &lt;strong&gt;you then just make a simple substitution using the data given, and calculate $U^*$ from the equation $U^* = \sqrt{\tau_0/\rho}$.&lt;/strong&gt; (see edit below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to the question of the boundary conditions, my understanding is that the equation supplied to you in the question is the logarithmic overlap law. These constants were not solved by using boundary conditions; they were solved by experiment. You can see this here &lt;a href=&quot;https://en.wikipedia.org/wiki/Law_of_the_wall#General_logarithmic_formulation&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Law_of_the_wall#General_logarithmic_formulation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are familiar with the Reynold's stresses equation, it can be reduced in 2D to yield:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau + \tau_{turbulence} = \tau_{wall}$ (or in your notation $\tau_0$)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here it can be non-dimensionalised and then boundary conditions are used with THIS equation in order to generate the logarithmic overlap law, and other valid solutions to describe the flow. In other words, the logarithmic overlap law is the outcome from the Reynold's stresses equation, and the constants in the logarithmic overlap law have been determined experimentally. I hope this helps, and sorry again for the lousy formatting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; I made a significant error in the third paragraph. The question asks you to find the shear stress, but &quot;make a simple substitution using the data given&quot; won't help you do that, since you know neither U* nor $\tau_o$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, for the equation to be the logarithmic overlap law, then &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ y/R = (y*U^*)/\nu$, where $\nu$ is the kinematic viscosity. --&gt; If you're not sure where this comes from read this: &lt;a href=&quot;https://en.wikipedia.org/wiki/Law_of_the_wall&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Law_of_the_wall&lt;/a&gt;. It is simply using the normalised equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, you can calculate $\tau_o$ in terms of R, as per the question. To do this, equate U* in terms of R, and then use $U^* = \sqrt{\tau_o/\rho}$. This will allow you to solve for tau in terms of R.&lt;/p&gt;&#xA;" OwnerUserId="5469" LastEditorUserId="5469" LastEditDate="2018-03-02T14:35:21.550" LastActivityDate="2018-03-02T14:35:21.550" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19663" PostTypeId="2" ParentId="19644" CreationDate="2018-02-28T20:12:44.613" Score="0" Body="&lt;p&gt;You can go various ways to solve your case.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You can just use many gears to create a huge rpm reduction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can PWM control your motor to go slow, but you'd still need a gearing of multiple gears; it'd be hard to find a motor that can steadily be rotated at 60rpm or lower. You can put a friction load on the motor so you'll have no/less stuttering. See, to drive the motor slow, power should be low(stuttering), or it should be kept at a low speed by a friction load. The latter gives less stuttering behaviour.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's not clear to me what kind of application you are intending, but you could just drive the motor with pulses. Kind of a ultra low frequency PWM. If you have a controller than powers the motor for a very short time, and then pauses for a longer time. E.g power the motor at 20% power for 0.25sec, stop power for 2.75sec. Repeat. You effectively have created a slow turning disk.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If your application is not power demanding, you can get a quartz clock mechanism and use that.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Recommended option: Use belts and pulleys or sprockets instead of gears. You can easily create large reductions cheaply. You can also create them yourself from wood and rubber bands.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-02-28T20:12:44.613" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19664" PostTypeId="2" ParentId="19636" CreationDate="2018-02-28T20:40:29.137" Score="2" Body="&lt;p&gt;Basically, materials have a tensile strength, and a yield strength. The latter describes how much it elongates under stress, the former how much stress it can have before breaking. It is always given in (M)Pa, so force per surface. This means it doesn't matter what shape or size your fibers are, it remains the same.&#xA;All the material properties remain the same at the same temperature, and before plastic deformation has been made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force you can put on it before it breaks, is determined by the surface/cross-section. So two times as much surface, two times as much tensile strength. If you know the cross-sectional area of your fiber, you can hang weights on it until it breaks, divide found force(Newton) by known surface(m2) and you have experimentally determined the tensile strength.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-02-28T20:40:29.137" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19665" PostTypeId="1" AcceptedAnswerId="19671" CreationDate="2018-03-01T00:42:30.667" Score="3" ViewCount="289" Body="&lt;p&gt;Taking a 1750RPM 110VAC appliance motor down to about 60RPM, maybe a little less. It's a home project on the cheap, so dropping a few hundred dollars on a ready-made gear box is to be avoided, and I'm working with the kids as a school project and so creating by hand as much as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/2&quot; plywood for an initial prototype/proof of concept, with a planned move to 3D printed gears afterward. Assuming a tooth pitch of 18-25mm, and that we play with the tooth count between the gears to spread out the wear and tear, which would be better for durability, chatter, noise? Fewer gears with larger reduction ratio at each step, or more gears with a smaller ratio? &lt;/p&gt;&#xA;" OwnerUserId="15115" LastActivityDate="2018-03-01T10:58:55.830" Title="~121:1 speed reduction; are more gears better?" Tags="|gears|mechanisms|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19666" PostTypeId="2" ParentId="19647" CreationDate="2018-03-01T02:21:44.110" Score="1" Body="&lt;p&gt;A photo resistor or solar panel make for pretty easy/low cost light measuring solutions, but they are not super accurate and will be much more complicated than the primary project. Hook the solar panel directly to a volt meter. It will measure the amount of green light but will not be linear. For the photo resistor hook a battery in series with it and the volt meter. It will likely not be linear either; but I doubt you need too much accuracy here. If you do, you will have to get a calibrated light meter like Fred suggested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though measuring the light intensity is possible, I would recommend just using a volt meter and/or amp meter directly like Chris Johns suggested to keep your project simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, for a more extendable hands-on measurement you could put a known resistor on each circuit and put each in a known volume of water containing a simple thermometer. The circuit with the larger diameter wire will have less resistance and deliver more power to the equal sized resistor and the temperature of the water would increase quicker. Then with some &lt;a href=&quot;https://en.wikipedia.org/wiki/Calorimetry#Practical_constant-volume_calorimetry_(bomb_calorimetry)_for_thermodynamic_studies&quot; rel=&quot;nofollow noreferrer&quot;&gt;simple calorimetry math&lt;/a&gt; you can calculate the resistance of each wire.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-03-01T02:21:44.110" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19667" PostTypeId="2" ParentId="19665" CreationDate="2018-03-01T02:34:51.247" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Gear&quot; rel=&quot;nofollow noreferrer&quot;&gt;one source&lt;/a&gt;, the tooth pitch is the distance on the circumference of the gear for one tooth and one gap. If your value is correct, you'll have quite a large diameter gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you are referring to pitch diameter, that brings the gears down to almost a tiny level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the 1750 rpm of the source, a destination rpm of 40 would mean a 35:1 ratio, a nice integer ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your starting point is a 25 mm drive gear, the driven gear will be almost a meter in diameter (875 mm). Obviously, if you go larger with the drive gear, the driven gear has to go up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A compound gear arrangement may be a bit better suited if you have the vertical space to stack gears. Another possible option would be to create and use a planetary gear set. 35:1 is well within the capabilities of such a mechanism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &quot;bunches&quot; of &lt;a href=&quot;https://www.thingiverse.com/search?q=planetary%20gear&amp;amp;sa=&amp;amp;dwh=985a9762ec103b0&quot; rel=&quot;nofollow noreferrer&quot;&gt;planetary gear models on Thingiverse&lt;/a&gt; for 3D printing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently working on a project involving a much lower ratio, only 1.5:1 which means an 80T sun gear and a 120T ring gear. Planet gear sizes are meaningless in this application. In order to get the greater reduction you require, you would have to stack at least two stages and that gets out of my current range of understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thingiverse has a few multi-stage designs, &lt;a href=&quot;https://www.thingiverse.com/thing:1727833&quot; rel=&quot;nofollow noreferrer&quot;&gt;one that looks particularly attractive&lt;/a&gt;. The design is based on a 6:1 reduction, therefore two stages would provide for 36:1, nearly the aforementioned 35:1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W8DcZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W8DcZ.jpg&quot; alt=&quot;multi-stage planetary gear reduction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The output of this design is not shown, but the crank drive could be adapted to take a motor shaft. If you desire to construct this of plywood, you could take the tooth counts for the components and enter them into one of the &lt;a href=&quot;http://geargenerator.com/#200,200,100,360,1,3,14536.200000000017,4,1,8,2,4,27,-90,0,0,16,4,4,27,-60,1,1,12,1,12,20,-60,2,0,60,5,12,20,0,0,0,2,-584&quot; rel=&quot;nofollow noreferrer&quot;&gt;many gear generator programs&lt;/a&gt; available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above linked site shows how to accomplish a greater reduction with a compound gear set rather than a planetary gear set, but will still allow you to construct the necessary drawings (and downloads) to build a model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0VYsz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0VYsz.jpg&quot; alt=&quot;36 rpm gear reduction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I did this properly, the above site and the link provided will appear with a 1750 rpm input and a 36 rpm output. A bit lower than you requested, but I had to change only one figure and the result is suitable for demonstration purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To attempt to answer your primary question, fewer gears means less wear, less noise, but much more space used to accomplish this. Planetary gear sets are not known for being quiet, especially when input is so high, but a compound gear set may not be much quieter, although there will be fewer gear components in the latter design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Design for the largest size you can fit to your application, to spread wear out over a greater area. Lubrication, even on wood, will be of great value at the speed you've provided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Larger gears will be in the range of hand tools such as a coping saw, unless you have the great resources of a laser cutter. I think I'd have had great fun in your class if I had such opportunities in my youth.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-03-01T02:34:51.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19668" PostTypeId="2" ParentId="19650" CreationDate="2018-03-01T03:56:55.650" Score="0" Body="&lt;p&gt;In reality there is underground water and also all variety of oil and mixture of other liquids, such as bicarbonates and other minerals dissolved in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the geology of soils, and the pore pressure of it, we get a wide range of situations. sometimes due to flow of  underground water there is even hydraulic head pressure or negative pressure to capillary and osmosis actions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We consider an ideally simplified case where the whole is large and forces are static and balanced and there is one liquid and no trapped air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a case like this if we could bore a long infinitesimally small radius hole without disturbing the layers of rock and water the pressure at any given depth is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P = \sum \rho.da.h $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total pressure is the weighted contribution of different material added together from grade level down to depth under consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pressure in the soil layer is transferred laterally by the factors such as passive lateral pressure, bedding angle and structural properties of the soil. &#xA;When the test bore gets to the level of water/liquid the pressure is equally transferred in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The soil surrounding this reservoir will react in complex ways, again depending on its structural properties and its content of aggregates. These properties can be determined by the soils engineer using a wide selection of tools. Such as sampling the soils, doing ultrasound and satellite imaging. etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finer aggregates and clayed strata would act as a container of the water and the pressure of the water would be equal to the pressure of surrounding soil. The larger aggregate will diffuse the pressure and can potentially lead to caving of the reservoir.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-01T03:56:55.650" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19669" PostTypeId="1" CreationDate="2018-03-01T07:06:50.320" Score="4" ViewCount="188" Body="&lt;p&gt;Instead of using steel and concrete to build towers why not use mostly light weight, high tensile materials and then suspend the four corners from four giant helium balloons? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The four main cables would be anchored to the ground at the bottom, and held apart at the top by the force of the balloons pushing off each other, as they are pulled inward by the tension required to suspend each floor. So the four corner cables would essentially form a parabola, like the top cables of a suspension bridge, except vertically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could a skyscraper be constructed in this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big would the balloons need to be to suspend the building?&lt;/p&gt;&#xA;" OwnerUserId="7205" LastEditorUserId="14479" LastEditDate="2018-03-01T12:52:54.963" LastActivityDate="2018-09-02T14:40:47.483" Title="Why don't towers use buoyancy to reduce weight?" Tags="|civil-engineering|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19670" PostTypeId="2" ParentId="19669" CreationDate="2018-03-01T07:54:42.277" Score="9" Body="&lt;p&gt;The engineering design of towers is not limited in a practical sense by their weight, meaning that the scheme you propose solves no problems in that field. Truly gigantic balloons would be necessary and the cost of a &quot;floating tower&quot; suspended from them would be huge- probably dominated by the price of the helium necessary to fill them and keep them full in the face of leakage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also not want to be in one of them when a windstorm hits because large lighter-than-air structures respond badly to wind shear and turbulence.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-03-01T07:54:42.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19671" PostTypeId="2" ParentId="19665" CreationDate="2018-03-01T10:03:46.197" Score="2" Body="&lt;p&gt;Gear chatter/noise is caused by all manner of things, but tooth count isn't a primary driver. Tooth count will impact the bending and pitting resistance of your gear teeth, meaning you could make them last longer, although if torque is minimal you probably don't need to worry about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To reduce noise you need to reduce transmission error (have involute gears as close to geometrically perfect as possible). Having a good surface finish on the gear teeth will help, 3D printing sounds fine for both accuracy and surface finish in your application. (Might be worth checking the material strength of whatever you choose to ensure it can handle the torque from the motor)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were you (assuming the torque from the motor is low enough that you essentially have infinite fatigue life) I would go for the bigger reduction over a single gear (to save space) and focus on making the teeth as accurately as possible, with smooth surface finish. Also look at reducing the backlash, in a low speed application this could be a primary source of noise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outside of that, I recommend reading &lt;a href=&quot;http://khkgears.net/gear-knowledge/gear-technical-reference/how-to-reduce-gear-noise/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; - it's quite a nice summary for reducing gear noise. There are some sources of noise you won't be able to do anything about. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If you are unsure about durability, go for more gears to prevent one big ratio step. For noise, improve the accuracy of geometry and surface finish on all gears where possible, reduce backlash (but not too much!).&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: I worked as a gear designer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lXZcC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lXZcC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14498" LastEditorUserId="14498" LastEditDate="2018-03-01T10:18:53.223" LastActivityDate="2018-03-01T10:18:53.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19672" PostTypeId="2" ParentId="19665" CreationDate="2018-03-01T10:58:55.830" Score="1" Body="&lt;p&gt;Why use 3d printed gears ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3D printing is a terrible way to make gears and you can just buy standard ones with proper tooth profiles off the shelf. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that sort of high reduction ratio a planetary gearbox is generally preferable to long compound gear trains. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However it sounds like you might be better of with a belt drive. Vee belt pulleys are fairly cheap, can achieve the required ratio without too much compounding and are reasonably quiet and importantly lend themselves well to low-tech fabrication as they are reasonably tolerant of misalignment and have a lot of inherent scope for adjustment. Unlike gearboxes which require pretty good manufacturing tolerances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For all of these reasons belt drives tend to be the preferred option for stationary tools driven by AC motors. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-03-01T10:58:55.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19675" PostTypeId="2" ParentId="19669" CreationDate="2018-03-01T15:54:13.397" Score="4" Body="&lt;p&gt;I think there are multiple aspects to this question.&#xA;If I'm not mistaken, your concept would basically mean, that every floor is anchored to those 4 cables in the corners, thus basically &quot;floating&quot; in the air?&#xA;As a consequence, the upward force generated by those balloons would be (greater) equal to the weight of the floors. Something like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iZbc9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iZbc9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this graphic I have neglected the tension due to the dead load of the cables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;stability and safety&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the biggest problem aside from statics would be the stability and safety of the structure; leaks, failure of a balloon, wind, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;statics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In normal conditions, Helium has a density of $\rho_{He}\approx0.179kg/m^3$, while air's is around $\rho_{air}\approx1.29kg/m^3$.&#xA;If we neglect the weight of the hull of the balloon as well, we can come up with a net buoyancy force $f_B$ per cubic meter of Helium.&#xA;$$ f_{B}=(\rho_{air}-\rho_{He})g=10.9N/m^3_{He} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take a $10m\times10m$ concrete floor, with a thickness of $t=0.2m$, and a density $\rho_c=2'500kg/m^3$, you would already need a balloon volume&#xA;$$ V_{He}\approx45'000m^3 $$&#xA;which is a sphere of radius $r=22m$. An that is for one single floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;availability of helium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Though being the most abundant element in the universe, helium is relatively rare on the earth. Currently, most helium is extracted from natural-gas, but those fossil fuel resources are limited. I'm not sure whether we would want to risk losing such a precious resource.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="8632" LastEditDate="2018-09-02T14:40:47.483" LastActivityDate="2018-09-02T14:40:47.483" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="19676" PostTypeId="1" AcceptedAnswerId="19682" CreationDate="2018-03-01T17:02:45.513" Score="3" ViewCount="1743" Body="&lt;p&gt;I am deciding between two design ideas for a project in which I have to build a balsa wood tower that can maximize load weight and minimize structural weight (maximize efficiency). I've included a picture with the two designs I am deciding between. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U20AL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U20AL.jpg&quot; alt=&quot;(Left, tower with height 40 cm, width 5 cm. 10 cross-braces with height 4 cm each. Right, height and width 40 cm and 5 cm. 5 cross-braces with height 8 cm each. Both towers are supported by pin supports on the right and left sides.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering which design would be able to hold greater weight, whether increasing the number of braces changes the load capacity, and in which way it changes (linear correlation, exponential correlation, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, I'm considering a load weight of &lt;strong&gt;145 N&lt;/strong&gt;, and I am looking for the values of the internal member forces, especially those before buckling (assuming a static system). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of this subject only extends to truss calculations, considering diagonal bracing (e.g. Howe Truss), which is why I am posting on this forum. Further, if you'd be willing to give a brief explanation of these diagrams from &lt;a href=&quot;https://www.aisc.org/globalassets/aisc/research-library/bracing-for-stability.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Bracing for Stability&lt;/em&gt;&lt;/a&gt;, I'd really appreciate it.  &lt;a href=&quot;https://i.stack.imgur.com/erzUE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/erzUE.png&quot; alt=&quot;Design recommendation for discrete bracing systems for columns, pg. 6&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/hzMAO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hzMAO.png&quot; alt=&quot;Figure 6 on pg. 6, the accompanying graph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15125" LastActivityDate="2018-03-01T23:26:16.057" Title="How does the number of braces in a balsa wood tower affect the load capacity?" Tags="|structural-engineering|design|statics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19677" PostTypeId="1" AcceptedAnswerId="19679" CreationDate="2018-03-01T18:00:08.887" Score="20" ViewCount="7314" Body="&lt;p&gt;Modern cars use cogwheel to transfer power from the engine to the wheels.&#xA;Steam locomotives &lt;a href=&quot;https://i.imgur.com/ou5yWsJ.gifv&quot; rel=&quot;noreferrer&quot;&gt;used some kind of bars&lt;/a&gt; (sorry, I'm not a native speaker) to transfer the power to the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did the engineers not use cogwheels?&#xA;Would steam locomotives had been faster if they had used cogwheels?&lt;/p&gt;&#xA;" OwnerUserId="15126" LastActivityDate="2018-03-04T05:54:44.393" Title="Why did steam locomotives not transfer power by cogwheels?" Tags="|steam|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19678" PostTypeId="2" ParentId="19677" CreationDate="2018-03-01T18:32:06.083" Score="5" Body="&lt;p&gt;for a while, steam locomotives actually used gears and cylinder/piston sets which drove crankshafts. These were called geared locomotives and they were used to haul heavy loads up particularly steep inclines at low speeds. this made them popular for timber logging operations in the western US during the days of steam power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for higher speed use on more gradual slopes, the direct-drive method (in which the connecting rod engages the driven wheel(s)) is simpler and furnishes an adequate match between the impedance of the load and that of the engine. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13363" LastEditDate="2018-03-02T02:36:35.053" LastActivityDate="2018-03-02T02:36:35.053" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19679" PostTypeId="2" ParentId="19677" CreationDate="2018-03-01T18:35:13.023" Score="12" Body="&lt;p&gt;I would like to point out that modern cars do not use cogwheels for transmission, they use shafts. Cogwheels are used for the gearing and differential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But bar mechanisms were used mostly because they didnt have the kind of manufacturing facilities as we do today. Bar mechanisms are easy to make, they are flexible and field maintainable. In any case in this particular design also because the entire mechanism would have had to turn the power transmission direction 2 times. See the piston is directly connected to the front wheel and transfering that to the subsequent wheel is pretty simple with a bar while a shaft coupling would have needed more parts, which again were hard to manufacture.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-03-03T13:13:44.397" LastActivityDate="2018-03-03T13:13:44.397" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="19680" PostTypeId="2" ParentId="19676" CreationDate="2018-03-01T18:51:22.520" Score="0" Body="&lt;p&gt;You need to find out structural properties of the cross-section of balsa member you use. namely at what length it is considered a short column. A short column will crush in compression rather than buckle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is ideally the joint height of the tower. However, practical constraints may force you to chose a compromise near that joint span. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-01T18:51:22.520" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19681" PostTypeId="2" ParentId="19677" CreationDate="2018-03-01T19:59:22.457" Score="37" Body="&lt;p&gt;Steam piston engines can generate a lot of torque from stationary and the pistons can be physically remote from the boiler, so in most cases it is most convenient to have the pistons directly drive the wheels via a crank. Equally as trains don't have a steering mechanism as such and have conical section wheels, you don't need a differential either. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, internal combustion engines need to be turning at a fairly moderate RPM to generate useful torque and produce most of their torque and power in a fairly  narrow rev range so they need both a means of disengaging drive (clutch or viscous torque converter) and a selectable ratio gearbox in order to provide useful torque at a wide range of road speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, IC engines tend to work better with multiple cylinders as this smooths out the power delivery over the various different stages of the working cycle and so need a crankshaft with a common output shaft. Steam engines are essentially pneumatic actuators so you can make the working stroke as long as is convenient and get a reasonably consistent linear force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The external connecting rods on a steam locomotive are a direct analogue of the connecting rods which link the piston of an IC engine to the crankshaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is that the torque characteristic of a steam engine simply means that a gearbox is unnecessary, as torque is more or less independent of RPM for its normal working speed range. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-03-02T19:33:46.213" LastActivityDate="2018-03-02T19:33:46.213" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19682" PostTypeId="2" ParentId="19676" CreationDate="2018-03-01T23:26:16.057" Score="5" Body="&lt;p&gt;Elements under compression such as the vertical columns in your tower can collapse in two very different ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is via simple crushing of the member. This happens when the applied load generates an internal stress in the member which is higher than the member's strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is via buckling. In this case, infinitesimal imperfections in the structure make it &quot;easier&quot; for the member to bow away from the load. Get a plastic straw or a piece of paper and try to place either under compression. You'll notice they just &quot;jump&quot; to the side. However, once the compression is removed, the straw or piece of paper will &quot;jump&quot; right back into its original shape as if nothing'd happened. The theoretical equation for the buckling load (often called Euler buckling) is&#xA;$$P_e = \dfrac{\pi^2EI}{(kL)^2}$$&#xA;where $E$ is the member's modulus of elasticity, $I$ is its second moment of area (aka, moment of inertia), and $kL$ is the member's unbraced length ($k$ is a coefficient which depends on the member's boundary conditions; in your case, you could conservatively assume $k=1$). Obviously, the real-world buckling load is much lower than $P_e$, since real-world imperfections are actually quite significant, not infinitesimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Braces aren't meant to carry any of the applied load. Their purpose is only to guarantee that the principal members (in this case, the vertical columns) do not buckle under compression. So a properly braced structure will instead collapse due to crushing or global buckling (where the entire structure buckles as if it were one member, which the braces can't help against).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the unbraced length for your columns, you could rework the Euler's buckling load equation to give you $L$ given the other values (which you can find online). However, given how that equation is highly theoretical (and not used directly in actual engineering), it is probably best to merely create prototypes to see at which length the parts start to buckle and then adopt that.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-03-01T23:26:16.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19686" PostTypeId="1" AcceptedAnswerId="19812" CreationDate="2018-03-02T15:01:47.640" Score="1" ViewCount="101" Body="&lt;p&gt;In the answer to this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/10341/difference-between-saturated-liquid-and-saturated-steam&quot;&gt;Difference between Saturated Liquid and Saturated Steam&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can energy be added with constant temperature? Isn't the energy input itself a change of temperature? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the repeat, but I don't have enough reputation to post a comment. &lt;/p&gt;&#xA;" OwnerUserId="13823" LastEditorUserId="14498" LastEditDate="2018-03-09T10:54:17.870" LastActivityDate="2018-03-09T11:36:15.133" Title="How can energy be added at constant temprature?" Tags="|thermodynamics|heat-transfer|steam|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19687" PostTypeId="2" ParentId="19024" CreationDate="2018-03-02T15:07:06.237" Score="0" Body="&lt;p&gt;Robin's statements are valid, I would only recommend to use open source libraries like &lt;a href=&quot;http://dual.sphysics.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;DualSPHysics&lt;/a&gt;, which is a quite fast, robust and well-documented SPH solver for a wide range of applications. Being open-source, it helps you to implement and configure your own or other's theories concerning the boundary conditions for moving gears. Although I never tried any of them, I don't believe that there are major benefits of using commercial SPH software today.&lt;/p&gt;&#xA;" OwnerUserId="15139" LastActivityDate="2018-03-02T15:07:06.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19688" PostTypeId="2" ParentId="19677" CreationDate="2018-03-02T15:08:52.910" Score="14" Body="&lt;p&gt;Here is a picture of the crankshafts inside a modern internal combustion engine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZJIob.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZJIob.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of these is to convert the back-and forth motion of the piston into rotational motion. It's very much the same mechanism as used on old steam engines:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6MMvC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6MMvC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference is that in the internal combustion engine the power is not transferred directly to the wheels but to a shaft. The reasons for this difference are discussed in other answers --- I just wanted to point out that the underlying mechanism is the same.&lt;/p&gt;&#xA;" OwnerUserId="15141" LastActivityDate="2018-03-02T15:08:52.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19689" PostTypeId="1" CreationDate="2018-03-02T16:00:43.117" Score="5" ViewCount="258" Body="&lt;p&gt;When using public transport, we often need to change routes in the middle of the journey. These changes have costs – I can think of the following, but there may be more.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If we're reading, working, etc. on board, we need to take a break, losing flow.&lt;/li&gt;&#xA;&lt;li&gt;We spend time moving from one route to the next.&lt;/li&gt;&#xA;&lt;li&gt;We spend time waiting for the next departure.&lt;/li&gt;&#xA;&lt;li&gt;There are lots of passengers moving between routes. They take up space and cause queues when on- and off-boarding. This also increases the risk of accidents.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As far as I can understand, we could decrease the necessity of route changes by having multiple routes to the terminals (final stations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've reviewed the subway maps of London, New York and Oslo, finding that none of them do this to any significant degree – the terminals usually have just one route.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, let's look at the simplest of the three – Oslo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4OOP2.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4OOP2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that you're travelling from Østerås (orange, left) to Ellingsrudåsen (orange, right). No changes required – route 2 takes you all the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you're travelling from Østerås to Mortensrud (purple, bottom right), you would have to change trains from route 2 to route 3 at some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine that we use some of the capacity used for routes 2 and 3 to serve routes 6 and 7. Route 6 (gray) serves Østerås-Mortensrud, and route 7 (brown) serves Kolsås-Ellingsrudåsen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AtDvE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AtDvE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this design, some route changes can be avoided. If we added enough routes, many more route changes could be avoided. Granted, we would have to wait longer for a route that went exactly to the station we want to go, but we would always have the option of taking the first train and then changing like before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I've been asking myself, why don't public transport authorities do this? These are the reasons I can think of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Simplicity for passengers: When only one route serves a particular station, you don't have to think about which route you'll board. Also, the maps would be more chaotic.&lt;/li&gt;&#xA;&lt;li&gt;Simplicity for transport providers: More routes would mean more difficult (and therefore costly) planning and management.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From my (admittedly ignorant) perspective, the above reasons don't seem sufficient to outweigh the advantages of fewer changes. Have I overlooked some problems with adding more routes? Or am I underestimating the costs and/or overestimating the benefits of more routes?&lt;/p&gt;&#xA;" OwnerUserId="15140" LastActivityDate="2018-03-06T09:54:25.893" Title="Why don't public transport networks have more routes to each terminal?" Tags="|design|transportation|architecture|infrastructure|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19690" PostTypeId="2" ParentId="19689" CreationDate="2018-03-02T17:13:16.930" Score="6" Body="&lt;p&gt;The main reason is economics. Who is going to pay for all the additional lines? The other part to this is, will the additional lines have enough passengers to make them profitable and sustainable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the lines are on the Earth's surface how much land will they occupy and could the land be better used for something else? If the lines are underground, or above ground, using elevated tracks, all this adds to the capital cost of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how much extra disruption would the additional lines create in terms of noise, vibration, aesthetic degradation to the environment?&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-03-02T17:13:16.930" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19691" PostTypeId="1" CreationDate="2018-03-02T18:26:59.737" Score="1" ViewCount="32" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZzXaV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZzXaV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I really need some help identifying this diode.  I contacted ON Semiconductor's technical support and they couldn't tell me what it is, even though it is a Fairchild Semiconductor diode (ON owns Fairchild now I guess).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am attaching a picture of this diode.  It has a Fairchild logo &quot;F&quot; followed by three lines of code &quot;523&quot; &quot;IC&quot; and &quot;F8.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can help me out with this, or point me in the direction of someone who can, I would be very thankful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="15146" LastActivityDate="2018-03-02T18:26:59.737" Title="Need help identifying a mystery diode!" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19692" PostTypeId="1" CreationDate="2018-03-02T18:28:16.800" Score="3" ViewCount="164" Body="&lt;p&gt;During world war two the British had plans for an aircraft carrier made of pykrete (a mixture of 14% sawdust or wood pulp and 86% ice.) the idea was to have a refrigeration system to stop the ice from melting however I can't help but think that since the temperature would still be very near to the melting temperature of ice that creep would have been a problem. does anyone have any knowledge about the properties of ice regarding creep?&lt;/p&gt;&#xA;" OwnerUserId="15147" LastActivityDate="2018-03-02T19:35:25.830" Title="Would creep have been a major problem for pykrete designs?" Tags="|materials|temperature|creep|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19693" PostTypeId="2" ParentId="19692" CreationDate="2018-03-02T18:58:32.840" Score="2" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: see &lt;a href=&quot;https://www.igsoc.org/journal/1/3/igs_journal_vol01_issue003_pg95-104.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for a contemporaneous study of this exact question: Perutz, &quot;A description of the iceburg aircraft carrier,&quot; &lt;em&gt;Journal of Glaciology&lt;/em&gt; (1947).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dominant creep mode and rate can be read from a deformation mechanism map. Shouji and Higashi, in &quot;&lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.474.5289&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;A deformation mechanism map for ice&lt;/a&gt;&quot;, &lt;em&gt;Journal of Glaciology&lt;/em&gt; (1978) included the following map for ice with a 2 mm grain size, based on experimental data from studies in the 1960s and 1970s:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z6AZN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z6AZN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Michael Ashby, who contributed substantially to developing and popularizing these types of maps, performed a &lt;a href=&quot;http://engineering.dartmouth.edu/defmech/Chapter_19.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;case study on Antarctic &#xA;ice&lt;/a&gt; that included the following map for ice with a grain size of 10 mm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nr0Ke.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nr0Ke.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trends shown in these maps emphasize the importance of ensuring low stresses if ice is to be used as an engineering material near its melting point without substantial creep deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these diagrams are for pure ice; the presence of a binder can be expected to alter the dominant failure model and strain rate. I haven't found a corresponding map for pykrete.  However, the first link (given in the edit comment) suggests that the relevant parameters are a normal stress of approximately 70 MPa—presumably corresponding to a shear stress of approximately 35 MPa—and a temperature of -15°C. It's also emphasized that the addition of the wood pulp/sawdust can strongly suppress long-term ice creep. The other links here provide an introduction for the various modes of creep that apply and a general idea of the strain rates that can be expected.&lt;/p&gt;&#xA;" OwnerUserId="10123" LastEditorUserId="10123" LastEditDate="2018-03-02T19:35:25.830" LastActivityDate="2018-03-02T19:35:25.830" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19694" PostTypeId="2" ParentId="19689" CreationDate="2018-03-02T19:25:24.777" Score="1" Body="&lt;p&gt;Best guess to the problem of adding more: Logistics and economics. First of all, never underestimate the cost of public construction project such as tunnel buldin&lt;code&gt;n&lt;/code&gt;stuff. The last sizeable expansion was the Løren-banen which was finished a couple of years ago. The short stretch between Sinsen and Økern ended up costing more than 1 billion NOK (Google, 2018). An aspect regarding the logistical side of this can arise when you cross lines. F.X. there will probably be complications that will result in longer queues in the tunnels between the stations when you have to change tracks, specially when a train is already running late which is something that is known to happen every now and then. But I completely agree that there are great benefits for doing this. An example would be the mentioned Løren-Line which made life a lot easier for those of us that live on the wrong side of the Akersriver&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Litteraturliste:&#xA;Google. (2018). Google search engine.&lt;/p&gt;&#xA;" OwnerUserId="15151" LastActivityDate="2018-03-02T19:25:24.777" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19695" PostTypeId="2" ParentId="17839" CreationDate="2018-03-02T19:59:20.170" Score="0" Body="&lt;p&gt;I would not find out the sum of average force acting on the glass. But a more appropriate way would be to find the most critical stress which happens near the bottom of the your panel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would pick the pressur at the bottom and use that multiplied by vertical projection of you surface to get the force downward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P= \rho.d =\rho.3 \space$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ Assuming \  rho=1 $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you get $P =3000 kg/m^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you can plug this pressure into the equation for plate tress for a horizontal plat of 4by 2 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one such formula is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GQ22h.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GQ22h.gif&quot; alt=&quot;the link:http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b is the width=2, a is the length=4, and t is thickness.&#xA;Off course there are finite element softwares that are more precise, but in this case you will be fine!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source of the formule:&lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-03-02T23:07:30.180" LastActivityDate="2018-03-02T23:07:30.180" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19696" PostTypeId="2" ParentId="19677" CreationDate="2018-03-02T22:09:45.933" Score="4" Body="&lt;p&gt;Steam generates full torque at zero speed as has been mentioned elsewhere, so like in an electric car (which has very much the same sort of characteristic), there is little to be gained from a gearbox, may as well drive the wheels directly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is also why the vast majority of diesel locomotives above the very small, are really diesel-electric, it makes the near zero speed part of the performance less annoying, and removes the need to try to cool a very high power clutch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally a steam loco does have sort of &quot;gearing&quot;, in that the driver can control the valve timing to vary the volume of steam admitted per stroke and thus the torque available, this is subtly separate (but interacts with) from varying the steam pressure.... You see this when a steam train pulls away as initially there will be powerful blasts of steam from the stack because the driver has the valve gear set such that there is still significant pressure in the cylinder as the exhaust valve opens (To maximise torque), as the speed comes up the fraction of a cycle the intake valve is open is reduced to improve efficiency and the exhaust note evens out as the exhaust becomes closer to atmospheric pressure. &#xA;These variable valve linkages were one of the more patent infested waters back then with fights between all the big players.&lt;/p&gt;&#xA;" OwnerUserId="10072" LastActivityDate="2018-03-02T22:09:45.933" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19698" PostTypeId="2" ParentId="19633" CreationDate="2018-03-03T02:52:31.313" Score="0" Body="&lt;p&gt;The vendor McMaster has MXL timing belts at the following page:&#xA;&lt;a href=&quot;https://www.mcmaster.com/#timing-belts/=1bswega&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mcmaster.com/#timing-belts/=1bswega&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Timings belts should have no or minimal slop. An easy way to remove slop is mount the motor using slots, position motor so belt is snug, and tighten screws holding motor in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would try for 120 degree teeth engagement or more, but for a low load I don't thing the pulleys being close together will be an issue.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-03-03T02:52:31.313" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19699" PostTypeId="1" CreationDate="2018-03-03T03:13:55.520" Score="0" ViewCount="2226" Body="&lt;p&gt;I am creating the 3d Design in Autodesk Autocad 2017. I need to specify distance between two parallel lines, I could successfully defined the parallel constraint but not able to add constraint that would restrict the distance between them. As per documentation &lt;a href=&quot;https://knowledge.autodesk.com/support/autocad/learn-explore/caas/CloudHelp/cloudhelp/2016/ENU/AutoCAD-Core/files/GUID-F22AD919-40DF-4AC3-8339-CD212D79EB94-htm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; we can use &quot;Aligned&quot; constraint to restrict the distance between point-point/point-line/line-line.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Aligned Constrains the length of a line or the distance between two&#xA;  lines, a point on an object and a line, or two points on different&#xA;  objects.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Point &amp;amp; Line Selects a point and a line object. The aligned constraint&#xA;  controls the distance between a point and the closest point on a line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But when I try this constraint I am not able to select the line but the closest point on line gets auto selected. Now these parallel lines end at different points on sketch and in turn results in Constraint which just defines the distance between line end points which is not desirable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need perpendicular distance between two parallel lines to be constrained. How can I achieve this?&lt;/p&gt;&#xA;" OwnerUserId="8900" LastActivityDate="2018-10-30T09:00:32.137" Title="Autocad &quot;Aligned&quot; Constraint - distance between two parallel lines" Tags="|autocad|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19702" PostTypeId="2" ParentId="19689" CreationDate="2018-03-03T05:55:27.147" Score="0" Body="&lt;p&gt;If I were working at the public transport authorities office, my main concern would be the added complexity for travellers that you already mention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your suggestion would require more from travellers in terms of planning and understanding and would (I suspect) increase the probablity of some personality types not getting to their desired destination on time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see the benefit of more routes for travellers more inclined to plan and reason, but for others I'm not entirely sure that the reduced cost of less changes would outweigh the costs related to confusion, backtracking and/or delays in getting to the desired destination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that this might just be me underestimating travellers. But then again, the same might the case for the transport authorities.&lt;/p&gt;&#xA;" OwnerUserId="15157" LastActivityDate="2018-03-03T05:55:27.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19709" PostTypeId="2" ParentId="19689" CreationDate="2018-03-03T18:59:14.280" Score="1" Body="&lt;p&gt;Given that the six parallel lines in the centre of town actually consist of only two tracks (one eastbound and one westbound) running through &quot;the tunnel&quot; (Oslo T-bane or Oslo Tunnelbane) and that points or switches exist at the appropriate points (which they must to allow the existing route options) then your idea is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2Uwto.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Uwto.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Stortinget station, Oslo. Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Stortinget_(station)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;. Note that there are only two tracks.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines can handle 24 trains per hour in each direction giving a 2.5 minute headway (separation between trains). This is the system bottleneck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The convenience of the multiple direct route options must be contrasted with the simplicity of the single-route system. &quot;Take the red line to X and then take the blue line to Y&quot; will always work. With a 2.5 minute headway the timetable could be designed around a repeating pattern so that the offset between trains is predictable. This may allow passengers to commit the pattern to memory and know that there is a train at 2, 17, 32 and 47 minutes past the hour, for example, and that the journey will always take the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rolling stock is all the same type so there should not be any physical reasons why trains are restricted to one route.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-03-03T18:59:14.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19710" PostTypeId="2" ParentId="19677" CreationDate="2018-03-04T05:43:43.180" Score="2" Body="&lt;h2&gt;Steam locomotives use steam &lt;em&gt;pistons&lt;/em&gt;, not steam &lt;em&gt;turbines&lt;/em&gt;.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Gears/cogs would be pointless since there is no rotary source of power on steam locomotives.  They use steam pistons, which go back and forth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the physics worked out, direct-drive worked out really well with achievable values of piston diameter, stroke/eccentric, and wheel size.  &lt;em&gt;Until it didn't.&lt;/em&gt; And what got them was the curves.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Mainline pullers stuck with rods: way too big for gears&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As fully superheated boilers became very powerful, fast passenger locomotives used this power at higher speeds. For them, the side-rod design was perfect.  But slow lugging freight locomotives needed more weight on the rail to transfer the power at low speeds.  This required more driving axles to spread the weight. That made a single rigid group of driving axles too long for curves.  So they split into two (rarely, three) groups of driving axles. Power transfer was done with an engine on each group, usually simple, sometimes compound.  Union Pacific's Big Boy had 8 drive axles in two groups (each with a simple engine, still avoiding gears), handling curves like a 4-drive-axle locomotive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T6oa4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T6oa4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:28884.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;src&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Taken to absurdity.  The Virginian Railway finally gave up and electrified.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At these power levels, 4000-6000 horsepower, gear drive was out of the question: it was an order of magnitude too much power for gears. Even the electric GG1 of the era used &lt;em&gt;twelve&lt;/em&gt; massive pinions to transfer a similar amount of power to six axles. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Much smaller engines could be geared&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Mountain railroads used low power, lightweight locomotives which had to slither up fairly tight curves.  Even a very modest side-rod steam engine was too stiff for the curves. They also wasted a lot of precious weight on non-drive wheels, e.g. the pilot truck and the tender.  Ephraim Shay solved this problem with, indeed, geared locomotives.  Keep in mind these are small locomotives: the largest, Western Maryland #6, has boiler pressure of 200 psi and a top speed of 23 mph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ephraim Shay put a drive shaft along one side of the locomotive, gearing to each wheel.  The pistons directly cranked the drive shaft. Note the elaborate telescoping drive shafts, most especially important due to its off-center location.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qGmqA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qGmqA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SgqZm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SgqZm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Note the gears. &lt;a href=&quot;https://en.wikipedia.org/wiki/Geared_steam_locomotive&quot; rel=&quot;nofollow noreferrer&quot;&gt;sources&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Charles Heisler put the drive shaft down the locomotive centerline, and used a &quot;vee-twin&quot; piston arrangement.  Note the side rods: that means only one of the two axles is geared to the drive shaft, the side rods transfer power to the other axle. Side rods like that imply perhaps 100 horsepower per axle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wpxBk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wpxBk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Climax Manufacturing Co. took Heisler's centerline-shaft arrangement and added a cross-shaft and more gearing to put the steam pistons in an &lt;em&gt;almost&lt;/em&gt; conventional location. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UwmLF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UwmLF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having seen these geared-locomotive arrangements, you can see where they would not &quot;scale&quot; up to multithousand horsepower outputs.&lt;/p&gt;&#xA;" OwnerUserId="11673" LastEditorUserId="11673" LastEditDate="2018-03-04T05:54:44.393" LastActivityDate="2018-03-04T05:54:44.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19711" PostTypeId="2" ParentId="19689" CreationDate="2018-03-04T06:17:33.880" Score="6" Body="&lt;p&gt;Keep drawing combinations.  You forgot  Østerås to Vestli,  Østerås to Bergkrystallen, and every other combination of possible origin and destination.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This breaks down in a hurry.  By the time you split up all the trains to hit every destination, you end up with only a few trains a &lt;em&gt;day&lt;/em&gt; going where you want to go. Since you don't want to travel then, you wind up taking a 33  Østerås-Vestli then transferring at Majorstuen to a 21 Frognersetern-Mortensrud run.  &lt;em&gt;You still don't avoid the transfer, only now everything's more complicated.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Transfers are the better way&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Maybe you haven't tried this run, but when you have a binary pair of lines like the ones  you are looking at -- the far-and-easist way to provide all the routes you want it the &lt;strong&gt;timed, across-the-platform transfer&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Line 2 and 3 trains pull up at the same time to an island platform - the kind with tracks on each side.  They sit there, 30 feet apart, with their doors hanging open for 20-30 seconds.  Then when all the shuffling is done, the trains depart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That might happen at the first place they meet, Smestad, or it might happen at one of the downtown stations where there's platform space to do this.  You just walk 30 feet and sit down in the other train and done.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that designated transfer station, transfers are not guaranteed. They &lt;em&gt;might&lt;/em&gt; roll into Hellerud abeam. You &lt;em&gt;might&lt;/em&gt; be able to bolt across. I've done it.  But it's not a guaranteed timed transfer - if the other train slams the door in your face, tough beans.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only downside to this is when the routes are coincident for a good length, as in this diagram.  This means people traveling &lt;em&gt;within&lt;/em&gt; the overlapping section have twice the wait for two redundant trains.  It makes sense &lt;em&gt;for them&lt;/em&gt; if the two parallel lines (4 in your diagram actually) stagger themselves so the core zone gets very frequent service if  you don't want to go past Brynseng.  &lt;/p&gt;&#xA;" OwnerUserId="11673" LastActivityDate="2018-03-04T06:17:33.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19712" PostTypeId="1" CreationDate="2018-03-04T14:23:15.820" Score="0" ViewCount="76" Body="&lt;p&gt;I’m reading “Theory of Machines and Mechanisms” by Joseph Shigley. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page No. 118 - The author states that   A mechanism has as many instant centres as there are ways of pairing the link numbers. Thus, for an n-link Mechanism there are N = n(n-1)/2  instantaneous centres. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the following image having 3 links (one of them being the fixed reference frame), two of them being connected by a revolute joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IRfop.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IRfop.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I see it, there should be 2 instantaneous centres. One, point A itself about which each link can rotate. Second, some instantaneous centre in the plane about which this linkage system, as a whole, describes rotational motion at any given instant. &#xA;But by the formula the no. Should be 3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, where am I going wrong ?&lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2018-03-04T14:23:15.820" Title="No. Of Instantaneous centres" Tags="|mechanical-engineering|kinematics|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19715" PostTypeId="1" AcceptedAnswerId="19722" CreationDate="2018-03-05T02:16:18.867" Score="1" ViewCount="322" Body="&lt;p&gt;Suppose I have a 5kg framed picture that I want to hang on hooks, each of which can hold 2kg. Is there some arrangement of hooks that will make this safe? How many hooks would I need? I thought that I could arrange the hooks as something like a catenary to distribute the load, but I'm having trouble calculating the force on each hook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AWMnv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AWMnv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the techniques I need to use to solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="15182" LastActivityDate="2018-03-05T12:34:34.150" Title="Hang heavy picture with multiple hooks" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19716" PostTypeId="1" AcceptedAnswerId="19729" CreationDate="2018-03-05T02:21:19.327" Score="2" ViewCount="244" Body="&lt;p&gt;This question goes right back to the first principles of mechanics, so to speak. I'm confused as to the nature of forces applied to rigid bodies. Mathematically, a force acts on a particle of a known mass, but I don't see the mathematical connection that allows a rigid body of the same mass to behave similarly under the influence of the same force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that a rigid body can be obtained by extending the idea of a system with only a finite number of particles to a system with its number of particles tending towards infinity. Thus each particle in a rigid body will have a differential mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, what I don't understand is how a force applied at a particular location on a rigid body doesn't lead to an infinite acceleration due to the infinitesimal mass of the point at which the force is applied. Obviously this is not the case in real life, however this is what occurs when I try to take the definitions that surround systems of particles and try to build up rigid body mechanics around it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does a force applied to a rigid body simply imply that this force is spread over the body? This would mean that the definition of a force on a rigid body is different to that of a force on a system of particles (since if it is acting on the rigid body as a whole it is not acting at a single point), right? What am I missing? &lt;/p&gt;&#xA;" OwnerUserId="15183" LastEditorUserId="15183" LastEditDate="2018-03-05T03:05:14.990" LastActivityDate="2018-03-05T17:42:42.890" Title="Mathematical connection between systems of particles and rigid bodies" Tags="|mechanical-engineering|mathematics|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19717" PostTypeId="2" ParentId="19716" CreationDate="2018-03-05T07:07:06.810" Score="0" Body="&lt;p&gt;You are right to some degree. Two clarifications though, there is no such thing as a rigid object, nor is there a real zero area point to carry the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real life, a force applied to an object at a certain point means applied through a sharp enough object such as a tooltip, small sharp hard object which has a surface area, start to accelerate and move that molecules at that point, however, due to the existing bond between the molecules of the object that point will immediately radiate the stress and start a dynamic shock wave which will distribute the stress throughout the part. Ther are many ways of measuring the propagation of stress such as imaging or dying the surface with special paints. One of the causes of fatigue in metals is this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In solids: A force applied to a point in a solid object if strong enough and the impact is fast enough so that the stress cannot be spread fast will, compress a small part of the impacted area and push it into an indentation hole similar to a bullet hole. Or the stress wave will propagate through the part and may crash it along the way. Modern car frames take advantage of this as crumple zones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the part is brittle like rock or glass it will break. These are addressed in studies on the continuum medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In liquids: The same force applied faster than the viscosity and elasticity of the liquid can accommodate it will cause turbulence and create a shock wave opening into concentric circles or shooting out some of the liquid as a projectile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In gases: A sharp point load such as the tip of a wing will create vortices or shock waves that can be detrimental to the wing or propeller.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-05T07:07:06.810" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19719" PostTypeId="1" CreationDate="2018-03-05T08:44:05.613" Score="1" ViewCount="3558" Body="&lt;p&gt;I am designing a watch case and would like to know how to make an annular snap fit for the backplate. The material used is stainless steel. I could find references for Metal-plastic snap fit but I need both parts to be metal. I have seen a similar design in a Withings watch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PFA image link: &lt;a href=&quot;https://www.wareable.com/media/images/2015/01/img-6044-1421770737-9RIG-full-width-inline.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.wareable.com/media/images/2015/01/img-6044-1421770737-9RIG-full-width-inline.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For plastics I used : &lt;a href=&quot;http://fab.cba.mit.edu/classes/S62.12/people/vernelle.noel/Plastic_Snap_fit_design.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://fab.cba.mit.edu/classes/S62.12/people/vernelle.noel/Plastic_Snap_fit_design.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15186" LastActivityDate="2018-03-05T10:14:07.207" Title="How to make an annular metal to metal snapfit?" Tags="|mechanical-engineering|design|manufacturing-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19720" PostTypeId="2" ParentId="19719" CreationDate="2018-03-05T10:14:07.207" Score="1" Body="&lt;p&gt;An annular snap fit using inflexible material such as stainless steel (or aluminum!) will require a yield point in the ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allowing that any ring you can design will be a non-zero thickness cross section of a cylinder, the resulting item will resist compression (internal snap component) or will resist stretching (external snap component) and require excessive force to assemble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such assembly will likely damage one or both components. Disassembly will likely damage one or both components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider to provide a slot in one or both components, allowing for compression of the inside portion or expansion of the outside component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A common implementation of such a design is called a snap ring or circlip. This is a C-shaped piece of metal designed to engage a groove in another part. The metal is often a stamped out (cut/sheared) ring which will distort in the plane of the sheet when &quot;stretched&quot; open to engage the groove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would therefore have to allow for such distortion in your design by providing for such flex, although I am uncertain of the direction of flex if your C-shape uses cylindrical cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below comes from a &lt;a href=&quot;https://www.rotorclip.com/retaining_rings.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;commerical web site&lt;/a&gt; for this type of product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wcsWs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wcsWs.png&quot; alt=&quot;snaprings&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the rings have a proportionally large gap, while others are designed to close completely but allow for spread during operation.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-03-05T10:14:07.207" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19722" PostTypeId="2" ParentId="19715" CreationDate="2018-03-05T12:26:18.343" Score="3" Body="&lt;p&gt;I couldn't resist, so: First, you should note:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The tension in the string will be the same along its entire length (assuming frictionless conditions)&lt;/li&gt;&#xA;&lt;li&gt;To use the hooks optimally, you should arrange them that they carry equal load&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Taking this into consideration, the angle that the string bends at each hook  should be the same (the wall hooks should be spaced equally on the arc of a circle between the picture hooks, not a catenary as noted).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The picture below illustrates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ø = \frac{2θ}{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or more generally:&#xA;$$ø = \frac{2θ}{n}$$&#xA;where n = the number of hooks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d8Zv6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d8Zv6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the FBD at one of the frame hooks:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$F_t$ is the tension in the string.&lt;/li&gt;&#xA;&lt;li&gt;$F_p$ is the vertical force on the&#xA;frame hook.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5qnCP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5qnCP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, the tension in the string can be calculated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_t = \frac{F_p}{sin(θ)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The FBD at one of the hooks is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $F_h$ as the resultant force on the hook.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IsiJD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IsiJD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_h$ can be calculated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_h = 2F_tsin(0.5ø)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be generalised as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_h = \frac{2F_p}{sin(θ)}*sin(\frac{θ}{n})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this you can see that as $θ$ approaches $0$, $F_h$ approaches $2F_p/n$, but $F_t$ approaches infinity. Thus you should consider the tensile strength of the string and the forces that the frame and frame hooks can handle as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can hang your picture with 3 hooks with $θ = 62.11$ degrees, giving you $F_t = 28.285N$ and $F_h = 19.999N$. This will be &quot;safe&quot; assuming that the gravitational constant is 10, wihch it isn't, but that's just nit picking.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastEditorUserId="14440" LastEditDate="2018-03-05T12:34:34.150" LastActivityDate="2018-03-05T12:34:34.150" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19723" PostTypeId="1" CreationDate="2018-03-05T12:29:47.243" Score="1" ViewCount="81" Body="&lt;p&gt;We have a plastic part that is a receptacle for small ( &amp;lt; 1mm ) particles. It is a disk with a shallow cone shaped indent made using a drill. When a particle is placed in the receptacle it should fall to the centre, but sometimes it stays on the side due to friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The particles can be spherical, globular or circular flat, and are made of calcium carbonate :) They are dropped in one by one and the slope of the cone is about 20 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fix this, we are thinking the receptacle should be metal with teflon coating. However, is there any other coating we can apply to the plastic to increase its slipperiness? Perhaps rubbing graphite or a pencil lead in there? &lt;/p&gt;&#xA;" OwnerUserId="5001" LastEditorUserId="5001" LastEditDate="2018-03-05T13:34:20.993" LastActivityDate="2018-03-05T13:34:20.993" Title="How to make machined plastic part slippery for small dry particles" Tags="|friction|plastic|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19724" PostTypeId="1" CreationDate="2018-03-05T14:30:42.750" Score="4" ViewCount="195" Body="&lt;p&gt;I am looking at the specifications for thread lockers, and henkel/loctite specify &quot;Torque break&quot; and &quot;Torque prevail&quot; in in./lb. as shown in the link below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.na.henkel-adhesives.com/industrial/product-search-1554.htm?nodeid=8797951950849&amp;amp;msdsLanguage=EN_US&amp;amp;selectedTab=technical&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.na.henkel-adhesives.com/industrial/product-search-1554.htm?nodeid=8797951950849&amp;amp;msdsLanguage=EN_US&amp;amp;selectedTab=technical&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had expected to see the torque required to break the &quot;lock&quot;, and to continue turning in in.⋅lb. (or N⋅m) as described on wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thread-locking_fluid&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thread-locking_fluid&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does this &quot;Torque break&quot; value mean, and how is it calculated?&lt;/p&gt;&#xA;" OwnerUserId="15189" LastActivityDate="2018-03-05T16:35:59.117" Title="What is &quot;Torque Break&quot;?" Tags="|mechanical-engineering|torque|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19725" PostTypeId="2" ParentId="19716" CreationDate="2018-03-05T14:31:26.400" Score="0" Body="&lt;p&gt;The answer for most commonly used definition of rigid bodies is quite banal: It does not include any such thing. Instead an assumption (simplification) is made that the body is fully rigid, as in no part can move in relation to other parts. Hence the name rigid body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this the investigation goes further asking how such a element would be minimally defined. This eventually leads to a definition where the entire body is reduced to 6 degrees of freedom (3 in 2D), and they come up with moment of inertia to reduce the rotational energy to a higher order particle etc etc. So rigid bodies work like particles with different formulas governing their motion. Mainly handling rotation is added.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-03-05T17:07:43.200" LastActivityDate="2018-03-05T17:07:43.200" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19726" PostTypeId="1" AcceptedAnswerId="19793" CreationDate="2018-03-05T15:13:25.240" Score="2" ViewCount="252" Body="&lt;p&gt;I need to design a pivot arm which is pulled up by a float (buoyancy force). In the upper positions of the lever arm, there's a mass trying to pull down. &#xA;The reason that I'm presenting this problem is to consider all the variables (most significant) in order to build a low-cost prototype.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/APHbe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/APHbe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variables that I consider are:&#xA;1.- Volume of the float, the material is polypropylene. I need a low density but high volume float in order to generate more buoyancy force, is that correct?&#xA;2.- The distance between the force applied and the pivot &quot;d1&quot;. If I generate more torque, more will be the force generated by the buoyancy, right?&#xA;3.- The angle of the arm A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my first attempt presenting this problem, so any suggestion will work.  The goal is to achieve a bigger force than the 95 N generates by the &quot;object x&quot; in order to reach the 'sealed position' denoted by &quot;d2&quot;.&lt;/p&gt;&#xA;" OwnerUserId="14265" LastActivityDate="2018-03-09T08:21:36.127" Title="Buoyance+torque generated force" Tags="|mechanical-engineering|fluid-mechanics|torque|experimental-physics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19728" PostTypeId="1" CreationDate="2018-03-05T17:38:49.997" Score="0" ViewCount="40" Body="&lt;p&gt;I am trying to make use the CD4081B AND logic gate for starting basic Logic Gate circuits and I am trying to find the input current at 5 volts for the chip. I was wondering how to find it in the datasheet because I cannot. If anyone could help find it or show me how to calculate it myself that would be great. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the link to the CD4081B datasheet:&#xA;&lt;a href=&quot;http://www.ti.com/lit/ds/symlink/cd4081b.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ti.com/lit/ds/symlink/cd4081b.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15191" LastEditorUserId="1832" LastEditDate="2018-03-05T18:18:44.913" LastActivityDate="2018-03-05T18:56:30.803" Title="Where on the data sheet is the supply current at 5 volts for the CD4081B?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19729" PostTypeId="2" ParentId="19716" CreationDate="2018-03-05T17:42:42.890" Score="1" Body="&lt;p&gt;Suppose you have body with mass M experiencing a force F. If you model the force as being applied to the whole body, then this gives an acceleration F/M. Now suppose you cut this body into n equal parts. The part that the force is directly applied to will experience force F over mass M/n. So if you model the force as being applied just to one part, that appears to give an acceleration of nF/M. Contradiction? No. We're modeling this as a rigid body. So if one part accelerates, it must take the rest of the body with it. The rest of the body has mass M(n-1)/n. If this mass is accelerating at F/M, then it is experiencing a force of (F/M)(M(n-1)/n) = F(n-1)/n. So if we're modeling the force as applying only to one part, the rest of the body isn't being accelerated by this force, at least not directly. Instead, it must be being accelerated by the part that has the force applied to it. So that part is applying a force of F(n-1)/n to the rest of the body, and by Newton's Third Law, the rest of the body is applying a force of F(n-1)/n to that part. So the total resultant force for the part with an external force is the external force F minus the force F(n-1)/n from the rest of the body, which gives F/n. A force of F/n applied to a mass of M/n gives an acceleration of F/M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want, you can break this analysis down even further. Suppose you have a object of n sections, and you apply a force to a section. Then that section will experience the exterior force plus forces from the adjacent sections. Those sections will experience a force from that original section, plus forces &lt;em&gt;their&lt;/em&gt; adjacent sections. And so on. You can then set up a system of equations involving all these forces, relate the acceleration of each part to the net force it experiences, apply the constraint that the acceleration on each part is equal to the acceleration of each other part, and solve that system of equations. Intuitively, we should find that the acceleration for each part is F/M, regardless of how we divide the body into &quot;parts&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or we can simply recognize that we are dealing with a model to begin with, and different aspects of the model apply to different situations. We can treat the body as being made up of differential masses for such purposes as calculating moments of inertia, but a single mass for calculating acceleration from force. We can treat the body as being a point mass for purposes of calculating what gravitational force it creates, and yet treat forces as being displaced from the center of mass for purposes of calculating torque.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This would mean that the definition of a force on a rigid body is different to that of a force on a system of particles (since if it is acting on the rigid body as a whole it is not acting at a single point)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The definition of force is the same. What &lt;em&gt;model&lt;/em&gt; we use depends on the context. The term &quot;rigid body&quot; does not refer to any actual physical object. Forces are transmitted through an object at the speed of sound in that object, which cannot be higher than c. No object is truly &quot;rigid&quot;. When we talk about a &quot;rigid object&quot;, that is simply shorthand for &quot;an object that, in the scenario currently under consideration, deviates so little from the theoretical ideal of a rigid body that it can be modeled as a rigid body to the degree of accuracy required&quot;. This is similar to how we talk about applying the ideal gas law to gases, even though there is no such thing as an ideal gas in the real world. So when we model something as a rigid body, we treat all forces applied to it as being applied to its whole mass. When we model something as a system of particles, we model each force as being applied to a specific particle. The actual physical laws are the same in both cases, we're simply applying different &lt;em&gt;models&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that treating a force as being applied to a particular point is itself a simplification used in particular models. A full quantum mechanical analysis would have the particle not localized to a point, so the force would not be localized to a point. In fact, in quantum mechanics, forces are themselves carried by particles, and those particles have some positional spread.&lt;/p&gt;&#xA;" OwnerUserId="12911" LastActivityDate="2018-03-05T17:42:42.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19730" PostTypeId="2" ParentId="19642" CreationDate="2018-03-05T17:45:03.727" Score="1" Body="&lt;p&gt;I was able to reproduce the error: basically you cannot assign any NumPad key to AutoCAD 2015 or newer (2018 included) and in any flavour (Full, LT, OEM, etc.) when using a keyboard different from English (mine was Italian).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZBq6p.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZBq6p.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to &lt;strong&gt;workaround&lt;/strong&gt; this bug, change/add an &lt;strong&gt;English keyboard BEFORE starting your AutoCAD product&lt;/strong&gt; and then you'll be able to assign the correct Numpad key in the &lt;strong&gt;Customize User Interface&lt;/strong&gt; dialog. It will appear as &lt;strong&gt;NUMPAD8&lt;/strong&gt; and it will work even after deleting the English keyboard (if not used).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3GCAu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3GCAu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15102" LastActivityDate="2018-03-05T17:45:03.727" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19731" PostTypeId="2" ParentId="19728" CreationDate="2018-03-05T18:56:30.803" Score="1" Body="&lt;p&gt;If you mean the current drawn by the chip in quiescent / static conditions then it is given in an alternative &lt;a href=&quot;http://www.futurlec.com/4000Series/CD4081.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;datasheet by Fairchild&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TyZZX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TyZZX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Quiescent device current is very low. (1) The device current. (2) The input current.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally the input resistance is so high that we ignore it. The situation changes when we feed it with a high frequency signal. The input capacitance can become a significant factor as it has to be charged and discharged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EUsyA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EUsyA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 2. (3) Input capacitance.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-03-05T18:56:30.803" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19732" PostTypeId="1" AcceptedAnswerId="21215" CreationDate="2018-03-05T22:59:30.893" Score="2" ViewCount="575" Body="&lt;p&gt;What kind of motor is suitable for the below requirements?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;2700 rpm, needs to be accurate, not +- few rpms. I think the best approach is to use a hall effect sensor and adjust current/voltage on the fly? Or a better idea?&lt;/li&gt;&#xA;&lt;li&gt;Be as noiseless as possible. The noise 12V square fan motors make is perfect.&lt;/li&gt;&#xA;&lt;li&gt;Torque is not much of an issue as even though it is rotating a 100g load on top of it the load is supported by 4 bearings.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;What motor am I looking for?&lt;/p&gt;&#xA;" OwnerUserId="14841" LastEditorUserId="8055" LastEditDate="2018-03-06T19:19:56.107" LastActivityDate="2018-04-06T01:52:07.030" Title="a small noiseless motor with accurate RPM control?" Tags="|motors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19733" PostTypeId="2" ParentId="2774" CreationDate="2018-03-06T07:05:57.883" Score="2" Body="&lt;p&gt;Here is the formula: &lt;a href=&quot;https://www.researchgate.net/post/how_to_calculate_cutting_force_required_by_hacksaw_for_particular_work_piece_for_particular_hardness&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.researchgate.net/post/how_to_calculate_cutting_force_required_by_hacksaw_for_particular_work_piece_for_particular_hardness&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fc = zc&lt;em&gt;kz&lt;/em&gt;A*f&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fc = cutting force in kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zc=number of contact teeths (this number is maximun when the saw reach the diameter of the bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;kz=especific pressure of cutting in kg/mm2 (this value depend of the cutting material).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A=width of saw in mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = feed per stroke in mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visit website for more info - &lt;a href=&quot;https://www.acealloysllp.com/stainless-steel-bending.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.acealloysllp.com/stainless-steel-bending.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps&lt;/p&gt;&#xA;" OwnerUserId="14994" LastActivityDate="2018-03-06T07:05:57.883" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19734" PostTypeId="1" CreationDate="2018-03-06T08:28:11.060" Score="2" ViewCount="71" Body="&lt;p&gt;Below I attached two BLDC motor commonly used for skate boards, I am confused with the calculation of stall torque for these two motors,What i need to know,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the maximum weight lifting capacity in the condition fig 1 and 2(x and y in kg)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is maximum lifting capacity is just below stall torque?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In figure 1 and 2 they mention torque as 1-2N in yellow motor and 1-2N/N in red motor what they are?Is it a mistake?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In red motor speed is RPN is it a typing mistake?&lt;a href=&quot;https://i.stack.imgur.com/slE1t.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/slE1t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15193" LastEditorUserId="9" LastEditDate="2018-03-06T09:42:05.773" LastActivityDate="2018-03-06T09:42:05.773" Title="Mechanical engineering, Please help me finding the torque and answering other doubts too" Tags="|mechanical-engineering|motors|automotive-engineering|torque|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19735" PostTypeId="2" ParentId="19689" CreationDate="2018-03-06T09:54:25.893" Score="0" Body="&lt;p&gt;Imagine this simplified scenario:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You have 4 stations: A, B, C and D.&lt;/li&gt;&#xA;&lt;li&gt;There are 2 trains, one is travelling only on line A-B and the other on line C-D.&lt;/li&gt;&#xA;&lt;li&gt;There are 2 people at station A – one wants to go to B and the other wants to go to D.&lt;/li&gt;&#xA;&lt;li&gt;There are 2 people at station C – one wants to go to B and the other wants to go to D.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ODUaX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ODUaX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The passengers at A both hop on train A-B and those at C hop on train C-D. When the trains meet at the central line, the passenger on train A-B wanting to go to D switches to train C-D and the passenger on train C-D wanting to go to B switches to train A-B. Then the trains continue to their respective stations and drops everyone off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If each passenger is lazy and doesn’t want to switch trains, You’ll either need to invest in double the amount of trains, or each passenger will need to wait twice as long for his/her train because the trains need to switch routes between each cycle to satisfy the passengers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing that mandatory stops are made on the central line where people will hop on and off the trains in anyway, why not use this opportunity to optimise the rest of the system?&lt;/p&gt;&#xA;" OwnerUserId="14440" LastActivityDate="2018-03-06T09:54:25.893" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19736" PostTypeId="2" ParentId="19732" CreationDate="2018-03-06T11:46:29.207" Score="0" Body="&lt;p&gt;I suggest a synchronous motor in order to obtain desired the rpm, it runs at a constant speed irrespective of and load there are various methods to control speed. The sound produced is dependent on the design-friction and others things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In synchronous motors the stater and rotor rotates at the same speed. However, you may need to use a feed back system, because even it works in magnetic locking it drops rpm when loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure about the power supply nature either AC or DC, this was not mentioned in the question?&lt;/p&gt;&#xA;" OwnerUserId="15193" LastEditorUserId="90" LastEditDate="2018-03-06T15:37:34.600" LastActivityDate="2018-03-06T15:37:34.600" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19737" PostTypeId="2" ParentId="3616" CreationDate="2018-03-06T13:17:02.077" Score="4" Body="&lt;p&gt;There are good answers here, but just wanted to clarify some points.  It's important to consider what the liquid (in this case water) is used for in the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin with, it helps to see that the majority of pumps are made of two main components - an electric motor, and the actual pump (impeller and volute) itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the pump itself, the water is used to lubricate the seal (normally a mechanical seal these days not gland packing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric motor will require some form of cooling.  On most motors a fan is fitted to the NDE (non-drive end) to blow air along the cooling fins on the side of the motor.  Where the motor is required to be fully submerged, the motor is fully sealed (to IP66 or above) and the liquid is required to transfer heat away from the motor.  In general you can ascertain whether a motor is air-cooled or liquid-cooled by the presence of a fan - if it has one it is probably air cooled.  Therefore, a submerged motor should not be run for more than a few seconds without being immersed to prevent over-heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point made earlier that a dry pump will over-run and cause damage to the rotating elements (bearings, etc) is also valid.&lt;/p&gt;&#xA;" OwnerUserId="15195" LastActivityDate="2018-03-06T13:17:02.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19739" PostTypeId="1" AcceptedAnswerId="19753" CreationDate="2018-03-06T14:34:56.807" Score="2" ViewCount="2752" Body="&lt;p&gt;Simple one. Some extension springs are wound with an initial tension which acts to pull the coils together, even when not under external loading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When calculating the preload in the spring, should the initial tension be added to the spring force (e.g. initial length x spring rate)? I have seen a few examples and some people include it, others don't. Which is correct? &lt;/p&gt;&#xA;" OwnerUserId="14498" LastActivityDate="2018-07-03T13:29:38.673" Title="Calculating pretension of a spring with an initial tension" Tags="|springs|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19742" PostTypeId="1" AcceptedAnswerId="19747" CreationDate="2018-03-06T20:29:39.503" Score="2" ViewCount="71" Body="&lt;p&gt;I'm not entirely sure if this is the right place to ask this, so apologies in advance if it doesn't belong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a curved surface with many small, irregular holes in it that I would like to measure. The surface is in a location that makes it awkward to work with, and is effectively immobile. Are there any technologies or techniques that could let me take a 1:1 imprint or copy that is more portable? Being able to 'flatten' the copy would be convenient but not required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface is thin and the reverse is accessible (though again with some difficulty) if at all applicable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Specifically, i'm trying to gauge the accuracy of another measuring process, so computer vision techniques are undesirable.&lt;/p&gt;&#xA;" OwnerUserId="15200" LastEditorUserId="15200" LastEditDate="2018-03-06T21:40:16.447" LastActivityDate="2018-03-06T23:19:42.257" Title="Measurement of immobile curved surface" Tags="|measurements|data|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19743" PostTypeId="2" ParentId="18699" CreationDate="2018-03-06T20:32:43.527" Score="1" Body="&lt;p&gt;One way to do this would be using permanent magnets arranged in a circle spaced out on a base plate with a central pivot point for a top plate (your said wheel) that has a ferrous disc mounted in it. Use the magnets with either pole facing up. As you rotate the wheel, the ferrous disc will be attracted to whichever magnet it is closest to, you can space the magnets to give you any detent pattern you like.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2018-03-06T20:32:43.527" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19745" PostTypeId="2" ParentId="19742" CreationDate="2018-03-06T21:26:28.693" Score="0" Body="&lt;p&gt;You can mark a masking tape in inches or centimetres and tape it along the visual prominent features, such as along the ridge, or length or any axis that would yield more info as to the dimension and geometry of your shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using those measurement you can build a 3d model in a drafting program like Autocad or free wares such as  google Sketchup. Then you have a model of your shape which you can measure in any direction or print a 3d print-out  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-06T21:26:28.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19746" PostTypeId="1" CreationDate="2018-03-06T22:01:05.357" Score="1" ViewCount="25" Body="&lt;p&gt;Consider the system in the (drawn in Paint) image. Assume that there is an input acceleration at the ground defined by a PSD. This acceleration can be in any axis. Points A and B are initially aligned and just offset vertically. You are shining a laser through them and need to know what the maximum possible misalignment is to make sure your optics can handle it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Evih.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Evih.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if I have an FEM model, I can find the displacement PSD's in each direction at both points. However, these PSD's are still a stochastic representation of the displacement. I have a few questions because of that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do the output PSD's represent some confidence interval? Or are they the same level of confidence as the input? (i.e. if the input is an envelope describing the maximum possible input acceleration, this describes the maximum possible output displacement)&lt;/li&gt;&#xA;&lt;li&gt;What features of the PSD can be used to decide a maximum? Does the RMS describe this? And in that case, would it be $rms_{A} + rms_B = disp_{max}$ because each could be off by that much in the worst case? Then, you'd do some vector addition to account for the multiple directions of misalignment to get an absolute misalignment. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9073" LastActivityDate="2018-03-06T22:01:05.357" Title="How to find the maximum misalignment from vibration?" Tags="|modeling|vibration|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19747" PostTypeId="2" ParentId="19742" CreationDate="2018-03-06T23:19:42.257" Score="1" Body="&lt;p&gt;If the holes are small enough and you don't care about spacing measurements, you could use what are called pin gauges. These will obviously not work on non-circular holes (irregular?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QV5Ty.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QV5Ty.jpg&quot; alt=&quot;pin gauges&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are available in sets of various ranges. The precision is quite good, often measured in tenths of a thousandths of an inch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is excessive in terms of precision and especially in terms of cost, there's a much less elaborate method you can consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I initially planned to suggest modeling clay, but it has limits with respect to handling. You would flatten a sheet of the stuff and press it against the surface with sufficient force to create bumps as it enters the holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once peeled off, you should be able to measure the bumps to get the hole sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it would be difficult to handle, however, without breaking it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an alternative, especially as you have access to the inside/reverse, is to use a silicone molding compound of some high Shore value (hardness). The clay might be useful to create an inner dam as well as a container of sorts for the silicone on the outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicone molding material, especially the high hardness stuff, is quite durable. This should allow the portability and handling you require. Some compounds have minimal shrinkage, allowing for reasonable precision. It may be necessary to test sample holes to determine exact shrinkage levels, but it would be consistent across a specific range of diameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expanding from the above suggestion would be to seal off the holes with adhesive tape rather than modeling clay or other dam material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ideal situation would be that your curved surface has a horizontal orientation, minimizing the damming requirement and the amount of silicone used.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-03-06T23:19:42.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19748" PostTypeId="1" CreationDate="2018-03-07T00:52:30.943" Score="-1" ViewCount="83" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7clz0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7clz0.jpg&quot; alt=&quot;Image of circuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wire is attached to the bottom of each D battery.&#xA;I already know that I have to put the bottom of the light bulb on the positive side of the battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My multimeter looks like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ovSOQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ovSOQ.jpg&quot; alt=&quot;Multimeter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure where to put the testing probes. I don't think this is important, but the wire is made of steel wool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I can get some help.&lt;/p&gt;&#xA;" OwnerUserId="15105" LastEditorUserId="7083" LastEditDate="2018-03-28T02:18:13.913" LastActivityDate="2018-03-28T02:18:13.913" Title="How would I measure this battery circuit with a multimeter?" Tags="|electrical-engineering|measurements|circuits|battery|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19750" PostTypeId="2" ParentId="19748" CreationDate="2018-03-07T01:21:53.957" Score="2" Body="&lt;p&gt;If you trying to measure voltage, connect the black connector probe (Terminal) to COM. Move the rotatory switch 20 DCV. The dot should point to 20 DCV. Then place the black exposed probe to negative and red expose probe to positive button on the battery. You should read somewhere around 1.5V for the D size battery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WaUHc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WaUHc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Checkout some of the links below. This should help you with your questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://learn.sparkfun.com/tutorials/how-to-use-a-multimeter/measuring-voltage&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to Use a Multimeter?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=GZX3MyBkMvA&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to Use a Multimeter: Measuring Voltage&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-03-07T01:41:21.380" LastActivityDate="2018-03-07T01:41:21.380" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19751" PostTypeId="1" CreationDate="2018-03-07T01:42:07.540" Score="2" ViewCount="1997" Body="&lt;p&gt;I need to seal a rotating shaft that protrudes through a wall in a 3D-printed enclosure I am designing for a school project. The enclosure is for what will be a small generator that will be used in all weather conditions, so I'd like to make the shaft waterproof. To clarify, it is not the motor shaft that needs to be sealed, it's just a 1/4&quot; diameter metal shaft which connects to some gears in the enclosure. Do I need anything more than a waterproof bearing (which I will be buying, not 3D printing) press fit in the enclosure wall? Such as, do I need some type of rotary seal as well? Is a press fit setting for the bearing enough to prevent water intrusion between the bearing's outer diameter and the enclosure wall? And how should I design the enclosure wall to hold the bearing (i.e. Is all I have to do make the wall the same thickness as the bearing? Does it need some sort of lip to hold the bearing?)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have searched for a long time online about how to do this, but nothing I've found is very helpful or is too confusing for me to understand. Any help is greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="15205" LastActivityDate="2018-03-07T16:30:06.823" Title="Sealing a rotating shaft in a 3D-printed part" Tags="|bearings|3d-printing|waterproofing|seals|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19752" PostTypeId="1" AcceptedAnswerId="19781" CreationDate="2018-03-07T03:10:22.963" Score="3" ViewCount="91" Body="&lt;p&gt;I am making a tea nitrogen laser based off the rimstarorg youtube channel. I have a perfect setup as seen below&#xA;&lt;a href=&quot;https://i.stack.imgur.com/h4Ypp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h4Ypp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I use 11, 1kv diodes to convert my 7,500 volt lug on my neon sign transformer to DC. My dielectric, well, i tried everything from transparency paper, to mylar, to acrylic and i tried multiple layers for each one.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8dYyW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8dYyW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fUHLg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fUHLg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Yeah thats what happened. So I eventually just tried setting up a piece of aluminum foil on the bottom, my dielectric, and then one aluminum sheet with my spark gap, and for some reasonthe spark jumps between the bottom of my 2 angle pieces before it jumps between the TWO CLOSEST POINTS AKA THE SCREW THATS 1mm AWAY!!&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KoyiC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KoyiC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I should also mention when i used just one piece of transparency paper this happened:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/NSQJm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NSQJm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Please help me on where to even go from here, i don't even know what I'm doing wrong at this point&lt;/p&gt;&#xA;" OwnerUserId="15207" LastActivityDate="2018-03-07T22:37:49.387" Title="Tea nitrogen laser debugging" Tags="|lasers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19753" PostTypeId="2" ParentId="19739" CreationDate="2018-03-07T09:25:17.917" Score="2" Body="&lt;p&gt;After being pointed in the correct direction by an acquaintance I think I have the answer now: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8JZFF.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8JZFF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that initial tension should be included when calculating load in an extension spring, as shown in the formulae above. However, in some applications the initial tension may be insignificant compared with the load at a given extension. People either ignore it due to it being negligible, or in error. &lt;/p&gt;&#xA;" OwnerUserId="14498" LastActivityDate="2018-03-07T09:25:17.917" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19755" PostTypeId="2" ParentId="19751" CreationDate="2018-03-07T10:41:46.260" Score="2" Body="&lt;p&gt;Being able to create a waterproof seal on a shaft would also have to take into consideration the water pressure on the outside of the assembly. In this case, waterproofing is more weatherproofing, reducing that aspect by a considerable amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest that your waterproof bearing, unless otherwise specified, is not a water-sealing bearing. If the specifications reference resistance to water pressure, you may have a good solution, otherwise no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the boat world, one will find shaft drive systems in which the motor/engine is inside the hull and drives the propeller by a shaft which passes through the hull. The device which prevents the boat from sinking due to water ingress is called a stuffing box or a packing box. A search for those terms results in many hundreds of links.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general design of a stuffing box is a series of seals restricting the flow and passage of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that your shaft passes through a metal tube slightly larger in diameter and that tube is affixed in the wall of your 3D printed enclosure. One would choose an inside diameter for the tube to match that of a selected O-ring size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The O-ring would be sized for a 1/4&quot; inside diameter, and for example's sake, a 3/8&quot; outside diameter. Your tubing would be also 3/8&quot; inside diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carefully lubricate the O-rings with a silicone based grease and guide them over the shaft and press the shaft into the tube. Using at least two O-rings will provide for reasonably good protection from water ingress due to weather.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WTuZg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WTuZg.jpg&quot; alt=&quot;stuffing box&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above image is for a boat and is larger and more complex than you need for your project. The packing rings in the diagram represent the O-rings on your shaft.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-03-07T10:41:46.260" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19756" PostTypeId="1" AcceptedAnswerId="19757" CreationDate="2018-03-07T11:16:01.540" Score="2" ViewCount="82" Body="&lt;p&gt;I see tractors with really high wheels and with the carriage placed really high. Similar to this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YE4gN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YE4gN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are the wheels powered? Is there is an electric motor for every wheel powered from diesel generator somewhere in tractor?&lt;/p&gt;&#xA;" OwnerUserId="15210" LastEditorUserId="7083" LastEditDate="2018-03-10T23:22:41.690" LastActivityDate="2018-03-10T23:22:41.690" Title="How do tractors with a highly placed carriage power their wheels?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19757" PostTypeId="2" ParentId="19756" CreationDate="2018-03-07T11:19:40.663" Score="5" Body="&lt;p&gt;The wheels are powered by hydraulic motors, one for each wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pipes for supply and return can be seen on the left side wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;It could be possible to provide an &quot;ugly&quot; (term as used in the comment below) solution with driveshafts, but, since driveshafts either with Hardy-Spicer joints or Constant Velocity joints have a maximum working angle then they would eat into the clearance available under the machine which is necessary for the crop to pass under without damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one did use the driveshaft solution this would increase the height of the machine and, if the wheelbase stays the same, would increase the chances of it falling over when used on sloping terrain - not all countries are blessed with large flat fields...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't they use electric motors? Well, a good guess is that most agricultural mechanics are already well-versed in hydraulics (all the ones I have met and worked with) and are more than capable of sorting most, if not all, hydraulic problems : spool valve blocks, oneway valves, pumps, motors etc. Going to electric would give them a completely new ball game and they would probably avoid purchasing machines and stick with what they know...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-03-07T16:45:34.700" LastActivityDate="2018-03-07T16:45:34.700" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19758" PostTypeId="1" AcceptedAnswerId="19759" CreationDate="2018-03-07T12:05:36.320" Score="42" ViewCount="63081" Body="&lt;p&gt;I found this &lt;a href=&quot;https://engineering.stackexchange.com/a/295/15211&quot;&gt;answer&lt;/a&gt; to a related question. The part of the answer that's confusing me is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Transmitting DC power over a long distance is inefficient. Thus AC supply is a far more efficient to transmit power.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to Siemens it's quite the &lt;a href=&quot;https://www.energy.siemens.com/hq/en/power-transmission/hvdc/applications-benefits/configurations/long-distance-power-transmission.htm&quot; rel=&quot;noreferrer&quot;&gt;opposite&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Whenever power has to be transmitted over long distances, DC transmission is the most economical solution compared to high-voltage AC.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also, from &lt;a href=&quot;https://en.wikipedia.org/wiki/High-voltage_direct_current#Advantages&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;HVDC transmission losses are quoted as less than 3% per 1,000 km, which are 30 to 40% less than with AC lines, at the same voltage levels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is the posted &lt;a href=&quot;https://engineering.stackexchange.com/a/295/15211&quot;&gt;answer&lt;/a&gt; correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;- - EDIT - -&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Chris H&lt;/strong&gt; made a very important observation (see his comment below): The context of the post I mentioned was of &lt;strong&gt;low voltage&lt;/strong&gt; whereas I was blindly thinking of high voltage. Indeed I learnt loads by the answers and comments. Thanks.  &lt;/p&gt;&#xA;" OwnerUserId="15211" LastEditorUserId="15211" LastEditDate="2018-03-08T15:05:36.947" LastActivityDate="2023-04-18T06:43:41.807" Title="Transmitting power over long distances what is better AC or DC?" Tags="|power-transmission|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19759" PostTypeId="2" ParentId="19758" CreationDate="2018-03-07T13:12:32.423" Score="45" Body="&lt;p&gt;It is more efficient to &lt;i&gt;transmit&lt;/i&gt; DC using about the same infrastructure.  This is because of several effects:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Skin effect experienced with AC.  There is no skin effect with DC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;Higher voltage allowed with DC for the same transmission lines.  The lines have to withstand the peak voltage.  With AC, that is 1.4 times higher than the RMS.  With DC, the RMS and peak voltages are the same.  However, the power transmitted is the current times the RMS, not peak, voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;No radiation loss with DC.  Long transmission lines act as antennas and do radiate some power.  That can only happen with AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;No induction losses.  The changing magnetic field around a wire carrying AC current causes induced voltage and current in nearby conductors.  In effect, the transmission line is the primary of a transformer, and conductors near it are secondaries.  With DC current, the magnetic field doesn't change and therefore doesn't transfer power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage of DC is that it doesn't require synchronization between grids.  Two AC grids need to be phase-synced to be connected together.  This gets tricky when the distances are large enough to be significant fractions of a cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flip side is that AC is easier to convert between voltages.  Converting DC back to AC to dump it onto the local grid at the receiving end is not a trivial process.  It takes a large plant to do this, which means significant expense.  That expense is only worth it if the transmission distance is long enough so that the efficiency savings outweigh the cost of the DC-AC conversion plant over its lifetime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a example of what it takes to convert high voltage DC back to AC:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Iq1qJ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iq1qJ.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DC power from large dams in Quebec enters at top right.  This plant converts that to AC and dumps the power onto a large regional AC transmission line in Ayer Massachusetts at &lt;a href=&quot;https://www.openstreetmap.org/#map=17/42.5702/-71.5242&quot; rel=&quot;noreferrer&quot;&gt;42.5702N 71.5242W&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expense of building and operating this plant is worth it due to the significant power savings of transmitting DC instead of AC.  Synchronization was also a factor in using DC.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="6927" LastEditDate="2018-03-07T23:28:43.380" LastActivityDate="2018-03-07T23:28:43.380" CommentCount="16" ContentLicense="CC BY-SA 3.0" />
  <row Id="19762" PostTypeId="1" AcceptedAnswerId="19797" CreationDate="2018-03-07T14:27:55.693" Score="0" ViewCount="363" Body="&lt;p&gt;this is my first project that i am working on. I want to make a IR sensor mount using stepper motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SrDyc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SrDyc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seen this on youtube and would like to make a similar design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Wn8H1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wn8H1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plan on using aluminium mounting hubs to transmit the torque from the shafts to the aluminium frame as shown above. My question is how is the torque transmitted from the motors to this camera mount? Is there a bearing and if so, what kind? &lt;/p&gt;&#xA;" OwnerUserId="15061" LastActivityDate="2018-03-08T14:33:25.417" Title="Pan and tilt mechanism" Tags="|motors|torque|friction|bearings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19763" PostTypeId="2" ParentId="19686" CreationDate="2018-03-07T14:33:08.077" Score="2" Body="&lt;p&gt;At one atmosphere of pressure water boils at a constant 212 F as you add heat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat is used for the state change from liquid to vapor.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-03-07T14:33:08.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19765" PostTypeId="2" ParentId="19758" CreationDate="2018-03-07T14:46:09.407" Score="47" Body="&lt;p&gt;I actually worked on HVDC schemes, back in the mid-to-late 90s.  Olin Lathrop's answer is partially right, but not quite.  I'll try not to repeat too much of his answer, but I'll clear up a few things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The losses for AC primarily come down to the inductance of the cable.  This creates &lt;a href=&quot;https://en.wikipedia.org/wiki/Electrical_reactance&quot; rel=&quot;noreferrer&quot;&gt;reactance&lt;/a&gt; for AC power transmission.  A common misconception (repeated by Olin) is that this is due to transferring power to things around it.  It isn't - a coil of wire halfway between here and the Magellanic Cloud will have precisely the same reactance and cause precisely the same electrical effects sat on your desk.  For this reason, it's called &lt;a href=&quot;https://en.wikipedia.org/wiki/Inductance&quot; rel=&quot;noreferrer&quot;&gt;self-inductance&lt;/a&gt;, and the self-inductance of a long transmission cable is really significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cable does not lose any significant power from &lt;a href=&quot;https://en.wikipedia.org/wiki/Inductive_coupling&quot; rel=&quot;noreferrer&quot;&gt;inductive coupling&lt;/a&gt; with other metalwork - this is the other half of that common misconception.  The effectiveness of inductive coupling is a function of the AC frequency and the distance between the cables.  For AC transmission at 50/60Hz, the frequency is so low that inductive coupling at any kind of distance is utterly ineffective; and unless you want to get electrocuted, those distances have to be several metres apart.  This just does not happen to any measurable extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edited to add one thing I forgot) For cables running underwater, there are also very high cable capacitances due to their construction.  This is a different source of reactive losses, but is significant in the same way.  These may be the  dominant cause of losses in underwater cables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Skin_effect&quot; rel=&quot;noreferrer&quot;&gt;Skin effect&lt;/a&gt; does cause higher resistance for AC power transmission, as Olin says.  In practise though, the need for flexible cables makes this less of an issue.  A single cable thick enough to transmit significant power would generally be too inflexible and unwieldy to hang from a pylon, so transmission cables are assembled from a bundle of wires held apart with spacers.  We'd need to do this anyway, whether we were using DC or AC.  The result of this though is to put the wires within the skin effect zone for the bundle.  Clearly there is engineering involved in this, and there will still be some losses, but by this happy coincidence we can make sure they're a lot lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buried and submarine cables are a single thick cable, of course, so in principle they could still get bitten by the skin effect.  Heavy-duty cable construction though will generally use a strong central core which provides structural integrity for the cable, with other connectors wound onto that core.  Again, we can use that to our advantage to reduce skin effect in AC, and even HVDC cables will be built the same way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big win in power transmission though is eliminating reactive losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Olin says, there is also an issue with joining two power grids together, because they will &lt;em&gt;never&lt;/em&gt; be exactly the same frequency and phase.  Clever use of filters in the mid-20th century did allow connection of grids, but designing these was as much art as science, and they were inherently inefficient.  Once you've got your power transmitted in DC though, you can reconstruct AC with the exact same frequency and phase as the destination grid, and avoid the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only that, but it is much more efficient to convert from AC to DC and back to AC again, instead of trying to use filters to compensate for phase and frequency.  Grids these days are commonly joined with &lt;a href=&quot;https://en.wikipedia.org/wiki/High-voltage_direct_current#Back_to_back&quot; rel=&quot;noreferrer&quot;&gt;back-to-back schemes&lt;/a&gt;.  These are essentially both halves of an HVDC link next to each other, with an enormous busbar between the two instead of kilometres of transmission cable.&lt;/p&gt;&#xA;" OwnerUserId="15214" LastEditorUserId="15214" LastEditDate="2018-03-07T18:16:26.717" LastActivityDate="2018-03-07T18:16:26.717" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="19766" PostTypeId="1" CreationDate="2018-03-07T15:54:11.883" Score="2" ViewCount="70" Body="&lt;h2&gt;What I have learned&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The detectors of the PET actually measure the photons, which are caused by the radioactive tracer injected into the patient. Then, with the help of backprojection, we get the photon count for each voxel. I hope I understood these processes right, please correct me if this is not the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as final result, we get the radioactivity concentration in kBq/mL. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;How can the scanner convert the counted photons into radioactivity concentration? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess would be, that a phantom of known volume is used, with an inserted radiotracer of known radioactivity concentration.&lt;/p&gt;&#xA;" OwnerUserId="15224" LastEditorUserId="615" LastEditDate="2018-03-07T16:17:06.143" LastActivityDate="2023-03-23T08:04:50.570" Title="Positron Emission Tomography: How to get from photon count to radioactivity?" Tags="|biomedical-engineering|medical-devices|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19767" PostTypeId="2" ParentId="19751" CreationDate="2018-03-07T16:30:06.823" Score="0" Body="&lt;p&gt;If your shaft only needs to be fluid-sealed during operation, consider the approach taken in the Wankel engine as produced by Mazda in their RX-series cars.  They essentially reverse-threaded the shaft just inside the bulkhead point. Once the engine is running, fluid (engine oil in the Mazda case) is hurled away from the bulkhead by the threads -- more or less Archimedes' Spiral.   They (Mazda, or anyone using this approach for an internal combustion engine) can get away with this because the oil level is well below the crankshaft when the engine is not running. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-03-07T16:30:06.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19770" PostTypeId="1" CreationDate="2018-03-07T16:53:43.640" Score="0" ViewCount="28" Body="&lt;p&gt;I have made a CNC machine from CD drive and want it to use for engraving purpose. which laser would be best for this job?&lt;/p&gt;&#xA;" OwnerUserId="15229" LastActivityDate="2018-04-06T19:24:20.420" Title="Suitable laser for Engraving" Tags="|lasers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19771" PostTypeId="1" AcceptedAnswerId="19777" CreationDate="2018-03-07T17:11:54.707" Score="3" ViewCount="3612" Body="&lt;p&gt;Today while driving I had to make a quick shift to a lower gear that was not as smooth as it should've been. I've been taught that while shifting to a lower gear you have to be more careful when releasing the clutch than while shifting a gear up. Why is this? I understand the basic mechanism of transmission and clutch. When shifting up, the car can &quot;bump&quot; slightly if the rotational speeds of the engine and wheels are far apart. But why does this effect seem more noticeable when shifting &lt;strong&gt;down&lt;/strong&gt;?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastEditorDisplayName="user4139" LastEditDate="2018-06-10T05:46:03.650" LastActivityDate="2018-06-10T05:46:03.650" Title="Manual transmission: Why downshifting is less smooth than shifting up?" Tags="|automotive-engineering|engines|transmission|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19772" PostTypeId="2" ParentId="19771" CreationDate="2018-03-07T17:40:06.803" Score="-1" Body="&lt;p&gt;One factor is that engine braking is stronger in lower gears - releasing the clutch suddenly into a lower gear has the effect of applying a sudden braking force to the car, causing a jolt that you can feel as the engine spins up to match the speed of the wheels. When moving into a higher gear, the engine braking is reduced compared to your previous gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be a jolt due to the sudden engine braking effect whenever the engine is spinning slower than ideal, and has to be 'dragged up to speed'. This is much more likely to happen when downshifting than upshifting.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-03-07T17:40:06.803" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19773" PostTypeId="2" ParentId="19726" CreationDate="2018-03-07T17:40:24.140" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DZKjj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DZKjj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to increase the buoyancy force over than the force applied by the weight of the object x, I'm considering increasing the distance &quot;d1&quot; or maybe decrease distance &quot;d2&quot;, which is the distance between the force applied by the weight and the pivot point. &#xA;So, the momentum in point A (pivot) generated by force of 95.6 N will be lower than the buoyancy force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other suggestion?&lt;/p&gt;&#xA;" OwnerUserId="14265" LastActivityDate="2018-03-07T17:40:24.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19774" PostTypeId="2" ParentId="19758" CreationDate="2018-03-07T17:47:58.877" Score="7" Body="&lt;h2&gt;They're talking about complexity and costing (\$\$\$\$\$)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The people saying &quot;DC is less efficient&quot; are using the word &quot;efficiency&quot; to talk about design factors like complexity of conversion hardware, and more critically, &lt;strong&gt;its cost&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have a Santa Claus machine that can pop out DC/DC converters as cheap and reliable as comparable transformers, then DC wins.  (on skin effect alone).  However out in the practical world, once your boots are laced up and lineman's gloves are on, you run into a few other hitches. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In AC, the speed of light creates phasing issues as loads move around - particularly a problem on electric railways, which is why they like ultra-low frequencies like 25 Hz or 16-2/3 Hz.  &lt;strong&gt;This problem goes away with DC&lt;/strong&gt;.  &lt;/li&gt;&#xA;&lt;li&gt;You can't increase current.  Current is limited by wire heating, and wire heating is already based on RMS of AC.  &lt;/li&gt;&#xA;&lt;li&gt;Most of the installed base of transmission and distribution towers are made for 3-phase &quot;delta&quot;, so they have 3 conductors.  It's hard to use all 3 wires effectively in DC, so &lt;strong&gt;DC will reduce the effective capacity&lt;/strong&gt; of these lines significantly by wasting a wire.  How much?  DC carries the same as single-phase AC, and 3-wire 3-phase carries sqrt(3) (1.732) times as much.  Ouch. &lt;/li&gt;&#xA;&lt;li&gt;You &lt;em&gt;could well&lt;/em&gt; increase voltage.  AC lines are insulated for peak voltage [peak = RMS * sqrt(2)] so you could hypothetically up DC voltage to that.  However...&lt;/li&gt;&#xA;&lt;li&gt;Once DC power strikes an arc, it is very difficult to extinguish because it never stops (unlike AC, where every zero-crossing gives the arc a chance to extinguish).  This may be addressable with arc-fault detection.  AC lines already have reclosers that will auto-reconnect after a trip; a DC recloser could retry after as little as a few milliseconds, replicating the effect of the AC zero crossing.    &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11673" LastActivityDate="2018-03-07T17:47:58.877" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19775" PostTypeId="2" ParentId="19771" CreationDate="2018-03-07T18:03:12.597" Score="0" Body="&lt;p&gt;Say 3rd gear is 3:1&lt;br&gt;&#xA;And 2nd gear is 2:1      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are at 4000 rpm in 3rd and down shift you take the rpm to 6000.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly spinning an engine to 6000 rpm will have more resistance than spinning an  engine to 4000 rpm.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-03-07T18:03:12.597" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19776" PostTypeId="2" ParentId="19770" CreationDate="2018-03-07T18:16:47.133" Score="-1" Body="&lt;p&gt;If the Wattage of semiconductor blue laser will be about 1 W (as those laser from blue-ray CD) then plastic and wood may be carved and mild aluminium may be scratched  (matted). But in the metal will no deep hole. For such activity is needed at least 10W CO2. If the cutting is aim, then at least one order more, but if the cut target is sth like railways rails then one more order of magnitude stronger.&lt;/p&gt;&#xA;" OwnerUserId="15231" LastActivityDate="2018-03-07T18:16:47.133" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19777" PostTypeId="2" ParentId="19771" CreationDate="2018-03-07T18:23:18.143" Score="5" Body="&lt;p&gt;That's because when you upshift, you select a lower ratio, so your clutch speed drops. Since you took your foot of the gas, the engine rpm also dropped, and now the clutch and engine speed are close to synced, and hence little force is felt when you engage the clutch. The best way to upshift is to relieve the throttle a little(not fully), so the rpm will drop a bit as soon as you declutch. Meanwhile, shift up and clutch. After practice, no force will be felt. Every gearshift will in/-decrease rpm by about 20-25%. So aim with the throttle at a rpm 20-25% higher or lower when resp. down- or upshifting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you downshift, you select a higher ratio, so the clutch speed increases. You should apply a little throttle so the engine spins up and matches the increased clutch speed. Again, you'll feel no force when engaging the clutch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't do this, the clutch will have to make up for the difference in speed from the engine and the clutch. That costs time and gives you the annoying feedback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to downshift, is to keep your foot the same on the throttle as you were while cruising. As soon as you disengage the clutch, the engine rpm will increase. Now quickly downshift and re-engage the clutch. While you briefly disengaged the clutch, the rpm rose just enough to match the clutch in the lower gearing, if you did it right. But you have to shift and clutch quickly. It will require some practice, but you'll have supersmooth shifts and impress your passengers.. :p On top of this, you'll save on some wear on the clutch.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-03-07T18:23:18.143" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="19778" PostTypeId="2" ParentId="19644" CreationDate="2018-03-07T18:33:35.860" Score="0" Body="&lt;p&gt;Cheap price method one: at the scrap-vehicle &lt;code&gt;autoschrott&lt;/code&gt; buy the cranckshaft flywheel. Normally it not be over 20 USD, but older cars may be even 10 USD. Buy too the compassing alternator only small sprocket. If find a totally broken alternator, it may cost simbolically, probably 1 USD. &#xA;Second option - take the backellyte plywood plate, about inch thick, process it completely round and with good glue attach a perimeter with CNC specialized rubber belt. This kind of belt has practically no elasticity (thread inside) and has SINUS teeth profile. It means it gives zero gap between teeth and belt or just superior accuracy. The belt of course must be glued with teeth outside. And over it put the identical belt with teeth inside, what somewhere gets around a sprocket wheel of standard 12mm CNC NEMA17 mass product sprocket. This sprocket costs below 1 USD and belt cost about 1 USD per meter of length. It is sold in &lt;code&gt;cut yourself&lt;/code&gt; rolls.&lt;/p&gt;&#xA;" OwnerUserId="15232" LastActivityDate="2018-03-07T18:33:35.860" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19779" PostTypeId="2" ParentId="19766" CreationDate="2018-03-07T19:26:59.047" Score="2" Body="&lt;p&gt;I happen to be a cancer patient and have had the PET-CT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way they explained it to me is after giving you a rapidly decaying radioactive sugar with a half-life of about one hour which has one carbon switched with fluoride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the hungry cancer cells break the sugar, positrons are released which in a deadly collision with electrons release two gamma rays which are colinear and in opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine is counting only these colinear gamma rays and will trace them back to their origin's x,y,z. This point will be shown as a red dot superposed on a regular ct of patients body that is used as a reference to his/her anatomy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem is because of ethnicity and nutrition and diet some false positive shadows may show up.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-03-08T15:58:24.623" LastActivityDate="2018-03-08T15:58:24.623" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19780" PostTypeId="1" CreationDate="2018-03-07T20:26:20.577" Score="0" ViewCount="174" Body="&lt;p&gt;O-rings are very common components in all kinds a machinery used to, say, keep a fluid from leaking from one side of a piston to another inside a cylinder. But why exactly are O-rings tori? Why are they not designed so that the contact side of the ring to the cylinder wall is flat? Wouldn't this give a more secure seal?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2018-03-07T20:26:20.577" Title="O-rings: Why a torus?" Tags="|seals|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19781" PostTypeId="2" ParentId="19752" CreationDate="2018-03-07T22:37:49.387" Score="0" Body="&lt;p&gt;I finally fixed it, to people who are confused please listen. In some videos you see that there is a spark at the spark gap every 2 seconds or so. This is not gonna be true all the time, in fact there was hardly anything wrong with my laser at all! I just had to leave it plugged in. Yes its loud as hell and it looks like its just sparking with no let up but if you look left at your piece of paper to see the beam, you will be surprised to see that everything has worked!&#xA;Check out my laser working&#xA;&lt;a href=&quot;https://youtu.be/lU-eeZtHVeE&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/lU-eeZtHVeE&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15207" LastActivityDate="2018-03-07T22:37:49.387" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19783" PostTypeId="1" AcceptedAnswerId="19786" CreationDate="2018-03-08T00:04:10.663" Score="2" ViewCount="129" Body="&lt;p&gt;Recently came across the following 2-minute YouTube video that claims &lt;a href=&quot;https://www.youtube.com/watch?v=KhHrJ45bSVE&quot; rel=&quot;nofollow noreferrer&quot;&gt;A Balanced Rig Saves Fuel&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While nicely done, it is a marketing video.  Does anyone know of any unbiased studies that support the claim?&lt;/p&gt;&#xA;" OwnerUserId="8718" LastEditorUserId="7083" LastEditDate="2018-03-10T23:21:32.817" LastActivityDate="2018-03-10T23:21:32.817" Title="does payload weight distribution affect rolling resistance?" Tags="|applied-mechanics|fuel-economy|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19784" PostTypeId="1" CreationDate="2018-03-08T01:34:45.763" Score="0" ViewCount="27" Body="&lt;p&gt;Why does the efficiency of all hydraulic machines increase initially with increase in discharge and then decrease on further increase in discharge?&lt;/p&gt;&#xA;" OwnerDisplayName="user15212" LastActivityDate="2018-03-08T01:54:52.513" Title="The variation in efficiency of hydraulic machines" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19785" PostTypeId="2" ParentId="19784" CreationDate="2018-03-08T01:54:52.513" Score="1" Body="&lt;p&gt;Consider the limiting cases, as follows. the limit of the low flow condition is blocked flow, where the pump is churning the fluid up to maximum pressure but no work is being done on the load because there is &lt;strong&gt;zero flow&lt;/strong&gt;. Efficiency will be low in this case. The limit of the high flow condition is the no-load state, where the pump is moving the fluid at maximum flow rate but no useful work is performed because &lt;strong&gt;no pressure&lt;/strong&gt; is being applied to any load. Efficiency tends to zero in this case too. Between these cases, &lt;strong&gt;both&lt;/strong&gt; fluid flow rate &lt;strong&gt;and&lt;/strong&gt; pressure are simultaneously nonzero and the efficiency will go through a maximum. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-03-08T01:54:52.513" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19786" PostTypeId="2" ParentId="19783" CreationDate="2018-03-08T02:04:15.937" Score="3" Body="&lt;p&gt;Consider to simplify the model to two wheels in line as in a bicycle. For nice round figures, the tires are inflated to 100 pounds per square inch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load is two hundred pounds and is placed over the rear tire in such a way as to fully load the rear axle and completely unload the front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The footprint of the rear tire is now two square inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to accomplish that footprint, the tire has to deflect along the sidewalls in such a manner as to substantially distort the rubber. This distortion absorbs energy that would otherwise be used to propel the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar situation is in place when a tire is underinflated. I have a cart similar to that shown in the video. When it is heavily loaded and the tires are at 45 psi, the cart rolls easily. One can allow that the tire footprint matches more appropriately the engineering of the tire and that sidewall flex is at a minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the tire is underinflated (often!) at say, 20 psi, the footprint is greater, the flex is greater and it sure is much more difficult to pull when loaded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a vehicle such as a semitrailer rig, heavy loads over one axle would not necessarily cause visible sidewall flex, but it could be mechanically significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above assessment is a simplification, but appears in the video to be substantiated by the rolling cart test process. When you see someone pulling an airliner with his teeth, you can bet the load is evenly distributed and the tires are either properly inflated or are over-inflated for the purpose of the demonstration.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-03-08T02:04:15.937" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19787" PostTypeId="1" AcceptedAnswerId="19789" CreationDate="2018-03-08T03:22:34.097" Score="3" ViewCount="173" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HdEkH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HdEkH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what these things on the ceiling of this building are called?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am referring to those objects with semicircle shape that act as a ceiling on the &#xA; top of the building.&lt;/p&gt;&#xA;" OwnerUserId="15244" LastActivityDate="2018-03-08T09:11:38.447" Title="What are those objects of the top of this building called?" Tags="|civil-engineering|concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19788" PostTypeId="1" CreationDate="2018-03-08T03:58:25.533" Score="1" ViewCount="104" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bar_(unit)&quot; rel=&quot;nofollow noreferrer&quot;&gt;I don't see anything in Wikipedia&lt;/a&gt; about it, but (from &lt;a href=&quot;https://physics.stackexchange.com/a/114132/176&quot;&gt;this question&lt;/a&gt;) when I try to reproduce the equations in &lt;a href=&quot;https://asa.scitation.org/doi/abs/10.1121/1.1915608&quot; rel=&quot;nofollow noreferrer&quot;&gt;this 1933 paper&lt;/a&gt;, I get values 10&lt;sup&gt;6&lt;/sup&gt; away from the values in the paper.  When I reproduce this equation, for instance:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3BxQx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3BxQx.png&quot; alt=&quot;The r.m.s. pressure P in the interval between f1 and f2, is ... For example, if f1=1000 and f2=6000 c.p.s., P=5×10&amp;lt;sup&amp;gt;−5&amp;lt;/sup&amp;gt; bars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get &lt;a href=&quot;https://www.wolframalpha.com/input/?i=(8%20%CF%80%20*%201.2754%20kg%2Fm%C2%B3%20*%201.38065*10%5E-23%20J%2FK%20*%20300%20K%2F(3*340.27%20m%2Fs)*((6000%20Hz)%5E3%20-%20(1000%20Hz)%5E3))%5E(1%2F2)%20to%20bar&quot; rel=&quot;nofollow noreferrer&quot;&gt;5×10&lt;sup&gt;−11&lt;/sup&gt; bar&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, it says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l1Vp3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l1Vp3.png&quot; alt=&quot;At 1000 c.p.s the average M.A.F observed is 1.9×10&amp;lt;sup&amp;gt;−16&amp;lt;/sup&amp;gt; watts per cm², corresponding to a pressure 71 db below 1 bar.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sound intensity and pressure are related by acoustic impedance of air &lt;i&gt;Z&lt;/i&gt;&lt;sub&gt;0&lt;/sub&gt; = 400 N·s/m&lt;sup&gt;3&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1.9×10&lt;sup&gt;−16&lt;/sup&gt; W/cm² ↔ &lt;a href=&quot;https://www.wolframalpha.com/input/?i=sqrt(1.9%C3%9710%5E-16+W%2Fcm%C2%B2+*+400+N%C2%B7s%2Fm%C2%B3)+to+Pa&quot; rel=&quot;nofollow noreferrer&quot;&gt;2.7×10&lt;sup&gt;−5&lt;/sup&gt; Pa&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;71 dB below 1 bar = 2.8×10&lt;sup&gt;−4&lt;/sup&gt; ⋅ 1×10&lt;sup&gt;5&lt;/sup&gt; Pa = &lt;a href=&quot;https://www.wolframalpha.com/input/?i=10%5E(-71%2F20)%20%E2%8B%85%201%20bar%20to%20Pa&quot; rel=&quot;nofollow noreferrer&quot;&gt;28 Pa&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="448" LastEditorUserId="448" LastEditDate="2018-03-08T12:32:43.670" LastActivityDate="2018-03-09T01:12:16.487" Title="Did the pressure unit &quot;bar&quot; change definitions since the 1930s?" Tags="|pressure|acoustics|unit|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="19789" PostTypeId="2" ParentId="19787" CreationDate="2018-03-08T04:01:27.443" Score="3" Body="&lt;p&gt;This is Kuwait National assembly building designed by Jorn Utzon, Danish architect. In your image the building is under construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objects are permanent formwork for roof concrete slab. They create the grooves on top that is going to be populated with reinforcement and mold the stem of T-beam Roof slab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time, they are nicely arched on the bottom and will play as the dramatic voluted ceiling of the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He utilizes this type of architectural feature in many of his designs.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="255" LastEditDate="2018-03-08T09:11:38.447" LastActivityDate="2018-03-08T09:11:38.447" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19791" PostTypeId="1" CreationDate="2018-03-08T07:49:20.170" Score="2" ViewCount="906" Body="&lt;p&gt;I'm trying to simulate a model which has one inlet and three outlets, and I'm passing air through it. Now, I want one of the outlets to have exhaust fan. I see options of exhaust fan and outlet-vent in the boundary conditions. So how should I simulate for this condition? And previously, I had done the simulation with all the three outlets which were given the outflow factors as 0.4,0.3&amp;amp;0.3, and the case I'm referring to, what outflow factor I need to give and simulate while one being used as exhaust fan/ outlet vent and the rest being the regular outflows?&lt;/p&gt;&#xA;" OwnerUserId="15249" LastActivityDate="2018-03-08T07:49:20.170" Title="How to simulate an exhaust fan condition in ANSYS Fluent?" Tags="|simulation|cfd|ansys-workbench|fluent|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19792" PostTypeId="1" CreationDate="2018-03-08T08:04:38.820" Score="0" ViewCount="39" Body="&lt;p&gt;This is the situation: We have a flat concrete roof with (from bottom to top) bituminous sheet, insulation, bituminous sheet. There's a blower to be set onto the roof, blower has a stand with a broad flange (only one side drawn):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vp29N.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vp29N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how to connect the upper sheet to the blower stand? Due to thermal expansion, there will be movement between the insulating layer and the blower, so simply glueing the sheet to the blower will tear soon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to do the connection?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-03-08T21:06:50.603" Title="How to integrate air duct into roof insulation &amp; bituminous sheeting?" Tags="|civil-engineering|hvac|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19793" PostTypeId="2" ParentId="19726" CreationDate="2018-03-08T08:46:56.057" Score="2" Body="&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: &lt;em&gt;Added the requested drawing + corrected formulae for the 4 bar setup&lt;/em&gt;&lt;/p&gt;&#xA;&lt;br/&gt;&#xA;&lt;p&gt;As Carl Witthoft mentioned, the buoyancy force &lt;span class=&quot;math-container&quot;&gt;$F_1$&lt;/span&gt; from the block is proportional to the submerged/displaced volume (&lt;span class=&quot;math-container&quot;&gt;$V_{sub}$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F_1=\rho gV_{sub} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let's the distance between the Arm A pivot and the intersection of Arms A and B be &lt;span class=&quot;math-container&quot;&gt;$d_3$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Summing the moments around Pivot A (for the 4 bar setup, see below):&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_1 sin( \theta )  \cdot d_1  + F_{objectx}sin(\theta)\cdot d_3=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F_{objectx}= \dfrac{-\rho g V_{sub}\cdot d_1}{d_3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;(I have negated the weight of the bars here)&lt;/p&gt;&#xA;&lt;p&gt;So you can move &lt;em&gt;object x&lt;/em&gt;, reducing the length &lt;span class=&quot;math-container&quot;&gt;$d_3$&lt;/span&gt;, or to increase the length &lt;span class=&quot;math-container&quot;&gt;$d_1$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Other than that, you can only increase &lt;span class=&quot;math-container&quot;&gt;$V_{sub}$&lt;/span&gt;, which will increase anyway as the water rises. But you will have to adjust your geometry to make sure you reach the sealing force at your desired water height.&lt;/p&gt;&#xA;&lt;br/&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Adjusting for the desired water height:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate the required &lt;span class=&quot;math-container&quot;&gt;$V_{sub}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$F_{objectx} $&lt;/span&gt; is the force that we  need to both suspend the weight of &lt;em&gt;object x&lt;/em&gt; (95.6N?) and seal the container.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Make sure that your block volume &amp;gt; &lt;span class=&quot;math-container&quot;&gt;$V_{sub} $&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Decide how high you want the water to go in order to create a seal (&lt;span class=&quot;math-container&quot;&gt;$h_{wmax}$&lt;/span&gt;). Adjust the geometry to ensure that when &lt;em&gt;object x&lt;/em&gt; is in it's sealed position and the water is at &lt;span class=&quot;math-container&quot;&gt;$h_{wmax}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$V_{sub} $&lt;/span&gt; is as calculated.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;br/&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I would suggest using a 4 bar linkage instead of your current set-up, so that &lt;em&gt;object x&lt;/em&gt; does not rotate, and only transfers force vertically, assuming this was your design intention.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Add an ARM C parallel to ARM A. It should be connected to the wall on a pivot above point A, and connected to ARM B with a pin, above ARM A.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Change the connection between ARM A and ARM B to be a pin.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;br/&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8zwEh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8zwEh.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15247" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-03-09T08:21:36.127" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19794" PostTypeId="1" CreationDate="2018-03-08T09:57:44.447" Score="2" ViewCount="57" Body="&lt;p&gt;Is there a pneumatic actuator capable to achieve 5m/s and will be good idea for constant work? I know SMC have up to 3m/s, but I can't find faster.&lt;/p&gt;&#xA;" OwnerUserId="15251" LastActivityDate="2018-03-10T02:40:51.640" Title="Pneumatic actuator capable to achieve 5m/s?" Tags="|pneumatic|actuator|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19795" PostTypeId="1" CreationDate="2018-03-08T12:22:57.587" Score="0" ViewCount="47" Body="&lt;p&gt;Do we have an online utility which can be used to ping by selecting locations?&lt;/p&gt;&#xA;" OwnerUserId="4960" LastActivityDate="2018-03-08T12:41:22.957" Title="How to we ping to check accessibility of a website or a server from different locations?" Tags="|networking|" AnswerCount="1" CommentCount="4" ClosedDate="2018-03-08T16:05:47.020" ContentLicense="CC BY-SA 3.0" />
  <row Id="19796" PostTypeId="2" ParentId="19795" CreationDate="2018-03-08T12:41:22.957" Score="1" Body="&lt;p&gt;Try this. There is plenty of test locations. &lt;a href=&quot;https://www.dotcom-tools.com/ping-test.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.dotcom-tools.com/ping-test.aspx&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15251" LastActivityDate="2018-03-08T12:41:22.957" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19797" PostTypeId="2" ParentId="19762" CreationDate="2018-03-08T14:33:25.417" Score="0" Body="&lt;p&gt;I don't think there's any bearings in there. The mounting hub held rigidly onto the motor shaft with one or two grub screw(s), and the hub itself is rigidly screwed onto the frame with the 4 threaded holes you can see in the photo. As long as the motor can provide enough torque, when the shaft turns, the rest (hub &amp;amp; frame) will follow. Of course, backlash is quite likely to be present, and the motor control is of paramount importance if you want a smooth, jerk-free, motion of the frame, but in essence it's pretty simple, mechanically at least.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2018-03-08T14:33:25.417" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19798" PostTypeId="2" ParentId="19758" CreationDate="2018-03-08T15:52:00.023" Score="2" Body="&lt;p&gt;All else being equal DC transmission is more efficient than AC transmission at the same nominal voltage due to the elimination of reactive losses.&lt;/p&gt;&#xA;&lt;p&gt;However all else is rarely equal.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;At a given voltage DC is far more prone to sustaining arcs than AC.&lt;/li&gt;&#xA;&lt;li&gt;It is only relatively recently that we have developed the ability to convert between DC voltages with reasonable cost and efficiency. At high power levels it is still more expensive and less efficient than transformers.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;The result is that DC systems have tended to operate at lower voltages than AC systems and this is what has got DC it's reputation for being inefficient.&lt;/p&gt;&#xA;&lt;p&gt;Voltage has a massive effect on cost and/or efficiency of transmission. If you halve the voltage then to maintain the same level of resistive losses you have to quadruple the size of the conductors. Alternatively you have four-times the losses for the same size of conductors. So the ability to easilly convert voltage allowed efficient and economical AC transmission systems to be built. Even today AC remains the default option for most transmission and distribution needs.&lt;/p&gt;&#xA;&lt;p&gt;The exception to this is very high power point to point power transmission over long distances, over undersea cables or between non-synchronised grids. In these cases the costs and hazards involved with converting the AC used on the grid to high voltage DC become more justified.&lt;/p&gt;&#xA;" OwnerUserId="3377" LastEditorUserId="3377" LastEditDate="2021-08-01T22:13:52.327" LastActivityDate="2021-08-01T22:13:52.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19799" PostTypeId="1" AcceptedAnswerId="19802" CreationDate="2018-03-08T18:11:32.127" Score="1" ViewCount="1155" Body="&lt;p&gt;I have a Structures question about determining strain energy that I am unable to solve. &lt;strong&gt;I have attempted it and I have the answer, I would appreciate if if someone can just tell me where I am going wrong.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q: A solid conical bar of circular cross-section is suspended vertically. The length of the bar is $L$, the diameter at the base is $D$ and the weight per unit volume is $\gamma$ (equivalent to density x gravity). Determine the strain energy of the bar due to its own weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Answer} = U = \dfrac{\pi D^2 \gamma^2 L^3}{360E}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My work is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;U &amp;amp;= \dfrac{\gamma((1/3)\pi r^2 L)^2)L}{2E\pi r^2} \\&#xA;U &amp;amp;= \dfrac{\gamma(\pi^2r^4(1/9)L^2)L}{2E\pi r^2} \\&#xA;U &amp;amp;= \dfrac{\gamma\pi r^2L^3}{9\cdot2E} \\&#xA;U &amp;amp;= \dfrac{\gamma\pi(d^2/4)L^3}{18E} \\&#xA;U &amp;amp;= \dfrac{\gamma\pi d^2L^3}{18\cdot4E} \\&#xA;U &amp;amp;= \dfrac{\pi d^2 \gamma L^3}{72E} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am incorrect by a factor of 1/5. Where in my working/logic did I go wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$U = \dfrac{(\rho gAL)^2L}{2EA}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$AL = \text{Volume (of a cone)} = (1/3)\pi r^2 h$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused as to whether the A in the denominator is the base of the&#xA;cone or the volume divided by the height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can derive the correct numerator, however I am not sure how my&#xA;lecturer arrived at 360E for the denominator.&lt;/p&gt;&#xA;" OwnerUserId="11039" LastEditorUserId="1832" LastEditDate="2018-03-08T22:57:48.803" LastActivityDate="2018-03-08T23:26:23.150" Title="Determine the strain energy of a bar" Tags="|mechanical-engineering|structural-engineering|civil-engineering|structures|energy|" AnswerCount="1" CommentCount="13" ContentLicense="CC BY-SA 3.0" />
  <row Id="19800" PostTypeId="2" ParentId="19792" CreationDate="2018-03-08T20:54:54.127" Score="2" Body="&lt;p&gt;You should attach a galvanized sealed flange to the duct like an apron hanging down 3 inches going around the duct like a ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you use a 2x2 angle galvanized flashing pasted to your roofing by roofing mastic. The vertical lip of this flashing will runn inside the apron skirt which is hanging down, and it will be free to flex a little there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the typical detail for moderated climates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fzdu3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fzdu3.jpg&quot; alt=&quot; flashing detail.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-03-08T21:06:50.603" LastActivityDate="2018-03-08T21:06:50.603" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19801" PostTypeId="1" AcceptedAnswerId="24239" CreationDate="2018-03-08T21:17:49.567" Score="4" ViewCount="687" Body="&lt;p&gt;I try to design a shaft with yet unknown diameter, i know the distribution of forces and stresses (shear, torsion and normal) in/ along the shaft. I need just one more parameter nl. the tensile strength of martial. I use Rolloff-Matek machine elements table book. In the table i choose a material with tensile stress say $1000 Mpa$, but all the values in the table are normalised for a rod with diameter $d_N=16 mm$. &#xA;May i use the tensile stress as mentioned in the tabel ? or i should find a correction factor somewhere else ? &lt;a href=&quot;https://i.stack.imgur.com/gdK0T.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gdK0T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-03-08T21:41:20.107" LastActivityDate="2018-10-31T12:45:04.630" Title="How to determine tensile strength while the normalised tensile strength is known?" Tags="|mechanical-engineering|materials|design|stresses|strength|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19802" PostTypeId="2" ParentId="19799" CreationDate="2018-03-08T21:18:56.743" Score="2" Body="&lt;p&gt;We assume that the wide end of the cone is up, of course—otherwise the stress would go to infinity as the full weight of the cone is applied to a cross-sectional area that tapers to zero. (I incorrectly specified the opposite configuration in my comment above.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measure the coordinate $z$ from the bottom (i.e., from the tip of the cone). Then, the diameter at any location $z$ is $$D(z)=\frac{Dz}{L}.$$ Therefore, the corresponding area is $$A(z)=\frac{\pi D(z)^2}{4}.$$ The volume hanging underneath location $z$ is $$V(z)=\frac{\pi D(z)^2z}{12},$$ with corresponding weight $W=\gamma V(z)$ applied across the cross section. Therefore, the stress on any horizontal infinitesimal slice of thickness $dz$ at location $z$ is $$\sigma(z)=\frac{W}{A(z)},$$ and the volumetric strain energy within that slice is thus $$dU=\left(\frac{\sigma(z)^2}{2E}\right)A(z)dz,$$ where I've applied the linear elasticity result that the differential strain energy &lt;em&gt;per unit volume&lt;/em&gt; can always be expressed as $du=\sigma\,d\epsilon=\frac{\sigma}{E}d\sigma$ by Hooke's Law, or, integrated, $u=\frac{\sigma^2}{2E}$. Now, integrate $dU$ from $z=1$ to $L$.&lt;/p&gt;&#xA;" OwnerUserId="10123" LastEditorUserId="10123" LastEditDate="2018-03-08T23:26:23.150" LastActivityDate="2018-03-08T23:26:23.150" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19803" PostTypeId="1" AcceptedAnswerId="19811" CreationDate="2018-03-08T22:37:05.570" Score="2" ViewCount="377" Body="&lt;p&gt;I'm currently taking classical control theory and have some questions regarding basic concepts related to closed loop systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have a simple servo motor controlled by an electric current. Using a PID controller, my block diagram is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CmdCa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CmdCa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm still struggling to understand some things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;At what point is my loop actually producing a signal that my servo can use? And how would I extract it from my control loop?&lt;/li&gt;&#xA;&lt;li&gt;Why can I simply take my PID'ed error signal and plug it into a plant requiring current as an input? If I had a set current I could calculate the torque on my motor, from which I could then determine the accerleration and thus get the position by integrating that value twice. How do I do such a calculation if I only have a value that tells me to either increase or decrease my current by some factor?&lt;/li&gt;&#xA;&lt;li&gt;Why is the positional output calculated by my plant reconnected to my sensor in the feedback loop? Why is there an output in the first place? Shouldn't the sensor simply deliver continous data and the loop respond according to my error term?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would be very glad if somebody could answer these basics questions for me, I just can't quiet wrap my head around the basic mechanics of a control loop like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="8674" LastEditorUserId="8674" LastEditDate="2018-03-08T23:08:12.407" LastActivityDate="2018-03-09T11:41:37.150" Title="How does a closed loop control system actually work?" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19805" PostTypeId="1" AcceptedAnswerId="19809" CreationDate="2018-03-09T04:51:48.610" Score="0" ViewCount="115" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/di8nb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MAYKd.jpg&quot; alt=&quot;enter image here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/axuin.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In question 3-37,we have &#xA;P-p(ATM)=density * g* h and we get h as 8.3 m&#xA;But the answer is 1.53 m.How and why?What is wrong in my approach?&lt;/p&gt;&#xA;" OwnerUserId="15268" LastEditorUserId="15268" LastEditDate="2018-03-09T05:49:16.320" LastActivityDate="2018-03-09T07:43:39.037" Title="Pressure measurement" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19808" PostTypeId="2" ParentId="19803" CreationDate="2018-03-09T06:54:19.333" Score="1" Body="&lt;p&gt;This is two nested control loops. In the current loop, the motor drive voltage is varied in to correct any errors in motor current, relative to  &lt;span class=&quot;math-container&quot;&gt;$I_C$&lt;/span&gt;. In the position loop, which the diagram shows, &lt;span class=&quot;math-container&quot;&gt;$I_C$&lt;/span&gt; is varied to correct any errors in position.&lt;/p&gt;&#xA;&lt;p&gt;I hope I've understood your questions correctly.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The servo is all the electronics shown in the diagram, i.e. the motor and controls. The plant in this case is the Motor and Load, and it's input signal is the output signal of the current loop.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The motor has resistance, so we can't simply supply an amount of current and that will be the current in the motor. We need to power the motor by giving it a  voltage that we control to give the desired current.&lt;/p&gt;&#xA;&lt;p&gt;You could use the integration you mentioned to simulate the plant. What the position loop does is take the real position, wherever it is at a given time, and correct to the desired position over time. It does this by feeding back the measured position information and using it to vary &lt;span class=&quot;math-container&quot;&gt;$I_C$&lt;/span&gt;. This allows it to respond to any real life situation, and without any inaccuracies caused by using a simulated model.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The output in this system is just the real life position that the plant creates. The sensor (resolver or encoder) provides continuous data on this which is fed back via the position loop, and the system responds to the error.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15247" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-03-09T11:41:37.150" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19809" PostTypeId="2" ParentId="19805" CreationDate="2018-03-09T07:24:55.243" Score="2" Body="&lt;p&gt;Pressure due to mercury is:&#xA;Pm = 13600*9.81*0.12&#xA;   = 16010&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, that is the pressure created by the blood and it will then rise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hb = 16010/(1040*9.81)&#xA;   = 1.57m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, I have used constants I am aware of - subtle differences will give 1.53m...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit the density of mercury is 13560kg/m3 so that is where my error comes in...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-03-09T07:43:39.037" LastActivityDate="2018-03-09T07:43:39.037" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19810" PostTypeId="1" AcceptedAnswerId="19824" CreationDate="2018-03-09T07:29:40.143" Score="3" ViewCount="924" Body="&lt;p&gt;I have a structures problem I cannot solve, I have the answer and I have attempted the question. I just need to know where I am going off track. Any assistance would be highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7yhgw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7yhgw.png&quot; alt=&quot;Instructions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My work is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At point B:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = Fd$, where $d = 300\text{ mm}$ and $F = 100\text{ N}$, therefore $T = 30,000\text{ Nmm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From torsion, shear stress $\tau = Tr/J$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$J = \dfrac{\pi}{2}\cdot(r_1^4 - r_2^4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore $\tau = \dfrac{30,000 \cdot 30 \cdot 2}{\pi(30^4 - 25^4)} = 1.37\text{ MPa}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first answer listed, &lt;strong&gt;which leads me to believe my lecturer listed the answers for (B,A) instead of (A,B)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However then for point A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = Fd$, I believe $d$ now changes to 330&amp;nbsp;mm as you have to add the outer&#xA;radius of the bar so $T = 33,000\text{ Nmm}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following the same process used for B:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$t = \dfrac{33,000 \cdot 30 \cdot 2}{\pi(30^4 - 25^4)} = 1.50\text{ MPa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is clearly not the answer my lecturer gave. I am a confused as to&#xA;where I went wrong.&lt;/strong&gt; I am aware that I did not use the value 250mm but I&#xA;am unsure where it applies. &lt;strong&gt;What makes points A and B different, aside&#xA;from their distance from the force?&lt;/strong&gt; I would really appreciate any assistance in the matter, thank you so much for reading.&lt;/p&gt;&#xA;" OwnerUserId="11039" LastEditorUserId="1832" LastEditDate="2018-03-09T12:13:21.690" LastActivityDate="2018-03-11T13:03:43.330" Title="Caulculating maximum shear stress at certain points" Tags="|mechanical-engineering|structural-engineering|structures|steel|stresses|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19811" PostTypeId="2" ParentId="19803" CreationDate="2018-03-09T08:30:24.650" Score="2" Body="&lt;h2&gt;Your system:&lt;/h2&gt;&#xA;&lt;p&gt;You have a PID connected to a servo. So you have a closed system connected to a closed system. A servomotor already makes a control loop, that's what being a servo means. So more than likely the servomotor already has a PID inside it (or some other controller), so you don't really need a PID to connect to the servo. you can just connect the &lt;span class=&quot;math-container&quot;&gt;$P_C$&lt;/span&gt; to the servo if you know how it is mapped by the controller in the servo and it is driving by the value you want.&lt;/p&gt;&#xA;&lt;p&gt;However, it is possible you want to drive some other property than the servo drives. For example lets suppose the servo drives position instead of speed. Or the motor system after the servo is somewhat unknown/hard/problematic and you don't want to drive the value by a measurement at the engine then you want to have a second controller to dive your servo.&lt;/p&gt;&#xA;&lt;p&gt;Or your system is highly nonlinear and unstable and you want to stabilize it based on 2 parameters. Like say a inverse pendulum.&lt;/p&gt;&#xA;&lt;p&gt;Or just because you happened to have a servo available.&lt;/p&gt;&#xA;&lt;h2&gt;Ok for your questions:&lt;/h2&gt;&#xA;&lt;p&gt;Lets start with the simple one first.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Why can I simply take my PID'ed error signal and plug it into a plant requiring current as an input?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can, what you effectively then have is a P-Controller with no adjustable gain. It may work, but it can be too slow to react, or too fast. Also it does not have D and I terms so it can not react to sudden and persistent errors. See a P controller rarely reaches the target, it reaches just under or just above the target as the equilibrium of the computation is not spot on target. The I factor would take care of this. Most likely you dont need the D factor.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1a. At what point is my loop actually producing a signal that my servo can use?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;After the summing of the PID terms. The control signal inside the servo may be a bit harder to obtain without dismantling the servo.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;1b. And how would I extract it from my control loop?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Connecting a wire to the the node and measuring voltage?&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;3a. Why is the positional output calculated by my plant reconnected to my sensor in the feedback loop?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Its not (atleast not ususally). The sensor is connected to the real plant output. If your doing simulation than the calculation there to work as a stand in for the real deal so that you can test your understanding, the real plant may or may not work as expected.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;3b. Why is there an output in the first place?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Presumably your plant does something with the output.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-03-09T08:30:24.650" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19812" PostTypeId="2" ParentId="19686" CreationDate="2018-03-09T08:49:28.530" Score="1" Body="&lt;p&gt;Temperature is a measure &lt;strong&gt;proportional&lt;/strong&gt; to the average translational energy (bouncing around) of molecules in a substance, but not energy itself. It is an important measure because energy always transfers from high temperature to low temperature bodies. Heat is the energy change during the transfer. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For Example: When we heat a fluid at a constant pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ Q $ is our heat added&#xA;$ \Delta t $ is our temperature change, $m$ is our mass, $ C_p $ is our constant pressure specific heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ Q = mC_p\Delta t $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a body with a higher $m$ and $C_p$ can be heated by the same amount, but with a lower temperature increase. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Between the saturated liquid and saturated gas points, adding or losing heat does affect the temperature at all, but changes the relative proportions of liquid and gas in the mixture. &lt;/p&gt;&#xA;" OwnerUserId="15247" LastEditorUserId="15247" LastEditDate="2018-03-09T11:36:15.133" LastActivityDate="2018-03-09T11:36:15.133" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19813" PostTypeId="2" ParentId="19277" CreationDate="2018-03-09T10:10:53.640" Score="1" Body="&lt;p&gt;I can't tell about railway sensors, but for automotive (traffic lights).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where it comes to reliability of readout - &quot;false positives&quot; - no, they don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensors are inducing a pretty high own frequency (which means any naturally occurring mechanical vibrations are going to be at worst a distant harmonic), and perform a lot of post-processing after detection of frequency as caused by changed inductance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both timing and intensity of the change are taken into account. As for timing, both too short pulses (EMI noise, surges, semi-distant lightnings etc) and too slow changes (weather, temperature due to time of day) are ignored; the long ones are integrated into the baseline to compare against too. Similarly, too weak shifts are ignored even if their timing is right - to avoid false positives of vehicles in neighboring lanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do the vibrations affect the inductive sensors? Happening for long enough, with sufficient intensity, they cause material wear and break the induction loops, necessitating repairs. This is the primary disadvantage of inductive sensors and by far the most frequent mode of failure. Luckily such damage is easily detected by the circuitry, so the sensors can alert the services, fall back to backups, or enter &quot;failsafe mode&quot; e.g. producing &quot;always active&quot; signal, resulting in non-optimal but acceptable control, or switching the controller to a failsafe &quot;fixed time&quot; program.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2018-03-09T10:19:19.057" LastActivityDate="2018-03-09T10:19:19.057" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19814" PostTypeId="1" CreationDate="2018-03-09T11:51:21.867" Score="1" ViewCount="452" Body="&lt;p&gt;I have a part(let's name it as supporting plate) which is gonna be welded on two parallel sides of a rectangular channel. &#xA;The supporting plate has holes drilled at its end, the two supporting plates holes must be concentric within some dimension so as that the pin can be assembled without any problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I provide/represent such concentricity tolerance in assembly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attached a picture for your easier understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Advanced thanks for your help and support&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MUdt2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MUdt2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &amp;amp; Regards&#xA;Charles A&lt;/p&gt;&#xA;" OwnerUserId="15273" LastActivityDate="2018-03-09T16:44:03.783" Title="How to provide concentricity tolerance in assembly?" Tags="|mechanical-engineering|tolerance|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19815" PostTypeId="1" CreationDate="2018-03-09T16:27:24.780" Score="4" ViewCount="1083" Body="&lt;p&gt;I am creating a microcontroller on a custom PCB design for an MP3 player. I would like to consider integrating bluetooth into it. When I say integrating bluetooth into it, I specifically mean transmitting the chosen song on my custom MP3 player to any bluetooth speaker device that has been connected to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of what I see on the web specifically talks about receiving audio. AKA, audio from a cellphone would be provided to my system and music will be played. This is the opposite functionality of what I really want (although if the module supports both, that is fine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've taken a look at this module:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.mouser.com/ds/2/268/RN52%20DATASHEET-1.0r-1181113.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;RN52 Bluetooth Audio Module&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://learn.sparkfun.com/tutorials/rn-52-bluetooth-hookup-guide&quot; rel=&quot;nofollow noreferrer&quot;&gt;RN-52 Bluetooth Hookup Guide&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sparkfun.com/products/12849?_ga=2.256381579.1238042627.1520609890-762015303.1514738678&quot; rel=&quot;nofollow noreferrer&quot;&gt;SparkFun Audio Bluetooth Breakout - RN-52&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It looks promising, but all of the projects I see in reference to it are still the receiving design. Whenever it talks about audio out, it also specifically mentions the MIC IN pins. Since my data coming in isn't technically through a MIC, I didn't know if this functioned the way I would expect. From what I gather, MIC IN tends to only support 8Khz, while MP3 files are up to 48KHz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other one I was looking at:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://ww1.microchip.com/downloads/en/DeviceDoc/60001403C.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Microchips BM62/64 Bluetooth® 4.2 Stereo Audio Module&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This one seems to have a lot of the buzz words I've found while researching this, such PCM formatting, but the pictures on the page labeled 5 once again only show receiving data from a Bluetooth device such as a cellphone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All other projects that I see that involve transmitting to anything also involve having a second custom microcontroller and Bluetooth module on the receive end. Instead, I want the receive end to be any generic Bluetooth audio receiver; similar to how I can pair any Bluetooth device to my cellphone and output it from my cellphone to the headphones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any direction or confirmation of functionality would be great.&lt;/p&gt;&#xA;" OwnerUserId="10713" LastEditorUserId="6767" LastEditDate="2019-05-06T16:16:57.610" LastActivityDate="2022-02-19T13:51:13.890" Title="Making my microcontroller transmit audio to any bluetooth headphones/speakers" Tags="|wireless-communication|audio-engineering|microelectronics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="19816" PostTypeId="2" ParentId="19814" CreationDate="2018-03-09T16:44:03.783" Score="2" Body="&lt;p&gt;As you have not placed constraints on the project implementation, consider that most welding requires a jig of some form to maintain alignment of parts. The use of the generality &quot;most&quot; also refers to simple clamping of two pieces prior to welding, which is common in non-critical alignment projects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allowing for the specifications you've provided, consider to use a piece of square tubing of the same dimensions as that portion to which the ears have been added. This piece of square tubing would have a hole drilled to match those of the ears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bolt of the proper diameter would pass through the holes and through the jigged up square tubing, to maintain alignment and provide a stable structure for welding. If it is not going to negatively impact the primary part, a pair of tack welds to hold the square tube to the main part would provide additional stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the ears have been welded in place, the bolt should be removed and the tack welds ground clear, allowing for the desired pin placement without complications.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-03-09T16:44:03.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19817" PostTypeId="2" ParentId="19815" CreationDate="2018-03-09T18:31:30.003" Score="1" Body="&lt;p&gt;There are Bluetooth adapters available which will hook into audio system via 3mm jack or optical jack and broadcast the music to your headset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you want to receive the music into your phone, you need another adapter to hook into your phone audio jack. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have both and they cost me $20 each.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-09T18:31:30.003" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19819" PostTypeId="1" CreationDate="2018-03-09T22:38:36.220" Score="0" ViewCount="95" Body="&lt;p&gt;I have the following differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{x} = \frac{q}{m}\cdot I^2 - k_s (x+l)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to linearize it around an (unknown) point $x_0$. I know I can calculate $I_0$ by inputting $x_0$ (and $\ddot{x}=0$) into my differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 0=\frac{q}{m}\cdot I_0^2 - k_s(x_0+l) \Rightarrow I_0=\sqrt{\frac{k_s\cdot m}{q}(x_0+l)} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then take the partial derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Delta \ddot{x} = \frac{2*q}{m} I_0  \Delta I -k_s \Delta x = 2\sqrt{\frac{k_sq}{m}(x_0+l)}\Delta I -k_s \Delta x $$&#xA;$$ \Rightarrow  \Delta \ddot{x} +k_s \Delta x  = 2\sqrt{\frac{k_sq}{m}(x_0+l)}\Delta I$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now have a differential equation describing my equation of motion in terms of differences around my working point. This working point is not set though, and changes depending on the x value my system currently has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial approach to this problem was to use a sensor that reads my x value and then simply plug it into my control loop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JANOV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JANOV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For simplicity I've set $x=p_{is}$.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this does not feel like it's the correct way to do things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for $x_0=0.001\text{m}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N4lTL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N4lTL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I actually go about implementing this linearized transfer function into my loop? Are there even ways to let Simulink linearize my system directly, without me having to manually linearize around a working point?&lt;/p&gt;&#xA;" OwnerUserId="8674" LastEditorUserId="15287" LastEditDate="2018-03-10T21:23:55.933" LastActivityDate="2018-04-09T02:29:57.627" Title="Implementing a linearized differential equation in a simulink control loop" Tags="|control-engineering|control-theory|simulink|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19820" PostTypeId="2" ParentId="44" CreationDate="2018-03-10T01:18:43.493" Score="1" Body="&lt;p&gt;Is the steam trap installed on the batch heater/ vessel or in the steam main? Best practice for steam design is to install a drip trap before the process heater in the steam main ( thermodynamic is good for this application as pointed out by others) and a process steam trap for the process steam laos. Process or batch heater should use a steam trap for process loads, often a bucket or float type.&#xA;Thermodynamic traps are good for a drip leg designed to drain the main steam header. The drip leg is usually installed at the end of the header, at the header low point or right before the control valve if that is the low point.&lt;/p&gt;&#xA;" OwnerUserId="14901" LastEditorUserId="14901" LastEditDate="2018-03-10T01:38:18.687" LastActivityDate="2018-03-10T01:38:18.687" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19821" PostTypeId="2" ParentId="19794" CreationDate="2018-03-10T02:40:51.640" Score="1" Body="&lt;p&gt;Pneumatic probably not. Something that fast would probably be belt driven servo, or linear servo such as LinMot.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-03-10T02:40:51.640" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19824" PostTypeId="2" ParentId="19810" CreationDate="2018-03-10T17:47:28.747" Score="1" Body="&lt;p&gt;The reason is, that the pipe is also subject to transverse shear.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/15FeQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/15FeQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is due to the fact, that this lever doesn't apply a &quot;pure&quot; torque moment, but an eccentric force, which results in torque and shear.&#xA;Now you get two different shear stresses when you consider A and B.&#xA;As this pipe is subject to the shear force $V_z=100N$, the transverse shear &quot;flows&quot; from B over A towards the opposite side of B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(note: There would be a bending moment $M_y=100N\cdot L_{pipe}$ the base of the pipe as well, but it doesn't result in transverse shear.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can determine the transverse shear stress, using the following formula&#xA;$$ \tau=\frac{V_z\cdot Q_y}{I_y\cdot t} $$&#xA;Take a look at &lt;a href=&quot;https://engineering.stackexchange.com/questions/15006/shear-stress-formula/15015#15015&quot;&gt;this&lt;/a&gt; question for further explanation of transverse shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;further explanation&lt;/strong&gt; &lt;em&gt;[edited]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f7xIE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f7xIE.png&quot; alt=&quot;![enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at this graphic, especially at the transformed coordinated systems. (By convention, a dot in a circle depicts a force/direction facing away from you, and a cross in a circle is a force/direction facing towards you)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In sketch $II$ you can see, how the $100N$ force does create a bending moment, the orange line is a sketch of the deformation curve, that said force would cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, in sketch $III$ you can see two opposing eccentric forces, resulting in the same torque $T=2\cdot(50N\cdot 0.3m)=30Nm$, which would &lt;em&gt;not&lt;/em&gt; result in a bending moment, as they cancel each other out. $M_y=50N\cdot L_{pipe}-50N\cdot L_{pipe}=0$. Furthermore, there would be no transverse shear stress in the pipe as well.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="8632" LastEditDate="2018-03-11T13:03:43.330" LastActivityDate="2018-03-11T13:03:43.330" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19827" PostTypeId="1" CreationDate="2018-03-10T23:44:19.547" Score="0" ViewCount="66" Body="&lt;p&gt;I'm considering doing some experiments with 3D-printed surfaces and sound transmission with constrained layer damping.  I've also been thinking about using isogrid or honeycomb interior surfaces, which got me wondering: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I had a honeycomb sandwich that was cut through the middle back into two sandwich faces, each face having one side flat and the other side exposed honeycomb.  If I then sandwiched the faces back together with a thin layer of viscoelastic polymer in between the honeycomb sides, would I expect more or less damping than with the same constrained layer between solid faces of equal thickness?  Has there been any prior work on this?&lt;/p&gt;&#xA;" OwnerUserId="15295" LastEditorUserId="15295" LastEditDate="2018-03-11T02:22:29.603" LastActivityDate="2018-07-19T03:12:39.637" Title="How well would viscoelastic damping work between isogrid- or honeycomb-faced constraining layers?" Tags="|acoustics|sound-isolation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19828" PostTypeId="1" CreationDate="2018-03-11T04:56:28.200" Score="0" ViewCount="71" Body="&lt;p&gt;When the box is upside down and filled with water, is it possible to rotate the white wheel freely without water being leakage by any method or any existing technology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;White is wheel and green is axle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BDVdC.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15296" LastEditorUserId="1832" LastEditDate="2018-03-11T16:16:38.780" LastActivityDate="2018-03-11T18:57:19.783" Title="Can we rotate the given white wheel?" Tags="|mechanical-engineering|structural-engineering|waterproofing|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19829" PostTypeId="2" ParentId="19828" CreationDate="2018-03-11T10:34:19.357" Score="4" Body="&lt;p&gt;This sort of sliding seal is very difficult to achieve, especially when you have a square corner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotary engines have this problem and tend to accept a certain amount of leakage as inevitable and design around it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Changing the wheel profile to give it a more rounded shoulder would certainly help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One possible solution is an expanding seal, either an inflatable tube or a material which expands when wet, this tends to be the solution used for inboard propeller shafts on boats, even then you tend to accept a certain amount of leakage.  Indeed the new Royal Navy aircraft carrier had exactly this problem is sea trials which goes to show how difficult  a problem it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addling additional layers of seals ie a &lt;a href=&quot;https://en.wikipedia.org/wiki/Labyrinth_seal&quot; rel=&quot;nofollow noreferrer&quot;&gt;labyrinth seal&lt;/a&gt; may also help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;another potential issue is that if the wheel if open to the environment contamination by grit etc may be an issue, either disrupting the interface or damaging the seals themselves. Similarly the wheel would need good concentricity  and surface finish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not entirely sure what this is intend to achieve either, if you don't want water to escape why not just have the wheel fully enclosed on one side or the other ? The only similar application I can think of is something like the self loading rollers used for line painting on sports fields etc or indeed a ball point pen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to be more specific without more information about the intended application. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-03-11T18:57:19.783" LastActivityDate="2018-03-11T18:57:19.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19830" PostTypeId="2" ParentId="2557" CreationDate="2018-03-11T11:14:21.160" Score="0" Body="&lt;p&gt;There are a few fundamental difficulties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy stored in a flywheel is a combination of its moment of inertia and its speed. To get a higher moment of inertia you either need more mass or a larger diameter. The problem being that bicycles tend to be very sensitive to both weight and space. This leaves you with speed, firstly this will require quite a high gear ratio to get from the relatively low speeds of a bicycle wheel to the high speeds required for an effective flywheel as well as a mechanism to engage and disengage it smoothly and transfer between extracting and imputing energy, this adds weight and complexity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also a flywheel with a useful amount of energy will have a string gyroscopic effect which is a particular issue on a bicycle unless you have an even more complex system to mitigate it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally high speed flywheels pose materials challenges for bearings and the flywheel itself which is trying to explosively pull itself apart and obviously even a light, low speed flywheel is going to need to be fully enclosed for safety. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is that with current technology an electrical system with batteries is always going to be more attractive, the technology is mature and well developed and for anything like reasonable cost is going to provide much better energy density and is much easier to package on a bicycle as well as allowing for a much greater degree of control. For example a digitally controlled electrical system can have many different modes of operation without any additional mechanical parts. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-03-11T11:14:21.160" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19836" PostTypeId="2" ParentId="9092" CreationDate="2018-03-11T18:05:13.577" Score="-1" Body="&lt;p&gt;We prepared a bike propelled on acetylene for studying such problems. We found it was possible to do do, but some of the factors that govern the operation of acetylene powered engines are: cooling of the gas, proper mixing of gas and air and satisfactory cooling of the engine.&lt;/p&gt;&#xA;" OwnerUserId="15304" LastEditorUserId="90" LastEditDate="2018-03-12T01:47:46.983" LastActivityDate="2018-03-12T01:47:46.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19837" PostTypeId="2" ParentId="16022" CreationDate="2018-03-11T19:27:21.830" Score="1" Body="&lt;p&gt;I know this is a while back, but my answer is generally applicable. From experience, these sort of experiments can make use of reaction time as a &quot;free&quot; variable. Before the DOE is even planned, take samples of the reaction at the three temperatures 20, 35 and 50 degrees at regular intervals. In this way you'll get a feel for the reaction rates. In fact, if you use an internal standard,  you could generate actual rates. &#xA;This reaction time information can be plugged into the DOE.  My advice is to take more than three time samples because you can then expand your statistical analysis if necessary e.g. Design Expert allows you to add extra lines to your DOE or you can use a &quot;historical data&quot; approach.&lt;br&gt;&#xA;So long as the timepoints are appropriately spaced,  you should get a good model.&#xA;When you have factors that are time bound or hard to change, then a split plot method can be used if that factor is not randomised to check there isn't a false analysis. &lt;/p&gt;&#xA;" OwnerUserId="11590" LastActivityDate="2018-03-11T19:27:21.830" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19838" PostTypeId="1" AcceptedAnswerId="19841" CreationDate="2018-03-11T21:01:48.703" Score="1" ViewCount="376" Body="&lt;p&gt;How do I analyse for, or what do I need to consider, when comparing gravity-opposed vs gravity assisted flow in a tube containing a porous bed that partially shrinks (through partial mass loss) over a fixed time period? &lt;/p&gt;&#xA;" OwnerUserId="11590" LastActivityDate="2018-03-12T00:04:24.740" Title="Up flow vs down flow through a porous fixed bed" Tags="|chemical-engineering|porous-medium|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19839" PostTypeId="1" AcceptedAnswerId="19840" CreationDate="2018-03-11T21:17:07.213" Score="-1" ViewCount="103" Body="&lt;p&gt;What I've known in my robotics course is that control resolution is given by this equation:&#xA;$Control Resolution = Total Range/2^n$&#xA;I'm actually confused whether $2^n$ is the number of bits or $n$ is the number of bits, and if the first case is the right one, what does $n$ stand for then, the same for the 2nd case, what does $2^n$ stand for? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="15306" LastActivityDate="2018-03-11T23:23:47.553" Title="Robots Control Resolution" Tags="|robotics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19840" PostTypeId="2" ParentId="19839" CreationDate="2018-03-11T23:23:47.553" Score="0" Body="&lt;p&gt;$2^n$ is the number of increments and $n$ is the number of bits&lt;/p&gt;&#xA;" OwnerUserId="15306" LastActivityDate="2018-03-11T23:23:47.553" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19841" PostTypeId="2" ParentId="19838" CreationDate="2018-03-12T00:04:24.740" Score="1" Body="&lt;p&gt;With gravity assisted flow, you have a higher potential of channel flow through the media, so sections of your plant packing won't be wetted by the solvent. Making sure you have the optimal flow pattern is one of the primary factors in packed column performance. You indicated a 10 to 15% volume drop in the plant matter, so this poses an issue that over time the solvent flow in will squish the plant matter against one end of the container; the magnitude of this problem will depend on when the extraction of the molecules you want takes place. It's also possible that you could fluidize this plant material after the volume starts to decrease. Fluidizing could be beneficial to get high surface area but you have to control it to avoid the bed compaction effect mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main thing to measure is the pressure drop through the column. Too low a pressure drop means you have channel flow and too high a pressure drop can indicate plant matter compaction. If you can use a clear container to hold the plant matter, you can tune your setup to keep the flow pattern you want - provided you can control inlet pressure of your solvent.&lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2018-03-12T00:04:24.740" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19842" PostTypeId="1" AcceptedAnswerId="19843" CreationDate="2018-03-12T03:31:54.563" Score="3" ViewCount="202" Body="&lt;p&gt;As the title suggests, why not use high-powered air pumps instead of ceiling fans to inject hot air from the ceiling down to the cold, bottom floor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand it may seem silly (especially since it may require a lot of energy), but wouldn't it work better than a series of fans in a large, tall warehouse for example? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't the pressure differential in air pumps be sufficient to force the hot air at the ceiling down to the floor further and faster than ceiling fans?&lt;/p&gt;&#xA;" OwnerUserId="15308" LastActivityDate="2018-03-12T07:17:22.187" Title="Why is it more common to use ceiling fans instead of air pumps?" Tags="|mechanical-engineering|design|heat-transfer|hvac|building-design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19843" PostTypeId="2" ParentId="19842" CreationDate="2018-03-12T07:17:22.187" Score="7" Body="&lt;p&gt;Mainly cost difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also, a fan is sufficient for the job. The problem being solved is thermal stratification. That's caused by a lack of vertical mixing; the warm air rises to the top, the cold air sinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, there's quite a bit of movement in the first two metres or so from the ground - thermal stratification starts when the ceiling gets to around 3 metres or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fans introduce enough vertical mixing, that when combined with general activities at ground-level, the thermal stratification gets sufficiently disrupted, for pretty low capital and operating costs, and with a device that's very easy to fit and to maintain, has a long life, doesn't have filters that need cleaning, and is generally quite quiet in operation.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2018-03-12T07:17:22.187" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19844" PostTypeId="1" CreationDate="2018-03-12T10:14:55.307" Score="1" ViewCount="753" Body="&lt;p&gt;The steel type &lt;a href=&quot;http://www.m-woite.de/en/materials/14571.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;1.4571 / (X6CrNiMoTi17-12-2, ASTM: more or less 316Ti)&lt;/a&gt; is commonly used in wastewater applications due to it's good corrosion resistance. An alternative is &lt;a href=&quot;http://www.m-woite.de/en/materials/14404.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;1.4404 / (X2CrNiMo17-12-2, ASTM: 316L)&lt;/a&gt;, which is often said to be inferior. &lt;a href=&quot;https://www.montanstahl.com/de/magazin/edelstahl-werkstoffgueten-1-4404-1-4571/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Several&lt;/a&gt; steel &lt;a href=&quot;http://www.edelstahl-rostfrei.de/downloads/ISER/4571_4404_DEW.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;plants&lt;/a&gt; published &lt;a href=&quot;https://www.montanstahl.com/de/magazin/edelstahl-werkstoffgueten-1-4404-1-4571/&quot; rel=&quot;nofollow noreferrer&quot;&gt;information&lt;/a&gt; materials to the following effect (don't have an english source, so I summarize the key points):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat treatment of carbon steels would lead to the formation of Chrome-Carbides, these are attack angles for corrosion. Ni and Ti are added to form 1.4571 because these elements form carbides and so prevent the formation of Chrome-Carbides.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Current (since a few decades ...) manufacturing technology allows lower carbon content in steel production, so that addition of Ni &amp;amp; Ti are no longer neccessary to have corrosion resistance as good or better as 1.4571 in many cases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;1.4571 is clearly superior in applications with T &gt; 350°C&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All information about the comparative corrosion resistance of these steels I found so far comes from steel plants and traders who have a vested interest in pushing 1.4404: 1.4571 is commonly used in Germany, Austria and some eastern European countries, apparantly the rest of the world moved on to 1.4404 et. al. This means german steel traders need to warehouse two types of steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is, according to an independent assessment, the corrosion resistance of 1.4404 same or better than 1.4571 in most conditions?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="10902" LastEditDate="2018-03-12T12:57:33.717" LastActivityDate="2018-03-13T12:30:52.940" Title="Do the steels 1.4404 and 1.4571 have the same corrosion resistance?" Tags="|materials|steel|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19849" PostTypeId="2" ParentId="19844" CreationDate="2018-03-12T19:04:59.357" Score="2" Body="&lt;p&gt;That is some serious hair splitting. They have essentially the same corrosion resistance. One would need testing in a particular environment to POSSIBLY find some difference. The 316 L has low carbon. The Ti addition in 4571 combines with the carbon  to make titanium carbide also resulting in very low C in solution. When welded ,the carbon in solution forms chromium carbide reducing chrome content in the grain boundaries ,thereby reducing corrosion resistance in the grain boundaries  : So the result is very similar corrosion resistance. Both alloys would be poor choices for elevated service. ASME did not have allowable high temperature stresses for &quot;L&quot; grades , when I was working. In addition , Mo has a &quot;volatile &quot; oxide the can promote oxidation at high temperatures in some situations. American mills tend to make L grades while European mills added Ti or Nb (Cb), at least in the past. Also , Ni is not a factor in this consideration . Comment ; American mills add smaller amounts of Ti to deoxidize the melt, so metallography will find the orange Ti carbonitides in both alloys &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-03-13T02:52:29.370" LastActivityDate="2018-03-13T02:52:29.370" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19850" PostTypeId="1" CreationDate="2018-03-12T19:24:54.480" Score="0" ViewCount="169" Body="&lt;p&gt;I am building an air-hockey table. Essentially, a rectangular wooden box with holes on the top surface.  I am fitting a blower fan on the bottom and wondering how affects the airflow, if at all.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my understanding, air is sucked from under the table and pushed in the box, increasing pressure and forcing air out of the holes on top.  Is pressure uniform within the box with air flowing in? Or will the area around the fan have a higher pressure compared to the corners, for example?&lt;/p&gt;&#xA;" OwnerUserId="15316" LastEditorUserId="1832" LastEditDate="2018-03-12T20:10:03.827" LastActivityDate="2018-07-12T04:57:46.377" Title="Fan placement effect on pressure in chamber" Tags="|fluid-mechanics|pressure|airflow|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19851" PostTypeId="1" AcceptedAnswerId="19854" CreationDate="2018-03-12T21:28:31.083" Score="0" ViewCount="133" Body="&lt;p&gt;I want to convert my son's pedal car to electric. The car is meant for 3-6 year old children.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I got the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;50W, 3300 RPM &lt;a href=&quot;https://www.dunkermotoren.com/en/products/brushed-dc-motors/detail/8844201221/&quot; rel=&quot;nofollow noreferrer&quot;&gt;electric motor&lt;/a&gt; with 8mm round shaft (no keyway or shaft groove)&lt;/li&gt;&#xA;&lt;li&gt;10mm threaded rod made of some fairly soft metal to use as a shaft&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I was thinking to connect the motor to the shaft with XL timing pulleys. With a 10 teeth pulley connected to the motor and a 100T one on the shaft, I would get a 10X RPM reduction and with 10-inch wheels, I calculate that the max speed of the car will be ~15KM/h. Probably still a bit to high but I could do further electronic limiting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have a few questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Would there be issues if I combine the smallest pully in the range with the largest one? It seems XL-10T is the smallest and XL-100T is the largest.&lt;a href=&quot;https://i.stack.imgur.com/4P7mh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4P7mh.jpg&quot; alt=&quot;Timing pulleys&quot;&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Considering these easily available pulleys are almost always mounted with set screws would the grip be sufficient for this application?&lt;/li&gt;&#xA;&lt;li&gt;What is the required belt length if the distance between shafts is ~90mm?&lt;/li&gt;&#xA;&lt;li&gt;What is a practical way to tension the belt for something like this?&lt;/li&gt;&#xA;&lt;li&gt;Is 10mm thick enough for this belt?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15318" LastActivityDate="2018-10-15T13:43:37.540" Title="timing pulleys setup for children's pedal car" Tags="|mechanical-engineering|pulleys|mechanical|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19852" PostTypeId="2" ParentId="19850" CreationDate="2018-03-12T22:05:28.713" Score="0" Body="&lt;p&gt;The area near the fan usually has higher pressure.  But by adding fins and or partitions you can try to achieve a more even pressure distribution for high flow situations.&#xA;If you just have a deep box and fan at the bottom near the center and the dimensions of the holes are small, you can get a better balanced distribution of pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if the holes are rather large the box will act as a manifold and the holes nearer to the fan will carry an unproprtionally large share if flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are numerical simultiin softwares that can plot a very good estimate if flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to have a head pressure consistently one order higher than the total drain. Like a Scottish bagpipe.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-05-12T06:24:05.653" LastActivityDate="2018-05-12T06:24:05.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="19854" PostTypeId="2" ParentId="19851" CreationDate="2018-03-13T03:34:20.530" Score="0" Body="&lt;p&gt;The belt length can be calculated using the formula in the end below. It comes out from geometry.  &lt;a href=&quot;https://i.stack.imgur.com/rUtfG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rUtfG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bbA6o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bbA6o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just try plugging in the values for R1, R2 and d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the thickness of the belt required depends on the amount of tension it will be handling which depends upon the torque it has to transfer. Now, we know the torque to be transferred. So, we know the maximum tension it has to handle. What remains then is to identify the designed elastic tensile limit for belts of differing thicknesses. Now, I suppose that isn’t mentioned. So we can try searching online for this material’s modulus of elasticity E. Would be best if there is a stress-strain curve as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the elastic limit stress and multiply it by the cross-sectional area of the belt. That should give us then it’s max. Limit. &lt;/p&gt;&#xA;" OwnerUserId="10075" LastEditorUserId="10075" LastEditDate="2018-03-13T03:44:06.923" LastActivityDate="2018-03-13T03:44:06.923" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19855" PostTypeId="1" CreationDate="2018-03-13T06:17:48.560" Score="0" ViewCount="67" Body="&lt;p&gt;Here is link of research paper in which spectrum efficiency is given in terms of dB/s/Hz. How can we define it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://jwcn-eurasipjournals.springeropen.com/articles/10.1186/s13638-016-0792-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://jwcn-eurasipjournals.springeropen.com/articles/10.1186/s13638-016-0792-0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15324" LastEditorUserId="9" LastEditDate="2018-03-13T14:59:57.430" LastActivityDate="2018-12-08T20:01:25.273" Title="dB/s/Hz in spectrum efficiency" Tags="|wireless-communication|signal-processing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19856" PostTypeId="1" CreationDate="2018-03-13T08:32:35.530" Score="1" ViewCount="235" Body="&lt;p&gt;Hi I'm designing a pneumatic can crusher for my course, and came across a problem.  How many allowable holes would be ideal to relieve pressure on the crushing plate of the device?  Are there ways to go about it formula wise that may be useful?  The clearances between the crushing plate and the housing would be enough to alleviate pressure built up I'd imagine but I was curious if there was a way of figuring it out anyways.  Thanks in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JjWTZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JjWTZ.jpg&quot; alt=&quot;just so we&amp;#39;re clear this is what I&amp;#39;m talking about&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15327" LastActivityDate="2018-06-11T13:47:46.540" Title="Pneumatic Can Crusher (Number of holes allowed for crushing plate)" Tags="|fluid-mechanics|structural-analysis|manufacturing-engineering|flow-control|lathe|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19857" PostTypeId="2" ParentId="19856" CreationDate="2018-03-13T08:56:07.283" Score="0" Body="&lt;p&gt;One hole minimum to match the can opening... that is assuming that this is slow... not an instantaneous operation.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-03-13T08:56:07.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19859" PostTypeId="2" ParentId="19844" CreationDate="2018-03-13T12:30:52.940" Score="0" Body="&lt;p&gt;Background related to the first answer: Roughly 50 years ago Amoco Chemicals used 317 L , essentially the same as 316 L but with more Mo ( 3.5 %). Laboratory testing in the specific environment ( concentrated acetic plus halides) showed a small advantage for European 317 with the traditional Ti / Nb additions. The advantage of the European grades was found not to be the Ti / Nb . The advantage was that they had Mo content in the high end of range while American mills had Mo in the low end of range. And European mills also used a higher solution anneal temperature than the American mills. I doubt this was published , but if it was, the author was Dr. M R Berry , Amoco Chemicals. Amoco changed to Titanium equipment as time passed and no longer used 317 SS. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is the same ; very similar corrosion resistance. But testing in some specific environments might show differentiation. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-03-13T12:30:52.940" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19860" PostTypeId="2" ParentId="19855" CreationDate="2018-03-13T14:20:32.360" Score="1" Body="&lt;p&gt;As with any physical equation, the units of a constant, in this case &quot;spectrum efficiency,&quot; are chosen so that the final result is in the units desired.  So, if you're going for an answer in dB, then somewhere in this equation you'll multiply this constant by time (s , seconds) and a bandwidth in Hz.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-03-13T14:20:32.360" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19861" PostTypeId="1" CreationDate="2018-03-13T17:57:06.697" Score="2" ViewCount="85" Body="&lt;p&gt;I am trying to see if I can generate a plasma from an induction coil that is wrapped around a glass jar under vacuum. Tons of induction coil heatering circuits are on ebay for forging that can supply up to 1000W and operate at a frequency of 100khz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am feeding in 30V at 5A to my circuit with the copper coil wrapped around my bell jar under vacuum (~100mTorr) though I am not getting any plasma. Any suggestions? I should be able to ignite that with that pressure and the power in the coil? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;induction coil link: &lt;a href=&quot;https://www.ebay.com/itm/1000W-20A-ZVS-Low-Voltage-Induction-Heating-Module-Flyback-Driver-Heater-Board/322396719383?_trkparms=aid%3D555019%26algo%3DPL.BANDIT%26ao%3D1%26asc%3D20150817211623%26meid%3D45911a2f1fa74911b1b46e10b5efcd20%26pid%3D100505%26rk%3D1%26rkt%3D1%26%26itm%3D322396719383&amp;amp;_trksid=p2045573.c100505.m3226&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15332" LastActivityDate="2018-03-13T17:57:06.697" Title="Generating a plasma via induction coil" Tags="|pressure|vacuum|vacuum-pumps|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19863" PostTypeId="1" AcceptedAnswerId="21061" CreationDate="2018-03-13T20:29:32.190" Score="1" ViewCount="63" Body="&lt;p&gt;How can I find a region of attraction for given system? $k&amp;lt;0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}  &#xA;x' =-kx+x^3\\&#xA;   \end{align}&#xA;    $$&lt;/p&gt;&#xA;" OwnerUserId="9772" LastEditorUserId="9" LastEditDate="2018-03-13T22:10:33.800" LastActivityDate="2018-03-26T18:59:16.803" Title="A region of attraction analysis using invariant sets" Tags="|electrical-engineering|control-engineering|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19866" PostTypeId="1" CreationDate="2018-03-14T08:26:18.393" Score="6" ViewCount="746" Body="&lt;p&gt;It is common to find this kind of detailing and patterning on engine blocks and related parts. Older engine blocks display much more flatter and simple geometry and this kind of mold work must come with a considerable production cost. What is it's purpose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3IDEY.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3IDEY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15340" LastActivityDate="2018-03-19T13:13:43.703" Title="What is the purpose of grid-like details on engine blocks?" Tags="|automotive-engineering|metallurgy|casting|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19867" PostTypeId="2" ParentId="19866" CreationDate="2018-03-14T08:55:34.857" Score="12" Body="&lt;p&gt;This pattern is to provide sufficient strength while minimising the mass of the block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These &quot;webs&quot; are designed to prevent any vibration, if the block wall was made thin and the full length and width it would buckle or fail under the loads / stresses applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This design allows the wall to be thin in-between the webs so reducing the mass and helping to reduce the overall mass of the vehicle and help fuel consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, you can see how the bolt holes or fixing points are re-inforced with extra material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mention the &quot;extra work&quot; necessary, but also consider the amount of material saved per block multiplied by the number of blocks produced...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-03-14T08:55:34.857" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19869" PostTypeId="2" ParentId="19866" CreationDate="2018-03-14T13:50:02.640" Score="4" Body="&lt;p&gt;As mentioned in other answers these patterns are structural stiffeners to give strength to a lighter engine block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they also have the task of distributing stresses due to internal vibrations of &#xA; moving parts and concentrated stress of attached accessories such as exhaust manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is a big help in increasing longevity of the entire engine assembly, reducing fatigue cracks in the long run, and bringing the maintenance cost down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These fins are also designed to make the engine run quieter and smoother.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-14T13:50:02.640" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19870" PostTypeId="1" CreationDate="2018-03-14T14:40:26.880" Score="1" ViewCount="255" Body="&lt;p&gt;I am currently doing a solar tracker project. Can I use the LDR as position sensor by detecting the sunlight intensity? I mean if light intensity on LDR east is greater than LDR west, the LDR east will send a data to microcontroller and tell the motor where the sun now and rotate to that position.&lt;/p&gt;&#xA;" OwnerUserId="15345" LastEditorUserId="7083" LastEditDate="2018-07-31T22:56:23.537" LastActivityDate="2018-07-31T22:56:23.537" Title="Can LDR acts as a position sensor?" Tags="|electrical-engineering|sensors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19871" PostTypeId="2" ParentId="19870" CreationDate="2018-03-14T16:00:17.550" Score="2" Body="&lt;p&gt;You already answered your own question: yes it is possible. The question that remains is whether it is the best option. LDR's tend to foul and get less reliable, especially when used outdoors.&#xA;Also the object that you are tracking (the sun) tends to be very predictable. At noon it will be at the same point in the sky each day. If you have a clock you do not need to actively track the sun with sensors. Depending on your geographical location the height of the sun over the horizon may vary each day, but also this is very predictable.&lt;/p&gt;&#xA;" OwnerUserId="14479" LastActivityDate="2018-03-14T16:00:17.550" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="19873" PostTypeId="1" AcceptedAnswerId="21015" CreationDate="2018-03-14T17:27:30.277" Score="0" ViewCount="105" Body="&lt;p&gt;I am looking for a replacement bearing for an old lathe.  The lathe spindle assemble features a collet-like bronze bearing with a nut for tightening the inner diameter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/umA5G.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/umA5G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I cannot find this style bearing at the required size.  Is it possible to use a bearing adapter sleeve inside a plain bronze bearing?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WFCiw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WFCiw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, are there any suitable alternatives that would allow me to reduce shaft runout?&lt;/p&gt;&#xA;" OwnerUserId="2248" LastActivityDate="2019-12-14T12:03:15.557" Title="Can a sleeve bearing adapter be used with a plain bronze bearing?" Tags="|mechanical-engineering|bearings|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19874" PostTypeId="2" ParentId="19866" CreationDate="2018-03-14T19:54:55.860" Score="3" Body="&lt;p&gt;There are two closely related reasons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is that in terms of pure geometry webs tend to be more efficient in terms of strength and stiffness to weight ratios compared to a solid block of similar dimensions. You also see this a lot in fabricated and pressed structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is that in castings it is generally desirable to keep section thickness as consistent as possible and within a certain range to reduce the potential for casting defects related to shrinkage. It is fairly rare to see quite such intricate webbing in an engine block as it somewhat complicates mould design  and in many cases an over engineered approach is preferred in terms of stiffness and vibration damping and  where weight is critical it tends to be easier to change materials rather than produce highly optimised castings. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-03-19T13:13:43.703" LastActivityDate="2018-03-19T13:13:43.703" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19875" PostTypeId="2" ParentId="19866" CreationDate="2018-03-14T21:22:28.713" Score="-3" Body="&lt;p&gt;I have never seen that, very interesting. I expect the webs are stiffeners as mentioned. But I expect thin flat sheet of metal will cover the surface shown to enclose a water jacket. There are not enough fins for air cooling. so there must be a water jacket.Is it iron or aluminum ?  It looks like aluminum but I don't see steel/iron cylinder liners.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-03-14T21:22:28.713" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19876" PostTypeId="1" AcceptedAnswerId="19881" CreationDate="2018-03-15T07:31:05.310" Score="3" ViewCount="698" Body="&lt;p&gt;I am having trouble conceptualizing this problem. I have attempted it but am confused about a few things, I would highly appreciate any insight or advice for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ImAX0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ImAX0.png&quot; alt=&quot;image question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know you need to resolve P into its Y and Z components and that you can move them to the end of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand Py causes bending in the XZ plane and Pz causes bending in the YX plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that Pz contributed to σzz and Py contributed to  σyy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, my lecturer said that both Py and Pz contribute to σxx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm a little confused as to how this is possible when they are orthogonal to each other.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both cause bending on the beam (Py causes torsion) but do they not bend the beam in different planes therefore &lt;strong&gt;shouldn't they contribute to different stresses?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="11039" LastActivityDate="2018-03-16T16:22:50.447" Title="Determining σxx on a 3d system with resolved forces (stress, strain, shear strain energy)" Tags="|mechanical-engineering|structural-engineering|materials|structures|stresses|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19877" PostTypeId="1" CreationDate="2018-03-15T07:49:18.163" Score="1" ViewCount="124" Body="&lt;p&gt;I am currently working on designing a teeterboard for the circus (a seesaw that people jump up and down on the ends), and need some advice on my methodology for wood selection. The primary wood will be Vic ash (a.k.a Tasmanian Oak), with a possible second layer of pine underneath. Currently i plan to do 3 point bend tests on each wood standalone, and with various thicknessses as a double or triple layer (ash/pine, pine/ash, ash/pine/ash, pine/ash/pine etc). This will be used to get material data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My primary concerns are with scalability, i.e can the results stress values obtained be applied to the full size modeldue to the somewhat unpredictable and irregular nature of wood. My other concern is incorporating the glue into analysis. I do not know how to get a reasonable theoretical calculation for the multi-layer board, unlike the single board for which the calculations are extremely simple.&lt;/p&gt;&#xA;" OwnerUserId="6304" LastActivityDate="2018-09-12T03:00:54.457" Title="Design of multiple wood layers" Tags="|design|statics|wood|flexures|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19878" PostTypeId="2" ParentId="19877" CreationDate="2018-03-15T10:06:07.460" Score="1" Body="&lt;p&gt;The results will be scalable, but you will be assuming the quality of the wood to be uniform throughout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Building trade there are standards that specify the size, number and location of defects such as knots etc - these reduce the carrying capacities etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to either make sure the wood has no or few defects or oversize the material to cover the effects...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-03-15T10:06:07.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19879" PostTypeId="1" CreationDate="2018-03-15T15:10:12.980" Score="2" ViewCount="96" Body="&lt;p&gt;Please indicate if this question would be better asked on other fora, I'm not an engineer but computer scientist, and have very little experience with FPGA's / ASICs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand from what I have read online, FPGA's can take up the role of compiled C/C++ and even OpenCL programs and simulate output of a GPU. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if it would be possible to connect a normal desktop PC GPU (NVidia) to an FPGA such as from Xilinx / Lattice Crosslink such that the video signal can be re-processed afterwards? Would this be possible with a DisplayPort or HDMI connection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, if the processinig is entirely moved to the FPGA, is it common to have high bandwith connections (order of &gt; 10Gbps) into an FPGA?&lt;/p&gt;&#xA;" OwnerUserId="15357" LastEditorUserId="7083" LastEditDate="2018-03-28T02:18:05.080" LastActivityDate="2018-03-28T02:18:05.080" Title="Connect a GPU to a mobile FPGA using" Tags="|electrical-engineering|embedded-systems|computer-engineering|computer-hardware|" AnswerCount="0" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19880" PostTypeId="1" AcceptedAnswerId="19896" CreationDate="2018-03-15T17:45:01.977" Score="2" ViewCount="661" Body="&lt;p&gt;I need to use this to find a motor that will be able to pull at most 2250&amp;nbsp;kg at a 1.5&amp;nbsp;m/s pace. At peak acceleration, I would like this to accelerate at 0.5&amp;nbsp;m/s&lt;sup&gt;2&lt;/sup&gt;. The area that the load will be pulled is a concrete sidewalk with, at its most extreme, a 20° incline.&lt;/p&gt;&#xA;" OwnerUserId="15360" LastEditorUserId="1832" LastEditDate="2018-03-15T18:44:37.473" LastActivityDate="2018-03-16T16:54:11.270" Title="How much torque will I need to pull a 2250Kg load?" Tags="|motors|applied-mechanics|torque|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19881" PostTypeId="2" ParentId="19876" CreationDate="2018-03-15T18:15:12.150" Score="2" Body="&lt;p&gt;The confusion seems to arise from a misconception about how $\sigma$ is defined. For example, $\sigma_{yy}$ corresponds to a force in the $y$ direction applied to a surface facing the $y$ direction. Both $\sigma_{yy}$ and $\sigma_{zz}$ are zero essentially everywhere in the shaft (idealizing it as long and narrow) because there's no distributed region where a $y$-or $z$-direction force is internally or externally applied to a $y$- or $z$-direction surface, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I'm not including the point loads at either end, which are localized. It's implicit in these problems that stress concentrations can be ignored and that failure doesn't arise directly in the attachment, though it may occur within the shaft &lt;em&gt;adjacent&lt;/em&gt; to the attachment or point of load application.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, $\sigma_{xx}$ and $\tau$ are the key parameters (and the only nonzero stress components in this idealized problem), and the first varies strongly with the location on the shaft. For example, the $y$-direction component of $P$ causes a location-dependent tensile stress $\sigma_{xx}$ (i.e., locations with positive $\sigma_{xx}$) on the top of the shaft as a result of the bending moment). The $z$-direction component causes a location-dependent compressive stress $\sigma_{xx}$ on the left side of the shaft, also arising from the bending moment. The torque causes a shear $\tau$ that's maximized on the surface of the shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The key point is that a single stress component is not necessarily uniform in an object&lt;/strong&gt;; here, for example, $\sigma_{xx}$ varies across the surface and interior of the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very important point, so I suggest reviewing a good mechanics of materials text (e.g., Beer and Johnston) until it's familiar.&lt;/p&gt;&#xA;" OwnerUserId="10123" LastEditorUserId="10123" LastEditDate="2018-03-16T16:22:50.447" LastActivityDate="2018-03-16T16:22:50.447" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19883" PostTypeId="1" CreationDate="2018-03-15T19:27:37.330" Score="1" ViewCount="79" Body="&lt;p&gt;this is my first time posting on here and I have a question regarding the structural properties a beam will need in order to fulfill some requirements we have for our robotics team. The scenario is as follows: Our robot has two wings that drop to our sides in order to support two other robots in order to simultaneously lift all three. The situation results in the lifted robot applying a large amount of force on the top of two beams supported only from the side of the wing held against the robot frame. The current wings we have are made out of 1/16in aluminum tubing, which has ended up being heavy and not sturdy enough for our needs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have been doing research on potentially using fiberglass or carbon fiber rods. We think that 3/4in OD solid fiberglass rods will be strong enough for what we need, but are not sure. I understand that fiberglass is very flexible, and while this normally would be good we only have room for about an inch of bending. This is because all three robots must clear 12 inches while being lifted, and we lift ~15 inches up. This only leaves 3 inches of clearance, and we would like to keep as much wiggle room as we can for any tilting/unbalancing after lifting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this, we looked to hollow carbon fiber tubes. We can afford .625in OD .515in ID Pultruded rods, but after doing research we believe these will be ineffective, as they focus on tensile and compression strength, while the force we are dealing with is transverse on top of the tube. As roll wrapped rods are quite expensive, the largest tube we can afford is .422in OD .372in ID twill rods. I understand because the way the fibers are directed, roll wrapped rods are much stronger in dealing with perpendicular or twisting forces, but believe the size we will need to support the robot is outside our price range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this quite a lot and I dont expect anyone to completely solve our problems. I have presented all this information in hopes someone can provide the methods we can use to calculated the type of rod we will need and its amount of deformation, bending, breaking point, etc. We are currently planning on using the previously mentioned 3/4in OD solid fiberglass rods from goodwinds composites, held by an aluminum holding piece that attaches the rods to the frame. Below I will attach some pictures and diagrams to better express the forces involved and our setup. Keep in mind that weight is a huge factor as well, but we think we any composite used will be lighter than our original wing. Currently each one is around 5lbs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks guys!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kdumK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kdumK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CNOyo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/CNOyo.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15362" LastActivityDate="2018-03-15T19:27:37.330" Title="Calculating the Structural Forces Required to Support a Robot on an Overhanging Beam" Tags="|mechanical-engineering|structural-engineering|structural-analysis|structures|stresses|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19884" PostTypeId="1" CreationDate="2018-03-15T19:40:57.400" Score="0" ViewCount="135" Body="&lt;p&gt;I have a linear actuator with 52 millimeters of travel. I have a set of linkages with specific dimensions as shown in the images below, to which this actuator will be attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tfAVM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tfAVM.png&quot; alt=&quot;bones image one&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The REF dimension is one of the constrained parameters and is supposed to be 219 mm, but I can't get that level of granularity with Solvespace at this zoom level and it's close enough to use in this reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 50 mm dimension on the left is variable and is one of the values I seek via mathematics. The 180° angle was necessary to get Solvespace to move the components in the manner needed to create these images, although the linkage composed of the 50 and 120 figures will eventually be one piece with the measurements determined by this query.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cbLaj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cbLaj.png&quot; alt=&quot;bones image two&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above image shows the objective after the travel of the actuator has reached the maximum extension. The REF figure of 109.91 (110) is provided by Solvespace as the distance when the two links (120 and 116) are co-linear. As before, the 50 mm measurement is arbitrary and variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2YTF5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2YTF5.png&quot; alt=&quot;bones image three&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image shows my first attempt to model the actuator. The previous 50 mm dimension has been shortened to 40 mm and two additional links have been added, both of them at 25 mm length. These are also arbitrary and is an attempt to get in the ballpark without knowing the correct mathematical workflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actuator is represented by the REF 99.45 figure on the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realized after viewing this image for the purpose of posting that one of the links could have been omitted for clarity. As the actuator extends, the REF 99.45 figure would approach 152 at full travel and the H 25 line also extends downward. The H 25 line could have been omitted, but I'm also working on this design in such a way as to mirror the linkage. The H 25 line will be duplicated to the left in the real-world construct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 120 mm of space between the left and right linkage assemblies, which would be fine if the 50 mm remained and allows even more room with the current 40 mm dimension. Shorter lengths are not a complication for that portion of the linkage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c9XZb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c9XZb.png&quot; alt=&quot;bones image four&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the previous linkage approximation when positioned for the objective. The goal is achieved when the actuator reaches REF 126.43, but the desired figure is 152, starting at the previously noted 100 mm drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As best as I can determine, the 40 mm figure and the angled 25 mm figure (in the third image) are the two dimensions to be adjusted to achieve the travel within the constraints. The angled 25 mm figure becomes vertical in the final image, but that's coincidental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did well in school in trigonometry, even did well in calculus, but can't apply my mathematical abilities after all these years and determine the workflow to solve this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping this doesn't come across as a homework problem, although it's a problem I'm trying to solve from home. It's part of a makerspace construction for extensible wing bones:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NEmal.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NEmal.png&quot; alt=&quot;bones panel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The center panel will hold the actuator. When the dimensions are determined the lower &quot;bones&quot; will be changed out with newly cut pieces and the intermediary links will also be installed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to have the actuator travel of 52 mm as a critical part of this build. The actuator has built-in limit switches, allowing the operator to flip a DPDT switch to either side and not fear for destroying the build or burning out the motor. Too little extension of the bones at full actuator extension is undesirable for cosmetic reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping the math is going to be simple enough to understand, but the brain cells just aren't cutting it lately. The original plan was to use bicycle cables and housing to create movement, but in such a small area the friction and necessary sharp bends would not work well. I then switched my alleged mind over to using pulleys, but managed to squeeze out this design above as a preferable solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 225 pound load capacity on the actuator, I believe there will be enough power to move the linkage without an excessive load on the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, how can I calculate the lengths of the links (currently at 40 mm and 25 mm) in order to get the desired movement from 52 mm of actuator travel?&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2018-03-15T20:38:47.527" LastActivityDate="2018-03-15T20:38:47.527" Title="What is the correct mathematical approach to this linkage problem?" Tags="|actuator|linkage|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="19885" PostTypeId="1" CreationDate="2018-03-15T20:13:30.343" Score="1" ViewCount="532" Body="&lt;p&gt;I need Some help in Checking the stability of a masonry vault (Voute in french) or even an arch.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fv0qq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fv0qq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hand Calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone know any article or reference that can help me to check the stability of this model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Software:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know any software where this type of structure (compression only elements since the masonry wall and arch does not take any flexure)I tried to model it in robot but their was no method to make the panels or bars take only compression and the results were not correct (although if anyone knows a way to do it in robot is very much welcome to correct me). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any further information can be provided.&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2018-03-16T19:56:48.157" Title="Stability Check of Masonry or Brick Vault (or arch)" Tags="|structural-engineering|civil-engineering|structural-analysis|masonry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19886" PostTypeId="2" ParentId="19885" CreationDate="2018-03-15T21:39:30.317" Score="3" Body="&lt;p&gt;There are equations for calculating normal and shear forces and moments on thin shell arches. They are typically a part of an engineering course, late undergrad or early graduate level curriculum. You can search for thin shell arches stresses and will find many answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thick and heavy gravity arches have too many degrees of indeterminacy and most of the time are legacy structure made by ancient artisans, masons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your sketch looks not stable for a pure gravity arch by first inspection. It would buckle under its own weight and push the wall laterally out. Semicircles are not optimal shapes for compression arches. There are known geometry for arches that will work under compression and have passed the test of time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one reference for calculating stresses in arches, &lt;a href=&quot;https://pdfs.semanticscholar.org/6dd2/d37ad84aaa39a0e6cdb77af32ada75c6fbd5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Analitacal stress estimates on brick or masonry arches.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-15T21:39:30.317" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19887" PostTypeId="2" ParentId="19877" CreationDate="2018-03-15T22:41:35.153" Score="0" Body="&lt;p&gt;Laminated would properties have to be calculated and then analyzed by conventional methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a substitute I, second moment of area you can substitute $ \sum da. dy^2  $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With $\sum da_1dy^2+ da_2dy^2 + da_ndy^2.. $ But with the difference that you multiply the areas of different laminates by their Young's modules proportions. So if a layer's E is 1.3 times the other layer it's area is multiplied by 1.3 but with the same center of geometry, Dy is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For shear you can use composit beam shear equations. All these are easily doable on a spreadsheet that you save for alternative materials and dimensions. So questions of scalability is mute.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-15T22:41:35.153" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19888" PostTypeId="1" CreationDate="2018-03-16T00:15:59.267" Score="3" ViewCount="155" Body="&lt;p&gt;In a story about &lt;a href=&quot;https://www.nytimes.com/2018/03/15/us/fiu-bridge-collapse.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;today's Miami bridge collapse&lt;/a&gt;, I saw this photo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://static01.nyt.com/images/2018/03/16/us/16bridge/16bridge-superJumbo-v2.jpg&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the upper left, there is a white box truck which seems to say &quot;MIAMI DADE FIRE RESCUE / TECHNICAL RESCUE / COLLAPSE UNIT,&quot; even though the start of each line is obscured by a pole.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First responders, including another fire truck, are visible on the scene and one of the crushed passenger vehicles' headlights are still on. (Reports indicate the crushed vehicles were stopped for a red light).  The disaster scene looks pretty fresh, though clear of bloodied victims and easily clearable bystanders.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, collapses like this are supposed to be quite rare.  The impressive observation that such a specialized unit got there so quickly got me wondering, &lt;strong&gt;how many of these units are there?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recognizing that collapses might be more common in countries with weaker building codes, and thus Collapse Units like this might be more necessary there, I'm primarily restricting the scope of this to the US. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Photo by Joe Raedle, Getty. I do not have original image credit/permission to relicense this under CC-BY and thus am not re-uploading it, but asking the question seems like a fair use exemption.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3697" LastActivityDate="2018-03-16T00:15:59.267" Title="How many Technical Rescue: Collapse Units are there in the US?" Tags="|structural-engineering|civil-engineering|systems-engineering|construction-management|infrastructure|" AnswerCount="0" CommentCount="6" ClosedDate="2018-03-16T19:46:27.680" ContentLicense="CC BY-SA 3.0" />
  <row Id="19889" PostTypeId="1" AcceptedAnswerId="19890" CreationDate="2018-03-16T00:22:42.397" Score="2" ViewCount="345" Body="&lt;p&gt;Is there any benefit to using triaxial weave pattern instead of common twill?&lt;/p&gt;&#xA;" OwnerUserId="10080" LastActivityDate="2018-03-16T00:46:46.437" Title="Is triaxial carbon fiber gimmick?" Tags="|carbon-fiber|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19890" PostTypeId="2" ParentId="19889" CreationDate="2018-03-16T00:46:46.437" Score="3" Body="&lt;p&gt;No, it is not a gimmick. Triaxial weaving has been around for a long time, and can be seen in baskets, hats, etc. etc. It often exhibits superior structural properties such as greater resistance to shear forces, tearing, bursting etc. The sheet will also be generally more isotropic, so if you don't know which direction your forces will be coming from, it may be a better choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether there is any &lt;em&gt;benefit&lt;/em&gt;, however, depends entirely on the specific application. You may not need any of the above properties, making the increased cost not worth it. Further, if you are exclusively loading the fibres in one orientation, then a square weave might be a better choice. What has prompted this question?&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-03-16T00:46:46.437" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19891" PostTypeId="1" AcceptedAnswerId="19902" CreationDate="2018-03-16T04:28:12.683" Score="8" ViewCount="2057" Body="&lt;p&gt;This is a pretty simple problem, but I'm just not sure how to go about it. I want to find the moment of inertia of a discontinuous hollow circle. But since the shape is so irregular, I'm not sure how to go about solving for it. I've posted the shape in mind, so hopefully that'll clear up any confusion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/122mM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/122mM.jpg&quot; alt=&quot;discontinuous hollow circle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would an equivalence be needed to calculate the moment of inertia? For example, would I assume that the &quot;new&quot; diameter to calculate with for the outer circle is (0.625 - 0.375)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: All these numbers have inches as units.&lt;/p&gt;&#xA;" OwnerUserId="15308" LastEditorUserId="15308" LastEditDate="2018-03-19T22:35:46.857" LastActivityDate="2018-03-19T22:35:46.857" Title="Calculating moment of inertia for a hollow discontinuous circle?" Tags="|statics|moments|geometry|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="19892" PostTypeId="2" ParentId="19851" CreationDate="2018-03-16T04:39:32.817" Score="0" Body="&lt;p&gt;Comments: You want to use solid bar stock instead of a threaded rod. The websites McMaster and Misumi have shafting, pulleys, and belts. Also, a flat for a setscrew can be easily added to a shaft to increase torque transmitted. A belt length calculator is at: &lt;a href=&quot;https://www.bbman.com/belt-length-calculator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.bbman.com/belt-length-calculator/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest computing the torque the motor can produce or use an online timing belt calculator to see if a larger pitch belt such as L is needed. If the belt is tensioned by using a tensioner on the belt return side, or mounting the motor on slots, the ratio is okay.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-03-16T04:39:32.817" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19893" PostTypeId="2" ParentId="19891" CreationDate="2018-03-16T05:40:42.597" Score="1" Body="&lt;p&gt;Assuming you want the moment of inertia, &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$Ixx$&lt;/span&gt;&amp;quot;, about the horizontal axis, you do as follows:&lt;/p&gt;&#xA;&lt;p&gt;You find the &amp;quot;I&amp;quot; of a complete ring about xx axis first and that is done by calculating the &amp;quot;I&amp;quot; of the big circle: R = 0.625 subtracted by the &amp;quot;I&amp;quot; of small inner disk: R = 0.5.&lt;/p&gt;&#xA;&lt;p&gt;Now you subtract the &amp;quot;I&amp;quot; of two missing parts from the top and bottom.&#xA;These can be broken into segments and small triangles on their sides. one may need the help of Matlab, or Mathematica or such tools.&lt;/p&gt;&#xA;&lt;p&gt;But generally for each part you find its &amp;quot;I&amp;quot; about its own CG then add that to its area times the square of its CG height from xx axis. Wolfram has the equation for these &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$ y^-$&lt;/span&gt;&amp;quot;s, of a segment as well as its area. You can search for CG and area of a circle segment.&lt;/p&gt;&#xA;&lt;p&gt;Similar routine can be applied to find:&lt;span class=&quot;math-container&quot;&gt;$Iyy$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;p&gt;As mentioned in other answers another way of calculating Ixx is two subtract the I of smaller segment from that of bigger segment.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Iyy = R^4/24 ( 3\theta -3sin\theta -2sin\theta sin^2(\theta/2)  $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Based on this equation there is an online calculator which you can plug in first the big segment and find its Iy then the small one and subtract the results to find the I of your section. Finaly you need to multiplay it by 2, Because you have 2 sides.&#xA;Hear is the link to online calculator, the blue bottom on top of the page turns the calculator on/off. &lt;a href=&quot;http://structx.com/Shape_Formulas_003.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;circle segment geometric properties&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;You may want to bookmark this page, It has many calculators and shapes!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-03-16T16:30:46.287" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19894" PostTypeId="2" ParentId="19880" CreationDate="2018-03-16T06:10:04.563" Score="0" Body="&lt;p&gt;Your question contradicts itself. Acceleration will mean infinitely increasing speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we disregard that and assume a steady speed of 1.5m/s up a ramp of 20 degrees for a mass of 2250kg, you need:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P= mgh/s= 2250(9.8)sin(20)1.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And force = P/vertical velocity= P/sin(20)1.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the force needed. From here depending on the diameter of the winch's pulley you can calculate the torque needed. Just divide the power by the diameter.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="10902" LastEditDate="2018-03-16T07:26:23.177" LastActivityDate="2018-03-16T07:26:23.177" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="19896" PostTypeId="2" ParentId="19880" CreationDate="2018-03-16T10:02:31.717" Score="1" Body="&lt;p&gt;Ignoring friction, because you have not provided any details regarding this in your question, and assuming that the object starts from rest on a $20°$ incline, and accelerates at a peak rate of $0.5m/s^2$ on it's way to a max speed of $1.5m/s$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force required to stop a static object from rolling down a hill is the same as that required to keep it at a constant speed on that hill, $F_1=mg\sin\theta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The additional resultant force required to provide an acceleration of $0.5m/s^2$ can then simply be added using $F=ma$, therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=m(g\sin\theta+a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which equates to a peak force in your case of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=2250(9.81\sin20+0.5)=8674=8.7kN$$&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-03-16T16:54:11.270" LastActivityDate="2018-03-16T16:54:11.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19898" PostTypeId="2" ParentId="17822" CreationDate="2018-03-16T10:21:00.747" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The issue is that, i know trusses cannot carry load on their members&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not quite right. &lt;strong&gt;&lt;em&gt;Idealised&lt;/em&gt;&lt;/strong&gt; trusses can't carry load applied within the length of the member, and can only carry load applied at nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality an individual truss member can act as a beam. You need to design it as such.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-03-16T10:21:00.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19900" PostTypeId="1" CreationDate="2018-03-16T12:34:08.620" Score="1" ViewCount="28" Body="&lt;p&gt;I am trying to develop a field by using water injection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am facing a problem that the average reservoir pressure is dropping down and the BHP of the production wells dropping down as well until it reach the limit and stop producing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, my injection is not achieving its purpose by maintaining the reservoir pressure and the production. The injectors are in the layers 4, 5 , 6 , 7 and 8 While producers are in the layers 1, 2 and 3 Would you please suggest to me how can i maintain the pressure and the production&lt;/p&gt;&#xA;" OwnerUserId="15371" LastEditorUserId="15371" LastEditDate="2018-03-16T14:42:47.277" LastActivityDate="2018-03-16T14:42:47.277" Title="Injection and lose of fluid injected and pressure drop down?" Tags="|fluid-mechanics|petroleum-engineering|injection|" AnswerCount="0" CommentCount="8" ClosedDate="2018-03-17T07:37:29.050" ContentLicense="CC BY-SA 3.0" />
  <row Id="19901" PostTypeId="1" AcceptedAnswerId="19905" CreationDate="2018-03-16T12:39:49.210" Score="3" ViewCount="62" Body="&lt;p&gt;I recently got interested in an electronic paper.&#xA;One design that I liked was the one from Mirasol, &lt;a href=&quot;https://en.wikipedia.org/wiki/Interferometric_modulator_display&quot; rel=&quot;nofollow noreferrer&quot;&gt;Interferometric modulator display.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5EF4Q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5EF4Q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what I don't understand, is the following:&#xA;&lt;strong&gt;Their subpixels are either ON (color) or OFF (black). &#xA;Therefore, I am wondering how they can produce lighter and darker colors?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By lighter and darker colors, I mean that the color intensity is tuned. For instance for red: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2zN4G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2zN4G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9039" LastEditorUserId="7083" LastEditDate="2018-03-18T22:54:26.960" LastActivityDate="2018-03-19T12:51:06.040" Title="Different Color Intenensities from simple ON/OFF RGB subpixels" Tags="|electrical-engineering|experimental-physics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19902" PostTypeId="2" ParentId="19891" CreationDate="2018-03-16T13:17:54.777" Score="7" Body="&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt;&#xA;There seems to be a much easier way I overlooked, which I'll explain. My first answer is kept below for reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your assembly consists of a small sector subtracted from a larger sector as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Co9cb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Co9cb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You calculate the moment of inertia of the sector about the horizontal axis as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\frac{R^4}{24}(3\phi-3sin(\phi)-2sin(\phi)sin^2(\frac{\phi}{2}))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you will have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{xx}=2(I_{out}-I_{in})$$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Old Answer:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will take some effort, but it can be done analytically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jfgTl.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jfgTl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You first calculate the moment of inertia of the complete annulus with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{ann}=\frac{\pi}{4}(R^4-r^4)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you subract twice the moment of inertia, $I_{rec}$, of the lxb rectangle calculated with $\frac{lb^3}{12}$ and the parallel axis theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have subtracted the red hatched part which shouldn't be, but not yet the green part, which should be, so lets fix this quickly by adding the red moment twice, and subtracting the green moment twice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below explains how to calculate the moment of inertia of the sectors about the centreline:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QsMSd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QsMSd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{sect}=\frac{r^4}{8}(\phi_i-sin(\phi_i)+2sin(\phi_i)sin^2(\frac{\phi_i}{2}))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi_i$ is measured in radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the end you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I_{xx}=I_{ann}-2(I_{rec}+I_{sect-green}-I_{sect-red})$$&lt;/p&gt;&#xA;" OwnerUserId="14440" LastEditorUserId="14440" LastEditDate="2018-03-16T15:36:06.607" LastActivityDate="2018-03-16T15:36:06.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19903" PostTypeId="2" ParentId="19885" CreationDate="2018-03-16T13:48:31.633" Score="3" Body="&lt;p&gt;Yes there are many ways to assess arches, of varying theoretical rigour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the UK Highways situation, you could use the methods set out in the highways standard (technically an advice note) BA16/97, which is freely available &lt;a href=&quot;http://www.standardsforhighways.co.uk/ha/standards/dmrb/vol3/section4/ba1697.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fastest method is the ‘modified MEXE’, which is modified from a military method intended for use in the field (i.e., to answer the question, can I move this artillery piece over that bridge now).  Chapter 3 in the standard tells you how to do it.  It’s not so useful except for vehicle loads, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Annex E discusses various assessment methods, however, and references a number of elements of software.  They are all getting somewhat long-in-the-tooth now.&#xA;The software I use most often is either &lt;a href=&quot;http://www.obvis.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Archie&lt;/a&gt; (which is referenced in BA16/97, though it has evolved considerably) or &lt;a href=&quot;http://www.limitstate.com/ring&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ring&lt;/a&gt;.  I have a natural affinity for Archie (having been using it intermittently since the early 90s), but in some respects (notably treatment of barrel separation) Ring is superior.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analyses are limit analyses, and not about assessing moments, shears etc. but rather solve for a collapse mechanism recognising the particulate nature of masonry.  In the case of Archie (which is the most clearly expressed method), it solves for a zone of thrust solution in which there is a discrete zone of thrust within the thickness of the barrel that satisfies all the necessary constraints (geometrical, masonry strength etc.).  In the screenshot on the Obvis front page you can see this – the orange lines are a zone of thrust that satisfies the constraints and are in equilibrium with factored loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, Ring analyses the arch as a system of interactions between rigid blocks meeting constraints and determines that an equilibrium solution exists, determining the additional factor that could be applied to all loads and still determine an equilibrium solution for every position of load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would theoretically be possible to analysis arches as a non-linear (compression-only) material within an FEA code.  Indeed, CIRIA C656 (‘Masonry arch bridges: condition appraisal and remedial treatment’, a publication that is unfortunately not freely available) discusses this as the third class of analysis (after empirical, e.g. MEXE, and collapse mechanisms e.g. Archie and Ring).  I’ve never done it in practice – if I need an analysis of an arch I’d use one of the two specialist analysis programs noted.&lt;/p&gt;&#xA;" OwnerUserId="902" LastEditorUserId="1832" LastEditDate="2018-03-16T19:56:48.157" LastActivityDate="2018-03-16T19:56:48.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19905" PostTypeId="2" ParentId="19901" CreationDate="2018-03-16T14:17:40.310" Score="2" Body="&lt;p&gt;If I understand that picture correctly, the brightness is a function of how many subpixels are turned ON in a given image frame.  Your eye integrates all the photons it receives, so if all 16 red subpixels are ON, you'll perceive &amp;quot;red&amp;quot; but also perceive &amp;quot;very bright.&amp;quot;  If only one is ON, you still perceive &amp;quot;red&amp;quot; but will see &amp;quot;very dark.&amp;quot;&lt;/p&gt;&#xA;&lt;h2&gt;edit to add:&lt;/h2&gt;&#xA;&lt;p&gt;It occurs to me that the perceived color could in theory be adjusted by toggling the pixels' states at, e.g., 240Hz to be ON for only part of a video frame.  However, this has not been achieved in actual devices, and in any case would defeat one of the desired features of this technology, which is to maintain a color image without drawing additional power (the pixels hold state until addressed).&lt;/p&gt;&#xA;&lt;p&gt;For those who don't know much about the retina, some info:&lt;/p&gt;&#xA;&lt;p&gt;There are color cones, typically 3 types, each of which is sensitive to a different band of frequencies.  The combination of signals from these cones is what the brain processes to sense a color.  Along with these are rods, which are more sensitive detectors but are monochromatic (only report total intensity).  Combine the intensity signal with the color signal to get teh final perceived shade.&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-03-19T12:51:06.040" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19906" PostTypeId="2" ParentId="19891" CreationDate="2018-03-16T16:08:21.147" Score="3" Body="&lt;p&gt;very close to 1E-3 in^4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More precisely: 997E-6 in^4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't (didn't) use either of the other algebraic proposals.  It's relatively easy to write an equation for the width of cut face at a defined offset from the centre:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R.o = outer radius = 0.3125&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R.i = inner radius = 0.25&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = gap = 0.375&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;width across the outer circle (whether it's actually there or not) at y from centre = 2*sqrt(R.o^2 - y^2) by pythagoras&lt;/p&gt;&#xA;&#xA;&lt;p&gt;width across the inner circle = 2*sqrt(R.i^2 - y^2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cut width = max(0, width across outer-max(G, width across inner)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's trivial (5 mins) on excel with a row calcuating the width of each of (say) 500 slices from -0.3125&quot; to 0.3125&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I actually did it in a numerical computer algebra system to get 997E-6, when I do it on Excel and 500 slices I get 996E-6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hand check - by eye, it's similar to two rectangles each 0.0625&quot; wide and 0.5&quot; deep, which would have I 1.3E-3, so I probably trust my answer.&lt;/p&gt;&#xA;" OwnerUserId="902" LastActivityDate="2018-03-16T16:08:21.147" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19910" PostTypeId="1" CreationDate="2018-03-16T20:18:55.253" Score="3" ViewCount="317" Body="&lt;p&gt;If I have a grid array of LED's surrounded by roughly 0.7mm thickness plastic walls. What material at this thickness would prevent light bleed side to side so that individual LED's can light up but not light up their neighbor? &lt;a href=&quot;https://i.stack.imgur.com/VW2ep.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VW2ep.png&quot; alt=&quot;grid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15383" LastActivityDate="2019-02-08T18:07:12.627" Title="What plastic material blocks the most amount of light?" Tags="|mechanical-engineering|materials|plastic|lighting|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="19911" PostTypeId="1" CreationDate="2018-03-17T01:22:15.733" Score="5" ViewCount="184" Body="&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm building a steel target for long-range rifle shooting. For those unfamiliar, the purpose of the steel is to produce a loud &quot;&lt;strong&gt;PING&lt;/strong&gt;&quot; when the bullet strikes the target. It's very fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I have a variety of materials to choose from, I'd like to know the ideal shape and size of a target to produce the loudest possible noise upon impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Materials&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steel grade is AR500 (very hard but not too brittle). Its density is ~0.281 lb/in&lt;sup&gt;3&lt;/sup&gt; (7.78 g/mL).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can choose from three thicknesses: 1/4&quot;, 3/8&quot; or 1/2&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metal can be cut into any shape, and I'd like it to be somewhere in the neighborhood of 24&quot;-36&quot; in any dimension. The most convenient shapes would be rectangle, square or circle. It could be bent along one axis if there's a &lt;em&gt;very&lt;/em&gt; good reason to, but preferably kept flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Parameters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The target will be placed 1000 yards from the shooting position (see why it needs to be loud?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the sake of discussion, the projectile will be a &quot;6.5 Creedmoor&quot; round: 6.5mm diameter, 140 grain (9 grams), 2700 fps (830 m/s) muzzle velocity, copper jacketed with a lead core. Ballistics calculations give us ~1200 fps (366 m/s) velocity at 1000 yds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I build my target to produce the most audible sound upon impact from 1000 yards away?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bonus Nachos:&lt;/strong&gt;  How should I mount the steel plate to a wooden post to reduce dampening? Options include bolts, metal hooks, chain, rubber cables, or nylon straps.&lt;/p&gt;&#xA;" OwnerUserId="15385" LastActivityDate="2020-08-30T21:04:50.560" Title="Optimal size and shape for a steel target?" Tags="|steel|vibration|metals|acoustics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="19912" PostTypeId="1" AcceptedAnswerId="19915" CreationDate="2018-03-17T03:30:26.780" Score="0" ViewCount="34" Body="&lt;p&gt;I want to find a suitable AC or DC gear motor for concrete paste mixing. The application details are given below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Material type - concrete paste&#xA;RPM - 30-200rpm&#xA;Qty of material - 30Liters&#xA;Mixing style - Planetary&#xA;To understand the type of mixer, you may consider bakery mixer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody help me how to calculate the capacity of gear motor that I have to select for this application?&lt;/p&gt;&#xA;" OwnerUserId="15386" LastEditorUserId="15386" LastEditDate="2018-03-17T04:38:03.103" LastActivityDate="2018-03-17T05:25:15.557" Title="How to find required size of AC gear motor for planetary mixing application?" Tags="|motors|gears|" AnswerCount="1" CommentCount="0" ClosedDate="2018-03-18T12:20:31.903" ContentLicense="CC BY-SA 3.0" />
  <row Id="19913" PostTypeId="2" ParentId="556" CreationDate="2018-03-17T04:18:18.600" Score="1" Body="&lt;p&gt;Energy delivered to the primary goes to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The secondary load, off coarse, zero if no load,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Copper losses: both primary and secondary IR losses of winding resistance.&#xA;If the secondary has no load, that part of the loss is zero. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Iron loss: A. To spin magnetic flux one way and the other the iron needs a magnetizing current. This current generates part of the IR loss in primary loss,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;3B. Magnetic properties of iron are &quot;sticky&quot; in that residual magnetism remains when magnetized, and energy has to be spent to remove it before it reverses its orientation. The cycle is hysteresis loss, becoming heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3C. Magnetic flux induces 'eddy currents' circulating along the circumference of iron core ending as IR loss, the R being resistance of iron along the cross-section.  Lamination of the core increases the effective resistance, since now the induce voltage on 'thin' laminate is smaller, the path of flow is longer.&lt;/p&gt;&#xA;" OwnerUserId="15387" LastEditorUserId="90" LastEditDate="2018-03-17T11:00:08.103" LastActivityDate="2018-03-17T11:00:08.103" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="19915" PostTypeId="2" ParentId="19912" CreationDate="2018-03-17T05:25:15.557" Score="1" Body="&lt;p&gt;Concrete mixers usually have 3-4 blades attached to the walls of the mixer and rotating with the mixer tilted 60-70 degrees. Let's assume the diameter of the mixer 0.5 meters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the blades lifting up carry half of the weight of the mix,&#xA;$$w=30/2 (2.5)\space with 2.5 = density\space of\space mix. $$&#xA; w = 37.5 kg. And the work mixer has to do at 100rpm is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$W = (37.5)9.8.100( 0.5)/60seconds/mins= 306.25 watts$$&#xA;This is a very rough approximation. One would need to consider the clumping of the concrete, potentially clutching of the mixer, etc. So for a 200rpm mixer, you'd need a 612-watt motor, say one horsepower.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-17T05:25:15.557" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="19917" PostTypeId="2" ParentId="16325" CreationDate="2018-03-17T13:11:49.223" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BQVMM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BQVMM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. The solution is to ensure that the pail fill pipe is above the maximum liquid level.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-03-17T13:11:49.223" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20917" PostTypeId="1" AcceptedAnswerId="20933" CreationDate="2018-03-17T15:39:56.610" Score="3" ViewCount="156" Body="&lt;p&gt;I always thought 20 has to be used for voltage or field strength and 10 for electric power. Now I found this formula which uses power of antennas and they use the factor 20 instead of 10, why?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Friis_transmission_equation#cite_note-AEH-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friis transmission equation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14595" LastEditorUserId="7083" LastEditDate="2018-03-18T20:17:01.423" LastActivityDate="2018-03-19T16:21:52.407" Title="Decibel calculations, when to use 20 and when 10?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20918" PostTypeId="1" AcceptedAnswerId="21008" CreationDate="2018-03-17T17:17:46.193" Score="2" ViewCount="367" Body="&lt;p&gt;I am trying to model water gas shift reactor on Aspen using an Rgibbs reactor. Unfortunately whenever i pass the stream into the reactor(HTS) the amount of CO and H2 both are increasing and C in the system is not balanced. Aspen is not prompting an error regarding elemental balance not being intact. I am confused what should I do&lt;/p&gt;&#xA;" OwnerUserId="8535" LastActivityDate="2018-03-26T21:57:08.240" Title="Problems regarding aspen simulation" Tags="|chemical-engineering|simulation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20920" PostTypeId="1" CreationDate="2018-03-17T18:12:17.360" Score="1" ViewCount="682" Body="&lt;p&gt;I have noob question,I dont understand why are parts made from woven carbon fiber composite stiff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that carbon fibers are strong and have high Young modulus,its stiff and strong when the applied force is tensile type parallel to the direction of the fibre,basically its strong and doesnt stretch when we use it as rope and hang some weight from it,but otherwise,its very flexible yarn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The woven twill cloth without resin is not much stiffer when it comes to holding complex 3D shape,like car mirror for example,than t shirt or woven shopping bag.Its still massively strong and stiff under tensile load in the direction of fibre,but if force is applied in any other direction,its extremly weak and soft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when epoxy resin is applied,and its cured properly,I dont understand how the carbon fiber can improve strenght or stiffness over the values provided by the epoxy which have lower tensile strenght and Young modulus than carbon fiber,but is isotropic so  its strong and stiff more equaly in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example when making car mirror,there is not much tensile load,I dont understand how could carbon fiber improve it,why not just make it from epoxy resin only and skip the CF.When car goes fast it blows air on the mirror,its pressure/compressive force perpendicular to the woven carbon fiber surface,when it comes to perpendicular compressive force,as when crushing in hand carbon fiber cloth in 3D shape,its about as strong and stiff as pajamas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I dont understand how this carbon fiber twill cloth that is soft and weak in every possible direction except when trying to stretch it becomes super stiff once resin was cured.I cant imagine reason why it would be any stiffer in ability to hold complex 3D shape than the stiffness of the epoxy resin alone,since when force is applied perpendicular to fibers,the fibers have practicly zero strenght and stiffness,so they cant contribute to the stiffness of the part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe car mirror is vague and unclear example,sphere is better.If I imagine carbon fiber composite sphere,I dont see how it will be stiffer against crushing force than cured resin alone without CF.Even if you tried to stretch it by grabbing left and right side of sphere,the carbon fiber will not begin to be loaded in tensile kind of way until the sphere collapsed into pancake disk shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some &quot;mechanism&quot;,some fundamental factor that I dont understand, that helps the cured resin to &quot;translate&quot; the high tensile stiffness of carbon fiber to other directions,so the cured part,such as mirror,sphere or car wheel are stiff and hold their shape against crushing,twisting forces?&lt;/p&gt;&#xA;" OwnerUserId="10080" LastEditorUserId="10080" LastEditDate="2018-03-17T18:25:31.547" LastActivityDate="2018-03-17T20:41:04.373" Title="Why are parts from carbon fiber stiff?" Tags="|carbon-fiber|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20921" PostTypeId="2" ParentId="20920" CreationDate="2018-03-17T20:41:04.373" Score="2" Body="&lt;p&gt;a fiber-composite like carbon &amp;amp; epoxy works as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tensile strength is furnished by the fibers and the shear strength is furnished by the resin. Furthermore, since the resin encases the fibers and prevents them from buckling in compression, it also enhances the compressive and bending strength of the part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fine-diameter fibers possess a lot of exposed surface area to which the resin can bind, and since adhesion between the fiber and the resin is key to obtaining the benefits of a composite, this enhances the strength of the part as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nature uses this same scheme in wood, where the fibers are cellulose and the resin is lignin. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-03-17T20:41:04.373" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20922" PostTypeId="1" CreationDate="2018-03-18T01:39:02.203" Score="3" ViewCount="99" Body="&lt;p&gt;I've got what is basically a hammock frame design, made from tubing, that I'm trying to make as light as possible, while meeting a necessary strength safety factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've hired a few different mechanical engineers to tell me if the design is feasible and, if so, what the best combination of aluminum/steel and diameter/wall thickness would be optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before hiring anyone, I went through a few iterations of a physical prototype and have been able to see some of the problems when a load is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that some of those problems aren't showing up on the FEAs, and, additionally, there seems to be a lack of familiarity with the different aluminums and steels (comparative strengths, weldability, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this more of a job for a structural engineer, or am I just not hiring good enough mechanical engineers?&lt;/p&gt;&#xA;" OwnerUserId="15392" LastActivityDate="2018-03-18T14:55:13.480" Title="Looking for the right kind of engineer" Tags="|structural-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20924" PostTypeId="1" AcceptedAnswerId="23534" CreationDate="2018-03-18T09:17:48.050" Score="1" ViewCount="326" Body="&lt;p&gt;I am looking at a Flood Risk Assessment that uses HEC-RAS to model the predicted levels of flooding on a piece of land. As shown in the attached image, the report lists a number of variables with abbreviated names and I would like to know briefly what each one means and which one of those values shows the predicted height of flood water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g5ndW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g5ndW.png&quot; alt=&quot;My HEC-RAS data&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this document: &lt;a href=&quot;https://www.novaregion.org/DocumentCenter/Home/View/286&quot; rel=&quot;nofollow noreferrer&quot;&gt;HEC-RAS Output Variables&lt;/a&gt; which gives the full names of the abbreviated variable names. But the actual significance of each variable still make little sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for any help you can give me.&lt;/p&gt;&#xA;" OwnerUserId="15395" LastEditorUserId="33" LastEditDate="2018-10-30T01:52:18.190" LastActivityDate="2018-10-30T01:52:18.190" Title="Explain each of the values output by HEC-RAS" Tags="|fluid-mechanics|modeling|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="20925" PostTypeId="1" CreationDate="2018-03-18T13:59:51.463" Score="1" ViewCount="31" Body="&lt;p&gt;i am trying to make a small, simple 3d printed robot with humanoid characteristics. i am working on its arm at the moment. it is being driven by a small sg90 servo motor which pulls the forearm about a pivot towards the upper arm using a pulley made from fishing wire. this works well enough. however i am having trouble getting it to go the other way. i have tried adding an elastic band from point A to point B on the diagram. this works but it is not very smooth. i have researched into springs and it seems that the torsion variety are exactly what i need but they are jaw dropingly expensive(whats the deal with this? other spring types are quite cheap). could anyone offer an alternative solution to my problem? i am aware that this is a very basic question so hopefully it wont be too difficult for you folks to solve. thanks!&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8gahm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8gahm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15398" LastActivityDate="2018-03-18T13:59:51.463" Title="Creating torsion for a small robot arm" Tags="|torque|servo|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20926" PostTypeId="2" ParentId="20922" CreationDate="2018-03-18T14:55:13.480" Score="2" Body="&lt;p&gt;It sounds like what  you really need is some one with a good mix of theoretical knowledge and practical fabrication experience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice relatively few things are designed from scratch using purely theoretical analysis. even with FEA you need a bit of practical knowledge to set the right parameters and judge what is a genuine issue and what is an artefact of the simulation for example FEA tends to over estimate the significance of stress raisers like holes unless you have a well tuned high resolution model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually  your approach of optimising based on physical testing and reinforcing the bits which break is not terrible, especially if you have an experience fabricator on hand who has a good intuitive sense of what is likely to work and the ability to accurately diagnose the root cause of failures. For example if a welded joint fails is it inherent bad design or a bad weld ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally it is sometimes better to start with a conservative design and then lighten it rather than the other way around. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally I woudl look at small companies rather than individual consultants custom or small volume motorcycle or sports/racing car builders may be a good place to start as they tend to have a lot of expertise in designing and building lightweight space frames. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you probably need to accept that there may not be a single optimal solution and you will help any technical collaborators a lot of you can provide them with a concise list of priorities going from essential through desirable to nice to have if possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eg &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Minimum weight capacity &lt;/li&gt;&#xA;&lt;li&gt;Target weight &lt;/li&gt;&#xA;&lt;li&gt;target price &lt;/li&gt;&#xA;&lt;li&gt;any additional features, easy of transport and disassembly etc. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-03-18T14:55:13.480" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20929" PostTypeId="1" AcceptedAnswerId="20953" CreationDate="2018-03-18T21:41:14.747" Score="1" ViewCount="120" Body="&lt;p&gt;I am trying to calculate the static yield stress of a specimen using the&#xA;0.2% offset line on a stress strain curve. I found the static yield stress of the specimen to be the &lt;strong&gt;number in bold (104.1239)&lt;/strong&gt;. Given the shape of the&#xA;graph and how close the static yield stress appears to be to the UTS, I was wondering if this value actually looked correct? Or could I have accidentally calculated the dynamic yield stress instead? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material I used in this test was a &quot;low grade aluminum&quot;... which is throwing my thought process off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would really appreciate any insight on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jj0mK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jj0mK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11039" LastEditorUserId="1832" LastEditDate="2018-03-20T01:29:09.777" LastActivityDate="2018-03-21T04:58:19.970" Title="Stress yield of a specimen, result confusion" Tags="|mechanical-engineering|structural-engineering|materials|structures|stresses|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20932" PostTypeId="1" AcceptedAnswerId="20934" CreationDate="2018-03-19T14:57:02.650" Score="0" ViewCount="91" Body="&lt;p&gt;I'm building a mount for a small IR sensor which will look very similar to this, minus the large camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J91Gk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J91Gk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering whether or not it would be necessary to do calculations to ensure that the structure is safe and will not collapse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential problems I see are the bending moment due to the weight of the top motor. Would I calculate this and compare to the yield stresses in the bolts of the bracket? Potential buckling? Would it be worthwhile to use a software such as Ansys to run a stress simulation? The weight of the camera is very small and would not effect any calculations. I am using aluminium. Thanks&lt;/p&gt;&#xA;" OwnerUserId="15061" LastActivityDate="2018-03-20T19:18:33.477" Title="Stresses in a camera mount" Tags="|mechanical-engineering|stresses|aluminum|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20933" PostTypeId="2" ParentId="20917" CreationDate="2018-03-19T16:21:52.407" Score="2" Body="&lt;p&gt;Look carefully:  the equation in linear units has $(\frac{\lambda}{4\pi d})^2$ .&#xA;Take the log (as part of the &quot;dB = 10log(x)&quot;), and notice that $log(x^2) = 2log(x)$ &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-03-19T16:21:52.407" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20934" PostTypeId="2" ParentId="20932" CreationDate="2018-03-19T16:47:50.253" Score="1" Body="&lt;p&gt;If it’s not just a hobby project, you should at least hand check the major stresses/forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While doing that instead of using the weight of the motor, the mass times max expected max acceleration must be considered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration depends on your operational limit cases, I.e. drop, touchdown, or crash scenarios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not foresee any reasonable scenarios, still, you could take rough factors to weight, (the more serious it is, the higher the factor). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally a 1.25 or 1.5 for safety must be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without any checks, if it’s a hobby project, you can of course go ahead and experience the world, and tweak the design in time. &lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2018-03-19T16:47:50.253" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20936" PostTypeId="1" CreationDate="2018-03-19T21:45:53.770" Score="3" ViewCount="39" Body="&lt;p&gt;Although ink waste (mixed colors of different inks for printing) is 90% water, separating the inks appears to be problematic. We have tried membranes, RO, electrolysis, boiling, and solidifying with kitty litter. Membranes do not trap the dyes, electrolysis is slow and expensive, boiling is expensive and messy, and litter makes too much waste. I would like to separate out the water and decimate (reduce by 90%) the waste. Any other process ideas? I have worked on precipitating the ink but I am not much of a chemist and there are all kinds of different components.&lt;/p&gt;&#xA;" OwnerUserId="15411" LastActivityDate="2018-07-18T18:58:36.847" Title="Efficient processing waste water with pigments and dyes to reduce cost" Tags="|waste-water-treatment|waste-disposal|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20938" PostTypeId="2" ParentId="20929" CreationDate="2018-03-20T03:51:26.500" Score="0" Body="&lt;p&gt;In the old days before computers, we just drew the 0.2 % offset or the 0.5 % strain lines , and where they intersected the stress / strain curve was the yield strength. Of course materials are strain rate sensitive but it does not appear that this data contains such information.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-03-20T20:27:33.783" LastActivityDate="2018-03-20T20:27:33.783" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20939" PostTypeId="2" ParentId="20936" CreationDate="2018-03-20T05:32:45.157" Score="1" Body="&lt;p&gt;many dyes and pigments can be decomposed by buffering the ink to either acid or base, depending on their molecular structure. you can then dewater the resulting effluent with the right RO membrane. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-03-20T05:32:45.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20940" PostTypeId="1" CreationDate="2018-03-20T09:37:33.807" Score="4" ViewCount="610" Body="&lt;p&gt;I am trying to build flexible, visual tool to calculate flow distribution through the piping network (mobile hydraulics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have root node of the network which has defined flow and pressure inputs.&#xA;Each node (of course including root one) can have multiple children - lines, connectors or outputs. Each branch ends up with outlet to atmospheric pressure.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QV8mC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QV8mC.png&quot; alt=&quot;This is how the example network could look like&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I know that I have to calculate losses at each node going from the bottom of the tree, but losses are flow dependent... When I would have losses at each node, I could somehow (how?) calculate flow distribution going from the root node I would get flow to each child, then for each child flow to each of its children etc. Not sure if it helps, but I know the target flow for each output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main issue is probably the fact that there is no particular example as it suppose to be generic tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I though it is pretty simple (I still believe it is) but I got stuck and cannot find the way out.&lt;/p&gt;&#xA;" OwnerUserId="15414" LastEditorUserId="2015" LastEditDate="2018-06-08T16:42:01.973" LastActivityDate="2020-04-28T23:05:50.733" Title="Fluid mechanics - how to calculate flow distribution in piping network?" Tags="|mechanical-engineering|fluid-mechanics|hydraulics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20941" PostTypeId="1" AcceptedAnswerId="20942" CreationDate="2018-03-20T11:43:06.153" Score="6" ViewCount="3973" Body="&lt;p&gt;(I'm not sure if this question belongs to this stack exchange, but it seems appropriate to me)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a very practical situation and our knowledge is lacking to be able to solve our debate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The situation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our country, and I'm guessing everywhere in the world, people seem to always fill their drinking glasses with about one or even half a page of newspaper. I'm talking about regular 20 or 30 cl drinking glasses or mugs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When asked why they do this, it is to &quot;protect the glass&quot;. The most scienty of them told me &quot;the pressure given by the paper inside the glass helps hold its integrity in case the glass undergoing opposite pressure from the outside (which it most likely is considering it's packed in the same box with many other glasses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My argument is that the glass being a solid and pretty much not flexible at all, you'd have to put a huge amount of paper in each glass to actually make a difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, not adding too much pressure from the inside to not break it that way, but filling it so that there is as little air as possible seems to me that this is the safest option. But that is by far the worst in terms of using paper and time consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I told them to do is focus on the outside of the glass when packing. Adding a few layers of very scrunched paper to act as some kind of &quot;spring&quot; all outside the glass, therefore spreading the pressure coming from the outside into the paper, instead of helping the glass sustain said pressure by adding pressure from the inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The questions :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What solution works best to protect the glass ?&lt;/li&gt;&#xA;&lt;li&gt;What solution is actually viable, not costing 20 minutes per glass and thousands of pages of newspapers.&lt;/li&gt;&#xA;&lt;li&gt;Is there anything else to this?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that I am not a native english speaker and there might be false friends between our languages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By &quot;pressure&quot; I mean the act of applying a force onto a surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Scrunched&quot; was google translated, it should mean &quot;the extreme opposite of flat, like when you make a ball with paper and then expand it again&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading and for your time :)&lt;/p&gt;&#xA;" OwnerUserId="15417" LastEditorUserId="15417" LastEditDate="2018-03-20T12:32:26.737" LastActivityDate="2018-03-21T12:11:10.623" Title="Is it useful to stuff my glasses with newspaper to protect them?" Tags="|structural-engineering|design|pressure|structural-analysis|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="20942" PostTypeId="2" ParentId="20941" CreationDate="2018-03-20T12:07:48.313" Score="12" Body="&lt;p&gt;In itself putting paper &lt;em&gt;inside&lt;/em&gt; a glass will make very little difference. Glass is a brittle material and tends to fail by shock and point loading. Its static tensile strength is actually pretty good. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will help is packing paper between the glasses as it will help prevent them coming into contact with each other and dampen vibration and impact forces. Without going into actual numbers you want it crumpled so it has a bit of stiffness but still with enough give that it is fairly springy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where putting paper inside the glasses may help it that it is filling space so that there is less chance for the packaging to move around and create gaps by working its way into the empty space inside the glasses.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4970" LastEditDate="2018-03-21T00:52:07.770" LastActivityDate="2018-03-21T00:52:07.770" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20943" PostTypeId="1" AcceptedAnswerId="20945" CreationDate="2018-03-20T14:28:16.080" Score="1" ViewCount="3539" Body="&lt;p&gt;For a given 2D print, is it recommended to number section views and details views with the same alphabetical list, i.e. Section views A,B and C followed by detail views D, E and F, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or go with different alphabetical list, i.e. Section views, A, B and C, followed by detail views, A, B and C?&lt;/p&gt;&#xA;" OwnerUserId="14168" LastActivityDate="2018-03-20T15:52:09.763" Title="How to number section and detail view for a same drawing" Tags="|technical-drawing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20944" PostTypeId="2" ParentId="20941" CreationDate="2018-03-20T15:19:12.587" Score="8" Body="&lt;p&gt;Filling inside the cup with paper is just to lock in the wrapping of the same paper around the glass.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It also offers coverage and protection of the rim of the glass.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because of the integrated, all-around coverage the whole package remains intact in the box with no chance of clinking against each other and it renders strength to the box as well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It is even easier and faster to wrap glasses this way, because you can cut your paper random in size and just stick the left-over in the glass.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-20T15:19:12.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20945" PostTypeId="2" ParentId="20943" CreationDate="2018-03-20T15:37:41.193" Score="2" Body="&lt;p&gt;Your first suggestion is correct - there should not be two different views (of any type) with the same letter on any one drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the quote below from ASME Y14.3-2003&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;1.7.3 Identifying Removed Views.&lt;/strong&gt; To relate the viewing&#xA;  plane or cutting plane to its removed view, capital&#xA;  letters such as A, B, C, etc., are placed near each arrowhead.&#xA;  The corresponding removed views are identified&#xA;  as VIEW A-A, VIEW B-B, VIEW C-C, etc. View letters&#xA;  should be used in alphabetical order excluding I, O, Q,&#xA;  S, X, and Z. When the alphabet is exhausted, additional&#xA;  removed views shall be identified by double letters in&#xA;  alphabetical order, as in AA-AA, AB-AB, AC-AC, etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This applies to &lt;em&gt;all&lt;/em&gt; &quot;Removed Views&quot;, of any type.&#xA;I, O, Q, S, X, and Z are all excluded due to their visual similarity to 1, 0, 0, 5, [blank that needs filling in], and 2, respectively.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-03-20T15:52:09.763" LastActivityDate="2018-03-20T15:52:09.763" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20946" PostTypeId="2" ParentId="20936" CreationDate="2018-03-20T16:32:23.533" Score="0" Body="&lt;p&gt;Low pressure desalination is a known technique to avoid high temperatures, the same problem that appears to hinder you in boiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't aim for a single process, though. As you noted, boiling will concentrate the result and give you sticky solution to work with. At that point you may want to switch to that litter-based solution - most of your intial volume will have been boiled off as pretty clean water vapor. (If your ink contains alcohols, they'll boil off preferentially and can be condensed out before the water vapor.)&lt;/p&gt;&#xA;" OwnerUserId="669" LastActivityDate="2018-03-20T16:32:23.533" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20947" PostTypeId="1" CreationDate="2018-03-20T16:42:02.410" Score="5" ViewCount="116" Body="&lt;p&gt;Is cauchy's stress tensor symmetric always?Does this symmetricity  hold in accelerating and non-zero cross velocity gradient flow field?&lt;/p&gt;&#xA;" OwnerUserId="4690" LastActivityDate="2018-05-22T22:11:13.057" Title="Fluid mechanics cauchy's stress tensor" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20948" PostTypeId="2" ParentId="20932" CreationDate="2018-03-20T19:18:33.477" Score="0" Body="&lt;p&gt;In this sport of situation requirement for stiffness tend to dominate so for any halfway acceptable design actual yield probably won't be an issue as a design which is acceptably stiff will be fairly lightly stressed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;FE analysis may help you to optimise a design but its not foolproof and you still need a reasonable degree of knowledge and experience to set up the model properly and interpret the results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just by eyeballing the example image I would make a couple of points. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The corner joints will tend to be the limiting factor, you're probably better off using short lengths of equal angle extrusion a L shaped brackets than the small blocks shown. &lt;/li&gt;&#xA;&lt;li&gt;With aluminium is it generally better to bolt through with a nut and bolt then to drill and tap the aluminium itself. &lt;/li&gt;&#xA;&lt;li&gt;Putting four bolts per side rather than two will greatly improve stiffness, you want to avoid putting any bending load on bolts. &lt;/li&gt;&#xA;&lt;li&gt;Adjust the dimensions of the structure to keep cantilever lengths to a minimum, for a beam supported at one end only the maximum deflection is proportional to the &lt;em&gt;cube&lt;/em&gt; of the unsupported length. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-03-20T19:18:33.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20951" PostTypeId="2" ParentId="20941" CreationDate="2018-03-20T22:59:28.693" Score="3" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As you have correctly guessed, putting crumpled paper inside the glass would not have any protective benefits. Glass itself is a hard, but brittle substance and thus would need to be protected from impact forces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I moved my household stuff between two cities a month back and have realized that wrapping the glasses in flat, uncrumpled is not the best strategy. Wrapping a glass in crumpled paper is better, but still might leave a couple of relatively under-protected resulting in chips. Even though I do not like this approach, this is the easiest one.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The best protection would be offered by styrofoam or similar materials. While getting individual foam cases would be a pain and a huge waste resources, you could always toss small foam cubes between your paper-packed glasses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15427" LastActivityDate="2018-03-20T22:59:28.693" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20953" PostTypeId="2" ParentId="20929" CreationDate="2018-03-21T04:58:19.970" Score="2" Body="&lt;p&gt;The mechanical properties of an aluminum alloy definitely vary with the various series designations. As I am not totally sure which series you are working with and how you are doing your calculations, I cannot say whether your actual strength values are correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is possible that your yield strength and ultimate tensile strength are not very far apart. For example, in aluminum 1100, the difference between yield strength and ultimate tensile strength is only 5 MPa at room temperature. Your alloy may be similar in nature.&lt;/p&gt;&#xA;" OwnerUserId="12571" LastActivityDate="2018-03-21T04:58:19.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20954" PostTypeId="2" ParentId="20940" CreationDate="2018-03-21T05:09:50.383" Score="1" Body="&lt;p&gt;Please refer to EPANET from &lt;a href=&quot;https://www.epa.gov/water-research/epanet&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.epa.gov/water-research/epanet&lt;/a&gt; &#xA;It is a open source software for solving flow networks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps. &#xA;Thanking you&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-03-21T05:09:50.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20955" PostTypeId="2" ParentId="20947" CreationDate="2018-03-21T06:30:48.890" Score="1" Body="&lt;p&gt;The symmetric property of stress tensor is generally proved through Moment of momentum on a fluid element&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it holds as long as momentum conservation and moment equilibrium are satisfied for that element.&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-03-21T06:30:48.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20956" PostTypeId="1" CreationDate="2018-03-21T08:18:48.927" Score="4" ViewCount="4095" Body="&lt;p&gt;There is a rain pipe in a house. I want to know how many cubic meter rain water can flows through the given pipe in a minute. I need to know the &lt;strong&gt;maximum capacity&lt;/strong&gt; of the pipe. In other words: How many liter water can pass through this pipe in one minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Data:&lt;/strong&gt;&lt;br&gt;&#xA;Pipe diameter: &lt;strong&gt;70mm&lt;/strong&gt;&lt;br&gt;&#xA;Pipe length: &lt;strong&gt;3000mm&lt;/strong&gt;&lt;br&gt;&#xA;Pipe angle: &lt;strong&gt;90°&lt;/strong&gt;&lt;br&gt;&#xA;Height of the top hole: &lt;strong&gt;3000mm&lt;/strong&gt;&lt;br&gt;&#xA;Height of the bottom hole: &lt;strong&gt;0mm&lt;/strong&gt;&lt;br&gt;&#xA;Pipe material: &lt;strong&gt;Copper&lt;/strong&gt;&lt;br&gt;&#xA;Time: &lt;strong&gt;1 minute&lt;/strong&gt;&lt;br&gt;&#xA;Fluid: &lt;strong&gt;Water&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone describe me or suggest me an online tool to calculate that?&lt;/p&gt;&#xA;" OwnerUserId="15434" LastEditorUserId="15434" LastEditDate="2018-04-06T11:00:06.270" LastActivityDate="2018-06-02T18:07:38.263" Title="How to calculate the volume capacity flow rate of a vertical pipe?" Tags="|fluid-mechanics|fluid|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20957" PostTypeId="1" AcceptedAnswerId="20958" CreationDate="2018-03-21T09:08:23.667" Score="3" ViewCount="328" Body="&lt;p&gt;My pulleys are going to be rotated with a 1000-2000 RPM motor, with a constant RPM controlled by a hall effect sensor, there is going to be almost no torque as one of the pulleys just has a 100 gram optical element attached to it and all the pulleys are on bearings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions on choosing v belt or circular belt? Price is pretty similar and v seems to reduce risk of slipping and has better grip. Cant find a disadvantage of v belts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just assume slippage might become in issue with round belts at these RPMs if there isn't enough grip, no other reason to not choose them over v belts, really.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd imagine tooted belts wouldn't provide much advantage here besides any backlash or slippage when the system begins rotating but I don't see how it would be more precise after the speed is reached and the speed and load remain constant during operation.&lt;/p&gt;&#xA;" OwnerUserId="14841" LastActivityDate="2018-03-21T10:06:53.830" Title="choosing between v belt and round belt for fast RPM, low torque application" Tags="|mechanical-engineering|motors|pulleys|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20958" PostTypeId="2" ParentId="20957" CreationDate="2018-03-21T10:00:40.290" Score="1" Body="&lt;p&gt;Depends on many things. For example I have sometimes used round belts for these applications simply because i happened to have suitably sized O- rings which i could use as belts and didnt have V- belts around. This means i could have a working version today, instead of in a day or two. *&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The groove for a round belt may be minimally easier to manufacture with a manual lathe. Also the round belt is slightly easier to install if you do not use a tensioner mechanism as its possible to roll it into place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember that your feedback should come from the final axis thisway it hardly matters if your belt slips. Slippage may be minimal anyway, but if you are really concerned use a toothed belt instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* dont underestimate this, as you may end up doing some higher priority work in future. But then if you do not need to concern yourself with assembly and manufacturing choose what you think is best.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-03-21T10:06:53.830" LastActivityDate="2018-03-21T10:06:53.830" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20959" PostTypeId="1" CreationDate="2018-03-21T10:13:02.593" Score="0" ViewCount="158" Body="&lt;p&gt;About Matlab (a software very useful for engineerinng, too), Is &quot;Matlab 2008&quot; or &quot;Matlab 2012a&quot; usable on laptop with core &quot;i3&quot; with &quot;Win 10 Home&quot;?&lt;/p&gt;&#xA;" OwnerUserId="4916" LastEditorUserId="4916" LastEditDate="2018-03-21T15:06:24.447" LastActivityDate="2018-03-21T15:06:24.447" Title="Is Matlab old versions usable with Win 10 on core i3" Tags="|matlab|" AnswerCount="1" CommentCount="7" ClosedDate="2018-03-21T22:04:35.313" ContentLicense="CC BY-SA 3.0" />
  <row Id="20961" PostTypeId="1" AcceptedAnswerId="20973" CreationDate="2018-03-21T10:17:29.213" Score="0" ViewCount="218" Body="&lt;p&gt;I have a piece of mahinery with a grease fitting in an awkward to reach place. The fitting is of this shape ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/m860K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m860K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... but not screwed into place but pressed. So we can't have it removed and replaced with a pipe leading to a more accessible place. &#xA;My question is: Can I mount a tube to the fitting in a way that it stays there for the foreseeable future (a few years)? Grease will be applied twice a year. I have doubts that simply pushing a tube over the fitting and applying a clamp will suffice, as greasing will put some pressure on the thing. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-03-24T01:50:53.907" Title="How to mount a tube to a grease fitting?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20962" PostTypeId="2" ParentId="20959" CreationDate="2018-03-21T10:34:31.167" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;&quot;Windows 2010&quot; is not a Windows version so i'm guessing that you are referring to Windows 10.&lt;/li&gt;&#xA;&lt;li&gt;In general all software that came out since Windows Vista (2007) also runs fine on newer Windows versions like Windows 7, 8 or 10.&lt;/li&gt;&#xA;&lt;li&gt;Since Matlab 2008 requires only a &lt;a href=&quot;https://www.mathworks.com/support/sysreq/previous_releases.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pentium 4 with 512MB memory&lt;/a&gt; it will probably run fine on any current system&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14479" LastActivityDate="2018-03-21T10:34:31.167" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20964" PostTypeId="1" CreationDate="2018-03-21T10:48:35.277" Score="0" ViewCount="1313" Body="&lt;p&gt;This is about several diesel tanks that need to be monitored in terms of diesel volume.&#xA;I want to measure and monitor the volume of Diesel liquid inside horizontally cylindrical tanks with standard dimensions, however the safety rules are strict, where no electrical devices or wires has to exist in the tank zone, which limits the solution to few methods or devices.&#xA;What i want to achieve is to measure the diesel volume inside the tank without having to use an ultrasonic sensor, as it will require some modification to the tank, not mentioning the wires connecting from the transducer to the display unit. the question here is whether we can use a sensor that can be attached to the output pipe of the tank and calculate the diesel volume. a sensor to measure pressure or flow, will it work for such requirements ? If yes, how do I calculate the volume based on those values ? and is there's any device in the market that can provide the volume of liquid inside the tank straight forward without the need to make any further calculations using the outlet pipe ?&#xA;The tank is similar to the one in the picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qlusz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qlusz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="15435" LastActivityDate="2018-05-21T11:55:03.197" Title="Calculating diesel volume in a horizontally-cylindrical tank" Tags="|fluid-mechanics|fluid|" AnswerCount="3" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20965" PostTypeId="2" ParentId="8362" CreationDate="2018-03-21T10:59:04.770" Score="1" Body="&lt;p&gt;Let's find the concepts of acceleration/velocity response and the pseudo counterparts from beginners' perspective.&lt;/p&gt;&#xA;&lt;p&gt;The derivation lies in the dynamic equilibrium equation of a single-degree-of-freedom oscillator, characterized by natural circular frequency &lt;span class=&quot;math-container&quot;&gt;$\omega_\mathrm{n}$&lt;/span&gt; and damping ratio &lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt;. Assume it is under the action of ground motion, represented by the ground acceleration &lt;span class=&quot;math-container&quot;&gt;$\ddot{u}_{\rm g}$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ddot{u}+2\zeta\omega_{\rm n}\dot{u}+\omega_{\rm n}^2u+\ddot{u}_{\rm g}=0.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here, the response of the oscillator &lt;span class=&quot;math-container&quot;&gt;$u(t),\dot{u}(t),\ddot{u}(t)$&lt;/span&gt; are defined relative to the shaking ground, not to a fixed point. Hence, the absolute acceleration is denoted as superposition of ground and the relative part:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$a_{\rm abs}(t)=\ddot{u}_{\rm g}(t)+\ddot{u}(t).$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Given a ground acceleration &lt;span class=&quot;math-container&quot;&gt;$\ddot{u}_{\rm g}$&lt;/span&gt;, the differential equation could be solved either analytically or numerically. Upon further analysis, it is evident that:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$a_{\rm abs}=\ddot{u}+\ddot{u}_{\rm g}=-(2\zeta\omega_{\rm n}\dot{u}+\omega_{\rm n}^2u),$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;since the damping ratio is usually very small, e.g. 5% for most structures, the term &lt;span class=&quot;math-container&quot;&gt;$2\zeta\omega_{\rm n}\dot{u}$&lt;/span&gt; can be ommited compared to &lt;span class=&quot;math-container&quot;&gt;$\omega_{\rm n}^2u$&lt;/span&gt;, especially in lightly-damped and high-frequency cases. Thus, the approximate acceleration is proportional to the relative displacement &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt;, termed &lt;strong&gt;pseudo-acceleration&lt;/strong&gt;. In many cases, only the value (not the sign) of the response is of interest. Therefore, the pseudo-acceleration is defined as:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$a_{\rm psu}=\omega_{\rm n}^2u.$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Pseudo-acceleration can be computed straightforwardly by multiplying &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$\omega_{\rm n}^2$&lt;/span&gt;, eliminating the need for double differentiation of &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt;. Notably, while pseudo-acceleration serves as an approximate estimate of absolute acceleration, it remains significant as it correlates with the base shear of the SDOF system( the term &lt;span class=&quot;math-container&quot;&gt;$2\zeta\omega_{\rm n}\dot{u}$&lt;/span&gt; associated with velocity having minimal impact on the base).&lt;/p&gt;&#xA;&lt;p&gt;Similarly, pseudo velocity, &lt;span class=&quot;math-container&quot;&gt;$v_{\rm psu}=\omega_{\rm n}u,$&lt;/span&gt;&#xA;is usually used as approximation of relative velocity.&lt;/p&gt;&#xA;&lt;p&gt;If we have calculated the responses a series of sdof oscillators within a scope of natural periods/frequencies, to the same ground motion and then draw periods versus the maximum responses in a diagram, a response spectrum is produced.&lt;/p&gt;&#xA;&lt;p&gt;The type of maximum response distingushes the type of response spectrum. If maximum absolute acceleration response is taken, it is an acceleration response spectrum, same for pseudo-acceleration, velocity, pseudo-velocity.&lt;/p&gt;&#xA;" OwnerUserId="15437" LastEditorUserId="15437" LastEditDate="2024-03-26T01:30:55.763" LastActivityDate="2024-03-26T01:30:55.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="20967" PostTypeId="2" ParentId="20941" CreationDate="2018-03-21T12:11:10.623" Score="0" Body="&lt;p&gt;I think of a few benefits of adding the paper inside:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I think the glass is weaker from the inside than the outside, so any hard object somehow falling towards the inside of the glass would have a cushion.&lt;/li&gt;&#xA;&lt;li&gt;In case the glass breaks, the paper inside might also be a better (or at least an extra) barrier for shards, possibly avoiding a chain reaction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15440" LastActivityDate="2018-03-21T12:11:10.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20968" PostTypeId="1" CreationDate="2018-03-21T12:22:35.357" Score="2" ViewCount="234" Body="&lt;p&gt;Wanted to press punching of 0.22mm dia (Bottom) and on (top) 0.6mm dia &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest how can I make such die?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7WDni.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7WDni.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15441" LastActivityDate="2024-03-26T02:20:20.447" Title="How Can I do punching of 0.22mm in stainless steel sheet (Sheet thickness 0.5mm)" Tags="|hydraulics|metals|power-engineering|cnc|drilling|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20970" PostTypeId="1" CreationDate="2018-03-21T13:53:28.880" Score="1" ViewCount="381" Body="&lt;p&gt;First of all, this is &lt;em&gt;not&lt;/em&gt; a homework question.&#xA;I simply have a hard time finding relevant sources or help. Therefore, I am asking this question here. Hopefully, someone can help me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to calculate the analytic solution of the deflection curve of the L-Shaped cantilever beam as shown in the following image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/utxuk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/utxuk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is difficult, because the force acts out of the plane, and will therefore induce an out of plane bending of the cantilever. The cantilever will exhibit torsinal and bending stress. &lt;/p&gt;&#xA;" OwnerUserId="9039" LastActivityDate="2018-03-21T15:41:27.663" Title="Calculate Out of Plane Deflection Curve" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="20971" PostTypeId="1" AcceptedAnswerId="20984" CreationDate="2018-03-21T14:26:35.027" Score="1" ViewCount="78" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GIJTr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GIJTr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion was that they are for Tamper-evident band, but there isn't one beneath the cap.&lt;a href=&quot;https://i.stack.imgur.com/fIVMQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fIVMQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15443" LastEditorUserId="15443" LastEditDate="2018-03-21T15:00:20.183" LastActivityDate="2018-03-22T04:04:51.080" Title="Why there are grooves at inner side of plastic nasal spray cap?" Tags="|plastic|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="20972" PostTypeId="1" CreationDate="2018-03-21T15:10:59.260" Score="2" ViewCount="245" Body="&lt;p&gt;Let's say I have a fish tank with a total volume of 0.25m3 (100x50x50cm) filled with 160l of water. There's no lid and the thing is placed in a room with sea level atmospheric pressure. The whole thing hasn't been moved for a while so the water is absolutely calm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if I were to lift the tank by 3cm in a seesaw motion in order for the water to start flowing and forming a wave, how would I go about calculating the maximum pressure the water puts on the walls? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course I'd wait for a fixed amount of time before lifting the other side of the tank so the wave stays the same in height. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I could do the calculations myself but I have no idea where to start or even where to look for a hint (piston pressure? mass inertia? &#xA;Bernoulli's equation?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I need this (or at least I think so) is to calculate the wall thickness. It's basically like a wave motion machine, but open and filled with water: &lt;a href=&quot;http://www.waynesthisandthat.com/images/wmm%201.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Photo&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15444" LastEditorUserId="15444" LastEditDate="2018-03-21T16:40:38.110" LastActivityDate="2024-02-19T22:07:40.883" Title="Calculate max. pressure of liquid in motion" Tags="|fluid-mechanics|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="20973" PostTypeId="2" ParentId="20961" CreationDate="2018-03-21T15:18:25.587" Score="2" Body="&lt;p&gt;Your problem appears to have a commercial solution. Using the Google, I searched for &quot;locking grease coupler&quot; and found a number of products with guaranteed no-leak connections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://locknlube.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://locknlube.com/&lt;/a&gt; appears to be using fittings with a release mechanism that can be separated from the attached hose, or perhaps the other end of the hose can be modified for your specific application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yTRNp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yTRNp.png&quot; alt=&quot;locking coupler&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.gurtech.co.za/locknlube-grease-coupler/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.gurtech.co.za/locknlube-grease-coupler/&lt;/a&gt; references a locking seal up to 690 bar / 10,000 psi with no release or leaking. It also appears to have a threaded joint to which other hoses could be attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cl4Gs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cl4Gs.png&quot; alt=&quot;locking fitting&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's feasible to consider that the locking lever can be removed if access is tight/restricted, as long as the spring forces are not modified in the process.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-03-21T15:18:25.587" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20974" PostTypeId="2" ParentId="20970" CreationDate="2018-03-21T15:35:42.200" Score="2" Body="&lt;p&gt;For the L2 part the deflection is $$PL^3/3EI $$ considering it a simple cantilever beam with  L= L2-W1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The L1 with the L = L1+ W2 has two deflections. first acting as a cantilever under load P. Second twisting under the torque P(L2-W1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is simplifying the stresses in the small corner rectangular W1.W2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Roark Formulas hanbook,&#xA;twist per inch length = T/(KG)&#xA;$$K = ab^3*[16/3-3.36*(b/a)*(1-((b/a)^4)/12)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = long side, W ; b = short side, T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T =Torque in-lb&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L= Length in inches&lt;/p&gt;&#xA;&#xA;&lt;p&gt;G = Modulus of rigidity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is twist per inch length, you multiply by L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now add the deflection of the L1 under cantilever action to the axis of L1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can modify theese results by considerin the small rectangula W1.W2 strains too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is kind of a rough first estimate. Just to give you preliminary numbers.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-03-21T15:41:27.663" LastActivityDate="2018-03-21T15:41:27.663" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20975" PostTypeId="1" CreationDate="2018-03-21T15:42:54.867" Score="0" ViewCount="64" Body="&lt;p&gt;Steels 316L, 420, 17-4PH and 316LVM are often chosen for medical products and labelled as &quot;surgical grade&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious as to how these types of steel are used for the production of common medical tools, for example ball-tipped tools such as &quot;dental ball burnishers&quot;. Are these cast? If so, I suppose the parting line would be visible along the tool. Or alternatively, could the steel ball, rod and handle be manufactured separately and then welded? I guess this doesn't make much economic sense. As a third approach, I have noticed piercing parts being manufacture by screwing threaded steel balls on threaded rods. Again more labour intensive than casting, but less setup costs (no molds needed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XUyLX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XUyLX.jpg&quot; alt=&quot;ball-tip dental burnisher&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15445" LastActivityDate="2018-04-21T00:10:52.990" Title="Surgical steel and part production" Tags="|steel|manufacturing-engineering|biomedical-engineering|casting|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20976" PostTypeId="2" ParentId="20964" CreationDate="2018-03-21T16:17:37.813" Score="0" Body="&lt;p&gt;Use the formula $$ A = (R^2/2)*(\theta - sin\theta) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $$R-H = cosa$$ and $$a =\theta/2$$&#xA;A is the surface of Diesel liquid inside the tank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;R is the radius of the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H is height of fuel in the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\theta$ is the central angle between the 2 sides of section made by fuel to center of the tank.&#xA;Once you find A the volume is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V= A.L$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L is length of tank.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-21T16:17:37.813" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20979" PostTypeId="1" CreationDate="2018-03-21T19:04:10.003" Score="0" ViewCount="48" Body="&lt;p&gt;My lab uses a home-built biomedical instrument that acts as a ventilator for test-subjects. In other words: our device combines nitrogen and oxygen gases and pumps them down the line at a specified rate. The actuation is controlled with three parameters: breath rate (breaths/min), inhalation duration, and breath hold duration. While it is running, it considers these parameters and actuates accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to calibrate our device, we have regulators on each gas line that control their pressures. This obviously changes the gas volume output at the end of the line. &lt;strong&gt;What is an easy way for us to measure the volume of gas flow out of the system?&lt;/strong&gt; I'm &lt;em&gt;trying&lt;/em&gt; to avoid buying any devices if I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This device is very low output, approximately 0.25 mL/breath or ~40 mL/min. I had originally thought to use a small syringe, but the backpressure was too much for the ventilator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our current method for measuring volume output is to fill a bucket with water, place a water-filled graduated cylinder on the surface, and then have the ventilator actuate several times to displace the water in the cylinder, measure that, and then divide by the number of actuations to get volume/breath.&lt;/p&gt;&#xA;" OwnerUserId="15449" LastEditorUserId="15449" LastEditDate="2018-03-22T15:36:45.173" LastActivityDate="2018-03-22T15:36:45.173" Title="Measuring gas flow from a low-output device" Tags="|pressure|gas|biomedical-engineering|medical-devices|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="20980" PostTypeId="1" CreationDate="2018-03-21T19:47:26.017" Score="1" ViewCount="56" Body="&lt;p&gt;By “scratch” I mean, if someone with all the knowledge needed were put in the middle of a plot of land containing all the BASE COMPONENTS (e.g. iron, copper, whatever other materials you would need) needed to build a computer, would they be able to feasibly create a &lt;em&gt;modern computer&lt;/em&gt; (such as your average desktop pc) by themselves within their lifetime? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I haven’t seen this &lt;strong&gt;specific&lt;/strong&gt; question asked anywhere so not sure if this is the right topic to be posting in. Just took my best guess.&lt;/p&gt;&#xA;" OwnerUserId="15450" LastActivityDate="2018-03-21T19:47:26.017" Title="Could a single person build a single modern computer from scratch" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="0" CommentCount="10" ClosedDate="2018-03-23T01:01:45.193" ContentLicense="CC BY-SA 3.0" />
  <row Id="20981" PostTypeId="2" ParentId="20975" CreationDate="2018-03-21T21:30:50.750" Score="1" Body="&lt;p&gt;When I was a boy; We made a eight shapes ( as I remember) of parts like that for Ritter. They were produced by investment casting using 17 - 4 PH. There was a die witness line that carried over from the plastic pattern; it was polished away during tumble polishing. Medical equipment also uses 440 A, B and C for cutting tools. Vitalium may still have some use also.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-03-21T21:30:50.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20982" PostTypeId="2" ParentId="20972" CreationDate="2018-03-22T00:25:09.690" Score="0" Body="&lt;p&gt;Assuming you time your rocking motion in a way that you get a sinusoidal wave going nicely back and forth you may just add 3*2= 6 centimeter to the height of water. This is without slushing force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For estimating rough numbers for slushing, I would consider the potential energy of a test rectangular sliver of water standing half the 3 cen at the middle of the tank width as a tide turning into kinetic energy at the wall, without friction breaking into wall of tank. Then you pick another test point along the width and the height and valume corresponding to this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 3-4 trial pints you add the maximum value to the prior value you got for hydrostatic pressure. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-22T00:25:09.690" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20984" PostTypeId="2" ParentId="20971" CreationDate="2018-03-22T04:04:51.080" Score="3" Body="&lt;p&gt;that part was injection molded in an unscrewing mold which allows fully-formed threads to be molded into a part. Ordinarily, the resulting undercuts would prevent the part from being extracted from the mold but in an unscrewing mold, a motorized shaft inside the mold unscrews the finished part from the mold after the plastic has solidified and the two halves of the mold are being separated. those features in the cap are engagement splines that mate into the rotating shaft that unscrews the finished part out of the threaded half of the mold cavity. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-03-22T04:04:51.080" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20985" PostTypeId="1" CreationDate="2018-03-22T04:22:13.390" Score="1" ViewCount="37" Body="&lt;p&gt;. . A thought recently crossed my mind which I cant seem to find any studied material on. It occurred while hearing a story about a potential future landslide in Kauai causing a large tsunami aimed at Japan; similar to this July 2017 landslide in Greenland &lt;a href=&quot;https://www.nature.com/news/huge-landslide-triggered-rare-greenland-mega-tsunami-1.22374&quot; rel=&quot;nofollow noreferrer&quot;&gt;[1]&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story made me wonder, can a breakwater/seawall built around any potential landslide area prevent the future buildup of a tsunami; or maybe a significant portion of its energy transfer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, I am &lt;strong&gt;not&lt;/strong&gt; writing about defensive measures that articles mostly seem to speak of &lt;a href=&quot;https://www.independent.co.uk/news/world/asia/japans-sea-wall-storm-brews-over-plans-to-construct-giant-5bn-barrier-against-tsunamis-a6914781.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;[2]&lt;/a&gt;. Instead, I'm interested in preventative measures for controllable instances. Thanks for any insight.&lt;/p&gt;&#xA;" OwnerUserId="15454" LastEditorUserId="1832" LastEditDate="2018-03-22T08:54:52.263" LastActivityDate="2018-03-22T15:48:19.463" Title="the possibility of a breakwater/seawall preventing the beginnings of a landslide-formed tsunami" Tags="|fluid-mechanics|civil-engineering|design|dynamics|" AnswerCount="1" CommentCount="1" ClosedDate="2018-03-24T09:48:19.573" ContentLicense="CC BY-SA 3.0" />
  <row Id="20986" PostTypeId="2" ParentId="20979" CreationDate="2018-03-22T04:31:43.503" Score="0" Body="&lt;p&gt;You can attach the system's outlet to the top of a bucket through a valve on the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This bucket at it's bottom and on the other side is connected to an open riser pipe going up just to the top of container to provide a gravity seal. The riser will allow overflow water to discharge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you open the valve for one minute, or the time count of your choice and shut it back. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of water drained out through riser is the volume of output of your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your system generates large volume of gas outlet use more buckets connected on parallel network.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-22T04:31:43.503" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20988" PostTypeId="1" CreationDate="2018-03-22T08:46:18.313" Score="0" ViewCount="38" Body="&lt;p&gt;There is a wall-opposed flame monitor detecting coal, oil and gas combustion in the burners, attached to the furnace walls in a coal-fired power plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for suggestions to clear the dust or particulates from the lens, as it is affecting the signal and leads to tripping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, it is cleaned by supplying purged air from the core air fan. However, this has been found to not be effective as the contaminants still exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the cause of such an ineffective system? Could the purge air system be improved by using higher pressure air from other components?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much appreciated&lt;/p&gt;&#xA;" OwnerUserId="15458" LastActivityDate="2018-03-22T09:53:14.113" Title="Ineffective purging of a flame monitor lens" Tags="|mechanical-engineering|power-engineering|boilers|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="20990" PostTypeId="1" AcceptedAnswerId="20992" CreationDate="2018-03-22T10:50:33.193" Score="0" ViewCount="1925" Body="&lt;p&gt;I'm trying to fit a shaft into a hole in an acrylic piece that is 5mm thick to cause an interference fit. The shaft has a max. diameter of ⌀5mm and a min. diameter of ⌀4.988mm. I have access to a laser cutter with a precision of 0.2mm. This basically means that i have to cut the hole in the acrylic to a diameter of less than ⌀4.788mm, to ensure an interference fit right? Lets say I cut the hole in the acrylic, using a diameter of ⌀4.788mm, the diameter could be as low as ⌀4.588mm. Would the shaft of ⌀5mm even fit in this hole? Given the material and uncertainty in the laser cutter, what diameter do you think is suitable? Thanks&lt;/p&gt;&#xA;" OwnerUserId="15061" LastEditorUserId="15061" LastEditDate="2018-03-22T16:07:22.510" LastActivityDate="2018-03-22T16:07:22.510" Title="Interference fit in acrylic/plexiglass (PMMA)" Tags="|mechanical-engineering|materials|tolerance|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="20991" PostTypeId="2" ParentId="20990" CreationDate="2018-03-22T14:01:02.823" Score="1" Body="&lt;p&gt;Presuming laser cutter and not laser printer, the project you suggest would be most effectively approached by creating test pieces. With 5 mm thick acrylic, there are enough variables to make it difficult to provide certain answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cast acrylic behaves differently from extruded acrylic. High power lasers are able to use higher travel speeds than lower powered lasers, changing the characteristics of the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 5 mm diameter hole is small enough that high travel speeds are not likely regardless of the performance of the laser. Laser cutters typically slow for circles and curves and especially for right angle corner cuts. This changes the profile of the cut and the surface rendered by such a cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your project allows for creating test pieces, consider to use a variety of hole sizes and cutting profiles, testing your holes with pin gauges or with the shaft to determine correct settings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kerf settings in the software that create the laser job can be adjusted to provide more precision for inside dimensions, but I suspect the tests would give you simpler results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With such a thick material, a low power laser may have some divergence as it nears the lower edge of the material and may create a non-cylindrical hole. The variation may not be a factor in your project, but the above suggested tests will ascertain that consideration.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-03-22T14:01:02.823" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20992" PostTypeId="2" ParentId="20990" CreationDate="2018-03-22T14:11:18.063" Score="1" Body="&lt;p&gt;There are a couple of issues. Firstly laser cutting may well not provide good enough tolerances or surface finish to reliably produce the fit you need and you woudl probably be better off drilling the holes, possibly using the laser to create a reference mark or pilot hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly poly-carbonate and acrylic don't really like this sort of interference fit and the advice is generally to give holes decent clearance as the stresses involved tend to cause cracks, especially if the holes aren't polished and chamfered. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would probably be better off drilling a hole with adequate clearance and using an adhesive to secure the shaft. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-03-22T14:11:18.063" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20993" PostTypeId="2" ParentId="20979" CreationDate="2018-03-22T14:44:52.367" Score="1" Body="&lt;p&gt;You can use a float type flowmeter:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Cv5xx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cv5xx.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;A float sits in a conically expanding tube, the higher the flow rate the higher the float hovers. In the closest catalog I have at hand, the smallest meter is for flow rates  in 0.2 m³/h - 2.5 m³/h range (gas applications).&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-03-22T14:44:52.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20994" PostTypeId="2" ParentId="20985" CreationDate="2018-03-22T15:48:19.463" Score="0" Body="&lt;p&gt;As a minimum, you'd need a catch-basin roughly the volume of the anticipated landslide.  That is a very large volume.  But that's not enough.  The reason a tsunami wave is formed is that the landslide mass transfers most of its energy to creating a near-plane-wave front, which means there's not much directional dissipation.  And what that means is that your proposed seawall would have to be strong enough and mechanically designed well enough to force energy dissipation in the plane perpendicular to the direction of motion of the wave.  That's not going to be possible, basically.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;[if only what-if  were still active, I'm sure Randall would jump all over the numerical analysis here]&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-03-22T15:48:19.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20995" PostTypeId="2" ParentId="14129" CreationDate="2018-03-22T16:45:40.123" Score="4" Body="&lt;p&gt;For a MIMO system $y(s) = G(s)d(s)$, with $m$ inputs and $l$ outputs.&#xA;Consider a fixed frequency $\omega$ where $G(j\omega)$ is a constant $l \times m$ complex matrix.&#xA;For the sake of simplicity the matrix $G(j\omega)$ is written as $G$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the singular value decomposition (SVD) states that any matrix $G$ may be decomposed into an input rotation $V$, a scaling matrix $\Sigma$ and an output rotation $U$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G = U \Sigma V^H$$&#xA;where $V^H$ denotes the conjugate transpose or Hermitian transpose of matrix $V$.&#xA;Furthermore,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$U$ is an $l \times l$ unitary matrix of output singular vectors, $u_i$; &lt;/li&gt;&#xA;&lt;li&gt;$V$ is an $m \times m$ unitary matrix of input singular vectors, $v_i$;&lt;/li&gt;&#xA;&lt;li&gt;$\Sigma$ is an $l \times m$ diagonal matrix with $k =&#xA;   \min\lbrace l,m\rbrace$ non-negative singular values, $\sigma_i$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The gain of matrix $G$ in the $i$'th direction (from input $v_i$ to output $u_i$) is given by the singular value $\sigma_i$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, for a SISO system, the singular value $\sigma$ describes the amplification from the input to the output on frequency $\omega$.&#xA;This information can also be obtained from the Bode magnitude response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why the MATLAB documentation of the function &lt;em&gt;sigma&lt;/em&gt; states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The singular value response of a SISO system is identical to its Bode magnitude response.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2018-03-22T16:45:40.123" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="20998" PostTypeId="1" CreationDate="2018-03-22T19:01:21.143" Score="3" ViewCount="170" Body="&lt;p&gt;I would like to know what test can be done to quickly measure the modulus of elasticity of soil with varying parameters (density, moisture content etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method I know is using the oedometer test. By varying the vertical stress and measuring the variation of strain, we can measure the modulus of elasticity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the oedometer test can take a lot of time. (48 hours). Is there a way that is quicker?&lt;/p&gt;&#xA;" OwnerUserId="15462" LastActivityDate="2019-12-26T05:19:33.967" Title="What methods exist to measure the modulus of Elasticity of Soil?" Tags="|civil-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="20999" PostTypeId="1" CreationDate="2018-03-22T19:20:59.757" Score="1" ViewCount="32" Body="&lt;p&gt;I am trying to use Ansys Workbench (Transient Thermal probably coupled with Static Structural analysis) to simulate behavior of a fire protection sleeve. It is used to seal the fire sectors in place where the combustible installations go through walls. Basically, as the temperature goes up, the sleeve expands (its volume goes up rapidly), it cuts out for example a plastic pipe and seals the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to model this in Ansys?&lt;/p&gt;&#xA;" OwnerUserId="12057" LastActivityDate="2018-03-22T19:20:59.757" Title="Modeling high thermal expansion under fire" Tags="|civil-engineering|ansys|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21000" PostTypeId="2" ParentId="18334" CreationDate="2018-03-22T19:56:34.827" Score="1" Body="&lt;p&gt;Most SAGD facilities inject steam at much higher temperatures and pressures (often over 300°C).  Downhole temperatures may be 180°C at 2 MPa.&lt;/p&gt;&#xA;" OwnerUserId="2277" LastActivityDate="2018-03-22T19:56:34.827" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21001" PostTypeId="2" ParentId="20998" CreationDate="2018-03-22T22:28:51.870" Score="2" Body="&lt;p&gt;You could use cone penetrometer tests and compare the results to local data or geotechnical reports. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can give a rough estimate to begin with. In many urban areas in the U.S. there's is enough tests and reports to use.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="16486" LastEditDate="2019-12-22T19:10:45.810" LastActivityDate="2019-12-22T19:10:45.810" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21002" PostTypeId="1" AcceptedAnswerId="21004" CreationDate="2018-03-23T03:11:30.910" Score="1" ViewCount="5512" Body="&lt;p&gt;I have a friend who isn't properly visualizing energy loss in a moving car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;His idea:&lt;/strong&gt; Cars that can run forever without being recharged  - - &gt; while the engine rotates the front wheels the move the car forward, there is a device at the back wheels using the back wheels' rotation to generate and store power. The engine draws this restored power to move the car forward and the cycle continues. Now, this is a &quot;perfect&quot; scenario where there's zero energy loss through friction, heat, sound, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I first told him that something similar was already being done: regenerative braking. He said his idea was to generate energy **while*  the car was moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My counter to why that was rubbish:&lt;/strong&gt; &#xA;I reminded him the amount of energy in a given system remains constant. Nothing created or destroyed, just transferred form one form to another. Thus he could not &quot;recover&quot; energy that was being spent to move the car forward. He moved on to elaborate that there would be two engines, one rotating the front wheels with an attached dynamo to recover energy, and another rotating the back wheels. I tried to explain he had added another energy consumer, with no way to recover this extra energy being spent...and that's how it would go. You add an extra engine to overcome the dynamo's resistance and move the car, &lt;em&gt;you are spending more than you are recovering&lt;/em&gt;, and will eventually have to charge up. To help him visualize, I made a sketch, trying to be more realistic: &lt;a href=&quot;https://i.stack.imgur.com/IQsSF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IQsSF.jpg&quot; alt=&quot;How I imagine movement/expenditure of energy in vehicle&quot;&gt;&lt;/a&gt; (please ignore my use of battery capacity, it's only there for the concept, not to be realistic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also mentioned to him that this scenario was even ignoring the other loads of problems and dangers that present themselves with having wheels rotating at different speeds. Plus the wear on the slower rotating ties which are effectively being dragged against the road. For some never-to-be-discovered reason, he simply could not  understand that to get the slower (front) tires to NOT be a problem, you have to increase the front engine power output for to match the wheel rotation speed of the rear wheels. Again, only recovering a fraction of that energy (where did the rest of it go? He can't understand and I can't figure out how to explain it to him). To counter my logic, he proceeded to mention that trucks don't have an engine attached to all their wheels. Hence, some energy can be captured from those &quot;free-spinning&quot; wheels. &lt;em&gt;Oh my God&lt;/em&gt;. &#xA;Newton's first law of motion applies here: A car with the engine rotating the wheels towards a given direction goes nowhere if say, a dynamo is offering equivalent or greater resistance in the opposite direction to capture that energy (again, assuming a &quot;perfect&quot; scenario with no energy loss). I tried to elaborate that to move the vehicle, you first exert a force (engine power output) great enough to move the vehicle from its static position (resistance? gravity? I don't understand physics enough to know what to call this), and anything like a dynamo (friction, resistance) serves to make that more difficult, so you can't simultaneously &quot;recover&quot; energy you are spending. I proceeded to explain that regenerative braking worked because the generated energy was being recovered from the moving vehicle's momentum through offering resistance, hence why the car slows down and why you can't sustain-ably use it while the car is in motion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This guy is super stubborn to being corrected once a very dim bulb lights up in his mind and he believes it a genius idea and can't be dis-proven. I am horrendous at teaching constructively, and would appreciate someone breaking down why his idea is a victim of basic laws of physics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I even stated if that were possible, it would already have popular applications in industry like electric vehicles.&lt;/p&gt;&#xA;" OwnerUserId="15465" LastEditorUserId="15465" LastEditDate="2018-03-23T03:19:46.543" LastActivityDate="2018-12-30T17:25:15.553" Title="Why can't energy be entirely recovered from a moving car's rotating wheels?" Tags="|automotive-engineering|energy|regenerative-braking|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21003" PostTypeId="2" ParentId="21002" CreationDate="2018-03-23T04:24:59.147" Score="5" Body="&lt;p&gt;There are some people who cannot be told. They think they know everything and that they are never wrong. My advice is do not try to convince your friend that your friend is wrong. You will always lose the argument. State what you think and if your friend rejects that, leave it at that and talk about something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing you can do is to tell him to &quot;prove&quot; his ideas by making such devices himself. Hopefully your friend has enough intelligence to eventually realize that your were correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your ideas are sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What your friend is assuming is that is possible to construct a perpetual motion device. Which sensible people know cannot be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason why your friend is wrong is that your friend is not considering energy loss via friction, electrical resistance and the generation of heat and noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these forms of energy loss may be reduced, but never eliminated.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-03-23T04:24:59.147" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21004" PostTypeId="2" ParentId="21002" CreationDate="2018-03-23T06:07:49.290" Score="7" Body="&lt;p&gt;Get a small DC motor from a cheap toy or something and put a gear or a flywheel or something on the shaft to make it easy to spin with your fingers. You should find these in the toy as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask your friend to spin the motor while nothing is attached to the terminals. He will find that it spins easily and some momentum keeps it going for a short while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, connect a small light to the terminals and ask him to spin the motor again. He will see the bulb light up but also feel that it is much harder to rotate the shaft and it brakes immediately after he stops spinning it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way you can show him that he has to personally put in extra effort (energy) to generate electricity to light up the bulb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking this concept back to the vehicle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tell him that if you connect that motor to the free-spinning wheels, the wheels won't be free spinning anymore, they will feel the same resistance he felt. Thus, to keep the rear wheels spinning at the same rate, you need to put more energy into the front (driving) wheels in order to overcome that resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to add extra energy to the driving wheels, it is defeating the purpose of generating energy from the rear, &quot;not-anymore-free-spinning&quot;, wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Fred said, he will probably still not change his mind, but I think this is a good illustration for someone that is open for persuasion or are just starting to learn these concepts.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastActivityDate="2018-03-23T06:07:49.290" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21005" PostTypeId="2" ParentId="21002" CreationDate="2018-03-23T07:17:18.587" Score="2" Body="&lt;p&gt;The problem with his logic is once you attach a dynamo you no longer have a &quot;free-spinning&quot; wheel. The dynamo produces resistance (braking). Even if the dynamo has no loss it only produces the power needed to spin it. In the perfect scenario on blocks could spin the wheels but the car has no net horse power.  The engine consumes all the power produced and the car is not moving. The car has no power to propel itself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say this was a generator on a dam.  If there was zero loss you could pump the water back up into the dam but you produce no net power.  All the power is used to pump the water.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-03-23T16:01:48.847" LastActivityDate="2018-03-23T16:01:48.847" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21006" PostTypeId="2" ParentId="20998" CreationDate="2018-03-23T09:32:27.143" Score="3" Body="&lt;p&gt;A plate loading test is relatively quick and will give you a modulus of elasticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a section on plate loading tests in EN 1997-2 (eurocode for geotechnical investigations and tests), and they are also covered by a specific execution standard, EN ISO 22476 part 13.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2018-03-23T09:32:27.143" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21008" PostTypeId="2" ParentId="20918" CreationDate="2018-03-23T20:50:59.960" Score="1" Body="&lt;p&gt;Without knowing too much about Aspen (I've been looking for a chemical engineering modeling software, so thank you), the documentation data says &lt;strong&gt;RGibbs&lt;/strong&gt; &lt;em&gt;doesn't pay attention to stoichiometry&lt;/em&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The (Rgibbs Reactor) provides reaction calculations without the need for detailed stoichiometry or yield. The calculations are based on minimizing the Gibbs energy for the system as discussed in page 98&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This reactor is intended for systems where the exact reactant composition is unknown but the energy is, such as &quot;crude oil&quot; or &quot;natural gas&quot;.  As such, this reactor won't notice a change in C concentration.  It is probably adding the additional thermal energy to change the concentration of the higher thermal energy products, CO and H2, as it is ignoring the Redox mechanism that would force the consumption of CO to generate CO2 to form H2.  A better reactor would be the &lt;strong&gt;REquil&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When one or more reactions involved are equilibrium reaction, the REquil block can be used. The block requires knowledge of the reaction stoichiometry, and performs chemical and phase equilibrium reactions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using this method would be a better model as the equilibrium constant's dependence on temperature is well documented:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$K_{eq} = exp(-2.4198 +0.0003855T + \frac{2180.6}{T})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $T$ is in Kelvins, from the range of 600K - 2000K.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-03-26T21:57:08.240" LastActivityDate="2018-03-26T21:57:08.240" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21009" PostTypeId="2" ParentId="20947" CreationDate="2018-03-23T21:03:05.973" Score="1" Body="&lt;p&gt;As @mustang says, the symmetry of the Cauchy stress is a direct consequence of the conservation of angular momentum.  However, angular momentum is not conserved in a non-inertial frame and one has to use an inertial frame of reference to take advantage of stress symmetry.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-03-23T21:03:05.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21014" PostTypeId="2" ParentId="20961" CreationDate="2018-03-24T01:50:53.907" Score="0" Body="&lt;p&gt;Did you also consider a 90 or 60 degree grease fitting. It would be a cheaper solution to piping to a remote manifold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mcmaster.com/#grease-fittings/=1c3ojei&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mcmaster.com/#grease-fittings/=1c3ojei&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-03-24T01:50:53.907" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21015" PostTypeId="2" ParentId="19873" CreationDate="2018-03-24T02:04:10.157" Score="0" Body="&lt;p&gt;I would check McMaster and Misumi to see if a a stock bearing or flange bearing would fit. If not get one sizer smaller and (hand) ream to inside diameter needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mcmaster.com/#bronze-bearings/=1c3onde&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mcmaster.com/#bronze-bearings/=1c3onde&lt;/a&gt;&#xA;&lt;a href=&quot;https://us.misumi-ec.com/vona2/mech/M0100000000/M0107000000/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://us.misumi-ec.com/vona2/mech/M0100000000/M0107000000/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-03-24T02:04:10.157" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21018" PostTypeId="1" CreationDate="2018-03-24T08:39:41.753" Score="0" ViewCount="592" Body="&lt;p&gt;I have a modified pico video projector that I need to add an autofocus to.&#xA;The focusing ring already has a fast stepper motor added to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the focus had to be done on the middle of the projection frame I suppose I could just put a laser distance sensor in the optical path of the projection by having a 45 degree IR reflect, visible spectrum pass dichro attached after and maybe before the final focusing and zooming lens. But I want the ability to focus to any point in the frame. This requires the sensor to be able to be rotated, or if it goes through lenses, tilted in two axis. I could do that with a small gantry system with stepper motors and the distance sensor attached to it, or galvonometer mirrors that would reflect the distance sensor beam to a given point/angle, but there may be a better way or a better sensor for this I am not aware of which wouldn't require so many parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Requirements are ability to measure distance up to 4 meters, accuracy about at least +-5cm up to 1 meter, +-10cm after that, which I believe is not much to ask from a sensor and acceptable accuracy for a video projection focus. Measuring on shiny or reflective surfaces don't need to be supported as projectors don't project properly on those.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having a camera and software which focuses until software finds the sharpest result would work if it was fast enough (200 ms). As I understand with this method the lens is focused until the sharpest image is detected which is slow.&lt;/p&gt;&#xA;" OwnerUserId="14841" LastEditorUserId="14841" LastEditDate="2018-03-25T18:50:11.737" LastActivityDate="2019-12-16T17:32:32.187" Title="how to implement autofocus for video projector" Tags="|sensors|optics|lighting|photonics|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21019" PostTypeId="1" CreationDate="2018-03-24T10:02:31.293" Score="1" ViewCount="309" Body="&lt;p&gt;I would like to calculate the spring constant of a frame structure.&#xA;I thought that I could do it like this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I apply a point force on the structure:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WXMGo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WXMGo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then I calculate the displacement at that same point:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/R0YJa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R0YJa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Finnaly, I asume that the material bends within its elastic limit:&#xA;$k=\frac{F}{\delta_y}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question1: Do you agree with this method ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question2: If I have a different type of loading, for instance a distributed load, would the spring constant change ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="9039" LastActivityDate="2018-03-25T02:24:46.473" Title="Find spring constant of frame structure" Tags="|mechanical-engineering|materials|statistics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21021" PostTypeId="1" CreationDate="2018-03-24T14:59:31.330" Score="0" ViewCount="65" Body="&lt;p&gt;Imagine I was working in the woods and dropped my cell phone somewhere in a large area where it is now hidden beneath dense vegetation.  By calling it I confirm that either it has powered off or its battery is dead.  How can I find it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A consumer metal detector (induction magnetometer) will easily find it &lt;em&gt;if&lt;/em&gt; I can get within a few feet of it.  But there must be more efficient means of finding electronics in a wide wilderness area given their unique and unnatural characteristics.  For example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;They contain dense, refined metals.  There is easily more copper, nickel, and perhaps a few other elements in a phone than in an acre of wilderness almost anywhere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They contain antennas.  Even when powered off, the electronic structures are there to receive and transmit signals efficiently in a particular spectrum.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can radar be used to discriminate either of these characteristics and provide a practical means of scanning a large area to zero in on the device?  Are there existing systems that do this?&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2018-03-25T10:15:30.017" Title="Find electronics in wilderness; or, &quot;I dropped my cell phone in the woods!&quot;" Tags="|electrical-engineering|metals|rf-electronics|radar|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21022" PostTypeId="1" CreationDate="2018-03-24T15:33:49.063" Score="1" ViewCount="4001" Body="&lt;p&gt;My professor gave us yesterday a problem to calculate the equivalent spring of the system shown in the attached figure. The problem: a cantilever beam supported by two springs $k_1$ and $k_2$ located at distances $a$ and $b$, respectively, from the left end. The beam has length $l$ and flexural rigidity $EI$. The idea is to approximate the motion of the mass in the vertical direction by a simple one degree of freedom system, so the first step on it is to calculate the equivalent stiffness coefficient of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rojcn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rojcn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having some trouble to understand his solutions, because I don't agree entirely with what he said. His approach to the problem:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Given the deflection equation of a cantilever beam with a load $P$ applied at the tip,&#xA;$$&#xA;v=\frac{-Px^2}{6EI}(3l-x)&#xA;$$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;he says an equivalent spring $k_{eq12}$ substitutes the springs $k_1$ and $k_2$ at the right-end of the beam. To calculate $k_{eq12}$, the total potential energy is&#xA;$$\begin{align}&#xA;U&amp;amp;=\frac{1}{2}k_{eq_{12}}y_l^2=\frac{1}{2}k_{1}y_a^2+\frac{1}{2}k_{2}y_b^2\\&#xA;k_{eq_{12}}&amp;amp;=k_1\left(\frac{y_a}{y_l}\right)^2+k_2\left(\frac{y_b}{y_l}\right)^2&#xA;\end{align}$$&#xA;where $y_a$, $y_b$ and $y_l$ are the deflections of the beam for a load located at $a$, $b$ and $l$. He claims $y_a$ and $y_b$ are the deflection of the springs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, for a unit load,&#xA;$$\begin{align}&#xA;y_a&amp;amp;=\frac{a^2}{6EI}(3l-a) \tag1\\&#xA;y_b&amp;amp;=\frac{b^2}{6EI}(3l-b) \tag2\\&#xA;y_l&amp;amp;=\frac{l^3}{3EI}       \tag3&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if $k_{eq_{12}}$ is an equivalent spring at the right-end of the beam, then the equivalent spring of the beam is in series with this equivalent spring, so the equivalent spring of the entire system supporting the mass is, &#xA;$$&#xA;\frac{1}{k_{eq}}=\frac{1}{k_{eq_{12}}}+\frac{1}{k_{beam}}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the deflections $y_a$, $y_b$, $y_l$, and the equivalent stiffness of the beam being $k_{beam}=3EI/l^3$, we get&#xA;$$&#xA;k_{eq}=\frac{\left( k_1\left(\frac{y_a}{y_l}\right)^2+k_2\left(\frac{y_b}{y_l}\right)^2 \right)k_{beam}}{k_1\left(\frac{y_a}{y_l}\right)^2+k_2\left(\frac{y_b}{y_l}\right)^2+k_{beam}}&#xA;$$&#xA;which can be simplified, substituting $(1)$, $(2)$, $(3)$ and the $k_{beam}$, to&#xA;$$&#xA;k_{eq}=\frac{3EIk_1a^4(3l-a)^2+3EIk_2b^4(3l-b)^2}{l^3\left[k_1a^4(3l-a)^2+k_2b^4(3l-b)^2+12l^3EI\right]}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the answer he gave us. Particularly, I don't agree with a few steps along the way, but the key point here for me is that we cannot use the deflection equation for a cantilever beam a tip load at the end, because this beam is not exactly a cantilever beam, but it has two extra supports, the springs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach to this problem would be to, given an applied load $P$ at the left end, find the deflection equation for the system with the springs and the approximate the equivalent stiffness with the new equation. The deflection equation now would have to be calculated from the $EI\frac{d^4v}{dx^4}=w(x)$ elastic curve. A much harder work, because it is a cantilever beam supported by two springs, and it is a statically indeterminate beam. Is this line of though correct or was his approach correct?&lt;/p&gt;&#xA;" OwnerUserId="15487" LastEditorUserId="1832" LastEditDate="2018-03-25T02:53:52.550" LastActivityDate="2018-03-25T02:53:52.550" Title="Equivalent spring of beam supported by springs" Tags="|mechanical-engineering|beam|vibration|springs|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21023" PostTypeId="1" CreationDate="2018-03-24T16:39:34.467" Score="0" ViewCount="88" Body="&lt;p&gt;I've been studying for a while now topics related to &lt;strong&gt;power quality&lt;/strong&gt; issues, such as &lt;strong&gt;harmonics&lt;/strong&gt; caused by &lt;strong&gt;non-linear loads&lt;/strong&gt; and &lt;strong&gt;power converters&lt;/strong&gt;, and I've been working on Active Power Filters, but since I live in developing country sadly all we have is theoretical knowledge, &lt;em&gt;So I appreciate if you add any professional knowledge to those conversation&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the commercial name of an &lt;strong&gt;active power filter&lt;/strong&gt; (if I write&#xA;that under electronics in Amazon, it doesn't give me any result)&lt;/li&gt;&#xA;&lt;li&gt;Same question goes to &lt;strong&gt;Unified Power Quality Controller&lt;/strong&gt; ? (I can't&#xA;find those in Amazon too)&lt;/li&gt;&#xA;&lt;li&gt;What are the main &lt;strong&gt;manufacturers&lt;/strong&gt; of those devices in the world?&lt;/li&gt;&#xA;&lt;li&gt;Could you recommend any &lt;strong&gt;good readings&lt;/strong&gt; about this topic to have an&#xA;idea about the industrial and marketing field&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15488" LastEditorUserId="7083" LastEditDate="2018-03-28T08:30:30.873" LastActivityDate="2018-03-28T08:30:30.873" Title="what is the commercial name of active power filters?" Tags="|electrical-engineering|power-electronics|quality-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21025" PostTypeId="1" CreationDate="2018-03-24T20:03:56.347" Score="0" ViewCount="145" Body="&lt;p&gt;I'm working on a small project where an armature is actuated by a servo via a couple feet of tendon. The servo+tendon needs to be able to pull about 5-10 lbs. There will be springs at the endpoints to allow some flexibility and shock absorption, so I don't need the tendon to incorporate much flexibility of its own. What's the appropriate type of wire to use for the tendon? Should I use Nylon? Or stranded steel? I see a lot of projects using tendons, but I don't see much description of what's the best material to use to avoid snapping and breakage.&lt;/p&gt;&#xA;" OwnerUserId="175" LastActivityDate="2018-03-24T20:45:10.090" Title="Wire type for robotic tendon?" Tags="|mechanical-engineering|pulleys|servo|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21026" PostTypeId="2" ParentId="21025" CreationDate="2018-03-24T20:45:10.090" Score="2" Body="&lt;p&gt;I have successfully used a non-stretch polymer known by various names such as Dyneema, GelSpun or Spectra for a similar purpose. It is easily purchased from most fishing supply stores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material is extremely strong and non-elastic - some care is needed with termination since it is slippery. Crimps, wire loops and model aircraft clevises are all good terminations.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2018-03-24T20:45:10.090" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21028" PostTypeId="2" ParentId="21019" CreationDate="2018-03-25T02:24:46.473" Score="1" Body="&lt;p&gt;Yes it would change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example same load if applied at quarter span would cause less deflection and a different curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence bigger K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;basically for the type of loading you display applying the load at center corrolates to smallest K.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-25T02:24:46.473" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21029" PostTypeId="1" CreationDate="2018-03-25T02:28:47.280" Score="2" ViewCount="29" Body="&lt;p&gt;Is there an alternative for the expensive HDPE Piping for long range sea water transport?&lt;/p&gt;&#xA;" OwnerUserId="15497" LastActivityDate="2018-03-25T10:31:50.060" Title="Alternative to HDPE Piping for long range sea-water + water transport" Tags="|pipelines|water-resources|piping|waste-water-treatment|" AnswerCount="0" CommentCount="0" ClosedDate="2018-03-25T21:52:20.900" ContentLicense="CC BY-SA 3.0" />
  <row Id="21030" PostTypeId="1" CreationDate="2018-03-25T06:44:24.490" Score="3" ViewCount="5888" Body="&lt;p&gt;What are the advantages of using X-form rudders in comparison to cross-form rudders?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do some submarines, torpedos and underwater autonomous vehicles use X-stern control?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X-stern:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZT8ev.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZT8ev.jpg&quot; alt=&quot;X-stern picture&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cross-form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vphtZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vphtZ.jpg&quot; alt=&quot;Cross-form&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15499" LastEditorUserId="6790" LastEditDate="2020-05-04T01:48:53.320" LastActivityDate="2020-05-04T01:48:53.320" Title="X-form and cross-form rudders in underwater control" Tags="|mechanical-engineering|fluid-mechanics|marine-engineering|naval-engineering|submarines|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21032" PostTypeId="2" ParentId="21030" CreationDate="2018-03-25T07:59:26.177" Score="0" Body="&lt;p&gt;Effectively none, the result is you can control up, down, left, right but with the x-form a combination of fins is necessary while with cross (second image) only a pair of fins is needed.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-03-25T07:59:26.177" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21036" PostTypeId="2" ParentId="21023" CreationDate="2018-03-25T09:42:33.267" Score="-1" Body="&lt;p&gt;An UPS can be used for filtered power but if still hum/harmonix /.... you can use an Op-Amp as a 'Notch' filter to filter out undesirable frequencies&lt;/p&gt;&#xA;" OwnerUserId="15500" LastActivityDate="2018-03-25T09:42:33.267" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21037" PostTypeId="2" ParentId="21018" CreationDate="2018-03-25T09:48:11.393" Score="1" Body="&lt;p&gt;U can use a mirror for feedback of the projected image and a SLR as in cameras (a piece of glass at an 45` angle to measure the difference between projected and image onthe screen and approp. adjust the focus.&lt;/p&gt;&#xA;" OwnerUserId="15500" LastActivityDate="2018-03-25T09:48:11.393" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21047" PostTypeId="1" CreationDate="2018-03-25T17:34:51.253" Score="2" ViewCount="38" Body="&lt;p&gt;I have two wind turbine blades, one that that consists of NACA 44xx airfoils and the other NACA 63-2xx airfoils. Some sections of these blades have airfoils whose maximal thickness are specified by three digits, such as .247c (where 'c' is the chord length). Is there any program that will allow me to plot such an airfoil?&lt;/p&gt;&#xA;" OwnerUserId="15350" LastActivityDate="2018-03-25T17:34:51.253" Title="Generating NACA airfoils with three digit maximal thickness numbers" Tags="|airfoils|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21049" PostTypeId="1" CreationDate="2018-03-25T23:11:57.063" Score="3" ViewCount="65" Body="&lt;p&gt;What's the longest lasting tubing for use in low pressure air systems?  I'm looking at building a greenhouse with double or triple layer ETFE panels and the tubing would be run through the metal pipe structure, so it would experience exposure to air an either steel or aluminum with low (a couple PSI or so at most) air pressures.  It would be a much bigger pain to replace the tubing than the ETFE panels so I'd like something which goes significantly beyond the 30-40 lifespan of the ETFE panels.&lt;/p&gt;&#xA;" OwnerUserId="9597" LastActivityDate="2018-05-29T16:32:40.243" Title="What's the longest lasting tubing in air?" Tags="|structural-engineering|piping|air|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21050" PostTypeId="2" ParentId="21023" CreationDate="2018-03-26T01:33:08.890" Score="1" Body="&lt;p&gt;I have procured &quot;active harmonic filters&quot; for an engineering project. We needed them to compensate for the harmonics caused by a large number of variable speed drives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Active harmonic filters are not a consumer product, so you won't find them on Amazon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try looking in the catalogues of Danfoss or Siemens, who are two manufacturers of these type of units. Other manufacturers of VSDs, i.e. ABB, are also likely to offer an active harmonic filter.&lt;/p&gt;&#xA;" OwnerUserId="1001" LastActivityDate="2018-03-26T01:33:08.890" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21053" PostTypeId="1" AcceptedAnswerId="21062" CreationDate="2018-03-26T07:23:39.533" Score="3" ViewCount="57" Body="&lt;p&gt;I have written a piece of code solving the linear elasticity problem on transversely isotropic materials (before that, the code only supported isotropic materials).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to validate it before using it for other applications, have you got any idea how to do it ? I would like to be sure that the 5 independents coefficients : E1, E3, nu12, nu13, G are well integrated in the code.&lt;/p&gt;&#xA;" OwnerUserId="15505" LastActivityDate="2018-03-26T19:37:32.517" Title="How to properly validate transverse isotropic elasticity Finite Element Code" Tags="|mechanical-engineering|finite-element-method|elastic-modulus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21054" PostTypeId="1" AcceptedAnswerId="21055" CreationDate="2018-03-26T07:34:00.590" Score="1" ViewCount="193" Body="&lt;p&gt;I am trying to design a lifting mechanism and I would love some insight/opinions as to how I can shed some weight off it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current weight I have for my mechanism is 187.5kg, however, this is supposed to be something that is &quot;easily transportable&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using stainless steel. Originally we had 20.3m of steel with a cross section of 40mm x 40mm and 10mm thick. Seeing as 187.5kg is quite high I am looking to reduce the weight. I will admit the material and cross section were chosen relatively arbitrarily, but we wanted to be sure that our structure could hold any load without actually worrying about the calculations for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main stress that will be exerted on the beam is bending, as the structure is built like a table, and will have a weight suspended from the middle. I looked up that the yield stress for stainless steel is 193&amp;nbsp;MPa (I took the lowest value I could find for safety reasons) and I have applied various different cross sections for the steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my original dimensions 40 by 40 by 10mm thick:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma\text{ (bending)} = \dfrac{My}{EI}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming my table top is completely made of a hollow rectangle of stainless steel:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;M = 200\cdot9.81\cdot500\text{ (lifting 200kgs 50cm away from it)} = 98100\text{ Nmm} \\&#xA;I = \dfrac{40^4 - 20^4}{3} = 800000\text{ mm}^4 \\&#xA;y = 20\text{ mm} \\&#xA;E = 193\text{ MPa} \\&#xA;\therefore \sigma_b = 0.0127\text{ MPa}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I worked in mm because 1&amp;nbsp;kg/(mm*s&lt;sup&gt;2&lt;/sup&gt;) = 1&amp;nbsp;MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reduced, solid cross section of 20mm gave a bending stress of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;M &amp;amp;= 98100\text{ Nmm} \\&#xA;I &amp;amp;= \dfrac{20^4}{3} = 53333\text{ mm}^4\\&#xA;y &amp;amp;= 10\text{ mm}\\&#xA;E &amp;amp;= 193\text{ MPa}\\&#xA;\therefore \sigma_b &amp;amp;= 0.0953\text{ MPa}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I know stainless steel is a very strong metal, which is why I leaned towards using it originally. I would just really like to know if the logic behind all of my calculations make sense? Am I on the right track for this or have I done something horribly wrong that I am unaware of?&lt;/strong&gt; I am aware I did not include self weight, but did not see the need for it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for reading, I would appreciate any thoughts on this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: this is for a university project, I am mainly looking for pointers on my conceptual design and work.&lt;/p&gt;&#xA;" OwnerUserId="11039" LastEditorUserId="1832" LastEditDate="2018-03-26T19:12:50.803" LastActivityDate="2018-03-26T19:29:14.920" Title="Trying to shed some weight on my design, any thoughts?" Tags="|mechanical-engineering|structural-engineering|materials|structural-analysis|structures|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21055" PostTypeId="2" ParentId="21054" CreationDate="2018-03-26T08:41:00.027" Score="5" Body="&lt;p&gt;Your equation for bending is wrong. Force x distance is appropriate for a cantilever. For a point load applied to a &quot;table&quot; you are either talking simply supported (in which case Mmax = Force x span / 4) or fixed-ended (in which case Mmax = Force x span / 8). &lt;em&gt;Note that you have not provided enough info to show whether a calc this simple is appropriate or not.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your equation for second moment of area is wrong. For a square section it's b^4/12.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your equation for stress is wrong. Stress = My/I. No E. E is Young's modulus, not yield stress. You just need to calculate Stress = My/I and make sure that the stress is less than the yield stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you've said, you've neglected self-weight. What's your justification for this? You have a self-weight of 187.5kg and a live load of 200kg; self-weight is a significant proportion of your load!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You haven't explicitly included any safety factors on your load or on your yield. Are there any? Do you need any? (I can't answer this for you, I work in bridge design not small mechanical products, so I have no idea of applicable standards / codes of practice. But I suspect you'll need to make allowance with some sort of factor.)&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-03-26T08:41:00.027" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21056" PostTypeId="2" ParentId="21054" CreationDate="2018-03-26T15:29:39.410" Score="3" Body="&lt;p&gt;Structural shapes are your friend with this.  Look at I-beams, channels, and struts...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3IbZx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3IbZx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You won't believe how much more weight this piece can carry than a table that is much heavier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stainless will add cost to your project with no structural advantage over carbon steel.  The additional elements in stainless prevent corrosion, which you haven't mentioned as a design issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminum comes in a zillion flavors, but a stronger aluminum alloy is pound-for-pound much stronger than typical stainless or carbon steel.  Aluminum resists corrosion like stainless steel, too... it won't rust outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A long horizontal span of a table increases the stress in the top without increasing the capacity.  Is there any way to reduce the horizontal span?  Also, if you adopt a truss or A-frame design, rather than a cantilever beam, you will achieve much greater strength per weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Casters!  Put wheels on the thing and let it roll around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quick assembly/disassembly.  Can you use toggle clamps or quick-release fasteners, like the wheels on a bike, to disassemble it?  Then you transport it in pieces rather than the whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't rule out wood.  Use 4 X 4 or 4 X 6 douglas fir or redwood... wood is really light compared with metal for many applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do the statics on your design by making a free-body diagram, and remember the basic truss premise: a truss shape is light and strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J7k5Z.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J7k5Z.gif&quot; alt=&quot;Truss&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15506" LastEditorUserId="15506" LastEditDate="2018-03-26T19:29:14.920" LastActivityDate="2018-03-26T19:29:14.920" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21057" PostTypeId="2" ParentId="21030" CreationDate="2018-03-26T16:59:44.347" Score="2" Body="&lt;p&gt;×-form surfaces result in a shallower draft.  ×-form surfaces are more convenient for landing on the floor.  With +-form   the bottom fin which stick down below the bottom of the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I've noticed, the littoral submarines have ×-form surfaces, because they are intended for shallow waters.&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2018-03-26T16:59:44.347" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21058" PostTypeId="2" ParentId="21030" CreationDate="2018-03-26T18:00:24.033" Score="3" Body="&lt;p&gt;People, people people (by which I mean the unreferenced other answers at this time) !  look it up!   Here is info from a &lt;a href=&quot;http://www.clausiuspress.com/conferences/ACSS/ICCSC%202017/BSES270.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chinese research paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;. Compared with the cross rudder, X-form rudder has the following&#xA;  advantages:  Firstly, X-form rudder has the higher rudder efficiency&#xA;  and its manipulate surface area is smaller about 10% than cross&#xA;  rudder, so that submarine could avoid the collision problem when it is&#xA;  landing. Secondly, X-form rudder reduces the serious consequences by&#xA;  rudder-blocked and improves the security and the underwater dynamic&#xA;  unsinkability. Besides, X-form rudder reduces the heeling moment,&#xA;  weight and stern heaviness when cross-rudder submarine turning, and&#xA;  having an advantage 13 2017 3rd International Conference on&#xA;  Computational Systems and Communications (ICCSC 2017) Published by CSP&#xA;  © 2017 the Authors to sail steadily when it is attacked. Finally,&#xA;  X-form rudder reduces interaction between propeller and rudder, and&#xA;  reduces the noise[4].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And a page I found quotes the following from a &lt;a href=&quot;https://books.google.com/books?id=cN4cdjU5KTcC&amp;amp;pg=PA184&amp;amp;lpg=PA184&amp;amp;dq=x%20submarine%20rudder%20configuration&amp;amp;source=web&amp;amp;ots=8lC7YEEB3F&amp;amp;sig=EiS6D3LjGjEtixvulRvY7p5K3fo#v=onepage&amp;amp;q=x%20submarine%20rudder%20configuration&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google Book reference&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most modern military submarines have a hull form that at least&#xA;  approximates an axisymmetric body of revolution. Most of these have&#xA;  four control surfaces at the stern for steering the vessel, that is,&#xA;  for making it turn left or right--the rudder--or rise or dive--diving&#xA;  plane--or a combination of both. In turn, in most modern submarines&#xA;  these control surfaces are in cruciform. That is, the rise-dive&#xA;  surfaces are generally in the same plane as the horizontal plane&#xA;  through the centerline of the vessel, and the turning surfaces are in&#xA;  the same plane as the vertical plane through the centerline. Thus, the&#xA;  control surfaces are generally in the form of a Greek cross. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In most cases the two rudder planes are yoked together, and the two&#xA;  diving planes are yoked together. Because of this yoking, each pair of&#xA;  control surfaces is operated by a single actuating rod. Thus, one rod&#xA;  turns the ship, and the other rod causes the ship to rise or dive. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;It is known that arranging the control surfaces or planes of a&#xA;  submarine in an X configuration has certain advantages. In this form,&#xA;  the control surfaces are in the form of an X. Unlike cruciform&#xA;  designs, X-stern designs utilize all four planes as part of any&#xA;  maneuver. Therefore, an X-stern design enjoys more maneuvering force&#xA;  per unit of control surface area than cruciform designs. X-stern ships&#xA;  can be designed with smaller control surfaces while maintaining&#xA;  maneuvering envelopes comparable to cruciform ships with larger&#xA;  control surfaces. Smaller control surfaces obviously have less drag,&#xA;  but may also be quieter--a very important factor today for a&#xA;  submarine. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The submarine USS ALBACORE had an X-stern configuration where the&#xA;  opposite control surfaces were yoked together. Australian submarines&#xA;  of the recent COLLINS class have X-stern configurations, but the&#xA;  control surfaces are not yoked together and each of the four surfaces&#xA;  has its own actuator. These are two examples of the current known&#xA;  methods of actuating X-sterns. In both cases, the control system for&#xA;  the operating rods is more complicated than that aboard a cruciform&#xA;  ship. In a cruciform ship, if the helmsman wants to turn the ship, the&#xA;  control system commands the rudder operating rod to extend or retract.&#xA;  If a change in depth is required, the control system commands the&#xA;  diving operating rod to extend or retract. In both X-stern designs,&#xA;  the control system commands every operating rod to move in one&#xA;  direction or the other, for any maneuver. Controlling these&#xA;  coordinated operating rod movements is a complex task that can be&#xA;  accomplished with a computer. However, manual coordination of the&#xA;  operating rods, in the event of a computer casualty, is difficult. &#xA;  The proven X-rudder configuration was chosen for a&#xA;  high manoeuvrability. The rudders are laid out in a way that they&#xA;  press a preswirl on the propeller inflow which homogenizes the wake&#xA;  flow field and increases the propeller efficiency on the one hand and&#xA;  reduces the noise signature on the other hand. The propelling power is&#xA;  provided by a Skewback propeller whose extremely low rotation speed&#xA;  prevents cavity and which generates thrust silently even at high&#xA;  speeds.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-03-26T18:00:24.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21059" PostTypeId="1" AcceptedAnswerId="21076" CreationDate="2018-03-26T18:20:01.603" Score="-1" ViewCount="689" Body="&lt;p&gt;I submitted the edited inp. file in abaqus. I edited the thickness of shell elements&#xA;then after I run the edited inp. file, I am suffering from this error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, there is another error:&#xA;&quot;There are fewer than 2 items on a data line for which a generation sequence (start id, end id, optional increment number) is expected. The keyword in question is related to *ELSET. Please check input data&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;does anyone know how to solve these error?&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-03-28T12:39:36.623" Title="in keyword *SURFACE, file &quot;FSECase1cor.inp&quot;, line 79025: Unknown assembly element set_M_SURF-1016_E3 (What does this error mean?)" Tags="|finite-element-method|abaqus|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21060" PostTypeId="2" ParentId="16603" CreationDate="2018-03-26T18:25:12.750" Score="1" Body="&lt;p&gt;I taught about it a lot and I think I have come to an answer .&#xA;the housing of the cable is actually a tube, that tube pushes &lt;strong&gt;down&lt;/strong&gt; the lever .&#xA;suppose the pad break from the other side of the wheel is already pressing the rim so now when you pull- the magic starts to happen - the lever from the side of the cable is pushed down by the &lt;strong&gt;cablehouse&lt;/strong&gt;&#xA;. now when you squeeze the break - where does the cable come from ? (the cable that is attached to the lever doesn't move, so how come you succeed pulling the cable ? ) answer--&gt; the cablehouse is curved -  is not a strait line between the break handle and the lever.if the cable was bear with no cablehouse you would think that there is a lot of redundant cable .now you are pulling that &quot;redundant&quot; cable - now the cable house dosnt have enough space so it pushes the lever- its like trying to insert a 25 cm spring(not streched spring )to a 20 cm gap.the string has a &quot;belly&quot; trying to push the spring to make it without a belly will cause a pressure on the gap to open&#xA;. I hope this is helpfull ....yoav Israel&lt;/p&gt;&#xA;" OwnerUserId="15508" LastEditorUserId="15508" LastEditDate="2018-03-27T16:11:27.657" LastActivityDate="2018-03-27T16:11:27.657" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21061" PostTypeId="2" ParentId="19863" CreationDate="2018-03-26T18:59:16.803" Score="1" Body="&lt;p&gt;If $k &amp;lt; 0$ and the system is $x' = -kx+x^3$ it can be rewritten to $x'=kx+x^3$ with $k &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the equilibrium is $x = 0$, there are two interesting regions, $x &amp;lt; 0$ and $x &amp;gt; 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Region 1, $x &amp;lt; 0$:&#xA;$$x'=\underbrace{kx}_{&amp;lt;0} + \underbrace{x^3}_{&amp;lt;0} &amp;lt; 0.$$&#xA;Region 2, $x &amp;gt; 0$:&#xA;$$x'=\underbrace{kx}_{&amp;gt;0} + \underbrace{x^3}_{&amp;gt;0} &amp;gt; 0.$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarizing the results in a &quot;phase portrait''&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tF8g6m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tF8g6m.png&quot; alt=&quot;Phase portrait 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it can be seen that the equilibrium point $x = 0$ is repulsive (unstable), therefore there is no region of attraction (i.e. the region of attraction is an empty set).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the question makes more sense for $k &amp;gt; 0$ and $x'=-kx+x^3$.&#xA;This system has three equilibria,  $x = \pm \sqrt {k}$ and $x = 0$, therefore there are 4 interesting regions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;$x &amp;lt; -\sqrt{k}$&lt;/li&gt;&#xA;&lt;li&gt;$-\sqrt{k} &amp;lt; x &amp;lt; 0$&lt;/li&gt;&#xA;&lt;li&gt;$0 &amp;lt; x &amp;lt; \sqrt{k}$&lt;/li&gt;&#xA;&lt;li&gt;$x &amp;gt; \sqrt{k}$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Region 1, $ x &amp;lt; -\sqrt{k} $:&#xA;$$ x'=-kx+x^3 &amp;lt; 0 $$&#xA;Region 2, $ -\sqrt{k} &amp;lt; x &amp;lt; 0 $:&#xA;$$ x'=-kx+x^3 &amp;gt; 0 $$&#xA;Region 3, $ 0 &amp;lt; x &amp;lt; \sqrt{k} $:&#xA;$$ x'=-kx+x^3 &amp;lt; 0 $$&#xA;Region 4, $ x &amp;gt; \sqrt{k} $:&#xA;$$ x'=-kx+x^3 &amp;gt; 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summarizing the results in a &quot;phase portrait''&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4etqRm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4etqRm.png&quot; alt=&quot;Phase portrait 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it can be seen that the equilibrium point $x = 0$ is attractive (stable), while the equilibria $x = \pm \sqrt {k}$ are repulsive (unstable).&#xA;Hence the region of attraction is $-\sqrt{k} &amp;lt; x &amp;lt; \sqrt{k}$.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2018-03-26T18:59:16.803" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21062" PostTypeId="2" ParentId="21053" CreationDate="2018-03-26T19:37:32.517" Score="3" Body="&lt;h1&gt;Definitions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Before we can answer your question, let us look at two standard definitions of terms (from &lt;a href=&quot;https://www.asme.org/products/codes.../v-v-10-2006-guide-verification-validation&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASME  Guide for Verification and Validation in Computational Solid Mechanics&lt;/a&gt;) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;strong&gt;Verification&lt;/strong&gt;: The process of determining that a computational model accurately represents the underlying mathematical model and its solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;strong&gt;Validation&lt;/strong&gt;: The process of determining the degree to which a model is an accurate representation of the real world from the perspective of the intended uses of the model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably mean &quot;verification&quot; rather than &quot;validation&quot; in your question.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Verification of anisotropic elasticity&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Assuming you want to verify your code for three-dimensional problems, there is a sequence of steps that's needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;strong&gt;Frame indifference&lt;/strong&gt;: Rotate your elements and confirm that stresses are not developed due to pure rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;strong&gt;Exact solutions&lt;/strong&gt;:  Search for exact solutions for 3D anisotropic elasticity in the literature and confirm that your code can reproduce those. (see, e.g., &lt;a href=&quot;http://www-personal.umich.edu/~jbarber/Ting.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www-personal.umich.edu/~jbarber/Ting.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;strong&gt;Manufactured solutions&lt;/strong&gt;:  Create exact manufactured solutions and verify that your code can match those. (see, e.g., &lt;a href=&quot;http://www.eng.utah.edu/~banerjee/Notes/MMS.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;www.eng.utah.edu/~banerjee/Notes/MMS.pdf&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can pass these tests for a wide range of loading paths, you can be reasonable sure that your implementation is OK.  However, I don't think one can prove correctness of an implementation in a mathematical sense.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-03-26T19:37:32.517" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21063" PostTypeId="2" ParentId="20968" CreationDate="2018-03-26T19:43:31.060" Score="0" Body="&lt;p&gt;You could &quot;micro metal etch&quot; the pattern either onto a stamp or onto a die.  This kind of thing is done by these companies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.microetchtech.com/guidelines-chemically-etching-part-dimensions&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.microetchtech.com/guidelines-chemically-etching-part-dimensions&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.micrometal.de/en.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.micrometal.de/en.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15506" LastEditorUserId="15506" LastEditDate="2018-03-26T19:48:52.960" LastActivityDate="2018-03-26T19:48:52.960" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21067" PostTypeId="2" ParentId="151" CreationDate="2018-03-27T10:02:54.133" Score="1" Body="&lt;p&gt;An almost-critical sphere of fissionable material would do. Add some neutron reflecting material that can be rotated to increase or decrease the neutron reflection. Or you could have two halves of a critical sphere that are moved closer together or further apart to control the reactivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A moderator is not necessary if the reactor operates on &lt;a href=&quot;https://en.wikipedia.org/wiki/Fast-neutron_reactor&quot; rel=&quot;nofollow noreferrer&quot;&gt;fast neutrons&lt;/a&gt;, thermal (slow) neutron reactors are just easier/cheaper to operate and more proliferation resistant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only operate it at low power levels you would not need a lot of cooling (though the reactor would probably not be very useful). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest amount of mass would be the shielding. So as long as you don't care about that, you're fine (and in your question you don't mention it). There has been research into nuclear powered &lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_propulsion#Cars&quot; rel=&quot;nofollow noreferrer&quot;&gt;cars&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear-powered_aircraft&quot; rel=&quot;nofollow noreferrer&quot;&gt;aircraft&lt;/a&gt;, and trains, but the size and weight of the shielding was the main technical problem for such applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The output of nuclear reactions is heat, if you want to convert it to electricity or another more useful form of energy, you will need some kind of heat engine, which (depending on the temperature) needs to dump 75% of the heat input as low temperature waste heat. So you will need some kind of heat sink for that. But again that is not part of the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;edit:&lt;/em&gt; To be more specific, the output of a nuclear reactor is some heat in the reaction products, and most of the energy in the neutron radiation. Reactors usually also capture those neutrons in order to convert their energy to heat, so that is anoher reason why you might want to use neutron reflectors. The most difficult to shield radiation are the gamma rays, which typically require several meters of concrete or a thick slab of lead to block.&lt;/p&gt;&#xA;" OwnerUserId="12424" LastEditorUserId="12424" LastEditDate="2020-03-10T16:10:35.613" LastActivityDate="2020-03-10T16:10:35.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21068" PostTypeId="2" ParentId="17328" CreationDate="2018-03-27T13:18:38.610" Score="3" Body="&lt;p&gt;It is actually quite common in Scandinavia in my experience, using steel beams and columns with concrete hollowcore slabs. &lt;a href=&quot;http://www.stalforbund.no/uploads/source/files/Materialet/Hulldekker/Veileder_Hulldekker_2008_2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This PDF&lt;/a&gt; describes one example of such a system.&lt;/p&gt;&#xA;" OwnerUserId="15514" LastEditorUserId="368" LastEditDate="2018-03-27T15:11:58.510" LastActivityDate="2018-03-27T15:11:58.510" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21069" PostTypeId="1" CreationDate="2018-03-27T13:22:32.713" Score="1" ViewCount="3793" Body="&lt;p&gt;Can somebody help me find the equation for the deformation of a simply supported beam under trapezoidal load (with intensity Q1 and Q2, at starting point X1 and end point X2)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find it myself using double integration, but without luck. I already have a numerical solution - but it's too slow (i'm doing thousands of calculations). &lt;/p&gt;&#xA;" OwnerUserId="15514" LastActivityDate="2018-05-06T05:15:40.237" Title="Beam deformation under trapezoidal load" Tags="|steel|beam|deformation|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21070" PostTypeId="1" CreationDate="2018-03-27T17:24:39.117" Score="1" ViewCount="4943" Body="&lt;p&gt;Suppose I know the mass and volume of an object, as well as its exact shape and density distribution as a function of space (x,y,z).  Without submerging it to determine the amount of water displaced, how can I determine the buoyant force?  Is there a way to determine the amount of displaced water based on knowing the density of the object as well as its shape?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, I'd like to use this information to predict whether a particular design I have made will sink or float, but I need to know the buoyant force in order to do this.  How can I predict this force a priori without experimentation?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="15506" LastEditDate="2018-03-27T19:13:07.583" LastActivityDate="2018-03-28T18:07:26.987" Title="How can I calculate the buoyant force on an object without knowing in advance the amount of water displaced?" Tags="|fluid-mechanics|statics|naval-engineering|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="21071" PostTypeId="2" ParentId="21070" CreationDate="2018-03-27T18:50:42.830" Score="2" Body="&lt;p&gt;The buoyancy is equal to weight of water of displaced volume of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's heavier than water it will sink and as long as you know the volume you know the buoyancy!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the object is not heavier than water, it will float.  A part of the object will submerge, displacing (in the static case) exactly enough water to equalize the water's buoyant force with the object's weight.  The level of water at which this occurs is named the &lt;em&gt;draft line.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to find the object's CG, and its center of geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ CG = (\sum \Delta v.\rho_v)/v $$ and center of geometry is a analytical process depending on its shape or combination of shapes and surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axis of float is vertical from CG to center of geometry and draft volume is such that $$\Sigma \Delta v.\rho\space \text{submerged volume of water} = \text{weight of object}$$&#xA;  ** Edit**&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The submerged volume is a given and is equall to the weight of the object. However to calculate the draft line one needs to compile the geometey  of the object as a function of the vertical axis say Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by equating the volume integral of  submerged volume by the weight we get the Z of draft-line. Z is the upper limit of integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process for many nicely behaving volumes is rather easy for instance for a submerged cone the volume is $$ V=2/3 Z. \pi .sin(theta)^2$$ and $\theta$ is half the cone angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Z = 2/3 . \pi.sin(\theta)^2/V  $$&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2018-03-28T08:33:23.147" LastActivityDate="2018-03-28T08:33:23.147" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21074" PostTypeId="2" ParentId="19029" CreationDate="2018-03-28T09:22:58.970" Score="0" Body="&lt;p&gt;I have resolved this problem. in my case, it was a mistake of a non-uniform unit in abaqus. so I strongly recommend to check the uniformity of your unit and boundary condition, the insufficient boundary condition can lead this error to occur&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-03-28T09:22:58.970" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21075" PostTypeId="2" ParentId="21070" CreationDate="2018-03-28T12:07:53.227" Score="3" Body="&lt;p&gt;Don't refer to the volume as the volume of material.  Use volume as the outer dimensions of the object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume of the water will be equal to the mass of the object (it if floats).  Just divide the mass of the object by the density of water for displaced water.  That will also be the exact volume of your object that is below the water line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not matter to the water if the object is hollow or not.  If the mass / volume of the object is greater than that of water it will sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Buoyancy&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;with the clarifications that for a sunken object the volume of&#xA;  displaced fluid is the volume of the object, and for a floating object&#xA;  on a liquid, the weight of the displaced liquid is the weight of the&#xA;  object&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-03-28T15:57:54.577" LastActivityDate="2018-03-28T15:57:54.577" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21076" PostTypeId="2" ParentId="21059" CreationDate="2018-03-28T12:39:36.623" Score="0" Body="&lt;p&gt;I have solved this problem.&#xA;I have made typing mistakes when editing the inp. file.&#xA;so it is highly recommended to write with high accuracy, otherwise, it can cause a lot of errors.&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-03-28T12:39:36.623" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21079" PostTypeId="1" CreationDate="2018-03-28T21:20:38.373" Score="1" ViewCount="69" Body="&lt;p&gt;Let's say I have the geometry of the rotor: the number of blades, the airfoils that make up each blade, their position, chord length and their twist angle. Would it be possible to use this information to determine the tip-speed ratio?&lt;/p&gt;&#xA;" OwnerUserId="15350" LastActivityDate="2018-03-28T21:56:28.087" Title="Is it possible to determine tip-speed ratio from rotor geometry?" Tags="|mechanical-engineering|turbines|wind-power|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21080" PostTypeId="2" ParentId="21079" CreationDate="2018-03-28T21:56:28.087" Score="1" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would point you to the 2014 paper by M. Ragheb titled Optimal Tip Speed Ratio&#xA;&lt;a href=&quot;http://mragheb.com/NPRE%20475%20Wind%20Power%20Systems/Optimal%20Rotor%20Tip%20Speed%20Ratio.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mragheb.com/NPRE%20475%20Wind%20Power%20Systems/Optimal%20Rotor%20Tip%20Speed%20Ratio.pdf&lt;/a&gt; which includes a fairly clear description of the factors affecting turbine efficency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rotor with a variable pitch system, the rotational speed vs. power generated can be controlled by feathering the blades to accommodate varying wind speeds and therefore keep the system operating at the best TSR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a fixed pitch system, some method of governing the maximum RPM and will be required. &lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2018-03-28T21:56:28.087" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21081" PostTypeId="2" ParentId="10486" CreationDate="2018-03-28T22:15:12.460" Score="1" Body="&lt;p&gt;Look at the size of the nonlinear terms relative to the linear ones; then decide for yourself whether you can throw them away. This is greatly simplified using nondimensional analysis contrived so that all coefficients have a magnitude on the order one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So set up reference quantities for the terms that appear in the governing equations. The reference mass might be the starting mass or the mass of a half filled bucket. So the mass in the governing equation is the actual mass divided by the reference mass. You can constrain the frequency using nondimensional time. For instance, you might divide the actual period by the natural period of the reference mass. Instantaneous flow rate out the bucket is now expressed in reference masses per reference cycle. Now you can decide under what conditions you are willing to disregard higher order terms (and justify the decisions) because you don't have to worry about huge coefficients counteracting some product of otherwise small terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a big part of exploiting series expansions such as Taylor series. To do so, you need to know the higher order terms get smaller at a decent rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This problem would be a lot more fun if the bucket was supplying fuel to a rocket engine and the spring represented a vibration mode. Because now you really do want to get the terms on the order of one.) &lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-03-28T22:15:12.460" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21082" PostTypeId="1" CreationDate="2018-03-29T05:15:43.523" Score="3" ViewCount="236" Body="&lt;p&gt;I am designing an NSAM crane. I have encountered the following issue when designing the vehicle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally the pivot(or the point about which the NSAM crane rotates) is in the center. But from two online sources(&lt;a href=&quot;https://patents.google.com/patent/US8002074&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://patents.google.com/patent/US8002074&lt;/a&gt;, &lt;a href=&quot;https://www.tandfonline.com/doi/abs/10.7158/M12-108.2014.12.1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.tandfonline.com/doi/abs/10.7158/M12-108.2014.12.1&lt;/a&gt;) I have found that if the pivot is in the center, then the NSAM crane has easy maneuverability and tip over stability. If the pivot is away from the center and towards the rear axle, the roll over stability is more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though some calculations have been given in the sources, I am not able to understand why this is the case and which stability is more important for the NSAM crane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also please help me with any Simply supported beam type of calculations for the stability if any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any references would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="15531" LastActivityDate="2018-04-28T22:50:27.530" Title="Distance of Pivot from the Rear axle" Tags="|mechanical-engineering|automotive-engineering|structures|strength|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21083" PostTypeId="1" CreationDate="2018-03-29T07:32:23.887" Score="0" ViewCount="230" Body="&lt;p&gt;I was wondering, what are the factors to consider or how to find the required technical specifications for the new machine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we consider a design of horizontal centrifugal casting machine with orbitary dimensions, Let suppose I do not know the dimensions of the machine. How do I calculate&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;centrifugal force required?&lt;/li&gt;&#xA;&lt;li&gt;Torque of the Motor?&lt;/li&gt;&#xA;&lt;li&gt;Power required to rotate the rotating tube?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To calculate the Centrifugal force, $F_c=mv^2 / r$ or $F_c=M\omega^2r$.&#xA;where, angular velocity, $\omega=2\pi N/60$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, in order to calculate the centrifugal force we need to have the Speed ($N$) of the rotating tube. But, how do we know that whether the assumed Speed is enough for the casting process? Using this $F_c$, torque and power can be calculated, $T=F_c*r$ and $P=2\pi NT/60$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming $N=200\text{ rpm}$, I have carried out some calculations which are as follows, $\omega=2\pi N/60=20.94\text{ rad/s} = 3.33\text{ rps}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let, part to be cast is Aluminium and has $\phi_{out}=20\text{ cm}$, $\phi_{in}=10\text{ cm}$ and $L=50\text{ cm}$. Then, $V=3750\pi\text{ cm}^3$ and $m=31.83\text{ kg}$. Hence, from $F_c=M\omega^2r=173.13\text{ N}$ (considering inner radius of the cast part as the Minimum distance need to achieve which 5&amp;nbsp;cm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimensions of rotating tube is assumed as $\phi_{out}=30\text{ cm}$ and $\phi_{in}=20\text{ cm}$ and $L=50\text{ cm}$. Torque required $T=173.13\cdot0.15=25.96\text{ Nm}$ (considering radius as the outer radius of the rotating tube that is 15&amp;nbsp;cm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power, $P=543.70\text{ W}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the steps I have performed are correct? If so, How do I know that assumed $N$ is the suitable value for the present machine or what is the usual Speed of the rotating tube in horizontal casting machine considered?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, how do I approach when there are several unknowns or dependencies present in the system, for example centrifugal force depends on angular velocity which in turn depended on speed of the rotating tube.&lt;/p&gt;&#xA;" OwnerUserId="15526" LastEditorUserId="1832" LastEditDate="2018-03-29T08:24:59.587" LastActivityDate="2019-01-07T18:31:56.730" Title="How to design a casting machine?" Tags="|mechanical-engineering|design|torque|manufacturing-engineering|power|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21084" PostTypeId="1" CreationDate="2018-03-29T09:50:03.863" Score="1" ViewCount="413" Body="&lt;p&gt;I am reading B. P Lathi, &lt;em&gt;Modern Analog and Digital Communication&lt;/em&gt;.&#xA;In it, there is a topic where the author says that maximum information rate is two pieces of information per second per hertz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am a little confused as according to the sampling theorem, we need a minimum frequency of 2B samples per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why does then the author say that only a maximum of 2*(maximum frequency) independent pieces of information per second can be transmitted, error free, over a noiseless channel of bandwidth Bhz.&lt;/p&gt;&#xA;" OwnerUserId="15533" LastEditorUserId="90" LastEditDate="2018-03-29T10:47:17.700" LastActivityDate="2018-04-01T15:55:01.483" Title="Maximum Information Rate" Tags="|digital-communication|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21085" PostTypeId="2" ParentId="21083" CreationDate="2018-03-29T11:38:46.330" Score="1" Body="&lt;p&gt;Don't make a centrifugal machine for producing parts of your dimensions ; Length 50 cm : Use static casting or possibly extrusion for aluminum. Centrifugal casting is used to produce about 10 m lengths  of iron, steel and stainless steel ( and concrete) , I never saw one for aluminum. Steel oil well casing couplings are short but they cast as a long tube which is cut to short lengths.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-03-29T11:38:46.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21086" PostTypeId="2" ParentId="21084" CreationDate="2018-03-29T11:58:25.163" Score="1" Body="&lt;p&gt;Take a sinusoid of frequency B.  Within a single cycle, the signal will hit its maximum and minimum amplitudes once each.  Said another way, if the signal ranges from 0 to 1, then we'll measure the values of 0 and 1 once each during a single cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those maximums and minimums represent the pieces of information that Lathi is referring to.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we get one maximum and and one minimum per cycle, we have two pieces of information per cycle.  And the maximum information rate is then 2 * B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sampling doesn't come into play here as we're using simple detectors to determine if maximum or minimum were reached.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In a follow up comment you had asked:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But why did you consider a sinusoidal signal only. The signal could as well be a non periodic one, right? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I picked a sinusoid because it's easier to visualize and draw observations.  A non-periodic signal yields the same results, but isn't as clear to observe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With an analog signal, we use the maximum and minimum amplitudes of the signal because we can't determine if intermediate values are intended as valid signal points or if they are incidental values while in the process of traveling to a different value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we have a sinusoid with an amplitude of +/- 1, then we'll cross the zero point twice.  The receiver has now way of knowing if those zero values were meaningful or just incidental.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's probably worth reviewing the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nyquist_rate&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nyquist Rate&lt;/a&gt;, which is the source of this information rate.  Nyquist was working in an analog world and trying to figure out how much information could be packed into a telegraph signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, look into the &lt;a href=&quot;https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shannon-Hartley theorem&lt;/a&gt;, which builds upon the work that Nyquist did and lends a more nuanced perspective of channel capacity.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2018-04-01T15:55:01.483" LastActivityDate="2018-04-01T15:55:01.483" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21087" PostTypeId="2" ParentId="21069" CreationDate="2018-03-29T13:29:08.560" Score="1" Body="&lt;p&gt;(edit 3/31/2018. Completed all equations)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would split the trapezoidal load&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XV1zQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XV1zQ.png&quot; alt=&quot;trapezoidal load&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;into the sum of a uniform load&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3yeIY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3yeIY.png&quot; alt=&quot;constant load&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and triangular load&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j35On.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j35On.png&quot; alt=&quot;triangular load&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactions $R_1$, $R_5$, $R_4$, and $R_8$ can be calculated from statics.&#xA;$$R_4=Q_1L_{23}\frac{L_{12}+\frac 12L_{23}}{L_{14}}$$&#xA;$$R_1=Q_1L_{23}-R_4$$&#xA;$$R_5=\frac 12Q_2L_{67}(\frac 13L_{67}+L_{78})\frac{1}{L_{58}}$$&#xA;$$R_8=\frac 12Q_2L_{67}-R_5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uniform and triangular loads can be integrated section by section to get the shear, moment, slope, and displacement. In each of these equations, x is measured from the previous point, not the beginning of the beam. Using the following equations, it is easy to create a spreadsheet to calculate the value at each point $V_1, V_2, \ldots, \delta_7, \delta_8$ or at any points in between using the equations that are functions of (x).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uniform load, shear: $V(x)=V_i+\int load(x)\,dx$ where i is at the start of the segment and load(x)=$Q_1$ between points 2 and 3.&#xA;$$V_{1}=R_1$$&#xA;$$V_{12}(x)=V_1$$&#xA;$$V_{2}=V_1$$&#xA;$$V_{23}(x)=V_2-Q_1x$$&#xA;$$V_3=V_2-Q_1L_{23}$$&#xA;$$V_{34}(x)=V_3$$&#xA;$$V_4=V_3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uniform load, moment: $M(x)=M_i+\int V(x)\,dx$&#xA;$$M_{1}=0$$&#xA;$$M_{12}(x)=M_1+V_1x$$&#xA;$$M_2=M_1+V_1L_{12}$$&#xA;$$M_{23}(x)=M_2+V_2x-\frac 12Q_1x^2$$&#xA;$$M_{3}=M_2+V_2L_{23}-\frac 12Q_1L_{23}^2$$&#xA;$$M_{34}(x)=M_3+V_3x$$&#xA;$$M_{4}=M_3+V_3L_{34}$$&#xA;$M_4$ should equal 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uniform load, slope: $\theta(x)=\theta_i+\int \frac{M(x)}{EI}\,dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_{1}=unknown$$&#xA;$$\theta_{12}(x)=\theta_1+\frac{1}{EI}(M_1x+\frac 12V_1x^2)$$&#xA;$$\theta_{2}=\theta_1+\frac{1}{EI}(M_1L_{12}+\frac 12V_1L_{12}^2)=\theta_1+c_{s2}$$&#xA;$$\theta_{23}(x)=\theta_2+\frac{1}{EI}(M_2x+\frac 12V_2x^2-\frac 16Q_1x^3)$$&#xA;$$\theta_{3}=\theta_2+\frac{1}{EI}(M_2L_{23}+\frac 12V_2L_{23}^2-\frac 16Q_1L_{23}^3)=\theta_2+c_{s3}$$&#xA;$$\theta_{34}(x)=\theta_3+\frac{1}{EI}(M_3x+\frac 12V_3x^2)$$&#xA;$$\theta_{4}=\theta_3+\frac{1}{EI}(M_3L_{34}+\frac 12V_3L_{34}^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uniform load, displacement: $\delta (x)=\delta_i+\int \theta(x)\,dx$&#xA;$$\delta_1=0$$&#xA;$$\delta_{12}(x)=\delta_1+\theta_1x+\frac{1}{EI}(\frac 12M_1x^2+\frac 16V_1x^3)$$&#xA;$$\delta_2=\delta_1+\theta_1L_{12}+\frac{1}{EI}(\frac 12M_1L_{12}^2+\frac 16V_1L_{12}^3)=\delta_1+\theta_1L_{12}+c_{d2}$$&#xA;$$\delta_{23}(x)=\delta_2+\theta_2x+\frac{1}{EI}(\frac 12M_2x^2+\frac 16V_2x^3-\frac{1}{24}Q_1x^4)$$&#xA;$$\delta_3=\delta_2+\theta_2L_{23}+\frac{1}{EI}(\frac 12M_2L_{23}^2+\frac 16V_2L_{23}^3-\frac{1}{24}Q_1L_{23}^4)=\delta_2+\theta_2L_{23}+c_{d3}$$&#xA;$$\delta_{34}(x)=\delta_3+\theta_3x+\frac{1}{EI}(\frac 12M_3x^2+\frac 16V_3x^3)$$&#xA;$$\delta_4=\delta_3+\theta_3L_{34}+\frac{1}{EI}(\frac 12M_3L_{34}^2+\frac 16V_3L_{34}^3)=\delta_3+\theta_3L_{34}+c_{d4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last equation gives $\delta_4=0=\theta_1+known\; values$ after you make all of the substitutions, so this gives the result for $\theta_1$ as follows (where $c_s$ and $c_d$ are simplification of constant terms defined above):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_1=-\frac{\delta_1+c_{d2}+c_{d3}+c_{d4}+c_{s2}L_{23}+(c_{s2}+c_{s3})L_{34}}{L_{14}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All values are now known for the beam with a uniform load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar equations are written for the triangular load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;triangular load, shear: $V(x)=V_i+\int load(x)\,dx$ where load(x)=$Q_2\frac xL$ between points 6 and 7.&#xA;$$V_{5}=R_5$$&#xA;$$V_{56}(x)=V_5$$&#xA;$$V_{6}=V_5$$&#xA;$$V_{67}(x)=V_6-\frac 12Q_2\frac{x^2}{L_{67}}$$&#xA;$$V_7=V_6-\frac 12Q_2L_{67}$$&#xA;$$V_{78}(x)=V_7$$&#xA;$$V_8=V_7$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;triangular load, moment: $M(x)=M_i+\int V(x)\,dx$&#xA;$$M_{5}=0$$&#xA;$$M_{56}(x)=M_5+V_5x$$&#xA;$$M_6=M_5+V_5L_{56}$$&#xA;$$M_{67}(x)=M_6+V_6x-\frac 16Q_2\frac{x^3}{L_{67}}$$&#xA;$$M_{7}=M_6+V_6L_{67}-\frac 16Q_2L_{67}^2$$&#xA;$$M_{78}(x)=M_7+V_7x$$&#xA;$$M_{8}=M_7+V_7L_{78}$$&#xA;$M_8$ should equal 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;triangular load, slope: $\theta(x)=\theta_i+\int \frac{M(x)}{EI}\,dx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_{5}=unknown$$&#xA;$$\theta_{56}(x)=\theta_5+\frac{1}{EI}(M_5x+\frac 12V_5x^2)$$&#xA;$$\theta_{6}=\theta_5+\frac{1}{EI}(M_5L_{56}+\frac 12V_5L_{56}^2)=\theta_5+c_{s6}$$&#xA;$$\theta_{67}(x)=\theta_6+\frac{1}{EI}(M_6x+\frac 12V_6x^2-\frac{1}{24}Q_2\frac{x^4}{L_{67}})$$&#xA;$$\theta_{7}=\theta_6+\frac{1}{EI}(M_6L_{67}+\frac 12V_6L_{67}^2-\frac{1}{24}Q_2L_{67}^3)=\theta_6+c_{s7}$$&#xA;$$\theta_{78}(x)=\theta_7+\frac{1}{EI}(M_7x+\frac 12V_7x^2)$$&#xA;$$\theta_{8}=\theta_7+\frac{1}{EI}(M_7L_{78}+\frac 12V_7L_{78}^2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;triangular load, displacement: $\delta (x)=\delta_i+\int \theta(x)\,dx$&#xA;$$\delta_5=0$$&#xA;$$\delta_{56}(x)=\delta_5+\theta_5x+\frac{1}{EI}(\frac 12M_5x^2+\frac 16V_5x^3)$$&#xA;$$\delta_6=\delta_5+\theta_5L_{56}+\frac{1}{EI}(\frac 12M_5L_{56}^2+\frac 16V_5L_{56}^3)=\delta_5+\theta_5L_{56}+c_{d6}$$&#xA;$$\delta_{67}(x)=\delta_6+\theta_6x+\frac{1}{EI}(\frac 12M_6x^2+\frac 16V_6x^3-\frac{1}{120}Q_2\frac{x^5}{L_{67}})$$&#xA;$$\delta_7=\delta_6+\theta_6L_{67}+\frac{1}{EI}(\frac 12M_6L_{67}^2+\frac 16V_6L_{67}^3-\frac{1}{120}Q_2L_{67}^4)=\delta_6+\theta_6L_{67}+c_{d7}$$&#xA;$$\delta_{78}(x)=\delta_7+\theta_7x+\frac{1}{EI}(\frac 12M_7x^2+\frac 16V_7x^3)$$&#xA;$$\delta_8=\delta_7+\theta_7L_{78}+\frac{1}{EI}(\frac 12M_7L_{78}^2+\frac 16V_7L_{78}^3)=\delta_7+\theta_7L_{78}+c_{d8}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last equation gives $\delta_8=0=\theta_5+known\; values$ after you make all of the substitutions, so this gives the result for $\theta_5$ as follows (where $c_s$ and $c_d$ are terms defined above):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta_5=-\frac{\delta_5+c_{d6}+c_{d7}+c_{d8}+c_{s6}L_{67}+(c_{s6}+c_{s7})L_{78}}{L_{58}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All values are now known for the beam with a triangular load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results for the trapezoidal load are thus&#xA;$$R_A=R_1+R_5$$&#xA;$$R_D=R_4+R_8$$&#xA;$$V_A=V_1+V5$$&#xA;$$V_B=V_2+V6$$&#xA;$$\ldots$$&#xA;$$\delta_C=\delta_3+\delta_7$$&#xA;$$\delta_D=\delta_4+\delta_8$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example which I verified with FEA is as follows:&#xA;$$L_{AB}=20$$&#xA;$$L_{BC}=30$$&#xA;$$L_{CD}=40$$&#xA;$$Q_1=10$$&#xA;$$Q_2=15$$&#xA;$$E=1E6$$&#xA;$$I=0.75$$&#xA;gives results of&#xA;$$\theta_1=-0.1976\;radians$$&#xA;$$\delta_3=-5.2370$$&#xA;$$\theta_5=-0.1513\;radians$$&#xA;$$\delta_7=-4.2267$$&#xA;The point of maximum displacement occurs at $\theta_{23}(x)+\theta_{67}(x)=0$ which is a solution to a 5th order equation which I did not attempt to solve. An approximate solution is at x=23.283 from B to C which gives a maximum displacement of (-5.412)+(-4.333)=(-9.745).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should get the same results if I did not make any typos.&lt;/p&gt;&#xA;" OwnerUserId="12984" LastEditorUserId="12984" LastEditDate="2018-03-31T21:58:55.547" LastActivityDate="2018-03-31T21:58:55.547" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21088" PostTypeId="1" CreationDate="2018-03-29T16:14:37.833" Score="1" ViewCount="3915" Body="&lt;p&gt;What is meant by shock load in gears?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are steady, light, medium and heavy shock loads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Tb48e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tb48e.jpg&quot; alt=&quot;I found this table in a design data handbook&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14355" LastEditorUserId="10902" LastEditDate="2018-03-30T19:23:16.263" LastActivityDate="2018-03-30T19:23:16.263" Title="What is meant by &quot;shock load&quot; in gears?" Tags="|gears|kinematics|power-transmission|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21089" PostTypeId="2" ParentId="21088" CreationDate="2018-03-29T16:35:42.677" Score="2" Body="&lt;p&gt;a shock load on gearing is imposed when the gear train is forced to start or stop turning very suddenly, or when a heavy load is suddenly applied to a set of rotating gears. A good example of this is when the driver of a car with a manual transmission slams or &quot;bangs&quot; the gear changes without skilled use of the clutch, causing the car to suddenly lurch. This imposes high stresses on the gear teeth for over short timescales. Gears subjected to shock loads may fail due to metal fatigue of the gear teeth rather than wearout of the contact faces on the teeth themselves. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-03-29T16:35:42.677" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21090" PostTypeId="1" CreationDate="2018-03-29T16:59:50.070" Score="0" ViewCount="23" Body="&lt;p&gt;I use a piezoelectric actuator on an experimental set up. At low temperatures (Lower than 0 Celsius), the piezo will freeze. Meaning, it will actuate about half way, and then stop moving. Other than the temperature, other conditions have changed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that this sticking is in fact some lubricant in the device. I &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone comment upon any possible physical origin for a piezo to stick like this?&lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2018-03-29T16:59:50.070" Title="Why does my piezoelectric actuator stick at low temperatures?" Tags="|design|temperature|actuator|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21093" PostTypeId="2" ParentId="21088" CreationDate="2018-03-29T17:57:04.187" Score="1" Body="&lt;p&gt;In gear design for different machinery there is a  combination of loads that causes damage and excessive wear and the equation for that load:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F*   =   Ka Kv Km Ft &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ka is application factor inherent in connecting to moving parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kv is dynamic factor caused by imbalance and deficiency in gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Km is load distribution factor on gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ft is service design load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part your question refers to is application factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These factors combine with the other above mentioned are used to define the shock load and service need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JOGLK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JOGLK.jpg&quot; alt=&quot;Applicatin load factor for machinery gears.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://www-mdp.eng.cam.ac.uk/web/library/enginfo/textbooks_dvd_only/DAN/gears/failure/failure.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www-mdp.eng.cam.ac.uk/web/library/enginfo/textbooks_dvd_only/DAN/gears/failure/failure.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-03-29T18:10:26.790" LastActivityDate="2018-03-29T18:10:26.790" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21094" PostTypeId="1" CreationDate="2018-03-29T19:07:16.783" Score="3" ViewCount="200" Body="&lt;p&gt;I have a simple cruise control closed-loop system with negative feedback and have no problem using the Nyquist stability criterion to find its stability. However, I have another function that describes the fuel consumption of the throttle in the closed loop. There is no feedback from the fuel consumption to the throttle or the speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, I have an additional transfer function outside the loop from the plant to the fuel consumption and want to assess the stability of the fuel consumption.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Since the fuel consumption is just a function of the throttle which is inside the feedback loop, am I guaranteed stability if the closed loop system is stable?&lt;/li&gt;&#xA;&lt;li&gt;Is it possible for the fuel consumption function to essentially add a gain that makes the consumption unstable even if the speed control is stable?&lt;/li&gt;&#xA;&lt;li&gt;Should I treat the fuel consumption as a purely open loop problem from the throttle to the fuel?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="15540" LastActivityDate="2018-05-23T20:11:25.950" Title="Finding stability of open loop system connected to closed loop system" Tags="|mechanical-engineering|electrical-engineering|stability|feedback-loop|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21095" PostTypeId="1" CreationDate="2018-03-29T22:19:24.037" Score="0" ViewCount="186" Body="&lt;p&gt;I am looking for some mirror setup where rotation one or more elements in the mirror setup will allow to shift the beam along its axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two mirrors in a galvanometer scanner type setup work well for controlling the angle of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotating risley prisms can be used to only offset the beam without changing its angle but the possible positions are limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dove prisms are great for rotating the beam along itself or rotate in a circular fashion if prism center is shifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as for offsetting (shifting) a beam along its axis from a (-1,-1) position to (1,1) without changing the angle of the beam, I can't find any setup which involves a rotating mirror rather than shifting mirror. But I feel there may be something missing from my limited optics knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 rotating galvo scanner mirrors may do this but this may be an overkill and the size of the mirrors on the last two will increase to impractical sizes for decent 45 degree FOVs or more. So I'm crossing this one out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case not clear, &lt;a href=&quot;https://i.stack.imgur.com/ITNSL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/ITNSL.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14841" LastEditorUserId="14841" LastEditDate="2018-03-29T23:14:14.360" LastActivityDate="2018-03-30T09:24:34.973" Title="rotating mirror system for shifting a beam?" Tags="|optics|photonics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21096" PostTypeId="2" ParentId="21082" CreationDate="2018-03-29T22:27:25.607" Score="2" Body="&lt;p&gt;There's a &lt;a href=&quot;https://www.researchgate.net/publication/269755529_A_general_articulation_angle_stability_model_for_non-slewing_articulated_mobile_cranes_on_slopes&quot; rel=&quot;nofollow noreferrer&quot;&gt;rather exhaustive dissertation&lt;/a&gt; regarding this particular type of crane, Non Slewing Articulated Mobile, based on a study of apparent out-of-balance situations in Australia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formulae are provided as well as diagrams of center of gravity calculations on flat and sloped surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JVHEE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JVHEE.png&quot; alt=&quot;nasm crane identification&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TyhIi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TyhIi.png&quot; alt=&quot;more nasm info&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above link also includes a download of the entire document in PDF. There is a substantial section containing the formulae and more diagrams that may be of use to your research.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-03-29T22:27:25.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21097" PostTypeId="2" ParentId="3863" CreationDate="2018-03-29T23:46:37.100" Score="0" Body="&lt;p&gt;Have look at these two:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YcCuf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YcCuf.jpg&quot; alt=&quot;clock1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sk6Ez.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sk6Ez.jpg&quot; alt=&quot;clock2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are medieval clockwork mechanisms.  You could make them with a smithy and hammer.  Cast and bash was the principle technique.  And then a bit of heavy bashing followed by more light bashing.  To be honest, the greater skill lies in the mechanical arrangement of parts. They had torture racks and stuff and rudimentary locks including ridiculously large padlocks.  But they had to invent and evolve these sophisticated mechanisms to keep some sort of decent time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A great report of the restoration of the Salisbury cathedral clock (1386!) is &lt;a href=&quot;https://en.wikipedia.org/wiki/Salisbury_cathedral_clock&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.  Look at the &lt;a href=&quot;https://en.wikipedia.org/wiki/Salisbury_cathedral_clock#1956_restoration&quot; rel=&quot;nofollow noreferrer&quot;&gt;1956 restoration&lt;/a&gt; bit which details making parts by hand forging (bashing).&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2018-03-29T23:46:37.100" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21098" PostTypeId="2" ParentId="21095" CreationDate="2018-03-30T00:30:52.103" Score="2" Body="&lt;p&gt;You could use a slab of glass or other transparent medium with a different refractive index then air. The slab should have two surfaces which are as parallel as possible and be mounted on a two axis gimbal. Namely the more you tilt the slab in any direction relative to axis of the beam the more the beam will shift in that direction, while maintaining the beam direction, as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3QqvQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3QqvQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However you will get more losses due to reflections the further you tilt the slab. And depending on how large of a displacement or the beam you want you might also need a quite thick slab. In order to get your hand on a thick slab of transparent material you could also use a liquid filled container.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The lateral displacement of the beam $l$ can be expressed as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;l = d\,\sin\theta\left(1 - \frac{\cos\theta}{\sqrt{n^2 - \sin^2\theta}}\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $d$ is the thickness of the slab, $\theta$ the angle between the beam and the normal of the slab and $n$ the refractive index of the slab.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastEditorUserId="1765" LastEditDate="2018-03-30T09:24:34.973" LastActivityDate="2018-03-30T09:24:34.973" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21099" PostTypeId="1" CreationDate="2018-03-30T04:43:31.510" Score="2" ViewCount="136" Body="&lt;p&gt;I've been trying to find the boundary conditions for a fourth order DE equation for a bending circular plate. I'm assuming angular symmetry and an even pressure distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Displacement Equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ D \nabla^2\nabla^2w(x,y,t) + h\rho \frac{\partial^2{w}}{\partial{t}^2}(x,y,t) = P(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D: Flexural Rigidity (Constant)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h: Thickness&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\rho$: Density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(t): Pressure (Evenly applied across diaphragm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conversion to Polar Coordinates&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \nabla^2\nabla^2w(x,y,t) = \frac{1}{r^3}\frac{\partial{w}}{\partial{r}} - \frac{1}{r^2}\frac{\partial^2{w}}{\partial{r}^2} + \frac{2}{r}\frac{\partial^3{w}}{\partial{r}^3} + \frac{\partial^4{w}}{\partial{r}^4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I need to solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D(\frac{1}{r^3}\frac{\partial{w}}{\partial{r}} - \frac{1}{r^2}\frac{\partial^2{w}}{\partial{r}^2} + \frac{2}{r}\frac{\partial^3{w}}{\partial{r}^3} + \frac{\partial^4{w}}{\partial{r}^4}) + h\rho\frac{\partial^2{w}}{\partial{t}^2}(r,t) = P(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The boundary conditions I currently have&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial{w}}{\partial{t}}(r=a,t) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$w(r=a,t) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{\partial{w}}{\partial{t}}(r=0,t) = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to discretize it with N divisions and transform the system into State Space so that I can use P(t) as a controller as well as create a Bode Plot and look at the system's resonance frequencies. I believe that I need one additional boundary condition due to the spacial derivative being of order 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also plan on using central and forward/backward finite differencing of order 4 to replace the spacial differences. Possibly order 2 because that would simplify the problem.&lt;/p&gt;&#xA;" OwnerUserId="15542" LastActivityDate="2018-03-30T04:43:31.510" Title="Boundary Conditions for a circular clamped plate" Tags="|mechanical-engineering|control-engineering|applied-mechanics|mathematics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21101" PostTypeId="1" CreationDate="2018-03-30T11:59:30.920" Score="1" ViewCount="109" Body="&lt;p&gt;I'm working on a pipe design problem using Bernoulli's equation, from first principles. The pipe is from a water storage reservoir (point 1) and discharges into a river (point 2). These are my two reference points respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I know that the velocity at the reservoir is zero (point 1) as the water level is assumed to be stationary, but is the velocity at point 2 equal to the speed of the river?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to make sense from my understanding as the movement of water in the river would carry some energy which must be considered. &lt;/p&gt;&#xA;" OwnerUserId="15545" LastEditorUserId="10902" LastEditDate="2018-03-30T19:21:31.160" LastActivityDate="2024-03-29T17:06:05.713" Title="Application of Bernoulli's equation between stationary and moving reference points" Tags="|design|piping|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21102" PostTypeId="1" CreationDate="2018-03-30T12:32:45.690" Score="0" ViewCount="744" Body="&lt;p&gt;I'm looking to construct a something with hollow steel tubing (e.g. 10ga 2&quot;x3&quot; rectangular tubing) suspended horizontally welded to an outer frame on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already calculated the deflection given the loads it'll experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I know there'll be a point as I increase the load where the sidewalls will buckle/crinkle, and this point could be lower than the load where the metal bends too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate this load?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done lots of searches, but nearly everything I can find regarding buckling loads is for metal oriented as a column and compressed, and that's not the case for this metal.&lt;/p&gt;&#xA;" OwnerUserId="1177" LastActivityDate="2018-05-30T15:23:54.497" Title="Hollow Steel Tubing: Critical Load for Sidewall Buckling" Tags="|mechanical-engineering|mechanical-failure|buckling|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21103" PostTypeId="1" CreationDate="2018-03-30T13:10:44.533" Score="1" ViewCount="73" Body="&lt;p&gt;What is the type of and principle of operation of the fitting connector circled in this photo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RvWrt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RvWrt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few other similar joints depicted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like this connector is tightened &quot;circumferentially&quot; as a hoop, possibly / probably over the top of the mating flanges of the two fittings. So that makes the fitting ends universal but its &lt;em&gt;not obvious how it would provide reaction against the pressure at the joint&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Contrast to the bolted flanges just below the circled area - its more obvious that they press the fittings together, maybe with a gasket, which would resist internal pressure forcing the joint apart).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note: photo snippet from &lt;a href=&quot;https://engineering.stackexchange.com/questions/15736/what-is-this-plumbing-apparatus&quot;&gt;What is this plumbing apparatus?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6489" LastActivityDate="2018-08-27T18:01:05.330" Title="Type and mechanism of piping fitting" Tags="|pressure|piping|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21104" PostTypeId="2" ParentId="21103" CreationDate="2018-03-30T13:22:48.973" Score="2" Body="&lt;p&gt;The flange on the pipes are beveled and the groove in the clamp is V shaped. There is also a seal between the 2 pipes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the clamp is screwed down it will force the 2 ends together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimensions are such that the clamp will not bottom out without providing enough pressure to ensure water tightness.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2018-03-30T13:22:48.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21105" PostTypeId="2" ParentId="21102" CreationDate="2018-03-30T13:54:20.317" Score="0" Body="&lt;p&gt;It depends on your load and how concentrated it is. A distributed load will not cause local buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually for a case like yours buckling or plastic deformation start at the fixed ends for the greatest bending moment is there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the complex geometry of the connection there there is no analytical way to calculate failure load, only tests and or manufacturer charts, if available, could help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, the ends were not welded and simply supported you could, as a first estimate, consider the plastic moment at the center and reduce it by a factor of safety.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crushing under a point load can happen in different forms and shapes, all of which can bear in the final strength. There is a random pattern in the actual shape of the buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But usually the process ends by developing a plastic groove on top which will extend beyond the cross section's sidewalls and deform them by pushing them out, creating a plastic hinge, hence a mechanism.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="10902" LastEditDate="2018-03-30T19:19:53.367" LastActivityDate="2018-03-30T19:19:53.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21106" PostTypeId="2" ParentId="21103" CreationDate="2018-03-30T14:57:32.770" Score="1" Body="&lt;p&gt;These are rigid grooved clamps and are frequently used in fire protection and sprinkler systems plumbing because they allow fast replacement of valves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one 2/12 inch example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://m.made-in-china.com/product/2-1-2-Size-Grooved-Rigid-Coupling-and-Plumbing-Clamps-for-Fire-Protection-Piping-705008162.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://m.made-in-china.com/product/2-1-2-Size-Grooved-Rigid-Coupling-and-Plumbing-Clamps-for-Fire-Protection-Piping-705008162.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-30T14:57:32.770" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21107" PostTypeId="2" ParentId="21101" CreationDate="2018-03-30T16:12:19.057" Score="0" Body="&lt;p&gt;If the water discharges there is a flow, then it has a velocity depending on head and pipe specification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these types of plumbing of gravity pipes they leave a gap at the discharge so as not to create a siphone. So no consideration of flow of river. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-03-30T16:12:19.057" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21108" PostTypeId="2" ParentId="10624" CreationDate="2018-03-30T17:03:01.550" Score="1" Body="&lt;p&gt;I know this is a bit late but an Arduino should be capable of interpreting pulse train or positional transducers. These are used in industrial machines so are really rugged. This would allow the software to map the signal onto all sorts of things.&lt;/p&gt;&#xA;" OwnerUserId="15548" LastActivityDate="2018-03-30T17:03:01.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21109" PostTypeId="1" AcceptedAnswerId="21121" CreationDate="2018-03-30T17:53:23.303" Score="6" ViewCount="114" Body="&lt;p&gt;This may sound rather vague on its own so I will explain with examples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solid booster rockets used by NASA space shuttle used a central bore design because it had a certain characteristic on burn rate and thrust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M8rMT.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M8rMT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this data was to be acquired only form experimentation like burning test rockets and not deriving any mathematical relation between bore design and thrust then would it be considered sound engineering?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are any designs based solely on data from trial and error used in  critical mainstream engineering? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like we use it just because it works and do not investigate why it works like that because it's very complex to understand but still easier to implement.&lt;/p&gt;&#xA;" OwnerUserId="15549" LastEditorUserId="90" LastEditDate="2018-03-31T07:56:23.250" LastActivityDate="2018-04-04T08:26:27.327" Title="Does every engineering design needs to have a complete theoretical backing or may experimental data suffice?" Tags="|aerospace-engineering|mathematics|manufacturing-engineering|process-engineering|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21110" PostTypeId="2" ParentId="18187" CreationDate="2018-03-30T18:14:09.667" Score="0" Body="&lt;p&gt;If the size of your vents is consistent, you could construct a section of duct that hooks up to a vane anemometer. As long as your setup doesn't restrict flow, you can get fairly accurate, repeatable measurements. The duct needs to capture all of the flow, but not restrict it. This might not be possible in your setup, but probably worth looking at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical measurements are in flow rate, not just velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HjHk4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HjHk4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11151" LastActivityDate="2018-03-30T18:14:09.667" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21111" PostTypeId="1" AcceptedAnswerId="21112" CreationDate="2018-03-30T18:28:28.070" Score="1" ViewCount="2365" Body="&lt;p&gt;I have a 3d printed (ABS or PLA) case which houses small single board computer. I want to glue it to glass (car's windshield from the inside) so that it sticks nicely and does not vibrate. However, there should be a chance to remove the case from the glass later if wanted without damaging the glass in any way. Will someone recommend suitable adhesive and removal method (heat etc.)?&lt;/p&gt;&#xA;" OwnerUserId="414" LastActivityDate="2018-03-30T22:14:20.977" Title="Glue plastic to glass with chance for easy removal later" Tags="|plastic|adhesive|glass|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21112" PostTypeId="2" ParentId="21111" CreationDate="2018-03-30T22:14:20.977" Score="2" Body="&lt;p&gt;Because the surface is a windshield and because you wish to have it somewhat removable, consider Solar Mike's suggestion of double sided tape, but add in a parameter for outdoor use. &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B00004Z4BV&quot; rel=&quot;nofollow noreferrer&quot;&gt;3M makes a great high density foam tape&lt;/a&gt; that I've been using for many years to attach miscellaneous items to the windshield. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surfaces have to be clean, of course, and the bond between the tape and the plastic should be secure. Because you're considering the plastic to be 3D printed, you may want to sand the surface with fine sandpaper, even to the point of getting it glossy for the best bond. If you construct the part with ABS, performing acetone-vapor smoothing will give you a good surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest to attach the tape to the plastic surface and allow it to bond for a day, then bond the assembly to the windshield. You can see how well the tape is bonding by looking at the glass, pushing against the box to work any visible bubbles out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've recently noted that Eclipse Sunshades switched adhesive pads on their product to this 3M product (grey foam adhesive) perhaps because the other verions previously used would release. I'd clean them off, put this 3M stuff in place and be assured of a solid attachment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removal means careful slicing with a knife or razor, then additional mechanical removal (fingers scraping) followed by any convenient solvent (acetone - use carefully) or alcohol, although I've not used that, and windows cleaner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OxKNx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OxKNx.jpg&quot; alt=&quot;mounting tape&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-03-30T22:14:20.977" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21113" PostTypeId="1" CreationDate="2018-03-31T00:52:57.833" Score="0" ViewCount="264" Body="&lt;p&gt;Is the formula $T=2qA$ valid for both open thin-walled and closed thin-walled members, where $T$ is torque, $q$ is shear flow, and $A$ is the area?&lt;/p&gt;&#xA;" OwnerUserId="10571" LastEditorUserId="1832" LastEditDate="2018-03-31T11:12:11.083" LastActivityDate="2018-04-01T19:47:31.807" Title="Torsion on thin walled members" Tags="|structural-engineering|civil-engineering|structures|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21114" PostTypeId="2" ParentId="21113" CreationDate="2018-03-31T03:11:38.870" Score="2" Body="&lt;p&gt;No , the statement is true only for closed thin wall sections. &#xA;$$ T = \oint_s\text{d}t = \oint q(h\text{d}s) = q \oint h \text{d}s = q \oint 2 \text{d}A = q(2A) = 2qA $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is the area of the section; $T$, the torque; $q$, the shear flow; and $h$, the distance from the center of the section to the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque is much smaller in open wall sections.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2018-03-31T11:15:05.877" LastActivityDate="2018-03-31T11:15:05.877" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21115" PostTypeId="1" CreationDate="2018-03-31T07:37:41.780" Score="2" ViewCount="44" Body="&lt;p&gt;What's the best way to control the following type of platform so that it remains balanced and upright?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gHv7x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gHv7x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a box supported by a single leg that can pivot at the foot (F) and at the hip (H). The foot and hip joints are designed so that they can only rotate in the xz plane (where x goes left-to-right and z goes up-and-down). It's similar in concept to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverted_pendulum&quot; rel=&quot;nofollow noreferrer&quot;&gt;inverted pendulum&lt;/a&gt;, except the base (F) is fixed, and the actuation is done by shifting the mass at the &quot;top&quot; of the pendulum. I've seen some academic papers refer to this as the &quot;cart-table model&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hip joint is powered by a servo motor which can report its angle of deflection relative to the box (ϴⱼ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the center-of-mass (COM), there's an accelerometer that reports the box's angle of deflection relative to gravity (ϴᵢ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've constructed this platform and I'm now trying to program a microcontroller to actuate the servo so that the platform remains both upright and level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure depicts the four basic states that the platform will experience. At t₀, it's perfectly level, but in an unstable leg configuration so that its center-of-mass is off-center and will cause it to begin toppling over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At t₁, the platform has tipped forward slightly. I've been able to program my controller to detect this and start actuating the servo to rotate the box counter to the angle of deflection so that the center-of-mass shifts to the other side of the leg, taking me to state t₂.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is where I run into problems. I'm using a simple PID controller that's being told the target value of ϴᵢ, while being fed the current value of ϴᵢ, and it's attempting to output a servo control signal to achieve this correction. After some testing, I realized this doesn't work because ϴᵢ isn't the only angle that needs to be controlled. By only looking at ϴᵢ, I was seeing massive oscillations of the servo all while the box topples over, because as soon as the servo tilted the box over the foot, ϴᵢ was then horribly unbalanced in the opposite direction, causing the PID controller to again correct it by shifting the box back over the foot...in the direction that the platform is falling, which only hastens the fall. I'm not sure how to tell the controller to &quot;ignore&quot; the unbalanced ϴᵢ once the center-of-mass has been shifted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I think the solution is that I need to either directly measure ϴₖ, or infer it by using ϴᵢ and ϴⱼ, and then feed both ϴₖ and ϴᵢ into my controller. However, I'm only familiar with simple PID controllers that take a single input. What type of controller would I use to take two inputs and output a servo signal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I input ϴₖ into my PID controller, I could probably get my platform to stop toppling over, but the box wouldn't be level. It would achieve a state close to that shown at t₂, where the box is skewed but its center-of-mass is balanced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would another solution be to re-locate the accelerometer to point H? If it were there, I could use the raw acceleration readings to get a reliable sense of when the platform has stopped shifting in the wrong direction, and use that period to temporarily disable the PID controller so it doesn't erroneously un-shift itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to get the platform to the state depicted in tₙ, where the box is level and the leg is angled so that the center-of-mass is evenly distributed so that the platform can remain upright. What type of controller would I use to accomplish this? Is there anything else I'm missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Updated the figure to depict the state described by @fibonatic.&lt;/p&gt;&#xA;" OwnerUserId="175" LastEditorUserId="175" LastEditDate="2018-03-31T17:58:41.350" LastActivityDate="2018-03-31T17:58:41.350" Title="How to control an underactuated platform with a single servo" Tags="|control-engineering|control-theory|dynamics|pid-control|signal-processing|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21116" PostTypeId="1" CreationDate="2018-03-31T10:10:57.867" Score="0" ViewCount="196" Body="&lt;p&gt;I am using a method proposed by R. G. Reynolds to estimate attitude based on two vector measurements, taken from: &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19990052720.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19990052720.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we measure 2 unit vectors $b_{1}$ and $b_{2}$ in the body frame, and we know the components in the inertial frame are $r_{1}$ and $r_{2}$. The attitude quaternion can be calculated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ c = |(b_{1} - r_{1}) \times (b_{2} - r_{2})|^2 + ((b_{1} + r_{1})\cdot (b_{2} - r_{2}))^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q = \frac{1}{\sqrt{c}} [b_{2} \cdot r_{1} - b_{1} \cdot r_{2}, (b_{1} - r_{1}) \times (b_{2} - r_{2})] $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $c$ is a normalisation factor and $q$ is in the form $[w, \vec{v}]$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying this method out and it works half the time, but occasionally all the elements of the quaternion are their negative. Is there a way around this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there is also a singularity problem when the two vectors are parallel but I don't think this is the problem?&lt;/p&gt;&#xA;" OwnerUserId="15557" LastActivityDate="2018-04-01T22:55:15.257" Title="Calculate attitude quaternion from 2 vector measurements" Tags="|aerospace-engineering|geometry|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21119" PostTypeId="1" AcceptedAnswerId="21153" CreationDate="2018-03-31T12:56:53.623" Score="0" ViewCount="918" Body="&lt;p&gt;I have a solids problem I cannot solve, I have the answer and I have attempted the question. I just need to know where I am going off track, (perhaps there was an issue setting up my deflection equations?). Any assistance would be highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RDqX2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RDqX2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My work is as follows:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Castigliano's theorem: deflection = partial derivative of strain energy U wrt a force F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I split the beam into two parts, A - B and C - B (So I don't have to&#xA;deal with the reaction forces at C)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;X is the distance from the free end for Mab and Uab&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And it is the distance from F for Mbc and Ubc&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My first moment:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mab = Fx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2nd moment:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mbc = F(x-b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U = integral from o to x of (M^2/(2EI))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uab = integral from 0 to b of ((f^2*x^2)/2EI) = (f^2/2EI)*[x^3/3] (0-b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uab = (f^2/2EI)*(b^3/3) = ((F^2*b^3)/6EI)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ubc = integral from 0 to b of ((F^2*(x-b)^2)/2EI) = (F^2/2EI)*[(x -&#xA;b)^3/3] (L-b)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ubc = ((F^2*(L^3 - (3b(L^2))+(3L(b^2)) - b^3))/6EI)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Therefore&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Utot = ((F^2*b^3)/6EI) + ((F^2*(L^3 - (3b(L^2))+(3L(b^2)) - b^3))/6EI)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Utot = ((F^2*(b^3+L^3 - (3b(L^2))+(3L(b^2)) - b^3))/6EI)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Utot = ((F^2*(L^3 - (3b(L^2))+(3L(b^2))))/6EI)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deflection = dUtot/dF = ((2*F*(L^3 - (3b(L^2))+(3L(b^2))))/6EI)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Deflection = ((F*(L^3 - (3b(L^2))+(3L(b^2))))/3EI)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Clearly not the answer,&lt;/strong&gt; I am not sure but maybe I went wrong somewhere in the first&#xA;couple of steps, perhaps either setting up my moment equations or setting them&#xA;up in the integral. But I can't figure it out. Can someone please help ID my error and let me know what I did wrong?&lt;/p&gt;&#xA;" OwnerUserId="11039" LastEditorUserId="11039" LastEditDate="2018-03-31T13:22:26.040" LastActivityDate="2018-04-03T00:43:44.710" Title="Using strain energy to calculate deflection" Tags="|mechanical-engineering|structural-engineering|materials|structures|stresses|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21121" PostTypeId="2" ParentId="21109" CreationDate="2018-03-31T14:44:58.860" Score="7" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are any designs based solely on data from trial and error used in critical mainstream engineering?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Usually not. And the reason is that trial and error is expensive and time consuming. As engineers, we are always working on projects with a budget and a deadline. Take your rocket example. Rockets are expensive. For sake of argument, let's just say it's $1 million per rocket test. You can't afford to build rocket after rocket after rocket solely by trial and error. You'll go through your budget very quickly. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does every engineering design needs to have a complete theoretical backing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, usually not.  And the reason is that developing &lt;strong&gt;complete&lt;/strong&gt; theoretical understanding is also expensive.  You could hire army of PhD researchers to come up with an incredibly detailed model and buy giant supercomputers to simulate it.  But an engineer with a PhD will cost $100k/year just in salary. If your model is so detailed that you need a team of 20 researchers working for years to come up with the model, it would be cheaper to just run a test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a balance.  You try to come up with a model that is sophisticated enough that it explains the majority of the behavior, but simple enough that you don't break the bank coming up with it. Your model won't be perfect, so you run a few experiments to fill in the holes in the theory, but not so many that you kill the budget.  The key decision is deciding the tradeoff between model and experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, if I had to describe what engineers do in exactly one word, that word is &quot;tradeoffs&quot;.  In every engineering decision, there is always a balance between multiple competing objectives, and the engineers job is to make the best tradeoff.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-03-31T14:44:58.860" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21123" PostTypeId="1" CreationDate="2018-03-31T15:36:22.470" Score="1" ViewCount="134" Body="&lt;p&gt;I have an ultrasonic testing equipment that can only measure the direct and semi-direct ( (a) and (b) in the image below):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VXOps.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VXOps.jpg&quot; alt=&quot;image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can measure compression wave velocities fairly easily, but I would like to know if it is possible to measure the shear wave velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to measure the shear wave velocity in a lab setting using this? What should the distance between the two sensors be to provide accurate measurments?&lt;/p&gt;&#xA;" OwnerUserId="15561" LastEditorUserId="1832" LastEditDate="2018-04-01T23:06:48.340" LastActivityDate="2018-04-02T16:15:04.723" Title="Is it possible to calculate the shear wave velocity in lab using ultrasonic test?" Tags="|civil-engineering|vibration|ultrasound|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21125" PostTypeId="2" ParentId="16661" CreationDate="2018-03-31T15:58:59.330" Score="0" Body="&lt;p&gt;You can think of capacity of the total system as being proportional to the mass flow times the temperature jump the compressor puts on the refrigerant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot Q \propto \dot{m}\cdot\left(T_{discharge}-T_{suction} \right)$$&#xA;or&#xA;$$ \dot Q = \epsilon \cdot k \cdot \dot{m}\cdot\left(T_{discharge}-T_{suction} \right)$$&#xA;If you keep everything else the same, and heat up the suction gas, then that heat has to be rejected by the system using the outdoor/condensing coil above and beyond the heat from the conditioned space, and above and beyond the heat from the compressor motor.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All things are not equal.  This is an irony of science, we say &lt;em&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ceteris_paribus&quot; rel=&quot;nofollow noreferrer&quot;&gt;ceteris paribus&lt;/a&gt;&lt;/em&gt;, but it never works out that way.  If you change the suction line temperature, also called the evaporator temperature, then the compressor will try to make the discharge hotter.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you could look at the &quot;compressor equations&quot; for your compressor and refrigerant in the operating envelope that operate on the evaporator(suction) and condenser (discharge) temperatures and give power, amps, mass flow, and capacity.  These sorts of things should be published, public information.  &lt;/p&gt;&#xA;" OwnerUserId="1953" LastActivityDate="2018-03-31T15:58:59.330" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21126" PostTypeId="1" CreationDate="2018-03-31T17:37:34.907" Score="-3" ViewCount="86" Body="&lt;p&gt;How come you can pour molten titanium with a melting point of 3291 degrees into a steel funnel with a melting point of 1300 degrees and it doesn't melt the funnel?&lt;/p&gt;&#xA;" OwnerUserId="15562" LastEditorUserId="1832" LastEditDate="2018-04-01T01:16:29.830" LastActivityDate="2018-04-01T01:16:29.830" Title="Why doesnt steel melt when pouring titanium through it" Tags="|finite-element-method|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21127" PostTypeId="2" ParentId="21126" CreationDate="2018-03-31T17:50:24.187" Score="5" Body="&lt;p&gt;Firstly you're mixing units &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The melting point of titanium is 1668C (3000F)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The melting point of steel starts at around 1300C up to around 1500C depending on carbon content. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For brief contact where the mass of the steel is reasonably high you would probably get away with this but generally foundry equipment which comes into contact with molten metal is lined with a refractory of some sort. Which both provides thermal insulation and prevents cross-contamination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather more of an issue is the reactivity of titanium which will burn in air some time before it gets to its melting point and needs inert gas shielding to be worked hot. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-03-31T17:50:24.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21128" PostTypeId="1" CreationDate="2018-03-31T21:33:26.957" Score="3" ViewCount="80" Body="&lt;p&gt;I have an electronic assembly that is heat sensitive. Due to the incorporation of a CR3032 coin cell battery, it can survive being heated to a maximum of 60 degrees C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to coat this assembly in a biocompatible, waterproof plastic. This layer will be the ultimate barrier to moisture between my device and a biological environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I am using parylene-C, which performs admirably, but is very expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various species of thermoplastic promise to perform just as well, at dramatically lower cost, but I don't know how I'd apply them without damaging my battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question to you is, how could I coat my circuit in a thermoplastic without thermally damaging it, and without spending more than a few dollars per unit?&lt;/p&gt;&#xA;" OwnerUserId="15564" LastActivityDate="2019-10-04T09:01:50.303" Title="Molding thermoplastic over sensitive components" Tags="|plastic|molding|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21130" PostTypeId="1" CreationDate="2018-04-01T10:12:45.473" Score="1" ViewCount="64" Body="&lt;p&gt;I am using concentrated NaCl(10%) to elute Cobalt (II) ions from a cationic exchange column. How would I determine the volumetric flowrate of regenerant needed to completely elute the cobalt ions from the resin?&lt;/p&gt;&#xA;" OwnerUserId="12385" LastActivityDate="2024-04-01T17:04:55.583" Title="Elution from an Ion Exchange Column" Tags="|chemical-engineering|waste-water-treatment|columns|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21131" PostTypeId="1" CreationDate="2018-04-01T16:05:55.670" Score="1" ViewCount="17" Body="&lt;p&gt;I am exploring whether a diamond based capacitor could be made for powering a drone. From my initial research, it has the right material properties (thermal conductivity, strength to weight ratio, dielectric strength), but I know from experience that doesn't mean it can be made. I found that the only practical way of making the near flawless diamonds are through chemical vapor deposition techniques. Since the capacitor would need to have conductive plates that sandwich the diamond, it would make the part a composite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am here to ask the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would make a composite chemical vapor deposition diamond hard to make?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my initial thinking about it, it seems that mating the two layers chemically would be difficult, as diamond is quite inert and doesn't want to bond to the conductive metals needed for the electrodes. But, if carbon, like graphite or graphene is used, I think it could work since the whole capacitor would essentially be the same element, just configured differently. The size of the capacitor would be 0.1 cc, so I think most vacuum chambers already in existence would be able to make the part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this doable with current technology, or are we missing something that makes this not possible currently?&lt;/p&gt;&#xA;" OwnerUserId="15570" LastActivityDate="2018-04-01T16:05:55.670" Title="What would make a composite chemical vapor deposition diamond hard to make?" Tags="|electrical-engineering|materials|chemical-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21135" PostTypeId="2" ParentId="11121" CreationDate="2018-04-01T19:38:19.140" Score="2" Body="&lt;p&gt;Continuous beams are safer than simply supported ones in general. They have greater degree of static undetermination. If one section fails, it will form a plastic hinge and will redistribute to the other sections. Then it will form another hinge and another until the whole beam fails. For simply supported beams, we need just one section to fail and it is done. This is in case you design both as needed but something unpredicted happens later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, you can make any beam safer than any other, regardless the static scheme, by simply increasing the size and reinforcement.&lt;/p&gt;&#xA;" OwnerDisplayName="user15573" LastActivityDate="2018-04-01T19:38:19.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21136" PostTypeId="1" AcceptedAnswerId="21138" CreationDate="2018-04-01T22:49:17.270" Score="0" ViewCount="3717" Body="&lt;p&gt;I'm trying to figure out how to construct a CNC XY table, ala this instructable: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.instructables.com/id/Low-Cost-Hobby-Servo-XY-Table/#discuss&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.instructables.com/id/Low-Cost-Hobby-Servo-XY-Table/#discuss&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a noob with mechanical design/mechanical parts, so I don't understand a lot of what this guy is saying. One of those things though, is that you can use (paraphrasing here, with limited understanding) a 10 turn potentiometer with a servo to ... make a motor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to know is, what is a 10-turn potentiometer? Is that somehow different from regular potentiometers? How do you modify a servo with a potentiometer, and how does that allow you to turn something more than the 180 or 90 degrees that a servo allows?&lt;/p&gt;&#xA;" OwnerUserId="15574" LastEditorUserId="1832" LastEditDate="2018-04-01T22:54:40.997" LastActivityDate="2018-04-01T23:39:21.820" Title="What is a 10 turn potentiometer" Tags="|motors|servo|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21137" PostTypeId="2" ParentId="21116" CreationDate="2018-04-01T22:55:15.257" Score="1" Body="&lt;p&gt;In this answer I will represent a quaternion as $q = (q_r,q_v)$, with $q_r$ the real part and $q_v$ the complex vector part of the quaternion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rotating a 3D vector $x$ by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotation represented by a unit quaternion&lt;/a&gt; can be done as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(0,x') = (q_r,q_v)\,(0,x)\,(q_r,-q_v) =q\,(0,x)\,\bar{q}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with $\bar{q}$ the complex conjugate of $q$. Note that the complex conjugate of $-q$ is equal to $-\bar{q}$. So if the quaternion is negated $q^*=-q$ then the rotation with this gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(0,x') = q^*\,(0,x)\bar{q^*} = (-q)\,(0,x)\,(-\bar{q}) = q\,(0,x)\,\bar{q})&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the same end result. So $q$ and $-q$ represent the same rotation.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-04-01T22:55:15.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21138" PostTypeId="2" ParentId="21136" CreationDate="2018-04-01T23:39:21.820" Score="4" Body="&lt;p&gt;it's not unusual to find a single turn potentiomenter (aka pot) that provides a change in resistance over the range of the turning component. The benefit of a single turn pot is that you can get a large change in resistance with minimal movement. Volume controls are a good example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A multi-turn potentiometer is just the opposite. It provides a similar range of resistance, but requires that you perform multiple rotations of the control to &quot;travel&quot; from one end of the range to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, a volume control may provide 10K resistance from one point to the other. You can select a 5K resistance (5000 ohms) by approximately turning to the halfway point. Due to the large range of resistance over such a small movement, it is difficult to pinpoint the 5000 ohm figure. For something as objective as volume control, such a coarse level of resistance is acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, with a ten turn pot, as in your example, of the same 10K value, you can easily guess at five turns to get close to the 5000 ohm level, but more important, you can move a quarter turn and have substantially less change than with the single turn pot and be in a better position to get the figure you require.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows one to provide for fine adjustment of various devices, such as the servo mentioned in the article you've linked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had a circuit to control a servo, using a single turn pot, you may get 180° movement over the full turn of the pot, but with the ten turn pot, each turn would possibly provide only 10° of movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a generalization that I'm hopeful will assist your understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask how to modify a servo with a potentiometer. A servo is a small motor with a potentiomenter within the housing. The potentiomenter sends a feedback signal to the controller to inform the controller of the location of the servo. What modification do you require of this internal pot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some servos are engineered to turn more than 90° or 180° and some can be modified to do so. This would be considered outside the scope of this answer and also can be found by using search terms such as &quot;modify servo rotation&quot; or similar. You may also discover such a reference in the world of &lt;a href=&quot;https://www.instructables.com/id/How-to-Modify-Micro-Servo-for-360-Rotation/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Instructables&lt;/a&gt; such as this provided link.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-04-01T23:39:21.820" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21142" PostTypeId="2" ParentId="19801" CreationDate="2018-04-02T12:24:44.003" Score="2" Body="&lt;p&gt;The tensile strength is diameter dependent. How greater the diameter, how longer it takes to discharge the heat, so, the time it takes to cool down the material (here is steel) has an impact on the microstructure and consequently the mechanical characteristic of the material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tensile strength would be lower than what is already mentioned in the table, because the shaft has a bigger diameter than $16mm$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correction factors are located at the manufacturing catalogues and not in academic table books.  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-04-02T12:24:44.003" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21143" PostTypeId="1" AcceptedAnswerId="21147" CreationDate="2018-04-02T12:51:16.207" Score="0" ViewCount="1320" Body="&lt;p&gt;Consider, while building a new machine which has different elements in it. For instance, a machine having a driving assembly, main body, chassis, crucible etc.&#xA;Then, how would I calculate the center of gravity and how would I know the calculated value is an exact value or Cg value gives a balanced system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I calculate a center of gravity of each element or the whole system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I need to calculate the center of gravity of each element, how would I calculate? What are the factors to be considered other than weight and distance from the reference plane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance&lt;/p&gt;&#xA;" OwnerUserId="15526" LastActivityDate="2018-04-02T16:31:10.700" Title="How to find the center of gravity(Cg) of machine?" Tags="|structural-engineering|" AnswerCount="2" CommentCount="1" ClosedDate="2018-04-03T17:37:22.340" ContentLicense="CC BY-SA 3.0" />
  <row Id="21144" PostTypeId="1" AcceptedAnswerId="21151" CreationDate="2018-04-02T13:02:03.607" Score="1" ViewCount="4511" Body="&lt;p&gt;I'm designing a shaft, and i have to choose a material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I deal with cyclic stress in my machine, so i set up an expression for all the possible stresses and combined it with von Mises criteria for plastic deformation, it gives me the so called &lt;strong&gt;comparison stress&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This comparison stress should be lower than the endurance limit of the material. I don't have the endurance limit, but instead the tensile strength. &#xA;Is there any connection between endurance limit and the tensile strength ?  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-04-03T03:05:31.283" Title="Is there any link between tensile strength and endurance limit for steel ?" Tags="|mechanical-engineering|stresses|fatigue|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21145" PostTypeId="2" ParentId="21143" CreationDate="2018-04-02T14:56:40.430" Score="2" Body="&lt;p&gt;You can use the a finite element-like method like you're describing, chop up the object in box shaped pieces and combine their individual weight and gravity centre(the exact middle with boxes), but that's rather hard and time-consuming to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the opbject at your disposal, you can determine it by just using a scale. If you take three or four points, the front, back and side, you know the weight distribution and location of the gravity centre in the horizontal plane. If you tilt the object at a known angle and measure one of the points again, you can calculate the height of the gravity centre with trigonometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're working with a vehicle or dynamic object, determining the rotational inertia may also be interesting.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-04-02T14:56:40.430" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21146" PostTypeId="2" ParentId="21123" CreationDate="2018-04-02T16:15:04.723" Score="1" Body="&lt;p&gt;Well, nothing is impossible. But, lets see what you are trying to do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iP40q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iP40q.png&quot; alt=&quot;Diagram of waves&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram is simplified; in reality, there are 3 different P (compression) waves and 2 different S (shear) waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measurement of the first P wave is very easy because it depends on the first arrival time (it is the fastest wave).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to measure the S wave, you first need to isolate the noise, the reflections caused by P waves, the refraction and then determine if you have the correct S wave or not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing so is very hard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you can use specific transducers called shear wave transducer that will generate shear waves directly so you can measure them. &lt;/p&gt;&#xA;" OwnerUserId="25" LastActivityDate="2018-04-02T16:15:04.723" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21147" PostTypeId="2" ParentId="21143" CreationDate="2018-04-02T16:31:10.700" Score="3" Body="&lt;p&gt;You have to know the mass and CG of each individual part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then once assembled you set your machine against a corner wall which you will use as xy axis. measure the Dx and Dy, the distance from CG of parts to corresponding X and Y walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you calculate CG_x as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ CG_x = (\Sigma Mn. Dxn )/W  $$ and CG_y as:$$ CG_y = (\Sigma Mn.Dyn)/ W $$ Then you can transfer this X and Y to a more convenient axis, such as one parallel with a prominent axis on your machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many CAD programs that can calculate CG by the specs and wire-mesh of the machine.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-04-02T16:31:10.700" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21148" PostTypeId="1" CreationDate="2018-04-02T17:21:41.063" Score="1" ViewCount="81" Body="&lt;p&gt;Looking for advice on how to turn off a motor when there is an item blocking a sensor. I found &lt;a href=&quot;http://store.ezautomation.net/product-p/ez-m12-pnpno-fmw.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;an inductive proximity sensor&lt;/a&gt; that I think could work. I am trying to make a brass case annealer &lt;a href=&quot;https://www.youtube.com/watch?v=tBUBOdxHgho&amp;amp;t=10s&quot; rel=&quot;nofollow noreferrer&quot;&gt;similar to this one&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15587" LastActivityDate="2018-04-03T13:56:16.767" Title="DC motor disconnect sensor" Tags="|motors|electrical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21149" PostTypeId="2" ParentId="21144" CreationDate="2018-04-02T17:32:09.253" Score="0" Body="&lt;p&gt;per wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Typical values of the limit for steels are 1/2 the ultimate tensile strength, to a maximum of 290 MPa&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fatigue_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Fatigue_limit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you want to rely on such a rule of thumb depends on your application.&lt;/p&gt;&#xA;" OwnerUserId="7461" LastActivityDate="2018-04-02T17:32:09.253" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21150" PostTypeId="2" ParentId="21144" CreationDate="2018-04-02T18:52:40.183" Score="0" Body="&lt;p&gt;I suspect not, based on a table provided &lt;a href=&quot;http://www.chegg.com/homework-help/endurance-limit-fatigue-life-steel-approximately-one-half-ul-chapter-5-problem-37p-solution-9780136081685-exc&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; .  It would appear that the relationship is strongly material-dependent and not analytically calculable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9ouf0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ouf0.png&quot; alt=&quot;table&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-04-02T18:52:40.183" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21151" PostTypeId="2" ParentId="21144" CreationDate="2018-04-02T19:34:15.870" Score="0" Body="&lt;p&gt;Yes, to a degree. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The endurance limit is normally expressed as a fraction of the yield stress and for steel a typical value is 1/2 (50%). This is is certainly a reasonable assumption for common low alloy, medium carbon steels ie the sort of steels that you might reasonably select for  a shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact for many practical purposes the factor of safety that you use to account for dynamic loading will be similar or greater than this anyway so it's often not a limiting factor in steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally for high performance, highly optimised designs you might not consider endurance stress as an absolute limitation and just design the part with a fatigue life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for really critical parts you relay need to consider the very detailed design in terms of surface finish, surface treatments and manufacturing processes as well as things like minimum radii when calculating fatigue life, for which tables of coefficients exist. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general in engineering design optimisation is a process of getting increasingly confident about the actual real world loads a part will see, a process which is very much subject to diminishing returns and for general purpose deign it is often a case of selecting a sensible and pragmatic factor of safety in which you can be confident. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Response to comment by OP &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is mostly empirical. The difficulty with fatigue is that it is highly dependant on local stress raisers. As a bulk material steel is not very susceptible to fatigue, but you can easily get local stress concentrations eg around a surface defect in the wrong place which exceed the yield stress of the material and consequent work hardening over multiple cycles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't &lt;em&gt;exactly&lt;/em&gt; the same mechanism as fatigue in say aluminium which has a very low endurance limit, even in ideal conditions.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with many things in practical engineering it boils down to an empirical coefficient whcih is known to work for a given context. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-04-02T21:15:15.747" LastActivityDate="2018-04-02T21:15:15.747" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21152" PostTypeId="2" ParentId="21148" CreationDate="2018-04-02T20:14:44.373" Score="1" Body="&lt;p&gt;What you are looking for is a normally closed relay which is sometimes abbreviated as NC relay. You should place the relay in you circuit so that switch side has the ability to interrupt the motor when the sensor is tripped. Here is a possible schematic of how your circuit could work.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1fsIy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1fsIy.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;When an object is placed in front of the sensor it closes the sensor which opens the relay which cuts power to the motor. This is one way to achieve control of your motor when an object is placed in front of the sensor.&lt;/p&gt;&#xA;" OwnerUserId="9656" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-04-03T13:56:16.767" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21153" PostTypeId="2" ParentId="21119" CreationDate="2018-04-03T00:43:44.710" Score="3" Body="&lt;p&gt;Castigliano's theorem says that we can apply a dummy load $P$ at any point of interest; the deflection in that location is $$y=\int_L\frac{M}{EI}\frac{\partial M}{\partial P}\,dx$$ letting $P\to 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply the dummy load at the free end of the beam (call this $x=0$). I'm only going to consider the region of the beam to the right of the applied force $F$, since the other part stores no real strain energy. The bending moment in this region is $M=Px+F(x-b)$; thus, $\frac{\partial M}{\partial P}=x$. So we have $$y=\int_b^L\frac{[Px+F(x-b)]x}{EI}\,dx$$ which can be verified for $P\to 0$ to give $y=\frac{F}{6EI}(2L^3-3L^2b+b^3)$, which matches the answer given in the problem.&lt;/p&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2018-04-03T00:43:44.710" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21154" PostTypeId="2" ParentId="21130" CreationDate="2018-04-03T01:55:05.097" Score="0" Body="&lt;p&gt;If you have lab data that tells you how long it takes for the Co(II) to be eluted, you can scale up keeping the residence time the same (and then add is some margin for slower mass transfer at the larger scale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow rate can't be too fast otherwise the capacity and resolution of the elution will be low and you can damage the column beads. You can go slow, but you sacrifice column up-time.&lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2018-04-03T01:55:05.097" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21155" PostTypeId="2" ParentId="21144" CreationDate="2018-04-03T03:05:31.283" Score="0" Body="&lt;p&gt;Steels are relatively unique among metals with a very reliable endurance limit equal to 1/2 of tensile strength. Cast irons definitely do not have this feature.  Nickel and cobalt alloys might have it but there is limited data. Fatigue test coupons are smooth with no stress concentration ; not to say that they never contain an inclusion. So fatigue tests give the best possible value; Real world components normally have various stress concentrators. It  is the duty of the builder to determine the severity of these stress concentrators. By &quot;endurance limit&quot; I mean the number of stress cycles approach infinity.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-04-03T03:05:31.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21156" PostTypeId="2" ParentId="16942" CreationDate="2018-04-03T05:20:20.570" Score="2" Body="&lt;p&gt;That is a wrench flat, presumably used when installing a nut on the conventional threaded portion at the left end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall enlargement into which the wrench flat is machined probably faces either a bearing or possibly some short of pulley that would installed between it and the threaded portion to the left.&lt;/p&gt;&#xA;" OwnerUserId="15129" LastActivityDate="2018-04-03T05:20:20.570" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21157" PostTypeId="2" ParentId="19699" CreationDate="2018-04-03T06:10:44.463" Score="1" Body="&lt;p&gt;With your &lt;em&gt;aligned constraint&lt;/em&gt; command active, press the down arrow on your keyboard and you will see a dropdown with options like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vs0MY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vs0MY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure that &lt;em&gt;Dynamic Input&lt;/em&gt; is switched on to see the feedback at your cursor instead of the command bar at the bottom.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastActivityDate="2018-04-03T06:10:44.463" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21158" PostTypeId="2" ParentId="15798" CreationDate="2018-04-03T11:10:52.550" Score="2" Body="&lt;p&gt;I been told by a number a  fridge maint. People Their rule of thumb is how ever long it lays on its side,,  leave it upright that long before plugging it in . Will give oils time to drain back down to where they are supposed to be. I am sure I didn't word it the exact same way but I am sure  you get what I am saying. if it is on its side  for 12 hours  upright 12 hours before you plug it in. I myself wait as long as I can&lt;/p&gt;&#xA;" OwnerUserId="15590" LastActivityDate="2018-04-03T11:10:52.550" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21159" PostTypeId="2" ParentId="18380" CreationDate="2018-04-03T13:22:54.477" Score="0" Body="&lt;p&gt;According to &lt;a href=&quot;https://knowledge.autodesk.com/search-result/caas/CloudHelp/cloudhelp/2016/ENU/AutoCAD-Customization/files/GUID-3FF72BD0-9863-4739-8A45-B14AF1B67B06-htm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Autodesk's website&lt;/a&gt;, an ARX file is basically a DLL but with a different extension. It is a plugin developed with the ObjectARX API. I don't really know if there is a significant distinction from a DLL, I have only ever written plugins using DLLs and loaded them using the NetLoad command.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A CUIx file is just a &lt;a href=&quot;https://knowledge.autodesk.com/support/autocad-lt/learn-explore/caas/CloudHelp/cloudhelp/2018/ENU/AutoCAD-LT/files/GUID-55B0E62E-9F40-4A95-BCE6-180EB47C9DDD-htm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;user interface customisation file&lt;/a&gt;. When you customise you UI elements and workspace, you can export this file to use on other PCs or as a backup for future use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can, however, &lt;a href=&quot;https://knowledge.autodesk.com/search-result/caas/CloudHelp/cloudhelp/2017/ENU/AutoCAD-NET/files/GUID-71554E76-8FD5-4853-82CD-3587764CBCAC-htm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;control the CUI programatically&lt;/a&gt; in order to provide contextual UI updates to augment the user experience of your plug-ins.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastActivityDate="2018-04-03T13:22:54.477" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21160" PostTypeId="1" CreationDate="2018-04-03T13:56:54.533" Score="1" ViewCount="76" Body="&lt;p&gt;Can someone explain what this sentence means? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The mean power produced in this plant is 100 MW with a mean productivity of 5100 kWh/tcoal.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically what does &lt;strong&gt;5100 kWh/tcoal&lt;/strong&gt; and &lt;strong&gt;100MW&lt;/strong&gt; mean?&lt;/p&gt;&#xA;" OwnerUserId="15591" LastEditorUserId="9656" LastEditDate="2018-04-03T17:38:17.563" LastActivityDate="2018-04-03T18:07:35.283" Title="Meaning of kWh/tcoal" Tags="|chemical-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21162" PostTypeId="2" ParentId="21160" CreationDate="2018-04-03T15:53:43.960" Score="1" Body="&lt;p&gt;FWIW,  Wikipedia quotes coal's energy content as on the order of 6.7 kWh per kilogram.   Data provided &lt;a href=&quot;https://www.wou.edu/las/physci/GS361/Energy_From_Fossil_Fuels.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here &lt;/a&gt;  cites anthracite as being nearly 1.5 X as energy-dense as bituminous,  30 vs 20 kJ/g .&#xA;If you take those numbers you can now calculate the efficiency of the power plant in question.  -- With the caveat that I'm not sure whether that's an English or metric ton referred to. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-04-03T15:53:43.960" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21163" PostTypeId="2" ParentId="21160" CreationDate="2018-04-03T18:07:35.283" Score="4" Body="&lt;p&gt;On average, the power plant produces 100 MW of electricity, or 100 000 kW of electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per tonne of coal burned, the plant produces 5100 kWh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the average amount of coal the plant burns per hour to generate this energy is 100 000/5100, which is 19.608 tonnes per hour.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-04-03T18:07:35.283" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21164" PostTypeId="1" CreationDate="2018-04-03T18:14:19.250" Score="0" ViewCount="147" Body="&lt;p&gt;I have a table flat surface made from aluminum, and the table legs are made from medium carbon steel. What are 3 welding processes that could be used to weld the legs to the flat surface.&lt;/p&gt;&#xA;" OwnerUserId="13130" LastActivityDate="2018-04-08T18:58:36.357" Title="Welding Aluminum to Medium Carbon Steel" Tags="|welding|" AnswerCount="1" CommentCount="5" ClosedDate="2018-04-09T02:32:07.293" ContentLicense="CC BY-SA 3.0" />
  <row Id="21165" PostTypeId="2" ParentId="19205" CreationDate="2018-04-03T19:24:14.827" Score="0" Body="&lt;p&gt;You can use FlightGear as your image generation (animation) software. What you have to do is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;model your system dynamics (forces, moments)&lt;/li&gt;&#xA;&lt;li&gt;write code to solve your system dynamics in time (use Equations of Motion libraries, or write your own)&lt;/li&gt;&#xA;&lt;li&gt;use your code to start and run the simulation and feed the states to FlightGear.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can replace FlightGear with any other tool that you might find. It is mentioned for availability, being free and open source and it has lots of documentation and tutorials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your code can be in any language that you are comfortable with: Python, Matlab, C, C++, Java, Assembly, etc.&lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2018-04-03T19:24:14.827" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21166" PostTypeId="1" CreationDate="2018-04-03T19:50:44.397" Score="1" ViewCount="36" Body="&lt;p&gt;I have a question regarding fluid flow and the change in viscosity of the fluid flow as the field parameter changes from one step to another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am modelling fluid flow, followed by fluid and proppant flow inside a cohesive fracture. In step 2, I just pump in normal fluid (that has low viscosity of 1 cP) and at the beginning of step 3, I am assigning to a particular node a field variable number so that the viscosity of the fluid will change from 1 cP to 3cP at the start of step 3. Thus, what I am doing is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under *Gap Flow keyword,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am entering the following lines:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;*Gap Flow, dependencies=1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;0.001, ,   0.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;0.00122408, , 0.05&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;0.00153288, ,  0.1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;0.00558678, ,  0.3&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;0.0187778, ,  0.4&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Under *Step keyword (when it is step 3),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am entering the following line:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;*FIELD, VARIABLE=1&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Part-1-1.inlet_Node, 0.273&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;Part-1-1.inlet_Node&quot; is the node where I am injecting the fluid, using the command *cflow at the beginning of step 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When step 3 will start, I want Abaqus to change the fluid viscosity because I am introducing the field variable at this node. Basically, I am trying to tell Abaqus that I am injecting fluid with proppants mixed with them at the beginning of Step 3, which will result in the higher pressure gradient accross the fracture length!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I do this, for some reason, this change is not taking place! The pressure change, due to injecting viscous fluid, is not taking place! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please help me with this? Maybe I am doing this wrong! &lt;/p&gt;&#xA;" OwnerUserId="15597" LastEditorDisplayName="user16" LastEditDate="2018-04-24T22:58:20.753" LastActivityDate="2018-04-24T22:58:20.753" Title="How does one activate viscosity change, in gap flow, as the dependency (field variable) is specified at a particular node in Abaqus Standard?" Tags="|abaqus|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21167" PostTypeId="2" ParentId="21164" CreationDate="2018-04-03T20:17:08.113" Score="2" Body="&lt;p&gt;Can't be done . Soldering and brazing physical properties would be poor. I think there is one possibility , which you can not afford, explosion bonding . A piece of aluminum is forced against steel by explosive force. Even that would be a challenge because of the very large differences in properties.  Then aluminum would be welded to the aluminum side and steel welded to the steel side ; still I see a lot of problems. Think epoxy or bolts.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-04-03T20:17:08.113" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21169" PostTypeId="1" AcceptedAnswerId="21174" CreationDate="2018-04-04T04:33:16.817" Score="0" ViewCount="118" Body="&lt;p&gt;We have steel hollow tubes of 80 mm length and 0.5 mm cross section, that we need to cut in smaller pieces of 10.1 and 10.5 mm length. We have been using utility knifes with very poor results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a tool that can do the job. Something like scissors or pliers, to be handled manually. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every answer or suggestions is welcomed.  &lt;/p&gt;&#xA;" OwnerUserId="15600" LastActivityDate="2018-04-04T14:53:52.523" Title="Cutting tool for hollow tubes" Tags="|mechanical-engineering|applied-mechanics|cutting|tools|" AnswerCount="2" CommentCount="8" ClosedDate="2018-04-04T17:15:48.973" ContentLicense="CC BY-SA 3.0" />
  <row Id="21170" PostTypeId="2" ParentId="21169" CreationDate="2018-04-04T06:55:16.837" Score="0" Body="&lt;p&gt;Can you flatten them and use a hole punch, at least there are leather punches used for punching holes in leather...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or put a mandrel inside and use the punch...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-04-04T06:55:16.837" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21171" PostTypeId="1" CreationDate="2018-04-04T07:09:50.157" Score="1" ViewCount="121" Body="&lt;p&gt;My graduation project involves designing several buildings with flat and solid slabs. I already know that ETABS can design flat slabs but is it possible to design solid slabs? &lt;/p&gt;&#xA;" OwnerUserId="15601" LastActivityDate="2018-04-04T07:09:50.157" Title="Is it possible to design solid slabs on ETABS 2016?" Tags="|structural-engineering|finite-element-method|" AnswerCount="0" CommentCount="2" ClosedDate="2018-04-09T02:32:24.277" ContentLicense="CC BY-SA 3.0" />
  <row Id="21172" PostTypeId="2" ParentId="21109" CreationDate="2018-04-04T08:26:27.327" Score="0" Body="&lt;p&gt;For complex and expensive systems you usually want to have an understanding of its behavior, cast in models. To model a system (or part of it), a &lt;strong&gt;complete theoretical understanding&lt;/strong&gt; is usually not required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should, however, have a &lt;strong&gt;complete knowledge&lt;/strong&gt; of the phenomena that act on your system. Then you identify the key phenomena and, if it does not yet exist, a theoretical understanding of those phenomena to a degree required to describe your system accurately enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual work is in finding out which phenomena are important and which are not, what effects to include, what effects to neglect, what depth of understading is actually required, which effects need to be understood, which effects are too expensive to understand now but can be compensated by overly conservative design decisions, etc. Those are some of the tradeoffs that @DanielKiracofe mentions in his answer.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2018-04-04T08:26:27.327" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21173" PostTypeId="1" CreationDate="2018-04-04T08:36:27.687" Score="1" ViewCount="71" Body="&lt;p&gt;Are there any commercial solutions for electric rotary motors with integrated torque sensor and torque control?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm only aware of www.sensodrive.de which are essentially the drives used in the DLR/Kuka Light Weight Robot. But those use harmonic drives (gear ratio of 1/100, &amp;lt; 10 RPM). I need something with much higher RPM and lower torque range (less than 0.01 Nm resolution would be good). Up to 300-1000 RPM but no more than 5 Nm torque is required. So something with a small gear ratio or no gear at all.&lt;/p&gt;&#xA;" OwnerUserId="15602" LastActivityDate="2018-04-04T08:36:27.687" Title="Commercial electric motor with integrated torque sensor and torque control" Tags="|sensors|torque|actuator|" AnswerCount="0" CommentCount="1" ClosedDate="2018-04-04T15:27:58.960" ContentLicense="CC BY-SA 3.0" />
  <row Id="21174" PostTypeId="2" ParentId="21169" CreationDate="2018-04-04T08:55:45.113" Score="4" Body="&lt;p&gt;Probably a skilled metalworker will come along and lough at me for this, but I don't think you can cut to 0.1 mm accuracy with many hand tools. I see these ways:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;using a small saw (your tubes are fairly small) and then file down to the desired length. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I think pliers or bolt cutters could cut through your tube, but will likely chinch it shut&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Or get a tube cutting device - rollers opposite a blade, with a thread to push the blade into the tube. &lt;a href=&quot;https://www.homedepot.com/b/Plumbing-Plumbing-Tools-Pipe-Tube-Cutting/N-5yc1vZc4ft&quot; rel=&quot;nofollow noreferrer&quot;&gt;Look here for these look like&lt;/a&gt;. Maybe you can get sub-mm accuracy with these.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whatever tool you use, if at all possible use internal support (since your dimensions indicate a very small tube this may be hard to implement), a wood dowel or plug could work (hattip to @fred_dot_u)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2018-04-04T14:53:52.523" LastActivityDate="2018-04-04T14:53:52.523" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21176" PostTypeId="1" CreationDate="2018-04-04T14:19:10.847" Score="0" ViewCount="652" Body="&lt;p&gt;diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HnNIx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HnNIx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) the scanner can angle a beam anywhere on the circular off-axis mirror and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the off-axis mirror can be rotated as an the diagram to any angle needed, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;can the beam shooting at any point on it from the scanner be made parallel to the middle beam bounced off of the off-axis mirror? &lt;/p&gt;&#xA;" OwnerUserId="14841" LastActivityDate="2024-01-04T16:07:10.863" Title="galvo scanner + rotating off-axis mirror for beam offseting?" Tags="|optics|photonics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21177" PostTypeId="2" ParentId="21176" CreationDate="2018-04-04T14:45:16.140" Score="0" Body="&lt;p&gt;No. The final mirror on the scanner has a smaller surface area than the 'rotating off-axis mirror'. Consequently, for a beam of light to hit near the edge of the larger mirror, it must be diverging from the centreline connecting the centre of those two mirrors. As the light paths are diverging, they will not be parallel after the off-axis mirror, unless that mirror is not flat.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-04-04T14:45:16.140" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21178" PostTypeId="2" ParentId="21176" CreationDate="2018-04-04T14:50:07.137" Score="0" Body="&lt;p&gt;Yes, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WsA5J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WsA5J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you see, the outgoing part of the yellow beam is parallel to the reflected part of the red beam, just as you wished. But for that to happen, the galvo must be located in such a way that it will obscure a fragment of the reflected field - the green beam, reflected from the mirror, will strike the structure of the galvo, and the reflected image will have a big black blob of shadow of the galvo somewhere off to the side.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-04-04T14:50:07.137" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21179" PostTypeId="1" AcceptedAnswerId="21196" CreationDate="2018-04-04T14:51:54.037" Score="1" ViewCount="67" Body="&lt;p&gt;Yesterday in class this question came up by one of the students in the class. Does anyone know why there is a minus sign in front of the W2 in the figure?&lt;a href=&quot;https://i.stack.imgur.com/AJ1gC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AJ1gC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5393" LastActivityDate="2018-04-05T00:10:31.750" Title="	Skogestad &amp; Postlethwaite, Multivariable Feedback Control: Analysis &amp; Design, Wiley, 2005 Figure 9.10 Why a negative sign on W2?" Tags="|control-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21180" PostTypeId="1" AcceptedAnswerId="21181" CreationDate="2018-04-04T15:00:17.417" Score="1" ViewCount="38" Body="&lt;p&gt;in most of car cabin filters (for example MANN filter) I see that I need to place it according the &quot;Air flow&quot; label? &#xA;But why I need to do that? &#xA;What will be wrong if I will put it in opposite side of &quot;Air flow&quot; label? &lt;/p&gt;&#xA;" OwnerUserId="15607" LastActivityDate="2018-04-04T15:02:41.637" Title="Why do I need to place car cabin filter according Air flow label" Tags="|car|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21181" PostTypeId="2" ParentId="21180" CreationDate="2018-04-04T15:02:41.637" Score="3" Body="&lt;p&gt;Usually because the strengthening in the filter is optimised for one direction and the layers may also be optimised so the oncoming air is met with a coarse layer first, followed by successively finer layers.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-04-04T15:02:41.637" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21183" PostTypeId="1" AcceptedAnswerId="21191" CreationDate="2018-04-04T15:53:15.760" Score="5" ViewCount="284" Body="&lt;p&gt;(please excuse my bad English. I don't speak English)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to deal with an old hydraulic machine, recovered from a deposit. I have no knowledge on mechanical engineering. Only civil engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine is for doing traction tests on steel bars (for example ribbed bars). The machine is old, from the 70’s or 80’s I guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has a system for plotting the stress-strain/deformation chart on paper, but also has digital instrumentation added, which is being read by an old PC still running Windows XP, and it only provides the raw data from the instruments (as a text file).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The digital system seems to be added, I guess, in the 90’s, and comes with a handbook with some instructions (as for example, how to calibrate the machine), but no explanation on the internal workings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to algorithmically process the raw data produced by the tests, to extract various parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I had been taught theoretically, the type of chart I should get, must look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vqyN0.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vqyN0.png&quot; alt=&quot;Theorical result&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But off course, this is a real world physical test, so I get this kind of chart&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c0FzM.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c0FzM.png&quot; alt=&quot;Real data&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had been told that it starts with an accommodation phase which I should ignore, and I need to extract various information from the rest of the chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, here is a zoom of the “elastic phase”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/td7tH.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/td7tH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no knowledge on why it behaves that way, but I conjecture (guess) that the traction force is increased in discrete steps, and between each force increase, the machine waits some amount of time for the steel bar slowly elongating until it reaches a stress-strain/deformation static equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ts8cP.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ts8cP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that conjecture is correct, then I should isolate only the “static equilibrium” points to do calculations, and discard the other, because the other points are transient states, and the theoretical applications of the test are only for static behaviors (civil engineering applications).&#xA;If the conjecture is correct, I should discard the transient states and retain only the static data for purposes of static mechanic applications of the traction test, which are the only theoretical applications of the test in civil engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3HZED.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3HZED.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But off course, I have no knowledge on dynamic mechanics, and I may be making some huge mistake. I’m not a mechanical engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is:&#xA;Is the conjecture correct?&#xA;Should I make some additional considerations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit] I would welcome if you can give me a book title, for reference, on the dynamical behavior of this type of machine.&lt;/p&gt;&#xA;" OwnerUserId="15593" LastEditorUserId="15593" LastEditDate="2018-04-04T16:08:07.883" LastActivityDate="2018-04-05T05:55:34.147" Title="Should I discard part of the data (steel traction test)" Tags="|steel|elastic-modulus|traction|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21184" PostTypeId="1" CreationDate="2018-04-04T16:23:03.583" Score="1" ViewCount="1378" Body="&lt;p&gt;I'm looking at building a sustainable energy project that will convert biogas (methane) into electricity. I haven't found any biogas or methane electric generators for sale. It seems that there are some projects that are using generators to convert the methane gas into electricity, however, I can't tell if this is special kind of generator and I just don't know what it's called or if it is a natural gas or propane generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So will off the shelf natural gas or propane generators also run off of methane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how would I go about trying to calculate how much methane it would take to run the generator to produce enough electricity to say charge a cell phone or another small appliance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="15609" LastEditorUserId="3505" LastEditDate="2024-03-05T22:45:48.407" LastActivityDate="2024-03-07T00:46:10.973" Title="Will natural gas or propane generators also run on methane?" Tags="|mechanical-engineering|chemical-engineering|gas|biogas|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21185" PostTypeId="1" CreationDate="2018-04-04T17:01:01.293" Score="0" ViewCount="464" Body="&lt;p&gt;If I have a water pump positioned horizontally where the diameter of the inlet and the outlet of the water is equal: Din = Dout. If the volume flow rate is know experimentally by measuring the time taken to fill a container with a known volume. Is it possible to calculate thrust force of the water pump based on these information? If possible how? If not why? &lt;/p&gt;&#xA;" OwnerUserId="15610" LastActivityDate="2018-07-08T22:39:34.633" Title="Water Pump Thrust Calculation" Tags="|open-channel-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21186" PostTypeId="1" AcceptedAnswerId="21204" CreationDate="2018-04-04T17:52:37.083" Score="1" ViewCount="1200" Body="&lt;p&gt;I want to torque a nut to 320 ft lbs, but my torque wrench measures only to 250 ft lbs.  Since I know I can get it to 250 ft lbs, is it possible to use that value to calculate the additional degree of rotation it would take to get it to approximately 320 ft lbs?  For example... torque to 250, switch to another ratchet to avoid damaging the torque wrench, then rotate the nut an additional 45 degrees.&lt;/p&gt;&#xA;" OwnerUserId="15612" LastActivityDate="2018-04-05T11:47:59.390" Title="Torquing a nut beyond a known value using degree of rotation?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21187" PostTypeId="2" ParentId="21184" CreationDate="2018-04-04T18:06:35.707" Score="3" Body="&lt;p&gt;natural gas is methane, so no problems there. an internal combustion engine that is equipped to operate on gasoline can be converted easily to run on gaseous fuels and the conversion kits usually contain a little fitment which allows the user to run the engine on their choice of either propane or methane.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-04-04T18:06:35.707" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21189" PostTypeId="2" ParentId="21183" CreationDate="2018-04-04T18:14:50.507" Score="2" Body="&lt;p&gt;all tensile testers exhibit this accomodation range to a slight degree, but in your case the large accomodation range is most likely caused by worn parts in the drive mechanism that pulls on the bar- most likely the mechanical &quot;grabbers&quot; that dig into and engage the ends of the bar under test. this might be the reason that this machine was discarded in the first place. when analyzing the test data, the accomodation &quot;foot&quot; is cut from the data by shooting the straight portion of the elastic slope down to the x-axis and using the resulting intercept as the true &quot;beginning&quot; of the test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I recommend inspecting the gears inside the machinery and the teeth in the grabbing fixtures to determine if repair is possible.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-04-04T18:14:50.507" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21191" PostTypeId="2" ParentId="21183" CreationDate="2018-04-04T19:31:25.363" Score="1" Body="&lt;p&gt;Edit after AgentP's comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As AgentP pointed out, you have swapped your stress-strain axes in the zoomed image. This fact alone refutes your conjecture... Your graph should look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n0FnY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n0FnY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see vertical increases in stress, and sudden jumps in strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reason the steps might occur:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steps in your data occur at very specific displacement/strain intervals, this says to me that the displacement read-out has a too low resolution compared to the force read-out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What to do with the data:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, this data and machine won't cut it if you would like to certify any materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the resolution of the displacement read-out is too low, you would be better off using the points indicating a change in displacement. Judging by the spacing of the force read-out points, their resolution seems much higher, therefor, if you read a change in the displacement read-out, it is more likely that that point will be the most representative of the stress-strain relationship of the specimen at that point. Below is a picture illustrating:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y7DvX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y7DvX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;(Pardon my Paint skills, I'm more of a CAD guy)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the accommodation zone, as niels nielsen suggests, extrapolate the elastic zone to the x-axis and disregard the data to the left of that line. Below is an image illustrating:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6kXtb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6kXtb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S. Why your conjecture would be wrong even if you hypothetically did not swap your axes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine wouldn't apply a force on the bar and wait for the bar to elongate to &quot;catch up&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, you would have to apply the force &lt;strong&gt;very&lt;/strong&gt; quickly to see any significant effect, I don't think it would be possible in the elastic zone. And if that was even possible, your graph won't have steps with a horizontal plateau, but it will be a series of bumps, going up in force and elongation, and when the force is stopped being applied, you will see a decrease in stress as the bar elongates to catch up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Properly built traction testers are built to be strain-controlled, i.e. the machine would have a PID controller or something similar that controls the force applied to ensure that the strain increases at a linear rate w.r.t. time. This reduces the effects of potentially catastrophic failure, especially when testing brittle materials and prevents the jacks from &quot;running away&quot; once the specimen yielded.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastEditorUserId="14440" LastEditDate="2018-04-05T05:55:34.147" LastActivityDate="2018-04-05T05:55:34.147" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21193" PostTypeId="1" CreationDate="2018-04-04T20:26:28.143" Score="0" ViewCount="161" Body="&lt;p&gt;dear StackExchange colleagues;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I want to find the fresh air rate in kg/s and the supply rate in kg/s during winter and summer for the case below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DATA:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Indoor condition all year        =       22oC dB temperature, 50% saturation. &lt;/li&gt;&#xA;&lt;li&gt;Outdoor condition summer         =  28oC dB temperature, 80% saturation. &lt;/li&gt;&#xA;&lt;li&gt;Outdoor condition winter          =     -3oC dB temperature, 80% saturation. &lt;/li&gt;&#xA;&lt;li&gt;Maximum occupancy                  =   1000 people. &lt;/li&gt;&#xA;&lt;li&gt;ADP of the cooling coil                    =       8oC &lt;/li&gt;&#xA;&lt;li&gt;Fresh air requirement                  = 12 l/s/person &lt;/li&gt;&#xA;&lt;li&gt;Supply air rate                         =      8 air changes per hour. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Fresh Air Flow Rate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fresh air rate = 12 l/s/p x 1000 people =         12,000 l/s =         12.0 m3/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific volume at the outside condition may be determined from a psychrometric chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is approximately 0.88 m3/kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass flow rate = Volume flow rate / specific volume&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass flow rate     (Fresh Air)   = 12.0 / 0.88      =       13.64 kg/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supply Air Flow Rate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ventilation supply air rate is taken from table B2.3 and is given as 8 air changes per hour; the mass flow rate can be calculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume flow rate (m3/h)         =       Volume of room (m3) x air change rate (ac/h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume of room (m3)             =       40 x 20 x 8          =       6400 m3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume flow rate (m3/h)         =       6400 (m3) x 8 (ac/h)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume flow rate (m3/h)         =       51,200 m3/h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume flow rate (m3/s)         =       51,200 / 3600               =       14.22 m3/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass flow rate = Volume flow rate / specific volume&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific volume at the supply condition may be approximated at this stage from a psychrometric chart. It can be taken as 0.834 m3/kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass flow rate (Supply Air) = 14.22 / 0.834                =       17.05 kg/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;right this procedure is for the air flows during summer, I did all the calculations for winter and found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fresh Air Flow Rate: 12/0.77 [0.77 from the chart with -3oC and 80%]&#xA;Supply air Flow Rate: 14.22/0.88 [ from chart with room temp conditions plus 8, 30oC and 50%]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this example that is very close to my doubt:&#xA;&lt;a href=&quot;http://www.arca53.dsl.pipex.com/index_files/psy21.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.arca53.dsl.pipex.com/index_files/psy21.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it contradicts my thinking, he uses the same flow for winter and summer, can I do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks a lot&lt;/p&gt;&#xA;" OwnerUserId="14565" LastActivityDate="2018-05-10T17:56:55.030" Title="Defining fresh air rates for sizing heating coils" Tags="|fluid-mechanics|airflow|hvac|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21195" PostTypeId="2" ParentId="14658" CreationDate="2018-04-04T22:06:59.877" Score="1" Body="&lt;p&gt;I would recommend   a slightly less technical approach and use double sided sticky tape.  It's wonderful stuff, and you can get what's called &lt;em&gt;removable&lt;/em&gt; tape.  Probably makes it a bit easier, but you can always use a heat gun /hair dryer to warm the PCB and lift it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should have enough gooey shear strength to resist the slight lateral milling forces, especially if you get enough of it down.  Clean up with solvents. &lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2018-04-04T22:06:59.877" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21196" PostTypeId="2" ParentId="21179" CreationDate="2018-04-05T00:10:31.750" Score="0" Body="&lt;p&gt;If you work out the transfer functions from $d$ to $z_1$ and $z_2$ it can be shown that you get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;z_1 = W_1 \frac{1}{1+K\,G} d,&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;z_2 = -W_2 \frac{-K}{1+K\,G} d.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part of the transfer function of $z_1$ without $W_1$ is often called the sensitivity function. For $H_\infty$ you often choose $W_1$ to be a semi-inverse of a desired sensitivity function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part of the transfer function of $z_2$, after cancelling the two minuses, without $W_2$ is often called the control sensitivity function (however &lt;a href=&quot;https://www.google.nl/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;url=https://www.cds.caltech.edu/~murray/courses/cds101/fa02/caltech/astrom-ch5.pdf&amp;amp;ved=2ahUKEwi3q-iC56HaAhUpsaQKHUd-DE8QFjAAegQIBxAB&amp;amp;usg=AOvVaw3qbmrbZGHBAw5tYwyn1kXI&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; they also call it the noise sensitivity function). For $H_\infty$ you again often choose $W_2$ to be semi-inverse of a desired control sensitivity function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order for $W_2$ to have a more meaningful meaning they added a minus sign into its definition, such that the control sensitivity function is weighted by $W_2$ instead of $-W_2$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: by semi-inverse I am roughly referring to a transfer function which is causal/proper, stable (all poles have a nonzero negative real part) and is an approximation of the inverse of a given transfer function.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-04-05T00:10:31.750" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21198" PostTypeId="1" CreationDate="2018-04-05T00:13:43.687" Score="1" ViewCount="125" Body="&lt;p&gt;I'm working on a project to make a Heads Up Display in my car.  I need a way to present my project in front of my class, for I obviously cannot bring my car into the presenting room.  My professor told me to look into an OBD-II simulator, and since then everything I looked into has a VGA connection(like this one: &lt;a href=&quot;https://www.sparkfun.com/products/9555&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sparkfun.com/products/9555&lt;/a&gt;).  I've tried looking into this, but I was not able to find anything.  Everything that I read talks a lot about VGA, but never explains why OBD-II used VGA.  It seems to me that an OBD-II connection is completely different from a VGA connection, and a VGA is for old computer monitors. So my question is: what are the similarities between an OBD-II connection and a VGA connecting, and why do these simulators use a VGA connection?&lt;/p&gt;&#xA;" OwnerUserId="15614" LastActivityDate="2018-04-05T19:10:46.237" Title="Similarities between OBD-II and VGA" Tags="|electrical-engineering|computer-engineering|computer-hardware|car|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21200" PostTypeId="1" CreationDate="2018-04-05T06:02:33.523" Score="1" ViewCount="1823" Body="&lt;p&gt;My application needed to measure human height on a standing kiosk. I need to attach the sensor on kiosk ceiling type top. I used different kind of ultrasonic sensor for this purpose but nothing proving reliable results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, general ultrasonic sensor have deadband within which the sensor doesn't work and second human head and should varies and thus the reflecting pattern. Can anyone suggest the best method and sensor to use for this purpose? I didn't use laser till now because of eye safety reasons.&lt;/p&gt;&#xA;" OwnerUserId="5973" LastActivityDate="2019-03-29T18:02:21.840" Title="What is best sensor &amp; method to measure human height?" Tags="|measurements|sensors|lasers|distance-measurement|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21201" PostTypeId="1" CreationDate="2018-04-05T06:40:12.133" Score="0" ViewCount="74" Body="&lt;p&gt;What is the advantage of a shell type transformer over core type and which one has smaller flux leakage?&lt;/p&gt;&#xA;" OwnerUserId="15617" LastEditorUserId="90" LastEditDate="2018-04-05T09:03:37.357" LastActivityDate="2018-04-05T09:03:37.357" Title="Differences between shell type and core type transformers" Tags="|electrical-engineering|electromagnetism|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21202" PostTypeId="1" AcceptedAnswerId="21224" CreationDate="2018-04-05T09:46:07.947" Score="1" ViewCount="100" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c0BQ2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c0BQ2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;Say I have a ring with the center of mass G. Also I rotate the ring around a point P on the ring, and I have a &quot;anything&quot; moving along the ring at the same time with velocity v.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would one explain the motion of the &quot;anything&quot;? Would one say that it's absolute velocity is the velocity of G + the relative velocity (velocity measured from P) + the rotational velocity of the entire system? Or is some of these terms zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="14441" LastEditorUserId="14441" LastEditDate="2018-04-05T14:47:44.933" LastActivityDate="2018-04-06T16:42:15.473" Title="Rotating ring with &quot;thing&quot; moving on it" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21203" PostTypeId="1" CreationDate="2018-04-05T11:31:03.430" Score="1" ViewCount="131" Body="&lt;p&gt;I understand how a pin and tumbler lock stops you from opening the door unless you have the key that aligns all the pins with the edge of the plug, allowing it to turn. But I can't find anywhere how the opposite side of the lock works!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does turning the lock from the inside unlock it if the pins are blocking the plug?&lt;/p&gt;&#xA;" OwnerUserId="15618" LastActivityDate="2018-04-06T09:38:24.660" Title="What allows a lock to be unlocked from the inside without a key?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21204" PostTypeId="2" ParentId="21186" CreationDate="2018-04-05T11:47:59.390" Score="2" Body="&lt;p&gt;The key to solving your problem is to understand what the units of torque are and how to apply them to your particular situation. Torque is defined as force per unit length. In every torque problem there is an arm that rotates around a fixed point and force that acts on the end of that arm. The equation that describes the resultant torque is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\tau = r * F * cos\theta$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a graphic that discribes the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EXgDD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EXgDD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This general equation takes into account that the force acting on the arm may not be perpendicular to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to asnwer your question. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;is it possible to use that value to calculate the additional degree of rotation it would take to get it to approximately 320 ft lbs?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. It is much easier to calculate torque directly. You can do this by placing your wrench on the nut so that is parallel to the ground and then placing a weight on the end. The resultant torque is the length($r$) of the arm times the weight on the end($F$). Since the arm is parallel to the earth and the force of gravity is perpendicular to the earth, $\theta$ is at $90\unicode{xb0}$ and thus $cos\theta$ equals one and does not change the outcome of the equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, divide 320 by the length of your bar in feet and place that many pounds of force on the end of your bar at a 90 degree angle and the resultant torque will be 320 foot pounds.&lt;/p&gt;&#xA;" OwnerUserId="9656" LastActivityDate="2018-04-05T11:47:59.390" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21205" PostTypeId="2" ParentId="21198" CreationDate="2018-04-05T12:00:28.867" Score="1" Body="&lt;p&gt;The link you posted is not an ODB-II simulator, it is an ODB-II interface you can plug into your car. If you have read the &lt;a href=&quot;https://learn.sparkfun.com/tutorials/obd-ii-uart-hookup-guide?_ga=2.47668391.1816384797.1522928605-744497538.1522928605&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hook-up Guide&lt;/a&gt; on the website, you'll have noticed that the interface port is a DB9 connector, usually used with serial connections, this one uses the UART protocol it seems. Furthermore, I haven't read anything about VGA in your link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ODB-II &lt;em&gt;emulator&lt;/em&gt; you're looking for will be something like the &lt;a href=&quot;https://freematics.com/pages/products/freematics-obd-emulator-mk2/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Freematics emulator&lt;/a&gt; (I literally typed into Goole: &lt;em&gt;odb-ii simulator&lt;/em&gt; and it was the first link Google gave me.) It can be controlled via your PC to simulate the signals as if it were a real car and sends the signals over an ODB-II connector, exactly as you would find in your car, which you could use for your demo. And the whole thing fits in your trouser pocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you may perhaps have come across might be some ODB-II interfaces with it's own proprietary firmware and OS with a VGA output you can directly connect to a screen without requiring additional software. I haven't seen something like this, but I can think that it should exist.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastEditorUserId="14440" LastEditDate="2018-04-05T19:10:46.237" LastActivityDate="2018-04-05T19:10:46.237" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21210" PostTypeId="2" ParentId="6015" CreationDate="2018-04-05T18:37:24.787" Score="1" Body="&lt;p&gt;There is a phase shift that makes it work. Acceleration is 90 degrees out of phase of velocity. The balancing mass must have non Newtonian charateoristics.&lt;/p&gt;&#xA;" OwnerUserId="15630" LastActivityDate="2018-04-05T18:37:24.787" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21211" PostTypeId="1" AcceptedAnswerId="21221" CreationDate="2018-04-05T19:25:51.420" Score="3" ViewCount="990" Body="&lt;p&gt;I am sorry if this is a strange question. Sometimes I like to visit parliaments from around the world and compare their buildings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qP0JV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qP0JV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After looking at the British house of commons chamber, I noticed that there was sun light coming out of ceiling of the chamber. At first I thought that the ceiling was made of glass and it was just sun light. However after checking the building from outside there seems to be at least another floor on top of the chamber. So it can't be sun light. Then I thought that it must be just some electric light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lsFQK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lsFQK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, after checking some old images of the chamber(before electricity was used) there is still light coming from ceiling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know how this happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6KsKL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6KsKL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15631" LastEditorUserId="15631" LastEditDate="2018-04-06T02:47:34.887" LastActivityDate="2022-07-14T20:53:29.183" Title="Where does the light in the ceiling of the British house of commons chamber come from?" Tags="|civil-engineering|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21212" PostTypeId="1" CreationDate="2018-04-05T19:30:08.570" Score="2" ViewCount="58" Body="&lt;p&gt;Suppose I have a rope that is capable of lifting 100 Kg without breaking. But I would like the rope to break if it is exposed to a sudden change in acceleration (jerk). So if the rope was lifting 10 Kg but was instantaneous exposed to &lt;em&gt;700 N impulse force&lt;/em&gt;, it must break. The numbers here are just examples.&#xA;[that edit should make the physics clearer]&#xA;So perhaps such ropes or similar exist? Or the rope could be divided in two with a hinge or something in between? Anything would work - just as long as it has these properties of breaking at some predefined change of acceleration...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does something like this exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="15632" LastEditorUserId="615" LastEditDate="2018-04-06T17:43:47.600" LastActivityDate="2018-04-06T17:43:47.600" Title="Hinge or similar that breaks at a certain change in acceleration?" Tags="|mechanical-engineering|applied-mechanics|mechanical|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21213" PostTypeId="2" ParentId="21212" CreationDate="2018-04-05T20:26:01.983" Score="1" Body="&lt;p&gt;It is fairly common to put deliberate break points in rope or cable to ensure that they fail in a controlled and predictable manner when overloaded. Typically this is done in tings like fishing rigs or oil well wire-line winches to try to ensue that any overload condition results in a least bad scenario for example by preventing overload of expensive bits of kit or leaving making disconnected equipment potentially recoverable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is often preferred to put in a specific breakaway linkage rather than design it into the rope itself as this makes it simpler to engineer and manage as you can add it to an existing system and change to breakaway characteristic fairly easily and also gives you a predicable point of failure. Plus it potentially saves you from scrapping an expensive length of cable if it does fail as you can be reasonably sure that the cable itself wasn't overloaded. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as in-line linkages whcih can be placed anywhere in the system there are also various other systems such as torque limiting clutches and various types of tear failure linkages which are also good at absorbing and mitigating impact forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also use inertial and/or  damping systems to engage or disengage clutches or couplings a bit like an inertial reel seat-belt in reverse. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the simplest system is just a shear pin as part of a clevis type coupling . For failure under jerk conditions a brittle material with a notch might be a good option. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-04-05T20:26:01.983" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21214" PostTypeId="2" ParentId="21211" CreationDate="2018-04-05T21:24:19.793" Score="2" Body="&lt;p&gt;I suspect that its is sunlight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No part of the palace of Westminster is that tall and the commons chamber itself is pretty tall with several tiers of seating in both the MPs benches and the public galleries. It's also often difficult to judge the height of Gothic buildings at the best of times and the sort of wide angle lenses you would need to get a full interior shot distort perspective a lot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also see form the images in the OP that the roof is pitched which strongly suggests that the ceiling is the exterior roof. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would also not be at all surprising, given the style of Gothic architecture if there were false exterior walls for either structure or cosmetic purposes plus false windows in the lower story. So you could well have three stories of exterior windows but one level or real windows which actually admit light to the main chamber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;this also fits well with the general scheme of Gothic ecclesiastical whcih tended the be based on a pattern of 4 main levels (regardless of overall height)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-aisle : the ground level and actual working art of the building, usually bounded by corridors under buttresses&lt;br&gt;&#xA;- Gallery/triforium : mid level with narrow galleries overlooking the ground floor, sometimes with small windows and the bases exterior flying buttresses &#xA;- Clerestory : structurally 'light' with most of the windows often supported on the exterior by flying buttresses &#xA;- Vaulted roof &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EGln8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EGln8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-04-05T21:24:19.793" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21215" PostTypeId="2" ParentId="19732" CreationDate="2018-04-06T01:52:07.030" Score="1" Body="&lt;p&gt;A brushless motor controlled by an electronic speed controller can produce an accurate and relatively powerful system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most BLDC motors used for hobby r/c equipment would be suitable and these are capable of supporting a significant axial and tangential load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When operating at low RPMs (2700 RPM is slow) good quality motors are virtually silent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed controller (ESC) can be controlled by pwm or a coded signal such as sBus which would also provide an interface for a feedback system via a microcontroller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BLDC motors are specified by the can size and the revs per volt rating. For example a 2216-900kv motor would be approximately 22mm dia, with 16mm magnets, rotating at 900 rpm per volt unloaded.  Small motors operate on 2S-4S LiPo batteries or 7v to 14v.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2018-04-06T01:52:07.030" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21216" PostTypeId="2" ParentId="18740" CreationDate="2018-04-06T02:31:20.053" Score="1" Body="&lt;h3&gt;Yes&lt;/h3&gt;&#xA;&lt;p&gt;The reason that weathering steel was developed was so that it would have better weathering characteristics than regular carbon steel. It was specifically developed so that it creates a patina of rust that is durable and keeps the remaining steel from rusting further.&lt;/p&gt;&#xA;&lt;h3&gt;When not to use it?&lt;/h3&gt;&#xA;&lt;p&gt;In certain environments, the protective patina will not form. In these locations, additional protective coatings are recommended.  These can be paint or zinc coatings.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Weathering_steel#Disadvantages&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; lists more specific details about locations where uncoated weathering steel is not recommended.&lt;/p&gt;&#xA;&lt;p&gt;In general, locations that are corrosive (e.g. pollution or swimming pools) or locations that do not experience wetting and drying cycles are not good locations for weathering steel.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-04-06T02:31:20.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21217" PostTypeId="1" CreationDate="2018-04-06T03:27:43.817" Score="1" ViewCount="30" Body="&lt;p&gt;I'm searching for 3D MRI images for generating an mesh to be used in a Finite Element Modeling project. My group has access to a low-resolution CT scan and cadaver knees, but we don't have any good MRI scans. I'm hoping that there might be a public repository of MRI anatomical images somewhere. Any pointers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, how acceptable is it to ask to loan out CT time. This is an unfunded graduate project, so the scan would have to be pro-bono.&lt;/p&gt;&#xA;" OwnerUserId="14465" LastActivityDate="2018-04-06T03:27:43.817" Title="Searching for Anterior Cruciate Ligament MRI Images for Finite Element Model." Tags="|finite-element-method|biomedical-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21219" PostTypeId="2" ParentId="21203" CreationDate="2018-04-06T09:38:24.660" Score="2" Body="&lt;p&gt;The parts responsible are the clutch and cam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UsMG1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UsMG1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the key bottoms out, besides aligning tumblers, it engages the clutch. As result, as you turn the key (and the core) - the cam turns too, moving the rest of the lock mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With dual-core locks (openable with a key from either side) the clutch is just friction-based, so once a key pushes it into &quot;engaged&quot; position, it stays in it until pushed to the position engaged for the other side once a key from the other side is inserted. For locks that just have a simple knob for opening from inside, the clutch is actuated with a spring - in absence of a key it engages the knob side and the core (key side) is disengaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side effect, it's a slight improvement of lock safety - when picking such lock, even if all tumblers are latched, all you achieve is turning the core uselessly while the deadbolt doesn't move. You need to depress the clutch to engage the cam and transfer the rotation to the rest of the mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you want to lock someone in, with the lock only lockable from your side, but openable with the knob from the other, after locking it, insert a &lt;em&gt;wrong&lt;/em&gt; key. This way the clutch engages for the key side, and the tumblers block the core - the lock can't be opened from any side.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-04-06T09:38:24.660" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21220" PostTypeId="2" ParentId="21200" CreationDate="2018-04-06T10:11:24.797" Score="3" Body="&lt;p&gt;Hair is mostly transparent to ultrasonic, so this is still the reliable&amp;amp;convenient approach - any EM spectrum readouts (3D, video, laser) will be completely falsified by hair. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just need a narrow cone ultrasonic sensor - one with readout area restricted to top of the head. It's not something to be found in robotics/industrial equipment supplies, but ones qualified for medical use should work. This is still not 100% reliable, but allows for quick, easy, contactless measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only 100% reliable but way less convenient method is mechanical - like a stadiometer. Electronic determination of the position of the slider should be vastly easier, e.g. using a linear encoder with Grey code encoded positioning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vn5sR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vn5sR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-04-06T10:11:24.797" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21221" PostTypeId="2" ParentId="21211" CreationDate="2018-04-06T10:22:33.473" Score="6" Body="&lt;p&gt;&lt;strong&gt;It is not sunlight. The lights in the roof are artificial&lt;/strong&gt;. Your 'old photo before electricity' is presumably from before the war, when the roof &lt;em&gt;did&lt;/em&gt; have real windows, in 'direct contact with the sky'. See this aerial view from 1919 for reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pzhvdm.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pzhvdm.png&quot; alt=&quot;Aerial view of Commons Chamber 1919&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is recorded evidence of resistance to the building of offices etc. above the chamber, as members worried about the lack of daylight.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before the Debate goes any further I would like to express my disapproval of the suggestion of the hon. Member for West Walthamstow (Mr. McEntee) and certain other hon. Members that when we come to rebuilding the House of Commons special accommodation should be made available for other purposes above it. If that were done, we would have no daylight whatever in the new Chamber, and having sat, like many others, for two years in this artificial illumination, I certainly look forward to the day when we can have a Chamber very much better lit than the former House of Commons used to be.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://api.parliament.uk/historic-hansard/commons/1943/oct/28/house-of-commons-rebuilding#S5CV0393P0_19431028_HOC_307&quot; rel=&quot;noreferrer&quot;&gt;Sir Alfred Beit (St. Pancras, South-East), 28/Oct/1943&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.parliament.uk/about/living-heritage/building/palace/architecture/palace-s-interiors/commons-chamber/&quot; rel=&quot;noreferrer&quot;&gt;parliment.uk website&lt;/a&gt; states that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the Chamber was rebuilt after 1945 at the cost of £2 million, Sir Giles Gilbert Scott designed a steel-framed building of five floors (two taken by the Chamber), with offices both above and below.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&quot;The Construction of the House of Commons&quot; by Oscar Faber notes further that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The elaborate wall and ceiling linings to the frame replicate the original timber detailing by Pugin. The frame steps back along its length and addtional height above the Chamber has been allowed for a two-tier roof over the whole rebuilt House of Commons.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An elaborate network of branched ductwork sent the conditioned air to numerous outlets set into the ornate detail of the Chamber’s gallery structure, and at the upper level of the spring point of the replica roof trusses. Extract ducts were positioned at regular centres in the apex of the roof lining of the Chamber.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i.e. the new chamber internal roof was designed to look like the old one, but also had extraction ducts etc. added, in keeping with there being in-use office space above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phrase &quot;two tier roof&quot; does not refer to the internal roof of the commons, since that is a ceiling, not a roof. Rather, the new roof of the commons simply has two tiers, but this is not accurately picked up by Google Maps. See here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5118U.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5118U.jpg&quot; alt=&quot;Close up of Commons Roof showing two-tiers above chamber&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, a friend of mine who is currently at work inside the House of Commons &lt;em&gt;right now&lt;/em&gt; has confirmed both that the look/feel of the roof-lights is artificial, and that the offices on the floor above do not follow a long corridor around the void which would be required to direct daylight down to the commons chamber roof two floors below.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-04-06T10:22:33.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21222" PostTypeId="1" CreationDate="2018-04-05T00:26:04.280" Score="6" ViewCount="98" Body="&lt;p&gt;Last night I was reading some fascinating &lt;a href=&quot;http://web.archive.org/web/20130121073127/http://www.mri.jhu.edu/~edelstei/Edelstein.Aberdeen.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;history of the first full-body MRI scanner&lt;/a&gt; and noticed that it's fundamentally different from modern scanners in one particular way: the static B field is vertical, while all the modern machines I've seen pictures of strongly suggest a static B field parallel to the patient's axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a reason why most modern machines orient their static field along the patient's axis and not through their frontal plane - was that first scanner's vertical field a non-fatal &quot;mistake&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there implications for image quality depending on which slice is selected, or is there a limitation on which slices &lt;em&gt;can&lt;/em&gt; be selected? Are axial slices (which I imagine to be what modern machines optimize for) perhaps more commonly useful than frontal slices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I don't know why this was migrated without conferring with me. This question is about the &lt;em&gt;application&lt;/em&gt; of these scanners, i.e. what &lt;strong&gt;medical&lt;/strong&gt; diagnostic reasons are there to prefer one slice orientation over another. Clearly engineering issues impose constraints on what medical professionals can do, and I'm interested in those too, but I want this to be primarily about what these machines &lt;em&gt;do&lt;/em&gt; (specifically with humans in the B field) rather than about how they achieve that.&lt;/p&gt;&#xA;" OwnerUserId="15650" OwnerDisplayName="Bernd Jendrissek" LastEditorUserId="15650" LastEditDate="2018-04-07T10:19:07.203" LastActivityDate="2024-02-21T03:05:35.160" Title="Why do modern MRI scanners use an axial static field?" Tags="|medical-devices|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21223" PostTypeId="1" CreationDate="2018-04-06T16:12:15.667" Score="2" ViewCount="1593" Body="&lt;p&gt;Higher compression ratios in internal combustion engines generally improve the engine's fuel efficiency. But diesel engines don't seem to go above a ratio of 25 or so. What is the limiting factor for the compression ratio in diesel engines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For anyone who's wondering about gasoline engines, there the compression ratio is limited by the undesired auto ignition of the fuel, which does not apply to diesel engines.)&lt;/p&gt;&#xA;" OwnerUserId="12424" LastActivityDate="2018-04-06T18:35:20.473" Title="What are the limiting factors on diesel engine compression ratios?" Tags="|thermodynamics|automotive-engineering|energy-efficiency|diesel|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21224" PostTypeId="2" ParentId="21202" CreationDate="2018-04-06T16:42:15.473" Score="1" Body="&lt;p&gt;The Anything path is a circle with G at its center. So in that reference frame&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x= \sqrt{1-y^2}  $$&#xA;But G is turning with its $$ X_p = \sqrt{1-Y^2_p}  $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you have to transfer the X1 and Y1 to new frame by substituting X1, with (X1-Xp) , and same for y, Y1 substitute with Y1-Yp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final path will look like a Lotus pattern of a round flower with oval petals. The radius of this pattern will be twice the radius of your ring and the number of petals depends on the ratio of the to rotations.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-04-06T16:42:15.473" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21227" PostTypeId="2" ParentId="2612" CreationDate="2018-04-06T17:33:48.167" Score="0" Body="&lt;p&gt;The RFD77402 (Qwiic) uses an infrared VCSEL (Vertical Cavity Surface Emitting Laser) module to measure the amount of time it takes to bounce off a target. This allows low-cost millimeter readings up to two meters.&lt;/p&gt;&#xA;&lt;p&gt;&lt;s&gt;SparkFun offers a breakout board for like 15 USD.&lt;/s&gt; The sparkfun board is not available as of 2022 (or earlier)&lt;/p&gt;&#xA;" OwnerUserId="15639" LastEditorUserId="15639" LastEditDate="2022-08-31T15:37:46.607" LastActivityDate="2022-08-31T15:37:46.607" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21228" PostTypeId="2" ParentId="21223" CreationDate="2018-04-06T18:35:20.473" Score="1" Body="&lt;p&gt;A  &lt;a href=&quot;https://www.reddit.com/r/askscience/comments/37e7ig/how_well_would_a_diesel_engine_with_a/&quot; rel=&quot;nofollow noreferrer&quot;&gt;redditor&lt;/a&gt;  states,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Now for the down sides, as you guessed higher compression gives higher&#xA;  delta T therefore higher peak pressures and this means higher loads on&#xA;  all the components. There is one other major downside, as we increase&#xA;  compression our cylinder temperature increases just before our&#xA;  combustion process from the compression doing work on the gas. So at&#xA;  some point mechanically it’s not practical to go higher compression.&#xA;  (This increase in temperature is used to ignite the diesel fuel, but&#xA;  as incorrectly noted earlier this ignition temperature is reached far&#xA;  before TDC (top dead center) or before the fuel is in the cylinder.&#xA;  Fuel ignites instantly upon injection into the cylinder, so timing is&#xA;  controlled from the injector and when it injects fuel) It’s also&#xA;  interesting to note high pressure injectors are needed for direct&#xA;  injection and seen on all diesels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://www.researchgate.net/post/What_is_the_maximum_compression_ratio_achieved_so_far_for_a_compression_ignition_engine&quot; rel=&quot;nofollow noreferrer&quot;&gt;NASA employee&lt;/a&gt; offers thusly:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Typically, about 25:1 for production type 4 stroke engines using&#xA;  diesel fuel. Materials, and thus costs, are probably the main&#xA;  limitation. Emissions are another factor, NOx goes up with compression&#xA;  ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-04-06T18:35:20.473" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21229" PostTypeId="2" ParentId="12421" CreationDate="2018-04-06T19:43:45.323" Score="0" Body="&lt;p&gt;Another reason that hasn't been mentioned yet is the 'network effect'. One US state or European country can't sensibly increase its octane requirements. The reason you want higher octane ratings is because it allows for higher compression ratios in engines, which increases the fuel efficiency. In most engines the compression ratio is fixed, and is optimised for the fuel that is available in the area where the car is sold. (Compression ratios sometimes differ for EU and US variants of the same car, due to the higher octane ratings of fuel in the EU.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If octane requirements were increased somewhere, that would only allow new cars sold there to utilise the higher octane fuel by having a higher compression ratio. But if only one US state or EU country had this higher octane requirements, cars using a corresponding higher compression ratio would not work well when driven to another state/country using the lower octane rating, so car manufacturers would not start selling cars with a higher compression ratio unless a large area (e.g. the entire US or EU) would change its fuel octane. And changes that influence a continent are very slow with lots of stakeholders, if they happen at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So once octane ratings are set, they are difficult to change. And the focus on fuel efficiency is probably relatively recent (though I don't know when these octane ratings were standardised).&lt;/p&gt;&#xA;" OwnerUserId="12424" LastActivityDate="2018-04-06T19:43:45.323" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21231" PostTypeId="1" AcceptedAnswerId="21235" CreationDate="2018-04-07T03:49:47.067" Score="0" ViewCount="73" Body="&lt;p&gt;I read the paper &quot;Bilateral Control of Teleoperators with Time Delay&quot;,which is written by Robert J. Anderson and Mark W. Spong. In that paper, they defined passivity of an n-port flow as&lt;br&gt;&#xA;$\int^\infty_0 F^T (t) v(t) dt \geqslant 0$&lt;br&gt;&#xA;$F$ means effort and $v$ means flow.&lt;br&gt;&#xA;And they said if a two-port system is passive, the system is stable. They didn't said that directly, but they said an system can be unstable by being nonpassive. They said that without proof, so I want to know why an passive system is stable.&lt;/p&gt;&#xA;" OwnerUserId="15645" LastActivityDate="2018-04-07T08:08:46.033" Title="Passivity and stability" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21235" PostTypeId="2" ParentId="21231" CreationDate="2018-04-07T08:08:46.033" Score="0" Body="&lt;p&gt;I didn't read the entire paper but only checked the statement you mentioned.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The two-port communication circuit for this system is nonpassive, and is the cause of the instability.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;And later they write again&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;[...] it was shown that the instability [...] is due to a nonpassive communication block.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;It looks to me like you mixed up sufficient and necessary conditions here.&#xA;They write in the paper, as you already mentioned in your question, that &lt;em&gt;nonpassive&lt;/em&gt; implies &lt;em&gt;unstable&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;This makes passivity a &lt;strong&gt;necessary&lt;/strong&gt; condition for stability in this case.&lt;/p&gt;&#xA;&lt;p&gt;They did &lt;strong&gt;not&lt;/strong&gt; say that it is a &lt;strong&gt;sufficient&lt;/strong&gt; condition, which would be how you understood it, that &lt;em&gt;passive&lt;/em&gt; implies &lt;em&gt;stable&lt;/em&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Necessity_and_sufficiency#Sufficiency&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; explains it a bit more compact than me.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;If P is sufficient for Q, then knowing P to be true is adequate grounds to conclude that Q is true; however, knowing P to be false does not meet a minimal need to conclude that Q is false.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I hope I could explain it somewhat understandable, don't hesitate to comment if not.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-04-07T08:08:46.033" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21236" PostTypeId="2" ParentId="8193" CreationDate="2018-04-07T10:10:48.250" Score="2" Body="&lt;p&gt;The difference of few large cylinders or many small cylinders has less to do with running on steam or diesel, but more with the era in which the engines were built. Early engines (both steam and diesel) had few, large cylinders, probably because that is how engineers tried to make more powerful engines: make everything bigger. The realization that multiple smaller cylinders gives a smoother running engine and more power output (because small pistons can run at higher rpm) came later, and by that time steam cylinder engines were being replaced with either diesels or steam turbines. But smaller cylinder &lt;a href=&quot;https://en.wikipedia.org/wiki/Steam_motor&quot; rel=&quot;nofollow noreferrer&quot;&gt;steam motors&lt;/a&gt; also existed. And if you search for videos of people running antique diesel engines, those are also often single or few cylindered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Length of the stroke is not really relevant, what matters is the ratio between the minimum and maximum volume, and you can increase that by having a long stroke with a large maximum volume, or a very small minimum volume. Modern engines use short strokes because those allow for higher rpms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;addition:&lt;/em&gt; Another factor is that a single cylinder engine is easier to build than a multi cylinder engine. In the 19th century the technology for building engines (both steam or internal combustion) was less advanced, and the simpler designs were preferred. As tools and construction methods improved, and as more uses for light high speed engines became apparent, engine designs moved from single large cylinders to multiple smaller cylinders. This too has little to do with the engine type, but much more with the era in which the engine was built.&lt;/p&gt;&#xA;" OwnerUserId="12424" LastEditorUserId="12424" LastEditDate="2019-08-15T19:33:53.333" LastActivityDate="2019-08-15T19:33:53.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21237" PostTypeId="2" ParentId="18408" CreationDate="2018-04-07T10:29:06.697" Score="2" Body="&lt;p&gt;There are &lt;a href=&quot;https://en.wikipedia.org/wiki/Carbureted_compression_ignition_model_engine&quot; rel=&quot;nofollow noreferrer&quot;&gt;model diesel engines&lt;/a&gt;, so technically it is possible. My guess in the case of car engines would be that it is about economics. Diesel engines are more expensive to manufacture (because they need to be stronger to withstand higher compression), but you earn that back in their better fuel efficiency and lower fuel taxes if you make enough kilometers. People who buy small cars are more likely to not make a lot of kilometers per year, and more likely to care more about the car's initial sale price compared to fuel costs.&lt;/p&gt;&#xA;" OwnerUserId="12424" LastActivityDate="2018-04-07T10:29:06.697" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21239" PostTypeId="1" CreationDate="2018-04-07T12:29:48.223" Score="2" ViewCount="1547" Body="&lt;p&gt;I can't wrap my head around it so if someone could explain this to me. According to &lt;a href=&quot;http://www.vncbearing.com/roller-bearings-vs-ball-bearings-major-differences/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.vncbearing.com/roller-bearings-vs-ball-bearings-major-differences/&lt;/a&gt; &#xA;ball bearings can handle thrust loads but roller bearings can't. Now I understand that ball can rotate around all axis and roller around only one, but still... If you put a shaft on those bearings it's meant to rotate, you don't allow a shaft to go through a bearing, it's always fixated in one way or another. So what's really the point of that statement? &lt;/p&gt;&#xA;" OwnerUserId="8851" LastActivityDate="2018-08-20T15:23:36.390" Title="What does it mean that a ball bearing can handle thrust load but a roller bearing can't?" Tags="|mechanical-engineering|design|bearings|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21240" PostTypeId="2" ParentId="21239" CreationDate="2018-04-07T17:52:41.097" Score="3" Body="&lt;p&gt;if you look at a ball bearing in cross-section, you will see that the balls are assembled into the two races in such a way that the races have side walls against which the balls can press as they roll. this means that a thrust load on one race will be transmitted through the balls to the opposite race without tending to push the races apart and open up the bearing. note that to assemble the bearing, one of the sidewalls on one of the races has to be missing its sidewall so the balls can be inserted between the races, so the bearing's ability to sustain thrust loads in one direction will be limited by that. But a roller bearing's races do not have sidewalls that can carry thrust loads without scuffing the ends of the rollers against other parts of the bearing and thereby galling and/or grinding those parts into failure.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13363" LastEditDate="2018-04-07T19:56:21.300" LastActivityDate="2018-04-07T19:56:21.300" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21242" PostTypeId="1" AcceptedAnswerId="21271" CreationDate="2018-04-08T02:45:46.347" Score="0" ViewCount="53" Body="&lt;p&gt;I'm not sure if this is an appropriate place to ask this, but I figure the worst that's going to happen is my question will just get closed, so here we go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just got my first home defense handgun, a Glock 19 gen 4, and I was wanting to store it within my combination safe. I timed myself at roughly 25 seconds to open the safe and retrieve the gun. This is fast, however, I was wondering if I could make it faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally you would need to turn the dial like this(Please note i don't count the first 4 turns because that should be muscle memory when opening a safe anyway):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4 times to the left, and end on the first number&#xA;3 times to the right, and end on the second number&#xA;2 times to the left and end on the third number&#xA;1 time to the right and end on 0. (This last step seems to be universal)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What I would like to do is reconfigure my safe to do something like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;4 times to the left, and end on the first number&#xA;1 time to the right, and end on the second number&#xA;1 time to the left, and end on the third number&#xA;1 time to the right and end on 0.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm essentially cutting out the extra turns of the tumbler. The concept seems solid, and I can indeed make it to the back of the safe tumbler to see the internals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is it possible to change not only the combination, but the way the combination is entered, and skip additional turns? The only downside I can see to this is by overriding the functionality of the safe, I may be decreasing the number of permutations for the combination. I can live with that if it means cutting down the retrieval time of my gun to around 5 seconds, which my concept will likely do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GSG2v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GSG2v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/85rR3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/85rR3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15652" LastEditorUserId="15652" LastEditDate="2018-04-08T03:05:09.843" LastActivityDate="2018-04-09T22:24:57.230" Title="Reconfiguring a Safe to only need Four turns" Tags="|mechanisms|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21243" PostTypeId="2" ParentId="18408" CreationDate="2018-04-08T04:02:21.147" Score="0" Body="&lt;p&gt;Ignition lag in diesels makes it difficult to run them over a range of rpms that peak out above about 4500 rpm. You can tinker with them to run just at higher rpms, but not to run over the wide range of rpms required of a vehicle diesel. The little 1 liter gas jobs can be perfectly driveable at 7000 rpms or more. Hence they don't need to have as large a displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honda's new little &lt;a href=&quot;https://www.topgear.com/car-reviews/honda/civic-type-r/16-i-dtec-sport-5dr/spec&quot; rel=&quot;nofollow noreferrer&quot;&gt;aluminum diesel&lt;/a&gt; apparently tops out at 4000 rpm.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-04-08T04:02:21.147" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21244" PostTypeId="1" AcceptedAnswerId="21259" CreationDate="2018-04-08T04:25:20.443" Score="2" ViewCount="1100" Body="&lt;p&gt;I have a schematic for some gas lines in a wafer processing unit.  One of the symbols which occurs is a circle with an x in it that has the left and right quarters filled in to form two black triangles.  (I've attached a drawing.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From context, i can tell that these represent valves to allow or block flowing of the gas.  However i would like to find this symbol in some guide to give a more exact interpretation of the meaning, and just to have an authority that i can cite to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not found the symbol in the context of a gas line schematic in any online reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, and how, can i look?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5TrFA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5TrFA.png&quot; alt=&quot;circle with two black sectors&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15653" LastActivityDate="2018-04-10T05:55:36.163" Title="What is a reference that i can use to find the interpretation of a symbol in a gas line schematic?" Tags="|gas|pipelines|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21245" PostTypeId="1" AcceptedAnswerId="21246" CreationDate="2018-04-08T08:42:40.753" Score="0" ViewCount="429" Body="&lt;p&gt;in the following picture, we should find the viscosity of fluid and we have to have the Area surface of cylinder that is in touch with the fluid.&#xA;the area surface depends on the bottom of the shape.&#xA;but i have no idea what's the shape of this cylinder. what is that curve on the bottom of cylinder? does it mean that cylinder is empty inside or what?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WdNA9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WdNA9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the solution of this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/m1Yn7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m1Yn7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14854" LastActivityDate="2018-04-08T11:37:59.723" Title="shape of cylinder in the fluid mechanics" Tags="|fluid-mechanics|fluid|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21246" PostTypeId="2" ParentId="21245" CreationDate="2018-04-08T09:14:24.780" Score="0" Body="&lt;p&gt;There is no info about the shape of that curve in your, ie the book's, question, also, the torque shearing stress is calculated for the gap between the wall and cylinder which is small and much smaller than the gap between the base and the end of the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;After a re-read, the question does also say neglect end effects.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-04-08T11:37:59.723" LastActivityDate="2018-04-08T11:37:59.723" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21251" PostTypeId="1" AcceptedAnswerId="21252" CreationDate="2018-04-08T19:58:54.757" Score="0" ViewCount="61" Body="&lt;p&gt;I am looking at Phase Change Materials as a building fabric option and from reading up on the topic theres something I don’t understand. Hopefully somebody can help explain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read that when a substance melts it stores latent energy, otherwise charging the material. When the material freezes again it releases this ‘stored’ energy and this is what is utilised for heating or cooling. However, I’ve also read that some substances can store energy during the freezing process and this is where it loses me...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I misunderstanding the process or is this specific to certain materials?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind Regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L&lt;/p&gt;&#xA;" OwnerUserId="14630" LastEditorUserId="14630" LastEditDate="2018-04-10T17:07:59.050" LastActivityDate="2018-04-10T17:07:59.050" Title="Question on Phase Change Materials" Tags="|mechanical-engineering|materials|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21252" PostTypeId="2" ParentId="21251" CreationDate="2018-04-08T20:18:20.850" Score="1" Body="&lt;p&gt;Let's take water as an example. to freeze water into ice, you must extract from it a certain amount of heat energy per unit of mass, which is called the specific heat of fusion. This same amount of heat must be added to the resulting ice in order to melt it back into water again. In this sense, the work it took to draw the heat of fusion out of the liquid water and freeze it is &quot;stored&quot; in the resulting chunk of ice, and is available to perform refrigeration work for you at a later time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, the heat that the chunk of ice pulled out of its surroundings as it melted is similarly &quot;stored&quot; in the resulting body of melt water. A device called a heat pump can then be used to pull that absorbed heat back out of the liquid water and exhaust it into a space where heating is needed, thereby turning the water back into ice. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-04-08T20:18:20.850" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21253" PostTypeId="1" AcceptedAnswerId="21255" CreationDate="2018-04-08T21:41:33.387" Score="3" ViewCount="198" Body="&lt;p&gt;How does one determine the equilibrium height/altitude of an enclosure of helium, given payload, weight of enclosure, and net lift?  What is the formula that will help calculate roughly at what altitude the enclosure will reach equilibrium with the surrounding atmosphere?&lt;/p&gt;&#xA;" OwnerUserId="12987" LastEditorUserId="1832" LastEditDate="2018-04-08T22:08:12.343" LastActivityDate="2018-06-08T08:35:14.810" Title="How do you determine equilibrium altitude of helium enclosure?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21255" PostTypeId="2" ParentId="21253" CreationDate="2018-04-09T02:16:16.473" Score="1" Body="&lt;p&gt;At balance altitude the weight of helium and its container is equal to the weight of displaced air at that level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume you let free a semi-deflated balloon, which can potentially get inflated by diminished surrounding air pressure to 100 cubic meters, containing 10 cubic meters of helium weighing at sea level 1.786 kg and your balloon is 4 kg thin plastic, total of 5.786kg. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to find out at what altitude the 100 cubic meters of air weighs 5.786 kg. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence we need to find at what altitude the atmosphere density is $$5.786/100 = 0.05786$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By checking International Standard Atmospheric table in Wikipedia &lt;a href=&quot;https://en.wikipedia.org/wiki/International_Standard_Atmosphere&quot; rel=&quot;nofollow noreferrer&quot;&gt;The link&lt;/a&gt; and very rough curve fitting, we find the balance altitude at, 21000 meters MSL, Stratosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the reality is more complicated and requires refined calculations and considering the event's weather forecast.   &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="10902" LastEditDate="2018-04-09T07:38:25.537" LastActivityDate="2018-04-09T07:38:25.537" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21256" PostTypeId="2" ParentId="19819" CreationDate="2018-04-09T02:29:57.627" Score="1" Body="&lt;p&gt;By including $x_0$ as feedback (making $x_0 = x(t)$) you are not linearizing this system. Instead you've created a Linear Parameter-Varying (LPV) system (more specifically a quasi-LPV system since your parameter varies according to your state). Such systems do not obey the same stability properties as Linear Time-Invariant systems so if you're looking to apply a linear control strategy you may find your system is unexpectedly unstable. If you're dead-set on using this type of implementation, see Herbert Werner's &lt;a href=&quot;https://www.tuhh.de/ics/teaching/lecture-courses/advanced-topics-in-control.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Advanced Topics in Control&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you linearize your system you need to pick a single operating point ($x_0$ has to be constant) and when you do so your controller will only be effective in a small region about that operating point. You can optionally pick multiple operating points and design a controller for each point, having your system switch between operating points as it changes state. However, you may still encounter some instability at the switching points, especially if you operate near a switching point.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2018-04-09T02:29:57.627" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21259" PostTypeId="2" ParentId="21244" CreationDate="2018-04-09T07:46:53.343" Score="1" Body="&lt;p&gt;Normally you look at the relevant industrial code. I believe in most parts of the world this would be DIN EN &lt;a href=&quot;https://www.iso.org/standard/51841.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISO 10628-2&lt;/a&gt; - &lt;em&gt;Diagrams for the chemical and petrochemical industry -- Part 2: Graphical symbols&lt;/em&gt;. Your sombol is not included.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next step would be to look at the relevant ASME code (I beleive the ASME provides the relevant code here, not working in the americas I'm not sure). I don't have that available and I'm not sure valves are drawn that differently in ASME P&amp;amp;IDs. &lt;a href=&quot;https://hardhatengineer.com/pid-pfd-pefs-pfs-drawing-symbols-legend-list/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This page references some non-ISO 10628 compliant symbols&lt;/a&gt;, but the valves look like in the ISO code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the last step is to ask wether the company providing the drawing has an internal standard they use, for whatever reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to paraphrase Mark's comment:&lt;/strong&gt; Sometimes, the first sheet of the set of P&amp;amp;IDs has a symbol legend and explanation of the numbering system. &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2018-04-10T05:55:36.163" LastActivityDate="2018-04-10T05:55:36.163" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21260" PostTypeId="2" ParentId="21244" CreationDate="2018-04-09T08:33:05.233" Score="0" Body="&lt;p&gt;From my experience in bridges, I would expect that to represent a borehole: &lt;a href=&quot;https://www.google.co.uk/search?biw=770&amp;amp;bih=742&amp;amp;tbm=isch&amp;amp;sa=1&amp;amp;ei=KSTLWoCyDcvNwAKV3LKYDA&amp;amp;q=borehole%20symbol&amp;amp;oq=borehole%20symbol&amp;amp;gs_l=psy-ab.3..0l2j0i8i30k1.20959.22743.0.23128.9.9.0.0.0.0.79.479.9.9.0....0...1c.1.64.psy-ab..0.9.478...0i10k1j0i13k1j0i7i30k1j0i7i5i10i30k1j0i8i13i30k1j0i13i5i10i30k1.0.wx_0zbo6uy0#imgrc=Lqp9DkRjKgQ8wM:&quot; rel=&quot;nofollow noreferrer&quot;&gt;google image result&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this isn't a pipeline-specific symbol, but a general engineering symbol. I can well believe that boreholes would have been taken to sample the soil/strata before a pipeline is constructed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my googled image, it appears with the letters BH next to it to, to signify BoreHole. In my experience there are often also numbers to correlate it with a table of borehole results, e.g. it might say &quot;BH 10436&quot;. Are there any letters/numbers near your symbol which could be part of it?&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-04-09T08:33:05.233" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21264" PostTypeId="1" AcceptedAnswerId="21270" CreationDate="2018-04-09T19:19:16.840" Score="2" ViewCount="4099" Body="&lt;p&gt;I'm attaching a metal pipe to a metal tube as shown in the below diagram. The bolt is threaded through one wall of the pipe and does not pass through the opposite wall of the pipe, but is tightened to the point of bottoming out against it. The bolt passes through unthreaded holes in the tube. It needs to support a load in the direction of the red arrow. I need a bit of a gap between the pipe and tube, so I'll be relying on the ability of the bolt to resist bending under this force. Assuming that nothing else will give out first, what metric would I use to estimate how much weight it can support before flexing the bolts? Is it &lt;a href=&quot;https://en.wikipedia.org/wiki/Bending_stiffness&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;bending stiffness&quot;&lt;/a&gt; rather than &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;shear strength&quot;&lt;/a&gt;? Bolts don't typically advertise bending stiffness, so how would I estimate that? I know distance between tube and pipe is a factor because it increases the leverage the force has to bend the bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PQjkl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PQjkl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd think that filling the void between tube and pipe with nuts/washers would strengthen this arrangement when two flat surfaces are being attached, but since the pipe is round, is there any point? I don't expect a precise answer I just need a way to get a ballpark estimate without bending the bolts before selecting them. Hardware stores don't care for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's for an improvised rail if anyone's wondering. I need upwards of 80% of the circumference of the pipe for the slider to grip it, which is why I'm attaching it in such a dumb way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like this (not my image) will be rolling along it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YqdMQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YqdMQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't found a way to manufacture a proper cradle for the pipe. (EDIT, a U channel would probably work if I can find one the right size) I know that it would be much stronger if done vertically, but this has some advantages given my situation.&lt;/p&gt;&#xA;" OwnerUserId="15672" LastEditorUserId="15672" LastEditDate="2018-04-09T19:25:30.763" LastActivityDate="2018-04-09T22:04:58.453" Title="Determining when bolt will bend" Tags="|bolting|shear|stiffness|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21265" PostTypeId="1" CreationDate="2018-04-09T21:08:49.940" Score="4" ViewCount="351" Body="&lt;p&gt;This is a strange question for me because this is so extremely common, but I have never heard a name to describe exactly what it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EmhkL.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EmhkL.png&quot; alt=&quot;Image of object&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is commonly used to provide water to animals from a trough or tube at a constant water level without dripping. As the water in the trough is reduced in height from drinking, air bubbles enter the reservoir which then releases sufficient water from the container to refill the trough and stop further air from entering the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining liquid is held inside the container by the vacuum above the water, and atmospheric pressure holding the water up in the evacuated space, even though the bottom remains open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this reservoir container is formally called a.... ?&lt;/p&gt;&#xA;" OwnerUserId="3126" LastActivityDate="2020-11-22T10:00:25.063" Title="What is the technical name for a water reservoir that retains its contents via vacuum and ambient air pressure?" Tags="|terminology|hydrology|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21266" PostTypeId="1" CreationDate="2018-04-09T21:12:16.553" Score="4" ViewCount="325" Body="&lt;p&gt;My project is about finding the efficiency of solar trough (cylindrical parabola) concentrator  that will boil water in a pipe located at the focus (check picture please). In the process of doing so (of course I made many assumptions, but I think they are irrelevant to my question now) I am trying to solve for heat loss from the pipe referred to as Q_loss (please always refer to my picture). I started by assuming Tco in equation3, and knowing T_ambient, T_sky I could find h_w and Q_loss. Using those result in eq2 I could find Tci(internal temperature of the glass envelope cover around the pipe). Now I want to substituting Tci in eq1 to find Q_loss which I will compare against the Q_loss I got from eq3 to see if my assumption about Tco was right, but the problem is that I don't have Tr(temperature on the pipe's surface which i assumed constant). How I would go about finding Tr please? I read many papers, but none of them pointed how to find it. They assume it is known or something!&#xA;I have being thinking about this issue for days now. Please advise. Appreciate any guidance&lt;img src=&quot;https://i.stack.imgur.com/gQqaY.jpg&quot; alt=&quot;![enter image description here&quot;&gt;]&lt;a href=&quot;https://i.stack.imgur.com/gQqaY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13830" LastActivityDate="2018-10-10T14:09:48.657" Title="Solving heat loss equation associated with solar trough concentrators" Tags="|mechanical-engineering|heat-transfer|environmental-engineering|solar-energy|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21268" PostTypeId="2" ParentId="21264" CreationDate="2018-04-09T21:52:23.270" Score="0" Body="&lt;p&gt;If you picture your washers as a solid mass and remove from the resulting cylinder a shape to match the tubing to which it is going to be joined, you will have what is called a pipe saddle or tubing saddle. In the world of hang gliding of many years gone by, these saddles were placed between two segments of tubing, with a bolt holding the assembly together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One saddle would mate to the tubing in your image, the other would mate to the pipe and the contacting surfaces of the saddles, being flat, transfer the forces more uniformly to the two/four pieces via the bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a quick Google search for tubing saddle as searching for pipe saddles was non-productive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/crKmv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/crKmv.jpg&quot; alt=&quot;tubing saddle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not too surprising that a &lt;a href=&quot;http://www.aircraftspruce.com/catalog/appages/0400930.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;product from Aircraft Spruce and Specialty&lt;/a&gt; would be priced at US$4.20, but it is aluminum, rather than hang glider plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/78YKh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/78YKh.jpg&quot; alt=&quot;another set of saddles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less expensive at &lt;a href=&quot;http://aircraftproducts.wicksaircraft.com/viewitems/tubing-aircraft-tools/tubing-saddle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wicks Aircraft Supply&lt;/a&gt;, one can get the plastic version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one is even slightly handy and has appropriate tools, one could create a set by drilling aluminum or plastic block to the required diameters and slide of sections to create saddles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key feature of this device is spreading the forces over a greater surface area, reducing the stresses that would result if the join were done by bolt only.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-04-09T21:52:23.270" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21269" PostTypeId="2" ParentId="21265" CreationDate="2018-04-09T22:04:00.833" Score="0" Body="&lt;p&gt;Some people refer to this arrangement as a bubbler. You will also see it referred to as a watering tube. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="1832" LastEditDate="2018-04-10T01:28:34.343" LastActivityDate="2018-04-10T01:28:34.343" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21270" PostTypeId="2" ParentId="21264" CreationDate="2018-04-09T22:04:58.453" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Assuming that nothing else will give out first, what metric would I use to estimate how much weight it can support before flexing the bolts? Is it &quot;bending stiffness&quot; rather than &quot;shear strength&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The metric that you should be interested in is &quot;bending stiffness&quot; instead of &quot;shear strength&quot;. Shear strength has to do with opposing forces that are very close to each other. This is the principle that scissors work on. Instead of cutting the paper like a knife the two blades cause the material to exceed its shear strength and you get a nice clean cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are worried about the bolts bending then you should consider the area of elastic bodies that deals with deflection in beams. In your particular case you have a cantilever beams with a single force acting on it. This is similar to &quot;bending stiffness&quot;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bolts don't typically advertise bending stiffness, so how would I estimate that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can estimate the deflection of a cantilever beam with a single point load with this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma = (Fa^3/(3EI))(1 +3b/2a)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma = $ deflection at the end of the beam&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = $ force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E = $ the modulous of elasticity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I = $ the moment of inertia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a = $ distance from the beginning of the beam to where the force is applied&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$b = $ the distance from the point where the force is applied to the end of the beam&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://www.engineeringtoolbox.com/cantilever-beams-d_1848.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt; for more info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will let you determine how much you bolt or &quot;beam&quot; will deflect for a certain force applied. However I would guess that in you situation you do not have easy access to accurate figures for the modulus of elasticity or the moment of inertia for your bolts.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'd think that filling the void between tube and pipe with nuts/washers would strengthen this arrangement when two flat surfaces are being attached, but since the pipe is round, is there any point? I don't expect a precise answer I just need a way to get a ballpark estimate without bending the bolts before selecting them.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are two ways of reducing the the deflection at the end of your bolt while holding the force and geometry the same. By looking at the above equation we see that we can only change $E$ and $I$. In fact one of your suggestions falls under the moment of inertia category. By installing a larger bolt or adding nuts to a smaller bolt you are increasing the moment of inertia which causes the deflection to grow smaller. You worry that this won't do anything because the pipe is round but it will cause a part of the bolt to flex less which means less deflection at the roller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second option you have is to use materials with a high modulus of elasticity. By using a bolt made out of a stiffer material the roller will deflect less. By maximizing these two values you can significantly strengthen the bolts in this system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for a quick note of warning. Per your request of &quot;Assuming that nothing else will give out first&quot; this answer does not take into account many other variables that could lead to structural failure. For example, stiffer bolts are more prone to sudden failure and by adding nuts you can possibly add higher stress to the area in between the nuts and the pipe. Please use common sense when you finalize your design and test it beyond what it would normally handle to ensure it is safe to use.&lt;/p&gt;&#xA;" OwnerUserId="9656" LastActivityDate="2018-04-09T22:04:58.453" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21271" PostTypeId="2" ParentId="21242" CreationDate="2018-04-09T22:24:57.230" Score="2" Body="&lt;p&gt;You cannot do this with this mechanism. I will now explain in detail how mechanical rotary knob combination lock mechanisms actually work rather than just tell you that you can't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These kinds of locks are extremely simple, using a series of stacked disks, with a protruding rod on both sides of each disk, and a notch at some random location on each disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you rotate the knob 360 degrees in one direction, the first disk will at some point have its protruding rod make contact with the rod of the second disk. The two disks now move together as a group.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you rotate the knob a second 360 degrees in the same direction, the knob of the second disk will make contact with the knob of the third disk. The three disks now move together as a group. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process continues until all disks are moving together as a group. The number at which you are to stop rotating the dial after four turns, will align the notch in the fourth disk with the release lever. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You now need to turn the dial in the opposite direction, so that the protruding pins make contact on the opposite side of each disk, until you are moving all but the fourth disk. You now carefully move the dial so that the notch of the third disk is aligned with the previous fourth disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You back out of the number of turns because you are gradually aligning the slots of each disk with the release mechanism, without disturbing the previously positioned disks in the stack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no other way to position the disks other than to gradually back out 4 turns, 3 turns, 2 turns, 1 turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also due to how these disk combination locks work, there are actually two ways to position the disks that will open the lock, depending on if you start by turning the knob to the left or to the right. Though you will normally only be provided instructions for one of these methods, and only by starting the turning in one direction.&lt;/p&gt;&#xA;" OwnerUserId="3126" LastActivityDate="2018-04-09T22:24:57.230" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21272" PostTypeId="2" ParentId="12155" CreationDate="2018-04-10T02:07:44.053" Score="1" Body="&lt;p&gt;You forge weld it. I watch Forged in fire and this is done regularly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.bladesmithsforum.com/index.php?/topic/31392-forge-welding-stainless-and-fire-pattern-damascus-wip/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.bladesmithsforum.com/index.php?/topic/31392-forge-welding-stainless-and-fire-pattern-damascus-wip/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ax0WY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ax0WY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-04-10T02:07:44.053" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21273" PostTypeId="2" ParentId="21265" CreationDate="2018-04-10T04:09:28.287" Score="1" Body="&lt;p&gt;A Torricelli vacuum, and barometers with mercury apply the same principle.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-04-10T04:09:28.287" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21274" PostTypeId="1" CreationDate="2018-04-10T05:52:50.467" Score="1" ViewCount="341" Body="&lt;p&gt;A liquid mixture containing 35.0 mole% acetone and 65.0 mole% water is to be partially vaporized to produce a vapor that is 70.0 mole% acetone and a residual liquid that is 30.0 mole% acetone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a) Suppose the process is to be carried out continuously, and at steady state the feed rate to the evaporator is 10.0 kmol/h. Calculate the molar flow rates of the product streams and the temperature (°C) and pressure (mm Hg) of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you calculate the temperature and the pressure?&lt;/p&gt;&#xA;" OwnerUserId="15676" LastActivityDate="2018-06-21T07:35:44.877" Title="Chemical Engineering Question using Raoult's Law" Tags="|chemical-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21277" PostTypeId="2" ParentId="21266" CreationDate="2018-04-10T10:17:59.933" Score="0" Body="&lt;p&gt;Sketch answer, may be utterly wrong:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do an energy balance for the pipe. At each point along the pipe, you have:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$Q_{in}$: Radiation heating the pipe element&lt;/li&gt;&#xA;&lt;li&gt;$Q_{loss}$ as per your question, function of $T_r$ among other things&lt;/li&gt;&#xA;&lt;li&gt;$Q_{trans}$ - power conducted to the medium, function of $T_r$, $T_{medium}$ and flow conditions&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assume short length elements so $T_{medium}$ is constant per length element. $Q_{loss}$ will of course not be constant along the whole length. $T_{medium}$ of course depends on conditions in the preceding element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also assumes that the pipe is thin, meaning the temperature across the wall of the pipe is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll ingore convective heat transfer (space around pipe is evacuated, right?) and heat transfer along the pipe as negligible. With $Q_{in}$ known, $Q_{in} =  Q_{loss} + Q_{trans}$ can maybe be solved numerically for each length element, using equation 1. As far as I can tell, $T_r$ should be the only free variable here. I have however little experience with numerical approaches to equations like these, so I'm not certain it's solvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, you'd solve for the first elemt in flow direction, find $T_{medium}$ for the second and so on.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2018-04-10T11:18:28.917" LastActivityDate="2018-04-10T11:18:28.917" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21280" PostTypeId="1" CreationDate="2018-04-10T20:59:09.810" Score="0" ViewCount="831" Body="&lt;p&gt;I am planning to build my own sports car when I am older and am very eco-friendly. I was wondering if a lighter car with everything else the same would be better than a heavier car. I am not expecting it to go up to 120 mph or something really fast, I just want it to have a really high mpg. I was thinking the lighter car would be better than a heavier car at stopping, and a heavier car would be better at accelerating, but not too heavy because it wouldn't be able to accelerate as well. So I was thinking a car with weight in the middle would be best.&lt;/p&gt;&#xA;" OwnerUserId="15686" LastActivityDate="2018-06-15T19:21:09.190" Title="Is a heavier car better than a lighter one?" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="9" ClosedDate="2018-04-16T00:42:01.363" ContentLicense="CC BY-SA 3.0" />
  <row Id="21281" PostTypeId="2" ParentId="21280" CreationDate="2018-04-10T21:48:10.760" Score="4" Body="&lt;p&gt;For  any car lighter is better and this goes double for a sports car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is true that having less weight over the drive wheels can reduce traction but the gain from less overall mass will always be greater. While car designers will use weight &lt;em&gt;distribution&lt;/em&gt; to tune handling there are almost no circumstances where you would deliberately add weight, except in say a racing formulas where there is a minimum weight in whcih case you would place ballast (usually lead) to tune weight distribution and meet the minimum weight requirements, assuming your design is under it in the first place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration is force/mass so for a given amount of torque less mass always means better acceleration, braking and handling, certainly up to the point where the tyres start to slip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While adding weight over the drive wheels will improve traction you are always going to lose more than you gain and it is much better to look at tyres, suspension and even aerodynamic down-force to add grip at minimum cost in extra weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally if your primary goal is high mpg then you won;t want to be accelerating very aggressively in any case. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-06-15T19:21:09.190" LastActivityDate="2018-06-15T19:21:09.190" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21282" PostTypeId="1" AcceptedAnswerId="21346" CreationDate="2018-04-10T21:53:43.927" Score="-1" ViewCount="30" Body="&lt;p&gt;I have a backplane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A backplane is a PCB which more or less serves as a hub for other PCBs. The other PCBs connect to it through its &quot;backplane connectors&quot; and it routes the numerous quantity of signals between those connected PCBs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On my backplane, the connector ports (backplane headers) were designed too close for testing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is anyone aware of backplane-connector right angle adapters such that I could plug into the original header connector, adapt at a right angle, and then&#xA;lay a board on its side to plug in at that right angle?&lt;/p&gt;&#xA;" OwnerUserId="4169" LastEditorUserId="1832" LastEditDate="2018-04-11T01:28:06.443" LastActivityDate="2018-04-16T21:02:01.390" Title="Backplane connector adapter" Tags="|mechanical-engineering|electrical-engineering|product-testing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21283" PostTypeId="1" AcceptedAnswerId="24989" CreationDate="2018-04-10T22:28:58.707" Score="2" ViewCount="175" Body="&lt;p&gt;I changed one laptop keyboard on the new one and while changing it I noticed that it has 30 line bus.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bwxrM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bwxrM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But from my VLSI practice, I know that keyboard can use only 1 data line and 1 clock line (PS/2) or 2 data lines on (USB).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k1CZ4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k1CZ4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is why it has such number of lines? Is it because it doesn't have a microcontroller in it and it is presumably located on the motherboard? Please explain the structure in detail if you can.&lt;/p&gt;&#xA;" OwnerUserId="15687" LastEditorUserId="282" LastEditDate="2018-12-04T22:17:26.103" LastActivityDate="2018-12-04T23:21:54.707" Title="Why laptop keyboard has so many bus lines?" Tags="|electrical-engineering|computer-engineering|computer-hardware|data|computer|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21284" PostTypeId="1" CreationDate="2018-04-11T02:27:42.143" Score="2" ViewCount="447" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L3qIa.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L3qIa.gif&quot; alt=&quot;Levitron toy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ntIc7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ntIc7.png&quot; alt=&quot;Magnetic field of Levitron &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above picture is a basic concept of how a levitron toy works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could the top floating peace stay entrapped in the magnetic field without having to spin if the edges of the balloon keep the top piece from flipping over?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balloon is neutrally buoyant with the small piece inside and secured not to flip over. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do know the spinning causes gyroscopic stabilization and keeps it from flipping over to attracting poles. The opposing poles of the 2 magnets creates the lift. The shape of the base magnet keeps the levitation entrapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=b-MSiQTXIG0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Youtube video&lt;/a&gt;&#xA;In other words could I replace the spinning with a balloon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to find support to that theory using current models and I am blown out the water wrong because those model were electronically stabilized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would a balloon make a difference?? In this model the difference is the helium balloon may make it weigh less and allow the top to be pushed higher by the base. Like a sail ship mast the buoyancy of the helium and the weight of the secured to balloon top would be kept up right entrapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will not know until I try it. Unless it has been tried some where else this way? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c9v40.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c9v40.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/7528/how-does-a-levitron-work?rq=1&quot;&gt;How does a Levitron work?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://earthscience.stackexchange.com/questions/13854/where-on-earth-is-the-magnetic-field-intensity-stronger/13857?noredirect=1#comment26171_13857&quot;&gt;Where on Earth is the magnetic field intensity stronger?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/26140/could-a-satellite-levitate-above-the-magenetic-fields-of-earth#26140&quot;&gt;Could a satellite levitate above the magenetic fields of Earth?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://worldbuilding.stackexchange.com/questions/108896/could-a-city-be-built-out-of-balloons&quot;&gt;Could a city be built out of Balloons?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UGCHx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UGCHx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorUserId="7083" LastEditDate="2018-06-10T20:00:52.703" LastActivityDate="2018-06-10T20:00:52.703" Title="Can a neutrally bouyant helium balloon be suspended in air using magnet entrapment?" Tags="|mechanical-engineering|structural-engineering|control-engineering|magnets|building-physics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21285" PostTypeId="1" CreationDate="2018-04-11T03:50:15.963" Score="1" ViewCount="76" Body="&lt;p&gt;I actually need a magnetic proximity switching circuit such that&#xA;In the case of water in fuel sensor, whenever the water level in water in fuel sensor increases the magnetic float shall move and come in forward in the proximity of hall sensor which in turn switches the magnetic reed position and the output has to given to the ECU is this concept correct. Can we do this with a timer circuit? Will timer circuit do the purpose? Does timer circuit need any extra protection circuit when used in industrial applications?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ECMeZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15690" LastEditorUserId="1586" LastEditDate="2018-10-08T19:28:03.627" LastActivityDate="2018-10-08T19:28:03.627" Title="Could we use NE 555 Timer for industrial magnetic proximity switching circuit?" Tags="|electrical-engineering|automotive-engineering|sensors|magnets|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21288" PostTypeId="2" ParentId="21284" CreationDate="2018-04-11T09:46:33.720" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could the top floating peace be place inside a helium balloon buoyantly neutral and stay entrapped in the magnetic field &lt;strong&gt;without having to spin&lt;/strong&gt;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;No.&lt;/strong&gt;  The system relies on the gyroscopic stabilization from the spin of the upper magnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia page for &lt;a href=&quot;https://en.wikipedia.org/wiki/Earnshaw%27s_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earnshaw's theorem&lt;/a&gt; explains:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Earnshaw%27s_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Earnshaw's&#xA;  theorem&lt;/a&gt; has no&#xA;  exceptions for non-moving permanent ferromagnets. However, Earnshaw's&#xA;  theorem does not necessarily apply to moving&#xA;  ferromagnets,&lt;a href=&quot;https://en.wikipedia.org/wiki/Earnshaw%27s_theorem#cite_note-AJOP-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;sup&gt;[3]&lt;/sup&gt;&lt;/a&gt;&#xA;  certain electromagnetic systems, pseudo-levitation and diamagnetic&#xA;  materials. These can thus seem to be exceptions, though in fact they&#xA;  exploit the constraints of the theorem.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Spinning ferromagnets (such as the Levitron) can—while&#xA;  spinning—magnetically levitate using only permanent&#xA;  ferromagnets.&lt;a href=&quot;https://en.wikipedia.org/wiki/Earnshaw%27s_theorem#cite_note-AJOP-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;sup&gt;[3]&lt;/sup&gt;&lt;/a&gt;&#xA;  Note that since this is spinning, this is not a non-moving&#xA;  ferromagnet.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Switching the polarity of an electromagnet or system of electromagnets&#xA;  can levitate a system by continuous expenditure of energy. Maglev&#xA;  trains are one application&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Any product advertised as a levitron that does not appear to require spinning of the levitated item must be powered by an external power source which can supply energy to a polarity-switching electromagnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is impossible for a Levitron with no external power source to maintain levitation without spinning.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-04-13T09:39:56.863" LastActivityDate="2018-04-13T09:39:56.863" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21289" PostTypeId="1" CreationDate="2018-04-11T11:18:45.597" Score="-4" ViewCount="96" Body="&lt;p&gt;My question is why are tires inflated? Wouldn't it just be better to make a rim out of a light but strong material, like fiber glass, then cover it with a thick coat of rubber, then it wouldn't pop if you run over a nail. I think my idea would be a lot better and would help a lot of people who otherwise would have got stranded in the middle of a desert. &lt;/p&gt;&#xA;" OwnerUserId="15686" LastActivityDate="2018-04-12T03:45:09.423" Title="Why are tires inflated?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="21290" PostTypeId="2" ParentId="21280" CreationDate="2018-04-11T13:43:12.490" Score="0" Body="&lt;p&gt;If you're building an efficient sports car in the near future, it will probably have some electric drive system, whether it's a hybrid or all-electric car.  If your car has batteries, it's likely to be on the heavier end of sports cars.  A heavier car will put more weight on your tires, increasing the force that your tires can apply before slipping, but F=ma, meaning that a greater force will be required to accelerate your heavier car.&lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2018-04-11T13:43:12.490" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21291" PostTypeId="2" ParentId="19547" CreationDate="2018-04-11T13:47:32.463" Score="3" Body="&lt;p&gt;I think it is similar to what it is called by &lt;a href=&quot;http://web.mit.edu/2.75/fundamentals/FUNdaMENTALs%20Book%20pdf/FUNdaMENTALs%20Topic%2010.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;MIT Fundamentals&lt;/a&gt;: &lt;strong&gt;Flexural Rolling Bearing&lt;/strong&gt; or &lt;strong&gt;Rolling Contact Flexure&lt;/strong&gt; in section 10.14.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As is often the case, a hybrid system can combine the best of both&#xA;  worlds. So it is with rolling contact flexural bearings. These bearings allow&#xA;  for compact rolling element joints to be designed. These joints can sustain&#xA;  large normal compressive forces due to the rolling contact, and respectable&#xA;  shear and tensile forces due to the flexible bands. This also make them inherently&#xA;  preloaded. Most importantly, they allow for rolling motion with a minimal&#xA;  number of elements and no lubrication which makes them particularly&#xA;  well-suited for precision instruments medical device applications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=XIA04EasAFM&quot; rel=&quot;nofollow noreferrer&quot;&gt;Youtube link for a similar bearing&lt;/a&gt; which is called &lt;strong&gt;Compliant Rolling-Contact Element&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="15699" LastEditorUserId="15699" LastEditDate="2018-04-11T14:17:01.573" LastActivityDate="2018-04-11T14:17:01.573" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21292" PostTypeId="2" ParentId="21280" CreationDate="2018-04-11T14:57:03.587" Score="0" Body="&lt;p&gt;More mass equals to more inertia. More inertia means more work to accelerate. On the otherhand more mass means more tire friction. More friction means more power from wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these dont work the same way more mass only gets you so much more friction and the more mass you add the more diminishing your returns are. Also you can artificially increase the downforce independently of things. You can use different rubber etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is noting other than making things lighter you can do to inertia. There is no other technological way to combat it. Purposefully adding weight makes no sense, but since things can not wheigh nothing you have to add it to gain other properties of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For energy consumption less is definitely more, but you might want to optimize other factors too. &lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2018-04-11T14:57:03.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21293" PostTypeId="1" CreationDate="2018-04-11T16:09:54.253" Score="4" ViewCount="2624" Body="&lt;p&gt;I essentially have a piston in a tube that is sealed with an O-Ring and actuated with a solenoid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friction of the O-Ring however, is higher than estimated (guessed as far as I can tell) by the designer who doesn't have any calculations to show me as reference. I have searched a little with no luck. The piston is pushed back to it's original position (after the solenoid is deactivated) by a spring, but the 5 Newton force the spring provides is not enough to overcome the friction of the seal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an idea of the force required I attached a bolt and added weight until the piston moved, but &lt;strong&gt;theoretically, is there a simple way of calculating the force required to overcome the friction of a rubber O-Ring seal?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A paper or two exists with relatively complex mathematical models, which i don't have the time to reproduce. Given how common rubber O-Rings are I'd be surprised if there weren't a commonly used formula. &lt;/p&gt;&#xA;" OwnerUserId="14498" LastActivityDate="2024-01-18T19:06:45.813" Title="How Can You Calculate Force to Overcome Friction From an O-Ring Seal?" Tags="|mechanical-engineering|design|friction|pistons|seals|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21295" PostTypeId="1" CreationDate="2018-04-11T17:47:06.493" Score="0" ViewCount="229" Body="&lt;p&gt;3D source potential flow is given by:&#xA;$\phi = \frac{-\lambda}{4\pi r}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laplace's equation in cylindrical coordinates is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{r}\frac{\partial }{\partial r}\left(r \frac{\partial \phi}{\partial r}\right) + f(\theta,z) = 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f(\theta,z) = 0$ since $\phi$ is independent of $\theta$ and $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{r}\frac{\partial }{\partial r}(r \frac{\partial \phi}{\partial r}) = \frac{1}{r}\frac{\partial }{\partial r} \left(\frac{\lambda}{4\pi r}\right) = &#xA;     \left(\frac{-\lambda}{4\pi r^3}\right) \neq 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, the 3D potential flow is not a solution to Laplace's equation in cylindrical coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Laplace's equation in spherical coordinates is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{r^2sin(\theta)}\left[\frac{\partial }{\partial r}\left(r^2sin(\theta)\frac{\partial \phi}{\partial r}\right) + f(\theta,\Phi)\right] = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $f(\theta,\Phi) = 0$ since $\phi$ is independent of $\theta$ and $\Phi$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\frac{1}{r^2sin(\theta)}\frac{\partial }{\partial r}&#xA;(r^2sin(\theta)\frac{\partial \phi}{\partial r}) = &#xA;\frac{1}{r^2sin(\theta)}\frac{\partial }{\partial r}&#xA;(r^2sin(\theta)\frac{\lambda}{4\pi r^2})=\frac{1}{r^2sin(\theta)}\frac{\partial }{\partial r}(f(\theta))=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, source potential flow is a solution to Laplace's equation in spherical coordinates. As I understand it, a solution to Laplace's equation should be independent of coordinate system, so what is going on here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All equations for this question have come from &lt;em&gt;John D. Anderson Jr., Fundamentals of Aerodynamics Fifth Edition.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="9226" LastActivityDate="2018-08-15T11:02:06.930" Title="Why does 3D source potential flow only satisfy Laplace's equation in spherical coordinates and not cylindrical coordinates?" Tags="|fluid-mechanics|aerodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21297" PostTypeId="1" CreationDate="2018-04-11T18:50:15.417" Score="2" ViewCount="220" Body="&lt;p&gt;Is there a way to refill an evacuated water supply reservoir from an elevated water supply tank, without using any valves, but only the power of gravity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know that a siphon is needed to start the flow from the elevated supply tank, but I do not know how to push the air out of the lower reservoir, send water into it, and then retain the water without admitting air through the water fill tube or air communication tubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BE5N2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BE5N2.png&quot; alt=&quot;siphon refill of evacuated water reservoir&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, where would be a place to look for possible solutions to these questions? I have researched the technology of &lt;a href=&quot;https://www.google.com/search?q=intermittent%20dosing%20siphon&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;intermittent dosing siphons&lt;/strong&gt;&lt;/a&gt; for sewer flush systems from the 1880's before electricity was widespread, and I know that air communication tubes can be extremely complex and dynamic in operation, but there is apparently no formal way to describe it.&lt;/p&gt;&#xA;" OwnerUserId="3126" LastEditorUserId="9504" LastEditDate="2019-05-18T03:07:20.540" LastActivityDate="2024-02-22T13:00:35.070" Title="Refill evacuated water reservoir from elevated tank, without valves?" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21298" PostTypeId="2" ParentId="21289" CreationDate="2018-04-12T03:45:09.423" Score="1" Body="&lt;p&gt;By filling the tire with air, we obtain a means by which shocks and bumps in the road surface can be prevented from reducing ride quality in the vehicle that rolls upon those tires. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, an air-filled tire smooths out shocks which would otherwise be transmitted into and then damage the suspension mechanisms and the frame of the vehicle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air-filled tires also reduce the amount of damage to the roadway surface which would otherwise be inflicted by solid tires, especially at high speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These days, tire technology is so good that it is rare to pop or burst a tire during normal operation unless the tire is underinflated. The occasional burst tire is a relatively small price to pay for the practical benefits of air-filled tires.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-04-12T03:45:09.423" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21300" PostTypeId="1" CreationDate="2018-04-12T07:40:36.437" Score="4" ViewCount="917" Body="&lt;p&gt;in linear optimal control,linear quadratic regulator,we have a system of the form: x=Ax+Bu,the optimal control law U is a state feedback,it's a function of the riccati equation solution and the state vector.I'm asking about stability,is the optimal control law U always stable?is the system is always stable? Can you give me some references about the demonstration?&lt;/p&gt;&#xA;" OwnerUserId="15547" LastActivityDate="2018-04-12T13:28:33.633" Title="Stability of the optimal control law" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21302" PostTypeId="2" ParentId="21300" CreationDate="2018-04-12T13:28:33.633" Score="4" Body="&lt;p&gt;So, just to formally repeat your question, we consider an infinite horizon continuous-time optimal control problem&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J^*(x_0) = min \int_0^{\infty} x(t)^TQx(t)+u(t)^TRu(t)dt$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is subject to some constraints, basically just the system dynamics and the initial condition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot x (t) = A x(t) + B u(t) \\ x(0) = x_0  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assume the cost matrices $Q = C^TC$ to be positive semidefinite and $R$ to be positive definite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, if the pair $(A,B)$ is controllable and the pair $(A,C)$ is observable, we can find a &lt;strong&gt;unique, positive definite&lt;/strong&gt; solution $P^*$ for the ARE (algebraic riccati equation) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P^* A+A^TP^* -P^*BR^{-1}B^TP^* + Q = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three nice statements we can make with this solution $P^*$:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The optimal cost is simply $J^*(x_0)=x_0^TP^*x_0$&lt;/li&gt;&#xA;&lt;li&gt;The optimal input is $u^*(t)=-R^{-1}B^TP^*x^*(t)$&lt;/li&gt;&#xA;&lt;li&gt;The closed loop $\dot x^* = (A-BR^{-1}B^TP^*)x^*$ is asymptotically stable.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The idea for the proof of statement 3 is, to take the optimal cost as a Lyapunov function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$V(x) = x^TP^*x$, which is positive definite (because $P^*$ is positive definite)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we still need to show that $\dot V$ is negative semidefinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\dot V \ = 2x^TP^*\dot x \\ \quad = 2x^TP^*(A-BR^{-1}B^TP^*)x \\ \quad = 2(x^TP^*Ax-x^TP^*BR^{-1}B^TP^*x) \\ \quad = x^T(P^*A+A^TP^*)x - 2x^T(P^*BR^{-1}B^TP^*)x \\ \quad = -x^TQx - x^TP^*BR^{-1}B^TP^*x$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both $Q$ as well as $P^*BR^{-1}B^TP^*$ are positive semidefinite, so $\dot V$ is negative semidefinite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This proves stability at the origin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we now consider the biggest invariant set $\{x \in \mathbb{R}^n|\dot V(x)=0\}$ and consider the output $y = Cx$. Since both $x^TQx$ as well as $x^TP^*BR^{-1}B^TP^*x$ are nonnegative, they must both independently be zero. With $x^TP^*BR^{-1}B^TP^*x=x^TP^*BR^{-1}RR^{-1}B^TP^*x = u^TRu \equiv 0$ we know that $u \equiv0$. We also know that $y^Ty \equiv 0$, which leads to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$y = Cx = 0 \\ \dot y = CAx = 0 \\ \ddot y = CA^2x = 0 \\ \vdots $ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We made the initial assumption that (A,C) is observable, so the matrix $\begin{bmatrix}{\ C \\CA\\ \ \ \vdots}\end{bmatrix}$ has full rank and $x \equiv 0$ is the only invariant solution in the biggest invariant set $\{x \in \mathbb{R}^n|\dot V(x)=0\}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This (finally) yields that the origin is asymptotically stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I didn't make a mistake somewhere, I'll look through it again later. For further reference, you might want to read for example &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/0005109877900176&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; paper.  &lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2018-04-12T13:28:33.633" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21303" PostTypeId="1" AcceptedAnswerId="21327" CreationDate="2018-04-12T13:57:52.367" Score="0" ViewCount="62" Body="&lt;p&gt;This should be a pretty straight forward question but I can't seem to find the type of material that is used..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am making my own heating coil out of Nickel Chromium wire. I plan on using two bolts in a frame that hold the entire thing together. Obviously the bolts themselves are conductive so my plan is to wrap the bolts in the &quot;heat resistant cardboard&quot; that I see inside of hair dryers and other heating devices that use the same wire. Does anyone know which type of material this is, as I am sure normal cardboard would simply ignite...&lt;/p&gt;&#xA;" OwnerUserId="14534" LastEditorUserId="14534" LastEditDate="2018-04-16T13:57:13.860" LastActivityDate="2018-04-16T13:57:13.860" Title="Heat resistant material for Nickel Chromium Wire?" Tags="|materials|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 3.0" />
  <row Id="21307" PostTypeId="1" AcceptedAnswerId="21315" CreationDate="2018-04-13T03:03:56.780" Score="0" ViewCount="124" Body="&lt;p&gt;If we are able to see an object in a certain color,that means it reflects the color.A blackbody is one which has zero reflectivity.Then how come it is not black always?&lt;/p&gt;&#xA;" OwnerUserId="15268" LastActivityDate="2018-06-15T14:53:59.787" Title="Blackbody radiation" Tags="|heat-transfer|radiation|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21308" PostTypeId="1" AcceptedAnswerId="21332" CreationDate="2018-04-13T07:10:54.547" Score="2" ViewCount="284" Body="&lt;p&gt;In this question I’m comparing non-plug-in hybrid fuel/electric vehicles with fuel-only vehicles. Also, I’m not disputing that hybrid cars are more fuel efficient, I’m just trying to understand how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I remember two things from high school physics, it’s:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Creating energy from nothing is against the law.&lt;/li&gt;&#xA;&lt;li&gt;Converting energy from kinetic to stored energy and back again is always less that 100% efficient - you end up with less kinetic energy than you put in, the rest being “lost” to the environment in the form of heat or whatever.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both a fuel-only car and a hybrid derive all their locomotive energy from the fuel (ok, and gravity, if you’re going downhill, but let’s ignore that). But the hybrid involves an extra conversion to/from stored energy on the way, i.e. charging and discharging the battery, which according to the second principle above should make it less efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the battery entails additional mass, which, if it were dead weight, would reduce the power/weight ratio of the car, reducing efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to achieve greater efficiency, the hybrid engine must somehow make use of energy that would ordinarily be lost in the engine of the fuel-only car. In order for the hybrid to be more fuel efficient than the fuel-only engine, it needs to do this in a way that overcomes the above inefficiencies and then some.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, what broad physical/mechanical techniques does a hybrid engine use to achieve this, and (in rough numbers) how much of the efficiency gains come from the different techniques?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there’s regenerative braking, but how efficient is this in practice? I.e. how much of a car’s kinetic energy that would otherwise be lost through braking is actually captured and re-used as locomotive energy? And is this basically it, or are there other techniques that significantly contribute to the efficiency of a hybrid?&lt;/p&gt;&#xA;" OwnerUserId="15716" LastEditorUserId="15716" LastEditDate="2018-04-13T07:23:10.277" LastActivityDate="2018-04-16T16:04:45.647" Title="How do hybrid cars achieve greater fuel efficiency?" Tags="|automotive-engineering|car|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 3.0" />
  <row Id="21309" PostTypeId="1" AcceptedAnswerId="21312" CreationDate="2018-04-13T08:20:35.677" Score="5" ViewCount="17462" Body="&lt;p&gt;I've been attempting this fundamental shear force diagram problem for several days, but can't seem to get the correct result. I'm trying to calculate the shear force diagram in terms of $x$, but I'm unsure about the intensity $w(x)$ of the triangular load distribution between $0m \le x \lt 3m$. I am able to calculate the correct result for the latter section $3m \lt x \le 6m$, so I'm a little confused as to what the correct intensity of the triangular load distribution is and how to calculate the correct shear force using the correct intensity $w(x)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below I've attached the problem and calculated the support reactions, which are $A_y=15kN$ and  $B_y=15kN$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s64b9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s64b9.jpg&quot; alt=&quot;Problem and Free Body Diagram with support reaction solutions&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I've attached my free body diagram of the first section between $0m \le x\lt 3m $ and indicated the positive sign convention for this beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rbpa4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rbpa4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then proceeded to find the shear force in terms of $x$ as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum F_y=0:$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$15-w(x)·x·\frac12 - v_1 = 0 \quad (eq\ 1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $w(x)·x·\frac12$ is the area of the triangular load distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I get confused. My understanding of triangular load distribution in terms of the intensity $w(x)$ is that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w(x)=\frac{w_0x}{L}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $w_0 = 10$ and $L=3$ for this problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But substituting these values into the intensity $w(x)$ and back into $(eq\ 1)$ gets me the wrong result of:&#xA;$$v_1=15-\frac53 x^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading multiple textbooks and watching several videos, I finally found out that if the maximum load of a triangular load distribution is at the initial point $x=0$ then the following formula should be applied: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$w(x)=\frac{w_0x}{L}-w_0$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now understand this a bit, but I am wondering where I could get a good explanation as to why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm struggling to find a good explanation as almost every example I've found in textbooks/videos use triangular load distributions that increase from the initial point and not decrease. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, after utilising this formula, I still get the wrong solution. My working out is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum F_y=0:$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$15-\Bigl(w(x)·x·\frac12 \Bigr) - v_1 = 0$$&#xA;$$15-\biggl(\Bigl(\frac{10x}{3}-10\Bigr)·x·\frac12 \biggr)- v_1 = 0$$&#xA;$$15-\biggl(\Bigl(\frac{10x}{6}-\frac{10}{2}\Bigr)·x\biggr) - v_1 = 0$$&#xA;$$15-\biggl(\Bigl(\frac{5x}{3}-5\Bigr)·x \biggr)- v_1 = 0$$&#xA;$$15-\frac{5x^2}{3}+5x - v_1 = 0$$&#xA;$$\Rightarrow v_1=15-\frac{5x^2}{3}+5x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual solution is:&#xA;$$v_1=15+\frac{5x^2}{3}-10x$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm not sure whether I'm using the correct intensity $w(x)$ and/or whether the triangle area has been correctly calculated using this intensity $w(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second section $3m\le x\lt6m$ I am able to calculate the correct shear force in terms of $x$, this solution is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_2=-15-\frac{5x^2}{3}+10x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plotting a diagram of the correct shear forces $v_1$ and $v_2$ in terms of $x$ looks the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ljlOB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ljlOB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your reference, this problem (F11.6) can be found in chapter 11 of Statics and Mechanics of Materials (4th Ed. SI edition) by Hibbeler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate if someone could explain intensity loads for situations similar to above and where I went wrong in my calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading a few examples, I found that if I calculate the shear force from the left end I am able to get the correct shear force using my initial intensity $w(x)=\frac{w_0x}{L}$ and not the latter intensity $w(x)=\frac{w_0x}{L}-w_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I'm unsure why I can't calculate this from the right end? Does it have something to do with the left support $A_y=15kN$ creating a discontinuity? If I calculate from the left end am I correct in changing the section's range to $0m \lt x \le 3m$ to not include the left support $A_y$? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My working out is as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum F_y=0:$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\Bigl(w(x)·x·\frac12 \Bigr) + v_1 = 0$$&#xA;$$-\biggl(\Bigl(\frac{10}{3}(3-x)\Bigr)·(3-x)·\frac12 \biggr)+ v_1 = 0$$&#xA;$$-\biggl(\Bigl(10-\frac{10x}{3}\Bigr)·(3-x)·\frac12 \biggr)+ v_1 = 0$$&#xA;$$-\biggl(\bigl(30-10x-10x+\frac{10x^2}{3}\bigr)·\frac12 \biggr)+ v_1 = 0$$&#xA;$$-\biggl(\bigl(30-20x+\frac{10x^2}{3}\bigr)·\frac12 \biggr)+ v_1 = 0$$&#xA;$$-\bigl(15-10x+\frac{10x^2}{6}\bigr)+ v_1 = 0$$&#xA;$$-15+10x-\frac{5x^2}{3}+ v_1 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\Rightarrow v_1=15+\frac{5x^2}{3}-10x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the correct solution. &lt;/p&gt;&#xA;" OwnerUserId="15715" LastEditorUserId="15715" LastEditDate="2018-04-14T01:35:24.860" LastActivityDate="2018-04-14T12:29:49.607" Title="Shear Force Diagram of a Simply Supported Beam with triangular load distribution" Tags="|statics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21311" PostTypeId="1" CreationDate="2018-04-13T11:13:41.120" Score="0" ViewCount="4707" Body="&lt;p&gt;is it possible to calculate the flow of water out of a pipe where gravity is the only force acting upon the water and therefore generating the flow. The pipe diameter is X and there is minimal fall (say 6 inches?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real life scenario here is that I am trying to calculate the flow of water out of a washing machine waste when the filter plug is removed at the bottom?&lt;/p&gt;&#xA;" OwnerUserId="15717" LastEditorDisplayName="user16" LastEditDate="2018-04-16T11:29:47.253" LastActivityDate="2018-04-16T22:01:13.240" Title="Water flow under pressure of gravity" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21312" PostTypeId="2" ParentId="21309" CreationDate="2018-04-13T12:40:32.340" Score="3" Body="&lt;p&gt;Your procedure is correct, but you have made a mistake with the sign convention.&#xA;Apparently you are using the same convention as I do, where a vertical load/force is negative, and an upward force is positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consequently, as you've written correctly&#xA;$$ w(x)=\frac{w_0}{L}x-w_0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your mistake happens as you formulate the force equilibrium equation.&#xA;With this definition of $w(x)$ you already comply to the sign convention.&#xA;If you now formulate the shear force equation and write&#xA;$$ V_1=15-\int w(x)dx $$ &#xA;you basically reverse the sign convention again.&#xA;To formulate the force equilibrium equation you have to &lt;strong&gt;sum&lt;/strong&gt; all forces, not &lt;strong&gt;subtract&lt;/strong&gt; them, thus&#xA;$$ V_1=15+\int w(x)dx $$ &#xA;which leads to&#xA;$$ V_1=15+\frac{5}{3}x^2-10x $$&#xA;which is the correct result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sign convention&lt;/strong&gt; &lt;em&gt;[edited]&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at your $w(x)$. It's a force pointing downwards, so it should be negative.&#xA;You have written it as&#xA;$$ w(x)=\frac{w_0}{L}x-w_0 \qquad \mbox{for}\qquad x=\{0...3\}$$&#xA;Thus $w(0)=-w_0$, which means your load $w(x)$ is already defined in the coordinate system you specified. If you now sum or integrate and add a minus sign in front of $w(x)$ you basically turn the downward load into an upwards facing load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider, e.g. $w(x)=const.=-w_0$, i.e. an evenly distributed downwards load, thus a negative value. It's easy to figure out, that the reaction at the bearing is $A_y=\frac{1}{2} w_0L $.&#xA;Now to find the shear force distribution, if we use your method, we'd write:&#xA;$$ V_1=A_y-\int_0^L (-w_0)dx=\frac{1}{2} w_0L - (-w_0)L=\frac{3}{2}w_0L $$&#xA;This is the wrong result, because for the shear force distribution calculation we suddenly switch sign convention (subtracting during integration instead of adding).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first part you said $w(x)$ was $w(x)=-w_0+\frac{x}{L}w_0$, which goes linearly from $-w_0$ to $0$, which in your example would be $w(x)=\frac{10}{3}(x-3)$. During your second calculation you managed to get the right result because you changed the sign convention of your $w(x)$, as you inserted $w(x)=\frac{10}{3}(3-x)$ (which is an upwards load), but then you turned it downwards again by adding a negative sign in front of the whole term.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really recommend sticking to &lt;em&gt;one&lt;/em&gt; sign convention. Either you say you &lt;strong&gt;sum&lt;/strong&gt; all forces, and defined downward loads as negative, or you &lt;strong&gt;subtract&lt;/strong&gt; downward loads from upward reactions.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="8632" LastEditDate="2018-04-14T12:29:49.607" LastActivityDate="2018-04-14T12:29:49.607" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21313" PostTypeId="2" ParentId="21311" CreationDate="2018-04-13T13:11:46.140" Score="1" Body="&lt;p&gt;The Bernoulli equation will give you a pretty good estimate.  It says that:&#xA;$$ P_{1} + \frac{1}{2}\rho v_{1}^{2} + \rho g h_{1} = P_{2} + \frac{1}{2}\rho v_{2}^{2} + \rho g h_{2} $$&#xA;You pick 2 points in a flow (1 and 2).  In this case, you can ignore the pressure terms, $P$, and focus on velocity, $v$, height, $h$, and gravitational acceleration, $g$. &#xA;In this case, you can use the free surface of the water in the machine as point 1, meaning that  $P_{1} = v _{1} = 0$, $h_{1}= 0.15 m$, then point 2 is at the outlet of your pipe, and $P_{2} = h_{2} = 0$.  Then: $$ v_{2} = \sqrt{2gh_{1}} $$  gravitational acceleration is $9.81m/s^2$.&#xA;If you want to add a (probably small) degree of accuracy to your calculations , the so-called head-loss calculations are useful.  You can use this calculator:&#xA;&lt;a href=&quot;https://apps.engineeringtoolbox.com/head-loss-water-pipe-a_15.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://apps.engineeringtoolbox.com/head-loss-water-pipe-a_15.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which will give you a pressure loss in kPa, and velocity loss in m/s.  You can adjust your prediction as:&#xA;$$ v_{adjusted} = v_{2} - v_{loss} $$&#xA;so your average flow velocity will &#xA;You'll notice, though, that you need to know the flow rate:$$Q_{2} = v_{2}\pi r_{pipe}^{2}$$&#xA;before you can calculate the head loss.&lt;/p&gt;&#xA;" OwnerUserId="4058" LastEditorUserId="4058" LastEditDate="2018-04-16T22:01:13.240" LastActivityDate="2018-04-16T22:01:13.240" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21314" PostTypeId="1" CreationDate="2018-04-13T17:19:04.987" Score="1" ViewCount="111" Body="&lt;p&gt;The electrolyser used in the process at 25°C and 10.0 bar exhibits a global yield of 82% with a potential of 1.481 V. Determine the electrical energy that can be electrolyzed by year in order to get molar flow of hydrogen: $F(H_2) = 12.016 \times 10^6\ mol/h$.&#xA;So, can anyone help with the calculation of energy required for this electrolysis ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;br&gt;&#xA;Yes, it's homework. I actually calculated the needed energy but i don't know if it's true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrolysis: $\ce{2H_2O + Electricity -&amp;gt; 2H_2 + O_2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The half-reaction is: $\ce{4e^- + 4H_2O(l) \to 2H_2(g) + 4OH^-(aq)}$ .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the molar flow of electrons is: $F(e) = 2F(\ce{H_2})/0.92 = 12.016 \times \frac{10^6}{0.92} = 14.653\cdot 10^6 \pu{mol/h}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, $1F(faraday\ constant) = q(e)\ Na = 1.602 \times 10^{-19}\times 6.023\times10^{23} = 96488.46 C/mol$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F(e) \ F = 14.653 \times 10^6 mol/h \times 96488.46 C/mol = 1413838.6 \times 10^6 C/h$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$I(current) = q/t = 1413838.6 \times 10^6 C/h \times 1h/3600s = 392.74 \times 10^6 A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$E(energy) = UIt= 1.481 \times 392.74\times 10^6 \times 31536000s  (1 year) = 1.834\times 10^{16} J$&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your formatting was hard to read, so I formatted with MathJax.  Please check it out to make sure I got the correct syntax.&lt;/p&gt;&#xA;" OwnerUserId="15592" LastEditorUserId="-1" LastEditDate="2018-04-15T16:37:56.953" LastActivityDate="2018-06-21T09:28:42.873" Title="Eletrolysis of water" Tags="|chemical-engineering|chemistry|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21315" PostTypeId="2" ParentId="21307" CreationDate="2018-04-13T17:21:23.493" Score="3" Body="&lt;p&gt;Your assumption is incorrect. The color we see is the combination of reflected light, if any, and radiated light.   All objects radiate photons of one wavelength or another.  All objects absorb photons of one wavelength or another.  These two spectral curves (absorption and radiation) are the same, but the spectral power curve of the irradiance obviously depends on external sources, not on the material or temperature of the black body.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-04-13T17:21:23.493" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21316" PostTypeId="2" ParentId="18834" CreationDate="2018-04-13T18:31:34.837" Score="3" Body="&lt;p&gt;The most serious and comprehensive information about whistles is provided by the folks who work on pipe organs: expensive instruments which endure critique by serious listeners, and which require maintenance where they are installed because of their size.  A pipe organ's &lt;strong&gt;Flue Pipes&lt;/strong&gt; are in fact no different from whistles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A whistle is a noisemaker that uses no reed.  In this sense, we might say that a brass player's lips and a singer's vocal cords are &lt;em&gt;reeds&lt;/em&gt;, while a flute player is playing a kind of &lt;em&gt;whistle&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A whistle has two components: a resonator and an air supply.  The resonator's opening may be called its mouth, and the air supply must flow freely past the resonator's mouth in order for it to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air flows over the resonator's &lt;em&gt;mouth&lt;/em&gt; and sucks out a bit of pressure from the resonator by the Bernoulli principle.  Once this pressure is lower inside the resonator, the free air supply flow changes course and starts packing into the resonator, increasing its pressure.  This sends a high pressure wave through the resonator which may either bounce back (for a closed resonator) or depart (for an open resonator).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some moment in time while the air is flowing into the resonator, the pressure inside increases enough to divert the airflow away from the mouth of the resonator.  Once the flow is diverted away from the resonator, it begins to remove air from inside and lower the pressure, creating another pressure wave inside.  Now the cycle repeats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sound you hear originates both from the opening at the top of the resonator and an opening at the bottom (for open resonators) where there is this fluctuating air pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sound is at a tuned pitch (or frequency) that is the result of the pressure wave bouncing from top to bottom, bottom to top in the resonator.  The frequency is a simple matter of length with flue pipes, but I can't even explain what's going on in an ocarina.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How can I reduce the pressure of my whistle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In order to make a sound, your instrument must initiate the aforedescribed cycle, and also convert the air source's energy into acoustic energy.  Improving either (by initiating sound earlier or converting energy more efficiently) may enable your instrument to use less pressure.  Here are some conditions which may affect either, in order of my guess for the most important first:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Velocity of air supply&lt;/strong&gt; (high velocity = earlier initiation of tone)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Cross-sectional area of resonator near its mouth&lt;/strong&gt; (smaller cross-sectional area = earlier initiation of tone)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;The resonator being stopped or open&lt;/strong&gt; (stopped = earlier initiation of tone)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Volumetric flow of air supply&lt;/strong&gt; (high flow = more volume)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Length along the flow direction of the resonator &quot;mouth&quot;&lt;/strong&gt; (must be optimized)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Length across the flow direction of the resonator &quot;mouth&quot;&lt;/strong&gt; (must be optimized)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Distance that the air supply travels freely&lt;/strong&gt; (must be optimized)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Laminarity of the air supply&lt;/strong&gt; (unknown effect by me whether more or less sound)&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You might increase the laminarity by passing the air supply through a long, straight tube, for instance.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Intended pitch of the resonator&lt;/strong&gt; (higher pitch = more sound, to a point)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Length of the resonator&lt;/strong&gt; (shorter length = more sound, to a point)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Atmospheric condition&lt;/strong&gt; (ambient wind may &quot;kill&quot; the free flow of the air supply)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Cross-sectional area of the resonator's other opening for open resonators&lt;/strong&gt; (bigger rear opening = more volume)  &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There's a reason tubas, gramophones, Alpenhorns, and megaphones have a flared end: the smaller volumetric velocity produced by the large cross-sectional area enables more sound energy to transfer into the open air, making the best use of the player's effort.  This is called &lt;strong&gt;impedance matching&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As suggested elsewhere, dividing your airflow among multiple resonators may solve your issue.  Good luck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://hardmanwurlitzer.com/pipes/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Further reading on pipe organ flue pipes (&quot;whistles&quot;)&lt;/em&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15506" LastEditorUserId="15506" LastEditDate="2018-04-13T20:42:45.943" LastActivityDate="2018-04-13T20:42:45.943" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21318" PostTypeId="1" CreationDate="2018-04-14T03:21:03.133" Score="1" ViewCount="267" Body="&lt;p&gt;It seems that flexible objects just don't model well in many popular constructive solid geometry (CSG) CAD programs. Most CSG modelers seem to assume exclusively rigid bodies with sharply defined and constrained motions. Actually having fully dynamic unconstrained components seems mind-blowingly out of scope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there is a &quot;collapsing string toy&quot; from decades ago, that seems to be far beyond the ability of most CSG CAD to model and then dynamically simulate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Google image search: collapsing string toy&lt;/strong&gt;  (also known as a &quot;thumb puppet&quot;)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.google.com/search?tbm=isch&amp;amp;q=collapsing+string+toy&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/search?tbm=isch&amp;amp;q=collapsing+string+toy&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;It is similarly exceptionally difficult for most CSG modeling software to accurately model and simulate the motion of a solid rectangular cube supported by two eye bolts, attached to two compound block and tackle rope pulleys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zx35u.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zx35u.png&quot; alt=&quot;two compound pulley blocks supporting a dynamic load&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very limiting for a mechanical engineer to be locked into such an inflexible and rigid understanding of modeling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what CSG modeling software is actually capable of simulating such dynamic complexity of motion?&lt;/p&gt;&#xA;" OwnerUserId="3126" LastEditorUserId="3126" LastEditDate="2018-04-14T03:28:11.647" LastActivityDate="2018-04-14T03:28:11.647" Title="What constructive solid geometry CAD programs can accurately model and simulate structural rope, cable, pulleys, and dynamic motion elements?" Tags="|mechanical-engineering|dynamics|simulation|computer-aided-design|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21321" PostTypeId="1" CreationDate="2018-04-14T11:20:57.133" Score="0" ViewCount="713" Body="&lt;p&gt;I downloaded a 3D  model of a planetary gearbox from &lt;strong&gt;Rossi&lt;/strong&gt; website, so the 3D model is reliable. As you can see, there is no keyway on the opening of the gearbox, thus i can't connect my shaft with traditional key designs. &lt;a href=&quot;https://i.stack.imgur.com/f3TNm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f3TNm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is axial section view of the gearbox, the shaft sits inside the left side. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/OpoHJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OpoHJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What woud be the best way to connect the shaft with gearbox ? should the shaft has exactly the form of inside gearbox i mean can the shaft come directly into the gearbox? or i should have consider a soort of coupling ? &#xA;Here is the suggested output side details:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7IJmx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7IJmx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jb7O8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jb7O8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-04-14T12:30:30.997" LastActivityDate="2018-04-14T12:30:30.997" Title="How to connect shaft to gearbox?" Tags="|mechanical-engineering|design|gears|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21323" PostTypeId="2" ParentId="3174" CreationDate="2018-04-14T14:41:36.033" Score="1" Body="&lt;p&gt;I just came from a heavy rain in a cold day, and wanted to know if any body had posted about this.&#xA;My facts are that using only the heater, is more delayed and besides that all other car windows get fogged,&#xA;There after I left the heater on, and turned on the AC, and the result is amazing, not only the windshield cleared, but all the windows around, to the point that I didn´t need to use the rear defogger.&lt;/p&gt;&#xA;" OwnerUserId="15727" LastActivityDate="2018-04-14T14:41:36.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21324" PostTypeId="1" CreationDate="2018-04-14T19:26:56.040" Score="2" ViewCount="82" Body="&lt;p&gt;Resolution of MRI can be improved if FOV is reduced keeping other factors constant, but this reduces voxel volume,  which in turn reduces SNR. How much SNR is enough for good imaging? &lt;/p&gt;&#xA;" OwnerUserId="15729" LastActivityDate="2024-01-16T20:06:55.790" Title="SNR for good MRI imaging" Tags="|biomedical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21325" PostTypeId="1" CreationDate="2018-04-14T20:04:04.947" Score="2" ViewCount="133" Body="&lt;p&gt;I have some plungers made of 38CrNiMoAlA. Their required hardness to be achieved via quenching/tempering is 51 HRC. I've scoured our technical order for heat treatment of steel alloys but the material is not there. I even checked the internet, it doesn't seem to exist at all. The closest I could get was 38CrMoAlA. Please help if you know this material or something close to it.&lt;/p&gt;&#xA;" OwnerUserId="15731" LastActivityDate="2019-02-04T10:51:41.233" Title="38CrNiMoAlA: Does it even exist?" Tags="|materials|metallurgy|metals|alloys|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21327" PostTypeId="2" ParentId="21303" CreationDate="2018-04-15T11:30:23.093" Score="1" Body="&lt;p&gt;The &quot;Heat resistant cardboard&quot; that you are describing is likely '&lt;a href=&quot;https://en.wikipedia.org/wiki/Mica#Built-up_mica&quot; rel=&quot;nofollow noreferrer&quot;&gt;Built-up Mica&lt;/a&gt;'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly this is not &quot;cardboard&quot; (as specified in the original question - now edited to say &quot;material&quot;!), since it is not paper-based, but, I hope it can be used for your application.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-04-15T15:22:05.607" LastActivityDate="2018-04-15T15:22:05.607" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21331" PostTypeId="1" CreationDate="2018-04-16T05:09:33.660" Score="1" ViewCount="198" Body="&lt;p&gt;I am a telecommunication and network engineer working as a programmer and data analyst. One of my dreams was and still to be a control system engineer and to design and program a smart robots and systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a background in programming and matlab. Where should I start ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get a book from Amazon for &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/8126537280&quot; rel=&quot;nofollow noreferrer&quot;&gt;Norman S. Nise&lt;/a&gt;. The book is very technical and there is lots of things that I can't understand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where should I start and what electronic kit should I buy to start with and do experiments. &lt;/p&gt;&#xA;" OwnerUserId="15647" LastActivityDate="2018-04-16T18:47:19.623" Title="Where should I start to be a control system engineer?" Tags="|control-engineering|control-theory|pid-control|robotics|" AnswerCount="2" CommentCount="6" ClosedDate="2018-04-16T22:32:27.480" ContentLicense="CC BY-SA 3.0" />
  <row Id="21332" PostTypeId="2" ParentId="21308" CreationDate="2018-04-16T06:48:27.673" Score="5" Body="&lt;p&gt;The difference is the regenerative braking.  It take as much work to decelerate a car as to accelerate.  I don't know the actual efficiency but even 50% recovers a lot of the energy that would have been lost as brake heat and brake dust. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electric assist control is also tuned to let the gasoline engine operate in a range where it is most efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to this &lt;a href=&quot;https://en.wikipedia.org/wiki/Regenerative_brake&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; the DOE puts efficiency at 60%.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-04-16T12:26:54.883" LastActivityDate="2018-04-16T12:26:54.883" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21333" PostTypeId="1" CreationDate="2018-04-16T08:28:56.840" Score="0" ViewCount="296" Body="&lt;p&gt;As what the title says&#xA;When I connect my designed preamplifier,which is in a breadboard,&#xA;unto a power amp kit (Acoustic 50W Mono) with a speaker, there is a farting sound, as if it is a pulsating DC. And if I directly connect my music or mic to the power amp the output of the speaker is fine but very weak since there is no preamp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the farting sound is always present, meaning with or without an input of music or mic the farting sound is there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I could think of right now is that it is due to the bias also going to the poweramp or too much low frequency that I'm not aware. And I need help in solving this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is an image of the preamp.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/PxyHm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PxyHm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15741" LastEditorDisplayName="user16" LastEditDate="2018-04-16T11:30:26.837" LastActivityDate="2018-05-16T19:18:58.993" Title="Farting sound when connecting power amp with to the pre amp" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21336" PostTypeId="2" ParentId="21331" CreationDate="2018-04-16T14:30:52.900" Score="1" Body="&lt;p&gt;If you want to get into control theory and the math behind it, I would recommend to roughly follow these steps:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Understand the math basics&lt;/strong&gt; from linear algrebra as @fibonatic already mentioned, but also solving differential equations.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Understand the basics of system dynamics and system analysis&lt;/strong&gt;, including but not limited to &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Bode diagrams and Nyquist diagrams&lt;/li&gt;&#xA;&lt;li&gt;stability analysis&lt;/li&gt;&#xA;&lt;li&gt;frequency domain and state space&lt;/li&gt;&#xA;&lt;li&gt;linearizing systems&lt;/li&gt;&#xA;&lt;li&gt;robustness&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Beyond analysing given system equations, you should also learn how to derive these equations from a given real system, which includes system identification.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Learn the basic approaches and concepts of controller design&lt;/strong&gt;, like tuning PI/PID controllers, state feedback with pole placement, state observation, cascade control, anti-windup and possibly the linear-quadratic regulator (a rather simple optimal control approach).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From this point on, the following control theory concepts get a bit more sophisticated, before you continue you should first look at the concept of &lt;strong&gt;Lyapunov stability&lt;/strong&gt; which might just be the most important concept in control theory, from here on you can for example look at nonlinear control and sophisticated stuff like Model Predictive Control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned the basics of control theory in german, so I don't know any english introduction level books, but &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/1783267798&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; &lt;em&gt;seems&lt;/em&gt; to cover most of what I mentioned in my list.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to @fibonatic for some more ideas, anyone feel free to make further suggestions when I forgot something.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastEditorUserId="14629" LastEditDate="2018-04-16T18:47:19.623" LastActivityDate="2018-04-16T18:47:19.623" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21337" PostTypeId="1" CreationDate="2018-04-16T14:39:38.427" Score="0" ViewCount="41" Body="&lt;p&gt;I planned to do a solar tracker for my project and I am going to make a RTC solar tracker(actually still considering). So, do I still need photosensor like LDR for tracking the sun?&lt;/p&gt;&#xA;" OwnerUserId="15345" LastActivityDate="2018-05-16T18:18:57.823" Title="Did a RTC solar tracker need any photosensor?" Tags="|solar-energy|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21338" PostTypeId="2" ParentId="21337" CreationDate="2018-04-16T15:22:54.853" Score="1" Body="&lt;p&gt;Do not put a photosensor or sun-finder on your array.  Your real-time clock may use the astronomy algorithms provided by Sandia National Lab to be very certain about where the sun is at any given time.  (See link below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please install a GPS, however, so that your time will be accurate as well as your exact location on the Earth's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find out how much performance your solar panels lose when they're off by a few degrees--- allowing some error could save motor energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, check shading row-to-row.  If there's shading, your perfect sun-tracking won't actually be helping during sunrise and sunset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://pvpmc.sandia.gov/modeling-steps/1-weather-design-inputs/sun-position/solar-position-algorithm-spa/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://pvpmc.sandia.gov/modeling-steps/1-weather-design-inputs/sun-position/solar-position-algorithm-spa/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15506" LastActivityDate="2018-04-16T15:22:54.853" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21339" PostTypeId="2" ParentId="21308" CreationDate="2018-04-16T16:04:45.647" Score="5" Body="&lt;p&gt;Paparazzo pretty much nailed it, but I'd like to add a few things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regenerative braking is certainly a large part of it, &lt;a href=&quot;https://www.geotab.com/blog/why-hybrid-vehicles-are-more-efficient/&quot; rel=&quot;noreferrer&quot;&gt;but there are other synergies when utilizing multiple engines in an automobile, too.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ICE can be smaller and sized closer to what is needed for normal operation instead of peak power since the electric motor helps the ICE during the less common acceleration events&lt;/li&gt;&#xA;&lt;li&gt;The electric motor moves the vehicle at lower speeds where the ICE is less efficient, specifically when accelerating from a standstill&lt;/li&gt;&#xA;&lt;li&gt;The electric motor helps move the vehicle at speeds at which the ICE would have to operate further away from its most optimal point, helping to keep the ICE within its optimal range&lt;/li&gt;&#xA;&lt;li&gt;The ICE can be turned off during idling. This is called stop start and is not limited to hybrids&lt;/li&gt;&#xA;&lt;li&gt;The ICE can be designed to be more efficient within its narrower operating range since it does not have to be a master of all trades&lt;/li&gt;&#xA;&lt;li&gt;The batteries and the electric drive allow for regenerative braking which helps recoup some of the energy lost during normal operation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In addition to this list from geotab, here are a few more insights to energy balance around an automobile:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;in certain hybrid setups (unlike through-the-road setups) the motor can charge the batteries during times of idle, utilizing the best load/speed for the motor&lt;/li&gt;&#xA;&lt;li&gt;being able to undersize the ICE is inherently beneficial to thermodynamic efficiency (similar to geotab point #1). ICE's operate best under high load. Ergo, if you can undersize the motor and force it to constantly be under high load, then kick the electric motor in when the peak load exceeds the capabilities of the ICE&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Jl231.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jl231.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other considerations: &#xA;I don't have any data for this and at the moment do not have time to research it, but I think that there could easily be a human factor to consider as well. I've driven with many individuals, and I don't think that any one will argue that different people have different driving styles. Considering that some people accelerate slowly,  while other accelerate hard and brake hard. Considering the synergies of the hybrid electric-ICE power train, it's possible that the hybrid power train can more robustly serve the two different drivers under equally efficient conditions. Consider that the aggressive driver would lose substantial energy in the braking system - and to get the acceleration they desire would require a larger ICE (think big V8s). they would spend more time NOT under peak load, which is what an ICE needs to be efficient. The hybrid drive train would deliver their desired acceleration, while maximizing thermal efficiency AND recouping the energy spent at the above quoted 60% efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eRSRd.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eRSRd.png&quot; alt=&quot;Power curves of ICE and electric drive trains&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then imagine that slow driving individual gets in the same vehicle; they would likewise be penalized for their slow driving in a large ICE. Considering the hybrid drivetrain, this individual would no longer be penalized for their tame driving habits. In essence, the hybrid drive train could better service larger extremes in driving style, and is over all more robust in terms of energy efficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just a thought :) &lt;/p&gt;&#xA;" OwnerUserId="8369" LastActivityDate="2018-04-16T16:04:45.647" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21340" PostTypeId="2" ParentId="21337" CreationDate="2018-04-16T16:46:08.367" Score="0" Body="&lt;p&gt;People building solar trackers are common, as a financial gain : not really, but as a project fun : great if you are the tinkering type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two examples below, and the one I hoped to find used two small (3 inch sq) pv panels with a jar of water as a lens to drive a small motor in either direction...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.mdpub.com/suntracker/&quot; rel=&quot;nofollow noreferrer&quot;&gt;tracker 1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.motherearthnews.com/diy/solar-tracker-zmaz77ndzgoe&quot; rel=&quot;nofollow noreferrer&quot;&gt;tracker 2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-04-16T16:46:08.367" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21341" PostTypeId="1" AcceptedAnswerId="21347" CreationDate="2018-04-16T17:27:46.923" Score="0" ViewCount="228" Body="&lt;p&gt;What is the maximum annual capacity factor that can be expected of a CCGT plant? I see in US in 2015 the &lt;em&gt;average&lt;/em&gt; was about &lt;a href=&quot;https://www.eia.gov/todayinenergy/detail.php?id=25652&quot; rel=&quot;nofollow noreferrer&quot;&gt;0.56&lt;/a&gt;, but can it be higher? I guess it needs some time for maintenance but I don't know how much. Any guidance on where I can get such information?&lt;/p&gt;&#xA;" OwnerUserId="9049" LastActivityDate="2018-04-16T22:40:29.660" Title="maximum capacity factors of combined cycle gas power plants" Tags="|power|gas|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21342" PostTypeId="2" ParentId="18045" CreationDate="2018-04-16T17:30:14.373" Score="1" Body="&lt;p&gt;I'm assuming that your equation is only used to define the spacing from an end to the middle, after which you mirror it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of dx1, write L / N, where L is the total length.  Now you have &lt;/p&gt;&#xA;&#xA;&lt;p&gt;x(i) = (L / N) * r ^ x(i-1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...and when you change N, your x(i)s will also scale.  I don't think you want r to be a function of N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if this helps!&lt;/p&gt;&#xA;" OwnerUserId="15506" LastEditorUserId="15506" LastEditDate="2018-04-16T19:13:43.453" LastActivityDate="2018-04-16T19:13:43.453" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21343" PostTypeId="2" ParentId="21331" CreationDate="2018-04-16T17:51:53.773" Score="1" Body="&lt;p&gt;A robot is anything with a sensor, an actuator, and some smarts.  Build anything you want!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest looking at the website for some college engineering classes on mechatronics to see what they have been building and controlling.  For instance, UC Berkeley, Cal Tech, or MIT.  All you have to do is see the student projects to get an idea of what to build yourself.  Maybe you want to get a degree in controls!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the MAKE: webpage.  One peek into MAKE: magazine, the web site, or a Maker Faire and you'll get a zillion ideas for getting started.  These folks never stop at the algorithm: they want something that &lt;em&gt;moves!&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the US, you may be interested in getting the controls P.E. stamp.  Take the Engineer-In-Training exam right away, then try to get a job where somebody has the controls P.E. stamp, such as at any infrastructure plant like sewage, water treatment, electrical grid, etc.  Then after a few years you can take the test to be a controls PE yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Learn an appropriate controls code language... I've heard that C (or was it C#?) is a good one for this purpose.  Try it out on a Raspberry PI or Arduino.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="15506" LastActivityDate="2018-04-16T17:51:53.773" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21344" PostTypeId="2" ParentId="21333" CreationDate="2018-04-16T18:14:18.720" Score="1" Body="&lt;p&gt;one source of the farting sound is low-frequency oscillation, called &quot;motorboating&quot; in the trade. this occurs when there is a feedback path inside the amplifier which couples the output to the input at low frequencies. The most common cause is when the different stages of the power supply are poorly decoupled, so that a power surge drawn by one stage causes the rest of the supply voltages to droop. the upstream stages of the amp then interpret the supply droop as a signal and the process continues. Check your power amp supply wiring against the schematic and make sure the series dropping resistors are the correct value, and that the filter caps are wired with the correct polarity.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-04-16T18:14:18.720" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21346" PostTypeId="2" ParentId="21282" CreationDate="2018-04-16T21:02:01.390" Score="0" Body="&lt;p&gt;After more searching and asking,&lt;br&gt;&#xA;I think the answer is that&lt;br&gt;&#xA;this does not exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes sense,&lt;br&gt;&#xA;since the connection isn't particularly firm.&lt;/p&gt;&#xA;" OwnerUserId="4169" LastActivityDate="2018-04-16T21:02:01.390" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21347" PostTypeId="2" ParentId="21341" CreationDate="2018-04-16T22:40:29.660" Score="2" Body="&lt;p&gt;It's on the link you provided.  Look at the middle graph, titled &quot;Distribution of annual capacity factors for natural gas combined-cycle plants (2005, 2015)&quot; and look at the Y axis, ranging from 0-10% up to 90-100%.  You can see that there were about 20 plants with a CF greater than 90% in 2015, and about 70 plants with a CF less than 10% that same year.&lt;/p&gt;&#xA;" OwnerUserId="15506" LastActivityDate="2018-04-16T22:40:29.660" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21348" PostTypeId="1" CreationDate="2018-04-17T03:05:42.990" Score="2" ViewCount="447" Body="&lt;p&gt;I'm dealing a C code here which is run on MSP430 LaunchPad. What I need to do here is modifying the program, so that at the push of the button, the micro cycles through 4 different modes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;LED1 flashes while LED2 off&lt;/li&gt;&#xA;&lt;li&gt;LED2 flashes while LED1 off&lt;/li&gt;&#xA;&lt;li&gt;LED1 and LED2 flash simultaneously &lt;/li&gt;&#xA;&lt;li&gt;LED1 and LED2 flash alternatively&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I quite don't understand here what I can do to make both LED flash simultaneously and alternatively. LED1 flashes and LED2 off if SW2 is HIGH and vice versa. How do the other two work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Down here is what has been done so far:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;msp430.h&amp;gt; &#xA;#include &quot;msp430g2553.h&quot;&#xA;int sw2=0;&#xA;&#xA;int main(void) &#xA;{&#xA;    WDTCTL = WDTPW | WDTHOLD;      //stop watchdog timer&#xA;&#xA;    P1DIR  = 0x00;         //port 1 all inputs&#xA;    P1DIR |= (BIT0 | BIT6);        //set P1.0 and P1.6 as outputs (LED1, LED2)&#xA;    P1REN |= BIT3;         //activate resister on P1.3&#xA;    P1OUT |= BIT3;         //make it pull up because SW2 is active low&#xA;&#xA;    for(;;)&#xA;    { sw2 = P1IN;          //read values from P1&#xA;      sw2 &amp;amp;= BIT3;         //mask out only BIT3 where SW2 is connected&#xA;      if (sw2 == BIT3)               &#xA;         {             //if SW2 is high&#xA;            P1OUT &amp;amp;= ~BIT6;    //turn LED2 off &#xA;            P1OUT ^= BIT0;     //toggle LED1&#xA;          __    delay_cycles(50000);   //delay 50,000 micro seconds&#xA;         }&#xA;      else             &#xA;         {             //else (SW2 is low) &#xA;            P1OUT &amp;amp;= ~BIT0;    //turn LED1 off&#xA;            P1OUT ^= BIT6;     //toggle LED2&#xA;            __delay_cycles(200000);   //delay 200,000 micro seconds&#xA;         }&#xA;     }&#xA;// end of infinite loop&#xA;}&#xA;//end of main&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15749" LastEditorUserId="7083" LastEditDate="2018-04-18T11:34:47.853" LastActivityDate="2018-06-17T13:03:13.903" Title="MSP430 - Semaphore - C language" Tags="|electrical-engineering|computer-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21349" PostTypeId="1" CreationDate="2018-04-17T03:12:23.523" Score="2" ViewCount="265" Body="&lt;p&gt;If you watch &lt;a href=&quot;https://www.youtube.com/watch?v=8OBHUZetYIU&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt;, you can get a sense of how an elevator can operate in two dimensions. But, what I can't figure out is how it is driven. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider a conventional elevator, there is a car and a counterweight connected by steel rope. The motor only has to oppose or resist the difference in weight between them (at most half the load, because it can be counterweighted to the car plus half its rated load). This limits the work the motor has to do when lifting the imbalance, and limits the energy it has to remove when lowering an imbalance. Also, the work is done by fixed equipment (typically at the top of the shaft) which doesn't contribute to the weight of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the two-dimensional ropeless system, there is no counterwieght, the motor is part of the car (and adds to its weight), so has to drive all that weight going up, and get rid of all that energy going down. It also has to do it without risk of runaway/freefall. And, it shouldn't just burn off all that energy as heat (terribly inefficient). So my question is, how is it done?&lt;/p&gt;&#xA;" OwnerUserId="15169" LastActivityDate="2020-07-25T10:02:10.913" Title="How does a two dimensional elevator work?" Tags="|motors|mechanical|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21351" PostTypeId="1" CreationDate="2018-04-17T06:57:48.550" Score="2" ViewCount="2437" Body="&lt;p&gt;The extrusion shape is a simple square tube. If I set it down on a perfectly flat surface with its curvature upwards, the middle raises up by around 5mm over the length of 1m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I tried to just put a weight on it while its laying on a glass pane (my perfectly flat surface) but after 24h that did just about nothing.. maybe if I left it there for a year? I'm not that patient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other options for me to straighten it without specialized tools? I could heat it up with a torch if that helps..? &lt;/p&gt;&#xA;" OwnerUserId="15752" LastActivityDate="2022-06-17T00:03:30.370" Title="How to straighten an aluminium extrusion?" Tags="|aluminum|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21352" PostTypeId="2" ParentId="21295" CreationDate="2018-04-17T10:03:53.077" Score="1" Body="&lt;p&gt;The appropriate potential (from irrotational condition) for 2d would give you, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\phi = \Lambda ~ln~r$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$u_r = \frac{\lambda}{r}$  .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, check your definition of potential function again. &lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-04-17T10:03:53.077" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21353" PostTypeId="2" ParentId="18240" CreationDate="2018-04-17T11:55:50.650" Score="4" Body="&lt;p&gt;Bamboo in its natural shape is very prone to splitting. Having any kind of hole will reduce its high bending strength resistance significantly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, bamboo culm does not have a constant cross-section in the longitudinal direction, makes it harder to create a reliable bond. &lt;strong&gt;J.F. Correal&lt;/strong&gt; &quot;&lt;em&gt;&lt;a href=&quot;https://doi.org/10.1016/B978-0-08-100038-0.00014-7&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bamboo design and construction&lt;/a&gt;&lt;/em&gt;&quot; has few suggestions for bamboo connections. the following are few of the typically used and suggested connections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pUy1O.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pUy1O.jpg&quot; alt=&quot;orthogonal connections&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WmxGd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WmxGd.jpg&quot; alt=&quot;longitudinal (sliced) connections&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15755" LastActivityDate="2018-04-17T11:55:50.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21354" PostTypeId="2" ParentId="15048" CreationDate="2018-04-17T13:11:35.547" Score="0" Body="&lt;p&gt;As touched on in other answers: a car starter is in a harsh environment, so it is pretty well sealed against dirt and moisture, and has no provision for cooling. In operation, a starter like all motors will generate heat due to electrical resistance in the windings. Its normal job as a starter requires it to run for a few seconds and then has minutes, hours, or days to get rid of the heat that builds up as turns the engine. In continuous operation, it will generate heat much faster than it can get rid of it, eventually causing something to fail (maybe melting or even ignition of insulation). Any motor suitable for continuous operation will have provision for cooling - the windings will be visible through ports in the housing and there will almost certainly be a fan to move air through it.&lt;/p&gt;&#xA;" OwnerUserId="15169" LastActivityDate="2018-04-17T13:11:35.547" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21355" PostTypeId="2" ParentId="15625" CreationDate="2018-04-17T13:17:18.747" Score="0" Body="&lt;p&gt;The wiper arm is on a shaft, which passes through the car's bodywork. What you don't see is another arm (perhaps only an inch or two in length) on the backside of the shaft, with the crank rocker arrangement attached to it.&lt;/p&gt;&#xA;" OwnerUserId="15169" LastActivityDate="2018-04-17T13:17:18.747" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21356" PostTypeId="2" ParentId="21348" CreationDate="2018-04-17T13:30:31.480" Score="1" Body="&lt;p&gt;I can only see that you've programmed the switch to &quot;if high - do one thing, if low - do something else&quot;. You need to include a counter for the switch (and debounce). then 4 if statements, one for every mode you're cycling through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pseudo-ish code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;int sw2 = 0;&#xA;int counter = 0;&#xA;flipbit = 0;&#xA;&#xA;Int main(void) {&#xA;&#xA;   (whatever ports/bits you need to set up here)&#xA;&#xA;   If (sw2 == 1) {                    //if the switch is depressed..&#xA;      delay_cycles(200,000);          //Wait 200ms (lazy debounce) as long as you click a button quicker than 200ms no worries&#xA;      counter == counter + 1;         //increment the counter&#xA;   }&#xA;   if (counter == 1) {&#xA;      P1OUT ^= BIT0;                  //turn led1 on&#xA;      __delay_cycles(200000);         //wait 200ms&#xA;      P1OUT ^= BIT0;                  //turn led1 off&#xA;      __delay_cycles(200000);         //wait 200ms&#xA;   }&#xA;   if (counter == 2) {&#xA;      P1OUT ^= BIT6;                  //turn led2 on&#xA;      __delay_cycles(200000);         //wait 200ms&#xA;      P1OUT ^= BIT6;                  //turn led2 off&#xA;      __delay_cycles(200000);         //wait 200ms&#xA;   }&#xA;   if (counter == 3) {&#xA;      P1OUT &amp;amp;= ~BIT0;                 //turn led 1 on&#xA;      P1OUT ^= BIT6;                  //turn led 2 on&#xA;      delay_cycles(200,000);          //wait 200ms&#xA;      P1OUT &amp;amp;= ~BIT0;                 //turn led 1 off&#xA;      P1OUT ^= BIT6;                  //turn led 2 off&#xA;      delay_cycles(200,000);          //wait 200ms&#xA;   }&#xA;   if (counter == 4) {                //now it gets exciting...&#xA;      if (flipbit == 0) {             //this statement lets us run a statement once per loop&#xA;         P1OUT &amp;amp;= ~BIT0;              //turn led1 on&#xA;         flipbit == flipbit + 1       //make flipbit = 1 so it doesn't run again this cycle&#xA;         }&#xA;      P1OUT &amp;amp;= ~BIT0;                 //turn led 1 off&#xA;      P1OUT ^= BIT6;                  //turn led 2 on&#xA;      delay_cycles(200,000);          //wait 200ms&#xA;      P1OUT &amp;amp;= ~BIT0;                 //turn led 1 on&#xA;      P1OUT ^= BIT6;                  //turn led 2 off&#xA;      delay_cycles(200,000);          //wait 200ms&#xA;   }&#xA;   if (counter == 5) {                //you clicked the button again, to return to the first mode&#xA;      P1OUT &amp;amp;= ~BIT0;                 //turn led1 back off, ready to start the cycle again&#xA;      flipbit = 0;&#xA;      counter = 1;                     //set counter to 1, so it starts on first mode&#xA;   }&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As you're 'toggling' bits rather than definitively setting them as 1 or 0, i've made sure to keep track of what state they're in by making them do a complete cycle within each statement, i.e turning them on and BACK off before it can move on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This 'flipbit' is there to offset the pattern by turning led1 on first so we can use the same pattern as mode 3, and the fifth step of the counter is to turn that led back off and to reset that flipbit to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a few ways of implementing this. This one was the easiest i could think that was 'readable'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just looked it up and the '__delay_cycles' is actually similar to arduino whereas it will halt code execution. This is not great as your button presses will go unnoticed for a large part of the program. There's 2 ways to go about it, either use an interrupt for the switch or use a timer instead of delay. &lt;a href=&quot;https://www.embeddedrelated.com/showarticle/182.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;This link&lt;/a&gt; shows you how.&lt;/p&gt;&#xA;" OwnerUserId="12187" LastEditorUserId="12187" LastEditDate="2018-04-17T13:46:29.130" LastActivityDate="2018-04-17T13:46:29.130" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21357" PostTypeId="1" CreationDate="2018-04-17T17:53:57.980" Score="1" ViewCount="97" Body="&lt;p&gt;How is a TXV supposed to behave in a no evaporator thermal load situation (for example, evaporator fan completely off)? Would it completely close, not letting any liquid refrigerant get through (and reach the pump still in the liquid phase), would it still let some liquid refrigerant leak past into the evaporator coil, eventually overcome by the pressure building up in the high pressure line, but not enough to manage to reach the evaporator's outlet without having turned into superheated vapor, or would it behave differently? Would the evaporator coil still be able to develop frost in that situation (and potentially end up making the evaporator completely unable to turn the liquid into superheated vapor)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="15424" LastActivityDate="2018-04-17T17:53:57.980" Title="Does a TXV completely close when there's no thermal load whatsoever?" Tags="|heat-transfer|hvac|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21358" PostTypeId="2" ParentId="11177" CreationDate="2018-04-17T18:49:54.457" Score="1" Body="&lt;p&gt;I have posted some pictures and links of wind power with tires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mjYxv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mjYxv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://www.google.com/url?sa=i&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=images&amp;amp;cd=&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=2ahUKEwiZ45-8-cHaAhXOneAKHVu1A-IQjhx6BAgAEAM&amp;amp;url=https%3A%2F%2Fphys.org%2Fnews%2F2012-07-blackbird-cart-fast.html&amp;amp;psig=AOvVaw39dFSGPy0w7NEXZdG-zHDA&amp;amp;ust=1524076299543377&quot; rel=&quot;nofollow noreferrer&quot;&gt;vehicle&lt;/a&gt;  proved it's possible to sail downwind faster than the wind is up for auction on Ebay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xFQUY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xFQUY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YHb8v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YHb8v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the wind to turn a generator then turns the drive motor or as a direct sail.&#xA;It would work like the diagram but with tires.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/orYim.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/orYim.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-04-17T18:49:54.457" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21359" PostTypeId="1" AcceptedAnswerId="21360" CreationDate="2018-04-17T19:50:13.077" Score="3" ViewCount="1629" Body="&lt;p&gt;I want to find the Poisson's ratio of a 3D model, not the material but the model. How can I do this in ANSYS workbench?&lt;/p&gt;&#xA;&lt;p&gt;I am using this formula, but I can't get it to work with different models. I am not sure what strain is the correct to use, can someone point me in the correct direction?&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\nu=-\frac{\text{Strain in direction of load}}{\text{Strain at right angle of load}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\nu=-\frac{\varepsilon_{lateral}}{\varepsilon_{axial}}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="15757" LastEditorUserId="9" LastEditDate="2020-12-16T11:33:01.277" LastActivityDate="2020-12-16T11:33:01.277" Title="Finding Poisson's ratio of 3D model in ANSYS" Tags="|ansys|ansys-workbench|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21360" PostTypeId="2" ParentId="21359" CreationDate="2018-04-17T21:11:51.073" Score="1" Body="&lt;p&gt;Are you asking which of the two equations you give to use? In the context of a continuous material, the first one is flipped; the second is correct. However, since you're trying to apply a materials concept to a distributed structure, you can't use the strain that ANSYS will output, because this will be the material strain. Better to use $$\nu=-\frac{\frac{\mathrm{displacement\; between\; two\; points\;A\;and\;B\;spaced\;at\;a\;right\;angle\;to\;the\;load}}{\mathrm{original\;distance\; between\; points\;A\; and\;B}}}{\frac{\mathrm{displacement\; between\; the\;two\; points\;of\;load\;application\;C\;and\;D}}{\mathrm{original\;distance\; between\; points\;C\;and\;D}}}$$&lt;/p&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2018-04-17T21:11:51.073" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21362" PostTypeId="2" ParentId="15931" CreationDate="2018-04-18T04:30:47.407" Score="3" Body="&lt;p&gt;use big rubber inflatable wheel barrow tires for they are IMO the best for pushing or pulling through soft loose soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eifQe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eifQe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorUserId="10902" LastEditDate="2018-06-17T07:05:31.273" LastActivityDate="2018-06-17T07:05:31.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21363" PostTypeId="2" ParentId="12688" CreationDate="2018-04-18T05:09:36.873" Score="1" Body="&lt;p&gt;I used to make candles in the wet sand by drawing out what I wanted it to look like. I guess it would work with plastic and then grind the sand out of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jplKL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jplKL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-04-18T05:09:36.873" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21365" PostTypeId="1" CreationDate="2018-04-18T10:11:49.077" Score="0" ViewCount="28" Body="&lt;p&gt;I have a database of locations. Each location is a latitude and longitude pair. Now for each query, I want to get top &quot;X&quot; nearest locations around a given latitude longitude point. How do I actually calculate the distance between two lat long points and is there any efficient algorithm to do this instead of a simple for loop?&lt;/p&gt;&#xA;" OwnerUserId="15765" LastActivityDate="2018-04-18T11:25:51.383" Title="How to find the nearest locations around a given latitude and longitude point?" Tags="|geometry|" AnswerCount="1" CommentCount="3" ClosedDate="2018-04-20T14:32:35.503" ContentLicense="CC BY-SA 3.0" />
  <row Id="21366" PostTypeId="1" CreationDate="2018-04-18T10:15:39.703" Score="0" ViewCount="1089" Body="&lt;p&gt;In a coupled tanks system, what is the control strategy to simultaneously control the level in multiple tanks. In this case the actuators influence multiple control variables. The system is as shown in this diagram&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YJu3C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YJu3C.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The variables to be controlled are the liquid levels in each of the tanks. The actuators are the valves on the pipes as well as the inlet flow q0. The problem is that the same actuator e.g. V1 influences both h1 and h2. I tried to 3 different PIDs for each tank. Each controller receives the height in the corresponding tank and sends the control signal to the corresponding outlet. So, PID1 receives input h1 and controls only V1; PID2 receives input h2 and controls only V2...&lt;/p&gt;&#xA;" OwnerUserId="14241" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-04-20T00:17:13.560" Title="How to control the level in multiple connected tanks at the same time?" Tags="|control-engineering|pid-control|flow-control|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 3.0" />
  <row Id="21367" PostTypeId="2" ParentId="21365" CreationDate="2018-04-18T11:25:51.383" Score="1" Body="&lt;p&gt;Your question comes down to a basic math problem.  You have two points in polar coordinates you want to find the distance between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, it is probably acceptable to model the earth as a sphere.  Convert each polar coordinate to cartesian.  Now you subtract them and compute the square root of the sum of the squares of the components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above method gives you straight line distance.  If all your &quot;near here&quot; regions are small enough, then that is a valid approximation.  If your distance are large enough so that you want to take the curvature of the earth into account, then you need to find the angle between the two points as seen from the center of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the vectors from the center of the earth to each point, unitize these vectors, then take the dot product.  The angle between the points is the arc-cosine of this dot product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beware of precision.  On a computer, you will probably want to use &quot;double precision&quot; for any cartesian components.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-04-18T11:25:51.383" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21368" PostTypeId="2" ParentId="12499" CreationDate="2018-04-18T11:32:03.927" Score="2" Body="&lt;p&gt;It all depends on what is in the waste water, and how much is allowed to be left after treatment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treatment can mean anything from a settling tank, to a filter or screen, to oxygenating, to centrifuging, to active biological break down, and more.  In some cases, treatment means it flows thru a &quot;magic pipe&quot; and is simply discharged when no inspectors are looking.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-04-18T11:32:03.927" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21370" PostTypeId="1" AcceptedAnswerId="21385" CreationDate="2018-04-18T15:54:10.357" Score="1" ViewCount="872" Body="&lt;p&gt;We do inspections using a robotic camera in the pipeline. It’s tractor type camera. (Ref. photo)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gobo1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gobo1.jpg&quot; alt=&quot;Pipeline Inspection camera&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QnjGE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QnjGE.jpg&quot; alt=&quot;camera inspection video&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While driving the camera through the pipeline it gives us the camera tilt degree output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The camera System gives us output as below,&#xA;&lt;a href=&quot;https://i.stack.imgur.com/g4mEv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g4mEv.jpg&quot; alt=&quot;camera tilt output degree against mar&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;could please help me to calculate the elevation/altitude of each meter. Using camera tilt degree value&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As below. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/9urX0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9urX0.jpg&quot; alt=&quot;example elevation graph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Awaiting for your answers. If you need any further info. Let me know. &lt;/p&gt;&#xA;" OwnerUserId="15769" LastActivityDate="2018-04-19T16:04:47.213" Title="Elevation / pipeline Elevation calculation" Tags="|civil-engineering|pipelines|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21373" PostTypeId="2" ParentId="4206" CreationDate="2018-04-18T18:42:55.793" Score="1" Body="&lt;p&gt;Yes, it is the spring that twists as its contracts and expands due to temperature. Change the setup, no other solution. There may be one or two reason, one is the shape of the spring on an isotropic expansion the second is because the spring is made out of a straight wire, it may not be to isotropic anymore. Actually, both effects are acting together.&lt;/p&gt;&#xA;" OwnerUserId="15771" LastEditorUserId="9" LastEditDate="2018-04-18T19:45:11.393" LastActivityDate="2018-04-18T19:45:11.393" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21374" PostTypeId="1" CreationDate="2018-04-19T06:26:49.197" Score="0" ViewCount="113" Body="&lt;p&gt;For my school project, I am trying to measure the effect of a mass damper by changing the length of the string (that carries the mass), my setup is similar to the following video:&#xA; &lt;a href=&quot;https://youtu.be/f1U4SAgy60c?t=4m4s&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/f1U4SAgy60c?t=4m4s&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find a relationship between the length of the string and the oscillation of the &quot;building&quot;. But I don't know exactly what the guy in the video is trying to measure and if it is related to what I'm trying to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to use an accelerometer to map the oscillation of the &quot;building&quot; and if so, is there a relationship between the accelerometer data and the displacement of the oscillation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My aim is to find the maximum amplitude for different lengths of string, so does a larger amplitude from the accelerometer data correspond to a larger amplitude in the displacement time graph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am desperate for help as I would much rather use accelerometer data rather than use a ruler in the background to measure the displacement of the oscillation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance!!!&lt;/p&gt;&#xA;" OwnerUserId="15776" LastActivityDate="2023-12-18T11:07:40.937" Title="How do I measure max oscillation of building using accelerometer?" Tags="|building-physics|acceleration|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21376" PostTypeId="1" CreationDate="2018-04-19T07:20:18.573" Score="0" ViewCount="19" Body="&lt;p&gt;Are there any fabrics available in today’s time which can stretch in one direction only and behave completely rigid in the perpendicular direction ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gPVKK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gPVKK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does any Materials Science expert know about any on-going research in this area ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I particularly wanted to know the way in which one might go about manufacturing something like this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if such a fabric is imagined to be made up of perpendicularly intersecting fibres, one of which is stretchable while the other is not, like criss-crossing rubber &amp;amp; fibreglass fibres, for instance, what would be the way to make something like this ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yvVYC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yvVYC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10075" LastEditorUserId="10075" LastEditDate="2018-04-20T04:51:13.727" LastActivityDate="2018-08-08T09:27:47.680" Title="Uni-Directional Stretchable Elastomers" Tags="|materials|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21377" PostTypeId="2" ParentId="21366" CreationDate="2018-04-19T07:22:51.553" Score="0" Body="&lt;p&gt;I'm not a control systems person, so forgive my not very systematic approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first thing you should do is to try to adjust the valves manually for a constant q, to reach the levels you want. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would control q from the average of the  height deviations from the wanted height in each tank. If the water level overall is lower than wanted, increase q, and vice versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control variable for the valves should be the height difference between two consecutive tanks: V1 is controled for h1-h2, these height differences you compute from the wanted heights in all tanks. Only valve 3 is controlled for h3 directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also make it so that the first valve is quicker than the second and the second quicker thant the third. This should help you avoid oscillations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite that I promised a non systematic approach, here's a bit of reasoning: For a given q, there's a relationship between height difference and how open or closed the valve is. This is likely why your strateg didn't work.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-04-19T07:22:51.553" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21378" PostTypeId="1" AcceptedAnswerId="21592" CreationDate="2018-04-19T07:40:43.100" Score="0" ViewCount="642" Body="&lt;p&gt;I need to simulate an external pressure load that varies sinusoidally along a circular arc. The sinusoidal variation is with arc length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to create a uniform pressure load in Patran but do not know how to create a non uniform pressure load. I can't seem to find any guides, tutorials or articles on it either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone knows how to do this or has a useful guide they could link, your help will be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using MSC Nastran/Patran student version 2017.0.1&lt;/p&gt;&#xA;" OwnerUserId="15777" LastEditorUserId="1832" LastEditDate="2018-04-21T10:36:42.340" LastActivityDate="2018-05-04T06:36:09.997" Title="How to create a non uniform pressure load in MSC Patran" Tags="|structural-engineering|finite-element-method|nastran|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21380" PostTypeId="1" CreationDate="2018-04-19T09:24:56.077" Score="4" ViewCount="828" Body="&lt;p&gt;I have a scenario where I have water flowing through a gravel filled pipe, and I need to find the velocity. The pipe is gravity fed (no pumps). My approach is to find the seepage velocity ($V_s$) which is the Velocity divided by the porosity. My understanding is that the seepage velocity is defined as the apparent velocity through the bulk of the porous medium. So that is what I need to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the velocity using the Darcy Weisbach equation and porosity of gravel from online research, however I am getting a very high value for the seepage velocity (about 56&amp;nbsp;m/s) even when using the highest porosity value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method that was suggested to me would be to use $Q = KIA$ where $k$ = hydraulic conductivity of gravel, $I$ = hydraulic gradient and $A$ = cross sectional area. However, this condition is only for laminar flow conditions (very slow velocities - such as groundwater under an aquifer. So I don't think I can use this equation. This method gives me a very low velocity which also seems incorrect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both methods seem to give me odd values so could anyone assist me with this problem?&lt;/p&gt;&#xA;" OwnerUserId="15545" LastEditorUserId="1832" LastEditDate="2018-04-21T10:24:18.553" LastActivityDate="2024-03-21T03:06:00.717" Title="Finding the velocity of water flowing through a gravel filled pipe" Tags="|fluid-mechanics|civil-engineering|design|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21383" PostTypeId="2" ParentId="21374" CreationDate="2018-04-19T13:48:57.977" Score="0" Body="&lt;p&gt;If you insist on using the accelerometer data, there are probably libraries, searching &quot;accelerometer tool&quot; (on the google play store returns several results that allow exporting of data etc), depending on your platform that do most of the maths for you.&#xA;but seeing as acceleration is rate of change of velocity which in turn is rate of change of distance. you should be able to numerically integrate the the acceleration to get a value for displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMHO: At the end of the day though this sounds like a school/lab experiment and you are much more likely to get good results using the ruler.&lt;/p&gt;&#xA;" OwnerUserId="12804" LastEditorUserId="12804" LastEditDate="2018-04-19T13:57:17.410" LastActivityDate="2018-04-19T13:57:17.410" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21384" PostTypeId="2" ParentId="7239" CreationDate="2018-04-19T15:49:20.447" Score="1" Body="&lt;p&gt;You do not need a license in any state to be an engineer; I was one in IL IN and TX. However you need a license in every state to be a &quot;Professional Engineer&quot;. Which will require a combination of formal education and experience. Many states do have reciprocity agreements with other states , so one license will let you practice in several states. Apologies; I believe I answered the question he wanted to ask , not the question that was asked.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-04-19T17:24:25.020" LastActivityDate="2018-04-19T17:24:25.020" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21385" PostTypeId="2" ParentId="21370" CreationDate="2018-04-19T16:04:47.213" Score="0" Body="&lt;p&gt;Assuming you know the original starting position, you can use trigonometry to work out the profile of the pipeline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you take coordinates $(x_n,y_n)$, and with a measured 'inside pipe travel distance' of $d_n$ meters inside the pipe, down a slope of $\theta_n°$, then that coordinate can be caluclated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x_n=x_{n-1}+\left(d_n-d_{n-1}\right)\sin\left(\theta_n\right)$$&#xA;$$y_n=y_{n-1}-\left(d_n-d_{n-1}\right)\cos\left(\theta_n\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to note, that while the negative incline of the pipe is increasing, this approach will slightly over-estimate the altitude of each point, and slightly under-estimate as it is levelling out. Over a long run of pipe, this should even out, and the total change in altitude from end to end will be fairly accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image of the graph that I was able to generate using the data that you supplied above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Wkcw0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wkcw0.png&quot; alt=&quot;Graph generated from supplied data&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-04-19T16:04:47.213" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21388" PostTypeId="1" CreationDate="2018-04-19T17:31:31.463" Score="-1" ViewCount="291" Body="&lt;p&gt;Given the mass flow rate in lbs/min, the specific gravity of a substance, the temperature in F, and the density in lbs/ft^3, what is the equation to derive the volumetric flow in Gallons/Min?&lt;/p&gt;&#xA;" OwnerUserId="15783" LastActivityDate="2018-04-21T00:07:50.583" Title="How to derive volumetric flow rate?" Tags="|chemical-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21389" PostTypeId="2" ParentId="21388" CreationDate="2018-04-19T18:30:33.903" Score="1" Body="&lt;p&gt;v lbs / min * w ft^3 / lbs * x gal / ft^3 = z gal / min  &lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-04-20T08:49:28.353" LastActivityDate="2018-04-20T08:49:28.353" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21392" PostTypeId="2" ParentId="17567" CreationDate="2018-04-20T00:14:43.310" Score="1" Body="&lt;p&gt;I'm sorry to disappoint you but I have tried numerous times to contact the makers that created that video and I have yet to get even one reply from them.  It seems they use those terms like, &quot;white opaque special laser colour&quot; in a manner that is deliberately vague; perhaps they have a proprietary paint or formula for mixing they don't want to share with the general community... sorta like the recipe for KFC original recipe chicken. They are a German based firm and I've translated most of what is contained on their website and the terms like &quot;laser colour&quot; have a more literal translation of enamel or varnish.  So it could be that their paint contains a laser reflective varnish they mix into the white paint. It's a bit confusing because they also call that special laserpaint white translucent, not opaque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are links to their site:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.simparts.de/en/faq.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.simparts.de/en/faq.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=vf7tQNYqliw&amp;amp;t=60s&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=vf7tQNYqliw&amp;amp;t=60s&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15788" LastEditorUserId="9" LastEditDate="2018-04-20T00:31:20.690" LastActivityDate="2018-04-20T00:31:20.690" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21393" PostTypeId="2" ParentId="21366" CreationDate="2018-04-20T00:17:13.560" Score="1" Body="&lt;p&gt;You are dealing with a &lt;a href=&quot;https://en.wikibooks.org/wiki/Control_Systems/MIMO_Systems&quot; rel=&quot;nofollow noreferrer&quot;&gt;multi-input multi-output system&lt;/a&gt; (MIMO) system in which case you can't in general apply the same techniques as for single-input single-output (SISO) systems. If you are dealing with a linear time invariant MIMO system, then you could use something called the &lt;a href=&quot;http://folk.ntnu.no/skoge/vgprosessregulering/papers-pensum/ch3-ch10-2ndedition.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;relative gain array&lt;/a&gt; to see how well a SISO control approach would work. However in this case I believe you are dealing with a nonlinear system, but it might be possible to do input-output decoupling, namely by controlling the netto inflow into each tank. For this I assume for now that you can choose the flow rates $q_0$, $q_1$, $q_2$ and $q_3$ and that the dynamics can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\dot{h}_i = \alpha_i (q_{i-1} - q_i) \tag{1}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the combines dynamics can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{\vec{h}} = &#xA;\begin{bmatrix}&#xA;\alpha_1 &amp;amp; -\alpha_1 &amp;amp; 0 &amp;amp; 0 \\&#xA;0 &amp;amp; \alpha_2 &amp;amp; -\alpha_2 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; \alpha_3 &amp;amp; -\alpha_3&#xA;\end{bmatrix} \vec{q} \tag{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\vec{h} = \begin{bmatrix}h_1 &amp;amp; h_2 &amp;amp; h_3\end{bmatrix}^\top$ and $\vec{q} = \begin{bmatrix}q_0 &amp;amp; q_1 &amp;amp; q_2 &amp;amp; q_3\end{bmatrix}^\top$. Now by defining a virtual input $\vec{v}$ such that $\dot{\vec{h}} = \vec{v}$, then every input and output pair is decoupled. For this we need&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}&#xA;\alpha_1 &amp;amp; -\alpha_1 &amp;amp; 0 &amp;amp; 0 \\&#xA;0 &amp;amp; \alpha_2 &amp;amp; -\alpha_2 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; \alpha_3 &amp;amp; -\alpha_3&#xA;\end{bmatrix} \vec{q} = \vec{v} \tag{3}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be solved for $\vec{q}$ using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse&quot; rel=&quot;nofollow noreferrer&quot;&gt;pseudo inverse&lt;/a&gt; of the matrix on the left hand side. There are multiple ways this could be done, but the solution which minimizes $\|\vec{q}\|$ yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\vec{q} = &#xA;\begin{bmatrix}&#xA;\frac{3}{4\,\alpha_1} &amp;amp; \frac{1}{2\,\alpha_2} &amp;amp; \frac{1}{4\,\alpha_3} \\&#xA;\frac{-1}{4\,\alpha_1} &amp;amp; \frac{1}{2\,\alpha_2} &amp;amp; \frac{1}{4\,\alpha_3} \\&#xA;\frac{-1}{4\,\alpha_1} &amp;amp; \frac{-1}{2\,\alpha_2} &amp;amp; \frac{1}{4\,\alpha_3} \\&#xA;\frac{-1}{4\,\alpha_1} &amp;amp; \frac{-1}{2\,\alpha_2} &amp;amp; \frac{-3}{4\,\alpha_3}&#xA;\end{bmatrix} \vec{v}. \tag{4}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now you can use normal SISO control methods, like PID, for each $h_i$-$v_i$ pair which would drive $\vec{h}$ to a desired value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is probably not a linear relation between $q_i$ and $V_i$, but more likely something like $q_i \propto \sqrt{V_i(h_i - h_{i+1})}$ or some other nonlinear state depended function. If you know this relation more accurately you could come up with a nonlinear controller, otherwise you could perform sequential loop closing. For the sequential loop closing you use $V_i$ to control $q_i$ to its desired value defined by $\vec{v}$ from equation $(4)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have some more challenging reference signal for the heights, then you might have to resort to even more advance techniques, such as model predictive control. For example when the reference heights are such that $h_{r,1} &amp;lt; h_{r,2}$ and $h_{r,2} &amp;gt; h_{r,3}$ but your system starts at $h_1 = h_2 = h_3 &amp;lt; h_{r,2}$, then it will initially not be possible to have a positive inflow into the second tank and to reach the desired heights $h_1$ actually has to be bigger than $h_2$ for a while in order for $h_2$ to reach its goal.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-04-20T00:17:13.560" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21397" PostTypeId="1" AcceptedAnswerId="24038" CreationDate="2018-04-20T07:52:54.503" Score="0" ViewCount="400" Body="&lt;p&gt;I want to run a coupled simulation in Ansys - RF heating and deformation of waveguide. Pictures of project layout and settings are &lt;a href=&quot;https://i.stack.imgur.com/rAmdl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I calculated RF fields and surface losses in HFSS; imported them to steady-state thermal, calculated temperature of the waveguide; calculated structure deformation in static structural. Now what is needed is to run this problem until it converges. But for some reason HSFF does not import calculated deformed mesh from static structural. What else should I do to make it import mesh? Can I then use feedback iterator to automatically recalculate this problem?&lt;/p&gt;&#xA;" OwnerUserId="15792" LastActivityDate="2018-11-09T10:05:09.907" Title="Ansys multiphysics import deformed mesh to HFSS" Tags="|ansys|ansys-workbench|radio|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21399" PostTypeId="1" CreationDate="2018-04-20T10:08:54.457" Score="1" ViewCount="512" Body="&lt;p&gt;Generally, I'm interested in simple (DIY) electric vehicle design problems. I saw some photos of electric vehicles as in the following pic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tO8oN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tO8oN.jpg&quot; alt=&quot;4dw diy vehicle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or also you could check &lt;a href=&quot;https://www.youtube.com/watch?v=6Z6ebDRXbxI&quot; rel=&quot;nofollow noreferrer&quot;&gt;this youtube video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vehicle is using an &lt;a href=&quot;https://en.wikipedia.org/wiki/Individual-wheel_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;independent wheel drive&lt;/a&gt;, so it has 4 bicycle hub motors - one for each wheel. &lt;strong&gt;How is possible to regulate/sync the motor's RPM/torque in the turns?&lt;/strong&gt; In the classic one-motor design this is solved by using a differential but how it is solved in an independent wheel drive scenario?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also wondering what happens when one wheel temporarily leaves the ground (e.g. in some off-road scenario). Wouldn't it start spinning very fast in the air, without contact with a surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a programmer, I could imagine some Arduino-based &quot;software differential&quot; for the front wheels, which will regulate the RPM on the wheels based on the &quot;steering wheel&quot; angle, but I don't know how to do the same on the rear wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, using an Arduino in such projects doesn't sound very &quot;bullet-proof&quot; but a car, especially such an important thing as engine RPM/torque regulation, needs to be &quot;bulletproof&quot; even as a DIY project, so I am not convinced that this would be a reasonable solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How it is possible to solve the independent wheel drive problems in the &lt;strong&gt;simple DIY world?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="866" LastEditorUserId="368" LastEditDate="2018-04-20T15:23:47.077" LastActivityDate="2018-04-20T15:23:47.077" Title="How to regulate torque/RPM on an IWD vehicle" Tags="|motors|automotive-engineering|torque|wheels|electric-vehicles|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21400" PostTypeId="1" AcceptedAnswerId="21414" CreationDate="2018-04-20T13:09:16.787" Score="2" ViewCount="3850" Body="&lt;p&gt;How do you do double interpolation for a large set of data in Excel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The user is required to input an x value and y value then the program has to find the points and do interpolation if necessary. For example, if the x is 3.5 and y value is 4.5  then I'll have 2 x value and 2 y values so I need to do the double interpolation then final interpolation between the remaining values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used a &lt;em&gt;vlookup&lt;/em&gt; function and &lt;em&gt;forecast&lt;/em&gt; function  but it doesn't seem to be working out. I have attached a picture of the data set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dXeJO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user15793" LastEditorUserId="3505" LastEditDate="2018-04-22T11:57:10.017" LastActivityDate="2018-04-22T11:57:10.017" Title="Double interpolation for a large set of data in Excel" Tags="|statistics|" AnswerCount="2" CommentCount="5" ClosedDate="2018-05-01T18:45:07.170" ContentLicense="CC BY-SA 3.0" />
  <row Id="21401" PostTypeId="2" ParentId="21400" CreationDate="2018-04-20T14:41:55.997" Score="1" Body="&lt;p&gt;I have a spreadsheet that I wrote years and years ago for interpolating. It generates points that follow the same sort of profile as the 'smoothed line' function when you add a line to an X/Y Scatter Plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image below, the known data is inputted on the left, and trend values are calculated to give an idea of the gradient at the outer edges of the data. The desired X values are put into the right hand table, and the Y values are calculated accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the graph, the Large Orange blobs are the original 'known data', the orange line is the 'smoothed line' generated by excel's graph functions, and the blue dots are the interpolated X/Y values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yY2Pv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yY2Pv.png&quot; alt=&quot;Spreadsheet Screenshot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can &lt;a href=&quot;http://www.filedropper.com/excelsmoothedcurveinterpolationspreadsheet&quot; rel=&quot;nofollow noreferrer&quot;&gt;download the spreadsheet here&lt;/a&gt;, but I know not everybody likes downloading unknown files from the internet, so I've included enough information to recreate the file from scratch below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for the 'trend' boxes are simple. For Cell D5 as shown, it is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=TREND(D6:D7,,0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The formulae for the 'Interoplated' boxes are less than simple. For Cell I5 as shown, it is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=SUM((1+1/IRR(MMULT({0,0,2,0;0,1,0,-1;-1,4,-5,2;1,-3,3,-1},INDEX(D$5:D$13,MATCH(H5,D$5:D$13)+N(IF(1,{-1;0;1;2})))-H5&#xA; )))^-{0;1;2;3}*MMULT({0,2,0,0;-1,0,1,0;2,-5,4,-1;-1,3,-3,1},INDEX(E$5:E$13,MATCH(H5,D$5:D$13)+N(IF(1,{-1;0;1;2})))))/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you recreate the spreadsheet exactly as in the image, and pay attention to how the formulae link the ranges, you can understand how to expand this for any size of dataset.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-04-20T14:41:55.997" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21402" PostTypeId="1" CreationDate="2018-04-20T16:26:18.573" Score="1" ViewCount="123" Body="&lt;p&gt;This is  real life issue I am having currently and I am puzzled as to why it happened. I have lived over the subway lines (two actually each going in the opposite direction, two sets of tracks) for two years. The noise has always been a soft background hum, with an occasional louder train (that usually needs maintenance), hardly noticeable. One early morning near the end of January I was jolted out of bed by a loud noise. Turns out it is the subway. It is as if someone turned the volume up from 3 to 8 all of a sudden. Noise much louder (both tracks), last longer and there is shaking indoors too. Some small variances in the loudness of each train, but all much louder. The quietest train now is much loudest then the loudest one was before. I know that no changes were made to the tracks, trains, schedules or my building. What else could have caused it - noise from subway became much louder overnight? &lt;/p&gt;&#xA;" OwnerUserId="15801" LastEditorUserId="15801" LastEditDate="2018-04-22T02:33:46.750" LastActivityDate="2018-04-22T02:33:46.750" Title="Sudden increase in noise levels - subway noise -happend overnight" Tags="|structural-engineering|civil-engineering|rail|acoustics|" AnswerCount="2" CommentCount="4" ClosedDate="2018-04-21T18:34:42.733" ContentLicense="CC BY-SA 3.0" />
  <row Id="21403" PostTypeId="2" ParentId="21402" CreationDate="2018-04-20T16:46:38.267" Score="0" Body="&lt;p&gt;If you were to stand next to a long train going by at speed, you would likely notice that some cars are incredibly noisy while others are relatively quiet. the most likely answer to your question is that the railroad cars now in service on that portion of track are different from those previously used. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-04-20T16:46:38.267" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21404" PostTypeId="1" AcceptedAnswerId="21405" CreationDate="2018-04-20T17:25:53.933" Score="-1" ViewCount="242" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PXBOv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PXBOv.jpg&quot; alt=&quot;I don’t get how to get the answer to part e. Should we assume reactions at A and B in all three figures are same? If so, why? &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don’t get how to get the answer to part e. Should we assume reactions at A and B in all three figures are same? If so, why?&lt;/p&gt;&#xA;" OwnerUserId="15802" LastActivityDate="2018-04-21T16:13:45.530" Title="Engineering mechanics(statics)" Tags="|mechanical-engineering|structural-engineering|civil-engineering|statics|beam|" AnswerCount="2" CommentCount="0" ClosedDate="2018-05-11T11:43:32.903" ContentLicense="CC BY-SA 3.0" />
  <row Id="21405" PostTypeId="2" ParentId="21404" CreationDate="2018-04-20T18:53:00.413" Score="0" Body="&lt;p&gt;Despite this being an apparent homework question, there is merit in that the answer is not the OP's objective. The question posed appears to be about the form of the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the first sentence is a typographical error, as it is beam AC not beam AD, but that's not in consideration here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you read the problem correctly, you'll see that &quot;reactions&quot; at A and B are support pillars. It is valid to assume that A and B in all three figures are the same. This logic is based on (e) stating that 'a' is being replaced by 'b' in one figure and is being replaced by 'c' in the other figure. If A and B were not the same, there would be reference to that in (e).&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-04-20T18:53:00.413" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21406" PostTypeId="2" ParentId="21388" CreationDate="2018-04-21T00:07:50.583" Score="0" Body="&lt;p&gt;Go metric, man. Go metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v = \rho m $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $ v $ is volume (litre), $ \rho $ is the density and $ m $ is the mass (kg).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given time, $ t $ the formula could be re-written as $ \frac {v}{t} = \rho \frac {m}{t} $ but the 't' terms cancel out and the numerical answer is the same.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-04-21T00:07:50.583" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21407" PostTypeId="2" ParentId="20975" CreationDate="2018-04-21T00:10:52.990" Score="0" Body="&lt;p&gt;I suspect that these are stamped to a rough blank and then the ball and tapered stem is finished with  some sort of axial grinding process, probably by hand with the assistance of a jig. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ball ends can be machined fairly easily with a rotating jig and it would be much easier to form the ball in one piece with the stem then try to weld it on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the kink on the end if probably just bent near the end of the process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It actually looks like something whcih was prototypes on a lathe but the knurling and shoulder radius have a lack or radial symmetry whcih suggest a stamping. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-04-21T00:10:52.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21408" PostTypeId="1" CreationDate="2018-04-21T00:30:16.757" Score="2" ViewCount="797" Body="&lt;p&gt;I wrote a UEL for one element and compared to a reference UEL (FEM), where the results agreed satisfactorily, except the reaction force. The stress, strain, strain energy all were pretty accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to understand this, how does Abaqus calculate reaction force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am simply restraining the element in the bottom(two nodes) and applying displacement(incrementally) at the top(two nodes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, the STIFFNESS matrix is used to calculate reaction force in these cases, the stiffness values are quite similar in my UEL and the FEM UEL. Also if this was the case, the other parameters wouldn't have had the same values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is probably leading to convergence issues when I am using my UEL for more than one element, where the solutions cannot even converge for the first increment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would be grateful if anybody could help me figure this out.&lt;/p&gt;&#xA;" OwnerUserId="15803" LastActivityDate="2018-04-21T00:30:16.757" Title="How does Abaqus calculate reaction force?" Tags="|mechanical-engineering|applied-mechanics|finite-element-method|abaqus|numerical-methods|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21409" PostTypeId="2" ParentId="21380" CreationDate="2018-04-21T00:32:06.023" Score="0" Body="&lt;p&gt;Theoretically, assuming your pipe is completely full of gravel and fluid this &lt;a href=&quot;http://www.deq.idaho.gov/media/60177882/rpa-lesson-plan-1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.deq.idaho.gov/media/60177882/rpa-lesson-plan-1.pdf&lt;/a&gt; should provide the material you need. If it is not completely full of fluid, and is a sewer/channel with gravel in it, use the manning’s equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some answers a professional may give depending on the application:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Calibrate – You seem to know the answer you want, reverse engineer to find the friction co-efficient of the material you are using.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For drainage – Replace, augment or clean the pipe;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For filtering – Can’t change filter material, gravel is about as course as it gets, increase pipe diameter or apply an increased pressure – either through a pump, static head, or increased slope.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Practically, depending on your situation, do one of the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Seek professional advice; the implications of your answer can have real world consequences such as flooding either upstream or downstream.  Such risk requires monetary reward, if only to pay insurance. A “perfect” (always correct) engineer still faces a high risk of their design being incorrectly applied, and to prove you’re correct in a court is expensive. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you are a professional either tell the client you are not qualified, or seek the advice of a principal engineer in your company.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user15789" LastActivityDate="2018-04-21T00:32:06.023" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21410" PostTypeId="1" CreationDate="2018-04-21T02:44:20.600" Score="0" ViewCount="96" Body="&lt;p&gt;I have a portable band saw, which I plan to add lubricant fluid to (this is getting ahead of myself, as I haven't even constructed a portable band saw stand). The main problem with this and the reason why manufacturers do not advise using lubricant fluid is because the band saw blade might slip off the tires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that got me wondering what rubber and cutting fluid pairings would be compatible. I don't mean just chemical compatibility in which the rubber would NOT BE degraded by the fluid, but &lt;em&gt;fluid that will not cause rubber to slip&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I've been wondering... Why do portable band saws use a tire wheel (grips) rather than a train wheel (with ridges) type of wheel? &lt;/p&gt;&#xA;" OwnerUserId="1915" LastEditorUserId="1915" LastEditDate="2018-04-21T02:53:53.043" LastActivityDate="2024-03-23T00:08:10.983" Title="Lubricant fluid and Rubber compatibility" Tags="|mechanical-engineering|materials|cooling|tools|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21411" PostTypeId="2" ParentId="20964" CreationDate="2018-04-21T08:10:10.983" Score="1" Body="&lt;p&gt;There is an explicit formula to calculate the cross-sectional area $A$ of a segment of a circle, and therefore the volume $V$. of the liquid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ASK82.png&quot; height=&quot;500&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tank is filled to the height $x$ (measured from the bottom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross-sectional area is given by:&#xA;$$ A=\mbox{arccos}\left(\frac{R-x}{R}\right)R^2-(R-x)\sqrt{2Rx-x^2} $$&#xA;The volume $V$ is just:&#xA;$$ V=A\cdot L $$&#xA;Now, I assume you would probably rather measure $d$, the distance of liquid level from the top, thus you would have to replace $x$:&#xA;$$ x=2R-d $$&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2018-04-21T08:10:10.983" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21412" PostTypeId="2" ParentId="20964" CreationDate="2018-04-21T10:55:02.717" Score="1" Body="&lt;p&gt;This is completely standard tank measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, diesel is hardly a very flammable liquid. Second, electrical atmosphere in flammable installation is a well known thing. We have a standard for it - ATEX. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get a compliant pressure sensor, or float sensor, and mount on a flange on the outlet pipe, or a flange on the tank if available. &lt;a href=&quot;https://www.danfoss.com/en/products/sensors-and-transmitters/dcs/industrial-pressure-transmitters/mbs-3000-series/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Danfos MBS3000&lt;/a&gt; can for instance be used. Correctly installed this is perfectly safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be surprised if ultrasounds are not made with ATEX specification as well.&lt;/p&gt;&#xA;" OwnerUserId="12173" LastActivityDate="2018-04-21T10:55:02.717" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21413" PostTypeId="2" ParentId="21349" CreationDate="2018-04-21T11:03:57.367" Score="1" Body="&lt;p&gt;Modern VFDs can feed energy back to the grid when braking a motor. Some energy will be lost, but 70-90% may be recovered, depending on sizing and available loads.&lt;/p&gt;&#xA;&lt;p&gt;The alternative to feeding power back to grid is to burn it off in braking resistors. This is a common practice in applications where feeding it back to grid is not feasible.&lt;/p&gt;&#xA;&lt;p&gt;You could even do it without a VFD or active braking. The grid is rather stiff; it wants to stay at whatever frequency it is. If you power up a standard asynchronous three phase motor, it will rotate at the same frequency as the grid, minus some slip. If you however power the axle, so it want's to move faster than grid, the slip will become slightly positive.&lt;/p&gt;&#xA;&lt;p&gt;But as the grid provides a large inertia, it will not change the frequency of the grid. Hoists using normal three phase motors without further braking will go slightly faster down than up, but will hold load perfectly fine.&lt;/p&gt;&#xA;" OwnerUserId="12173" LastEditorUserId="12173" LastEditDate="2020-07-25T10:02:10.913" LastActivityDate="2020-07-25T10:02:10.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21414" PostTypeId="2" ParentId="21400" CreationDate="2018-04-21T13:01:21.330" Score="3" Body="&lt;p&gt;It appears that your dataset has integer row and column header values starting at 0. You can try the following code.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Turn on Excel's &lt;em&gt;Developer&lt;/em&gt; ribbon if it's not already on.&lt;/li&gt;&#xA;&lt;li&gt;Click the &lt;em&gt;Visual Basic&lt;/em&gt; button.&lt;/li&gt;&#xA;&lt;li&gt;Insert | Module.&lt;/li&gt;&#xA;&lt;li&gt;Paste the code below into the module.&lt;/li&gt;&#xA;&lt;li&gt;Adjust the rowOffset value to suit.&lt;/li&gt;&#xA;&lt;li&gt;Enter the formula =Interpol2D(y, x) into your spreadsheet where you want the interpolation result displayed.&lt;/li&gt;&#xA;&lt;li&gt;Save as .xlsm (with macro).&lt;/li&gt;&#xA;&lt;li&gt;Every time you change one of the referenced values the calculation will be run again.&lt;/li&gt;&#xA;&lt;li&gt;You can monitor the calculated values in the Visual Basic &lt;em&gt;Immediate&lt;/em&gt; window.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Code:&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Function Interpol2D(r, c)           'Row and column.&#xA;'By Transistor.&#xA;'https://engineering.stackexchange.com/questions/21400/double-interpolation-for-a-large-set-of-data-in-excel/21414#21414&#xA;&#xA;    rowOffset = 10                  'The row number for the header line of the table.&#xA;    r1 = Int(r + rowOffset)         'The table row number for the first parameter&#xA;    r2 = Int(r) + rowOffset + 1     'The next row.&#xA;    c1 = Int(c) + 2                 'The table column number for the second parameter&#xA;    c2 = Int(c) + 3                 'The next column.&#xA;&#xA;    'Layout of the four adjacent cells and the interpolated result.&#xA;    '    (r1, c1) *---p-------* (r1, c2)&#xA;    '                 |&#xA;    '                 |&#xA;    '                 r (result)&#xA;    '                 |&#xA;    '    (r2, c1) *---q-------* (r2, c2)&#xA;&#xA;    r1c1 = ActiveSheet.Cells(r1, c1)&#xA;    r2c1 = ActiveSheet.Cells(r2, c1)&#xA;    r1c2 = ActiveSheet.Cells(r1, c2)&#xA;    r2c2 = ActiveSheet.Cells(r2, c2)&#xA;&#xA;    'Interpolate by multiplying the difference between the two cells by the fractional&#xA;    'part of the parameter and then add in the first cell value.&#xA;    p = (r1c2 - r1c1) * (c - Int(c)) + r1c1    'The interpolation along the first horizontal.&#xA;    q = (r2c2 - r2c1) * (c - Int(c)) + r2c1    'The interpolation along the second horizontal.&#xA;    r = (q - p) * (r - Int(r)) + p             'The interpolation along the vertical.&#xA;&#xA;    'Results will be printed out in the Immediate window.        &#xA;    Debug.Print &quot;r1c1:&quot;, r1c1, p, r1c2, &quot;:r1c2&quot;&#xA;    Debug.Print &quot;r: &quot;, , r&#xA;    Debug.Print &quot;r2c1:&quot;, r2c1, q, r2c2, &quot;:r2c2&quot;&#xA;    Debug.Print &quot;================================================================================&quot;&#xA;&#xA;    Interpol2D = r                          'Return the result.&#xA;&#xA;End Function&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q9n2A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q9n2A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Test data.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Table 1. Debug window result.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r1c1:          12            15            18           :r1c2&#xA;r:                           16.5 &#xA;r2c1:          14            17.5          21           :r2c2&#xA;================================================================================&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2018-04-21T16:27:10.793" LastActivityDate="2018-04-21T16:27:10.793" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21416" PostTypeId="2" ParentId="21404" CreationDate="2018-04-21T16:13:45.530" Score="0" Body="&lt;p&gt;No, they are not the same. I suppose you should calculate distances b and c for same condition as you calculated a in question d) and compare results.&lt;/p&gt;&#xA;" OwnerUserId="13748" LastActivityDate="2018-04-21T16:13:45.530" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21418" PostTypeId="2" ParentId="21402" CreationDate="2018-04-21T17:38:34.763" Score="1" Body="&lt;p&gt;It is possible that because of percolation, the moisture content of the layer of dirt between you and subway has changed to near optimal compaction, which is in the range of 12 to 24 % in siltyclay, assuming your strata is CL, ML, SC. These are sensitive to moisture ratio and at optimal compaction transmit sound better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some change in moisture is seasonal, therefor it could go back to initial condition. But if a main sewer or water line has cracked, there is room for concern.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-04-21T17:38:34.763" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21419" PostTypeId="1" AcceptedAnswerId="21421" CreationDate="2018-04-21T20:39:41.530" Score="0" ViewCount="51" Body="&lt;p&gt;I choose an AC motor (IEC) with frame number 71. The key on the shaft of the motor is $5mm,$, $4.23 mm$ and $6 mm$ in diameter. At other hand i cant't find a safety coupling (torque limiter) with exactly the same dimensions as the key way on the  motor. They are all a bit smaller or bigger but they have the same  diameter as the motor. for instance $4 mm$, $2.23 mm$ and $6 mm$ in diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what can i do ? Is it possible to use a slightly greater safety coupling or the dimensions of the both, should exactly be the same ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o8DrO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o8DrO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-04-21T21:10:09.537" LastActivityDate="2018-04-21T22:44:12.257" Title="Motor coupling connection" Tags="|motors|torque|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21420" PostTypeId="2" ParentId="19603" CreationDate="2018-04-21T21:18:32.740" Score="1" Body="&lt;p&gt;Have 2 color dyes with a 0 reactivity like yellow and blue. Once you have ideal diffusion it should appear green. Liquid steel a type of glue that you have to kneed to activate and are composed of 2 halves of white and grey. When mixed properly there would be no visible traces of either color but a solid light grey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When dealing with transparent liquids a color chart of can be made to cross reference if you need more of one or the other fluid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consistency of the dye is important but can be tested before use by measuring the weight of the dye and how much it changes the liquid's color.  &lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-04-21T21:18:32.740" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21421" PostTypeId="2" ParentId="21419" CreationDate="2018-04-21T22:32:23.140" Score="0" Body="&lt;p&gt;This extract from a &lt;a href=&quot;http://www.electricmotorservice.net/iechart.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Baldor IE Chart&lt;/a&gt; may help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nnhwt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nnhwt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. The Frame 71 keyway dimensions are highlighted.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As everyone else has pointed out, keyways and keys are rectangular so we can't figure out where all your diameters are coming from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your drawing is a little strange as it seems to have a shaft entry with keyway rather than a protruding shaft.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it possible to use a slightly greater safety coupling or the dimensions of the both, should exactly be the same?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The keyway should be a snug fit. Any play or slop will result in backlash during starting and stopping and this will chew up the key or shaft or both.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-04-21T22:32:23.140" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21422" PostTypeId="2" ParentId="21419" CreationDate="2018-04-21T22:44:12.257" Score="0" Body="&lt;p&gt;Once you find a coupling that fits the shaft diameter and the other constraints such as size etc, then all you need to do is to get a machine shop to cut a matching keyway to match the one in the motor shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They will also machine a key to match f necessary. This is a small easy job...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-04-21T22:44:12.257" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21423" PostTypeId="2" ParentId="21349" CreationDate="2018-04-21T23:01:14.143" Score="1" Body="&lt;p&gt;Further to vidarlo's answer, these lifts will be controlled by some sort of variable frequency drives (VFD). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a typical VFD deceleration causes regeneration and current is fed back onto the DC bus. This will cause the bus voltage to rise and at some point the voltage will reach the safe bus limit and the energy will have to be burnt off in a braking resistor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where multiple motors are accelerating, running and decelerating asynchronously it is common to use a common DC bus for a group of drives. This is relatively simple compared with feeding back into the grid which requires an inverter built into the drive. The idea is that the regenerated power is consumed by the driving motors and the overall effect is to reduce overall power consumption.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It also has to do it without risk of runaway/freefall. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct. This implies the use of brakes or motors with zero-speed holding torque. The least-wear / least power option would be to decelerate to zero speed and then apply the mechanical brake. The safety brake is, presumably, a mechanical centrifugal or standard lift / elevator brake.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And, it shouldn't just burn off all that energy as heat (terribly inefficient). So my question is, how is it done?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think we've covered that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks for the video link. It all looks horrendously complicated with a lot of maintenance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to think of an application for the system. A couple come to mind:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;With a simple two-car / two-shaft system one could have the cars travelling in a loop. The left door is always going up. The problem is that there is some lost time while the car traverses horizontally.&lt;/li&gt;&#xA;&lt;li&gt;It may be possible, as I think is shown in the video, to have multiple cars in each shaft. If the shafts operate as one-way streets with linking cross-roads every so-many floors it seems that a much higher capacity system could be created for a given pair of lift / elevator shafts. In a hundred-story building each lift shaft saved could create another 100 rooms. The economics then start to look rather different!&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-04-21T23:01:14.143" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21425" PostTypeId="1" CreationDate="2018-04-22T09:34:49.410" Score="0" ViewCount="46" Body="&lt;p&gt;I used a 2N3055 powerTrans with a collector transformer (6-220v)load with an audio input base bias from a  audio source , I am getting 160vAC from 6v DC , i want 220 v AC out , Can anyone suggest a solution&lt;/p&gt;&#xA;" OwnerUserId="15500" LastActivityDate="2018-09-15T05:25:47.557" Title="Electrical Engineering" Tags="|power-electronics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21427" PostTypeId="1" AcceptedAnswerId="21432" CreationDate="2018-04-22T21:50:33.260" Score="1" ViewCount="713" Body="&lt;p&gt;Back in the early 60s, Chrysler built a &lt;a href=&quot;https://en.wikipedia.org/wiki/Chrysler_Turbine_Car&quot; rel=&quot;nofollow noreferrer&quot;&gt;turbine car&lt;/a&gt; powered by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Chrysler_turbine_engines&quot; rel=&quot;nofollow noreferrer&quot;&gt;turbine engine&lt;/a&gt; it had under development.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting feature of the engine was a regenerator system to recover heat from the gases leaving the power turbine, using this heat to preheat the air on its way from the compressor to the combustor. The benefits of this were to improve fuel efficiency and reduce the final exit temperature of the exhaust. The regenerators involved some sort of rotating disks which somehow transferred the heat as they rotated. There are diagrams/animations depicting the gas flow through the engine and (generally) heat path through the regenerators, but in some ways they don't seem to make much sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LBwz6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LBwz6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: how did the regenerator setup actually work? How was the compressor output air (60 psi?) put into contact with a rotating(?) hot surface without the pressure leaking air away to the exhaust? Why would a regenerator need moving parts at all? Could it not just be a system of tubes within tubes?&lt;/p&gt;&#xA;" OwnerUserId="15169" LastActivityDate="2018-04-23T09:25:13.047" Title="Why regenerator &quot;wheels&quot; in a turbine engine?" Tags="|energy-efficiency|turbomachinery|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21431" PostTypeId="1" CreationDate="2018-04-23T08:13:58.880" Score="0" ViewCount="50" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/77GlX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/77GlX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) What is the area of the steel plate interface (in $m^2$) with the fluid conduit if the heat enters the steel plate at a temperature (T1) of 980K? Assume rest of the conduit surface (besides the steel plate interface) is well insulated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) What is the Carnot efficiency of the heat engine (in %) based on the heat entering the engine at T2 if the same amount heat is input to the heat engine? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Attempt&lt;/strong&gt;&#xA;I'm a bit confused as to how to approach both (a) and (b). Since this is a heat engine, the $\Delta U = 0$ so $Q_{in} - Q_{out} = W$. So is the correct way of approaching (a) is to conclude that $Q_{in}$ at $T_2$ is $500kW$ and then utilizing the fact that $500kW = h(A)(1000K - 980K)$ to solve for $A$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that reasoning, the Carnot efficiency for (b) would be 20%.&lt;/p&gt;&#xA;" OwnerUserId="15823" LastActivityDate="2018-04-23T22:02:26.990" Title="Area of Heat Transfer and Carnot Efficiency" Tags="|mechanical-engineering|heat-transfer|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21432" PostTypeId="2" ParentId="21427" CreationDate="2018-04-23T09:25:13.047" Score="3" Body="&lt;p&gt;The main advantage of a rotary regenerator is that you can pack a lot more surface area into a  given volume than with a shell and tube type device important for gas-gas heat exchange where heat capacities and heat transfer coefficients tend to be low compared to liquid-liquid. This obviously has advantages for packaging a turbine in a car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially they have a fine matrix of eg wire mesh which gets heated as it passed through the hot stream and cooled as it passes through the colds stream so you are basically heating up one part and moving that hot part to the cold side. A good analogy is if you have a bucket of hit water and a bucket of cold water you stick a copper rod in the hot bucket untill it gets hot and then take it out and put in in the cold bucket and repeat. A rotary regenerator is the same principal but continuous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You do get some cross contamination but in a turbine where you are just mixing intake with exhaust this doesn't matter much and isn't that significant in terms of pressure loss and there is no direct path from one side  to the other just the inherent limitations of rotary seals &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-04-23T09:25:13.047" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21433" PostTypeId="2" ParentId="21410" CreationDate="2018-04-23T09:41:07.417" Score="0" Body="&lt;p&gt;The problem is that, by definition, a lubricant is going to reduce friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some bandsaws do use steel or aluminium drive wheels but even then a flange doesn't remove the need for the blade to grip the wheel as it still needs to provide drive and also the blade still needs to be tracked accurately, relying on a flange will put a side load on the blade whcih could cause it to twist and will wear both the blade and wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with trains the flanges aren't the main thing which keeps the wheel on the track. The wheels stay on the track because they have cone section (indeed tracking wheels on belt drives etc are crowned for the same reason) the flanges are just a backstop in case the normal alignment mechanism gets pushed too far. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a portable band saw the tracking and blade tensioning system probably isn't that sophisticated  and thus more sensitive to sipping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Industrial band saws can and do just fuid cooling and blade lubrication stems but they are designed to do so. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-04-23T09:41:07.417" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21435" PostTypeId="1" CreationDate="2018-04-23T10:53:53.437" Score="0" ViewCount="504" Body="&lt;p&gt;At times, the Swagelok SS Ferrules gets crimped tight into the SS pipe. In those cases, can these ferrules be forcefully removed and reused?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or was this a result of faulty installation?&lt;/p&gt;&#xA;" OwnerUserId="14771" LastEditorUserId="14771" LastEditDate="2018-04-23T12:02:13.463" LastActivityDate="2018-04-23T12:02:13.463" Title="Can Swagelok SS ferrule be reused?" Tags="|hydraulics|pipelines|connections|" AnswerCount="0" CommentCount="11" ContentLicense="CC BY-SA 3.0" />
  <row Id="21436" PostTypeId="2" ParentId="14143" CreationDate="2018-04-23T13:58:50.037" Score="1" Body="&lt;p&gt;&lt;strong&gt;Yes, it is possible to consider Pade approximants for delays in linear state space systems.&lt;/strong&gt; The simplest approach is to construct the transfer function for the delay and convert this to a state space model. As an example, the second order Pade approximation for a time delay of $T$ is&#xA;$$&#xA;H_{\text{delay}}(s) = \frac{s^2-6/T s+12/T^2}{s^2+6/Ts+12/T^2}&#xA;$$&#xA;which corresponds to a state space system with matrices&#xA;$$&#xA;A_{\text{delay}} = \left[ \begin{array}{cc} 0 &amp;amp; -12/T \\ 1 &amp;amp; -6/T \end{array} \right], \quad&#xA;B_{\text{delay}} = \left[ \begin{array}{c} 1 \\ 0 \end{array} \right], \quad&#xA;C_{\text{delay}} = \left[ \begin{array}{c} -12/T &amp;amp; 72/T^2 \end{array} \right]&#xA;$$&#xA;where I have used the companion state space form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Supposing we want to apply this delay at the input to another linear state space system with matrices $A_{\text{sys}}$, $B_{\text{sys}}$, $C_{\text{sys}}$, $D_{\text{sys}}$, we can achieve this by forming the larger system with matrices:&#xA;$$&#xA;A = \left[ \begin{array}{cc} A_\text{delay} &amp;amp; 0 \\ B_\text{sys} &amp;amp; A_\text{sys} \end{array} \right], \quad&#xA;B = \left[ \begin{array}{c} B_{\text{delay}} \\ 0 \end{array} \right], \quad&#xA;C = \left[ \begin{array}{c} D_\text{sys} &amp;amp; C_\text{sys} \end{array} \right], \quad&#xA;D = 0&#xA;$$&#xA;In practice, most of this can be carried out easily in Matlab by using the pade() and ss() functions.&lt;/p&gt;&#xA;" OwnerUserId="15784" LastActivityDate="2018-04-23T13:58:50.037" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21437" PostTypeId="1" CreationDate="2018-04-23T17:06:13.033" Score="2" ViewCount="4978" Body="&lt;p&gt;I used a bucket to find out that at 1 feet high (a garden faucet) the flow rate from an half inch pipe is 1 gallon in 26 seconds&#xA;At 3 feet high (face basin) the flow rate is 1 gallon in 30 seconds&#xA;at 11 feet high (upstairs face basin) the flow rate is 1 gallon in 1 minute&#xA;at 15 feet high (upstairs showerhead) the flow rate is 1 gallon in 2 minutes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphing the data I see that this flow won't make it to the 23 feet I need it to fill a rooftop water tank. Is there an equation I can use to estimate the water flow I would need, measurable from the one foot pipe (It changes daily from the water company these days, some days it is more than strong enough) to reach 23 feet? I want to know if i'll have to just save money for a pump after all, or a bigger tank. Keep in mind I'm not an engineer, tho I did try to solve myself this with a natural number e or ln in there somewhere lol. It's been much too long since college math.    &lt;/p&gt;&#xA;" OwnerUserId="15826" LastActivityDate="2021-12-04T17:04:35.437" Title="I want to know the water flow rate from a certain height knowing the flow rate at near the ground level" Tags="|dynamics|fluid|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21438" PostTypeId="1" CreationDate="2018-04-23T17:40:38.337" Score="1" ViewCount="38" Body="&lt;p&gt;I don´t know if this is the rifght place to ask this question, but I would like to know if after the remediation of a soil contaminated with hydrocarbon you can reuse or even sell the oil recovered. And if it's possible, do you know any companie that can do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help and sorry for my bad english.&lt;/p&gt;&#xA;" OwnerUserId="15829" LastActivityDate="2018-05-23T20:15:29.610" Title="Is it possible to reuse or the oil recovered from a spill?" Tags="|soil|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21439" PostTypeId="2" ParentId="21438" CreationDate="2018-04-23T19:27:10.880" Score="1" Body="&lt;p&gt;Assuming you mean petroleum; crude and products are collected as much as possible , and taken to the refinery for processing. In much of the world , what can not be collected is burned in situ, this was formerly done in the US also. Today, in the US , regulations create a problem and contaminated soil is dug up and hauled someplace for cleaning . The burned soil is sterile but recovers - builds up organic matter- in a year +/-. Of course there are many degrees of variation. For example , I understand there pools of oil laying on the ground in Russia where the logistics of moving equipment into the site are difficult/expensive.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-04-23T19:27:10.880" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21440" PostTypeId="1" AcceptedAnswerId="21443" CreationDate="2018-04-23T20:21:04.713" Score="0" ViewCount="310" Body="&lt;p&gt;My machine design textbook states that the fulcrum pin of a band brake during braking is subjected to a double shear. How do you say it is subjected to a double shear and not to a single one. &#xA;By double shear the textbook means:&lt;br&gt;&#xA;  Double shear stress = 2 * shear stress&lt;/p&gt;&#xA;" OwnerUserId="14355" LastEditorUserId="14355" LastEditDate="2018-04-25T15:20:27.017" LastActivityDate="2018-04-25T15:20:27.017" Title="Why does the fulcrum pin of a band brake experience &quot;double&quot; shear stress" Tags="|mechanical-engineering|stresses|mechanical|shear|" AnswerCount="1" CommentCount="4" ClosedDate="2018-04-26T20:51:08.253" ContentLicense="CC BY-SA 3.0" />
  <row Id="21443" PostTypeId="2" ParentId="21440" CreationDate="2018-04-23T21:29:31.513" Score="1" Body="&lt;p&gt;From the image you should see that the pin on the left has two &quot;lines&quot; of shear stress while the one on the right only one, bending is of course an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KJ2nG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KJ2nG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-04-23T21:29:31.513" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21444" PostTypeId="2" ParentId="21437" CreationDate="2018-04-23T21:34:40.187" Score="0" Body="&lt;p&gt;The flow rate at a particular height would depend on the pressure and flow rate at the supply end. You may use the simple formula (v^2 /2g) + h = constant (assuming supply pressure to be constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v=flow rate/ area of cross section of the pipe.(use SI units)&#xA;g= 9.8m/s^2&#xA;h= height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First find the value of the relation at h=1 ft and find the constant. Then equate this value to the relation with h=23ft and find v.&#xA;Then flow rate= area of cross section of pipe× v&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B: Since the supply pressure is variable, as you have mentioned you may take the average of the value calculated on every day in a week.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2018-04-23T21:34:40.187" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21445" PostTypeId="2" ParentId="21431" CreationDate="2018-04-23T22:02:26.990" Score="1" Body="&lt;p&gt;I think your thought on part a) is correct. &#xA;But for part b), if we assume the plate to be of uniform temperature, then the efficiency may turn out to be 1- (300/980)= 69.4%.&#xA;If we consider the finite conductivity of steel, then also I dont think the given values could be obtained.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2018-04-23T22:02:26.990" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21446" PostTypeId="1" CreationDate="2018-04-23T22:04:57.610" Score="0" ViewCount="351" Body="&lt;p&gt;How can the DOM MILD Steel ASTM A513 Type 5 Tube be improved for corrosion resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it has a low but good welding and machinery ability but if we want to improved the material for corrosion resistance what element will we apply to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can we reduce carbon added to make it excellent welding ability and add more magnesium to make more corrosion resistance.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or we can use Application of zinc coating – steel galvanization it’s a process which creates a coating of zinc over metals surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;need more formal answer on this subject.&lt;/p&gt;&#xA;" OwnerUserId="12830" LastEditorUserId="12830" LastEditDate="2018-04-24T03:21:10.537" LastActivityDate="2018-04-24T12:47:55.220" Title="DOM MILD Steel ASTM A513 Type 5 Tube" Tags="|mechanical-engineering|materials|steel|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21448" PostTypeId="2" ParentId="15798" CreationDate="2018-04-24T08:30:41.470" Score="0" Body="&lt;p&gt;I agree with Craig below, in addition though, my understanding is, old refrigerators will accumulate some wear sludge in the bottom of the compressor - this is what will permanently stop a fridge from working.&lt;/p&gt;&#xA;" OwnerUserId="15833" LastEditorUserId="15833" LastEditDate="2018-04-24T08:40:01.257" LastActivityDate="2018-04-24T08:40:01.257" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21449" PostTypeId="1" AcceptedAnswerId="21455" CreationDate="2018-04-24T08:59:45.503" Score="0" ViewCount="43" Body="&lt;p&gt;When analyzing machines, how to tell if a point that was attached to another part will have only an $x$ or $y$ component or both?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PCrDy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PCrDy.png&quot; alt=&quot;here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;why wouldn't point $B$, $C$ and $A$ have an $x$ component? &lt;/p&gt;&#xA;" OwnerUserId="15099" LastActivityDate="2018-04-24T22:19:09.023" Title="Machines components?" Tags="|structural-analysis|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21451" PostTypeId="1" CreationDate="2018-04-24T06:02:13.110" Score="1" ViewCount="694" Body="&lt;p&gt;I am working on disturbance rejection technique in a vehicle of a wheel. Suppose our vehicle moves with a constant speed 15 km/h. Since roads are uneven in nature, suddenly vehicle got a bump or jerk. Is there any method or relationship to find out the force or torque applied by the jerk or bump?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume mass of a vehicle is M (kg) and radius of wheel is r (cm).&lt;/p&gt;&#xA;" OwnerUserId="15908" OwnerDisplayName="aman2909" LastActivityDate="2019-04-22T02:02:25.820" Title="How much force/torque act on a wheel of a vehicle when vehicle moves a constant speed and suddenly got a bump or jerk?" Tags="|design|mechanical|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 3.0" />
  <row Id="21452" PostTypeId="2" ParentId="21451" CreationDate="2018-04-24T08:33:05.290" Score="1" Body="&lt;p&gt;Not at all trivial problem! A vehicle is not rigid. More or less elastic wheels are the the first defence line against hits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More complexities:&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;The vehicle does not fall down vertically, it flies horizontally at the same time. This makes difficult to calculate how the wheels meet the next obstacle&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Different wheels receive different hits. The vehicle does not fall and rise, it also rotates. It's in 3D.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You need a realistic model of the road&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Rolling wheel (assuming it does not fly nor modify the road) meet the road profile. Solving the route of the wheel axis is a numerical equation solving problem. In practice the road can be smooth and the vehicle sometimes flies.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As a coarse approximation assume at first some road profile. Assume the wheel rises as high as is needed to prevent the wheel to sink into the road (=solve the route of the axis at first). If your wheel moves horizontally (= to X)  15km/h, present Y as function of the time. The second derivative is the acceleration input to your wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wheel elasticity, vehicle mass and frictional losses make a low pass filter. They make the effective hit input smoother. The second derivative should be calculated from the filtered signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculations like this are coarse. Make some measurements or use proper physics simulation software. Building the simulation model is a major task and the needed software can cost $10k. Making acceleration measurements give to you realistic info much easier.&lt;/p&gt;&#xA;" OwnerDisplayName="user33233" LastActivityDate="2018-04-24T08:46:16.300" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21453" PostTypeId="2" ParentId="21446" CreationDate="2018-04-24T12:47:55.220" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;Yellow Chromating over zinc for white rust from moisture    &lt;/li&gt;&#xA;&lt;li&gt;High Performance Passivation for red rust from salt spray        &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8847" LastActivityDate="2018-04-24T12:47:55.220" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21454" PostTypeId="1" CreationDate="2018-04-24T14:16:56.043" Score="0" ViewCount="59" Body="&lt;p&gt;I'm developing a machine with a reciprocating motion (mass: 50-100 kg, period: 1.35 s, accelerations: 1 g). This machine is part of a process line, and I don't want the resulting &quot;shaking&quot; to interfere with other machines. How do I go about doing this? A very stiff frame is an option, but a more elegant solution is preferable. Basically I'm building a frame around the reciprocating motion, and I have to connect that frame to the process line.&lt;/p&gt;&#xA;" OwnerUserId="15840" LastEditorUserId="11869" LastEditDate="2018-04-27T12:01:00.870" LastActivityDate="2018-05-27T22:44:20.913" Title="Isolating vibrations originating from my machine" Tags="|mechanical-engineering|structural-engineering|stresses|dynamics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21455" PostTypeId="2" ParentId="21449" CreationDate="2018-04-24T16:26:32.103" Score="2" Body="&lt;p&gt;Whether a reaction has both x and y components is going to depend on the geometry of the machine components and the orientation of the applied load. In the machine example you provided it appears all the components are pin connected, which means they &lt;em&gt;potentially&lt;/em&gt; have both x and y components. With experience, you may be able to determine by inspection if one of these components is zero. In the meantime, I find it easiest to simply assume nonzero components for both the x and y direction. Solving the equations of equilibrium will generally reveal if one of the components is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, looking at the machine you provided, we can determine by inspection that there is no x reaction at A or D because for link A-D to be in equilibrium there cannot be forces in the x direction. F_Ax and F_Dx would need to be equal and opposite to satisfy horizontal force equilibrium and that would produce a moment on the link (not allowed by equilibrium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With no x reaction at D, then for equilibrium of the bottom jaw, there can be no x reaction at E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the lower handle, with no x reaction at E and only vertical load applied, there can be no x reaction at C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the upper handle, with no x reaction at C and only vertical load applied, there can be no x reaction at B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In equilibrium we trust.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The image below shows an example of y-equilibrium. (I didn’t actually work out magnitudes and directions...just the principle of how forces might flow.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/of6sRl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/of6sRl.png&quot; alt=&quot;Assuming y-reactions only&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking just at the upper handle it’s possible to imagine that component remaining in equilibrium if we assumed x reactions at B and C (equal and opposite). However, when we follow those reactions throughout the machine it leads to a link AD that is not in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S4HMGl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S4HMGl.png&quot; alt=&quot;Assuming x-reactions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we have to assume no x reactions at link AD. When we follow that assumption throughout the machine we end up with no x reactions at all.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2018-04-24T22:19:09.023" LastActivityDate="2018-04-24T22:19:09.023" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21456" PostTypeId="1" AcceptedAnswerId="21461" CreationDate="2018-04-24T16:37:16.583" Score="4" ViewCount="214" Body="&lt;p&gt;I’ve some small U groove rollers I’m having made, 60mm diameter, 60mm wide, 28mm diameter at ceter of U profile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G4OoA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G4OoA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rollers need to be silent and have some friction on the rails. Polyurethane seems the most suitable material as it’s the highest load capacity to friction I can find but I’m open to suggestion if anyone can think of something else? I want to maximise how much weight they can take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can not harden the polyurethane much as they need to be soft for the friction and to run silent. Also I can’t thicken the diameter of the rollers though I can make the hub smaller to thicken the PU a little. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do polyurethane wheels fail under pressure? I’m presuming that the PU is displaced and shears of the bonded surface eventually splitting in the centre as in drawing A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I though of putting support on ether side as in picture B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I changed the profile of the U groove to have a dip in the centre. The groove should redistribute the pressure away from the centre and into the sides, acting against the side support and on the thickest part of the PU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1l0mF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1l0mF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was also looking a whether different shaped larger hubs and thinner PU or smaller hubs and thicker PU would be better under strain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l9ysw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l9ysw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally I was looking at the possibility of support within the PU to increase it’s load capacity and make it less likely to deform, something like a 3D printed Nylon mesh reinforcement. I’m going to talk to the company who makes the wheels tomorrow but I suspect that making a PU mould with a mesh inside will have problems with getting rid of bubbles and cavities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k1St9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k1St9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there is anyone out there with any experience or opinion on how PU behaves under stress, any advise how these wheels fail? Are there any other methods of increasing the load capacity of PU that you may know of? Or is there another material that may be more suitable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for taking the time Chris, I can’t see a button to answer you apart from the commet one which has a limit to the letters? So I did it here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight capacity is ok but it’s right there on the edge of ok. Unfortunately I can’t add rollers or increase the outside turning diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need no slippage, so I need some friction and I need a little give so as not to translate tiny imperfections up the shaft and into the mechanism. 85A hardness PU would do the job, maybe 90A, I don’t know, I’m waiting on harder samples to try them out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue of the wedging effect you brought up is something that has been on my mind, I’ll have a look at the conical idea though I’m hoping that by having the PU bonded to the two side supports will go a long way towards mitigating any wedging effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PU can move and flex but can it be compressed? I’m unsure but I suspect that it can not or of it can its only by a tiny amount. So the failure point of the roller is down to the displacement only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By bonding the PU to the hub and also the two side supports I hoped to have the weight acting on the PU where it has the two bonded surfaces to reduce the displacement that can happen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My natural reaction has been to increase the thickness of the PU to increase it’s load capacity but the more PU there is and the further it is from the bonded surface surely that means the more it can be displaced? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind I was thinking that it may be better to make the hub the same shape as the roller and have only a thin layer of PU as in the 3rd drawing down. This would increases the diameter of the hub from 16mm to 25mm at the thinnest point. Do you think the thick hub with the thin layer of PU would be better or the thinner hub with as much PU as possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there another materiality that would be more suitable? I was trying to think of something like layers of leather or rubber or PU coated fabric or something like this but couldn't think of anything.&lt;/p&gt;&#xA;" OwnerUserId="8597" LastEditorUserId="8597" LastEditDate="2018-04-25T13:03:18.150" LastActivityDate="2018-05-25T13:22:11.860" Title="Looking for advice on design of some rollers" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21457" PostTypeId="1" CreationDate="2018-04-24T22:42:59.053" Score="1" ViewCount="471" Body="&lt;p&gt;I work in Abaqus 2017 a lot. I have noticed that when I submit any abaqus job (that does not contain any subroutine!) via parallel processing (aka cpus=4 or 8), it runs just fine! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, when i submit an abaqus job that contains a user subroutine with it via parallel processing, it won't run at all. It will not abort, or terminate, or do anything. The job will freeze. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to run an abaqus job that has a user subroutine with it, is to run the job with single processing. This takes up a lot of time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do use the allocatable arrays for the job, and define them in my subroutine. Is there something I should include in my subroutine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please tell me why is it that the abaqus jobs (with user subroutines) get completely stuck with cpus of 4 and higher? Please? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!!!&lt;/p&gt;&#xA;" OwnerUserId="15597" LastActivityDate="2018-04-24T22:42:59.053" Title="If I want to run an abaqus job (in Linux redhat) with user subroutine on parallel processors, how do I do that?" Tags="|abaqus|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 3.0" />
  <row Id="21459" PostTypeId="1" AcceptedAnswerId="21460" CreationDate="2018-04-25T05:07:12.603" Score="0" ViewCount="193" Body="&lt;p&gt;Given a car configuration with a length of L and width of W. If you can give the top left wheel an angle of A to turn A degree how do you derive the angle of top right wheel for turning . I know that the bottom left angle will just be complimentary of the top left wheel and bottom right wheel is complimentary of top right wheel. This is a swerve drive system so i can control each wheel angle. &lt;/p&gt;&#xA;" OwnerUserId="15850" LastEditorUserId="15850" LastEditDate="2018-04-25T05:15:10.020" LastActivityDate="2018-04-25T09:43:18.673" Title="Wheel angle for turning" Tags="|mechanical-engineering|control-engineering|design|mechanisms|car|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21460" PostTypeId="2" ParentId="21459" CreationDate="2018-04-25T09:43:18.673" Score="0" Body="&lt;p&gt;Most road vehicle steering is based on &lt;a href=&quot;https://en.wikipedia.org/wiki/Ackermann_steering_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ackerman geometry&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put simply the plane of each steered wheel should be perpendicular to a straight line from the centre of rotation the wheel to the centre of the tuning circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally this is achieved by a mechanical linkage where the steering arms point along a line from the wheel pivot point to the centre of the rear axle. With independent steered wheels you need to translate that geometry to whatever control system you are using. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice there are several factors which might make you deviate from pure Ackerman steering, suspension often doesn't  travel in a perfectly vertical line and you may want non-zero camper, castor and toe-in in the suspension geometry for various reasons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also pneumatic tyres tend to operate with a small slip angle on turning as a result of the finite stiffness of the side wall of the tyre, which is how the turning force is generated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modeling and analysing this in detail get very complex very quickly and there is no single 'correct' solution as it will depend on a whole variety if interrelated factors and will often end up being some compromise. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-04-25T09:43:18.673" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21461" PostTypeId="2" ParentId="21456" CreationDate="2018-04-25T10:18:06.330" Score="0" Body="&lt;p&gt;In of the first issues you will get is that the area in contact with the rail will flatten, increasing the rolling resistance. In practice this is what tends to limit that capacity and they will become unusable deformed before they actually fail as such. So its really the Youngs Modulus rather than the tensile strength which is the issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most obvious way to improve the capacity is to increase the diameter. If you look at industrial castors the main difference between different capacities is the wheel diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that isn't possible then the next solution is just to use more rollers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also an issue that concave rollers on rails don't actually work that well, especially under moderate load as the shoulders tend to rub on the rails and wedge  you are probably much better off with a conic section like train wheels or using a set of guide wheels like a roller-coaster. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-04-25T10:18:06.330" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21464" PostTypeId="2" ParentId="16344" CreationDate="2018-04-25T13:52:44.247" Score="2" Body="&lt;p&gt;I think your question actually might be missing a key difference between the two lining systems you mention. Your top image shows a segmental lining. This lining is constructed using the stop and start manner that you suggest. There are variations on a theme to this but for a typical segmental lining the machine boring the tunnel (TBM) pushes off the leading edge of a completed ring to create the driving force to push the machine into the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am understanding your spiral proposal correctly then this is a different approach to lining, doing something similar to slip forming behind the lining. There are linings that are built in a similar manner to this but you have to get the driving force to shove the machine forward from a different source. A &lt;a href=&quot;https://www.herrenknecht.com/en/products/core-products/tunnelling/gripper-tbm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;gripper TBM&lt;/a&gt; would give you an indication of the type of machine you might address this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering a spiral cast in-situ lining vs. a segmental lining to resist earthquakes typically (but not always) I'd suggest the segments are better because they have additional flexibility due to the joints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For circular vs., horseshoe shaped linings, again it depends, but typically the circular lining will be stronger. &lt;/p&gt;&#xA;" OwnerUserId="15855" LastActivityDate="2018-04-25T13:52:44.247" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21465" PostTypeId="2" ParentId="16414" CreationDate="2018-04-25T13:54:56.810" Score="0" Body="&lt;p&gt;Within the tunnelling environment I work in we refer to that sort of image as a 'developed elevation'. &lt;/p&gt;&#xA;" OwnerUserId="15855" LastActivityDate="2018-04-25T13:54:56.810" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21466" PostTypeId="1" AcceptedAnswerId="21473" CreationDate="2018-04-25T15:16:48.157" Score="0" ViewCount="1543" Body="&lt;p&gt;Consider a spring like this (this one CoC hand gripper):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aWDfK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aWDfK.jpg&quot; alt=&quot;CoC hand gripper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They claim on their web site, that the spring will not degrade during time:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Do your grippers get weaker with time?&lt;/strong&gt;&#xA;  No, so don’t worry if your Captains of Crush No. 1 gripper is feeling easier to you now—it’s because you’re getting stronger! What is mistakenly called “seasoning” by some people is actually a weakening with use, and it’s a reflection of an under-designed gripper with a spring that is bending, which is why it’s not uncommon for low-quality grippers to get narrower and easier as they are used. On the other hand, Captains of Crush grippers hold their line for a lifetime of steady training.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is also a lot of discussion on this topic, referencing some re-measurements, but none of that seems really trustable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My questions, from engineering point of view:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is that possible/probable? I mean, can the degradation be small enough (for the given brand) so the user should not care? How likely it is true?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is possible what they say about the &quot;low-quality grippers&quot; that they degrade over time significantly. How probable is that?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I can imagine that both options (degradable and also stable springs) are possible. But how probable it is. Note that such a product cost usally $10-25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is kind of opinion based question, however the experiment would take few years, so any theory and/or experience based answer can help.&lt;/p&gt;&#xA;" OwnerUserId="15210" LastActivityDate="2019-12-14T22:14:38.870" Title="Spring degradation in hand grip trainer" Tags="|steel|springs|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21467" PostTypeId="1" CreationDate="2018-04-25T15:30:02.493" Score="2" ViewCount="461" Body="&lt;p&gt;I'm trying to find the formula for calculating the tensile stress area of a bolt. According to page 1528 of &lt;em&gt;The Machinery's Handbook (29th Ed.)&lt;/em&gt;, that formula is given as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A_S=\frac{\pi}{4}\left(\frac{d_m+d_p}{2}\right)^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the pitch diameter is $d_p=d-0.649519\times P$, and the minor diameter is $d_m=d-1.299038\times P$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was trying to understand where the definitions of $d_p$ and $d_m$ come from. The Wikipedia page for the Unified Thread Standard (&lt;a href=&quot;https://en.wikipedia.org/wiki/Unified_Thread_Standard&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;) was very helpful for understanding that. However, it is not clear where the equation for $d_m$ comes from. The value 1.299038 is equivalent to $\frac{3\sqrt{3}}{4}$. Translating this to be in terms of $H$ instead of $P$ results in $d_m=d-2\cdot\frac{3}{4}\cdot H$. In other words, it would appear that the boxed dimension in the drawing below (versions of which appear on pages 1807, 1912, and 1973 of the handbook mentioned above) should be $.75H$ instead of $.625H$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W35xi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W35xi.png&quot; alt=&quot;Basic Profile of Screw Thread&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming I must be missing something obvious here because I'm seeing variations of the formula for $A_S$ listed above everywhere. The best I can figure is that the equation for $A_S$ assumes that the outermost 1/8th of the V is included in the nominal diameter of the bolt. However, I can't find any justification for this. If someone could explain it to me, I would greatly appreciate it.&lt;/p&gt;&#xA;" OwnerUserId="3463" LastActivityDate="2018-04-25T15:30:02.493" Title="What is the minor diameter of a thread in terms of its nominal diameter and pitch?" Tags="|threads|technical-drawing|drawings|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21468" PostTypeId="1" CreationDate="2018-04-25T15:39:31.577" Score="0" ViewCount="134" Body="&lt;p&gt;Ive seen some images of band brakes and I notice there are usually two rivets (or screws) on the drum casing (encircled in the image below). What are the rivets riveted to? And why are they riveted?&#xA;Are these rivets present only in simple band brakes or are they also present in other types of drum brakes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xsNMP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xsNMP.jpg&quot; alt=&quot;Band brake&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image from Wikipedia. Usage licence: CC BY-SA 3.0&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="14355" LastActivityDate="2018-04-25T17:04:36.377" Title="Why are there rivets on the drum case of a band brake" Tags="|mechanical-engineering|automotive-engineering|mechanical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21469" PostTypeId="2" ParentId="18153" CreationDate="2018-04-25T16:00:45.557" Score="0" Body="&lt;p&gt;Depending on the degree of damage resistance, whether you want the temperature to be sacrificial or not and the duration of the applied temperature then something like Fendolite might be an option. &lt;/p&gt;&#xA;" OwnerUserId="15855" LastActivityDate="2018-04-25T16:00:45.557" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21471" PostTypeId="2" ParentId="21468" CreationDate="2018-04-25T17:04:36.377" Score="3" Body="&lt;p&gt;The circled portions of the image provided are not rivets. They are locking adjustment screws. According to information provided in an &lt;a href=&quot;https://bicycles.stackexchange.com/questions/8456/how-can-i-stop-a-belt-brake-band-brake-from-squealing&quot;&gt;answer (number 2) on the bicycles stack exchange&lt;/a&gt;, the primary purpose of those screws is to provide a uniform gap between the brake band and the drum when the brake lever is released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one long rivet in the 4 o'clock position securing the band from rotating and another rivet in the 3 o'clock position at which point the actuating lever is attached. Those with digital time keeping devices, disregard the above references.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-04-25T17:04:36.377" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21472" PostTypeId="1" CreationDate="2018-04-25T17:07:00.823" Score="0" ViewCount="813" Body="&lt;p&gt;I really would like to run my abaqus jobs (that have user subroutines attached to them) very quickly. I have read some sections in documentation about calling a utility subroutine called &quot;MutexInit&quot;. I didin't understand when i could use it so that there will be no errors!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone please direct me to anywhere where I can have an idea of what to type in my subroutine files so that I can use cpus 4 or higher to run my abaqus jobs super fast?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please and thank you very mch!&lt;/p&gt;&#xA;" OwnerUserId="15597" LastActivityDate="2024-01-15T03:07:50.997" Title="Does anyone know how would one use &quot;mutexes&quot; in fortran user subroutine so that I can run abaqus job via parallel processing?" Tags="|abaqus|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21473" PostTypeId="2" ParentId="21466" CreationDate="2018-04-25T17:15:15.950" Score="5" Body="&lt;p&gt;for a properly-designed (right materials, right size, right manufacture) spring of that type, the applied stresses are far too small to cause any plastic deformation (permanent bending) at all. Furthermore, there is no physical mechanism active at room temperature and low applied stress in spring alloys that causes them to &quot;season&quot; or take a set on initial flexure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature at which an otherwise properly-designed spring would lose strength and take a set upon flexure is so high that you'd get 3rd degree burns of your hand while using it, so don't fall for the false claim that storing one of these in a hot place will ruin the spring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A low-quality spring could become softer with use by the progressive growth of fatigue cracks in its surface but for a hand exerciser this is very unlikely because those cracks require a very large number of stress cycles to grow to significant size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spring alloy design and spring manufacture are both quite mature technologies and the materials used in them are not a significant cost-driver in a business model; the most probable way that a &quot;low quality&quot; spring would fail is due to a flaw hidden inside the spring (like a slag or scale inclusion) - but these would cause prompt and catastrophic failure of the spring within tens or hundreds of flexure cycles out-of-the-box. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-04-25T17:15:15.950" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21477" PostTypeId="1" CreationDate="2018-04-25T22:44:08.800" Score="2" ViewCount="138" Body="&lt;p&gt;Not sure if this is a Physics related or Engineering related question, but imagine that we would take 100's of small hobby telescopes and somehow managed to point them all at the same point in the sky at the same time. We would than digitally process this image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would we be able to combine these images and get image quality of a much larger telescope? We assume that all of them would point at a same part of sky at the very same time. Do you think we could get a significantly improved image compared to individual telescopes? And if so how much of a better image would we get?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2018-04-26T11:51:59.407" Title="Would connecting 100's of small telescopes into one huge telescope make for a significantly better telescope?" Tags="|design|measurements|optics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21478" PostTypeId="2" ParentId="21477" CreationDate="2018-04-25T22:54:31.027" Score="3" Body="&lt;p&gt;coupling together a large number of small telescopes to emulate the behavior of a single very large telescope is an area of active research in the telescope field. to properly combine a large number of optical images into one requires phase-matching all of the separate signal sources, which is a difficult job when dealing with light. it is easier in radio astronomy because the wavelengths involved are much longer; for this reason, the combination trick you describe has been in use for many years in that arena, where it is capable of producing resolution equivalent to that of a radio telescope &quot;aperture&quot; many miles across.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-04-25T22:54:31.027" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21479" PostTypeId="1" AcceptedAnswerId="21654" CreationDate="2018-04-26T05:03:22.883" Score="0" ViewCount="41" Body="&lt;p&gt;If I have a 4 inch PVC pipe connected to a rotary union, that I want to spin at 200-300 RPM -how do I do it? It appears like the rotary unions themselves have some lateral give -- So, I would need bearings? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there rotary unions with built-in motors? Are there motors that can have hollow, 4 inch cores? Does it make sense to get a custom gear machined for this purpose?&lt;/p&gt;&#xA;" OwnerUserId="15863" LastActivityDate="2018-05-08T08:21:15.807" Title="How can you spin a pipe connected to a rotary union?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21480" PostTypeId="1" AcceptedAnswerId="21577" CreationDate="2018-04-26T08:31:43.450" Score="1" ViewCount="83" Body="&lt;p&gt;Given that we have mechanisms of performing photolithography at 10nm, why are metal filters often meshes? Wouldn't it make a lot more sense to buy a single chunk of metal, and use the UV lasers to poke lots of N micron holes in it?&lt;/p&gt;&#xA;" OwnerUserId="15863" LastActivityDate="2018-05-03T05:13:12.650" Title="Why are fine metal filters made of meshes?" Tags="|lasers|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21482" PostTypeId="2" ParentId="21480" CreationDate="2018-04-26T10:54:46.843" Score="1" Body="&lt;p&gt;When using a laser to drill holes you are melting or sublimating the material, The liquefied/evaporated metal has to go somewhere before it solidifies again where you don't want it to and possible plugging the hole again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building the filter by only adding material makes sure there is nothing extra where you don't want it (the holes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making them out of meshes also means they can be (partially) disassembled and cleaned.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2018-04-26T10:54:46.843" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21483" PostTypeId="2" ParentId="21480" CreationDate="2018-04-26T11:23:26.277" Score="1" Body="&lt;p&gt;Using a mesh means that, not only are they cheap relatively, but they can be scaled easily to suit the task required, so a mesh for tea leaves or a mesh to allow concrete to be poured through etc etc.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-04-26T11:23:26.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21484" PostTypeId="2" ParentId="21477" CreationDate="2018-04-26T11:51:59.407" Score="4" Body="&lt;p&gt;There are two issues here, noise and resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining the optical images from multiple telescopes would reduce the overall noise.  This is because each image is independent of the others, so all the noise would be uncorrelated.  Averaging multiple images attenuates uncorrelated noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You wouldn't really get any enhancement of resolution.  There are techniques for combining the data from multiple telescopes to get higher resolution, but you need to capture phase information for that to work.  This is essentially past our capability for separate optical telescopes.  However, this is being done rather routinely at lower frequencies where capturing phase is possible.  There are radio telescope arrays that were specifically built to act as one large-aperture receiver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There have even been cases where multiple radio telescopes around the world have cooperated in imaging the same distance object.  The data from each telescope is carefully recorded and time-stamped.  The result requires lots of computing to decipher into a image.  If done right, the effective aperture can be a good fraction of the diameter of the earth.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-04-26T11:51:59.407" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21487" PostTypeId="2" ParentId="21480" CreationDate="2018-04-26T12:34:21.277" Score="2" Body="&lt;p&gt;Photo-lithography is expensive and is generally more useful in etching away some surface layer than drilling holes. Generally the light source doesn't actually remove material but selectively cures a coating on the surface which masks off teh pattern to be etched. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to weaving which was one of the first ever processes to be mechanised and automated this is an expensive and time-consuming process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want very fine filters they are often made my sintering fine metal particles produced by vacuum deposition. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-04-26T12:34:21.277" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21488" PostTypeId="2" ParentId="15483" CreationDate="2018-04-26T13:24:49.890" Score="1" Body="&lt;p&gt;I think the important difference to consider in a typical excavation is that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A strut will span the whole excavation and only be supported at the ends. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For a typical waling it will potentially span the whole excavation but it would typically be rigidly connected frequently to the excavation walls. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;By rigidly connecting the waling to the wall it would behave in a composite manner with the wall. Any development of axial strain the waling would be counteracted by restraining actions such as the restraint of the wall itself and friction in the surrounding ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't however mean that you do not need to do a check on the waler for high temperature conditions. If the strut heats up and additional forces are introduced due to the temperature rise then these additional forces would probably end up in the walers potentially introducing additional bending into the waler which would need to be checked. &lt;/p&gt;&#xA;" OwnerUserId="15855" LastActivityDate="2018-04-26T13:24:49.890" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21489" PostTypeId="2" ParentId="7252" CreationDate="2018-04-26T13:36:24.280" Score="1" Body="&lt;p&gt;P-Δ obviously only need be considered in members with high axial forces. There are two elements of P-Δ to consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The effect of end restraints etc. inducing bending into a strut meaning that the axial force is no oriented along the centreline of the member. Many struts I see are designed with large physical pinned connections at the end to avoid this. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The effect of direct forces on a strut causing the strut to bend and moving the centre of thrust away from the centre of the member. The obvious examples of this could either bending due to differential heating of a strut (top vs. bottom) or the self weight of the strut causing the mid span to sag (I think this is covered by a French design standard for props). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In all cases the basic P-Δ approach can apply. Create your structural model of the system. Run the model and find the various deflections and axial forces. Then add in additional moments into the system based on the deflection multiplied by the axial force. Then check whether the new defections are sufficiently similar to the previous deflections. If not then repeat. The exact process for adding the additional moments will vary depending on the structural model but a lot of typical frame analysis software has this capability built in. &lt;/p&gt;&#xA;" OwnerUserId="15855" LastActivityDate="2018-04-26T13:36:24.280" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21490" PostTypeId="2" ParentId="5474" CreationDate="2018-04-26T13:55:14.490" Score="0" Body="&lt;p&gt;I think by &lt;em&gt;intended member stiffness&lt;/em&gt; this refers back to your structural analysis that you undertook to determine the forces in the member. The member in that analysis would have had a certain stiffness and the bolts need to ensure that the same stiffness is maintained across the joint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your bolts are not in tension in any design case (and are not pretensioned) because of the magnitude of the axial load then the bolts will not contribute to the stiffness of the joint. I think this however unusual for all necessary load cases to result in no load combination that results in bolt tension at a joint like this. &lt;/p&gt;&#xA;" OwnerUserId="15855" LastActivityDate="2018-04-26T13:55:14.490" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21492" PostTypeId="2" ParentId="15691" CreationDate="2018-04-26T18:30:36.040" Score="1" Body="&lt;p&gt;I believe you could have also considered rock bolts/anchors. I don't know how thick your raft foundation was going to be, but sometimes making the slab thick enough to resist uplift forces can cause problems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It may be unnecessarily thick, and as such quite expensive to oversize&lt;/li&gt;&#xA;&lt;li&gt;It may not be possible to dig deeper to pour a thicker foundation&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Rock bolts also will most likely be concealed, so it won't affect architectural elements. They can be pricey however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Local economics/prices/availability may have a big effect on the viability of any of these solutions.&lt;/p&gt;&#xA;" OwnerUserId="15871" LastActivityDate="2018-04-26T18:30:36.040" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21494" PostTypeId="1" AcceptedAnswerId="21497" CreationDate="2018-04-27T10:07:18.447" Score="0" ViewCount="360" Body="&lt;p&gt;I have recently come across the concept of &lt;a href=&quot;https://en.wikipedia.org/wiki/Compact_linear_Fresnel_reflector&quot; rel=&quot;nofollow noreferrer&quot;&gt;CLFRs&lt;/a&gt;, and I want to build one myself, as a private experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a rudimentary knowledge of optics (which I'm sure I'll have to refresh on) - any useful links would be appreciated on bringing me up to speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of the technology is as follows (I stand to be corrected if this knowledge is incorrect).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;A linear Fresnel reflector is constructed by taking equal width &quot;strips&quot; out of a parabolic mirror and &quot;translating&quot; the &quot;strips&quot; on to a horizontal surface in such a way that all the strips have bases with the same Y coordinate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A compact linear fresnel reflector superimposes two parabolic mirrors (facing oposing directions) and decomposes them into two sets of alternating, equal width &quot;strips&quot; lying on the same horizontal line.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Is my understanding of the construction of CLFR correct? If yes, is there an algorithm I can use to &quot;slice&quot; a parabola into slices of width &lt;em&gt;w&lt;/em&gt;, whilst preserving the reflecting angle of each &quot;slice&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think of a way of generating an algorithm, using geometry and elementary calculus - BUT, I don't profess to be an expert in this area, so I want to know what specialists in this field suggest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A schematic of a CLFR is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GEWN2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GEWN2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, how may I build a CLFR by &quot;decomposing&quot; two parabolas into two alternating rows of strips of width &lt;em&gt;w&lt;/em&gt; ?&lt;/p&gt;&#xA;" OwnerUserId="932" LastEditorUserId="8055" LastEditDate="2018-04-27T20:01:11.180" LastActivityDate="2018-04-27T21:35:10.853" Title="How to construct a compact linear Fresnel reflector" Tags="|design|optics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21497" PostTypeId="2" ParentId="21494" CreationDate="2018-04-27T17:25:36.423" Score="1" Body="&lt;p&gt;Nope, it's not that simple. You are forgetting that as you move each section back that its focal point will move with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gGyQE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gGyQE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. If you move a section of the parabola back to the base line the focus will move with it.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never seen one but a Fresnel parabolic mirror would have to be made of strips of &lt;em&gt;different&lt;/em&gt; parabolas.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;To answer @fred_dot_u the angle for a flat section at radius &lt;strong&gt;r&lt;/strong&gt; with a focal point at &lt;strong&gt;h&lt;/strong&gt; above the mirror plane is given by the equation $ \frac{1}{2}tan^{-1} \frac{r}{h} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zm8Lu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zm8Lu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 2. Calculating the angle for a flat spiral strip Fresnel parabolic mirror.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2018-04-27T21:35:10.853" LastActivityDate="2018-04-27T21:35:10.853" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21498" PostTypeId="2" ParentId="21494" CreationDate="2018-04-27T19:08:29.557" Score="0" Body="&lt;p&gt;If the idea of a flat surface cut into a spiral qualifies for your objective, you may be able to make something work based on &lt;a href=&quot;https://omnivorenz.wordpress.com/2012/06/09/spiral-fresnel-reflector-a-simple-solar-concentrator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rick Steenblik's design&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the 1982 Popular Science magazine, the article described how he cut the shape and aligned marks placed on the spiral to attach the reflective material to the cross members. As each segment was &quot;pushed&quot; into position, the outside edge lifted at the angle necessary to provide a common focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allowing for some error, the continuous strip of reflective foil bent progressively greater as the spiral reached the edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zPprL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zPprL.png&quot; alt=&quot;flat spiral reflector&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was a young adult when I purchased the template for this item and did not have the resources to build it. I've long since given up on trying to find the template, although if I did, I'd be building one in a flash. I also recall attempting to contact Mr. Steekblik a number of years ago and having the envelope returned to me, probably using the address shown in the article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another &quot;dead-end&quot; link can be found at a &lt;a href=&quot;http://solarcooking.wikia.com/wiki/Georgia_Tech_Spiral_Solar_Cooker&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar cooking wiki page&lt;/a&gt;. One of the photographs on that site shows clearly the spiral as well as the pairs of mounting holes necessary to accomplish the bend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great if someone could perform the analysis and create a parametric file for creating this reflector in varying sizes and widths.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-04-27T19:08:29.557" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21499" PostTypeId="2" ParentId="21454" CreationDate="2018-04-27T21:24:18.923" Score="1" Body="&lt;p&gt;The most elegant, but sometimes impractical solution, is a tuned mass damper. There are excellent &lt;a href=&quot;https://m.youtube.com/watch?v=f1U4SAgy60c&quot; rel=&quot;nofollow noreferrer&quot;&gt;video resources&lt;/a&gt; available to discuss this concept, but the basic idea is that instead of attempting to stressfully contain vibrations, you cause a single part to move significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This single mass on a spring, attached to the moving component, can cause the vibrations to come to a practical standstill. It works well in systems where the driving frequency is well known and the resonance of the first system is also known. An experienced mechanical engineer can usually design such a system to handle this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The less elegant, more practical approach is to reduce the stiffness of the parts attached to the vibrating component. These vibration isolator will then allow the vibrating component to vibrate, without causing the rest of the building to vibrate.  These systems are well known and have more availability to unknown frequencies and unknown natural frequencies, but can cause unwanted vibrations nevertheless.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-04-27T21:24:18.923" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21500" PostTypeId="1" CreationDate="2018-04-27T21:41:53.530" Score="2" ViewCount="34" Body="&lt;p&gt;I have done a permeability experiment on LDPE in the presence of air using a 25mm circular specimen. At 30'C, the permeability was found to be 1.846 barrer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wGaqG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wGaqG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am trying to apply this permeability value to a calculation whereby I am trying to find what the change in pressure of a dunnage bag is (it is a bag that has the shape of a pillow) after 5 minutes. I have the following info of the dunnage bag:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Volume of bag = 0.2064&amp;nbsp;m&lt;sup&gt;3&lt;/sup&gt; (This was calculated using the Paper Bag Problem)&lt;/li&gt;&#xA;&lt;li&gt;Approximate area = 1.35&amp;nbsp;m&lt;sup&gt;2&lt;/sup&gt; (this is just a quick approximate)&lt;/li&gt;&#xA;&lt;li&gt;Thickness of LDPE = 0.042&amp;nbsp;x&amp;nbsp;10&lt;sup&gt;-3&lt;/sup&gt; m&lt;/li&gt;&#xA;&lt;li&gt;Time = 5&amp;nbsp;x&amp;nbsp;60&amp;nbsp;s = 300&amp;nbsp;s&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is it correct if I convert barrer to SI units in the following manner: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1&amp;nbsp;barrer = 7.5005&amp;nbsp;x&amp;nbsp;10&lt;sup&gt;-18&lt;/sup&gt; m&lt;sup&gt;2&lt;/sup&gt;s&lt;sup&gt;-1&lt;/sup&gt;Pa&lt;sup&gt;-1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using the following formula, I was hoping to calculate the change in pressure. However, my values don't make sense at all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\text{Permeability} = \dfrac{\text{Volume}\cdot\text{thickness}}{\text{Area}\cdot\text{time}\cdot\text{Change in Pressure}}$$&lt;/p&gt;&#xA;" OwnerUserId="15882" LastEditorUserId="1832" LastEditDate="2018-04-28T00:27:43.233" LastActivityDate="2018-04-28T00:27:43.233" Title="How to apply the permeability value found from an experiment?" Tags="|mechanical-engineering|chemical-engineering|experimental-physics|pressure-vessel|polyethylene|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21501" PostTypeId="1" AcceptedAnswerId="21516" CreationDate="2018-04-27T23:19:59.597" Score="0" ViewCount="32" Body="&lt;p&gt;I'm looking for something like this air valve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cIjJK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cIjJK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(as discussed in &lt;a href=&quot;https://engineering.stackexchange.com/a/12404/12007&quot;&gt;https://engineering.stackexchange.com/a/12404/12007&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I need to tube to be wider (3/4 in), the tube to be shorter (the shorter the better), and I would like it if the mechanism for opening and closing the valve wasn't so bulky. This is a picture of an air valve at a much larger scale:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y3OUC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y3OUC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.homedepot.com/p/Suncourt-12-in-Normally-Closed-Automated-Damper-ZC112/204268743?cm_mmc=Shopping%7cVF%7cG%7c0%7cG-VF-PLA%7c&amp;amp;gclid=CjwKCAjwlIvXBRBjEiwATWAQIp6R6KVXc7l3PrZILy2NjFr2LsexUXmn6eohfsIsHXPDVXo3r9prZBoC_QgQAvD_BwE&amp;amp;dclid=CIXgsZqh2toCFQN6fgodAd8ElA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.homedepot.com/p/Suncourt-12-in-Normally-Closed-Automated-Damper-ZC112/204268743?cm_mmc=Shopping%7cVF%7cG%7c0%7cG-VF-PLA%7c&amp;amp;gclid=CjwKCAjwlIvXBRBjEiwATWAQIp6R6KVXc7l3PrZILy2NjFr2LsexUXmn6eohfsIsHXPDVXo3r9prZBoC_QgQAvD_BwE&amp;amp;dclid=CIXgsZqh2toCFQN6fgodAd8ElA&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping to find something like that at a 3/4 in scale, where the mechanism is proportionally smaller also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only need the valve to be able to withstand the pressure of a person exhaling, which I think is like 3 psi max. So I'm hoping something less bulky than the solenoid in the first link could be used.&lt;/p&gt;&#xA;" OwnerUserId="12007" LastEditorUserId="12007" LastEditDate="2018-04-28T08:11:11.367" LastActivityDate="2018-04-28T23:27:30.717" Title="where to find / what to call a 3/4 inch air valve that can be opened/closed electrically?" Tags="|airflow|valves|electrical|" AnswerCount="1" CommentCount="4" ClosedDate="2018-04-30T12:44:58.893" ContentLicense="CC BY-SA 3.0" />
  <row Id="21505" PostTypeId="1" AcceptedAnswerId="21506" CreationDate="2018-04-28T05:35:47.683" Score="0" ViewCount="476" Body="&lt;p&gt;How do I get a rough estimate of how many litres per minute a motor can push?&lt;/p&gt;&#xA;" OwnerUserId="12586" LastActivityDate="2018-05-03T04:47:30.603" Title="Motors, Horsepower, RPM and Litres per minute" Tags="|motors|hydraulics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21506" PostTypeId="2" ParentId="21505" CreationDate="2018-04-28T05:48:01.153" Score="0" Body="&lt;p&gt;Formulating a relation will depend on the type of the pump. But if you find out the volume delivered in one revolution and multiply it by rpm , you can get volume output per minute.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastEditorUserId="11114" LastEditDate="2018-04-28T09:15:59.417" LastActivityDate="2018-04-28T09:15:59.417" CommentCount="2" ContentLicense="CC BY-SA 3.0" />
  <row Id="21509" PostTypeId="1" AcceptedAnswerId="21510" CreationDate="2018-04-28T16:20:22.397" Score="0" ViewCount="168" Body="&lt;p&gt;I just don't get the concept of Boolean Geometry... Can you help me understand how to solve these homework problems? &#xA;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2v5BN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2v5BN.jpg&quot; alt=&quot;HomeworkProblems&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15890" LastActivityDate="2018-04-28T16:44:18.823" Title="Boolean Geometry Help" Tags="|mechanical-engineering|geometry|" AnswerCount="1" CommentCount="2" ClosedDate="2018-05-11T11:39:29.397" ContentLicense="CC BY-SA 3.0" />
  <row Id="21510" PostTypeId="2" ParentId="21509" CreationDate="2018-04-28T16:44:18.823" Score="5" Body="&lt;p&gt;&quot;Boolean&quot; just means logically true or false. For this exercise you have to figure out which edges exist when the parts intersect or cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xCr5G.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xCr5G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Partial completion of exercise 1.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exercise 1 requires you to create the combined solid consisting of A and B but with a cutout C. This means that you have the box and wedge combined and a hole cut through with shape C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've drawn in a few edges for you to get you started. Note that the top edge of the box has a gap in it where the wedge juts through. If you were to run your finger across the left face you would feel no edge there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you continue from there?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the &lt;strong&gt;∩&lt;/strong&gt; symbol is the 'intersect' operation as used in set theory. The result of that boolean operation will be the parts that are common to both objects when merged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be nice to use a different colour to darken the 'hidden' edges - those you can't see from the chosen viewing angle.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-04-28T16:44:18.823" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21511" PostTypeId="1" AcceptedAnswerId="21513" CreationDate="2018-04-28T16:49:52.940" Score="2" ViewCount="154" Body="&lt;p&gt;Or alternately what is the name of the device I'm looking for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought is to make a narrow V shaped chamber by gluing 1/4&quot; acrylic together with 2 face plates and a mid plate with a triangular hole removed.  The liquid comes in at the bottom and lifts a slightly heavier than water bead.  Faster flow has to get further into the V chamber before it spreads far enough to match the net gravitational down on the bead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 10 liters per hour, flow through a 1/4&quot; pipe is about 9 cm/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems like such a simple way to do it, I'm sure there are off the shelf ones available for a few bucks.  But often searching depends on knowing what it's called. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/88pux.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/88pux.jpg&quot; alt=&quot;flowmeter&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one that I found on ebay, but intended for gasses, not liquids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this sort of design with 2-4&quot; plumbing is used for pneumatic seed sorters.  However in those, you are indifferent to turbulence.  For this application to have something reasonably repeatable, I think I need non-turbulent divergence.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Purpose:  I'm trying to make a fertilizer injector with minimal pressure drop.  Overall the notion is to put a slight constriction in a 3/4&quot; pipe creating a small (few oz) pressure drop.  Two small diameter hoses connect to the bottom and top of a chunk of PVC pipe with the fertilizer inside it.  Since the pipe is pressurized to the line pressure, the differential across the constriction will move water through the device.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small valve on the 1/4&quot; line can be used to adjust the flow to put the bead at about the same place each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thought would be to use something like the IV drip chamber.  This is a comparable flow rate.  For long duration, I don't know if you could maintain air in the chamber.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other effects I could use?&lt;/p&gt;&#xA;" OwnerUserId="8768" LastEditorUserId="8768" LastEditDate="2018-05-03T17:04:01.533" LastActivityDate="2021-01-13T16:05:39.780" Title="How can I make an uncalibrated flow meter for water flows of a few liters per hour" Tags="|fluid-mechanics|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21512" PostTypeId="2" ParentId="5950" CreationDate="2018-04-28T16:52:51.650" Score="0" Body="&lt;p&gt;Addressing laminar internal drag within a pipe; there is a new developing technology on the horizon called STIC; that utilizes methods to break up the straight line boundary layer drag and then providing a method to bleed portions of that turbulence into a separate chamber to be cooled and reintroduced back into the outer surface area. This process employs the &quot;conservation of energy&quot; principles to speed up the central flow; continuously bleeding the boundary layer drag and turbulence into a separate processing cavity. this new method does not require moving parts. recent testing in the Bakken Oil fields is able to triple the flow rate of oil with high levels of Hydrogen Sulfide. The new system originally designed to improve the liquid flow rates in various devices and applications. Utilizing this method of unloading oil tankers; the rate of the download was increased from 300 barrels per hour to +900 barrels per hour.              &lt;/p&gt;&#xA;" OwnerUserId="15891" LastActivityDate="2018-04-28T16:52:51.650" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21513" PostTypeId="2" ParentId="21511" CreationDate="2018-04-28T16:58:40.717" Score="2" Body="&lt;p&gt;I think you’re looking for a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Rotameter&quot; rel=&quot;nofollow noreferrer&quot;&gt;rotameter&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2018-04-28T16:58:40.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21514" PostTypeId="2" ParentId="21511" CreationDate="2018-04-28T21:18:28.033" Score="2" Body="&lt;p&gt;They make these for liquid's also. Uually the bob looks like an upside down top hat riding on an rod inside the glass or plastic V tube. As mentioned above a rotometer will work but requires a voltage PS to convert the electromagnetic pulses into a Vdc output. -AG &lt;/p&gt;&#xA;" OwnerUserId="15892" LastActivityDate="2018-04-28T21:18:28.033" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21515" PostTypeId="2" ParentId="21511" CreationDate="2018-04-28T21:42:55.953" Score="2" Body="&lt;p&gt;Try &lt;a href=&quot;https://www.google.ie/search?rlz=1C1PRFC_enIE653IE653&amp;amp;biw=1050&amp;amp;bih=1614&amp;amp;tbm=isch&amp;amp;sa=1&amp;amp;ei=C-rkWoCXJ6XOgAajxaz4Ag&amp;amp;q=variable+area+flowmeter&amp;amp;oq=variable+area+flowmeter&quot; rel=&quot;nofollow noreferrer&quot;&gt;variable+area+flowmeter&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kdtt7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kdtt7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Various types of variable area flowmeters.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-04-28T21:42:55.953" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21516" PostTypeId="2" ParentId="21501" CreationDate="2018-04-28T23:27:30.717" Score="1" Body="&lt;p&gt;You are looking for a 3/4&quot; &lt;strong&gt;magnetic&lt;/strong&gt; valve or &lt;strong&gt;solenoid&lt;/strong&gt; valve. Google these terms, then talk to a supplier. As you want the mechanism inside the duct (or so it appears), you could also look for a magnetic or electrical ball valve. However I beleive at this small size, the electrical componentes outside the duct will be about the same size as the physical body of the valve.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-04-28T23:27:30.717" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21517" PostTypeId="1" CreationDate="2018-04-29T16:43:56.600" Score="4" ViewCount="152" Body="&lt;p&gt;Given ist the following framework:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hHy1P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hHy1P.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E=210 \frac{\text{kN}}{\text{mm}^2}$$&#xA;$$ EA \to \infty $$&#xA;$$ I_1 = 80000\text{cm}^4 $$&#xA;$$ \frac{I_1}{I_2} = \frac{3}{4} $$&#xA;$$ I_3=I_1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the principle of virtual work, I'm suppossed to figure out by what angle $\varphi_i$ beam 2 and 3 are skewed. Shear forces can be omitted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my virtual system, I apply a moment at the joint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jhbno.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jhbno.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine the reaction forces needed in my virtual system? I assume that the moments in 1 and 4 need to be 0, but there is no way to cancel out the moments at the ends of 2 and 3 and still have a force equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is my understanding of a balanced virtual system flawed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is greatly appreciated. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case someone wants to work through the problem themselves, here is the solution and what I've calculated so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution:&#xA;$$\begin{align*}&#xA;\varphi_2&amp;amp;=0.009\text{rad}\\&#xA;\varphi_3&amp;amp;=0.007\text{rad}&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EI_1 =EI_3= 210\cdot 80000\cdot 10^4\text{kNmm}^4=168\cdot 10^9\text{kNmm}^4 = 168\cdot 10^3\text{kNm}^4$$&#xA;$$ EI_2=\frac{4}{3}EI_1=224\cdot 10^3 \text{kNm}^4$$&#xA;Reaction Forces:&#xA;$$ \begin{align*}&#xA;A_x &amp;amp;= -126\text{kN}\\&#xA;A_y&amp;amp;=-42\text{kN}\\&#xA;B_x&amp;amp;=-42\text{kN}\\&#xA;B_y&amp;amp;=\hphantom{-}\text{42kN}\\&#xA;G_x&amp;amp;=\hphantom{-}42\text{kN}\\&#xA;G_y&amp;amp;=-42\text{kN}&#xA;\end{align*} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we ignore shear forces and $EA\to \infty$ we only need to consider moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moments can be calculated to:&#xA;$$\begin{align*}&#xA;1:&amp;amp;-q\frac{x^2}{2}+126x=-\frac{21}{2}x^2+126x\\&#xA;2:&amp;amp;-42x+336\\&#xA;3:&amp;amp; -42x\\&#xA;4:&amp;amp; 42x-168&#xA;\end{align*}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fxEuH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fxEuH.png&quot; alt=&quot;Moment Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8674" LastEditorUserId="8674" LastEditDate="2018-04-30T19:10:23.653" LastActivityDate="2018-04-30T19:10:23.653" Title="Skewing of joints using virtual work" Tags="|statics|beam|moments|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21518" PostTypeId="1" CreationDate="2018-04-29T19:31:30.967" Score="2" ViewCount="69" Body="&lt;p&gt;I can imagine a steam engine driving multiple regular old vacuum pumps, which each raise water ~10m.  I would think that someone would combine the multiple pumps into a single piece of machinery for raising water &gt;10m without high pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was such a thing in use before the Newcomen steam engine (driven by horses or something)?  How was it designed?  Was the same pumping system used with the Newcomen steam engine?&lt;/p&gt;&#xA;" OwnerUserId="15901" LastActivityDate="2018-04-29T21:13:54.777" Title="How were multiple vacuum pumps used to raise water over 10m (e.g. by a steam engine)?" Tags="|pumps|engineering-history|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21519" PostTypeId="2" ParentId="21518" CreationDate="2018-04-29T21:13:54.777" Score="2" Body="&lt;p&gt;pumps called mechanical lift pumps existed before then. these had the piston/cylinder and check valve assembly deep in the hole. actuating the piston with a long rod lifted one cylinder's worth of water upwards in the well pipe after which a check valve closed and held the water in place. pushing the piston down opened another valve and refilled the cylinder, etc. such a pump can lift water higher than a vacuum pump can but it requires a long actuating rod to connect the handle with the piston. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-04-29T21:13:54.777" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21521" PostTypeId="1" AcceptedAnswerId="21524" CreationDate="2018-04-30T00:46:12.460" Score="0" ViewCount="4947" Body="&lt;p&gt;I have a motor thats rated for 220V single phase. My house has 240V single phase. Im trying to run this motor but it doesn't budge and I don't even hear a hum. I know power is getting to it from checking with a multimeter. Could it possibly have burned from too high of voltage? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, if wired improperly, say, ground was accidentally switched with one of the hot wires running to the motor, could this somehow burn the motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure whats going on with this thing, and I don't see any reset buttons either or fuses either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="15903" LastActivityDate="2021-08-17T12:49:26.787" Title="Running 220V single phase motor on 240V, and did I burn it?" Tags="|mechanical-engineering|electrical-engineering|motors|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21523" PostTypeId="2" ParentId="21521" CreationDate="2018-04-30T03:47:20.043" Score="0" Body="&lt;p&gt;Which country are you in? As UK used to be 240 but dropped to 230 to match with France, France and all EU moved from 220 to 230 for same reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 240 did not fry your motor, something else is wrong / broken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The variation of +10% -6% permitted on the actual voltage received at a property covers this variation anyway.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-04-30T03:47:20.043" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21524" PostTypeId="2" ParentId="21521" CreationDate="2018-04-30T05:43:53.993" Score="2" Body="&lt;p&gt;for short periods, 240VAC will not burn out a 220VAC motor. furthermore, switching the polarity of the wires will not blow up the motor. here are the things to check:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to begin with, disconnect the 240 VAC power and apply an ohmmeter to the SAME motor terminals to which you had applied the 240VAC power. your ohmmeter should read almost ZERO ohms. If it reads high ohms, either the windings of the motor are burnt open (throw the motor away) OR the terminals you chose are not those used to power the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can this be? in a single-phase motor, two of the wire terminals leading to the motor go to the stator winding and the third terminal (the green ground wire) is connected to the motor housing. connecting one of the 240VAC leads to one of the stator terminals and the other to ground will not power up the motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;furthermore, in a single-phase motor there is a switch which temporarily patches in a capacitor (the start capacitor) which furnishes a phase lag to one set of stator coils in order to get the motor to start from zero RPM and spin up to speed, after which the capacitor is patched out of the circuit. if this switch is bad, applying power to the motor terminals will not start the motor. If your motor is of the &quot;capacitor-start&quot; type and it won't start, you must check the proper operation of the start switch as well. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-04-30T05:43:53.993" CommentCount="4" ContentLicense="CC BY-SA 3.0" />
  <row Id="21526" PostTypeId="1" CreationDate="2018-04-30T10:55:50.987" Score="0" ViewCount="60" Body="&lt;p&gt;I have some difficulty to solve this question:&lt;a href=&quot;https://i.stack.imgur.com/21MOw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/21MOw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the condition for real eigen values other than symmetric matrix?&lt;/p&gt;&#xA;" OwnerUserId="15908" LastActivityDate="2018-05-03T01:22:11.357" Title="What is the condition of real eigen values other than symmetric matrix?" Tags="|control-engineering|control-theory|mathematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21529" PostTypeId="1" CreationDate="2018-04-30T13:47:27.317" Score="2" ViewCount="662" Body="&lt;p&gt;I am interested in mass producing a reusable coffee cup (à la &lt;a href=&quot;https://uk.keepcup.com/?country=United%20Kingdom#&quot; rel=&quot;nofollow noreferrer&quot;&gt;KeepCup&lt;/a&gt;) with a scannable id on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cup will be plastic or glass, ideally glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ID code will need to survive being washed regularly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking it could be put on by whatever method is used to put beer names on pint glasses, but I really have no idea how any of this is done. Onte thing I was wondering about was if I need to print in two colours (black and white for example) or if frosting some lines on it would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I am interested in any of this is I want to assess the feasibility of giving out takeaway coffee cups which can then be returned to a 'cup bin'. The barcodes are needed to identify which cups have been returned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the codes have to be unique for each glass.&lt;/p&gt;&#xA;" OwnerUserId="15910" LastEditorDisplayName="user16" LastEditDate="2018-05-11T11:32:09.910" LastActivityDate="2018-05-11T11:32:10.120" Title="How would I put a barcode of other scannable code on a coffee cup?" Tags="|plastic|glass|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21530" PostTypeId="1" CreationDate="2018-04-30T14:29:16.587" Score="-1" ViewCount="55" Body="&lt;p&gt;If I wanted to wash 10,000 identical plastic or glass cups a day, would I use a regular dishwasher or a specialised product designed to work with those specific cups.&lt;/p&gt;&#xA;" OwnerUserId="15910" LastActivityDate="2018-05-30T18:04:44.600" Title="Specialised Cup Washer" Tags="|engineering-economics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21531" PostTypeId="2" ParentId="21529" CreationDate="2018-04-30T15:09:07.887" Score="6" Body="&lt;p&gt;A few different marking processes are available to accomplish your objective. Glass is a good choice, for reasons of durability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Etching&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One creates a mask or stencil with the desired graphic. In the case of a complex design, vinyl cutters perform the task quite well. The stencil is applied using transfer tape, after which etching chemicals are applied. After a specific period of time for the chemical, it is washed off, leaving behind a permanent image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will not work on plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Con: requires creation of vinyl stencil for each unique image, labor intensive to weed image, apply chemicals, clean glass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Engraving&lt;/strong&gt; (mechanical)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular method requires a computer controlled engraving device, which spins a diamond bit at high speeds. The computer directs the cutting point and removes glass from the surface. Different cutting bits provide some flexibility in design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will work on plastic, results are varied depending on plastic composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Con: I have no direct experience or observation of hobbyist level rotary devices for CNC routers capable of this work. Commercial versions with rotary capability are huge and expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Engraving&lt;/strong&gt; (laser) *note below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method also uses a computer controlled device, typically a CO2 laser equipped with a rotary accessory. The glass is placed in the chuck or on rollers and rotates under the laser nozzle as it moves laterally. Microscopic pieces of glass are removed by the laser, similar to mechanical engraving. Power settings determine depth of engraving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will work with acrylic plastic, others may be possible. Vinyl is prohibited due to creation of chlorine gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Con: can be expensive (no budget referenced) and sometimes frustrating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Marking&lt;/strong&gt;  (laser applied coating)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substance exists (Laser Bond Tech) which when applied to glass (or other surfaces) and activated (engraving settings) with the laser, bonds to the glass. The substance creates dark, almost black marking on the surface, resistant to wear, very durable. Images created in the same manner as engraving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Con: Additional steps involved in applying, drying and cleaning after marking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Marking&lt;/strong&gt; (silk screen)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silk screen marking deposits a layer of ink on the surface of the glass (or plastic). In the case of a curved surface, special screens and squeegees are necessary to provide proper coverage. The screen is typically created using a photographic process allowing for quite good detail results. Ink is selected for durability for obvious reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will work on plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Con: requires creation of silk screen for each image, also labor intensive process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Marking&lt;/strong&gt; (transfer ink)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This process uses a flexible rubber surface and a raised surface containing the image to be created (think rubber stamp but in metal or more durable material). The image has ink applied by a roller, which is then transferred to the rubber surface. It helps to consider a sports ball with a smooth surface contacting the inked surface. The ball then moves over the surface to be inked and is pressed into contact, transferring the image. Durability of the image depends on the ink selected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will work on plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Con: very labor intensive and/or expensive to create unique image block, better suited to repeat marking of same image.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;note here: These suggestions are provided as a reference for further research and are not to be considered exhaustive resource guides. Much of this material is presented from a hobbyist perspective.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As I own a laser cutter/engraver, I see this method as the most easily accomplished, although initial expense can be high. I also believe that the labor involved in quantity production of unique bar codes is lower with a laser engraver than any of the other methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image generation for bar codes is about the same, but for all of the previously noted methods, one more step is involved to create a physical device to create the image on the glass. With a laser engraver, the image of the bar code is transferred to the software and the engraving begins, removing that intermediate step and attendant costs of time and money. The mechanical engraving (CNC router) is the closest to the laser process in that the image also is directly transferred to the machine without an intermediate device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Inexpensive&quot; Chinese-made CO2 lasers abound on eBay, although some sources have a high failure rate, my purchase included!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: testing completed, results below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not surprisingly, the bare naked engraved UPC-A bar code is unreadable. I tried many combinations of illumination, but none of them resulted in a read by my iPhone scanner app.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lQCJI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lQCJI.jpg&quot; alt=&quot;glass engraved upc code&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image is of the engraving directly on the glass. The irregular marking below is the adhesive remnants of the label.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aforementioned Laser Bond Tech 100 purports to bond to glass, and my tests show that this is true. Excessive power levels destroy the material and the markings wash off with water. Not-excessive power levels do a good job of bonding the material. It is resistant to abrasion, although I did not take an abrasive pad to my test item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bar code generated on the glass was readable IF there was a white background. The scanner snapped up the code in less than one second during the positioning!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vt3a4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vt3a4.jpg&quot; alt=&quot;marked bar code with paper background&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image is approximately 25 mm x 50 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that the OPs reference is aimed at use by the general public, I envision someone scratching away at the marking material. I think with a metal device such as a knife or sharp edged spoon, one could damage the marking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving right along, I figured that an inverse pattern on glass, engraved rather than marked, might work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L9R6n.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L9R6n.jpg&quot; alt=&quot;dark engraved bar code&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above image was taken with ambient light and the iPhone scanner snapped the image in about a second. I was surprised, considering the darkness of the area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q6fy1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q6fy1.jpg&quot; alt=&quot;light engraved bar code&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This photo was taken with the iPhone with the built-in light turned on. The scanner app also caught the code quickly with its light turned on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used an online free UPC generator, which created a mess of the SVG image I needed to perform this test. It was easily corrected, but time consuming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the OP, ten thousand unique codes would be better managed by someone fluent in SVG creation, with the ability to spit out those files (after testing) rather than use an online resource of questionable quality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, I'd say that the marking spray has promise, but an inverted engraving gives good durability with the least amount of labor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case presented by the OP, however, I would suggest to contract to an outside business specializing in glass marking technology. Alternatively, he will spend many hours producing these codes on glasses, a few at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could produce perhaps fifty per day and likely lose my sanity after the first week.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2018-05-02T21:53:44.333" LastActivityDate="2018-05-02T21:53:44.333" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21532" PostTypeId="2" ParentId="21530" CreationDate="2018-04-30T15:15:55.050" Score="0" Body="&lt;p&gt;If you mean by &quot;regular&quot; dishwasher, a unit designed for home use, it seems impractical to consider that device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commercial dish washing equipment can handle high volume cleaning and are used in large food service establishments such as restaurants. I viewed a television show recently that showed both fixed military bases and federal prisons using large capacity dish washing machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conveyor is loaded with the dishes to be cleaned. The conveyor has pins or panels upon which the dishes are placed. This ensures proper angles for the wash water to clean correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Solar Mike suggests, such machinery will also dispose of the waste water, as well as dry the dishes, which have to be removed from the conveyor at the exit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on your previous question posted, I suspect these glass cups are not identical, but the difference is inconsequential in this context.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-04-30T15:15:55.050" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21533" PostTypeId="2" ParentId="21530" CreationDate="2018-04-30T17:48:40.597" Score="1" Body="&lt;p&gt;This is not an engineering problem.  It's an Economics problem.   Compare the cost of a specialized machine, including maintenance and all that, vs. the cost of unskilled labor to perform the work manually.   Pick the cheaper option. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-04-30T17:48:40.597" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21535" PostTypeId="2" ParentId="21517" CreationDate="2018-04-30T18:48:30.083" Score="2" Body="&lt;p&gt;Let's assume that in equilibrium the links 1 and 4 will be at angles $\theta_1$ and  $\theta_4$ with the vertical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bYQw0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bYQw0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between these angles can be determined from the fact that the horizontal length between points A and B is $8+4=12$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ 8 \sin \theta _1+8 \cos \theta _1+4 \sin \theta _4+4&#xA;   \cos \theta _4 =12 \ \ \  (1)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The external forces acting on the links are their masses $m_i$ and $q$. The masses can be determined from the inertias. (The question does not make clear about what point the inertias are calculated, and also the units seem wrong. So I will just leave the equations I terms of the masses.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The virtual work by $q$ can be calculated as&#xA;   $$ \delta W_q=\int _0^8q \ \delta (l \ \sin (\theta_1 ))dl = 32 q \cos \theta_1  \delta \theta_1=672000 \cos \theta_1  \delta \theta_1 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The virtual work due to masses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta W_1= m_1 g\  (-\delta(4 \cos \theta_1))= 4 m_1 g \sin\theta_1 \delta \theta_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta W_2= m_2 g (-\delta(8 \cos \theta_1-4 \sin \theta_1)))=m_2 g (8  \sin \theta_1+4  \cos&#xA;   \theta_1)\delta\theta_1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta W_3= m_3 g\  (-\delta(2 \cos \theta_1))= 2 m_2 g \sin\theta_4 \delta \theta_4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta W_4= m_4 g (-\delta(4 \cos \theta_4-2 \sin \theta_4)))=m_2 g (4  \sin \theta_4+2  \cos&#xA;   \theta_4)\delta\theta_4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From $(1)$ we have&#xA;$$ 8 \cos \theta_1 \delta\theta_1-8  \sin \theta_1 \delta\theta_1+4  \cos&#xA;   \theta_4 \delta\theta_4-4 \sin \theta_4 \delta\theta_4=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be solved for $\delta\theta_4$ in terms of $\delta\theta_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we sum all the virtual works and set them to zero, and substitute for $\delta\theta_4$ in terms of $\delta\theta_1$ we get the second equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta W_q+\delta W_1+\delta W_2+\delta W_3+\delta W_4=0 \ \ \ (2)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two equations $(1)$ and $(2)$ have two unknowns $\theta_1$ and $\theta_4$ which must be solved with the values of $m_i$ and $g$. &lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2018-04-30T18:48:30.083" CommentCount="5" ContentLicense="CC BY-SA 3.0" />
  <row Id="21536" PostTypeId="2" ParentId="21526" CreationDate="2018-04-30T19:09:25.070" Score="2" Body="&lt;p&gt;A graphical method is to find the characteristic equation and then determine the threshold values using the root locus plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the calculations using Mathematica. The largest value is 5.75.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YlWij.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YlWij.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2018-04-30T19:09:25.070" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21538" PostTypeId="1" AcceptedAnswerId="21540" CreationDate="2018-04-30T20:34:48.330" Score="0" ViewCount="609" Body="&lt;p&gt;I'm trying to build my own weatherstation for my home, I want to measure temperature humidity and wind strenght and direction. For the wind I have a question. I want to measure the wind with a ball on top of a stick (schematic), and measure the bending of the stick with some strain gauges. Is this possible or just a crappy idea. And what material would I need for the stick that it bends for low wind forces too?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MO7dO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MO7dO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15916" LastActivityDate="2023-05-15T05:34:56.907" Title="Wind measure with strain gauge" Tags="|measurements|wind-power|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21539" PostTypeId="1" AcceptedAnswerId="21542" CreationDate="2018-04-30T20:59:51.150" Score="0" ViewCount="1305" Body="&lt;p&gt;I am designing a lifting device for practical use which is powered by a dc motor with worm gear. The lifting device should be able to hold the load in place at a certain height.however, i read that wormgears are self locking only in situations where shock and vibration are not present, in other words, if the lifting device is not perturbed in any way, the self-locking feature will probably perform as intended. This could present a problem for security reasons and functionality if the lifting device is intended for use in varied conditions. My question is, should I consider others options, like different types of gears or persist with the worm gear&lt;/p&gt;&#xA;" OwnerUserId="12046" LastActivityDate="2018-04-30T21:35:07.587" Title="Worm gear for a lifting device" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21540" PostTypeId="2" ParentId="21538" CreationDate="2018-04-30T21:12:04.320" Score="0" Body="&lt;p&gt;There may be other designs from which you can select, but I would consider a mechanism that does not depend on a specific material bending, especially at low air speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest this, because a light, easily bent material may snap at higher velocities, while a stronger one may exhibit insufficient bending at lower speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a design in which a material of sufficient strength, perhaps aluminum has a hole drilled through it. This plate can be of any outer shape but should have a thickness to allow pressure sensors to be attached to the inside surface of the hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of a plate, one could use a strong piece of thick walled aluminum tubing of a length that would support these pressure sensors. Placed at eight equally spaced locations on the inside wall of the tube, they would provide pressure information to your device, which I envision is something like an Arduino or Raspberry Pi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pressure sensing tube would be placed as near as possible to your sphere, while the sphere is mounted on another piece of lightweight tubing nearly the same diameter as the sensing tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposite end of the sphere tube should be some distance away to reduce the angle of contact between the two tubes. Consider it as a lever. Long levers have small angle changes compared to short levers. The best circumstances should arise if the inner tube is a fraction of a millimeter smaller than the diameter of the sensors within the outer tube. This reduces point contact on the sensors, reducing the possibility of damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I7jiO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I7jiO.png&quot; alt=&quot;pressure sensor cylinder&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By having many sensors inside the tube, you can do more accurate direction locating and pressure measurement. If your sensors are not strips but are instead short button types, the surrounding tube can be shorter, of course, but should still remain as tightly around the sphere tube as possible.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-04-30T21:12:04.320" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21541" PostTypeId="1" CreationDate="2018-04-30T21:29:30.940" Score="2" ViewCount="25" Body="&lt;p&gt;Given ist the following system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jX79R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jX79R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EI=100000\frac{\text{kN}}{\text{m}^2}$$&#xA;$$EA\to \infty$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm suppossed to find the horizontal and vertical shift of joint g. My first approach was to use the principle of virtual work using $$f_i=\int_0^l\frac{M\bar{M}}{EI}dx$$. (As per syllabus, shear forces can be neglected.) In order to determine the moments I need to figure out the reaction forces. However, the system is kinematically indeterminate, despite being statically deteminate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can easily be seen by determining the moments around the respective bearings in each system:&#xA;$$1: G_x+G_y+24=0$$&#xA;$$2: G_x + G_y + 48=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, it can be seen that there are no conflicting moment poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My next approach was the principle of virtual shifts, which (unfortunately as expected) led to the same equations as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the correct way to approach this problem? Any help would be appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case someone wants to know the solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\delta_x = 29.97\cdot10^{-3}\text{m}$$&#xA;$$\delta_y = 7.29\cdot10^{-3}\text{m}$$&#xA;$$\delta_{\text{ges}}=30.84\cdot 10^{-3}\text{m}$$&lt;/p&gt;&#xA;" OwnerUserId="8674" LastActivityDate="2018-04-30T21:29:30.940" Title="Shift of joint in kinematically indeterminate system" Tags="|statics|moments|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21542" PostTypeId="2" ParentId="21539" CreationDate="2018-04-30T21:35:07.587" Score="2" Body="&lt;p&gt;I've seen from simple observation that many lifting devices using winch mechanisms use worm gears. One advantage of using this mechanism is the high gear ratio (reduction) created with few components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my search to validate this reference, I found the answer to a &lt;a href=&quot;https://engineering.stackexchange.com/questions/7230/designing-non-self-locking-worm-gear?rq=1&quot;&gt;post in the Engineering Stack Exchange&lt;/a&gt; in which the replying party suggests that a worm gear design will be self-locking even if such a feature is not desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer also suggests that the angle on the worm gear (lead angle) will reduce or increase the self-locking capacity. This makes sense in that a nearly perpendicular gear engagement will have substantial friction resisting back-drive, while a more oblique angle will allow the teeth to slide more easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with some vibration and shock, a perpendicular lead angle may not have movement beyond a degree or so of rotation. If more mechanically complex construction is not a problem, one worm gear driving another, driving the winch would increase the self-locking capacity. I attempted to find an image of this type of assembly and failed. It may be excessive and impractical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X1RcC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X1RcC.jpg&quot; alt=&quot;worm gears x 3&quot;&gt;&lt;/a&gt;&#xA;Pick two!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the process, I discovered a worm gear being driven by a conventional reduction gear set, which increases the self-locking by requiring that the worm gear and the gear set be back-driven. Using a multi-level planetary gear set as a reduction gear set would fall into the overkill category, I believe, as planetary gear sets of multiple rings have substantial resistance to being back-driven.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-04-30T21:35:07.587" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21543" PostTypeId="2" ParentId="2300" CreationDate="2018-04-30T22:32:16.697" Score="0" Body="&lt;p&gt;Here's how I like to think of it. lbf is the force acting upon the mass. This is what, for example, your bathroom scale is measuring. lbm is the actual mass of the object. So the F =m*a in English units, lbf = lbm * a (aka gravity 32.2 ft/s2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's at least how I've always looked at it.&lt;/p&gt;&#xA;" OwnerUserId="15917" LastEditorUserId="15917" LastEditDate="2018-04-30T22:39:14.033" LastActivityDate="2018-04-30T22:39:14.033" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21544" PostTypeId="2" ParentId="21538" CreationDate="2018-04-30T22:47:22.910" Score="-2" Body="&lt;p&gt;I like the idea above, however I also like the idea of a bending poll. It could work. You'll need to pump electricity into the poll and measure resistance. Essentially, as the poll bends, resistance would increase. Setting the limits might be a bear. Never tried something like this, but you could theoretically gather the information needed. At 0 mph u see 0 deflection, with Cu possibly, u see 0 resistance. At 300mph the rod is at 90 degree, and you see resistance at half. Just an estimate of course, again this is theory. Could work however.&lt;/p&gt;&#xA;" OwnerUserId="15917" LastActivityDate="2018-04-30T22:47:22.910" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21545" PostTypeId="2" ParentId="21526" CreationDate="2018-05-01T00:03:36.973" Score="2" Body="&lt;p&gt;At the bifurcation between all real eigenvalues and one pair of complex conjugate eigenvalues there has to be a duplicate real eigenvalue. So the characteristic polynomial of the matrix at the bifurcation should be of the following form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;(\lambda - a)^2(\lambda - b)(\lambda - c) = &#xA;\lambda^4 - \left(2\,a + b + c\right)\,\lambda^3 + \left(2\,a\,(b + c) + b\,c + a^2\right)\,\lambda^2 - \left(a^2\,(b + C) + 2\,a\,b\,c\right)\,\lambda + a^2\,b\,c.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual characteristic polynomial of the matrix can be shown to be equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\det(\lambda\,I - A) = \lambda^4 + 0\,\lambda^3 - 15\,\lambda^2 + 10\,\lambda + 24 + \alpha.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order for these two to be equal the following is required&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;2\,a + b + c &amp;amp;= 0 \\&#xA;2\,a\,(b + c) + b\,c + a^2 &amp;amp;= -15 \\&#xA;a^2\,(b + c) + 2\,a\,b\,c &amp;amp;= -10 \\&#xA;a^2\,b\,c &amp;amp;= 24 + \alpha&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;simplifying these equations gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;b + c &amp;amp;= -2\,a \\&#xA;b\,c &amp;amp;= 3\,a^2 - 15 \\&#xA;2\,a^3 - 15\,a + 5 &amp;amp;= 0 \\&#xA;3\,a^2\,(a^2 - 5) &amp;amp;= 24 + \alpha&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for $b$ and $c$ to be real it can be shown that it is required that $2\,a^2 &amp;lt; 15$ and for $\alpha$ to be positive it is required that $2\,a^2 &amp;gt; 5 + \sqrt{57}$. By checking the roots of the cubic equation in $a$ against these constraints and using those can be used to solve for $\alpha$.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-05-01T00:03:36.973" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21548" PostTypeId="1" AcceptedAnswerId="21566" CreationDate="2018-05-01T13:59:06.233" Score="0" ViewCount="48" Body="&lt;p&gt;I am attempting some convection analysis for research I am doing. I was wondering someone could kindly confirm whether I am on the right track. The calculation is based up air being passed over a flat surface and the heat exchange taking place between the 'air' and ' surface'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following issues...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read books that say that hc for air by forced convection is normally within 10 - 10^3 range, however all my calculations seem to result in around 4-5 is this correct as per my example below?. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example i have based this upon is a 1m2 surface 1m x 1m dimensions and I want to know how long it would take to change the temperature of the surface material. My approach using the example below shows that if I had to add 460 kJ to change the temperature and then calculate the time, would it be the following to determine time to add the energy..... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;460000 J / 66.42 j.s = 6925 secs = 1.92 hours, is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kHhfI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kHhfI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14630" LastActivityDate="2018-05-02T16:14:13.377" Title="convection query" Tags="|mechanical|convection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 3.0" />
  <row Id="21551" PostTypeId="1" CreationDate="2018-05-01T18:09:26.003" Score="2" ViewCount="966" Body="&lt;p&gt;I'm posting this question on this stack-exchange site, because it appeared to be the one made for this kind of questions. If this happens to not be the case, I'm sorry, I didn't know better. Also forgive me any issues with terminology (or point them out to me), I'm not an engineer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to build a music instrument (I won't go into details here, because I think it's not necessary). In this instrument there will be keys (in the picture I drew, the keys are the blue boxes, as seen from above. The Black arrows indicate the directions in which the keys are movable). If a key is pressed (which means &quot;moved up&quot; in the picture), I want a little string to be pulled. &#xA;The rope is drawn with black color, and one can see the way it is attached to the keys. The red circles are supposed to be the bars that will press the string out of its equilibrium position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture, the middle key is pressed, and the string is elongated because of this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is:&lt;/strong&gt; What materials should I use for the string AND the bars, to make the string glide on the bar as friction-less as possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some additional information:&lt;br&gt;&#xA;There will be 20 keys, and the distance they move, if they are pressed, is 5 millimeters. The string can't be thicker than a thick guitar string. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial thoughts were to use fishing line, and make the bars out of brass. But maybe there are better options to use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The string isn't vibrating, and it won't be the thing to be responsible for the sound. Maybe I should have written &quot;rope&quot; instead of string. The purpose of the strings seen is to activate a little switch, when the key is pressed. To accomplish this, as the keys are pressed (and move), the string will be elongated, as it slides over the red bars. The ellongated string then pulls a little switch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is to say that there is some tension on the string (the switch effectively raises one of the vibrating strings, so the force acting on the non-vibrating strings is roughly the same force it takes to pull a guitar string out of its equilibrium position). &#xA;&lt;a href=&quot;https://i.stack.imgur.com/grXoJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/grXoJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15922" LastEditorUserId="15922" LastEditDate="2018-05-02T16:14:53.930" LastActivityDate="2018-08-31T06:00:31.807" Title="Materials to use for a low friction cable-pull" Tags="|materials|friction|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="21552" PostTypeId="1" CreationDate="2018-05-01T20:49:09.103" Score="1" ViewCount="669" Body="&lt;p&gt;Recently I learned a new word for what I'm looking for -- rotameter -- a device that uses a weight in a conical bore to measure fluid flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/21511/how-can-i-make-an-uncalibrated-flow-meter-for-water-flows-of-a-few-liters-per-ho&quot;&gt;How can I make an uncalibrated flow meter for water flows of a few liters per hour&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found many for gasses, but not many for liquids, and none for the range I'm looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To first order approximation do I need to do more than compare the viscosity of air to that of water to use an air calibrated one for water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The viscosities differ by about a factor of 50.  Would a 5 liter/minute gas rotometer be a 6 liter per hour water rotameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caveats:  In a gas meter the bead may be plastic, which has no buoyancy effect in air, but a substantial one in water.  Indeed the plastic might float, and I'd have to use the thing upside down.  Plastic with a specific gravity of .9 would make the device 10 times more sensitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on what's dissolved in the water, the viscosity can change somewhat.  I will be using more or less pure tapwater with about 600 ppm of assorted hardwater crud.&lt;/p&gt;&#xA;" OwnerUserId="8768" LastActivityDate="2018-05-01T22:23:41.720" Title="Can a gas rotameter be used for liquids?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 3.0" />
  <row Id="21554" PostTypeId="2" ParentId="21552" CreationDate="2018-05-01T22:23:41.720" Score="1" Body="&lt;p&gt;There are water ones that exist, what flow range are you attempting to use?  A very low flow rate (in the Liter / Hour range) would be difficult to size, as the uncertainty would vary substantially for the thing to be readable.  This would be the main reason why they would not be offered in low flow rates - the uncertainty.  What good is a measuring device if you can't read it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty part works as this:  In order to have consistent drag coefficient of a spherical float, the Reynolds number needs to be pretty near constant, and in the range of 10,000.  Otherwise the drag coefficient isn't constant, and you have high uncertainty:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sr751.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sr751.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have a Reynolds number around 10,000, at say 6L/hr with water we'd need a cross sectional area of the ball to be around 0.029 mm^2.  Then there are the tolerances that would be associated with ensuring the tube linearly expands, and the ball stays within the proper cross sectional area.  All of this then would vary drastically if the tube expands too drastically as that would cause compressible flow effects to emerge and the temperature of the water would change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, use a positive displacement meter or variable area meter for low flowrates.  There is a reason why there are different flowmeters for different applications. &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-01T22:23:41.720" CommentCount="6" ContentLicense="CC BY-SA 3.0" />
  <row Id="21555" PostTypeId="1" CreationDate="2018-05-01T23:05:42.490" Score="-1" ViewCount="12674" Body="&lt;p&gt;My Answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Benefits: The major advantage of aluminum in ship building is its strength to density ratio. Aluminum has high tensile strength with less weight, which results in lowering the weight of the overall structure and thus reduce the fuel cost. &#xA;When aluminum is exposed to air, a positive layer of oxide is formed on the surface which makes its corrosion resistant.&#xA;Also, it is easily available in many standard forms as third most abundant metal, and can be transformed to all shapes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disadvantages: i am seeking formal statements on what are some possible disadvantages of aluminum alloy as a replacement of steels in ship building. &lt;/p&gt;&#xA;" OwnerUserId="12830" LastEditorUserId="6790" LastEditDate="2020-05-02T19:34:22.040" LastActivityDate="2020-05-02T19:34:22.040" Title="Benefits and disadvantages of aluminium alloys as a replacement for structural steels in ships" Tags="|structural-engineering|steel|aluminum|alloys|naval-engineering|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 3.0" />
  <row Id="21556" PostTypeId="2" ParentId="21555" CreationDate="2018-05-02T00:10:40.097" Score="0" Body="&lt;p&gt;Aluminum is much more expensive than steel. Note here that the dead weight of a ship is not an important factor in its cost of operation, which means that the weight savings achieved by buying aluminum instead of steel are not paid for by fuel savings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only arena in which the weight savings associated with building the hull of a boat out of aluminum are important are those in which the boat is loaded and unloaded from a trailer by muscle power.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-05-02T00:10:40.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21557" PostTypeId="2" ParentId="21551" CreationDate="2018-05-02T02:19:16.173" Score="0" Body="&lt;p&gt;What sort of tension would be on the strings.  For fishing line, I would suggest high density polyethylene.  If you want to reduce friction further, keep it dusted with talc.  (I used to use talc to lubricate my slide rules.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moderator please delete this answer and make it a comment.&lt;/p&gt;&#xA;" OwnerUserId="8768" LastEditorUserId="8768" LastEditDate="2018-05-03T03:47:31.203" LastActivityDate="2018-05-03T03:47:31.203" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21558" PostTypeId="2" ParentId="21555" CreationDate="2018-05-02T02:36:27.430" Score="-1" Body="&lt;p&gt;Aluminium corrodes in seawater that is why it is used as sacrificial anodes on steel. It is more difficult to weld than steel and almost impossible to develop good mechanical properties in large weldments. Aluminium may not have ever been tested for toughness in the heavy sections (eg. 6 in.) needed for ships and I expect low toughness would result in ships breaking in half like the WW2 Liberty Ships....And it would be very difficult/expensive to cut thick plate sections for hull fabrication; You can shear or saw it but the logistics would be much more expensive than the torch cutting used for steel.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-05-02T14:17:24.893" LastActivityDate="2018-05-02T14:17:24.893" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21559" PostTypeId="2" ParentId="21538" CreationDate="2018-05-02T02:37:53.157" Score="0" Body="&lt;p&gt;Here's another approach:  &lt;a href=&quot;https://i.stack.imgur.com/GQ4WH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GQ4WH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While just a bit over the top, using something like this with somewhat lower density materials may be effective.  Doesn't turn in to electronic measurements readily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen simple wind guages that had a flap pivoted on one edge, and the angle from vertical was calibrated with wind speed.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A comment above is certainly correct:  A round pole will shed vortices and will require some post measurement tricks to get a reasonable number out of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to derive wind speed from the oscillations set up by the vortex shedding.  Another method that is fraught with calibration issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also may find that at certain wind speeds you get resonances that turn your flexible pole into so much debris, or give wild answers.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If your goal is to make a wind measuring device  without moving parts, you might be able to rig something with timing ultrasonic pulses on a round trip cross wind and up/down wind.  I bet you could get within a few percent with two sets at 45 degrees to each other.&lt;/p&gt;&#xA;" OwnerUserId="8768" LastActivityDate="2018-05-02T02:37:53.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21562" PostTypeId="1" AcceptedAnswerId="21563" CreationDate="2018-05-02T09:18:21.583" Score="3" ViewCount="6918" Body="&lt;p&gt;I am building a machine, and one of it's components is a mechanism that will convert rotational energy to a reciprocating horizontal movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of various mechanisms such as the Scotch Yoke shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ukWZe.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ukWZe.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which perform this translation. In fact, I would have used the Scotch yoke since it is mechanically simpler - however, the problem with the Scotch Yoke is that the &lt;em&gt;extent of the lateral motion is constrained to the diameter of the wheel&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want is to be able to create a lateral reciprocating movement of extent &lt;em&gt;NxD&lt;/em&gt; where &lt;em&gt;N&lt;/em&gt; is a positive real number, and &lt;em&gt;D&lt;/em&gt; is the diameter of the wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an existing mechanism that generates lateral displacements that are several times larger than the diameter of the rotating circle/wheel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZoYlT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZoYlT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/ZoYlT.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="932" LastEditorUserId="110" LastEditDate="2019-10-09T15:04:33.717" LastActivityDate="2022-12-15T22:47:18.510" Title="Mechanism to translate rotation into reciprocating horizontal movement" Tags="|mechanical-engineering|mechanisms|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21563" PostTypeId="2" ParentId="21562" CreationDate="2018-05-02T09:58:42.930" Score="3" Body="&lt;p&gt;You may find Nguyen Duc Thang's '2100 animated mechanisms' useful. There is a .zip containing a useful PDF index of these - you are interested in 'Part 2', pages 57-114 &lt;a href=&quot;http://wiki.dtonline.org/index.php/2100_Animated_Mechanical_Mechanisms&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wiki.dtonline.org/index.php/2100_Animated_Mechanical_Mechanisms&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two mechanisms I have picked out that you may find useful are: &lt;a href=&quot;https://www.youtube.com/watch?v=56fJjeeEo38&amp;amp;feature=youtu.be&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=56fJjeeEo38&amp;amp;feature=youtu.be&lt;/a&gt; and &lt;a href=&quot;https://www.youtube.com/watch?v=ITYKygWmD9Q&amp;amp;feature=youtu.be&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=ITYKygWmD9Q&amp;amp;feature=youtu.be&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but without knowing the full design constraints of your problem, I can't comment further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LAnh9.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LAnh9.gif&quot; alt=&quot;Slider-Crank Mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-05-02T09:58:42.930" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21564" PostTypeId="1" AcceptedAnswerId="21565" CreationDate="2018-05-02T15:18:51.520" Score="0" ViewCount="38" Body="&lt;p&gt;I'm doing a study on surfaces and I would like to describe all of the surfaces that I expect to encounter in the field and determine which ones will be the most difficult to detect. To do this, I need to determine what attributes in a surface I need to consider. I have no experience in optics so I'm asking this question here in hopes that someone could save me from overlooking important factors.&lt;/p&gt;&#xA;" OwnerUserId="7503" LastActivityDate="2018-05-02T17:32:15.403" Title="What attributes of a surface are relevant to light sensors?" Tags="|sensors|optics|" AnswerCount="2" CommentCount="1" ClosedDate="2018-05-05T18:37:22.073" ContentLicense="CC BY-SA 4.0" />
  <row Id="21565" PostTypeId="2" ParentId="21564" CreationDate="2018-05-02T15:42:04.353" Score="2" Body="&lt;p&gt;Surface roughness, absorptivity, transmittivity, absorption spectrum, etc might be some attributes that may help you.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2018-05-02T15:42:04.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21566" PostTypeId="2" ParentId="21548" CreationDate="2018-05-02T16:14:13.377" Score="3" Body="&lt;p&gt;You have the right general approach, and the numbers look reasonable. Your airspeed isn't very fast, so I'm not surprised that the convection coefficient you calculate is less than 10 W m&lt;sup&gt;-2&lt;/sup&gt; K&lt;sup&gt;-1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, however, that you're applying the heat flux you calculated at a plate temperature of 288 K for all other temperatures as well. In reality, the convective flux will decrease as the plate heats up to the airflow temperature. Try using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lumped_element_model#Solution_in_terms_of_object_heat_capacity&quot; rel=&quot;nofollow noreferrer&quot;&gt;lumped capacitance model&lt;/a&gt;. You end up getting a time constant (not a total time) of 6925 s; that is, every 6925 s, the temperature different is reduced by (1-1/e) = 73%. Because the heat flux magnitude affects the plate temperature but also depends on the plate temperature, this temperature approaches the airflow temperature asymptotically. People often assume that the process is essentially complete after 2-3 time constants have passed. Does this make sense?&lt;/p&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2018-05-02T16:14:13.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21567" PostTypeId="2" ParentId="21564" CreationDate="2018-05-02T17:32:15.403" Score="1" Body="&lt;p&gt;Your question lacks important parameters: in addition to the qualities of the surface or material that Pranav mentions, you need to be able to define what &quot;detect&quot; means, what sort of light sensors you think you want to use, what the interfering noise sources are, and more.&lt;br&gt;&#xA;If your target surface is not active (like a lightbulb, generating its own output) then you need to know as a minimum what the illuminating sources are.  Then you can calculate, based on surface qualities such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Bidirectional_reflectance_distribution_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;BRDF&lt;/a&gt; the expected signal magnitude at your sensor location.   But there's a huge difference between saying &quot;Look out! you're going to hit that thing!&quot;  and saying &quot;Hey, there's a chunk of rock with 20% silicates and 1% manganese at about 240 Kelvins over there&quot;&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-05-02T17:32:15.403" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21568" PostTypeId="2" ParentId="21551" CreationDate="2018-05-02T18:18:49.170" Score="1" Body="&lt;p&gt;Fishing line and brass sounds pretty sensible as a starting point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brass has a good balance of wear properties and ease of machining which makes it popular for this sort of application and it is also reasonably easy to get hold of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the string there are a few options Dyneema has very high tensile strength and is very slippery and other polymers have comparable properties, kevlar, polyester, nylon etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polymers have the advantage of being relatively soft and so will have less issues with wearing the contact surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably a key characteristic will be the tensile stiffness of the cord. Steel is significantly stiffer than most polymers, especially braided cords and this will almost certainly affect the feel and responsiveness of the keys but what is 'good' in this context is hard to judge. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-05-02T18:18:49.170" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21572" PostTypeId="1" CreationDate="2018-05-02T21:37:15.730" Score="0" ViewCount="59" Body="&lt;p&gt;I'm trying to match up two lines on a drawing I am doing, without changing the length of either line.  There is a perfect set of angles to do so- I just don't know what it each.  Is there a built in functionality to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what it currently looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H7H5w.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H7H5w.png&quot; alt=&quot;AutoCAD_Endpoint_Align&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="15934" LastActivityDate="2018-06-02T09:33:26.060" Title="AutoCAD 2019: How do I Find the Perfect Angles to Match Line Endpoints" Tags="|design|autocad|technical-drawing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21574" PostTypeId="2" ParentId="21555" CreationDate="2018-05-03T04:04:24.480" Score="2" Body="Advantages:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Makes a lighter craft.  However this is only significant for small fishing boats, canoes, punts, where they are lifted out of the water by hand.  I can see possible uses in warfare where weight may be critical.  In normal use the main energy cost of shipping is water friction.  A ship's weight is small compared to it's cargo's weight.  Reducing a ship's weight to zero wouldn't change the operating costs.  (You see references to ships traveling in ballast -- part of the cargo compartments are filled with water so that the ships handles properly.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;Disadvantages:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Expensive raw material.&lt;/li&gt;&#xA;&lt;li&gt;Expensive fabrication.  Requires special techniques to weld.&lt;/li&gt;&#xA;&lt;li&gt;Flamnable.  (See HMS Sheffield in the Falkland's war)&lt;/li&gt;&#xA;&lt;li&gt;Corrodes rapidly in salt water.  (In air or in fresh water aluminum skins over with an oxide layer.  In salt water this layer doesn't form, or is removed as fast as it forms.&lt;/li&gt;&#xA;&lt;li&gt;Softer than steel.  Parts exposed to casual friction (freight moving) would wear faster.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8768" LastActivityDate="2018-05-03T04:04:24.480" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21575" PostTypeId="2" ParentId="21505" CreationDate="2018-05-03T04:47:30.603" Score="1" Body="&lt;p&gt;First order approximation:  Do it as a work problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power = energy/time  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so Energy = power * time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy = Work = force * distance = pressure * volume.  (Do the dimensional analysis...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So power * time = pressure * volume&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rearranging again&lt;/p&gt;&#xA;&#xA;&lt;p&gt;power/pressure = volume/time&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest is picking a compatible set of units so you don't end up with something weird like board-feet per fortnight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given a 5 horse motor, and a 100 foot head, how much water can I move?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 hp = 746 watts&#xA;5 hp = 3730 watts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100 foot head = 43 psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you leave it in psi you will end up with a bizarre unit.  I think it would be kg&lt;em&gt;m/lb&lt;/em&gt;ft*s which gives me a headache trying to visualize.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Easier to conver 43 psi to about 280 kPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 3730 W/280 kPa =.013 cubic meters per second or about 13 liters per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add 5% losses to the pump.  10% pump loss.  10% hydraulic losses and you get about 10 liters per second.&lt;/p&gt;&#xA;" OwnerUserId="8768" LastActivityDate="2018-05-03T04:47:30.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21577" PostTypeId="2" ParentId="21480" CreationDate="2018-05-03T05:13:12.650" Score="1" Body="&lt;p&gt;There are MANY ways to make filters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A metal mesh is one.  Works well from gravel down to fine sand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your coffee pot you use a paper filter.  Will take out sand, but will pass suspended clay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Landscape fabric is often needle punched -- a 50 to 100 micro hole every 5 mm. As a filter it's not very efficient.  But it only has to let through a foot or  two of water a year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another not so common filter are a stack of rough disks.  The liquid passes btween the disks.  This has fairly low back pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make filters out of spun plastic that is heat fused.  This is basically what tyvek housewrap is.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The waterproofing in Goretex garments is basically teflon that is stretched so that liquid water had too much surface tension to squeeze through the holes, but water vapour passes thorugh.  There is heavy magic in getting the right plastic cross linking and the right stretching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One type of oil filter consisted of a frame that would hold a roll of toilet paper.  Oil pass through the roll lengthwise.  &lt;/p&gt;&#xA;" OwnerUserId="8768" LastActivityDate="2018-05-03T05:13:12.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21578" PostTypeId="1" AcceptedAnswerId="21583" CreationDate="2018-05-03T07:56:01.880" Score="2" ViewCount="236" Body="&lt;p&gt;Is there any scientific comparison between linear and nonlinear systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I often hear that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Nonlinear control is more sluggish than linear control.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which makes sense. But is there any research or any claim based on practical experience which supports this claim?&lt;/p&gt;&#xA;" OwnerUserId="15940" LastEditorUserId="6790" LastEditDate="2020-05-04T11:28:29.640" LastActivityDate="2020-05-04T11:28:29.640" Title="Is nonlinear control slower than linear control?" Tags="|control-engineering|control-theory|linear-control|nonlinear-control|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21579" PostTypeId="2" ParentId="21572" CreationDate="2018-05-03T08:52:46.410" Score="1" Body="&lt;p&gt;It's a really simple problem with a simple solution and I doubt that there's a built-in solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need to do is to draw a circle on each line with its radius equal to the length of the line and its centre on the stationary end of the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will encounter three possible cases:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The circles don't intersect - the lines won't meet&lt;/li&gt;&#xA;&lt;li&gt;The circles are tangential to each other - the lines will end up co-linear&lt;/li&gt;&#xA;&lt;li&gt;The circle intersect at to points - two solutions exist.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This may be tedious if you have a lot of lines, but considering the multiple possible solutions, a built-in function will also require you to inspect that the correct solution has been accepted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an image showing the two-solution case. The red lines are the initial condition and the blue lines are the two possible solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hj15w.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hj15w.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14440" LastActivityDate="2018-05-03T08:52:46.410" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21580" PostTypeId="1" AcceptedAnswerId="21590" CreationDate="2018-05-03T10:20:08.133" Score="0" ViewCount="95" Body="&lt;p&gt;Assume we have a cylindrical fin which has the effective length of &lt;strong&gt;L&lt;/strong&gt; and its efficiency is given by the equation: $$η=exp(-0.32mL)$$ where $$m=\sqrt{\frac{hP}{kA}}$$ where &lt;strong&gt;P&lt;/strong&gt; is circumference and &lt;strong&gt;A&lt;/strong&gt; is the cross sectional area of the fin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If the volume of the fin remains constant, which of the following statements is true?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;By increasing the length of the fin ...&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat transfer increases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat transfer decreases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat transfer increases then decreases.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat transfer remains constant because the volume is constant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="15941" LastEditorUserId="15941" LastEditDate="2018-05-03T14:38:20.707" LastActivityDate="2018-05-04T04:15:04.490" Title="Heat transfer in a fin with constant volume and given efficiency" Tags="|mechanical-engineering|heat-transfer|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21582" PostTypeId="2" ParentId="21580" CreationDate="2018-05-03T12:44:44.987" Score="0" Body="&lt;p&gt;The solution for the problem you posed would depend on the relation between P and A. This relation would be specific for each type of cross sectional area(circle, rectangle etc.). So you need to specify the exact shape of cross sectional area for this to be answerable.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2018-05-03T12:44:44.987" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21583" PostTypeId="2" ParentId="21578" CreationDate="2018-05-03T13:18:26.770" Score="9" Body="&lt;p&gt;&quot;Linear&quot; imposes a set of restrictions.  &quot;Non-linear&quot; simply means there are no restrictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many non-linear control schemes can be faster than linear ones.  Linear control schemes are restricted to &quot;smoothly&quot; transitioning.  Non-linear control can be implemented by suddenly slamming a digital value, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good example of a fast-responding non-linear control scheme is a typical thermostat.  When the room is below the set point, the heater is turned on fully.  When above the set point, the heater is turned off fully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously the fastest way to get to the desired temperature.  Such a system may be less &quot;good&quot; than a linear controller in other metrics, like overshoot, but that's not what you asked about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason linear control schemes are sometimes chosen is not for their speed or other control property, but because they can be mathematically analyzed more easily.  There has been much theory and techniques developed around linear systems that don't apply to non-linear systems.  Consider Laplace transforms and S-parameter analysis, for example.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-05-03T13:18:26.770" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21584" PostTypeId="1" AcceptedAnswerId="21585" CreationDate="2018-05-03T13:39:52.643" Score="0" ViewCount="897" Body="&lt;p&gt;I have no experience with actual CAD drawings and CNC machines so please forgive me if this is a really stupid question...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway I have a part which I have 3D printed and it contains some internal screw thread for an ISO metric screw. I modelled the threading myself so it will get 3D printed which took some trial and error. Now since I need it in high quality metal I want to CNC it. But I am wondering whether I have to let the screw thread stay in my model or indicate somehow this is actually ISO metric screw thread of this spec? Since I imagine there is special machinery/settings to make the threading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is appreciated&lt;/p&gt;&#xA;" OwnerUserId="15945" LastActivityDate="2018-05-08T13:12:51.250" Title="How to indicate standard screw thread in drawing" Tags="|cad|threads|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21585" PostTypeId="2" ParentId="21584" CreationDate="2018-05-03T17:51:54.803" Score="2" Body="&lt;p&gt;Most CAD software has the facility to automatically generate screw threads in  a model. Often this will be represented on the model as a texture to reduce the complexity of the model but the file will contain the specification for the thread. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally for machining you want to specify the tread rather than modeling it because, as you surmise the process for machining a thread is typically different to machining some arbitrary shape. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all else fails the best thing to do is probably just mark have the holes and specify the treads with annotation. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-05-03T17:51:54.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21588" PostTypeId="2" ParentId="16720" CreationDate="2018-05-04T03:18:00.803" Score="1" Body="&lt;p&gt;By default Abaqus decides time increments adaptively. The initial time increment is typically a guess. If the guess is too large to solve the equations for the first time increment, it guesses smaller and repeats the process. If the guess could be larger and the equations still solve well, it increases the guess. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your simulations run in a reasonable time frame, don't worry about the time increment. If you have to run a bunch of similar simulations and would like to speed things up a little, open up the monitor while your simulation is running and watch how Abaqus adaptively sets the first increment. Write down what it settles on, and use that as the first increment size for future simulations. This will save time because Abaqus can jump straight to the right answer without a ton of guessing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have to choose a fixed time increment scheme, the proper time increments depend on a lot of factors, including the physics you are trying to simulate and the non-linearity of your problem with respect to time. &lt;/p&gt;&#xA;" OwnerUserId="15951" LastActivityDate="2018-05-04T03:18:00.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21589" PostTypeId="2" ParentId="16836" CreationDate="2018-05-04T03:59:08.737" Score="0" Body="&lt;p&gt;For spark ignition methods, I would recommend using an already existing product if possible. You can modify the length of the high voltage wire(s) and spark points into something more attractive as needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often there is only one wire and the chassis is used for the return. This would be favorable in your design, since one wire is less clutter than two. They make some good &lt;a href=&quot;https://www.mcmaster.com/#8296k12/=1cot0ud&quot; rel=&quot;nofollow noreferrer&quot;&gt;small diameter high voltage wire&lt;/a&gt;, that you could probably chase up the inside of one of your pipes. Just make sure the voltage rating of the wire is higher than the voltage required for the spark gap. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Dielectric_strength#Breakdown_field_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;dielectric strength of air is 3MV/m&lt;/a&gt;, so for 15kv wire the spark gap needs to be less than 5mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make your own spark points you will want them to be out of high temperature oxide resistant metal, or over time (100+ ignitions) the surfaces of the sparking materials will erode to the point of failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B007VAXQGM&quot; rel=&quot;nofollow noreferrer&quot;&gt;pizeo igniter&lt;/a&gt; generates a high voltage spark by a push button spring striking a pizeo crystal. This obviously requires mechanical input, but wires can be extended to where you need them and no batteries are required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B007VAXQF8&quot; rel=&quot;nofollow noreferrer&quot;&gt;electric bbq igniter&lt;/a&gt; uses a AA or AAA battery, some internal electronics to create a pulse through a transformer to get the high voltage. Beyond the unit, the high voltage wire and ceramic insulated spark points are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option to consider is a nicrome hot wire (especially if your device turns on and off frequently or has variable flow rates). You run low voltage DC electricity through the hot wire till it glows red and it will ignite the gas. Looks like &lt;a href=&quot;http://www.instructables.com/id/NiCrome-%22hotwire%22-ignition-for-propane-flame-effec/&quot; rel=&quot;nofollow noreferrer&quot;&gt;someone on instructables&lt;/a&gt; used it for a similar application. This requires access to more electricity than the other options, but it makes your controls simpler since you would only need to modulate the gas. It is silent and doesn't make spark noises. It would also keep low gas flows ignited beyond the point of stable flame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would only need 5 mm of &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B0799QTFRW&quot; rel=&quot;nofollow noreferrer&quot;&gt;20gauge nicrome&lt;/a&gt; and 5 volt at a couple amps would probably do the trick. You may need to put it just to the side of the gas stream so it doesn't cool down too much in a rapid on/off situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty cool project, good luck as you move forward on it!&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-05-04T03:59:08.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21590" PostTypeId="2" ParentId="21580" CreationDate="2018-05-04T04:15:04.490" Score="0" Body="&lt;p&gt;The math for this problem goes like this:&#xA;The fin has a circular cross section and a constant volume. We can write a relationship between volume, length and radius: $V = L*pi*r^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving that for radius: $r = \sqrt{\frac{V}{L \pi}}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Efficiency is the ratio of the effectiveness of a fin divided by how effective that fin would be if 100% of its area were at the temperature of the base. Total heat transfer is proportional to efficiency times area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = hA\eta = hL\pi r = hL\pi\sqrt{\frac{V}{L\pi}} = h\sqrt{L\pi V} \eta$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can do similar substitutions for efficiency.&#xA;$\eta = exp(-0.32\sqrt{\frac{hP}{kA}}) = exp(-0.32\sqrt{\frac{h\pi r}{k\pi r^2}}) = exp(-0.32\sqrt{\frac{h}{k r}}) = exp(-0.32\sqrt{\frac{h}{k \sqrt{\frac{V}{L \pi}}}}) = exp(-0.32\sqrt{\frac{h}{k \sqrt{\frac{V}{\pi}}}}L^{1/4})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we put these two terms together we can ignore everything that isn't L for the purpose of this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Q = things*L^{1/2}*exp(-things*L^{1/4})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term with L increases with L, the second decreases and definitely not at the same rate. Put in some dummy values for &quot;things&quot; and you can get this graph with L on the horizontal and Q on the vertical. It clearly goes up and then down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d9K8i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d9K8i.png&quot; alt=&quot;L vs Q&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of this math was not really necessary; Common sense could suffice. The fin equation does not account for heat transfer from the fin tip. You need an additional term for that. If we have a very short fin, it will be $\sqrt{very}$ big around, and have very little area (very &gt; $\sqrt{very}$). Even though it will have 100% efficiency, there will be very little heat transfer. If we have a very long fin, the efficiency will be terrible because most of the fin is very far from the heat source and is cold. At some point in the middle there is an optimum.&lt;/p&gt;&#xA;" OwnerUserId="15951" LastActivityDate="2018-05-04T04:15:04.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21591" PostTypeId="1" CreationDate="2018-05-04T05:08:43.223" Score="-1" ViewCount="31" Body="&lt;p&gt;i didn't understand it how the air pressure cancel out.my eye is so sensitive,and why is the air pressure that comes towards my face doesn't hurt my eye? why is our body doesn't feel a thing while the air pressure being exerted?  &lt;/p&gt;&#xA;" OwnerUserId="15005" LastActivityDate="2018-05-04T11:47:17.120" Title="how is the air pressure that exerted to our body cancels out?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21592" PostTypeId="2" ParentId="21378" CreationDate="2018-05-04T06:36:09.997" Score="0" Body="&lt;p&gt;The best way to do it is with fields. In this specific case I had to set up two additional coordinate systems. These were cylindrical coordinate systems the were located at the centre of the two circular arcs which defined the surfaces to which the pressure was applied. &#xA;The two fields were then defined for in their respective coordinate frame using a PCL function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure is then applied like a normal constant pressure but the input is the field and not some static value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sorry I don't have any screenshots to better illustrate this but I hope it helps someone.&lt;/p&gt;&#xA;" OwnerUserId="15777" LastActivityDate="2018-05-04T06:36:09.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21594" PostTypeId="1" CreationDate="2018-05-04T10:46:25.020" Score="6" ViewCount="2845" Body="&lt;p&gt;Compressor is hot due to gas compression, and the gas then cooled in condenser. AC installed in tropical climate. Why compressor have insulation barrier?&lt;img src=&quot;https://i.stack.imgur.com/tSAZo.jpg&quot; alt=&quot;AC outdoor units, compressor&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2022-08-03T08:59:40.940" Title="Why home Air Conditioning AC compressor insulated" Tags="|hvac|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="21595" PostTypeId="2" ParentId="21594" CreationDate="2018-05-04T11:38:21.897" Score="0" Body="&lt;p&gt;Just guessing, but some possibilities:&lt;ol&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;To reduce noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;li&gt;To reduce heat exchange in places other than where it is wanted/used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;/ol&gt;&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-05-04T11:38:21.897" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21596" PostTypeId="2" ParentId="21591" CreationDate="2018-05-04T11:47:17.120" Score="1" Body="&lt;p&gt;Think about a sheet of paper held loosely horizontally.  There are about 14.7 pounds of pressure pushing down on every square inch of the top surface.  For a 8&amp;frac12; x 11 inch paper, that comes out to about 1,400 pounds.  The reason that paper isn't ripped out of your hand and slammed to the ground is because the same pressure is also pushing up on the bottom side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The various parts of your body, and any other objects lying around, work the same way.  The air pressure is pushing all around, and the net force is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, almost 0.  Air pressure decreases with altitude.  For large enough objects, the force pushing down on the top is noticeably less than the force pushing up at the bottom.  If you do all the math, you find that the net upward force is directly proportional to the volume of the object.  Blimps and helium-filled party balloons exploit this principle and are actually able to rise thru the air against gravity.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-05-04T11:47:17.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21597" PostTypeId="1" AcceptedAnswerId="21608" CreationDate="2018-05-04T13:32:03.660" Score="1" ViewCount="2975" Body="&lt;p&gt;We all know the typical streamline shape which looks like a tear. &lt;strong&gt;But what is the optimum shape (for speeds 0-200 km/h)?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the front a half sphere or is it an ovoid? How are the sides and the end formed? Can it be described in a formula or in a bezier curve? What is the drag coefficient compared to standard tear drops? Image with explanations how the shape must be built to minimize the drag are perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And no, I don't accept flat plates and perfectly laminar airflow :). The shape must be built around a pipe with diameter R and must not be longer than 10R.&lt;/p&gt;&#xA;" OwnerUserId="1137" LastActivityDate="2018-10-21T23:07:07.683" Title="What is the optimal streamlined shape?" Tags="|aerodynamics|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="21598" PostTypeId="1" CreationDate="2018-05-04T14:08:18.677" Score="0" ViewCount="1333" Body="&lt;p&gt;I am little bit curious about, is hall sensor also used in PMSM motor drive? Since in PMSM motor the voltage applied is sinusoidal, all 3 phase are active, so a rotating flux is introduced, which is not in BLDC motor. In BLDC motor only two phase are active at any instant. So it is necessary in switching in BLDC motor. But due to rotating flux in PMSM there is no any requirement of position sensor. Is my view correct or not?&#xA;Please help me.   &lt;/p&gt;&#xA;" OwnerUserId="15908" LastEditorUserId="15908" LastEditDate="2018-05-05T03:06:20.280" LastActivityDate="2018-09-04T15:57:56.370" Title="Is hall sensor also used in PMSM drive scheme as like used in BLDC motor?" Tags="|electrical-engineering|motors|power-electronics|electrical|electric-vehicles|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21599" PostTypeId="1" AcceptedAnswerId="21607" CreationDate="2018-05-04T14:08:33.783" Score="1" ViewCount="181" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nsZJ0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nsZJ0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Source: &lt;a href=&quot;http://www.cbc.ca/news/canada/new-brunswick/friday-flood-new-brunswick-2018-1.4647979&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cbc.ca/news/canada/new-brunswick/friday-flood-new-brunswick-2018-1.4647979&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to calculate how much weight a vessel can carry (in fresh water) before it submerges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The water is not disturbed (no waves or wind) and the cargo does not move&lt;/li&gt;&#xA;&lt;li&gt;The vessel weighs 200 lbs when empty&lt;/li&gt;&#xA;&lt;li&gt;The vessel's volume is 60 cubic ft&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="10536" LastEditorUserId="10536" LastEditDate="2018-05-04T14:14:31.513" LastActivityDate="2018-05-04T18:40:41.593" Title="How much weight can a floating vessel carry before submerging?" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21601" PostTypeId="1" CreationDate="2018-05-04T14:48:29.300" Score="-1" ViewCount="59" Body="&lt;p&gt;I have motor 1 1000 which rpm &amp;amp; motor 2 which 1500 rpm connected with shaft via gear ratio of 1:2 and what will be the shaft speed ? how can i calculate shaft speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am considering the shaft gear ratio as 3 ,then what will be the speed of shaft ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the shaft speed will be combined speed of 2 motor ?&#xA;or it will be maximum speed of any of this motor ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i am looking for suggestion to make a model to simulate &lt;/p&gt;&#xA;" OwnerUserId="15954" LastActivityDate="2018-05-04T19:34:41.290" Title="shaft speed when 2 motor connected" Tags="|mechanical-engineering|electrical|" AnswerCount="1" CommentCount="1" ClosedDate="2018-05-06T11:56:22.043" ContentLicense="CC BY-SA 4.0" />
  <row Id="21603" PostTypeId="2" ParentId="21599" CreationDate="2018-05-04T15:02:08.070" Score="0" Body="&lt;p&gt;Damn customary units, its equal to the weight of the displaced water minus own weight. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;60*0.3048m^3 * 1000 kg/m^3 -0.45kg*200=1600kg = 3500 lbs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would load it with max 0.8, so it is 1280 kg= 2800 lbs.&lt;/p&gt;&#xA;" OwnerUserId="1137" LastActivityDate="2018-05-04T15:02:08.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21604" PostTypeId="2" ParentId="21599" CreationDate="2018-05-04T16:45:18.887" Score="1" Body="&lt;p&gt;To show you how simple this in metric units:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Your boat is 91 kg.&lt;/li&gt;&#xA;&lt;li&gt;Your boat is 1700 litres.&lt;/li&gt;&#xA;&lt;li&gt;Water weighs 1 kg/L.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As Archimedes pointed out a couple of thousand years ago, the buoyancy will be equal to the mass of the water displaced.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Your boat can displace up to 1700 L. The buoyancy at that load will be 1700 kg.&lt;/li&gt;&#xA;&lt;li&gt;The boat weighs 91 kg. The maximum load will be 1700 - 91 = 1609 kg.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Go metric, man. Go metric. It's so much simpler.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-05-04T16:45:18.887" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21606" PostTypeId="2" ParentId="21597" CreationDate="2018-05-04T17:01:59.900" Score="1" Body="&lt;p&gt;The optimal shape of a streamlined wing, boat, wind mill blade, is a balance between several factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The speed, hence the variable wing angle, both in delta wings and angle the wing is attached to the fuselage.&lt;/li&gt;&#xA;&lt;li&gt;The density and viscosity of the fluid: e.g., different altitudes of flying require different shapes and configuration.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The utility of the airfoil or rudder or boat, how much volume it should hold, how wide it should be? If you want a streamlined bus it could be too long for the road!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The cost of manufacturing: precision and efficiency are expensive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The training of the pilot or operator: fighter pilots can handle a razor thin wing tip with minimal dynamic balance and thin envelope of authority, to the advantage of agility and speed of aerobatics; commercial planes have bull-nose wing tip to afford more versatility and better lift at angles near stall.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Basically the teardrop shape is practical in most cases, but as speed increases the shape approaches that of a thin blade with sharp tip to cut its way like a knife.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="77" LastEditDate="2018-10-21T23:07:07.683" LastActivityDate="2018-10-21T23:07:07.683" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21607" PostTypeId="2" ParentId="21599" CreationDate="2018-05-04T18:20:53.350" Score="1" Body="&lt;p&gt;A problem is that volume is not exactly displacement.  Typically a boat will be a little lower in the middle and / or more weighted in the rear.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretend this is a perfect boat and volume = displacement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buoyancy is exactly displaced water * water density.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density of water is nominally 62 lbs per cubic foot.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;60 ft^3 * 62 lb / ft^3 = 200 lbs + x lbs&lt;br&gt;&#xA;   3720 lb - 200 lbs = x lbs&lt;br&gt;&#xA;   x = 3520 lb&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-05-04T18:40:41.593" LastActivityDate="2018-05-04T18:40:41.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21608" PostTypeId="2" ParentId="21597" CreationDate="2018-05-04T19:03:45.967" Score="3" Body="&lt;p&gt;There's a lot more to it than just minimum drag shape. At 200 kph, you want it to be aerodynamically stable as well. If the pipe is supporting the fairing, and you want to limit the load on the pipe, then the pipe has to be forward of the center of resistance of the fairing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape will depend on temperature, altitude, humidity, the size of the pipe, turbulence and other non-uniform features in the freestream, and manufacturing ability. Getting to 95% of the ideal is pretty easy to achieve for say a 6&quot; pipe. Getting that last 5% can be tricky, requiring very stiff and strong materials and very high tolerances, like +/- .0005 inch overall and no wavyness along the cord. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some common ones. Plots are for Reynolds numbers of 500,000 and 1,000,000. You can adjust as necessary.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://airfoiltools.com/airfoil/details?airfoil=e862-il&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eppler 862&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://airfoiltools.com/airfoil/details?airfoil=e863-il&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eppler 863&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://airfoiltools.com/airfoil/details?airfoil=e864-il&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eppler 864&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-05-04T23:00:46.020" LastActivityDate="2018-05-04T23:00:46.020" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21609" PostTypeId="2" ParentId="21601" CreationDate="2018-05-04T19:34:41.290" Score="2" Body="&lt;p&gt;If you couple together two motors of unequal speed through a gear system which allows the faster motor to match speed with the slower one, then the rotating speed will equal that of the slower one and their torque outputs will add. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you couple them in a manner where their output speeds do not match, then the more powerful motor will drive the other one in reverse and their torque outputs will subtract.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not advisable to couple together motors whose operating speeds are different.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-05-04T19:34:41.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21610" PostTypeId="2" ParentId="21597" CreationDate="2018-05-04T20:25:26.657" Score="2" Body="&lt;p&gt;The general rule of thumb for laminar flow is that the tail needs to be about 4 times as long as the nose because it is more difficult to slow down a flow than to speed it up. Equally you want to avoid any geometric discontinuities so a sensible way to do this is treat it as circular arcs at a tangent to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZeQVH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZeQVH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The grey cube is 1x1 aligned with the major axes to give a sense of orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are a few caveats to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, scale matters in fluid flow and you get to a point where the onset of turbulence is inevitable at a given scale and speed whatever you do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally: when you talk about optimisation this, by definition, implies that you are seeking a useful balance between conflicting requirements. Reducing drag is only one requirement. In reality and even in in fluid dynamics the body under consideration has some purpose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:  Aircraft need to generate lift, thrust, control and carry some payload, so in practice a lot of aerodynamics is managing drag and turbulence rather than attempting to eliminate it entirely. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="77" LastEditDate="2018-10-21T23:05:03.920" LastActivityDate="2018-10-21T23:05:03.920" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21611" PostTypeId="2" ParentId="21594" CreationDate="2018-05-04T21:09:56.037" Score="2" Body="&lt;p&gt;Knowledgeable service man said it was for noise reduction. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-05-04T21:09:56.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21612" PostTypeId="1" CreationDate="2018-05-05T03:31:55.323" Score="1" ViewCount="100" Body="&lt;p&gt;Does anyone knows what is the name of this rod? are these available in the market?&lt;a href=&quot;https://i.stack.imgur.com/a57Sc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a57Sc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15958" LastActivityDate="2018-05-05T17:12:02.470" Title="rod technical name" Tags="|mechanical-engineering|product-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21613" PostTypeId="2" ParentId="21612" CreationDate="2018-05-05T05:20:39.750" Score="1" Body="&lt;p&gt;it looks like a piece of plain 1/2&quot; or 5/8&quot; steel rod with locking collars on its ends.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-05-05T05:20:39.750" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21614" PostTypeId="2" ParentId="21612" CreationDate="2018-05-05T06:30:59.120" Score="1" Body="&lt;p&gt;This could be called a « locking pin » or « locating pin » or « sliding stop » , depends on the function but most would understand the function and form.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-05-05T06:30:59.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21615" PostTypeId="2" ParentId="21612" CreationDate="2018-05-05T17:12:02.470" Score="1" Body="&lt;p&gt;This is part of work-out bench called &quot; Rogue Adjustable Bench.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see in the photo this Pin is to adjust the angle of the bench. This is a fitness company's brochure photo I used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there is more info &lt;a href=&quot;https://www.roguefitness.com/rogue-adjustable-bench-2-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sNY44.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sNY44.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-05T17:12:02.470" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21616" PostTypeId="1" CreationDate="2018-05-05T17:40:49.583" Score="1" ViewCount="696" Body="&lt;p&gt;The peak expiratory flow meter is usually used to manage asthma. I wonder what's inside it and how can it measure peak flow?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/md09K.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/md09K.jpg&quot; alt=&quot;peak expiratory flow meter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the difference between this and an anemometer? Can an anemometer be used to do this work? Or is there something can be used to DIY one (like small computer fan)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dadZc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dadZc.jpg&quot; alt=&quot;anemometer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15965" LastActivityDate="2018-05-05T17:40:49.583" Title="What's inside a peak expiratory flow meter, and what's the difference with an anemometer?" Tags="|airflow|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21618" PostTypeId="1" AcceptedAnswerId="21619" CreationDate="2018-05-05T19:02:36.307" Score="1" ViewCount="40" Body="&lt;p&gt;Are there any simplified formulas that could be used to generate plausible airplane, spaceplane/SSTO, and spacecraft performance specs for a game given some basic vehicle and contextual attributes? By plausible, I mean within 25% of the value a full analysis might achieve. For example, could I roughly determine maximum acceleration and top speed from the following attributes:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;VEHICLE&#xA;thrust (in real units)&#xA;mass (in real units)&#xA;aerodynamics (any intuitive scale: e.g. 0.0 = horrible, 1.0 = ideal, Boeing &#xA;747 set arbitrarily at 0.50)&#xA;&#xA;&#xA;CONTEXT&#xA;atmosphere (density? pressure? friction? How best to measure simply?)&#xA;gravity (real units)&#xA;angle of travel (0 = parallel to planet, 90 = perpendicular to planet)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="15966" LastActivityDate="2018-05-05T22:04:45.940" Title="simplified acceleration and max speed calculations for a game?" Tags="|modeling|simulation|" AnswerCount="1" CommentCount="2" ClosedDate="2018-05-06T22:26:04.493" ContentLicense="CC BY-SA 4.0" />
  <row Id="21619" PostTypeId="2" ParentId="21618" CreationDate="2018-05-05T22:04:45.940" Score="0" Body="&lt;p&gt;The maximum acceleration can be calculated simply by comparing your vehicle's mass and thrust values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=m*a$$&#xA;$$acceleration = \frac{thrust}{mass}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terminal velocity can be estimated by balancing your thrust values and the drag equation. When drag balances thrust, you cannot travel any faster&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F_D=\frac{1}{2}\rho u^2C_DA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$terminal\ velocity=\sqrt{\frac{2*thrust}{air \ density* drag\ coefficient*frontal\ area}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have also included 'angle of travel' as a factor - this will affect the total driving force in the direction of travel, and consequently the terminal velocity. If you are flying vertically upwards, the force due to gravity ($F=mg$) must be subtracted from the trust. If you are flying downwards at an angle of $20°$, then the component in the direction of travel ($F=mg\sin(\theta)$) should be added to the thrust.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-05-05T22:04:45.940" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21620" PostTypeId="2" ParentId="21069" CreationDate="2018-05-06T05:15:40.237" Score="1" Body="&lt;p&gt;I found a better way to solve this problem by using singularity functions. Not only is the solution simple and elegant, one equation can be written for the entire beam! Also, the format for each equation is similar, so it is easy to copy the formula in a spreadsheet or program for one result (such as for the shear) and revise the equation for other results (such as the displacement) -- just change the multiplier and exponent. Furthermore, you do not need to calculate all of the variables (shear, moment, slope) to calculate any of the the other equations (such as displacement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not to go into too much detail, but singularity functions are like a switch to turn a load on. Furthermore, they can be integrated just like a normal equation. The singularity functions work as follow: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\langle x-a \rangle ^{-1} &amp;amp;=&#xA;  \begin{cases}&#xA;  0, &amp;amp; x \neq a \\&#xA;  1, &amp;amp; x=a&#xA;  \end{cases} \\&#xA;\langle x-a \rangle ^0 &amp;amp;=&#xA;  \begin{cases}&#xA;  0, &amp;amp; x \neq a \\&#xA;  1, &amp;amp; x=a&#xA;  \end{cases} \\&#xA;\text{if n&amp;gt;0, } \langle x-a \rangle ^n &amp;amp;=&#xA;   \begin{cases}&#xA;   0,  &amp;amp; \text{x&amp;lt;a} \\&#xA;   (x-a)^n, &amp;amp; \text{x&amp;gt;a}&#xA;\end{cases} \\&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;&amp;amp; \int \langle x-a \rangle ^{-1} dx = \langle x-a \rangle ^0 \\&#xA;&amp;amp; \int \langle x-a \rangle ^0 dx = \langle x-a \rangle ^1 \\&#xA;&amp;amp; \int \langle x-a \rangle ^n dx = \frac{1}{n+1} \langle x-a \rangle ^{(n+1)}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect12.d/IAST.Lect12.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect12.d/IAST.Lect12.pdf&lt;/a&gt; is a good example of using singularity functions to solve beam deflections.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the singularity functions, the beam to solve&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ghjQH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ghjQH.png&quot; alt=&quot;trapezoidal load&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is represented as follows:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/m58li.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m58li.png&quot; alt=&quot;singularity equivalent&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $slope=(Q_C-Q_B)/(x_C-x_B)$, the load beginning with $Q_B$ continues for the remaining length, and the load beginning with $Q_C$ turns the load &quot;off&quot;. (The singularity function turns a load on but not off, so any load that ends before the end of the beam requires a second load and singularity function to remove the load.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Letting $x_A=0$ and $x_D=L$, the equation for the load $q(x)$ is $$&#xA;\bbox[yellow]&#xA;{&#xA;\begin{align}&#xA;q(x)=R_A\langle x \rangle ^{-1} + R_D\langle x-L \rangle ^{-1}&amp;amp;-Q_B\langle x-x_B \rangle ^0-slope\langle x-x_B \rangle ^1 \\&#xA;&amp;amp;+Q_C\langle x-x_C \rangle ^0+slope\langle x-x_C \rangle ^1&#xA;\end{align}&#xA;\tag 1&#xA;}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for the shear is $V(x)=\int q(x)dx$, $$&#xA;\bbox[yellow]&#xA;{&#xA;\begin{align}&#xA;V(x)=R_A\langle x \rangle ^{0} + R_D\langle x-L \rangle ^{0}&amp;amp;-Q_B\langle x-x_B \rangle ^1-\frac{1}{2}slope\langle x-x_B \rangle ^2 \\&#xA;&amp;amp;+Q_C\langle x-x_C \rangle ^1+\frac{1}{2}slope\langle x-x_C \rangle ^2&#xA;\end{align}&#xA;\tag 2&#xA;}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for the moment is $M(x)=\int V(x)dx$, $$&#xA;\bbox[yellow]&#xA;{&#xA;\begin{align}&#xA;M(x)=R_A\langle x \rangle ^{1} + R_D\langle x-L \rangle ^{1}&amp;amp;-\frac {1}{2}Q_B\langle x-x_B \rangle ^2-\frac{1}{6}slope\langle x-x_B \rangle ^3 \\&#xA;&amp;amp;+\frac {1}{2}Q_C\langle x-x_C \rangle ^2+\frac{1}{6}slope\langle x-x_C \rangle ^3&#xA;\end{align}&#xA;\tag 3&#xA;}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here on, singularity functions that begin at x=0 can be converted to regular functions, and singularity functions that begin at X=L can be dropped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for the slope is $EI\cdot\theta(x)=\int M(x)dx$, $$&#xA;\bbox[yellow]&#xA;{&#xA;\begin{align}&#xA;EI\cdot\theta(x)=\frac {1}{2}R_Ax^2+C_1 &amp;amp;-\frac {1}{6}Q_B\langle x-x_B \rangle ^3-\frac{1}{24}slope\langle x-x_B \rangle ^4 \\&#xA;&amp;amp;+\frac {1}{6}Q_C\langle x-x_C \rangle ^3+\frac{1}{24}slope\langle x-x_C \rangle ^4&#xA;\end{align}&#xA;\tag 4&#xA;}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for the deflection is $EI\cdot\delta(x)=\int \theta(x)dx$, $$&#xA;\bbox[yellow]&#xA;{&#xA;\begin{align}&#xA;EI\cdot\delta(x)=\frac {1}{6}R_Ax^3+C_1x+C_2 &amp;amp;-\frac {1}{24}Q_B\langle x-x_B \rangle ^4-\frac{1}{120}slope\langle x-x_B \rangle ^5 \\&#xA;&amp;amp;+\frac {1}{24}Q_C\langle x-x_C \rangle ^4+\frac{1}{120}slope\langle x-x_C \rangle ^5&#xA;\end{align}&#xA;\tag 5&#xA;}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction force $R_A$ can be determined from the moment equation. At x=L, M(x)=0,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;0=R_AL + R_D(0) &amp;amp;-\frac {1}{2}Q_B(L-x_B)^2-\frac{1}{6}slope(L-x_B)^3 \\&#xA;&amp;amp;+\frac {1}{2}Q_C(L-x_C)^2+\frac{1}{6}slope(L-x_C)^3\\&#xA;\tag 6&#xA;\end{align}&#xA;\begin{align}&#xA;R_AL =\frac {1}{6}[&amp;amp;+3Q_B(L-x_B)^2+slope(L-x_B)^3 \\&#xA;&amp;amp;-3Q_C(L-x_C)^2-slope(L-x_C)^3] \\&#xA;\tag 7&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\bbox[yellow]&#xA;{&#xA;\begin{align}&#xA;\text{thus }R_A =\frac {1}{6L}[&amp;amp;+3Q_B(L-x_B)^2+slope(L-x_B)^3 \\&#xA;&amp;amp;-3Q_C(L-x_C)^2-slope(L-x_C)^3]&#xA;\end{align}&#xA;\tag 8&#xA;}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\text{and }R_D &amp;amp;= \text{total load }-R_A \\&#xA;&amp;amp;= \frac 12(Q_B+Q_C)(x_C-x_B)-R_A&#xA;\tag 9&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At x=0, $\delta=0$ from which $C_2=0$.&#xA;At x=L, $\delta=0$ from which $C_1$ is calculated as follows:&#xA;\begin{align}&#xA;EI\cdot(0)=\frac 16R_AL^3+C_1L &amp;amp;-\frac{1}{24}Q_B(L-x_B)^4-\frac{1}{120}slope(L-x_B)^5\\&#xA;&amp;amp;+\frac{1}{24}Q_C(L-x_C)^4+\frac{1}{120}slope(L-x_C)^5\\&#xA;\tag{10}&#xA;\end{align}&#xA;\begin{align}&#xA;-C_1L =\frac {1}{120}[20R_AL^3 &amp;amp;-5Q_B(L-x_B)^4- slope(L-x_B)^5\\&#xA;&amp;amp;+5Q_C(L-x_C)^4+slope(L-x_C)^5]\\&#xA;\tag{11}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\bbox[yellow]&#xA;{&#xA;\begin{align}&#xA;\text{thus }C_1 =\frac {1}{120L}[-20R_AL^3 &amp;amp;+5Q_B(L-x_B)^4+ slope(L-x_B)^5\\&#xA;&amp;amp;-5Q_C(L-x_C)^4-slope(L-x_C)^5]\\&#xA;\end{align}&#xA;\tag{12}&#xA;}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All values are now known, so any of the first 5 equations can be used to calculate the quantity at any position $x$ long the length of the beam $0&amp;lt;x&amp;lt;L$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some example results for $x_B=20,\, x_C=50,\, x_D=L=90,\, E=1\times10^6,\, I=8.5$:&#xA;\begin{array}{cc|cccc }&#xA;Q_B &amp;amp;Q_C &amp;amp;\theta_A &amp;amp;\theta_D &amp;amp;\delta_B &amp;amp;\delta_C \\&#xA;\hline&#xA;20 &amp;amp;35 &amp;amp;-0.0482 &amp;amp;+0.0429 &amp;amp;-0.8872 &amp;amp;-1.2971 \\&#xA;35 &amp;amp;10 &amp;amp;-0.0388 &amp;amp;+0.0325 &amp;amp;-0.7075 &amp;amp;-0.9958 \\&#xA;100 &amp;amp;-100 &amp;amp;+0.0036 &amp;amp;-0.0140 &amp;amp;+0.0987 &amp;amp;+0.3516&#xA;\end{array}&lt;/p&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2018-05-06T05:15:40.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21621" PostTypeId="2" ParentId="21598" CreationDate="2018-05-06T12:54:54.960" Score="0" Body="&lt;p&gt;Yes, Hall sensors can be used as position feedback with PMSMs (permanent magnet synchronous motors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many applications you can't just blindly apply a frequency ramp and know that the motor will follow.  Most of the time you need some position feedback to know where the rotor is within the magnetic cycle.  This then dictates the phase of the drive signals.  Hall sensors are one means to provide this position feedback.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-05-06T12:54:54.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21622" PostTypeId="1" CreationDate="2018-05-06T13:30:25.240" Score="1" ViewCount="1069" Body="&lt;p&gt;Suppose I have a bolt that goes through a hole in a plate and is under tension. If I place a washer between the bolt head and the plate, does static frictional force (torque actually) really double? I'm assuming the bolt, washer, and plate are all of the same material for simplicity.&lt;/p&gt;&#xA;" OwnerUserId="15971" LastActivityDate="2018-05-07T02:15:21.410" Title="Does a washer really increase static friction force?" Tags="|statics|friction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21624" PostTypeId="2" ParentId="21622" CreationDate="2018-05-06T16:51:40.733" Score="6" Body="&lt;p&gt;No it doesn't, and it isn't meant for that either. A standard flat washer, that is. It's meant to give the nuts and bolts a better support, and it protects the material against the nut which may otherwise scratch it upon tightening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A washer may even reduce the friction on the nut, since the fastened material may in many cases have a higher friction against the nut than the washer does. That's why we either use lock washers, or nylock nuts, flange serrated nuts, or other types of self securing nuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your good understanding, it doesn't matter what kind or how many washers or nuts you apply, the one with the least friction determines the total friction. Like the weakest link determines the strength of a chain.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-05-06T16:51:40.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21625" PostTypeId="1" AcceptedAnswerId="21628" CreationDate="2018-05-06T18:02:19.410" Score="2" ViewCount="610" Body="&lt;p&gt;Do homes increase in temperature due to solar gain with curtains open if it's overcast?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, Does having curtains closed reduce the temperature of the home on an overcast day?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, Does an overcast day prevent the warming effect of the sun hitting a homes window and raising it's temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, Closing the curtains on a sunny clearly keeps a room cooler. Does closing the curtains on an overcast day have the same effect? Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="15975" LastEditorUserId="15975" LastEditDate="2018-05-06T21:52:39.130" LastActivityDate="2018-05-06T22:30:46.097" Title="Is a Home's Temperature Affected by Solar Gain Through Windows on Cloudy Days?" Tags="|thermodynamics|temperature|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="21626" PostTypeId="1" CreationDate="2018-05-06T20:40:14.483" Score="0" ViewCount="58" Body="&lt;p&gt;I have an electronic assembly and part of what it does is to measure and report temperature. The electronics are housed is a plastic enclosure, and there is a small fan blowing out of the enclosure. There are also intake openings in the enclosure wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a simple fact that the temperature inside the enclosure is above ambient temperature seeing as the electronics itself is a heat source and the enclosure naturally has thermal insulating properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said I want to report ambient temperature without my temperature sensor being outside the enclosure, and I'm willing to perform a calibration step to get there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions are,  (1) is it sufficient to model the transform as a simple offset? (2) What tools can I use to understand the relationship between the internal and external temperature in my enclosure,  over a range of external temperatures?  (3) What are the most important variables I need to control in order for my calibration to hold in practice? &lt;/p&gt;&#xA;" OwnerUserId="15976" LastEditorUserId="15976" LastEditDate="2018-05-06T21:50:38.890" LastActivityDate="2018-05-06T21:50:38.890" Title="Thermal Modeling" Tags="|mechanical-engineering|modeling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21627" PostTypeId="2" ParentId="21626" CreationDate="2018-05-06T21:31:40.417" Score="1" Body="&lt;p&gt;It's not clear why you wouldn't put the temperature sensor close to one of the points where outside air is drawn in and shield it from the inside temperature. However, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provided the power dissipated in the box (or the energy consumed inside the box) is constant then the temperature rise, $ \Delta T $ of the inside relative to the outside will be governed by the thermal resistance of the box inside to outside. This would be analogous to Ohm's law, $ V = IR $ but now we have $ \Delta T = QR $ where &lt;em&gt;Q&lt;/em&gt; is the heat flow and &lt;em&gt;R&lt;/em&gt; is the thermal resistance which in your case should be a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If power consumed inside the box is not a constant then Q will vary so you would have to measure power consumption and correct for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simple system a fixed offset should work and it should be possible to verify with a pair of probes or sensors such as the &lt;a href=&quot;http://www.ti.com/lit/ds/symlink/lm35.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;LM35&lt;/a&gt; which give 10 mV/°C output and, so, can be measured using a decent multimeter.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-05-06T21:31:40.417" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21628" PostTypeId="2" ParentId="21625" CreationDate="2018-05-06T22:30:46.097" Score="1" Body="&lt;p&gt;Closing the curtains will help even on a cloudy day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming there is minimal heat transfer from the curtain to the room due to infrared radiation of the curtain and assuming minimal convection of hot air trapped between the curtain and the glass, overflowing the curtain rail into the room, the curtain will help by reflecting some of the sun's rays that have penetrated through the clouds and also some of the infrared radiation caused by the clouds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between a cloudy day and a sunny day is some of the solar radiation energy reflects back into space by the clouds, but not all of it. It depends on reflection index of the cloud. The part of radiation which is not reflected partially heats up the cloud causing infrared radiation to earth and partially reaches the surface of the air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So having the shades closed still helps even during cloudy days.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-06T22:30:46.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21629" PostTypeId="2" ParentId="21622" CreationDate="2018-05-07T02:15:21.410" Score="1" Body="&lt;p&gt;You have made a very common mistake - if you want to increase the static friction, you have to &lt;em&gt;decrease&lt;/em&gt; the torque! So yes, a well designed fastener system that includes washers can significantly reduce the torque needed to achieve a given tension on the bolt. And it is the tension, not the torque applied to get there, that provides the friction and stiction. This is done by controlling the radius of the pressure. It is common practice to want the blind side to stick more than the side being turned. Easy to do if you think about it. Also, for torque critical applications, washers provide a more consistent torque to tension relationship.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-05-07T02:15:21.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21630" PostTypeId="1" CreationDate="2018-05-07T03:28:39.997" Score="1" ViewCount="197" Body="&lt;p&gt;I want to compare those values for different materials, e.g. wood (pine, cork) and plastic polymers (HDPE,PP). So I want to compute the max load for a given material and cross section. Can I say that the max load is prportional to the cross area? and is this also true when I use hollow poles?&#xA;Another question is what happen when the pole is loaded for limited amount of time. Can I tell when a pole that is loaded with, say, 20% more then its max load, will crush when loaded for 30 minutes every day?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a question I'm very interested in is which cheap material will give me max strength per its density?&lt;/p&gt;&#xA;" OwnerUserId="15977" LastEditorUserId="15977" LastEditDate="2018-05-07T03:39:59.550" LastActivityDate="2018-05-07T04:13:08.800" Title="How can I compute the thickness of a pole of a given material that is needed to carry a given load?" Tags="|mechanical-engineering|structural-engineering|materials|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21632" PostTypeId="2" ParentId="21630" CreationDate="2018-05-07T04:13:08.800" Score="1" Body="&lt;p&gt;There are 4 main things you should consider when thinking about this problem: normal stress due to axial loading, normal stress due to bending moment, shear stress due to axial loading and shear stress due to bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal stress due to axial loading states $\sigma=P/A$ where P is value of the load and A is the normal cross sectional area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for normal stress due to bending moment states $$\sigma=My/I$$which can be elaborated as saying that normal stress depends on the moment (M), the thickness of the bar (y) and moment of inertia (I).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear stress due to axial loading states $\tau_{ave}=P/A$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear stress due to bending moment has formula $$\tau_{ave}=VQ/It$$ where V is the load, Q is first moment of area, I is moment of inertia and t is thickness of the bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining all these different kinds of stress we can conclude that the pole will fail in &lt;strong&gt;at least one&lt;/strong&gt; of those scenarios. To make sure the pole won't break you need to make sure that the &lt;strong&gt;maximum yield stress&lt;/strong&gt; is greater than maximum stress experienced in all four of these categories $$\sigma_{max}&amp;gt;\sigma$$ and $$\tau_{max}&amp;gt;\tau$$&#xA;Simple calculations for 4 of these stresses will tell you which material to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the material you choose depends on the Load itself, thickness, cross sectional area, first moment of area and moment of inertia of the pole you are using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time also matters since all materials are prone to fatigue which weakens it. Some materials are more likely to experience higher fatigues than others but it requires very complex analysis to identify the correct choice.&lt;/p&gt;&#xA;" OwnerUserId="15978" LastActivityDate="2018-05-07T04:13:08.800" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21634" PostTypeId="1" CreationDate="2018-05-07T06:57:05.180" Score="1" ViewCount="47" Body="&lt;p&gt;I have some trouble to understand the relationship between internal stress tensor and the external traction force (in this example, the traction force is a volumic centripetal force). Consider a cube made of a very high elastic modulus homogeneous material (which can be approximated to a quasi-rigid body), it is rotating about axis-z in a constant spin speed $\Omega$. For sake of simplicity, the three axes are attached to the cube.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9vcrY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9vcrY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In its stationnary configuration, with respect to the attached rotating reference, the centrifugal force on any point in this cube should be equilibrated by an internal elastic force, in order to keep the cube from falling apart. Given the coordinate of a generic point $P$ in this cube as $\boldsymbol{x} = [x\;y\;z]^\text{T}$, the centripetal force density can be given by&#xA;$$\boldsymbol{f} = \rho [\boldsymbol{\omega}]^2 \boldsymbol{x}$$&#xA;where&#xA;$$[\boldsymbol{\omega}] = &#xA;\begin{bmatrix} &#xA;0 &amp;amp; -\Omega &amp;amp; 0 \\ \Omega &amp;amp; 0 &amp;amp; 0 \\ 0&amp;amp; 0 &amp;amp; 0&#xA;\end{bmatrix}$$&#xA;My question is how to express the corresponding true/cauchy stress tensor on this point for equilibrating the centripetal force density as&#xA;$$\boldsymbol{\sigma}_{\boldsymbol{f}} = &#xA;\begin{bmatrix} &#xA;\sigma_{xx} &amp;amp; \sigma_{xy} &amp;amp; \sigma_{xz} \\ \sigma_{xy} &amp;amp; \sigma_{yy} &amp;amp; \sigma_{yz} \\ \sigma_{xz}&amp;amp; \sigma_{yz} &amp;amp; \sigma_{zz}&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for taking a look!&lt;/p&gt;&#xA;" OwnerUserId="7441" LastActivityDate="2018-05-07T06:57:05.180" Title="Internal stress tensor associated with centrifugal/centripetal load" Tags="|stresses|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21635" PostTypeId="1" CreationDate="2018-05-07T08:06:17.457" Score="-1" ViewCount="95" Body="&lt;p&gt;While pressure measurement in manometers is independent of size,why do we use a reservoir that has a larger diameter than the other limb in inclined manometer?&lt;/p&gt;&#xA;" OwnerUserId="15268" LastActivityDate="2019-08-19T04:03:45.787" Title="Measurement of pressure using inclined manometer" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21636" PostTypeId="2" ParentId="21635" CreationDate="2018-05-07T08:14:50.607" Score="0" Body="&lt;p&gt;The q in your title does not match the q in the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for an lnclined mano it is to give a more precise reading as the working distance is longer but the range is shorter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in diameters does the same thing.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-05-07T08:14:50.607" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21637" PostTypeId="2" ParentId="280" CreationDate="2018-05-07T09:25:22.827" Score="1" Body="&lt;p&gt;If you still looking for answer, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could keep a well-designed wedge, with static holes on the wedge surface, eighther 1. the surface of the wedge is aligned with the flow axis or 2. aligned symmetric line with the flow axis. you will have pitot pressure from Raleigh pitot-tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you could measure the static pressure directly in case 1. or 2. with Oblique shock relations having known values of wedge angle($\theta$), $P_0$ &amp;amp; $P_{\infty}$ you will get oblique shock angle($\beta$) and Mach number($M$).&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-05-07T09:25:22.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21639" PostTypeId="1" CreationDate="2018-05-07T09:50:22.547" Score="1" ViewCount="24" Body="&lt;p&gt;It's my first time posting on this specific forum. I hope I'm doing this right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently modeling fire spreading through different buildings via FDS, but I lack the properties of the materials modeled, such as heat of combustion and heat of vaporization. I've searched the internet for days and some information is still missing. And I'm not sure what to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does any of you happen to know where I can find an extended combustion properties of materials table?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="15981" LastActivityDate="2018-05-07T09:50:22.547" Title="Heat of Combustion, Vaporization tables" Tags="|materials|thermodynamics|combustion|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21640" PostTypeId="1" CreationDate="2018-05-07T11:07:02.447" Score="2" ViewCount="155" Body="&lt;p&gt;I have a system with a motor connected to the trapezoidal lead screw. The lead screw is connected to the tube which is connected to the piston. The piston has to pump 1000cc oil against a pressure of 20 bar into the bladder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well , first I found out the force on the piston at 200m which is 10054 N. I calculated the torque considering 20 bar+1000 Pa of extra pressure to push the oil into the bladder at 20 bar. The resulting torque was calculated as 12.6 Nm (this seems a lot to me and hence asked the question on this forum).&#xA;Once I know the RPM of the motor, I can calculate the piston velocity (using stroke length ,lead and RPM)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this approach correct? Am I doing something wrong? If not, then which motor should I use?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Piston diameter: 80 mm&lt;/li&gt;&#xA;&lt;li&gt;Pitch of the screw: 3mm&lt;/li&gt;&#xA;&lt;li&gt;Lead angle: ø = 9.38˚&lt;/li&gt;&#xA;&lt;li&gt;Helix angle: α = 6.8˚&lt;/li&gt;&#xA;&lt;li&gt;Time to pump the oil into the bladder at 20 bar: 12s ± 3s (This can increase        but I am trying to keep it as low as possible) &lt;/li&gt;&#xA;&lt;li&gt;The orifice diameter is 17.77mm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.imgur.com/YYdcarw.png&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15938" LastEditorUserId="15938" LastEditDate="2018-05-09T05:52:26.973" LastActivityDate="2018-05-11T11:30:42.857" Title="How do I calculate the velocity of the piston?" Tags="|mechanical-engineering|fluid-mechanics|motors|systems-engineering|power-transmission|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="21642" PostTypeId="1" AcceptedAnswerId="21647" CreationDate="2018-05-07T11:40:02.103" Score="-2" ViewCount="170" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/o6Fs1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;What is wrong with totally replacing the inlet of a piston engine with a surgical mask?As surgical masks get rid of about 80% of air impurities and dusts.&lt;/p&gt;&#xA;" OwnerUserId="15982" LastActivityDate="2018-05-07T15:23:00.147" Title="Inlet of an ICE" Tags="|engines|" AnswerCount="3" CommentCount="13" ClosedDate="2018-05-11T11:42:06.553" ContentLicense="CC BY-SA 4.0" />
  <row Id="21643" PostTypeId="2" ParentId="21642" CreationDate="2018-05-07T11:50:12.153" Score="2" Body="&lt;p&gt;How about the volume of air going through it?&#xA;For a human with assuming 10 breaths a minute and 2 litres per breath that is 40 litres a minute...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An engine will use that is seconds so the life of that filter is going to be very short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most racing engines don't have air filters as the servicing and rebuilds are so much more frequent, but for a family car then longevity is a higher priority.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-05-07T11:50:12.153" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21645" PostTypeId="2" ParentId="21642" CreationDate="2018-05-07T12:58:54.533" Score="2" Body="&lt;p&gt;There are 2 ways to increase the airflow through a filter: bigger holes and more surface area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bigger holes is usually not possible because that would defeat the purpose of the filter. The face mask doesn't seal all that well to the face resulting in a few large holes at the side where air can escape and enter unfiltered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More surface area ban be achieved by pleating the filter using a mesh to support it. You can buy ready made performance filters made for high airflow for not all that much money (a quick google found several for &amp;lt; $70) and that will be a better deal than trying to make your own.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2018-05-07T13:42:48.737" LastActivityDate="2018-05-07T13:42:48.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21647" PostTypeId="2" ParentId="21642" CreationDate="2018-05-07T15:23:00.147" Score="1" Body="&lt;p&gt;Air impurities are no issue for an ICE, only dust and other bigger particle matter are. The material of a surgical mask has little flow capacity, so a bigger filter is needed than normal, without good reason. You could use them for an ICE, but as is mentioned before, their flow capacity is way too low. You'd need to glue lots of them together, and fold them into a saw pattern to efficiently package them into a filter box. Then they could properly function as an airfilter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it won't weigh less, so it won't contribute in your search to lower the weight of your engine. If you don't live in a dusty area, rather remove the entire filter if it's really important for the engine to weigh less. The engine will wear faster though.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-05-07T15:23:00.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21649" PostTypeId="1" CreationDate="2018-05-07T15:43:06.603" Score="1" ViewCount="106" Body="&lt;p&gt;I once read on a web page I can no longer find that corundum is a covalent network. I'm guessing that's because despite the fact that the electronegativity difference between aluminum and oxygen is more than 1.7, corundum has exactly 2 valence electrons per bond so its valence orbitals hybridize into one orbital for each bond each of which is filled with 2 electrons making the substance really strong like a covalent network. Since it's a covalent network, if you replace each aluminum atom with a silicon atoms, would you have excess electrons some of which random walk to the surface atoms filling their outer shells and making nothing adhere very strongly to the surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There could be so many practical applications for it when it's etched nanosmooth. According to &lt;a href=&quot;http://iopscience.iop.org/article/10.1088/0022-3727/25/10/006/meta&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;, diamond has high friction with itself in an ultra high vacuum and according to &lt;a href=&quot;https://physics.stackexchange.com/questions/15876/why-making-a-surface-super-smooth-increases-the-coefficient-of-friction&quot;&gt;this question&lt;/a&gt;, super smooth surfaces have high friction with each other so I'm guessing that because of diamond's strong adhesion to itself from forming bonds between the surfaces, even when the normal force is tiny, the force of friction per area between the surfaces is almost as high or higher than the sheer modulus of diamond so the two diamond surfaces undergo the stick-slip phenomenon contributing to wearing out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, when you have 2 smooth spheres of corundum with some of its aluminum atoms replaced with silicon atoms sliding against each other even in an ultra high vacuum, the force of friction between them per area of contact will be much smaller than the shear modulus of the material so they should slide smoothly and not wear out. Also, a nanosmooth frying pan made of that material would be so revolutionary. It would not chemically degrade over time. All sorts of food would have so little adhesion with it. Maybe soft solids like squash though won't roll off because they have some adhesion to it and lack the ability to flow like a liquid and roll off and would have to be rinsed off. We could make tongs out of the exact same substance for those pans and have the food not stick to the tongs either and the tongs would not scratch the pan because the pan is made of such a hard material and they're both smooth and the tongs would not undergo the stick-slip phenomenon with the pan. Also, according to the Wikipedia article &lt;a href=&quot;https://en.wikipedia.org/wiki/Fracture_mechanics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fracture mechanics&lt;/a&gt;, brittle materials are weak because they have surface cracks and it takes a smaller amount of tension to magnify the tension to the breaking point at the tip of the crack and the longer the crack is, the less tension it takes to crack the material but if it's etched nanosmooth, its strength will probably be a significant fraction of its sheer modulus but according to &lt;a href=&quot;https://physics.stackexchange.com/questions/305400/why-is-glass-so-breakable/305401#305401&quot;&gt;this answer&lt;/a&gt;, the strength a material needs in order to not break on collision varies as its sheer modulus to the power of $\frac{4}{5}$ so it should totally resist breaking from colliding as long as it doesn't hit something rough in the collision like a rock and get scratched. Also, because it's so nonstick, it could be etched smooth and when it's pulled out of the etching liquid, the liquid will not cling to it and evaporate redepositing the material it dissolved away as a rough surface so it will remain nanosmooth after it's etched.&lt;/p&gt;&#xA;" OwnerUserId="6419" LastEditorUserId="6419" LastEditDate="2018-06-12T16:15:06.783" LastActivityDate="2018-06-12T16:15:06.783" Title="Is it possible to create a hard slick non-stick non-degradable substance?" Tags="|mechanical-engineering|materials|ceramics|" AnswerCount="0" CommentCount="17" ClosedDate="2018-05-07T23:19:19.637" ContentLicense="CC BY-SA 4.0" />
  <row Id="21651" PostTypeId="1" CreationDate="2018-05-07T21:46:37.540" Score="2" ViewCount="92" Body="&lt;p&gt;I'm designing an electronic speed controller for an electric bike and I recently mounted my wheel and performed some tests. The wheel (an old 24V brushed DC motor) works pretty much exactly how I expected it to EXCEPT there is a large amount of mechanical resistance both when I attempt to turn it by hand and when I hook up my ESC. Here are the values of a preliminary test with the bike mounted (no load):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nKOnM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nKOnM.png&quot; alt=&quot;Preliminary Test Results&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weird part is that there is a little bit of &quot;give&quot; (low resistance) when the wheel is turned by hand for about 6 inches or so. I uploaded a &lt;a href=&quot;https://youtu.be/SKhiWkke_1I&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; to youtube of me moving the motor by hand to show this clearly (sound on!). The video doesn't exactly show this but the housing of the wheel is not rubbing anywhere on the fork of the bike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know if the mechanical resistance of the motor might have something to do with the brushes? Or maybe how the wheel itself is mounted? I appreciate any and all answers!&lt;/p&gt;&#xA;" OwnerUserId="15994" LastActivityDate="2018-05-08T12:38:02.403" Title="Where is the resistance in my DC hub motor coming from?" Tags="|motors|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="21652" PostTypeId="2" ParentId="21584" CreationDate="2018-05-08T01:34:57.967" Score="2" Body="&lt;p&gt;It’s not a hard no, but do not model threads. They certainly look a lot prettier, but they are mostly unnecessary. They can consume more memory and require more stress on the computer to render them. This is a big factor with larger models with many parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, as @Chris Johns stated, simply indicate the standard thread on the drawing. The machine shop should do the rest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, if you have specific requirement, you could specify the way the threads are to be manufactured(cut or formed)&lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2018-05-08T13:12:51.250" LastActivityDate="2018-05-08T13:12:51.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21654" PostTypeId="2" ParentId="21479" CreationDate="2018-05-08T08:21:15.807" Score="0" Body="&lt;p&gt;You can isolate the other side of the rotary union, and use a belt, or magnetic gears. Traditional gears will put lateral stresses on the rotary joint, which may result in buckling. &lt;/p&gt;&#xA;" OwnerUserId="15863" LastActivityDate="2018-05-08T08:21:15.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21655" PostTypeId="1" CreationDate="2018-05-08T08:21:37.513" Score="0" ViewCount="653" Body="&lt;p&gt;I am aware this is a vague question but it is something which has been on my mind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would anybody know the steps used to calculate the draught loss across a burner in the furnace of a boiler?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it involve using parameters from the forced or induced draft fan?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other values would need to be taken into consideration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="15458" LastActivityDate="2018-05-08T08:21:37.513" Title="How to calculate the register draught/draft loss (RDL) of a burner?" Tags="|pressure|power-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21656" PostTypeId="1" CreationDate="2018-05-08T10:01:37.807" Score="1" ViewCount="22" Body="&lt;p&gt;We have porous ceramic platelets of ~50x5x1 mm³ thickness infiltrated with &lt;a href=&quot;https://en.wikipedia.org/wiki/Ammonium_nitrate&quot; rel=&quot;nofollow noreferrer&quot;&gt;ammonium nitrate&lt;/a&gt; solution (NH4NO3). We want to decompose and evaporate it w/o damaging the fragile substrate (decomposition should take place at ~210 °C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is part of a repeated experimental procedure, it shouldn't take longer than ~15 minutes for heating, decomposition and cooling back to max ~50 °C for a batch of 10 to 20 samples at once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about a &lt;strong&gt;microwave furnace&lt;/strong&gt; since it would heat only the samples, or maybe even only the ammonium nitrate within the samples (not sure how well ceramics (ZrO2) couples in the µwave), but since it will heat the material uniformly, I'm concerned about explosive evaporation and damage of the fragile samples. Might still be suitable if the power is tuned well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What also came to my mind was &lt;strong&gt;inductive heating&lt;/strong&gt; of a small metal sample carrier or sample holding capsule. Could also be quick. However I have no idea if lab devices like this are available, at least I've found nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other ideas?&lt;/p&gt;&#xA;" OwnerUserId="15989" LastActivityDate="2018-05-08T10:01:37.807" Title="Safely evaporate electrolyte from fragile porous substrate" Tags="|heating-systems|chemistry|heat-treatment|evaporation|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21657" PostTypeId="2" ParentId="19277" CreationDate="2018-05-08T12:18:59.083" Score="1" Body="&lt;p&gt;Coils by themselves are not susceptible to mechanical vibrations.  However, relative motion between a coil and a external magnetic field can cause a signal out of the coil.  This is because the changing magnetic field at the location of the coil causes a signal.  After all, what coils sense is the change in the magnetic field around them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the coil and the source of the magnetic field were to vibrate together, then there would be no signal from the coil.  That is because relative to the coil, the magnetic field is not changing.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-05-08T12:18:59.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21658" PostTypeId="2" ParentId="21651" CreationDate="2018-05-08T12:38:02.403" Score="2" Body="&lt;p&gt;Instead of looking at a bunch of numbers, look at a plot to understand the data better:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/166l6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/166l6.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that torque is proportional to current, shown a amps on the X axis.  This shows that it take considerable torque to overcome something to initially get going.  After that, the speed is roughly proportional to the torque, as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like this is showing there is something in the motor causing significant friction.  It may be something is mis-aligned, causing rubbing, a bad bearing, etc.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-05-08T12:38:02.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21660" PostTypeId="1" CreationDate="2018-05-08T15:26:37.430" Score="2" ViewCount="49" Body="&lt;p&gt;I have water being diverted from underneath my foundation in one specific area from an unknown source. It's causing my foundation to settle and I have already had it leveled with piers.  I have confirmed that it is not a plumbing leak.  It only happens when it rains really hard.  I spoke to a plumber who told me he ran across this kind of issue before and after his customer investigated where their utility lines were, AT&amp;amp;T's lines were causing the problem and they ultimately were responsible for installing a sump pump to resolve the issue.  I know absolutely nothing about how utility lines run underground and how this could cause water to be diverted under a slab.  Does this make any sense to anyone?&lt;/p&gt;&#xA;" OwnerUserId="16002" LastActivityDate="2018-05-08T16:46:53.620" Title="Unknown Water Source" Tags="|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2018-05-11T11:37:12.553" ContentLicense="CC BY-SA 4.0" />
  <row Id="21661" PostTypeId="1" AcceptedAnswerId="21667" CreationDate="2018-05-08T16:02:14.327" Score="2" ViewCount="142" Body="&lt;p&gt;In structural mechanics, we often approximate a partial differential equation by a system of ordinary differential equations in matrix form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\boldsymbol{M}\ddot{\boldsymbol{q}}(t)+\boldsymbol{B}\dot{\boldsymbol{q}}(t)+\boldsymbol{K}{\boldsymbol{q}}(t)=\boldsymbol{f}(t),$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which $\boldsymbol{M}$ is the mass matrix, $\boldsymbol{B}$ is the damping matrix and $\boldsymbol{K}$ is the stiffness matrix. The vector $\boldsymbol{q}(t)$ is the displacement vector which contains the displacements at specific positions and the force vector $\boldsymbol{f}(t)$ does specify the forces applied at the same positions at which the displacements are specified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to deduce the mass matrix $\boldsymbol{M}$ and the stiffness matrix $\boldsymbol{K}$ by introducing finite elements. The damping matrix $\boldsymbol{B}$ is not as easy to determine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A possible model is Rayleigh’s damping model. It is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\boldsymbol{B}=\alpha \boldsymbol{M} +\beta \boldsymbol{K}.$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what I do not understand is the problem of choosing $\alpha$ and $\beta$. Are they obtained by system identification, while testing measured data with simulated data?&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="7627" LastEditDate="2018-05-09T07:00:58.220" LastActivityDate="2018-05-09T07:00:58.220" Title="Rayleigh damping in structural mechanics" Tags="|structural-analysis|applied-mechanics|vibration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21662" PostTypeId="2" ParentId="21660" CreationDate="2018-05-08T16:46:53.620" Score="2" Body="&lt;p&gt;There are imaging companies that can read off X-rays or Radar and try to locate a broken utility housing or other underground trenches and natural or man made crevices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that you have a soils engineering issue which will be uncovered by a soils investigation by an engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you can release the hydrostatic pressure by installing a drain and sump pump at a location upstream to this intermittent flow.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-08T16:46:53.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21663" PostTypeId="1" AcceptedAnswerId="21664" CreationDate="2018-05-08T20:43:08.560" Score="0" ViewCount="38" Body="&lt;p&gt;I grow plum trees from pits using &lt;a href=&quot;http://www.canadiantire.ca/en/pdp/mckenzie-21-cell-seed-starting-tray-3-in-0591680p.html#srp&quot; rel=&quot;nofollow noreferrer&quot;&gt;seed starting trays&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TPCsF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TPCsF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I have a problem where the roots of the seedlings grow in circles around the bottom of the cavities of the trays. The circling roots cause problems when the trees grow to maturity (they strangle each other out). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of the roots growing in circles, I need to train them downwards towards the hole (where they will be air-pruned).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ka5MO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ka5MO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The problem is normally solved by using &lt;a href=&quot;http://www.amaplas.com/ViewProduct.aspx?GroupID=0&amp;amp;InventoryID=5053&amp;amp;CategoryID=52&quot; rel=&quot;nofollow noreferrer&quot;&gt;seed starting trays that have specialized vertical fins&lt;/a&gt; that prevent the root from growing in circles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ep4Ab.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ep4Ab.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to &lt;strong&gt;retrofit&lt;/strong&gt; my existing trays (I have several dozen) so that they function like the anti-root circling trays. I think this might be achieved by adding some sort of vertical fin in each cavity that would act as a physical barrier/re-router to the roots (but I'm not sure how to do this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I retrofit my existing trays to prevent root circling?&lt;/p&gt;&#xA;" OwnerUserId="10536" LastActivityDate="2018-05-08T22:21:42.603" Title="How to retrofit seed starting trays to prevent root circling" Tags="|manufacturing-engineering|environmental-engineering|prototyping|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21664" PostTypeId="2" ParentId="21663" CreationDate="2018-05-08T20:54:17.363" Score="1" Body="&lt;p&gt;Attach some L-shaped metal sheets or plastic parts in between holes? Attaching can be accomplished by, gluing, nailing, silicone, screwing, etc. depending on the size of the plates and your pots. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or use straws to extend the roots radially/downwards a bit more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cutting any circling root is another method that I’ve seen being employed by bonsai hobbyists. However it takes time to repot the plants every month or so. &lt;/p&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2018-05-08T20:54:17.363" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21665" PostTypeId="1" CreationDate="2018-05-08T22:04:17.473" Score="0" ViewCount="31" Body="&lt;p&gt;How would one determine if a fluidised bed (with liquid flow) is determined to be a stable fluidised bed, which subsequently doesn't form a plug or bed at the top of a column (bounded by frits), when under pressure?&lt;/p&gt;&#xA;" OwnerUserId="11590" LastActivityDate="2018-05-09T11:44:55.743" Title="Determination of solid bed fluidisation" Tags="|fluid-mechanics|chemical-engineering|compressible-flow|multiphase-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21666" PostTypeId="2" ParentId="21663" CreationDate="2018-05-08T22:21:42.603" Score="2" Body="&lt;p&gt;Coffee (stirring) sticks might work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ScpEK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ScpEK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After potting is complete these could be inserted vertically with the narrow edge against the side of the pot in multiple positions around the edge. Seeing as your pots are conical at the bottom you may insert one stick vertically down to the start of the cone and another one on the other side at an angle to form a breaker in the conical section.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-05-08T22:21:42.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21667" PostTypeId="2" ParentId="21661" CreationDate="2018-05-09T00:38:32.880" Score="3" Body="&lt;p&gt;If you know the damping ratios ($\zeta$) at two frequencies, those 4 values determine $\alpha$ and $\beta$ by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\zeta = {1 \over 2}{({\alpha \over \omega} + \beta \omega)}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this graph, the blue curve is $\zeta$, and it's the sum of the other two curves which correspond to the two terms in brackets in the formula. If you know two points on the blue curve, that uniquely defines its parameters $\alpha$ and $\beta$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the special case where damping ratio is proportional to frequency, $\alpha = 0$, the mass matrix $\boldsymbol M$ is ignored, and the formula is exact. Otherwise, it's usually an approximation that's only correct at the two frequencies used to fit it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iequQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iequQ.png&quot; alt=&quot;Damping ratio vs angular frequency&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="4437" LastEditDate="2018-05-09T01:36:36.373" LastActivityDate="2018-05-09T01:36:36.373" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21669" PostTypeId="1" CreationDate="2018-05-09T06:57:17.700" Score="1" ViewCount="211" Body="&lt;p&gt;I am having trouble interpreting Ashby plots. Usually, I am given the plots with the shapes but I am asked to identify where each material (metal, ceramic, composite...) fits. Can someone provide me with a complete list of all the properties I must know of these basic materials in order to be able to fill in any usual Ashby plot?&lt;/p&gt;&#xA;" OwnerUserId="16008" LastActivityDate="2018-05-09T16:55:54.360" Title="What properties can be plotted on an Ashby diagram?" Tags="|materials|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21670" PostTypeId="2" ParentId="21669" CreationDate="2018-05-09T07:15:50.210" Score="2" Body="&lt;p&gt;A google search shows that any two properties of interest (or more) can be plotted - based on work by Michael Ashby, so there are many answers possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://www.grantadesign.com/download/pdf/teaching_resource_books/2-Materials-Charts-2010.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;material charts pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of several interesting or relevant results.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-05-09T07:15:50.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21671" PostTypeId="1" AcceptedAnswerId="21675" CreationDate="2018-05-09T08:00:59.633" Score="1" ViewCount="337" Body="&lt;p&gt;I am bit confused in considering the methods to apply for designing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a machine, which as several parts mounted on the frame (frame is of any sectional shape for exampe: ´I´ section).  Let the length and width of the frame is 2m and 0.6m respectively. Here the Distribution of the masses of the different components on the frame are eccentric in nature. And the machine supports at the four cornes which is similar to the support of the chair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how do I calculate the deflection, bending moment, shear stress etc of the Frame or whole machine due to the Forces acting on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the Forces acting here are eccentric and not exactly on the beam, I think it is not possible to solve it by using Beam theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HKsL9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HKsL9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YPrg0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YPrg0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, could anyone suggest me how to carry out this problem to solve it analytically. Or any advices on the study of any related topics are appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="15526" LastEditorUserId="15526" LastEditDate="2018-05-09T10:20:02.290" LastActivityDate="2018-05-10T02:54:38.427" Title="How to analyse a frame of a machine?" Tags="|mechanical-engineering|structural-engineering|structural-analysis|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="21672" PostTypeId="2" ParentId="21665" CreationDate="2018-05-09T08:21:08.200" Score="1" Body="&lt;p&gt;Use flow (which is a product of pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a single solid will be entrained at the flow rate then it will plug.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the bed has a packing factor of 50% then flow velocity in the bed is twice pipe flow velocity. As the bed fluidizes the packing factor goes down as does flow velocity.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;VelocityPacking = VelocityPipe / (1 - packing)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller particles will be lost (plug) first.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-05-09T11:44:55.743" LastActivityDate="2018-05-09T11:44:55.743" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21673" PostTypeId="1" AcceptedAnswerId="21680" CreationDate="2018-05-09T09:00:16.777" Score="1" ViewCount="447" Body="&lt;p&gt;I've been trying to solve this problem for a while with no success.  I think I figured out that $m_1$ is $420 kg/hr$ and that $n_{3,A} = 3 kmol/h$ using the mass balance equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$input-output+generation-consumption=accumulation$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where steady state conditions mean there will be no accumulation and the lack of a reaction means that there will be no generation or consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything else I've tried to solve for the remaining values has led me to a dead end (420=420, 16=16, etc.)  Any help would be greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/W2Oxv.png&quot; alt=&quot;Question&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="16009" LastEditorUserId="12804" LastEditDate="2018-05-09T11:31:43.910" LastActivityDate="2018-05-09T14:17:45.900" Title="How do I solve for the mole and mass flow rates in a system?" Tags="|materials|energy|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21675" PostTypeId="2" ParentId="21671" CreationDate="2018-05-09T10:23:16.110" Score="0" Body="&lt;p&gt;This case can be easily considered using standard beam theory, though analytical software will make things much easier by letting you use offsets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load from your equipment will be naturally transferred to the frame. This can be understood by imagining the equipment (or its own support structure) as additional transversal beams on the frame, such as below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MfIXL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MfIXL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that the equipment (or its support structure) suffers bending moment, which may or may not be transferred to your frame (as torsion) depending on how the connection is made (whether the connection can be considered pinned or fixed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, the vertical load will be transferred to the frame, which will lead to bending moments on the frame. Depending on the connection, this vertical load might itself generate torsion on the frame, in which case the connection would have to be modelled with offsets to guarantee this effect is modelled correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by &lt;a href=&quot;https://engineering.stackexchange.com/questions/21671/how-to-analyse-a-frame-of-a-machine#comment38902_21671&quot;&gt;@mart's comment on the OP&lt;/a&gt;, there's also the issue of the supports being off-center. This will undoubtedly lead to torsion on the frame, but this can also be easily modelled with offsets.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-05-09T10:23:16.110" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21676" PostTypeId="1" AcceptedAnswerId="21691" CreationDate="2018-05-09T11:17:02.243" Score="1" ViewCount="155" Body="&lt;p&gt;I am trying to figure out how I can assemble multiple finite elements to obtain the system mass matrix $\boldsymbol{M}$ and the stiffness matrix $\boldsymbol{K}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation is depicted in a picture (see below, &lt;strong&gt;assume the left side is not fixed&lt;/strong&gt;). Now, I thought about splitting the body into a cantilever beam and a disc. The beam has four degrees of freedom in a planar analysis. I will refer to $q_1$ as the vertical displacement at the left side of the beam and $q_2$ is the deflection angle at the left side of the beam. At the right side of the beam, the vertical displace is denoted by $q_3$ and the deflection is denoted by $q_4$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the disc, I assume the degrees of freedom as depicted in the picture below. Hence, $q_{\text{S},1}$ and $q_{\text{S},3}$ for displacements and $q_{\text{S},2}$ and $q_{\text{S},4}$ for the angles of deflections (double headed arrow). It is obvious that $q_3=q_{\text{S},3}$ and $q_4 =q_{\text{S},4}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I set up the mass matrix $\boldsymbol{M}_\text{B}$ for the beam I see that it has the format $4\times 4$. The same goes for the stiffness matrix $\boldsymbol{K}_\text{B}$. Assume that the mass matrix of the disc is given by $\boldsymbol{M}_\text{D}$ and the stiffness matrix is given by $\boldsymbol{K}_\text{D}$. How can I assemble the system mass matrix $\boldsymbol{M}$ and the stiffness matrix $\boldsymbol{K}$ from the previous matrices? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to write down the procedure on a sheet of paper, that would be good enough :). I also appreciate a resource that is explaining this procedure.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Ts1Sx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ts1Sx.png&quot; alt=&quot;whole system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7627" LastEditorUserId="4437" LastEditDate="2018-05-12T15:05:42.600" LastActivityDate="2018-05-12T15:05:42.600" Title="Cantilever beam with disc at end (structural mechanics)" Tags="|structural-analysis|applied-mechanics|finite-element-method|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21677" PostTypeId="2" ParentId="21673" CreationDate="2018-05-09T11:51:01.893" Score="0" Body="&lt;p&gt;I'm not going to solve this question numerically, but I believe you've been given enough information. Look at it piece by piece:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You have the composition of stream 1. Assume the divider maintains the composition ratios and just splits up the total flowrates. This means that the ratio of streams S2/S1 = S2_A/S1_A = 1/4. You can use that factor to solve for S2_B and S2_C.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;That also tells you that S3/S1 = 3/4. Use that to determine the molar flowrates of S3_A,B,C.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use the MW values to calculate total mass flows.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3717" LastActivityDate="2018-05-09T11:51:01.893" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21678" PostTypeId="2" ParentId="2612" CreationDate="2018-05-09T12:32:31.353" Score="5" Body="&lt;p&gt;Was looking for other distance solutions, and came across this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sensor that I've found works well is the Sharp GP2Y0E02B (digital version) (newer version is the GP2Y0E03).  Configuring it as an I2C sensor, using an Arduino, I've been able to get &lt;em&gt;sub-millimeter&lt;/em&gt; resolution (0.156 mm - Range is ~630 mm, with 12 bit resolution). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost didn't believe it when I read the spec, so I put the sensor on a Bridgeport with a test target, and used this as 'ground truth'.  Out to 50 mm on a simple test, the error was less than 0.2 mm; out at 100 mm, I got about a 0.4 mm error.&lt;/p&gt;&#xA;" OwnerUserId="16011" LastActivityDate="2018-05-09T12:32:31.353" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="21680" PostTypeId="2" ParentId="21673" CreationDate="2018-05-09T14:17:45.900" Score="2" Body="&lt;p&gt;First, applying your equation to your case, $n_1-(n_2+n_3) + 0 - 0 = 0$ or $n_1 = n_2+n_3$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$n_{1,a}=n_{2,a}+n_{3,a} \space (1) \\&#xA;n_{1,b}=n_{2,b}+n_{3,b} \space (2)\\&#xA;n_{1,c}=n_{2,c}+n_{3,c} \space (3)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and since there's nothing that separates/filters the components in different proportions, the output will contain exactly the same fraction as input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f n_{2,a} + (1-f) n_{3,a} = n_{1,a} \space (4)\\&#xA; f n_{2,b} + (1-f) n_{3,b} = n_{1,b} \space (5)\\&#xA; f n_{2,c} + (1-f) n_{3,c} = n_{1,c} \space (6)\\&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $f$ is the ratio at which the streams are split into output - the same for all components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the table, we have $n_{1,A}$ and $n_{2,a}$. From (1) we can quickly find $n_{3,a}$ : 4-1=3. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these 3 we can find $f$ from (4): 0.25&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remainder of $n$ is trivial as we have column $1$ - multiply the $1$ column value by 0.25 for column $2$ and 0.75 for column $3$, e.g. $n_{2,b} = 10\cdot0.25 = 2.5$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For totals in kg/h, simply take the units given to find the factor by which to multiply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$total_{c} = \sum_i^{a,b,c}{MW_i \cdot n_{c,i}} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(remembering the 1000 factor of $kmol$ vs $mol$ - multiply $n$ by 1000 first): For column $1$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4*10 + 10*20 + 6*30)*1,000 = 40+200+180 = 420,000 kg/h.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-05-09T14:17:45.900" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21681" PostTypeId="1" CreationDate="2018-05-09T14:27:01.967" Score="0" ViewCount="44" Body="&lt;p&gt;I am using an encoder having 1000 pulse in one revolution in both edges configuration. I used one method in which number of pulse is counted. What is the error in this case?&#xA;Another method is alternative approach to Quadrature encoder module where pulse duration is measured. But here we have only 69 pulse per revolution. Here clock cycle frequency is 7.38 MHz. In this case what is the error?&lt;/p&gt;&#xA;" OwnerUserId="15908" LastEditorUserId="15908" LastEditDate="2018-05-09T16:18:38.830" LastActivityDate="2018-06-08T16:44:48.093" Title="Maximum error in Encoder when discrete sample interval is 2 ms" Tags="|control-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21683" PostTypeId="2" ParentId="21681" CreationDate="2018-05-09T15:51:09.683" Score="1" Body="&lt;p&gt;Depending on readout type of readout, for the 1000 pulse, if you read both edges (x2), you have position error of at worst 1/2000 of revolution, or 0.18 degree for sufficiently low speeds. For speeds up to 15 RPM. When exceeding that speed, your readout becomes completely worthless - error value becoming arbitrarily big, as your sample readouts start skipping gaps. Also, obviously, this is an unidirectional solution; rotation can't reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If instead of sampling/polling you use interrupts on one edge (x1) at much higher polling frequency but with max 2ms per interrupt, you can double the RPM, halving the accuracy - 1/1000 of revolution up to 30RPM. At least exceeding that will cap the error to sane values (readouts will be between 15 and 30RPM and consistent enough that with smart software that detects exceeding the 30RPM you can even keep increasing the speed, halving the accuracy at each 30RPM increment). You may use interrupts on both edges, which would improve readouts, but the edge detection unit must operate faster than your readout subsystem - to retrigger on two subsequent identical edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quadrature encoder can be polled in 4x mode or use interrupts in 1x and 2x modes. So, 1/138, 1/69 and 1/276 rotation or 2.6 (x4), 5.2 (x1) or 1.3 (x2) degree &lt;em&gt;at worst&lt;/em&gt;; linearly dropping with speed reduced from the cap. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But are you sure you didn't confuse units? 7.38 Hz is something terribly low. At the fastest you'd have ridiculous caps of 6.4 RPM at 1x, 3.2 at 2x and 1.6RPM at 4x where your solution breaks down. And you may either try supporting higher speeds at cost of accuracy, or bidirectionality, but not both.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-05-09T15:51:09.683" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="21685" PostTypeId="2" ParentId="21669" CreationDate="2018-05-09T16:55:54.360" Score="1" Body="&lt;p&gt;All an Ashby Plot is, is a plot with one independent variable (X-axis) and two or three dependent variables.  One is plotted on the Y-axis, and the remaining identified by color, marker size, etc.  There's nothing special or unusual about this sort of graph.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-05-09T16:55:54.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21686" PostTypeId="1" AcceptedAnswerId="21689" CreationDate="2018-05-09T18:30:43.907" Score="3" ViewCount="105" Body="&lt;p&gt;I took this photo behind my place; there was this group of workers for about an hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have a small hose, from the black spool, they were pushing in and out of the manhole and a large hose that didn't seem to move at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truck's engine was quite noisy, so I'm assuming they have a large pump. I'm curious what kind of truck it is and what they could be doing with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dNvtI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dNvtI.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8789" LastEditorUserId="624" LastEditDate="2018-05-10T11:53:22.627" LastActivityDate="2018-05-10T11:53:22.627" Title="What is this kind of sewer maintenance work?" Tags="|civil-engineering|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21687" PostTypeId="2" ParentId="21686" CreationDate="2018-05-09T20:52:03.980" Score="1" Body="&lt;p&gt;It seems they are city or municipality maintenance crew.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This truck could be a tanker that has two reservoirs. Water, and collected sewage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have a variety of tools that can be attached to the hose and inserted into the sewer line through clean-out access.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They clean the sewer lines which are clogged and remove the sewage if needed from septic tanks. The have root and hard sludge breaking drill tips and expandable bladders to temporarily seal the downstream line and contain the mess.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-09T20:52:03.980" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21688" PostTypeId="2" ParentId="21686" CreationDate="2018-05-09T21:45:04.227" Score="3" Body="&lt;p&gt;The hose probably has a head on it similar to the one pictured below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uJEAi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uJEAi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. A &lt;a href=&quot;https://www.youtube.com/watch?v=_aMBFyVsI6o&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jetter Depot twister nozzle&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The head is attached to a pressurised hose (coming from the right of the photo). Notice that there are multiple jets pointing backwards and these are thrusting the head to the left causing it to advance through the pipe. Meanwhile a smaller (in thrust) jet is shot out the front to cut through the blockage in the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had occasion to call in a drain cleaning service for my father's sewers which had become blocked - possibly through collapse. I was very impressed with this. The operator fed the very flexible hose into the sewer, powered it up and the head pulled the hose in and with gentle shaking of the hose it rapidly progressed up to the blockage. The forward jet dislodged the obstruction and the reverse jets assist in breaking it up and flushing it quickly down the pipe. On pulling the hose back further flushing took place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably the big hose will collect the sludge from the sewer or sump for disposal.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-05-09T21:45:04.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21689" PostTypeId="2" ParentId="21686" CreationDate="2018-05-09T21:46:09.563" Score="5" Body="&lt;p&gt;The small black hose is a high pressure water hose, with a nozzle attached to its end. Something like that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5cCYD.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5cCYD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is used to flush down sand and other sediments&#xA;The nozzle usually features some backwards facing jets, so the hose gets pushed inside a pipe. The hose can be pulled back with the windlass.&#xA;There's a variety of differenz nozzles, as you can see in the pictures, I think nozzle 2 and 3 have a rotating bit and the chains on nozzle 4 are probably to knock free solid sediments like chalk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The large hose is a suction tube to either empty the manhole or pump out the sediments flushed down by the pressure hose.&#xA;Usually the tanks of those types of vehicles have a bulkhead to separate waste water from fresh water, and a methode to return the sediments inside, while returning the sediment-free waste water to the pipes.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2018-05-09T21:46:09.563" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21690" PostTypeId="1" CreationDate="2018-05-09T23:39:36.317" Score="0" ViewCount="84" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e3d4i.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e3d4i.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a driver in a headphone. These can produce every sound and beyond that humans can hear. I understand that there are full range speakers that are sometimes used, but they work in a very different way. The drivers that are used in loudspeakers use rubber surrounds, rather than an integrated plastic surround with the ridges seen here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone explain to me why drivers like these can't be up sized to produce a more durable full range speaker?&lt;/p&gt;&#xA;" OwnerUserId="16018" LastActivityDate="2018-05-10T17:14:14.183" Title="Why aren't these kind of drivers used in loudspeakers?" Tags="|mechanical-engineering|acoustics|consumer-electronics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21691" PostTypeId="2" ParentId="21676" CreationDate="2018-05-10T00:01:00.307" Score="1" Body="&lt;p&gt;The procedure for assembling the system mass and stiffness matrices are the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat steps 1 and 2 below for each element matrix, $K_D$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Expand the element matrix to have the same structure as the system matrix, putting zeros where there's no corresponding value in the element matrix. Do this by inserting each element matrix element into the expanded element matrix row with the same DOF that it came from and the column with the same DOF that it came from. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the DOFs in the system matrix $K$ are in this order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;qs2, q1, q2, q3, q4, qs1&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the DOFs in the element matrix $K_D$ are in this order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;qs1, qs2, q3, q4&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;K_D =\begin{bmatrix}a &amp;amp; b &amp;amp; c &amp;amp; d \\ e &amp;amp; f &amp;amp; g &amp;amp; h \\ i &amp;amp; j &amp;amp; k &amp;amp; l \\ m &amp;amp; n &amp;amp; o &amp;amp; p\end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the expanded element matrix would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{bmatrix}f &amp;amp; 0 &amp;amp; 0 &amp;amp; g &amp;amp; h &amp;amp; e \\ 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\ j &amp;amp; 0 &amp;amp; 0 &amp;amp; k &amp;amp; l &amp;amp; i \\ n &amp;amp; 0 &amp;amp; 0 &amp;amp; o &amp;amp; p &amp;amp; m \\ b &amp;amp; 0 &amp;amp; 0 &amp;amp; c &amp;amp; d &amp;amp; a \end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Add (matrix addition) the expanded element matrix to the system matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real life, you wouldn't actually build the expanded matrix explicitly but just add the non-zero elements.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="4437" LastEditDate="2018-05-10T00:07:16.667" LastActivityDate="2018-05-10T00:07:16.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21692" PostTypeId="2" ParentId="21671" CreationDate="2018-05-10T02:54:38.427" Score="1" Body="&lt;p&gt;Lets handle fore the sake of illustration just the horizontal cylinder being supported by four rollers and later we can add the stresses caused by other parts and superpose all of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume the cylinder weighs 400lbs and is touching the bearings at 45 degrees angle, and is not imposing an dynamic loading, or else we need to refer to its manual for dynamic load factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bearings horizontal and vertical reaction is each $$ R_h = 100\sqrt(2) \space = 141lbs \space , \space and \space R_v =100lbs  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say the arm of roller bearing is 6inch high and 2 inch in. So we have $$ 6\times 141 = 846 \space lb.in  \space - 2\times 100 =646lb.in$$ &#xA;total clockwise torque to the right beam at each bearing and symmetrical torque at the left beam supports.  This causes torsional stress at 2 points on the beam, which we can calculate, given the beam's cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the vertical loads which create moment and shear we have for point loads of 100lbs each and again we can calculate the bending moment and shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we go back to vertical cylinder on the back of diagram. That part can slide back and forth so we have to consider the influence line momentum and shear. But in this case for simplicity we consider the bending moment when the part is at most critical position at mid span and shear when it is at either support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we can apply all the loads at their proper offset from the start of the beam and calculate bending moment and shear and torsion for a two span continuous beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I usually use factor of safety of 2.8 for these situations in the absence of a code load factor, but as we said this was just an illustration. Most of the compressors and moving machines have recommended design charts for foundation and support.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-10T02:54:38.427" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21694" PostTypeId="1" CreationDate="2018-05-10T06:15:31.880" Score="1" ViewCount="26" Body="&lt;p&gt;We want to know how can we determine the pressure of the liquid at the delivery end of the pump. We have a centrifugal pump and a 2&quot; pipeline at the delivery end. Pump delivery performance as per catalog is 510LPM @20 mtr head to 348LPM @36mtr head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The delivery pipe is connected to a manifold on which spray nozzles are connected around 70 Nos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total length of pipe from the pump outlet to the spray manifold will be approx 5 metres of which 1.5mtr is vertical &amp;amp; remaining horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have mounted a pressure gauge on the line which gives a constant reading of 0.6-0.8 Kg/cm2 which is lower is lower than our expectations from a 5HP pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we block the line just before the manifold, the gauge shows a reading of 3.5-4 kg/cm2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kindly help us determine the process of finding out the pressure in the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, If I have to select a pump that constantly delivers liquid at 3 Kg/Cm2, how do I go about the selection? &lt;/p&gt;&#xA;" OwnerUserId="16020" LastActivityDate="2018-05-10T06:15:31.880" Title="Determination of delivery pressure from pump in my Parts Washer" Tags="|fluid-mechanics|pressure|hydraulics|fluid|liquid|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21695" PostTypeId="1" AcceptedAnswerId="21714" CreationDate="2018-05-10T06:26:13.020" Score="0" ViewCount="126" Body="&lt;p&gt;I need  to find heat loss from the greenhouse. I know how to find heat loss through the walls and roof , but what about soil ? how to find amount heat conductivity loss to the soil ? I would be gratefull for the any reference/method/ article or etc.  &lt;/p&gt;&#xA;" OwnerUserId="8686" LastEditorUserId="10902" LastEditDate="2018-05-11T12:48:40.330" LastActivityDate="2018-05-11T23:16:11.107" Title="Heat loss from the greenhouse to the soil" Tags="|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="21696" PostTypeId="1" CreationDate="2018-05-10T09:16:51.063" Score="2" ViewCount="49" Body="&lt;p&gt;According to &lt;a href=&quot;https://www.theverge.com/2017/9/25/16359966/microsoft-facebook-transatlantic-cable-160-terabits-a-second&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;, Microsoft and Facebook have completed a 160 terabits cross-Atlantic cable. Based on my calculations, this doesn't seem probable, so what did I miss in my calculation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum frequency that a mono-mode fiber can reach is roughly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{max}=(1/\lambda)*\sqrt{\frac{c}{DL}}$     $(1)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$D$ is the dispersion parameter.&#xA;$L$ its length. &#xA;$\lambda$ is the central wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $\lambda$=1500 nm , $D$ = 100 ps/nm/km and $L$ = 10 km. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I find: 10 GHz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming 400 wavelength channels, 2 polarizations,  8 fibers, without relays under the ocean and a bit of magic I cannot reach 160 THz.  What am I missing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(1)$ is obtained with the &lt;a href=&quot;https://www.rp-photonics.com/group_velocity_dispersion.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;definition of $D$&lt;/a&gt; assuming that the dispersion of a pulse of angular frequency $\omega_0$ , angular line-width width $\delta \omega_0$, duration $\delta \tau_0=2\pi/\delta \omega_0$ is $\delta \tau_1=L(1/Vg(\omega_0+\delta \omega_0 /2)- 1/Vg(\omega_0-\delta \omega_0 /2)=LD\lambda^2/(\delta \tau_0)$. $Vg$ is the group velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum frequency is such that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{max}=1/\delta \tau_0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_{max}=1/\delta \tau_1=\frac{\delta \tau_0}{LD\lambda^2}$&lt;/p&gt;&#xA;" OwnerUserId="16022" LastEditorDisplayName="user16" LastEditDate="2018-05-11T17:24:40.030" LastActivityDate="2018-05-11T17:24:43.987" Title="How can the Microsoft and the Facebook Fibers reach 160 Tbps between Virginia and Bilbao?" Tags="|telecommunication|data|fiber-optics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21698" PostTypeId="2" ParentId="21690" CreationDate="2018-05-10T17:14:14.183" Score="2" Body="&lt;p&gt;that thing is indeed an loudspeaker- just a small one. it operates on exactly the same principles as a full-size loudspeaker. because it doesn't take much movement of air in a headphone to fry your eardrums, these little guys don't require soft surrounds to accomodate large cone excursions. the flexure of the material itself is adequate for use as a surround. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;note the ribs molded into the cone: these stiffen the bulk of the cone without adding much weight. this trick allows the cone material to serve both as the radiating surface (which needs to be stiff) and as the flexible surround. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-05-10T17:14:14.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21699" PostTypeId="2" ParentId="21193" CreationDate="2018-05-10T17:56:55.030" Score="2" Body="&lt;p&gt;Using the same volume flow rate for winter and summer is ok, since the fan will push the same amount of volume. It is true that the specific volume will change at different moist air conditions, so to be extremely exact you could adjust for this change in winter and summer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, HVAC uses approximations because being extremely exact has diminishing returns when you are buying equipment that has nominal sizes. (No sense being accurate out to the third decimal (ie. 4.567 tons) when nominal tonnages are 5,10,20 etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct that using slightly different mass flow rates for winter and summer would be more exact, but I am sure that you would arrive at similar answers in the end, or at least close enough to size your equipment.  &lt;/p&gt;&#xA;" OwnerUserId="16027" LastActivityDate="2018-05-10T17:56:55.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21700" PostTypeId="1" AcceptedAnswerId="21701" CreationDate="2018-05-10T21:40:07.687" Score="1" ViewCount="340" Body="&lt;p&gt;My strength of materials book provides the following equation to determine the x-component of the normal stress acting on all possible planes of a two dimensional element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\sigma_{x^{\prime}} = \frac{1}{2}(\sigma_x + \sigma_y ) + \frac{1}{2}(\sigma_x - \sigma_y&#xA;)\cos(2\theta) + \tau_{xy}\sin(2\theta)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is -- in what orientation are the non-primed variables assumed? My instinct is that the initial orientation that the non-primed quantities are given in is irrelevant, only the numerical value of the components matter and the angle from which you measure the new components (primed quantities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you are given a 2-D element that is initially inclined at 60 degrees and you have the values of $\sigma_x$, $\sigma_y$, and $\tau_{xy}$ in that orientation and you want to find $\sigma_{x^{\prime}}$ at the 0 degree orientation you would compute the equation above using a value of $\theta = -60$ because you have to measure &lt;em&gt;relative&lt;/em&gt; to where you &quot;start&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct or do I have this completely confused? &lt;/p&gt;&#xA;" OwnerUserId="9732" LastActivityDate="2018-05-11T00:04:46.103" Title="2-D Plane Stress Transformations" Tags="|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21701" PostTypeId="2" ParentId="21700" CreationDate="2018-05-11T00:04:46.103" Score="2" Body="&lt;p&gt;The transformation relation for 2D stresses in matrix form is&#xA;$$&#xA;  \boldsymbol{\sigma}' = \mathbf{Q}^T\, \boldsymbol{\sigma}\, \mathbf{Q}&#xA;$$&#xA;where&#xA;$$&#xA; \boldsymbol{\sigma} = \begin{bmatrix} \sigma_{xx} &amp;amp; \tau_{xy} \\ \tau_{xy} &amp;amp; \sigma_{yy} \end{bmatrix} ~,~~ \mathbf{Q} = \begin{bmatrix} \cos\theta &amp;amp; -\sin\theta \\ \sin\theta &amp;amp; \cos\theta\end{bmatrix}&#xA;$$&#xA;In writing the stresses in matrix form, we assume that the stress components have been expressed in terms of a coordinate system with base vectors ($\mathbf{e}_x, \mathbf{e}_y$) and that the angle $\theta$ is measured counter-clockwise with respect to the $\mathbf{e}_x$ vector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the angle $\theta$ is always zero when you are looking along the $\mathbf{e}_x$ vector, by definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you work out the algebra, you will find that (assuming I've made no mistakes)&#xA;$$&#xA; \begin{bmatrix} \sigma_{xx}' \\ \sigma_{yy}' \\ \tau_{xy}' \end{bmatrix}&#xA; = \begin{bmatrix} \cos^2\theta &amp;amp; \sin^2\theta &amp;amp; \sin2\theta \\&#xA;                   \sin^2\theta &amp;amp; \cos^2\theta &amp;amp; -\sin2\theta \\   &#xA;                   -\tfrac{1}{2}\sin 2\theta &amp;amp; \tfrac{1}{2}\sin 2\theta &amp;amp; \cos 2\theta \\  &#xA;\end{bmatrix}&#xA;\begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \tau_{xy} \end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inverse relation is&#xA;$$&#xA;  \mathbf{Q} \,\boldsymbol{\sigma}' \, \mathbf{Q}^T=  \boldsymbol{\sigma}&#xA;$$&#xA;In expanded form,&#xA;$$&#xA; \begin{bmatrix} \sigma_{xx} \\ \sigma_{yy} \\ \tau_{xy} \end{bmatrix}&#xA; = \begin{bmatrix} \cos^2\theta &amp;amp; -\sin^2\theta &amp;amp; -\sin2\theta \\&#xA;                   \sin^2\theta &amp;amp; \cos^2\theta &amp;amp; \sin2\theta \\   &#xA;                   \tfrac{1}{2}\sin 2\theta &amp;amp; -\tfrac{1}{2}\sin 2\theta &amp;amp; \cos 2\theta \\  &#xA;\end{bmatrix}&#xA;\begin{bmatrix} \sigma_{xx}' \\ \sigma_{yy}' \\ \tau_{xy}' \end{bmatrix}&#xA;$$&#xA;If the initial stress components are with respect to a set of initial axes that are not aligned with ($\mathbf{e}_x, \mathbf{e}_y$) and you want to compute the components along $\mathbf{e}_x$ etc., you will have to use the inverse relationship.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-05-11T00:04:46.103" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="21702" PostTypeId="1" AcceptedAnswerId="21706" CreationDate="2018-05-11T00:36:45.613" Score="0" ViewCount="1879" Body="&lt;p&gt;How can I determine horizontal force reactions in a fixed on both ends beam like this one?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ItzGY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ItzGY.png&quot; alt=&quot;Fixed on both ends beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16029" LastActivityDate="2018-05-11T08:34:29.950" Title="How can I determine horizontal force reactions in a fixed on both ends beam" Tags="|mechanical-engineering|structural-engineering|civil-engineering|structural-analysis|statics|" AnswerCount="2" CommentCount="1" ClosedDate="2018-05-11T11:40:13.173" ContentLicense="CC BY-SA 4.0" />
  <row Id="21703" PostTypeId="1" CreationDate="2018-05-11T01:12:41.693" Score="4" ViewCount="1268" Body="&lt;p&gt;Suppose I have a dynamical system with a continuous time state-space description given by the matrices A, B, C, and D, under full-state feedback u=Kx (where x is the state and x is the control input). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do I compute the Bode plot of the open-loop system, as I would do in traditional frequency-domain analysis?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a simple, concrete example, consider the problem where the plant is a mass on a spring, on which I can exert a force $F_{ext}$. Suppose I use a PID controller to control the position of the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A state space description of the plant (with an adjoined integrator state to be used by the controller) is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A = [0, 1, 0; -k, -g, 0; 1, 0, 0];&#xA;B = [0; 1/m; 0];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And the PID controller can be realized through full-state feedback with the matrix:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;K = [kp, kd, ki];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, in the frequency domain, the plant has transfer function $G(s)$, and the controller has transfer function $H(s) = k_p + \frac{1}{s} k_i + s k_d$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In frequency domain analysis, the transfer function $GH$, the &lt;strong&gt;open loop transfer function&lt;/strong&gt; is of interest. This is what we example, for example, to determine gain margin and phase margin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I obtain $GH$ given $A, B, C, D$ and $K$?&lt;/p&gt;&#xA;" OwnerUserId="508" LastEditorUserId="508" LastEditDate="2018-05-11T16:09:52.373" LastActivityDate="2018-05-14T09:20:07.133" Title="How do I compute/plot the open loop transfer function of a system given its state-space description?" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21704" PostTypeId="2" ParentId="21703" CreationDate="2018-05-11T02:56:54.357" Score="5" Body="&lt;p&gt;In order for the open loop to be meaningful the input to the controller has to have the same dimensions and units as the output of the plant, since normally when doing control in the frequency domain using unity negative feedback the input to the controller is defined as the error signal (reference signal minus the output of the plant). But if the full state is not measured (in the output of the plant) then you can not recreate the 'state' from the error signal directly. Instead you would have to use an &lt;a href=&quot;https://en.wikipedia.org/wiki/State_observer#Typical_observer_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;observer&lt;/a&gt;, which normally is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{\hat{x}} = A\,\hat{x} + B\,u + L\,(y - C\,\hat{x} - D\,u)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\hat{x}$ is the estimate of the state, $L$ is a matrix such that $A-L\,C$ is Hurwitz and $y$ the output of the actual plant. When using full state feedback then $u = -K\,\hat{x}$. However in the case of the open loop $y$ will be replaced by minus the error signal $-e = y-r$, such that when $r=0$ then it is equivalent as the equation above. Combining this white the dynamics of the plant yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\begin{bmatrix}&#xA;\dot{x} \\ &#xA;\dot{\hat{x}}&#xA;\end{bmatrix} &amp;amp;= &#xA;\underbrace{&#xA;\begin{bmatrix}&#xA;A &amp;amp; -B\,K \\&#xA;0 &amp;amp; A - B\,K - L\,C + L\,D\,K&#xA;\end{bmatrix}}_{A_{ol}}&#xA;\begin{bmatrix}&#xA;x \\ &#xA;\hat{x}&#xA;\end{bmatrix} + &#xA;\underbrace{&#xA;\begin{bmatrix}&#xA;0 \\ -L&#xA;\end{bmatrix}}_{B_{ol}} e \\&#xA;y &amp;amp;= &#xA;\underbrace{&#xA;\begin{bmatrix}&#xA;C &amp;amp; -D\,K&#xA;\end{bmatrix}}_{C_{ol}}&#xA;\begin{bmatrix}&#xA;x \\ &#xA;\hat{x}&#xA;\end{bmatrix}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the way the observer is defined the 'controller' (the effective transfer function which gets multiplied by the plant) will always have a strictly &lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_transfer_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;proper transfer function&lt;/a&gt;. So even if a constant gain might be sufficient for control, this general method will always add some sort of low-pass filter to the controller as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This state space model can be converted to an equivalent transfer function &lt;a href=&quot;http://lpsa.swarthmore.edu/Representations/SysRepTransformations/TF2SS.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;using&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;G_{ol}(s) = C_{ol}\,(s\,I - A_{ol})^{-1}\,B_{ol}.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also calculate the points of the Bode plot directly by substituting $s = j\,\omega$, with $\omega$ the desired frequencies (in radians per second) for your Bode plot.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It can be noted that in the case of your example system the integral state will not be observable. But it can be an option to first filter the error signal with $\begin{bmatrix}1 &amp;amp; 1/s\end{bmatrix}^\top$ before passing it through $G_{ol}(s)$, and use for the observer in $G_{ol}(s)$ that both the position and the integral are measured.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastEditorUserId="1765" LastEditDate="2018-05-14T09:20:07.133" LastActivityDate="2018-05-14T09:20:07.133" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21705" PostTypeId="2" ParentId="21702" CreationDate="2018-05-11T06:04:21.287" Score="2" Body="&lt;p&gt;This seems like a hw question so I'm not going to give you the straight up answer, but the following should help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to this problem is to look at elongation and compression of the beam. To the left of where force F is applied , the beam is in tension and &quot;wants&quot; to elongate. To the right of where force F is applied the opposite is true and the beam is in compression and &quot;wants&quot; to shrink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Since the beam is constrained we know that the total elongation/deformation is 0&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By applying that constraint we know that the elongation of the left side of the beam is equal to the compression of the right side of the beam, and we can solve for our reactionary forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To develop intuition ask yourself this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would happen if $a=0$? $b=0$? or $a=b$? Does my answer reflect this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best of luck&#xA;-Brian&lt;/p&gt;&#xA;" OwnerUserId="14961" LastEditorUserId="9" LastEditDate="2018-05-11T08:34:29.950" LastActivityDate="2018-05-11T08:34:29.950" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21706" PostTypeId="2" ParentId="21702" CreationDate="2018-05-11T08:29:48.417" Score="0" Body="&lt;p&gt;The reaction at either end is simply equal and opposite to the axial load in the beam adjacent to it. So what you need to work out is the axial force each side of where &lt;strong&gt;F&lt;/strong&gt; is applied.&lt;/p&gt;&#xA;&lt;p&gt;To work this out you need the plea formula:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;d = PL/EA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;where d is extension, P is axial force, L is the original length, E is Young's modulus and A is cross-sectional area.&lt;/p&gt;&#xA;&lt;p&gt;Using subscript 1 for the left hand side and 2 for the right hand side, we then get two equations:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;d&lt;sub&gt;1&lt;/sub&gt; = P&lt;sub&gt;1&lt;/sub&gt;a/EA&lt;/p&gt;&#xA;&lt;p&gt;d&lt;sub&gt;2&lt;/sub&gt; = P&lt;sub&gt;2&lt;/sub&gt;b/EA&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We also know that for equilibrium:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;P&lt;sub&gt;2&lt;/sub&gt; + P&lt;sub&gt;1&lt;/sub&gt; = F&lt;/p&gt;&#xA;&lt;p&gt;d&lt;sub&gt;2&lt;/sub&gt; = d&lt;sub&gt;1&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We can then solve all of these simultaneous equations (I'll leave that step to you), and we'll find:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;P&lt;sub&gt;1&lt;/sub&gt; = F * b/(a+b)&lt;/p&gt;&#xA;&lt;p&gt;P&lt;sub&gt;2&lt;/sub&gt; = F * a/(a+b)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;em&gt;NB The plea formula works equally well in tension and compression (assuming no buckling). A tensile force leads to elongation, a compressive force leads to shortening. All my workings are on absolute values, if you want you can make P&lt;sub&gt;1&lt;/sub&gt; and d&lt;sub&gt;1&lt;/sub&gt; negative; this is technically more correct but it adds a layer of complexity that I don't feel is necessary.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-05-11T08:29:48.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21707" PostTypeId="1" AcceptedAnswerId="21743" CreationDate="2018-05-11T09:10:09.683" Score="1" ViewCount="35" Body="&lt;p&gt;I've been stuck on this for an hour. (Probem is below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is with part b. I understand its the basic mass flow in - mass flow out with air. The equation being $$\rho_{air}A_4V_4 = \rho_{air}A_0V_0$$ where we are trying to solve for $V_4$ or the speed of air coming in through pipe 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We rearrange to $$V_1=\frac{\rho_{air}A_0V_0}{\rho_{air}A_4}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything should cancel to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_1=\frac{A_0V_0}{A_4}$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue, what on earth is $V_0$???! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In part b, $\frac{dh}{dt}=0.1910\;\frac{ft}{s}$ and is the value I ASSUMED would be used for $V_0$ in this case but no, it gives a value of $27.5 \: \frac{ft}{s}$ which is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value supposedly used for $V_0$ is 0.1484&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this value?&#xA;Where does it come from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing I can think of is that it has something to do with 'average velocity' as stated in the part b question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet I'm drawing a complete blank. And have been for more time than I care to admit. Especially considering the solution is right in front of me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help is VERY appreciated at this point.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yir7q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yir7q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16036" LastActivityDate="2018-05-15T07:26:26.753" Title="Unknown value in solution of continuity problem?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21708" PostTypeId="1" CreationDate="2018-05-11T12:06:48.750" Score="0" ViewCount="176" Body="&lt;p&gt;i.e. a battery powered device which presses a digital cameras capture button at some constant interval of ~60 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be an easy project for someone with little experience with electronics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The camera will be in a water-tight housing (a large housing with space for somethings as big as the camera itself) a few meters underwater. Hence why i dont think a wireless remote control device will work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seems to be two devices I could just buy which would do the job, the MicroBot Push and the Tempus All  &lt;a href=&quot;http://bymac.org/new/photography/14-tempus-all-mechanical-intervalometer.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;TEMPUS ALL Mechanical Intervalometer&lt;/a&gt; but I don't want to pay 150 dollars for something so simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice appreciated!&lt;/p&gt;&#xA;" OwnerUserId="16039" OwnerDisplayName="BarnOwl" LastEditorUserId="8055" LastEditDate="2018-05-11T16:54:27.900" LastActivityDate="2018-05-11T16:54:27.900" Title="How would I most easily create a mechanical intervalometer for a digital camera?" Tags="|mechanical|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="21709" PostTypeId="2" ParentId="21708" CreationDate="2018-05-11T14:41:39.980" Score="2" Body="&lt;p&gt;You may find your objective obtainable via the world of Arduino in concert with radio control servos. The Arduino will support direct connection to a single servo without requiring an external driver circuit, while the programming can be custom tailored to your camera's requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanical aspect is going to be limited only to your fabrication skills. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick brain flash resulted in a search for &quot;arduino servo shutter release&quot; and discovered that others have created this type of mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=FhRnf90AYJU&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube video of one project&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also an &lt;a href=&quot;https://www.instructables.com/id/Automatic-Camera-Shutter-Switch/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Instructable project&lt;/a&gt; that covers camera types with cable releases, electronic &quot;cable&quot; releases, and mechanical releases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last link covers many aspects quite comprehensively and may be your best initial resource. With Arduino, if someone else's program doesn't meet your requirements, it can be easily modified.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-05-11T14:41:39.980" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21710" PostTypeId="1" CreationDate="2018-05-11T15:35:01.113" Score="1" ViewCount="55" Body="&lt;p&gt;I am recreating an engineering drawing package from old quotes for a large assembly. In the quote they say &quot;All length edges not to exceed 1/32&quot; deep feathering&quot;. For some context this is for a bunch of graphite blocks which were cut to a constant cross-section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First what exactly does feathering mean in this machining context? Secondly what is the best way to denote this tolerance on an engineering drawing?&lt;/p&gt;&#xA;" OwnerUserId="16043" LastActivityDate="2018-06-27T12:42:12.410" Title="Document Feathering Tolerance on Drawing" Tags="|manufacturing-engineering|machining|technical-drawing|drawings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21711" PostTypeId="1" CreationDate="2018-05-11T17:47:50.137" Score="1" ViewCount="1986" Body="&lt;p&gt;I've been trying to revise for an upcoming final so I am solving problems from the previous chapters.I've been wracking my head on this for a while (about 1-2hrs) but I just can't seem to get it.&lt;a href=&quot;https://i.stack.imgur.com/CWjWC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CWjWC.png&quot; alt=&quot;This is a simple problem about deflection using energy method&quot; /&gt;&lt;/a&gt;&#xA;For some stupid reason I can't figure out why Member CE is 0 in the virtual loading, shouldn't that pin at C have a vertical reaction of 1. That would mean the summation of the vertical forces at joint C would be something like this Fy=1-FCEsin(26.565)=0. Failing at method of joints is very depressing to think about with the upcoming finals :/. Of course, the solution manual is no help it already assumes any scrub can solve method of joints. Can you help me out :|&lt;/p&gt;&#xA;" OwnerUserId="16044" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-05-12T16:10:32.617" Title="Finding the displacement in a truss using energy method Question" Tags="|civil-engineering|structural-analysis|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21712" PostTypeId="1" CreationDate="2018-05-11T19:27:58.800" Score="1" ViewCount="900" Body="&lt;p&gt;I basically need a block on a shaft to rotate between 90 degrees (vertical) and approximately 180 degrees (horizontal) counterclockwise on a shaft. I don't want the part to fall below 180 degrees. At 90 degrees it will clip into a stand so clockwise rotation does not need a part preventing rotation. Please advise on a way to prevent clockwise rotation after a certain angle. Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="16047" LastActivityDate="2018-06-12T08:15:30.393" Title="How to prevent an object doing a full rotation on a shaft" Tags="|mechanical-engineering|design|mechanical|systems-design|prototyping|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21713" PostTypeId="1" CreationDate="2018-05-11T22:21:00.520" Score="3" ViewCount="127" Body="&lt;p&gt;In my project I've been using galvo scanners for rotating small mirrors but they seem like an overkill at this point for several reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Even though I use the slowest and smallest scanners out there, they are still over 100 grams each, in a device where every gram matters,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;on top of that, their drivers are rather large (8cm x 6cm) and need special dual rail +-12V power supplies and +-5V DAC circuits to run. These all require space of their own and I need everything take up as much little space as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) While they can achieve the rotation speeds I need they seem to exceed it. The cons of using galvos which is space for all the electronics seems to outweight the pro of the high speeds which aren't used to the fullest anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since steppers just need one 3.3, 5 or 12V PSU and one stepper driver I thought I'd give them a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest steppers I am aware of which are readily available are NEMA 14, which is huge for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The requirements are simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) torque is extremely low as a small 10x10mm 1mm thick mirror is rotated which is glued onto it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) acceleration and rotation speeds are important and demanding though: 500 degree/second&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found a stepper but the rotation speed is abysmal for this. The 28BYJ-48&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Can small enough steppers meet this criteria (low torque but high speed requirement) at all?&#xA;Bigger motors seem to achieve the desired result &lt;a href=&quot;https://youtu.be/KgPyR0Rg3FQ?t=1m17s&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/KgPyR0Rg3FQ?t=1m17s&lt;/a&gt;&#xA;2) If yes, what stepper(s) should I check.&#xA;3) can they not make those loud sqeaky noises like in CNCs and 3d printers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/pyFZKc356eQ?t=2m1s&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/pyFZKc356eQ?t=2m1s&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14841" LastEditorDisplayName="user4139" LastEditDate="2018-06-06T14:07:30.980" LastActivityDate="2018-11-03T17:00:39.583" Title="smallest highspeed stepper motors?" Tags="|electrical-engineering|motors|stepper-motor|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21714" PostTypeId="2" ParentId="21695" CreationDate="2018-05-11T23:09:06.460" Score="0" Body="&lt;p&gt;&lt;strong&gt;Easy answer:&lt;/strong&gt; Net heat loss averaged over day/night is zero because the heat that's transmitted into the soil when the greenhouse is hot is stored in the soil and returned to the greenhouse when the greenhouse becomes colder than the soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hard answer suggestion:&lt;/strong&gt; This problem is called &quot;One-dimensional transient heat conduction in semi-infinite body&quot;. Semi-infinite because the ground is infinitely deep in the downwards direction but not in the upwards direction. It's time dependent. The rate of heat loss will go down as it progresses because the ground becomes hotter. You can write and solve the DEs by hand if you wanted a hard exercise in calculus. Here's somebody's working:&#xA;&lt;a href=&quot;https://www.thermalfluidscentral.org/encyclopedia/index.php/One-dimensional_transient_heat_conduction_in_semi-infinite_body&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.thermalfluidscentral.org/encyclopedia/index.php/One-dimensional_transient_heat_conduction_in_semi-infinite_body&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whiteboard lecture video: &lt;a href=&quot;https://www.youtube.com/watch?v=X8ofOO-iHhk&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=X8ofOO-iHhk&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Practical answer:&lt;/strong&gt; I would use Finite Element Analysis (FEA) which does all that math for you. Download some FEA software and use time dependent or transient heat transfer analysis. This will also allow you to model 2D and 3D effects like heat flow around the edges. The two 1D suggestions above are not strictly correct because of lateral heat flow under the ground but the bigger the greenhouse, the more accurate the 1D approximation is.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="4437" LastEditDate="2018-05-11T23:16:11.107" LastActivityDate="2018-05-11T23:16:11.107" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21715" PostTypeId="2" ParentId="21711" CreationDate="2018-05-12T07:17:34.793" Score="1" Body="&lt;p&gt;Consider the geometry. Horizontal forces on the bearings can not be determined with forces and moment equilibrium, but you can determine the vertical forces!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, calculate those forces in the beams, which you are able to calculate.&#xA;&lt;br&gt;Note: Those little green triangles you see are the forces in the truss members, split up to horizontal and vertical. That way you can avoid using nasty decimal point numbers. (To get the absolute value, you can just use our fellah Pythagoras' formula)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KApbx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KApbx.jpg&quot; alt=&quot;![enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers in orange specify the order in which my calculations were made:&lt;br&gt;&#xA; 0. Calculate horizontal forces on bearing (moment equilibrium)&lt;br&gt;&#xA; 1. $AE_V=1$, as $AB$ cannot take the vertical force attacking at $A$. Therefore $AE_V=2$ (geometry)&lt;br&gt;&#xA; 2. Therefore, the $AB_H=-1$ (only has a horizontal part)&lt;br&gt;&#xA; 3. External horizontal force $=2$. Therefore $DE_H=2$, thus $DE_V=1$&lt;br&gt;&#xA; 4. $DE$ and $AE$ cancel each other out at node $E$, so $CE$ and $BE$ must do so as well. As only $CE$ has a horizontal part, both $CE=BE=0$.&lt;br&gt;&#xA; 5. see 4.&lt;br&gt;&#xA; 6. only horizontal part, so $BC=2$&lt;br&gt;&#xA; 7. from point 3 you can determine, that $D_V=1$ (see the triangle), and consequently $C_V=0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;conventions:&lt;br&gt;&#xA; 1. $AB_H$ is the horizontal part of the force in member AB, $AB_V$ the vertical one...&lt;br&gt;&#xA; 2. $-$: compression, $+$: tension&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2018-05-12T07:17:34.793" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21716" PostTypeId="2" ParentId="12455" CreationDate="2018-05-12T11:42:34.993" Score="1" Body="&lt;p&gt;We apply a few forces for it. The forces are likely to do two different things. To begin with, they will create a rigid body acceleration in accordance with F=ma. I.e. the middle of mass of this item will accelerate dependent on the amount of their forces. However, since we've got a flexible figure, these forces will also be likely to trigger localized deformations and stresses. We do not care about the stiff body acceleration of the middle of mass. We just care about the neighborhood strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can address a transient energetic solution (i.e. sol 109 at NASTRAN), for both the stiff body acceleration as well as also the local deformation, then afterward subtract the stiff body acceleration to receive only the neighborhood deformation we care for. We're spending lots of CPU resources to calculate something which we're simply going to through out anyhow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we would like to do would be to resolve only a static answer (i.e. sol 101 at NASTRAN). However, there's an issue. We can't invert it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can add a bogus support to the construction someplace, just for the interest of constraining it to find an invertible stiffness matrix. However, the reaction forces in this restriction will lead to localized deformations and stresses which aren't real and will wreck the response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we've got a issue, we do not wish to address a transient dynamic difficulty since it might take a long time, but a static remedy does not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is what inertia relief will. We add a restriction. I.e. we create the construction fixed at the border rather than free. Then we put in a distributed body force to the whole arrangement that just balances out the stiff body acceleration. That usually means that the response force at our brand new dummy restriction is zero. However, the response force in the service is zero, therefore it will not have some influence on the regional deformations and stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're attempting to model buckling of a free figure that's quickening, then I do not understand why not. The additional fake constraints are just likely to constrain out rigid body movement that should not impact the buckling results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in case you've got a constrained structure that's not accelerating, then I really don't see how it is logical. I guess what you're doing is deleting the border conditions entirely (since they're tough to model), therefore fictitiously unconstraining the model, then utilizing inertia relief to place them back into... Inertia relief will attempt and create a body force on your complete structure so as to apply a zero response in the fictitiously generated limitations. However, your arrangement does have actual limitations, so there should be a non-zero response in the actual constraints. So the body force will probably be incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not quite certain how NASTRAN manages that body pressure. If it uses it together with all the additional loads in creating the differential stiffness matrix, and then you get the incorrect answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that has been a very long winded answer... bottom line, though it may work, unless there's something particular about how NASTRAN does so I am unaware of, I wouldn't suggest it. Spend a little excess time to acquire the bounds modeled properly.&lt;/p&gt;&#xA;" OwnerUserId="16050" LastEditorUserId="1832" LastEditDate="2018-05-12T15:49:09.950" LastActivityDate="2018-05-12T15:49:09.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21718" PostTypeId="2" ParentId="21711" CreationDate="2018-05-12T16:10:32.617" Score="1" Body="&lt;p&gt;You're asking about the virtual case which only has the 1&amp;nbsp;kip load, so that's what I'll answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Let's start at node A.&lt;/strong&gt; It has beams AB and AE. Only AE is capable of supporting the vertical load and we know that AE's slope is 1/2, so the horizontal component is equal to 2&amp;nbsp;kip, for a resultant of $\sqrt{1^2+2^2}=2.236\text{ kip}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since AE's horizontal component is 2&amp;nbsp;kip, we know that AB is also 2&amp;nbsp;kip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now for node B.&lt;/strong&gt; This has beams AB, BC and BE. We already know AB is 2&amp;nbsp;kip (horizontal), and since BE is vertical, it can't absorb any of this load (it therefore suffers 0&amp;nbsp;kip), leaving all of it for BC, which therefore is also 2&amp;nbsp;kip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now for node E.&lt;/strong&gt; This has beams AE, BE, CE and DE. AE we know has a resultant of 2.236&amp;nbsp;kip and BE is 0&amp;nbsp;kip. We therefore need to figure out how to split AE's force between CE and DE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is actually also trivial. Look back at what we did for node B: we saw that one of the beams (BE) could only resist vertical forces, of which there were none, and we could therefore conclude that it suffers 0&amp;nbsp;kip. This case is identical, other than that it is rotated. Imagine for this case that the X-axis follows the line A-D, and the Y-axis is perpendicular to that. What you'll see is that AE and DE only work on the X-axis, while CE has a Y component. Since it is the only one with this component (since we already know BE is 0&amp;nbsp;kip), it can't be cancelled out. Therefore, for this node to be in equilibrium, this component (and therefore CE's entire resultant force) must be equal to 0&amp;nbsp;kip.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-05-12T16:10:32.617" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21724" PostTypeId="1" CreationDate="2018-05-13T08:03:12.083" Score="-1" ViewCount="333" Body="&lt;p&gt;What's the difference between simple ball bearing and deep groove ball bearing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;  I read about them in my textbook of design. The book has only mentioned that there are teo types of ball bearings. One, simple and the other, deep groove. No images or explanation have been provided. When I search for their images on google, Images of both the types of ball bearings look the same. &lt;/p&gt;&#xA;" OwnerUserId="14355" LastEditorUserId="14355" LastEditDate="2018-05-13T08:49:59.603" LastActivityDate="2018-05-13T08:49:59.603" Title="What's the difference between simple and deep groove ball bearing" Tags="|mechanical-engineering|bearings|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21725" PostTypeId="2" ParentId="21712" CreationDate="2018-05-13T08:08:56.333" Score="1" Body="&lt;p&gt;Ok, so completely not the best image, but if you have the shaft with a hardened pin fitted - the angular width of the slot can be adjusted to control the angle of rotation and it only needs to match the width of the pin...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y5sMJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y5sMJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-05-13T08:08:56.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21726" PostTypeId="2" ParentId="21724" CreationDate="2018-05-13T08:17:10.630" Score="1" Body="&lt;p&gt;Without an idea of which bearings you are looking at, a deep-groove ball-bearing is able to control a higher axial load, but for larger loads a taper-roller bearing can or is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example is the wheel bearings on cars - some have roller bearings and are not adjustable, others use taper roller bearings that have nuts and locking washers to set the free-play.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-05-13T08:17:10.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21728" PostTypeId="1" CreationDate="2018-05-13T14:30:03.727" Score="4" ViewCount="49" Body="&lt;p&gt;In terms of personal productivity (single machine scheduling) when a task is interrupted before being done in order to switch to a different task, it incurres a 'tear down' time cost - putting tools, files, materials away etc. It also incurres a set-up cost when you pick it back up again - getting the tools, files, materials ready, reviewing the work that was previously done etc. Given a limited set of information about the task, how would one estimate the tear-down and set-up cost associated with interrupting a task?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that we have the following information:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Initial total estimated time to complete task.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Current estimated remaining time to complete task.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Task value (this could be an amount in $ or some other measure of importance)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there a good, recognized, heuristic to evaluate the time cost associated with interrupting a task? Maybe something as simple as &quot;10% of the task's total estimated time&quot;.&lt;/p&gt;&#xA;" OwnerUserId="16061" LastActivityDate="2018-05-13T22:00:01.820" Title="How to estimate the time cost of interrupting a task and picking it up again later" Tags="|modeling|manufacturing-engineering|project-management|timing|cost-estimation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21729" PostTypeId="2" ParentId="21728" CreationDate="2018-05-13T15:34:54.077" Score="1" Body="&lt;p&gt;No, because the number to be produced changes the percentage ie the setting up time can be longer than the machining time for 1 piece but 3% for 100 pieces and 0.003% for 1000 pieces...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-05-13T15:34:54.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21730" PostTypeId="1" CreationDate="2018-05-13T19:28:34.347" Score="1" ViewCount="1544" Body="&lt;p&gt;I'm building an outdoor shelter and I'm wondering what would be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The maximum beam span for the type of wood/material used such it can support the roof weight and not sag in the middle,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Suggestions on other materials (beams or otherwise) or construction techniques that you'd recommend I use instead.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The current beams as currently designed are two standard 2x10's sandwiched together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roof is 30°. The trusses are 2x6's which will have four blocking bars between each set of trusses. The roof trusses are currently 20&quot; apart OC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance I have it currently span is 9.5', but I'm interested in increasing that distance, and knowing about how far I can reasonably go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plan for the roof is 1/2&quot; plywood with tar paper and shingles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture to help illustrate what I had in mind. (Note:  I do have plans for diagonal braces between the columns and the beams, I just didn't bother to illustrate it.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zUafw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zUafw.png&quot; alt=&quot;Outdoor shelter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16064" LastEditorUserId="16064" LastEditDate="2018-05-14T14:02:40.380" LastActivityDate="2018-05-14T14:02:40.380" Title="Beam Span for Outdoor Shelter" Tags="|beam|" AnswerCount="1" CommentCount="2" ClosedDate="2018-05-15T18:48:54.620" ContentLicense="CC BY-SA 4.0" />
  <row Id="21731" PostTypeId="2" ParentId="21728" CreationDate="2018-05-13T22:00:01.820" Score="1" Body="&lt;p&gt;The amount of information you specified is too limited.  If you had more information, maybe you could do it, but not as stated. Machining tasks can vary greatly. I can imagine some parts where it might literally take hours to get work piece fixtured in the machine, but then the actual machining operation takes only a few minutes (e.g. a giant part for a ship or an airplane, where the part weights several tons and it takes big cranes to get it setup, but you only need to drill a few holes).  But there are other parts where it might only take a few minutes to get the part fixtured, but then you need to spend 10 or 12 hours of machining.  Without knowing any detail about the operation, it's impossible to estimate the overhead involved in interrupting it.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-05-13T22:00:01.820" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21732" PostTypeId="1" AcceptedAnswerId="21770" CreationDate="2018-05-14T00:59:17.813" Score="2" ViewCount="650" Body="&lt;p&gt;Currently, I am thinking that since electromagnetic waves loose energy over distance, you might be able to figure out the relative distance to each of three radio &quot;beacons&quot; and use this information to triangulate the position of a robot with a radio sensor on-board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What information each radio signal could I use to measure the distance to correlating beacon? Amplitude, perhaps? The entire system would be running indoors in a relatively small area, so it needs to be relatively precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this even a reasonable idea, or are there better methods I could use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Note:&lt;/strong&gt; I asked a &lt;a href=&quot;https://robotics.stackexchange.com/questions/15680/radio-triangulation-with-raspberry-pi&quot;&gt;similar question&lt;/a&gt; about radio triangulation on the Robotics SE, where someone asked me to look into the Engineering SE instead.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="16067" LastActivityDate="2018-05-16T15:38:58.593" Title="Radio triangulation with Raspberry Pi" Tags="|positions-measurement|radio|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21733" PostTypeId="1" AcceptedAnswerId="21736" CreationDate="2018-05-14T02:16:42.763" Score="3" ViewCount="1140" Body="&lt;p&gt;What is the definition of anti-windup? How does it impose the constraints?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the advantages of MPC and anti-windup over each other?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anti-windup guarantee the constraints or does it just try respecting them?&lt;/p&gt;&#xA;" OwnerUserId="16032" LastEditorUserId="16032" LastEditDate="2018-05-14T05:12:57.827" LastActivityDate="2018-05-17T11:02:06.420" Title="Advantage of anti-windup" Tags="|control-engineering|control-theory|pid-control|optimal-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21734" PostTypeId="2" ParentId="21730" CreationDate="2018-05-14T04:59:10.573" Score="1" Body="&lt;p&gt;This is an unstable frame and can not support any load for variety of reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rafters have a horizontal outward force over the 2-2x10 beams which is not capable of offering any lateral resistance. However this can be remedied if you provide horizontal tie beams connecting the rafters at their two bases like an A frame, say of 2x4 lumber with enough nails to counter the roofs horizontal force. Then the 2-2x10s can be supporting the rafters' vertical load easily. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other problem is the connection of columns to the roof. It needs to be laterally braced somehow. For example by a pair of 4x4 diagonal bracing at 2 feet off-centre from corners of the roof to columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it is advisable to use one single beam of 4x10 instead of sandwiching to 2x10s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use composite light weight roofing shingles and 1/2&quot; sheathing, this tool-shack can be as big as approximately 10x12 feet. Rough estimates indicate by using 4x12 beams instead of 2-2x20s the width of the shack can increase to 14 feet. Many cities' building departments have pre approved hand outs with plans and details for such structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same issues apply to other materials such as light gauge steel or alum sections.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-14T04:59:10.573" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="21735" PostTypeId="1" CreationDate="2018-05-14T05:04:57.883" Score="2" ViewCount="520" Body="&lt;p&gt;When I was studying the &quot;Electronic devices and circuits - Conventional current&quot; book floyd, I came across the following circuit and its KVL equations,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NwTMa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NwTMa.jpg&quot; alt=&quot;Voltage-divider bias circuit of an PNP BJT&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My doubt is when I apply KVL for the circuit (a), I end up with the &lt;span class=&quot;math-container&quot;&gt;$+V_{BE}$&lt;/span&gt; in the equation if I trace the voltage drop in the clock-wise direction. But in the equation, it is &lt;span class=&quot;math-container&quot;&gt;$-V_{BE}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the circuit (b), I could not understand how they got &lt;span class=&quot;math-container&quot;&gt;$-V_{EE}-V_{BE}-V_{TH}$&lt;/span&gt; in the equation?&lt;/p&gt;&#xA;" OwnerUserId="5108" LastEditorUserId="110" LastEditDate="2019-04-29T23:34:29.673" LastActivityDate="2019-04-29T23:34:29.673" Title="How can I derive the KVL equation for the voltage-divider bias circuit of PNP BJT?" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="5" ClosedDate="2018-05-15T17:42:26.147" ContentLicense="CC BY-SA 4.0" />
  <row Id="21736" PostTypeId="2" ParentId="21733" CreationDate="2018-05-14T08:17:47.313" Score="6" Body="&lt;p&gt;Anti-windup is a concept for feedback controllers with integral terms, e.g. PID, to keep the integral term from „overcharging“ when regulating a large set point error. It basically saturates the integral term to keep the system from overshooting the set point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;classic&lt;/em&gt; form of anti-windup, as described above, does not actually ensure satisfaction of input or state constraints of the system, it just enforces a somewhat „auxiliary constraint“ on the integral term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Anti-windup can also be used to ensure satisfaction of state and input constraints by exploiting the unwanted mechanic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classic saturation circuit gets extended by a predictor, that predicts the state that corresponds to the control input. A possible violation of constraints can then be detected with saturation. In that case a feasible control input, i.e. a control input such that the resulting state will not violate constraints, is calculated. The details of the approach are explained in &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S1474667015395100&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not familiar enough with the above approach to directly compare Anti-windup constraint satisfaction and MPC constraint satisfaction. I read once in some lecture slides about MPC (I'll link to them when I find them again)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If PID does the job, take PID, otherwise take MPC&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so now that you know how to ensure constraint satisfaction with the former, just take the one that does the job.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastEditorUserId="14629" LastEditDate="2018-05-17T11:02:06.420" LastActivityDate="2018-05-17T11:02:06.420" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21737" PostTypeId="2" ParentId="21732" CreationDate="2018-05-14T12:07:47.373" Score="3" Body="&lt;p&gt;While your concept is theoretically correct, it doesn't work well in practice except for rough locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is interference.  Radio waves get absorbed, reflected, diffracted, and in some cases even refracted.  All those mean that there isn't just a single point source from the receiver's point of view.  The various components of the signal bouncing around the environment can interfere constructively and destructively, making received amplitude only a very rough estimate of distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of that, you also have to consider the radiation pattern of the transmitting and receiving antennas.  If you know, for example, that everything is constrained to be in a horizontal plane and always at the same orientation, then vertical dipoles or whips would work.  However, as soon as one end changes orientation or goes out of the plane, there will be additional changes in received signal strength just due to that.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-05-14T12:07:47.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21738" PostTypeId="1" AcceptedAnswerId="22093" CreationDate="2018-05-14T20:26:17.153" Score="2" ViewCount="723" Body="&lt;p&gt;I have a temperature sensor in a -20C freezer.  The probe is going into a bottle filled with propylene glycol solution.  I want to seal the hole where the probe is inserted into the bottle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's a good sealant for this?&lt;/p&gt;&#xA;" OwnerUserId="16072" LastEditorDisplayName="user4139" LastEditDate="2018-06-06T08:45:04.690" LastActivityDate="2018-11-09T01:44:54.210" Title="How would you seal a hole in an HDPE container that's in -20C environment" Tags="|materials|temperature|seals|adhesive|" AnswerCount="6" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21739" PostTypeId="2" ParentId="21738" CreationDate="2018-05-14T20:47:01.383" Score="1" Body="&lt;p&gt;Try this stuff. It's about as good as it gets off the shelf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.3m.com/3M/en_US/company-us/all-3m-products/~/3M-Scotch-Weld-Structural-Plastic-Adhesive-DP8005/?N=5002385+3293241357&amp;amp;rt=rud&quot; rel=&quot;nofollow noreferrer&quot;&gt;3M™ Scotch-Weld™ Structural Plastic Adhesive DP8005&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-05-14T20:47:01.383" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="21740" PostTypeId="2" ParentId="21738" CreationDate="2018-05-14T23:19:21.287" Score="0" Body="&lt;p&gt;You may actually get your best bet by encasing the entire thing in &lt;a href=&quot;https://en.wikipedia.org/wiki/Shrink_wrap&quot; rel=&quot;nofollow noreferrer&quot;&gt;shrink wrap&lt;/a&gt; (depending upon the size of the bottle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shrink wrap is typically made of LDPE.  Due to the low density, the LDPE has many sites where it can still &lt;a href=&quot;https://en.wikipedia.org/wiki/Cross-link&quot; rel=&quot;nofollow noreferrer&quot;&gt;cross-link&lt;/a&gt; upon itself.  Heating the wrap begins this reaction, and as the material cross-links, it shrinks to form HDPE.  This shrinking action forms a good seal.  In the area of the hole, additional heating could be applied until the LDPE and HDPE entered a plastic state, welding the two materials together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other methods, such as acrylics, work well, but this would insure similar chemical resistance inside and outside the freezer.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-14T23:19:21.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21742" PostTypeId="1" CreationDate="2018-05-15T06:08:33.717" Score="1" ViewCount="84" Body="&lt;p&gt;I am attempting to solve the flow rate for a fixture and need to calculate the losses in the fixture that consist of ported holes. Please see picture below. Blue arrow signify air flow and red arrows signify capped holes.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oYTHO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oYTHO.png&quot; alt=&quot;Fixture&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my mechanical engineering schooling I figured this would be similar to flow in a pipe, however when calculating loses how should I account for the cross profiles for the porting (red arrows)? Essentially, I am trying to figure out the Kl, loss coefficient, variable in Hloss = Kl*(V^2)/2g. For pipes it is normally a corner component and thus I have a table of loss coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attempted to research prior to posting this, but I am not 100% sure what this corner profile is even called.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="7248" LastActivityDate="2018-05-15T06:08:33.717" Title="Fluid head loss coefficient for drilled hole cross profiles" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21743" PostTypeId="2" ParentId="21707" CreationDate="2018-05-15T07:26:26.753" Score="0" Body="&lt;p&gt;In my opinion, this is a typo and your calculation is correct. In (a), the velocity at the boundary water-air is denoted $dh/dt$. Therefore, when considering only the air volume, and considering the air as incompressible,&#xA;$$&#xA;\rho_{air}\,A_0\,\frac{dh}{dt}=\rho_{air}\,A_4\,v&#xA;$$&#xA;as you wrote (though I would use $v$ instead of $V$ for the velocity, since $V$ is often associated with volume).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $dh/dt\approx 0.1910\,\mathrm{ft/s}$ yields $v\approx 27.5\, \mathrm{ft/s}$. Using instead the 0.1484 (ft/s) for $dh/dt$ that given in the solution for (b), $v$ calculates to 21.4 ft/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Please, unlike the author of the problem, use units in your calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PPS: You might check if there is a newer version of the textbook available, or if there is an errata.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2018-05-15T07:26:26.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21745" PostTypeId="1" AcceptedAnswerId="21751" CreationDate="2018-05-15T08:13:49.810" Score="3" ViewCount="1702" Body="&lt;p&gt;I have a large boom of a radio antenna weighing $3,500 kg$ which is rotating at $0.5 RPM$. The antenna is a T shape and the boom radius is $14m$. The power to the motor is cut off suddenly and the motor acts as a rigid brake. The boom is not rigid and it continues to move/bend in the same direction for 1 second (at which points it bounces back and fourth until the forces damper out). I am trying to work out the max torque this would apply to the gearbox shaft, assuming there is no friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For angular movement, I have designated the centroid of each boom radius as a point load (so a point load at a radius of 7 m on both ends of the boom). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can work out the angular moment which is $m \cdot r \cdot v$, which works out to $1750 kg \cdot 7m \cdot 0.367m/s$ or $4489.86kg\frac{m^2}{s}$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I am not sure about though. I have read that torque is simply the change in angular momentum over time. So would that be $4489.86kg\cdot m $ $(44.1kN \cdot m)$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this be an accurate way to determine the torque in the shaft? This is for a real scenario so If I have missed any big considerations please let me know.&lt;/p&gt;&#xA;" OwnerUserId="13810" LastActivityDate="2018-05-16T17:27:55.300" Title="How to calculate torque needed to stop a rotating beam over x seconds" Tags="|torque|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21746" PostTypeId="1" AcceptedAnswerId="21763" CreationDate="2018-05-15T09:09:02.717" Score="2" ViewCount="171" Body="&lt;p&gt;I am interested on simulating the converse Piezoelectric effect of materials and I have given the mechanical (Young's modulus and Poisson's ratio) and electric material properties (Electric Permittivity, Piezoelectric coupling matrix etc). However from the literature I got double index notation, I am struggling to convert double index notation to triple index notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The double index notation from literature is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double index notation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}&#xA;d_{11} &amp;amp; d_{12} &amp;amp; d_{13} &amp;amp; d_{14} &amp;amp; d_{15} &amp;amp; d_{16} \\&#xA;d_{21} &amp;amp; d_{22} &amp;amp; d_{23} &amp;amp; d_{24} &amp;amp; d_{25} &amp;amp; d_{26} \\&#xA;d_{31} &amp;amp; d_{32} &amp;amp; d_{33} &amp;amp; d_{34} &amp;amp; d_{35} &amp;amp; d_{36}&#xA;\end{bmatrix} = &#xA;\begin{bmatrix}&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 21.3 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 21.3 &amp;amp; 0 &amp;amp; 0 \\&#xA;-2.6 &amp;amp; -2.69 &amp;amp; 3.65 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to convert them to Triple index notation used by some commercial software providers like Abaqus and MSC Marc etc and is written as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{bmatrix}&#xA;d_{11-1} &amp;amp; d_{11-2} &amp;amp; d_{11-3} \\&#xA;d_{22-1} &amp;amp; d_{22-2} &amp;amp; d_{22-3} \\&#xA;d_{33-2} &amp;amp; d_{33-2} &amp;amp; d_{33-3} \\&#xA;d_{12-2} &amp;amp; d_{12-2} &amp;amp; d_{12-3} \\&#xA;d_{23-2} &amp;amp; d_{23-2} &amp;amp; d_{23-3} \\&#xA;d_{31-2} &amp;amp; d_{31-2} &amp;amp; d_{31-3}&#xA;\end{bmatrix} = &#xA;\begin{bmatrix}&#xA;0 &amp;amp; 0 &amp;amp; -2.6 \\&#xA;0 &amp;amp; 0 &amp;amp; -2.6 \\&#xA;0 &amp;amp; 0 &amp;amp; 3.65 \\&#xA;0 &amp;amp; 21.3 &amp;amp; 0 \\&#xA;21.3 &amp;amp; 0 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; 0&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I doing the conversion correctly? Are the values placed in correct positions?&#xA;Double Index = Triple Index = Coupling Matrix Value&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{2}&#xA;d_{11} &amp;amp;= d_{11-1} &amp;amp;&amp;amp;= 0 \\&#xA;d_{21} &amp;amp;= d_{11-2} &amp;amp;&amp;amp;= 0 \\&#xA;d_{31} &amp;amp;= d_{11-3} &amp;amp;&amp;amp;= -2.6 \\&#xA;d_{12} &amp;amp;= d_{12-1} &amp;amp;&amp;amp;= 0 \\&#xA;d_{22} &amp;amp;= d_{22-2} &amp;amp;&amp;amp;= 0 \\&#xA;d_{32} &amp;amp;= d_{22-3} &amp;amp;&amp;amp;= -2.6 \\&#xA;\vdots \\&#xA;d_{16} &amp;amp;= d_{31-1} &amp;amp;&amp;amp;= 0 \\&#xA;d_{26} &amp;amp;= d_{31-2} &amp;amp;&amp;amp;= 0 \\&#xA;d_{36} &amp;amp;= d_{31-3} &amp;amp;&amp;amp;= 0 \\&#xA;\end{alignat}$$&lt;/p&gt;&#xA;" OwnerUserId="16076" LastEditorUserId="1832" LastEditDate="2018-05-15T10:35:17.940" LastActivityDate="2018-05-15T23:03:51.137" Title="Conversion of Piezoelectric coefficients" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21747" PostTypeId="2" ParentId="18972" CreationDate="2018-05-15T10:52:31.440" Score="0" Body="&lt;p&gt;There’s no mention of volume per dose, so for smaller applications, how about using a peristaltic pump which you can control in finite increments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;￼&lt;a href=&quot;https://i.stack.imgur.com/3sZFS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3sZFS.jpg&quot; alt=&quot;peristaltic pump&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2018-05-15T10:52:31.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21748" PostTypeId="1" AcceptedAnswerId="21749" CreationDate="2018-05-15T12:19:29.187" Score="1" ViewCount="102" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J0MkO.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J0MkO.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is deblistering machine by Stripfoil. I wounder to know the mechanism of blades movement ( equally ) with &lt;code&gt;one?&lt;/code&gt; screw. when the operator rotates the handle, all blades move with the same distance.&lt;/p&gt;&#xA;" OwnerUserId="10310" LastEditorUserId="10310" LastEditDate="2018-05-15T12:45:38.870" LastActivityDate="2018-05-15T19:35:33.787" Title="What is the mechanism of the blades movement" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21749" PostTypeId="2" ParentId="21748" CreationDate="2018-05-15T13:58:17.507" Score="4" Body="&lt;p&gt;One way to achieve this is by a series of 4 bar linkages &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f4gT4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f4gT4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is used for marking out equally spaced points on a line of arbitrary length.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty simple and elegant. By virtue of the geometry you can choose any two points as the start and end and the distance between them will always be equal fractions of the total. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously in this particular context you would need some extra parts to implement it properly and maybe make the whole thing a bit stiffer, but within  reasonable range of adjustment it should do exactly what you are looking for. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-05-15T19:35:33.787" LastActivityDate="2018-05-15T19:35:33.787" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21750" PostTypeId="1" CreationDate="2018-05-15T14:00:06.703" Score="2" ViewCount="40" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YdNYr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YdNYr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about resetting the position of the green diamond which acts as an anchor point for these text objects? &#xA;The idea is to be able to move the text box around with the green diamond in it's original position where it moves around with the text box, as it is by default.&lt;/p&gt;&#xA;" OwnerUserId="16079" LastActivityDate="2018-05-15T14:00:06.703" Title="How to put the green diamond back in it's original position in AutoCAD 2018" Tags="|cad|autocad|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21751" PostTypeId="2" ParentId="21745" CreationDate="2018-05-15T14:44:57.550" Score="2" Body="&lt;p&gt;You make 1 major false assumption, and that is that the torque is uniform over the 1 second period of deceleration of the beam. If the motor is truly a rigid break (doesn't back drive at all) then the bottom of the beam is stopping instantaneous (infinite torque) and the far end of the beam stops 1 second later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the shock load right as you stop may require testing or detailed analysis of the rigidity of your gear box, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if we assume your idealized scenario where all the mass is at a radius of 7 meters attached to the gear box by a massless beam in bending varies with the deflection. So zero deflection (time t=0) means 0 force on the beam, so 0 torque and maximum deflection would mean maximum force and maximum torque. I suggest using beam in bending calculations to give you the torque curve vs time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.engineersedge.com/beam_calc_menu.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineersedge.com/beam_calc_menu.shtml&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16080" LastActivityDate="2018-05-15T14:44:57.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21752" PostTypeId="1" CreationDate="2018-05-15T16:02:37.787" Score="1" ViewCount="165" Body="&lt;p&gt;I am attempting to create a ThermoElectric Cooler to Chill Water. I have thought of two options as to how I can cool the water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some limitations of the design: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The water is still, not moving. No &quot;PVC water pipes&quot; can be used.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;1) Running the still water through a copper water block, as shown below. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I would have the Peltier cool the waterblock, which would then take water from the top and cool. Nothing would be pumped in, water would freely exit and enter the holes from the top and exit without any assistance. &lt;strong&gt;(design limitations)&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OlcIh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OlcIh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Having the water touch an aluminum heatsink&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I would have the Petier cool the aluminum heatsink and simply have the water run through the grills of the cooled heatsink .&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jnPQ3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jnPQ3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the above designs, which would be more efficient at cooling water? Having water seep into the waterblock, or have water run along the grills of the aluminum heatsink? This is given the water is still, no pipes to funnel in water to the waterblock just gravity.&lt;/p&gt;&#xA;" OwnerUserId="16074" LastEditorUserId="16074" LastEditDate="2018-05-15T17:56:43.077" LastActivityDate="2018-05-15T17:56:43.077" Title="Using a Copper Waterblock or Aluminum Heatsink to Cool Water via Thermo Electric Cooler" Tags="|thermodynamics|heat-transfer|" AnswerCount="0" CommentCount="5" ClosedDate="2018-05-16T04:51:13.033" ContentLicense="CC BY-SA 4.0" />
  <row Id="21753" PostTypeId="1" CreationDate="2018-05-15T16:25:22.423" Score="1" ViewCount="79" Body="&lt;p&gt;I'm looking to do some really repetitive assembly that involves screw-driving a ton of M2 screws over and over again into threaded holes to put two plastic pieces together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it's a high volume job, I want to find the right tool. I thought it would be so awesome if there was a one handed electric screwdriver of some sort that would dispense the M2 screw as I hold the button, then once it's in the threaded hole/fully dispensed a new one pops out of the gun, and I can move onto the next hole, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Found some related stuff for nails obviously, which is good for woodworking but not exactly my application: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.aliexpress.com/item/Woodworking-decoration-chain-with-screw-gun-professional-automatic-screw-driver-decorating-tools-220V-780W-1PC/32832683890.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.aliexpress.com/item/Woodworking-decoration-chain-with-screw-gun-professional-automatic-screw-driver-decorating-tools-220V-780W-1PC/32832683890.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone confirm if what I'm looking for exists, and point me to some example products? Also down to modify similar existing tools to suit my application if need be. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="7764" LastActivityDate="2018-05-15T19:24:31.203" Title="Does this exist - 'screwdriver gun' with cartridge to automatically load new screws?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21754" PostTypeId="2" ParentId="21753" CreationDate="2018-05-15T16:33:56.550" Score="1" Body="&lt;p&gt;Home Depot and many hardware stores sell a variety of screw driving guns with different cartridges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure which will meet your need. &lt;a href=&quot;https://www.homedepot.com/p/RIDGID-3-in-Drywall-and-Deck-Collated-Screwdriver-R6791/100646501&quot; rel=&quot;nofollow noreferrer&quot;&gt;Home Depot selection her&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XcAnT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XcAnT.jpg&quot; alt=&quot;screw gun with screw belt&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-05-15T16:39:50.193" LastActivityDate="2018-05-15T16:39:50.193" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21755" PostTypeId="1" AcceptedAnswerId="21758" CreationDate="2018-05-15T16:45:09.290" Score="0" ViewCount="54" Body="&lt;p&gt;I just purchased an aluminum heatsink that I am using to &quot;wick&quot; away heat from a Peltier; however, I was wondering how I can pull heat from the hot side of the peltier quicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have purchased large, small, medium sized aluminum heatsinks and have not been able to explain why one works better than the other. This is the root of the problem: what determines how fast a heatsink can wick heat away from the hotside of the Peltier? &#xA;&lt;a href=&quot;https://i.stack.imgur.com/ymYtP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ymYtP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking adding copper heat pipes, such as the ones on a laptop heatsink-fan combinations; however, I would need to thermally attach the copper heatpipes to the same heatsink to cool them. I am afraid this might kill the efficiency of the heatsink. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you, and I am looking forward to hearing your suggestions&lt;/p&gt;&#xA;" OwnerUserId="16074" LastActivityDate="2018-05-15T19:06:51.023" Title="How to Pull Heat Quicker from a Thermo-Electric Cooler using an Aluminum Heatsink" Tags="|heat-transfer|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="21756" PostTypeId="2" ParentId="21753" CreationDate="2018-05-15T17:41:50.397" Score="2" Body="&lt;p&gt;There are systems. Your question made me do a search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interesting version is &lt;a href=&quot;https://www.pemnet.com/the-stickscrew-system/&quot; rel=&quot;nofollow noreferrer&quot;&gt;StickScrew&lt;/a&gt; and they do screws down to M2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/neFHX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/neFHX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. The StickScrews come in sticks. These are inserted through the rear of the tool and the screw snaps off when torqued. The screws have hex heads.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See a 30 s &lt;a href=&quot;https://www.youtube.com/watch?v=P4T3U6nB-RY&quot; rel=&quot;nofollow noreferrer&quot;&gt;StickScrew video&lt;/a&gt; on YouTube.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2018-05-15T19:24:31.203" LastActivityDate="2018-05-15T19:24:31.203" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21757" PostTypeId="2" ParentId="21745" CreationDate="2018-05-15T17:52:02.810" Score="1" Body="&lt;p&gt;Assuming your antenna's boom is solid then its moment of inertia along the long axis is$$ I = 1/12ml^2 $$ approximately which is less than you attributing the mass to the two ends.&#xA; The torque this boom causes while stopping and wobbling is &#xA;$$ \tau = I\alpha  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case is $$ 1/2-0 = 1/2 \space = \pi  $$&#xA;So your torque is $$1/12\times 3500\times28^2\times \pi = \pi/12\times 28^2\times 3500 \space Nm $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But We should consider the vibration of the boom and its dynamic whiplash effect two, which is dependant on the stiffness of the boom and its detailed construction such as bolted together or welded together or what not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;codes in similar situations have a dynamic load factor of 2.8.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-05-16T17:27:55.300" LastActivityDate="2018-05-16T17:27:55.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21758" PostTypeId="2" ParentId="21755" CreationDate="2018-05-15T19:06:51.023" Score="0" Body="&lt;p&gt;Conductivity, surface area, and air flow are the major factors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some have a plastic funnel so the fan can be larger than the sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume (hope) you get better with larger heat sinks and fans.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A case fan also helps.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That looks like a relatively small fan.  I get full size cases for better cooling.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-05-15T19:06:51.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21759" PostTypeId="1" AcceptedAnswerId="21761" CreationDate="2018-05-15T19:21:22.697" Score="3" ViewCount="772" Body="&lt;p&gt;I have a question regarding the efficiency of lean burn gas engines. Maybe it is just a confusion concerning the terminology of &quot;thermal efficiency&quot;, &quot;brake mean pressure&quot;, &quot;brake thermal efficiency&quot; etc. but let's see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stationary gas engines in combined heat and power plants are usually run at a very high air-fuel-ratio in order maximize efficiency and reduce NOx-emission at the same time. With the introduction of more air, thermal efficiency is increased due to a higher heat capacity ratio of the gas mixture. At the same time this also results in reduced temperatures in the combustion chamber which reduceds the oxidation of nitrogen and therefore lowers nitrogenoxide emissions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in literature a tradeoff between low NOx-emissions and a high engine efficiency is often mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I assumed that higher temperatures in a combustion chamber were indicative of an efficient combustion, which is also not in accordance with the first paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other papers that I have read also indicate a maximum of brake thermal efficiency at the same air-fuel-ratio as the maximum NOx-emissions, which is also not in accordance with the first paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone here have a good understanding of lean burn engine mechanisms and can help me understand this? I can also provide more details regarding my own misunderstandings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Follow up questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, thank you for your answer, Mark. Here are some further thoughts including the graphs that confused me initially. Most notably the graph &quot;otto cycle efficiency vs. compression ratio&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the beforementioned graph: With the increase of lambda, the isentropic exponent gamma is also increased, resulting in a higher thermal efficiency. Does this increase in efficiency also imply an increase of the available power at the crank shaft? My assumption is that the increase in efficiency also increases the available work in the p-V-diagramm of an Otto cycle (area between the 4 points of the different steps of an otto cycle). One thought experiment would be increasing lambda to infinity, therefore maximizing thermal efficiency as the isentropic exponent approaches that of air. However, this does not make much sense now as there will not be any power at the crank shaft.&#xA;Or to put this whole paragraph briefly: What do I gain by an increase in thermal efficiency? Where does the statement &quot;lean burn engines are efficient and reduce NOx-emission&quot; originate from, when the minimum of the brake specific fuel consumption is at the same lambda as maximum NOx-emissions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The argument of the efficiency of a lean burn engine is often used in literature, however the lowest brake specific fuel consumption occurs at a lambda only slightly higher than the stochiometric operation of the engine and quickly rises then. This is very confusing as the previos investigation of gamma would lead to the assumption that I can simply &quot;increase lambda to get more efficiency&quot;. What is the motivation behind an increase in thermal efficiency when it cannot be used as mechanical power at the crank shaft? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse that my questions seem a little naive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And one last very important question: When investigating lambda, is it correct that with an increase of lambda the mass of the actual fuel is reduced? Or is the mass of fuel usually held constant in such experiments and lambda is increased by adding more oxygen, effectively increasing the total mass of gas in the combustion chamber? (To make a long question short: Is the convention that total mass in the combustion chamber is held constant when varying lambda or is the fuel mass held constant and lambda is varied by adding more oxygen)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FZB3S.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FZB3S.png&quot; alt=&quot;Efficiency Increase due to increase in gamma&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8nKBv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8nKBv.png&quot; alt=&quot;Spark Ignition Engine Fuel Consumption, Power and NOx&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jdkSn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jdkSn.png&quot; alt=&quot;Brake Thermal Efficiency and NOx emission at different Air-Fuel-Ratios&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16082" LastEditorUserId="16082" LastEditDate="2018-05-16T18:12:01.683" LastActivityDate="2018-05-16T18:12:01.683" Title="Efficiency and NOx emission of lean burn gas engines" Tags="|automotive-engineering|energy-efficiency|biogas|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21760" PostTypeId="1" CreationDate="2018-05-15T20:26:44.913" Score="0" ViewCount="218" Body="&lt;p&gt;I'm trying to make a slider mechanism that is guided around a curved rail that consists of two different radii. As you can see in the image, the section of the rail with a diameter of 120 mm requires the 3 circles to be spaced 9.615 mm apart, and the section of the rail with a diameter of 196.615 mm requires the 3 circles to be spaced 9.686 mm apart. This difference of .071 mm causes too much play between the slider and the rail, and I really need the slider remain as stable as possible. Any suggestions of how to design the slider in order for the slider mechanism to slide along the rail smoothly with the least amount of play?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ldx4o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ldx4o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7869" LastActivityDate="2018-05-16T17:06:16.057" Title="Curved Slider Mechanism With Two Different Radii" Tags="|mechanical-engineering|mechanisms|cad|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21761" PostTypeId="2" ParentId="21759" CreationDate="2018-05-15T21:30:24.500" Score="2" Body="&lt;p&gt;This is a good, complex question, and the short answer is that it has entirely to do with terminology confusion.  To answer it, I have to explain the physics behind the equations, because a lot of single experiments are going into each of these terms, but the net result of the entire system fosters other terminology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start with the &lt;a href=&quot;https://www.thermalfluidscentral.org/encyclopedia/index.php/Heat_of_Combustion&quot; rel=&quot;nofollow noreferrer&quot;&gt;heat of combustion&lt;/a&gt; is determined in a laboratory where they have a fixed quantity of fuel in a closed chamber that is insulated so heat cannot escape.  In this case, we obtain that the reaction would release some amount of heat (in a unit called Joules, which is the amount of energy in 1 second of a 1 Watt power source) for every amount of fuel.  Typically this is done on a mass basis (J/kg), but other versions, such as a volume or per-molecule basis (J/L or J/mol) are used.  The heat released (in Joules), is a measure of energy.  To determine how much energy was released, we actually need to do some backwards calculations.  In an ideal world, where the experiment has the stoichiometric (chemically exact amount needed for the reaction - no more, no less) amount of pure oxygen and fuel mixed together, we would run the calculations as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Measure the temperature before the reaction&lt;/li&gt;&#xA;&lt;li&gt;Measure the temperature after the reaction&lt;/li&gt;&#xA;&lt;li&gt;Note the &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;specific heat&lt;/a&gt;s of the reaction products (Steam and Carbon Dioxide)&lt;/li&gt;&#xA;&lt;li&gt;Account for the phase change of water if this was done at room temperature.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Together, we see a temperature rise, and using properties taken in an earlier experiment (how the temperature of some substances changes when a specific amount of electrical heat was applied), we can find the amount of energy generated during the reaction.  Note that &lt;em&gt;your assumption&lt;/em&gt; in this experiment is completely accurate - a high temperature rise would indicate the reaction went to completion.  If it was low temperature, we would unfortunately see a lot of side reactions which resulted in poor combustion (such as NOx formation). We would call the highest temperature rise possible a value of 1, and anything that got to 90% of this value would be considered 90% efficient.  This is a form of thermal efficiency known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Fuel_efficiency&quot; rel=&quot;nofollow noreferrer&quot;&gt;fuel efficiency&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way the temperature in this thought experiment would be lowered is if it was full of an inert gas, such as Helium, that did not participate in the reaction.  The limited amount of fuel in the chamber would raise the temperature of the fuel and oxygen, but it would need to also heat up the helium.  Such a temperature rise would use some of the available energy.  These are both ways we can have low temperature - through poor combustion, and through dilution.  One still has a high thermal efficiency, the other is simply a lower temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final thought experiment, if we had an ideal engine in-taking air from a hot oven, it would be a lot less efficient.  The engine would need to be cooled a lot more, and it would have to run very slow.  This is partly due to the components over-heating, and partly due to the fuel knocking, but mostly due to the fact that it would be a lot more work to compress that hot air then it would be to compress colder air.  As a result, the engine is a lot less efficient.  In the early 1820s, it was proven that the most energy you could get using heat was a &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermodynamic_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermodynamic cycle&lt;/a&gt; called the Carnot cycle was the most efficient, and the efficiency depended entirely on how hot you could get the gas, and on how cold the gas started.  This Carnot Efficiency, between the two temperatures, could be called 1, and anything that is 90% of this power extraction would be a form of thermal efficiency we could call a &lt;a href=&quot;https://en.wikipedia.org/wiki/Carnot%27s_theorem_(thermodynamics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;brake thermal efficiency&lt;/a&gt;, because by physics no engine can be more efficient than that value.  It is called a brake efficiency or brake power because you measure the power that you are using at the brake to stop the engine, not the power (in terms of chemical energy flowing through the engine per second).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that brake thermal efficiency is different than thermal efficiency in this context - thermal efficiency of the fuel is how much of the fuel was converted from fuel to it's final end products, but the brake thermal efficiency simply refers to how warm the gas becomes.  If the gas does not get warm in the combustion, it can not efficiently extract power.  As a result, going to the though experiment with the helium, if we flood the engine with air, most of the air will remain unreacted, and act as helium.  However, the excess air will ensure all of the fuel would combust.  We would have excellent fuel efficiency, but terrible Carnot efficiency, since the gas wouldn't get warm enough for good Carnot efficiency.  This low temperature prevents the formation of NOx, but overall the system will not produce as much power as it could have produced if the excess air was not introduced into the system.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-15T21:30:24.500" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21762" PostTypeId="2" ParentId="18972" CreationDate="2018-05-15T21:59:00.023" Score="1" Body="&lt;p&gt;I am pretty sure most dosing/metering pump manufacturers provide some degree of what you are asking for.  I know that we have used Prominent before and they make metering pumps that can be controlled with a 4-20 mA signal, as do several other manufacturers (I think I have seen some from Grundfos also).&lt;br&gt;&#xA;Prominent also sells controllers that you can integrate with pumps to control via various means.&lt;br&gt;&#xA;I imagine most dosing pump manufacturers will also sell you a package setup that will have everything you need (Prominent has them here: &lt;a href=&quot;https://www.prominent.com/en/Products/Products/Metering-Systems/Metering-Systems-for-Liquids/p-dulcodos-modular.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.prominent.com/en/Products/Products/Metering-Systems/Metering-Systems-for-Liquids/p-dulcodos-modular.html&lt;/a&gt;), though you will likely need to spend a little time talking to the manufacturer or vendor you want to go with so that you get an appropriate system.  With pumps, there's not really a one-size-fits-all product, so you will need to have a little data before you start ordering anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[I am not affiliated with ProMinent, I just happened to have used them recently on a project, and it made it easy to quickly find their product data for my answer]&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2018-05-15T21:59:00.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21763" PostTypeId="2" ParentId="21746" CreationDate="2018-05-15T22:42:40.657" Score="1" Body="&lt;p&gt;&lt;strong&gt;Caveat&lt;/strong&gt;:  I don't know the convention used in Marc.  The following is based on what I can recall about Abaqus and Ansys.  &lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Constitutive relations&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The constitutive relations for &lt;em&gt;linear&lt;/em&gt; piezoelectric materials are typically expressed in two forms:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the strain and electric field are the independent variables (called the stress-charge or e-form), and&lt;/li&gt;&#xA;&lt;li&gt;the stress and electric field are the independent variables (called the strain-charge or d-form).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Case 1 : stress-charge or e-form&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The coupled equations in this case are&#xA;$$&#xA;\begin{align}&#xA;  \sigma_{ij} &amp;amp;= C_{ijkl} \varepsilon_{kl} - e_{mij} E_m \\&#xA;  q_i &amp;amp;= e_{ijk} \varepsilon_{jk} + D_{ij} E_j&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Abaqus uses this form&lt;/strong&gt; because solid mechanics finite element models are more easily solved in strain-driven form.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Case 2 : strain-charge or d-form&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The above equations can also be written in the form&#xA;$$&#xA;\begin{align}&#xA;  \varepsilon_{ij} &amp;amp;= S_{ijkl} \sigma_{kl} + d_{mij} E_m \\&#xA;  q_i &amp;amp;= d_{ijk} \sigma_{jk} + D_{ij} E_j&#xA;\end{align}&#xA;$$&#xA;&lt;strong&gt;Abaqus allows input of the coupling tensor $d$&lt;/strong&gt; because these numbers are easily converted into the $e$ tensor using the relation&#xA;$$&#xA;  e_{mij} = C_{ijkl} d_{mkl}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Converse piezoelectric effect&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The converse piezoelectric effect is modeled by the first of the d-form equations while the direct effect is modeled by the second.  In particular, in the absence of stresses and external forces, the converse effect is written as&#xA;$$&#xA;\varepsilon_{ij} = d_{mij} E_m&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Conventions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The symmetry of the strain tensor leads to various ways of representing the components in matrix form.  For example, the standard solid mechanics convention is&#xA;$$&#xA;  [\boldsymbol{\varepsilon}] = \begin{bmatrix} \varepsilon_{11} &amp;amp; \varepsilon_{22} &amp;amp; \varepsilon_{33} &amp;amp; \varepsilon_{23} &amp;amp; \varepsilon_{31} &amp;amp; \varepsilon_{12} \end{bmatrix}^T \rightarrow \begin{bmatrix} \varepsilon_{1} &amp;amp; \varepsilon_{2} &amp;amp; \varepsilon_{3} &amp;amp; \varepsilon_{4} &amp;amp; \varepsilon_{5} &amp;amp; \varepsilon_{6} \end{bmatrix}^T&#xA;$$&#xA;The &lt;strong&gt;Abaqus convention&lt;/strong&gt; is&#xA;$$&#xA;  [\boldsymbol{\varepsilon}] = \begin{bmatrix} \varepsilon_{11} &amp;amp; \varepsilon_{22} &amp;amp; \varepsilon_{33} &amp;amp; \varepsilon_{12} &amp;amp; \varepsilon_{31} &amp;amp; \varepsilon_{23} \end{bmatrix}^T \rightarrow \begin{bmatrix} \varepsilon_{1} &amp;amp; \varepsilon_{2} &amp;amp; \varepsilon_{3} &amp;amp; \varepsilon_{4} &amp;amp; \varepsilon_{5} &amp;amp; \varepsilon_{6} \end{bmatrix}^T&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Tensor d in matrix form&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;To express the coupling tensor in matrix form, let us expand out each term:&#xA;$$&#xA;\begin{align}&#xA;\varepsilon_{11} &amp;amp; = E_1 d_{111} + E_2 d_{211} + E_3 d_{311} \\&#xA;\varepsilon_{22} &amp;amp; = E_1 d_{122} + E_2 d_{222} + E_3 d_{322} \\&#xA;\varepsilon_{33} &amp;amp; = E_1 d_{133} + E_2 d_{233} + E_3 d_{333} \\&#xA;\varepsilon_{12} &amp;amp; = E_1 d_{112} + E_2 d_{212} + E_3 d_{312} \\&#xA;\varepsilon_{13} &amp;amp; = E_1 d_{113} + E_2 d_{213} + E_3 d_{313} \\&#xA;\varepsilon_{23} &amp;amp; = E_1 d_{123} + E_2 d_{223} + E_3 d_{323} &#xA;\end{align}&#xA;$$&#xA;Using the &lt;strong&gt;Abaqus convention&lt;/strong&gt;, we can write the above as&#xA;$$&#xA;\begin{align}&#xA;\varepsilon_{1} &amp;amp; = E_1 d_{11} + E_2 d_{21} + E_3 d_{31} \\&#xA;\varepsilon_{2} &amp;amp; = E_1 d_{12} + E_2 d_{22} + E_3 d_{32} \\&#xA;\varepsilon_{3} &amp;amp; = E_1 d_{13} + E_2 d_{23} + E_3 d_{33} \\&#xA;\varepsilon_{4} &amp;amp; = E_1 d_{14} + E_2 d_{24} + E_3 d_{34} \\&#xA;\varepsilon_{5} &amp;amp; = E_1 d_{15} + E_2 d_{25} + E_3 d_{35} \\&#xA;\varepsilon_{6} &amp;amp; = E_1 d_{16} + E_2 d_{26} + E_3 d_{36} &#xA;\end{align}&#xA;$$&#xA;In matrix form,&#xA;$$&#xA; \begin{bmatrix} \varepsilon_{1} \\ \varepsilon_{2} \\ \varepsilon_{3} \\ \varepsilon_{4} \\ \varepsilon_{5} \\ \varepsilon_{6} \end{bmatrix}&#xA; = \begin{bmatrix} d_{11} &amp;amp; d_{21} &amp;amp; d_{31} \\ d_{12} &amp;amp; d_{22} &amp;amp; d_{32} \\ d_{13} &amp;amp; d_{23} &amp;amp; d_{33} \\ d_{14} &amp;amp; d_{24} &amp;amp; d_{34} \\ d_{15} &amp;amp; d_{25} &amp;amp; d_{35} \\ d_{16} &amp;amp; d_{26} &amp;amp; d_{36} \end{bmatrix} \begin{bmatrix} E_1 \\ E_2 \\ E_3 \end{bmatrix}&#xA;$$&#xA;or&#xA;$$&#xA;  [\boldsymbol{\varepsilon}] = [\mathbf{d}]^T [\mathbf{E}]&#xA;$$&#xA;where&#xA;$$&#xA;  [\mathbf{d}] = \begin{bmatrix} d_{11} &amp;amp; d_{12} &amp;amp; d_{13} &amp;amp; d_{14} &amp;amp; d_{15} &amp;amp; d_{16} \\ d_{21} &amp;amp; d_{22} &amp;amp; d_{23} &amp;amp; d_{24} &amp;amp; d_{25} &amp;amp; d_{26} \\d_{31} &amp;amp; d_{32} &amp;amp; d_{33} &amp;amp; d_{34} &amp;amp; d_{35} &amp;amp; d_{36} \end{bmatrix} &#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Double to triple index map&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;From the above we see that, &lt;strong&gt;in Abaqus&lt;/strong&gt;, the matrix $[\mathbf{d}]$ is&#xA;$$&#xA;\begin{align}&#xA;  [\mathbf{d}] &amp;amp;= \begin{bmatrix} d_{111} &amp;amp; d_{122} &amp;amp; d_{133} &amp;amp; d_{112} &amp;amp; d_{113} &amp;amp; d_{123} \\ d_{211} &amp;amp; d_{222} &amp;amp; d_{233} &amp;amp; d_{212} &amp;amp; d_{213} &amp;amp; d_{223} \\d_{311} &amp;amp; d_{322} &amp;amp; d_{333} &amp;amp; d_{312} &amp;amp; d_{313} &amp;amp; d_{323} \end{bmatrix} \\&#xA; &amp;amp; = \begin{bmatrix} d_{11} &amp;amp; d_{12} &amp;amp; d_{13} &amp;amp; d_{14} &amp;amp; d_{15} &amp;amp; d_{16} \\ d_{21} &amp;amp; d_{22} &amp;amp; d_{23} &amp;amp; d_{24} &amp;amp; d_{25} &amp;amp; d_{26} \\d_{31} &amp;amp; d_{32} &amp;amp; d_{33} &amp;amp; d_{34} &amp;amp; d_{35} &amp;amp; d_{36} \end{bmatrix} &#xA;\end{align}&#xA;$$&#xA;Ansys (and possibly Marc) uses a different convention and the numbers will have to be changed accordingly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the constants in your initial matrix are arranged according to the &lt;strong&gt;standard mechanics convention&lt;/strong&gt;, the above map changes to&#xA;$$&#xA;\begin{align}&#xA;  [\mathbf{d}] &amp;amp;= \begin{bmatrix} d_{111} &amp;amp; d_{122} &amp;amp; d_{133} &amp;amp; d_{123} &amp;amp; d_{113} &amp;amp; d_{112} \\ d_{211} &amp;amp; d_{222} &amp;amp; d_{233} &amp;amp; d_{223} &amp;amp; d_{213} &amp;amp; d_{212} \\d_{311} &amp;amp; d_{322} &amp;amp; d_{333} &amp;amp; d_{323} &amp;amp; d_{313} &amp;amp; d_{312} \end{bmatrix} \\&#xA; &amp;amp; = \begin{bmatrix} d_{11} &amp;amp; d_{12} &amp;amp; d_{13} &amp;amp; d_{14} &amp;amp; d_{15} &amp;amp; d_{16} \\ d_{21} &amp;amp; d_{22} &amp;amp; d_{23} &amp;amp; d_{24} &amp;amp; d_{25} &amp;amp; d_{26} \\d_{31} &amp;amp; d_{32} &amp;amp; d_{33} &amp;amp; d_{34} &amp;amp; d_{35} &amp;amp; d_{36} \end{bmatrix} &#xA;\end{align}&#xA;$$&#xA;You have to make sure that the convention used in the input 2-index matrix is known before you can write down the correct mapping.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="3488" LastEditDate="2018-05-15T23:03:51.137" LastActivityDate="2018-05-15T23:03:51.137" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21766" PostTypeId="1" CreationDate="2018-05-16T07:42:31.750" Score="1" ViewCount="147" Body="&lt;p&gt;A building where three toilets (toilet + urinal each) are connected to the same sewage pipe under the building. The pipe is DN 100 and should have (according to old plans) a gradient of 1:50. Two floor drains are also connected to this pipe. It empties after a few meters into a DN125 sewage pipe that collects the water from a dozen showers. This empties after 6m or so in a larger pipe outside the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The branch of the network near the toilets hade a ventilation pipe straight up through the roof, that was errouneously removed during recent refurbishments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since then, at certain shift changes when the toilets and showers where heavily in use, the floor drain overflowed and sewage from the toilets flooded the area. The whole network was checked with cameras. There's no blockage, only a slight sag near the toilets. It was observed that when you flush a toilet, the urinal in the same stall by will empty partially. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same building, in a nother branch of the sewaer pipe network, the ventilation pipes where removed also. Here, we have a lot of pipe upstream of the toilet and so far no problems occured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, our assumption is that due to the removal of the ventilation, we have this backflow problem. I want to check if my reasoning here is correct &amp;amp; if we are not solving the wrong problem by installing a new ventilation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;When flushing the toilet, I have a plug of water in the pipe and an area of underpressure behind until air can pass over this plug.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When the, downstream of the toilets, the pipe is relativly full because the showers are in use, the water from the toilets flows off slower and the time till equalization is longer&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Hence backflow and overflowing floor drain&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-05-16T18:39:04.527" Title="Why does removal of ventilation of a sewage pipe lead to backflow from the toilet?" Tags="|fluid-mechanics|piping|drain-waste-vent|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21767" PostTypeId="2" ParentId="21766" CreationDate="2018-05-16T08:06:13.693" Score="2" Body="&lt;p&gt;Simply if you remove the ventilation then you create a closed tube where the « piston » ie waste is trying to create a vacuum behind it, that is, untill it can draw past the liquid in the u-bend...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-05-16T10:19:34.910" LastActivityDate="2018-05-16T10:19:34.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21768" PostTypeId="1" AcceptedAnswerId="21776" CreationDate="2018-05-16T08:21:18.667" Score="1" ViewCount="228" Body="&lt;p&gt;If you look at the screenshot you will see a composite auxetic material. How would you calculate the E-modulus for it? Is it as simple as the rule of mixtures, &lt;code&gt;E = V1 * E1 + V2 * E2&lt;/code&gt;? Or is it more complex because of its structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bTy05.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bTy05.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BMjgy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BMjgy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15757" LastEditorUserId="15757" LastEditDate="2018-05-16T10:47:28.620" LastActivityDate="2018-05-16T23:32:57.520" Title="Young's modulus for composite auxetic material" Tags="|materials|elastic-modulus|composite|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21770" PostTypeId="2" ParentId="21732" CreationDate="2018-05-16T15:38:58.593" Score="1" Body="&lt;p&gt;I believe there has been some research done on robot localization based on Wifi signal strength. For example: &lt;a href=&quot;http://robotics.usc.edu/~ahoward/projects_wifi.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://robotics.usc.edu/~ahoward/projects_wifi.php&lt;/a&gt; and &lt;a href=&quot;http://www.cs.cmu.edu/~mmv/papers/10icra-joydeep.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cs.cmu.edu/~mmv/papers/10icra-joydeep.pdf&lt;/a&gt;. But I believe accuracy is very rough (maybe meters of error).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more precise radio localization (cm level), you can use ultrawideband (UWB) radios which can measure the time of flight between endpoints. Yes, there are multipath issues indoors, but there are ways to mitigate this. There are two main players here: &lt;a href=&quot;https://www.decawave.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.decawave.com/&lt;/a&gt; and &lt;a href=&quot;http://bespoon.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://bespoon.com/&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also FYI, triangulation is when you know the angle to the fixed beacons, trilateration is when you know the range. See: &lt;a href=&quot;https://en.wikipedia.org/wiki/Trilateration&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Trilateration&lt;/a&gt;. Your proposed application would be trilateration.&lt;/p&gt;&#xA;" OwnerUserId="16098" LastActivityDate="2018-05-16T15:38:58.593" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21771" PostTypeId="2" ParentId="21760" CreationDate="2018-05-16T17:06:16.057" Score="1" Body="&lt;p&gt;You may need a five-wheel carriage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z2RKB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z2RKB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Illustration of a possible solution using a five-wheeled carriage.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ease of illustration I've used two radii - one of which is infinite.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On the larger radius the inner centre wheel provides the accurate location.&lt;/li&gt;&#xA;&lt;li&gt;On the smaller radius the alignment is done by the previously floating wheels.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;There are some details for you to work out:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What exactly happens at the transition.&lt;/li&gt;&#xA;&lt;li&gt;The centre of the carriage has moved inwards a little on the curve.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-05-16T17:06:16.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21772" PostTypeId="2" ParentId="21766" CreationDate="2018-05-16T17:15:28.530" Score="1" Body="&lt;p&gt;The size and slope of sewer plumbing has been optimized for gravity flow, meaning top part of pipe shall stays dry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the vent pipe is not helping to break the back stream vacuum, the flow may stagnate and either leave deposits clogging the sewer pipe or backing up. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-16T17:15:28.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21773" PostTypeId="2" ParentId="14611" CreationDate="2018-05-16T18:25:15.697" Score="1" Body="&lt;p&gt;&lt;strong&gt;Short Answer:&lt;/strong&gt; Castigliano gives rapid exact solutions at a few key points in a large complex structure, while virtual work gives usable approximate models to complex systems that otherwise would be unsolvable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Castigliano's Theorem and virtual work are two sides of the same mathematical coin.  Castigliano's Methods predates virtual work, but begins the fundamental principles of virtual work.  It forms the simpler half of virtual work, where the displacements could be solved via linear analysis, but we use virtual work to get the answer faster.  Virtual work is described for the latter half, where we could not solve the displacements with linear analysis (without solving differential equations and throwing in a bunch of coefficients), and we rely on virtual work instead to find a good approximate answer that fits many of the boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated above, inn most applications of what engineers use for Castigliano's method, the main principle is to use what is known from linear elastic beam or truss theory (it can be used in multiple aspects for these structures), and quickly solve for a structure subjected to highly unusual forces.  An &lt;strong&gt;&lt;em&gt;equation for the forces&lt;/em&gt;&lt;/strong&gt; is written in terms of many unknown forces of the statically determinate structure, and then the unknown forces are removed.  One of the unknown (or unusual, but known) forces are applied, and the &lt;a href=&quot;https://mechanicalc.com/reference/beam-deflection-tables&quot; rel=&quot;nofollow noreferrer&quot;&gt;linear models and tables&lt;/a&gt; used for single forces can quickly tell us the actual displacement at various points in the structure. The single force may result in 500 newtons of force at a reaction point per newton of original force, or 5 newtons. This is recorded.  The unknown force is removed, and a new force is added, and tested.  Once all of these reactions and forces are found, Castigliano's method can then solve for what the final deflection would be for the entire load-state, which may not be found in a table of solved load-states.  This is especially useful in the case where there are elastic supports, supports that deflect based upon how much force they apply, which happens in any real system.  The only limit on this approach is how detailed the tables are, and the principle of superposition.  So long as the system can be treated using superposition, Castigliano's method will produce exact results (given the model used) at various points within the structure without solving for all of the other points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principle of virtual work extends beyond this principle - the idea is instead to simply write an &lt;strong&gt;&lt;em&gt;equation for the displacements&lt;/em&gt;&lt;/strong&gt; with unknown coefficients.  It could be the solution to the governing DE, or it could be completely inaccurate, but it needs to be able to resolve all of the boundary conditions (at point A, the displacement is 0, etc).  For beams, taking the second derivative of the displacement equation results in the moment equation, taking the third results in the shear equation.  For plates and other continuum, the displacement is the strain times the length.  Any stress terms can be written as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hooke%27s_law#Matrix_representation_.28stiffness_tensor.29&quot; rel=&quot;nofollow noreferrer&quot;&gt;stiffness tensor&lt;/a&gt; times the strain, so the entire virtual work can be expressed simply in terms of our unknown displacement equation, in principle.  Hence, the work is simply to solve for those unknown coefficients such that they minimize the virtual work (both in potential energy for static systems, or the sum of potential energy and kinetic energy for dynamic systems).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this is frequently given with the equations used for finite element analysis, where instead of the normally quartic displacement equations, a cubic equation is used for the displacement.  This is because we have at most two degrees of freedom for rotations, and two degrees of freedom for displacements, so the most we can have is four unknown coefficients - a cubic equation.  Note this means that an FEA needs to then split a distributed load into point loads that allow the cubic equation to have the same the same deflections as the original quartic.  This is what causes single elements to not show the same mid-span deflections as the original quartic:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ywQXU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ywQXU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even without superposition, the principal of virtual work still applies, so long as your stiffness tensor accounts for the change in stress with regards to strain.  This may take an independent &lt;strong&gt;&lt;em&gt;unknown stress equation&lt;/em&gt;&lt;/strong&gt;, to use in lieu of the stiffness tensor.  Variations of this kind are used in many fields by the engineers who need to make mathematical models of their systems, which form the basis for practically all finite element methods.  In summary, Castigliano gives rapid exact solutions at a few key points in a large complex structure, while virtual work gives usable approximate models to complex systems that otherwise would be unsolvable.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-05-21T17:48:01.547" LastActivityDate="2018-05-21T17:48:01.547" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21774" PostTypeId="2" ParentId="21766" CreationDate="2018-05-16T18:39:04.527" Score="0" Body="&lt;p&gt;To go a bit deeper - you can have a situation where there is no ventilation pipe and the sewer still works, but you will be lucky.  In this case, having two ventilation pipes allowed the first building to work when one was removed - the remaining pipe allowed the building to continue to ventilate, but it worked twice as hard.  The other situation is when the sewer line is under vacuum, which can happen if you live on a hilly area and the sewage line runs down stream.  If you live in the valley, the opposite occurs, and the manway covers can have &lt;a href=&quot;https://sploid.gizmodo.com/why-the-hell-is-this-manhole-cover-levitating-1784883167&quot; rel=&quot;nofollow noreferrer&quot;&gt;so much pressure they come off&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many situations, if you have nuisance odors from the ventilation pipes, consider an odor control system.  Activated carbon systems are especially well suited for this application. &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-16T18:39:04.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21775" PostTypeId="1" CreationDate="2018-05-16T21:12:28.410" Score="1" ViewCount="262" Body="&lt;p&gt;I know how to find the power and the energy of a signal. What I have a problem with is knowing whether a signal is an energy signal or a power signal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my teacher, a signal is a power signal if its power is finite (less than infinity) and non-zero and a signal is an energy signal if its energy is finite (less than infinity). What he does not specify is what happens if the energy of a signal is 0 such as in the case of the signal x(t) = 0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy is obviously 0. Is this signal considered an energy signal or the signal x(t) is not considered a signal at all since its zero?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, is it possible for a signal to be neither a power signal nor an energy signal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerDisplayName="user12377" LastEditorUserId="-1" LastEditDate="2019-05-18T20:58:27.247" LastActivityDate="2022-06-02T03:01:40.177" Title="What are the conditions for a signal to be an energy signal and a power signal?" Tags="|electrical-engineering|power|energy|signal-processing|homework|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21776" PostTypeId="2" ParentId="21768" CreationDate="2018-05-16T21:19:58.910" Score="2" Body="&lt;p&gt;Rather than thinking about a Young's modulus directly, the problem becomes easier to grasp if you think in terms of a relationship between the load and displacement, and subsequently between stress and the strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the case where deformations are small, and just consider the stiff direction (tension/compression vertically and not horizontal shear in your first figure), you can apply a vertical load to the structure and compute the vertical deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P_y$ be the applied load and let $u_y$ be the resulting displacement.  For the linear part of the load-displacement curve you will get a relation&#xA;$$&#xA;  P_y = K u_y&#xA;$$&#xA;where $K$ is the slope of the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convert these into stresses and strains, you will need two quantities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the area ($A_{xz}$) over which the load is applied&lt;/li&gt;&#xA;&lt;li&gt;the initial length ($L_y$) of the unit cell.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The choice of these quantities, particularly the area, is not obvious and several options are possible.  For example, the area might be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;just the area of the cylinder on which the load is applied&lt;/li&gt;&#xA;&lt;li&gt;the area of the triangular domain of the unit cell&lt;/li&gt;&#xA;&lt;li&gt;a square area that tiles space in the horizontal plane, etc.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You have to explicitly define what the area is and justify that choice satisfactorily.  Once that choice has been made, you can compute the stress-strain relation using&#xA;$$&#xA;  \sigma_{yy} = \frac{P_y}{A_{xz}} = \frac{K L_y}{A_{xz}}\frac{u_y}{L_y} = \frac{K L_y}{A_{xz}} \varepsilon_{yy} =: E_{yy} \varepsilon_{yy}&#xA;$$&#xA;where your definition of modulus is&#xA;$$&#xA;  E_{yy} = \frac{K L_y}{A_{xz}}&#xA;$$&#xA;That's the preferred way of computing a modulus.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can compare that with your mixture relation (or any other continuum theory) to check what the error in the rule of mixtures is.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-05-16T21:19:58.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21777" PostTypeId="2" ParentId="21775" CreationDate="2018-05-16T22:27:49.620" Score="1" Body="&lt;p&gt;I don't think that this line of thought is going to help you much. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;signal&lt;/strong&gt; [sig-nl]&lt;br&gt;&#xA;&lt;em&gt;Electronics&lt;/em&gt;. an electrical quantity or effect, as current, voltage, or electromagnetic waves, that can be varied in such a way as to convey information. Source: &lt;a href=&quot;http://www.dictionary.com/browse/signal?s=t&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dictionary.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In most cases some power will be required to pass the information but that is inherent in the way electronics works and it doesn't mean that power is a signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast power suggests providing &lt;em&gt;energy&lt;/em&gt; to a device with no information transmitted.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The energy is obviously 0. Is this signal considered an energy signal or the signal x(t) is not considered a signal at all since its zero?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would think that zero indicates a lack of power or signal. It's nothing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Furthermore, is it possible for a signal to be neither a power signal nor an energy signal?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We're back to the definitions again. If it's a signal then it's a signal.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-05-16T22:27:49.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21778" PostTypeId="2" ParentId="21768" CreationDate="2018-05-16T23:32:57.520" Score="0" Body="&lt;p&gt;Related: &lt;a href=&quot;https://engineering.stackexchange.com/questions/4182/how-do-i-calculate-an-estimate-for-the-properties-of-a-composite-material&quot;&gt;How do I calculate an estimate for the properties of a composite material?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we actually have an unusual case of the fourth method.  We could look at each individual unit cell of the material and apply a unit force on the structure in the x, y, and z directions.  Each application of unit force would correspond to a certain deflection.  The net deflection divided by the bounding box dimensions of the unit cell would result in a strain, and the unit force divided by a stress would result in a stress.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this simple approach does not appreciate the 3D complexity of the problem.  Since we are calculating the response to a force, we are finding the individual relations of the &lt;a href=&quot;https://en.wikiversity.org/wiki/Introduction_to_Elasticity/Constitutive_relations&quot; rel=&quot;nofollow noreferrer&quot;&gt;stiffness tensor&lt;/a&gt;.  Note that for general anisotropic materials such as this one, we need to find 21 different relationships.  However, afterwards, the stiffness tensor is not usable, we need to invert the matrix to find the compliance tensor.  The elastic modulus would then be the inverse of the first term of the stiffness tensor, but the usable format would utilize the compliance tensor.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-16T23:32:57.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21781" PostTypeId="1" AcceptedAnswerId="21792" CreationDate="2018-05-17T08:38:40.787" Score="1" ViewCount="33" Body="&lt;p&gt;In CAD, there are multiple file formats only for the scope of saving polygonized models in three dimensions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, there is the &lt;a href=&quot;http://segeval.cs.princeton.edu/public/off_format.html&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Object FIle Format&quot;&gt;Object File Format&lt;/a&gt; and furthermore, there is the much more famous &lt;a href=&quot;http://www.fabbers.com/tech/STL_Format&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;STL&quot;&gt;STL&lt;/a&gt; format. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the different (most important) use cases and reasons to use especially one of them?&lt;/p&gt;&#xA;" OwnerUserId="14377" LastActivityDate="2018-05-17T21:42:20.030" Title="Why are there multiple ways to save polygonized 3D model data?" Tags="|cad|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21782" PostTypeId="1" CreationDate="2018-05-17T09:18:19.530" Score="0" ViewCount="72" Body="&lt;p&gt;If there is an 11kW wind turbine, and a 15kW wind turbine. Why is the 15 kw wind turbine more expensive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it seems like a basic question, but what exactly makes it more expensive? Is it only the generator? Or is there something less obvious that causes it to drive the turbine price higher, with higher power?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am hypothesizing that the weight of the turbine might be the major factor that determines the price of the turbine, and not its generator capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any additional thoughts are welcome.&lt;/p&gt;&#xA;" OwnerUserId="3538" LastEditorUserId="9" LastEditDate="2018-05-20T16:50:56.980" LastActivityDate="2018-05-20T16:50:56.980" Title="What determines the cost of the wind turbines?" Tags="|turbines|wind-power|cost-engineering|cost-estimation|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21783" PostTypeId="1" AcceptedAnswerId="23269" CreationDate="2018-05-17T09:18:33.580" Score="1" ViewCount="114" Body="&lt;p&gt;I have non-rigid beam attached to a gearbox and motor assembly. The gearbox outputs $1000 N \cdot m$ at the shaft. The beam is rotated at $3000 RPM$ around it's center. If the motor is switched off immediately and acts as a brake, is it possible that the torque generated from the momentum of the beam could be greater than the $1000N \cdot m$ that the motor originally produced, and if so how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to work out the max torque a beam could create at the gearbox mount when the power is cut suddenly, however we don't know any of the beam characteristics such as where the point loads are, how long it takes to stop etc. All I have is the gearbox and motor specifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This question is different from a similar one I asked &lt;a href=&quot;https://engineering.stackexchange.com/questions/21745/how-to-calculate-torque-needed-to-stop-a-rotating-beam-over-x-seconds&quot;&gt;here&lt;/a&gt;. In the other question I mentioned that the beam was uniformly distributed, however in this example I have no beam information. All I have to go off is the motor and gearbox specifications. &lt;/p&gt;&#xA;" OwnerUserId="13810" LastEditorUserId="13810" LastEditDate="2018-05-17T10:05:05.593" LastActivityDate="2018-08-16T15:17:03.823" Title="How to calculate the max torque due to angular momentum of a beam given motor specs" Tags="|dynamics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21784" PostTypeId="2" ParentId="21782" CreationDate="2018-05-17T09:59:04.100" Score="3" Body="&lt;p&gt;Where does the power for a wind turbine come from? Answer - the wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do you get 15kW from the wind rather than 11kW? Answer - you either need a location with a higher wind speed, or you need to pick up more wind, i.e. the blades need to be longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this affect cost? Answer - a longer blade needs more material for its extra length. But then you're also taking more force, so that needs extra material along the blade too. And the extra force means the turbine tower needs to be stronger, so more material for that too. It all adds up.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-05-17T09:59:04.100" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21786" PostTypeId="1" CreationDate="2018-05-17T13:06:47.160" Score="1" ViewCount="35" Body="&lt;p&gt;What's the purpose of the special railway transport wagons shown in the photograph?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DXein.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DXein.jpg&quot; alt=&quot;railway wagons&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16108" LastEditorUserId="1832" LastEditDate="2018-05-17T14:23:14.500" LastActivityDate="2018-05-17T14:23:14.500" Title="Purpose of special railway transport wagons?" Tags="|transportation|" AnswerCount="0" CommentCount="2" ClosedDate="2018-05-18T15:55:42.107" ContentLicense="CC BY-SA 4.0" />
  <row Id="21788" PostTypeId="1" CreationDate="2018-05-17T14:52:10.453" Score="1" ViewCount="79" Body="&lt;p&gt;Various sources (e.g., &lt;a href=&quot;https://en.wikipedia.org/wiki/Barricade_tape#OSHA_and_ANSI_requirements_for_barricade_tape&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; and &lt;a href=&quot;http://www.worksafely.com/safety-equipment/barricade-tape.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;WorkSafely&lt;/a&gt;) claim that the seven color combinations commonly used for barricade tape (a.k.a. caution tape) and safety markings in North America (yellow and black for physical hazards, green and white for first aid supplies, etc.) are specified by the US Occupational Safety and Health Administration (OSHA). OSHA standard &lt;a href=&quot;https://www.osha.gov/pls/oshaweb/owadisp.show_document?p_table=STANDARDS&amp;amp;p_id=9793&quot; rel=&quot;nofollow noreferrer&quot;&gt;1910.144&lt;/a&gt; (&quot;Safety color code for marking physical hazards&quot;) is often mentioned. However, 1910.144 states only the following regarding color:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The color red should be used to identify fire protection equipment, sources of danger, and emergency stop controls.&lt;/li&gt;&#xA;&lt;li&gt;The color yellow should be used to mark physical hazards.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are the seven color combinations specified in another OSHA standard? (If so, I haven't been able to find it.) If not, where did they originate?&lt;/p&gt;&#xA;" OwnerUserId="11066" LastActivityDate="2018-05-17T14:52:10.453" Title="Origin of barricade tape color combinations (allegedly OSHA)" Tags="|safety|standards|regulations|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21789" PostTypeId="2" ParentId="21783" CreationDate="2018-05-17T14:55:26.530" Score="1" Body="&lt;p&gt;Depending on the elasticity of your beam and its mass, it will vibrate in a complex harmonic pattern advancing the 3000rpm ad retarding it, in a pattern similar to that of the winded spring of mechanical hand watches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the absence of more detail on the beam we assume this simplified harmonic motion of the beam rotation above and below 3000rpm, while its center is attached to motor at constant 3000 rpm. Lets call the difference in rotation, or swinging like a pendulum about center, delta rpm, DN .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$  DN_{(t)} = Acos(\omega t)$$ and $$\space \omega = \sqrt {k/m} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the torque follows the same harmonic pattern, it can be more or less than 1000N.m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may also be higher frequency modes on the beam where it can vibrate on 2nd harmonic mode, meaning the is vibration along the length of the beam, while it advances and retards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These all have impact on the torque!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-17T14:55:26.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21791" PostTypeId="2" ParentId="21782" CreationDate="2018-05-17T15:48:46.763" Score="0" Body="&lt;p&gt;There's manufacturing cost, and then there's optimum sale price.  You charge what the market will bear, and people tend to pay more for &quot;bigger&quot; things.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Historically, this law of economics caused major headaches for American car manufacturers in the mid-70s.  Fuel prices led to demand for small cars, and consumers expected that a car half the size of typical American sedans would cost half as much.  But since material costs are a small fraction of total manufacturing cost,  this didn't happen.  &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-05-17T15:48:46.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21792" PostTypeId="2" ParentId="21781" CreationDate="2018-05-17T15:51:28.307" Score="1" Body="&lt;p&gt;&quot;The great thing about standards is there are so many to choose from&quot; attr. Andrew Tannenbaum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some CAD formats were deliberately developed to be proprietary so as to enhance sales of someone's CAD software.  Some were developed with certain applications in mind.  Others were developed ad hoc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what it's worth, there are probably 10X as many &quot;standard&quot; image file formats as there are CAD formats.  Something we all have to live with -- and a gold mine for those who sell format-conversion software tools. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-05-17T15:51:28.307" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21794" PostTypeId="2" ParentId="21781" CreationDate="2018-05-17T21:42:20.030" Score="0" Body="&lt;p&gt;The formats have different functions. While it is true that some of the formats duplicate functionality it would be a mistake to think all the formats are interchageable and serve same purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example stl is the file format intended for 3D printing. All it does is store a bunch of sepaarate triangles. It is meant to be extremely easy to dump. But it does not contain any other reliable info. The fileformat forces you to rebuild all semantic info that may have been in the mesh. The format is nearly unusable for anything other it was intended for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually you then have one or more formats meant for rendering like the obj format which resembles OFF you linked but has info on rudimentary material properties, uv maps normals, connectivity and so on. This makes the format much more usable. But alas unlike STL there is no binary version so these can become really big for highly tesselated meshes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you usually have the intermediate format needed to make 3D PDF files, you get it for free if you intend to support 3D PDF, its nearly useless outside the context.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2018-05-17T21:42:20.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21795" PostTypeId="1" AcceptedAnswerId="21799" CreationDate="2018-05-17T22:55:13.543" Score="0" ViewCount="245" Body="&lt;p&gt;For a plant G(s), the transfer function is given by:&#xA;$$G(s) = \frac{(10s+50)}{s(s^2+20s+2500)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Ziegler-Nicholas method to get the parameter: $k_p = \dfrac{-1000}{3}$ and $w =-28.868j$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct? what should I do in case of negative $k_p$ and how do I calculate the PID parameters?&lt;/p&gt;&#xA;" OwnerUserId="16116" LastEditorUserId="1832" LastEditDate="2018-05-18T01:03:29.783" LastActivityDate="2018-05-18T08:07:27.503" Title="Negative gain in Ziegler-Nicholas Oscillation Method?" Tags="|electrical-engineering|control-engineering|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21796" PostTypeId="1" CreationDate="2018-05-17T23:20:35.897" Score="1" ViewCount="18" Body="&lt;p&gt;Can I use the delay of different paths of a RAKE receiver for orthogonal channels. The idea is that if I have a square law detector for a packet of bits there will be no channel interference because the delay effectively acts as a pad of zeros. It's a technique that I want to use? If there are any papers on this I would greatly appreciate it.&lt;/p&gt;&#xA;" OwnerUserId="16118" LastActivityDate="2018-05-17T23:20:35.897" Title="Orthogonal Channels in a RAKE receiver?" Tags="|electrical-engineering|wireless-communication|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21799" PostTypeId="2" ParentId="21795" CreationDate="2018-05-18T08:07:27.503" Score="1" Body="&lt;p&gt;The Ziegler-Nichols cycling method is not appropriate for this system. Because of the zero located in -5, the closed loop system is always stable and will not oscillate for proportional compensator (poles will never be conjugated on top of the $j\omega$ axis), as can be evidenced by the root locus plot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LME2P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LME2P.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you still wish to use the Ziegler-Nichols table for this system you should use a different approach, like the step response method.&lt;/p&gt;&#xA;" OwnerUserId="16125" LastActivityDate="2018-05-18T08:07:27.503" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21800" PostTypeId="1" CreationDate="2018-05-18T09:33:48.030" Score="0" ViewCount="2275" Body="&lt;p&gt;I have a gearbox adapter that bolts onto a flange and I need to work out how much clamping force is needed to prevent the faces from slipping and keep the bolts in tension rather than shear when torque is applied. However I can't use the straightforward friction equation here $(F_f =\mu \cdot F_n)$ because I don't know how to work out how the force from the applied torque is distributed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, the torque being applied to the flange is $9123N \cdot m$ and the flange has an inside diameter of $185 mm$ and and outside diameter of $275 mm$. There are $12$ x $16 mm$ bolt holes at a diameter of $196 mm$. The two faces bolted together are made of $1020$ mild steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured that while friction doesn't have anything to do with surface area, in this case it would be better to have the flange face diameters as wide as possible because it would require less frictional force to resist the torque at a wider radius. However this has just confused me as I don't know what/where the forces are acting on the faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find to force from torque that is distributed over a surface such as this?&lt;/p&gt;&#xA;" OwnerUserId="13810" LastActivityDate="2018-05-18T13:39:30.050" Title="How to find required clamping force of bolts to prevent flange faces from twising due to torque?" Tags="|torque|friction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21801" PostTypeId="1" CreationDate="2018-05-18T10:40:32.797" Score="4" ViewCount="1019" Body="&lt;p&gt;I've got a structural engineer ready to provide calcs on this job, but there is a specific aspect where I'd like a heads-up what the normal approach would be, before I raise it with him, in case he says it's a matter for builders to decide (which it probably isn't!).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The building&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The property is a pretty typical London UK suburban 2 bedroom terrace, early 20th century. The wall is 2 x 100mm old London brick, with a 50mm unfilled cavity. The aim is to install a steel 203x203x46kg steel moment-resisting frame on the ground floor (USA:1st floor), as part of sorting out inadequate support for the rear wall as part of an extension. The rear wall is 6m wide and the opening will be about 2800 H x 6000 W resting on the original trench foundations, so there will be a horizontal load spreader, and 3 vertical columns (both sides + middle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've sketched one side of the frame and related wall:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P0zdF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P0zdF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty confident about installing the horizontal load spreader (I've cut out below a cavity wall and retrofitted lintels/RHS below it, elsewhere). But for this frame, I can't mount the vertical columns against the wall - I need to cut a slot through both leaves of the rear cavity wall, and fit the UC into the slot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The space for the slot isn't wide (maybe 220mm wide - the width of a single brick), but it is quite tall.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The problem&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;My concern is the lateral bracing shown in the top-right sketch. While most of the load on the removed brickwork is vertical and removing 200mm width isn't an issue, the masonry probably also provides lateral/torsion/racking support to the adjacent rear wall and party wall. I'd be happy if the slot was only a metre or so high, but I'm worried that a 2.8m slot might be too much to trust without appropriate lateral bracing, while the UC is being installed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;This is surely a very common situation for domestic extensions. What should I expect to be told as the usual/proper way to brace the slot, to make absolutely sure there can be no adverse lateral freedom or movement affecting the adjacent wall, during the few days it'll take for the 2.8m slot to be cut and emptied of brickwork, and the UC being installed and mortared in place?&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;My thoughts so far&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I have a couple of ideas, sketched below, but I can't find a definitively good answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/06Y5t.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/06Y5t.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own ideas revolve around using short (200mm vertical) steel sections to brace the wall laterally, and hold the slot open properly while the column is installed. Once the column is in, the braces become redundant and can be cut away or left in place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of either using RHS and slotting the new UC inside it - then it just needs moving sideways to fit - or creating heavy duty &quot;U&quot; shaped sections with 20mm steel bracing at their &quot;open&quot; end, which can be selectively opened and closed one at a time, to allow the UC to be installed. But I've never seen a description of a &quot;proper&quot; or safe way to do this, so these are just guesses.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Questions on my mind:&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;What is the realistic risk?&lt;/strong&gt;&lt;br/&gt;Realistically, how significant is the lateral movement/racking/twisting risk in this situation? How much/how likely is it that there really is a problem?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;How severe and what direction are the non-vertical forces/moments likely to be?&lt;/strong&gt;&lt;br/&gt;Assuming &quot;normal/ordinary&quot; construction, what should I expect for typical/relevant non-vertical forces/moments and their directions, to help anticipate what the engineer/builder needs to counter?&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;What is the usual way this is done safely, and are my ideas any good?&lt;/strong&gt;&lt;br/&gt;What is the usual way to install UCs without exposing this risk? How heavy duty is this bracing typically, if bracing is needed (Are my ideas adequate, or are there better/cheaper ways?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2018-05-18T10:55:15.367" LastActivityDate="2018-05-18T16:10:47.967" Title="How to provide lateral support for adjacent masonry during installation of steel UC in an old cavity wall?" Tags="|structural-engineering|civil-engineering|structures|strength|reinforcement|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21802" PostTypeId="2" ParentId="21800" CreationDate="2018-05-18T13:39:30.050" Score="4" Body="&lt;p&gt;The simplest way by far is to select a prospective bolt, calculate the clamping force this creates when it is torqued to rated torque, and use the bolt circle as an approximation of the moment arm. Compare this against the &lt;em&gt;rated&lt;/em&gt; torque spec and duty type for the gearbox. Grade 5 bolts are ordinarily adequate for industrial uses. Some fussier machines or harsh environments may need grade 8. Most fussy ones would have the receiving flange tapped for a particular bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real problem here is for shock load tolerance, torque pulses from diesels, vibration tolerance, brake and clutch engagement torques, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since many of these issues are balled into &lt;em&gt;load factor&lt;/em&gt; coefficients, and can be as high as 300% or 400% in severe cases, getting the effective flange pressure radius right to within 5% isn't very important to the big picture. Repeatability with new and to-spec hardware using a calibrated torque wrench is only 10%. Five years from now, some mechanic is going to replace a bolt with one pulled from a coffee can and tighten it with pliers. That's what you need to design for in most cases, and is why typical flanges are more than adequate.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-05-18T13:39:30.050" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21803" PostTypeId="2" ParentId="21801" CreationDate="2018-05-18T15:30:40.997" Score="3" Body="&lt;p&gt;&lt;em&gt;Disclaimer - No contract exists between us. This post is provided as information and opinion only and does not form &quot;advice&quot; in any way. Although I am a Chartered Civil Engineer I have zero professional experience with masonry. My views here should be treated as those of a keen DIYer (which I also am).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I suspect it'll probably be ok without any lateral support. The vast majority of lateral loading in the UK comes from wind - as long as your builder doesn't cut out the existing wall directly before the highest winds for the past 25 years, I can't see it failing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you could do is ask your structural engineer whether they've designed for a positive lateral connection between the UC and the existing brickwork. My guess is not, and if that connection isn't needed for support long term, then it won't be needed during construction either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way of looking at it is to consider patio / bi-fold doors. The vast majority of frames for these are non-structural, and yet there isn't a problem with houses' rear walls falling over from lateral loading. Your hole might be wider than most of these, but I don't think it's higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The practical solution for you is of course to just talk to your structural engineer. Construction (Design &amp;amp; Management) regulations (otherwise known as CDM Regs) require the designer to come up with at least one safe method of constructing their design. So your engineer &lt;em&gt;will&lt;/em&gt; have already considered a method that is safe. And if that method is &quot;just cut out without support - it's not required&quot; then they'll be able to tell you that.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-05-18T15:30:40.997" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21805" PostTypeId="2" ParentId="21801" CreationDate="2018-05-18T16:10:47.967" Score="1" Body="&lt;p&gt;I would recommend a temporary shoring of the wall by attaching a vertical 3&quot; x 6&quot; board to the wall near the location of new column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This board will be attached to the wall by 5x 1/4 inch anchor bolts, and will be laterally braced by a few diagonal 2x4&quot; braces which are secured to the ground. later this board shall be removed with the anchors, and any holes patched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would make shore my Engineer has details on positive connection between the new column and adjacent masonry, such as epoxy dowels or other means.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-18T16:10:47.967" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="21807" PostTypeId="1" AcceptedAnswerId="21811" CreationDate="2018-05-18T20:59:37.927" Score="1" ViewCount="840" Body="&lt;p&gt;How would one work out the force required to compress flat, using a flat bar on top pressing down, a thick walled (15mm OD 10mm ID) piece of flexible plastic (PVC) tubing i.e. so reducing the x-sectional area inside the tube to 0?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just looking for any way to approximate the real situation without having to use an FEA simulation.&lt;a href=&quot;https://i.stack.imgur.com/1Q8H2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Q8H2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="11394" LastEditorUserId="11394" LastEditDate="2018-05-18T21:20:49.487" LastActivityDate="2018-05-19T00:26:46.170" Title="Force to compress flexible elastic tubing" Tags="|compressors|elastic-modulus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21808" PostTypeId="2" ParentId="2935" CreationDate="2018-05-18T21:20:40.413" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/4182/how-do-i-calculate-an-estimate-for-the-properties-of-a-composite-material&quot;&gt;Related - how do i calculate an estimate for the properties of a composite material&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The reference to &lt;a href=&quot;https://www.library.ucdavis.edu/wp-content/uploads/2017/03/HDBK17-3F.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mil Handbook 17F&lt;/a&gt;, p. 213 is summarized here:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J7Rbi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J7Rbi.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Computation of effective elastic moduli is a very difficult problem in&#xA;elasticity theory and only a few simple models permit exact analysis.&#xA;One type of model consists of periodic arrays of identical circular fibers, e.g., square periodic arrays or hexagonal periodic arrays ... These models are analyzed by numerical finite difference or finite element procedures. Note that the square array is not a suitable model for the majority of Uni-Directional Composites since it is not transversely isotropic.&lt;/p&gt;&#xA;&lt;p&gt;The composite cylinder assemblage (CCA) model permits exact analytical determination of effective elastic moduli ... Consider a collection of composite cylinders, each with a circular fiber core and a concentric matrix shell. The size of the cylinders may vary but the ratio of core radius to shell radius is held constant.  Then...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LOxId.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LOxId.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;(Where &lt;span class=&quot;math-container&quot;&gt;$V_f$&lt;/span&gt; is the volume fraction of fibers to the total amount of material.&#xA;&lt;span class=&quot;math-container&quot;&gt;$X_{m}$&lt;/span&gt; is a property of the matrix, &lt;span class=&quot;math-container&quot;&gt;$X_{f}$&lt;/span&gt; is a property of the fiber, and &lt;span class=&quot;math-container&quot;&gt;$E, G, k$&lt;/span&gt; are the elastic modulus, shear modulus, and bulk modulus properties.  The bulk modulus, k, can be computed for isotropic materials as &lt;span class=&quot;math-container&quot;&gt;$\frac{E}{2(1-\nu-2\nu^2)}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\nu$&lt;/span&gt; is the Poisson's ratio.  The G without a subscript is a typo, and should be replaced with &lt;span class=&quot;math-container&quot;&gt;$G_m$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A preferred alternative is to use a method of approximation which has been called the Generalized Self Consistent Scheme (GSCS). According to this method, the stress and strain in any fiber is approximated by embedding a composite cylinder in the effective fiber composite material. The volume fractions of fiber and matrix in the composite cylinder are those of the entire composite. Such an analysis ... results in a quadratic equation for the shear modulus...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kRVB9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kRVB9.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The net algorithm is to compute the effective bulk modulus &lt;span class=&quot;math-container&quot;&gt;$k^*$&lt;/span&gt;, &lt;em&gt;12&lt;/em&gt; poisson's ratio &lt;span class=&quot;math-container&quot;&gt;$\nu_{12}^*$&lt;/span&gt;, and young's modulus &lt;span class=&quot;math-container&quot;&gt;$E_{1}^*$&lt;/span&gt; first, then use the quadratic formula listed to calculate the second shear modulus, &lt;span class=&quot;math-container&quot;&gt;$G_2^*$&lt;/span&gt;.  Using &lt;span class=&quot;math-container&quot;&gt;$G_2^*$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$E_2^*$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\nu_{23}^*$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$G_1$&lt;/span&gt; can be calculated.  These are in the local coordinate system of the fiber.  To translate to global coordinates:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AHLti.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AHLti.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;We can then rotate the fiber to find the properties of the uni-directional composite to find the properties in an arbitrary direction:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wkSOI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wkSOI.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;where Qbar is the rotated matrix, and Q is the original inverse matrix.  For a stochastic model, the angle of the fiber and the volume fraction can be the inputs, and the outputs would be the resulting properties.  Note that for a uniform random distribution, it is possible to integrate the Qbar matrix as theta varies from 0 to &lt;span class=&quot;math-container&quot;&gt;$2\pi$&lt;/span&gt;, then divide by &lt;span class=&quot;math-container&quot;&gt;$2\pi$&lt;/span&gt; to obtain a symmetrical matrix.  The results from this method match well with data on random fiber materials in the fiberglass industry.&lt;/p&gt;&#xA;&lt;p&gt;As you asked about a differential equation, we'd need to review the appropriate theory from this point on.  For example, the classical plate equation, &lt;span class=&quot;math-container&quot;&gt;$$\nabla^2\nabla^2 = \frac{q}{D}$$&lt;/span&gt;, works partly.  We have to include another stoichastic variable, the height of the fiber inside of a block of concrete.  The closer the fiber is to the top, the stiffer the block will be against the bending load.  The block can be divided into arbitrary segments of uniform thickness, and the volume of the fibers in each segment is added, generating different Qbars.  A different distribution would result in different properties of the block:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6fBoD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6fBoD.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This matrix, called the ABD matrix, would then redefine the plate equation as follows:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$D_{11}\frac{\partial^4 w}{\partial x^4} + 2(D_{12}+2 D_{66})\frac{\partial^4 w}{\partial x^2 \partial y^2} + D_{22}\frac{\partial^4 w}{\partial y^4} = q(x,y)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;for the simplest of cases (B matrix irrelevant, no transverse loading, etc...).  The cases go stranger from there, but can be derived from the original derivations, but stopping when the model says to assume the stress is proportional to the stain.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-05-21T17:46:11.073" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21809" PostTypeId="1" AcceptedAnswerId="21819" CreationDate="2018-05-18T21:41:16.000" Score="2" ViewCount="260" Body="&lt;p&gt;How might one calculate or approximate the negative pressure applied to the outer surface of a cylinder required to expand it a certain distance, when held open rigidly at each end?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cylinder ID = 20*the wall thickness and the material is very elastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LTr3w.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LTr3w.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for your help!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oli&lt;/p&gt;&#xA;" OwnerUserId="11394" LastEditorUserId="11394" LastEditDate="2018-05-19T09:59:30.597" LastActivityDate="2018-05-19T21:14:02.630" Title="Calculate or approximate force required to expand a thin-walled elastic cylinder held rigidly at both ends?" Tags="|pressure|elastic-modulus|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="21810" PostTypeId="2" ParentId="21809" CreationDate="2018-05-18T23:49:19.360" Score="2" Body="&lt;p&gt;This is a link to a solved problem similar to your question.&#xA;&lt;a href=&quot;http://nptel.ac.in/courses/105106049/lecnotes/mainch7.html#x43-1260007.4&quot; rel=&quot;nofollow noreferrer&quot;&gt;inflation of an annular cylinder&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In their example the two ends are fixed, no traction of cylinder, meaning the length of cylinder doesn't change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a partial deferential boundary value problem solved as part of their online course. NPTEL is one of India's online university institutes.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-18T23:49:19.360" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21811" PostTypeId="2" ParentId="21807" CreationDate="2018-05-19T00:26:46.170" Score="1" Body="&lt;p&gt;At the crushed configuration we can assume we are dealing with a flat beam. &#xA;Width= L of pipe, span = 15pi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Height = 5mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for simplicity we assume three plastic hinges ( we will refine this by assuming more hinges, such as one at either edge of pressure block if need be), one at either crushed edge of the pipe, one at the center of horizontal flatly deflected top of the pipe right under the middle of back block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we assume this ultimate failure configuration? because we think this is the lowest energy needed, of course we later can try other collapse models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two plastic hinges on the side need to be enough to counter the resistance of these hinges M_p.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M_p = L5^2/4 \times \sigma_{plastic yield}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the force needed is $$M_p/7.5\times \pi$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to add to above the force needed to crush the pipe at the center which is equal to previous force/2: effective length is double.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first collapse mechanism. now we can assume other modes and compare the results, in a few iterations we get close to required precision.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-19T00:26:46.170" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21812" PostTypeId="2" ParentId="2852" CreationDate="2018-05-19T04:23:57.777" Score="2" Body="&lt;p&gt;Measuring water in open channels is a basic element of water conservation. With the increasing demand for improved water management techniques, there is a serious need for low-cost and accurate water-measuring devices like flow meters &amp;amp; &lt;a href=&quot;https://www.iconprocon.com/product-category/levelpro/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Liquid Level sensors.&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ever since the development of the Parshall flume, attempts have been made to simplify the construction and improve the accuracy of water-measuring devices in open channels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circular flume is an appropriate device for measuring flow through furrows because its circular shape fits to the natural shape of a furrow, reducing the possibility of lateral flow around the flume. The device also has been used successfully in lined and unlined canals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High costs have prevented the use of water measuring flumes by growers. However, recently a practical water measuring device has been designed that can be used by growers at low cost: the circular flume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just my raw information which i've shared with you, further you can also study about it&lt;/p&gt;&#xA;" OwnerUserId="16133" LastActivityDate="2018-05-19T04:23:57.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21813" PostTypeId="1" CreationDate="2018-05-19T05:02:47.487" Score="0" ViewCount="51" Body="&lt;p&gt;I have been recommended not to run the two-wheeler (Suzuki Access) at a constant speed (I give a constant throttle in a way that I run essentially at a constant speed of about 40-45 kmph). In fact, I have been told to keep changing the acceleration (smoothly though) for lesser fuel consumption. I want to understand the technical reason behind this claim.&lt;/p&gt;&#xA;" OwnerDisplayName="user16060" LastActivityDate="2018-05-20T08:50:58.650" Title="Should I run 2-wheeler at constant speed?" Tags="|mechanical-engineering|energy-efficiency|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21814" PostTypeId="2" ParentId="21813" CreationDate="2018-05-19T05:15:30.770" Score="0" Body="&lt;p&gt;The optimum speed for obtaining maximum efficiency varies is not a constant with the load. That might be the reason for the advice.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastActivityDate="2018-05-19T05:15:30.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21815" PostTypeId="1" AcceptedAnswerId="21816" CreationDate="2018-05-19T16:44:08.327" Score="-1" ViewCount="26" Body="&lt;p&gt;A sealed ball is not connected to anything externally, but can have any complex mechanism inside (mechanical, chemical, or any other). Can a wireless trigger from a remote or any other wireless mechanism make the ball's mass heavier? Take into account that the ball is sealed and therefore only internal changes can occur, no external objects should increase the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Someone suggested that the ball should create a Higgs field inside it which is impractical but theoretically correct.&lt;/p&gt;&#xA;" OwnerUserId="16137" LastActivityDate="2018-05-19T19:23:02.420" Title="A sealed ball is not connected to anything externally, but can have any complex mechanism inside. Can the mass change?" Tags="|applied-mechanics|" AnswerCount="1" CommentCount="0" ClosedDate="2018-05-20T01:07:08.760" ContentLicense="CC BY-SA 4.0" />
  <row Id="21816" PostTypeId="2" ParentId="21815" CreationDate="2018-05-19T18:29:25.930" Score="1" Body="&lt;p&gt;Within the realms of practical engineering (the limits of this SE), &lt;strong&gt;No.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Conservation_of_mass&quot; rel=&quot;nofollow noreferrer&quot;&gt;Law of Conservation of Mass&lt;/a&gt; applies, and as such, if the ball is truly sealed, it cannot become heavier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even trying to cheat by using physics, &lt;strong&gt;No.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mass-energy equivalence&lt;/a&gt; ($E=mc^2$) shows that an object that a bodies relativistic mass is greater than its rest-mass, that is to say, if a body is moving, it will be heavier than if the same body were stationary. Adding energy to an object adds mass, but, without approaching the speed of light (difficult, to say the least), you would struggle to measure the change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But... The ball would have to be accelerated by an external energy source for this to work. Any mechanism inside the ball used to make it roll or spin (e.g. a motor) would require its own energy source, and, by the same equation shown above, a &lt;a href=&quot;https://physics.stackexchange.com/questions/34421/does-the-mass-of-a-battery-change-when-charged-discharged&quot;&gt;full battery weighs more than an empty battery&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I make the ball lighter, then? &lt;strong&gt;Yes, but you won't be able to measure it&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you allow the ball to radiate heat to it's surroundings, therefore, loosing energy, you could make the ball &lt;em&gt;lighter&lt;/em&gt; (e.g. by running a heater off some batteries inside) without loosing any matter/atoms, but there is no way you could measure a change this small (see the excellent answer on the weight of batteries that I linked above for discussion on this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One other thought:&#xA;If an electromagnet were sealed inside, and the ball were near some iron filings, then the ball could &lt;em&gt;pick up&lt;/em&gt; mass from the outside world, without breaking its seal, but, that seems like cheating your rules, rather than a solution.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-05-19T18:52:59.770" LastActivityDate="2018-05-19T18:52:59.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21818" PostTypeId="1" AcceptedAnswerId="21824" CreationDate="2018-05-19T19:25:22.610" Score="0" ViewCount="170" Body="&lt;p&gt;The shaft is the most important element in my machine, so i rather prefer forging for production technology, the circumscribed circle of the shaft is about 140 mm and the length is about 1400 mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of dimensions should i specify on my drawing to help the operator to manufacture it ? The diameter of a circumscribed circle here does't help. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-05-20T08:35:03.080" Title="Production technology for hexagonal shaft ?" Tags="|product-engineering|forging|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="21819" PostTypeId="2" ParentId="21809" CreationDate="2018-05-19T21:14:02.630" Score="1" Body="&lt;p&gt;It would be indetrminant structure if we release the axial constraints. In a very crude first estimate one could compare it to a syspended bridge with flexible end supports, or a balloon with built-in geometric constraints. I guess if you don't want to use FEM then, you would want to guess a curve for the expanded shape and solve by energy methods. It would be interesting to start from trig functions for the cross section deformation an see where it goes.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-19T21:14:02.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21822" PostTypeId="1" AcceptedAnswerId="21823" CreationDate="2018-05-20T00:56:40.900" Score="5" ViewCount="386" Body="&lt;p&gt;I am working on this question on my statics assignment:&lt;br/&gt;&#xA;Approximately at point B there are two forces, 20kN and 50kN.So far, I have only calculated global equilibrium and got the following values:&#xA;&lt;br/&gt;&#xA;$D_y=(\frac{80}{7})kN$&lt;br/&gt;&#xA;$D_x=(\frac{20}{\frac{\sqrt{13}}{2}})kN$&#xA;&lt;br/&gt;&#xA;$A_y\approx89.094 kN$&#xA;&lt;br/&gt;&#xA;My question here is, when I'm drawing the shear force diagram do I sum up the 20kN force and the 50kN force into one resultant and graph it? Or, do I draw them separately. I need help.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GwI3d.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GwI3d.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15848" LastActivityDate="2018-05-20T01:35:50.767" Title="How to draw a shear force diagram when two forces are acting at the same point" Tags="|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21823" PostTypeId="2" ParentId="21822" CreationDate="2018-05-20T01:21:23.783" Score="5" Body="&lt;p&gt;To answer your question, you just need to break the inclined force into its component X and Y forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{4}&#xA;f_x &amp;amp;= \dfrac{20}{\sqrt{1^2+1.5^2}}&amp;amp;&amp;amp;=11.09\text{ kN} \\&#xA;f_y &amp;amp;= -1.5f_x&amp;amp;&amp;amp;=-16.64\text{ kN} \\&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and then add the vertical component to the 50&amp;nbsp;kN force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, your reactions (other than $D_x$) are incorrect:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\sum M_A &amp;amp;= -66.64\times2 - 6\times4\times\left(2+\dfrac{4}{2}\right) + 200 + 9D_y - 2\times2\times\left(9 + \dfrac{2}{2}\right) - 200 = 0 \\&#xA;\therefore D_y &amp;amp;= 29.92\text{ kN}\\&#xA;\sum F_y &amp;amp;= A_y + 29.92 - 66.64 - 6\times4 - 2\times2 = 0 \\&#xA;\therefore A_y &amp;amp;= 64.72\text{ kN}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then draw the diagram as normal, remembering to add the inclined force's vertical component to the vertical force at B:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9JCs7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9JCs7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Diagram obtained with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis program.&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2018-05-20T01:35:50.767" LastActivityDate="2018-05-20T01:35:50.767" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21824" PostTypeId="2" ParentId="21818" CreationDate="2018-05-20T08:35:03.080" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zt3hZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zt3hZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Hex AF (across-flats) dimensioning.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's how I'd do it (but I'm an electrical engineer so what would I know!).&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-05-20T08:35:03.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21825" PostTypeId="2" ParentId="21813" CreationDate="2018-05-20T08:50:58.650" Score="1" Body="&lt;p&gt;Your question doesn't make much sense. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For a given set of conditions - selected gear, gradient, load, wind and road resistance - there is a required power to maintain constant speed. &lt;/li&gt;&#xA;&lt;li&gt;Since power is a function of throttle position and this increases with throttle rotation then there must be only one throttle position that generates that power.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RE4SM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RE4SM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph and topic is discussed further in &lt;a href=&quot;https://physics.stackexchange.com/questions/57888/why-does-higher-acceleration-minimize-a-cars-fuel-consumption&quot;&gt;https://physics.stackexchange.com/questions/57888/why-does-higher-acceleration-minimize-a-cars-fuel-consumption&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-05-20T08:50:58.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21826" PostTypeId="1" CreationDate="2018-05-20T12:39:14.313" Score="2" ViewCount="183" Body="&lt;p&gt;I am building a robot (2 powered wheels and one ball bearing). The problem is that I can't seem to make it drive straight. I literally find it impossible, I have been trying for weeks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I am able to rely on rotations (of both motors) or the gyro readings(I also have two gyros, each near the two tyres)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way I can fuse those together, giving me a more accurate way to determine which motor I need to speed up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My motors accept a value from 0-900 (although the speed should be determined by me and not fixed). Also if an algorithm exists, I'd like some directions of what I'd need to swap if I make the motors go backwards.&lt;/p&gt;&#xA;" OwnerUserId="16146" LastEditorUserId="16146" LastEditDate="2018-05-20T13:17:46.147" LastActivityDate="2022-07-29T19:04:20.353" Title="How do I 'fuse' together two different kinds of data to get a final result?" Tags="|control-engineering|control-theory|robotics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21827" PostTypeId="1" CreationDate="2018-05-20T13:03:53.630" Score="-1" ViewCount="1381" Body="&lt;p&gt;One of the &amp;quot;rules&amp;quot; for sketching the root locus is that for k&amp;gt;0 the RL exists for every part where the number of poles and zeros is odd (&amp;quot;to the right of it&amp;quot;).&#xA;If we take the simple transfer function &lt;code&gt;P(S) = 1/(S^2)(S+1)&lt;/code&gt;&lt;br&gt;&#xA;There are even number of poles to the right of s=0 thus i thought the graph should exists there, but i see from wolfram that it does&#xA;&lt;a href=&quot;https://i.stack.imgur.com/xoAZd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xoAZd.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;what am i missing?&lt;/p&gt;&#xA;" OwnerUserId="16147" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-05-21T14:47:17.620" Title="root locus of a system with two poles in the same place" Tags="|electrical-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21828" PostTypeId="1" CreationDate="2018-05-20T13:06:53.257" Score="3" ViewCount="3335" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/A4LTL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My electrician made a cut in the wall and the beam as in the image in my absence. The cutting is done with hammer and chisel and it is around 1.5inch deep..I'm really worried that it might impact the structural strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beam appears to be larger than 9inch by 9inch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it safe to do it? If it is not, what could be done to fix this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot! &lt;/p&gt;&#xA;" OwnerUserId="16148" LastEditorUserId="16148" LastEditDate="2018-05-20T13:20:37.660" LastActivityDate="2021-10-06T12:08:00.663" Title="Is it safe to slightly cut the edge of RCC beam for passage of AC Vent and wiring?" Tags="|structural-engineering|structural-analysis|structures|building-design|" AnswerCount="2" CommentCount="6" ClosedDate="2021-10-08T15:06:29.313" ContentLicense="CC BY-SA 4.0" />
  <row Id="21829" PostTypeId="1" CreationDate="2018-05-20T13:59:04.743" Score="0" ViewCount="353" Body="&lt;p&gt;In &lt;a href=&quot;http://au.mathworks.com/help/mpc/ug/adaptive-mpc-control-of-nonlinear-chemical-reactor-using-successive-linearization.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt;, two methods have been referred for linearization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- offline linearization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- online linearization&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not understand what is the role of &quot;successive linearization&quot; in the online method. What is the difference between successive and typical linearization?&lt;/p&gt;&#xA;" OwnerUserId="16032" LastActivityDate="2024-03-20T07:02:55.320" Title="What is the advantage of successive linearization?" Tags="|control-theory|modeling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21830" PostTypeId="2" ParentId="21782" CreationDate="2018-05-20T14:43:08.620" Score="1" Body="&lt;p&gt;There are many points which directly affects the cost of wind turbine.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Before installing any wind turbine plant for 1 to 2 years of data analysis of wind density , direction and speed are done by engineer for success of plant which require money and resources.&lt;/li&gt;&#xA;&lt;li&gt;Plants needs lands for installing turbines also land cost depends on market scenario.&lt;/li&gt;&#xA;&lt;li&gt;Height plays major role in wind turbine which make's it costly to install at height.&lt;/li&gt;&#xA;&lt;li&gt;Transportation of large wind blades, large poles, motors, machineries and many more materials from production line to the installation area determine it's cost.&lt;/li&gt;&#xA;&lt;li&gt;Labours are one of the another factor which determine the cost of turbine, we need skilled labour which is directly proportional to cost.&lt;/li&gt;&#xA;&lt;li&gt;Technological aspect of wind turbine like which type of motor is used and how long turbine blades is used so that it can generate power per rotation.&lt;/li&gt;&#xA;&lt;li&gt;The major consideration of wind turbine is power storage device used to store generated power so that it can transmit when needed ,so larger storing device is needed which affect cost of wind turbine.          &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16144" LastActivityDate="2018-05-20T14:43:08.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21832" PostTypeId="2" ParentId="21829" CreationDate="2018-05-20T15:15:00.173" Score="0" Body="&lt;p&gt;In the section &amp;quot;Implement Adaptive MPC Control of CSTR Plant in Simulink (R)&amp;quot;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The model includes three parts:&lt;/p&gt;&#xA;&lt;p&gt;[...]&lt;/p&gt;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;The &amp;quot;Successive Linearizer&amp;quot; block [...] that [...] provides the linear plant model to the &amp;quot;Adaptive MPC Controller&amp;quot; block &lt;strong&gt;at each control interval&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If you look closely, they don't make a difference between &lt;em&gt;typical&lt;/em&gt; and &lt;em&gt;successive&lt;/em&gt; linearization, but call &lt;em&gt;successive&lt;/em&gt; linearization and &lt;em&gt;online estimation&lt;/em&gt; the two typical forms of online linearization. Read the second section, &amp;quot;About Adaptive Model Predictive Control&amp;quot; again, it is actually all explained there.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-05-20T15:15:00.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21833" PostTypeId="2" ParentId="21828" CreationDate="2018-05-20T21:50:44.153" Score="-2" Body="&lt;p&gt;If the beam is not at a critical position, you can move the conduit lower out of the way and patch the damaged area with a 28&quot; long steel, L 4&quot;x 4&quot; x 3/8&quot; running on the corner of the beam and saddling the exposed area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This L needs to bee attached with 1/4&quot; x 4&quot; long concrete anchors at 6&quot; O.C. on both sides of the L to the beam, total of 8 anchor bolts. The bolts should be placed at 1&quot; near to the edge of L.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-20T21:50:44.153" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21834" PostTypeId="1" CreationDate="2018-05-21T01:35:37.397" Score="1" ViewCount="723" Body="&lt;p&gt;What type of pneumatic solenoid valves would represent the valves in the schematic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, the lower valves could be represented by a 5 port 4 way 2 position valve. This is because there is only one port with bidirectional flow for each side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm not sure how to realize the upper valve system since there are 2 ports with bidirectional flow. Is there a way to realize this system with multi-port valves or does the upper system need to utilize direct acting (not pilot spool) valves?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yYyJu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yYyJu.png&quot; alt=&quot;Pressure Swing Adsorption&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16152" LastActivityDate="2024-04-06T03:07:55.917" Title="What type of pneumatic valves would represent the valves in the schematic?" Tags="|pneumatic|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21835" PostTypeId="2" ParentId="21826" CreationDate="2018-05-21T04:38:38.287" Score="1" Body="&lt;p&gt;The robot is unable to drive straight because no two motors are exact. Given the same voltage and target speed, both motors will have slight difference from one another which will accumulate over time to make the robot move in a non-straight direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of feedback control are you implementing? A simple PID control should work fine for adjusting the speed since you have rotation and speed feedback from both the motors. &#xA;A very good explanation and example is given at &lt;a href=&quot;http://www.inpharmix.com/jps/PID_Controller_For_Lego_Mindstorms_Robots.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; website. While it is for lego, but the concept is the same for the PID control. You can also see the code implementation from my code at &lt;a href=&quot;https://github.com/Atif-Anwer/QRCode_Following_LegoEV3_Robot/blob/master/LeCartv2.m&quot; rel=&quot;nofollow noreferrer&quot;&gt;GitHub&lt;/a&gt; (se function &lt;code&gt;calculateTurnDirDistandSpeed&lt;/code&gt;). &lt;/p&gt;&#xA;" OwnerUserId="8778" LastActivityDate="2018-05-21T04:38:38.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21836" PostTypeId="2" ParentId="15511" CreationDate="2018-05-21T05:51:15.020" Score="1" Body="&lt;p&gt;Try the experiment in a darkened room and set it up so you can see the actual contact point in profile. If the value of the sense resistor is too low, it will pass too much current and you'll create a spark upon rebound which means the setup is conducting for a certain time AFTER the rod and plate have bounced out of contact. This will make it appear like they are in contact for longer than they actually are. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used this same arrangement to trigger an oscilloscope upon impact but I used a much lower source voltage (1.5 volts instead of 9) and I recommend a sense resistor bigger than 1K ohm. Both of these things will minimize arcing. Turn up the vertical gain on the y-axis of the scope to catch the smaller signal, and adjust the trigger threshold accordingly. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-05-21T05:51:15.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21838" PostTypeId="2" ParentId="19472" CreationDate="2018-05-21T14:11:36.180" Score="0" Body="&lt;p&gt;The simplest explanation I can give is that a forward biased diode will start to transmit a high level of current with only a small increase in voltage, once a certain threshold is reached. This means that if you increase the rate at which electrons are generated with more sunlight, then instead of the voltage increasing in line with this, the current will go up instead. There are numerous resources explaining the mechanisms behind this elsewhere on Electronics SE, and the wider internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you would like a higher voltage for a particular project, you can simply connect multiple solar cells in series.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-05-21T14:11:36.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21840" PostTypeId="2" ParentId="21827" CreationDate="2018-05-21T14:47:17.620" Score="1" Body="&lt;p&gt;Just read through my old notes, you are not missing anything! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I've always learned the rule is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;'The loci are on the real axis to the &lt;strong&gt;left&lt;/strong&gt; of an odd number of poles and zeros'&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as a proof, I have looked up my Control theory book &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Feedback Control of Dynamic Systems, 7th edition&quot; by Gene F. Franklin, J. David Powell and Abbas Emami-Naeini. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you happen to have that book laying around somewhere, see page 263, expressed after &quot;RULE 2.&quot;&lt;/p&gt;&#xA;" OwnerUserId="16158" LastActivityDate="2018-05-21T14:47:17.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21841" PostTypeId="1" CreationDate="2018-05-21T16:10:59.420" Score="1" ViewCount="223" Body="&lt;p&gt;When I generate Mesh elements in Ansys Fluent, they aren't generated as expected with the edge sizing control.&#xA;&lt;br&gt;I've set the Edge Sizing control of this edge with these settings: (Number of Divisions: 20 and Biased):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M5r7V.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M5r7V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these settings it's expected to see on the edge: &lt;em&gt;20 mesh elements&lt;/em&gt; and with their &lt;em&gt;size biased&lt;/em&gt;.&#xA;But it's not the case: Their sizes are &lt;strong&gt;not biased&lt;/strong&gt; and they are &lt;strong&gt;not 20 mesh elements&lt;/strong&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BBNb3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BBNb3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;These are the Mesh general settings:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/n0MJn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n0MJn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16161" LastEditorUserId="16161" LastEditDate="2018-05-21T16:16:40.983" LastActivityDate="2018-05-21T16:16:40.983" Title="Edge sizing control is applying number of divisions and biasing unexpectedly" Tags="|fluid-mechanics|ansys|ansys-workbench|fluent|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21842" PostTypeId="1" AcceptedAnswerId="21843" CreationDate="2018-05-21T16:27:36.507" Score="3" ViewCount="780" Body="&lt;p&gt;On a datasheet, sensitivity temperature drift (TCs) is specified in %/K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read this as &quot;the percent change per kelvin&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this is simple as this sounds, or is there a reason for using kelvin over celsius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e if TCs=0.01 and the temperature changes from 20C to 30C, or from 293K to 303K, then the total percentage change is 0.1%?&lt;/p&gt;&#xA;" OwnerUserId="16159" LastEditorUserId="16159" LastEditDate="2018-05-23T09:56:01.293" LastActivityDate="2019-10-31T02:00:44.747" Title="What does &quot;+- %/K&quot; mean?" Tags="|electrical-engineering|temperature|nomenclature|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21843" PostTypeId="2" ParentId="21842" CreationDate="2018-05-21T17:45:06.370" Score="2" Body="&lt;p&gt;This is as simple as it sounds - if the reading was 5mA current when the temperature was 20C, and it changed to 30C, with TCs = 0.01, then the reading could be off by as much as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$(0.01 \frac{\%}{K})*(30^\circ C-20^\circ C)*\frac{0.01}{\%}*5mA = 0.005mA$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While ISO convention dictates the use of kelvin, inverse kelvin are a derived unit and there is no standard when choosing between inverse degrees Celsius and inverse kelvin - except to avoid having to place a unicode degree symbol (kelvin &lt;strong&gt;are&lt;/strong&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Kelvin&quot; rel=&quot;nofollow noreferrer&quot;&gt;standalone without the symbol&lt;/a&gt;).  Another reason for the notation discrepancy is that some measurement devices, such as thermocouples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vcjNb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vcjNb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;have a semi-non-linear response with respect to temperature.  As such, these typically will denote their average change over a large range as per degree Celsius, to hint that it is more of an &quot;average&quot; amount and that more precise thermal drift values can be obtained by reviewing the exact temperature range that is being measured.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-05-23T04:07:04.720" LastActivityDate="2018-05-23T04:07:04.720" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21844" PostTypeId="2" ParentId="21842" CreationDate="2018-05-21T17:46:36.087" Score="3" Body="&lt;p&gt;Yes, this is the percentage change per kelvin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most engineers work in kelvin as some formulae need the absolute temperature such as when calculating the total internal energy, but we sometimes don’t bother such as when dealing with temperature difference, but it was good practice when we were taught it so we continue to use K...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="3505" LastEditDate="2018-05-22T02:02:43.563" LastActivityDate="2018-05-22T02:02:43.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21845" PostTypeId="1" CreationDate="2018-05-21T21:25:15.763" Score="1" ViewCount="131" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tv8pN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tv8pN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An object with the weight of $80N$ is balanced with the force of $F$  as seen on the diagram. What is the magnitude of $F$? (Ignore the weight of pulley and friction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum F_y = 0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_y \sin 37 - 80 = 0 \implies F_y \sin 37 = 80 \implies F_y = 133.33N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum F_x = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F_x \cos 37 = 0 \implies F_x = 0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$F = 133.33N$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right? &lt;/p&gt;&#xA;" OwnerUserId="12679" LastActivityDate="2018-05-24T20:18:18.927" Title="Finding the magnitude of F" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21846" PostTypeId="2" ParentId="21845" CreationDate="2018-05-22T02:53:00.167" Score="4" Body="&lt;p&gt;Your calculations are incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two sides of the cable around the pulley that helps to support the 80 N load. See the diagrams below. So the vertical force from the cable around the pulley is $F+F\sin(37)$, not $F_y\sin(37)$. This gives the sum of vertical forces as $F+F\sin(37)-80=0$ from which F=49.9 N. (Note that $F_y\sin(37)$ does not make sense.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal force from the cable around the pulley is $F\cos(37)$, not $F_x\cos(37)$. Furthermore, there is a force in the horizontal &quot;rope&quot; of magnitude T, so the sum of forces in the horizontal forces is $F\cos(37)-T=0$ from which T=39.9N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a free body diagram showing all of the forces acting on the pulley. The statement to neglect the friction in the pulley indicates that the tension in the cable is the same value, F, on each side of the pulley.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qhI8m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qhI8m.png&quot; alt=&quot;free body diagram&quot;&gt;&lt;/a&gt; or like this &lt;a href=&quot;https://i.stack.imgur.com/h8jQR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h8jQR.png&quot; alt=&quot;FBD 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12984" LastEditorUserId="12984" LastEditDate="2018-05-24T20:18:18.927" LastActivityDate="2018-05-24T20:18:18.927" CommentCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="21848" PostTypeId="1" CreationDate="2018-05-22T08:37:17.967" Score="1" ViewCount="575" Body="&lt;p&gt;I'm not sure I'm in the right place for this, if not, Apologies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my education material under Construction Steel, in its application it lists Masking Details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under Tool Steel it list an application as Maskin Parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But nowhere I look or ask have I been able to find an answer as to the difference in meaning between Part and Detail has here. &#xA;And from my other education material the word 'detail' is used for (as far as I can confirm) anything on a schematic and I've never encountered the word 'part' in such connotation. On the other hand in the workshop I hear the word 'detail' used to refer to anything made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, and the language I'm taking the education in is Swedish, so I've translated the terms.&lt;/p&gt;&#xA;" OwnerUserId="16167" LastEditorUserId="2559" LastEditDate="2018-05-22T12:45:08.510" LastActivityDate="2018-05-22T12:45:08.510" Title="Difference between a Part and a Detail" Tags="|terminology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21849" PostTypeId="1" AcceptedAnswerId="21855" CreationDate="2018-05-22T08:51:37.170" Score="1" ViewCount="1419" Body="&lt;p&gt;I'm working on the statics problem below:&#xA;&lt;br/&gt;&lt;a href=&quot;https://i.stack.imgur.com/FYh6q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FYh6q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I'm tasked to draw the shear and bending moment diagram for the frame. I'm trying to work out the reactions and I'm taking the moment about pin E.Here's my working so far:&lt;br/&gt;&#xA;$$\Sigma M_E=0$$&#xA;$$(10)(5)-100-(2*10)(\frac{1}{2}*10+5)+(10)(15)-100-(\frac{30}{\sqrt{2}})+(0.5*5*4)(\frac{1}{3}*5)+A_y(20)=0$$&#xA;My question here is, for the triangular UDL, is the distance one-third or two thirds? My lecturer did a similar problem and he said that the distance is two thirds but I don't quite get why it is two thirds. Someone help.&lt;/p&gt;&#xA;" OwnerUserId="15848" LastEditorUserId="15848" LastEditDate="2018-05-22T10:57:06.873" LastActivityDate="2018-05-22T21:29:15.727" Title="What is the distance at which the triangular UDL is acting on in this problem?" Tags="|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21850" PostTypeId="1" CreationDate="2018-05-22T11:28:27.510" Score="4" ViewCount="283" Body="&lt;p&gt;Here is a schematic cross-section of a water jet cutter nozzle design:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MWred.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MWred.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is shown, the thin water jet grabs abrasive particles in the mixing chamber and the mix flows through the nozzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what makes the water flow through the nozzle despite the loss of kinetic energy due to the abrasive material and the friction inside the nozzle instead of filling up the mixing chamber and running up in the ambient pressured abrasive feeding tube?&lt;/p&gt;&#xA;" OwnerUserId="15139" LastActivityDate="2018-05-22T19:00:27.007" Title="How does water flow in the water jet cutter nozzle?" Tags="|cutting|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21851" PostTypeId="2" ParentId="21850" CreationDate="2018-05-22T11:56:22.743" Score="2" Body="&lt;p&gt;It's called &lt;i&gt;Bernoulli's principle&lt;/i&gt;.  It works because the fluid has momentum, so doesn't cause equal outward pressure at all points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This principle can be used to convert a high-speed low-volume flow into a high-volume low-speed flow.  Since the output flow has more volume than the input flow, additional material is sucked in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter of section 4 is a little larger than that of section 2 for this process to work.&lt;/p&gt;&#xA;" OwnerUserId="624" LastEditorUserId="624" LastEditDate="2018-05-22T15:17:56.347" LastActivityDate="2018-05-22T15:17:56.347" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21852" PostTypeId="2" ParentId="21848" CreationDate="2018-05-22T12:41:36.590" Score="1" Body="&lt;p&gt;I've heard both terms used and in my experience have different meanings depending on the context and industry. It's nothing to get hung up on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, sometimes a &quot;detail&quot; can refer to an entire part and it's typically an item that goes into an assembly. Some assembly drawings will actually refer to the parts that make the assembly as &quot;details&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it's pretty common in a shop to refer to parts as details as they typically are &quot;details&quot;. &lt;strong&gt;Therefore &quot;detail&quot; and &quot;part&quot; are synonymous.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Also, a &quot;detail&quot; can be literal&lt;/strong&gt;. Commonly used in architecture/construction which would refer to how a particular area or elements come together with each aspect being defined with the &quot;detail&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;It is used as both a noun and verb&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If drafting, there are &quot;detail&quot; views which are typically used to &quot;zoom&quot; on more complex areas to show features that need to be described.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2018-05-22T12:41:36.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21853" PostTypeId="2" ParentId="4009" CreationDate="2018-05-22T14:18:31.717" Score="1" Body="&lt;p&gt;The present technology is all about IOT i. e with millions of devices connected over internet. Zigbee and BLE are two types of wireless communication which is used for interacting between two or more things or devices. Both have their own protocol and both have their merits and demerits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BLE(Bluetooth low energy)  is a wireless personal area network with low power and low cost. It has high dataa rate. It is specially used in Healthcare application like glucose monitoring,blood pressure monitoring of patients which is connected by IOT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ZIGBEE is a personal are network with a range of about 100m this is not conventional method even though it uses low Power and has low cost because it does not support  all types of operating systems (eg. Microsoft, Linux, Android, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BLE is preferred over Zigbee, because it supports all operating systems in general, and mainly because it is supported by mobile. This is a conventional method as we can update the data using mobile sitting anywhere without the physical presence. &lt;/p&gt;&#xA;" OwnerUserId="16169" LastEditorUserId="9" LastEditDate="2018-05-22T16:44:03.267" LastActivityDate="2018-05-22T16:44:03.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21855" PostTypeId="2" ParentId="21849" CreationDate="2018-05-22T18:34:04.260" Score="1" Body="&lt;p&gt;It is 2/3 and also if we assume the moment direction positive clockwise, to be consistent with the rest of your positive signs, then that vertical loading's moment is negative, the correct sign and factor is (-0.5*5*4) (2/3*5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the distance is 2/3 is because C.G. of a the vertical 5m load triangle is at 1/3 from its base, which here is 2/3s from E. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the loading triangle was reversed increasing up your assumption would be correct.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-05-22T18:43:26.790" LastActivityDate="2018-05-22T18:43:26.790" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21856" PostTypeId="2" ParentId="21850" CreationDate="2018-05-22T19:00:27.007" Score="0" Body="&lt;p&gt;As has been mentioned in other answers it is Bernoulli's principle. Part of the pressure in the water feed line turns into momentum and eject the jet. This jet now has a high velocity and a large momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When this jet in part 3 picks up the abrasive sand and accelerates it to mix and carry it, there is some loss of momentum of water in nuzzle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have calibrated the feed of the abrasive material such that its inertia and friction is kept well below what can cause clogging of the nuzzle or backing up.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-22T19:00:27.007" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21857" PostTypeId="2" ParentId="21849" CreationDate="2018-05-22T21:29:15.727" Score="3" Body="&lt;p&gt;As for the title of your question, any statics textbook should show you the centroid of a right triangle. &lt;a href=&quot;https://en.m.wikipedia.org/wiki/List_of_centroids&quot; rel=&quot;nofollow noreferrer&quot;&gt;This wikipedia article &lt;/a&gt;gives the centroids of common shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When replacing a distributed load with an equivalent point load, that point load will act through the centroid of the distributed load. In this case, the equivalent point load for the triangular distributed load will act (1/3)*5 up from Point A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When summing the moments about a point it is important to indicate your sign convention (clockwise positive or counterclockwise positive). Either convention is acceptable so long as you are consistent throughout. Looking at the work you showed, there appear to be some inconsistencies. For example P1 and M1 (see sketch below for notation) should both produce clockwise rotation about Point E, however your work shows them producing moments with opposite signs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won’t write out the whole equation for summing moments about Point E because I don’t want to do the homework for you. You’re on the right track though!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1osH6l.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1osH6l.jpg&quot; alt=&quot;Free Body Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2018-05-22T21:29:15.727" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21858" PostTypeId="1" CreationDate="2018-05-22T23:34:30.237" Score="2" ViewCount="115" Body="&lt;p&gt;Pulsejet expels pulse of hot gas (unlike turbojet engine) which create some thrust(Not good enough for ). These gases are hot, fast and pulsed. Is it possible to convert kinetic energy of these gases to mechanical energy using Turbines? (Assuming hot gases are not too hot for turbine blades to melt down). My concern is about Pulsed gases. What kind of problems/challenges it may cause in an engine perspective?&lt;/p&gt;&#xA;" OwnerUserId="16175" LastEditorUserId="16175" LastEditDate="2018-11-05T14:56:42.087" LastActivityDate="2018-11-05T14:56:42.087" Title="Is it possible to extract mechanical energy using Turbines behind PulseJet engine?" Tags="|mechanical-engineering|turbines|aerodynamics|propulsion|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21859" PostTypeId="1" CreationDate="2018-05-23T01:16:40.823" Score="1" ViewCount="27" Body="&lt;p&gt;I had this thought with me for quite some time, that you can't use the usual formula for estimating carbon content on a Nital etched hypoeutectoid sample (DarkArea*0.77) if the steel obviously has more than around 0.6% carbon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure I read that somewhere, but I can't find that, and I might have been operating in a false premise for some time now.&lt;/p&gt;&#xA;" OwnerUserId="11240" LastEditorUserId="11240" LastEditDate="2018-05-23T01:23:50.073" LastActivityDate="2018-05-23T01:23:50.073" Title="Why shouldn't dark area analysis be used to estimate carbon content on high carbon hypoeutectoid steels?" Tags="|mechanical-engineering|materials|steel|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21862" PostTypeId="1" CreationDate="2018-05-23T08:51:52.760" Score="1" ViewCount="108" Body="&lt;p&gt;Honeycomb sieves are starting to see &lt;a href=&quot;https://www.aqseptence.com/app/en/products/wts-noggerath-centre-flo-cf/&quot; rel=&quot;nofollow noreferrer&quot;&gt;applications in wastewater screening&lt;/a&gt;. What is the flow resistance of a honeycomb screen?&lt;/p&gt;&#xA;&lt;p&gt;I would expect something similar to the Kirschmer formula (used for bar screens):&lt;br /&gt;&#xA;Headloss = constant, dependent on bar shapes X function of square of velocity&lt;/p&gt;&#xA;&lt;p&gt;(Actually, the Kirschmer formula would be corrected for angle and for loading of the screen)&lt;/p&gt;&#xA;&lt;p&gt;where velocity means in the screen, so the free survece is taken into account. While a honeycomb can have a far larger free surface compared to a screen with holes, I would expect the first factor to be different to, as the thinner bars in the honeycomb allow for a different flow regime immediatly downstream.&lt;/p&gt;&#xA;&lt;p&gt;What is the headloss for flow through a honeycomb screen?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-05-23T08:51:52.760" Title="What's the flow resistance of a honeycomb sieve?" Tags="|fluid-mechanics|civil-engineering|fluid-filtration|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21863" PostTypeId="2" ParentId="21858" CreationDate="2018-05-23T11:05:44.687" Score="1" Body="&lt;p&gt;Two challenges (not necessarily &quot;problems&quot; so long as the device is designed to deal with them) are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Vibration. If the pulse frequency matches a resonant frequency of the turbine, the vibrations may reach a high enough amplitude to break something, or significantly reduce the life of the components. The waveform of the pulse will probably contain a large number of harmonics that are multiples of its fundamental frequency, and a typical turbine also has a large number of vibration modes - not just the modes of a single blade on its own, but coupled modes where all the blades and the turbine disk vibrate as a unit. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Putting a turbine in the exhaust jet stream will change the aerodynamics of the jet itself. If the speed of the jet is subsonic, some energy will be &quot;reflected&quot; from the turbine and travel back upstream through the jet, in effect partially blocking the flow of the jet compared with the original pulse-jet design.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Even apparently small issues here can be catastrophic. True story of an anecdote when testing a conventional turbojet engine in a different test bed from the standard design: the gap between the back of the engine and the entry into the test bed exhaust silencer (muffler) happened to be just the right length to set up a standing wave in the gas flow, at the same frequency as a vibration mode of the last turbine stage in the engine. That mode would never have been excited in normal engine operation and therefore nobody had paid much attention to it - until 5 minutes into the first test run, when there was a very  loud bang as the turbine disk and blades disintegrated!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-05-23T11:05:44.687" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21866" PostTypeId="1" CreationDate="2018-05-23T14:24:46.530" Score="4" ViewCount="1982" Body="&lt;p&gt;I aim on building a silent, completely passively cooled PC, with no moving parts at all (I am aware that it is cheaper to go for an inaudible computer instead, using low speed fans. This is not the point.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came up with the idea to install not one but two CPU coolers, of which in theory one alone should keep a high performance CPU cool only using natural convection, but not in practice. For this I thought of using a triangular shaped piece of copper, with thermal compound between each surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will this improve heat dissipation compared to a single heat sink or is the thermal resistance of the copper block a problem?&#xA;I sketched the idea below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eLgXz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eLgXz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16181" LastActivityDate="2019-02-15T13:20:17.123" Title="Use two passive CPU coolers on one single CPU?" Tags="|cooling|computer-engineering|thermal-conduction|computer-hardware|computer|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="21867" PostTypeId="5" CreationDate="2018-05-23T18:05:00.077" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-05-23T18:05:00.077" LastActivityDate="2018-05-23T18:05:00.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21868" PostTypeId="4" CreationDate="2018-05-23T18:05:00.077" Score="0" Body="Structural analysis is the determination of the effects of loads on physical structures and their components. Structures subject to this type of analysis include all that must withstand loads, such as buildings, bridges, vehicles, furniture, soil strata, prostheses and biological tissue." OwnerUserId="7006" LastEditorUserId="1832" LastEditDate="2018-05-23T19:24:29.873" LastActivityDate="2018-05-23T19:24:29.873" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21869" PostTypeId="1" AcceptedAnswerId="21871" CreationDate="2018-05-23T18:22:32.907" Score="2" ViewCount="84" Body="&lt;p&gt;I have the following system (at P and Q there is a ball joint; the length of L beam is l; the PB and PA forces are applied at l/2 and are known; Fz is not zero and is known):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pykcv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pykcv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system has 0 degrees of freedom because each ball joint removes three degrees of freedom (6-2*3=0). &#xA;I replaced each ball joint with three unknown forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n72HL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n72HL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the second cardinal equation along x gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-F_z \, l = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I solve the problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much for your time.&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="7006" LastEditDate="2018-05-24T06:50:20.000" LastActivityDate="2018-05-24T06:50:20.000" Title="L beam - Strange result" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="21870" PostTypeId="1" CreationDate="2018-05-23T18:38:36.670" Score="3" ViewCount="253" Body="&lt;p&gt;Suppose that I have a state-space system &lt;span class=&quot;math-container&quot;&gt;$A, B, C, D$&lt;/span&gt; and a gains matrix &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; computed via LQR. I would like to track a reference signal &lt;span class=&quot;math-container&quot;&gt;$x_{cmd}$&lt;/span&gt;. I do so by using the control law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ u = -K(x - x_{cmd})$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is the system state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I develop a feed-forward term as a function of &lt;span class=&quot;math-container&quot;&gt;$x_{cmd}$&lt;/span&gt; to encourage the system to follow the reference more closely, particularly in the case of time-varying &lt;span class=&quot;math-container&quot;&gt;$t_{cmd}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="508" LastEditorUserId="18238" LastEditDate="2020-05-05T18:33:25.327" LastActivityDate="2020-05-05T18:33:25.327" Title="How to improve reference tracking using state feedback?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21871" PostTypeId="2" ParentId="21869" CreationDate="2018-05-23T19:29:51.163" Score="1" Body="&lt;p&gt;The system is still statically indeterminate, but not due to lack of DOF.  You have a singularity.  Having correct DOFs only assures you have a chance of being statically determinate, but it does not guarantee this condition.  In this case, the singularity is because the beam is allowed to rotate about the x axis unless $F_z$ is an unknown reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, in this case, $F_zl = 0$ means $F_z = 0$.  Here are your other equations, in no particular order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{4}&#xA;\sum&amp;amp; F_x &amp;amp;&amp;amp;= A + PB_x + D + PA_x &amp;amp;&amp;amp;= 0 \\&#xA;\sum&amp;amp; F_y &amp;amp;&amp;amp;= B + PB_y + E + PA_y &amp;amp;&amp;amp;= 0 \\&#xA;\sum&amp;amp; F_z &amp;amp;&amp;amp;= C + G + (F_z = 0) &amp;amp;&amp;amp;= 0 \\&#xA;\sum&amp;amp; M_{y@Q} &amp;amp;&amp;amp;= Cl &amp;amp;&amp;amp;= 0 \\&#xA;\sum&amp;amp; M_{x@Q} &amp;amp;&amp;amp;= F_zl &amp;amp;&amp;amp;= 0 \\&#xA;\sum&amp;amp; M_{z@Q} &amp;amp;&amp;amp;= -Bl + PB_yl/2 - PA_xl/2 &amp;amp;&amp;amp;= 0&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can immediately see $C = F_z = 0$. Because of $\sum F_z$, $G = 0$.  We now have four unknowns and three equations. We can solve for $B$ immediately using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum M_{z@Q} = -Bl + PB_yl/2 - PA_xl/2 = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging, we can use this result for $B$ to solve for $E$ using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum F_y = B + PB_y + E + PA_y = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the final equation is singular:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum F_x = A + PB_x + D + PA_x = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This cannot be resolved using statics, but require static indeterminate methods.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-05-23T20:43:38.883" LastActivityDate="2018-05-23T20:43:38.883" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21872" PostTypeId="2" ParentId="21094" CreationDate="2018-05-23T19:52:51.630" Score="1" Body="&lt;p&gt;Since you're worried about an open-loop system, in theory all you need to do is perform standard stability analysis on the open-loop system alone. The source of the input signal to the open loop system is irrelevant if you can prove that it is stable for a known/expected set of input signals. For example, for a linear system you would need to demonstrate its stability for an expected range of input magnitudes and frequencies. A nonlinear system would be more complicated, but methods such as Lyapunov stability analysis and passivity can be applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have a counter-question for you: what does it mean to have &quot;stable&quot; fuel consumption? Stability usually implies that given a certain initial state, the state does not move beyond a certain distance from the equilibrium point for all time (Lyapunov stability). Since the state in a fuel system would be the volume remaining in the tank, and we know that the tank has a maximum volume ($V_{max}$) and a minimum volume (0, no fuel) then for any initial state,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 \leq V(0) \leq V_{max} \implies 0 \leq V(t) \leq V_{max} \text{.}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note this conclusion is independent of the input. If we further consider that, in the absence of refueling, $\frac{dV}{dt} \leq 0$, we can conclude that $0 \leq V(t) \leq V(0)$. Either result leads to the conclusion that the fuel consumption is stable (in the Lyapunov sense) regardless of input.&lt;/p&gt;&#xA;" OwnerUserId="6695" LastEditorUserId="6695" LastEditDate="2018-05-23T20:11:25.950" LastActivityDate="2018-05-23T20:11:25.950" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21873" PostTypeId="1" CreationDate="2018-05-23T21:23:11.557" Score="0" ViewCount="49" Body="&lt;p&gt;Assume the water bottle is the most sturdily built standard water bottle one could acquire in an american supermarket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first question I've asked, so I'm sorry if I'm not in the right place or asking it the right way. If I may I'd like to throw in these follow up questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a good way to approximate how good an object would be at serving as a long term floatation device, like a pontoon or bouy? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to calculate what the optimum size or shape of a 'water bottle like structure/material' (pocket of air encased in plastic) would be for displacing weight lifted out of the water by the buoyancy?&lt;/p&gt;&#xA;" OwnerUserId="16186" LastActivityDate="2018-05-24T06:21:29.367" Title="How deep would a water bottle have to be submerged in the sea to burst?" Tags="|compressed-air|waterproofing|" AnswerCount="2" CommentCount="4" ClosedDate="2018-05-30T18:44:46.270" ContentLicense="CC BY-SA 4.0" />
  <row Id="21874" PostTypeId="2" ParentId="21873" CreationDate="2018-05-23T21:55:22.883" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How deep would a water bottle have to be submerged in the sea to burst?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It won't burst. It will be crushed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure increases by 1 bar approximately every 10 m in salt water.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a good way to approximate how good an object would be at serving as a long term floatation device, like a pontoon or bouy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. 1 L of displaced water gives 1 kg (10 N) lift.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a way to calculate what the optimum size or shape of a 'water bottle like structure/material' (pocket of air encased in plastic) would be for displacing weight lifted out of the water by the buoyancy?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. Use finite element analysis on a model of the shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strongest structure for a given weight of material and will probably be a sphere or egg-shape. Cylindrical shapes with vertical axis may be easier to manufacture.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-05-23T21:55:22.883" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21875" PostTypeId="1" CreationDate="2018-05-23T22:49:23.350" Score="1" ViewCount="23" Body="&lt;p&gt;For scrubbing with activated carbon, air is generally brought through a dry adsorption media, where the contaminant is adsorbed onto the media.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what mathematical model is used for this process.  I presume the rate of adsorption is proportional to the concentration of the contaminant, and can utilize a &lt;a href=&quot;https://en.wikipedia.org/wiki/Plug_flow_reactor_model&quot; rel=&quot;nofollow noreferrer&quot;&gt;plugged flow reactor model&lt;/a&gt; - but the carbon has a finite amount of capacity it can adsorb.  So, how does this change the equation?  After t = 5 days if the first 0.1 mm is completely saturated, how would it take for a 1m long bed to reach breakthrough (i.e. detectable concentrations)?  There has been rules of thumb used in industry (such as if the first 2/3 are saturation, change out the media), but I would like to see a more thorough mathematical derivation.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-23T22:49:23.350" Title="Adsorption - what models are used for dry scrubbing?" Tags="|mechanical-engineering|chemical-engineering|environmental-engineering|mass-transfer|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21876" PostTypeId="2" ParentId="6624" CreationDate="2018-05-23T23:10:34.387" Score="0" Body="&lt;p&gt;The main reason is likely due to Wolfram Alpha applying the inverse Fourier transform as a second Fourier transform.  In fact, doing so &quot;flips time&quot; - as can &lt;a href=&quot;https://en.wikipedia.org/wiki/Fourier_transform#Invertibility_and_periodicity&quot; rel=&quot;nofollow noreferrer&quot;&gt;be shown mathematically&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Defining the '''flip-time operator''' $\mathcal{P}$ that inverts time, $\mathcal{P}[f(t)] ↦ f(−t)$&#xA;$$\begin{align}&#xA;  \mathcal{F}^0 &amp;amp;= \mathrm{Id}, \quad&#xA;  \mathcal{F}^1  = \mathcal{F}, \\&#xA;  \mathcal{F}^2 &amp;amp;= \mathcal{P}, \quad&#xA;  \mathcal{F}^4  = \mathrm{Id}, \\&#xA;  \mathcal{F}^3 &amp;amp;= \mathcal{F}^{-1} = \mathcal{P} \circ \mathcal{F} = \mathcal{F} \circ \mathcal{P}&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the fourier transform 3 times to the system will get you the version in normal time.  Since waves are time consistent, it normally does not matter.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-23T23:10:34.387" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21877" PostTypeId="2" ParentId="21873" CreationDate="2018-05-24T06:21:29.367" Score="0" Body="&lt;p&gt;As Transistor already wrote, according to &lt;a href=&quot;https://en.wikipedia.org/wiki/Archimedes%27_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Archimedes' principle&lt;/a&gt;, the lift is identical to the weight of the superseded material, reduced by the weight of the superseding material. Since &lt;a href=&quot;https://water.usgs.gov/edu/density.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;water is nearly incompressible&lt;/a&gt;, you can use 1 kg per liter (or 1000 kg/m³) lift for most cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main failure mode of a submerged standard supermarket plastic bottle is &lt;a href=&quot;https://en.wikipedia.org/wiki/Buckling&quot; rel=&quot;nofollow noreferrer&quot;&gt;plate buckling&lt;/a&gt;. The corresponding &lt;a href=&quot;https://de.wikipedia.org/wiki/Beulen&quot; rel=&quot;nofollow noreferrer&quot;&gt;German wikipedia article&lt;/a&gt; gives a pressure limit of &#xA;$$&#xA;p_{\mathrm{buckling}} = 0.65 \cdot E \cdot \frac{r}{l} \cdot \left(\frac{t}{r}\right)^{\frac 52}&#xA;$$&#xA;as maximum external pressure $p_{\mathrm{buckling}}$ for buckling of a cylinder, with $E$ being &lt;a href=&quot;https://en.wikipedia.org/wiki/Young%27s_modulus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Young's modulus&lt;/a&gt;, $r$ the mean radius and $l$ the length of the cylinder and $t$ the wall thickness. Please note that in this equation there are no safety factors included, e.g. against a sudden local change of external pressure due to waves or contact with other solids (ground, ships, whatever) and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more complex shapes, as Transistor already wrote, finite elements are a suitable way to estimate the buckling resistance.&lt;/p&gt;&#xA;" OwnerUserId="4122" LastActivityDate="2018-05-24T06:21:29.367" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21878" PostTypeId="1" CreationDate="2018-05-24T06:40:17.680" Score="0" ViewCount="34" Body="&lt;p&gt;I am wondering if there is a procedure to verify that the constraints of a system of rigid bodies are independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here there is an example of a system with dependent constraints: &lt;a href=&quot;https://engineering.stackexchange.com/questions/21869/l-beam-strange-result&quot;&gt;L beam - Strange result&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your willingness.&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="7006" LastEditDate="2018-05-24T07:17:42.357" LastActivityDate="2018-05-24T07:17:42.357" Title="Independent constraints - Procedure to verify it" Tags="|mechanical-engineering|structural-engineering|civil-engineering|structural-analysis|structures|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21879" PostTypeId="1" CreationDate="2018-05-24T10:50:06.307" Score="0" ViewCount="260" Body="&lt;p&gt;From my understanding,when there is  no change in volume due to pressure applied on a fluid,the fluid can be called as incompressible.As there can be still be change in volume due to other properties,can I say that flow may be compressible?&lt;/p&gt;&#xA;" OwnerUserId="15268" LastActivityDate="2024-01-18T17:01:55.250" Title="If a fluid is incompressible,is it correct to say that the flow may or may not be incompressible?" Tags="|mechanical-engineering|fluid-mechanics|compressible-flow|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21880" PostTypeId="1" AcceptedAnswerId="21890" CreationDate="2018-05-24T10:51:32.800" Score="1" ViewCount="1044" Body="&lt;p&gt;Hardening leads to a material being more brittle.However I can't understand what is strain hardening or strain hardening exponent.Please explain why hardening leads to brittleness also in this answer itself.I would be happy to get a detailed answer of this.&lt;/p&gt;&#xA;" OwnerUserId="12558" LastEditorUserId="12558" LastEditDate="2018-05-29T14:15:29.433" LastActivityDate="2018-05-29T14:15:29.433" Title="What is strain hardening exponent?" Tags="|mechanical-engineering|materials|metallurgy|metals|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21881" PostTypeId="2" ParentId="21879" CreationDate="2018-05-24T11:11:17.377" Score="0" Body="&lt;p&gt;No, the fluid is still incompressible, but the container it is in can change volume...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-05-24T11:11:17.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21882" PostTypeId="1" AcceptedAnswerId="22025" CreationDate="2018-05-24T11:21:02.247" Score="2" ViewCount="192" Body="&lt;p&gt;Please bear with me - I'm a lapsed mathematician and I'm self-studying these concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A question asks the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Water flows in a pipe of diameter 5 m at a velocity of 10 m/s. It then&#xA;  flows down into a smaller pipe of diameter 2 m. The height between the&#xA;  centre of pipe sections is 5 m. The density is assumed to be uniform&#xA;  over the cross sections. The gauge pressure at Boundary 1 is 120 kPa.&#xA;  Calculate the velocity at the smaller pipe section.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no reason I can see to assume mass flow continuity doesn't apply, and using $v_1A_1 = v_2A_2$ one obtains $v_2 = 62.5$ m/s. However, using the Bernoulli equation while assuming atmospheric pressure at the smaller section, one gets&#xA;$$&#xA;\frac{p_1}{\rho} + \frac{1}{2}v_1^2 + + gz_1 = \frac{p_2}{\rho}+\frac{1}{2}v_2^2 + gz_2&#xA;$$&#xA;$$&#xA;120+ 50 + 5g = 0 + \frac{1}{2}v_2^2 + 0&#xA;$$&#xA;$$&#xA;v_2 = 20.93,&#xA;$$&#xA;and in fact this is what the textbook answer gives. I am quite confused as to why mass continuity applies in other situations, even with changes in pressure, but doesn't seem to apply here.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;is this textbook question then ill-posed?&lt;/strong&gt; I feel as though by providing too much information about the pipe section without checking the calculations, the question is bound to create a contradiction. The 5m/2m diameters don't actually make it into the final answer.&lt;/p&gt;&#xA;" OwnerUserId="16190" LastEditorUserId="16190" LastEditDate="2018-05-24T11:26:29.130" LastActivityDate="2018-06-10T19:30:19.323" Title="Can the continuity equation and Bernoulli contradict each other?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21884" PostTypeId="2" ParentId="21879" CreationDate="2018-05-24T11:54:53.980" Score="0" Body="&lt;p&gt;When you use the incompressible fluid and apply pressure to it then the fluid´s density will not Change i.e.,volume of the incompressible fluid does not changes even with the application of external pressure. May be the outer part within which it is stored might undergo volume Change.&lt;/p&gt;&#xA;" OwnerUserId="15526" LastActivityDate="2018-05-24T11:54:53.980" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21885" PostTypeId="1" CreationDate="2018-04-02T20:48:50.490" Score="4" ViewCount="84" Body="&lt;p&gt;My friend has two young Pokemon-addicted sons who carry their 60-card playing decks with them everywhere, typically in a cargo pants pocket.  Unfortunately, they have either managed to destroy every card box they've tried, or been bruised by sharp corners/edges during other activities.&lt;/p&gt;&#xA;&lt;p&gt;I thought I could design and print a better one.  I started with an inner rectangular box surrounded by a shell having curved edges and corners for safety, and with slightly bulging sides for strength.&lt;/p&gt;&#xA;&lt;p&gt;Then I researched a bit on getting more crush resistance from the shape itself (rather than adding ever more infill) and found that catenaries are very crush-resistant, as are parabolas.&lt;/p&gt;&#xA;&lt;p&gt;Then I stumbled into &lt;a href=&quot;https://en.wikipedia.org/wiki/Squircle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Squircles&lt;/a&gt;, the more general &lt;a href=&quot;https://en.wikipedia.org/wiki/Superellipse&quot; rel=&quot;nofollow noreferrer&quot;&gt;Super-Ellipses&lt;/a&gt;, and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Superellipsoid&quot; rel=&quot;nofollow noreferrer&quot;&gt;3D Super-Ellipsoid&lt;/a&gt;.  This seems to be exactly what I'm looking for!  A closed 3D surface with continuously varying curvature.&lt;/p&gt;&#xA;&lt;h3&gt;[the above edited for clarity by CGW; the next section added in the hopes it will focus answers]&lt;/h3&gt;&#xA;&lt;p&gt;Given that an ideal sphere, and even an idealized ellipsoid of revolution, won't fit nicely in a pants pocket, what general engineering guidelines should [the OP] follow in designing a card box?  Rounded corners and edges are a must, to protect fabric and flesh. Are internal ribs on near-flat surfaces a good option?&lt;/p&gt;&#xA;&lt;h3&gt;[end of added material]&lt;/h3&gt;&#xA;&lt;p&gt;The generic form of the 3D Super-Ellipsoid is:&lt;/p&gt;&#xA;&lt;p&gt;(x/a)^n + (y/b)^n + (z/c)^n = 1&lt;/p&gt;&#xA;&lt;p&gt;Where:&#xA;a, b and c are axis scale factors,&#xA;n is the exponent controlling the shape (2..8, typ. 5-6), and&#xA;x, y and z are bounded to +/-1&lt;/p&gt;&#xA;&lt;p&gt;But the above isn't directly usable to create a point cloud.&lt;/p&gt;&#xA;&lt;p&gt;The parametric form would iterate over three angles.  But the problem I'm having is with powers greater than two.&lt;/p&gt;&#xA;&lt;p&gt;Anyone have some clues to share?&lt;/p&gt;&#xA;&lt;p&gt;I promise to include a full write-up (including credits from this thread) when I post the final box on Thingiverse.  (Perhaps using OpenSCAD?)&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;BTW, after finding the Squircle I detoured to write a &lt;a href=&quot;https://bobtechnits.blogspot.com/2018/03/playing-with-equation-for-circle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;blog post&lt;/a&gt; about the characteristics of generalized equations of the form X^n + Y^n = 1&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16210" OwnerDisplayName="BobC" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-10-31T19:20:31.730" Title="Designing a Safe and Crush-proof Pokemon playing card box using a 3D Super-Elipsoid" Tags="|mechanical-engineering|stresses|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21889" PostTypeId="1" CreationDate="2018-05-24T16:29:10.230" Score="2" ViewCount="1169" Body="&lt;p&gt;True Airspeed (TAS) is the vehicle speed relative to the surrounding air. My assumption is that if I am in orbit say at 120 km altitude with an instantaneous velocity of 8 km/s my TAS would be fairly close to 8 km/s even though my Indicated Airspeed (IAS) shows zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the above holds true, the &quot;ram rise&quot; (RR) formulas I find are all based on Mach (which, in turn, is based on TAS) or TAS directly. They do not appear to take into account the static pressure or atmospheric composition, only the adiabatic index. Therefore, my TAT (OAT plus ram rise) is in the thousands of degrees Kelvin ... which makes no sense because there are practically no molecules of air to hit!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are these RR=TAS²/87 formulas assuming to be used below a specific altitude? I can't find anything for orbit or reentry. I also can't find any space shuttle telemetry to corroborate the TAS assumptions I made above. I am aware there is also a Mach-based RR formula and it has the same problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your help is sure appreciated!&lt;/p&gt;&#xA;" OwnerUserId="16202" LastEditorUserId="16202" LastEditDate="2018-05-25T22:02:33.043" LastActivityDate="2018-05-25T22:02:33.043" Title="Formula for &quot;ram rise&quot; that works all the way to orbit" Tags="|thermodynamics|aerodynamics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21890" PostTypeId="2" ParentId="21880" CreationDate="2018-05-24T16:37:09.693" Score="0" Body="&lt;p&gt;Some metals like steel go through a large range of hardening, while still ductile and under going plastic yield, before getting to the rupture point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the entire area under the strain/stress curve up to fracture point it's called toughness module and in steell it is much larger the resilience (elastic) module.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Sw7Jn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sw7Jn.jpg&quot; alt=&quot;mild steel stress/ strain graph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cmrp.com/blog/faq/analysis-design/exploring-stress-strain-curve-mild-steel.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source of graph here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-05-24T16:37:09.693" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21891" PostTypeId="1" AcceptedAnswerId="21893" CreationDate="2018-05-24T16:51:03.310" Score="1" ViewCount="805" Body="&lt;p&gt;I recently had a machine shop machine some aluminium panels (3mm thickness) which I then sent to an anodizing shop where the panels were bead blasted and anodized. When they came back they were seriously bent (dimensions of the panel are about 310x130mm). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did specify 6061-T6 for the alloy but I suspect the machine shop might have used a cheaper non-tempered aluminium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my thinking correct or does this problem come from the anodisation process? &lt;/p&gt;&#xA;" OwnerUserId="16203" LastEditorUserId="16203" LastEditDate="2018-05-24T18:21:51.130" LastActivityDate="2020-12-22T22:39:06.797" Title="deformation / bending problem with anodized aluminium panels" Tags="|mechanical-engineering|machining|aluminum|deformation|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21893" PostTypeId="2" ParentId="21891" CreationDate="2018-05-24T19:11:21.420" Score="1" Body="&lt;p&gt;Very difficult to think that anodizing could cause distortion. I think there is a possibility that bead blasting could cause distortion with high pressure and large beads. Bead blasting is similar to shot peening which is used to put residual stresses and distortion into metals. Lower strength materials would be more likely to distort under extreme bead blast conditions. However , I would look for another cause.   &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-05-24T19:11:21.420" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21894" PostTypeId="2" ParentId="18866" CreationDate="2018-05-24T21:27:17.613" Score="2" Body="&lt;p&gt;Backing off of micro-crack fracture theory, what your engineering introduction book is trying to hint at is the implications of &lt;a href=&quot;https://en.wikipedia.org/wiki/Mohr%27s_circle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mohr's circle&lt;/a&gt;.  To illustrate, from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8PqLw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8PqLw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, when you have a stress in x-direction, a stress in y-direction, and a shear stress, you can convert the stress to a new arbitrary axis (x' and y') with a different normal and shear stress.  Basic statics can resolve these forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_n = \frac{1}{2} ( \sigma_x + \sigma_y ) + \frac{1}{2} ( \sigma_x - \sigma_y )\cos 2\theta + \tau_{xy} \sin 2\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau_n = -\frac{1}{2}(\sigma_x - \sigma_y )\sin 2\theta + \tau_{xy}\cos 2\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a shaft under pure torsion, the equations simplify:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_n = \tau_{xy} \sin 2\theta$$&#xA;$$\tau_n = \tau_{xy}\cos 2\theta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the primary failure method is shear failure, such as the case for most ductile materials, then this occurs when we look at the axis where $\theta = 0$, and the stress is entirely in shear.  However, if we look at the axis where $\theta = 45$°, then the stress is all in tension normal stress.  In this case, when the material fails easily in tension, it will fail in this direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real world is actually more complicated with micro-crack fracture theory, stoichastic methods, and fatigue stress.  However, when running designs, this is practice to keep in mind - just because the shear stress is low doesn't mean that the brittle material can handle the tensile forces.  As the introduction to machine design book, this is simply to make you aware of this phenomenon and know there should be a difference.  However, to handle this in the real world, design criteria methods such as &lt;a href=&quot;https://en.wikipedia.org/wiki/Mohr%E2%80%93Coulomb_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mohr-Coulomb Theory&lt;/a&gt; would be utilized for brittle materials, while &lt;a href=&quot;https://en.wikipedia.org/wiki/Von_Mises_yield_criterion&quot; rel=&quot;nofollow noreferrer&quot;&gt;von Mises yield criteria&lt;/a&gt; are used on ductile materials.  Creep is used in plastics.  A nice summary of other methods is listed at &lt;a href=&quot;https://www.engr.colostate.edu/~dga/mech325/handouts/failure_theories.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Colorado State&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-24T21:27:17.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21895" PostTypeId="1" CreationDate="2018-05-24T21:28:52.017" Score="6" ViewCount="2285" Body="&lt;p&gt;Suppose I have a system described in state-space form. Can I invert the system (swapping the roles of the input $u$ and output $y$)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we start with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\dot{x} &amp;amp;= A\,x + B\,u \\&#xA;y &amp;amp;= C\,x + D\,u&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and $D$ is invertible, then we can use the second equation to solve for $u$ in terms of $x$ and $y$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;u = D^{-1} (y - C\,x)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then substitute that back into the original state-space description to get the inverse system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;\dot{x} &amp;amp;= (A - B\,D^{-1}\,C)\,x + B\,D^{-1}\,y \\&#xA;u &amp;amp;= -D^{-1}\,C\,x + D^{-1}\,y&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in typical systems $D$ is zero. What is the significance of requiring $D$ to be invertible to inverse the system? Is there a work-around allowing us to invert systems where $D=0$?&lt;/p&gt;&#xA;" OwnerUserId="508" LastEditorUserId="1765" LastEditDate="2018-05-30T09:06:05.380" LastActivityDate="2018-06-22T19:30:53.143" Title="How do I invert a system in state space form?" Tags="|control-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21896" PostTypeId="2" ParentId="21891" CreationDate="2018-05-24T23:46:14.287" Score="0" Body="&lt;p&gt;Assuming the pieces were formed to shape before anodizing: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the aluminum alloy was a high-strength variant, then bending it would result in a significant amount of strain energy being stored in the bend zone- and the sharper the bend, the more concentrated will be that energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anodization is a chemical etching process in which the deposition of reaction products on the surface of the workpiece exceeds their dissolution rate, and it requires that the reaction product film be porous enough to support the transport of charge and reactants through the film in order to sustain the reaction until such time as the film has grown to the intended thickness. After that, the anodized workpiece is removed from the chemical bath and &lt;strong&gt;baked&lt;/strong&gt; in an oxidizing atmosphere furnace long enough for the porosity in the film to be closed off by the creation of fresh oxide in those pores. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your parts are probably being stress-relief annealed during the post-anodization bake, and warping as a result. To prevent this, they should be stress-relieved after bending, and then anodized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming instead the pieces were flat when they went into the bath, then any anisotropic stresses left in the sheet by the rolling &amp;amp; finishing process would get released during the bake, and leave the parts warped out-of-flat. Same fix: stress-relieve before anodize.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13363" LastEditDate="2018-05-25T05:10:42.583" LastActivityDate="2018-05-25T05:10:42.583" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21897" PostTypeId="2" ParentId="21858" CreationDate="2018-05-25T00:20:04.367" Score="0" Body="&lt;p&gt;Yes, you can extract turbine energy from the exhaust of a pulse jet, but you need to evaluate the consequences before you think it worthwhile.  I assume the pulse jet is on an aircraft, in which case, its exhaust normally jets into the atmosphere, and so the turbine needs to be attached to the aircraft and placed in the pulse jet exhaust stream, or a better idea may be to attach it directly by means of a short duct to the pulse jet exhaust. Regardless, one can see immediately that such a scheme greatly affects the operation of the pulse jet.  If the turbine is mounted downstream in the free jet, the jet will cause a force that pushes on the turbine apparatus, in a direction opposite to the thrust of the pulse jet, thus reducing the net forward force on the aircraft.  If the turbine is mounted to the exhaust of the pulse jet, the pressure drop of the turbine will increase the pulse jet exhaust pressure, thus reducing the performance of the pulse jet.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the turbine is mounted downstream in the free jet, the resulting pressure disturbances in the exhaust stream can possibly travel upstream, resulting in an increase in the exit pressure of the pulse jet, which consequently reduces both the jet exit velocity and thrust of the jet.  However, this upstream migration can happen only if the exhaust jet velocity is subsonic.  Wherever this jet velocity is supersonic, the disturbance cannot travel upstream. The fact that the jet velocity is pulsed greatly complicates estimation of the magnitude of this reduced thrust, and also the magnitude of the backward-directed reaction force transmitted to the aircraft.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the simplest arrangement would be to mount the turbine in a short duct attached to the back of the pulse jet.  But then, I wonder why not substitute this kind of contraption with a turbojet?&lt;/p&gt;&#xA;" OwnerUserId="14623" LastActivityDate="2018-05-25T00:20:04.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21898" PostTypeId="1" AcceptedAnswerId="21907" CreationDate="2018-05-25T01:16:50.927" Score="2" ViewCount="68" Body="&lt;p&gt;If I do a port scan from the internet, do I hit a bundle of wires that are physical ports into the motherboard (like a PCE port)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or do I actually go into the operating system and interact with the virtual or 'logical' ports of the OS? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand what the difference is between these ports, and why, if they are logical, I would be able to scan &lt;em&gt;any&lt;/em&gt; of them from the internet, because it implies my bytes are passing over hardware to reach these 'logical' ports inside the CPU and cache, rather than on the peripheral hardware (before they are rejected)… This feels wrong...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if they are actually physical, then where are they? And, if I have a daemon listen on a port that is &quot;opened&quot;, does this mean that the program is listening to bytes that pass over a physical component on the motherboard, exterior to the cache-ram-cpu area?&lt;/p&gt;&#xA;" OwnerUserId="11577" LastActivityDate="2018-05-25T12:04:37.270" Title="Computer &quot;Ports&quot;, port scans and daemons" Tags="|computer-engineering|computer-hardware|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21899" PostTypeId="1" AcceptedAnswerId="21912" CreationDate="2018-05-25T03:22:26.117" Score="2" ViewCount="736" Body="&lt;p&gt;I have a crane boom that is rotating and when it is stopped suddenly, the boom flexes back and fourth several times. I have measured the angles between each deflection and have $3.16 ^\circ$, $2.42 ^\circ$, $1.77 ^\circ$ and $1.31 ^\circ$, which if plotted in Excel forms a logarithmic curve equal to $$y = -1.33ln(x) + 3.2221$$ The goal now is to find how to plot this as a sin curve with the period of 3.5 seconds so that it it looks like a deflection vs time chart (or damping curve), similar to the one in red below. Can someone please explain how I can do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JJcG2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JJcG2.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13810" LastActivityDate="2018-05-25T23:09:58.880" Title="How to work out the damping or deflection curve from deflection angles?" Tags="|mathematics|vibration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21902" PostTypeId="1" AcceptedAnswerId="21909" CreationDate="2018-05-25T06:53:29.187" Score="4" ViewCount="11578" Body="&lt;p&gt;How do we calculate the deflection of a cantilever with two or three cross-sections, and different forces acting at the end of each section?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/x27LV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x27LV.png&quot; alt=&quot;image of three-sectioned cantilever&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16206" LastEditorUserId="255" LastEditDate="2018-05-25T08:02:18.883" LastActivityDate="2021-07-05T15:21:31.357" Title="Multiple Varying cross section beam deflection" Tags="|structural-engineering|beam|aircraft-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21904" PostTypeId="2" ParentId="21902" CreationDate="2018-05-25T08:07:29.933" Score="2" Body="&lt;p&gt;Let's give the name &quot;section 1&quot; to the section with length $\ell_1$, similarly for sections 2 and 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section 1 can be treated as a cantilever with shear force $A+B+C$ applied at its end, and a moment $B\ell_2 + C(\ell_2+\ell_3)$ also applied at its end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section 2 can then be treated as a cantilever from Section 1, with shear force $B+C$ applied at its end, and a moment $C\ell_3$ also applied at its end; but remember that the deflection and rotation at the left end are set by the calculation for Section 1, so you'll need to add this on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section 3 can then be treated as a cantilever from Section 2, with shear force $C$ applied at its end; but remember that the deflection and rotation at the left end are set by the calculation for Section 2, so you'll need to add this on.&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="1832" LastEditDate="2018-05-25T10:39:06.960" LastActivityDate="2018-05-25T10:39:06.960" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21906" PostTypeId="1" CreationDate="2018-05-25T10:21:49.973" Score="1" ViewCount="24" Body="&lt;p&gt;This morning I had a presentation on &lt;a href=&quot;https://www.phpspec.net/en/stable/&quot; rel=&quot;nofollow noreferrer&quot;&gt;PHPSPec&lt;/a&gt; which the official website describes as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A php toolset to drive emergent design by specification.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As defined within an adjective context the word &quot;emergent&quot; is defined as follow:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the process of coming into being&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On the other hand &lt;a href=&quot;https://en.wikipedia.org/wiki/PHPUnit&quot; rel=&quot;nofollow noreferrer&quot;&gt;PHPUnit&lt;/a&gt; is said to focus on quick code commits and asserting that no code regression occurs in other parts of the code. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that being said the argument that came out of presentation is that PHPSpec is still a unit test framework which I did, in no way disputed that, but mine I think was more on what the tool really focuses on which is design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The differences among these two tools is succinctly explained in this &lt;a href=&quot;https://laracasts.com/discuss/channels/testing/whats-the-difference-between-phpunit-and-phpspec&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; on Laracasts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes Agile Methodology  which is explained &lt;a href=&quot;https://blog.azendoo.com/agile-project-management-methodology-explained/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://www.infoworld.com/article/3237508/agile-development/what-is-agile-methodology-modern-software-development-explained.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. The latter article on Agile Methodology highlights mentions why this methodology is better (&lt;strong&gt;&lt;em&gt;see below for an excerpt&lt;/em&gt;&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When you take the aggregate of agile principles, implement them in an&#xA;  agile framework, leverage collaboration tools, and adopt agile&#xA;  development practices, you usually get better-quality applications,&#xA;  faster-developed applications, and better technical practices (aka&#xA;  hygiene).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The view is that PHPSpec offers nothing that PHPUnit does not. However, if we look into the Engineering discipline and what it is about we can almost see the mention of tools (&lt;a href=&quot;https://www.unr.edu/cse/prospective-students/what-is-software-engineering&quot; rel=&quot;nofollow noreferrer&quot;&gt;see below excerpt&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It encompasses concepts, principles, theories, techniques and &lt;strong&gt;tools&lt;/strong&gt;&#xA;  that can be used for developing high-quality professional software.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Looking at all this within the perspective of Continuous Integration tools include, in that context, things like PHPSpec, PHPUnit, Behat, Codeception and so forth for achieving varying test strategies. Within an engineering context all these tools have roles to play. However, it does seem that one cannot use the other, particularly PHPSPec, within an Agile environment. What I think lacks is a real-world context where one can really drive Test-Driven and Behavior-Driven Design using these two tools in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might be wrong in how I have been seeing things within the context of Agility coupled with Software Engineering strategies to deliver high quality code and so forth. I do not claim to be Mr Know-it-all mine is to try and become better and knowledgeable within my profession and thus improve on my skills.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody please help me understand Agile Methodology when mixed with Software Engineering within the context of PHP and how one can go about implementing a Software Product Development process taking advantage of readily available testing tools and strategies out there? Thank you in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question has been moved into the &lt;a href=&quot;https://softwareengineering.stackexchange.com/questions/371560/what-constitutes-agility-where-behavior-driven-and-unit-testing-are-concerned&quot;&gt;Software Engineering&lt;/a&gt; space. I apologise for posting it here I did not know any better. Thanks&lt;/p&gt;&#xA;" OwnerUserId="16208" LastEditorUserId="16208" LastEditDate="2018-05-25T15:16:06.497" LastActivityDate="2018-05-25T15:16:06.497" Title="What constitutes AGILITY where Behavior-Driven and Unit Testing are concerned" Tags="|industrial-engineering|tools|research|" AnswerCount="0" CommentCount="3" ClosedDate="2018-06-01T18:01:21.837" ContentLicense="CC BY-SA 4.0" />
  <row Id="21907" PostTypeId="2" ParentId="21898" CreationDate="2018-05-25T12:04:37.270" Score="1" Body="&lt;p&gt;The ports referred to in a &quot;port scan&quot; are logical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the fact that all network communication to a computer typically is over a single physical connection, usually Ethernet.  There are multiple simultaneous conversations happening routinely.  Think about how the network traffic for fetching email in one window, a web browser in another window, various apps accessing servers elsewhere, and all those viruses sending your passwords and bank account details over to my servers all work independently without getting each other's traffic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the TCP and UDP level, this is done with virtual ports.  Servers are waiting on &quot;well known ports&quot; to accept connections.  For example, port 25 is usually reserved for requesting new connections of a SMTP server.  Once a connection is established, it is assigned a free port.  Subsequent traffic then references that port to identify the connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A port scan essentially (this is over-simplified, but that's all that's needed for the basic concept) sends a connection request to every port (they are in a 16 bit namespace) to see which ones the server machine reacts to.  Some OS services of particular operating systems and versions are known to have servers on particular ports.  Some of there are know to have vulnerabilities.  One purpose of a port scan is to find out what ports your machine reacts to, and if there are any you might know how to exploit in doing unintended things with the machine.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-05-25T12:04:37.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21909" PostTypeId="2" ParentId="21902" CreationDate="2018-05-25T17:04:20.887" Score="1" Body="&lt;p&gt;Castigliano's method can simplify your calculations to an extent - instead of computing the deflection at every point on the beam, you can calculate the deflection at the points you care about only, and ignore the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we need to write a moment equation in terms of x.  This is fairly straightforward.  Defining:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ &amp;lt;x&amp;gt; = \left\{\begin{matrix}&#xA;0&amp;amp;\text{if}&amp;amp;x &amp;lt; 0 \\&#xA;x&amp;amp;\text{if}&amp;amp;x&amp;gt;= 0&#xA;\end{matrix}\right.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then we have the moment equation for the entire beam as simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M(x) = A&amp;lt;l_1-x&amp;gt; + B&amp;lt;l_2+l_1-x&amp;gt; +C&amp;lt;l_3+l_2+l_1-x&amp;gt;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also can identify the moment of inertia of each section as &lt;span class=&quot;math-container&quot;&gt;$I_1,I_2,I_3$&lt;/span&gt; respectively.  Using the &lt;a href=&quot;http://mathworld.wolfram.com/HeavisideStepFunction.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heaviside Step function&lt;/a&gt; (which acts as the basis for the &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;x&amp;gt;$&lt;/span&gt; convention - i.e. this convention is simply shorthand for &lt;span class=&quot;math-container&quot;&gt;$(x)*H(x)$&lt;/span&gt;) the Section modulus of each section becomes simply:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$EI(x) = EI_1*H(l_1-x)+EI_2*H(x-l_1)*H(l_1+l_2-x)+EI_3*H(x-l_1+l_2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we differentiate the moment equation with respect to the forces (A,B,C); for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial M}{\partial A} = &amp;lt;l_1-x&amp;gt;$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial M}{\partial B} = &amp;lt;l_1+l_2-x&amp;gt;$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial M}{\partial C} = &amp;lt;l_1+l_2+l_3-x&amp;gt;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, we have everything we need.  To find the deflection at A, we simply take Castigliano's theorem regarding the energy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$U = \int^L_0 \frac{M(x)^2}{2EI(x)}dx$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And therefore the deflection is equal to &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial U}{\partial P}$&lt;/span&gt;, where P is the load at the point you want to test.  If there isn't a load, you can just add it into the moment equation, and run through the integral.  In this case, you asked for the deflection at the three points, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta_j = \frac{\partial U}{\partial P_j} = \int^L_0 \frac{M(x)}{EI(x)}\frac{\partial M(x)}{\partial P_j}dx $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Where &lt;span class=&quot;math-container&quot;&gt;$P_j = A, B, C$&lt;/span&gt; respectively, and &lt;span class=&quot;math-container&quot;&gt;$\delta_j = x_1, x_2, x_3$&lt;/span&gt;, the points under the loads A, B, and C respectively.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substitute the different partial derivatives of the moment equation, plug into the integral.  If the Heaviside step integral becomes difficult, break the integral into the different lengths.  For example, running with the first deflection, since &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial M}{\partial A} = &amp;lt;l_1-x&amp;gt;$&lt;/span&gt;, we only need to compute the integral from &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$l_1$&lt;/span&gt;.  This simplifies the result immensely:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta_1 = \int^{l_1}_0 \frac{A*(l_1-x) + B*(l_2+l_1-x) +C*(l_3+l_2+l_1-x)}{EI_1}(l_1-x) = \frac{l_1^2*(2Al_1+2Bl_1+3Bl_2+2Cl_1+3Cl_2+3Cl_3)}{6EI_1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar applications utilizing the properties of the &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;x&amp;gt;$&lt;/span&gt; convention can reduce the problem to computing a few additional integrals, which are easily resolved with most computer algebra systems.  For a quick double check, the second deflection is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta_2 = \frac{A(2l_1^3+3l_2^2)}{6EI_1}+\frac{B(2l_1^3+6l_1^2l_2+6l_1l_2^2)}{6EI_1}-\frac{B(2l_1^3-2l_2^3)}{6EI_2}+\frac{C(2l_1^3+6l_1^2l_2+3l_3l_1^2+6l_1l_2^2+6l_1l_2l_3)}{6EI_1}-\frac{C(l_1^3+3l_1^2l_3-2l_2^3-3l_2^2l_3)}{6EI_2}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2019-06-05T19:37:16.103" LastActivityDate="2019-06-05T19:37:16.103" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="21910" PostTypeId="1" CreationDate="2018-05-25T17:26:35.627" Score="3" ViewCount="1313" Body="&lt;p&gt;I am doing some research on impacts related to natural gas compressor stations. In some states, I've received data for compressor stations alongside other facilities in the oil and gas world. These booster stations and pump stations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before filtering out these other types of facilities, I'd like to make sure that I understand the difference between the facilities. Could someone outline key differences across these facilities?&lt;/p&gt;&#xA;" OwnerUserId="16213" LastActivityDate="2022-12-01T20:07:50.917" Title="What are the differences between compressor stations, pump stations, and booster stations?" Tags="|gas|pipelines|energy|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="21911" PostTypeId="2" ParentId="19230" CreationDate="2018-05-25T20:45:43.463" Score="0" Body="&lt;p&gt;In the perfectly elastic spectrum, response to the &lt;strong&gt;dampening&lt;/strong&gt; coefficient of the second order differential equation of vibration behaves very similar to how you have shown, but not quite.  It is well documented by reviewing how a system responds when subjected to a forced time-varying load at different speeds.  Notice how it doesn't get wider for the frequency shifts as the dampening increases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RfEYv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RfEYv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;x(t) = X_0 \sin(\omega t + \phi) \\&#xA;X_0 = \dfrac{KF_0}{\sqrt{\left(1-\dfrac{\omega^2}{\omega_n^2}\right)^2+\left(2\varsigma\dfrac{\omega}{\omega_n}\right)^2}} \\&#xA;\phi = \tan^{-1}\left(\dfrac{-2\varsigma\dfrac{\omega}{\omega_n}}{1-\dfrac{\omega^2}{\omega_n^2}}\right)&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as time goes on, the natural frequency of a material experiencing &lt;em&gt;creep&lt;/em&gt; will vary.  While creep adds to the dampening, it adds some other time-dependent side effects. From &lt;a href=&quot;https://en.wikipedia.org/wiki/Creep_(deformation)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Creep (sometimes called cold flow) is the tendency of a solid material to move slowly or deform permanently under the influence of mechanical stresses. It can occur as a result of long-term exposure to high levels of stress that are still below the yield strength of the material.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As a result of creep, the natural frequency of the curve can become wider over time, but that's somewhat speculative until the actual creep mechanism is understood and the governing PDEs are resolved.  However, this could give results similar to your temperature plot.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="1832" LastEditDate="2018-05-25T20:57:15.850" LastActivityDate="2018-05-25T20:57:15.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21912" PostTypeId="2" ParentId="21899" CreationDate="2018-05-25T23:09:58.880" Score="2" Body="&lt;h1&gt;The model&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The equation for a damped periodic oscillation is&#xA;$$&#xA;  y(t) = A e^{-\lambda t} \cos(\omega t + \phi)&#xA;$$&#xA;where $A$ is the amplitude, $\lambda$ is the decay constant, $\omega = 2\pi/\tau$ where $\tau$ is the period, and $\phi$ is the phase angle at $t=0$.  In your figure, $\phi = -\pi/2$, and therefore your damping curve has the form&#xA;$$&#xA;  y(t) = A e^{-\lambda t} \sin(\omega t)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Your mistake&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Instead of using&#xA;$$&#xA;  y(t) = A e^{-\lambda t} \quad \quad \text{(Correct)}&#xA;$$&#xA;you have fit the portion of the curve where $\sin(\omega t) = 1$ using the relation &#xA;$$&#xA;  y(t) = B \ln(t) + C \quad \quad \text{(Wrong)}\,.&#xA;$$&#xA;This model predicts an infinite displacement $y$ at time $t=0$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Data needed&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;You have provided angles but not displacements.  We have to guess what those are from your fit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we fit the correct model to data extracted from your log model, we get the following curves (note that we can calculate the full deflection-time curve after fitting $A$ and $\lambda$ because $\omega$ can be calculated from the period):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VUDF3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VUDF3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-05-25T23:09:58.880" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21913" PostTypeId="2" ParentId="21910" CreationDate="2018-05-25T23:59:55.683" Score="1" Body="&lt;p&gt;Compressor stations are the main gas movers. Pumps would be for liquids. I guess &#xA;&quot;Booster&quot; could be for either . Depending on situation , one location could have both a compressor and a pump for liquid and gas going the same area. There are many variations like one liquid line carrying diesel and gasoline to a producing area and a gas line and liquid line taking production out. Some locations may pipe out produced water to a location for treatment; significant When onshore US wells produce roughly 90% water ( not counting the production from fracking ). Although most produced water is reinjected in the same general area. And, the biggest compressor station I have seen, compressed gas for reinjection. Reading the question again, this sounds like the beginning of some more computer generated misinformation.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-05-25T23:59:55.683" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21914" PostTypeId="1" AcceptedAnswerId="22661" CreationDate="2018-05-26T02:20:58.393" Score="5" ViewCount="803" Body="&lt;p&gt;If a bridge has a parapet that is nota fully solid block and resembles railings with posts, as the following image&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BdN3C.png&quot; alt=&quot;bridge parapet&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is this type of parapet called?&lt;/li&gt;&#xA;&lt;li&gt;What are the top, bottom, post, and hole of the railing-like structure called?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16216" LastActivityDate="2018-10-12T17:09:16.323" Title="Bridge terminology - railing-like parapet" Tags="|bridges|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21915" PostTypeId="1" AcceptedAnswerId="21921" CreationDate="2018-05-26T06:37:21.110" Score="1" ViewCount="3759" Body="&lt;p&gt;What is the difference between a blade and a vane in a turbine? When constructing velocity triangles, there are two angles at inlet and outlet (α &amp;amp; θ, β &amp;amp; φ), respectively. My textbook defines α &amp;amp; β as blade angles and θ &amp;amp; φ as vane angles. What is the difference between them?&lt;a href=&quot;https://i.stack.imgur.com/SU8Cf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SU8Cf.jpg&quot; alt=&quot;Velocity triangle of a curved vane.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16219" LastActivityDate="2018-05-26T12:07:14.050" Title="Difference between 'vanes' and 'blades'" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21917" PostTypeId="1" CreationDate="2018-05-26T07:08:36.453" Score="2" ViewCount="492" Body="&lt;p&gt;I am designing a roof rack for my car that attaches to my canopy and extends over the roof of the cab. I am deciding what size of aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would need to support me walking on it ideally (80kg).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be a 1800x3600mm rectangle that is welded onto the canopy for 2400mm from one end, leaving the 1200mm overhanging the cabin. There will be 8 cross supports including the ends. Options are all in box section:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;32 x 32 in 2 and 3mm&lt;/li&gt;&#xA;&lt;li&gt;40 x 40 in 2 and 3mm&lt;/li&gt;&#xA;&lt;li&gt;50 x 50 in 2,3 and 5mm.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;if needed, I can add a central support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/63848m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/63848m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KX4d8m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KX4d8m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16220" LastEditorUserId="9" LastEditDate="2018-05-27T22:50:35.337" LastActivityDate="2018-05-27T22:50:35.337" Title="How to calculate the swl of aluminium?" Tags="|materials|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21920" PostTypeId="1" CreationDate="2018-05-26T10:17:42.410" Score="2" ViewCount="106" Body="&lt;p&gt;I am investigating what would be the most appropriate compressed air source (fan/blower/compressor) for designing a separate air compressor system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the compressed air would be supplied to the lens of an industrial sensing instrument, which is mounted onto a pipe leading into a boiler's furnace&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;there are 12 instruments on each row and there are four rows in total (first row starts at 20m and there is an increase of 5m per row). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Purpose:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;to provide sufficient pressure in order to prevent accumulation of dust on an instrumentation's lens &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;to provide sufficient flow to keep the entire pipe free of contaminants in the first place&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Demand requirements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;flow (800&amp;nbsp;m&lt;sup&gt;3&lt;/sup&gt;/hour) for all 84 instruments (100&amp;nbsp;m&lt;sup&gt;3&lt;/sup&gt;/hour for one row of instruments) (&lt;strong&gt;8.6&amp;nbsp;m&lt;sup&gt;3&lt;/sup&gt;/hour for one instrument)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;pressure (40&amp;nbsp;mbar)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know that the pressure demand is very little, it simply needs to be slightly above the furnace pressure. However, my concern is that the fans we have been using have not been able to provide sufficient pressure. Meanwhile, the flow is quite large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, I am contemplating between a blower (more powerful fan) or a compressor (not sure if an air receiver would be necessary). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which one would be more suitable (for both 100&amp;nbsp;m&lt;sup&gt;3&lt;/sup&gt; per hour and 800&amp;nbsp;m&lt;sup&gt;3&lt;/sup&gt;/hour)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are the pros and cons of either (in terms of cost and space required and efficiency).?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a compressor, &lt;strong&gt;which type of compressor&lt;/strong&gt; (positive displacement, rotary, reciprocating) would be suggested?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the duty cycle is 100% (all the tube needs to be cleaned at all times). If a compressor is used, is a receiver necessary?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many instruments could be fed with one compressor?&lt;/p&gt;&#xA;" OwnerUserId="15458" LastEditorUserId="15458" LastEditDate="2018-05-26T11:22:04.733" LastActivityDate="2018-06-08T09:14:31.677" Title="Recommended compressed air source and location for separate air compression system" Tags="|mechanical-engineering|compressed-air|compressors|industrial-engineering|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="21921" PostTypeId="2" ParentId="21915" CreationDate="2018-05-26T12:07:14.050" Score="0" Body="&lt;p&gt;I don't understand why the textbook uses the terms &quot;blade angle&quot; and &quot;vane angle&quot; here, but the point is that you need to consider two sets of angles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One set ($\alpha$ and $\beta$) describe the fluid flow direction relative to the fixed turbine structure. The other set ($\theta$ and $\phi$) describe the flow direction relative to the rotating blades. The vectors $u1$ and $u2$ in the velocity triangles come from the velocity $u$ of the blade as it rotates.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-05-26T12:07:14.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21922" PostTypeId="2" ParentId="21917" CreationDate="2018-05-26T18:45:00.203" Score="1" Body="&lt;p&gt;Depending on what series of aluminum section you use the tensile strength varies greatly. But generally these are the steps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I of a square aluminum section is $$  5\times 5^3 /12 - 4.7\times 4.7^3/12 = ~ 12 cm^4 $$ for 3mm thickness profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $$ S = 2I/H = 4.5 cm^3 $$  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You multiply your weight by 2.8 for dynamic load factor 80. 2.8 = 224 kg.&#xA;Moment of your weight is 224. 1.2 = 268 kgm. = 26800 kgcm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now we have $$ \sigma = m/S = 26800/4.5 = ~6000kg/ cm^2 $$ &#xA;I did not check my calcs but basically this is the stress your aluminum profile should be able to take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to check the data sheet of the section.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-05-26T18:51:18.467" LastActivityDate="2018-05-26T18:51:18.467" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21923" PostTypeId="1" CreationDate="2018-05-27T01:28:02.787" Score="3" ViewCount="3527" Body="&lt;p&gt;I need to calculate the moment of Inertia Iyz but I find 0 information online about how to get it or solve it in a real problem, only that is equal to &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar{I}_{yz}=\int yz\text{ d}A$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I solve this in the case of an antisymmetric beam section?&lt;/p&gt;&#xA;" OwnerUserId="12596" LastEditorUserId="2015" LastEditDate="2018-05-31T16:07:03.643" LastActivityDate="2018-06-30T16:50:57.370" Title="Second Moment of Inertia (or Second Moment of Area) of Beam Section" Tags="|structural-engineering|civil-engineering|mathematics|inertia-matrix|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21924" PostTypeId="2" ParentId="21923" CreationDate="2018-05-27T02:22:18.640" Score="-1" Body="&lt;p&gt;Second moment of inertia with respect to Z axis is in the form$$ \bar{I}_{yz}=\int yz^2\text{ d}A&#xA;   $$ By this we mean I-z but when the beam's base is considered Y. Alternatively we could seek I_xz which would have x as base and z as beam height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise if we are looking for diagonal I_yz we know it is $ \space Iyz = \sqrt{[(I_z)^2 + (I_y)^2]}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you have shown is first moment of area, not second. Second moment of area is the sum of all small parts of a beam's surface, multiplied by their moment arm WRT the beam's neutral axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example for a rectangular cross section beam with dimensions$$ X = 6cm,\space Y =12 cm\space and\space Z =180 cm $$ which could be a log of timber the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \bar{I}_{yz}=\int yz^2\text{ d}A = 12 \times 180^3 /12  = 5832000 cm^4 $$&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-05-27T18:14:34.147" LastActivityDate="2018-05-27T18:14:34.147" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="21925" PostTypeId="1" AcceptedAnswerId="21926" CreationDate="2018-05-27T07:32:59.990" Score="1" ViewCount="292" Body="&lt;p&gt;I'm a engineering student currently taking Thermodynamics and we have this project for our finals which is a sort of reeling /towing device, the assignment consists in a reeling motor that has to tow a small car up along a slope, loaded with a certain weight, the team that is able to reel in the most weight in the shortest time will win the competition and have the maximum grade and extra credits for the semester.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rules are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can only use 1 Peltier thermoelectric generator module that will deliver 12 volts to the motor of my choosing. No capacitors or energy storage devices. I could use a voltage regulator to rise the voltage but speaking to one of my professors he said that due to the fact the power is just a relationship between current and voltage, rising the voltage wouldn't really make a difference as the current would drop, so my best shot was to shop for a motor with good torque that would surely work under 12 volts in case the peltier module does not deliver its full capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had 2 options, and because I need to move the biggest load in the shortest amount of time, I had to look for a motor with a good speed/torque relationship that would work under 12 volts in the worst case scenario, but I read somewhere you can make a speed reduction arrangement to help torque like the one below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=LHPduoql93A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reducing Engine&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I went and bought two different motors,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One with high speed and comparably low torque (to see if I the reducing arrangement can help me get a better torque in the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B01DVHAW6A&quot; rel=&quot;nofollow noreferrer&quot;&gt;BEMONOC Small DC Motor 12V High Speed 12000 RPM High Torque PMDC Motor for DIY Parts &amp;nbsp;(0.135 N.m)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And one with low speed and high torque, to test both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B071GTTSV3&quot; rel=&quot;nofollow noreferrer&quot;&gt;Greartisan DC 12V 200RPM Gear Motor High Torque Electric Micro Speed Reduction Geared Motor Centric Output Shaft 37mm Diameter Gearbox (0.2157463 N.m)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I was wondering if you have any suggestions as to how to improve this torque/speed relationship and get the best out of it, I really want to win this competition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would truly appreciate any information you can share with me in contribution to my project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance and sorry for my grammar, English is not my first language. &lt;/p&gt;&#xA;" OwnerUserId="15504" LastActivityDate="2018-05-27T13:38:08.593" Title="What is your best recommendation to optimize the torque/speed relationship of a motor?" Tags="|mechanical-engineering|motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21926" PostTypeId="2" ParentId="21925" CreationDate="2018-05-27T13:38:08.593" Score="1" Body="&lt;p&gt;With such a limited power supply, the key metric that you should be concerned with, is efficiency, i.e. how much of the energy supplied to the motor can &lt;em&gt;actually be used&lt;/em&gt; for lifting the weight. For a typical Permanent Magnet DC Motor, peak efficiency occurs at at approximately 90% of the motor's 'no load speed', and 10% of its 'stall torque'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QVXTD.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QVXTD.gif&quot; alt=&quot;Typical PMDC Motor Performance Graph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you will find that using a &quot;low speed and high torque motor&quot; which is able to reel in the load without 'struggling' at all (i.e. the motor itself, before the gearbox, is able to turn at approximately 90% of it's no-load speed), then your system will run more efficiently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say that the winners will be able to lift &quot;the most weight in the shortest time&quot; - this is an ambiguous scoring system. Is it:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the most weight lifted over a certain distance wins, but if there are two teams lifting the same weight, then the faster team wins?&lt;/li&gt;&#xA;&lt;li&gt;Teams get a score base on the weight they lifted and the time it took - something like $\frac{Weight}{Time}$?&lt;/li&gt;&#xA;&lt;li&gt;Something else?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Without that information, you can't make an informed decision regarding which motor is likely to be best.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-05-27T13:38:08.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21929" PostTypeId="2" ParentId="17508" CreationDate="2018-05-27T22:30:42.240" Score="2" Body="&lt;p&gt;The best vibration correlation to damage I have seen is &quot;Fatigue Damage Spectrum&quot;.  An excellent explanation of this is in the text, &lt;strong&gt;Mechanical Vibration and Shock Analysis, Fatigue Damage (Mechanical Vibrations and Shock Analysis) (Volume 4)&lt;/strong&gt;, Christian Lalanne; ISBN 978-1848216471.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the technique estimates the modal response of a structure to a particular base excitation vibration across a range of natural frequencies and the damage is assumed to be proportional to the maximum displacement of the object.  &lt;/p&gt;&#xA;" OwnerUserId="16229" LastActivityDate="2018-05-27T22:30:42.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21930" PostTypeId="1" CreationDate="2018-05-27T22:56:26.033" Score="1" ViewCount="39" Body="&lt;p&gt;I'd like to know how can I estimate the maximum speed that can be achieved by a&#xA;&quot;portion of wind&quot; accelerated upward from rest on the surface of&#xA;a place by pushing forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've thinking in the relation between lift and speed (for an object such as an air plane), &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$Lift=\frac{1}{2}ρv^2 × C_l × area$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But not sure if I can use this expression for &quot;a portion of air&quot;, for air moving specifically.&lt;/p&gt;&#xA;" OwnerUserId="16230" LastEditorUserId="110" LastEditDate="2018-05-28T01:25:12.430" LastActivityDate="2018-05-28T01:25:12.430" Title="Relation between airespeed and pushing force" Tags="|fluid-mechanics|fluid|air|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21932" PostTypeId="2" ParentId="5908" CreationDate="2018-05-28T10:02:38.620" Score="0" Body="&lt;p&gt;Wall function solves buffer layer and outer layer, but not viscous sublayer. see &lt;a href=&quot;https://en.wikipedia.org/wiki/Law_of_the_wall&quot; rel=&quot;nofollow noreferrer&quot;&gt;law of the wall&lt;/a&gt; explanation from wiki.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For flow separation modelling, sub-viscous Boudnary layer (yplus&amp;lt;5) need to be resolved, thus yplus &amp;lt; 2, use SST k-omega RANS, or more advanced models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y* is almost identical to Y+, detail see  &quot;wall function and near wall treatement&quot; in theory guide of fluent.&lt;/p&gt;&#xA;" OwnerUserId="16231" LastEditorUserId="9" LastEditDate="2018-05-28T10:47:15.687" LastActivityDate="2018-05-28T10:47:15.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21933" PostTypeId="1" AcceptedAnswerId="21953" CreationDate="2018-05-28T14:21:19.753" Score="2" ViewCount="114" Body="&lt;p&gt;I'm currently trying to design a bulk-handling system which minimizes the distance from the bulk material pile(s) where the material is loaded (the volume) to the unloading platform(s) (the point), but I have many possible combinations on the volume shape and the number of unloading point. I would ideally like to iterate through various designs in order to find the most feasible one, and a useful parameter to get from each design would be to get this average travel distance from the pile to the platform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have toyed around with the idea of using the centroid as the average distance to a point, but the problem is that the directional component of the centroid makes it such that it doesn't truly represent the average distance. For example, a disk with a radius of 10 meters might have the same centroid as a disk with a radius of 100 meters, but the average distance from the center would be different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_{V}r\text{ d}V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A slightly better representation would be the moment of inertia, since the squared component better represents average distance (so that opposite directions don't cancel out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_V r^2\text{ d}V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to truly get the average distance, I would need something that looks more like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\int_V \lvert r\rvert\text{ d}V$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any simple way of getting this value from a CAD software package, or is it derivable from any other existing mathematical definitions like the centroid or the moment of inertia? Similarly, if this average distance is obtained only relative to the centroid, how could I extrapolate it to another arbitrary point?&lt;/p&gt;&#xA;" OwnerUserId="12823" LastEditorUserId="1832" LastEditDate="2018-05-28T16:31:23.670" LastActivityDate="2018-06-01T15:33:57.937" Title="Is there any defined mathematical parameter that can give me the average distance of a volume to a point?" Tags="|design|mathematics|computer-aided-design|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="21935" PostTypeId="1" CreationDate="2018-05-28T17:53:43.083" Score="1" ViewCount="19" Body="&lt;p&gt;I am trying to measure the axial induction factor of a small aeromodelling propeller.&#xA;By axial induction factor I call the ratio of the velocity of the wind far from the propeller $V_{\infty}$ over the velocity through the actuator disk $V_{d}$ as introduced in the Rankine-Froude actuator theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us call $$a=\frac{V_{d}}{V_{\infty}}$$ the induction factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The experimental setup is the following : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The propeller is mounted on the axis of a DC motor&lt;/li&gt;&#xA;&lt;li&gt;Those are placed into a wind tunnel&lt;/li&gt;&#xA;&lt;li&gt;A resistance $R$ (of a value really close from the internal engine resistance) is connected to the poles of the DC motor and the voltage drop $U$ across the resistance is measured.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From the actuator theory we know that the mechanical power transmitted to the propeller can be expressed as :&#xA;$$P_{mech}=2\rho_{\infty}SV_{\infty}^2a^2(1-a)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that $S$ is the area of the actuator disk.&#xA;From the experimental setup supposing that the efficientcy of the DC engine is $\eta$ I have :&#xA;$$P_{elec}=2\frac{U^2}{R}=\eta P_{mech}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this I can find :&#xA;$$a^2(1-a)=\frac{U^2}{R\eta\rho_{\infty}SV_{\infty}^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is that the experiments do not provide a satisfying value for $a$, the values I find are really close to $a=1$ which wolud mean that the propeller does not have any impact on the flow even if it rotates really fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any flaws either in my experimental setup or in my theory ?&lt;/p&gt;&#xA;" OwnerUserId="8192" LastActivityDate="2018-05-28T17:53:43.083" Title="How can I mesure the axial induction factor of a small propeller using a DC motor?" Tags="|mechanical-engineering|electrical-engineering|fluid-mechanics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21937" PostTypeId="1" AcceptedAnswerId="21942" CreationDate="2018-05-28T21:25:39.233" Score="1" ViewCount="780" Body="&lt;p&gt;If the output of a diaphragm pump is impeded in any way, or meant to be at higher pressure than the input of the diaphragm pump, will it work, or will it stall out and fail to pump? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to this, are there reversible diaphragm pumps?&lt;/p&gt;&#xA;" OwnerUserId="15863" LastActivityDate="2018-05-29T05:00:08.570" Title="Can the same diaphragm pump be used to pressurize a container?" Tags="|pumps|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21939" PostTypeId="2" ParentId="21891" CreationDate="2018-05-29T00:55:22.203" Score="0" Body="&lt;p&gt;Your plate is relatively long and thin, and Aluminum (even with T6 temper), is not particularly stiff or strong.  That said, typical bead blasting or shot peening should not deform it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When anodizing, a bunch of electrical current is being pumped into your part, since it is relatively large surface area.  In order to prevent arcing of the electrical current, it needs to be clamped or otherwise held securely to the rack used for the anodizing process.  Depending how it was clamped (and what kind of warping you're seeing), this could be the cause of your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how badly warped it is, because I'm not sure what your tolerances on the flatness of the plate were, and therefore by what margin it is warped.  If you're holding relatively tight tolerances for flatness (less than, say, .010&quot;), shot peening or bead blasting prior to anodizing is a good idea.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, I'm thinking that whatever non-flatness was in place prior to anodizing, plus any additional warpage that was incurred in retaining the plate to the rack caused some bending or warping.  When the anodizing was applied, it 'locked in' this warpage, and now your part is more warped than before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from the obvious possibility of them simply mishandling your parts, I don't think the machine shop or the anodizing shop did anything out of the ordinary.  You may just have to make your plate a bit thicker if possible, or control the flatness tighter to begin with (again, not sure how badly out of tolerance we're talking here), or have a discussion with the anodizer about your problem to see if they can process it differently next time.&lt;/p&gt;&#xA;" OwnerUserId="14912" LastActivityDate="2018-05-29T00:55:22.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21941" PostTypeId="2" ParentId="7884" CreationDate="2018-05-29T03:15:22.783" Score="0" Body="&lt;p&gt;You would need a brake when the motor is pushing a car. The reversing of the polarities in the magnets to break are currently used in Prius cars and are also to recoup the kinetic energy down hills called synergy. This has also been done to reverse engineer ceiling fans to make them into wind generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After talking to the author of the picture below he said this was also used in trains since the 1900 to collect kinetic energy from going downhill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tZiLz.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tZiLz.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;You have my permission to use or link to illustrations from my tutorial on stepping motors at&#xA;  --&#xA;or any mirror of that web site, on the condition that you credit those illustrations with something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Illustration from&#xA;  -- &lt;a href=&quot;http://www.cs.uiowa.edu/~jones/step/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cs.uiowa.edu/~jones/step/&lt;/a&gt;,&#xA;  -- Control of Stepping Motors&#xA;  -- by Douglas W. Jones&#xA;  -- used with permission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual wording is your choice -- make it fit gracefully into what you're doing.  All I care is that the final 4 elements listed above are included in the credit.  If you are crediting multiple sources, it reads better if you use a similar form of credit for all of them, but if not, just go with this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-- Illustration from &lt;a href=&quot;http://www.cs.uiowa.edu/~jones/step/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.cs.uiowa.edu/~jones/step/&lt;/a&gt;, Control of Stepping Motors by Douglas W. Jones, used with permission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you may not do is imply any endorsement by me of what you are writing, since I have no idea what it is.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;            Doug Jones&#xA;            jones@cs.uiowa.edu&#xA;            (aka douglas-w-jones@uiowa.edu)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://homepage.divms.uiowa.edu/~jones/step/typesf/2anim.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Original image retrieved from here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-05-30T00:00:21.123" LastActivityDate="2018-05-30T00:00:21.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21942" PostTypeId="2" ParentId="21937" CreationDate="2018-05-29T05:00:08.570" Score="1" Body="&lt;p&gt;Yes, a diaphragm pump is able to pump against a pressure gradient. Typically, positive displacement pumps are able to generate a higher pressure difference, but at a lower flow rate than other technologies, such as centrifugal pumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are not reversible diaphragm pumps, since they rely on one-way-valves for their intrinsic operation.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-05-29T05:00:08.570" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="21944" PostTypeId="2" ParentId="21933" CreationDate="2018-05-29T08:49:02.690" Score="1" Body="&lt;p&gt;Having been involved with a similar situation many, many years ago, I'm not aware of a simple equation that will solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are good, or you know someone who good, at computer programming, or you could use a spreadsheet, you can use an imaginary grid. Fix either the volume location or loading location in the grid and then iteratively place the other location at points on the grid and calculate the distance between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this you will get a 2D array of distance data from which you can then develop contours of distance between the two locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then have to repeat the procedure for different loading location or volume locations.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-05-29T08:49:02.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21947" PostTypeId="2" ParentId="21049" CreationDate="2018-05-29T16:32:40.243" Score="1" Body="&lt;p&gt;In general, for air life-time, no tubing manutcurer will recommend a useful lifetime beyond 20 years, no matter the service.  This isn't really due to shoddy craftmanship, but due to creep and test data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All plastics experience &lt;a href=&quot;https://en.wikipedia.org/wiki/Creep_(deformation)&quot; rel=&quot;nofollow noreferrer&quot;&gt;creep&lt;/a&gt; - a phenomenon where the material will continue to deform so long as it is being stressed.  No matter how low the pressure of the air, the tubing will be stressed, and the material will deform, and at some point it will rupture because it cannot deform anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some plastics resolve this phenomenon with reinforcement - either steel reinforced on the outside, or in the case of &lt;a href=&quot;https://en.wikipedia.org/wiki/Fibre-reinforced_plastic&quot; rel=&quot;nofollow noreferrer&quot;&gt;fiber-reinforced plastic&lt;/a&gt;, fibers of a material that will not creep.  In this case, 100% of the stress can be diverted to the fibers and the material can last forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there is a catch to this reinforcement phenomenon - the testing for such materials requires strict reproduction to ensure it's final life - and this rarely happens.  As such, most manufacturers will simply stick to the plastic life - 20 years, as they cannot provide a better number for the fiber reinforced version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this scenario, most polyethylene tubing would work wonderfully, with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Double-walled_pipe&quot; rel=&quot;nofollow noreferrer&quot;&gt;double-containment pipe&lt;/a&gt; around the tube consisting of either steel or PVC.  However, I won't even be able to promise you it would last longer than the panels.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-29T16:32:40.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21948" PostTypeId="2" ParentId="18536" CreationDate="2018-05-29T18:46:29.650" Score="0" Body="&lt;p&gt;For a localized plastic loading, you may want to consider the &lt;a href=&quot;https://en.wikipedia.org/wiki/Plastic_hinge&quot; rel=&quot;nofollow noreferrer&quot;&gt;plastic hinge model&lt;/a&gt;.  In this model, it is assumed that the moment the beam reaches yield stress, it becomes a &lt;strong&gt;plastic hinge&lt;/strong&gt;.  Here is a freebody diagram of the plastic hinge:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZfXD6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZfXD6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the shear at the wall ($V.w$) and the shear at the hinge ($V.b$) may be different, but the plastic moment is still the same - the plastic stresses have redistributed the loading so that less load occurs at the wall, and more of the load can be supported at the hinge point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plastic hinge model doesn't use the normal second moment of area, but utilizes the &lt;a href=&quot;https://en.wikipedia.org/wiki/Section_modulus#Plastic_section_modulus&quot; rel=&quot;nofollow noreferrer&quot;&gt;plastic section modulus&lt;/a&gt; as well as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Yield_(engineering)&quot; rel=&quot;nofollow noreferrer&quot;&gt;yield stress&lt;/a&gt; to find the ultimate allowable plastic moment.  As an example, while a normal rectangle would be $\frac{bh^2}{6}$, in plastic sections it would be $\frac{bh^2}{4}$.  Multiply by a yield stress of a hypothetical $100 MPa$ would generate a final moment.  Using these methods, we would generate a free-body diagram of each side of the hinge, solve for the different values of $M_p$, and find out which loading causes $M_p$ to reach the ultimate value determined above.  This would be the final allowable forces, and the point of plastic rupture.  &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-29T18:46:29.650" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21950" PostTypeId="1" CreationDate="2018-05-30T02:35:55.420" Score="0" ViewCount="245" Body="&lt;p&gt;Given a triangle with nodes 0, 1, and 2 where each node is located at $r_i = x_i,y_i,z_i$ for $i = 0:2$ and has an associated scalar value $\phi_i$ at the node, how would I evaluate the directional derivative $\nabla_u \phi$ at the centroid of the triangle at the centroid of the triangular element given the direction $u$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've figured out how to evaluate $\nabla_u \phi$ on each edge separately but not sure how to get the value at the centroid.&lt;/p&gt;&#xA;" OwnerUserId="8225" LastEditorUserId="8225" LastEditDate="2018-05-30T04:11:04.730" LastActivityDate="2024-04-03T03:01:50.180" Title="Calculate numerical directional derivative on triangular element" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21952" PostTypeId="2" ParentId="21895" CreationDate="2018-05-30T07:45:43.043" Score="3" Body="&lt;p&gt;When $D$ is invertible means that an equivalent transfer function has the same order numerator as denominator. So the inverse also has the same order numerator as denominator. The stated state space model can only represent transfer functions which are &lt;a href=&quot;https://en.wikipedia.org/wiki/Proper_transfer_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;proper&lt;/a&gt;, the degree of the numerator does not exceed the degree of the denominator. When $D$ is not invertible means that the order of the numerator is less then the degree of the denominator, so for the inverse the degree of the numerator does exceed the degree of the denominator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is still possible to represent an 'improper' transfer function as a state space model, however you then have to introduce an $E$ matrix as well. This is called a &lt;a href=&quot;https://nl.mathworks.com/help/control/ref/dss.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;descriptor state space model&lt;/a&gt; and is defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;E\,\dot{x} &amp;amp;= A\,x + B\,u \\&#xA;y &amp;amp;= C\,x + D\,u&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has an improper transfer function when $E$ is not full rank, so not invertible. finding the inverse of a state space model with a not invertible $D$ is more complicated, since it is not directly clear what the degree difference would be between numerator and denominator, thus how much rank deficiency $E$ needs to have.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-05-30T07:45:43.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21953" PostTypeId="2" ParentId="21933" CreationDate="2018-05-30T08:17:44.067" Score="3" Body="&lt;p&gt;First, let's consider a Cartesian plane with the loading platform at the origin. It is obvious that the distance to any point is on the plane is: $l = \sqrt{x^2+y^2}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you extend this to the $3^{rd}$ dimension by plotting $l$ on the $z$-axis, you get an inverted cone with a pitch of 90 degrees with its tip at the origin. Your material piles will be projected onto this cone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total distance to all points on the pile from the loading bay will be the volume under the cone bounded by the area of the piles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$D=\int\int{\sqrt{x^2+y^2}}dydx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the average distance per trip, you need to divide the total distance by the number of trips. The non-discreet version of this equates to the volume under the cone divided by the plan area of the pile:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d=\frac{D}{A}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve your real problem, the obvious answer is to have as much material as possible as close as possible to your loading platforms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Just some additional info that might help with your problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously you can translate the functions to other points than the origin if you have to plot multiple loading platforms. Multiple cones will intersect each other at some point in space, and viewed in plan, these intersections will form lines enclosing the platforms in &quot;cells&quot;, with any point within a cell being closest to the origin of the cell. This is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Voronoi_diagram&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voronoi Diagram&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T8X33.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T8X33.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Image from Mathworks&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this as a graphic aid to assist with the placement and shape of the piles. I'll leave the determination of the ideal shapes and placements as an exercise to the reader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that we don't know any of the restrictions, there won't be one optimal solution. (For all intents and purposes, with the info we have we could just as well put the stockpile right on the loading platform in a column of infinite height, thus the average distance would be 0).&lt;/p&gt;&#xA;" OwnerUserId="14440" LastEditorUserId="14440" LastEditDate="2018-05-30T08:50:36.577" LastActivityDate="2018-05-30T08:50:36.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21954" PostTypeId="1" CreationDate="2018-05-30T09:01:01.230" Score="0" ViewCount="64" Body="&lt;p&gt;I am looking for some pretty specific tasks to be carried out by a software program.  I’m currently using Civil 3D, but I can’t help but think there is a better program (or at least a better Autodesk based workflow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an overview of my project, followed by the requirements I’m looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overview:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Approx. 40 locations/sites &lt;/li&gt;&#xA;&lt;li&gt;Use hi-res raster aerial images (most obtained via UAV) to analyze properties&lt;/li&gt;&#xA;&lt;li&gt;Looking to identify obvious problems such as poor drainage, inadequate parking, deteriorated pavements, potential future growth areas&lt;/li&gt;&#xA;&lt;li&gt;The properties are mostly commercial or industrial&lt;/li&gt;&#xA;&lt;li&gt;Based on the owner’s vision and property maintenance budget, develop a rudimentary 5-10 year outlook (i.e. install underground storm detention and adjacent pond in 2020, add a loading dock in 2022, etc.)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Needs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Create polygons around problem areas or items of interest&lt;/li&gt;&#xA;&lt;li&gt;Ability to assign a polygon to a layer or tag (there will be 10-15 total layers or tags that every markup will fall under&lt;/li&gt;&#xA;&lt;li&gt;Automatically change the appearance (color, transparency) depending on the layer or tag&lt;/li&gt;&#xA;&lt;li&gt;Quantify square footages, # in each layer or group&lt;/li&gt;&#xA;&lt;li&gt;Generate a report in .csv or .txt format for each location&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note: Final deliverable would hopefully be web-map based (think google maps JavaScript APIs).&lt;/p&gt;&#xA;" OwnerUserId="16252" LastEditorUserId="1832" LastEditDate="2018-06-29T15:20:21.027" LastActivityDate="2018-06-29T15:20:21.027" Title="I’m a civil engineering consultant, and I am analyzing different properties across US to develop long-term property maintenance plans" Tags="|civil-engineering|software|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21956" PostTypeId="2" ParentId="21870" CreationDate="2018-05-30T09:35:23.153" Score="1" Body="&lt;p&gt;In the case that there are no other inputs acting on the system and the system starts at $x=x_\mathrm{cmd}$, then in order to ensure that you keep perfect tracking you also need that $\dot{x}=\dot{x}_\mathrm{cmd}$. This can be achieved by solving for $u_\mathrm{cmd}$ in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{x}_\mathrm{cmd} = A\,x_\mathrm{cmd} + B\,u_\mathrm{cmd}. \tag{1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does require that $x_\mathrm{cmd}$ is a feasible trajectory as a function of time, namely $\dot{x}_\mathrm{cmd} - A\,x_\mathrm{cmd}$ should lie in the span of $B$. If this is the case, then the feedforward term can be found using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Inverse_element#Matrices&quot; rel=&quot;nofollow noreferrer&quot;&gt;pseudo inverse&lt;/a&gt; if $B^\top\,B$ is full rank&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;u_\mathrm{cmd} = \left(B^\top\,B\right)^{-1} B^\top\left(\dot{x}_\mathrm{cmd} - A\,x_\mathrm{cmd}\right). \tag{2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total control law would become&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;u = u_\mathrm{cmd} - K\,(x - x_\mathrm{cmd}). \tag{3}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here by defining $e = x - x_\mathrm{cmd}$, under the assumption that no other inputs are acting on the system, then it can be shown that the dynamics of $e$ becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{e} = (A - B\,K)\,e. \tag{4}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if $A-B\,K$ is Hurwitz, then as time goes to infinity $e$ should go to zero. And when $e=0$ by definition you also have $x = x_\mathrm{cmd}$.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastEditorUserId="1765" LastEditDate="2018-05-30T09:43:48.967" LastActivityDate="2018-05-30T09:43:48.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21958" PostTypeId="2" ParentId="5627" CreationDate="2018-05-30T11:13:59.877" Score="2" Body="&lt;p&gt;In Abaqus, there are two primary analysis methods used for solving structural problems, namely Abaqus/Standard and Abaqus/Explicit (includes dynamic Explicit). Abaqus/Standard has static implicit and dynamic implicit and is suitable for solving smooth nonlinear problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the solution of these problems may converge with difficulty, because of the contact or material complexities which can cause a large number of iterations. Such analyses are more expensive in Abaqus/Standard due to the iteration which needs many sets of linear equations to be solved. Meanwhile, Abaqus/Explicit conducts the solution without iterating by explicitly advancing the kinematic state from the previous increment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, Explicit stands for explicit time integration. Although a given analysis may require a large number of time increments by using the explicit method, the analysis requires much less disk space and memory compared with Abaqus/Standard which requires many iterations. Hence, for problems in which the computational costs of the two programs are comparable, the disk space and memory savings of Abaqus/Explicit make it more attractive.&lt;/p&gt;&#xA;" OwnerUserId="16256" LastEditorUserId="9" LastEditDate="2018-05-30T12:04:01.633" LastActivityDate="2018-05-30T12:04:01.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21959" PostTypeId="2" ParentId="21954" CreationDate="2018-05-30T11:39:00.873" Score="1" Body="&lt;p&gt;I don't see why Civil 3D doesn't meet your needs. It can accomplish and automate everything you've specified apart from visually identifying and marking the problem areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Drawing Setup&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can start by creating a template and setting up all the layers and line work etc for your areas and for every task you create your drawing starting with the template. Civil 3D has hundreds of default layers you might not want to use, so perhaps you can create a barebones template from AutoCAD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you start to draw you regions, make sure everything is set to &lt;em&gt;ByLayer&lt;/em&gt; and pick the appropriate layer before sketching. This way the appearance is added as you draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Data Export&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Autocad has a command, &lt;em&gt;&lt;a href=&quot;https://knowledge.autodesk.com/support/autocad/learn-explore/caas/CloudHelp/cloudhelp/2019/ENU/AutoCAD-Core/files/GUID-30D85568-2BCF-4F6C-A95E-FC6F394C5D3A-htm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;EATTEXT&lt;/a&gt;&lt;/em&gt;, which you can use to export almost any data of almost any object to a csv file, and you can do it in bulk. This has aparently bee deprecated and replaced by &lt;em&gt;Attribute Extraction wizard&lt;/em&gt; in version 2018.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Export to Maps&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Civil 3D has a built-in feature to export data to a kml/kmz file for Google Earth. Just make sure that you start off with a properly georeferenced file, preferably in WGS84.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also easily write scripts to automate some repetitive tasks if you need to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Possible (probably better) Alternative&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Autodesk Revit is in general more user friendly than any AutoCAD based product. You can create areas, rooms and sites and assign specific metadata to them and automatically annotate them as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there you can easily create schedules of quantities with all the info you need and export to csv if you need to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Revit just doesn't have an option to export to Google maps I know of, but perhaps a plug-in exists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're not familiar with Revit, do yourself a favour and check it out.&lt;/p&gt;&#xA;" OwnerUserId="14440" LastActivityDate="2018-05-30T11:39:00.873" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21960" PostTypeId="2" ParentId="7884" CreationDate="2018-05-30T14:13:51.910" Score="1" Body="&lt;p&gt;Your solution fits a rather narrow gap of the market. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of the market is where a brake is not needed - the motor propelling devices where standard operation friction is enough to stop them when unpowered. It's a very broad domain: motor being non-critical, say, powering a fan which is harmless when allowed to spin down slowly after switching power off; high-torque (heavy industrial) machines of moderate inertia, that will stop within 1-2 turns by themselves without constant power input, machinery where duty cycle of motor operation is so long comparing to switch on/off times these become negligible (pumps operating on large reservoirs) and many others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other side of the market requires braking far more efficient than a couple resistors can provide. These will use mechanical clutches or external mechanical brakes that can stop the rotation within much less than several rotations, and inhibit slow free rotation (your brake does nothing against that!) - this will be safety equipment stopping machines that can be dangerous to life and health: saw blades, elevators, heavy machinery of high inertia that would take a long time to sink all the stored momentum through the resistors, vehicles and transport devices (cranes/gantries), also production lines where a fault at a point could lead to serious accident if allowed to propagate down the production line with damaged product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're in a narrow segment where a weak brake like yours is sufficient. It may be budget machinery which allows a delay at braking at expense of safety; it may be very heavy, high-inertia machinery where rapid mechanical braking is simply impossible due to forces involved. It may be also a comfort/efficiency feature for devices that don't require brakes, but presence of one allows faster change of operation - say, service/regulation access to a section of a machine that is locked out for duration of operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And even in these cases a segment of your market is stolen by controllers that have your feature integrated, and provide it as a part of control, not just extra safety. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-05-30T14:13:51.910" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21961" PostTypeId="1" CreationDate="2018-05-30T15:09:07.517" Score="0" ViewCount="297" Body="&lt;p&gt;I would like to weld a piece of hardened steel but i don't want to ruin the treatment on the tip of the piece.&#xA;Can i put some ice in the middle of the shaft so the heat generated from the welding (SMAW DCEN) doesn't temper the tip?&#xA;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="15000" LastActivityDate="2018-12-26T19:01:05.247" Title="Can I keep hardened steel cool with ice while welding?" Tags="|cooling|welding|heat-treatment|hss|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21962" PostTypeId="2" ParentId="21961" CreationDate="2018-05-30T15:17:17.360" Score="1" Body="&lt;p&gt;It will reduce the weld HAZ some , but not much. It depends on the dimensions and the heat input of the weld, the alloy , the original hardness , and the original heat treatment. The primary cooling is by conduction which is affected by the temperature difference. So the temperature difference will be 75 F / 2800 F , or 32 F /2800 F , not much of a difference.  And, if the steel has much carbon ( eg. over 0.15% ) , the weld HAZ will contain some hard brittle zones with a significant chance of cracks. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-05-30T15:28:12.257" LastActivityDate="2018-05-30T15:28:12.257" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21963" PostTypeId="2" ParentId="21102" CreationDate="2018-05-30T15:23:54.497" Score="1" Body="&lt;p&gt;I am not sure if I understand your question correctly but to determine whether your steel sections compression flange will buckle before it will fail by some other mechanism it might be helpful to look up &lt;strong&gt;Section classification&lt;/strong&gt; in following standards:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EN 1993-1-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AISC Steel Manual&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on size and material of your section it will place it into one of 4 classes (EN) and that will determine if you have to worry about buckling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my belief that most of standard sections are designed in a way that the section fails by some other mechanism rather than buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jakub&lt;/p&gt;&#xA;" OwnerUserId="16257" LastActivityDate="2018-05-30T15:23:54.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21964" PostTypeId="2" ParentId="21961" CreationDate="2018-05-30T16:20:03.573" Score="0" Body="&lt;p&gt;Second, more planned answer . Hard steel ( Rockwell C 35 +) is not, can not be welded. In an unfortunate situation where it must be attempted ; 1 - Soften at about 1300 F , one hour . 2 - Preheat to 400  F ( I have recommended up to 600 F) . 3 - Weld. ( maintain preheat ). 4 Inspect for cracks. 5 - Soften the weld HAZ  at 1300 F. 6 Heat-treat , quench and temper, according to the alloy. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-05-30T16:20:03.573" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21965" PostTypeId="2" ParentId="17252" CreationDate="2018-05-30T17:45:49.110" Score="2" Body="&lt;p&gt;The thought process is not entirely correct in the work section. The heat fluxes are not additive, rather they should be equal due to conservation of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in the $y$ direction, $q_y=q_{yA}=q_{yB}$.&#xA;Since $q=k\frac{\Delta T}{L}$, &#xA;\begin{align}&#xA;q_{yA}&amp;amp;=k_A\frac{\Delta T_A}{L_A}\\&#xA;q_{yB}&amp;amp;=k_B\frac{\Delta T_B}{L_B}.&#xA;\end{align}&#xA;Therefore,&#xA;\begin{align}&#xA;q_{y}&amp;amp;=\frac{k_A}{L_A}(T_0-T_1)\\&#xA;q_{y}&amp;amp;=\frac{k_B}{L_B}(T_1-T_2).&#xA;\end{align}&#xA;Rearranging the equations and replacing in for $T_1$, &#xA;$$&#xA;T_0-\frac{L_A}{k_A}q_y=T_2+\frac{L_B}{k_B}q_y,&#xA;$$&#xA;from which&#xA;$$&#xA;T_0-T_2=q_y\left(\frac{L_A}{k_A}+\frac{L_B}{k_B}\right).&#xA;$$&#xA;Defining $\Delta T=T_0-T_2$,&#xA;$$&#xA;q_y=\frac{1}{\frac{L_A}{k_A}+\frac{L_B}{k_B}}\Delta T.&#xA;$$&#xA;Since $G_y=-\frac{\Delta T}{L_A+L_B}$,&#xA;$$&#xA;q_y=\frac{L_A+L_B}{\frac{L_A}{k_A}+\frac{L_B}{k_B}}\frac{\Delta T}{L_A+L_B}=-\frac{L_A+L_B}{\frac{L_A}{k_A}+\frac{L_B}{k_B}}G_y.&#xA;$$&#xA;Rearranging and using the definition $\phi=\frac{L_A}{L_A+L_B}$,&#xA;$$&#xA;q_y=-\frac{k_Ak_B}{(1-\phi)k_A+\phi k_B}G_y.&#xA;$$&#xA;Since $q_y=-k_{yy}G_y$,&#xA;$$&#xA;k_{yy}=\frac{k_Ak_B}{(1-\phi)k_A+\phi k_B}.&#xA;$$&#xA;For $k_{xx}$ and $k_{zz}$,&#xA;the paths through $A$ and $B$ are parallel, so $\frac{\Delta T}{L}$ is the same for $A$ and $B$, and $G_x=G_z=-\frac{\Delta T}{L}$ is the same for $A$ and $B$.&#xA;Since the paths are parallel, &#xA;\begin{align}&#xA;q_x&amp;amp;=\frac{L_A}{L_A+L_B}q_{xA}+\frac{L_B}{L_A+L_B}q_{xB}\\&#xA;q_z&amp;amp;=\frac{L_A}{L_A+L_B}q_{zA}+\frac{L_B}{L_A+L_B}q_{zB}.&#xA;\end{align}&#xA;Now,&#xA;\begin{align}&#xA;q_{xA}&amp;amp;=q_{zA}=k_A\left(\frac{\Delta T}{L}\right)_A\\&#xA;q_{xB}&amp;amp;=q_{zB}=k_B\left(\frac{\Delta T}{L}\right)_B,&#xA;\end{align}&#xA;but&#xA;$$&#xA;\left(\frac{\Delta T}{L}\right)_A=\left(\frac{\Delta T}{L}\right)_B=-G_x=-G_z,&#xA;$$&#xA;so&#xA;\begin{align}&#xA;q_x&amp;amp;=\phi k_A(-G_x)+(1-\phi)k_B(-G_x)=-\left[\phi k_A+(1-\phi)k_B\right]G_x\\&#xA;q_z&amp;amp;=\phi k_A(-G_z)+(1-\phi)k_B(-G_z)=-\left[\phi k_A+(1-\phi)k_B\right]G_z.&#xA;\end{align}&#xA;Since&#xA;\begin{align}&#xA;q_x&amp;amp;=-k_{xx}G_x\\&#xA;q_z&amp;amp;=-k_{zz}G_z,&#xA;\end{align}&#xA;$$&#xA;k_{xx}=k_{zz}=\phi k_A+(1-\phi)k_B.&#xA;$$&#xA;Therefore,&#xA;$$&#xA;\mathbf{k}=\left(&#xA;\begin{array}{ccc}&#xA;k_{zz}=\phi k_A+(1-\phi)k_B&amp;amp;0&amp;amp;0\\&#xA;0&amp;amp;\frac{k_Ak_B}{(1-\phi)k_A+\phi k_B}&amp;amp;0\\&#xA;0&amp;amp;0&amp;amp;k_{zz}=\phi k_A+(1-\phi)k_B&#xA;\end{array}&#xA;\right).&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part b, in order for two vectors $\mathbf{a}$ and $\mathbf{b}$ to be parallel, the ratio $a_i/b_i$ must be constant for all $i$.&#xA;Here,&#xA;\begin{align}&#xA;\mathbf{Q}&amp;amp;=-k_{xx}G_x\hat{x}-k_{yy}G_y\hat{y}-k_{zz}G_z\hat{z}\\&#xA;-\mathbf{G}&amp;amp;=-G_x\hat{x}-G_y\hat{y}-G_z\hat{z}.&#xA;\end{align}&#xA;\begin{align}&#xA;\frac{-k_{xx}G_x}{-G_x}&amp;amp;=k_{xx}=k_{zz}=\frac{-k_{zz}G_z}{-G_z}\\&#xA;\frac{-k_{yy}G_y}{-G_y}&amp;amp;=k_{yy},&#xA;\end{align}&#xA;but unless $k_A=k_B$, $k_{yy}\neq k_{xx}$, so the ratio does not hold and the vectors are not parallel.&lt;/p&gt;&#xA;" OwnerUserId="13036" LastActivityDate="2018-05-30T17:45:49.110" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21969" PostTypeId="2" ParentId="4230" CreationDate="2018-05-30T18:42:34.710" Score="2" Body="&lt;p&gt;The simple answer is both - flow never really cares about travelling &quot;out of a high density&quot; or &quot;into a low density&quot; zone, it just travels from areas where there are density differentials.  However, your intuition is right - and I personally consider it more of an expansion wave than a compression wave.  The reasons are complicated, as you would imagine for a question that remained unanswered for two years.  The short version is you consider the expansion of the pipe in the first half while the fluid compresses.  In the second half, when the pipe compresses back to it's unstressed state, the fluid expands backwards to the source to refill the reservoir from whence it started flowing.  Sometimes that's a pump, not a reservoir, which can cause additional problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A source I have has a rigid mathematical model to water hammer, as well as a few other situations.  For more, see &lt;a href=&quot;https://www.google.com/search?q=Analysis%20and%20Design%20of%20Energy%20Systems,%203rd%20Edition,%20by%20BK%20Hodge%20and%20Robert%20P%20Taylor&amp;amp;ei=-uAOW5LmAoaOsQXo77GIAQ&amp;amp;start=0&amp;amp;sa=N&amp;amp;biw=1455&amp;amp;bih=735&quot; rel=&quot;nofollow noreferrer&quot;&gt;Analysis and Design of Energy Systems, 3rd Edition, by BK Hodge and Robert P Taylor&lt;/a&gt; - ISBN 978-0135259733, specifically chapter 7.  The simplest model is a water filled tank, with water at height h, connected via a long pipe of length l, diameter D, with a valve at the end of the pipe.  We begin with fluid in the pipe, and open the valve.  Working through the continuum mechanics, we find the equation for the fluid beginning to flow out through the valve follows along the lines of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$-\frac{\partial H}{\partial x} - \left(\frac{\partial z}{\partial x} = 0 \right) -\frac{fV|V|}{2gD} = \frac{1}{g}\frac{dV}{dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the weight per volume, $\gamma$ is treated as a constant - this is considered a rigid body of water.  Integrating across the pipe, pretending $f$ (the friction factor) is constant, and dividing the pressure by the specific weight to treat this as &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydraulic_head&quot; rel=&quot;nofollow noreferrer&quot;&gt;fluid head&lt;/a&gt;, we wind up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$H-\frac{fL}{D}\frac{V^2}{2g} = \frac{L}{g}\frac{dV}{dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple differential equation to solve for flow development velocity $V$ with respect to time.  For closing the valve, the water is treated as compressible, but &lt;strong&gt;&lt;em&gt;so is the pipe&lt;/em&gt;&lt;/strong&gt; and as a result the pipe bulges with stored potential energy.  As a result, the speed of water hammer is a combination of the speed of sound of water and the pipe, and slightly less than the speed of sound in water:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ a =\left(\frac{Kg}{\rho(1+(K/E)c)}\right)^\frac{1}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $K$ is the bulk modulus of the fluid, $E$ is the modulus of the pipe, and $c$ is an empirical factor that varies from 0 to $\infty$, but is on the order of the ratio of the Diameter of the pipe to the pipe wall thickness.  Note that plastic and flexible pipes will have a slower hammer speed because of a higher $(K/E)$ ratio.  The water hammer is modeled with a second wave equation coupled to the original:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{a^2}{g}\frac{\partial V}{\partial x} + \frac{\partial H}{\partial t} = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The source goes on with methods of using these equations together to get useful finite-element based numerical results.  In this, water is returned back to the original tank, the pipe compresses excessively, and as the pipe returns back to the original unstressed state, the liquid fills back up under pressure from the reservoir and the second cycle repeats, dampened by the friction factor only.  As such, the main results to be seen is that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Depending upon what you look at, the fluid compresses, followed by the pipe compressing&lt;/li&gt;&#xA;&lt;li&gt;Or ... the pipe expands from a low stress state to a high stress state, followed by the fluid expanding to form a high head.&lt;/li&gt;&#xA;&lt;li&gt;Both are valid ways to look at it, but the second seems more intuitive, and captures the main point of the danger of water hammer - it will damage your pipes if you fail to account for it!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-30T18:42:34.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21970" PostTypeId="1" CreationDate="2018-05-30T23:18:17.910" Score="0" ViewCount="21" Body="&lt;p&gt;I am able to find that the class for materials that can be in contact with wine (~10-15% alcohol by volume) is called Class VI-A and Class VI-C, but I am not able to find a list of materials that are approved for those classes. Does anyone have experience with navigating the FDA for a list of materials that are approved for a certain class of use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.fda.gov/RegulatoryInformation/Guidances/ucm081818.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.fda.gov/RegulatoryInformation/Guidances/ucm081818.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15383" LastActivityDate="2018-05-30T23:18:17.910" Title="What FDA compliant plastics can have contact with alochol (wine)?" Tags="|plastic|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21971" PostTypeId="2" ParentId="11018" CreationDate="2018-05-31T01:22:49.140" Score="0" Body="&lt;p&gt;In heat exchanger design, typically a maximum rise is specified by the end user - be it the environmental control groups, or the plant designer itself.  The heat exchanger is then designed to remove the required amount of heat, which is known, by having the output have a certain temperature fall of the first stream, and a maximum temperature rise.  The actual temperatures are then simulated in various selected heat exchangers, until a desired heat exchanger that matches the constraints is determined.  Usually this comes with a desired flowrate on both the hot line and the cool line.  The cool line flowrate would usually be a minimum to remain open, and conservatively additional flow would be designed into the system (larger pumps, larger pipes, etc).&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-31T01:22:49.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21976" PostTypeId="1" CreationDate="2018-05-31T11:09:24.060" Score="1" ViewCount="47" Body="&lt;p&gt;I was thinking about interesting powder metallurgy method, hot isostatic pressing. Often thick steel walled vessel is required increasing costs and decreasing the prevalence of this technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This led me to consider say several metal spherical concentric tanks. I wonder if the inner most tank is at very high pressure more so than the metal wall could handle but was counter acted mostly by a second concentric tank pressurized nearly to the same pressure but slightly less, and so on so forth until finally the outer most tank was low enough pressure to be contained.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not really sure of the net forces? Would there just be large compression forces on the tanks? Would there be a large net force inwards or outwards? I ask because it would be interesting if say I had two balloons and everything worked out such that say I had a 99% chance of rupturing balloon A at pressure P_1 but it balloon B had less than 1% chance of rupturing at pressure P_2 and I found both had low chance of rupturing at respective pressures with balloon A contained inside balloon B. Maybe this help to understand the idea I wonder about, perhaps it is a fallacy I'm not sure. &lt;/p&gt;&#xA;" OwnerDisplayName="user10585" LastActivityDate="2018-05-31T11:09:24.060" Title="multiple wall, pressurized gas pressure vessel" Tags="|pressure|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="21977" PostTypeId="1" CreationDate="2018-05-31T12:25:28.253" Score="4" ViewCount="89" Body="&lt;p&gt;I need to send out some PRT files for tooling that have been made using Creo. I do not have a windows PC or mac available at the moment, only linux. Does anyone have some advice how to go from here?&lt;/p&gt;&#xA;" OwnerUserId="16270" LastEditorUserId="16270" LastEditDate="2018-05-31T12:47:07.497" LastActivityDate="2018-05-31T12:47:07.497" Title="Converting Creo files to STP on linux" Tags="|cad|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="21978" PostTypeId="2" ParentId="21923" CreationDate="2018-05-31T12:31:55.877" Score="2" Body="&lt;p&gt;The products of inertia aren’t something that is encountered often. This is mainly because we can break up most cross section into small pieces and combine them with the parallel axis theorem. As such, we simply skip the integration step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The product comes into play when a load is applied at a strange angle or you are dealing with a non-symmetrical cross section. In the case of an equal legged angle shape, for example, one can compute the product of inertia is 0 for each of the legs. However, when the legs are joined, the centroid of the shape moves. As a result, since the centroid moves in both x and y axis, the net change in the product inertia would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A*x*y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where x would be the distance the centroid moved in the x direction and y would be the distance the centroid moved in the y direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This becomes especially relevant in the Mohr’s circle of inertia to determine the primary axis of inertia. You can rotate the reference frame by an angle $\theta$ to remove the product of inertia, where $\theta$ is found as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\theta = \frac{1}{2} atan\left(\frac{-2I_{xy}}{I_x-I_y}\right)$$&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-05-31T12:31:55.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21979" PostTypeId="1" CreationDate="2018-05-30T19:17:35.640" Score="0" ViewCount="48" Body="&lt;p&gt;I apologize in advance if this is the wrong site for this question.  If this is the wrong site, would some kind soul point me towards the correct site?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a couple of Vision Engineering TS4 microscopes in my lab (both purchased used from eBay).  One came with a camera attachment with a completely unknown camera mounted.  No documentation, no cables, no power supply.  Luckily, the microscope has a standard &quot;C-mount&quot; attach point.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The TS4 microscope has an optical zoom range from x6 to x40.  This is a great range when working with surface-mount components on circuit boards.  These microscopes project the image onto a view screen rather than forcing the user to sit hunched over a pair of eyepieces.  Current listing on eBay: &lt;a href=&quot;https://www.ebay.com/itm/Vision-Engineering-TS4-Stereo-Dynascope-30-Day-No-Hassle-Warranty/322914386626&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vision Engineering TS4 Stereo Dynascope&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I removed the existing camera and instead installed a 2.0 MP camera, also purchased from eBay.  &lt;a href=&quot;https://www.ebay.com/itm/Wireless-Control-1080P-HDMI-HD-Industrial-Lab-C-mount-2-0MP-Microscope-Camera-US/311759645479&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wireless Control 1080P HDMI HD Industrial Lab C-mount 2.0MP Microscope Camera US&lt;/a&gt;.  I've used these cameras previously when modifying simple lens-based microscopes with great success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem that I'm having is that the image on the monitor driven by the camera has far too much magnification as compared to the optical screen mounted on the microscope.  So much so that the resulting image simply can't get small enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I know about optics and lenses might fit on the head of a very small pin.  I don't have any idea how or what to change to bring the image size seen by the camera down to something similar to what is displayed on the native view screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guidance appreciated.&lt;/p&gt;&#xA;" OwnerUserId="686" OwnerDisplayName="Dwayne Reid" LastActivityDate="2018-05-31T16:19:14.427" Title="How to adjust magnification when mounting camera on microscope" Tags="|optics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21980" PostTypeId="2" ParentId="21979" CreationDate="2018-05-31T04:15:31.307" Score="1" Body="&lt;p&gt;I'm not familiar with this particular microscope, but usually, the magnification is determined by the objective lens of the microscope, and less so by the camera (although changing the camera position with respect to the tube lens will vary the magnification slightly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;more precisely, the magnification is determined by the ratio of focal lengths between the tube lens and the objective lens (in an infinity-corrected microscope, which is pretty much the standard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest you do some reading on geometric optics and infinity-corrected microscopes to expand your understanding. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerDisplayName="Yuval Weissler" LastActivityDate="2018-05-31T04:15:31.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21982" PostTypeId="1" AcceptedAnswerId="21983" CreationDate="2018-05-31T18:27:54.340" Score="0" ViewCount="3102" Body="&lt;p&gt;How would I find the second moment of area of a parallelogram, as shown?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dp2Pt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dp2Pt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2020-05-27T02:42:25.383" Title="How would I find the Second Moment of Area of a Parallelogram?" Tags="|mechanical-engineering|structural-engineering|statics|inertia-matrix|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21983" PostTypeId="2" ParentId="21982" CreationDate="2018-05-31T18:27:54.340" Score="2" Body="&lt;p&gt;While this could be resolved using rectangle and triangle formulas, a generalized approach could be applied that would work for any shape using integration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$A = \int^c_0\int^{x\:=\:\frac{b}{c}y+a}_{x\:=\:\frac{b}{c}y}1\:dx\:dy = ac$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the origin, we evaluate the first moment of the areas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Q_x = \int^c_0\int^{x\:=\:\frac{b}{c}y+a}_{x\:=\:\frac{b}{c}y}x\:dx\:dy = \frac{ac(a+b)}{2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q_y = \int^c_0\int^{x\:=\:\frac{b}{c}y+a}_{x\:=\:\frac{b}{c}y}y\:dx\:dy = \frac{ac^2}{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here, the centroid of the parallelogram could be evaluated, as &lt;span class=&quot;math-container&quot;&gt;$\overline{x} = \frac{Q_x}{A}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\overline{y} = \frac{Q_y}{A}$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\overline{x} = \frac{Q_x}{A} = \frac{a+b}{2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\overline{y} = \frac{Q_y}{A} = \frac{c}{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can then evaluate the integrals.  We need to change our origin, and set ourselves new limits of integration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$c-\overline{y} = \frac{c}{2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$0-\overline{y} = -\frac{c}{2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{b}{c}y+a-\overline{x} = \frac{b}{c}y+\frac{a-b}{2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{b}{c}y-\overline{x} = \frac{b}{c}y-\frac{a+b}{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, we can evaluate &lt;span class=&quot;math-container&quot;&gt;$I_{xx}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$I_{xy}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$I_{yy}$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_{xx} = \int^\frac{c}{2}_{-\frac{c}{2}}\int^{x\:=\:\frac{b}{c}y+\frac{a-b}{2}}_{x\:=\:\frac{b}{c}y-\frac{a+b}{2}}y^2\:dx\:dy = \frac{ac^3}{12}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$I_{xy} = \int^\frac{c}{2}_{-\frac{c}{2}}\int^{x\:=\:\frac{b}{c}y+\frac{a-b}{2}}_{x\:=\:\frac{b}{c}y-\frac{a+b}{2}}xy\:dx\:dy = \frac{abc^2}{12}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$I_{yy} = \int^\frac{c}{2}_{-\frac{c}{2}}\int^{x\:=\:\frac{b}{c}y+\frac{a-b}{2}}_{x\:=\:\frac{b}{c}y-\frac{a+b}{2}}x^2\:dx\:dy = \frac{ac(a^2+4b^2)}{12}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use of the perpendicular axis theorem would give the polar moment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_{zz} = I_{xx}+I_{yy} = \frac{ac(a^2+4b^2+c^2)}{12}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the product of inertia is not zero, the Mohr's circle of Inertia will be needed to generate the Inertia-Matrix.  The principal moments of Inertia are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_{\overline{xx}} = \frac{I_{xx}+I_{yy}}{2} \pm \sqrt{\left(\frac{I_{xx}-I_{yy}}{2}\right)^2 + I_{xy}^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This yields a principal moment of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_{\overline{xx}} = \frac{ac}{24}\left[a^2+4b^2+c^2 \pm \sqrt{(a-c)^2(a+c)^2+4b^2(2a^2+4b^2-c^2)}\right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This principle moment would be found by rotating the reference frame by an angle &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\theta = \frac{1}{2} atan\left(\frac{-2I_{xy}}{I_{xx}-I_{yy}}\right) = \frac{1}{2} atan\left(\frac{2bc}{c^2-4b^2-a^2}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this can be brought into tensor format:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{bmatrix}&#xA;    I_{xx} &amp;amp; I_{xy} &amp;amp; I_{xz} \\&#xA;    I_{xy} &amp;amp; I_{yy} &amp;amp; I_{yz} \\&#xA;    I_{xz} &amp;amp; I_{yz} &amp;amp; I_{zz} &#xA;\end{bmatrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is useful in some tensor-based applications.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="-1" LastEditDate="2020-04-13T17:41:05.300" LastActivityDate="2020-04-13T17:41:05.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21984" PostTypeId="1" CreationDate="2018-05-31T21:04:27.487" Score="2" ViewCount="69" Body="&lt;p&gt;Shown in the images below I have designed a small stainless steel apparatus with accompanying hollow metal cylinders ontop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s0Hf5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s0Hf5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/EGu0K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EGu0K.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking into larger quantity production and simply CNC'ing these would be cost prohibitive. I was thinking that pressing the steel into shape would be more cost effective, but I'm unsure how I'd go about adhering the hollow pipes to the top side. Would I do some type of welding to get these on with a solid airtight connection all the way around? &lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2018-06-01T17:46:54.887" Title="Best way to manufacture stainless steel apparatus?" Tags="|manufacturing-engineering|metallurgy|welding|" AnswerCount="1" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="21986" PostTypeId="2" ParentId="16616" CreationDate="2018-06-01T05:51:02.643" Score="1" Body="&lt;p&gt;With out taking any of the comments in consideration the OP is asking what insulating ration per area without asking what material is used. This question should be is it possible to get a 100% insulation? No, just like you cannot get a 100% reflection rate from photons from a mirrors. This does pose the question what is the best insulation material available? I would think Space.SE would be a good home fore this question. The OP said wall space was not a restriction for insulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is that silica aerogels the best insulator in the world right now. Silica aerogels has thermal conductivities of less than 0.03 W/m*K in the atmosphere. Aerogel  can prevent ice from melting on a hot plate at 80 degrees Celsius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;www.aerogeltechnologies.com&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RrKgc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RrKgc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/RTgiq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTgiq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This material would take the least space for insulation.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-06-01T05:51:02.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21988" PostTypeId="1" CreationDate="2018-06-01T08:49:35.483" Score="1" ViewCount="42" Body="&lt;p&gt;In the following &lt;a href=&quot;https://pdfs.semanticscholar.org/e0a7/5b33eff3f0116ec5f7321dbe7b51e95347b3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Kuhne, Felipe, Walter Fetter Lages, and JM Gomes da Silva Jr. &quot;Model predictive control of a mobile robot using linearization.&quot; In Proceedings of mechatronics and robotics, pp. 525-530. 2004.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They have linearized the system using Jacobian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\boldsymbol{\dot x}_r=\boldsymbol{f}(\boldsymbol{x}_r,\boldsymbol{u}_r)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DIHWL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DIHWL.png&quot; alt=&quot;linearization&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I am wondering&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Is it the only way to linearize the system or are there many methods to linearize and this is only one of them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- If there is a disturbance $\boldsymbol{d}$, then using this method, should we calculate the Jacobian around $\boldsymbol{d}_r$ or should we calculate it around zero disturbance? Which one is more common?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Is linearization of disturbance in the similar way to the linearization of states and input?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4- Is there any special linearization specific to robust control?&lt;/p&gt;&#xA;" OwnerUserId="16032" LastActivityDate="2018-06-01T08:49:35.483" Title="Robust linearization" Tags="|control-engineering|control-theory|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21989" PostTypeId="1" AcceptedAnswerId="21993" CreationDate="2018-06-01T09:01:38.070" Score="0" ViewCount="164" Body="&lt;p&gt;What is error in action? And why is it called error &quot;in action&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ij38O.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ij38O.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14355" LastActivityDate="2018-06-01T14:50:14.997" Title="What is &quot;error in action&quot; in gears?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="21990" PostTypeId="2" ParentId="21984" CreationDate="2018-06-01T09:03:22.060" Score="1" Body="&lt;p&gt;The tooling to make part-specific press dies is going to be very expensive to set up, especially for stainless steel and I suspect that you would have to modify the design somewhat to make it a viable pressing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By far the best solution for anything other than pretty large volumes is going to be having them fabricated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might be able to find the main body as an off the shelf pressing otherwise just weld a cap onto a length of tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plate on the base could be laser cut and the tubes on top are just lengths of tube, both of whcih could be TIG welded on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is all pretty straightforward. The only slight complication is the rounded shoulder at the top of the main cylinder. The simplest solution would be to just simplify the design to a flat plate cap on the end of a tube. If it absolutely needs the radius then you could get it machined or pressed. It should be simple enough not to be too cost prohibitive. For example a simple dished shape for the cap would be vastly easier to tool than a press die for the whole part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To expand on what I've said in the comments this looks like a fairly straightforward welding job, especially as the volume would certainly justify making a fairly sophisticated jig but you can let the manufacturer worry about the fine detail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the only possible issue I can see is that it is a bit tight for space in the gap between the two small tubes it get in to weld it but that doesn't look like a major issue. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-06-01T17:46:54.887" LastActivityDate="2018-06-01T17:46:54.887" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="21992" PostTypeId="1" CreationDate="2018-06-01T13:44:11.643" Score="2" ViewCount="12" Body="&lt;p&gt;Is there an intuitive way to approximate/estimate missing cells in a (symmetric) distance matrix?&#xA;In this case, each unknown pair has several &quot;neighbors&quot; for which the distance is known.&lt;/p&gt;&#xA;" OwnerUserId="16279" LastActivityDate="2018-06-01T13:44:11.643" Title="Complete symmetric distance matrix" Tags="|mathematics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="21993" PostTypeId="2" ParentId="21989" CreationDate="2018-06-01T14:50:14.997" Score="2" Body="&lt;p&gt;From Manual of Gear Design: Spur and Internal gears, volume II:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The load carrying ability of a pair of gears may be limited by either the beam strength of the gear teeth, or the surface endurance limit of the material. The lower of these two values will be used to establish the load carrying ability of any pair, and must be greater than the maximum dynamic load.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Dynamic Load -  whenever there is an error in action between a pair of gears, one of&#xA;  two reactions must take place: the connected masses must be&#xA;  accelerated and decelerated to compensate for this error, or the&#xA;  system must be elastically deformed the amount of the error. In&#xA;  practice a combination of the two will occur. &lt;strong&gt;Even with perfect gears,&#xA;  there will be an error in action due to the deformations under load.&lt;/strong&gt;&#xA;  The load caused by these reactions is called the dynamic load.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.globalspec.com/reference/68707/203279/gear-tooth-loads&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.globalspec.com/reference/68707/203279/gear-tooth-loads&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's called &quot;in action&quot; because the error is not present when the gears are not under load.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-06-01T14:50:14.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21994" PostTypeId="2" ParentId="21933" CreationDate="2018-06-01T15:33:57.937" Score="1" Body="&lt;p&gt;Based off of ChP and Fred's answers, I ended up coding the 2-D solution for the problem. I might eventually want to extrapolate it to 3-D, but it shouldn't be too complicated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used a &quot;clipped&quot; Voronoi diagram, code examples of which are readily available online, which I used to separate an arbitrary polygon (representing the pile) into different sections, each representing the area closest to one of the given points (representing the unloading/loading points). Then, I placed a uniform point grid inside of the polygon, and solved for the distance of each of these with respect to the closest loading point. After getting an array of these values, I averaged them to get the average distance. It's rudimentary, but it's accurate enough for my particular purpose.&lt;/p&gt;&#xA;" OwnerUserId="12823" LastActivityDate="2018-06-01T15:33:57.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21995" PostTypeId="1" CreationDate="2018-06-01T16:30:33.860" Score="4" ViewCount="202" Body="&lt;p&gt;I am trying to find an expression for the natural frequencies of an orthotropic bending beam with a circular cross-section. I have tried searching for some literature on this, but have so far been unsuccessful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an isotropic bending beam, I believe the natural frequencies can be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_n = \frac{n}{2L^2}\sqrt{\frac{EI}{m}}, $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $L$ is the length, $E$ is the Young's modulus, $I$ is the area moment of inertia of the cross-section about the neutral axis, and $m$ is the mass per unit length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an equivalent expression for an orthotropic beam? i.e. one that has a different Young's modulus in the $x$ and $y$ direction say. Therefore it would have different values $E_x$ and $E_y$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is I am considering a &lt;strong&gt;circular beam made out of wood&lt;/strong&gt;. Hence, it has different values of stiffness along the grain and across the grain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is an expression for the natural frequencies of a circular orthotropic beam, I want to be able to show that if $E_x$ and $E_y$ are equal, the expression should reduce to the isotropic case. &lt;/p&gt;&#xA;" OwnerDisplayName="user16281" LastActivityDate="2018-09-06T14:24:38.850" Title="Natural frequencies of an orthotropic bending beam with circular cross-section" Tags="|mechanical-engineering|beam|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21996" PostTypeId="2" ParentId="8279" CreationDate="2018-06-01T16:33:40.367" Score="1" Body="&lt;p&gt;Measure the water and distilling container separately and together. Subtract the added weight left behind in the distiller after distilling all the water out. You could count the hydrocarbons in a drop of water under a microscope and multiply. Get a microscope.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-06-01T16:33:40.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21997" PostTypeId="1" AcceptedAnswerId="22000" CreationDate="2018-06-01T18:23:53.010" Score="2" ViewCount="135" Body="&lt;p&gt;I have a question on the Bauschinger effect. Let's say that we have a material with yield stress $400 \,\mathrm{MPa}$, and let's say that the actual applied stress is tensile for starters. If we load the material to the point where the stress is, for example, $500 \,\mathrm{MPa}$, then that means that the loading has surpassed the material's yield stress by $100 \,\mathrm{MPa}$ and that the material has entered the inelastic region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the direction of stress is now reversed, from tensile to compressive, the Bauschinger effect tells us that the new compressive yield stress will be lower than the initial one and will be $-400+100=-300 \,\mathrm{MPa}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to know is: What happens if in my initial loading of the material I don't surpass the yield strength? In other words, if yielding occurs at $400 \,\mathrm{MPa}$, I apply less than that ($350 \,\mathrm{MPa}$, for example). If I now unload and start loading with compressive stress, does the Bauschinger effect apply? Or is the yield stress for compressive loading still $-400 \,\mathrm{MPa}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&#xA;Note: I don't know what tags to add, so if you know of more suitable ones, please add them.&lt;/p&gt;&#xA;" OwnerUserId="16282" LastEditorUserId="10123" LastEditDate="2018-06-03T21:06:19.790" LastActivityDate="2018-06-03T21:06:19.790" Title="The Bauschinger effect" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="21999" PostTypeId="1" AcceptedAnswerId="22003" CreationDate="2018-06-01T19:23:54.797" Score="2" ViewCount="591" Body="&lt;p&gt;I'm using two stepper motors (datasheet below) to pan and tilt a platform. The bottom motor spins fine without the breadboard glued to the top platform. However, it behaves unpredictably and skips steps when the breadboard is glued in place (shown below). My question is why does this happen? Is it related to rotor inertia? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3ejdB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3ejdB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2kI5v.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2kI5v.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15061" LastEditorUserId="7083" LastEditDate="2018-06-26T01:21:27.873" LastActivityDate="2018-06-26T01:21:27.873" Title="Stepper motor skipping steps" Tags="|motors|torque|stepper-motor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22000" PostTypeId="2" ParentId="21997" CreationDate="2018-06-01T19:41:34.140" Score="1" Body="&lt;p&gt;Bauschinger effect is based on plastic strain , so there would be no effect for stresses in the elastic strain ranges. I learned this from Drs N.H.Polakowski and S  Mostovoy before there were computers so no tags. However; I am sure they have references in the literature.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-06-01T19:41:34.140" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22001" PostTypeId="1" AcceptedAnswerId="22002" CreationDate="2018-06-01T22:19:38.663" Score="2" ViewCount="34" Body="&lt;p&gt;I have been designing a prototype for a keychain that sprays perfume. I have a metal keychain piece which has a female thread and the plastic sprayer screws right into it and has a male thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to get these manufacturers ASAP but I need to be ensured it wont leak. I think I will switch my straight thread to a tapered thread, will that help? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can I do to ensure that it will not leak? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!!&lt;/p&gt;&#xA;" OwnerUserId="16285" LastActivityDate="2018-06-01T23:15:50.610" Title="making a leak proof thread for my keychain idea" Tags="|threads|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22002" PostTypeId="2" ParentId="22001" CreationDate="2018-06-01T23:15:50.610" Score="1" Body="&lt;p&gt;tapered threads are intended to be leakproof, but the degree to which they resist leaks depends on how tightly they are screwed together. This generally requires the use of a tool of some sort, as it is usually not possible to tighten a tapered-thread joint enough by hand to make it stop leaking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason you should instead use straight threads and a gasket like a soft O-ring which is trapped between the two pieces when they are screwed together. Unlike the tapered thread joint, these can be tightened enough by hand to be leak-proof. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-01T23:15:50.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22003" PostTypeId="2" ParentId="21999" CreationDate="2018-06-01T23:25:03.447" Score="1" Body="&lt;p&gt;I assume the bottom stepper shaft is affixed directly to the clear plastic platform that supports the top motor and the breadboard, is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, there are two possible causes of your problem. first, the rotary inertia of the assembly consisting of the top motor and the PC board on its lever arm may be too much for the bottom stepper to rotate at the speed with which you are sending it pulses. try slowing down the pulse rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;second, the combination of board + support arm + top motor may have a torsional resonance close to the step signal frequency. trying to rotate the assembly with a string of discrete pulses may then excite the resonance and kick the torque impulses back to the bottom motor, causing it to jump between cog positions and lose steps. try hitting the bottom motor with one, two, three, ... pulses in a string and watch the assembly closely to see if it has been set into vibration by the end of the pulse string. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-01T23:25:03.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22005" PostTypeId="2" ParentId="19422" CreationDate="2018-06-02T01:55:43.387" Score="0" Body="&lt;p&gt;The average magnetization exiting the tube in steady state should just be the product not the convolution.&#xA;$$\int_0^\infty E(t)\, E2(t) \, dt$$&#xA;$$\int_{\frac{\tau}2}^\infty \left(1-e^{-t\frac{t}{k}}\right)\, \frac{\tau^2}{2\,t^3} \, dt$$&#xA;$$-\frac{\tau^2 Ei\left(-\frac{\tau}{2 k}\right)}{4 k^2} - \frac{\tau \, e^{-\frac{\tau}{2 k}}}{2 k} + e^{-\frac{\tau}{2 k}}-1$$&#xA;Where $Ei$ is the &lt;a href=&quot;http://mathworld.wolfram.com/ExponentialIntegral.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Exponential Integral&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks like it comes out to a bit less than your CFD results:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/leK16.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/leK16.png&quot; alt=&quot;Plot of fraction of flow magnetized&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The x axis in this case is the average residence time which should be proportional to the length along the tube, making our plots comparable.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2018-06-02T01:55:43.387" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22007" PostTypeId="1" AcceptedAnswerId="22009" CreationDate="2018-06-02T05:52:07.553" Score="2" ViewCount="55" Body="&lt;p&gt;What is symmetric?Please explain geometrically and in terms of displacements. I have encountered this while looking up for engineering strain and true strain values have totally different compression and expansion parameters for their respective strains.&lt;/p&gt;&#xA;" OwnerUserId="12558" LastActivityDate="2018-06-02T08:38:24.290" Title="What is symmetric?" Tags="|mechanical-engineering|structures|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22009" PostTypeId="2" ParentId="22007" CreationDate="2018-06-02T08:38:24.290" Score="2" Body="&lt;p&gt;My guess would be that talking about symmetry in the context of material science, the meaning of &quot;symmetry&quot; is probably describing direction-dependant properties of a material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel for example is an isotropic material (excluding drawn/stretched steel), meaning that its properties are not dependent on the direction in which you measure them.&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note, this does &lt;strong&gt;not&lt;/strong&gt; mean that compressive and tensile value must be the same. It basically means that if you have for material sample, you will measure the same properties, whether you measure them in $x$, $y$ or $z$ direction.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, wood, reinforced concrete and other composite materials are usually not isotropic. Wood for example has a much higher strength when loaded in axial direction, compared to a radial or tangential load.&lt;br&gt;&#xA;This makes wood an material with orthotropic symmetry, because its properties mainly vary along the three orthogonal directions (ax/rad/tang).&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a formal explanation on the whole subject on &lt;a href=&quot;https://en.wikipedia.org/wiki/Orthotropic_material#Condition_for_material_symmetry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, but it requires some understanding of linear algebra.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2018-06-02T08:38:24.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22010" PostTypeId="1" AcceptedAnswerId="22013" CreationDate="2018-06-02T10:23:26.343" Score="3" ViewCount="113" Body="&lt;p&gt;What are the advantages and disadvantages (technical, financial etc.) of having battery swapping system over charging stations for electric cars? As per my reading, battery swapping seems feasible for electric buses and 2-wheelers but not for cars. Is that really the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it possible for the battery to be housed under the bonnet instead of having it under the floor of the car so as to do away with underground swapping (like the one showcased by Tesla)? I know that the battery is placed there so as to lower the center of gravity. However, if there aren't any other significant benefits, can the aforementioned be done and battery swapping be made easier?&lt;/p&gt;&#xA;" OwnerDisplayName="user16060" LastActivityDate="2018-06-02T16:30:57.047" Title="Battery Swapping in Electric Cars" Tags="|mechanical-engineering|electrical-engineering|battery|electric-vehicles|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22012" PostTypeId="1" CreationDate="2018-06-02T16:04:24.137" Score="2" ViewCount="1923" Body="&lt;p&gt;&lt;strong&gt;First disclaimer&lt;/strong&gt;: I have a background in math, but I'm no expert in engineering or physics. This is the first question I've ever posted and I think it belongs here, but please let me know if belongs somewhere else and I will be happy to move it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second disclaimer&lt;/strong&gt;: Realistically I am aware that the maximum acceleration from rest (or from any given vehicle speed, for that matter) is limited by the force of static friction between the tires and the road, but for this question I'd like to assume &quot;infinite traction&quot; just as a thought experiment. This is mostly a theoretical question anyway. So, here goes:&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BACKGROUND&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I love cars and I love math so I'm working on a very simple model of vehicle dynamics for fun (at this point it is just modeling basic kinematics). Now &lt;strong&gt;please excuse the wordiness, but I'm going to explicitly go through the key points in my thought process leading up to where I'm stuck so that you can correct any inaccurate assumptions of mine&lt;/strong&gt;...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate an acceleration, we need a force and we need a mass. We can easily calculate the mass of the car by looking up its weight, but what about the force? Well, let's say the engine produces 267 lb*ft of torque at 6000 RPM and 328 HP at 7000 RPM. We can use the torque to calculate the force at the driven wheels, $F_{tract}$, but since this is the force at the wheels, we need to know the torque at the wheels, not the torque at the engine crankshaft. Therefore, one could argue that the quoted torque figure is meaningless, since it doesn't tell us anything useful unless we also know how the car is geared and, moreover, engine torque is going to change as a function of engine RPM. Theoretically, then, let's suppose the engine's torque curve is roughly parabolic and we (maybe) know how the car is geared. Ignoring drag for now, we can calculate the acceleration based on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{car}=\frac{F_{tract}}{m_{car}}=\frac{\tau_{wheel}}{r_{wheel}m_{car}}=\frac{\tau_{eng}\gamma_{trans}\gamma_{FD}}{r_{wheel}m_{car}}=\frac{\tau_{eng}\omega_{eng}\omega_{trans}}{\omega_{trans}\omega_{wheel}r_{wheel}m_{car}}=\frac{\tau_{eng}\omega_{eng}}{v_{car}m_{car}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\gamma_{trans}$ and $\gamma_{FD}$ are the transmission and final drive gearing ratios, respectively. We want to maximize wheel torque to maximize acceleration, so intuitively we would like a lot of engine torque and a large transmission gear ratio. There is a limit, though, to how high we can feasibly make the gear ratio because, while it amplifies the torque to the wheels, it also means that the engine is spinning faster for any given vehicle speed. The engine can't rev above its redline and, moreover, for a given vehicle speed, we see from the above equation that maximizing wheel torque equates to maximizing the product of engine torque and engine rotational speed, which might not necessarily be maximized at the engine's redline anyway if the engine torque drops off enough from the peak after a certain engine RPM. So how do we maximize this quantity? Well, luckily this quantity is exactly equal to the definition of power, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a_{car}=\frac{\tau_{eng}\omega_{eng}}{v_{car}m_{car}}=\frac{P}{v_{car}m_{car}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, to maximize acceleration at any given speed, we need to maximize engine power (which is the same as wheel power, ignoring losses due to friction). This result has important theoretical implications. First, it proves that, although the peak engine torque number is essentially meaningless, the peak engine power number is actually enough to, theoretically, calculate the maximum possible acceleration the car can achieve at any given speed. So it sort of ends the age-old &quot;power vs. torque&quot; debate. Second, in a car with a stepped transmission, it tells you exactly when to shift to maximize acceleration. You want to shift when the power (or, equivalently, wheel torque) in the next gear up equals that in the current gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, the equation provides some insight into an interesting question, which really starts to get at the point where I become confused: when does my car accelerate the fastest? Well, let's suppose it has a stepped transmission. You can simulate the wheel torque curves for each gear as functions of vehicle speed, and you'll find that the highest value for wheel torque and hence, acceleration, occurs at the point where the engine is making maximum torque in first gear. But, wait! I thought the peak engine torque number was meaningless and only the peak power number mattered in terms of acceleration! Well, sort of. Before we were talking about theoretical maximum acceleration, and now we are talking about actual (more or less) acceleration, which is limited by the available gear ratios. Within each gear, maximum acceleration occurs when the engine produces peak torque, because this is when maximum wheel torque is realized in that gear. The key is that before we assumed the gear ratio and, hence, the engine rotational speed, could vary for any given vehicle speed, so we could play with both quantities to maximize their product (i.e. power). Now everything is fixed except for engine torque, so that's all we can maximize. When the engine reaches its maximum power output within this same gear, it is now accelerating the car as fast as theoretically possible at that new given vehicle speed. So, in other words, technically the car accelerates fastest in each gear at maximum engine torque, but if it isn't simultaneously making maximum power, this implies that it &lt;em&gt;could be accelerating faster if&lt;/em&gt; it were geared differently such that the wheel torque (i.e. power) were maximized. A CVT can be designed to constantly adjust the gear ratio such that, at any given vehicle speed, wheel torque (i.e. power) and, thus, acceleration, is always maximized.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MY CONFUSION/QUESTIONS&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;AT LAST&lt;/strong&gt; we get to my point of confusion. This is all well and good, and my simulations work just fine as long as I start my car at idle engine RPM in first gear (i.e. idle vehicle speed). The problem is that my grasp on the theory breaks down when I start the car from rest. Technically if the car is at rest ($v=0$) and is physically connected to the engine, the engine must also not be rotating $(RPM_{eng}=0)$. But that's why manual transmission cars have clutches and automatic transmission cars have transmission fluid (I think that is what an automatic car uses to ensure that the engine can spin while the wheels do not, right?) So if I have the brake on in my automatic car or the clutch in in my manual car, how can I calculate the maximum acceleration from rest? If I appeal to the maximum acceleration in terms of power equation (i.e. what would a CVT do?) it's not defined for $v=0$. The maximum acceleration grows arbitrarily large for smaller and smaller vehicle speed, presumably because you'd need an infinitely high gear ratio to amplify the zero torque the engine produces at 0 RPM in order to make the product equal maximum engine power. I guess that makes sense, but this equation only works when the engine and wheels are physically coupled, right? If I have the clutch in / brake on, there has to be an optimal RPM to rev the engine up to before transferring the power to the wheels to achieve maximum acceleration, but I don't know how to find it. Intuitively it's either going to be where the engine makes maximum torque or where it makes maximum power, but I don't know how to figure out which one it should be. It seems that maximum power is always the answer, but if the transmission is stepped then in first gear it makes the most wheel torque when the engine makes the most torque, so wouldn't I launch from here? But then instantaneously I would want the engine to produce maximum power again, not maximum torque?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the meat of my theoretical question, but I also have a more practical question about how to implement this into my model. Let's say I've decided what RPM to launch at. When I dump the clutch, if I assume I have traction, do I assume the wheels instantaneously match that speed and continue with the simulation? This doesn't seem right. How could I model the transfer of speeds (preferably in the simplest way)? I know that, in reality, when you dump the clutch the engine RPMs drop as, I assume, the inertia of the system increases massively and then the wheel and engine rotational speeds &quot;synchronize.&quot; &#xA;This was a long question and I apologize. If anyone actually takes the time to read through it and provide some insight I would really appreciate it. I love thinking about this stuff but I'm stuck on this issue and could really use the help! &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SUMMARY/tl;dr&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, maximum acceleration of a car for a given vehicle speed is achieved by maximizing engine power. This is the principle a CVT uses to maximize acceleration. However, what about the case when $v=0$? The equation for acceleration is undefined, so how can I calculate the maximum theoretical acceleration from rest, assuming infinite traction and given the engine's torque curve and the car's mass? What if I have a stepped transmission and I know the first gear ratio? Do I want to launch the car at the RPM where engine torque is highest, or where engine power is highest?&lt;/p&gt;&#xA;" OwnerUserId="16286" LastEditorUserId="16286" LastEditDate="2018-06-02T22:11:45.647" LastActivityDate="2018-06-04T12:38:48.883" Title="Calculate maximum acceleration of a car from rest (optimum launch)" Tags="|mechanical-engineering|automotive-engineering|torque|power|acceleration|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="22013" PostTypeId="2" ParentId="22010" CreationDate="2018-06-02T16:30:57.047" Score="1" Body="&lt;p&gt;This idea has been suggested often, but there's way more to it than you'd think.&#xA;One big problem is that batteries are heavy, and there are multiple problems arising because of this. The main thing being the packaging in the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want a decent range in a car, so the battery will be large and heavy. If you put it in the front, the car will be front heavy, and it'll be understeered. The front will break out in the corner. Same problem when putting it in the back, it'll be oversteered, which can be even more dangerous for ignorant drivers. You'll notice that a disturbed weight distribution causes bad driving behaviour in general, from a car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Placing it too high will also cause problems, the car will dive under braking and accelerating, and it'll roll in the corners. Long story short, batteries are annoyingly big and heavy and awkaward things we have to get stuffed inside a car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you make people use batteries on exchange basis, you can only have a few standard battery sizes and configurations, or it'll be unfeasible. This means the battery will never be optimal for a specific car. And this is why exchange batteries aren't a thing. (yet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Others often mention that different batteries will have a different quality, but I don't think this really matters. If you you have a lease contract, or if you pay per kilometer, the battery quality doesn't matter anymore. Bad ones will just be taken outof the cylce and be recycled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be feasible if only a few different cars were used on the road, so the number of batteries exchange stations need to have in stock is also limited. Or, we'll have to wait for batterytechnology to advance a lot, so we can have a decent range with small and light batteries. Then the packaging would be less critical and standard battery sizes could be a solution. Until then, i don't think we'll see exhange batteries being offered at petrolstations.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-06-02T16:30:57.047" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22015" PostTypeId="2" ParentId="22012" CreationDate="2018-06-02T16:44:06.893" Score="0" Body="&lt;p&gt;Not really an answer but too much for a comment.  I think you are making this harder than it needs to be.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you start at like 200 rpm and shift at redline. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think max acceleration will actually be at max horse power (not max torque).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque and horsepower are both functions of rpm so this becomes an integral equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to get force is the question as neither torque or horsepower are the correct units.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to account for air friction.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-06-02T18:31:13.187" LastActivityDate="2018-06-02T18:31:13.187" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="22016" PostTypeId="2" ParentId="20956" CreationDate="2018-06-02T18:07:38.263" Score="2" Body="&lt;p&gt;To estimate the frictional losses you can use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Darcy–Weisbach equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h_f = f_D \frac{L}{D} \frac{V^2}{2\,g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This head-loss ($h_f$) can be added to Bernoulli's equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h - h_f = \frac{V^2}{2\,g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the length ($L$) is the same as your height ($h$) so these equations combine like so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$h - f_D \frac{h}{D} \frac{V^2}{2\,g} = \frac{V^2}{2\,g}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V=\sqrt{\frac{2\,g\,h}{1+f_D\frac{h}D}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the friction factor ($f_D$) you need the Reynolds number:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Re=\frac{V\,D}{\nu}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you can look it up on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Moody_chart&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moody Chart&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging our equation for $V$ into our Reynolds number equations we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Re = \frac{D}{\nu}\sqrt{\frac{2\,g\,h}{1+f_D\frac{h}D}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Re=\frac{70mm}{1.0035\frac{m^2}s}\sqrt{\frac{2 \bullet 9.8\frac{m}{s^2} \bullet 3 m}{1+f_d\frac{3m}{70mm}}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Re=\frac{81707}{\sqrt{0.0233+f_d}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need two more things to use the Moody Chart, a roughness value ($1.3 \mu m$ new to $30\mu m$ used) and an initial guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets pick an initial guess of $Re= 10^5$. Looking at the Moody Chart that would give us a friction factor of about 0.023 for a new pipe.&#xA;$$Re_{new}=\frac{81707}{\sqrt{0.0233+0.023}}\approx 380000$$&#xA;Looks like we were too low. So when we look up the friction factor for the new Reynolds number we get about 0.0205.&#xA;$$Re_{new}=\frac{81707}{\sqrt{0.0233+0.0205}}\approx 390000$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is about as much accuracy as you will get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we can solve for flow rate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q=\frac{\pi}4 \nu \, D\,Re \approx 1300 \frac{L}{min}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeating the procedure for the rougher pipe yields:&#xA;$$960 \frac{L}{min}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One big warning of this analysis however, is that many pipes can get debris in them severely limiting their flow capacity. Rain gutters are typically as large as they are, not just to pass high volumes of water, but to allow debris to pass through them.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2018-06-02T18:07:38.263" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22017" PostTypeId="2" ParentId="19568" CreationDate="2018-06-02T18:13:29.730" Score="0" Body="&lt;p&gt;You're equations of motion are correct. Your book is either wrong, or is making the assumption that $m_1 \gg m_2$ in which case the terms contributing from the interaction with $m_2$ would be negligible.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2018-06-02T18:13:29.730" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22018" PostTypeId="1" CreationDate="2018-06-02T19:15:10.717" Score="2" ViewCount="146" Body="&lt;p&gt;It seems to me that pump vibration isolators need to support:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The weight of the pump&lt;/li&gt;&#xA;&lt;li&gt;The weight of the water in the pump&lt;/li&gt;&#xA;&lt;li&gt;The weight of the water (only) in the drop to the pump&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The supplier of vibration isolators on my project does not seem to share this view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that pump manufactures want none of the pipe weight to bear on their connections. The pipe hangers can be adjusted to prevent that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A braided flex connector can prevent the thrust force due to pressure in the pipe from being an issue at the vibration isolators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the water weight in the drop to the pump seems to have no support other than what is provided by bearing on the bottom of the pump’s volute. This weight is transferred to the vibration isolators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram below (considering only on the discharge side of the pump to explore the concept), the vibration isolators need to support 8 feet of water in the 14” diameter drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water weight is 59.75 lb/ft in the chart here:&#xA;&lt;a href=&quot;https://www.engineersedge.com/calculators/std_pipe_weights/std_pipe_weights.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pipe weights per foot&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the water weight (due to only the discharge pipe) on the vibration isolators seems to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(59.75 lb/ft) * (8 feet) = 478 lbs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? Or are the vibration isolators not subjected to the water-weight in the drop for some reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lXJyZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lXJyZ.png&quot; alt=&quot;Centrifugal pump on vibration isolators&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16293" LastActivityDate="2018-06-02T19:15:10.717" Title="Determining the load on a pump vibration isolator" Tags="|mechanical-engineering|statics|vibration|hvac|deflection|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22020" PostTypeId="1" AcceptedAnswerId="22026" CreationDate="2018-06-02T22:21:05.890" Score="4" ViewCount="276" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Dear engineers,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have to make an engineering project which is a portable solar phone charger. One thing I need to do in it is to somehow charge the phone (android) from the battery (which is charged by solar panels) and simultaneously send data about the battery level to the phone through the same usb connection. Is this even possible to do? If so, could you please advise a way to do it? Would arduino be helpful?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that usb has separate lines for data and power, but I don’t know how a phone can receive this data and whether it will work while charging. I was thinking about using arduino, but it doesn’t seem to be well suited for charging a phone (maximum current output is rather low). Would it work if I made separate circuits, one for charging a phone and another for reading the data from voltmeter and sending it to the phone through the other pins on usb?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m sorry for such a tremendous question, I just don’t have much experience with electronics (not even a year at college), and I must somehow do all this stuff. &lt;/p&gt;&#xA;" OwnerUserId="16295" LastEditorUserId="16295" LastEditDate="2018-06-02T22:28:25.857" LastActivityDate="2018-06-03T11:17:03.177" Title="Simultaneously charging and transmitting data to an android phone using usb" Tags="|electrical-engineering|power-electronics|circuits|battery|circuit-design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22021" PostTypeId="2" ParentId="12511" CreationDate="2018-06-02T22:58:10.223" Score="2" Body="&lt;p&gt;I don't know if you're still interested in this topic, but it sort of relates to a question I had about modeling acceleration. I might be able to clear up some confusion, though, because I had the same question when I was learning how to build a similar model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It turns out that you don't need to know &quot;how fast the engine revs up&quot; while the car is accelerating at maximum capacity -- at least, you don't need to know it directly; it sort of &quot;falls out&quot; of the simulation.  The key is that as long as the engine and wheels are physically connected, all you need is an initial condition: either the initial speed of the car or the initial RPM of the engine in a particular gear. The two are &quot;linked&quot; via the gear ratios of the car, so you can always use one to calculate the other, then check to see what the acceleration is at that speed, then update one, then figure out the other, etc. until you decide to shift.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic procedure for a simple computer simulation goes like this:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Start in first gear at some vehicle speed, $v_{car}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Figure out what the engine speed is for these conditions using: $RPM_{eng}=\frac{v_{car}*\gamma_{FD}*\gamma_{trans}*(5280ft/mi)}{2*(60min/hr)*\pi * r_{wheel}}$ where I am using $\gamma_{FD}$ and $\gamma_{trans}$ to represent the final drive and transmission gearing ratios, and I'm assuming the vehicle speed is in MPH and the radius of the tire is in feet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Now you need to look at the engine's torque curve. Figure out, based on the curve, what the engine torque, $\tau_{eng}$ is at the current $RPM_{eng}$.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(4) Use the engine torque and the current gear to calculate the force at the wheels and, thus, the acceleration: $a_{car}=\frac{\tau_{eng}*\gamma_{trans}*\gamma_{FD}}{r_{wheel}*m_{car}}$ where I've left out the conversion factors this time because I don't know how you want to express the acceleration (if torque is in lb*ft then this formula would give you acceleration in terms of pounds per whatever units you used for mass).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(5) Now just use a small timestep $dt$ and Euler's method of numerical integration to calculate the new vehicle speed: $v_{car}=v_{car}+a_{car}*dt$. It doesn't matter what $dt$ is exactly; smaller values will provide a slightly more accurate theoretical simulation but the code will also run slower. I think I used 0.005 seconds. Just make sure all the units agree -- that's the easiest way to mess up! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No go back to step (2) and repeat until you decide to shift (or the engine hits the redline). For maximum acceleration you want to shift exactly when the wheel torque (or, equivalently, engine power) provided at the same vehicle speed in the next gear up exactly equals that in the current gear. Hope this helps! &lt;/p&gt;&#xA;" OwnerUserId="16286" LastActivityDate="2018-06-02T22:58:10.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22022" PostTypeId="1" CreationDate="2018-06-02T23:53:33.983" Score="2" ViewCount="35" Body="&lt;p&gt;My question refers to a typical bog standard normally-aspirated 4-cylinder petrol/gasoline engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say it's a 2.0 litre engine (10:1 compression ratio, firing order 1-3-4-2) operating at 2000 rpm with 30% throttle open. Ambient conditions are 1000 millibars of pressure at 12 degrees C (285 degrees K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When cylinder 1 is on its compression stroke (increasing pressure), cylinder 2 is on its power/expansion stroke (decreasing pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, for most of the period in question, the pressure in cylinder 2 is higher than that in cylinder 1, but there is a cross-over point at which the pressures in both cylinders are the same. My question is, at what crank angle should I expect to find this equality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If necessary, and for simplicity, assume that 100% of the petrol/gasoline has fully combusted into CO2 &amp;amp; H2O, but if you'd like to assume more realistic figures please do so. I doubt that it would affect the equality point too much though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the equal point is obviously fairly late in cylinder 1's compression stroke, inlet valve timing is probably irrelevant. The exhaust valve timing though might be relevant, so please assume that it (they) open after the equal pressure point.&lt;/p&gt;&#xA;" OwnerUserId="16296" LastEditorUserId="16296" LastEditDate="2018-06-03T02:06:06.623" LastActivityDate="2018-06-03T02:06:06.623" Title="When is the Equal Pressure Point in a Typical ICE?" Tags="|automotive-engineering|vapor-pressure|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22024" PostTypeId="2" ParentId="21950" CreationDate="2018-06-03T10:29:46.713" Score="0" Body="&lt;p&gt;The easiest way is to fit a linear function a + b x + c y. To find a, b and c solve the system of equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a + b&lt;em&gt;x0 + c&lt;/em&gt;y0 = phi0&#xA;a + b&lt;em&gt;x1 + c&lt;/em&gt;y1 = phi1&#xA;a + b&lt;em&gt;x2 + c&lt;/em&gt;y2 = phi2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If (u1,u2) is the unit vector in the direction u, the the directional derivative is: b&lt;em&gt;u1 + c&lt;/em&gt;u2&lt;/p&gt;&#xA;" OwnerUserId="16299" LastActivityDate="2018-06-03T10:29:46.713" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22025" PostTypeId="2" ParentId="21882" CreationDate="2018-06-03T10:42:56.973" Score="2" Body="&lt;p&gt;I think this question is ill posed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both pipes are flowing full, the correct answer is 62.5 m/s from continuity.&#xA;If there is an abrupt transition from the large to the small pipe, there will be a total pressure loss and Bernoulli's equation will not apply without a loss factor.&lt;/p&gt;&#xA;" OwnerUserId="16299" LastActivityDate="2018-06-03T10:42:56.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22026" PostTypeId="2" ParentId="22020" CreationDate="2018-06-03T11:05:38.813" Score="0" Body="&lt;p&gt;It is assume that the PV panel is charging a battery such as a LiPo or a Li-Ion battery Cell. A battery Gas gauge circuit such a &lt;a href=&quot;http://www.st.com/content/ccc/resource/technical/document/datasheet/ea/b5/01/6e/dd/f0/49/3e/DM00105047.pdf/files/DM00105047.pdf/jcr:content/translations/en.DM00105047.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;STC3117 (Gas gauge IC with battery charger control for handheld applications)&lt;/a&gt; and micro-controller. To use a STC3117, micro controller with I2C such a Arduino DIY platform is suitable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1DJ11.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1DJ11.jpg&quot; alt=&quot;STC3117 I2C&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cFJrb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cFJrb.jpg&quot; alt=&quot;Arduino I2C pins&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are correct USB has a data lines which can be used to communicate the status of the portable solar charger. Upon connecting the phone an application on the phone could read the status of the mobile solar charger display the status. This link on &lt;a href=&quot;https://www.allaboutcircuits.com/projects/communicate-with-your-arduino-through-android/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Communicate with Your Arduino Through Android&lt;/a&gt; should provide a starting point to communicate to a phone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this will help you get started.&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-06-03T11:17:03.177" LastActivityDate="2018-06-03T11:17:03.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22028" PostTypeId="1" AcceptedAnswerId="22032" CreationDate="2018-06-03T11:48:56.277" Score="1" ViewCount="65" Body="&lt;p&gt;Based on the following publication:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Kouro, S., Perez, M.A., Rodriguez, J., Llor, A.M. and Young, H.A., 2015. Model predictive control: MPC's role in the evolution of power electronics. IEEE Industrial Electronics Magazine, 9(4), pp.8-21.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Available &lt;a href=&quot;https://www.researchgate.net/publication/287796890_Model_Predictive_Control_MPC%27s_Role_in_the_Evolution_of_Power_Electronics?enrichId=rgreq-963aef38d7d14b76effad0c8f661a507-XXX&amp;amp;enrichSource=Y292ZXJQYWdlOzI4Nzc5Njg5MDtBUzozNzgyOTIwNzYwNzI5NjBAMTQ2NzIwMzI2NzM4MA%3D%3D&amp;amp;el=1_x_3&amp;amp;_esc=publicationCoverPdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, it is said that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;MPC can be also used with &lt;strong&gt;nonminimum&lt;/strong&gt; phase plants, which usually appear in rectifiers connected to the grid, avoiding &lt;strong&gt;linear&lt;/strong&gt; controllers, which, in these cases, have a very restricted performance [76], [79], [83].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Dose it mean that linear controllers are not adequate for nonminimum phase systems? If yes, why?&lt;/p&gt;&#xA;" OwnerUserId="16032" LastEditorUserId="110" LastEditDate="2018-06-03T16:37:51.990" LastActivityDate="2018-06-03T20:18:55.860" Title="Are linear controllers inadequate for nonminimum phase systems?" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22029" PostTypeId="2" ParentId="14840" CreationDate="2018-06-03T16:31:36.743" Score="0" Body="&lt;p&gt;Using Matlab R2018a with minreal function from Robust control package and inputting original A, B, C, D matrices I got: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{array}{c|c}&#xA;A^*_m &amp;amp; B^*_m \\ \hline C^*_m&#xA;\end{array} = &#xA;\begin{array}{cc|cc}&#xA;1 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 \\&#xA;\sqrt2 &amp;amp; 1 &amp;amp; 0 &amp;amp; \sqrt2 \\&#xA;\hline&#xA;1 &amp;amp; \sqrt2&#xA;\end{array}.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One state was removed.&lt;/p&gt;&#xA;" OwnerUserId="5393" LastActivityDate="2018-06-03T16:31:36.743" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22031" PostTypeId="1" CreationDate="2018-06-03T17:40:14.867" Score="1" ViewCount="2433" Body="&lt;p&gt;I used to calculate the power needed by a pump using Bernoulli equation,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ W =(p_2-p_1) + {1 \over 2}\rho(u_2^2-u_1^2) + g\rho(h_2-h_1) + R $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ is fluid density (and assuming it is constant), &lt;em&gt;subscript&lt;/em&gt; &lt;em&gt;1&lt;/em&gt; is inlet, &lt;em&gt;2&lt;/em&gt; is outlet, $p$ is pressure, $u$ is speed, g is acceleration of gravity, $h$ is and $R$ are pressure drops (see &lt;strong&gt;EDIT2&lt;/strong&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this formula fluid properties are represented by $\rho$. What will happen if I change the flowing fluid? e.g. water for &lt;a href=&quot;https://wiki.anton-paar.com/en/engine-oil/&quot; rel=&quot;nofollow noreferrer&quot;&gt;mineral oil&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With water, I need more power than to flow the same amount of mineral oil, due to its &lt;a href=&quot;https://wiki.anton-paar.com/en/engine-oil/&quot; rel=&quot;nofollow noreferrer&quot;&gt;lower density&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How this assumption could be right/complete!?&lt;/strong&gt; Since does not appear the property of viscosity of the fluid anywhere and &lt;a href=&quot;https://en.wikipedia.org/wiki/Viscosity&quot; rel=&quot;nofollow noreferrer&quot;&gt;that's&lt;/a&gt; just the resistance of a fluid to the deformation/movement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT1:&lt;/strong&gt; fixed the Bernoulli formula; add $\rho$ assumption;&lt;br&gt;&#xA;&lt;strong&gt;EDIT2:&lt;/strong&gt; clarify $ R$ term:  &lt;em&gt;since we are talking about a pump, losses are lumped, not distributed, we are not talking about losses in a pipe, thus they are not Reynolds dependants. Thus $R$ does not depend on fluid properties but relies only on inlet/outlet pump geometry&lt;/em&gt;;&lt;/p&gt;&#xA;" OwnerUserId="2157" LastEditorUserId="2157" LastEditDate="2018-06-03T20:58:12.910" LastActivityDate="2022-09-12T12:13:22.703" Title="Pump power and fluid viscosity" Tags="|fluid-mechanics|pumps|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="22032" PostTypeId="2" ParentId="22028" CreationDate="2018-06-03T20:18:55.860" Score="3" Body="&lt;p&gt;A LTI system is nonminimum phase if it has one or more zeros in the right half plane. When using a LTI controller for feedback also then the bandwidth of this controller will be limited to roughly to smallest magnitude of zeros in the right half plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because a right half plane zero gives a phase drop and slope increase. But near the bandwidth a slope of minus one (time 20 dB per decade) with a phase of -90° is desired. A minimum phase zero, so located in the left hand plane, could be counteracted by a left hand plane pole in order to ensure the desired asymptote at the bandwidth. If this could also be done for a nonminimum phase zero, in which case a right half plane pole would need to be added. According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nyquist_stability_criterion#The_Nyquist_criterion&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nyquist criterion&lt;/a&gt; this would then require a counterclockwise encirclement of the minus one point of the openloop Nyquist diagram. The magnitude and phase of such counterclockwise encirclement look very similar to a that of a right half plane pole and zero, however when encircling the minus one point you already cross the zero dB line and therefore also give an upper bound for the bandwidth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no, a nonminimum phase LTI system can be controller using a LTI controller for feedback, but the bandwidth will be limited.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-06-03T20:18:55.860" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22033" PostTypeId="2" ParentId="22012" CreationDate="2018-06-03T22:30:44.457" Score="1" Body="&lt;p&gt;The first thing you need to note is that when your engine is rotating freely, i.e. your clutch is disengaged, then it &quot;produces&quot; little to no torque, regardless of the current RPM. The torque rating of the engine at a certain RPM is the torque you need to apply to the output shaft to bring it to an instantaneous standstill. Thus, to mobilise the torque, you need something to resist it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, with an instantaneous clutch and infinite traction, if you drop the clutch the engine will come to a standstill immediately and stall, and the car will barely move. You'd have the full torque of the engine but only for a split second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine hitting a punching bag with a baseball bat as hard as you can. It will barely move, but with the same force gradually applied you can move it further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second thing to note is that power is the rate at which the engine is generating energy, measured in Watt. 1 Watt is 1 joule per second, and a Joule is the energy needed to move 1 newton a distance of 1 meter (Nm, but not to be confused with torque, which is just a rotational force).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although your peak power is at a point where the torque is below peak, you are adding energy into the system more quickly, thus &quot;distance increases more quickly&quot; i.e. you accelerate faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine hitting the punching bag repeatedly with full force; it will move further and faster with each hit. Now imagine decreasing the force a bit, but hit it more quickly, you'll burn more calories and move the bag quicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you launch your car, you'd like to keep the engine at max power. So start at the RPM for peak power and as you release the clutch, you'll want to increase the gas to compensate for the drop in RPM. You want to do this as quickly as possible without having your wheels spin, so practice by doing this quicker with every launch unit your wheels spin, then turn it down a notch. Obviously the surfacing of the road will also have a big impact on how quickly you can do this. &lt;/p&gt;&#xA;" OwnerUserId="14440" LastEditorUserId="14440" LastEditDate="2018-06-04T12:38:48.883" LastActivityDate="2018-06-04T12:38:48.883" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22034" PostTypeId="1" AcceptedAnswerId="22052" CreationDate="2018-06-04T00:23:45.437" Score="3" ViewCount="56" Body="&lt;p&gt;I recently installed a VFD to reduce the speed of my pump, however, occasionally the configuration will stop pumping liquid through the system. The pump is a three phase 60 Hz and this issue has happened when running the system at 13 Hz and 40 Hz. Any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="4674" LastActivityDate="2018-06-04T15:44:31.493" Title="Why does my VFD and pump configuration stop pumping?" Tags="|mechanical-engineering|electrical-engineering|pumps|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22035" PostTypeId="1" AcceptedAnswerId="22039" CreationDate="2018-06-04T01:14:43.557" Score="1" ViewCount="153" Body="&lt;p&gt;I've tried to find some rules of thumb about this without any luck. I'm designing a part with a blind cartridge bearing bore. It's a 624 bearing (13x4x5) and the fit size guides point me to a tolerance of 13.0-13.026mm for the bore for my application, so slip fit on the housing. FWIW the shaft is the rotating part of the design and will be pressed into the inner race. Do I need to create some kind of lip or shelf inside the bore so that the inner race doesn't come into contact with the housing and create friction? It seems like this would obviously be a concern if both the inner race and outer race are the same thickness.&lt;a href=&quot;https://i.stack.imgur.com/0XXda.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0XXda.png&quot; alt=&quot;housing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11211" LastEditorUserId="11211" LastEditDate="2018-06-04T03:28:01.010" LastActivityDate="2022-02-24T00:36:45.997" Title="Blind bearing bore design" Tags="|machining|bearings|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22039" PostTypeId="2" ParentId="22035" CreationDate="2018-06-04T05:28:30.077" Score="1" Body="&lt;p&gt;Yes, this is normally the solution, both to avoid the inner race touching or the shaft it supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method is to fit a spacer / washer to do the same thing - possibly cheaper but can be left out when taken apart.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-06-04T05:28:30.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22040" PostTypeId="1" CreationDate="2018-06-04T06:35:31.297" Score="5" ViewCount="1144" Body="&lt;p&gt;I have an option to either attend &lt;strong&gt;Middle East Technical University - North Cyprus Campus&lt;/strong&gt; (non-ABET accredited) or &lt;strong&gt;Mapua University&lt;/strong&gt; (ABET accredited) in the Philippines to study Chemical Engineering as an undergraduate. I would personally prefer to study in (METU NCC) than in Mapua, but have been reticent due to the Chemical Engineering program there not being ABET accredited. I have a feeling, however, that ABET accreditation mainly pertains to engineers in the United States. &lt;em&gt;How important is the ABET accreditation if one wants to pursue a job in the EU or other countries NOT in North America after their bachelors in engineering?&lt;/em&gt; &lt;em&gt;How will this affect my job prospects?&lt;/em&gt; If you have any experience in this, please don't hesitate to share them. There seems to be a lack or contradictory information in the internet regarding this matter. Thank you so much!&lt;/p&gt;&#xA;" OwnerUserId="16310" LastActivityDate="2018-06-04T09:35:24.560" Title="How important is an ABET-accredited Engineering Degree in the EU (like Germany) or other countries that is not in North America?" Tags="|chemical-engineering|licensure|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22041" PostTypeId="1" CreationDate="2018-06-04T06:38:09.610" Score="2" ViewCount="38" Body="&lt;p&gt;&lt;a href=&quot;https://link.springer.com/content/pdf/10.1007%2F978-3-642-33926-4_49.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Planar Evasive Aircrafts Maneuvers Using Reinforcement Learning&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$d=\sqrt{(x_m-x_e)^2+(y_m-y_e)^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{d} = V_e \cos (\lambda-\Psi_e) - V_m \cos(\lambda-\Psi_m)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\lambda = \arctan\frac{x_e - x_m}{y_e - y_m}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{\lambda} = \frac{V_m\sin(\lambda - \Psi_m) + V_e\sin(\lambda-\Psi_e)}{d}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the state equations, $d$ is the distance between the objects $m$ and $e$. $\lambda$ is the line of sight angle between $m$ and $e$. $V_e$ and $V_m$ are constant scalars. $\Psi$ is the heading angle of an object. The picture of the geometry can be seen below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eWbkG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eWbkG.jpg&quot; alt=&quot;geometry&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/97Rye.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/97Rye.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in the discrete time state update $x(k+1) = x(k) + \delta t \times x(k)$. I have solved $d$ dotdot and $\lambda$ dotdot above. I can just plug in $\dot{psi}$ = acceleration / velocity and other variables except for $\psi_m = arctan(\dot{y_m}/\dot{x_m})$ and $\psi_e = arctan(\dot{y_e}/\dot{x_e})$, which I cannot solve using the four state variables. Please advise me how to solve for heading angles $\psi_m$ and $\psi_e$.&lt;/p&gt;&#xA;" OwnerUserId="16308" LastEditorUserId="16308" LastEditDate="2018-06-26T23:46:48.803" LastActivityDate="2018-06-27T01:31:10.860" Title="aircraft pursuit dynamic system state update" Tags="|mechanical-engineering|mathematics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22042" PostTypeId="2" ParentId="22031" CreationDate="2018-06-04T08:59:39.537" Score="2" Body="&lt;p&gt;Strictly speaking, the energy content of the moving fluid is a function of pressure/head and velocity, so that vscostiy does not show up in your equation is no wonder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The viscosity mostly changes the behavior in the pipe system, you will have more friction losses, so for the same flow rate you need a higher head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, with higher viscostiy the efficiency of your pump falls. This nomogram, cribbed from a planning handbook from KSB &amp;amp; based on the methodology of the Hydraulics Insititue, shows how you can find the correct efficiency + correction factors for other values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xZGPU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xZGPU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you know the values for water for flow rate $Q_w$, head $H_w$ and efficency $\eta_w$, you find the correction values $k_{Q}$, $k_{H}$ and $k_{\eta}$ from the nomogram. This relation holds true from Q between 0.8 - 1.2 time Q at best efficiency point. So do the conversion at BEP and one operating point higher and one lower than BEP and you should be able to draw a new pumping curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for power, I believe your formula is wrong because of the way it accounts for efficiency. Power is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P=\rho g H Q\frac{1}{\eta}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;obviously using the corrected values!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: KSB, Auslegung von Kreiselpumpen (apparently no longer available online)&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-06-04T08:59:39.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22043" PostTypeId="1" AcceptedAnswerId="22051" CreationDate="2018-06-04T09:02:02.290" Score="2" ViewCount="2481" Body="&lt;p&gt;How much could this motor lift and how fast? If I connect a belt to its axis about 1cm away or even 10cm away from its tip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque: 1.5Nm&#xA;Peak Torque: 10Nm&#xA;Power: 9-30A, to 1000W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.alibaba.com/product-detail/86BLF40-24v-48v-1000W-big-brushless_60644013087.html?spm=a2700.7724838.2017115.56.f5f862211YuM1C&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.alibaba.com/product-detail/86BLF40-24v-48v-1000W-big-brushless_60644013087.html?spm=a2700.7724838.2017115.56.f5f862211YuM1C&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of motor would I need to be able to lift 50kg about 1m high, in say 4 seconds? And to control finely the exact position within the 1m, sometimes it should lift to 20cm sometimes to 30cm. Is a stepper or servo motor capable of this task? Would a linear actuator be better, but I read those have low duty cycle, and they are slow so they would lift it slowly and wait for next time?&lt;/p&gt;&#xA;" OwnerUserId="16311" LastActivityDate="2018-09-02T20:01:58.280" Title="How much could this motor lift and how fast?" Tags="|motors|robotics|stepper-motor|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22044" PostTypeId="2" ParentId="22040" CreationDate="2018-06-04T09:28:53.080" Score="2" Body="&lt;p&gt;One of benefits of accreditation is when dealing with governments bodies. When one has to get education credential verified such in case EU having a international recognized credentials make the process less painful. I have a ABET accredited credentials and found it value when I have had to get my qualification verified and accepted by government bodies and Professional Engineering bodies other than the domicile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also take a look at the &lt;a href=&quot;http://www.ieagreements.org/accords/washington/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Washington Accord&lt;/a&gt;. This will give alternative view points. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;References:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/6/what-is-the-process-of-transferring-an-engineering-license-from-the-united-state?utm_medium=organic&amp;amp;utm_source=google_rich_qa&amp;amp;utm_campaign=google_rich_qa&quot;&gt;What is the process of transferring an engineering license from the United States to Germany?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/7244/110&quot;&gt;Which US states require a bachelors in Engineering to become licensed?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-06-04T09:35:24.560" LastActivityDate="2018-06-04T09:35:24.560" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22050" PostTypeId="2" ParentId="5826" CreationDate="2018-06-04T14:35:54.807" Score="0" Body="&lt;p&gt;Yes and no. The basic difference is carbon content, which is very important. You did not define heat treatment which makes a great difference with the carbon. If both are hardened and tempered at high temperature (like 1300°F), they will have almost identical properties. Traditionally, 420 with 0.15% min. C is used for cutting instruments and cutlery. It is generally not a good choice for torsion and fatigue stresses. The 410 (0.15% C max.) is a much better choice, depending on hardness/strength. 410 can have low toughness, 420 is expected to have low toughness. Also, 410 with 0.04% C is a very different material from 410 with 0.14% C. It sounds like you need a metallurgist; I would say the original choice of 420 was poor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: a 13% Cr with 0.150% C could be both 410 and 420.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="1832" LastEditDate="2018-06-06T00:22:31.977" LastActivityDate="2018-06-06T00:22:31.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22051" PostTypeId="2" ParentId="22043" CreationDate="2018-06-04T15:26:26.927" Score="0" Body="&lt;p&gt;Torque is measured in N M (Newton Meters)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Newton is kg m / s^2  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So torque kg m^2 / s^2  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque = r x F = r x m x a &lt;/p&gt;&#xA;&#xA;&lt;p&gt;s is distance&lt;br&gt;&#xA;t is time&lt;br&gt;&#xA;a is acceleration&lt;br&gt;&#xA;r is radius = 1/10 meter   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;s = 1/2 a t^2 = 1 m &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 / 4^2 = a = 1/8  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;plug it in    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.5 = 1/10 m 1/8&lt;br&gt;&#xA;m = 1.5 x 10 x 8 = 1200 kg    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;that seems too high to me but that is what I get&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-06-04T15:57:04.610" LastActivityDate="2018-06-04T15:57:04.610" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22052" PostTypeId="2" ParentId="22034" CreationDate="2018-06-04T15:44:31.493" Score="1" Body="&lt;p&gt;may be its not getting the enough rotating flux , as you know V/F ratio needs to be maintained constant for motor to run at low speeds... may be your VFD is not providing the appropriate ratio of Votlage to Freq so that its gets right rotating flux... &lt;/p&gt;&#xA;" OwnerUserId="16317" LastActivityDate="2018-06-04T15:44:31.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22054" PostTypeId="1" AcceptedAnswerId="22055" CreationDate="2018-06-04T16:07:34.237" Score="5" ViewCount="152" Body="&lt;p&gt;I have been doing some research on valve systems for sample analyzers. In many of the valve systems' product documentation they refer to a flow path as 'fully swept'. From the context I can conclude that it increases the rate of purging a previous gas or liquid from the system. My best guess is that it's a path that minimizes 'dead zones' in the flow, where gas or liquid may get caught. Is it simply marketing jargon, or is there a fluid mechanics definition? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.swagelok.com/downloads/webcatalogs/en/ms-02-326.pdf&quot; rel=&quot;noreferrer&quot;&gt;An example usage of the term (page 7)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16318" LastActivityDate="2018-06-04T16:37:35.583" Title="What is a 'swept path' in a valve system?" Tags="|mechanical-engineering|fluid-mechanics|chemical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22055" PostTypeId="2" ParentId="22054" CreationDate="2018-06-04T16:26:58.243" Score="2" Body="&lt;p&gt;You are correct; as described in product literature, a &quot;fully swept&quot; path &quot;&lt;a href=&quot;https://www.designworldonline.com/diaphragm-valve-withstands-rugged-semiconductor-environment/&quot; rel=&quot;nofollow noreferrer&quot;&gt;minimize[s] entrapment areas, facilitate[s] purging and maximize[s] flow capacity&lt;/a&gt;&quot; (SwageLok) and ensures &quot;&lt;a href=&quot;http://ppl.saint-gobain.com/uploadedFiles/SGppl-global/Documents/High%20Purity%20Systems.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;low shear&lt;/a&gt;&quot; with &quot;&lt;a href=&quot;http://ppl.saint-gobain.com/uploadedFiles/SGppl-global/Documents/High%20Purity%20Systems.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;no hold-up volume&lt;/a&gt;&quot; (Saint Gobain) and &quot;&lt;a href=&quot;http://www.indev-cheaper.it/Brochure/28_trasduttori_ds-pr-solidsense_ii_atex-pt-eng0.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;with no hidden recesses or dead spaces&lt;/a&gt;&quot; (SolidSense). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether this capability is valuable to you depends on your application. It's certainly industry-specific jargon. Since it's rigorously defined (there either are dead spaces or there aren't) and can be easily verified, on the spectrum of marketing fluff to an objective industry term, it's firmly on the latter side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own research, for example, involved sending live biological cells in solution through PEEK tubing and a variety of valves. Increasing the degree of sweeping in the fluid path meant the difference between a quick post-experiment clean with 10% bleach followed by deionized water vs. the possibility of bacterial contamination flourishing in cracks filled with dead cells and the possibility of residual bleach leaking out of crevices and killing my live cells.&lt;/p&gt;&#xA;" OwnerUserId="10123" LastEditorUserId="10123" LastEditDate="2018-06-04T16:37:35.583" LastActivityDate="2018-06-04T16:37:35.583" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22056" PostTypeId="1" CreationDate="2018-06-04T17:18:19.567" Score="-1" ViewCount="181" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Synthesis_of_precious_metals&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Synthesis_of_precious_metals&lt;/a&gt;, precious metals can be synthsized using fission. I think that if we found a way to do it with fusion, it would be more efficient and more of the ingredients would turn into the desired element with less of the ingredients getting wasted. According to &lt;a href=&quot;https://physics.stackexchange.com/questions/203757/why-can-gold-be-drawn-out-finer-than-light/203786#203786&quot;&gt;this answer&lt;/a&gt;, gold cannot support crack propagation at all. If we synthesized gold using fusion, maybe we could produce so much of it and use it as a building material without worrying that it will suddenly and unpredictably fracture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also disadvantages to using gold. Gold is so weak compared to other materials. Also according to this answer, when two spheres of a given material collide with each other at a given speed, the strength they must have in order to resist forming a crack or denting varies as their shear modulus to the power of $\frac{4}{5}$ but the yield strength of gold might be much smaller than its shear modulus so two spheres of gold might not withstand collisions with a speed of 5 m/s without denting.&lt;/p&gt;&#xA;" OwnerUserId="6419" LastEditorUserId="6419" LastEditDate="2018-06-04T20:04:24.930" LastActivityDate="2018-06-04T22:29:29.733" Title="Can fusion be used to synthesize precious metals?" Tags="|mechanical-engineering|materials|nuclear-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2018-06-07T12:34:34.307" ContentLicense="CC BY-SA 4.0" />
  <row Id="22057" PostTypeId="2" ParentId="22056" CreationDate="2018-06-04T18:24:59.400" Score="3" Body="&lt;p&gt;To produce precious metals via fission, you need a starting metal with about twice the atomic weight of the desired end product. In practical terms this means the starting metal is more precious than the end product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the fission process releases neutrons and gamma rays in deadly quantities, and the fission products are themselves often violently radioactive for periods of time ranging from hours to centuries. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fission process is also &quot;unclean&quot; from the standpoint that you do not get two clean halves of the raw material as the product; you get a variety of fission products from which you then must sort out the one you wanted in the first place. Remember that since most of the fission products are to one degree or another radioactive, this implies wet chemical extraction processing in a highly radioactive environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fusion processes are even more difficult to master for the purposes of making your own precious metals, for a variety of reasons that I will summarize below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take two heavy elements and endeavor to press them together hard enough to fuse them into one heavier element, we discover that the amount of pressure required to get the nuclei close enough together to fuse is truly gigantic because of how hard they electrostatically repel one another. we have been trying for almost 70 years to invent some way of doing this and we are not there yet. This process is so hard to trigger that the only way it happens in bulk is  in the collapsing core of a supermassive star during a supernova.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we instead take one heavy element and try to add neutrons or protons to it one or two at a time, we find that the most likely outcome is that as soon as we add them, the resulting nucleus is highly unstable and quickly decays or fissions or spits out a small lump of neutrons and protons and the net yield of heavier nuclei is tiny. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the only environment in which this process produces heavier nuclei in bulk quantities occurs in the extended envelope of a red giant star and takes hundreds of thousands of years. Small quantities of heavy elements (specifically, isotopes of plutonium) can be made this way by starting with uranium and subjecting it to extremely intense neutron bombardment- which is how we make plutonium for use in fission bombs. This process is extraordinarily expensive, hazardous, and produces huge quantities of highly radioactive waste which require foolproof deep storage for tens of thousands of years. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13363" LastEditDate="2018-06-04T22:29:29.733" LastActivityDate="2018-06-04T22:29:29.733" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22059" PostTypeId="1" AcceptedAnswerId="22061" CreationDate="2018-06-04T19:18:09.227" Score="3" ViewCount="78" Body="&lt;p&gt;I got a mini-project: I want to automatically get rid of the water produced by an indoor air conditioner unit. Since the unit is underground, the water needs to climb ~2m to get to the outside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a cheap pump online that has a max head of 3m and flow rate of 240l/h, the pipes on it have 6mm on the inside diameter. I also have some tubing left over from a gardening project but its 4mm on the inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can figure out how to fit the two together mechanically but since the tubing is tad too small, wouldn't that increase the pressure requirement and thus decrease the max head? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really don't mind if it affects the flow rate, the amount of water I need to pump out is ~2L/day max so I'm adding a microcontroller to turn it on for just a few minutes out of a day anyways.&lt;/p&gt;&#xA;" OwnerUserId="15752" LastActivityDate="2018-06-04T21:38:36.630" Title="Would too small tubing reduce the max head of a pump?" Tags="|pumps|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22060" PostTypeId="1" AcceptedAnswerId="22062" CreationDate="2018-06-04T20:56:38.313" Score="2" ViewCount="74" Body="&lt;p&gt;We have an open container sitting at 1000mm off the ground and a receiver tank sitting one metre above it. Container holds 400l of water and is connected to the receiver by 80mm id pipe. I was planning on using extraction fans sitting in the open end of the receiver to create a vacuum to remove the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is a formula I could use to work out how inHg needed to be produced by the fans to remove water in x amount of time?&lt;/p&gt;&#xA;" OwnerUserId="16322" LastEditorUserId="16322" LastEditDate="2018-06-04T21:09:18.110" LastActivityDate="2018-06-05T16:38:53.223" Title="What is the formula to work out time to remove water from an open vessel with vacuum?" Tags="|vacuum|vacuum-pumps|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22061" PostTypeId="2" ParentId="22059" CreationDate="2018-06-04T21:38:36.630" Score="1" Body="&lt;p&gt;Related: &lt;a href=&quot;https://engineering.stackexchange.com/questions/16481/system-curve-for-use-in-determining-pump-operating-point&quot;&gt;System curve for use in determining pump operating point&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smaller lines will increase the resistance to flow, which will decrease the flow rate.  The use able head is reduced, but this wouldn't be a problem with such a low flow rate. However, the lines will still flow, just at a reduced flow rate.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-06-04T21:38:36.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22062" PostTypeId="2" ParentId="22060" CreationDate="2018-06-04T23:24:31.100" Score="1" Body="&lt;p&gt;So if your fan doesn't produce a minimum of 1m of water (which is about 2.9 inHg), there would be no flow.  Afterwards, it depends upon your fan curve.  You would find that the system curve would likely equal 2.9 inHg + K*Q^2, where Q is the flowrate and K is a factor that depends upon how you move the flow - note with smooth curves K is small, and with sharp turns and disruptive contours, K is large.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wherever your pump curve intersects this system curve is the flow rate.  Plain and simple.  Once you know the flowrate, divide the volume to move by the flowrate to gain the time estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a wet/dry vacuum style compressor, you're most like to see a curve similar to this one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5BmmQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5BmmQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the 40&quot; of water (approx. 1 meter), the vacuum would remove 300 cfm of air (and therefore water) a minute.  At less than 100 cfm removal, the vacuum would stall.  So, you should expect removal between 100-300 cfm, or between 3-9 seconds.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-06-05T16:38:53.223" LastActivityDate="2018-06-05T16:38:53.223" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22063" PostTypeId="1" AcceptedAnswerId="22076" CreationDate="2018-06-05T09:22:34.877" Score="4" ViewCount="214" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SQPmL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SQPmL.jpg&quot; alt=&quot;triangular load single support&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From equilibrium: $$C_y = \frac{w_0L}{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the load is given by: $$w(x)  = \frac{w_0}{L}x$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the expressions for shear and moment are given by: $$V(x) =\int{-w(x)dx}=-\frac{w_0}{2L}x^2 + C_1\\M(x) = \int{V(x)} = -\frac{w_0}{6L}x^3 + C_1x + C_2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are $C_1$ and $C_2$ determined?.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="16325" LastEditorUserId="2015" LastEditDate="2018-06-05T17:11:22.690" LastActivityDate="2018-06-06T10:35:27.030" Title="Shear force expression for singly supported triangular load" Tags="|mechanical-engineering|structural-engineering|structural-analysis|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22064" PostTypeId="1" CreationDate="2018-06-05T09:36:46.730" Score="1" ViewCount="216" Body="&lt;p&gt;I am working on a HVAC project where lizard in going inside the electrical unit and causing wire damage and short circuit. I need to find some solution to detect or repel lizard before any catastrophic situation. Since lizard are cold blooded species, I cannot use IR head index measurement methods or PIR sensor. Also, lizard are very small creature so method need to be foolproof. &lt;/p&gt;&#xA;" OwnerUserId="5973" LastActivityDate="2020-08-23T23:04:15.467" Title="What are some robust method to detect or repel lizard?" Tags="|electrical-engineering|sensors|positions-measurement|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22065" PostTypeId="2" ParentId="22064" CreationDate="2018-06-05T10:33:24.373" Score="1" Body="&lt;p&gt;A camera with provided lighting (IR is okay) within the duct (area where no stray moving objects should appear) would be perfectly okay for detection. Lizards aren't invisible. Now, stopping the lizards might be trickier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think though, that the best approach would be to make the intakes/exhausts to face downwards and grease the walls - or include a T-shape section, with a service hatch on the short downwards facing segment, thoroughly greased. Detection no longer required. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-06-05T10:33:24.373" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22066" PostTypeId="1" AcceptedAnswerId="22067" CreationDate="2018-06-05T10:57:37.123" Score="-1" ViewCount="57" Body="&lt;p&gt;The discs must remain in flat contact to the surface at all times but you can assume no friction while pushing them about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No using magnets from the other side of the surface, for all intensive purposes assume it is infinitely thick on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must be able to move the discs &quot;freely&quot; meaning that normal robotic arms won't cut it due to getting tangled after moving each disc in a circle around the centre maybe once at most.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I imagine the answer could be very useful if there is one, I couldn't find one after scouring the internet for a few hours but a solution would open a few doors, especially when it comes to &quot;omnidirectional treadmills&quot;, if you replace each disc with a vertical cylinder you could crudely mimic any terrain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this isn't too theoretical, apologies if it is, I do really think there are numerous practical applications though and for that reason any working answer at all would be useful.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit: Since it was apparently ambiguous, let me clarify, no using magnets, obviously if a solution uses a machine that has a magnet internally for some reason required for that machine to function then that's fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface is arbitrary, it could be any material within reason and cannot be changed, assume the area above the surface cannot be altered in any way.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit 2: My mistake, I completely forgot to mention the surface has a finite size, although the shape could be practically anything as long as no corridors thinner than the sum of both discs' radii are within it, assume it is convex if you cannot find a solution for concave, or just assume it is a rectangle!&lt;/p&gt;&#xA;" OwnerUserId="16314" LastEditorUserId="16314" LastEditDate="2018-06-05T12:11:37.593" LastActivityDate="2018-06-05T17:47:15.363" Title="Move two discs freely and quickly on the underside of a surface?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="9" ClosedDate="2018-06-06T10:10:49.287" ContentLicense="CC BY-SA 4.0" />
  <row Id="22067" PostTypeId="2" ParentId="22066" CreationDate="2018-06-05T11:36:42.643" Score="0" Body="&lt;p&gt;Use a single robotic arm or other mechanism providing 2 degrees of freedom, to move one disc. Said disc is attached to the arm through an axis of a motor. The stator/chassis of the motor would attach to another arm with a linear actuator moving the other disc closer/farther (and the motor turning would change the angle relative to the first) - turning and extending/retracting in a plane that lies below the plane of movement of the robotic arm. You'll need a rotary slip ring connector to provide control/power to the stator and actuator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, attach both one disc and the base of a second robotic arm (moving the second disk) to the end of the first robotic arm.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2018-06-05T12:03:52.657" LastActivityDate="2018-06-05T12:03:52.657" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="22068" PostTypeId="2" ParentId="22063" CreationDate="2018-06-05T12:13:12.587" Score="0" Body="&lt;p&gt;The constants of integration are determined from the known conditions. In this case, there is no load at A, so the shear and moment at A are 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your equation for w(x), it appears that x=0 at A, so the equations for V(x) and M(x) result in C1=0, C2=0.&lt;/p&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2018-06-05T12:13:12.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22069" PostTypeId="2" ParentId="44" CreationDate="2018-06-05T13:04:27.130" Score="-1" Body="&lt;p&gt;You have to go for automatic pumping trap, as a normal float trap does not get enough differential pressure, and hence you have to waste lot of live steam by opening bypass valves. &lt;/p&gt;&#xA;" OwnerUserId="16329" LastEditorUserId="2015" LastEditDate="2018-06-05T17:14:37.723" LastActivityDate="2018-06-05T17:14:37.723" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22072" PostTypeId="1" AcceptedAnswerId="22074" CreationDate="2018-06-05T15:33:03.213" Score="18" ViewCount="1471" Body="&lt;p&gt;While browsing a local news site, I stumbled upon this picture of a broken bus door. The glass was fractured due to a rock impact. As far as i know there are no heating elements on the window section in the picture. Now, what interests me is the apparent periodicity and structure of the fracture pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are periodic fractures something commonly encountered in tempered glass?&#xA;Is the mechanism that generates such fractures understood?&#xA;Is it a consequence of the manufacturing process?&#xA;Perhaps the way the glass is affixed to the door?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fYYK9.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fYYK9.jpg&quot; alt=&quot;Glass fracture&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16330" LastActivityDate="2018-06-05T16:23:10.643" Title="Fracture patterns in tempered glass" Tags="|glass|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22074" PostTypeId="2" ParentId="22072" CreationDate="2018-06-05T16:23:10.643" Score="21" Body="&lt;p&gt;It's an artifact of production - in particular, the construction of the conveyor belt of the machine performing the quenching process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pattern can be observed through a polarizing filter in the undamaged glass:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FRmQy.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FRmQy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/b/b2/CarWindowPolarization.jpg/800px-CarWindowPolarization.jpg&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and it's a result of the structure of the conveyor belt through which the surface of the glass is cooled to generate the stress that gives it the special properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iCdZZ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iCdZZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=GCnQVVfjabk&quot; rel=&quot;noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The contact area of the glass with the conveyor is small enough that it doesn't negatively impact the process - all throughout the surface the stress is introduced in sufficient amount, but the amount of stress applied differs by the &quot;openness&quot; of given area, creating this pattern.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-06-05T16:23:10.643" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22075" PostTypeId="1" CreationDate="2018-06-05T16:25:26.860" Score="3" ViewCount="26" Body="&lt;p&gt;I am quite new to this so please forgive me for any mistakes.&#xA;I am working on a simulation project and have very little clue as to what to do and I hope someone can point me to the right direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the summary of my problem. (Please keep in mind that this is a simulation, no real equipment or IMU is involved)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an object O and a stereo camera C. Both have an 9dof IMU attached and are moving freely in space. O and C are 3m in range. O's IMU broadcast acceleratometer info periodically. C is equipped with software to extract the 3D position of O. Find the best way to fuse the data and predict O's position in C's reference in the very near future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought on solving the problem is as follows: (please point out any mistakes I have because I am honestly super new at this...)&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Initialize the distance between O and C with an error distribution&lt;/li&gt;&#xA;&lt;li&gt;Every time I receive O's imu data, I need to calculate where it is&#xA;approximately with an error distribution&lt;/li&gt;&#xA;&lt;li&gt;then use a nonlinear Kalman filter to fuse the 2 distributions&#xA;together and find where O might be.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My question is this&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How do I calculate where O is given approximate initial pos, and&#xA;9dof IMU data?&lt;/li&gt;&#xA;&lt;li&gt;How do I combine C's IMU and O's IMU so that the measurement is done in C's reference frame?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If anyone can point me towards the right direction I will be very very grateful. Thank you so much!!&lt;/p&gt;&#xA;" OwnerUserId="16323" LastEditorUserId="7083" LastEditDate="2018-07-31T22:56:16.463" LastActivityDate="2018-07-31T22:56:16.463" Title="Predictive object tracking with 2 IMUs and a stereo camera" Tags="|electrical-engineering|sensors|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22076" PostTypeId="2" ParentId="22063" CreationDate="2018-06-05T17:10:57.157" Score="1" Body="&lt;p&gt;The original analysis is incorrect, because you failed to account for the sudden load at point C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(x) = -\frac{w_0}{2L}x^2 + C_1 + C_y*H(x-\frac{2L}{3})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $$H(x) = \lim_{k\to\infty} \frac{1}{1+e^{-2kx}}$$ is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heaviside_step_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;heaviside step function&lt;/a&gt;, As such, M(x) is also different:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M(x) = -\frac{w_0}{6L}x^3 + C_1x + C_2 + C_y(x-\frac{2L}{3})H(x-\frac{2L}{3})$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can determine the constants.  In V(x), at x=0, V(x) is 0, so $C_1$ is 0.  In M(x), at x=0, M(x) is 0, so $C_2$ is 0.  However, integrating further for deflection and rotation, it will be seen that the remaining constants will not be 0, because we will need to balance out the evolving step function with the growing polynomial.  We will evaluate those at x = 2L/3 where the rotation must be 0 and the deflection must be 0.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-06-05T17:10:57.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22078" PostTypeId="2" ParentId="22066" CreationDate="2018-06-05T17:47:15.363" Score="0" Body="&lt;p&gt;Given your multi-edited question, this seems as reasonable an answer as any other: mount each disk on top of a blimp, thus providing the necessary anti-gravity force to keep the disk on the surface.  Use the blimp's guidance system (propellors, ailerons, etc) to move wherever you please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, this is absurd, but your scenario is unrealistic. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-06-05T17:47:15.363" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22079" PostTypeId="1" CreationDate="2018-06-05T18:45:44.400" Score="2" ViewCount="43" Body="&lt;p&gt;I'd like to calculate the viscosity of a liquid in a container systematically (and feed the results back to a computer). I only have basic knowledge of physics and mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that a motor with a tiny &quot;windmill&quot; constantly turning inside the liquid would be enough because someone one could measure the current and voltage being supplied to the motor and then &lt;em&gt;somehow&lt;/em&gt; link that back to the force which the motor is putting in. I'm hoping that there's a formula to link force and volume of liquid moved to viscosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this sound like a good approach and if not, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is a good approach, how would I go about calculating liquid viscosity from the original data and what sorts of measurements would need to be taken?&lt;/p&gt;&#xA;" OwnerUserId="11904" LastActivityDate="2018-06-05T18:45:44.400" Title="Finding liquid viscosity" Tags="|measurements|liquid|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="22081" PostTypeId="1" CreationDate="2018-06-05T20:21:06.573" Score="3" ViewCount="207" Body="&lt;p&gt;I prototyped this attractive wooden sit/stand laptop stand&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v6BI0m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v6BI0m.jpg&quot; alt=&quot;this attractive wooden sit/stand laptop stand&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but ran into a problem keeping the legs tight enough to hold their position. There are 1/16&quot; thick EPDM washers between the oak legs - the washers are almost 1.25&quot; in diameter with a .25&quot; hole for a threaded tension rod. If I want to resist about 15 lbs of force on a 8&quot; lever arm, that's only about 10 ft-lb of torque that I need to resist. What the hell, let's say 20 ft-lb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to find a toothed washer that I could use in pairs, each one epoxied to one of the oak legs, but couldn't find a suitable design. (For example the Nord Lock wedge locking washers won't work because they are directional. They will work on one side, but on the other the rotation will be in the wrong direction.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What alternative material(s) should I use which might resist that rotation? &lt;/p&gt;&#xA;" OwnerUserId="16334" LastEditorDisplayName="user4139" LastEditDate="2018-06-06T08:44:55.673" LastActivityDate="2018-11-03T13:00:33.630" Title="Resist rotation with friction or teeth?" Tags="|mechanical-engineering|design|gears|structural-analysis|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22082" PostTypeId="2" ParentId="21738" CreationDate="2018-06-06T02:31:12.493" Score="1" Body="&lt;p&gt;JB Water weld works very well for me in many applications like this. It sets under water and is not brittle. I built a lip for a water pump that eroded away. Please let us know what you went with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G128G.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G128G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-11-08T21:33:27.040" LastActivityDate="2018-11-08T21:33:27.040" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22083" PostTypeId="2" ParentId="21713" CreationDate="2018-06-06T02:35:51.953" Score="1" Body="&lt;p&gt;This is a good question you could modify 12volt a 3-5 watt brush less water pump commonly found on ebay from China to run a gear instead of impellers . You can add a dimmer switch to control the revolutions. You may have to change out the motor more often but for 3 dollars each. These motors run fine dry and is about 2-4 inches cubed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NIsFx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NIsFx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remove the screws to expose the drive shaft and impellers. &lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-06-06T02:47:40.647" LastActivityDate="2018-06-06T02:47:40.647" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22084" PostTypeId="2" ParentId="22081" CreationDate="2018-06-06T02:59:31.827" Score="1" Body="&lt;p&gt;I think this old lawn chair gear would work for you well. It should be old enough where the patient on it has expired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/999uv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/999uv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-06-06T02:59:31.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22087" PostTypeId="1" AcceptedAnswerId="22095" CreationDate="2018-06-06T08:51:49.497" Score="12" ViewCount="41499" Body="&lt;p&gt;My AC outdoor unit gets direct sunlight (no shading whatsoever) from morning to noon. Temperature normally soars up to 40 degree Celsius (I am from India). I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Is this bad for the AC's refrigerant, does it lose efficiency over time?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does an AC compressor located in full sun work harder than one located in the shade?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0ZgMq.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0ZgMq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="12665" LastEditorUserId="12665" LastEditDate="2018-06-06T10:34:57.623" LastActivityDate="2021-06-14T14:33:44.477" Title="Should we protect our AC's outdoor unit from sunlight?Does it increases the durability of an AC and its refrigerant if it is well shaded?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|refrigeration|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22088" PostTypeId="1" CreationDate="2018-06-06T10:12:57.600" Score="1" ViewCount="86" Body="&lt;p&gt;Is it possible to define a finite element that isn't planar? For example the shell element shown in the picture below...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kNgZY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kNgZY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14531" LastActivityDate="2020-09-05T08:38:54.010" Title="Do finite elements have to be planar?" Tags="|finite-element-method|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22089" PostTypeId="2" ParentId="22063" CreationDate="2018-06-06T10:35:27.030" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/22068/1832&quot;&gt;JohnHoltz's answer&lt;/a&gt; is correct in that you determine $C_1$ and $C_2$ by looking at the boundary conditions. Looking at the left-hand side of the beam, we know that the cantilever has zero shear and bending moment, which let's us determine (note the use of the auxiliary variable $w$ which describes the load's variation per unit length):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;w &amp;amp;= \dfrac{w_0}{L} \\&#xA;Q &amp;amp;= \int q\text{ d}x \\&#xA;&amp;amp;= \int wx\text{ d}x\\&#xA;&amp;amp;=\dfrac{1}{2}w x^2 + C_1 \\&#xA;Q(0) &amp;amp;= C_1 = 0 \\&#xA;\therefore Q &amp;amp;=\dfrac{1}{2}w x^2 \\&#xA;M &amp;amp;= \int Q\text{ d}x \\&#xA;&amp;amp;= \int \dfrac{1}{2}w x^2\text{ d}x \\&#xA;&amp;amp;= \dfrac{1}{6}w x^3 + C_2 \\&#xA;M(0) &amp;amp;= C_2 = 0 \\&#xA;\therefore M &amp;amp;= \dfrac{1}{6}w x^3 \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by &lt;a href=&quot;https://engineering.stackexchange.com/a/22076/1832&quot;&gt;Mark's answer&lt;/a&gt;, however, this only describes the beam from the left cantilever until the support. The discontinuity caused by the support means the equations are no longer valid and must be recalculated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use Mark's heaviside function, or you can just repeat the process above for the other side as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's start by pretending that we only have the right-hand side, so that $x=0$ is at point C, and therefore define a new length $\ell$ which describes this cantilever, and $w_c$ as the value of the distributed load at point C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\ell &amp;amp;= \dfrac{L}{3} \\&#xA;w_c &amp;amp;= \dfrac{2}{3}w_0 \\&#xA;Q &amp;amp;= \int q\text{ d}x \\&#xA;&amp;amp;= \int \left(wx + w_c\right)\text{ d}x\\&#xA;&amp;amp;=\dfrac{1}{2}wx^2 + w_c x + C_1 \\&#xA;Q(\ell) &amp;amp;= \dfrac{1}{2}w\ell^2 + w_c \ell + C_1 = 0 \\&#xA;\therefore C_1 &amp;amp;= -\dfrac{1}{2}w\ell^2 - w_c \ell \\&#xA;&amp;amp;= -\dfrac{1}{18}w_0 L - \dfrac{2}{9}w_0 L \\&#xA;&amp;amp;= -\dfrac{5}{18}w_0 L \\&#xA;\therefore Q &amp;amp;=\dfrac{1}{2}wx^2 + \dfrac{2}{3}w_0 x -\dfrac{5}{18}w_0 L \\&#xA;M &amp;amp;= \int Q\text{ d}x \\&#xA;&amp;amp;= \int \left(\dfrac{1}{2}wx^2 + \dfrac{2}{3}w_0 x -\dfrac{5}{18}w_0 L\right)\text{ d}x \\&#xA;&amp;amp;= \dfrac{1}{6}wx^3 + \dfrac{1}{3}w_0 x^2 -\dfrac{5}{18}w_0 Lx + C_2 \\&#xA;M(\ell) &amp;amp;= \dfrac{1}{6}w\ell^3 + \dfrac{1}{3}w_0 \ell^2 - \dfrac{5}{18}w_0 L\ell + C_2 = 0 \\&#xA;\therefore C_2 &amp;amp;= -\dfrac{1}{6}w\ell^3 - \dfrac{1}{3}w_0 \ell^2 + \dfrac{5}{18}w_0 L\ell \\&#xA;&amp;amp;= \dfrac{1}{162}w_0 L^2 + \dfrac{1}{27}w_0 L^2 -\dfrac{5}{54}w_0 L^2 \\&#xA;&amp;amp;= -\dfrac{4}{81}w_0 L^2 \\&#xA;\therefore M &amp;amp;= \dfrac{1}{6}wx^3 + \dfrac{1}{3}w_0 x^2 -\dfrac{5}{18}w_0 Lx - \dfrac{4}{81}w_0 L^2 \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we have these values, we just need to shift the origin back to $x=0$ at point A to keep these results compatible with what we got for the left cantilever, so we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;Q &amp;amp;=\dfrac{1}{2}w\left(x-\dfrac{2}{3}L\right)^2 + \dfrac{2}{3}w_0\left(x-\dfrac{2}{3}L\right) -\dfrac{5}{18}w_0 L \\&#xA;M &amp;amp;= \dfrac{1}{6}w\left(x-\dfrac{2}{3}L\right)^3 + \dfrac{1}{3}w_0 \left(x-\dfrac{2}{3}L\right)^2 -\dfrac{5}{18}w_0 L\left(x-\dfrac{2}{3}L\right) - \dfrac{4}{81}w_0 L^2&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-06-06T10:35:27.030" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22090" PostTypeId="2" ParentId="22087" CreationDate="2018-06-06T11:08:41.097" Score="4" Body="&lt;p&gt;Direct sunlight increases power usage and wear of the unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cooling efficiency depends directly on temperature gradient - difference between temperature of the coolant and ambient air pushed through it. With direct sunlight the chassis heats up, the protective mesh on the fan heats up, surrounding walls and floors heat up from sunlight, the air passes, it also heats up from these, entering the radiator at significantly higher temperature than had the unit been shaded. Also, the radiator and coolant heat up through conduction from chassis, requiring more cooling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that means less heat is drawn from the coolant, the coolant returns warmer to the indoor unit, the AC will need to work longer before it reaches the temperature set on the thermostat, and in consequence it will use up more electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the extra workload put on the device results in faster wear and shorter lifetime.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2018-06-06T12:41:35.857" LastActivityDate="2018-06-06T12:41:35.857" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22091" PostTypeId="2" ParentId="22088" CreationDate="2018-06-06T12:21:28.443" Score="1" Body="&lt;p&gt;Finite elements can have curved edges or faces. See the pictures of the element library for almost any FE program.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It probably doesn't make much sense to try to define a curved shell element with straight edges as in your picture. Curved elements usually have mid-side nodes as well as corner nodes, to give &quot;smooth&quot; curved shapes.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-06-06T12:21:28.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22092" PostTypeId="1" CreationDate="2018-06-06T12:36:56.007" Score="3" ViewCount="81" Body="&lt;p&gt;It seems dynamic magnetic field may damages electronics components like diode,  transistor etc, probably via electromagnetic induction.  How much strength and what frequency magnetic field is dangerous? Surprisingly there is hardly any info on Google on this. &lt;/p&gt;&#xA;" OwnerUserId="16343" LastActivityDate="2018-06-06T12:36:56.007" Title="Dynamic magnetic field Vs electronics" Tags="|electrical-engineering|electromagnetism|" AnswerCount="0" CommentCount="2" ClosedDate="2018-06-25T19:28:31.827" ContentLicense="CC BY-SA 4.0" />
  <row Id="22093" PostTypeId="2" ParentId="21738" CreationDate="2018-06-06T18:15:07.577" Score="2" Body="&lt;p&gt;Bonding to HDPE is very difficult. A few adhesives developed fairly recently do ok in some applications, but that kind of cooling will likely cause them to fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, I'd recommend a mechanical seal. I would (and often have for this exact purpose) use a &lt;a href=&quot;https://www.google.com/search?q=bulkhead%20fitting&amp;amp;oq=bulkhead%20fitting&amp;amp;aqs=chrome..69i57j0l5.2617j0j4&amp;amp;sourceid=chrome&amp;amp;ie=UTF-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;bulkhead fitting&lt;/a&gt; to seal against the wall and an instrument fitting, aka &lt;a href=&quot;https://www.google.com/search?ei=USQYW8zOMKiS0gLz-4rQDg&amp;amp;q=compression%20tube%20fitting&amp;amp;oq=compression%20tube%20fitting&amp;amp;gs_l=psy-ab.3..0j0i30k1l9.4720.5061.0.5935.5.5.0.0.0.0.176.278.0j2.2.0....0...1c.1.64.psy-ab..4.1.176....0.A-ShHQH8myE&quot; rel=&quot;nofollow noreferrer&quot;&gt;compression tube fitting&lt;/a&gt; to seal against the temperature sensor.&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2018-06-06T18:15:07.577" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22094" PostTypeId="1" AcceptedAnswerId="22102" CreationDate="2018-06-06T19:38:58.423" Score="0" ViewCount="74" Body="&lt;p&gt;Note, as I have my degree in chemistry, I denote &quot;=&gt;&quot; as the next step in the equation. Also, sorry for any obvious mistakes that I made - as I mentioned, my degree is not in engineering. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am considering a design for a PET spherical toy ball. The ball will have a radius of 0.07&amp;nbsp;m that will have pressurized air in it at a pressure of 5.81&amp;nbsp;atm (588698.3&amp;nbsp;Pa). Also, the ball will include a 35&amp;nbsp;nm coating of Reduced Graphene Oxide (RGO) as a durable coating on the outside of the ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the thickness of the walls of the ball, I know that the ultimate tensile strength of PET is 55&amp;nbsp;MPa. Because I want a 15% safety factor for the ball, I do the following calculation:&#xA;$$(1-15\%)\cdot55 = 46.75\text{ MPa} = 46750000\text{ Pa}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then use the reduced ultimate tensile strength to find the thickness of the wall using the formula for a thin-walled sphere where $p$ is pressure, $t$ is thickness, $r$ is radius, and $\sigma$ is stress:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;\sigma = \dfrac{pr}{2t} \\&#xA;46750000 = \dfrac{588698.3 \cdot 0.07}{2t} \\&#xA;\therefore t = 4.41\times 10^{-4}\text{ m}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can calculate the burst pressure of the sphere $P$ for pressure (gauge) inside sphere, $FS$ for factor of safety, $\sigma$ for allowable stress and additionally $R_i$ for inner radius, $R_o$ for outer radius:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \dfrac{(R_o^2-R_i^2)\sigma}{(R_i^2)FS}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For calculation of bursting pressure, we take $\sigma$ as ultimate stress for a given material and put $FS=1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note here that I chose the express the ultimate tensile strength of PET as $5.5 \times 10^7$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$P = \dfrac{((0.07 + 4.41 \times 10^{-4})^2 - 0.07^2)\cdot 5.5 \times 10^7}{0.07^2 \times 1} = 695183.67\text{ Pa} = 6.86\text{ atm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now want to calculate the volume of PET needed to create a sphere of the desired dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;V &amp;amp;= \dfrac{4}{3}\pi \left(\left(\dfrac{d_o}{2}\right)^3 - \left(\dfrac{d_i}{2}\right)^3\right) \\&#xA;&amp;amp;= \dfrac{4}{3}\pi \left(\left(\dfrac{0.140 + 4.41 \times 10^{-4}}{2}\right)^3  - \left(\dfrac{0.140}{2}\right)^3\right) \\&#xA;&amp;amp;= 1.36 \times 10^{-5}\text{ m}^3&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now want to calculate the volume of Reduced Graphene Oxide (RGO) to apply to the surface of the sphere:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;V &amp;amp;= \dfrac{4}{3}\pi \left(\left(\dfrac{d_o}{2}\right)^3 - \left(\dfrac{d_i}{2}\right)^3\right) \\&#xA;&amp;amp;= \dfrac{4}{3}\pi \left(\left(\dfrac{0.140 + 4.41 \times 10^{-4} + 35 \times 10^{-9}}{2}\right)^3 - \left(\dfrac{0.140 + 4.41 \times 10^{-4}}{2}\right)^3\right) \\&#xA;&amp;amp;= 1.084366 \times 10^{-9}\text{ m}^3&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you see any mistakes in my calculations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please do not ask why I am using PET as a material or other design decisions like that. There are good reasons for them.&lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="1832" LastEditDate="2018-06-06T23:00:08.260" LastActivityDate="2018-06-06T23:13:59.657" Title="Difficult Toy Ball Analysis" Tags="|mechanical-engineering|pressure|structural-analysis|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22095" PostTypeId="2" ParentId="22087" CreationDate="2018-06-06T19:53:01.583" Score="1" Body="&lt;p&gt;Direct sunshine heats up the enclosure of the air conditioner to much higher levels than ambient temperatures, greenhouse effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the UV Rays will break up plastic and paint and rubber molecules in such component as washers, hoses, brackets and clips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compressor must work much longer to cool off the room because it has to deal with a much higher initial temperature in the housing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This causes extra wear and tear, higher utility bills, loss of refrigerant liquid more rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very helpful to choose the window or location of the unit judiciously, where you get mostly shade from trees or have minded the angle of sun in the afternoons and have a wall or other part of building blocking it, when the daily temperatures are at their peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is advisable to cover the unit in a nice bamboo or similar indigeneus shade lake a small cabinet with enough room around the air condition to allow free ventilation.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-06-06T22:26:17.790" LastActivityDate="2018-06-06T22:26:17.790" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22097" PostTypeId="2" ParentId="22087" CreationDate="2018-06-06T20:29:20.417" Score="4" Body="&lt;p&gt;An AC uses a large volume of air for cooling.  Air in the shade is the same temperature as air in the sun.  Yes sun will heat the outside of the unit but it will have a small effect on the cooling air.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-06-06T20:29:20.417" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22098" PostTypeId="1" AcceptedAnswerId="22099" CreationDate="2018-06-06T20:51:44.573" Score="3" ViewCount="245" Body="&lt;p&gt;I am wanting to produce custom mass plates for a project, very similar to those found in gyms. Here's a rough idea of what each one should look like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jf8aW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jf8aW.jpg&quot; alt=&quot;Rough concept of custom mass plate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is essentially &lt;em&gt;identical&lt;/em&gt; to mass plates one can buy online, however I am needing to produce my own, with custom specifications, sizes, etc. Most notably each plate will need a thread through the center hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where would I even start looking for someone who could do this? Looking online, I get confused by all the metallurgical terms. What type of professional am I looking for? I feel clueless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="16345" LastEditorUserId="2015" LastEditDate="2018-06-06T23:12:57.710" LastActivityDate="2018-06-06T23:12:57.710" Title="Where does one go to produce custom metal objects?" Tags="|mechanical-engineering|metallurgy|project-management|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22099" PostTypeId="2" ParentId="22098" CreationDate="2018-06-06T21:08:06.603" Score="0" Body="&lt;p&gt;The best place to look with just a drawing is a machine shop.  Machine shops work by contacting metal suppliers for them, then creating the items as you need.  For a small batch such as this, a machine shop is ideal.  They will take metal already pre-made and shape it to the shape you need.  For large batches with pre-made work, the machine shop may refer you to a cast-iron shop.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-06-06T21:08:06.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22100" PostTypeId="1" AcceptedAnswerId="22104" CreationDate="2018-06-06T22:03:55.797" Score="2" ViewCount="50" Body="&lt;p&gt;Cement heats up like crazy. I live in the top floor and the heat is unbearable. Also it has a high carbon footprint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why still cement is vastly used? When there are so many better alternatives? &lt;/p&gt;&#xA;" OwnerUserId="12665" LastActivityDate="2018-06-07T04:53:20.077" Title="Why don't we use other substitute of cement while building a roof for thermal proofing?" Tags="|mechanical-engineering|civil-engineering|heat-transfer|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22101" PostTypeId="2" ParentId="22098" CreationDate="2018-06-06T23:08:41.973" Score="2" Body="&lt;p&gt;At higher volumes, these weight plates are generally cast iron. Here's a link to some more info on that: &lt;a href=&quot;http://www.metals-china.com/cast-iron-weight-plate-iron-casting-foundry.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.metals-china.com/cast-iron-weight-plate-iron-casting-foundry.html&lt;/a&gt; . You'll want to search for &quot;cast iron manufacturer.&quot; Adding the country to the end of that search term will help you find someone based on location. Note: Generally, it may be less expensive to manufacture it in Asia, however due to the weight, shipping it back may make it more expensive than making it in the US. Get quotes from both markets to compare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now however, casting isn't ideal for having threads. In this case, after casting, the weight plates need secondary processing via machining to get the threads. Afterwards, a third aditional process would be added to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Most plates are coated with enamel paint or hammertone to resist corrosion; more expensive varieties may be coated with chrome, rubber, or plastic &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;PS. I'd recommend adding some fillets to your sharp edges for comfort and to prevent cuts to the user of your product.&lt;/p&gt;&#xA;" OwnerUserId="15383" LastActivityDate="2018-06-06T23:08:41.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22102" PostTypeId="2" ParentId="22094" CreationDate="2018-06-06T23:13:59.657" Score="1" Body="&lt;p&gt;I don't understand what you're trying to achieve, whether a 15% factor of safety is correct, whether having a factor of safety only on the material strength (instead of a combination of strength-minimizing and load-maximizing factors) or any such thing is adequate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also didn't bother checking your calculations (as in, punching numbers into a calculator). But just looking at your reasoning, it seems fine enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only mistake I spotted was in the volume calculations, where you calculate the volume using diameters, but then add the wall thicknesses only once. If you're using diameters, you should add the thicknesses twice, once for each side of the ball.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or just simplify your life and use radii instead (as you did for the rest of the question):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{align}&#xA;V &amp;amp;= \dfrac{4}{3}\pi \left(r_o^3 - r_i^3\right) \\&#xA;&amp;amp;= \dfrac{4}{3}\pi \left(\left(0.07 + 4.41 \times 10^{-4}\right)^3 - 0.07^3\right) \\&#xA;&amp;amp;= 2.73\times 10^{-5}\text{ m}^3\text{ of PET} \\&#xA;V &amp;amp;= \dfrac{4}{3}\pi \left(\left(0.07 + 4.41 \times 10^{-4} + 35\times 10^{−9} \right)^3 - \left(0.07 + 4.41 \times 10^{-4}\right)^3\right) \\&#xA;&amp;amp;= 2.18\times 10^{-9}\text{ m}^3\text{ of RGO}&#xA;\end{align}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I did not check that your calculations for the values needed for the above (such as 4.41x10&lt;sup&gt;-4&lt;/sup&gt;&amp;nbsp;m needed for the wall thickness) were correct.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-06-06T23:13:59.657" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22103" PostTypeId="2" ParentId="21995" CreationDate="2018-06-07T00:01:50.173" Score="1" Body="&lt;p&gt;I don't believe there are general (longitudinal, transverse, torsion) closed form solutions even for the case where the cylinder is of infinite length.  However, you can use the review article Kostas Soldatos (Soldatos, Kostas P. &quot;&lt;a href=&quot;http://appliedmechanicsreviews.asmedigitalcollection.asme.org/article.aspx?articleid=1395384&quot; rel=&quot;nofollow noreferrer&quot;&gt;Review of three dimensional dynamic analyses of circular cylinders and cylindrical shells&lt;/a&gt;.&quot; Applied Mechanics Reviews 47.10 (1994): 501-516.) as a starting point and solve the eigenvalue problem numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic idea is to start from the three-dimensional equations of elasticity in cylindrical coordinates and then assume appropriate harmonic displacement fields.  Plugging in the displacement fields leads to an eigenvalue problem that can be solve numerically (or analytically in a few cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also look at Grigorenko and Efimova's work (Grigorenko, A. Ya, and T. L. Efimova. &quot;&lt;a href=&quot;https://link.springer.com/article/10.1007/s10778-009-0126-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;Using spline-approximation to solve problems of axisymmetric free vibration of thick-walled orthotropic cylinders.&lt;/a&gt;&quot; International Applied Mechanics 44.10 (2008): 1137-1147.) for a different approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a more recent overview (haven't looked at it closely) you might want to see &lt;a href=&quot;http://14.139.56.14/library/phd/TH188.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://14.139.56.14/library/phd/TH188.pdf&lt;/a&gt; and Grigorenko, A. Ya, and T. L. Efimova. &quot;Free vibrations of continuously inhomogeneous cylindrical bodies.&quot; Shell Structures: Theory and Applications 3 (2013): 293.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-06-07T00:01:50.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22104" PostTypeId="2" ParentId="22100" CreationDate="2018-06-07T04:53:20.077" Score="3" Body="&lt;p&gt;as pointed out by others here, it is hard to beat concrete for inexpensive strength and ease of construction. In the dorm building I inhabited in 1970 at the University of California at Davis (a hot place!), the structure was entirely concrete but the roof of the top floor (where I lived) was covered with several inches of insulating foam topped with tar-covered fabric that was painted a light color. The insulation and light color kept the heat out while the concrete underneath it provided the necessary structural strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, the advantages of both materials were exploited in combination to produce an inexpensive and strong building that did not get roasted by the sun. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-07T04:53:20.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22106" PostTypeId="1" CreationDate="2018-06-07T08:59:16.030" Score="0" ViewCount="242" Body="&lt;p&gt;What is the highest sampling time ever applied in Model Predictive Control (MPC)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an HAVC (Heat, Air Ventilation, and Cooling) it can go to hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I am looking for a sampling interval in the range of months or years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does any one know any practical application?&lt;/p&gt;&#xA;" OwnerUserId="16032" LastActivityDate="2018-07-12T13:39:15.033" Title="MPC highest sampling time" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="1" ClosedDate="2018-07-15T21:31:29.650" ContentLicense="CC BY-SA 4.0" />
  <row Id="22108" PostTypeId="1" CreationDate="2018-06-07T10:24:53.007" Score="2" ViewCount="84" Body="&lt;p&gt;To clarify the title since it cannot be said within the word limit:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;A &quot;simulation&quot; exoskeleton, is an exoskeleton that doesn't exist to support a disabled user's movement or to help someone carry more. It is an exoskeleton that one wears in order to be better immersed into a virtual environment, either for recreation or training.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To explain further, the increased immersion is achieved using the exoskeleton for mechanical feedback, and for full body input, by matching the &quot;rag doll&quot; avatar's position/form to the exoskeleton you can both apply more diverse control to and experience the virtual environment in a far more immersed way. Allowing running, wrestling, climbing, carrying, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the theory at least, and products like &lt;a href=&quot;http://www.dextarobotics.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dexmo&lt;/a&gt; achieve it on a small scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there was a product in progress called Axon VR which would have been a full body experience, I can't find what happened but it seems they've decreased scope and are also doing &lt;a href=&quot;https://haptx.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;gloves&lt;/a&gt;. The shtick of these gloves compared to Dexmo is that there is better tactile feedback, such as heat/texture/etc. Axon VR even has an ex-CEO of twitter invest in it so obviously the concept isn't too out there.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Exoskeletons like these that can be mounted and so don't have to lift your or their own weight, nor have to worry about batteries since then can run off the mains, should be significantly easier to make economically viably right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I want to know&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Whether something is feasibly possible with a 100,000 dollar hardware budget per item at all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) If not what is limiting it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) If it is, then how much would it cost (ball park numbers), and a rough cost breakdown in terms of hardware per complete item.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I understand that software for it is hard as hell, and ensuring safety within legal limits is difficult, or perhaps there isn't a big enough market that even if they were $25,000 to buy each that they wouldn't sell. By answering the above questions I hope to narrow down on the &quot;bigger picture&quot; of it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From what I read online I could find upper limit of power outputs of humans reached around 2000kw (~2.5hp) or for a short period for a professional athlete, in this case cyclists. A 5hp pump can be bought for under \$1k off the &lt;a href=&quot;http://www.fostermfgcorp.com/page/electric/1hp_10hp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;first site I found&lt;/a&gt;, even if you had to go as high as 10hp to not only stop but also &quot;reverse&quot; any actions of a &quot;sprinting&quot; human, which probably wouldn't be safe anyway and wouldn't be a feature, as well as being extremely rare, that's sub \$2k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that you have material for a frame, ~40 cylinders/hydraulic motors (total), hoses, sensors, and (most expensive as far as I know) ~40 servo valves, and I'm sure some other things. (~40 estimated as 3 for each shoulder and hip joint (+12), 2 for each upper and lower arm and leg (+8), a few for each wrist/ankle, and 5 for each hand (+10)). Now I admit servo values can be PRICEY, but in bulk and many of the 40 not under much stress like the ones for fingers I figure the price should stay WAY under \$250 a valve + cylinder, which even then becomes \$10000 which for what you're getting isn't too high in my opinion. And that's an all out all motion one, I imagine most people could have lots of fun with just hands, feet, and waist &quot;simulation&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I mean there are already exoskeletons on the market that can walk people who can't walk! On battery, against the force of gravity, within a reasonable price range. I'm honestly mostly wondering if I'm crazy more than anything, wondering why we don't have simulation exoskeletons already, maybe I am just vastly over estimating their market value.&lt;/p&gt;&#xA;" OwnerUserId="16314" LastActivityDate="2018-06-07T11:21:02.670" Title="What (if any) are the engineering limitations for producing &quot;simulation&quot; exoskeletons" Tags="|mechanical-engineering|electrical-engineering|hydraulics|safety|actuator|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22109" PostTypeId="2" ParentId="22106" CreationDate="2018-06-07T10:42:20.200" Score="3" Body="&lt;p&gt;More of a comment, but a bit longer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sampling interval correlates with the time constant of the controlled system. You would need to find a system that is slow enough such that MPC with sampling time of a few months is still sufficient. Maybe you can control the movement of a glacier..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chemical industry has some of the slowest controlled processes as far as I know (please correct me if there are slower controlled processes) which is also why it was the origin of MPC. Of course the time constant of the system is only an upper bound for the sampling time, whoever designs the controller might still decide to pick a shorter sampling time.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2018-06-07T10:42:20.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22110" PostTypeId="2" ParentId="22108" CreationDate="2018-06-07T11:21:02.670" Score="1" Body="&lt;p&gt;The answer is fairly simple: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;for data &lt;em&gt;input&lt;/em&gt;, video recognition is far less cumbersome than an exoskeleton reading the user moves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For &lt;em&gt;output&lt;/em&gt; - force feedback - our senses are way too complex; the technologies currently provided are all snake oil and money grabs because they would be extremely lackluster had they came to fruition in current form.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The problem is that these devices provide force feedback at bone/joint level. This is not how humans perceive world - we have a huge number of touch receptors in our skin, and they are the primary means of giving us tactile feedback. When you push a wall with your hand, you expect to feel pressure on your palm and bottoms of fingers, not a pull on their topside. You feel the palm resisting as your wrist, elbow and shoulder strain against the resistance, not pressure of straps on your arm, forearm, and a pull on top of your hand, with again, no pressure on palm!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a completely different sensation, not even remotely similar to the actual experience. You'd need an exoskeleton that encloses your entire body tightly, and is able to apply resistance at any point while not obstructing any other - you couldn't use the same 'bones' of the exoskeleton to provide sensation to forearm and simultaneously support the bones of your palm, because if an obstruction blocks the palm, the skeleton must transfer the force all the way from the palm to where the equal and opposing force is applied - at foot soles(!) without affecting any other body part in the process! If it makes the elbow joint rigid, you're surely feel it in your arm and forearm, not just in your palm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The price and complexity of even a simple, flawed exoskeleton is huge, and the result so lackluster and disappointing in comparison, that not a whole lot of people are willing to invest into it - and most of these who do don't realize the flaws. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-06-07T11:21:02.670" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22111" PostTypeId="1" CreationDate="2018-06-07T12:29:49.657" Score="3" ViewCount="324" Body="&lt;p&gt;I have a water jet pump pushing water through a 6&quot; diameter open ended pipe 100 ft long, I need to get 12 bar gauge pressure on the pump discharge, what size (m3/hour) pump do I need ?&lt;/p&gt;&#xA;" OwnerUserId="16350" LastEditorUserId="16486" LastEditDate="2019-01-20T04:16:09.040" LastActivityDate="2020-02-14T10:01:55.247" Title="Flow metrics of water at 12 bar through a 6&quot; diameter open ended pipe 100 ft long" Tags="|mechanical-engineering|fluid-mechanics|piping|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22113" PostTypeId="1" CreationDate="2018-06-07T15:08:33.610" Score="4" ViewCount="136" Body="&lt;p&gt;I am tasked with designing a compressed air piping network,with  and would like to determine the pressure loss across the different layers of instruments&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-now here is the 3D layout of the network. The blue circles are instruments I want to deliver air to. They are in parallel with each other and connected in rows parallel to each other. &lt;strong&gt;L&lt;/strong&gt; connotes pipe length (and in some contexts also the vertical elevation) while &lt;strong&gt;d&lt;/strong&gt; is the diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JZhZ7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JZhZ7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-I would like to find the maximum possible head loss (from point 1 to point 2) given that a proportion of the flow is teed off at each branch.  I would also like to find consequently the proportion of flow teed off in each branch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in assuming points 1 and 2 (below) and going about the normal head loss calculation? How do I account for the tees and wyes branching off?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumptions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow is turbulent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure drop across each brand is designed to be constant (with regulator valves -otherwise they differ due to elevation - maximum between the bottom and the top being 35 meters).&lt;/p&gt;&#xA;" OwnerUserId="15458" LastEditorUserId="2015" LastEditDate="2018-06-07T19:08:48.943" LastActivityDate="2018-06-07T19:08:48.943" Title="Simplifying head loss across a piping network" Tags="|mechanical-engineering|fluid-mechanics|civil-engineering|pipelines|compressed-air|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22114" PostTypeId="1" CreationDate="2018-06-07T16:01:23.433" Score="1" ViewCount="32" Body="&lt;p&gt;I'm not sure if this is the right place to ask this, however:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know if it's possible to make videocam Eyewear. What I mean:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Take simple glasses without lens&lt;/li&gt;&#xA;&lt;li&gt;Instead of lens, use screens on which video from camera will be displayed&lt;/li&gt;&#xA;&lt;li&gt;I build two cameras, one on the left, one on the right,which will capture video to left and right screen.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can this be done? How can it be done? What is needed to create this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P. S. Is it also possible to add a zoom in and out feature through an exterior control? &lt;/p&gt;&#xA;" OwnerUserId="16352" LastEditorUserId="9" LastEditDate="2018-06-07T16:30:58.263" LastActivityDate="2018-06-07T17:43:47.160" Title="Making videocamera Eyewear (glasses)" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="5" ClosedDate="2018-06-13T14:41:12.133" ContentLicense="CC BY-SA 4.0" />
  <row Id="22115" PostTypeId="2" ParentId="22111" CreationDate="2018-06-07T16:12:33.907" Score="1" Body="&lt;p&gt;Related: &lt;a href=&quot;https://engineering.stackexchange.com/questions/16481/system-curve-for-use-in-determining-pump-operating-point/16488#16488&quot;&gt;System curve for use in determining pump operating point&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation, copied, is: $$\Delta P =f_D \frac{\rho V^2}{2}\frac{L}{D} = f_D \frac{8\rho Q^2}{\pi^2}\frac{L}{D^5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friction factor, $f_D$, is based on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Moody_chart&quot; rel=&quot;nofollow noreferrer&quot;&gt;moody chart&lt;/a&gt;.  $Q$ is the flowrate, $\rho$ is the density of the liquid your pumping, $L$ is the length of the pipe, $\Delta P$ is the pressure buildup and $D$ is the diameter.  The basis seems to be that you want a system that would build up 12 bar of gauge pressure on the pump discharge, and at the end of 100ft of pipe, you would discharge to jet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We rearrange as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{12\:bar \: \pi^2 (6\:in)^5 }{(8)\frac{1\:kg}{1\:L}\:100ft} = f_DQ^2 = 0.004 \frac{m^6}{s^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We estimate a relative roughness of 0.001 with relatively smooth cast iron pipes, then the friction factor would be 0.02.  This would make the flow rate run at about 0.5 cubic meters per second, with a water jet of 27.5 m/s.  The pump would need to put out 12 bar at 0.5 cubic meter per second, which would be difficult to source, but not impossible.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-06-07T18:53:11.367" LastActivityDate="2018-06-07T18:53:11.367" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22116" PostTypeId="1" CreationDate="2018-06-07T16:58:53.390" Score="-1" ViewCount="78" Body="&lt;p&gt;When an engineer optimizes a hoe to make an excavator, what do they actually do? I guess that when they see a room for improvement of the hoe, they start solving differential equations. But as the solutions can be transformed under Fourier transform/series, so it's just controlling oscillators. So even if the problem doesn't involve waves or vibrations, what they actually do is adding new oscillators and making old oscillators more or less resonated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct that everything they do involves differential equations? To quote Sidney Coleman: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia page &lt;a href=&quot;https://en.wikipedia.org/wiki/Mathematical_optimization&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mathematical optimization&lt;/a&gt; doesn't talk about this. I wonder how this is viewed in the engineering perspective.&lt;/p&gt;&#xA;" OwnerUserId="11004" LastEditorUserId="11004" LastEditDate="2018-06-08T02:08:19.537" LastActivityDate="2018-06-08T02:08:19.537" Title="Is engineering basically controlling oscillators?" Tags="|mathematics|" AnswerCount="1" CommentCount="3" ClosedDate="2018-06-08T13:07:10.377" ContentLicense="CC BY-SA 4.0" />
  <row Id="22117" PostTypeId="2" ParentId="22116" CreationDate="2018-06-07T17:39:05.707" Score="5" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, there is much much more to most engineering disciplines than analyzing the response of something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, even when that is part of the task, it isn't done by &quot;adding oscillators&quot;, whatever that even means.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-06-07T17:39:05.707" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22118" PostTypeId="2" ParentId="22114" CreationDate="2018-06-07T17:43:47.160" Score="1" Body="&lt;p&gt;What you describe is most of what a Virtual Reality Headset is.  It presents two video signals, one to each eye.  VR headsets are meant to display computer-generated video signals, but it would be a rather small modification to get the two video signals from cameras instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing you seem to be missing is that projecting a image in the same place where lenses of glasses are doesn't work if you expect to see the image.  It would be way too close for your eyes to focus on.  You need optics between the video screens and the eyes so that the images appear at infinity to the user.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-06-07T17:43:47.160" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22120" PostTypeId="2" ParentId="21995" CreationDate="2018-06-07T20:09:30.423" Score="1" Body="&lt;p&gt;If you're designing something, you don't need an analytical formula and can use finite element analysis instead to solve specific configurations. This problem fits very easily into the basic functionality of most FEA software.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2018-06-07T20:09:30.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22121" PostTypeId="1" AcceptedAnswerId="22123" CreationDate="2018-06-07T20:10:51.213" Score="3" ViewCount="375" Body="&lt;p&gt;If I had an aircraft that used propellers for movement and took the air from the front running it through a tube that gets very small towards the end, would it have any impact on the speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the aircraft get faster because the ejected air would be faster then when it got in, similar to a waterpipe that is half closed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be slower because the compressed air would act like a barrier for the new air that comes in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would the aircraft have the same speed regardless of whether we run the air thrusted by the propeller through our cone or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image with what I mean:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/D1xBX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D1xBX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16354" LastActivityDate="2019-04-13T03:57:52.443" Title="Compressing the air thrusted by an aircraft's propeller through a cone like tube" Tags="|airflow|aerodynamics|aircraft-design|propulsion|thrust|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22122" PostTypeId="1" AcceptedAnswerId="22125" CreationDate="2018-06-07T21:08:06.660" Score="0" ViewCount="77" Body="&lt;p&gt;Would an aircraft that uses propellers for movement but with a turbo jet like structure where the ignition chamber only heats the air a bit, similar to a hair dryer instead of releasing fuel and combusting it have any impact on the speed of the craft?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it just act like a normal propeller or will the aircraft get a little faster?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/F8bLv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F8bLv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16354" LastActivityDate="2018-06-08T00:01:20.177" Title="Heating the air inside a turbojet instead of combusting it" Tags="|airflow|aerodynamics|aircraft-design|propulsion|thrust|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22123" PostTypeId="2" ParentId="22121" CreationDate="2018-06-07T21:54:16.590" Score="2" Body="&lt;p&gt;&lt;strong&gt;It would be slower.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compressing the air takes energy - that will slow the propeller, and the plane as a whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a water pump, which is pumping water out of a pipe. Now, if you put your finger over the end of the pipe, the water speeds up, but the pump impeller will slow down, and, as a result, &lt;em&gt;the total mass-flow rate goes down&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of the plane is linked to the mass-flow rate of air going out the back, since it is propelled by the equal/opposite reaction force according to Newton's Third Law.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-06-07T21:54:16.590" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22124" PostTypeId="2" ParentId="22122" CreationDate="2018-06-07T23:45:03.297" Score="1" Body="&lt;p&gt;the amount of energy per second you would have to dump into the air passing through the &quot;heating chamber&quot; in order to make the turboprop engine work would have to be at least as great as that produced by the usual combustion of jet fuel in the heating chamber. Your homework is to do a quick energy balance on the system (using the two heat sources- jet fuel and electrical heat) to see what you would be up against. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-07T23:45:03.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22125" PostTypeId="2" ParentId="22122" CreationDate="2018-06-08T00:01:20.177" Score="2" Body="&lt;p&gt;As a reference, consider the &lt;a href=&quot;https://en.wikipedia.org/wiki/Meredith_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meredith effect&lt;/a&gt;, which applies (applied) in the construction of the P-51 Mustang radiator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the linked Wiki site:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Meredith effect occurs when air flowing through a duct is heated&#xA;  by a heat-exchanger or radiator containing a hot working fluid such as&#xA;  ethylene glycol. Typically the fluid is a coolant carrying waste heat&#xA;  from an internal combustion engine.&lt;a href=&quot;https://en.wikipedia.org/wiki/Meredith_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For the effect to occur, the duct must be travelling at a significant&#xA;  speed with respect to the air. Air flowing into the duct meets drag&#xA;  resistance from the radiator surface and is compressed due to the ram&#xA;  air effect. As the air flows through the radiator it is heated,&#xA;  raising its temperature slightly and further increasing its volume.&#xA;  The hot, pressurised air then exits through the exhaust duct which is&#xA;  shaped to be convergent, i.e. to narrow towards the rear. This&#xA;  accelerates the air backwards and the reaction of this acceleration&#xA;  against the installation provides a small forward thrust.[2] The air&#xA;  expands and decreases temperature as it passes along the duct, before&#xA;  emerging to join the external air flow. Thus, the three processes of&#xA;  an open Brayton cycle are achieved: compression, heat addition at&#xA;  constant pressure and expansion. The thrust obtainable depends upon&#xA;  the pressure ratio between the inside and outside of the duct and the&#xA;  temperature of the coolant.&lt;a href=&quot;https://en.wikipedia.org/wiki/Meredith_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; The higher boiling point of ethylene&#xA;  glycol compared to water allows the air to attain a higher temperature&#xA;  increasing the specific thrust.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If the generated thrust is less than the aerodynamic drag of the&#xA;  ducting and radiator, then the arrangement serves to reduce the net&#xA;  aerodynamic drag of the radiator installation. If the generated thrust&#xA;  exceeds the aerodynamic drag of the installation, then the entire&#xA;  assemblage contributes a net forward thrust to the vehicle.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's important to note that the primary purpose of this construction was to counteract the drag of the exposed radiator. It's certainly unlikely to generate sufficient thrust to power an aircraft.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-06-08T00:01:20.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22126" PostTypeId="1" AcceptedAnswerId="22130" CreationDate="2018-06-08T02:13:37.670" Score="0" ViewCount="497" Body="&lt;p&gt;I need a high voltage power supply for the homemade CO2 laser I am building. I understand that a 15kV 60mA transformer is a thing, but its very big and heavy and its out of phase meaning each lug is only 7,500V and the difference is 15,000V. This would work for most applications but not one where i need 15kV positive and GROUND. I have seen simple arc lighter things that generate a supposed 15kV from a small battery, but the amps must be low? How can i design a simple 15kV high amperage transformer that can be smaller than a neon sign transformer?&lt;/p&gt;&#xA;" OwnerUserId="15207" LastEditorUserId="90" LastEditDate="2018-06-08T14:46:29.280" LastActivityDate="2018-06-08T19:24:41.920" Title="why do 15kV 60mA transformers have to be so big?" Tags="|power-electronics|electrical|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22127" PostTypeId="1" CreationDate="2018-06-08T02:19:41.633" Score="1" ViewCount="45" Body="&lt;p&gt;My main constraint is that the wake pattern to be worked on is far downstream.&#xA;Which one will give me more accurate results?&#xA;Will these two work well for the far wake at all?&lt;/p&gt;&#xA;" OwnerUserId="16357" LastEditorUserId="-1" LastEditDate="2018-06-08T09:18:18.530" LastActivityDate="2018-06-08T09:18:18.530" Title="Is RotCFD or Ansys Fluent better to capture the wake pattern far downstream of a flow in the case of a rotating blade?" Tags="|fluid-mechanics|ansys|cfd|fluid|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22128" PostTypeId="1" CreationDate="2018-06-08T05:03:47.580" Score="1" ViewCount="87" Body="&lt;p&gt;I want to import initial state of a model using predefined fields in import analysis and change the general stiffness matrix and continue simulation. But by default model get the section(material) state of the job file specify in predefined fields. Is there any method to change general stiffness? &lt;/p&gt;&#xA;" OwnerUserId="16359" LastActivityDate="2018-06-08T05:03:47.580" Title="Abaqus Import Analysis" Tags="|abaqus|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22129" PostTypeId="1" CreationDate="2018-06-08T08:55:43.597" Score="1" ViewCount="52" Body="&lt;p&gt;I'm a software developer, who wants to create a monitoring system for an industrial airflow system. The airflow of the system is huge - between 15 and 20 cubic liters per minute. Here is my problem, the strongest airflow sensor which could be connected to a microcomputer, is with capacity 300 SLPM which is 0.3 cubic liters per minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary I have the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a way to use such a sensor to measure airflow which is much much bigger?&lt;/li&gt;&#xA;&lt;li&gt;Maybe there is a way to decrease the airflow in a parallel connection where to put the sensor, measure it, than do some calculations in order to infer the real airflow?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16360" LastEditorUserId="9" LastEditDate="2018-06-08T09:19:39.437" LastActivityDate="2018-06-08T09:19:39.437" Title="AirFlow Measurement" Tags="|measurements|airflow|experimental-physics|industrial-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22130" PostTypeId="2" ParentId="22126" CreationDate="2018-06-08T08:59:36.773" Score="1" Body="&lt;p&gt;Unfortunately, if you have the high-voltage coils, they must be isolated thickly so that you don't get a short-circuit between loops of the coil. Add a rather thick wire to conduct the current needed without excessive losses, and add plenty of coils, because with fewer coils you generate much less of magnetic field in the transformer (in relation to what goes out through supply wires and the rest of circuitry) - and you have a massive device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arc lighters are massively lossy devices. They take a relatively high current on the relatively short primary coiling, and can produce massive voltage on the secondary (which still needs to be isolated well, but can be now much shorter; fewer loops of coil) - but most energy goes 'poof' as heat, induction of wires, and losses in general; the energy efficiency is very poor - but since to create the spark you need very little energy, you can afford these losses, at benefit of small, simple, handy device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get reasonably good efficiency, you must retain the primary:secondary loops count ratio, but you need much more loops on the primary, and thicker wire on secondary. And you can't really afford low efficiency, because the puny 3.7V battery of a spark lighter will barely heat up your circuit, while massive losses with much higher current will melt the wires and set things on fire. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-06-08T08:59:36.773" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="22131" PostTypeId="2" ParentId="22129" CreationDate="2018-06-08T09:05:43.877" Score="1" Body="&lt;p&gt;To measure an airflow you can measure the pressure drop across a section or before and after a bend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That pressure difference can then be calculated into the flow rate.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-06-08T09:05:43.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22132" PostTypeId="2" ParentId="21920" CreationDate="2018-06-08T09:14:31.677" Score="1" Body="&lt;p&gt;From skimming a (sadly german language) &quot;blower bible&quot;, it seems that high pressure radial blowers are mostly used in the pressure range you need. Axial blowers appear to deliver a too low pressure, a compressor is certainly overblown for just 40mbar. However, I would suggest that due to your high volume flows you experience significant pressure losses and that may be why your existing blowers are not up to the task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand your requirement to deliver 100 or 800 m³/h. A radial blower is the wrong tool to adjust over such a wide range of flow rates when you need mostly contant pressure, you you'd need to mount several in paralell and switch them on/off as needed.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-06-08T09:14:31.677" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22134" PostTypeId="2" ParentId="21738" CreationDate="2018-06-08T13:25:26.887" Score="1" Body="&lt;p&gt;I suggest buying extra caps/stoppers and drill a close-fitting hole. If the caps (and your probe/wires can take it), perhaps use some hot-melt glue for a mechanical seal.&lt;/p&gt;&#xA;" OwnerUserId="13014" LastActivityDate="2018-06-08T13:25:26.887" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22135" PostTypeId="2" ParentId="22126" CreationDate="2018-06-08T14:51:17.200" Score="2" Body="&lt;p&gt;Voltage and Current multiply to give power. 15kV times 60mA is the same as 60Volt times 15 Amps (=900 Watts) You would not expect that to be a tiny thing, would you? And that power is without transformation losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that 900 watts a common size for an electric space heater. This is not trivial power you are dealing with.&lt;/p&gt;&#xA;" OwnerUserId="16248" LastEditorUserId="4953" LastEditDate="2018-06-08T19:24:41.920" LastActivityDate="2018-06-08T19:24:41.920" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22136" PostTypeId="1" CreationDate="2018-06-08T14:59:47.337" Score="9" ViewCount="3333" Body="&lt;p&gt;How does anything actually cut anything, since we all know stuff is made up of atoms and molecules, which can't be touched together (It Requires Binding Energy). So how come we see things all around us being cut so easily?&lt;/p&gt;&#xA;" OwnerUserId="16364" LastEditorUserId="-1" LastEditDate="2018-06-10T05:44:34.520" LastActivityDate="2018-06-10T05:44:34.520" Title="How can a knife cut through butter?" Tags="|mechanical-engineering|materials|cutting|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22137" PostTypeId="2" ParentId="22136" CreationDate="2018-06-08T15:50:58.997" Score="3" Body="&lt;p&gt;The knife breaks the bonds between the atoms and pushes the atoms to the side. Depending on the type of bond and density made when the atoms are in a solid state or moving slow atoms are harder to move with another set of atoms. it Less energy is need to move excited the atoms (like butter) creates weaker bonds to break. This is a good for Physics.SE as well.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-06-08T19:13:36.647" LastActivityDate="2018-06-08T19:13:36.647" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22138" PostTypeId="1" CreationDate="2018-06-08T15:57:42.603" Score="2" ViewCount="753" Body="&lt;p&gt;How does time needed for cooking pasta (or any other food) change depending on stove power output (for simplicity, let's say flame size)? If it does... And why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the point that water is &lt;strong&gt;boiling&lt;/strong&gt; and continues in it (thus water temperature is constant, let's say, 100 degrees Celsius),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is there a difference in cooking time:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if I set the biggest fire flame (water is boiling &lt;em&gt;strongly&lt;/em&gt;, many bubbles, water is really rough).&lt;/li&gt;&#xA;&lt;li&gt;I set the smallest fire flame (but enough to keep water boiling) and water is &lt;em&gt;quiet&lt;/em&gt; but still boiling.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&#xA;The simplest model I have in mind is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; Temperature is constant at 100°C;&lt;br&gt;&#xA; &lt;strong&gt;2.&lt;/strong&gt; The kind of pasta is irrelevant; for simplicity we can modelize it as a sphere;&lt;br&gt;&#xA; &lt;strong&gt;3.&lt;/strong&gt; Forget any kind of transport of water into pasta, amid compound into water, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I think the main issues are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;how do we define &lt;em&gt;cooked&lt;/em&gt; for this problem?&lt;/li&gt;&#xA;&lt;li&gt;does heat exchange coefficient change with fire size?&lt;br&gt;&#xA;larger fire -&gt; greater heat goes into water -&gt; much heat goes into pasta -&gt; pasta cooked in less time&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2157" LastEditorUserId="2157" LastEditDate="2018-06-08T21:15:52.540" LastActivityDate="2018-06-09T23:02:00.250" Title="Pasta cooking time" Tags="|heat-transfer|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22139" PostTypeId="2" ParentId="22138" CreationDate="2018-06-08T16:17:12.140" Score="0" Body="&lt;p&gt;More dark or low frequency infrared radiation created from the eye of the stove is present in the pasta on full blast which heat the food more directly then bringing pasta to a boil. Normally the radiation from the stove eye heat the pot then the pot atoms heats the water atoms then heats the food molecules. A UV cooker and light ovens skips those mediums of heat transference cooking the food directly and that is why they are so much faster than a conventional stove.  Chemistry.SE does cook and may want to ask there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP does not mention the heating element. Cooking time differs because of how the food is warmed by what type of element and the amount of radiation produced. When you put food on that is not the same temperature as the stove eye it cools the eye. Depending on the efficiency and robustness of the stove to bring the eye back up to temperature matters in cooking time. &lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-06-09T19:40:03.780" LastActivityDate="2018-06-09T19:40:03.780" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22140" PostTypeId="2" ParentId="22136" CreationDate="2018-06-08T16:18:38.947" Score="16" Body="&lt;p&gt;Muze had a good answer from the Physics side.  From a materials perspective, we see things have grain structure, which explains how different fractures occur.  Butter cuts smoothly because its micro-structure is rather loosely held.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Butter is primarily composed of triglycerides:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H0aoG.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H0aoG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These short heads (where all the Os are) bunch up, while the long tails spread out.  The tails find tails from other triglycerides and weave themselves into giant knots, in the sea of other cells, water, and other compounds in the butter.  This begins a process called crosslinking, which makes butter a polymer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iuhe8.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iuhe8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the butter solidifies into its final solid shape, it's a loosely held together knot surrounding bubbles of loosely solidified organic material.  The net result is an intergranular fracture as you split the various loosely bound knots apart from each other by simply pushing them away from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NnTB2.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NnTB2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorDisplayName="user4139" LastEditDate="2018-06-08T19:24:23.517" LastActivityDate="2018-06-08T19:24:23.517" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22142" PostTypeId="2" ParentId="22138" CreationDate="2018-06-08T18:17:30.993" Score="4" Body="&lt;p&gt;This is a domain where &lt;em&gt;experimental calibration&lt;/em&gt; is the correct approach. Simply because there are far too many variables to determine mathematically.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Pasta starts cooking at 80C, as starch starts absorbing water then. So the moment you add pasta matters.&lt;/li&gt;&#xA;&lt;li&gt;The absorption speed largely depends on mechanical aid of stirring - as water boils, motion of the bubbles stirs the pasta. This process is far too complex to model mathematically.&lt;/li&gt;&#xA;&lt;li&gt;Pasta of different cross-sections, composition, dryness will cook at different rate. This is again not something you determine through models, but experimentally.&lt;/li&gt;&#xA;&lt;li&gt;there are other engineering factors to consider: splashing water (and pasta) when boiling too intensely, ratio of heat transfer into water and modelling temperature in the pot (hotter by bottom, cooler by surface); wasting energy through heat escaping through the sides, pasta sticking to the bottom of the pot or forming clumps if not stirred, is the water salted? (boiling point rise up to 4 Celsius degrees), is the pot covered (higher temperature near the top, which may speed up cooking but will build pressure and splash worse).&lt;/li&gt;&#xA;&lt;li&gt;How do you define 'cooked' - the most accepted method involves taste test, that is experimental. For your model you'd need to develop a hardness function of pasta, determined through dedicated devices.&lt;/li&gt;&#xA;&lt;li&gt;How uniformly will the pasta cook? If you cook thick noodles in very hot water shortly, they will be overcooked on the surface and hard inside.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In short, that all depends. Generally boiling stronger will speed the process up, but may not necessarily provide preferable results than gentle simmering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've actually added some answerable questions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;how do we define &lt;em&gt;cooked&lt;/em&gt; for this problem?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The 'correct' baseline is organoleptic, 'tastes just right'. If you want this to be more 'scientific' you can go with tests spread across many people, and do statistics. This is not necessary though, because you're not discovering a new field; pasta science is long known and established, so you can stand on the shoulders of giants: apply standard cooking conditions to pasta with known recipe, and it will result in 'cooked pasta'. So you have the baseline 'cooked pasta' template. Now for further tests you will need a more simple test, based on properties less cumbersome to measure - &lt;a href=&quot;https://cooking.stackexchange.com/questions/66134/how-much-weight-does-pasta-gain-when-boiled&quot;&gt;for example, mass&lt;/a&gt;. If the mass of drained pasta rose by a factor of 2.25, you have 'al dente'. With a factor of 2.4, you have it thoroughly cooked. It's an easy, reliable test usable in practice.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;does heat exchange coefficient change with fire size?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Heat exchange coefficient is completely insignificant here. Pasta has a very low specific heat coefficient&lt;sup&gt;[citation needed]&lt;/sup&gt; comparing to water, and with it being usually thin-walled, reaches the temperature of surrounding water within seconds after immersion; especially that it soaks the hot water rapidly. So, no, except for very low heat where you dump a lot of pasta into little water significantly reducing the water temperature (and resulting in lousy clumped cooked pasta) you're not changing regarding heat exchange coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cooked pasta differs from raw, soggy pasta in content of water absorbed. (after all, if you cool down cooked pasta, it doesn't revert to uncooked...) And that is what changes with cooking intensity. Wet pasta secretes starch, which forms a weak barrier obstructing further absorption. Stirring washes it off, and allows water to penetrate deeper; motion moves water through pores and promotes penetrating cellular membranes. In short, water that boils hard stirs the pasta and lets it cook &lt;em&gt;a little&lt;/em&gt; faster. Temperature and heat transfer have little influence here: if you stir a slow-simmering pasta with a ladle, you'll get it cooked just as fast as if you cook it on high fire.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2018-06-09T23:02:00.250" LastActivityDate="2018-06-09T23:02:00.250" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="22143" PostTypeId="1" CreationDate="2018-06-08T18:58:46.863" Score="2" ViewCount="26" Body="&lt;p&gt;So, I know it's a long shot but my Mechanics exam is coming up in less than a week and my course is really vague and poorly structured. Also, it has all these exercises with no example at all, so I'm kinda stuck. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would anyone guide me through this problem please? The following ones are similar in structure, and I'm sure that if I understand this one, I'll be able to do the other ones alone.&lt;/p&gt;&#xA;" OwnerUserId="16368" LastActivityDate="2018-06-10T00:37:40.723" Title="Given a material point that goes on a circular trajectory, of radius R, in the z=0 plane, write the intrinsic components for the velocity" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ClosedDate="2018-06-13T14:41:53.703" ContentLicense="CC BY-SA 4.0" />
  <row Id="22144" PostTypeId="1" AcceptedAnswerId="22158" CreationDate="2018-06-08T19:19:23.907" Score="2" ViewCount="122" Body="&lt;p&gt;This mainly would concern Europe, since part of the infrastructure is already in place (trolleys overhead cables). What &lt;strong&gt;economic&lt;/strong&gt; reasons prevent electric autos sans &lt;em&gt;large&lt;/em&gt;  batteries from being feasible in European cities? Lithium is considered by most environmental unfriendly, both extraction, usage(flammable), and disposal. I would see Europe being forefront of electric autos without batteries &lt;em&gt;if&lt;/em&gt; it were feasible.&lt;/p&gt;&#xA;" OwnerUserId="16369" LastActivityDate="2018-07-10T10:05:16.773" Title="Overhead cable vs batteries for vehicles" Tags="|electric-vehicles|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22145" PostTypeId="2" ParentId="6317" CreationDate="2018-06-08T21:29:08.337" Score="0" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;What are the downsides for galvanized steel pipes + press fittings vs. welded steel pipes in a heating application? Diameter would be DN25/1&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Background: &lt;strong&gt;The customer specifies&lt;/strong&gt; only steel, only welded connections for heating pipes starting at this size. &lt;strong&gt;The contractor wants&lt;/strong&gt; to use press fittings, presumably because this way the installation will be easier. The rooms the installation will take place are rather full.&lt;/p&gt;&#xA;&lt;p&gt;Are there good technical reasons to avoid galvanized steel pipes + press fittings? Corrosion, longevity, others?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Technical reasons: Regulations: Health and Safety Executive - &lt;a href=&quot;http://www.hse.gov.uk/comah/sragtech/techmeaspipework.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Design Codes - Pipework&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Corrosion, longevity: Corrosion, yes and longevity, no. If it's for galvanized hot water radiators you should be OK, but it's up to the customer and regulations. For drinking and bathing water you wouldn't want galvanized.&lt;/p&gt;&#xA;&lt;p&gt;See these answers, people dislike galvanized - presumably so does the customer.&lt;/p&gt;&#xA;&lt;p&gt;Answers from our sites:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://diy.stackexchange.com/questions/89397/galvanized-pipe-life-expectancy?rq=1&quot;&gt;Galvanized pipe life expectancy&lt;/a&gt; - particularly &lt;a href=&quot;https://diy.stackexchange.com/a/89409&quot;&gt;this answer&lt;/a&gt; and &lt;a href=&quot;https://i.stack.imgur.com/NI5Yxm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;associated photo&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://diy.stackexchange.com/questions/13898/are-galvanised-steel-pipes-suitable-for-supplying-hot-water&quot;&gt;Are galvanised steel pipes suitable for supplying hot water?&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://diy.stackexchange.com/questions/94401/replacing-galvanized?rq=1&quot;&gt;Replacing galvanized&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;If the customer has spoken with an Engineer or spent a lot on drawings and inspections they may be unreceptive to changes. If it's an idea they came up with on their own they might appreciate &lt;strong&gt;unbiased&lt;/strong&gt; advice - the contractor would need to be certified to weld pipe, press-fitting is easier.&lt;/p&gt;&#xA;" OwnerUserId="14556" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-06-08T21:29:08.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22146" PostTypeId="1" CreationDate="2018-06-08T22:40:00.490" Score="1" ViewCount="73" Body="&lt;p&gt;The last time I welded was too many years ago in engineering school but a friend and I recently bought a welding machine and have set about trying to learn how to weld.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In researching what welding rods to buy, I found there are “sodium” rods such as AWS E6010 (High cellulose sodium) and “potassium” rods such as the AWS E6011 (High cellulose potassium). The web sites for example &lt;a href=&quot;https://www.lincolnelectric.com/en-us/support/process-and-theory/Pages/aws-classifications-detail.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;AWS Classifications Explained&lt;/a&gt; say E6010’s should be used for Electrode Positive DC (DC+) welding and the E6011’s can be used for any of AC, DC+ or DC-.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It makes sense that you would need charged particles to maintain the arc during zero crossings of the current but why should potassium work while sodium would not?  Sodium and potassium are chemically similar and everything else is pretty much the same so why should potassium based rods work with both AC and DC while sodium based rods only work with DC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insights will be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastEditorUserId="7083" LastEditDate="2018-06-11T09:01:33.533" LastActivityDate="2018-06-11T09:01:33.533" Title="Welding rods: Why are some DC and other AC also?" Tags="|mechanical-engineering|welding|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22147" PostTypeId="2" ParentId="8824" CreationDate="2018-06-08T22:53:20.410" Score="1" Body="&lt;p&gt;Steel is better than stainless steel, with the same kind of aluminum being even better; as far as corrosion is concerned. The difference between steel and stainless for strength won't matter as much for rivets as the potential for corrosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Galvanic_corrosion#Anodic_index&quot; rel=&quot;nofollow noreferrer&quot;&gt;anodic index&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The compatibility of two different metals may be predicted by consideration of their anodic index. This parameter is a measure of the electrochemical voltage that will be developed between the metal[s] &lt;s&gt;and gold&lt;/s&gt;. To find the relative voltage of a pair of metals it is only required to subtract their anodic indices.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For normal environments, such as storage in warehouses or non-temperature and humidity controlled environments, there should &lt;strong&gt;not be more than 0.25 V difference&lt;/strong&gt; in the anodic index. For controlled environments, in which temperature and humidity are controlled, 0.50 V can be tolerated. For harsh environments, such as outdoors, high humidity, and salt environments, there should &lt;strong&gt;not be more than 0.15 V difference&lt;/strong&gt; in the anodic index.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;$\begin{array}{lc} \\&#xA;\text{Metal} &amp;amp; \text{Index (V)} \\&#xA;\hline &#xA;\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \text{Most Cathodic} &amp;amp; \\&#xA;\hline &#xA;\text{18% chromium type corrosion-resistant steels} &amp;amp; −0.50 \\&#xA;\text{Chromium plated; tin plated; 12% chromium type corrosion-resistant steels} &amp;amp; −0.60 \\&#xA;\text{Tin-plate; tin-lead solder} &amp;amp; −0.65 \\&#xA;\text{Lead, solid or plated; high lead alloys} &amp;amp; −0.70 \\&#xA;\text{2000 series wrought aluminum} &amp;amp; −0.75 \\&#xA;\text{Iron, wrought, gray or malleable, plain carbon and low alloy steels} &amp;amp; −0.85 \\&#xA;\text{Aluminum, wrought other than 2000 series, or cast alloys (silicon type)} &amp;amp; -0.90 \\&#xA;\text{Aluminum, cast other than silicon type, cadmium, plated and chromate} &amp;amp; −0.95 \\&#xA;\hline &#xA;\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{Most Anodic} &amp;amp; \\&#xA;\hline &#xA;\end{array}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The difference between stainless and aluminum.&lt;/p&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2018-06-08T22:53:20.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22148" PostTypeId="2" ParentId="22138" CreationDate="2018-06-08T23:38:25.540" Score="2" Body="&lt;p&gt;If the temperature is constant 100 C that is all that matters.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting more heat into the pot does not put more heat into the pasta.  That heat just goes to vaporize more water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anything a rapid boil would be longer as heat transfer from a vapor is less than heat transfer from a liquid (same temperature).  Steam will also be 100 C.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-06-08T23:38:25.540" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22149" PostTypeId="2" ParentId="22144" CreationDate="2018-06-09T03:09:24.137" Score="0" Body="&lt;p&gt;Batteries are limited and can explode, short or need service vs while wires do not. The best batteries can only charge about 3000 times before they are depleted. Problems in the wire can be viewed easily vs where a battery has to be tested.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-06-09T03:09:24.137" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22152" PostTypeId="2" ParentId="22143" CreationDate="2018-06-09T08:15:36.470" Score="0" Body="&lt;p&gt;The trajectory you described can be represented as x^2+y^2=R^2.&#xA;So the velocity in x direction will be 2x and that in y direction will be 2y. &#xA;The resultant velocity at any point will be√[4x^2+4y^2]=2R.&lt;/p&gt;&#xA;" OwnerUserId="11114" LastEditorUserId="11114" LastEditDate="2018-06-10T00:37:40.723" LastActivityDate="2018-06-10T00:37:40.723" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22153" PostTypeId="1" CreationDate="2018-06-09T12:35:37.823" Score="0" ViewCount="51" Body="&lt;p&gt;Being an electrical engineer, I'm hoping someone will help me double-check my math. I have a small circuit that I want to power for years that consumes an average 10uA at 3V. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to power this circuit from a small water turbine with a magnetic coupling that can sustain at most 1 oz-in of torque before slipping.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the turbine runs 1hr/day at 100rpms (and 1 oz-in of torque), and assuming the generator and regulation circuit run at 100% efficiency, could it power that circuit. I think it's close.&lt;/p&gt;&#xA;" OwnerUserId="7015" LastActivityDate="2018-09-07T20:01:04.183" Title="Theoretical maximum generated power for a given torque" Tags="|torque|power|energy|generator|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22154" PostTypeId="1" CreationDate="2018-06-09T12:47:09.307" Score="1" ViewCount="84" Body="&lt;p&gt;From the &quot;meep meep&quot; of the smallest scooter to the &quot;BLLLAAAAAGGGGGHHH&quot; of the largest seagoing container vessel, vehicles seem to have horn sounds that, in an auditory way, seem to indicate their mass. Of course, nothing prevents a designer from putting a &quot;large&quot; sounding semi truck horn into a mini automobile, it seems one would only do this for a joke and indeed, I can imagine it would make people laugh because of the unexpected mismatch between the vehicle and its warning sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there actual engineering guidelines that drive such design decisions? Presumably, from a psychological acceptability standpoint or from safety or general human-machine interface principles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note to audience: I tried my best to find the most appropriate SE mini-site. We have &quot;Sound Design&quot; which seems to relate to music; and &quot;User Experience&quot; which doesn't say so but seems 100% related to software design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note to site moderators: A tag for usability engineering or human machine interface design would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="14775" LastEditorUserId="2015" LastEditDate="2018-06-11T16:47:52.370" LastActivityDate="2018-06-13T19:06:21.223" Title="Mapping of vehicle to its horn sound" Tags="|design|safety|environmental-engineering|systems-design|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22155" PostTypeId="1" AcceptedAnswerId="22161" CreationDate="2018-06-09T13:10:36.710" Score="1" ViewCount="136" Body="&lt;p&gt;in my aquarium Ive got a little pump for bubbling air through the water so it is substained with enough oxygen. Internaly the pump works fairly similar to old door bells, where the mains is flowing through a coil thus creating a magnetic field which moves an armature, which again presses air thorugh a hose and underneath the water's surface. (As from the German Wikipedia, &lt;a href=&quot;https://de.wikipedia.org/wiki/Schwingankerpumpe&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://de.wikipedia.org/wiki/Schwingankerpumpe&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now one day I looked at my pump as I found out that its housing  is very tight so that air cant come to the inside. I wrapped it up completely in plastic foil, but the air output didnt reduce. I followed the hose with my eyes, but it (obviously) didnt show any holes in it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Still, I can use the pump (without any possible air input) to fill up a balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is that possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greetings,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~Cr0n&lt;/p&gt;&#xA;" OwnerUserId="16380" LastActivityDate="2018-06-09T18:02:10.013" Title="How does a Vibratory Pump (&quot;Schwingankerpumpe&quot;) work?" Tags="|pressure|airflow|pumps|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22156" PostTypeId="1" AcceptedAnswerId="23179" CreationDate="2018-06-09T13:11:18.590" Score="0" ViewCount="88" Body="&lt;p&gt;When we can manually measure the discharge flow rate,what for have we developed the concept of venturi meter or orifice metre?&lt;/p&gt;&#xA;" OwnerUserId="15268" LastActivityDate="2018-08-10T23:23:46.180" Title="Why do we need flow rate measurement devices?" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22157" PostTypeId="2" ParentId="22154" CreationDate="2018-06-09T15:47:31.150" Score="1" Body="&lt;p&gt;Needs to be heard (volume) within vehicle stopping distance.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-06-11T17:16:01.907" LastActivityDate="2018-06-11T17:16:01.907" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="22158" PostTypeId="2" ParentId="22144" CreationDate="2018-06-09T15:59:23.720" Score="1" Body="&lt;p&gt;The main problem is the extremely peaky load. Your gas tank stores power for a week, batteries store power for days, and can be charged off peak. OLEs deliver on demand, and that sort of power is expensive to dispatch compared to more balanced loads. You really need to look at what this sort of recharging infrastructure consists of. How many combined cycle gas turbine peaker plants would you need to build to supply Munich, say? Maybe six of &lt;a href=&quot;https://www.ge.com/content/dam/gepower/global/en_US/images/h-class/Haveli%20COD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt; would do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a slightly brighter note, there are some limited uses that have been looked at. For longer distances, an intermittent charge-on-the-go system that dump charges supercaps which then discharge to batteries can reduce vehicle weight and improve the versatility of electric vehicles. Using intelligent vehicles to plan routes and use the resources evenly can help. Recharging on straightish uphill segments would be a good exploit.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-06-10T02:42:49.907" LastActivityDate="2018-06-10T02:42:49.907" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22159" PostTypeId="1" CreationDate="2018-06-09T16:39:28.060" Score="2" ViewCount="49" Body="&lt;p&gt;I'm working on a project for a printing company with the main goal to reduce their waste. Paper products are not the issue but they also use materials like plexiglass, polycarbonate and dibond.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if it is at all possible to recycle those materials and how. And if not, what happens with them?&lt;/p&gt;&#xA;" OwnerUserId="16381" LastActivityDate="2018-06-09T16:39:28.060" Title="Recycling materials like plexiglass, polycarbonate and dibond" Tags="|materials|recycling|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22160" PostTypeId="2" ParentId="18680" CreationDate="2018-06-09T17:04:47.680" Score="1" Body="&lt;p&gt;It is the rotational torch on the drive tires through the transmission, drive shaft, yoke and tires that lifts up the one tire not just the engine alone as stated in other answers. I have experience driving a semi-trucks to its limits loaded, unloaded and bobtail and unless the truck extremely modified the truck will never get both wheels in the air. The same why when you rev a car engine it lifts up on the direction of the rotation a little but with the transmission engaged its much more. This flexing is standard,does not hurt the frame and is designed that way to take loads on uneven terrain off road. I cannot say the same for the drive train.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-06-09T17:39:12.140" LastActivityDate="2018-06-09T17:39:12.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22161" PostTypeId="2" ParentId="22155" CreationDate="2018-06-09T17:53:56.333" Score="2" Body="&lt;p&gt;aluminum foil will not stop the seepage of air through its folds. Here is how your pump works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most vibratory aquarium pumps have air inlets of four types: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) underneath the bottom you may find a small tightly-squeezed-in plug of porous cloth (felt) which hides the air inlet and acts like a dust filter to keep dirt out of the air hoses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) cheaper air pumps use the small gap between the top and bottom halves of the plastic injection-molded pump housing as the air inlet. This does not act as a filter but it does allow air to enter the pump housing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) on your pump there is a place where the hose fitment (or &quot;barb&quot;) emerges from the side or top of the pump housing through a hole. The gap between the pump housing and that hose barb can act as the air inlet for the pump inside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) a similar gap exists between the pump housing and where the AC mains wires enter the housing. this gap can also act as an air inlet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know about this because in my previous lifetime, I had to make small vacuum pumps for lab use with very little money to spend. I used aquarium pumps bought in second-hand stores and had to learn how to seal their housings airtight and install a separate vacuum inlet. this required me to know all the different air inlet locations on those things!&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-09T17:53:56.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22162" PostTypeId="2" ParentId="22154" CreationDate="2018-06-09T17:56:40.520" Score="1" Body="&lt;p&gt;the mismatch between the horn sound and the size of the vehicle is a very useful way to gain attention. My Suzuki GS1000 carries a pair (!) of Italian sports car horns, tuned to two different notes, which sound like those horns used on large diesel trucks. VERY effective! &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-09T17:56:40.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22163" PostTypeId="2" ParentId="22155" CreationDate="2018-06-09T18:02:10.013" Score="0" Body="&lt;p&gt;It may by creating small vacuum bobbled by just creating turbulence of enough intensity to create foam bubbles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Agitating water can produce vacuum cavities and foam. In some cases it is taken advantage of, such as submarine torpedoes. They have been designed to create a shroud of small bubbles to reduce the skin friction drastically and move so fast as to hit the target before it can defend itself.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-06-09T18:02:10.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22164" PostTypeId="2" ParentId="22153" CreationDate="2018-06-09T18:02:33.493" Score="1" Body="&lt;p&gt;You're supplying a power of only 30 microwatt, so in a day 2.6 joules will be generated. That's very little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 oz-in is about 0.0071 newton meter, 100rpm is about 10.5 rad/s.&#xA;0.0071Nm at 10.5rad/s multiplied is 0.074 Watt. With 2.6 Joule you'll be able to do that for about 35 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's assumed there are no losses, anywhere. But there will be. Depending on how you'll store the energy, i think you you'll be able to run it for 5 seconds, if at all. So nowhere near an hour, which you're targeting. You just need more power for that, and tell us how the application is designed, only then guesses can be made about losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One tip: get familiar with the metric system, converting an calculating with units is a lot easier.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-06-09T18:02:33.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22165" PostTypeId="2" ParentId="22153" CreationDate="2018-06-09T18:05:28.247" Score="1" Body="&lt;p&gt;your circuit requires (3 volts x 10 microamps) = 30 microwatts to operate. in a 24 hour day it will consume (30 uW x 24hr) = 720 uW-hours of energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;your pump produces (1 oz-in x 100RPM) units of power (we'll worry about the unit conversion later). in a day it will produce (1 oz-in x 100RPM x 60 minutes) of mechanical energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now go to &quot;Unit Conversions&quot; on the web and find the conversion factors necessary to get the pump output into watt-hours, and equate.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-09T18:05:28.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22166" PostTypeId="2" ParentId="22154" CreationDate="2018-06-09T18:07:50.980" Score="3" Body="&lt;p&gt;Ships have loud horns because they need to be heard from far away. Because they can't stop abruptly. For cars it is less so, for scooters even less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A car's engine isn't powerful enough to drive a shiphorn, a scooter not powerful enough to power a car horn. You can go a louder than standard, but regulations prevent manufacturers from doing so, and fines prevent people from fitting loud horns to their vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually horns with alternating tones are also forbidden. Very boring indeed.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="9920" LastEditDate="2018-06-13T19:06:21.223" LastActivityDate="2018-06-13T19:06:21.223" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22167" PostTypeId="2" ParentId="21882" CreationDate="2018-06-09T18:47:22.557" Score="1" Body="&lt;p&gt;I think @sam is correct. To provide some intuition here, imagine that we have a pipe that does what is described in the problem but that had constant diameter throughout. In this case, the pressure at point 2 is 169 kPa gauge. If we increase the diameter of the pipe at point 2, the pressure increases. You can think of this as some of the velocity head of the fluid at point 1 becoming pressure head at point 2. If the diameter decreases, the pressure must go down. That is, some of the head of the fluid at point 1 gets converted to velocity head at point 2. As the problem is posed, there is not  enough head at point 1 to achieve the velocity at point 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To fix this we would either have to increase the pressure at point 1 or increase the height difference. &lt;/p&gt;&#xA;" OwnerUserId="1031" LastEditorUserId="1031" LastEditDate="2018-06-10T19:30:19.323" LastActivityDate="2018-06-10T19:30:19.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22169" PostTypeId="1" CreationDate="2018-06-09T20:23:25.457" Score="2" ViewCount="63" Body="&lt;p&gt;Ques: I have a question regarding solid mechanics and plate theory,is 2 layer plate means laminated plate or simply refers to as a plate?&lt;/p&gt;&#xA;" OwnerUserId="16383" LastActivityDate="2018-06-11T16:33:20.737" Title="Structural mechanics" Tags="|structural-engineering|applied-mechanics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22170" PostTypeId="1" CreationDate="2018-06-09T23:43:15.570" Score="0" ViewCount="114" Body="&lt;p&gt;I have a high sleeper in my room (width 3,70m, length 4,50m, height 3,30m) consisting of one large plane (yellow) and one window (darker grey). The plance leaves about 80cm space to sleep. The door (not visible) is under the plane and should be irrelevant because it remains closed over night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KYmdK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KYmdK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The space between plane and ceiling tends to get hot relatively to the room temperature because warm air rises up. Opening the window (tilt or completely) is not a problem, but the cooling only has a small effect close to the floor of the room - without moving air/wind -, and as good as none under the ceiling for hours (when I wake up at night an go down from the bed and up again I feel a significant temperature difference).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore I'd like to transport cold air from outside inside the room to create a stronger/faster cooling effect. I'm considering a system with multiple small vents. Which direction of airflow would be advantageous? Blowing warm air outside? Sucking cold air in? Laying out airflow in a line (e.g. from window to bed) or in a circle? Directing the airflow close to the ground or the ceiling?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm renting the room and can't change the door, window or floor. The door has no joints and leaves some mm gaps at all sites. The ceiling in the build are old wood constructions which at least lets warm air pass very well in winter (I don't have to heat in the early winter because my neighbors below do that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construction of supports, ducts or else is not a problem because I have access to carpenter tools and expertise in the neighborhood - nor is the look of the installation/making the installation look nice. I want to figure out the engineering aspects of the setup.&lt;/p&gt;&#xA;" OwnerUserId="9780" LastEditorUserId="9780" LastEditDate="2018-07-31T10:44:49.540" LastActivityDate="2018-07-31T10:44:49.540" Title="Ventilation setup high sleeper night cooling" Tags="|airflow|temperature|building-physics|air|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22171" PostTypeId="2" ParentId="22170" CreationDate="2018-06-10T00:36:07.207" Score="1" Body="&lt;p&gt;From a simple physics perspective, please consider the following. A fan or blower located in or near the window will pull cooler air into the room. Ideally, you have to have a means to exhaust the ambient air, allowing such airflow. This would require an exhaust portion in the window partially separated from the fan. As you've noted, hot air is in the upper portion of the room, making the top of the window a better location for the exhaust feature, either passive or another fan. This means the answer to your question about which direction is better is that both are required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty arises in splitting the airflow. It is practical to consider that the incoming air will shed some momentum and reverse course and exit the upper portion of the window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your best solution may lie in the ability to create baffling to direct the airflow for your comfort. A baffle could be something as simple as a lightweight wooden frame covered with paper, easily moved, not particularly expensive. Use the baffle(s) to split the incoming air, specifically around the area of the sleeping plank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your budget allows, small (personal) fans attached to the baffle frame can be used to further guide the airflow to provide assurance of comfort from the cooling air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a living space, it seems impractical to create and install ducting from the window to the sleeping area, although such a construction would provide for the greater temperature management.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your living space allows, suspending paper frame diverters from the ceiling on light rope or strong cord would also allow for vertical management of the airflow. Such an installation could be &quot;dressed up&quot; as an art piece improving the overall appearance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you lack the materials or tools to construct a light wood frame for paper, you should be able suspend from lines some paper panels to help direct airflow. Folding the edges over and applying glue will create some stiffness. Adding a layer of inexpensive cardboard would improve this, especially if the cardboard was attached edge-on to the paper with the corrugations appropriately aligned to enhance stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if you are unable to suspend from the ceiling any of the constructions, another method must be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foundation of all this mindwork is that you have to have a means of moving the air in and moving the air out and keeping the two paths separated and aimed in the desired directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: additionally, you can have a bit of fun at the same time as determining the usually invisible airflow. Purchase a helium balloon of mylar construction. Ordinary plastic balloons have a too-short lifespan for this aspect of the project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attach to the balloon string as closely as possible a paper cup. Either cut the excess string or place it in the cup. Add easily removable weight to the cup until the balloon neither rises nor falls. In the past, I have used balled up pieces of newspaper. Also with a foam coffee cup, one can break off small sections and return some to the cup as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the balloon in a neutral buoyancy state, it will follow airflow in the room quite well. The above configuration should be performed when the windows and doors are closed. When your selected airflow director system is in place, the balloon should be manually placed in the incoming air. It will show you clearly whether the airflow will reach the sleeping area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you discover, for example, that the airflow diverts under the plank or to one side or another, you will know with more certainty where to place a diverter.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2018-06-10T10:41:16.827" LastActivityDate="2018-06-10T10:41:16.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22172" PostTypeId="2" ParentId="22170" CreationDate="2018-06-10T02:05:48.713" Score="2" Body="&lt;p&gt;Air hates to flow into corners, even if there is a fan. A thermally stratified box is about the most obstinate thing there is to make comfortable. The best solution would be to run a small duct against the wall from the floor up through the sleeping loft floor near your feet and on up another 40cm. Something like 5cm x 20cm. Use a fan to push air up the duct. It would be best not to slam the intake right in the corner of the room. Then use a decent window fan to create air exchange through the window. Decent window fan is 50 dollars. Decent duct fan is 35 dollars. The duct can be made out of scrap, or 10 dollars worth of thin plywood. &lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-06-10T02:05:48.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22173" PostTypeId="2" ParentId="10952" CreationDate="2018-06-10T04:02:46.523" Score="2" Body="&lt;p&gt;Use a hanging scale like the ones in the grocery store and attach it to the steering wheel as a medium to see the pounds of force needed to turn the wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JVuNU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JVuNU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-06-10T04:02:46.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22174" PostTypeId="1" CreationDate="2018-06-10T04:16:10.797" Score="2" ViewCount="4287" Body="&lt;p&gt;The processes sound really similar; I'm asking with respect to Martensite processing. Essentially the difference between Tempered martensite and Aged martensite for Steels&lt;/p&gt;&#xA;" OwnerUserId="16385" LastActivityDate="2018-06-13T22:43:07.950" Title="What's the difference between tempering and aging?" Tags="|materials|steel|metals|alloys|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22175" PostTypeId="2" ParentId="10887" CreationDate="2018-06-10T04:21:39.357" Score="2" Body="&lt;p&gt;This jack allow planes to use the air based head sets or use the electrical based headsets. It is an upgraded seat and jack that meet either audio requirements. Older planes use air power headsets normally. Headsets on planes are odd, robust and almost fit nothing to deter theft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The golden pieces are negative positive and right left speaker but I could be wrong and the magnets conduct electricity and the golden pieces are Left negative/positive and Right negative/positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T4fV6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T4fV6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4Ym42.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Ym42.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-06-11T03:46:11.887" LastActivityDate="2018-06-11T03:46:11.887" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22178" PostTypeId="1" CreationDate="2018-06-10T06:22:34.983" Score="1" ViewCount="248" Body="&lt;p&gt;When simulating systems of ODEs, I'm used to always using a numerical integration scheme to propagate the equations forward in time, such as simple Euler integration or Runge-Kutta methods. However, with model predictive control (MPC), every textbook I see gives the following form for propagating a linear system forward in time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$x(t+1)=Ax(t)+Bu(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if we see how this system evolves in time using arbitrary inputs $u(t)$, the results are extremely inaccurate. I would have thought that MPC would incorporate some form of integration scheme while performing the control optimization. So, for example, if we consider Euler integration we would have,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x}(t)=Ax(t)+Bu(t)$$&#xA;$$\frac{x(t+1)-x(t)}{\Delta t}=Ax(t)+Bu(t)$$&#xA;and then our dynamics used as linear equality constraints in an MPC algorithm would be given by,&#xA;$$x(t+1)= x(t) + \Delta t[Ax(t)+Bu(t)]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I've never seen a form like this used before. I know I must be wrong, but I don't know what I'm missing. Any clarification of why general MPC algorithms use $x(t+1)=Ax(t)+Bu(t)$ without any integration scheme would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="9755" LastActivityDate="2018-06-10T10:27:36.527" Title="Model Predictive Control and Numerical Integration Schemes" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22179" PostTypeId="2" ParentId="22178" CreationDate="2018-06-10T08:21:55.917" Score="2" Body="&lt;p&gt;The first equation you wrote is called &lt;strong&gt;discrete-time system&lt;/strong&gt;. They are often also written as&#xA;$$x_{k+1} = Ax_k + Bu_k$$&#xA;Recall that MPC solves an optimization problem &lt;em&gt;every time-step&lt;/em&gt;, so you need your system to have discrete time-steps in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, you can use your Euler integration scheme to discretize a continuous time-system of the form&#xA;$$\dot x = Ax + Bu$$&#xA;before applying an MPC scheme, you basically already did it right in your question without realizing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider, as you already mentioned, the forward Euler equation, you get&#xA;$$\frac{x(t_{k+1})-x(t_k)}{\Delta t} = Ax(t_k) + Bu(t_k),$$ then you can solve this for $x(t_{k+1})$, which yields the discrete-time system&#xA;$$x(t_{k+1}) = (I+A \Delta t)x(t_k) + B\Delta tu(t_k)$$ where $I$ is the identity matrix.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastEditorUserId="14629" LastEditDate="2018-06-10T10:27:36.527" LastActivityDate="2018-06-10T10:27:36.527" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22181" PostTypeId="2" ParentId="10952" CreationDate="2018-06-10T11:51:54.690" Score="2" Body="&lt;p&gt;It heavily depends on the caster that the suspension is set at. Caster makes the wheels go back to neutral position. The more caster, the more torque is needed to keep the wheel at a position. Scrub radius will also make a difference. Google these terms and get familiar with them to get the idea. It will wildly vary per car. Use a scale like Muze mentioned to actualy measure the force needed to keep the steering wheel. Multiply by the radius of the steering wheel to get the needed torque.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-06-10T11:51:54.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22182" PostTypeId="2" ParentId="3348" CreationDate="2018-06-10T13:40:52.303" Score="4" Body="&lt;p&gt;From the accelerator sensor data, you can only calculate pitch and roll.&#xA;The bellow document from Freescale explains with plenty of information what you need:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://cache.freescale.com/files/sensors/doc/app_note/AN3461.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;AN3461 - Tilt Sensing Using a Three-Axis Accelerometer&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the sayings of the document,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan \phi_{xyz} = \frac{G_{py}}{G_{pz}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tan \theta_{xyz} = \frac{-G_{px}}{G_{py}\sin \phi + G_{pz}\cos \phi} = \frac{-G_{px}}{\sqrt{G_{py}^2 + G_{pz}^2}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which equates to:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;roll = atan2(accelerationY, accelerationZ)&#xA;&#xA;pitch = atan2(-accelerationX, sqrt(accelerationY*accelerationY + accelerationZ*accelerationZ))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Of course, the result is this only when the rotations are occurring on a specific order (Rxyz):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Roll around the x-axis by angle $\phi$&lt;/li&gt;&#xA;&lt;li&gt;Pitch around the y-axis by angle $\theta$&lt;/li&gt;&#xA;&lt;li&gt;Yaw around z-axis by angle $\psi$&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Depending on the rotations order, you get different equations.&#xA;For the $R_{xyz}$ rotation order, you can not find the angle $\psi$ for the Yaw around z-axis.&lt;/p&gt;&#xA;" OwnerUserId="16386" LastEditorUserId="9" LastEditDate="2018-06-10T14:11:04.593" LastActivityDate="2018-06-10T14:11:04.593" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22183" PostTypeId="1" AcceptedAnswerId="22188" CreationDate="2018-06-10T14:01:35.580" Score="5" ViewCount="1690" Body="&lt;p&gt;Scooters and skateboards are obviously meant for slower speeds, below 5 MPH. Typically if you go faster than that, it becomes harder and harder to control the scooter, and you will fall off. An example of this is the speed wobble, examples easily found on youtube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the deal? Why is it that comparatively, bicycles can handle higher speeds with more stability and comfort?&lt;/p&gt;&#xA;" OwnerUserId="16388" LastActivityDate="2021-11-22T01:22:01.570" Title="What makes scooters unstable at high speeds?" Tags="|wheels|geometry|bicycles|" AnswerCount="7" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22184" PostTypeId="2" ParentId="22174" CreationDate="2018-06-10T14:14:45.437" Score="4" Body="&lt;p&gt;Although the time and temperatures may be the same, different things are happening. Tempering generally reduces hardness/strength, but improves toughness. Aging martensite is done for a group of specialty steels; PH-precipitation hardening. 17-4 PH is the most common. During aging, hardness/strength and toughness increase. Precipitation hardening is more common in metals other than steels, like aluminum. What is happening on an atomic and molecular scale needs a book to explain.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="1832" LastEditDate="2018-06-12T21:17:59.543" LastActivityDate="2018-06-12T21:17:59.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22185" PostTypeId="2" ParentId="22183" CreationDate="2018-06-10T17:46:23.620" Score="0" Body="&lt;p&gt;The smaller the wheel the less gyroscopic effect is created. Death wobbles can occur at slower speeds with smaller tires. Taller tires is what gives motorcycles and bicycles their stability compared to the smaller tires of a scooter as the speed increases. &lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user16" LastEditDate="2018-06-10T21:06:33.957" LastActivityDate="2018-06-10T21:06:33.957" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22187" PostTypeId="2" ParentId="22183" CreationDate="2018-06-10T18:26:46.560" Score="2" Body="&lt;p&gt;There is less gyroscopic effect but this is not as much a factor.   It is camber of two wheeled vehicle that is self correcting.  If you push a bicycle backwards it will not stay up.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the smaller radius the rpm is higher.  If the tire is out of round it will wobble.  They often put cheaper tires on a scooter and / or do not balance them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller wheel base makes them less stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.ivespa.com/building-a-winning-race-scooter/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Racing scooter&lt;/a&gt; attain high speeds.  &lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-06-10T18:26:46.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22188" PostTypeId="2" ParentId="22183" CreationDate="2018-06-10T20:50:57.150" Score="6" Body="&lt;p&gt;As has been mentioned above the size of tires plays a major role.&#xA;Let's assume the manufacturer standards for a wheel after improving the shaft allows 1/100 mm of play. This much play would encourage a small wheel to resonated while shimmying much more than a big wheel which would need much mor time to complete one turn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 10 cm diameter wheel at 20 km per hour rotates roughly 10 turn/ second which is much closer to its natural frequency of vibration around the shaft versus a large wheel which takes ~1 seconds at the same speed.&#xA;Also, the small tire has a very small I, moment of inertia WRT the shaft axis so it will bounce around easier. $$ \omega_p = \frac {mgr}{ Iw}   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precession is inversely proportional to the spin of the wheel speed meaning the small wheel will precess much faster and because it needs much less play at the shaft to wobble, it will resonate and shimmy. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-06-11T03:11:00.880" LastActivityDate="2018-06-11T03:11:00.880" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22190" PostTypeId="1" CreationDate="2018-06-10T22:47:08.467" Score="4" ViewCount="2164" Body="&lt;p&gt;I'm looking at a soil bearing capacity report, which has a column labeled &quot;kips&quot; and the values are followed by &quot;kps&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;kip&quot; is 1000 pounds, but I'm not sure how to convert from &quot;kps&quot; to psf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data has values in the 17 to 45 kps range. If I assume that there are 1000 psf in one kps, then the soil bearing capacity would be 17,000 pounds per square foot to 45,000 pounds per square foot. This seems wrong to me, as I would expect the values to be in the 1,700 to 4,500 range.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What does the abbreviation &quot;kps&quot; actually mean?&lt;/li&gt;&#xA;&lt;li&gt;How do I convert kps to psf?&lt;/li&gt;&#xA;&lt;li&gt;If a &quot;kip&quot; is 1000 pounds, why does a &quot;kps&quot; seem to be only 100 psf?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16394" LastActivityDate="2018-06-11T08:45:04.170" Title="How do you convert &quot;kps&quot; to Pounds per Square Foot (psf)?" Tags="|structural-engineering|measurements|geotechnical-engineering|unit|" AnswerCount="1" CommentCount="4" ClosedDate="2018-06-25T20:01:20.580" ContentLicense="CC BY-SA 4.0" />
  <row Id="22191" PostTypeId="2" ParentId="21128" CreationDate="2018-06-10T23:29:32.710" Score="0" Body="&lt;p&gt;Use wax the kind people use to pull hair it is durable and doesn't crack unlike candle wax and has a lower melting point. I us this on match sticks to water proof them. It doesn't burn the skin so it should be cold enough not harm your battery. The down side it is hard to get off once one. You can also use rubber cement since it has a 0 reactivity but is flammable in the liquid state and does not dry as hard as the wax, but is easier to get off. I have not tried to mix the to together. I know you where looking for an out of the box solution and hope this helps. Please let me know.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-06-10T23:29:32.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22193" PostTypeId="1" CreationDate="2018-06-01T18:24:01.763" Score="0" ViewCount="36" Body="&lt;p&gt;What are the advantages and disadvantages (technical, financial etc.) of having battery swapping system over charging stations for electric cars? As per my reading, battery swapping seems feasible for electric buses and 2-wheelers but not for cars. Is that really the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it possible for the battery to be housed under the bonnet instead of having it under the floor of the car so as to do away with underground swapping (like the one showcased by Tesla)? I know that the battery is placed there so as to lower the center of gravity. However, if there aren't any other significant benefits, can the aforementioned be done and battery swapping be made easier?&lt;/p&gt;&#xA;" OwnerDisplayName="user35995" LastEditorUserId="33" LastEditDate="2018-12-16T01:14:38.567" LastActivityDate="2018-12-16T01:14:38.567" Title="Battery Swapping in Electric Cars" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2018-06-11T11:34:19.500" ContentLicense="CC BY-SA 4.0" />
  <row Id="22194" PostTypeId="2" ParentId="22010" CreationDate="2018-06-01T19:31:36.770" Score="2" Body="&lt;p&gt;Advantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Minimal turnover time&lt;/li&gt;&#xA;&lt;li&gt;Simple, standardized connection&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Disadvantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Varying &quot;remaining life&quot; in batteries&lt;/li&gt;&#xA;&lt;li&gt;Overhead of disposing of spent batteries&lt;/li&gt;&#xA;&lt;li&gt;Needs standardization (or better, automation) of battery replacement&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For example regarding varying age/use and recycling, I could start a business &quot;recycling&quot; batteries from BE (British Electricity) by putting them in my car and driving to the Cell station for replacement.  I think that to minimize that kind of fraud, you'd have to &quot;subscribe&quot; to a particular vendor, and let them worry about swapping batteries with their competitors (kind of like roaming on phones, a premium service at first but eventually becoming an expected free service).&lt;/p&gt;&#xA;" OwnerUserId="35515" OwnerDisplayName="Cristobol Polychronopolis" LastActivityDate="2018-06-01T19:31:36.770" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22195" PostTypeId="2" ParentId="16301" CreationDate="2018-06-11T01:50:52.317" Score="1" Body="&lt;p&gt;I know of the plastic pellets that you talk about. I would use a aqueduct system to carry them similar how rivers are used to transport logs down stream. Unlike conveyor system since there is much less to break down and can be a closed system using 1 pump to drive the water but I would keep a spare pump tied in. Then at the end of the track use a screened or mesh conveyor belt to catch, dry and preheat the pellets.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-06-11T01:58:58.953" LastActivityDate="2018-06-11T01:58:58.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22197" PostTypeId="2" ParentId="7303" CreationDate="2018-06-11T03:38:40.857" Score="0" Body="&lt;p&gt;The Pratt &amp;amp; Whitney Precision Tools catalog says these had a Jarno #4 Taper in the tailstock.&#xA;[url]&lt;a href=&quot;https://archive.org/details/precisiontools00pratrich[/url]&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://archive.org/details/precisiontools00pratrich[/url]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16396" LastActivityDate="2018-06-11T03:38:40.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22198" PostTypeId="1" CreationDate="2018-06-11T05:50:06.333" Score="3" ViewCount="447" Body="&lt;p&gt;I am trying to work out fatigue life of a part using Miner's rule, however I am having some trouble working out which stress values to use. I have seen some examples but there seems no rhyme or reason to the values they have used in the accumulative Miner's rule, where fatigue failure occurs when :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sum_{j=1}^{j=k}\frac{n_j}{N_j}=1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $n_1, n_2...n_k$&#xA; represents the number of cycles at specific overstress levels and $N_1, N_2...N_k$ represent the life (in cycles) at these overstress levels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example below just used the maximum positive values, with no regard for negative values, even if they were nearly the same as that maximum positive value. Surely this negative stress would have some impact on fatigue, more so than a zero based loading?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b646H.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b646H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I have a stress vs time plot (shown below) that varies from positive to negative quite randomly, where do I get these overstress levels from? Are they the alternating stresses, mean stresses, or max/min stresses that are over the endurance limit of the material?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FkouI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FkouI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13810" LastEditorUserId="13810" LastEditDate="2018-06-11T07:11:35.313" LastActivityDate="2024-01-14T20:07:45.957" Title="Do I use alternating, mean or max/min stresses for Miner's rule?" Tags="|fatigue|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22199" PostTypeId="2" ParentId="22190" CreationDate="2018-06-11T08:45:04.170" Score="3" Body="&lt;p&gt;I've done some searching, according to this &lt;a href=&quot;https://sciencing.com/how-to-convert-kps-to-psi-12399615.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;converter&lt;/a&gt; &lt;em&gt;kps&lt;/em&gt; or &lt;em&gt;KPS&lt;/em&gt; stands for &lt;em&gt;Kilo Pascal&lt;/em&gt;, but don't use that abbreviation.&#xA;The actual abbreviation for Pascal is &lt;em&gt;Pa&lt;/em&gt;, and for kilo a small &lt;em&gt;k&lt;/em&gt;, combined &lt;em&gt;kPa&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus:&#xA;$$ 1\hspace{1mm}\mbox{psf} = 47.88\hspace{1mm}\mbox{Pa} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that assumption, are your values in a reasonable range?&lt;br&gt;&#xA;I assume that &lt;em&gt;kips&lt;/em&gt; in the column title is an error, because usually the tested values need to be in some pressure form, not force itself.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2018-06-11T08:45:04.170" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22200" PostTypeId="1" CreationDate="2018-06-11T10:00:34.660" Score="2" ViewCount="42" Body="&lt;p&gt;I am trying to bond solid wood with with clear acrylic, with requirement to able to withstand shearing force, not more than &lt;strong&gt;6kg over an area of 100 sq.cm&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have tried to bond with result of the Acrylic side getting shear off. Has anyone used better epoxy for the job? Thank you in advance!&lt;/p&gt;&#xA;" OwnerUserId="12799" LastActivityDate="2018-06-11T10:00:34.660" Title="Epoxy/Adhesive to bond solid wood and acrylic" Tags="|mechanical|epoxy|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22201" PostTypeId="2" ParentId="22198" CreationDate="2018-06-11T13:45:11.030" Score="0" Body="&lt;p&gt;I used Miners rule to summarized the fatigue life of different stress cycles for various times . As : stress A for B cycles + stress C for D cycles + stress E for F cycles. I don't think it applies to single stress applications.   &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-06-11T13:45:11.030" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22202" PostTypeId="2" ParentId="19856" CreationDate="2018-06-11T13:47:46.540" Score="2" Body="&lt;p&gt;The number isn't as important as total surface area of the holes - it doesn't need to be greater than the can opening for the remaining pressure resistance to be entirely negligible. Decision on the number should be a trade-off between uniform distribution of the load and possibility of having the holes sealed by dirt or whatever contaminants could be ejected along with the air. You'd need a complex air flow analysis if you want to crush at high speed, where pathways of the air matter but in general case like yours even using corrugated piston surface instead of smooth will be more than sufficient (and also help against the can slipping sideways).  &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-06-11T13:47:46.540" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22203" PostTypeId="2" ParentId="19850" CreationDate="2018-06-11T13:54:18.630" Score="1" Body="&lt;p&gt;In case of air hockey, the holes should be tiny enough, that the difference in pressure should be negligible - the differential between different parts of the surface should be much lower than differential between sub-table and surface/outside. (otherwise, your puck will just plug the holes it's over, instead of floating on air pushed out) I'm not sure if common fan is sufficient for that; the large open area between the fins makes any pressure build-up very difficult. you might want to use a radial turbine, the sort used in a vacuum cleaner. &lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2018-07-12T04:57:46.377" LastActivityDate="2018-07-12T04:57:46.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22204" PostTypeId="1" CreationDate="2018-06-11T14:41:58.363" Score="1" ViewCount="79" Body="&lt;p&gt;I'm in Cambodia where I don't speak the lingo and I'm hoping to do a little DIY project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to buy two different off the shelf steel tubes am I right in thinking that the most common nylon and rubber bush sizes are always between these common tube sizes and I should have no problem finding say top hat bushes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is there more common lengths that these bushes come in, and if one can't find the right length is there generally long lengths of nylon bushing that they can cut down for you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="16403" LastEditorDisplayName="user4139" LastEditDate="2018-06-12T05:44:25.277" LastActivityDate="2018-06-12T19:43:16.863" Title="Do nylon and rubber bushes usually come in sizes between steel pipe sizes?" Tags="|steel|pipelines|connections|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22205" PostTypeId="2" ParentId="6317" CreationDate="2018-06-11T15:23:05.810" Score="1" Body="&lt;p&gt;Galvanized coatings tend to chip and flake after a while. Regardless of the coating, a plain steel pipe WILL corrode/rust after a while. Compression fittings will also eventually loosen up over time, depending on the conditions, sooner or later. Welded stainless (316 or so) while the most difficult to install, would be the most dependable setup for the long haul. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-06-11T15:23:05.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22206" PostTypeId="2" ParentId="22169" CreationDate="2018-06-11T16:33:20.737" Score="1" Body="&lt;p&gt;A 2-layered plate of different materials would be best handled by means of laminated plate theory.  This is due to the strange side affects of bending-stretching coupling - where bending the material can cause it to pull, and vice-versa.  This is because the centroid is not in the geometric center, and as a result pulling on the two materials will cause vastly different strains that will result in bending-like behavior (as can be seen with a bimetalic strip).&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-06-11T16:33:20.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22207" PostTypeId="1" CreationDate="2018-06-11T18:53:05.060" Score="1" ViewCount="49" Body="&lt;p&gt;Let's say that we have a radar that is not an OTH on a boat, and it is placed 30 meters above the ground. Is there a way for it to detect an offshore platform 152 kilometers away from it, since the Earths curvature should prevent it from doing so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I'm asking is that I saw a flat earth video where the guy is in the north sea and this is the supposed setup of his radar. Because the curvature of the earth the offshore platform must be 1376 meters high for it to be detected by the radar. Is there any way for a radar to extend its range over the horizon?&lt;/p&gt;&#xA;" OwnerUserId="16404" LastActivityDate="2018-08-12T17:00:58.103" Title="A radar that is not an OTH detection distance" Tags="|telecommunication|electromagnetism|radar|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22208" PostTypeId="2" ParentId="22183" CreationDate="2018-06-11T18:59:21.810" Score="1" Body="&lt;p&gt;I agree that wobble magnitude vs. wheel size is a significant source of instability.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My suggestion is that another major instability source is the relative lateral acceleration in a turn compared with your ability to compensate. The center of mass is very high on a skateboard, compared with the wheel axes, so a little centripetal force leads to the CM going &quot;outboard&quot; and causing a crash.  By comparison, not only is the CM of a bicycle much closer to the wheel axes (in height), but you can force a far greater out-of-plane tilt on the wheels to correct against that centripetal force. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-06-11T18:59:21.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22209" PostTypeId="2" ParentId="22170" CreationDate="2018-06-11T19:03:09.980" Score="3" Body="&lt;p&gt;I would recommend a modification Phil's answer: airflow is almost always better when the fan is exhausting to the outside - fans don't &quot;like&quot; back pressure and work most efficiently when dumping to an open area.  So, use a fan in the window to push air out, and fill the rest of the window with an inlet duct which is routed up to your loft.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way, there's no other (I hope...) inlet source so all the replacement air is directed at your loft volume, and the fan operates at max efficiency in removing the warm air from the room. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-06-11T19:03:09.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22210" PostTypeId="1" CreationDate="2018-06-11T19:35:57.223" Score="0" ViewCount="68" Body="&lt;p&gt;I'm trying to model, in MATLAB, a membrane in 2D which has an initial triangle-like shape.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nV1VP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nV1VP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This membrane has pressure acting on it from the inside. There are also some pulling forces in random directions on the inner side. I want to analyse the behavior of this membrane over time. The way I want to model this is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the set of membrane points. I define the total energy of the membrane in terms of the pressure and the forces acting on it. Since this is in 2D, my total energy is actually energy per some unit length. So, the pressure energy can be modeled as $-\text{p}(\text{area under the membrane})$, and I can use trapezoidal rule to write area as $\sum_{i=1}^{n-1}(y_i+y_{i+1})(x_{i+1}-x_i)$, where $(x_i,y_i)$ is $i$th point of the membrane, and $n$ is the total number of points of the membrane. (Although it doesn't look like it, the base)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unaware how to model the Bending Energy of the membrane in a similar way. It has to be in the form of $x_i,y_i$ and some constant. A reference to an appropriate bending theory would also be helpful. I seem to remember some formula like $\int \chi H^2 dA$, but I can't really find a source online. Once I have the total energy of the membrane, I can solve some differential equations to get $x_i(t),y_i(t)$. How do I model the bending energy term?&lt;/p&gt;&#xA;" OwnerUserId="11008" LastActivityDate="2018-12-09T05:01:10.227" Title="Modeling the stiffness of a membrane" Tags="|modeling|simulation|energy|matlab|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22211" PostTypeId="2" ParentId="16181" CreationDate="2018-06-11T19:38:04.340" Score="1" Body="&lt;p&gt;I'm going to take the formal question statement as &amp;quot;Is evaporative cooling for a pool effective?&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;There are a handful of factors that need to be considered. First and foremost though, is &lt;strong&gt;wet bulb temperature&lt;/strong&gt;. The wet bulb temperature is &lt;strong&gt;theoretically&lt;/strong&gt; the coldest temperature that you will ever be able to achieve utilizing a wet cooling system (a system that evaporates water to reject heat). Taking this into consideration, the absolute coldest that you will be able to make the pool given your stated data points and method is the wet bulb temperature of 80 Fahrenheit. Realistically, you will probably only be able to come within 2-5 degrees F of this (82-85 Fahrenheit) due to mass transfer limitations and lack of crossflow packing &amp;amp; drift elimination in a small residential unit. For sake of the argument, assume that 80 F is the best you can do using wet cooling.&lt;/p&gt;&#xA;&lt;p&gt;If 80 F is suitable and you would like to pursue this, the next step is to determine total &lt;strong&gt;capacity&lt;/strong&gt; required. &lt;strong&gt;Capacity&lt;/strong&gt; when talking about heat rejection is measured in MegaWatts, kiloWatts, Watts, or BTU/hr. To help understand this concept, think about the total energy coming into the pool that's heating it up; likely the predominant source is sunlight being absorbed into the pool (there may be some heat transfer from the dry bulb temp of the air and the water, but odds are it's negligible, or possibly even cooling the pool off a small amount via some natural evaporation). So if we want to keep the pool at a constant, cooler temperature, we must have enough &lt;strong&gt;capacity&lt;/strong&gt; to remove all of the heat that the sun is putting into the pool. Because we have a very small amount of knowledge about your pool, I'm going to make some considerable assumptions &lt;a href=&quot;https://www.troublefreepool.com/threads/9205-Water-Absorption-and-Heating-from-Sunlight&quot; rel=&quot;nofollow noreferrer&quot;&gt;utilizing previously calculated values&lt;/a&gt;. So we're going to assume a crude measurement of a peak solar energy absorption that induces a 0.75 F / hr (0.41 C / hr) increase in pool temperature. Note that this is NOT accurate if your pool is considerably different geometry, or has a different level of reflection on the pool floor. Additionally I will use a per-unit-of-water calculation so that you can extrapolate this to your pool size as this was not provided. An example was provided for a pool size of 10,000 US gallons&lt;/p&gt;&#xA;&lt;p&gt;Heat capacity of water: 4.184 kJ/kg*k;&#xA;Q = heat flow rate;&#xA;1 US gallon water = 3.785 kg&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2Oebc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Oebc.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, what this is saying: for a 10,000 US gallon pool, at the same conditions described in the link, we should expect to need at least 18 kW of heat removal ability to keep the pool cool.&lt;/p&gt;&#xA;&lt;p&gt;Remember that the coldest we can get is going to be the wet bulb temp; but to get to that wet bulb temp, we have to be able to reject the same amount of heat that's coming into the pool from the sun that we previously calculated (18 kW). To do this, you need to size your evaporative cooler. Unfortunately, evaporative coolers are sized and sold on a &amp;quot;CFM&amp;quot; basis, as their BTU performance is variable depending on inlet conditions (namely temperature and humidity). You can find &lt;a href=&quot;http://www.breezair.com/us/help-me-choose/model-specifications&quot; rel=&quot;nofollow noreferrer&quot;&gt;some manufacturers that will rate their units&lt;/a&gt; based on very tight design conditions; but note that a 15 kW (50,000 BTU/hr) unit at 0.3 IWG conditions may not be able to achieve 15 kW regularly. Accurate sizing of the evaporative cooler will require measurement of heating degree days, historic wet bulb temperatures, and then analysis of this against said design criteria. Alternatively, you could just way over shoot your kW target (if you need 18, maybe go with 30 kW worth, or two 50,000 BTU units). Note: if you try to size an air conditioner using BTU's or tons and translate that to a comparable evaporative cooler by using the CFM required to cool the same square footage/volume, you will undersize the evaporative cooler - The cooler is only designed to hit its design wet bulb temperature approach, NOT to match the same BTU output of an air conditioner.&lt;/p&gt;&#xA;&lt;p&gt;I do not know which model(s) you're looking at, but just looking at some home depot styles and some on the link above, it seems like most coolers are around &lt;span class=&quot;math-container&quot;&gt;$100-$&lt;/span&gt;200 USD/1000 CFM. If we extrapolate &lt;strong&gt;very crudely&lt;/strong&gt; based on the Australian design conditions used by breezair, you can achieve ~1.5 kW / 1000 CFM. To hit 18 kW for a 10,000 US gallon pool would require 12,000 CFM, or roughly &lt;span class=&quot;math-container&quot;&gt;$ 1200 to $&lt;/span&gt; 2400 in capital expense. This doesn't include the power to run the fan (~1kW * (12 hrs / day) * &lt;span class=&quot;math-container&quot;&gt;$.15/kWh =~ $&lt;/span&gt;2/day to operate at peak load), or water costs (&lt;span class=&quot;math-container&quot;&gt;$0.10/gal * ~100 Gal / 12 hrs operation =~ $&lt;/span&gt;10/day peak operation)&lt;/p&gt;&#xA;&lt;p&gt;One other concern with this is cooling water source. It would generally not be recommended to utilize pool water for the evaporative effect, as all the minerals in the water will concentrate up over time as you evaporate water out and recirculate it back into the pool - this will eventually cause scale buildup in the pool water, making it undesirable to swim in. If one discharges the water somewhere else as opposed to back into the pool, then you'd be wasting the treatment chemicals (bleach etc), which would be an additional cost.&lt;/p&gt;&#xA;&lt;p&gt;Alternatives to utilizing a swamp cooler / evaporative cooler is to simply remove the heat source in the first place. If all the heat is coming from the sun, then building a way to provide shade to the water may provide a substantial benefit.&lt;/p&gt;&#xA;" OwnerUserId="8369" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-06-11T19:38:04.340" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22212" PostTypeId="1" CreationDate="2018-06-11T19:59:52.603" Score="1" ViewCount="40" Body="&lt;p&gt;How would I determine if an upward flow of liquid at 100 bar forces a bed of porous solid upwards or flows through without affecting the same bed of material? Is it related to bulk density and pores?&lt;/p&gt;&#xA;" OwnerUserId="11590" LastActivityDate="2018-06-11T19:59:52.603" Title="Overcoming gravity with fluid pressure" Tags="|fluid-mechanics|pressure|liquid|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22213" PostTypeId="1" CreationDate="2018-06-11T21:18:41.007" Score="1" ViewCount="7752" Body="&lt;p&gt;a description for virbation monitoring of industrial bearings states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;peak amplitues at 1 x rpm&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;should be an indicator for a fault in my case. Can somebody explain me what I have to understand under 1 x rpm. My chart (FFT) shows Amplitude in m/s^2 on the y-axis and Frquency in [Hz] on the x-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I need to know how much rotations my bearings do in one minute? If somebody can give an example, also was 2 x rpm mean and how I calculate this value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="14138" LastActivityDate="2018-06-11T21:42:51.207" Title="What does 1 x rpm or 2 x rpm or 3 x rpm in vibration analysis mean?" Tags="|structural-analysis|vibration|bearings|failure-analysis|maintenance|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22214" PostTypeId="1" CreationDate="2018-06-11T21:21:33.873" Score="0" ViewCount="307" Body="&lt;p&gt;I would like to reverse the gearing of my Husaberg FS 650e motorcycle so that by pressing the shifting pedal the gearbox changes up instead of down to the first gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many solutions on the market consisting of linkages that redirect the movement of the shifting pedal. Unfortunately this is not possible on my bike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of there is no ready to buy solution, secondly there is not enough free space around the shaft that would make this approach feasable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am wondering if theoretically a mirrored/inversed shifting drum would accomplish this?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pnTX3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pnTX3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16405" LastActivityDate="2024-03-06T18:02:55.410" Title="Crafting a mirrored shifting drum for motorcycle gearbox" Tags="|mechanical-engineering|gears|cad|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22215" PostTypeId="1" AcceptedAnswerId="22308" CreationDate="2018-06-11T21:22:04.160" Score="1" ViewCount="40" Body="&lt;p&gt;Consider a dynamical system as bellow that is subjected to an impulse input:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sum_{n=0}^N {a_n x^{(n)}}= f_0 \delta(t-t_0)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the initial values are non-resting (all non-zero):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x(t_0^-)=x_0\hspace{0.2cm} ;\hspace{0.2cm} x'(t_0^-)=x_1\hspace{0.2cm} ;\hspace{0.2cm} ... \hspace{0.2cm} ;\hspace{0.2cm} x^{(n-1)}(t_0^-)=x_{(n-1)} $.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I write the correct initial values for $t=t_0^+$? This is ambiguous since integrating the differential equation above, yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ \sum_{n=1}^N a_n  \big[x^{(n-1)} (t_0^+)-x^{(n-1)} (t_0^-)\big] = \sum_{n=1}^N a_n  c_{n-1} = f_0$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $c_i$s are the jumps in initial conditions. In literature, it is said that only the highest order will undergo a discontinuity, yet it's not clear for me why.&lt;/p&gt;&#xA;" OwnerUserId="16406" LastActivityDate="2018-06-18T07:45:42.330" Title="impulse input and discontinuity in initial condition after" Tags="|control-engineering|control-theory|mathematics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22216" PostTypeId="2" ParentId="22213" CreationDate="2018-06-11T21:42:51.207" Score="4" Body="&lt;p&gt;RPM = revolutions per minute. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of vibration monitoring, 1 x rpm means a vibration frequency that is the same as the rotation speed (one vibration oscillation per revolution of the rotor), 2 x rpm means a frequency 2 x the rpm (two oscillations per revolution), etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of nonlinear effects in the mechanical system, vibrations can be excited at different multiples of the rotation speed (1x, 2x, 3x, etc) and even at fractional multiples (for example 0.5x). However the &quot;synchronous&quot; vibration at 1x RPM usually has the biggest amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact some faults in mechanical bearings can excite vibrations at a frequency which doesn't have any &quot;obvious&quot; relation to the rotor speed - for example problems with the bearing lubrication system can excite so-called &quot;oil whip&quot; vibrations at about 0.9 x the rotor speed. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-06-11T21:42:51.207" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22217" PostTypeId="2" ParentId="22210" CreationDate="2018-06-11T21:47:50.983" Score="1" Body="&lt;p&gt;The usual definition of &quot;a membrane&quot; is &quot;something sufficiently thin that the bending energy can be ignored&quot;. If you don't want to ignore it, you are talking about a &quot;thin shell,&quot; not a &quot;membrane&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The theory of thin shells is well known, but too long to describe in an answer here. Any good textbook on solid mechanics, or the finite element method, should have a section on the subject.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-06-11T21:47:50.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22219" PostTypeId="2" ParentId="22214" CreationDate="2018-06-12T01:21:56.913" Score="0" Body="&lt;p&gt;To answer this question you must furnish a complete kinematic model of the entire shifter mechanism, and then study it to determine exactly how to reverse the indexing action by recutting the shifting drum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Neither of these tasks will be straightforward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would be better off designing a simple 2-gear reverser mechanism which would bolt on to the shifter shaft and transmission housing, or by asking yourself just how important it really is to reverse the shifting action on that bike. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-12T01:21:56.913" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="22220" PostTypeId="2" ParentId="22204" CreationDate="2018-06-12T03:11:05.603" Score="1" Body="&lt;p&gt;The nylon bushings(below) that pull up first in google are 1/2 and 3/4 inch and seem the most common on other searches and are the cheapest of the bushings. They are used to make an air tight seal between metal pipes to transport fresh water. The 2 inch to 3/4th inch pipe bushing and pipes would be good for using for a bath tube and light irrigation, 1/4 to 1/2 would be good for a toilet and a 1/8th would be good for a refrigerator ice maker.  Depending on how much water flow you want will depend on what is the most common for that task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YIBeH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YIBeH.jpg&quot; alt=&quot;a&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-06-12T19:43:16.863" LastActivityDate="2018-06-12T19:43:16.863" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22221" PostTypeId="1" CreationDate="2018-06-12T03:16:25.577" Score="1" ViewCount="662" Body="&lt;p&gt;I am making a concrete sleeper wall and am unsure of the current structural integrity (I am not a civil engineer) specifically regarding the width of each terrace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My retaining wall will be made from concrete posts, and sleeper slats with rebar as a reinforcer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QbydF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QbydF.jpg&quot; alt=&quot;3d rendering of wall pieces&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that my rudimentary drawings provide enough information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 5 degree lean back on each terrace of the wall, and each post, and slat is reinforced. However, I do not know how to determine if 4' for each terrace will cause integrity issues. If possible, I'd like to go down to 3' per each terrace. Is there a way to calculate this, or a rule of thumb for these types of retaining walls? Is there a point where I have to use geo-grid reinforcement?&lt;/p&gt;&#xA;" OwnerUserId="16408" LastEditorUserId="16408" LastEditDate="2018-06-12T15:54:30.543" LastActivityDate="2018-06-12T17:16:34.347" Title="How to determine how large each terrace should be on a multi-level retaining wall?" Tags="|structural-engineering|civil-engineering|retaining-wall|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22223" PostTypeId="2" ParentId="22221" CreationDate="2018-06-12T04:12:40.673" Score="4" Body="&lt;p&gt;Design of retaining walls has to deal with many factors and there is no typical design but for heights of less than 4 feet and even then for soils of known competence, in California at least, which I am aware of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the factors are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type of the soil to be retained and its properties such as density, cohesion, compaction ratio, bearing capacity, underlying strata.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factors affecting the stability of the retained soil and general grade and geology supporting it. Even statistics of flooding or seismic activity or subterranean water levels.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, for any job other than short garden walls, the services of a soils engineer are needed. And they will test in situ conditions, assessment of site geology, define the type and height of the walls allowed and passive and active soil pressure and expected seismic load and allowable settlements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then a civil engineer using their numbers and recommendations will design the retaining walls. If there is provisions for tiered setbacks it will be in the soils engineer report, because they would define the influence angle of surcharge on the foundation.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-06-12T04:12:40.673" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22224" PostTypeId="2" ParentId="6654" CreationDate="2018-06-12T06:33:12.090" Score="2" Body="&lt;p&gt;RADAR (RAdio Detecting And Ranging) was first employed during World War Two to locate enemy ships and planes. Since that time radar has evolved into the navigation aid primarily used to avoid collisions. Today it's often referred to as ARPA or automatic radar plotting aid and is a combination of radar and computer technologies merged to augment watch standing. There are two basic marine radar frequencies commonly known as &quot;X&quot; and &quot;S&quot; band. &quot;X&quot; band, because of its higher frequency, 10 GHz provides a higher resolution and a crisper image while &quot;S&quot; band, at 3 GHz is less affected by rain and fog. In most situations larger vessels are fitted with both &quot;X&quot; and &quot;S&quot; band radars while smaller vessels will only have an &quot;X&quot; band. Vessels in excess of 300 gross tons are required to have two operational marine radars and one of those radars must be an ARPA. An industry misconception is that if a vessel is fitted with three radars that all three must be operational. This is not true; only two radars, one being an ATA (Automatic Tracking - like ARPA but requires one user click first!), are required. (Full ARPA [no manual target selection] is a requirement at 10,000GT)&lt;/p&gt;&#xA;" OwnerUserId="16409" LastEditorUserId="-1" LastEditDate="2020-01-06T22:41:51.480" LastActivityDate="2020-01-06T22:41:51.480" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22226" PostTypeId="1" CreationDate="2018-06-12T08:05:18.827" Score="1" ViewCount="120" Body="&lt;p&gt;often we encounter examples where specific energy consumption of a plant(i.e energy consumption/production rate) improves just by increasing the thoroughput or by &quot;pushing&quot; more load through a plant. I want to understand why. Why does the energy consumption of major equipment (say columns, reboilers etc.) increase non-linearly with mass feed rates?&lt;/p&gt;&#xA;" OwnerUserId="16411" LastActivityDate="2018-06-12T15:09:33.103" Title="specific energy consumption vs plant load" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2018-06-25T19:15:09.710" ContentLicense="CC BY-SA 4.0" />
  <row Id="22227" PostTypeId="1" CreationDate="2018-06-12T09:22:50.010" Score="3" ViewCount="245" Body="&lt;p&gt;Let's say that I have the following continuous system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G(s)= \frac{2}{1+s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could convert it to a discrete system using for example the Tustin approximation &lt;a href=&quot;https://en.wikipedia.org/wiki/Bilinear_transform&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Bilinear_transform&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I replace s with: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s \rightarrow{}  \frac{2(1-z^{-1})}{T_e(1+z^{-1})} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence I get the approx. discrete transfer function:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$G(z)= \frac{2}{1+\frac{2(1-z^{-1})}{T_e(1+z^{-1})}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is, &lt;strong&gt;how can I compute its frequency response ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, I would like to be able to compare the discrete approx. freqe. response with the  freq. response of the continuous original transfer function.&lt;/p&gt;&#xA;" OwnerUserId="9039" LastActivityDate="2018-06-13T07:09:44.353" Title="Frequency Response of a Discrete System" Tags="|mathematics|transfer-function|systems-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22228" PostTypeId="2" ParentId="22227" CreationDate="2018-06-12T11:06:24.653" Score="2" Body="&lt;p&gt;In matlab, using &lt;code&gt;c2d&lt;/code&gt; (&lt;a href=&quot;https://www.mathworks.com/help/control/ref/c2d.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;) and &lt;code&gt;bode&lt;/code&gt; (&lt;a href=&quot;https://www.mathworks.com/help/ident/ref/bode.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;) functions:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = tf('s');&#xA;G_c = 2/(1+s);&#xA;Ts = 1;&#xA;G_d = c2d(G_c,Ts,'Tustin');&#xA;bode(G_c), hold on, bode(G_d)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2mRSb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2mRSb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16125" LastActivityDate="2018-06-12T11:06:24.653" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22229" PostTypeId="1" CreationDate="2018-06-12T11:26:51.693" Score="1" ViewCount="156" Body="&lt;p&gt;I have to calculate the slope/tilt from a 6 axis (3D Accelero, 3D Gyro) IMU. Are there open source libraries that help one do this or should I build it on my own? How do I get started?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an open source library to calculate slope, tilt, collision, shock, turning intensity of a vehicle using an IMU? There are great FWs for flight control from which I can derive many things, but couldn't find one for automotives.&lt;/p&gt;&#xA;" OwnerUserId="16414" LastActivityDate="2018-06-12T11:26:51.693" Title="How to calculate slope/tilt from IMU" Tags="|electrical-engineering|automotive-engineering|sensors|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22230" PostTypeId="1" CreationDate="2018-06-12T12:22:00.940" Score="9" ViewCount="780" Body="&lt;p&gt;I am studying the thrust distribution of jet engines.&#xA;But I'm now a bit confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Rolls-Royce's &quot;The Jet Engine&quot; book,&#xA;&lt;a href=&quot;https://i.stack.imgur.com/J96zv.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J96zv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://aeromodelbasic.blogspot.com/2012/05/thrust-distribution-distribution-of.html&quot; rel=&quot;noreferrer&quot;&gt;http://aeromodelbasic.blogspot.com/2012/05/thrust-distribution-distribution-of.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At the start of the cycle, air is induced into the engine and is compressed. The rearward accelera- tions through the compressor stages and the resultant pressure rise produces a large reactive force in a forward direction. On the next stage of its journey the air passes through the diffuser where it  exerts a small reactive force, also in a forward direction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand the first part of the paragraph that the compressor is providing forward thrust, as it is pushing (so compressing) air rearward. But why is the diffuser also providing forward thrust?&#xA;And also why the nozzle is providing rearward thrust?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The similar conculsion is also shown here:&#xA;&lt;a href=&quot;http://www.pulse-jets.com/phpbb3/viewtopic.php?t=2183&quot; rel=&quot;noreferrer&quot;&gt;http://www.pulse-jets.com/phpbb3/viewtopic.php?t=2183&lt;/a&gt;, that the diffuser is providing positive thrust by calculating the pressure force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my understanding of basic fluid mechanics,&#xA;shouldn't a nozzle be providing forward thrust, like the sprinkler in the garden or a fire hose?&#xA;And shouldn't a diffuser be providing rearward thrust, as the outlet speed is slower than the inlet speed, and so m dot X (v - u) is negative?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with my understanding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like I don't know when I need to add a pressure force term P_out A_out - P_in A_in, in addition to the change of momentum term (m dot x (v-u)), while calculating the force of a control volume.&lt;/p&gt;&#xA;" OwnerUserId="16412" LastActivityDate="2020-07-27T20:05:25.477" Title="Thrust distribution of engine" Tags="|fluid-mechanics|propulsion|thrust|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22232" PostTypeId="2" ParentId="19266" CreationDate="2018-06-12T14:51:59.257" Score="1" Body="&lt;p&gt;Moore's law is not a scientific law, it is only a business model (a roadmap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has become very famous because of its accuracy especially during the time when microelectronics was making its first steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are a business man and you manage to predict accurately how technology will develop you will have a much better chance to also estimate how markets will develop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also since ~2009 the resolution of photolithography is not linked Moore's law anymore. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Here is a simplified chronology:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When microelectronics was making its first steps the most successful business men investing in microelectronics were those who followed this estimation from Moore on how microelectronics would develop.&lt;/li&gt;&#xA;&lt;li&gt;Following the success this estimation became widely known and got a name: the Moore's law. This law then started to also drive the investments and would serve as a kind of expectation/agreement between the companies and investors.&lt;/li&gt;&#xA;&lt;li&gt;there was several attempt to push the technology faster, especially as the microelectronic markets were continuously growing up. If someone could bring a technology a little faster there was a possibility to make a lot of money. All of these attempts somehow failed or nearly did. There was a small acceleration with the nodes 90nm-&gt;45nm but the technology limits then catched up, together with the Financial crisis of 2007–2008.&lt;/li&gt;&#xA;&lt;li&gt;Since the technology node 45nm (or 32nm depending on the opinions) the density of devices per area is not anymore linked with the patterning resolution but with the patterning technique. You can add to this a small set of submicron effects in the devices taking more importance: It is now more the investments that drive the technology than the other way around.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Nowaday, a new technology has become very expensive to bring on the market, and is therefore only possible for the richest companies and the smaller investors are looking toward a some end point for this advance. But no worry, In the meanwhile they created the 'More-than-Moore' to drive new investments.&lt;/p&gt;&#xA;" OwnerUserId="10684" LastActivityDate="2018-06-12T14:51:59.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22233" PostTypeId="2" ParentId="22226" CreationDate="2018-06-12T15:09:33.103" Score="4" Body="&lt;p&gt;This is a great question, though I feel it should be refined a bit and have the main question stated more explicitly. That being said, I will attempt to answer it with the assumption that the question is: &lt;strong&gt;Why does the energy consumption of major equipment increase non-linearly with mass feed rates in a processing facility?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any processing facility, you have different forms of input energy. In the vast majority of facilities, you have some sort of industrial heat source (coal, more often natural gas) and usually electricity. Considering this, there are three main areas where you'll notice what are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Economies_of_scale&quot; rel=&quot;nofollow noreferrer&quot;&gt;Economies of Scale&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The first is individual unit operations&lt;/strong&gt;. A great example is a typical distillation column utilizing sieve tray style separation, and attempting to separate a two-species mixture (Ethanol-Water as an example). The individual unit op (in this case distillation) has a design tray vapor and liquid flux rate. This flux rate is generally fixed based upon the physical characteristics of the binary mixture (determined by viscosity, A-B interactions, azeotropes etc). The total flow rate of the column then, is fixed based upon the cross sectional area of each tray (the surface area). In this instance, if you try to run the column too slow, you'll end up with &lt;a href=&quot;https://www.youtube.com/watch?v=6Olr_IliW-k&quot; rel=&quot;nofollow noreferrer&quot;&gt;weeping&lt;/a&gt;, which causes thermodynamic inefficiencies in the separation. Running the column too fast causes &lt;a href=&quot;https://www.youtube.com/watch?v=q7u3NkpeatY&quot; rel=&quot;nofollow noreferrer&quot;&gt;entainment&lt;/a&gt;, which also causes inefficiencies. When most unit ops are designed, they are designed to have some wiggle room for rate, but they usually are designed slightly larger than the intended rate. If we look at the Wikipedia provided image for economies of scale below, you can see if we lie on the left hand side of the curve as we would if the column was over-designed, then pushing the column harder would drive our specific unit cost (or specific unit energy) down. The caveat is that you aren't always on the left hand side of the curve. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/jwhYW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jwhYW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The second phenomenon is heat recovery&lt;/strong&gt;. While you cannot change the energy required to separate ethanol and water in the above example, you can recycle the heat from the distillate and the column bottoms. In most processing facilities this takes place in a heat exchanger. The governing limitations for heat transfer are the total temperature difference (Hottest temperature of the hot stream - coldest temperature of the cold stream), surface area of the heat exchanger, and the heat transfer coefficient. When pushing an existing production facility, you're likely not going to spend extra capital to up-size heat exchangers (at least not at first), and the process temperatures are limited to the physical requirements of the process, so changing delta T is out. But we CAN increase the heat transfer coefficient by running flow faster (and with more pressure drop across the exchanger). If you think about an example as the distillate being condensed from the top of the column exchanging with the feed flow to the column, having higher flow rates of either or both streams will produce better heat transfer coefficients and result in more heat recovery. This directly translates to a lower energy input into the process, and a lower specific energy for the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The final is fixed costs, or more specifically fixed energy costs&lt;/strong&gt;. In any production facility, you have hundreds if not thousands or even tens of thousands of electric motor driven pumps, and process control equipment being powered by compressed air. While some electric motors do have &lt;a href=&quot;https://en.wikipedia.org/wiki/Variable-frequency_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;Variable Frequency Drives&lt;/a&gt;, they can be prohibitively expensive to put on certain motors (4160 Volt VFD's can costs hundreds of thousands of dollars for just one which makes it hard to pay itself back). So if had a 500 kW motor running 24 hrs a day and you made 500 units, (500*24/500), you would use 24 kWh of energy to produce those units. But if you sped the plant up and produce 1000 units, you would have only used 12 kWh to produce those units. This is the fundamental working of economies of scale. &#xA;Similarly, compressed air driven control equipment operates like this. A control valve doesn't care whether you're flowing 10, 100 or 1000 GPM, it's going to use (roughly) the same amount of compressed air energy. This lends itself to the same economies of scale effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To expound further, there are also &lt;a href=&quot;https://en.wikipedia.org/wiki/Diseconomies_of_scale&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dis-economies of scale&lt;/a&gt;. This is the antithesis of an economy of scale, and manifests itself as the right hand side of the graph shown on the Wikipedia article. a Dis-economy of scale is occurring when increasing unit production actually causes an INCREASE in specific production cost (or energy cost in our analysis). This could be something like the distillation entrainment example, pushing phase change equipment (reboilers, evaporators etc) past their &lt;a href=&quot;https://en.wikipedia.org/wiki/Critical_heat_flux&quot; rel=&quot;nofollow noreferrer&quot;&gt;Critical heat flux&lt;/a&gt; and causes heat transfer inefficiencies, or not having enough retention time for chemical reaction unit ops (Fermentation, catalysis, polymerization etc). I've experience a real world example of this where attempting to push a distillation column too fast outpaced the reboiler system capacity and caused excessive live steam injection into the column, drastically increasing water consumption &amp;amp; total energy into the column. Dis-economies of scale are great opportunities for de-bottlenecking.&lt;/p&gt;&#xA;" OwnerUserId="8369" LastActivityDate="2018-06-12T15:09:33.103" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22235" PostTypeId="2" ParentId="22221" CreationDate="2018-06-12T17:16:34.347" Score="3" Body="&lt;p&gt;Please note that in many municipalities, you are required to apply for a building permit before undertaking the construction of anything which requires the moving or excavation of more than a certain quantity of earth on the site. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is true for construction of retaining walls and the like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you derive what you consider a safe design for this project, if the city or county discovers your project and you do not have a permit on file, it is likely they will require you to tear it down- especially if, for example, the drainage pipes you put in there are not visible to the building inspector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the process of getting the permit, you'll be required to furnish a signoff by a registered professional engineer, demonstrating conformance to the construction codes and general due diligence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall objective of this permission chain is to prevent the collapse of the structure and thereby minimize hazards to yourself and the general public. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-12T17:16:34.347" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22236" PostTypeId="2" ParentId="22227" CreationDate="2018-06-12T17:29:43.147" Score="5" Body="&lt;p&gt;If you want to evaluate a continues time transfer function at a specific frequency $\omega$ in rad/s you substitute $s$ with $j\,\omega$. For a discrete time transfer function you substitute $z$ with $e^{T_e\,j\,\omega}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to see why you have to substitute $z$ with $e^{T_e\,j\,\omega}$ you can consider the transfer function $z^{-1}$, which is a delay of $T_e$. So the response to $\sin(\omega\,t)$ would be $\sin(\omega\,(t-T_e))$ which is equivalent to $\sin(\omega\,t-\omega\,T_e)$. The frequency response function should therefore have a constant magnitude of one (or 0 dB), since the output sine wave always has the same amplitude as the input sine wave and the amount of phase shift the output has relative to the input is $-\omega\,T_e$. It can be shown that both the amplitude and phase are captured by $e^{-T_e\,j\,\omega}$, where the minus sign is because we are considering $z^{-1}$ instead of $z$.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastEditorUserId="1765" LastEditDate="2018-06-13T07:09:44.353" LastActivityDate="2018-06-13T07:09:44.353" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22237" PostTypeId="1" CreationDate="2018-06-13T02:12:44.177" Score="-1" ViewCount="124" Body="&lt;p&gt;Good day, I have assigned a task where I have to calculate efficiency per operator as shown in image below. But I have no idea how to get the correct formula. Any help would be greatly appreciated. Thank you&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5M2Ar.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5M2Ar.jpg&quot; alt=&quot;Operator Efficiency&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16420" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-06-13T04:08:50.673" Title="Operator's Efficiency" Tags="|manufacturing-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22239" PostTypeId="2" ParentId="22237" CreationDate="2018-06-13T04:08:50.673" Score="0" Body="&lt;p&gt;You could take the shift duration multiplied by the pieces per minute to get a maximum possible then compare with actual produced ie&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Efficiency is output / input...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the downtime may need to be included,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is downtime acceptable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so - how much?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then perhaps account for minimum allowed downtime by reducing the time for the maximum calculation ie (shift hours - downtime) * pieces per min...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-06-13T04:08:50.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22240" PostTypeId="1" AcceptedAnswerId="22241" CreationDate="2018-06-13T05:45:20.577" Score="1" ViewCount="103" Body="&lt;p&gt;In the following &lt;a href=&quot;http://fumblog.um.ac.ir/gallery/839/mpc_ug.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;book&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Model Predictive Control ToolboxTM (User's Guide)&#xA;&#xA;    Alberto Bemporad&#xA;    Manfred Morari&#xA;    N. Lawrence Ricker&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At page &lt;a href=&quot;http://fumblog.um.ac.ir/gallery/839/mpc_ug.pdf#page=18&quot; rel=&quot;nofollow noreferrer&quot;&gt;1-6&lt;/a&gt;, it has been mentioned that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the plant is open-loop unstable, the maximum possible p is the number of control intervals required for the plant’s open-loop step response to become infinite. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;code&gt;p&lt;/code&gt; means the prediction horizon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This claim does not look reasonable to me.&#xA;I have not understood the reason for this claim. &#xA;What will happen if I choose the prediction horizon longer?&lt;/p&gt;&#xA;" OwnerUserId="16032" LastActivityDate="2018-06-13T08:43:28.990" Title="Maximum MPC prediction horizon for an unstable plant" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22241" PostTypeId="2" ParentId="22240" CreationDate="2018-06-13T08:43:28.990" Score="1" Body="&lt;p&gt;I think the book is using &quot;infinite&quot; to mean &quot;a number that is too big for the computer software to represent&quot;, not in the strict mathematical sense. See near the top of page 1-6:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Open-loop unstable plants: if p*Ts is too large, such that the plant&#xA;  step responses become infinite during this amount of time, key&#xA;  parameters needed for MPC calculations become undefined, generating an&#xA;  error message.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, this is practical limitation of how the software works, not a theoretical limitation of the MPC method.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-06-13T08:43:28.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22243" PostTypeId="1" CreationDate="2018-06-13T11:21:53.237" Score="0" ViewCount="1315" Body="&lt;p&gt;In the equation for half band width given below, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;B = (D + 1)f&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;my textbook states D to be the maximum difference between node numbers in an element and all the elements are considered for its determination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what the above statement means, can someone please explain what D means and how to find its value.&lt;/p&gt;&#xA;" OwnerUserId="14355" LastActivityDate="2018-06-13T17:25:43.527" Title="FEM: What's D in the equation for half band width" Tags="|mechanical-engineering|finite-element-method|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22245" PostTypeId="1" CreationDate="2018-06-13T14:38:30.963" Score="1" ViewCount="71" Body="&lt;p&gt;I'm working on a project that involves supporting a distributed load on a bar that is held up by a moment at one end of a lever arm. My goal is to assess how much the bar bends under the distributed load and how much difference in displacement exists for the actuated and non-actuated lever arms. Here's the setup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lpqEU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lpqEU.jpg&quot; alt=&quot;Torque is purely in the x direction and only on the lever arm shown. Distributed load is entirely in the xz plane. Lever arms are pointed in the y direction.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having done only limited finite element analysis in school before, I'm not sure how best to add constraints. In reality, both lever arms will be free to rotate and will not transfer the moment load past the pivoting axis, where the dashed line is. I did a quick setup the other day where I fixed the bar, but that was more to get a feel for the software I'm using and gave inaccurate results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anywhere that can be fixed in this setup? The only place I can think of is the structure that I haven't shown in the drawing, which is what the pivot points interface with.&lt;/p&gt;&#xA;" OwnerUserId="16304" LastActivityDate="2018-06-13T14:38:30.963" Title="Where to fix part in moment setup for FEA" Tags="|mechanical-engineering|finite-element-method|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22246" PostTypeId="1" CreationDate="2018-06-13T14:44:10.003" Score="1" ViewCount="51" Body="&lt;p&gt;Generally if &lt;code&gt;UE&lt;/code&gt; is in &lt;code&gt;RRC Idle&lt;/code&gt; state, &lt;code&gt;ENB&lt;/code&gt; will inform to &lt;code&gt;UE&lt;/code&gt; via &lt;strong&gt;paging&lt;/strong&gt; message about CS domain i.e incoming call.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My doubt is in LTE, when &lt;code&gt;UE&lt;/code&gt; is in &lt;code&gt;RRC Connected&lt;/code&gt; State, how Network(EnB) will notify to &lt;code&gt;UE&lt;/code&gt; about incoming &lt;code&gt;MT call&lt;/code&gt; ? In which Information element(IE) or message ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help will be appreciated !!&lt;/p&gt;&#xA;" OwnerUserId="16425" LastEditorUserId="7083" LastEditDate="2019-08-07T20:33:59.020" LastActivityDate="2019-08-07T20:33:59.020" Title="How EnB will inform to UE about incoming call in RRC connected state in LTE?" Tags="|electrical-engineering|telecommunication|lte|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22247" PostTypeId="2" ParentId="22207" CreationDate="2018-06-13T15:24:35.747" Score="1" Body="&lt;p&gt;No.  Stating the obvious,  a line of sight (LOS) system won't detect things below the horizon.&lt;br&gt;&#xA;Now, if atmospheric conditions cause a waveguide effect, signals could propogate farther (and back), but that means intermittent dropouts at best.  In addition, since that signal path includes an unknown number of bounces (reflections), the true distance is rather difficult to calculate.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-06-13T15:24:35.747" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22248" PostTypeId="1" CreationDate="2018-06-13T15:47:12.660" Score="1" ViewCount="20" Body="&lt;p&gt;I want to build rotating arms of length around 2ft that can hold up to 30 lbs of weight and am wondering what kind of motor to get. I'm thinking of getting a step motor so that I can know exactly which angle the arm is at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key requirement is that it needs to be a high torque delivering system. Gears would be good to achieve this but as a beginner, I don't have a lot of experience with constructing my own gear-motor combinations.&lt;/p&gt;&#xA;" OwnerUserId="16430" LastActivityDate="2018-06-13T15:47:12.660" Title="What are some things to keep in mind when searching for a step motor where a lot of torque is required?" Tags="|stepper-motor|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22249" PostTypeId="2" ParentId="10305" CreationDate="2018-06-13T16:23:44.630" Score="1" Body="&lt;p&gt;&lt;strong&gt;Polyurethane is the better choice&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both polyurethane and polyester resins need UV inhibitors added to form UV-resistance.  The polyurethane will hold up fairly well to UV with the metal filings, and can handle the molding defects better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However, a composite of the two may be better&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Polyester can be handled with styrene addition.  This lowers the viscosity of the working laminate and allows for a longer cure time, to resolve any problems.  Additionally, polyester and other epoxy resins bond well to steel when using a methacrylate based adhesive.  It may be best to coat your foam with polyester first for structure, and bond it to the steel, then cover with the final layer of polyurethane and metal filings.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-06-13T16:23:44.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22250" PostTypeId="2" ParentId="22243" CreationDate="2018-06-13T17:25:43.527" Score="2" Body="&lt;p&gt;This would only apply for a simple FE program where the nodes in the model are always labelled from 1 to $n$, every node has the same number of degrees of freedom $f$, and the rows and columns of the global matrix are also ordered in groups of $f$ from 1 to $n$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, think about the case where there is only one degree of freedom at each node, or $f = 1$. If the smallest and largest notes in an element are $p$ and $q$, when you assemble element matrix into the global matrix all the terms fit within the square submatrix between rows and columns $p$ and $q$, so the maximum distance of the terms from the diagonal of the global matrix (i.e. the bandwidth) is $q-p+1$ or $D+1$ in your notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bandwidth of the global matrix is therefore the maximum value of $D+1$ for all the elements in the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are $f$ degrees of freedom at each node, the system matrix is $f$ times bigger, and the bandwidth is also multiplied by $f$ to give your formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 1, the value of $D$ in each element depends how you number the nodes in the model, and reducing the bandwidth of the global matrix will reduce the computer time required to solve the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2, in modern &quot;real world&quot; FE software this is not very important any more, since different equation solution methods would be used which don't depend on minimizing the bandwidth to make them efficient.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-06-13T17:25:43.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22251" PostTypeId="1" AcceptedAnswerId="22375" CreationDate="2018-06-13T18:00:42.117" Score="1" ViewCount="1256" Body="&lt;p&gt;I am modelling a plate as 3D deformable solid. My plate has a spatially dependent isotropic thermal expansion coefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have defined this spatial distribution as a Discrete Field in a tabular form (for the mesh elements). My question is how can I link this distribution to my material definition? Or is it not possible to have spatially dependent thermal expansion in Abaqus/CAE?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16434" LastEditorUserId="16434" LastEditDate="2018-06-14T21:04:51.683" LastActivityDate="2018-06-21T14:50:49.163" Title="Spatially dependent material property in Abaqus/CAE" Tags="|finite-element-method|abaqus|thermal-expansion|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22252" PostTypeId="1" AcceptedAnswerId="33197" CreationDate="2018-06-13T19:33:18.607" Score="1" ViewCount="1133" Body="&lt;p&gt;I have some coefficient values i.e. &lt;code&gt;74.9&lt;/code&gt;, &lt;code&gt;81.3&lt;/code&gt;, and &lt;code&gt;80.5&lt;/code&gt; at frequency centers &lt;code&gt;63&lt;/code&gt;, &lt;code&gt;160&lt;/code&gt; and &lt;code&gt;400&lt;/code&gt; Hz. The values are &lt;code&gt;1/3&lt;/code&gt; octave band values. I would like to convert them to the &lt;code&gt;1/1&lt;/code&gt; octave band values. Can someone suggest how to do that? &lt;/p&gt;&#xA;" OwnerUserId="16435" LastActivityDate="2023-09-16T18:59:35.797" Title="How do I convert coefficient values from 1/3 octave band to 1/1 octave band?" Tags="|acoustics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22253" PostTypeId="1" CreationDate="2018-06-13T20:15:01.020" Score="2" ViewCount="30" Body="&lt;p&gt;How to create a 3D model of a person's veins using ultrasound? How to do something like this, &lt;a href=&quot;https://www.biosensewebster.com/products/carto-3/cartosound-module.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;CARTOSOUND® Module&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This 3D ultrasound technology gives you all the benefits of traditional intracardiac echo  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why is most ultrasound 2D?&lt;/li&gt;&#xA;&lt;li&gt;Is 3D ultrasound a separate technology from 2D? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4214" LastEditorUserId="5899" LastEditDate="2018-06-14T14:35:29.900" LastActivityDate="2018-06-14T14:35:29.900" Title="How to create a 3D model of a person's veins using ultrasound?" Tags="|ultrasound|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22254" PostTypeId="1" CreationDate="2018-06-13T20:55:02.033" Score="1" ViewCount="149" Body="&lt;p&gt;I am trying to construct a window seat like this one:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0zqnT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0zqnT.jpg&quot; alt=&quot;window seat&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in Canada.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering what are the common construction code points to consider on this one?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Materials - does it matter if this is just a seat and not a bearing wall, I was planing 2x4s from Home Depot.&lt;/li&gt;&#xA;&lt;li&gt;Screws vs Nails - I was thinking screws with pre-drilling.&lt;/li&gt;&#xA;&lt;li&gt;Framing - should I follow the spacing guidelines for the wood frame or as long as it's sturdy it's all that matters?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electrical Heating Element&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have an electrical heating element attached to the wall right where the seat will go, so I was thinking to move it onto the face of the seat once completed. Is there any specific plating that has to be installed to drywall?&lt;/p&gt;&#xA;" OwnerUserId="6205" LastActivityDate="2018-07-14T07:20:13.870" Title="Common points to consider when building a window seat" Tags="|building-design|wood|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22255" PostTypeId="2" ParentId="22174" CreationDate="2018-06-13T22:30:12.910" Score="0" Body="&lt;p&gt;TL;DR:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tempering of martensite is trading strength and hardness for toughness amd ductility. Martempering is a heat treatment where a workpiece is held at constant temperature until through thickness temperature equilibrium is reached, to ensure through thickness martensite formation. Aging is elevated temperature precipitation hardening. Maraging is aging which involves the presence and maintenance of a martensite phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, you may want to search for the terms &quot;martempering&quot; and &quot;maraging&quot;. They are contractions (portmanteaus?) of martensite and the respective process name. It is worth noting these processing methods are somewhat different from traditional tempering and aging. Different enough to warrant a special name.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First a bit of background. When a steel is heated to above its A3 temperature, the iron changes phases from BCC to FCC. The FCC lattice has larger free spaces, or interstitial sites, and the carbon enters these sites, forming a solid solution with the iron. If the steel is quenched rapidly enough and to a low enough temperature, a speed-of-sound diffusionless process occurs where the dissolved carbon is trapped in the iron BCC lattice. The carbon doesnt fit well in the BCC lattice interstitial sites, distorting the latticr into an elongated BCT lattice. The BCT lattice is under high strain, which in turn causes stresses in the bulk steel, resulting in work hardening by formation of a very high density of dislocations. The dislocations are a large part of what makes martensitic steel hard, strong, and brittle. Dislocations are what allows metals to undergo plastic deformation, by gliding. Its only when there are so many of them that they get tangled up, as in the BCT distortion, or caught on the interstitial carbon atoms, that they increase strength and brittleness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Martempering is the process of quenching a martensitic steel workpiece to below the martensite start temperature, and holding it there until thermal equilibrium, to achieve a martensitic microstructure throughout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditional tempering of martensitic pieces exchanges hardness and strength for toughness by allowing dislocation density to decrease at the elevated tempering temperature. Dislocations are metastable, so they eventually work their way out of the lattice, given enough time. Higher temperatures make the process go exponentially faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maraging is a special process which involves precipitation hardening. It requires large quantities of alloying elements like Ni, Co, Ti, and Mn. These elements have high affinity for carbon, nitrogen, sulfur, and phosphorus, and will form intermetallic phases given enough time at the right temperatures. Finely distributed intermetallic particles harden metals by blocking dislocation motion, preventing glide, and trading toughness for more strength and hardness. To achieve this the workpiece is quenched to achieve a martensitic structure, then aged at a slightly elevated temperature long enough for the intermetallic phases to precipitate out of solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditional aging is very similar, but does not generally involve a martensitic phase.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2018-06-13T22:43:07.950" LastActivityDate="2018-06-13T22:43:07.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22256" PostTypeId="1" AcceptedAnswerId="22270" CreationDate="2018-06-13T23:53:25.043" Score="0" ViewCount="174" Body="&lt;p&gt;Most undergraduate majors require 10-12 courses in the major, and maybe up to three courses from other departments as &quot;co-requisites.&quot; But most engineering majors require 15-18 courses in the department itself, plus 10 -15 courses in math, chemistry, physics and computer science as &quot;co-requisites.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one engineering program I looked at, a &quot;minor&quot; consisted of six engineering courses, plus 14 courses in &quot;co-requisites.&quot; I &quot;get&quot; that much, particularly the co-requisites (basically enough for another minor). But the major consisted of 18 engineering courses, three times the &quot;six&quot; for the minor, whereas other majors, require twelve courses (double the six for the minor) as well as co-requisites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's 25-33 courses, that is two thirds to four fifths of the whole curriculum, versus about one-third for other majors. Why is that?&lt;/p&gt;&#xA;" OwnerUserId="62" LastEditorUserId="62" LastEditDate="2018-06-14T05:15:40.257" LastActivityDate="2018-06-14T16:25:45.040" Title="Why are engineering programs so course intensive?" Tags="|education|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="22257" PostTypeId="1" CreationDate="2018-06-13T23:53:28.443" Score="1" ViewCount="77" Body="&lt;p&gt;I'm new to all of this and am working on a project with someone. They mentioned using this kind of a rotary table, &lt;a href=&quot;https://www.velmex.com/Products/Rotary_Tables/Motorized-Rotary-Tables.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Velmex - Motorized Rotary Tables&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XFj5x.png&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Motorized Rotary Tables&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XFj5x.png&quot; alt=&quot;Motorized Rotary Tables&quot; title=&quot;Motorized Rotary Tables&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and mounting it on a flat horizontal surface using a mounted bearing of some kind - like this but larger, &lt;a href=&quot;https://www.grainger.com/product/36UZ04?cm_mmc=PPC:%20Google%20PLA&amp;amp;s_kwcid=AL!2966!3!50916755157!!!g!109230485397!&amp;amp;ef_id=Wh_ziQAAAsT13BME:20180613235028:s&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pillow Block Bearing, Number of Bolts: 2, Ball Bearing Type, 7/8&quot; Bore Dia.&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4RwDl.png&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Pillow Block Bearing&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4RwDl.png&quot; alt=&quot;Pillow Block Bearing&quot; title=&quot;Pillow Block Bearing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The descriptions that I'm reading are saying that the rotary table comes with bearing of its own? So I'm confused if this is something that's possible. I'm going to look through some more diagrams of the rotary table and see if I can see how a shaft would go through &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't need to tell me how it can be done, just if it is something that's possible because my other idea was using an L-shaped clamp, something like this, &lt;a href=&quot;https://hacktronics.co.in/mechanical-parts/large-l-shaped-clamp-with-extra-holes-for-mounting&quot; rel=&quot;nofollow noreferrer&quot;&gt;Large L Shaped Clamp with Extra Holes for Mounting&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7zOYX.png&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Large L Shaped Clamp with Extra Holes for Mounting&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7zOYX.png&quot; alt=&quot;Large L Shaped Clamp with Extra Holes for Mounting&quot; title=&quot;Large L Shaped Clamp with Extra Holes for Mounting&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16433" LastEditorUserId="5899" LastEditDate="2018-06-14T14:35:14.630" LastActivityDate="2018-06-14T16:25:29.600" Title="How to go about attaching a rotary table to a mounted bearing?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22258" PostTypeId="2" ParentId="22256" CreationDate="2018-06-14T05:12:57.800" Score="2" Body="&lt;p&gt;Some of the difference is that some other courses have a lot of individual reading time and engineers get more class / lab / experiment time - which also means that the Uni needs expensive equipment that needs to be maintained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Uni had sub &amp;amp; supersonic wind tunnels , water flow stuff axial and radial, water channel etc etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineering courses are not cheap and some Universities do concentrate on law and marketing and humanities as only books are required with less contact time... Each of our lab experiments were 3 hours ....&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-06-14T05:12:57.800" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22259" PostTypeId="2" ParentId="22254" CreationDate="2018-06-14T05:20:31.817" Score="1" Body="&lt;p&gt;I don't think there is any problem with the idea, with these considerations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat register should be relocated to somwhere that is not likely going to be exposing someone to excessive heat or burn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2x4 at 16&quot; o.c. type 5 nailing is ok with with 16 d nails, unless you have decking screws with data sheet so you can adequately provide for 100 lbs/ sqr foot of bench load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bench covers and fabrics should meet locale fire retardant code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glazing should be tempered and meet local impact resistance codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your local buulding department should be a good source of information.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-06-14T06:35:14.983" LastActivityDate="2018-06-14T06:35:14.983" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="22260" PostTypeId="1" AcceptedAnswerId="22276" CreationDate="2018-06-14T06:12:34.083" Score="2" ViewCount="2221" Body="&lt;p&gt;I want to use a DC 12/24V motor to rotate an assembly of glass tubes(image attached) through a hollow shaft of 25mm OD and 22mm ID, the shaft is supported on bearings on both the sides(assume minimum friction), two 3mm thick circular plates hold the four tubes together at both the end. The shaft is attached to the center of the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Data:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Glass tube: 58&amp;nbsp;mm OD, 45&amp;nbsp;mm ID 1.8&amp;nbsp;m long&lt;/li&gt;&#xA;&lt;li&gt;Disk: 180&amp;nbsp;mm OD, 3&amp;nbsp;mm thick attached at 25&amp;nbsp;mm from both the edges of the glass tube.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Even a general procedure/idea of calculating torque requirement is appreciated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l3XKY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l3XKY.jpg&quot; alt=&quot;Close up of the assembly&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nrls4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nrls4.jpg&quot; alt=&quot;Complete assembly&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/giCp3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/giCp3.jpg&quot; alt=&quot;Front view of the assembly&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16440" LastEditorUserId="1832" LastEditDate="2018-06-14T11:19:22.040" LastActivityDate="2018-06-15T07:39:16.750" Title="Calculating the torque requirements of the motor for a specific case" Tags="|motors|applied-mechanics|torque|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22261" PostTypeId="1" CreationDate="2018-06-14T06:44:56.233" Score="4" ViewCount="123" Body="&lt;p&gt;An existing sewer treatment plant is experiencing maintenance issues with their first stage (bar screens) where grit which is just barely larger than the apperture (bar spacing) of the bar screens gets jammed in between the bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been tasked to solve the problem. It was suggested to install a grit separator upstream from the bar screens. The conventional approach is to provide  grit separation downstream however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could not find a solution in my normal literature (Metcalf &amp;amp; Eddy &amp;amp; MWH) and google, so am asking the question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If grit separation can be done upstream, what type would be used? I assume that it would probably employ a vortex type system.&lt;/p&gt;&#xA;" OwnerUserId="2637" LastEditorUserId="2637" LastEditDate="2018-06-14T07:33:40.543" LastActivityDate="2020-08-02T21:31:34.223" Title="What type of grit separator can be installed between the inlet works and bar screen of a wastewater treatment plant?" Tags="|civil-engineering|waste-water-treatment|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22264" PostTypeId="1" AcceptedAnswerId="22265" CreationDate="2018-06-14T11:03:43.353" Score="2" ViewCount="228" Body="&lt;p&gt;I have a high speed roller bearing rated at 3000rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know in the electrical industry that tolerances of around 10-15% are built into components so that they continue working in fluctuating electrical environments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similar tolerance in the bearing industry ?&lt;/p&gt;&#xA;" OwnerUserId="16443" LastEditorUserId="1031" LastEditDate="2018-06-14T14:35:36.373" LastActivityDate="2018-06-14T14:35:36.373" Title="What is the typical rpm variance-tolerance of a roller bearing?" Tags="|bearings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22265" PostTypeId="2" ParentId="22264" CreationDate="2018-06-14T13:11:35.377" Score="3" Body="&lt;p&gt;The maximum speed for a bearing recommended by the manufacturer can be due to one of two limiting factors, Thermal, or Mechanical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermal speed rating (often called &quot;reference speed&quot;) is calculated using ISO 15312 standardised operating and cooling conditions, and as such, if you are able to provide improved cooling in your application, then, you will be able to exceed this rating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanical speed limit of a particular bearing should not be exceeded except in very rare cases. If you need a higher speed, then you should look to find a more appropriate bearing specification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;http://www.skf.com/uk/products/bearings-units-housings/principles/bearing-selection-process/operating-temperature-and-speed/speed-limitations/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;SKF website&lt;/a&gt; says this about operating above the published speeds:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is possible to operate a bearing at speeds above its reference speed, its adjusted reference speed, or even the limiting speed. Before doing so, first make a detailed thermal analysis, and take whatever further measures may be required, such as use of special cage executions, or consider using high precision bearings. Regarding management of the effects of increased speed, consider the following options:&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Control the resulting increase in bearing temperature by additional    cooling.&lt;/li&gt;&#xA;  &lt;li&gt;Compensate for any reduction in bearing clearance resulting    from increased bearing temperature.&lt;/li&gt;&#xA;  &lt;li&gt;Revise the housing fitting    tolerance choice to ensure that the influence of increased bearing    temperature does not impair the&#xA;  axial displaceability of non-locating    bearing outer rings.&lt;/li&gt;&#xA;  &lt;li&gt;Revise the bearing tolerance class, together    with the geometrical precision of the shaft and housing seats, to    ensure&#xA;  these are sufficient to avoid excessive vibration&lt;/li&gt;&#xA;  &lt;li&gt;Consider    using an alternative cage execution that is suitable for higher speed    operation, in particular when approaching or&#xA;  exceeding the limiting    speed.&lt;/li&gt;&#xA;  &lt;li&gt;Ensure that the lubricant and lubrication method used are    compatible with the higher operating temperature and the cage&lt;br&gt;&#xA;  execution.&lt;/li&gt;&#xA;  &lt;li&gt;Check that the relubrication interval is still acceptable,&#xA;  particularly for grease lubricated bearings. Oil lubrication may be&#xA;  required.&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-06-14T13:11:35.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22266" PostTypeId="2" ParentId="22261" CreationDate="2018-06-14T13:41:18.860" Score="1" Body="&lt;p&gt;The OP wrote they use a Huber screen. Huber often uses bars that are hydraulically optimized bars with a profile like the left one in the picture:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3U6YI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3U6YI.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;These have a slight funnel effect, so if d = 7mm, a particle with ~8mm size in one direction can get stuck in the part between the rounded tips. I suggest talking to Huber if you can replace the bars with a different form, the wedge shape on the right. This removes the funnel. I have not seen the hydraulic coefficients for these wedges, but according to the what I'Ve read the downstream end is the biggest influence on flow resistance of the bar so the wedges could be okay too. I don't know how constrained you are hydraulically.&lt;/p&gt;&#xA;&lt;p&gt;I would also check the rakes: It should be so that one rake clears every other space (say space 1, 3, 5), alternating with the following rake (which cleans 2, 4 ,6 ...) so the bars can bend slightly. Also if the rake gets stuck, it should reverse and to try to dislodge the blockage. All of this is, AFAIK, standard with every rake manufactuer and certainly Huber so it should be implemented on your site.&lt;/p&gt;&#xA;&lt;p&gt;I beleive a modification to the screens should be the first thing to try, because - like you - I don't see a viable alternative. You can have a screen upstream, but then you might as well replace the screens you have. You can try to sediment the large particles, but then you build a second grit separator. I beleive cyclones to be too energy intensive for a wastewater stream and maybe not robust enough for unscreened wastewater - screen waste is mostly tissues (at least where I am), I could imagine those blocking a cyclone or its pump.&lt;/p&gt;&#xA;&lt;p&gt;The other idea I might consider would be to have a calm, low flow channel upstream of the screen to separate large grit &amp;lt;7mm - maybe just two wide channel with a small weir at the end, to be cleared by with a Bobcat every year or so. Not knowing the site and hydraulic conditions, I don't know if this is viable.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="3505" LastEditDate="2020-08-02T21:31:34.223" LastActivityDate="2020-08-02T21:31:34.223" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22267" PostTypeId="1" CreationDate="2018-06-14T15:28:47.803" Score="0" ViewCount="58" Body="&lt;p&gt;I'm currently looking for a tubing that can withstand 350F+, have a shore rating of less than 80A, and must be impermeable or very close to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found a tubing called &quot;Viton&quot; which is close but not quite impermeable; are there any other materials offered out there that fit these specifications? &lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2018-08-10T23:18:36.390" Title="Recommendations on impermeable high temp material selection?" Tags="|materials|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22268" PostTypeId="1" CreationDate="2018-06-14T15:28:58.543" Score="0" ViewCount="135" Body="&lt;p&gt;I am building a tree house in my backyard and it is going to be relatively simple. There are no parts that require drilling into the tree. The whole house is propped up by four 4x4 wood supports. I'm wondering if quikrete is the way to best support the whole structure or if there are better alternatives. &lt;/p&gt;&#xA;" OwnerUserId="16447" LastActivityDate="2022-04-03T07:39:39.530" Title="Base Support for a Tree house" Tags="|structural-engineering|design|stresses|beam|concrete|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22269" PostTypeId="2" ParentId="22257" CreationDate="2018-06-14T16:25:29.600" Score="3" Body="&lt;p&gt;On the &lt;a href=&quot;https://www.velmex.com/Products/Rotary_Tables/Motorized-Rotary-Tables.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;product page&lt;/a&gt; you linked, there is a drawing that shows how the rotary table should be mounted. On the right side of the drawing you can see the mounting holes. The annotation also tells you that you can mount either from the top or from the bottom and which bolts you should use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.velmex.com/images/Rotary_Tables/B4800-DWG-web-j.jpg&quot; alt=&quot;Drawing of rotary table&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There does not seem to be any need to devise some kind of bearing or other mechanism to mount that rotary table to a flat and horizontal surface.&lt;/p&gt;&#xA;" OwnerUserId="16448" LastActivityDate="2018-06-14T16:25:29.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22270" PostTypeId="2" ParentId="22256" CreationDate="2018-06-14T16:25:45.040" Score="5" Body="&lt;p&gt;I went with my daughter's class to the science museum and while we were looking at things, I commented on why it worked that way and gave occasional interpretation of the displays.  Impressed, my daughter's friend said, “Gee, Are you some sort of scientist of something?”  The answer I had ready was “No, something better. I am an engineer.  Scientists only report the news, engineers make it.”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The definition I got for an engineer when I was in high school was “Engineers are science and technology problem solvers.”  You can’t solve problems without knowledge of the underlying science, along with the specific technology you are dealing with.  When I was in school (BSME, 1981 (I am an old man)), they told us that they were teaching us everything they could fit in four years, it was not enough but it should be sufficient to allow us to function in the real world.  As I recall, I had a few “free” electives (I took more math), three (not very free) humanities electives and a few mechanical engineering electives.  Everything else was listed out in the requirements when I started.  The school knows that much of what they teach is going to be obsolete in a few years so they teach as much of the foundations as they can and how to learn and solve problems.  The problem is if you keep squeezing things in, there is not enough room for much else and you wind up with a &quot;course intensive&quot; engineering curriculum &lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2018-06-14T16:25:45.040" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22271" PostTypeId="2" ParentId="22268" CreationDate="2018-06-14T16:48:55.617" Score="1" Body="&lt;p&gt;What's the point of building a tree house if the house is not actually connected to the tree?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, yes concrete foundations should work. There are two points that I would suggest considering though:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The wood should not be in direct contact with the concrete or the ground. Use some kind of steel brackets and ensure that water can drain from the joint. A post bracket could be used. For example this &lt;a href=&quot;https://www.strongtie.com/nonstandoffpostbases_postbases/rcps_saddle/p/rcps&quot; rel=&quot;nofollow noreferrer&quot;&gt;rebar bracket&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Make sure to dig deep enough when pouring the concrete. You are going to put in a long lever and probably want to make sure that your house does not move when you bump against it.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16448" LastEditorUserId="16448" LastEditDate="2018-06-15T21:20:12.563" LastActivityDate="2018-06-15T21:20:12.563" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22273" PostTypeId="2" ParentId="14869" CreationDate="2018-06-14T19:41:18.117" Score="0" Body="&lt;p&gt;This is used to move dirty water in ponds. The cover is used to screen out anything that would jam the pump but it is also used to mix the particles evenly with the water on slower velocity pumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1pai9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1pai9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally in those gold mining high velocity it is a pump&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RhJdp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RhJdp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if it the rock will pass the hose it will not jam the pump and the high velocity keeps better but cost more. I would like to see what kind of pump you are using now.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorUserId="110" LastEditDate="2018-09-21T22:55:01.563" LastActivityDate="2018-09-21T22:55:01.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22274" PostTypeId="2" ParentId="22267" CreationDate="2018-06-15T02:22:36.490" Score="1" Body="&lt;p&gt;Check out ePTFE tubing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water impermeable and bio-inert. Strong, can be really thin, and can be softer than 80A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding odor impermeable, you'll have to test. &lt;/p&gt;&#xA;" OwnerUserId="16453" LastActivityDate="2018-06-15T02:22:36.490" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22276" PostTypeId="2" ParentId="22260" CreationDate="2018-06-15T07:39:16.750" Score="0" Body="&lt;p&gt;The torque required during the acceleration phase can be calculated using the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T=Ia$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: $T = \text{Torque (Nm)}\\ I = \text{moment of inertia (kg.m²)}\\ a = \text{angular acceleration (rad/s²)}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment of inertia of a solid disc is $I=\frac{1}{2}MR^2$, where $M=\text{mass (kg)}$, and $R=\text{radius (m)}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the moment of inertia will be lower than that of a solid disk, since the arrangement of the glass tubes concentrate the mass closer to the axis. Therefore, it is a safe estimate to use for calculation, and will ensure that the selected motor has sufficient power. Using the numbers provided in your question, the radius of the disc is found to be $0.09\text{ m}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$I=\frac{1}{2}(23)(0.09)^2=0.0932\text{ (kg.m²)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the numbers that you provided in your comment, the angular acceleration can be calculated by converting the steady-state speed $100\text{RPM}=10.47\text{rad/s}$, and dividing by the 'spin up time' in seconds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\frac{10.47}{2.5}=4.19\text{ (rad/s²)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combining these two calculated values in the original equation gives the peak torque during acceleration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$T=(0.0932)(4.19)=0.39\text{ Nm}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to note, however, that the torque generated by any motor is not constant in relation to angular velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fndYPm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fndYPm.jpg&quot; alt=&quot;Torque Speed Curves for different Motor Windings&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you specify a motor which is advertised as being capable of sustaining $0.4\text{ Nm}$ at $100\text{ RPM}$, then you should be OK, since the available torque will be higher at lower speeds. This corresponds to a power of $\approx 4.2\text{ W}$, which is certainly achievable with a small DC motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to limit the spin-up-speed to take $2.5\text{ s}$ in any case, then this should be done using e.g. PWM control of the motor.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-06-15T07:39:16.750" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22278" PostTypeId="2" ParentId="22156" CreationDate="2018-06-15T14:46:34.870" Score="1" Body="&lt;p&gt;Flow measurement device is used where you cannot just put the output of a fluid in to a tank where you can measure the time and volume filled, it is used in conditions like flow of refrigerant liquid through a pipe, or a petroleum gas flow inside a pipe. and also it is not easy to measure flow of gas directly.&lt;/p&gt;&#xA;" OwnerUserId="16457" LastActivityDate="2018-06-15T14:46:34.870" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22279" PostTypeId="2" ParentId="21879" CreationDate="2018-06-15T14:50:08.350" Score="0" Body="&lt;p&gt;Fluids includes gas, liquids and some solids which flows, as you know gas is compressible, in some case fluid flow is considered as compressible like in the case of a rocket design, aircraft etc...&lt;/p&gt;&#xA;" OwnerUserId="16457" LastActivityDate="2018-06-15T14:50:08.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22280" PostTypeId="2" ParentId="21307" CreationDate="2018-06-15T14:53:59.787" Score="0" Body="&lt;p&gt;There is no Perfect black body (including black hole see hawkings radiation for details) that is why you can see a small light coming up from every black body and hence no body have emissivity of 1. This means true black is not yet discovered. &lt;/p&gt;&#xA;" OwnerUserId="16457" LastActivityDate="2018-06-15T14:53:59.787" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22281" PostTypeId="1" AcceptedAnswerId="23745" CreationDate="2018-06-15T21:24:04.333" Score="0" ViewCount="183" Body="&lt;p&gt;I understand that there is such a metric as the Rayleigh number which governs convective cooling in a medium.  Assume for example, a fist sized heatsink like that below in free air:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tnvsI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tnvsI.png&quot; alt=&quot;heatsink&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the Rayleigh number mean that convection might not start between the fins?  So in effect, the finned heatsink might as well be a large solid lump? And insofar as thermal resistance goes, the thermal conductivity would only depend on the gross overall dimensions (height x width x length) and not the actual finned area?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2018-09-14T07:42:24.990" Title="What is the practical application of Rayleigh number to heat sinks?" Tags="|thermodynamics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22282" PostTypeId="1" AcceptedAnswerId="22287" CreationDate="2018-06-15T21:35:55.147" Score="1" ViewCount="63" Body="&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xA;   \sigma_{ij}=&#xA;  \left[ {\begin{array}{cc}&#xA;   -20 &amp;amp; 60 \\&#xA;   60 &amp;amp; 90 \\&#xA;  \end{array} } \right],\quad i,j=x,y&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the principle stress tensor $\sigma_{ij}^{pr}$.&#xA;Using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mohr%27s_circle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mohr's Cirlce&lt;/a&gt;, I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_{max}=116.39,\sigma_{min}=-46.39$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the points where the circle intersects with the $x(\sigma_{xx},\sigma_{yy})$ axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there, how do these points make up the principle stress tensor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ &#xA;   \sigma_{ij}^{pr}=&#xA;  \left[ {\begin{array}{cc}&#xA;   116.39 &amp;amp; 0 \\&#xA;   0 &amp;amp; -46.39 \\&#xA;  \end{array} } \right],\quad i,j=x,y&#xA;$$&#xA;correct?&lt;/p&gt;&#xA;" OwnerUserId="16459" LastActivityDate="2018-06-16T09:45:57.267" Title="Mohr's circle: maximum normal stress tensor" Tags="|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22283" PostTypeId="1" CreationDate="2018-06-15T22:45:11.217" Score="1" ViewCount="44" Body="&lt;p&gt;I'm looking at the option of running two blinds motors in one tube, running in alternate rotations so that they can help to lift more weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that the dominant motor will slow to the other motor but when under load (lifting blind) they will kick in together. These motors are 1.1Nm but struggle to lift the blind, we tried a hard wire running at 1.5Nm but still it struggled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I put two of the battery ones together it should run about 2Nm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="16460" LastActivityDate="2018-06-15T22:45:11.217" Title="Running two motors in one tube" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22284" PostTypeId="2" ParentId="21879" CreationDate="2018-06-16T00:46:52.210" Score="0" Body="&lt;p&gt;When you say &quot;incompressible&quot; that is what it is by definition. Other things can change and the density of the fluid can change but it is still incompressible. &#xA;Nothing in the real world is completely incompressible.  For example 40°F water at atmospheric pressure has a density of 62.4 lb/ft³ while at 500 psig, that goes up to 62.5 lb/ft³. That is not much of a difference so for the purposes of calculation, we assume cold water is incompressible.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2018-06-16T00:46:52.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22286" PostTypeId="2" ParentId="22281" CreationDate="2018-06-16T05:04:41.670" Score="0" Body="&lt;p&gt;Let's assume for a moment as per your hypothesis the convection doesn't start and the sink acts as a solid heated from bottom face and in contact with air at 5 other faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat will infuse this cube and propagate in gradients contoured similar to top half of an onion, core hottest and spherical layers gradually less warm to the surface which is least warm..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the air at the core is warmer it will rise like a cumulus cloud and leave a low pressure in zone in the base of this column. This low pressure zone will suck in air from around the base of the heat sink to fill in the vacuum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convection will stablish by circulating the colder air in between the teeth of the heat sink and getting it warmer and going up through the teeth and opening up in the shape of a canopy.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-06-16T05:04:41.670" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22287" PostTypeId="2" ParentId="22282" CreationDate="2018-06-16T09:45:57.267" Score="0" Body="&lt;p&gt;The values you determined for $\sigma_{\mathit{max}}$ and $\sigma_\mathit{min}$ are correct. However, your answer is only partially correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of $\sigma^\mathrm{pr}$ has to be expressed in the principal coordinate system. You can calculate the angle of the principal system to your original coordinate system using:&#xA;$$&#xA;tan(2\varphi) = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2 * 60}{-20-90}&#xA;$$&#xA;$$&#xA;\varphi_1 \approx 23.74°&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes the values are correct, but your coordinate system $x^\mathrm{pr}, y^\mathrm{pr}$ is rotated by $\varphi_1$ degrees, compared to your original coordinate system $x, y$.&lt;/p&gt;&#xA;" OwnerUserId="16448" LastActivityDate="2018-06-16T09:45:57.267" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22288" PostTypeId="1" AcceptedAnswerId="22289" CreationDate="2018-06-16T10:58:11.857" Score="1" ViewCount="3862" Body="&lt;p&gt;My professor says, the internal hinge can not transfer moment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UNZM7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UNZM7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But can we calculate $\Sigma M_A=0$ for whole beam ( without cutting it or something) this way?:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Sigma M_A=0  ===\Rightarrow M_A+(D_y \times 4m ) + 3KN.M =0$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with this equation we literally transferred the 3KN.M moment to point A while my professor says the internal hinge can not transfer moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something is wrong, and I don't know what it is.&lt;/p&gt;&#xA;" OwnerUserId="16461" LastEditorUserId="2559" LastEditDate="2018-06-18T10:49:49.610" LastActivityDate="2018-06-18T10:49:49.610" Title="Transfering moment through internal hinge" Tags="|structural-engineering|structural-analysis|beam|moments|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22289" PostTypeId="2" ParentId="22288" CreationDate="2018-06-16T11:31:17.407" Score="2" Body="&lt;p&gt;&quot;Hinges don't transfer moment&quot; is just a way of saying that hinges do not resist bending, and therefore suffer no bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can basically &quot;translate&quot; that term as &quot;Bending moment at hinges is always zero.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just note that in more complex structures such as frames there may be hinges which only affect some members but not others on the same node. In that case, there may still be moment on the node, but the hinged beam will have zero moment on that node, such as below (see the hinge on the central node only affecting the left-hand span):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E2jOY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E2jOY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-06-16T11:31:17.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22290" PostTypeId="2" ParentId="22288" CreationDate="2018-06-16T11:36:49.190" Score="1" Body="&lt;p&gt;The internal bending moment at the hinge must be zero, but the hinge can still transmit a force into the AB cantilever. If you consider a plain cantilever with point force at the end, you have not applied a moment, but there is a moment reaction force at the base - that’s what is happening here.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-06-16T11:36:49.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22291" PostTypeId="2" ParentId="22288" CreationDate="2018-06-16T18:05:59.130" Score="1" Body="&lt;p&gt;The moment 3KN creates two equal and opposite forces at supports B and D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each is equal to $ 3KN/(Length- of-BD). $ With Dy pushing up and By pushing down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D-y is countered by the D upward reaction and cancels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct moment at A is $$ M_a = (\frac {3KN}{length-of-BD} \times AB) $$ which is clockwise. &#xA;$$ M_a = force_B \times (length AB) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And your teacher is right, no moment is transferred, but shear has been transferred.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-06-16T18:29:15.840" LastActivityDate="2018-06-16T18:29:15.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22292" PostTypeId="1" CreationDate="2018-06-16T19:42:15.310" Score="1" ViewCount="57" Body="&lt;p&gt;After looking through dozens upon dozens of IEEE papers, it seems that no one has really developed a simplified formula for a surge arrester.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would like to be able to do is perform a transient voltage study, and would like to be able to determine how the transient voltage would be impacted with the addition of a surge arrester, purely mathematically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a method of developing a &quot;cutoff&quot; equation for a surge arrester given its V-I data from a manufacturer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;$$\begin{array}{c} &#xA; &amp;amp; \text{Vrated (kV)}\\ \hline&#xA;\ 15kV\\ \hline&#xA;\end{array}&#xA;\begin{array}{c} &#xA; &amp;amp; \text{I(kA) (8/20µs), (kV)} \\ \hline&#xA;\ 1 (33.1) &amp;amp; 2.5(37.8) &amp;amp; 5(41.3) &amp;amp; 10(45.6) \\ \hline&#xA;\end{array}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the table above, 15kV is the rating of the surge arrester, and to the right is the characteristic curve (V-I) of the arrester, where the number on the outside is the current in kA and the number in parenthesis is the residual voltage in kV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know how to take a curve, such as the one above, and transform it into a time-dependent equation that I would 'subtract' from a sinusoidal transient voltage,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(t) = [transient-waveform]-[surge -arrester-curve]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only formula I've come across that characterizes manufacturer data and is used in EMTP software is:&#xA;$$I=p(V/Vref)^q$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is not time dependent and absolutely no mentions how the 'p', 'q' and 'Vref' variables are calculated.&lt;/p&gt;&#xA;" OwnerUserId="16465" LastActivityDate="2018-06-16T19:42:15.310" Title="Deriving a Time-Dependent Equation from a Surge Arrester for Transient Voltage Analysis" Tags="|electrical-engineering|mathematics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22293" PostTypeId="1" CreationDate="2018-06-16T21:31:02.750" Score="1" ViewCount="19" Body="&lt;p&gt;I have only taken standard (basic) level math courses in high school and had learned only the fundamentals of wave and ray optics in physics classes. Therefore, Im asking only for a very general answer: Is it possible that lens consisting of a flat plate with a very high refractive index gradient (maximal n=2.7, minimal n=1) going across the plane of the lens, with thickness equal to 20 wavelengths be used to refract a laser beam, or is the scale too small for the behaviour of the light to be predicted from the law of refraction? &lt;/p&gt;&#xA;" OwnerUserId="16466" LastActivityDate="2018-06-16T21:31:02.750" Title="Gradient of refractive index at very short distances" Tags="|optics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22294" PostTypeId="1" CreationDate="2018-06-17T01:52:37.687" Score="1" ViewCount="188" Body="&lt;p&gt;I have been working on some acoustic synthesis software for modelling several acoustic instruments. I am using the principle of modal synthesis, where every mode of vibration is individually synthesized, generally using a resonant bandpass filter (or sine wave). I have used this approach to get great results on stringed instruments like guitar/cello where the relationship between modes is simple and reasonably predictable. However, I am struggling with drums.&lt;/p&gt;&#xA;&lt;p&gt;As a 2D circular membrane, the vibrational properties of a drumhead (and even more so, a two headed drum with a drum shell between) are much more complicated than a simple string. However, I still believe this should be possible to synthesize well given modern computing power. With the correct data, I can easily synthesize 500+ modes simultaneously to give a likely good sound.&lt;/p&gt;&#xA;&lt;p&gt;Imagine a simple 2D circular membrane excited by a strike of energy at its dead center. The information I ideally need for each mode is:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Frequency ratio relative to the fundamental (0,1)&lt;/strong&gt; - this is easily given in an ideal circular membrane by the Bessel zeros. Things would be different in a &amp;quot;nonideal&amp;quot; membrane, but this is easy to get from any program like ANSYS either way, and I would prefer an &amp;quot;ideal&amp;quot; simulation first.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Maximum amplitude of sound of each mode relative to the fundamental (0,1)&lt;/strong&gt;. If a point of reference is needed, let's say we are measuring sound at 2-3&amp;quot; from the center of the excited membrane.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Decay rate of each mode&lt;/strong&gt;, where decay rate is given by time to reach 1/e amplitude relative to that mode's highest initial amplitude, or alternatively in dB/s.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Time delay from excitation to beginning of a mode's oscillation&lt;/strong&gt; in milliseconds or in radians/degrees for that mode's frequency.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;If I have a table of that data for the first 500 significant resonances of a circular membrane (or full drum model), I can easily put that into synthesis to see what I get.&lt;/p&gt;&#xA;&lt;p&gt;ANSYS is okay even in simple usage for providing basic modal frequencies (#1). But I am uncertain if/how it or another program can possibly provide #2-4 on that list.&lt;/p&gt;&#xA;&lt;p&gt;Is this a very simple or challenging set of data to try to get? How would you approach it, ie. with what program or modelling technique?&lt;/p&gt;&#xA;&lt;p&gt;Alternatively, the biggest component of the &amp;quot;sound&amp;quot; besides the frequency ratios is the decay rates. If you are aware of any equation that can, using an arbitrary damping coefficient &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, express the theoretical decay rate of any (m,n) mode from a given decay rate of (0,1), that would likely work well enough too. It was easy enough in strings to work in this way, but I don't think it will be so easy in 2D. I'm sure such an equation can be derived, but I don't know it, and I'm not sure anyone else does either. I am hoping to be able to work that relationship out from modeled data if I can get modeling working.&lt;/p&gt;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="16468" LastEditorUserId="6790" LastEditDate="2020-08-06T13:03:47.803" LastActivityDate="2021-05-03T15:03:55.473" Title="Modelling acoustic instruments — predicting frequencies, decays, levels, and phases of modes of a circular membrane (drum head)" Tags="|finite-element-method|modeling|vibration|acoustics|modal-analysis|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="22295" PostTypeId="1" AcceptedAnswerId="22321" CreationDate="2018-06-17T03:32:50.713" Score="4" ViewCount="7262" Body="&lt;p&gt;What is the difference between a state observer and a Kalman filter? Having implemented various types of Kalman filters, I'm still a bit confused, mainly because state observers require the selection of poles similar to a control system setup, whereas Kalman filters don't. Any clarification on the two and their differences would be great.&lt;/p&gt;&#xA;" OwnerUserId="9755" LastEditorUserId="6790" LastEditDate="2020-05-03T18:13:37.697" LastActivityDate="2021-02-02T21:41:43.580" Title="Kalman filters vs. state observers" Tags="|control-engineering|kalman-filters|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22296" PostTypeId="1" AcceptedAnswerId="22307" CreationDate="2018-06-17T04:17:05.247" Score="3" ViewCount="327" Body="&lt;p&gt;If you were to use a centrifugal axial fan instead, won't you be able to achieve higher flow rates and better pressures? It seems like the efficiency of these motors are very low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wzZvA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wzZvA.jpg&quot; alt=&quot;Typical centrifugal fan&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16469" LastActivityDate="2018-06-18T00:48:21.250" Title="Why do most centrifugal fans found in laptops have tiny blades?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22298" PostTypeId="1" CreationDate="2018-06-17T08:39:12.480" Score="2" ViewCount="82" Body="&lt;p&gt;What are the requirements to call a controller real-time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the entire standards for that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not think that just rendering the output quickly before the sampling interval passes is sufficient for the real-time certification.&lt;/p&gt;&#xA;" OwnerUserId="16032" LastActivityDate="2018-06-18T20:20:09.447" Title="Requirements for realtime control" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22299" PostTypeId="1" CreationDate="2018-06-17T13:35:03.610" Score="2" ViewCount="229" Body="&lt;p&gt;I have two disks to bolt with. 4 centrally located holes on each with Pitch circle diameter P and one hole at center of disc.&#xA;If standard bolt diameter is B +/- b, what should be each hole diameter if tolerance of drilling hole is +/- h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the load rating I got M6 bolts. Their dimensions are 5.85+/- 0.1 mm. And PCD is 10 cm, and I have to bolt it via 4 bolts symmetrically placed and one in center. Drilling of holes on pitch circle is of tolerance 0.5 mm, while central hole tolerance is 0.1 mm. Now, what should be the hole diameters?, so as all four bolts get into and proper fitment occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12092" LastEditorUserId="12092" LastEditDate="2018-06-17T19:25:13.823" LastActivityDate="2018-10-16T03:00:31.670" Title="How to find diameter of hole for correct assembly?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22300" PostTypeId="2" ParentId="22298" CreationDate="2018-06-17T13:58:11.410" Score="2" Body="&lt;p&gt;A real time controller is any controller that can control a system as it is operating.  This is opposed to, for example, simulating control after the fact.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-06-17T13:58:11.410" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22304" PostTypeId="2" ParentId="21472" CreationDate="2018-06-17T21:57:28.373" Score="-1" Body="&lt;p&gt;If you are using CAE then you can specifiy the number of CPUs or GPUs in the Job Manger, when creating a job click the Parallelization tab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also do it from the command line &lt;code&gt;abaqus job=myJob cpus=4&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how successful you will be depends on your Abaqus license and how many tokens you can use. It will throw an error if you dont have enough tokens or the wrong type of license.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It cant be done just from the usermat. The usermat is compiled into the solver it cant just spawn new threads because the solver is locked to only use a certain number of cores. So unless you have a license to unlock that it cant be done. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition when executing the job the number of cores needs to be explicitly declared. If you ask for 4 cores Abaqus will split the whole analysis across 4 cores. If you then try &amp;amp; spawn more cores from your usermat it will throw an error. Sorry but there is no way around this using Abaqus.&lt;/p&gt;&#xA;" OwnerUserId="14531" LastEditorUserId="14531" LastEditDate="2018-06-18T16:01:31.817" LastActivityDate="2018-06-18T16:01:31.817" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="22305" PostTypeId="2" ParentId="22299" CreationDate="2018-06-18T00:27:21.463" Score="1" Body="&lt;p&gt;Let's take worst case:  5.85+0.1mm = 5.95 bolt, 0.5mm offset off the center when drilling on one plate, -0.5mm offset on the other plate, and -h error in dimension of the hole. For the plates to be guaranteed to align, the hole would need to be at least 6.95+h mm diameter. And that leaves at least a millimeter of play, which is &lt;em&gt;bad&lt;/em&gt;. For the central hole you'd get away with 6.15mm+h but that's still over 0.3mm of play. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A much better approach is to align and temporarily bind (clamp) the disks together, and drill the holes through both at the same time. That way whatever error in hole placement is made on one, it's duplicated exactly on the other so there's no misalignment. Make the holes 5.75mm-h (assuming h is in the ballpark of 0.1mm or less) and expand them using an M6 thread cutter (tap) to fit the bolt's thread and hold the bolt snugly. And as soon as you have one hole complete (drilled and threaded) install the bolt and tighten it, before drilling another hole; this way any play/error introduced initially by the clamps will be nullified in consecutive bolts. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-06-18T00:27:21.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22307" PostTypeId="2" ParentId="22296" CreationDate="2018-06-18T00:48:21.250" Score="6" Body="&lt;p&gt;Lower pressure differential and lower speed per blade translates to lower noise levels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, for this sort of applications, higher speed would mean higher torque (scales with square of airspeed) and that converts to higher power usage; many blades allow to keep a decent flow while keeping the pressure, and resulting torque low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these factors are important in laptops. And while better flow rate would be desirable, higher pressure is absolutely not needed. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-06-18T00:48:21.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22308" PostTypeId="2" ParentId="22215" CreationDate="2018-06-18T07:45:42.330" Score="0" Body="&lt;p&gt;I asked the question &lt;a href=&quot;https://physics.stackexchange.com/questions/411213/non-resting-initial-value-problem-with-impulsive-input&quot;&gt;here&lt;/a&gt; and I found the answer myself. This is really a simple one really. Here is my answer for this question:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;[...] I revisited this problem with another point of view which gives the answer right away. Since the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is at least &lt;span class=&quot;math-container&quot;&gt;$N-1$&lt;/span&gt; times differentiable, it's derivatives (up to order &lt;span class=&quot;math-container&quot;&gt;$N-2$&lt;/span&gt;) and itself must be continuous. So no shift until the last derivative. Using the my question's notation,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$c_n =0\hspace{1cm} \forall n&amp;lt;N-1$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$c_{N-1} =f_0/a_N$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16406" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-06-18T07:45:42.330" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22309" PostTypeId="1" CreationDate="2018-06-18T09:55:39.487" Score="0" ViewCount="121" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Nig2L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Nig2L.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;My target is to create a valve with multiple inlet capillaries. And my output is a pipe of a larger diameter. I do not want to use any electrical systems, and at the same time, I want to maintain enough negative pressure inside the valve to influence the flow of fluid to be from capillaries to the pipe and not the other way round. The pipe-capillary system is evacuated and the fluid inside is at a higher than atmospheric pressure (we can assume 4-5 atm). I also wish the flow to remain laminar and avoid turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked into kinds of pumps which, in principle, achieve this... But I could not find any self-controlling system which does not require external power to maintain the pressure force required. Pneumatic systems like the flow valve looked promising but I could not figure out a way to achieve the requirement I am aiming for. &#xA;I was thinking of inserting a high pressured section inside the valve which would push a diaphragm with constant force. And that force can be used to maintain the negative pressure that I want. Another idea with having a vacuum instead of a high-pressure fluid to achieve the same (effectively the inverse mechanism). But from a design perspective, I could not figure out a way to avoid leakage of the high pressure over time. My idea was inspired by the action of a syringe. I tried to design a CAD model for the same to simulate it... But could not figure out how to implement all that I read on this topic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a valve or pneumatic system exist that achieves this (am I missing some information that I should know)? And if not, how can I make such a valve? &#xA;I need to manufacture this as well, so simplicity/availability of resources is important. But any theoretically sound solution will also be greatly appreciated :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance &lt;/p&gt;&#xA;" OwnerUserId="16476" LastEditorUserId="16476" LastEditDate="2018-06-18T13:23:30.310" LastActivityDate="2018-08-10T23:13:05.833" Title="How to maintain a static negative pressure in a valve?" Tags="|fluid-mechanics|pipelines|valves|building-physics|pneumatic|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="22310" PostTypeId="1" AcceptedAnswerId="22311" CreationDate="2018-06-18T10:33:14.667" Score="3" ViewCount="61" Body="&lt;p&gt;I am trying to calculate the running cost of my clothes dryer and have bought myself an energy meter to check its usage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plugged the meter in, reset it, and then turned it over to the aggregate kWh mode and did a single drying session with wet clothes from the washing machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the load had finished the meter read 1.828 kWh and the total time was 1h 13m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as I understand it, a single load used 1.828 kWh, and as we are doing about 7 loads per week that would be a total of 665 kWh / year. At the current rate of 33c per kWh the annual running cost is therefore about $219.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this to me is a bit strange as just about every website I have seen says that dryers should be costing upwards of $400 per annum for this kind of usage. In addition, my dryer is only a 2 out of 5 stars for energy efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone see a problem with my math or logic?&lt;/p&gt;&#xA;" OwnerUserId="16477" LastActivityDate="2018-06-18T11:15:28.577" Title="Trying to understand how much it costs to run my clothes dryer" Tags="|energy|cost-estimation|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22311" PostTypeId="2" ParentId="22310" CreationDate="2018-06-18T11:15:28.577" Score="3" Body="&lt;p&gt;You measured one load to take 1.8 kWh.  That implies that 7 loads would take 12.6 kWh.  If these 7 loads are repeated 52 times per year, then a whole year of operating takes 655 kWh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a sanity check, 1.828 kWh over 1 hour and 13 minutes comes out to 1.5 kW, which is plausible for some type of clothes dryer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\$0.33 per kWh sounds very high, but going with that, 655 kWh would cost $216.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems your calculations are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Web sites that claim some appliance costs some amount per year to operate are a very rough guide at best, complete fabrication at worst.  These sites usually have some agenda, so push the number towards whatever makes their argument look better.  They also rarely state assumptions.  Even then, usage patterns for something like a clothes dryer vary so widely, as do electricity prices, that they can pick from a wide range of values and then claim a case that justifies what they picked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep monitoring the electricity usage of your dryer.  It probably varies quite a bit from load to load.  There will be inevitable variations in amount of clothes that are being dried, the ambient humidity, the wetness of the clothes, and other factors that are hard to predict.  Over a whole year, you'll get a much better idea of the expected power usage of the dryer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you really sure that you pay $0.33 incrementally per kWh?  That's probably correct for some part of the world at least some of the time, but sounds rather high.  If your area has time of day metering, it might be worth your trouble to arrange to run the dryer over night when the rates are lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to save money, try using a plain old clothes line, at least during the warm parts of the year.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-06-18T11:15:28.577" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22313" PostTypeId="1" AcceptedAnswerId="22318" CreationDate="2018-06-18T16:46:07.800" Score="4" ViewCount="1187" Body="&lt;p&gt;I am planning a prop for a marching band show and I want to determine how to best address the challenge of the wind blowing it over.  This is definitely not my area of study and I am hoping for some guidance here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The panel is 10' high and 4' wide.  My working assumption is that we won't be dealing with winds any stronger than 15 mph.  That may be wishful thinking but I needed to throw out a number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The panel will be mounted on a small platform (5' x 3.5') and angled at roughly 80 degrees.  Something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/91zgd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/91zgd.png&quot; alt=&quot;Side View of prop&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The panel will be mounted top and bottom to the support and base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to get an idea about how much weight it would take to keep it from flipping over, I found a generic formula for calculating wind load:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F=AxPxCd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this formula and plugging in 0.00256xV^2 for P&#xA;0.00256 x 15^2 = 0.576&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And based on some very cursory reading, I am guessing at the Cd value for a rectangular panel, setting it to 2.0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I arrive at F = 40ft^2 x 0.576 psf x 2.0&#xA;= 46.08 pounds&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not sure what this is telling me or if it is even correct for my situation.  I'm primarily concerned with the wind coming from that support side and leading to it lifting or flipping the panel.  I'm limited on the footprint but could add weights to the base and/or behind the base attached to the support poles and/or the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="16480" LastActivityDate="2018-07-11T19:46:44.727" Title="Wind load of panel" Tags="|structural-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22315" PostTypeId="2" ParentId="22313" CreationDate="2018-06-18T19:46:32.010" Score="1" Body="&lt;p&gt;So, &lt;em&gt;I am not a professional engineer&lt;/em&gt;, but: that 46 lbs. is a horizontal, equivalent force at the center of the panel.  To figure out how much weight you need to keep it from tipping over, you will need to use statics (think levers).  Put in a factor of safety of, say 2 -&gt; you need to be able to balance a 100lb force acting on the middle of the panel (5' from the ground).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5' x 100 lb = 500 ft-lb of torque.  Looking at your base, it looks like you have a lever of about 3 ft in length.  Assuming you have the weight evenly distributed on the base (this will provide the same resistance to tipping in both directions), your weight needs to act on a 1 1/2 ft lever arm and provide that 500 ft-lb of torque.  This translates to about 340 lbs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will also need to make sure that the method you use to secure the panel to the base can withstand transmitting that force: if the sign just rips off, it doesn't matter how much the base weighs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also make sure that the sign doesn't deflect too much when the wind is blowing: imagine some suitable person sitting in the middle of the sign when it is laid flat on some sawhorses.  If werf ends up on the ground, you need to improve your design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving the brace to connect to the sign 2/3rds of the way up could help.&lt;/p&gt;&#xA;" OwnerUserId="16283" LastActivityDate="2018-06-18T19:46:32.010" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22316" PostTypeId="2" ParentId="22298" CreationDate="2018-06-18T20:20:09.447" Score="1" Body="&lt;p&gt;I don't think your question calls for pointing to any of the various official standard definitions of &quot;real-time&quot; (e.g., ANSI/IEEE POSIX 1003.1b-1993 and POSIX 1003.1i-1995). And the real-time computing research and practitioner communities have little consensus on a &quot;definition,&quot; instead having a great many ad hoc interpretations. The following actually captures the essence of &quot;real-time&quot; and its relationship to control systems (among others):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people implicitly have an informal mental model that considers information or an action as being &quot;real-time&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;•   if, or to the extent that, it is manifest to them with some expected delay that can be related to its perceived currency&lt;/p&gt;&#xA;&#xA;&lt;p&gt;•   i.e., in an expected time frame that the information or action has satisfactory usefulness (or at least interest) to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That mental model is natural despite its informality. However, it must be formalized to create one for theoretical and any specific application purposes --such as for maintaining a control loop's dynamics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That requires reasoning about “real-time” from a set of first principles,  including &quot;timeliness,&quot; &quot;predictability of timeliness,&quot; and &quot;uncertainty,&quot; to formulate a precise model which is satisfactorily useful for designing and measuring the behavior of the (say) control system.&lt;/p&gt;&#xA;" OwnerUserId="15671" LastActivityDate="2018-06-18T20:20:09.447" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22317" PostTypeId="1" AcceptedAnswerId="22323" CreationDate="2018-06-18T21:33:42.697" Score="-1" ViewCount="605" Body="&lt;p&gt;The pads are used in air coolers to cool the air but make it more humid.&#xA;Is it possible to use copper / aluminum coils instead of pads to get cold and dry air?&lt;/p&gt;&#xA;" OwnerUserId="11138" LastActivityDate="2018-06-19T14:31:39.173" Title="is it possible to use copper coil instead of pads in air cooler" Tags="|cooling|air|" AnswerCount="2" CommentCount="3" ClosedDate="2018-06-25T19:24:47.440" ContentLicense="CC BY-SA 4.0" />
  <row Id="22318" PostTypeId="2" ParentId="22313" CreationDate="2018-06-18T22:17:19.747" Score="2" Body="&lt;p&gt;In US there are zones of maximum wind velocity and its pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Majority of the zones assume (15 to 35) lbs/sqr ft pressure for heights of up to 60 feet, depending on combination of factors such as zone, exposure, utility of structure. Lets use just for illustration pick wind pressure of 20 #/sf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will have an overturning momentum of &#xA;$$  m_{overturning} = (20\times 4 \times 10)_{total\space p} \times (10/2)_{arm} = 4000 \space lbsft \space moment$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to counter this by adding enough width and weight to the base to resist this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a first guess say we pick the base to be 3 feet deep. then 4000/3= 1660lbs so you need a ballast of 2\times 1660 =3300 lbs at the center of base, which is not practical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We add the depth to 5 feet and get the $ballast= 4000/2.5 = 2 \times 800 lbs.$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there are reduction factores for temporary uses such as yours but this is just to get a feeling of the magnitude and order of wind pressure. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-06-18T22:24:27.893" LastActivityDate="2018-06-18T22:24:27.893" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22319" PostTypeId="2" ParentId="22317" CreationDate="2018-06-19T01:00:44.613" Score="0" Body="&lt;p&gt;Copper or aluminum coils are used not in evaporative coolers (which add humidity); they are used in heat pump-style coolers that have compressors, pumps and freon refrigerant (and subtract humidity). These two types of coolers are completely different in how they work and their cooling components cannot be interchanged. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-19T01:00:44.613" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22321" PostTypeId="2" ParentId="22295" CreationDate="2018-06-19T07:21:01.473" Score="4" Body="&lt;p&gt;They are both different forms of the same thing. I will try to clarify this a bit without a lot of buzzwords, which many people seem to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: The Kalman filter is an optimal observer similar to how LQR is an optimal controller. Both together form an optimal state feedback controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bit longer:&#xA;You already noted that an observer is nothing more than a kind of controller controlling your estimate of a system towards the real system based on your sensor output, so just think of it as a controller. You can design a controller using frequency domain techniques giving you a transfer function with poles and zeros, using a state space representation and providing a control matrix yourself, or by solving a Riccati equation to find an optimal controller (LQR), requiring you to define cost matrices penalizing control effort and error. (And in many many other ways.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Kalman filter is just an optimal observer/estimator similar to how an LQR controller is an optimal controller. You can even design them using the same Riccati equation. (You just change your weighting matrices to the covariance matrices of process and observation noise and so on, but you are familiar with this.) The Kalman filter or Linear Quadratic Estimator (LQE) is a way of selecting the observer gains. You did this manually using pole placement to stabilize your error dynamics (see the controller analogy?). The Riccati equation gives you stabilizing gains by definition, but you lose control over where poles are placed exactly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you (or anyone else reading this) want to know more about this, the wikipedia page about Linear Quadratic Gaussian Control (LQG), which is nothing more than the combination of a Kalman Filter and a LQR controller might be a good starting point.&lt;/p&gt;&#xA;" OwnerUserId="16413" LastActivityDate="2018-06-19T07:21:01.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22323" PostTypeId="2" ParentId="22317" CreationDate="2018-06-19T14:31:39.173" Score="0" Body="&lt;p&gt;Reverse cycle air conditioners, heat pumps by another name, are closed system coolers whereby heat is transferred form one side of a divide to another, ie from inside a room to outside the building, when cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Evaporative air coolers operate on a totally different principle. Water is evaporated and dispersed into the air to cool the air by absorbing heat. In doing so they increase the humidity of the air inside the room. Because of this, such coolers are not suited to environments with humid air, such as tropical regions and coastal regions. They are best suited to areas with dry air, such as deserts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pads of an evaporative air cooler were to be replaced with a coil, as you ask, there is nothing to evaporate. The only way for such a system to work is to turn the evaporative air cooler into a refrigerative air conditioner (also known as a reverse cycle air condition or a heat pump). This defeats the purpose of having an evaporative air cooler. You may as well just acquire a reverse cycle air conditioner.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-06-19T14:31:39.173" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22324" PostTypeId="1" AcceptedAnswerId="22327" CreationDate="2018-06-19T16:54:30.687" Score="1" ViewCount="22" Body="&lt;p&gt;I know that heat affects material permeation (increasing it); but I have not been able to find much in the way of data other than biologically, on how it effects materials such as plastics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does heat increase permeation exponentially, linearly, etc..? &lt;/p&gt;&#xA;" OwnerUserId="3970" LastActivityDate="2018-06-19T18:26:41.713" Title="Heats affect on material permeation?" Tags="|materials|heat-transfer|temperature|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22325" PostTypeId="1" CreationDate="2018-06-19T17:22:00.247" Score="1" ViewCount="816" Body="&lt;p&gt;I’m in a situation where a multi unit structure under construction will receive its water from local municipality via three metered 5/8” lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We’d like to combine all of the lines into a 3” horizontal main. The question is, can we assume that all three lines would have equal pressure, since they originate from the same water main? The concerb is that our volume would be reduced if any of the lines have lower pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="16487" LastActivityDate="2018-09-20T09:02:23.743" Title="Will multiple lines off same water main have equal pressure?" Tags="|fluid-mechanics|water-resources|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22326" PostTypeId="2" ParentId="22325" CreationDate="2018-06-19T17:37:08.257" Score="0" Body="&lt;p&gt;If there is no flow then pressure will be equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is flow and the lines are different lengths or have extra bends or other loss causing changes then the delivery pressures will be different but in some countries you must not allow reverse flow back into the supply so you may have to fit one-way valves to prevent that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But feeding a 3” pipe from three 5/8 supplies seems odd... I suppose you have what you have.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-06-19T17:37:08.257" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22327" PostTypeId="2" ParentId="22324" CreationDate="2018-06-19T18:26:41.713" Score="2" Body="&lt;p&gt;Because diffusion processes go like e^kT, the apparent permeabililty will rise exponentially. The picture is a bit more complex than this for plastics, because they expand significantly with temperature, thereby lowering their bulk density and enhancing their permeability.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-19T18:26:41.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22328" PostTypeId="1" CreationDate="2018-06-19T20:02:21.823" Score="3" ViewCount="7952" Body="&lt;p&gt;I'm trying to clarify my understanding about the requirements [read: best practices] for measurement resolution and measurement confidence.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Rule of 10s&quot; is a commonly tossed around guideline for measurement resolution that, to the best of my understanding and searching, suggests the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;That your measurement systems should be able to resolve to a level of 10% of your total tolerance. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, regardless of whether you consider &quot;tolerance&quot; to refer to the product or process tolerance, there still seems to be some confusion among those I've talked with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, take the following dimension:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;125mm +/- 10mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the total tolerance range is 20mm.  Ten percent of 20mm would be 2mm.  Therefore your measurement system should be able to resolve down to a 2mm level which would get you roughly 10 distinct &quot;categories&quot; into which your results fall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternate theory that I've come across however, is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;That the measurement resolution should be &quot;ten times greater than your specification/tolerance as written. &lt;/li&gt;&#xA;&lt;li&gt;This is the &quot;move the decimal place&quot; approach and is more a &quot;significant figures&quot; discussion. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Taking the same previous example where the tolerance is +/- 10mm, increasing the resolution 10X would imply that measurements should be obtained at the 1mm level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't believe this latter approach is the intent of the &quot;10 to 1&quot; rule, but I'm looking for some general consensus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To complicate this question slightly, consider a scenario with an engineering drawing which defines a length as follows:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;468.2mm +/- 6.48mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(THIS is where the team arguments begin.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total tolerance is 12.96mm (6.48mm x 2), of which 10% would be 1.296mm.  Being a good inspector, and dropping the decimal places to utilize a 1mm resolution measurement method, this still seems to be contradictory to the intent of the dimension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the dimension or tolerance is specified out to multiple decimal places, should the &quot;move the decimal place&quot; method be invoked? If a dimension is specified out two decimal places, it would only make sense to be able to resolve it to 3 decimals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Appreciate anyone who takes the time to read through this and give your thoughts!&lt;/p&gt;&#xA;" OwnerUserId="4897" LastActivityDate="2019-08-29T15:56:11.550" Title="Rule of Tens/Ten to 1 Rule in Measurement and Inspection" Tags="|measurements|technical-drawing|quality-engineering|metrology|calibration|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22329" PostTypeId="1" AcceptedAnswerId="22330" CreationDate="2018-06-19T21:48:39.510" Score="0" ViewCount="658" Body="&lt;p&gt;I am working on a project where I must find the maximum principal stresses on an object subjected to impact loading condition. I know the velocity of the object (the one which will hit the test object) and its weight. My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;For analysis should I apply force or pressure to the area of impact?&lt;/li&gt;&#xA;&lt;li&gt;How do I calculate the force/pressure?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16489" LastEditorUserId="1832" LastEditDate="2018-06-19T22:05:00.397" LastActivityDate="2018-06-19T23:01:28.743" Title="Should I apply Force or Pressure, Ansys Static Structural" Tags="|mechanical-engineering|structural-engineering|structural-analysis|finite-element-method|ansys|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22330" PostTypeId="2" ParentId="22329" CreationDate="2018-06-19T22:54:44.870" Score="0" Body="&lt;p&gt;Your problem is ill-posed as stated.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a zeroth-order estimate of the impact force can be calculated using a one-dimensional balance of energy.  The kinetic energy of the projectile is&#xA;$$&#xA;  \text{KE} = \tfrac{1}{2} m v^2&#xA;$$&#xA;where $m$ is the mass of the projectile and $v$ is its velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strain energy of the target, assuming linear elasticity, is&#xA;$$&#xA;  \text{U} = \tfrac{1}{2} \sigma \varepsilon V&#xA;$$&#xA;where $\sigma$ is the stress in the target, $\varepsilon$ is the strain in the target, and $V$ is the volume of the target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation the two, we have&#xA;$$&#xA;  m v^2 = \sigma \varepsilon V = \frac{F}{A} \varepsilon A L = F \varepsilon L&#xA;$$&#xA;where $F$ is the impact force, $A$ is the impact area and $L$ is the length of the target normal to the impact direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The impact force estimate is, therefore,&#xA;$$&#xA;  F = \frac{m v^2}{\varepsilon L} = \frac{m v^2}{d}&#xA;$$&#xA;where $d$ is the depth of penetration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, instead, you would like to use the pressure,&#xA;$$&#xA; p = \sigma = \frac{m v^2}{\varepsilon A L} = \frac{m v^2}{A d}&#xA;$$&#xA;Either can be used, but note that the force has to be applied over the contact area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is ill-posed because, even if you know $A$, you don't know $\varepsilon$ and $L$.  People usually assume that $L$ is the thickness of the object (which may be infinite in some cases) and that $\varepsilon$ is small (e.g., 0.001). On the other hand, you may know a value of $d$ and get a reasonable estimate that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only accurate way to compute the impact forces is to do a full-physics dynamic simulation.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="3488" LastEditDate="2018-06-19T23:01:28.743" LastActivityDate="2018-06-19T23:01:28.743" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22332" PostTypeId="1" CreationDate="2018-06-20T01:35:26.957" Score="1" ViewCount="801" Body="&lt;p&gt;I am doing polyurethane tube simulation in Solidworks Simulation. I am using material PUR for polyurethane already defined in Solidworks. I have a question about the model type to be selected for material? As polyurethane is plastic, should I choose Non-Linear elastic or plastic or hyperelastic? Any help would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot. &lt;/p&gt;&#xA;" OwnerUserId="16491" LastActivityDate="2018-06-20T01:35:26.957" Title="Polyurethane tube simulation in Solidworks" Tags="|mechanical-engineering|structural-analysis|finite-element-method|" AnswerCount="0" CommentCount="5" ClosedDate="2018-07-07T22:18:26.013" ContentLicense="CC BY-SA 4.0" />
  <row Id="22333" PostTypeId="1" CreationDate="2018-06-20T02:14:24.470" Score="0" ViewCount="369" Body="&lt;p&gt;Consider a pipe in which flow is taking place.A textbook says,when there is a valve or heat exchanger or resistance to flow between two points in this pipe,pressure decreases from one point to another.It occurts to me that when there is resistance to flow, velocity decreases and hence kinetic head decreases which implies pressure head increases (according to Bernoulli's equation).Could someone clarify?&lt;/p&gt;&#xA;" OwnerUserId="15268" LastActivityDate="2018-08-19T10:01:40.273" Title="How does pressure drop occur across a pipe" Tags="|fluid-mechanics|flow-control|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22334" PostTypeId="2" ParentId="22333" CreationDate="2018-06-20T04:09:51.953" Score="0" Body="&lt;p&gt;The pressure losses in a pipe are due to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity squared&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length, material and surface roughness ie pipe friction (Thanks Fred)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And any losses due to bends, restrictions etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has a good explanation of the phenomena : &lt;a href=&quot;http://my.me.queensu.ca/People/Sellens/LossesinPipes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Losses in pipes - Queens University&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-06-20T04:43:41.737" LastActivityDate="2018-06-20T04:43:41.737" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22338" PostTypeId="2" ParentId="22333" CreationDate="2018-06-20T08:10:44.147" Score="1" Body="&lt;p&gt;The Bernoulli Equation can be considered to be a statement of the conservation of energy, i.e. the internal energy of the fluid remains constant. This can be useful for certain calculations e.g. the Venturi effect&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a long pipe run, however, the internal energy is not constant, as some is lost. &#xA;Any frictional effects (due to surface roughness of the pipe, going around bends etc. as noted by Solar Mike) will cause the pipework to warm up slightly, and consequently that energy is lost to the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the diameter of the pipe is equal at the two points that you're considering, and the internal energy has been reduced, then the fluid must be at a reduced pressure, since the constant mass-flow rate fixes it to being at the same velocity.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-06-20T08:10:44.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22339" PostTypeId="1" AcceptedAnswerId="22341" CreationDate="2018-06-20T09:50:00.463" Score="2" ViewCount="188" Body="&lt;p&gt;i'm doing a simulation in matlab which uses a 3D matrix as input, i.e. the matrix has the form (xDim, yDim, zDim).&#xA;With a simple STL export and a voxelization of this STL file i can get only a matrix with all elements be either one or zero.&#xA;But my part has four different materials which i need to distinguish in matlab.&#xA;One single slice in the matrix should look like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 0 0 0 1 1 1 1 2 2 2 0 ...&#xA;0 0 0 1 1 1 1 2 2 2 0 0 ...&#xA;0 0 0 1 1 1 1 1 2 2 2 2 ...&#xA;.&#xA;.&#xA;.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;As a first step: in which format should i export my part to preserve the materials as some value?&lt;/p&gt;&#xA;" OwnerUserId="16494" LastActivityDate="2018-06-20T12:18:13.460" Title="What is a method to export a SolidWorks part (with distinguishing between materials) to a voxel cloud?" Tags="|solidworks|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22340" PostTypeId="2" ParentId="22328" CreationDate="2018-06-20T11:51:07.683" Score="1" Body="&lt;p&gt;Correct. The rule of thumb is multiply your tolerance range by 10% (or divide by 10) to get the &lt;em&gt;minumum&lt;/em&gt; resolution of your measuring device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the 12.96 tolerance range because more precision is applied to the dimension, two decimal places, or .01mm precision, you need a measuring device 10x that precision to be accurate. So a device with .001mm precision would be correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my opinion on the +-6.48 tolerance is that it lacks consideration for design for manufacturing and inspection. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2018-06-20T11:51:07.683" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22341" PostTypeId="2" ParentId="22339" CreationDate="2018-06-20T12:18:13.460" Score="0" Body="&lt;p&gt;If you are stuck with using STL, here's a crude way to do it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rpuS1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rpuS1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Create two references which extend beyond the object in the x, y, z space (I used the cylinders shown on the upper right and lower left corners)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For each material (body), export STL along with the references&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Voxelise using the same parameters&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The references are used to ensure that the binary voxel arrays are generated in a consistent space between the different bodies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can multiply each separate voxel array using a different number (2,3,4 ...) and then add them together. If you want to get rid of the references, you can use Matlab's &lt;code&gt;bwlabel&lt;/code&gt; &lt;a href=&quot;https://au.mathworks.com/help/images/ref/bwlabel.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;function&lt;/a&gt; to detect unconnected regions and remove them.&lt;/p&gt;&#xA;" OwnerUserId="15586" LastActivityDate="2018-06-20T12:18:13.460" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22342" PostTypeId="1" CreationDate="2018-06-20T12:49:01.387" Score="-1" ViewCount="134" Body="&lt;p&gt;I played some video games with beautiful environments filled with buildings which can't be found in the real world. And so I'm wondering, would such buildings be even realistically possible to exist in the real world?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is that they are made from the most likely material - stone. If it is impossible, would it be possible to make it from some other material while being able to maintain the aesthetics of the exterior (painted steel)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At some point, the materials (marble, concrete) under certain conditions must start deforming, correct? So how high could a gothic cathedral possibly be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, I will enclose a couple of pictures from Dark Souls 3 which will hopefully give you an idea of what I have on my mind. If you ever played this game, you will know how colossal the buildings are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4NgDi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;IMGUR&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your convenience, I added red lines to represent the height of a human in comparison to these structures. If you can't find them, I put them into yellow circles. Additionally, these two pictures are showing the same castle, and I made sure that these representations of the human height are on the same sea level. So the heights add up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, for a better sense of how high it actually is, I am also enclosing a video where the player travels from the location of the second human height representation. Hopefully, it will give you a better sense of the size:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=Fd7PMzSw5lM&quot; rel=&quot;nofollow noreferrer&quot;&gt;From the outside&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a second video, so that you can grasp how large the castle is when you stand on its outer wall:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=diNeFY3M7OY&amp;amp;t=135&quot; rel=&quot;nofollow noreferrer&quot;&gt;From the inside&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16495" LastActivityDate="2018-08-20T02:00:20.880" Title="What are the limits on building very high buildings using certain materials and architectural style?" Tags="|building-design|building-physics|architecture|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="22345" PostTypeId="1" AcceptedAnswerId="22348" CreationDate="2018-06-20T13:47:18.067" Score="0" ViewCount="116" Body="&lt;p&gt;As &lt;strong&gt;roadway&lt;/strong&gt; &lt;em&gt;consists of a width of road on which a vehicle is not restricted by any physical barriers or separation to move laterally&lt;/em&gt; &lt;sup&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Carriageway&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/sup&gt;, what is a similar term which encompasses the width of the entire road's infrastructure - barriers, drainage ditches, sound screens, generally the entirety of terrain allocated for construction as a road is built?&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-03-19T17:36:13.477" Title="What is the entirety of breadth of road called?" Tags="|civil-engineering|terminology|highway-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22347" PostTypeId="2" ParentId="22342" CreationDate="2018-06-20T14:12:16.203" Score="3" Body="&lt;p&gt;The tallest masonry building erected is the &lt;a href=&quot;https://weburbanist.com/2008/01/19/the-tallest-masonry-building-in-the-world-philadelphias-beautiful-load-bearing-city-hall/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Philadelphia City Hall&lt;/a&gt; . It is 548 feet high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue with building masonry structures is the walls have to support the weight of the building and the weight of the wall itself. Bricks &amp;amp; stone are heavy, so for a masonry wall to be tall the width of the wall at its base needs to wide - it can taper to a thinner wider up the wall. This can place limitations on the area of usable space inside such buildings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue is, with so much weight there will be a lot bearing pressure on the ground and the ground needs to be strong enough to support the structure. Soils are not as strong as bed rock, and some bedrock can be weak, depending on the bedrock. Sometimes masonry wall need to have foundations set in bedrock.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-06-20T14:12:16.203" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="22348" PostTypeId="2" ParentId="22345" CreationDate="2018-06-20T14:20:17.707" Score="2" Body="&lt;p&gt;A number of online sources use the term road reserve to encompass the road, associated infrastructure, such as drainage and spare land for additional lanes in the future. The width of this would be the road reserve width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lFbqt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lFbqt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-06-20T14:20:17.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22350" PostTypeId="1" CreationDate="2018-06-20T17:41:04.260" Score="1" ViewCount="120" Body="&lt;p&gt;Using an old copy of an automobile magazine, these are the cars with hp and lb ft I am trying to work out the RPM for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2007 Honda Accord 2.4 sedan, 2.4-liter 187hp 4-cylinder, 164 lb ft pulling power&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The math I did to work out RPM:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;187 x 5252/164 = 5,988.56 (rounded up to 6000rpm, how far should i round up in significant figures or decimal places)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2007 Peugeot 207 1.6 HDi 110 GT, 1.6-liter 110hp 4-cylinder, 177 lb ft pulling power&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The math:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;110 x 5252/177 = 3,263.95 (rounded up to 3,264 rpm, or should it be 3000 rpm?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used these formulas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque and horsepower relations:&#xA;T = HP * 5252 / RPM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HP = T * RPM / 5252&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RPM = HP * 5252 / T&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my two examples, have I got the math correct for car engines? With RPM, when do I need to round up correctly and what significant figure level do I need?&lt;/p&gt;&#xA;" OwnerUserId="16501" LastActivityDate="2018-07-24T11:22:39.337" Title="Working out RPM from car engine's lb ft?" Tags="|mechanical-engineering|automotive-engineering|mathematics|car|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22351" PostTypeId="1" CreationDate="2018-06-20T18:43:26.320" Score="1" ViewCount="30" Body="&lt;p&gt;I'm trying to build a filament extruder with some basic available tools and materials. Simply the main principle of the machine is to insert some amount of plastic raw material ( ABS in our Case ) to a hopper , move them inside a barrel with the help of helical screw which is derived by a motor, heating the material and extruding it from a predefined hole size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets assume that these are our givens:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Tube inner diameter: 20 mm&lt;/li&gt;&#xA;&lt;li&gt;Helical screw outer diameter: 18 mm&lt;/li&gt;&#xA;&lt;li&gt;Helical screw body diameter: 15 mm&lt;/li&gt;&#xA;&lt;li&gt;Helical screw pitch: 30 mm&lt;/li&gt;&#xA;&lt;li&gt;ABS density @ melting: 0.97 gr/cc&lt;/li&gt;&#xA;&lt;li&gt;Motor speed: 30 RPM&lt;/li&gt;&#xA;&lt;li&gt;Material output hole: 3 mm&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now what is the proper power of the motor to drive the machine?&lt;/p&gt;&#xA;" OwnerUserId="16505" LastEditorUserId="9" LastEditDate="2018-06-20T18:48:05.233" LastActivityDate="2018-06-20T18:48:05.233" Title="Calculating the proper power of a motor" Tags="|design|motors|power|equipment-selection|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22353" PostTypeId="2" ParentId="18149" CreationDate="2018-06-20T20:34:03.740" Score="0" Body="&lt;p&gt;3/16 BSF (3/16in dia 32 turns per inch)&lt;/p&gt;&#xA;" OwnerUserId="16502" LastActivityDate="2018-06-20T20:34:03.740" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22354" PostTypeId="2" ParentId="22350" CreationDate="2018-06-20T21:12:03.990" Score="1" Body="&lt;p&gt;As I already mentioned in my comment, the problem in your calculations is that maximum torque and maximum power are at different RPM. The picture below is an example for a torque-power-diagram (excuse the sloppy quality, I just took a photo of old lecture notes with badly edited translations) of a combustion engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kxi4x.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kxi4x.jpg&quot; alt=&quot;exemplary torque-power-diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solid line is the torque, the dashed line is the power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can clearly see that the maximum torque and maximum power are at least 1000 RPM apart. To calculate the RPM at which maximum torque is provided, you cannot use the maximum power from the data sheet. You need to use the power that corresponds to the RPM you are searching for, so in the diagram, the power where the dashed line intersects with the red line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this, usually car data sheets tell at which RPM the maximum power and torque are delivered. I hope I could make it a bit clearer why your calculations give &lt;em&gt;some&lt;/em&gt; value, but not a meaningful RPM value.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2018-06-20T21:12:03.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22355" PostTypeId="1" CreationDate="2018-06-20T23:27:55.797" Score="0" ViewCount="163" Body="&lt;p&gt;A majority of semi-automatic pistols (revolvers excluded) have a spring-loaded component called &lt;a href=&quot;https://en.wikipedia.org/wiki/Pistol_slide&quot; rel=&quot;nofollow noreferrer&quot;&gt;slide&lt;/a&gt; as a part of their firing mechanism. Although historically there are designs without a slide, (notably &lt;a href=&quot;https://en.wikipedia.org/wiki/Mauser_C96&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mauser C96&lt;/a&gt;), those are, if not non-existent, very rare after the World War II.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It probably is safe to conclude that the displacement of other designs are due to some advantages in the slide mechanism. What are those advantages?&lt;/p&gt;&#xA;" OwnerUserId="9953" LastEditorUserId="1832" LastEditDate="2018-06-21T00:08:32.323" LastActivityDate="2020-11-12T09:19:35.717" Title="What are the advantages of the slide in semi-automatic pistol design?" Tags="|mechanical-engineering|firearms|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="22357" PostTypeId="1" CreationDate="2018-06-21T00:40:50.963" Score="2" ViewCount="740" Body="&lt;p&gt;I have done my project on alcohol detection for drivers, but I don't know the minimum and maximum distances between the alcohol sensor and the driver.&#xA;What are the maximum and minimum distances?&lt;/p&gt;&#xA;" OwnerUserId="16503" LastEditorUserId="16624" LastEditDate="2018-07-02T09:05:44.260" LastActivityDate="2018-07-02T09:05:44.260" Title="MQ3 Alcohol sensor" Tags="|electrical-engineering|sensors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22358" PostTypeId="2" ParentId="22328" CreationDate="2018-06-21T01:59:02.350" Score="2" Body="&lt;p&gt;Remember that a rule of thumb is just that, a rule of thumb.  It is not a hard specification, so in some sense there really is no right answer here.  It is a good starting point, but to determine what is acceptable for your system you need to balance three things&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is the cost of a false positive?  i.e. what is the consequence if a perfectly acceptable +9 mm part is flagged as +11 mm and thrown away?  Does that cost you 2 cents or a million dollars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What is the consequence of a missed negative?  i.e. what is the consequence if an unacceptable +11 mm part is marked as acceptable?  Does it cost you 2 cents or a million dollars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) What is the cost of various levels of resolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if missed negatives are costing your company \$1,000 per year, and upgrading from 2 mm resolution to 1 mm resolution costs $1 million, well then just stick with 2 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, on the other hand, if missed negatives are costing you a million dollars per year, and and upgrading resolution only costs $10,000 then you should upgrade as quick as you can.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-06-21T01:59:02.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22360" PostTypeId="2" ParentId="15040" CreationDate="2018-06-21T03:35:26.493" Score="0" Body="&lt;p&gt;The problem mentioned that support A and C are both pins, therefore you should use the modified slope-deflection equation.&lt;/p&gt;&#xA;" OwnerUserId="16509" LastActivityDate="2018-06-21T03:35:26.493" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22361" PostTypeId="2" ParentId="22355" CreationDate="2018-06-21T04:04:22.003" Score="4" Body="&lt;p&gt;A simple mechanism, rapid ejection of spent cartridge, rapid loading of fresh cartridge and some of the recoil energy is absorbed in driving the slide helping improve accuracy.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-06-21T04:04:22.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22363" PostTypeId="1" AcceptedAnswerId="22369" CreationDate="2018-06-21T05:56:53.667" Score="-2" ViewCount="112" Body="&lt;p&gt;How do &amp;quot;engineering leaders&amp;quot; work with large-scale &amp;quot;trends&amp;quot;?&lt;/p&gt;&#xA;&lt;p&gt;Like matching against predicted requirements in renewable energy etc.&lt;/p&gt;&#xA;&lt;p&gt;These seem like abstractly unknowable future states. Yet, &amp;quot;engineering leaders&amp;quot; seem to be able to ride the trends and collect the requirements needed to work on these trends.&lt;/p&gt;&#xA;&lt;p&gt;Also, nowadays the projects are quite large-scale already, because they may touch even millions of people, or take tens of years to fully implement. It seems like black magic, because humans and human organizations have a lot of variables. One cannot e.g. control the supply of labour market.&lt;/p&gt;&#xA;&lt;p&gt;So how do they progress in such?&lt;/p&gt;&#xA;" OwnerUserId="2643" LastEditorUserId="2643" LastEditDate="2022-06-19T11:07:36.573" LastActivityDate="2022-06-19T11:07:36.573" Title="How do &quot;engineering leaders&quot; work with large-scale &quot;trends&quot;?" Tags="|engineering-managment|" AnswerCount="2" CommentCount="9" ClosedDate="2018-06-28T12:50:20.040" ContentLicense="CC BY-SA 4.0" />
  <row Id="22364" PostTypeId="2" ParentId="22363" CreationDate="2018-06-21T06:31:57.350" Score="1" Body="&lt;p&gt;They don't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you call a trend, has usually been around for ages. Engineering on a large commercial scale does not really start until the fundamentals are known. This means that there has already been quite a lot of basic research going on, for years sometimes decades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If nothing else, then the limitations and pitfalls of current tech is well known. Through mathematical modeling they can say what the limits of different technological methods are even before they overtake the competition. Therefore people in the R&amp;amp;D departments know what to look for, and revisit what is available from time to time. This is why photographing inside factories is a no-no as a person in the know can near instantaneously be able to spot the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big companies however, deliberately create demand. They don't just passively make things happen.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-06-22T19:37:49.133" LastActivityDate="2018-06-22T19:37:49.133" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22369" PostTypeId="2" ParentId="22363" CreationDate="2018-06-21T12:12:26.347" Score="2" Body="&lt;p&gt;The strategic planning of long term projects isn't solely an engineering problem but also involves business, financial and political planning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take the example of renewable energy and to generalise somewhat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First the technology needs to be developed and proven. This is usually an iterative process, initially on a relatively small scale. You first have prototypes then pilot schemes then full commercial roll-out   Infrastructure technology often involves a degree of collaboration between business, technology and government. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example government may stimulate demand by offering grants, tax breaks etc or by legislating to reduce the attractiveness of alternatives. For example if renewable energy is taxed at a lower rate than coal power, then that makes renewable a more attractive investment proposition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any new technology is a business risk and investors will calculate whether the potential return on investment justifies that risk. This calculation will involve both engineering feasibility studies and economic modelling to try to predict future economic conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally while you can't completely control future markets you can trade in them, energy wholesale contracts may be agreed decades in advance and often energy infrastructure projects will include complex contracts which set pricing structures. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="624" LastEditDate="2018-06-21T12:19:23.950" LastActivityDate="2018-06-21T12:19:23.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22370" PostTypeId="1" CreationDate="2018-06-21T12:32:33.897" Score="3" ViewCount="38" Body="&lt;p&gt;The storage coefficient (s) for artesian basin can be determined by the following equation (Theis equation): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$s=\frac{Q}{4\pi T}\int_{\frac{r^2S}{4Tt}}^\infty\frac{e^{-u}}{u}du$$&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;p&gt;Does this equation work for confined aquifers with an arbitrary &lt;em&gt;t&lt;/em&gt; or there are some limits for time (t) to be used in this equation?&lt;/p&gt;&#xA;" OwnerUserId="16515" LastEditorUserId="9" LastEditDate="2018-06-21T12:39:28.580" LastActivityDate="2018-06-21T12:39:28.580" Title="The use of Theis equation based on time" Tags="|fluid-mechanics|civil-engineering|hydraulics|geotechnical-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22371" PostTypeId="1" CreationDate="2018-06-21T13:19:27.900" Score="1" ViewCount="214" Body="&lt;p&gt;I received these fittings as part of a package, but now we need more of them. Can anyone identify this assembly, or even better help me learn how to recreate it? My coworkers and I are stumped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pictures: &lt;a href=&quot;https://i.stack.imgur.com/38qb9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/38qb9.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sSmeE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sSmeE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j2feD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j2feD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ask9J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ask9J.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vgQYA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vgQYA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16517" LastEditorUserId="7083" LastEditDate="2018-07-02T22:20:41.703" LastActivityDate="2018-10-03T14:28:16.343" Title="I have an unidentified press-fit nut that I need to reproduce, can anyone tell me how to find this?" Tags="|mechanical-engineering|machining|fasteners|connections|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22373" PostTypeId="1" CreationDate="2018-06-21T14:06:38.083" Score="3" ViewCount="9922" Body="&lt;p&gt;I'm looking at this &lt;a href=&quot;http://www.disensors.com/product_details.aspx?id=1390&quot; rel=&quot;nofollow noreferrer&quot;&gt;inclinometer&lt;/a&gt; and in the features it says &quot;12 bit resolution.&quot; I'm looking for a resolution of up to a 1/10th of a degree but don't understand what the 12 bit resolution means.&lt;/p&gt;&#xA;" OwnerUserId="16433" LastActivityDate="2018-06-22T16:54:37.977" Title="What does 12 bit resolution mean?" Tags="|measurements|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22374" PostTypeId="2" ParentId="22373" CreationDate="2018-06-21T14:17:26.387" Score="6" Body="&lt;p&gt;With 12-bit resolution, $2^{12}= 4096$ different numbers can be represented, this means that the measuring range of a given sensor gets split into 4096 pieces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, if your sensor can measure from 0V to 409.6V (just for nicer math), then with 12-bit resolution, you can resolve the measured value in increments of 0.1V.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastEditorUserId="14629" LastEditDate="2018-06-21T16:27:09.013" LastActivityDate="2018-06-21T16:27:09.013" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22375" PostTypeId="2" ParentId="22251" CreationDate="2018-06-21T14:50:49.163" Score="0" Body="&lt;p&gt;It turns out that defining predefined field variables in CAE is a feature that was added in the 2018 version. The steps that one needs to follow when creating a field-dependent material property are the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) In the Property module, make your Material dependent on a Field Variable(s) by specifying the number of field variables. In the table you will have to map the material property to the field variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Create the Field Variable: In the Load module, go to &quot;Predefined fields&quot;-&gt;Create-&gt;Other-&gt;Field and specify the desired distribution, which can be one of several options, including an Analytical Field. This analytical field in my case is the spatial distribution of the expansion coefficient - a function of X,Y,Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you have an earlier version of Abaqus/CAE, apparently you could use a temperature definition for the Field, and modify your input file to reroute to the field variable. I haven't tried this approach, but it should work just as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple people suggested parts of this and then I finally figured the rest out through some trial and error. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16434" LastActivityDate="2018-06-21T14:50:49.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22377" PostTypeId="1" CreationDate="2018-06-21T18:15:59.250" Score="1" ViewCount="2413" Body="&lt;p&gt;I have these two parts shown below, the bottom cylindrical part is stainless steel, the rectangular upper piece is PTFE (Teflon). The SS prongs from the lower part need to insert into the holes of the PTFE part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p2PRx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p2PRx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to create an airtight connection between the two; but I'm unsure of how to go about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I'd like to avoid adhesives all together as this will undergo high temps.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3970" LastEditorUserId="3970" LastEditDate="2018-06-21T19:18:40.687" LastActivityDate="2018-11-08T03:42:09.600" Title="How to create an airtight connection between Stainless Steel and PTFE?" Tags="|mechanical-engineering|materials|metals|plastic|connections|" AnswerCount="6" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22378" PostTypeId="2" ParentId="22377" CreationDate="2018-06-21T18:44:15.380" Score="1" Body="&lt;p&gt;If you can modify the stainless steel prongs with groves or indentations to accept either O-rings or Compression brass or plastic fittings you would not need any adhesive!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plastic compression connectors can withstand high temperatures and pressures. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-06-21T18:44:15.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22379" PostTypeId="2" ParentId="22377" CreationDate="2018-06-21T19:04:34.000" Score="1" Body="&lt;p&gt;A few design improvement notes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;PTFE is not Nylon.&lt;/em&gt;  I'll assume from here on out you really mean PTFE here, but it's better known as Teflon.  Nylon would be done entirely differently.&lt;/li&gt;&#xA;&lt;li&gt;You can't just snap these two parts together for an airtight seal.  You can get mostly air tight, but you will not make an airtight seal between stainless and PTFE with just a snapped connection.&lt;/li&gt;&#xA;&lt;li&gt;PTFE is slippery stuff.  It's well known for this property.  You will need some kind of mechanical way to keep it on (such as end flanges that can be screwed onto the ends of the SST parts).  Otherwise most mechanical stresses will break the very fragile airtight seal of the passive method, and up the power requirements of the active method.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There are two methods for joining these two materials.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Passive Method&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In the passive method, you would roughen the stainless steel's surface to a white metal blast finish (SSPC.SP5 standard or NACE #1).  Then the steel would be coated with a powder coat of PFA (or PFA/FEP combination), then baked in a powder coating application.  Additional layers of PFA could be added, and baked on.  A minimum of three would be recommended.  Finally, heat both the stainless and PTFE to 260 Celsius, and join together at that temperature.  The parts will have a fairly weak bond, but so long as the stress around the joint can be mitigated mechanically as discussed above, it will be workable.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Active Method&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is more for robust applications and with thin, relatively flexible PTFE linings.  In this method, you still treat the steel surface as mentioned above, but instead of all the thermal work, there is a manifold that would supply active compressed air to a thin layer of PTFE surrounding the joint.  So long as the air pressure does not exceed the compressed air pressure, the mechanical push forces will maintain an active seal.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Testing&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You should make samples ahead of time between the proposed materials and test them with a mechanical peel test (Such as an ASTM D1781 &lt;a href=&quot;https://www.youtube.com/watch?v=QOxlNpQ_s7w&quot; rel=&quot;nofollow noreferrer&quot;&gt;climbing drum test&lt;/a&gt;).  A 50 lb/in peel is a bare minimum for pass/fail, 100 lb/in is more suitable.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-06-21T23:12:13.420" LastActivityDate="2018-06-21T23:12:13.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22380" PostTypeId="2" ParentId="22377" CreationDate="2018-06-21T20:13:57.423" Score="4" Body="&lt;p&gt;You're going to struggle to get a reliable gas tight fit with that design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a start it would be better if the two connections were separate so they aren't fighting against each other in terms of fit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally rather that trying to reinvent the wheel you would be better using off the shelf connectors which are known to work. It would be fairly trivial to replace the plain tubes with threaded or compression fit connectors in a suitable material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also hard to offer advice without more detail of what you are ultimately trying to achieve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If for example you wanted to be able to 'plug-in' the stainless steel section you might be ok with a taper fit if you could keep the joint compressed eg with an external spring or cam but again you could just use off the shelf quick release fittings.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-06-23T18:30:18.577" LastActivityDate="2018-06-23T18:30:18.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22381" PostTypeId="2" ParentId="2934" CreationDate="2018-06-21T20:49:06.687" Score="0" Body="&lt;p&gt;You can use a SPRAG type clutch from Altramotion.com &lt;/p&gt;&#xA;&#xA;&lt;p&gt;or a backstop that would spin the opposite direction which could act like a engagement gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i still believe you can get a all in one unit from them for a SPRAG clutch that will be useful for you. &lt;/p&gt;&#xA;" OwnerUserId="15027" LastActivityDate="2018-06-21T20:49:06.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22383" PostTypeId="2" ParentId="22350" CreationDate="2018-06-22T00:04:11.377" Score="0" Body="&lt;p&gt;Max HP will always lag max torque&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HP = torque x rpm  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless torque drops off massively fast HP peak will lag (higher rpm).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;HP peak cannot be a lower rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be proved mathematically but this site does not support Latex.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-07-24T11:22:39.337" LastActivityDate="2018-07-24T11:22:39.337" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22384" PostTypeId="2" ParentId="22325" CreationDate="2018-06-22T00:23:58.877" Score="1" Body="&lt;p&gt;The pressure drop over feet from a main water line is near zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A main line is over powered to maintain pressure.  They have to deliver pressure to the last person on the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3&quot; is way bigger than 3 x 5/8&quot;.   It is closer to a 1&quot; line.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-06-22T00:23:58.877" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22386" PostTypeId="2" ParentId="15040" CreationDate="2018-06-22T00:36:46.210" Score="1" Body="&lt;p&gt;If you look at the structure (ignoring the loading), it is symmetrical: two spans of equal length, with pins on the extremities and a roller in the middle. It is also a hyperstatic (or statically indeterminate) structure, with more unknowns than static equilibrium equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might therefore be tempted into simplifying this model into a single fixed-and-pinned beam. After all, a symmetrical load on both spans will cancel out the rotation at B, and a point with bending and no rotation is equivalent to a fixed support. So why not simplify the model into a single span? Sure, it's still hyperstatic, but it's a classic condition with known reactions as given by your tables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, obviously the problem is that, in this case, the loading &lt;strong&gt;&lt;em&gt;isn't&lt;/em&gt;&lt;/strong&gt; symmetric. So what do you do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ignore that little detail and momentarily pretend that you are in fact dealing with two fixed-and-pinned spans. You then calculate the moment reaction at the &quot;fixed&quot; point B for each span. You then use slope-deflection equations to figure out what the &lt;strong&gt;&lt;em&gt;actual&lt;/em&gt;&lt;/strong&gt; rotation around B is and use that to recalculate your reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's take this one step at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume AB and BC are pinned-and-fixed beams and calculate the moment reaction at B in each case using your tables:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f5Zmt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f5Zmt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{4}&#xA;M_{B,AB} &amp;amp;= \dfrac{P}{L^2}\left(b^{2}a + \dfrac{a^{2}b}{2}\right) &amp;amp;&amp;amp;= 52.5\text{ kNm} \\&#xA;M_{B,BC} &amp;amp;= \dfrac{3PL}{16} &amp;amp;&amp;amp;= -30\text{ kNm}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that $M_{B,BC}$ used the top-right case from your table since the load was centered, while $M_{B,AB}$ used the next one below since the force is off-center. Also note that the structure in both cases is the same: a fixed-and-pinned beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that the results for $M_{B,AB}$ and $M_{B,BC}$ aren't equal, which tells you that the assumption that point B was the same as a fixed support with no rotation was incorrect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You therefore use the slope-deflection equations to figure out the relationship between bending moment and rotation for each span, use them to calculate the actual rotation around B, and then use that to calculate the actual bending moment around B:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{4}&#xA;M_{B,AB} &amp;amp;= \dfrac{3EI}{8}\theta_B + 52.5 \\&#xA;M_{B,BC} &amp;amp;= \dfrac{3EI}{8}\theta_B - 30 \\&#xA;M_{B,AB} &amp;amp;= M_{B,BC} \\&#xA;\dfrac{3EI}{8}\theta_B + 52.5 &amp;amp;= \dfrac{3EI}{8}\theta_B - 30 \\&#xA;\therefore \theta_B &amp;amp;= \dfrac{-30}{EI} \\&#xA;\therefore M_B &amp;amp;= \dfrac{3EI}{8}\theta_B + 52.5 &amp;amp;&amp;amp;= -41.25\text{ kNm} \\&#xA;&amp;amp;= \dfrac{3EI}{8}\theta_B - 30 &amp;amp;&amp;amp;= -41.25\text{ kNm}&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I just calculated $M_B$ twice to show that you can use either of the equations to find its value, obviously)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that you have the actual moment at B and have solved the problem.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-06-22T00:36:46.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22387" PostTypeId="1" AcceptedAnswerId="22389" CreationDate="2018-06-22T00:56:20.247" Score="0" ViewCount="293" Body="&lt;p&gt;Convention says to measure at the bottom of the meniscus, but I think the cohesion/adhesion balance of mercury results in it &quot;bulging out&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you measure at the top of the bulge  (near the center) or the bottom (near the edges)?&lt;/p&gt;&#xA;" OwnerUserId="6488" LastActivityDate="2018-06-22T02:34:29.510" Title="How do I measure mercury in a graduated cylinder?" Tags="|measurements|liquid|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22389" PostTypeId="2" ParentId="22387" CreationDate="2018-06-22T02:34:29.510" Score="2" Body="&lt;p&gt;You need to measure from the top of mercury column, in tubes that are wide enough to allow a flat area of mercury on top and ignore the lower level along the perimeter where mercury is repelled by glass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This page has enough detailed instruction on calibration of barometers. &lt;a href=&quot;http://www.instruction.greenriver.edu/kmarr/Chem%20161/Instrument-%20Glassware%20Videos/How%20to%20Zero%20and%20Read%20a%20Barometer/How%20to%20zero%20and%20read%20a%20barometer.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Green river U link.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-06-22T02:34:29.510" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22390" PostTypeId="1" CreationDate="2018-06-22T03:06:49.080" Score="2" ViewCount="166" Body="&lt;p&gt;Referring to the  document here on &lt;a href=&quot;http://www.buildingsmart-tech.org/mvd/IFC4x1/Alignment/1.0/html/schema/ifcgeometryresource/lexical/ifcclothoidalarcsegment2d.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;IfcClothoidalArcSegment2D&lt;/a&gt;, we have &lt;code&gt;ClothoidConstant&lt;/code&gt;, which is defined as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The constant $A$ (not $A^2$), that determines the rate of curvature&#xA;  change along the clothoid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wonder what it actually is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the wiki page about &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler_spiral#Symbols&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler Spiral&lt;/a&gt;, I found a symbol $a$, which is calculated to be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$a={1\over \sqrt{2R_cs_o}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has the same dimension of the above &lt;code&gt;ClothoidConstant&lt;/code&gt;, $A$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they actually the same? If not, how should the &lt;code&gt;ClothoidConstant&lt;/code&gt; be defined?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="3353" LastEditDate="2018-06-22T03:42:07.897" LastActivityDate="2018-06-22T03:42:07.897" Title="How to define the Clothoid Constant in Clothodial Arc ( Spiral Curve)?" Tags="|civil-engineering|rail|highway-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22391" PostTypeId="1" CreationDate="2018-06-22T04:41:44.323" Score="1" ViewCount="1664" Body="&lt;p&gt;I have a stepper motor (&lt;a href=&quot;https://www.adafruit.com/product/858&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;). Its shaft is of type &quot;oval&quot; or whatever, so like a circle but with 2 flat sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the motor shaft pointing straight up, in the z-axis. I want a 6-inch rod, or shaft, (honestly doesn't really matter, I just want a rigid metal stick of some sort) to stick out from it. I.e. So that when the motor spins, this rod/shaft is creating a circle in the x-y plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gathered that what I want is a &quot;coupler&quot;, possibly a &quot;flexible coupler&quot; which connects shafts which are NOT parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is there a standard term, or language, for describing my motor's shaft, as well as the coupler which would connect my shaft to a metal rod?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Am I even choosing the right things for this task of spinning a rod perpendicular to the shaft's axis?&lt;/p&gt;&#xA;" OwnerUserId="16528" LastActivityDate="2018-06-26T12:39:25.180" Title="Coupler to connect motor shaft to perpendicular rod/shaft" Tags="|mechanical-engineering|motors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22392" PostTypeId="1" CreationDate="2018-06-22T05:41:02.067" Score="0" ViewCount="105" Body="&lt;p&gt;Anyone who has owned a car has ultimately dealt with failure of a sensor (parking), or A/C failure, battery shutdown or a tyre puncture. Part breakdowns like these are pretty common, at least relatively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a car however, the most stress (mechanical and thermal) is faced by the engine and engine-parts. If you think about it a little more, you could see that the most stress is faced by the piston-rod and connecting shaft. Those and related parts face constant tensions and compressions, on the order of many times a second, and not to mention the higher temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite that, I rarely hear of such failures (in proportion to other common failures mentioned above). What could be the reason for this? Are those engine parts over-engineering to never fail (almost, 'cause nothing is certain), over the lifetime of the vehicle?&lt;/p&gt;&#xA;" OwnerUserId="7057" LastActivityDate="2018-07-22T13:36:36.407" Title="Why don't car engine piston-rods (or similar parts) fail as often as other car parts?" Tags="|automotive-engineering|stresses|mechanical-failure|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22393" PostTypeId="2" ParentId="22391" CreationDate="2018-06-22T05:47:06.717" Score="2" Body="&lt;p&gt;yes, it is called a flexible coupling or flex coupling. they come in a wide variety of designs and sizes. Any engineering supply house (for instance, McMaster-Carr) offers these for sale off-the-shelf. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your step motor shaft is described as a round shaft with two setscrew flats. the couplings you are interested in are affixed to shafts using setscrews and if the shaft flats are symmetrically opposite one another, you want to specify &quot;2 setscrews at 180 degrees&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that some types of flex couplings accomodate radial and/or axial misalignment via springy or rubbery components. these perform poorly when driven by stepper motors because they introduce torsional compliance, which can then resonate off the rotary inertia of your actuator arm and then oscillate uncontrollably when driven by step impulses. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-22T05:47:06.717" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22394" PostTypeId="2" ParentId="22392" CreationDate="2018-06-22T07:13:29.477" Score="6" Body="&lt;p&gt;The main reason from a designer's point of view is the &lt;em&gt;consequence&lt;/em&gt; of a failure. Almost always, something like battery failure or a puncture doesn't cause any collateral damage - you just replace the failed part. (Even a tyre blowout when driving at high speed rarely causes much secondary damage - and that is a rare type of &quot;tyre failure&quot; compared with the more common slow puncture, which doesn't even permanently damage the tyre if you regularly check the tyre pressures.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, failure of a piston or conrod usually destroys the whole engine, or damages it beyond economic repair compared with the cost of replacing it - much more expensive than plugging a hole in a tyre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &lt;em&gt;could&lt;/em&gt; design a tyre that would have the same life as the car (say 15 years and 250,000 miles) but the cost would be huge, and many car users would never get the benefits simply because they would trade in the car long before &quot;wearing out&quot; such a tire. Would a car buyer want to pay say \$10,000 for such a tyre, compared with say \$100 for a &quot;normal&quot; one? For most people, the answer is rather obvious.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-06-22T07:20:45.283" LastActivityDate="2018-06-22T07:20:45.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22396" PostTypeId="1" CreationDate="2018-06-22T08:45:42.253" Score="1" ViewCount="50" Body="&lt;p&gt;I'm gutting and redoing an older house (early 20th C, ordinary suburban terrace house, London UK).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd planned beam and (dense) block for the upstairs structural floor, for a few reasons, and already got structural calculations for load spreading on the old foundations for the extra load this would involve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beams would be supported on engineering brick in some places, or on the stiffened flanges of 203x203x46kg beams across openings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planned floor above would then be neoprene pads with battens resting on them, and floorboards on those (for a nice effect and also ensures space for services and exceptional acoustic isolation - almost &quot;box within box&quot; by design, stopping most vibration through floors and around through the terrace walls). The wider design takes care of acoustic isolation of the walls which isn't relevant here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My reasons are a bit diverse, in no special order - beam and block guarantees no squeaking underfoot, guarantees very high quality acoustic resilience, excellent fireproofing, and makes it dead easy to support a wide (4m) span or to support battening anywhere needed. Also it's quick,cheap, and easy, no pump either, just some heavy lifting.  I know this is probably way way overkill, but I'm comfortable with that and don't consider it a factor in any decision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final, and perhaps most important, reason is that the building is space challenged. By which I mean, I'm fighting for every vertical cm I can save on the underfloor thickness. Every cm is a cm less of ceiling height, and also either a steeper or a larger stairwell. I can afford a couple of cm thicker for the 1st floor structure, but more than about 5-8cm would really hurt due to knock-on effects. Beam and block gets my structural floor thickness down to 150mm and nothing else except steel can do that, and I need that space badly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I had this all sorted out. But the job got passed to a different engineer in the same company, and he's urging a different design - engineering wood beams with 22mm plywood and 60-80mm of poured screed with mesh. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;His reason is load and age. He agrees the original plan will be fine, and the work done on walls and load spreading will be ample to accommodate the extra weight both within the walls and on the original foundations, but from what I can gather its a feeling that the building was originally designed for timber floors and just wasn't designed to have this kind of extra load or structure. He also cites building cost but I think that's more like a carrot dangling on a string - its a nice thing but in the scheme of things any cost difference will be tiny and I discount it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He agrees that screed (1800kg/m3) and full-on concrete (2400 kg/m3) are similar enough density to use the latter instead of screed if I wanted, and I feel slightly pressured, although if I say no, it will be no. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apart from considering whether or not to accede to his desire to lighten the load, my other more serious issue is that the depth would go up a lot. The original plan's depth would be 150mm floor beams, plus pads, battens and flooring. A version done as he suggests would be 200mm engineering timber beam, 20-25mm plywood, and 60-80mm screed/concrete for a total of almost 300mm - almost 150mm more... and all to save at most 110kg/m3 (beam = 250kg/m3 max, 75mm screed = 135kg/m3 min) that we've already calculated our way around anyhow. It's just not worth it as far as I can see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I say no, is there anything important here &lt;strong&gt;from a purely structural engineering, rather than general home improvement, perspective&lt;/strong&gt; that I might be missing or need to take into account, when I consider this new option against the existing one they have presented?&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2018-06-22T08:52:20.867" LastActivityDate="2018-06-22T08:52:20.867" Title="Will screed on plywood do instead of beam and block, to replace a floor?" Tags="|structural-engineering|concrete|construction-management|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22398" PostTypeId="2" ParentId="22377" CreationDate="2018-06-22T10:57:44.153" Score="0" Body="&lt;p&gt;Thread (or make otherwise irregular, like machine circular indentations) in the steel part's holes to a size that fits snugly on the PTFE pipe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then get two  cylinders of steel or other strong material that will be able to withstand whatever the pipes will carry, threaded on the outside, of outer diameter a bit larger than the inner diameter of the PTFE pipe; radius larger at least by height of the thread; they aren't just to fit snugly, they must squash the pipe a bit. Length about the width of the metal opening. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Screw them into the pipe, stretching it and squeezing the material into the threads of the steel holes. It won't be easy, because teflon is quite hard, but under sufficient pressure it starts flowing; the action of screwing the pipe in must cause it to flow into the gaps.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-06-22T10:57:44.153" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22400" PostTypeId="1" CreationDate="2018-06-22T13:46:14.677" Score="1" ViewCount="264" Body="&lt;p&gt;I am simulating a spherical grooved bearing….e.g. A bearing that has spiral grooves on the ball. Please see the following figures.&#xA;There is a kind of lubricant between ball and cup, so as to prevent metal-to-metal contact (Full film lubrication theory). The dimension of groove and gap between ball and cup are in (mm) and (micron). For example, Value of groove depth is equal to 0.006 mm and also, areas of some surfaces are too small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r7Sot.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r7Sot.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UTI2o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UTI2o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now in Ansys meshig, when I tried to mesh, I got following error(Fig 3,4,5,6). I want to mesh be structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yQvLV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yQvLV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PXwYQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PXwYQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig5 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gVSPr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gVSPr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FvOjl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FvOjl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the following Fig, I increase number of cell in edges and change dropped option to kept option in mesh method and the error is only as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ku91Y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ku91Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Fig5 is explained about narrow surface. What do these errors happen? I am grateful that guide me in this case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="16485" LastActivityDate="2018-06-22T13:46:14.677" Title="How to solve mesh error for narrow surfaces in Ansys Meshing?" Tags="|mechanical-engineering|ansys|cfd|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22401" PostTypeId="2" ParentId="22373" CreationDate="2018-06-22T16:54:37.977" Score="0" Body="&lt;p&gt;In computers a lot of stuff is measured in bits.  A bit is binary base 2.     $2^{12}= 4096$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That unit measures +- 90 for a total of 180 for a resolution of about 1/23 of a degree.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastActivityDate="2018-06-22T16:54:37.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22402" PostTypeId="1" CreationDate="2018-06-22T18:11:13.240" Score="1" ViewCount="88" Body="&lt;p&gt;I used an Instron 5581 Tensile Testing machine to test the tension of several bamboo samples that were sanded into dogbone shapes. They kept breaking at the grips and I am not sure why?&lt;a href=&quot;https://i.stack.imgur.com/IayIz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IayIz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16533" LastEditorUserId="16533" LastEditDate="2018-06-22T19:02:42.363" LastActivityDate="2018-06-22T19:12:30.150" Title="My dogbone bamboo sample's grips keep breaking" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22403" PostTypeId="1" AcceptedAnswerId="22423" CreationDate="2018-06-22T18:32:40.187" Score="1" ViewCount="2904" Body="&lt;p&gt;I have a half cylinder shape, and another half cylinder sketch perpendicular to that first cylinder. I wish to create a &quot;half-dome&quot; between these two surfaces (picture what should look like almost a pill sliced in half, but hollow). Here is my current design:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mqUb2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mqUb2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to create the dome between the sketch pictured and the top face of the half-cylinder. I've tried using dome and loft, but am having trouble doing so. I tried using revolve, but I cannot seem to pick a good axis so that it does what I want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to using Solidworks, but I've used Inventor in a class before and a lot of the features seem fairly similar. Any help would be appreciated. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried adding a center axis to the cylinder and then revolving 180$^\circ$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Bidd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Bidd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly something isn't right; the error says the axis cannot intersect the drawing, but the shape of the revolve seems to be in the right spot. I created an axis on the back of the half-cylinder so it didn't intersect, and the revolve command called it a not acceptable axis.&lt;/p&gt;&#xA;" OwnerUserId="10715" LastEditorUserId="10715" LastEditDate="2018-06-22T21:55:12.930" LastActivityDate="2018-06-23T16:12:38.980" Title="Creating a &quot;half dome&quot; in solidworks" Tags="|cad|solidworks|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22404" PostTypeId="2" ParentId="22402" CreationDate="2018-06-22T19:12:30.150" Score="1" Body="&lt;p&gt;Have the necked section slightly shorter and make sure the radiusing is smooth...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, I would suggest that you should consider either natural defects (that may not be visible) causing failure or you are just hitting the tensile strength - did they break at around the same value?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many have you tested so far - if it is only those two then you need to do more to confirm the cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should do a test on a piece that does not have any reduction in csa but perhaps just a bit thinner to see as well.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-06-22T19:12:30.150" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22406" PostTypeId="2" ParentId="22377" CreationDate="2018-06-22T19:24:08.547" Score="3" Body="&lt;p&gt;If you have to use the Teflon for some reason(s), you can bond the Teflon part to the SS by first etching the Teflon with a special etchant designed for Teflon and then you can use a high temp epoxy adhesive to get a good bond (seal) between the two assemblies.MasterBond Inc. has a large selection of epoxies for all kinds of applications. I would sandblast the SS also to get a good bonding surface. I have used this method of etching and bonding Teflon myself, the etchant turns the Teflon brownish, so if there is a problem with looks , you can't do it. The epoxy bonds nicely to the Teflon, I was amazed the first time I tried it. That's why we love engineering. &lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2018-06-22T19:24:08.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22407" PostTypeId="2" ParentId="21895" CreationDate="2018-06-22T19:30:53.143" Score="0" Body="&lt;p&gt;Suppose I know the trajectory $x(t)$ and $\dot{x}(t)$ that I want the system to follow. If the B matrix is invertible, then the inverse system is simply $u=B^{-1}\left(\dot{x}-A x\right)$.&lt;/p&gt;&#xA;" OwnerUserId="508" LastActivityDate="2018-06-22T19:30:53.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22408" PostTypeId="2" ParentId="22403" CreationDate="2018-06-22T23:08:53.967" Score="1" Body="&lt;p&gt;The basic problem is that it should be a half arc; then it should (and in my case did) work properly with the revolve tool and the center axis.&lt;/p&gt;&#xA;" OwnerUserId="10715" LastActivityDate="2018-06-22T23:08:53.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22409" PostTypeId="1" CreationDate="2018-06-23T00:39:17.670" Score="2" ViewCount="467" Body="&lt;p&gt;I am a novice who is trying to make some gears. I have followed the formula's but sadly something seems to go wrong and the gear's don't come out right. Below is some dimensions and pictures.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Gear 1&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Module(Diametral Pitch)(P) = 3,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Teeth(N)                   = 12,&lt;/p&gt;&#xA;&lt;p&gt;Pitch Diameter(PD)         = 36,&lt;/p&gt;&#xA;&lt;p&gt;Addendum(A) = MODULE       = 3,&lt;/p&gt;&#xA;&lt;p&gt;Deddendum(B)               = 32.529,&lt;/p&gt;&#xA;&lt;p&gt;Outside Diameter(OD)       = 39,&lt;/p&gt;&#xA;&lt;p&gt;Root Diameter(RD)          = 28.5,&lt;/p&gt;&#xA;&lt;p&gt;Base Circle(BC)            = 33.82884,&lt;/p&gt;&#xA;&lt;p&gt;Circular Pitch (CP)        = 9.4248,&lt;/p&gt;&#xA;&lt;p&gt;This produces a gear that looks &amp;quot;right&amp;quot; (small gear on right)&#xA;here is a link to my CAD file&#xA;&lt;a href=&quot;https://cad.onshape.com/documents/e67e4eedf7e94a76c439bef0/w/4a5cb0732a0c5ca585fb5d87/e/2a98076fc77593239bb0eaa7&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cad.onshape.com/documents/e67e4eedf7e94a76c439bef0/w/4a5cb0732a0c5ca585fb5d87/e/2a98076fc77593239bb0eaa7&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Now when i apply the same formula's to gear 2 i get a issue.&lt;/p&gt;&#xA;&lt;p&gt;Gear 2&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Module(Diametral Pitch)(P) = 3,&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Teeth(N)                   = 40,&lt;/p&gt;&#xA;&lt;p&gt;Pitch Diameter(PD)         = 120,&lt;/p&gt;&#xA;&lt;p&gt;Addendum(A) = MODULE       = 3,&lt;/p&gt;&#xA;&lt;p&gt;Deddendum(B)               = 116.529,&lt;/p&gt;&#xA;&lt;p&gt;Outside Diameter(OD)       = 123,&lt;/p&gt;&#xA;&lt;p&gt;Root Diameter(RD)          = 112.5&lt;/p&gt;&#xA;&lt;p&gt;Base Circle(BC)            = 112.7628,&lt;/p&gt;&#xA;&lt;p&gt;Circular Pitch (CP)        = 9.4248,&lt;/p&gt;&#xA;&lt;p&gt;Pressure Angle             = 20 (for both gears)&lt;/p&gt;&#xA;&lt;p&gt;Here is link to the CAD file&#xA;&lt;a href=&quot;https://cad.onshape.com/documents/e67e4eedf7e94a76c439bef0/w/4a5cb0732a0c5ca585fb5d87/e/2a98076fc77593239bb0eaa7&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cad.onshape.com/documents/e67e4eedf7e94a76c439bef0/w/4a5cb0732a0c5ca585fb5d87/e/2a98076fc77593239bb0eaa7&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The base circle and root diameter are practically the same ?!&lt;/p&gt;&#xA;&lt;p&gt;The gear teeth look different but i have used the same formula's for both gears.&lt;/p&gt;&#xA;&lt;p&gt;Here are the formula's i have used, just in case there is a massive error!!!!&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Module(Diametral Pitch)(P) = 3 &lt;strong&gt;(Saw this number used so stuck with it!)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Teeth(N)                   = 40, &lt;strong&gt;(Wanted to try a bigger number)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Pitch Diameter(PD)         = 120, &lt;strong&gt;(Teeth x Module) (40 X 3)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Addendum(A) = MODULE       = 3,  &lt;strong&gt;(I hope this is true !)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Deddendum(B)               = 116.529, &lt;strong&gt;(1.157 x P(Module) = 1.157 x 3 = 3.471 =120 - 3.471 )&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Outside Diameter(OD)       = 123, &lt;strong&gt;(Pitch diameter + Addendum)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Root Diameter(RD)          = 112.5 &lt;strong&gt;(Pitch Diameter - (2.5 X Module) = 120 - 7.5)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Base Circle(BC)            = 112.7628, &lt;strong&gt;(Pitch Diameter x COS (20pressure angle)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Circular Pitch (CP)        = 9.4248, &lt;strong&gt;(PI x Module)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Here are some pictures&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0ymmY.png&quot; alt=&quot;gear 1 and 2 close up meshing &quot; /&gt;&lt;img src=&quot;https://i.stack.imgur.com/oWk2p.png&quot; alt=&quot;same picture but further out &quot; /&gt;&lt;img src=&quot;https://i.stack.imgur.com/mFzja.png&quot; alt=&quot;view of both gears &quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;The diameters you can see on the pictures are in order from the top of the tooth to the centre:&lt;/p&gt;&#xA;&lt;p&gt;Outside diameter,&#xA;Pitch circle(pitch diameter),&#xA;Base circle,&#xA;Root diameter,&lt;/p&gt;&#xA;&lt;p&gt;The images show the bigger gear with its base circle and root diameter practically on the same diameter. The smaller gear seems more in proportion&lt;/p&gt;&#xA;&lt;p&gt;I set the Module(diametral pitch), which in this case is 3. I only used this number because I saw it used in a example Involving 12 and 24 teeth gears.&lt;/p&gt;&#xA;&lt;p&gt;Does anyone think I should have change it as one gear is quite big (gear 1 is 12 teeth and gear 2 is 40 teeth) or is it just personally preference?&lt;/p&gt;&#xA;&lt;p&gt;Did I use the module of 3 incorrectly for these gears?&lt;/p&gt;&#xA;&lt;p&gt;I sadly forgot to mention in the original question the centre distance between the gears, which is 78mm. I am not sure if you would set the module (diametral pitch) based on the centre distance.&lt;/p&gt;&#xA;" OwnerUserId="16538" LastEditorUserId="-1" LastEditorDisplayName="user16" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2024-03-25T00:01:30.267" Title="Trouble with Base circle and meshing with spur gears" Tags="|mechanical-engineering|gears|meshing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22410" PostTypeId="1" CreationDate="2018-06-23T03:44:41.747" Score="1" ViewCount="13" Body="&lt;p&gt;Rock-solid guide specifications are the ultimate pipe-dream of consulting engineers because they minimize the amount of time that needs to be repeatedly spent on the minutiae of tiny details from project to project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In specifications for items that are commercially available, I often see specific models listed in the specifications.  On one hand, this is great because if an alternate item is proposed by a contractor during the submittal process, the similarities can easily be compared.  Putting model numbers in a specification, however runs the risk that by the time someone uses the specification again, that model might have been discontinued.  Plumbing fixtures are the absolute worst for this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every single manufacturer seems to have products that run the gamut from low end to high end products, and they ALL seem to change model numbers constantly.  Is there a brand of plumbing fixtures (for any or all fixture types) that maintains consistent models for long periods of time or is there a better way to avoid spending time constantly updating specs without contractors trying to sneak the cheapest builder-grade fixtures through submittals?&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2018-06-23T03:44:41.747" Title="How to specify plumbing fixtures so models are not immediately out of date" Tags="|construction-management|specifications|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22411" PostTypeId="1" CreationDate="2018-06-23T06:06:17.200" Score="-1" ViewCount="10208" Body="&lt;p&gt;I want to find out how much torque will be needed from my motor if I want to pull a mass. So imagine a car towing a trailer.How would I calculate the torque need to move the trailer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the friction acting on the wheels and then did sum of the forces in the x direction to find the tension on the hitch. &lt;/p&gt;&#xA;" OwnerUserId="16542" LastActivityDate="2020-10-26T17:51:48.813" Title="How would I calculate the torque need to move the trailer?" Tags="|torque|dynamics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22413" PostTypeId="5" CreationDate="2018-06-23T09:04:45.677" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-06-23T09:04:45.677" LastActivityDate="2018-06-23T09:04:45.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22414" PostTypeId="4" CreationDate="2018-06-23T09:04:45.677" Score="0" Body="Questions related to energy produced by renewable systems such as solar, wind, water and geothermal." OwnerUserId="2157" LastEditorUserId="2157" LastEditDate="2018-06-24T10:47:05.500" LastActivityDate="2018-06-24T10:47:05.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22415" PostTypeId="2" ParentId="22409" CreationDate="2018-06-23T09:50:04.037" Score="0" Body="&lt;p&gt;Firstly it is correct that the gears are &lt;strong&gt;not&lt;/strong&gt; in proportion. For a pair of meshing gears the teeth on both need to be the same size ie you don't just scale a standard gear by some factor to get different diameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as you have noticed, on a larger gear the pitch circle diameter and root diameter will be proportionally closer to each other than on a smaller gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what the Module number is for, its a ratio between the diameter and number of teeth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also looks like there is a problem resolving the tooth profiles. It looks like this may be an issue with the Addendum and Dedendum values you have used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are, conventionally the difference between the pitch circle diameter and the tip and root respectively ie addendum plus dedendum = the height of the tooth above the root circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The A and D values are what determines the overall shape the individual teeth &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this may have confused your software and generated weird tooth profiles. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-06-23T09:50:04.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22416" PostTypeId="1" CreationDate="2018-06-23T12:12:35.183" Score="0" ViewCount="58" Body="&lt;p&gt;I learned in high school science that everything has a hardness index, that metal has a relatively low hardness compared to many kinds of stones, and that something cannot be scratched by a substance that is less hard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then why does the disc on an angle grinder wear down? Why can't it be made of a substance that can stand up to metal?&lt;/p&gt;&#xA;" OwnerUserId="7205" LastActivityDate="2022-12-23T05:29:03.947" Title="Why are angle grinder discs consumable?" Tags="|tools|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22417" PostTypeId="2" ParentId="22416" CreationDate="2018-06-23T12:36:23.540" Score="1" Body="&lt;p&gt;Scratching is hardly the only thing that destroys objects. For example: Glass is harder than cast iron, a cast iron anvil thus can not scratch a glass pane. However if you drop the anvil on the pane it is very likely to shatter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the individual particles on the grinding wheel are being hit repeatedly by a impulse force. This causes the ginding grains to fracture. More generally it causes material fatigue after enough cycles. Heat can also cause problems to the surface and certainly makes fatigue worse.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2018-06-23T12:36:23.540" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22419" PostTypeId="1" CreationDate="2018-06-23T07:42:41.910" Score="2" ViewCount="91" Body="&lt;p&gt;Resist means without catastrophic corrosion here.&#xA;We assume that there is no coating or anodic protection. &#xA;The steel can be mirror polished and passivated.&#xA;To make it more difficult, the shape is not a perfect sphere but a mechanism with some very bad water stagnation.&#xA;Since all seas are different let's say that the part is a few miles away from San Francisco seashore. &#xA;We are 50 meters underwater. There is no source of heat in the mechanism. &lt;/p&gt;&#xA;" OwnerUserId="36406" OwnerDisplayName="Thibd" LastEditorUserId="90" LastEditDate="2018-06-24T09:18:07.910" LastActivityDate="2018-06-24T09:18:07.910" Title="What type of stainless steels can resist 6 months under the sea?" Tags="|materials|corrosion|waterproofing|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22420" PostTypeId="2" ParentId="22419" CreationDate="2018-06-23T08:17:05.343" Score="4" Body="&lt;p&gt;For this type of use and to make it simple we will calculate the Pitting resistance equivalent number. This is based upon the material composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PREN = 1 x %Cr + 3.3 ( %Mo + 0.5 x %W ) + 16 x %N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, I would not test any steel that have a PREN inferior to 40.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we are left with Super Duplex, Hyper Duplex and Super Austenitic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I would only choose steels made with very high end manufacturing processes in order to avoid any pollution. The heat treatment is also very important. Electropolishing can help to clean the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical Super Austenitic include 904L, 254 SMO, AL-6XN, F44...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prefer Duplex over Austenitic because it has better mechanical properties as long as you don't encounter very high temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical Super Duplex include F53, F55... These could be very interesting to test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Super Duplex does not corrode easily. It usually happens at high temperatures or pressures. But let's say that there are very very bad conditions in this mechanism... We would be left with Hyper Duplex like 2707 or 3207. The good point of these steels is that these have the best mechanical characteristics available for this type of use.&lt;/p&gt;&#xA;" OwnerUserId="36406" OwnerDisplayName="Thibd" LastEditorUserId="10902" LastEditDate="2018-06-23T15:48:25.450" LastActivityDate="2018-06-23T15:48:25.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22421" PostTypeId="2" ParentId="22416" CreationDate="2018-06-23T14:52:02.910" Score="2" Body="&lt;p&gt;as a traditional grinding wheel is used, the abrasive particles get shed in the process of removing material. this exposes fresh and sharp abrasive particles so the wheel cuts like new throughout its life. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(an exception: when grinding gummy materials like aluminum, the wheel will fill with galled-up metal which has to be removed by the user with a wheel-dressing tool. this also removes some of the wheel itself, but in the end, fresh cutting surfaces are exposed.) &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-23T14:52:02.910" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22422" PostTypeId="2" ParentId="22411" CreationDate="2018-06-23T15:40:48.947" Score="0" Body="&lt;p&gt;Most of car makers talk about torque at engine flywheel. And most have for the first gear approximately 1 to 3.6 gear ratio and then 2 , 1.3, 1, etc. You find these in car specs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets assume you have 16 inch diameter tires, meaning 8 inch radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have 12/8 = 1.5 mechanical advantage at wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You divide F towing demand by 1.5 to gat torque at wheels, then you divide that by gear ratio, say 3.6 to get engine torque required to tow your trailer.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-06-23T15:40:48.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22423" PostTypeId="2" ParentId="22403" CreationDate="2018-06-23T16:12:38.980" Score="1" Body="&lt;p&gt;Whenever you are sketching a profile to be revolved in any CAD package, you should start by drawing your axis of revolution into that sketch as a construction line. Then, keep &lt;em&gt;every&lt;/em&gt; other line to one side of that axis. In SolidWorks, this gives you the added benefit of being able to dimension diameters, rather than radii, which may be more appropriate. I've shown a 'best practice' sketch below, showing how you can generate the object you're looking for, with just four dimensions, and one revolve feature in the tree. Simply revolve this profile by 180°&#xA;&lt;a href=&quot;https://i.stack.imgur.com/X4L55.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X4L55.png&quot; alt=&quot;Best Practice Revolve Sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, there are of course several other ways to generate the 'half dome', which you are after. I'll go through one of these below. Please note that this is not recommended in this situation but I've included it to give an idea of how to use the tools involved more generally, and to show how many options are present in all situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original question included a question on how to define the axis. If you wanted to revolve your dome in the other direction, then you need to create a 'master' sketch to not only define the axis, but also the plane on which the sketch should be created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by extruding the half-cylinder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5Zjy0m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Zjy0m.png&quot; alt=&quot;Half Cylinder Extrude&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then on the Right Plane, create a master sketch, which defines the shape of the dome that you would like. This should only contain one driving dimension, You should also add driven dimensions that show the radius of the dome, and the angle of the revolve to come&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OK9m2m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OK9m2m.png&quot; alt=&quot;Master Sketch Demo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now define a new plane for your sketch. It needs to intersect the centerpoint of the arc in your master sketch. I used the line from that sketch (coincident) and the Right Plane (perpendicular) as my references&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lBY8fm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lBY8fm.png&quot; alt=&quot;Plane Definition&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, start your sketch for the dome on the new plane. All the dimensions in this sketch should be equation driven, since you've already defined all the geometry elsewhere. Again, &lt;strong&gt;remember to put a centreline into your sketch and keep everything else on one side of it&lt;/strong&gt;. The diameter of the circle should be twice the radius you measured in the master sketch, and the thickness should be equal to that set in your original extrude sketch. This means that if you ever wish to modify the height of the dome, or the diameter of the tube etc., this sketch will automatically rebuild to the correct dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VGjUFm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VGjUFm.png&quot; alt=&quot;Equation Driven Sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you are ready to revolve your dome. Use the angle dimension from your master sketch to define the revolve angle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/scSOKm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/scSOKm.png&quot; alt=&quot;Equation controlled revolve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will now be left with a shape which is correct on the inside, but has some undesired overhangs on the outside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4sEAsm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4sEAsm.png&quot; alt=&quot;Dome Overhangs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you could simply run a cut-extrude around the outside, but another way to achieve this (in the spirit of explaining more complex tools for future use) is the 'delete face' tool. Simply delete the face underneath the overhang, with &quot;delete and patch&quot; selected, and SolidWorks will extend the Cylinder face up to meet the dome. You will have to do this once for each overhang. Deleting both at once will not work due to the way both overhang faces meet at a single vertex causing the maths to fall over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aTgOZm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aTgOZm.png&quot; alt=&quot;Finished Item&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part has now been completed, and is dimensionally identical to what was achieved with a single revolve previously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental concept of master modelling (defining geometry in a sketch that isn't actually used to build the model, and referring to that using equations in order to allow automatic updating of the built geometry) is extremely powerful, however, and in other more complicated examples, it is absolutely the best way to work.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-06-23T16:12:38.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22424" PostTypeId="2" ParentId="22419" CreationDate="2018-06-23T16:14:51.733" Score="0" Body="&lt;p&gt;You could chrome plate the steel and it would work fine. Steel rims that are plated with chrome last for a very long time on cars on roads that are heavily covered in salt during the winter.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-06-23T16:14:51.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22425" PostTypeId="1" CreationDate="2018-06-23T18:00:33.620" Score="0" ViewCount="3034" Body="&lt;p&gt;I saw a &lt;a href=&quot;https://m.youtube.com/watch?v=XOQOW2Gqs7Q#&quot; rel=&quot;nofollow noreferrer&quot;&gt;youtube video&lt;/a&gt; of an ultralight airplane with a 5.5hp engine that has been modified to produce 30hp. So how do I modify a 5.5hp engine to make 15hp+ of output?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GDRKH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="15982" LastEditorUserId="2015" LastEditDate="2018-06-25T22:47:21.353" LastActivityDate="2018-06-25T22:47:21.353" Title="What to do to increase the hp of a gx-160 engine?" Tags="|mechanical-engineering|torque|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22426" PostTypeId="2" ParentId="22419" CreationDate="2018-06-23T18:20:13.317" Score="2" Body="&lt;p&gt;I would say Thibd's answer excellent. However, one should look at the cost (including fabrication) on Monel, Incoloy 825, any Inconel or Hastelloy or Titanium . Although not &quot;steel&quot; they are metals that would answer the intent of the question. Also in &quot;deep&quot; water, corrosion of even carbon steel is not a problem because of low oxygen; but I think testing would be necessary to determine if the the subject location falls into that situation. And, although not exotic, attaching an aluminum , magnesium or zinc anode to a carbon steel object would certainly work and be the cheapest option. One would just need to check literature for the size and type of anode.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-06-23T18:28:02.503" LastActivityDate="2018-06-23T18:28:02.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22427" PostTypeId="1" CreationDate="2018-06-23T19:49:15.027" Score="1" ViewCount="3344" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jdCe7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jdCe7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a picture of a scuffed liner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What causes the presence of vertical scratches?&lt;/p&gt;&#xA;" OwnerUserId="5825" LastActivityDate="2018-06-24T07:34:44.253" Title="Why scuffed liners or piston rings always have vertical lines?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22428" PostTypeId="2" ParentId="22357" CreationDate="2018-06-24T03:01:24.470" Score="1" Body="&lt;p&gt;MQ-3 Alcohol Sensor is base the amount ethanol molecules that is present near the electrodes. Therefore minimum and maximum distance had no or little bearing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a little insight in to how the MQ3 sensor works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensing element as shown in the picture below is a cuboid structure made out of an Alumina (Yellow) tube covered with Tin Dioxide (Light Brown). The electrode which is made out of Aurum is shown in Black. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mqU1l.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mqU1l.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pin out and a drive circuit for the sensor are shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uxhH9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uxhH9.jpg&quot; alt=&quot;MQ3 Sensor pin out&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mXApU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mXApU.jpg&quot; alt=&quot;MQ3 Circuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above circuit activate the heating element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JbRbb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JbRbb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result electrons begin to move because Tin Oxide ceramics has become a semi-conductor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ixRLV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ixRLV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when air with alcohol elements enter the cuboid structure the circuit generate additional current because the ethanol burns into Acetic acid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hmvii.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hmvii.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the minimum or maximum distance has no bearing, but the presence of alcohol will have significant effect on the sensor. For more detail information review the references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://sensorworkshop.blogspot.com/2008/04/sensor-report-mq3-gas-sensor.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sensor Report - MQ3 Gas sensor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.elprocus.com/mq-135-alcohol-sensor-circuit-and-working/&quot; rel=&quot;nofollow noreferrer&quot;&gt;MQ135 Alcohol Sensor Circuit And Working&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://cdn.sparkfun.com/datasheets/Sensors/Biometric/MQ-3%20ver1.3%20-%20Manual.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alcohol Gas Sensor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2018-06-24T03:01:24.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22429" PostTypeId="1" CreationDate="2018-06-24T03:48:17.417" Score="3" ViewCount="109" Body="&lt;p&gt;Most of what I was able to find pointed towards thermoelectric generators which are very inefficient, stirling engines which have a low torque output at low temperatures, or generators that use the organic rankine cycle. What are other methods of getting power from a low temperature heat source? And are any of them currently viable or potentially viable in the near future?&lt;/p&gt;&#xA;" OwnerUserId="16548" LastActivityDate="2018-06-26T23:30:57.060" Title="What methods are used to extract energy from low energy heat sources?" Tags="|electrical-engineering|thermodynamics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22430" PostTypeId="2" ParentId="22425" CreationDate="2018-06-24T04:06:06.473" Score="4" Body="&lt;p&gt;to triple the output of a 5.5HP engine, you must either triple the displacement or triple the rotating speed. there is no way to triple the displacement by boring out the block and fitting a bigger piston and/or by installing a crank with a longer throw to increase the stroke n that engine. Now, tripling the rotating speed requires a long list of modifications to achieve which do not make economic sense for an engine this small, compared to simply buying a different engine with triple the cylinder displacement. Here is why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engine you picture has a &quot;comfortable&quot; design speed of about 3600RPM. It has valves operated by pushrods. Inertia in the valve-actuating mechanism limits its rotating speed to no more than about 2X 3600RPM or 7200RPM. To go faster than that requires re-engineering the engine for an overhead cam system (with reduced inertia) for a start, but there are other issues that must be dealt with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the induction system which supplies fuel and air mixture to the cylinder is proportioned to support a flow rate which matches the design speed of the engine. Tripling the design speed requires re-engineering this with a much-shortened manifold tube and a carburetor which can supply 3X the flow rate so the engine does not choke out at the new, higher RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;3X&quot; engine will be generating heat at 3X the rate of the original, which must be gotten rid of. Since the fan that cools this engine runs at the same speed as the crankshaft, one would think that heat rejection is not a problem, but the finning pattern in the engine must be modified so as to pull heat out of the head (where the valves and valve seats reside) much faster than that required to support a 3600RPM engine, to prevent the valves and seats from overheating and getting burnt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar comments apply to the lubrication system in the engine as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we come to final drive and reliability issues. Tripling the engine speed requires a redesign of the reduction system in the final drive, to maintain the prop tips subsonic. It also requires a prop redesign, to absorb 3X the power of the original engine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your &quot;3X&quot; engine will wear itself out in 1/3rd the time as the original design since its parts are moving 3X faster- but even if that were acceptable, what won't be is the overall reduction of reliability which accompanies an engine that fundamentally runs hot. So while racing motorcycles can use engines that run at 16,000 RPM, it's OK for them to blow up after one season's worth of races since they get torn down and completely rebuilt after ~tens of hours at 16,000RPM. For high-reliability applications like aircraft, you want a two-thousand-hour engine to statistically ensure an acceptably low failure rate in 1,000 hours of operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is this: it is &lt;em&gt;possible&lt;/em&gt; to re-engineer a 5.5HP lawnmower engine to run at 10,800RPM and it is &lt;em&gt;possible&lt;/em&gt; to bore out the cylinder to get 1.2X the displacement and combine that with 1.2X the stroke with a new crank and it is &lt;em&gt;possible&lt;/em&gt; to put a carb on that engine pulled from a 15HP engine, but it generally makes little sense. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-24T04:06:06.473" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22432" PostTypeId="2" ParentId="22429" CreationDate="2018-06-24T06:05:06.947" Score="1" Body="&lt;p&gt;All the approaches to designing a heat engine that runs off a relatively low-temperature heat source suffer from severe limits on their thermodynamic efficiency which cannot be circumvented by clever design. For any of them to become viable requires that energy costs rise to much higher levels than they are at present.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-24T06:05:06.947" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22434" PostTypeId="2" ParentId="22427" CreationDate="2018-06-24T07:11:47.057" Score="1" Body="&lt;p&gt;When piston rings are fitted, let us assume 3 rings (2 compression &amp;amp; 1 scraper) they are fitted with the gaps with a 120 degree offset to avoid blow-by (where combustion gasses leak past the rings - an issue that can happen when the gaps are all in line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the temperature of the piston increases due to the combustion temperature, the gaps close up due to the expansion of the ring, The rings may rotate, but that rotation will be very slow - however the linear movement of the piston up and down will be much greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scoring can be due to combustion particles getting trapped between the ring and the bore surface or the skirt of the piston coming into contact with the bore surface. Once this has started then it just gets worse over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gudgeon pin (aka &quot;wrist-pin&quot;) in the piston is located fractionally off-centre to make sure that the piston is always tilted as it moves so that it does not &quot;wobble&quot; during the stroke, this wobbling can cause a noise called &quot;piston-slap&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Another source of particles that can start scoring is the oil itself - if it is not changed regularly or other contaminants have entered the engine / sump...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-06-24T07:34:44.253" LastActivityDate="2018-06-24T07:34:44.253" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22437" PostTypeId="2" ParentId="9" CreationDate="2018-06-24T10:59:19.160" Score="0" Body="&lt;p&gt;Check the coolant temperature (It should be below 28 degree Centigrade)   which used for cooling the heat where it is being generated during machining operation due to shearing force as well as reduce the tool diameter meter in order to reduce the the amount of cutting force  &lt;/p&gt;&#xA;" OwnerUserId="16552" LastActivityDate="2018-06-24T10:59:19.160" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22439" PostTypeId="2" ParentId="16537" CreationDate="2018-06-24T12:31:11.563" Score="1" Body="&lt;p&gt;Assuming that following &lt;a href=&quot;https://cdn.sparkfun.com/datasheets/Sensors/Biometric/MQ-3%20ver1.3%20-%20Manual.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;MQ3 Alcohol Gas Sensor&lt;/a&gt; is the sensor referenced in this question. Per the specification the &lt;strong&gt;&lt;em&gt;detection range&lt;/em&gt;&lt;/strong&gt; is &lt;strong&gt;&lt;em&gt;25-500 ppm&lt;/em&gt;&lt;/strong&gt; of Alcohol. The sensitivity is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{R_{o(in \ air)}}{R_{s(125ppm\ C_{2}H_{5}OH)}} \ \leq \ 5$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively take a look at &lt;a href=&quot;http://www.winsensor.com/products/MEMS/MEMSjjqtcgqGM_302B_661.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;MEMS Alcohol Gas Sensor GM-302B&lt;/a&gt; (note the link is in Chinese, so please use translate) which has a &lt;strong&gt;detection range&lt;/strong&gt; of &lt;strong&gt;1-500 ppm&lt;/strong&gt; of alcohol. Also the sensitivity is 1.2 to 50,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U0bTj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U0bTj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/22428/110&quot;&gt;MQ3 Alcohol sensor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://media.digikey.com/pdf/Data%20Sheets/Austriamicrosystems%20PDFs/CCS803.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ultra-Low Power Gas Sensor for Ethanol Detection&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-06-24T12:39:44.710" LastActivityDate="2018-06-24T12:39:44.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22440" PostTypeId="1" AcceptedAnswerId="22460" CreationDate="2018-06-25T02:01:06.970" Score="2" ViewCount="140" Body="&lt;p&gt;Anyone who's spent much time in a machine shop will be familiar with the hardware commonly used for work-holding and fixtures on a machine tool. In particular, the &lt;em&gt;flange nut&lt;/em&gt; is a common item used with strap clamps and other work-holding devices. TeCo is a well-known manufacturer of these items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wgfON.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wgfON.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just happened to be shopping for some of these flange nuts for a fixture and noticed that &lt;em&gt;some&lt;/em&gt; of these are specified as being made from 12L14 (leaded steel). This detail is not buried in the data sheet, but appears where I would interpret it as a &lt;em&gt;feature&lt;/em&gt;, meaning a purchaser might have reasons for choosing one in this material, as opposed to one in a stronger steel for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why 12L14 might be a &lt;em&gt;preferred&lt;/em&gt; material in this application. I've understood leaded steels to be free-machining, which would make manufacturing easier (although making the part non-weldable), but I can't imagine that would be a driving factor in high-volume parts like this and in any case wouldn't rise to the level of a feature to be advertised if that were all it was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two theories:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The lead in 12L14 somehow increases the lubricity of the fastener, making it less likely to bind (or perhaps gall) in a frequently used fastener like one in a manufacturing fixture.&lt;/li&gt;&#xA;&lt;li&gt;Lower tensile strength (although not &lt;em&gt;that&lt;/em&gt; much lower) is a plus, because if over-tensioned, it's better for the fastener to give than the t-slots machined into an expensive machine table.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can anyone enlighten me?&lt;/p&gt;&#xA;" OwnerUserId="3937" LastActivityDate="2018-06-26T13:47:35.267" Title="Why would 12L14 material be a &quot;feature&quot; of a work-holding flange nut?" Tags="|mechanical-engineering|materials|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22441" PostTypeId="2" ParentId="22440" CreationDate="2018-06-25T02:18:31.237" Score="1" Body="&lt;p&gt;The lead improves machineability with some sacrifice in toughness and ductility. It does nothing to improve serviceability. The lead may be present as stringers which are undesirable , the mill tries to have it as isolated globs. The shape and size of the mill product determine the lead. Industrial applications do not permit free machining ( lead or sulfur) in fasteners. I think it is OK for consumer fasteners as they generally have lower stresses. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-06-25T02:18:31.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22442" PostTypeId="1" CreationDate="2018-06-25T02:51:32.923" Score="2" ViewCount="68" Body="&lt;p&gt;I have designed an LED driver with SEPIC converter for 220 volts line. I hear  starting current for 3 or 4 cycle is very high in the branch of switch and second inductor(18 A). I want to reduce it to less than 1 amp. What is the way that I can apply to limit startup current of SEPIC?&lt;/p&gt;&#xA;" OwnerUserId="16558" LastEditorUserId="110" LastEditDate="2021-04-11T17:51:30.793" LastActivityDate="2023-12-27T22:04:15.513" Title="What is the current limiter for SEPIC LED driver?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22443" PostTypeId="1" CreationDate="2018-06-25T06:21:16.997" Score="-2" ViewCount="107" Body="&lt;p&gt;Does ground conduct heat or it acts like an insulator ? If yes then whats the conduction co-efficient or convection co-efficient ( in terms of convection heat transfer ) ?&lt;/p&gt;&#xA;" OwnerUserId="14271" LastActivityDate="2018-06-25T12:49:16.303" Title="Does ground conduct heat?" Tags="|heat-transfer|" AnswerCount="1" CommentCount="3" ClosedDate="2018-06-26T05:34:18.100" ContentLicense="CC BY-SA 4.0" />
  <row Id="22444" PostTypeId="1" AcceptedAnswerId="22445" CreationDate="2018-06-25T09:10:31.213" Score="1" ViewCount="94" Body="&lt;p&gt;The history of computers seems somewhat pervasive since there are so many steps in modern manufacturing and so many changes in trends with transistors over many years. Grasping how we got to our current level of processing power seems somewhat like a chicken/egg situation. How exactly were we able to create such precise processors when we only could have had more primitive tools to craft them with? For instance, take the UV stencil used on a silicon wafer to make a CPU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could engineers have designed a machine capable of etching out so many millions of transistors in the stencil given that they didn't originally have access to that level of processing that they are trying to achieve? Or in other words, how is it possible for computers to evolve in the process of making more precise machines from less precise machines? &lt;/p&gt;&#xA;" OwnerUserId="15677" LastEditorUserId="15677" LastEditDate="2018-06-25T09:56:07.343" LastActivityDate="2018-06-25T09:56:07.343" Title="How did scientists develop the stencils that made even more precise processors than they had previously?" Tags="|manufacturing-engineering|computer-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22445" PostTypeId="2" ParentId="22444" CreationDate="2018-06-25T09:22:14.660" Score="1" Body="&lt;p&gt;Incrementally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first computers were made in discrete parts. At first vacuum tubes, then transistors. Then the first integrated circuits that had only a couple dozen transistors and were based on manually projected stencils. The computers would help a bit, but most of the work was done with pen on paper. What is now done by a single tiny chip was occupying a large board with a hundred integrated circuits - each designed by hand. Never mind precision of the process wouldn't allow more elements in an IC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At certain point, computers were advanced enough to implement CAD software. This allowed moving the design from paper to circuitry, applying common cut&amp;amp;paste to stuff several chips into one, then plot a large, complex mask on paper to be shrunk by photo-techniques and made into more complex chips. As these entered use, increasingly less of the process was done &quot;by hand&quot; and increasingly more was automated. Specialized software that allowed use of a high-level programming language to &quot;write&quot; the IC architecture, that compiled to paths and transistors without human interaction, robotic assembly lines that could prepare increasingly more complex masks etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of it that way: you think the mask can't be made using anything less than it produces... are you sure? Can't it be made using something 5% less than it produces? And that one can't be made with something another 5% less complex?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ps. you refer to 'cutting'. That's not how these stencils were made. They were about universally made using photo-etching.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Make a (big) printout of a layer of the IC.&lt;/li&gt;&#xA;&lt;li&gt;Take a photo. Now your printout is the size of a microfiche.&lt;/li&gt;&#xA;&lt;li&gt;Spray silicone wafer with conductor, copper vapor or such.&lt;/li&gt;&#xA;&lt;li&gt;Spray that with photo-hardening emulsion.&lt;/li&gt;&#xA;&lt;li&gt;Shine UV light through the microfiche stencil with optics focusing it to the size of the IC. It hardens the emulsion where paths are to go.&lt;/li&gt;&#xA;&lt;li&gt;Apply acid bath, etching the conductor and emulsion away where they aren't needed. &lt;/li&gt;&#xA;&lt;li&gt;Wash the emulsion off. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;...and so on. Different substances, different emulsions, different masks, layer by layer the chip was built in an entirely analog process. The only digital part was design and plotting of the first stage on huge sheets of paper.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2018-06-25T09:29:36.957" LastActivityDate="2018-06-25T09:29:36.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22446" PostTypeId="2" ParentId="22443" CreationDate="2018-06-25T12:49:16.303" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://www.geothermal-energy.org/pdf/IGAstandard/WGC/2010/2952.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a paper on measuring soil thermal conductivity, with quite a few values.  As noted by Fred, the value depends on the soil composition and moisture content.  It varies over quite a range, with values in this paper ranging from 0.15 to 5.03 W/m K.  Depending on the type of soil, the actual conductivity may be outside this range.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2018-06-25T12:49:16.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22447" PostTypeId="1" CreationDate="2018-06-25T13:22:00.670" Score="2" ViewCount="37" Body="&lt;p&gt;I'd like to ask an environemental question which has to do with waste packaging and the EU regulations that enclosure this subject&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to package loads of different hazardous wastes in compliance with ADR. So, depending on the waste, I use the labels that are pictured in this link. &lt;a href=&quot;https://conceptdraw.com/a1276c4/preview&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://conceptdraw.com/a1276c4/preview&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I label the packages of some wastes that are not applicable to ADR? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These wastes are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;brake fluids (16 01 13*)&lt;/li&gt;&#xA;&lt;li&gt;antifreeze (16 01 14*) &lt;/li&gt;&#xA;&lt;li&gt;synthetic engine, gear and lubricating oils (13 02 06*)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A possible answer is to label them in compliance with the GHS (Globally Harmonized System of Classification and Labelling of Chemicals) which is the regulation for hazardous products but there is no a certain answer if this system is applied for hazardous wastes too.&lt;/p&gt;&#xA;" OwnerUserId="16563" LastEditorDisplayName="user16" LastEditDate="2018-06-26T12:04:40.180" LastActivityDate="2018-06-26T12:04:40.180" Title="Hazardous waste packaging" Tags="|environmental-engineering|waste-disposal|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22448" PostTypeId="1" AcceptedAnswerId="22450" CreationDate="2018-06-25T14:49:00.813" Score="0" ViewCount="63" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/GHBz0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm a hobbiest who loves doing and learning about different subjects. Often I run into difficult problems that I normally can find solutions to however this one stumped me, I have reviewed many physics journals on the problem but either they are two complex or tackle the much easier version of 'viscous' friction.(p 2/54)&#xA;Essentially when I even begin a solution I yield to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ma = sign(v) * mg*mu - kx&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would be simple if the friction didn't change direction. Does anyone have a suggestion on how to approach this problem? I'm not looking for someone to solve it I just want an idea on how to start it. Thank you&lt;/p&gt;&#xA;" OwnerUserId="7652" LastActivityDate="2018-06-25T22:55:30.123" Title="Damped harmonic motion with dry friction" Tags="|dynamics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22450" PostTypeId="2" ParentId="22448" CreationDate="2018-06-25T18:35:54.373" Score="1" Body="&lt;p&gt;The sign change problem can be avoided slightly - by substituting a different equation for sign(v).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ma = mg\mu*sign(v) - kx$$&#xA;$$ma = mg\mu\lim_{n\to\infty}tanh(nv)-kx$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring the limit for now, we have a real differential equation!  Since we don't depend on t explicitly, we'll make an immediate autonomous substitution.  $v(x) = v$.  Then, $a(t) = v*\frac{dv}{dx} = v*v'$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$vmv' = mg\mu\lim_{n\to\infty}{tanh(nv)}-{kx}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a first order equation, this is easier, but not by much.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-06-25T18:35:54.373" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22451" PostTypeId="2" ParentId="22371" CreationDate="2018-06-25T19:23:45.883" Score="1" Body="&lt;p&gt;There's really not enough information to be certain, but the images seem to match TR Fastenings' &quot;&lt;a href=&quot;https://www.trfastenings.com/Products/Catalogue/nuts/Hank-Self-Clinch-Nuts/Non-Locking-Mini-Squeezed&quot; rel=&quot;nofollow noreferrer&quot;&gt;Self Clinch Non-Locking Mini Squeezed Nut&lt;/a&gt;&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B. It also looks to have been improperly installed, as the 'spiked' diameter should be fully embedded in the bulk material.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-06-25T19:23:45.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22452" PostTypeId="1" CreationDate="2018-06-25T20:27:52.163" Score="4" ViewCount="423" Body="&lt;p&gt;Assume the following setup:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Input -&amp;gt; First    ----------&amp;gt; Second    ----------&amp;gt; etc.&#xA;Air      heatpump   hot air   heatpump    hot air&#xA;T_0        |        T_h1&amp;gt;T_0     |        T_h2&amp;gt;T_h1&#xA;           v                     v&#xA;        cold air              cold air&#xA;        T_c0&amp;lt;T_0              T_c1&amp;lt;T_h1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mZeLG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mZeLG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;T&lt;/code&gt;s are temperatures; the darker the red, the warmer. The barriers between the areas are isolating materials. The throughput is smaller for nested pumps so that the most inner, hottest area is quite small and heated up slowly in comparison to the out ones. The heat pumps would operate with different fluids suited for the temperature differences. The output temperature (after &lt;code&gt;etc.&lt;/code&gt;) is limited by the availability of the right fluid and the construction of a pump which can withstand the heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only question to decide whether such a machine could operate efficiently, i.e. provide more energy than the pumps consume, is how much do the heat pumps consume? I'm searching for figures to estimate the consumption of thermal heat pumps (let's assume they're electric).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure that it's not a perpetuum mobile of the second kind because the system is fed with the energy taken out of the cold air which leaves the first (left) heat pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figures for the energy consumption would show whether such a machine can be constructed and provide (heat) energy rather than consume more than it provides. Example use cases would be geo-/climate-engineering and production of electricity or heat for heavy engineering.&lt;/p&gt;&#xA;" OwnerUserId="9780" LastEditorUserId="90" LastEditDate="2022-08-03T00:36:51.533" LastActivityDate="2022-11-29T01:18:25.260" Title="Can a chain of nested thermal heat pumps be used to to gain thermal energy out of the environment?" Tags="|thermodynamics|heat-transfer|pumps|temperature|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="22454" PostTypeId="2" ParentId="22448" CreationDate="2018-06-25T22:55:30.123" Score="1" Body="&lt;h2&gt;Method 1&lt;/h2&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This would be simple if the friction didn't change direction&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Indeed.  So here is one way to deal with it.  First, determine the initial velocity at t=0.  E.g. let's just say v(0) is positive.  Then instead of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ma =sign(v) * mg*\mu - kx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ma =mg*\mu - kx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, solve that equation, which should be pretty easy, and plot velocity versus time.  You'll note that at some point, the velocity is no longer positive, and has changed to negative. Let's just call that time $t_1$.  i.e. $v(t_1)=0$.  So at that point, &quot;stop&quot; that solution, and write a new equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ma =-mg*\mu - kx$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this equation &quot;starts&quot; at time $t=t_1$ with initial conditions $v(t_1)=0$ and $x(t_1)$ determined from the ending position of the first solution.  This equation should also be easy to solve.  Continue the process, switching back and forth as necessary.  At each point, you will also need to check if the spring force is enough to overcome the static friction.  At some point it will not be, and then you stop.  At the end, stitch together all of the partial solutions to each segment into one big piecewise solution.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Method 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Use numerical integration. E.g. apply Euler's method for solving differential equations.  Over each time step, you can assume that the friction force acts in a constant direction.  Assuming the time steps are small enough, you'll get about the same answer as above.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-06-25T22:55:30.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22456" PostTypeId="2" ParentId="22442" CreationDate="2018-06-26T11:37:35.303" Score="0" Body="&lt;p&gt;Since the current stored in the second inductor is added to the current supplied by the first conductor when the switch &quot;S1&quot; is closed &quot;open circuit&quot;. Reducing the inductance &quot;amount of current stored in them&quot; of one of the inductors will result in the decrease of the total current supplied to the load.&lt;/p&gt;&#xA;" OwnerUserId="16505" LastActivityDate="2018-06-26T11:37:35.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22458" PostTypeId="2" ParentId="22391" CreationDate="2018-06-26T12:39:25.180" Score="0" Body="&lt;p&gt;You can get your perpendicular, rotary motion from a miter gear or worm gear drive. Generally, miter for 1:1 ratio, worm for a reduction or heavy load. If your going for an &quot;elbow&quot; go miter, for a &quot;T&quot; or &quot;cross&quot; go worm.&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-06-26T12:39:25.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22459" PostTypeId="2" ParentId="3176" CreationDate="2018-06-26T12:46:25.327" Score="0" Body="&lt;p&gt;Do the pipes need to be &quot;keyed&quot; such that they cant rotate with respect to each other? If not, you can maybe seal the inner pipe like a piston(inner pipe) in a cylinder( outer pipe)  shaft/T/U/oring/seal etc. Not sure how those seals work with gases as they do with liquids, but air is a gas and pneumatic cylinders work the same!&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-06-26T12:46:25.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22460" PostTypeId="2" ParentId="22440" CreationDate="2018-06-26T13:47:35.267" Score="1" Body="&lt;p&gt;It is likely made from 12L14 for machine-ability of the threaded/tapped hole in the nut.   Manufacturers can make them faster and probably have better tool life. Reduced cost. Simply that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those nuts should be case hardened. That protects from wear from friction and compression when spinning the nut down when clamping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m going to disagree with your theories. There’s no galling or anti-friction happening from the leaded steel. Lead primarily works as a lubricant during cutting. This works because of lead’s lower melting temperature and heat generated at the cutting contact point. Liquid Lead acts as a lubricant to the cut. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your machine table is likely some heat treated cast iron. It will likely take a few dings and dents from impacts, but your clamps will almost always yield before your table does. The t slots have much more stiffness and material than your t-slot clamping System. If your clamps damage the t-slot, something else is wrong. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2018-06-26T13:47:35.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22461" PostTypeId="1" CreationDate="2018-06-26T13:54:04.870" Score="1" ViewCount="141" Body="&lt;p&gt;For a nonlinear system,&#xA;$$&#xA;\begin{align}&#xA;&amp;amp;{\boldsymbol x}(k+1)={\boldsymbol f}({\boldsymbol x}(k),{\boldsymbol u}(k),{\boldsymbol w}(k)) \\&#xA;&amp;amp;{\boldsymbol y}(k)={\boldsymbol h}({\boldsymbol x}(k),{\boldsymbol v}(k))&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to find the Extended Kalman Filter (EKF), having a look at &lt;a href=&quot;https://en.wikipedia.org/wiki/Extended_Kalman_filter&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt;, I have reached&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;&amp;amp; \hat {\boldsymbol x}(k+1|k)={\boldsymbol f}(\hat {\boldsymbol x}(k|k),{\boldsymbol u}(k|k)) \\&#xA;&amp;amp; {\boldsymbol P}(k+1|k)={\boldsymbol F}(k){\boldsymbol P}(k|k){\boldsymbol F}^T(k)+{\boldsymbol L}(k) {\boldsymbol Q}(k) {\boldsymbol L}^T(k)&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;&amp;amp;{\boldsymbol S}(k+1)={\boldsymbol H}(k+1){\boldsymbol P}(k+1|k){\boldsymbol H}^T(k+1)+{\boldsymbol M}(k+1){\boldsymbol R}(k+1){\boldsymbol M}^T(k+1) \\&#xA;&amp;amp; {\boldsymbol K}(k+1)={\boldsymbol P}(k+1|k) {\boldsymbol H}^T (k+1){\boldsymbol S}^{-1}(k+1) \\&#xA;&amp;amp; \hat{\boldsymbol x}(k+1|k+1)=\hat{\boldsymbol x}(k+1|k)+{\boldsymbol K}(k+1)[y(k+1)-{\boldsymbol h}(\hat {\boldsymbol x}(k+1|k))] \\&#xA;&amp;amp; {\boldsymbol P}(k+1|k+1)k=[{\boldsymbol I}-{\boldsymbol K}(k+1){\boldsymbol H}(k+1)]{\boldsymbol P}(k+1|k)&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;&amp;amp; {\boldsymbol F}(k)=\frac{\partial \boldsymbol f}{\partial \boldsymbol x}|_{\hat {\boldsymbol x}(k|k),{\boldsymbol u}(k)} \\&#xA;&amp;amp; {\boldsymbol H}(k+1)=\frac{\partial \boldsymbol h}{\partial \boldsymbol x}|_{\hat {\boldsymbol x}(k+1|k)}\\&#xA;&amp;amp;{\boldsymbol L}(k)=\frac{\partial\boldsymbol f}{\partial\boldsymbol w}|_{\hat {\boldsymbol x}(k|k),{\boldsymbol u}(k)}\\&#xA;&amp;amp;{\boldsymbol M}(k+1)=\frac{\partial\boldsymbol h}{\partial\boldsymbol v}|_{\hat {\boldsymbol x}(k+1|{\color{red}{k+1}})}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this obtained formulation correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main concerns are about the whole timing indexes especially the last one.&lt;/p&gt;&#xA;" OwnerUserId="16032" LastEditorUserId="6790" LastEditDate="2020-05-04T08:02:22.320" LastActivityDate="2024-03-25T03:09:41.133" Title="Extended Kalman Filter formulation" Tags="|control-engineering|control-theory|optimal-control|kalman-filters|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22463" PostTypeId="1" CreationDate="2018-06-26T14:15:08.293" Score="2" ViewCount="102" Body="&lt;p&gt;I have an interlocking set of parts (&lt;a href=&quot;https://ca.vwr.com/store/product/en/4548753/evaporating-flasks-heavy-wall-chemglass&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; and &lt;a href=&quot;https://ca.vwr.com/store/product/en/80066-014/adapter-universal-inlet-with-hose-connection&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;) which are used under vacuum. The last time I used it, the interface was wet with a mixture of water and ethanol. When the vacuum was removed, the two parts stayed firmly fused and no amount of force that I can apply can pull them apart. I suspect I will break the glass neck before I break the connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone suggest some tricks I could try to use to unstick them? I have tried heating the interface, which had no effect, and I have tried lubricating it, but the connection is too tight to allow the lubricant to penetrate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a crudely drawn picture of the fused pieces:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yxhTg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yxhTg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16575" LastEditorUserId="90" LastEditDate="2022-06-20T09:27:36.673" LastActivityDate="2022-06-20T09:27:36.673" Title="Unsticking two glass elements fused by vacuum" Tags="|vacuum|glass|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22464" PostTypeId="2" ParentId="22463" CreationDate="2018-06-26T14:42:06.163" Score="1" Body="&lt;p&gt;Vacuum bagging is used in the creation of various items, including and especially amateur built aircraft. A suitably strong surface holds the item being bagged. Plastic film is placed over the item and taped to the surface, along with a port for the vacuum source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, a layer of dacron material is placed over the fiberglass, as it does not adhere to epoxy resin, but also serves as a porous space for air to be removed. You won't be applying resin, but you would want the spacer. In your case, the material would be useful below the item as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many resources when using &quot;vacuum bagging composites&quot; as a search term. One source I've used in the past is &lt;a href=&quot;https://www.fibreglast.com/product/vacuum-bagging-equipment-and-techniques-for-room-temp-applications/Learning_Center&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;fibreglast.com&quot;&gt;fiberglast.com&lt;/a&gt; for supplies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following image is from their site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RmKuy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RmKuy.jpg&quot; alt=&quot;bagging image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But most of it would not apply to your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the tricky part would be moving it after the vacuum is in force. Two opposing strings? It would slip under the sealing tape and yet be more easily applied than trying to slide the two parts with hands through the bagging film. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, bunch up the film where you need movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the film tears, you would not be breaking glass!&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-06-26T14:42:06.163" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22465" PostTypeId="2" ParentId="22463" CreationDate="2018-06-26T17:18:31.790" Score="4" Body="&lt;p&gt;Perhaps, depending on the materials, you could try gently heating one piece while cooling (or at least heat-sinking) the other piece, and hoping that the expansion mismatch will separate them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You haven't identified the glass types nor their surface quality-- if the interface areas are porous, your contaminants may well have formed physical bonds which are stronger than the glass.  In which case, you're stuck (pun intended)&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-06-26T17:18:31.790" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22466" PostTypeId="1" CreationDate="2018-06-26T18:05:18.977" Score="1" ViewCount="59" Body="&lt;p&gt;I have been doing research into artificial Intelligence applications for protection relays. I have read a lot about theory and the possibility of them being used, but have failed to find evidence of it ever being implemented. Can anyone point me to some information on actual implementations and statistics on its usage.&lt;/p&gt;&#xA;" OwnerUserId="16580" LastActivityDate="2018-06-26T18:05:18.977" Title="AI in protection relays" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="1" ClosedDate="2018-07-02T06:39:46.040" ContentLicense="CC BY-SA 4.0" />
  <row Id="22469" PostTypeId="1" AcceptedAnswerId="22471" CreationDate="2018-06-26T19:17:15.697" Score="2" ViewCount="1448" Body="&lt;p&gt;I'm hoping this is the correct StackExchange for this question/request for advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking to buy a welder to introduce me a little to welding, my main use-case right now is the hobby projects I do with chainmail and soldering is just not good enough in my book. I'm working with stainless steel jump rings with a diameter ranging from 0.8mm to around 2mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen on &lt;a href=&quot;http://theringlord.com/cart/shopdisplayproducts.asp?page=1&amp;amp;id=164&quot; rel=&quot;nofollow noreferrer&quot;&gt;TheRingLord&lt;/a&gt; and &lt;a href=&quot;https://www.ganoksin.com/article/pulse-arc-welder-closing-chains/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ganoksin&lt;/a&gt; that this can be done with an arc welder. As far as I've seen arc welders have the lowest prices and to me also seem like the least maintenance heavy since you don't need to stock up on gasses. So arc welders have my focus for now. My budged is around €200,- tops preferably from a dutch reseller, and I have no clue what to look for in a welder (my only clue so far is probably something that can be activated for short amounts). Would anyone be able to give me some advice on what would be a good welder and if I'm even looking in the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;please do note that I'm not just looking for something fast and easy, I really don't mind getting my hands dirty learning how it works and getting some terrible welds in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some sites that are available for me (you might not be able to read the descriptions, but the welder device-names should still be the same):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.beslist.nl/products/klussen/klussen_486173_3621539/?sortby=price_min&quot; rel=&quot;nofollow noreferrer&quot;&gt;CoolBlue&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.allesvoorlassen.nl/lasapparatuur/elektrode-lassen&quot; rel=&quot;nofollow noreferrer&quot;&gt;AllesVoorLassen.nl&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.beslist.nl/products/klussen/klussen_486173_3621539/?sortby=price_min&quot; rel=&quot;nofollow noreferrer&quot;&gt;Beslist.nl&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16583" LastEditorUserId="2015" LastEditDate="2018-06-26T21:56:52.667" LastActivityDate="2018-06-26T22:06:52.077" Title="Arc welder? Cheap welding device for chainmail/chainmail jewelry" Tags="|materials|manufacturing-engineering|metallurgy|welding|strength|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="22471" PostTypeId="2" ParentId="22469" CreationDate="2018-06-26T22:06:52.077" Score="1" Body="&lt;p&gt;So from a fellow chain mail enthusiast (who uses split rings due to this), the spot welder is still the premier choice for this, and the ones on ring lord really are ideal for most carbon steels and even stainless steels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you want to get into aluminum, silver, or beyond, you'll want to invest into TIG welding.  TIG welding setups can be found on say McMaster Carr for around $300 - $400, not sure on the conversion rates or if the british government allows you to have one of those without a license (you will also need a grinding wheel to grind down the tungsten electrode, plus a supply of inert gas).  You'll find for ring welds, this will simply consist of heating up the ring at the gap, then tapping with weld rod like soldering.  Unlike soldering, the inert gas keeps it sealed from oxidation, and the temperatures are hot enough to melt the base materials, so as a result you get a TIG weld.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you get more advanced, you can use this method to weld large steel plates - simply chaining dots like this in a line until you get a full weld.  It becomes quite satisfying to weld quite precious metals like aluminum and titanium.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-06-26T22:06:52.077" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22472" PostTypeId="2" ParentId="22429" CreationDate="2018-06-26T23:30:57.060" Score="1" Body="&lt;p&gt;On the &lt;a href=&quot;https://en.wikipedia.org/wiki/Carnot%27s_theorem_(thermodynamics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carnot Theorem&lt;/a&gt;, which is based on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Second_law_of_thermodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Second Law of the Thermodynamics&lt;/a&gt;, the maximal effectivity of any heat engine is $1-\frac{T_C}{T_H}$. $T_C$ and $T_H$ are the temperatures of the cold and warm heat tanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if we burn gas ($1000K$ burning temperature) in a $\approx 300K$ environment, then the theoretical maximum is $70\%$. If you want to power your watch with your body heat, then $T_H \approx 20C^\circ$, $T_C \approx 36C^\circ$, thus the maximal effectivity is $\approx 3\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the engineering practice, there is no try to extract too much energy from little temperature differences. Because it is practically impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is unavoidible, then the rule of thumb is, that the thermomechanical conversions are the best (for example, Stirling-engine), but they have many disadvantage (cost, complexity, high and unpredictable fault rate of the moving parts). Here very complex (=costly) solutions aren't too far away from the theoretical limit. On obvious reasons, there is still &lt;a href=&quot;http://www.powermag.com/long-form-stories/2016-POWER-Plant-of-the-Year-Award/&quot; rel=&quot;nofollow noreferrer&quot;&gt;intensive development&lt;/a&gt; here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermochemical, thermoelectrical conversions are far away in effectivity, but they are simple and cheap with a better predictable wear rate.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2018-06-26T23:30:57.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22473" PostTypeId="2" ParentId="22041" CreationDate="2018-06-26T23:52:11.733" Score="1" Body="&lt;p&gt;So far there is no solution for updating the states directly. I suspect the authors of the paper update the states $x_m, y_m, \dot{x}_m, \dot{y}_m, \psi_m$ at each step using the acceleration input and then recompute the states $d, \lambda$, etc every time.&lt;/p&gt;&#xA;" OwnerUserId="16308" LastActivityDate="2018-06-26T23:52:11.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22474" PostTypeId="2" ParentId="22041" CreationDate="2018-06-27T01:31:10.860" Score="0" Body="&lt;p&gt;Following on from phoque, go back to you first four equations and rewrite them with each and every variable expressed one step forward in terms of the current state plus delta. You will need to do a few simple calcs to get some of the deltas. And you may want to think about how you want to update position. For instance do you want to use a delta based on the old velocity, the new one, or an average?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can avoid averages by using two different time step phases. Calc. Velocities at 1.0, 2.0, 3.0 delta t's and positions at 0.5, 1.5, 2.5 delta t's etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pro tip. The sketch that conveys the problem the clearest in terms of the variables is never the one that has the variables in the most convenient arrangement for computation. Get used to this. Draw two diagrams. The first should have all angles based on the common reference frame - Earth in this case, XYZ = ENU). All angles must have 1 and only 1 arrow. And they go from +X to +Y in the XY plane, +Y to +Z in the YZ plane, and +Z to +X in the ZX plane. This is for all Right Hand Rule coordinate systems. For instance, the angle labna evader should be from the +X and run ccw to the line of sight segment. If what you really want in the end is that minus phi, you can assign that a new name in the second sketch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do this and your autopilot won't unexpectedly turn the plane upside down when it crosses the equator.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-06-27T01:31:10.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22475" PostTypeId="1" CreationDate="2018-06-27T01:33:17.173" Score="1" ViewCount="620" Body="&lt;p&gt;Does the module of 2 meshing spur gears have to be the exact same value ? What is the allowable difference? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically my module is off by 0 01 and my coworker made a model and it works. So I'm just wondering what the allowable difference  is ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ps these are custom made gears. &lt;/p&gt;&#xA;" OwnerUserId="16585" LastActivityDate="2018-06-27T01:33:17.173" Title="Does the module of 2 meshing spur gears have to be the exact same value ?" Tags="|gears|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22477" PostTypeId="1" CreationDate="2018-06-27T05:31:56.563" Score="3" ViewCount="2792" Body="&lt;p&gt;I am working on a project that requires a wing to be rotated and I need the rotation mechanism to be fail-safe -- if power fails, the wing will stay in place and not move around.  My thought was to use a worm drive since they are not reversible.  However, worm drives typically reduce speed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there be anyway to make this reduction in speed from the motor to the gear as little as possible using a worm drive?&lt;br&gt;&#xA;Should I just forget the idea of a worm drive and use a motor that can hold its position even if turned off (if so what type of motor could do this?)?&lt;/p&gt;&#xA;" OwnerUserId="16586" LastEditorDisplayName="user16" LastEditDate="2018-06-27T11:52:06.083" LastActivityDate="2018-07-04T03:07:22.290" Title="Is a 1:1 Worm Drive possible?" Tags="|mechanical-engineering|gears|servo|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22478" PostTypeId="1" CreationDate="2018-06-27T05:34:40.153" Score="2" ViewCount="40" Body="&lt;p&gt;Do Oxygen lambda sensors (used in automotive applications) work in trace gas environments? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My requirements for a trace Oxygen gas sensor:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The sensor must be capable of working in an inert gas environment of either Nitrogen or Argon.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sensor's operating temperature is not more than 40-50 degrees Celsius.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sensor is capable of sensing an Oxygen gas concentration as low as 10 ppm to an Oxygen gas concentration as high as (19-25%, that is, an Oxygen concentration of regular air).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Could anybody who is working with such sensors suggest some sensor that I can use in such environments? &lt;/p&gt;&#xA;" OwnerUserId="16513" LastEditorUserId="7083" LastEditDate="2018-07-02T22:20:58.377" LastActivityDate="2018-07-02T22:20:58.377" Title="Oxygen sensors in trace gas environments" Tags="|electrical-engineering|sensors|gas|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="22479" PostTypeId="1" CreationDate="2018-06-27T06:05:34.350" Score="1" ViewCount="517" Body="&lt;p&gt;What is a mesh topology? I have read this term being used in a lot of articles dealing with meshing the computational domain, but could not understand its intuitive meaning. And why is topology called a global property? Also, what does topology mean wrt CFD analysis anyways? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2018-07-29T10:01:07.710" Title="What does mesh topology mean?" Tags="|cfd|meshing|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22480" PostTypeId="2" ParentId="22479" CreationDate="2018-06-27T07:01:00.440" Score="1" Body="&lt;p&gt;Depending on the context, mesh topology is the cell or mesh density and how it changes across the whole calculation field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get detail then the computational area is divided into cells - more cells means better detail / results but at a cost of more calculations ie  it takes longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, engineers look to “target” extra cells in areas of importance or relevance by increasing the mesh density in those areas and leaving other areas with fewer cells to limit or reduce the total number of cells so controlling the total time to calculate .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some jobs I did ran for 40 days ie 40 * 24 hours - any reduction due to optimisation of the calculation field was welcome and looked for...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-06-27T08:11:32.547" LastActivityDate="2018-06-27T08:11:32.547" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="22481" PostTypeId="2" ParentId="22477" CreationDate="2018-06-27T09:29:52.720" Score="1" Body="&lt;p&gt;A worm drive presents a reduction with a ratio of one to the number of teeth on the driven gear. This means it would be necessary to have a gear with one tooth and yet remain circular. Obviously not in our universe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to Phil Sweet, I've become &lt;a href=&quot;https://khkgears.net/new/worm_gear.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;aware of multi-start worm gears&lt;/a&gt;, also called multi-threaded worms. This would improve the mechanism by reducing friction and increasing efficiency. With the typical maximum multi-start of 4, you would still need a 4 tooth gear, also impractical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it becomes easier to implement the secondary gearset. If a 12 tooth driven gear is used, the ratio become 1:3 requiring a 3:1 secondary gearset. That's easier than having to deal with larger numbers, such as 40:1 or worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to accomplish your objective, you would have to select the smallest gearing mechanism of your resources and use it to drive another set of gears to increase the gear ratio.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2018-06-27T19:52:33.237" LastActivityDate="2018-06-27T19:52:33.237" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22482" PostTypeId="2" ParentId="14636" CreationDate="2018-06-27T12:28:37.320" Score="2" Body="&lt;p&gt;I always cheat the thread engagement percentage in tough materials. When allowable of course, going from say 75% to 65% will make it much easier to tap. So for a M12, chart says letter &quot;Y&quot; drill at .404&quot; will give you 75% engagement. I'd try going up to letter &quot;Z&quot; drill at .413&quot;. ONLY if you can! Otherwise, it sounds like your problem is chip extraction, the chip formed from cutting the threads packing in the bottom of the hole, making it hard to get that last bit of thread. A lighter weight cutting oil and a tap that aids in chip removal will help. You can also get a tap with a coating that aids in chip removal. Check Travers tool and Mcmaster, they have decent descriptions of types of taps and coatings and what there good for.&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-06-27T12:28:37.320" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22484" PostTypeId="2" ParentId="21710" CreationDate="2018-06-27T12:42:12.410" Score="1" Body="&lt;p&gt;Could they possibly be referring to surface finish? If they are, its called surface roughness, denoted as a value &quot;Ra&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lxlbU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lxlbU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-06-27T12:42:12.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22486" PostTypeId="2" ParentId="22479" CreationDate="2018-06-27T15:31:40.540" Score="0" Body="&lt;p&gt;Quoting from one &lt;a href=&quot;https://pdfs.semanticscholar.org/520b/a3e0ce4a2d17493ab9bde7adbc4f2efc4034.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt; related to FEA,  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The output of a mesh generator consists of a set of nodes, and a set&#xA;  of elements (which represents the mesh  topology).  [...] The nodes&#xA;  are created first, and they are then connected to form triangular or&#xA;  quadrilateral elements. This can be divided into two subcases. If the&#xA;  vertexes of the object are assumed to be the only nodes in the mesh, a&#xA;  triangulation algorithm can be applied to create a minimum number of&#xA;  nonoverlapping triangles that cover the object (see Figure 10). This&#xA;  minimal set of triangles is determined mainly by the topology of the&#xA;  object. The complex topology of the object has been decomposed into&#xA;  the simple topology of the triangles, so this approach will be called&#xA;  the topology decomposition approach. The mesh thus created has&#xA;  distorted and mostly coarse elements, and cannot be used for analysis&#xA;  purposes. Mesh refinement, possibly with element rearrangements by the&#xA;  diagonal transpose technique, must be applied to produce a reasonable&#xA;  mesh.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, colloquially, &quot;mesh topology&quot; may be used to indicate node density variations, face orientation variations, etc.  (The term &quot;topology&quot; itself comes straight from mathematics, where it means the study of sets and surfaces. )&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-06-27T15:31:40.540" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22487" PostTypeId="1" CreationDate="2018-06-27T15:57:58.267" Score="-1" ViewCount="445" Body="&lt;p&gt;I've come across two equations that relate power transmitted with the belt tensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;P &amp;amp;= 10^{-3}(T_1 - T_2)(k)(v) \\&#xA;P &amp;amp;= 10^{-3}(T_1 - T_c)(k)(v)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see one equation is based on slack side tension ($T_2$) and the other is based on centrifugal tension ($T_c$).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is when should I use any one of those two equations to find power.&lt;/p&gt;&#xA;" OwnerUserId="14355" LastEditorUserId="14355" LastEditDate="2018-06-27T17:35:58.590" LastActivityDate="2022-12-10T16:04:45.563" Title="Equations for power transmitted by a flat belt drive" Tags="|mechanical-engineering|power-transmission|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22489" PostTypeId="1" CreationDate="2018-06-27T19:05:55.457" Score="1" ViewCount="34" Body="&lt;p&gt;My question is very simple, I need to know why the minimum reinforcement for slabs is less than the minimum reinforcement required in Shear walls?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the criteria and reasons to set minimum steel reinforcement in the two structural elements. (if a code reference is needed plz refer to ACI and EC)&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="1832" LastEditDate="2018-06-27T20:23:13.930" LastActivityDate="2018-06-27T20:23:13.930" Title="Difference Between minimum steel required in Slabs and Shear Walls" Tags="|structural-engineering|concrete|reinforced-concrete|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22490" PostTypeId="2" ParentId="22477" CreationDate="2018-06-27T19:07:11.550" Score="1" Body="&lt;p&gt;You can use a one to one worm drive, even though I don't see the advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also as has been mentioned in other answers there are many cluching- locking mechanisms that need either a mechanical or electrical switch to release.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aviation industry commonly uses all manner of cluch and release in applications such as the hinge connection of propellers to shaft in adjustable pitch propellers. You can google Cessna 180 and Mooney, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o4tU7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o4tU7.jpg&quot; alt=&quot;one to one worm-drive&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-06-27T22:42:30.843" LastActivityDate="2018-06-27T22:42:30.843" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22493" PostTypeId="1" AcceptedAnswerId="22582" CreationDate="2018-06-28T02:42:48.293" Score="0" ViewCount="234" Body="&lt;p&gt;I am planning to build a large scale 3D printer that will utilise 2 1000mm linear motion shafts that each have a young modulus of 210 GPA. The ends of the shafts will be secured in a block of plastic and will not be able to move and the distance from one usable end to the other will total 900mm. Due to the large size of the parts being used I have concerns over possible deflection that could occur. Therefore I need to calculate the minimum required diameter necessary for 2 shafts to support a 2kg weight in the centre while deflecting less than 0.05mm (my layer height is going to be 0.1mm). Thanks for any help I am not an engineer so I have not the slightest clue how to do the calculations to determine this.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GPLvh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GPLvh.jpg&quot; alt=&quot;Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16593" LastActivityDate="2018-07-04T03:59:39.923" Title="What is the minimum diameter necessary for 2 linear motion shafts to deflect less than 0.05mm when a 2kg weight is in the centre?" Tags="|deflection|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22494" PostTypeId="1" AcceptedAnswerId="22498" CreationDate="2018-06-28T05:30:45.403" Score="2" ViewCount="251" Body="&lt;p&gt;I have a drive way which has knick at  point ot smoothen is out i leveled it by putting loose soil &lt;a href=&quot;https://i.stack.imgur.com/ZQPNX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZQPNX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But its monsoon season and rain washed away all my soil as water comes running down slope.&#xA;In pic red is soil i am putting to level road, IS there way to stabilize this loose soil , so that water does not wash away soil &lt;/p&gt;&#xA;" OwnerUserId="10977" LastActivityDate="2018-07-04T00:21:11.767" Title="Save loose soil from erosion on a slope" Tags="|civil-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22495" PostTypeId="2" ParentId="22493" CreationDate="2018-06-28T06:22:03.077" Score="0" Body="&lt;p&gt;There are equations to calculate the static deflection of your rods and from there design their diameter. But that will give incorrect answers because your print-heads are moving back and forth fast and depending on their movement, can impact the beams in a dynamic resonate mode and create huge vibrations and deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a start let's look at static deflection. &#xA;$$ Deflection = Y = \frac { pl^3} {48EI} =&amp;lt; 0.05mm $$ &#xA; l is 900mm and E = 290 GPA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I = \pi d^4/64 $$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;plugging I in Y we get $$ Y = \frac { pl^3} {48E\times \pi d^4/64} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation gives us $$ 0.05 \times D^4 = \frac {64 pl^3}{48E\pi} $$ And $$ D = ( 80pl^3/E\pi)^{1/4}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this D is an underestimate by a large margin. One would need to have data on the range of acceleration and deceleration impact of the print head and then design the shafts accordingly. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="16486" LastEditDate="2018-07-04T03:59:39.923" LastActivityDate="2018-07-04T03:59:39.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22496" PostTypeId="2" ParentId="22494" CreationDate="2018-06-28T06:58:31.173" Score="0" Body="&lt;p&gt;First you need to plant some fast-growing, root generating plants and cover the loose soil with landscape netting and shout the whole top with green mulch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have to create a concrete swale on the top-slope or at least create a berm by sandbags. The swale and or berm have to be leading the rain down in a gentle slope not blocking it like a dam.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-06-28T06:58:31.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22497" PostTypeId="1" CreationDate="2018-06-28T07:17:03.817" Score="1" ViewCount="564" Body="&lt;p&gt;I'm using the Ansys ACT Api with C# to create an extension for Ansys (Mechanical).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When developing the extension I need to build a dll file from C# code (using Visual Studio). My problem is that every time I want to test my extension with a new build, I need to restart Ansys Workbench so that the dll of my extension can correctly be replaced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is rather time consuming, and I assume there must be a work around, but can't find anything in the documentation or online.&lt;/p&gt;&#xA;" OwnerUserId="16326" LastEditorUserId="33210" LastEditDate="2022-03-28T07:46:19.250" LastActivityDate="2023-12-18T15:07:00.900" Title="Debugging Ansys ACT C# Extensions" Tags="|finite-element-method|ansys|ansys-workbench|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22498" PostTypeId="2" ParentId="22494" CreationDate="2018-06-28T09:50:16.950" Score="3" Body="&lt;p&gt;The key to preventing erosion and soil loss is to slow the speed of the water traveling down the slope during rain events and to keep the soil in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that could be done is to construct a retaining wall at the bottom of the slope. This can be expensive and depending on the amount of soil that needs to be retained may require and engineer to design it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, the slope surface of the soil should be &quot;moonscaped&quot;. This involves creating dips and bumps up and down and along the full length of the soil slope. The bumps should be staggered and not line up with each other. This will slow down the speed of surface water on the slope and reduce erosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing that needs to be done is to lock the grains of soil in place. This is best done with roots of plants and you may need to go through a staged serious of plantings: grasses, shrubs and trees. Don't just plant trees and expect them to do everything. The roots of the grasses will lock soil near the surface of the slope. The shrubs will lock the soil a little deeper into the slope and tress and even deeper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After having done all of this, covering the slope with heavy mulch, that will not be carried away by water will also assist in slowing the speed of surface water and help retain moisture in the soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geotextiles and landscape netting can be used to prevent the top portion of the soil from being washed away. Or the slope surface could be sprayed with a mixture of bitumen and grass seed. Once the grass seeds have established themselves, the bitumen will eventually degrade and be absorbed by the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construction of a drainage ditch, just above the deposited soil will also help reduce the amount of water the soil slope needs to deal with. This could be a dug earth ditch of a concrete/paved ditch, as long as the run off water is directed away from the slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other things that can be done is to place hay bales, or rocks, on the soil slope and on the slope above the deposited soil. These can help to reduce the speed of surface water running down the slope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If hay bales are used they should be placed in a staggered, off-set pattern, so that long drainage channels, which would lead to the formation of erosion gullies, are not created by the bales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moonscaping of the upper natural slope, above the deposited soil slope would also help in preserving the deposited soil slope.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2018-06-28T13:51:30.557" LastActivityDate="2018-06-28T13:51:30.557" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22499" PostTypeId="1" AcceptedAnswerId="22506" CreationDate="2018-06-28T10:48:24.123" Score="1" ViewCount="98" Body="&lt;p&gt;I need a check valve ( one way valve ) for my project which has less actuation force (IDK real term. shouldn't require high pressure for inflow of fluid) and I searched in market for the valve but couldn't find one. Also I am not in place where I can easily order online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw an hydraulic (can hold 2 tons ~ 7k psi. I need about 500 psi only though.) with ball valve that lifts when inflow and gets down to its seat due to gravity. It is really simple with no spring and also has no rubber or anything like that to prevent seepage of oil but it works. So I was thinking if I can make one in lathe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two problems for that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;how to make a spherical hole.&lt;/li&gt;&#xA;&lt;li&gt;despite making the structure how can I be sure that it won't leak.&#xA;Like theoretically it is possible but mostly somehow the products&#xA;don't turn out okay as planned.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am also thinking instead of ball, a tapered one will be easier to manufacture but not sure if it will work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;" OwnerUserId="14520" LastActivityDate="2018-06-28T18:59:33.460" Title="Can I make check valve in simple workshop?" Tags="|mechanical-engineering|fluid-mechanics|hydraulics|valves|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22500" PostTypeId="2" ParentId="22499" CreationDate="2018-06-28T12:10:27.747" Score="0" Body="&lt;p&gt;Spring-loaded ball check valve, you have described it, now you need to make it and that is down to the precision you are capable of.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-06-28T12:10:27.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22502" PostTypeId="2" ParentId="22479" CreationDate="2018-06-28T13:06:06.223" Score="0" Body="&lt;p&gt;Mesh topology, just like the mathematical definition of a topology, defines how things are connected or related. The meshing of a computational domain is made up of nodes and elements(take here line elements ie line connecting one node to another). Topology means how these nodes are connected by elements as the same nodes could be connected in other ways too and this would be another topology. If the mesh is considered as a set, and the nodes and elements are considered the part of this set then topology is how all these parts are connected or related to each other. There are different topologies on the same set.&lt;/p&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="16275" LastEditDate="2018-06-29T09:51:23.130" LastActivityDate="2018-06-29T09:51:23.130" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22503" PostTypeId="2" ParentId="22452" CreationDate="2018-06-28T14:19:11.220" Score="2" Body="&lt;p&gt;Here is the diagram that I was trying to lead you to in the comments&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;       Energy In E1          Energy out E2&#xA;           |                     ^&#xA;           V                     |&#xA;Input -&amp;gt; First    ----------&amp;gt;  Heat    ----------&amp;gt; etc.&#xA;Air      heatpump   hot air    engine    warm air&#xA;T_0        |        T_h1&amp;gt;T_0     ^       T_c1&amp;lt;T_h2&amp;lt;T_h1&#xA;           v                     |&#xA;        cold air              cold air&#xA;        T_c0&amp;lt;T_0              T_c1&amp;lt;T_h1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;First thing, is that heat pumps require external energy input to run.  If you don't believe me, go look at your refrigerator or air conditioner.  Both of these are heat pumps, and I guarantee that they are plugged into an external electricity source, and will cease to function if unplugged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second thing is that just generating hot air does not automatically get you energy.  You need to extract the energy from that hot air using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_engine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heat engine&lt;/a&gt;.  This could take many forms. E.g. a stirling engine might work well here.  Note also that the heat engine must have not only a source of hot air, but also a source of cold air too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I know what you are thinking, that you can just use the energy E2 extracted from the heat engine in order to drive the heat pump.  But the problem is that E2 will always be less than E1.  You need to put more energy into this system than you will get out of it.  Therefore, this system does not produce any net energy. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-06-28T14:19:11.220" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22504" PostTypeId="2" ParentId="22493" CreationDate="2018-06-28T14:42:42.073" Score="0" Body="&lt;p&gt;The design you show is not going to work reliably. In any machine predictability is required and your design will produce an unpredictable machine because the load and deflections are unconstrained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct way to look at the problem is to understand the nature and magnitude of the forces and the direction they can be applied. In your tubular design, any side loading will also produce deflections. So the better way to build the machine is to constrain the deflections in the vertical and lateral directions by using a fully supported rail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Search for linear rails and look for products that have a solid rectangular mounting rail such as this rail from Misumi Corp.&lt;a href=&quot;https://i.stack.imgur.com/McHPP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/McHPP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2018-06-28T14:42:42.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22506" PostTypeId="2" ParentId="22499" CreationDate="2018-06-28T18:59:33.460" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cpYaa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cpYaa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. A sprung check-valve using a ball and a taper. Source: &lt;a href=&quot;http://crosscontrols.com/product/walworth-api-602-ball-check-valve/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cross Controls&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm no machinist but it seems to me that a purchased ball (a bearing, for example) should make a good fit into a well machined tapered hole. If operating with gravity assist the spring may be omitted.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-06-28T18:59:33.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22508" PostTypeId="2" ParentId="18536" CreationDate="2018-06-28T22:20:55.360" Score="0" Body="&lt;p&gt;Plastic yield stress is the stress above the linear part of stress strain curve and varies by the class of steel. When a member is stressed beyond this limit it will strain by larger amounts and it won't relax back to original geometry. However while in deforming in plastic range the section can absorb huge amounts of energy and provide the critical short relief safety from disasters such as earthquake or explosion before total collapse of the structure .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In structural engineering this property of steel is taken advantage of by designing the connections and members of structure to encourage plastic hinge at prescribed locations such as to save a column at the expense of a beam by tapering the flanges of the beam just before the joint to a narrower width, thus providing a ductile behavior under above strength limits of loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though a section can be stressed in such a way as to experience both elastic and plastic stresses at the same time on different parts of profile common practice to calculate plastic module is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the neutral axis of the section, in I beams and typical symmetrical sections it's the horizontal axis of symmetry. Find the center of geometry of the top and bottom parts. Multiply top part area by its moment arm, which is the distance from its center of geometry to the neutral axis,  then do the same for bottom part and add the result. This is the section's plastic module. Now you can multiply this by the Fy to get the plastic strength of the section.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-06-28T22:31:26.807" LastActivityDate="2018-06-28T22:31:26.807" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22510" PostTypeId="1" AcceptedAnswerId="22523" CreationDate="2018-06-29T01:09:07.460" Score="1" ViewCount="267" Body="&lt;p&gt;I’m a electrical engineer but an amateur in mechanics. I’ve been checking some tools and machines in the market for small works which would allow me to manufacture small but precise pieces, such as a 5mm H6 tolerance shaft to fit a bearing bore under interference. The thing is this shaft should have a tolerance of +-4um, maximum. Ok, so usual lathe setups allow you to position the cutting tool in a scale of 50um marks, with god knows which accuracy. I’d guess no less than +- 50um. But even if it was perfectly match, 50um is rather large near H6 tolerances for my shaft.&#xA;Ok, let’s assume I could iteratively remove material using a lathe and use a top precision micrometer to check if my shaft is within required dimension. Still it seems to be a very non-professional way to do it, basically a try-and-error approach, moreover I could accidentally remove more material than I should and end up losing all the work and also money.&#xA;Could anyone tell me if there’s a more efficient method to produce such precise pieces?&lt;/p&gt;&#xA;" OwnerUserId="7086" LastActivityDate="2018-06-29T11:48:22.287" Title="H6 tolerance shaft manufacturing" Tags="|tolerance|metrology|lathe|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22511" PostTypeId="1" CreationDate="2018-06-29T01:27:50.100" Score="1" ViewCount="69" Body="&lt;p&gt;If the power $P = C\omega$, so how a DC motor powered by $P = 10\text{ W}$ can work as $\omega$ is of course 0&amp;nbsp;rad/s at first, because if $\omega = 0$, then $P = C\omega = C\cdot0 = 0$?&lt;/p&gt;&#xA;" OwnerUserId="14830" LastEditorUserId="14830" LastEditDate="2018-06-29T13:52:35.660" LastActivityDate="2018-06-29T13:52:35.660" Title="Power torque and rotational speed of a not running yet DC motor" Tags="|electrical-engineering|motors|power-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22512" PostTypeId="2" ParentId="22511" CreationDate="2018-06-29T05:57:13.700" Score="0" Body="&lt;p&gt;Because Power is also &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P = I * V&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;That is current (amps) and voltage, which is not zero.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-06-29T05:57:13.700" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22513" PostTypeId="1" CreationDate="2018-06-29T06:46:32.380" Score="0" ViewCount="56" Body="&lt;p&gt;I am doing research on biomedical device development, and focusing on devices that work on grasping tissues in the abdominal area. I need some help with applying the concepts of material strength to human tissues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simple language, let's assume that the stiffness constant (or the spring constant, K) of a given tissue is known, and we assume that it is a constant and hence the compression of the tissue is linearly proportional to the compressive force applied on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that we know K, is there a way to figure out the upper limit of the Force or the upper limit of the compression I can allow, before the tissue gets damaged? i.e. can we estimate the compressive yield strength of the tissue given that we know its K?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another small confusion: If we are talking about the point at which tissue damages, is that the point of yield strength or ultimate strength?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm familiar with basic solid mechanics and strength of materials on a very beginner level and I'm getting stuck in applying those concepts in the context of biomechanics. Any help in this direction would be really helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: This is not Homework help, I work as an RA and I'm stuck at this issue in my research :)&lt;/p&gt;&#xA;" OwnerUserId="16601" LastActivityDate="2018-12-31T05:01:15.220" Title="Need advice: Estimating the compressive yield strength of abdominal tissues from the data of its rigidity or stiffness constant" Tags="|mechanical-engineering|strength|biomechanics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22514" PostTypeId="2" ParentId="22511" CreationDate="2018-06-29T08:01:48.853" Score="1" Body="&lt;p&gt;When $\omega = 0$ the motor is not doing any mechanical work, or producing any mechanical power, because nothing is moving. (If this is a long term situation, then the motor is &lt;em&gt;stalled&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when $\omega = 0$ the motor is still producing a &lt;em&gt;torque&lt;/em&gt; (proportional to $I$) which causes a mechanical &lt;em&gt;angular acceleration&lt;/em&gt; (by Newton's second law) and starts the system moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the motor is stalled, the power $P = I * V$ mentioned by Solar Mike is converted into &lt;em&gt;heat,&lt;/em&gt; not mechanical work - i.e. the motor is behaving exactly the same as an electrical resistor.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-06-29T08:01:48.853" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22515" PostTypeId="2" ParentId="22513" CreationDate="2018-06-29T08:12:16.913" Score="1" Body="&lt;p&gt;There is no general relation between the stiffness and strength of different materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, be aware that in general, the mechanical behaviour of biological materials &lt;em&gt;much&lt;/em&gt; more complicated than the &quot;simple&quot; linear-elastic isotropic materials typically used in structural engineering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a medical expert, but common sense suggests that for a living biomaterial, &quot;damage&quot; is a much more complicated notion than &quot;how much force does it take to break it!&quot;&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-06-29T08:12:16.913" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22516" PostTypeId="1" CreationDate="2018-06-29T08:38:48.873" Score="1" ViewCount="159" Body="&lt;p&gt;I'm looking to design a system that can control a solar panel through an electrical actuator for a range of motion of -45 to +45 degrees for maximum solar exposure. The panel will be mounted on a 10 feet pole and will be attached at the center of that pole. This means that either the mechanism will be at the pole's top or at least some kind of bearing to allow for the movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The caveat in this case (because the panel is attached high up) is that due to security reasons the system must be designed in such a way that the panel will be held stable (mustn't rotate and mustn't fall off) &lt;strong&gt;without relying on the actuator itself&lt;/strong&gt;, in case of power failures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've started looking around for actuator mechanisms but haven't yet found anything that talks specifics about similar security requirements. I'm looking for pointers / ideas as I'm not yet experienced in such cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: For anyone who would be interested in numbers the panel weighs about 150 kg and will possibly receive a horizontal wind force of 300N. Also a torque of 100Nm will need to be applied in order to change its facing direction.&lt;/p&gt;&#xA;" OwnerUserId="16602" LastActivityDate="2023-12-31T07:04:15.480" Title="Solar panel actuator control" Tags="|actuator|stability|systems-design|photovoltaics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22518" PostTypeId="1" CreationDate="2018-06-29T09:14:02.457" Score="1" ViewCount="61" Body="&lt;p&gt;I was learning about PCV valve, when one of the online resources pointed out that the PCV valve will be closed during  high engine load to prevent the lean-mixture (blow-by) in the crank case from entering the combustion chamber, as a lean mixture is not desirable during high engine load.&#xA;But my basic automobile engineering textbook (local author) points out that during higher engine load, the blow-by increases and causes the crank case pressure to rise thereby increasing the opening of the PCV valve and letting more blow-by enter the combustion chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm confused wether the valve actually closes or opens during high engine load.&lt;/p&gt;&#xA;" OwnerUserId="16605" LastActivityDate="2018-06-29T09:14:02.457" Title="Is the PCV valve open during high engine load" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22519" PostTypeId="1" CreationDate="2018-06-29T09:31:24.400" Score="0" ViewCount="181" Body="&lt;p&gt;I have a shaft with a bearing in each end, and a large eccentric mass covering the entire &quot;free length&quot; of the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LgcKs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LgcKs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The attachment of the eccentric mass is done using heat shrinking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without bearings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mDFN4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mDFN4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without bearings and eccentric masses:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8uBk9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8uBk9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to calculate deflection equations $y_{ab}$ ($x\in\{0,a\}$, from outmost left to left hand side of flywheel), $y_{BC}$ ($x\in\{a,b\}$, from left hand side of flywheels to right hand side of flywheels), and $y_{CD}$ ($x\in\{b,l\}$, from right hand side of flywheels to outmost right):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PiRsD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PiRsD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is similar to Shigley's Mechanical Engineering Design problem 4-19, 10$^{th}$ ed.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FBD:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F4Roo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F4Roo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7873" LastEditorUserId="7873" LastEditDate="2018-07-25T04:54:25.673" LastActivityDate="2018-07-25T04:54:25.673" Title="How does the superposition principle relate to Bernoulli-Euler and Timoshenko beam theory?" Tags="|mechanical-engineering|beam|modeling|" AnswerCount="1" CommentCount="3" ClosedDate="2018-07-02T15:59:03.847" ContentLicense="CC BY-SA 4.0" />
  <row Id="22520" PostTypeId="1" AcceptedAnswerId="22529" CreationDate="2018-06-29T10:44:22.740" Score="2" ViewCount="904" Body="&lt;p&gt;Solar furnaces using mirrors to concentrate sunlight on a small area do not produce much power - perhaps 1kW per square metre of mirror - but can reach very high temperatures. From the earliest metalworking through to the beginnings of the industrial revolution, metal refining and working was limited by the difficulty of supplying enough heat to the furnace. For a small scale operation - the sort of activity carried out before the industrial revolution - could a solar furnace have been used instead of one using a fuel (usually charcoal) and air mixture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm imagining a sunny hillside with perhaps 10 to 100 polished metal mirrors, perhaps a square metre apiece, aligned by humans running about, all pointing at a small crucible or forge inside as much insulation as could be wrapped around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to use that for refining or forging?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to ask, over on history stack exchange, whether such an arrangement was ever used, but I thought I'd make a fool of myself asking a question over here before I make a fool of myself asking one over there. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I did find these:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://books.google.co.uk/books?id=-bQ9DwAAQBAJ&amp;amp;pg=PT106&amp;amp;lpg=PT106&amp;amp;dq=small%20solar%20forge&amp;amp;source=bl&amp;amp;ots=z5BeB-pO3K&amp;amp;sig=rVtQ_MdLaQ-O5GcIEdohsG7g0sY&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwiOp5zb-vjbAhXQJVAKHYZ3AUUQ6AEIrAEwFg#v=onepage&amp;amp;q=small%20solar%20forge&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fresnel solar forge&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://woodwaredesigns.com/energy/solarforge.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Woodward solar forge&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm guessing it would have been hard to make a large Fresnel lens until relatively recently. Mirrors are much easier, they have lovely shiny steel armour in many museums, and child labour to polish the metal and align the mirrors would have been cheap in the past. The Woodward forge looks good but the blog says that it hasn't been used for metalwork. Has anyone made this work on a small scale with mirrors?&lt;/p&gt;&#xA;" OwnerUserId="16606" LastEditorUserId="16606" LastEditDate="2018-06-29T13:25:13.480" LastActivityDate="2019-08-22T11:33:25.160" Title="Is it practical to make a solar furnace for small scale metal production?" Tags="|heat-transfer|metallurgy|solar-energy|forging|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22521" PostTypeId="2" ParentId="22520" CreationDate="2018-06-29T10:59:20.813" Score="1" Body="&lt;p&gt;Well, I think the French built a large one...&#xA;See &#xA;&lt;a href=&quot;https://anglophone-direct.com/mont-louisfont-romeu-odeillo-via/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://anglophone-direct.com/mont-louisfont-romeu-odeillo-via/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for historical, the issue may have been the quality of the mirrors...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, 1000 years ago - understanding that the power was available as well as reliability ie not weather dependant.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-06-29T11:09:59.733" LastActivityDate="2018-06-29T11:09:59.733" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22522" PostTypeId="2" ParentId="22516" CreationDate="2018-06-29T11:33:49.267" Score="0" Body="&lt;p&gt;There are lots of home-brew projects completed and written up in this magazine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One used a water bucket with a controlled fill to follow the sun and a controlled tap to empty the bucket with a return spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another used a small 3v motor with two small pv panels and a « lens » made from a jar filled with water...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.homepower.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.homepower.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-06-29T11:33:49.267" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22523" PostTypeId="2" ParentId="22510" CreationDate="2018-06-29T11:48:22.287" Score="0" Body="&lt;p&gt;If you truly need that type of tolerance and repeatability, then you are way out of the hobby equipment territory. You’ll need high precision calibrated measuring equipment, temperature stability, and specialty manufacturing equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likely an O.D. cylindrical grinding machine will get you there. If the bars are long, they need to be supported while being processed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to get close at the hobby scale, patience, scrap material and some decent micrometers will get you close. You can turn(machine) your diameter oversize and then you could use some abrasives(Emory cloth and scotch bright to polish the diameter to size. You could also make a homemade toolpost grinder. Essentially mount a grinding wheel and motor to your carriage. Trial and error. Cut and measure. Adjust, re-cut. Possible, but seems like an academic effort. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start, I suggest you purchase pre-made shafts to that tolerance and focus on the lower precision items you can actually produce. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2018-06-29T11:48:22.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22524" PostTypeId="1" AcceptedAnswerId="22944" CreationDate="2018-06-29T12:19:14.420" Score="4" ViewCount="360" Body="&lt;p&gt;Roughness measurement instruments output those two values, where Rt is the distance  between the top of the highest peak and the bottom of the lowest valley, and Ra is the average distance of every point to the mean line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In almost all technical specifications I see the Ra value being reported, so when is the Rt value used/why is it measured?&lt;/p&gt;&#xA;" OwnerUserId="16607" LastActivityDate="2022-08-08T13:45:54.617" Title="In what cases are the roughness values Rt and Ra used?" Tags="|mechanical-engineering|materials|measurements|surface-preparation|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22525" PostTypeId="2" ParentId="22520" CreationDate="2018-06-29T15:52:58.413" Score="0" Body="&lt;p&gt;I think the first step is to decide if you are refining or forging. If refining you are pretty much limited to making steel from pig iron : certainly not practical on a small scale; it is more about blowing hot air + oxygen into the bottom of a ladle than a heat source. Forging sounds possible ; build a insulated chamber that the beam can enter and a hatch for access. Then why or what are going to forge ? An almost practical objective would be to melt copper alloys to pour castings . Start by melting zinc for castings , low melt temperature and fairly e.asy to handle&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-06-29T15:52:58.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22527" PostTypeId="2" ParentId="22519" CreationDate="2018-06-29T17:31:09.467" Score="2" Body="&lt;p&gt;Your mechanism is rigid between the points B-C. Using w as uniform load is meaningless; uniform load Is best imagined as the pressure of a liquid to the bottom of its container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this case you need to check both vertical and torque shear at the start and end of eccentric mass. And you need to plot the shear stresses on a Mohr circle to calculate the principal stresses and axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This mechanism looks like its designed to counter another off balance flywheel. If so, then you need to calculate the dynamic vibration and dangerous resonance frequencies of the system and its response spectra.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many FEM systems that can do that. Important point is you are looking at the problem in a totally wrong angle.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-06-29T17:36:28.613" LastActivityDate="2018-06-29T17:36:28.613" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22529" PostTypeId="2" ParentId="22520" CreationDate="2018-06-29T21:51:52.803" Score="1" Body="&lt;p&gt;I'm not aware of any small scale historical versions, but it seems more-or-less possible. There is &lt;a href=&quot;https://www.aalborgcsp.com/business-areas/solar-district-heating/csp-parabolic-troughs/history-of-csp/&quot; rel=&quot;nofollow noreferrer&quot;&gt;a good myth&lt;/a&gt; about the ancient Greek's using shields to focus sunlight and light Roman ships on fire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see a few engineering challenges. You either have to have a nice roughly parabolic shaped mirror, which was really tough to manufacture until the last century, or approximate one with many small flat mirrors. The small flat mirrors (check out &lt;a href=&quot;https://www.google.com/search?q=linear%20fresnel%20csp&amp;amp;oq=linear%20fresnel%20csp&amp;amp;aqs=chrome..69i57j0.2855j0j7&amp;amp;sourceid=chrome&amp;amp;ie=UTF-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;linear fresnel concentrated solar collectors&lt;/a&gt;) quickly become a controls nightmare without automation. Turns out, when you're trying to focus the sun with a mirror, the sun actually moves annoyingly fast (well, in your frame of reference). Modern CSP heliostats adjust position on a one minute type timescale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature control and solar flux are also challenging problems. If you work through the geometry to collect enough sun to produce the requisite heat, you'll find the focal distance between not hot and burned-a-hole-in-your-work is not very big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On an only sort of related, but really interesting, note, someone in maybe the 70s decided it would be fun to make a concentrated solar power plant using a liquid sodium/potassium alloy as the heat transfer fluid. As you likely guessed based on this answer and a little bit of knowledge about liquid sodium, it burned down, killing an operator. The moral is to only forge non-flammable metals, I guess?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2018-06-29T21:51:52.803" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22530" PostTypeId="1" CreationDate="2018-06-30T05:26:05.223" Score="0" ViewCount="125" Body="&lt;p&gt;I’m wondering if anybody can help me by critiquing my project idea of having a failsafe power train 4x4 for remote touring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of having one big engine, I’m toying with the idea of running two small engines, parallel to each other, which would turn a mitre gearbox (90* bevel gears), which would both then have output shafts entering a differential to mate them into a single output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would run a differential lock so that I could disconnect a blown motor and drive turn the differential off a single engine (instead of just having the opposite side of the differential spin in reverse).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only major hurdle I have thought of so far, is that the shafts would be spinning in opposite directions as they enter the diff which would be relatively ineffective and make me ever so slightly sad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m thinking to overcome this I’ll need a 180* bevel bear to reverse the drive rotate of one of the shafts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What say you?&lt;/p&gt;&#xA;" OwnerUserId="16553" LastEditorUserId="90" LastEditDate="2018-06-30T09:20:39.177" LastActivityDate="2024-04-02T12:03:00.373" Title="Will this work for single output twin engine" Tags="|gears|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22532" PostTypeId="1" AcceptedAnswerId="22551" CreationDate="2018-06-30T14:29:51.960" Score="1" ViewCount="114" Body="&lt;p&gt;During the era of the typewriter there were a few kinds of basic designs. The (afaik) most well known is the type bar design, where each key is connected to an individual type bar that hits the paper when the key is pressed, each type bar carries the uppercase and lowercase shapes of a single letter. The other variant is the type wheel or type ball design, where a single ball or wheel shaped element contains all the letters. The mechanism rotates and/or tilts the wheel/ball before it strikes the paper so that the right letter is imprinted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Type wheel/balls have some advantages over type bars: They cannot get stuck, and the entire wheel/ball can be replaced to change the font. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Blickensderfer_typewriter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blickensderfer&lt;/a&gt;, one of the early wheel operated typewriters, is often described as being much simpler than other contemporary (often type bar operated) machines, containing only a quarter of the moving parts of other contemporary typtewriters, so the type wheel mechanism appears to not necessarily be more complex than type bars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the last decades of the typewriter era, the electric type ball operated &lt;a href=&quot;https://en.wikipedia.org/wiki/IBM_Selectric_typewriter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Selectric&lt;/a&gt; machines where very popular as higher end typewriters, but I have the impression that type bars were more common in manually operated lower end machines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were there technical reasons for type bars being more popular in simpler typewriters? If so what were they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This question is based on a few assumptions, which may be wrong. Please tell me if that is the case. And it may be that nontechnical reasons were more important in why type bar designs were more popular. I'm also interested in those, but in that case this is not the right site.)&lt;/p&gt;&#xA;" OwnerUserId="12424" LastActivityDate="2018-07-02T21:47:58.627" Title="What were the technical reasons for type bar designs being more common than type wheels for manual typewriters?" Tags="|engineering-history|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="22533" PostTypeId="2" ParentId="22530" CreationDate="2018-06-30T17:36:13.567" Score="0" Body="&lt;p&gt;While that which you propose is possible from a purely technical standpoint, each extra gizmo required to implement it introduces its own failure modes into the overall reliabililty equation for your vehicle: the more things that can break, the better the chances are that SOMETHING will break. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your overall objective is to furnish redundancy in case of failure, then you would most likely be better off studying the existing failure modes of your 4X4 and their frequency of occurrence, and beefing up their weak points rather than doubling up on your drivetrain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If on the other hand your objective is to get out your welding rig and a pile of tools and have fun, then by all means go for it. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-06-30T17:36:13.567" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22536" PostTypeId="1" CreationDate="2018-07-01T08:57:16.553" Score="1" ViewCount="153" Body="&lt;p&gt;I am thinking to develop a remote-sensing applications for vehicle pollutions. In such cases generally expensive optical IR and UV devices are used because more precise. These devices are generally ready-to-use and have also a graphical interfaces and other commands plus all the algorithms already implemented while I am interested only in the sensor part of such devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary electrochemical gas sensors are cheaper but much less precise. Generally these sensors have a response time from 20 to 60 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I can use a less precise estimate: can environmental polluting gas (electrochemical) sensors detect emissions of a low speed passing vehicle in a semi-closed environment such as a garage to have an approximate estimation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My biggest fear is that these sensors are too slow to detect anything, even for a low-speed vehicle in load conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively do you now any optical IR/UV open-path emitter/receiver sensor to analyze polluting gases from vehicle emissions and whose (electrical) output can be analyzed by embedded devices such as microcontrollers or micro-computers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: The pollution gases due to road vehicles are typically CO, NO, NO2 but also CO2 should be monitored. These are the gases maybe with more concern but also SO2 might be important.&#xA;The goal is to have an estimate on how much each vehicle is polluting: this estimate is not precise but it becomes important having large enough data to see for example which brand is most polluting.&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=hsKihzZxIhg&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; is an example of device but it is used for industrial applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT2&lt;/strong&gt; Let me reply to Bart's answer because it seems to me that the topic is not really clear here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has already being done (from late 80s-early 90s) using NDIR and UV. You do not measure the concentration directly but rather the ratio between a pollutant and CO2 because it is assumed to remain constant with the gases dispersion (so the wind effect should be minimized).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is not to get a single precise costly and time-requiring measurement but to get many cheap less precise fast measurements. In this way it is possible to guess for example which brand or model is more polluting, and also to guess if a vehicle is polluting too much. With on-board expensive tests you confirm or reject those results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the analysis of emissions using extra data such as model/brand, category diesel or not you can provide also other estimations.&lt;/p&gt;&#xA;" OwnerUserId="16624" LastEditorUserId="16624" LastEditDate="2018-07-03T09:32:09.833" LastActivityDate="2024-04-03T02:06:05.807" Title="Which type of sensors to analyze polluting gas from vehicle emissions" Tags="|electrical-engineering|sensors|embedded-systems|gas|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22537" PostTypeId="2" ParentId="11018" CreationDate="2018-07-01T13:00:09.777" Score="1" Body="&lt;p&gt;One way is to use massive pond systems for evaporate cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turkey Point nuclear plant. Closed loop system explained here - &lt;a href=&quot;https://youtu.be/qt9dtYC7RDE&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/qt9dtYC7RDE&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tTlS0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tTlS0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d2afM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d2afM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-07-01T13:10:01.993" LastActivityDate="2018-07-01T13:10:01.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22540" PostTypeId="2" ParentId="12108" CreationDate="2018-07-01T21:20:02.780" Score="0" Body="&lt;p&gt;From the materials perspective you need to look seriously at NH3, not on Stack Exchange. Of course no copper,brass or bronze and monel should be checked. Steam turbines can be hot , like 1000 F , at this temperature you will nitride carbon steel with ammonia. GE condensing steam turbines had/have a problem with hydrogen stress cracking and/ or corrosion  fatigue of hard 13 chrome blades ( GE has denied this).  But one should be very careful with any hydrogen compounds around hard ( Rockwell C 23 ) steel . And , unrelated, anhydrous ammonia can stress corrosion crack moderate strength steels at ambient temperatures ( Re: Dr Phelps , USSteel, NACE Corrosion 1969-? ). This is prevented by addition  of a few hundred ppm of water, so most engineers are unaware of it today. And ,furthermore, unrelated , is high temperature hydrogen attack ( Re: API 941 ,Nelson curves).This occurs above 500 F in high pressure hydrogen ; However ,corrosion can produce hydrogen with high fugacity ( relates chemical activity to pressure - No , I do not understand it). So hydrogen attack has occurred in steam boilers with essentially no hydrogen pressure.So, using ammonia as a working fluid  does introduce potential materials problems. .PS , I love hydrogen it basically provided me a good career.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-07-01T22:15:11.800" LastActivityDate="2018-07-01T22:15:11.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22541" PostTypeId="2" ParentId="12108" CreationDate="2018-07-01T21:49:09.190" Score="0" Body="&lt;p&gt;The worst complication of the ammonia vapor is that its lethal dose is around 0.015 $\frac{mg}{kg}$. As comparison, cyan has 0.0015 $\frac{mg}{kg}$. &lt;sup&gt;(&lt;a href=&quot;https://en.wikipedia.org/wiki/Ammonia&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref1&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrogen_cyanide&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref2&lt;/a&gt;)&lt;/sup&gt;&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2018-07-01T21:49:09.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22542" PostTypeId="2" ParentId="19129" CreationDate="2018-07-01T22:56:20.497" Score="0" Body="&lt;p&gt;You are missing a huge number of guide curves in this scenario. I've taken a screenshot showing what happens when you have an internal and external diameter guide curve for just one 'tooth' on your profile - the other vertices aren't guided, attempt to go in a straight line, and produce self-intersecting geometry that can be previewed, but not built.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xwb2j.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xwb2j.png&quot; alt=&quot;Two of 8 guide curves&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am, therefore, going to suggest another way to model your part, since adding all of those curves would be exceptionally time consuming...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that your &quot;Flow Axis&quot; is a straight line, this would make &lt;em&gt;much&lt;/em&gt; more sense if it were built using a revolve, rather than a sweep. This gives you full control over the cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have modelled your part using the following procedure:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Create a 'master sketch', showing the internal profile, external profile, and the root of the grooves. I've assumed that the wall thickness should scale down as the diameter reduces (what would happen if the sweep/guide curve procedure you were trying had 'worked'), but if you wanted constant wall thickness, then simply edit the 'master sketch' to use 'offset' curves from the outer diameter, rather than the curves shown.&lt;/li&gt;&#xA;&lt;li&gt;Revolve the inner area, the solid segment of the body through 360 degrees&lt;/li&gt;&#xA;&lt;li&gt;Revolve the outer area, the 'ribs', through 8 degrees (my geometry is an arbitrary guess, of course)&lt;/li&gt;&#xA;&lt;li&gt;Circular Pattern the rib revolve into 24 instances&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EK7Fb.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EK7Fb.gif&quot; alt=&quot;Demo Gif&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lo3Uf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lo3Uf.png&quot; alt=&quot;Finished Item&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-07-01T22:56:20.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22543" PostTypeId="1" CreationDate="2018-07-02T03:13:36.777" Score="1" ViewCount="40" Body="&lt;p&gt;Several questions, regarding traffic signal preemption at a grade crossing &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the preempt relay, which triggers the preemption sequence,&#xA;located inside the bungalow which houses the grade crossing&#xA;predictor, or is it located inside the cabinet which houses the&#xA;traffic signal controller?&lt;/li&gt;&#xA;&lt;li&gt;It is my understanding that the speed of an approaching train&#xA;determines the length of time that the preemption sequence will be&#xA;provided for. Is that true? And if so, does the crossing predictor&#xA;calculate the train's speed, determine how long the preemption&#xA;should be provided for and then tell the traffic signal controller&#xA;to provide it for that amount of time?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16630" LastEditorUserId="7083" LastEditDate="2018-07-02T22:20:49.833" LastActivityDate="2018-07-02T22:20:49.833" Title="Specific Questions about Traffic Signal Preemption at Grade Crossings" Tags="|rail|traffic-intersections|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22544" PostTypeId="1" AcceptedAnswerId="22548" CreationDate="2018-07-02T09:47:31.137" Score="0" ViewCount="452" Body="&lt;p&gt;I am trying to make this&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hQRWp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hQRWp.png&quot; alt=&quot;Goal&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to make the base feature but I am facing problems with the secondary features. So could anyone shed some light on how to make the datum planes on cones. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made one feature using this &lt;a href=&quot;https://i.stack.imgur.com/0KnhH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0KnhH.png&quot; alt=&quot;Plane&quot;&gt;&lt;/a&gt;&#xA;as was instructed from where I am learning.&#xA;However, I am unable to make the plane for the other feature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have made this &lt;a href=&quot;https://i.stack.imgur.com/TZrGq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TZrGq.png&quot; alt=&quot;Model&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="16633" LastEditorUserId="16633" LastEditDate="2018-07-02T10:03:23.350" LastActivityDate="2018-07-02T18:40:37.353" Title="How to make datum planes on a cone in Creo?" Tags="|mechanical-engineering|design|modeling|cad|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22545" PostTypeId="1" CreationDate="2018-07-02T11:09:29.813" Score="1" ViewCount="23" Body="&lt;p&gt;Is there a way to approximate the charge and discharge time of a PCM. I understand that this is a transient process as a material melts from the outside in, but I was wondering it there a means of estimating this time in order to determine the charge and discharge cycles. I am looking for a good source that shows different melting times, steady state calcs or published data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking at different vendors of these materials but their tech sheets rarely include this info. I am particularly interested in Eutectic salts if anyone has a good source on this sort of info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P&lt;/p&gt;&#xA;" OwnerUserId="14630" LastActivityDate="2018-07-02T11:09:29.813" Title="Phase Change Materials Charge and Discharge" Tags="|mechanical-engineering|materials|chemical-engineering|" AnswerCount="0" CommentCount="1" ClosedDate="2018-07-08T09:52:19.970" ContentLicense="CC BY-SA 4.0" />
  <row Id="22548" PostTypeId="2" ParentId="22544" CreationDate="2018-07-02T18:40:37.353" Score="1" Body="&lt;p&gt;You dont actually need a datum plane, you can do it with a revolve. But yes you can do this much more easily than you think. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When creating a plane click on he cone surface quilt, then change the option from &lt;em&gt;Through&lt;/em&gt; to &lt;em&gt;Tangent&lt;/em&gt;. Yeah the GUI is a bit sleazy the option is a drop down even if it has no affordance. Bad UX designer!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2mjxC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2mjxC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then add another constraint and choose a plane or line that defines the rotation around the central axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TnKvN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TnKvN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And finally&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Op9w2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Op9w2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also note extrusions can define different extrusion values to both directions so no need to create the second set of planes offset (or the extra axes for that matter). Personally I would make the planes internal to the feature creation as it clutters my view and model tree less.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2018-07-02T18:40:37.353" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22549" PostTypeId="2" ParentId="15042" CreationDate="2018-07-02T19:27:34.247" Score="1" Body="&lt;p&gt;Outside of the obvious (different R value, Cd, Omega), you get a system that allows for higher ductility and thus less shear transfer requirement. Reinforcement requirements for the intermediate wall is noted in TMS to adhere to section 1.17.3.2.3.1 and also not exceed vertical reinforcement spacing of 48&quot; o.c. (section 1.17.3.2.5).&lt;/p&gt;&#xA;" OwnerUserId="16636" LastActivityDate="2018-07-02T19:27:34.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22550" PostTypeId="1" CreationDate="2018-07-02T20:47:05.940" Score="1" ViewCount="53" Body="&lt;p&gt;I'm looking through this &lt;a href=&quot;https://www.velmex.com/Downloads/Documents/Velmex%20catalog.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;catalog&lt;/a&gt; at the motorized rotary tables and it says on page 114 that the maximum input shaft torque is 150 oz-in.The rotary table doesn't have a shaft so I'm confused as to how I should interpret this. Does it just mean that the maximum torque exerted by the motor is 150 oz-in or is there something else I need to be considering?&lt;/p&gt;&#xA;" OwnerUserId="16433" LastActivityDate="2018-07-03T12:13:06.180" Title="Understanding input shaft torque for a motor without a shaft" Tags="|motors|torque|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22551" PostTypeId="2" ParentId="22532" CreationDate="2018-07-02T21:47:58.627" Score="4" Body="&lt;p&gt;The reason turns out to be that type bar typewriters are faster. A type wheel is much heavier than a single type bar and needs to both rotate and move forward, so operating it with a finger requires more effort. For an electric typewriter that is not a problem as the motor provides the additional force. See &lt;a href=&quot;https://history.stackexchange.com/a/46955/32441&quot;&gt;this answer&lt;/a&gt; for more details and references.&lt;/p&gt;&#xA;" OwnerUserId="12424" LastActivityDate="2018-07-02T21:47:58.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22552" PostTypeId="2" ParentId="22536" CreationDate="2018-07-02T22:05:45.157" Score="0" Body="&lt;p&gt;It's very inaccurate to measure for emissions of a vehicle passing by, in a non hermetically closed environment. The slightest breeze will make the measurements incoherent. Depending on the sensor and type of emission, you may not even detect it. Ideally, you measure in the exhaust of the vehicle, which can be considered a closed environment. For emission tests, they accumulate exhaust gases in 'bags' and then measure them, either accurately diluted with air or not. Constant volume flow systems are also used. Google for them to get an idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding emissions, most electrochemical sensors are inferior to NDIR sensors, and most of them get consumed by measuring. They have limited shelf and usage life, and measurements can be spoiled by cross interfering gases. For instance, a CO sensor may see CO2 as CO, and thus reports a higher CO concentration than in reality. This also happens with NDIR sensors though. Using them directly in the exhaust gives them a too concentrated sample, and they'll get contaminated with sooth very quickly. So find a way to dilute the exhaust gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sulphur based emissions are barely present, and thus hard to detect. NOx is also rather hard to detect, HC and CO are easily detected, CO2 is really easily detected because of it's abundance in exhaust gas.(roughly 15%) I'd recommend you to look into the typical emission concentrations in exhaust gas. It may give you a better idea about what and how to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to compare various fuels or vehicles, make sure they're operating in the exact same way, so the fuel or vehicle is the only thing that makes the difference. Otherwise the tests are meaningless.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-07-02T22:05:45.157" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22553" PostTypeId="2" ParentId="22530" CreationDate="2018-07-02T22:22:08.707" Score="0" Body="&lt;p&gt;With just two normal gears, the direction of rotation can be reversed. That's not your major hurdle. Please understand that when using a diff, one engine can only bring you to half the speed, in comparison to using both engines. That's more of a problem, eg. when choosing the total gearing ratio's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK, the Prius uses a planetary gear system to achieve what you're trying. It may be a better option. Otherwise, you could just have one main gear to the transmission, where two gears are driving it, one for each engine, and disconnectable from the main gear when one engine is shut off. The engines will run synchronised though, so the throttles have to be connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to go the Citroën 2CV way, use one engine for the front wheels, the other for the rearwheels. It has two engines, two transmissions, two fuel tanks... Typical French invention..&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-07-02T22:22:08.707" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22554" PostTypeId="2" ParentId="22550" CreationDate="2018-07-02T23:28:39.810" Score="0" Body="&lt;p&gt;This is a complex and apparently mill spec piece of machinery. I would recommend you check with manufacturer. But as a first guess it says its out-put torque and speed are those limits. Now if in the previous sections it has defined other restrictions on configuration and placement of the load or its geometry, it has to be checked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually they set up a short course with appropriate checklists for the operators of these type of machinery. You'd want to look into it.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-07-02T23:28:39.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22555" PostTypeId="1" CreationDate="2018-07-03T01:44:33.297" Score="0" ViewCount="82" Body="&lt;p&gt;I seem to be having an issue with my little vacuum chamber I am building where I can only reach a base pressure of 6,000mTorr which is pretty high and am trying to figure out what is wrong. I have a new, clean viton gasket; a brass base plate, I have tried thick plastic and glass chambers. All the ports are sealed with either teflon tape and/or dow corning vacuum grease. I am using just water valves with the internal ball and rubber seals to vent or pump the chamber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions on what is wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4xTYi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4xTYi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15332" LastEditorUserId="282" LastEditDate="2018-07-03T21:29:24.347" LastActivityDate="2018-10-02T11:12:49.323" Title="Vacuum Pump Issues" Tags="|vacuum|vacuum-pumps|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22556" PostTypeId="1" CreationDate="2018-07-03T03:14:44.347" Score="1" ViewCount="29" Body="&lt;p&gt;Our setup &lt;a href=&quot;https://www.youtube.com/watch?v=uKuzshqjkYA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=uKuzshqjkYA&lt;/a&gt; &#xA;The diesel generator exhaust is routed via our test setup. (notice the difference between ON/OFF states). Generator in test is similar spec as a 20kVA &lt;a href=&quot;https://www.americasgenerators.com/20-kVA-Perkins-Diesel-Generator-Non-EPA.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.americasgenerators.com/20-kVA-Perkins-Diesel-Generator-Non-EPA.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a project i've been pursuing for the past few months. This is to capture fumes from particle laden gas sources. (A lot of generators in developing countries aren't maintained properly, and their soot traps don't work too well). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows nanoparticles/pm 2.5 to be captured and segregated much more efficiently, without using any mechanical obstruction. We've also put in a self cleaning mechanical assembly inside to clean the collected particulate matter continually.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a lab scale build. Would like to understand how to go about building larger versions of this setup that can work on generators much larger which are a major source of urban pollution such as &lt;a href=&quot;https://www.hindustantimes.com/gurugram/gurugram-will-get-new-air-quality-monitoring-station-to-keep-eye-on-pollution/story-xWgfUGHgLRLWjJD1FmLatJ.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.hindustantimes.com/gurugram/gurugram-will-get-new-air-quality-monitoring-station-to-keep-eye-on-pollution/story-xWgfUGHgLRLWjJD1FmLatJ.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;What to keep in mind while scaling our mini setup? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Should we attempt it ourselves, our rather talk to companies who might be interested in something like this and already comfortable with large fabrications.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We want to make 4-5 such setups and install it for a small pilot in Southern India where generators are used heavily.&lt;/p&gt;&#xA;" OwnerUserId="12071" LastActivityDate="2021-09-23T07:06:02.467" Title="Fabrication strategy to build larger version of our ESP" Tags="|manufacturing-engineering|power-electronics|diesel|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22557" PostTypeId="2" ParentId="16836" CreationDate="2018-07-03T06:55:27.100" Score="0" Body="&lt;p&gt;Go to a home &amp;amp; garden supply store that sells replacement pushbutton piezoelectric igniter units for propane BBQ grills and buy one. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-07-03T06:55:27.100" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22558" PostTypeId="2" ParentId="16836" CreationDate="2018-07-03T11:46:30.360" Score="1" Body="&lt;p&gt;How about a car ignition coil and sparkplug?  They are intended to do almost exactly what you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ignition coil is really a auto-transformer.  You apply 12 V between the tap and the low end (the primary).  This builds up current over a few milliseconds.  When you suddenly interrupt that current, the energy in the magnetic core is released by forcing current out the top of the coil (the secondary).  That can be 10s of kV if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd keep the ignition coil near the sparkplug, like it is in a car.  The two wires to the ignition coil primary can be long, with the power source and switch at the remote end.  All you need at the remote end is a 12 V source and a pushbutton that can handle the current of 12 V divided by the DC resistance of the ignition coil primary winding.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-07-03T11:46:30.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22559" PostTypeId="2" ParentId="22550" CreationDate="2018-07-03T12:13:06.180" Score="1" Body="&lt;p&gt;Seems like the maximum allowable input torque to the rotary table is 150 oz-in. From the catalog description, tables come standard with hollow shafts and mounting for NEMA 23 stepper motors. Pretty much, make sure the motor you install does not exceed the torque and RPM specs allowed by the rotary table.&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-07-03T12:13:06.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22562" PostTypeId="1" AcceptedAnswerId="22576" CreationDate="2018-07-03T13:54:18.537" Score="1" ViewCount="467" Body="&lt;p&gt;My lab is evaluating the liquid, plastic, and shrinkage limits of some soil samples and we are following &lt;a href=&quot;https://youtu.be/EcXJ961qjGA?t=3m31s&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video provided by the Missouri University of Science and Technology&lt;/a&gt;. The video indicates we should conduct at least 3 tests where it takes 15-20, 20-25, and 25-30 hits of the Casagrande Apparatus to close the gap that we make via the grooving tool. And this video also explains that the liquid limit is defined as the moisture content in which it takes 25 hits of the Casagrande Apparatus to close the divide made by the grooving tool.  We have been doing our best to turn our soil sample into a &quot;uniform paste&quot; through adding water as the video indicates, and even when that is the case, on a consistent basis, it takes fewer than 10 hits for the gap to be closed. &lt;strong&gt;Bottom line: My question is could there be soil samples where their liquid limits are lower than what the Atterberg Limit and Casagrande Apparatus method could detect?&lt;/strong&gt; I do recognize the possibility of user error on our part, that we can continue to refine our technique on making the soil sample a &quot;uniform paste.&quot; Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="16645" LastEditorUserId="90" LastEditDate="2022-08-03T00:51:48.560" LastActivityDate="2022-08-03T00:51:48.560" Title="Liquid Limit of Soil through Atterberg Tests" Tags="|geotechnical-engineering|soil|soil-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22563" PostTypeId="1" AcceptedAnswerId="22570" CreationDate="2018-07-03T14:33:09.767" Score="2" ViewCount="427" Body="&lt;p&gt;I am interested in the challenges that Solar Balloons face when flying.  For example, the high altitude could cause drastic changes in:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Wind speed&lt;/li&gt;&#xA;&lt;li&gt;Temperature&lt;/li&gt;&#xA;&lt;li&gt;Moisture&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How does the unique design of a solar balloon address these challenges?&lt;/p&gt;&#xA;" OwnerUserId="12800" LastEditorUserId="2015" LastEditDate="2018-07-03T21:41:34.510" LastActivityDate="2018-07-04T04:03:51.917" Title="How do solar balloons work at high altitudes?" Tags="|mechanical-engineering|thermodynamics|measurements|aerospace-engineering|instrumentation|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22564" PostTypeId="2" ParentId="14748" CreationDate="2018-07-03T14:37:21.243" Score="0" Body="&lt;p&gt;Regarding efficiency:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Fit ceramic bearings. There are many videos on YouTube which demonstrate how much less friction these bearings provide.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you have a large budget; Make use of active magnetic bearing technology which provides zero friction as the shaft effectively hovers within the bearing. You will also need to factor in the power usage of the active magnetic bearing controller into this, as such, this would more likely pay off in a large motor setup. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Place the motor within a strong vacuum. This will eliminate drag. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use a permanent magnet motor which is approximately 20% more efficient than an exciter based motor.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Regarding speed:&#xA;You could increase the voltage within tolerance which should increase the RPM speed. For example, if you run 220V 50hz motor, you could supply 240V or 250V to get a slight boost in speed. Note, that the motor would draw more current under load and you could damage the motor.&lt;/p&gt;&#xA;" OwnerUserId="16443" LastActivityDate="2018-07-03T14:37:21.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22565" PostTypeId="1" AcceptedAnswerId="22586" CreationDate="2018-07-03T14:38:44.373" Score="1" ViewCount="3107" Body="&lt;p&gt;I have a typical small electric induction motor given to me in pieces. &#xA;It has two &lt;strong&gt;wave washers&lt;/strong&gt; which I am told should sit between the bearing and the assembly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure which bearing needs the wave washers.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do I place on one wave washer before the front and rear bearings ?&lt;/li&gt;&#xA;&lt;li&gt;Both wave washers on the rear bearing? &lt;/li&gt;&#xA;&lt;li&gt;Both wave washers on the front bearing?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please advise.&lt;/p&gt;&#xA;" OwnerUserId="16443" LastActivityDate="2024-03-22T07:10:32.363" Title="Do wave washers need to be on either end of bearings in double bearing induction motor" Tags="|motors|bearings|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22566" PostTypeId="1" AcceptedAnswerId="22567" CreationDate="2018-07-03T14:43:39.323" Score="0" ViewCount="179" Body="&lt;p&gt;Can someone help identifying the electrical connectors in this picture? These connectors are from an aftermarket igniter. Also where can one purchase replacement connectors. I have looked &lt;a href=&quot;https://www.digikey.com/en/resources/connectors/index&quot; rel=&quot;nofollow noreferrer&quot;&gt;connectors from digikey&lt;/a&gt;. Any help towards looking for replacement connectors is appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kmCXX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kmCXX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8055" LastEditorUserId="1586" LastEditDate="2018-07-03T21:29:43.723" LastActivityDate="2021-11-24T15:58:50.720" Title="How to identify electrical connectors?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22567" PostTypeId="2" ParentId="22566" CreationDate="2018-07-03T15:04:16.327" Score="1" Body="&lt;p&gt;These are known as “Spade Connectors”, or “Blade Terminals”. Here are some:&#xA;&lt;a href=&quot;http://www.electronix.com/insulated-female-push-terminal-awg-red-p-13642.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;INSULATED FEMALE PUSH ON TERMINAL 22-18 AWG RED&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="1586" LastEditDate="2018-07-03T21:29:35.887" LastActivityDate="2018-07-03T21:29:35.887" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22569" PostTypeId="2" ParentId="22563" CreationDate="2018-07-03T16:30:07.333" Score="1" Body="&lt;p&gt;As long as the density of the air inside the solar balloon is less than the density of the air outside the balloon, the balloon will stay afloat. This is the principle of buoyancy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding wind speed, if the balloon has been made strong enough wind will not be a issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise with atmospheric moisture. The moisture only affects the atmospheric air and will only change the density of the atmospheric air. Increases in atmospheric moisture will increase the density of the atmosphere which will assist buoyancy of the balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2011, French scientists &lt;a href=&quot;https://en.wikipedia.org/wiki/Solar_balloon&quot; rel=&quot;nofollow noreferrer&quot;&gt;launched a solar balloon from the French base in Antarctica&lt;/a&gt;. The balloon rose to an altitude of 14 000 m (46 000 ft). Rising from one cold location to an even colder location was not an issue.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2018-07-04T04:03:51.917" LastActivityDate="2018-07-04T04:03:51.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22570" PostTypeId="2" ParentId="22563" CreationDate="2018-07-03T16:54:52.833" Score="2" Body="&lt;p&gt;I think you are asking about how the blimp is maneuvered or functions? Wind currents are widely used and is not that turbulent not to cause damage unless you guide it into a lightning storm or hurricane even then it could get high enough to be safe. Here is a picture of the altitude and wind direction:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1sB9M.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1sB9M.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://earth.nullschool.net/#current/wind/isobaric/250hPa/orthographic&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://earth.nullschool.net/#current/wind/isobaric/250hPa/orthographic&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To change direction the blimp in many cases the blimp would only have to change altitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X02KD.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X02KD.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not Photo shopped Airship Yuanmeng.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MhUYl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MhUYl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: The helium in the balloon only needs to be heated to raise the altitude in some models. This model not built yet suggest to do this by having a transparent side of the balloon/blimp facing the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1B0Uo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1B0Uo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Stratobus_artiste.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://commons.wikimedia.org/wiki/File:Stratobus_artiste.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I deduct that a parabolic shape could be designed to heat an inner dome filled with a breathable oxygen/helium atmosphere supplying warm air to the passengers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of heating the hydrogen have an air compressor to compress the air to land and regulate pressure and use heat to gain more height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make the whole balloon transparent an put everything outside it that can be inside and make those part like the framework on the outside on the inside with a wider surface area that gets hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zvPLZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zvPLZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related: &lt;a href=&quot;https://space.stackexchange.com/questions/26429/can-air-pressure-be-accumulated-this-way-for-a-biodome-or-spacesuit&quot;&gt;https://space.stackexchange.com/questions/26429/can-air-pressure-be-accumulated-this-way-for-a-biodome-or-spacesuit&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-07-03T19:21:29.907" LastActivityDate="2018-07-03T19:21:29.907" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22571" PostTypeId="2" ParentId="22566" CreationDate="2018-07-03T17:01:22.983" Score="1" Body="&lt;p&gt;These connectors are also called Quick-Disconnect Terminals. Here is a link from Master Carr. &lt;a href=&quot;https://www.mcmaster.com/#spade-connectors/=1djynbs&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quick-Disconnect Terminals&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PknAm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PknAm.png&quot; alt=&quot;Quick Connects&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1586" LastActivityDate="2018-07-03T17:01:22.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22572" PostTypeId="1" AcceptedAnswerId="22575" CreationDate="2018-07-03T17:32:17.880" Score="0" ViewCount="415" Body="&lt;p&gt;I want to work out how much I can expand a piece of rubber tubing before it ruptures. The wall is 0.75mm and the ID 5mm. Hence I am working out the stress/strain from thick-walled cylinder theory using &lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Mechanics/Cylinders.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt; &lt;a href=&quot;http://courses.washington.edu/me354a/Thick%20Walled%20Cylinders.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;pages&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Lame's equations (derived on those pages), are the radiuses in the formula the original radiuses before expansion, or those resulting from expansion? Does the same go for the radius under investigation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;many thanks for your help&lt;/p&gt;&#xA;" OwnerUserId="11394" LastEditorUserId="282" LastEditDate="2018-07-03T19:48:15.643" LastActivityDate="2018-07-04T01:11:17.527" Title="Thick walled cylinder theory - Stress/strain in ballooned out tube" Tags="|stresses|elastic-modulus|strength|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22573" PostTypeId="1" CreationDate="2018-07-03T18:10:01.957" Score="1" ViewCount="37" Body="&lt;p&gt;I'm considering a topology optimization problem where the design domain is a rotating annulus (in $\mathbb{R}^2$). One of the adjoint equations of the Navier-Stokes system whose solution is to be used in the sensitivity analysis has a weak form with righthand side: $$\int_{\Gamma_{\text{out}}} \rho(r\times u) \cdot \omega)n\cdot v + \int_{\Gamma_{\text{out}}} \rho(r\times \omega \times r)\cdot \omega)n \cdot v + \int_{\Gamma_{\text{out}}} (\rho(\omega\times r)\bigotimes u)n\cdot v,$$ where $v$ is a test function, $r$ is &quot;perpendicular to the axis of rotation&quot; (I've assumed that this is a position vector, see below), $\omega$ is the angular velocity (which is constant, say $|\omega| = 500$), $\Gamma_{\text{out}}$ is the outflow boundary (the outer boundary of the annulus), $n$ is the outward normal to $\Gamma_{\text{out}}$ (i.e. $n(x) = \frac{x}{||x||}$), $u$ is the velocity computed by solving the Navier-Stokes equations, and $\rho$ is the mass density (scalar). I'm wondering how I can formulate this problem for a 2D problem. What I'm doing now is writing $$r(x) = \begin{bmatrix} x_1 &amp;amp; x_2 &amp;amp; 0\end{bmatrix}^\top,\\\omega(x) = \frac{500}{||x||}\begin{bmatrix} -x_2 &amp;amp; x_1 &amp;amp; 0\end{bmatrix}^\top,$$ and so on, treating every 2D vector as a 3D vector in the $xy$-plane, doing the computations by hand and then coding them explicitly (since of course the PDE software I'm using doesn't know how to interpret a cross product of vectors in $\mathbb{R}^2$ either). I'm wondering if this approach is correct or there's some other way of formulating this problem that I'm missing. I'm also not sure that I'm interpreting $r$ or $\omega$ properly, so if there's anything I'm missing there I'd be grateful for some feedback. I'm not an engineer or physicist by training, so I'm not aware of the conventions for these things!&lt;/p&gt;&#xA;" OwnerUserId="16649" LastActivityDate="2018-07-03T18:10:01.957" Title="How to formulate a 3D version of the Navier-Stokes equations in 2D for numerical solution?" Tags="|numerical-methods|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22574" PostTypeId="2" ParentId="18710" CreationDate="2018-07-03T20:19:11.070" Score="1" Body="&lt;p&gt;Just count the cogs teeth on both sides that is your ratio. Keep dividing by 2. You can cross check it by counting the teeth on the housing and dividing it by the number of teeth on the cog or spindle. &lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-07-03T20:19:11.070" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22575" PostTypeId="2" ParentId="22572" CreationDate="2018-07-03T22:10:36.747" Score="2" Body="&lt;p&gt;A good place to start is Ogden's book &lt;strong&gt;Nonlinear elastic deformations&lt;/strong&gt;, section 5.2.3: &quot;&lt;em&gt;Extension and inflation of a circular cylindrical tube&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a material with a strain energy function:&#xA;$$&#xA;  W = F(I_1) - A I_2 + B I_3&#xA;$$&#xA;where $I_1$, $I_2$, $I_3$ are the three principal invariants of the stress tensor, Ogden shows that for the situation where a stretch $\lambda_3$ is applied along the cylindrical axis and there are no internal or external pressures, the final radius ($r$) of the cylinder is given by&#xA;$$&#xA;  r(R) = \tfrac{1}{2} (\alpha - \lambda_3) R + \frac{\beta}{R}&#xA;$$&#xA;where $R$ is the initial location of a point in the radial direction and&#xA;$\beta = 0$.  The term $\alpha$ is obtained by solving the equation&#xA;$$&#xA;  F'(\alpha) - A\alpha + \tfrac{1}{2}(A + B\lambda_3)(\alpha - \lambda_3) = 0&#xA;$$&#xA;The tube fails when $r(R_o) - r(R_i)$ reaches a critical value ($R_i, R_o$ are the initial inner and outer radii of the cylinder).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can follow the approach used by Ogden to solve the problem for other strain energy functions, boundary conditions, and constraints such as incompressibility.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-07-03T22:10:36.747" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22576" PostTypeId="2" ParentId="22562" CreationDate="2018-07-03T22:26:40.690" Score="1" Body="&lt;p&gt;You have to run the test at various moisture contents.  Plot the number of hits vs. moisture content and then extrapolate to 25 hits to find the moisture content at the liquid limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ASTM D4318 standard should be followed.  However, there is always some operator-dependence and an alternative is the cone penetration test described in BS 1377-1/2.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-07-03T22:26:40.690" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22577" PostTypeId="1" AcceptedAnswerId="22584" CreationDate="2018-07-03T22:54:05.970" Score="3" ViewCount="88" Body="&lt;p&gt;Sulfur concrete only uses sulfur as a binding agent then an aggregate. Why does it not use cement?&lt;/p&gt;&#xA;" OwnerUserId="16653" LastActivityDate="2018-07-04T03:03:36.930" Title="Why does sulfur concrete require no cement?" Tags="|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22578" PostTypeId="2" ParentId="22494" CreationDate="2018-07-04T00:21:11.767" Score="2" Body="&lt;p&gt;In Zimbabwe we experience flash flooding due to tropical thunderstorms. To prevent soil erosion, a major problem to farmers on sloping ground, channels are dug on roads diagonally across the surface to channel water away to drainage ditches that open up into fields of crops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method was proven most successful by Rhodesian farmers during the flood decade of the 1970's. The specific angle depends on the vertical gradient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The channel begins as a narrow 10cm deep v in the road centre and expands into a much deeper 1m wide V about 30cm deep to the roadside. The trailing edge angle of the cut is usually 30° to the horizontal. This method of alternating cuts in the surface slows down the surface flew of water and diverts it to specific drainage channels calculated to carry maximum capacity storm water+30%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this method helps you solve your problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am alternative drainage method is to create large diagonal speed humps that are backfilled with 12-15mm gravel stone across the drive way. The water flows downhill and then slows while being diverted away from the drive slope. This method is employed in many African game reserves on steep gradient dirt roads and is longer term more effective.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-07-04T00:21:11.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22579" PostTypeId="2" ParentId="22572" CreationDate="2018-07-04T01:11:17.527" Score="1" Body="&lt;p&gt;According to J. E. Gordon's book &quot;Structures or why things don't fall down&quot; he explains The concept of Work of Fracture. During his exposition he explains that cylindrical shapes have a specific relationship of tensile strength to the thickness of the cylinder weigh is what a tube is. The general rule is that the tensile strength of any material is determined by the surface area perpindicular to the plane in which the strain exists multiplied by the length of the cylinder. On page 77 he states rubber has about 300% working strain. On p122 a diagram displays an image of circumferential stress in a cylinder with the formula of $s_2=\frac{rp}t$ where s=stress, r=radius, p=pressure and t=cylinder thickness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In almost all cases where internal pressure exceeds the strain capacity of the cylinder, a longitudinal rupture occurs. Since your question asks about rubber specifically there is another issue in that the cylindrical thickness decreases with pressure increases proportionally to the radial expansion. In simple terms, the rubber is thinnest where the diameter is greatest but the strain energy is identical throughout the tube because the internal pressure is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, the thickness of the tube is 0.75mm with an internal diameter of 5mm. Since we know rubber had approximately 300% working strain, the pressure required to break it must therefore be&#xA;$$ s=\frac {rp}t$$&#xA;$$p=\frac {st}{r}$$&#xA;$$p=\frac {3 \cdot 0.75}{5}$$&#xA;$$p=0.45$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So pressure must exceed $0.45 units of pressure per {mm^2}$ to rupture the tube by which the length of the unexpanded section to be expanded must be multiplied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gqDNT.jpg&quot; alt=&quot;book cover&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book is a great read and humorous in many places. As well as being an invaluable source of education, it is a recommended course book by the university of Wolverhampton for civil engineering honours students.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-07-04T01:11:17.527" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22580" PostTypeId="2" ParentId="22513" CreationDate="2018-07-04T01:32:01.297" Score="0" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hKPGy.jpg&quot; alt=&quot;An excellent research book&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above book ought to educate you on structural materials whilst giving you the mathematical resources to complete your research. It covers materials such as bone, tendon, muscle, steels, rock and polymeric origin types also. It is not nearly as dry as you might expect from a non-fiction book and worth the effort. It explains tensile stresses, strain, Work of Fracture, energy storage capacity of materials etc. You really ought to be able to find many answers derived from Mr Gordon's knowledge. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-07-04T01:32:01.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22581" PostTypeId="2" ParentId="22520" CreationDate="2018-07-04T02:15:48.510" Score="0" Body="&lt;p&gt;You asked if it is practical. The answers will be subjective because it depends on the pal solar performances in relation to the latitude in which the work is being done and under what weather conditions allow it. Whilst mirrors seem obvious, focal distances are imperative and relative to surface curvature of the reflective surface. In my engineering view, sustaining a temperature in excess of 1746°C will require a flat mirror surface area of about 400 {m^2} or an equal parabolic surface area of that divided by the depth of the parabola. A Fresnel lense of 1 square meter can transmit enough energy to heat a local object up to 800°C but that takes time. You only have 120° of daylight and of that 45° is optimum power meaning 3 hours of sunlight a day IF weather conditions are clear . It may take a couple hours to obtain smelting temperatures but sustaining them is going to be the challenge. You may be better off using solar to generate power for arc furnace use through parabolic solar heated steam turbines instead, in my opinion. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-07-04T02:15:48.510" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22582" PostTypeId="2" ParentId="22493" CreationDate="2018-07-04T02:52:24.040" Score="0" Body="&lt;p&gt;I think that @Donald Gibson answered well. There is no such thing as &quot;will not be able to move&quot; in engineering because everything moves; we just cannot humanly observe those movements in most cases without specialist equipment. Plastic deforms substantially but its elastic modulus returns it to near-original shape whilst marginally deforming permanently under each repetitive stress action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cylindrical bars may seem a good idea but your problem is the variable average 19.6133N static force. When that carriage stops dead centre, the inertia will cause a moment of force to develop in excess of the gravitational force of a static object. To counter that, an I beam support would be better and must be in permanent tension to counter deflection in dynamic operation or the supports must be designed as curved rails with some sort of kinetic dampers to absorb inertial acceleration &amp;amp; deceleration forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not unusual for printer heads to travel up to 30 m.s. if that is combined with gravity, then maybe you need to consider a vertical arrangement instead of horizontal to negate deflection issues better. Use of a third rail would also help somewhat if it acted as a moment energy deflection rail to transfer inertial energy laterally outside the scope of the two cylindrical support rails. Vibration is another issue that may cause harmonic distortions which is why a tubular section under compression may actually be better than a solid cylinder shape. Elliptical bars may actually be the best answer in the application of printer design since they are likely to resist deformation far more readily than cylindrical shapes while being able to withstand greater deflection over the same distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This kind of precision engineering is not going to be cheap which is why large printers are so expensive. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-07-04T02:52:24.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22583" PostTypeId="2" ParentId="22487" CreationDate="2018-07-04T03:00:17.753" Score="0" Body="&lt;p&gt;The first equation is a static system or an instant of time when the machine is being operated. Usually this would be as the machine starts at maximum force before centrifugal force begins to take effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second equation is when the machine is dynamically being operated and accounts for centrifugal force also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's my opinion. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2018-09-02T04:13:49.553" LastActivityDate="2018-09-02T04:13:49.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22584" PostTypeId="2" ParentId="22577" CreationDate="2018-07-04T03:03:36.930" Score="2" Body="&lt;p&gt;The molten sulfur is the &quot;cement&quot;.  I understand that hundreds of years ago sulfur concrete was occasionally used. In the 70's , Amoco experimented with sulfur + asphalt concretes. They had excellent properties but were too expensive.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-07-04T03:03:36.930" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22585" PostTypeId="2" ParentId="22477" CreationDate="2018-07-04T03:07:22.290" Score="0" Body="&lt;p&gt;I would use a stator motor with a sliding shaft key to lock position as needed. A ratchet gear system may not be a bad idea in this case to lock final position as long as fatigue is not going to be an issue longer term. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-07-04T03:07:22.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22586" PostTypeId="2" ParentId="22565" CreationDate="2018-07-04T05:34:36.973" Score="1" Body="&lt;p&gt;I would assume one wave washer for each bearing - this allows pre-load to limit any play and expansion control as it get hot in normal operation.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-04T05:34:36.973" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22589" PostTypeId="2" ParentId="22565" CreationDate="2018-07-04T07:42:18.897" Score="2" Body="&lt;p&gt;I intuitively agree with Solar Mike, there should be one washer per bearing. I wanted to find a source/reference, however:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image appears to show wave washers in front of each bearing &#xA;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/a/a9/Rotterdam_Ahoy_Europort_2011_%2814%29.JPG&quot; alt=&quot;motor cross section 1&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this image doesn’t appear to show any wave washers at all:&lt;img src=&quot;https://i.stack.imgur.com/PZWrE.jpg&quot; alt=&quot;motor cross section 2&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the short answer is: probably 1 washer per bearing, but it will of course depend on the exact motor design. We’ll need more information from you if you want us to comment further.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-07-04T07:49:10.587" LastActivityDate="2018-07-04T07:49:10.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22590" PostTypeId="1" AcceptedAnswerId="22595" CreationDate="2018-07-04T08:02:05.350" Score="0" ViewCount="567" Body="&lt;p&gt;I need to duct a PVC pipe (DN80) through a stainless steel sheet metal covering. The hole thing needs to be tight against stormwater. There's tons of products to duct pipes through thick walls, but I'm at loss on how to do that here. The cover is horizontal, upside checkered. Ideas I've considered:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Welding a short stainless steel pipe stub onto the sheet and use a ring seal&lt;/li&gt;&#xA;&lt;li&gt;Hole with ID of pipe in steel sheet, flange the pipe through the sheet&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Both of these appear overblown to me.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-07-04T12:39:41.663" Title="Simple way to duct a PVC pipe through a stainless steel sheet cover" Tags="|civil-engineering|pipelines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22591" PostTypeId="1" AcceptedAnswerId="22594" CreationDate="2018-07-04T09:54:50.260" Score="0" ViewCount="237" Body="&lt;p&gt;I have a large and unwieldy Solidworks assembly and I would like to save it as an &quot;inert&quot; geometry file such as an STL or STEP format.  Then I want to import this into another project assembly and move it into a specific location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that I am unable to move the STL version from its default insertion point in the new project.  I have tried opening the STL file and saving it as a SLDPRT file and bringing this into the assembly but I am unable to move that either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I want to do this is so that there are no dependency links between one project and the other, in order to avoid any possibility of design corruption while I'm experimenting with the new assembly.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In a nutshell, how can I import and position an inert STL (or STEP or whatever) into an existing SLDASM file and be able to move it around?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastActivityDate="2018-07-04T13:39:01.343" Title="Solidworks 2014 - Convert a complex assembly into a single &quot;inert&quot; geometry file and import it" Tags="|solidworks|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22592" PostTypeId="1" CreationDate="2018-07-04T11:12:17.940" Score="2" ViewCount="28" Body="&lt;p&gt;I have a small shaft, 2.5m wide by 4m deep, sunk into the ground. Due to explosion protection I should ensure &quot;natural ventilation&quot; - this is the termgiven in the relevant ordinance, no air flow specified. As the shaft is sunk into the ground, in the summer months it will always be colder than the air above. I see no way to ensure passive ventilation, so I will need a small blower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or am I wrong and there is way to ensure natural ventilation?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-07-04T11:12:17.940" Title="How to ensure natural ventilation through a shaft?" Tags="|civil-engineering|building-physics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22593" PostTypeId="2" ParentId="22591" CreationDate="2018-07-04T11:51:56.433" Score="0" Body="&lt;p&gt;&lt;strong&gt;EDIT: I misread your question, and the below instructions are for inserting into an existing .SLDPRT file, not a .SLDASM file as requested - I'm leaving it visible for reference, as I believe the information is still useful. I have submitted a separate answer to your actual question!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you insert the part (Insert --&gt; Part), you can either check the box labelled &quot;Locate part with Move/Copy feature&quot;, or simply use the Move/Copy Bodies Tool (Insert --&gt; Surface --&gt; Move/Copy) to move the object afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also choose to break the link to the part that you inserted at this point, so for a .SLDPRT file, you don't need to do any conversion. For assemblies, you will need to save as a .SLDPRT. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure you select &quot;All Components&quot; while saving, if you want all the data to be present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4g7hWm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4g7hWm.png&quot; alt=&quot;All components option&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WuF2sm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WuF2sm.png&quot; alt=&quot;Insert Part Move Copy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once in the Move/Copy Tool, you can click the &quot;Constraints&quot; button to get a rudimentary version of the Assembly Constraints environment, but, I'd stick to translation where possible, since this then will not be affected by geometry changes, giving fewer errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CUGLVm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CUGLVm.png&quot; alt=&quot;Constraints Button&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-07-04T12:03:33.567" LastActivityDate="2018-07-04T12:03:33.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22594" PostTypeId="2" ParentId="22591" CreationDate="2018-07-04T12:02:30.960" Score="0" Body="&lt;p&gt;In order to insert a 'dumb' copy of an assembly into another higher level assembly, you can save the assembly as a .SLDPRT File. Make sure to select the correct amount of geometry to save, according to your needs from the options in the 'Save As' dialogue.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4g7hW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4g7hW.png&quot; alt=&quot;Save as Part Options&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This .SLDPRT will be completely independent from the original assembly, being made up of internally saved .IGES components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then simply insert this new part into your high level assembly, and mate it as normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B., if you ever use the 'Edit Component' Function inside the Assembly environment, you can toggle &quot;No External References&quot;, which will prevent you from accidentally creating references to the 'dumb assembly', e.g. by accidentally snapping a sketch point to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NVN4a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NVN4a.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-07-04T13:39:01.343" LastActivityDate="2018-07-04T13:39:01.343" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22595" PostTypeId="2" ParentId="22590" CreationDate="2018-07-04T12:39:41.663" Score="2" Body="&lt;p&gt;The way this would typically be achieved, is with an off-the-shelf &quot;Tank Connector&quot;. This essentially replaces the welding operation in your first idea. Simply cut the appropriately sized hole, and install the connector. There are face seals on both sides of the sheet metal, which are clamped tight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n1Tzw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n1Tzw.png&quot; alt=&quot;Tank Connector&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've shown a brass connector, since I thought it was the clearest image - but needless to say, these are available in PVC, Stainless Steel, and with any number of diameters or Male/Female/Etc. configurations, according to your exact needs.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-07-04T12:39:41.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22596" PostTypeId="1" CreationDate="2018-07-04T15:15:15.183" Score="0" ViewCount="117" Body="&lt;p&gt;I have been looking a lot into shape memory polymers and alloys recently and have been wondering wether they have an application in headphones. By this I mean that when plugged into the phone the warming current is sufficient to ‘untangle’ them back to their original, straight shape, avoiding the time consuming process of manually untangling the knot. Would this be possible, or is the current provided by the headphone jack insufficient. Any ideas would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="16657" LastActivityDate="2018-07-04T15:58:17.990" Title="Shape memory alloy for headphones" Tags="|electrical-engineering|materials|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22597" PostTypeId="2" ParentId="22596" CreationDate="2018-07-04T15:58:17.990" Score="1" Body="&lt;p&gt;This would not work, simply because SMA's cannot untie themselves. If you bend a SMA wire, then it can return to 'straight' (within reason, the return is rarely perfect), but if you tie a knot in the same wire, it's prevented from becoming straight by the knot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Headphones can become very tangled with multiple loops crossing/interweaving - the fact that the power that can be provided by an average headphone jack is not sufficient (but it could be augmented by an in-headphone battery pack, I suppose), isn't the biggest problem with this idea...&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-07-04T15:58:17.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22598" PostTypeId="1" CreationDate="2018-07-04T17:09:04.787" Score="1" ViewCount="1456" Body="&lt;p&gt;I'm trying to build a hovercraft of sorts, and I thought I'd take advantage of the ground effect. I've done some searches, but I can't seem to find any guides on the actual physics. How do I go about calculating the additional lift my hovercraft will receive from the ground effect?&lt;/p&gt;&#xA;" OwnerUserId="16660" LastActivityDate="2019-06-02T13:01:55.223" Title="How to calculate added lift from ground effect" Tags="|airflow|compressed-air|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22599" PostTypeId="2" ParentId="22598" CreationDate="2018-07-04T18:09:24.320" Score="1" Body="&lt;p&gt;Hovercraft lift is a totally different dynamics then the airplane lift. Hovercraft basically rides on a cushion of high pressure air trapped in a skirt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ground effect is due to the earth constraining free flow of air stream aft the wing, forcing it to build up extra pressure and extra resistance, and hence providing extra lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically you would need to fly your Hovercraft much faster like the range of 80kt, and then design its aerodynamics like a flying saucer or a teardrop shape, and test it in a wind tunnel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be possible to come up with a low altitude hybrid flying a few inches above ground. But at those speeds it would need a flat runway or smooth surface of a lake. It can not possibly navigate the bumps and natural obstacles as hovercraft can.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-07-05T20:19:17.863" LastActivityDate="2018-07-05T20:19:17.863" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22601" PostTypeId="1" CreationDate="2018-07-04T20:13:40.370" Score="0" ViewCount="476" Body="&lt;p&gt;I have two parallel shafts, 3 millimetres in diameter and offset by 8 millimetres, both anchored in surrounding housing by ball bearings. When one shaft is rotated (via gear), I want the other shaft to rotate with it. I have less than a cubic centimetre of space between the shafts in which to place the coupling mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have been using a stiff orthodontic elastic looped around high-friction rollers fixed onto the shafts. However, this introduces backlash, resistance, and unpredictability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A timing belt would work well but the pitch length would have to be no greater than 1.07&quot;, which is smaller than any timing belt I can find on the market.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions for coupling these shafts precisely in a space- and preferably cost-efficient way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; I believe the offset is too large for Oldham coupling. Here's an image of the shafts without surrounding housing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6ckk6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6ckk6.png&quot; alt=&quot;Shafts 8.79 mm apart&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13317" LastEditorUserId="13317" LastEditDate="2018-07-05T13:18:00.273" LastActivityDate="2018-07-05T13:18:00.273" Title="How can I couple two small parallel shafts offset a short distance from each other?" Tags="|mechanical-engineering|gears|manufacturing-engineering|pulleys|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="22602" PostTypeId="2" ParentId="22601" CreationDate="2018-07-04T22:13:25.227" Score="2" Body="&lt;p&gt;The first option is Oldham coupling, you say &lt;strong&gt;' I have less than a cubic centimetre of space between the shafts in which to place the coupling mechanism.',&lt;/strong&gt; so i recommend Oldham coupling. They require less space in compare to the other options. Another name for Oldham is cross disc coupling (German). Friction is unavoidable, so you should consider a way to lubricate it now and then. This type of coupling is way old and they are not popular anymore in industry.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second option is Schmidt coupling, i don't recommend it here, because they demand much more room than Oldham coupling, and they are designed to compensate extreme radial deviations think about rolling machines.There are also some issues(start up of machines), that make the application of this type of coupling less interesting.    &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-07-04T22:13:25.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22603" PostTypeId="1" CreationDate="2018-07-05T02:47:14.140" Score="1" ViewCount="41" Body="&lt;p&gt;I need to measure the width of a bar made out of glowing red steel (900-1200 degrees Celsius) using computer vision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to ensure resolution of a 0.5 millimeters/px, the camera is needed to have enough pixels and, considering the field-of-view, to be at a certain distance from the bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody think of a way to use a camera at a distance of 0.5m from the glowing hot bar? Air refrigeration? Refractory enclosure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's talk about it?&lt;/p&gt;&#xA;" OwnerUserId="16663" LastEditorUserId="90" LastEditDate="2018-07-05T09:02:49.637" LastActivityDate="2018-08-04T18:01:39.500" Title="How to use consumer electronics for industrial application on high temperature" Tags="|thermodynamics|consumer-electronics|microelectronics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22604" PostTypeId="1" CreationDate="2018-07-05T05:30:53.033" Score="3" ViewCount="55" Body="&lt;p&gt;I was talking to someone about brickmaking and he said doing a &quot;hot casting&quot; of a brick requires more energy and there are alternatives that use less energy. What does that mean?&lt;/p&gt;&#xA;" OwnerUserId="16653" LastActivityDate="2018-07-05T05:30:53.033" Title="What does it mean to do a &quot;hot casting&quot; of a brick?" Tags="|civil-engineering|" AnswerCount="0" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="22605" PostTypeId="1" AcceptedAnswerId="22607" CreationDate="2018-07-05T09:34:29.580" Score="0" ViewCount="1046" Body="&lt;p&gt;I have 2 files in prt, and I need to render this parts but my software can't open this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone export this files for me to step?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.dropbox.com/s/m9ysajdp6vo89od/pokretlo1.prt.1?dl=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.dropbox.com/s/m9ysajdp6vo89od/pokretlo1.prt.1?dl=0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.dropbox.com/s/nkgqe9scxhjqsw0/pokretlo2.prt.1?dl=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.dropbox.com/s/nkgqe9scxhjqsw0/pokretlo2.prt.1?dl=0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16665" LastEditorUserId="9" LastEditDate="2018-07-05T10:51:52.503" LastActivityDate="2018-07-05T10:51:52.503" Title="Can someone export file from prt.1 to step?" Tags="|software|" AnswerCount="1" CommentCount="0" ClosedDate="2018-07-06T12:27:12.853" ContentLicense="CC BY-SA 4.0" />
  <row Id="22606" PostTypeId="1" AcceptedAnswerId="22609" CreationDate="2018-07-05T10:06:28.777" Score="1" ViewCount="2133" Body="&lt;p&gt;I hope this is the right place to ask since I couldn't find a control systems SE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say we have the transfer function $P(s) = \frac{1}{s(s+15)(s+10)} = \frac{1}{s(s^2+25s+150)}$(completely made up numbers) of some sort of device. This is without any controller or feedback attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I look at the $\frac{1}{s}$ that's in there? Do I treat it as a separate multiplier or do I handle this as a third order system? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without the integrator this is a simple 2nd order system with clearly defined relative damping and natural frequency because of $\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically I could treat the 1/s as an I controller with the value of $P_i$ value of 1. Would this system even have a steady state error with a simple unity P feedback? Technically it would automatically be a PI controlled system right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the questions don't seem clear it's because I don't really know what to ask here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: I managed to read something about types of a transfer function(0, 1 or 2) which only modifies the error you will have. The transfer function shown here would be considered a type 1 because it has 1 pole in the origin(2 means type 2 and 0 means type 0 I guess). Type 1 means that the steady state error will be 0 for a step input, which confirms my speculation about the error. I still don't know if it has any other effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit2: &lt;a href=&quot;https://i.stack.imgur.com/XoEhU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XoEhU.png&quot; alt=&quot;equivalent&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this picture shows the exact same thing in 2 forms. They're are functionally the same. The PID controller is C(s) in this case. Well it's a PI controller with K = 1, without me actually adding anything to the system. I've just added a negative feedback loop.&lt;/p&gt;&#xA;" OwnerUserId="16666" LastEditorUserId="16666" LastEditDate="2018-07-05T14:35:16.193" LastActivityDate="2018-07-05T14:35:16.193" Title="Order of a transfer function and its properties" Tags="|control-engineering|transfer-function|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22607" PostTypeId="2" ParentId="22605" CreationDate="2018-07-05T10:40:36.420" Score="2" Body="&lt;p&gt;You can remove the &quot;.1&quot; from the extension, so they are just .prt files - the .1 is to do with sequential numbering of backups, I believe. They can then be opened / converted using Autodesk Fusion 360, which is free for non-commercial use.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-07-05T10:40:36.420" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22608" PostTypeId="1" AcceptedAnswerId="22682" CreationDate="2018-07-05T10:57:49.583" Score="1" ViewCount="322" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tham_Luang_cave_rescue&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tham Luang cave rescue&lt;/a&gt; incident is taking combined effort of Engineers throughout the world. Various news media is telling currently it is not possible to pass more than one person through some very narrow portions of the cave tunnel (so the boys have to learn diving within cave). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XBVwx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XBVwx.png&quot; alt=&quot;BBC MAP&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a Map by BBC, perhaps from &lt;a href=&quot;https://www.bbc.com/news/world-asia-44695232&quot; rel=&quot;nofollow noreferrer&quot;&gt;here of BBC&lt;/a&gt;; showing the narrow places in cross sections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now; my question is&lt;/strong&gt;; why the engineers are &lt;strong&gt;not attempting&lt;/strong&gt; to use &lt;a href=&quot;https://en.wikipedia.org/wiki/Abrasion_(mechanical)&quot; rel=&quot;nofollow noreferrer&quot;&gt;abrasion&lt;/a&gt; (with a &lt;a href=&quot;https://en.wikipedia.org/wiki/Grindstone&quot; rel=&quot;nofollow noreferrer&quot;&gt;grinding wheel&lt;/a&gt; or such thing) and rock cutting wheels to broaden these &quot;too narrow&quot; parts so that later on 2 people (a diver carrying a boy) can come out at a time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any hidden risk of attempting so? Is that the cause for not atempting so? &lt;/p&gt;&#xA;" OwnerDisplayName="user8831" LastEditorUserId="16486" LastEditDate="2018-07-11T09:25:17.837" LastActivityDate="2018-07-11T11:21:00.133" Title="Thai Cave Rescue- Why the narrow portions are not being broadened?" Tags="|civil-engineering|geotechnical-engineering|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="22609" PostTypeId="2" ParentId="22606" CreationDate="2018-07-05T12:49:20.100" Score="0" Body="&lt;p&gt;Okay, something indeed does not seem very clear to you.&#xA;Let me try to clarify that ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $P_s = \frac{1}{(s+15)(s+20)}$ denote the second order part you do understand. I don't know what you mean by type 2 system, but this can represent any mass-spring-damper system as you already stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where you go wrong is that you view the integrator $I = \frac{1}{s}$ as a controller. You recognized that $P = IP_s$, which is true. A controller, however, would introduce (negative) feedback, which is not present here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Controller interpretation - wrong!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let $Y$ denote the Fourier transform of the output of your plant and $E$ that of the error. (Without loss of generality we set the reference $R=0$).&#xA;Then: $$Y = P E = P(R-Y),$$ from which we can derive the well-known expression for the complementary sensitivity: $$T = \frac{Y}{R} = \frac{P}{1+P}.$$ (In literature, often $L$ is used instead to denote the open-loop transfer function $CP$, where $C$ is the controller, but let's keep using your notation instead.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$T = \frac{1}{s^3 + 25s^2 + 150s+1}$, is the real transfer function of your second order system with your integrator as negative feedback controller from input $R$ to output $Y$. Note that this system indeed has no steady state error as you correctly noted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Correct interpretation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to your system and the correct interpretation: the integrator is nothing more than a ''multiplier'' as you stated. A physical interpreation could be a valve controlling a water flow into a vat on an old-fashioned mechanical scale. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your input corresponds to the valve setting $u$. The output of the integrator, $Us$ or $\int u \mathrm dt$ in the time domain corresponds to the amount of water in a vat. Then your second order transfer function $P_s$ takes the amount of water as input (force induced by the weight in this case) and outputs movement of the scale $Y$. Where this analogy goes wrong is that $m$ obviously changes as the vat fills, and inertia of the water flow is ignored, but hopefully you get my point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another analogy would be an ideal capacitor in-between your input and what actually goes into your second-order system.&lt;/p&gt;&#xA;" OwnerUserId="16413" LastEditorUserId="16413" LastEditDate="2018-07-05T13:17:47.437" LastActivityDate="2018-07-05T13:17:47.437" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="22610" PostTypeId="1" CreationDate="2018-07-05T15:51:58.207" Score="1" ViewCount="7470" Body="&lt;p&gt;This question is regarding the safest way of installing a heavy punching bag of 100 lbs from the basement ceiling. Above the basement ceiling is a wood floor. The goal is to sleep soundly at night knowing that we don't have to worry about damaging the ceiling/floor structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;QUESTION 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is one of these methods, or another method, the best and safest for hanging and using a 100 lbs punching bag and not having to worry about it damaging the floor structure or the floor above over time?:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;(a)&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Adding 3 or 4 beams parallel between the existing basement ceiling joists&lt;/li&gt;&#xA;&lt;li&gt;Adding a perpendicular beam that will screw into those parallel beams (so that the weight is distributed amongst those beams)&lt;/li&gt;&#xA;&lt;li&gt;Attaching the ceiling mount &lt;em&gt;(Note 1)&lt;/em&gt; to that perpendicular beam&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;(b)&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Adding a beam parallel between the basement ceiling joists&lt;/li&gt;&#xA;&lt;li&gt;Attaching the ceiling mount &lt;em&gt;(Note 1)&lt;/em&gt; to that one parallel beam&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;(c)&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: The ceiling mount is designed to absorb the shock produced by hitting the bag. The bag hangs from the &lt;a href=&quot;https://rivalboxing.us/products/copy-of-rival-ceilling-mount-system-150-lbs&quot; rel=&quot;nofollow noreferrer&quot;&gt;ceiling mount&lt;/a&gt;, which is screwed into the ceiling beam.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;QUESTION 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to secure/install the parallel beams?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;QUESTION 3&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is that hanging anything from the existing joists, or even making holes in the existing joists, is best avoided so as not to weaken them or tamper with the existing floor structure. Is this a valid assumption?&lt;/p&gt;&#xA;" OwnerUserId="16669" LastEditorUserId="1832" LastEditDate="2018-07-05T17:04:43.697" LastActivityDate="2018-07-06T02:40:59.677" Title="Ceiling mounted punching bag without compromising structure or floor above" Tags="|structural-engineering|civil-engineering|beam|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22611" PostTypeId="2" ParentId="22603" CreationDate="2018-07-05T17:26:56.913" Score="1" Body="&lt;p&gt;Do you need a single camera, or could you build a precision mount with 2 smaller cameras, each covering one end of the bar?  Saves you a ton of pixels.  Or, for that matter, a single camera with an external  set of path-folding mirrors &amp;amp; prisms so that what the camera &quot;sees&quot; is just the ends of the bars.  This is a common approach taken in various laboratory work when only two small non-adjacent regions are of interest. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-07-05T17:26:56.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22613" PostTypeId="2" ParentId="22610" CreationDate="2018-07-05T19:12:18.790" Score="1" Body="&lt;p&gt;Its good that you are thinking through a structure change like this. The world would be much safer if people took the time to ask. That said, I would hope that the floor structure has enough safety factor to handle a small load like a 100lb punching bag. House floors are designed to handle water beds, refrigerators, and other large loads. The 100lb punching bag would easily fall under the live loading the floor was designed for. Remember that 100lb is a very small load as far as your house is considered. Most likely you do not need to add anything to the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your house is very old (1950 or older) and had questionable if any engineering, you may want to have the structure evaluated (bigger problem than just the punching bag). If issues are found, you would want to have the whole structure evaluated as the roof would probably even have bigger problems. You would need to hire a licensed Structural Engineer to do this evaluation for you. If you are concerned, I recommend you submit a new question specifically asking if the beams(floor joists) in your floor are adequate. Include a drawing with dimensions of the joist and some photos of the floor joists and connections. The civil/structural engineers on here would be able to recommend if you need to get the structure professionally evaluated or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When attaching your punching bag mount, I recommend attaching to two joists. This is to keep the bag from twisting the joist as it is impacted. The spacing on the mount is probably something standard like 16&quot;. If it doesn't line up correctly, you will need to box in between the joists with 2x6 or larger so no twisting will occur. For attaching the mount itself, predrill a hole the size of the root (the solid inside part) of the lag you have selected. Make sure the hole is centered in the bottom of the joist. This small hole and lack of splitting, ensures the floor joist's strength stays above 100% of its rated strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without seeing your exact installation location, I would recommend using four &lt;a href=&quot;https://www.homedepot.com/p/Everbilt-3-8-in-x-3-in-Zinc-Plated-Hex-Lag-Screw-801566/204645650&quot; rel=&quot;nofollow noreferrer&quot;&gt;3/8&quot; x 3&quot; lag bolts&lt;/a&gt; with a 1/4&quot; predrill.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-07-05T19:12:18.790" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22614" PostTypeId="2" ParentId="8246" CreationDate="2018-07-05T19:33:02.387" Score="1" Body="&lt;p&gt;Call local machine shops, and ask them, if they have a sonic mill. It uses special bits with coolant and ocelats to make the cuts. It can cut accurately through 1/2 inch in about 45 minutes. It shouldn't be too expensive, basically it vibrates the the material.&lt;/p&gt;&#xA;" OwnerUserId="16674" LastEditorUserId="9" LastEditDate="2018-07-05T20:46:15.243" LastActivityDate="2018-07-05T20:46:15.243" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22615" PostTypeId="1" CreationDate="2018-07-05T20:42:06.307" Score="3" ViewCount="63" Body="&lt;p&gt;I need a pressure switch for an electric circuit that has to be closed once a piston is compressing at a pressure of (approximately as we haven't specified how much force we have to use, and maybe it will be greater) 5 tons/cm^2 (75000 psi). The problem is that I can't find any model that satisfies these characteristics due to either&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;them not being as common as to be talked about in the sites I can access&lt;/li&gt;&#xA;&lt;li&gt;my abilities at intelligent Google search are not good enough or&lt;/li&gt;&#xA;&lt;li&gt;them not existing.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I can't believe a component like that doesn't exist and I'm very confident about my intelligent Google search skills so, can someone tell me if there's a chance I can get to know about components like that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be very thankful if anyone could give me a model or a brand or any reference or recommendation to find such information (or about why I won't).&lt;/p&gt;&#xA;" OwnerUserId="16673" LastEditorUserId="16673" LastEditDate="2018-07-06T23:36:10.727" LastActivityDate="2018-08-10T22:42:18.787" Title="What kind of pressure switch can I use for pressures up to 5-10 tons/cm^2?" Tags="|electrical-engineering|pressure|hydraulics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22616" PostTypeId="2" ParentId="22615" CreationDate="2018-07-05T22:50:58.470" Score="4" Body="&lt;p&gt;You do this by monitoring the hydraulic pressure rather than the press tonnage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;strong&gt;&lt;em&gt;force = pressure x area&lt;/em&gt;&lt;/strong&gt; we can rearrange so that &lt;strong&gt;&lt;em&gt;pressure = force / area&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Using your quaint units of measure, 5 ton = 5 x 2,240 lbs* = 11,200 lbs.&lt;/li&gt;&#xA;&lt;li&gt;Let's say your hydraulic cylinder has a cross-sectional area of 2 sq. in. then &lt;strong&gt;&lt;em&gt;pressure = 11,200 / 2 = 5,600 psi&lt;/em&gt;&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In that case you would be looking for a pressure switch that can be set to 5,600 psi. You would tee this into the drive line of your press.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ph5Gm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ph5Gm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. An image search for 'digital hydraulic pressure switch' throws up an impressive array of options.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 imperial ton = 2,240 lbs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-07-05T22:50:58.470" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22617" PostTypeId="2" ParentId="22610" CreationDate="2018-07-06T02:40:59.677" Score="0" Body="&lt;p&gt;I assume you are living in US. My answer is in reference to conventional construction with light framing and what is known as Type v construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you open a small 16&quot; wide by 6&quot; high opening in a place near the top of the wall where you expect the floor joist will meet and be supported by the double-plate to check for moisture leaks, dry rot, termite damage or an unexpected sewer pipe which has eaten out most of the structure, basically general soundness of the structure, then if all seems ok proceed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to hang the anchor within 1/3 span of the floor joists to minimize booming sound throughout the house, and the extra momentum to the beam. Do not attach the anchor directly to the joist but to a blocking of the same dimension as the joists saddling and connecting the two joists with appropriate joist fasteners which you can readily find at big box warehouses. Most of the anchors have their own fasteners and relevant instructions, try to follow them as best. Paint the ceiling in some kind of square chess pattern to give you a sense of aim and perspective. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally try to leave a half an inch gap at the ceiling entrance point( or what you judge the play of anchor would be)  filled with caulking  so as you don't worry about future cracks in the ceiling.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-07-06T02:40:59.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22618" PostTypeId="1" CreationDate="2018-07-06T03:43:20.817" Score="1" ViewCount="243" Body="&lt;p&gt;Our company builds rather large video walls that are driven by computers next to them. In one of our current project the size of the installation is 21 55&quot; displays side by side in portrait mode and 4 computers. The total heat dissipation would be maximum 8000 W, distributed evenly along the housing. This whole setup is in a standalone, about 15 meters long installation. Both the displays and the computers implement forced heating internally. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interior decoration company that builds the housing asked the question: How big and what kind of ventilation holes should they plan for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was searching the net and found certain solutions and online calculators (&lt;a href=&quot;https://www.engineersedge.com/heat_transfer/convection.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineersedge.com/heat_transfer/convection.htm&lt;/a&gt;), also a related question here on Engineering (&lt;a href=&quot;https://engineering.stackexchange.com/questions/13837&quot;&gt;How to design a cooling chimney for a computer server?&lt;/a&gt;) but these are concerned with surfaces being cooled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case the surfaces are irrelevant -I would think- as all devices have forced cooling, therefore I should care about the warm air they produce and push out of their individual chassis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give me a handle, what is the best approach to calculate the venting hole requirements.&lt;/p&gt;&#xA;" OwnerUserId="16678" LastActivityDate="2018-07-06T03:43:20.817" Title="Ventilation requirements of a closed housing of a video wall" Tags="|heat-transfer|airflow|cooling|convection|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22620" PostTypeId="1" CreationDate="2018-07-06T08:19:00.600" Score="0" ViewCount="95" Body="&lt;p&gt;How do I prove that during a loading session of a glass container, between two failure modes, overstressing or buckling, which will happen first via FEA? Is there any scientific way of approaching this problem? Possible references will also be appreciated. &lt;/p&gt;&#xA;" OwnerUserId="16489" LastEditorUserId="255" LastEditDate="2019-01-08T11:39:42.563" LastActivityDate="2020-02-02T13:01:55.217" Title="How to prove that during loading session of glass container, which will happen first - overstressing or buckling - via FEA?" Tags="|structural-engineering|finite-element-method|ansys|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22621" PostTypeId="1" CreationDate="2018-07-06T09:54:20.527" Score="2" ViewCount="428" Body="&lt;p&gt;What are the changes specific to mecanum wheels when calculating torque requirements for a vehicle. I have the data for the vehicles mass, top speed and top acceleration.&lt;/p&gt;&#xA;" OwnerUserId="16680" LastEditorUserId="16680" LastEditDate="2018-07-07T19:14:50.947" LastActivityDate="2024-01-30T06:02:50.303" Title="How to calculate torque requirement for vehicle with Mecanum wheels?" Tags="|motors|torque|mechanical|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22622" PostTypeId="1" AcceptedAnswerId="22623" CreationDate="2018-07-06T10:47:44.280" Score="5" ViewCount="2388" Body="&lt;p&gt;I’ve been working on an electric Dirtsurfer project that has many parts to be made from folded sheet aluminium. &lt;a href=&quot;https://i.stack.imgur.com/LyYjI.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LyYjI.jpg&quot; alt=&quot;screenshot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the sheet metal needs holes in it for bolts. The sheet aluminium will be either CNC laser cut or CNC water jet cut. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve got three options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Add holes to my design at the exact dimensions that I need them to be. I am placing myself at the mercy of the accuracy of the CNC cutter to ensure I have no slop which might let the bolts move. But no drilling!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Leave the holes out of the design. I measure the cut patterns to find the right places for the holes and drill them myself. Using a drill will give me a clean hole that I can fully trust the dimensions of. This is what I’ve done in the past, but I haven’t been accurate enough with my placement. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Undersize the holes in my design, and drill them out to the correct size later. I don’t have to measure anything, but still get a clean accurate drill hole.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any practical advice from someone who’s done this before?&lt;/p&gt;&#xA;" OwnerUserId="16681" LastActivityDate="2018-07-06T17:08:13.663" Title="How should I go about adding holes to laser or water jet cut sheet metal?" Tags="|mechanical-engineering|cutting|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22623" PostTypeId="2" ParentId="22622" CreationDate="2018-07-06T11:13:23.387" Score="8" Body="&lt;p&gt;This answer only refers to laser cutting&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your main problem will not be the accuracy of the CNC machine, which will be way better than what you can achieve when measuring yourself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem with clean edges is the hole size, if they are too small, say under 6mm diameter (this also depends on the thickness of the metal), the initial puncturing for each hole will leave some residue that will cause unclean edges of the holes. The laser basically drills a small hole and then afterwards cuts out the shape of the actual hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small holes, I would suggest an adaption of your 3rd idea. Some CNC machines can basically do punch marking, marking the center of the hole with a single short pulse. You dont have a lot of residue because very little material is removed, but the hole is marked accurately and you can afterwards drill exactly how you need it.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastEditorUserId="14629" LastEditDate="2018-07-06T11:19:24.837" LastActivityDate="2018-07-06T11:19:24.837" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22624" PostTypeId="1" CreationDate="2018-07-06T11:46:20.690" Score="1" ViewCount="30" Body="&lt;p&gt;Water will be circulated in the water pipe and liquid nitrogen tanks heat going to cool the water pipe,by the way to the water too. And fan is going to blow this cold air and heat is going to be decreased. Is it possible ?&lt;/p&gt;&#xA;" OwnerUserId="16682" LastEditorUserId="16682" LastEditDate="2018-07-06T12:53:09.557" LastActivityDate="2018-07-06T12:53:09.557" Title="How can I develope a water loop cooler system sustaining with liquid nitrogen?" Tags="|cooling|experimental-physics|air|cryogenics|" AnswerCount="0" CommentCount="3" ClosedDate="2018-07-06T12:25:51.030" ContentLicense="CC BY-SA 4.0" />
  <row Id="22625" PostTypeId="2" ParentId="22622" CreationDate="2018-07-06T13:24:28.730" Score="4" Body="&lt;p&gt;Similar to OpticalResonator's answer, I would suggest using the laser/waterjet cutter to mark the hole locations (although you can always mark the holes  manually after cutting). You should probably use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Punch_press&quot; rel=&quot;nofollow noreferrer&quot;&gt;punch press&lt;/a&gt; to cut the holes, if one is available in your machine shop. Then it's just a matter of selecting the exact die size for the hole you want to make. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a small, manually-operated punch press in the machine shop that I work in. Check with your machinist to see if you have something similar. A drill isn't the best tool for cutting holes in thin sheets, but it will get the job done. A punch press is intended for punching holes in sheet metals, but bear in mind that there is an upper limit to the thickness of material that you can cut with one.&lt;/p&gt;&#xA;" OwnerUserId="7430" LastActivityDate="2018-07-06T13:24:28.730" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22626" PostTypeId="1" CreationDate="2018-07-06T13:46:14.503" Score="1" ViewCount="444" Body="&lt;p&gt;For an at-grade intersection of multi-lane roads, the &quot;Turborotonde&quot; (&lt;a href=&quot;https://nl.wikipedia.org/wiki/Turborotonde&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dutch language Wikipedia article&lt;/a&gt;) or &quot;Turboplein&quot; (&lt;a href=&quot;https://www.wegenwiki.nl/Turboplein&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dutch language wegenwiki article&lt;/a&gt;) is an alternative to the multi-lane roundabout:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Turborotonde.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/e/e4/Turborotonde.svg/749px-Turborotonde.svg.png&quot; alt=&quot;Turborotonde&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;em&gt;Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Turborotonde.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Juerd, Wikimedia Commons, CC-BY-SA 4.0&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or the larger turboplein:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wegenwiki.nl/Bestand:Turboplein_N213.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://www.wegenwiki.nl/images/Turboplein_N213.jpg&quot; alt=&quot;Turboplein&quot;&gt;&lt;/a&gt; &#xA;&lt;em&gt;Source: &lt;a href=&quot;https://www.wegenwiki.nl/Bestand:Turboplein_N213.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rijkswaterstaat / Joop van Houdt, via wegenwiki&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compared to a classic multi-lane roundabout:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:NonUK_Roundabout_8_Cars.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/4/41/NonUK_Roundabout_8_Cars.gif&quot; alt=&quot;Roundabout&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;em&gt;Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:NonUK_Roundabout_8_Cars.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ferdrik, Wikimedia Commons, CC-BY-SA 4.0&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an at-grade intersection of multi-lane roads, what are the (dis)advantages of a &quot;turborotonde&quot; compared to a classical roundabout?  It would appear that the turborotonde should be safer, but it may be that there are disadvantages too.  For example, it appears not (legally) possible to turn around (full circle).  What are other disadvantages compared to a classic roundabout?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also: &lt;em&gt;&lt;a href=&quot;https://engineering.stackexchange.com/q/22627/3599&quot;&gt;Are there any turborotonde / turboplein designs with more than four exits?&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3599" LastEditorUserId="3599" LastEditDate="2018-07-06T14:00:06.757" LastActivityDate="2018-07-06T16:02:14.197" Title="For an at-grade intersection of multi-lane roads, what are the (dis)advantages of a &quot;turborotonde&quot; or &quot;turboplein&quot; compared to a classical roundabout?" Tags="|highway-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22627" PostTypeId="1" CreationDate="2018-07-06T13:51:36.917" Score="0" ViewCount="83" Body="&lt;p&gt;For an at-grade intersection of multi-lane roads, the &quot;Turborotonde&quot; (&lt;a href=&quot;https://nl.wikipedia.org/wiki/Turborotonde&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dutch language Wikipedia article&lt;/a&gt;) or &quot;Turboplein&quot; (&lt;a href=&quot;https://www.wegenwiki.nl/Turboplein&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dutch language wegenwiki article&lt;/a&gt;) is an alternative to the multi-lane roundabout:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Turborotonde.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/e/e4/Turborotonde.svg/749px-Turborotonde.svg.png&quot; alt=&quot;Turborotonde&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;em&gt;Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Turborotonde.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Juerd, Wikimedia Commons, CC-BY-SA 4.0&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or the larger turboplein:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wegenwiki.nl/Bestand:Turboplein_N213.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://www.wegenwiki.nl/images/Turboplein_N213.jpg&quot; alt=&quot;Turboplein&quot;&gt;&lt;/a&gt; &#xA;&lt;em&gt;Source: &lt;a href=&quot;https://www.wegenwiki.nl/Bestand:Turboplein_N213.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rijkswaterstaat / Joop van Houdt, via wegenwiki&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both examples have exactly four exits.  In principle, I can imagine that such a design can be extended to more than four exits.  Are there any examples of a &lt;em&gt;turborotonde&lt;/em&gt; or &lt;em&gt;turboplein&lt;/em&gt; design with more than four exits?  What would it look like?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;See also: &lt;a href=&quot;https://engineering.stackexchange.com/q/22626/3599&quot;&gt;For an at-grade intersection of multi-lane roads, what are the (dis)advantages of a &amp;quot;turborotonde&amp;quot; or &amp;quot;turboplein&amp;quot; compared to a classical roundabout?&lt;/a&gt;&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3599" LastEditorUserId="3599" LastEditDate="2018-07-06T14:00:35.273" LastActivityDate="2018-07-06T14:00:35.273" Title="Are there any turborotonde / turboplein designs with more than four exits?" Tags="|highway-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22628" PostTypeId="2" ParentId="22622" CreationDate="2018-07-06T15:25:53.680" Score="2" Body="&lt;p&gt;TLDR: Ask someone and do tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've worked with hobby level CNC mills (480 watt spindle) and CNC laser cutters (80-100 watt), and I've found that after doing a couple of test cuts, I know what I can and can't do.  As ConjuringFrictionForces mentions, talk to your machinists to see what you're likely to get for a result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're likely going to have to do a couple (several) prototypes, so looking at the results, with measuring, will help you figure out what you need to do.  It might be cutting a 6mm hole at 5mm and punching/drilling it out, or it might be cutting it at 6.5mm to allow room for the bolt to slip through easily during assembly.  (BTW, you usually need &lt;em&gt;some&lt;/em&gt; &quot;slop&quot; for bolts, otherwise they end up being screws.  Besides, not all bolts are created equal, unfortunately.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As much time as I've had to use my machines, I still rely on actual tests to see what result I'm going to get with a material and machine combo.  Your machinists probably have a better idea of what their machines can do.  I don't use my machines 8 hours a day, and I'm still learning what my home built CNC mill can do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already up-voted the other 2 answers and a comment, so this is just my 2 cents with the experience I've earned the hard way.&lt;/p&gt;&#xA;" OwnerUserId="16686" LastActivityDate="2018-07-06T15:25:53.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22629" PostTypeId="2" ParentId="22626" CreationDate="2018-07-06T16:02:14.197" Score="1" Body="&lt;p&gt;For a turbo roundabout traffic has to commit to a direction in advance of the intersection. This means having a long approach and adding clear signs above the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on where you put it people will be unfamiliar with the concept and do odd maneuvers, creating dangerous situations. Though this is temporary and extra signs can help with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can add a U-turn path should the flow be big enough to require it.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2018-07-06T16:02:14.197" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22631" PostTypeId="1" CreationDate="2018-07-06T16:29:48.040" Score="-1" ViewCount="113" Body="&lt;p&gt;There are red, green, and blue lasers out there but i can't find a white laser? how hard is to make one? Is it possible?&lt;/p&gt;&#xA;" OwnerUserId="16688" LastEditorUserId="10902" LastEditDate="2018-07-07T06:38:06.193" LastActivityDate="2018-07-07T06:38:06.193" Title="Why is making a white laser so difficult?" Tags="|lasers|" AnswerCount="1" CommentCount="1" ClosedDate="2018-07-08T02:32:30.687" ContentLicense="CC BY-SA 4.0" />
  <row Id="22632" PostTypeId="2" ParentId="22631" CreationDate="2018-07-06T16:44:56.810" Score="3" Body="&lt;p&gt;Its not hard- it bends the idea of what a laser is. Lasers have a single (or very narrow) wavelength of light emitted. There is no &quot;white&quot; wavelength- its just what your eye perceives from a (relatively) full spectrum of wavelengths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is generally considered by the scientific community to be an advantage, not a disadvantage. You can do many cool things with a single wavelength that you can't do with a broad spectrum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could potentially make a white-ish laser by syncing up many different lasers to make a combination beam that your eye perceives as white. This would be a very difficult engineering problem though, if you actually wanted a spatially coherent laser. Thanks to OpticalResonator for pointing out &lt;a href=&quot;https://www.nature.com/articles/nnano.2015.149&quot; rel=&quot;nofollow noreferrer&quot;&gt;some folks who managed to do it in a single crystal&lt;/a&gt;, which is pretty wild.&lt;/p&gt;&#xA;" OwnerUserId="2781" LastEditorUserId="2781" LastEditDate="2018-07-06T17:07:50.873" LastActivityDate="2018-07-06T17:07:50.873" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22633" PostTypeId="2" ParentId="22622" CreationDate="2018-07-06T17:08:13.663" Score="1" Body="&lt;p&gt;What I have done before from a very short period of time, I have used a CNC laser cutter to cut 12mm sheet of stainless steel, which I'm sure that you are going to use a thinner sheat metal. My design was something about 135 x 18 mm and the final result was completely nice. But you have to take in consideration that laser has its own cutting diameter &quot;pierce&quot; ( some thing near 0.2 mm - according to the machine power and the sheet metal thickness ) . Think of it as the tool used to cut metal in CNC router. Ensure that the size of this pierce is taken in consideration while cutting to get your desired hole diamter. Machine operators can handle this issue without any problem. You just have to manage this issue with them.&lt;/p&gt;&#xA;" OwnerUserId="16505" LastActivityDate="2018-07-06T17:08:13.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22634" PostTypeId="1" CreationDate="2018-07-06T17:38:38.123" Score="0" ViewCount="74" Body="&lt;p&gt;I want to design a custom 3D printed fan that would be made of several axial fans that would be connected (fixed) to their cylindrical casing. I am planning to rotate the whole assembly (the cylinder with fixed - nonrotating fans inside) using one engine on one axle only around it’s axis. I am hoping to achieve a higher static pressure (sum of individual static pressures of fans). Do you think this will work?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2018-08-06T07:01:58.420" Title="Will a fan that is composed of several fans fixed inside rotating tube work?" Tags="|fluid-mechanics|airflow|aerodynamics|fluid|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22635" PostTypeId="2" ParentId="9039" CreationDate="2018-07-06T18:39:35.837" Score="1" Body="&lt;p&gt;I know this is an old question, but I'd like to expand on the general info alephzero stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my research for building my own metal cutting CNC laser, I've found that it takes at least a 150 watt CO2 laser to cut through even thin metals.  This is also coupled with using a gas (N2 or O2) to either shield the material cut or to help it be cut.  Thickness can also be a factor in whether a shield gas or a cutting gas is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://am.co.za/laser/thickness&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://am.co.za/laser/thickness&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.fabricatingandmetalworking.com/2012/11/nitrogen-vs-oxygenwhich-should-you-use-to-cut-steel/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.fabricatingandmetalworking.com/2012/11/nitrogen-vs-oxygenwhich-should-you-use-to-cut-steel/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the different reflective nature of different metals, not all metals can be cut at the same power.  Thin steel and stainless steel can be cut with a 150 watt laser, but much more powerful lasers are needed for more highly reflective materials, like aluminum and silver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ND or ND:YAG lasers are commonly used for metal cutting, although they are much more expensive laser systems, and therefore much more expensive machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Laser_cutting&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Laser_cutting&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my own machine, I've made it large (4'x8' capacity) and found that a 180 watt CO2 laser tube is between 6.5' and 7' long.  Since my CNC frame is so big, I don't have to worry about that, though.  The tube, if gotten directly from China can be as &quot;little&quot; as \$900-\$1000, but a more local USA dealer has them for \$1500-\$1600.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've spent around \$8000 on my own machine and still don't have the tube yet, but that's not the £100,000 to £500,000 alephzero mentions.  I've seen machines (from China) that were around \$40,000 with power ratings to do metal, but that's still not something you are likely to get with the change in your couch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing about the reflections is that they definitely need to be managed.  In too small of a laser cutter, the reflections will go back into the tube and basically set up a standing wave that eventually burns out the tube, since not enough of the light is absorbed or dispersed, exceeding the rated power of the tube for too long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a laser that can cut through metal, the laser can still be reflected into the tube, but only until it pierces the material.  At that point, the angle of reflection moves any reflected laser to a different angle.  That usually means the laser is reflected into the bottom the of laser cabinet and handled/absorbed there, if not used to heat the metal.  By the time the laser hits something at that point, it is usually not focused enough to do any damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, many laser cutters have a full cabinet with windows that absorb the laser wavelength.  In the one I have access to, it uses specially coated glass windows to absorb the laser.  In the other one, it's Plexiglas/acrylic windows.  Acrylic goggles/glasses are used when a laser cutter doesn't have a cabinet.&lt;/p&gt;&#xA;" OwnerUserId="16686" LastActivityDate="2018-07-06T18:39:35.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22637" PostTypeId="2" ParentId="22634" CreationDate="2018-07-06T21:10:46.397" Score="0" Body="&lt;p&gt;I think this wouldn't work because of the following: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- the amount of power needed to rotate the whole mechanism is going to be much more than the required power to drive each fan individually due to the added weight of the cylinder ( casing ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- the eccentricity ( offset ) between the casing axial axis and one of your fans axis, will cause that fan to work inefficiently due to that that the required time to complete one revolution per fan is going to be longer, thus less output power from fan since power = torque x omega ( rotational speed ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- also fan blade design will no longer be effective because the change in the blade angle of attack and other specs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider using the total lost power to drive the fan on their own axis, I think you will get a better results rather than as expected.&lt;/p&gt;&#xA;" OwnerUserId="16505" LastActivityDate="2018-07-06T21:10:46.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22639" PostTypeId="1" CreationDate="2018-07-06T23:44:45.840" Score="0" ViewCount="5639" Body="&lt;p&gt;The common knowledge these days seems to be that positive air pressure pumped into the common hallways of highrise buildings will allow smokers and non-smokers to live in the same highrise.  Good seals between units is still needed, but the positive pressure prevents smoke from going into the hallway, and then into neighbouring units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the summer, many people open their windows.  In fact, I like to put a fan in front of the window.  Often, this blows cigarette smoke from neighbouring units into my unit.  There is simply not enough ventilation to mitigate the heating up of the unit without a fan.  Otherwise, it might possible that the positive pressure alone would ensure that smoke doesn't come in from the outside (assuming that the air from the corridor is bearably cool, which it often isn't).  For the time being, I try to choose a window that is as far from the smoking apartment as possible, and place the fan in front of that window.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the spring and fall, however, it might be possible to do without a fan in front of the window, and let the apartment be ventilated by the positive pressure.  I don't know where the intake for the positive pressure is, however, and the air quality smells really &quot;non-ideal&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another complication is that the windward side of the building will experience a high pressure outside, and I wonder it is enough to counteract the positive pressure from the inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seem to be so many variables that can upset the conditions for positive pressure to work.  I wonder &lt;strong&gt;how well understood is the benefit of positive pressure?&lt;/strong&gt;  Or is it just one of those things that sound reasonable, but difficult to study empirically based on actual highrises?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I eventually want to move, but in the mean time, I am trying various coping mechanisms based on whatever I can understand about positive pressure.  A complicating factor seems be the eddies on the leeward side of the building, making it hard to gauge whether to open the windows or which windows to open.&lt;/p&gt;&#xA;" OwnerUserId="16693" LastActivityDate="2022-11-15T06:27:30.560" Title="Effectiveness of positive pressure in highrise buildings?" Tags="|hvac|building-design|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22640" PostTypeId="2" ParentId="22634" CreationDate="2018-07-07T06:02:45.587" Score="1" Body="&lt;p&gt;The fan blades ideally should work like a wing getting lift (or giving lift, like a helicopter) out of 'flying' in a smooth stream or wind, not in a turbulent flow already agitated by previous layers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by adding more fans and turning the cylinder you may end up wasting the torque energy by just mostly twisting and turning the entire volume of air inside the cylinder with the blades, not producing much lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you compare the old windmill fabric sales to the new wind-energy propellers you see the difference. The old design worked based on direct wind pressure on the sails, like a parachute. But the new blades act like a wing in the stream of wind, creating much more energy efficiently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to create a lot of speed and pressure in an enclosed multi-layered turbine, but that has its walls purposely stationary, has vanes and gears, and the chamber itself narrows along as air accelerates.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-07-07T06:02:45.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22641" PostTypeId="1" CreationDate="2018-07-07T09:16:19.590" Score="1" ViewCount="1660" Body="&lt;p&gt;I'm attempting to convert a fridge compressor into a vacuum pump, but it got me confused. There should be one inlet and one outlet I believe, but instead I have three, with the third being connected to the second (the white one) through a thin copper tube with the bulky thing at the end of it (see picture). The white tube is the suction, it has a label on it. How would I hook it up, what are the three tubes for? (Sorry if it's in the wrong forum)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vdyVm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vdyVm.jpg&quot; alt=&quot;The three pipes are on the right&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16695" LastEditorUserId="3505" LastEditDate="2018-07-10T03:13:31.007" LastActivityDate="2018-07-11T20:07:37.367" Title="Inlet/outlet of a fridge compressor" Tags="|vacuum|vacuum-pumps|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22642" PostTypeId="2" ParentId="22641" CreationDate="2018-07-07T09:32:20.897" Score="0" Body="&lt;p&gt;One inlet (suction), one outlet (delivery) and the original filling point.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-07T09:32:20.897" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22646" PostTypeId="1" CreationDate="2018-07-07T16:17:59.167" Score="1" ViewCount="4077" Body="&lt;p&gt;In wind turbine, Is the number of revolution taken in 1s is called 1P frequency? If not, is it possible to find the 1P rpm from rpm measured for 1 hour (that is, I have 60 rpm data points from tachometer)?&lt;/p&gt;&#xA;" OwnerUserId="16697" LastEditorUserId="16697" LastEditDate="2018-07-07T18:16:19.447" LastActivityDate="2020-01-21T12:18:21.363" Title="what is 1P frequency in rotor (wind turbine)?" Tags="|measurements|signal-processing|wind-power|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22647" PostTypeId="2" ParentId="22646" CreationDate="2018-07-07T18:24:14.083" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In wind turbine, is the number of revolution taken in 1 s is called 1P frequency?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://epubs.surrey.ac.uk/790106/1/ICE-GE-Special_Issue_2013.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dynamics of offshore wind turbines&#xA;supported on two foundations&lt;/a&gt; you are correct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If not, is it possible to find the 1P rpm from rpm measured for 1 hour (that is, I have 60 rpm data points from tachometer)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Average the RPM readings and divide by 60 (seconds in one minute).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: I looked up the first part. I am an EE but never heard of 1P before.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2018-07-07T18:24:14.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22648" PostTypeId="2" ParentId="17291" CreationDate="2018-07-07T22:39:37.617" Score="0" Body="&lt;p&gt;We rented one back in 1988 to cool our computer room so they definitely make them. The coolant tubes were about 30 feet long and as flexible as hose pipes. I don't know if they used gas or water as the coolant. But the outdoor side was hung out of an open window. &lt;/p&gt;&#xA;" OwnerUserId="16700" LastActivityDate="2018-07-07T22:39:37.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22649" PostTypeId="1" CreationDate="2018-07-07T23:06:25.050" Score="1" ViewCount="75" Body="&lt;p&gt;One of the potential ideas that was suggested for rescuing the Thai soccer team trapped in the cave is to locate a &quot;chimney&quot; or secondary entrance to the chamber where the team is trapped.  However, crews searching for this have not been able to find one that is suitable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose diver crews were to bring a rare chemical or isotope into the cave that is easy to detect in minute concentrations, and let it diffuse through the cave network.  Then, searchers on the surface could be supplied with equipment or animals that can detect this gas, and would have a way of knowing how close they are to entrances that are connected to the main cave.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, 2,4-dithiapentane is found in truffles, which pigs can sniff out.  So the surface team could be supplied with already-trained pigs.  However, its high mass density might make it unsuitable for diffusing through the cave network in a reasonable amount of time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this be a reasonable approach to locating entrances?  What chemicals or detection methods would be suitable? Please cite a specific chemical/isotope and detection method with your answer.&lt;/p&gt;&#xA;" OwnerUserId="16699" LastEditorUserId="16699" LastEditDate="2018-07-11T07:16:49.623" LastActivityDate="2018-07-11T07:16:49.623" Title="Thai Cave Rescue-Using trace gas detection to locate chimney entrances" Tags="|civil-engineering|chemical-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2018-07-15T20:25:11.710" ContentLicense="CC BY-SA 4.0" />
  <row Id="22651" PostTypeId="1" CreationDate="2018-07-08T04:44:04.203" Score="1" ViewCount="361" Body="&lt;p&gt;What is the cross-sectional shape in blue called? The shape, used on the shafts, is used for torque transfer, gears or similar. The three arcs are same diameter as the ball bearing ID, so bearings rest centered perfectly, but flat parts are used for engagement of sliding gears. Because of the shape and surface of engagement is much stronger than standard cut out keyway.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZBwwe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZBwwe.jpg&quot; alt=&quot;cross-sectional shape&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16701" LastEditorUserId="16486" LastEditDate="2018-07-11T16:01:32.560" LastActivityDate="2018-07-11T16:01:32.560" Title="The name of this cross-sectional shape" Tags="|mechanical-engineering|gears|torque|mechanisms|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22652" PostTypeId="1" AcceptedAnswerId="22653" CreationDate="2018-07-08T09:02:53.757" Score="0" ViewCount="62" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bh7XQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bh7XQ.jpg&quot; alt=&quot;Image 1 - what type of nut it is? what tool I have to use to open it?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UgbNb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UgbNb.jpg&quot; alt=&quot;Image2 - what type of lock it is? what tool I have to use to open it?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see the two images. I marked a nut and locking part. what type of nut and lock mechanism it is? and what is the tool I have to use to open it?&lt;/p&gt;&#xA;" OwnerUserId="15386" LastActivityDate="2018-07-08T11:01:08.820" Title="What is the type of nut and tool to be used to unscrew?" Tags="|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22653" PostTypeId="2" ParentId="22652" CreationDate="2018-07-08T09:12:50.057" Score="3" Body="&lt;p&gt;You need to hold the shaft (in a vise with protective jaws) and turn that nut with a &quot;C&quot; spanner - which looks like a C with a single protruding tooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EVGd9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EVGd9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A poor method is to use a drift or punch, but this always damages the nut - perhaps fine for a temporary repair...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit as Fred says that is a circlip, actually in internal circlip which means that the nut above needs to be removed first then that circlip will come out.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-07-08T11:01:08.820" LastActivityDate="2018-07-08T11:01:08.820" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22654" PostTypeId="2" ParentId="22649" CreationDate="2018-07-08T09:45:39.260" Score="2" Body="&lt;p&gt;Most underground mines in developed countries use a &lt;a href=&quot;http://www.minearc.com/wp-content/uploads/2015/09/Stench-Gas-System-Brochure.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;stench gas system&lt;/a&gt; to alert the underground workforce of a danger, such as an underground fire, so that those who are underground can report to emergency safety bases (also known as fresh air bases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gas used is ethyl mercaptan, which smells of rotten onions. It's the same chemical that is added to methane, natural gas, supplies so that people can detect leaks of natural gas in domestic or industrial situations. Normally methane is odorless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This chemical could be used in the cave in Thailand but it would be very unpleasant for everyone trapped underground because the smell lingers for a very long time, which could lead to people experiencing nausea and/or headaches if exposed to the gas for a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another flaw with using this method is that air flows through any underground system will be variable with most of the air and any stench gas following the path of least resistance. So not all, so called chimneys, may have stench gas go through them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other flaws include: if so called chimneys do exists they may be too narrow, or too steep, or have too many changes of direction (dog legs), or their ground conditions may not be safe for anyone to use.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-07-08T09:45:39.260" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22655" PostTypeId="2" ParentId="22651" CreationDate="2018-07-08T13:07:01.680" Score="2" Body="&lt;p&gt;I would call it a “Tri-Lobe”. It comes in many forms, but that’s the general concept. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here’s a similar object which uses a tri-lobe shape which is a “capto” style tool holder&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lwo1J.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lwo1J.jpg&quot; alt=&quot;tri-lobe tool holder&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2018-07-08T13:07:01.680" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22656" PostTypeId="2" ParentId="22641" CreationDate="2018-07-08T15:02:26.640" Score="3" Body="&lt;p&gt;The small tube I believe is for the freezer coil, or it could be a capillary tube (thank you Swift). Regardless, simply solder it shut and use the two larger lines for pumping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to note, this style of compressor cycles some of its lubrication through the refrigeration system, so keeping the crankcase full of oil is important since any oil lost during pump down has no chance of being recovered since this will no longer be a closed system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, it is possible to use these as a good pump. I used to run (and still have) an old refrigerator pump that had a separate oiling system sight tube that made for a great vacuum pump. These style of compressors are easy to spot (although not terribly common).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running power to the unit needs to be done correctly. Typically black is hot, white is neutral and there is usually a green screw for ground. If you are unsure how to do this, find someone who is rather than messing with it as the wires need to be tied correctly to avoid risk of electric shock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck with the project!&lt;/p&gt;&#xA;" OwnerUserId="553" LastEditorUserId="553" LastEditDate="2018-07-11T20:07:37.367" LastActivityDate="2018-07-11T20:07:37.367" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22657" PostTypeId="2" ParentId="22555" CreationDate="2018-07-08T15:35:07.460" Score="1" Body="&lt;p&gt;6 Torr isn't too bad really and that vacuum pump looks a bit small and I am guessing is not a high quality unit? Also, as said in the comments, that water valve may be leaking as well as it isn't rated for vacuum but instead for pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, here is a simple leak detecting technique that can be of help. When we have a vacuum problem in the lab we start with a squirt bottle of acetone and lightly apply around the connection points. If there is a leak where the acetone is applied it will quickly enter the chamber, vaporize, and increase pressure on your gauge letting you know where the leak is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps, and good luck with the project!&lt;/p&gt;&#xA;" OwnerUserId="553" LastActivityDate="2018-07-08T15:35:07.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22659" PostTypeId="2" ParentId="21185" CreationDate="2018-07-08T22:39:34.633" Score="2" Body="&lt;p&gt;This is enough information to determine thrust provided you know the actual value of output diameter Dout because it is the only given distance measure.&lt;br&gt;&#xA;Cross sectional area m^2 = Pi * (Dout m / 2)^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets reduce everything to what happens during 1 second of continuous flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;initial velocity at input m/s = 0 m/s&lt;br&gt;&#xA;final velocity at output m/s = volumetric flow rate m^3/sec / cross sectional area m^2&lt;br&gt;&#xA;mass kg = volumetric flow rate m^3/sec * density kg/m^3 * 1 sec&lt;br&gt;&#xA;acceleration m/s^2 = (final velocity m/sec - initial velocity m/sec) / (1 sec - 0 sec)&lt;br&gt;&#xA;thrust N = mass kg * acceleration m/s^2  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thrust in Newtons N, is the force required to accelerate 1 kg of mass to 1 meter per second per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the pump is producing a constant acceleration if the flow rate is uniform.&lt;/p&gt;&#xA;" OwnerUserId="9532" LastActivityDate="2018-07-08T22:39:34.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22661" PostTypeId="2" ParentId="21914" CreationDate="2018-07-09T15:25:15.073" Score="4" Body="&lt;p&gt;I do a lot of bridge inspections in Ontario, Canada following OSIM (Ontario Structural Inspection Manual).  Based on your photo, I would hazard to guess this is an older style barrier system.  I would break what you are looking at the side of the bridge into 3 separate components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the bottom you have a barrier or safety curb.  That approximately 150-200 mm square protrusion at the base of the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wall itself I would call a parapet wall.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part at the top I would call a post and railing or hand rail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would avoid calling it a post an open railing system as the wall is really solid and looks to be a solid connection with the deck.  Meaning in a collision the load would be transferred down and directly to the deck.  In a post and railing system the load gets transferred to the posts and the posts transfer it to the deck.&lt;/p&gt;&#xA;" OwnerUserId="6993" LastEditorUserId="6993" LastEditDate="2018-10-12T17:09:16.323" LastActivityDate="2018-10-12T17:09:16.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22666" PostTypeId="1" AcceptedAnswerId="22668" CreationDate="2018-07-09T18:01:31.450" Score="0" ViewCount="43" Body="&lt;p&gt;Suppose I have a rough 3d model of a building created it some CAD software. Now I want to change some portion of the model based on the following factors&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Lighting&lt;/li&gt;&#xA;&lt;li&gt;Energy consumption&lt;/li&gt;&#xA;&lt;li&gt;Structural Load&lt;/li&gt;&#xA;&lt;li&gt;Fire Safety&lt;/li&gt;&#xA;&lt;li&gt;Heating &amp;amp; cooling load&lt;/li&gt;&#xA;&lt;li&gt;etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is there any Software that can provide these feature. I know there are some good software for each of the feature mentioned above, but how can we integrate their results into the 3d model. I am from a Computer science background so don't have much idea about this, Any suggestion would be of great help.&lt;/p&gt;&#xA;" OwnerDisplayName="user16716" LastActivityDate="2018-07-09T20:20:12.857" Title="How buildings 3D model is created" Tags="|structural-engineering|cad|building-design|" AnswerCount="1" CommentCount="1" ClosedDate="2018-07-10T18:00:53.187" ContentLicense="CC BY-SA 4.0" />
  <row Id="22667" PostTypeId="1" CreationDate="2018-07-09T18:56:18.673" Score="1" ViewCount="41" Body="&lt;p&gt;I'm working on a project that involves pressing a softened thermoplastic sheet in a two-part mold to form a part 0.090&quot; thick. None of us at our small company are experts in matched molds, and we've come up against a question that must be answered before we put more financial resources into the project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape that we are working with can be approximated as an arc, with the open side up. That makes a female mold with a cupping shape and a male mold with a rounded shape. Our question resides in the pressures that we can achieve along the side walls that approach vertical. In my mind, this is a simple statics problem. Assuming there are no other significant effects besides the normal force between the mold faces, then if $F$ is the pressing force and $\theta$ is the angle of the sidewall, measured from horizontal, the sidewall normal force $N$ will simply be $N=F\sin(\theta)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my assumptions are good and shear doesn't play a significant effect on the final geometry, then $N$ is reduced to below 20% of $F$ when $\theta=10^{\circ}$. That has potential to heavily influence the quality of part in those regions. My question is, as someone with no prior molding experience, &lt;strong&gt;is the normal force the only significant player when it comes to pressing a part at steep angles?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16304" LastEditorUserId="16304" LastEditDate="2018-07-12T15:02:44.533" LastActivityDate="2018-07-12T15:02:44.533" Title="Influence of shear force on pressed part with steep walls" Tags="|manufacturing-engineering|shear|molding|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22668" PostTypeId="2" ParentId="22666" CreationDate="2018-07-09T20:20:12.857" Score="0" Body="&lt;p&gt;A good Starting point for what you need is REVIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our structural engineers do not use REVIT. They use RAM Steel to perform their structural analysis (Gravity only) and then give the work to CAD operators to document in AutoCAD or Revit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of the building Components can be stress tested with  multiple scenarios in INVENTOR per individual components, but they also can be done in revit if you know the know how. (this is much more complex)&lt;/p&gt;&#xA;" OwnerUserId="15027" LastActivityDate="2018-07-09T20:20:12.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22670" PostTypeId="2" ParentId="22620" CreationDate="2018-07-09T21:37:10.183" Score="1" Body="&lt;p&gt;Many Finite Element programs (such as ANSYS) will comes with several modes for their model.  ANSYS comes with a buckling mode as well as a normal stress mode.  Simply run the same model in both modes and review.  You should find that your answers will tell you the maximum stress (for the stress model) and the maximum load for the buckling.  Then you can run factors of safety on both modes.  Bear in mind that buckling in FEA is &lt;strong&gt;&lt;em&gt;highly&lt;/em&gt;&lt;/strong&gt; sensitive to flaws, smaller than can usually be detected.  As such, the results are almost always taken to be about 80% lower for most models and it is suggested that a scale model be built and crushed in a test lab to ensure accuracy.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-07-09T21:37:10.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22671" PostTypeId="2" ParentId="21866" CreationDate="2018-07-10T07:08:39.600" Score="3" Body="&lt;p&gt;Well, I would &lt;em&gt;think&lt;/em&gt; that two smaller ones rigged up in an &quot;amateurish&quot; fashion (apologies, but that's we both are, right) are going to be worse than just one (sufficiently large) properly mounted heatsink. That giant triangular piece closest to the CPU would probably delay heat transfer -- even if you have excellent cooling on the other side of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have done was to use just one cooler, and that &lt;em&gt;is&lt;/em&gt; sufficient if you choose one that's large enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have one giant case (Antec 182) with a giant heat sink (Thermalright HR-1) which never even gets lukewarm under load of a 3.something GHz cpu.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an HTPC case (Streacom FC10) in which the sides themselves are giant heat sinks connected to the CPU and PSU by way of heat pipes. This also works excellently, and is fully enclosed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Here's what the case-as-heatsink solution looks like: &lt;a href=&quot;https://i.stack.imgur.com/lfZ1P.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lfZ1P.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1011" LastEditorUserId="19376" LastEditDate="2019-02-15T13:20:17.123" LastActivityDate="2019-02-15T13:20:17.123" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22672" PostTypeId="1" CreationDate="2018-07-10T08:26:17.817" Score="0" ViewCount="163" Body="&lt;p&gt;The paddles are pulled counter-clockwise, using a muscle-like mechanism that attaches between the distal part of the paddle, and the inner part of the paddle next to it, similar to the arrow in the image below, and the hypothesis is that it would cause the device to rotate in a clockwise direction. What are some thoughts on the hypothesis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SHOq2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SHOq2.jpg&quot; alt=&quot;3d model image stationary&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/18CL8.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/18CL8.gif&quot; alt=&quot;slowly rotating image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/byfMk.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/byfMk.gif&quot; alt=&quot;rapidly rotating image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16720" LastEditorUserId="13297" LastEditDate="2018-07-10T09:28:28.943" LastActivityDate="2018-07-10T15:04:22.720" Title="A device like this, submerged in water, would it rotate?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22673" PostTypeId="2" ParentId="22144" CreationDate="2018-07-10T10:05:16.773" Score="1" Body="&lt;p&gt;One thing about using overhead trolley wires to power vehicles is traffic congestion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can remember riding on trolley buses when I was very young &amp;amp; if a trolley bus had to pull over at a bus stop to drop off or pick up passengers it blocked the vehicles behind it. All vehicles behind it had to wait until the bus moved again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any vehicle connected to an overhead trolley wire, whether it be bus or car, has restricted movement. It cannot deviate significantly from the line of the overhead trolley wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, if something is blocking the path of a self powered vehicle whether it be a combustion engine of a battery the vehicle can go around the obstacle. But vehicles powered by an overhead trolley wire cannot move around. They have to stay where they are until the obstacle is removed.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-07-10T10:05:16.773" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22674" PostTypeId="1" CreationDate="2018-07-10T12:20:57.820" Score="1" ViewCount="110" Body="&lt;p&gt;In a system I am modeling for personal research, I have a sphere (S) submerged in air (A), initially at the same temperature Ts = Ta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the temperature of air is slowly increased (or decreased), by something like 0.5 °C/h. In these conditions the Delta-T = Ts-Ta is necessarily low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, would &lt;em&gt;h&lt;/em&gt; be so low that the main heat transfer mode is not convection anymore?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE: It is a sphere of say 20 mm radius, made of steel, and exposed to ambient air - which means the temperature of air will change with the diurnal cycle; 0.5 °C/h is a rough estimate based on the observed diurnal excursion in a certain locality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In light of the informative comments &amp;amp; answers below, I would then like to ask how is the conductive heat transfer coefficient defined for this case.&lt;/p&gt;&#xA;" OwnerUserId="16721" LastEditorUserId="2015" LastEditDate="2018-07-17T17:29:19.157" LastActivityDate="2018-07-19T15:23:31.427" Title="What about the natural convection heat transfer coefficient (*h*) at low Delta-T?" Tags="|mechanical-engineering|heat-transfer|chemical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22675" PostTypeId="2" ParentId="22672" CreationDate="2018-07-10T15:04:22.720" Score="2" Body="&lt;p&gt;The way the cylinders are set up ( all parallel to vertical axis) it seems it will more jerk back and forth rather than rotating.&#xA;However if you install the cylinders in a slanting axis so that they roughly follows a spiral around the center, then it would become similar to a pump and would turn and gently pump the water down along the central axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if you hold it laying flat in a stream of water it would turn into a turbine and generate power.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-07-10T15:04:22.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22677" PostTypeId="2" ParentId="556" CreationDate="2018-07-10T16:33:53.823" Score="0" Body="&lt;p&gt;If a transformer has no load on the secondary, there is no current consumption. Maybe some leakage but this is minuscule. If you see the transformer as an inductor, this will imply that the transformer winding blocks AC and passes DC. Versus capacitance that blocks DC and passes AC. So an inductor is simply an AC resistor. If you do the ohms law math, your voltage is constant so the resistance of the coil is what changes when you add a load to the secondary winding. It is like completing the circuit allowing more current to flow.  &lt;/p&gt;&#xA;" OwnerUserId="16724" LastActivityDate="2018-07-10T16:33:53.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22678" PostTypeId="2" ParentId="22674" CreationDate="2018-07-10T20:05:17.320" Score="2" Body="&lt;p&gt;We can't say without knowing more information about the system. Luckily, we have a concept for evaluating this problem: &lt;a href=&quot;https://en.wikipedia.org/wiki/Rayleigh_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Rayleigh number&lt;/a&gt;. As outlined in the reference, if the number is low, convection is unimportant relative to conduction. If its high, convection is dominant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect finding correlations for the Grashof number (to plug into the Raleigh number equation) in your application will be difficult. If you have specific questions about that though, please update the question with more details. Otherwise, I suspect it will be easier to use this concept to do some rough estimates to determine how wide the range of possible outcomes is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update about conduction:&#xA;If conduction is dominant, you unfortunately can't treat the air as a bulk material and solve with nodes, but rather you'd need to solve for the temperature profile/gradient. You'd need to solve &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_conduction#Fourier&amp;#39;s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fourier's Law of Conduction&lt;/a&gt;, either explicitly or numerically. Different equations time!&lt;/p&gt;&#xA;" OwnerUserId="2781" LastEditorUserId="2781" LastEditDate="2018-07-19T15:23:31.427" LastActivityDate="2018-07-19T15:23:31.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22679" PostTypeId="2" ParentId="4159" CreationDate="2018-07-11T02:02:09.083" Score="2" Body="&lt;p&gt;Another advantage is to counter mechanical thrust which is why the bearings are usually conical&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2018-07-11T11:23:08.190" LastActivityDate="2018-07-11T11:23:08.190" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22681" PostTypeId="2" ParentId="2561" CreationDate="2018-07-11T02:09:42.057" Score="0" Body="&lt;p&gt;You need to understand that whilst the flow rates may remain static, the mass will not be so because warmer air is lighter and less dense whilst chilled aircon air is substantially heavier. The difference between them determines the power required to transfer the air by the machine to the zone your looking into. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in calculations is to answer two different questions: power consumption or airflow {m^3 \cdot s}&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-07-11T02:09:42.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22682" PostTypeId="2" ParentId="22608" CreationDate="2018-07-11T02:22:33.413" Score="0" Body="&lt;p&gt;Any underground work requires special research which takes time. Modifying the tunnel would mean increasing flow rate of water and therefore current strength. With larger volumes comes additional erosion issues such as deposition. As the diameter of an opening increases the flow velocity decreases causing deposition of material and narrowing of the vertical space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Risk of collapse is probably high in wet ground which is already causing problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Removing material means it has to be manually transported a very long distance underground and delay rescue before flooding expected widely at any moment. Rather than remove cave material, remove the problem: the children. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2018-07-11T11:21:00.133" LastActivityDate="2018-07-11T11:21:00.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22683" PostTypeId="1" AcceptedAnswerId="22688" CreationDate="2018-07-11T06:11:09.273" Score="0" ViewCount="65" Body="&lt;p&gt;There is a gate which is to be controlled by 2 motors, but one should work and when one fails the other should work. AND THE CONTROLLER USED IS A PLC&lt;/p&gt;&#xA;" OwnerUserId="16727" LastEditorUserId="7083" LastEditDate="2018-07-15T20:20:13.190" LastActivityDate="2018-07-15T20:20:13.190" Title="How to control a roller gate using 2 motors - when one fails the other works" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22686" PostTypeId="1" CreationDate="2018-07-11T09:46:36.113" Score="0" ViewCount="478" Body="&lt;p&gt;What are the parameters required to calculate the overall efficiency of a steam turbine and how to calculate. Is there are certain assumptions to be made while calculating the overall efficiency of a steam turbine.&lt;/p&gt;&#xA;" OwnerUserId="16732" LastActivityDate="2018-09-09T12:02:04.993" Title="How to calculate actual efficiency of a steam turbine" Tags="|energy-efficiency|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22687" PostTypeId="2" ParentId="22686" CreationDate="2018-07-11T11:40:35.633" Score="-1" Body="&lt;p&gt;The objective of a stream turbine is to extract all the available latent heat in steam and convert it to mechanical power. In any energy transfer system it is nigh impossible to achieve 99.999% efficiency because of friction, inertia and uncontrollable energy creep due to entropy. A simplistic observation would suggest that energy input=energy output in a direct system but a turbine is a mechanical obstruction that introduces friction and inertia issues. To compound the initial issues are the aerodynamics of the turbine according to the methods of kinetic energy extraction from the steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am example would be an initial pressure of 2700 psi and post extraction pressure of 0.17 psi. The thermal gradient may well drop from 300°C to ambient temperature before the fluid is cycled again. The formulae for the calculations you speak of are complex, requiring specific measurements and parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, turbines only extract about 36% to 68% of their throughput. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-07-11T11:40:35.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22688" PostTypeId="2" ParentId="22683" CreationDate="2018-07-11T12:56:49.693" Score="0" Body="&lt;p&gt;Probably the simplest, most efficient solution would be powering the mechanism by both motors simultaneously, connected on the low-torque side of any mechanics - e.g. both propelling the same linear actuator, or the same first gear of a gearbox. Have them with enough surplus power that each can move the gate and idle second motor simultaneously. The gate will open slower with one motor, but it will open just fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can expand on that idea by having the PLC read the current drawn by either of the motors when active, and switch it off (to idle, open contact) if detecting either the current is zero (burnt) or way over the top (short circuit). If the current in both is way higher than normal, but still not at short circuit level, the PLC should switch both motors off and signal gate blocked / mechanism jammed. (similarly, current way low; gearbox broken, motor running without load.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That way you're keeping the mechanics simple and move most of complex decisions into software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you want to save up on electricity, and use cheaper, weaker motors, you'll need either electromagnetic clutches (...and there go all your savings on cheaper motors), or unidirectional (free-wheel) clutches (so that an idle motor isn't pulled by the active one), and forgo simple switching of direction by reversing polarity, in favor of electromechanical gearbox that allows switching direction of the output rotation.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-07-11T12:56:49.693" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="22690" PostTypeId="1" CreationDate="2018-07-11T16:49:33.280" Score="0" ViewCount="158" Body="&lt;p&gt;I am trying to find the easiest way to cut a thick glass pattern (glass thickness of 1-3 mm thick) without breaking. I am considering laser cutting but I don't know if I can cut glass of this thickness. Can anyone suggest an easy way to cut curved shapes precisely into glass?&lt;/p&gt;&#xA;" OwnerUserId="16736" LastActivityDate="2019-06-09T08:01:11.610" Title="Methods for cutting glass" Tags="|materials|manufacturing-engineering|glass|ceramics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22691" PostTypeId="2" ParentId="22313" CreationDate="2018-07-11T18:44:28.197" Score="1" Body="&lt;p&gt;&lt;strong&gt;I am not a qualified civil engineer.&lt;/strong&gt; &lt;em&gt;This advice is theoretical and not to be used for any actual design in the real world.&lt;/em&gt; It is for informational purposes only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In civil engineering we always design things for worst case scenarios for good reason. Were that panel to fall over and injure a child as happened with &lt;a href=&quot;https://www.mirror.co.uk/news/uk-news/hugo-boss-fined-12-million-6382709&quot; rel=&quot;nofollow noreferrer&quot;&gt;a Hugo Boss store&lt;/a&gt; in Bicester (pronounced Bister) Village, England in 2015, there may be a criminal case to answer and also legal issues that cost vast sums of money. No design of any structure is taken on without a full assessment of its possible environmental impact which means how that structure is going to perform in its position and the effects it might have in failure situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To formuate any coherent design we must make educated assumptions that take into account safety factors, environmental factors, mitigation of loading extremeties, alternative design issues to overcome any unforseen issues.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In this case we need to assume that the maximum wind velocity is not only possible, but is likely and then use something like the Eurocodes safety factor margins to improve the safety of the structure.&lt;/li&gt;&#xA;&lt;li&gt;The panel must be able to withstand a unifomly distributed load from wind at a far higher load than 15 mph for good reason: Wind has a tendency to NOT be uniform and GUSTS may easily be up to 80% times higher than median velocity. This means that 15 mph is possibly gusted up to 27 mph.&lt;/li&gt;&#xA;&lt;li&gt;A safety factor must be incorporated to withstand the moment of lateral forces and usually it is an additional 5 to 30% additional force tolerance.&lt;/li&gt;&#xA;&lt;li&gt;Altitude also plays a major factor in designing for wind load. Although wind is a dynamic load, all building structures are usually designed for a uniform load on account of safety and for simplified material strength calculations. The higher the board above ground level, the greater will be the unrestricted wind verlocity. There is often a vast difference in wind velocity between 4' and 40' depending on surrounding environmental fators.&lt;/li&gt;&#xA;&lt;li&gt;Depending on the location of the panel, the force of the wind may be substantially greater in certain sitings such as between buildings 2' to 12' above the ground compared to open area of a playing field.&lt;/li&gt;&#xA;&lt;li&gt;Aerodynamics also must be accounted for since it is flat and to be used specifically as an aerofoil or fro want of a better description, a wing.&lt;/li&gt;&#xA;&lt;li&gt;There is no indication as to whether this wing is going to be fixed or in motion on a float in a parade.&lt;/li&gt;&#xA;&lt;li&gt;According to the diagram provided, it is not going to be pinned (fixed) at all. It has two rollers meaning it can be moved mechanically. If this is true, then we need to consider using cable stays for a fixed device an mass-moment reaction balances for motion designs.&lt;/li&gt;&#xA;&lt;li&gt;There is no indication as to what the material is to be used because we do not know the magnitude of forces that may be exerted upon the wing for now.&lt;/li&gt;&#xA;&lt;li&gt;Self-weight of the entire system must be accounted for to calculate accurately what is a safe working design.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Imperial (Metric) vertical Panel Dimensions&lt;/strong&gt;&#xA;$$&#xA;10'(3.048m)&#xA;\times&#xA;4'(1.2192m)&#xA;=40'^2(3.7161216m^2)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparent inclined panel dimensions:&#xA;$$&#xA;10'\sin 80^\circ \times 4'&#xA;\approx 39.39231012 \text{ ft}^2 \text{ or } 3.659665363 \text{ m}^2&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Wind Velocity:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Client estimated maximum velocity: 15 mph&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feasible velocity: $30 \text{mph}$ with gust velocity estimate: $\approx &amp;lt;60\%$&#xA;$$&#xA;30 \text{mph}+(30 \times 6)=48 \text{mph}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Safety factor: 25%&#xA;$$&#xA;48 \text{mph}\times 1.25 = 60 mph&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we have a design velocity for the wind force of 48 mph for which &lt;a href=&quot;https://www.physicsforums.com/threads/what-psi-can-be-achieved-from-a-60mph-wind.698861/&quot; rel=&quot;nofollow noreferrer&quot;&gt;another website has calculations&lt;/a&gt; we can use to determine pressure and therfore a lateral force measure that may be exerted on the panel with a safety margin for error of 60 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that air density differs according to its thermal state which means we also need to account for the temperature of the air as more dense air has greater inertia and hence a greater pressure against the panel. Wind pressure is proportional to air density multiplied by the square of the velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\rho = \frac{p}{R_{specific} \times K}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;R_{specific}=287.058 J\cdot(kg·K)^{-1} \text { or } 53.35 ft·lbf(lb·°R)^{-1}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When taking account of density, it is unlikely that anyone will sensibly march their band in temperatures below $5^\circ$C ($41^\circ$F) for which the density of air at sea level is going to be &lt;a href=&quot;https://en.wikipedia.org/wiki/Density_of_air#Temperature_and_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;$ \rho=1.269 kg.m^3@100$KPa&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that our surface area is $3.659665363 m $^2$ and therefore the wind force will calculates as 1671N according to &lt;a href=&quot;https://www.engineeringtoolbox.com/wind-load-d_1775.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Engineering Toolbox&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see from the result that that board would have to withstand a dead load of 170.4 Kg.f or 375.66 lb.f at the top vertical edge of the panel.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Design Brief and Recommendations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use a few different ways to prevent the board being blown over as enumerated below:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Deadweight welded to frame attached platform base;&lt;/li&gt;&#xA;&lt;li&gt;Cable type stays like tent rope &amp;amp; peg systems;&lt;/li&gt;&#xA;&lt;li&gt;Ground spike systems used to hold down groundsheets;&lt;/li&gt;&#xA;&lt;li&gt;Equilateral triangle systems with internal deadweight.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Looking at your design, a single sheet of wood would theoretically be strong enough as long as it was able to withstand a minimum weight of 9.536 lb.ft$^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that it is only a wood sheet, you might want to consider using &lt;a href=&quot;https://en.wikipedia.org/wiki/Voile&quot; rel=&quot;nofollow noreferrer&quot;&gt;voile&lt;/a&gt; over a framed support as an alternative or canvas with eyes. These options are to save on the weight of the device and work as a windbreak also. This is an important considerations since your wooden support's bending moments are going to require a great mass in a small area to prevent the system from toppling over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider that the total lateral force is effectively applied as a combined moment of force at the top edge of the panel in a lateral direction, you will require an equal but opposing force to counter that force in two opposing directions. This is a major concern because wind arrives from 540$^\circ$ of direction at ground level. To understand this, you need to consider a circular direction in plan and add 180$^\circ$ in elevation which is effectively an hemisphere of wind action towards the centroid mass of the force system which needs to have a very low centre of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A centroid is where the centre of mass combined with gravitation and wind inertia is situated. An example is that the entire construction is going to be in motion so we cannot use cable stays that are not also mobile in tandem with the panel and supporting structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, if the platform is going to be transported, use of ratchet straps that pull the panel downwards against the base of the support would be a good idea. Four straps of one inch width that wrap around the panel should hold it in position as long as the base and the apex are securely fixed to each other.&#xA;In reference to previous ideas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1. Deadweight welded frame and base&lt;/strong&gt;&#xA;While this is probably not a bad idea if transporting heavy loads above human carriage is not an issue, it requires financial expenditure if you are not a welder and do not have an encapsulating frame. On cost efficients grounds this may have to be a forgone conclusion and obsolete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2. Cable type stays&lt;/strong&gt;&#xA;Feasible if using a vehicle like a pickup but will require some lateral moment opposition of a wider base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3. Ground spike system&lt;/strong&gt;&#xA;This is clearly not going to help at all. Dropped into &lt;em&gt;File 13&lt;/em&gt; for trash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4. Equilateral triangle system with deadweight.&lt;/strong&gt;&#xA;This is a modification of option 1 but with a design change to your initial proposal in the next section because frankly, my recomendation.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Design considerations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Moments of force&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic formula for calculating moments around a point is $M_O = Fd$ where F is the total force multiplied by the combined distance from the point of rotation. Since the greatest force is accumulated at the top of the system, we combine the uniformly distributed load into a single point of force at the maximum point of rotation to calculate the rotation forces that need to be opposed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;M_O = 1671 \times 3.048&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$&#xA;M_O = 5093.208 N.m \text { (1144 lbf)}&#xA;$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The platform base should be cut into 4 lengths of equal width. Your proposed vertical panel is going to become the new platfrom base. Your vertical support will be supported by the 4 platform pieces as diagonals attached to the new base and upright support at angles of $60^\circ$ to the horizontal and $90^\circ$ to each other being pinned to the vertical support edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we need to ensure is that the wind pressure does not blow the system over and the height is the issue because The force at the top edge multiplied by the height of the system will explain what magnitude of downward force we need to use to compensate for the turning moment. Fortunately, this design is quite simple and we can compensate for lifting forces whilst remembering that the design load is for 48 mph winds as a maximum and the extra 14 mph safety is to prevent anyone being hurt by the construction.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2018-07-11T19:46:44.727" LastActivityDate="2018-07-11T19:46:44.727" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22694" PostTypeId="2" ParentId="22690" CreationDate="2018-07-11T20:14:35.240" Score="1" Body="&lt;p&gt;Waterjet cutters work like a charm up to 5mm. I have cut 'fireproof' glass of 4mm and it came out very well.&lt;/p&gt;&#xA;" OwnerUserId="16663" LastActivityDate="2018-07-11T20:14:35.240" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22695" PostTypeId="1" AcceptedAnswerId="22699" CreationDate="2018-07-11T22:39:15.283" Score="1" ViewCount="44" Body="&lt;p&gt;With 9 blades that are in a slanting axis so that they roughly follows a spiral around the center, would this device be similar to a pump, and turn (rotate) and gently pump the water down along the central axis as it contracts and squeezes itself together?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SULld.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SULld.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16739" LastActivityDate="2018-07-12T06:49:28.313" Title="If this is submerged in water, and contracts and squeezes itself together, does it rotate?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22696" PostTypeId="1" CreationDate="2018-07-12T00:48:29.473" Score="1" ViewCount="489" Body="&lt;p&gt;A company called Devecitech has just created what it calls the 'ENLIL' vertical axis wind turbine, that sits at the side of a highway generating electricity from the wind-power created by passing cars&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gK6ms.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gK6ms.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Video here: &lt;a href=&quot;http://devecitech.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://devecitech.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: does the presence of this turbine 'steal' energy ('negligible' or otherwise) from passing cars by creating drag where there would otherwise be free space?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, will it only ever generate less energy than it 'steals' from the passing cars?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it purely harnessing energy that would otherwise be wasted?&lt;/p&gt;&#xA;" OwnerUserId="16740" LastEditorUserId="90" LastEditDate="2018-07-12T05:14:11.480" LastActivityDate="2018-07-12T20:21:14.227" Title="ENLIL road-side vertical turbine - creating drag?" Tags="|aerodynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22699" PostTypeId="2" ParentId="22695" CreationDate="2018-07-12T06:49:28.313" Score="0" Body="&lt;p&gt;If it is primed so there is a stablished flow of water down with enough momentum, yes it will rotate while contracting and expanding acting like a fish swimming and will generate either pressure down to pump a central flux of water or will start to move up if free to move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fish swinging its tail to left and right, or up and down, creates propulsion and moves forward, by riding its own wake, as opposed to steering.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-07-12T06:49:28.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22700" PostTypeId="2" ParentId="22696" CreationDate="2018-07-12T07:12:14.133" Score="1" Body="&lt;p&gt;Depends, If it is in such a location as to interfere and incur into smooth streamlined envelope of air around the bus, yes it will disrupt the easy glide of the bus and lose part of the stolen energy to efficiency factor( around 50% at best). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is far in a safe zone, then it receives minimal benefit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can imagine a wind tunnel with openings on the side walls, hardly any wing will create lift in it.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-07-12T07:12:14.133" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22703" PostTypeId="1" CreationDate="2018-07-12T11:14:16.410" Score="2" ViewCount="97" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v84nK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v84nK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Problem 2/54, I have come up with a solution only one decimal place off from the books results. I am posting to see if someone can look over my solution to see if it is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin the equation given:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;a = u*g - (k/m)x&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only works in cases where it only moves in a direction where the friction force is strictly positive (i.e. the block is moving leftward).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the case of motion in the opposite direction must be considered, the equation for motion opposite this direction is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;a = -u*g - (k/m)x&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now using the relationship in the book:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;vdv = ads&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating and noting that each starting and stopping point has zero velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) &lt;strong&gt;0 = u*g(xf - xi) -(k/2m)(xf^2 - xi^2)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) &lt;strong&gt;0 = -u*g(xf - xi) -(k/2m)(xf^2 - xi^2)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now using (1) the first stop start point occurrs at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;xf = .23866m&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the way back however the motion is reveresed and equation (2) is used to find:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;xf = .022666m&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution in the book was: 23.3mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did I make a mistake or why is there such a seemingly slight difference in the answer?&lt;/p&gt;&#xA;" OwnerUserId="16746" LastEditorUserId="16746" LastEditDate="2018-07-12T11:22:27.330" LastActivityDate="2023-12-17T15:06:45.807" Title="Linearly Damped Harmonic Motion with Sliding Friction" Tags="|dynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22704" PostTypeId="1" AcceptedAnswerId="22705" CreationDate="2018-07-12T12:04:47.890" Score="3" ViewCount="6325" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kishF.png&quot; alt=&quot;illustration&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My dad has a tub by a lake.&#xA;He has a water pump in the lake to pump up water into a tub.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He claims that if he use a flexible hose, the pressure in the hose at the top will be lower than if he use a rigid hose. The pump has to keep a flexible hose stretched, causing a pressure drop, he says.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say that if the diameter of the hose doesn’t change while the pump runs, there will be no pressure drop, so once the pump is fully running, it doesn’t matter if the hose is flexible or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Who is right?&lt;/p&gt;&#xA;" OwnerUserId="16747" LastActivityDate="2023-03-05T15:57:31.643" Title="Do flexible water hose cause pressure drop with constant flow?" Tags="|pressure|statics|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22705" PostTypeId="2" ParentId="22704" CreationDate="2018-07-12T12:17:01.143" Score="1" Body="&lt;p&gt;There is no difference once the flow is stabilized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a dynamic situation, the hose may yield under increase of pressure; the hose stretches, more water is accumulated in it, outflow is lower than inflow, and outflow pressure is lower. But the hose can't stretch indefinitely - at certain point its tension offsets the pressure, amounts of flow stabilize, pressure and flow rate at outlet rise, and the situation becomes equivalent to rigid hose in all respects. Whether the hose is rigid or not, the forces acting on its walls don't move them, so no work is done on the hose - the pressure is not affected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And of course, the moment you switch the pump off, the flow may continue for a while in the soft hose, as the pressure drops and the hose contracts against it; that wouldn't happen with a rigid one. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-07-12T12:17:01.143" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22706" PostTypeId="2" ParentId="22106" CreationDate="2018-07-12T13:39:15.033" Score="2" Body="&lt;p&gt;NMPC (nonlinear MPC) of oil recovery processes is an ongoing field of research, and it involves control intervals on the scale of months. It is also referred to as closed-loop reservoir management (CLRM).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www2.imm.dtu.dk/~jbjo/oilproduction.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This website&lt;/a&gt; has an introduction to NMPC for oil reservoirs and CLRM. It also contains links to other research groups and organizations, publications, and software.&lt;/p&gt;&#xA;" OwnerUserId="16750" LastActivityDate="2018-07-12T13:39:15.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22707" PostTypeId="1" CreationDate="2018-07-12T13:41:15.457" Score="0" ViewCount="6064" Body="&lt;p&gt;Pump action seems to be a faster way to cycle a weapon compared to bolt action and it's a rather simple action, unlike straight-pull bolt which are more prone to malfunction. It almost seems like a direct upgrade to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we have magazine-fed pump action shotguns nowadays, It's probably possible to do the same with rifles. Even if we only had tubular magazines, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lebel_Model_1886_rifle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lebel rifle&lt;/a&gt; actually had a tubular magazine with spitzer (pointy) bullets that was designed to prevent cartridges from firing the one in front of it in the magazine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that nowadays this would be a novelty because of the availability of semi-auto rifles but why couldn't it replace bolt-action rifles?&lt;/p&gt;&#xA;" OwnerUserId="16749" LastActivityDate="2024-03-01T00:42:49.723" Title="Why are pump action rifles so uncommon compared to bolt-action rifles?" Tags="|firearms|" AnswerCount="5" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="22710" PostTypeId="2" ParentId="22704" CreationDate="2018-07-12T16:19:40.767" Score="1" Body="&lt;p&gt;Yes hoses or pipes always cause some pressure drop when flowing, it may be small. For example ; part of a test given to industrial painters is &quot; how much pressure drop is there for each of different lengths of hose when using an airless paint sprayer? &quot; . That is, they must be able to adjust for the pressure drop between the pump and the nozzle for various flow rates and hose lengths  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-07-12T16:19:40.767" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22711" PostTypeId="2" ParentId="15307" CreationDate="2018-07-12T17:23:37.647" Score="6" Body="&lt;p&gt;This is an excellent question, and one that has not yet been fully answered by people working in this area of research. Let's look at some of the bigger projects and see what their proposals are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ITER, the tokamak (giant doughnut) being built in France is supposed to be the first reactor to ever beat the break even point, unless someone either beats them to it beforehand or it doesn't work like we saw with the hotly anticipated NIF (National Ignition Facility) fusion experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their goal is to simply beat the break even energy barrier, there are no good plans, that I have seen at this time, on how to easily extract energy for power production, but it will likely use a steam process. Their current best shot is to use the high energy neutrons resulting from the fusion reactions to lose energy in the walls, thereby heating it and having that heat removed by a high pressure water blanket on the outside. This will be extraordinarily challenging, but not as difficult an engineering problem as solving fusion in the first place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The NIF (National Ignition Facility) uses small spherical capsules of super cooled deuterium/tritium in a hohlraum (precision made tube) that converts the energy of the worlds largest laser into x-rays in order to compress the fusion fuel to a density of around 1000g/cm^3 causing fusion to occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their plan was to make a system where these capsules would be loaded into the chamber in quick succession to maintain a &quot;constant&quot; energy output. Problems with this are cost, complexity, and inability (so far) to be able to get past the break even threshold. They likely would implement a steam cycle for power generation if ever successful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there are some private sector ventures. Most are not doing much in my opinion (they can't even manage basic fusion reactions, let alone anything near the break even barrier), although there are a couple exceptions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;General Fusion has probably the second best idea in fusion for electrical production. They are using a massive sphere full of molten lead that swirls around to create a vortex that they shoot a sub fusion temperature plasma into. Once the plasma is in the center of the swirling lead, a series of pistons SMACK hammers that in turn produce a massive shock wave that then collapses the plasma resulting in fusion reactions. This in turn would heat the lead which has heat exchangers built into the reactor which in turn boils water for a steam cycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several advantages to their design. First, they don't have to worry about melting down the containment walls (like with a tokamak) as they are already molten. Second, the high energy neutrons from the deuterium/tritium reactions can be moderated by the large volume of lead (it's a poor moderator compared to other elements, but having such a large physical cross sectional area it works). Third, the lead can be made into an amalgam with Li6 to absorb the now thermal neutrons to produce more tritium (most expensive component of their fuel at ~$30,000.00/g) while also providing a nice 'boost' in power. Fourth, the pistons could be directly driven by the steam created from the reactor for greater efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another company, Tri Alpha Energy (although they are pushing for being called TAE now) is trying proton/boron 11 fusion to create an unstable C12 that blows itself apart into 3 helium 4 nuclei that are referred to as alpha particles, hence the name 'Tri Alpha' Energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being in the industry, I have had ample opportunity to speak with some of their people. It seems their big focus (no pun intended), as I understand it, is on trying a new technique for compressing and maintaining a plasma through self generated magnetic fields (as opposed to shoving it together externally). The issue from what they have told me is their equipment is off by a minimum of an order of magnitude in order to hit fusion temperatures, even after 20 years of development (fusion is hard after all). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mention them because they did have a unique idea on electrical production which ties directly to your question. They claimed to be working on a 'reverse cyclotron' where they would input the alpha particles 'backwards' into a cyclotron to generate electrical power. There are about a million challenges to this problem, e.g. how do you get the alpha particles to enter the device precisely enough to generate a magnetic field to produce electrical current and do it without cooling the plasma to such a degree that the plasma can no longer maintain fusion reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put simply, the big goal right now is just try to figure out how to create excess energy, by any means necessary, and work out power production after.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this answers your question!&lt;/p&gt;&#xA;" OwnerUserId="553" LastActivityDate="2018-07-12T17:23:37.647" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22712" PostTypeId="1" CreationDate="2018-07-12T19:22:23.337" Score="0" ViewCount="70" Body="&lt;h2&gt;The problem domain&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You have a shelter that you use in a cold climate, and you want to provide it with some heat to take the edge off the chill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You already have a structure in a location which gets cold winters, let's say in the -15C to -30C range. When it gets down to -30C you are used to turning on a heater, whose fire usually increases the temperature by approximately 10 degrees (to -20C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to construct a similar structure - exactly like the first, or as close as possible - except this one will be in an even colder area near one of the planetary poles and getting winters that might reach temperatures as low as -80C. Similar to before, you plan to turn on the heater if you wake up to the lowest expected temperature of -80C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to figure out how much the heater will warm the structure in the colder area. It very well may just heat the structure by 10 degrees as before up to -70C, but I do not know for sure.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This question does not seem to be as straightforward as I originally thought it would be, and there have been other comments and posts probing some specifics here and not getting me to a satisfactory answer, &lt;strong&gt;so the question is this...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are all the non-negligible forces at play here which should be considered to decide if the secondary structure will be heated by the same amount (10 degrees) as the first structure, and in what way does each of these affect the result?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it matters: The building will likely be either a cold-weather tent or a human-made cave made from snow which is designed with the floor higher than the doorway to trap the warmer air inside. If the specific choice matters, feel free to either say so in comment or assume one or the other for your answer.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;See also&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I think the primary concern might be the &lt;code&gt;heat capacity&lt;/code&gt; (ie: the number of heat units needed to raise the temperature of a body by one degree) of air. See also my question &lt;a href=&quot;https://physics.stackexchange.com/q/416337/155598&quot;&gt;Physics.SE: If two objects of different temperatures have the same heat source applied, do they heat up by the same amount?&lt;/a&gt; for the Q&amp;amp;A which led me to this (hopefully correct) assumption. One of the answers even seemed to suggest that the heat capacity change may be significant, assuming I read the complicated answer correctly (maybe not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to Google up whether the heat capacity of air changes with temperature. I found an interesting &lt;a href=&quot;https://www.ohio.edu/mechanical/thermo/property_tables/air/air_cp_cv.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;table labeled &quot;Specific Heat Capacities of Air&quot;&lt;/a&gt;, though that does not go down far enough, &lt;a href=&quot;https://www.engineeringtoolbox.com/air-specific-heat-capacity-d_705.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;and a calculator&lt;/a&gt; which does. I get 1.005 kJ/(kg&lt;em&gt;K) for 0C and 1.009 kJ/(kg&lt;/em&gt;K) for -80C. In my naivety, I would guess that means the heat source might work 0.4% better at the colder temperature (making this tree I'm barking up a negligible one), but I would not be surprised if I am way off base.&lt;/p&gt;&#xA;" OwnerUserId="16752" LastActivityDate="2018-07-13T19:43:38.277" Title="How do I correctly model a building to determine effectiveness of a heat source?" Tags="|heat-transfer|heating-systems|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22714" PostTypeId="2" ParentId="22696" CreationDate="2018-07-12T20:21:14.227" Score="1" Body="&lt;p&gt;If you assume that there is normally some cross-wind, then the rotor will be downstream and operating in the wake of vehicles in the windward lane that are powering it. Then this is basically free energy. Also, it might lower the buffeting felt by the lee-side lane's vehicles, netting a further global gain.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-07-12T20:21:14.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22716" PostTypeId="1" CreationDate="2018-07-12T23:22:15.190" Score="2" ViewCount="109" Body="&lt;h1&gt;This is not a StackOverflow question because the software engineers do not have appropriate math skills to answer an IEEE 754 question and thus must be answered by an IEEE tutor.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I had to implement my own string-to-float and float-to-string algorithms for &lt;a href=&quot;https://www.cs.tufts.edu/~nr/cs257/archive/florian-loitsch/printf.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Grisu&lt;/a&gt; and Unicode using C++ templates. I used the following algorithm found in the Wikipedia &lt;a href=&quot;https://www.wikiwand.com/en/Single-precision_floating-point_format&quot; rel=&quot;nofollow noreferrer&quot;&gt;Single-precision Floating Point&lt;/a&gt; Format article:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;consider a real number with an integer and a fraction part such as 12.375&lt;/li&gt;&#xA;&lt;li&gt;convert and normalize the integer part into binary&lt;/li&gt;&#xA;&lt;li&gt;convert the fraction part using the following technique as shown here&lt;/li&gt;&#xA;&lt;li&gt;add the two results and adjust them to produce a proper final conversion&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am unsure what to do when the string being converted to float has more digits precision then I can store in a 32-bit (for example) float but it does not hit +/- infinity, and the proper way to round the imprecise digits/decimals. For 32-bit floating-point numbers, there is only 7 digits of precision. For 64-bit floating-point numbers, it's 16 digits of precision. Any more numbers than that must be rounded using the &lt;a href=&quot;https://www.wikiwand.com/en/IEEE_754#/Rounding_rules&quot; rel=&quot;nofollow noreferrer&quot;&gt;IEEE 754 rounding rules&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Example&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Base 2 Option&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We convert the integer portion first, like the algorithm in the &lt;a href=&quot;https://www.wikiwand.com/en/Single-precision_floating-point_format&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki entry&lt;/a&gt;, by scanning to the least significant digit then parse back to the most significant digit by multiplying each digit by a power of 10 (i.e. the standard string-to-integer algorithm) and use unsigned multiplication wrap-around to detect bit overflow, round in base 2; which doesn't sound right to me and I think the Wikipedia article is not the best algorithm.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Base 10 Option&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;We convert every number to into exponent notation with a single most significant digit, such as converting 123456780123456789.012. to 1.234567e20 using the &lt;a href=&quot;https://www.wikiwand.com/en/IEEE_754#/Rounding_rules&quot; rel=&quot;nofollow noreferrer&quot;&gt;IEEE 754 rounding rules&lt;/a&gt;. Sounds like the best choice because we only have to parse the text once and we have to scan for the exponent and we can avoid the string-to-integer algorithm. Rounding in base 10 sounds the sanest due to the number still being in base 10 and you only need to read the digits and decimals once without multiplying by a power of 10. If this is the preferred method then wiki entry should be updated.&lt;/p&gt;&#xA;" OwnerDisplayName="user16755" LastEditorDisplayName="user16755" LastEditDate="2018-07-20T14:22:20.677" LastActivityDate="2018-07-20T14:22:20.677" Title="What is the proper way to handle string-to-float decimal truncation in IEEE 754?" Tags="|computer-engineering|" AnswerCount="0" CommentCount="8" ClosedDate="2018-07-13T09:25:38.850" ContentLicense="CC BY-SA 4.0" />
  <row Id="22717" PostTypeId="2" ParentId="22712" CreationDate="2018-07-13T03:59:05.933" Score="1" Body="&lt;p&gt;If we look at this as realistic experiment, the distribution of warmer air throughout the shelter is mainly due to convection of warm air from the walls and object near to the flow of warm air to colder surfaces and circulating back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Convection in a space potentially partitioned by walls and openings and crowded by articles and furniture of deferent material and shape will be very complex. strata of layers of warm air with different temperature will create an union like geometry starting from warmer at center zones to colder at outer zones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However one major component is the heat source. If it is going to be the same in both cases, meaning same surface temperature of heat element such as electrical heater or furnace or similar, because of the larger difference in heat source and ambient temperature, convection will happen in narrower, relatively warmer, and faster rising plums and the gradient of layers temperatures will be steeper. One would need to imagine a time laps image of the shelter's occupants movements, and usage of space density, then design the distribution of warm air efficiently to cover that usage density pattern closely.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-07-13T03:59:05.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22718" PostTypeId="2" ParentId="22704" CreationDate="2018-07-13T05:54:48.307" Score="2" Body="&lt;p&gt;Going around corners does induce a pressure drop, so if the rigid hose has fewer tight bends, it will perform (slightly) better in this application. Realistically, however, the gain is marginal, and you’re better off making the decision on financial/availability criteria.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-07-13T05:54:48.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22719" PostTypeId="1" AcceptedAnswerId="22776" CreationDate="2018-07-13T07:42:06.403" Score="1" ViewCount="107" Body="&lt;p&gt;I am going to store used li-ion batteries for some long time &lt;code&gt;&amp;gt;1&lt;/code&gt; year, and &lt;a href=&quot;http://batteryuniversity.com/index.php/learn/article/how_to_store_batteries&quot; rel=&quot;nofollow noreferrer&quot;&gt;I have read&lt;/a&gt; that optimal storage conditions are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;cool ambient temperatures, like &lt;em&gt;2-5 degrees celsium&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;3.82V voltage per cell.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I cannot find is what is correct operation sequence to prepare a &lt;em&gt;used&lt;/em&gt; battery for cold storage (which is initially in &lt;em&gt;random&lt;/em&gt; state)?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Should I charge it to 4.2V then &lt;strong&gt;discharge&lt;/strong&gt; to 3.82V? &lt;/li&gt;&#xA;&lt;li&gt;Should I discharge it to 3.6V then &lt;strong&gt;charge&lt;/strong&gt; to 3.82V? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16760" LastEditorUserId="7083" LastEditDate="2018-08-14T00:10:55.260" LastActivityDate="2018-08-14T00:10:55.260" Title="Li-ion batteries storage" Tags="|electrical-engineering|battery|energy-storage|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22720" PostTypeId="1" CreationDate="2018-07-13T08:28:24.233" Score="1" ViewCount="32060" Body="&lt;p&gt;Free Air Delivery (FAD) is a common measure of the capacity of an air compressor. I have a question that comes at the end here. My question is a little different from what you might be expecting. So first, let me establish some basics so that we don't just get replies giving a version of the standard answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;FAD is a flow rate of air derived from measurement at the outlet of the compressor, defined at the corresponding outlet pressure, e.g. 300 litres per minute @ 700kPa. However, FAD is not simply the raw flow rate measured at the discharge at this raised pressure. The measured flow at the outlet first has to be converted back using a formula, to be the equivalent flow rate that would be obtained if the pressure and temperature were the same as at the intake. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put in other words, here is a quote from a guide put out by the German compressor manufacturer Kaeser:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The air delivery of a compressor (known also as the free air delivery or FAD) is the expanded volume of air it forces into the air main (network) over a given period of time. The correct method of measuring this volume is given in the following standards: ISO 1217, annex C and DIN 1945, Part 1, Appendix F. ... Proceed as follows to measure FAD: the temperature, atmospheric pressure and humidity must first be measured at the air inlet of the compressor package. Then, the maximum working pressure, temperature and volume of compressed air discharged from the compressor are measured. Finally, the volume V2 measured at the compressor outlet is referred back to the inlet conditions using the following equation ... The result is the free air delivery (FAD) of the compressor package. This figure is not to be confused with the airend delivery.&#xA;  $$ V_1= \frac{V_2 \cdot P_2 \cdot T_1}{T_2 \cdot F_1} $$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://us.kaeser.com/Images/Compressed%20Air%20Engineering%20Guide-tcm9-775853.pdf/&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Kaeser &amp;#39;Compressed Air Engineering: Basic principles, tips and suggestions&amp;#39;.&quot;&gt;Source: Kaeser &quot;Compressed Air Engineering: Basic principles, tips and suggestions.&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that F&lt;sub&gt;1&lt;/sub&gt; is a factor where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F1= p_1- (p_p \cdot F_{rel}) $$&#xA;where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;p&lt;sub&gt;1&lt;/sub&gt; = inlet pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p&lt;sub&gt;p&lt;/sub&gt; = vapour pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F&lt;sub&gt;rel&lt;/sub&gt; = inlet humidity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For teaching purposes, that is a quite a lot to take in at once. It would be good to state an initial, approximate, concept that can then be qualified and refined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to my question. Is this not virtually the same as saying: &quot;&lt;em&gt;Aside from leakage losses, FAD is equivalent to the flow rate of air being sucked into the intake of a compressor that is working against a certain backpressure&lt;/em&gt;.&quot; Is that a legitimate statement?&lt;/p&gt;&#xA;" OwnerUserId="16759" LastEditorUserId="16759" LastEditDate="2018-07-13T11:16:53.773" LastActivityDate="2022-04-27T03:27:20.717" Title="Define Free Air Delivery for an air compressor" Tags="|compressed-air|compressors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22721" PostTypeId="2" ParentId="22720" CreationDate="2018-07-13T10:22:18.670" Score="0" Body="&lt;p&gt;I would suggest that since p, v and T are changing you should modify that statement to be about the mass flow input being equal to the mass flow output.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-13T10:22:18.670" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22722" PostTypeId="2" ParentId="22720" CreationDate="2018-07-13T10:49:55.240" Score="1" Body="&lt;p&gt;Build up to the formula (is $F_2$ a typo for $P_2$?) one step at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the simplest situation, $V_1 = V_2$ (which should be obvious to everybody in the room). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for a real compressor test, the inlet and outlet pressures will be different, so you need to correct for that using the ideal gas law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also the inlet and output temperatures will be different, so you need to correct for that as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's all there is to it! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good way to reinforce the formula is to actually &lt;em&gt;measure&lt;/em&gt; the flow rate, temperatures, and pressures, and show that the &quot;correction factors&quot; do matter in practice. For a small compressor, you should be able to find a way to measure $V_1$ directly at the inlet to verify the formula. Even if that slightly affects the compressor operation (e.g. attaching a pipe with a flow meter to the inlet), it shouldn't matter for the purpose of demonstrating what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the mass flow in does equal the mass flow out, but that is just a different way of deriving the &quot;complete&quot; formula.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-07-13T10:56:04.323" LastActivityDate="2018-07-13T10:56:04.323" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22723" PostTypeId="2" ParentId="22704" CreationDate="2018-07-13T11:21:51.360" Score="0" Body="&lt;p&gt;For Same size and same length of pipe lines, pressure drop will depend only on number of bends and internal smoothness of the pipe line. So depending upon above factors you may find out out in which case pressure drop will be higher.&lt;/p&gt;&#xA;" OwnerUserId="16732" LastActivityDate="2018-07-13T11:21:51.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22724" PostTypeId="1" CreationDate="2018-07-13T12:09:12.887" Score="1" ViewCount="22" Body="&lt;p&gt;Disclaimer: I am new in the design of any fluid systems, however the selection of these units is critical for the electrical design I am responsible for this system, so please excuse me if my approach is naive. We are running the system @ 480V 60Hz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have an application where we are using Fuji Ring Compressors (&lt;a href=&quot;https://americas.fujielectric.com/files/Fuji%20Electric%20VFC90.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://americas.fujielectric.com/files/Fuji%20Electric%20VFC90.pdf&lt;/a&gt;) in a vacuum mode to hold aluminum panels up with vacuum belts. The outlet of the ring compressor goes into 3 chambers connected to the belt track. The belts have holes along the middle which provide enough suction force to hold up the panel. This method has proven to be successful in the past but we are expanding the number of belts, and would like to avoid simply adding more ring compressors (we would need 9 more units to expand the system in the same way, and we are short on space). What I am investigating is if we can use a larger compressor to match the outputs of two of these old units. This is the highest capactiy model that Fuji can offer in the same range (&lt;a href=&quot;https://americas.fujielectric.com/wp-content/uploads/2017/05/VFB2000-7Wa-Single-Stage-Ring-Compressor.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://americas.fujielectric.com/wp-content/uploads/2017/05/VFB2000-7Wa-Single-Stage-Ring-Compressor.pdf&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach was to look at this from a black box perspective, and for now not worry about the force required to hold the panels, figuring out # of holes covering etc, since I know the singular ring compressor could handle it. My thought is that since the pressure is inversely proportional to area, I can simply compare the outlets of the units and the pressure they can supply at a given airflow. So I imagine a y-connector at the 4&quot; outlet of the larger VFB unit splitting into two 3&quot; lines representing two of the VFC units. If the vacuum generated at the end of those two 3&quot; lines was greater than then what the VFC's could output, I can use the singular VFB unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The old VFC units have a 3&quot; diameter outlet for an area of ~7.07 inches^2. The new VFB units have a 4&quot; diameter outlet for an area of ~12.57 inches^2. If I am trying to match the output of two VFC units, I double the areas for those and compare the ratio (14.14/12.57 = 1.125). At any given point on the Flowrate/Pressure curves given in the datasheet, the pressure supplied by the VFB is greater than that ratio multiplied by the pressure supplied by the VFC by about 10-15%. I also anticipate less head loss in the lines because running one 4&quot; line and splitting it should be more efficient then two 3&quot; lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still want to do the analysis back to the force required to hold panel, and required pressure at each vacuum hole etc, but I want to know if my idea is even remotely feasible before I spend too much time investigating. My main concern is the drop in vacuum flow at each hole, but I am not sure if this is a concern if I satisfy the vacuum requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help and I would be glad to offer more information.&lt;/p&gt;&#xA;" OwnerUserId="16762" LastActivityDate="2018-07-13T12:09:12.887" Title="Matching Output of Two Old Ring Compressors using Larger Single Unit" Tags="|mechanical-engineering|fluid-mechanics|vacuum|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22725" PostTypeId="1" AcceptedAnswerId="22729" CreationDate="2018-07-13T15:06:59.147" Score="14" ViewCount="4341" Body="&lt;p&gt;I'm studying Euler's work on structural engineering from a book out of curiosity and it is mentioned that he developed a mathematical theory describing the buckling of columns under a parallel load (the weight-force of the load is directed down along the column). The theory is covered quickly without much motivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this got me thinking; why does a column &quot;buckle&quot; in the first place? If the load presses the column down, why does the column even start deflecting sideways? I know this happens in real life since this fact is easily confirmeable with household objects, but theoretically, why do objects start deflecting sideways instead of just compressing under loads? This might be something obvious and maybe I'm just overthinking but I find this curious nonetheless.&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2021-08-16T09:38:05.477" Title="Why does column buckling occur when the load is parallel to the column?" Tags="|structural-engineering|civil-engineering|" AnswerCount="7" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22726" PostTypeId="1" CreationDate="2018-07-13T15:17:30.903" Score="1" ViewCount="96" Body="&lt;p&gt;I am trying to make a &lt;strong&gt;MOSFET&lt;/strong&gt; switching circuit where its gate input is 12V clock pulse. Now when I put the load between (Vdd) and (Drain), there is no drop voltage VDS, but when the gate input falls down to 0, there still a voltage between (Vdd) and (Drain) (across the load).  - - - - - - case (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I put the load between (Source) and (Ground), the situation is reversed so that there is a voltage drop VDS, but when the input falls down to 0 the load voltage also becomes 0 (pulled down) . - - - - - - case (2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what is better, putting the load on case 1 or case 2 ?&lt;/p&gt;&#xA;" OwnerUserId="16767" LastEditorUserId="1832" LastEditDate="2018-07-14T00:32:21.160" LastActivityDate="2018-07-14T00:32:21.160" Title="Mosfet switching circuit" Tags="|electrical-engineering|power-electronics|circuits|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22728" PostTypeId="2" ParentId="22725" CreationDate="2018-07-13T15:30:31.360" Score="0" Body="&lt;p&gt;If the load is applied through the centre-line of the column then there is no side force, but if the load is offset, but parallel, then there is a sideforce which leads to buckling.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-13T15:30:31.360" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22729" PostTypeId="2" ParentId="22725" CreationDate="2018-07-13T15:36:53.873" Score="15" Body="&lt;p&gt;Euler buckling occurs because the world isn't perfect. So that theory assumes that there is an initial infinitesimal deviation along the column (assuming the column is in fact not perfectly vertical*). This deviation causes a bending moment along the beam, which increases the deviation, which increases the bending moment, which increases the deviation...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For loads lower than the Euler load, this vicious cycle eventually stabilizes and the beam doesn't buckle. For the Euler load and above, the cycle never stabilizes and the deflection goes to infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the real world has initial deviations and other problems which are much higher than &quot;infinitesimal&quot;. So in the real world, columns buckle with loads far lower than the theoretical Euler load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* This is the assumption for Euler buckling, but another possible deviation is that the load is actually not perfectly centered on the column. In the real world, both cases probably happen simultaneously&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-07-13T15:36:53.873" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22730" PostTypeId="2" ParentId="22725" CreationDate="2018-07-13T16:29:23.113" Score="4" Body="&lt;p&gt;Think about a &quot;thin&quot; beam, for example a strip of springy steel. It is very easy to bend the strip into a curve, compared with stretching or compressing it along its length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it is bent into a curve, the length of the strip &lt;em&gt;measured around the curve&lt;/em&gt; does not change significantly, and that means the straight-line  distance between the two ends becomes smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try this experimentally with something you can bend easily with your hands, you will find that a graph of the force against distance between the two ends is &lt;em&gt;not&lt;/em&gt; a straight line - the effective stiffness reduces as the load increases and the beam curves more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the stiffness when compressing the beam along its length without bending it is constant (and equal to $EA/L$, as shown in any strength of materials textbook).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since it is impossible to make a perfectly straight beam in the real world, the beam will buckle when the end load reaches the point where the stiffness in &quot;bending sideways&quot; becomes less than the stiffness in &quot;perfect compression&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Euler's formula gives a fairly good approximation to that load, though it makes a few more assumptions (for example, about the shape of the beam when it bends sideways) which are not completely accurate. But since the tolerances in the beam geometry are also unknown, Euler's formula is good enough to be useful in practice, even though it usually &lt;em&gt;over-estimates&lt;/em&gt; the actual buckling load by a factor of a few times (say between 2 and 5 times) compared with real life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the beam becomes more flexible after it buckles, if you apply a &lt;em&gt;constant&lt;/em&gt; end load (e.g. the weight of something pressing on the end of the column) the buckling results in catastrophic failure, as the beam curves more and more until it breaks. On the other hand, if you apply a controlled &lt;em&gt;displacement&lt;/em&gt; to the end, the process is reversible and when the load is removed the beam will return to its (nominally) straight shape, with no permanent damage.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-07-13T16:29:23.113" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22731" PostTypeId="1" AcceptedAnswerId="22735" CreationDate="2018-07-13T17:01:33.743" Score="5" ViewCount="533" Body="&lt;p&gt;Lets say you want to build a bridge out of carbon fiber. This has never been never been done before, and it's super expensive stuff. So your goal is to use as little material as possible, but obviously your bridge still needs to have enough strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To start you look up how strong is carbon fiber, and the online data sheets tell you that a 10 oz cloth has a tensile strength of 800KSI. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After running the math this sounds perfect and should give your bridge enough strength for it's uses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, before you start a super expensive project that's never been done, based on nothing more than datasheets you found online and equations that might not fully apply to your material you run a sanity check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you start building your full sized bridge, you build a small one that's 1/10th the size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming this small bridge is exactly the same as the large one, just using 1oz fiber instead of 10 oz fiber would your small bridge endure 1/10th the load of the full sized one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't mean to apply this question to carbon fiber alone. With anything in engineering if you scale up a model perfectly using the same materials would the strength scale up proportionally, or is there some kind of real life nuance that blows all this mathematical beauty out of the water?&lt;/p&gt;&#xA;" OwnerUserId="16726" LastEditorUserId="16726" LastEditDate="2018-07-13T19:43:01.197" LastActivityDate="2018-08-06T15:06:22.733" Title="Do scaled models reflect strength characteristics of full size" Tags="|structural-engineering|structural-analysis|stresses|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22732" PostTypeId="2" ParentId="22725" CreationDate="2018-07-13T17:28:45.237" Score="2" Body="&lt;p&gt;Not all the columns fail under compression by buckling. In steel columns shorter than slenderness ration of 50 they fail by direct compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the principal of stability bifurcation and it appears not only in columns but in failure mode of many other shapes, such as beams, trusses, vessels, and the buckling pattern could be quite complex. Fro example if you cut the cap and bottom of a can of coke and put it under a micro control press, it will buckle along diamond pattern on its wall, twisting around the vertical axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In columns it happens because of the elastic behavior of the material leading to bifurcation, be it steel or aluminum, wood, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not due to residual imperfection in manufacturing of the column, neither due to load not applied at perfect centre, although those conditions will affect the reaction of column but that belongs to another topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you increase the load applied to the column compression stress develops on the area of the cross section. This stress is applied evenly over the surface of the section, $$ \sigma = P/A$$&#xA;But this stress is constantly seeking out ways to force the column to curve so as to release the stress by creating small variation in intensity distribution over the surface area while the total stress is constant, hence creating a lateral momentum, but up to buckling force this virtual stress is just not enough to force the buckling.&#xA;When the load reaches the buckling level the column will fail by randomly bending on either of two sides which have bigger slenderness ratio.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-07-13T20:04:45.667" LastActivityDate="2018-07-13T20:04:45.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22733" PostTypeId="2" ParentId="22703" CreationDate="2018-07-13T18:41:12.810" Score="0" Body="&lt;p&gt;I believe my solution is correct based on the solution here: &lt;a href=&quot;https://aapt.scitation.org/doi/10.1119/1.1976111&quot; rel=&quot;nofollow noreferrer&quot;&gt;American Physics Journal Solution&lt;/a&gt; The phycist uses a different more general method to solve the same problem, following his approach and using his derivations I arrived at the final position being at -22.6mm (in compression). Therefore I will assume the books answer to be a little off from the actual solution. Hope this helps someone later on down the line!&lt;/p&gt;&#xA;" OwnerUserId="16746" LastActivityDate="2018-07-13T18:41:12.810" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22734" PostTypeId="2" ParentId="22731" CreationDate="2018-07-13T18:50:16.957" Score="0" Body="&lt;p&gt;According to J. E. Gordon in his book on structures, the answer is no. The reasons are simple: structural strength is a property related to material density. As you decrease the size of the structure you must proportionately decrease its material structural strength. This means you need to use a material that is $\frac 1{10}$ in scale strength also. When you scale size, scale material strength also. This means that its appearance may differ in the model but in performance metrics be similar to the full scale structure. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2018-08-06T15:06:22.733" LastActivityDate="2018-08-06T15:06:22.733" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22735" PostTypeId="2" ParentId="22731" CreationDate="2018-07-13T18:53:07.487" Score="9" Body="&lt;p&gt;Things scale, but not always in a nice simple way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example consider the units of &lt;em&gt;stiffness&lt;/em&gt; of part of the structure are $E$ (Young's Modulus) times some length, but the units of &lt;em&gt;weight&lt;/em&gt; are $\rho$ (density) times some length cubed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a scaled-down model structure will deform much less under its own weight than the original. (This explains why elephants have thicker legs than ants, relative to the size of their bodies.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same sort of thing happens with other types of loading, for example wind loads would probably be proportional to a surface area, i.e. length squared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real life the joins between the various parts will be important for the strength of the complete structure, and you might not be able to scale those down by a factor of 10 easily. Think about the practical size of bolts and rivets or the thickness of adhesive films, welds, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geometrical tolerances usually don't scale the way you would like either. Neither do things like defects in the surface finish that might act as sites for crack propagation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above list is far from complete. All this doesn't mean that scale models are useless, but you have to think through what you are really trying to do when you use them, and they aren't a &quot;magic solution&quot; to validating a design.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-07-13T18:53:07.487" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22736" PostTypeId="2" ParentId="22712" CreationDate="2018-07-13T19:43:38.277" Score="1" Body="&lt;p&gt;Like others have mentioned, this can be complex depending on your conditions. Providing additional information helps us determine not only complicated things, but also determine if a simple method is satisfactory or not. I will provide some simple intuition on the topic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_conduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thermal conduction&lt;/a&gt; is a linear equation. If all of your heat transfer in this shelter was via conduction, your 10C temperature increase in both shelters would be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_radiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thermal radiation&lt;/a&gt; is an exponential equation that result in exponentially more heat being lost from a higher temperature source. In your case it will probably make no difference at a small 10C differential. (considering the 37C human in the shelter, it will mater a lot)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Convective_heat_transfer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thermal convection&lt;/a&gt; is more complicated and will result in exponentially more heat being lost from a higher temperature source. This will depend on the size and shape of the shelter, design of the heater (forced or natural convection), as well as the location of the human and how much heat they are giving off. Since it is not possible to calculate without significantly more information; your assumption of 10C increase in both locations is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what your goal is in knowing this information, but if you plan to heat such a shelter I recommend just using the shelter to eliminate the outside wind (forced convection) and create a protected space. A temperature increase from -80C to -70C is kindof a waste of energy. It would be better to use this heat energy to heat the inside layer of an occupants sleeping bag or insulated clothing. This reduces the area the heat can be dissipated over, and the clothing sounds like it will have much better insulation properties than your structure. It is important to stay at least a few degrees below body temperature or sweating will create a new problem.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-07-13T19:43:38.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22737" PostTypeId="2" ParentId="22707" CreationDate="2018-07-13T22:08:42.073" Score="4" Body="&lt;p&gt;I found this &quot;answer&quot; on Quora:&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pump rifles are not popular because a pump action excels at quick shots rather than precise ones. Moving the forend to reload throws off the shooting position. This is fine for shotguns, but kind of defeats the point of a rifle.  For people who might need a quick follow-up shot, autoloading (semi-automatic) rifles fit the bill a lot better, because the shooter does not need to move between shots. Also, by the time pump rifle designs became practical, autoloaders cost similar to them, so the only benefit they might have had was gone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;I'd have to say that the main reason why pump action rifles just aren't that popular is that there really isn't a niche market for them. Some manufactures do make them&lt;sup&gt;1&lt;/sup&gt;, but as far as I can tell from articles I've seen in various publications, they're mostly available in small calibers. It's not so much a matter of technology, but of marketing.&lt;br&gt;&lt;br&gt;And there might be some issues with the locking mechanism. With any single-shot rifle over, say, 30-30, there has to be a reliable bolt-locking mechanism to keep the breech closed until the projectile exits the muzzle and the muzzle blast dissipates. The strength needed to operate the pump on such a rifle would even further cause the shooter to completely lose their original sight picture further obviating the advantage of a pump action mechanism.&lt;br&gt;&lt;br&gt;In short, there just isn't a market for them.&lt;br&gt;&lt;br&gt;_______________&lt;br&gt;1. &lt;a href=&quot;https://www.remington.com/rifles/pump-action&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.remington.com/rifles/pump-action&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12535" LastActivityDate="2018-07-13T22:08:42.073" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22738" PostTypeId="2" ParentId="22731" CreationDate="2018-07-13T22:14:16.043" Score="0" Body="&lt;p&gt;There has been some valid comments on the other answers, at the expense of being repetitive I mention just a few more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your analogy of the bridge for example. The cables' strength scales to the surface area of the cable, L^2 but the seismic forces are related to mass of the structure which is volume L^3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wind testing of a scaled model is very complex and depending on geometry and openings has to be individually element by element analyzed and scaled back up&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In lab tests, say to test the shear strength of a manufacturer's proprietary assembly of a shear wall they do have scaled down models occasionally, but they are aware and calculate all these disparities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;City of Los Angeles accepts sometimes theses test results for performance of some products( LA RR reports) but they expect careful analysis and logic used to justify scaling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However thanks to powerful simulation algorithms many projects can be directly simulated for analysis from the digital drawings.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-07-13T22:14:16.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22739" PostTypeId="1" CreationDate="2018-07-14T05:46:24.963" Score="2" ViewCount="65" Body="&lt;p&gt;I have a design for a gasoline engine cylinder that is open at the top, closed at the bottom (a single pipe cylinder) that operates on opposed force (the pistons move lateral to the up-down direction of the pipe.  I believe it's an extremely efficient engine design that would produce -no disorganized cloud of vapor explosion-, absolutely none, but a single diamond shaped concentrated vapor burst between the two pistons, and at the top the flame profile turns into a string before disintegrating.  The description is, it says top torus, actually those are the pistons, and they're both half hemispheres.  The ring around the base of the pistons are repeated metallic X's that control the flame profile, and the cylinders move in a horizontal position laterally, not up and down as this is an opposed force design.  Here is a youtube video of an open pipe explosion: &lt;a href=&quot;https://www.youtube.com/watch?gl=UG&amp;amp;hl=en-GB&amp;amp;v=2--CWLW_TvE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Open Pipe Explosion #2&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, since this a nonlinear explosion in an engine that should be the most efficient, how would somebody go about making the engine, or convincing an automaker to build one.  I don't have a patent on it, but I designed it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9aiX3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9aiX3.jpg&quot; alt=&quot;cylinder&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16774" LastEditorUserId="7083" LastEditDate="2018-07-15T21:30:05.913" LastActivityDate="2018-07-15T21:30:05.913" Title="Explosion in an open pipe" Tags="|automotive-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22740" PostTypeId="1" AcceptedAnswerId="22759" CreationDate="2018-07-14T20:17:31.717" Score="0" ViewCount="139" Body="&lt;p&gt;I was going through the wikipedia page on polymer concretes. It seems like a good type of building material. I was wondering which types of polymers are used in the concrete, since I cannot find any online besides epoxy.&lt;/p&gt;&#xA;" OwnerUserId="16653" LastActivityDate="2018-08-15T08:01:34.053" Title="Which polymers are used to make polymer concrete?" Tags="|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22746" PostTypeId="1" CreationDate="2018-07-15T07:43:34.860" Score="0" ViewCount="81" Body="&lt;p&gt;In a polymer production process, there are a lot of manipulated variables(MV) such as temperature ,pressure , or level of reactor that affect the quality of my polymer product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it usually has &quot;dead-time&quot; problems in a process especially when the scale of the system is quite large. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already collected empirical data from the process for 6 months. Now, I want to find out the &quot;dead-time&quot; of each MV that has significantly importance on the final quality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, if I increase the temperature for 1 degree C, after 1 hour, we'll see the viscosity of polymer also increase for 100 cP. The time 1 hour is the answer I'd like to figure out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I do if I can only analyze the empirical data on hand without doing experiment on the production lines?&lt;/p&gt;&#xA;" OwnerUserId="16779" LastActivityDate="2024-04-04T18:00:55.143" Title="How to identify the time delay of manipulated variables which are important to the final quality of polymer product" Tags="|control-engineering|chemical-engineering|process-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22752" PostTypeId="1" AcceptedAnswerId="22755" CreationDate="2018-07-15T12:23:10.210" Score="1" ViewCount="48" Body="&lt;p&gt;Are electron photographs showing electrical charges? Do electrically charged parts show up more (darker in the photo) ?&lt;/p&gt;&#xA;" OwnerUserId="16780" LastActivityDate="2018-08-07T06:51:17.020" Title="Electron micrographs, is it showing electrical charges?" Tags="|optics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22755" PostTypeId="2" ParentId="22752" CreationDate="2018-07-15T13:30:34.293" Score="0" Body="&lt;p&gt;Yes, because the electrons interact mainly electromagnetically, all of their other interactions are negligible to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electronmicroscope generate a 2D plot, essentially a matrix, mapping the picture we made. It depends on the imaging software, how it maps the values to greyscale colors.&lt;/p&gt;&#xA;" OwnerUserId="9" LastEditorUserId="9" LastEditDate="2018-08-07T06:51:17.020" LastActivityDate="2018-08-07T06:51:17.020" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22759" PostTypeId="2" ParentId="22740" CreationDate="2018-07-16T03:04:10.880" Score="2" Body="&lt;p&gt;There are many categories of binders and more are being added due to research and development in high strength binders. The two categories of binders or resins are thermosetting and thermoplastic. Each has sub categories as shown below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermosetting resins: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Epoxy&lt;/li&gt;&#xA;&lt;li&gt;Polyester&lt;/li&gt;&#xA;&lt;li&gt;Acetone formaldehyde&lt;/li&gt;&#xA;&lt;li&gt;Phenol formaldehyde&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thermoplastic resins&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Methyl methacrylate&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Indencumaron&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com/books?id=NiQtAgAAQBAJ&amp;amp;lpg=PA3&amp;amp;ots=BLCaU29OOc&amp;amp;dq=Indencumaron&amp;amp;pg=PA3#v=onepage&amp;amp;q=Indencumaron&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google book research, click here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="10902" LastEditDate="2018-07-16T05:08:56.537" LastActivityDate="2018-07-16T05:08:56.537" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22760" PostTypeId="1" CreationDate="2018-07-16T10:17:53.330" Score="5" ViewCount="1898" Body="&lt;p&gt;During tensile test, why does the specimen's(let's say mild steel) volume changes during elastic deformation and does not change during plastic deformation?&lt;/p&gt;&#xA;" OwnerUserId="11070" LastActivityDate="2018-07-16T12:02:53.657" Title="Volume change during tensile test" Tags="|civil-engineering|materials|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22761" PostTypeId="1" AcceptedAnswerId="22766" CreationDate="2018-07-16T11:57:08.640" Score="0" ViewCount="20" Body="&lt;p&gt;The problem is illustrated in the picture below. A viewer looks through a slit and into a room. The problem is to find the maximum height of the wall in the room that the viewer can see if they are able to move 'right in front' of the slit and up to 166 cm away from the slit 'in the right'. (If it makes things harder or anything.. the slit is 3.5 cm wide)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In other words, what distance must the viewer be from the slit too see most of the wall which is 280 cm tall&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/PMgsb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PMgsb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a problem for fun. I am not sure how one might go about finding this optimal distance from the slit. While it can be done by experimental trial and error, I have a guess that one can see more of the wall if the viewer is closer (right next) to the slit.&lt;/p&gt;&#xA;" OwnerUserId="16789" LastActivityDate="2018-07-16T18:04:40.373" Title="Maximum height view-able in the box when looking through slit" Tags="|measurements|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22762" PostTypeId="2" ParentId="22760" CreationDate="2018-07-16T12:02:53.657" Score="5" Body="&lt;p&gt;In short, elastic deformation of crystalline structures (like steel) happen on an atomic scale, where the bonds of the atoms in the lattice are stretched. This allows for a change in how close the individual atoms are packed together.&lt;br&gt;&#xA;If you remove the load, the energy stored within these bonds can be reversed (comparable to a spring), therefore it's only elastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, plastic deformation happens along gliding planes, for example between individual crystals (inter-crystalline) or within the crystal itself (intra-crystalline). During plastic deformation crystals &quot;slide&quot; along that plane, they perform a translation which is permanent, therefore plastic.&#xA;The crystalline structure itself remains more or less unaffected during that translation, therefore no volumetric change occurs.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2018-07-16T12:02:53.657" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22763" PostTypeId="2" ParentId="22497" CreationDate="2018-07-16T12:28:53.030" Score="0" Body="&lt;p&gt;As I discussed about Ansys about this problem, the only solution so far is to copy dll to some location and import this copy instead of the one from you debug/release folder...&lt;/p&gt;&#xA;" OwnerUserId="16791" LastActivityDate="2018-07-16T12:28:53.030" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22764" PostTypeId="1" CreationDate="2018-07-16T15:14:31.957" Score="1" ViewCount="40" Body="&lt;p&gt;I am simulating flow around a model. This model have an inlet and an outlet. Flow is incompressible. I know that value of pressure at the outlet is very much different from ambient pressure(Operating pressure). But, I don't know real value of pressure at the outlet. Now, I applied value of zero gauge pressure at the outlet (Is this OK? Why?) and this means that pressure at the outlet is equal to operating pressure while this is not. How do i find correct boundary condition for outlet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But some people say that when flow is incopressible and boundary condition at inlet is velocity, we can use pressure-outlet as a boundary condition at the outlet because flow is not dependent on outlet pressure (so, we can use zero gauge pressure at the outlet). Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2018-07-16T15:14:31.957" Title="How to unserstand Correct boundary condition at outlet for incompressible flow?" Tags="|cfd|outlet|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22766" PostTypeId="2" ParentId="22761" CreationDate="2018-07-16T18:04:40.373" Score="0" Body="&lt;p&gt;You have it essentially correct.  The farther away from the slit, the smaller your angular field of view and thus the less of the far wall you can see.   When your eye is in the aperture (that's the optics name for the slit in the near wall), your angular field of view is limited only by the eyeball itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when you move away, draw the lines from the center of your eye to the top and bottom edges of the slit, then continue those lines to the far wall. You will quickly see that, the farther away you are, the smaller angle of view is made by those two limiting lines. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-07-16T18:04:40.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22767" PostTypeId="2" ParentId="22752" CreationDate="2018-07-16T18:09:28.373" Score="0" Body="&lt;p&gt;If you read the excellent discussions on the wikipedia page, you'll see that a transmission electron microscope works on the same principle as an x-ray machine: the image shows how many particles/photons made it through the subject.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other techniques analyze the various secondary products produced by electron collisions with the subject, such as heat or X-ray emission.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The charge or lack thereof on the subject material has essentially zero bearing on the final image.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-07-16T18:09:28.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22770" PostTypeId="1" CreationDate="2018-07-17T10:22:44.603" Score="1" ViewCount="841" Body="&lt;p&gt;What will be the result if the pipe losses is greater than head produced by the pump? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will be the result if outlet is closed?&lt;/p&gt;&#xA;" OwnerUserId="16798" LastEditorDisplayName="user14407" LastEditDate="2018-08-16T22:03:33.240" LastActivityDate="2018-08-16T22:03:33.240" Title="What happens when pumps are blocked off?" Tags="|mechanical-engineering|pressure|pumps|hydraulics|temperature|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22771" PostTypeId="2" ParentId="22770" CreationDate="2018-07-17T11:08:04.260" Score="2" Body="&lt;p&gt;If losses exceed head then there won’t be any flow or if the head exceeds losses at start and then the losses increase then the flow will fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why the losses are calculated then the pump size chosen ...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-17T11:08:04.260" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22772" PostTypeId="1" CreationDate="2018-07-17T18:34:04.473" Score="1" ViewCount="15" Body="&lt;p&gt;In designing an atomizer which generates a spray of liquid, one can adopt atomisation through shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear in the following cases are to be considered:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;stagnant Gas- moving Liquid: in this gas case the gas is stagnant but the liquid is injected at high velocity. This encountered in daily diesel IC engines.&lt;/li&gt;&#xA;&lt;li&gt;moving gas - moving liquid: this is used in aircraft engines wherein the fast-moving gas atomises the liquid due to shear&lt;/li&gt;&#xA;&lt;li&gt;moving liquid - moving liquid: this is a not-so-often encountered area but shear does occur when two liquid layers move one over the other. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Comparing the atomization/shear intensity, which of the following will be the highest?&lt;/p&gt;&#xA;" OwnerUserId="14771" LastActivityDate="2018-07-17T18:34:04.473" Title="Shear between which two fluids is better for atomisation - liquid &amp; liquid or gas &amp; liquid?" Tags="|fluid-mechanics|shear|liquid|injection|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22773" PostTypeId="1" CreationDate="2018-07-17T19:24:05.840" Score="1" ViewCount="26" Body="&lt;p&gt;The simple orifice and a pipe have well-documented pressure loss calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when an injector has complicated flow patterns, is it possible to break it down into simpler sections by which the pressure loss can be calculated? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or are there procedures to estimate the injector pressure loss?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is empirical the only way to measure the pressure loss?&lt;/p&gt;&#xA;" OwnerUserId="14771" LastActivityDate="2018-07-17T19:24:05.840" Title="How to analytically estimate the pressure loss through an liquid injector?" Tags="|fluid-mechanics|pressure|airflow|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22774" PostTypeId="1" CreationDate="2018-07-17T22:25:20.893" Score="1" ViewCount="271" Body="&lt;p&gt;I'm looking for a circulation pump in a very high pressure loop and having trouble finding one. I'm hoping I just don't know the correct search term and someone can point me to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The absolute pressure (what the housing needs to be rated to) is 100 bar / 1500 psi, but the pressure rise created by the pump is less than 1 bar /  15 psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a centrifugal pump would be appropriate, it would just need an extra beefy housing and appropriate seals. Is there a common name for this type of pump?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2018-12-18T02:01:15.233" Title="Pump with high pressure, low delta pressure" Tags="|pumps|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22775" PostTypeId="1" CreationDate="2018-07-18T04:04:02.570" Score="2" ViewCount="12273" Body="&lt;p&gt;I want to measure a tube's diameter with a precision of about 0.1 mm. I have a caliper but sadly the tube is wider than my tool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what methods exist to precisely measure a diameter. Ideally, without buying any (expensive) tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few ideas I have are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tape measure the circumference. Then divide by &lt;code&gt;3.1416&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Roll the tube once and measure the distance traveled. Divide by &lt;code&gt;3.1416&lt;/code&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if I would be able to achieve the desired accuracy with one of those methods. Any ideas will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Why I can't use my caliper:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4B2CS.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4B2CS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16804" LastEditorUserId="16804" LastEditDate="2018-07-18T21:08:00.673" LastActivityDate="2021-04-23T22:23:21.350" Title="Methods to measure diameter" Tags="|measurements|" AnswerCount="4" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="22776" PostTypeId="2" ParentId="22719" CreationDate="2018-07-18T04:26:45.013" Score="1" Body="&lt;p&gt;You should first &quot;balance charge&quot; the battery to full capacity using a suitable charger. Batteries do vary between manufacturers, and between battery types so check the manufacturer's rated voltage. Most modern chargers are smart enough to recognize the change of resistance as the battery charges to prevent over charging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the battery cells maintain the same voltage, the battery is considered to be balanced and fully charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now discharge the battery gently so it does not warm up significantly, until the rated cell &quot;storage voltage&quot; is reached. Older batteries may take several cycles to achieve a stable and even cell voltage for each cell or may not balance at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an individual cell voltage/internal resistance is significantly different than the other cells, the battery can still be stored, but of course not at the manufacturer's optimum voltages. The cell to cell difference should be less than 0.1v.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can keep the battery in a cool space at the storage voltage for more than 12 months.&lt;/p&gt;&#xA;" OwnerUserId="7021" LastActivityDate="2018-07-18T04:26:45.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22778" PostTypeId="1" CreationDate="2018-07-18T10:03:56.220" Score="1" ViewCount="149" Body="&lt;p&gt;I'm building a roller for a &lt;a href=&quot;http://www.plastipack.co.uk/downloads/ENGLISH_solguard.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;pool&quot; cover&lt;/a&gt; I got at home. The cover weighs 10kg (8kg but I'm considering some weigh added by the water present on it). I have a bosch profile structure where I will support it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I need to choose a pipe. I don't know which pipe to choose. Will a steel 35mm diameter with 1.5mm thickness pipe does the job without deflection? Should I increase the thickness or the diameter? The tube lenght will be 4m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J1PNW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J1PNW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16806" LastEditorUserId="16806" LastEditDate="2018-07-18T16:52:33.560" LastActivityDate="2018-07-18T16:52:33.560" Title="Which diameter and thickness pipe to choose" Tags="|mechanical-engineering|deflection|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22779" PostTypeId="2" ParentId="22778" CreationDate="2018-07-18T11:47:08.320" Score="1" Body="&lt;p&gt;10kg of cover on roughly 10kg worth of pipe? Never mind moist environment where rust will be a serious problem... never mind if you intend it to roll by crank (like in the PDF attached), in the beginning you'll be rolling about 10cm of the cover per revolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go with a much larger diameter PVC pipe - 10cm or more. Rust-free, can be easily worked with lightweight tools - no need for welding or similar hard metalworking; a set of drills and good epoxy glue will suffice; pre-made plastic parts (like end-caps) should be easy enough to attach to the axles, it will suffice to hold the weight of the cover and roll it in/out in far fewer turns of the crank, and it won't be as dangerous as pool-side device (leg against pipe breaks pipe, not leg.)&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-07-18T11:47:08.320" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="22781" PostTypeId="1" CreationDate="2018-07-18T14:19:03.307" Score="0" ViewCount="119" Body="&lt;p&gt;I'm designing an electronic musical instrument and need to injection mold the mouthpiece.  This is my project: &lt;a href=&quot;https://hackaday.io/project/159481-multiwind&quot; rel=&quot;nofollow noreferrer&quot;&gt;Multiwind&lt;/a&gt;.  The mouthpiece needs to be safe to go in the mouth, opaque (preferably black), durable enough not to wear excessively if the user rests their top teeth on it while playing, and maintain good dimensional tolerances in molding.  Other manufacturers such as Yamaha and Roland make mouthpieces that appear to have these properties.  What resin should I specify for the molding?&lt;/p&gt;&#xA;" OwnerUserId="16810" LastActivityDate="2018-07-18T21:17:15.860" Title="What type of resin should be used to injection mold an instrument mouthpiece?" Tags="|materials|manufacturing-engineering|plastic|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22782" PostTypeId="2" ParentId="22775" CreationDate="2018-07-18T15:01:04.780" Score="0" Body="&lt;p&gt;The time-honored method is to attach a string, or wire, or straightedge, to one point on the perimeter.  Hold the string taut and sweep it across the tube. Take note of where the greatest length of string is needed to reach the perimeter.  That's a diameter. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-07-18T15:01:04.780" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22783" PostTypeId="1" AcceptedAnswerId="22803" CreationDate="2018-07-18T15:13:01.413" Score="2" ViewCount="80" Body="&lt;p&gt;Given a system dynamics &#xA;\begin{equation}&#xA;\dot{x}=A(p)x+Bu,&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A(p)$ is the uncertain system state matrix with the nominal matrix being $A(p_0)$. The uncertainty in $A(p)$ satisfies the matching condition, i.e. &#xA;\begin{equation}&#xA;A(p)-A(p_0)=B\phi(p)&#xA;\end{equation}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now from the book &quot;Robust control design (an optimal control approach)&quot; I encountered an exercise problem which goes as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If $(A(p_0),B)$ is controllable, show that $(A(p),B)$ is also controllable for any $\phi(p)$ where $A(p)=A(p_0)+B\phi(p)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My approach was to use the rank condition of the controllability matrix for the nominal case and extend that result to the original system but it does not give me the result I expected. Any hint will be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="16813" LastActivityDate="2018-07-19T17:05:12.753" Title="Does the controllability of nominal system imply the controllability of the actual uncertain system?" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22784" PostTypeId="2" ParentId="22781" CreationDate="2018-07-18T17:40:52.690" Score="1" Body="&lt;p&gt;I would imagine that the professional mouthpieces are compression molded with Ebonite (see wikipedia article - &lt;a href=&quot;https://en.wikipedia.org/wiki/Ebonite&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Ebonite&lt;/a&gt;).  My late father worked on a customer's mouthpiece with this material focusing on the vibrational properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you must injection mold this piece, then you are limited to less rigid and durable materials, because they need to flow down the channels of the injection molding device.  Plastic utensils are made with polystyrene and a pigment can be added to turn it black.&lt;/p&gt;&#xA;" OwnerUserId="8533" LastActivityDate="2018-07-18T17:40:52.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22785" PostTypeId="2" ParentId="22781" CreationDate="2018-07-18T21:17:15.860" Score="0" Body="&lt;p&gt;I think I found a material.  Protomold suggested looking at their list of medical grade injection molding resins.  The best one for my requirements seems to be Makrolon grade 2458-550115.  It's clear but can be tinted black and remain medical grade, and it's tough and molds with good accuracy.  This is probably what I'll go with.  If anyone thinks this is not a good option please let me know.&lt;/p&gt;&#xA;" OwnerUserId="16810" LastActivityDate="2018-07-18T21:17:15.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22787" PostTypeId="1" CreationDate="2018-07-18T22:40:21.557" Score="0" ViewCount="330" Body="&lt;p&gt;Hoping this is the best area to ask this, mostly in the DIY and hobby world I've seen 2 engines coupled together to make a twin engine some so this for the zen of doing, but others do it to align the pistons I've noticed particularly in RC aircraft rather than buy a twin engine where the pistons operate at exactly the same time and I supposed matched this  Introduces a reciprocating vibration that causes issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been hearing if you take 2 two stroke engines and couple them together with 1 at TDC and the other at BDC they cancel each other out and a guy on YouTube showed a video of one with a glass of water next to it barely moving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the name for such an arrangement? Is there any disadvantage of doing this? Links to more information? I shall search for YouTube video as described above. &lt;/p&gt;&#xA;" OwnerUserId="16818" LastActivityDate="2018-07-19T00:09:11.543" Title="Twin cylinder engine opposed cylinders" Tags="|mechanical-engineering|automotive-engineering|vibration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22788" PostTypeId="2" ParentId="22787" CreationDate="2018-07-18T22:49:39.747" Score="2" Body="&lt;p&gt;Well, the arrangement used for Volkswagen (air-cooled) engines was known as &lt;em&gt;horizontally-opposed&lt;/em&gt; when I worked on them. The cylinders were slightly offset from each other and each cylinder had its own crankshaft journal, so not &lt;em&gt;exactly&lt;/em&gt; opposed, but perhaps along the lines of what you were looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also heard them called &lt;em&gt;pancake&lt;/em&gt; engines, which might turn up some other search results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On blacksmith37's excellent comment, here's what the BMW motorcycle version looks like inside (these days at least).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2Lt20.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Lt20.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason I never discovered, these were sometimes called &quot;boxer&quot; engines, which might provide another fruitful search term.&lt;/p&gt;&#xA;" OwnerUserId="3937" LastEditorUserId="3937" LastEditDate="2018-07-19T00:09:11.543" LastActivityDate="2018-07-19T00:09:11.543" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22790" PostTypeId="2" ParentId="19827" CreationDate="2018-07-19T02:32:58.940" Score="1" Body="&lt;p&gt;Currently, honeycomb cores are sometimes faced with an adhesive film between the Prepreg skin and the core.   But you seem to be describing a viscoelastic midlayer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This arrangement could be simulated parametrically with FEA and the polymer’s FEA Support Test Data. (Parametric FEA model: adjusting thickness and surrounding core/facesheet assembly material and geometry)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could simulate the behavior you are interested in by performing a DMA temperature and frequency sweep on a viscous polymer you are interested in.  Then, using time-temperature superposition principle, use WLF curve shifting to get the polymer’s dynamic properties in the acoustic frequency range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply these properties to your FEA model and adjust (polymer thickness, core/facesheet geometry and material) as necessary to arrive at your desired design goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawback to this honeycomb modification:&#xA;One drawback to breaking the out-of-plane cell wall direction with a viscous polymer is the loss of in-plane Shear stiffness that honeycomb cores are particularly good at providing.   The interface at the “hard” outer walls would need to move in unison or risk delaminating the midlayer and compromising your engineered benefit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manufacturing alternative:&#xA;I was thinking about a way to not detract from the core’s structural purpose and achieve something similar:&#xA;What if “polymeric bubbles” were blown through hexagonal injectors into each cell.  Blown in such a fashion to adhere to the cell walls and cure in place (likely with some inherent curvature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please contact me if you’d be interested in further research (toby_samples@ardl.com).  I’m quite interested in this topic since reading your question.&lt;/p&gt;&#xA;" OwnerUserId="8533" LastEditorUserId="8533" LastEditDate="2018-07-19T03:12:39.637" LastActivityDate="2018-07-19T03:12:39.637" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22791" PostTypeId="1" CreationDate="2018-07-19T02:36:40.227" Score="1" ViewCount="63" Body="&lt;p&gt;In the following publication &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Arce, A., Alejandro, J., Bordons, C., &amp;amp; Ramirez, D. R. (2010).&#xA;&quot;Real-time implementation of a constrained MPC for efficient &#xA;airflow control in a PEM fuel cell.&quot;&#xA;IEEE Transactions on Industrial Electronics, 57(6), 1892-1905.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There is a claim that&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In fact, as the fuel cell is open-loop stable and the input signal is bounded, the closed-loop system is stable, at least in the bounded-input–bounded-output sense.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, I doubt this claim is true for a &lt;strong&gt;Model Predictive Control (MPC)&lt;/strong&gt;. But, I do not have a counter example too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone please inform if this claim is correct or any counter example exists.&lt;/p&gt;&#xA;" OwnerUserId="16032" LastActivityDate="2018-07-19T02:36:40.227" Title="Does open loop stability guarantee an MPC stability?" Tags="|control-engineering|control-theory|optimal-control|stability|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22792" PostTypeId="2" ParentId="357" CreationDate="2018-07-19T02:38:46.490" Score="2" Body="&lt;p&gt;The chemical inside the battery begins to heat up, which causes the degradation of the separator. The battery can reach a temperature of more than 1000 degree F. At this point flammable electrolytes can ignite or even explode when exposed to oxygen.&lt;/p&gt;&#xA;" OwnerUserId="16819" LastEditorUserId="90" LastEditDate="2018-07-19T04:32:32.797" LastActivityDate="2018-07-19T04:32:32.797" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22793" PostTypeId="1" AcceptedAnswerId="22794" CreationDate="2018-07-19T05:07:34.963" Score="0" ViewCount="45" Body="&lt;p&gt;Let us assume that the an object collides into me in one of two scenarios, either I’m standing on shore, or else we’re submerged underwater. The object doesn’t change it shape, and possesses the same velocity prior to the collision. Would the consequences be the same? more severe? less severe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Points to take into account:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The &quot;wet&quot; weight is larger than &quot;dry&quot; weight, thus the hitting body has more momentum prior to collision. This is true especially if for instance the hitting body has some cavity which may be filled with water thus giving it larger effective mass during collision.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;My another concern is the effect of the medium. Since water are incompressible, I should start &quot;feeling&quot; the body &quot;pushing&quot; water towards me earlier (prior to impact), thus the effective duration of the collision is longer, which should imply smaller forces.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Aside from that, since the collision in water cannot be thought of as sudden, how should I acount for drag forces during the process.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16820" LastActivityDate="2018-08-18T19:01:06.503" Title="Collision of submerged bodies" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22794" PostTypeId="2" ParentId="22793" CreationDate="2018-07-19T07:27:17.840" Score="1" Body="&lt;p&gt;Let us not consider the wet weight, because we don't at this stage have enough info on the size and location of cavities and how fast the can shed the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the duration of the impact it can be very short, crashing impact. Consider an sledge hammer hitting a glass panel under the water, assuming the same impact velocity, it will impart same force. I don't think there will be cushion of pressure wave in front of the hammer to break the blow, not significantly, water is incompressible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An obvious difference is the way our body responds. On the air our body will roll with punch and take the blow in a more flexible way. If you hit me with a fast ball in my torso my rib cage will bend in under the impact and my longs will dampen the blow, may be by bulging out around and acting like shock absorber, but if they are pre-pressured with the hydrostatic pressure they will resist the punch in a less flexible way and chances of tissue injury are higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the viscosity of water added as a shear friction to the entire surface of our body, it will be that much harder to recoil, hence it will receive the impact more forcefully. Although if wear some wetsuit with lubricated skin it may help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All these, of course if we assume we somehow manage to move the ball or projectile with the same speed below the water as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** EDIT **&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After OP's adding more detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something that is semi autonomic and follows you has to have fault safety algorithm built into it. Such as it will default to a fast left turn and stop, or suddenly dives down and slows in a circle ready to be towed back. You may be able to attach a soft cone to the tip as a sacrifice soft structure to break the impact, etc., etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We my google about design of torpedoes and submarines.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-07-19T18:55:53.870" LastActivityDate="2018-07-19T18:55:53.870" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22795" PostTypeId="1" CreationDate="2018-07-19T08:18:31.300" Score="1" ViewCount="66" Body="&lt;p&gt;The most important factor for mainspring is, as far as I understand the matter, is their yield point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this regard, Kevlar beats steel by at least factor of two. So why aren't we using it instead of steel in watches, well their mainsprings specifically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could understand the costs factor, but even premium watches seem to avoid it.&#xA;As a matter of fact, finding any non-steel mainspring is pretty rare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason for that?&lt;/p&gt;&#xA;" OwnerUserId="16821" LastActivityDate="2018-07-19T08:18:31.300" Title="Why don't we make mainsprings out of Kevlar?" Tags="|mechanical-engineering|mechanisms|springs|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="22796" PostTypeId="1" CreationDate="2018-07-19T09:34:27.023" Score="1" ViewCount="53" Body="&lt;p&gt;I have a simple thread joint between two objects:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J7avn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J7avn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and want to find a size of those threads so that after rotating male part it stands at the specific angle. For those who is curious I made it in Fusion 360 and added joints to these parts, also specified limits of sliding upward and downward. The problem is when it totally rotates till the end, the upper part stands in the wrong direction. I am going to 3d print it, so I need it to be as precise as possible.&lt;/p&gt;&#xA;" OwnerUserId="16822" LastActivityDate="2019-07-12T22:49:31.680" Title="How to calculate properties of a thread?" Tags="|mechanical-engineering|cad|threads|joining|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22797" PostTypeId="2" ParentId="22796" CreationDate="2018-07-19T10:07:02.573" Score="3" Body="&lt;p&gt;Well I would suggest that you don’t thread one of the items - just have a clearance hole then a locking nut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, to get a precise position have a hole with a locating pin - this will ensure the positioning.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-19T10:07:02.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22798" PostTypeId="1" CreationDate="2018-07-19T10:28:43.743" Score="1" ViewCount="521" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h0ayB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h0ayB.png&quot; alt=&quot;Sectional View of thread&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have a large piece that I want to thread (the entire region in blue), but I am not sure how much space to leave at the bottom for retraction of the blade (region in red), see above image.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if I have a M55 x 2.0 thread, should the little box the in red be the size of one thread? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meaning: Width = pitch , Height = (Major Diameter - Minor Diameter of thread).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or can it be smaller while ensuring that the entire blue region is threaded?&lt;/p&gt;&#xA;" OwnerUserId="9186" LastEditorUserId="33" LastEditDate="2018-07-20T15:44:59.147" LastActivityDate="2018-07-20T15:44:59.147" Title="Tool Clearance at Bottom of Milled Thread" Tags="|mechanical-engineering|cad|threads|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22801" PostTypeId="1" CreationDate="2018-07-19T16:35:01.290" Score="1" ViewCount="643" Body="&lt;p&gt;Trough is 1.87 meters long and 0.5 meter wide and 0.22 meter deep.  Water only flows over the 1.87 meter long side and from both sides (so x2).  The water overflow is 0.012 meters tall (ruler positioned on edge of trough and water hits 1.2 cm up ruler, this is similar to depth of a waterfall at the ledge).  It is consistent across trough length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the flow rate of water over the edge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I modeled water molecules like balls thrown upward and being decelerated by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{2 \cdot 0.012}{t^2} = 9.81$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming final velocity of rising water molecule is zero, then initial velocity of 0.485 m/s can be calculated (as it leaves trough edge traveling upward). [sqrt (2*gravity*height)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The average is therefore half of that at 0.243 m/s, since we are averaging with zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross sectional area over which water flows is 1.87 m * 0.012 m and multiplying that by 0.243 m/s gives 0.00545 m^3/s which converts to 86.4 gpm.  There are two sides of the trough so multiply by 2 to get 172.8 gpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I derived this naively based on physics knowledge, so please tell me where I'm wrong regarding a simple model of the system.&lt;/p&gt;&#xA;" OwnerUserId="16828" LastEditorUserId="9" LastEditDate="2018-07-19T17:34:46.683" LastActivityDate="2018-08-02T01:50:56.440" Title="Easy model to calculate water flow rate over trough ledge?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22802" PostTypeId="1" CreationDate="2018-07-19T16:40:33.913" Score="2" ViewCount="311" Body="&lt;p&gt;We are specifying pipes for a water injection system (high pressure - Rating 1500) with a physiochemical analysis of 36400 ppm chloride (CL). We need to use Carbon Steel pipe and to limit the corrosion,we use a corrosion allowance of 0.125 in, however, we think that this ppm of chloride are excesive. How we can calculate the corrosion allowance for this kind of fluid? Is it neccesary include some internal coating material or change the carbon steel to Alloy Steel?&lt;/p&gt;&#xA;" OwnerUserId="16829" LastActivityDate="2018-07-19T19:53:25.467" Title="Carbon Steel vs CL" Tags="|steel|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22803" PostTypeId="2" ParentId="22783" CreationDate="2018-07-19T17:05:12.753" Score="3" Body="&lt;p&gt;You can prove it using the PBH test for controllability. It states that $rank \left(&#xA;\begin{array}{cc}&#xA;  \lambda I -A &amp;amp; B \\&#xA;\end{array}&#xA;\right) = n$ for all values of $\lambda$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the uncertain system this becomes $\left(&#xA;\begin{array}{cc}&#xA;  \lambda I -A -B \phi  &amp;amp; B \\&#xA;\end{array}&#xA;\right) = \left(&#xA;\begin{array}{cc}&#xA;  \lambda I -A &amp;amp; B \\&#xA;\end{array}&#xA;\right).\left(&#xA;\begin{array}{cc}&#xA; I_n &amp;amp; 0 \\&#xA; -\phi  &amp;amp; I_m \\&#xA;\end{array}&#xA;\right)$ which also has full rank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Here $n$ is the number of states, and $m$ the number of inputs.)&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2018-07-19T17:05:12.753" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22804" PostTypeId="2" ParentId="22802" CreationDate="2018-07-19T19:53:25.467" Score="2" Body="&lt;p&gt;in the presence of that much chloride, carbon steel pipe would be a bad choice. The idea that you could somehow accomodate its use by specifying 1/8&quot; worth of allowance is poor practice because it assumes the corrosion rate is perfectly uniform, which is usually not the case in practice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you &lt;em&gt;must&lt;/em&gt; use carbon steel in this application, it would be imperative to deoxygenate the feed water, buffer it to an alkaline pH, and spike the water with a silicate, chromate or tungstate corrosion inhibitor; even then, you would have no guarantees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you cannot condition the feed water, then you cannot use plain steel in this application.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-07-19T19:53:25.467" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22805" PostTypeId="1" CreationDate="2018-07-19T20:00:02.717" Score="-1" ViewCount="77" Body="&lt;p&gt;I have two 20W solar panels (each Voc = 22.3, Isc = 1.22) in series connected directly to an axial fan driven by an EC motor (rated voltage 48V). Here the maximum operating voltage when very sunny has been about 43 V. This configuration has worked well in the past but I need a bit more airflow. I want to add in series another panel, this one is a 10W, same voltage, about half the current. I have read elsewhere that this not need to be a problem for the panels since the operating current will fall to that of the smallest panel. That is fine to me, at the moment I am not aiming for high efficiency. What worries me is that the EC motor might get too much voltage out of the three panels. From the panel curves it appears that with full sun (around 1000 W/m2) the motor might get 60-62 V. The operating voltage range of the fan according to its data sheet is 28-56 V. The rated current is 0.5 A, however when I connected the fan to 56V I measured 0.4 A. Is it possible to connect a power resistor in series in the circuit so that with full sun the motor &quot;sees&quot; about 56V instead of 60-62V? Is there a way to calculate the required resistance? Unfortunately for me the fan does not have much information on its properties to help with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate some guidance with this. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="16831" LastActivityDate="2019-04-17T23:03:25.707" Title="Resistor to limit maximum voltage of solar panels" Tags="|motors|photovoltaics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22807" PostTypeId="1" CreationDate="2018-07-19T21:45:13.677" Score="1" ViewCount="48" Body="&lt;p&gt;I'm looking to conduct water level monitoring in streams and rivers for water resource engineering purposes. My plan is to use Arduinos and Raspberry Pis in conjunction with a pressure transducer, as Pis and Arduinos work with the telemetry system and water quality sensors that I have chosen for the project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the standards in British Columbia, Canada (where I work) require that the sensor must have a 2 mm accuracy to collect &quot;Grade A&quot; data. Grade B data requires 5 mm accuracy, and as it turns out a pressure transducer with these accuracies are hard to find unless you spend &gt;$500 per pressure transducer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know if a low cost ($300 max, but preferably $50 to $150) pressure transducer designed for measuring water with 2 mm or 5 mm accuracy exist? I'm also open to other technologies as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page 4 in the attached in BC's &lt;a href=&quot;https://www2.gov.bc.ca/assets/gov/environment/air-land-water/water/science-data/man_bc_hydrometric_stand_v10.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;standards doccument&lt;/a&gt; is attached if anyone is interested in checking it out.&lt;/p&gt;&#xA;" OwnerUserId="16832" LastEditorUserId="16832" LastEditDate="2018-07-20T23:51:57.563" LastActivityDate="2018-07-20T23:51:57.563" Title="2 mm accuracy in an affordable water level sensor (pressure transducer) - Does one exist?" Tags="|pressure|water-resources|hydrology|" AnswerCount="0" CommentCount="7" ClosedDate="2018-07-20T01:41:52.503" ContentLicense="CC BY-SA 4.0" />
  <row Id="22808" PostTypeId="1" CreationDate="2018-07-20T03:52:11.713" Score="1" ViewCount="181" Body="&lt;p&gt;I must clarify that I am not an engineer and I am not an engineering student. I am only doing a personal project in which I am stuck for months with the following problem, since my knowledge is not enough to answer it on my own! So I appeal to you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Issue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the following system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RN4dd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RN4dd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be seen in the drawing, what is sought is to transfer the blue fluid (water) to tank 2. For this the device consists of two very heavy pistons that will be used to transfer the fluid to tank 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is this: what weight must the pistons have to transfer water to tank 2 in a certain time? What equations should I use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the data is in the drawing.&lt;/p&gt;&#xA;" OwnerUserId="16834" LastEditorUserId="1832" LastEditDate="2018-09-22T22:03:18.353" LastActivityDate="2018-11-21T23:02:30.473" Title="Calculating piston mass, for transfering fluid through a length of pipe" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22809" PostTypeId="1" AcceptedAnswerId="22810" CreationDate="2018-07-20T03:58:10.097" Score="3" ViewCount="126" Body="&lt;p&gt;I am designing a system to have solar panels track the sun, but only with one axis of rotation that will tilt the panels from east to west. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to connect the panels together so that one motor could rotate multiple panels. I was thinking of a something like a bicycle roller chain and sprockets. Where a long chain goes around, say, 5 sprockets that each rotation a solar panel, and wraps around again forming a loop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if the chain will have enough contact with the sprockets, especially on panels that are in the middle of the chain system. The chain would basically be resting on those sprockets from the weight of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a good design choice for such a system? Is there more cost effective way to accomplish the same thing? Should I supply tension to system so it has good contact with all the sprockets? &lt;/p&gt;&#xA;" OwnerUserId="4819" LastActivityDate="2018-07-25T09:03:16.477" Title="Rotating Multiple Solar Panels with One Motor" Tags="|gears|photovoltaics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22810" PostTypeId="2" ParentId="22809" CreationDate="2018-07-20T04:09:58.537" Score="1" Body="&lt;p&gt;You could consider idler sprockets to cause the chain to wrap around the drive sprockets more to increase the number of teeth in contact. the panels may move a bit due to play in the chain - tension and a brake may help...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could have a single long drive shaft with worm gears and gears to each panel. The advantage is that the worm/gear will “self-lock” when stationary ie wind load won’t cause any rotation except due to play between the teeth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit re comments : have a shaft with worm gears that slide on and clamp - easy to position accurately and a method to &quot;disconnect&quot; or provide a &quot;neutral&quot; leaving the others working when you need to move one panel to work on it.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-07-20T08:44:52.237" LastActivityDate="2018-07-20T08:44:52.237" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="22812" PostTypeId="1" AcceptedAnswerId="22813" CreationDate="2018-07-20T08:50:44.043" Score="-2" ViewCount="35" Body="&lt;p&gt;This is what I'm talking about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Ampyx_Power&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=lykCHdMNvQM&quot; rel=&quot;nofollow noreferrer&quot;&gt;Youtube video 1&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=uCHk4TRr5hE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Youtube video 2&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.ampyxpower.com/technology&quot; rel=&quot;nofollow noreferrer&quot;&gt;Official site&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Can you please explain me on how this plane is generating energy? Reeling back the aircraft should consume the energy produced and maybe more, as now you also fight the wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some energy may be from gravity when the plane descends, but I don't see how that will be enough.&lt;/p&gt;&#xA;" OwnerUserId="16837" LastActivityDate="2018-07-20T09:10:57.173" Title="How is the Ampyx Power plain generating energy?" Tags="|energy-efficiency|wind-power|" AnswerCount="1" CommentCount="1" ClosedDate="2018-07-23T19:21:47.013" ContentLicense="CC BY-SA 4.0" />
  <row Id="22813" PostTypeId="2" ParentId="22812" CreationDate="2018-07-20T09:10:57.173" Score="1" Body="&lt;p&gt;You've pretty much worked it out: when the wind is lower, or when the plane's controls reduce the plane's lift, it drops, and the winch is wound in. Then the lift increases again, and that reels out the winch, which is where the generator sits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its generation is therefore very similar system to Kite Power Systems, which is explained in &lt;a href=&quot;https://www.youtube.com/watch?v=Ap_B-9o7uV8&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2018-07-20T09:10:57.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22814" PostTypeId="2" ParentId="22798" CreationDate="2018-07-20T12:41:21.653" Score="3" Body="&lt;p&gt;You need clearance at the bottom of the threaded hole or shorten the depth of the thread. This is to allow for the machining process and clearance for the cutting tool. Otherwise your thread will not be a long as you specified(because the machinist needed clearance, or the machinist will break a tool or drag the tool at the bottom of the hole if they are not careful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I interpret your picture correctly, then I think 4mm(equal to 2 x thread pitch) would be sufficient/safe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, talk to the machine shop and ask what tool will be used to cut the thread. A single point tool or a thread-mill? That will show you how much clearance below the thread you need. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/GPbt7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GPbt7.jpg&quot; alt=&quot;sketch&quot;&gt;&lt;/a&gt;&#xA;See below what a single point thread mill looks like and you will see how the tool requires clearance at the bottom due to geometry of the tool&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7xVDe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7xVDe.jpg&quot; alt=&quot;single point thread mill&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2018-07-20T12:48:48.317" LastActivityDate="2018-07-20T12:48:48.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22815" PostTypeId="5" CreationDate="2018-07-20T17:38:09.560" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-07-20T17:38:09.560" LastActivityDate="2018-07-20T17:38:09.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22816" PostTypeId="4" CreationDate="2018-07-20T17:38:09.560" Score="0" Body="For questions about words, phrases and definitions that are specific to engineering." OwnerUserId="6767" LastEditorUserId="6767" LastEditDate="2018-07-21T12:50:26.090" LastActivityDate="2018-07-21T12:50:26.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22817" PostTypeId="2" ParentId="22809" CreationDate="2018-07-20T17:46:00.737" Score="1" Body="&lt;p&gt;I might go with a  set-up use an offset arm to translate a long beam. Said beam has a link at each solar panel to a pole which push/pulls the panel to the desired angle.  Kind of like the strings in a window's Venetian blind.&lt;br&gt;&#xA;The drawback here may be in the amount of mechanical parts required at each panel.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-07-20T17:46:00.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22819" PostTypeId="1" AcceptedAnswerId="22843" CreationDate="2018-07-20T20:59:49.720" Score="2" ViewCount="2904" Body="&lt;p&gt;So lets compare my dirt bike and my daily driver:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My dirtbike is a single 350cc cylinder 4 stroke engine. It has a redline at 12.1k RPM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My daily driver is a six cylinder 3.6L V6. The redline on it is 6.5k RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On my daily driver, and without much mechanical knowledge, I understand that at 6000 RPM, each cylinder would be moving at 1/6th of the total RPM as its a &quot;team effort&quot; &#xA;My motorcycle piston would be working alone, so it has to reach the 12.1k on its own. &#xA;Is the bottleneck not the piston, but in the crankshaft? How can a single cylinder work faster alone as opposed to working with other cylinders?&lt;/p&gt;&#xA;" OwnerUserId="16843" LastEditorUserId="9" LastEditDate="2018-07-20T21:08:38.510" LastActivityDate="2018-07-30T09:03:08.083" Title="Why do small displacement engines have higher RPM ranges?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22820" PostTypeId="2" ParentId="22774" CreationDate="2018-07-20T21:30:29.447" Score="1" Body="&lt;p&gt;What you are looking for is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Circulating_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;circulation pump&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Types commonly found in central water heating systems where the system is maintained at pressure (to detect leaks in heating or supply flow for hot water consumption) and then a pressure differential is applied with the pump to circulate the water around the loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_coupling&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetically driven&lt;/a&gt; pumps are popular for sealed fluid circuits due to pressure, corrosive, toxic or radioactive nature of the fluids or need to maintain cleanliness, these systems may have specifications in common with your needs and a magnetically driven pump is often robustly manufactured for the other demands.&lt;/p&gt;&#xA;" OwnerUserId="3497" LastActivityDate="2018-07-20T21:30:29.447" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22821" PostTypeId="2" ParentId="22819" CreationDate="2018-07-21T01:12:16.993" Score="2" Body="&lt;p&gt;Smaller engines have smaller pistons, connecting rods, and crankshafts. This means these components weigh less than their counterparts in a larger engine. This in turn means that the dynamic forces they experience while whirling around and moving up and down are less than those experienced by those parts in a larger engine running at the same speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This then means that a smaller engine can rev up to a higher speed before the stresses generated by the dynamic forces cause the engine to fly apart. This is why smaller engines can withstand higher revs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, the smaller engine is likely to wear out faster because it is spinning faster, and a larger, slower-turning engine with the same power rating will last longer before it wears out. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-07-21T01:12:16.993" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="22825" PostTypeId="1" CreationDate="2018-07-21T10:42:36.833" Score="1" ViewCount="2550" Body="&lt;p&gt;this is probably a beginner's question. I have a simple solid material, lets say wood wall. I am performing transient-thermal analysis. On one side, I have elevated temperatures (after 30 minutes, I already have some 750 degrees Celsius). On this side of wall I defined convection and radiation. Film coefficient is constant, the ambient temperature I defined using a table with values. The same I did for the radiation (in this case I defined constant emissivity and used same temperature table as for convection).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other side I have only ambient temperature. So, the temperature at time 0 is 25 degrees Celsius and should go up over the time. This is defined under Initial temperature. I also specified convection and radiation of the wood on the other side, and here is the problem. I defined convection again (this time the film coefficient and temperature is constant, 25 degrees Celsius) and radiation as well (again, the temperature and emissivity are constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the elevated temperatures of the first side are getting pretty high over the time, it is logical to expect higher temperatures on the other side of wood wall (on its surface at least). However, when I got results, the temperature stays at 25 degrees Celsius whole time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to define just the initial ambient temperature on the &quot;cold&quot; side of wall? I do not want to limit the temperature on the cold side, on contrary, I want to see how high the temperature can get on that side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12057" LastActivityDate="2018-07-29T19:52:31.960" Title="Ansys Workbench, defining ambient temperature correctly" Tags="|ansys|ansys-workbench|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22826" PostTypeId="1" AcceptedAnswerId="22839" CreationDate="2018-07-21T17:51:00.383" Score="-1" ViewCount="212" Body="&lt;p&gt;How to represent numbers 10 to 15 in 2421 bcd?&#xA;The original question is this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cnKcC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cnKcC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="609" LastActivityDate="2018-07-22T16:05:21.573" Title="How can I convert (AF36) hexadecimal to 2421 BCD?" Tags="|computer-engineering|computer|" AnswerCount="1" CommentCount="2" ClosedDate="2018-07-23T19:20:36.460" ContentLicense="CC BY-SA 4.0" />
  <row Id="22827" PostTypeId="2" ParentId="22805" CreationDate="2018-07-21T20:38:19.873" Score="0" Body="&lt;p&gt;Get 6 volt panels that match the amps of the other panel and run them in series. 2 Get another match to the one you have to increase the amperage and run them parallel. Picture below is in series.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x15vr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x15vr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-07-21T20:44:44.313" LastActivityDate="2018-07-21T20:44:44.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22830" PostTypeId="1" CreationDate="2018-07-22T04:05:26.880" Score="4" ViewCount="811" Body="&lt;p&gt;I recently finished a project that simulated the dynamics and control of a 6DOF quadcopter model using a state-space LQR control approach, but I had a few questions that I wanted to ask that might help me develop the model further. From the literature, if we have a general nonlinear dynamical model of the form,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x}(t) = f(x(t),u(t))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then the &quot;standard&quot; approach to control engineering is to linearize the nonlinear system dynamics into the form,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{x}(t) = Ax(t) + Bu(t)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where,&#xA;$$A = \frac{\partial f}{\partial x}, \space B = \frac{\partial f}{\partial u}$$&#xA;are &lt;em&gt;constant&lt;/em&gt; matrices that have been evaluated at an equilibrium state $x = x_{eq}$ and $u=u_{eq}$ (for a quadrotor this equilibrium state is a hover with pitch, roll and yaw equal to zero and control inputs being enough to counteract gravity). Once this is done, we can then use our $A$ and $B$ matrices to find the &quot;optimal&quot; LQR gain matrix $K$ by solving the Riccati equation (or equivalently using MATLAB's &lt;code&gt;lqr&lt;/code&gt; function), and use this gain in a control law of the form&#xA;$$u(t) = -K(x(t)-x_d(t))$$&#xA;to find our control inputs $u(t)$, that allow us to track a desired state trajectory $x_d(t)$ by applying $u(t)$ to our full nonlinear system dynamics $\dot{x}(t)=f(x(t),u(t))$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach worked pretty well for the quadrotor simulation, but I was wondering why we're required to linearize our system dynamics about an equilibrium point. Wouldn't it also be possible to have a time-varying state-transition matrix $A(t)$ and time-varying control matrix $B(t)$, that would be computed using the same linearization process above, except that instead of linearizing about the hover equilibrium state we would linearize about our current state $x(t)$, which would then allow us to re-compute the LQR gain matrix at each time-step? This would then give us a time-varying gain matrix $K$, and would presumably give us better tracking capabilities since our next state $x(t+1)$ would be close to our previously linearized state. Any advice would be much appreciated, since I'm struggling to find out why linearization always occurs around static equlibrium points, and not about current states.&lt;/p&gt;&#xA;" OwnerUserId="16851" LastEditorUserId="16851" LastEditDate="2018-07-22T18:34:31.687" LastActivityDate="2019-08-16T22:01:00.100" Title="LQR control and system dynamics linearization" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22831" PostTypeId="2" ParentId="22325" CreationDate="2018-07-22T07:59:07.287" Score="0" Body="&lt;p&gt;As mentioned in other answers, hydrostatic pressure with no flow remains the same.&#xA;Now, to estimate changes in pressure head with flow, you can use the &lt;em&gt;Darcy-Weisbach&lt;/em&gt;-equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{equation} (1)\qquad\Delta z_E=\lambda\frac{L}{D}\frac{v^2}{2g} \end{equation}&#xA;$L$ is the length of the pipe, $D$ its diameter, $v$ the flow velocity within the pipe, $g$ the gravitational acceleration and $\lambda$ the friction coefficient.&#xA;Note, that $v$ is needed to solve this equation, thus this usually is an iterative process, where you calculate the velocity without any losses, then calculate the losses with &lt;em&gt;Darcy-Weisbach&lt;/em&gt; and so forth...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an empiric formula for $\lambda$, but its quite complicated and on an iterative basis as well. I think it will suffice to analyse the equations qualitatively. As you can see in equation (1), losses in pressure head $\Delta z_E$ due to friction are reduced, with an increasing pipe diameter.&#xA;The same is generally true for the friction coefficient $\lambda$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, switching from multiple pipes to one with the same total cross-sectional area is beneficial, as there are fewer losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the initial total cross-sectional area is&#xA;$$ A_0=3\cdot\left(\frac{5}{8}\mbox{in}\right)^2\frac{\pi}{4}=0.92 \mbox{sq in}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Afterwards it will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A_1=(3\mbox{in})^2\frac{\pi}{4}=7.1\mbox{sq in} $$&#xA;Which is more than seven times the initial cross-sectional area, resulting in a decreased velocity for the same flow, thus fewer losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore it should be safe to say that there will be no drop in pressure compared to the system you had before.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2018-07-22T07:59:07.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22833" PostTypeId="1" CreationDate="2018-07-22T10:39:26.107" Score="0" ViewCount="1168" Body="&lt;p&gt;I have a transfer function (From Ogata's Modern Control Engineering)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{s+2.5}{(s+2.5)(s-1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the theory says the system has a pole zero cancellation and is uncontrollable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They said that a state space rep of this transfer function has A and B matrix of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A = \begin{bmatrix} 0 &amp;amp; 1 \\ 2.5 &amp;amp; -1.5 \end{bmatrix} \ \text{and} \ B =  \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;code&gt;rank(ctrb(A,B))&lt;/code&gt; in MATLAB, the result is not equal to the dimension of the state space so it not controllable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I got curious and used tf2ss in MATLAB and got another state space rep:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ A = \begin{bmatrix} -1.5 &amp;amp; 2.5 \\ 1 &amp;amp; 0 \end{bmatrix} \ \text{and} \ B =  \begin{bmatrix} 1 \\ 0 \end{bmatrix} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;code&gt;rank(ctrb(A,B))&lt;/code&gt; in MATLAB, I got a value equal to the dimension of the state space so it is controllable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What have I misconceived? (and could someone teach me on how to make matrices in Markdown for the above?)&lt;/p&gt;&#xA;" OwnerUserId="16854" LastEditorUserId="1832" LastEditDate="2018-07-22T13:33:12.947" LastActivityDate="2020-04-06T18:34:04.793" Title="Transfer function with cancellable zero pole and controllability" Tags="|mechanical-engineering|control-engineering|control-theory|optimal-control|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22834" PostTypeId="2" ParentId="22830" CreationDate="2018-07-22T11:38:14.150" Score="3" Body="&lt;p&gt;The idea that you describe is basically Model Predictive Control with successive linearization (the optimization problem that is solved in MPC is finite-horizon while LQR is infinite-horizon, but the concept is still the same). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This requires a lot of computational power, because you need to be able to linearize the system and solve the optimization problem repeatedly at runtime, and every time faster than the time-step duration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the contrary, an LQR-controller is meant to be computed beforehand to give you an explicit control law that requires no big computations at runtime.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2018-07-22T11:38:14.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22835" PostTypeId="2" ParentId="22392" CreationDate="2018-07-22T12:10:32.107" Score="-4" Body="&lt;p&gt;The failure occur in piston rod or connecting rod is less as compare to the other parts (battery, tyre alternator, etc) because the material used for manufacturing a connecting rod are usually steel which has lower thermal conductivity  and the other properties like yield strength (659&amp;nbsp;MPa), tensile strength (855&amp;nbsp;MPaa), elastic modulus (201&amp;nbsp;GPa) are high and the hardness value of connecting rod is 276 vh due to which failure occur less in connecting rod. Now days aluminium alloys or titanium is used as a material for connecting rod so that it becomes light in weight and the performance of the engine increases. The main aim of connecting rod is to connect piston to a crank shaft because of the connecting rod the motion in piston and crank shaft occur so if connecting rod fails engine will not work that's why the designing of connecting rod is done like that the chances of failure occur less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its not like that connecting rod does not fails. The connecting rod is under tremendous stress from the reciprocating load represented by the piston, actually stretching and being compressed with every rotation, and the load increases as the square of the engine speed increase. Failure of a connecting rod usually called throwing a rod which is a most common cause of catastrophic engine failure in a car and the failure occur on connecting rod due to many reason like at the end of exhaust stroke it fails because the tensile force being at the highest at tdc. You can know the connecting rod fails by the knocking noise from engine low oil pressure. Similarly for the other parts of engine like piston, push rod, valves etc failure is less compare to the other parts.&lt;/p&gt;&#xA;" OwnerUserId="16856" LastEditorUserId="1832" LastEditDate="2018-07-22T13:36:36.407" LastActivityDate="2018-07-22T13:36:36.407" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22837" PostTypeId="2" ParentId="22830" CreationDate="2018-07-22T13:33:12.577" Score="0" Body="&lt;p&gt;Sometimes linearisation about a non-equilibrium point leads to a result that does not conform with the nonlinear model. Thus if at some x(t) that is non-equilbrium we try to linearise, we may get incorrect answers.&lt;/p&gt;&#xA;" OwnerUserId="16854" LastActivityDate="2018-07-22T13:33:12.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22838" PostTypeId="2" ParentId="22833" CreationDate="2018-07-22T13:50:23.017" Score="1" Body="&lt;p&gt;Normally state space models who are equivalent to the same transfer function are also equivalent to each other, such that there exists a &lt;a href=&quot;http://lpsa.swarthmore.edu/Representations/SysRepTransformations/SS2SS.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;similarity transformation&lt;/a&gt; between them. However if the considered transfer function has pole zero cancellations then an equivalent state space model would be a rank deficient &lt;a href=&quot;https://en.wikibooks.org/wiki/Control_Systems/Controllability_and_Observability#Controllability_Matrix&quot; rel=&quot;nofollow noreferrer&quot;&gt;controllability-&lt;/a&gt; or &lt;a href=&quot;https://en.wikibooks.org/wiki/Control_Systems/Controllability_and_Observability#Observability_Matrix&quot; rel=&quot;nofollow noreferrer&quot;&gt;observability matrix&lt;/a&gt;, but it is not fixed which of the two has to be rank deficient (those ranks are conserved under a similarity transformation). But the lowest rank of either the  controllability- or observability matrix will be the same for every state space model which is equivalent to the transfer function. So when checking pole zero cancellation you have to check for both controllability and observability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: One might argue that there still exists a similarity transform between those state space models. Only the transform should be rank deficient itself (not invertible) but still satisfy $\hat{A}\,T = T\,A$ and $\hat{C}\,T = C$.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastEditorUserId="1765" LastEditDate="2018-07-22T14:08:59.263" LastActivityDate="2018-07-22T14:08:59.263" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22839" PostTypeId="2" ParentId="22826" CreationDate="2018-07-22T14:53:32.743" Score="0" Body="&lt;p&gt;To Convert a hexadecimal to BCD first you have to convert the hexadecimal to decimal then we can convert decimal to BCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now first take a look at how to convert decimal to BCD:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is fix set of rule for this it is like&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0 in decimal = 0000 BCD(2421), &lt;/li&gt;&#xA;&lt;li&gt;1 in decimal = 0001 BCD(2421),&lt;/li&gt;&#xA;&lt;li&gt;2 = 0010&lt;/li&gt;&#xA;&lt;li&gt;3 = 0011&lt;/li&gt;&#xA;&lt;li&gt;4 = 0100&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Until here it is same as binary but it changes from here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;5 = 1011, &lt;/li&gt;&#xA;&lt;li&gt;6 = 1100, &lt;/li&gt;&#xA;&lt;li&gt;7 = 1101, &lt;/li&gt;&#xA;&lt;li&gt;8 = 1110, &lt;/li&gt;&#xA;&lt;li&gt;9 = 1111&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So how BCD(2421) is converted lets see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take an example of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$8 = 1110 = 1\times2+1\times4+1\times2+1\times0 = 8$&lt;/li&gt;&#xA;&lt;li&gt;$9 = 1111 = 1\times2+1\times4+1\times2+1\times1 = 9$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the numbers here are multiplied by (2421) that's why it is called 2421 BCD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now then how to write 10?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;10 = combination of 1&amp;amp;0 i.e. 00010000, &lt;/li&gt;&#xA;&lt;li&gt;11 = 00010001 likewise.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So now to convert hexadecimal to binary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\text{(AF36) hexa} &amp;amp;= 10\times16^3+15\times16^2+3\times16^1+6\times16^0 \\&#xA;                   &amp;amp;= 40960+3840+48+6 \\&#xA;                   &amp;amp;=44854&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now convert this decimal no. 44854 into 2421 BCD&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;4=0100, &lt;/li&gt;&#xA;&lt;li&gt;4=0100, &lt;/li&gt;&#xA;&lt;li&gt;8=1110, &lt;/li&gt;&#xA;&lt;li&gt;5=1011, &lt;/li&gt;&#xA;&lt;li&gt;4=0100, &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Combine them: 01000100111010110100 (option c)&lt;/p&gt;&#xA;" OwnerUserId="16859" LastEditorUserId="1832" LastEditDate="2018-07-22T16:05:21.573" LastActivityDate="2018-07-22T16:05:21.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22843" PostTypeId="2" ParentId="22819" CreationDate="2018-07-23T11:51:44.177" Score="0" Body="&lt;p&gt;Before I answer your actual question, I'd like to correct a point you made here, &quot;I understand that at 6000 RPM, each cylinder would be moving at 1/6th of the total RPM&quot;. Each cylinder will not be moving at 1/6th the total RPM, all the pistons are connected to a common crankshaft, via their respective connecting rods, so they are all moving at the same speeds. They are all generating 1/6th of the total power, not the total rpm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coming to your question, the redline is not decided by the number of pistons but by the bore and stroke measurements of the engine. The bore of the engine is basically the diameter of the piston and stroke is the distance the piston moves from Top Dead Centre (TDC) to Bottom Dead Centre (BDC). I hope this image helps you understand that better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S9VaY.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S9VaY.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that you know what the bore and stroke length is, there are three different types of piston-cylinder configurations. Over-square, square and under-square. Over-square is when the bore:stroke ratio is over 1, square is when the ratio is equal to 1 and under-square is when the ratio is less than one. The image below will help you understand the 3 different configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vIpEF.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vIpEF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the over-square engine has a smaller stroke length than an under-square engine. So the over-square piston has to move a smaller distance every stroke, that's the reason why over-square engines have a higher redline than under-square engines. Over-square engines are usually more efficient at higher RPMs, that's why they are found in vehicles that are designed to run at higher RPMs, such as your Dirtbike. Your daily driver is designed to be used at lower RPMs, so as to achieve peak torque at lower RPMs, hence has an under-square engine.&lt;/p&gt;&#xA;" OwnerUserId="16864" LastEditorUserId="10902" LastEditDate="2018-07-26T08:17:46.313" LastActivityDate="2018-07-26T08:17:46.313" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22845" PostTypeId="2" ParentId="6071" CreationDate="2018-07-23T13:57:32.323" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;firstly you have ensue that the mesh quality is good.&lt;/li&gt;&#xA;&lt;li&gt;try to reduce Under Relaxation Factors to values less than defaults.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16867" LastActivityDate="2018-07-23T13:57:32.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22846" PostTypeId="2" ParentId="22808" CreationDate="2018-07-23T14:53:44.347" Score="0" Body="&lt;p&gt;You can solve this problem by dividing the problem in three parts. We can consider &lt;a href=&quot;https://www.khanacademy.org/science/physics/fluids/fluid-dynamics/a/what-is-bernoullis-equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bernouli's equation&lt;/a&gt; for the horizontal parts, and for the vertical one we can use the formula $P = \rho \cdot g \cdot h$.&lt;/p&gt;&#xA;" OwnerUserId="16868" LastEditorUserId="9" LastEditDate="2018-07-23T14:59:41.777" LastActivityDate="2018-07-23T14:59:41.777" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22847" PostTypeId="1" CreationDate="2018-07-23T15:41:07.290" Score="0" ViewCount="51" Body="&lt;p&gt;How would I create a stress curve using this given data: Time(s),Extension(mm),Load(N,Tensile strain (Extension)(mm/mm),Tensile extension(mm)&lt;/p&gt;&#xA;" OwnerUserId="16533" LastActivityDate="2018-07-23T22:49:05.587" Title="How to create a Stress Curve" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22848" PostTypeId="1" AcceptedAnswerId="22850" CreationDate="2018-07-23T16:03:34.507" Score="19" ViewCount="4348" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pu91N.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pu91N.jpg&quot; alt=&quot;Enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just by taking a train ride across my home city I can see truss bridges like the one in above picture everywhere. There are numerous variations, but the most common design seems to be this. But why are they built specifically this way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can intuitively kind-of see why such a design probably is strong, but is there any kind of in-depth reason? I would be interested to know the answer as much from the physics side of things as possible. Googling didn't help much; I could find information on the different variations and many examples, but none really covered what is it about this design specifically that makes it so popular.&lt;/p&gt;&#xA;" OwnerUserId="13372" LastEditorUserId="330" LastEditDate="2018-07-24T14:19:50.913" LastActivityDate="2023-04-16T05:03:40.677" Title="Why are truss bridges the way they are?" Tags="|structural-engineering|civil-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22849" PostTypeId="2" ParentId="22847" CreationDate="2018-07-23T16:14:17.970" Score="1" Body="&lt;p&gt;to calculate the stress, you divide load by the surface area of your section be it a wire or rod or square bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember $$\sigma = \frac{F}{Area}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want to graph a stress/strain curve, you divide all the strains to initial length of your sample. you get numbers starting at 1 and increasing to say 1.0, 1.03 corresponding to increasing loads. You drop the 1 and plot the decimal parts on x axis and stresses on the y axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;Explaining why i did not consider the change in surface area of the wire or bar.&#xA;Usually we ignored narrowing of the specimen before we reached plastic region in our lab. And bottle necking happens just before the rapture. You can estimate 30% of tensile strain but it doesn't add up to a significant amount. Say you have 6% tensile strain , you get 0.018 radial shrinking which raised to the power of two is 0.0036 which is smaller than tolerance of many of the jacks you use to test the tensile strain.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-07-23T22:49:05.587" LastActivityDate="2018-07-23T22:49:05.587" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22850" PostTypeId="2" ParentId="22848" CreationDate="2018-07-23T17:13:40.607" Score="25" Body="&lt;p&gt;That looks like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Truss_bridge#Pratt_truss&quot; rel=&quot;noreferrer&quot;&gt;Pratt truss&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These trusses have diagonals which go from the outer-top nodes to the inner-bottom nodes (i.e. they connect to the top chord on the node furthest from the center of the span, and to the bottom chord on the node closest to the center). This design means that the diagonals are under tension and the verticals are under compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another famous design is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Truss_bridge#Allan_truss&quot; rel=&quot;noreferrer&quot;&gt;Allan truss&lt;/a&gt;, which is the exact opposite: the diagonals go from the inner-top nodes to the outer-bottom nodes, which means that the diagonals are under compression and the verticals are under tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason the Pratt truss is so common in steel bridges is because it tends to be more economical. This is because steel works better under tension than under compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under tension, steel can theoretically operate very close to its yield stress. Under compression, however, there's the risk of buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buckling is a behavior of slender elements under compression to effectively collapse at loads far below their yield stress (think of the classic &quot;compress a ruler from both ends&quot; experiment). &quot;Slender&quot; here means beams which are very long and with relatively small cross-sections (see &lt;a href=&quot;https://en.wikipedia.org/wiki/Slenderness_ratio&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia's slenderness ratio page&lt;/a&gt;). Steel beams are frequently slender and therefore buckle under compression (as opposed to simply being crushed). The longer the element, the smaller the buckling stress and, therefore, the larger the beam's cross-section must be to resist buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, with the Pratt truss, the verticals are under compression and the diagonals are under tension. As can be clearly seen in the image (or derived from geometry), the diagonals are longer than the verticals. Therefore, the diagonals' buckling load is smaller than the verticals'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in an Allan truss, the longer diagonals will have a larger cross-section and the shorter verticals will have a smaller cross-section.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a Pratt truss, however, the longer diagonals can have a smaller cross-section while the verticals will have a larger cross-section.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the advantage of the Pratt truss is that material tends to be used more efficiently: the longer elements have as small (and therefore light and cheap) a cross-section as possible by &quot;sacrificing&quot; the shorter elements. This works because those shorter elements require a smaller &quot;upgrade&quot; to resist buckling than the longer ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* Note that when I say above that, for example, &quot;the verticals will have a larger cross-section&quot;, I don't mean that the verticals' cross-section will be larger than the diagonals'. I mean only that it will be larger than if buckling weren't an issue.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-07-23T17:13:40.607" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22851" PostTypeId="1" CreationDate="2018-07-23T17:28:57.680" Score="1" ViewCount="28" Body="&lt;p&gt;I am very much new in this field and trying to get some information on adulteration of cow milk. I have searched and saw that mostly &lt;strong&gt;urea&lt;/strong&gt; and &lt;strong&gt;hydrogen per oxide&lt;/strong&gt; are mixed with cow milk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that to quantify adulterants in milk, Fourier-transform infrared spectroscopy (&lt;strong&gt;FTIR&lt;/strong&gt;) can be used. Normal cow milk contains &lt;strong&gt;(20-70)mg/100ml&lt;/strong&gt; urea. So, urea content above &lt;strong&gt;70mg/100ml&lt;/strong&gt; in milk is considered as added urea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the exact proportion/amount of urea and hydrogen per oxide are mixed in the cow milk as adulterants. I need to carry out the experiment by adding urea/hydrogen per oxide in pure cow milk to get the desired spectrum in &lt;strong&gt;FTIR&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it in between 71-100mg or more or in which range actually?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone point me out any reference? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="16869" LastEditorUserId="16869" LastEditDate="2018-07-23T17:31:32.243" LastActivityDate="2018-07-23T17:31:32.243" Title="How to quantify adulteration of urea and hydrogen per oxide in cow milk?" Tags="|chemical-engineering|chemistry|spectrometry|" AnswerCount="0" CommentCount="5" ClosedDate="2018-07-23T23:59:30.093" ContentLicense="CC BY-SA 4.0" />
  <row Id="22852" PostTypeId="1" CreationDate="2018-07-23T18:42:07.463" Score="0" ViewCount="124" Body="&lt;p&gt;I'm wondering if anyone in the community has experience with foil strain gauges used to measure strain components that are normal to the plane of the sensor? If the strain gauge were pressed between two large flat plates, for example, would there be enough deformation of the conductive path to cause a measurable change in resistance?&lt;/p&gt;&#xA;" OwnerUserId="16304" LastActivityDate="2024-03-26T13:01:15.183" Title="Can foil strain gauges measure strain normal to the plane of the sensor?" Tags="|measurements|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22853" PostTypeId="2" ParentId="22852" CreationDate="2018-07-23T19:21:13.840" Score="0" Body="&lt;p&gt;Mostly, strain gauges are pasted on a surface and give indications about the in-plane strains. &#xA;A Load cell is a device that uses strain-gauges  and can measure the &lt;em&gt;force&lt;/em&gt; perpendicular to the surface. &lt;a href=&quot;https://en.wikipedia.org/wiki/Load_cell&quot; rel=&quot;nofollow noreferrer&quot;&gt;Load cells in wiki&lt;/a&gt;&#xA;Can you please describe what specifically do you want to measure?&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2018-07-23T19:21:13.840" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22854" PostTypeId="1" CreationDate="2018-07-23T20:08:36.400" Score="1" ViewCount="112" Body="&lt;p&gt;I want to calculate drag force at special angle on a sphere . There are some grooves on the sphere and these groove have a special angle on the sphere. In other words, I want to calculate drag force on the these grooves. How can I do it in Fluent?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xGTNz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xGTNz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sYotb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sYotb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2985" LastEditorUserId="2985" LastEditDate="2018-07-25T20:02:09.650" LastActivityDate="2018-07-25T20:02:09.650" Title="How can i calculate drag on an Angled area?" Tags="|fluid-mechanics|aerodynamics|fluent|drag|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22855" PostTypeId="1" CreationDate="2018-07-23T23:18:37.873" Score="2" ViewCount="772" Body="&lt;p&gt;My limited understanding of hydraulic systems is that they aren't nearly as efficient as electrical systems, and that wire is cheaper than high pressure lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a zero turn mower, each of the drive wheels has a separate pump and hydraulic motor geared down with a planetary to the actual wheel shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It occurs to me that doing this with the engine running a generator and putting electric motors to drive the wheels would be a simpler design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am faced with replacing an Eaton Gemini drive unit to the tune of $1500 for the unit, plus labour.  Seems to me that there should be a simpler way to make systems like these?)&lt;/p&gt;&#xA;" OwnerUserId="8768" LastActivityDate="2018-07-25T01:26:43.033" Title="What are the tradeoffs between using hydraulic motors and electric motors?" Tags="|hydraulics|electric-vehicles|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22856" PostTypeId="2" ParentId="22855" CreationDate="2018-07-24T04:51:04.667" Score="3" Body="&lt;p&gt;you will find that a generator capable of delivering, say, 10HP (electrical equivalent, 750 watts = 1HP) is very big and heavy and a hydraulic pump with a 10HP rating is smaller and lighter. Similarly, an electric motor good for 5HP will be much heavier than a hydraulic motor with the same rating. So if hydraulic power is available, it makes sense to use hydraulic motors and live with the inefficiencies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exception to this general rule is when using high-voltage motors (480V) and three-phase AC power, which result in a smaller generator and a smaller motor- but motor speed control becomes more complex in this case. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-07-24T04:51:04.667" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22859" PostTypeId="1" CreationDate="2018-07-24T11:06:14.377" Score="1" ViewCount="82" Body="&lt;p&gt;This is supposed to be the locating bearing in my design, the outer ring is supported by the housing and a taconite seal (taconite seal is hidden). So it is fixed in place and can't move at all.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no significant axial forces working on the bearing system. I want to know if the inner ring can move to the left side due to thermal expansion of the shaft while it is not supported axially by anything else ? (same direction as the red arrow) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the conical shape of the sleeves adapter prevents any movement of inner ring anyway am i right? The length of the shaft is 50 inches and the diameter is 4.5 inches  &lt;a href=&quot;https://i.stack.imgur.com/BG2rc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BG2rc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-07-24T20:48:25.720" LastActivityDate="2024-03-16T12:05:20.603" Title="Dose adapter sleeve hinder the axial displacement of inner ring of spherical roller bearings?" Tags="|mechanical-engineering|design|bearings|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22861" PostTypeId="2" ParentId="22859" CreationDate="2018-07-24T18:33:07.907" Score="0" Body="&lt;p&gt;Expansion coefficient of steel is 7.2* 10^-6 for each Fahrenheit degree of change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calling your shaft's length L and assuming the range of design temperature is 100, we get the expansion: delta L = &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \delta L = L \times 100\times 7.2^{-6} =L\times 7.2^{-4} $$&#xA;Or very roughly you have your shaft elongated by to L* 1.001.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your bearing can not tolerate this much play it has to be able to contain the shaft back with a force of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = E*e =29*10^6(0.001)psi= approx. 29000 psi. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a first estimate but gives some idea about the order of forces involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;Ok now having the length of 50 inches we assuming the block is of the same family of steel with more or less same expansion coefficient there is no need to be that concerned about differential expansion unless we are talking extreme work temperature changes which may cause contortion and warping  in the block. Otherwise assuming a difference in temperature between the block and shaft of +/- 50 degrees we will have thermal expansion of the shaft, &#xA;E = 50 * 0.0005 *25mm/inch = .625 mm elongation or shrinkage. Now you need to check with your bearing manufacturer to see if the bearing can take this much displacement. Another thing I recommend to check is systems natural frequency and its spectra of vibration to check for destructive resonance modes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One cautionary note: your shaft area of 1/4* 4.5^2*pi = 15.9 sq. inch can produce a considerable compressive load, need to check for that. All the temps are preliminary hypothetical numbers, you want to use correct data in your estimates.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-07-24T23:29:18.937" LastActivityDate="2018-07-24T23:29:18.937" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22862" PostTypeId="2" ParentId="3321" CreationDate="2018-07-24T19:11:41.850" Score="0" Body="&lt;p&gt;The calculation perfectly right, bu the normal electric vehicle weights just 1/10th on the weight that you have considered. Hence the results are scaled in that ratio (10x more than what the real world EV could generate)... Hope that helps!&lt;/p&gt;&#xA;" OwnerUserId="16893" LastActivityDate="2018-07-24T19:11:41.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22863" PostTypeId="2" ParentId="22859" CreationDate="2018-07-24T20:22:08.900" Score="0" Body="&lt;p&gt;We would be able to better estimate the lateral expansion if you edit your question and add the length and radius of the shaft. But it seems it will be okay at a first estimate, say your shaft is 10 inches. The expected variance in its length will be +/- 10* 0.0005= 0.005 inches or 0.125 mm.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-07-24T20:22:08.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22864" PostTypeId="2" ParentId="22854" CreationDate="2018-07-24T20:58:34.853" Score="1" Body="&lt;p&gt;I believe under &lt;em&gt;Results&gt;Reports&gt;Forces&lt;/em&gt;, you can calculate the force on a chosen surface in a particular direction vector (by providing x,y,z coordinates). Does that help?&lt;/p&gt;&#xA;" OwnerUserId="3307" LastActivityDate="2018-07-24T20:58:34.853" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="22867" PostTypeId="1" CreationDate="2018-07-25T00:36:55.397" Score="1" ViewCount="39" Body="&lt;p&gt;Assume that a gas obeys the following equation of state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\nu = {\frac{RT}{P}}-{\frac{a}{T}} + b$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $a , b $ are constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Derive an equation for the temperature at the Joule-Thompson inversion state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor gave us the following hint:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ T = {\frac{\mu_{JT}c_p + \nu}{\frac{\partial \nu}{\partial T}}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where ${\frac{\partial \nu}{\partial T}} $ is held constant at P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempt at the solution starts here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I differentiate the equation of state wrt T, I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ {\frac{\partial \nu}{\partial T}} = {\frac{RT}{P}} + {\frac{a}{T^2}}  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging this back into the equation for T leads to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$ T = {\frac{\mu_{JT}c_p + \nu}{\frac{RT}{P}+{\frac{a}{T^2}}}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the correct approach to take to solve for the temperature at the Joule - Thompson inversion state? It seems I have need to take a cubic root for the temperature so I was not sure if this is what the intended answer is.&lt;/p&gt;&#xA;" OwnerUserId="10703" LastActivityDate="2018-07-25T00:36:55.397" Title="Equation for temperature of gas at Joule-Thompson Inversion State" Tags="|thermodynamics|gas|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22868" PostTypeId="2" ParentId="22855" CreationDate="2018-07-25T00:56:25.447" Score="3" Body="&lt;p&gt;You get more free gearing with the hydrostat, and much smaller package and less weight. Take a 60hp diesel and hook it to a hydraulic drive servo. It makes up to 4500 psi peak pressure, or 8 gpm peak flow. It gives infinite gearing from max reverse to max forward speed. The servo weighs about 30 pounds and is the size of a loaf of bread. The wheel motors are the size of a coffee mug and sit inside an axle tube. They might have a vol/rev that is ten times or more greater than the servo, giving 2X10 = 20:1 mechanical reduction. There is no torque differential with speed differential when turning unless you build this in (which you normally do in bigger machines). You can drive as many wheels as you like with relative ease (just some plumbing and a block with some check valves). So there is a huge savings in getting rid of mechanical multispeed transmissions. (Some equipment will still have road gears - a mechanical tranny between the engine and servo). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrostats make a lot of heat, but they carry it all away in the return line to a fluid cooler. E-motors don't have this feature. You would need a sealed motor with liquid cooling circuit to fit it in an axle tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydrostats have most of the control computer executed as plumbing hardware. This can be good or bad. You don't need an electrical genius to troubleshoot the system. You do need a trained hydraulics tech. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So show me the engine/genny/electric motor combo you had in mind that can match the maximum torque capability of a hydrostat system set at 4500 psi with a 20:1 vol/turn ratio from servo to wheels. Or at lease one that can spin all four wheels of a machine on dry concrete, and still perform at speed. In other words, calculate the flow rate of 4500 psi at 60 hp. And calculate the pressure at 8 gpm and 60 hp, and try to duplicate that sort of versatility with an electric motor controller that can be pressure washed twice per day for 20 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest downside by far is that leaking hydraulic fluid kills grass better than glyphosate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For zero turns, its pretty much the same. You need two servos with separate actuators. But you can still get 1000:1 direct reduction from the hydrostats. Often, the transport function is sort of a pto load, with the deck drive hooked (belted) to the engine shaft mechanically. So if you hang a genset off the crankshaft, you need to figure out how you will drive the deck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't to say electrics are bad. The lunar rover went with electrics, but it wasn't mowing grass; and it had some pretty fancy hub reducers that probably wouldn't hold up too well where I work.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-07-25T01:26:43.033" LastActivityDate="2018-07-25T01:26:43.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22869" PostTypeId="1" AcceptedAnswerId="22871" CreationDate="2018-07-25T05:57:04.503" Score="0" ViewCount="94" Body="&lt;p&gt;I am a 3d designer trying to build a simple device for my project and I have a problem. As you can see in schema below, point G has weight on it and it needs to be a controlled, vertical rotation. I am afraid that when I build this, it will go faster or slower on sides. Rotation needs to be stable and I need rotation to stop where it is, as soon as motor stops.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can a mini motor achieve this and what kind?&lt;/li&gt;&#xA;&lt;li&gt;Do I need to add extra parts to the device and what are they?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance and I am looking forward to answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JyrPX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JyrPX.jpg&quot; alt=&quot;schema&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16896" LastActivityDate="2018-07-28T10:09:16.390" Title="Establish a controlled rotation on gears" Tags="|mechanical-engineering|motors|gears|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22871" PostTypeId="2" ParentId="22869" CreationDate="2018-07-25T09:39:01.263" Score="1" Body="&lt;p&gt;The easiest, to implement would be to use a stepper motor with enough hold torque or possibly detent torque if you need the thing to hold even unenergized. The benefit of this is also that you do not need a closed loop controller as the stepper either moves a step or skips it. Which means as long as your stepper is strong enough its not a issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your post mentions gears. While i think involving gears is a bit premature as you havent done motor selection, remember that gears change the effective torque of your motor. So it also affects the effective hold  and detent torque.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-07-28T10:09:16.390" LastActivityDate="2018-07-28T10:09:16.390" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22873" PostTypeId="2" ParentId="22801" CreationDate="2018-07-25T12:57:21.183" Score="1" Body="&lt;p&gt;At each location along the vertical distance $h$, the velocity is given by $\sqrt {2gh}$, where $g$ is the acceleration due to gravity, and this velocity of flow varies as $h$ changes along the vertical. So, you must integrate over the vertical distance since the relationship is nonlinear and you can't simply take the flow rate at the center. The first integral of the equation above is $\dfrac{2}{3}\sqrt{2g}\cdot h^{3/2}$. I could find the equation but not the derivation online (surprisingly), so this is what I did on my own. &lt;/p&gt;&#xA;" OwnerUserId="16828" LastEditorUserId="1832" LastEditDate="2018-08-02T01:50:56.440" LastActivityDate="2018-08-02T01:50:56.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22875" PostTypeId="1" AcceptedAnswerId="22877" CreationDate="2018-07-25T14:43:49.543" Score="2" ViewCount="2289" Body="&lt;p&gt;Before the yield point, the deformation is elastic, and then it is plastic. But what happens to the elastic 'part' of strain after the load is removed? Is it retracted fully or partly, and how does this relate to the extent of plastic deformation?&lt;/p&gt;&#xA;" OwnerUserId="16863" LastActivityDate="2018-07-25T19:00:13.210" Title="Springback after a plastic deformation" Tags="|materials|strength|deformation|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22876" PostTypeId="2" ParentId="22875" CreationDate="2018-07-25T14:53:43.437" Score="1" Body="&lt;p&gt;If you want to bend a piece of metal to a particular angle then you have to bend it slightly more than the angle as when it is released it will spring back a bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try it in a vise and see how it works.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-25T14:53:43.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22877" PostTypeId="2" ParentId="22875" CreationDate="2018-07-25T15:20:11.083" Score="5" Body="&lt;p&gt;When the load is removed, the elastic strain is recovered completely (by definition); the plastic strain is not recovered at all (by definition):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aj7XI.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aj7XI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the load is reapplied, the curve picks up where it left off (i.e., the system moves back up the unloading line and resumes plastic deformation). This point is explained &lt;a href=&quot;https://engineering.stackexchange.com/questions/8045/does-unloading-beyond-yield-point-also-affect-tensile-strength&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2018-07-25T15:20:11.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22878" PostTypeId="1" CreationDate="2018-07-25T15:42:42.857" Score="1" ViewCount="23" Body="&lt;p&gt;I am a student learning about centrifugal pumps in my studies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve learnt about the 3 different type of flows - axial, radial and mixed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve also learnt about different impeller designs - mainly - opened, semi-enclosed, enclosed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering, how does these 2 link together ? For the types of flow classifications, when does these different flows occurs ? And does the impeller design have a specific cause on how the fluid will flow ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also which flow is ideal to operate under ? &lt;/p&gt;&#xA;" OwnerUserId="15037" LastActivityDate="2018-07-25T15:42:42.857" Title="Does impeller design have a specific cause on how the fluid will flow in a centrifugal pump?" Tags="|fluid-mechanics|pumps|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22879" PostTypeId="1" AcceptedAnswerId="41038" CreationDate="2018-07-25T17:01:42.310" Score="2" ViewCount="737" Body="&lt;p&gt;I am trying to optimize the heat control of a test bench at our facility. The test bench is basically a pipe with an air flow that can be heated up to desired temperature. To heat the air flow up, three heat zones are available as shown in the sketch below. Measurements of the temperature are available for &lt;span class=&quot;math-container&quot;&gt;$T_1, T_2, T_3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T_4$&lt;/span&gt;. The three zones are each equipped with there own PID controller, so it is possible to set target temperatures for each zone. However the single temperature states obviously affect the values of each other, especially in the direction of the flow, but also slightly in upstream direction via conduction in the pipe material.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QUlhe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QUlhe.png&quot; alt=&quot;Sketch of test bench&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The ultimate goal is to set a target temperature for &lt;span class=&quot;math-container&quot;&gt;$T_4$&lt;/span&gt; and control the three heat zones accordingly.&lt;/p&gt;&#xA;&lt;p&gt;Right now, basically, only the PID controller of the second heating zone is used for the temperature control. Ideally I would like to use all three heating zones. The PID controllers of the heating zones could either be used to do so or just set to a static gain and building a control on top of this.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;What is the general strategy to get such a MISO control set up?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10513" LastEditorUserId="24822" LastEditDate="2021-03-18T14:44:15.150" LastActivityDate="2021-03-18T16:28:31.527" Title="How can we control the output temperature of a multi-input single-output pipe air-flow system?" Tags="|control-engineering|control-theory|airflow|pid-control|temperature|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22880" PostTypeId="1" AcceptedAnswerId="23108" CreationDate="2018-07-25T17:31:48.700" Score="1" ViewCount="73" Body="&lt;p&gt;How can I quickly and accurately form batches (10 to 100) of parts from mild steel wire about 3mm diameter ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parts are 3D with bends of several radii,  about 200mm in the longest direction with a 0.2&quot;diam 270 degree loop in one end for a bolt.&lt;/p&gt;&#xA;" OwnerUserId="16810" LastEditorUserId="1097" LastEditDate="2018-07-31T22:56:08.377" LastActivityDate="2021-11-16T22:42:27.350" Title="CNC or other method for Wire Forming" Tags="|manufacturing-engineering|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22882" PostTypeId="2" ParentId="22875" CreationDate="2018-07-25T18:46:14.207" Score="1" Body="&lt;p&gt;In ductile materials there is a well defined  straight area at the beginning of the stress/strain curve which is purely elastic, the slope of this part of the curve is Young's modulus. If you remove the load the sample will completely relax to its original length and can give back all the work put into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you keep loading the specimen beyond this elastic region and then remove the load it will not fully recover to its original length, but it will shrink with the same slope as Young's modulus. The area under this loading and unloading is the work gone into the system and never recovered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be a huge amount of energy in structural frames and connections. It is employed to provide the last precious extra minutes of safety in case of a big earthquake, by absorbing the energy of earthquake.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-07-25T19:00:13.210" LastActivityDate="2018-07-25T19:00:13.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22883" PostTypeId="2" ParentId="22879" CreationDate="2018-07-26T00:31:50.700" Score="2" Body="&lt;p&gt;There might be so many methods to perform that. But, the premium method is to use a nonlinear Model Predictive Control (MPC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a multi-input/multi-output control method which predicts the future based on the available model. Then it optimizes a series of control inputs to minimize a forecast cost function with respect to the constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nonlinear MPC is like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;\begin{cases}&#xA;J(\Delta \boldsymbol U|t)=\sum_{k=1}^{N_p} ||T_4(t+kT_s|t)-T_4^{\text{ref}}||^2+\sum_{k=0}^{N_c-1}||\Delta\boldsymbol U||^2 \\&#xA;\Delta \boldsymbol U^{\star}(t) =\text{argmin}_{\Delta \boldsymbol U} ~~J(\Delta \boldsymbol U|t) \\&#xA;\text{Optional constraints}&#xA;\end{cases}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x(t'|t)$ means $x$ predicted of time $t'$ at the current time of $t$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term in the cost function is penalizing the error. The second term minimizes the control actions. The constraints could be maximum or minimum temperatures. $N_p$ is called the prediction horizon and $N_c$ is the control horizon and $T_s$ is the sample time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\boldsymbol U$ is the series of control actions such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\boldsymbol U=\begin{bmatrix}&#xA;T_1(t)\\&#xA;T_2(t)\\&#xA;T_3(t)\\&#xA;\dots\\&#xA;T_1(t+(N_c-1)T_s)\\&#xA;T_2(t+(N_c-1)T_s)\\&#xA;T_3(t+(N_c-1)T_s)\\&#xA;\end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Delta \boldsymbol U$ is the difference of control input&#xA;$$\Delta \boldsymbol U(t)=\boldsymbol U(t)-\boldsymbol U(t-T_s)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control input is updated at each sample time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no constraint and if the problem is convex the problem becomes far easier to solve. You can use gradient descent method. If it is even linear, it will reduce to Linear Quadratic Regulator (LQR) problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information about MPC, have a look at my papers in Researchgate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, MATLAB has MPC toolbox as well as system identification. You can easily use them.&lt;/p&gt;&#xA;" OwnerUserId="14288" LastEditorUserId="14288" LastEditDate="2018-07-27T00:14:01.727" LastActivityDate="2018-07-27T00:14:01.727" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22884" PostTypeId="1" AcceptedAnswerId="22887" CreationDate="2018-07-26T03:47:19.990" Score="0" ViewCount="35" Body="&lt;p&gt;Let us consider a ceramic material, $\mathrm{Mg_2SiO_4}$ - Forsterite, with a sphere shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At room temperature, the ceramic sphere is &quot;trapped&quot; in a matrix which does not expand (null thermal expansion coefficient).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I raise the temperature of the system (matrix + ceramic), the sphere will not be able to expand and internal stress will grow due to the resulting compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you estimate the maximum temperature, as well as the maximum internal stress associated before the ceramic sphere fails ?&lt;/p&gt;&#xA;" OwnerUserId="16903" LastEditorUserId="9" LastEditDate="2018-07-26T09:54:43.590" LastActivityDate="2018-07-26T09:54:43.590" Title="How to determine the maximum internal stress for a material under compression before brittle failure?" Tags="|mechanical-engineering|materials|failure-analysis|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22885" PostTypeId="1" CreationDate="2018-07-26T07:22:25.873" Score="2" ViewCount="100" Body="&lt;p&gt;I have a spanof 9m that I bridge with a (as yet unspecified) beam. From this beam hangs a load (a pipeline), right in the middle, of 4.295 kN. The beam is simply supported. I want to use an IPE steel beam. How do I go from here to specifiying the exact size? I'd appreciate a specific answer, but more importantly a walkthrough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand I have a bending moment of 5.163 kNm. Now I need to divide this moment by the moment of inertia (that I can find in the right table) and I arrive at a stress. So my question is, which specific of the many moments of inertia my Lölner Stahlbautabellen list is the right one, and what is a decent assumption for  max stress, and what margin of safety to apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of this question is a) to arrive at a size that's no too far off, for puprpose of a drawing and maybe cost estimates, and b) to educate myself a bit. Before the structure is built, we will run this by a structural engineer, which brings me to purpose c) not to look ridiculous with a horribly under- or oversized beam.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-07-26T15:31:30.443" Title="How to find a beam for my (simple) application" Tags="|mechanical-engineering|structural-engineering|beam|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22887" PostTypeId="2" ParentId="22884" CreationDate="2018-07-26T09:09:51.120" Score="0" Body="&lt;p&gt;I assume here, the linear coefficient of thermal expansion of the matrix is lower than that of the ceramic so we land in compression, otherwise the sphere tries to expand in all directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fracture criterium of isotrope materials and anisotrope materials are not the same but in general, fractures occurs: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the greatest principal stress is equal to the fracture stress as measured in axial tensile test. This is true for &lt;strong&gt;isotrope&lt;/strong&gt; materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what can we do, theoretically, first we should construct the stress tensor and then find the eigen values of this matrix, the greatest eigen values indicates the greatest principal stress. &#xA;Again all this is true if the material is isotope. For anisotrope materials you need a comprehensive knowledge of fracture mechanics, and it ain't as easy as decibed above.      &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-07-26T09:09:51.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22888" PostTypeId="1" CreationDate="2018-07-26T09:13:16.047" Score="0" ViewCount="896" Body="&lt;p&gt;What are the materials we can use for the long-term storage of 30~liters of liquid nitrogen with composition and grade, along with the thickness of the sheet and storage capacity? Can we use any composites for the same? And which is the best among all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nitrogen is for rocket cooling, but the device doesn't need to be a thermo-flask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please provide references also.&lt;/p&gt;&#xA;" OwnerUserId="16906" LastEditorUserId="1832" LastEditDate="2018-07-26T10:22:05.587" LastActivityDate="2019-01-08T05:02:55.687" Title="Which materials we can use for storing liquid nitrogen?" Tags="|materials|metallurgy|metals|liquid|cryogenics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22889" PostTypeId="2" ParentId="18567" CreationDate="2018-07-26T13:09:13.170" Score="1" Body="&lt;p&gt;You'd need two clutches by the looks of it.&#xA;One to engage and disengage the generator and one for the propeller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise you'd be always generating power with the generator which will add a load to your motor and when you try to spin the electric motor it will spin the diesel engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question is how you'd want the clutch to be activated, by hand or should it be powered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way in the range of the torque you want, large motorcycle clutches are your best bet I think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or if you have access to tools you could attempt to make one, but that would require things like a lathe and you'd need to buy springs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also another issue will be storing the energy the generator will produce, but I won't get into that for now.&lt;/p&gt;&#xA;" OwnerUserId="16556" LastActivityDate="2018-07-26T13:09:13.170" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22890" PostTypeId="2" ParentId="22885" CreationDate="2018-07-26T13:25:18.363" Score="1" Body="&lt;p&gt;You could use this to calculate moment of inertia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.amesweb.info/SectionalPropertiesTabs/SectionalPropertiesIbeam.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amesweb.info/SectionalPropertiesTabs/SectionalPropertiesIbeam.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you put the beam like an I, you use the Ixx in your stress calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factors of safety are application specific so I cannot reasonably suggest one.&#xA;Important factors determining a factor of safety are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How sure you are the load is the stated load, temperature, dynamic loads that cause fatigue,stress concentrations,how the beam was made and many other factors can change material properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can be really conservative and put a factor of safety of 4-6 times if you don't know what you are doing and chances are, it &quot;might&quot; be okay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another thing to think about is that the beam will also need to support it's own weight, so that should also be in your calculations in the form of a load at the center of mass of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also learn from &quot;structurefree&quot; in youtube who teaches these kinds of things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And at the end of the day these calculations are &quot;ballpark&quot; calculations, so they aren't absolute, it might work or it might not.&lt;/p&gt;&#xA;" OwnerUserId="16556" LastActivityDate="2018-07-26T13:25:18.363" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22891" PostTypeId="2" ParentId="18153" CreationDate="2018-07-26T13:37:13.533" Score="1" Body="&lt;p&gt;Without knowing the application it would be difficult to provide you with a correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Super alloys might be what you are looking for, materials usually used for stuff like jet engines are made to retain their mechanical properties at high temperatures.&lt;/p&gt;&#xA;" OwnerUserId="16556" LastActivityDate="2018-07-26T13:37:13.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22892" PostTypeId="2" ParentId="17855" CreationDate="2018-07-26T13:54:47.637" Score="0" Body="&lt;p&gt;It might be easier to add a part that has teeth machined on to both sides of it from a hard material that under the clamping pressure of the bolt will dig the teeth into the titanium parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might also be cheaper than trying to make properly toleranced grooves on both parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution I'm proposing is from a type of washer that digs into the material as you tighten the bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MTZIS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MTZIS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KOWUA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KOWUA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16556" LastEditorUserId="16556" LastEditDate="2018-07-26T16:35:51.670" LastActivityDate="2018-07-26T16:35:51.670" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22893" PostTypeId="1" CreationDate="2018-07-26T15:19:05.400" Score="3" ViewCount="416" Body="&lt;p&gt;I'm trying to determine the effect of shearing force on the deflection of a quasistatic cantilever beam.  In his book, Theory of Elasticity, Timoshenko added the following term to the deflection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{Pc^2}{2IG} * (l-x)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the Wikipedia article on Timoshenko beam theory added the term&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{P(l-x)}{kAG}$$&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Timoshenko_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Timoshenko_beam_theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these reduce to the Euler-Bernoulli form when shear is disregarded. &#xA;What is the difference between these two forms?  Should both be included in the deflection?&lt;/p&gt;&#xA;" OwnerUserId="16908" LastEditorUserId="16908" LastEditDate="2018-07-26T23:41:45.383" LastActivityDate="2018-07-26T23:41:45.383" Title="Timoshenko Beam Theory for Quasistatic Cantilever Beam: Shear Term" Tags="|mechanical-engineering|structural-engineering|statics|beam|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22894" PostTypeId="1" CreationDate="2018-07-26T15:30:18.997" Score="2" ViewCount="125" Body="&lt;p&gt;It is a known fact that beams and spheres have the ability to carry relatively high loads.&#xA;Beams are seen in bridges and other structures even since the days of ancient roman architecture.&#xA;Spheres have even more impressive abilities. Eggs, for instance, can deal with great compression loads without breaking while their shell is ridiculously thin. Pressure chambers also usually have domes at their ends - being able to resist much higher loads than plane surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the trick is that there are almost no shear stresses - most of the stress is tensile due to the shapes geometry. Generally, materials can withstand loads more easily this way (bending is more destructing than stretching).&#xA;Can someone shed some light about this mechanism? How the stresses are arranged in a tension-compression pattern? If find it quite challenging to visualize it.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2018-10-26T00:50:04.470" Title="strength of eggshell" Tags="|mechanical-engineering|structural-engineering|structures|stresses|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22895" PostTypeId="2" ParentId="22885" CreationDate="2018-07-26T15:31:30.443" Score="2" Body="&lt;p&gt;Depending upon your region, codes, safety factors can change.  In essence, someone else typically comes up with the safety factor, while we come up with the maximum stress.  However, you are performing one of the many functions of newly graduated engineers who do not have a license and need to perform work outside school before they can be licensed.  The order of operations here is simple.  You will usually want a spreadsheet.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Guess a safety factor. For steels, use 2.2 for this step.  &lt;/li&gt;&#xA;&lt;li&gt;Take the load and multiply the load by a safety factor before beginning (due to non-linear effects, it is considered better practice to increase the load instead of reducing the allowable stress, though for your beam it will not matter).&lt;/li&gt;&#xA;&lt;li&gt;Calculate the moment for this setup as described.  You calculated incorrectly, the maximum moment in the middle would normally be $\frac{PL}{4}$, where L is the span of the entire beam (9m), so the moment is 9,663.75 Nm.  With 2.2 multiplies it is 21,260.25 N*m (round in last step).&lt;/li&gt;&#xA;&lt;li&gt;Divide by the yield stress for the steel you intend to use (290 MPa is typical for this kind of beam).&lt;/li&gt;&#xA;&lt;li&gt;Glance at your list of beams that are near this value (the IPE 160 and the IPE 180 are close.  The IPE 160 is a bit light, but we had a lot of safety.)&lt;/li&gt;&#xA;&lt;li&gt;Record in your spreadsheet the 3-4 sizes that work closely.  Go a bit above and a bit below.  You also need to record the kg/m for the beam.  &lt;/li&gt;&#xA;&lt;li&gt;Calculate the load from the dead weight.  This works out to $\frac{9.81*w*L^2}{8}$, where w is the kg/m for the beam, L is the full span.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Using this information, now calculate the actual factor of safety when using this beam.  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Multiply the dead load by 1.1 and the pipe weight by a safety factor in a cell.  &lt;/li&gt;&#xA;&lt;li&gt;Determine the stress by dividing by the actual beam section and compare to the yield strength (290 MPa).  &lt;/li&gt;&#xA;&lt;li&gt;Setup a solver in the spreadsheet to change the safety factor until it finds a safety factor such that the stress in the beam is equal to the yield stress.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;With the various factors of safety, you now know how these beams perform.  1.6 can be used for light loading, 1.8 is typical for most loading, and some people will ask for factors of safety of 2 or greater in unusual circumstances.  For this application, finding a beam that meets a minimum of 1.8 will probably be satisfactory.  Show all work, including the spreadsheet printout to your structural engineer, who can best advise on what the code says is allowed and what yield strength to use.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-07-26T15:31:30.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22896" PostTypeId="1" AcceptedAnswerId="22915" CreationDate="2018-07-26T15:37:13.853" Score="0" ViewCount="179" Body="&lt;p&gt;I'm talking about IEC motors-gearbox alignement. Here are an example: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/LkW3o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LkW3o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YNJ8P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YNJ8P.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H9vZT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H9vZT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found the drawing in cross hydraulic website, so it's quite reliable. &#xA;What i don't understand, is, how this design makes sure the perfect alignement of gearbox coupling and the motor?   &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-07-26T16:21:24.720" LastActivityDate="2018-07-27T18:47:17.053" Title="How does bell housing guarantee the concentricity of the shafts?" Tags="|mechanical-engineering|pumps|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22897" PostTypeId="2" ParentId="22896" CreationDate="2018-07-26T15:59:22.837" Score="0" Body="&lt;p&gt;Because they are both machined to a specified tolerance - you have the diagrams showing the information to choose the correct one for your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are not the same as the manufacturing drawings,,,&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-26T15:59:22.837" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22898" PostTypeId="2" ParentId="17855" CreationDate="2018-07-26T16:39:38.460" Score="0" Body="&lt;p&gt;You can also try adding a second bolt next to the first one if possible to increase the clamping force if possible, this will be simpler than what I propose in my other answer.&#xA;If not possible, changing the grade of the bolt to a higher one to provide higher strength might work.&lt;/p&gt;&#xA;" OwnerUserId="16556" LastActivityDate="2018-07-26T16:39:38.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22899" PostTypeId="2" ParentId="22893" CreationDate="2018-07-26T17:31:19.363" Score="5" Body="&lt;p&gt;Timoshenko originally solved the equation with a $k$ factor.  The $c^2$ factor came as a result of Timoshenko's analysis.  The goal was to keep $k$ as a constant, which would make the equations easier.  Unfortunately, $k$ is actually a function of frequency, which is unknown.  This can be seen from a finite element or other elasticity analysis of the beam cross section with dynamic loading, and analyzing the shear stress in the cross sectional area.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most low-frequency applications (like we see today), the effects from k can be ignored, so Timoshenko ignored these values.  Instead, he focused on when the Euler equations began to deviate.  For example, the Euler equations could not predict the phenomenon of a longitudinal wave.  This occurs when a wave travels along the axis of a beam - like when stretching a spring:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j40VI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j40VI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental longitudinal wave frequency, therefore was considered highly important.  Since this is a property that is based upon the material itself, as the $c$ in $c^2$ is the &quot;speed of sound in the material&quot;, it made sense to find the value for $k$ such that Timoshenko's equations predicted this frequency.  In other words, since $k$ is a function of frequency, when Timoshenko's equations were set to describe the longitudinal wave, the value for $k$ would correctly predict the first fundamental frequency.  The net result is the $c^2$ term, which is accurate for this desired value of $k$.  Other values of k can occur.  For more information, there is a declassified resource from the Army Services I was able to consult - &lt;a href=&quot;http://www.dtic.mil/dtic/tr/fulltext/u2/013061.pdf&quot; rel=&quot;noreferrer&quot;&gt;AD013061&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-07-26T17:31:19.363" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22900" PostTypeId="1" AcceptedAnswerId="22903" CreationDate="2018-07-26T17:52:38.427" Score="3" ViewCount="6178" Body="&lt;p&gt;What does it mean with respect to calculating stresses on a beam undergoing loading and when should it be applied?&lt;/p&gt;&#xA;" OwnerUserId="16912" LastActivityDate="2018-07-27T02:42:39.090" Title="What exactly is the thin wall assumption in regards to beam cross sections?" Tags="|structures|stresses|beam|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22901" PostTypeId="2" ParentId="22900" CreationDate="2018-07-26T18:27:38.147" Score="-1" Body="&lt;p&gt;I would guess (please provide a reference for your question) that it means you can assume all the nice simplifications which allow the problem to be solved by hand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumtions/simplifications &lt;a href=&quot;http://structureanalysis.weebly.com/thin-beam-theory.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;are&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The structure must be thin (width: length ≈ 1:100)&lt;/li&gt;&#xA;&lt;li&gt;Bending deflection doesn't exceed the structure's thickness.&lt;/li&gt;&#xA;&lt;li&gt;Perpendicular lines in the structure before bending will stay perpendicular on the neutral axis after bending&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can also look for &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bernoulli/Euler&lt;/a&gt;. A more advanced beam theory is provided by &lt;a href=&quot;https://www.researchgate.net/deref/http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FTimoshenko_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Timoshenko&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over at &lt;a href=&quot;https://www.researchgate.net/post/Whats_the_basic_difference_between_Euler-Bernoulli_and_Timoshenko_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;rearchgate a similar&lt;/a&gt; question was asked. &lt;/p&gt;&#xA;" OwnerUserId="1339" LastEditorUserId="1339" LastEditDate="2018-07-26T18:39:13.587" LastActivityDate="2018-07-26T18:39:13.587" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22903" PostTypeId="2" ParentId="22900" CreationDate="2018-07-27T02:42:39.090" Score="2" Body="&lt;p&gt;The thin-wall assumption is relevant when calculating stresses due to shear and torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because shear and torsion generate &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_flow&quot; rel=&quot;nofollow noreferrer&quot;&gt;shear flow&lt;/a&gt; through the cross-section, as seen below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sgKdd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sgKdd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;&lt;a href=&quot;https://slideplayer.com/slide/9308724/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shear flow is a representation of how the reaction to the applied shear force (or shear force binary pair is equivalent to the applied torsion) &quot;flows&quot; through the section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any given segment of the section (for example, each of the sides of the box section above), it's obvious that the total shear stress &quot;going in&quot; is the same as the stress &quot;going out&quot;. However, how much of that stress is found near the outer face, near the inner face and around the middle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honestly, that gets kind of messy to calculate, so instead the &quot;thin-walled assumption&quot; is made instead. We just assume that the thickness of each of the walls is so small that whatever variance there is through the wall can be ignored. Put another way, it assumes that the shear flow is uniform throughout each of the walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know, there is no formal definition of when a wall can be considered thick or thin, but a common one is $d/t&amp;gt;20$ (seen &lt;a href=&quot;https://en.wikipedia.org/wiki/Cylinder_stress#Thin-walled_assumption&quot; rel=&quot;nofollow noreferrer&quot;&gt;here on Wikipedia&lt;/a&gt;), where $d$ is the section's width or diameter, and $t$ is the wall's thickness. That being said, I've seen the thin-wall assumption used in sections with $d/t=10$ as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That &lt;a href=&quot;https://en.wikipedia.org/wiki/Cylinder_stress#Thin-walled_assumption&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; actually has a good description of how much easier thin-walled sections are to calculate than thick-walled ones. Thin-walled sections have closed-form solutions, while thick-walled ones require integration to solve. Fun times!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example I gave above is of a closed thin-walled section, but the assumption can also be made with open sections (I, L and T sections, for example). Obviously the solution for the shear flow is dependent on the cross-section, but closed sections perform drastically better under torsion than open ones.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-07-27T02:42:39.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22905" PostTypeId="1" CreationDate="2018-07-27T05:30:30.087" Score="1" ViewCount="91" Body="&lt;p&gt;I had a transfer function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{10s+20}{s^3+10s^2+24s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I found the state space representation of the above using MATLAB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using $place(A,B,[poles])$, I found a gain matrix K that corresponds to the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that when I tried to do this using an equivalent alternate state space representation of the transfer function (e.g. using the controllable canonical form), I got a different gain matrix K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this the case? Since the system's transfer function is one and the same, shouldn't the K matrix be the same? Does this mean that there are multiple K matrices that when multiplied with the states, would give us a desired closed loop characteristic? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this then be an advantage over classical control, where here we can use different kinds of controllers with different gains, whereas in classical control we can only have 1 gain value that can get us to the desired closed loop pole characteristic?&lt;/p&gt;&#xA;" OwnerUserId="16854" LastActivityDate="2018-07-27T15:08:55.640" Title="Using place command in MATLAB using different state representation" Tags="|mechanical-engineering|electrical-engineering|control-engineering|control-theory|optimal-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22906" PostTypeId="1" CreationDate="2018-07-27T05:34:17.943" Score="0" ViewCount="65" Body="&lt;p&gt;NPSHa formula - h(sp) +/- h(s) - h(f) - h(vp)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h(sp)- static pressure head (absolute) applied to the fluid &lt;/p&gt;&#xA;&#xA;&lt;p&gt;h(s)- elevation difference between fluid level in reservoir to pump inlet &lt;/p&gt;&#xA;&#xA;&lt;p&gt;h(f) - friction loss in suction piping &lt;/p&gt;&#xA;&#xA;&lt;p&gt;h(vp)- cap or pressure of the liquid at a stated temperature &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I determine whether h(s) is a positive or negative ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve researched that- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pump below reservoir , h(s) is a positive and vice versa but I don’t know the reason behind it ... I’ve been told that if it’s not good for the pump, u have to substract away from the NPSHa formula which makes sense for h(f) and h(vp). and if it’s good for the pump, you have to add. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;am I right to say it’s because the fluid loses more energy as it needs to fight against gravity , that is why h(s) is negative if the pump is above the reservoir, this means more work for the pump. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if pump is below reservoir, the fluid doesn’t lose much energy. &lt;/p&gt;&#xA;" OwnerUserId="15037" LastActivityDate="2020-06-19T16:19:16.243" Title="Calculating NPSHa for a pump" Tags="|mechanical-engineering|pumps|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22907" PostTypeId="2" ParentId="22906" CreationDate="2018-07-27T08:29:05.857" Score="1" Body="&lt;p&gt;The reason h(s) is positive when the pump is below the reservoir fluid level is that there is a pressure created by the height difference which contributes to the net work done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the pump is above the fluid level, then the pump has to apply work to get the fluid into the pump before delivering on it so that work reduces the net work done.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-27T08:29:05.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22909" PostTypeId="1" CreationDate="2018-07-27T14:34:45.307" Score="0" ViewCount="137" Body="&lt;p&gt;We have a tank which has a constant inflow of water (Qi). The level (h) of water in tank is controlled by opening/ closing (l) a valve downstream of the tank. Overflow height of the tank is 4 meter. I want to find the position (ls) of the valve, mathematically, such that inflow and outflow through tank becomes constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also how can this be done using simulink. &lt;/p&gt;&#xA;" OwnerUserId="16924" LastEditorUserId="8055" LastEditDate="2018-07-27T17:04:17.460" LastActivityDate="2020-02-04T05:15:04.230" Title="How to find the balance position of valve downstream a tank?" Tags="|mathematics|simulation|process-engineering|valves|simulink|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22910" PostTypeId="2" ParentId="22905" CreationDate="2018-07-27T15:08:55.640" Score="2" Body="&lt;p&gt;Using a different state space model, which is equivalent to the same transfer function, means that you are applying a &lt;a href=&quot;http://lpsa.swarthmore.edu/Representations/SysRepTransformations/SS2SS.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;similarity transformation&lt;/a&gt;. The new state vector will therefore have a different meaning. For example when given a state space model $(A,B,C,D)$ and applying a similarity transformation $\hat{x}=T\,x$ then the new state space model $(\hat{A},\hat{B},\hat{C},\hat{D})$ can be defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\hat{A} = T\,A\,T^{-1} \\&#xA;\hat{B} = T\,B \\&#xA;\hat{C} = C\,T^{-1} \\&#xA;\hat{D} = D&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and there can be infinitely many different similarity transformation $T$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A state feedback, based on the first model, $u=-K\,x$ can also be transformed into the coordinates of the new state space model using $u=-K\,T^{-1} \hat{x}=-\hat{K}\,\hat{x}$. The open- and closed loop poles are invariant under a similarity transformation. So from this it can be concluded that when using a different equivalent state space model the feedback matrix should most likely change as well.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-07-27T15:08:55.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22911" PostTypeId="1" AcceptedAnswerId="32425" CreationDate="2018-07-27T15:44:20.223" Score="1" ViewCount="1020" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cltOJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cltOJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I am trying to apply an existing blower curve in a hydraulic model for a single stage blower with variable inlet diffuser vanes.  The variety of curves shown on the plot does not seem to make sense to me.  In particular, I don't know which of the blower curves I should be looking at.  The legend at the top-middle of the plot shows IGV: 10 etc. but then those same line-types are used in the plot with 12, 10, 08 etc. and no units are shown anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked up the ASME document referenced (PTC-10) but did not find anything in it to shed any light on how this curve was developed or how it is meant to be read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know how to interpret this or know of a resource that explains this style curve in detail?&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2019-12-18T21:37:06.863" Title="How do you read a blower/compressor curve with multi-curves for inlet guide vanes?" Tags="|design|modeling|compressed-air|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22912" PostTypeId="2" ParentId="22911" CreationDate="2018-07-27T16:24:47.320" Score="1" Body="&lt;p&gt;This is just four separate charts drawn on one piece of paper, to save space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For IGV:0, there are two curves (02 and 04) and the working line at the bottom left. Presumably, a higher flow rate than 04 would stall the compressor with this IGV setting (but it uses less power at low flow rates)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For IGV:4, there are four curves (02, 04, 06, and 08) because you can push the flow rate higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly for IGV:6 and IGV:10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For IGV:10 you can use flow rates up to 12, but the device is less efficient for flow rates less than 06 than at the other IGV settings. The curves for low flow rates at IGV:10 are not shown because nobody would use them in practice, and they would make the chart look &lt;em&gt;really&lt;/em&gt; confusing by overlapping the &quot;useful&quot; curves on the chart.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-07-27T16:24:47.320" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22913" PostTypeId="1" AcceptedAnswerId="22918" CreationDate="2018-07-27T16:33:54.843" Score="0" ViewCount="67" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xlny8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xlny8.jpg&quot; alt=&quot;load disp curves comparison results between FEM and Experimental results&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my results, I can have this curves.&#xA;And somehow in this figure, it is not pretty much similar, &#xA;What can I say if it is asked why I still want to use the FEM results? &#xA;Or how can I validate the FEM results from this figure? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help?&#xA;I really appreciate it&#xA;Thank you very much in advance&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-07-28T15:31:30.847" Title="What is the good reason if the curves between two results are not similar?" Tags="|structural-engineering|abaqus|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22914" PostTypeId="2" ParentId="22913" CreationDate="2018-07-27T16:42:50.600" Score="1" Body="&lt;p&gt;Well, two categories of questions come to mind straight away:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) what values were used for the FEM? Are they based on the actual material tested? What was the source of those values and the confidence level?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) what accuracy were the primary results measured to? Change in length for example - how was it measured? A plastic ruler or a digital vernier with 3 digit resolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you establish an error for the primary values, then you can work out an error for the result...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-27T16:42:50.600" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="22915" PostTypeId="2" ParentId="22896" CreationDate="2018-07-27T18:36:35.580" Score="0" Body="&lt;p&gt;Bell housings are often fixed with bolts or studs, these are convenient and take the required loads but do not generally provide adequate precision in location. Threaded fasteners really need a certain amount of clearance otherwise there is a significant risk of damaging the threads or the through hole during assembly or at least making assembly much more difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead additional matching sets of features will be machined onto both mating parts which are designed to provide alignment to the required tolerances. In some cases fine adjustment may be provided by shims or eccentric pins etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fairly generic method is to use dowel pins, unlike bolts or studs, which generally require a certain amount of clearance to avoid damaging the threads these will be machined with a slight interference fit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other options include machining matching rings, steps or tapers onto both parts. Taper fits tend to be especially good at ensuring concentrically on standardised parts as they tend to be self aligning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases final machining with be carried out with both parts assembled together making them essentially a matched pair. This tends to give the best possible accuracy but makes parts non-interchangeable.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-07-27T18:47:17.053" LastActivityDate="2018-07-27T18:47:17.053" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22916" PostTypeId="2" ParentId="22909" CreationDate="2018-07-28T00:20:12.170" Score="0" Body="&lt;p&gt;Let's say your valve output q is a function of hydrostatic head and valve's position between 100% to 0% open as $f_{valve}$ with n as its independent variable and, Y the dependant, the percentage of opening of its flow, plotted along q curves. And you need the water level stable at n meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$Q_i =  Y(n)   $$ at the q = Qi curve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have the valve's graph, you can create one by setting some heads and positions of valve and by curve fitting stablish your graphs for range of heads.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-07-28T00:42:21.823" LastActivityDate="2018-07-28T00:42:21.823" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22917" PostTypeId="2" ParentId="22913" CreationDate="2018-07-28T06:47:41.833" Score="3" Body="&lt;p&gt;These look like some sort of tensile test on a ductile material which is loaded until it breaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the experiment, it looks like the test rig has some &quot;slack&quot; which gets taken up when the load reaches about 25 kN. There may be some flexibility in the test rig that makes the slope of the elastic deformation different from the theoretical model of a perfectly rigid test rig, or you might simply have a different values of Young's modulus from the test material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quite likely the load is applied to the test piece in a different way in the FE model and the test. For example the test may be applying a relatively concentrated load via a pin through a hole in the specimen (and there may be local plastic deformation around the hole, the pin may be bending slightly, etc, etc), but the FE model may by applying a uniform tension across all the cross section. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elastic limit is slightly different between the test and FE model, but the difference of 10% is nothing to be worried about. The experiment is done in the real world, not in some idealized environment where everything is known to 16 decimal places! If you are going to design some &quot;real life&quot; structure based on these results, you would probably have &quot;safety factors&quot; much bigger than 10% in any case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At large displacements, presumably the FE model has some effect (material properties again?) which makes the load curve slope downwards, after some value of strain is reached. The test does the same thing but to a smaller amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't say what FE software you used, but you might have only defined the material properties up to the strain that corresponds to the deflection of 35mm. Make sure that Young's modulus doesn't suddenly jump to zero for higher strains in your FE model, or something similar and unrealistic!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The test specimen then suddenly broke. There is probably nothing in the FE model to represent that explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general agreement between test and model is very good here IMO. You just need to sort out a few details. Either correct the test data for the defects in the test rig, or make a more detailed FE model that includes the same defects.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-07-28T15:31:30.847" LastActivityDate="2018-07-28T15:31:30.847" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22918" PostTypeId="2" ParentId="22913" CreationDate="2018-07-28T10:39:06.200" Score="2" Body="&lt;p&gt;The tail end of the curve is probably not a massive issue as that is well into the region which would normally be considered failure in any case and FE systems often don't really look at failure mode in detail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of more concern is the elastic region where the FEM results suggest it is  a lot stiffer than it is in the experiment. The experimental curve also doesn't follow a straight line at the start of the experiment which suggest that there is some slack being taken up somewhere. This could either be an issue with the experimental apparatus or the sample itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing the details of the part its is hard to speculate in detail but it could be say that you have bolted joints with inadequate preload and they are slipping under the initial loading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference in Youngs modulus implies a more fundamental issue with the FE model though, perhaps an incorrect value for E or invalid assumptions about things like joint stiffness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A likely possibility is that you have some elements of the model whcih are constrained in too many degrees of freedom compared to the experiment eg you have a rigid joint where you should have a pinned joint.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-07-28T10:39:06.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22919" PostTypeId="1" CreationDate="2018-07-28T11:18:18.990" Score="2" ViewCount="191" Body="&lt;p&gt;Currently, I have an actual entire corroded bridge model in the FEM analysis in Abaqus&#xA;I want to estimate the residual strength of this bridge due to corrosion&#xA;from the output results in the Abaqus, I can have stress and strain values &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I was wondering how to calculate the residual strength of this corroded bridge,? is there any possibility to derive an equation to estimate the residual strength of this bridge using the results from Abaqus model?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you very much in advance&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-08-27T17:01:47.070" Title="How to calculate residual strength of a corroded structure from their stress values?" Tags="|structural-analysis|stresses|finite-element-method|abaqus|" AnswerCount="1" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="22922" PostTypeId="1" AcceptedAnswerId="22928" CreationDate="2018-07-28T14:13:42.653" Score="-2" ViewCount="327" Body="&lt;p&gt;I have a fan on a stand, to circulate air in my home. The three fab blades are somewhat &quot;D&quot;-shaped, with one edge being rounded and one being straight-ish or even curving inward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, from my sailing, as well as my understanding of wing and propeller (edit: and low-noise computer fan) design, I would expect the outward-curving edge to be the leading one, but it's not. In the photo*, the fan spins clockwise to push air towards me.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pi4u6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pi4u6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;*excuse potato quality, the fan was spinning so this is with reduced shutter time and high ISO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, &lt;a href=&quot;https://images-na.ssl-images-amazon.com/images/I/61XFMzZLFkL._SY355_.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this low noise pc fan&lt;/a&gt; also spins clockwise, yet its blades &lt;em&gt;do&lt;/em&gt; curve &quot;backwards&quot; in relation to their direction of spin (as I'd expect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't my fan be less noisy if it spun the other way, &lt;strong&gt;at least theoretically&lt;/strong&gt;? Obviously, including inverting the blade tilt, in order to keep airflow direction.&lt;/p&gt;&#xA;" OwnerUserId="1011" LastEditorUserId="1011" LastEditDate="2018-07-28T17:51:15.270" LastActivityDate="2018-07-29T00:52:37.820" Title="Doesn't this fan spin the wrong way?" Tags="|airflow|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22923" PostTypeId="2" ParentId="22922" CreationDate="2018-07-28T14:32:04.460" Score="0" Body="&lt;p&gt;Most people like to have the air blown towards them - a direct flow of air feels like it helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The noise is mostly (probably) due to the motor noise - it is unlikely that the blade tips are going supersonic... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit based on comment - the shape of the blades will have been considered in terms of solidity (efficiency of moving the air) and also, but less so, in terms of flicker ie the gap between the blades as they rotate.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-07-28T15:03:55.743" LastActivityDate="2018-07-28T15:03:55.743" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="22924" PostTypeId="2" ParentId="22919" CreationDate="2018-07-28T16:52:39.920" Score="2" Body="&lt;p&gt;The strength of the steel has not changed unless there is some factor you have not mentioned. You can get a good estimate of the remaining strength of the structure by putting in the minimum remaining thickness of the various sections into the program. This technique is accepted by AGA ( American Gas Association) , and I think ASME or ANSI, for determining the remaining strength of corroded line pipe. That is , if the original crossection area was 5 sq. in. and corrosion has removed 1 sq. in. of steel in the most corroded point of that member ; The strength of that member is reduced by 20 %.   &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-07-28T16:52:39.920" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22925" PostTypeId="1" AcceptedAnswerId="22926" CreationDate="2018-07-28T21:02:28.223" Score="3" ViewCount="1745" Body="&lt;p&gt;I was playing with a spring, and realized that I've never wondered why a spring gets harder and harder to compress the more you try to compress it.&lt;/p&gt;&#xA;&lt;p&gt;It seems easy at first though. What is the scientific explanation behind this?&lt;/p&gt;&#xA;" OwnerUserId="16934" LastEditorUserId="90" LastEditDate="2022-01-20T17:02:01.703" LastActivityDate="2022-01-20T17:02:01.703" Title="Why do springs get harder to compress?" Tags="|springs|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22926" PostTypeId="2" ParentId="22925" CreationDate="2018-07-28T21:06:14.340" Score="2" Body="&lt;p&gt;Because for every inch or centimeter that you compress the spring an additional force has to be applied to make it compress an additional inch or centimeter, which is why spring rates are given as 245lb/in etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit, just to note some compression coil springs are designed so that the initial travel is at one rate , then further travel is at a higher rate - commonly used in suspension systems on cars to prevent or reduce “bottoming” ie the spring to become so compressed it is solid.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-07-29T07:26:33.507" LastActivityDate="2018-07-29T07:26:33.507" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22927" PostTypeId="1" CreationDate="2018-07-28T23:33:26.057" Score="2" ViewCount="165" Body="&lt;p&gt;I finished taking my control theory course, and I want to build an RC car that has cruise control. So I started off with the force analysis, and I got external forces that are constant. For example if the road is sloped, then the gravitational force is holding the car back. Another force is rolling resistance. I believe these are the only two considerable forces that will resist the car's movement; however, this does not provide a stable system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing a little research online, I found that most models, if not all, say that the external forces are proportional to the velocity of the car. I understand how that can be related to the spring-damper model, but I don't see why the external forces are proportional to velocity when all the forces are constant. For instance, rolling resistance only depends on the weight/normal force of the car--not velocity. So how can I say they are variably dependent on velocity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, let's assume that the forces are proportional to velocity, then how would I figure out that proportionally constant? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to create a model in simulink with some sort of controller, probably PID, then later transfer it into reality. And I am already stuck with the main system. Thanks for any help!&lt;/p&gt;&#xA;" OwnerUserId="16935" LastActivityDate="2020-12-15T18:04:15.373" Title="Cruise Control Project on Toy Car" Tags="|mechanical-engineering|electrical-engineering|control-engineering|control-theory|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22928" PostTypeId="2" ParentId="22922" CreationDate="2018-07-29T00:52:37.820" Score="3" Body="&lt;p&gt;I believe the main reason for the different blade shapes is ducted vs non-ducted fan design and tip vorticity. PC fans are essentially ducted fans with a fan-blade shape that produces tip vortices. These vortices hit the edge of the duct, which dissipates (or weakens) the vortex. OTOH, the portable fan in your picture has no duct, so the shape of the fan blade (forward swept) is designed to direct the span-wise flow towards the inside of the fan rather than the edge. This allows the blade to produce more airflow, since less energy is used in producing a tip vortex.&lt;/p&gt;&#xA;" OwnerUserId="12535" LastActivityDate="2018-07-29T00:52:37.820" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22929" PostTypeId="2" ParentId="22925" CreationDate="2018-07-29T01:03:20.747" Score="5" Body="&lt;p&gt;There are different designs of springs, with different properties:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://springipedia.com/constant-force-about.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Constant Force Springs&lt;/a&gt; - &lt;a href=&quot;http://springipedia.com/constant-force-design-theory.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Constant Force Spring Design Theory&lt;/a&gt; (2nd &lt;a href=&quot;https://www.engineersedge.com/spring_material/constant-force-spring.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GmOWD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GmOWD.png&quot; alt=&quot;Constant Force Spring&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The extension type of constant force spring represents the most basic, yet most versatile, type of constant force spring. It is a pre-stressed flat strip of spring material which is formed into virtually constant radius coils around itself or on a drum. When the strip is extended (deflected) the inherent stress resists the loading force, the same as a common extension spring, but at a nearly constant (zero) rate (change in load per unit deflection, generally given in pounds per inch).&lt;/p&gt;&#xA;&lt;p&gt;A constant torque is obtained when the outer end of the spring is attached to another spool and caused to wind in either the reverse or same direction as it is originally wound. The full rated load of the spring is reached after being deflected to a length equal to 1.25 times its diameter. Thereafter, it maintains a relatively constant force regardless of extension length. Load (force applied to a spring that causes a deflection) is basically determined by the thickness and width of the material and the diameter of the coil.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://springipedia.com/compression-about.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Compression Springs&lt;/a&gt; - &lt;a href=&quot;http://springipedia.com/compression-general-design.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Compression Spring Design General Considerations&lt;/a&gt; (2nd &lt;a href=&quot;https://www.engineersedge.com/spring_comp_calc_k.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UeixC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UeixC.jpg&quot; alt=&quot;Compression Spring&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A compression spring is an open-coil helical spring that offers resistance to a compressive force applied axially. Compression Springs are the most common metal spring configuration and are in fact one of the most efficient energy storage devices available.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://springipedia.com/extension-about.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Extension or Tension Springs&lt;/a&gt; - &lt;a href=&quot;http://springipedia.com/extension-design-theory.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Extension Spring Design Theory&lt;/a&gt; (2nd &lt;a href=&quot;https://www.engineersedge.com/spring_tension_calc_k.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fyjgz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fyjgz.jpg&quot; alt=&quot;Extension or Tension Springs&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Helical extension springs are similar to helical compression springs, but are loaded in tension. Hooks or loops are provided to allow a pull force to be applied. Usually, extension springs are attached at both ends to other components. When these components move apart, the spring tries to bring them together again.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://springipedia.com/torsion-about.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Helical or Torsion Springs&lt;/a&gt; - &lt;a href=&quot;http://springipedia.com/torsion-design-theory.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Torsion Spring Design Theory&lt;/a&gt; (2nd &lt;a href=&quot;https://www.engineersedge.com/spring_torsion_calc.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PkgLx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PkgLx.jpg&quot; alt=&quot;Helical or Torsion Spring&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Helical springs used to apply a torque or store rotational energy are commonly referred to as torsion springs. The ends of torsion springs are attached to other components, and when those components rotate around the center of the spring, the spring tries to push them back to their original position. Although the name implies otherwise, torsion springs are subjected to bending stress rather than torsional stress. They can store and release angular energy or statically hold a mechanism in place by deflecting the legs about the body centerline axis.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Comparison of the force applied vs. amount of movement.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H2R4Z.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H2R4Z.gif&quot; alt=&quot;Spring Deflection Force vs. Type&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14556" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-07-29T01:03:20.747" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22931" PostTypeId="1" AcceptedAnswerId="22941" CreationDate="2018-07-29T03:05:04.187" Score="1" ViewCount="55" Body="&lt;p&gt;I  am comparing the stress results between original or non-corroded model to corroded model.&#xA;I found that the stress value in non corroded model is lot more bigger than the corroded model, &#xA;is it correct?&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2018-07-29T12:21:57.453" Title="Is it true that if the structure damaged or corroded, the stress will be lower?" Tags="|structural-analysis|stresses|abaqus|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22933" PostTypeId="1" CreationDate="2018-07-29T06:05:45.520" Score="0" ViewCount="28" Body="&lt;p&gt;Can someone help me with the dimensioning convention used here?Usually the coordinates are given as cartesion X,Y,but I am unable to understand this convention.&#xA;Also each individual hole size is not mentioned,so how does one find the hole size if they are close in diameter.Here the sizes are quite disparate,so it is visually obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers &amp;amp; TIA&lt;a href=&quot;https://i.stack.imgur.com/bVyLS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bVyLS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16939" LastActivityDate="2018-07-29T06:50:09.913" Title="Dimensioning Confusion" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22934" PostTypeId="2" ParentId="22933" CreationDate="2018-07-29T06:46:53.467" Score="2" Body="&lt;p&gt;Top left is given as zero origin and all centres are measured from that point horizontal and vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two holes have diameter details with one saying 6 off and the smaller 25.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-29T06:46:53.467" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22935" PostTypeId="2" ParentId="22933" CreationDate="2018-07-29T06:50:09.913" Score="0" Body="&lt;p&gt;Ok , I figured it out. 10 4x means there are 4 holes at 10mm X axis.Similarly for Y axis.Hence you can get the coordinate of each hole individually.&lt;/p&gt;&#xA;" OwnerUserId="16939" LastActivityDate="2018-07-29T06:50:09.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22939" PostTypeId="2" ParentId="22927" CreationDate="2018-07-29T10:30:15.947" Score="1" Body="&lt;p&gt;SimuLink is a great environment for learning about and developing a cruise control system, since it provides real life/time testing while observing every element in the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resistances of the vehicle are certainly not proportional with the vehicle's speed, but it may come close at low speeds, ie. less than 10km/h. Both air and rolling fricion are not linear. In your case, rolling friction probably prevails because of the low speed I presume an RC car will maintain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to design a model of your vehicle in SL to be able to quickly and accurately design a cruise control for it. There are SL models available on the web for free that allow you to simply fill in the vehicle's specs: mass, air drag coefficient, rolling resistance curve etc. I'd highly recommend to make use of this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have created a model for your vehicle that you think quite accurately fits the real vehicle, start setting up a PID for the cruise control. The only input is the vehicle speed, the only output is the power demanded from the motor. Try to make the PID output and engine power proportional to get a quick and stable system. Same goes for the speed and input signal. In the beginning, make use of the saturation blocks to keep things stable, it allows for easier troubleshooting.&#xA;Begin playing with the P, then I and then maybe D. There are methods for manually tuning the PID, like the Ziegler-Nichols method. They can be a good start.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have created a good and stable PID, you can make the PID values dependant on other variables like eg. speed, gearing ratio or battery charge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best of luck.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-07-29T10:30:15.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22941" PostTypeId="2" ParentId="22931" CreationDate="2018-07-29T12:21:57.453" Score="2" Body="&lt;p&gt;Maybe it is correct, maybe it isn't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose you &quot;corrode away&quot; a small radius that was causing a stress concentration, and it becomes a bigger radius. Quite likely the maximum stress in the  structure (at that concentration) will &lt;em&gt;decrease&lt;/em&gt; because of the corrosion. For example an &quot;average&quot; stress of 1.0 and a stress concentration of 10.0 might become an &quot;average&quot; stress of 1.1 and a stress concentration of 9.0. Whether you count that as a &quot;decrease from 10 to 9&quot; or an &quot;increase from 1.0 to 1.1&quot; is your choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if you reduce the cross section of a structural member like a beam, or the thickness of a plate, the stress will probably increase - i.e. stress = load/area, and you decreased the area.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-07-29T12:21:57.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22943" PostTypeId="1" CreationDate="2018-07-29T14:31:03.620" Score="2" ViewCount="538" Body="&lt;p&gt;I've been trying to produce a cheap panel using fiber cement and zincalume sheet. There is a &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/0143749682901257&quot; rel=&quot;nofollow noreferrer&quot;&gt;research paper&lt;/a&gt; that I read about bonding steel plate to concrete to increase their tensile strength by reaching a fully composite state when they are bonded with epoxy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've calculated the cost of bonding those two materials with epoxy and mechanical fasteners. I found out that if I cover the entire surface with epoxy the cost is too high. I wonder can I just make a Square and X shape ( [X] ) with the epoxy adhesive and reach the same benefit as covering the entire surface? Or is there an alternative solution to this?&lt;/p&gt;&#xA;" OwnerUserId="16942" LastEditorUserId="16942" LastEditDate="2018-07-29T18:14:38.707" LastActivityDate="2018-07-29T18:14:38.707" Title="Bonding steel plate to fiber cement sheet effectively" Tags="|steel|concrete|reinforced-concrete|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22944" PostTypeId="2" ParentId="22524" CreationDate="2018-07-29T18:48:33.273" Score="1" Body="&lt;p&gt;Theoretically talking, the peak to peak value let us to estimate the worst case scenarios. Here is an example, A shaft fixed inside a conical coupling.&lt;a href=&quot;https://i.stack.imgur.com/dtA6b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dtA6b.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;How we make sure if the coupling can transmit the required torque? We need to calculate it using this pk-pk value. I'm not going further into details.&lt;/p&gt;&#xA;&lt;p&gt;In production series of a specific component, people are less interested about how much really this peak to peak value is, but rather they are interested in the average. This average is not sensitive to strong deviations, and is an indicator of your machining tools health and lifetime. Strong deviations in average value means worn cutters ... .&lt;/p&gt;&#xA;&lt;p&gt;But that's not all, sometimes the average value plays very important role in theoretical calculations as well, for instance, designing the thrust bearing requires always the average value, and the peak-to-peak value here doesn't make sense.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorUserId="37432" LastEditDate="2022-08-08T13:45:54.617" LastActivityDate="2022-08-08T13:45:54.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22946" PostTypeId="2" ParentId="22825" CreationDate="2018-07-29T19:52:31.960" Score="1" Body="&lt;p&gt;the possibility of the simulation going wrong can be in two places. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Firstly, you chose the material to be wood, which is not a good conductor. This means heat is more likely to dissipate from the face rather than distributing itself in the entire part. &lt;/li&gt;&#xA;&lt;li&gt;Secondly, if you entered the face temperature on the other side to be 25°C, it will be constant throughout the time period. So you might was well check your setup again.&lt;/li&gt;&#xA;&lt;li&gt;Personally, I suggest you to change the material to a good conductor (say a metal) and repeat the simulation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="16944" LastActivityDate="2018-07-29T19:52:31.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22949" PostTypeId="1" AcceptedAnswerId="22958" CreationDate="2018-07-30T08:14:26.323" Score="3" ViewCount="539" Body="&lt;p&gt;Load sequence in phases can be considered applying &lt;a href=&quot;https://forums.autodesk.com/t5/robot-structural-analysis-forum/load-sequencing-construction-stages/m-p/4570879/highlight/true#M18259&quot; rel=&quot;nofollow noreferrer&quot;&gt;the self-weight load only to the story added in the appropriate phase&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YEMyz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YEMyz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that the material will not go into yield stage ( will always be at the elastic stage), is there any need to consider load sequence in construction stage for structural analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there is &lt;strong&gt;no need&lt;/strong&gt;, because no matter what, the final finished structure should always have the maximum loads at each floor/element, whereas during the construction, each floor/element loading is only partial. So if we consider the final structure analysis, we already consider everything adequate for design. No?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why code of practice still recommend it?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="3353" LastEditDate="2018-07-30T10:34:37.960" LastActivityDate="2018-08-02T01:33:06.250" Title="Is there any need to consider loading sequence in construction stage for structure analysis?" Tags="|structural-engineering|structural-analysis|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22950" PostTypeId="2" ParentId="22949" CreationDate="2018-07-30T08:25:56.860" Score="3" Body="&lt;p&gt;If all the material stays elastic at every step in building the structure, you are right that the final analysis is all you need to consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But … &lt;em&gt;how do you know all the material will stay elastic, unless you analyse all the steps in the construction???&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you DO need to do the analysis, in order to prove that you DON'T need to do it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern materials make it less likely that half-constructed buildings fall down, but it does happen. On a small scale, one of my neighbours was building an extension to his house, and over one weekend a storm blew down a 10-foot high section of brick wall which wasn't yet &quot;built in&quot; to the original structure. On a much bigger scale, the history of building medieval cathedrals in Europe contained many examples of half-finished buildings that collapsed, and had to be re-built after changing the design.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-07-30T08:25:56.860" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="22951" PostTypeId="2" ParentId="22819" CreationDate="2018-07-30T09:03:08.083" Score="0" Body="&lt;p&gt;Go back to first principles. Engine horsepower depends upon:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.) Cubic capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ii.) Rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;iii.) Compression ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also the inevitable ‘ proportionality constant ‘ , which tidies up all other constants relating to interchanging units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i. &amp;amp; ii. together give you the amount of air passing through the engine in 1 minute, &amp;amp; hence the amount of fuel that can be burnt to generate power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a small &amp;amp; compact, lightweight, engine, keep the cubic capacity small, &amp;amp; hence the rpm will need to increase to achieve the same total ‘flow’.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more you can compress the charge prior to ignition, the greater the relative expansion, &amp;amp; the more power that can be absorbed from the process. There is, of course, a practical limit. Due to the heating effect of compressing gases, petrol will tend to ignite prematurely, before the ideal point in the cylinder cycle - detonation, also known as ‘pinking’. This limits the available power, and damages the engine into the bargain. The upper limit is about 8.5 -9 :1. Above this, you have to put additives into the fuel, to deter pre-ignition. The most famous of which was tetra-ethyl lead, which is now banned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because diesel oil is basically a much heavier, less volatile hydrocarbon than petrol [ gasoline], and it is only injected directly into the cylinder at the point of almost maximum compression, much higher compression ratios are possible, a factor of maybe 2–3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the diesel engine has to be much more robust, &amp;amp; hence heavier to survive these stresses. I’ve never heard of a practical diesel motorbike!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can also be slower revving, &amp;amp; longer stroke, which gives greater torque. I’ve driven trucks with 13–14 litre [ 780 -840  ] that produce 450 bhp, in a power band that runs from 1000–1500 rpm. The torque is truly awesome. My road car has a 1.9 litre twin turbo diesel, connected to a 6 speed gearbox, &amp;amp; redlined at 4500rpm, to yield 180bhp. It does not have the same long stroke, hence low[ish] torque, &amp;amp; cannot really pull away in 2nd gear.&lt;/p&gt;&#xA;" OwnerUserId="16923" LastActivityDate="2018-07-30T09:03:08.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22952" PostTypeId="1" CreationDate="2018-07-30T09:38:14.653" Score="0" ViewCount="102" Body="&lt;p&gt;I am looking to use some load cells (foil based type) to detect presence for home automation within in a room. I was looking to mount some loadcells to the joists for the floor (UK Based house so around 6x2&quot; Joists around 13ft long on 400mm (1.3ft Centres) to try and measure deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not bothered for Exact measurements, but knowing the difference between a small dog and a human may be useful or determining more that one person.. but again not a real issue..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What size loadcell would people recommend, I've seen very small sensors used for deflection in steel components such as suspension components on cars..  this makes me think that I dont need large sensors on timber joists that 'should' move more?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mounting I was thinking of bonding / gluing to the side of the joist, near the centre of the span and sanding it smooth first.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thoughts, suggestions, help? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="16951" LastActivityDate="2021-04-15T18:47:48.613" Title="Loadcell / strain gauge sizing for a joist" Tags="|structures|stresses|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22953" PostTypeId="2" ParentId="22952" CreationDate="2018-07-30T09:46:57.870" Score="0" Body="&lt;p&gt;Well, centre of the span is good as that is where the greatest change will be - but at the edge, because if you put it on the centre-line then the change in length will be minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for size - you need to estimate the actual change in length you are likely to see for the heaviest person or persons ( lifts always make me laugh as the average weight is about half my weight., but at over 6&quot;...), then select a sensor which covers that range - usually best not at the extremes of its working range.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-07-30T09:46:57.870" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22954" PostTypeId="1" CreationDate="2018-07-30T10:42:02.060" Score="0" ViewCount="561" Body="&lt;p&gt;Where is it best to place a strain gauge to measure the twist of a beam? And what type of gauge is best?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my application, I have a small metal part which experiences a twisting and bending force. The beam part is 8mm wide, 9mm long, and 1.3mm thick. It is fixed at the two screw holes, and a force is applied to the little flat rectangular patch at the end of the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wMJS2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wMJS2.png&quot; alt=&quot;Twisting beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see in the image, most of the stress occurs around the root of the beam, but I'm not quite sure which areas on the surface are experiencing compression, and which are experiencing tension.&lt;/p&gt;&#xA;" OwnerUserId="16953" LastActivityDate="2018-07-31T00:08:50.640" Title="Strain gauge placement to measure a twisting beam" Tags="|structural-analysis|measurements|sensors|finite-element-method|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22955" PostTypeId="1" CreationDate="2018-07-30T11:51:38.150" Score="2" ViewCount="968" Body="&lt;p&gt;I have 2 plates that are fillet welded to each other. How can I determinate the max force the weld can take before it cracks?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plates are only welded on one side because the other side is not reachable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added a image to clarify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bt4L3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bt4L3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8464" LastEditorUserId="33" LastEditDate="2018-07-31T12:46:45.243" LastActivityDate="2018-10-09T08:12:29.037" Title="Fillet weld strength calculation" Tags="|design|welding|strength|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22956" PostTypeId="2" ParentId="22952" CreationDate="2018-07-30T12:24:15.100" Score="1" Body="&lt;p&gt;As &lt;a href=&quot;https://engineering.stackexchange.com/a/22953/1832&quot;&gt;mentioned by Solar Mike&lt;/a&gt;, you'll need to calculate the deflection at midspan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One's first instinct would be to make use of the fact that joist structures tend to be heavily standardized and there are handbooks around which give you all the information you need (I personally don't have any because I'm from Brazil, which doesn't have a tradition of joists).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this first instinct won't do you much good. As standardized as joists should be, reality isn't as kind. The problem isn't really the interaction with the flooring &lt;a href=&quot;https://engineering.stackexchange.com/questions/22952/loadcell-strain-gauge-sizing-for-a-joist#comment41614_22953&quot;&gt;you've mentioned&lt;/a&gt;, but the simple fact that wood is a highly variable material, with significantly different Elastic Modulus for every specimen of the same species. Given that the Elastic Modulus is a key variable in calculating deflection, that makes the calculation very imprecise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there's the fact that buildings aren't always built exactly to code and construction imperfections exist everywhere. And calculations are always made in a world of theory, where supports are perfectly pinned or fixed, when reality is much messier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you really need to calculate the deflection, you'll need to take a core sample of the joist (even though removing a beam sample is perfectly fine, good luck telling the owner that!) and do a pretty detailed survey of the structural system. And you'll probably still be significantly wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest solution here is to forget about actually calculating the deflection, and just use the strain gauge results directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;a href=&quot;https://engineering.stackexchange.com/questions/22952/loadcell-strain-gauge-sizing-for-a-joist#comment41614_22953&quot;&gt;you've apparently already bought some&lt;/a&gt;, stick one to your house and then see what the strain gauge measures when you walk around the room, when your dog walks around, when you move furniture around, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See if you can distinguish that signal. If you can, then that strain gauge is sufficient (feel free to buy cheaper ones and try them). If you can't, then you'll need to buy a more sensitive one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this will require significant calibration, but you should probably get used to that anyways, since you'll probably need to calibrate the results for each of your customers' locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll then have to figure out how to handle moving furniture (which will alter the &quot;empty room baseline&quot;) and long-term effects like creep.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-07-30T12:24:15.100" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22957" PostTypeId="2" ParentId="22949" CreationDate="2018-07-30T12:58:48.720" Score="1" Body="&lt;p&gt;You are right that there should be no issue if the loads during construction never exceed the design loads.  There are situations where construction loads may exceed the final loads, though. Here are a few examples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Wind or other lateral forces before braces are installed.&lt;/strong&gt; You may need to do analysis so that you can define when in the construction sequence these braces need to be installed. Otherwise the contractor may decide to install them last.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Construction equipment.&lt;/strong&gt; Sometimes contractors use equipment that applies much greater loads during construction than during use. Is your office building floor able to support a crane?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Construction stockpiles.&lt;/strong&gt; Sometimes contractors store material in piles on the built floors. Is your floor designed to handle a pile of columns for the next floor?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="33" LastActivityDate="2018-07-30T12:58:48.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22958" PostTypeId="2" ParentId="22949" CreationDate="2018-07-30T13:15:21.640" Score="2" Body="&lt;p&gt;That's because the finished structure with full loading &lt;strong&gt;&lt;em&gt;is not&lt;/em&gt;&lt;/strong&gt; the same as the finished structure as a composite of the loading of each construction phase. And the latter is the correct solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be seen with the trivial structural model below. The first two structures represent the two construction phases, and the last one is the complete structure with all the loading. Here are its axial results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MknvH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MknvH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the compression at the very top of the right-most column is different between the second phase (47.7&amp;nbsp;kN) and the total structure (50.6&amp;nbsp;kN). Indeed, every single one of the values is different between the total structure and the addition of the first and second phases. For example, the base of the central column has 395.6&amp;nbsp;kN in the total-structure-total-loading case, while the addition of both phases gives us $209.3+197=406.3\text{ kN}$.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, the reason is that the completed structure is applying the loading of the first phase on the whole structure, when in fact those loads are only being supported by the first phase. Compare the results below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WTWg0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WTWg0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The existence of the second level is causing the loads to be redistributed. The central column is under less loading, while the outer ones are supporting a greater portion of the load. Hell, the second phase is even presenting forces even though it didn't actually exist when those loads were applied.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, why does the existence of the second phase modify the results from the loading of the first phase? To get an intuitive understanding, let's look at the structure's deflection. But let's do so while telling a story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The story starts by assuming the second phase is only composed of the external columns, with no central column and no second-phase beams. In this case, the loads from the first phase cause an &quot;inwards&quot; rotation on the top-external nodes of the first phase. Since we're assuming the second-phase columns are also there, those columns also get rotated. However, this is a solid-body motion, with no deformation and no internal forces or moments generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ISH54.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ISH54.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the real second phase obviously has the beams, so let's include them (while still leaving the central column out). The current deflection of the external columns will obviously cause a shortening of the beams, so they'll be under compression (as can be seen in the axial results above). However, every force has an equal and opposite reaction, so the compressed beams will also push the external columns outwards, counteracting the vast majority of the inwards motion we see above. Indeed, the beams in my model are so axially stiff that they barely budge horizontally at all, such that the columns feel as if the node is almost fixed in the horizontal direction. In this case, the top of the second-phase columns ends up rotating outwards, and the beams suffer negative bending moment (as can be seen by their deflection which causes tension on the upper fiber).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CFB0r.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CFB0r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the loads have caused a compression and shortening of the first phase's central column. However, the second phase's beams are presenting uplift at midspan, where the second phase's central column should be. Therefore, that column will obviously be under tension as it needs to stretch to resist that motion. That tension will need to be transmitted down to the ground support, so it actually represents an overall reduction in the compression of the first phase's central column (it therefore stands to reason that the external columns end up taking that additional load).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that if the axial stiffness of the second-phase beams were very low compared to the columns' bending stiffness (and I mean &lt;strong&gt;&lt;em&gt;low&lt;/em&gt;&lt;/strong&gt;, given elements' natural inclination to be much stiffer axially than under bending), they wouldn't offer much resistance to the external columns' desire to move inwards, and would actually present positive bending moment (tension on the inferior fiber):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c72ME.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c72ME.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, it's possible that the structural behavior is different, with the central column absorbing loads from the external ones (the unloaded second phase presents tension in the external columns and compression in the central one). However, for this particular model I had to use unreasonably low axial stiffnesses to achieve that reversed result.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;An argument could be made that over time, due to long-term redistribution effects, the loads will in fact approach the &quot;total structure under total loading&quot; results. But that's beyond the scope of this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, depending on the precise type of structure, the phases might be independent from one another (due to the existence of hinges, for example). In that case, yeah, the results should be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anything, I'd suggest doing an envelope of results between &quot;sum of each phase&quot; and &quot;total structure&quot; and designing members for the greater of the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this analysis is obviously only valid for permanent and construction loads. Once the structure is totally built, all variable loads are supported by the total structure.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2018-08-02T01:33:06.250" LastActivityDate="2018-08-02T01:33:06.250" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22960" PostTypeId="1" CreationDate="2018-07-30T14:47:59.570" Score="1" ViewCount="20" Body="&lt;p&gt;I have a vertical load (F) acting at a distance from the edge of the cylinder. The load is acting on the cylinder via chunk of steel sliding across the surface of the roller. The coefficient of friction between the two objects is 0.25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shaft is fixed at both ends. The cylinder has seized and has become a part of the shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know the torque and the force at the yellow point.&#xA;You must state the variables you use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MDAdQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MDAdQ.png&quot; alt=&quot;Image of Problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also like an explanation of where the frictional force from the load is going. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="16955" LastActivityDate="2018-07-30T14:47:59.570" Title="Torque on a siezed part" Tags="|torque|" AnswerCount="0" CommentCount="3" ClosedDate="2018-07-31T09:35:47.513" ContentLicense="CC BY-SA 4.0" />
  <row Id="22961" PostTypeId="2" ParentId="22949" CreationDate="2018-07-30T15:20:42.663" Score="1" Body="&lt;p&gt;There is definitely a need to look at loading sequence through staged construction.  Modern analysis usually has a section for staged construction.  I normally run across this in bridge construction.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The piers can be constructed before the tops are stabilized by the bridge deck.  &lt;/li&gt;&#xA;&lt;li&gt;The bridge decks themselves can be simply supported between piers before the concrete over the piers cures and the deck can subsequently be treated as a continuous member&lt;/li&gt;&#xA;&lt;li&gt;Construction equipment and materials can generate abnormal loading conditions&lt;/li&gt;&#xA;&lt;li&gt;Due to the size of your bridge deck and location, you may have limits on how much deck you can place and finish in a given time frame.  The wet concrete will provide dead load in the poured sections, and there will be no strength from the deck in the incomplete sections (or the wet concrete sections)  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6993" LastActivityDate="2018-07-30T15:20:42.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22962" PostTypeId="1" AcceptedAnswerId="22964" CreationDate="2018-07-30T17:14:50.977" Score="0" ViewCount="76" Body="&lt;p&gt;I'm about to start learning about control engineering. But am I advised to learn first Thermodynamics? (Since control engineering has applications in Thermodynamics)&lt;/p&gt;&#xA;" OwnerDisplayName="user16958" LastActivityDate="2018-07-30T18:17:06.097" Title="Learn thermodynamics before beginning with control engineering ?" Tags="|thermodynamics|control-engineering|" AnswerCount="1" CommentCount="9" ClosedDate="2018-07-31T13:02:38.483" ContentLicense="CC BY-SA 4.0" />
  <row Id="22963" PostTypeId="2" ParentId="19007" CreationDate="2018-07-30T18:11:59.383" Score="2" Body="&lt;p&gt;The draft will decrease as you move up the radiant section as the elevation is reducing with very little pressure drop. The convection section on the other hand will have significant pressure drop for the flue gas flow with little elevation change to move through it. The negative pressure above the convection section exit must be at a higher negative pressure than the inlet due to this pressure drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more helpful way to think about this might be to envision a 35 foot tall radiant section operating at 1800F. The base draft is -0.40 In. W.C. and with the elevation change a rough decrease in draft expected is 0.10 In. W.C. for every 10 feet in elevation change so lets say the draft at the top of the radiant section or bridgewall is now -0.05 In. W.C. (-0.40 minus 0.10/10 feet * 35 feet). Just above this starts the convection section with several rows of close tolerance tubes. At the operating conditions the flue gas must all flow through these spaces which will increase its velocity and cause several changes in direction to the flow. The convection section is only 10' deep here but contains 20 tube banks which causes a differential pressure of 0.30 In. W.C. from inlet to exit. Here we know what the pressure is just below the tubes (-0.05 In. W.C.) and we know the final stack pressure on the downstream side of the convection tubes must be enough to overcome the loss in draft across the tubes. The loss in draft across the tubes is the elevation change plus the flowing pressure drop or 0.40 In. W.C. (0.30 + 0.10/10 feet *10 feet). The downstream side of the convection section or breeching section must then be -0.35 In. W.C. (-0.05 + 0.40 = -X). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do take issue with the stack damper listing. The draft at the end or top of the stack is close to zero however the draft at the stack damper is very unlikely to be close to zero unless the damper is at the top of the stack which is essentially never done. The damper adds pressure drop to the exiting flue gas, the elevation and flue gas temperature is where the draft is created. If the breeching section is at -0.35 In. W.C. and the stack damper is fully open the stack must be around 35 feet (0.10 In W.C./10 feet * 35 feet). If the damper were to be closed down to say only 25% open the downstream side of the damper would still be at -0.35 In W.C. however the upstream side would now be lower due to the pressure drop added by the damper ( a rough estimate might be for the upstream damper draft to now be -0.2 In W.C. with the damper 25% closed).&lt;/p&gt;&#xA;" OwnerUserId="16960" LastActivityDate="2018-07-30T18:11:59.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22964" PostTypeId="2" ParentId="22962" CreationDate="2018-07-30T18:17:06.097" Score="0" Body="&lt;p&gt;It is difficult to understand the aspect at play on the control side of processes before first understanding how the process behaves.  Thermodynamics, fluid mechanics/dynamics, unit operations, kinetics, etc. will all improve your process knowledge and make what you are learning in controls much easier to implement appropriately.  The first two listed would be very important in the learning phase in order to understand the why behind most control operations.&lt;/p&gt;&#xA;" OwnerUserId="16960" LastActivityDate="2018-07-30T18:17:06.097" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22965" PostTypeId="2" ParentId="22955" CreationDate="2018-07-30T18:26:10.633" Score="1" Body="&lt;p&gt;Depending on the type of electrode and the properties of the base material, we get a wide range of possible answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical electrodes tensile strength is ranging between 60 -70- 80- 90 ksi. They ultimately fail at shear which we call f_w at ultimate level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we consider factor of safety of 75% let's assume a= width of the weld and L =length of the weld.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$φ Rn = 0.75 f_w \times 0.707\times a \times L$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to simplify, you plug in 3mm=1/6 inch and 11mm = 15/32 inch length of the weld in the formula and input the f_w from the electrode's spec sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the same time you need to check the base material shear strength and factor of safety which is usually 0.60.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="33" LastEditDate="2018-08-31T01:08:23.190" LastActivityDate="2018-08-31T01:08:23.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22966" PostTypeId="1" AcceptedAnswerId="22992" CreationDate="2018-07-30T18:28:48.523" Score="1" ViewCount="598" Body="&lt;p&gt;According to wikipedia there are 3 types of steel based on the crystal structure: austenitic, ferritic and martensitic. Austenitic steels are non-magnetic where 300 series stainless steels belong to. Ferritic steels have magnetic properties. But I didn't manage to find any information about martensitic steel magnetic properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular I am interested in stainless steel alloy 455. According to the data I have this steel is magnetic (though didn't manage to find the source of this data). But when I obtained a sample of this steel It appeared to be non-magnetic. This brings to possibilities: either my source is wrong or the sample I got is not steel 455.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Upd.&#xA;Turned out that the samples I got were “accidentally” made of austenitic steel 304 instead of requested 455.&lt;/p&gt;&#xA;" OwnerUserId="16961" LastEditorUserId="16961" LastEditDate="2018-08-03T10:07:21.653" LastActivityDate="2018-08-03T10:07:21.653" Title="Is martensitic steel 455 magnetic?" Tags="|steel|magnets|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22967" PostTypeId="1" AcceptedAnswerId="22970" CreationDate="2018-07-30T18:55:49.123" Score="3" ViewCount="154" Body="&lt;p&gt;On &lt;a href=&quot;https://youtu.be/mqPMANOEG7Y?t=3m26s&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; from Captain Joe, you can see that the landing gear of the airplane has got some sort of v-shaped strut connecting the two moving parts.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0tiNr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0tiNr.jpg&quot; alt=&quot;v-shaped strut&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that I saw the same thing on some vespas with a suspension in the front wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are they for ?&lt;/p&gt;&#xA;" OwnerUserId="16962" LastActivityDate="2018-07-30T19:35:26.610" Title="What is this V-shaped strut on the landing gear for?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22968" PostTypeId="2" ParentId="22967" CreationDate="2018-07-30T19:21:51.747" Score="1" Body="&lt;p&gt;I have a small plane and also  fly with other small single engine prop airplanes, Such as Cessna, Cherokee, Moony and the like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They do have a similar structure providing alignment and resisting shimmying of the wheel when the plane hits a bump on the tarmac.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is likely the same structure except possibly doing some more sophisticated things in addition to that.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-07-30T19:21:51.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22970" PostTypeId="2" ParentId="22967" CreationDate="2018-07-30T19:29:17.107" Score="4" Body="&lt;p&gt;This is the mechanism that allows the piston to move so absorbing the landing forces, while making sure that there is no rotation of the wheel assembly which would create a huge amount of side loading and possibly cause the landing gear to fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have had the opportunity to visit Airbus and stand next to some of the units used to test these... serious engineering, but then 230 tonnes does take some controlling...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-07-30T19:35:26.610" LastActivityDate="2018-07-30T19:35:26.610" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22971" PostTypeId="2" ParentId="8001" CreationDate="2018-07-30T19:41:03.680" Score="2" Body="&lt;p&gt;Carlton is mistaken.  Look at the exponent carefully.  It should be 3/2, and not 2/3, this will obviously make a huge difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://engineering.stackexchange.com/a/22873/1832&quot;&gt;another answer&lt;/a&gt;, I give an explanation for why this equation is the case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At each location along the vertical distance $h$, the velocity is given by $\sqrt {2gh}$, where $g$ is the acceleration due to gravity, and this velocity of flow varies as $h$ changes along the vertical. So, you must integrate over the vertical distance since the relationship is nonlinear and you can't simply take the flow rate at the center. The first integral of the equation above is $\dfrac{2}{3}\sqrt{2g}\cdot h^{3/2}$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I neglect the C term which is experimentally derived.&lt;/p&gt;&#xA;" OwnerUserId="16828" LastEditorUserId="1832" LastEditDate="2018-08-02T01:51:30.060" LastActivityDate="2018-08-02T01:51:30.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22972" PostTypeId="1" CreationDate="2018-07-30T19:46:58.357" Score="2" ViewCount="187" Body="&lt;p&gt;Hi I am making a beam model using Abaqus. There are two beams created. I wanted to join two beams rigidly in the connection to transfer moments between beams. However, I do not know what connector or constraint should be used to complete such a task.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the model I used, I created chain wire at the same point and created the following as a connector section to be assigned to this wire set. However, it does not work and give error:&quot;Node set assembly_Part-1_Part-refpt_ has no members and will be ignored. Nodes defined in this set may have been deleted because they were not connected to any elements.&quot;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1vRvA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1vRvA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fncIV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fncIV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7st2J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7st2J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help to solve this connection problem. Thank you!!&lt;/p&gt;&#xA;" OwnerUserId="16965" LastActivityDate="2018-07-30T19:46:58.357" Title="How to join two beams rigidly?" Tags="|mechanical-engineering|structural-analysis|modeling|abaqus|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22974" PostTypeId="1" AcceptedAnswerId="22980" CreationDate="2018-07-30T20:48:42.617" Score="1" ViewCount="2625" Body="&lt;p&gt;I am thinking of building a house with basement next to some existing structures. Are there rules of thumb on how closely/deeply you can excavate next to an existing structure? By moving farther away from the existing structure do I reduce the potential need for temporary retaining?&lt;/p&gt;&#xA;" OwnerUserId="16967" LastActivityDate="2018-08-21T03:08:37.390" Title="Excavation near existing structure" Tags="|structural-engineering|civil-engineering|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22975" PostTypeId="2" ParentId="22974" CreationDate="2018-07-30T21:36:41.147" Score="0" Body="&lt;p&gt;There are various factors that need to be considered.&#xA;Amongst them are soil strength, depth of the excavation, distance, weight (i.e. pressure exerted) of the neighbouring structure as well as the depth of its foundation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without any numbers it‘s not really possible to make an estimation about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the further away the new excavation site, the lower the risk of settling of the neighbouring structure.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2018-07-30T21:36:41.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22976" PostTypeId="1" CreationDate="2018-07-30T22:57:59.853" Score="1" ViewCount="33" Body="&lt;p&gt;I have a 12v 300 watt power supply.  I replaced 13 20watt halogen bulbs with 13 3watt LED bulbs. Questions: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;should I downgrade power supply since I'm now drawing 39 watts as opposed to the 260watts I was pulling with the old halogen bulbs?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Am I correct in assuming that power usage is based on the total wattage of the lights and not the rated wattage of the power supply? If true, then my power consumption has dropped from a rough equivalent of 3 100watt light bulbs to approximately a single 40watt bulb, correct?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16968" LastEditorUserId="288" LastEditDate="2018-07-31T12:02:40.577" LastActivityDate="2018-08-10T22:19:19.140" Title="Outdoor 12v power supply" Tags="|electrical|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22977" PostTypeId="2" ParentId="22954" CreationDate="2018-07-31T00:08:50.640" Score="1" Body="&lt;p&gt;Generally you are going to want to place the strain gages in the locations with the highest strain, i.e. the red spots.  In addition to knowing whether the locations are in tension or compression, you will also want to know the direction of the stress.  i.e. which direction should you align the gage?  For this you will need to plot directional stresses.  Alternatively, depending on your analysis package you could ask it to plot the vector principle stress.  In which case you will want to align the gage in the direction of 1st principle stress (or maybe 3rd).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A caveat though is that the locations with the highest stress usually also have the highest gradient.  That means if you place the gage a tiny bit left or right of where you wanted it, or if your analysis was off a little bit, then you aren't going to get quite what you expected.  You could also look for regions with medium high strain, but low gradients.  i.e. the orange areas.  That should get you a value that is easier to compare to the analysis.  Whether this matters to you depends on your application.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-07-31T00:08:50.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22979" PostTypeId="2" ParentId="22976" CreationDate="2018-07-31T00:22:12.387" Score="2" Body="&lt;p&gt;1) You could if you wanted to, but it's probably not necessary.  Drawing less than the rated power should not hurt it.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) yes and yes.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-07-31T00:22:12.387" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22980" PostTypeId="2" ParentId="22974" CreationDate="2018-07-31T00:38:25.513" Score="0" Body="&lt;p&gt;Usually your building department is the first source. They will recommend what needs to be done. &#xA;If your neighbor is a conventional building with few stories of wood or light framing, and the soil is sound and firm without high level of subterranean water table, by staying outside of a 45 degree projection plane from the bottom of his foundation you should be ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even then it is recommended to excavate near the neighbor in sections of 4 to 6 feet wide and build either a temporary shoring or part of permeant retaining wall and then dig and build next section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many Soils engineers for a small fee do a pre review of the site and give you a professional advice as to whether you need a soils report or no.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the basement walls you need to control and drain any moisture caused by rains or underground water from standing behind walls, through insulation and drainage system via a possible sump pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally your retaining walls should be designed to be strong enough for combination of worst conditions, both gravity and soils pressure.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-07-31T00:38:25.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22981" PostTypeId="1" CreationDate="2018-07-31T01:37:02.433" Score="3" ViewCount="109" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2bpa8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2bpa8.jpg&quot; alt=&quot;&amp;quot;gear style component&amp;quot; - unknown name&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You loosen a screw (knob) and then move to the angle you want and tighten. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know what the engineering term for this part is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance&lt;/p&gt;&#xA;" OwnerUserId="16969" LastEditorUserId="13311" LastEditDate="2018-07-31T22:54:32.407" LastActivityDate="2018-08-30T23:01:14.557" Title="What is the engineering term for the gear style articulated arm" Tags="|structural-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="22982" PostTypeId="2" ParentId="22911" CreationDate="2018-07-31T05:49:12.943" Score="0" Body="&lt;p&gt;I finally found someone from the OEM that could concisely explain what is going on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IGV curves listed at the top are the inlet guide vanes I mentioned. The numbers are positions of those vanes and are used to control efficiency by effectively imparting spin on the incoming air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The groups of lines that were not labeled are vane diffuser positions and are used to control capacity by opening or restricting flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By changing the combinations of these two features, you can hit the various curves shown on the plot. The top of the individual curves are the surge pressures. &lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2018-07-31T05:49:12.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22983" PostTypeId="2" ParentId="15643" CreationDate="2018-07-31T06:51:43.753" Score="1" Body="&lt;p&gt;In a normal NPN transistor we are dealing with three states (seen comment section on Josemi post). &lt;a href=&quot;https://i.stack.imgur.com/bHEqB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bHEqB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red area, indicates the saturation the BC junction is forward polarised, the the green area is called linear, the BC junction is inverse polarised and the transistor works as an amplifier and when $I_B = 0$ we land in cut-off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prefere to work with the simplified version, so i take the threshold voltage of NP or PN junction as &lt;strong&gt;$0.7_v$&lt;/strong&gt;. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/fzf1d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fzf1d.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let investigate all the three area a little bit more, starting with cut-off:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V_{BE} &amp;lt; 0.7_v \Rightarrow I_B = I_C = I_E = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear area:$$&#xA; \left\{&#xA;    \begin{array}{ll}&#xA;        V_{BE} = 0.7_v &amp;amp;  \\&#xA;        V_{CE} &amp;gt; 0 &#xA;    \end{array}&#xA;\right. \Rightarrow   \left\{&#xA; \begin{array}{ll}&#xA;I_B&amp;gt;0 &amp;amp;  \\&#xA;  I_C = \beta I_B &#xA; \end{array}&#xA;\right.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;saturation: &#xA;$$&#xA; \left\{&#xA;    \begin{array}{ll}&#xA;        V_{BE} = 0.7_v &amp;amp;  \\&#xA;        V_{CE} = 0&#xA;    \end{array}&#xA;\right. \Rightarrow   \left\{&#xA; \begin{array}{ll}&#xA;I_B&amp;gt;0 &amp;amp;  \\&#xA;  I_C &amp;lt; \beta I_B &#xA; \end{array}&#xA;\right.&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see there is no limitation on $I_B$. However, this just a theory, in reality, i don't think so they come in handy when the bias current is about some femtoamps, otherwise why, using CMOS then? &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-07-31T08:14:46.647" LastActivityDate="2018-07-31T08:14:46.647" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22984" PostTypeId="2" ParentId="22411" CreationDate="2018-07-31T09:30:28.677" Score="1" Body="&lt;p&gt;Torque of the engine does not matter. What matters when pulling is the force exterted by the car at the base of the wheel pushing the car forward. This force depends on many things but the  two most important things are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The current speed of the vehicle&lt;/li&gt;&#xA;&lt;li&gt;The power of the engine&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In fact ultimately, assuming a perfect transmission and no losses, you can write an equation for the force which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ f_p = {P \over v}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$f_p$ - is the pulling force of your vehicle,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$P$  - is the power of the engine (at the current rpm/throttle) and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v$ - is the current speed of the car&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have the current pulling force you can compare it to the force required to pull your trailer. &#xA;Note that the pulling force tends to infinity as speed approaches 0, this is correct. In fact even a very low power tractor can still pull tonnes on a trailer because of a very advantageous transmission ratio and consequently very low speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to answer your question: to pull any trailer you need a torque greater than zero at an angular speed greater than zero. This give us a power which is greater than 0 which will be enough to pull any trailer at a certain speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed ultimately depends on many other forces which have to be balanced with the initial pulling forces but the most important are air drag, road slope losses, and tires losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air drag depends on the size and shape of the vehicle and increases roughly with the square of the speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slope losses depends on the mass of the vehicle and the slope of the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tires losses depends on the weight of the vehicle and increase roughly linearly with the speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more 'correct' question should be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At which speed $v$ can I pull a trailer of mass $m_t$ with a car of mass $m_c$ and power $P$ on a road with a slope of $s$ if the air drag coefficient is $k_a$ and the friction coefficient is $k_f$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then a very rough answer would be obtained by solving for v the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$${P \over v} - {{k_a}v^2 } - {((m_t + m_c)sin(s))} - ((m_t + m_c){k_f}v)  = 0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${P \over v}$  is the pulling force of the car at speed v&lt;/p&gt;&#xA;&#xA;&lt;p&gt;${{k_a}v^2 }$ is the air drag force acting on the whole vehicle at speed v&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$((m_t + m_c)sin(s))$ is the gravity force acting on the vehicle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$(m_t + m_c){k_f}v$ are the friction losses at speed v&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is a very &quot;simple&quot; model and does not take into consideration many things, for example even friction changes as the slope of the terrain changes.&lt;/p&gt;&#xA;" OwnerUserId="13837" LastEditorUserId="13837" LastEditDate="2018-07-31T12:00:58.653" LastActivityDate="2018-07-31T12:00:58.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22987" PostTypeId="1" CreationDate="2018-07-31T11:41:34.073" Score="1" ViewCount="64" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hCakZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hCakZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I am building a Project that can pinpoint location of a pen or any object (something as small as 0.5 mm radius sphere) placed on a flat screen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using a ultrasound sensor (initially but i can switch to a different sensor if the situation demands) attached to one side of the flat screen ( on top edge). I'll be attaching a metal ( aluminium or copper) film on the object whose location I want to detect. I'll be using two ultrasound sensors to measure the x and y axis distance and pinpoint the location&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find a way wherein I can measure the intensity and direction of the reflected wave from the metal film to detect location. However I can't find a way to do so yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to do so and if it is will ultrasound suffice ?&lt;/p&gt;&#xA;" OwnerUserId="14887" LastEditorUserId="14887" LastEditDate="2018-07-31T16:52:02.237" LastActivityDate="2018-08-30T19:00:33.403" Title="Measuring intensity of a radio wave reflected from a metal plate" Tags="|experimental-physics|distance-measurement|ultrasound|radio|transmission|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="22989" PostTypeId="2" ParentId="22981" CreationDate="2018-07-31T15:02:36.753" Score="2" Body="&lt;p&gt;There isn't a 'universally recognised' name for this. &quot;Toothed Clamping Element/Plate&quot; was the most common that I could find, with multiple sellers online using this terminology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/koAD9m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/koAD9m.png&quot; alt=&quot;Toothed Clamping Element&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/NRyqrm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NRyqrm.png&quot; alt=&quot;Clamping Element Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are often used in conjunction with spring-loaded clamping lever/screws, which allow you to disengage the lever by pressing down on the screw head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MrSCBm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MrSCBm.png&quot; alt=&quot;Clamping Lever&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-07-31T15:02:36.753" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22991" PostTypeId="2" ParentId="22987" CreationDate="2018-07-31T18:16:09.483" Score="1" Body="&lt;p&gt;Typical distance sensors depend on time-of-flight.  If you're trying to hack the distance from return power level, you will not be happy (to put it mildly).  It's next to impossible to calibrate any system that way.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to direction, it should be obvious that you cannot possibly do so with a single receiver. As a minimum you'd need two sensors aka two pixels, and then either compare the incoming phases (as done with radar) or the location of peak energy of a focussed input ( as is done with optics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a trivial problem .&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-07-31T18:16:09.483" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22992" PostTypeId="2" ParentId="22966" CreationDate="2018-07-31T20:34:22.977" Score="1" Body="&lt;p&gt;&lt;strong&gt;Yes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any type of martensitic steel will be ferro magnetic. However the magnetic susceptibility may not be the same as that of a material like iron. This could make a sample look like it is not magnetic or very weakly magnetic when compared to a material with high magnetic susceptibility like iron. There is also the possibility that the sample has been worked in such a way that the magnetic susceptibility was reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would first make sure you are checking the sample with a strong magnet like a neodymium magnet. Otherwise, the manufacturer of the sample will know what has been done to the steel and how to best answer you question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://webcache.googleusercontent.com/search?q=cache:7ciwBO5U08MJ:ispatguru.com/martensitic-stainless-steels/+&amp;amp;cd=14&amp;amp;hl=en&amp;amp;ct=clnk&amp;amp;gl=us&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; for more information about martensitic stainless steel magnetic properties and &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_susceptibility&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia article&lt;/a&gt; for information on magnetic susceptibility.&lt;/p&gt;&#xA;" OwnerUserId="9656" LastEditorUserId="9656" LastEditDate="2018-08-01T13:51:50.870" LastActivityDate="2018-08-01T13:51:50.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22993" PostTypeId="2" ParentId="18844" CreationDate="2018-07-31T22:03:36.497" Score="2" Body="&lt;p&gt;I've done a lot of sailing (won a couple national championships) and the answer is yes. But you wouldn't want the fan blowing forward from the stern.  Instead you would want it blowing towards the sails from the side. Even though the fan would want to pull the boat sideways, sailboats are equipped with centerboards or keels which function like a wing underwater. Since water is much more dense then air, you'll have a hell of a time pulling the boat sideways.  Instead, the air will wrap around the sail thus creating a high pressure zone on the &quot;windward&quot; side and a low pressure zone on the &quot;leeward&quot; side and propelling the boat forwards. You'll also get a bit of drive from &quot;deflecting&quot; air backwards (equal and opposite reaction). &lt;/p&gt;&#xA;" OwnerUserId="16979" LastActivityDate="2018-07-31T22:03:36.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22995" PostTypeId="1" CreationDate="2018-07-31T23:49:07.943" Score="0" ViewCount="65" Body="&lt;p&gt;I'm a headphone enthusiast, and once I tried to open up my headphones to see how much they move when I play my music.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I was really surprised to see that the sounds that I could &quot;feel&quot; actually came from a driver that didn't have any visible motion at all! The only way I could see even the smallest motions was by turning the volume up to extremely high levels that I would never listen to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to use the calculator I found &lt;a href=&quot;http://www.baudline.com/erik/bass/xmaxer.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; to calculate the motion that my headphones would make, and the results I got seemed to make sense, but this calculator was made for large speakers, not small headphones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone tell me if this calculator should scale down to a headphone level (45 mm)? If not, what is an estimate for how much motion a headphone driver should be making at moderate volumes in terms of mm traveled?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="16018" LastActivityDate="2018-08-01T07:39:59.507" Title="How can I calculate headphone driver excursion?" Tags="|mechanical-engineering|electrical-engineering|acoustics|linear-motion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22996" PostTypeId="1" CreationDate="2018-07-31T23:58:15.480" Score="3" ViewCount="496" Body="&lt;p&gt;I enjoy creating mechanical designs but I am not a trained mechanical engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a technical issue with a joint design that needs to be permanently retained using a press (snap) fit. Any idea on the size and tolerances for these two 304 SS parts to be pressed together using an arbour press without permanent deformation of either part? I will use a little lithium grease to help with the press fitting which will also serve as general lubrication for the shock absorber joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EpG1Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EpG1Z.jpg&quot; alt=&quot;press fit size mm&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lteDF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lteDF.jpg&quot; alt=&quot;sample with rubber surround&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ilkaK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ilkaK.jpg&quot; alt=&quot;timing station&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16980" LastEditorUserId="9" LastEditDate="2018-08-01T00:06:47.940" LastActivityDate="2021-06-08T08:01:25.113" Title="press fit ball joint" Tags="|mechanical-engineering|design|applied-mechanics|manufacturing-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="22997" PostTypeId="2" ParentId="1" CreationDate="2018-08-01T01:31:42.900" Score="0" Body="&lt;p&gt;Yes you can pump powders using a diaphragm pump. The above comments all focus on alternate ways to transport powder they are not talking about pumping. Check out powder pumping with a diaphragm pump on youtube.&lt;/p&gt;&#xA;" OwnerUserId="16982" LastActivityDate="2018-08-01T01:31:42.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="22998" PostTypeId="2" ParentId="22966" CreationDate="2018-08-01T03:50:59.143" Score="1" Body="&lt;p&gt;This is a precipitation hardening martensitic so will be ferromagnetic unless it has been mistreated. To anneal, it is quenched from high temperature and makes a soft martensite = ferromagnetic. It is then machined and/or cold formed . And then final heat-treatment is an age at roughly 1000 F when it will harden. There may be minor changes in ferromagnetic properties depending on condition but I expect it is always similar to conventional steel. I never worked with it but the high nickel - 9%, makes it prone to have retained austenitic ( very low ferromagentism) structure after quenching: If so, a sub zero cooling ( like dry ice and alcohol) may get it to transform to martensite. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-08-01T18:24:46.210" LastActivityDate="2018-08-01T18:24:46.210" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="22999" PostTypeId="2" ParentId="22995" CreationDate="2018-08-01T07:39:59.507" Score="1" Body="&lt;p&gt;I doubt that the calculator you referenced is relevant for headphones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fundamental difference between headphones and loudspeakers is that headphones transmit energy to the small, (almost) sealed, volume of air trapped between the headphone and the structure of your ear. That is completely different from a loudspeaker which is designed to transmit energy into a (theoretically) infinite region of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your ear responds to pressure variations in the air, for a small &quot;sealed&quot; volume of air the transducer movement to produce the same pressure change is roughly &lt;em&gt;independent&lt;/em&gt; of the frequency - you are changing the &lt;em&gt;volume&lt;/em&gt; of the trapped air by the same amount, and thus changing its pressure, independent of the frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is very different from a loudspeaker radiating into free air, where you need (approximately) the same &lt;em&gt;velocity&lt;/em&gt; of the speaker cone to create the same pressure difference, and the displacement to produce that velocity is (approximately) inversely proportional to frequency - which is why high power bass speaker cones may move with an amplitude of several inches at low frequency, to produce the required sound pressure level.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-08-01T07:39:59.507" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23000" PostTypeId="1" CreationDate="2018-08-01T11:27:03.880" Score="1" ViewCount="322" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JTQv4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JTQv4.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/AlOGy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AlOGy.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Case 1&lt;/h2&gt;&#xA;&lt;p&gt;The development of Case 1 of this problem in which the axes of the prisms intersect at right angles, is shown in Diagram 1. Note that inter-edge lines are required for the development of both solids, their positions being found by projectors drawn from the different views of the drawing to the full view of the end.&lt;/p&gt;&#xA;&lt;h2&gt;Case 2&lt;/h2&gt;&#xA;&lt;p&gt;Case 2 of the problem in which the axes intersect at 45°, is developed in a similar manner. The projections for this are shown on the plate at Fig 1(a). In order to avoid confusion of lines, projectors are sometimes drawn as shown on this plate, that is, only their starting points are indicated, as between the two views of the octagonal prism in this drawing. First, reproduce the projections. Next, develop the stretch outs for the solids on the lines MN and M'N', and draw edge and inter-edge lines through their respective points. The positions of the inter-edge lines are found by projecting the point A in the plan across to the full view, as shown in (d) afterwards locating the points x and y at x' and y' in (d'); then they are projected to the elevation and carried to the development in the usual manner, the resulting figures at (b) and (c) completing the development of the solids.&lt;/p&gt;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&lt;p&gt;What I need to know is, should both drawings be drawn in an elevation and plan way? And where exactly is the problem in both cases that should be developed?&lt;/p&gt;&#xA;" OwnerUserId="16974" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-09-30T19:01:09.440" Title="Two intersecting prisms" Tags="|mechanical-engineering|manufacturing-engineering|mathematics|drafting|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23001" PostTypeId="1" CreationDate="2018-08-01T13:51:54.910" Score="1" ViewCount="3568" Body="&lt;p&gt;I am designing a scissor lifting table and I have an issue with the linear actuator. &lt;a href=&quot;https://i.stack.imgur.com/GjdUq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GjdUq.png&quot; alt=&quot;table&quot;&gt;&lt;/a&gt; This table should be lifting 100 kg. I studied the problem in two ways : a static study and a power study. My table is a 3-level table. Here is a schema of a one level table : &lt;a href=&quot;https://i.stack.imgur.com/JvbC2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JvbC2.png&quot; alt=&quot;lifting table sketch&quot;&gt;&lt;/a&gt; (1) and (4) are V-rails that guide wheels on points A and F. (5) is the linear actuator. (2) and (3) are scissors, and all the mechanical connections between the different parts are pivots (revolute).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://drive.google.com/open?id=0B_UWwmc5Pc4uZTdQdmU3aWprNV9hcU5SbkZtN2RSd1VLV3ZN&quot; rel=&quot;nofollow noreferrer&quot;&gt;this static study&lt;/a&gt;, I found (page 3) that the two linear actuators have to lift a load of minimum 2158.2 N. So each linear actuator has to be lifting 1079.1N. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;a href=&quot;https://drive.google.com/open?id=16jXbSmludoOoTBbopUWp0tn63H-8GgmE&quot; rel=&quot;nofollow noreferrer&quot;&gt;this power study&lt;/a&gt;, I found (page 2) that the two linear actuators have to lift 12988 N (6494 N without the safety factor). So each linear actuator has to be lifting 6494 N (3247 N without the safety factor). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between the two results, I have a ratio of 3. I guess the second study is false, but I'm struggling finding why, and I still have no solution...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you can help me, and I thank you for your time :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Muthu&lt;/p&gt;&#xA;" OwnerUserId="16986" LastActivityDate="2018-08-02T05:16:01.577" Title="Linear actuator on a scissor lifting table" Tags="|mechanical-engineering|statics|power|power-transmission|linear-motion|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23003" PostTypeId="1" AcceptedAnswerId="23004" CreationDate="2018-08-01T15:30:55.077" Score="1" ViewCount="3534" Body="&lt;p&gt;I have been trying to understand the torque curve and power curve of an engine. From what I have read and understood I know that the torque curve and power curve is has a peak at a particular RPMs(RPM is different for torque curve and power curve). I also understand that when the engine is running at a particular RPM it will be producing a particular torque(drive torque) and if we increase the load on the engine(increase the load torque) then the RPM will start to decrease. To keep the RPM from decreasing, we can open up the throttle( which will increase the fuel burnt and so the energy produced at the power stroke) and so, increase the drive torque being produced to match the increased load torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am not able to understand is if the torque value at a particular RPM shown in the torque curve is the maximum possible torque at that RPM (i.e throttle valve fully open at that RPM) or if that torque curve is given for a particular external load(i.e we can still increase the load and to keep the RPM constant we can still open up the throttle valve). I am guessing that the curve shows the maximum possible torque at a particular RPM but I am not sure.   &lt;/p&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="2015" LastEditDate="2018-08-01T17:08:14.170" LastActivityDate="2021-08-29T20:54:44.563" Title="What does the torque curve of an engine mean?" Tags="|mechanical-engineering|automotive-engineering|torque|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23004" PostTypeId="2" ParentId="23003" CreationDate="2018-08-01T15:52:24.863" Score="2" Body="&lt;p&gt;to produce a torque curve, the throttle is opened, the load increased, the throttle is opened more, the load is increased more, etc., etc. until the throttle is wide open- then the torque measurement is made. then the load is decreased slightly, which causes the RPM to increase slightly, and the torque is measured again. the process of decreasing the load incrementally and measuring the torque after the engine's RPM has increased is continued until the engine's redline RPM is reached. then the process is reversed: the load is increased, which brings the RPM down, torque is measured; then load is increased some more to lower the RPM some more and the torque is measured, and so on until the engine is finally lugged down to the minimum speed at which it can still run smoothly. This produces a torque curve at every point of which the throttle is wide open.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-08-01T15:52:24.863" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23005" PostTypeId="2" ParentId="23000" CreationDate="2018-08-01T16:58:06.593" Score="1" Body="&lt;p&gt;The problem that needs to be solved and the method you described to solve that problem would be how to cut the octagonal prism along it's circumference in such a way that it would smoothly fit with the kite shaped prism.  The plausible, theoretical shape described by the drawing is useless if it cannot be manufactured.  This method allows one to mark special points along the circumference of the prism, trace their path, and cut along that path in such a way to generate a smooth surface that will allow the two pieces to be joined with minimal gap.  This development before computer aided drafting was critical in making many important cuts with minimal aid.  Especially for pipe fitters, which frequently need to be joined pipes together with a &quot;saddle-cut&quot; all the time:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KqXSy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KqXSy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to start with flat edge prisms and then improve the technique to circles.  However, I found the most complicated of these in real life involve non-prismatic shapes, such as a cone intersecting a pipe at 90 degrees.  Other complications involve intersections off-center, as is shown in case 3, where further complications are found that need to be resolved via the drafter to derive a complete solution.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-08-01T17:06:40.417" LastActivityDate="2018-08-01T17:06:40.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23006" PostTypeId="1" AcceptedAnswerId="23138" CreationDate="2018-08-01T17:02:53.997" Score="3" ViewCount="101" Body="&lt;p&gt;I am solving problems through Fox and McDonald.Here is the problem &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/nPj2G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried by finding out the torque due to viscous forces which is $$\tau_1=\mu(2\pi Rh)\frac{R\omega}{a}$$ and the torque due to mass m1 as $$\tau_2=m_1gR$$ Writing into equation $$\tau_2- \tau_1=m_2R^2 \frac{d\omega}{dt}$$ integrating this and using boundary condition $\omega=0$ at $t=0$ I got $$\omega=\frac{m_1ga}{2\pi Rh\mu}[1-exp(\frac{-2\pi \mu htR}{am_2})]$$ . However I am missing $m_1+m_2$ instead of $m_2$ in the exponential part.Any ideas? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastEditorUserId="2015" LastEditDate="2018-08-09T12:24:57.547" LastActivityDate="2018-08-09T21:51:51.840" Title="Finding out torque in a viscosimeter" Tags="|mechanical-engineering|fluid-mechanics|dynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23007" PostTypeId="1" AcceptedAnswerId="23008" CreationDate="2018-08-01T21:07:28.703" Score="3" ViewCount="1137" Body="&lt;p&gt;I'm new to part design/mechanical engineering and I have the following question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HRMUW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HRMUW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Blue: steel shaft, 5/8&quot; diameter, connected to a motor&lt;/li&gt;&#xA;&lt;li&gt;Green: aluminum disk, 10&quot; diameter, 1/2&quot; thick, 3&quot; hole&lt;/li&gt;&#xA;&lt;li&gt;Goal: the shaft should rotate the disk&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Question:&lt;/strong&gt; I'm wondering how can the shaft be connected to the disk?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried searching for bushings, various forms of couplings, etc, but couldn't find anything that can connect the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean I would have to machine my own coupling, and there are no standard fittings/parts that would do the job?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FYI, the project needs high torque, so I'm guessing (please correct me if I'm wrong) that a coupling needs to have a large diameter as to apply the torque at a greater distance from the center.&lt;/p&gt;&#xA;" OwnerUserId="16991" LastEditorUserId="16991" LastEditDate="2018-08-01T21:25:07.877" LastActivityDate="2018-08-02T04:19:23.663" Title="How to couple/connect shaft to disk?" Tags="|mechanical-engineering|motors|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23008" PostTypeId="2" ParentId="23007" CreationDate="2018-08-02T04:19:23.663" Score="2" Body="&lt;p&gt;&lt;a href=&quot;http://www.masterdrives.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Masterdrives.com&lt;/a&gt; (.PDF - Catalog: &lt;a href=&quot;http://www.masterdrives.com/bushings/QD%2520Bushing%2520Pages.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;QD Bushings&lt;/a&gt;) sells bushings of a suitable size but you may or may not want to do some machining to reduce the &quot;E&quot; dimension. Apparently you want bushing type SF, the E length with that taper is bound to fit. 5/8&quot; bore size is available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Catalog page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PpCgA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PpCgA.jpg&quot; alt=&quot;QD Bushings&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another view of one of those T-bearings:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yHrCu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yHrCu.jpg&quot; alt=&quot;QD Bushing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively make your own or have one made for you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mOSkt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mOSkt.jpg&quot; alt=&quot;Homemade #1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another design:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P8OjF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P8OjF.jpg&quot; alt=&quot;Homemade #2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2018-08-02T04:19:23.663" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23010" PostTypeId="2" ParentId="22639" CreationDate="2018-08-02T04:51:16.477" Score="0" Body="&lt;p&gt;I think that I can say that, at least for my building, the benefit of positive pressure is a myth.  I can hear the wind whistling into the hallway from the vent, so I expect positive pressure in the hallways.  If I open my windows, and I open my apartment door a crack, I should be able to do the tissue test to indicate which way the air is flowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The empirical result is that it flows in the same direction as the external wind.  My apartment faces West, and if the wind is coming moderately from the West, then air flows from the apartment to the hallway.  If the wind is easterly, the air flows from the hallway into the apartment.  Essentially, the positive pressure scheme does very little to control the flow of air between my apartment and the hallway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My building is just one data point, but if positive pressure works most of the time, then what I observed would have been quite unlikely.&lt;/p&gt;&#xA;" OwnerUserId="16693" LastActivityDate="2018-08-02T04:51:16.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23011" PostTypeId="2" ParentId="23001" CreationDate="2018-08-02T05:16:01.577" Score="1" Body="&lt;p&gt;Basically it works like a lever with the force or weight arm length as AC and resisting arm which is connected to the actuator as BC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of mechanical advantage remains constant as the jack lifts the load because the ratio of the horizontal projection of the arms which is proportional to their respective lengths multiplied by the cosine of angle Alpha is the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the demand on actuator is $ \space =\frac {( F\times AC)}{BC}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By roughly scaling your sketch the force on actuator is $ = 100\times 9.8 \times 5/2 = 2450 N $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for the two actuators, thus we dived by 2 and get 1225 N per actuator. Now we have to multiply this by the safety factor. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-08-02T05:16:01.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23012" PostTypeId="1" CreationDate="2018-08-02T05:55:01.613" Score="3" ViewCount="1756" Body="&lt;p&gt;I was looking for some shear strength properties of common polymers/plastics.... I have not found any usefull information yet. I know polymers aren't that linear elastic, but I expected to find some range of shear properties. I would like to use it for a stress evalution of a general cross section at shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me with some numbers or tell me why it is impossible to give such figures?&lt;/p&gt;&#xA;" OwnerUserId="8134" LastEditorUserId="2015" LastEditDate="2018-08-07T18:06:36.757" LastActivityDate="2020-01-09T15:09:00.153" Title="Shear strength polymers" Tags="|mechanical-engineering|structural-engineering|materials|structural-analysis|plastic|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23013" PostTypeId="1" CreationDate="2018-08-02T05:59:15.230" Score="0" ViewCount="51" Body="&lt;p&gt;In our Factory we have a Compressor which rates 660 CFM (Kirloskar Make), which is a VFD Compressor.  And next to compressor, we have a Receiver tank.  Next to Receiver Tank, we have a Dryer. The Consumption of the Total Plant is around 350-400 CFM. The Flow in the Dryer Outlet was around 400-450 CFM, whereas the flow is uniform for 4 to 5 min.  Then the flow drops down to 50-70 CFM and remains in that for 40 Secs. But still all the machines were in the same running State. And after 40 secs, the Flow in the Dryer Outlet was about 400 to 450 CFM.  Why this kind of low CFM is happening in the Dryer Outlet?&lt;/p&gt;&#xA;" OwnerUserId="16994" LastActivityDate="2019-05-05T16:01:50.477" Title="Compressed Air Piping Network and Flow" Tags="|mechanical|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23014" PostTypeId="1" CreationDate="2018-08-02T11:10:12.803" Score="0" ViewCount="673" Body="&lt;p&gt;What would be the best ceiling shape for inducing condensation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing it would a curve ︶ or V shape, but those are just guesses. &#xA;(I don't have any background on thermal physics)&lt;/p&gt;&#xA;" OwnerUserId="16999" LastEditorUserId="90" LastEditDate="2018-08-02T13:50:46.743" LastActivityDate="2019-06-07T06:02:50.700" Title="what's the best shape for inducing condensation?" Tags="|fluid-mechanics|thermodynamics|heat-transfer|" AnswerCount="5" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="23015" PostTypeId="1" CreationDate="2018-08-02T12:51:14.237" Score="1" ViewCount="57" Body="&lt;p&gt;I'm designing a coordinate measuring machine for my FYP. I need to know what is that black zigzag flexible cover usually made of. Is that leather, rubber or some other material? And how is it fabricated?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/j3QUo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j3QUo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;It is called a ‘Ball Screw Cover’, they are available in just nylon or fiber coated plastic-&#xA;Sometimes, you just need the term for the component to get going down the right path: “Ball Screw Cover”&lt;/p&gt;&#xA;" OwnerUserId="17001" LastEditorUserId="8533" LastEditDate="2018-08-08T14:09:49.997" LastActivityDate="2018-08-08T14:09:49.997" Title="What material is it?" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="1" ClosedDate="2018-08-07T18:05:16.290" ContentLicense="CC BY-SA 4.0" />
  <row Id="23016" PostTypeId="1" AcceptedAnswerId="26035" CreationDate="2018-08-02T15:49:06.600" Score="1" ViewCount="114" Body="&lt;p&gt;I recently started working with a new team doing detailed energy modeling for all flavors of building projects -- commercial, industrial, residential, new construction, renovations, additions. As the team has been growing, we've been discussing three particular problems that I believe version control could help with.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Our challenges&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Coordinating multiple people working on a single energy model.&lt;/strong&gt; Each building model involves different components -- geometry, envelope, HVAC, lighting, etc. For time-crunched projects different people may work on different components simultaneously. To bring each person's work together at the end can be complicated and time-consuming.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Keeping track of &quot;known good versions&quot; for a particular application.&lt;/strong&gt; To streamline our work we create templates for various building components. Each template may only be ready to use for certain building types (say, &lt;code&gt;HVAC template #n&lt;/code&gt; is working for commercial buildings but not tested for residential). When starting a new model we review release notes to make sure we apply the appropriate template, until our periodic review when we test/update templates for all use-cases. Both the tracking of historic versions, and the periodic integration of updates, is complicated and time-consuming. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Reproducing results from a report sent to a client.&lt;/strong&gt; During model development we periodically prepare reports for clients. Each report is tied to a version of the model and templates which is archived on our server. This way we can re-open the old model to address any questions the client has, even as model development has continued. At times we also need to change aspects of the old model to answer specific client questions, before a model update is ready. At this point, the task of integrating two separate streams of model changes becomes... complicated and time consuming.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;All of these processes can be improved with version control -- but nobody here has ever used version control!&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if others here have been in a similar position, and implemented a version control system. What did you use, and how did it go? What best practices can you share?&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Some details about our team and our work&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All engineers use Windows 10 &lt;/li&gt;&#xA;&lt;li&gt;All of our modeling tools are Win32 applications (eQuest, Open Studio, TRNSYS), but modeling source files are stored as text-based files (not binary)&lt;/li&gt;&#xA;&lt;li&gt;We're considering Git, Mercurial, and Bazaar&lt;/li&gt;&#xA;&lt;li&gt;We do not currently have a server where we could run something like SVN, so we'd prefer a distributed system which could store files on a shared drive (such as a networked drive, SharePoint, DropBox, etc).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17004" LastEditorUserId="17004" LastEditDate="2018-08-03T21:44:29.327" LastActivityDate="2019-02-18T16:32:36.820" Title="Git, Mercurial, others -- what's the best system for an engineering team new to version control?" Tags="|modeling|tools|" AnswerCount="4" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="23017" PostTypeId="2" ParentId="23015" CreationDate="2018-08-02T16:16:04.617" Score="0" Body="&lt;p&gt;probably synthetic rubber. the most commonly-used one is ethylene propylene diene monomer, or EPDM. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-08-02T16:16:04.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23019" PostTypeId="2" ParentId="23014" CreationDate="2018-08-02T16:58:13.023" Score="1" Body="&lt;p&gt;Assuming the outside temperature is colder, then a heat sink slopped gently to drain the water droplets, installed upside down on the ceiling would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;depending on the construction of your space, this heat sink could be some kind of finned galvanized plate that works both as a roof and ceiling. There is a wide selection on the market. The function of heat sink is to lower the temperature to get below due point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If outside temperatures are not colder, then the same heat sink should be attached vertically to the walls, to avoid heat accumulated near ceiling by convection in the room. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-08-02T17:10:58.780" LastActivityDate="2018-08-02T17:10:58.780" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23021" PostTypeId="1" CreationDate="2018-08-02T19:47:35.210" Score="2" ViewCount="552" Body="&lt;p&gt;What is a Butterworth filter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have raw X, Y, Z acceleration sensor data that I need to smooth them from noise. I have read that a Butterworth filter is one of the most appropriate for the specific kind of job I am working on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that filter being applied directly on the raw data (time-amplitude domain) or I need to translate the data to the frequency-amplitude domain (using FFT)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is IIR any kind of a Butterworth filter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do I make with the result from that filter? Do I multiply it on my data in order to filter them out?&lt;/p&gt;&#xA;" OwnerUserId="16386" LastActivityDate="2019-03-17T22:17:02.787" Title="What is a Butterworth filter?" Tags="|sensors|signal-processing|electronic-filters|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23022" PostTypeId="1" CreationDate="2018-08-02T20:48:20.990" Score="4" ViewCount="297" Body="&lt;p&gt;I'm learning how to control a double integrator with $H_\infty$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my model is simply &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;\dot{r} = v \\&#xA;\dot{v} = F/m \\&#xA;r(t_0) = 0\text{ m}, $v(t_0) = 0\text{ m/s}, m = 1000\text{ kg}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so I want to be able to track a step command. I have noise on measurements of position, velocity and force, assumed with a noise having a std of 0.02&amp;nbsp;m, 0.01&amp;nbsp;m/s and 0.2&amp;nbsp;N. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to have closed-loop bandwidth equal to 0.2 Hz with a steady-state error of 0.1&amp;nbsp;m, and a sensitivity peak at $f_{p} = 1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scheme I implemented is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zXddC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zXddC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weigthing functions are the following. Since I want to track low-frequency changing signals, I imposed&#xA;$$W_{ref,r} = \frac{1}{s/\omega_{lpt} +1 }$$&#xA;with the low-pass tracking function $f_{lpt}$ equal to $2\pi f_{lpt}$, and $f_{lpt}$ equal to 0.2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The noise weighting functions are constants corresponding to the values mentioned above, while there is no feedforward contribution, so $W_{ref_F}$ is equal to 1, and so does $W_{ctrl_inn}$ (perfect inner dynamics). If I understand the theory correctly, the functions $W_{p_r}$ and $W_u$ play a similar role as the matrices $Q$ and $R$ in LQR, except that we can shape them frequency-wise, and that we are minimizing the $\infty$ norm instead of the Euclidean one. So, as recommended by Skogestad in his wonderful book I specified &#xA;$$W_{p_r} = \frac{s/M+2\pi f_{p}}{1+2\pi f_{p} A}$$&#xA;with $A = 0.1$, $f_p = 1$, and the peak for the sensitivity transfer function M equal to 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transfer function for control performance is a high-pass filter needed to penalize high-frequencies such that the controller does not waste efforts in trying to control high-frequency dynamics (in my case &gt; 10&amp;nbsp;Hz)&#xA;$$W_u = \frac{3+ 2 \pi f_{hpf}/2}{1+2 \pi f_{hpf}}$$&#xA;with the high pass frequency $f_{hpf}$ equal to 10&amp;nbsp;Hz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get this bode plot of the inverses of $W_{p_r}$ and $W_u$, so at small frequency the sensitivity tf $S$ is small, and at large frequency $KS$ is small, that is, no big control efforts.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XpWAs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XpWAs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I synthesize the Hinf with matlab I get a $\gamma$ equal to 10. I would expect a small value because we want to make the z output small for the expected exogenous inputs. Can someone tell me what I'm doing wrong? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm getting the generalized plant by using linmod on the above specified Simulink model to get A,B,C,D and I transform it into P by doing&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P = ss(A,B,C,D)&#xA;P = minreal(P)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;P.P.S. The step response also diverges.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/R4oDZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R4oDZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17008" LastEditorUserId="1832" LastEditDate="2018-08-02T21:30:11.070" LastActivityDate="2024-04-04T15:04:50.593" Title="what's wrong with this robust control scheme?" Tags="|control-engineering|control-theory|pid-control|optimal-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23023" PostTypeId="1" AcceptedAnswerId="23027" CreationDate="2018-08-02T23:24:45.010" Score="2" ViewCount="63" Body="&lt;p&gt;I was originally planning to as this on Physics Stack exchange, but found &lt;a href=&quot;https://physics.stackexchange.com/questions/243555/focus-correcting-screen/243583#243583&quot;&gt;this question&lt;/a&gt; while searching, which directed the asker to here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While my question is not a duplicate of that, it still falls in the same classification, so I believe it's on topic here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I recently had a &lt;a href=&quot;https://en.wikipedia.org/wiki/Fluorescein_angiography&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fluorescein angiography&lt;/a&gt; and the following thought occurred to me about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume there's a piece of graph paper in front of me, the job of the lens in my eye is to project a sharp image of the paper on my retina.  I have astigmatism, in both eyes as it happens, making glasses with a cylindrical correction necessary to read anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, during the angiogram, the exact process is reversed.  The camera involved projects a sharp image of my retina on a photographic sensor.  Except I don't have my glasses on, so I would expect the astigmatism would lead to an imperfect image on the sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How then is the camera able to get a sharp image?  When reviewing the results afterwards with the doctor, I was quite impressed with the detail and clarity of the images.&lt;/p&gt;&#xA;" OwnerUserId="17011" LastEditorUserId="17011" LastEditDate="2018-08-03T01:27:48.823" LastActivityDate="2018-08-03T06:07:17.810" Title="What lens is needed for a retina scan?" Tags="|optics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23024" PostTypeId="1" CreationDate="2018-08-03T01:44:21.727" Score="1" ViewCount="18" Body="&lt;p&gt;Is detailed experimental information on the distribution of energy dissipation in a stirred tank(45o Pitch blade) available? I require this variation in all directions in a standard stirred tank configuration. &lt;/p&gt;&#xA;" OwnerUserId="17013" LastActivityDate="2018-08-03T01:44:21.727" Title="Distribution of turbulent energy dissipation in a stirred tank" Tags="|chemical-engineering|turbulence|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23025" PostTypeId="1" CreationDate="2018-08-03T03:51:43.350" Score="1" ViewCount="146" Body="&lt;p&gt;The question reads&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Design a horizontal pressure vessel with hemispherical head supported by saddle support and fitted with 2 nozzles welded to the head. It's used for storing and delivering fuel oil with API gravity 70. Internal pressure is 140/150 Psi and internal temperature 190-200 deg C.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't figure out where to start.There seems no way to find out the thickness or radii. How do I approach this problem?&lt;/p&gt;&#xA;" OwnerUserId="10255" LastActivityDate="2018-08-03T17:17:24.840" Title="Designing a horizontal pressure vessel" Tags="|mechanical-engineering|design|chemical-engineering|pressure-vessel|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23026" PostTypeId="1" CreationDate="2018-08-03T04:17:58.360" Score="1" ViewCount="217" Body="&lt;p&gt;I am building a temporary structure in a windy area (this is for Burning Man).  We are essentially using 4x4s that are embedded into the ground (hard clay floor) that are spaced apart by 2x6 rafters.  The purpose of this structure is to hold some shade sails which will effectively catch a reasonable amount of wind (though we try to take them down if the wind gets too strong), and otherwise the structure is not load bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sketch of the structure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w6tnP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w6tnP.jpg&quot; alt=&quot;shade structure&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that this year we want the long rafters to reach about 27'.  Previous years were 24' long, a lumber length that is reasonably easy to find.  Furthermore, we previously only had rafters across the angular spans, there were no horizontal beams for support, and the shade structure was more than able to handle the wind load (to give an idea of what forces we are dealing with).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can't find 28' 2x6 lumber in our area (or order it in time to get here), so I am considering my options, and so far I've come up with:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Using I-Joists (most expensive)&lt;/li&gt;&#xA;&lt;li&gt;Sistering 2x8s together and figuring out a hanger solution&lt;/li&gt;&#xA;&lt;li&gt;Sistering 2x6s together and hoping for the best (least expensive).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By &quot;sistering&quot; I mean to sandwich two 2x6s or 2x8s together (in offset lengths) to reach the full 27' using wood glue and then heavy duty screws on both sides (or perhaps bolting them together in a number of places if we wanted easier storage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a feeling that #3 is enough, considering we'd also be adding the horizontal beams on the outside rafters and collar ties (or maybe a &quot;half collar tie?) on the inner rafters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on whether this would be good enough, considering what has worked for us before?&lt;/p&gt;&#xA;" OwnerUserId="17012" LastActivityDate="2018-08-03T04:17:58.360" Title="How to span a 28' gap, preferably with wood" Tags="|structural-engineering|wood|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23027" PostTypeId="2" ParentId="23023" CreationDate="2018-08-03T06:07:17.810" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I recently had a Fluorescein angiography and the following thought occurred to me about it.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I have astigmatism, in both eyes as it happens, making glasses with a cylindrical correction necessary to read anything.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The camera involved projects a sharp image of my retina on a photographic sensor. Except I don't have my glasses on, so I would expect the astigmatism would lead to an imperfect image on the sensor.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;How then is the camera able to get a sharp image? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From the American Academy of Ophthalmology: &lt;a href=&quot;https://www.aao.org/eye-health/diseases/what-do-astigmatism-measurements-mean&quot; rel=&quot;nofollow noreferrer&quot;&gt;Astigmatism&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Astigmatism is measured in diopters. &lt;strong&gt;A perfect eye with no astigmatism has 0 diopters. Most people have between 0.5 to 0.75 diopters of astigmatism&lt;/strong&gt;. People with a measurement of 1.5 or more typically need contacts or eyeglasses to correct their astigmatism in order to maintain clear vision.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most people have some astigmatism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Topcon Medical Systems &quot;&lt;a href=&quot;http://www.topconmedical.com/products/trc50dx.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;TRC-50DX Mydriatic Retinal Camera&lt;/a&gt;&quot; webpage lists their &lt;strong&gt;TRC-50DX Datasheet&lt;/strong&gt; (&lt;a href=&quot;http://www.topconmedical.com//_assets/cf/downloadFile.cfm?section=productLit&amp;amp;file=TRC-50DX%20Datasheet.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;.PDF&lt;/a&gt;), which says the device has these settings (and autofocus):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;&lt;strong&gt;Diopter Compensation Range For Patient’s Eye&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; 0 Setting:  -10 to +6 Diopter&#xA; - Setting:  -9 to -23 Diopter&#xA; + Setting:  +5 to +23 Diopter&#xA; + Setting: +22 to +41 Diopter (Ocular Anterior Photography)&quot;.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The machine has its own &lt;em&gt;glasses&lt;/em&gt; built-in, which the doctor can adjust to match the eyes of the patient (whether they own glasses or not). Glasses are not worn since there could be smudges, scratches or dirt; it would be more glass in-between the sensor and your eye, and more to adjust for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9PRT9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9PRT9.jpg&quot; alt=&quot;TRC-50DX Mydriatic Retinal Camera&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2018-08-03T06:07:17.810" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23031" PostTypeId="1" AcceptedAnswerId="23041" CreationDate="2018-08-03T10:50:48.880" Score="0" ViewCount="941" Body="&lt;p&gt;A 3&amp;nbsp;m deep basement has a two story timber frame house above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They share a side, however both have parts &quot;outside&quot; of the other, as can be seen in the sketch below. The red dash is the basement, the blue is the house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BFae7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BFae7.png&quot; alt=&quot;rough diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a basement form part of the foundations for the house in conjunction with other foundations not part of the basement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any risks or known issues with this concept?&lt;/p&gt;&#xA;" OwnerUserId="16967" LastEditorUserId="1832" LastEditDate="2018-08-03T20:37:53.277" LastActivityDate="2018-08-03T23:50:24.570" Title="irregular shape basement as part foundation for two story timber frame house" Tags="|structural-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23032" PostTypeId="2" ParentId="22043" CreationDate="2018-08-03T17:05:57.637" Score="1" Body="&lt;p&gt;The height has nothing to do with the selection of the motor. &#xA;Whichever motor type you choose (servo or step motors will allow you control the exact position while simple induction motor just cannot) - you should verify it has sufficient torque capabilities for the desired lifting velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The needed torque in your case is calculated as follows:&#xA;1. for a constant speed: the load mass * g * the distance of the weight from the motor axis center (in meters)&#xA;2. for accelerating lifting: the load mass * the linear acceleration (in m/t^2) * the distance of the weight from the motor axis center (in meters)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the results for both cases are in Nm units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, you need to convert the desired lift speed to the motor angular velocity. Basically, all you need to do is to divide to linear speed (m/sec) by pulley radius (m) to have it in rad/sec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each motor has a specific torque-speed curve. You should make sure your motor can supply the needed torque for the desired speed.&#xA;Furthermore, you should relate to the motor peak torque only when this torque is applied for a short period of time. Otherwise, the motor will heat-up and wouldn't be able to supply enough torque.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastEditorUserId="16860" LastEditDate="2018-08-03T18:28:09.573" LastActivityDate="2018-08-03T18:28:09.573" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23033" PostTypeId="2" ParentId="23025" CreationDate="2018-08-03T17:17:24.840" Score="1" Body="&lt;p&gt;I'm going to combine two criteria to find the wall thickness (the only unknown here). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first criterion is based on yielding of the wall material prior to to failure as a result of crack formation and propagation, so the inspection team can observe the plastic deformation and the pressure within the tank be released before total failure of the tank. So materials with large critical crack length are suitable, what you can do, is, rank the materials with higher &lt;strong&gt;fracture toughness&lt;/strong&gt;. how ? See further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second criterion  is &lt;strong&gt;Leak before break&lt;/strong&gt;. The cracks may grow and penetrate throughout the wall thickness, but, they never propagate rapidly, again inspector sees the leaking of liquid, and the catastrophic failure can be prevented. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, i assume the condition of plain stress, and i assume the pressure vessel is in one whole piece, and i neglect what ever shape, the nozzles introduce on the tank. So a simple cylindrical vessel with spherical heads and no stress risers. The volume of the tank and the pressure are known.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stress cause by pressure inside are: $$\begin{bmatrix}&#xA;    \sigma_{rrp}       &amp;amp; 0 &amp;amp; 0 &amp;amp;\\&#xA;    0       &amp;amp; \sigma_{\theta\theta p} &amp;amp; 0 &amp;amp; \\&#xA;    0       &amp;amp; 0 &amp;amp; 0 &amp;amp; &#xA;\end{bmatrix}&#xA;= \frac{pr}{2t}&#xA;\begin{bmatrix}&#xA;    1      &amp;amp; 0 &amp;amp; 0 &amp;amp;\\&#xA;    0       &amp;amp; 2 &amp;amp; 0 &amp;amp; \\&#xA;    0       &amp;amp; 0 &amp;amp; 0 &amp;amp; &#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the $\sigma_{rr p}$ and  $\sigma_{\theta\theta p}$ by solving the ODE (you can find it by writing the equilibrium equations for free body diagram ) or simply using the Lamé equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thermal effect, is also important, you didn't specify how many degrees are outside, so i can't jump into details now, but it will introduces new stresses  $\sigma_{rrt}$ and  $\sigma_{\theta\theta t}$. The subscript t stands for thermal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we are dealing with a complex stress condition, i choose to work with &lt;a href=&quot;https://en.wikipedia.org/wiki/Von_Mises_yield_criterion&quot; rel=&quot;nofollow noreferrer&quot;&gt;von Mises criterium&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I call the result of &lt;em&gt;von Misses yield criterium&lt;/em&gt; $\sigma_v$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's use our criteria to find the wall thickness.&#xA;The first design criterion (Yielding before failure):&#xA;I borrow this simple equation from fracture mechanics: $$K_{Ic} = Y\sigma \sqrt{\pi a}  \qquad (1)$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$K_{Ic}$ is called &lt;strong&gt;plain strain fracture toughness&lt;/strong&gt;. $Y$ is a dimensionless parameter $Y \cong 1.1$, $a$ is the half of the crack length, and $\pi$ is the mathematical constant. &#xA;Before we use the expression above, i want to make a small modification, i replace the $\sigma$ with $\sigma_y$ (yield strength) and divid it by $N$ safety factor. &#xA;Now solving for $a$:$$a_c = \frac{N^2}{\pi Y^2} (\frac{K_{Ic}}{\sigma_y})^2$$. Now we find the critical crack length, we can find the material using the ratio $\frac{K_{Ic}}{\sigma{y}}$. (it's a material properties). How to find $a_c$? That's not something you can exactly specify the value at this moment, but we should estimate it, it's practically the wall thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After choosing a material, we use the $\sigma_v$ ( it contains the wall thickness), and replace $a$ in equation (1)  with $t$ the wall thickness and $\sigma$ with $\sigma_v$. and solve it for the wall thickness. The wall thickness, that we have so far found should be equal or slightly bigger than that we already estimated. If not got back there, improve it and repeat the last step until you get it right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the naivest, and the most simplified version of design the pressure vessel, based on the information provided in the question.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-08-03T17:17:24.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23034" PostTypeId="2" ParentId="22746" CreationDate="2018-08-03T17:18:39.463" Score="0" Body="&lt;p&gt;Consider that you categorize your data by parametric variations. Suppose you have three $MVs$: $p, T, z$. You might separate the data as such ...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;runs where only $p$, $T$, or $z$ are varied (three different sets of data streams)&lt;/li&gt;&#xA;&lt;li&gt;runs where $p, T$ or $p, z$ or $T, z$ are varied (three different sets of data streams)&lt;/li&gt;&#xA;&lt;li&gt;runs where all three values are varied&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Analyze the first three sets. Does anything happen when only $p$ changes? Does anything happen when only $T$ changes? Does anything happen when only $z$ changes? Make notes. Check for consistency across each $MV$. The results are the influences of the fundamental or first order control factors. Derive whatever metrics you want from this analysis (e.g. &quot;delay time&quot; or other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continue the analysis with the other sets. The results are the cross-correlation influences (e.g. how the combined set $p,T$ effects the output).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analysis may be done manually by visual inspection of the data sets you have. Alternatively, you may find statistical programs such as R to aid in culling through the noise.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-08-03T17:18:39.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23035" PostTypeId="2" ParentId="23022" CreationDate="2018-08-03T18:32:07.810" Score="0" Body="&lt;p&gt;Where's the rest of your block diagram? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like you have a polarity problem - You give a step of +1 and your &lt;code&gt;pos_meas_Hinf&lt;/code&gt; goes down instead of up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing you've shown in your diagram looks like a controller. You have a position reference that doesn't appear to go &lt;em&gt;anywhere&lt;/em&gt;, except to a position error block, and you have a force reference that just feeds directly into an actuator model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where is your controller in that diagram?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically a feedback controller uses &lt;code&gt;reference - feedback&lt;/code&gt; as the error signal to drive the controller, but it looks like you're adding your &lt;code&gt;control force&lt;/code&gt; with the &lt;code&gt;ref_F&lt;/code&gt;. But again, I don't see where you're actually doing any control. You make a position error signal... do you actually use it? &lt;em&gt;How&lt;/em&gt; are you using it? How does that signal get back to... &lt;code&gt;ref_F&lt;/code&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is &lt;code&gt;control force&lt;/code&gt;, if you're using &lt;code&gt;ref_F&lt;/code&gt; to control? Why would a control force be an input to an actuator? Wouldn't the force be the &lt;em&gt;output&lt;/em&gt; of the actuator? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In re-reading the question, it's not even clear to me &lt;em&gt;what&lt;/em&gt; you're stepping. You have &lt;em&gt;two&lt;/em&gt; things labeled as references in your block diagram - &lt;code&gt;ref_Pos&lt;/code&gt; and &lt;code&gt;ref_F&lt;/code&gt;. Your step response just has &lt;code&gt;command&lt;/code&gt;, which is a pretty meaningless label. Are you stepping position reference or force reference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the state space matrices you're using?&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2018-08-03T18:32:07.810" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23037" PostTypeId="2" ParentId="23021" CreationDate="2018-08-03T18:50:32.567" Score="2" Body="&lt;p&gt;This.... this is the content of most of a semester on digital signal processing. You said, &quot;I found [a book on signal processing], but it is a book of 700 pages with plenty of math.&quot; Well, yes. Welcome to digital signal processing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then said, &quot;I was just expecting some kind of answer for dummies.&quot; Okay, &lt;a href=&quot;https://en.wikipedia.org/wiki/Butterworth_filter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia says&lt;/a&gt; it's a type of filter &quot;designed to have a frequency response as flat as possible in the passband. It is also referred to as a maximally flat magnitude filter.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Filter design is typically done in the frequency/&lt;a href=&quot;https://en.wikipedia.org/wiki/Laplace_transform&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laplace domain&lt;/a&gt;. If you want something you can use, then you need to convert back to the time domain. That would leave you with a continuous time filter; if you want a discrete time filter (you do), then you would use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Bilinear_transform&quot; rel=&quot;nofollow noreferrer&quot;&gt;discrete time approximation&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2018-08-03T18:50:32.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23038" PostTypeId="2" ParentId="23016" CreationDate="2018-08-03T18:51:31.457" Score="0" Body="&lt;p&gt;It does not matter. Any version control is better than none. What you choose depends on your needs. Git has a more elborate system that allows editing of history afterwards. Personally i find that mercurial is more straightforward for people who haven't done much version control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this really depends on what system you are on and whether you expect to run the system from commandline or a GUI frontend of somekind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway even on a distributed system you still need a central server for synching. Even if it is just a shared drive.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2018-08-03T18:51:31.457" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23039" PostTypeId="2" ParentId="23016" CreationDate="2018-08-03T20:05:33.950" Score="1" Body="&lt;p&gt;Source control isn't going to fix your problems. Let's review:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;We don't have a way to allow multiple people to work on different components of a model simultaneously.&lt;/strong&gt; Why do you have multiple people working on the same thing at the same time? Either (1) they're duplicating each others work, or (2) you have your stuff improperly written/stored as some monolithic &lt;em&gt;thing&lt;/em&gt; instead of as sub-components, sub-assemblies, etc. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;For some projects we want to use older versions with known issues, rather than newer versions with unknown issues. This often applies when a change has been made to fix a problem in one building type but hasn't been tested for others.&lt;/strong&gt; Why is your team not testing the work that it's producing? &lt;a href=&quot;https://en.wikipedia.org/wiki/Unit_testing#Advantages&quot; rel=&quot;nofollow noreferrer&quot;&gt;Unit testing&lt;/a&gt; is a method where you have a list of inputs that you know could be provided to your software/tools, and then you also have a list of what the outputs that you expect your [stuff] to produce, given those inputs. As your team finds things/inputs that breaks your [stuff], you add that to the list of tests. There is &lt;em&gt;no reason&lt;/em&gt; why your team should release work that would contain previously fixed bugs.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;During model development we periodically prepare reports for clients. [...the toolchain configuration can be lost, which...]  can result in hours spent figuring out how to make the model produce the previous results, which is about the biggest waste of time imaginable.&lt;/strong&gt; Why are you generating reports without including information about &lt;em&gt;how&lt;/em&gt; the report was generated?? Your report should have a section or appendix on &quot;methods,&quot; which would generally detail &lt;em&gt;how&lt;/em&gt; you're arriving at the results. &lt;strong&gt;Any time you use a reference, you cite it.&lt;/strong&gt; If you are using a program/tool you or your company wrote, &lt;strong&gt;you cite it.&lt;/strong&gt; &lt;a href=&quot;http://blog.apastyle.org/apastyle/2015/01/how-to-cite-software-in-apa-style.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Software citations should include the version number&lt;/a&gt;. If you're writing a report for a customer (non-profit or otherwise), and you're presenting numbers/facts with no explanation or reproducible source, they might as well come from your butt. &lt;em&gt;Nobody has a means to verify anything you say.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So, to summarize, it sounds like your team is working on some monolithic project, where people are overwriting each others work, and then you're using this to generate reports with no citations to the software version you're using to generate that report. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Version control doesn't fix any of this.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Git doesn't structure your project. Git doesn't assign work flows. Git doesn't (automatically) know when you are at a stable release. Git archives your work. That's it. Any version control just archives your work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't have time to setup a central repository? Then how do you expect each individual developer to share work? Don't have anyone responsible for maintaining the archive? Then whose job is it to merge branches?&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2018-08-03T20:05:33.950" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23040" PostTypeId="2" ParentId="23031" CreationDate="2018-08-03T21:07:54.317" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Are there any risks or known issues with this concept?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is excess overhang on the left side, but that leaves plenty of room to bore three holes and fill them with concrete; thus fully supporting that side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o79fw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o79fw.png&quot; alt=&quot;Design 1 - Three Pillars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another technique is to invert your plans, centering the floors over the basement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hOPwx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hOPwx.png&quot; alt=&quot;Design 2 - Inverted Plans&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;The drawing would be better if it were not done on a mobile phone APP.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2018-08-03T21:07:54.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23041" PostTypeId="2" ParentId="23031" CreationDate="2018-08-03T23:50:24.570" Score="0" Body="&lt;p&gt;There are some problems in your sketch that can be fixed. Other than that there is no fundamental issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A house or structure can be fully or partially supported by the walls of a basement, provided the bearing walls of the house are supported by bearing walls in the basement and the parts of the house that are outside of the perimeter of the basement have their own foundation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your sketch you have not shown the houses bearing walls, but you have to make sure all bearing walls especially the house's exterior walls are directly supported by the walls in the basement. Of course you can have interruptions in those walls such as doors or partial open wall but they have to be engineered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most of the US. these are called vertical discontinuities and there are special &quot;penalty&quot; load factors that apply to the structure, sometimes as large as 300% over load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is not practical to have bearing walls under all the house walls in the basement You can have a solid structural concrete slab that runs under the entire house and also covers the roof of basement where it projects out of the foot print of the house, designed to carry the load of the walls. This is actually structurally the preferred method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other Issue of concern is the detailed design of decking, roof of basement, water proofing, slope of it and practical flashing, specially at the joints of the house with the roof of basement.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-08-03T23:50:24.570" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23043" PostTypeId="1" CreationDate="2018-08-04T04:05:57.840" Score="4" ViewCount="5513" Body="&lt;p&gt;Looking for a durable material you can walk on that will attract pet hair from shoes without sticky or adhesive qualities.  Preferably won’t need active electric current but can consider that.&lt;/p&gt;&#xA;" OwnerUserId="17022" LastEditorUserId="6264" LastEditDate="2020-08-03T09:23:11.367" LastActivityDate="2020-08-03T09:23:11.367" Title="Need electrostatic material to attract pet hair" Tags="|adhesive|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23044" PostTypeId="1" AcceptedAnswerId="23057" CreationDate="2018-08-04T11:30:02.343" Score="2" ViewCount="144" Body="&lt;p&gt;I am working on an assignment and I have a doubt in a step in my problem. Could someone please help me understand it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question and its solution is:&lt;a href=&quot;https://i.stack.imgur.com/91xFV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/91xFV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&#xA;My doubt is in the part which is underlined. How did the 2nd term become zero? Also. how did they calculate external moment vector. If someone can explain these steps it would be very helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: If this is not the correct forum to ask such question, I apologize.&lt;/p&gt;&#xA;" OwnerUserId="17025" LastEditorUserId="17025" LastEditDate="2018-08-04T12:52:34.743" LastActivityDate="2018-08-05T15:35:55.330" Title="Rigid circular disc equation of motion" Tags="|mechanical-engineering|vibration|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23045" PostTypeId="2" ParentId="23016" CreationDate="2018-08-04T12:26:39.507" Score="1" Body="&lt;p&gt;Bearing in mind all of the points that &lt;a href=&quot;https://engineering.stackexchange.com/a/23039/16&quot;&gt;Chuck's answer&lt;/a&gt; brings up, it still sounds like you have more people problems than technical problems. No piece of technology can resolve people issues, but based upon the comments it sounds like you need something to facilitate the process improvement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The modeling files you describe don't sound like they'll merge gracefully, so I believe you'll need to use a version control system that allows &lt;a href=&quot;https://en.wikipedia.org/wiki/Comparison_of_version_control_software&quot; rel=&quot;nofollow noreferrer&quot;&gt;locking instead of just merging&lt;/a&gt;.  If Ashmi needs to wait until Jane is done working on a component before Ashmi can make her changes, then locking is the way to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also doesn't sound like you need a distributed VCS.  Everyone needs a client regardless of what you pick, but you don't need the ability (or complexity) of running your own local VCS.  DVCS's are merge only for concurrency and are more useful when running detached from the main repository.  Branching and merging are (potentially) more sophisticated with a DVCS, but again, I don't foresee you needing that capability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend looking into SVN.  Yes, you'll need to setup a &quot;server&quot; to provide a central repository.  You need a central repository regardless of what VCS you choose, and SVN is comparatively lightweight to setup.  And &quot;server&quot; is in scary quotes because in reality all you need is a spare desktop to act as the server.  VisualSVN may be worth looking into if the Apache SVN route appears too daunting.  TortoiseSVN is a pretty reasonable client to consider as well, but there are several others to consider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage of SVN is that it's a lot easier to verify that &quot;really-old-project-with-known-issue&quot; has remained in that state by simply examining the file system hosting that branch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if for whatever reason that SVN is an absolute non-starter, then take a look into a private GitHub account for your main repository and git for your local clients.  You'll end up with a lot more headaches this way, but it is an option.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2018-08-04T12:26:39.507" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23046" PostTypeId="1" CreationDate="2018-08-04T13:50:56.313" Score="1" ViewCount="543" Body="&lt;p&gt;How can I scale a 3D object only on 2 dimensions with Autocad?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, I would like to scale a complex object whose maximum dimensions are 50x30x10 (on axes &lt;code&gt;x&lt;/code&gt;, &lt;code&gt;y&lt;/code&gt; and &lt;code&gt;z&lt;/code&gt;, respectively) obtaining an object whose maximum dimensions are 75x45x10, i.e. I would like to scale it up to 150% on &lt;code&gt;x&lt;/code&gt; and &lt;code&gt;y&lt;/code&gt; axes only, leaving unchanged the &lt;code&gt;z&lt;/code&gt; dimension.&lt;/p&gt;&#xA;" OwnerUserId="7271" LastActivityDate="2018-08-04T17:43:44.583" Title="How to scale a 3D object on 2 dimensions only?" Tags="|autocad|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23047" PostTypeId="2" ParentId="23046" CreationDate="2018-08-04T17:43:44.583" Score="1" Body="&lt;p&gt;This method, directly pasted from &lt;a href=&quot;https://knowledge.autodesk.com/support/autocad/learn-explore/caas/CloudHelp/cloudhelp/2016/ENU/AutoCAD-Core/files/GUID-E96829D6-A58A-48B6-8B65-A7C99C879262-htm.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Autodesk Knowledge Network&lt;/a&gt; would appear to be the solution you seek. Using a plane for the two dimensions you wish to scale should exclude the third from the changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use this method to resize the objects only along a plane that you specify.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Click Home tab Modify3D Scale. &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Select the objects and subobjects you want to scale using the following methods:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Press and hold Ctrl to select subobjects (faces, edges, and vertices).&#xA;Release Ctrl to select entire objects.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When you have selected all objects, press Enter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Scale gizmo is displayed at the center of the selected object or objects.&#xA;Do one of the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;To scale along a plane: Hover the cursor over one of the bars found between each of the axes of the gizmo until it turns yellow. Click the yellow bar.&#xA;To scale uniformly: Hover the cursor over the triangular area closest to the center point of the gizmo until it turns yellow. Click the yellow area.&#xA;To scale along an axis: Hover the cursor over one of the axes of the gizmo until it turns yellow. Click the yellow axis.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fu8Y6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fu8Y6.png&quot; alt=&quot;axis display&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To resize the selection, drag and release, or enter a scale factor while holding the mouse button. &lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-08-04T17:43:44.583" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23048" PostTypeId="2" ParentId="23014" CreationDate="2018-08-04T19:42:52.530" Score="2" Body="&lt;p&gt;Let's call the rate of condensation of a gas on to a surface as $r_{cond}$ in units of kg/m$^2$ s (i.e. it is a flux). The value of $r_{cond}$ in the case of heterogeneous nucleation depends on the surface temperature and the contact angle of the liquid on the surface. See these links:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://pubs.acs.org/doi/pdf/10.1021/acs.jpcc.5b07709&quot; rel=&quot;nofollow noreferrer&quot;&gt;ACS Article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=2&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=2ahUKEwjwj7D6ktTcAhWRxFkKHa87BkMQFjABegQICBAC&amp;amp;url=http%3A%2F%2Fwww.eng.utah.edu%2F~lzang%2Fimages%2Flecture-12.pdf&amp;amp;usg=AOvVaw2ALPI_VTEmKTHvC9b6GV0B&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lecture Notes&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nature.com/articles/s41598-017-16692-9&quot; rel=&quot;nofollow noreferrer&quot;&gt;Scientific Reports Article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The macroscopic shape of the object has no bearing on this result.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-08-04T19:42:52.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23050" PostTypeId="2" ParentId="23043" CreationDate="2018-08-04T20:56:51.823" Score="1" Body="&lt;p&gt;Installing enough electrostatic material on your floor will be significantly more work than having a handheld device that you can use to brush off your shoes.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.prnewswire.com/news-releases/innovative-new-chomchom-roller-promises-easier-pet-hair-removal-than-ever-before-without-the-need-for-adhesives-or-awkward-vacuum-attachments-190185361.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;This device&lt;/a&gt; appears to use electrostatics to hasten the removal of pet hair:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The ChomChom Roller is handheld and requires no power source. While it looks similar to the traditional lint roller, the similarities stop there. Moving the roller back and forth over any fabric surface creates an electrostatic charge, attracting nearby hair and dust. Hair and dust are then deposited in a small compartment behind the roller. Users simply empty the compartment in the trash and the ChomChom Roller is ready for more.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can buy this device (or significantly cheaper generic versions) on Amazon. Interestingly, the branded device's &lt;a href=&quot;https://www.pethairgone.com/how-it-works/&quot; rel=&quot;nofollow noreferrer&quot;&gt;website&lt;/a&gt; mentions that it is &amp;quot;internationally patented,&amp;quot; but I can find no evidence to support this claim. The strongest legal protection I can find relating to their company is their &lt;a href=&quot;https://trademarks.justia.com/873/95/chomchom-87395211.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;trademark registration&lt;/a&gt;, so I filed a support request on their &amp;quot;Contact Us&amp;quot; page:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/txzOS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/txzOS.png&quot; alt=&quot;ChomChom patent request&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;They mention a 24-hour turnaround policy on these customer support inquiries, so I'll edit this answer within the next two days if they respond with supporting documentation.&lt;/p&gt;&#xA;&lt;h3&gt;Documented patent applications for adhesive-free hair removers:&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://patents.google.com/patent/US20150342426A1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;US20150342426A1 &lt;em&gt;(Hair remover)&lt;/em&gt;&lt;/a&gt; describes a suction-driven cleaner but contains a brush head unit reminiscent of the branded device.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;It realizes the extraction of pet hair and bubbles through a motion of translation on fabric, which can be linear or circular as regards the fabric.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Classifications... Suction cleaners adapted to the cleaning of brushes&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://patents.google.com/patent/US20150257361A1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;US20150257361A1 &lt;em&gt;(Static electricity pet hair remover)&lt;/em&gt;&lt;/a&gt; is mechanistically similar to the branded device but is constructed of large freestanding brush-covered posts for grooming.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The brushes are made of a material having a tendency to attract large numbers of electrons and hold a negative static charge. As a pet rubs itself against the posts, the pet's hair becomes positively charged. The hair is attracted to the negatively charged brush...&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Classifications... Rubbing-posts&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://patents.google.com/patent/US5957090&quot; rel=&quot;nofollow noreferrer&quot;&gt;US5957090A &lt;em&gt;(Electrostatic pet hair remover)&lt;/em&gt;&lt;/a&gt; is a hilarious electrostatic generator that you're supposed to feed your pets/hair-covered household items through. I won't even include text from this patent in favor of this incredible figure:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a78dG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a78dG.png&quot; alt=&quot;Electrostatic cat&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16783" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-08-04T21:22:29.947" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="23051" PostTypeId="2" ParentId="23021" CreationDate="2018-08-04T23:10:36.530" Score="2" Body="&lt;p&gt;A Butterworth filter is just a type of filter (there are lots of types) with a particular shape to its response. In your situation, a Butterworth filter would be used as a low-pass filter (LPF) to cut off higher-frequencies that contain noise. The filter can be implemented either on time-domain samples or on the frequency domain (after FFT), but if you want to use the data quickly then you probably need to process it in the time domain. You correctly refer to IIR (infinite impulse response - another option is FIR or Finite Impulse Response). There are filters that are constructed as a series of 'taps' (or 'weights') that are multiplied by the present sample and a series of previous samples, then summed and averaged. You will need to learn a bit about FIR/IIR implementations of a low-pass filter to calculate tap values to give you a cut-off above your signal frequency (but low enough to cut out the noise). This might seem complicated so here is a simple example of a 'tap' based filter - add the present sample to the previous 4 and divide by 5. The output of that filter is a low-pass, because it is simply a moving average - all the tap values are simply '1'. Your next step is.... check out how to implement a simply low-pass FIR filter, and try to implement it! Good luck :)&lt;/p&gt;&#xA;" OwnerDisplayName="user17030" LastActivityDate="2018-08-04T23:10:36.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23052" PostTypeId="1" CreationDate="2018-08-04T23:27:24.870" Score="2" ViewCount="72" Body="&lt;p&gt;I have a 3DOF platform that I'm trying to make more stable. It's difficult to explain in words what the issue is so I've added a GIF below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sPEHP.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sPEHP.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas of how I can get rid of this wobble? The DOF arm tends to cave inwards. I know it's being caused by the very small gap between the screw and the channel but is there any way to prevent this? Since the servo's I'm using don't have a dual shaft going through and through I can't modify the servo arm to be a inverted-U attached on both ends of the servo.&lt;/p&gt;&#xA;" OwnerUserId="17031" LastActivityDate="2021-01-07T06:13:41.587" Title="How to prevent movement in this 3 DOF setup?" Tags="|mechanical-engineering|design|mechanisms|fasteners|servo|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23053" PostTypeId="1" AcceptedAnswerId="23074" CreationDate="2018-08-05T03:25:15.547" Score="0" ViewCount="517" Body="&lt;p&gt;For a small 30 square meter home (8 x 3.75) is it feasible to have plinth beam 2 feet below ground? Technically, is it better than or worse off than standard procedure of plinth beam at ground level?&#xA;Also, please consider the fact that it is an earthquake prone zone.&lt;/p&gt;&#xA;" OwnerUserId="4569" LastActivityDate="2018-08-06T18:27:17.103" Title="Can we have plinth beam below ground level for small homes?" Tags="|civil-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23054" PostTypeId="2" ParentId="23043" CreationDate="2018-08-05T07:06:25.907" Score="0" Body="&lt;p&gt;Actually if you don't mind using active electronics, it is trivial to make a floor that can attract all the dust and hairs you want. Just coat your floor with an electrically conductive paint like a nickel filled acrylic. Then insulate this conductive layer with a good insulative coating like an epoxy paint that is two part, resin and hardener. Make sure you double coat the edges of the conductive layer because that is where the electrostatic charge will be greatest. Make a contact point on the conductive layer and make an attachment to a high voltage charging circuit like the ones designed to charge up electrostatic loudspeaker diaphragms, just a low power high voltage power supply that feeds the conductive layer through a high value resistor, somewhere in the range of 5 megohms or so. This is the way to safely generate a strong electric field without the chance of getting a bad shock. You can apply any high voltage you want to the conductive layer, you are only limited by the dielectric strength of your epoxy insulating layer. &lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2018-08-05T07:06:25.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23056" PostTypeId="1" AcceptedAnswerId="23060" CreationDate="2018-08-05T09:37:17.800" Score="2" ViewCount="3693" Body="&lt;p&gt;Patrols have higher self-ignition temperature than diesel, but still, petrol engines have lower compression ratios than diesel engines. As the self-ignition temperature of petrol is shouldn't we be able to compress it more. I am not saying compress petrol to the point of auto-ignition which can cause knocking, but just have a higher compression ratio than diesel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because when the combustion of petrol-air mixture starts the moving flame front starts to compress the remaining air-fuel mixture which could reach a point of self-ignition when using higher CR?&lt;/p&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2018-08-05T21:22:53.583" Title="Why do petrol engines have lower compression ratio than diesel engine?" Tags="|automotive-engineering|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23057" PostTypeId="2" ParentId="23044" CreationDate="2018-08-05T15:35:55.330" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;How did the second term become zero?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Recall that the cross product of vectors $\vec{a}$ and $\vec{b}$ is a vector representing the area of the parallelogram spanned by $\vec{a}$ and $\vec{b}$ in the direction perpendicular on the parallelogram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\begin{align}&#xA;\vec{\omega}(t) \times \mathbf{J}\vec\omega(t) &amp;amp;= &#xA;\begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix} &#xA;\times&#xA;\begin{pmatrix} J_x &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; J_y &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; J_z \end{pmatrix}&#xA;\begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix}  \\&#xA;&amp;amp;=&#xA;\underbrace{\begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix}}_{\vec{a}}&#xA;\times&#xA;\underbrace{\begin{pmatrix} 0 \\ J_y\omega_y \\ 0 \end{pmatrix}}_{\vec{b}}&#xA;\\&#xA;&amp;amp;= \vec{0}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that within this case the vectors $\vec{a}$ and $\vec{b}$ are parallel, hence the area of the parallelogram they span is $0$ and therefore the cross product is also zero.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;How did they calculate the external moment vector?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is only one external force present in this system, the bearing force&#xA;$$\vec{F} = \begin{pmatrix} A_x \\ 0 \\ A_z \end{pmatrix}.$$&#xA;The moment this forces causes on the center of gravity is&#xA;$$\begin{align}&#xA;\vec{M}_C&#xA;&amp;amp;= \vec{r}_C \times \vec{F}\\&#xA;&amp;amp;= \begin{pmatrix} r_x \\ r_y \\ r_z \end{pmatrix} \times&#xA;\begin{pmatrix} A_x \\ 0 \\ A_z \end{pmatrix}\\&#xA;&amp;amp;=&#xA;\begin{pmatrix} A_z r_y -  0 r_z \\ A_x r_z - A_z r_x \\ 0 r_x - A_x r_y \end{pmatrix}&#xA;\end{align}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that $r_y = 0$ result in&#xA;$$ \vec{M}_C = &#xA;\begin{pmatrix} 0 \\ A_x r_z - A_z r_x \\ 0 \end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the rotation $\varphi_y$ is clockwise and not counter clockwise as it would have been by using a right handed coordinate system.&#xA;Therefore the external moment should be defined in the opposite direction&#xA;$$ \vec{\mathbf{I}} = -\vec{M}_C =&#xA;\begin{pmatrix} 0 \\ -A_x r_z + A_z r_x \\ 0 \end{pmatrix}.&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2018-08-05T15:35:55.330" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23059" PostTypeId="2" ParentId="23056" CreationDate="2018-08-05T19:53:46.343" Score="1" Body="&lt;p&gt;Gas engines are spark ignited. &#xA;Diesel engines have no spark,  but achieve ignition through compression alone. &#xA;It's really that simple. &lt;/p&gt;&#xA;" OwnerUserId="14245" LastActivityDate="2018-08-05T19:53:46.343" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23060" PostTypeId="2" ParentId="23056" CreationDate="2018-08-05T21:22:53.583" Score="4" Body="&lt;p&gt;It's not because of the fuel, but because of the process.&#xA;The diesel process differs fundamentally from the otto(petrol/gasoline) cycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the otto cycle, where fuel is present in the cylinder while compressing, the compression is limited by the auto ignition temperature of the fuel, whatever fuel is used. The compression warms up the mixture, ideally just before auto-ignition. The spark plug adds the needed flame source to start the combustion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diesel cycle, fuel is added only when compression has already taken place, the temperature in the combustion chamber is way higher than the auto-ignition point of the fuel, which is why the fuel combusts as soon as it's injected into the cylinder. This removes the limit set by the auto-ignition temperature when choosing a compression ratio. Thus, the ratio can be higher, up to where materials begin to form a problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compression is the very reason diesels are more efficient; the compression ratio can be higher, and they also always run at 'wide open throttle' giving high compression.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastActivityDate="2018-08-05T21:22:53.583" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23061" PostTypeId="1" CreationDate="2018-08-05T23:07:25.190" Score="0" ViewCount="330" Body="&lt;p&gt;i have a 300kg motorcycle with a 1630mm wheelbase. I have a lazy susan 1000kg bearing, which will be centrally mounted on the steel plate. what is the minimum thickness steel plate at 350mm wide i need in order to support the motorcycle with both wheels off the ground.&#xA;I am only 5’ tall and need to keep the turntable as low as possible. I initially considered using wood but the thickness required would mean my short legs wouldn’t touch the ground! &#xA;Thanks for the framework suggestions. I can maybe now combine steel and wood into the design to reduce height and weight.&#xA;Thanks to all you very knowledgable engineers who have given me quite a bit to think about! I will post a photo when....or if, I manage to make a workable turntable. &lt;/p&gt;&#xA;" OwnerUserId="17041" LastEditorUserId="17041" LastEditDate="2018-08-06T19:39:27.433" LastActivityDate="2020-08-26T06:03:45.293" Title="what thickness steel sheet for a motorcycle turntable" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23062" PostTypeId="1" AcceptedAnswerId="23069" CreationDate="2018-08-06T02:44:07.903" Score="0" ViewCount="128" Body="&lt;p&gt;I am attempting to design a &quot;flying saucer.&quot; The saucer is an ellipsoidal-ish body with a RC drone inside. I plan to allow for air access to the rotors as shown in the rudimentary diagram below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0481D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0481D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am pretty sure that the drag on the air flowing through the intake and through the body of the flying saucer will reduce the thrust capacity of the drone. Can anyone confirm this? Are there any other considerations I should keep in mind (maybe there's a reason helicopters don't have a big tube surrounding the lift rotor)? I want the flying saucer to behave like a slow, mellow RC drone (I don't really want high speeds or elevations).&lt;/p&gt;&#xA;" OwnerUserId="1174" LastActivityDate="2018-08-06T18:45:10.333" Title="RC drone rotor air access" Tags="|aerodynamics|propulsion|thrust|drag|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23063" PostTypeId="1" AcceptedAnswerId="23064" CreationDate="2018-08-06T03:35:19.333" Score="5" ViewCount="142" Body="&lt;p&gt;I'm designing a ducted fan, and I'm deciding on the airfoil profile. NACA 16-series profiles are well-suited to M &gt; .5 or .6, but they're not particularly efficient at lower Mach numbers. On the other hand, older profiles, like Clark YM, work quite well at lower speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why do propeller designers generally only use one profile (with scaled chord and thickness) throughout the propeller? Is there some reason not to use, for example, a Clark YM for 2/3 of the blade and a NACA 16-709 for the outer 1/3?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this being a ducted fan, laminar profiles will not work, so they aren't under consideration.&lt;/p&gt;&#xA;" OwnerUserId="17040" LastActivityDate="2018-08-06T06:12:03.970" Title="Why don't propeller designers use multiple profiles?" Tags="|aerospace-engineering|aerodynamics|propulsion|airfoils|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23064" PostTypeId="2" ParentId="23063" CreationDate="2018-08-06T06:12:03.970" Score="6" Body="&lt;p&gt;There are propellers with different aerofoil sections. &#xA;For example, the propeller attached to the Rolls Royce engine kept in my graduate college had symmetric aerofoil sections roughly about 25 % and smoothly varying to bottom flat airfoil sections at the rest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aerofoil profile selections are based on the performance requirement, structural rigidity, and modes of operation. &#xA;You can go for any aerofoil sections with high ($C_l / C_d$) but, in the end,  it should be structurally rigid to produce high thrust at high RPM. When you have variable pitch propellers you don't need different aerofoil sections since the required performance achieved by adequately rotating the blade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my point of view having the same aerofoil sections will be advantageous in the below-given ways. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Matching the aerodynamic centres of the same aerofoils sections would be easier &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;for example, NACA 0006 have the $a.c$ at $\sim c/4$ and NACA 2412 have at 0.3$c$, then locations of $a.c$'s will not be smooth along the span of the propeller. This would be critical in structural stress distribution on the propeller. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Downwash of the propeller sections will have smooth variations. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For similar sections, the downwash induced velocity component will be predictable and the twist of the blades could be compensated easily while designing. And the twist will be smooth for similar profiles from the structural point of view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-08-06T06:12:03.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23066" PostTypeId="2" ParentId="23061" CreationDate="2018-08-06T09:34:24.850" Score="1" Body="&lt;p&gt;A steel sheet that diameter which is reasonably stiff would be pretty thick. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most efficient solution would be  to constructs a frame to support in to steel angle or rectangular tube with a steel or plywood plate on top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JpLgC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JpLgC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would probably also be a good idea to put castors and screw jacks around the circumference to give it a bit more stability, reduce the bending loads on the bearings and give you a way to lock it in place. Having support at the edge as well as the middle will also make the whole thing a lot stiffer&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-08-06T09:34:24.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23067" PostTypeId="1" CreationDate="2018-08-06T09:50:09.413" Score="2" ViewCount="55" Body="&lt;p&gt;In Printed circuit board/Plywood lamination Hydraulic Press machines (&lt;a href=&quot;https://www.youtube.com/watch?v=bz8bu_vZM6c&quot; rel=&quot;nofollow noreferrer&quot;&gt;seen here&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ramming action comes from bottom in all the machines I saw so far. For such action from bottom, excavation has to be done. Is such excavation mandatory in ramming from bottom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will there be any difference if we do such ramming from top ?&lt;/p&gt;&#xA;" OwnerUserId="17042" LastEditorUserId="1832" LastEditDate="2018-08-06T13:44:56.343" LastActivityDate="2018-08-06T13:44:56.343" Title="Difference between direction of ramming in hydraulic press?" Tags="|structural-engineering|civil-engineering|hydraulics|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="23068" PostTypeId="1" CreationDate="2018-08-06T14:28:48.863" Score="1" ViewCount="134" Body="&lt;p&gt;I'am trying to implement this open access research article &lt;a href=&quot;https://www.hindawi.com/journals/mpe/2016/8573235/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Attitude Optimal Backstepping Controller Based Quaternion for a UAV&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not find the same result in this paper Can anyone tell me where my error is in this code&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Case 1&lt;/strong&gt;. Let the initial condition of the quaternion error be $\mathrm{Q}_{e0}= (0.8224, 0.2226, 0.4397, 0.3604)^{T}$, which corresponds to an angle $\phi(0) = 34.67^{\circ}$; the matrices $P_{1}$ and $P_{2}$ are computed through Riccati equation and $\gamma_{2}$ is chosen to achieve good performances. For that case $q_{0}(0) &amp;gt;0$ and will be stabilized to 1. Figures 1, 2, and 3 show the attitude responses and the control inputs, which gives a convergence in short time. These results are nearby the normal reflect when analyzing the initial location of $\phi$ which lies in the first quadrant.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function xdot = test(x)&#xA;R1=0.1*eye(3,3);&#xA;v=[0.0098,0.0098,0.0176];&#xA;J=diag(v);&#xA;R2=0.03*eye(3,3);&#xA;gamma= 150;&#xA;Q1=eye(3,3);&#xA;Q2=eye(3,3);&#xA;W_d=[0.56;0.61;0.42];&#xA;%--------------------------------------------------------------------------&#xA;R_T=[ x(1)^2+x(2)^2-x(3)^2-x(4)^2, 2*(x(2)*x(3)+x(1)*x(4)), 2*(x(2)*x(4)-x(1)*x(3))   &#xA;          2*(x(2)*x(3)-x(1)*x(4)), x(1)^2-x(2)^2+x(3)^2-x(4)^2, 2*(x(3)*x(4)+x(1)*x(2))   &#xA;          2*(x(2)*x(4)+x(1)*x(3)), 2*(x(3)*x(4)-x(1)*x(2)), x(1)^2-x(2)^2-x(3)^2+x(4)^2];&#xA;W_star=R_T*W_d;&#xA;q0=x(1);&#xA;qv=[x(2);x(3);x(4)];&#xA;W_aux=[x(5);x(6);x(7)];&#xA;&#xA;B1=0.5*[0,x(4),-x(3);-x(4),0,x(2);x(3),-x(2),0]+q0*eye(3,3);&#xA;B2=inv(J);&#xA;A2=(-inv(J)*[0,x(7),-x(6);-x(7),0,x(5);x(6),-x(5),0]*J-inv(J)*[0,W_star(3),-W_star(2);-W_star(3),0,W_star(1);W_star(2),-W_star(1),0]*J);&#xA;G2=inv(J);&#xA;B22=inv(J);&#xA;z1=[x(8);x(9);x(10)];&#xA;z2=[x(11);x(12);x(13)];&#xA;p1=[x(14) x(15) x(16);x(17) x(18) x(19);x(20) x(21) x(22)];&#xA;p2=[x(23) x(24) x(25);x(26) x(27) x(28);x(29) x(30) x(31)];&#xA;W2=1/(2*gamma^2)*B22'*p2*z2;&#xA;Ksi2=-inv(R2)*B2'*p2*z2;&#xA;&#xA;v1=-inv(R1)*inv(B1)*inv(R1)*p1*qv*sign(q0);&#xA;q0dot=-0.5*qv'*W_aux;&#xA;qvdot=B1*W_aux;&#xA;B1dot=0.5*[0,-qvdot(3),qvdot(2);qvdot(3),0,-qvdot(1);-qvdot(2),qvdot(1),0]+q0dot*eye(3,3);&#xA;p1dot=p1*B1*inv(R1)*B1'*p1-Q1;&#xA;v1dot=inv((B1^2*R1^2))*(B1*p1*qvdot + B1*p1dot*qv - p1*B1dot*qv)*sign(q0);&#xA;%v1dot=-inv(R1)*invB1dot*inv(R1)*p1*x1+(-inv(R1)*inv(B1)*inv(R1))*(p1dot*x1+p1*x1dot);&#xA;v2=inv(R2)*B2'*p2*z2+inv(B2)*v1dot*sign(q0)-inv(B2)*A2*v1*sign(q0)+inv(B2)*inv(p2)*B1'*p1*z1-inv(B2)*(1/2*gamma^2)*B22*(B22')*p2*z2;&#xA;W_auxdot=A2*W_aux+B2*v2+G2*W2;&#xA;z1dot=B1*z2-B1*inv(R1)*B1'*p1*z1;&#xA;z2dot=A2*z2+B2*Ksi2-inv(p2)*B1'*p1*z1;&#xA;p1dot=p1dot(:);&#xA;T2=1/(2*gamma^2)*G2*(G2')-B2*inv(R2)*B2';&#xA;p2dot=-A2'*p2-p2*A2-p2*T2*p2-Q2;&#xA;p2dot=p2dot(:);&#xA;xdot=[q0dot;qvdot;W_auxdot;z1dot;z2dot;p1dot;p2dot];&#xA;xdot=xdot(:);&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;execute function&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;% test for function deri&#xA;clc;clear ALL;close all;&#xA;[t,x]=ode45(@(t,x)test(x),[0 1],[0.8224 0.2226 0.4397 0.3604 0 0 0 1 0.5 3 2 1 1.2 0.1 0.2 0.3 0.4 0.5 0.1 0.2 0.3 0.1 0.01 0.02 0.3 0.4 0.2 0.03 0.02 0.001 0.4])&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="17032" LastEditorUserId="110" LastEditDate="2018-09-05T22:55:25.520" LastActivityDate="2021-12-19T10:05:10.510" Title="simulation of an attitude optimal Backstepping controller based quaternion for a UAV in MATLAB" Tags="|control-engineering|aerospace-engineering|simulation|matlab|optimal-control|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23069" PostTypeId="2" ParentId="23062" CreationDate="2018-08-06T14:55:54.213" Score="2" Body="&lt;p&gt;In the 50's the DARPA designed such a craft but due to aerodynamic instability it was a complete failure and abandoned. The problem you have is maintaining 4 separate intakes which have a greater volumetric capacity than each propeller can handle at maximum power in coldest conditions. You cannot combine the intakes and expect stability, it's just not possible. Each airscrew creates a pressure differential with the air cone immediately above and below it. This creates a spinning torus vortex when stationary as part of the stabilising effect of air wash. When that wash is excluded by ducting you introduce other aerodynamic effects. One of these effects is what causes a Dyson fan with a hollow centre to blow a larger volume of air. Your idea will cause the same effect external to the structure which will destabilise it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An image is available here: &lt;a href=&quot;https://www.nationalmuseum.af.mil/Upcoming/Press-Room/News/Article-Display/Article/198892/usafs-flying-saucer-on-display-at-national-museum-of-the-us-air-force/&quot; rel=&quot;nofollow noreferrer&quot;&gt;USAF's 'flying saucer' on display at National Museum of the U.S. Air Force&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="9" LastEditDate="2018-08-06T18:45:10.333" LastActivityDate="2018-08-06T18:45:10.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23070" PostTypeId="2" ParentId="23068" CreationDate="2018-08-06T15:35:31.550" Score="1" Body="&lt;p&gt;That is quite the wall of text. You don't need to rewrite an entire academic paper to ask a question (&lt;strong&gt;please&lt;/strong&gt; don't). You failed to &lt;em&gt;show&lt;/em&gt; what it is you thought you were supposed to get or what you actually got. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I think your quaternion to rotation matrix equation is not correct. It looks like maybe it only works for unit quaternions. The quaternion errors you're passing in are not. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2018-08-06T15:35:31.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23072" PostTypeId="2" ParentId="22639" CreationDate="2018-08-06T17:07:08.627" Score="1" Body="&lt;p&gt;It is more accurate to think of these as negative pressure systems. But the apartment is the thing that is negative. There is typically a &quot;whole house&quot; ventilation fan that is constanctly exhausting air from the apartment and keeping it at a negative presure relative to outdoors. This draws in air via windows or permamant openings (sometimes trickle vents) for ventilation purposes. The positive pressure in the hallway is to ensure that air does no migrate from the apartment to the hallways, not so that air does not come in the window. Air coming in the window is actually by design. Mitigating odors from outdoors is very difficult and would require a much more complex system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another important reason for the positive pressure (other than odors) is for fire protection. In the event of a fire in an apartment, the airflow from the corridor should be keeping the smoke contained to that unit, keeping the path of egress clear. There are typically fire/smoke barriers between units to mitigate the impact of a fire and keep it contained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this quick sketch helps. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/sTCBM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sTCBM.png&quot; alt=&quot;residential vent sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16027" LastActivityDate="2018-08-06T17:07:08.627" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23073" PostTypeId="2" ParentId="23061" CreationDate="2018-08-06T18:11:46.520" Score="0" Body="&lt;p&gt;You have to think of a positive lock-in mechanism for a bracket or support of some sort, depending on what angle of slant you want to present the bike or do you want to offer a kind of ballast counter weight suspended mechanism so the bike can be easily angled on the display for variety of perspectives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even at the most basic configuration you need to have a rough estimate of how the weight will be distributed between the jack and suspension and wheels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As has been mentioned in other answers it would be better to have a light truss or some base under the plate to make it lighter. But the most important question is the entire tray and the lazy susan have to be strong enough to support both the weight and off-center loading and more importantly it is going to resist toppling ( overturning moment), even during the time of loading the bike on it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise the thickness of the plate sounds right at about 1/8&quot; to 3/16&quot;. We can approximate the design using steel column bas plate equations.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-08-06T18:11:46.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23074" PostTypeId="2" ParentId="23053" CreationDate="2018-08-06T18:27:17.103" Score="1" Body="&lt;p&gt;It is better to have the plinth beam above the ground level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For two reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You keep the floor above rain run off or subterranean water and you reduce the height of your building or the length of the columns which helps them better perform in an earthquake.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-08-06T18:27:17.103" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23075" PostTypeId="1" CreationDate="2018-08-06T21:54:15.830" Score="1" ViewCount="58" Body="&lt;p&gt;I am trying to develop my own spring steel version of a carbon fiber running blade that you might see in the paralympics.  Except mine is for someone who still has their foot/leg.  Nothing like that exists and out of necessity I am trying to develop one that is similar and I think i might be able to do it using a less expensive material too.   I really need help calculating what steel to use, thickness, width, angles, etc.  Carbon fiber is ideal but very expensive.  Would spring steel work?  Any concerns with that choice other than a bit heavier?  It would have to be formed in the shape of a &quot;J&quot;This would need to support a 215 pound person walking, running etc.  Any help is greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="17048" LastActivityDate="2018-08-07T05:16:43.210" Title="What steel to use? What thickness and width? What angles to use?" Tags="|steel|carbon-fiber|molding|" AnswerCount="2" CommentCount="1" ClosedDate="2018-08-07T23:10:29.253" ContentLicense="CC BY-SA 4.0" />
  <row Id="23076" PostTypeId="2" ParentId="23075" CreationDate="2018-08-06T23:38:02.753" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What steel to use? What thickness and width? What angles to use?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;An inexpensive source of spring steel for experimenting with are leaf springs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GHfxO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GHfxO.jpg&quot; alt=&quot;Leaf Springs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have serious competition, named &lt;a href=&quot;https://twitter.com/bionicboots?lang=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Keahi Seymour&lt;/a&gt; and his 25 mph 'bionic boots'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vytGW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vytGW.jpg&quot; alt=&quot;Bionic Boots&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also &lt;a href=&quot;http://half-life.wikia.com/wiki/Long_Fall_Boot&quot; rel=&quot;nofollow noreferrer&quot;&gt;fictional designs&lt;/a&gt; to draw from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A7dpA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A7dpA.jpg&quot; alt=&quot;Long Fall Boots&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2018-08-06T23:38:02.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23077" PostTypeId="1" CreationDate="2018-08-06T23:49:22.357" Score="2" ViewCount="363" Body="&lt;p&gt;I'm trying to calculate the deflection of a stack of metallic disks, when subjected to a force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the figure below, representing a valve for a shock absorber, the force is due to a pressure differential. The stack of disks (also called &quot;shim&quot;) is represented in orange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fnKbr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fnKbr.jpg&quot; alt=&quot;Deflection of a stack of disks in a valve, due to a pressure differential&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a couple of papers that delve in this theme, but they all seem to be a direct citation of the following paper: &quot;&lt;em&gt;An Experimentally Validated Physical Model of a High-Performance Mono-Tube Damper&lt;/em&gt;&quot;, by &lt;em&gt;Michael S. Talbott&lt;/em&gt; and &lt;em&gt;John Starkey&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They briefly exemplify for a 3 disk stack, but actually explain very little of the method of computing the deflection of the stack as a function of force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using Roark's formulas for disk plates and the superposition method, one can obtain a system of equations to be solved, akin to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;y_1 = z_{2,R12} + z_{2,R23} \\&#xA;y_2 = z_{3,R23} + z_{3,R34} \\&#xA;\vdots&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $y_i$ is the deflection at the outer radius of disk $i$, and $z_{i,Rkj}$ is the deflection of disk $i$ due to the reaction force between disks $k$ and $j$ (this is better explained in the papers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble however in determining how the disks transfer forces between each others. It is assumed that two different disks only touch at the tip of the smallest one, the pressure difference is known and by using Roark's formulas one can calculate the reaction force at the base of one isolated disk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm however unsure on what amount of force is transferred between disk 1 and disk 2, disk 2 and disk 3, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any papers or books that described this problem or a similar one (stacked disks, stacked beams, stacked plates, etc.) and that illustrate the solution method in greater detail?&lt;/p&gt;&#xA;" OwnerUserId="6364" LastEditorUserId="1832" LastEditDate="2018-08-07T00:17:19.787" LastActivityDate="2020-04-23T06:09:57.237" Title="Deflection of a stack of disks under applied force/pressure" Tags="|mechanical-engineering|structural-engineering|valves|deflection|deformation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23078" PostTypeId="2" ParentId="22639" CreationDate="2018-08-07T02:31:32.153" Score="0" Body="&lt;p&gt;Properly designed HVAC systems in multi-storey buildings introduce Make-Up air which compensates for air leaving the system thru exhaust air dampers.&#xA;A mixing box section accepts return air from the conditioned space,  make-up air from the outdoor air dampers,  and the outdoor air intake for make-up air is tied to the exhaust air damper position. In some modes of enthalpy control,  one hundred percent outdoor air is taken in and the entire return air stream is exhausted (when the OA stream enthalpy is lower than the return air enthalpy). Dedicated exhaust air leaving the conditioned space is incorporated into proper design,  wherein the outdoor air damper minimum position is set to makeup the fixed exhaust air (from bathrooms, etc.)&lt;/p&gt;&#xA;&lt;p&gt;Additionally,  core and perimeter systems are separate entities.  Building pressure control is achieved,  with varying degrees of complexity,  by outdoor pressure sensors on all four sides of the building,  and integrated with discrete indoor air pressure sensors  with the computation output fed into the HVAC unit. In the case of VAV (variable air volume) systems,  the supply and/or return air fans are speed or vane controlled to match the computed requirements,  thereby controlling the system, and the served space pressure. Bottom line is without knowledge of the systems in the building,  all you can do is hunt and peck for your specific situational &amp;quot;fix&amp;quot;, in this case regarding smokers' smoke. Fire smoke is typically handled by purge fan systems coupled with the unit mixing box dampers to maintain relative pressures which ensure smoke is drawn out of the building. It's complicated and every building is an individual in and of itself. Hallways and stairwells are always kept at positive pressure to mitigate the introduction of fire-generated smoke from entering areas of egress from the building.  Yup,  HVAC for multi story buildings is a craft,  requiring professional design engineering training and skills.&lt;/p&gt;&#xA;" OwnerUserId="14245" LastEditorUserId="90" LastEditDate="2022-11-15T06:27:30.560" LastActivityDate="2022-11-15T06:27:30.560" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23079" PostTypeId="1" CreationDate="2018-08-07T03:05:44.837" Score="0" ViewCount="102" Body="&lt;p&gt;I have a small table-like platform with two legs. Underneath each, I need to measure the amount of force. I want to use the difference between the two force measurements to estimate the platform's center of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's a good sensor to use for this? Looking through suppliers Digikey, all the sensors that can handle a large amount of sustained force are either enormously expensive or way too large. Is there another way to measure force without bulky discrete sensors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a cheaper alternative, I was thinking of placing the platform on a low-profile seesaw, with a rotary potentiometer at the fulcrum, and springs under either end. The seesaw would tilt in proportion to the offset of the platform's center of gravity, and this could be measured with the potentiometer. It could do the same measurement as a $200 force sensor at a fraction of the cost. The only problem is that most potentiometers distribute their range of values across 360 degrees, whereas I need all those values within the range of 1 degree.&lt;/p&gt;&#xA;" OwnerUserId="175" LastActivityDate="2018-09-07T07:16:51.453" Title="High impact force sensor?" Tags="|measurements|sensors|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23080" PostTypeId="2" ParentId="23075" CreationDate="2018-08-07T05:16:43.210" Score="0" Body="&lt;p&gt;Have a look at the spring steel used in cultivator tines for the agricultural industry - lighter than leaf springs and already designed in the vertical orientation you are looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link to some examples &lt;a href=&quot;https://www.spaldings.co.uk/replacement-parts/website/replacement-parts/cultivation/heavy-and-light/conventional-spring-s-tines&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-08-07T05:16:43.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23081" PostTypeId="2" ParentId="23079" CreationDate="2018-08-07T06:57:57.177" Score="0" Body="&lt;p&gt;You can attach a low price Laser pointer to your table and read angles off a board positioned on one side as far as is needed to give you resolution in the range you work with, similar triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amazon sells a pack of three for $8.03. in different colors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just need to attach them secure and approximately level, because of your small angle variation anticipation, it doesn't matter if they are not calibrated to be perfectly level. But you only need to wrap a tape around the switch to keep it on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B07BQDSKQ7&quot; rel=&quot;nofollow noreferrer&quot;&gt;3 pieces Laser Torch &lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-08-07T06:57:57.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23082" PostTypeId="2" ParentId="23077" CreationDate="2018-08-07T07:20:54.240" Score="1" Body="&lt;p&gt;The Radius of curvature on a beam bent under moment is $1/r = m/EI  $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use this just as a rough model the m affecting each disk gradually becomes smaller as we go up and radius of curvature bigger. So it is safe to assume only the outer edge of each top disk touches the lower disk.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-08-07T07:20:54.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23084" PostTypeId="1" AcceptedAnswerId="24729" CreationDate="2018-08-07T08:47:32.210" Score="1" ViewCount="1231" Body="&lt;p&gt;I am attempting to write a user model in Fortran that can be connected to the Aspen Plus process simulator. Can Fortran 90 be used, or does Aspen only work with the older Fortran 77? I have checked the Aspen Plus documentation, but can find nothing about which versions of Fortran it is compatible with.&lt;/p&gt;&#xA;" OwnerUserId="6968" LastEditorUserId="6968" LastEditDate="2018-11-19T06:32:30.467" LastActivityDate="2018-11-19T06:32:30.467" Title="Can Fortran 90 be used with Aspen Plus, or only Fortran 77?" Tags="|chemical-engineering|simulation|process-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23085" PostTypeId="2" ParentId="23014" CreationDate="2018-08-07T11:05:09.020" Score="1" Body="&lt;p&gt;Let's imagine condensation as the process of nucleation and growth. &#xA;The nucleation is a process of forming the embryos and growth will give stable condensed particles. &#xA;There are two ways you could trigger nucleation events namely, homogeneous and heterogeneous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In homogeneous nucleation, you will not have any wall/ impurities etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This depends on the following parameters, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;undercooling $\Delta T$&lt;/li&gt;&#xA;&lt;li&gt;density of condesed (liquid phase) particles $\eta_{l}$ &lt;/li&gt;&#xA;&lt;li&gt;surface energy (tension) $\gamma_{sl}$&lt;/li&gt;&#xA;&lt;li&gt;frequency of molecular collision $\nu$&lt;/li&gt;&#xA;&lt;li&gt;probabality of the collision  $\sigma$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;$N_{homo}$ = $f(\Delta T, \gamma_{sl}, \nu, \sigma)$ $ -&amp;gt;$ a big equation (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your case is heterogeneous since you have a wall to grow nucleation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, including the geometric parameters, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$N_{hetro}$ = $N_{homo} * g(\theta)$ $-&amp;gt;$ another equation (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given undercooling ($\Delta T$), heterogeneous nucleation will be more than the homogeneous nucleation. Also based on the $g(\theta)$ function, the $V$ shape or inverted V - ($\wedge$) will trigger more nucleation compared to the concave or convex shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that the cavity side of the shapes (V or $\wedge$ ) will be facing the medium containing condensing particles.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-08-07T11:05:09.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23086" PostTypeId="1" CreationDate="2018-08-07T11:38:15.593" Score="1" ViewCount="735" Body="&lt;p&gt;I am trying to build pitch, roll and yaw from sensor readings of a 9 degree of freedom sensing device.&#xA;The data look like that (in a .csv file):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TIME,ACC_X,ACC_Y,ACC_Z,GYRO_X,GYRO_Y,GYRO_Z,MAG_X,MAG_Y,MAG_Z&#xA;2313992204,-1.694118,-2.053042,-8.886943,0.006272,-0.004091,0.000682,-48.200001,-68.400002,13.000000&#xA;2313992204,-1.674976,-2.029114,-8.886943,0.006545,-0.003682,0.000682,-48.500000,-67.599998,11.300000&#xA;2313992204,-1.674976,-2.033899,-8.882157,0.006136,-0.004091,0.000273,-47.799999,-67.500000,12.300000&#xA;2313992204,-1.703690,-2.033899,-8.882157,0.006818,-0.003682,0.000409,-47.700001,-68.099998,11.800000&#xA;2313992204,-1.694118,-2.053042,-8.877372,0.006409,-0.004500,0.000682,-47.900002,-67.400002,11.200000&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I am not familiar with building this measures from these data and also I am not familiar with quaternions. &#xA;This question is the only thing I have found to be related to mine questions (&lt;a href=&quot;https://engineering.stackexchange.com/questions/3348/calculating-pitch-yaw-and-roll-from-mag-acc-and-gyro-data?newreg=30ff31f79f5a48fcb7598581deca4754&quot;&gt;Calculating pitch, yaw, and roll from mag, acc, and gyro data&lt;/a&gt;), however, it is not using all the axis.&#xA;Ideally, I want to build this vector and be able to visualize them in 3D (like cube animation that is made live from 9dof sensors and Arduino). Moreover, I want to be able to compute the heading from this data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me, I have all my pipeline in python so it would be helpful to have advice on how to do that with python?&lt;/p&gt;&#xA;" OwnerUserId="17054" LastEditorUserId="7083" LastEditDate="2018-08-14T00:10:49.017" LastActivityDate="2018-08-14T00:10:49.017" Title="Build and visualize Pitch, Roll and Yaw from 9dof data offline" Tags="|mechanical-engineering|electrical-engineering|sensors|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23087" PostTypeId="2" ParentId="23084" CreationDate="2018-08-07T12:43:33.623" Score="0" Body="&lt;p&gt;I'm not 100% on this, but I don't think that AspenPlus itself cares.  This is because AspenPlus calls a Fortran compiler to compile-and-link your code, then it consumes your code as a &lt;code&gt;.dll&lt;/code&gt; (dynamic link library).  The details behind your Fortran and how the compiler compiles it seem to beyond the scope of AspenPlus's concern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, AspenPlus is probably happy as long as the compiler can compile your code into a library that links properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, it'd seem like the version of Fortran accepted would be a function of your compiler.  As long as your compiler turns the Fortran into a &lt;code&gt;.dll&lt;/code&gt; with the call convention that AspenPlus expects, it's probably sufficient.&lt;/p&gt;&#xA;" OwnerUserId="10678" LastActivityDate="2018-08-07T12:43:33.623" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23088" PostTypeId="1" CreationDate="2018-08-07T14:22:50.310" Score="2" ViewCount="126" Body="&lt;p&gt;I don't understand how bending moment is being calculated in the problem mentioned in the above link. I understand the parts after that i.e. calculating elastic potential and partial derivatives but I am just not able to understand the bending moment part. I would be glad to receive some responses on this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YoBGX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YoBGX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16862" OwnerDisplayName="MechNovice" LastEditorUserId="282" LastEditDate="2018-11-02T16:40:38.047" LastActivityDate="2018-11-02T16:40:38.047" Title="Bending Moment Doubt" Tags="|kinematics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23089" PostTypeId="2" ParentId="23088" CreationDate="2018-08-07T16:17:08.710" Score="1" Body="&lt;p&gt;When you constrain a beam end to point in the same direction as the clamped end (which is the case with the lower beams), then you have to apply a moment equal to $Ql$ (where $Q$ is the end load and $l$ is the length) to maintain this angle. You'll note that this results in an end-moment magnitude of $Ql$ at each end, which produces symmetric curvature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, though, $Q_1$ is split between two beams, so the applied end moment for each is $Q_1l_1/2$, and the moment along each is $-Q_1x+Q_1l_1/2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No such end moment is needed in the free beam on top, so the moment along that beam is simply $-Q_2x$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this what you're asking about?&lt;/p&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2018-08-07T16:17:08.710" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23091" PostTypeId="1" CreationDate="2018-08-07T17:45:07.740" Score="2" ViewCount="83" Body="&lt;p&gt;I was wondering why carbon nanotubes are not used as artificial muscles yet? I mean they behave similar to real muscles, only that they expand horizontally. What are the current challenges that researchers are facing?&lt;/p&gt;&#xA;" OwnerUserId="17057" LastActivityDate="2024-01-02T05:39:52.287" Title="Carbon Nanotubes as muscles?" Tags="|carbon-fiber|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23092" PostTypeId="2" ParentId="23012" CreationDate="2018-08-07T18:32:35.453" Score="4" Body="&lt;p&gt;The reason it is impossible to give you figures is that nobody has run the tests to verify the shear strength.  Why?  Wouldn't it be easy to get a block of polypropylene and run a test?  Well, that test would only be valid for that supplier of polypropylene, and only for that thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This uncertainty in testing results from several interesting factors:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Unlike steels, who's unique properties can be determined by specifying a set additive concentration, a polymer's properties depends on chain lengths as well as cross linking, which come down to manufacturing methods.&lt;/li&gt;&#xA;&lt;li&gt;This means that a &quot;standard&quot; polymer is never - standard.  For example, polypropylene is usually broken into homopolymer, copolymer, and random-block copolymer.  All of these can be made in long and short chain.  And the co-polymer block added can greatly influence the degree of cross-linking.  Each specifically manufactured chain-length polymer is typically called a resin, with a manufacturer's trade name.&lt;/li&gt;&#xA;&lt;li&gt;Even these wildly varying property differences can be subtly influenced by the extrusion method, not unlike how annealing changes the property of a steel.  &lt;/li&gt;&#xA;&lt;li&gt;A manufacturer will not run unnecessary tests to bring a resin to market. If a resin will not be needed to operate in shear, there is no need to run a test on shear strength.  &lt;/li&gt;&#xA;&lt;li&gt;Any engineers concerned with shear have learned to convert the stress via Mohr's circle to tensile stress and estimate using a very conservative factor of safety using those methods.  &lt;/li&gt;&#xA;&lt;li&gt;A shear strength estimate would typically be made under the presumption that the ratio of shear strength to tensile strength would be similar to the ratio between shear modulus and tensile modulus, and utilize the Poisson's ratio (which typically remains constant regardless of cross-linking or chain length).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-08-07T20:27:41.277" LastActivityDate="2018-08-07T20:27:41.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23093" PostTypeId="2" ParentId="21325" CreationDate="2018-08-07T18:53:23.900" Score="1" Body="&lt;p&gt;The closest to this grade is AMS 6470 steel, but it does not contain nickel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore 51 HRC is easily possible via surface diffusion process like Nitriding to improve wear resistance.&lt;/p&gt;&#xA;" OwnerUserId="17058" LastEditorUserId="1832" LastEditDate="2018-08-07T21:30:10.770" LastActivityDate="2018-08-07T21:30:10.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23094" PostTypeId="1" AcceptedAnswerId="23611" CreationDate="2018-08-07T20:34:54.473" Score="3" ViewCount="336" Body="&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; I am currently encountering Hard Drive Memory space issue to store file on a National Instruments based CDAQ Tester. I am using Labview  2013 SP1 and have to all the software load in the CDAQ Controller.  Some of the basic specifications for the CDAQ are as follows&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Processor: Intel Atom 1.33GHz Dual Core&lt;/li&gt;&#xA;&lt;li&gt;Controller OS: Windows Embedded Standard 7&lt;/li&gt;&#xA;&lt;li&gt;Hard Drive Memory Size: 16 GB&lt;/li&gt;&#xA;&lt;li&gt;Number of Slots: 8&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OKYrP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OKYrP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Background:&lt;/strong&gt; I am trying to control and acquire electrical measurement data from a Keysight 34461A digital multimeter via GBIP using Labview and CDAQ. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZQl2a.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZQl2a.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="90" LastEditDate="2018-11-02T16:18:01.843" LastActivityDate="2018-11-02T16:18:01.843" Title="How to resolve file space issue on CDAQ Controller with Labview Test System?" Tags="|electrical-engineering|product-testing|labview|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23095" PostTypeId="1" CreationDate="2018-08-07T21:59:18.840" Score="0" ViewCount="50" Body="&lt;p&gt;At a very low-tech, rustic forest camp, I've built a 'water tower' connected to a &lt;a href=&quot;https://international.sawyer.com/products/sawyer-pointone-filter-bucket-adapter-kit-1-ft-hose/&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravity-fed water filter&lt;/a&gt; to provide something like a clean water faucet. It's basically a 5 gallon bucket on top of a ~6' high platform, with a 1' long hose running down from it connected to a filter. It is idle for long periods of time and then used for a period of 1 to 9 days, receiving apx. 3 gallons of moderately turbid, acidic water input each day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the end of the filter (i.e. the clean water spout) to be about waist high (way easier to work with than a spout only a couple inches above the ground). To accomplish this I could either shorten the filter's hose or raise the bucket, and thinking that height/gravity was important for a gravity fed filter, I opted to raise the bucket. This presented two challenges: I need a taller platform, and I need a way to fill the bucket when it's at or above head height. The taller platform was easy enough to build, but a sturdy enough ladder to carry water up to the bucket is harder to craft. How can I get water from ground level up in to a bucket that's rim is about 7' high?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conditions on site are very rustic and remote. It's a long drive to get there, then a bit of a walk to camp (lugging gear in a backpack and by hand), and currently the only dry storage is makeshift via tarps (not reliably dry). I don't always go there with a large vehicle, so transporting a ladder back and forth isn't a great option, and if I leave a steel ladder there it will be exposed to extreme temperature swings &amp;amp; moisture. I'm envisioning some relatively primitive technology like a pulley, an Archimedes screw, or more compact easier-to-transport modern technology like a small hand-pump with a hose, but I know very little about the details of those options. So I ask Engineering.SE: what practical rugged method can I use to elevate water 7 feet high? If some kind of pump, what specs to look for and what other equipment will I need to make it work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus if the method can be scaled or setup in series so to move water further distances. As of now, to get water to the ground next to the water tower, one has to carry water by bucket up a steep ~100' slope that has ~75% grade.&lt;/p&gt;&#xA;" OwnerUserId="5315" LastActivityDate="2018-08-10T22:00:18.133" Title="Practical rugged method for elevating water 7 feet high?" Tags="|mechanical-engineering|hydraulics|" AnswerCount="2" CommentCount="8" ClosedDate="2018-08-14T23:09:49.380" ContentLicense="CC BY-SA 4.0" />
  <row Id="23096" PostTypeId="1" AcceptedAnswerId="23101" CreationDate="2018-08-08T01:47:08.597" Score="1" ViewCount="408" Body="&lt;p&gt;I am currently involved with a project using a mechanism similar to that seen in 'table tilters' and we are struggling to work out how to begin resolving the forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image shows the state we are considering in equilibrium to resolve the forces. The plate (blue) is in a horizontal position and is 350mm in length&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linkages (red) are 270mm long and are at an angle of 20 degrees to the horizontal at this position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The green component is restricted to motion in the vertical direction and is being pulled downwards with a force of 3000N. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to work out the downward force the linkages exert on the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Oaw8H.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oaw8H.jpg&quot; alt=&quot;ISO view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VATO7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VATO7.jpg&quot; alt=&quot;Side View&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started to try and break the system down into free body diagrams (below), but am struggling to make a start resolving the forces. Any advice would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XfcP8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XfcP8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vdWBW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vdWBW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17061" LastEditorUserId="17061" LastEditDate="2018-08-08T07:13:55.970" LastActivityDate="2018-08-08T15:15:07.170" Title="Calculating force exerted on plate by link" Tags="|mechanical-engineering|statics|linkage|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23097" PostTypeId="1" CreationDate="2018-08-08T02:48:49.663" Score="0" ViewCount="571" Body="&lt;p&gt;Assuming a notional 4.5&quot; disc, on a common 4.5&quot; angle grinder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which abrasive cuts faster, a flap disc, a sanding disc or a grinding wheel of common grit (I believe 36 or 40)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume, at least for the sanding disc, and the flap disc that the abrasive material are similar, for example, aluminum oxide.  Ditto for the grinding disc, however they do not seem to be as clearly marked as to the abrasive material. &lt;/p&gt;&#xA;" OwnerUserId="17062" LastActivityDate="2018-08-08T12:26:51.130" Title="Flap disc vs sand paper vs grinding disc" Tags="|cutting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23098" PostTypeId="1" AcceptedAnswerId="23109" CreationDate="2018-08-08T04:09:57.077" Score="1" ViewCount="567" Body="&lt;p&gt;I am trying to calculate the increase in pressure caused by liquid nitrogen at the moment it changes from liquid to vapor within a closed, constant volume at atmospheric pressure. Do I need to include the heat of vaporization? How can this be done?&lt;/p&gt;&#xA;" OwnerUserId="17063" LastActivityDate="2018-08-08T22:29:06.997" Title="How to calculate change in pressure from LN2 state change (constant volume)" Tags="|pressure|liquid|vaporization|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23099" PostTypeId="1" CreationDate="2018-08-08T05:20:04.800" Score="1" ViewCount="139" Body="&lt;p&gt;Does anyone know how this works? I hypothesize that the reason that the bullet shatters is that boron carbide is much harder than the copper of the bullet. Does anyone think that this may be made into a conventional armor design as some people are saying or is boron carbide too expensive for use in this application? &#xA;Link:  &lt;a href=&quot;http://www.slate.com/articles/video/video/2016/04/metal_foam_armor_can_shatter_bullet_on_contact_video.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.slate.com/articles/video/video/2016/04/metal_foam_armor_can_shatter_bullet_on_contact_video.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="7538" LastEditorUserId="7538" LastEditDate="2018-08-08T14:35:36.667" LastActivityDate="2018-11-22T15:06:00.633" Title="Boron Carbide Shattering a Bullet" Tags="|materials|design|strength|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23100" PostTypeId="2" ParentId="21376" CreationDate="2018-08-08T05:58:01.193" Score="1" Body="&lt;p&gt;If you calender a rubber with some thin steel cords in the calendered direction you could achieve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;For those who are unfamiliar with &lt;a href=&quot;https://en.wikipedia.org/wiki/Calender&quot; rel=&quot;nofollow noreferrer&quot;&gt;calenders&lt;/a&gt;, they are &quot;a series of hard pressure rollers used to finish or smooth a sheet of material such as paper, textiles, or plastics&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do not confuse calenders (with an ers ending), which are rollers, with calendars (with an ars ending) which is a time keeping device.&lt;/p&gt;&#xA;" OwnerUserId="8533" LastEditorUserId="90" LastEditDate="2018-08-08T09:27:47.680" LastActivityDate="2018-08-08T09:27:47.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23101" PostTypeId="2" ParentId="23096" CreationDate="2018-08-08T06:58:43.583" Score="0" Body="&lt;p&gt;your sketches misses a bit of information. where is the resisting element?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a post, or leg/legs under the table. Assuming the center of resisting forces falls right under the red bar on the point of contact. then the force on plate is same as 3 kN downward. If the resisting forces are not directly under the red bar then the downward force becomes bigger or smaller, depending on the location of resisting force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reviewing additional info, your table is receiving 3kN downward force. But you need to be mindful of horizontal forces at points green and yellow on the vertical; black bar. Those are $Fh = cos(20)/ Sin(20)\times 3kN = approx.\space 9 kN$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two work as a couple trying to topple your table.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-08-08T15:15:07.170" LastActivityDate="2018-08-08T15:15:07.170" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23102" PostTypeId="1" AcceptedAnswerId="23120" CreationDate="2018-08-08T10:23:01.693" Score="-1" ViewCount="1006" Body="&lt;p&gt;I have a TEC-12706 thermoelectric peltier module. I am making a mini refrigerator (simple box refrigerator). I'm using this peltier and heatsink of equal size as compared with the peltier. I have given the picture of the heatsink attached with the hot side of the peltier. I'm facing a very big problem. While I connect this peltier with a DC source, it works properly for some time, i.e., the cold side of the peltier gets cool and the heatsink starts becoming hot. But just after a few time the peltier starts working in a different way. That is, the cold side of the peltier also gets some heated and the heatsink stops or reduces absorbing heat from the hot side of the peltier. I think the heatsink compound (or thermal paste) which I'm using is not proper. What do you think, what may be the problem? Please help me, I'm struck and am not able to make my refrigerator.!See this picture &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FMpEh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FMpEh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12570" LastEditorUserId="8055" LastEditDate="2018-08-08T14:09:56.080" LastActivityDate="2018-08-10T22:05:47.320" Title="Thermoelectric Module Not Working Properly!" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23103" PostTypeId="1" AcceptedAnswerId="23111" CreationDate="2018-08-08T12:09:59.653" Score="2" ViewCount="324" Body="&lt;p&gt;I'm looking for a name of a part. It looks like a cable inside another cable. The idea is that you can just pull on the inside cable and get the force at the other end. You can just attach the outside cable somewhere, and don't have to bother about pulleys and such. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this exists, and what is it called?&lt;/p&gt;&#xA;" OwnerUserId="17072" LastEditorUserId="1324" LastEditDate="2018-08-10T13:18:28.690" LastActivityDate="2018-08-13T14:22:29.377" Title="Looking for the name of a part" Tags="|mechanical-engineering|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23104" PostTypeId="2" ParentId="23103" CreationDate="2018-08-08T12:23:39.387" Score="2" Body="&lt;p&gt;Sounds like your talking about bicycle cable and housing. Search for &quot;brake cable&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also different types of these cables for different applications. Teflon lined inner for smoother operations, stainless for corrosion resistance, linear wound, opposed to radially, outer housings that can withstand turning/mild kinking without tightening the inner cable, etc. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastEditorUserId="11327" LastEditDate="2018-08-08T14:13:18.877" LastActivityDate="2018-08-08T14:13:18.877" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23105" PostTypeId="2" ParentId="23097" CreationDate="2018-08-08T12:26:51.130" Score="1" Body="&lt;p&gt;There are a lot of variables to say which cuts &quot;faster&quot; , but generally IMHO, grinding wheel &gt; flap disk &gt; sanding disk as far as material removal in lets say mild steel. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-08-08T12:26:51.130" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23106" PostTypeId="1" CreationDate="2018-08-08T12:35:16.657" Score="1" ViewCount="4374" Body="&lt;p&gt;(Disclaimer: To make things clear I'm talking about adhesion railways and excluding rack railways as those are an entirely different technology and requires specially equipped vehicles. Also I'm looking for examples as how this is engineered.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was surprised recently as I visited the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wutach_Valley_Railway&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wutachtal railway&lt;/a&gt; in southern Germany. A particular section of this railway makes extremes detours with multiple viaducs and tunnels (including a circular tunnel) in order to climb 231m while having a gradient of only 1%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the museum's explanation, the gradient of 1% was required because they wanted to allow heavy military freight trains to circulate on this railway. However, steeper railways are very common, including rails which allows heavy freight trains. For instance, according to German language Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Gotthard railway has a max steepness of 2.8%&lt;/li&gt;&#xA;&lt;li&gt;The Simplon railway as well as the Neuchâtel–Pontarlier railway a max steepness of 2.5%&lt;/li&gt;&#xA;&lt;li&gt;The Arlberg railway in Austria has a steepness of 3%.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some trains which only allows passenger train and have narrow gauge have much steeper rails:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The Uetli train in Zürich has a steepness of almost 8%&lt;/li&gt;&#xA;&lt;li&gt;The Pöstlingbergbahn has a steepness of 11.6%, (narrow gauge)&lt;/li&gt;&#xA;&lt;li&gt;Lisbon tramway has a steepness of 13.5%&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Since the maximum allowed steepness is a major factor when designing a train line, I really wonder what factors determined it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What kind of problems appears when the rail is too steep for a given train, is it a problem when going uphill (train slip and can't advance) or when going downhill (train slips and can't brake).&lt;/li&gt;&#xA;&lt;li&gt;Does the weight of the train, or the power of the locomotive play a role&lt;/li&gt;&#xA;&lt;li&gt;Did technology (i.e. more powerful motors, motors and/or braking on more wheels) allow steeper trains as time passed, or is there a hard physical limit that can't be overcome ?&lt;/li&gt;&#xA;&lt;li&gt;Why could Switzerland and Austria allows freight trains on 3% steep tracks while Germany needed to limit itself 1% ?&lt;/li&gt;&#xA;&lt;li&gt;When an international freight train circulates, how do we know which routes it can or cannot take due to its weight ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;EDIT: Since I asked the question, I've learn about a very interesting thing on the subject : In Geneva, 1904 a tramway line with a steepness of 11.8% was ready for usage but an &lt;a href=&quot;https://www.snotpg.ch/site/tram-cite/&quot; rel=&quot;nofollow noreferrer&quot;&gt;accident when doing tests, where the tramway slipped and was unable to brake, cancelled the line&lt;/a&gt;. So apparently the major problem is being able to brake when going down. But the Lisbon tramway seems even steeper and didn't have similar accident as far as I'm aware.&lt;/p&gt;&#xA;" OwnerUserId="17069" LastEditorUserId="3505" LastEditDate="2020-05-31T17:54:08.130" LastActivityDate="2022-03-18T00:13:56.243" Title="What determines the maximum possible slope of rail transportation (without rack)?" Tags="|rail|transportation|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23107" PostTypeId="2" ParentId="23013" CreationDate="2018-08-08T12:36:01.457" Score="1" Body="&lt;p&gt;Does this happen repeatedly, when you turn it on, after a while, etc? If your load is around 50-75% of your compressors output and your operating in VFD, if it happens at start up, perhaps its the PID trying to figure out what it wants to tell your VFD to do. If that's the case, after a while running at your plants normal load, the cfm drop and time in between should taper off. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And your sure the system is plumbed COMP &gt; TANK &gt; DRYER ?&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-08-08T12:36:01.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23108" PostTypeId="2" ParentId="22880" CreationDate="2018-08-08T12:41:08.583" Score="1" Body="&lt;p&gt;A CNC bender would be the way to go for quality and repeat ability. Otherwise, you can try building some bending jigs to try and replicate some of the bends between pieces better. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-08-08T12:41:08.583" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23109" PostTypeId="2" ParentId="23098" CreationDate="2018-08-08T12:47:18.670" Score="0" Body="&lt;p&gt;Assume ideal gas conditions in a container at constant $T_o, V_o$. The change in pressure by adding an amount $\Delta n$ of gas is $\Delta p = \Delta n R T_o / V_o$.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-08-08T12:47:18.670" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23110" PostTypeId="1" AcceptedAnswerId="23139" CreationDate="2018-08-08T12:51:23.300" Score="1" ViewCount="470" Body="&lt;p&gt;I already got great help on this topic, but I'd like to make sure that I get it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, take a beam of length 6 meters, left side fixed support and right side rolled support. There is a load of 30 kN-m that starts at 3 meters and ends at 6 meters with a magnitude of 60 kN-m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The function of the load is easy, f(x)=10x. The centroid is given by &#xA;$$\frac{\int_3^6 10x² dx}{\int_3^6 10x dx} = 4.67m$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the force translated into a point load is &#xA;$$ \int_3^6 10x dx   = 135 kN $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the sum of forces as seen from a is 0, and can be used to calculate Mb&#xA;$$ \sum Ma =0 = 4.67*135 -6*Mb \Rightarrow Mb = 105 kN \Rightarrow Ma=35kN$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then to find the shear forces I integrate the equation from 3 to x taking into account Ma&#xA;$$ \int_3^x 10x dx -Ma   = 5x² -75 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to find the Bending moments I just integrate again, this time for the whole equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \int_0^x (5x²-75 )dx = \frac{5x³}{3} -75x $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alas this does not give the complete answer, I have many uncertainties, for the shear forces I notice that it only gives the correct answer for x between 3 and 6; for x under 3 it becomes Ma, and I imagine if the length of the beam would be greater than the distributed load's length, it would be Mb for x&gt;6. If I would add a point load of 3kN at l=2, do I need to replace Ma with Ma'- 3kN in the shear force equation? What would happen with a point load after 6 meters (eg lenght of beam = 9 meters, point load at 7m)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the bending moments, I kinda get the right answer for x between 3 and 6, but it is 90 kN too much, which is the BM at the start of the distributed load. For x=2 for example I get completely wrong answers, and I need to revert to calculating BM by using trigonometry to calculate the actual area under the shear force curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean I can solve them, but I don't get the why - why can't I use the BM function to calculate all of x, considering the boundary condition should have been solved in the first integral (from 3 to 6 ). Why do I need to substract the BM at the beginning of the distributed load,... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would greatly appreciate if someone would take the time to enlighten me - as it is I'm not really confident for the exam (I can't afford to fail) &lt;/p&gt;&#xA;" OwnerUserId="17073" LastActivityDate="2018-08-09T14:00:05.167" Title="Shear force and bending diagram un-uniformly distributed load" Tags="|mechanical-engineering|moments|shear|diagram|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23111" PostTypeId="2" ParentId="23103" CreationDate="2018-08-08T12:53:23.877" Score="3" Body="&lt;p&gt;It's called a &lt;strong&gt;Bowden cable.&lt;/strong&gt; See &lt;a href=&quot;https://en.wikipedia.org/wiki/Bowden_cable&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Bowden_cable&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Corey said, they are often used for bike brake cables, but they have many other uses as well.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-08-08T12:53:23.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23112" PostTypeId="2" ParentId="23103" CreationDate="2018-08-08T12:54:44.203" Score="2" Body="&lt;p&gt;One name is “Bowden cable”, which is the inner cable and outer sheath / cover providing a force by the movement of one relative to the other - usually the inner moves relative to the outer...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-08-08T12:54:44.203" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23113" PostTypeId="2" ParentId="23106" CreationDate="2018-08-08T12:55:39.343" Score="1" Body="&lt;p&gt;A combination of the pulled weight, the weight of the movers (the locomotives) and the coefficient of friction between wheel and rail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more weight on each driven wheel the more force it can exert before it starts slipping. Being able to start uphill on the slope is also important especially when considering the military needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more weight you need to tow the more force you need to get it moving and keep it moving. &lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2018-08-08T12:55:39.343" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23114" PostTypeId="1" CreationDate="2018-08-08T13:19:58.203" Score="1" ViewCount="148" Body="&lt;p&gt;I've been reading into air bearing recently and keep coming across the term &quot;pre-load&quot;. References online and wikipedia both state that a pre-load helps stiffen the bearing such that their is less variation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing the pre-load ensures the air-gap is maintained/little variation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm having a difficult understanding how adding some force (pre-load) would actually help with this. &lt;/p&gt;&#xA;" OwnerUserId="7664" LastActivityDate="2019-06-03T11:46:43.893" Title="How does an air-bearing &quot;pre-load&quot; work?" Tags="|bearings|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23115" PostTypeId="1" CreationDate="2018-08-08T13:21:21.500" Score="0" ViewCount="3273" Body="&lt;p&gt;While testing the toughness of brittle material by izod or charpy test why do we provide v notch?&lt;/p&gt;&#xA;" OwnerUserId="17075" LastActivityDate="2018-08-09T13:55:21.270" Title="What is the use of v notch in Izod charpy test" Tags="|materials|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23116" PostTypeId="2" ParentId="3547" CreationDate="2018-08-08T13:27:19.000" Score="1" Body="&lt;p&gt;Something important to consider is that if the corridor taking you down to the underground is narrower then the width of the way turning in, then there are certain sizes of cars that are able to go in but are unable to go out from the underground garage. So these cars can only go out in reverse.&lt;/p&gt;&#xA;" OwnerUserId="17071" LastEditorUserId="2015" LastEditDate="2018-08-08T20:16:42.380" LastActivityDate="2018-08-08T20:16:42.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23118" PostTypeId="2" ParentId="23115" CreationDate="2018-08-08T13:47:39.587" Score="2" Body="&lt;p&gt;The point is to promote the condition of brittle fracture. The notch here acts as a stress riser and favours the crack propagation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This test is designed for any material, if you already know that the material is brittle then why do you put it through the test ? The absolute values and numbers here are less important and they are not reliable. The impact engineers want to know when a material favours brittle fracture, some materials are ductile in a specific range of temperature, but they still fail like brittle material, why ? presence of stress risers, high rate of strain ... , by bringing the notch they promote the condition of brittle fracture and the notch provides three dimensional stress condition.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-08-09T13:55:21.270" LastActivityDate="2018-08-09T13:55:21.270" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23119" PostTypeId="2" ParentId="23114" CreationDate="2018-08-08T14:18:45.307" Score="1" Body="&lt;p&gt;The reason is to prevent instability. What is instability? Let's have look at a simple shaft-journal bearing system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an ideal system wherein a shaft is rotating clockwise inside a journal bearing. As you can see the eccentricity is zero. Notice the gap between the shaft and the journal is filled with oil, or air in case of air bearings.&lt;a href=&quot;https://i.stack.imgur.com/f8b4x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f8b4x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the whole system rotates unloaded in the beginning. Now duo to some irregularities (bumps, impacts ...) the eccentricity between the shaft and the geometrical center of the journal bearing increases. Here you can see indeed the shaft moves upward to the right, the direction of the eccentricity is &lt;span class=&quot;math-container&quot;&gt;$\theta = 38.78°$&lt;/span&gt;:&lt;a href=&quot;https://i.stack.imgur.com/EgMJY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EgMJY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the pressure distribution tends to rotate the shaft in counterclockwise direction, the pressure in narrow split is the highest. The resultant of pressure in narrow split is &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;. If the bearing is not preloaded then this force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; makes the system unstable by disrupting the natural rotational motion of the shaft. By preloading the bearing in all directions of course (because this force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; happens to have random directions because the irregularities happen randomly) compensate the force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; and the system stays stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the angular speed of the shaft is low enough and the viscosity is high enough then the disruption goes away automatically and no preload is needed, but that's not the case for high angular velocities in combination with low viscosity. The viscosity of air is low enough and the air bearing usually are used in high rpm, so the preloading is indispensable. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2019-06-01T18:43:39.237" LastActivityDate="2019-06-01T18:43:39.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23120" PostTypeId="2" ParentId="23102" CreationDate="2018-08-08T14:42:01.140" Score="0" Body="&lt;p&gt;It is typical of a Peltier device to require an active method of dissipating the heat. You should have a fan on the hot-side capable of moving cooler air over the fins, in order to keep from destroying the device. The same aspect applies to the cold side. Use a fan to move cooled air to the desired location.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-08-08T14:42:01.140" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23121" PostTypeId="2" ParentId="23103" CreationDate="2018-08-08T15:02:37.267" Score="1" Body="&lt;p&gt;Also known as a &quot;Push-Pull Cable&quot; or &quot;(mechanical) Control Cable&quot;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2018-08-13T14:22:29.377" LastActivityDate="2018-08-13T14:22:29.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23123" PostTypeId="2" ParentId="14323" CreationDate="2018-08-08T17:53:02.967" Score="1" Body="&lt;p&gt;The entire point of a &quot;pin&quot; connection is that it does not resist a moment.  If you move two parts connected via a pin connection, the parts are supposed to move freely, but rotate relative to each other at the pin.  As such, the moments are always considered 0 at a pin connection.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider classical static analysis frames from second year.  These always have the loads applied directly to the pins, never on the members.  As such, the moment arm is zero, and therefore the moment equation is irrelevant.  Now let's consider real world analysis, when a load is applied to the middle of a member. Since the member is considered to have pinned connections, where the moment at each pin is zero, we do not need to count on other members transmitting that load.  Instead, that single member that is loaded could be treated as a pinned connection beam.  Once the load is resolved, the pins still have - no moments.  Again, no additional equations to resolve the indeterminacy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note this applies only for static frames.  When frames move dynamically, the members can rotate, and this rotation is caused by moments from dynamic movements in the members.  &lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-08-08T19:11:32.993" LastActivityDate="2018-08-08T19:11:32.993" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23124" PostTypeId="2" ParentId="23095" CreationDate="2018-08-08T18:02:47.350" Score="2" Body="&lt;p&gt;Winch the bucket down, fill the bucket, then winch the bucket up. You're not trying to move this some huge distance, and it's not some huge reservoir to move. A pump is overkill. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could, alternatively, just use a couple empty milk jugs to move one gallon at a time. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2018-08-08T18:02:47.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23125" PostTypeId="2" ParentId="13663" CreationDate="2018-08-08T18:04:02.163" Score="3" Body="&lt;p&gt;For concrete and asphalt, you would be more interested in the properties of creep, which is always a function of temperature.  For example, there is an entire &lt;a href=&quot;https://en.wikipedia.org/wiki/Creep_and_shrinkage_of_concrete&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt; page about this at this time, which mainly references ACI Committee 209&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For polymers, there are many models for &lt;a href=&quot;https://en.wikipedia.org/wiki/Creep_(deformation)&quot; rel=&quot;nofollow noreferrer&quot;&gt;creep&lt;/a&gt;.  Due to the extensive range of polymer chains, it would be difficult to model a general version without testing some asphalts specifically.  &lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-08-08T18:04:02.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23126" PostTypeId="2" ParentId="23106" CreationDate="2018-08-08T18:06:29.043" Score="3" Body="&lt;p&gt;The actual maximum attainable slope is determined by the weight of the locomotive, the total weight of the train, the rolling friction of the carriages, and the kinetic(sliding) friction between two materials; the locomotive's wheels and road surface(rails).&lt;/p&gt;&#xA;&lt;p&gt;The static friction(present when things aren't slipping yet) allows for steeper hills, but it'd be unsafe to go that far, one slip of the wheel and the train won't be able to regain grip.&lt;/p&gt;&#xA;&lt;p&gt;The friction coefficient of steel on steel(dry) is around 0.6. That means that a locomotive of 100 ton is able to give 60 ton of tractive force on a flat rails.&#xA;It is less so on a hill, since gravity is not working perpendicular on the train, here its 'weight' is cos(4.6°)*100 = 99.7 ton, so it can pull 59.8 ton.&#xA;At a slope of 8%(4.6°), that means the total weight of the train can be 59.8/sin(4.6°) = 746 ton. No rolling friction is assumed here.&lt;/p&gt;&#xA;&lt;p&gt;Of course, there is rolling friction, safety margins have to be used, and in bad weather traction will be less, so the total weight of the train must be less. If a locomotive only has to drive itself up the hill, then the hill could be as steep as 30 degrees and it wouldn't slip yet.&lt;/p&gt;&#xA;&lt;p&gt;The slope of the rails is just a choice, steeper just means less load can be hauled by one locomotive.&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="90" LastEditDate="2022-03-18T00:13:56.243" LastActivityDate="2022-03-18T00:13:56.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23127" PostTypeId="2" ParentId="14323" CreationDate="2018-08-08T20:16:09.580" Score="1" Body="&lt;p&gt;To be clear, just as you consider global bending moment for beams, you also consider it for trusses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, note that this is the &lt;strong&gt;&lt;em&gt;global&lt;/em&gt;&lt;/strong&gt; bending moment. Once you've calculated that once, there's no reason to calculate it again. Sure, you choose some arbitrary point $A$ from which to calculate $\sum M_A=0$, but the point chosen is completely immaterial: no matter which point you select, the result will be the same. Doing the work again for another point $B$ will grant you absolutely no new information, since both $\sum M_A=0$ and $\sum M_B=0$ are representing the exact same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why, when calculating beams, we don't calculate the sum of moment around each support: if $\sum M_A=0$ is satisfied, and so are the global force equations $\left(\sum F_x=\sum F_y=0\right)$, then the global bending moment from the other support is &lt;strong&gt;&lt;em&gt;guaranteed&lt;/em&gt;&lt;/strong&gt; to be satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so, just as with beams, you &lt;strong&gt;&lt;em&gt;do&lt;/em&gt;&lt;/strong&gt; need that global bending moment calculation. But you only need it once.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also don't need to bother calculating the internal bending moment at any of the truss' pins because, well, pins are defined as having zero bending moment. And, fundamentally, that's why each joint's bending moment isn't counted when thinking about a truss' determinacy: determinacy is about the balance between the number of global equilibrium equations we have available and the number of &lt;strong&gt;&lt;em&gt;unknowns&lt;/em&gt;&lt;/strong&gt; in the system. The bending moment on a truss' joint isn't an unknown; it is &lt;strong&gt;&lt;em&gt;defined&lt;/em&gt;&lt;/strong&gt; as zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why the static determinacy equation for trusses is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{alignat}{4}&#xA;\text{if 2D: }&amp;amp; 2j − b &amp;amp;&amp;amp;= 3 \\&#xA;\text{if 3D: }&amp;amp; 3j − b &amp;amp;&amp;amp;= 6 \\&#xA;\end{alignat}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $j$ is the number of joints and $b$ is the number of bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the equation is satisfied, the structure is statically determinate. If $2j-b&amp;lt;3$ (or the 3D equivalent), then it is indeterminate. If $2j-b&amp;gt;3$ (or the 3D equivalent), then it is unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The constant ahead of $j$ represents the number of internal forces existing in the structure: each joint has $F_x$ and $F_y$ (and $F_z$ in 3D). The more joints you have, the more internal forces you need to solve for. Sure, each joint also has $M_z$ (and $M_x$ and $M_y$ in 3D), but we know it's zero so we don't need to solve for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then why is the right-hand side equal to 3 (or 6 in 3D)? Well, that's precisely because that's how many global equilibrium equations we have: $\sum F_x=\sum F_y=\sum M_z=0$ (and $\sum F_z=\sum M_x=\sum M_y=0$ in 3D). So there's the global moment equation you were looking for.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-08-08T20:16:09.580" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23128" PostTypeId="2" ParentId="14323" CreationDate="2018-08-08T20:23:32.043" Score="1" Body="&lt;p&gt;For a stable and determinant truss we can say that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(# External Reactions) + (# Members) = 2*(# Joints)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why we only have two equations per joint and not three (i.e. why we ignore moment) is largely due to the assumptions we made when dealing with a truss to begin with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical assumptions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Members are straight and carry only axial load&lt;/li&gt;&#xA;&lt;li&gt;Joints are frictionless pins (cannot carry moment)&lt;/li&gt;&#xA;&lt;li&gt;Loads are applied only at the joints&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As a result of these assumptions, each joint represents a concurrent force system. That is, all forces pass through a common point. If we consider our handy statics potato, we can see that this mean moment equilibrium is automatically satisfied. We can only write the trivial equation ‘sum of moments = 0’. None of our unknowns of interest participate in this equation and therefore the ‘sum of the moments’ equation cannot help us solve for our member forces. Therefore we only have two equations available per joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1TxqNl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1TxqNl.jpg&quot; alt=&quot;Concurrent Force System&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going back to that initial equation then. When we say “determinant” all we mean is the structure under consideration can be solved with the equations of statics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this, we must have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Number of things we want to solve for = Number of equations we can write&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We’ve established that the number of equations we can write = 2*No. Joints. The things we want to solve for are the external reactions and the internal member forces (axial only because of our initial assumption).&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2018-08-08T20:23:32.043" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23130" PostTypeId="1" AcceptedAnswerId="23144" CreationDate="2018-08-08T20:43:38.307" Score="1" ViewCount="128" Body="&lt;p&gt;I have a 2D lidar mounted to a plate on a robot. I am looking for a method to adjust the pitch and roll of the plate so the plane of the laser is parallel to the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I am using thin metal shims, but I want to find an &lt;em&gt;easier&lt;/em&gt; method that will last. I was thinking about using set screws, but I've been told that it's a bad idea because they will inevitably come loose due to vibrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other methods exist for adjusting the roll and pitch of a mounting plate? Are there any specially engineered solutions for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, I'm using a Hokuyo 20Lx lidar.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VpSfn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VpSfn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7503" LastActivityDate="2018-08-09T19:03:49.670" Title="Method for permanent adjustment of pitch and roll of a mounting plate" Tags="|mechanical-engineering|calibration|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23131" PostTypeId="2" ParentId="23098" CreationDate="2018-08-08T22:29:06.997" Score="0" Body="&lt;p&gt;You have some assumptions, so it's important to clarify the mechanism you are describing.  I'll be assuming that you are describing a container filled with LN2 at the boiling point of LN2 under 1 atmosphere of pressure.  Then this container is then left out in a room temperature environment while sealed off.  The end result is - your container explodes.  Proving it is complicated, but you wind up at pressures that far exceed most container's withholding pressure.  While I know some thermodynamics, it's not my strongest area, so this may not be technically correct, but it gets a good ballpark estimate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We begin by considering the phase diagram of nitrogen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KMQuy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KMQuy.jpg&quot; alt=&quot;Phase Diagram of Nitrogen showing critical point at 5 MPa and -150C&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZPMTT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZPMTT.jpg&quot; alt=&quot;ST Diagram of LN2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Link &lt;a href=&quot;https://www.researchgate.net/figure/The-temperature-pressure-phase-diagram-for-nitrogen_fig1_315888614&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Enthalpy#Diagrams&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; because images are broken as I write this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would begin our journey at point &quot;g&quot; on the ST diagram.  The two phases can be considered by considering a fraction x at point &quot;h&quot; on the diagram, at gas phase.  We put a permeable barrier between the two phases and begin a set amount of heat flow into the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the internal energy increases, some portion at &quot;g&quot; crosses the barrier into &quot;h&quot;.  The result increases in pressure and entropy.  As such, we move further up the dark red lines, since each phase can cross the barrier.  After enough energy, the system reaches the critical point, at top of the dark red peak.  I'm not sure how to write the equations to calculate how much energy or what the time it would take, but once we reach this point, we've reached supercritical nitrogen between 20-50 bars.  This difficulty is resolved easily in practical engineering via the use of a &lt;a href=&quot;https://en.wikipedia.org/wiki/Steam_drum&quot; rel=&quot;nofollow noreferrer&quot;&gt;steam drum&lt;/a&gt;, a piece of process equipment that safely separates the phases while preventing &lt;a href=&quot;https://en.wikipedia.org/wiki/Bumping_(chemistry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;bumping&lt;/a&gt;.  However, advanced techniques today allow the resolution of multi-phase flow and multi-phase heat transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point, the process would follow an &lt;a href=&quot;https://en.wikipedia.org/wiki/Isochoric_process&quot; rel=&quot;nofollow noreferrer&quot;&gt;isochoric&lt;/a&gt; process.  While hard to describe using a TS diagram, a &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/6/63/Temperature-entropy_chart_for_steam%2C_US_units.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;similar diagram&lt;/a&gt; for steam will confirm that isochoric processes gain relatively little entropy compared to the pressure they gain with a rise in temperature.  Doubling the temperature nearly has a 10-fold increase in pressure.  With such a significant rise, the only conclusion I can arrive at is that you would be at a pressure that would be best measured in units of &lt;a href=&quot;https://en.wikipedia.org/wiki/Orders_of_magnitude_(pressure)&quot; rel=&quot;nofollow noreferrer&quot;&gt;GigaPascals (GPa)&lt;/a&gt;.  In short, this would most likely destroy your container.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-08-08T22:29:06.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23132" PostTypeId="1" CreationDate="2018-08-08T22:32:05.790" Score="1" ViewCount="104" Body="&lt;p&gt;I'm planning on building a plant growth chamber inside a sealed/insulated 55&amp;nbsp;g rain barrel, and I'm trying to figure out how much of the heat from my lights will enter the chamber, and how much will leave through the heat sinks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plan on attracting 3x3000k 80cri Vero 29 gen 7 COB LEDs to the lid and running around 120-150&amp;nbsp;W/h total through them. According to their &lt;a href=&quot;https://www.bridgelux.com/sites/default/files/resource_media/DS93%20Vero%2029%20Array%20Data%20Sheet%20Rev%20A%2020160622.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;datasheet&lt;/a&gt;, their efficiency runs around 130-190&amp;nbsp;lumens/W depending on how hard you drive them which I have read is around 40-55% efficient (although Wiki seems to disagree with that). I believe this means that they will produce 54-82.5&amp;nbsp;W/h of heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be attaching both COBs to their own Arctic apline 11 plus active heatsinks rated for 100&amp;nbsp;W (see &lt;a href=&quot;https://www.arctic.ac/us_en/alpine-11-pro-rev-2.html#techspecs&quot; rel=&quot;nofollow noreferrer&quot;&gt;datasheet&lt;/a&gt;). They are aluminium and the fan moves 37.8cfm over its fins, but surface area and mass values for the aluminum portions are not provided by the manufacturer. That being said it’s total mass is 428&amp;nbsp;g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really would like to know what fraction of heat will be diverted into the growth chamber vs what fraction of heat will exit through the heat sink, so I can figure out if using DIY thermoelectric cooling is a viable option to combat this heat and keep the inside of the chamber no warmer than 85°F/30°C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so this will allow me to build a sweet 55 gallon growth chamber that can utilize CO2 injection while being temperature controlled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other variables:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Max ambient temperature 35°C&lt;/li&gt;&#xA;&lt;li&gt;Inside walls of container have an Emissivity value of 0.044&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17082" LastEditorUserId="1832" LastEditDate="2018-08-08T22:46:43.090" LastActivityDate="2018-10-11T21:00:23.833" Title="how would one calculate the % of heat not absorbed by a heat sink?" Tags="|thermodynamics|heat-transfer|energy-efficiency|building-physics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23133" PostTypeId="1" AcceptedAnswerId="23229" CreationDate="2018-08-09T01:03:32.013" Score="-1" ViewCount="62" Body="&lt;p&gt;I was looking at some papers related to moving bed Biofilm reactor and it says that it can be used both in anoxic and oxic condition. So, I want to know what are the advantages of treatment plants that can work in both anoxic and oxic conditions?&lt;/p&gt;&#xA;" OwnerUserId="17083" LastActivityDate="2018-08-13T20:58:40.590" Title="What are the advantages of wastewater treatment technology that can work in both anoxic and oxic condition?" Tags="|waste-water-treatment|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23135" PostTypeId="1" AcceptedAnswerId="23136" CreationDate="2018-08-09T08:30:34.147" Score="6" ViewCount="10815" Body="&lt;p&gt;There are two square tubes of the same length, material and wall thickness, one being 40x40mm, the other 20x20mm. How does their strength against bending compare?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the one hand, the cross-sectional area of the 40mm tube is 4 times bigger. On the other hand, I guess it is mostly the two walls that are in the same plane as the bending force that work against it. The other two walls provide very little resistance comparing to the first two. And because the surface of the &quot;working&quot; walls of the 40mm tube is twice as much bigger as those of the 20mm one, I am inclined to think that the 40mm tube is roughly twice as much stronger. How does this conclusion stand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, this question is about galvanized steel tubes with 1.6mm thick walls:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ewW7N.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ewW7N.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking at using them as two or more horizontal beams 2.4m long which will take some evenly distributed load. The vertical space is limited so that I would rather use four/eight/sixteen 20mm tubes than two 40mm ones. But how many 20mm tubes would be roughly equal to two 40mm ones?&lt;/p&gt;&#xA;" OwnerUserId="17085" LastEditorDisplayName="user14407" LastEditDate="2018-08-09T08:36:08.740" LastActivityDate="2018-08-15T03:16:36.467" Title="How does the strength of a square tube depend on the side size?" Tags="|structural-engineering|materials|steel|strength|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23136" PostTypeId="2" ParentId="23135" CreationDate="2018-08-09T09:51:20.257" Score="10" Body="&lt;p&gt;Flexural capacity is based on the stress at the extreme fiber (the point farthest vertically from the neutral axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma = \frac{My}{I}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, rearranging&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M = \sigma \frac{I}{y} = \sigma S$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M = moment capacity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\sigma$ = allowable stress&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S = I/y = section modulus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the relative capacity of the tubes will be a function of their section moduli.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a rectangular tube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$S = \frac{BH^2}{6} - \frac{bh^3}{6H}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DiHczm.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DiHczm.jpg&quot; alt=&quot;rectangular tube&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2018-08-09T09:51:20.257" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23138" PostTypeId="2" ParentId="23006" CreationDate="2018-08-09T12:52:29.060" Score="4" Body="&lt;p&gt;You didn't account for the acceleration of $m_1$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Setting up a free body diagram on the weight shows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$m_1g - T = m_1a_y$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where $T$ is the tension in the rope, $a_y$ is the acceleration of the block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to the following corrections:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau_1=\mu(2\pi Rh)\frac{R\omega}{a}*R$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The original had a value for force, whereas we need a torque.)  This is because the entire viscous force would operate at a distance R from the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau_2=TR = (m_1g-m_1a_y)R=m_1gR-m_1R^2\frac{d\omega}{dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To account for the tension in the rope properly, the acceleration of $m_1$ must be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau_2- \tau_1=m_2R^2 \frac{d\omega}{dt}$$&#xA;$$m_1gR-m_1R^2\frac{d\omega}{dt} - \mu(2\pi Rh)\frac{R\omega}{a}*R = m_2R^2 \frac{d\omega}{dt}$$&#xA;$$m_1g - \mu(2\pi Rh)\frac{R\omega}{a} = (m_1+m_2)R \frac{d\omega}{dt}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is easily solved into:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \frac{m_1ga}{2\mu\pi R^2h}(1-exp(\frac{-2\mu\pi Rht}{a(m_1+m_2)}))$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly when the exponential disappears, the maximum speed will settle as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\omega = \frac{m_1ga}{2\mu\pi R^2h}$$&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-08-09T12:52:29.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23139" PostTypeId="2" ParentId="23110" CreationDate="2018-08-09T14:00:05.167" Score="1" Body="&lt;p&gt;I am a bit confused here, so I went under the assumption you have a statically undetermined beam with a fixed support (moment resisting) on the left and a roller support (non-moment resisting) on the right.  In either case, this method still applied, but the constants become significantly easier in the case where it is a pinned support (non-moment resisting, but fixed in x-axis) on the left side.  I'll describe the more complicated case and you can see how it can simplify to the less complicated case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing how you want a method that gives the complete picture for all of x, I will suggest what I personally use, and what books such as &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/0071742476&quot; rel=&quot;nofollow noreferrer&quot;&gt;Roark's formulas for stresses and strains&lt;/a&gt; uses, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Macaulay_brackets&quot; rel=&quot;nofollow noreferrer&quot;&gt;Macaulay Brackets&lt;/a&gt;.  These are defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\langle x-a \rangle^n = \begin{cases} 0, &amp;amp; x &amp;lt; a \\ (x-a)^n, &amp;amp; x \ge a. \end{cases}(n \ge 0)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, you can ignore what is in the bracket when x is less than a, and then start from 0 when x is greater than a.  It is even defined at n=0 to be equal to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heaviside_step_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;heaviside step function.&lt;/a&gt;  This forms &lt;a href=&quot;https://en.wikipedia.org/wiki/Macaulay%27s_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Macaulay's method.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's redefine your load function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ w(x) = 30 \frac{kN}{m} \langle x-3 \rangle^0 + 10 \frac{kN}{m^2} \langle x-3 \rangle^1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks a lot more complicated then it did before.  But the same theory applies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The centroid of the load is still at $\frac{\int w(x)*x dx}{\int w(x) dx} $ and has a value equal to $\int w(x) dx$.  We then need to work on the remaining equations.  The reaction forces at each end are - actually unknown if this is a statically indeterminate beam as described above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can determine the shear and bending moment in the beam throughout the section simply by integrating w(x):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(x) = 30\langle x-3 \rangle^1 + 5\langle x-3 \rangle^2 + F_a$$&#xA;$$M(x) = 15\langle x-3 \rangle^2 + \frac{5}{3}\langle x-3 \rangle^3 + F_a*x + M_a $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then can divide by $EI$, the beam property, and continue to integrate to determine the slope and deflection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EI\theta(x) = 5\langle x-3 \rangle^3 + \frac{5}{12}\langle x-3 \rangle^4 + \frac{F_a}{2}x^2 + M_ax + C_1 $$&#xA;$$EIy(x) = \frac{5}{4}\langle x-3 \rangle^4 + \frac{1}{12}\langle x-3 \rangle^5 + \frac{F_a}{6}x^3 + \frac{M_a}{2}x^2 + C_1x + C_2 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see immediately that $C_2 = 0$ because the deflection at a is 0.&#xA;We then see that $C_1 = 0$ because the slope at a is 0.  Finally, we can solve for the reactions.  We can use the fact that the moment at b is 0 on M(x) to get a single equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$M(6) = 0 = 15\langle 6-3 \rangle^2 + \frac{5}{3}\langle 6-3 \rangle^3 + F_a*6 + M_a $$&#xA;$$ 0 = 180 + 6*F_a + M_a $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We then use the deflection equation for the last equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$EIy(6) = 0 = \frac{5}{4}\langle 6-3 \rangle^4 + \frac{1}{12}\langle 6-3 \rangle^5 + \frac{F_a}{6}6^3 + \frac{M_a}{2}6^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$0 = 101.25 + 20.25 + 36F_a + 18M_a$$&#xA;$$0 = 121.5 + 36F_a + 18M_a$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can then easily be resolved into $F_a = -43.3125$, $M_a = 79.875$.  $F_b$ follows shortly thereafter.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-08-09T14:00:05.167" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23140" PostTypeId="2" ParentId="23135" CreationDate="2018-08-09T14:26:40.513" Score="1" Body="&lt;p&gt;The second moment area of a hollow square section, I, can be calculated by subtracting the I of hole from the I of square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We call the outer side ,a and inner side b.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ I = \frac {(a^4-b^4)}{12} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And $$ S = \frac {(a^4-b^4)}{6a} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So roughly, you need three of the smaller size beams for each 40x40 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since the deflection of the smaller beam will be THREE times more under the same load, we need to check on that not to become too excessive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4EksO.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4EksO.gif&quot; alt=&quot;square hollow section&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-08-09T14:32:31.717" LastActivityDate="2018-08-09T14:32:31.717" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23142" PostTypeId="2" ParentId="23133" CreationDate="2018-08-09T16:41:49.737" Score="0" Body="&lt;p&gt;the two processes are used to degrade two different classes of contaminants in the wastewater stream. the oxic/activated sludge process is nearly universal but depending on the discharge regulations, the anoxic process may be required to meet the regs. this means the plant would have to contain both processes. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-08-09T16:41:49.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23143" PostTypeId="2" ParentId="23130" CreationDate="2018-08-09T16:49:44.203" Score="1" Body="&lt;p&gt;You could order two large washers that are cut off a tapered plane, called beveled Washers. by rotating them over each other you get a level platform with ease.&#xA;&lt;a href=&quot;https://shop.stafast.com/washers/square-beveled-washers/malleable-square-beveled-washer&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-08-09T16:49:44.203" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23144" PostTypeId="2" ParentId="23130" CreationDate="2018-08-09T19:03:49.670" Score="1" Body="&lt;p&gt;A simple solution would be :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Two matching square plates with an M8 hole in each corner.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sandwich a bit of moderately thick rubber, 10mm  thick maybe,   with matching holes between the two plates &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;arrange whatever mounting points you need for the lidar and vehicle in the top and bottom plates &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bolt through the holes with bolts any nyloc nuts &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;adjust the bolt tension to level the top plate, the rubber will take up the slack and &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A slightly more sophisticated version is to drill a hole in the centre of the plates whcih you use to locate a ball bearing to act as a pivot point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also drill through some (soft) ball bearings and use them as sort of spherical washers in conjunction with countersunk holes on the top plate to get better location for the bolt heads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a very large range of adjustment you could use something like this (made from  modified pipe flanges) possibly substituting a bolt any nyloc for the Bristol lever, in fact with a smaller ball bearing (or no ball and a robber block or springs that is similar to what I described above.  :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Leolc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Leolc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-08-09T19:03:49.670" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23145" PostTypeId="2" ParentId="23132" CreationDate="2018-08-09T19:36:29.417" Score="1" Body="&lt;p&gt;I believe that you have two possible approaches to an energy balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JvYgy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JvYgy.png&quot; alt=&quot;picture of system in two cases&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;System A&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The walls allow heat flow. The defining equations are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;P_i = q_x + q_a + P_o \\&#xA;q_x = \epsilon_{HS} (1 - \epsilon_L) P_i \\&#xA;q_a = (T_c - T_a) / R_T \\&#xA;R_T = R_c + R_w + R_a \\&#xA;R_c = 1/h_c A_w\ \ R_w = w_w / A_w k_c\ \ R_a = 1/h_a A_w&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balance here is with power input, heat flow out of the fins, heat flow across the container, and cooling power output. The heat flow from the fins is the heat sink efficiency times the heat produced by the lamps as determined from the lamp efficiency (for light production). The heat flow across the wall is a thermal resistance calculation. In this case, you assume / calculate the overall thermal resistance $R_T$. You define the fin efficiency $\epsilon_{HS}$ and lamp efficiency $\epsilon_L$. You set the air temperature $T_a$ and power input $P_i$. What is left is the cooling power that you need $P_o$. The maximum is when $\epsilon_{HS} = 0$ (all lamp heat is dumped to the container) and $\epsilon_L = 0$ (the lamp produces only heat and no light). This approach allows you to size the cooling unit for the maximum required cooling rate $P_{o,max}$ (W) at the output.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;System B&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The system is adiabatic. The defining equations are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;q_L = q_c \\&#xA;q_L = (1 - \epsilon_{HS}) (1 - \epsilon_L) P_i \\&#xA;q_c = (T_c - T_x) / R_c \\&#xA;R_c = 1/h_c A_c&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balance here is heat flow from the lamps and heat flow to the cooling unit. Assume that all heat produced by the lamps is dumped perfectly to the system. The heat flow to the cooling unit is defined by the thermal resistance. This approach allows you to design the temperature, efficiency (convection coefficient), and area of the cooling unit. This will again give a maximum design when $\epsilon_{HS} = 0$ and $\epsilon_L = 0$.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Questions?&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Here are the questions that I find are left to address:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Which system model is more appropriate, conducting walls or insulated walls (System A or System B)?&lt;/li&gt;&#xA;&lt;li&gt;How much heat will the lamps dump to the container (what are $\epsilon_{HS}$ and $\epsilon_L$)?&lt;/li&gt;&#xA;&lt;li&gt;How effective is the cooling unit (what are its nominal values of $h_c$ and $A_c$)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Notes&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;System A assumes the chamber is hotter than the surrounding air. System B does not care.&lt;/li&gt;&#xA;&lt;li&gt;Neglect radiation, especially in System A. Unless something in the system is above a few hundred oF, radiation is of third order consequence.&lt;/li&gt;&#xA;&lt;li&gt;This assumes that all heat produced by the lamps is dumped perfectly to the system. When the lamps are hot above the system, convection is nearly non-existent. The chamber will have a hot zone at the top and a cool zone at the bottom. You must adjust the placement of hot and cold elements accordingly.&lt;/li&gt;&#xA;&lt;li&gt;This analysis only gives a picture of how to find the maximum design parameters. The next step may be to decide how much control you want / need on the cooling unit.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h1&gt;Edit (After the Fact)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The original question was ...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I really would like to know what fraction of heat will be diverted into the growth chamber vs what fraction of heat will exit through the heat sink,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The variable you want to determine is $\epsilon_{HS}$. Based on the system picture, I can suggest a simple experiment. Set up the chamber with everything as desired but exclude the cooling system. Put a thermocouple in the chamber. Run the lamps until the internal temperature stabilizes. Use the equations to back-calculate a value or estimate for $\epsilon_{HS}$.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-08-12T19:52:27.393" LastActivityDate="2018-08-12T19:52:27.393" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23146" PostTypeId="2" ParentId="23006" CreationDate="2018-08-09T21:51:51.840" Score="3" Body="&lt;p&gt;Mark has already pointed out what you are missing, and this is the Lagrangian approach which also should give the same result - and it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of $m_1$  is $x_1 = R \ \theta(t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of $m_1$: $KE_1=\frac{1}{2}m_1x_1'(t)^2 =\frac{1}{2} m_1 R^2 \theta '(t)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The potential energy of $m_1$: $PE_1=-m_1 g x_1=-m_1 g R \ \theta (t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of the $m_2$: $KE_2=\frac{1}{2}I\theta '(t)^2=\frac{1}{2} m_2 R^2 \theta '(t)^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lagrangian: $L=KE_1+KE_2-PE_1=\frac{1}{2} m_1 R^2 \theta&#xA;   '(t)^2+\frac{1}{2} m_2 R^2 \theta '(t)^2+m_1 g R \theta (t)$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The generalized force is the torque $\tau_2$: $Q=-\mu(2\pi Rh)\frac{R\omega}{a}R$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations of motion are $\frac{d }{d t}\frac{d L}{d&#xA;   \theta'(t)}-\frac{d L}{d \theta(t)}=Q$ which turn out as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ m_1 R^2 \theta ''(t)+m_2 R^2 \theta ''(t)- m_1 g R=-\frac{2 \pi  h \mu  R^3 \theta '(t)}{a} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting $\theta '(t)=\omega (t)$ and simplifying we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(m_1+m_2\right) R \omega '(t)+ \frac{2 \pi  h \mu  R^2 \omega (t)}{a}-m_1 g=0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the solution with $\omega(0) = 0$ is $$\omega(t) =\frac{ m_1 g a}{2 \pi  h \mu &#xA;   R^2} \left(1-e^{-\frac{2 \pi  h \mu  R t}{a \left(m_1+m_2\right)}}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which agrees with the free-body diagram approach.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2018-08-09T21:51:51.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23148" PostTypeId="2" ParentId="16301" CreationDate="2018-08-10T04:42:35.803" Score="2" Body="&lt;p&gt;There are several companies out there that can assist with silo/hopper design, which would allow easy offloading of truck to silo,, but that can also offer a solution to use vacuum pneumatic conveying to each machine as you currently have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there are many technical things to consider throughout the system, therefore it will be best to get a company that has delivered similar system to provide a solution. Each of these components are a science in itself and you can spend lots of money on a system that in the end could be useless if not done correctly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these include:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silo design:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Material flow out of the Silo: typical you would want a mass flow system rather than a funnel flow system (this will ensure you use the older material first and reduce the risk of buildup in the silo).&lt;/li&gt;&#xA;&lt;li&gt;Do you want level measurement or weight of the silo to know when you are running low and have to order a new truck. You can add load cells in the supports of the silo to its main structure (typical suupliers will be mettler toledo for load cell and Endress Hauser or Vega for level measurement)&lt;/li&gt;&#xA;&lt;li&gt;Also, the silo must be earthed to ensure you don't have static buildup (typical for plastic pellets)&lt;/li&gt;&#xA;&lt;li&gt;A bag filter for when you fill the silo (assuming you will use a supplier with pneumatic pump fitted to the truck) - a typical supplier will be Donaldson&lt;/li&gt;&#xA;&lt;li&gt;an over under valve or bursting disk for the silo to ensure you don't over pressurise it or accidentally get a vacuum generated within the silo that could cause it to implode&lt;/li&gt;&#xA;&lt;li&gt;Depending on the flow ability of the product, the size of the silo you are considering you will have to consider whether you need devices to assist with the material flow out of the silo. This will be as the material will be under its own pressure and dependent on the shape of the plastic pallets. Typical devices that might get added is an inverted cone to break material pressure and air pads to assist the material to flow down the chute of the silo. alternative materials could also be used to assist with the material flow - stainless steel has better flow than mild steel, ultra-high molecular polyethylene (UHMWPE) sheets can be used to line the chute.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Vacuum pump system:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What type of system do you want to have - Lean or dense phase (dense phase will be at a slightly higher pressure and slower velocity but will be at lower running cost but with higher upfront costs)&lt;/li&gt;&#xA;&lt;li&gt;You will need a feeder that slowly feeds the material to your vacuum line (this could be a star feed or a rotofeeder, which will gauge the amount of material that gets fed).&lt;/li&gt;&#xA;&lt;li&gt;The vacuum pump sized according to the pressure drop of your system and the material flow rate you want to achieve&lt;/li&gt;&#xA;&lt;li&gt;Air inlet filter with valve at you silo&lt;/li&gt;&#xA;&lt;li&gt;Piping sized based on your route and pump sizing&lt;/li&gt;&#xA;&lt;li&gt;Receiving cyclones/hoppers at your units as what you currently have&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Typical global suppliers for these systems will be Flexicon and Schenck Process, but there are smaller companies that can offer similar solutions in every country.&lt;/p&gt;&#xA;" OwnerUserId="17092" LastActivityDate="2018-08-10T04:42:35.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23149" PostTypeId="1" AcceptedAnswerId="23156" CreationDate="2018-08-10T06:31:40.357" Score="0" ViewCount="227" Body="&lt;p&gt;&lt;a href=&quot;https://www.multitran.ru/c/m.exe?a=4&amp;amp;MessNum=349503&amp;amp;l1=1&amp;amp;l2=2&quot; rel=&quot;nofollow noreferrer&quot;&gt;A fellow translator is translating a list of parts into Russian&lt;/a&gt;, and can't understand the meaning of &quot;female eye&quot;. Here's the table: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2mWhw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2mWhw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Maybe someone here could come up with some ideas about the possible meaning of this &quot;female eye&quot;. &lt;/p&gt;&#xA;" OwnerUserId="349" LastActivityDate="2018-08-12T14:54:18.423" Title="Meaning of &quot;female eye&quot; in a list of parts" Tags="|terminology|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23151" PostTypeId="1" CreationDate="2018-08-10T08:11:28.387" Score="1" ViewCount="447" Body="&lt;p&gt;Is it possible to calculate torque required to rotate wheel but weight of wheel is unknown?&lt;/p&gt;&#xA;" OwnerUserId="17097" LastActivityDate="2018-08-10T08:55:56.253" Title="How can we calculate torque required to rotate the wheel (mass unknown)" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23153" PostTypeId="1" CreationDate="2018-08-10T08:15:21.413" Score="4" ViewCount="108" Body="&lt;p&gt;I'm building a Wien Bridge oscillator. The oscillation part works just fine, but it currently limits on the supply rails. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to add amplitude control, so I've added a precision rectifier which feeds an integrator that controls an LDR to change the gain of the circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This setup makes the output amplitude oscillate. To get a better understanding of what happens I want to construct a block diagram of the amplitude control and use control theory to improve the design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm already a bit stuck on building the block diagram. I've the following:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ahUQG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ahUQG.png&quot; alt=&quot;Block diagram of amplitude control&quot;&gt;&lt;/a&gt;&#xA;Where the 3 is the ideal gain, to compensate the $\frac{1}{3}$ of the frequency selective network. The $\delta R$ is the change in the gain resistor due to the LDR. Since the frequency selective network has 0 phase shift at the oscillation frequency I believe there is no need to put in an actual transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've modelled the rectifier with a gain of 1, as it doesn't really change the amplitude it just presents it in a different form. The $\frac{1}{sC}$ is the integrator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this somewhat correct? How can I now add the fact that the integrator output alters the $\delta R$?&lt;/p&gt;&#xA;" OwnerUserId="17095" LastActivityDate="2018-11-11T20:01:25.283" Title="Constructing a block diagram for amplitude control of an oscillator" Tags="|control-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23154" PostTypeId="1" AcceptedAnswerId="23166" CreationDate="2018-08-10T08:43:41.383" Score="-1" ViewCount="70" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hzopt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hzopt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;A homogeneous rigid circular disc (radius R, mass ms, center point S) rolls without a slip on a plain with a small angle of inclination alpha against the horizontal. A point mass MG is fixed on the disc in a distance e to the center point S. The motion of the disc can be described by the angle phi between the vertical and the connection line SMG. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see the figure for reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to calculate Kinetic Energy and Potential energy of the given problem (pic #2) and I have considered both masses as one system but I am not sure if that's correct or not. Also, I am not sure if the h term in potential energy (mgh) is correct here. So I would be glad if someone could throw some light on this. Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="16862" LastEditorUserId="16862" LastEditDate="2018-08-10T16:29:18.283" LastActivityDate="2018-08-10T20:36:20.800" Title="Analytic expressions of the kinetic and potential energy" Tags="|mechanical-engineering|applied-mechanics|vibration|mechanical|linear-motion|" AnswerCount="1" CommentCount="1" ClosedDate="2018-08-13T14:34:35.053" ContentLicense="CC BY-SA 4.0" />
  <row Id="23156" PostTypeId="2" ParentId="23149" CreationDate="2018-08-10T11:34:27.727" Score="3" Body="&lt;p&gt;&quot;Eye&quot; is very non specific, but I expect it to be an item with a female thread (like a nut). This could take any number of forms, however. I've illustrated two below, but more information from context etc. would be required for a more specific answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;Female Eye Bolt&quot; - Yes, it gets called '... bolt' despite being more like a nut. You would use this if, for example, you had studding welded to something.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/54qmtm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/54qmtm.png&quot; alt=&quot;female eye bolt&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one is a female &quot;Heim Eye Bolt&quot;, sometimes called a &quot;Rod End&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WZbC1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WZbC1.png&quot; alt=&quot;Female Rod End&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully that give you a bit of an idea, and a route down which you can continue your own research.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-08-10T11:34:27.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23157" PostTypeId="2" ParentId="18144" CreationDate="2018-08-10T12:12:48.537" Score="0" Body="&lt;p&gt;It's not possible to give a discrete answer only by knowing what the geometries are. There are many factors affecting penetration. You still need to give more data about your analysis to get a better answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should first make sure of Geometries, that are well assembled and are input with a good format. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good practice is adjust to touch and pinball radius. If not you should then check other contact parameters.&lt;/p&gt;&#xA;" OwnerUserId="17070" LastEditorUserId="90" LastEditDate="2018-08-10T15:07:47.370" LastActivityDate="2018-08-10T15:07:47.370" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23158" PostTypeId="2" ParentId="21239" CreationDate="2018-08-10T16:54:35.913" Score="2" Body="&lt;p&gt;Your main assumption, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If you put a shaft on those bearings it's meant to rotate, you don't&#xA;  allow a shaft to go through a bearing, it's always fixated in one way&#xA;  or another&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;does not hold true in many cases. For example the only thing that keeps a bicycle wheel from sliding around are the bearings (if anything but the ball bearing, you'd have a sliding point of contact which would mean more friction and wear). Another example is the free spinning carriage wheel axle set that was the need fulfilled by the &lt;a href=&quot;http://web.mit.edu/2.75/fundamentals/FUNdaMENTALs%20Book%20pdf/FUNdaMENTALs%20Topic%2010.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;Timken roller bearing&lt;/a&gt; (p 18). Rollers can take a much heavier load than balls. Until then that required a set of radial bearings and another set of thrust bearings. If simple roller bearings take an axial load the ends of the cylinders are sliding against the cage with force. Timken angled the profile of the bearing to make it be used with another like a pair of opposing angular ball bearings, but the key to that was making them &lt;em&gt;tapered&lt;/em&gt;. A plain cylinder can't roll in a circle on a cone without sliding a little over much of it's length. So the races and the cylinders all tapered to the same point allowing all the parts to roll together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when you throw an engine in the middle of that axle (a car) that's connected to the frame as is the outer race of your wheel bearing you don't want all the axial forces from the wheels taken up by expensive parts in the transmission. You still want the bearing to handle that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deep groove radial ball bearings can take axial force. The balls can roll slightly in the groove which takes the axial force. Angling the groove allows it to take more thrust in that direction. There are simple diagrams in this &lt;a href=&quot;http://www.nachiamerica.com/content/downloads/Nachi_Bearing_Training_Manual_NAB1202-3_rev_May_2017_small.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nachi training guide&lt;/a&gt; showing the relative loads different bearing designs can handle.&lt;/p&gt;&#xA;" OwnerUserId="17099" LastEditorUserId="17099" LastEditDate="2018-08-20T15:23:36.390" LastActivityDate="2018-08-20T15:23:36.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23159" PostTypeId="2" ParentId="23149" CreationDate="2018-08-10T17:11:16.720" Score="2" Body="&lt;p&gt;On some hydraulic cylinders there are different attachments on the ends called mounts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kY78D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kY78D.png&quot; alt=&quot;Female Eye&quot;&gt;&lt;/a&gt;&#xA;Female Eye&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SgqOK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SgqOK.jpg&quot; alt=&quot;Pin Eye&quot;&gt;&lt;/a&gt;&#xA;Pin Eye&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ebay.com/itm/New-Clevis-Bracket-Fits-40-mm-Bore-Cylinders-/181951183859&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nMFOf.jpg&quot; alt=&quot;Clevis&quot;&gt;&lt;/a&gt;&#xA;Clevis&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These go on the ends of some hydraulic cylinders like on this rack and pinion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FqnE5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FqnE5.jpg&quot; alt=&quot;Rack and Pinion&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14556" LastEditorUserId="14556" LastEditDate="2018-08-12T14:54:18.423" LastActivityDate="2018-08-12T14:54:18.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23160" PostTypeId="1" AcceptedAnswerId="23298" CreationDate="2018-08-10T17:46:18.170" Score="0" ViewCount="692" Body="&lt;p&gt;I need some help with a statics design problem. I am tasked with designing a solar panel frame to mount on a building and am the only engineer on staff so I don't have anyone else to ask the simple dumb questions. Since I have no real experience doing wind load calculations I figured I would assume a load normal to the solar panel face. My idea is to use two square aluminum tubes mounted along the long side of the three solar panels stacked vertically along the tube. My analysis involves the support structure for one side, of one array column. As of right now I am hoping the solar panels themselves will suffice as cross-member connections. The details really don't matter, in fact the wind loading really doesn't matter either. It is the method I am having trouble with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Want it to be able to withstand 125 MPH gusts. That appears to equate to 30 lbf/square foot. Given 18.47 sq. ft./ panel * 30PSF * 3 Panels + 135 lb panel wt = 1797 lbf. That's 1800 lbf total distributed load accross all three solar panels in one array column. Since there will be two support structures per array column that's 900 lbf distributed load per support structure or 4.48 lbf/inch. (Not sure where I got 6.2 on my drawing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I need to determine the reactions at each pin joint, do shear and moment diagrams and then I will do a buckling failure analysis on each support strut member. Any guidance would be appreciated, started doing some calcs on my own and things just weren't coming out right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DFSVZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DFSVZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16668" LastEditorUserId="16668" LastEditDate="2018-08-10T18:30:38.803" LastActivityDate="2018-08-17T14:45:07.920" Title="Solar Array Frame Design / Statics Design Problem" Tags="|structural-engineering|structural-analysis|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23161" PostTypeId="1" AcceptedAnswerId="23181" CreationDate="2018-08-10T18:28:44.153" Score="-1" ViewCount="69" Body="&lt;p&gt;I understand that the lead ions can be rendered into a precipitate with chloride ions in water treatment, but why on earth does the U.S. still use lead pipes at all when contamination poses an enormous risk? &lt;/p&gt;&#xA;" OwnerUserId="17102" LastEditorDisplayName="user14407" LastEditDate="2018-08-10T21:50:52.600" LastActivityDate="2018-08-11T00:35:06.750" Title="Why does the U.S. still use lead pipes?" Tags="|waste-water-treatment|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23162" PostTypeId="1" CreationDate="2018-08-10T18:33:28.430" Score="2" ViewCount="68" Body="&lt;p&gt;Would it be best to place the earth protection (the copper spike) for a building on a slope downhill (A) or uphill (B) from the structure? Assume that lightning is most likely to strike at (C), the hill which is also much higher than the lightning rod on top of the building. Assume the soil is rocky and there is no way to drive the stake in at B far enough to get BELOW the foundation.&lt;a href=&quot;https://i.stack.imgur.com/Fm3my.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fm3my.png&quot; alt=&quot;building on a slope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17101" LastActivityDate="2018-08-11T19:30:07.777" Title="For a building on a slope, should the grounded earth lightning protection be downhill or uphill from the structure?" Tags="|electrical-engineering|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23163" PostTypeId="2" ParentId="23162" CreationDate="2018-08-10T18:34:19.297" Score="2" Body="&lt;p&gt;I got this response from Bud at lightning.org:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I can give you my opinion by answering the question “Does it matter?” with a qualified NO. The nature of lightning is that it wants to reach earth ground to neutralize the charge imbalance created in the clouds. Once the lightning reaches earth ground at any point, it spreads out over the surface depending on the electrolytic content of the soil strata until neutralized. I’m not sure it makes a lot of difference, all things being equal, where you place it into the earth, the effect will be the same.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;On the other hand the soil itself makes a difference. Lightning will normally seek the lower resistance ground path. If you have a ground in good clay/loam soil with moisture versus a ground driven into rock or sand with little to no moisture content, the lightning will seek the better path to ground. In system design, we normally try to split the lightning along multiple paths to minimize the charge on any one lead and avoid side-flashing to interior systems. I’m aware of a project I worked on many years ago, similar to what you have drawn, with a square building and all four corners grounded – 2 along line B and 2 along line A. Along line B they had blasted out of a cliff for the project making very rocky conditions on this side. Line A was good dirt backfill. Lightning struck this building nearer the B roofline and the majority charge started down those leads, but at basement level it came into close proximity with the electrical grounding inside – flashed through the basement wall and onto the electrical grounding which had been interconnected with the lightning protection grounds on the A building side only. This caused much damage to equipment connected electrically in the basement. The problem was resolved by installing a ground loop, interconnecting all the grounds to maintain equal potential on all downleads and system elements. This is why I think the grounding medium makes more difference than the uphill/downhill side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17101" LastEditorUserId="1832" LastEditDate="2018-08-11T13:50:04.510" LastActivityDate="2018-08-11T13:50:04.510" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23164" PostTypeId="1" AcceptedAnswerId="23180" CreationDate="2018-08-09T22:03:35.710" Score="0" ViewCount="75" Body="&lt;p&gt;I have a peltier (TEC) device capable of 400W Qmax, a 160W thermal load that I'd like to keep at ambient, and &lt;a href=&quot;https://customthermoelectric.com/water-block-3-0-x-3-0-x-0-85.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this water block&lt;/a&gt; running at about 1.5 gpm w/ room temp water. I think this system is right on the border of what is possible, but I'm struggling to determine the exact efficiency of my peltier unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The datasheet for &lt;a href=&quot;https://customthermoelectric.com/19911-5m31-28cz-thermoelectric-peltier-module.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;my peltier&lt;/a&gt; indicates a Qmax of 400W and claims that maximum applied power is about 690W -- does this mean that I have a 57% efficiency? To me that seems extraordinarily high for a TEC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I don't believe 57% is right, let's assume a 30% efficiency. In that case, my peltier will need to be operated at around 533W in order to dissipate the 160W load + the load from the peltier. And now, I have 693W total being dumped into my water block. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I doing this calculation correctly? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other notable factors:&#xA;- This system will be operated in a vacuum environment so convective cooling is a no-go. &#xA;- There is a thin (~1/8&quot;) aluminum plate between the heat load and the peltier.&lt;/p&gt;&#xA;" OwnerUserId="17105" OwnerDisplayName="Austin Prater" LastActivityDate="2018-08-10T23:47:14.300" Title="Peltier Calculation Conundrum" Tags="|thermodynamics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23165" PostTypeId="2" ParentId="23164" CreationDate="2018-08-09T22:54:15.117" Score="0" Body="&lt;p&gt;TEC efficiency (COP=$Q_C/P_{el}$) depends on many factors, including temperature difference, and can exceed 1 (&lt;a href=&quot;https://www.meerstetter.ch/compendium/peltier-element-efficiency&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.meerstetter.ch/compendium/peltier-element-efficiency&lt;/a&gt;). So one cannot say that 57% is too high without a more detailed analysis.&lt;/p&gt;&#xA;" OwnerUserId="1463" OwnerDisplayName="akhmeteli" LastActivityDate="2018-08-09T22:54:15.117" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23166" PostTypeId="2" ParentId="23154" CreationDate="2018-08-10T20:36:20.800" Score="1" Body="&lt;p&gt;The calculations (in Mathematica) with minimal explanations. Hopefully you can pick it up from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D0c2E.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D0c2E.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2018-08-10T20:36:20.800" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23167" PostTypeId="2" ParentId="22690" CreationDate="2018-08-10T21:49:00.633" Score="0" Body="&lt;p&gt;1-3 mm is perfectly doable with a cheap-o, off the shelf, carbide glass cutter. So long as your shapes are generally large sweeping curves (i.e. few acute corners), you can use something like &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B078YS9KJY&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; to score then glass, then something like &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B005I67ZMW&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; to cleanly snap the piece. &lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-10T21:49:00.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23168" PostTypeId="2" ParentId="23095" CreationDate="2018-08-10T22:00:18.133" Score="1" Body="&lt;p&gt;I will echo what Chuck said -- this is a pretty simple/ common problem. A basic shop winch will suffice just fine. There are lots of cheap ones on the market (some even come with fancy remotes!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check out this one --&gt; &lt;a href=&quot;https://www.northerntool.com/shop/tools/product_200672565_200672565?cm_mmc=Google-pla&amp;amp;utm_source=Google_PLA&amp;amp;utm_medium=Winches%20%3E%20AC%20Powered%20Winches&amp;amp;utm_campaign=Ironton&amp;amp;utm_content=54129&amp;amp;gclid=EAIaIQobChMI2cC6ub3j3AIVFMpkCh3rJguCEAQYBCABEgJjKPD_BwE&quot; rel=&quot;nofollow noreferrer&quot;&gt;1500 shop winch, for example&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-10T22:00:18.133" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23169" PostTypeId="2" ParentId="23102" CreationDate="2018-08-10T22:05:47.320" Score="0" Body="&lt;p&gt;Peltier modules are extremely inefficient and therefore need excessive heat dissipation. What you are experiencing is that the peltier module cannot dissipate all of the heat that it is moving (either from a heat load or from itself) and as a result it cannot maintain an appropriate delta-Temp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest adding a larger heat sink, a fan blowing over the heat sink (good 'ol convective cooling), or a water block. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See some options below:&#xA;Fan: &lt;a href=&quot;https://www.newegg.com/Product/Product.aspx?Item=N82E16835230010&amp;amp;ignorebbr=1&amp;amp;nm_mc=KNC-GoogleAdwords-PC&amp;amp;cm_mmc=KNC-GoogleAdwords-PC-_-pla-_-CPU%20Cooling-_-N82E16835230010&amp;amp;gclid=EAIaIQobChMIgJCd8L7j3AIVDttkCh3MTw2lEAYYAiABEgJRHvD_BwE&amp;amp;gclsrc=aw.ds&quot; rel=&quot;nofollow noreferrer&quot;&gt;Click me!&lt;/a&gt;&#xA;Water Block: &lt;a href=&quot;https://customthermoelectric.com/water-block-3-0-x-3-0-x-0-85.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Click me!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-10T22:05:47.320" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23171" PostTypeId="2" ParentId="23014" CreationDate="2018-08-10T22:15:32.973" Score="1" Body="&lt;p&gt;In short:&#xA;What you want is to MAXIMIZE surface area and utilize a proper environment for condensation (humidity, temp, and dew point). Environment is everything and surface area will provide the most material for water vapor to condense onto. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look into designs for common heat sinks and you'll find a solution.&lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-10T22:15:32.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23172" PostTypeId="2" ParentId="22976" CreationDate="2018-08-10T22:19:19.140" Score="0" Body="&lt;p&gt;What Daniel said is the best, correct answer. To add my two cents, keep in mind that a device will &lt;strong&gt;only ever pull as much current as it needs&lt;/strong&gt;. Most devices have a constant, non-changing impedance -- just a simple case of &lt;em&gt;V=IR&lt;/em&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-10T22:19:19.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23173" PostTypeId="2" ParentId="22888" CreationDate="2018-08-10T22:23:18.683" Score="1" Body="&lt;p&gt;If you like living, do not attempt to build your own high-pressure LN2 tank if you have never designed commercial-grade pressure vessels before. &lt;a href=&quot;https://www.thermofisher.com/order/catalog/product/8121?SID=srch-srp-8121&quot; rel=&quot;nofollow noreferrer&quot;&gt;Use this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-10T22:23:18.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23174" PostTypeId="1" AcceptedAnswerId="25228" CreationDate="2018-08-10T22:23:22.430" Score="3" ViewCount="738" Body="&lt;p&gt;I have recently been working on a gearbox for a robot.  I have this pattern of 4 gears with each gear connected to its 2 neighbors.  It looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ry2HY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ry2HY.png&quot; alt=&quot;Early Drawings&quot;&gt;&lt;/a&gt;&#xA;You will probably notice something though.  Large gear on the right and the small one in the center do not properly mesh.  I made sure that the center to center distance of the all the gears was the sum of their respective pitch radii so that is not the issue:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1iHns.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1iHns.png&quot; alt=&quot;The issue&quot;&gt;&lt;/a&gt;&#xA;There only seems to be a few positions where the gears engage one another properly.  By moving things around and eyeballing it I got a set of constraints where everything appears to mesh properly but it is unlikely that it is perfect:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aeQy3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aeQy3.png&quot; alt=&quot;eyeballed fix&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lUMAz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lUMAz.png&quot; alt=&quot;sketch&quot;&gt;&lt;/a&gt;&#xA;So my question is how do I calculate distance between the small gear in the center and the large one on the bottom so the gears will all mesh properly.  I don't like eyeballing it because I don't think the gears will mesh perfectly. &#xA; Also, it makes it harder to improve my design if I can't make these calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;br&gt;&#xA;Ben&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some specs on the gears:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Large Gears:  44 tooth, 2.2&quot; pitch diameter, 2.3&quot; outer diameter, 14.5 deg pressure angle, 20 Diametrical Pitch.&lt;br&gt;&#xA;Small Gears:  12 tooth, 0.6&quot; pitch diameter, 0.7&quot; outer diameter, 14.5 deg pressure angle, 20 Diametrical Pitch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar question that got no answer: &lt;a href=&quot;https://engineering.stackexchange.com/questions/2637/how-to-find-angle-of-timing-gears?rq=1&quot;&gt;How to find angle of timing gears&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17104" LastEditorUserId="17104" LastEditDate="2018-08-10T22:57:46.843" LastActivityDate="2018-12-24T01:08:37.660" Title="4 Gears in a Circular Pattern - Calculating the Proper Distances for Meshing" Tags="|mechanical-engineering|gears|robotics|mechanical|kinematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23175" PostTypeId="2" ParentId="22615" CreationDate="2018-08-10T22:42:18.787" Score="2" Body="&lt;p&gt;Let me just start by saying... &lt;em&gt;what an awful unit of measure&lt;/em&gt;. Tons/cm^2?? Whew... that's gut wrenching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyways, this range of pressure transducer certainly exists, albeit they are expensive. These are easiest to find by using the keywords &quot;ultra high pressure transducer&quot;. Here is an example --&gt; &lt;a href=&quot;https://www.omega.com/pptst/PX91-5VC.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Click me!&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-10T22:42:18.787" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23176" PostTypeId="2" ParentId="22555" CreationDate="2018-08-10T22:59:27.840" Score="1" Body="&lt;p&gt;eatscrayons suggestion for leak detection is great. However, instead of acetone, I would use isopropyl alcohol (or even better, a helium gas source since He is such a small molecule). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is certainly a small roughing pump so I would not be surprised if you're hitting a floor of 6 Torr. Time to start looking into larger roughing pumps -- those which use oil tend to be able to achieve harder vacuum levels! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're trying to reach pressures &amp;lt;50 mTorr (dubbed &quot;high vaccum&quot;) you will almost certainly need to introduce a cryo-pump or a turbo-molecular pump. &lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-10T22:59:27.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23177" PostTypeId="2" ParentId="22309" CreationDate="2018-08-10T23:13:05.833" Score="1" Body="&lt;p&gt;Sounds like you're looking for some sort of &lt;strong&gt;regulator&lt;/strong&gt;, not a valve. You did not give any indication to the size or scale of this device, so a link to a specific PN would likely not be useful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, regulators are pressure-reducing devices; most of which operate passively, without electromechanical components. I think that this video, which describes the basic principle of a reg, will help you. &lt;a href=&quot;https://www.youtube.com/watch?v=Kf3xc1BGYGo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Single Stage Regulator: Working Principle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-10T23:13:05.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23178" PostTypeId="2" ParentId="22267" CreationDate="2018-08-10T23:18:36.390" Score="1" Body="&lt;p&gt;Viton is about as close to &quot;impermeable&quot; as it gets. Viton gaskets are what are commonly used for ultra-high vacuum chamber systems which can achieve vacuum levels on the order of 10E-12 Torr. &lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-10T23:18:36.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23179" PostTypeId="2" ParentId="22156" CreationDate="2018-08-10T23:23:46.180" Score="1" Body="&lt;p&gt;Flow meters can also be used for high-precision applications. Imagine wanting to measure flow of gas, on the order of 0.1 sccm through a very small tube. &lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-10T23:23:46.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23180" PostTypeId="2" ParentId="23164" CreationDate="2018-08-10T23:47:14.300" Score="1" Body="&lt;p&gt;Your water is at 20 C and the block is at -15 C.  Temperature differential is 35 C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a pretty big gradient for the TEC. The plot of Q vs I says that you can shift no more than about 125 W but you'll need 28 A and that will cost you about 675 W. (this is from the plots on the data sheet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the source of heat is 25 W or so it will require 10 A and cost you about 100 W (10 V * 10 A). This is where I would use the TEC you specified.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to indicate that you'll need either:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;To settle for a lower amount of heat shifted &lt;em&gt;or&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;More TECs in series to reduce the temperature gradient of each one &lt;em&gt;or&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;More TECs in parallel to increase the heat pumped &lt;em&gt;or&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;Settle for a smaller temperature differential&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;or a combination of the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: You'll need to do option 3 when you apply option 2 because now you need to get rid of the energy used to pump the heat from the colder stage.&lt;/p&gt;&#xA;" OwnerUserId="16544" LastActivityDate="2018-08-10T23:47:14.300" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23181" PostTypeId="2" ParentId="23161" CreationDate="2018-08-11T00:35:06.750" Score="1" Body="&lt;p&gt;Lead pipes have not been used in new construction in the USA for many decades. The problem is with old buildings and old water lines which may still be equipped with lead pipes if they have not been upgraded since new. Copper pipes for residential and commercial construction were common for the last 50 years or so and since they were assembled on-site using lead solder, it is possible to produce lead contamination of water flowing through them even if the copper pipes themselves contain no lead. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-08-11T00:35:06.750" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23182" PostTypeId="1" AcceptedAnswerId="23183" CreationDate="2018-08-11T00:41:09.713" Score="1" ViewCount="994" Body="&lt;p&gt;From wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Before opening either door, the air pressure of the airlock—the space between the doors—is equalized with that of the environment beyond the next door to open&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Anyone knows how this work? What the mechanism make possible &quot;equalize&quot; the air pressure between the doors? For example, in the ISS (International Space Station).&lt;/p&gt;&#xA;" OwnerUserId="16964" LastActivityDate="2018-08-11T05:37:51.293" Title="How an airlock works?" Tags="|pressure|air|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23183" PostTypeId="2" ParentId="23182" CreationDate="2018-08-11T05:37:51.293" Score="2" Body="&lt;p&gt;each &quot;end&quot; of the airlock has a valve in it which, when opened, allows the pressure between the inside of the airlock chamber and the pressure inside the adjacent chamber to equalize. Once this is done, the equalizing valve is shut and then the door to the adjacent chamber can safely be opened. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-08-11T05:37:51.293" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23186" PostTypeId="1" AcceptedAnswerId="23188" CreationDate="2018-08-11T15:37:40.647" Score="2" ViewCount="54" Body="&lt;p&gt;Below the handrail of a staircase is some kind of gridwork to prevent people or objects from falling off the staircase into the space below. What is the term for this gridwork? - perhaps &quot;handrail grating&quot;?&lt;/p&gt;&#xA;" OwnerUserId="17113" LastActivityDate="2018-08-11T18:41:28.350" Title="What is the protective part of a staircase handrail called?" Tags="|terminology|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23187" PostTypeId="1" AcceptedAnswerId="23190" CreationDate="2018-08-11T15:40:39.023" Score="0" ViewCount="97" Body="&lt;p&gt;I currently have &lt;a href=&quot;https://www.omega.com/pressure/pdf/LC703.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; load cell, which has a 10V excitation voltage. However, there is a very nice/convenient load cell amplifier that works with an excitation voltage of 5 VDC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the consequences of operating a cell at a voltage lower than the normal excitation? Just increased noise in the signal?&lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2018-08-11T18:03:53.867" Title="Operating a load cell below excitation voltage" Tags="|signal-processing|instrumentation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23188" PostTypeId="2" ParentId="23186" CreationDate="2018-08-11T16:58:48.070" Score="1" Body="&lt;p&gt;If the handrail is not mounted on wall then it must be mounted on a guardrail with 42 inch height and no opening which would allow any ball greater then or equal to 4 inch pass through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is LA code.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-08-11T16:58:48.070" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23190" PostTypeId="2" ParentId="23187" CreationDate="2018-08-11T18:03:53.867" Score="1" Body="&lt;p&gt;The load cell you linked to has a line item on the specs that says &quot;Output: 2 mV/V nominal&quot;.  Let's just say that you picked the version with 100 lbf full scale output.  What that means is this: if you use a 1 V excitation, you'll get a response something like this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Load    Output&#xA;  0 lbf   0 mV&#xA; 50 lbf   1 mV&#xA;100 lbf   2 mV&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now if you use a 10 V excitation, you'll get this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Load    Output&#xA;  0 lbf   0 mV&#xA; 50 lbf  10 mV&#xA;100 lbf  20 mV&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So if you want to use a 5V excitation, it will work perfectly fine.  Just do the math on the expected output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your comment on noise. The signal to noise ratio will be likely be better with a higher excitation voltage.  So yes, if you use 5 V instead of 10V, you probably have a slightly worse signal to noise ratio. Depends on your application and the precision required as to whether that will be acceptable.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-08-11T18:03:53.867" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23191" PostTypeId="2" ParentId="23186" CreationDate="2018-08-11T18:41:28.350" Score="3" Body="&lt;p&gt;The first and last are called newel posts and the intermediate ones are called spindles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But whether these names apply to materials other than wood is a different question.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-08-11T18:41:28.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23192" PostTypeId="2" ParentId="23162" CreationDate="2018-08-11T19:30:07.777" Score="0" Body="&lt;p&gt;Regardless of the site topography, it is best to follow local codes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But usually depending on the size of the building and its roof shape they install lightning rods on the highest part of the roof or on the ends of the ridge beams. They have to be connected to ground bar or the ground line of electrical network. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The connections and cables should be approved and the grounding has to be planned to get connected at approved number of loops in ground.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-08-11T19:30:07.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23194" PostTypeId="2" ParentId="23149" CreationDate="2018-08-12T03:51:05.483" Score="1" Body="&lt;p&gt;The operative engineering term is &quot;screw&quot; (no pun intended).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The screw itself is the &quot;male&quot; part, and the shaft into which a screw is inserted and rotated is a female part. The &quot;eyes&quot; referred to are &quot;receivers&quot; for screws, and are therefore categorized as &quot;female&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terminology is derived, of course, from animal anatomy, in which females are the recipients of &quot;screw&quot;-like actions. &lt;/p&gt;&#xA;" OwnerUserId="62" LastActivityDate="2018-08-12T03:51:05.483" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23195" PostTypeId="1" AcceptedAnswerId="23263" CreationDate="2018-08-12T04:06:59.500" Score="1" ViewCount="71" Body="&lt;p&gt;As far as I know a treatment plant's job is to treat black water to become effluent. Part of the process is removing dangerous microbes, grease and other environment damaging substances. However, is it possible that the odor remains in the effluent? Or that safe effluent can still contain odorous compounds unless scrubbed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: I live few kms away from one that discharges into a river that has moderate odor but at that distance away from residences, I guess it doesn't matter.&lt;/p&gt;&#xA;" OwnerUserId="11378" LastEditorUserId="61" LastEditDate="2018-08-16T09:19:17.893" LastActivityDate="2018-09-15T11:01:00.150" Title="Is the sewer outfall discharge supposed to have foul odor?" Tags="|waste-water-treatment|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23196" PostTypeId="1" CreationDate="2018-08-12T05:12:55.753" Score="1" ViewCount="432" Body="&lt;p&gt;While visiting Vancouver, Canada recently, I noticed there is a lot of construction for apartments and condos.  Attached is a photo of one of many of those sites, and at these sites, I noticed something unusual to me: temporary columns on lower floors appeared to support floors above, and all floors are made of concrete. In my photo, three of the uppermost floors were supported by floors beneath them. This seems fairly consistent across several construction sites.  So my question presumes that this practice is structurally sound, Canada is not a country that is devoid of competent architecture design.  I just don't know why it is sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that concrete has very little tensile strength, and so perhaps the concrete flooring used in these buildings are reinforced - presumably with rebar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, why only 3 floors?  If the concrete is not reinforced, wouldn't all floors from the ground up need to support the floors above it?  And if the concrete is reinforced, why is there a need at all to support the floors above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem to me that the 4th floor in the photo is bearing the brunt of all of the support above it, as it is not being supported by anything underneath it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm further curious about what happens next: will each of the floors be given a permanent support?  If so, why hasn't it been done already?  If not, what will happen when those temporary supports are removed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SMZiY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SMZiY.jpg&quot; alt=&quot;Building site&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17118" LastActivityDate="2018-08-12T07:10:53.130" Title="Why is this structurally sound?" Tags="|structural-engineering|materials|concrete|reinforced-concrete|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23198" PostTypeId="2" ParentId="23196" CreationDate="2018-08-12T06:41:11.830" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;Why is this structurally sound?&lt;/p&gt;&#xA;&lt;p&gt;In my photo, three of the uppermost floors were supported by floors beneath them. This seems fairly consistent across several construction sites. So my question presumes that this practice is structurally sound, ...&lt;/p&gt;&#xA;&lt;p&gt;why only 3 floors? If the concrete is not reinforced, wouldn't all floors from the ground up need to support the floors above it? And if the concrete is reinforced, why is there a need at all to support the floors above?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Generally it's approximately one floor per week, setup, laying rebar, in-floor conduit and pipes, then the pour. The supports stay in while it's hardening. As soon as it's hard they get the supports out and return them for use at the next site - everything costs money.&lt;/p&gt;&#xA;&lt;p&gt;That you saw the same thing at a few places and nothing collapsed says: &amp;quot;It works&amp;quot;.&lt;/p&gt;&#xA;&lt;p&gt;Since you didn't say where that is (exact address) I can't Google more photos so I had to supply my own pick. You need side, inside and downview to get a glimpse of the technique.&lt;/p&gt;&#xA;&lt;p&gt;These photos are from this webpage: &amp;quot;&lt;a href=&quot;http://www.highriseglass.ca/projects.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;HRG Projects&lt;/a&gt;&amp;quot;, for the apartment titled: &amp;quot;Cielo (2007) - 1205 West Hastings Street, Vancouver, BC&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v2q3d.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v2q3d.jpg&quot; alt=&quot;Supports pulled from lower floors&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Supports pulled from lower floors.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AU36h.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AU36h.jpg&quot; alt=&quot;Supports left on most recently poured floors&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Supports left on most recently poured floors.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vnHkA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vnHkA.jpg&quot; alt=&quot;Inside view - look at ceiling above worker's head - notice the marks from the supports&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Inside view - look at ceiling above worker's head - notice the marks from the supports.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/poAlO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/poAlO.jpg&quot; alt=&quot;Finished - No supports - Still standing&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Finished - No supports - Still standing.&lt;/p&gt;&#xA;" OwnerUserId="14556" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-08-12T06:41:11.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23199" PostTypeId="2" ParentId="23196" CreationDate="2018-08-12T07:10:53.130" Score="4" Body="&lt;p&gt;What you see is temporary support, only till the time the floors attain their design strength. Then all these will be removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rate of progress of the project must be faster than the time required for the concrete to be strong enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as using rebars, just looking at the spans and proportion of cantilevered area's it seems they are using some of the most advanced and strongest technologies. I read in some engineering magazines the are using welded/bolted rebars as opposed to overlapping, extensively.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thin slabs indicate they are using pre and post-tensioned high tensile strength strands or pre-tensioned cables, this as well as using the very high strength concrete (in the range of 4-8 k psi).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combination of these high strength reinforcing and thin lightweight slab and shear walls lead to a leaner, more ductile structure, better prepared to withstand potential earthquakes.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-08-12T07:10:53.130" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23203" PostTypeId="1" CreationDate="2018-08-12T14:20:14.560" Score="1" ViewCount="34" Body="&lt;p&gt;I am working on monitoring research. During it I found a lot of papers use Hall effect sensors to get spindle current. However some of them are accessed internally to spindle driver current. I will be thankful if any cnc professional can guide me how to access internally to spindle current and which types of cnc have this ability? &lt;/p&gt;&#xA;" OwnerUserId="17124" LastEditorUserId="7083" LastEditDate="2018-08-14T00:12:41.053" LastActivityDate="2018-08-14T00:12:41.053" Title="How can I get the spindle driver current internally?" Tags="|manufacturing-engineering|machining|cnc|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23204" PostTypeId="1" AcceptedAnswerId="23208" CreationDate="2018-08-12T14:24:16.143" Score="1" ViewCount="400" Body="&lt;p&gt;I am using Bernoulli's equation in order to calculate a relative area between two points in a pipe. Problem is, based on the solution I have obtained, I end up with imaginary numbers and not sure what to make of this practically. Here is my math:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;P_1 + \dfrac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \dfrac{1}{2}\rho v_2^2 + \rho gh_2 \\&#xA;v_1 A_1 = v_2 A_2 \\&#xA;\dfrac{A_1}{A_2} = \sqrt{\dfrac{v_2^2}{v_2^2+\dfrac{2}{\rho}\left(P_2-P_1\right)}}&#xA;\end{gather}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know my velocity at location 2, density, and pressures. I have no change in height so I got rid of third element in Bernoulli's equation. I am trying to obtain an area ratio between both points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very possible in my model that I will have a low P2 and high P1 leading to a negative denominator and an imaginary solution. I feel like I am missing something practical or mathematical here that is simple but my mind has drawn a blank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, I am assuming non-compressible airflow. Cannot assume laminar flow. &lt;/p&gt;&#xA;" OwnerUserId="17123" LastEditorUserId="1832" LastEditDate="2018-08-12T14:53:59.187" LastActivityDate="2018-08-13T18:57:51.957" Title="Area ratio using Bernoulli's equation" Tags="|pressure|airflow|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23206" PostTypeId="1" CreationDate="2018-08-12T14:49:09.333" Score="0" ViewCount="276" Body="&lt;p&gt;The thermal oil Feroline U in &lt;a href=&quot;https://www.siekerkotte.de/en/flange-heaters/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Boiler&lt;/a&gt; is not gaining the desired temp of 240 Degrees. We have physically measured the temp using digital thermometer at a distance of 20 cm where the boiler temp sensor is installed. 40 degrees are lost in 20cm insulated area.&#xA;The boiler is well insulated with Rockwool &amp;amp; oil quality is good (recently tested). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could be the reasons? Any solutions please?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZrDIX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZrDIX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17079" LastEditorUserId="17079" LastEditDate="2018-08-12T18:01:29.757" LastActivityDate="2022-04-25T23:04:20.430" Title="Thermal oil in Boiler not reaching desired temperature" Tags="|heat-transfer|heating-systems|heat-exchanger|thermal-conduction|boilers|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23207" PostTypeId="2" ParentId="16705" CreationDate="2018-08-12T17:34:58.417" Score="1" Body="&lt;p&gt;Take a system picture as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j6kH2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j6kH2.png&quot; alt=&quot;picture of hen house&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$\dot{V}$ - air flow (m$^3$/s)&lt;/li&gt;&#xA;&lt;li&gt;$\rho$ - air density (kg/m$^3$)&lt;/li&gt;&#xA;&lt;li&gt;$\tilde{C}_p$ - air specific heat (J/kg $^o$C)&lt;/li&gt;&#xA;&lt;li&gt;$h_a$ - air convection coefficient (W/m$^2$ $^o$C)&lt;/li&gt;&#xA;&lt;li&gt;$T_a$ - air temperature (K)&lt;/li&gt;&#xA;&lt;li&gt;$A$ - area of container (m$^2$)&lt;/li&gt;&#xA;&lt;li&gt;$w$ - wall thickness (m)&lt;/li&gt;&#xA;&lt;li&gt;$k$ - wall thermal conductivity (W/m K)&lt;/li&gt;&#xA;&lt;li&gt;$\hat{\dot{q}}_I$ - internal heat generation (W/m$^3$)&lt;/li&gt;&#xA;&lt;li&gt;$V$ - volume of container (m$^3$)&lt;/li&gt;&#xA;&lt;li&gt;$T_f$ - unknown final equilibrium temperature&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assume that heat generation is uniform throughout the container, the container is (mostly) air, and the air is well mixed. The energy balance equation gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\hat{\dot{q}}_I V = \left(\dot{V}\rho\tilde{C}_p + (A/R_T)\right)\left(T_f - T_a\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $R_T$ is a thermal resistance defined as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R_T = \left(\frac{w}{k} + \frac{1}{h_a}\right)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cast this into the following form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{T_f}{T_a} - 1 \right) = \left[\alpha\ \hat{\dot{q}}_I\right] \left[1 + \alpha \left(\frac{A}{V}\right)\left(\frac{1}{R_T}\right) \right]^{-1}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\alpha = \frac{\tau}{\rho \tilde{C}_p T_a}$$&#xA;$$ \tau = \frac{V}{\dot{V}}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the left side as a relative difference in the temperature inside the container versus the air (this can be expressed as a percentage when multiplied by 100). Consider $\tau$ as a turnover time (how often does one entire volume of air in the container get exchanged).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take two extremes.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Case A&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In the limit that you do not ventilate, $\tau \rightarrow \infty$. For large $\alpha$, the expression approximates as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{T_f}{T_a} - 1 \right) \approx \left[\frac{A}{V}\right]^{-1}\left[R_T\right] \left[\hat{\dot{q}}_I\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep the inside of the container closest to the external air temperature, have the greatest wall area to container volume and have a low thermal resistance. As an extreme analogy, using a thin walled metal sphere that houses all the chickens would be the best option.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Case B&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In the limit that you ventilate well, $\tau \rightarrow 0$. For small $\alpha$, the expression approximates as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\left(\frac{T_f}{T_a} - 1 \right) \approx \left[\tau\right]\left[\rho\tilde{C}_p T_a\right]^{-1} \left[\hat{\dot{q}}_I\right]$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To keep the inside of the container closest the external air temperature, have a high air flow (low $\tau$) and replace the air with a denser gas that has a higher heat capacity. So again by extreme analogy, do not ventilate the container with helium (for other reasons as well I can imagine :-)).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use a container with the greatest area to volume possible&lt;/li&gt;&#xA;&lt;li&gt;Use a container that has low thermal resistance (alternatively blow the air around the outside of the container as well to increase $h_a$)&lt;/li&gt;&#xA;&lt;li&gt;Ventilate well&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Interestingly, Case B suggests that, all else being equal, cooling will be more efficient on cooler days than on hot days.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-08-12T17:34:58.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23208" PostTypeId="2" ParentId="23204" CreationDate="2018-08-12T20:02:28.343" Score="1" Body="&lt;p&gt;Unless you are putting &quot;randomly chosen numbers&quot; into your equation, or you have measured the numbers but your assumptions about Bernoulli and compressibility don't apply to your real-world flow situation, you shouldn't have a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at it this way: for given values of $P_2$ and $v_2$, physically $P_1$ must always be less than some maximum value - namely, the value when $v_1 = 0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That follows directly from your first equation (your statement of Bernoulli's principle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That maximum value of $P_1$ that is physically possible will make the denominator of your square-root equal to zero, or in other words, $A_1$ will be &quot;infinite&quot; - which of course is correct, if $v_1$ is zero!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-08-12T20:02:28.343" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23209" PostTypeId="1" CreationDate="2018-08-12T20:43:29.663" Score="0" ViewCount="38" Body="&lt;p&gt;So I need two modules to give each other data with the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They are relatively short range. They need to be near each other to transmit the data, say around 5 cm.&lt;/li&gt;&#xA;&lt;li&gt;They are relatively small.&lt;/li&gt;&#xA;&lt;li&gt;Battery powered, preferably ready to transmit the moment its turned on with a switch.&lt;/li&gt;&#xA;&lt;li&gt;The data they receive should be sent via bluetooth to another device (There is a reason I can't use bluetooth directly)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17129" LastEditorUserId="1586" LastEditDate="2018-10-08T19:28:34.063" LastActivityDate="2018-10-08T19:28:34.063" Title="Best way to communicate data wirelessly with specific conditions" Tags="|electrical-engineering|battery|waves|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23210" PostTypeId="1" CreationDate="2018-08-12T21:36:11.340" Score="1" ViewCount="384" Body="&lt;p&gt;I want to write to file data from XY-Plot with temperature depending on the location. In Axis menu I set a max precision (9) for X and Y. However, Fluent writes data like this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;293.094 0.000364067&#xA;293.094 0.000361604&#xA;293.094 0.000359141&#xA;293.095 0.000356678&#xA;293.095 0.000354215&#xA;293.096 0.000351752&#xA;293.096 0.000349289&#xA;293.096 0.000346826&#xA;...&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I need a data about temperature as precise as possible (cause I want to use it in another program) but Fluent writes to xy-file only 3 digits after the decimal point despite my settings (for example, 293.194 instead of 293.093621..). I set precision this way for a many times for another physical quantities (velocity, heat flux at walls etc.), and there was always good precision exporting to in xy-files. But not for temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to google this problem but didn't find any special information for Fluent about this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot for any help!&lt;/p&gt;&#xA;" OwnerUserId="17130" LastActivityDate="2018-08-12T21:36:11.340" Title="Ansys Fluent: how to change the temperature digit precision in writing to file XY-plot?" Tags="|fluid-mechanics|ansys|ansys-workbench|fluent|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23212" PostTypeId="1" CreationDate="2018-08-12T23:59:52.743" Score="1" ViewCount="193" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xX4hp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find out what type of hinge or mechanism is used in fold down car TV's. I need to buy one for a project I'm working on but dont know the name. See picture above.&lt;/p&gt;&#xA;" OwnerUserId="17133" LastActivityDate="2023-12-16T05:08:06.010" Title="What is the type of hinge used in fold-down car tv's?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23214" PostTypeId="2" ParentId="23206" CreationDate="2018-08-13T05:06:03.503" Score="0" Body="&lt;p&gt;There are a couple things that could be going on, but most likely it has to do with how and where the temperature is measured rather than some sort of unknown heat loss. People are being critical in the comments, because the two measurements (based on the info given) should be nearly identical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seeing your replies in the comments, I am assuming both your temperatures were measured with the same measuring device, placed between the insulation and the metal pipe/vessel, and given time to reach equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume A is your vessel and B is your outlet. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Is B in fact the Outlet and not the Inlet? It is sometimes easy to mistake the flow through a plumbing system. Things are also not necessarily labeled correctly. It is very possible B is the Inlet to the heater and this temperature difference is the heat that is being delivered to users as planned. This is the most likely scenario with the info I have.&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Have you checked temperature B in different locations about the diameter? Maybe the flow is all to one side or there is rust/scale near the measuring location.&#xA;&lt;br&gt;&lt;br&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Have you checked temperature A in different locations on the vessel?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;How close is point A to the heating element? (not sure of the energy input type, but the location/proximity of the electric heating element or heat exchanger will make a big difference. This is also a likely scenario.)&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How close is A to the inlet? (this is where the cool oil returns in a closed loop system) (this is more likely to make A colder than B, but something to think about as well) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;Other things to consider is that the steel thickness at B may be thicker or the insulation may be thinner. This results in your measurement being closer to ambient temperature than the the measurement at A. When we place a thermometer below insulation we are making the assumption that the insulation is much much more resistant to heat flow than the metal. Most the time this assumption is true, but it is something to consider when looking for accuracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If none of those items help you track down the issue please include info like flow-rate, energy source, energy users, etc.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-08-13T05:06:03.503" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23216" PostTypeId="2" ParentId="23204" CreationDate="2018-08-13T05:36:15.317" Score="0" Body="&lt;p&gt;Adding to alephzero, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you are getting a negative denominator, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$v_{2}^2 + \frac{2}{\rho}(P_2 - P_1) &amp;lt; 0 $, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow   \frac{1}{2}{\rho}v_{2}^2 +(P_2 - P_1) &amp;lt; 0 $ ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow   \frac{1}{2}{\rho}v_{2}^2 +P_2  &amp;lt; P_1 $ ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$\Rightarrow P_{02} &amp;lt; P_1 $ , This violates the Bernoulli's principle. So you will never get an imaginary velocity value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As alephzero clearly mentioned, $P_1 \rightarrow P_0$ when $v_1 \rightarrow 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:  $P_0$ is the total pressure and it will be constant along the streamline on wihch you apply the Bernoulli's equation.&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-08-13T05:36:15.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23217" PostTypeId="2" ParentId="23160" CreationDate="2018-08-13T05:58:09.073" Score="0" Body="&lt;p&gt;Often times your assumptions make a bigger difference than the rigor of your mathematical analysis. Remember you should select assumptions that favor an easier analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember that the guys your hire/employ to make a simple structure like this are not going to measure things to the .1 degrees; their tools don't even support that. These are probably pinned connections, so only lengths are needed, but keep that in mind and leave those numbers off the mfg print.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't put your structure right up to the corner of the building; give yourself a standoff. If there are any errors in length they will not be easily corrected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not sure if your location has snow, but often times snow load will be greater than wind load. Wind loads are horizontal except when around other structures like in your situation. The two highest loads left to right and right to left could actually be lifting the panels up due to the captured air pocket; so look at that scenario also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having built solar hot water and solar PV mounts, I do not recommend using the panels to hold square or take any load between the other supports. The structure should bear all the load and the panels should be attached per the manufacturer recommended method (that also covers you if they do break).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were my design, I would select a large enough primary beam (201&quot;) such that it can handle the chosen load with just two supports. One at a standoff, say 6&quot; up from corner of the building, and another selected such that it is a symmetrical connection point based on your desired mounting configuration. Basically the only optimization you have to do is figuring out the amount of cantilever to amount of span. Simple designs are easier to calculate, easier to optimize without a computer, and easier to manufacture. Unless you need to split hairs for low mass designs like in aerospace, your materials are much much cheaper than labor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you still wish to solve your design and assumptions as is, I would rotate the whole system mathematically so that the distributed load is either in just x or y. That gives you load components that are either in just x or y. Then use the angle of each member to compute the total load on each.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2018-08-14T22:05:17.240" LastActivityDate="2018-08-14T22:05:17.240" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23218" PostTypeId="2" ParentId="23212" CreationDate="2018-08-13T06:41:38.107" Score="0" Body="&lt;p&gt;These are not usually simple hinges - they are often designed with cams etc to give a specific movement - so are usually specific to the item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of kitchen cupboard hinges - the simple ones are easy to replace, but some have extending movements so doors can move out to clear other doors...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find a hinge that does the movement you require and then you will need to re-manufacture it to cope with the load you will be applying...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-08-13T06:41:38.107" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23220" PostTypeId="1" AcceptedAnswerId="23242" CreationDate="2018-08-13T10:26:32.957" Score="2" ViewCount="1562" Body="&lt;p&gt;I have a beam, simply supported with 2 points loads and 2 UDLs. The loads are symmetric about the centre of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to derive an expression for the moment along the beam and then via 2 integrations obtained expressions for slope and deflection respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please let me know whether I'm close with what I have or help me to create an overall moment equation, which i can plot in MathCAD preferably, and also help me with finding the maximum deflection of the beam? Attached is the FBD for the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xBXVm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xBXVm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is my attempt at a solution for deflection (In the case where x = L).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zVRH3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zVRH3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="17139" LastEditorDisplayName="user14407" LastEditDate="2018-08-14T23:26:18.673" LastActivityDate="2018-08-14T23:26:18.673" Title="Bending equation and max deflection of UDL and multiple point loads" Tags="|structural-engineering|statics|beam|deflection|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23222" PostTypeId="1" AcceptedAnswerId="23225" CreationDate="2018-08-13T13:14:53.690" Score="0" ViewCount="794" Body="&lt;p&gt;It's just a matter of curiosity, but I find hard to justify why every single watch keep the hour hand below the minute one; from a user perspective, it seems a poorly design decision for those times when they overlap the hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some manufacturers – notably Rolex with its Mercedes-like symbol – even added larger stuffs to the hour hand to be seen when such occurrences happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose it may be done to simplify the internal gears, but I also suppose that by placing the hour-minute-seconds hands in a reverse order shouldn't be a problem…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IMO, a more reasonable user experience should show from top-to-bottom: seconds, hour, [GMT], minute hands. Isn't this achievable ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone satisfy my curiosity ? :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="17142" LastActivityDate="2018-08-13T14:52:28.850" Title="Why all watches hands are stacked in that specific order ? Can it be changed?" Tags="|mechanical-engineering|design|gears|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23223" PostTypeId="2" ParentId="23204" CreationDate="2018-08-13T13:55:30.770" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As a side note, I am assuming non-compressible airflow. Cannot assume laminar flow.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The assumptions that $\dot{V}_1 = v_1 A_1 = v_2 A_2 = \dot{V}_2$ and that $\rho =$ constant are only valid for incompressible flow. Air is compressible, not incompressible. Use the ideal gas law with $\rho = M_{air}(p/RT)$ and $v = \dot{V}/A = \dot{n}RT/pA$. This gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dot{n}_1 = \dot{n}_2 = \dot{n}$$&#xA;$$\frac{RT_1}{M_{air}} + \frac{1}{2}\left(\frac{\dot{n}RT_1}{p_1 A_1}\right)^2 = \frac{RT_2}{M_{air}} + \frac{1}{2}\left(\frac{\dot{n}RT_2}{p_2 A_2}\right)^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The proper analysis for a compressible (ideal) gas must consider pressure and &lt;em&gt;temperature&lt;/em&gt; not pressure and &lt;em&gt;velocity&lt;/em&gt;. Corrections for non-ideality may be applied depending on how drastically the gas is compressed and/or how accurate the answer really must be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of areas as a function of velocities for ideal gas flow is expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \frac{A_1}{A_2} = \frac{\dot{V}_1 v_2}{\dot{V}_2 v_1} =  \frac{T_1 v_2 p_2}{T_2 v_1 p_1}$$&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-08-13T18:57:51.957" LastActivityDate="2018-08-13T18:57:51.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23225" PostTypeId="2" ParentId="23222" CreationDate="2018-08-13T14:52:28.850" Score="2" Body="&lt;p&gt;At current time - probably only tradition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The origin would be historical, when original single-hand clocks gained the minute hand. There would be existing, prior clockwork near the (back side of the) face of the clock, just where it was originally, and then new clockwork would be built, to propel the new hand, placed further back from the face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you propel an axis-inside-axis transmission, the inner axis must extend further from both ends of the outer (tubular) axis and can be propelled on the extended part. So you have the tube reaching to the hour work, and the inner axis reaching into the new gear on top. And so, the hands on the opposite side follow, hour hand attached to the shorter tube axis, minute hand on top, on the longer inner axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since there was never really any significant reason to challenge this, and the clockmaking industry is quite conservative and traditional, it wasn't changed. A minor argument might be the amount of friction (and so, frequency of need to wind the clock/watch) - the faster hand only moves against one surface, the slower is working between two - but that's a really minor issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sure it could be done. It would most likely involve mirroring the clockwork against the plane of the clock face (or just moving the face to the opposite side) and switching the roles of the two axis (plus obvious minor diameter adjustments. At current time though... an LCD screen displays both hands on the same plane.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-08-13T14:52:28.850" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23227" PostTypeId="1" CreationDate="2018-08-13T20:03:20.473" Score="0" ViewCount="139" Body="&lt;p&gt;I am new to the field, most of the papers I have searched directly conduct CFD simulation but skipped the calculation. I hope to get an estimation by calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am designing a small scale horizontal overhead air curtain covering a 0.5m x 1m (Width x Height) area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume no temperature, air density difference between inside and outside air, no winds or other kinds of draught. Considering an ideal case, and the air curtain is blowing vertically downward,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$b\times h$ = outlet cross-sectional area&lt;/li&gt;&#xA;&lt;li&gt;$L$ = throw distance&lt;/li&gt;&#xA;&lt;li&gt;$v_0$ = initial velocity&lt;/li&gt;&#xA;&lt;li&gt;$v_L$ = velocity at $L$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there an equation or other numerical methods to determine the relationships between $b$, $h$, $L$, $v_0$ and $v_L$?&lt;/li&gt;&#xA;&lt;li&gt;Also, is it possible to find $b_L\times h_L$ (cross-sectional area of the air stream at $L$)?&lt;/li&gt;&#xA;&lt;li&gt;How to determine optimal outlet cross-sectional area $b \times h$?&lt;/li&gt;&#xA;&lt;li&gt;How to determine optimal air speed vL? (to minimize leakage but also turbulence) (2.5-3.5&amp;nbsp;m/s for most of the currently existing products)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Are there any relevant considerations I should consider?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What references/sources would be good to read?&lt;/p&gt;&#xA;" OwnerUserId="17143" LastEditorUserId="1832" LastEditDate="2018-08-13T20:20:12.803" LastActivityDate="2024-03-18T01:06:05.820" Title="Design of a small scale air curtain" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23228" PostTypeId="1" AcceptedAnswerId="23230" CreationDate="2018-08-13T20:40:22.183" Score="2" ViewCount="61" Body="&lt;p&gt;Is Solidworks Motion capable of finding the torque of the planet carrier (output) in a planetary gear set when in steady state? The sun and ring gears are both inputs and all gears are rotating at a constant angular velocity.&lt;/p&gt;&#xA;" OwnerUserId="17148" LastActivityDate="2018-08-13T21:02:13.913" Title="Solidworks Motion capabilities" Tags="|automotive-engineering|simulation|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23229" PostTypeId="2" ParentId="23133" CreationDate="2018-08-13T20:58:40.590" Score="0" Body="&lt;p&gt;&lt;strong&gt;TLDR: Treatment technologies that can function in oxic and anoxic conditions are more flexible and can be applied in more circumstances.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I think clearing up some terminology would be helpful. These terms are not super specific, but this is how they are generally used in my experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;Treatment plant&lt;/strong&gt; is a facility which receives wastewater, treats it, and discharges it somewhere. It is not oxic or anoxic, but it has &lt;strong&gt;treatment processes&lt;/strong&gt; which can be oxic or anoxic. Most plants have numerous processes, including both oxic and anoxic (and anaerobic) processes for treatment of different things. A treatment process uses a specific &lt;strong&gt;treatment technology or technologies&lt;/strong&gt; to accomplish its goals. A &lt;strong&gt;moving bed biofilm reactor (MBBR)&lt;/strong&gt; is a specific technology involving suspended media to host biofilm growth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MBBR tanks can be aerated (oxic conditions) or un-aerated (anoxic conditions) depending on the goals of the treatment process. The fact that the technology can be applied in both kinds of processes doesn't usually* benefit the process itself. It does benefit the manufacturer because their potential market is larger however. This does not mean this is just marketing hype; it serves to clarify for customers the potential applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I say doesn't usually benefit the process, but there are some plants that have &quot;swing&quot; zones that can be operated either oxically or anoxically depending on the influent conditions. I don't know of an MBBR that is operated this way, but I don't see why it couldn't be. In this case, the flexibility helps the plant adapt to the current conditions.&lt;/p&gt;&#xA;" OwnerUserId="17150" LastActivityDate="2018-08-13T20:58:40.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23230" PostTypeId="2" ParentId="23228" CreationDate="2018-08-13T21:02:13.913" Score="1" Body="&lt;p&gt;Yes, it is possible, if you define all physical properties (mass, moment of inertia ...) of the rotating object. You need to run the simulation first, notice put the motor at the component which you want to find the torque of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Click on results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- Under property manger (under results) select 'forces'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Select 'motor torque' under subcategory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4- Select 'magnitude' of whatever component you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5 click on the green check mark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll see the plot in function of time. &#xA;This is also true for finding power, forces, acceleration ... . &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-08-13T21:02:13.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23231" PostTypeId="1" AcceptedAnswerId="23234" CreationDate="2018-08-13T22:18:21.543" Score="0" ViewCount="28" Body="&lt;p&gt;I'm driving a small, benchtop vibe table (&lt;a href=&quot;https://controlledvibration.com/product-item/12-inch-platform-shaker/#tab-id-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;shaker&lt;/a&gt;) with two Crown DSI-1000 PA's, and also cleaning my signal with a DBX 231s EQ. Currently running the system at 6.8 GRMS, but I'm looking to increase my overall driving level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the test article mass/ test levels I'm shooting for, I'm conflicting with the specified peak capability of my shaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be more precise, I would like to drive a ~5.5 lb object to ~10 grms (random vibe). According to my back-of-the-envelope calcs (Eqn 1), I don't &lt;em&gt;think&lt;/em&gt; there is much I can do besides purchasing a larger shaker. This shaker tops out at 200 lbf. However I figured it would be worthwhile to probe the community for ideas in case there is some sort of &lt;em&gt;engineering voodoo&lt;/em&gt; I can employ. Will also post in EE stack exchange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;TL:DR, Is there any way for me to increase the peak driving force/ capabiltiy of my shaker table without the risk of damaging it?&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My PA's and EQ are also getting close to their maximums, at just 6.8 grms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Desired profile: NASA GEVs STD-7000a: &lt;a href=&quot;https://i.stack.imgur.com/ss5E2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ss5E2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eqn 1: &lt;em&gt;GRMS = [PEAK_SHAKER_FORCE / (ARTICLE_MASS + ARMATURE_MASS)] / 2&lt;/em&gt; . --&gt; GRMS = (200 lbs/ 13.5 lbs) / 2 ...................................................................... --&gt; GRMS_MAX = 7.4&lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2018-08-13T23:09:19.663" Title="Vibe Table: Increase Peak Force Capability?" Tags="|aerospace-engineering|vibration|power|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23232" PostTypeId="1" CreationDate="2018-08-13T22:51:58.463" Score="0" ViewCount="672" Body="&lt;p&gt;I'm trying to determine whether it would be possible to use multiple power outlets in vehicle to increase the wattage that can be obtained. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is driven from a desire to install an DC to AC inverter without having to connect directly to the vehicle battery. Typically, the units sold that plug into the cigarette lighter outlets in a car max out around 150 watts.  Some simple math suggests that, even with a 20 amp fuse, the maximum wattage you can draw from the outlet -before blowing a fuse - would be 240 Watts (12V x 20A = 240 W).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if there are multiple, separately fused outlets available, can they be connected in parallel to increase the available power?  For example, could two outlets be used to supply up to 480 watts of power?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems simple in theory, but I'm certain there are some more complicated aspects to be considered.  I realize that the length and resistance of the supply wires are a factor, and that if the voltage isn't balanced, then one leg will supply more amperage than the other. However, I believe this would really only be an issue as you started to approach the upper end of the amperage limit for one of the legs.  Once one of the legs maxed out, the fuse would blow, transferring the entire load to the second leg which would also blow quickly thereafter. But up until that point, both legs would share some level of the load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've included a simple schematic that I sketched for a visual aid.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any educated feedback or explanation would be greatly enjoyed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bNHEK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bNHEK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4897" LastActivityDate="2018-08-13T23:03:30.897" Title="Utilize multiple vehicle power outlets to increase wattage capacity of inverter" Tags="|electrical-engineering|circuits|circuit-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23233" PostTypeId="2" ParentId="23232" CreationDate="2018-08-13T23:03:30.897" Score="1" Body="&lt;p&gt;It is possible that the outlets you want to pair together may have the same supply : relay and / or fuse...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should connect a set of wires with appropriate fuse directly to the battery, if this is for short periods of use otherwise you will end uo with a flat battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best solution is to fit a second battery, run the inverter from that and have a split-charge system to charge the second battery when the engine is running.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-08-13T23:03:30.897" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23234" PostTypeId="2" ParentId="23231" CreationDate="2018-08-13T23:09:19.663" Score="1" Body="&lt;p&gt;No, there is not really much you can do.  If you want to get 10 Grms, you'll need a bigger shaker.  The only trick you could play would be to decrease the bandwidth of your test spectrum.  i.e. instead of testing 20 - 2000 Hz, just test 20 - 1000 Hz.  The g^2/Hz stays the same, but the Hz range is smaller, so now you don't have to drive 10 Grms anymore, but a smaller number.  You could perhaps split your testing into multiple segments.  Segment 1 test 20 - 1000 Hz, and then segment 2 test 1000 - 2000.  Depending on the intent of the test, this may or may not be acceptable.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-08-13T23:09:19.663" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23235" PostTypeId="1" CreationDate="2018-08-11T19:43:00.530" Score="2" ViewCount="54" Body="&lt;p&gt;I’ve been looking for a temperature sensor for my PI3 home thermostat. I’m to the point where my app is stable and running for over 20 days with the only issues relating to the temperature sensors. I have dual DHT22 sensors and those will freeze up occasionally, for which the power needs to be reset. I will try to handle this reset in code, but because this controls my home HVAC system, I would like to get sensors that will not freeze up and have greater reliability. Recommendations from someone with a current PI3 thermostat being used to control their HVAC would be appreciated. I would like to be able to rely on the sensors.&lt;/p&gt;&#xA;" OwnerDisplayName="Jesse" LastActivityDate="2018-09-13T09:26:11.200" Title="Temperature sensor for PI3 thermostat" Tags="|sensors|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23240" PostTypeId="1" CreationDate="2018-08-14T08:49:03.177" Score="2" ViewCount="49" Body="&lt;p&gt;If you would search for open source software projects, you'd probably run into GitHub and BitBucket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a similar site/community around open hardware to share &lt;em&gt;and version&lt;/em&gt; CAD files, curcuits, etc. (not modeling software but an open hardware product oriented)? &lt;/p&gt;&#xA;" OwnerUserId="11687" LastEditorUserId="11687" LastEditDate="2018-08-14T17:54:21.370" LastActivityDate="2018-08-14T17:54:21.370" Title="Go-to resource for open hardware engineering resources?" Tags="|technical-drawing|" AnswerCount="0" CommentCount="8" ClosedDate="2018-08-14T12:46:11.507" ContentLicense="CC BY-SA 4.0" />
  <row Id="23241" PostTypeId="2" ParentId="10978" CreationDate="2018-08-14T14:30:18.010" Score="2" Body="&lt;p&gt;I realize this is an old thread, and I am not sure how deep of a dive you finally took on this, but one thing I don't see accounted for in your equations is drum/cable friction. This will be small, and like the accumulated mass of the wound steel wire rope you did not include, it may not be on your list. The cable could be pre-stretched and pre-loaded, however any movement between the cable and drum due to cable stretch will also encounter friction. In my industry (theater rigging, stage machinery design), the groove contacts a greater area than a flat drum application, and we usually have additional friction along the redirect sheaves and mules in the lineset to account for especially in 2:1 or 4:1 mechanical advantage systems. &lt;/p&gt;&#xA;" OwnerUserId="17162" LastActivityDate="2018-08-14T14:30:18.010" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23242" PostTypeId="2" ParentId="23220" CreationDate="2018-08-14T17:38:34.050" Score="1" Body="&lt;p&gt;Note: I'll be using the following notation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$q_1, q_2$ are the distributed loads (where you use $w$, which I reserve for deflections)&lt;/li&gt;&#xA;&lt;li&gt;$L = c + 2(a+b)$, the total length of the span&lt;/li&gt;&#xA;&lt;li&gt;$\ell = a + b$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'll also be making use of the structure and loading's symmetry, and adopting the following model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0XLPm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0XLPm.png&quot; alt=&quot;half of beam modelled by placing Rz support at midspan&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let's get the reactions. Since the structure and the loading is symmetrical, this is trivial, with each support getting half of each load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$R = -\dfrac{q_1(L-2a)}{2} - \dfrac{q_2L}{2} - P$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's then use singularity functions to describe the loading on the beam and integrate our way up to the deflection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\newcommand{\a}[1]{\langle #1 \rangle}&#xA;q &amp;amp;= R\a{x}^{-1} + q_1\a{x-a}^0 + q_2\a{x}^0 + P\a{x-\ell}^{-1} \\&#xA;Q &amp;amp;= R\a{x}^0 + q_1\a{x-a}^1 + q_2\a{x}^1 + P\a{x-\ell}^0 \\&#xA;M &amp;amp;= R\a{x}^1 + \dfrac{q_1\a{x-a}^2}{2} + \dfrac{q_2\a{x}^2}{2} + P\a{x-\ell}^1 \\&#xA;EI\theta &amp;amp;= \dfrac{R\a{x}^2}{2} + \dfrac{q_1\a{x-a}^3}{6} + \dfrac{q_2\a{x}^3}{6} + \dfrac{P\a{x-\ell}^2}{2} + C_1 \\&#xA;EIw &amp;amp;= \dfrac{R\a{x}^3}{6} + \dfrac{q_1\a{x-a}^4}{24} + \dfrac{q_2\a{x}^4}{24} + \dfrac{P\a{x-\ell}^3}{6} + C_1x + C_2 \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's use the boundary conditions to find those constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;\theta\left(\dfrac{L}{2}\right) = EI\theta ={}&amp;amp; \dfrac{R}{2}\left(\dfrac{L}{2}\right)^2 + \dfrac{q_1}{6}\left(\dfrac{L}{2}-a\right)^3 + \dfrac{q_2}{6}\left(\dfrac{L}{2}\right)^3 + \dfrac{P}{2}\left(\dfrac{L}{2}-\ell\right)^2 + C_1 = 0 \\&#xA;\therefore C_1 ={}&amp;amp; q_1\left(\dfrac{(L-2a)}{4}\left(\dfrac{L}{2}\right)^2 - \dfrac{1}{6}\left(\dfrac{L}{2}-a\right)^3\right) \\&#xA;&amp;amp;+ q_2\left(\dfrac{L}{4}\left(\dfrac{L}{2}\right)^2 - \dfrac{1}{6}\left(\dfrac{L}{2}\right)^3\right) \\&#xA;&amp;amp;+ P\left(\dfrac{1}{2}\left(\dfrac{L}{2}\right)^2 - \dfrac{1}{2}\left(\dfrac{L}{2}-\ell\right)^2\right) \\&#xA;={}&amp;amp; q_1\left(\dfrac{a^3}{6} - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\&#xA;&amp;amp;+ \dfrac{q_2L^3}{24} \\&#xA;&amp;amp;+ \dfrac{P\ell(L-\ell)}{2} \\&#xA;w(0) = C_2 ={}&amp;amp; 0&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We therefore have the full description of the entire beam. Personally, I prefer splitting the functions up for each beam segment. Therefore, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x\in[0,a]$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;Q ={}&amp;amp; R\a{x}^0 + q_2\a{x}^1 \\&#xA;={}&amp;amp; \dfrac{q_1(2a - L)}{2} \\&#xA;&amp;amp;+ q_2\left(x - \dfrac{L}{2}\right) \\&#xA;&amp;amp;- P \\&#xA;M ={}&amp;amp; R\a{x}^1 + \dfrac{q_2\a{x}^2}{2} \\&#xA;={}&amp;amp;\dfrac{q_1(2a - L)}{2}x \\&#xA;&amp;amp;+ q_2\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x\right)  \\&#xA;&amp;amp;- Px \\&#xA;EI\theta ={}&amp;amp; \dfrac{R\a{x}^2}{2} + \dfrac{q_2\a{x}^3}{6} + C_1 \\&#xA;={}&amp;amp;q_1\left(\dfrac{2a - L}{4}x^2 + \dfrac{a^3}{6} - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\&#xA;&amp;amp;+ q_2\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{L^3}{24}\right) \\&#xA;&amp;amp;+ P\left(-\dfrac{1}{2}x^2 + \dfrac{\ell(L-\ell)}{2}\right) \\&#xA;EIw ={}&amp;amp; \dfrac{R\a{x}^3}{6} + \dfrac{q_2\a{x}^4}{24} + C_1x \\&#xA;={}&amp;amp;q_1\left(\dfrac{2a - L}{12}x^3 + \left(\dfrac{a^3}{6} - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right)x\right) \\&#xA;&amp;amp;+ q_2\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{L^3}{24}x\right) \\&#xA;&amp;amp;+ P\left(-\dfrac{1}{6}x^3 + \dfrac{\ell}{2}(L-\ell)x\right)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For $x\in(a,\ell]$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;Q ={}&amp;amp; R\a{x}^0 + q_1\a{x-a}^1 + q_2\a{x}^1 \\&#xA;={}&amp;amp;q_1\left(x - \dfrac{L}{2}\right) \\&#xA;&amp;amp;+ q_2\left(x - \dfrac{L}{2}\right) \\&#xA;&amp;amp;- P \\&#xA;M ={}&amp;amp;R\a{x}^1 + \dfrac{q_1\a{x-a}^2}{2} + \dfrac{q_2\a{x}^2}{2} \\&#xA;={}&amp;amp;q_1\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x + \dfrac{a^2}{2}\right) \\&#xA;&amp;amp;+ q_2\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x\right)  \\&#xA;&amp;amp;- Px \\&#xA;EI\theta ={}&amp;amp;\dfrac{R\a{x}^2}{2} + \dfrac{q_1\a{x-a}^3}{6} + \dfrac{q_2\a{x}^3}{6} + C_1 \\&#xA;={}&amp;amp;q_1\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{a^2}{2}x - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\&#xA;&amp;amp;+ q_2\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{L^3}{24}\right) \\&#xA;&amp;amp;+ P\left(-\dfrac{1}{2}x^2 + \dfrac{\ell(L-\ell)}{2}\right) \\&#xA;EIw ={}&amp;amp;\dfrac{R\a{x}^3}{6} + \dfrac{q_1\a{x-a}^4}{24} + \dfrac{q_2\a{x}^4}{24} + C_1x \\&#xA;={}&amp;amp;q_1\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{a^2}{4}x^2 + \left(\dfrac{L^3}{24} - \dfrac{a^2L}{4}\right)x + \dfrac{a^4}{24}\right) \\&#xA;&amp;amp;+ q_2\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{L^3}{24}x\right) \\&#xA;&amp;amp;+ P\left(-\dfrac{1}{6}x^3 + \dfrac{\ell}{2}(L-\ell)x\right)&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, for $x\in(\ell,\frac{L}{2}]$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;Q ={}&amp;amp; R\a{x}^0 + q_1\a{x-a}^1 + q_2\a{x}^1 + P\a{x-\ell}^0 \\&#xA;={}&amp;amp;q_1\left(x - \dfrac{L}{2}\right) \\&#xA;&amp;amp;+ q_2\left(x - \dfrac{L}{2}\right) \\&#xA;&amp;amp;+ 0 \\&#xA;M ={}&amp;amp;R\a{x}^1 + \dfrac{q_1\a{x-a}^2}{2} + \dfrac{q_2\a{x}^2}{2} + P\a{x-\ell}^1 \\&#xA;={}&amp;amp;q_1\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x + \dfrac{a^2}{2}\right) \\&#xA;&amp;amp;+ q_2\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x\right)  \\&#xA;&amp;amp;-P\ell \\&#xA;EI\theta ={}&amp;amp;\dfrac{R\a{x}^2}{2} + \dfrac{q_1\a{x-a}^3}{6} + \dfrac{q_2\a{x}^3}{6} + \dfrac{P\a{x-\ell}^2}{2} + C_1 \\&#xA;={}&amp;amp;q_1\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{a^2}{2}x - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\&#xA;&amp;amp;+ q_2\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{q_2L^3}{24}\right) \\&#xA;&amp;amp;+ P\left(-\ell x + \dfrac{L\ell}{2}\right) \\&#xA;EIw ={}&amp;amp;\dfrac{R\a{x}^3}{6} + \dfrac{q_1\a{x-a}^4}{24} + \dfrac{q_2\a{x}^4}{24} + \dfrac{P\a{x-\ell}^3}{6} + C_1x \\&#xA;={}&amp;amp;q_1\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{a^2}{4}x^2 + \left(\dfrac{L^3}{24} - \dfrac{a^2L}{4}\right)x + \dfrac{a^4}{24}\right) \\&#xA;&amp;amp;+ q_2\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{L^3}{24}x\right) \\&#xA;&amp;amp;+ P\left(-\dfrac{\ell}{2}x^2 + \dfrac{L\ell}{2}x - \dfrac{\ell^3}{6}\right)&#xA;\end{align}$$&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-08-14T17:38:34.050" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="23244" PostTypeId="1" CreationDate="2018-08-14T19:07:28.653" Score="0" ViewCount="3948" Body="&lt;p&gt;I'm an EE trying to interpret a pneumatic schematic and have a few questions regarding the symbols being used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've come across two different regulator symbols (or so I believe them to be regulator symbols). I'm curious if someone could break down the difference between each.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my interpretations and questions for each symbol:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I believe this symbol represents a pressure regulator. The spring indicates that it is a pressure regulator valve. However, I'm somewhat confused as to what the dashed line is meant to represent. Is it just meant to represent a feedback line for regulation purposes?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This symbol is similar to symbol 1, however, it differs in how the arrow is oriented. What is the significance of the arrow direction?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0a5Sj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0a5Sj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7664" LastActivityDate="2023-04-17T13:09:04.067" Title="How do you interpret a pneumatic regulator symbol?" Tags="|hydraulics|compressed-air|pneumatic|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23245" PostTypeId="2" ParentId="23244" CreationDate="2018-08-14T19:40:47.340" Score="1" Body="&lt;p&gt;Both of symbols represent 'relief valves', to my knowledge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of arrow in the first schematics represent the pilot operated relief valve and the second one is direct operated relief valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first schematic, the pressure valve is in its normal state, no flow is passing, once the pressure exceeds the pre-defined value, it will change its state to the second schematic. The second schematic also has a pressure indicator, the circle with an arrow inside. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dashed line are signal lines, no flow passe through the dashed line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct operated relief valve: &lt;a href=&quot;https://i.stack.imgur.com/SDVoe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SDVoe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pilot operated relief vale: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t0u6Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t0u6Z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-08-14T20:44:56.157" LastActivityDate="2018-08-14T20:44:56.157" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23246" PostTypeId="1" CreationDate="2018-08-14T19:52:36.570" Score="1" ViewCount="179" Body="&lt;p&gt;I have this water rower (&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B075MB7DM8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sunny Health &amp;amp; Fitness Water Rowing Machine w/LCD Monitor&lt;/a&gt;) :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bj9Jd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am going to make a monitor for it (to calculate time/500m and watts). I am going to attach a magnet to the tip of one of the blades and a sensor on the inside of the tank. I was wondering what the physic equations to calculate the minutes/500m were. Please ask any clarifying questions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further information :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The monitor provided with the machine is very inaccurate and cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The handle is attached to a fabric cord. The cord goes into the machine and goes around a pulley (it is now parallel with the cord coming from the handle). It then goes around a circle and is wrapped around many times. When the handle is pulled, the circle moves, which it attached to a shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also attached to that pole/shaft is the fan. The fan has three blades at the end of three poles. The blades are a couple inches wide and are about 1/4 in water. There is also a bungie cord wrapped around the pole to bring the handle back. I don't think that the bungie cord changes the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some images:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hsq4r.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hIVFq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/JanRf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0WquD.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note : there are several images of better quality on the Amazon website linked in the 1st paragraph.&lt;/p&gt;&#xA;" OwnerUserId="17164" LastEditorUserId="17164" LastEditDate="2018-08-16T19:06:32.743" LastActivityDate="2018-08-16T19:06:32.743" Title="Equation for water rower" Tags="|experimental-physics|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23249" PostTypeId="2" ParentId="23135" CreationDate="2018-08-15T03:16:36.467" Score="1" Body="&lt;p&gt;Contrary to popular opinion, the smaller tubes will provide greater strength under load than the larger tubes. Here's why:&#xA;Four 20x20mm tubes have a greater cross sectional surface area than a single 40×40mm tube. In fact, the 20×20 occupy the same external volume but have treble the strength according to the physical size. This issue is directly related to the square-cube law. Because the work of fracture required to cause structural failure in all 4 of the 20×20 tubes is higher, it is therefore more reliable. If the 4 20×20 tubes are spot welded together, their structural strength may be decreased but the rigidity of the whole resultant 40×40 tube will be 3 times stronger because of the internal cross of material measuring 3.2mm thick in the shape of a &quot;+&quot;. Obviously the 40×40mm tube does not have that rigidity or ability to resist deflection as well as 4 20×20mm tubes welded together as a 40×40mm tube. Another advantage of the 20×20mm is their ability to resist torsion better when welded together as a box tube.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-08-15T03:16:36.467" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23250" PostTypeId="2" ParentId="63" CreationDate="2018-08-15T03:41:34.920" Score="2" Body="&lt;p&gt;In any design regarding public traffic your biggest problem is going to be managing the human factors of competence, confidence and error. The entire purpose of traffic management is to control human behaviour. With automated traffic it is simply an issue of coordination but humans are not well coordinated. Some are slow, others are nervous, other people are frustrated while even more are driving along on autopilot or putting on make-up...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These factors are what make a junction of roads work or fail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cons of traffic lights are electricity consumption and telecommunication with the rest of the network. A good system will work at all times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A modular traffic circle will always work as long as nobody fearful turns up and flawless around each stage of turns. As capacity is reached they can become a gridlocked bottleneck but are excellent on one way street systems. Traffic lights work best at minor junctions to regulate traffic flow. In this respect the UK has ubiquitous installations of traffic lights on traffic circles which is an oxymoron contrary to traffic flow behaviours. The UK problem would be reduced drastically of drivers actually obeyed the 3 second rule of following distance but they steadfastly do not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume of traffic is a major consideration also. Where volume is low and speed is low, a painted circle should suffice but why bother when a 4 way stop does exactly the same thing? Intersections are always an opportunity for an hazard to be amplified or reduced. You just have to work out which hazard needs reduction and the opposite hazard increased. Whatever the design, it is to manage human behaviour in that location. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-08-15T03:41:34.920" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23253" PostTypeId="1" CreationDate="2018-08-15T15:03:18.913" Score="0" ViewCount="623" Body="&lt;p&gt;I want to embark on an acoustic project of designing an enclosure for some 40mm microspeakers, identical to ones found in a typical on-ear headphone solution. Are microspeaker enclosures treated in the same way as a typical loudspeaker enclosure, Just with smaller dimensions? I ask this because every acoustic/loudspeaker design book including ones with chapters on headphones in, I have read does not mention anything on enclosure design for headphones and earbuds and google is nearly a complete deadend for the level of information I am looking for... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If headphone/earbud enclosures are not treated as loudspeaker enclosures with reduced dimensions, what is the design procedure/equations for the desired acoustic &amp;amp; frequency response from the enclosure? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance&lt;/p&gt;&#xA;" OwnerUserId="17169" LastActivityDate="2018-08-15T21:55:02.653" Title="Microspeaker enclosure design for headphones" Tags="|mechanical-engineering|electrical-engineering|acoustics|audio-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23254" PostTypeId="2" ParentId="23253" CreationDate="2018-08-15T20:06:15.193" Score="1" Body="&lt;p&gt;the equations scale linearly for the linear dimensions of enclosures. However, here is why micro-enclosures don't get designed according to the same laws and equations used for full-sized speaker boxes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the primary objective of full-sized speaker enclosure design is to manage the resonances of the box/driver system so as to prevent large response peaks in the bass range while extending bass response to the lowest frequencies possible. This is a solved problem in the field and results in the so-called Theile-Small parameters for ported enclosures, which nearly everyone uses when designing hi-fi speakers.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a tiny little 40mm speaker is going to have a response peak between 5 and 10kHz and except in the so-called near field (speaker is closer to your ear than one speaker diameter), that speaker is going to put out very, very small amounts of audible bass energy. So there is no need to vent or horn-load the driver to smooth out the bass response. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the bump in the response curve at, say, 7.5kHz can be managed simply by turning down the treble knob on the amp that is driving the little speaker. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why speakers that small are usually mounted in equally small enclosures (like the ear cups of old-school Koss headphones, for example) that are just stuffed with a little loose cotton. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-08-15T20:06:15.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23255" PostTypeId="2" ParentId="8544" CreationDate="2018-08-15T20:59:22.547" Score="0" Body="&lt;p&gt;The CDV surface may be more slippery than the untreated block, but the ramp surface is the same. Adding lube to the interface will make the tilting ramp more slippery, reducing the overall coefficient of friction. Unless both surfaces are already super slippery, then the dry powder lube could conceivably cake up and increase the CoF. I think if the CoF of the dry surfaces is considerably less than the lube, the lube may increase the CoF. Ie, Mud makes it easier to drag your canoe onshore, but it plays havoc with your trailer wheel bearings.&#xA;I am a Mech, so no math out of me.&lt;/p&gt;&#xA;" OwnerUserId="17172" LastActivityDate="2018-08-15T20:59:22.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23257" PostTypeId="1" AcceptedAnswerId="23479" CreationDate="2018-08-15T21:21:18.190" Score="3" ViewCount="4025" Body="&lt;p&gt;For &lt;a href=&quot;https://en.wikipedia.org/wiki/Pressure_vessel#Stress_in_thin-walled_pressure_vessels&quot; rel=&quot;nofollow noreferrer&quot;&gt;pressure vessels in the shape of circular cylinders&lt;/a&gt;, we can use $\sigma_{hoop}=\frac{pr}{t}$ to find the minimum skin thickness by setting the hoop stress the maximum allowed value, and then solving for t. But, what about (the more complicated) &lt;a href=&quot;https://barbaramatthews.files.wordpress.com/2012/08/ellipsetank.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;elliptical cylinders&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since ellipses have a &lt;a href=&quot;https://en.wikipedia.org/wiki/Semi-major_and_semi-minor_axes&quot; rel=&quot;nofollow noreferrer&quot;&gt;semi-major axis and a semi-minor axis&lt;/a&gt; (as opposed to circles with just a single radius), it stands to reason that an elliptically cylindrical pressure vessel would have two minimum thickness (instead of just one). Is this correct? If so, how does one go about finding them?&lt;/p&gt;&#xA;" OwnerUserId="17023" LastEditorUserId="2015" LastEditDate="2018-08-26T04:01:51.943" LastActivityDate="2018-08-27T21:30:06.250" Title="Minimum thicknesses of an elliptical cylinder pressure vessel" Tags="|mechanical-engineering|structural-engineering|structural-analysis|pressure-vessel|" AnswerCount="3" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="23258" PostTypeId="2" ParentId="23253" CreationDate="2018-08-15T21:55:02.653" Score="0" Body="&lt;p&gt;Earbuds do &lt;em&gt;not&lt;/em&gt; work the same way as loudspeakers, because an earbud is working into a small &lt;em&gt;sealed&lt;/em&gt; volume of air (the space between the earbud and the wearer's eardrum) while a speaker is designed to radiate sound into an (infinite) volume of &lt;em&gt;free&lt;/em&gt; air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bass response for the two systems is completely different. For the earbuds, the motion of the driver is changing  the volume of the sealed space, and hence changing the air pressure in the space. At low frequencies, the transmission of the pressure waves at the speed of sound is irrelevant - the pressure changes can be assumed to be instantaneous. On the other hand, a loudspeaker radiates sound by creating a &lt;em&gt;travelling wave&lt;/em&gt; in the air, and for low frequencies the amplitude of the speaker cone must be (approximately) &lt;em&gt;inversely proportional to the frequency&lt;/em&gt; to generate the same sound pressure variation. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-08-15T21:55:02.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23259" PostTypeId="1" CreationDate="2018-08-16T03:55:16.227" Score="0" ViewCount="357" Body="&lt;p&gt;I have an STL file, all I want is to be able to measure dimension from point A to point B on the model in the file. How can I do this? I do have access to Solidworks but don't know if that the best software to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="17174" LastActivityDate="2018-09-15T10:00:45.387" Title="Dimensions on a STL model" Tags="|solidworks|3d-printing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23260" PostTypeId="1" CreationDate="2018-08-16T05:53:18.177" Score="1" ViewCount="39" Body="&lt;p&gt;I have a RaspberryPi that will control 4 DC motors. I will be using 2 L293d ICs. The 2 ICs will be powered by a 24v 5A Power Adapter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;  What other components do I need? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine I would be needing resistors so the ICs don't burn but I am not sure what power rating. I am also worried that I might damage my RaspberryPi. Also, the motors will be connected through a long wire (about 1.5m long) and I am worrying about too much power being lost due to the long wire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2vRQ0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2vRQ0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17176" LastEditorUserId="110" LastEditDate="2018-10-18T18:05:11.887" LastActivityDate="2018-10-18T18:05:11.887" Title="How do I make sure I don't damage electrical components (L293D, RaspberryPi, &amp; DC Motors)?" Tags="|electrical-engineering|microelectronics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23262" PostTypeId="2" ParentId="23259" CreationDate="2018-08-16T09:03:17.390" Score="1" Body="&lt;p&gt;'Evaluate&gt;Measure'. Now a dialogue box appears. now select two points to know the distance. This tool can be used for measuring area, diameter, perimeter etc &lt;/p&gt;&#xA;" OwnerUserId="16206" LastActivityDate="2018-08-16T09:03:17.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23263" PostTypeId="2" ParentId="23195" CreationDate="2018-08-16T09:30:03.893" Score="0" Body="&lt;p&gt;Where I live (Germany), WWTP are required to clean the water to a degree so it does not harm the receiving water. This is far from drinking water quality. In your area, restrictions may be more lax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience, WWTP stink, I know one you can smell kilometers away! Especially the larger ones, even when most tanks are covered and there's an odor control. The stinkiest parts are usually near the inflow, rakes etc. Sewage sludge pre-fouling is also very stinky. Secondary clarifiers should not be smelly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it may be hard to tell apart if the whole WWTP is stinking or only the effluent, if the &lt;em&gt;effluent&lt;/em&gt; stinks this could prompt further questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What are the (legal) requirements for WWTPs in your locale?&lt;/li&gt;&#xA;&lt;li&gt;How is the ecological health of the receiving water body?&lt;/li&gt;&#xA;&lt;li&gt;Does the effluent smell on all days, or only after strong rain events? Or only ater longer dry spells?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-08-16T09:30:03.893" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23264" PostTypeId="1" CreationDate="2018-08-16T11:30:31.443" Score="1" ViewCount="556" Body="&lt;p&gt;Let’s say this odd design, the arrow is the direction of water flow (don’t ask why because I’m just curious about it)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drew this out and just wanted to ask if water were to flow into the pipe like that, will there be entrance losses ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know if water were to flow the opposite direction, and the exit pipe goes into the water as a submerged exit, there are exit losses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I reverse the direction of flow and now, so is there entrance losses ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dVyd8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dVyd8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15037" LastActivityDate="2018-09-01T03:59:19.733" Title="Is there any entrance losses in this particular Pipe design?" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23265" PostTypeId="2" ParentId="23264" CreationDate="2018-08-16T11:49:48.050" Score="2" Body="&lt;p&gt;Whichever end you use as the intake will have entry losses - a suitable entrance is a “bell end” or rounded.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-08-16T11:49:48.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23266" PostTypeId="2" ParentId="23227" CreationDate="2018-08-16T13:05:21.777" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Imagine the air curtain as a 2D laminar jet ($b/h$ &gt; 20),&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Then the velocity along the centre line decays as $L^{-1/3}$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More clearly, $v_L = \bigg( \frac{3 \rho~ v_0^4~ h^2}{32\nu~ L} \bigg)^{1/3}$ ; where $\nu$ is the kinematic viscosity of air. &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Yes, possible. But you have to decide what $\%v_L$ defines the edge/boundary of the jet at $L$ for your curtain. Since the jet diffuses laterally by $\sim sec^2y$.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;3 &amp;amp; 4. You decide what is your need on $v_L$ or constraint on $v_0$, Then the first point will give you the answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need a more detailed answer or if your jet is turbulent rather laminar, I would suggest the great book &quot;Boundary-Layer Theory&quot; by  Hermann Schlichting&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-08-16T13:05:21.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23268" PostTypeId="2" ParentId="22770" CreationDate="2018-08-16T14:34:48.287" Score="2" Body="&lt;p&gt;Related: &lt;a href=&quot;https://engineering.stackexchange.com/questions/16481/system-curve-for-use-in-determining-pump-operating-point&quot;&gt;System curve for use in determining pump operating point&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Frictional pipe losses&lt;/h2&gt;&#xA;&lt;p&gt;Hydraulic systems are more complex devices then you are picturing with this question.  Frictional losses in a fluid system are a function of the flow rate - the higher the flow rate, the higher the frictional loss.  The lower the flow rate, the lower the frictional loss, in general.&lt;/p&gt;&#xA;&lt;p&gt;As such, a system with a very long pipe simply develops weaker and weaker flow, but the flow will continue.  For example, if the system were to flow, and you added more frictional pipe losses by closing a throttling valves, the system would throttle, obviously.  The flow rate would diminish.  In theory, so long as the valve was still open, the flow rate would be non-zero for a system that only cared about frictional loads.  But there are also gravitational loads/losses.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VW0uT.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VW0uT.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Gravity-induced pressure&lt;/h2&gt;&#xA;&lt;p&gt;Gravitational loads, on the other hand, are not a function of the flow rate - to raise the fluid level 10 meters, the pump needs to supply enough pressure to counteract the gravity-induced pressure (called head) at the other end of the pipe.&lt;/p&gt;&#xA;&lt;p&gt;We can then see that if the pump cannot overcome the gravitational load, then the pump will simply raise the height of liquid in the pipe until it stops going higher.  If the inlet was a reservoir and the outlet was simultaneously a reservoir, the difference in height would remain unchanged while the pump was running.  This would be a system in a quasi-static equilibrium, and as such pump manufacturers call this system a &amp;quot;Total static head&amp;quot; (though others call it a Total dynamic head), and is a useful test for defining the pump curve:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yQQLq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yQQLq.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Dead head&lt;/h2&gt;&#xA;&lt;p&gt;While this is useful for testing, it can be impractical to have a reservoir several hundred meters above the pump (for some pump types).  In these cases they can substitute the reservoir with a pressure vessel and a pressure gauge.  This leads to your final question - what happens when the outlet is closed - it's the same thing.  However, it is unwanted in applications, and in actual project design, it is given a more ominous name of &amp;quot;dead head&amp;quot; in this situation.  This is because the impeller is simply twirling in a non-flowing fluid, and any heat generated by the impeller due to inefficiencies and entropy generation simply stays close to the impeller, unable to flow out into the fluid via convection.  This can damage most pumps and is usually avoided at all costs.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-08-16T14:34:48.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23269" PostTypeId="2" ParentId="21783" CreationDate="2018-08-16T15:17:03.823" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;is it possible that the torque generated from the momentum of the beam&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm going to stop you right there. Momentum doesn't generate torque. Momentum is moment of inertia times speed. Torque is moment of inertia times acceleration. Just like speed and acceleration aren't the same thing, momentum and force aren't the same thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you apply brakes, the torque that results is whatever the braking torque is. The load doesn't dictate what the braking torque is. The load will determine how quickly it decelerates for a given braking torque, but the brake is what sets the torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, momentum is $p = mv$, or $L = I\omega$. Torque affects acceleration. It changes the speed, so it changes momentum $\tau = dL/dt$, but that's just because again $L = I\omega$, so really what that's saying is that $\tau = (I)d\omega/dt$, or the more common expression for torque, $\tau = I\alpha$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So again, to reiterate, the load doesn't set the torque. The brakes set the torque. The load reacts to torque. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am trying to work out the max torque a beam could create at the gearbox mount when the power is cut suddenly, however we don't know any of the beam characteristics such as where the point loads are, how long it takes to stop etc.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you want to calculate the load, but you don't know anything about the load, you're pretty well out of luck.&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2018-08-16T15:17:03.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23270" PostTypeId="1" CreationDate="2018-08-16T15:37:51.070" Score="1" ViewCount="24" Body="&lt;p&gt;Ethylene &lt;a href=&quot;https://bmcbiol.biomedcentral.com/articles/10.1186/s12915-016-0230-0&quot; rel=&quot;nofollow noreferrer&quot;&gt;works&lt;/a&gt; as a gaseous &quot;aging&quot; hormone for many fruits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of papers on ethylene receptor inhibitors (e.g. 1-MCP) to delay this process. I also found out that &lt;a href=&quot;https://pubs.acs.org/cen/coverstory/85/8544cover.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;1-MCP has been EPA-approved in 2002&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In 2002, 1-MCP was approved by the Environmental Protection Agency for use in the U.S. as a tool for delaying ripening. The molecule has since been approved in 27 countries, including some in the European Union, for fruits as diverse as apples, kiwis, avocados, and persimmons. [...] Fruits are treated in an enclosed room for several hours, and the treatment blocks the receptor for up to 12 days. [...] The molecule has also been particularly successful in apples—in 2006, it received two gold medals from French apple industry trade groups. In bananas, however, this molecule's impact can sometimes be misleading [...]. 1-MCP may prevent banana skin from undergoing ripening-induced yellowing, but it doesn't always stop ripening of the fruit inside, possibly because it can't pass beyond the skin. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I can't find out is how widespread such ethylene-inhibiting products are used in practice in the industry (of fruit packaging &amp;amp; shipping). Are there any statistics on this, even on small sub-sectors?&lt;/p&gt;&#xA;" OwnerUserId="480" LastEditorUserId="480" LastEditDate="2018-08-16T18:16:59.357" LastActivityDate="2018-08-16T18:16:59.357" Title="How widely used are ethylene receptor inhibitors in fruit packaging &amp; shipping?" Tags="|chemical-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23271" PostTypeId="1" AcceptedAnswerId="23472" CreationDate="2018-08-16T15:47:35.127" Score="2" ViewCount="369" Body="&lt;p&gt;$\newcommand{\a}[1]{\langle#1\rangle}$&#xA;As you can see, the left end of the beam is fixed, and a distributed load is applied to the beam. I calculated the shear forces by conventional means, but when I calculate it with singularity function I get weird results. Here are my calculations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SiTTe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SiTTe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$q(x)=-3000\a{x-0}^{-2} + 3000\a{x-0}^{-1} - \frac{2000}{3}\a{x-0}^1 + \frac{2000}{3}\a{x-0}^0$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term comes from reaction torque and the second term comes from reaction normal force. The shear force is just the integral of the $q(x)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v(x) = 3000\a{x-0}^{-1} - 3000\a{x-0}^{0} + \frac{2000}{6}\a{x-0}^2 - \frac{2000}{3}\a{x-0}^1$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see the results are false. I'll appreciate any help. &lt;/p&gt;&#xA;" OwnerUserId="17181" LastEditorUserId="1832" LastEditDate="2018-08-16T17:10:57.440" LastActivityDate="2019-03-12T16:43:09.600" Title="Shear forces along a beam using singularity functions" Tags="|structural-engineering|beam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23274" PostTypeId="1" AcceptedAnswerId="23279" CreationDate="2018-08-16T18:17:32.757" Score="1" ViewCount="60" Body="&lt;p&gt;I am looking to classify ECG signal using with neural network. I have never done signal analysis before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the samples of data I have, I have noramlised, transformed using the FFT , get the pics, amplitudes and the frequencies. &#xA;Now, How do I feed those values, to my ANN (Artificial Neural Network)? This looks like a tuple(pair of value) for each row.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry. I have no much experience in engineering but I am recently challenge myself on area that is not particularly mine.&lt;/p&gt;&#xA;" OwnerUserId="17184" LastActivityDate="2018-08-16T20:09:53.737" Title="classify ECG signal using neural network" Tags="|electrical-engineering|matlab|signal-processing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23275" PostTypeId="1" CreationDate="2018-08-16T18:17:53.140" Score="1" ViewCount="110" Body="&lt;p&gt;We know that time varying magnetic field produces voltage in a coil. But what if the wire forms a semicircle or an arc instead of a complete loop,  will such magnetic field produce voltage between the ends of the wire?  If yes,  what is the formula to calculate the induced voltage?&lt;/p&gt;&#xA;" OwnerUserId="17188" LastActivityDate="2024-03-18T14:07:20.860" Title="Induced voltage in an arc of wire" Tags="|electrical-engineering|electromagnetism|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23276" PostTypeId="2" ParentId="23275" CreationDate="2018-08-16T18:55:57.067" Score="0" Body="&lt;p&gt;Yes, it induces the voltage across the both ends, but no current can flow. This is similar to motion of conductor inside a magnetic filed, or motional EMF. The motional EMF is described as: $$\oint ( \vec{v}^{\,}\times \vec{B}^{\,})d \vec{l}^{\,}&#xA; $$&#xA;Here $\vec{v}^{\,} &#xA; $ is the speed, $\vec{B}^{\,}$ is the magnetic field, and the integral should be evaluated along the conductor curve. This situation is not exactly a time varying magnetic filed, but everything here is relative if you know what i mean. The best example is antenna. The equation doesn't provide an immediate quantitive values to your question, i try here to make it clear that the EMF can be induced in an open loop circuit.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-08-16T19:23:56.040" LastActivityDate="2018-08-16T19:23:56.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23277" PostTypeId="1" CreationDate="2018-08-16T19:15:32.193" Score="1" ViewCount="210" Body="&lt;p&gt;I am trying to model a test which has grips that can translate in X and Y directions, as well as rotate. I have been trying to model this using constraints and a reference point but have not been able to model a situation where the coupon is rotated and translated, using the top surface of the coupon as the surface being moved, the rotation is about the Z axis.&lt;a href=&quot;https://i.stack.imgur.com/dl3iG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dl3iG.jpg&quot; alt=&quot;Coupon&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="17189" LastActivityDate="2018-08-16T19:15:32.193" Title="Modeling Rotation and Translation of surface B.C. in Abaqus" Tags="|mechanical-engineering|finite-element-method|abaqus|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23278" PostTypeId="1" CreationDate="2018-08-16T20:00:12.500" Score="2" ViewCount="43" Body="&lt;p&gt;Diode and transistor have reverse breakdown voltage. Are these devices permanently destroyed under such voltage? Also,  do these devices have breakdown voltage when AC voltage is applied? How much AC voltage and frequency these can tolerate,  say $1V$ at $50$ Hz,  $1V$ at $1$ MHz?&lt;/p&gt;&#xA;" OwnerUserId="17190" LastEditorDisplayName="user14407" LastEditDate="2018-08-16T22:03:42.697" LastActivityDate="2018-08-16T22:03:42.697" Title="Breakdown voltage of electronic devices" Tags="|electrical-engineering|ac|semiconductor|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23279" PostTypeId="2" ParentId="23274" CreationDate="2018-08-16T20:09:53.737" Score="0" Body="&lt;p&gt;Since you haven't mentioned exactly what you want to classify or how you're building your network I'll give a more general explanation of how you prepare data for a classification task and the general shape of the network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important part of any classification problem using a neural network is to figure out what your classes are going to be. Sometimes you know the classes in advance (i.e. you can label your data), sometimes you need to identify the classes after collecting your data using a clustering algorithm or some other prcess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you're dealing with ECG FFT data (I assume you're referring to an electrocardiogram?) some examples of classes you might have would be specific people (i.e. heartbeat info for person A, person B, and person C) and you want your ANN to classify which person it is based on heartbeat. Another case could be a binary classification between a &quot;healthy&quot; heartbeat and an &quot;unhealthy&quot; heartbeat. For these sorts of examples you would probably be able to label the data in advance (ex. you know that dataset 17 was measured from participant #27, who was healthy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have labelled data you can think about how to use it in your network. Let's use the example of having three people who you want to identify based on heartbeat (person A, person B, and person C). You have $n$ frequency data per test, and $N$ total tests (with multiple tests per person I hope, otherwise you won't be able to do anything meaningful). We need to express the classification as some vector, so with three classes you should have a three-element vector ([1 0 0] means person A, [0 1 0] means person B, and [0 0 1] means person C). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you know your input (a vector containing $n$ magnitudes corresponding to a set of frequencies) and your output (a three-element vector indicating the class). Therefore you need a network with $n$ input nodes (one for each data point in a given test set) and three output nodes (one for each class). When you input a given frequency set to your network you'll get a 3-element output, for example [1 0 1]. You compare that to your label (let's say this set was taken from person A, so your label should be [1 0 0]) and take the difference between the two as your error, which is fed back to the network to help it learn. This process is repeated many times until the network gets good at classifying the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most packages handle the minutia of the learning process for you, so all you need to do is specify the size of your network, give it an input set and the corresponding output set (labels).&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2018-08-16T20:09:53.737" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23280" PostTypeId="1" CreationDate="2018-08-16T21:07:16.357" Score="2" ViewCount="138" Body="&lt;p&gt;Some diodes have reverse breakdown voltage of 50v and some 1000v. What determines this difference? &lt;/p&gt;&#xA;" OwnerUserId="17192" LastActivityDate="2021-11-29T10:05:10.427" Title="What determines the breakdown voltage of diodes?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23281" PostTypeId="2" ParentId="23280" CreationDate="2018-08-16T22:20:41.167" Score="1" Body="&lt;p&gt;Avalanche multiplication and quantum mechanical tunneling of carriers through the bandgap are two methods that impact breakdown voltage. Generally breakdown voltage is dependent of the struct of the PN junction. Usually the breakdown voltage is dependent on the doping density. Check out some attached references. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://ecee.colorado.edu/~bart/book/book/chapter4/ch4_5.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reverse bias breakdown&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2018-08-16T22:20:41.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23282" PostTypeId="1" CreationDate="2018-08-17T01:56:36.123" Score="2" ViewCount="103" Body="&lt;p&gt;Is there a database showing how many grams of a particular fuel generates how many Kilowatt hours of electricity?&lt;/p&gt;&#xA;" OwnerUserId="17193" LastEditorUserId="9" LastEditDate="2018-08-17T03:11:22.033" LastActivityDate="2020-10-23T07:04:10.440" Title="Converting grams of fuel per kWh of energy" Tags="|energy|energy-efficiency|renewable-energy|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23283" PostTypeId="2" ParentId="23282" CreationDate="2018-08-17T03:06:12.703" Score="1" Body="&lt;p&gt;The burning of the fuel generates thermal energy, i.e. hot gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy what it generates is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_of_combustion&quot; rel=&quot;nofollow noreferrer&quot;&gt;heat of combustion&lt;/a&gt;. Its ideal value is well known, depends mainly in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Binding_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;binding energy&lt;/a&gt; of the of the fuel and the combustion products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here comes the first problem: the burning is typically not perfect and it depends significantly on the circumstances (mainly temperature, pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the motors, generators, power plants convert this energy to electrical energy. It happens typically by the conversion to mechanical energy (typically rotating parts), and then the mechanical is converted to electrical energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All energy conversion using a different path have typically a very bad efficiency, typically below $20\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The conversion of the thermal energy to mechanical or electrical energy is subject of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Second_law_of_thermodynamics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Second law of thermodynamics&lt;/a&gt;. It is the same Law which says that&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;heat moves always from the warmer body to the colder, never back&lt;/li&gt;&#xA;&lt;li&gt;you can't build a ship which would cool the water of the sea and using its energy to move&lt;/li&gt;&#xA;&lt;li&gt;more technically, the maximal efficiency is $\frac{T_2-T_1}{T_2}$, for example if the $1000K$ burning products are cooled down to $300K$, then the maximal efficiency is $\approx 70\%$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The conversion of the mechanical energy to electrical energy is much easier, because (ideally) there is no heat exchange, it happens by coils in generators, the efficiency is typically over $90\%$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practical generators have an efficiency between $30\%$ and $70\%$, depending on the circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the burning heat of the fuel (anywhere, it is public data), and find the efficiency of the device which burns it. After that, you need only some multiplications.&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2018-08-17T03:06:12.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23284" PostTypeId="1" CreationDate="2018-08-17T04:15:51.040" Score="0" ViewCount="159" Body="&lt;p&gt;Clearly if it were easy to produce attractive looking cars for cheap, there would be a lot more 'modern' and good looking cars on the road.  While beauty certainly is in the eye of the beholder in many respects, I would rather have a car that looks more like a Porsche than a Prius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why don't most automotive companies manufacture cars that look as good as a sports car?  Specifically what is the major cost driver for the production of a sports car's body?&lt;/p&gt;&#xA;" OwnerUserId="14040" LastActivityDate="2018-12-23T19:35:57.260" Title="What are the challenges in building a 'sports car' frame and body?" Tags="|mechanical-engineering|automotive-engineering|manufacturing-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23286" PostTypeId="1" AcceptedAnswerId="23305" CreationDate="2018-08-17T05:42:37.287" Score="4" ViewCount="2715" Body="&lt;p&gt;Near where I live there's a construction site for a new supermarket. The last week or so they did things I don't quite understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some background:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The ground in our area (Ruhrgebiet) is rather clayey&lt;/li&gt;&#xA;&lt;li&gt;The site is near a small hilltop, the area generally drains into the Ruhr&lt;/li&gt;&#xA;&lt;li&gt;The site was excavated maybe 2m deep (below ground level) and filled in with 1.5m of sand gravel mix.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now the part that I don't understand:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;An excavator with a drill drilled through the sand-gravel and the soil beneath, maybe 3m total depth (below ground level), I guess 0.5 m diameter&lt;/li&gt;&#xA;&lt;li&gt;The holes are set in rows, maybe 2m apart&lt;/li&gt;&#xA;&lt;li&gt;A huge machine that on first glance looks a bit like the ones they set sheet pile walls with drives up to each hole, inserts a pipe and (appears to) fill the hole with gravel - not the sand gravel mix, but fairly large, roundish individual stones with no small infill material&lt;/li&gt;&#xA;&lt;li&gt;I see no way that a binder or water could be added to the gravel - there's just a funnel for the gravel and a loooong pipe that is entered into the hole. I don't think it's for a pile foundation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Picture of the machine (not from &quot;my&quot; site, but &lt;a href=&quot;https://www.bauforum24.biz/forums/topic/15880-keller-raupe-110/&quot; rel=&quot;nofollow noreferrer&quot;&gt;from here&lt;/a&gt; - sadly no explanation what the ultimate purpose is):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CmJrQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CmJrQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;The funnel can be hoisted up near the end of the pipe, or lowered to be filled by front loader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My theory is that this is for surface water drainage - water will percolate through the sand-gravel mix and then into lower ground layers though the highly porous gravel. There are a few other sites close by dedicated to infiltrate storm water into the ground, so maybe this is a thing in my locale. The image source has someone describing the machine purpose built for &quot;Kieseleinbausäulen&quot; (Gravel infill piles) but I don't find that term anywhere else. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately I don't know (and the only way to really find out is to ask one of the guys on site, but so far they've been to far away from the fence when I came by), my question is: To what end does one drill holes and fill them with gravel? Is my hunch correct, are there other explanations?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="5411" LastEditDate="2022-01-12T14:06:07.040" LastActivityDate="2022-01-12T14:06:07.040" Title="Drilling holes in ground and filling with gravel - to what ends?" Tags="|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23287" PostTypeId="1" CreationDate="2018-08-17T07:18:18.380" Score="2" ViewCount="38" Body="&lt;p&gt;I have just learned that &lt;a href=&quot;https://www.welt.de/finanzen/immobilien/article147422556/Werden-Haeuser-bald-vom-3D-Drucker-gebaut.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;in Germany you wouldn't be able to &lt;em&gt;print&lt;/em&gt; a house&lt;/a&gt; because of existing rules, especially regarding energy efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't it possible to print an insulating inlayer to fulfil this standard?&lt;/p&gt;&#xA;" OwnerUserId="11687" LastActivityDate="2018-08-17T10:42:04.937" Title="3D printed houses and energy efficiency in Germany" Tags="|3d-printing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23288" PostTypeId="1" CreationDate="2018-08-17T07:19:31.367" Score="1" ViewCount="36" Body="&lt;p&gt;Will a 1 kW induction cook top damage a transistor which is kept on its surface? If yes,  what is the mechanism? Is it electromagnetic induction? &lt;/p&gt;&#xA;" OwnerUserId="17194" LastEditorUserId="90" LastEditDate="2018-08-17T09:26:23.250" LastActivityDate="2019-10-31T01:39:51.217" Title="Transistor and diode on induction cookware" Tags="|electrical-engineering|electromagnetism|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23291" PostTypeId="1" CreationDate="2018-08-17T08:38:25.587" Score="3" ViewCount="42" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In addition to the machine parts we manufacture in our foundry we also produce high quality steel abrasive including low carbon round corn, cut wire, steel grit amongst others. We will be happy to advise you of the best abrasive solution for your requirements.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here what does &quot;corn&quot; imply? Is it powder?&lt;/p&gt;&#xA;" OwnerUserId="17198" LastEditorDisplayName="user16" LastEditDate="2018-08-21T23:34:48.640" LastActivityDate="2018-08-23T22:48:31.133" Title="About Shot blast machine" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23292" PostTypeId="1" AcceptedAnswerId="23314" CreationDate="2018-08-17T09:24:38.197" Score="0" ViewCount="386" Body="&lt;p&gt;If a rod is turned around it's own axis, force/torque is needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume the rod it self has a diameter $d$, a length $l$, the mass $m$ and should be turned by a motor on one side. The other side is laying on a support, that prevents it from moving sideways which is not frictionless $\mu_{rod-supp}$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/I74m7.jpg&quot; alt=&quot;Sketch&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How strong must this motor be?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I tried around a bit with Moment of Inertia. But I don't know how to go from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a real life problem I need to solve. The rod needs to turn once in 3-8 minutes and I have to find a fitting stepper motor for that, that's why I need the torque.&lt;/p&gt;&#xA;" OwnerUserId="17199" LastEditorUserId="1832" LastEditDate="2018-08-20T14:56:39.500" LastActivityDate="2018-08-20T14:56:39.500" Title="Calculate the torque needed to turn a rod it's own axis" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="1" CommentCount="4" ClosedDate="2018-08-22T14:34:14.343" ContentLicense="CC BY-SA 4.0" />
  <row Id="23293" PostTypeId="1" CreationDate="2018-08-17T10:26:56.440" Score="4" ViewCount="245" Body="&lt;p&gt;The United Kingdom has a housing crisis (see &lt;a href=&quot;https://www.ft.com/content/4b2e6e9a-a07a-11e8-85da-eeb7a9ce36e4&quot; rel=&quot;nofollow noreferrer&quot;&gt;Financial Times&lt;/a&gt;, &lt;a href=&quot;http://england.shelter.org.uk/campaigns_/why_we_campaign/the_housing_crisis/what_is_the_housing_crisis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shelter&lt;/a&gt;, &lt;a href=&quot;https://www.independent.co.uk/news/uk/home-news/housing-homeless-crisis-homes-a8356646.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Independent&lt;/a&gt;, &lt;a href=&quot;https://www.telegraph.co.uk/housing-crisis/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Telegraph&lt;/a&gt;) and a housing quality crisis (killing &lt;a href=&quot;https://www.independent.co.uk/news/business/news/cold-weather-uk-winter-deaths-europe-polar-vortex-a8224276.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;3000&lt;/a&gt;–&lt;a href=&quot;https://www.independent.co.uk/news/uk/home-news/fuel-poverty-killed-15000-people-last-winter-10217215.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;15000&lt;/a&gt; per year), due to &lt;a href=&quot;https://www.bbc.co.uk/news/business-39107973&quot; rel=&quot;nofollow noreferrer&quot;&gt;poor insulation&lt;/a&gt;.&lt;br&gt;&#xA;Cash-strapped local authorities &lt;a href=&quot;https://politics.stackexchange.com/q/10989/130&quot;&gt;aren't building&lt;/a&gt;.  Despite this,  &lt;a href=&quot;https://www.telegraph.co.uk/men/thinking-man/11092924/Why-is-Britain-so-terrible-at-domestic-architecture.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;new homes appear to be mostly built using traditional masonry&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that &lt;a href=&quot;https://en.wikipedia.org/wiki/Prefabrication&quot; rel=&quot;nofollow noreferrer&quot;&gt;prefabrication&lt;/a&gt; has higher quality and lower cost.  What are advantages of masonry over prefabrication?  Why would housebuilders in the UK choose the former?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://s0.geograph.org.uk/geophotos/05/75/35/5753519_37323f13.jpg&quot; alt=&quot;masonry house&quot;&gt;&lt;br&gt;&#xA;New British masonry house.  Source: &lt;a href=&quot;https://www.geograph.org.uk/photo/5753519&quot; rel=&quot;nofollow noreferrer&quot;&gt;geograph&lt;/a&gt; (cc-by-sa 2.0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/c/ca/Bricklayer_J5.jpg/447px-Bricklayer_J5.jpg&quot; alt=&quot;bricklayer&quot;&gt;&lt;br&gt;&#xA;Bricklayer.  Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Bricklayer_J5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikimedia Commons&lt;/a&gt; (cc-by-sa 3.0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/4/44/Flatlands_Linwood_Avs_Spring_Creek_06_-_Nehemiah.jpg/1024px-Flatlands_Linwood_Avs_Spring_Creek_06_-_Nehemiah.jpg&quot; alt=&quot;pre-fab house&quot;&gt;&lt;br&gt;&#xA;&lt;a href=&quot;http://www.nydailynews.com/life-style/real-estate/spring-creek-nehemiah-affordable-housing-success-story-east-new-york-article-1.1123089&quot; rel=&quot;nofollow noreferrer&quot;&gt;Affordable&lt;/a&gt; pre-fab houses Spring Creek, Nehemiah, New York.  Source: &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Flatlands_Linwood_Avs_Spring_Creek_06_-_Nehemiah.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikimedia Commons&lt;/a&gt; (cc-by-sa 3.0)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://www.bbc.co.uk/news/resources/idt-3cca82c0-af80-4c3a-8a79-84fda5015115&quot; rel=&quot;nofollow noreferrer&quot;&gt;flatpack skyscrapers&lt;/a&gt;, or &lt;a href=&quot;https://www.alamy.com/stock-photo-apartment-blocks-hafnarfjordur-greater-reykjavik-area-iceland-32858015.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;those apartment blocks near Reykjavík, Iceland&lt;/a&gt; (I don't know if the latter are prefab, but they look to me like they could be).&lt;/p&gt;&#xA;" OwnerUserId="3599" LastEditorDisplayName="user4139" LastEditDate="2018-08-17T21:00:28.940" LastActivityDate="2018-08-24T11:03:52.543" Title="What are some advantages of masonry over prefab for homebuilding, or, why are prefab homes so rare in the United Kingdom?" Tags="|materials|building-design|" AnswerCount="2" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="23294" PostTypeId="2" ParentId="23287" CreationDate="2018-08-17T10:42:04.937" Score="1" Body="&lt;p&gt;Of course it is possible to meet the energy efficiency standards - the article itself suggests that 3D printed houses have the &lt;em&gt;potential&lt;/em&gt; to exceed traditional building methods for efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point that the article is making under the &quot;Deutschland tut sich schwer mit dem 3-D-Druck&quot; heading, is that there are just &lt;em&gt;a lot&lt;/em&gt; of regulatory hurdles. Paperwork. Not only does your potential 3D printed house need to be suitably efficient, but you need to &lt;em&gt;prove to the authorities&lt;/em&gt; that it will be, before you've built it, or you won't get a permit to start the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For such a new technology with little existing data to draw upon, convincing those in charge that your design meets &lt;em&gt;all&lt;/em&gt; of the standards (energy efficiency is just one example) will be extremely difficult, and explains why research into this area is not a priority for the established house builders in the country.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-08-17T10:42:04.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23295" PostTypeId="1" CreationDate="2018-08-17T13:08:00.797" Score="0" ViewCount="287" Body="&lt;p&gt;Sometimes the dropping goes on so long(10-15 minutes) that the test is thrown out and the valve is opened and filled with water for a longer period of time. Pressure testers claim that it is because of air, but when I ask why does air in the valve cause a bad test, they can't answer me. Some technicians seem to think that it is because air is more compressible, but this doesn't make a lot of sense to me because some of our products are tested with nitrogen. I feel like it is possible that the system is not in equilibrium and air bubbles are sort of crawling around making their way to gather with a larger pocket of air in a stable pocket of the liquid.&lt;/p&gt;&#xA;" OwnerUserId="13451" LastActivityDate="2023-07-21T13:06:45.867" Title="Why does the water pressure initially drop then stabilize when pressure testing a valve?" Tags="|fluid-mechanics|pressure|valves|air|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23298" PostTypeId="2" ParentId="23160" CreationDate="2018-08-17T14:35:42.743" Score="0" Body="&lt;p&gt;Just FYI I was able to solve my design using RISA 3D. This is an excellent software! Ended up designing for a 125 MPH / 40 PSF wind load. Tested the frame in both horizontal and 50 degrees from horizontal orientations. Test loads were applied normal to the panel face and parallel to the ground in two different scenario groups. Loads were applied downward on all panels at 40 PSF, upward on all panels at 40 PSF, combination load: upward on the top panel / downward on the two lower panels, and the reverse of that combination load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the moment I am neglecting snow loads based on the fact that as you move north the required solar panel tilt angle becomes greater up to about 45 degree slope. And where the solar panel array would be most vulnerable (near horizontal) is in regions with zero snow load. But I will have to revisit this topic later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ended up using 6061-T6 3x2.03 I-beams for the long side members. Was even able to eliminate two support struts in this configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vortex shedding may be a problem however depending on how the assembly of frames is arranged. Structure frequency is 7 to 77 Hz for the first five modes. Based on the Strouhal number vortex shedding frequencies could range from 11 to 26 Hz. However when I analyze the shedding frequency across the assembly of all six planned frames the range drops to 2.2 to 2.7 Hz. Some type of vortex suppression devices may be required on the long struts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KwVqq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KwVqq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16668" LastEditorUserId="16668" LastEditDate="2018-08-17T14:45:07.920" LastActivityDate="2018-08-17T14:45:07.920" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23299" PostTypeId="1" CreationDate="2018-08-17T15:09:30.477" Score="1" ViewCount="35" Body="&lt;p&gt;Household supplies are 110v AC.  Is it possible to series connect two plug points and get 220v AC? &lt;/p&gt;&#xA;" OwnerUserId="17203" LastActivityDate="2018-08-17T15:23:20.073" Title="Series connection of household AC" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23300" PostTypeId="2" ParentId="23299" CreationDate="2018-08-17T15:23:20.073" Score="3" Body="&lt;p&gt;No, when go connect 2 plug points in series you will connect the live from one directly to the neutral of the other. That is a short and will trip the breaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some circuits are wired in split-phase where there are 2 lives and a neutral. Live to live will be 220V and either live to neutral will be 110V. So in that specific case you could get 220V from 2 110V outlets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But &lt;strong&gt;DON'T DO THAT&lt;/strong&gt; especially when you don't know how electricity works. Instead pay an electrician to install a 220V outlet where you need it.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2018-08-17T15:23:20.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23303" PostTypeId="2" ParentId="23295" CreationDate="2018-08-17T18:32:42.567" Score="1" Body="&lt;p&gt;I can think of two reasons why this might happen, though I don't know for sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, air trapped in the test piece could be slowly dissolving into the water, which will change the overall density and hence the pressure. This doesn't happen when pure nitrogen is used because the nitrogen is not dissolving into anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the mechanical parts of the valve could be getting adjusted to the pressure and might creep for a while until they balance the water pressure. Things like rubber valve seats and gaskets will do this.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2018-08-17T18:32:42.567" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23304" PostTypeId="2" ParentId="283" CreationDate="2018-08-17T19:10:55.007" Score="0" Body="&lt;p&gt;I know this is an old post, however there is a fairly simple way to determine the gas pressure required for launching things from a tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume isentropic expansion, so that the expression $\frac{p}{p_0}=\left(\frac{V_0}{V} \right)^\gamma$ holds. Here, $p_0$ and $V_0$ are the initial pressure and volume behind the projectile, $p$ and $V$ are the pressure and volume after the projectile has moved some amount, and $\gamma$ is the specific heat capacity ratio of the driver gas (air = 1.4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now integrate the pressure on the projectile over the volume it covers in the tube to get its kinetic energy at exit (i.e. &quot;PV&quot; work):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;KE=\frac{1}{2}mv^2 = \int_{V_0}^{V_e}p\cdot dV&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $m$ is projectile mass, $v$ is exit velocity, and $V_e$ is the volume of gas behind the projectile just as it's exiting. Now just substitute the isentropic equation into the integral and solve it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{1}{2}mv^2 = \int_{V_0}^{V_e} p_0 \cdot \left(\frac{V_0}{V} \right)^\gamma dV&#xA;$$&#xA;$$ = p_0 V_0^\gamma \int_{V_0}^{V_e} \frac{dV}{V^{\gamma}}&#xA;$$&#xA;$$ = \frac{p_0 V_0^\gamma}{1-\gamma} \left( V_{e}^{1-\gamma} - V_{0}^{1-\gamma} \right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can just plug in values for everything else and solve for $p_0$.&lt;/p&gt;&#xA;" OwnerUserId="1560" LastActivityDate="2018-08-17T19:10:55.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23305" PostTypeId="2" ParentId="23286" CreationDate="2018-08-17T20:25:36.743" Score="5" Body="&lt;h1&gt;You might be right&lt;/h1&gt;&#xA;&lt;p&gt;Your first thought about using the holes for drainage could be right. This would be similar to a &lt;a href=&quot;https://en.wikipedia.org/wiki/French_drain&quot; rel=&quot;noreferrer&quot;&gt;French Drain&lt;/a&gt; or a &lt;a href=&quot;https://en.wikipedia.org/wiki/Dry_well&quot; rel=&quot;noreferrer&quot;&gt;Dry Well&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;French drains are usually horizontal and dry wells aren't usually installed on hills.&lt;/p&gt;&#xA;&lt;h1&gt;The likely option&lt;/h1&gt;&#xA;&lt;p&gt;What is more likely is that the contractor is installing a form of compacted aggregate pier. See this for a description from a company that does them &lt;a href=&quot;https://www.haywardbaker.com/solutions/techniques/vibro-aggregate-piers&quot; rel=&quot;noreferrer&quot;&gt;(vibro aggregate pier)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The main idea is that a hole is drilled and then filled with compacted aggregate. This provides a firmer foundation without treating the entire area or installing steel or concrete piles. The aggregate piers are sometimes used instead of micropiles, but they aren't as good of replacement for large concrete piers or driven steel piles.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-08-17T20:25:36.743" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23306" PostTypeId="2" ParentId="22775" CreationDate="2018-08-17T22:23:01.903" Score="1" Body="&lt;p&gt;Accurate circumference measurements by string of tape ted to be difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest buying or making some simple external cailpers &lt;a href=&quot;https://i.stack.imgur.com/ySGtH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ySGtH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And use them to transfer the measurement you your vernier calipers using the external jaws on the back  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will lose a bit of accuracy in the transfer but not that much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some skill is required to use jaw calipers as you need to feel for the actual diameter by rocking them and adjusting untill you just feel very light touch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other option is to measure the end of the tube ie rotate the vernier 90 degrees, again you need to feel for the max measurement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having aid that 0.1mm accuracy is perhaps a bit ambitious without the right tools and ERW tube can easily vary by more than that depending on where you measure it and there are only a finite number of nominal  diameters standard tube can possibly be. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-08-17T22:23:01.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23307" PostTypeId="2" ParentId="23293" CreationDate="2018-08-17T22:25:01.133" Score="0" Body="&lt;p&gt;One size doesn't fit all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Putting the US prefab houses mentioned by Muze and blown away by stronger wind, if you go with the european prefab, you can quickly end up with architecture like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dCF5z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dCF5z.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.sas1946.com/main/index.php?topic=41786.0&quot; rel=&quot;nofollow noreferrer&quot;&gt;photo credit&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ask anyone living there to list the disadvantages, you won't hear the end of the list. One could say the entire list of advantages consists of &quot;it's affordable to build and maintain,&quot; &quot;they can be built fast&quot; and &quot;one can live in there.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest part of the prefab house problem, I think, is lack of individualism. If you make &quot;short runs&quot; of individualized prefabs you're losing the economy of scale; price goes up, list of other disadvantages remains. If you go large scale, it's hard not to end with monolithic, dull, depressing architecture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brick offers full flexibility. Fancy curved walls, levels that don't correspond to the integer 'floor#', small quirks, fancy decorative styles, and a house that is tailored to your specific needs. It also doesn't require the sort of heavy machinery heavy prefab construction does. It's extensible, modifiable, can be arbitrarily individualized; plays well with other technologies - wood, steel, glass, aluminum. In short, a brick house can be pretty much anything, from a multi-floor block of flats, to a little fancy castle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aHe6G.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aHe6G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prefabs don't offer anything near this flexibility. Even room sizes come in small multiples of slab sizes: 1-wide, 2-long bedroom; 2-wide, 2-long living room, 1-wide 1-long bathroom and kitchen, plus half-slab wide, 2-long corridor... it gets depressing. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-08-17T22:25:01.133" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23308" PostTypeId="2" ParentId="63" CreationDate="2018-08-17T22:33:29.287" Score="5" Body="&lt;p&gt;People using the road make mistakes (like running stop signs and red lights), always have and always will. Crashes will always be with us, but they need not result in fatalities or serious injury. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern roundabouts are the safest form of intersection in the world - the intersection type with the lowest risk of fatal or serious injury crashes - (much more so than comparable signals).  Modern roundabouts require a change in speed and alter the geometry of one of the most dangerous parts of the system - intersections.  The reduction in speed and sideswipe geometry mean that, when a crash does happen at a modern roundabout, you usually need a tow truck, not an ambulance.  Visit the Insurance Institute for Highway Safety for modern roundabout FAQs and safety facts.  Roundabouts are one of several proven road safety features (FHWA).&lt;br&gt;&#xA;The life saved may be your own.&#xA;&lt;a href=&quot;https://safety.fhwa.dot.gov/provencountermeasures/roundabouts/&quot; rel=&quot;noreferrer&quot;&gt;https://safety.fhwa.dot.gov/provencountermeasures/roundabouts/&lt;/a&gt; &#xA;&lt;a href=&quot;https://safety.fhwa.dot.gov/intersection/innovative/roundabouts/&quot; rel=&quot;noreferrer&quot;&gt;https://safety.fhwa.dot.gov/intersection/innovative/roundabouts/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern, slow and go, roundabout intersections have less daily delay than a stop light or stop sign, especially the other 20 hours a day people aren’t driving to or from work (it’s the #2 reason they’re built).  Average daily delay at a signal is around 12 seconds per car.  At a modern roundabout average daily delay is less than five seconds. Signals take an hour of demand and restrict it to a half hour, at best only half the traffic gets to go at any one time.  'At best' because traffic signals must have the yellow and all red portion (6+ seconds per cycle) for safety, and modern roundabouts do not.  At a modern roundabout, drivers entering from different directions can all enter at the same time.  Don’t try that with a signalized intersection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First cost is the wrong way to compare projects. It would be like buying a car without knowing the fuel economy or safety of the thing, just its price to buy.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; Present Value Life Cycle Cost Analysis (LCCA) is the best way to compare two or more choices.  When comparing modern roundabouts to signals for a 20-year life cycle (the standard period), modern roundabouts usually cost less.  Costs to compare include: first cost (design/land/construction), operation and maintenance (electricity, re-striping, upgrades, etc.), crash reduction (what’s your/your family’s safety worth?), daily delay (what’s your time worth?), daily fuel consumption (spend much on gas?), point source pollution (generated by stopped vehicles = health cost), area insurance rates (this costs more where it is less safe to drive).  Each of these things, and others, can be estimated for any two choices and everyone near or using the project area will pay some portion of all these costs (and gain the benefits).    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Many people confuse other and older styles of circular intersections with modern roundabouts. High speed, east coast rotaries, large multi-lane traffic circles (Arc D’Triomphe, Dupont Circle), and small neighborhood traffic circles are not modern roundabouts. &#xA;The Brits even call a merry-go-round a kid’s roundabout.&#xA;Go to  &lt;a href=&quot;http://www.k-state.edu/roundabouts/photos.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.k-state.edu/roundabouts/photos.htm&lt;/a&gt; to see pictures.&lt;br&gt;&#xA;What is, and is not, a modern roundabout:&#xA;WA DOT:  &lt;a href=&quot;http://www.youtube.com/watch?v=vsCoI7lERGE&quot; rel=&quot;noreferrer&quot;&gt;http://www.youtube.com/watch?v=vsCoI7lERGE&lt;/a&gt; &#xA;NJ traffic circles: &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_traffic_circles_in_New_Jersey&quot; rel=&quot;noreferrer&quot;&gt;https://en.wikipedia.org/wiki/List_of_traffic_circles_in_New_Jersey&lt;/a&gt; &#xA;NJ wins award for building roundabout: &#xA;&lt;a href=&quot;https://www.prnewswire.com/news-releases/national-roadway-safety-award-winners-announced-300556007.html&quot; rel=&quot;noreferrer&quot;&gt;https://www.prnewswire.com/news-releases/national-roadway-safety-award-winners-announced-300556007.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, safety projects traditionally qualify for 80% Federal reimbursement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mini-roundabouts are less common in North America, but frequently used in the UK.  Their footprint is smaller, making them suitable for retrofit situations with right of way constraints (and lower cost).  They are also commonly used where truck U-turns are not needed, so achieve all the safety benefits of compact modern roundabouts at a much lower cost.  They are all truck apron, and in the UK are sometimes just paint on the road.&lt;br&gt;&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=-wok-PZ4pGI&amp;amp;feature=youtu.be&quot; rel=&quot;noreferrer&quot;&gt;https://www.youtube.com/watch?v=-wok-PZ4pGI&amp;amp;feature=youtu.be&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Single-lane modern roundabouts (50-120 feet in diameter) can handle intersections that serve up to 20,000 vehicles per day with peak-hour flows between 2,000 and 2,500 vehicles per hour.  Two- and three-lane modern roundabouts (150-220 feet in diameter) can serve up to 60,000 vehicles per day and handle 2,500 to 5,500 vehicles per hour.  Right-turn slip lanes can increase those numbers if needed (just like for signal intersections).  Much depends on how balanced the entries are, but only in determining how many lanes are needed for each movement – just like at a signal controlled intersection.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="17206" LastActivityDate="2018-08-17T22:33:29.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23309" PostTypeId="1" CreationDate="2018-08-18T07:09:30.590" Score="1" ViewCount="74" Body="&lt;p&gt;I'm designing a rolling machine for reduction of steel ingots of $5 mm$ to $1.96 mm$ thick. I can convince myself using equations why increasing the coefficient of friction (in case of lubrication system fails) results in evenly distributed load in rolling machines, but i can't describe intuitively why this happens. &lt;a href=&quot;https://i.stack.imgur.com/KpJgf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KpJgf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the friction coefficient increase, normal forces increase as wel and i expect the peak to move upward, but why it moves to the right as well ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal axis, represents the angle of contact zone, and the vertical axis shows the dimensionless stress. $\frac{\theta}{\theta_0} = 0$ means exit angle(reduced) and $\frac{\theta}{\theta_0} = 1$ means entry angle.  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-08-18T07:14:39.097" LastActivityDate="2018-08-18T07:14:39.097" Title="Why increasing the coefficient of friction results in evenly distributed load in rolling machines?" Tags="|mechanical-engineering|steel|friction|rolling|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23310" PostTypeId="1" CreationDate="2018-08-18T07:21:55.790" Score="1" ViewCount="539" Body="&lt;p&gt;This is not a real system but rather something I made up to understand how the placement of pressure gauge in a tank takes its readings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here’s a diagram of it- &lt;a href=&quot;https://i.stack.imgur.com/rEp4y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rEp4y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure gauge placed in the middle of the tank. Let’s say I’m working in gage pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which point in the tank does this pressure gauge measures until ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it measure till the middle point only ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example - pressure at the level where pressure gauge is placed - &#xA;P (vapor) + P (liquid at h=1m) = $120 + \rho g h$ = ____ kpa &lt;/p&gt;&#xA;" OwnerUserId="15037" LastEditorUserId="15037" LastEditDate="2018-08-18T07:58:17.837" LastActivityDate="2022-04-14T07:07:10.760" Title="How does placement of a pressure gauge in a tank of fluid affect the pressure readings?" Tags="|fluid-mechanics|pressure|vapor-pressure|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23312" PostTypeId="2" ParentId="22775" CreationDate="2018-08-18T12:51:40.277" Score="0" Body="&lt;p&gt;This should work in theory, but the accuracy would decrease with increasing diameter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your image shows measuring a chord of the circle at an unknown distance from the center of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you place a gauge block (of known thickness) between the beam of the caliper and the circumference of the tube, you can measure the length of a second chord of the circle at a known radial displacement from the first chord.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter of the circle should be obtainable from the lengths of the two chords and the known radial distance between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The illustration at the top of this page is applicable:&#xA;&lt;a href=&quot;http://jwilson.coe.uga.edu/emt725/ParallelChords/ParallelChords.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://jwilson.coe.uga.edu/emt725/ParallelChords/ParallelChords.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result would be affected by the roundness of the tube, surface roughness on the circumference, the perpendicularity of the caliper to the axis of the tube, and the accuracy with which the thickness of the gauge block is known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't done this, worked out the formula, or determined how sensitive the result would be to the variables listed above, but it might be an adequate and inexpensive approach in some cases.&lt;/p&gt;&#xA;" OwnerUserId="13553" LastEditorUserId="13553" LastEditDate="2018-08-18T13:27:54.507" LastActivityDate="2018-08-18T13:27:54.507" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23313" PostTypeId="1" CreationDate="2018-08-18T14:39:59.253" Score="1" ViewCount="1397" Body="&lt;p&gt;In welding, we have different types of grooves like V-shaped groove, U-shaped groove, Double V, Double U, J and Double J. What is the advantage of each kind over the other? &lt;/p&gt;&#xA;" OwnerUserId="17075" LastActivityDate="2018-10-08T13:39:50.923" Title="What are the advantages of different types of groove in welding?" Tags="|manufacturing-engineering|welding|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23314" PostTypeId="2" ParentId="23292" CreationDate="2018-08-18T14:42:29.560" Score="1" Body="&lt;p&gt;If you assume that 100% of the weight is carried by the support, then the normal force is just $mg$, so the friction force is $mg\mu$, and the torque required to overcome friction is $mg\mu d/2$.  If you want to be able to acceleration the rod quickly, then the polar moment of inertia will come into play.  However, since you indicated a rate of 1 rev every few minutes, the accelerations involved here will be trivial. The friction force will dominate.  Take the calculation for friction and add a little margin and call it a day.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-08-18T14:42:29.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23315" PostTypeId="1" AcceptedAnswerId="23330" CreationDate="2018-08-18T15:53:11.623" Score="1" ViewCount="225" Body="&lt;p&gt;I read that it &quot;is a positive pressure&#xA;against the environment&quot;. But I am not too sure what that means.&lt;/p&gt;&#xA;" OwnerUserId="17210" LastActivityDate="2018-08-20T21:26:19.680" Title="Why is the saturation pressure of the refrigerant desired to be slightly greater than atmospheric pressure in an evaporator?" Tags="|thermodynamics|refrigeration|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23316" PostTypeId="2" ParentId="22775" CreationDate="2018-08-18T18:43:37.483" Score="0" Body="&lt;p&gt;One other possible and simple solution would be to tape or twist-wire a couple of pieces of flat aluminum to the jaws of the caliper. The diameter of the pipe would be the measurement minus twice the thickness of the flat aluminum. Pretty straightforward.&lt;br&gt;&lt;br&gt;There's also &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B007OJVYIC&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; item on Amazon, if you're willing to spend fewer than 10 bucks.&lt;/p&gt;&#xA;" OwnerUserId="12535" LastActivityDate="2018-08-18T18:43:37.483" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23317" PostTypeId="2" ParentId="23313" CreationDate="2018-08-18T19:36:59.960" Score="0" Body="&lt;p&gt;Mostly a function of thickness. It is expensive to fill a groove with weld metal so try to use a low volume but simple groove. Such a ( simple) V groove at 1/2 &quot; thickness and a U groove at 3&quot; thickness.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-08-18T19:36:59.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23318" PostTypeId="1" CreationDate="2018-08-18T21:38:24.870" Score="0" ViewCount="140" Body="&lt;p&gt;The process of punching a soft material uses a hallow punch with beveled and sharpened edges that is driven into the material. The backing material is usually a soft metal or wood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on such a process and have a few questions for the community.&#xA;I am punching 3 mm discs of thickness 0.2 mm from a elastomer material with tensile modulus of 220 MPa and a max tensile strength of 35 MPa. The disc remains within the hallow punch and is extracted afterwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have measured the force required to punch using a force gauge and found that 140 N was required to punch and retain the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end process will use a pneumatic cylinder to drive the punch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is the punch force calculation the following - &quot;Punch Force = perimeter x thickness x shear stress&quot;? Can the shear stress be equated to the materials tensile strength? Using the above, I get a punch force of 66 N. This is well below what is measured using the gauge. Why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What is the effect of a blunt or dull blade on the punch force? Hows does the blades increase in contact area influence the punch force (shear)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) In the above example, what is the relationship between the normal force/pressure to the punch force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) How would you select a backing material? Softer than the punch's material (HSS Steel), but not too soft as to deform when struck by the punch?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="17212" LastEditorUserId="17212" LastEditDate="2018-08-18T21:43:58.970" LastActivityDate="2018-09-18T04:30:55.927" Title="Punching of soft materials" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23319" PostTypeId="2" ParentId="23318" CreationDate="2018-08-18T22:22:01.307" Score="1" Body="&lt;p&gt;The force requires will indeed depends on the sharpness of the tool and indeed its overall geometry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is complicated by the fact that, as the punch blade as finite thickness you are also pushing material out of the way and have friction between the tool and the sides of the cut. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally 'sharpness' isn't really a very well defined property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cutting a soft material with a wad punch is a somewhat different situation from punching say sheet metal where you are shearing the material. Resistance to cutting with a blade isn't straightforwardly related to tensile strength. Rate effects may be significant too. So I don't think that formula you quote is very appropriate in this context. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a few sensible options for the backing board, hard rubber, particle board, end-grain hardwood, polypropylene etc, whatever you use you are probably going to want to be able to change it fairly frequently. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="1324" LastEditDate="2018-09-18T04:30:55.927" LastActivityDate="2018-09-18T04:30:55.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23320" PostTypeId="1" AcceptedAnswerId="23322" CreationDate="2018-08-18T22:43:54.013" Score="0" ViewCount="3665" Body="&lt;p&gt;I found a technical drawing of a popular guitar shape online and I'm having trouble to understand a certain annotation on some arcs in the drawing. It basically says stuff like &quot;0.43R 2PL&quot; or &quot;3PL&quot; (see attached).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any idea what this could mean? My current guess is that it means that there are X PL other curves in the loop that have the same radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3uquT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3uquT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.tdpri.com/threads/d-size-tele-body-blueprint-files-here.74504/page-30#post-6860876&quot; rel=&quot;nofollow noreferrer&quot;&gt;Drawing Source&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17213" LastEditorUserId="17213" LastEditDate="2018-08-18T23:24:38.483" LastActivityDate="2018-08-19T00:35:18.473" Title="2PL / 3PL annotation on arc in technical drawing" Tags="|technical-drawing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23322" PostTypeId="2" ParentId="23320" CreationDate="2018-08-19T00:35:18.473" Score="1" Body="&lt;p&gt;Your guess is correct. &quot;0.430R 2PL&quot; means there are two places where there is a curve with the given radius. Usually it is obvious from looking at the complete drawing where all the curves are - for example in the OP's complete drawing, &lt;a href=&quot;http://www.tdpri.com/attachments/tele_body_drawing_reve-1-pdf.347484/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.tdpri.com/attachments/tele_body_drawing_reve-1-pdf.347484/&lt;/a&gt;, the part being referred to is symmetrical (though it is oriented at an angle to the symmetry axes of the complete object).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is more commonly used for things like small holes, where you might see something like &quot;0.1R 36PL&quot; rather than each one of the 36 holes being dimensioned individually. There are some examples of this on the OP's complete drawing.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-08-19T00:35:18.473" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23323" PostTypeId="1" CreationDate="2018-08-19T02:36:54.947" Score="1" ViewCount="953" Body="&lt;p&gt;I am trying to build a super-efficient electric boat. The dilemma I am coming across is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In normal long range running mode the boat will go at a slow 2.5 knots and only require 400 W power. This is the optimally efficient setting through the water. However for safety and practicality reasons like getting around in high winds etc I still need about 5kw max power available on the shaft. From my understanding any electric motor with a 5 kW capacity will be very inefficient at 400 W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my proposed solution is to put 2 motors on one shaft, a small one optimized for 400 W and a big one, the small one will be rigidly connected to the shaft and the big one via a clutch. I will then manually manually engage and disengage the large motor with a switch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please advise if this is a normal approach of if there is a better way to handle my situation? Am I misunderstanding something or reinventing the wheel?&lt;/p&gt;&#xA;" OwnerUserId="17214" LastEditorUserId="90" LastEditDate="2018-08-19T04:22:16.927" LastActivityDate="2018-08-19T05:00:32.510" Title="Multiple Electric Motors on single shaft to improve efficiency" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23324" PostTypeId="2" ParentId="23323" CreationDate="2018-08-19T05:00:32.510" Score="1" Body="&lt;p&gt;While the mechanism you described will produce greater motor efficiency than turning down a 5kW motor to less than 10%, you will lose overall system reliability IMHO with the clutch/2 motor system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clutch/2 motor system should work as described, but why not just use a VFD?&lt;/p&gt;&#xA;" OwnerUserId="17215" LastActivityDate="2018-08-19T05:00:32.510" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23325" PostTypeId="2" ParentId="17803" CreationDate="2018-08-19T05:23:14.763" Score="1" Body="&lt;p&gt;Since the composition is  given in mole % then your calculation basis must be in moles or kilo moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start with a Kg basis then you need to know the weight composition then calculate the mole composition.&lt;/p&gt;&#xA;" OwnerUserId="17216" LastEditorUserId="17216" LastEditDate="2018-08-19T20:21:22.557" LastActivityDate="2018-08-19T20:21:22.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23326" PostTypeId="1" CreationDate="2018-08-19T11:23:11.600" Score="2" ViewCount="135" Body="&lt;p&gt;Specifically, does in pairs mean that there are 2 C20019 members per indicated line on a technical drawing?&lt;/p&gt;&#xA;" OwnerUserId="17220" LastActivityDate="2018-08-19T11:23:11.600" Title="What does C20019 IN PAIRS AT 900 MAX CTS WITH 2 ROWS OF BRIDGING AT 1/3 POINTS mean in a technical drawing?" Tags="|civil-engineering|technical-drawing|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23327" PostTypeId="1" CreationDate="2018-08-19T14:21:08.617" Score="1" ViewCount="124" Body="&lt;p&gt;Imagine a source of oscillating magnetic field, say 10 Ghz.  Now change in a  magnetic field propagates at the speed of light,  assuming a vacuum. Consider point B to be 30 cm from the source. The change in magnetic field will take 1 ns to reach B.   Does this mean that B will experience changes at 10 Ghz but delayed by 1 ns? Is my inference correct or am I wrong? &lt;/p&gt;&#xA;" OwnerUserId="17222" LastEditorUserId="90" LastEditDate="2018-08-19T15:01:02.873" LastActivityDate="2018-10-19T01:00:24.770" Title="Propagation delay of magnetic field" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23328" PostTypeId="1" AcceptedAnswerId="23329" CreationDate="2018-08-19T14:22:30.980" Score="0" ViewCount="9432" Body="&lt;p&gt;I am trying to interpret the hole specifications in the attached diagram. What does each component mean? I understand what 'Thru All' means, does 4x mean 4 holes? What is the diameter of the hole? is it 5mm or M6 = 6mm? What is 6H?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kRGPm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kRGPm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17221" LastActivityDate="2018-08-19T14:55:45.773" Title="Understanding hole specifications" Tags="|mechanical-engineering|technical-drawing|drawings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23329" PostTypeId="2" ParentId="23328" CreationDate="2018-08-19T14:55:45.773" Score="3" Body="&lt;p&gt;4 x means it applies to the four instances that look similar on the drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ø 5 means the hole diameter (drill size) is 5mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Thru all” is added here to remove the requirement for a section view to show hole depth. You can’t tell by the outlines whether it’s through, or you can see the bottom of the hole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M6x1.0 - 6H is the thread and tolerance to be tapped into the drilled hole - again, through all. With blind holes it’s often the case that a minimum thread depth is specified that’s less than the hole depth to avoid multiple tapping operations using a bottoming tap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This drawing would likely be better using a centreline and symmetry to dimension the hole positions. Even if distances to the bottom right corner are critical, it still needs some sort of ‘construction lines’ to e.g. show that the top left hole is at the same vertical height as the top right. This is a fairly simple part, but in more complex geometry you can’t assume things like that unless they are explicitly shown. Adding centrelines would do this.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-08-19T14:55:45.773" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23330" PostTypeId="2" ParentId="23315" CreationDate="2018-08-19T14:56:25.100" Score="1" Body="&lt;p&gt;Saturation pressure is vapor pressure $p_{vap}$. The statement is saying to have vapor pressure greater than external (atmospheric) pressure $p_{vap} &amp;gt; p$. In a hypothetical system where a liquid is evaporating with no boundaries, we could evaporate with $p_{vap} = p$. In an actual evaporator with physical boundaries (i.e. a container), the evaporating liquid must &quot;push out&quot; existing vapor. To do so, the existing vapor must &quot;flow out&quot; of the container. Flow requires a pressure difference. Hence $p_{vap} &amp;gt; p$ for effective operation in real evaporators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One point of confusion remains in the original statement. The phrase &quot;atmospheric pressure&quot; should be replaced by &quot;applied pressure&quot;. Consider the applied pressure to first order as the pressure felt in the tank before evaporation starts.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-08-20T21:26:19.680" LastActivityDate="2018-08-20T21:26:19.680" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23331" PostTypeId="1" AcceptedAnswerId="23351" CreationDate="2018-08-19T17:05:52.503" Score="0" ViewCount="77" Body="&lt;p&gt;I know that on the moon, while the daytime surface temperature can be above 270° F, in the shade but a few feet above the surface temperature can get quite low.  Wondering whether a similar phenomenon can be observed if a vaccum is created around an object which is suspended in he middle of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say for example, a steel ball is suspended in the middle of a  plexiglass cube, by 3 plexiglass threads, and the interior volume of the cube is 1 cubic meter,  and the ball has a diameter of 10cm, the box walls are 5cm thick.  The cube has been evacuated to the practical limit of modern vacuum pumps.   Let's say the steel ball is chilled to a temperature of -20 c (measured at the time the evacuation is complete).   Let's say it's 70° F outside, and the box is sitting on a plexiglass tripod in the shade, shielded from direct sunlight.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the temperature reading of the steel ball, approximately, after say 2 weeks? (i.e. at equilibrium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I don't mean to be overly prescriptive, but trying to be sufficiently specific to avoid answers like &quot;it depends on how thick the walls are&quot; and similar)&lt;/p&gt;&#xA;" OwnerUserId="17223" LastEditorDisplayName="user14407" LastEditDate="2018-08-19T20:04:29.683" LastActivityDate="2018-08-20T21:50:16.263" Title="How cold for lack of heat?" Tags="|thermodynamics|thermal-conduction|thermal-insulation|thermal-radiation|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="23332" PostTypeId="2" ParentId="22342" CreationDate="2018-08-19T18:36:48.633" Score="0" Body="&lt;p&gt;The tallest brick building is the Monadnock building in Chicago at 215 ft. A solid brick structure could be made taller ; with no interior space. All very tall buildings are steel . The stone or glass facing you see is hanging on the steel; &quot;curtain wall&quot; design. Foundations that I know of are based on steel pilings drilled hundreds of feet into bed rock; Criteria are based on the strength of bedrock. Correction ; Since the Sears Tower .they are concrete/steel composites, not steel. I hate when I need to learn something new. I worked for Amoco when they built the SOB ( Standard Oil Bldg ) , aka Aon Tower, so that was my reference point.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-08-20T02:00:20.880" LastActivityDate="2018-08-20T02:00:20.880" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23333" PostTypeId="2" ParentId="2158" CreationDate="2018-08-19T21:17:50.997" Score="1" Body="&lt;p&gt;15-30 mins under any vacuum pump will be fine. If there are any air bubbles still there at the surface I recommend to burst it with pressurized air gun.&#xA;Make sure that your vacuum chamber lid is sealed well. Some time it is useful to use vacuum grease to achieve that.&lt;/p&gt;&#xA;" OwnerUserId="17216" LastActivityDate="2018-08-19T21:17:50.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23334" PostTypeId="2" ParentId="23315" CreationDate="2018-08-19T21:46:10.323" Score="0" Body="&lt;p&gt;This isn't always true. Industrial low pressure ammonia vapor refrigeration was usually operated below external pressure so you didn't poison people or contaminate stuff in the event of a leak. It was relatively easy to remove atmospheric leakage into the systems, and it didn't pose any great threat to the system. With modern equipment, you don't want anything getting into the system. The systems are designed to be hermetically sealed and they don't suffer contamination well. Nobody wants to purge leaked atmosphere out of there refer system once a day like was common practice in the past. These old systems basically used to require an operator to run them, like old boilers and everything else back then. Old icemakers used to be a problem because many operated near atmospheric pressure on the low side, and if there was a high side leak, the low side would suck air and you lost your cool. This is why they all had little bubble windows in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See table 2.9 in the following FAO report on seawater ice making for fisheries.&#xA;&lt;a href=&quot;http://www.fao.org/docrep/006/Y5013E/y5013e05.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.fao.org/docrep/006/Y5013E/y5013e05.htm&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as a practical matter, detecting a leak is easier if the low side is below atmospheric, but locating the leak is easier with positive pressure, and except for residential comfort air, you won't often be shutting a system down to go on a leak hunt.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-08-19T22:42:57.850" LastActivityDate="2018-08-19T22:42:57.850" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23335" PostTypeId="2" ParentId="23327" CreationDate="2018-08-20T00:56:29.273" Score="1" Body="&lt;p&gt;In general, yes. Note that in the design of extra-high frequency comm links and circuitry, the propagation time delays and associated signal phase shifts have to accounted for, which makes the circuit designer's job much harder. This is especially true when designing computer component layouts for machines that run at GHz clock rates but less of a worry when dealing with wireless signal transmission.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-08-20T00:56:29.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23336" PostTypeId="1" AcceptedAnswerId="23338" CreationDate="2018-08-20T06:32:54.580" Score="0" ViewCount="347" Body="&lt;p&gt;In Oldham coupling,is it possible that the same angular velocity is transferred from one shaft to the other when the centre disc is not sliding?If so,why do we want the centre disc to slide?I know that the shafts are misaligned but it is occurring to me that the same motion is possible when the centre disc is held back from sliding?&lt;/p&gt;&#xA;" OwnerUserId="15268" LastActivityDate="2018-08-20T09:25:01.640" Title="Transfer of angular velocity in Oldham's coupling with the centre disc not sliding" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23338" PostTypeId="2" ParentId="23336" CreationDate="2018-08-20T09:25:01.640" Score="0" Body="&lt;p&gt;If you plot the center of the third disc in one revolution you'll see it moves in a circle with diameter $δk$  with $δk$&#xA; the radial deviation of the two shafts. So if you fix the disc, it can't compensate the radial deviation at all. If i remember the frequency of the disc is about twice as it frequency of the shafts.&lt;a href=&quot;https://i.stack.imgur.com/M2J9A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M2J9A.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fAAOP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fAAOP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Oldham coupling is a kind of cross disc coupling (German). The second figure shows the general principal behind this soort of couplings.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-08-20T09:25:01.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23339" PostTypeId="1" CreationDate="2018-08-20T11:47:36.043" Score="2" ViewCount="7465" Body="&lt;p&gt;One of the assumptions we make when analyzing truss structures (method of joints for example) is that all the components are assumed to experience only compressive and tensile forces, not bending moments. &lt;strong&gt;But why can we make this assumption?&lt;/strong&gt; This is not made clear in any texts that I've read on the subject. I would assume that if we put a load on what is essentially a set of beams positioned over a gap, the beams would experience bending as well (for example the horizontal beams in Warren truss).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read some answers on another topic on Quora: &lt;a href=&quot;https://www.quora.com/Why-do-we-not-use-fixed-supports-in-a-truss&quot; rel=&quot;nofollow noreferrer&quot;&gt;Why truss bridges use a non-fixed support?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the answers was that if we used fixed supports on both ends, then the truss bridge would experience bending moments and the whole purpose of using a truss network would be lost. Could somebody shed some light on this? Why is this fact true?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastEditorUserId="1832" LastEditDate="2018-08-20T15:02:48.173" LastActivityDate="2021-09-18T20:30:14.933" Title="Why aren't there moments in truss bridges?" Tags="|structural-engineering|civil-engineering|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23341" PostTypeId="1" AcceptedAnswerId="23393" CreationDate="2018-08-20T13:26:07.347" Score="1" ViewCount="362" Body="&lt;p&gt;I know how to translate torques and vertical forces into singularity functions, but may i also express the horizontal forces with singular function? if yes then what would be that expression?  &lt;a href=&quot;https://i.stack.imgur.com/K6Z4o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K6Z4o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the green beam is fixed at point A, and a distributed load acts on the beam, and a horizontal force acts at the upper edge of the beam.&lt;/p&gt;&#xA;" OwnerUserId="17181" LastActivityDate="2018-08-22T03:26:17.327" Title="Is is possible to use singularity function to expresse the horizontal forces ?" Tags="|structural-engineering|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23342" PostTypeId="1" CreationDate="2018-08-20T13:40:20.293" Score="2" ViewCount="101" Body="&lt;p&gt;&lt;br&gt;&#xA;I am currently designing a simple pen for drawing with different types of liquid varnishes. My current design has a tip at the bottom side of the pen (blue) and a tank at the top side where the varnish is filled in (red). This varnish flows through a 0.7 mm hole in the tip. The drawing works so far. But the problem is, that it flows too quickly such that I have to draw extremely fast to prevent it from dripping. The main goal is now to reduce the flow/pressure in the tip. Unfortunately, the tip is not manufactured by me so that I cannot change it. But everything else is printed with a 3D printer and can be changed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about two different design changes. Option 1 includes a small channel between the tank and the tip. I printed this option with a channel size of 1mm, but I didn't help. Maybe the channel was too big. Option 2 added cap on the top side of the pen. I hoped that it would reduce the pressure by letting less air in, but again this did not really help. The size was 1mm which is maybe still too big. My question is if these changes help to reduce the pressure on the tip. Are there any other physical tricks to reduce the pressure/flow? Would it help to make the tank wider/narrower? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, I wanted to ask if it is possible and how to compute the pressure on the tip with these designs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nNBkN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nNBkN.png&quot; alt=&quot;Pen designs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17227" LastActivityDate="2023-03-28T07:06:51.010" Title="Reduce and compute flow in self designed pen" Tags="|fluid-mechanics|pressure|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23343" PostTypeId="2" ParentId="23339" CreationDate="2018-08-20T13:53:32.117" Score="2" Body="&lt;p&gt;Real-life truss structures are often constructed from individual &quot;rods&quot; of material which are bolted or riveted together at the joints. Since bolts and rivets can work lose over time, the most pessimistic assumption is that the bolts &lt;em&gt;are&lt;/em&gt; loose, and therefore can &lt;em&gt;not&lt;/em&gt; transmit any moments across them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, you assume that all the joints are actually &lt;em&gt;hinges&lt;/em&gt;, not rigid connections. That is why truss structures usually contain so many &quot;triangles&quot; - a triangle made from three rods is still a rigid structure even when the three vertices are only hinged together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another reason is that it is much easier to find the forces in the members of the truss if you assume the joints don't carry any moments. If they &lt;em&gt;did&lt;/em&gt; carry moments, the only way to analyse the structure would be to make a computer model (e.g. a finite element model) and run it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you don't really know how &lt;em&gt;much&lt;/em&gt; moment each joint can carry (see the first paragraph above!) a more &lt;em&gt;detailed&lt;/em&gt; computer model is not necessarily a more &lt;em&gt;accurate&lt;/em&gt; model. A simple model that you can analyse by hand, plus a reasonably large safety factor, is all you need to design a truss structure that will be safe to withstand the required loads.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-08-20T14:00:36.067" LastActivityDate="2018-08-20T14:00:36.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23344" PostTypeId="2" ParentId="23339" CreationDate="2018-08-20T13:59:24.027" Score="1" Body="&lt;p&gt;Recall that trusses have been developed incrementally over the last thousand years or so. The advantage of using a structure without rotational constraints at the joints and without midpoint connections is that the integrity of the structure depends only on the strength and stability of axially loaded members. In practice, this means you need to worry only about the tensile strength of the members (easily measured) and the chance of buckling (easily characterized).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach lets you avoid having to devise a joint that's strong enough to prevent revolute motion (which is challenging, considering the large bending moments occurring in a member fixed in place—moments that could arise from loads, thermal expansion, misfit, wind, failure elsewhere, etc.). Such a joint would require a weld, or some type of adhesive, or rivets, or a thousand other solutions that were not as well understood at simple tension and buckling. And if you don't need such revolute constraints (because the fundamental triangular unit of a truss is stable), why use them? It's simpler just to design for and (and thus assume) a lack of bending moment in the members&lt;/p&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2018-08-20T13:59:24.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23345" PostTypeId="1" AcceptedAnswerId="23347" CreationDate="2018-08-20T14:11:47.627" Score="1" ViewCount="573" Body="&lt;p&gt;Is the length of a steel dowel pin standardized to a certain tolerance? McMaster-Carr provides tolerances on the diameter of dowel pins but not the length. In the event where I want a dowel pin to fit into two blind locating holes, I'd like to know if I need to leave a little extra &quot;wiggle room&quot; in the hole depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have two parts that need to meet as close together as possible. One part has a 5 mm deep locating hole for a dowel pin. If I have a 10 mm dowel, can I specify a 5 mm deep locating hole on the second part (assuming the dowel is actually &amp;lt;10 mm long)? Or should I choose something deeper (5.5 or 6 mm perhaps)?&lt;/p&gt;&#xA;" OwnerUserId="6695" LastActivityDate="2018-08-20T14:27:27.297" Title="Tolerance on dowel pin length" Tags="|mechanical-engineering|design|machining|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23346" PostTypeId="1" AcceptedAnswerId="23349" CreationDate="2018-08-20T14:20:58.463" Score="1" ViewCount="78" Body="&lt;p&gt;I am wondering why turbines are not implemented with controllable pitch blades (on the rotor system)?  I assume the reason is that the coupling between the stator and rotor systems posed by typical controllable pitch mechanisms poses an unacceptable frictional loss for such a high speed/ low torque application.&lt;/p&gt;&#xA;" OwnerUserId="17228" LastEditorUserId="17228" LastEditDate="2018-08-20T15:01:19.997" LastActivityDate="2019-06-10T00:22:47.863" Title="Why aren't turbos/turbines controllable pitch?" Tags="|fluid-mechanics|turbines|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23347" PostTypeId="2" ParentId="23345" CreationDate="2018-08-20T14:27:27.297" Score="1" Body="&lt;p&gt;Usually the nominal length of the dowel is the total overall including the actual ground, pin section + the rounded end and + beveled end. Usually lengths are not toleranced so you should have a deeper hole then the nominal length as they run slightly larger. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ALSO....If you are using dowel pin to locate two blind holes, make sure to use a pull dowel and grind a &quot;vent&quot; on the side on the side of the pin so that it can be removed. Better if one of those holes doesn't have to be blind so the pin can be pushed out. A regular dowel pin, pressed into a blind hole is a real chore to try and remove. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-08-20T14:27:27.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23348" PostTypeId="2" ParentId="23339" CreationDate="2018-08-20T14:37:37.790" Score="6" Body="&lt;p&gt;In an ideal truss, this is actually true. As all members in an ideal truss are pin-connected (meaning the nodes cannot support any moments) the members themselves can only be loaded in compression or tension, not shear. Thus, they do not experience any moment either. There are 4 assumptions made for ideal trusses:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;joints are friction-less&lt;/li&gt;&#xA;&lt;li&gt;members are assumed to be weightless&lt;/li&gt;&#xA;&lt;li&gt;nodes are only at the ends of members&lt;/li&gt;&#xA;&lt;li&gt;all loads attack at nodes&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These are some severe simplifications, but this can be accounted for by other means.&#xA;The main reason why idealised trusses are/were used is because they facilitate calculations &lt;em&gt;a lot&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have done some calculations on beams, you may know how complex things can get, as soon as you get into statically indeterminate structures.&lt;br&gt;&#xA;Before computer allowed to solve complex statical problems with methods like FEM, simplified systems like trusses were of great help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It must be noted that e.g. the nodes of truss bridges are not really constructed with pin supports, as this is not really practical for larger structures. Often the members are actually welded or bolted together in a way that doesn't allow rotations like an ideal truss would. But with large enough safety factors, as well as a stiff enough design (steel comes in quite handy with an elasticity of $200$ $GPa$) the necessary rotation in the truss nodes is minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning points 2 and 4 of the &quot;ideal truss rules&quot;, this seems to be counterintuitive at first. But firstly, the normal forces modern I-beams for example can take is a multiple of its own weight. Furthermore, methods have been in use to approximate the weight distribution, such as the following:&#xA;(This is just an example)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kyaYP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kyaYP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the mass of the entire truss $m$ is approximated by a uniformly distributed load $$q=\frac{m\cdot g}{L}$$, which is then replaced by point loads $Q$ at the nodes. $$Q=q\frac{L}{5}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;addendum:&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fixing both supports of the truss creates the same problem as with a regular beam: It becomes statically indeterminate&lt;sup&gt;i&lt;/sup&gt;. Consider a beam, fixed at both ends. When this beam experiences a change in temperature it would want to contract or expand, however the two fixed supports prevent that from happening. Thus, resulting in tensile or compressive stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the same for trusses. Note how the truss basically consists of triangles. Imagine one of the members experiencing an increase in temperature, therefore expanding. To allow this, the other two members of said triangles just rotate to allow for that expansion. Consequently no thermal stresses occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In statically indeterminate trusses this is however prevented by the fixed supports, or by the shape of the truss itself (see footnote &lt;sup&gt;i&lt;/sup&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;i&lt;/sup&gt; Note that trusses can be statically indeterminate even when there are not two fixed supports. Basically any truss where you could remove $n$ members without the system collapsing is $n$-fold statically indeterminate.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastEditorUserId="1832" LastEditDate="2018-08-21T11:37:51.760" LastActivityDate="2018-08-21T11:37:51.760" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23349" PostTypeId="2" ParentId="23346" CreationDate="2018-08-20T14:39:04.167" Score="1" Body="&lt;p&gt;There are turbos with variable geometry, in fact my car is fitted with one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obvious with the actuator visible which, when the ignition is switched on, the actuator is cycled from min to max and back to min.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two classic issues that put the engine error light on - one is the actuator fails (usually soldered joints apparently) and two is that the vanes seize due to lots of low speed use clogging the vanes (a good regular &quot;italian service&quot; on the motorway is one solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.dieselnet.com/tech/air_turbo_vgt.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a link to some further info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To add the info based on the comments, turbos for cars tend to be around 2&quot; to 2.5&quot; max and the turbine blades very thin (rotating around 180000rpm), trying to put a control system to control the pitch of those blades would be a challenge due to size and the effects of play in the mechanisms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The images below are from the Jaguar x type manual pages 1129 and 1130, showing low and high engine speed operation and the effects of the vane position:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gGAAB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gGAAB.png&quot; alt=&quot;Low speed&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sg8g1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sg8g1.png&quot; alt=&quot;High speed&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-08-20T15:27:08.050" LastActivityDate="2018-08-20T15:27:08.050" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23350" PostTypeId="1" CreationDate="2018-08-20T15:45:30.223" Score="2" ViewCount="246" Body="&lt;p&gt;A foam which I have performed compression tests on shows results similar to the following hysteresis cycle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fWxY8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fWxY8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the first bend in the graph is not the yield point; the hysteresis cycle returns to the origin.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to characterize the compressibility of the material using a commonly used statistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Young's modulus is an obvious candidate (see dotted line), but it doesn't really capture the characteristics of the material when you compress it further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead I had the idea of using the dashed line: the minimal value of stress divided by strain over all values along the curve of the compression. However, I don't know whether that is a commonly used statistic to describe compression of foams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What commonly used statistic best describes the compression behavior of a foam at higher strain values?&lt;/p&gt;&#xA;" OwnerUserId="17229" LastActivityDate="2018-08-20T17:58:43.073" Title="Stiffness of highly compressible materials" Tags="|elastic-modulus|stiffness|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23351" PostTypeId="2" ParentId="23331" CreationDate="2018-08-20T16:46:06.893" Score="1" Body="&lt;p&gt;Imagine the system below at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ERaVc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ERaVc.png&quot; alt=&quot;ball in box&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy balances are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the ball ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon_b T_b^4 =\epsilon_w T_w^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the wall ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\epsilon_w A_w \sigma T_w^4 = h_a A_w (T_a - T_w) + \epsilon_a A_w \sigma T_a^4$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these balances, $\epsilon_j$ are emissivities, $A_j$ are areas, $\sigma$ is the Stefan-Boltzmann constant, $T_j$ are temperatures, and $h_a$ is the convection coefficient of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose all materials and sizes to specify all $A_j$ and $\epsilon_j$ values. Choose a still or windy day to specify $h_a$. Define the air temperature. This leaves two unknowns ($T_b$ and $T_w$) with two equations. The problem can be solved.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Take the case where the ball and walls are the same material so their emissivities are the same. We find that $T_b = T_w$. The radiation heat flows inside the box are balanced. Now assume the air has no radiation and the walls are perfect black bodies ($\epsilon_w = 1$) to obtain $\sigma T_w^4 = h_a (T_a - T_w)$. Take stagnate air with $h_a = 5$ W/m$^2$ K and $T_a = 275$ K. The wall and ball are at 238 K. In this case, the air is pumping heat to the walls by convection, and the (black body) walls are radiating back to the air. As the walls (and ball) go toward gray bodies (as $\epsilon_w \rightarrow 0$), the wall temperature increases. In the limit where the walls emit no radiation, we end with the simple case that $T_b = T_w = T_a$. No heat flows from anything at equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-08-20T21:50:16.263" LastActivityDate="2018-08-20T21:50:16.263" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23352" PostTypeId="2" ParentId="23342" CreationDate="2018-08-20T17:20:55.550" Score="2" Body="&lt;p&gt;you will need some way to apply a constant very slightly negative pressure on the ink reservoir to prevent the pen tip from drooling due to gravity. this is a solved problem in the field of inkjet printhead design. I recommend you obtain a spent HP model 45 printhead, which contains a flexible bag preloaded with a stainless steel spring. when properly filled and primed, this arrangement will produce the required negative backpressure over a discharge volume of almost 40 cc's. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would use this by first sealing off the nozzle array and then using the fill port as the insertion point for your pen tip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contact me if you need more help with this. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-08-20T17:20:55.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23353" PostTypeId="2" ParentId="23350" CreationDate="2018-08-20T17:58:43.073" Score="3" Body="&lt;p&gt;Material science can almost always be broken down into &quot;generally useful&quot; and &quot;specifically useful&quot;.  The link in the comments demonstrates a commonly used statistic that describes behavior of foams at higher strain values - the three dimensional &lt;a href=&quot;https://en.wikipedia.org/wiki/Strain_energy_density_function&quot; rel=&quot;nofollow noreferrer&quot;&gt;strain energy density function&lt;/a&gt;.  But this statistic is &quot;generally useful&quot;.  It generalizes to all foams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In specific circumstances, you may find that you design your own tests to reflect a specific loading you need to test for your design.  These later become specific ASTM procedures that people can utilize to check the quality of foams and guarantee similar performance.  For example, one method may simply be to run an integral of the area under the stress-strain curve.  This energy is not the same as the stress energy density function, because your anisotropic material is not charachterized in all three dimensions, but the resulting energy difference would nevertheless be a useful mechanic to determine how much energy the foam can absorb from an impact.  Such an energy is typically called the &lt;a href=&quot;https://en.wikipedia.org/wiki/Toughness&quot; rel=&quot;nofollow noreferrer&quot;&gt;toughness&lt;/a&gt;, and the full value is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Resilience_(materials_science)&quot; rel=&quot;nofollow noreferrer&quot;&gt;modulus of resilience&lt;/a&gt;.  A specific test of a specific sized sample of foam could be utilized to find the toughness as a function of compression force, and compare it to a known good sample.  Such a test would easily be characterizeable and usable for the industry.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-08-20T17:58:43.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23355" PostTypeId="1" CreationDate="2018-08-20T23:44:21.040" Score="0" ViewCount="369" Body="&lt;p&gt;I am working on a linear actuator for a valve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wondering if anyone out there can point me towards the tolerances that are used for valve lift in a typical internal combustion engine using a camshaft? &#xA;i.e. what is the plus/minus range that the valve lifts to? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1175" LastActivityDate="2018-08-21T22:20:18.900" Title="Typical tolerances for valve lift using a camshaft?" Tags="|mechanical-engineering|valves|mechanical|tolerance|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23356" PostTypeId="1" CreationDate="2018-08-21T01:15:17.420" Score="1" ViewCount="151" Body="&lt;p&gt;I'm a high school student who hasn't had the pleasure of taking physics yet, so forgive me if the answer is obvious. The problem I have involves an earth anchor drilled vertically into the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lka19.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lka19.jpg&quot; alt=&quot;https://static.gemplers.com/img/auger-style-earth-227706-lrg.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to figure out the equation that would give the highest amount of permissible force uniformly distributed on the exposed top part before the anchor would rotate. Just a force perpendicular to the vertical auger. What I mean by rotate is that it would pivot at a point halfway down the auger, like a lever with a fulcrum in the middle (x axis), and not unscrew itself from the earth (z axis). Of course this would be easy if it were in air, but I'm lost at how to calculate it in dirt. I came here from Physics Stack Exchange. &#xA;The auger would be 9 inches long and immersed in 1.52 g/cm consistently dry soil. Any help would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="17232" LastEditorUserId="17232" LastEditDate="2018-08-21T17:23:42.430" LastActivityDate="2024-03-16T10:04:55.563" Title="How would you calculate the force needed to rotate along the x axis vertical earth anchor in dirt?" Tags="|civil-engineering|torque|soil|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23357" PostTypeId="2" ParentId="23310" CreationDate="2018-08-21T01:52:00.703" Score="-4" Body="&lt;p&gt;From what I have learned at university, pressure in a  vessel is equal at all points of the interior surface. This means it does not matter where the measuring gauge is placed. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2018-09-01T21:13:24.387" LastActivityDate="2018-09-01T21:13:24.387" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23358" PostTypeId="1" CreationDate="2018-08-21T02:04:56.623" Score="1" ViewCount="307" Body="&lt;p&gt;I am using a sheet of a certain material to dampen a product. Now, I have been looking at different soft dampening materials, and they're all defined by their durometer scale and other factors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which factor should I look at and how should I use it to determine how thick the sheet should be to withstand a certain amount of vertical force or acceleration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="17233" LastActivityDate="2018-08-21T02:04:56.623" Title="How to calculate the required thickness of a material to withstand a certain force or acceleration?" Tags="|materials|stresses|" AnswerCount="0" CommentCount="2" ClosedDate="2018-08-22T14:30:02.410" ContentLicense="CC BY-SA 4.0" />
  <row Id="23359" PostTypeId="2" ParentId="23293" CreationDate="2018-08-21T02:13:27.740" Score="3" Body="&lt;p&gt;In UK, building codes are unusual. There is something called Mobile Homes Act which specifies particular types of homes as being covered under different protections and building laws than masonry buildings. Basically put, politics had made prefab homes unattractive legally because you have to meet different ownership requirements related to land occupation than masonry homes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This issue arose because government wanted to legislate against random land occupation by gypsies and the traveller communities which resulted in a stilted and broken recognition of what legally and politically constitutes a &quot;fixed abode&quot; without singling out a specific group or ethnicity of people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cnNga.jpg&quot; alt=&quot;Mobile home&quot;&gt;&#xA;Mobile Homes are not covered under rent regulations and insurance companies are reluctant to insure them unless they are located in very specific conditions in their environment. If they are on a &quot;trailer park&quot; or &quot;holiday park&quot; the rates are quite different to ones with wheels removed. Masonry buildings are considered to be fixed assets whereas prefabs are categorised under moveable assets in the same way as caravans. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can argue the point but the laws are so convoluted and Machiavellian that prefab homes are too expensive to build in quantity, politically or legally. They tend to be classed as park homes which can only be placed in certain areas under specific conditions if English Heritage agree to allow them to be built... such are planning laws in UK that a charity determines who build what and where it when they can build it. When it comes down to cost, it is not the material film which the home is constructed that accounts for the price but price is largely attributed to demand. These mobile homes cost about the same as a masonry home which negates the saving on materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whilst a landlord will order a mobile home to be built for about £45k, they will charge between £80k &amp;amp; £140k to the prospective buyer. In addition to this, no landlords of mobile home parks will allow you to bring your own unit on-site because they don't make a pretty penny off you otherwise. One your home arrives, you start paying ground rent at around £70 a week in addition to buying the overpriced unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better to buy a brick and mortar home that costs buyers about the same because you don't have to fork out an additional £36400 over 10 years in addition to the sale price of £120000. Both types of homes incur the same council taxes. Both cost around £160000. You cannot usually get a mortgage for prefab homes either which means only richer people can afford prefab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cost of a mobile home single unit is about £35k and sells for about ££90k.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2018-08-24T11:03:52.543" LastActivityDate="2018-08-24T11:03:52.543" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23362" PostTypeId="2" ParentId="23257" CreationDate="2018-08-21T02:55:52.520" Score="0" Body="&lt;p&gt;I would say that calculating for the thickness with the least stress would be the best thickness for the whole container. A variable thickness container is going to cost more to construct and be difficult to consistently manufacture to specifications. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-08-21T02:55:52.520" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23363" PostTypeId="2" ParentId="22974" CreationDate="2018-08-21T03:08:37.390" Score="0" Body="&lt;p&gt;Without a competent geotechnical analysis by a qualified civil engineer, you are likely to face difficulties and lawsuits. Generally speaking, forces exerted by buildings act at 33° to the horizontal in soil. This does not account for hogging in soil or loss of compression against neighbouring structures of the soil. Fluid content aints for much of the rigidity or fluidity of soil strength. Opening it up to air will change its characteristics and behaviour messing something unexpected could happen.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-08-21T03:08:37.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23364" PostTypeId="1" AcceptedAnswerId="23368" CreationDate="2018-08-21T05:12:31.460" Score="1" ViewCount="158" Body="&lt;p&gt;In books of dynamics, I normally see topics in kinematics and then in kinetics, and finally a last chapter on mechanical vibrations dealing with SDOF and MDOF systems. But I do not see any book mentioning how the topic of vibration fits in dyynamics? It feels natural to say to mechanical vibrations would be an extension of kinetics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the correct classification of mechanical vibrations?&lt;/p&gt;&#xA;" OwnerUserId="15972" LastActivityDate="2018-08-21T07:52:42.393" Title="Mechanical vibrations: kinetics of kinematics?" Tags="|dynamics|vibration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23365" PostTypeId="1" CreationDate="2018-08-21T05:44:47.580" Score="0" ViewCount="56" Body="&lt;p&gt;Please check this newspaper article and its pictures and the video in it: &lt;a href=&quot;http://ilpiccolo.gelocal.it/trieste/cronaca/2016/08/17/news/panico-sul-tram-di-opicina-9-feriti-nello-scontro-frontale-1.13972540#gallery-slider=undefined&quot; rel=&quot;nofollow noreferrer&quot;&gt;Article&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It contains photos and a video of a frontal crash between two vehicles of a trolley line. From the photos, it seems like one vehicle suffered much more damage the the other. Judging by this photos, what can be said about the speed of the two vehicles? Can the difference in damage be attributed to much different speeds of the two vehicles?&lt;/p&gt;&#xA;" OwnerUserId="17234" LastActivityDate="2018-08-21T14:51:13.783" Title="Damage differences in frontal collision of two vehicles and their speed" Tags="|dynamics|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23366" PostTypeId="2" ParentId="23365" CreationDate="2018-08-21T06:27:50.093" Score="1" Body="&lt;p&gt;The imact energy is the sum of the individual energies just prior to impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The damage to the fastest is not greater and vice versa - the relative damage is down to things like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Older or newer designs ie crush zones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The age ie effects of corrosion or previous damage on strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle of impact&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point of impact - can be a stronger impact point on one cf the other..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many factors influence the level of damage.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-08-21T06:27:50.093" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23368" PostTypeId="2" ParentId="23364" CreationDate="2018-08-21T07:52:42.393" Score="1" Body="&lt;p&gt;&quot;Kinetics&quot; is just a synonym for &quot;dynamics,&quot; and most English books use the term &quot;dynamics&quot; rather than &quot;kinetics.&quot; There are plenty of mechanical engineering textbooks with &quot;dynamics&quot; in the title, but few with &quot;kinetics&quot; (I'm ignoring books on topics like &quot;kinetics of chemical reactions&quot;, etc, which are not about &lt;em&gt;mechanical&lt;/em&gt; engineering)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between &quot;kinetics&quot; and &quot;kinematics&quot; is that kinematics deals with the motion of object &lt;em&gt;without considering what causes that motion&lt;/em&gt; (for example the mathematical relationships between position, velocity, and acceleration of a particle moving along a given curved path in space), but kinetics/dynamics &lt;em&gt;does&lt;/em&gt; consider the forces causing the motion (i.e. Newton's laws of motion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since a fundamental topic in studying mechanical vibrations is the fact that for linear elastic materials which obey Hooke's law, Newton's laws mean that the motion for small displacements about an equilibrium position &lt;em&gt;is&lt;/em&gt; simple harmonic motion, I would classify most of the subject of &quot;mechanical vibrations&quot; to be part of dynamics/kinematics, rather than kinetics.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-08-21T07:52:42.393" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23369" PostTypeId="1" AcceptedAnswerId="23388" CreationDate="2018-08-21T09:47:38.257" Score="0" ViewCount="141" Body="&lt;p&gt;I am an economist, so please keep replies simple ;)&#xA;I am searching for the relationship between heat rate and capacity for gas-fired power plants. Here is what I refer to as heat rate, i.e. the inverse of efficiency: &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_rate_(efficiency)&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Heat_rate_(efficiency)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From an IEA publication I got the following approximate relationship (for a CCGT plant):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;maximum heat rate = 1.1 * minimum heat rate&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Maximum heat rate holds whenever the power plant is operated at minimum capacity, and&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;minimum capacity = 0.4*maximum capacity&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'd like to know what the heat rate is when I run the power plant at a medium capacity of let's say:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;medium capacity = minimum capacity + (maximum capacity-minimum capacity)/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;medium heat rate = ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is the relationship linear (i.e. = minimum hr+ (maximum hr-minimum hr)/2) or how would you estimate the heat rate here?&lt;/p&gt;&#xA;" OwnerUserId="11639" LastEditorUserId="11639" LastEditDate="2018-08-21T13:18:54.463" LastActivityDate="2018-08-21T22:37:46.227" Title="Relationship heat rate with capacity" Tags="|power|gas|power-engineering|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23370" PostTypeId="2" ParentId="3183" CreationDate="2018-08-21T09:48:18.950" Score="4" Body="&lt;p&gt;Typically, the max is 100mW (20dBm). You can refer to an example wifi chip datasheet (This one is used in many mobile phones with Snapdragon processor) - &#xA;&lt;a href=&quot;https://developer.qualcomm.com/download/sd410/wcn3620-wireless-connectivity-ic-device-specification.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://developer.qualcomm.com/download/sd410/wcn3620-wireless-connectivity-ic-device-specification.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17235" LastActivityDate="2018-08-21T09:48:18.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23373" PostTypeId="2" ParentId="23365" CreationDate="2018-08-21T13:48:53.387" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can the difference in damage be attributed to much different speeds of the two vehicles?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which speed? If the cars hit each other, then there is only one speed that matters - the speed of car A relative to car B. This also means that the speed for both cars is the &lt;strong&gt;same&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one car is moving at 30 km/h, and the other is moving the other direction at 60 km/h, then they collide at a relative speed of 90 km/h. You &lt;strong&gt;cannot&lt;/strong&gt; analyze the collision with individual speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the alternative - if you yourself are standing still (0 km/h), and a truck hits you going 100 km/h, do you suffer any damage? Your speed was zero, correct? Was this a low speed collision for you?&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2018-08-21T14:51:13.783" LastActivityDate="2018-08-21T14:51:13.783" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23374" PostTypeId="1" CreationDate="2018-08-21T14:12:43.733" Score="1" ViewCount="103" Body="&lt;p&gt;Why transition frequency of power transistor is typically below 30 mhz?  Detailed search on internet doesn't reveal any power transistor which can handle 5 Amp or more and has transition frequency of more than 30 mhz.  While signals transistors have frequency up to 300 ghz. &lt;/p&gt;&#xA;" OwnerUserId="17239" LastActivityDate="2018-08-21T14:12:43.733" Title="Transition frequency of power transistor" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23375" PostTypeId="2" ParentId="23365" CreationDate="2018-08-21T14:19:55.840" Score="0" Body="&lt;p&gt;As Solar Mike said, from theoretical physics point of view, there's no difference: A impacts B with exactly the same force as B impacts A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are finer details though, that come out from minor asymmetries. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Due to wear and aging, no two parts are identical.&lt;/li&gt;&#xA;&lt;li&gt;Material that started buckling will continue to buckle. So if the weaker parts gives in first, it will give in all the way before the stronger part starts sustaining damage; the split isn't equal or proportional to strength of materials - even slightly stronger wins by a large margin.&lt;/li&gt;&#xA;&lt;li&gt;Material reinforced by frame is tougher than separated from the frame.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here, it seems like the cars have very solid undercarriage; their frames didn't buckle, but one lifted above the other, and plowed into the structure of the other car, ripping the chassis away, and supporting own chassis which continued to plow into the weakened counterpart. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-08-21T14:19:55.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23377" PostTypeId="1" CreationDate="2018-08-21T16:40:04.707" Score="0" ViewCount="60" Body="&lt;p&gt;I need to solve the next equation and to find $x_1$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\dfrac{d(x_0x_1)}{dz}=-\beta\dfrac{dx_0}{dz}\dfrac{1}{r-z(r-1)}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $x_0$, $x_1$, $z$ are variables, and $r$, $\beta$ are constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point is it possible to multiply whole equation with $dz$, will I lose some information because of the therm $\dfrac{1}{r-z(r-1)}$?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other side, I have expression for $x_0=f(z)$, it is dependent on $z$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$x_0=\left(1+\dfrac{\beta}{r-1}\left(  1-\dfrac{1}{(r-z(r-1))^3} \right)\right)^{0.5}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On which side I need to go to integrate this equation?&lt;/p&gt;&#xA;" OwnerUserId="14507" LastEditorUserId="3488" LastEditDate="2018-08-22T03:13:31.327" LastActivityDate="2018-08-22T03:13:31.327" Title="How to solve analytically this equation?" Tags="|fluid-mechanics|fluid|" AnswerCount="1" CommentCount="3" ClosedDate="2018-08-22T09:11:55.227" ContentLicense="CC BY-SA 4.0" />
  <row Id="23378" PostTypeId="1" AcceptedAnswerId="23386" CreationDate="2018-08-21T17:36:42.193" Score="0" ViewCount="1053" Body="&lt;p&gt;I'm attempting to understand the difference between a pressure relief valve and a pressure regulating valve. From my research far, I've come to the following conclusion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Pressure relief valves are primarily used for limiting pressure build ups and are often safety related. Pressure relief valves vent the fluid when the pressure exceed some setpoint.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Pressure regulator valves on the other hand are used to regulate a pressure via step-down. These regulator are typically non-venting.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So I'm somewhat confused as to whether or not I should be using a pressure relief valve or pressure regulator in my application. It seems like a pressure relief valve would do the same exact thing as a regulator except it would vent the gas. Do pressure regulator offer better regulation than a relief valve? &lt;/p&gt;&#xA;" OwnerUserId="7664" LastActivityDate="2018-08-21T21:30:05.387" Title="Is it standard practice to use a pressure relief valve as a pressure regulator?" Tags="|pressure|pneumatic|compressed-gases|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23380" PostTypeId="2" ParentId="23378" CreationDate="2018-08-21T17:50:51.987" Score="0" Body="&lt;p&gt;Id say a regulator is more optimised for attaining a given setpoint with precision and lack of chatter, whereas a relief valve is more concerned with reliability and less with finesse. But you are correct on an abstract level they are both flow control feedback mechanisms of a very similar kind; though in practice you are unlikely to confuse the two application wise. Without regretting it that is.&lt;/p&gt;&#xA;" OwnerUserId="14899" LastActivityDate="2018-08-21T17:50:51.987" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23381" PostTypeId="2" ParentId="23341" CreationDate="2018-08-21T17:53:07.800" Score="0" Body="&lt;p&gt;No, it is not possible. The singularity functions are defined only for vertical forces.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-08-21T17:53:07.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23382" PostTypeId="1" CreationDate="2018-08-21T18:00:53.690" Score="0" ViewCount="67" Body="&lt;p&gt;Optical IR sensors are able to measure the surface temperature of solids. But is there a range of the EM spectrum that can accomplish the same for measuring the temperature of nitrox (dry nitrogen, oxygen mixtures)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so what are the constraints?&lt;/p&gt;&#xA;" OwnerUserId="820" LastActivityDate="2018-08-22T18:15:12.523" Title="Is there an optical technology able to measure the temperature of dry air?" Tags="|gas|temperature|metrology|air|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23383" PostTypeId="2" ParentId="2841" CreationDate="2018-08-21T18:01:06.453" Score="0" Body="&lt;p&gt;The total flow resistance is the sum of its parts. Pipes are long, valves are short. Making a pipe wider is relatively inexpensive, and making a valve wider is relatively expensive. As you make the opening wider, everything about the valve has to be beefed up, as the forces involved will scale along with the size of the opening. So on balance, youd always expect valve openings to be smaller relative to pipe dimensions, in a cost optimized system.&lt;/p&gt;&#xA;" OwnerUserId="14899" LastActivityDate="2018-08-21T18:01:06.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23385" PostTypeId="2" ParentId="23382" CreationDate="2018-08-21T18:58:21.167" Score="0" Body="&lt;p&gt;You can use a sensor such as an &quot;&lt;a href=&quot;https://www.sparkfun.com/products/9570&quot; rel=&quot;nofollow noreferrer&quot;&gt;Infrared Thermometer - MLX90614&lt;/a&gt;&quot; and a simple circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &quot;&lt;a href=&quot;https://www.analoxsensortechnology.com/sda-temperature-and-humidity-saturation-diving.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;SDA Temperature and Humidity monitors&lt;/a&gt;&quot; available for making measurements in diving chambers and dive systems. They are certified to be accurate and used where safety is a concern, as such they won't be an inexpensive solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To measure using sound see: &quot;&lt;a href=&quot;https://www.edn.com/design/analog/4369636/An-introduction-to-acoustic-thermometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;An introduction to acoustic thermometry&lt;/a&gt;&quot; (Apr 21 2011), by Jim Williams and Omar Sanchez-Felipe, of Linear Technology. Also a DIY solution.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;Acoustic thermometry is an arcane, elegant technique that measures temperature using the temperature-dependent transit time of sound in a medium. The medium can be a solid, a liquid, or a gas. Acoustic thermometers function in environments, including extreme temperatures, destructive physical abuse, and nuclear reactors, that conventional sensors cannot tolerate. Sonic speed in air varies predictably as the square root of temperature. The sonic transit time in a gas-path thermometer is almost entirely insensitive to pressure and humidity. Gas-path acoustic thermometers respond quickly to temperature changes. They have essentially no thermal mass or lag.&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2018-08-21T18:58:21.167" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23386" PostTypeId="2" ParentId="23378" CreationDate="2018-08-21T21:30:05.387" Score="2" Body="&lt;p&gt;Sort of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly things like pressure cookers and turbochargers use blow off vales as a routine way to regulate pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally this is to make sure that internal pressure doesn't exceed some design maximum rather than for fine control and it is a bit contextual as it tends to imply that whatever working fluid you are dealing with goes to 'waste'. Obviously in both of the above examples this doesn't really matter much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a turbo charger both the fluid and the energy to compresses it are more or less 'free' once you have the physical plant in place so a blow off valve is a simple and reliable way to regulate pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast if you have something like industrial argon as a welding shielding gas you don't want to vent that to atmosphere if you don't need to so you would use a more sophisticated regulator to control the flow-rate indeed y ou mmight even use two, one for course pressure regulation and one for fine volume flow-rate control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally in a more general case if you are using bottled gas the unregulated pressure is directly related to the mass of gas in the cylinder so are often forced to use  a regulator unless you can accept  a drop-off in  available pressure with usage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside several Robot Wars Robots have used direct bottle pressure (CO2) ie just a solenoid valve and no regulator  to fire weapons, flippers etc. This tends to give fairly  fierce performance while the bottle is full and is simple to implement, especially  by remote control but every time you use it the available pressure drops and so can be a bit inefficient. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-08-21T21:30:05.387" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23387" PostTypeId="2" ParentId="23355" CreationDate="2018-08-21T22:20:18.900" Score="1" Body="&lt;p&gt;Broadly speaking opening a valve further reduces flow resistance, at least up to  point but more travel means greater acceleration and thus greater forces on parts, which is magnified as RPM increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed with mechanical valves you get to a point where the return springs can't act fast enough and you starts  to get harmonic effects where the valves themselves are no longer directly coupled to the cams ie 'valve float' which can be pretty serious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally it is entirely possible in high compression engines that the valves and piston occupy the same space so if timing isn't right they can collide. In mechanically driven systems this should only happen if something breaks or the design rpm is exceeded (see valve float above) although even then many pistons have small crescent shaped cutouts to allow valve clearance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bdrfG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bdrfG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the short answer is that the most important tolerance  that the valves and pistons absolutely can't occupy the same space at the same time, everything else im more or less negotiable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F1 engines have for a while now have used pneumatic valves as camshafts and springs just can't  respond fast enough to 15k+ rpm. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-08-21T22:20:18.900" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23388" PostTypeId="2" ParentId="23369" CreationDate="2018-08-21T22:37:46.227" Score="1" Body="&lt;p&gt;I will take a stab at an answer in what I hope might be simple and general terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is defined as heat rate is akin to a &lt;em&gt;relative inefficiency&lt;/em&gt; or &lt;em&gt;relative loss&lt;/em&gt; rather than &lt;em&gt;heat rate&lt;/em&gt;. The value itself has no units. Multiple by 100 and you get a percentage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You first statement says, the greatest relative loss we expect in a plant is 10% over its minimum. The minimum relative loss occurs when the plant is operating at it greatest efficiency. So, a plant that is 85% efficient in the best case has a heat rate (relative loss) of $1/0.85 = 1.2$ or 120% in the best case. We expect this plant to have a heat rate (relative loss) of $1.1 * 1.2 = 1.3$ or 130% at its worst case operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your next statements say, the worst case (130% heat rate) happens when the plant operates at its lowest capacity. Capacity is akin to the actual amount of electricity that is produced. So, a plant will be least efficient (have the highest heat rate) when it produces the lowest amount of actual electrical energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your next statement says, the lowest capacity of a plant is 40% of its maximum capacity. Basically, this says, we do not operate power plants to produce less than 40% of their maximum rated output of electricity. Let's suppose the maximum capacity of the example power plant from above is $C_{max} = 100$ (with whatever units). The lowest capacity is $C_{min} = 0.4C_{max} = 40$. At 40, the plant operates with a heat rate of 130%. Now for the tricky part ... &lt;em&gt;We apparently have no information about the relationship between increasing capacity and decreasing heat rate&lt;/em&gt;. We do not know whether the minimum heat rate of 120% is at maximum capacity ($C = 100$) or perhaps even before. All we know is, as we increase capacity, we can expect that heat rate decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The finding leaves us with this conclusion. Absent any further information, we are free to assume any model we want for a relationship between increasing capacity and decreasing heat rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real life, power plants are not operated continuously at their physical maximum capacity. They are operated below this physical maximum limit, by example continuing from above perhaps at $C_{operation} = 80 - 90$. In real life, inefficiencies also kick in above this operational design &quot;set point&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instinct to tackle this absent any other information would be to do the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Set the maximum heat rate at the lowest capacity. This is defined by reference.&lt;/li&gt;&#xA;&lt;li&gt;Set the minimum heat rate somewhere at 80% - 90% of maximum capacity. This &lt;em&gt;can&lt;/em&gt; be defined by further research (i.e. look up typical operating capacities of power plants relative to their rated maximum capacity).&lt;/li&gt;&#xA;&lt;li&gt;Define the relationship between increasing capacity and decreasing heat rate as linear over the above range. This is &lt;em&gt;your&lt;/em&gt; model. Publish it.&lt;/li&gt;&#xA;&lt;li&gt;Do not model the behavior of systems below their minimum capacity.&lt;/li&gt;&#xA;&lt;li&gt;Do not model the behavior of systems above the 80% - 90% their rated maximum capacity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-08-21T22:37:46.227" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23389" PostTypeId="2" ParentId="23377" CreationDate="2018-08-21T23:03:52.467" Score="1" Body="&lt;p&gt;Define $\alpha := r - 1$.  Then, the original equation can be written as &#xA;$$&#xA; (1) \quad \dfrac{dx_1}{dz} = \left[x_1+\dfrac{\beta}{r-\alpha z}\right]\left[-\dfrac{1}{x_0}\dfrac{dx_0}{dz}\right]&#xA;$$&#xA;where&#xA;$$&#xA; (2) \quad x_0^2 = \left[1 + \dfrac{\beta}{\alpha}\right] -\left[\dfrac{\beta}{\alpha}\dfrac{1}{(r-\alpha z)^3} \right]&#xA;$$&#xA;We want to find $x_1$ analytically by solving (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiation of (2) gives&#xA;$$&#xA; (3) \quad \dfrac{dx_0}{dz} = -\dfrac{1}{2x_0}\dfrac{3\beta}{(r-\alpha z)^4}&#xA;$$&#xA;Therefore, the last factor in (1) can be written as&#xA;$$&#xA; (4) \quad -\dfrac{1}{x_0} \dfrac{dx_0}{dz} = \dfrac{3}{2}\dfrac{\beta^2}{x_0^2(r-\alpha z)^4}&#xA;$$&#xA;Now, using (2),&#xA;$$&#xA;  \begin{align}&#xA;  \frac{x_0^2(r-\alpha z)^4}{\beta^2}  &amp;amp; = \frac{1}{\beta^2} \left(1 + \dfrac{\beta}{\alpha}\right)(r-\alpha z)^4 - \dfrac{\beta}{\alpha}(r-\alpha z) \\&#xA;  &amp;amp; = \dfrac{r-\alpha z}{\alpha\beta^2} \left[\left(\alpha + \beta\right)(r-\alpha z)^3 - \beta\right]&#xA;  \end{align}&#xA;$$&#xA;We can now write (4) as&#xA;$$&#xA;(5) \quad -\dfrac{1}{x_0} \dfrac{dx_0}{dz} = \left(\dfrac{3\alpha\beta}{2}\right) \left(\dfrac{\beta}{r-\alpha z}\right) \left[\dfrac{1}{\left(\alpha + \beta\right)(r-\alpha z)^3 - \beta}\right]&#xA;$$&#xA;Plug (5) into (1) to get&#xA;$$&#xA;(6) \quad \dfrac{dx_1}{dz} = \dfrac{3\alpha\beta}{2}\left[x_1+\dfrac{\beta}{r-\alpha z}\right] \left(\dfrac{\beta}{r-\alpha z}\right) \left[\dfrac{1}{\left(\alpha + \beta\right)(r-\alpha z)^3 - \beta}\right]&#xA;$$&#xA;Do a change of variables by defining $y := r - \alpha z$.  Then&#xA;$$&#xA;  \frac{dx_1}{dz} = \frac{dx_1}{dy} \dfrac{dy}{dz} = -\alpha \frac{dx_1}{dy}&#xA;$$&#xA;We can now write (6) as&#xA;$$&#xA;(7) \quad \dfrac{dx_1}{dy} = -\dfrac{3\beta}{2}\left[x_1+\dfrac{\beta}{y}\right] \left[\dfrac{\beta}{\left(\alpha + \beta\right)y^4 - \beta y}\right]&#xA;$$&#xA;or&#xA;$$&#xA;(8) \quad \dfrac{dx_1}{dy} = -\dfrac{3\beta^2}{2}\left[\dfrac{ x_1}{\left(\alpha + \beta\right)y^4 - \beta y} +  \dfrac{\beta}{\left(\alpha + \beta\right)y^5 - \beta y^2}\right]&#xA;$$&#xA;You can now solve the ODE using standard approaches, e.g., &lt;a href=&quot;http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx&lt;/a&gt;.  The solution is a complex mix of square roots, logs, and arctans.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-08-21T23:03:52.467" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23390" PostTypeId="2" ParentId="23094" CreationDate="2018-08-22T00:03:54.440" Score="2" Body="&lt;p&gt;If the built-in storage is fully utilized, there are three other options for saving test data:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Save to a network share&lt;/li&gt;&#xA;&lt;li&gt;Save to a USB flash drive&lt;/li&gt;&#xA;&lt;li&gt;Save to an SD card&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;All of these connections are on the &lt;a href=&quot;http://www.ni.com/documentation/en/compactdaq-controller/latest/cdaq-9135/pinout/&quot; rel=&quot;nofollow noreferrer&quot;&gt;main front panel&lt;/a&gt; (items 3, 15, and 10) of the cDAQ controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zWrN5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zWrN5.png&quot; alt=&quot;cDAQ Controller Front Panel Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the &lt;a href=&quot;http://www.ni.com/pdf/manuals/374668b.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;specifications&lt;/a&gt; report that the USB controller supports USB 2.0 devices and that the SD controller supports 32 GB cards.&lt;/p&gt;&#xA;" OwnerUserId="2009" LastActivityDate="2018-08-22T00:03:54.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23391" PostTypeId="2" ParentId="23235" CreationDate="2018-08-22T02:22:34.340" Score="2" Body="&lt;p&gt;Try a DS18B20. They are readily and cheaply available packaged in water proof, stainless steel enclosures.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use lots of them and they operate very reliably in field conditions that are often extreme. Some have been operating continuously for 18 months. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be aware though that they are a 5v unit (I use them on arduinos that then connect through to a RPI). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get best results with the 3x wire config (+5v, GND, signal) with different pull up resistors on the signal depending on cable length and how many are on a single signal bus (you can have many on a single signal). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Libraries are readily available for arduinos and I recall also finding libraries for RPI a while back.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This looks sensible: &#xA;&lt;a href=&quot;https://www.modmypi.com/blog/ds18b20-one-wire-digital-temperature-sensor-and-the-raspberry-pi&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.modmypi.com/blog/ds18b20-one-wire-digital-temperature-sensor-and-the-raspberry-pi&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example of the stainless steel enclosed component:&#xA;&lt;a href=&quot;https://www.altronics.com.au/p/z6386-stainless-steel-housing-waterproof-ds18b20-temperature-probe/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.altronics.com.au/p/z6386-stainless-steel-housing-waterproof-ds18b20-temperature-probe/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Extended discussion on solutions for an anomonly using these sensors in industrial conditions FYI (problem not with the sensor):&#xA;&lt;a href=&quot;https://iot.stackexchange.com/questions/2859/ds18b20-oscillating-rapidly/2929#2929&quot;&gt;https://iot.stackexchange.com/questions/2859/ds18b20-oscillating-rapidly/2929#2929&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17245" LastEditorUserId="17245" LastEditDate="2018-08-22T02:35:09.783" LastActivityDate="2018-08-22T02:35:09.783" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23393" PostTypeId="2" ParentId="23341" CreationDate="2018-08-22T03:26:17.327" Score="1" Body="&lt;p&gt;Yes, it is possible to describe horizontal loading with singularity functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in order to do so you must change the loading equation from scalar to vector-form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\{q\} = \left\{\matrix{q_x \\ q_y}\right\}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you'll handle the horizontal and vertical load components separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, how you then interpret the singularity functions changes, since beams behave differently under transversal and axial loads (the fourth integral of transversal loading is the deflection, which isn't true for axial loading).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, I don't know how useful singularity functions are when dealing with horizontal loading. However, your question is whether it is possible to express the horizontal loading with singularity equations and the answer to that question is &quot;yes&quot;.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-08-22T03:26:17.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23395" PostTypeId="1" AcceptedAnswerId="23403" CreationDate="2018-08-22T08:36:53.953" Score="0" ViewCount="2268" Body="&lt;p&gt;I'm reading the derivation of the equation of the hanging catenary from Wikipedia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Catenary&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Catenary&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same article the parabolic equation governing the shape of the main cable of a suspension bridge is discussed, and its equation is obtained as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$y={\frac {w}{2T_{0}}}x^{2}+\beta$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $w$ is the density of the deck and $T_{0}$ is the horizontal tension on the cable at lowest point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I'm designing a suspension bridge with the length and density of the deck and its other features are known. I would be interested in knowing the equation of the main cable to calculate the length of the secondary cables (the vertical ones). &lt;strong&gt;But how do I get $T_{0}$?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2018-08-22T21:16:21.060" Title="Equation of the parabolic suspension bridge cable when the deck mass is known" Tags="|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23397" PostTypeId="1" CreationDate="2018-08-22T13:36:51.047" Score="2" ViewCount="2390" Body="&lt;p&gt;On regular conveyor belts, motors are usually placed on the head shaft where material is discharged, and tension is highest at the entrance of this head shaft. However, if the motor were to be simply reversed, it seems as if the load-carrying side would lose tension, turn slack, and the material handled could fall out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to reverse a conveyor belt without having two separate motors, and simply having a reversible one, what sort of system do you need to maintain the tension?&lt;/p&gt;&#xA;" OwnerUserId="12823" LastActivityDate="2022-11-15T12:08:08.683" Title="What components are needed to reverse the direction of a belt conveyor using one reversible motor, and how do they change the tension on the belt?" Tags="|applied-mechanics|mechanisms|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23398" PostTypeId="1" CreationDate="2018-08-22T14:32:40.733" Score="3" ViewCount="92" Body="&lt;p&gt;Hello I am a high school student and I am curious about how street light is arranged along the road. Is there government laws to install street light on the road or is the arangement of the street light that is the most efficient to its road? If it is, how are the lights arrangement? Is it by some sort of software?&lt;/p&gt;&#xA;" OwnerUserId="17240" LastActivityDate="2019-03-19T17:26:33.497" Title="Street Light Arrangement Method" Tags="|civil-engineering|highway-engineering|lighting|roadway|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23399" PostTypeId="2" ParentId="17692" CreationDate="2018-08-22T14:34:34.690" Score="0" Body="&lt;p&gt;As niels said, they have all induction motors, some advanced kitchen equipments has also a safety coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance if you want crush nuts on your smoothie but the nuts are too hard to crush by the baldes, safety coupling (automatic engagement) disengage the motor and the driven shaft to protect sensitive components of motor, baldes or ... . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also have another coupling in serie with the safety coupling, it is called the magnetic coupling (only in recent applications) to couple the motor and the driven shaft, as far as you keep pressing the button, friction disc of magnetic coupling get engage with driving shaft (motor) and the driven shaft (blades). Old food processors use another type of conventional coupling. If you can drive the motor manually by rotating it with your hand it's not a magnetic coupling.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the driven end of magnetic coupling there is a crown wiel (negative or female) and a male part at the top of driven shaft which is connected via a polymer rod to the baldes, this maple and female part fit together to complete the the torque transmission. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There exist many variants of mechanisms, but they share the same principal. All of them have at least an induction motor, a coupling and blades.   &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-08-22T14:34:34.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23400" PostTypeId="2" ParentId="23356" CreationDate="2018-08-22T16:56:52.043" Score="0" Body="&lt;p&gt;I am going about explains this question at the level of high school physics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume you auger blade covers 2 full circles at length of embedment of 1.5&amp;nbsp;feet. And assuming the radius of blades 6&amp;nbsp;inches the diameter of the bore as 1&amp;nbsp;foot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the surface of the earth resisting torsion is either &lt;span class=&quot;math-container&quot;&gt;$1.5\pi\times1 = 1.5\pi$&lt;/span&gt; feet or a expanded length of &lt;span class=&quot;math-container&quot;&gt;$2\times2\pi\times1.41 = 5.64\pi$&lt;/span&gt; times shear strength of your soil, typically at approximately 100&amp;nbsp;lbs/ft&lt;sup&gt;2&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So picking the bigger shear as the one on the surface of the blade versus the bore, we get &lt;span class=&quot;math-container&quot;&gt;$564\times 0.5= 282\text{ lbs ft}$&lt;/span&gt; torque. This is dislodging of the soil's resistance which is not large enough to create a hinge in the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we look at the other end and try to see how much force the shaft can take before it yields and creates a plastic hinge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the diameter of the shaft is 1 inch and it's made of steel with 25 p.s.i. torsion strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torsion strength of a shaft is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$z\tau= \dfrac{\pi}{16}d^3\times25000 = \dfrac{\pi}{16} \times 25000 \approx 5000\text{ lb in}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to apply 5000 &lt;span class=&quot;math-container&quot;&gt;$lb.in^1$&lt;/span&gt; total torque to the top part before it starts to develop a plastic hinge regardless of the soil it is boring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After new comment by OP, here I attach my &lt;a href=&quot;https://engineering.stackexchange.com/a/11464/4816&quot;&gt;previous answer to a similar question&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="16486" LastEditDate="2018-11-23T08:20:41.843" LastActivityDate="2018-11-23T08:20:41.843" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23401" PostTypeId="2" ParentId="23397" CreationDate="2018-08-22T17:59:33.810" Score="1" Body="&lt;p&gt;You might consider adding a clutch so that, upon reversing the motor direction, it's freed from the local (head) shaft and drives, either via secondary belt or geared shaft, the far-end head shaft.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I think Mr. R. Goldberg would be proud of me)&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-08-22T17:59:33.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23402" PostTypeId="2" ParentId="23382" CreationDate="2018-08-22T18:15:12.523" Score="1" Body="&lt;p&gt;The fundamental problem is: how do you reject the optical signal from solids or liquids &quot;behind&quot; the air mass you want to measure?   The esoteric methods reported in : &lt;a href=&quot;https://pdfs.semanticscholar.org/9ccd/6df082576a620a2a89f8e6b7203186e6b08c.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://pdfs.semanticscholar.org/9ccd/6df082576a620a2a89f8e6b7203186e6b08c.pdf&lt;/a&gt; , &lt;a href=&quot;http://www.mdpi.com/1424-8220/18/1/72/pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.mdpi.com/1424-8220/18/1/72/pdf&lt;/a&gt; &#xA;may work, because they indirectly measure air temperature rather than trying to sense the spectral power distribution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, you essentially need a cold plate behind your sample volume to act as a near-zero-emission (not emissivity) baffle. So the answer is that it's really not a good way to go.  Is there a reason you can't work with thermocouples?&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-08-22T18:15:12.523" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23403" PostTypeId="2" ParentId="23395" CreationDate="2018-08-22T18:27:04.083" Score="0" Body="&lt;p&gt;Since $x=0$ marks the midpoint of the parabola, let $y(0)=0$ at this point to eliminate $\beta$. Then, we only need to know the sag $s$ of the cable (which is typically set during the design process). The sag is equivalent to the value of $y$ at one of the symmetric towers (where $x=L/2$, where $L$ is the span or horizontal distance between the towers).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, if we know the sag $s$ and span $L$ (and the distributed weight $w$), then we can immediately obtain $T_0$ from $y=\frac{w}{2T_0}x^2$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we know the length $l$ of the cable but not the sag, then we can use the formula for the &lt;a href=&quot;http://mathworld.wolfram.com/ParabolicSegment.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;length of a parabola&lt;/a&gt; $$l=\sqrt{\left(\frac{L}{2}\right)^2+4s^2}+\frac{L^2}{8s}\sinh^{-1}\left(\frac{4s}{L}\right)$$ to obtain the sag $s$. (We'll probably have to solve this equation numerically.)&lt;/p&gt;&#xA;" OwnerUserId="10123" LastEditorUserId="10123" LastEditDate="2018-08-22T21:16:21.060" LastActivityDate="2018-08-22T21:16:21.060" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23404" PostTypeId="1" AcceptedAnswerId="23405" CreationDate="2018-08-22T20:30:38.350" Score="4" ViewCount="292" Body="&lt;p&gt;When widening a road, why does construction seem to inevitably include a &quot;smoothing&quot; out of the curves on the road, essentially making it more straight? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen this personally on roads where I travel, and you can even catch a glimpse on Google satellite imagery of roads that are in the widening process: Google overlays a &quot;ghost&quot; lane to show where the new lanes of the road will be. It's faint, but you can see an example here on a road near Flowery Branch, GA:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xO3WZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xO3WZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the engineers' way of correcting previous mistakes, or a response to devleopment that's sprung up around the road since its initial construction, or safety reasons? Or something else entirely?&lt;/p&gt;&#xA;" OwnerUserId="17255" LastActivityDate="2018-08-22T21:24:27.710" Title="Why are roads straightened when widened?" Tags="|highway-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23405" PostTypeId="2" ParentId="23404" CreationDate="2018-08-22T21:20:03.337" Score="6" Body="&lt;p&gt;A wider road will have faster average traffic speeds, because faster vehicles will be hindered less by slow ones, with more lanes available for overtaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So longer visibility distances are needed for safety, and hence more gradual curves and more gradual changes of gradient (no &quot;hidden spots&quot; over the brow of a hill, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, for most &quot;fast&quot; roads traffic speeds are not limited at all by the radius of the curves, except perhaps in very severe weather conditions (snow and ice, standing rainwater, etc). Aside from breaking the law and the hazards posed by other vehicles, there would be no problem in driving a modern vehicle along the road way above the legal speed limit.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-08-22T21:20:03.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23406" PostTypeId="2" ParentId="23404" CreationDate="2018-08-22T21:24:27.710" Score="1" Body="&lt;p&gt;I am familiar with this effect; with careful observation you can count at least three rounds of road-straightening on highway 17 between the cities of San Jose and Santa Cruz in California.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first road was built over 100 years ago and carried horse-drawn wagons. At the time, moving large volumes of earth and rock was more expensive than surfacing the road and so it followed sharp curves in the mountainsides instead of being extensively built on fill from nearby road cuts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then came automobiles and more and faster traffic, and when widening the original stage road the sharpest turns were rounded off to carry faster vehicles and cut-and-fill was used so the mountain contours didn't have to be closely followed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At each step in this decades-long process, the cost of earthmoving fell and the cost of pavement rose, and so it became economical to use more and more cut-and-fill in conjunction with straighter and straighter road paths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stage road is now a 4-lane freeway which still has a few somewhat sharp turns in it but much of it is built by cut-and-fill and can be traversed at 50 to 60 MPH.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-08-22T21:24:27.710" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23407" PostTypeId="1" AcceptedAnswerId="23408" CreationDate="2018-08-23T00:26:59.840" Score="3" ViewCount="72" Body="&lt;p&gt;Reading an article about multistage rockets (not educated at all on the topic), and from the get go it seems to assume they are used/need to be used.&lt;/p&gt;&#xA;" OwnerUserId="17257" LastActivityDate="2018-08-23T09:45:19.367" Title="What is the purpose of building multistage rockets, rather then packing more fuel into a single stage?" Tags="|aerospace-engineering|rocketry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23408" PostTypeId="2" ParentId="23407" CreationDate="2018-08-23T01:02:11.037" Score="4" Body="&lt;p&gt;As the rocket is propelled upwards, it expends fuel. So there is no need to carry half empty fuel tanks. By splitting it up into separate stages, you can simply drop off unneeded mass.&lt;/p&gt;&#xA;" OwnerUserId="16325" LastActivityDate="2018-08-23T01:02:11.037" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23409" PostTypeId="1" AcceptedAnswerId="23429" CreationDate="2018-08-23T01:11:52.797" Score="1" ViewCount="235" Body="&lt;p&gt;If I have a square plate of height h supported along one side (an L-bracket, modeled as fixed support), and there is an evenly distributed force acting over the face of the plate. From load analysis I am able to determine the resulting force (acting in the centre of the plate) to be P, thus the distributed load over the plate is P/A = P/(L^2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if we are considering the bending of the plate, can we model this as a cantilever beam with UDL of P/L or will it be something else? I am leaning to the former, and my calculations with such an assumption seems to yeild a reasonable result, but I wanted to double check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="16325" LastActivityDate="2018-08-24T09:00:23.073" Title="How to model a UDL over a plate for bending?" Tags="|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23410" PostTypeId="1" CreationDate="2018-08-23T01:16:52.217" Score="1" ViewCount="302" Body="&lt;p&gt;I am having trouble understanding the difference between positive and negative adjustable voltage regulators (LDO's). There are circuit diagrams for the positive LDO's such as the following: &lt;a href=&quot;https://i.stack.imgur.com/3GfnP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3GfnP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not able to understand what a circuit diagram of a negative voltage regulator looks like or how different it would be from a positive voltage regulator. &lt;/p&gt;&#xA;" OwnerUserId="10432" LastActivityDate="2018-08-23T01:16:52.217" Title="Difference between a positive and negative adjustable voltage regulator (LDO's)" Tags="|electrical-engineering|circuit-design|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23411" PostTypeId="1" AcceptedAnswerId="23412" CreationDate="2018-08-23T02:09:27.597" Score="6" ViewCount="345" Body="&lt;p&gt;Looking at the teeth of a chain saw, I believe the chain moves from &lt;em&gt;right to left&lt;/em&gt; while cutting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has two wedges &lt;strong&gt;A&lt;/strong&gt; and &lt;strong&gt;B&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've never owned a chain saw, so I've been trying to visualize how these two wedges work.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I feel stuck because the wedge &lt;strong&gt;A&lt;/strong&gt; seems unnecessary. Also it doesn't look that sharp and it seems to block the sharper wedge &lt;strong&gt;B&lt;/strong&gt;. I'm dying to know the purpose of wedge &lt;strong&gt;A&lt;/strong&gt; here. Appreciate any help. Thanks!&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cf4J2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cf4J2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17258" LastEditorUserId="33" LastEditDate="2018-08-23T13:42:01.793" LastActivityDate="2018-08-23T13:42:01.793" Title="Shape of chain saw teeth" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23412" PostTypeId="2" ParentId="23411" CreationDate="2018-08-23T05:47:40.427" Score="5" Body="&lt;p&gt;The part you label A is the depth control to stop B cutting so deep it blocks. This depth control only works on a hard material like wood, on flesh the blade cuts down to the base of the chain - experience speaking here - I had a blade do its final stop in my leg...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of B is dependent on the purpose of the blade - some have a flat top and angled face with no wedge - also the size of the blade is a factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you sharpen your own blades you understand how they work and the blades do have the angle information required in or on the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice the difference in the cutting edge A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HuRHA.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HuRHA.png&quot; alt=&quot;Speedcut blade shape from Oregon&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://www.oregonproducts.co.uk/en-gb/products/forestry-chainsaw.html&quot; rel=&quot;noreferrer&quot;&gt;http://www.oregonproducts.co.uk/en-gb/products/forestry-chainsaw.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-08-23T06:02:27.683" LastActivityDate="2018-08-23T06:02:27.683" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23413" PostTypeId="2" ParentId="23407" CreationDate="2018-08-23T09:45:19.367" Score="2" Body="&lt;p&gt;Another reason is that each motor stage can be designed for different purposes to match the flights characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, a relatively short high thrust burn to escape gravity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a longer or continuous burn to accelerate over a course / trajectory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can save weight and complexity so adjustable nozzles are not always needed.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-08-23T09:45:19.367" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23414" PostTypeId="2" ParentId="23411" CreationDate="2018-08-23T10:08:03.840" Score="5" Body="&lt;p&gt;The first issue: the most important cutting edge is actually on top of the link, a segment curved towards the middle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pmk4t.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pmk4t.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blade on the side is secondary, just for clearing any remains of whatever might have sprung out due to material flexibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These links alternate the sides, so that the overhang isn't excessively long but the cut provides a gap wide enough for the entire chain to fit. Otherwise you'd be stuck with a very shallow cut and unable to get anywhere deeper than what the blade sticks above the chain!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next: Cutting wood with a chainsaw is subject to the same set of rules as any machining; lathe, CNC mill, drill on a boring machine - all the same rules:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5uIKj.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5uIKj.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's the blade angle, speeds&amp;amp;feeds, material properties - and there's the cutting depth, which is an important parameter that accounts for power, material durability, chip properties and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the blade can't be right above the chain, because there would be no room left for the chip! It must be offset from the chain (by the distancer &quot;A&quot;) to fit all the material it removes before it exits the cut and can be ejected!&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-08-23T10:08:03.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23416" PostTypeId="2" ParentId="23397" CreationDate="2018-08-23T11:16:12.897" Score="2" Body="&lt;p&gt;There would be a few ways of doing it, but one the simplest I've seen is having a separate drive roller, and idlers somewhere on the underside of the conveyor. These idlers can be moved for tensioning the belt, whilst the ends of the conveyors would be used for tracking the belt. Here's a diagram in paint. My paint skills are somewhat lacking....&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3sbJQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3sbJQ.png&quot; alt=&quot;drive roller U shape beneath main conveyor&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12187" LastEditorUserId="9448" LastEditDate="2022-11-15T12:08:08.683" LastActivityDate="2022-11-15T12:08:08.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23417" PostTypeId="2" ParentId="23397" CreationDate="2018-08-23T12:13:00.853" Score="1" Body="&lt;p&gt;All installations I have seen with a reversible conveyor belt had a seperate tensioning system for the belt (as Solar Mike pointed out in a comment) and the motor at one end. Non reversible belts typically have a tensioning system:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XZKy6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XZKy6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The belt is enclosed, but you see the roller. The long threaded bolt (red cloud) is used to tension the belt. This conveyor is non reversible.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-08-23T12:13:00.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23419" PostTypeId="1" CreationDate="2018-08-23T14:31:53.380" Score="1" ViewCount="42" Body="&lt;p&gt;I have a complicated cylindrical geometry with a speaker mounted on somewhere middle of the rig. The one end is closed, the other is open. Sweeping the speaker frequency from 0 to 1000 Hz, I tried to find the acoustic resonance frequency of this geometry from several sensors at different location. It is found that there are some frequencies that give the local peak amplitudes. I thought those frequencies are the resonance frequencies, but when I compared the signals from two different sensors, the phase difference was neither in phase nor out of phase, but like 120 degree.  This result is counter-intuitive to me because I thought the resonance always happens with the standing wave, which gives only in phase or out of phase between two sensors. If the end boundary conditions are hard or soft BCs, the resonance theoretically occurs with the standing wave pattern. So I think my open end is not actually sound soft BC, but rather have low impedance BC.&lt;br&gt;&#xA;Anyway, the below is my questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does resonance always happen with the standing wave?&lt;/li&gt;&#xA;&lt;li&gt;What properties affect the end impedance. Is it function of temperature? Frequency?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17265" LastActivityDate="2018-08-23T16:34:06.883" Title="Acoustics resonance question" Tags="|acoustics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23420" PostTypeId="1" AcceptedAnswerId="23435" CreationDate="2018-08-23T15:38:59.700" Score="3" ViewCount="3562" Body="&lt;p&gt;When browsing through pneumatic parts, I notice that some vendors specify different components for pressure regulator and vacuum regulators. &lt;a href=&quot;https://www.smcusa.com/products&quot; rel=&quot;nofollow noreferrer&quot;&gt;SMC for example does this&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious why vendors would actually need to differentiate vacuum regulators from pressure regulators. It seems like a vacuum regulator is simply a pressure regulator whose inlet port is connected to something wished to be vacuumed, and the outlet port is connected to a vacuum supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there more going on here?&lt;/p&gt;&#xA;" OwnerUserId="7664" LastActivityDate="2018-08-28T08:25:34.843" Title="What's the difference between a vacuum regulator and pressure regulator?" Tags="|compressed-air|vacuum|pneumatic|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23421" PostTypeId="2" ParentId="23419" CreationDate="2018-08-23T16:34:06.883" Score="4" Body="&lt;p&gt;The idealized &quot;open boundary condition&quot; that is given in a first course on sound  waves in pipes doesn't exist in the real world. The reason is that the pipe radiates sound into the environment, (otherwise you would not be able to hear any wind instrument playing a note!) and the sound wave &lt;em&gt;outside&lt;/em&gt; the pipe is a traveling wave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, the &quot;end correction&quot; that you have to apply to simple experiments on resonance in pipes is a way to model the effect of this sound radiation on the &lt;em&gt;frequency&lt;/em&gt; of the resonance, but simple experiments don't attempt to measure the phase of the sound waves - they just assume there is &quot;a resonance&quot; and don't encourage the beginning students to ask too many questions about what is going on!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the wave inside the pipe is some combination of a standing wave and a pure travelling wave, and the phase angle is not constant everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can visualize the wave inside the pipe as a travelling wave moving towards the &quot;open&quot; end, but the reflected travelling wave in the opposite direction has a smaller amplitude, because some of the sound energy has been radiated away outside the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the sound radiation out of the pipe also creates damping of the resonance response - which is why you need a continuous excitation source, otherwise the sound energy inside the pipe would decay to zero very quickly.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-08-23T16:34:06.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23422" PostTypeId="1" CreationDate="2018-08-23T18:11:25.383" Score="0" ViewCount="59" Body="&lt;p&gt;I have a 3 inch line with a 1.5 inch flow orifice. If I take out the flow orifice and put in a 1.5 inch coriolis meter will the pressure drop be the same or better with the new coriolis meter than with the flow orifice?&lt;/p&gt;&#xA;" OwnerUserId="17269" LastActivityDate="2018-10-19T00:07:26.683" Title="If I replace a flow orifice with a Coriolis flow meter of the same size do I need to worry about pressure drop with the new Coriolis meter?" Tags="|fluid-mechanics|control-engineering|chemical-engineering|process-engineering|instrumentation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23425" PostTypeId="2" ParentId="23310" CreationDate="2018-08-24T05:30:54.580" Score="1" Body="&lt;p&gt;Yes, If you remove mercury manometer, then the pressure measured will be $P = 120000 + \rho *9.81 * (1m)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to measure the pressure at the bottom of the vessel with the mercury manometer, make sure the pipe connecting the limb and vessel is filled with air. (just to avoid hydrostatic pressure by the fluid inside the connecting pipe). This would be difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best method I could think is, go for single limb manometer and keep the bottom of the limb and the bottom of the vessel are at the same level (head). &lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-08-24T05:30:54.580" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23427" PostTypeId="1" CreationDate="2018-08-24T06:52:32.587" Score="0" ViewCount="58" Body="&lt;p&gt;Say we have a required mass flow rate, in order to have equivalent volume flow rate, we need to consider density of the operating fluid (compressible fluid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To measure the  volume flow rate in LPM, I'm using a rotameter. I'm controlling the flow using a needle valve upstream of the rotameter. The regulated pressure upstream of the needle valve is 20&amp;nbsp;bar and the pressure downstream the rotameter is ambient (1&amp;nbsp;atm).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I consider the density of the operating fluid (say oxygen) relating to the upstream pressure (20&amp;nbsp;bar) or the downstream (1&amp;nbsp;bar ambient)&lt;/p&gt;&#xA;" OwnerUserId="17275" LastEditorUserId="1832" LastEditDate="2018-08-24T22:05:50.633" LastActivityDate="2018-09-23T23:01:13.497" Title="What should be the density of a compressed gas while measuring the flow rate using Rotameter?" Tags="|mechanical-engineering|fluid-mechanics|applied-mechanics|compressible-flow|flow-control|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23428" PostTypeId="2" ParentId="23427" CreationDate="2018-08-24T07:02:26.793" Score="0" Body="&lt;p&gt;Gas density can be calculated from: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\rho = \dfrac{P}{RT}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\rho$ is the density, $P$, the pressure, $R$, the gas constant, and $T$, the temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should evaluate the change in density between the two pressures and decide what to do - this will be based on the fluid you are using...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="1832" LastEditDate="2018-08-24T22:07:24.473" LastActivityDate="2018-08-24T22:07:24.473" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23429" PostTypeId="2" ParentId="23409" CreationDate="2018-08-24T09:00:23.073" Score="1" Body="&lt;p&gt;A plate supported along one side, subject to a uniformly distributed load, will behave exactly the same as a cantilever beam. You are correct that the UDL will be P/L, as it will be P/A*w where w is the width.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a non-uniform load, treating it as a beam would give you the &lt;em&gt;average&lt;/em&gt; stress, but miss the fact that there will be regions of higher stress and regions of lower stress.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-08-24T09:00:23.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23431" PostTypeId="1" CreationDate="2018-08-24T09:52:57.350" Score="2" ViewCount="62" Body="&lt;p&gt;Recently I encountered a problem of oil leakage from different driving systems SERVO gearboxes. &#xA;The transmitted mechanical power ranges from 2 to 8 kW, while the motor speed (the gearbox input speed) varies from 1440 to 3000 RPM.&#xA;Most of the units are 1 stage planetary gearboxes although some of theme are helical-bevel gears.&lt;br&gt;&#xA;Despite the gearbox manufacturers (several European companies) are well-known and having significant heritage in this field, these gearboxes all begin to leak from their input seal before reaching 3000 hours of operation.&#xA;Replacing the oil to grease improves the amount of leakage, but the phenomena still exists.&#xA;Adding breathers / pressure relief valves, hardening the gearbox input shaft (which acts against the input seal), and replacing the seal material to viton - all made just a small improvement.&#xA;Furthermore, the same happens even in our air-conditioned lab - so the problem doesn't seem to be related to harsh environmental conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect the root cause is the continuous operation profile (the system runs for 24/7) - which kills the seal many hours before the units operation life.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspecting the seal and the shaft of a leaking gearbox did not show strong indications of wear. As suggested here, it may also be a vibration problem, probably coming from the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas how to improve it? &#xA;Are there instances out there for SERVO gearboxes running continuously for more than 3000 hours before leakage?&lt;/p&gt;&#xA;" OwnerUserId="16860" LastEditorUserId="16860" LastEditDate="2018-08-25T07:24:07.237" LastActivityDate="2018-08-25T07:24:07.237" Title="Gearbox oil/grease leakage" Tags="|mechanical-engineering|gears|servo|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23433" PostTypeId="2" ParentId="23427" CreationDate="2018-08-24T11:09:21.597" Score="1" Body="&lt;p&gt;In, compressible flow, the mass flow rate remains constant throughout the flow. But volume flow rate is changed according to the local velocity and area of cross-section. So one can calculate the volume flow rate locally by diving the mass flow rate by local density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let say, If you need volume flow rate at the exit, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the exit, the pressure is 1 bar and the ambient temperature is 27$^\circ$C which gives ~ 1.88 kg/m$^3$. (refer equation given by solar mike).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your volume flow rate at the exit is  $\dot V =  \dot m$/1.88 m$^3$/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: It looks like your flow is a choked flow since the pressure comes down from 20 bar to 1 bar. So keep this in mind while calculating and interpreting the values.&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-08-24T11:09:21.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23434" PostTypeId="2" ParentId="23420" CreationDate="2018-08-24T11:27:13.713" Score="0" Body="&lt;p&gt;Just the operating range differs in these regulating valves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure regulators work from 1 bar to the high pressure with respect to atm. &#xA;Vacuum regulators work from 1 bar to low pressure with respect to atm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally, the vacuum regulators are attached to the vacuum chamber where you need to maintain the subatmospheric pressure. For example, in the sputtering process, you have to provide vacuum pressure with some inert gas as the medium. &lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-08-24T11:27:13.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23435" PostTypeId="2" ParentId="23420" CreationDate="2018-08-24T16:34:34.587" Score="2" Body="&lt;p&gt;The difference is the side on which pressure is regulated, and the reaction curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical pressure regulator takes arbitrary input pressure (from compressor) and provides specified (lower) pressure on output, locking the valve as output pressure rises; not allowing it to drop below preset level. The vacuum regulator assures the input pressure is as specified, while accepting arbitrary output (vacuum pump), by closing as the input pressure drops to preset. In the regulator feedback loop, the &quot;readout&quot; is performed on different sides of the valve. Since the regulators are usually unidirectional devices, not allowing for backflow, you can't just flip one around, and expect it to work like the other - if e.g. you put a vacuum pump in place of the compressor connected to a tank through a standard pressure regulator - it won't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can imagine a tank that needs to maintain a pressure in specific range; the fill input will have a pressure regulator that allows filling the valve to preset pressure. But if for other reasons (e.g. the tank being heated by sunlight) the pressure rises, the relief valve will disengage, releasing the excess build-up; a vacuum regulator works similar to the relief valve, except its set point is somewhere below 1 bar, while a typical relief (safety) valve will open when internal pressure rises considerably above atmospheric.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2018-08-28T08:25:34.843" LastActivityDate="2018-08-28T08:25:34.843" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23437" PostTypeId="1" AcceptedAnswerId="23438" CreationDate="2018-08-24T16:40:11.570" Score="-2" ViewCount="1758" Body="&lt;p&gt;Gears are called positive gears and belt drives are called negative drives. Explain in detail.&lt;/p&gt;&#xA;" OwnerUserId="17284" LastEditorDisplayName="user14407" LastEditDate="2018-08-24T17:28:15.673" LastActivityDate="2018-08-24T17:28:15.673" Title="What are positive drive gears?" Tags="|mechanical-engineering|automotive-engineering|gears|" AnswerCount="1" CommentCount="2" ClosedDate="2018-08-25T00:48:31.600" ContentLicense="CC BY-SA 4.0" />
  <row Id="23438" PostTypeId="2" ParentId="23437" CreationDate="2018-08-24T16:50:25.053" Score="1" Body="&lt;p&gt;The negative drive is an informal expression to describe 'friction drive'. The belt drive is based on friction between the disc and the belt, however positive drive is based on movements without slipping. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not all belt drive system are 'negative drive'. We also have a positive drive belt system, in this special case, the belt doesn't slip over the driving disc but it rolls over teeth. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-08-24T16:50:25.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23439" PostTypeId="1" CreationDate="2018-08-24T18:42:45.207" Score="1" ViewCount="73" Body="&lt;p&gt;Machining a metal part will create residual stresses on its surface due to the deformation that the process introduces to it, is there any model to estimate how thick will this stress layer may be?&lt;/p&gt;&#xA;" OwnerUserId="17286" LastEditorUserId="17286" LastEditDate="2018-08-27T13:19:10.283" LastActivityDate="2018-08-27T13:19:10.283" Title="How to calculate the thickness of the residual stress layer generated by a machining process" Tags="|mechanical-engineering|stresses|machining|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23440" PostTypeId="2" ParentId="16882" CreationDate="2018-08-24T18:54:54.783" Score="1" Body="&lt;p&gt;I'll take your fill options one at a time:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Sand&lt;/strong&gt; - This isn't going to do anything. You could argue that filling the space between the tube and the H-pile would add bracing between one and the other. This could help if you were having issues with the unbraced length part of axial capacity. The sand will not transfer shear though, so it isn't a composite section.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Concrete&lt;/strong&gt; - Concrete will bond to (or at least have greater friction with) the steel elements. This is the same as reinforcing in concrete. Also, just like reinforced concrete this will cause a composite section to be produced. But just like reinforcing, there will be some concept of development length. Data on this has been hard to find in my experience. AISC has a publication specifically about this: &lt;a href=&quot;https://www.aisc.org/Design-Guide-6-Load-and-Resistance-Factor-Design-of-W-Shapes-Encased-in-Concrete-Print&quot; rel=&quot;nofollow noreferrer&quot;&gt;Design Guide 6: Load and Resistance Factor Design of W-Shapes Encased in Concrete&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Grout&lt;/strong&gt; - This is generally the same as concrete. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="33" LastActivityDate="2018-08-24T18:54:54.783" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23441" PostTypeId="2" ParentId="14508" CreationDate="2018-08-24T19:59:13.100" Score="2" Body="&lt;p&gt;As has been mentioned in the comments, there is no specific combination of dimensions that mean that punching shear is not a problem. An Engineer may make a judgement that a check isn't required based on experience, but a check would always be the safe option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACI (American Concrete Institute) even has a similar question in its Frequently Asked Questions: &lt;a href=&quot;https://www.concrete.org/tools/frequentlyaskedquestions.aspx?faqid=876&quot; rel=&quot;nofollow noreferrer&quot;&gt;Punching shear check for shear wall foundations&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2018-08-24T19:59:13.100" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23442" PostTypeId="2" ParentId="16961" CreationDate="2018-08-24T21:02:25.680" Score="1" Body="&lt;p&gt;These appear to be different types of &lt;strong&gt;Depression Contour Lines&lt;/strong&gt;.  These lines are typically used on topographic maps to show where the ground slopes down. Sometimes pure contours are hard to read (or impossible) if elevations aren't marked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This site has some &lt;a href=&quot;http://academic.brooklyn.cuny.edu/geology/leveson/core/linksa/depression.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;good examples&lt;/a&gt; of where they are found on topographic maps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These same contour line types are also used in other situations just to show slope directly. It doesn't necessarily mean that the surface wasn't triangulated. It is just a helpful display method.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2018-08-24T21:02:25.680" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23443" PostTypeId="1" CreationDate="2018-08-24T21:50:54.403" Score="4" ViewCount="612" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ykwZF.jpg&quot; alt=&quot;Rotatable Composting bin &quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to insert couple of moisture, temp and humidity sensors in this bin to take readings from the compost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do i install it without turning/tangling the wires? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me figure a solution I am not an engineer but I am trying hard to solve. &lt;/p&gt;&#xA;" OwnerUserId="17287" LastEditorUserId="1832" LastEditDate="2018-08-24T22:44:04.233" LastActivityDate="2018-08-27T19:54:34.610" Title="How to install a wired sensor in a rotating bin without tangling the wires?" Tags="|electrical-engineering|environmental-engineering|" AnswerCount="8" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23444" PostTypeId="2" ParentId="23443" CreationDate="2018-08-24T22:06:54.423" Score="1" Body="&lt;p&gt;Put a disc on the side of the drum centered around the shaft it rotates on and have two, three or four concentric copper tracks for carbon brushes to rub on, spring loaded and they should work fine.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-08-24T22:06:54.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23445" PostTypeId="2" ParentId="19147" CreationDate="2018-08-24T22:24:07.473" Score="0" Body="&lt;p&gt;Temperature indicator consists of two parts: the first part is  the temperature element itself and the second part is the electronic part that conditions the signal and convert it to 4 to 20  and then send it to control system .&lt;/p&gt;&#xA;" OwnerUserId="17248" LastActivityDate="2018-08-24T22:24:07.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23446" PostTypeId="2" ParentId="23443" CreationDate="2018-08-24T23:15:36.383" Score="3" Body="&lt;p&gt;There is a device called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Slip_ring&quot; rel=&quot;nofollow noreferrer&quot;&gt;slip ring&lt;/a&gt; which does what @Solar Mike describes. They can be fairly cheap if obtained on the &lt;a href=&quot;https://www.surplusshed.com/pages/item/R3567.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;surplus market&lt;/a&gt; or even purchased new from &lt;a href=&quot;https://www.sparkfun.com/products/13063&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;https://www.adafruit.com/product/736&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. They can be either hollow or solid shafts. Solid shaft models can be used as part of the axle (if mechanically strong enough) while the axle can pass through the middle of the hollow shaft varieties.&lt;/p&gt;&#xA;" OwnerUserId="9682" LastActivityDate="2018-08-24T23:15:36.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23447" PostTypeId="2" ParentId="23443" CreationDate="2018-08-25T02:10:41.693" Score="0" Body="&lt;p&gt;This &lt;a href=&quot;https://www.appliancepartspros.com/whirlpool-brush-8182500-ap3777488.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;clothes dryer&lt;/a&gt; has conducting bands around the drum with wire brushes (#16 in diagram) &quot;in order to transfer the inputs from the sensors in the baffles to the control board&quot;.&lt;/p&gt;&#xA;" OwnerUserId="17291" LastActivityDate="2018-08-25T02:10:41.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23448" PostTypeId="2" ParentId="23443" CreationDate="2018-08-25T02:29:38.880" Score="2" Body="&lt;p&gt;Slip rings are a straightforward solution to the problem, so long as the electrical noise they generate is small compared with the signals you want to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are already using an Arduino, you might consider an alternative, which is to have a second, battery powered, Arduino attached to the rotating drum itself, and use that to transmit the data to an external computer wirelessly, for example using a Bluetooth data link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming the composting bin only rotates slowly, to mix up the contents, so there would not be any mechanical problems with a relatively large unbalanced mass (i.e. the Arduino) attached to it. But this &quot;rotating transmitter&quot; concept is used successfully in situations where the environmental conditions are very hostile (e.g. rotational speeds of the order of 10,000 RPM, high temperatures, and a corrosive atmosphere which would quickly destroy slip rings) - using purpose-built transmitters though, not an off-the-shelf Arduino.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-08-25T02:29:38.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23449" PostTypeId="2" ParentId="2349" CreationDate="2018-08-25T05:49:26.750" Score="2" Body="&lt;p&gt;Rephrasing your question:&lt;br&gt;&#xA;1. How much energy does a car use at 60mph doing 34mpg?&lt;br&gt;&#xA;2. What is its efficiency if it requires 20hp to maintain constant speed?&lt;br&gt;&#xA;&lt;br&gt;&#xA;In SI units:&lt;br&gt;&#xA;Car at 26.8m/s doing 14.4km/L requires 14.9kW&lt;br&gt;&#xA;&lt;br&gt;&#xA;Empirical data:&lt;br&gt;&#xA;* HHV for gasoline approx 47MJ/Kg&lt;br&gt;&#xA;* Density of gasoline approx 0.74 Kg/L&lt;br&gt;&#xA;             =&gt; ~ 35MJ/L&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculations:&lt;br&gt;&#xA;PowerIn = (35MJ/L) * (26.8m/s) / (14km/L)&lt;br&gt;&#xA;       = (35kJ) * (26.8/s) / (14), NB: MJ=kkJ&lt;br&gt;&#xA;       = 67kJ/s&lt;br&gt;&#xA;       = 67kW&lt;br&gt;&#xA;Efficiency = 14.9/67&lt;br&gt;&#xA;        = 22.2%&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB1: The quoted 240hp is the maximum OUTPUT power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB2: The % load in the diagram refers to maximum power at that particular speed. It is roughly proportional to volumetric efficiency of the engine, which depends on throttle position, valve timing and manifold resonance characteristics.&lt;/p&gt;&#xA;" OwnerUserId="17292" LastEditorUserId="17292" LastEditDate="2018-08-25T11:15:34.783" LastActivityDate="2018-08-25T11:15:34.783" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23450" PostTypeId="2" ParentId="23443" CreationDate="2018-08-25T07:15:08.470" Score="0" Body="&lt;p&gt;Commutating slip rings will work but another option is to use a slew ring for the main bearing of the drum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FXWGt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FXWGt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensor wires and mounting can then extend to outside of the bin and not therefore not have to rotate at all... (there is no shaft in the way)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other end can be a traditional shaft and bearing if required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're expensive new (but so can be slip-rings) - suggest you look at surplus/old spare stock on ebay etc.&lt;/p&gt;&#xA;" OwnerUserId="3231" LastActivityDate="2018-08-25T07:15:08.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23451" PostTypeId="1" CreationDate="2018-08-25T12:11:58.237" Score="0" ViewCount="381" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/igNJb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Here in this book of Theory of machines by S.S. Rattan it is written that diametral pitch is not used in S I units. Why is that so?&lt;/p&gt;&#xA;" OwnerUserId="17284" LastActivityDate="2019-12-18T18:03:05.663" Title="Why diametral pitch is not used in S.I units" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23452" PostTypeId="2" ParentId="23451" CreationDate="2018-08-25T13:00:09.787" Score="1" Body="&lt;p&gt;European standards differ from US standards. US standards re not generally SI based. European standards have defined module, then therefore there is no need to define a second set of definitions. &lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2018-08-25T13:00:09.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23453" PostTypeId="2" ParentId="23443" CreationDate="2018-08-25T15:17:21.830" Score="0" Body="&lt;p&gt;Additional to the slip rings solution, you can easily find wireless sensors on the market. Which I personally find much more easy to implement.&lt;/p&gt;&#xA;" OwnerUserId="17286" LastActivityDate="2018-08-25T15:17:21.830" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23455" PostTypeId="1" AcceptedAnswerId="23465" CreationDate="2018-08-25T20:10:18.967" Score="2" ViewCount="3177" Body="&lt;p&gt;I have Sketch 1. It's on Plane 1. There's a different Plane 2, and I want to make Pketch 2 which is simply the entities in Sketch 1 copied to Sketch 2. I don't want to copy-paste, because I want to have their constrains preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing &quot;Convert Entities&quot; doesn't help, because the shapes become different, since they are projected to the plane. &lt;/p&gt;&#xA;" OwnerUserId="11510" LastActivityDate="2019-01-08T10:41:46.720" Title="Solidworks: Make a copy of a sketch on a different plane" Tags="|solidworks|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23456" PostTypeId="1" AcceptedAnswerId="23463" CreationDate="2018-08-25T20:14:20.660" Score="-1" ViewCount="65" Body="&lt;p&gt;I am a freshman student and I am very interested in using solar energy to break water into hydrogen and oxygen. So, I want to know which areas are important in this regard..&lt;/p&gt;&#xA;" OwnerUserId="17210" LastEditorUserId="17210" LastEditDate="2018-08-26T05:28:45.057" LastActivityDate="2018-08-26T14:31:31.013" Title="what classes should I take so that I can perform research into finding catalysts to split water into oxygen and hydrogen?" Tags="|chemical-engineering|photovoltaics|" AnswerCount="2" CommentCount="6" ClosedDate="2018-08-26T14:32:09.040" ContentLicense="CC BY-SA 4.0" />
  <row Id="23457" PostTypeId="1" AcceptedAnswerId="23468" CreationDate="2018-08-25T21:13:53.227" Score="1" ViewCount="262" Body="&lt;p&gt;I have the following differential equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\ddot{\psi_t} + \omega^2 \psi_t = \omega R_s i_f - \omega u_f + \dot{u}_t - R_s \dot{i}_t$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $w$ is the electrical speed of the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have considered R.H.S. as $\Delta\omega$ and would be feed-forwarded, so I obtain the following transfer function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\frac{\psi_t}{\Delta\omega} = \frac{1}{s^2 + \omega^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any tips for controlling the above undamped second order system? Can I use PI controller for this?&lt;/p&gt;&#xA;" OwnerUserId="17294" LastEditorUserId="1832" LastEditDate="2018-08-25T23:50:12.570" LastActivityDate="2018-08-27T00:06:05.187" Title="Control of second order undamped system" Tags="|electrical-engineering|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23458" PostTypeId="2" ParentId="23456" CreationDate="2018-08-25T22:47:30.947" Score="0" Body="&lt;p&gt;you'll need to be expert in (at least) these fields:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;photocatalysis&lt;/li&gt;&#xA;&lt;li&gt;electrochemistry&lt;/li&gt;&#xA;&lt;li&gt;photovoltaics &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In addition, you'll have to become intimately familiar with the published efforts others have made in this field over the last 45 years or so. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="1832" LastEditDate="2018-08-26T01:27:08.673" LastActivityDate="2018-08-26T01:27:08.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23459" PostTypeId="1" CreationDate="2018-08-26T07:43:01.157" Score="1" ViewCount="116" Body="&lt;p&gt;I was ruminating over &lt;em&gt;A Connecticut Yankee in King Arthur's Court&lt;/em&gt;. In the book, a trained American engineer goes back in time and creates guns. The book itself predates tanks, but one could imagine shoddy homemade technicals rolling around in medieval England. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is: is that possible? Let's imagine we have the full support of an entire kingdom, and we're working purely from memory. We still have bad quality steel, primitive tools, and no CAD. Can we make an engine at all, even a terribly leaky one? &lt;/p&gt;&#xA;" OwnerUserId="17298" LastActivityDate="2018-08-26T08:18:16.573" Title="How difficult would it be to build an entire engine (block and all) in 16th century conditions?" Tags="|engines|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23460" PostTypeId="2" ParentId="23459" CreationDate="2018-08-26T08:01:26.243" Score="1" Body="&lt;p&gt;Yes. The principle of the steam engine was known already in the 16th century. Their problem was that they had a terrible efficiency, thus they were weak. James Watt's invention was not the steam engine, but that he used first the condensating water steam to convert heat to mechanical energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CAD significantly improves the engineering work, but it is not really needed for building engines (consider, for example, the cathedrals - they were built purely on experimental knowledge).&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2018-08-26T08:01:26.243" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23461" PostTypeId="2" ParentId="23459" CreationDate="2018-08-26T08:18:16.573" Score="1" Body="&lt;p&gt;The casting and forging are two ancient technics to deform metals. Even these days, manufacturers still make use of casting to form engine blocks. For other parts, usually smaller than engine blocks and for large series man invented (3000 B.C  in southwest Asai) the 'Investment casting' which was a high precision method to manufacture parts (metals). As peterh pointed, CAD gives us an insight into more complex mechanical structures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More than 4000 B.C man found out, how to form gold into desire forms by striking on its surface, so the forging born. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-08-26T08:18:16.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23462" PostTypeId="2" ParentId="23443" CreationDate="2018-08-26T12:40:53.727" Score="0" Body="&lt;p&gt;A rotary transformer &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Rotary_transformer&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Rotary_transformer&lt;/a&gt; could be an option. You may use one channel to take some AC power supply to the moving side, and conversely  one or more channels to bring back some kind of serial data. Some simple electronics should be installed in the rotating part to condition power supply and communications. Rotary transformer itself could probably be a relatively low specs one and hence could be a home made one.&lt;/p&gt;&#xA;" OwnerUserId="7668" LastActivityDate="2018-08-26T12:40:53.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23463" PostTypeId="2" ParentId="23456" CreationDate="2018-08-26T13:36:14.130" Score="0" Body="&lt;p&gt;At your level, start with the two semesters of general chemistry. You will need a solid basis in chemical dynamics, redox chemistry, and chemical thermodynamics. Next take physical chemistry, especially when it includes electrochemistry. This will expand your understanding of the three core topics. Finally, take a course that covers aspects of heterogeneous catalysis, for example as offered in chemical engineering departments. This will explain  how to translate the core topics from bulk systems to surfaces. As you move to graduate school, you should take courses in catalysis, solid state chemistry/physics, and electrochemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Undergrad curricula in chemistry, chemical engineering, or materials science (not materials engineering) should provide what you need. Graduate curricula would be in the same fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do &lt;strong&gt;not&lt;/strong&gt; neglect to take the associated laboratory courses as an undergraduate. You &lt;em&gt;must&lt;/em&gt; learn how to do experiments in general chemistry and physical chemistry. You &lt;em&gt;must&lt;/em&gt; learn how to keep certifiable notebooks, review literature, write well-structured technical reports, and prepare publishable-level documents on your research. These skills are built from the groundings in the undergraduate laboratories, they do not just magically appear because you show up one day fully enthused and excited to do your research work in a laboratory.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-08-26T14:31:31.013" LastActivityDate="2018-08-26T14:31:31.013" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23465" PostTypeId="2" ParentId="23455" CreationDate="2018-08-26T21:38:11.307" Score="0" Body="&lt;p&gt;Step 1: &#xA;Select the entity you want to copy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2: Select the plane you want to copy the sketch to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3: Go to Insert $\rightarrow$  Derived Sketch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you'll see it makes the exact copy of the entity in the second plane.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-08-26T21:38:11.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23468" PostTypeId="2" ParentId="23457" CreationDate="2018-08-27T00:06:05.187" Score="1" Body="&lt;p&gt;So let's just start with a pure proportional controller.  What does success (i.e. error=0) look like for the controller?  Basically its when $\psi=0$ (or equal to a setpoint, but let's just say 0 for simplicity).  And a proportional controller can drive the system to $\psi=0$.  When $\psi=0$, the control effort will be zero, but $\dot\psi$ will not, in general, equal zero. Let's just call it $\dot\psi=v$. So your system will not stop at $\psi=0$. It's going to overshoot.  The controller will notice this, and correct, and drive the system back to $\psi=0$.  But at this point, now $\dot\psi=-v$, so the system is going to undershoot.  It will keep oscillating like this forever.  Now, if you had a damped system, the damping would take out a little energy each time, so that you don't come back to $\dot\psi=v$, but maybe $0.9v$, and eventually the oscillations will die out and you reach steady state.  But if the damping is zero, the oscillations never die, they will continue at the same amplitude forever. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you have a PI controller, it's probably even worse.  When the system hits $\psi=0$, you probably have some integrated error built up, so the control effort is not zero, but non-zero.  So the overshoot is even higher than with a P controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to overcome this is to use a PD or PID controller. The D term essentially acts on $\dot\psi$. Now when $\psi=0$, the error term and thus control effort will be zero only if $\dot\psi=0$ too.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably a better way, however, is to use two nested controllers.  One controller acts on $\psi$ and the other controller acts on $\dot\psi$.  The inner controller compares $\psi$ to the setpoint and generates a desired $\dot\psi$ that will achieve that (e.g. if $\psi&amp;gt;0$ then the $\dot\psi$ setpoint is negative).  The outer controller takes the setpoint for $\dot\psi$ as an input, and applies control effort (in your case $\Delta\omega$) in order to drive $\dot\psi$ to the desired value.  Both controllers could be P or PI.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-08-27T00:06:05.187" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23471" PostTypeId="2" ParentId="22909" CreationDate="2018-08-27T08:18:06.413" Score="0" Body="&lt;p&gt;Becoming an engineer is learning when to trust your gut feeling &amp;amp; experience, and when to trust the math.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;i have done some math with the mass balance and have derived the differntial equation. i have also simulated it but the problem is as follows: the level becomes stable at any given valve opening but most of the time at more height than overflow height. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This time, it seems your math checks out - congrats! - and you should trust it. It is indeed the case that for any Qin and valve setting, the level of the tank will stabilize seomwhere (unless there's an overflow): For any given valve setting, flow Q will result in a specific head loss. If the water level is high enough to provide that head for Qin, you have reached a stable level.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;there should be one value of valve opening at which the level stabelizes. i want to find that opening&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not to sound harsh, but I think you confused yourelf here a little bit. You need to question &lt;em&gt;why&lt;/em&gt; you think there sould be only one stable valve setting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to exercise your math-muscles by searching specific valve positions, find the valve position for a specific stable tank level (say 3.99 m). &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2018-08-27T11:15:23.393" LastActivityDate="2018-08-27T11:15:23.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23472" PostTypeId="2" ParentId="23271" CreationDate="2018-08-27T08:40:03.673" Score="2" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\newcommand{\a}[1]{\langle#1\rangle}$&lt;/span&gt;&#xA;&lt;strong&gt;Loading&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the last two parts of your loading equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{2000}{3}\a{x-0}^0 - \frac{2000}{3}\a{x-0}^1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XzppF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XzppF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to superimpose the two graphs they would not look like the triangular loading you have in your picture. If we wanted that loading, we would use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$-2000\a{x-0}^0 + \frac{2000}{3}\a{x-0}^1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would make sure that we had -2000kN/m at x=0m, and 0kN/m at x=3m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Integration&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\int \langle x-a\rangle ^{n}dx={\begin{cases}\langle x-a\rangle ^{{n+1}},&amp;amp;n\leq 0\\{\frac  {\langle x-a\rangle ^{{n+1}}}{n+1}},&amp;amp;n\geq 0\end{cases}}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;q(x)=-3000\a{x-0}^{-2} + 3000\a{x-0}^{-1} - 2000\a{x-0}^0 + \frac{2000}{3}\a{x-0}^1&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;v(x) = - 3000\a{x-0}^{-1} + 3000\a{x-0}^{0} - 2000\a{x-0}^1 + \frac{1000}{3}\a{x-0}^2&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is continuous over the entire length, we can simplify to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;v(x) =  3000 - 2000x + \frac{1000}{3}x^2&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is exactly what we would expect, since there is 3000kN at x=0m, and 0kN at x=3m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we can also plot in &lt;a href=&quot;http://www.wolframalpha.com/input/?i=y%20%3D%203000-2000x%2B(1000%2F3)x%5E2%20from%20x%20%3D%200%20to%20x%20%3D%203&quot; rel=&quot;nofollow noreferrer&quot;&gt;wolfram alpha&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17306" LastEditorUserId="1832" LastEditDate="2019-03-12T16:43:09.600" LastActivityDate="2019-03-12T16:43:09.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23473" PostTypeId="1" AcceptedAnswerId="23476" CreationDate="2018-08-27T10:38:57.380" Score="1" ViewCount="481" Body="&lt;p&gt;Question is on the meaning of input variable TID within material model 163 of LS-Dyna 971.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Definition of TID variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TID: Table ID defining yield stress versus volumetric strain at different strain rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remark of TID: Rate effects are accounted for by defining a table of curves using *DEFINE_TABLE. Each curve defines the yield stress versus volumetric strain for a different strain rate. The yield stress is obtained by interpolating between the two curves that bound the strain rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I understand of the yield stress is a single point on a stress vs volumetric strain curve for each individual strain rate. So wouldn't that give me a yield stress vs volumetric strain point for each individual strain rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the yield stress referring to just the stress. Having trouble understanding the inputs, any clarification would be helpful.&lt;/p&gt;&#xA;" OwnerUserId="16825" LastActivityDate="2018-08-27T21:17:02.173" Title="LS DYNA Material Model 163, Modified Crushable Foam" Tags="|mechanical-engineering|materials|applied-mechanics|stresses|simulation|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23476" PostTypeId="2" ParentId="23473" CreationDate="2018-08-27T13:16:15.703" Score="2" Body="&lt;p&gt;While the yield stress is a well defined point in materials such as steel or other metals, that is not the case when we talk about foams.&#xA;Here they are calling &quot;yield stress&quot; to the whole volumetric strain-stress curve.&#xA;So they are basically asking for a number of strain-stress curves at different strain rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can use material 63 if strain rate effects are not important.&lt;/p&gt;&#xA;" OwnerUserId="17286" LastEditorUserId="17286" LastEditDate="2018-08-27T13:34:58.667" LastActivityDate="2018-08-27T13:34:58.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23479" PostTypeId="2" ParentId="23257" CreationDate="2018-08-27T18:09:48.657" Score="6" Body="&lt;h2&gt;NOTE ON OBROUND VESSELS&lt;/h2&gt;&#xA;&lt;p&gt;While not a direct answer to your question, it should be quickly noted that &lt;strong&gt;ASME BPVC Section VIII, Mandatory Appendix 13, Section 10&lt;/strong&gt; has a section to design obround vessels (circular half sections with flat plate walls, see image below).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lLRvG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lLRvG.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;While not the same, these can typically optimize the available area compared to elliptical vessels (for a given dimension A and B).  As such, these are almost always the preferred design choice compared to elliptical, and more commercially available.  The equations, while handling a membrane stress discontinuity, are easier.&lt;/p&gt;&#xA;&lt;h2&gt;Thin wall assumption basis&lt;/h2&gt;&#xA;&lt;p&gt;We can easily derive a thin wall version of this.  It can be noted that the ellipse shown above can be expressed as all points&#xA;&lt;span class=&quot;math-container&quot;&gt;$$(\frac{b}{2}\cos(\phi),\frac{a}{2}\sin(\phi))$$&lt;/span&gt;&#xA;with a parameter &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; that varies from &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$2\pi$&lt;/span&gt;.  Note this isn't the same as the angle from the center, &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;, but is related to this angle by the relationship&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\tan(\theta) = \frac{b}{a} \tan(\phi)$$&lt;/span&gt;&#xA;For a given &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt;, the stress in a thin-wall vessel can be easily seen.  Similar to a circular vessel, we take a cross section to a point at a specific &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; through to the center.  Using a similar thin wall approximation as cylindrical vessels, we presume the tangential (membrane) stress is uniform across this cut of a thickness &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;.  We presume this is uniform regardless of &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$t&amp;lt;&amp;lt;a ; t&amp;lt;&amp;lt;b$&lt;/span&gt;.  The distance &lt;span class=&quot;math-container&quot;&gt;$D(\phi)$&lt;/span&gt; is a function based upon &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; from the edge, through the center to the other edge would be easily&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D(\phi) = \sqrt{b^2\cos^2(\phi) + a^2\sin^2(\phi)}$$&lt;/span&gt;.&#xA;Since the wall is pushed uniformly across this cut with a pressure &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;, then it is clear that, similar to circles, for a load per unit length basis:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$PD(\phi) = 2t\sigma(\phi)$$&lt;/span&gt;.&lt;br /&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_t(\phi) = \frac{PD(\phi)}{2t}$$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;h2&gt;Presuming constant perimeter&lt;/h2&gt;&#xA;&lt;p&gt;Note that stress concentrations will definitely emerge, as the vessel will want to take on a cylindrical shape.  Running a cut through both the x and y axis and running a statics analysis of this section, we note that a much smaller tangential stress comes from one axis than the other.  The non-uniform strain will certainly cause bending moments in the thin wall material.&lt;/p&gt;&#xA;&lt;p&gt;Without running a thick wall analysis, these stress concentrations cannot be determined exactly.  For this thin wall presumption, we simply assume the perimeter will stay roughly the same before and after deflection.  The perimeter of an ellipse is determined by using elliptical integrals.  Defining:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$k^2 = 1-\frac{a^2}{b^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Then the perimeter for the ellipse is &lt;span class=&quot;math-container&quot;&gt;$2bE(k)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is the elliptical integral of the second kind.  We note that &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is a function of &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;.  Taking the derivative of this with respect to &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, and a lot of math, we find the beautiful result that:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{E(k)}{F(k)} = \frac{a^2}{b^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We approximate &lt;span class=&quot;math-container&quot;&gt;$E(k)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F(k)$&lt;/span&gt;, the elliptical integral of the first kind, as a constant for small changes in &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; due to changes in &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;.  As such, for a small contraction of &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Delta b$&lt;/span&gt;, we find an expansion of &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Delta a$&lt;/span&gt; as follows:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta a = - \frac{b^3}{a^3} \Delta b$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The cubic relation implies that highly eccentric ellipses will quickly snap from the elliptical shape into a cylindrical shape!  As such, the above approximations will quickly become invalid.  The negative implying the expansion is in converse to the contraction.  With this information, we can attempt to resolve the stress concentrations to a reasonable approximation.  The &amp;quot;radial&amp;quot; stress from a statics view indicates it should have a linear radial growth, while the perimeter analysis shows a cubic radial growth.  This deflection would present itself as a force, adding a bending concentration at the tips of edge b (worst case) of approximately:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M = \frac{Pb^2(b^2-a^2)}{4a^2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_b = \frac{3Pb^2(b^2-a^2)}{2t^2a^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This stress would have &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; swapped with &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; for the other axis, to obtain the two maximum stresses in the vessel.  Note the original tangential stress will need to be added to the bending stress.  As such, the total stress in the long-axis side would be:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_b = \frac{3Pb^2(b^2-a^2)}{2t^2a^2} + \frac{Pb}{2t}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Adding together with the axial stress of your case-by-case analysis (complicated, but could be assumed to be reasonable by dividing the area of an elliptical cap by the strange perimeter of an ellipse), and you have the stresses for a Von-Mises stress.  This, as a function of the thickness, would be set to maximum value, and the minimum thickness would be the result that allows that Von-Mises stress to reach the maximum value.  This portion is commonly utilized in pressure vessel design, as environmental loading adds more stress than your equation presumes.&lt;/p&gt;&#xA;&lt;p&gt;I would run some tests to verify, but this is my conclusion as a PE for 8 years.  Others can disagree as they see fit, and I'd be happy with a better analysis that shows better work.&lt;/p&gt;&#xA;&lt;h2&gt;Notes on thick wall vessel&lt;/h2&gt;&#xA;&lt;p&gt;I did attempt a thick wall version, but it quickly became apparent that I needed to invest more time than I was able to dedicate to it.  Note you did not ask for this - and since your question and equation related to the thin-wall case, I saw no reason to push forward.  Some notes, if you are familiar with elliptical integrals:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The practicalities are still an issue - obround works well even in thick-wall cases.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Sources are elusive for other works - Roark's was empty, as was ASME code.  As such, I again defer to obround as the better method of construction for all practical applications.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-08-27T19:47:33.583" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23481" PostTypeId="2" ParentId="23443" CreationDate="2018-08-27T19:54:34.610" Score="0" Body="&lt;p&gt;I suggest having a fixed shaft, with two bearing where it carries the drum. The drum can be driven from the outside with a roller, belt or similar. That way you don't need so many slip rings and no wireless connections.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-08-27T19:54:34.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23482" PostTypeId="2" ParentId="23257" CreationDate="2018-08-27T21:13:24.517" Score="1" Body="&lt;p&gt;You can use a finite element code to simulate the stress distribution in the wall.&#xA;Simulating elastic problems under pressure is straight forward to model and you get the advantage of getting the stress distributions in 2 or 3 dimensions in contrast with the other methods mentioned.&lt;/p&gt;&#xA;" OwnerUserId="17286" LastEditorUserId="17286" LastEditDate="2018-08-27T21:30:06.250" LastActivityDate="2018-08-27T21:30:06.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23484" PostTypeId="6" CreationDate="2018-08-27T21:36:41.693" Score="0" Body="&lt;p&gt;I do not think I am the most qualified poster for this position, and would happily endorse the optimum candidates.  But, I have high interest in this site, and I want to help it grow.  Put simply, I have made a hobby of solving problems using this site, something that makes me personally happy.  I would like to continue being happy.  So, I am taking a stab at this election and I will see what happens.  I want to help welcome new users into the community and I want to see this site take off from where it is today.  I want to encourage thought provoking physics in the engineering sense (we approximate everything but we get the job done, am I right?).  I want to encourage practical applications and help answer questions from real engineers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not the firm candidate.  In many vote to close queues, I find myself voting to leave open and maybe even proposing a strange answer to the question at hand.  If you feel the site needs firmer rules and more cleanup, then look to someone else.  I will encourage and propose alternates that engage and help new people discover what this site can be first before taking such drastic action.  &lt;strong&gt;I find solutions where others see problems.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="2015" LastEditDate="2018-08-27T21:36:41.693" LastActivityDate="2018-08-27T21:36:41.693" CommentCount="1" CommunityOwnedDate="2018-08-27T21:36:41.693" ContentLicense="CC BY-SA 4.0" />
  <row Id="23486" PostTypeId="1" CreationDate="2018-08-28T04:55:05.767" Score="1" ViewCount="44" Body="&lt;p&gt;I am currently developing an AHU for all weather conditions.&#xA;I need the design process for calculating coil capacity and air discharge rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Issue is that, while plotting the process on psychrometric chart, you can give only 1 outside air condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you would need to plot 12 different charts for a year (considering that the temp. is averaged out for a month, which I think is not the correct way to design) and choose the highest required capacity as design requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence for all year air conditioning how can I design coil capacity and air discharge?? Any other methods?&lt;/p&gt;&#xA;" OwnerUserId="17317" LastActivityDate="2018-08-28T04:55:05.767" Title="Air handling unit design" Tags="|hvac|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23487" PostTypeId="1" CreationDate="2018-08-28T08:26:53.270" Score="2" ViewCount="251" Body="&lt;p&gt;I'm looking for a formula to arrive at a pump curve for a centrifugal pump: $H(Q, n)$ - total head delivered as a function of flow and speed (rpm), and of course some pump specific parameters. I find no formula I can transform into this. For actual pump selection I would of course use the pump curves from the supplier, but a formula like this would be interesting for simulation purposes of pumping systems where VFD is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As both head and flow are mostly a function of speed, impeller size and shape  I think a non purely empirical fomrlua for a pump curve should exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. Alternativly, I considered using a simple polynomial or similar curve and fitting it numerically to an existing pump curve. Pump simulators, is this an actually useful approach lacking an analytically correct curve?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="2015" LastEditDate="2018-08-28T22:42:55.397" LastActivityDate="2018-08-28T22:42:55.397" Title="Analytical shape of a pump curve" Tags="|mechanical-engineering|pumps|hydraulics|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23488" PostTypeId="1" CreationDate="2018-08-28T13:36:39.677" Score="2" ViewCount="333" Body="&lt;p&gt;I have been trying to build an electric go kart, and power requirements (as always) have been the problem. I have been going through some formulas to find out how big of a motor I would need from how much torque is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Formulas used:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;$F=ma$&lt;/li&gt;&#xA;&lt;li&gt;$M=Fr$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My specs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Weight of vehicle (driver included): 200kg&lt;/li&gt;&#xA;&lt;li&gt;Top speed: 20m/s (45mph) acc. (a)= 2m/s^2(10 sec to reach top speed) radius(r)=0.2 (variable tho, this is the problem)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;From calculations, F=200kg*2m/s^2=400N, torque=400N*0.2m=80 Nm. If a smaller radius is used(say 0.1 m instead of 0.2 m), less torque is needed(400N*0.1=40Nm). Clearly, there is some misunderstanding, or we would all use smaller wheels if they resulted in less torque needed. What am I doing wrong? maybe I did not understand the formulas correctly? If I am correct, how would you explain using bigger wheel sizes?&lt;/p&gt;&#xA;" OwnerUserId="17322" LastEditorUserId="9" LastEditDate="2018-08-28T14:28:57.313" LastActivityDate="2018-08-28T17:47:31.860" Title="smaller radius requires less torque?" Tags="|torque|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23489" PostTypeId="1" CreationDate="2018-08-28T13:46:59.147" Score="2" ViewCount="120" Body="&lt;p&gt;Anybody knows a tutorial to make a finite element analysis for bolted joints like junker test(fastener vibration) in Solidworks or Ansys?&lt;/p&gt;&#xA;" OwnerUserId="17326" LastActivityDate="2018-08-28T13:46:59.147" Title="Bolted joint junker test with finite element method" Tags="|finite-element-method|simulation|ansys|solidworks|ansys-workbench|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23491" PostTypeId="1" CreationDate="2018-08-28T14:00:01.443" Score="4" ViewCount="2195" Body="&lt;p&gt;I'm trying to learn a bit about building kinematic mechanisms by building a small project. The goal is to be able to open and close a box programmatically via a servo motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a small box that I've reproduced in fusion 360 and I'm trying to design a crank rocker mechanism to open and close the box. After reading up on the crank rocker concept (I have no background in engineering so I'm brand new to all of these concepts) and building a couple of free standing proof of concepts (&lt;a href=&quot;https://www.thingiverse.com/thing:3068478&quot; rel=&quot;nofollow noreferrer&quot;&gt;I put them up on thingiverse&lt;/a&gt;) I tried to start fitting the linkages into the box model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue I'm running into is that, given the size constraints of the box and the length of the servo's horn, I'm not sure how to calculate the correct coupler link length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gmpOG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gmpOG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that according to grashof's law I need to make sure that the sum of the shortest and longest links are less than the sum of the other two links; $s + l &amp;lt; q + p$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started by modeling the rocker arm ($q$) first so that I could figure out the length of the sliding slot (not sure what to call this) I would have to add to the lid so that the arm could actually open the lid. I placed the points for the rocker arm on the box and the lid so that there was room for the rocker arm to move without hitting the back of the box when the lid was fully opened, so it was kind of an arbitrary placement choice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already know the length of my default servo horn so I was going to stick with that (though I can 3d print a new horn if necessary).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured that between:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The positioning of the servo and the bottom of the rocker arm (making the grounded link $p = 55.228mm$&lt;/li&gt;&#xA;&lt;li&gt;The length of the rocker arm $q = 60mm$&lt;/li&gt;&#xA;&lt;li&gt;The length of the crank arm $s = 25.53 mm$&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I could just calculate the length of the coupler arm ($l$). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$25.53 + x &amp;lt; 55.228 + 60$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part that I'm not sure how to figure out the correct X value. It seems like, if I understand all of this correctly, I have a range of values I can pick from and the top of that range is $89.698 mm$. I know that the bottom of somewhere around $68.081 mm$ (I don't actually know the proper way to calculate this, I just closed the lid on my model and measured the distance between the end of the servo horn and the top of the rocker arm :| ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I can figure it out by noodling around with the coupler arm length until I find one that will let me fully extend the lid, but I'm trying to use this all as a learning exercise/introduction into these concepts. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is there a way of determining the ideal coupler length? &lt;/li&gt;&#xA;&lt;li&gt;Am I even right on the way that I've approached it so far?&lt;/li&gt;&#xA;&lt;li&gt;Am I missing anything super obvious?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for the help!&lt;/p&gt;&#xA;" OwnerUserId="17325" LastEditorDisplayName="user14407" LastEditDate="2018-08-28T15:57:02.813" LastActivityDate="2018-08-28T15:57:02.813" Title="How to calculate the coupler length in a crank rocker mechanism" Tags="|mechanical-engineering|kinematics|servo|linkage|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23492" PostTypeId="2" ParentId="23488" CreationDate="2018-08-28T14:08:05.453" Score="1" Body="&lt;p&gt;The important requirement for accelerating the kart quickly is the &lt;em&gt;power&lt;/em&gt; (work done per second) that the motor can deliver. The relevant equations are $$\text{power}= \text{force} \times \text{velocity}$$&#xA;for motion in a straight line, or&#xA;$$\text{power}= \text{torque} \times \text{angular velocity}$$&#xA;for a rotating shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use smaller wheels, the angular velocity increases for the same linear speed of the kart, and the torque is therefore smaller for the same power, but the motor has to turn faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you changed the gear ratio to keep the motor speed the same, then the &lt;em&gt;motor&lt;/em&gt; torque would be the same, though (because of the different gearing) the torque at the &lt;em&gt;wheels&lt;/em&gt; would be less for the smaller wheels.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-08-28T14:08:05.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23493" PostTypeId="1" AcceptedAnswerId="23494" CreationDate="2018-08-28T14:36:55.343" Score="4" ViewCount="126" Body="&lt;p&gt;The common way to model or estimate gas (e.g. CO2) desorption rate from a liquid solution is by using an equation similar to this one&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$J_{CO_2} = -k_L a(H·C^G_{CO_2} - C^L_{CO_2})$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $J_{CO_2}$ is the molar (or mass) flowrate, $k_L$ is the mass transfer coefficient, $a$ is the specific contact surface, $H$ is one of the many expressions used for Henry's constant, and the G and L scripts stand for gas and liquid phase respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with this approach is that all the correlations I know used to estimate $k_L$, $a$, or most commonly $k_La$ imply that there is a gas stream with a given gas velocity (e.g. aeration or sparging).&#xA;How could I estimate $k_La$ for a still solution, for instance a glass containing beer where the gas bubbles are stripped slowly? Furthermore, if I wanted to assess the effect of stirring the beer with a spoon, how could I estimate the effect? &lt;/p&gt;&#xA;" OwnerUserId="3991" LastActivityDate="2018-09-06T00:51:06.500" Title="Modelling gas desorption from a still solution (e.g. a beer)" Tags="|chemical-engineering|modeling|process-engineering|multiphase-flow|mass-transfer|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23494" PostTypeId="2" ParentId="23493" CreationDate="2018-08-28T15:53:20.790" Score="3" Body="&lt;p&gt;In perfectly stagnate gas systems, transport of a component through a gas collapses to diffusion of the component through the gas. In this case, $k_L a \propto (D / \delta)^n$ where $D$ is diffusion coefficient, $\delta$ is a film thickness, and $n$ can be correlated or derived from first principles (typically $n \rightarrow 1$ in film theories). References can be taken from correlations of the Sherwood number to the Grashof and Schmidt numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A perfectly stirred liquid says $C^L$ is uniform throughout. A perfectly stagnate liquid requires that transport through the liquid also be modeled by diffusion. The approaches for diffusion in gas and diffusion in liquid are similar in concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transport models to review are film theory, penetration theory, and two-film theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you really are considering flow of a gas dissolved in a liquid, solubility can be an additional concern. The process of gas bubbles forming from dissolved gas is nucleation + growth. The bubbles become buoyant. This motion is &lt;em&gt;not&lt;/em&gt; modeled by diffusion, rather by a process that might be equivalent to free convection. The combined transport of gas bubbles from a dissolved gas in a stagnate liquid must be modeled to include nucleation + growth of bubbles + free convection of the bubbles. I would have to wonder whether the net result (nucleate + grow + rise out of liquid) tends to a homogenized bubble concentration in the container at some point. Considerations made in the absence of gravity will, as will stirring.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;References&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here are some links that I found in a search &quot;stagnate air, mass transfer rate&quot; with or without &quot;stirred liquid&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://doi.org/10.4319/lo.1995.40.2.0290&quot; rel=&quot;nofollow noreferrer&quot;&gt;Methane transfer across the water-air interface in stagnate wooded swamps ...&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://pubs.acs.org/doi/pdf/10.1021/es60111a012&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rates of evaporation of low-solubility contaminants from water bodies ...&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://pubs.acs.org/doi/pdf/10.1021/es00057a018&quot; rel=&quot;nofollow noreferrer&quot;&gt;Seasonal variations in air-water exchange of PCBs in Lake Superior&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://pubs.acs.org/doi/pdf/10.1021/ac960814r&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mass transfer characteristics of solvent extraction ...&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://pubs.acs.org/doi/pdf/10.1021/ie50498a055&quot; rel=&quot;nofollow noreferrer&quot;&gt;Significance of liquid-film coefficients in gas adsorption&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sciencedirect.com/science/article/abs/pii/S0255270103001004&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gas-liquid mass transfer coefficient in stirred tanks ...&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sciencedirect.com/science/article/abs/pii/S0255270104000388&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gas-liquid mass transfer coefficient in stirred tanks ... eddy structure ..&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-09-06T00:51:06.500" LastActivityDate="2018-09-06T00:51:06.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23495" PostTypeId="2" ParentId="23488" CreationDate="2018-08-28T17:47:31.860" Score="2" Body="&lt;p&gt;Welcome, Hey. a quick and easy way to get an estimate is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have you ever watched kart races in which the vehicles are powered by gasoline engines? Can you find out the horsepower rating for a gas kart whose speed and acceleration rate you wish to match? This is kind of a cheat (no equations required, woo hoo!) but it will get you started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next step is to decide on what you want the top speed of the kart to be; this then sets the overall gearing ratio for the drive system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The desired range tells you how heavy a battery pack you will need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my own experience and as a calibration point, a 1.5HP gas kart is fairly doggy, a 3HP kart is almost fun, a 5HP gas kart has more zing and can carry two people, and a 10HP gas kart (especially a shifter kart) can go like stink. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 746 watts to the HP (sorry about the equation), you can do the math!&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-08-28T17:47:31.860" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23496" PostTypeId="2" ParentId="18367" CreationDate="2018-08-28T18:26:08.353" Score="0" Body="&lt;p&gt;For thick wall assumption, the normal stress is also called the radial stress.  The thick wall assumption has thick walls and adds a small internal &quot;normal&quot; radial stress:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\sigma_r = \frac{r_i^2P_i-r_o^2P_o}{r_o^2-r_i^2} - \frac{(P_i-P_o)r_i^2r_o^2}{(r_o^2-r_i^2)r^2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note this stress is much smaller compared to the tangential and axial stress, which is why it is normally ignored.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="1832" LastEditDate="2018-08-28T18:41:58.307" LastActivityDate="2018-08-28T18:41:58.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23497" PostTypeId="2" ParentId="23346" CreationDate="2018-08-28T19:59:46.083" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://www.asme.org/engineering-topics/articles/turbines/turbine-fuel-efficiency-fitting-a-pitch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Turbine Fuel Efficiency: Fitting a Pitch&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an excellent answer to my question.&lt;/p&gt;&#xA;" OwnerUserId="17228" LastEditorUserId="110" LastEditDate="2019-06-10T00:22:47.863" LastActivityDate="2019-06-10T00:22:47.863" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23499" PostTypeId="1" AcceptedAnswerId="23503" CreationDate="2018-08-29T01:19:30.683" Score="4" ViewCount="85" Body="&lt;p&gt;I have a question about the use of a servo motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to lift a load of $130\ kg$ with two servo motors with $13\ Nm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameter of the shaft is $Ø40\ mm$, ie, $r=\ 20mm$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\ N = 1\ Kg\ *\ 1\  m/s^2$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1\ Nm = 1\ Kg\ *\ 1\  m/s^2*\ 1\  m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$25.506\ Nm = 130\ Kg\ *\ 9.81\  m/s^2*\ 0.02\  m$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By my reasoning, two servo motors could raise this load if the point of application of the load is exactly the diameter of the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use a cable that every time its diameter increases inside a pulley, at some point I will not be able to lift that load anymore?&lt;/p&gt;&#xA;" OwnerUserId="6100" LastActivityDate="2018-08-29T14:09:02.283" Title="Power decreases as the diameter increases" Tags="|dynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23500" PostTypeId="2" ParentId="23499" CreationDate="2018-08-29T01:40:03.677" Score="4" Body="&lt;p&gt;The downwards tension force in a cable with a $130\text{kg}$ mass hanging under gravity is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=mg=130*9.81=1275.3\text{N}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upwards force provided by a servo with arm radius $0.02\text{m}$ and torque $13\text{Nm}$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\frac{T}{r}=\frac{13}{0.02}=650N$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With two servos in your proposed system, the torque would be doubled, and so your theoretical lifting force would be 1300N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the maths, as long as the radius remains &lt;em&gt;under&lt;/em&gt; $\frac{13}{1275.3\ / \ 2}=0.02039\text{m}$, then you will be able to lift the mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your problem, however, comes at the speed of lifting. At your original radius, the resultant force (lift minus weight) is only $1300-1275.3=24.7\text{N}$. This means a maximum acceleration of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$a=\frac{F}{m}=\frac{24.7}{130}=0.19\text{m/s}^2$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is really very slow, and will only go down as the cable winds in. For any practical application you will find that the performance of the servos may not match the theoretical peak values from the data sheets. I'd be extremely surprised if the servo motors that you have in mind are able to lift the $130\text{kg}$ mass at all, let alone at a speed that is useful.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-08-29T01:40:03.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23502" PostTypeId="1" CreationDate="2018-08-29T13:48:09.453" Score="1" ViewCount="108" Body="&lt;p&gt;Numerous state regulations adopt the NCEES model language for the use of digital signatures.  This language says, among other things, that the signature must be &quot;capable of verification.&quot;  It is not clear to me what this means, though I can make some guesses.  Is there any authoritative answer as to what, exactly, must be verifiable when using a digital signature to sign an engineering document?&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorDisplayName="user16" LastEditDate="2018-08-30T12:35:29.523" LastActivityDate="2018-08-30T12:35:29.523" Title="When using digital signatures, what does &quot;capable of verification&quot; mean?" Tags="|licensure|seals|regulations|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23503" PostTypeId="2" ParentId="23499" CreationDate="2018-08-29T14:03:14.847" Score="4" Body="&lt;p&gt;In addition to what Jonathan has said, there is several other practical things to consider:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;not only torque is limited, but power also; since power is torque times angular velocity (P=T*omega), your motor specification will limit the maximum speed you will be able to lift the load at&lt;/li&gt;&#xA;&lt;li&gt;depending on your application, you will not necessarily have to apply all the weight force by your motors; in typical elevators there is a counterweight (usually the size of the average load) which balances the &lt;strong&gt;weight force&lt;/strong&gt;, so the motor will only have to apply the &lt;strong&gt;inertial force&lt;/strong&gt; (of load + counterweight, plus any deviations from the average weight)&lt;/li&gt;&#xA;&lt;li&gt;so, if you are building a kind of elevator, you don't have to wind the cable around a pulley either; because the counterweight pulls at the cable, you only need to take care that there is enough traction at the cable sheave&lt;/li&gt;&#xA;&lt;li&gt;if you really want to lift all the weight by the motors, you could also use a gear or a block and tackle mechanism, which can translate torque into a more manageable range (without resorting to ridiculously small shaft diameters)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So your pulley diameter of 20mm is not set in stone after all. But remember: if you use a mechanical advantage of some sort (like in bullet 4), you will only change force/torque, whereas power stays the same (or almost, considering friction effects). So the power limitation of your motors is usually the more severe one, and this depends on what ultimate speed you want to move the load at.&lt;/p&gt;&#xA;" OwnerUserId="16376" LastEditorUserId="16376" LastEditDate="2018-08-29T14:09:02.283" LastActivityDate="2018-08-29T14:09:02.283" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23504" PostTypeId="1" AcceptedAnswerId="23506" CreationDate="2018-08-29T15:02:50.963" Score="0" ViewCount="242" Body="&lt;p&gt;I need to thermally insulate two metal (say aluminum) parts, the first of which is at 200-400 °C (473-673 K), and the second one is at room temperature. It should be possible to screw the parts together or something, to form one solid assembly. The typical dimensions are like 10-40 mm. The insulator should be able to be created in a DIY setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The insulation material has to be reasonably stiff/strong to enable a rigid compound construction, so all plastic foams etc. are out and, due to the high temperatures, almost all (easily available) solid plastics are also out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing thermal conductivities I have found that all mineral-based materials are reasonably insulating at ~1 W/(mK) when compared to metals (~10^2 W/(mK)).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What has come to my mind (in order of ideation):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;porcelain&lt;/li&gt;&#xA;&lt;li&gt;glass&lt;/li&gt;&#xA;&lt;li&gt;gypsum&lt;/li&gt;&#xA;&lt;li&gt;fine-grained concrete&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Anything obvious I have missed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gypsum seems very practical in that it can be easily cast with very fine structures. However I am not so sure whether it resists the heat (crystal water? gets more brittle at higher temp?). Anyway it is certainly a bad material for threading a metal screw in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Glass is stiff and temperature resistant, but it is also sensitive to high temperature gradients and breaks easily under mechanic stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fine-grained concrete (don't know if this exists at all as a product) - probably similar to gypsum, but a little stronger. But the question remains, if it can resist repeated heating/cooling/dehydration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Porcelain (or other sintered minerals): sounds perfect, but can I reasonably expect to create a shaped part with threads/holes etc. at home?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any best industry practices for parts like that?&lt;/p&gt;&#xA;" OwnerUserId="16376" LastActivityDate="2018-08-29T19:16:27.160" Title="DIY thermally insulating construction material" Tags="|thermal-insulation|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23505" PostTypeId="1" CreationDate="2018-08-29T15:26:14.860" Score="1" ViewCount="44" Body="&lt;p&gt;Suppose that we have a $2$ by $2$ by 2 $cm^3$ closed box, made up of glass, with thickness $0.1 cm$. Yes, it is a very small box. Let's say in that box there is a ball (also made with glass) with diameter $0.5 cm$. If we shake the box as fast as we can for one minute, making the ball bouncing like crazy inside the box, how do we measure (exactly or approximately) how many times the ball bounces/makes contact with one out of the six sides of the cube?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I never heard of any device that can do this. The closest ones are vibration meter and drumometer, I believe? However, it looks like vibrameter does not measure the number of bounces and drumometer is, well, too huge? Perhaps we need a mini drumometer? I mean, whatever that device is, does that mean a very small sensor that is connected to the device must be put into that side of the cube?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since the stuffs are made with glass, meaning they produce sound, can we measure the bounce by measuring (from the outside) the sound made from the shaking? My guess is that a device called sound level meter is what we need but it still won't count the bounce?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Please correct my tags/title if needed as I have very little engineering background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional question: What if (instead of one particular face) we measure the number of impacts coming from any face of the cube? Well, in this case, we don't need cube, but it can be a ball-shaped glass large enough to contain the small one.&lt;/p&gt;&#xA;" OwnerUserId="17338" LastEditorUserId="17338" LastEditDate="2018-08-30T09:02:26.337" LastActivityDate="2018-08-30T09:02:26.337" Title="Measuring beats/bounce in a closed cubed glass" Tags="|electrical-engineering|measurements|vibration|experimental-physics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23506" PostTypeId="2" ParentId="23504" CreationDate="2018-08-29T15:32:10.347" Score="1" Body="&lt;p&gt;have you considered firebrick, a product designed to be used in high temperature kilns, for example? It's light weight and easily cut with ordinary hand tools. It may not tolerate severe compression, but should bond well with aluminum and high-temperature adhesive.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-08-29T15:32:10.347" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23507" PostTypeId="2" ParentId="23504" CreationDate="2018-08-29T15:37:16.013" Score="1" Body="&lt;p&gt;Don't consider this as an answer, but since i can't post a photo in comments, here i post it, i generated this diagram using CES, level two, i marked the most suitable and accessible materials. &lt;a href=&quot;https://i.stack.imgur.com/rKAVt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rKAVt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disadvantage of all the materials here, is that they are brittle.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-08-29T15:37:16.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23508" PostTypeId="2" ParentId="23502" CreationDate="2018-08-29T16:59:27.120" Score="3" Body="&lt;p&gt;The question mentions numerous states, but I will use Washington State as a defined example.&lt;/p&gt;&#xA;&lt;h2&gt;Law&lt;/h2&gt;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.dol.wa.gov/business/engineerslandsurveyors/docs/eDocsInterpretiveGuideline.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;state law interpreted guidance in Washington&lt;/a&gt; states this:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Digital Signature&lt;br /&gt;&#xA;A Digital Signature has the following properties:&lt;br /&gt;&#xA;A. It has a hardware or software cryptographic protection (such as a public/private key pair) that is unique to the licensee.&lt;br /&gt;&#xA;B. It can be independently verified by a third-party certification authority.&lt;br /&gt;&#xA;C. The Electronic Record to which it is attached is encrypted such that the recipient can verify that the document was uniquely signed by the licensee.&lt;br /&gt;&#xA;D. The Electronic Record to which it is attached is encrypted such that the recipient can verify if the document has been altered since the licensee signed it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;h2&gt;Implementation&lt;/h2&gt;&#xA;&lt;p&gt;I haven't seen very many variations in usage. All of the digital signature that I have seen are on Adobe PDFs. Their help goes through the details of how to actually use the signatures, but it also mentions how each part meets the law as stated above.&lt;/p&gt;&#xA;&lt;h3&gt;Certificate-based Signatures&lt;/h3&gt;&#xA;&lt;p&gt;On &lt;a href=&quot;https://helpx.adobe.com/acrobat/using/certificate-based-signatures.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;, the following is stated:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The digital ID contains a private key and a certificate with a public key and more. The private key is used to create the certificate-based signature. The certificate is a credential that is automatically applied to the signed document. The signature is verified when recipients open the document.&lt;/p&gt;&#xA;&lt;p&gt;When you apply a certificate-based signature, Acrobat uses a hashing algorithm to generate a message digest, which it encrypts using your private key. Acrobat embeds the encrypted message digest in the PDF, certificate details, signature image, and a version of the document when it was signed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This statement basically meets all of the items as described in the law.&lt;/p&gt;&#xA;&lt;p&gt;Adobe has a &lt;a href=&quot;https://helpx.adobe.com/acrobat/using/validating-digital-signatures.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;separate page that talks about how the verification works&lt;/a&gt;, but that is only more background information stating that it does was is described above.&lt;/p&gt;&#xA;&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&lt;p&gt;All of that above is a lot of text that basically says that the entire process works together, i.e. signing and verification are parts of the same general process. There is no one part that is verifiable; either the entire document (signature and document) are verified, or they aren't. If you think about it, there is no use to have one or the other.&lt;/p&gt;&#xA;&lt;p&gt;As I mentioned in the beginning, I'm sure that there are other available methods/programs, but I have only ever seen the Adobe version in practice.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-08-29T16:59:27.120" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23509" PostTypeId="2" ParentId="23504" CreationDate="2018-08-29T19:16:27.160" Score="2" Body="&lt;p&gt;The aluminum can barely support itself at 400 C.&quot;Lightweight castable&quot; refractory ( can be handled like concrete) should do the job; but you would need to find a supplier .It is what a petrochemical or power-plant might use . Portland cement/concrete type material will not tolerate temperatures of 400 C. Johns-Manville and Harbison-Walker made product like this in the past.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-08-29T19:16:27.160" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23510" PostTypeId="2" ParentId="23505" CreationDate="2018-08-29T19:38:25.863" Score="1" Body="&lt;p&gt;I would attach some kind of microphone or piezo pickup to the center of the side of the box that you're measuring.&lt;br&gt;&#xA;Now, the devil is in the calibration - you'll need to do something like tapping the &quot;test&quot; side and seeing what the piezo output amplitudes are, then tap the other 5 faces of the cube and compare.  If there's a reasonably large difference in amplitude, or even in shape (spike vs. wider pulse or different decay tail), then you may be able to analyse the results of shaking the box via pulse height discrimination and/or FFT signatures. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-08-29T19:38:25.863" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23511" PostTypeId="6" CreationDate="2018-08-30T02:30:58.470" Score="0" Body="&lt;p&gt;I am passionate about this site. My browser has eight pinned tabs which are always open. Four of them are for this site (my filtered question list, the review page, chat and meta). I usually check in here multiple times a day (though this election has been fantastically ill-timed for me, since I'm currently on vacation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As moderator, I'll try to keep a reasonably light touch on review items, acting only in obvious cases, though I might keep adding my &quot;this looks like a homework question&quot; comments in cases which I believe should be closed but would rather leave for the community to decide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mostly hang around the structural-engineering and mechanical-engineering tags, and will probably mainly continue to do so except for whatever shows up on the Review page, so keep flaggin'!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be very glad to be your moderator, and I hope you wish the same.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2018-08-30T02:30:58.470" LastActivityDate="2018-08-30T02:30:58.470" CommentCount="2" CommunityOwnedDate="2018-08-30T02:30:58.470" ContentLicense="CC BY-SA 4.0" />
  <row Id="23512" PostTypeId="1" CreationDate="2018-08-30T02:48:32.980" Score="1" ViewCount="1391" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bncoG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bncoG.png&quot; alt=&quot;AIR LIFT?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much pressure would I need or, better yet, how do I compute for the pressure required to push water out of a submerged tank? Do I use 1 meter as head or 7 meters? (Please just bear with the drawing :) )&lt;/p&gt;&#xA;" OwnerUserId="17339" LastActivityDate="2018-09-29T16:01:12.120" Title="How much pressure would I need to push water out of tank?" Tags="|pressure|pressure-vessel|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23514" PostTypeId="1" CreationDate="2018-08-30T05:38:29.323" Score="1" ViewCount="810" Body="&lt;p&gt;I couldnt get a good explanation differentiating phase and grain.From what I understood, a collection of unit cells arranged in the same orientation is called grain.A collection of grains with similar structure is called phase.Is this interpretation right?&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastActivityDate="2018-08-30T13:43:24.183" Title="Difference between phase and grain in a material" Tags="|materials|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23515" PostTypeId="1" CreationDate="2018-08-30T06:02:19.013" Score="0" ViewCount="42" Body="&lt;p&gt;Electric field inside a metal is zero , when it is placed in an external electric field. What about human body? If it is placed in an external electric field, will the field inside the body be zero? I am confused because human body is a conductor but not as good as metals.&lt;/p&gt;&#xA;" OwnerUserId="17342" LastActivityDate="2018-08-30T06:35:27.340" Title="Electric field inside human body" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23516" PostTypeId="2" ParentId="23515" CreationDate="2018-08-30T06:35:27.340" Score="2" Body="&lt;p&gt;No. Humans are not faraday cages. This is the reason such things as MRI scanners work. &lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2018-08-30T06:35:27.340" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23517" PostTypeId="2" ParentId="23514" CreationDate="2018-08-30T06:49:09.143" Score="2" Body="&lt;p&gt;A phase can be defined as as a homogeneous portion of a system that has uniform physical and chemical characteristics, so every pure material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;grain is defined as an individual crystal in a polycrystalline metal or ceramic.Let's clear the confusion with an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p9Jnw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p9Jnw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here there is SEM photo of a pearlite (multiphase). The white area represents iron and the black lines represent carbon. The grains are also visible in this microscopic image. As you can see, each grain consists of two phase nl. iron and carbon, but the orientation of lamellar carbon is not the same in each grain. Your first interpretation of grain concept is almost correct, but be carful  with the word 'structure' in the definition of phase.  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-08-30T06:49:09.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23518" PostTypeId="1" CreationDate="2018-08-30T09:17:25.350" Score="0" ViewCount="2741" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HnNb3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HnNb3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a pipe which is channeling air flow from a compressor with a known mass flow rate and temperature. I can adjust the mass flow rate (power setting) and measure the delta T for the temp of air coming out the nozzle for each setting. It's for a cooling application and I want to maximise mass flow rate and minimise temperature of the flow. How can I optimise this system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air is channeled perpendicular to the disc solid with a gap between the nozzle and the solid to allow the air flow to dissipate. The nozzle and the disc are fixed.&lt;/p&gt;&#xA;" OwnerUserId="17345" LastActivityDate="2018-09-04T03:07:45.377" Title="Calculation for heat transfer rate / cooling capacity for air cooled solid" Tags="|fluid-mechanics|thermodynamics|thermal-conduction|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23519" PostTypeId="2" ParentId="16616" CreationDate="2018-08-30T12:40:28.993" Score="3" Body="&lt;p&gt;No, it's not meaningful to express it as a ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best insulation is deep insulation everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to consider convection losses, and your ventilation strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The attic is usually targetted first, for a combination of several reasons. It's cheap, it's easy, and heat rises, so thoroughly insulating the attic cuts down a lot of conduction losses and convection losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the next-easiest intervention is, will depend on the individual property.&#xA;Upgrading from single-glazed windows to triple is often the next best, because not only does it cut conduction and convection losses, it brings improved security, and acoustic insulation too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get very cheap thermal camera clip-ons for mobile phones now, so you could do your own thermal survey on a particularly cold, windy day, to find out where the cold's getting in (and thus where the heat's getting out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interventions on floors and walls tend to be expensive and disruptive, unless you have masonry cavity walls that can be injected with insulation material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're going to insulate your home properly, then you're going to be doing all surfaces anyway. At this point, you'll want someone to come in, do a site survey, and offer you a full specification that treats the house as a whole system. Look for passivhaus-certified architects.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2018-08-30T12:40:28.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23520" PostTypeId="1" CreationDate="2018-08-30T12:49:10.293" Score="-1" ViewCount="124" Body="&lt;p&gt;Microwave oven generate around 1 KW of power. If this energy is dumped on a electronic device, say wifi router, gsm phone or fm receiver etc, what will happen? Will the devices be damaged? &#xA;By dumping I mean the magnetron at a distance of 1m or less from the electronic device and assume body cage absent.&lt;/p&gt;&#xA;" OwnerUserId="17342" LastEditorUserId="17342" LastEditDate="2018-08-30T13:07:52.267" LastActivityDate="2018-08-30T13:34:02.243" Title="Dumping microwave oven energy on electronic device" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23521" PostTypeId="2" ParentId="23512" CreationDate="2018-08-30T13:22:03.110" Score="-2" Body="&lt;p&gt;you need to supply no head pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is as per your diagram the tank and its drain duct are connected at the open bottom. The pressure is equal on the drain duct to that of the tank and equal to the water it is submerged into at all levels and all locations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given enough time the lake water will replace the tank's water. Any small pressure in the hose will expedite this process. if you want the pressure at the discharge of the hose is 6 meters.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-08-30T13:37:30.100" LastActivityDate="2018-08-30T13:37:30.100" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23522" PostTypeId="2" ParentId="23520" CreationDate="2018-08-30T13:34:02.243" Score="1" Body="&lt;p&gt;Yes, the device will be damaged. See &lt;a href=&quot;https://www.youtube.com/watch?v=vI8JTUql3bQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=vI8JTUql3bQ&lt;/a&gt; for reference... Sure, inside the body cage, the microwaves are bouncing around and may be more concentrated, but even with no cage, the effect is the same - microwaves hitting thin metal (as on a PCB) causes arcing, and consequently damage. You may not get an explosion, but you definitely won't get working devices.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-08-30T13:34:02.243" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23523" PostTypeId="2" ParentId="23514" CreationDate="2018-08-30T13:43:24.183" Score="1" Body="&lt;p&gt;Sam's response is good, but just to confirm your ideas:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;a collection of unit cells arranged in the same orientation is called grain&quot; - Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A collection of grains with similar structure is called phase&quot; - Yes.&lt;/p&gt;&#xA;" OwnerUserId="17286" LastActivityDate="2018-08-30T13:43:24.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23524" PostTypeId="1" CreationDate="2018-08-30T13:52:43.140" Score="2" ViewCount="148" Body="&lt;p&gt;My question relates to finding the vertical component of force due to a rigid wheel hitting a rigid step. I know the details of the step and I know the horizontal velocity of the wheel. I'm using the result to try and determine a damper of some sort to react this vertical force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With reference to the image below, the shallow angle of the step is, a = 9.1° with the sides of the triangle being, x = 62.4 mm, y = 10 mm.&#xA;Horizontal velocity of the wheel is, $V_x = 8,33 \frac{m}{s} [8330 \frac{mm}{s}]$&#xA;The unsprung mass on this wheel is, M = 150 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time taken to cover the distance of the x dimension:&#xA;$t = \frac{62,43mm}{8330\frac{mm}{s}} = 0,0075 s $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Acceleration, $A_x = \frac{V_x}{t} = \frac{8,33\frac{m}{s}}{0,0075s} = 1110,67\frac{m}{s^2}$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I reckon the horizontal force component is, $F_x = MA_x = 166 kN$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so the vertical force component should be, $F_y = F_x\sin(9,1°) = 26,25 kN$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please review my method and determine if this is a plausible method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I've made any errors I would appreciate any advice on how to correct this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QM3uO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QM3uO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17139" LastEditorUserId="2015" LastEditDate="2018-08-30T16:22:31.573" LastActivityDate="2018-08-30T16:59:28.047" Title="Vertical component of force for rigid wheel hitting a step" Tags="|mechanical-engineering|dynamics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23525" PostTypeId="6" CreationDate="2018-08-30T14:35:51.717" Score="0" Body="&lt;h3&gt;My Site History&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I have been a member of this site since it started (and Area 51 before that). Over the past 3.5 years, I have seen both good and bad members come and go. I have seen the site &lt;a href=&quot;https://engineering.meta.stackexchange.com/a/79&quot;&gt;grow greatly&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;About Me&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I am a licensed Professional Engineer (PE) and Structural Engineer (SE) in various states in the US. I have personally used this site to answer my own questions as well as to learn other interesting topics. Without this site I would not know anything about the details of how engineering is similar or different around the world.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Why I Would be a Good Moderator&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;I already check the review queue regularly. I want to welcome and teach more new users about the site. We need to grow and keep users in various fields and levels (curious to professional) while making sure that the site stays on topic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has been long enough since some basic site norms have been discussed. These topics could be discussed again to get more input from current users.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that the best moderators are active in discussing the site and their opinions, but accepting of the consensus that has been created. &lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2018-08-30T14:35:51.717" LastActivityDate="2018-08-30T14:35:51.717" CommentCount="1" CommunityOwnedDate="2018-08-30T14:35:51.717" ContentLicense="CC BY-SA 4.0" />
  <row Id="23526" PostTypeId="2" ParentId="23518" CreationDate="2018-08-30T15:11:43.057" Score="1" Body="&lt;h2&gt;Stagnate Gas&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When the gas is stagnate, view the system as though it is an extended surface (from compressor to outlet). Take the heat transfer inside the tube as having no radial component. Develop the differential equation along the tube $z$ as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d}{dz} k_G A \frac{dT}{dz} - \frac{P}{R'}\left(T - T_\infty\right) = 0 \\&#xA;\frac{1}{R'} = \left(\frac{1}{h} + \frac{w}{k_w}\right)^{-1}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $A$ is the cross-sectional area and $P$ is the perimeter length at a given $z$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allow that gas thermal conductivity $k_G$ is constant to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{d^2T}{dz^2} - m^2\left(T - T_\infty\right) = 0 \\&#xA;m^2 = \frac{P}{k_G R'A}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $P$ is the (outer) perimeter of the tube, $w$ is the tube thickness, $k_w$ is the thermal conductivity of the tube, and $A$ is (inner) circular area of the gas at a position $z$ along the tube. For this geometry&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\frac{P}{A} = \frac{2\pi r_o}{\pi r_i^2} = \frac{2r_o}{r_i^2}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fix the temperature at the compressor. Three solutions are obtained for the temperature along the (center of the) tube $T(z)$ depending on the boundary conditions. One is when the end temperature is fixed, one is when the end is insulated (no heat flow), and the final is when the end allows heat flow. Examples are found &lt;a href=&quot;http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node128.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;at this link to MIT course notes&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system you have is modeled as one where heat flows out of the end. To first order, the heat flow out the end is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{q}_L = -\left.k_G\frac{dT}{dz}\right|_L = \dot{m} \tilde{C}_p\left(T - T_\infty\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This adds complexity to the simple answers and requires analysis at an advanced level. In short though, as you increase $\dot{m}$, the temperature $T_L$ will become closer to $T_o$. The relationship has nothing that lends itself to be &quot;optimized&quot;. In short, you &lt;strong&gt;cannot&lt;/strong&gt; find a relationship to obtain a minimum $T_o - T_L$ based on setting a maximum $\dot{m}$ because $T_L \rightarrow T_o$ as $\dot{m} \rightarrow \infty$.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Gas Flow - Well Mixed Radial&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When the gas flows along $z$ and is well-mixed along $r$, we replace the term $-k_G A dT/dx$ by $\dot{m}\tilde{C}_p T$ to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{m}\tilde{C}_p\frac{dT}{dz} - \frac{A}{R'}\left(T - T_\infty\right) = 0 $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This first order differential equation can be solved for $T(z)$ with the boundary condition $T(0) = T_o$. Here again, the system has no optimization between $\dot{m}$ and $T_o - T_L$. The same limits apply as above.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Gas Flow - Non-Mixed Radial&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;When the gas is not well mixed in the radial direction, the temperature profile inside the tube will not be uniform along the radius. To continue with the model of an extended surface, we will have to include an internal convection coefficient. This gets ugly fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simpler system in this case is to model the tube as a heat exchanger. The first order relationship is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\dot{q} = \dot{m}\tilde{C}_p\left(T_o - T_L\right) = \varepsilon U A \left(T_o - T_\infty\right)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $\varepsilon$ is the efficiency of the exchanger, $U$ is its overall heat transfer coefficient, and $A$ is the (inner/outer) tube area. This is what is called &lt;a href=&quot;https://en.wikipedia.org/wiki/NTU_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;an NTU analysis&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in this case, one does not have a condition to optimize. As $\dot{m} \rightarrow \infty$, you will find that $T_L \rightarrow T_o$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this gives you a useful starting point. In summary, there is nothing to optimize in a balance of $\dot{m}$ and $T_o - T_L$.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-08-30T21:53:16.837" LastActivityDate="2018-08-30T21:53:16.837" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23527" PostTypeId="2" ParentId="23512" CreationDate="2018-08-30T15:49:09.960" Score="1" Body="&lt;p&gt;Taking a look at the tank initially:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nvsEU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nvsEU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have pressure at the top, and with air you've got about the same pressure.  Zooming in on the bottom, the picture is pretty clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hN34G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hN34G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To hold this position, the sum of forces must be equal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ P_{air}A_{tank} = P_{liquid}A_{liquid} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the area the air is pushing on is the same as the area the water is pushing on the air, the pressures are equal.  In this position, the pressure of the air must have the same head as the liquid, or 6 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9BCTF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9BCTF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We look into this second position.  Note this is the most the air can push out the liquid.  If the liquid level was lowered, the air would bubble out the tailpipe.  In this picture, the pressure of the air must be equal to 6 meters plus the height of the tailpipe, XX meters.  In the limit where the tail pipe reaches the bottom, the pressure would be equal to 7 meters.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-08-30T15:49:09.960" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23528" PostTypeId="1" AcceptedAnswerId="23531" CreationDate="2018-08-30T16:39:58.293" Score="0" ViewCount="28" Body="&lt;p&gt;In this solution &lt;a href=&quot;https://www.drdo.gov.in/drdo/English/index.jsp?pg=helicoptor-fire-control-system.jsp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Helicoptor Fire Control System&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CMA is used. For motion analysis we use computer vision and image processing. But what is special about CMA.&lt;/p&gt;&#xA;" OwnerUserId="2679" LastActivityDate="2018-08-30T17:48:46.220" Title="Contact Motion Analysis (CMA)" Tags="|sensors|aerospace-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2018-09-10T20:31:49.510" ContentLicense="CC BY-SA 4.0" />
  <row Id="23529" PostTypeId="2" ParentId="23524" CreationDate="2018-08-30T16:59:28.047" Score="3" Body="&lt;p&gt;Mark has indicated that he's going to write an answer relating to conservation of angular momentum - this is very likely the best way to approach the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to answer very specifically the &quot;Please review my method and determine if this is a plausible method.&quot; part of your question, and leave the final answer to Mark, since I'm in a rush!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of the wheel with mass $150\text{kg}$, and velocity $8.33\text{m/s}$, before it reaches the step is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_0=\frac{1}{2}mv_0^2=\frac{1}{2}*150 *8.33^2=5208.33\text{J}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work done against gravity to lift the wheel up the step of height $10\text{mm}=0.01\text{m}$ is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$E_g=mg\Delta h=150*9.81*0.01=14.72\text{J}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final horizontal velocity of the wheel, after the step, therefore, can be calculated using its new kinetic energy, $E_1=E_0-E_g=5208.33-14.72=5193.61\text{J}$:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$v_1=\sqrt{\frac{2E_1}{m}}=\sqrt{\frac{2*5193.61}{150}}=8.32\text{m/s}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly then, the horizontal speed is not constant, so much of your original working, which relies upon a time calculation for the climb up the step, is not valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth stressing, that the Forces in the $x$ and $y$ directions will not be constant during the 'climbing' phase, with each decaying sinusoidally until there is no force (other than reaction to the mass of the wheel), i.e. the step has been fully climbed.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-08-30T16:59:28.047" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23531" PostTypeId="2" ParentId="23528" CreationDate="2018-08-30T17:48:46.220" Score="1" Body="&lt;p&gt;From a quick skim of a few monographs and book excerpts (which, by the way, are easily found via web search -- something worth doing before posting a wide-open question), it appears that CMA is an active system - sonar or radar, for example, which provides a wealth of information on object range, size, and motion.  Further, any active system can operate regardless of local illumination, as opposed to a vision system which shuts down after sunset (roughly speaking).&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-08-30T17:48:46.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23532" PostTypeId="1" AcceptedAnswerId="23536" CreationDate="2018-08-30T22:42:52.153" Score="0" ViewCount="28" Body="&lt;p&gt;I have a planar object whose center of mass is not at the geometric centroid, and I want to roughly estimate the magnitude of four hypothetical point masses that at the four corners that would result in the center of mass. In summary, I want to backsolve the center of mass equation with reasonable assumptions. I currently have 2 equations for the center of mass in the x and y direction, but I need two more equations to be able to solve for all four. &lt;/p&gt;&#xA;" OwnerUserId="2804" LastActivityDate="2018-08-31T02:09:35.560" Title="Estimating planar mass distribution" Tags="|statics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23533" PostTypeId="2" ParentId="23264" CreationDate="2018-08-30T22:46:09.197" Score="1" Body="&lt;p&gt;There are most definitely entrance losses, and the configuration you've shown would have the greatest losses.  Basically, drawing streamlines that represent how a fluid would flow into the pipe gives you a good idea of how severe the losses would be.  In your example some of the fluid would have to flow down alongside your projecting pipe then do a U-turn to enter the pipe.  The horizontal velocity as it enters the pipe would cause a constriction in the flow, resulting in very high losses.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the flow entering the pipe was from the face of a smooth plate with a large radius leading to a gradual narrowing of the flow field, the streamlines would gradually merge and the losses would be much, much less, on the order of 20x less than the example you have shown.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, if you reverse the direction of the flow in your diagram, the entrance is now at the bottom and there would be a square edged entrance to the pipe flow.  The losses in this case would be intermediate between the two cases above.  &lt;/p&gt;&#xA;" OwnerUserId="9637" LastActivityDate="2018-08-30T22:46:09.197" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23534" PostTypeId="2" ParentId="20924" CreationDate="2018-08-30T23:39:01.160" Score="1" Body="&lt;p&gt;HEC-RAS stands for Hydrologic Engineering Center - River Analysis System.  It's an Army Corps of Engineers program for characterizing flow in river  or other large open channel systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the output for your question is the channel data at one particular cross section in an open channel.  In order from left to right:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q = total channel flow in cubic meters per second&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Min Ch El = elevation of the lowest point of the channel at that point (section)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W.S. Elev. = the elevation in feet of the water surface at that section&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Crit W.S. = this is the elevation of the water surface at critical flow;&#xA;  higher than this and the flow is subcritical, lower than this and the flow is supercritical.  Think of a slow meandering river (subcritical) and a rapid, shooting river (supercritical).  Compare this elevation to the actual WS Elev to determine which type of flow it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.G. Elev = The is the elevation of the Energy Grade Line, which is the sum of the actual water surface elevation and the additional head derived from the flow velocity.  This number represents the elevation if the flow velocity were cleanly directed upwards at this point in the channel.  It's important to determine factors of safety of nearby facilities if the flow should become obstructed at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.G. Slope = The is just the slope of the above Grade Line.  It's related to the slope of the channel bottom and the velocity together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vel Chnl = The average flow velocity for the channel in meters per second.  This is an &lt;strong&gt;average&lt;/strong&gt; just to get a feel for the overall flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow Area = Looking at the cross section of the channel, the area of the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Top Width = the width of the flow section measured at the top free surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Froude Number = This number characterizes the flow as subcritical or supercritical.  See above for the meanings.  If this number is greater than 1, the flow is supercritical, and less than 1 means subcritical.  If it's very near or equal to one, the flow can be unstable, varying (surging) back and forth between sub and super - critical.  In this case it's well below 1 so the flow will be well behaved, subcritical, meaning essentially gentle flowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given channel, there will be several of these cross sections with outputs for each section.  Put together, they form a picture of the overall flow characteristics for the entire length of the channel being evaluated.  Good results come from good inputs, and it takes some experience to know where to take your sections for these programs to yield good results.  &lt;/p&gt;&#xA;" OwnerUserId="9637" LastActivityDate="2018-08-30T23:39:01.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23536" PostTypeId="2" ParentId="23532" CreationDate="2018-08-31T02:09:35.560" Score="1" Body="&lt;p&gt;let's say the off-center of center of mass from geometrical centroid is dx and dy, and total mass is M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assume the masses from top left counting clockwise m1,m2,m3,m4,located with their CGs equally located on four corner of a vertical rectangular. First let's assume the object is symmetrical, with b as the base of the rectangle and a as its vertical long side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so we have $$\bar{x} =\frac {m1\times a/2 + m2\times a/2 +m3\times a/2 +m4\times a/2}{M} $$ and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\bar {y}= \frac {m1\times b/2 + m2\times b/2 +m3\times b/2 +m4\times b/2}{M}   $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now we back calculate dx and dy into the above EQs to figure the difference in mass of individual parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We observe that if we add  a very small differential amount dm1 to m1 the center of gravity will move up and left away from center of geometry by a factor of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ dx=\space  \thicksim \frac{dm1\times a/2} {M+dm1}\space and\space \space dy =\space\thicksim \frac{ dm1\times b/2}{M+dm1}  $$&#xA;This rough approximation should give us intuition to figure out the discrepancy in mass or if we have an indication about the geometry we can set up the differential discrepancy as Dx for that.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-08-31T02:09:35.560" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23538" PostTypeId="1" CreationDate="2018-08-31T07:25:00.100" Score="0" ViewCount="99" Body="&lt;p&gt;While architectural features are oftentimes more aesthetic than they are technically complex, this still seemed like a suitable question for here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on a project for an industrial type facility that will include one or more large exhaust air stacks. These stacks are part of an odor control system, so while the air will be clean, the final step is to get the air high over the area for the purposes of dispersion. &#xA;The facility will be in a relatively nice part of town so the residents don't really want to see the stack. Are there proven ways to hide such a structure or to make it more appealing similar to what is done with cell towers disguised to look like trees?&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2018-08-31T11:44:04.273" Title="Disguising tall exhaust stacks" Tags="|structures|building-design|architecture|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23539" PostTypeId="2" ParentId="8279" CreationDate="2018-08-31T08:49:55.377" Score="2" Body="&lt;p&gt;The only or vastly major source of &lt;a href=&quot;https://en.wikipedia.org/wiki/Chemical_oxygen_demand&quot; rel=&quot;nofollow noreferrer&quot;&gt;chemical oxygen demand (COD)&lt;/a&gt; should be your dissolved or suspended hydrocarbons, so a COD test could be the way to go. . I don't know hw confident you are in your analytical chemical and labwork skills. COD determintation requires handling of strong acids and oxidizers, would avoid without proper training. However, there exist micro vials for safe, quick test for people with little training. Search for micro vial COD quick test. These are marketed to WWTP operators without proper labs or labwork training.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need check the following: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chloride content of sample, Chloride can distort the result and would need to be masked.&lt;/li&gt;&#xA;&lt;li&gt;COD equivalent to your 20mg/l THC - either derived from stochiometry of your hydrocarbons (that you likely don't know exactly), or maybe you find a code you can reference. Or you measure the COD of samples with known THC and map these. &lt;/li&gt;&#xA;&lt;li&gt;Not that engine oils are often modified (-H substituted) heavily and you may underestimate the THC/COD ration if you take the stochiometry of straight petrol or Diesel as a baseline&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Either way, I suggest you define a COD equivalent to THC value way less than 20mg/l and then laboratory test only sample above this treshold to be sure.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-08-31T08:49:55.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23540" PostTypeId="2" ParentId="23538" CreationDate="2018-08-31T11:16:42.013" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Disguising tall exhaust stacks - The facility will be in a relatively nice part of town so the residents don't really want to see the stack. Are there proven ways to hide such a structure or to make it more appealing similar to what is done with cell towers disguised to look like trees?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is possible to make them look like trees but without branches and foliage they'll be a couple of sticks. &lt;a href=&quot;https://www.dictionary.com/browse/can-t-make-a-silk-purse-out-of-a-sow-s-ear&quot; rel=&quot;nofollow noreferrer&quot;&gt;You can't make a silk purse out of a sow's ear&lt;/a&gt;, but I have a workaround.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is brick, or build an artistic structure around them to make them an attraction instead of an eyesore. It depends on the height and the budget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GrL1W.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GrL1W.jpg&quot; alt=&quot;Cover up the stacks&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Companies like &lt;a href=&quot;https://trashpresso.com/trash-tech/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Trashpresso&lt;/a&gt; make tiles from waste plastic, part of the beautification of tower can be accomplished by using recycled materials; good for the environment and the view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s2Wxf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s2Wxf.jpg&quot; alt=&quot;TrashPresso Tile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2018-08-31T11:16:42.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23541" PostTypeId="2" ParentId="23538" CreationDate="2018-08-31T11:44:04.273" Score="0" Body="&lt;p&gt;You will have buildings wih ventilation, with exhaust air vents on the roof (usually). Duct your air from the odour control system there. Worst case you will need a slightly bigger blower.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-08-31T11:44:04.273" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23543" PostTypeId="1" CreationDate="2018-08-31T15:56:16.150" Score="-1" ViewCount="69" Body="&lt;p&gt;Intensity of RF wave at a distance from source can be calculated by using inverse square law. But imagine an RF energy travels a distance $d1$ before being perfectly reflected from a metal surface. How much is the energy intensity at a point B, which is at a distance $d2$ from the point of reflection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) B lies in the line of reflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) B is not located at the line of reflection. It forms an angle $\theta$ with the line of reflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My though process is following. Intensity must be much more along line of reflection. Probably , it should follow inverse square law along line of reflection, but i am not sure. &lt;/p&gt;&#xA;" OwnerUserId="17342" LastEditorDisplayName="user14407" LastEditDate="2018-08-31T17:24:58.030" LastActivityDate="2022-12-09T12:02:55.373" Title="RF energy intensity after reflection" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23544" PostTypeId="2" ParentId="23518" CreationDate="2018-08-31T16:28:21.193" Score="0" Body="&lt;h2&gt;Temperature of Disk&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Consider a flat, circular disk that is being cooled by convection only (ignore radiation). The configuration is illustrated below with symbols for geometry ($w$ and $R$), temperatures $T_j$, thermal conductivity $k$, density $\rho$, specific heat capacity $\tilde{C}_p$, and convection coefficient $h_a$. The problem is to find the temperature in the disk as a function of time $T(t)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4LOVi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4LOVi.png&quot; alt=&quot;image of heat flows for disk problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important number to determine first is &lt;a href=&quot;https://en.wikipedia.org/wiki/Biot_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Biot number&lt;/a&gt; $Bi = h_a (V/A) / k$. For the disk, $V/A = w\pi R^2 / (2\pi R^2 + 2\pi R w)$. For a thin disk $R &amp;gt;&amp;gt; w$, this approaches $w/2$. When $Bi &amp;lt; 0.1$, the disk can be handled as a lumped system. Basically, heat is conducted through the disk faster than heat is pulled out of the disk. The temperature profile inside the disk can be neglected. The resultant analysis for $T(t)$ is given by example &lt;a href=&quot;https://www.sfu.ca/~mbahrami/ENSC%20388/Notes/Transient%20Heat%20Conduction.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;at this link&lt;/a&gt;. When $Bi &amp;gt; 1$, the analysis must include the temperature profile inside the disk. The analysis leads to $T(t,z)$ solved by what are called &lt;a href=&quot;https://en.wikipedia.org/wiki/Heisler_chart&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heisler charts&lt;/a&gt;. A quick and dirty approach in this case is to pick the temperature either at the center point of the object or on its surface to follow as a function of time. With this approach, the $T(t,z=0)$ or $T(t,z=w)$ results are about the same as found by the lumped analysis. The temperature changes exponentially with time in either case.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Convection Coefficient Correlations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Convection coefficients are correlated to gas flow rate. The starting point is to appreciate that correlations are used to find the &lt;a href=&quot;https://en.wikipedia.org/wiki/Nusselt_number&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Nusselt number&lt;/a&gt; $Nu = h_a (V/A)/k$. The convection coefficient is calculated from $Nu$. Correlations in forced convection depend on the Reynolds number and the Prandtl number. See the same link as given for $Nu$ to see examples in various geometries. The flow rate of a gas over the disk appears as velocity in $Re = \rho_g v_g L_c/\mu_g$. For the system at hand with mass flow, the gas velocity can calculate at the end of the tube as $v_g = \dot{m} /(\rho_g \pi R_T^2)$. All else being equal, velocity is therefore proportional to mass flow rate. As a side note, $L_c$ is a characteristic length of flow over the disk. For flow over the disk laterally, $L_c \approx 2R$ at the opposite end(s) of the disk. For flow perpendicularly from the center line, $L_c = R$ at the edges.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Cooling Rate&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;At any instance in time, the cooling rate of the plate is $\dot{q} \approx 2 h_a A (T - T_a)$ (ignore loss from the edge of the disk). As $T_a$ decreases, the cooling rate increases linearly. To first order in laminar flow, $Nu \propto \sqrt{Re}$. Therefore, to first order in laminar flow over the disk, cooling rate will depend on the square root of mass flow. The correlations are complex for turbulent flow. The convection coefficient still increases with mass flow rate, however the correlation is never directly linear.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Summary&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Decreases in temperature $T_a$ will increase cooling rate linearly. Increasing mass flow $\dot{m}$ will increase cooling rate but less effectively (less than a linear increase).&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-09-04T03:07:45.377" LastActivityDate="2018-09-04T03:07:45.377" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23546" PostTypeId="2" ParentId="23543" CreationDate="2018-08-31T17:08:44.197" Score="0" Body="&lt;p&gt;Take a look at the image below: The Yellow side shows a point source, bouncing off a mirror. You can clearly see that the radiation intensity starts bright at the source, and then falls off according to the inverse square law. It's also visibly brighter in the area where the reflected radiation overlaps the incident radiation. At that point, there's more overall light, because it's coming from two directions. Of course, putting a detector in there can create shadows that messes with this - I'm assuming no shadows for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orange half of the image shows what's know as 'extended rays', or, the point from which the reflected light rays &lt;em&gt;appear&lt;/em&gt; to originate from, to an observer on the yellow side of the mirror. You can see here how the rays again start bright, and then fall off according to inverse square. To calculate the intensity at any point, then, you simply calculate the distance to the detector from the yellow and orange sources, work out intensity using inverse square, and subtract any shadows as necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GP6ND.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GP6ND.png&quot; alt=&quot;Point Source at Mirror&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-08-31T17:08:44.197" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23547" PostTypeId="1" AcceptedAnswerId="23548" CreationDate="2018-08-31T19:38:05.407" Score="1" ViewCount="5762" Body="&lt;p&gt;I need to save some engineering files that need the thicknesses in the filename. Windows hates &quot;/&quot; or &quot;\&quot;, so I can't use a simple fraction. Symbols like this...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;⅓, ⅔, ⅕, ⅖, ⅗, ⅘, ⅙, ⅚, ⅛, ⅜, ⅝, ⅞&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...are nice, but don't cover the full gamitt of dimensions (ie. 1/16&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, Windows hates &lt;strong&gt;&quot;&lt;/strong&gt; in the filename, so I assume &lt;strong&gt;in&lt;/strong&gt; is used instead?&lt;/p&gt;&#xA;" OwnerUserId="15383" LastEditorUserId="2559" LastEditDate="2018-08-31T20:49:24.003" LastActivityDate="2020-08-17T04:41:48.743" Title="What is the engineering standard for filenames with fractions and inches?" Tags="|measurements|standards|nomenclature|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23548" PostTypeId="2" ParentId="23547" CreationDate="2018-08-31T20:38:20.780" Score="5" Body="&lt;p&gt;I've seen numerous formats in my experience. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Per some of the comments, sometimes you'll see a fraction represented with dashes, underscores, or as a decimal. I find that most often, decimals are used.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1/2 = 1_2 ,1-2, 500&lt;/li&gt;&#xA;&lt;li&gt;5/16 = 5_16 , 5-16, 3125&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you have 1 1/2&quot; then I see 1_500 where the underscore would represent the decimal point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think there is a standard per se, but you should &lt;strong&gt;be consistent&lt;/strong&gt; with your naming. Also, make an info-graphic that describes your filename nomenclature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes for standard sizes or increments or catalog items, things are represented just by a simple number. E.g.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1/16&quot; = 01&lt;/li&gt;&#xA;&lt;li&gt;1/8&quot; = 02&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This graphic below is an excellent example(aside from the slash usage for a computer file name)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/RTHcP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTHcP.jpg&quot; alt=&quot;Nomenclature Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2018-08-31T20:45:28.523" LastActivityDate="2018-08-31T20:45:28.523" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23549" PostTypeId="1" CreationDate="2018-08-31T21:59:32.160" Score="1" ViewCount="33" Body="&lt;p&gt;I am using standard test blocks to find the uncertainty of a supermicrometer, which is capable of measuring down to 1/10 of a micron. The issue is that it measures the EXACT same length within a run of even 100 measurements. Understandably, it is suspicious to calculate uncertainty based on data with zero variation; however, the value that it consistently measures is different every time I zero the supermicrometer at a different position. For example, I may get a value of 5.0000 mm repeatedly during one run, zero the supermicrometer, and then get a value of 5.0003 mm repeatedly during the next run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is, how can I take into consideration variation between runs? Traditionally, one would use a single run to find uncertainty, but I don't think I can do that. &lt;/p&gt;&#xA;" OwnerUserId="17358" LastActivityDate="2018-08-31T21:59:32.160" Title="How can I calculate uncertainty based on measurements from multiple separate runs?" Tags="|statistics|quality-engineering|calibration|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23550" PostTypeId="6" CreationDate="2018-09-01T02:35:44.290" Score="0" Body="&lt;p&gt;On Stack Exchange, we believe the core moderators should come from the community, and be elected by the community itself through popular vote. We hold regular elections to determine who these community moderators will be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Community moderators are accorded the highest level of privilege on our community, and should themselves be exemplars of positive behavior and leaders within the community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our general criteria for moderators is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;patient and fair&#xA;leads by example&#xA;shows respect for their fellow community members in their actions and words&#xA;open to some light but firm moderation to keep the community on track and resolve (hopefully) uncommon disputes and exceptions&#xA;Every election has three phases:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nomination&#xA;Primary&#xA;Election&#xA;Please participate in the moderator elections by voting, and perhaps even by nominating yourself to be a community moderator!&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-09-01T02:35:44.290" LastActivityDate="2018-09-01T02:35:44.290" CommentCount="0" CommunityOwnedDate="2018-09-01T02:35:44.290" ContentLicense="CC BY-SA 4.0" />
  <row Id="23551" PostTypeId="2" ParentId="23264" CreationDate="2018-09-01T03:59:19.733" Score="1" Body="&lt;p&gt;&lt;strong&gt;Yes, there are entrance losses.&lt;/strong&gt;&#xA;Pretty significant entrance losses compared to some other entrance designs, in fact. @PJNoess gives some excellent information on why you should expect entrance losses in the situation you drew. To calculate the losses one method is to use the K method, described &lt;a href=&quot;https://neutrium.net/fluid_flow/pressure-loss-from-fittings-excess-head-k-method/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and summarized below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Headloss in the pipe is calculated as: $$h_p = f\frac{L}{D}\frac{V^2}{2g}$$&#xA;where $f$ is the friction factor found on a Moody diagram, $L$ is the length, $D$ is the diameter, $V$ is the velocity, and $g$ is the gravitational constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minor losses are from entrances, exits, valves, bends, nozzles, etc. They can be calculated as $$h_m = K\frac{V^2}{2g}$$ where $K$ is the sum of the coefficients for the minor loss. Some $K$ values for entrances can be looked up &lt;a href=&quot;https://neutrium.net/fluid_flow/pressure-loss-from-pipe-entrances-and-exits/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You drew an &lt;strong&gt;inward projecting entrance&lt;/strong&gt;, which has: $K=0.78$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;flush entrance&lt;/strong&gt; like the exit on your drawing would have $K=0.5$ if used as an entrance ($K=1$ for all exits).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;strong&gt;rounded entrance&lt;/strong&gt; with a large radius can have $K$ as low as $0.04$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lower $K$ values mean lower losses. An inward projecting entrance has high losses compared to many other designs.&lt;/p&gt;&#xA;" OwnerUserId="17151" LastActivityDate="2018-09-01T03:59:19.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23552" PostTypeId="1" CreationDate="2018-09-01T04:43:48.867" Score="0" ViewCount="413" Body="&lt;p&gt;I was reading an article on the impact of jet in vanes. In this article I found that the net torque on the unsymmetrical or radially curved vanes is non zero. &#xA;But, my question is that, why the torque on the symmetrical curved vane, when jet impinges tangentially at one end, is zero.&#xA;My thoughts: As the force on the symmetrical vane, for the case described above, is in the horizontal direction only and the net force in vertical direction will be zero, if no losses are considered. Hence, the line of action of horizontal force is finite distance apart from the axis of the vane. Therefore, net troque must be finite.&lt;img src=&quot;https://i.stack.imgur.com/IGXK4.png&quot; alt=&quot;symmetrical curved vane jet striking at one end&quot;&gt;&#xA;Note: this picture is only meant to define the sense of horizontal and vertical directions only and the axis of the vane and not the other terms which can vary according to the user.&lt;/p&gt;&#xA;" OwnerUserId="17361" LastActivityDate="2024-04-02T18:05:55.700" Title="Torque on the curved moving vane" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23553" PostTypeId="2" ParentId="23552" CreationDate="2018-09-01T05:39:23.420" Score="0" Body="&lt;p&gt;A symetrical blade imparts the torque to the shaft with no turning force beween its ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a non-symetrical blade will have an imbalance beween the torque one half compared to the other - which can cause one bearing to fail earlier for example or slight bending of the shaft causing clearance issues...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-09-01T05:39:23.420" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23555" PostTypeId="1" CreationDate="2018-09-01T14:59:22.977" Score="0" ViewCount="953" Body="&lt;p&gt;Electromagnetic pulse is said to detsroy electronics. Lightning is a type of EMP. As per my understanding lightning is basically a dielectric breakdown, it can be considered as DC current. Can it or any dielectric breakdown damage electronics from a distance, say 1m, 10 m , 100m or 1 km? If yes, how is that possible, because it cannot induce voltage in any wire as it is DC? Even if it is AC of few Hz, the length of any household wire will be very small as compared to the wavelength, then also there will be huge resistance to any volatge induced.&lt;/p&gt;&#xA;" OwnerUserId="17342" LastEditorUserId="17342" LastEditDate="2018-09-01T15:07:58.503" LastActivityDate="2019-04-12T20:01:15.210" Title="Damage to electronics from lightning strike at a distance" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="23556" PostTypeId="1" CreationDate="2018-09-01T15:40:44.393" Score="1" ViewCount="205" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uZGyY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uZGyY.jpg&quot; alt=&quot;representative image showing a drone from side view where the two parellel and horizontal lines represent thrust-producing propellers and at the left end is the torque-reaction propeller we&amp;#39;re talking about&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a unique idea of mine. In the above image, the torque reaction propeller is rotating clockwise producing a side-thrust of 40N. With its reaction torque alone, it has to balance the moment generated by the payload on the other side. So...&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vOgfF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vOgfF.jpg&quot; alt=&quot;the equation (eqn.1) applies equilibrium principle to the moments on both sides of the &amp;#39;pivot&amp;#39; point. The point of meeting of the two horizontal arms, where the vertical thrust acts upon in the frame, is taken to be the pivot point&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding this, I put my arguments as follows. &lt;strong&gt;Please choose one of these as correct one or create your own concept to solve the issue I'm having.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Argument 1&lt;/strong&gt;&#xA;I've also read somewhere that the yawing torque of a motor is the torque of the motor itself. If this is the case, the torque of the propeller will be:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WwJza.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WwJza.jpg&quot; alt=&quot;This equation (eqn.2) is what I&amp;#39;ve derived from the momentum theory. Here *tau* is the torque I&amp;#39;ve calculated.&quot;&gt;&lt;/a&gt;&#xA;This result is based on the assumption that the relation&lt;br&gt;&#xA;Rotational power=torque×angular velocity holds true in this case where a propeller involves.&#xA;The torque so obtained will not be enough to balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Argument 2&lt;/strong&gt;&#xA;If this torque is considered to be leveraged by the distance from the pivot upon which it acts, units on both sides of the equation doesn't balance. &#xA;&lt;em&gt;Refer eqn.1 in image 2 where the LHS would be in N.m and the RHS would be in N.m-squared if considered to be leveraged&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Argument 3&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vDIHQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vDIHQ.jpg&quot; alt=&quot;This, eqn.3, is assuming that the thrust is directly a force acting perpendicular to the axis, on the frame&quot;&gt;&lt;/a&gt;&#xA;In this case, at the mentioned distance of 8.75cm, the frame will be level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Argument 4&lt;/strong&gt;&#xA;Here, the torque we obtained from eqn.2 in image 3 is considered as a rotational force acting at a distance, which is the propeller's mean radius, from the centre of the hub. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/Ou9cB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ou9cB.jpg&quot; alt=&quot;Here, in eqn.4, the mean radius is approximated to half of the outer radius of the propeller&quot;&gt;&lt;/a&gt;&#xA;In this case too, like in the previous, the propeller torque reaction will be able to hold the frame level albeit at a greater distance from the pivot.&lt;/p&gt;&#xA;" OwnerUserId="17365" LastActivityDate="2018-09-01T15:40:44.393" Title="Can a propeller be used as a balancer by spinning it perpendicular to ground?" Tags="|mechanical-engineering|statics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23557" PostTypeId="1" CreationDate="2018-09-01T16:09:48.403" Score="2" ViewCount="1369" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nhDYj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nhDYj.png&quot; alt=&quot;Table Top&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm hoping someone can help me with some questions to know whether or not the table can be supported by this design and how to calculate the length of the support runners and the width necessary to prevent tipping/rocking along the width as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Table length and width ~90&quot; x 36&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angle of the cantilever arm (terminology?) either 130 degrees or 50 degree&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Which of the two designs from the elevation view would provide better support (could I use either as long as runners are long enough?) Which would require shorter runners [longer overhang]?&lt;/li&gt;&#xA;&lt;li&gt;What would the minimum width between the two support arms would I need to prevent horizontal tipping?&lt;/li&gt;&#xA;&lt;li&gt;How do I calculate tipping weight on the end of the table (opposite of the support)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Material is wood. I found some figures for &quot;E = modulus of elasticity&quot; and for the wood I'm using it'd be between 1.6-1.7 and the user posted this note:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;units of measure are psi, values must be multiplied by $10^6$.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is all beyond my math skills, but am willing to try and learn!&lt;/p&gt;&#xA;" OwnerUserId="17366" LastEditorUserId="1832" LastEditDate="2018-09-01T20:42:09.913" LastActivityDate="2018-12-31T09:02:25.487" Title="Help with cantilever table top questions" Tags="|structural-engineering|design|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23558" PostTypeId="1" CreationDate="2018-09-01T17:15:36.190" Score="1" ViewCount="102" Body="&lt;p&gt;I'm trying to describe a part that I'm designing that has a disc with incremental teeth, like what is pictured here; which mates with a similar part. What is this type mechanism called? And what are the &quot;teeth&quot; called (for some reason I thought they were called detents, but found that not to be the case)?  Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bcnzy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bcnzy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17367" LastActivityDate="2018-09-01T17:30:43.470" Title="What is a mechanism with teeth like this called?" Tags="|mechanical-engineering|cad|" AnswerCount="1" CommentCount="0" ClosedDate="2018-09-01T22:22:20.427" ContentLicense="CC BY-SA 4.0" />
  <row Id="23559" PostTypeId="2" ParentId="23558" CreationDate="2018-09-01T17:24:31.850" Score="0" Body="&lt;p&gt;I would call that a swivel mount with detents. if the teeth are sharp they are sometimes called ratchet detents. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-01T17:24:31.850" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23561" PostTypeId="2" ParentId="23555" CreationDate="2018-09-01T20:03:20.867" Score="0" Body="&lt;p&gt;The lightening can damage the facilities in two ways, by direct strikes or by electromagnetic waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the electromagnetic wave cause any damage ? yes, it's possible. The wave induces extra voltage across electronic components, if the components receive the right amount of voltage then you know what would happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your second question, the length of household wires are a way too small that's true, but the length of electrical networks across a continent can be big enough, for instance the electrical network between France, Belgium, Netherland, Germany, Austria and Czech, and they are all connected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another example is the phenomenon called 'coronal mass ejection', it's combination of materials and electromagnetic radiation, that caused some disturbances on the earth several years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further i refer you to this paper: &lt;a href=&quot;https://digitalcommons.trinity.edu/cgi/viewcontent.cgi?article=1027&amp;amp;context=physics_faculty&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://digitalcommons.trinity.edu/cgi/viewcontent.cgi?article=1027&amp;amp;context=physics_faculty&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-09-01T20:03:20.867" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23562" PostTypeId="1" CreationDate="2018-09-01T22:37:58.177" Score="1" ViewCount="94" Body="&lt;p&gt;I am building a fatigue testing machine that includes the ability to induce stress from torsion, however I am stuck at how to convert the linear motion from the linear actuators we have chosen to torque efficiently. Because the cycles of the machine will be in the millions and involve high accelerations, a rack and pinion mechanism is not an option, so we have looked at a four bar linkage slider crank mechanism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the slider crank is that we need to transmit the maximum torque at the extremes of the linear actuator position i.e. when the actuator is fully extended and fully retracted. The slider cranks transmit the most when the transmission angle is 90 degrees, so we have to try and create the motion in this &lt;a href=&quot;https://www.youtube.com/watch?v=PMejdaku4Do&amp;amp;feature=youtu.be&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand most slider cranks that have the slider as the input and rely on inertia to keep going in a certain direction. Our setup needs to be able to go very slow and can't rely on inertia to keep the crank spinning. Does anyone know how we can achieve this type of motion?&lt;/p&gt;&#xA;" OwnerUserId="13810" LastActivityDate="2018-09-02T05:17:29.747" Title="How to make slider driven crank go in a certain direction when transmission angle is zero?" Tags="|mechanisms|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23564" PostTypeId="2" ParentId="23562" CreationDate="2018-09-02T05:17:29.747" Score="1" Body="&lt;p&gt;a four-bar linkage is going to have a dead spot in its torque response where it will stall and stick at low speeds. My solution to this problem when I first encountered it  years ago was to add a toothed belt drive that would pull the mechanism out of the stall point. In this case, the belt drive was exercising the 4-bar mechanism which idled along while the belt did the work. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-02T05:17:29.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23565" PostTypeId="1" CreationDate="2018-09-02T06:37:27.117" Score="1" ViewCount="11775" Body="&lt;p&gt;I am working on a project for an electronics class, but need some help with one of the mechanical details.&#xA;My job is to program a stepper motor to move a conveyor belt back and forth as specified. The &quot;belt&quot; will have a chain and the shaft of the motor will have the sprocket. But before I choose my stepper, I need to know what the required torque for my stepper motor is. Unfortunately my physics knowledge isn't great, so I googled to find the equation I should use. This is what I came up with.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/04yJO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/04yJO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found this equation which comes out to be similar to the above equation except this one doesn't have the efficiency and gear ratio involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6TzPD.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6TzPD.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have 3 questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;what kind of force would I calculate for F? If the object is laying flat horizontally, what kind of force would be pushing against it east to west?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the two equations give me very different results due to the efficiency and gear ratio. when would I use one equation over the other?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;where would the required speed come into the picture? if I want to move my object quickly vs slowly, would this affect the torque needed?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you are still reading this, I thank you for your time, and hope you could help me with any or all of my questions...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="17371" LastActivityDate="2021-07-13T07:51:22.250" Title="Determining required torque for a chain and sprocket conveyor belt" Tags="|mechanical-engineering|electrical-engineering|torque|stepper-motor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23566" PostTypeId="2" ParentId="23557" CreationDate="2018-09-02T07:38:56.127" Score="1" Body="&lt;p&gt;let's say your table and the support weighs 100 lbs. and the support at 90-degree angle projects a rectangle that is 1 foot shorter on each side from the edge of the table, meaning it is a rectangle of the size 66&quot; by 12&quot; and a height of 21&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This support when tilted is $$ 21\times sin(50) = 16&quot;$$ offset horizontally, standing at the height of $$21\times cos(50) = 13.5&quot;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now immediately we observe that as far as stability both 130 and 50 degrees are a mirror image of each other and provide one favorite side, one unfavorite side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that CG of the table is at its center of geometry, which is a plausible assumption considering the weight of the two legs touching the floor is much smaller than the top part it is an acceptable assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we want to figure how much weight added right on the edge of the table will topple it. At the unfavorable tilt of the support considering the legs or angled out dislocating the CG by 16&quot;out, our restoring moment is the 100 lbs multiplied by 90/2- (16+ 12) = 1700 lbs.inch. now we divide this by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1700/(16+12 =28)= 60 lbs$...-which is so low that we do not further check for the strength of the legs. of course, this load can be increased as it moves toward the center of the table limited only by the material strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now checking for lateral stability, we have a restoring moment of 100 lbs multiplied by 6&quot;= 600 lbs.inch. we dived this by 12&quot; we get 50lbs. Anything larger than 50lbs will laterally topple the table.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-09-02T07:38:56.127" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23567" PostTypeId="1" CreationDate="2018-09-02T12:53:05.707" Score="0" ViewCount="807" Body="&lt;p&gt;ESD from human affects electronics. Why it does not affect the pacemaker? Is it beacuse it is much below the surface?&lt;/p&gt;&#xA;" OwnerUserId="17342" LastActivityDate="2021-07-20T09:06:25.490" Title="Why ESD does not affect pacemaker?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23569" PostTypeId="1" CreationDate="2018-09-02T13:36:13.193" Score="0" ViewCount="2706" Body="&lt;p&gt;Consider an electronic device, say wifi dongle. it is inside a wooden box. It may be a) powered up or b) without any power supply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to damage it to render it dysfunctional without damaging the box? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this can be done using 1)Electromagnetic waves, 2) Static electric field and 3) Oscillating electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider these options one by one. 1) Powerful em waves will induce large electric fields in the antenna and conductors. This may damage the subsequent electronics. What power and frequency is enough for the purpose? 2) Static electric field should induce voltage ascross conductors, resulting in damage. Is this correct? What field stregth is enough? 4) Oscillating electric field should be more effective than static one. What field strength and frequency are enough?&lt;/p&gt;&#xA;" OwnerUserId="17373" LastActivityDate="2024-03-17T00:05:06.003" Title="Damaging electronics from a distance" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23570" PostTypeId="2" ParentId="14165" CreationDate="2018-09-02T13:52:07.107" Score="0" Body="&lt;p&gt;Think of an egg. It is able to withstand substantial force die to its shape longitudinally yet laterally it is weak. The same idea applies to corrugation. The perpendicular forces exerted on a corrugated pipe are transformed into lateral forces due to the shape of the configuration of the corrugation. The different depth at which the compressive forces act on the pipe create tensile and compressive forces in the pipe material which means that the pipe wall can be thinner and therefore lighter for transportation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each peak resists compressive forces whilst each valley resists with tensile strength radially. This is why damage to the valley section in a corrugated pipe is more likely than at a peak and radial damage is far more likely than lateral damage. Once the pipe is buried, the tensile forces also balance out to become compressive forces due to the covering material being the opposing yet equal force acting on the structure. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-09-02T13:52:07.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23571" PostTypeId="2" ParentId="17457" CreationDate="2018-09-02T14:03:15.410" Score="0" Body="&lt;p&gt;Since the combined force (F) is 100 and 25% is in member OA, the balance must therefore be 75% in member OB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both member you need to convert the diagonal forces into vertical and horizontal components that provide the resultant sigma of forces. Since you have at last one angle and the final force (f), it is therefore possible to find all the other angles and forces, too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since alpha is 20°, the internal angle at A should be 90°-20° since we created a right angle triangle by separating the members into 2 force diagrams&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Angle A=70°&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need to find the distances for right angled triangle AO $\alpha$&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2018-09-02T14:11:46.873" LastActivityDate="2018-09-02T14:11:46.873" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23573" PostTypeId="1" CreationDate="2018-09-02T14:40:42.023" Score="0" ViewCount="1175" Body="&lt;p&gt;Microwave oven uses magnetron to generate hugh power em waves. How the power output of a microwave can be increased? It uses a 1-2 KVAC capacitor. Will doubling the voltage of capacitor double the power output? &lt;/p&gt;&#xA;" OwnerUserId="17373" LastActivityDate="2021-03-26T10:36:04.747" Title="Increasing power output of microwave oven" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23575" PostTypeId="2" ParentId="11554" CreationDate="2018-09-02T16:21:42.167" Score="-1" Body="&lt;p&gt;I guess if the design is like a wind turbine, like a propeller sat in the flow with no ducting then the betz limit would apply. See: &lt;a href=&quot;https://energyeducation.ca/encyclopedia/Betz_limit&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://energyeducation.ca/encyclopedia/Betz_limit&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17375" LastActivityDate="2018-09-02T16:21:42.167" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23576" PostTypeId="2" ParentId="23567" CreationDate="2018-09-02T19:23:39.597" Score="1" Body="&lt;p&gt;ESD problems are caused by buildup of static electricity on the surface of a human or other device. The damage is mainly caused by the sudden equalisation of potential. Like a very small lightning strike. Another is noise from other devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have experienced electric shock when touching a door handle or a friend. This is on the higher end of the for very small smd components can take. They may be damaged even by shocks humans can not feel. However its not like they will always get destroyed by such shocks, its more of a statistical thing for manufacturing while the component is still separated from the board.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the important bit is that this causes damage because it is localized, the shock traveled directly to the component before it starts to spread around. The pacemaker is inside a human so it has a lot of insulating and dissipating medium around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the pacemaker is also shielded by its casing. It is after all a device that sends shocks. Your remote control or computer is not very likely to break from a shock as the case shields the components from being in the direct line for the shock to ground. Similarily the case may protect from other esd sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But yes the pacemaker can fry if you have a big enough shock. But then the human would be dead regardless.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-09-02T19:46:11.270" LastActivityDate="2018-09-02T19:46:11.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23577" PostTypeId="1" CreationDate="2018-09-02T20:25:16.833" Score="0" ViewCount="76" Body="&lt;p&gt;I am simulating a furnace as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JoNtk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JoNtk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to capture temperature distribution in this furnace without designing intake fan? Which kind of boundary conditions can replace with intake fan?&#xA;In this model, there is a electrical element for generating heat (in fact, transferring heat to air flow). How can I set this element as a boundary condition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, There are four cylinder in this model and they rotate. I want to obtain temperature distribution around these cylinders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is unsteady an 3D. Firstly (t=0), temperature in furnace is 25 but when air flow cross air duct and electrical element, temperature in exit of duct is equal to 200 C. All of furnace walls are adiabatic. Therefore, We have two kind of work(electrical work and mechanical work) and Q=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What information (pressure, temperature, heat flux, power and ...) should I have for boundary conditions in parts of Fan and element?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Major goal in this simulation is to find that temperature distribution (inside the furnace) is uniform or not (i.e 200 C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am grateful that guide me and give me useful links and documents in this case. &lt;/p&gt;&#xA;" OwnerUserId="2985" LastActivityDate="2018-09-02T20:25:16.833" Title="Tempreture distribution in a furnace" Tags="|fluid-mechanics|heat-transfer|cfd|temperature|fluent|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23579" PostTypeId="2" ParentId="23573" CreationDate="2018-09-03T06:07:54.653" Score="2" Body="&lt;p&gt;No, it will not. the power output of a microwave oven is set by the power rating of the magnetron tube that generates the microwaves. If you want more microwave power, you need a more powerful magnetron tube, and a larger power supply to drive it as well. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-03T06:07:54.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23580" PostTypeId="2" ParentId="23569" CreationDate="2018-09-03T06:12:53.733" Score="0" Body="&lt;p&gt;the classic method to destroy a piece of solid-state electronics is with an electromagnetic pulse (EMP), as is generated by the explosion of an atomic or nuclear bomb, or a device called a disruptor- which is used by police to disable an explosive device with an electronic timer in it. It generates an EMP which is strong enough to burn out the electronics inside a bomb package without blowing up the city it is in. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-03T06:12:53.733" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23581" PostTypeId="2" ParentId="23569" CreationDate="2018-09-03T11:00:05.860" Score="1" Body="&lt;p&gt;The simplest approach is to put the box into a microwave oven. Ten seconds at nominal power won't even heat up the box, and will make a short story of the dongle. The box may sustain superficial damage &lt;em&gt;on the inside&lt;/em&gt; if the dongle spews sparks, the plastic catches fire, or some electronic parts explode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, a directional microwave emitter will easily damage the electronics. If you want to damage all electronic devices within a hundred meters or so, you should search the dark web for plans of EMP bomb. It involves a permanent magnet, a coil, and a pyrotechnic charge that causes change of the geometry of the device in such a way that a massive electric current is rapidly induced in the coil, which is then immediately transformed into a powerful electromagnetic pulse destroying all electronic devices within range.  &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-09-03T11:00:05.860" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23582" PostTypeId="1" AcceptedAnswerId="23885" CreationDate="2018-09-03T14:18:08.680" Score="2" ViewCount="64" Body="&lt;p&gt;I'm designing a diagnostic system that simulates an environment and user input for a dummy vehicle. All possible critical situations are simulated in order to find out whether any errors are triggered, or if the software contains any faults or bugs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a vehicle for state defense, it doesn't have OBD like normal cars, but a display with buttons with which you can seek and troubleshoot errors. I have to design a procedure for testing each new software release. I was wondering if any standard exists for testing/validating software. Just like driving cycles(NEDC, WLTP) exist for emission tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do any standards(SAE, DIN, etc.) exist for this kind of thing?&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="9920" LastEditDate="2018-09-06T06:51:28.247" LastActivityDate="2018-09-23T12:18:51.827" Title="What standards exist regarding diagnostic systems in automotive engineering?" Tags="|automotive-engineering|simulation|software|standards|car|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23583" PostTypeId="1" CreationDate="2018-09-03T14:27:48.730" Score="2" ViewCount="82" Body="&lt;p&gt;They are tightly connected as some concepts of feedforward is based on PID algorithms, and both of them are control systems. But when I see a phrase such as &quot;Feed forwards augment PID control&quot;, I'm confused. Can they be used at the same time?&lt;/p&gt;&#xA;" OwnerUserId="17353" LastActivityDate="2018-09-03T14:27:48.730" Title="What is the connection between PID and feedforward controllers?" Tags="|mechanical-engineering|electrical-engineering|control-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23587" PostTypeId="1" CreationDate="2018-09-03T21:34:14.637" Score="-2" ViewCount="391" Body="&lt;p&gt;My knowledge about methods of transmission of data is limited, but I believe that communication between transmitters and receivers often uses Morse code to send information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If say, a satellite wanted to send a message consisting of letters from the English Alphabet, wouldn't it be possible to send it via either Morse, or ASCII, i.e., 'A' may be sent as .- (which may then be translated to the most appropriate form to be transmitted over an EM Wave) or as 01000001?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reason why Morse is preferred?&lt;/p&gt;&#xA;" OwnerUserId="17381" LastActivityDate="2019-09-15T17:46:08.663" Title="Why is Morse code used in satellite communication compared to ASCII encoding" Tags="|telecommunication|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23588" PostTypeId="2" ParentId="23587" CreationDate="2018-09-03T23:17:33.300" Score="1" Body="&lt;p&gt;Morse isn't used to communicate with satellites. In its simplest form, satellite communications are done by digitizing the text, encoding it into blocks or &quot;packets&quot; (ASCII can be used here), tagging the block or packet with routing information headers and footers, and transmitting the resulting bit string. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-03T23:17:33.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23590" PostTypeId="1" CreationDate="2018-09-04T01:09:35.837" Score="1" ViewCount="48" Body="&lt;p&gt;Small background: for a project of mine I need to develop a (very light) conical spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While there are plenty resources out there for regular springs I really struggled to find anything on conical springs (non-linear and telescoping conical springs, to be precise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the main constraint for my spring are its dimensions, I tried to calculate the wire diameter needed that can accomodate for compression force of around &lt;code&gt;5 cN&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some digging I managed to find &lt;a href=&quot;http://www.mate.tue.nl/mate/pdfs/10229.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; and made some calculations according to it.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D_1 = 0.003 m    # upper mean diameter&#xA;D_2 = 0.011 m    # lower mean diameter&#xA;n = 10           # number of coils&#xA;H = 0.0055 m     # height of spring at rest&#xA;G = 79.3 GPa     # steel shear modulus&#xA;F_C = 0.05 N     # full compression force&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;$ F_C $ is the load for which the smallest active coil (with local&#xA;  spring diameter $ D_1 $) reaches its maximum deflection $ H $&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;$ F_C = \frac{Gd^4H}{8D_1^3n} $&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Hence, $ d = \sqrt[4]{\frac{8F_CD_1n}{GH}} $&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the inputs above, I calculated that wire diameter should equal approximately &lt;code&gt;d = 0.12544 mm&lt;/code&gt;. (By the way if you can see it's wrong, I'd greatly appreciate some directions)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I approached some manufacturers with these numbers, several of them said it could not be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you please tell me what kind of constraints are typically taken into consideration when manufacturing such springs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it that it's impossible to create a spring of this shape that will hold its properties or are they simply lacking proper machinery?&lt;/p&gt;&#xA;" OwnerUserId="17383" LastEditorUserId="17383" LastEditDate="2018-09-04T12:28:45.977" LastActivityDate="2018-09-04T12:28:45.977" Title="What are typical manufacturing constraints when producing (conical) springs?" Tags="|mechanical-engineering|springs|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23591" PostTypeId="1" AcceptedAnswerId="23681" CreationDate="2018-09-04T03:49:32.517" Score="0" ViewCount="150" Body="&lt;p&gt;While I'm reading the paper, it said&lt;br&gt;&#xA;&quot;The I control can makes the phase lag, so it can makes stability margin reduced.&quot;&lt;br&gt;&#xA;Words are not same as that, but the meaning is the same.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How the I control makes the phase lag in the PID controller?&lt;br&gt;&#xA;If the plant's transfer function is $G_p (s)$, the open-loop transfer function $G(s)$ is&lt;br&gt;&#xA;$G(s) = \left(\frac{k_i}{s} + k_p + k_d s\right)G_p (s)$&lt;br&gt;&#xA;when we use the PID for the controller. After substitute $s=j\omega$,&lt;br&gt;&#xA;$G(j\omega) = \frac{\left(k_i-k_d\omega^2  \right)j+jk_p\omega}{\omega} G_P (s)$&lt;br&gt;&#xA;So, how can I find that the I controller makes phase lag?  &lt;/li&gt;&#xA;&lt;li&gt;How the phase lag makes stability margin reduced?&lt;br&gt;&#xA;In the paper, it wrote &quot;stability margin&quot;. I think that's phase margin. Anyway, how can I get the relation between 'stability margin' and 'phase lag'?  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you for advising!&lt;/p&gt;&#xA;" OwnerUserId="15645" LastActivityDate="2018-09-09T10:42:26.193" Title="The relationship between the I control and the stability margin" Tags="|control-engineering|pid-control|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23592" PostTypeId="1" CreationDate="2018-09-04T05:18:53.277" Score="1" ViewCount="458" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HDDVQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HDDVQ.jpg&quot; alt=&quot;HELICAL 2 STAGE GEAR ARRANGEMENT&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here P, Q, R, S are the bearings and G1, G2, G3, G4 are four helical gears.&#xA;The gear forces tangential(Kp), Axial(Ka), Radial(Kn) are known.&#xA;All the distances are known. The forces acting on bearing are radial on non locating bearing and both radial and axial on locating bearing.  P, R are non locating bearings. Q, S are locating bearings.. Fh and Fv are components of radial force on bearings and Fa is the axial force... How to calculate Fh, Fv and Fa for each bearing shown??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated all the gear forces but i am confused about the calculation of the forces. will be solved using moment equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance..&lt;/p&gt;&#xA;" OwnerUserId="17385" LastEditorUserId="17385" LastEditDate="2018-09-04T06:34:44.573" LastActivityDate="2018-09-04T06:34:44.573" Title="How to calculate bearing loads in a two stage helical gear box?" Tags="|mechanical-engineering|design|gears|applied-mechanics|bearings|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23593" PostTypeId="1" CreationDate="2018-09-04T07:22:53.417" Score="1" ViewCount="24" Body="&lt;p&gt;I have some questions about overhead line and substation equipment in high speed line.&#xA;My first question is about if all the overhead voltage is the same for all the intervals of speed? I mean for 250, 300 and 350 Km/h is it 25 Kv?&#xA;My second question is about if we want to increase speed of line, for example, after 5 years, should we change all these equipment  related to its consumable power? Since when the speed increases, it needs more power? What can we do to have a cost-effective solution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance &lt;/p&gt;&#xA;" OwnerUserId="17387" LastActivityDate="2018-09-04T07:22:53.417" Title="Electrification in highspeed train" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23594" PostTypeId="1" CreationDate="2018-09-04T10:32:11.767" Score="2" ViewCount="631" Body="&lt;p&gt;Theses are my thoughts on the question. I can't find an answer to &quot;why not?&quot;. It there something I am missing, or are auto manufacturers simply too focussed on cars only?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The automotive industry has been developing internal combustion engines for many decades. They are now routinely capable of running for 20,000 miles service interval, which is a simple oil and filters change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;20,000 miles will be around 2000 litres of fuel, which is comparable to the oil required to heat a house in the UK for a year using an oil  boiler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air source heat pumps are only just capable of offering cost savings to a UK household, compared to an oil-fired boiler, because of the inefficiency of the power stations and electricity grid in delivering electricity to the house, compared to burning the fuel locally. A local heat-pump gain of 4+ has to be multiplied by electricity generation and distribution inefficiency (around 0.5? Not sure). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The auto manufacturers' investment in internal combustion engines in under threat from electric vehicles. They have made the investment to mass-produce engines, and ought to welcome another possible use for much the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why don't the auto manufacturers develop an domestic air source heat pump based on an auto engine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water cooling system would be the house's central heating system. Auto cooling normally runs at 90C. Domestic heating systems normally run at no more than 70C for safety reasons, and often as low as 30C for thermodynamic efficiency reasons. Would this be a problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what fraction of a car's fuel is lost as heat via the exhaust pipe. In any case, some (large?) paer if this could be reclaimed via a heat-exchanger, as with a condenser boiler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotary motion the engine generates could either directly run a compressor, or run a generator to provide electrical energy for the house with excess sold back to the electricity utility. Unlike solar panels, this energy would almost by definition be generated at times of maximum demand (in the UK, the coldest weather. In hot countries, the hottest weather, for cooling).  So this would automatically be a benefit to utility companies and they might offer a good price even without the government forcing their hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An auto engine can provide a range of power levels, so the system could easily modulate its output to match demand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd guess that an auto engine driving a heat pump, directly or indirectly, could have a heating efficiency of 2-3. Heat pump efficiency for driving a hot water central heating system is typically 3, so that lower figure allows for some losses. A modern boiler's efficiency is 0.85. I'm guessing that overall electricity utility plus heat pump is around 1.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Annual servicing would be an oil and filter change. By using a larger oil supply and larger filters, it might be straightforward to push this to 2, even 3 years. Costs not a lot more than oil boiler servicing. Also a fixed installation should be a more benign environment than a moving vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that a petrol engine can also run on natural gas. Presumably a diesel could be modified to run on kerosene. If not, this would require a household without a gas supply to use diesel as heating fuel. Governments might worry about leakage of tax revenue from the heating oil tank into a car's fuel tank. However, in any case, electric cars are going to destroy that revenue in coming decades. (In the UK, farmers use low-tax &quot;Red diesel&quot; for tractors and heating.  This fuel is dyed red so auto fuel tax evasion can be detected). &lt;/p&gt;&#xA;" OwnerUserId="17388" LastEditorUserId="17388" LastEditDate="2018-09-04T10:44:58.000" LastActivityDate="2023-04-24T23:05:18.233" Title="Why not drive an air source heat pump domestic heating system with an auto engine?" Tags="|thermodynamics|heat-transfer|automotive-engineering|heating-systems|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23595" PostTypeId="1" CreationDate="2018-09-04T11:14:23.307" Score="-1" ViewCount="47" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2AZFj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2AZFj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MjNNt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MjNNt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dear all, &#xA;as a novice I need some help from somebody with knowledge in electrical engineering. I need to supply a sensor with power, as shown on the left side of the picture. The sensor uses a plug as shown in the right section.&#xA;I would like the source to be a battery (standalone solution), can I use a 12V motorcycle battery or do I risk to destroy the sensor for any kind of reason ?&#xA;What is the name of this type of plug  ? &lt;/p&gt;&#xA;" OwnerUserId="10960" LastEditorUserId="10960" LastEditDate="2018-09-05T06:30:30.483" LastActivityDate="2018-10-05T11:00:15.767" Title="Sensor connectivity" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="23596" PostTypeId="2" ParentId="14636" CreationDate="2018-09-04T12:30:12.447" Score="1" Body="&lt;p&gt;Do a google search for HSSE VAP (high speed steel with cobalt and vaporised surface) threading taps. They have a 1400nm/mm2 cutting capacity. Speacialy made for stainless steel. I’ve used them and they are great.&lt;/p&gt;&#xA;" OwnerUserId="17390" LastActivityDate="2018-09-04T12:30:12.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23597" PostTypeId="1" CreationDate="2018-09-04T13:42:58.390" Score="2" ViewCount="101" Body="&lt;p&gt;I am designing a new, closed loop,chilled water system for our small injection molding outfit and can use some help where to begin. Here are some of my knowns:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Northeast NJ, USA, no air conditioning. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 hydraulic machines( 2x250ton, 1x310ton, 1x400 ton), total HP, 210hp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 molds, ~1 ton per mold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;material is HDPE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 2 existing, relatively new, air cooled chillers, a 4 ton unit, and a variable speed 10 ton unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4 machines are located in a square, total piping for a loop to all would be ~ 250',  25' up in the air, would need 3 machine &quot;drops&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My oil heat exchangers will have water savers @ ~ 105*F&#xA;My molds should be ~ 55-65*F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like a central system, say a 500 gal. tank or so, cooled with my existing chillers and pumped through the loop to each machine, returned to my tank. Capability of adding say 2 more machines @ another 10hp + another 2 molds @ 1 ton each would be nice so lets say, 40 ton in an non-central setup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not want a cooling tower&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I go about calculating my circulating pipe/pump sizes? My 2 chillers have their own pumps&#xA;In a non-central setup, I don't have enough chiller for my requirements ( 14 tons &amp;lt; 25 tons) Will 14 tons be enough in a central setup with 500 gal of water? If not, how do I calculate how much of a chiller i'd still need?&#xA;My TO PROCESS goal for water temp. is ~60*F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any guidance in any direction is appreciated&lt;/p&gt;&#xA;" OwnerUserId="11327" LastEditorUserId="11327" LastEditDate="2018-09-05T13:09:06.083" LastActivityDate="2018-09-05T14:12:33.240" Title="Designing a Chilled Process Water System for Small Plant" Tags="|mechanical-engineering|process-engineering|cooling|injection|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23599" PostTypeId="1" CreationDate="2018-09-04T13:54:01.343" Score="1" ViewCount="28" Body="&lt;p&gt;Consider a receiving dipole antennae of length $L$. If electromagnetic waves of power intensity $P$ W/m&lt;sup&gt;2&lt;/sup&gt; is incident on it, then open circuit voltage induced in the antenna is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V=\frac{L}{2}\sqrt{P\times 2\times 377}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V$ is independent of the wavelength of the electromagnetic waves. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another fact is that dipole antennae are resonant at half wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to reconcile these two facts? I mean to say a dipole of 6cm is resonant for 2.4GHz. But if one uses it for receiving 10 Mhz signal, what exactly will be problem as the open circuit voltage will be same irrespective of frequency (of course $P$ is constant)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding says following. While voltage induced will be same, what will depended on frequency is impedance of antenna. At half wavelength, the dipole will be resistive only. At lower frequencies capacitive reactance will be high while at high frequencies inductive reactance will be high, both these will reduce the current induced in the circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I wrong? or, am I missing some aspect?&lt;/p&gt;&#xA;" OwnerUserId="17373" LastEditorUserId="1832" LastEditDate="2018-09-04T14:56:33.240" LastActivityDate="2018-09-04T14:56:33.240" Title="Role of antenna length" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23600" PostTypeId="2" ParentId="23594" CreationDate="2018-09-04T14:02:00.007" Score="1" Body="&lt;p&gt;I had a similar thought some time ago, but my concept was to use an internal combustion engine that was modified to use the domestic gas supply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The use of domestic gas avoids issues with transporting and storing petrol (gasoline) to domestic premises and associated safety issues. Domestic gas is already supplied to most dwellings in certain regions, so the fuel is already there. All that is required to attach a pipe from the domestic gas supply to the engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having stated that, there are a number of reasons why using an internal combustion engine as a domestic energy supply is problematic:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The installation of the system would require skilled people, which&#xA;could be expensive.&lt;/li&gt;&#xA;&lt;li&gt;Car manufacturers do not make combustion engines for the domestic&#xA;market most likely because the market size would be small and there would&#xA;most likely would be regulations that would make it difficult.&lt;/li&gt;&#xA;&lt;li&gt;Neighbors could be easily upset by both noise and exhaust fumes.&#xA;Noise can be mitigated by enclosing the engine in a sound proofed&#xA;enclosure - such as a shed. The exhaust fumes may require a very tall flue, which may&#xA;not be approved by local municipal councils.&lt;/li&gt;&#xA;&lt;li&gt;In some jurisdictions, generating electricity requires permits and&#xA;has government regulations associated with it; which can be an issue&#xA;for some people with solar panels.&lt;/li&gt;&#xA;&lt;li&gt;Such systems would require periodic safety checks and the upkeep of&#xA;maintenance.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-09-04T14:02:00.007" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23605" PostTypeId="2" ParentId="8760" CreationDate="2018-09-04T15:51:52.037" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XC36W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XC36W.png&quot; alt=&quot;Tripartite Paper - Period on X-axis&quot;&gt;&lt;/a&gt;&#xA;This is the only place on the internet that I could find that actually gave decent resolution images for tripartite plotting paper.  However, it had some limits that affected use in our area: 1) the design earthquakes top out in our area at about 1g and 2) we generally are only concerned with periods between 0.01s and 10s, or 0.1 Hz to 100 Hz.  This resulted in plots of the earthquake crammed down in one corner, and the 144-yr operating basis earthquake barely showed up at all.  Also, we generally plot with the period, rather than the frequency, on the X-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took a weekend to learn just enough R to modify Biswajit Banerjee's excellent script to come up with a plot that would serve my purposes.  I'm going to throw it up in case it's useful for anybody else.  If you've not worked with R, you can install the basic R interpreter, cut and paste either Mr. Banerjee's or my code into a script, and run it to get a higher-quality PDF output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It didn't take long to learn enough to actually plot the earthquakes using relatively easy code, but the way I did it is extremely user-unfriendly, if effective.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;require(ggplot2)&#xA;require(data.table)&#xA;&#xA;# The constant value grid lines&#xA;periods = unique(c(seq(0.01, 0.02, by = 0.005), seq(0.02, 0.1, by = 0.01),&#xA;      seq(0.1, 0.2, by = 0.05), seq(0.2, 1.0, by = 0.1),&#xA;      seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0)))&#xA;vels = unique(c(seq(0.01, 0.02, by = 0.005), seq(0.02, 0.1, by = 0.01),&#xA;    seq(0.1, 0.2, by = 0.05), seq(0.2, 1.0, by = 0.1),&#xA;    seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0),&#xA;      seq(10.0, 20.0, by = 5.0)))&#xA;accs = unique(c(seq(0.0001, 0.0002, by = 0.00005),&#xA;    seq(0.0002, 0.001, by = 0.0001),&#xA;    seq(0.001, 0.002, by = 0.0005), seq(0.002, 0.01, by = 0.001),&#xA;      seq(0.01, 0.02, by = 0.005), seq(0.02, 0.1, by = 0.01),&#xA;      seq(0.1, 0.2, by = 0.05), seq(0.2, 1.0, by = 0.1),&#xA;      seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0),&#xA;      seq(10.0, 20.0, by = 5.0), seq(20.0, 100.0, by = 10.0)))&#xA;disps = unique(c(seq(0.00001, 0.00002, by = 0.000005), seq(0.00002, 0.0001, by = 0.00001),&#xA;          seq(0.0001, 0.0002, by = 0.00005), seq(0.0002, 0.001, by = 0.0001),&#xA;          seq(0.001, 0.002, by = 0.0005), seq(0.002, 0.01, by = 0.001),&#xA;          seq(0.01, 0.02, by = 0.005), seq(0.02, 0.1, by = 0.01),&#xA;          seq(0.1, 0.2, by = 0.05), seq(0.2, 1.0, by = 0.1),&#xA;          seq(1.0, 2.0, by = 0.50), seq(2.0, 10.0, by = 1.0),&#xA;          seq(10.0, 20.0, by = 5.0)))&#xA;&#xA;# Horizontal grid lines (const vel = in/s)&#xA;vel_grid = data.table()&#xA;periodmin = min(periods)&#xA;periodmax = max(periods)&#xA;for (vel in vels) {&#xA;  xx = c(periodmin, periodmax)&#xA;  yy = c(vel, vel)&#xA;  label = c(paste0(&quot;vel&quot;,vel), paste0(&quot;vel&quot;,vel)) &#xA;  color = &quot;vel&quot;&#xA;  local_dt = data.table(x = xx, y = yy, label = label, color = color)&#xA;  vel_grid = rbind(vel_grid, local_dt)&#xA;}&#xA;&#xA;# Vertical grid lines (freq = cycles/sec)&#xA;period_grid = data.table()&#xA;velmin = min(vels)&#xA;velmax = max(vels)&#xA;for (period in periods) {&#xA;  xx = c(period, period)&#xA;  yy = c(velmin, velmax)&#xA;  label = c(paste0(&quot;period&quot;,period), paste0(&quot;period&quot;,period)) &#xA;  color = &quot;period&quot;&#xA;  local_dt = data.table(x = xx, y = yy, label = label, color = color)&#xA;  period_grid = rbind(period_grid, local_dt)&#xA;}&#xA;&#xA;# 45 grid lines (constant acc = omega * vel = (2*pi/period) radians/sec * vel in/sec  &#xA;#                              = (2*pi/period)*vel in/sec^2 = (2*pi/period)*vel/g  Gs)&#xA;gravity = 32.2*12.0  # in/s^2&#xA;acc_grid = data.table()&#xA;for (acc in accs) {&#xA;  for (period in periods) {&#xA;    Ts = period&#xA;    vel = acc*gravity*period/(2*pi)&#xA;    if (vel &amp;lt; velmin) {&#xA;      vel = velmin&#xA;      Ts = vel*2*pi/(acc*gravity)&#xA;    }&#xA;    if (vel &amp;gt; velmax) {&#xA;      vel = velmax&#xA;      Ts = vel*2*pi/(acc*gravity)&#xA;    }&#xA;    xx = c(Ts)&#xA;    yy = c(vel)&#xA;    label = c(paste0(&quot;acc&quot;,acc)) &#xA;    color = &quot;acc&quot;&#xA;    local_dt = data.table(x = xx, y = yy, label = label, color = color)&#xA;    acc_grid = rbind(acc_grid, local_dt)&#xA;  }&#xA;}&#xA;&#xA;## The const acc=1 line is the disp axis&#xA;acc_1 = acc_grid[acc_grid$label == &quot;acc0.01&quot;]&#xA;disp_axis = data.table()&#xA;for (disp in disps) {&#xA;  acc = 0.012&#xA;  omega = sqrt(acc*gravity/disp)&#xA;  period = (2*pi)/omega&#xA;  vel = omega*disp&#xA;  xx = c(period)&#xA;  yy = c(vel)&#xA;  label = c(paste0(&quot;disp_axis&quot;,acc)) &#xA;  color = &quot;disp_axis&quot;&#xA;  local_dt = data.table(x = xx, y = yy, z = disp, label = label, color = color)&#xA;  disp_axis = rbind(disp_axis, local_dt)&#xA;}&#xA;&#xA;# -45 grid lines (const disp = vel/omega = vel/(2*pi/period) in)&#xA;disp_grid = data.table()&#xA;for (disp in disps) {&#xA;  for (period in periods) {&#xA;    Ts = period&#xA;    vel = disp*(2*pi)/period&#xA;    if (vel &amp;lt; velmin) {&#xA;      vel = velmin&#xA;      Ts = disp*2*pi/vel&#xA;    }&#xA;    if (vel &amp;gt; velmax) {&#xA;      vel = velmax&#xA;      Ts = disp*2*pi/vel&#xA;    }&#xA;    xx = c(Ts)&#xA;    yy = c(vel)&#xA;    label = c(paste0(&quot;disp&quot;,disp)) &#xA;    color = &quot;disp&quot;&#xA;    local_dt = data.table(x = xx, y = yy, label = label, color = color)&#xA;    disp_grid = rbind(disp_grid, local_dt)&#xA;  }&#xA;}&#xA;&#xA;## The const disp=0.1 line is the acc axis&#xA;disp_1 = disp_grid[disp_grid$label == &quot;disp0.1&quot;]&#xA;acc_axis = data.table()&#xA;for (acc in accs) {&#xA;  disp = 0.12&#xA;  omega = sqrt(acc*gravity/disp)&#xA;  period = 2*pi/omega&#xA;  vel = omega*disp&#xA;  xx = c(period)&#xA;  yy = c(vel)&#xA;  label = c(paste0(&quot;acc_axis&quot;,acc)) &#xA;  color = &quot;acc_axis&quot;&#xA;  local_dt = data.table(x = xx, y = yy, z = acc, label = label, color = color)&#xA;  acc_axis = rbind(acc_axis, local_dt)&#xA;}&#xA;&#xA;# Join vertical and horizontal&#xA;grid = rbind(vel_grid, period_grid, acc_grid, disp_grid)&#xA;&#xA;### Axis label points&#xA;disp = 0.08&#xA;acc = 0.002&#xA;omega = sqrt(acc*gravity/disp)&#xA;period = 2*pi/omega&#xA;vel = omega*disp&#xA;acc_axis_label = data.table(x = c(period), y = c(vel), label=&quot;Spectral acceleration (g)&quot;)&#xA;&#xA;disp = 0.006&#xA;acc = 0.008&#xA;omega = sqrt(acc*gravity/disp)&#xA;period = 2*pi/omega&#xA;vel = omega*disp&#xA;disp_axis_label = data.table(x = c(period), y = c(vel), label=&quot;Spectral displacement (in)&quot;)&#xA;&#xA;# Plot horizontal grid&#xA;xticks = periods[c(TRUE, FALSE)]&#xA;yticks = vels[c(TRUE, FALSE)]&#xA;&#xA;&#xA;plt = ggplot() +&#xA;      geom_line(data = grid, aes(x = x, y = y, group = label, color = color),&#xA;                size=0.25) +&#xA;      geom_line(data = acc_1, aes(x = x, y = y), color = 1, size=0.5) + &#xA;      geom_line(data = disp_1, aes(x = x, y = y), color = 1, size=0.5) + &#xA;      xlab(&quot;Period (s)&quot;) + &#xA;      ylab(&quot;Spectral velocity (in/sec)&quot;) + &#xA;      scale_color_discrete(guide = FALSE) +&#xA;      scale_x_log10(limits = c(periodmin, periodmax),&#xA;                    expand = c(0,0),&#xA;                    breaks=xticks,&#xA;                    labels= c(&quot;0.01&quot;,&quot;0.02&quot;,&quot;0.04&quot;,&quot;0.06&quot;,&quot;0.08&quot;, xticks[xticks &amp;gt;= 0.1])) +&#xA;      scale_y_log10(limits = c(velmin, velmax),&#xA;                    expand = c(0,0),&#xA;                    breaks=yticks) +&#xA;      coord_fixed() +&#xA;      theme_bw() +&#xA;      theme(panel.grid.major = element_blank(),&#xA;            panel.grid.minor = element_blank()) +&#xA;      annotate(&quot;text&quot;, x = acc_axis$x, y = acc_axis$y, label = as.character(acc_axis$z), &#xA;               size = 3, angle = 45) + &#xA;      annotate(&quot;text&quot;, x = disp_axis$x, y = disp_axis$y, label = as.character(disp_axis$z), &#xA;               size = 3, angle = -45)  +&#xA;      annotate(&quot;text&quot;, x = acc_axis_label$x, y = acc_axis_label$y, label = acc_axis_label$label,&#xA;               size = 4, angle = -45)  +&#xA;      annotate(&quot;text&quot;, x = disp_axis_label$x, y = disp_axis_label$y, label = disp_axis_label$label,&#xA;               size = 4, angle = 45)&#xA;&#xA;print(plt)&#xA;ggsave(&quot;TripartitePaper.pdf&quot;,width = 10.0, height = 10.0, units = &quot;in&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="17396" LastActivityDate="2018-09-04T15:51:52.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23606" PostTypeId="2" ParentId="21598" CreationDate="2018-09-04T15:57:56.370" Score="1" Body="&lt;p&gt;Short answer: neither BLDC nor PMSM requires a hall sensor, and both may need need it for some applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BLDC (brush-Less Direct Current) motors and PMSM (Permanent Magnet Synchronous Motor) are basically the same type of motor: three-phase synchronous motors, with permanent magnets to generate the rotor's magnetic field (it would work with other numbers than 3 phases, but it's analog to AC motors and allow 6-steps control). PMSM and BLDC acronyms are used to indicate for which control type the motor has been optimized. On PMSM, we try to get no magnetic saturation to get a sinusoidal back-EMF (electromotive force) and therefore an optimal control with sinusoidal current, while the magnetic saturation tends to compensate torque ripple on BLDC. The principle of both motor is still the one of a three-phase synchronous motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is the principle of a three-phase synchronous motor ? The stator produces a fixed amplitude (at a given speed and load) rotating magnetic field, and the rotor produces a fixed magnetic field that will align on the rotating field from the stator. To get the best efficiency, the stator's field should stay at 90° from the rotor's field. To adjust the stator field intensity, it is therefore required to know where the rotor is. Using a sensor (such as hall effect sensor) is always an option, but for cost reduction, and to mitigate the risk of sensor malfunction, sensor-less solutions exist:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;on BLDC, with a 6-step control, there always is a non-driven phase, on which reading the voltage and detecting zero-crossings is enough to know when to commute steps.&lt;/li&gt;&#xA;&lt;li&gt;on PMSM, with either sinusoidal or vector control, the three phases are driven, and it's a bit more tricky to rebuild the position of the rotor from the current measurement in phases&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Those sensor-less solutions use the fact that the motor produces a back-EMF which depends on the position of the rotor. The issue is that the motor produces back-EMF proportionally to its speed. Those methods don't work for position control. Also, the position computation from the phase currents is computationally more demanding than Zero-Crossing detection, and may be impossible to run at high speed on limited hardwares. Last but not least, sensor-less solutions start in &lt;a href=&quot;https://en.wikipedia.org/wiki/Open-loop_controller&quot; rel=&quot;nofollow noreferrer&quot;&gt;open-loop&lt;/a&gt;, with no-feedback from the motor, to make it reach the require speed to apply sensor-less control. This requires to know roughly the load applied on the motor when it starts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short, the choice to use a sensor depends more on your application than on your control method.&lt;/p&gt;&#xA;" OwnerUserId="2664" LastActivityDate="2018-09-04T15:57:56.370" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23608" PostTypeId="2" ParentId="23594" CreationDate="2018-09-04T16:44:03.710" Score="1" Body="&lt;p&gt;gas and diesel engine-driven heat pumps have been on the market for decades, for use in running cooling systems for refrigerated truck trailers and railroad cars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;they are noisy, and would require the same noise-abatement systems used to control noise in residential gas-powered electric generator installations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;something to consider here is the idea of using the waste heat of the engine as part of the useful output of the system, to provide space heat or hot water for domestic use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find that Honda is actually selling an engine-powered electric generator right now that runs on natural gas and furnishes both electricity and heat. the system is powerful enough to serve the needs of a small house and fits under  a kitchen counter, occupying the same space as a dishwasher. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-04T16:44:03.710" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23610" PostTypeId="2" ParentId="23594" CreationDate="2018-09-04T19:35:26.737" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;Automotive engines have their own inefficiencies. If the engine runs at its optimum speed, expect to convert ~30% of the energy in the fuel to motive power, 30% to heat in the cooling system and 30% to hot gases in the exhaust. At low loads, a petrol engine gets much worse figures, diesel scales a bit better. &lt;/li&gt;&#xA;&lt;li&gt;Reliability is not high enough. 20,000 miles takes about 500 hours, so you'd need to service the engine every month. &lt;/li&gt;&#xA;&lt;li&gt;Car engines are a bit too large for this application. A 75-kW engine produces ~70 kW of electricity (50 times as much as you need) and 75 kW of heat (a central heating furnace produces 25-40 kW of heat and only needs a small duty cycle), so you have to store the excess and run the engine intermittently. You can make the engine smaller, but that also makes it less efficient and noisier. &lt;/li&gt;&#xA;&lt;li&gt;&quot;The auto manufacturers' investment in internal combustion engines in under threat from electric vehicles&quot;, yes, and auto manufacturers are switching for three reasons: we're running out of fossil fuels, we can't go on producing as much CO2 as we do now and we've noticed exhaust products are bad for your health. All of these would count just as much for domestic use of ICEs, so any attempt to introduce this would face heavy legislation. &lt;/li&gt;&#xA;&lt;li&gt;It's very difficult to reduce the noise of an ICE to levels acceptable for domestic use. Car engines produce ~ 80 dB running at their most efficient point, you want the noise level outside the enclosure to be below 30 dB. 50 dB reduction requires heavy, expensive and large noise insulation. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="11645" LastActivityDate="2018-09-04T19:35:26.737" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23611" PostTypeId="2" ParentId="23094" CreationDate="2018-09-04T21:15:57.687" Score="3" Body="&lt;p&gt;@Mahendra, I had run into Hard Drive Memory Space issue few months ago on my NI cDAQ. I am using cDAQ-9132 with a 16 GB hard drive, similar to your cDAQ. I was running LabVIEW Professional 2017 SP1 with all softwares installed. The memory space got full when I needed to install some additional drivers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cDAQ-9132 is not meant to have a full LabVIEW system running on it. The best practice is to only have LabVIEW run-time engine on the cDAQ and run LabVIEW application which is developed on a full computer with complete LabVIEW system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid memory space issue, I developed my code on the laptop which has over 200 GB memory space and has a full-fledged LabVIEW system. Once I developed the code I created a standalone application (EXE file) using instruction from the following tutorial: &lt;a href=&quot;http://www.ni.com/tutorial/3303/en/#toc2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Distributing Applications with the LabVIEW Application Builder: Stand-Alone Applications&lt;/a&gt; . After that I created an installer for my application which I used to install the LabVIEW runtime engine and the drivers I needed on the cDAQ. You can find step by step instructions to create installer in the following link:&lt;a href=&quot;http://www.ni.com/tutorial/3303/en/#toc5&quot; rel=&quot;nofollow noreferrer&quot;&gt;Distributing Applications with the LabVIEW Application Builder:  Installers (Windows)&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now to cleanup space on the cDAQ, you need to uninstall LabVIEW completely. Go to the control panel in Windows 7 and follow this tutorial:  &lt;a href=&quot;http://www.ni.com/product-documentation/54335/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Uninstall or Repair National Instruments Software or Drivers in Windows&lt;/a&gt;. I have also used Temp File Cleaner which cleaned about 2 GB of space on cDAQ.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope it helps!&lt;/p&gt;&#xA;" OwnerUserId="10379" LastEditorUserId="7083" LastEditDate="2018-09-05T01:11:01.830" LastActivityDate="2018-09-05T01:11:01.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23612" PostTypeId="1" AcceptedAnswerId="23619" CreationDate="2018-09-05T05:47:25.223" Score="2" ViewCount="3268" Body="&lt;p&gt;In the case of a simple cantilever beam having a uniform shape and being loaded at its end - the calculation of the deflection at any point along the beam is well known.&#xA;What happens if we split it along its longitudinal axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Xiu2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Xiu2.jpg&quot; alt=&quot;cantilever beam composed of separate planks - taken from http://www.bu.edu/moss/mechanics-of-materials-bending-shear-stress/ &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize we are eliminating the shear forces along that axis between the beam segments, and this allows each segment to slide relatively to its neighbors. However, I'm not sure how this phenomena affects the beam deflection. How can I calculate it?&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2020-08-07T08:24:12.710" Title="Deflection of a cantilever beam composed of separate (not bonded) planks" Tags="|mechanical-engineering|structural-engineering|beam|deflection|shear|" AnswerCount="3" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23613" PostTypeId="1" AcceptedAnswerId="23617" CreationDate="2018-09-05T07:03:06.910" Score="3" ViewCount="44" Body="&lt;p&gt;Im talking about 10k 5k 2k 1k 800 400 grit papers, wax polishing bars and polish paste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do they have a universal color code which corresponds to grit? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many times the letters are either not on the paper or hard to read.&lt;/p&gt;&#xA;" OwnerUserId="17400" LastEditorUserId="17400" LastEditDate="2018-09-05T08:48:21.693" LastActivityDate="2018-09-05T12:04:23.760" Title="Do fine grit sand papers and wax polish bars and pastes have a universal color code?" Tags="|polishing|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23614" PostTypeId="1" AcceptedAnswerId="26320" CreationDate="2018-09-05T07:48:49.757" Score="0" ViewCount="1433" Body="&lt;p&gt;I ordered this model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ebay.com/itm/coil-winding-mechanical-tensioner-MT-01-wire-diameter-0-04-0-12-tension-device/181965432380?hash=item2a5dfc263c:g:KFIAAOSw~4NZt4lY&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ebay.com/itm/coil-winding-mechanical-tensioner-MT-01-wire-diameter-0-04-0-12-tension-device/181965432380?hash=item2a5dfc263c:g:KFIAAOSw~4NZt4lY&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I downloaded this diagram to get a better idea of the parts(This from a different model but they share a majority of components):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Pxs2y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pxs2y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just curious about the 1-10 knob on the back that puts a metal pad either closer or farther away from the device. I assume this varies the magnetic resistance&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also curious about the back-tension button. I see that when you tighten the screw the tension arm takes more pressure to pull down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All-in-all, I just want to know how the knob on the back works in conjunction with the back-tension&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, If you've used a Coil winding magnetic tensioner before, any info on its functions I would love to learn about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reaching out because there is basically nothing online about how these things work and all the labels on my tensioner are in Chinese so I'm clueless&lt;/p&gt;&#xA;" OwnerUserId="17402" LastActivityDate="2019-03-10T18:07:40.177" Title="Coil winding magnetic tensioner: How does it Work?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23616" PostTypeId="2" ParentId="23595" CreationDate="2018-09-05T10:55:19.167" Score="0" Body="&lt;p&gt;Based on the information provided I don't think you will have any issue using a 12V motorcycle battery. Attach below is information on 12V Motorcycle battery current draw which is over 2.5 Ah. The current requirement for you sensor module is 200mA. Therefore you should not have any issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dC3pK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dC3pK.jpg&quot; alt=&quot;12V Motorcycle Battery Specification&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.yuasabatteries.com/wp-content/uploads/2017/12/Catalog-2018-Final.x78244.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battery Applications and Specifications - Yuasa&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.motorcyclecruiser.com/tech-tip-for-recharging-your-motorcycle-battery-right-way&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tech Tip for Recharging Your Motorcycle Battery the Right Way&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2018-09-05T10:55:19.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23617" PostTypeId="2" ParentId="23613" CreationDate="2018-09-05T12:04:23.760" Score="3" Body="&lt;p&gt;No, they do not. I've got red, yellow, green, and grey 800 grit in a drawer in my garage!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have various different grit values that are all grey.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-09-05T12:04:23.760" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23619" PostTypeId="2" ParentId="23612" CreationDate="2018-09-05T13:39:52.590" Score="6" Body="&lt;p&gt;in a cantilever beam the deflection is $$\delta_{max} = \frac {PL^3}{3EI}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case assuming free sliding between the planks the load P is going to be supported equally between the 3 planks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the deflection will be $$ \delta_{max} = \frac {(P/3)L^3}{3EI_{\text{single board}}}  $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because $I$ is proportional to the cube of the board's height (in this case, its thickness), the single board's inertia will be $(1/3)^3=1/27$ that of the bonded boards. Therefore the unbonded deflection will be greater than the bonded boards by a factor of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{gather}&#xA;\dfrac{\left(\frac{1}{3}\right)}{\left(\frac{1}{27}\right)} = \frac{27}{3} = 9 \\&#xA;\therefore \delta_{unbonded}= 9\delta_{bonded}&#xA;\end{gather}$$&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2018-09-05T17:22:59.957" LastActivityDate="2018-09-05T17:22:59.957" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23620" PostTypeId="2" ParentId="23597" CreationDate="2018-09-05T14:12:33.240" Score="1" Body="&lt;p&gt;Disclaimer 1: I've dealt with heating circuits, never with cooling circuits&lt;br&gt;&#xA;Disclaimer 2: I don't feel like doing your math for you, I'll just show the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I would set it up (ignore the shaded areas, that's a scanner artifact):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aHAC5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aHAC5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have a cooling manifold with 4 branches (on of which is drawn) for the hydraulic power packs and 4 branches for the molds. Each branch gets it's own 3-way valve and pump and temperature control. Note that if you want to cool your molds to 55F, you need colder water than 55F - I've assumed 50F but check back with the documentation of your molds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for pump sizing you need to know your heat loads, your fluid (water or water + x% glycol?) and your delta T between flow and return. Volume flow equals heat load by delta T times heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then pick a pipe size so your flow velocity is around 1-1.5 m/s. Take this info, the layout of your piping and what you know about the molds and HPUs to a supplier of heating pumps, they will help you with sizing. At leat in my part o the world, high efficiency heating pumps that control themselves for head are standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the air chiller side, I've assumed one pump for each chiller. Install a hydraulic compensator as your volume flows on chilling and consumption side will be missmatched - this could be your 500 gal. tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But including a large tank only makes sense if you have serious discontinous operation and hope. Say you want to run the chillers 24/7, you plan operates 8/5 or so. Modelling this would be it's own question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want about 50F, that's colder than on a warm day. You will need some sort of refridgerator beyond a simple air chiller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, it mayturn out you need to higher an HVAC engineer to design a system for you, or get someone to do a turn-key installation.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-09-05T14:12:33.240" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23623" PostTypeId="1" AcceptedAnswerId="23628" CreationDate="2018-09-05T19:04:26.607" Score="4" ViewCount="6385" Body="&lt;p&gt;Gapless railroads have their joints welded together to reduce noise. The lack of a gap between rail sections also reduces wear on both the track and the wheels. However, because the weld essentially produces one very long track, the individual sections have no room to expand as they get warm on hot days. What prevents the track from warping beyond usability? (Yes, I have heard of tracks warping beyond safety margins on extremely hot days, but I'm wondering about just normal expansion between normal weather extremes.)&lt;/p&gt;&#xA;" OwnerUserId="12535" LastEditorUserId="4800" LastEditDate="2018-09-06T10:49:38.097" LastActivityDate="2018-09-06T10:49:38.097" Title="How do gapless rails deal with thermal expansion?" Tags="|rail|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23624" PostTypeId="2" ParentId="23623" CreationDate="2018-09-05T19:24:56.863" Score="7" Body="&lt;p&gt;The tracks are preheated and welded at a calculated temperature (that depends on the local weather) at the time of installation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the track cools to the ambient temperature, this leaves a tensile stress in them. This tensile stress increases on colder days, and decreases on hotter days. As long as the stress remains tensile there is no warping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On extremely hot days the stress may become compressive. This can lead to buckling of the rails.&lt;/p&gt;&#xA;" OwnerUserId="17409" LastEditorUserId="255" LastEditDate="2018-09-06T08:41:46.917" LastActivityDate="2018-09-06T08:41:46.917" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23625" PostTypeId="1" CreationDate="2018-09-05T19:47:28.787" Score="0" ViewCount="48" Body="&lt;p&gt;How to create an algorithms for fitting a network of tubes with a range of diameters into a cube, with maximum volume in the tubes? Assume:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;30 cm minimum square side&lt;/li&gt;&#xA;&lt;li&gt;1 cm minimum pipe diameter&lt;/li&gt;&#xA;&lt;li&gt;1 mm pipe wall thickness. &lt;/li&gt;&#xA;&lt;li&gt;Cube edge length $L$&lt;/li&gt;&#xA;&lt;li&gt;number of different diameters $n$&lt;/li&gt;&#xA;&lt;li&gt;volume Pipes $= \pi r^2\times\text{length}$.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The tubes form a network in the sense of a continuous network of connected Tubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tricky parts:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Finding the ideal number of different radiuses of pipes to use.&lt;/li&gt;&#xA;&lt;li&gt;accounting for curves in the pipes. May be worth assuming straight pipes with U-turns at the ends? Obviously square pipes would be ideal. But we are restricted to circular pipes.  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Best to break down the problem into components:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;assuming (one) 1&amp;nbsp;cm diameter pipe, how to fit most volume pipe into 30cm cube.&lt;/li&gt;&#xA;&lt;li&gt;assuming (multiple) diameter pipes: 1&amp;nbsp;cm and 5&amp;nbsp;cm diameter pipes, how to fit most volume pipe into 30&amp;nbsp;cm cube.&lt;/li&gt;&#xA;&lt;li&gt;how to determine the optimal number of diameters&lt;br&gt;&#xA;i. given limited number or radii,&lt;br&gt;&#xA;ii. given unlimited number of radii.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4214" LastEditorUserId="4214" LastEditDate="2018-09-06T01:08:48.140" LastActivityDate="2018-09-06T12:55:52.933" Title="How to create an algorithms for fitting a network of tubes into a cube?" Tags="|pipelines|" AnswerCount="1" CommentCount="7" ClosedDate="2018-09-07T12:10:48.263" ContentLicense="CC BY-SA 4.0" />
  <row Id="23626" PostTypeId="1" CreationDate="2018-09-05T20:40:21.623" Score="2" ViewCount="642" Body="&lt;p&gt;I am looking for a cable to be placed in a drag chain that is moved multiple times per minute. Therefore, the cable needs to have a flex lifetime of over 1M times. Since the space is also tight, the bend radius is of interest. However the datasheet mentions:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Flexing bending radius: $5\phi$&lt;/li&gt;&#xA;  &lt;li&gt;Static bending radius: $10\phi$&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where $\phi$ is the diameter of the cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't the flexing bending radius be larger than the static bending radius? I would expect that a cable flexing all the time would have more restrictions on the bending radius than one that never moves. &lt;/p&gt;&#xA;" OwnerUserId="17410" LastEditorUserId="1832" LastEditDate="2018-09-06T00:15:17.727" LastActivityDate="2018-09-06T00:15:17.727" Title="Flex and Static Bend Radius of a Cable: Which one should be larger?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23627" PostTypeId="2" ParentId="23625" CreationDate="2018-09-05T20:47:48.437" Score="1" Body="&lt;p&gt;If you assume that your pipes are straight, your problem reduces to 2D circle packing as Mohammad suggests. But unlike what he says, using the smallest pipes wont get you the best packing density: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For seven of these radius ratios a compact packing is known that&#xA;  achieves the maximum possible packing fraction (above that of&#xA;  uniformly-sized discs) for mixtures of discs with that radius ratio.&#xA;  The highest packing density is 0.911627478 for a radius ratio of&#xA;  0.545151042.&#xA;  &lt;a href=&quot;https://en.wikipedia.org/wiki/Circle_packing&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Circle_packing&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, in order to write a meaningful algorithm you need to have a minimum diameter constraint, otherwise you will be able to fill the gaps with increasingly smaller tubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find better responses for this question on &lt;a href=&quot;http://math.stackexchange.com/&quot;&gt;math.stackexchange.com&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17409" LastEditorUserId="17409" LastEditDate="2018-09-06T12:55:52.933" LastActivityDate="2018-09-06T12:55:52.933" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23628" PostTypeId="2" ParentId="23623" CreationDate="2018-09-05T21:35:50.070" Score="1" Body="&lt;p&gt;What I found looking a many web sites : The continuous rails are usually in 1/4 mile ( USA) lengths made by welding together standard rails .They are hydraulically pretensioned for a particular average temperature/length depending on location ( in cold weather construction the rail may be heated to increase length). They require very good roadbed and they are firmly anchored to it so that the thermal expansion/contraction strain is held at each cross tie ( avoiding a cumulative effect of a few feet of movement at the 1/4 mile joint ). So they have a tensile strain that is greater in cold weather and may go negative in very hot weather. And, some railroads will make welds at the 1/4 mile joints.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-09-05T21:35:50.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23631" PostTypeId="1" CreationDate="2018-09-06T07:01:09.210" Score="0" ViewCount="455" Body="&lt;p&gt;I am trying to understand the underlying process of computer circuits from the beginning. I am reading about punch cards. I have come across several youtube videos and articles which are useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.reddit.com/r/explainlikeimfive/comments/1t4r0g/eli5_how_did_punch_cards_work_specifically_what/&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference article: How did punch cards work? Specifically, what is the reading mechanism?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain me back in time when punch cards were used. Punch cards with holes were entered into a reader and the holes lead to completion of circuit, how does completion of circuit helps to compute something? How was that circuit designed? Please refer some article to understand some examples about this.&lt;/p&gt;&#xA;" OwnerUserId="17418" LastEditorUserId="7083" LastEditDate="2018-10-13T21:36:48.573" LastActivityDate="2018-12-13T08:24:40.960" Title="How does punched card computer circuit design work?" Tags="|computer-engineering|computer-hardware|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23632" PostTypeId="1" CreationDate="2018-09-06T07:10:20.010" Score="1" ViewCount="401" Body="&lt;p&gt;In gas power cycles, one mentioned that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Carnot cycle is both externally and internally reversible.&#xA;  Therefore making the Carnot cycles more efficient than other practical&#xA;  cycles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;h2&gt;Question:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;How do external and internal irreversibilities affect the efficiency of a Thermodynamic cycle?&lt;/p&gt;&#xA;" OwnerUserId="17210" LastEditorUserId="9" LastEditDate="2018-09-06T21:36:30.237" LastActivityDate="2018-10-11T22:24:44.560" Title="How do external irreversibilities decrease the efficiency of thermodynamic cycles?" Tags="|thermodynamics|engines|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23633" PostTypeId="1" CreationDate="2018-09-06T08:03:50.357" Score="0" ViewCount="57" Body="&lt;p&gt;I need to measure a torque on an 8mm shaft of an electrical motor on a prototypical device. The expected torque is ~15 Nm and for the data processing I intend to use an Arduino. One requirement which excludes commercially available torque sensors, is the available installation space in the device as well as weight limits and the price. Due to the small diameter, I am afraid that the curvature of the shaft could be too high for the strain gauges - is there any recommendation or best practice how to handle that?&lt;/p&gt;&#xA;" OwnerUserId="17406" LastActivityDate="2018-09-07T09:38:41.580" Title="Can strain gauges radially be mounted on 8 mm diameter shaft for torque measurement for effective result?" Tags="|measurements|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23634" PostTypeId="2" ParentId="23633" CreationDate="2018-09-06T08:42:51.893" Score="1" Body="&lt;p&gt;Many off the shelf torque meters are basically just a socket extension bar with a strain sensor (or several) wrapped around the outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you quanitfy how much space you have available, with pictures?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;15 Nm is a fairly small amount of torque to measure through the deformation of a metal shaft.  One way to solve this and also to increase the diameter (that your strain sensor has to fit around) would be as follows...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add a hard-rubber coupling between your motor and the shaft it needs to drive.  The rubber will deform a lot more than the metal would do, thus increasing the sensor's effectiveness.  If you make the rubber a larger diameter (25mm perhaps if available space allows) then that should be less problematic for the strain sensor to wrap around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rubber shock absorbing engine mount (from just about any car) could probably be modified rather easily to this purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One downside of this is that it would introduce some backlash into your motor drive.  This can be reduced by using a harder rubber compound).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without photos of your device it's rather hard to imagine what would work for you.&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastEditorDisplayName="user6335" LastEditDate="2018-09-07T09:38:41.580" LastActivityDate="2018-09-07T09:38:41.580" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23635" PostTypeId="2" ParentId="23631" CreationDate="2018-09-06T09:45:22.400" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Punched_card&quot; rel=&quot;nofollow noreferrer&quot;&gt;Punched cards&lt;/a&gt;, like &lt;a href=&quot;https://en.wikipedia.org/wiki/Punched_tape&quot; rel=&quot;nofollow noreferrer&quot;&gt;punched paper tape&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_tape_data_storage&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetic tape&lt;/a&gt; &amp;amp; &lt;a href=&quot;https://en.wikipedia.org/wiki/Drum_memory&quot; rel=&quot;nofollow noreferrer&quot;&gt;magnetic drums&lt;/a&gt; were all forms of data storage in the early development of digital computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They served a similar function as magnetic hard drives, &lt;a href=&quot;https://en.wikipedia.org/wiki/Solid-state_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;solid state drives&lt;/a&gt;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Memory_card&quot; rel=&quot;nofollow noreferrer&quot;&gt;memory cards&lt;/a&gt; &amp;amp; &lt;a href=&quot;https://en.wikipedia.org/wiki/USB_flash_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;USB memory storage&lt;/a&gt; devices currently do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They store data and software (programs, applications) that can be accessed and used by computers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With punched cards, each card represented one line of information - analogous to one line of a text file, one line of code in a computer program or one line of data from a data file. Each column on the punched cards represented characters, or numbers, depending on which holes in the column were created.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most computer cards had holes numbered from 0 to 9 going down the card &amp;amp; holes 11 and 12 at the top of the card (hole 12 being the topmost hole) - particularly those that used the &lt;a href=&quot;http://www.columbia.edu/cu/computinghistory/029.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;DEC 029 code&lt;/a&gt; for punching holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With punched cards, only capital letters were used - due to limitations imposed by computer memory and how many &lt;a href=&quot;https://en.wikipedia.org/wiki/Bit&quot; rel=&quot;nofollow noreferrer&quot;&gt;bits&lt;/a&gt; were in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Byte&quot; rel=&quot;nofollow noreferrer&quot;&gt;byte&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the DEC 029 code, the letter A was designated as holes 12 &amp;amp; 1 in a column. The letter B was holes 12 &amp;amp; 2. The letter J was holes 11 &amp;amp; 1. For special characters such as + - = &amp;lt; &gt;, three holes would be punched in a column, such as 0 3 8 (me reviving long dormant brain cells).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For numbers, the number 1 was hole number 1 and number 5 was hole number 5, etc. To resent a number like 24, one column would have only the number 2 hole punched and in the adjacent column to the right, the number 4 would be punched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way punched card readers worked is each card would be placed against a shiny metal plate and the card was illuminated by a strong light. The metal exposed by the holes would reflect the light which would then be read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For people who did not have access to a punched card machine to punch the holes into the card, the holes could be manually marked with a HB pencil - preferably using a large dot filling most of the rectangle. The graphite in the pencil lead reflected light in a similar way as the metal backing when the cards had holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit - several weeks later&lt;/strong&gt;&#xA;Computer cards had eighty columns of holes, which is why a lot of old computer programs, from the 1960s and 1970s was limited to having eighty characters per line.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2018-12-13T08:24:40.960" LastActivityDate="2018-12-13T08:24:40.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23636" PostTypeId="2" ParentId="23623" CreationDate="2018-09-06T10:34:08.520" Score="0" Body="&lt;p&gt;To take it back to the basic engineering principle: a change in temperature of something changes its stress-free length. That is, the length it would be if there was no stress on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For gapless rails (I'd call them Continually Welded Rails or CWR in the UK), the position of the rail is restrained by the connection of the rail to the sleeper or the slab below it. So the length is unable to change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you change the stress-free length without changing the &lt;em&gt;actual&lt;/em&gt; length, you create stress. If your rail has got colder, it wants to get shorter, and this causes tensile stress; if your rail gets hotter it wants to get longer, and this causes compressive stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failure occurs when the rail supports can't provide enough restraint against this tensile or compressive stress. On curved track this will manifest itself as the curve radius decreasing under tensile stress / cold weather, or radius increasing under compressive stress / hot weather. For a straight section of track this usually only happens under compressive stress / hot weather, where the track will buckle, snaking left and right, which makes the path of the rail longer, hence relieving the compressive stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what prevents the track warping is friction between the track and its connection to the sleeper, and friction between the sleeper and the ballast. When too excessive a temperature change occurs, the failure point is normally the friction between the sleeper and the ballast.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note that as per &lt;a href=&quot;https://engineering.stackexchange.com/a/23624/255&quot;&gt;user190081&lt;/a&gt; and &lt;a href=&quot;https://engineering.stackexchange.com/a/23628/255&quot;&gt;blacksmith37&lt;/a&gt;'s answers, the stress-free length at installation may not be the distance between sleepers, as the rail may either be mechanically stretched or heated. This is as failure in hot weather is far more of a problem than failure in cold weather, so a distance between the sleepers equal to the stress-free length at a relatively high temperature makes failure less likely. There is obviously a limit to this though, as failure at cold temperature extremes is still physically possible.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2018-09-06T10:34:08.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23639" PostTypeId="1" CreationDate="2018-09-06T14:00:43.237" Score="1" ViewCount="31" Body="&lt;p&gt;I am currently planing to build a consrtuction for a simulator.&#xA;My simulator seat should be able to fall like 30cm and afterwards move to the starting position .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The falling should be as natural as possible. For that my actuator has to be really fast.&#xA;I am searching for a good soultion to lift up the seat with a person like up to 100kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about a linear actuator or a linear motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;About the linear monitor I know it is really fast and would probably provide a fast motion.&#xA;I wasn't sure, whether it can hold a heavy person at one position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas, suggestion and solution or links to good linear motor or linear actuator would be really hepful.&#xA;Thanks in advance :-)&lt;/p&gt;&#xA;" OwnerUserId="17423" LastActivityDate="2018-09-06T14:00:43.237" Title="Searching for a High Speed Actuator for a simulator" Tags="|hydraulics|building-design|actuator|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23640" PostTypeId="1" AcceptedAnswerId="23643" CreationDate="2018-09-06T14:24:37.237" Score="0" ViewCount="219" Body="&lt;p&gt;If someone were to invent a device that uses the hydrodynamic energy contained within the city water supply to generate enough electricity to reduce their monthly electric bill (however unlikely this would be), would it legal to do so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, are there any existing federal/state laws that specifically state that a home owner cannot use any device that generates electricity from the hydrodynamic energy contained within the city water supply?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an idea for such a device but I don't want to invest the time and money to construct it if say sometime in the future my local water company finds out about it and then sends one of their lawyers to my home informing me that I have to stop using it or face heavy fines.&lt;/p&gt;&#xA;" OwnerDisplayName="user17424" LastActivityDate="2018-09-06T17:25:03.153" Title="Is it illegal to use the hydrodynamic energy contained within the city water supply to generate electricity?" Tags="|mechanical-engineering|electrical-engineering|hydraulics|electrical|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="23641" PostTypeId="2" ParentId="21995" CreationDate="2018-09-06T14:24:38.850" Score="0" Body="&lt;p&gt;The formula presented neglects shear effects - in other words, follows Euler-Bernoulli beam theory.  For design purposes this is typically conservative to ensure something does not vibrate, but has the drawback that it does not predict longitudinal vibrations at all.  Without a shear modulus, it would not predict torsional vibrations either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically for long beams (about 10 times longer than they are wide), this model is accurate, and you can demonstrate what you want to.  In the y-direction, the model uses $E_y$, and in the x-direction, the model uses $E_x$ for classical lamination theory (CLT) models utilizing ABD matrices.  If both were equal, then this model would show they have the same value in both directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note there are many flaws to this model - if the beam were struck at an angle, how would it vibrate?  The model essentially would split the force into the two components and run the oscillations in both directions at the natural frequency of each, which is highly counterintuitive.  Surely one would dominate after energy dissipation removes the other, but how quickly?  For these and other questions for specialty design, the original Euler-Bernoulli beam theory wouldn't have answers, and neither would CLT based Euler-Bernoulli beam theory.  These would come back to the isotropic case of their original theory, but most likely would not come back to the same equation you presented.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-09-06T14:24:38.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23642" PostTypeId="2" ParentId="23640" CreationDate="2018-09-06T14:34:54.090" Score="1" Body="&lt;p&gt;Someone already came up with this idea - but working with the water companies and it was fitted as part of the pressure regulation system so that any over-pressure situations were controlled and something useful generated. Can't find a link at the moment though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for you doing it privately, depending on the country and legal statutes, you may well be charged with waste of resource, misappropriation of resource or somesuch. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-09-06T14:34:54.090" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23643" PostTypeId="2" ParentId="23640" CreationDate="2018-09-06T17:25:03.153" Score="3" Body="&lt;p&gt;Interesting problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s see, if we have a cylinder 12 inches in diameter and 8 feet high that would be &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(pi*(0.5f)^2*8f)*7.48 g/ft3=47 gallons per operation.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;At 3 GPM, that would be about 15 minutes to fill the cylinder all the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it were used to lift a 6780 pound weight, 8 feet up, that would be 8*6780 foot pounds of work, in 15 minutes.  That would be &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(8 feet*6780 pounds/(15 minutes*60 seconds/minute) = 60 foot pounds per second.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Converting that to watts, we get &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;60 fps*1.36w/fps=81.7 watts in 15 minutes.  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Converting that to watt hours, which is how the electric company bills, we get &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;81.7W*0.25 hours=20.4 WH or 0.0204 KWH.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where I live, we pay about $0.12 per KWH so you are generating &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;(0.0204KWH*$0.12/KWH)=$0.00254 of electricity.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To do that, you need 47 gallons of water.  In my area, we pay about $0.008 per gallon of water. That comes out to &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;47g*$0.008/g=$0.38.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you use this 100 times, you will use 38 dollars of water to generate 25 cents of electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if your water is free? In that case, if you run this continuously, it will cycle 96 times in 24 hours (15 minutes per cycle*24 hours) and use 96*47, about 4500 gallons of water to produce about enough to power to run a few LED light bulbs.  If you can get the equipment for free and nobody is going to notice if your water consumption goes up to like 135,000 gallons per month, go for it.   In the meantime, I suggest that if you start using that much tap water, somebody is going to notice and if it was not illegal when you started, it would be after a short time.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2018-09-06T17:25:03.153" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23644" PostTypeId="2" ParentId="23612" CreationDate="2018-09-06T18:00:56.127" Score="3" Body="&lt;p&gt;Thanks to @kamran for his answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I simulated the problem in ANSYS student v19 to verify his approach.&#xA;In the pictures below, the upper beam is solid, the middle one is split into two segments and the lower one is split into 3 segments.&#xA;Each segment is allowed to slide in respect to its neighbors. It is clear that the deflection of the 3 segments beam is 9 times the full one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f5O8C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f5O8C.png&quot; alt=&quot;beam deflection ANSYS simulation results - for the case of frictionless contacts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case the segments are bonded together (i.e cannot slip in respect to each other) - We get the same results for all the case. All the beam act like a full solid body in bending:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IJULq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IJULq.png&quot; alt=&quot;beam deflection ANSYS simulation results - for the case of bonded contacts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2018-09-06T18:00:56.127" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23645" PostTypeId="1" AcceptedAnswerId="23664" CreationDate="2018-09-06T18:34:09.333" Score="1" ViewCount="505" Body="&lt;p&gt;Can two adiabatic process paths intersect on a plot? This question is for 2 cases-&#xA;1.) Can two irreversible adiabatic curves intersect.&#xA;2.) Can a reversible &amp;amp; an irreversible adiabatic curve intersect?&lt;/p&gt;&#xA;&lt;p&gt;There are some answers for this question on Quora, but they mostly address the intersection of two reversible adiabatic processes.&lt;/p&gt;&#xA;&lt;p&gt;Also if the answer to either of the above 2 questions is 'yes', then I would like to know that if we do the proof given in the first answer of Quora link below with a reversible &amp;amp; an irreversible adiabatic process (or even two irreversible one) instead of two reversible adiabatic processes, it appears to be in violation of Kelvin Planck's statement. That is if we take an initial state and expand the system reversible adiabatically to a temperature 'T', now we take same system with same initial state but this time expand it irreversible adiabatically to same  temperature 'T' (but with different P &amp;amp; V ofcourse) &amp;amp; join these two final  states by an isothermal curve, we get a cycle with 3 processes (diagram in Quora link below, first answer). Now considering this cycle, we get a net positive work since it's area on a P-V diagram is positive but the system interact with surrounding only in 1 of the process, i.e the system recieves/rejects heat in only one of the processes, i.e the isothermal one. So it appears that we have a device which  produces work continuously in a cycle just by interacting with a single thermal reservoir thus violating Kelvin Planck's statement of the second law of thermodynamics. how is that possible?&lt;/p&gt;&#xA;&lt;p&gt;Also I want a physical interpretation of the processes that if they don't intersect then why don't and what it would be physically for a system.&lt;/p&gt;&#xA;&lt;p&gt;Here are answers on Quora-&#xA;&lt;a href=&quot;https://www.quora.com/Why-do-two-adiabatic-curves-never-cut-each-other&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.quora.com/Why-do-two-adiabatic-curves-never-cut-each-other&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17427" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-04-14T13:21:18.353" Title="Adiabatic paths" Tags="|thermodynamics|heat-transfer|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23647" PostTypeId="1" AcceptedAnswerId="23650" CreationDate="2018-09-06T19:22:37.270" Score="2" ViewCount="1154" Body="&lt;p&gt;I’ve been trying to figure out what this connector is and everyone has no clue. There isn’t enough identifying information to find anything on google. All I see is a label “lwai”. It’s located in a panel with power outlets, rj45 jacks, and some audio jacks. I think I found that it might be related to some hvac system but I can’t find any description or picture that confirms that. &lt;a href=&quot;https://i.stack.imgur.com/5kgSk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5kgSk.jpg&quot; alt=&quot;mystery connector&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17429" LastActivityDate="2018-09-07T09:43:33.740" Title="What is this connector" Tags="|outlet|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23648" PostTypeId="2" ParentId="23647" CreationDate="2018-09-06T19:37:30.273" Score="1" Body="&lt;p&gt;It looks like a connection for a trailer picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n5l6N.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5l6N.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-09-06T19:37:30.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23649" PostTypeId="1" CreationDate="2018-09-06T19:48:30.327" Score="1" ViewCount="1449" Body="&lt;p&gt;I have an exhaust that drones with the vehicle. However, outside, it's not a loud exhaust. So my assumption is the vibrations are transmitted into vehicle interior via the points of contact - in this case the rubber exhaust hangers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exhaust hangers are used fit very tight and are somewhat stretched, so is my theory correct - does the pre-stretch on the hangers allows more vibration transmission into the vehicle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xc3Hg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xc3Hg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6205" LastActivityDate="2018-10-07T02:00:19.873" Title="Does the stress on exhaust hanger amplify sound transmission into the vehicle chassis?" Tags="|acoustics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23650" PostTypeId="2" ParentId="23647" CreationDate="2018-09-06T20:04:18.173" Score="8" Body="&lt;p&gt;Looking at the keys in the socket, I don't think that it is meant for a trailer power cable. It seems to be a dock connector to supply a docked boat with power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cCm8a.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cCm8a.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture comes from &lt;a href=&quot;http://www.downwindmarine.com/Dock-Connectors-Hubbell-Plugs-and-Receptacles-120-208V-p-90891076.html&quot; rel=&quot;noreferrer&quot;&gt;this&lt;/a&gt; page.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2018-09-06T20:04:18.173" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23651" PostTypeId="1" AcceptedAnswerId="29930" CreationDate="2018-09-06T23:41:33.870" Score="3" ViewCount="8331" Body="&lt;p&gt;The crux of  a PID controller is:-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;output = Kp * error + Ki * integral + Kd * derivative&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So typically there is proportionality and a continuously varying (analogue) output.  In the case of a simple process like a domestic gas boiler, the boiler is either fully on or fully off.  Or, say a cooling fan that can only be switched on or off to keep something cold.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these cases, can a PID controller be used at all? There is still an error term.  Or would an alternative form of control be necessary?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2023-04-19T21:37:50.697" Title="Can a PID controller have a simple on/off output?" Tags="|pid-control|feedback-loop|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23652" PostTypeId="2" ParentId="23649" CreationDate="2018-09-07T01:41:26.567" Score="1" Body="&lt;p&gt;Consider a metal rod. You put your ear to one end. Someone &quot;pings&quot; the other. You will hear the ping thru the metal. Now add a wad of cotton balls at your end and repeat. Press your ear firmly on the cotton. The sound will be muffled if it comes thru at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When your vibrating exhaust is attached directly to the car, the car picks up and can amplify the sound. To avoid the problem, you must decouple the vibrations generated by the exhaust from reaching the car. You have two approaches. You can add the equivalent of a non-transmitting medium between the exhaust and the car body. Rubber acts as a vibration isolation medium. An appropriately designed spring can do the same. The other approach is to reduce the contact area between the exhaust and the car. By example, at the point where the exhaust runs under the car, hang it by a strong-enough wire rather than attaching it with clamps to the car body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the theory. The best practice is likely found by researching trade journals or chatting with folks from car shops that do the installations.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-09-07T01:41:26.567" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23653" PostTypeId="2" ParentId="23647" CreationDate="2018-09-07T03:31:44.327" Score="2" Body="&lt;p&gt;That seems to be an industrial plug power, I saw something similar in a milk factory. That can provide highter voltage than a regular plug.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Industrial_and_multiphase_power_plugs_and_sockets&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Industrial_and_multiphase_power_plugs_and_sockets&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17430" LastActivityDate="2018-09-07T03:31:44.327" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23655" PostTypeId="2" ParentId="23647" CreationDate="2018-09-07T04:59:51.043" Score="3" Body="&lt;p&gt;It's an IEC pin and sleeve connector, based on color, pins and ground position it should be 3 phase 120/208V.  Google iec pin and sleeve and you'll find mfr guides.&lt;/p&gt;&#xA;" OwnerUserId="17431" LastActivityDate="2018-09-07T04:59:51.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23656" PostTypeId="2" ParentId="23651" CreationDate="2018-09-07T06:03:49.110" Score="0" Body="&lt;p&gt;Yes, a PID controller does not know of the system it is controlling. It is perfectly fine to turn the control signal to binary on/off that is beyond the scope of the controller itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the system does is unknown to the PID. There might be some lag, some predictive component or anything in the system between it and the controller. So even though its a PID controller does not mean it has to controll what you think is obvious. Hell, the PID may be inside or control another controller**.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, the description of PID allows for many implementations to occur. The formula above is more a abstraction instrument, to categorize rather than a this is how you implement a PID*. A lot of things are PID controllers even though they may do the terms differently than the formula says. For example the integrative term may just be a history dependent term, nowhere does it say it has to integrate the error over the lifespan of the machine. Instead it may be a time limited slice too implemented like say a running average. The values may be capped and have some of the later systems features reduced into them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* Although, since you can abstract to this then you might as well use it as a guide to implement. But then many things are PID controllers implicitly like spring/shock absorber systems and hydraulic pistons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;** So while one can be tuning a PID, does not mean the entire overall controller is a PID&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-09-07T06:15:24.467" LastActivityDate="2018-09-07T06:15:24.467" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23657" PostTypeId="2" ParentId="23079" CreationDate="2018-09-07T07:16:51.453" Score="1" Body="&lt;p&gt;get you a battery-operated digital (i.e.,electronic) bathroom scale at a goodwill store. Disassemble it; you will find several compressive strain gauges inside along with the associated electronics. the electronics are cleverly designed to determine the weight of the object on the platform even if that weight is off-center, by artfully polling each sensor , A-to-D converting each, and feeding the results into a clever algorithm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to hack this into something that might suit your purpose. Of course, the easy way would be to buy one bathroom scale for each leg and then hack the display driver on each to extract a decimal value for the weight based on which segments in the display are energized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;whatever you do, have fun! &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-07T07:16:51.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23658" PostTypeId="1" AcceptedAnswerId="23660" CreationDate="2018-09-07T08:51:52.007" Score="0" ViewCount="69" Body="&lt;p&gt;now I'm working on a project which involves a dc motor and high voltage. I'm developing a machine consists of a motor to rotate a platform disc and high voltage will be applied during the rotation. The problem is when I supply a positive high voltage at motor shaft during rotation, the motor stop working. Do you guys have any idea, how I can supply the high voltage properly the high voltage supply?&lt;a href=&quot;https://i.stack.imgur.com/eq3fh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eq3fh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1368" LastActivityDate="2018-09-09T08:43:02.590" Title="how to avoid a clash of magnetic charge and electric charge" Tags="|mechanical-engineering|electrical-engineering|motors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23659" PostTypeId="2" ParentId="23647" CreationDate="2018-09-07T09:43:33.740" Score="3" Body="&lt;p&gt;This is a 3 phase power connector with neutral and earth according to the IEC 60309. It's made for 120V/208-144V/250V 50-60Hz Systems. The contact in the top should be ground. The contact in the center could be a pilot contact which might indicate a 63A connector&lt;/p&gt;&#xA;" OwnerUserId="17433" LastActivityDate="2018-09-07T09:43:33.740" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23660" PostTypeId="2" ParentId="23658" CreationDate="2018-09-07T09:59:47.553" Score="2" Body="&lt;p&gt;Don't apply the high voltage to the motor shaft, isolate the disc and only apply the high voltage to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit based on comments:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;isolate ie insulate, electrically separate the disc from the motor shaft while keeping the mechanical drive to make it rotate. Slip rings can be used to apply the voltage to the, now insulated, disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here for an explanation of slip rings: &lt;a href=&quot;https://en.wikipedia.org/wiki/Slip_ring&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Slip_ring&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-09-09T08:43:02.590" LastActivityDate="2018-09-09T08:43:02.590" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="23661" PostTypeId="1" AcceptedAnswerId="23662" CreationDate="2018-09-07T14:42:06.660" Score="0" ViewCount="69" Body="&lt;p&gt;3d printing is a slow process and for my solution I need fast printing. So I print with no infill and I fill inside with a two part, 10 min drying polyurethane resin. I mix them in a disposable cup and use a syringe to put it inside. It gives super results as it is both sturdy and lightweight so I recommend this method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that if the print has a high volume, the mixture begins solidifying inside syringe. So I need a device which can mix parts and apply it on the spot. Below is a crude drawing of what I need. Is there a name for such a device readily avaible or will I have to build it myself? Or any other advise is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OBkvg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OBkvg.jpg&quot; alt=&quot;Mixer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have tried so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If I use a slower drying formula then most of the solution leaks from small holes in printed object. In faster drying formulas it is managable.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16896" LastEditorUserId="16896" LastEditDate="2018-09-07T16:31:18.633" LastActivityDate="2018-09-08T02:17:59.660" Title="Polyurethane mix-on-the-spot system" Tags="|fluid-mechanics|applied-mechanics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23662" PostTypeId="2" ParentId="23661" CreationDate="2018-09-07T19:44:33.400" Score="0" Body="&lt;p&gt;there is a device called a mixing tube or mixing nozzle which is used with industrial-duty epoxy dispensing systems. it clips onto the dual-nozzle outlets of a pair of syringe tubes that are molded together into a single unit and filled with the 2-part glue composition of your choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This assembly then snaps into a hand-operated ratchet dispenser which forces the two parts into the mixing tube, which contains a series of twisted baffles that fold together the two substances so by the time they emerge from the other end of the tube, they are thoroughly mixed and ready to apply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the tube is a single-use device as there is no way to clean out the mixed epoxy from the baffles and re-use it. However, the way you use it is by staging up a whole series of things to be glued and then dispensing the mixed glue until there is none left in the syringe tubes- at which point you remove the empty tubes and the mixing nozzle and throw them all away. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-07T19:44:33.400" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23663" PostTypeId="1" AcceptedAnswerId="25614" CreationDate="2018-09-07T22:23:47.623" Score="3" ViewCount="381" Body="&lt;p&gt;I have a medical device I am designing which will lie under the patient bed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to add a &quot;connector/joint&quot; on this device to permit the attachment of some consumable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a quick overview of the process:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cAC3s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cAC3s.png&quot; alt=&quot;basic assembly schema&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Blue part is low cost injected plastic disposable part that will be assembled and disassembled only 2-3 times before being disposed.&lt;/li&gt;&#xA;&lt;li&gt;Orange part is the device, can contain more complex parts (springs...).&lt;/li&gt;&#xA;&lt;li&gt;Orange part must last thousands of assembly/disassembly.&lt;/li&gt;&#xA;&lt;li&gt;Orange part is not movable, only blue part can be moved when assembling.&lt;/li&gt;&#xA;&lt;li&gt;When assembled, it must resist to at least 100N force in all direction, including rotation around the &quot;snap axis&quot;.&lt;/li&gt;&#xA;&lt;li&gt;Orange part must be around 3cm which is a good side to apply pressure with two fingers.&lt;/li&gt;&#xA;&lt;li&gt;Disassembly pressure force should be low as only two fingers will be use, other hand will retain blue part.&lt;/li&gt;&#xA;&lt;li&gt;It should be easy to assemble/disassemble with &quot;closed eyes&quot; as the device will be under a bed and there might be no visibility on the device itself.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As my drawing might be a bit confusing, here is a common device that is similar (aka a buckle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FJ1rE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FJ1rE.jpg&quot; alt=&quot;buckle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem with buckle design is the require pressure force for disassembly which is way too high, it's also too hard to assemble and not robust enough. But it is a good example for the &quot;general principle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am looking for is not for someone to design the assembly for me, but I am looking for exiting systems that are an alternative to a snap in joint and any pointer that would help me design it.&lt;/p&gt;&#xA;" OwnerUserId="17438" LastEditorDisplayName="user16" LastEditDate="2018-09-08T00:11:59.860" LastActivityDate="2019-01-20T03:44:12.073" Title="Overcoming limitations of a snap in joint" Tags="|mechanical-engineering|joining|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23664" PostTypeId="2" ParentId="23645" CreationDate="2018-09-08T01:49:11.920" Score="2" Body="&lt;p&gt;Intuition should tell you that it is possible because irreversible paths can result in an infinite array of end points sharing the same volume (or pressure or temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this: You have 2 identical systems. For the first system, reversibly compress the system to $\frac{1}{2}$ its volume. For the second system, &lt;em&gt;irreversibly&lt;/em&gt; compress the system to $\frac{1}{2}$ its volume. Intuitively it should made sense that you have imparted more energy into the system in the irreversible case, and thus, it should be at a greater temperature and/or pressure than the second system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, we can show this is true as well. I will use $$PV_m=RT$$ to represent the ideal gas system, where $V_m$ is the molar volume. $c_v = 3/2R$ for a monatomic gas (a simply explanation of this is Degrees of Freedom of a monatomic gas; see equipartition theorem for more information). Also, for simplicity, I will assume $n=1$, so I will use $V$ in place of $V_m$ for the rest of the derivation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us consider the following system: You 2 identical systems with $1$ $mol$ of a monatomic ideal gas with volume $1 m^3$ and pressure of $1$ $Pa$. It follows that the starting temperature of this system is $\frac{1}{R}K$. I will consider the compression to $\frac{1}{2}m^3$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Reversible&lt;/em&gt;&lt;/strong&gt;: Let's begin with the more restricted reversible case. Since $q=0$ we have $\Delta U=w$. We know pressure volume work is given by $w=-\int_{1}^{1/2} P dV$, but $P = \frac{RT}{V}$ and we can flip the integral bounds by multiplying by $-1$, giving $w=\int_{1/2}^{1} \frac{RT}{V} dV$. We also know (can easily derive) $dU = c_v dT$ for an ideal gas. This implies $\Delta U = \int_{T_i}^{T_f} c_v dT$. We then have $\int_{T_i}^{T_f} c_v dT = \int_{1/2}^{1} \frac{RT}{V} dV$. We can then &quot;carefully&quot; rearrange the equation to $\int_{T_i}^{T_f} \frac{c_v}{T} dT = \int_{1/2}^{1} \frac{R}{V} dV$, which gives $c_v \ln(\frac{T_f}{T_i}) = R \ln(\frac{1}{\frac{1}{2}})$. (I say &quot;carefully&quot; because such a rearrangement of dividing within integrals isn't always allowed, though I've never come across a case in a science where it is not allowed to be done. The theorem regarding it's use, however, is not trivial at all. To finish the problem up, we rearrange to obtain $\ln (\frac{T_f}{T_i}) = \ln (2^{\frac{R}{c_v}})$, raise both sides to the power of $e$, and solve $T_f = T_i \cdot 2^{\frac{R}{c_v}} = \frac{1}{R} \cdot 2^{\frac{2}{3}}$. Thus $T_f \approx 0.191K$. From the ideal gas law, we get $P \approx 3.175$ $Pa$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Irreversible&lt;/em&gt;&lt;/strong&gt;: For the irreversible case, we will undergo a 1-step compression by applying a constant pressure on the system. Since the system is still adiabatic, $q=0$ and $\Delta U=w$ still hold true. Expanding $\Delta U$ in the same way as above, we get $\int_{T_i}^{T_f} c_v dT = -\int_{1}^{1/2} P dV$; however since $P$ is now constant, we can easily simplify the equation down to $c_v \cdot (T_f-T_i) = P \cdot (1-\frac{1}{2})$. We don't know the $T_f$ of this path, so we replace $T_f$ with $\frac{PV_f}{R}$ by the ideal gas law. After distributing, we see that $P \cdot \frac{1}{2} = \frac{Pc_vV_f}{R} - c_vT_i$. We can then factor $P$ out and rearrange to get $P = \frac{c_vT_i}{\frac{c_vV_f}{R}-\frac{1}{2}}$. Plugging in numbers, we see that $P = \frac{\frac{3}{2}}{\frac{3}{2} \cdot \frac{1}{2} - \frac{1}{2}} = 6$ $Pa$. It follows that $T_f = 3 \cdot Ti \approx 0.361K$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, although these two systems started at the same place, they clearly diverged from each other, ending at a different pressure and temperature for the same volume. You can show this is true for expansion in the same manner. It follows naturally then, that you can have two systems cross in the middle of a path. Please check my math, as I may have made an error trying to format my thoughts into mathjax. Regardless, only reversible paths (of the same process) cannot intersect with each other. Irreversible paths can intersect with other irreversible paths (you can show this in a similar manner as with reversible paths) or with reversible paths. I hope this clears up the &quot;impossibility&quot; of  system paths crossing each other.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: The Kelvin-Planck statement is essentially a consequence (possibly an equivalent statement) of the 2$^{nd}$ Law of Thermo. It essentially says that the energy you put into a system cannot (as a consequence of probability) be entirely converted to [macroscopic] work. In other words, if you heat a system $100J$, the system cannot perform $100J$ of work. Why? Because this would mean that $\Delta S_{univ}$ does not increase. Why? In a cycle, the system will return to its original state ($\Delta S = 0$), and the entropy lost by the heat reservoir is $\frac{q}{T_{res}}$. This means the 2$^{nd}$ external body (the one you want to do work on) must have $\Delta S  \geq \frac{q}{T_res}$. If all the heat gained from the reservoir were converted to energy, however, $\Delta S  \ngeq \frac{q}{T_res}$, as such a system would have q = 0. Thus, a functioning heat engine cycle would need a second heat sink (a lower temperature heat reservoir) to restore the cycle to its initial state.  (Think about why a heat “sink” of the same temperature would cause the system to perform no work.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the significance of the Kelvin-Planck statement? &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It tells us that if the energy &lt;em&gt;imparted into&lt;/em&gt; a cycle is derived from &lt;em&gt;heat&lt;/em&gt; (such as from an external heat source reservoir), a second heat sink/reservoir is needed to restore the initial state of the cycle. What this &lt;em&gt;does not&lt;/em&gt; say, however, is that you cannot have &lt;em&gt;only one heat reservoir&lt;/em&gt; in a cycle. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is possible for a cycle to have only one heat reservoir in its path – it just must be a heat sink (absorbs heat from the cycle). This heat lost by the system to the heat sink must then be provided to the system from an isentropic* source (such as an adiabatic expansion/compression). In other words, the logic argued on Quora is flawed, or incomplete at best. (I find quora and Chegg are flawed all the time… Take everything you read/hear/etc. with a grain of salt. Think about whether or not what someone says makes sense with what you think.) In fact, if you complete the scenario I presented as a system such that you have 3 steps: 1. Irreversible adiabatic compression ($1m^3$ to $\frac{1}{2}m^3$; 2. Constant volume cooling ($.361K$ to $.191K$); 3. Reversible adiabatic expansion ($\frac{1}{2}m^3$ to $1m^3$), you can see that it indeed does not violate the 2$^{nd}$ Law of Thermo (the Kelvin-Planck statement). In fact, for one full cycle, $\Delta S_{univ} \approx 0+7.938+3.16 = 11.098$ $J/K$, a clearly positive value (assuming I didn't mess up my math).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the Kelvin-Planck statement often stated and misinterpreted as “a cycle can’t must have more than one heat reservoir”? This is because there’s little reason in applying work to a system to extract less work from the system (while this might be needed occasionally, there are better ways of going about this than a heat engine). Say you have a piston and you put in 200kJ of energy to compress the gas, the cylinder is then used to spin a turbine with 150kJ of energy. Why not just turn the turbine directly? (Obviously, real world scenarios are not this simple, but I think this gets the point across.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I say &lt;em&gt;isentropic&lt;/em&gt;, but I really mean any process that would decrease the entropy its source less than the entropy gained by the the sink. It is safe to ignore the system since you are considering a cycle. Note that while an irreversible adiabatic compression is not an isentropic process, the entropy of the source does not change.&lt;/p&gt;&#xA;" OwnerUserId="17439" LastEditorUserId="17439" LastEditDate="2018-09-08T21:00:06.533" LastActivityDate="2018-09-08T21:00:06.533" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="23665" PostTypeId="1" CreationDate="2018-09-08T02:09:45.477" Score="2" ViewCount="106" Body="&lt;p&gt;I need a -30 C freezer. Inside volume 5-30 l. I've seen videos of double Peltier stacks reaching -50 C with one Peltier running at 12V and the other at 5 or 7 V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How big a freezer could such a stack cool to -30 C? Either a cube with up to, say, 5 cm thick expanded/extruded polystyrene (EPS/XPS) walls, or a wide neck Dewar with EPS/XPS lid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot side cooling is probably with a waterblock, pump and heat exchanger with fan. T-hot ... does 40 C at 30 C ambient sound realistic?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking of something like standard TEC1-12709 or -06 modules (12 V, 6/9 A, 40x40 mm), if they do the job. Local ones are kinda expensive (20-30 EUR). Aliexpress has some for a few USD. Are they ok?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found some math to calculate Peltiers (&lt;a href=&quot;https://www.overclockers.com/peltier-stacking/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.overclockers.com/peltier-stacking/&lt;/a&gt;), but I was a bit overwhelmed (or maybe just lazy ;). EPS/EPX seems to have about 0.035 W/(m·K) thermal conductivity. I haven't found (or researched) Dewar thermal conductivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The freezer should reach -30 C in a finite time, e.g. 30 min, so remove a bit more heat at -30 C as is flowing in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you have experience I'd appreciate estimates. Calculations would be cool, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compressor coolers or Sterling cryocoolers are probably not an option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="17426" LastActivityDate="2018-09-08T02:09:45.477" Title="How much can a double Peltier stack cool to -30 C?" Tags="|thermodynamics|heat-transfer|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23666" PostTypeId="2" ParentId="23661" CreationDate="2018-09-08T02:17:59.660" Score="0" Body="&lt;p&gt;in the construction industry, there are many companies which sell two component epoxy and its applicator. which happens to be a double cylinder pump with a mixing nozzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These pumps do exactly what you need, they mix the two components at the final stage in a mixing nozzle. Some are hand operated, some have work with air pressure and some are electrical.&#xA;I add the battery operated epoxy gun by &quot;Read Head&quot; Co. &lt;a href=&quot;https://i.stack.imgur.com/L2mH8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L2mH8.jpg&quot; alt=&quot;here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nozzle is a plastic cone with spiraling inside blades designed to mix the two components and deliver to the fasteners.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-09-08T02:17:59.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23667" PostTypeId="1" AcceptedAnswerId="23669" CreationDate="2018-09-08T02:25:57.647" Score="2" ViewCount="138" Body="&lt;p&gt;A couple of books on elasticity, e.g. Gould and Soutas-Little, in the section on 2d elasticity in cylindrical coordinates, mention the case of &lt;em&gt;'quasi-axisymmetric'&lt;/em&gt; problems. These are problems where the stress and strain retain and axial symmetry (i.e., no dependence on the angular coordinate $\theta$), whereas the displacements $u_r$ and $u_\theta$ may carry $\theta$ dependence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, on writing the expression for $u_\theta$:&#xA;$$&#xA;u_\theta = A*r\theta + B*\cos\theta + C*\sin\theta + D*r&#xA;$$&#xA;with A,B,C,D constants, both mention that the last three terms rigid body motions, and hence do not contribute to any strain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This statement per se I understand. Rigid body motions cannot give rise to strains or stresses. However, while I can imagine that the $\cos\theta$ and $\sin\theta$ represent rotations, I do not see how the $r$ term cannot give rise to strains! If $u_\theta$ is (say) zero at the inner radius and finite at the outer radius, it must surely give rise to strains! What is going on? What am I misunderstanding?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Addition to answer below&lt;/strong&gt; --&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see my comment to the answer for an explanation of the $r\theta$ term as well.&lt;/p&gt;&#xA;" OwnerUserId="17440" LastEditorUserId="17440" LastEditDate="2018-09-10T10:36:50.620" LastActivityDate="2018-09-10T10:36:50.620" Title="Meaning of quasi-axisymmetric case in 2D elasticity" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23669" PostTypeId="2" ParentId="23667" CreationDate="2018-09-08T08:31:32.270" Score="1" Body="&lt;p&gt;Consider a (small) rigid rotation $\alpha$ about the axis of the cylindrical coordinate system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The displacement of a point at radius $r$ from the axis is&#xA;$$\begin{align} u_r &amp;amp;= 0 \\ u_\theta &amp;amp;= \alpha r. \end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's where your $D \star r$ term comes from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The $\cos\theta$ and $\sin\theta$ terms come from rigid body &lt;em&gt;translations&lt;/em&gt; in the $\theta = 0$ and $\theta = \pi/2$ directions, not from rotations. For example, consider a (small) rigid translation $a$ in the $\theta =0$ direction. Describing this is cylindrical coordinates gives&#xA;$$\begin{align} u_r &amp;amp;= a\cos\theta \\ u_\theta &amp;amp;= a\sin\theta. \end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rigid translation in the $\theta = \pi/2$ direction is similar, but with the $\cos$ and $\sin$ terms interchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the $A * r\theta$ term looks wrong. It doesn't make any sense, because it doesn't even represent a continuous displacement field. The values of $u_\theta$ when $\theta = 0$ and $\theta = 2\pi$ must be the same if the deformed shape of the structure &quot;fits together&quot; properly, but they are not!&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-09-08T08:38:33.307" LastActivityDate="2018-09-08T08:38:33.307" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23670" PostTypeId="1" CreationDate="2018-09-08T09:05:46.990" Score="1" ViewCount="315" Body="&lt;p&gt;Let´s pretend there is a source of homogenous noise. Up to which decibel level could active noise cancelling be viable to decrease the volume of this noise significantly? Is it like 10, 50, 100 dB?&lt;/p&gt;&#xA;" OwnerUserId="17444" LastEditorDisplayName="user16" LastEditDate="2018-09-08T12:30:14.563" LastActivityDate="2018-09-08T12:30:14.563" Title="What are the limitations of active noise cancelling?" Tags="|mechanical-engineering|sound-isolation|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23672" PostTypeId="1" CreationDate="2018-09-08T14:37:49.833" Score="1" ViewCount="116" Body="&lt;p&gt;Microwave ovens use oscillating electric field of microwaves to heat up the food. Consider following design of an oven, which runs on electric field, one may call it field oven. The oven is a rectangular box similar to microwave oven. However, the opening is not in the front but on the top in the form of a hole or cavity, not a door. Think of a 15 -20 cm dia of hole on top of box. the cookware sits in this hole, such that the brim of the cookware extends few cm above the top surface of the hole. material of cookware is chosen to suit the application. Front and back sides are parallel plates of a capacitor. A high frequency say few ghz or 100s of mhz of AC is applied to the capacitor plates. The required electronics is attached outside the box. Hence, in the oven (the box) high frequency AC electric field is generated. 1) Will this work for cooking the food? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the difference is that it uses AC electric field instead of em waves for cooking. Some em waves will be produced as by product because of AC source, it can be contained or directed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Microwave ovens suffer from skin depth issues, hence the cooking happens only few cm from surface. Will this new design also suffer from skin effect?&lt;/p&gt;&#xA;" OwnerUserId="17373" LastEditorUserId="17373" LastEditDate="2018-09-09T06:34:07.873" LastActivityDate="2018-09-09T22:38:11.543" Title="Electric field oven" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23673" PostTypeId="2" ParentId="23645" CreationDate="2018-09-08T14:41:10.503" Score="0" Body="&lt;p&gt;A simple approach considers only the first law for a system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;dU = \delta q + \delta w&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Internal energy change &lt;span class=&quot;math-container&quot;&gt;$dU$&lt;/span&gt; is caused by heat flow &lt;span class=&quot;math-container&quot;&gt;$\delta q$&lt;/span&gt; and work flow &lt;span class=&quot;math-container&quot;&gt;$\delta w$&lt;/span&gt; across a boundary between the system and surroundings. Heat leaving the system (an exothermic process) decreases internal energy. Work done by the system leaves the system (and is negative), thereby also decreasing the internal energy. Recall that &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; is a state function. For a closed system, it depends only on &lt;span class=&quot;math-container&quot;&gt;$T, V$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reversible process allows that any step that increases &lt;span class=&quot;math-container&quot;&gt;$dU$&lt;/span&gt; can be identically and exactly reversed to decrease &lt;span class=&quot;math-container&quot;&gt;$dU$&lt;/span&gt; just by changing the signs of &lt;span class=&quot;math-container&quot;&gt;$\delta q$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\delta w$&lt;/span&gt;. So, when heat leaves and work enters to increase &lt;span class=&quot;math-container&quot;&gt;$dU$&lt;/span&gt;, the reversible step is to have &lt;strong&gt;exactly&lt;/strong&gt; the same heat enter as left and &lt;strong&gt;exactly&lt;/strong&gt; the same work leave as entered. These heat and work flows are &lt;strong&gt;always&lt;/strong&gt; across a boundary between the system and surroundings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An irreversible process has an additional heat flow &lt;span class=&quot;math-container&quot;&gt;$\delta q_{irr}$&lt;/span&gt; that always leaves the system. That heat flow can be due to friction or internal interactions in the material in the system. That heat flow is &lt;strong&gt;never&lt;/strong&gt; returned during the reverse path. It also does NOT flow (leave or enter) through the boundary between the system and surroundings. It is either directly absorbed in the system or surroundings (as a consequence of having internal interactions of the fundamental particles -- i.e. as a consequence of using a &quot;real&quot; substance rather than an ideal gas) or it is absorbed at the boundary (i.e. as friction). Because we loose f &lt;span class=&quot;math-container&quot;&gt;$q_{irr}$&lt;/span&gt;, to return to the same internal energy &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; as the starting point after taking an irreversible step, we must supply more work during the irreversible path than during the reversible path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An adiabatic process has no &lt;em&gt;reversible&lt;/em&gt; heat flow. A reversible adiabatic process has no heat flow. An irreversible has irreversible heat flow &lt;span class=&quot;math-container&quot;&gt;$\delta q_{irr}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work done &lt;em&gt;by the system&lt;/em&gt; during a compression or expansion process is &lt;span class=&quot;math-container&quot;&gt;$\delta w = -p_{ext} dV$&lt;/span&gt;. We can allow that &lt;span class=&quot;math-container&quot;&gt;$p_{int} = p_{ext}$&lt;/span&gt; during the process regardless that it is reversible or irreversible (reversible processes must have &lt;span class=&quot;math-container&quot;&gt;$p_{int} = p_{ext}$&lt;/span&gt; by definition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a graph of &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, work is the area under the curve. During a reversible, adiabatic process, that area will be &lt;span class=&quot;math-container&quot;&gt;$dU$&lt;/span&gt; &lt;strong&gt;exactly&lt;/strong&gt;. During an irreversible, adiabatic process, that area will be &lt;span class=&quot;math-container&quot;&gt;$dU \pm \delta q_{irr}$&lt;/span&gt;. When the two curves are drawn on a graph, they can intersect either at one common starting point or at one common ending point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the total work done during a reversible, adiabatic process is never the same as the total work done during an irreversible, adiabatic process that either starts or ends at the same set of &lt;span class=&quot;math-container&quot;&gt;$T, V$&lt;/span&gt; conditions (i.e. starts or ends at the same &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;). Therefore, adiabatic curves for the two different processes, reversible and irreversible, will diverge from a common starting point or intersect at a common end point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any concerns about whether this violates the Kelvin-Planck form of the second law cannot be addressed here. This discussion is about the intersection of individual paths. The Kelvin-Planck statement is for cycles.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-04-14T13:21:18.353" LastActivityDate="2019-04-14T13:21:18.353" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23674" PostTypeId="1" CreationDate="2018-09-08T15:16:34.387" Score="0" ViewCount="100" Body="&lt;p&gt;I am looking to purchase a fan with high velocity. I am trying to make solid data driven decision based on specs.&#xA;I found that most fan's of the 20&amp;quot; diameter variety are around 3000 CFM; however found one that is stated at 6000 CFM  (2.83 m^3/s) with 162 watt and 1450 RPM.&#xA;Link to fan:&#xA;&lt;a href=&quot;https://www.homedepot.com/p/Commercial-Electric-20-in-3-Speed-High-Velocity-Floor-Fan-SFC1-500B/302845111&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.homedepot.com/p/Commercial-Electric-20-in-3-Speed-High-Velocity-Floor-Fan-SFC1-500B/302845111&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I am using equation efficiency = dP*flow rate/Power and using 50 Pa dP (not sure what is proper value there, but it is giving me reasonable efficiency ratings so I normalized to that) I am coming to an efficiency of 87% for this fan.&lt;/p&gt;&#xA;&lt;p&gt;It seems like the rest of marketplace is more like 30 to 50% efficiency (again using the assumed dP of 50 Pa).&lt;/p&gt;&#xA;&lt;p&gt;Can someone help me understand if this is possible efficiency for a floor fan or too good to be true? Am I using correct equations and dP estimate. This is not my area of expertise by any stretch, but from what I have found the efficiency seems implausible.&lt;/p&gt;&#xA;&lt;p&gt;Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="17446" LastEditorUserId="3505" LastEditDate="2021-03-22T09:59:56.853" LastActivityDate="2024-04-05T16:08:11.040" Title="Calculating typical floor fan efficiency or dP? (Spec sheet too good to be true?)" Tags="|mechanical-engineering|fluid-mechanics|airflow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23675" PostTypeId="2" ParentId="23663" CreationDate="2018-09-08T15:31:45.820" Score="0" Body="&lt;p&gt;For a circular attachment, then a groove and matching balls held with a spring loaded collar. Used on brake lines for trailer and also much higher pressure/flow rates in hydraulic fittings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from &lt;a href=&quot;https://hydraulicmegastore.com/product-category/hydraulic-products/quick-release/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://hydraulicmegastore.com/product-category/hydraulic-products/quick-release/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uQMFw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uQMFw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-09-08T20:55:39.503" LastActivityDate="2018-09-08T20:55:39.503" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23676" PostTypeId="1" CreationDate="2018-09-08T18:33:16.167" Score="0" ViewCount="2862" Body="&lt;p&gt;I need help in understanding the connection between Nyquist plot and root locus plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have a Nyquist plot, what can I infer about the root locus plot? and vise versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the root locus plot, can I draw the Nyquist plot?&lt;/p&gt;&#xA;" OwnerUserId="17447" LastEditorUserId="683" LastEditDate="2018-09-08T19:07:10.833" LastActivityDate="2018-09-10T17:53:46.523" Title="Control systems, Nyquist plot and root locus" Tags="|mechanical-engineering|control-theory|stability|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23678" PostTypeId="1" AcceptedAnswerId="23679" CreationDate="2018-09-08T23:42:48.040" Score="1" ViewCount="159" Body="&lt;p&gt;I am currently building myself a &quot;hobby&quot; vacuum chamber from 1 inch tickness acrylic sheets (I am very aware that acrylic is not the proper material). Inner dimensions of the the chamber is 40x40x40mm. After a few simulations in Solidworks, safety factor and displacements seemed okay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I need is to feed 8 cables through the sheets. After a research, I couldn't find any airtight cable gland or any low cost vacuum feedthrough. What is your suggestion for a low cost feedthough?&lt;/p&gt;&#xA;" OwnerUserId="17452" LastEditorUserId="9" LastEditDate="2018-09-09T01:03:13.227" LastActivityDate="2018-09-09T09:11:44.413" Title="Low cost feedthrough for an acrylic vacuum chamber" Tags="|vacuum|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23679" PostTypeId="2" ParentId="23678" CreationDate="2018-09-09T04:31:11.883" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://www.heyco.com/Liquid_Tight_Cordgrips/product.cfm?product=Liquid-Tight-Cordgrips-Pre-Assembled&amp;amp;section=Liquid_Tight_Cordgrips&quot; rel=&quot;nofollow noreferrer&quot;&gt;Liquid tight cable glans&lt;/a&gt; are cheap, and many manufacturers make them.  If you have a productive enough vacuum pump, it may be able to keep up with small air leakage through them.  You could also add grease inside the gland.&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2018-09-09T06:02:27.013" LastActivityDate="2018-09-09T06:02:27.013" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23680" PostTypeId="2" ParentId="23678" CreationDate="2018-09-09T09:05:16.940" Score="1" Body="&lt;p&gt;A sparkplug is air tight and provides an electrical feedthru for very high voltages.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They're also threaded and taps for the thread are common. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also they are cheap!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The body of the plug is resistive  ( between about 40 $\Omega$ and 1 k$\Omega$) and sometimes inductive - check the specs)&lt;/p&gt;&#xA;" OwnerUserId="16544" LastEditorUserId="16544" LastEditDate="2018-09-09T09:11:44.413" LastActivityDate="2018-09-09T09:11:44.413" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23681" PostTypeId="2" ParentId="23591" CreationDate="2018-09-09T10:42:26.193" Score="0" Body="&lt;p&gt;In this answer stability is covered first, thereafter the effect of the PID-controller on the stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the following simple control system&#xA;&lt;img src=&quot;https://i.stack.imgur.com/OvINx.png&quot; width=&quot;400&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has a closed loop transfer function of $$\Gamma(s)= \frac{C(s)G(s)}{1+C(s)G(s)}=\frac{L(s)}{1+L(s)}.$$&#xA;Here $L(s) = C(s)G(s)$  is called the loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is &lt;strong&gt;unstable&lt;/strong&gt; if at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|&amp;gt;1(=0$dB$)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the system is &lt;strong&gt;stable&lt;/strong&gt; if at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|&amp;lt;1(=0$dB$)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, neutral (or marginal) stability is at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|=1(=0$dB$)$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stability margin is ''the space'' you have until the system becomes unstable (i.e. robustness).&#xA;Since stability is determined by using two measures: the gain and the phase, consequently there are also two stability margins (the gain margin and the phase margin).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gain margin is ``the space'' you have at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$ until $|L(j\omega)| = 0$.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phase margin is ``the space'' you have at frequency $\omega$ where $|L(j\omega)| = 0$ until $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$ until.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gain, phase and corresponding margins can be computed by hand, however I would recommend using software (e.g. MATLAB), since they can be very extensive.&#xA;To find effect your controller $C(s)$, you can draw Bodediagram and see the effect of the various parameters (MATLAB command &lt;code&gt;bode()&lt;/code&gt;):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Fvt94.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fvt94.png&quot; alt=&quot;Bodediagram PID-controlers&quot;&gt;&lt;/a&gt;&#xA;The margins of the complete system can also be visualized using MATLAB (command &lt;code&gt;margin()&lt;/code&gt;), which should give more insight in the effect of the chosen controller and your stability margins.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lxeX9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lxeX9.png&quot; alt=&quot;Stability margins&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2018-09-09T10:42:26.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23682" PostTypeId="1" AcceptedAnswerId="23706" CreationDate="2018-09-09T16:16:07.773" Score="0" ViewCount="692" Body="&lt;p&gt;I'm trying to upgrade a multiple gas flow controller, which must be capable of controlling the mass flowrates of three gases independently through one outlet line to our system, and am in need of a bit of advice on programming it. I'm using an arduino and a few DACs to tell the &lt;a href=&quot;https://en.wikipedia.org/wiki/Proportioning_valve&quot; rel=&quot;nofollow noreferrer&quot;&gt;proportioning solenoid valves&lt;/a&gt; how open they should be. The current setup schematic is below, though I have more pressure sensors and mass flow sensors at my disposal if necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MScS8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MScS8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My colleagues' advice is to use PID loops for each of the gases. This would be easy, particularly as there's an &lt;a href=&quot;https://playground.arduino.cc/Code/PIDLibaryBasicExample&quot; rel=&quot;nofollow noreferrer&quot;&gt;arduino code already available&lt;/a&gt; - however I'm not convinced that's either necessary or the best option, because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There is hardly any inertia in the system, unlike the thermal inertia in an oven for example. If it were just one gas, with one inlet pressure and one outlet, just the P part of a PID would be enough I think. &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The flow rate of each gas is dependent on the pressure differential before and after the valve, which is then dependent on the mass flow of the other gases and the outlet pressure, hence a PID could reach the optimal flow for one gas but in doing so change the flow rates of the other gases - leaving oscillation in the system.  (the pressure in the system at the outlet may be changing slightly also) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know if its possible, or if so then how, to write a PID loop for simultaneous multiple input/outputs.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Hence I would appreciate any ideas, or solutions that exist which I was unaware of, for how my program should go to control the proportioning solenoid valves?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;P.S. I'm not asking anyone to actually write a code for me, but just a basic idea of how I should do it would be great please? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.P.S. (There is a formula for gas flow rate &lt;a href=&quot;https://www.pipeflowcalculations.com/gasleak/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, so I suppose you could try to work out theoretically what the optimal signal to the valves would be based on some kind of simultaneous solution of this formula for the three gases, taking into account friction factors for all the little fittings and parts in the system and expansion factors for the particular gas mixtures concerned (I don't think any of the gas cylinders are absolutely pure mixtures) etc. but I would think it's possible not to go that complex here!)&lt;/p&gt;&#xA;" OwnerUserId="11394" LastEditorUserId="11394" LastEditDate="2018-09-09T16:23:49.677" LastActivityDate="2022-06-02T13:42:53.647" Title="Programming a gas mass flow controller rig" Tags="|control-engineering|pressure|gas|pid-control|flow-control|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="23683" PostTypeId="2" ParentId="19357" CreationDate="2018-09-09T21:51:38.300" Score="1" Body="&lt;p&gt;An easy solution is to use a &lt;em&gt;Motorized Linear Slide Pot&lt;/em&gt; - they're not expensive and they have a motor and a pot - you still need to build the feedback loop to control the position and use a DAC to provide a voltage for the setpoint. A mechanical linkage can reduce the range of motion to what you require.&lt;/p&gt;&#xA;" OwnerUserId="16544" LastActivityDate="2018-09-09T21:51:38.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23684" PostTypeId="2" ParentId="23672" CreationDate="2018-09-09T22:38:11.543" Score="1" Body="&lt;p&gt;Microwaves are oscillating electromagnetic fields. They are absorbed by (frustrated) rotational modes of molecules. These articles explain the basics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/B9781845694203500019&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electromagnetic basis of microwave heating&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0065262808600828&quot; rel=&quot;nofollow noreferrer&quot;&gt;Basic principles of microwave and recent developments&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A capacitor generates an electric field between its plates. Oscillating fields induce (frustrated) rotations, ionic vibrations, and electronic oscillations in the material between the plates depending on the frequency of the field. These articles explain the modes of polarization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://nptel.ac.in/courses/113104005/lecture18a/18_2.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mechanisms of polarization&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://eng.libretexts.org/Textbook_Maps/Chemical_Engineering/Supplemental_Modules_(Materials_Science)/Optical_Properties/Dielectric_Polarization&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dielectric polarization&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www4.ncsu.edu/~franzen/public_html/CH736/lecture/Dielectric_Polarization.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dielectric polarization&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principles of dielectric absorption of an oscillating electric field at microwave frequencies are the same as those for the absorption of microwaves. Therefore, oscillating electric fields will (and do) heat the material when they are absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The practical limitations of replacing a basic microwave oven with a capacitor AC-field oven are likely to be that one will not be able to generate or sustain input powers that are comparable to a basic microwave oven (1 kW) across a parallel plate capacitor.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-09-09T22:38:11.543" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23685" PostTypeId="1" AcceptedAnswerId="23686" CreationDate="2018-09-10T04:55:30.933" Score="0" ViewCount="231" Body="&lt;p&gt;The project is to rotate a load using a motor with a varying speed between like &lt;em&gt;2000-10000 rpm&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using a &lt;strong&gt;flange coupling&lt;/strong&gt; to attach the cylindrical tube to the motor. Since I will rotate it to high speed, I think I need to tightly and safely attach the load to the motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any suggestion on what coupling or what is the best way to attach the cylindrical tube to the motor &lt;strong&gt;safely&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE: the shaft and load are the same lines. this machine is about centrifugal electrospinning which is a machine to fabricate a nanofiber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JsGkQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JsGkQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1368" LastEditorUserId="1368" LastEditDate="2018-09-10T09:37:27.677" LastActivityDate="2018-09-10T14:15:43.983" Title="How to attach a cylindrical load to a motor safetly" Tags="|mechanical-engineering|motors|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23686" PostTypeId="2" ParentId="23685" CreationDate="2018-09-10T05:35:15.763" Score="1" Body="&lt;p&gt;One can use the Flange Coupling with Key feature and Axis Support Fixed Seat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qtpBZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qtpBZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This design includes both. Also, one can find these off the shelf according to dimension requirements, or one can design one of them easily.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For example: &lt;a href=&quot;https://joycedayton.com/products/system-accessories/couplings&quot; rel=&quot;nofollow noreferrer&quot;&gt;JoyceDayton&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nX6DM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nX6DM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: &lt;a href=&quot;https://www.ebay.co.uk/itm/8mm-Flange-Coupling-Steel-Rigid-Flange-Coupling-Motor-Guide-Shaft-Axis-Bearing-F-/222683278447&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ebay&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vON9es.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vON9es.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Most importantly, after the calculation of the bolt torques, one can use &lt;strong&gt;lock tight&lt;/strong&gt; in order to make sure rigid connection.&lt;/p&gt;&#xA;" OwnerUserId="7468" LastEditorUserId="13311" LastEditDate="2018-09-10T14:15:43.983" LastActivityDate="2018-09-10T14:15:43.983" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23687" PostTypeId="2" ParentId="23685" CreationDate="2018-09-10T05:38:56.120" Score="3" Body="&lt;p&gt;The safest way to attach a non-symmetrical load (such as that shown in your photo) that spins that fast (up to 10,000rpm) will be to mount it  to a solid base, using bearings or similar, such that its weight is supported separately from the motor shaft. This will minimise vibrations/oscillations of the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, there will now be a slight misalignment between the vibrating motor, and the constrained load. Using a a semi-flexible coupling (jaw, Oldham, etc) to transfer the rotation will accomodate this misalignment safely without damaging the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your biggest worry should be wobble from any off-centre mass of the load, rather than the direct strength of the coupling, per se.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-09-10T14:07:21.887" LastActivityDate="2018-09-10T14:07:21.887" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23688" PostTypeId="2" ParentId="23632" CreationDate="2018-09-10T07:25:47.597" Score="1" Body="&lt;h2&gt;Irreversible process&lt;/h2&gt;&#xA;&lt;p&gt;In science, a process that is not reversible is called irreversible. This concept arises frequently in thermodynamics.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In thermodynamics, a change in the thermodynamic state of a system and all of its surroundings &lt;strong&gt;cannot be precisely restored to its initial state&lt;/strong&gt; by infinitesimal changes in some property of the system without expenditure of energy.&lt;/p&gt;&#xA;&lt;p&gt;A system that undergoes an irreversible process may still be capable of returning to its initial state. However, the &lt;strong&gt;impossibility&lt;/strong&gt; occurs in restoring the environment to its own initial conditions. (&lt;em&gt;Wikipedia&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then here it comes:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;All complex natural processes are irreversible.&lt;/strong&gt;&#xA;(&lt;em&gt;&lt;strong&gt;Lucia, U&lt;/strong&gt; (1995). &amp;quot;Mathematical consequences and Gyarmati's principle in Rational Thermodynamics&amp;quot;.&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;The Carnot Cycle&lt;/h2&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Carnot cycle is a &lt;strong&gt;theoretical&lt;/strong&gt; thermodynamic cycle proposed by Nicolas Léonard Sadi Carnot in 1824 and expanded upon by others in the 1830s and 1840s. It provides an upper limit on the efficiency that any classical thermodynamic engine can achieve during the conversion of heat into work, or conversely, the efficiency of a refrigeration system in creating a temperature difference (e.g. refrigeration) by the application of work to the system. It is not an actual thermodynamic cycle but is a &lt;strong&gt;theoretical&lt;/strong&gt; construct. (&lt;em&gt;Wikipedia&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which is a theoretical cycle in order to determine the upper limit of a practical thermodynamic cycle.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h2&gt;Answer&lt;/h2&gt;&#xA;&lt;p&gt;The External and Internal Irreversibilities, &lt;strong&gt;Decreases&lt;/strong&gt; the efficiency of a Thermodynamic Cycle.&lt;/p&gt;&#xA;&lt;p&gt;Here is how;&#xA;First I strongly suggest you to read this simple introduction from &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_efficiency&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;This limiting value is called the Carnot cycle efficiency because it is the efficiency of an &lt;strong&gt;unattainable&lt;/strong&gt;, &lt;strong&gt;ideal&lt;/strong&gt;, &lt;strong&gt;reversible&lt;/strong&gt; engine cycle called the Carnot cycle. &lt;strong&gt;No device converting heat into mechanical energy, regardless of its construction, can exceed this efficiency.&lt;/strong&gt; (&lt;em&gt;Wikipedia&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In thermodynamic cycle, the processes that are applied to the mass that runs the cycle is irreversible. That is the law of entropy.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Carnot efficiency formulas are based on simple idealized mathematical models of engines, with no friction and working fluids that obey simple thermodynamic rules called the ideal gas law.&lt;/p&gt;&#xA;&lt;p&gt;Real engines have many departures from ideal behavior that waste energy, reducing actual efficiencies far below the theoretical values given above. (&lt;em&gt;Wikipedia&lt;/em&gt;)&lt;/p&gt;&#xA;&lt;p&gt;Examples are:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;friction of moving parts&lt;/li&gt;&#xA;&lt;li&gt;inefficient combustion&lt;/li&gt;&#xA;&lt;li&gt;heat loss from the combustion chamber&lt;/li&gt;&#xA;&lt;li&gt;departure of the working fluid from the thermodynamic properties of an ideal gas&lt;/li&gt;&#xA;&lt;li&gt;aerodynamic drag of air moving through the engine&lt;/li&gt;&#xA;&lt;li&gt;energy used by auxiliary equipment like oil and water pumps.&lt;/li&gt;&#xA;&lt;li&gt;inefficient compressors and turbines&lt;/li&gt;&#xA;&lt;li&gt;imperfect valve timing&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Then for further reading, you can take a look at &lt;a href=&quot;https://engineering.stackexchange.com/a/11209/7468&quot;&gt;this topic&lt;/a&gt; too.&lt;/p&gt;&#xA;" OwnerUserId="7468" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-09-10T07:25:47.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23689" PostTypeId="2" ParentId="23685" CreationDate="2018-09-10T08:40:32.470" Score="0" Body="&lt;p&gt;Normally, you would use two grub screws at 90 degrees to each other, to secure the load onto the motor shaft. I have done this to attach a stainless steel flywheel on a motor spinning at 6,000 rpm and it's holding just fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give it extra strength, you can also use something like Loctite 638 to glue the flange coupling, or whatever coupling you should choose, to the shaft. This probably means the coupling needs to be metal, not plastic (although I'm sure there'll be other glues to bong plastic to metal). Note that Loctite 638 is pretty much permanent: once you've glued your coupling onto the shaft, it won't be coming off again (easily).&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2018-09-10T08:40:32.470" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23690" PostTypeId="2" ParentId="15788" CreationDate="2018-09-10T10:50:14.830" Score="3" Body="&lt;p&gt;The answer to this lies in the defining equation for the strains that you have supplied. The full displacement gradient $u_{i,j}$ (where the $\bullet_{,j}$ represents the $j^{\rm th}$ derivative) can be linearly decomposed into a symmetric and an anti-symmetric component: $u_{i,j} = \epsilon_{i,j} + \omega_{i,j}$, in the usual way as for all rank-2 tensors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symmetric part, $\epsilon_{i,j} = \frac{u_{i,j}+u_{j,i}}{2}$ represents the strain, the only thing that counts for elastic energy, whereas the anti-symmetric part $\omega_{i,j} = \frac{u_{i,j}-u_{j,i}}{2}$ represents rigid body rotation. And as you remarked, this costs no energy to the system. Hence, for elasticity, we only ever talk about $\epsilon_{i,j}$.&lt;/p&gt;&#xA;" OwnerUserId="17440" LastActivityDate="2018-09-10T10:50:14.830" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23691" PostTypeId="1" CreationDate="2018-09-10T12:34:27.423" Score="1" ViewCount="253" Body="&lt;p&gt;What can happen to a pump in a pit without water for many months, then the fall heavy rains start adding water to the pit, will the pump be efficient to do its job? Should water be added to the pit to exercise the pump? If yes, what would be the recommended frequency?&lt;/p&gt;&#xA;" OwnerUserId="17458" LastActivityDate="2018-12-11T19:04:00.930" Title="Submersible sump pump life in pit without water?" Tags="|pumps|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23692" PostTypeId="1" CreationDate="2018-09-10T14:15:39.150" Score="0" ViewCount="2970" Body="&lt;p&gt;I work with Abaqus CAE/Standard a lot!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever I have tried to put the number below the *Depvar keyword to be greater than 15 (and there can be any subroutines (UMAT or USDFLD) &lt;em&gt;with no coding in it&lt;/em&gt;, just the basic input parameters as provided by Abaqus User Subroutine Guide), the job always, without fail, aborts right away and gives me the error: &quot;Illegal memory reference (signal 11)&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind, there is no actual code anywhere in the user subroutine where there could be any mistake; I have tried this just as a curiosity, and just as expected, as soon as the number of dependent variable is higher than 15, it aborts right away!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the cause of this?&lt;/p&gt;&#xA;" OwnerUserId="15597" LastEditorDisplayName="user16" LastEditDate="2018-09-11T12:12:33.350" LastActivityDate="2018-09-12T00:46:30.537" Title="Is there any reason for Abaqus to give this error message &quot;Illegal memory reference (signal11)&quot; whenever the number of statev is greater than 15?" Tags="|abaqus|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23693" PostTypeId="2" ParentId="23691" CreationDate="2018-09-10T17:44:27.240" Score="1" Body="&lt;p&gt;Two issues I would check&#xA;1) can the pump prime okay without intervention. An air release valve in the discharge would help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Pumps left idle can suffer from corrosion which causes them to seize when they are operated.Suitable materials and clearances would be required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A final problem with not using machinery until it is needed is that you don't know if the pump has failed until you need it, at which point it is too late.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speak with the manufacturer regarding the frequency of exercising the pump. Monthly would seem reasonable&lt;/p&gt;&#xA;" OwnerUserId="8482" LastActivityDate="2018-09-10T17:44:27.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23694" PostTypeId="2" ParentId="23676" CreationDate="2018-09-10T17:53:46.523" Score="3" Body="&lt;p&gt;A root locus plot shows how the &lt;em&gt;poles&lt;/em&gt; of the transfer function &lt;em&gt;change&lt;/em&gt;, when some parameter of the system changes. The parameter is often the loop gain of the system, but it could be anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Nyquist plot shows how the &lt;em&gt;response&lt;/em&gt; of the system changes with &lt;em&gt;frequency,&lt;/em&gt; for one fixed value of the parameter and one fixed set of pole positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you can't draw a root locus plot from just &lt;em&gt;one&lt;/em&gt; Nyquist plot. Well, you &lt;em&gt;can&lt;/em&gt;, but it would just be a plot showing the positions of the poles of that response curve, which isn't very interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a root locus plot, in principle you could draw &lt;em&gt;many&lt;/em&gt; Nyquist plots for different values of the parameter, but in practice the curves on the root locus plot are unlikely to be labeled to show how the parameter changes along each curve, so you can't extract the data you need (i.e. the positions of all the poles for one value of the parameter) just from the plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize: a root locus plot and a Nyquist plot are two different ways of looking at the same underlying information (i.e. the system transfer function) but they show different things, and they are useful for different purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might say that a root locus plot is more useful for &lt;em&gt;designing&lt;/em&gt; a system (for example it tells you what values of the parameter will give a stable or unstable system response) while a Nyquist plot is more useful for &lt;em&gt;analysing&lt;/em&gt; a system where the parameter value has already been chosen (for example it tells you about the phase and gain stability margins of the system at different frequencies).&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-09-10T17:53:46.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23696" PostTypeId="1" CreationDate="2018-09-10T21:16:28.323" Score="7" ViewCount="686" Body="&lt;p&gt;Our hearts are peristaltic pumps, which has a periodic compression and expansion. The head developed can be taken as blood pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are situation when the blood pressure is normal while the pulse rate is high. What is the physical meaning of this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it mean?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Low flow rate per contraction?&lt;/li&gt;&#xA;&lt;li&gt;more pressure loss(an indication of blood vessel blockage)?&lt;/li&gt;&#xA;&lt;li&gt;contraction rate of heart muscle is low?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14771" LastActivityDate="2018-09-11T01:54:41.810" Title="What is the physical meaning of pulse rate and blood pressure, from a turbo machinery point of view?" Tags="|mechanical-engineering|pumps|biomechanics|turbomachinery|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23697" PostTypeId="2" ParentId="23696" CreationDate="2018-09-10T21:58:48.113" Score="12" Body="&lt;p&gt;high pulse rate + low pressure means low &lt;em&gt;pumping efficiency&lt;/em&gt; of the heart. this can be caused by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;leaking valves, especially the main aortic valve, due to disease or calcification;&lt;/li&gt;&#xA;&lt;li&gt;low blood volume, due to bleedout as in bodily injury or an aneurysm rupture;&lt;/li&gt;&#xA;&lt;li&gt;incomplete contraction of the heart muscle due to a breakage or damage in the nerves which carry the firing impulse around the heart, as in a heart attack;&lt;/li&gt;&#xA;&lt;li&gt;cardiac nerve misfire, as in fibrillation or a fault in the nerve cluster that generates the cardiac trigger signal;&lt;/li&gt;&#xA;&lt;li&gt;dead tissue in the heart muscle, as caused by a heart attack;&lt;/li&gt;&#xA;&lt;li&gt;a perforation between the ventricles, as in a congenital heart defect;&lt;/li&gt;&#xA;&lt;li&gt;breakdown of the elastic component of the heart tissue, causing ballooning of the heart, as in congestive heart failure;&lt;/li&gt;&#xA;&lt;li&gt;severe relaxation of the muscular walls of the arteries in the body.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="1832" LastEditDate="2018-09-11T01:54:41.810" LastActivityDate="2018-09-11T01:54:41.810" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23698" PostTypeId="2" ParentId="23674" CreationDate="2018-09-10T22:17:07.367" Score="0" Body="&lt;p&gt;Your confusion seems to be in assigning a dP at all.  On the back side of the fan, the air is at 1 atmosphere, and on the front side, the air is at 1 atmosphere.  There is no static pressure gain at 6,000 CFM.  Instead, you would consider the velocity pressure.  This is more challenging:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First, take the velocity at each point on the outlet of the fan.  These measurement points should be as many as possible for the highest accuracy.&lt;/li&gt;&#xA;&lt;li&gt;Square the velocity, and multiply by the area this point covers.  &lt;/li&gt;&#xA;&lt;li&gt;Add all of these together, and divide by the total area.  &lt;/li&gt;&#xA;&lt;li&gt;Multiply by the density and divide by 2.  &lt;/li&gt;&#xA;&lt;li&gt;Convert to Pascals.&lt;/li&gt;&#xA;&lt;li&gt;Use this to define the efficiency.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As someone who has experienced these fans, a significant fraction of the air at high speeds winds up heading out radially or even slightly backwards.  Thus, a different manufacturer may have wound up running such a measurement for CFM and including this additional airflow, whereas the others disregarded it, knowing it was useless vortexing.  As such, it may simply be marketing.&lt;/p&gt;&#xA;" OwnerUserId="2015" LastActivityDate="2018-09-10T22:17:07.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23699" PostTypeId="2" ParentId="15788" CreationDate="2018-09-11T01:39:47.550" Score="1" Body="&lt;p&gt;From your picture, your deformation map would be something like&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;\varphi(x) = &#xA;\left[\begin{array}{c}&#xA;xcos(\varphi) - ysin(\varphi) \\&#xA;xsin(\varphi) + ycos(\varphi) \\&#xA;z \\&#xA;\end{array}\right]&#xA;$$&#xA;Deformation gradient&#xA;$$&#xA;F = &#xA;\left[\begin{array}{ccc}&#xA;cos(\varphi) &amp;amp; -sin(\varphi) &amp;amp; 0 \\&#xA;sin(\varphi) &amp;amp; cos(\varphi) &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; 1\\&#xA;\end{array}\right]&#xA;$$&#xA;Green deformation tensor&#xA;$$&#xA;G = &#xA;\left[\begin{array}{ccc}&#xA;1 &amp;amp; 0 &amp;amp; 0 \\&#xA;0 &amp;amp; 1 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; 1\\&#xA;\end{array}\right]&#xA;$$&#xA;$$&#xA;\therefore&#xA;E = \frac{1}{2} [G - I] = 0&#xA;$$&#xA;Which is what we would expect, but in order to calculate this we needed to use higher order terms. Since:&#xA;$$&#xA;G = FF^T = (\nabla u + I)(\nabla u + I)^T = \nabla u + \nabla u^T + \nabla u \nabla u ^T + I&#xA;$$&#xA;If each component of:&#xA;$$&#xA;\nabla u &amp;lt; 1&#xA;$$&#xA;then each component of:&#xA;$$&#xA;\nabla u \nabla u ^T \ll 1&#xA;$$&#xA;We expect our $\varphi$ to be very small, so we can ignore our higher order terms. Let's linearize the deformation gradient:&#xA;$$&#xA;cos(\varphi) \approx 1, sin(\varphi) \approx \varphi \\&#xA;F = &#xA;\left[\begin{array}{ccc}&#xA;1 &amp;amp; -\varphi &amp;amp; 0 \\&#xA;\varphi &amp;amp; 1 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; 1\\&#xA;\end{array}\right] \\&#xA;G = &#xA;\left[\begin{array}{ccc}&#xA;1 + \varphi ^2 &amp;amp; 0 &amp;amp; 0 \\&#xA;0 &amp;amp; 1 + \varphi ^2 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; 1\\&#xA;\end{array}\right] \\&#xA;\therefore&#xA;E = \frac{1}{2}&#xA;\left[\begin{array}{ccc}&#xA;\varphi ^2 &amp;amp; 0 &amp;amp; 0 \\&#xA;0 &amp;amp; \varphi ^2 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; 0\\&#xA;\end{array}\right] \approx 0&#xA;$$&#xA;This makes since, because if $\varphi$ is small, $\varphi^2$ is much smaller and we can accept the error assuming geometrically linear behavior. We just need to watch out for non-small rotations that will cause our linearization to have a larger error.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Resource:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fundamentals of Structural Mechanics 2nd, Keith D. Hjelmstad&lt;/p&gt;&#xA;" OwnerUserId="17306" LastActivityDate="2018-09-11T01:39:47.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23700" PostTypeId="1" AcceptedAnswerId="23701" CreationDate="2018-09-11T04:48:09.740" Score="3" ViewCount="254" Body="&lt;p&gt;The only one I know is the three point bending test for flexural strength.&lt;/p&gt;&#xA;" OwnerUserId="16653" LastActivityDate="2018-09-11T13:39:57.450" Title="What methods are there to test the flexural, tensile and compressive strength of concrete?" Tags="|concrete|strength|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23701" PostTypeId="2" ParentId="23700" CreationDate="2018-09-11T06:04:37.697" Score="8" Body="&lt;p&gt;There are two methods to test flexural strength, three point and four point method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three point method: &lt;a href=&quot;https://i.stack.imgur.com/RFnzR.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RFnzR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the corresponding moment line: &lt;a href=&quot;https://i.stack.imgur.com/dA9WX.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dA9WX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;four point load: &lt;a href=&quot;https://i.stack.imgur.com/g5Gwm.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g5Gwm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and its moment line: &lt;a href=&quot;https://i.stack.imgur.com/c8j7N.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c8j7N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum flexural stress happens to concentrate only in a point in three point test, if you are testing a non-homogeouse material as it's always so in reality, it's not handy at all, brittle materials in three point test are prone to premature fracture, however the stresses in four point test, extended across the beam (construction ...), it means in case of not-homogenous material béton armé (Eng. Reinforcement concrete), we can have more accurate map of material behaviour.     &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually these two methodes are considered as destructive test. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another destructive test is &lt;strong&gt;splitting tensile strength test&lt;/strong&gt;,  &lt;a href=&quot;https://www.youtube.com/watch?v=gC7XB0A3F4M&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, you can find more about this test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pure axial (tensile, compression) tests (destructive tests) are quite popular. Universal testing machines or hydraulic presses are two familiar examples of instrument testing machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides hardness tests (semi-destructive) such as Brinell, Rockwel and Vickers are among the most popular tests these days in Europe. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-09-11T13:39:57.450" LastActivityDate="2018-09-11T13:39:57.450" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23702" PostTypeId="2" ParentId="23692" CreationDate="2018-09-11T12:16:50.663" Score="1" Body="&lt;p&gt;They likely use a nibble (aka. half a byte or 4 bits) to represent the counter that represents that value.  The largest unsigned value that a nibble can represent is 15, with a range of 0-15. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is likely logic in place to throw the error you mentioned whenever you try to use a value that exceeds that range.  A simple &lt;code&gt;if(0xF0 &amp;amp; num_states)&lt;/code&gt; check is sufficient to detect a value greater than 15.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And while I can't comment on the reasoning the Abaqus developers used, nibbles are typically used when either:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Available memory is a limiting resource, and they are trying to squeeze things as much as is possible.&lt;/li&gt;&#xA;&lt;li&gt;&quot;No one will ever need more than 15 of these!&quot; is declared&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2018-09-12T00:46:30.537" LastActivityDate="2018-09-12T00:46:30.537" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23703" PostTypeId="2" ParentId="23692" CreationDate="2018-09-11T16:31:24.740" Score="0" Body="&lt;p&gt;99.9 percent of the time signal 11 means there is a bug in the program. So in this case you have a bug that is only triggered when this variable is above 15. Either stay under 15 or ask the vendor to fix the bug&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-09-11T16:31:24.740" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23706" PostTypeId="2" ParentId="23682" CreationDate="2018-09-11T21:52:24.237" Score="1" Body="&lt;p&gt;My best stab at this would be to attach parallel inputs of:&lt;br&gt;&lt;pre&gt;&#xA;Gas1(100psi) -&gt; Flow Meter1 -&gt; Valve1 -&gt; Manifold Input1&lt;br&gt;&#xA;Gas2(100psi) -&gt; Flow Meter2 -&gt; Valve2 -&gt; Manifold Input2&lt;br&gt;&#xA;Gas3(100psi) -&gt; Flow Meter3 -&gt; Valve3 -&gt; Manifold Input3&lt;br&gt;&lt;/pre&gt;&#xA;...where:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I picked &quot;100psi&quot; to stress that the input pressure should be &gt;&gt; the output pressure. Could be 50psi, or whatever...&lt;/li&gt;&#xA;&lt;li&gt;Valve[n] is controlled by your spiffy Arduino code taking Flow Meter[n] as its input.&lt;/li&gt;&#xA;&lt;li&gt;All Valve[n] outputs go to a manifold that effectively connects together all inputs.&lt;/li&gt;&#xA;&lt;li&gt;The Manifold output (not shown) connects to your artificial lung.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In addition:&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&#xA;Manifold Output -&gt; Lung -&gt; Px Sensor&lt;br&gt;&#xA;'                    | Purge Valve&lt;br&gt;&lt;/pre&gt;&#xA;&#xA;&lt;h2&gt;Strategy:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Open purge valve.&lt;/li&gt;&#xA;&lt;li&gt;Control Proportional Valve[n] so each gas has its proportional mass    flow input. (This is where the spiffy software comes in, more on that    later). &lt;/li&gt;&#xA;&lt;li&gt;Allow purge to stay open while all 3 gasses are flowing, until steady state has been achieved. &lt;/li&gt;&#xA;&lt;li&gt;Close the purge valve. &lt;/li&gt;&#xA;&lt;li&gt;Monitor the Px sensor while the pressure builds up. While in the background, maintain proper flow proportions. &lt;/li&gt;&#xA;&lt;li&gt;Once the target Px is reached, close all Valve[n] valves simultaneously.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Control Algorithm&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I would suspect that you could get away with a straight P controller.  I assume that you are not too concerned with small errors (you didn't say). I do not think that you will need to work the pressure sensor data into any of your calculations.  However, if delta-p starts getting too small, then maybe you would need to build in a higher order control system, or perhaps the Px data to scale the valve outputs as delta-p goes down.  TBD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will spend 80% of your time prototyping 1 control system to work.  Then another 80% of your time integrating the hardware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting the proportional valve to work as you expect, well that's the golden goose.  Stability will be dependent on a few things, but macroscopically I would be concerned about:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Valve Cv value (Ahh, Cv numbers.  Mean different things to different people, and nothing to most.)  Are the valves ported properly for the desire flow rates?  Are the valve orifices a suitable size for the flow rates and control range?&lt;/li&gt;&#xA;&lt;li&gt;Arduino limitations.  We love Arduino's.  They're fun.  They also suck.  Especially when you want to do something real, and real fast.  I'd bet dollars-to-donuts that you could get this to work with some flavor of Arduino.  I'd also bet that you will end up working around some wonky limitation that you could have avoided by spinning your own electronics, and programming in native c.  But heck, it's fun. The parts, and more importantly your time, are really cheap - so why not.&lt;/li&gt;&#xA;&lt;li&gt;Is it all really stable under all conditions?  Everyone will scream &quot;Model it!&quot;.  Instead, just try it and find out.  A few hundred times.  It will be a good experience.  Even if you do take the modeling route, you will still need to try it out a few hundred times, so leave the expensive modeling software to the interns, where it'll do less damage.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Purchase a $30 adjustable pressure relief valve, in case something goes awry.  No need to blow the poor fella's lungs apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like a fun project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!&lt;/p&gt;&#xA;" OwnerUserId="10176" LastActivityDate="2018-09-11T21:52:24.237" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23707" PostTypeId="1" AcceptedAnswerId="23716" CreationDate="2018-09-12T04:45:44.323" Score="2" ViewCount="263" Body="&lt;p&gt;The curing process for regular cement concrete is sprinkling water fog or using a moisture retaining fabric. I was wondering what the process is for polymer concretes because they don't use any water.&lt;/p&gt;&#xA;" OwnerUserId="16653" LastActivityDate="2018-09-12T18:05:58.300" Title="What is the curing process for polymer concrete if it doesn't use any water?" Tags="|civil-engineering|concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23708" PostTypeId="1" AcceptedAnswerId="23719" CreationDate="2018-09-12T05:12:22.380" Score="2" ViewCount="2010" Body="&lt;p&gt;When I was younger, I had interned in a civil/structural engineering firm specializing in bridges that had many PE civil engineers on staff and a couple licensed mechanical engineers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as a soon-to-be licensed mechanical engineer, interested in starting his own business, I'm wondering what is limit of a mechanical PE. Can a mechanical PE subcontract from a civil firm? Can the mechanical PE only work on certain tasks/items of a civil project? Of the little I remember, the mechanical engineers were the experts on moving bridges (bascule, swing). My specialty is finite element analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd be curious if anybody knows where exactly this line is, and even better, if they can point to laws/regulations/policy documents that describe this. For specifics, I am located in California, but would be interested in all United States-based answers.&lt;/p&gt;&#xA;" OwnerUserId="17476" LastEditorUserId="17476" LastEditDate="2018-09-12T06:13:16.537" LastActivityDate="2018-09-12T18:37:19.417" Title="What can a licensed Mechanical Engineer do in the civil/structural field?" Tags="|mechanical-engineering|structural-engineering|civil-engineering|licensure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23709" PostTypeId="1" CreationDate="2018-09-12T05:35:18.107" Score="5" ViewCount="99" Body="&lt;p&gt;My mine is looking to install a 600 mm open borehole to drop particles up to 150 mm through.  It's a 320m shaft to drop ballast down to save time compared to using the drift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it gets blocked they typically take a water truck out and that works.  They have used explosives to clear a blockage before but that was a decade ago. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the seniors have rough rule of thumbs on the borehole diameter required to drop particles without causing a blockage but I'm unable to find any references. And it is based on experience and what they've always done. I'm unaware of any design work for this beyond &quot;this should be large enough&quot;, but believe there should be more behind it than that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a formula available to determine the borehole diameter required to safety drop in the particles?&lt;/p&gt;&#xA;" OwnerUserId="17477" LastEditorDisplayName="user16" LastEditDate="2018-09-14T01:08:18.213" LastActivityDate="2018-09-14T01:08:18.213" Title="What diameter of particles will result in a blocked borehole?" Tags="|geotechnical-engineering|soil|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23710" PostTypeId="1" CreationDate="2018-09-12T08:01:03.567" Score="-1" ViewCount="750" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7U2ps.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7U2ps.png&quot; alt=&quot;ASUS S550CM Back Panel &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a picture of the back of my laptop (An &lt;a href=&quot;https://www.asus.com/Laptops/ASUS_VivoBook_S550CM/specifications/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASUS S550CM&lt;/a&gt;) with a motherboard/HDD/RAM access panel removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The white connector connects the motherboard with the laptops back panel which in turns has contact with the battery. It is either the power cable for the motherboard or for the speakers which are battery-integrated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cable has not been disconnected in 5 years. I want to remove the back panel, clean the heatsink and fans of accumulated dust but am unable to remove the back panel since I am unable to disconnect this wire. My questions&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I tried using tweezers to disconnect this wire but was not sure as to how much pressure to apply. Since the connector has not been removed in 5 years, is there anything I can apply or spray to loosen it up and smoothly disconnect without risking damage to the motherboard ? &lt;/li&gt;&#xA;&lt;li&gt;What does this connector do - is it a power cable for the motherboard or for the speakers or something else entirely?&lt;/li&gt;&#xA;&lt;li&gt;What is the name of this sort of connector? (a general commercial name which will let me locate this on eBay if I happen to break it)&lt;/li&gt;&#xA;&lt;li&gt;What is the name for the connecting wire? Are these wires easily available in the market? (I live in Sydney, Australia)  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17470" LastEditorUserId="7083" LastEditDate="2018-10-13T21:35:08.337" LastActivityDate="2019-12-08T01:01:35.223" Title="What is the white colored connector circled in red in the picture of a laptop motherboard called?" Tags="|electrical-engineering|computer-hardware|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23711" PostTypeId="2" ParentId="23710" CreationDate="2018-09-12T08:43:40.160" Score="1" Body="&lt;p&gt;It looks like a Molex or AMP 4-pin power connector. Note the AMP and Molex versions are similar, but not interchangeable. See  &lt;a href=&quot;https://en.wikipedia.org/wiki/Molex_connector&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Molex_connector&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AMP connectors have one or two tabs which click into place and &quot;lock&quot; the parts together when you connect them. You need to bend the tab(s) to release them while you are pulling on the plug to remove it. The two &quot;ears&quot; on the ends of the connector in your picture may be these locking tabs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This video isn't very clear, but at about 1:30 it looks like the guy is using a small screwdriver to bend back the end tabs at the same time as pulling on the plug. Once the plug has started to move, the tab will be &quot;unlocked&quot; and you can then fully remove the plug without bothering about the tabs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=qhRuU7EAA_E&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=qhRuU7EAA_E&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These connectors can be a tight fit - even when you release the locking tabs, you may still need quite a bit of force to unplug it. Try to pull on the shell of the connector itself, not on the wires!&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-09-12T08:52:57.110" LastActivityDate="2018-09-12T08:52:57.110" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23712" PostTypeId="1" CreationDate="2018-09-12T12:51:54.090" Score="0" ViewCount="136" Body="&lt;p&gt;When I try to create a flat form of a archimedes screw using Solidworks it gives me a diameter different from what is calculated manually basically using a Pythagoras Theorem. Can anyone explain the reason as to what is used by any 3D Software in order to calculate the diameter and essentially what factors am I assuming when I calculate using the pythagoras theorem only.&lt;a href=&quot;https://i.stack.imgur.com/eUq3V.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eUq3V.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;I tried to calculate a screw with pitch- 400mm, OD - 400mm and Id - 268mm&#xA;Using Pythagoras theorem gives me a OD - 430mm and ID - 297mm circle&#xA;whereas Solidworks gives a OD - 450mm and ID - 318mm circle.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="17480" LastEditorUserId="17480" LastEditDate="2018-09-13T16:55:42.763" LastActivityDate="2019-12-07T20:03:20.500" Title="Difference in Diameter calculation using Solidworks and manually" Tags="|mechanical-engineering|fluid-mechanics|applied-mechanics|solidworks|geometry|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="23713" PostTypeId="1" CreationDate="2018-09-12T14:08:58.387" Score="1" ViewCount="419" Body="&lt;p&gt;Ansys Workbench 17.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a mesh that was imported from Abaqus into Ansys but the beam cross section data is missing. As a result the structural model omits the beam elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I access the beam cross section data in Ansys? I cant find any way to do it in the FE Modeler, Mechanical Model or Mesh modules so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: There appears to be no way of directly doing this in Ansys Work Bench or AIM. I had to write some code the alter the *inp files to include the missing information prior to importing to Ansys.&lt;/p&gt;&#xA;" OwnerUserId="14531" LastEditorUserId="14531" LastEditDate="2018-09-17T11:49:05.183" LastActivityDate="2018-09-17T11:49:05.183" Title="How to add Beam cross section properties to imported mesh in Ansys?" Tags="|beam|ansys|ansys-workbench|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23714" PostTypeId="1" AcceptedAnswerId="23720" CreationDate="2018-09-12T15:31:04.703" Score="0" ViewCount="62" Body="&lt;p&gt;I have seen many times that people gets the frequency response function using DTFT. Can I use the Fourier series, and when I do that?&lt;/p&gt;&#xA;" OwnerUserId="17484" LastEditorUserId="9" LastEditDate="2018-09-12T16:48:28.930" LastActivityDate="2018-09-12T18:49:01.530" Title="Fourier series with frequency response" Tags="|frequency-response|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23715" PostTypeId="2" ParentId="23710" CreationDate="2018-09-12T16:05:42.417" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;3) What is the name of this sort of connector?&lt;br&gt;&#xA;  (a general commercial name which will let me locate this on eBay if I&#xA;  happen to break it)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'll hazard a guess that this is &lt;a href=&quot;http://www.molex.com/molex/products/family?key=picoblade&amp;amp;channel=products&amp;amp;chanName=family&amp;amp;pageTitle=Introduction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Molex PicoBlade&lt;/a&gt;.&lt;br&gt;&#xA;If that's the case, then the pitch should be 1.25mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please measure the pitch (center-to-center distance between the pins) and update your question with that information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related threads on the Electrical Engineering stack.  &lt;a href=&quot;https://electronics.stackexchange.com/a/164791/7036&quot;&gt;Surface-mount PicoBlade header&lt;/a&gt; like yours.  &lt;a href=&quot;https://electronics.stackexchange.com/a/245257/7036&quot;&gt;Throughole header&lt;/a&gt;, same connector family.&lt;br&gt;&#xA;(Be advised, however, that repair and identification questions are frowned upon on the Electrical Engineering stack.  Please don't misconstrue these references as an invitation to post repair and identification questions to the Electrical Engineering stack.)&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2018-09-12T16:17:49.733" LastActivityDate="2018-09-12T16:17:49.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23716" PostTypeId="2" ParentId="23707" CreationDate="2018-09-12T18:05:58.300" Score="3" Body="&lt;p&gt;Note that the curing process for ordinary concrete is a chemical reaction between the ingredients of the cement mix and water which is stirred into that mix. Water is sprinkled onto the curing concrete to make sure that enough water is available to make the curing reaction go to completion and also to cool the concrete, since the reaction releases heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of polymer cement, &quot;curing&quot; is a chemical reaction between the resin and the curing agent which are mixed together before pouring. Sometimes the curing agent is air, which fuels the chemical reaction that causes the resin to turn into a solid, and in some cases the air-curing chemical reaction is assisted by moisture in the air. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-12T18:05:58.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23717" PostTypeId="2" ParentId="23691" CreationDate="2018-09-12T18:14:55.697" Score="1" Body="&lt;p&gt;a submersible pump in a sump pit is subject to being fouled by sedimentation i.e., the slow accumulation of sand and silt in the pit during extended periods of inactivity. If that silt gets into the impeller scroll, it will lock the impeller and stall the pump motor; if it fouls the float switch then the pump will not turn on and the sump pit will continue to accumulate more silt and sand. If the float switch is not fouled but the impeller is, then since the pump relies on water flow to cool it while it is running, a silted up pump will overheat and burn out its windings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to prevent this is by inspection of the pump in its pit and periodic cleanout of the silt and sand.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-12T18:14:55.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23718" PostTypeId="1" AcceptedAnswerId="23728" CreationDate="2018-09-12T18:27:13.360" Score="2" ViewCount="3026" Body="&lt;p&gt;I want to make a &lt;strong&gt;geodesic dome greenhouse&lt;/strong&gt; (16ft wide, type: 3V 5/8). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DXMt6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DXMt6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Source: &lt;a href=&quot;http://geodesicdomes.ca/wp-content/uploads/2009/12/domes-fullsize.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://geodesicdomes.ca/wp-content/uploads/2009/12/domes-fullsize.jpg&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would cover a geodesic dome frame with clear poly plastic sheeting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can purchase clear poly plastic in almost any rectangular dimension (either as &lt;a href=&quot;https://www.homedepot.ca/en/home/p.240-in-x-1200-in-cgsb-approved-vapour-barrier.1000166576.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;narrow house wrap&lt;/a&gt; or as wider greenhouse poly). &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I've researched this sort of setup online, I've found lots of photos of attempted wrapping jobs. Unfortunately, when the poly is wrapped around the dome, it ends up looking as &lt;strong&gt;ugly as sin&lt;/strong&gt;. There ends up being lots of overlap and/or loose areas, which have the added disadvantage of being noisy in wind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ssDXN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ssDXN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;Source: &lt;a href=&quot;https://www.instructables.com/id/20-Geodesic-Dome-Greenhouse/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.instructables.com/id/20-Geodesic-Dome-Greenhouse/&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What method of wrapping a geodesic dome would result in the least amount of overlapping and loose areas?&lt;/p&gt;&#xA;" OwnerUserId="10536" LastEditorUserId="10536" LastEditDate="2019-02-28T01:24:13.153" LastActivityDate="2024-02-27T04:17:43.743" Title="Wrap a geodesic dome with a poly sheet: Minimize overlap &amp; loose areas" Tags="|geometry|" AnswerCount="6" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23719" PostTypeId="2" ParentId="23708" CreationDate="2018-09-12T18:37:19.417" Score="1" Body="&lt;p&gt;I'm a licensed civil engineer in California, and I don't think the answer is a black and white yes or no.  My civil license allows me to supervise and/or stamp work that I have taken personal responsibility for.  It allows me to stamp only work that I have direct work experience in.  I have no specific seismic experience, for example, so I'm not legally allowed to stamp and work with seismic calculations, even though that work rightly falls under civil engineering.  Likewise, I'm not aware of any regulation in the Business and Professions Code that might prevent me from stamping work that might be considered mechanical engineering, if I can show relevant and recent experience in that field. In fact, I've done some pump station work in the past, which is just as much a mechanical discipline as a civil one.  (Ironically, my engineering degree is mechanical, not civil).  Most, if not all public agencies do require a registered civil engineer's stamp for public works type projects, but that is up to the agency, not part of the state regulations.  That might also be true for agency contracts for mechanical/structural type work as well.  But that would be a separate agency requirement, not a state regulation.  FEA is most certainly as much a mechanical discipline as anything else, so I suspect you'd be fine doing that work with a mechanical license.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advice would be to read the Business and Professions Code for Professional Engineers available at the state's website. This is a good place to start:  &lt;a href=&quot;https://www.bpelsg.ca.gov/laws/pe_act.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.bpelsg.ca.gov/laws/pe_act.pdf&lt;/a&gt;  If you need to ask a specific question to someone at the state level, you could start here:  &lt;a href=&quot;https://www.bpelsg.ca.gov/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.bpelsg.ca.gov/&lt;/a&gt;  There are also professional engineering associations around the state that might also give you good guidance.  All of the above is from my memory so take it with a grain of salt, and good luck!&lt;/p&gt;&#xA;" OwnerUserId="9637" LastActivityDate="2018-09-12T18:37:19.417" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23720" PostTypeId="2" ParentId="23714" CreationDate="2018-09-12T18:49:01.530" Score="0" Body="&lt;p&gt;Examining a physical phenomena (expressed as a mathematical function) in the frequency domain can be very fruitful sometimes.&#xA;In case you are capable to perfectly express your function mathematically, i.e. by writing down an analytical function, you can manually use a Fourier series to find the amplitudes of any specific frequency.&#xA;However, this is a mathematical approach which is quite irrelevant to the most of engineering tasks. Since you probably do not know exactly what you are looking for, trying to analyze what is going on will be very tedious this way.&#xA;More often, you find yourself with a lot of recorded digital data you would like to analyze. You will be interested to find the dominant frequencies (by which the amplitude is the largest) and will be glad to obtain a plot showing you all the data arrange along the frequency axis. This is exactly what DFT (or FFT and etc,,) does.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2018-09-12T18:49:01.530" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23721" PostTypeId="2" ParentId="23718" CreationDate="2018-09-12T20:27:13.370" Score="2" Body="&lt;p&gt;I've been fascinated with geodesic domes since the seventies (last century) and had given some thought to the problem you face. Unfortunately, there is no single-sheet solution to this puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have noted that a geodesic sphere from which the dome is built is constructed of pentagons. You can determine the frequency of a dome by counting from the center of one pentagon to the next the number of struts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, one can cut a sheet of plastic to match the individual pentagons, but it still means that you have to slice a single segment in order to either tape flush (butted) or overlap on the pentagon each sheet. You can get reasonably tight coverage in this manner, but one may consider the taped joints to be unsightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining segments that are not pentagons have to be attacked individually. Your selection of a 3v dome gives you a bit of an advantage in that the remaining segments are hexagons, with a center point, with radial struts of the same length, which can be approached in the manner of the pentagon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some geodesic designs use individual triangles for construction, which results in duplicated struts at every connection, effectively doubling the material along each path between vertices. This would allow you to pull the plastic into the joint rather than to have to tape or otherwise fasten the plastic to the outside. Think picnic tablecloths draped over the edges, while butting another table against it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of your EMT based product, this is impractical, especially as it doubles the cost of materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could create an add-on solution by attaching furring strips or similar panels to outline the pentagons and hexagons, using them as clamps to secure the plastic to the EMT. It would still be necessary to slice one sheet per pentagon and hexagon to pull the perimeter tight and secure with the clamping system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would hesitate to drill that many holes through what is relatively flimsy EMT and instead would create a hot dog bun or I-beam type of attachment, which would allow for the bolts securing the wood to the EMT to pass alongside the EMT rather than through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7nOEt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7nOEt.png&quot; alt=&quot;geodesic bolt block drawing attempt&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could pass the plastic sheet between the top wood block and the EMT or perhaps staple the plastic to the outside of the wood block or to the edges of the block, leaving the top surface of the wood exposed/painted/treated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One characteristic I recall from my years of reading of construction and pondering geodesic domes is that homespun solutions always had leaking problems!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicone caulk/sealer may be desirable to provide additional leak protection.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-09-12T20:27:13.370" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23722" PostTypeId="1" AcceptedAnswerId="23723" CreationDate="2018-09-12T20:57:20.827" Score="0" ViewCount="43" Body="&lt;p&gt;I am studying gauges for measurements and I can't find the answer to what device I would use to set angles in 2 plates.&lt;/p&gt;&#xA;" OwnerUserId="17486" LastActivityDate="2018-09-12T21:23:28.777" Title="What device would you use to set angles in two plates?" Tags="|measurements|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23723" PostTypeId="2" ParentId="23722" CreationDate="2018-09-12T21:23:28.777" Score="4" Body="&lt;p&gt;Although a photo or diagram or more detailed description would be helpful, you may find what you seek in a machinist's protractor. There are a few different designs, including a few electronic digital models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EDBlg.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EDBlg.jpg&quot; alt=&quot;digital protractor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the above model, but have yet to use it for anything requiring extreme precision. It has a tendency to flex a bit even after tightening the angle lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rPtKF.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rPtKF.jpg&quot; alt=&quot;protractor 320&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model appears often in searches and is much more accurate, according to my previous reading via a machinists' email list. Much if the discussion is that the typical home hobby machinist does not require this much precision, as these devices are rather expensive, upwards of US$300.00&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6RPML.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6RPML.jpg&quot; alt=&quot;simple protractor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Less precise and less expensive, the above type provides perhaps one-half degree accuracy, but does qualify as a possible solution to your question.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-09-12T21:23:28.777" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23724" PostTypeId="2" ParentId="23712" CreationDate="2018-09-12T22:37:52.860" Score="1" Body="&lt;p&gt;A flat helix (&lt;a href=&quot;https://en.wikipedia.org/wiki/Helicoid&quot; rel=&quot;nofollow noreferrer&quot;&gt;helicoid&lt;/a&gt;) is a ruled surface, but it is not a developable surface. So you are going to have a flattening problem. Try using the middle fiber radius and compare the hand calc to the software and see if they aren't a lot closer.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-09-12T22:37:52.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23725" PostTypeId="2" ParentId="13972" CreationDate="2018-09-13T00:24:36.887" Score="3" Body="&lt;p&gt;A 3Hz pulsating hum-- noise that is getting louder and softer at 3Hz.  Yes, I agree that would be quite annoying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is a residential unit in the US, both the compressor and the fan are run with single phase asynchronous motors.  Almost certainly, they are four pole with an 1800 RPM synchronous speed, running about 1750 RPM or so, depending upon load.  The peak you are seeing at 116.8 Hz is consistent with this since that corresponds to the second harmonic of a 1752 RPM rotation rate.  The big peak you see at 120Hz is interesting.  I wonder if this is due to a loose winding in one of the motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compressor is inside the A/C unit, in this case, on top of the house.  The fan has to be exposed to the outside.  My experience is the fan is what you hear most of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the graphs, it looks like everything is working correctly.  The compressor or fan is cycling in and out every 20 minutes which is to be expected.  The duty cycle is less than 50% which means it is properly cooling so I guess the coolant levels are adequate.  The sounds are probably caused by a combination of loose sheet metal and maybe a loose winding. If it were my unit, I would put my hand here and there to try to guess where the noises are coming from and make sure everything I can get to was bolted down securely.  Recognizing that it is not your unit and you don’t have access to it, much less permission to go monkeying around, this won’t do you much good.  If it is not running in heat pump mode and it is running at close to 50% duty cycle in February, (when this question was originally posted), it won’t be able to keep the house cool when it starts to get hot and your neighbor will get it fixed, hopefully with a replacement unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading the linked document, I think you should go over there in the dark of night and do a “power trip test” if for no other reason than to get some peace and quiet.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastEditorUserId="4953" LastEditDate="2018-09-14T18:12:53.690" LastActivityDate="2018-09-14T18:12:53.690" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23726" PostTypeId="1" AcceptedAnswerId="23727" CreationDate="2018-09-13T04:45:44.023" Score="1" ViewCount="1244" Body="&lt;p&gt;I'm designing a chair from scratch and I'd like it to have a swivel base. For this my initial idea was to press fit a bearing (thrust bearing) in the shaft of the base and connect the seat by press fitting a small shaft through the middle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm having my doubts if this can handle the load of a person sitting on the chair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking for a purpose built or better suited bearing for this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also opened to suggestions for a better way to build this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="1636" LastEditorUserId="1636" LastEditDate="2018-09-13T10:25:04.483" LastActivityDate="2018-09-13T10:25:04.483" Title="What is the type of bearing used on swivel base chairs?" Tags="|mechanical-engineering|bearings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23727" PostTypeId="2" ParentId="23726" CreationDate="2018-09-13T05:37:33.800" Score="2" Body="&lt;p&gt;Most 'Swivel Chairs' use simple thrust bearings, which consist of a low-friction layer (either ballbearings, or a flat disc of plastic), sandwiched between a lower disc attached to the swivel legs and an upper disc attached to the chair base. the shaft coming down out of the chair base goes through the sandwich and extends down to where the legs branch off, and somewhere in there is a bushing that holds the shaft centered in the tube which supports the bottom disc as described above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HG3gFm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HG3gFm.png&quot; alt=&quot;Swivel Chair Thrust Bearing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13311" LastEditDate="2018-09-13T09:29:09.177" LastActivityDate="2018-09-13T09:29:09.177" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23728" PostTypeId="2" ParentId="23718" CreationDate="2018-09-13T06:33:09.450" Score="3" Body="&lt;p&gt;Minimising overlap and loose areas at the same time will be very difficult. It may be possible to source a material that you can ‘shrink wrap’ on, eg using heat, but I don’t have specific pointers there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding laying a flat sheet, your key aim here should be to have &lt;em&gt;regular&lt;/em&gt; overlaps. See tutorials on wrapping circular objects &lt;a href=&quot;https://youtu.be/dH_gpSfaA6E&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/dH_gpSfaA6E&lt;/a&gt; or industrially wrapped spheres such as old golf balls, or Chocolate Oranges (if you’re from the UK!) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WUh2D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WUh2D.png&quot; alt=&quot;Wrapped Golf Ball&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-09-13T22:52:38.500" LastActivityDate="2018-09-13T22:52:38.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23729" PostTypeId="1" CreationDate="2018-09-13T06:40:04.273" Score="2" ViewCount="40" Body="&lt;p&gt;I'm currently an underground Geotechnical Engineer, don't spend a lot of time on the surface. I'm trying to find a reference or method for calculating the required diameter of a service hole. My Mine is looking to install a 600mm drop hole to transport ballast underground, and the ballast itself is up to 150mm. I don't believe this will be large enough and that at some stage it will become blocked, this is based on past experienced, but nothing more. I'm struggling to find a reference or method to back this up and to provide an alternative diameter. There are several &quot;rules of thumb&quot; from older Engineers, but no references to support them. Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="17491" LastActivityDate="2018-09-13T06:40:04.273" Title="Required diameter for ballast drop hole?" Tags="|civil-engineering|design|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ClosedDate="2018-09-14T01:08:56.193" ContentLicense="CC BY-SA 4.0" />
  <row Id="23730" PostTypeId="2" ParentId="23235" CreationDate="2018-09-13T09:15:14.490" Score="2" Body="&lt;p&gt;You've fallen into the trap of using a low cost first generation device because it got picked up by the hobbyist community. Whilst this is a great source of early-adopter information, it can be quite sticky, and not too concerned about reliability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the genuine 'safe' operating conditions for this sort of device can be hard since reliability was probably never a high priority for the original design, so you can expect to see dramatically better performance from a different part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need remote sensors, the digital wiring will be your greatest concern - you need something which is resilient to EMI and potentially has a built-in watchdog at the sensor end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your sensors are local, analogue inputs &lt;em&gt;might&lt;/em&gt; be more reliable than digital, but you can also think about building in some 'fault recovery' logic for the cases where your sensors fail and cease to return plausible data.&lt;/p&gt;&#xA;" OwnerUserId="6483" LastEditorUserId="6483" LastEditDate="2018-09-13T09:26:11.200" LastActivityDate="2018-09-13T09:26:11.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23731" PostTypeId="1" AcceptedAnswerId="23733" CreationDate="2018-09-13T11:24:02.270" Score="0" ViewCount="707" Body="&lt;p&gt;I have a 4-wheeled vehicle that is 10&amp;nbsp;x&amp;nbsp;10&amp;nbsp;cm. Each wheel has its own motor. The wheel diameter is 3.1&amp;nbsp;cm. The total weight of the vehicle is 500&amp;nbsp;grams. The friction coefficient of the rubber wheels to concrete is 0.6-0.85 (found this online, from the research I did I think it is needed for calculating torque). How much torque is needed on the wheel before the motor becomes too powerful and just cause itself to slip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone share the steps and formulas needed to solve the problem it would be of great help. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I only calculated the weight on each wheel (I'm not even sure if this is correct)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\begin{align}&#xA;W &amp;amp;= \text{wheel}_1 + \text{wheel}_2 + \text{wheel}_3 + \text{wheel}_4 \\&#xA;500\text{ g} &amp;amp;= 4\text{wheel} \\&#xA;125\text{ g} &amp;amp;= \text{wheel} \\&#xA;\end{align}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so 125&amp;nbsp;g per wheel (assuming weight is distributed equally).&lt;/p&gt;&#xA;" OwnerUserId="17493" LastEditorUserId="1832" LastEditDate="2018-09-13T11:52:09.883" LastActivityDate="2018-12-12T16:01:00.193" Title="How to calculate the torque of 4 wheeled vehicle before slipping" Tags="|torque|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23732" PostTypeId="1" AcceptedAnswerId="23735" CreationDate="2018-09-13T11:35:55.807" Score="3" ViewCount="316" Body="&lt;p&gt;There are thermo-shrinking plastics, commonly used in electronics, which stay shrunk after cooling down. There are thermally expanding foams. The closest to 'expanding when cooled' I know is water, with ice expanding, but only by a small margin, a scarce couple percent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a material that would start expanding significantly when cooled? May shrink back when heated, or may stay expanded. It doesn't need to be as forceful as ice (which can explode rocks) but it should have at least 30% expansion ratio, the more the better. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-09-13T18:45:07.200" Title="Are there any cold-expandable materials of significant expansion factor?" Tags="|materials|thermal-expansion|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23733" PostTypeId="2" ParentId="23731" CreationDate="2018-09-13T11:51:47.890" Score="2" Body="&lt;p&gt;Before start to solve the problem, i like to use your assumption about load distribution. And i assume you found the static coefficient of friction. Notice there is a difference between static and dynamic coefficient of friction. The dynamic coefficient is slightly smaller in amplitude than static coefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the free body diagram of one of the wheels: &lt;a href=&quot;https://i.stack.imgur.com/Vgtno.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vgtno.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the wheel move, you need to apply enough torque to cancel out the force $F_t$: &#xA;$$F_t = \mu F_N$$ &#xA;$$F_N = mg$$&#xA;Here $\mu$ is the static coefficient of friction, $F_N$ normal or reaction force on the wheel, $g$ is  constant, about $9.81 \frac {m}{s^2}$ and $m$ is the mass of the vehicle in this case $m$ is just a fraction of it nl. one fourth.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can calculate the torque: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$\tau = F_t \frac{d}{2} =\frac{d}{2} mg \mu $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here $d$ represent the diameter.  Notice this is enough to bring the vehicle in movement, but if you want to accelerate the vehicle then this amount of torque is not enough.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-09-13T13:25:03.997" LastActivityDate="2018-09-13T13:25:03.997" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23735" PostTypeId="2" ParentId="23732" CreationDate="2018-09-13T15:22:58.467" Score="4" Body="&lt;p&gt;While there's a list of (rare) materials with negative thermal expansion, that 30% expansion ratio excludes most of them and narrows down your search to polymers or structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Within the realm of polymers, &lt;strong&gt;two-way reversible shape memory polymers&lt;/strong&gt; might be your option. They can have relatively high reversible strains and can be thermally activated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wxslM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wxslM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;WM Huang, Thermo-Moisture Responsive Polyurethane Shape Memory Polymer for&#xA;Biomedical Devices, The Open Medical Devices Journal, 2010, 2, 11-19&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ovTII.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ovTII.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;C. Liu, H. Qinb, P.T. Mather, Review of progress in shape-memory polymers, Journal of Materials Chemistry&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to look for &lt;strong&gt;structures&lt;/strong&gt; which are made of positive thermal expansion materials, but when assembled they have a the opposite effect: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K9DA3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K9DA3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Christopher Spadaccini, Mechanical Metamaterials: Design, Fabrication, and Performance, Frontiers of Engineering: Reports on Leading-Edge Engineering from the 2015 Symposium (2016)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="17409" LastEditorUserId="17409" LastEditDate="2018-09-13T18:45:07.200" LastActivityDate="2018-09-13T18:45:07.200" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23736" PostTypeId="1" CreationDate="2018-09-13T16:37:29.500" Score="1" ViewCount="61" Body="&lt;p&gt;I don't know why they tried to ruin EES by doing this, but what people want to do is have the solver recognize an equation with a variable, and then take that same variable and plot a table of it with varying values in a separate parametric table. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is it the software states a conflict between assuming two definitions of the variable at once, so unlike competently designed software, users have to choose between either a table or the equation solver, not both. How can this be fixed? &lt;/p&gt;&#xA;" OwnerUserId="17498" LastActivityDate="2018-09-13T16:37:29.500" Title="How can Engineering Equation Solver solve both equations and a table?" Tags="|software|" AnswerCount="0" CommentCount="3" ClosedDate="2018-09-14T11:52:51.003" ContentLicense="CC BY-SA 4.0" />
  <row Id="23739" PostTypeId="1" CreationDate="2018-09-13T21:17:46.787" Score="1" ViewCount="2503" Body="&lt;p&gt;I found the following paper and am attempting to put Saturated Pressure and Saturated Temperature into excel. I got Saturated Pressure to work, but can't get saturated Temperature to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://core.ac.uk/download/pdf/81944678.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saturated Steam Equations Paper&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Saturated Pressure (working)&lt;/strong&gt;&lt;br&gt;&#xA;=(B18+273.15)/647.096 [cacluates Tr from C]&lt;br&gt;&#xA;=EXP(9.56756 + 5.39806*LN(B19) - 6.16183*(LN(B19))^2 + 1.49572*(LN(B19))^4 + 0.433*B19^5 )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Saturated Temperature (not working)&lt;/strong&gt;&lt;br&gt;&#xA;=B15/22064 [calculates Pr from kPa]&lt;br&gt;&#xA;=EXP((0.00937817 + 0.000498951*B16 + 0.0000111049*B16^2 + 0.000000334995*B16^3 + 0.0000000344102*B16^4)^(-0.4))&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought maybe it was outputting in K, but that did not solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either I didn't correctly translate the equation, or the equation is wrong. Thoughts? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2819" LastEditorUserId="2819" LastEditDate="2018-09-13T22:23:20.687" LastActivityDate="2023-12-17T22:08:00.940" Title="Saturated steam equations" Tags="|steam|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23740" PostTypeId="1" CreationDate="2018-09-13T22:44:09.910" Score="2" ViewCount="231" Body="&lt;p&gt;I have a wind tunnel (subsonic flow of air) which consists of a fan inside (and ahead of) a long cylinder of cross-sectional area of $A_e$, followed by a diffuser of $A_o$ c.s. area ($A_o&amp;gt;A_e$).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MM9SK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MM9SK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The fan maintains a velocity of $V_e$ inside the tunnel.&#xA;Assuming the flow inviscid and incompressible, I want to find the amount of force per unit area that the fan applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I chose the entire inside of the tunnel (fan included) as the fixed, non-deforming control volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying Conservation of Momentum for the c.v. (gage pressure used):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$F=\rho(V_o^2A_o-V_i^2A_e)=\rho V_e^2A_e(A_e/A_o)$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume integral was dropped since the flow is steady.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last transition was done with the help of Conservation of Mass which yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$A_eV_e=A_oV_o$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer should be $(\rho/2)V_e^2A_e(A_e/A_o)^2$ and in the available solutions to me, Bernoulli was used on the stream line that begins after the fan and ends outside the tunnel at atm' pressure, and Momentum Conservation was applied to a control volume enclosing the fan alone.&lt;/p&gt;&#xA;" OwnerUserId="11557" LastEditorUserId="11557" LastEditDate="2018-09-14T06:38:31.333" LastActivityDate="2018-09-16T13:37:46.530" Title="Force applied by fan under inviscid &amp; incompressible flow" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23741" PostTypeId="1" CreationDate="2018-09-14T00:21:54.033" Score="2" ViewCount="552" Body="&lt;p&gt;I feel as if I understand the conceptual difference between Lagrangian and Eulerian descriptions. But I can't seem to follow the math. I have one example that I would greatly appreciate an explanation or full calculation for&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The motion can be expressed as:&#xA;  $$\left\{\begin{align}&#xA;x_1(X_1,X_2,X_3,t) &amp;amp;= X_1 \\&#xA;x_2(X_1,X_2,X_3,t) &amp;amp;= X_2 + \dfrac{X_2 v_0 t}{h_0} \\&#xA;x_3(X_1,X_2,X_3,t) &amp;amp;= X_3 \\&#xA;\end{align}\right.$$&#xA;  velocity in Lagrangian description:&#xA;  $$v_i = \begin{bmatrix}&#xA;0 \\&#xA;\left(\dfrac{X_2v_0}{h_0}\right) \\&#xA;0&#xA;\end{bmatrix}$$&#xA;  in Eulerian description:&#xA;  $$v_i = \begin{bmatrix}&#xA;0 \\&#xA;\left(\dfrac{x_2v_0}{h_0 + v_0 t}\right) \\&#xA;0&#xA;\end{bmatrix}$$&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am entirely on board on the velocity in the Lagrangian description. I cannot for the life of me figure out how the Eulerian description is computed nor why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone show me how the Eulerian description of the velocity is calculated? &lt;/p&gt;&#xA;" OwnerDisplayName="user14808" LastEditorUserId="1832" LastEditDate="2018-09-14T00:44:24.997" LastActivityDate="2018-09-15T00:53:51.843" Title="Eulerian and Lagrangian descriptions of velocity" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23743" PostTypeId="2" ParentId="23739" CreationDate="2018-09-14T02:56:07.593" Score="0" Body="&lt;p&gt;For your temperature, the last term ^-0.4 needs to be applied to the full expression, not just the term to the power ^0.4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ie [exp(xxx + xxx + xxx + xxx)] ^-0.4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, so I realise that what Eric has is correct, I read it wrong, but I won't delete it as the comments have value.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-09-14T06:38:54.037" LastActivityDate="2018-09-14T06:38:54.037" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23745" PostTypeId="2" ParentId="22281" CreationDate="2018-09-14T07:31:33.297" Score="1" Body="&lt;p&gt;The Reynolds number is irrelevant in these situations because it is very small and does not deal with buoyancy and natural convection. There is instead a similitude parameter called the Grashof number that deals with balances between buoyant forces and viscuous forces which is relevant here.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13363" LastEditDate="2018-09-14T07:42:24.990" LastActivityDate="2018-09-14T07:42:24.990" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23747" PostTypeId="1" CreationDate="2018-09-14T13:02:30.133" Score="0" ViewCount="50" Body="&lt;p&gt;Before asking a question, I have no background knowledge of engineering. So I have no idea about these issues like control or signal engineering. (Also, I'm not good at using English. sorry for that.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attached the picture related to my question below.&#xA;It is time series data and I want to find the black circled points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) these data follows Gaussian distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The most important thing in this question is that I can't know future data at specific time point. I know only past data at specific time point. I want to recognize black circled points around local max/min points or right after local max/min points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) There is no model with these data. e.g. There is no equations like relations of velocity, acceleration and positions. Only these data are the only measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got some recommendations to use Kalman filter things, and It helps to make these lines differentiable. Now I'm studying Kalman filter but It seems I do need kinds of model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible or useful to use Kalman filter? or, do I have to another methods to filter these data?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DvoHZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DvoHZ.png&quot; alt=&quot;Question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17510" LastActivityDate="2018-09-14T13:27:11.397" Title="how to filter model-free observations (e.g. Kalman filter)" Tags="|control-engineering|signal-processing|electronic-filters|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23748" PostTypeId="2" ParentId="23747" CreationDate="2018-09-14T13:27:11.397" Score="1" Body="&lt;p&gt;The Kalman-filter is an observer, predicting the next system state based on an initial value. That prediction &lt;em&gt;needs&lt;/em&gt; to be made based on some kind of model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest solution that I can come up with is a learning-based adaptive control approach, to identify the model and later use the identified model to design an observer. This requires a rudimentary initial model. I never tried something like that for some arbitrary measured data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can specify a bit where the data comes from, for example if you can influence the measured system in some way (does it have actuators that a controller could use?), I might be able to give a clearer answer.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2018-09-14T13:27:11.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23749" PostTypeId="1" CreationDate="2018-09-14T16:19:09.943" Score="1" ViewCount="1220" Body="&lt;p&gt;I work in a university building designed in the 1930s, located in western New York State (not in NYC). To enter classrooms, offices or labs one passes through a single door. However, the restrooms ALL have two layers of door with a closet-sized compartment in between. This poses issues for people with mobility impairments. All the inner doors have transoms, with a ventilation grating above each transom. Both the men's and women's restrooms have these two-layer doors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two door system is awkward and wastes space. Surely there are better ways to prevent hallway users from peeping in. And the bathrooms are not noisy; I've been in classrooms that share a wall with a restroom and have never heard any water-related noise. Finally, the grating between the restroom and the compartment eliminates the argument that the reason is odor control. I suspect the reason relates to the building codes in force when the building was designed - but why would a building code require two layers of door on a restroom? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any insights!&lt;/p&gt;&#xA;" OwnerUserId="17512" LastActivityDate="2019-04-21T04:14:06.630" Title="Why would bathrooms have two layers of doors?" Tags="|architecture|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23751" PostTypeId="2" ParentId="23555" CreationDate="2018-09-14T17:57:07.190" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can it or any dielectric breakdown damage electronics from a distance, say 1m, 10 m , 100m or 1 km? If yes, how is that possible&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.  This is not an engineering answer, it is an empirical answer. &#xA;As a ham radio op, I had my station connected with antenna.  In my beginning years, I never disconnected the antenna.&#xA;The station consisted of the antenna, and an HW-16 (Heathkit) with external power supply (also Heathkit).  One day, there was a nearby lightning strike. Not direct, but within 500 feet (count of one second).&lt;br&gt;&#xA;Later, I found that the &quot;transmit&quot; did not function anymore.  I reached around the power supply to touch the HW-16, and got 800 volts in my arm.&lt;br&gt;&#xA;Seems that the nearby strike raised enough EMF in the antenna, which traveled down the feed line, which traveled down the &quot;ground&quot; lead to the power supply, which traveled down the actual ground wire to ground.  Burned out the rig to power supply ground lead.&lt;br&gt;&#xA;I replaced the wire, and it worked fine.&lt;br&gt;&#xA;But remember, the Heathkit HW-16 is an all tube rig.  If it had any transistors, I don't know what would have happened. &#xA;Just a little experience.   So now I always disconnect the antenna when not operating.  My friends say to ground the floating antenna too, but that would be too smart. &#xA;73, Joe W3TTT &lt;/p&gt;&#xA;" OwnerUserId="17513" LastActivityDate="2018-09-14T17:57:07.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23752" PostTypeId="1" CreationDate="2018-09-14T20:56:44.337" Score="1" ViewCount="3963" Body="&lt;p&gt;I hope this is a suitable question to ask here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I'm taking Numerical Methods this semester (also called 'Numerical Analysis' in some places).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The term-assignment is to find a real-life problem which is solvable by numerical methods. And then present two different methods to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't have to be something new, simply presenting someone else's solution is acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems like an incredibly easy thing but I'm having a hard time finding something of reasonable difficulty to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas or suggestions?&lt;/p&gt;&#xA;" OwnerUserId="13485" LastEditorUserId="13485" LastEditDate="2018-09-14T21:08:14.257" LastActivityDate="2018-09-14T22:50:25.203" Title="Example of real-life problem solved with numerical methods?" Tags="|modeling|mathematics|numerical-methods|" AnswerCount="2" CommentCount="3" ClosedDate="2018-09-15T01:16:52.520" ContentLicense="CC BY-SA 4.0" />
  <row Id="23753" PostTypeId="2" ParentId="23752" CreationDate="2018-09-14T21:34:44.087" Score="1" Body="&lt;p&gt;You could develop some saturated steam equations by running a regression on steam table data. Even if it was a giant equation, sure would be nice to not have to do a a table lookup in an otherwise automatic spreadsheet or script. I attempted to use some I found on the internet the other day and only one of them worked (&lt;a href=&quot;https://engineering.stackexchange.com/questions/23739/saturated-steam-equations&quot;&gt;saturated steam equations SE question&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, there is one idea. If you do it you should definitely share ;-)&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-09-14T21:34:44.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23754" PostTypeId="1" CreationDate="2018-09-13T11:27:39.447" Score="3" ViewCount="794" Body="&lt;p&gt;My friend and I were discussing this earlier: why don't escalators slow down when people get on? We are not interested in any fancy schmancy computers that might adjust their speed for capacity/traffic flow, we were simply wondering about the most basic possible escalator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our thoughts were: If an escalator is running at a constant speed, and then a single person (or more people) stand on it, it is obviously doing more work to lift that mass. Should the speed of the escalator drop? If something is keeping the speed from changing, does that mean the torque of the motor changes? Or does it simply draw more power but keep the same torque? Are we missing something obvious?&lt;/p&gt;&#xA;" OwnerDisplayName="flinnja" LastEditorUserId="33" LastEditDate="2018-09-15T01:27:17.500" LastActivityDate="2018-09-15T01:27:17.500" Title="Why don't escalators slow down when people get on?" Tags="|motors|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23755" PostTypeId="2" ParentId="23754" CreationDate="2018-09-13T11:55:18.617" Score="0" Body="&lt;p&gt;The response of the system you describe is of course dependent on the construction. Even if we do not allow &quot;fancy schmancy computers&quot;, &lt;a href=&quot;https://en.wikipedia.org/wiki/Centrifugal_governor&quot; rel=&quot;nofollow noreferrer&quot;&gt;mechanical RPM-controllers&lt;/a&gt; have been around since the steam era.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a first example we can assume that we have a simple escalator that is powered by an engine &lt;em&gt;with an RPM controller&lt;/em&gt;. This engine will likely have a maximum torque. The actual torque excerted by the engine is dependent on the load, and the decrease in engine RPM (i.e. elevator speed) that you are expecting will only occur if the load is greater than the maximum torque of the engine at the set RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, the excerted torque - and in normal cases, the consumed energy - of the engine will increase with increasing loads, but the speed will be constant (until the load is greater than the maximum torque).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a second example we can assume an engine running with a constant energy flow (e.g. a gas throttle) and &lt;em&gt;no RPM controller&lt;/em&gt;. In this system the load increase will directly incur a decrease in escalator speed - the work performed by the system is dependent on the inflow of energy to the engine, and as we stated that this is a constant, the work must also be constant. With more weight added, the lifting speed must be lowered.&lt;/p&gt;&#xA;" OwnerDisplayName="Sternerson" LastActivityDate="2018-09-13T12:23:59.397" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23756" PostTypeId="2" ParentId="23754" CreationDate="2018-09-13T12:09:24.870" Score="0" Body="&lt;p&gt;It depends on how your escalator is running.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like you want the simplest thing possible, and it seems to me the simplest thing is an escalator whose mechanism can only apply a single, constant, force/torque (I say force/torque because we aren't being specific about what is actually causing the escalator to move. From now on I will just say force.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider before anyone gets on the escalator. It is supplying enough force to move the mass of the escalator parts as well as resist internal frictions that are present (which for our discussion we will assume to be constant friction forces that do not depend on the velocity of the escalator. Having velocity dependent friction could change some things, but I will leave thinking about that to you). Let's assume at this point the stairs are moving at a constant speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now someone steps on. If we want to maintain a constant speed up the escalator, the thing supplying our force would need to increase its &quot;strength&quot; to counter the new &quot;resistance&quot; caused by the person stepping on. It needs to be able to lift the stairs, fight the friction, and now also lift the person. Since our force being supplied is constant, it cannot do this. Therefore the escalator would actually begin to slow down, and then eventually start to move backwards if the person does not get off in time (this could change with a velocity dependent friction force). This would be analogous to me lifting my hand at a constant speed, and then someone puts a mass in my hand while I supply the same upward force to my hand. The net force on the mass will be downwards, so it will have a downward acceleration.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If something is keeping the speed from changing, does that mean the torque of the motor changes? Or does it simply draw more power but keep the same torque?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So let's say our escalator can adjust its applied force/torque accordingly to keep the escalator moving at a constant rate. Then yes, the force would need to increase to keep people moving at the constant rate. Just how if someone puts that mass onto my hand I will need to push harder to keep my hand (and mass) moving at a constant speed. This also means more power is needed too. If we want the escalator to move at a constant rate, but it is doing more work to lift the people up, then it is doing more work in the same amount of time. Therefore, the rate of energy expenditure increases, i.e. more power needs to be supplied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, to answer your question title, escalators don't slow down because they are designed to be able to rotate the steps at a constant speed by adjusting the supplied force/torque.&lt;/p&gt;&#xA;" OwnerUserId="17515" OwnerDisplayName="Aaron Stevens" LastActivityDate="2018-09-13T12:27:41.553" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23757" PostTypeId="2" ParentId="23754" CreationDate="2018-09-13T15:47:31.753" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Should the speed of the escalator drop?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That would be undesirable behavior. A competent engineering team would not allow that to happen.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If something is keeping the speed from changing, does that mean the torque of the motor changes?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The steps form a chain. Weight on the steps is transmitted up the chain to the drive sprocket that keeps the chain moving.  Any change in weight on the steps equates to a change in torque on the sprocket's drive shaft. And, that torque must be transmitted through some transmission to the motor shaft.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or does it simply...keep the same torque?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That would not make sense. See above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Are we missing something obvious?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, I don't think so.  The system is designed to keep the motor running at a constant speed.  For some types of motor, that can be achieved simply by connecting the motor to a constant voltage power supply. For other types of motor, there may be a controller that monitors the speed and adjusts the voltage accordingly.  Changes in weight equate to changes in torque which cause changes in the amount of electric current (and therefore, the amount of electric power) drawn by the motor.&lt;/p&gt;&#xA;" OwnerDisplayName="user17527" LastActivityDate="2018-09-13T15:47:31.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23758" PostTypeId="2" ParentId="23752" CreationDate="2018-09-14T22:44:55.247" Score="0" Body="&lt;p&gt;A heat transfer problem might be suitable here. Temperature is an intuitive parameter and a simple scalar, and energy balances are easy to write down but often challenging to solve analytically because of temperature-dependent material properties, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote &lt;a href=&quot;http://john.maloney.org/Structural/parabolas_and_catenaries.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; about the parabolic and catenary temperature profiles that arise in a simple 1-D geometry with heat generation. One practical application is described &lt;a href=&quot;http://john.maloney.org/thermal_actuation.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;: a microfabricated suspended silicon beam that heats up from an applied current, expands, and deflects—thus, a microscale linear actuator:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rH4qI.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rH4qI.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the temperature profile only and forget about the motion. In the second link, I write about how the time-dependent analysis diverges from the experimental results because the analytical solution doesn't incorporate the temperature dependence of certain material properties. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you request, there are at least two ways to obtain a more accurate temperature distribution numerically: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) One could use a lookup table for the temperature-dependent material properties (for simplicity, maybe just one material property, say, the thermal conductivity of silicon) and perform a 1-D finite-difference heat transfer analysis by discretizing the beam into segments, each with a uniform temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) One could fit the temperature-dependent material property of interest by an analytical function and solve the resulting differential heat transfer equation (namely, $k(T)\frac{\partial^2T(x,t)}{\partial x^2}+J^2r=c\rho\frac{\partial T(x,t)}{\partial t}$, where $k$ is the thermal conductivity, $T$ is temperature, $x$ is the position, $t$ is time, $J$ is the current density, $r$ is the resistivity, $c$ is the heat capacity, and $\rho$ is the density) using a preferred numerical scheme. Or for simplicity, one could drop the time dependence and simply solve $k(T)\frac{\partial^2T(x,t)}{\partial x^2}+J^2r=0$. (Here, if $k$ were constant, we'd simply obtain a parabola.)&lt;/p&gt;&#xA;" OwnerUserId="10123" LastEditorUserId="10123" LastEditDate="2018-09-14T22:50:25.203" LastActivityDate="2018-09-14T22:50:25.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23760" PostTypeId="2" ParentId="23741" CreationDate="2018-09-15T00:53:51.843" Score="3" Body="&lt;p&gt;The motion is&#xA;$$&#xA;  x = \varphi(X, t) \quad \leftrightarrow \quad X = \varphi^{-1}(x, t)&#xA;$$&#xA;The Lagrangian velocity is&#xA;$$&#xA; V(X, t) = \dot{\varphi}(X, t) = \frac{\partial \varphi (X, t)}{\partial t}&#xA;$$&#xA;The spatial (Eulerian) velocity is defined as&#xA;$$&#xA;  v(x, t) = V(\varphi^{-1}(x, t),t) = \dot{\varphi}(\varphi^{-1}(x, t),t) = \frac{\partial \varphi ( \varphi^{-1}(x, t), t)}{\partial t}&#xA;$$&#xA;In your case&#xA;$$&#xA;  \varphi(X, t) = \begin{bmatrix} X_1 \\ \frac{h_0 + v_0 t}{h_0} X_2 \\ X_3 \end{bmatrix} \quad \implies \quad&#xA;  V(X, t) = \begin{bmatrix} 0 \\ \frac{v_0}{h_0} X_2 \\ 0 \end{bmatrix}&#xA;$$&#xA;and&#xA;$$&#xA;  \varphi^{-1}(x, t) = \begin{bmatrix} x_1 \\ \frac{h_0}{h_0 + v_0 t} x_2 \\ x_3 \end{bmatrix} \quad \implies \quad&#xA;  v(x, t) = V(\varphi^{-1}(x, t),t) = \begin{bmatrix} 0 \\ \frac{v_0}{h_0} \frac{h_0}{h_0 + v_0 t} x_2 \\ 0 \end{bmatrix}&#xA;$$&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-09-15T00:53:51.843" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23762" PostTypeId="2" ParentId="21425" CreationDate="2018-09-15T05:25:47.557" Score="-1" Body="&lt;p&gt;Respected sir/madam,&#xA;    we can get the desired 220V Ac Output by giving the arrived 160V AC output as input to the step up transformer.&#xA;                                        Thank you.&lt;/p&gt;&#xA;" OwnerUserId="17505" LastActivityDate="2018-09-15T05:25:47.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23763" PostTypeId="1" CreationDate="2018-09-15T08:44:37.127" Score="1" ViewCount="1270" Body="&lt;p&gt;&lt;strong&gt;In an adiabatic throttling process of ideal gas:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that mathematically lost work is given by Tsurr * entropy generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But&lt;/strong&gt; intuitively I can't understand if there is not heat exchange between system and surroundings how can the temperature of surrounding affect the work obtained by system.&lt;/p&gt;&#xA;" OwnerUserId="8213" LastActivityDate="2024-03-21T02:03:50.510" Title="Entropy change in a throttling process" Tags="|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23764" PostTypeId="1" CreationDate="2018-09-15T09:10:33.747" Score="2" ViewCount="104" Body="&lt;p&gt;I'm currently stuck on obtaining a description of the robot base velocities (xdot, ydot, thetadot) based on data obtained from 4 IMU sensors located, with an offset, on 4 pivot points. See picture below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mgQtt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mgQtt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And a closeup of 1 of the pivot wheels:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TrllS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TrllS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The IMU is located at the red dot and since this is a 2d problem, I'll only consider acceleration in the X,Y direction of the IMU frame and rotation around the Z axis of the IMU frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since I'm looking to obtain a description for the velocities xdot, ydot en thetadot, this all would result in something like the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[dx dy dtheta]' = [3x12 matrix] * [ax_1 ay_1 gz_1 .. .. .. etc.]'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I think that if a description of how the accelerations of the individual IMU's influence the robots odometry is known, I can average the individual components to come to the 'true' odometry. So what's first (I think) is obtaining a description of either dx or dy or dtheta as a function of the angle delta and the velocities of the IMU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to know the location of the IMU in the robot's frame, we would need translation matrices which look like the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;H_imu_p = [1 0 -d_s; 0 1 0; 0 0 1];&#xA;H_p_r = [cos(d) -sin(d) -d_px; sin(d) cos(d) -d_py; 0 0 1];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where d_s is the distance in X between the IMU and the pivot point and d_px and d_py are the distance between Robot Origin and pivot point expressed in the Robot's frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying these matrices would result in a matrix that is able to give the location of the IMU in the Robot's frame. Since the IMU is located wrt the IMU frame at:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;p_imu = [0 0 1]'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We can obtain the location of the IMU wrt the global frame by taking the last column of the multiplied translation matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differentiating this information twice obtains how the accelerations of the IMU are expressed in the Robot's frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is where the question part begins:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming all this information above is correct, how do I go from an expression that expresses xddot_IMU yddot_IMU and thetaddot_IMU in the robot's frame to that of determining the velocities of the robot's frame? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would integrating the measurement results (so obtain velocities of the IMU) and then having the expression of location of the IMU wrt the robot's frame differentiated once? deltadot could then be obtained from the gyroscopic data around the Z-axis. I'm having doubts about this method due to the complex kinematic relationships between the IMU and the robot. Essentially all the four pivot points could rotate without the robot ever moving (while still measuring values at [ax_1 ay_1 gz_1 .. .. .. etc.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not asking for a worked out solution but I'm quite stuck here and couldn't find the required literature to find a way out. If someone could point me towards the solution (or towards some examples in literature), that would be very much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="17519" LastEditorUserId="17519" LastEditDate="2018-09-15T10:34:12.977" LastActivityDate="2018-10-15T18:00:42.780" Title="Robot velocities based on IMU data" Tags="|mechanical-engineering|robotics|acceleration|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23765" PostTypeId="1" AcceptedAnswerId="23797" CreationDate="2018-09-15T11:02:13.700" Score="3" ViewCount="2313" Body="&lt;p&gt;I would like to calculate the deflection of the following geometry with a 3D load in Y direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/38eTt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/38eTt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know of any textbook, website etc. which discusses how to do that analytically ? I cannot find any suitable reference.&lt;/p&gt;&#xA;" OwnerUserId="9039" LastActivityDate="2018-09-21T01:05:20.073" Title="Deflection of L shaped plate" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23766" PostTypeId="2" ParentId="23740" CreationDate="2018-09-15T12:43:04.153" Score="1" Body="&lt;p&gt;I believe I figured it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we want the force that the fan applies on the air we use Conservation of linear momentum on a CV the encapsulates the fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we choose the CV, fan to outlet, excluding the tunnel itself, we must take into consideration $p{dA}\hat{n}\cdot\hat{x}$ that changes with the curve of the tunnel geometry. This we would like to avoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we choose to solve this issue by expanding our CV to include the tunnel itself, we cut through matter which experiences internal forces, which we don't have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining option is to choose the CV around the fan itself.&#xA;$$F+(p_i-p_e)A_e=\dot{m}(v_e-v_i)$$&#xA;Where $p_i$ and $p_e$ are the pressures just before and just after the fan, respectively, and $F$ is the force we are looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of mass yields $v_i=v_e$ (incompressible flow and constant cross-section), so the RHS disappears,&#xA;$$F=(p_e-p_i)A_e \tag{1}$$ &#xA;Now, since the flow is steady, inviscid and incompressible, we can use Bernoulli on a streamline that stretches :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) from &quot;far away&quot; before the inlet, where the pressure is atmospheric and the velocity is zero, up to the cross-section where we have defined $p_i$: $$\frac{1}{2}\rho v_e^2=p_a-p_i \tag{2}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) from the point where we have defined $p_e$ up to the outlet, where the pressure is atmospheric and the velocity is determined by Conservation of mass (fan to outlet),$$v_eA_e=v_oA_o$$ $$\frac{1}{2}\rho (v_e^2-v_o^2)=p_a-p_e$$&#xA;It can shown that the last equation may be rewritten as,$$\frac{1}{2}\rho v_e^2(1-(\frac{A_e}{A_o})^2)=p_a-p_e \tag{3}$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, subtract eq. (3) from (2) and we have the RHS of (1),$$p_e-p_i=\frac{1}{2}\rho v_e^2(\frac{Ae}{A_o})^2$$&#xA;Finally, the force per unit area is,$$\frac{F}{A_e}=\frac{1}{2}\rho v_e^2(\frac{Ae}{A_o})^2$$&lt;/p&gt;&#xA;" OwnerUserId="11557" LastActivityDate="2018-09-15T12:43:04.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23767" PostTypeId="2" ParentId="23765" CreationDate="2018-09-15T12:51:08.333" Score="1" Body="&lt;p&gt;I think, for the part attached to the wall, the applied load may be considered as a transverse shear load (which causes bending) and a twisting moment(which twists the part).For the part parallel to the wall, the load is just a transverse shear load(causes only bending).I think the stress concentration factors due to sharp edges or joints should be included.(I think the book by Timoshenko and Gere may be helpful)&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastActivityDate="2018-09-15T12:51:08.333" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23769" PostTypeId="1" CreationDate="2018-09-15T14:06:31.763" Score="1" ViewCount="31" Body="&lt;p&gt;Consider a series LCR circuit. Once power source is removed, the electrical energy oscillates for some time before damping out completely. I want a LCR circuit such that 1) it has high resonance frequency , say 1 GHz and 2) it oscillates for some time say 1 min.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider following numbers. To achieve 1 GHz, L and C must be very low. Let C=1 pF and L=25nH. This gives us resonance frequency around 1 GHz. At such high frequency the inductor wire will pose some resistance, let us assume R=0.001 Ohm, even though it is likely to be more. The attenuation factor $\alpha=R/2L$ is going to be high, which means the oscillations even if under-damped, will not sustain even for a second. Clearly there is a trade-off between high frequency and longer oscillation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to achieve both? It is ok even it is achieved without removing the power source, however, note that the voltage source is 10-100KV DC.&#xA;Just keeping the LCR connected to the power source will not result in oscillations, in my understabnding.&lt;/p&gt;&#xA;" OwnerUserId="17373" LastActivityDate="2018-09-15T14:06:31.763" Title="Achieveing sustained LCR oscillations" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="23771" PostTypeId="2" ParentId="23764" CreationDate="2018-09-15T16:48:04.677" Score="1" Body="&lt;p&gt;Unfortunately, the arrangement you show won't work very well. There are a whole raft of reasons, but they boil down to two classes - gimbal lock and singularities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to illustrate the problem with an example of an entire class of motion that this setup can't handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on your sketch, I'm assuming the IMU output is acceleration, and that you are looking to define instantaneous chassis acceleration based on the instantaneous accelerations of four IMUs. In general, you can't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose all IMU &lt;strong&gt;x&lt;/strong&gt; accelerations are zero and &lt;strong&gt;y&lt;/strong&gt; accelerations are all equal. The obvious solution is Thetadot = 0 and the robot doing orbits at constant speed with the casters all having equal angles and rates of rotation about the chasis. However, there are other solutions. Suppose the chasis is spinning about its center. Now &lt;strong&gt;X&lt;/strong&gt;,&lt;strong&gt;Y&lt;/strong&gt; are zero and Thetadot is nonzero. This can have exactly the same IMU outputs. Furthermore, you can combine these motions, where a retrograde spin and orbit combine to yield the same IMU outputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other problems occur when the IMU velocity is low relative to other factors such as caster angle rates. For instance, you can keep all four IMUs stationary and still make orbits of radius = caster offset without triggering any output. EDIT - this would produce a yaw rate, which I didn't know was available when I wrote the previous sentence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As formulated, even given a complete initial condition and IMU history, you still can't generate a trajectory in general, and an error budget analysis would also show serious problems in ordinary, smooth, well behaved trajectories.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use angular resolvers for a reason.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-09-15T17:08:01.437" LastActivityDate="2018-09-15T17:08:01.437" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23772" PostTypeId="2" ParentId="23740" CreationDate="2018-09-15T17:33:09.673" Score="1" Body="&lt;p&gt;Here is the proper approach for the starting point using air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the gas as compressible and ideal. Conservation of mass gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ \dot{n}_e = \dot{n}_i = \dot{n} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_e A_e \frac{p_e}{T_e} = v_i A_i \frac{p_i}{T_i} $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume constant area and isothermal flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ v_e = \frac{p_i}{p_e} v_i $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under isothermal flow of an ideal gas, the force of the fan is therefore set by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$ F = \left(p_e - p_i\right)A_e + mv_i \left(\frac{p_i}{p_e} - 1\right) $$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two terms are the work to move the fluid (ideal gas) across a pressure difference and the work done as the gas expands/contracts across that pressure difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we assume that $p_i/p_e \approx 1$, we obtain the answer found by the (incorrect) assumption of an incompressible fluid. To be blunt, a way can always be found in a physical system to validate an assumption that pressure change is small across a control volume. Nothing physically ever validates a blanket starting assumption that gases are incompressible!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This demonstrates, an analysis using a compressible fluid assumption for a gas will fail as $p_i/p_e$ deviates from unity, as mass flow increases, and as inlet velocity increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining derivation for a compressible, ideal gas is left as an exercise for the reader.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-09-16T13:37:46.530" LastActivityDate="2018-09-16T13:37:46.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23774" PostTypeId="1" CreationDate="2018-09-15T18:17:48.520" Score="0" ViewCount="514" Body="&lt;p&gt;The problem as well as my work so far are in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b0kWF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b0kWF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I begin with equation (2.23) and use the divergence theorem to express the second term as the surface integral part of the problem equation. From there I quickly end up with what appears to be an equation stating that the partial derivative of the scalar field at x is the same as the full derivative. This would imply that the integral of the convective part is zero. Since this is a general case I don't think I can make that statement. Either I've made a mistake somewhere or I am not seeing how to proceed from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would appreciate help in figuring out which of those two it is and how I should proceed from here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing that caught my eye repeatedly is the definition of the function for the problem. By definition and context I believe it is assumed that small x is always a function of time and initial position X, as in equation (2.20). Also, I do not fully understand why the second term of the final expression in (2.20) has to be evaluated at x. &lt;/p&gt;&#xA;" OwnerDisplayName="user14808" LastActivityDate="2018-09-28T09:12:24.123" Title="Reynolds transport theorem and eulerian material derivative" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23776" PostTypeId="1" CreationDate="2018-09-15T20:09:56.723" Score="2" ViewCount="253" Body="&lt;p&gt;Old hard drives are my favourite source of super-powered fridge magnets. I have a bunch of them holding up countless childrens' drawings. But the magnets themselves are stuck to base plates of various shapes, some with pins and protrusions which are sometimes unhelpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering what the best process would be to remove the magnets from those plates? They're definitely not stuck on solely by their own magnetism, so my best guess is &quot;some sort of glue&quot;. I don't want to take a saw or screwdriver to them, because they're probably way too firmly stuck and also too brittle -- I'd just break them (possibly spectacularly so). I have tried sawing off the odd parts of one base plate, but getting the metal particle sawdust away from (or off of) the magnet is quite a challenge. I have also tried to twist and pry those pins off, but they're stuck in their holes better than my measly DIY-ers toolbox can undo (I don't have access to a metalworking shop).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So ... Heat? Acid? Magic? Some other SE stack?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gBJLt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gBJLt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1011" LastActivityDate="2020-09-22T19:58:05.777" Title="How do I separate these powerful magnets from their base plates?" Tags="|magnets|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23777" PostTypeId="1" AcceptedAnswerId="23779" CreationDate="2018-09-15T21:33:32.117" Score="3" ViewCount="695" Body="&lt;p&gt;My limited knowledge in engineering tells me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Titanium is expensive, stronger and heavier than aluminium, lighter and weaker than high yield steel.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The most significant advantage of titanium is the good strength to weight ratio.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In case of submarines, weight is not really a big concern, as long as the weight is manageable (less than the displacement and bigger than displacement - maximum ballast mass).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The mentioned points exclude the use of titanium. Since the designers decided otherwise, and surely had some good reasons, I wonder where my misunderstanding is?&lt;/p&gt;&#xA;" OwnerUserId="13207" LastEditorUserId="6790" LastEditDate="2020-05-04T08:02:38.810" LastActivityDate="2023-11-16T06:28:26.417" Title="Why did the Soviet Alfa-class submarines have a titanium hull?" Tags="|materials|metals|marine-engineering|naval-engineering|submarines|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="23778" PostTypeId="2" ParentId="23776" CreationDate="2018-09-15T22:58:11.133" Score="2" Body="&lt;p&gt;You don't want to use heat on a neodymium magnet, as it will remove the magnetism. You are correct that adhesive is used to bond the magnets to the base. I've been successful in the same manner suggested by Solar Mike, with a screwdriver and hammer, but have managed to fracture more than one in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish a more controlled removal process, an arbor press or hydraulic press with a jig to secure the backing plate while applying pressure to the edge of the magnet may give the best results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've not tried a solvent, but acetone or MEK or similarly dangerous chemicals applied to the edges may help with release.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the screwdriver and hammer option is best, if you can afford to damage one or two. Alternative, a very sharp metal chisel (not wood chisel) or screwdriver with a sharpened edge will add a leverage moment to the shearing force, improving your chances.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-09-15T22:58:11.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23779" PostTypeId="2" ParentId="23777" CreationDate="2018-09-16T01:16:22.430" Score="6" Body="&lt;p&gt;the former soviet union had plentiful stores of titanium on hand, so its expense was not an issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being nonmagnetic, a titanium hull would be more difficult to detect with magnetometers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that in a properly-designed submarine pressure hull, the limiter is not tensile strength but compressive yield strength and elastic modulus, since the hull will fail by compressive collapse like a short column or buckling as for a long column.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-16T01:16:22.430" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23780" PostTypeId="1" CreationDate="2018-09-16T05:07:16.697" Score="1" ViewCount="691" Body="&lt;p&gt;Electronic components like transistors, diodes etc have maximum voltage ratings which when exceeded damages the respective component. Is voltage sufficient to damage a component or there must be a sufficient current? I mean one may place these components in an external electric field (static field not dynamic) such that the voltage between two points exceeds their rated voltage capacity. However, there is no current supply as the component is in an electric field(in air), and not connected to a power source. Will the component be damaged?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the electric field is strong enough to exceed components voltage rating but not high enough to cause dielectric breakdown of air or semiconductor material.&lt;/p&gt;&#xA;" OwnerUserId="17373" LastEditorUserId="17373" LastEditDate="2018-09-17T04:41:43.307" LastActivityDate="2021-01-15T10:47:29.603" Title="Will strong electric field damage electronic components?" Tags="|electrical-engineering|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23781" PostTypeId="2" ParentId="23780" CreationDate="2018-09-16T07:06:40.673" Score="0" Body="&lt;p&gt;If a high voltage is applied directly and incorrectly to a pin then yes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the field can induce a high enough voltage in components then yes they can be damaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The component data sheet will give information about relevant limits.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-09-16T07:06:40.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23782" PostTypeId="1" CreationDate="2018-09-16T12:07:51.937" Score="0" ViewCount="64" Body="&lt;p&gt;Transistors and diodes have (reverse) breakdown volatge ratings. What is not specified in datasheets are how long such voltages must be applied for breakdown to occur. Suppose a diode has reverse breakdown voltage of 100V. Will applying 100 V (appropriate bias) for duration of 1 ps, 1 ns, 1 us, 1ms or 1s cause breakdown ? How long breakdown voltages must be applied for breakdown to occur?&lt;/p&gt;&#xA;" OwnerUserId="17373" LastActivityDate="2024-03-17T06:07:05.857" Title="How long breakdown voltages must be applied for breakdown to occur?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23783" PostTypeId="2" ParentId="23782" CreationDate="2018-09-16T12:50:03.067" Score="0" Body="&lt;p&gt;Some data sheets are more comprehensive than others. The better ones will show this information if it is relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the very common 1N4148 diode for example, this data sheet from &lt;a href=&quot;http://www.farnell.com/datasheets/1934602.pdf?_ga=2.215960215.1266357674.1537026328-135853076.1471879212&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fairchild&lt;/a&gt; gives &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For the maximum repetitive reverse voltage,it doesn't give a duration, but note the word &quot;repetitive&quot;, which means the duration of a &lt;em&gt;single&lt;/em&gt; pulse is irrelevant. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the &lt;em&gt;non-repetitive&lt;/em&gt; peak forward surge current, the duration &lt;em&gt;is&lt;/em&gt; specified, and different currents are specified for two durations, 1 microsecond and 1 second.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the reverse breakdown voltage, the test conditions are again specified - a non-repetitive square pulse of length 8.3 milliseconds, and two different levels of breakdown current.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the data sheet for a far eastern clone might not be so explicit!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-09-16T12:50:03.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23784" PostTypeId="2" ParentId="8668" CreationDate="2018-09-16T14:42:58.803" Score="-2" Body="&lt;p&gt;In worst condition when pump are installed on top it cause cavitation in which affect both performance of pump and discharge also pump have to work more in this condition thus more consumption of electricity.&#xA;In your given situation there no such problem of cavitation it happen in worst condition but by the same means it affect performance and discharge of your pumb&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let it's understand by a life example of you are not a technical person then also you can easily understand this...&#xA;Suppose you have to drink, soft drink using straw for this you have to suck with your mouth by applying some negative pressure from your mouth and you achieve this easily, now in situation second you are on fifth floor and bottle is on ground floor you have to suck the drink with your mouth via a long length straw now it is next to impossible because suction need more vaccum pressure in reverse of this if you have to blow some drink from full bottle by applying blow of air from your mouth it is quite easier than sucking it to mouth &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In same manner bump have to do more work on water to suck this from sump than it pushes to top &#xA;There is another problem also if you apply that much of work on water to suck up to high head the water absorb that much amount of energy and start converting in vapour and forming of bubbles happen now you pump pumping bubbles that means very less amount of water only air and air (technically called cavitation and due to bubbles formation life of pump will suffer and discharge goes down)&lt;/p&gt;&#xA;" OwnerUserId="17525" LastActivityDate="2018-09-16T14:42:58.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23785" PostTypeId="1" CreationDate="2018-09-16T16:25:04.023" Score="1" ViewCount="21" Body="&lt;p&gt;I'm designing a BLDC motor with some unique geometry constraints (~250mm long, ~45mm diameter with a ~35mm bore down the center) and am considering using straight copper bars carrying high current instead of multiple turns around a core for the stater windings (motor needs to provide high torque at or close to stall for extended periods).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there existing motor designs that are similar that I could read up on? What would be the correct terminology to describe such a design so I can search for prior art on the subject? &lt;/p&gt;&#xA;" OwnerUserId="2201" LastActivityDate="2018-09-16T16:25:04.023" Title="Correct terminology for a zero turn stator winding on a BLDC?" Tags="|motors|terminology|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23786" PostTypeId="2" ParentId="23780" CreationDate="2018-09-16T17:05:13.683" Score="1" Body="&lt;p&gt;In theory if the component is in a circuit containing a loop then the wires of that loop could act as an antenna to receive the disturbance from an EMP and create an internal AC current without requiring the device with the circuit to be powered at the time. &lt;/p&gt;&#xA;" OwnerUserId="2201" LastActivityDate="2018-09-16T17:05:13.683" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23788" PostTypeId="2" ParentId="21374" CreationDate="2018-09-16T19:41:02.210" Score="0" Body="&lt;p&gt;The guy in the video actually used an accelerometer to measure the linear acceleration of the building. If you followed you noticed he did not pay any attention to the displacement. While comparing the different cases, he focused only in acceleration.&#xA;Actually it is very common to measure acceleration while experimenting with vibrating objects. Since it is directly related to force, you may deduce useful information about possible mechanical failure and may also find standardizations dealing with allowed/recommended vibration levels of different devices (I realize you are not interested in those right now). You may also plot the results in both time and frequency domains to obtain natural frequencies, damping coefficients, time constants and similar nice and handy stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess you would use MATLAB to manipulate the recorded data - in this case, all you have to do is to integrate the acceleration over time in order to get the velocity, and then integrate the velocity over time to have the displacement. You will easily find a lot of ready to use codes out there. If for some reason you are not able to find any, let us know and we will help.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2018-09-16T19:41:02.210" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23790" PostTypeId="1" CreationDate="2018-09-17T08:47:35.243" Score="0" ViewCount="69" Body="&lt;p&gt;Pressure inside a water droplet is $0.32 N/cm^2$ if surface tension of liquid is $0.095 N/m$. We have to find the radius of the bubble &lt;/p&gt;&#xA;&#xA;&lt;p&gt;since it's a water droplet the formula used is $$P_i-P_o=\frac{2T}{R}$$&#xA;The $P_i$ is internal pressure wehreas $P_0$ is external pressure so the $P_0$ should be taken in consideration but the solution of the question directly puts $0.32$ in place of pressure difference. where I am doing wrong in the question.&lt;/p&gt;&#xA;" OwnerUserId="8808" LastActivityDate="2018-09-17T19:27:07.443" Title="What pressure to take in surface tension question?" Tags="|fluid-mechanics|pressure|fluid|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23791" PostTypeId="1" AcceptedAnswerId="23792" CreationDate="2018-09-17T11:30:26.127" Score="1" ViewCount="39" Body="&lt;p&gt;I am new to the Underfloor heating system, and a bit confused on the terminology for Operative Temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the temperature that is measured by the sensor or is this the set-point temperature on the thermostat?&lt;/p&gt;&#xA;" OwnerUserId="17534" LastActivityDate="2018-09-17T11:53:31.600" Title="Operative Temperature in Underfloor Heating?" Tags="|heat-transfer|heating-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23792" PostTypeId="2" ParentId="23791" CreationDate="2018-09-17T11:53:31.600" Score="0" Body="&lt;p&gt;That may well depend on who did the writing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The set point is the desired temperature and the sensor should measure the ambient temperature so that it heats, or not, according to where the ambient temperature is wrt to the set point.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-09-17T11:53:31.600" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23793" PostTypeId="2" ParentId="23790" CreationDate="2018-09-17T12:55:39.197" Score="1" Body="&lt;p&gt;Have you considered that the solution is assuming that the pressure is gauge pressure ie above atmospheric...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that means if you have the bubble pressure as an absolute pressure reading and you subtract the external atmospheric pressure you end up with the difference which is the value given in the solution.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-09-17T19:27:07.443" LastActivityDate="2018-09-17T19:27:07.443" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23794" PostTypeId="2" ParentId="23777" CreationDate="2018-09-17T15:40:46.267" Score="2" Body="&lt;p&gt;It was more a political decision than engineering. The problems with fabricating thick titanium plate are enormous = expensive, and the Russian engineers did a remarkable job. There is much information on the web, but I could not find information on the particular alloy , thickness, strength, toughness, etc. An interesting summary is &quot;Unravelling a Cold War Mystery&quot;.So the titanium hulls  technical success and a functional failure, as all these subs have been decommissioned. The cost of the titanium does not justify possible advantages. Remember the political decision Russia made not to have containment for the Chernobyl class nuclear reactors ? Politicians should probably not make engineering decisions. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-09-17T15:40:46.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23795" PostTypeId="1" AcceptedAnswerId="24219" CreationDate="2018-09-17T16:28:04.107" Score="2" ViewCount="277" Body="&lt;p&gt;As an energy engineer focused on energy efficiency in the built environment, I am familiar with the standards and codes covering energy use in buildings; specifically &lt;a href=&quot;https://www.ashrae.org/technical-resources/bookstore/standard-90-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASHRAE 90.1&lt;/a&gt;, the &quot;Energy Standard for Buildings,&quot; and &lt;a href=&quot;https://basc.pnnl.gov/resources/2018-iecc-international-energy-conservation-code&quot; rel=&quot;nofollow noreferrer&quot;&gt;IECC 2018&lt;/a&gt;, the &quot;International Energy Conservation Code.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recently several areas of increasing concern are intersecting with this work, such as indoor air quality, but also &lt;strong&gt;stormwater management&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Stormwater management is the effort to reduce runoff of rainwater or melted snow into streets, lawns and other sites and the improvement of water quality, according to the United States Environmental Protection Agency (EPA). &lt;a href=&quot;https://www.eecenvironmental.com/what-is-stormwater-management/&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Source)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is &lt;strong&gt;what engineering codes and/or standards govern stormwater management in and around buildings?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some teams we work with are concerned about how to manage the water which falls on the roof of the buildings they are designing. Are there guidelines related to collecting and/or treating this water before diverting it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: my work is within the U.S., but I'm also interested to know of standards or codes that are in use in other countries. &lt;/p&gt;&#xA;" OwnerUserId="17004" LastEditorUserId="17004" LastEditDate="2018-10-18T19:24:32.563" LastActivityDate="2021-08-04T02:06:10.567" Title="What standard and/or code covers building stormwater management?" Tags="|standards|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23796" PostTypeId="2" ParentId="23790" CreationDate="2018-09-17T18:07:07.827" Score="0" Body="&lt;p&gt;I think , a good hint is, the given value is the gauge pressure value.&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastActivityDate="2018-09-17T18:07:07.827" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23797" PostTypeId="2" ParentId="23765" CreationDate="2018-09-17T18:10:06.770" Score="4" Body="&lt;p&gt;let's call the cross section width B, base and its height H.&#xA;And assume the length of the section supported by the wall L1 and the short elbow L2 and call the load P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the deflection at the end of L2 will be the sum of 3 components: the deflection of L2 as a simple cantilever loaded with P, the deflection of L2 due to torsional rotation of L1 under the torque P(L2), and the deflection of L1 loaded as a cantilever beam with load P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torsion strain is  &lt;span class=&quot;math-container&quot;&gt;$$\tau_{max} = \frac {P\times L2} { \alpha\times B\times H^2\times G }$$&lt;/span&gt;&#xA; &lt;span class=&quot;math-container&quot;&gt;$$\phi = \frac { TL1} {\beta BH^3G }  $$&lt;/span&gt;&#xA;where G is the section shear modulus, alpha and beta are constants depending on the proportion of the base to height of the section. In this case they can be approximated for 0.246 and 0.249 assuming b is twice the height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we know the deflection of a cantilever beam with a single load at the end &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta = PL^3/3 EI $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we calculate the 2 components of the deflection and add &lt;span class=&quot;math-container&quot;&gt;$\phi \times L2 $&lt;/span&gt; to it.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-09-21T01:05:20.073" LastActivityDate="2018-09-21T01:05:20.073" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23798" PostTypeId="1" CreationDate="2018-09-17T19:51:53.857" Score="0" ViewCount="41" Body="&lt;p&gt;can anybody tell me what subject in Mechanical engineering deals with the making of External cases for electronic gadgets? I know I need to use CAD/CAM but I am asking this question because I want to know about the materials - How to determine what to use, where and why. Suppose I want to design an IP 67 casing for an electronic gadget, How would I go about it, where do I start?&lt;/p&gt;&#xA;" OwnerUserId="16787" LastActivityDate="2018-09-17T22:37:33.560" Title="what subject in Mechanical engineering deals with the making of External cases for electronic gadgets?" Tags="|mechanical|computer-aided-design|product-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23799" PostTypeId="1" AcceptedAnswerId="23800" CreationDate="2018-09-17T20:36:12.367" Score="0" ViewCount="62" Body="&lt;p&gt;I would be grateful if someone could explain me what does the value of zeros of a transfer function tell about the dynamics of the system.&lt;/p&gt;&#xA;" OwnerUserId="17539" LastActivityDate="2018-09-17T21:08:44.263" Title="Zeros of a Trransfer Function" Tags="|control-engineering|control-theory|transfer-function|systems-design|regulations|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23800" PostTypeId="2" ParentId="23799" CreationDate="2018-09-17T21:08:44.263" Score="0" Body="&lt;p&gt;The zeros of a transfer function influence the dynamics of the system in multiple ways. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Consider the output of a given system for $u(t) = 0$ as a weighted linear combination $$y_a(t) = \sum_{i=1}^n R_i e^{P_it}$$ where $P_i$ are the poles of the system. The weights $R_i$ are influenced by the zeros.&lt;/li&gt;&#xA;&lt;li&gt;For a zero at $s = N_i$, a signal with the complex frequency $N_i$ is not transmitted anymore, i.e. input signals $u(t) = e^{N_i t}$ are &quot;blocked&quot; by the system.&lt;/li&gt;&#xA;&lt;li&gt;Zeros are invariant towards state feedback, i.e., while the poles of a system can easily be moved by a corresponding state feedback, that is not possible for the zeros. &lt;/li&gt;&#xA;&lt;li&gt;It can be shown that a system with a zero in the right half-plane will always be unstable for high feedback gains and that the possible control performance for such a system is limited.&lt;/li&gt;&#xA;&lt;li&gt;The zeros of the transfer function of the open loop describe an internal dynamic of the system, which is not observable. This internal dynamic was initially described by Isidori as &lt;strong&gt;zero dynamics&lt;/strong&gt;. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The zero dynamics describe the internal dynamic of a system for the special case of an input $u(t)$ and an initial condition $x(0)$ such that the output $y(t) \equiv 0$ for all times $t \geq t_0$. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further reading on this pretty interesting topic, I'd start right at the source with Isidori's book &quot;Nonlinear Control Systems: An introduction&quot;&lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2018-09-17T21:08:44.263" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23801" PostTypeId="2" ParentId="23798" CreationDate="2018-09-17T22:37:33.560" Score="1" Body="&lt;p&gt;the two main methods of making hard cases for electronic gadgets are 1) injection molding and 2) blow-molding, vacuum forming or sheet molding (these have similarities as described below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Injection molding is the squirting of molten plastic into a cavity in a metal mold, which is then opened up to remove the finished part. It is used when the part has to have a lot of fine details accurately included in it, which is easily done by appropriately machining the mold cavities. The housings for digital cameras and cell phones are made this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Blow-molding is done by pumping air into the center of a glob of molten plastic that is inside a mold cavity. The glob inflates and adopts the shape of the cavity. The mold is then split open to remove the finished part. Inexpensive plastic and glass bottles are made this way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vacuum-forming is done by heating up a thin sheet of plastic until it is soft and then draping it over a form. Small holes in the form are connected to a vacuum source, which then sucks the soft sheet into close contact with the form so the plastic sheet assumes the shape of the form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sheet molding is similar but may use air pressure on one side of the sheet instead of vacuum on the other side to push the soft plastic into contact with the form. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sheet molding, blow molding and vacuum-forming are used to make inexpensive carrying cases for small electrical appliances, the side pieces of brief cases, and hard-sided luggage. Grades of plastic sheet are available with reinforcing fiber in them which significantly increases the strength of the finished part. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-17T22:37:33.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23805" PostTypeId="2" ParentId="21826" CreationDate="2018-09-18T08:59:03.960" Score="0" Body="&lt;p&gt;I'm assuming you know what to do with the &quot;single input&quot; once you have it, so I'm not going to talk about control algorithms - just how to aggregate the inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your input is two gyros, providing in perfect conditions the same exact value; the values will differ though due to various 'inadequacies' of the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One simple method is averaging their data. Sum, divide by two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your input data is noisy, use a smoothing algorithm (weighted moving average, WMA). I like the ARX model which is computationally very lightweight and provides nice, well-tunable results: $ \text{WMA}_n = w \   p_n + (1-w) \   \text{WMA}_{n-1} $ where $p_n$ is the current sample, $\text{WMA}_n$ is the result ready for further use, $W_{n-1}$ is the previous result (obtained from previous sample), and $w$ is a smoothing factor, $0 &amp;lt; w ≪ 1 $, higher values like 0.1-0.2 provide faster reaction, lower values like 0.001 provide good smoothing. Just mix all these samples taking them from both inputs,  never caring whether current p_n comes from one gyro or the other, and it will work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can take a more advanced approach, e.g. evaluate the reliability of inputs. For that you need at least three values - but we can manufacture the third one, using e.g. the same ARX model as above. Keep calculating the moving average as above, but don't use it for control. Instead, compare the two inputs to it, and reject the farther one. This helps a lot with signal that has short high spikes, e.g. from EM pulse-type noise, static discharges - or terrain bumps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, independently, you can include fault detection and redundancy. Keep track of how much the two inputs differ from each other and if they do differ by much for a long time (again, WMA of the difference, $|p^A_i - p^B_i|$, with a long time constant), you have a condition of a sensor fault. You can estimate which sensor by keeping two WMAs, each separate for one sensor, summing the differences between readouts and average over a &quot;typical&quot; time and comparing to typical values (obtained experimentally). $ F = { 1 \over n-k} \sum_{i=k}^{n} |\text{WMA}_i - p_i| $. Very low $F$ means the sensor is stuck and producing a constant; very high $F$  means the sensor broke and produces white noise. Generally, the $F$ further from typical will indicate the faulty sensor. Report the fault and exclude the faulty sensor from further use until the error condition is gone.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-09-18T08:59:03.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23806" PostTypeId="2" ParentId="8668" CreationDate="2018-09-18T09:21:09.553" Score="2" Body="&lt;p&gt;Considering this is a &lt;strong&gt;centrifugal water pump&lt;/strong&gt;, it can provide very little &lt;strong&gt;air suction&lt;/strong&gt;. That means if your pump is airlocked, there's not a shade of a chance it will be able to produce nearly enough suction to pull the water up two floors. And it will be airlocked all the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a check valve preventing backflow into the mains, and then filling the water lines and the pump with a bucket, you might have chance to get the 'top pump' config started... until it gets airlocked one way or another again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;low pump&quot; config will get enough mains pressure to fill the pump with water and then the pump will have no trouble handling the height.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-09-18T09:21:09.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23807" PostTypeId="2" ParentId="23780" CreationDate="2018-09-18T10:32:20.870" Score="0" Body="&lt;p&gt;A constant strong electric field might have an adverse effect on elements that utilize magnetic field, especially moving - motors etc. But the electronics can work at pretty much any potential as long as it's a &lt;em&gt;constant&lt;/em&gt; potential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OTOH a field with a strong differential, or rapidly changing will fry your electronics quick. Yes, the voltage alone doesn't hurt, you need a current - but with sufficient voltage the current will happen all by itself easily. So, if half of the board happens to be on the negative side and the other on the positive side of the field, you can expect a malfunction; if it's allowed to discharge, expect damage. And if you rapidly induce a massive electromagnetic wave through e.g an EMP bomb, sparks will fly.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-09-18T10:32:20.870" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23808" PostTypeId="2" ParentId="8774" CreationDate="2018-09-18T10:33:41.683" Score="-1" Body="&lt;p&gt;Try this post from CNC Zone, where an explanation is given on making your own eccentric spacers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cnczone.com/forums/cnc-router-parts/208920-diy-eccentric-spacers-less-40c.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cnczone.com/forums/cnc-router-parts/208920-diy-eccentric-spacers-less-40c.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/9t0HL.jpg&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/EW7SZ.jpg&quot; alt=&quot;&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/fJnF5.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17546" LastEditorUserId="17546" LastEditDate="2018-09-18T10:40:15.110" LastActivityDate="2018-09-18T10:40:15.110" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23809" PostTypeId="2" ParentId="23763" CreationDate="2018-09-18T11:16:30.690" Score="0" Body="&lt;p&gt;If the system is isolated, Tsurr not going to affect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the system is isolated, (no mass and energy transfer between the system and the surrounding ) then the work could be extracted from the system will be still less than what it contains in terms of energy. This is due to the irreversible path of the work extraction from the system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate how much work could be extracted from the system, you could do exergy (not energy) analysis.  &lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-09-18T11:16:30.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23811" PostTypeId="1" CreationDate="2018-09-18T13:55:55.887" Score="1" ViewCount="946" Body="&lt;p&gt;I have a &quot;Micropump GB-P25.JVS.A&quot; gear pump. I pump in the range between 100 and 500 ml/min up to a column about 10 cm above the pump level. The column is filled from the top and has a bottom outlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have noticed lately that my pump allows the fluid to flow backwards when it is not on. I assume this is caused by the small pressure head caused in the column. The backflow is small, but it is there. This was not the case a couple of months ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, one can forcefully pump the fluid from the inlet side of the pump using a syringe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that either the gears or the O-ring are damaged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other possibilities I might be missing? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How hard is it to change an O-ring of this kind of pump?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other possible fixes to the situation? I have attached check-valves before and after the pump, but I doubt this is the ideal solution.&lt;/p&gt;&#xA;" OwnerUserId="6405" LastEditorUserId="6405" LastEditDate="2018-09-18T14:01:09.683" LastActivityDate="2018-09-19T15:59:15.367" Title="Gear pump has backflow and does not stop forced flow" Tags="|gears|pumps|flow-control|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23812" PostTypeId="1" CreationDate="2018-09-18T14:18:14.383" Score="2" ViewCount="155" Body="&lt;p&gt;As a graduation project, I was assigned to solve the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can the vehicle’s behaviour be simulated in the form of a testing programme, with which the VCU software can be fully tested for correct functioning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The VCU here is the central control unit that monitors the Engine control unit and the Transmission control unit. Additionally, it controls the Power Distribution Modules. These power and measure various sensors and systems around the vehicle. The VCU also contains the vehicle’s main interface to the driver by means of a display and buttons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The VCU is a complete display and controller package from Murphy, the manufacturer. It communicates with the connected devices via 2 CAN busses. The VCU is not directly programmed with code, but configured by PowerVision, a design program for the display, made available also by Murphy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In PowerVision, control and check loops, functions and systems are designed. Errors can be made during designing though, which makes them not having the intended effects on the connected hardware. This is what must be tested with the testing programme mentioned in the assignment above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since my experience with software testing is quite non-existent, I’m not sure how to approach this correctly, most efficiently, or most effectively. Hence my question: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How should the assignment to test software for a vehicle’s VCU be approached?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;VCU in question: &lt;a href=&quot;https://www.enovationcontrols.com/products/displays/powerview-780&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.enovationcontrols.com/products/displays/powerview-780&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9920" LastEditorUserId="9920" LastEditDate="2018-09-19T07:15:08.283" LastActivityDate="2018-09-22T16:41:02.653" Title="How should the assignment to test software for a vehicle’s VCU be aproached?" Tags="|automotive-engineering|simulation|software|product-testing|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23814" PostTypeId="1" CreationDate="2018-09-18T16:38:49.690" Score="2" ViewCount="25" Body="&lt;p&gt;Are there any visibility/height requirements with regards to plumbing air-gap design/installation?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're looking at using an air gap device to allow HVAC condensate drip to enter a buildings drainage system.  I've seen this before on residential applications where the washing machine drain and HVAC condensate drip are combined into a single box using a product such as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.siouxchief.com/products/supply/access/laundry-boxes/oxbox-washing-machine-outlet-box?template=P3Technical&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.siouxchief.com/products/supply/access/laundry-boxes/oxbox-washing-machine-outlet-box?template=P3Technical&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ferguson.com/product/lancaster-pump-34-x-1-12-in-fip-x-mip-multipurpose-air-gap-lg175/_/R-4376003?skuId=4376003&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ferguson.com/product/lancaster-pump-34-x-1-12-in-fip-x-mip-multipurpose-air-gap-lg175/_/R-4376003?skuId=4376003&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.freshwatersystems.com/p-328-mr-drain-air-gap-md-10a-15-or-2-standpipes.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.freshwatersystems.com/p-328-mr-drain-air-gap-md-10a-15-or-2-standpipes.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Customer is requesting that air-gap boxes be located below counter level with cutouts in the back of the installed cabinets.  Condensate from units on second floor too low for air gap to be installed in second flow.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the concern behind my question.  What happens if there is a system backup due to a clog and my air-gap is the lowest point?  Obviously it would result in an overflow.  If the air-gap is visible this would result in simple cleanup.  If the air-gap is hidden behind cabinets or in a wall then the overflow would go unnoticed for a period of time and result in a much larger (expensive) cleanup.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems like there should be code requirements that prevent hidden air-gaps or require them to be located above a nearby sink or other drain that is visible so that an overflow is &quot;contained&quot; in a visible area.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.  &lt;/p&gt;&#xA;" OwnerUserId="17549" LastActivityDate="2018-09-18T16:38:49.690" Title="Visibility/Height Requirements for Plumbing Air-Gaps (Commercial Building)" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23817" PostTypeId="2" ParentId="11664" CreationDate="2018-09-19T09:13:45.747" Score="0" Body="&lt;p&gt;I think, unless you are planning for mass production, you need not search for any standards.you can experiment and decide the values yourself.(Iam not sure about this suggestion.But from what I learnt IT grades are used in mass production of hole shaft assemblies because machining cannot be done precisely and identically for all the parts.In case of single components, I think its not needed)&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastActivityDate="2018-09-19T09:13:45.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23819" PostTypeId="1" AcceptedAnswerId="23822" CreationDate="2018-09-19T13:54:24.097" Score="0" ViewCount="1620" Body="&lt;p&gt;How would one describe the difference between measuring and gauging? As I understand it at my company, we use a fixed tool to &lt;code&gt;gauge&lt;/code&gt; something, but we will use a set of calipers to &lt;code&gt;measure&lt;/code&gt; something. Is this a correct understanding? &lt;/p&gt;&#xA;" OwnerUserId="17486" LastActivityDate="2018-09-19T15:51:48.257" Title="Measuring vs gauging" Tags="|measurements|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23820" PostTypeId="1" AcceptedAnswerId="23826" CreationDate="2018-09-19T13:54:42.010" Score="4" ViewCount="4423" Body="&lt;p&gt;Apologies if I've got the terminology wrong, I'm in now way involved in engineering, but I work in a high rise in a city with a lot of construction going on.  I see a lot of different types of cranes, and it's got me wondering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two most common types I see are (I think) both tower cranes, however Some have the classic T shape, where the arm stays level, whilst the others have an are that raises up and down.  Both types seem to be working on the same types of sites, so I've been wondering what the difference between the two types are, and why you'd pick one over the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are examples of what I'm talking about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QgGqB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QgGqB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zc8pN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zc8pN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17561" LastEditorUserId="1832" LastEditDate="2018-09-19T16:01:15.947" LastActivityDate="2018-09-20T08:00:46.907" Title="What's the difference/advantages/disadvantages to a tower with a moving jib, vs a flat jib" Tags="|building-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23821" PostTypeId="1" CreationDate="2018-09-19T14:03:30.630" Score="1" ViewCount="572" Body="&lt;p&gt;I try to implement a state observer (Luenberg Observer) to estimate the speed of my motor. Only the absolute position of the motor can be measured and I want to take into account also the target velocity as the input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my derived state space model looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The states are position, velocity and acceleration&#xA;$$&#xA;x = \begin{pmatrix}&#xA;p \\&#xA;v  \\&#xA;a&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Input is the velocity target vi minus the estimated velocity state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;u=(v_i-v)&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system matrices looks like this (T is the sampling time):&#xA;$$&#xA;A = \begin{pmatrix}&#xA;1 &amp;amp; T &amp;amp; \frac{T^2}{2} \\&#xA;0 &amp;amp; 1 &amp;amp; T  \\&#xA;0 &amp;amp; 0 &amp;amp; 0 &#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;B = \begin{pmatrix}&#xA;T \\&#xA;1  \\&#xA;\frac{1}{T} &#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;C = \begin{pmatrix}&#xA;1 &amp;amp; 0 &amp;amp; 0&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;D=0&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I move the input -v into the A matrix, it looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_{k+1} = \begin{pmatrix}&#xA;1 &amp;amp; 0 &amp;amp; \frac{T^2}{2} \\&#xA;0 &amp;amp; 0 &amp;amp; T  \\&#xA;0 &amp;amp; -\frac{1}{T} &amp;amp; 0 &#xA;\end{pmatrix} x_k + \begin{pmatrix}&#xA;T \\&#xA;1  \\&#xA;\frac{1}{T} &#xA;\end{pmatrix} v_i&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;y_k= \begin{pmatrix}&#xA;1 &amp;amp; 0 &amp;amp; 0&#xA;\end{pmatrix} x_k&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I tried to place the poles of the error dynamics with a Luenberg observer by choosen L for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;A-LC = \begin{pmatrix}&#xA;1-L_1 &amp;amp; 0 &amp;amp; \frac{T^2}{2} \\&#xA;-L2 &amp;amp; 0 &amp;amp; T  \\&#xA;-L3 &amp;amp; -\frac{1}{T} &amp;amp; 0 &#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I placed the poles in matlab as follows:&#xA;$$ |\lambda_i|&amp;lt;1$$&#xA; at [-0.5,-0.55,-0.6]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gave me for L:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;L = \begin{pmatrix}&#xA;2.65 \\&#xA;2.92 \times 10^3  \\&#xA;-1.9 \times 10^5&#xA;\end{pmatrix}&#xA;$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I embedded this observer in the real system, but the results are actually very bad. So I think I did something wrong with the state space model, especially the input u seems not to be a good idea. Has anyone an idea how else I can model this system, especially about the input u I'm very unsure. Should I use the actual current as input for u, since this is proportional to the torque and so the acceleration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; What i further see is, that state a is not observable (Observability matrix has determinante different from zero), but since I don't want to observe the acceleration (I give this directly as an input), I think this should not be a problem, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt; Well, my Luenberger gains are pretty high, so I think that might be a problem. I update the state as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$&#xA;x_{k+1}=(A-LC)x_k + B v_i + Ly_k&#xA;$$ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;y is my measured position at instance k, so let's say this is 4000 (12 bit encode resolution). Initially my states are zero. Then the acceleration will be directly 4000*(-1.9 * 10^5), huge! So  I think I really messed up something here.&lt;/p&gt;&#xA;" OwnerUserId="8784" LastEditorUserId="8784" LastEditDate="2018-09-19T15:31:06.070" LastActivityDate="2018-09-19T20:15:01.263" Title="State observer for speed estimation from encoder" Tags="|control-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23822" PostTypeId="2" ParentId="23819" CreationDate="2018-09-19T15:51:48.257" Score="3" Body="&lt;p&gt;I think you have it correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We used &quot;go / no go&quot; gauges to check some things and made measurements for others.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-09-19T15:51:48.257" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23823" PostTypeId="2" ParentId="23811" CreationDate="2018-09-19T15:59:15.367" Score="2" Body="&lt;p&gt;You will probably need to install a check valve (or manual valve) to prevent the backflow, and rebuild the pump heads periodically. Over time, the gears wear down and small gaps form. This both leads to reverse flow when the pump is off and reduces the liquid flow rate for a given pump speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used to work at a plant that used gear pumps that look similar to the one you have. When the pumps were in service, they were used continually, but when out of service they were manually valved out to prevent backflow like you describe. The pumps were connected to a flow meter and VFD to maintain the same flow rate despite pump wear. Additionally, we rebuilt the pump heads periodically (yearly if I remember right). I'm pretty sure the manufacturer sold rebuild kits for that purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side note: Eventually, we switched to peristaltic pumps which were much easier to control and maintain.&lt;/p&gt;&#xA;" OwnerUserId="17151" LastActivityDate="2018-09-19T15:59:15.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23824" PostTypeId="2" ParentId="23821" CreationDate="2018-09-19T20:15:01.263" Score="1" Body="&lt;p&gt;Acceleration shouldn't be part of the state unless you jerk would be your input. And tracking a target velocity seems more of a job for the controller using feedback. So your system input is more likely a force, which induces an acceleration, and the states position and velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inorder for an observer to work well you need to provide it with a good model of your system. Namely the motor will likely have some friction and maybe even some delay. For obtaining a model you could for example try to fit a &lt;a href=&quot;https://en.wikipedia.org/wiki/Autoregressive_integrated_moving_average&quot; rel=&quot;nofollow noreferrer&quot;&gt;ARIMA&lt;/a&gt; model onto measured data or use &lt;a href=&quot;https://en.wikipedia.org/wiki/Eigensystem_realization_algorithm&quot; rel=&quot;nofollow noreferrer&quot;&gt;ERA&lt;/a&gt;. The state space representation of a (minimal) LTI system is not unique, namely they are equivalent under a &lt;a href=&quot;http://lpsa.swarthmore.edu/Representations/SysRepTransformations/SS2SS.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;similarity transformation&lt;/a&gt;. So if it is desired to have a state vector with position and velocity, you could obtain this by using a similarity transformation.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-09-19T20:15:01.263" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23825" PostTypeId="1" CreationDate="2018-09-19T20:31:46.977" Score="1" ViewCount="262" Body="&lt;p&gt;Is there a general rule of thumb for how accurate a gauge should be in compared to the parts being checked? In my chemistry class, I remember making standards where the instrument needed 1 more significant figure in compared to the standard solution we were creating. Is it the same rule for gauges?&lt;/p&gt;&#xA;" OwnerUserId="17486" LastEditorUserId="90" LastEditDate="2018-09-21T13:54:34.377" LastActivityDate="2018-09-21T13:54:34.377" Title="How accurate must my gauge be?" Tags="|measurements|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23826" PostTypeId="2" ParentId="23820" CreationDate="2018-09-19T20:54:56.657" Score="2" Body="&lt;p&gt;&lt;strong&gt;The key benefit of Luffing Jibs, is their ability to park at night in a smaller radius&lt;/strong&gt;. When a crane is 'out of service', it's a requirement that it is able to 'weathervane'. This means that in high winds, the jib must always rotate such that the jib is facing downwind, and the ballast serves to counteract the rotation force of the wind. Consequently, it must be impossible for the jibs to hit one another, whatever direction they are facing. For multiple 'saddle jibs' (classic T shape), they get around this by setting the jibs at different heights, so that they overlap. They still can't have the towers close together than the length of the lower jib, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4OUa8m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4OUa8m.png&quot; alt=&quot;Overlapping Saddle Jibs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the increase in pre-fab, it's often desirable to fit more towers onto a single site. Furthermore, strict overhanging regulations in cities may prevent you from siting a saddle jib close to the boundary of your site, in case it swings over (or hits) neighbouring buildings when free-slewing at night. A luffing jib can be parked at a steep angle, and thus with this smaller out-of-service radius, you can fit more cranes onto the site, and stay within the overhanging regulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l2lDim.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l2lDim.png&quot; alt=&quot;Luffing jibs on site&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a trade-off, however. If you try to park your jib at &lt;em&gt;too&lt;/em&gt; steep an angle (over ~60degrees), the jib no longer has enough leverage to rotate the slew ring around under wind load. This is why you now sometimes see a solid 'rectangle' in the tip of a luffing jib. This is a 'windsail', and is literally a sheet of steel (sometimes fabric) designed to increase the surface area that the wind can push against, allowing an even smaller park radius compared to a traditional luffing jib.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AzCqim.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AzCqim.png&quot; alt=&quot;Fixed windsail on top of Shard in London&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to go tighter still (over ~75degrees), you need more area again, but, if you have a large fixed sail, that would make driving the crane in service very difficult under a light wind, and would limit the maximum wind speed under which it would be safe to operate the crane. The solution is to have a retractable windsail, which is deployed at night, and retracted when the crane is in use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MxRqym.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MxRqym.png&quot; alt=&quot;Retractable Windsail&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;his innovation was a key part in Select Plant Hire being offered the contract to supply cranes for the Shell Centre on London's south bank, which had 21 cranes on a single site, 17 of which required retractable windsails. This allowed them to offer a faster production schedule than other companies that couldn't get so many cranes into the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q8kIAm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q8kIAm.jpg&quot; alt=&quot;Shell Centre Site with multiple retractable windsails&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I'm aware, saddle jibs are just more simple, and consequently a bit cheaper to use, and are used where possible by default. When you see a mixture of cranes on site, it's likely that the luffing jibs are sited near the edges, where overhanging regulations are an issue, with saddles in the middle to save money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RTMB8l.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RTMB8l.png&quot; alt=&quot;Saddle jib in middle, luffing at edges&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-09-20T08:00:46.907" LastActivityDate="2018-09-20T08:00:46.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23827" PostTypeId="1" CreationDate="2018-09-19T22:11:33.587" Score="-2" ViewCount="238" Body="&lt;p&gt;I want to get this out there. What is a typical and common  range of a small hole gauge?&lt;/p&gt;&#xA;" OwnerUserId="17486" LastActivityDate="2022-07-31T08:03:00.163" Title="Typical range of a small hole gauge." Tags="|measurements|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23828" PostTypeId="2" ParentId="23827" CreationDate="2018-09-19T22:11:33.587" Score="0" Body="&lt;p&gt;Small hole gauges usually come in sets and their range will typically vary between 0.125&quot; to 0.500&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is according to NTMA Catelog Number 5023 measuring and gauging in the maching shop chapter 4.&lt;/p&gt;&#xA;" OwnerUserId="17486" LastActivityDate="2018-09-19T22:11:33.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23830" PostTypeId="2" ParentId="23825" CreationDate="2018-09-19T23:50:06.580" Score="2" Body="&lt;p&gt;A measuring device will have as two main contributions to its accuracy and precision, a calibration uncertainty and a scale uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a ruler. It expands and contracts as it is heated and cooled. Consider a beam that is being measured by that ruler. The beam does not expand and contract at the same extent as the ruler. The measurements at two different temperatures will give two different lengths. This is the calibration uncertainty. It causes readings to be inaccurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a ruler marked in millimeters and one marked in inches. The former has a scale (device) uncertainty of $\pm 0.5$ mm. The latter has a device uncertainty of $\pm 0.5$ in $\approx 13$ mm. Device uncertainties affect the precision of the measurements, &lt;em&gt;not&lt;/em&gt; the accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best approach to obtain a rule of thumb is to use relative uncertainties rather than absolute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that you will measure a beam that is EXACTLY 1 m long regardless of temperature. You measure on a cold day and on a hot day using a ruler that expands and contracts by ... say ... 100 microns. Your relative calibration uncertainty is $100/10^6 = \pm 100$~ppm. Using the millimeter ruler, your relative device uncertainty is $0.5/10^3 =  \pm0.05$%. Using the inch ruler, your relative device uncertainty is $13/10^3 = \pm 1.3$%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb, I routinely teach (in a colloquial manner) that better than $\pm 0.5$% is national labs standards, better than $\pm 1$% is calibration standards, better than $\pm 5$% is highly regarded, better than $\pm10$% is routinely reproducible, better than $\pm15$% is engineering seat of the pants, and anything higher than $\pm20$% is gossip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these rules codified in some resource? Yes, in my hand-written or print out notes for some of my classes. Otherwise, no. The requirements for measurement tolerance (overall uncertainty) &lt;em&gt;may&lt;/em&gt; be codified in a document that defines how the component is to be used. At that point, your job is to find a measuring device (gauge) that has no more than that overall uncertainty, presumably being better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, it is NOT about the &quot;extra&quot; number of significant digits. It &lt;em&gt;is&lt;/em&gt; about the extent that you accept that you will have an uncertainty and will need to define/control it.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-09-19T23:50:06.580" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23834" PostTypeId="1" CreationDate="2018-09-20T09:47:58.737" Score="5" ViewCount="206" Body="&lt;p&gt;I am trying to research the common methods on an underfloor heating system that are available in the market. &#xA;I know there are usually, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On/Off control to control the flow of water into the pipe&lt;/li&gt;&#xA;&lt;li&gt;Outdoor reset control to control the supply water temperature.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think the most common is ON/OFF control (simple PI controller), but not sure about it. &lt;/p&gt;&#xA;" OwnerUserId="17534" LastEditorUserId="33" LastEditDate="2018-09-20T17:57:23.550" LastActivityDate="2024-03-23T18:07:30.853" Title="What is the most common control method in Underfloor heating?" Tags="|control-engineering|heat-transfer|heating-systems|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23837" PostTypeId="1" CreationDate="2018-09-20T15:50:18.067" Score="0" ViewCount="76" Body="&lt;p&gt;Stepper motor should drive a frame to tilt, but not 360 degrees, but only within 20 to - 20 degree.&#xA;What simple machine elements do you use to restrict the freedom of movement?&#xA;It must be a mechanical solution, not a software one.&#xA;Can it be only done through gear elements and no further addition elements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like the following&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=w7v2lVsjH4g&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=w7v2lVsjH4g&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="17577" LastEditorDisplayName="user14407" LastEditDate="2018-09-20T17:56:41.833" LastActivityDate="2018-09-20T17:56:41.833" Title="Limiting rotation to +20 to -20 degree" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23838" PostTypeId="2" ParentId="23837" CreationDate="2018-09-20T15:52:23.533" Score="1" Body="&lt;p&gt;Have a slot with a pin, the slot limits the distance the pin can move, then make sure the slot is machined to give +/- 20 degrees.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-09-20T15:52:23.533" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23841" PostTypeId="2" ParentId="22328" CreationDate="2018-09-20T16:30:08.083" Score="1" Body="&lt;p&gt;The best metric of confidence in making measurements is the overall relative uncertainty of the measurement. Two factors contribute to uncertainty for a measuring device. One is the calibration uncertainty. This is an offset or bias. The other is the device or scale uncertainty (precision).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a force-gauge with a spring. The spring constant depends on temperature. Readings taken at different temperatures will be different. This is an example of a calibration uncertainty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Device uncertainties are $\pm$ half the tick mark spacings whether the ruler is a millimeter ruler (where such markings are impractical) or a football field marked in 10 yard increments (where the best we can then reliable measure is $\pm$ 5&#xA;yards).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By example, consider a perfect 1 m beam. A ruler that expands $\pm 100$ microns over the temperature range of measurements will have a relative calibration uncertainty of $100/10^6$ = 100 ppm. A ruler that is marked in millimeters will have a relative scale uncertainty of $0.5/10^3$ = 0.05%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Rule of 10ths&quot; is subjective. It is never to be used to define a consistent statement about the precision of a measurement. When you need a component to have a specific size at a specific tolerance level, use a measuring device that has at least that relative uncertainty or better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ACCURACY and PRECISION are two different terms. The former is not defined by the measurement device, the latter is.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-09-20T16:30:08.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23842" PostTypeId="1" AcceptedAnswerId="23849" CreationDate="2018-09-20T16:34:09.007" Score="0" ViewCount="34" Body="&lt;p&gt;I have been given a spherical drug particle that dissolves by surface erosion into the bloodstream. The rate of dissolution is proportional to the exposed surface area of the particle. I have to model the volume of the dissolving particle as a function of time, which I did, and I got&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$$V(t) = \left(\dfrac{3V_0^{1/3} - kt}{3}\right)^3$$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where $V_0 = V(t=0)$, the initial volume of particle. When I plot using Excel with $V_0=100$, I get a reasonable downward sloping curve which reaches 0 in about 3 sec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have to derive the rate of drug release as a function of time (g/s or mol/s) from the volume function. There is no other information given. How do I do this? Please help!&lt;/p&gt;&#xA;" OwnerUserId="17579" LastEditorUserId="1832" LastEditDate="2018-09-20T18:27:54.883" LastActivityDate="2018-09-20T19:00:07.167" Title="How can I convert volume as a function of time to mass or mole as a function of time for a dissolving drug particle?" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23843" PostTypeId="1" AcceptedAnswerId="23844" CreationDate="2018-09-20T16:53:29.830" Score="0" ViewCount="46" Body="&lt;p&gt;I am a sophomore in mechanical engineering and am unable to understand the purpose and functioning of Oldham's Coupling.&lt;/p&gt;&#xA;" OwnerUserId="17574" LastActivityDate="2018-09-20T19:41:04.463" Title="Which is the best tutorial on Oldham's Coupling?" Tags="|mechanisms|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23844" PostTypeId="2" ParentId="23843" CreationDate="2018-09-20T16:59:40.197" Score="1" Body="&lt;p&gt;Have a watch of this video: &lt;a href=&quot;https://www.youtube.com/watch?v=utEKKox2WHA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=utEKKox2WHA&lt;/a&gt; It's a marketing video for a specific make/model of oldham coupling, but shows good footage of the coupling running at a variety of RPM's, including failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oldham couplings are used to allow transmission of torque between two parallel shafts, which may be misaligned (non-coaxial) . The allowable misalignment will be shown on the data sheet for each specific coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This may be a static misalignment, or dynamic, such as a driven roller that moves up/down according to the thickness of stock that it is running on top of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In failure an Oldham Coupling will not transmit any torque, to the output shaft. This contrasts with flexible Jaw Couplings (which can also accommodate some misalignment, although not as much), which, if the insert fails, will continue to transmit torque, albeit with some backlash. Which of these is appropriate depends on which failure mechanism is safer for the system at hand.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="615" LastEditDate="2018-09-20T19:41:04.463" LastActivityDate="2018-09-20T19:41:04.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23845" PostTypeId="1" CreationDate="2018-09-20T17:56:36.980" Score="0" ViewCount="3113" Body="&lt;p&gt;For a temporary job I'm working, one of the tasks we are sometimes required to do is to unroll these giant spindells of paper (think rolls of toilet paper the size of a trash bin except the paper is tough) and pack the paper into a box for recycling. The goal, really is to remove the cardboard center piece. The way they've been doing it is to cut through it with a knife. But this is tiring and takes a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a more efficient way of doing this? I was thinking about maybe cutting in the middle to try and remove the center piece but it's just in there so tight.&lt;/p&gt;&#xA;" OwnerUserId="17580" LastActivityDate="2018-09-21T20:27:04.907" Title="What is the most efficient way to remove the cardboard center piece from a roll of paper?" Tags="|manufacturing-engineering|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23846" PostTypeId="2" ParentId="23845" CreationDate="2018-09-20T17:59:10.050" Score="1" Body="&lt;p&gt;How about a small battery powered circular saw with depth set to the thickness of the roll?   Google &quot;as seen on tv circular saw&quot; to get some ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut it twice, pull out the narrow part, gives the larger part room to collapse.&lt;/p&gt;&#xA;" OwnerUserId="17242" LastEditorUserId="17242" LastEditDate="2018-09-20T18:47:50.997" LastActivityDate="2018-09-20T18:47:50.997" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23847" PostTypeId="1" CreationDate="2018-09-20T18:03:39.000" Score="2" ViewCount="585" Body="&lt;p&gt;Can anyone point me to a calculation or formula that would tell me what percentage an 6061 T6 aluminum tube OD should be able to be flared to, using any or all of the following: OD size, Wall size, yield strength, ultimate tensile strength and elongation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 3&quot; long sample that has a 2&quot; outside diameter with a wall size of 0.090&quot;. The method of flaring is utilizing an Instron tensile testing machine. The part is being forced down onto a 60 degree cone with a flat surface at the top. There is not a predetermined speed or force that the cross-head must move. The T, Y &amp;amp; E properties fall into the middle of the tolerance for the alloy and temper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize the calculation couldn't be precise, but I'm wanting a ballpark figure to work with as my customers include a minimum flaring percentage that the part must make and I don't know where they're getting their data from. They are coming from steel so I want to make sure they are not just bringing that measurement straight across.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The typical part is anywhere from 2&quot; to 3.5&quot; OD with a wall in a range from .075&quot; to 0.100&quot;.. The customer's spec for minimum flare percentage is from 10% to 15% depending on the part.&lt;/p&gt;&#xA;" OwnerUserId="17581" LastEditorUserId="17581" LastEditDate="2018-09-21T15:32:49.110" LastActivityDate="2019-04-20T23:02:15.443" Title="Aluminum tube expansion (flaring)" Tags="|materials|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="23848" PostTypeId="2" ParentId="23845" CreationDate="2018-09-20T18:45:07.873" Score="0" Body="&lt;p&gt;you could also get a car jack and fixture it in such a way that when you crank the jack handle, it presses the core out of the roll sideways. A simple version of this is called an arbor press and can be bought from tool supply stores. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also try a log-splitter, which uses a motor to drive a wedge down into a log and split it into pieces. here you would fashion a ram pin to replace the wedge and set up the paper roll so that the ram would push the roll core out when you turned the thing on. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-20T18:45:07.873" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23849" PostTypeId="2" ParentId="23842" CreationDate="2018-09-20T18:45:29.343" Score="2" Body="&lt;p&gt;This is actually more of a calculus question than anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That function gives you the volume over time. You are looking for the rate of change in volume over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the definition of the derivative of that function. So all we need to do is get &lt;span class=&quot;math-container&quot;&gt;$\dfrac{\partial V}{\partial t}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that, we define &lt;span class=&quot;math-container&quot;&gt;$V(x) = f(g(x))$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$g(x) = 3V_0^{1/3}−kt$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$f(x) = \left(\dfrac{g(x)}{3}\right)^3 = \dfrac{1}{27}g(x)^3$&lt;/span&gt;. Via the &lt;a href=&quot;https://en.wikipedia.org/wiki/Chain_rule&quot; rel=&quot;nofollow noreferrer&quot;&gt;chain rule&lt;/a&gt;, we know that &lt;span class=&quot;math-container&quot;&gt;$\dfrac{\partial V}{\partial t} = f'(g(x))\cdot g'(x)$&lt;/span&gt;. Therefore, we have that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{alignat}{4}&#xA;\dfrac{\partial V}{\partial t} &amp;amp;= &amp;amp;&amp;amp;\dfrac{1}{9}(3V_0^{1/3}−kt)^2\cdot(-k) \\&#xA;&amp;amp;= -&amp;amp;&amp;amp;\dfrac{k}{9}(3V_0^{1/3}−kt)^2&#xA;\end{alignat}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result here is negative (assuming a positive &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; and that the result between parentheses is also positive), which indicates that the volume is dropping over time. Obviously, the volume lost is precisely what's released into the bloodstream. So if that latter interpretation (rate of release) is most relevant, just remove the negative sign.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2018-09-20T19:00:07.167" LastActivityDate="2018-09-20T19:00:07.167" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23851" PostTypeId="1" CreationDate="2018-09-21T06:08:36.347" Score="1" ViewCount="111" Body="&lt;p&gt;What books would you recommend for an inventor entrepreneur?&lt;/p&gt;&#xA;" OwnerUserId="17585" LastActivityDate="2018-09-21T06:08:36.347" Title="Books and resources for an entrepreneur inventor?" Tags="|mechanical-engineering|design|" AnswerCount="0" CommentCount="1" ClosedDate="2018-09-21T12:03:16.553" ContentLicense="CC BY-SA 4.0" />
  <row Id="23852" PostTypeId="2" ParentId="21437" CreationDate="2018-09-21T07:02:17.297" Score="1" Body="&lt;p&gt;Flow is a function of head (=pressure) and the piping system - the longer, the thinner, the more valves etc. the less flow. This much you know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you can't know from your little experiment alone if the head will suffice to reach your tank. I suggest running a hose up to your roof, if you have any flow at all the head in your water system suffices. At least sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;23 feet is 6.9m, you need more than 0.7 bar. Water mains &lt;em&gt;should&lt;/em&gt; deliver 4 bar but this will depend of course. I never took time when filling a bucket, but half a minute for a gallon does not seem too bad.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We don't know your plans etc. I suggest when you know if you can fill a tank on your roof, you sit down, think your system through, make a sketch or two and show this here to see if it will work to solve your problem.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-09-21T07:02:17.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23854" PostTypeId="2" ParentId="23834" CreationDate="2018-09-21T08:02:51.033" Score="0" Body="&lt;p&gt;The typical appraoch I see in heating systems is to have a three way valve on the suction side of the heating pump. This mixes cool return into the flow into the heater. The controling varialbe is the flow temeprature, the controlled varaiable the setting of the valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/c2ZxC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c2ZxC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then adjust the target flow temperature based on the room temperature you want. Check if such a system works with your heating in terms of hydraulics, I assume a normal heating manifold with hydraulic compensator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this is from large heating systems and may be overblown for a domestic purpose.  &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2018-09-21T09:26:37.190" LastActivityDate="2018-09-21T09:26:37.190" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23855" PostTypeId="1" CreationDate="2018-09-21T08:47:44.110" Score="0" ViewCount="292" Body="&lt;p&gt;I am new to fluid mechanics and am trying to understand the fluid mechanics of a simple valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume we have a simple tube, which is supplied with a constant stream of liquid using a pump.&#xA;At the end of the tube is a valve, which closes periodically.&#xA;What kind of changes, would I except in the tube due to the closing of the valve?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my naive understanding, I would assume, that the velocity should approach zero (since there is no way, where the fluid could go), however the pressure should stay constant - there is no fluid flow, but the pump provides still constant pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this understanding correct? Do you maybe have some papers / literature / videos, where this is explained in more (maybe mathematical) detail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!!&lt;/p&gt;&#xA;" OwnerUserId="17589" LastActivityDate="2020-12-09T23:39:33.743" Title="Fluid dynamics of a valve" Tags="|dynamics|fluid|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23856" PostTypeId="2" ParentId="23855" CreationDate="2018-09-21T09:33:09.203" Score="1" Body="&lt;p&gt;This is a very broad question in hydraulics but i try to clarify some points. &#xA;First the type of pump defines the parameter such as pressure ... . As you can see in the image bellow, the turbo pump (blue curve) and the volumetric pump (red curve) show totally different behaviours. The horizontal axis represent the pressure and the vertical axis is flow per unit time.&lt;a href=&quot;https://i.stack.imgur.com/MIinW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MIinW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the valve just blocks the stream for a while and then opens again. If you block the stream for a turbo pump, it means nothing flows through the pipe and pressure rises, but for a volumetric pump ( for instance double acting cylinder) the pressure rises even in constant flow, so the safety valve here is indispensable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the pressure eventually reaches a constant value for the turbo pump, and increases for the volumetric pump, notice in both cases the velocity is zero, nothing can flow through the hydraulic circuit, while the valve blocks the stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice, the red curve here is an idealised curve, in reality it decreases with pressure, but we can neglect this.  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-09-21T09:33:09.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23859" PostTypeId="1" CreationDate="2018-09-21T13:03:21.053" Score="1" ViewCount="220" Body="&lt;p&gt;I'm seeking a precise description of engine torque.&lt;/p&gt;&#xA;&lt;p&gt;We talk about torque in terms of the twisting force that the piston exerts on the crankshaft with a radius equivalent of the rod throw.&lt;/p&gt;&#xA;&lt;p&gt;Assume these two scenarios:&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Engine 1:&lt;/strong&gt; 6 cylinder, 2.5L diesel making 480 Nm torque&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Engine 2:&lt;/strong&gt; 4 cylinder, 2.0L diesel making 390 Nm torque&lt;/p&gt;&#xA;&lt;p&gt;Cylinder volume on eng 1 is 2.5 / 6 = 0.416L&lt;/p&gt;&#xA;&lt;p&gt;Cylinder volume on eng 2 is 2.0 / 4 = 0.5L&lt;/p&gt;&#xA;&lt;p&gt;So its apparent that the cylindrical volume on the smaller displacement engine is larger.&lt;/p&gt;&#xA;&lt;p&gt;How come the 6-cylinder engine produces more torque, instead of only smoothening out the performance due do less crank rotation per firing event?&lt;/p&gt;&#xA;&lt;p&gt;I'm thinking that torque is measured with two crank rotations in mind.&lt;/p&gt;&#xA;&lt;p&gt;Am I correct on this assumption, and if not - how is engine torque described precisely?&lt;/p&gt;&#xA;" OwnerUserId="3873" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-09-21T13:03:21.053" Title="Measurement of ICE torque" Tags="|motors|automotive-engineering|torque|power|combustion|" AnswerCount="0" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="23860" PostTypeId="2" ParentId="23845" CreationDate="2018-09-21T13:48:36.187" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Get a piece of dowel of similar diameter to the internal diameter of&#xA;the toilet roll and place it over a bin.&lt;/li&gt;&#xA;&lt;li&gt;Get a small rubber/friction wheel and secure it to a pin so the pin&#xA;can be placed into an electric drill.&lt;/li&gt;&#xA;&lt;li&gt;Unroll the paper into the bin using the electric drill and&#xA;rubber/wheel.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If possible attach a number of rubber wheels to an axle that goes into the electric drill and you can unravel several rolls of paper simultaneously.  &lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-09-21T13:48:36.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23861" PostTypeId="1" CreationDate="2018-09-21T16:20:25.627" Score="1" ViewCount="109" Body="&lt;p&gt;How are all LTI systems also dynamical systems?&#xA;I understand that LTI systems are linear and time invariant and that dynamical systems describe how one state develops into another, but how would this make all LTI systems dynamical systems?&lt;/p&gt;&#xA;" OwnerUserId="17595" LastActivityDate="2018-10-26T01:02:14.707" Title="How are all LTI systems also dynamical systems?" Tags="|control-engineering|dynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23863" PostTypeId="2" ParentId="23845" CreationDate="2018-09-21T18:42:47.140" Score="0" Body="&lt;p&gt;Drill a hole in the tube. Stick a blow-off nozzle in the hole with 140 psi of shop air and float the tube right out. Air bearings. Once you got it figured out, make a hard pipe handle long enough to keep your face out of harm's way. I used to fix high-end &lt;a href=&quot;http://stiffypushpoles.com/stiffy-one-piece-push-poles/graphite-push-pole&quot; rel=&quot;nofollow noreferrer&quot;&gt;push-poles&lt;/a&gt; that folks had broken. This is how I got the things off my mandrel.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-09-21T18:42:47.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23864" PostTypeId="2" ParentId="23845" CreationDate="2018-09-21T20:27:04.907" Score="0" Body="&lt;p&gt;I'd go down the route of trying to cut the core from the inside-out, after which you could roll it in on itself, and remove it without unrolling the reel, or generally making a mess as would be the case with cutting/sawing from the outside-in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the thickness of the cardboard, this might be achievable using a hot-wire cutter, or you may need to use a reciprocating saw. I have, on a past project involving a similar problem where a very long cut had to be made (with no access from the outside), re-purposed an old bandsaw blade for this task. It was threaded through the tube in question, and attached to a strong tensioning spring at the far end. A standard handheld reciprocating saw was then used on its lowest speed to pull the blade back and forth. You have to take it slow, but the cut can be made, with very little manual effort. If I can do it with Aluminium, you can do it with cardboard!&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-09-21T20:27:04.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23865" PostTypeId="1" CreationDate="2018-09-21T23:30:22.040" Score="5" ViewCount="634" Body="&lt;p&gt;I am looking for a way to convert rotary motion to linear, but it should take less energy to get back to initial position than when going from initial position to the end position.&lt;/p&gt;&#xA;&lt;p&gt;Specifically, a screw mechanism pushes a spring with multiple of full motor rotations. Then it stays in this ending position for a while as screw prevents the spring from pushing the mechanism back. Now, I'm looking for a mechanism that will retract back to the initial position with less work than before, f.i. with a single motor turn in the same or opposite direction.&lt;/p&gt;&#xA;&lt;p&gt;For example, a lipstick mechanism works with a cam and a pin. Imagine that this mechanism pushes a spring and you can draw the pin out of the cam so the spring will push back. The problem is how to do that with a single rotational motor.&lt;/p&gt;&#xA;" OwnerUserId="17597" LastEditorUserId="3505" LastEditDate="2022-04-05T08:25:13.937" LastActivityDate="2023-12-26T12:00:50.103" Title="Quick return mechanism for spring-loaded screw" Tags="|mechanical-engineering|motors|mechanisms|mechanical|linear-motion|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23866" PostTypeId="1" AcceptedAnswerId="23869" CreationDate="2018-09-21T23:32:00.943" Score="3" ViewCount="255" Body="&lt;p&gt;I was recently walking down a road when a car came my way (I'm an inveterate jaywalker). I moved over to the sidewalk, only to notice it clearly wasn't a sidewalk at all. The pavement looked like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RFDyd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RFDyd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The paving stones are embedded into and perpendicular to the ground, making it rather uncomfortable to walk on!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the purpose of such pavement? Is this layout meant to improve drainage in case of rain (this is in London, so it's always raining)? But then why pave the ground at all instead of simply using grass or greenery to anchor the soil and prevent erosion?&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2021-07-22T06:26:54.307" Title="What's sort of &quot;sidewalk&quot; pavement is this?" Tags="|civil-engineering|highway-engineering|pavement|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23867" PostTypeId="1" CreationDate="2018-09-22T00:03:16.567" Score="3" ViewCount="48" Body="&lt;p&gt;Can CVD Chambers be used to lay down thick (~1 mm) layers of metal?&lt;/p&gt;&#xA;" OwnerUserId="17599" LastEditorUserId="61" LastEditDate="2018-09-25T12:31:33.593" LastActivityDate="2018-09-25T12:31:33.593" Title="Can chemical vapor deposition chambers be used to lay down thick layers of metal?" Tags="|chemical-engineering|metals|experimental-physics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23868" PostTypeId="2" ParentId="23867" CreationDate="2018-09-22T02:35:12.807" Score="2" Body="&lt;p&gt;you do not need a CVD system to deposit metals, a planar magnetron sputtering system is more suitable. But for layers of order ~1mm in thickness you would be better off with an electroplating process. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-22T02:35:12.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23869" PostTypeId="2" ParentId="23866" CreationDate="2018-09-22T08:48:45.513" Score="9" Body="&lt;p&gt;The technical term for this (in the UK) is &quot;deterrent paving&quot; and its function is to discourage pedestrians, cyclists and drivers from being where they are not supposed to be. In the OP's photo, the paved area apparently is leading to a &quot;dead end&quot; which is divided from the roadway by a crash barrier - i.e. this is not a &quot;footpath&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the photo on their web page, this company supplies the same type as in the OP's image: &lt;a href=&quot;http://www.pavingexpert.com/special.htm&quot; rel=&quot;noreferrer&quot;&gt;http://www.pavingexpert.com/special.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design isn't intended to improve rainwater drainage - the paving blocks actually cover the entire surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This appears to be the exact product in the OP's picture: &lt;a href=&quot;https://www.aggregate.com/sites/aiuk/files/atoms/files/charcon-landscaping-paving-elite-tds.pdf&quot; rel=&quot;noreferrer&quot;&gt;https://www.aggregate.com/sites/aiuk/files/atoms/files/charcon-landscaping-paving-elite-tds.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-09-22T08:48:45.513" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23871" PostTypeId="2" ParentId="23812" CreationDate="2018-09-22T14:48:28.340" Score="2" Body="&lt;p&gt;I am not too familiar with the VCU, PV780, or PowerVision. Based on your description PV780 can be configured to be the Instrument Cluster Panel (ICP) for Engine Control Unit/Module (ECM) or Transmission Control Unit. To keep a response short I have selected a Temperature  gauge of the IPC. There is a very good chance this information communicated to the IPC from the ECM via the CAN bus in most automobiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KgVTC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kIyTe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the attach information you device supports CAN 2.0B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sDIqs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sDIqs.jpg&quot; alt=&quot;Can 2.0B&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is CAN bus message format. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TlbGb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TlbGb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming that a Temperature gauge is component in the PV780 base IPC, then to test the functionality it the Temperature gauge CAN messages need developed and injected to the PV780 per the about message format. The test cases will need to validate both nominal, boundary conditions and error conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is set of test to validate the fuel gauge on &lt;strong&gt;Toyota&lt;/strong&gt;. Note the CAN ID in this case is 0x07C0, and data length is 0x08 &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;&lt;pre&gt;&lt;code&gt;# Combo Meter Fuel Empty + beep&#xA;IDH: 07, IDL: C0, Len: 08, Data: 04 30 03 00 01 00 00 00&#xA;#Combo Meter Fuel Empty&#xA;IDH: 07, IDL: C0, Len: 08, Data: 04 30 03 00 02 00 00 00&#xA;#Combo Meter Fuel Empty&#xA;IDH: 07, IDL: C0, Len: 08, Data: 04 30 03 00 04 00 00 00&#xA;#Combo Meter Fuel 1/4 tank&#xA;IDH: 07, IDL: C0, Len: 08, Data: 04 30 03 00 08 00 00 00&#xA;#Combo Meter Fuel 1/2 tank&#xA;IDH: 07, IDL: C0, Len: 08, Data: 04 30 03 00 10 00 00 00&#xA;#Combo Meter Fuel 3/4 tank&#xA;IDH: 07, IDL: C0, Len: 08, Data: 04 30 03 00 20 00 00 00&#xA;#Combo Meter Fuel 4/4 tank&#xA;IDH: 07, IDL: C0, Len: 08, Data: 04 30 03 00 40 00 00 00&#xA;#Combo Meter Fuel Empty&#xA;IDH: 07, IDL: C0, Len: 08, Data: 04 30 03 00 80 00 00 00&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Below are some references that I used to develop this response. They expand on this summarized response.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next you will need a tool for data ingestion. The tool that I am most familiar is &lt;a href=&quot;http://www.csielectronics.com/saint/saintov.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Systems Analysis INterface (SAINT)&lt;/a&gt;. This uses Labview front end and can be used to analyze data packages as will as inject data packets. I am sure their other tool. I did some digging and found a few which I have listed under references. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.cars.com/articles/should-i-worry-about-how-hot-my-engine-is-running-1420680334271/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Should I Worry About How Hot My Engine Is Running?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://ioactive.com/pdfs/IOActive_Adventures_in_Automotive_Networks_and_Control_Units.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adventures in Automotive Networks and Control Units&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.csselectronics.com/screen/page/simple-intro-to-can-bus/language/en&quot; rel=&quot;nofollow noreferrer&quot;&gt;CAN BUS EXPLAINED - A SIMPLE INTRO (2018)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/FqLDpHsxvf8&quot; rel=&quot;nofollow noreferrer&quot;&gt;CAN BUS EXPLAINED - A SIMPLE INTRO (2018) Video&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://galileosky.com/products/gps-tracking-devices/galileosky-7.0-lite.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Galileosky 7.0 Lite&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://galileosky.com/assets/files/docs/manuals_en/can-bus.-receiving-data-on-fuel-level-from-the-can-bus-(281009-v3).pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;User’s manual. Receiving data on fuel level from the CAN-bus&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.microchip.com/Developmenttools/ProductDetails/APGDT002&quot; rel=&quot;nofollow noreferrer&quot;&gt;CAN BUS Analyzer Tool&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.csselectronics.com/screen/product/can-bus-logger-canlogger1000&quot; rel=&quot;nofollow noreferrer&quot;&gt;CL1000: CAN BUS LOGGER&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.slideshare.net/Acromag/introduction-to-can-bus-technology&quot; rel=&quot;nofollow noreferrer&quot;&gt;Introduction to CAN Bus Technology&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.csielectronics.com/saint/manuals/overview.ppt&quot; rel=&quot;nofollow noreferrer&quot;&gt;Systems Analysis INterface - (Saint) Presentation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ni.com/white-paper/2732/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Controller Area Network (CAN) Overview&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-09-22T16:41:02.653" LastActivityDate="2018-09-22T16:41:02.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23872" PostTypeId="1" AcceptedAnswerId="23873" CreationDate="2018-09-22T15:16:45.920" Score="0" ViewCount="102" Body="&lt;p&gt;In order to conduct study, I need to produce controlled vibrations with amplitude (maximum) approx 200 um (micro) at frequency between (100-150) Hz. Vibration amplitude must be controlled with input voltage (Battery operable) while frequency remain constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried many things :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Eccentric Motor - frequency and amplitude, both varied with voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Solenoid - difficult to control with voltage. (If any way exists, please suggest me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Voice Coil Motor - Its very costly, I need solution around 10-20 $ max. as project constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Linear Resonant Actuator - It is the best solution, but I am not able to find any LRA which can produce vibrations in um (micro metre).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest me any product exists or ways to produce controlled vibrations.&lt;/p&gt;&#xA;" OwnerUserId="17603" LastActivityDate="2020-04-14T19:01:45.243" Title="Need to produce vibration amplitude of 200 um (approx) at natural (resonant) frequency, can be any frequency between 100 to 150 Hz" Tags="|motors|vibration|linear-motors|linear-motion|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23873" PostTypeId="2" ParentId="23872" CreationDate="2018-09-22T15:39:39.810" Score="0" Body="&lt;p&gt;You can make an AC electromagnet pretty cheaply. Wrap some enamel coated wire around a steel bolt. Then get a steel cantilever beam with a resonant frequency equal to the frequency of interest. Place the magnet just under the tip of the beam and drive it with an ac voltage at the resonant frequency. 200 um should be achievable with a few volts if you have a sufficiently big magnet. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-09-22T15:39:39.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23874" PostTypeId="2" ParentId="23865" CreationDate="2018-09-22T15:40:02.370" Score="0" Body="&lt;p&gt;If you dont need the spring to stay at the ending position for a while with the help of the screw, you can design the screw such that it facilitates back driving (i.e. once the applied torque by the motor is removed, the spring force will be able to drive the screw back to its original position.)&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastActivityDate="2018-09-22T15:40:02.370" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23875" PostTypeId="2" ParentId="22808" CreationDate="2018-09-22T16:16:12.980" Score="1" Body="&lt;p&gt;From the diagram, the weight of both the pistons act in our favour(i.e. try to transfer water from the left to right tank).The opposing force is the weight of the vertical liquid column(60m) in the middle.So if the sum of the piston weights is more than the weight of the vertical liquid column, the water will be transferred from left to right.(This is neglecting the height of the water in the tanks since the height was not mentioned in the diagram).I think it is better to add a few known masses to the pison and relate the weight and the time taken to fill the tank istead of using analytical equations.&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastEditorDisplayName="user17332" LastEditDate="2018-09-22T17:18:16.967" LastActivityDate="2018-09-22T17:18:16.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23876" PostTypeId="1" AcceptedAnswerId="23877" CreationDate="2018-09-22T19:56:47.653" Score="0" ViewCount="132" Body="&lt;p&gt;The same geometry and the same free body diagram. However, if we replace the material from steel to glass, for example. How does that influence the analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things I think are&#xA;1. steel is ductile and glass is brittle. This means the cross section when it fails will be different.&#xA;2. glass has lower yield stress, which means it is much easier to fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything else that I omit? Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="17605" LastActivityDate="2018-09-22T21:22:18.903" Title="Stress analysis for the same configuration but different material" Tags="|structural-analysis|statics|stresses|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23877" PostTypeId="2" ParentId="23876" CreationDate="2018-09-22T21:22:18.903" Score="1" Body="&lt;p&gt;If you are dealing with a simple Back-of-the-envelope calculation, then it should not matter, from stress perspective, what type of material you chose. All you have to do is to compare the results to the material allowable strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you are talking about a FEA problem, you should take care closely of the material you choose. Tools like ANSYS calculate the object deformation (which is totally affected by the material Young's modulus and Poisson's ratio) and then derive the stresses out of it.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2018-09-22T21:22:18.903" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23878" PostTypeId="2" ParentId="23847" CreationDate="2018-09-22T21:55:39.050" Score="1" Body="&lt;p&gt;Though there are numerous rules of thumb used by manufacturers, a rigorous approach to finding an optimal amount of flare for a given geometry requires numerical solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest of these solutions will typical use von Mises plasticity (J2-plasticity) in conjunction with a damage model such as Johnson-Cook model.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An axisymmetric model (2D) will give you the plastic strains in the part but will not be able to predict tensile failures around the circumference. A random distribution of initial material parameters is needed to generate strain localization prior to failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often it is easier for the manufacturer to just test a few samples rather than do numerical simulations for which there may not be in-house expertise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below, I've attached a few images of what a simulation might produce.  Consider the axisymmetric case first.  The initial condition is shown below.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0Zz6J.png&quot; width=&quot;300&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a lot of friction between the punch and the cylinder, you may get a buckled part (see below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LAeNL.png&quot; width=&quot;300&quot;/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If friction is low, you will see flaring.  Typically, a plastic strain of 0.25 - 0.5 is enough to cause tensile failure in 6061-T6 Al.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/X2ubc.png&quot; width=&quot;300&quot;/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better estimate can be found by doing a 3D simulation. See the figure below for a feel of what you will see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/tPoTc.png&quot; width=&quot;300&quot;/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many studies in the literature on the subject.  A few recent ones are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &quot;A theoretical study of the expansion metal tubes&quot; by Liu and Qiu, International Journal of Mechanical Sciences 114 (2016) 157–165.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &quot;Towards the characterization of fracture in thin-walled tube forming&quot; by Centeno et al., International Journal of Mechanical Sciences 119 (2016) 12–22.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &quot;Numerical and experimental study of axial splitting of circular tubular structures&quot; by Rouzegar and Karimi, Thin-Walled Structures 105 (2016) 57–70.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) &quot;Expansion and reduction of thin-walled tubes using a die:Experimental and theoretical investigation&quot; by Almeida et al., International Journal of Machine Tools &amp;amp; Manufacture 46 (2006) 1643–1652.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will get a good idea about the state of the art in this field in these papers.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-09-22T21:55:39.050" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23879" PostTypeId="2" ParentId="12382" CreationDate="2018-09-22T22:10:14.407" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Q:&lt;/strong&gt; Can a blimp enter Mar's atmosphere at the orbital speeds or faster as a glider?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A:&lt;/strong&gt;  No.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A glider relies upon upward lift force in order to avoid crashing into the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity's sake, &lt;a href=&quot;https://www.engineersedge.com/calculators/air-density.htm&quot; rel=&quot;noreferrer&quot;&gt;Earth's air density&lt;/a&gt; at sea level is 1.225 kg/m3. &lt;a href=&quot;https://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html&quot; rel=&quot;noreferrer&quot;&gt;Mars's air density&lt;/a&gt; is much less than that amount at 0.020 kg/m3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/lifteq.html&quot; rel=&quot;noreferrer&quot;&gt;Lift force is directly related to density&lt;/a&gt;. So an atmosphere that has a fraction of the density of Earth's atmosphere is going to have a proportionately lower amount of lift force available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stated a bit more empirically - there's effectively no atmosphere for the wings of the glider to press against, so there's no lift force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a slightly more positive note, Mars's gravity is 38% that of the Earth's.  So the glider won't accelerate as fast towards the surface of Mars as it crashes down.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2018-09-23T13:18:23.690" LastActivityDate="2018-09-23T13:18:23.690" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23880" PostTypeId="1" AcceptedAnswerId="23881" CreationDate="2018-09-23T02:07:19.017" Score="1" ViewCount="71" Body="&lt;p&gt;What causes the pressure change in a horizontal pipe with no energy input. Is it the head loss? If so, how?&lt;/p&gt;&#xA;" OwnerUserId="17606" LastActivityDate="2018-09-23T02:31:56.477" Title="Pressure rise in horizontal pipe" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23881" PostTypeId="2" ParentId="23880" CreationDate="2018-09-23T02:31:56.477" Score="2" Body="&lt;p&gt;We take the case of a horizontal pipe open on one end and driven with a pressure source on the other, flow is subsonic, fluid is newtonian, and incompressible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure diminishes with length away from the source due to frictional dissipation from viscosity. This is commonly referred to as &quot;head loss&quot;. In the absence of energy input- that is, we stop the pump- then there is no flow, the viscosity terms are zero, no energy is dissipated, and there is no head loss.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-23T02:31:56.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23882" PostTypeId="1" CreationDate="2018-09-23T05:04:14.660" Score="0" ViewCount="105" Body="&lt;p&gt;Wind flow always generates naturally when I opening windows on the opposite sides of the building. &#xA;Is it possible to put turbine on the way and generate electricity from that?&#xA;It could be optimising by building a tunnel and protecting walls for better indoor/outdoor temperature differences.&lt;/p&gt;&#xA;" OwnerUserId="17608" LastActivityDate="2018-09-24T09:44:29.753" Title="Is it possible to generate energy from indoor draft by opening windows on opposite sides of the building?" Tags="|electrical-engineering|aerodynamics|wind-power|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23883" PostTypeId="1" CreationDate="2018-09-23T08:36:41.383" Score="2" ViewCount="344" Body="&lt;p&gt;I've drawn an FBD of one leg of the table the picture. I've a feeling I'm missing something. Am I?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/QByb9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4NqfJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17609" LastActivityDate="2023-11-24T16:01:50.223" Title="FBD of folding table leg" Tags="|structural-analysis|applied-mechanics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23884" PostTypeId="2" ParentId="23882" CreationDate="2018-09-23T08:48:25.740" Score="2" Body="&lt;p&gt;The two main considerations concerning wind power generation is the speed of the air and the duration of wind (air flow). The more of both, the better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to harness wind flow within a building will be difficult because the air flow will most likely be low (slow speeds) and the duration of flow may also be short, except on very windy days. Even a wind turbine with very low friction bearings would not be of much help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any electric generated this way would be small and for a short time period.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-09-23T08:48:25.740" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23885" PostTypeId="2" ParentId="23582" CreationDate="2018-09-23T12:18:51.827" Score="1" Body="&lt;p&gt;Since your project is defense software related I suggest taking a look at &lt;a href=&quot;https://en.wikipedia.org/wiki/MIL-STD-498&quot; rel=&quot;nofollow noreferrer&quot;&gt;MIL-STD-498&lt;/a&gt;. The one that is close related to your question is &lt;a href=&quot;https://web.archive.org/web/20110616201638/http://www.fisica.uniud.it/~cabras/swe/MIL-STD/498/STP-DID.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;SOFTWARE TEST PLAN (STP)&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2018-09-23T12:18:51.827" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23886" PostTypeId="2" ParentId="23883" CreationDate="2018-09-23T21:08:25.503" Score="-1" Body="&lt;p&gt;Your FBD is correct if we assume the legs are on a roller or the floor is very slippery and or if your table legs have received a few jerks enough to slide open to a relaxed position releasing the friction force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;otherwise the forces on the leg are indeterminate depending on the friction coefficient and the original base length the table is placed on the floor. Because the cable can not have compression stress the legs can be randomly positioned in different angles and &lt;span class=&quot;math-container&quot;&gt;$ friction= \mu\times sin\alpha \times weight$&lt;/span&gt; will be different.&#xA;Alpha is the angle of leg with floor, and meu the friction coefficient.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-09-23T21:08:25.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23887" PostTypeId="1" CreationDate="2018-09-23T21:21:35.707" Score="1" ViewCount="1271" Body="&lt;p&gt;What are some common ways that I can vaporize Metals into a vacuum chamber? How can I keep these Metals in a gaseous state?&lt;/p&gt;&#xA;" OwnerUserId="17599" LastActivityDate="2021-12-03T07:45:26.073" Title="How can I vaporize metals?" Tags="|electrical-engineering|chemical-engineering|experimental-physics|computer-engineering|chemistry|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23888" PostTypeId="2" ParentId="23887" CreationDate="2018-09-24T04:09:48.890" Score="1" Body="&lt;p&gt;vaporizing metal in a near-vacuum is a straightforward process; the simplest method is called sputtering and is one specific method of vacuum metallization as noted by William Hird above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keeping a vaporized metal in a gaseous state for more than a few milliseconds is a different proposition, however. For that you would need a furnace with tremendous power and extremely good insulation surrounding it.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-24T04:09:48.890" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23889" PostTypeId="2" ParentId="23882" CreationDate="2018-09-24T09:44:29.753" Score="0" Body="&lt;p&gt;Any natural 'wind' passing through the structure would be better harnessed by a turbine outside the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any generated air currents (e.g. due to thermal differences etc.) will rely on external energy input (e.g. from the heating system), so, if you are able to create an airflow by opening your windows, and put a turbine in it, you will certainly recoup less energy than the heating system used to make the airflow in the first place. See the Second Law of Thermodynamics...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;em&gt;is&lt;/em&gt; possible that in very specific circumstances, the heating effect of the sun could conspire to generate additional air movement without using up power on a heating system, but, in this case Solar-Thermal or PV Panels would be a &lt;em&gt;much&lt;/em&gt; more efficient way to be generating energy...&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-09-24T09:44:29.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23890" PostTypeId="1" CreationDate="2018-09-24T09:50:38.180" Score="-3" ViewCount="76" Body="&lt;p&gt;Do you float above the surface or you drown ?&lt;/p&gt;&#xA;" OwnerUserId="17616" LastActivityDate="2018-09-24T11:48:34.857" Title="What happens when you wear a hydrophobic suit and dive into water?" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="2" ClosedDate="2018-09-24T18:29:14.143" ContentLicense="CC BY-SA 4.0" />
  <row Id="23891" PostTypeId="2" ParentId="23890" CreationDate="2018-09-24T10:50:25.357" Score="2" Body="&lt;p&gt;Depending on your buyuoncy, you sink. Surface tension of water is not strong enough to hold something with the weight and shape of a human being afloat. This works for &lt;a href=&quot;https://en.wikipedia.org/wiki/Gerridae#Nature_of_the_ability_to_walk_on_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;water striders&lt;/a&gt; and maybe some tiny geckos. What will happen is that air bubbles cling to your suit longer than otherwise, maybe even forming a thin air layer until turbulence shears them away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hydrophobe&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydrophobic behavior&lt;/a&gt; does not - strictly speaking - mean that something repels water, it means that there's no attraction between the hydrophobic surface and individual water molecules so that the adhesion between water molecules is far stronger.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorDisplayName="user16" LastEditDate="2018-09-24T11:48:34.857" LastActivityDate="2018-09-24T11:48:34.857" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23892" PostTypeId="1" CreationDate="2018-09-24T17:06:18.700" Score="1" ViewCount="240" Body="&lt;p&gt;I'm looking to use an epoxy in a high-temp environment like 800C, and one that is optically transparent at 450nm (not below 90% transmission). But i'm having trouble finding one. It can be made of any material, as long as it can be cured in a mold.&lt;/p&gt;&#xA;" OwnerUserId="17619" LastActivityDate="2023-12-07T17:05:15.723" Title="Epoxy that can withstand up to 800C and is optically transparent?" Tags="|temperature|radiation|epoxy|photonics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23894" PostTypeId="2" ParentId="23892" CreationDate="2018-09-24T17:36:50.727" Score="0" Body="&lt;p&gt;since cured epoxy is a carbon-based polymer, you will have a very hard time finding one that can stand 800C. Have you considered a sintered ceramic? &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-24T17:36:50.727" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23897" PostTypeId="1" CreationDate="2018-09-24T19:50:04.873" Score="3" ViewCount="3193" Body="&lt;p&gt;I am doing some restauration on my old Vespa and since the fuel indicator doesn't work, I took the fuel gauge system out of the tank and tried to understand how it works to repair it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, 3 wires go to the tank, one is ground while the other two should be live (12V RMS voltage I think). Now, looking at the gauge, there is an empty plastic piece which stays on the surface of the gasoline and moves a potentiometer which can be also submerged by gasoline (when the fuel tank is full) but in practice it is often not submerged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tank is, of course, closed but air can go inside it (in order to let gasoline come out to the engine). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, at first I didn't even think about it but my father asked me:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Isn't it dangerous to have current (and possibly a spark) inside your fuel tank?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So far the number of Vespas I saw blowing up is zero so it probably isn't dangerous, but still, I don't understand why. My reasoning is the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The current passing through the potentiometer may be very low and therefore not cause much heat being dissipated.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The risk of a spark between contacts is probably low due to the low voltage and since the electrodes should be always in contact.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Needless to say I'm not satisfied with my explanation and I'd like someone who knows about physics to explain me why the risk of explosion is very low (or non existent hopefully).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you are wondering, the gauge is the following&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Lrnzz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lrnzz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="17623" LastActivityDate="2018-09-24T21:57:18.520" Title="Can current passing through a wire potentially ignite gasoline? If not, why?" Tags="|electrical-engineering|chemical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23898" PostTypeId="2" ParentId="23897" CreationDate="2018-09-24T21:57:18.520" Score="6" Body="&lt;p&gt;For a combustible mixture to ignite, it must be within its &lt;a href=&quot;https://www.mathesongas.com/pdfs/products/Lower-(LEL)-&amp;amp;-Upper-(UEL)-Explosive-Limits-.pdf&quot; rel=&quot;noreferrer&quot;&gt;flammability limits&lt;/a&gt;. &lt;a href=&quot;https://www.2carpros.com/questions/chevrolet-impala-worried-safety-and-electric-fuel-pumps&quot; rel=&quot;noreferrer&quot;&gt;This forum&lt;/a&gt; provides an anecdotal explanation for gasoline's inflammability in an automobile tank:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;...when the tank is empty the vapor level is above the HEL or UEL (upper explosive limit). If the flammable vapors are higher in concentration than this level it will not ignite. We used to weld inside the diesel tanks in the Navy because it was above the UEL and therefore was safe for hot work.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;This anecdote is correct: Chemical engineers are careful to use the &lt;a href=&quot;http://www.roberts-roberts.com/documents/Using%20Properties%20to%20Manage%20Flammable%20Liquid%20Hazards.pdf&quot; rel=&quot;noreferrer&quot;&gt;physical and chemical properties of flammable liquids&lt;/a&gt; to ensure their safe storage. However, can we be certain that gasoline will always be safe in your &lt;a href=&quot;https://auto.howstuffworks.com/fuel-gauge1.htm&quot; rel=&quot;noreferrer&quot;&gt;gas tank&lt;/a&gt;?&lt;/p&gt;&#xA;&lt;p&gt;The answer is yes – unless you're driving under very, very specific conditions. To be within gasoline's flammability limits, you would need a very low fill level (to allow  enough headspace for proper gas mixing) and very low temperatures (around -20°C). I've included more information about an experimental investigation into these conditions below, but this should be enough to answer your question – if you're really worried about it, keep your tank more than ~1/8th full, and you'll be just fine.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The National Renewable Energy Laboratory (NREL) investigated gasoline flammability in a series of publications titled &lt;em&gt;An Experimental and Modeling&#xA;Study of the Flammability of Fuel Tank Headspace Vapors from High Ethanol Content Fuels.&lt;/em&gt; &lt;a href=&quot;https://www.nrel.gov/docs/fy09osti/44040.pdf&quot; rel=&quot;noreferrer&quot;&gt;The first volume&lt;/a&gt; concluded:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Gasoline is so volatile at most ambient temperatures that the headspace vapors in the tank are actually too rich to burn, as long as some liquid fuel remains. However, as temperature drops, or as the liquid fuel level goes down, the volatility of the fuel decreases. As liquid level drops, there is less fuel vapor mixed with the air in the tank.&#xA;If the ambient temperature is cold enough and the tank is nearly empty, then the fuel-air mixture in the tank becomes flammable and can pose an explosion hazard if ignited. Similarly, when refueling a nearly empty tank at very cold temperatures, the headspace vapors are expelled as a plume of fuel vapor and air as the liquid enters the tank, producing a flammable plume. An ignition source within that plume could lead to a flame traveling back down into the tank and causing an explosion. This has always been the case with gasoline, but the temperatures and fill levels needed to produce a hazard are rarely encountered, so fires and tank explosions are very unlikely, although not impossible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;After generating their model in volume one, NREL examined &amp;quot;field samples&amp;quot; (gasoline from common fill stations) in &lt;a href=&quot;https://www.nrel.gov/docs/fy10osti/47819.pdf&quot; rel=&quot;noreferrer&quot;&gt;volume two&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;...at least some of the ethanol fuels that are currently available when and where Class 3 conditions exist (-5ºC and below) are likely to produce flammable vapors within the ambient temperature range where they are used.&lt;/p&gt;&#xA;&lt;p&gt;...results indicate that mid-level blends...are unlikely to significantly increase the risk of producing flammable vapors over that of the base gasoline used for the blends.&lt;/p&gt;&#xA;&lt;p&gt;The vapor pressure data for the blends indicated that a reduction in ethanol content to about 65% would be needed to achieve the minimum vapor pressure specification. The flammability data trends suggest that blends with ethanol content in this region would provide a favorable trade-off between avoiding the formation of flammable vapors and maximizing the ethanol content of the fuel.&lt;/p&gt;&#xA;&lt;p&gt;Comparisons of the flammability results of the field and laboratory samples indicate that&#xA;vapor pressure (DVPE) alone does not adequately predict the flammability of the fuel tank&#xA;vapors formed at low temperatures. The properties of the hydrocarbon portion of the fuel&#xA;must be considered as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Finally, &lt;a href=&quot;https://www.nrel.gov/docs/fy11osti/52043.pdf&quot; rel=&quot;noreferrer&quot;&gt;volume three&lt;/a&gt; focused on in-depth experimental study:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The experimental results presented in this report are, to some extent, unique to the&#xA;apparatus and test conditions that were used for the experiments. Important factors that would&#xA;have affected the observed behavior included the use of a fill level of 5%, the use of a strong&#xA;spark for ignition, and the use of test chambers with a relatively small internal volume. Thus, the&#xA;combustion pressures and flammability limit temperatures presented in this report would be&#xA;expected to differ from those that might be encountered in other situations. For example, a higher&#xA;fill level or a weaker ignition source would lower the upper flammability limit temperature (the&#xA;headspace vapors would have to be colder to be flammable), while a larger headspace volume&#xA;would likely lead to higher maximum pressures and higher rates of pressure rise.  Therefore, the&#xA;results presented in this report should be interpreted primarily as a relative indication of the&#xA;headspace vapor flammability of the test fuels. It should also be noted that results were obtained&#xA;with blends made from specific gasoline samples. The quantitative results at a given vapor&#xA;pressure level might not be exactly the same if different gasoline samples with different&#xA;hydrocarbon compositions were used.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="16783" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-09-24T21:57:18.520" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23900" PostTypeId="2" ParentId="23887" CreationDate="2018-09-24T23:32:37.690" Score="5" Body="&lt;p&gt;The easiest way to vaporize a metal is to create a filament of the metal. Heat the filament in the vacuum to its sublimation (vaporization) temperature. Alternatively, when the metal cannot be shaped into a filament, the metal can be encased in or painted on a &quot;wrapper&quot; metal. Commercial evaporators for alkali metals are designed this way. The wrapper is a thin tungsten foil encasing the alkali and having a small slit or hole in it. Heat the wrapper, and the alkali evaporates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to keeping a metal as a vapor, consider how to confine a metal within chamber walls AND keep it as a gas. When the walls are colder than the sublimation temperature of the metal, the instant the metal gas hits the walls, it will condense. So, as already noted, the chamber walls will have to be at least as hot as the sublimation temperature of the metal itself, if not hotter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative method to keep metal &quot;gaseous&quot; is to ionize the metal immediately as or immediately after it is vaporized. The metal ions can be confined within a magnetic field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The methods to vaporize metals are relatively cost effective, well-documented, and reliable for small amounts of material. Vaporizing more than nano-gram amounts of metal heads to expensive, professional-level designs. The principles remain the same.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-09-24T23:32:37.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23901" PostTypeId="1" CreationDate="2018-09-25T07:57:53.923" Score="2" ViewCount="11" Body="&lt;p&gt;I've found data related to the deformation and shear performance of typical PIR (polyisocyanurate) foams such as Celotex/Kingspan, but I'm not sure how to use these figures in my own situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am refitting a solid floor building in the UK. The existing oversite slab is intact and uncracked, but never had insulation. So the plan is to lay appropriate DPM/VCL, about 100-120mm Celotex (or equivalent), lay A242 mesh, and pour about 120 - 130mm of either dense concrete or screed. It won't be bearing any load other than usual UK floors (~ 2kN/m^2), and the area is &amp;lt;= 4 x 6 m. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately due to the way the original slab was laid and the topology of the immediate area, there is a slight chance that moisture could seep under these, above the original slab. Realistically it's unlikely/no evidence, but it is possible, so I'd like to rest the Celotex/DPM/VCL layers on some kind of front-to-back orientedbattening, and provide airbricks, to create a shallow ventilated void between the original oversite slab and the new flooring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is done, the batten widths and spacing would be quite crucial - if the battens are too narrow, or too widely spaced, the Celotex will deform excessively or even shear in the ~&amp;lt; 24 hours that the concrete/screed is fluid. (After it's set, of course, there will be no deformation beyond the equilibrium it was at when it set.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can imagine that narrow battens might have a limit on safe spacing different from wider battens, as shear and deformation are different phenomena. So there might be a range of &quot;safe limit&quot; combinations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't find anywhere, the appropriate formulae and safety factors that should be used, to calculate the limits of batten width/spacing beyond which shear and deformation would become issues.&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2018-09-25T08:34:31.557" LastActivityDate="2018-09-25T08:34:31.557" Title="What degree of open span/narrow support can rigid PIR thermal insulating foam sustain under the short term load of setting concrete?" Tags="|concrete|construction-management|shear|deformation|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23904" PostTypeId="1" AcceptedAnswerId="23908" CreationDate="2018-09-25T12:11:31.413" Score="0" ViewCount="34" Body="&lt;p&gt;I am currently looking ways to control or siwtch a max 220v 35A load. Simple magnetic relay rated for such load would work, but i am exploring the other alternatives too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here i stumbled into triacs, in which solid state relays are made of. I stumbled into this ic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.st.com/resource/en/datasheet/t16t.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.st.com/resource/en/datasheet/t16t.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and i could not get the grasp on how a small triac such as this one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is able to switch a high load of 220v 16amps&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="17493" LastActivityDate="2018-09-25T15:50:57.150" Title="How are small triacs able to &quot;switch&quot; or control high power AC loads" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23907" PostTypeId="1" CreationDate="2018-09-25T15:49:57.747" Score="0" ViewCount="73" Body="&lt;p&gt;Diesel electric locomotives seem popular, even these days.  The UK's Type 45 destroyers have tried integrated electric propulsion, and I think that there are hybrid motor-generator cars that can recharge their battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There clearly seems a steady move towards transforming more and more power train components to electricity in all sorts of vehicles. Assuming that we don't attach over head wires and a pantograph system to Volvos, the next best thing seems fully diesel electric power. No?&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2018-09-25T15:49:57.747" Title="Why are there no fully diesel electric cars?" Tags="|automotive-engineering|power-transmission|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23908" PostTypeId="2" ParentId="23904" CreationDate="2018-09-25T15:50:57.150" Score="0" Body="&lt;p&gt;35A needs a heavy duty contactor. It's a relay but designed for switching the high current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A triac latches on until the current drops to 0. This means that the actual switching off happens at the zero crossing point. Switching on is still done whenever the gate voltage exceeds the turn on threshold. This can result in a inrush current spike but as long as that isn't sustained the triac will be able to deal with it. The one you linked can handle 120A for one cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the triac is off the isolation is enough to stop current flow.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2018-09-25T15:50:57.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23911" PostTypeId="1" CreationDate="2018-09-25T20:13:24.097" Score="1" ViewCount="160" Body="&lt;p&gt;I'm working on a device that relies on data from a magnetometer. Unfortunately, the device also needs to have a fairly powerful magnet on it. Given that the design is fairly flexible, and within manufacturing tolerance I can know precisely where the magnet is and the layout of the magnetic field it should produce, is there a way to compensate for having a magnet in the device?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some mathematical means of cancelling out the contribution of the inbuilt magnet, or perhaps shielding the magnet from the remainder of the device somehow, as the magnet is only needed for mounting the device to metal surfaces and nothing else?&lt;/p&gt;&#xA;" OwnerUserId="11618" LastActivityDate="2024-01-06T00:08:15.993" Title="Compensating for a magnet in a device with a magnetometer" Tags="|electrical-engineering|sensors|magnets|electromagnetism|consumer-electronics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="23913" PostTypeId="2" ParentId="23911" CreationDate="2018-09-26T01:56:39.737" Score="0" Body="&lt;p&gt;Some fluxgate magnetometers have their sensor mounted inside Helmholtz coils and the coils can be used to compensate for errors in the sensor caused by external (or in your case, internal) magnetic fields or ferrous materials which disturb the natural flux lines of the earth's magnetic field. So yes, you can compensate for the effect of the magnet using Helmholtz coils but you may find that you have to run a lot of DC current through them to cancel the magnetic field of the magnet.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2018-09-26T01:56:39.737" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23916" PostTypeId="1" CreationDate="2018-09-26T05:14:45.927" Score="-1" ViewCount="46" Body="&lt;p&gt;Working on a new prototype and I would need the answer to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the cheapest and easiest way to cool 2 liters of water without using ice or a fridge?&lt;/p&gt;&#xA;" OwnerUserId="17642" LastActivityDate="2018-09-26T14:44:27.197" Title="What is the cheapest and easiest way to cool 2 liters of water without using ice or a fridge?" Tags="|cooling|" AnswerCount="2" CommentCount="3" ClosedDate="2018-09-26T12:15:45.540" ContentLicense="CC BY-SA 4.0" />
  <row Id="23918" PostTypeId="2" ParentId="23916" CreationDate="2018-09-26T06:37:28.247" Score="1" Body="&lt;p&gt;-Fill in a vase with clean cool silt and let the water drain through it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Spill it on a cool stone slab sloping down to a container.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Run it through a car radiator.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-09-26T14:44:27.197" LastActivityDate="2018-09-26T14:44:27.197" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23920" PostTypeId="2" ParentId="23916" CreationDate="2018-09-26T07:37:02.983" Score="1" Body="&lt;p&gt;Use a &quot;Zeer pot&quot;, commonly used for keeping wine chilled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ceramic pot which is soaked in water and as the water evaporates from the surface it cools the interior - only issue can be that if the ambient humidity is very high it does not work so well...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also used in hot climates to cool rooms...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A link to one description : &lt;a href=&quot;https://practicalaction.org/zeer-pot-fridge&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://practicalaction.org/zeer-pot-fridge&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-09-26T07:37:02.983" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23926" PostTypeId="1" AcceptedAnswerId="23937" CreationDate="2018-09-26T14:31:10.217" Score="1" ViewCount="155" Body="&lt;p&gt;I am attaching the model part as an Image below for your reference:-&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7SS81.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7SS81.png&quot; alt=&quot;bent plate model so far&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the pattern like this on the above model.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ikoc1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ikoc1.jpg&quot; alt=&quot;desired embossed pattern&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the model shown above, and want to put embossed lines 0.2 mm thick with 0.2 mm spacing in between on the top, sides, and the bends. The feature should be parallel to the horizontal section. I tried using emboss but I don't know how to emboss all three sides.&lt;/p&gt;&#xA;" OwnerUserId="16787" LastEditorUserId="13311" LastEditDate="2018-09-27T08:12:05.860" LastActivityDate="2018-09-27T08:12:05.860" Title="How do you implement a pattern over a bent plate in Autodesk Inventor?" Tags="|cad|software|autodesk-inventor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23928" PostTypeId="1" CreationDate="2018-09-26T15:40:35.580" Score="0" ViewCount="289" Body="&lt;p&gt;I am building a pneumatic shear to cut .135&quot; C1018 round wire. We will be holding the wire in a hardened thimble and using a heat treated beveled cutting knife.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much force do we need to be able to produce?&lt;/p&gt;&#xA;" OwnerUserId="17649" LastActivityDate="2018-09-26T16:20:22.963" Title="Shear force requirement" Tags="|shear|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23929" PostTypeId="1" AcceptedAnswerId="23931" CreationDate="2018-09-26T16:13:20.240" Score="0" ViewCount="74" Body="&lt;p&gt;Do you have any idea on why the &lt;code&gt;Calculation 1&lt;/code&gt; below doesn't match?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://developer.mozilla.org/en-US/docs/Web/API/Web_Audio_API/Basic_concepts_behind_Web_Audio_API#Audio_buffers_frames_samples_and_channels&quot; rel=&quot;nofollow noreferrer&quot;&gt;Audio buffers: frames, samples and channels&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;VLC - Media Information:&#xA;   num channels: 2 (stereo)&#xA;    sample rate: 44100 Hz&#xA;bits per sample: 32 =&amp;gt; bytes per sample: 4&#xA;        bitrate: 192 kb/s&#xA;       duration: 7 mins, 23 secs =&amp;gt; 443 secs&#xA;&#xA;Windows - File Properties:&#xA;           size: 10,642,422 bytes =&amp;gt; 10393 KB&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculation 1 (without using bitrate) (using rounded values for simplicity)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1 second: 44100 samples =&amp;gt; 44100 * 4 = 176400 bytes =&amp;gt; 172 KB (looks a lot for just 1 sec)&#xA;This way: 443 secs =&amp;gt; 443 * 172 KB = 76196 = 74 MB, but the file just 10 MB (!!! Contradiction !!!)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Calculation 2 (using bitrate)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; bitrate: 192 kb/s =&amp;gt; 24 KB/s&#xA;1 second: 24KB =&amp;gt; 443 seconds: 443 * 24 KB = 10632 KB = 10393 KB (like above) + [some metadata] (!!! Match !!!)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="17650" LastEditorUserId="8055" LastEditDate="2018-09-26T22:32:28.637" LastActivityDate="2018-09-26T22:32:28.637" Title="Using the bitrate value on a calculation is causing contradiction, why?" Tags="|audio-engineering|sound-isolation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23930" PostTypeId="2" ParentId="23928" CreationDate="2018-09-26T16:20:22.963" Score="1" Body="&lt;p&gt;I use to build a very simple model first, then try to expand it. Here i try to find the minimum, force require to cut the wire, by using Tresca criterion: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yielding starts when the maximum shear stress in the material &lt;span class=&quot;math-container&quot;&gt;$\tau_{max}$&lt;/span&gt; equals the maximum shear stress at yielding in a simple tension test &lt;span class=&quot;math-container&quot;&gt;$\tau_x$&lt;/span&gt;. You can verify the final result with the Mohr's circle: &lt;span class=&quot;math-container&quot;&gt;$$\tau_{max} = \frac{\sigma_y}{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here &lt;span class=&quot;math-container&quot;&gt;$\sigma_y$&lt;/span&gt; is the yield strength of your material. A quick google search, i find the yield strength  of &lt;span class=&quot;math-container&quot;&gt;$370 MPa$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the force: &lt;span class=&quot;math-container&quot;&gt;$$\tau_{max} = \frac{F}{A}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\tau_{max} = 185 Mpa$&lt;/span&gt; and the section area of the wires are about is about &lt;span class=&quot;math-container&quot;&gt;$37 m^2$&lt;/span&gt;. The require force is about &lt;span class=&quot;math-container&quot;&gt;$6845 MN$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I worked so far conservative, by using other criteria, you may find much lower values for the force, i think, but you have to make sure if the force is enough to cut through the wires.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-09-26T16:20:22.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23931" PostTypeId="2" ParentId="23929" CreationDate="2018-09-26T16:41:29.613" Score="2" Body="&lt;p&gt;Because the file is compressed. The sample rate and bits per sample are parameters of playback after decompression. The file size is for the compressed format, the amount of data that needs to be downloaded, not the amount of data that reaches the digital-to-analog converter.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2018-09-26T17:16:11.430" LastActivityDate="2018-09-26T17:16:11.430" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23932" PostTypeId="2" ParentId="2405" CreationDate="2018-09-26T17:14:34.553" Score="0" Body="&lt;p&gt;I'd like to add my understanding, in case it helps anyone comprehend the reason. The reason there is pressure from all sides in these situations is due to the properties of fluids in equilibrium. In the atmosphere, for example, the air molecules being &quot;weighted down on&quot; from above would squeeze out the sides of the air column, if that were possible. It is not, of course, because the air column adjacent is under &lt;em&gt;the same force&lt;/em&gt; and thus they are no better off. Gas molecules are energetic &lt;em&gt;in every direction&lt;/em&gt;, or in other words, fluid pressure cannot exist in one direction &lt;em&gt;if&lt;/em&gt; in equilibrium, as any difference in pressure would yield motion (wind).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason that you use the weight of the fluid above you (air, ocean, etc.) to know the horizontal pressure you would feel is because you are assuming you are in an area under equilibrium and thus you know from the above reasons that the &quot;horizontal&quot; pressure is equal to the &quot;vertical&quot; pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another intuition I like is the idea of a pneumatic piston. The cylinder that contains the fluid needs to be strong to keep from bursting. If you replaced the fluid with a metal rod and put piston force on that instead, the cylinder walls wouldn't feel anything.&lt;/p&gt;&#xA;" OwnerUserId="17652" LastActivityDate="2018-09-26T17:14:34.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23933" PostTypeId="1" CreationDate="2018-09-26T18:39:22.030" Score="0" ViewCount="265" Body="&lt;p&gt;I am looking for a mini freewheel, one that I could mount on a standard NEMA 17 stepper motor (or similar stepper). I want to use it with a pulley, such that it starts the motion of a wheel, which can then continue to rotate on its own. &#xA;Does anybody have an idea where I could get such a freewheel? Most of what I have seen is either for bikes or cars - I just need a very small version.&#xA;Any tips are appreciated!&lt;/p&gt;&#xA;" OwnerUserId="17654" LastActivityDate="2018-09-26T20:26:43.020" Title="Mini freewheel / overrunning clutch for stepper motor" Tags="|gears|stepper-motor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23934" PostTypeId="1" CreationDate="2018-09-26T18:59:24.827" Score="1" ViewCount="2251" Body="&lt;p&gt;... The sphere is placed over a hole, of radius a, in the tank bottom. Develop a general expression for the range of specific gravities for which the sphere will float to the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've attached the diagram for the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an attempt at the solution, I've summed the forces on the sphere, or the buoyant force &lt;span class=&quot;math-container&quot;&gt;$F_b$&lt;/span&gt; - the weight of the sphere. Buoyant force is equal to the weight of the displaced water by the sphere, &lt;span class=&quot;math-container&quot;&gt;$\frac{4}{3}\pi R^3\rho_{water}\times32\frac{ft}{s^2}$&lt;/span&gt;. The weight of the sphere is &lt;span class=&quot;math-container&quot;&gt;$\frac{4}{3}\pi R^3\rho_{sphere}\times32\frac{ft}{s^2}$&lt;/span&gt;. Based on force equation, in order for the ball to float the buoyant force would have to be greater than the weight of the sphere, or &lt;span class=&quot;math-container&quot;&gt;$F_b-W&amp;gt;0$&lt;/span&gt;. After some algebra you can say that the specific gravity of the sphere has to be &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;1$&lt;/span&gt;, which seems too simple to be the actual answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure I'd have to take the hole that the sphere is resting atop into consideration, but I'm not sure how to go about that. My only ideas on how to include that would be to include the force from the ambient pressure on the bottom of the ball pushing upward in the force equation, but even if that effects the specific gravity at which the ball will float, once it started to float and that force was no longer there, the ball would simply sink again, unless &lt;span class=&quot;math-container&quot;&gt;$S.G.&amp;lt;1$&lt;/span&gt;. My other thought was that I would have to consider the volume of the sphere not submerged due to the hole and thus not contributing to the buoyant force, but I have no idea how to calculate what that volume would be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my first solution correct, that &lt;span class=&quot;math-container&quot;&gt;$S.G.&amp;lt;1$&lt;/span&gt; would cause the ball to float? That again feels too simple for the given parameters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any direction would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1niCF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1niCF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17655" LastEditorDisplayName="user14407" LastEditDate="2018-09-26T19:29:12.653" LastActivityDate="2019-05-25T06:03:10.453" Title="A sphere of radius R, made from material of specific gravity, SG is submerged in a tank of water" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23935" PostTypeId="1" CreationDate="2018-09-26T19:01:02.593" Score="3" ViewCount="505" Body="&lt;p&gt;I have an old bend/twisted wooden door. It has some character to it so I want to keep it and add square metal bars to straighten it. I have provided a sample drawing of the door and where it is twisted. I have drilled a hole in the upper most open part and if I pull with a scale I am able to straighten it with about 20kg (I haven't used a scale yet, but I will)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BIAEu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BIAEu.png&quot; alt=&quot;Twisted door, arrows show twists&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I intend to put square metal bars on the bent sides (maybe even inside and out) and straighten the wood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F1XQ9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F1XQ9.png&quot; alt=&quot;straightening bars&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what is the formula that I can use to calculate the size of the metal bar? For example I see there is a square profile bar 40x40x3mm that I will be able to buy and attach. So, is this going to be enough for the long side 2m and for the short side 1m?&#xA;I can also order this square profile bar to be slightly bent and attach it so the two bends - the one of the wood and the one of the metal counter each other. But I am not really willing to go that way. I think its way too complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I think a visible rectangle bar on one or both faces will be stronger and more simple to attach, so I will go with that.&lt;/li&gt;&#xA;&lt;li&gt;No bend is acceptable - I want it straight.&lt;/li&gt;&#xA;&lt;li&gt;So the next question following the comment is how much of a bend do I need for what type of bar for each of the sides - short(1m) and long (2m)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17656" LastEditorUserId="17656" LastEditDate="2018-09-26T20:15:25.010" LastActivityDate="2018-12-26T01:01:10.663" Title="How to calculate the bending resistance of a square metal bar?" Tags="|materials|metals|wood|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23936" PostTypeId="2" ParentId="23933" CreationDate="2018-09-26T20:26:43.020" Score="1" Body="&lt;p&gt;You may be able to locate the desired component if you search for &quot;one way bearing for rc helicopter.&quot; I also used the term &quot;sprag clutch&quot; but came back with industrial grade, industrial sized items, along with the poorly spelled &quot;sprague&quot; clutch bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One such return shows an 8 mm inside diameter with a 22 mm outside diameter over a 9 mm depth for the bearing. A bit more than US$12.00 found on &lt;a href=&quot;https://www.vxb.com/CSK8-One-way-Sprag-Freewheel-Backstop-Clutch-p/Kit8176.htm?gclid=EAIaIQobChMI1IfO9r_Z3QIVh4zICh0X3gpQEAQYAiABEgJBovD_BwE&quot; rel=&quot;nofollow noreferrer&quot;&gt;vxb.com web site.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3dBoV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3dBoV.jpg&quot; alt=&quot;sprag bearing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-09-26T20:26:43.020" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23937" PostTypeId="2" ParentId="23926" CreationDate="2018-09-26T21:02:40.833" Score="1" Body="&lt;p&gt;The solution to this, is simply to add the &quot;embossed lines&quot; &lt;em&gt;before&lt;/em&gt; the bend. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at the modelling process below:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Extrude a flat plate. Because this model will be symmetrical this is only half the size of the final object.&lt;/li&gt;&#xA;&lt;li&gt;Extrude a 'master rib'. This is 0.1mm away from the YZ plane, so that there's an 0.2mm gap once it's mirrored&lt;/li&gt;&#xA;&lt;li&gt;Pattern the rib to cover the full plate (using the 'rectangular pattern' tool, with one direction defined&lt;/li&gt;&#xA;&lt;li&gt;Bend the plate, using the 'Bend Part' tool. You will need to make a sketch on the surface of the plate, and draw a line at the point where you want the bend to start. The line defines the &quot;end of the flat top&quot;, so if you want the overall width to be 20mm, with an internal bend radius of 2mm, and a plate thickness of 5mm, you would need the line 13mm from the YZ plane.&#xA;N.B. You may have to click the word &quot;Modify&quot; on the ribbon to reveal this command if you have it hidden, or just use the &quot;Search Help &amp;amp; Commands&quot; window at the top right&lt;/li&gt;&#xA;&lt;li&gt;Use the Extrude Tool with Cut Boolean to make the 'angled cutoff', as required&lt;/li&gt;&#xA;&lt;li&gt;Finally mirror the part around the YZ plane to complete your component.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vBmEl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vBmEl.png&quot; alt=&quot;Modelling with embossed lines&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-09-26T21:02:40.833" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23938" PostTypeId="1" CreationDate="2018-09-26T21:15:39.600" Score="2" ViewCount="1953" Body="&lt;p&gt;I need to know the difference between a direct and indirect reading instrument. Would a caliper be a direct and a gauge block be indirect? Is a direct instrument something with a variable measurement and indirect something with just a fixed measurement?&lt;/p&gt;&#xA;" OwnerUserId="17486" LastEditorUserId="110" LastEditDate="2018-09-28T10:13:54.390" LastActivityDate="2018-09-28T14:24:20.817" Title="What is the difference between direct and indirect reading instruments" Tags="|measurements|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23940" PostTypeId="2" ParentId="23935" CreationDate="2018-09-26T22:18:50.687" Score="1" Body="&lt;p&gt;If you estimate force needed to straighten the door applied at the corner is 20kg, it would be equivalent to the reaction of a uniformly loaded beam (your square bar) at one end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So lets design the bar and for simplicity only the top bar, we can do the side later if we need to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say the door stays out of door jamb 2cm at the corner.&#xA;So now we have all the data to design a bar or strap that has a deflection, built in bend, of 1/2 times 2cm =1cm and the 2 end reactions of 20kg, which for a 1m long strap mean omega is 40kg/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deflection in a simply supported strap:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta = \frac {5 }{384 } \frac {\omega L^3}{EI} $$&lt;/span&gt; This means &#xA;&lt;span class=&quot;math-container&quot;&gt;$$0.01 = 5/384 \times40\times 1^3/(200GPa\times I)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ 2GPa\times I = 5/384\times 40  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Gives the second moment, &quot;I&quot; of your strap and from there you can select the section of the bar that gives you optimal looks and required strength. Again this bar need to have a camber of 1cm built into it, so that after you attach it the door will be pressed into straight line. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-09-26T22:18:50.687" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23942" PostTypeId="2" ParentId="23934" CreationDate="2018-09-27T00:49:43.047" Score="1" Body="&lt;p&gt;The net force on the sphere due to the water around it is obtained by integrating the pressure as a vector at each point around the area of the sphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F_w = \iint \vec{p}(\theta, \phi)\bullet\vec{n}\ d\theta\ d\phi $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allow that water is incompressible and the pressure at the top of the tank is &lt;span class=&quot;math-container&quot;&gt;$p_o$&lt;/span&gt;. The magnitude of the pressure depends on distance from the bottom of the tank &lt;span class=&quot;math-container&quot;&gt;$d_b$&lt;/span&gt; as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ p = p_o + \rho_w g \left(H - d_b\right) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between &lt;span class=&quot;math-container&quot;&gt;$d_b$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\theta, \phi, R$&lt;/span&gt; can be developed by geometry. The vector relationship for &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; can be developed based on position &lt;span class=&quot;math-container&quot;&gt;$(r, \theta, \phi)$&lt;/span&gt;. Assume the pressure is otherwise isotropic (it only depends on &lt;span class=&quot;math-container&quot;&gt;$d_b$&lt;/span&gt; not on &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt;. For a free-floating sphere, the double integral becomes &lt;span class=&quot;math-container&quot;&gt;$4\pi R^2$&lt;/span&gt;. But ... the segment of the sphere at the bottom where no water exists must be dropped from the integral in this system. Alternatively, the accurate approach is to replace the pressure over only that segment with an integral where &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is constant (at &lt;span class=&quot;math-container&quot;&gt;$p_o$&lt;/span&gt;). The double integral does not easily become just &lt;span class=&quot;math-container&quot;&gt;$4\pi R^2$&lt;/span&gt; because of the change in limits on &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The request was for a hint at the right direction, not for a complete answer. In due respect to this (and indeed in great appreciation that you want to do the work yourself once you know how to start), I leave the grunt work, as one might say, as an exercise for the reader.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-09-27T00:54:53.103" LastActivityDate="2018-09-27T00:54:53.103" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23943" PostTypeId="1" CreationDate="2018-09-27T03:04:14.550" Score="3" ViewCount="1617" Body="&lt;p&gt;Consider a free falling tank of fluid. The goal is to find the pressure distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition says that there should be no pressure distribution; the pressure should be uniform since the container is also accelerating at the same rate and direction as gravity, so there is no upward reacting force on the fluid to create a hydrostatic pressure distribution. This is also previously supported by others: &lt;a href=&quot;https://physics.stackexchange.com/questions/4619/water-pressure-in-free-fall&quot;&gt;https://physics.stackexchange.com/questions/4619/water-pressure-in-free-fall&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Previous questions and answers, however, are entirely handwavy and also based on intuition. How can we prove this mathematically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start with Euler's equation in the downward direction of acceleration, let's say the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; direction:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\rho a_z = -\frac{\partial P}{\partial z} - \rho g$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the fluid is accelerating downward with acceleration &lt;span class=&quot;math-container&quot;&gt;$a_z = -g$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting into Euler's equation yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \frac{\partial P}{\partial z} = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thus proving that the pressure in a free-falling fluid is uniform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? Can I apply &lt;span class=&quot;math-container&quot;&gt;$a_z = -g$&lt;/span&gt; in the fluid equation of motion even though gravity is already accounted for?&lt;/p&gt;&#xA;" OwnerUserId="17504" LastActivityDate="2023-02-06T08:08:21.030" Title="Free falling fluid pressure distribution" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23946" PostTypeId="1" CreationDate="2018-09-27T12:10:05.957" Score="2" ViewCount="1446" Body="&lt;p&gt;To get realistic results from a FEM calculation, it is common (or so I've been taught by someone who may not be up to date) to use at least four or five elements through the thickness of a beam or plate. I am now modelling a composite plate bending problem, with two or three materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I use four or five elements through the thickness of &lt;em&gt;every&lt;/em&gt; material, or can I reduce my calculation time by using one or two elements per material, as long as the total adds up to four or five?&lt;/p&gt;&#xA;" OwnerUserId="946" LastEditorUserId="946" LastEditDate="2018-09-27T12:34:31.813" LastActivityDate="2020-04-19T20:01:40.157" Title="Should one use multiple elements through the thickness with multiple materials?" Tags="|mechanical-engineering|beam|finite-element-method|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="23947" PostTypeId="1" CreationDate="2018-09-27T12:42:05.633" Score="1" ViewCount="769" Body="&lt;p&gt;What is the fundamental difference between the flute geometry of an end mill as opposed to a drill bit? &#xA;Obviously drill bits are not intended to cut horizontally so what difference about them makes them unable to do this efficiently?&#xA;Also, since end mills can cut vertically, why not just manufacture drill bits with the same flute geometry as end mills?&lt;/p&gt;&#xA;" OwnerUserId="9273" LastActivityDate="2018-09-28T19:29:32.420" Title="End mill vs Drill Bit Flute Geometry" Tags="|machining|cnc|drilling|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="23948" PostTypeId="1" CreationDate="2018-09-27T12:47:04.527" Score="0" ViewCount="905" Body="&lt;p&gt;We have water cooling chillers and crahs in our data center. What I want to learn is what will happen if I increase the chilled water temperature. For example, the indoor units (CRAH) set temperature is 23 degrees Celsius and the chiller's outlet set water temperature is 16 degrees Celsius. I want to set the chiller's outlet temperature to 21 Celsius to have more efficiency in terms of electrical usage. Are there any pros or cons to do this?&lt;/p&gt;&#xA;" OwnerUserId="17667" LastEditorUserId="90" LastEditDate="2018-09-27T13:52:04.470" LastActivityDate="2018-09-28T23:03:07.997" Title="How to determine optimum chilled water outlet temperature?" Tags="|mechanical-engineering|cooling|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23949" PostTypeId="2" ParentId="23948" CreationDate="2018-09-27T13:10:09.607" Score="1" Body="&lt;p&gt;Find the component cooled by the water that is closest to it's overheat temperature and that will tell you how much margin you have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do this test when under sustained load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could monitor the load such that under idle you set the water chiller to a higher temperature, and when you need high performance the chiller gets set to a lower temperature. (Of course this requires that you can monitor the load and can sett he temperature automatically)&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="90" LastEditDate="2018-09-27T13:56:47.903" LastActivityDate="2018-09-27T13:56:47.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23950" PostTypeId="2" ParentId="23946" CreationDate="2018-09-27T13:52:39.720" Score="1" Body="&lt;p&gt;This kind of a rule of thumb is valid only for specific cases. I assume the person who shared with you this rule was investigating a very particular case using the same FEA tool for many times.&#xA;The number one rule, however, is to always use your own brain.&#xA;In your case, are the stresses or the deflections changing along the thickness direction? If not, you should not worry much about having a good resolution there. One fatty element will do the work.&#xA;If you ARE interested in the phenomena going on along the thickness of the material, you have a lot of weapons in your arsenal. You might use 2nd order or more elements (this way the deflection behavior between the nodes of each element is assumed to be polynomial rather linear), refine the mesh to as many layers as needed, and most importantly - check the convergence of the results to realize there is no need to improve the mesh even more.&#xA;This way you will obtain your own rules for this specific problem.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastEditorUserId="16860" LastEditDate="2018-09-27T16:31:17.053" LastActivityDate="2018-09-27T16:31:17.053" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="23951" PostTypeId="2" ParentId="23774" CreationDate="2018-09-27T13:58:20.883" Score="1" Body="&lt;p&gt;Using 2.24,&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{array}{l}\frac{d}{{dt}}\int_{CV}^{} {\phi d{\bf{x}} = } \int_{CV}^{} {\left( {\frac{{d\phi }}{{dt}} + \phi \frac{{\partial {v_i}}}{{\partial {x_i}}}} \right)d{\bf{x}}} \\ = \int_{CV}^{} {\frac{{d\phi }}{{dt}}d{\bf{x}}}  + \int_{CV}^{} {\phi \frac{{\partial {v_i}}}{{\partial {x_i}}}d{\bf{x}}} \end{array}&lt;/span&gt;&#xA;From 2.20,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$= \int_{CV}^{} {\left( {\frac{{\partial \phi }}{{\partial {x_i}}}\frac{{\partial {x_i}}}{{\partial t}} + \frac{{\partial \phi }}{{\partial t}}} \right)d{\bf{x}}}  + \int_{CV}^{} {\phi \frac{{\partial {v_i}}}{{\partial {x_i}}}d{\bf{x}}}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{array}{l} = \int_{CV}^{} {\frac{{\partial \phi }}{{\partial {x_i}}}\frac{{\partial {x_i}}}{{\partial t}}d{\bf{x}}}  + \int_{CV}^{} {\frac{{\partial \phi }}{{\partial t}}d{\bf{x}}}  + \int_{CV}^{} {\phi \frac{{\partial {v_i}}}{{\partial {x_i}}}d{\bf{x}}} \\ = \int_{CV}^{} {\frac{{\partial \phi }}{{\partial {x_i}}}{v_i}d{\bf{x}}}  + \int_{CV}^{} {\phi \frac{{\partial {v_i}}}{{\partial {x_i}}}d{\bf{x}}}  + \int_{CV}^{} {\frac{{\partial \phi }}{{\partial t}}d{\bf{x}}} \\ = \int_{CV}^{} {\left( {\frac{{\partial \phi }}{{\partial {x_i}}}{v_i} + \phi \frac{{\partial {v_i}}}{{\partial {x_i}}}} \right)d{\bf{x}}}  + \int_{CV}^{} {\frac{{\partial \phi }}{{\partial t}}d{\bf{x}}} \end{array}&#xA;$$&lt;/span&gt;&#xA;From differentiation of multiplication,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ = \int_{CV}^{} {\frac{\partial }{{\partial {x_i}}}\left( {\phi {v_i}} \right)d{\bf{x}}}  + \int_{CV}^{} {\frac{{\partial \phi }}{{\partial t}}d{\bf{x}}} &#xA;$$&lt;/span&gt;&#xA;Finally, use the divergence theorem,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ = \int_{CS}^{} {\phi {v_i}d{\bf{s}}}  + \int_{CV}^{} {\frac{{\partial \phi }}{{\partial t}}d{\bf{x}}} &#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="11557" LastEditorUserId="11557" LastEditDate="2018-09-28T09:12:24.123" LastActivityDate="2018-09-28T09:12:24.123" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23952" PostTypeId="1" CreationDate="2018-09-27T14:32:33.427" Score="1" ViewCount="165" Body="&lt;p&gt;I have a Wemo Smart WiFi Smart Dimmer with me that can control CFL, Incandescent and LED lights. (&lt;a href=&quot;https://www.belkin.com/us/p/P-F7C059/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.belkin.com/us/p/P-F7C059/&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious to know how one device can handle all type of bulbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I understand that,&#xA;In order to dim the light, usually, Triac/Solid State switches are used. They are generally called phase-cut dimmers. They can be either leading edge, trailing edge or universal dimmers. These usually go well with CFL and Incandescent lights. These dimmers circuits accept 230V 50Hz input for operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if I want to control a LED light, I probably will need a LED driver. These LED drivers work on PWM on either constant voltage or constant current. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Wemo Smart dimmer works for all kinds of the bulb, they will have to have a LED driver inbuilt and an AC/DC converter to convert the 230V AC to DC for LED lights. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am not sure If I am right. &#xA;What do you think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S - A shout out to the explanation Rod Elliott (&lt;a href=&quot;http://sound.whsites.net/lamps/dimmers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://sound.whsites.net/lamps/dimmers.html&lt;/a&gt;). It's a very well written piece of information on LED dimmers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I had a bit of misunderstanding. I think that the Wemo Smart dimmer can only control LED bulbs that are rated 230V. Bulbs like these( &lt;a href=&quot;https://www.philips.co.id/c-p/8718696485583/led-bulb&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.philips.co.id/c-p/8718696485583/led-bulb&lt;/a&gt;). And then you will need to a LED driver to control LED bulbs like MR16 LED bulbs (&lt;a href=&quot;http://www.lighting.philips.com/main/prof/led-lamps-and-tubes/led-spots/mr16-gu10-par16&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.lighting.philips.com/main/prof/led-lamps-and-tubes/led-spots/mr16-gu10-par16&lt;/a&gt;). They are rated 12V...24V DC-DC with either constant current or constant voltage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But again I want to make sure that my direction of thinking is right. I am no expert on LED lightning.&lt;/p&gt;&#xA;" OwnerUserId="17534" LastEditorUserId="17534" LastEditDate="2018-09-28T07:28:24.730" LastActivityDate="2018-09-28T07:28:24.730" Title="LED Lights Dimmer Electronics" Tags="|electrical-engineering|power-electronics|electrical|consumer-electronics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23953" PostTypeId="1" CreationDate="2018-09-27T15:09:13.517" Score="1" ViewCount="309" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;A uniform pipeline, &lt;span class=&quot;math-container&quot;&gt;$5km$&lt;/span&gt; in length, &lt;span class=&quot;math-container&quot;&gt;$100mm$&lt;/span&gt; in diameter and with a&#xA;  roughness size of &lt;span class=&quot;math-container&quot;&gt;$0.03mm$&lt;/span&gt;, conveys water between two reservoirs. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The difference in water level between the reservoirs is &lt;span class=&quot;math-container&quot;&gt;$50m$&lt;/span&gt;. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In addition to the entry loss of &lt;span class=&quot;math-container&quot;&gt;$0.5\frac{v^2}{2g}$&lt;/span&gt;, a valve produces&#xA;  a head loss of &lt;span class=&quot;math-container&quot;&gt;$10\frac{v^2}{2g}$&lt;/span&gt;. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Determine the steady discharge&#xA;  between the reservoirs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know that the first stage should be to find a value for velocity &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; using the Colebrook-White Formula:&lt;span class=&quot;math-container&quot;&gt;$$v=-2\sqrt{2gDs_{f}}log(\frac{k_{s}}{3.7D}+\frac{2.51u}{D\sqrt{2gDs_{f}}})$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$v=$&lt;/span&gt; velocity, &lt;span class=&quot;math-container&quot;&gt;$g= 9.81$&lt;/span&gt; m/s, &lt;span class=&quot;math-container&quot;&gt;$D=0.1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$s_{f}=0.01$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$k_{s}=0.03$&lt;/span&gt; x &lt;span class=&quot;math-container&quot;&gt;$10^{-3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u=1$&lt;/span&gt; x &lt;span class=&quot;math-container&quot;&gt;$10^{-6}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subbing these values in, I get velocity &lt;span class=&quot;math-container&quot;&gt;$v=2.312$&lt;/span&gt; m/s (3 d.p.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I assume I need to sum the entry and head losses &lt;span class=&quot;math-container&quot;&gt;$h_{l}$&lt;/span&gt;  &lt;span class=&quot;math-container&quot;&gt;$$h_{l}=0.5\frac{2.312^2}{2\times 9.81}+10\frac{2.312^2}{2\times 9.81}=2.861m$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now adjust &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; to get &lt;span class=&quot;math-container&quot;&gt;$h_{f}=50-2.861=47.139$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then find a new value for &lt;span class=&quot;math-container&quot;&gt;$s_{f}=47.139/5000=9.4278\times 10^{-3}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally recalculate for &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; using the Colebrook-White Formula:&lt;span class=&quot;math-container&quot;&gt;$$v=-2\sqrt{2gDs_{f}}log(\frac{k_{s}}{3.7D}+\frac{2.51u}{D\sqrt{2gDs_{f}}})$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$v=$&lt;/span&gt; velocity, &lt;span class=&quot;math-container&quot;&gt;$g= 9.81$&lt;/span&gt; m/s, &lt;span class=&quot;math-container&quot;&gt;$D=0.1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$s_{f}=9.4278\times 10^{-3}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$k_{s}=0.03$&lt;/span&gt; x &lt;span class=&quot;math-container&quot;&gt;$10^{-3}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u=1$&lt;/span&gt; x &lt;span class=&quot;math-container&quot;&gt;$10^{-6}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subbing these values in, I get velocity &lt;span class=&quot;math-container&quot;&gt;$v=2.240$&lt;/span&gt; m/s (3 d.p.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now recalculating &lt;span class=&quot;math-container&quot;&gt;$$h_{l}=0.5\frac{2.240^2}{2\times 9.81}+10\frac{2.240^2}{2\times 9.81}=2.685m$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$h_{l}+h_{f}=2.685+47.139=49.82\approx 50m$&lt;/span&gt; which is sufficiently accurate to be acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$Q=vA \implies Q=2.240\times 0.05^{2}\pi=0.01759$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$m^{3}/s$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I answered this correctly? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="17668" LastActivityDate="2018-09-27T15:09:13.517" Title="Determining the Steady Discharge Between Two Reservoirs" Tags="|fluid-mechanics|hydraulics|pipelines|open-channel-flow|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23954" PostTypeId="2" ParentId="23947" CreationDate="2018-09-27T15:50:03.027" Score="2" Body="&lt;p&gt;Take this all as very general....Flutes of a drill bit are ground only for chip extraction, where as end mills take the chip extraction flute and grind an actual cutting edge on it ( b - c ) and some relief (c - d). Picture is of a 4 flute end mill, the one on the left being proper. There's MUCH much more to cutting tool geometry than this brief explanation, a cutting tool maker or a book on tool &amp;amp; cutter grinders will do a much better job explaining it. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/dAVnL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dAVnL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as far as grinding cutting edges on the flutes of a drill, for one it would make drills much more expensive to manufacture. In manufacturing environments, end mills are generally re-sharpened, for me, any drill under 1/2&quot; gets replaced. Drills are also generally held in drill chucks, end mills in collets or mill holders. Drill chucks are not designed to withstand the radial cutting forces of side milling so a cutting edge on a drill flute would give a false sense of acceptable practice. And one more...the ratio of diameter of the tool to the length of cut is also much less with an endmill, IE, a 1/2&quot; endmill LOC is usually around 1&quot; where a 3/16&quot; drill can cut 2&quot; or so deep. That added LOC of a drill substantially weakens it in the radial direction. So standard, jobbers length drills would deflect far too much to be a usable side miller. Just some examples for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Corey&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some drills have somewhat of a cutting edge ground into the flute, but from my experience, its for helping cut materials that tend to &quot;close up&quot; on you as you drill deep into a hole, bronze/copper/etc. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastEditorUserId="11327" LastEditDate="2018-09-28T19:29:32.420" LastActivityDate="2018-09-28T19:29:32.420" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23956" PostTypeId="2" ParentId="23948" CreationDate="2018-09-27T20:07:10.040" Score="1" Body="&lt;p&gt;View this as a heat exchanger at steady state. Your components generate &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_g$&lt;/span&gt; of heat flow. The water takes in that heat flow as &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_w = \dot{m}_w \tilde{C}_{pw} (T_{wo} - T_{wi})$&lt;/span&gt;, with mass flow, specific heat, and temperatures defined. The remaining heat &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_r = \dot{q}_g - \dot{q}_w$&lt;/span&gt; heats the components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to increase &lt;span class=&quot;math-container&quot;&gt;$T_{wi}$&lt;/span&gt;. One of three things can happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) With constant mass flow of water, &lt;span class=&quot;math-container&quot;&gt;$T_{wo}$&lt;/span&gt; will increase. This can happen when the water leaving the heat exchanger was initially cooler than the components that are to be cooled. In this case, you had some room to increase the inlet temperature because you have not yet over-run the outlet temperature of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) With constant mass flow of water, &lt;span class=&quot;math-container&quot;&gt;$T_{wo}$&lt;/span&gt; will not change and &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_w$&lt;/span&gt; will decrease. This will happen when the water leaving the exchanger is already as hot as the components it is cooling. This is actually a case of a maximum heat exchange. In this case, when you increase &lt;span class=&quot;math-container&quot;&gt;$T_{wi}$&lt;/span&gt;, less heat is taken from the components. They will correspondingly get hotter as they reach a new equilibrium state. The output temperature of the water will then also get hotter, meaning less heat taken by the water. The result of this case will be that your components will get hotter, perhaps even hotter than you might have expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) You will increase the mass flow, &lt;span class=&quot;math-container&quot;&gt;$\dot{m}_w$&lt;/span&gt; to account for the increase in &lt;span class=&quot;math-container&quot;&gt;$T_{wo}$&lt;/span&gt;. The heat flow &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_w$&lt;/span&gt; will stay the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all cases, you will decrease the electrical demand in the chiller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's do a back of the envelope calculation. Assume the chiller runs at its thermodynamic efficiency. What is the difference in &lt;a href=&quot;https://en.wikipedia.org/wiki/Coefficient_of_performance&quot; rel=&quot;nofollow noreferrer&quot;&gt;CoP&lt;/a&gt; for a refrigerator unit that must cool water at 23 oC to 16 oC versus one that must cool water at 23 oC to 21 oC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ CoP = \frac{\dot{q}_C}{\dot{w}} = \frac{T_C}{T_H - T_C} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ CoP_o = \frac{289}{7} = 41 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ CoP_{new} = \frac{294}{2} = 147 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the same mass flow of water (case 1 and case 2). The ratio of cooling needs is &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_{new} / \dot{q}_o = \Delta T_{w,new} / \Delta T_{w,o} = 2/7$&lt;/span&gt;. The ratio of work loads is therefore found as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{\dot{w}_{new}}{\dot{w}_o} =  \frac{\dot{q}_{new}}{\dot{q}_o} \frac{CoP_o}{CoP_{new}} = \frac{2}{7} \frac{41}{147} = 0.080$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We find the power demand to a perfect chiller unit in the new case is predicted to be 8% of the power demand to a perfect chiller in the starting case. Inefficiencies in the chiller will increase this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, you can decrease the electricity load at the chiller unit by running as per the first or second cases. Before you move forward, you should test the temperatures at the components that are being cooled relative to the output temperature of the cooling water. When the temperatures are the same, your savings will come at a cost that your components will get hotter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third case has to consider that &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_{new} = \dot{q}_o$&lt;/span&gt;, so the maximum theoretical improvement will be 28%. The demand to the pump for the water will go up. Whether it offsets the gain is to be determined.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-09-28T23:03:07.997" LastActivityDate="2018-09-28T23:03:07.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23957" PostTypeId="2" ParentId="7451" CreationDate="2018-09-28T09:30:15.363" Score="0" Body="&lt;p&gt;I'd like to mention something else that will improve the retention of your data. If you want your data to be really safe - make a few copies. To be safe even if parts of the data are corrupted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have multiple physical copies of your data (say: 3 memory cards holding the same files) you can use a tool like &lt;a href=&quot;https://www.gnu.org/software/ddrescue/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ddrescue&lt;/a&gt; to heal problems in one copy with the other one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way even if some of the data gets corrupted - you have a very high chance of recovering it all as random errors will most likely not corrupt the same bytes in all of your copies. The more copies you have - the more recoverable your data is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you still have some holes in your ddrecovered LZIP archive - you can use &lt;code&gt;lziprecover&lt;/code&gt; to fix any residual errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach could be to create a ZFS pool on multiple USB drives or memory cards and set them to work in mirror mode. ZFS is a filesystem/volume manager designed to prevent data loss used in data centers and NAS deployments around the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'd need to use Linux or Solaris for that though.&lt;/p&gt;&#xA;" OwnerUserId="17676" LastActivityDate="2018-09-28T09:30:15.363" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23958" PostTypeId="1" AcceptedAnswerId="23966" CreationDate="2018-09-28T14:22:22.627" Score="0" ViewCount="862" Body="&lt;p&gt;I'm planning to build an air-dropper for a small drone - a device to release small payloads hanging under the drone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a commercial example of such a system, used with a much bigger drone - for heavier payloads, pretty heavy itself, and using the drone's built-in systems for control:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=e81eQ0pYLCs&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=e81eQ0pYLCs&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(7:18 - system in action. 5:44 - the part, for which counterpart I'm asking.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm planning to build something similar, but much smaller. Payloads of up to 100-150 gram (but often smaller too), a completely independent control system. I already have working remote control electronics capable of outputting 3.3V at 40mA (driving a tiny coreless motor like &lt;a href=&quot;https://www.alibaba.com/product-detail/DC-3-7V-dia-4mm-micro_60459075751.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;; I also have such motors), under 10 gram together with the battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I still have no idea what sort of actuator I could use to release the payload. I'd prefer it to remain unpowered when 'latched', needing only power to open (drop the payload). It should be as lightweight as possible, because any of its weight takes away from payload capacity of the drone. And not being very expensive would be a big plus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the ideas I had so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Most servos I've seen are rather on the heavy side, or provide so little torque I'm afraid any latch I come up with won't suffice. &lt;/li&gt;&#xA;&lt;li&gt;I considered pairing a permanent magnet and an electromagnet that would cancel out the magnet's field, but first I'm not sure if I can output a field strong enough, and then the cancellation won't be perfect and if my payload is very light, it won't drop.&lt;/li&gt;&#xA;&lt;li&gt;I don't think I have the current to drive a linear electromagnetic actuator to unlatch a hook (not sure about that though). &lt;/li&gt;&#xA;&lt;li&gt;I don't have means to build more complex gearboxes to convert the high RPM of the tiny motor to torque - at least from scratch; if something of that kind is commercially available and small enough, it might be good. &lt;/li&gt;&#xA;&lt;li&gt;...also, it would be great if it was fully reusable and safe. I seriously considered just a small relay to put full current from a battery through a resistive spiral to burn through the line with the payload. I'm not enthusiastic about this idea.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm not specifying the housing, 'hook', line, attachment methods etc here - these are to be made such as required for it to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(also worth considering: a fully mechanical connector with no electronics; disengaging on a maneuver not encountered in normal flight, e.g. a rapid drop. I can't generate negative-G, but I can make the drone -almost- drop out of the sky. If this could disengage the hook, that would also be a workable solution. This is not as preferable as it requires considerable vertical space and is rather risky.)&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-09-28T20:37:17.320" Title="air-dropping actuator for a small drone" Tags="|mechanical-engineering|actuator|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23959" PostTypeId="2" ParentId="23938" CreationDate="2018-09-28T14:24:20.817" Score="2" Body="&lt;p&gt;Direct measures the physical property directly. So if you want to measure length you would measure length. Both calipers and gauge blocks do that. But say optical measurement unit would not directly measure length, it would measure something else. Such as time for bounce in case of time of flight laser rangefinder, intensity of reflection for infrared distance units or angle for triangulation with a camera.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2018-09-28T14:24:20.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23960" PostTypeId="2" ParentId="23958" CreationDate="2018-09-28T15:23:45.923" Score="0" Body="&lt;p&gt;Latching relays in both the NO and NC state are &lt;a href=&quot;http://pbresource.com/relay-BLY.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;common&lt;/a&gt; enough.  Granted,  those will be a bit larger than what you need, but the principle is easy enough: you manually engage the latch, and an electrical pulse to the relay pulls the latch into the open state, releasing your cargo.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-09-28T15:23:45.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23964" PostTypeId="2" ParentId="23958" CreationDate="2018-09-28T19:31:36.457" Score="1" Body="&lt;p&gt;I've designed and built using a 3D printer a release mechanism to drop one kilogram payloads from a DJI Phantom 3 copter. It used the same principle as the &lt;a href=&quot;https://en.wikipedia.org/wiki/3-ring_release_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;3-ring release mechanism patented by Bill Booth&lt;/a&gt; for cutting away parachute canopies in the event of an emergency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concept behind this system is that the first &quot;fold-over&quot; takes the primary load close to the fulcrum. This allows the reduction of force at the end of the lever (virtual or real) which is then integrated into another &quot;fold-over&quot; at a much lower force. This force is applied close to the fulcrum, which is reduced at the end as before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the system I built, I used cotter keys on pivots. The first pivot point allowed the load to be applied close to the pivot. The end of the cotter key was then captured by another cotter key with the end being close to the pivot, but with the second cotter key 90° to the first. This was repeated again. With the dimensions I used, there was a one thousand to one reduction of force, based on the mathematics of leverage. It might have been a bit greater than that, but rounding down reduces the exaggeration of pride.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end result was that the final release required so little force to open that the camera integrated into the Phantom was able to push the release lever. This caused the cascade of the three links to open in sequence, dropping the payload.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &lt;a href=&quot;https://youtu.be/CcMh9ZbPzB4&quot; rel=&quot;nofollow noreferrer&quot;&gt;video link&lt;/a&gt; shows the release in action. The wheel on a pin is a locking feature, as the force is so small for release that it slides open without much load. The teardrop on the end of the release bears against the camera mechanism and is press-fit on the pin, allowing for rotational adjustment once the bracket is in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between corners 3 and 4, there is a small folded piece of soft drink bottle providing &quot;ejection force&quot; to the last pin, effectively applying additional force to keep friction between the last pin and the release pin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found that a tiny bit of lube causes the release to open with vibration, while a tiny bit of debris or debris will prevent opening. This is directly related to the extremely small force available from the camera acutator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one created a 3-link or greater release in the manner described, one could use extremely light actuator mechanisms, electro-magnetic or mechanical. My objective was to have zero interconnection with the electronics of the copter. If you can wire into the lights as shown in the video, you have more power for a solenoid release or similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism in the video shows a fairly poor use of mechanical advantage. The freely-pivoting bar is short, and there's a short distance to the loop gap (good) but an equally short distance to the release (bad). If the pivoting bar is longer, the amount of force on the release end is reduced.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-09-28T19:31:36.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23966" PostTypeId="2" ParentId="23958" CreationDate="2018-09-28T20:37:17.320" Score="1" Body="&lt;p&gt;I would recommend using servo (or other actuator) to pull a pin that is holding the load in shear. That way, as long as the friction of your system is low, the actuator will only need a fraction of the force (like a mousetrap). You would attach the pin to the servo arm such that it is as close to the rotation point as possible to give you the most torque. Also, make sure the attachment point can rotate for proper actuation and release. On the other end you will want to use something thin (or with a shallow hole) to support the pin. This way the stroke does not have to be very long.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Here is a &lt;a href=&quot;https://hobbyking.com/en_us/turnigytm-tgy-1440a-analog-servo-v2-0-8kg-0-10sec-4-4g.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;4.4 gram servo&lt;/a&gt; that should work well.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://hobbyking.com/en_us/radio-servos/servos.html?price=0-12&amp;amp;wrh=2%2C3&amp;amp;unit_weight_filterable=0-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hobbyking&lt;/a&gt; has a pretty good specifications filter now.&lt;/li&gt;&#xA;&lt;li&gt;The pin can be made out of &lt;a href=&quot;https://hobbyking.com/en_us/push-rod-and-sleeve-set-5pcs-bag.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;rc push rod material&lt;/a&gt; so you don't have to reinvent the wheel.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is a cartoon image of the proposed idea. Things are a bit exaggerated; you will want to reduce the distance between your servo and the pin support so that the pin as is short as possible. A shorter pin means it can be a thinner rod without bending or buckling under the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nFiUE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nFiUE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also note that the servo will need pwm control. You can get controllers for this, or if you want discrete input you can cut open the servo and just power the dc motor directly. Make sure you have a delay or a limit switch (possibly in its circuitry already) so you don't fry the motor.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-09-28T20:37:17.320" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23967" PostTypeId="1" AcceptedAnswerId="23972" CreationDate="2018-09-28T20:43:13.650" Score="0" ViewCount="888" Body="&lt;p&gt;I am aiming to program a quadcopter flight controller as a personal project.&#xA;To be able to do this I understand one of the  fundamentals is a PID control system . So my question is does anyone have any recommendations for books to read or will google be good enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found the following books , has anyone read these?:&#xA;&quot;Advanced PID  control&quot; - &lt;a href=&quot;https://www.amazon.co.uk/Advanced-PID-Control-Karl-Astrom/dp/1556179421/ref=sr_1_4?s=books&amp;amp;ie=UTF8&amp;amp;qid=1538160392&amp;amp;sr=1-4&amp;amp;keywords=pid+control&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&#xA;or ,&#xA;&quot;PID Control: New Identification and Design Methods&quot; - &lt;a href=&quot;https://www.amazon.co.uk/PID-Control-Identification-Design-Methods/dp/1849968977/ref=sr_1_14?s=books&amp;amp;ie=UTF8&amp;amp;qid=1538160392&amp;amp;sr=1-14&amp;amp;keywords=pid+control&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17684" LastEditorUserId="17684" LastEditDate="2018-09-28T21:55:55.400" LastActivityDate="2018-09-29T06:46:36.627" Title="Books for learning PID control?" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="2" CommentCount="2" ClosedDate="2018-09-30T09:47:59.820" ContentLicense="CC BY-SA 4.0" />
  <row Id="23968" PostTypeId="2" ParentId="23943" CreationDate="2018-09-28T23:32:03.500" Score="1" Body="&lt;p&gt;In free fall, any segment of a control volume for &lt;em&gt;any&lt;/em&gt; material experiences absolutely no gravity. As noted, Euler's equation yields a statement that pressure is identically isotropic at that point. In essence, there is no &quot;down&quot; to the control volume. Apply this throughout the control volume, and you find that pressure is identically isotropic throughout the entire &lt;em&gt;body&lt;/em&gt; of the control volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before we continue, the above statements are the founding principles behind &lt;a href=&quot;https://en.wikipedia.org/wiki/Drop_tube&quot; rel=&quot;nofollow noreferrer&quot;&gt;Drop Tube Experiments&lt;/a&gt; to measure phase transformations without the influence of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a control volume with a surface. That surface has a surface tension. The interesting note here happens when the external surroundings have a static fluid. We find that each point on the surface of the drop should theoretically experience a different pressure because the external fluid is static. In essence, the external surface knows which way is &quot;down&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue we face is to account for the drag affects of the external fluid on the falling body. A pressure difference will arise top to bottom of the body from this factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The escape out of having to consider the surface tension and drag affects is to invoke a perfect vacuum around the falling body.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-09-28T23:32:03.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23969" PostTypeId="2" ParentId="23967" CreationDate="2018-09-29T01:16:30.027" Score="0" Body="&lt;p&gt;The engineering problem of designing from first principles an optimal PID control system is a topic about which entire textbooks have been written and is the subject of semester-long introductory courses at the upper-division university level. Google will most assuredly not be enough. If it were, mechanical engineering students would not have to spend four years learning the basics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To even begin the process of designing a control system requires you first to furnish a dynamical model of the system you wish to control, which is a discipline all unto itself. Unless your model is accurate, your subsequent efforts to devise a suitable control system for it are doomed.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-09-29T01:16:30.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23970" PostTypeId="1" CreationDate="2018-09-29T05:35:39.917" Score="1" ViewCount="2704" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qgYz4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qgYz4.png&quot; alt=&quot;Motor Electrical Data&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a motor for a submersible pump with a Rated Current of 8.4A as given in the motor's electrical data. However, when it is running, I record a current of 3.4-3.9A. Is there a fault I am supposed to look into, or is this normal? &#xA;Why is the rated current so much more than the measured current.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/X5P9r.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X5P9r.jpg&quot; alt=&quot;Measure Current&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11147" LastActivityDate="2018-09-29T10:02:55.610" Title="Why is the Current to Motor less than Motor Rated Current?" Tags="|mechanical-engineering|electrical-engineering|motors|pumps|current|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23971" PostTypeId="2" ParentId="23970" CreationDate="2018-09-29T05:47:14.750" Score="1" Body="&lt;p&gt;This can be due to the rated current is showing the initial start current which can be greater than the running current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supply needs to be fused to cope with the start current as, if you put a fuse matched to the running current, it would blow on start-up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You probably won’t be able to measure the start ( or in-rush) current with that meter, you would need something more accurate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did an experiment about the current through an incandescent bulb, measuring the current v time - the in-rush current is much higher for initially and reduces rapidly as the bulb heats up - talking 20 to 30ms here iirc.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-09-29T05:47:14.750" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23972" PostTypeId="2" ParentId="23967" CreationDate="2018-09-29T06:40:31.993" Score="1" Body="&lt;p&gt;As @niels nielsen  mentions the topic is very broad and successful implementation and design require additional techniques not necessarily directly associated with control design. Depending on your specific application this might include dynamic modeling, system identification, model validation and updating, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are numerous books that deal with PID control and, given the above, it should be clear that it is impossible to provide an exhaustive list of references. I have not used the books you mention and although their table of contents looks decent their applicability will depend on your particular application. Some alternative references are mentioned below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get a quick overview of what a PID controller actually does the &lt;a href=&quot;https://en.m.wikipedia.org/wiki/PID_controller?wprov=sfti1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page on this subject&lt;/a&gt; is actually quite decent. However, I would not consider this sufficient information for any serious implementation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For SISO control systems I’ve used both “Feedback Control of Dynamic Systems” by Franklin and “Feedback Systems: An Introduction for Scientists and Engineers” by Astrom and Murray. The former is somewhat outdated but covers the basics well and also includes some material on dynamic modeling. The latter is more up to date, covers some more advanced materials and is also available &lt;a href=&quot;http://www.cds.caltech.edu/~murray/amwiki/index.php/Second_Edition&quot; rel=&quot;nofollow noreferrer&quot;&gt;online&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For control of MIMO systems I personally like “Multivariable Feedback Control” by Skogestadt. However, if you decide to go into this direction, be prepared for the mathematical approach that is inherent to understanding the design and analysis of MIMO control systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good reference, which approaches PID control from an implementation point of view, is “Applied Control Theory for Embedded Systems” by Tim Wescott. I think this is SISO only but it does discuss PID tuning based on measured transfer functions and how to measure these transfer functions. If you are looking for rigorous mathematical detail this is not the book for you.&lt;/p&gt;&#xA;" OwnerUserId="11869" LastEditorUserId="11869" LastEditDate="2018-09-29T06:46:36.627" LastActivityDate="2018-09-29T06:46:36.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23974" PostTypeId="1" CreationDate="2018-09-29T09:19:41.417" Score="0" ViewCount="223" Body="&lt;p&gt;I'm looking for an explicit numerical example of a parametric surface (u, v) |-&gt; (x(u, v), y(u, v), z(u, v)) that models a fan blade like that in the attached picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling results in NURBS/PDE approaches but with no numerical explicit data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to find a more simple model to be used in a engineering class for upper high school students.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, Humberto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cH0Z6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cH0Z6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17687" LastActivityDate="2018-09-29T17:53:42.263" Title="Parametric equations for fan blades" Tags="|modeling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23975" PostTypeId="2" ParentId="23970" CreationDate="2018-09-29T10:02:55.610" Score="3" Body="&lt;p&gt;The &quot;rated&quot; value is the maximum safe value for each parameter, not necessarily the actual value for any particular working condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current when the motor is running will depend on the load. If you were testing it with no load, the current will be lower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably the pump specification (not the motor spec) will tell you the conditions (flow rate, pressure head, etc) that make the motor work hardest. You could try simulating those conditions and measure the current again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW the numbers from the data sheet are consistent with each other. For the minimum voltage (200V), 8.4A current, and a power factor of 1.0, the calculated power input is 1.68 kW compared with the data sheet value of 1.65kW.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-09-29T10:02:55.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23977" PostTypeId="1" CreationDate="2018-09-29T10:22:31.680" Score="0" ViewCount="57" Body="&lt;p&gt;I have currently been experimenting with a phase change material to absorb the heat given off by the condenser in a small refrigeration system (about 1/5 hp compressor) and while it does absorb a substantial amount, it often requires a huge volume of the phase change material (many many kilograms of the stuff) which is not practical due to the heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any other alternatives that can absorb heat without a huge increase in weight? &lt;/p&gt;&#xA;" OwnerUserId="12174" LastActivityDate="2018-10-29T13:00:37.810" Title="Is there any practical way to partially absorb the heat from a refrigeration condenser" Tags="|heat-transfer|refrigeration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23979" PostTypeId="2" ParentId="23977" CreationDate="2018-09-29T12:44:11.477" Score="1" Body="&lt;p&gt;Many systems are designed to avoid condensors and matched to feed directly into other systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example is refrigeration units that pre-heat water on its way to a boiler - any temperature rise is less work the boiler has to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to consider what systems you can feed heat into and how “direct” you can achieve that - the more conversions the poorer the efficiency....&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-09-29T12:44:11.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23981" PostTypeId="2" ParentId="23974" CreationDate="2018-09-29T16:56:58.727" Score="3" Body="&lt;p&gt;For propellers, one usually has blade reference lines called &lt;em&gt;generator lines&lt;/em&gt;. They can be parametrically defined in terms of blade number, skew, rake, hub dia, tip dia. If the generator lines are the midpoint of the blade chord, they are called &lt;em&gt;blade center lines&lt;/em&gt; BCL. Next, you have a parametric expression for the pitch progression along the generator (often the deviation from a nominal pitch progression). Next, you have the blade area distribution along the generator. Basically, this is the blade chord  associated with each diameter (or projected arc length, choose one). Finally, you have a foil development for each diameter (this has to be consistent with the pitch definition). So it's a bit more complicated than what you envision. The point of parametric expressions in engineering is to be able to isolate performance variables and manufacturing variables and cost variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a fan is built by choosing the number of blades, hub diameter, and tip diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, define the blade generator - you need Skew(D), Rake(D).&lt;br&gt;&#xA;Next, define the geometric pitch expression, P(D)&lt;br&gt;&#xA;Next, develop the chordlines on the generator using (projected) blade arc and P(D).&lt;br&gt;&#xA;Next, define the blade foil's camber shape as a series of offsets from the chordline.&lt;br&gt;&#xA;Finally, find the surfaces by applying a thickness expression T(C,D) to the foil's camber line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You usually use cylindrical coordinate system for props - X (axial), R (radial), and Theta (angle from reference point)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Michigan Wheel PDF of terms and geometry systems. &lt;a href=&quot;http://navalex.com/downloads/Michigan_Wheel_Propeller_Geometry.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://navalex.com/downloads/Michigan_Wheel_Propeller_Geometry.pdf&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for your fan, lets say rake is zero. You need an expression for the skew(D) to create a generator. Let this be your &quot;u&quot;. Skew(D) = Theta = f(r).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take geometric pitch to be constant (meaning uniform geometric advance along the blade span.) Pitch angle(r) = Arctan(pitch/2*Pi*r). Given a uniform thickness, you now have everything needed to define face and back surfaces with respect to the generator ( X, R, Theta in terms of u(r), constant pitch, and constant thickness). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arc length of the chord at each point is your expression &quot;v&quot;. chord length(r) = g(r). This completes the boundary of the surface (along with hub dia, tip dia, if chord is nonzero at those radii.) The expression for chord length can be given in Expanded Coordinates, which is a shell of a cylinder unrolled into 2D and conventionally expressed in dimensions X and Z'. Or it can be expressed in the length projected onto the propeller plane and then unrolled. The two are related by the cosine of the local geometric pitch angle. So the trick now is you have a boundary value problem. One of your variables defines the surface and the other defines the boundary. This is why I don't think this isn't a very good example for your purpose.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-09-29T17:53:42.263" LastActivityDate="2018-09-29T17:53:42.263" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23983" PostTypeId="1" CreationDate="2018-09-29T22:13:27.280" Score="0" ViewCount="369" Body="&lt;p&gt;Suppose we do nonlinear structure mechanics computation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In each time step, we can get the results and can get Von Mises Stress at each node of the mesh. Can we say this number also means its residual stress?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, is there any way to get residual stress?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="17605" LastActivityDate="2018-09-29T22:54:50.587" Title="Relation between residual stress and von mises stress calculated by FEA in comercial software" Tags="|structural-engineering|computer-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23984" PostTypeId="2" ParentId="23983" CreationDate="2018-09-29T22:54:50.587" Score="2" Body="&lt;p&gt;The Von Mises Stress and the residual stress are not the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the residual stress, do another time step where all the applied loads are zero. If there is material nonlinearity (plasticity or creep) the stresses and deflections will not return to zero when you remove all the loads.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-09-29T22:54:50.587" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23987" PostTypeId="1" CreationDate="2018-09-30T17:31:58.940" Score="6" ViewCount="295" Body="&lt;p&gt;Based on the description of friend I drew this small example of a specific pan-tilt mechanism I'd like more information on, but as I'm not an engineer I don't really know what terms to use and can't seem to find anything.&#xA;I think I might have seen it somewhere related to nautical devices once, but I'm not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system I'm looking for uses two cylinders with a slanted face. Both are able to rotate around their central axis. As part B (top part) is connected to part A (bottom part) at the center of their slanted faces, it moves together with part A but is also able to rotate around it's own axis, combining both their movement and creating a lot of freedom in achievable positions.&#xA;See the image below for a 3d impression of the shapes and a 2 schematic of how they rotate (where the dotted line would be their rotational axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea what this is called, or what to look for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YfXvm.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YfXvm.png&quot; alt=&quot;3d image&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rwsHL.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rwsHL.png&quot; alt=&quot;2d image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17699" LastEditorUserId="17699" LastEditDate="2018-09-30T17:39:57.073" LastActivityDate="2018-12-26T23:49:19.660" Title="What is this pan-tilt system called?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23988" PostTypeId="1" CreationDate="2018-09-30T21:59:49.660" Score="2" ViewCount="1717" Body="&lt;p&gt;I have a BMI088 and am trying to get an accelerometer reading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data sheet and specs for Bosch Sensortec are &lt;a href=&quot;https://www.bosch-sensortec.com/bst/products/all_products/bmi088_1&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensitivity per range is this (its a 16 bit signed integer):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;± 3 g: 10920 LSB/g&lt;/li&gt;&#xA;&lt;li&gt;± 6 g: 5460 LSB/g&lt;/li&gt;&#xA;&lt;li&gt;± 12 g: 2730 LSB/g&lt;/li&gt;&#xA;&lt;li&gt;± 24 g: 1365 LSB/g&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Yet, the formula they give to calculate the force in mg is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mg = accel_int16 / 32768 * 1000 * 2^((range register value) +1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the range register has the following values:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;± 3g  = 0x0&lt;/li&gt;&#xA;&lt;li&gt;± 6g  = 0x1&lt;/li&gt;&#xA;&lt;li&gt;± 12g = 0x2&lt;/li&gt;&#xA;&lt;li&gt;± 24g = 0x3&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we do the math on that and plug in the highest value (+32768) and highest range register value (0x3), we get a force of 16000mg or 16g, so a range of -16g to +16, not -24g to +24g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what is the correct conversion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we do simply, accel_int16 LSB / 1365 LSB/g we get a maximum value of 24g for 32768 which is intuitively what one would expect. Going by the same logic of a linear response, we might also say, each value (LSB) represents a LSB/mg, which would be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;32768*x = 24000mg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x = 0.7324&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case we can take a reading, multiply by 0.7324 to get our mg.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong or not understanding? Why does the data sheet formula not give me the correct range?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part to that question, is, I  want to have a scale factor to convert from LSB to to meters/second/second, so intuitively one would think it would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;scale = 0.7324 * 0.001 * ~9.8m/s/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, if we follow the datasheet formula we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;scale = 0.4883 * 0.001 * ~9.8m/s/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I misunderstanding something fundamental here? Or does a 24g accelerometer only measure in a 16g range for some reason? &lt;/p&gt;&#xA;" OwnerUserId="17702" LastEditorUserId="282" LastEditDate="2019-05-30T23:35:01.490" LastActivityDate="2021-04-15T15:42:03.670" Title="Conversion from LSB to mg not making sense for accelerometer" Tags="|sensors|embedded-systems|acceleration|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="23989" PostTypeId="2" ParentId="7598" CreationDate="2018-10-01T00:50:23.430" Score="3" Body="&lt;h2&gt;Yes, this is permissible&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;NFPA 13 7.1.3 explicitly permits this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;7.1.3 Auxiliary Systems.&lt;/strong&gt; A wet pipe system shall be permitted&#xA;  to supply an auxiliary dry pipe, preaction, or deluge system,&#xA;  provided the water supply is adequate.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="964" LastActivityDate="2018-10-01T00:50:23.430" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23990" PostTypeId="1" AcceptedAnswerId="23992" CreationDate="2018-10-01T10:10:30.490" Score="0" ViewCount="72" Body="&lt;p&gt;Nowadays deep vacuum pumps are fairly expensive, so I'm wondering: how could people reach the deep vacuum required for &lt;a href=&quot;https://en.wikipedia.org/wiki/Crookes_tube&quot; rel=&quot;nofollow noreferrer&quot;&gt;crookes tube&lt;/a&gt; or &lt;a href=&quot;https://en.wikipedia.org/wiki/Geissler_tube&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geissler tubes&lt;/a&gt; in the 19th century ? (I'd be interested in building one myself)&lt;/p&gt;&#xA;" OwnerUserId="5926" LastActivityDate="2018-10-01T11:36:43.970" Title="deep vacuum in the 19th century" Tags="|vacuum|vacuum-pumps|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23991" PostTypeId="2" ParentId="16828" CreationDate="2018-10-01T10:41:22.643" Score="1" Body="&lt;p&gt;I'm not sure I totally understand your question, but I give a try.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have an actual situation and a calculation (model or analytic formula). The calculation resulted in a gate height for the desired flow rate distribution. After that you tried the gate height in the actual situation and it resulted in an unexpected flow rate distribution (2.7 m&lt;sup&gt;3&lt;/sup&gt;/s instead of 3 m&lt;sup&gt;3&lt;/sup&gt;/s). Probably you checked the other flow rate (it should be 2.3 m&lt;sup&gt;3&lt;/sup&gt;/s) and the total (5 m&lt;sup&gt;3&lt;/sup&gt;/s) as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose the calculation has been made with the assumption of stationair flow conditions and the actual situation can also be described as stationair. Then the friction losses by the gate are most likely the cause of the deviation. In your calculation you had to make some assumptions for the gate structure. If you want to reduce this deviation you can calibrate the structure with different flow rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall a calculation is &lt;em&gt;never&lt;/em&gt; this accurate because you always have to make some simplifications and assumptions. These cause a deviation in the actual situation compared to the calculation or model. In laboratory experiments it is possible to control the total situation as much as possible to minimalize these deviations. In the real world, you have to take some deviation into account and act on that. &lt;/p&gt;&#xA;" OwnerUserId="17698" LastActivityDate="2018-10-01T10:41:22.643" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23992" PostTypeId="2" ParentId="23990" CreationDate="2018-10-01T11:36:43.970" Score="4" Body="&lt;p&gt;Geissler invented the so-called &quot;liquid piston&quot; pump using mercury for the piston. The basic idea of the design is to entrap some of the air between two volumes of mercury in a pipe, and then force the mercury along the pipe to release the air into the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geissler achieved 0.1 Torr with that design of pump in the 1850s. Sprengel invented an improved version of the same idea. The first models were capable of &lt;span class=&quot;math-container&quot;&gt;$10^{-2}$&lt;/span&gt; Torr which was subsequently improved &lt;span class=&quot;math-container&quot;&gt;$10^{-5}$&lt;/span&gt; Torr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://cds.cern.ch/record/455984/files/p281.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cds.cern.ch/record/455984/files/p281.pdf&lt;/a&gt; has more history, and diagrams (but unfortunately, not of very good quality).&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-10-01T11:36:43.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23993" PostTypeId="1" CreationDate="2018-10-01T12:28:51.753" Score="2" ViewCount="165" Body="&lt;p&gt;Sorry if this is the wrong exchange, if it is could you point me in the right direction that'd be awesome!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a bit of a problem locally that I think there could be a simple solution that we're not seeing. I'm part of the local volunteer group that cleans the river running through the town but a lot of rubbish is thrown in. We have litter pickers to get the close bits and what is in the weeds, nets to get the closer litter that floats on the surface, and hooks to try and get the heavy items out from under water, but that's where the problem is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of the items that get thrown in take a lot of energy and time to get out. They may include things like shopping trolleys (carts), we recently pulled out part of a car exhaust, car and bike tires, large metal or wooden poles, children's toys, chairs, wrought iron bench ends, and so much more! On the edge of the water there are thick reeds to promote biodiversity in the river, and wooden beams to support the bank. The hook very often gets caught on these and gets stuck, meaning we have to put our selves in danger to free it. I would happily jump in the water, because hey its just a bit of water, but the medical team that follows us has strongly advised us not to as there are a lot of rodents around and we could contract diseases such as weil's disease. The hook we use is attached to about 5 meters of rope which allows us to throw the hook in and attempt to catch the items, but if we miss then we just have to keep trying.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a model cross section of the river that may help you visualise the area.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KKwQs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KKwQs.jpg&quot; alt=&quot;Cross Section&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have thought of a potential solution to this that may help. We use one or two people either side of the river who hold a guide rope which the hook is freely attached to. Each side then has a rope attached to the rope so that they can pull the hook across the guide rope. The guide is lowered into the water to hook the item and then the attached ropes can be pulled to pull the item out. If the hook or item becomes trapped or free we would simply pull the hook in the other direction to reattach or free the hook again. A model of this is also attached below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uTTyD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uTTyD.jpg&quot; alt=&quot;Solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are your thoughts on this? Could this be a valuable solution or am I just getting it completely wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advanced,&#xA;Shaun.&lt;/p&gt;&#xA;" OwnerUserId="17706" LastActivityDate="2019-09-02T07:02:25.293" Title="Engineering Solution for river cleaning" Tags="|design|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="23994" PostTypeId="2" ParentId="23993" CreationDate="2018-10-01T19:54:41.707" Score="0" Body="&lt;p&gt;I would recommend something different:  change the hook you are using.  No matter how you are suspending it, at some point it'll foul.  What would be ideal, then, is a grappling end with a locking hinge you can remotely release to change from, roughly, an L-shape to straight shape, thus freeing it from whatever is in the muck.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took a quick look thru some camping and climbing sites, but their grappling hooks are all designed to open upon deployment, not release remotely. You may have to search deeper, or design your own &quot;inverted&quot; version. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-10-01T19:54:41.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23996" PostTypeId="1" AcceptedAnswerId="23997" CreationDate="2018-10-01T22:30:10.377" Score="0" ViewCount="3370" Body="&lt;p&gt;I have this diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ChVQm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ChVQm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I have done these calculations:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;R_x&amp;amp;=-20\sin(30)+30\cos(35)+80\cos(45)=71.14\text{ lb} \\&#xA;R_y&amp;amp;=20\cos(30)+30\sin(35)-80\sin(45)=-22.04\text{ lb}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude of this resultant force is then:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$|R|=\sqrt{71.14^2+(-22.04)^2}=74.48\text{ lb}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the part I need help with. I can get a theta angle for where the resultant force's angle &lt;em&gt;is,&lt;/em&gt; but I want to figure out how one would get this angle measured counterclockwise from the positive x-axis:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\theta=\tan^{-1}\left(\frac{-22.04}{71.14}\right)=-17.21°$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I get this angle measured counterclockwise from the + x-axis? I don't get this and would really appreciate some help. This comes up a lot and I do not understand it.&lt;/p&gt;&#xA;" OwnerUserId="13219" LastEditorUserId="1832" LastEditDate="2018-10-02T01:43:02.997" LastActivityDate="2018-10-02T01:43:02.997" Title="Statics problem: Finding resultant force angle measured counterclockwise from positive x-axis" Tags="|statics|mathematics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="23997" PostTypeId="2" ParentId="23996" CreationDate="2018-10-02T01:38:48.427" Score="1" Body="&lt;p&gt;I haven't bothered checking your math since that doesn't seem relevant to your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, assuming that your answer of -17.21° is correct, that is in fact the counter-clockwise angle. It means that you need to go negative 17.21° counterclockwise, which is equivalent to rotating 17.21° clockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if what you actually mean is how to get the positive angle equivalent to -17.21°, well, that's just basic geometry: a circle has 360°. So to go from a negative angle to the equivalent positive angle, just add 360°.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$-17.21 + 360 = 342.79°$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-10-02T01:38:48.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="23999" PostTypeId="1" AcceptedAnswerId="24000" CreationDate="2018-10-02T08:23:20.397" Score="0" ViewCount="222" Body="&lt;p&gt;I am trying to perform static structural analysis of a turbine rotor which is rotating at a given angular speed, say 1000 rad/s. I also know the pressure distribution from CFD analysis. What are the boundary conditions for static structural analysis? I can use angular velocity as inertial load, pressure distribution from CFD as applied load. Do I have to restrain the  translation/rotation of surface 1, shown in figure? If yes, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mZVYV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mZVYV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2018-10-02T10:22:34.320" Title="Boundary Conditions fro Static Structural Analysis of a Turbine Rotor Using FEM" Tags="|finite-element-method|nastran|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24000" PostTypeId="2" ParentId="23999" CreationDate="2018-10-02T10:17:10.167" Score="2" Body="&lt;p&gt;If your CFD analysis for the pressure loads was correct, the resultant of all the pressure loads should contain axial load and a torque on the complete assembly. If there is no torque, the turbine is useless! The axial load is not useful for the complete machine, but you can't avoid it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether it is rotating or not, the turbine wheel can move in space as a rigid body. So you need to restrain it to prevent that, and create reaction forces that balance the applied axial load and torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to do this is to restrain it at the location where it would be attached to the turbine shaft. You will then get the &quot;correct&quot; stress distribution in the disk. If you restrain it at some other arbitrary point or points, you will probably get an unrealistic high stress at those points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The turbine disk is probably &lt;em&gt;not&lt;/em&gt; attached to the shaft at your &quot;surface 1&quot;, unless it is a shrink fit on the shaft - for example there is no practical way that you could bolt it to the shaft through that surface.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, a much more efficient way to model this would be to model a sector of the turbine wheel containing just one blade, and derive the correct boundary conditions from the fact that the deflections of all the other sectors are identical (when measured in cylindrical polar coordinates) - but going into the details of how to do that is outside the scope of your actual question.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-10-02T10:22:34.320" LastActivityDate="2018-10-02T10:22:34.320" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24001" PostTypeId="2" ParentId="22555" CreationDate="2018-10-02T11:12:49.323" Score="0" Body="&lt;p&gt;-I have 2 of the cheap chinese vacuum pump from the brand &quot;Value&quot; , they look very much like yours. They often go up to 3 Torr when the oil is fresh then 6 Torr when the oil is contaminated with water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-1st step is to test the max vacuum with simply a jauge connected directly to the pump, top see the max you can get without potential leaks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-then check the oil, if it's cloudy, there is water in it, it has to be changed with proper vacuum oil&lt;/p&gt;&#xA;" OwnerUserId="5926" LastActivityDate="2018-10-02T11:12:49.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24002" PostTypeId="2" ParentId="23993" CreationDate="2018-10-02T12:23:52.373" Score="1" Body="&lt;p&gt;Instead of trying to hook floating debris from shore, I'd look into grabbing it from a canoe or other suitable watercraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some items may be too large to lift into a canoe without risking capsizing during the process.  Since you are willing to deploy two people, you could have one on shore and one in the canoe.  The one in the canoe can attach the hook easily so that the one on shore can haul in the large item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small items are just lifted onto the canoe directly, and only need one person.  A single person could do a sweep with a canoe, then come back to work with someone else on shore for the larger and more difficult items.&lt;/p&gt;&#xA;" OwnerUserId="624" LastActivityDate="2018-10-02T12:23:52.373" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24004" PostTypeId="1" AcceptedAnswerId="24025" CreationDate="2018-10-02T16:22:51.447" Score="1" ViewCount="2920" Body="&lt;p&gt;I wonder, whether an equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$R = 2P + S$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;should give a geometrically sound design, where &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the tooth count for the ring; &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;, for the planets; and &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, for the sun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to model a planetary gears with &lt;span class=&quot;math-container&quot;&gt;$R = 47$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$P = 16$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$S = 15$&lt;/span&gt;. They do look well as for diameters, but the teeth do not match - it cannot be assembled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hIUZf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hIUZf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the equation above guarantees, that the gears will fit each other? I now tend to think, that it does not take into account assemble-ability. Or perhaps I made a mistake somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Module is 0.9. Centre circle of planets was calculated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$D = \frac{m(Z_1 + Z_2)}{2} = \frac{0.9(15 + 16)}{2} = 13.95$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I double checked everything and didn't find any mistake yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gears should be precise, I used &quot;Involute Gears&quot; macro in FreeCAD, which should provide a proven correct gears. The carriage is based on precise fully constrained sketch and the parts are assembled in fully constrained assembly. Thus, there should be no imprecisions here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I would look for possible mistake (but didn't findany) is the position of planets against outer ring and sun (calculated as shown above), imperfections of outer ring, which is produced as subtract shape of involute gear with 47 teeth from a disk. And starting rotations of the gears for assembly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2R2Sr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2R2Sr.png&quot; alt=&quot;Assembly&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZGsP0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZGsP0.png&quot; alt=&quot;Carrier&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what I'm thinking is, if it is even that easy. The outer ring has 47 teeth, which is a &lt;strong&gt;prime number&lt;/strong&gt; (and selected on purpose, because correct gears should employ prime numbers and have to care about a common divider), meaning that one third (we have 3 planets) has &lt;strong&gt;15,6 teeth&lt;/strong&gt;. And that means, that if you set the carrier sou that it points into the middle of the tooth with one of it's points, the other two will be offset from a center of a tooth by 6/10 of a tooth. While in the same time, the Sun has 15 teeth which are perfectly dividable by 3 and we can have it set with all three i.e. gaps pointing in the direction of all three carrier lines. This is confirmed by creating three lines constrained at 120° and rotating with one (other two follow), I cannot find any position at which would each line point at the same place on a tooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm thinking, that this prohibits same condition for every planet. Meaning this is not possible to assemble (as is). But perhaps I'm missing something; the pointed simulator does depict same configuration assembled, but then it's only infographics which could cheat... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sTPDv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sTPDv.png&quot; alt=&quot;Gear Simulator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So issue is still open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also good to point out what should be possible to assemble then. Well, one configuration which should work is &lt;span class=&quot;math-container&quot;&gt;$S = 15$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$P = 15$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R = 45$&lt;/span&gt;, but then it is a &lt;em&gt;very bad&lt;/em&gt; design due to &lt;span class=&quot;math-container&quot;&gt;$S = $&lt;/span&gt;P (having the worst common divisor possible). Adding 3 to 5 I get 48, which won't fit the initial equation, but adding another 3 I get &lt;span class=&quot;math-container&quot;&gt;$S = 15$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$P = 18$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R = 51$&lt;/span&gt; which doesn't seem to be that bad. The greatest common divisor of a Sun and Planet is 3, meaning sun will turn 5 times before the same tooth will meet same tooth on the other gear.&lt;/p&gt;&#xA;" OwnerUserId="17716" LastEditorUserId="17716" LastEditDate="2018-10-03T21:43:18.603" LastActivityDate="2021-10-28T02:34:05.743" Title="Functionally Matching Gears (teeth count) in Planetary Gearbox" Tags="|gears|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24006" PostTypeId="1" AcceptedAnswerId="24010" CreationDate="2018-10-02T18:16:53.503" Score="0" ViewCount="183" Body="&lt;p&gt;I'm not sure if the upper question is clear enough to understand but here's what I've got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made a model of excavator in Adams View and got trajectory of it's motion (boom, stick, bucket) using trace marker.&#xA;This image shows pretty much what I've done {not mine, from Wikipedia}:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ncdlm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ncdlm.png&quot; alt=&quot;what I mean&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my question is how I possibly can get the surface area of it's uhh.. trajectory in Adams View? Or the area that is circled in purple (dashed line) - I'm not sure what to call that.&lt;/p&gt;&#xA;" OwnerUserId="17719" LastActivityDate="2018-10-02T22:35:05.750" Title="How to get the surface area of trajectory [MSC Adams]?" Tags="|mechanical-engineering|modeling|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24007" PostTypeId="2" ParentId="24004" CreationDate="2018-10-02T18:27:34.977" Score="3" Body="&lt;p&gt;EDITED following removal of typo in original question and my mistype after counting teeth in the photo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've counted the teeth in your image, and you appear to have modelled it with R = 47. P=16 and S=15, which should work!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue seems to be in your definition of positions inside the modelling software - I've pulled your image into photoshop and done some minor adjustment, and it all seems to fit OK. Have a look at the highlighted blue line. Try arranging your gears like that to start, and then see if the other two drop in!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zWym0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zWym0.png&quot; alt=&quot;fixed positions of gears&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at the excellent simulator here: &lt;a href=&quot;http://www.thecatalystis.com/gears/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.thecatalystis.com/gears/&lt;/a&gt; for more insight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can take free control of the teeth on one of the three gears, and then adjusting one of the remaining values will automatically calculate the final number. Simply select your preferred configuration from the drop-down menu.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-10-03T02:11:09.213" LastActivityDate="2018-10-03T02:11:09.213" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24009" PostTypeId="1" CreationDate="2018-10-02T20:53:23.010" Score="0" ViewCount="635" Body="&lt;p&gt;I have to study several industrial pulleys made for lifting heavy weights. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important to select one which has a low coefficient of friction because the system is based in a hoist and several corner angle transmissions under heavy load. So that there is lots of strength lost by friction. I also have many other constraints of course. I'm not searching for pulley brand but for a test mechanism I could use to check the already selected candidate pulleys' coefficient of friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could test the static coefficient of friction quite easily using a weight at one side and a dynamometer at the other side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However my application wile involve several lifts a day with up to 20 meters of rope running under heavy load (1~2 metric tons per pulley). The speed is relatively low. 1m/s in the rope would be the absolute maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if the static coefficient of friction will be enough to make a relevant selection. What do you think about it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had several ideas to build a testing machine. All seems to be quite complicated : &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I firstly imagined 2 big hydraulic jacks with a dynamometer at each side. I'm not sure of the minimum length of the jacks but I had in the idea that 2m would be a minimum to get something representative.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I then imagined a relatively compact system with an endless rope running under load between a motor and pulley. I would measure the torque between the motor axis and the wheel actuator. I'm not sure of what I would exactly be measuring here but I could make some comparisons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I also imagined a system with 2 opposite pulleys under tension. I would measure the strength needed to move the rope with a dynamometer. I'm quite sure I should not do it manually but with a constant speed or constant strength system. I'm afraid to build a labyrinthine system. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Do you have better ideas or insights on how to do it ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards &lt;/p&gt;&#xA;" OwnerUserId="17720" LastActivityDate="2018-10-02T23:04:52.640" Title="How can I measure the coefficients of friction of pulleys?" Tags="|mechanical-engineering|measurements|mechanisms|mechanical|product-testing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24010" PostTypeId="2" ParentId="24006" CreationDate="2018-10-02T22:35:05.750" Score="1" Body="&lt;p&gt;I've used a feature in &lt;a href=&quot;https://inkscape.org/en/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Inkscape&lt;/a&gt; that allows one to select a polygon or series of joined paths which will return the area enclosed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd had to use this feature in the recent past and had found the answer in the &lt;a href=&quot;https://graphicdesign.stackexchange.com/questions/16312/quantify-the-area-of-a-shape-in-inkscape&quot;&gt;Graphics StackExchange.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7OGQ0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7OGQ0.png&quot; alt=&quot;area of polygon image 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first image, I've roughly traced the path with the Bezier tool, then selected Extensions, Visualize, Measure Path. In the drop down box for Measurement Type, I selected Area. Be sure to select the item you wish to measure. Click Apply.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program works slowly, presents (Not Responding) on a regular basis, but works!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0yDsi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0yDsi.png&quot; alt=&quot;area of polygon image 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, my units are set to millimeters. It would be necessary to have a reference scale object in order to have an accurate representation.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-10-02T22:35:05.750" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24011" PostTypeId="2" ParentId="24009" CreationDate="2018-10-02T23:04:52.640" Score="-1" Body="&lt;p&gt;The parameters are what you would experience on a sail boat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to measure your dynamic friction run your line through the a block mounted on the ceiling (at your desired speed and load) and measure the force between the block and the ceiling.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(on a boat: block = pulley)&lt;/p&gt;&#xA;" OwnerUserId="16544" LastActivityDate="2018-10-02T23:04:52.640" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24013" PostTypeId="1" AcceptedAnswerId="24019" CreationDate="2018-10-03T05:24:05.110" Score="0" ViewCount="93" Body="&lt;p&gt;Is understanding tensors a prerequisite for understanding plastic deformation (metal forming) equations, plane stress, plane strain conditions, etc..? If so which book should I use to understand tensors?. Because on the internet people suggest that a good resource to learn tensors are the physics books that deal with relativity, etc. So I am not sure whether that perception will be suitable for studying the stress tensors.&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastEditorUserId="90" LastEditDate="2018-10-03T09:13:44.610" LastActivityDate="2018-10-03T10:48:19.440" Title="Understanding plastic deformation" Tags="|structural-analysis|deformation|metal-folding|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24018" PostTypeId="2" ParentId="23993" CreationDate="2018-10-03T09:48:59.500" Score="0" Body="&lt;p&gt;Seems reasonable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact you probably don't need a guide rope  if you just tie the hook into the centre of one long rope you can just take in slack on the appropriate side. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A further refinement would be to bring some lightweight shear legs or a &lt;a href=&quot;https://en.wikipedia.org/wiki/Gin_pole&quot; rel=&quot;nofollow noreferrer&quot;&gt;gin pole&lt;/a&gt;. Either would provide a relatively lightweight and versatile way to allow you to lift rather than drag submerged objects. Both methods offer a reasonably efficient way to convert pulling force from ground level to a lifting force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially like a giant fishing rod or pole. Equally you have the option to use a pole with a rope fixed to the end as a large lever or with a pulley.&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-10-03T09:48:59.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24019" PostTypeId="2" ParentId="24013" CreationDate="2018-10-03T10:48:19.440" Score="0" Body="&lt;p&gt;In a mechanical engineering curriculum at university, you would probably take a series of 3 classes.  First would be called &quot;Strengths of Materials&quot; (or maybe &quot;mechanics of materials&quot;).  This would be a sophomore level class. No tensors, just linear algebra (matrices).  The second class would be &quot;theory of elasticity&quot;.  This is going to involve tensors. Third class will be &quot;Continuum mechanics&quot;.  This will be very heavy on tensors.    After that, there would probably be a specialized class on plasticity. Depending on your comfort level with the math and your background, pick a text book on one of these subjects.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-10-03T10:48:19.440" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24020" PostTypeId="1" CreationDate="2018-10-03T14:16:18.487" Score="0" ViewCount="142" Body="&lt;p&gt;I have a situation where I have a rough granite roadway about 15 feet wide and 50 feet long. The granite is solid granite. The surface variability is on the order of a maximum of about 4 inches. In other words at no place on the surface is one segment of the granite more than 4 inches higher than the granite adjacent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to smooth this into a crown. A crown is a gentle slope that allows water to shed off to the side of the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any kind of mechanized equipment that can do this kind of work?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2018-10-03T20:58:20.637" Title="How to grind smooth hard rock to create a road surface" Tags="|masonry|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24021" PostTypeId="2" ParentId="22371" CreationDate="2018-10-03T14:28:16.343" Score="0" Body="&lt;p&gt;Those are &quot;press fit&quot; or &quot;captive&quot; nuts. Mcmaster Carr should have whatever your looking for.&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-10-03T14:28:16.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24022" PostTypeId="2" ParentId="24020" CreationDate="2018-10-03T17:18:24.040" Score="2" Body="&lt;p&gt;Granite is hard a rock. On the &lt;a href=&quot;https://en.wikipedia.org/wiki/Mohs_scale_of_mineral_hardness&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mohs scale&lt;/a&gt; of hardness, which rates talc (being soft) at 1 and diamond (being the hardest material) at 10, granite is rated at being &lt;a href=&quot;https://www.reference.com/science/hard-granite-mohs-hardness-scale-f6ecbd0cce3a459a&quot; rel=&quot;nofollow noreferrer&quot;&gt;6 to 7&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The old way to do what you want to achieve is to drill and blast the area of interest to create a trench deeper than you require. This would then be back filled with finer grained fill to produce the smooth profiled surface you require.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding rock grinding, some companies offer &lt;a href=&quot;http://hardrockgrinding.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;rock grinding services&lt;/a&gt;. The pictures this company provides on its website show soft material being ground. The picks on the grinding drum/head are large and are spaced far apart. What you need is a grinding drum with the picks closer together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It a case of getting the correct gear for the application, if it is available and whether it is affordable.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-10-03T17:18:24.040" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24024" PostTypeId="1" CreationDate="2018-10-03T21:14:32.020" Score="1" ViewCount="95" Body="&lt;p&gt;I want to know, what the best way to connect an electric motor to a differential in a tight space. I first thought of a drive shaft, but the angle is to steep. The diff would be of a BMW E46 320d and the shaft of the motor has an outer diameter of 19.75mm and min 24.5mm depth and DIN 5480 teething.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the side view of the setup&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6aAAJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6aAAJ.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TkB9h.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TkB9h.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;And the top view of the setup&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.hribitec.com/si/attachment.php?id_attachment=10&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's more info about the electric motor&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17731" LastActivityDate="2018-10-03T21:14:32.020" Title="Best way to connect electric motor and differential in a tight space" Tags="|automotive-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24025" PostTypeId="2" ParentId="24004" CreationDate="2018-10-03T22:05:38.637" Score="1" Body="&lt;p&gt;It seems there are two conditions required to produce a valid involute planetary gear set. First was already written:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R = 2$&lt;/span&gt; x &lt;span class=&quot;math-container&quot;&gt;$P + $&lt;/span&gt;S&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the 2nd priciple is, that &lt;strong&gt;&lt;em&gt;the Ring and The sun have to have a common divisor equal to number of planets&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I.e. &lt;span class=&quot;math-container&quot;&gt;$S = $&lt;/span&gt;15, &lt;span class=&quot;math-container&quot;&gt;$P = $&lt;/span&gt;18 and &lt;span class=&quot;math-container&quot;&gt;$R = $&lt;/span&gt;51 have a common divisor 3 which is also a number of planets and it gives a correct result and assembly is easy to assemble:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FiAx9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FiAx9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17716" LastActivityDate="2018-10-03T22:05:38.637" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24027" PostTypeId="1" CreationDate="2018-10-04T01:54:33.070" Score="1" ViewCount="295" Body="&lt;p&gt;How to convert Joules(kinetic energy) plot chart into watts?&#xA;Anyone know how?&lt;a href=&quot;https://i.stack.imgur.com/VAC1z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VAC1z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17732" LastActivityDate="2018-10-04T13:14:47.030" Title="How to convert Joules(kinetic energy) plot chart into watts?" Tags="|mechanical-engineering|simulation|solidworks|energy|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24028" PostTypeId="1" CreationDate="2018-10-04T02:48:10.333" Score="2" ViewCount="70" Body="&lt;p&gt;&lt;strong&gt;Context:&lt;/strong&gt;&#xA;I have a paper roll of about 8 in (20 cm) diameter mounted in a 3 inches core (7.62cm). Its length is about 1.2 m. The paper is bond of 95 g/m2. I have to cut slices of it of about 10-18 cm in length, down to the paper core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As this is a very eventual task, I can't afford a log slitter, which is the right tool for the task, so I'm trying to do it in either a lathe or a cutoff machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I've tried:&lt;/strong&gt;&#xA;In the lathe I've tested with two different options: a fixed blade (a knife and a machete) and a rotating one (a lightweight circular saw with wich I've used ceramic and wood cutting discs). The machete produces the best cut, but as it's fixed, it dulls very quickly. The circular saw produces a lot of small paper chips that cause problems in the process that follows after the cutting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've experimented with different RPMs in the lathe, specially in the trials with the fixed blade, but haven't quite controlled for the feed of the blade into the roll. With the circular saw the RPMs of it are so high (5800), that I'm afraid the relative speeds (even if I have rotation of both pieces in the same direction) are too high (diameter of the cutting disc is about 19cm or 7.5 inches).&#xA;In the lathe I've mounted the roll both between points (as standard steel would be) and on a central bar (a core inside the core) to avoid tightening of the cutting area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the cutoff machine I've tried with discs of different tooth sizes, but all of them end up burning the paper in the cutting face (I don't have the RPM of the cutoff machine, but I can provide it in a few hours).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have searched for available information on speeds and feeds for paper in log-slitting operations, but have found nothing so far, and the available tables of speeds and feeds are just for metals and some plastics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What my expected result is:&lt;/strong&gt;&#xA;I'll appreciate if someone can provide an adequate speed and feed value for cutting paper, or if can point me in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the cut of the paper roll: I need a cut that is as clean as possible, without significant burning, that requires as less sharpening of the tool as possible, and that can be considered safe for the operator.&lt;/p&gt;&#xA;" OwnerUserId="17733" LastActivityDate="2018-10-07T14:36:56.607" Title="Speed and Feed values for log slitting of paper" Tags="|mechanical-engineering|machining|cutting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24029" PostTypeId="1" CreationDate="2018-10-04T11:07:55.337" Score="2" ViewCount="1412" Body="&lt;p&gt;I'm trying to control my 12V fan with a 5V fan controller.&#xA;With this circuit the fan run at full speed.&#xA;When I measure the voltage on PWM and ground I obtain 5V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which circuit can I use to adapt the 5V PWM on the 12V PWM ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cM8pB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cM8pB.png&quot; alt=&quot;schema&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17737" LastEditorUserId="8055" LastEditDate="2018-10-04T22:17:03.420" LastActivityDate="2023-12-09T17:02:55.357" Title="How can I use a 5V pwm with a 12V pwm" Tags="|electrical-engineering|pwm|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24030" PostTypeId="1" AcceptedAnswerId="24031" CreationDate="2018-10-04T11:16:18.493" Score="1" ViewCount="82" Body="&lt;p&gt;Sometimes we need a strong threaded joint in some part that is produced using molding. Say, this part itself is plastic. Then we design the part with a hole and insert there a.. &quot;threaded bushing&quot;? I don't know the proper name for this element in English. In Russian it is &quot;закладная втулка&quot; (insertable bushing/cylinder/etc - втулка/vtulka is a catch-all term for all things cylindrical that are used in an &quot;axial manner&quot; - something is inserted into them, or they are inserted somewhere, or both). I'm looking for a proper term in English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I googled for &lt;strong&gt;закладная втулка&lt;/strong&gt; and found this example relevant to my text: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mHNOzm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mHNOzm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In the text that I'm translating into English, these brass insertable threaded elements in a plastic part produced using molding look like this: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dX4Zj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dX4Zj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found a DIN on &lt;a href=&quot;http://www.fastbolt.ru/din16903&quot; rel=&quot;nofollow noreferrer&quot;&gt;one Russian page&lt;/a&gt; describing &quot;закладная втулка&quot; and this DIN led me to English terms &quot;&lt;a href=&quot;https://www.linn-praezision.de/en/products/din-16903.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;threaded bushing&lt;/a&gt;&quot;, &quot;threaded insert&quot;. Are these okay? &lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="349" LastEditDate="2018-10-04T11:28:47.253" LastActivityDate="2019-02-23T04:45:02.150" Title="Looking for translation of &quot;закладная втулка&quot; (threaded bushing?)" Tags="|terminology|molding|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24031" PostTypeId="2" ParentId="24030" CreationDate="2018-10-04T11:56:52.130" Score="3" Body="&lt;p&gt;As they are going into blind holes, then threaded insert would be better.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-10-04T11:56:52.130" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24032" PostTypeId="2" ParentId="24027" CreationDate="2018-10-04T13:14:47.030" Score="4" Body="&lt;p&gt;To convert a plot of J versus s to a plot of J/s vs s, take the derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my quick-and-dirty answer after digitizing the data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/erofy.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/erofy.png&quot; alt=&quot;derivative plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-10-04T13:14:47.030" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24033" PostTypeId="1" AcceptedAnswerId="24036" CreationDate="2018-10-04T15:50:24.237" Score="0" ViewCount="162" Body="&lt;p&gt;Attempting this kinematics question but not really understanding if what I'm doing is right or completely wrong. Should I be starting by finding Vp? Any help appreciated. Not an assignment, just practicing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YYfPY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YYfPY.png&quot; alt=&quot;Question&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LhGeB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LhGeB.jpg&quot; alt=&quot;Attempt&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user13170" LastActivityDate="2018-10-05T04:43:12.557" Title="Two Dimensional Kinematics" Tags="|mechanical-engineering|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24034" PostTypeId="2" ParentId="4226" CreationDate="2018-10-04T17:55:11.277" Score="4" Body="&lt;p&gt;The reason behind the twist is that the absolute inlet velocity (in your notation &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;) is constant from hub to shroud. However, the linear velocity (in your notation &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;) varies from hub to shroud. This means the relative velocity (in your notation &lt;span class=&quot;math-container&quot;&gt;$V_r$&lt;/span&gt;) should change to accommodate a constant absolute velocity. The modification in &lt;span class=&quot;math-container&quot;&gt;$V_r$&lt;/span&gt; is enforced through blade angle &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same velocity triangle principle shown in the original post is applied to multiple planes likes the hub plane and shroud planes to get a distribution of &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;; hence, a twist appears whenever a considerable change in radius is present. This is usually the case for a pump or compressor inlet, a Francis turbine outlet, a propeller, a wind turbine, etc. It should be mentioned that a 3D design considers many other factors and often uses a mixture of analytical, empirical and numerical relations to define the final blade twist. &lt;/p&gt;&#xA;" OwnerUserId="8742" LastEditorUserId="8742" LastEditDate="2018-10-05T20:12:21.120" LastActivityDate="2018-10-05T20:12:21.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24035" PostTypeId="2" ParentId="24030" CreationDate="2018-10-05T00:43:14.383" Score="4" Body="&lt;p&gt;This is called &quot;&lt;em&gt;threaded insert&lt;/em&gt;&quot; in English.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more context, here are product page for threaded inserts for plastic: &lt;a href=&quot;https://www.pemnet.com/fastening-products/si-inserts-for-plastic/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://www.stanleyengineeredfastening.com/fasteners/inserts/heat-ultrasonic-inserts/dodge-ultrasert-iv-flanged&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.  Both of are major manufacturers based in US.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional examples of plastic parts with threaded inserts: &lt;a href=&quot;https://www.flickr.com/photos/46038819@N07/15722721032&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; and &lt;a href=&quot;https://www.flickr.com/photos/46038819@N07/22177155596&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="63" LastEditorUserId="63" LastEditDate="2019-02-23T04:45:02.150" LastActivityDate="2019-02-23T04:45:02.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24036" PostTypeId="2" ParentId="24033" CreationDate="2018-10-05T04:43:12.557" Score="1" Body="&lt;p&gt;Let us work out the geometric relations first and then do the kinematics.  I will follow a purely algorithmic approach to show the details.  You can arrive at the same result from trigonometric relations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us choose a fixed coordinate frame that has its origin at A, with the three unit vectors &lt;span class=&quot;math-container&quot;&gt;$\mathbf{E}_1 = (1, 0, 0)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathbf{E}_2 = (0, 1, 0)$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\mathbf{E}_3 = (0, 0, 1)$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\mathbf{E}_1$&lt;/span&gt; is along AB, &lt;span class=&quot;math-container&quot;&gt;$\mathbf{E}_2$&lt;/span&gt; is perpendicular to AB (in the plane ABP), and &lt;span class=&quot;math-container&quot;&gt;$\mathbf{E}_3$&lt;/span&gt; is perpendicular to the plane ABP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position vectors of the three points are&#xA;&lt;span class=&quot;math-container&quot;&gt;$&#xA;  \mathbf{x}_A, \mathbf{x}_B, \mathbf{x}_P \,.&#xA;$&lt;/span&gt;&#xA;Then the direction vectors of interest are&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \mathbf{r}_{BA} = \mathbf{x}_B - \mathbf{x}_A ~,~~&#xA;  \mathbf{r}_{PA} = \mathbf{x}_P - \mathbf{x}_A ~,~~ &#xA;  \mathbf{r}_{PB} = \mathbf{x}_P - \mathbf{x}_B \,.&#xA;$$&lt;/span&gt;&#xA;From the figure, if &lt;span class=&quot;math-container&quot;&gt;$\theta_A, \theta_B$&lt;/span&gt; are the angles made by PA and PB with AB, and &lt;span class=&quot;math-container&quot;&gt;$r_A, r_B$&lt;/span&gt; are the lengths of PA and PB, we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \begin{align}&#xA;  \mathbf{r}_{BA} &amp;amp;= b \mathbf{E}_1 \\&#xA;  \mathbf{r}_{PA} &amp;amp;= -r_A \cos\theta_A \mathbf{E}_1 + r_A \sin\theta_A \mathbf{E}_2 \\&#xA;  \mathbf{r}_{PB} &amp;amp;= -r_B \cos\theta_B \mathbf{E}_1 + r_B \sin\theta_B \mathbf{E}_2&#xA;  \end{align}&#xA;$$&lt;/span&gt;&#xA;Since &lt;span class=&quot;math-container&quot;&gt;$\mathbf{r}_{PB} = -\mathbf{r}_{BA} + \mathbf{r}_{PA}$&lt;/span&gt;, we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  -r_B \cos\theta_B \mathbf{E}_1 + r_B \sin\theta_B \mathbf{E}_2 = -b \mathbf{E}_1 - r_A \cos\theta_A \mathbf{E}_1 + r_A \sin\theta_A \mathbf{E}_2&#xA;$$&lt;/span&gt;&#xA;or&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \begin{align}&#xA;  -r_B \cos\theta_B &amp;amp; = -b - r_A \cos\theta_A \\&#xA;  r_B \sin\theta_B &amp;amp; = r_A \sin\theta_A&#xA;  \end{align}&#xA;$$&lt;/span&gt;&#xA;In matrix form&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \begin{bmatrix}&#xA;    \cos\theta_A &amp;amp; -\cos\theta_B \\&#xA;    \sin\theta_A &amp;amp; -\sin\theta_B&#xA;  \end{bmatrix}&#xA;  \begin{bmatrix} r_A \\ r_B \end{bmatrix} = &#xA;  \begin{bmatrix} - b \\ 0 \end{bmatrix}&#xA;$$&lt;/span&gt;&#xA;The determinant of the &lt;span class=&quot;math-container&quot;&gt;$2 \times 2$&lt;/span&gt; matrix is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  D := -\cos\theta_A \sin\theta_B + \sin\theta_A \cos\theta_B = \sin(\theta_A - \theta_B)&#xA;$$&lt;/span&gt;&#xA;Therefore, inverting the matrix, we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \begin{bmatrix} r_A \\ r_B \end{bmatrix} = \frac{1}{\sin(\theta_A -\theta_B)}&#xA; \begin{bmatrix}&#xA;    -\sin\theta_B &amp;amp; \cos\theta_B \\&#xA;    -\sin\theta_A &amp;amp; \cos\theta_A&#xA;  \end{bmatrix}\begin{bmatrix} - b \\ 0 \end{bmatrix}&#xA;$$&lt;/span&gt;&#xA;or&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\boxed{&#xA; \begin{align}&#xA;   r_A &amp;amp;= \frac{b\sin\theta_B}{\sin(\theta_A -\theta_B)} \\&#xA;   r_B &amp;amp;= \frac{b\sin\theta_A}{\sin(\theta_A -\theta_B)}&#xA; \end{align}&#xA;}&#xA;$$&lt;/span&gt;&#xA;Plugging in &lt;span class=&quot;math-container&quot;&gt;$b = 300$&lt;/span&gt; mm, &lt;span class=&quot;math-container&quot;&gt;$\theta_A = 60^\circ$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\theta_B = 20^\circ$&lt;/span&gt;, we have &lt;span class=&quot;math-container&quot;&gt;$r_A = 159.6$&lt;/span&gt; mm and &lt;span class=&quot;math-container&quot;&gt;$r_B = 404.2$&lt;/span&gt; mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that we know the geometry, we can do the kinematics.  Choose a rotating frame that has origin&#xA;at point A, and the coordinate axes&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \begin{align}&#xA;   \mathbf{e}_1 &amp;amp;= \sin\theta_A \mathbf{E}_1 + \cos\theta_A \mathbf{E}_2 \\&#xA;   \mathbf{e}_2 &amp;amp;= -\cos\theta_A \mathbf{E}_1 + \sin\theta_A \mathbf{E}_2 = \mathbf{r}_{PA}/||\mathbf{r}_{PA}||\\&#xA;   \mathbf{e_3} &amp;amp;= \mathbf{E}_3.&#xA; \end{align}&#xA;$$&lt;/span&gt;&#xA;or, inverting the relation,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \begin{align}&#xA;   \mathbf{E}_1 &amp;amp;= \sin\theta_A \mathbf{e}_1 - \cos\theta_A \mathbf{e}_2 \\&#xA;   \mathbf{E}_2 &amp;amp;= \cos\theta_A \mathbf{e}_1 + \sin\theta_A \mathbf{e}_2 \\&#xA;   \mathbf{E_3} &amp;amp;= \mathbf{e}_3.&#xA; \end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the position vector of P in that coordinate system is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \mathbf{r}_{PA} = r_A \mathbf{e}_2&#xA;$$&lt;/span&gt;&#xA;and the velocity of P relative to the rotating frame is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \mathbf{v}_{P/R} = \dot{r}_A \mathbf{e}_2&#xA;     = \dot{r}_A \left[-\cos\theta_A \mathbf{E}_1 + \sin\theta_A \mathbf{E}_2\right]&#xA;$$&lt;/span&gt;&#xA;The time derivative of &lt;span class=&quot;math-container&quot;&gt;$r_A$&lt;/span&gt; is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \dot{r}_A = \frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} \dot{\theta}_B +&#xA;  \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}&#xA;    (\dot{\theta}_A - \dot{\theta}_B)&#xA;$$&lt;/span&gt;&#xA;or, using &lt;span class=&quot;math-container&quot;&gt;$\omega_A = \dot{\theta}_A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\omega_B = \dot{\theta}_B$&lt;/span&gt;, &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \boxed{&#xA;  \dot{r}_A = \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}&#xA;    \omega_A +&#xA;    \left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} -  &#xA;  \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right]&#xA;    \omega_B&#xA;   }&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{\Omega}_A = \omega_A \mathbf{E}_3 = \omega_A \mathbf{e}_3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{\Omega}_B = \omega_B \mathbf{E}_3 = \omega_B \mathbf{e}_3$&lt;/span&gt; are the angular velocities at A and B of AP and BP, respectively. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in the values that we know, we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \dot{r}_A = -190.24 \omega_A + 628.81 \omega_B \,.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At P, the velocity is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \mathbf{v}_P = \boldsymbol{\Omega}_A \times \mathbf{r}_{PA} +  \mathbf{v}_{P/R}&#xA;               = \boldsymbol{\Omega}_B \times \mathbf{r}_{PB}&#xA;$$&lt;/span&gt;&#xA;Therefore, we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \begin{align}&#xA;  &amp;amp; (\omega_A \mathbf{e}_3) \times (r_A \mathbf{e}_2) + \\&#xA;  &amp;amp; \left[\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}&#xA;    \omega_A +&#xA;    \left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} -  &#xA;  \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right]&#xA;    \omega_B\right]\mathbf{e}_2 \\&#xA;  &amp;amp;= (\omega_B \mathbf{e}_3) \times (-r_B \cos\theta_B \mathbf{E}_1 + r_B \sin\theta_B \mathbf{E}_2)&#xA;  \end{align}&#xA;$$&lt;/span&gt;&#xA;Now&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \begin{align}&#xA;  \mathbf{e}_3 \times \mathbf{e}_1 &amp;amp;= \mathbf{e}_2 \\&#xA;  \mathbf{e}_3 \times \mathbf{e}_2 &amp;amp;= -\mathbf{e}_1 \\&#xA;  \mathbf{e}_3 \times \mathbf{E}_1 &amp;amp;= \sin\theta_A \mathbf{e}_3 \times \mathbf{e}_1 -&#xA;                                      \cos\theta_A \mathbf{e}_3 \times \mathbf{e}_2&#xA;                                    = \sin\theta_A \mathbf{e}_2 + \cos\theta_A \mathbf{e}_1\\&#xA;  \mathbf{e}_3 \times \mathbf{E}_2 &amp;amp;= \cos\theta_A \mathbf{e}_3 \times \mathbf{e}_1 +&#xA;                                      \sin\theta_A \mathbf{e}_3 \times \mathbf{e}_2&#xA;                                    = \cos\theta_A \mathbf{e}_2 - \sin\theta_A \mathbf{e}_1&#xA;  \end{align}&#xA;$$&lt;/span&gt;&#xA;Therefore, we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \begin{align}&#xA;  &amp;amp; -\omega_A r_A \mathbf{e}_1 + \\&#xA;  &amp;amp; \left[\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}&#xA;    \omega_A +&#xA;    \left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} -  &#xA;  \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right]&#xA;    \omega_B\right]\mathbf{e}_2 \\&#xA;  &amp;amp; = \omega_B r_B \left[-\cos\theta_B(\sin\theta_A \mathbf{e}_2 + \cos\theta_A \mathbf{e}_1)&#xA;                         +\sin\theta_B(\cos\theta_A \mathbf{e}_2 - \sin\theta_A \mathbf{e}_1)\right]\\&#xA;  &amp;amp; = -\omega_B r_B \left[\cos(\theta_A-\theta_B) \mathbf{e}_1 + \sin(\theta_A-\theta_B) \mathbf{e}_2 &#xA;                    \right]&#xA;  \end{align}&#xA;$$&lt;/span&gt;&#xA;Comparing the components along &lt;span class=&quot;math-container&quot;&gt;$\mathbf{e}_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathbf{e}_2$&lt;/span&gt;, respectively, we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;   -\omega_A r_A = -\omega_B r_B \cos(\theta_A-\theta_B)&#xA;$$&lt;/span&gt;&#xA;and&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}&#xA;    \omega_A +&#xA;    \left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} -  &#xA;  \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right]&#xA;    \omega_B = -\omega_B r_B  \sin(\theta_A-\theta_B)&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in &lt;span class=&quot;math-container&quot;&gt;$\omega_A = 10$&lt;/span&gt; and the previously computed values of &lt;span class=&quot;math-container&quot;&gt;$r_A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$r_B$&lt;/span&gt; into the first equation above, we have &lt;span class=&quot;math-container&quot;&gt;$\omega_B =$&lt;/span&gt; 5.1555 rad/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the relative velocity of the slider block is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \dot{r}_A = -190.24 \omega_A + 628.81 \omega_B  = 1339.4~\text{mm/s} \,.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-10-05T04:43:12.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24038" PostTypeId="2" ParentId="21397" CreationDate="2018-10-05T12:16:30.080" Score="0" Body="&lt;p&gt;Problem was with workbench setup. Static structural node should be placed &lt;strong&gt;over&lt;/strong&gt; thermal node, not near it. More detailed description with pictures is avaliable &lt;a href=&quot;https://i.stack.imgur.com/RuHHZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15792" LastActivityDate="2018-10-05T12:16:30.080" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24039" PostTypeId="2" ParentId="24029" CreationDate="2018-10-05T12:32:22.437" Score="0" Body="&lt;p&gt;Based on provide information I suggest taking a look at &lt;a href=&quot;https://www.onsemi.com/pub/Collateral/ADT7467-D.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;ADT7467 dbCOOL Remote Thermal&#xA;Monitor and Fan Controller&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below should help on to configure the circuits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wBK6a.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wBK6a.gif&quot; alt=&quot;ADT7467 circuit configuration -1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the FAN specification a pullup to 3.3V, 5V or 12V might also be required. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oyQLd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oyQLd.png&quot; alt=&quot;ADT7467 circuit configuration Pull up&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively I would recommend looking at something like the following. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.newegg.com/Product/Product.aspx?Item=9SIACJF6HF8769&amp;amp;cm_re=fan_speed_controllers-_-9SIACJF6HF8769-_-Product&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newest Black PC 1 to 10 PWM 4Pin Cooler Cooling Fan Hub Splitter Cable Power by 12V SATA Power Supply Adapter For PC Mining&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bkacP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bkacP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With reference to 4-wire Specifications, I found a 2004 CPU fans specification from Intel. It is assumed that the your fan specification are similar. The full specification is attached in the reference section. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t2nP6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t2nP6.jpg&quot; alt=&quot;Intel 4-wire Fan Specification&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://superuser.com/questions/1126136/three-wire-and-four-wire-cpu-fans-whats-the-difference/1126155&quot;&gt;Three-wire and four-wire CPU fans, what's the difference?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/a/176061/50587&quot;&gt;Troubleshooting PCB of an ATX power supply?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://folk.uio.no/kyrrens/diverse/viftekontroller/developer-specs-REV1_2_Public.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;4-Wire Pulse Width Modulation (PWM) Controlled Fans&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-10-06T01:07:53.683" LastActivityDate="2018-10-06T01:07:53.683" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24041" PostTypeId="1" AcceptedAnswerId="24045" CreationDate="2018-10-05T13:58:30.133" Score="1" ViewCount="250" Body="&lt;p&gt;I have a task to create a software that calculates how much evaporative coolers are needed to maintain a certain temperature in a given room. I need to create a simple form, where variables are next :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - Location (i can fetch dry and wet bulb temperatures) 2 - Desired Temperature 3 - Air changes per hour 4 - Area of the room (m2) 5 - Number of people in the room&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working with AOLAN evaporative coolers, and i have all the specification from them for the devices used. I just can't wrap my head around the calculations. So my question(s) are rather simple : How do i calculate the amount of coolers needed to cool down an area with all the given parameters up there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would love if someone could give me guidelines into how to achieve this.&lt;/p&gt;&#xA;" OwnerUserId="17746" LastActivityDate="2018-10-07T19:08:57.540" Title="Calculate amount of evaporative coolers needed to cool down an area" Tags="|cooling|ac|evaporation|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24043" PostTypeId="2" ParentId="12636" CreationDate="2018-10-05T14:42:31.463" Score="2" Body="&lt;p&gt;Pop Rivets are a fastener composed of two portions: the mandrel and the hat. To install a pop rivet a hole is drilled into the materials, and then the hat is placed into the hole. A rivet installation too is then slid onto the exposed mandrel end. As it begins pulling the mandrel, it deforms the back of the hat creating a strong and quick hold. Once a certain pressure is reached, the mandrel snaps leaving the hat deformed and in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, and using the terminology from your diagram, the &lt;strong&gt;pulling stem&lt;/strong&gt; is pulled until exceeding its capacity. During this pulling period the &lt;strong&gt;rivet shell&lt;/strong&gt; deforms and creates a &lt;strong&gt;header&lt;/strong&gt; on the rear of the installation. Once the pulling stem is pulled too tightly it will snap resulting in the &lt;strong&gt;stem break point&lt;/strong&gt; being just below the surface of the rivet shell. This creates a firm secure hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pop rivets are most commonly used for their quick installation process and because you only need access to one side of the installation materials earning them the name blind rivets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structural rivets are a bit different as there is a built in locking mechanism that ensure the mandrel portion that stays inside the rivet does not fall out but actually creates a secondary wall. This increases both the rivets strength and hold over time.&lt;/p&gt;&#xA;" OwnerUserId="17751" LastActivityDate="2018-10-05T14:42:31.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24044" PostTypeId="1" CreationDate="2018-10-05T20:42:07.667" Score="0" ViewCount="1868" Body="&lt;p&gt;55 HRC is a very high hardness, even for steels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally the corrosion resistance of the steel would have to match AISI 316L or be even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qenched stainless steel can go to 55HRC. Even to 65 HRC. But this is at the price of a high Carbone content. Carbone react with free Chromium and makes it unavailable for corrosion resistance. So that qenched stainless steel often have a poor corrosion resistance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most basic option is AISI 420. It has the minimum amount of Carbone needed and show a relatively good corrosion resistance. It is available at a very good price. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the other hand we have the Cronidur30/Bohler N360. It is basically like if the 420 had been highly alloyed with everything available. It has 15% Chromium,  1% Molybdenium, 0,4% Nitrogen and some Vanadium. This gives it a really higher corrosion resistance but it actually only poorly match AISI 304, the most basic Austenitic Stainless Steel. It cost something like $30 per Kg. That's 6 time the price of the 420. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Precipitation Hardening stainless steel like 13-8 Mo PH would only reach 45 HRC. That's already high. I'm wondering if there is specific versions that can do better. This type of steel has a better corrosion resistance than quenched steels. I know that Myodo produce a steel called H1 but it is only available in Japan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also some maraging stainless steels that can reach 55HRC. I have no idea of the price and actual corrosion resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Highly strained by cold working classical stainless steel like 301 or Springflex can reach very high hardness. I don't know for what plate thickness 55 HRC could be reached. The other point is that cold working produce Martensite and reduce corrosion resistance. I don't know how good is final corrosion resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about high temperature Nitrocarburation. However this process reduce corrosion resistance. I don't know by what amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another solution is to use a surface treatment like chemical Nickel. However I'm not totally confident in this type of solution because the corrosion resistance is reduced to 0 locally where the coating is breached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have any idea. I would be greatly interested. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit : to make it short&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a steel that : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;is as corrosion resistant as 316L or better&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;reach a hardness of 55 HRC&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The steel will be used in a sort of bearing that is in contact with salty and a bit of citric acidic mixture in an industrial environment.&lt;/p&gt;&#xA;" OwnerUserId="17720" LastEditorUserId="17720" LastEditDate="2018-10-06T07:48:08.890" LastActivityDate="2018-10-06T07:51:43.317" Title="What is the most corrosion resistant 55HRC stainless steel?" Tags="|metallurgy|metals|corrosion|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24045" PostTypeId="2" ParentId="24041" CreationDate="2018-10-05T20:50:02.150" Score="1" Body="&lt;h1&gt;Foundations (Energy Balance)&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;For the Room&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Consider the system shown in the picture below. Air flows through a room volume (area and height) with a residence time. Heat is provided by people inside the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zHum0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zHum0.png&quot; alt=&quot;picture of room&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steady state energy balance equation becomes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{m}_a\left( \tilde{H}_{a,out} - \tilde{H}_{a,in} \right) = \dot{q}_p \\&#xA;\left(\frac{A h}{t_{TO}}\right) \left(\frac{M_a p}{RT_{in}}\right)  \tilde{C}_{p,a}\left(T_{out} - T_{in} \right) = N_p\hat{\dot{q}}_p&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says, the heat generated causes an enthalpy change in the air flow (out - in). The enthalpy change of an ideal gas causes it to undergo a temperature change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the room area. You also need its height. This gives volume. Divided by residence time gives volumetric flow. With ideal gas density you get mass flow. The specific heat and molar mass of air are known. The pressure of the inlet air is known. The output temperature of your AC unit is the inlet temperature to the room. You might also adjust the flow rate of air in to the room. This is the residence time. You know the number of people and the heat per person. The remaining factor is the outlet temperature. This is the desired room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your goal is to have a set room temperature. As more people enter the room, you will need either to increase the flow rate (decrease residence time) at the same inlet air temperature or you will need to decrease the inlet air temperature to the room (the outlet temperature of your AC unit) at the same air flow rate.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Sizing the AC Unit(s)&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The minimum cooling load on the AC units is &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_p$&lt;/span&gt;. You must remove at least the heat generated by the people in the room. The CoP of an AC is &lt;span class=&quot;math-container&quot;&gt;$CoP = \dot{q}/\dot{w}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\dot{w}$&lt;/span&gt; is the work required. This is to first order the electrical power input multiplied by an efficiency factor for the heat pump. Each AC takes &lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt; watts power at an efficiency of &lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt;. The net result to establish the minimum number of units is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{q}_p = N_u\ CoP\ \epsilon\ W = N_p \hat{\dot{q}}_p&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The approach might be as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine the minimum number of AC units to meet the cooling load based on the efficiency, CoP, and power load of a unit as well as the number of people in the room and their heat output.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Determine whether the AC units will meet the demand to maintain a desired room temperature based on the rated output temperature and air flow rate of the units.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-10-07T19:08:57.540" LastActivityDate="2018-10-07T19:08:57.540" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24046" PostTypeId="1" CreationDate="2018-10-05T23:59:19.983" Score="1" ViewCount="32" Body="&lt;p&gt;The literature on the existence of limit cycles is replete with complex and abstract mathematical analysis. Or for specific systems of specific structure, specific mathematical criteria that determine the existence or non-existence of limit cycles. But (for many years) I've not been able to find any references that address, in more general, &lt;em&gt;qualitative&lt;/em&gt; (non-mathematical) language, what minimum physical elements are necessary in a system for a limit cycle to exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts are a system must be structured (1) to contain at least two different states in which energy can be stored and which is able to flow periodically between them, and (2) to contain a non-linearity that facilitates regular orbits without either growth or decay and (3) sufficient initial amount of energy or else constant net energy such that input balances loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nonlinearity is a key element since you cannot practically realize an oscillator without decay or growth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these the essential key elements or can the necessary elements be described more in a more fundamental way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone know of references that treat this question?&lt;/p&gt;&#xA;" OwnerUserId="820" LastActivityDate="2018-10-06T09:23:32.583" Title="What are the necessary qualitative criteria for the existence of limit cycles" Tags="|energy|systems-engineering|stability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24047" PostTypeId="2" ParentId="24044" CreationDate="2018-10-06T03:03:28.593" Score="1" Body="&lt;p&gt;There is not a steel . 440A will make the hardness and is a &quot;stainless&quot; , but generally not as corrosion resistant as 316 ( except 316 won't tolerate hot chlorides) . You need to define your corrosive environment. If by &quot;bearing&quot; you mean a rolling element bearing , a materials engineer may be able to do something with tungsten carbides  , or ceramics, maybe a cobalt 6 ( aka Stellite 6) raceway. I assume money is no object. To answer our specific question , 440A is the most corrosion resistant steel with HRC 55.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2018-10-06T03:10:26.660" LastActivityDate="2018-10-06T03:10:26.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24051" PostTypeId="1" CreationDate="2018-10-06T08:01:44.410" Score="0" ViewCount="111" Body="&lt;p&gt;In most geometry inverse problem, information like the temperature measurement on the surface is use to detect the shape and location of a defect. My question is: is it possible to measure the temperature distribution or temperature gradient of a material up to certain depth? What kind of device is used for this kind of problem? Can you recommend a study that deals with such kind of problem?&lt;/p&gt;&#xA;" OwnerUserId="17756" LastActivityDate="2021-09-20T20:06:30.640" Title="A question on non-destructive testing and evaluation" Tags="|thermodynamics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24052" PostTypeId="2" ParentId="24051" CreationDate="2018-10-06T08:32:15.200" Score="2" Body="&lt;p&gt;Yes, in material engineering they call it &lt;strong&gt;Thermography&lt;/strong&gt;. By Radiating heat on the surface or infrared radiation we can map the temperature distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat transport throughout the material is affected by defects (holes, fracture ...). Also the heat that produce during the fracture of material can be detected in this technic.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually a laser puls is used to send the energy throughout the material and a thermographic camera to visualise the temperature gradient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This technic is very popular to find delimitation in fiber re-infocred plastics, or defects in welding connections between metals and composites. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a brief course in college, named &lt;strong&gt;fracture mechanics&lt;/strong&gt;, which deals with this kind of problem but not concentrated. I think material engineering and science is the right field. Also the basic knowledge of heat transport and thermodynamic is indispensable to understand what is happening.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-10-06T08:32:15.200" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24053" PostTypeId="2" ParentId="24046" CreationDate="2018-10-06T09:07:35.260" Score="1" Body="&lt;p&gt;Your (1) and (3) are along the right lines, but I'm not sure what (2) really means and some limit cycles don't look much like &quot;regular orbits&quot; anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is always a danger in trying to produce a very general explanation of something - hence the (probably apocryphal) quote &quot;This theory is so perfectly general that no particular application of it is possible&quot; - but let's try anyway ;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can state a very general and useful way to think about limit cycles like this. It is not completely &quot;qualitative&quot; since it needs some way to &quot;measure&quot; what is going on in the system - but in some sense, if you can't measure &lt;em&gt;anything&lt;/em&gt; about the system, you can't describe its behaviour at all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, identify some single parameter that measures the &quot;amplitude&quot; of the system's behaviour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;amplitude&quot; could measure things like displacement, velocity, gas pressure, etc, in a mechanical system, a voltage or current in an electrical system, the concentration of some component of a chemical reaction, etc, etc. It could also measure something that is not what an engineer or physicist might call an &quot;amplitude&quot; - for example the &lt;em&gt;frequency&lt;/em&gt; of vibration of a system. (Consider questions about limit cycles like &quot;why does the sound of brakes squealing have (almost) a constant pitch, independent of the speed of the vehicle&quot;?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, describe how the amount of &quot;stuff&quot; going into and out of the system depend on the &quot;amplitude&quot;. I use the word &quot;stuff&quot; rather than &quot;energy&quot; because this might not be exactly what a physicist would describe as &quot;energy&quot; - though often it will be exactly that. In some systems, it might be literally &quot;physical stuff,&quot; like the rate of flow of a material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, plot two lines on a graph, showing the amount of &quot;stuff going in&quot; and &quot;stuff coming out&quot; of the system, as functions of the &quot;amplitude&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two lines might be any shape of curve, depending on the particular system, but to get a limit cycle they must have these three properties:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;the graphs need to intersect at some point&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;for amplitudes smaller than the intersection point, more &quot;stuff&quot; is going in than is coming out&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;for amplitudes above the intersection point, more &quot;stuff&quot; is coming out than is going in.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In fact the graphs might have more than one intersection point, which can describe complex behaviour and several different limit points with different amplitudes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At any particular amplitude, if &quot;stuff out&quot; is &lt;em&gt;bigger&lt;/em&gt; than &quot;stuff in&quot;, the amplitude will tend to &lt;em&gt;decrease&lt;/em&gt; over time, but if &quot;stuff out&quot; is &lt;em&gt;less&lt;/em&gt; than &quot;stuff in&quot;, the amplitude will tend to &lt;em&gt;increase&lt;/em&gt;. Limit cycles can occur at the points where the two flows of &quot;stuff&quot; are equal.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-10-06T09:23:32.583" LastActivityDate="2018-10-06T09:23:32.583" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24054" PostTypeId="1" AcceptedAnswerId="24065" CreationDate="2018-10-06T09:45:11.187" Score="0" ViewCount="78" Body="&lt;p&gt;I have an Aluminium square cross sectional rod of &lt;span class=&quot;math-container&quot;&gt;$20mm$&lt;/span&gt; X &lt;span class=&quot;math-container&quot;&gt;$20mm$&lt;/span&gt;. Length 700mm.&#xA;At the end of this rod there is a mass of &lt;span class=&quot;math-container&quot;&gt;$660$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; attached.&#xA;My question is what torque do i need to lift this mass up assuming my pivot point is at the other extreme of the rod? And how do i consider the mass of the rod in the calculations!&lt;/p&gt;&#xA;" OwnerUserId="17758" LastEditorDisplayName="user14407" LastEditDate="2018-10-06T09:47:49.497" LastActivityDate="2018-10-07T05:05:56.563" Title="Understanding Motor Torque calculations" Tags="|mechanical-engineering|motors|torque|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24057" PostTypeId="1" CreationDate="2018-10-06T13:58:36.097" Score="3" ViewCount="561" Body="&lt;p&gt;I am building an Arcade cabinet with a 3-sided rotating control panel. I want the rotation of the control panel will be automated. Here's a render:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cv98x.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cv98x.gif&quot; alt=&quot;render&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have a problem: locking the control panel in place. It needs to be able to hold in place against the force of players leaning on the control panel (and angry players pounding on it). It also needs to have no play and feel very solid. It can't be rocking while players are pushing the joystick up and down (some people push somewhat hard). The front panel prevents it from rotating forward, but there is nothing preventing it from rotating backwards. Note that this will be made from wood in my shop, so exact tolerances can not be expected (may be off by up to 1/8&quot;-1/4&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I created a test setup for my original idea that used a tension latch to secure the control panel to the front. That worked well, but I would like the whole thing to be automated. Here's a video of that: &lt;a href=&quot;http://www.youtube.com/watch?v=KK2Xa4psLp0&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.youtube.com/watch?v=KK2Xa4psLp0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have had a couple ideas but I am not sure how feasible they are or exactly how to implement them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One idea was to use an indexing/dividing head like the one in my lathe. The rod that locks it in place could be automated via an actuator or something. But, I am not sure how to obtain one. I don't have the tools to machine anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another was to use a worm screw to drive the rotation of the control panel as I have read they are supposed to provide holding strength. However, I am not sure how much holding strength and how much play will be present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about an electromagnetic brake, but I think one that would provide enough force would be way too big (and could interfere with the CRT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also figured solutions could be combined. One component could be responsible from preventing the control panel from turning freely under large forces (like leaning/pounding) and another component could be used to prevent the control panel from rocking due to the smaller forces (like pushing on the joystick).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the problem of holding a rotating joint in place is one that has been readily solved by robotics. But I am a babe in the woods with this stuff and I have not been able to find a good solution in my research. I am completely new to automation/ME. Please let me know if there is a better place to post this question and sorry if this is not the correct place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; While posting this I had another idea: is there a way to automate the tension latch?&lt;/p&gt;&#xA;" OwnerUserId="17754" LastEditorUserId="17754" LastEditDate="2018-10-06T14:53:53.747" LastActivityDate="2018-10-06T20:16:33.213" Title="Rotating Control Panel for Arcade Cabinet" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24058" PostTypeId="2" ParentId="24057" CreationDate="2018-10-06T15:30:40.420" Score="5" Body="&lt;p&gt;The force you need to rotate the panel is tiny, compared with the restraining forces to hold it in place while it is being used (and abused) by the players.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would separate the two functions and devise some &lt;em&gt;passive&lt;/em&gt; locking features, for example pins that can extend and retract from the end panels and fit into holes in the ends of the rotating part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force required to &lt;em&gt;move&lt;/em&gt; the pins would also be small, so you could use small electrical devices (linear actuators, small motors driving a rack and pinion, or whatever) to deploy and retract them.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-10-06T15:30:40.420" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24060" PostTypeId="2" ParentId="24057" CreationDate="2018-10-06T20:16:33.213" Score="1" Body="&lt;p&gt;The class of device is called an intermittent. It is used, for example, in a film projector to advance the film one frame at a time. It is also used in the packaging industry to fill bottles and such from a stationary fill head. The gadget you are wanting is probably the cammed geneva wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=8ix2eGVs1U8&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=8ix2eGVs1U8&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2018-10-06T20:16:33.213" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24062" PostTypeId="2" ParentId="21325" CreationDate="2018-10-07T00:27:20.303" Score="1" Body="&lt;p&gt;It looks like a Russian standard ( GOST ). If so ,it is a low alloy steel with 0.38% C and can be nitrided ( indicated by Al).  Similar to SAE 4340 with additional Al which can be hardened up to HRC 60. What is the component ; could it be a nickel base high alloy ?&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-10-07T00:27:20.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24063" PostTypeId="2" ParentId="23993" CreationDate="2018-10-07T00:31:44.763" Score="1" Body="&lt;p&gt;The main problem I see here is the force you need to pull anything up. If the margins are suficiently higher than the river that will be ok, but if the margins are at almost same level you will need to do a lot of force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if θ is 10°, each person will need to do 2.9 times the force one would do alone if they were right above the object. And I am ignoring the weight of the rope which is probably not negligible specially if its get moist by dipping into the river.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the river is narrow (like 15 feet) I think it will work but with 50 feet you will start to get tired real quick. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mdthY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mdthY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8696" LastActivityDate="2018-10-07T00:31:44.763" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24064" PostTypeId="1" AcceptedAnswerId="24076" CreationDate="2018-10-07T02:59:01.597" Score="1" ViewCount="1026" Body="&lt;p&gt;Simple bending equation is derived assuming pure bending(when a couple is applied onto a beam with no shear stress).We tend to use the above derived equation when moment is caused due to shear stress?How far is this correct?&lt;/p&gt;&#xA;" OwnerUserId="15268" LastActivityDate="2018-10-07T16:28:10.873" Title="Why do we use simple bending equation when bending moment is caused due to shear force?" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24065" PostTypeId="2" ParentId="24054" CreationDate="2018-10-07T05:05:56.563" Score="0" Body="&lt;p&gt;The angular momentum of a rod or beam rotating about one end is its angular momentum about the center of mass of the rod plus mass times (l/2)^2, in this case, parallel axis momentum theory.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$L = m\frac{l^2}{12} +m\times (l/2)^2 = ml^2/3  $$&lt;/span&gt;&#xA;Where m is the mass and l the length of the aluminum rod, 700mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the torque needed is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T= 1/3(ml^2) + 660l$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-10-07T05:05:56.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24067" PostTypeId="5" CreationDate="2018-10-07T10:08:47.600" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-10-07T10:08:47.600" LastActivityDate="2018-10-07T10:08:47.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24068" PostTypeId="4" CreationDate="2018-10-07T10:08:47.600" Score="0" Body="This is a special tag that is designed to be used only by the system for questions that have had all of their other tags removed. Do not add this tag to existing questions, it is reserved for use by the system to identify posts with no valid tags." OwnerUserId="6767" LastEditorUserId="6767" LastEditDate="2018-10-07T12:25:53.573" LastActivityDate="2018-10-07T12:25:53.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24069" PostTypeId="1" AcceptedAnswerId="24070" CreationDate="2018-10-07T11:09:59.537" Score="1" ViewCount="2661" Body="&lt;p&gt;How do we get the moment when there's a distributed load?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the picture below has 9[(60)(18)]. I get that 60 is the distributed load and 18 is the total length of the load, my question is how did it get the 9?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pTXhS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pTXhS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3703" LastActivityDate="2018-10-07T13:37:05.093" Title="How to solve for moment with uniform distributed load" Tags="|moments|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24070" PostTypeId="2" ParentId="24069" CreationDate="2018-10-07T12:05:50.313" Score="2" Body="&lt;p&gt;The resultant of distributed loads always acts on the centroid of the distributed load geometry, here the distributed load is uniform so its centroid lies half the way. If the distributed load varies linearly from zero at one end to a maximum value at the other end, then its centroid would lie at &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{3} L$&lt;/span&gt; from the &quot;max load&quot; end and &lt;span class=&quot;math-container&quot;&gt;$\frac{2}{3}L$&lt;/span&gt; from the &quot;zero load&quot; end, with &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; the side length.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorUserId="683" LastEditDate="2018-10-07T13:37:05.093" LastActivityDate="2018-10-07T13:37:05.093" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24071" PostTypeId="2" ParentId="24028" CreationDate="2018-10-07T12:47:12.030" Score="1" Body="&lt;p&gt;If a knife is working but has poor edge retention your bet bet might be to look at a harder blade. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also possible that, if you are getting hot enough to scorch the paper, the temper of the blade is being affected. Similarly applying a dry lubricant (eg PTFE spray) to the blade may also help if that if compatible with the rest of the process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might be worth trying High Speed Steel for the blade, something like a parting tool blank may work as they are already a suitably flat, relatively narrow section. With care you should be able to grind it to a cutting edge without needing to heat treat it afterwards. Alternatively a section of a scrap HHS disc saw may work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The edge geometry will also be crucial. I woudl suggest that something similar to a wood chisel would be a good starting point ie a fairly acute angle on one side only with a small secondary bevel. Even a slight hollow grind on the primary bevel may help and give some clearance to help reduce heat from friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other Tools steels  may also work, something like D2 is an obvious contender, also O1 and A2. there woudl probably need to be custom made and heat treated but of you can find a small knife-maker  (of which there are many) you might be able to get a simple utilitarian blade fairly cheaply. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something else whcih might be worth trying is a chainsaw chain., Set the work up in the lathe and mount the chain so it wraps around part of the circumference with something to apply a bit of tension. Just draping it over teh top with a weight on one end may work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can get &lt;a href=&quot;https://www.amazon.co.uk/dp/B06ZYHNC17/ref=asc_df_B06ZYHNC1756083755/?tag=googshopuk-21&amp;amp;creative=22146&amp;amp;creativeASIN=B06ZYHNC17&amp;amp;linkCode=df0&amp;amp;hvadid=205188093108&amp;amp;hvpos=1o4&amp;amp;hvnetw=g&amp;amp;hvrand=9700879027246799415&amp;amp;hvpone=&amp;amp;hvptwo=&amp;amp;hvqmt=&amp;amp;hvdev=c&amp;amp;hvdvcmdl=&amp;amp;hvlocint=&amp;amp;hvlocphy=9045554&amp;amp;hvtargid=pla-350045836783&quot; rel=&quot;nofollow noreferrer&quot;&gt;'pocket chainsaws'&lt;/a&gt; whcih should be cheap enough to at least test the idea. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-10-07T14:36:56.607" LastActivityDate="2018-10-07T14:36:56.607" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24073" PostTypeId="1" AcceptedAnswerId="24075" CreationDate="2018-10-07T14:43:35.513" Score="4" ViewCount="339" Body="&lt;p&gt;I have observed that when a 4-wheeled vehicle takes a turn, one of it rear tire moves with more speed as compared to the other. For example A car is taking right turn, its right rear tire will be slower as compared to left rear tire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is that done?&lt;br&gt;&#xA;Without this system, the vehicle wont take a turn and collapse.&lt;/p&gt;&#xA;" OwnerUserId="17768" LastEditorUserId="17768" LastEditDate="2019-08-17T14:31:09.040" LastActivityDate="2019-08-17T14:31:09.040" Title="Proccess of a 4-wheeled vehicle taking turn" Tags="|car|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24074" PostTypeId="2" ParentId="24064" CreationDate="2018-10-07T14:45:38.537" Score="2" Body="&lt;p&gt;The equation you're talking about is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma = \dfrac{My}{I}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason this equation only considers bending moment even though &quot;moment is caused [by] shear stress&quot;, is because, well, that fact is irrelevant. There is no relationship between a given shear force and the resultant bending moment, since different beams (with different spans, support conditions, etc) might have the same shear force at a given point but wildly different bending moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bending moment causes stresses according to the equation above. Shear stress causes other stresses which follow a different behavior. The effects are completely independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, while bending moment generates longitudinal stresses (parallel to the beam's central axis), shear stress is transversal (perpendicular to the central axis... or at an angle to it).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also choose to separate these effects because it makes life easier. For example, look at this beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZBELQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZBELQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Shear diagram&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xTxMS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xTxMS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bending moment&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WHZKo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WHZKo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can separate this beam into two parts: between the supports and the loads, and the midspan between the loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the midspan, we have bending moment but no shear force. So whatever stresses are felt by the beam in that region are entirely due to the bending moment. So just use the equation above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, looking at the area between the supports and the loads, we have both bending moment and shear force. So the beam's internal stress state in this region is a combination of both forces. But still, you can just calculate the bending moment component using the equation above and others for the shear force component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, you can't just add those results since they operate in different orientations, but you could combine them to obtain the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cauchy_stress_tensor#Principal_stresses_and_stress_invariants&quot; rel=&quot;nofollow noreferrer&quot;&gt;principal stresses&lt;/a&gt;, we just &lt;a href=&quot;https://engineering.stackexchange.com/q/8043/1832&quot;&gt;don't bother doing so&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, it is easier to work with each force component in isolation than simultaneously. And given that there's no benefit to doing it all at once, we choose the easier route. (Obviously, in cases where the components actually influence each other, we take that into consideration, such as cases of flexo-compression or -tension).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-10-07T14:45:38.537" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24075" PostTypeId="2" ParentId="24073" CreationDate="2018-10-07T15:04:13.323" Score="4" Body="&lt;p&gt;It depends. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The driven wheels need to be connected by a &lt;a href=&quot;https://en.wikipedia.org/wiki/Differential_(mechanical_device)&quot; rel=&quot;nofollow noreferrer&quot;&gt;differential&lt;/a&gt; whcih allows power to be transferred to both wheels but still allowing them to turn at different speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not that easy to comprehends exactly how this works mechanically without having a physical model to play with but it does work , at least within certain limits.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with a gear based differential is that if one wheel loses traction the opposite wheel doesn't get any power either whcih is a big issue on slippery surfaces like snow, ice, mud or gravel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple fix is to lock the diff which improves traction but means that the wheels need to slip on cornering on slipper surfaces this may be acceptable but is a problem on dry pavement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many 4x4 vehicles have the ability to lock and unlock their differentials according to the conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more sophisticated solution is a &lt;a href=&quot;https://en.wikipedia.org/wiki/Limited-slip_differential&quot; rel=&quot;nofollow noreferrer&quot;&gt;limited slip differential&lt;/a&gt; whcih allows the drive wheels to rotate at different speeds but ensures that both get at least some torque in all conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various ways to implement LSDs but some sort of viscous coupling (a bit like the torque converter in an automatic gearbox) is common). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are often used in conditions where vehicle performance is traction limited ie some wheel slip is frequent but not consistent eg in rallying and racing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern systems allow the degree of slip to be controlled either by directly by the driver or by a computer with reference to some pre-set parameters. Indeed diff settings are one of the important driver-controlled parameters in a modern F1 car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a 'loose' setting any loss of traction on a driven wheel will reduce overall torque so you want  ta tight diff for exiting a low speed corner conversely a tight diff will tend to push the car in more of a straight line and induce under-steer (at least in general). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is further complicated by that fact that cornering at any speed tends to induce roll which unloads the inside wheel to the inside driven wheel will tend to lose traction before the outside one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It isn't &lt;em&gt;impossible&lt;/em&gt; to turn without a differential most go-karts don't have diffs  and they are more essential on fwd than rwd cars. A fws car with a simple Limited Slip Diff tends to have fairly epic torque steer. But in rwd you can use engine torque to deliberately slide the rear and break traction  to assist  turn-in so the tyres provide the slip rather than the transmission. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very obvious in 'drift' racing, whcih tends to use locked or relatively tight LSD.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-10-07T15:25:19.830" LastActivityDate="2018-10-07T15:25:19.830" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24076" PostTypeId="2" ParentId="24064" CreationDate="2018-10-07T16:17:50.617" Score="0" Body="&lt;p&gt;We only can ignore the shear when the effects it has on shape and displacement  are small enough to not be a concern. In general, we &lt;em&gt;can't&lt;/em&gt; ignore them. If we ignore shear, then we are saying that the beam ends (and all cross sections) are constrained to remain perpendicular to the beam's deflected surfaces. Real shear stresses don't let that happen. But for a long beam relative to the thickness, the shear strain can't develop very quickly along the length and the perpendicular assumption is mostly true. But for short, thick beams, shear strain has a large impact on the deformed shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the first comprehensive approaches to handle this was &lt;a href=&quot;https://en.wikipedia.org/wiki/Timoshenko_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Timoshenko beam theory&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-10-07T16:28:10.873" LastActivityDate="2018-10-07T16:28:10.873" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24079" PostTypeId="1" AcceptedAnswerId="24081" CreationDate="2018-10-08T03:09:49.643" Score="2" ViewCount="70" Body="&lt;p&gt;Take a loom at this diagram&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ANglD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ANglD.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;You have probably heard of this before but im just curious what laser they are using to do this. They do t appear to be ionizing air. It seems the one laser is just for color on the particle but what is the other? I would like to try this at home and make a dot for fun but i dont quite understand whats going on here. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="15207" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-10-08T11:28:01.287" Title="How 2 lasers interacting causes a dot of light" Tags="|lasers|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24080" PostTypeId="1" CreationDate="2018-10-08T10:56:12.553" Score="0" ViewCount="78" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/v75bJ.jpg&quot; alt=&quot;enter image description here&quot;&gt; Can someone explain how to get the Fx for the support A and E&lt;/p&gt;&#xA;" OwnerUserId="17776" LastActivityDate="2018-10-08T11:32:33.650" Title="Someone please explain how to get the FAx and FEx because both support are pin.. I only got the FAy and FEy" Tags="|structural-analysis|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24081" PostTypeId="2" ParentId="24079" CreationDate="2018-10-08T11:28:01.287" Score="2" Body="&lt;p&gt;The full letter is here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nature.com/articles/nature25176&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nature.com/articles/nature25176&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have 4 lasers: an RGB beam (3 lasers) that provides colour and an invisible 405nm beam that traps the particle. The RGB beam is pretty self-explanatory: you shine a beam of the colour you want the line to be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trapping beam is more complicated, by using lenses with particular imperfections you can focus the beam so that there is a dark volume near the focal point. Particles in this volume will be kept there by the photophoretic force. There is more information in this paper: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.osapublishing.org/oe/abstract.cfm?uri=oe-19-18-17350&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.osapublishing.org/oe/abstract.cfm?uri=oe-19-18-17350&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Playing with invisible lasers can be pretty dangerous &lt;/p&gt;&#xA;" OwnerUserId="17777" LastActivityDate="2018-10-08T11:28:01.287" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24082" PostTypeId="2" ParentId="24080" CreationDate="2018-10-08T11:32:33.650" Score="1" Body="&lt;p&gt;As always, let's start with the global equilibrium equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= A_x + E_x = 0 \\&#xA;\therefore A_x &amp;amp;= -E_x \\&#xA;\sum F_y &amp;amp;= A_y + E_y - 3\times40 = 0 \\&#xA;\therefore A_y &amp;amp;= 120 - E_y \\&#xA;\sum M_A &amp;amp;= 5E_x - 40(5 + 10 + 15) = 0 \\&#xA;\therefore E_x &amp;amp;= 240\text{ kN} \\&#xA;\therefore A_x &amp;amp;= -240\text{ kN}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's the horizontal reactions, trivially found by using global equilibrium equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For completeness, we can also directly find the vertical components since &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is only connected to one truss member, therefore &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;'s resultant reaction must be parallel to that member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\dfrac{E_y}{E_x} &amp;amp;= \dfrac{5}{15} \\&#xA;\therefore E_y &amp;amp;= \dfrac{E_x}{3} = 80\text{ kN} \\&#xA;\therefore A_y &amp;amp;= 40\text{ kN}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-10-08T11:32:33.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24083" PostTypeId="1" CreationDate="2018-10-08T12:42:33.850" Score="4" ViewCount="226" Body="&lt;p&gt;Problem: I have been asked to help determine whether a waterwheel is able to generate enough power to run 5 x 15kW pumps in a remote location in Zambia where there is no constant power supply. I work in the mining industry so this question is a bit out of my field to say the least.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The  pumps as mentioned are 15kW pulling 34 Amps (max speed is 2900RPM)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water flow to the waterwheel is 7m3/s in a channel that is 3mW and 1mH. The maximum head from the top of the channel is 3m. Please see attached picture (left is top view right is side view)&lt;a href=&quot;https://i.stack.imgur.com/AO0P9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AO0P9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have to do basic calculations to determine:&#xA;- The speed of the water wheel  (RPM)&#xA;- Dimensions of the water wheel &#xA;- Will the water wheel be able to produce that amount of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please guide me in any way regarding what calculations are required, where to find these calculations? What additional parameters are required if necessary?&lt;/p&gt;&#xA;" OwnerUserId="7092" LastActivityDate="2018-10-08T12:42:33.850" Title="Water wheel Calculations required" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24084" PostTypeId="2" ParentId="23313" CreationDate="2018-10-08T13:25:47.180" Score="2" Body="&lt;p&gt;Edge preparation is required for joining thick plates for proper welding joint. Its always preferable less volume of groove size. Less volume of groove gives less HAZ and having good mechanical properties. Double V, Double U joints are there but we need to follow certain welding sequence like first layer on one side and second layer on second groove which leads to less distortions. &#xA;Single V can be used most of the cases because of easy edge preparation instead of U.&#xA;J joint can be used in specific purposes like one material is weldable another is not weldable. Then due to J joint we have prepared J on only non weldable plate another plate is without edge preparation. It is also depends on the thermal conductivity of the material&lt;/p&gt;&#xA;" OwnerUserId="17778" LastEditorUserId="17778" LastEditDate="2018-10-08T13:39:50.923" LastActivityDate="2018-10-08T13:39:50.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24087" PostTypeId="1" CreationDate="2018-10-08T14:32:39.327" Score="1" ViewCount="1070" Body="&lt;p&gt;I dont know how to approach this problem. Suppose that a machine has a probability of failure of 43% at 90 hours of operation. How can you calculate the failure rate and the probability of surviving x hours without having failures?&lt;/p&gt;&#xA;" OwnerUserId="17782" LastActivityDate="2020-10-04T16:40:16.373" Title="How to calculate the failure rate of a machine with probability and mean failure time?" Tags="|failure-analysis|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24088" PostTypeId="1" AcceptedAnswerId="24865" CreationDate="2018-10-08T14:51:39.253" Score="1" ViewCount="781" Body="&lt;p&gt;I have to do a durability approval of feather key (shaft-hub connection) and the formulas are dependent if the shaft respectively the hub is ductile or brittle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what is the decision about to say it is ductile or brittle.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;My norm for feather key isn’t saying anything how to distinguish if the metal is ductile or brittle.&lt;/li&gt;&#xA;&lt;li&gt;The norm for pressfits uses the parameter fracture elongation and reduction of area.&lt;/li&gt;&#xA;&lt;li&gt;And the literature is talking about brittle metal when the Maximum Stress Theory is valid &lt;span class=&quot;math-container&quot;&gt;$(\sigma_{limit}/\tau_{limit}=1)$&lt;/span&gt; and ductile metal when the Shear Stress Hypothesis is valid &lt;span class=&quot;math-container&quot;&gt;$(\sigma_{limit}/ \tau_{limit}=2)$&lt;/span&gt;.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\sigma_{limit}$&lt;/span&gt; is tensile strength&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\tau_{limit}$&lt;/span&gt; is the strength when plasticity begins&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Okay, there are many possibilities… But which one is the right possibility? Or is there a connection between my findings that I’m not aware of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help.&lt;/p&gt;&#xA;" OwnerUserId="17780" LastEditorUserId="1832" LastEditDate="2018-10-08T19:42:57.653" LastActivityDate="2018-11-26T11:19:29.357" Title="With which parameter can you specify whether a metal is ductile or brittle" Tags="|materials|steel|metals|strength|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="24090" PostTypeId="2" ParentId="24087" CreationDate="2018-10-08T17:56:05.357" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\large{R(t)=e^{\lambda t}}\\&#xA;\begin{align}&#xA;R&amp;amp;=\text{Reliability}\\&#xA;t&amp;amp;=\text{time}\\&#xA;\lambda&amp;amp;=\text{failure rate}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;At &lt;span class=&quot;math-container&quot;&gt;$t=0$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;R(t=0)=1\\&#xA;\implies100\%&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I think, if the reliability follows exponential distribution, you can use the formula given above. Reliability is the probability that a machine will function normally during a period of time under proper working conditions. So, &lt;span class=&quot;math-container&quot;&gt;$$\text{reliability}=1-\text{probability of failure}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastEditorUserId="28917" LastEditDate="2020-10-04T16:40:16.373" LastActivityDate="2020-10-04T16:40:16.373" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24092" PostTypeId="1" CreationDate="2018-10-08T20:33:56.530" Score="2" ViewCount="71" Body="&lt;p&gt;I want to build a strong electro magnet. I came to the conclusion that the smartest arrangement would be something like hundreds of thin sheets of 0.2 mm copper with PTFE isolation. This reduces resistance (diameter) and allows maximally compact stacking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I struggle a little bit, because actually no one seems to produce something like this. Instead industry prefers wires, even if they are not compact enough to get a good performance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is: What is the smartest, cheapest and easiest way to get sheets like in the image below? Diameter would be around 4 to 8 cm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lt593.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lt593.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17788" LastEditorUserId="2559" LastEditDate="2018-10-09T15:34:33.903" LastActivityDate="2018-10-09T15:34:33.903" Title="Cheapest way to produce 200-1000 pieces of 0.2 mm copper plates" Tags="|manufacturing-engineering|electromagnetism|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24093" PostTypeId="1" AcceptedAnswerId="24495" CreationDate="2018-10-08T21:28:07.780" Score="1" ViewCount="294" Body="&lt;p&gt;So upon watching videos and reading articles i understand few things. For one when the laser beam is focused to a tight enough point when placing a material in there (people use crayola markers) it will trap and burn that particle for a little while.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/RjgWn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RjgWn.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hSkqC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hSkqC.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;My question is how can i do this without  having to touch a marker to it? Can I achieve a particle effect through something else or atleast something that wont burn off the particle right away. Also is this effect as easy as talking a 405nm laser, focusing it with some sharp lenses and then having that particle get trapped. I have seen a few diagrams but still it is unclear all the equipment they use and then in a video it is just a laser focused to a point.&#xA;Such as these:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TIaBg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TIaBg.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KB65D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KB65D.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;UPDATE: Upon reading more i see the mW required is only around 5mW which isnt bad at all. I also see people use graphite but im not sure how ling that particle lasts? Forever until the beam is broken or only for a certain amount of time. To use graphite could i simple take a block of it and wave it in front of the laser at its lowest density point to capture the particle? My other confusion is the lens required what is so special? Is there maybe an equation that includes a class of particles, wavelength, power, and focal length of lens to see where i stand with this. I know this isnt what stack overflow is for but i just want to set this up and see it and understand it rather then just read so maybe a good diagram that i or any member coming here could understand in order to setup this experiment.&lt;/p&gt;&#xA;" OwnerUserId="15207" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2022-01-12T03:03:33.573" Title="Trapping a particle in a laser beam (optical tweezers)" Tags="|lasers|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24095" PostTypeId="1" CreationDate="2018-10-09T04:04:23.527" Score="0" ViewCount="1929" Body="&lt;p&gt;How do I make this highlighted part of drawing a solid body so that I can than boolean / subtract it later?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these faces / planes are connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jJRds.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jJRds.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13701" LastActivityDate="2018-10-09T12:00:49.837" Title="Fusion360 - How do I fill these sketch planes into a body?" Tags="|cad|drawings|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24098" PostTypeId="2" ParentId="22955" CreationDate="2018-10-09T08:12:29.037" Score="2" Body="&lt;p&gt;For fillet weld only axial tensile/compressive load is possible. Due to this load shear will produce on weld joint. We know that shear stress is equal to (applied load/minimum cross-section area). Minimum cross section is 0.707XSXL&lt;/p&gt;&#xA;" OwnerUserId="17778" LastActivityDate="2018-10-09T08:12:29.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24099" PostTypeId="2" ParentId="24095" CreationDate="2018-10-09T11:50:52.643" Score="1" Body="&lt;p&gt;It looks like you are in patch mode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fusion has three main ways of creating solid bodies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the &lt;strong&gt;Model&lt;/strong&gt; environment you can either create a solid from scratch using Create Form which creates various different solid shapes with faces whcih can be manipulated by push/pull tools. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or also in the &lt;strong&gt;model&lt;/strong&gt; environment you can start with a sketch and &lt;em&gt;extrude&lt;/em&gt; it. as well as the basic extrude tool you can &lt;em&gt;sweep&lt;/em&gt;  the shape along a path, &lt;em&gt;revolve&lt;/em&gt; it around an axis or &lt;em&gt;loft&lt;/em&gt; between two or more different shapes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively in the &lt;strong&gt;Patch&lt;/strong&gt; environment (orange coloured tool icons) you start with 2 dimensional shapes and &lt;em&gt;thicken&lt;/em&gt; them. This tends to be most useful when creating complex hollow shapes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like you need to go into the &lt;strong&gt;model&lt;/strong&gt; environment (from the drop down menu in the top left where it says PATCH in your screenshot). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are trying to create a solid pyramid the easiest way might be to &lt;em&gt;loft&lt;/em&gt; one of the faces to a point on one of the vertices. Alternatively you could extrude one face to the appropriate length to create a prism and then use the other faces as cutting planes for the split tool or possibly extrude them in inwards to create an &lt;em&gt;intersection&lt;/em&gt; &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-10-09T12:00:49.837" LastActivityDate="2018-10-09T12:00:49.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24102" PostTypeId="1" AcceptedAnswerId="24105" CreationDate="2018-10-09T14:15:11.647" Score="0" ViewCount="99" Body="&lt;p&gt;Some days ago, I took a plane (Airbus A330) for a pretty long flight - around 8hrs. During the flight, we experienced some turbulences and I thought : what if the turbulences last the whole flight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would a plane react to 8+ hrs under mid to heavy turbulences during the whole flight?&lt;/p&gt;&#xA;" OwnerUserId="17794" LastActivityDate="2018-10-12T05:04:55.093" Title="For how long could a plane handle turbulences?" Tags="|aircraft-design|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24103" PostTypeId="2" ParentId="24102" CreationDate="2018-10-09T15:24:52.447" Score="1" Body="&lt;p&gt;I can't point to a reference, but I'd say that, as long as you're not talking about catastrophic turbulence, the plane can resist for basically the entire flight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So long as each of the &quot;waves&quot; of turbulence are individually safe, the entire turbulence is also safe. This is because the turbulence is unlikely to be at the plane's resonant frequency, and therefore you don't get magnifying effects over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other relevant failstate would be fatigue, however that happens over countless cycles. So unless you have a very old plane (20+ years, which has consumed most of its safe cycles), fatigue also likely wouldn't be reached in a single very-turbulent flight.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-10-09T15:24:52.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24104" PostTypeId="2" ParentId="24092" CreationDate="2018-10-09T15:33:36.083" Score="1" Body="&lt;p&gt;Your &quot;best&quot; options would be waterjet or laser cutting as long as the cutting tolerances are acceptable for your design. There are virtually no tooling costs. You can make a 2D cad drawing and get multiple quotes very quickly for this type of manufacturing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, continuous die cutting/stamping from a foil roll would be more suited for high volume, higher precision manufacturing.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2018-10-09T15:33:36.083" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24105" PostTypeId="2" ParentId="24102" CreationDate="2018-10-09T16:54:43.113" Score="4" Body="&lt;p&gt;You may want to look for &lt;strong&gt;aircraft fatigue loads&lt;/strong&gt;, which are the typical loads that an airplane may experience during a fly (some sets include data for harsh conditions). They look like this (depending where are you measuring):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ds6Zq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ds6Zq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O3SQO.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O3SQO.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And note that airplanes are designed to withstand thousands of those cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as long as the turbulence is not catastrophic (that one gust of wind is so strong that it causes immediate failure - which I think is unlikely) and you are not already near the end of the plane's life design, as Wasabi said, one long turbulent fly wont cause failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source for the first image: &lt;a href=&quot;http://www.dtic.mil/dtic/tr/fulltext/u2/762718.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fatigue life prediction for aircraft structures and materials&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17409" LastActivityDate="2018-10-09T16:54:43.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24107" PostTypeId="2" ParentId="24102" CreationDate="2018-10-09T21:30:03.927" Score="1" Body="&lt;p&gt;Duration of turbulence is not a factor, intensity is!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All modern airplanes have been designed with a flexible structure ready to take punishing weather and vibrate with the impact of the shock wave to dissipate the energy of turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have been designed to vibrate longitudinally along the fuselage and rotationally to take in the torque induced by wing shock. The wings are designed to flex and vibrate to behave in a ductile mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even the empennage, the end of the fuselage including the rudder and elevators, is designed to twist and shake safely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the pilot during an intense turbulence or even in the absence of it can subject the plane to stresses above the safe levels, do maneuvers that are not safe for that weather and cause damage to structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fatigue which is the first culprit to look into happens mostly by the routine stresses of flight and engine vibration and has to be checked at routine service intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FAA crash reports are a good source to start researching about effect of bad weather on flights, planes, and pilots' decision making. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some search on Youtube and found this on flutter of the airplanes in calm weather. &lt;a href=&quot;https://youtu.be/egDWh7jnNic&quot; rel=&quot;nofollow noreferrer&quot;&gt;Flutter&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-10-12T05:04:55.093" LastActivityDate="2018-10-12T05:04:55.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24108" PostTypeId="2" ParentId="21266" CreationDate="2018-10-10T01:34:24.943" Score="1" Body="&lt;h1&gt;Foundations&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Energy Balance&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Consider the picture below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cE40G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cE40G.png&quot; alt=&quot;picture of tube system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows a &lt;span class=&quot;math-container&quot;&gt;$dz$&lt;/span&gt; segment of the tube. Neglect radiation to/from the air and sky. The energy balance becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dot{q}_r + \dot{q}_{fi} = \dot{q}_h + \dot{q}_{fo} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand this as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;A_{go} f_r \varepsilon \sigma T_r^4  = h_a A_{go} (T_{go}(z) - T_a) + \dot{m}_w \tilde{C}_{pw}\Delta T_w(z) \\&#xA;2 \pi r_{go} \Delta z f_r \varepsilon \sigma T_r^4  = h_a 2\pi r_{go} \Delta z (T_{go}(z) - T_a) + \dot{m}_w \tilde{C}_{pw}\Delta T_w(z)&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Allow this to go to a differential as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;2\pi r_{go} f_r \varepsilon \sigma T_r^4  = 2 \pi r_{go} h_a (T_{go}(z) - T_a) + \dot{m}_w \tilde{C}_{pw}\frac{dT_w}{dz}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Tube Wall as Heat Exchanger&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Write the heat flow from the water through the metal and glass as a heat exchanger expression with the overall heat transfer coefficient of the water + metal + glass (the air is already accounted in the energy balance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dot{q}_h = U A_{go} \left(T_{w}(z) - T_{go}(z)\right) = h_a A_{go}\left(T_{go}(z) - T_a\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives an expression for &lt;span class=&quot;math-container&quot;&gt;$T_{go}(z)$&lt;/span&gt; as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;T_{go}(z) = R_U T_w(z) + R_h T_a \\&#xA;R_U = U / (U + h_a)\ \ \ R_h = h_a / (U + h_a)\ \ \ R_U + R_h = 1&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; is a combination of convection at the water/metal interface (&lt;span class=&quot;math-container&quot;&gt;$h_w$&lt;/span&gt;) +  conductive resistances &lt;span class=&quot;math-container&quot;&gt;$k_j$&lt;/span&gt; for metal tube and glass.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Tube Wall as Series Thermal Resisters&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Write the heat flow from the water through the metal and glass as a series of thermal resisters for water + metal + glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \left(T_w(z) - T_{go}(z)\right) = \dot{q}_h\left(R_w + R_t + R_g\right) \\&#xA;R_w = \frac{1}{2\pi r_{ti} \Delta z h_w} \hspace{0.5cm} R_j = \frac{\ln(r_{jo}/r_{ji})}{2\pi k_j \Delta z}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to an expression similar to the above with &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\left(T_w(z) - T_{go}(z)\right) = \frac{r_{go} h_a}{\pi} \left(T_{go}(z) - T_a\right) \left(\frac{1}{r_t h_w} + \frac{\ln(r_{to}/r_{ti})}{k_t}  + \frac{\ln(r_{go}/r_{gi})}{k_g}\right)&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relationship between &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\sum R_j$&lt;/span&gt; can be derived using the two expressions for &lt;span class=&quot;math-container&quot;&gt;$\left(T_w(z) - T_{go}(z)\right)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Evaluation&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Final Expression&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Combine the energy balance and the heat flow through the tube walls (as a heat exchanger) to obtain the first order differential equation for the temperature of the water as a function of position along the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{m}_w \tilde{C}_{pw}\frac{dT_w}{dz} + 2 \pi r_{go} h_a R_U T_w(z) - \left(R_U T_a + 2\pi r_{go} \sigma f_r \varepsilon T_r^4\right)  = 0&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression has only one unknown &lt;span class=&quot;math-container&quot;&gt;$T_w(z)$&lt;/span&gt;. Only one boundary condition is required. The BC is &lt;span class=&quot;math-container&quot;&gt;$T_w(0) = T_{wi}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Interpretation&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The first term is the enthalpy change of the water flowing through the &lt;span class=&quot;math-container&quot;&gt;$dz$&lt;/span&gt; cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term is the heat flow out of the water through the tube walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last term is the heat flow to the air and the heat flow from the radiator.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-10-10T14:09:48.657" LastActivityDate="2018-10-10T14:09:48.657" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24109" PostTypeId="1" AcceptedAnswerId="24133" CreationDate="2018-10-10T01:56:17.813" Score="1" ViewCount="147" Body="&lt;p&gt;I'm prototyping a hardware that involves many thin walls to bend around curves and such, so it needs to be flexible, cheap, and also transparent. It may also be the same plastic used in a production version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only example I can think of is the plastic you find around the inside of a new dress shirt collar. (McMaster Carr doesn't seem to have anything like it, except polycarbonate, which seems expensive and overkill; surely I'm thinking of a much cheaper plastic. Maybe PP, but not finding enough info to confirm, or find out where to get some in that form.) I almost went with transparency rolls for transparency projectors, but it's too thin/flexible (5 mils).&lt;/p&gt;&#xA;" OwnerUserId="9230" LastEditorUserId="9230" LastEditDate="2018-10-10T13:52:13.600" LastActivityDate="2018-10-12T11:47:06.627" Title="What type of plastic is a good clear, flexible plastic for prototyping? (~0.5mm - 1.5mm thick)" Tags="|materials|plastic|prototyping|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24111" PostTypeId="1" CreationDate="2018-10-10T08:03:47.747" Score="0" ViewCount="44" Body="&lt;p&gt;Let's say for a steel bar, you have the frequency information on each spatial location along the steel bar, could you make a decision that a defect/crack is exist on the specific location without having to inspect the steel bar first?&lt;/p&gt;&#xA;" OwnerUserId="17802" LastActivityDate="2018-10-10T16:43:43.393" Title="Can you make a decision on where is defect if you have the frequency spatially?" Tags="|mechanical-engineering|civil-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24115" PostTypeId="2" ParentId="24111" CreationDate="2018-10-10T16:43:43.393" Score="2" Body="&lt;p&gt;The answer is yes. Having information on natural frequencies and mode shapes, such as you could gather from a series of accelerometers, could allow you to detect defects. However the resolution might not be all that great. I.e. you may be able to detect a large defect but miss a small defect. Depends on what size defect is considered acceptable. If you need the best resolution, there are probably better methods such as ultrasonic. Also mounting a lot accelerometers will be time consuming. Ultrasonic would be much quicker. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-10-10T16:43:43.393" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24117" PostTypeId="2" ParentId="16475" CreationDate="2018-10-10T18:20:46.767" Score="0" Body="&lt;p&gt;In Mobile communication (unlike in satellite communication where source doesn't seems to have much battery (similar to UE in mobile communication)there are 2 devices one is mobile &amp;amp; other is mobile tower here mobile tower is in continue/adequate supply of power while mobile power will goes zero soon so here as per the requirement load can be put on on mobile tower hence high power(some one needs to be more) high frequency higher frequency higher throughput (throughput is secondary issue)&#xA;hence concluded&lt;/p&gt;&#xA;" OwnerUserId="17807" LastActivityDate="2018-10-10T18:20:46.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24118" PostTypeId="1" CreationDate="2018-10-10T21:38:09.903" Score="2" ViewCount="369" Body="&lt;p&gt;I am trying to calculate the maximum torque that a piece of cypress wooden beam can handle before it snaps or becomes structurally unsound. I know this will depend on the thickness and width of the wooden beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know where I can lookup the property value (coefficient or some other value) that will allow me to calculate the max tolerable torque given the thickness and width? I am looking for the value for cypress wood, but if there is a database for other types of woods, that'd be even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, I am trying to design a bathtub bench for sitting that will hold up to 300 pounds. The current plank is 1 inch thick x 3.5 inches wide x 30 inches long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="17808" LastActivityDate="2019-04-09T19:50:57.030" Title="Max torque allowed on cypress wooden beam" Tags="|structural-engineering|materials|torque|wood|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24119" PostTypeId="1" CreationDate="2018-10-10T22:17:26.557" Score="2" ViewCount="109" Body="&lt;p&gt;I have a CAD file in SketchUp that I would like to edit using code (ie: its dimensions, properties, etc.) What is the best way to do this? If I export the file as an STL, is there a program where I can edit the STL using code? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="17809" LastActivityDate="2018-10-10T22:17:26.557" Title="CAD - Using code to edit a CAD file" Tags="|cad|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24120" PostTypeId="1" AcceptedAnswerId="26795" CreationDate="2018-10-11T01:20:29.757" Score="2" ViewCount="1272" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QLYjT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QLYjT.png&quot; alt=&quot;\\this is the qeustion i am facing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, I thought the neutral axis would just be the y bar which can be obtained from the centroid of the cross-section but is different. I have tried letting the net moment at the wall = 0 and the reaction force I got is 1kn * 1m. But I am kinda lost on how to continue.&lt;/p&gt;&#xA;" OwnerUserId="17811" LastActivityDate="2019-04-09T18:46:49.140" Title="Neutral axis in canteliver beam at support" Tags="|structural-analysis|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24121" PostTypeId="2" ParentId="24120" CreationDate="2018-10-11T02:52:26.637" Score="-1" Body="&lt;p&gt;The stress is the superposition of two stresses: tensile due to bending moment and tensile due to 200kN tension.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \sigma_{bending}= m/S =\frac{1kN.1m}{(50\times 100^2)/6} =6kN.1000mm/500000mm^2 =6000N/500mm^2 = 12N/mm^2$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \sigma_{tension}= 200kMN/100^2mm^2 = 20Nmm^2  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So maximum tension is 20 +12= 32Nmm^2 on the upper face of the beam at support.&#xA;And neutral axis can be calculated by finding the slope of stress surface vertically and locating where it turns to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slope of stress surface from the bottom face to top face is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ [(20+12) -(20-12)]/100_{mm height}= 0.24Nm\space$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and Neutral axis is at the height from the bottom face of &lt;span class=&quot;math-container&quot;&gt;$$\frac{8_N}{0.24} =33.33mm \space or\space\frac{H}{3}  $$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-10-11T15:01:49.103" LastActivityDate="2018-10-11T15:01:49.103" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24123" PostTypeId="2" ParentId="14615" CreationDate="2018-10-11T11:39:26.117" Score="0" Body="&lt;p&gt;It is desirable but as far as Software concern, It is not possible. Something went to wrong with &quot;Response Spectrum Function Defination&quot; I.e. Define period 0 - 0.036, 0.1 - 0.09, and so on.&#xA;Also, check Load type is acceleration type or not.&#xA;&lt;img src=&quot;https://i.stack.imgur.com/0uCt1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17816" LastActivityDate="2018-10-11T11:39:26.117" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24125" PostTypeId="2" ParentId="24118" CreationDate="2018-10-11T16:13:17.767" Score="0" Body="&lt;p&gt;You can always try engineering toolbox online, it may take some digging but they would have something to give you an idea. If you are a student you can use Fusion 360 to test various materials, using the simulation environment. It is not exact, but it should have a material you can use for wood. I've used this method to calculate steel pipe strengths. Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="17818" LastActivityDate="2018-10-11T16:13:17.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24126" PostTypeId="2" ParentId="24118" CreationDate="2018-10-11T18:48:15.953" Score="0" Body="&lt;p&gt;The angle of twist of a wood beam in Radians is &#xA;&lt;span class=&quot;math-container&quot;&gt;$$ θ = \frac{TL}{&#xA;GK}\space and\space torsion\space stress\space is f= T/(\beta\times h\times b^2) $$&lt;/span&gt;&#xA;T applied torque, L member length, G shear modulus which could be approximated by E/16, K is cross-section shape factor for a rectangular beam and beta is approximately 3 for your member dimensions, 1 and 3.5 inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the stress is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_{shear} = \frac {T}{\beta\times h\times b^2}  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h is the long side and b is the short side of your member section.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-10-11T18:48:15.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24127" PostTypeId="2" ParentId="23632" CreationDate="2018-10-11T22:24:44.560" Score="0" Body="&lt;p&gt;A Carnot cycle is a theoretical limit. The irreversibility of a real cycle is always expressed in addition to the expression of the Carnot efficiency of the cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this regard, inefficiencies, whether internal (gas interactions) or external (loss of heat to friction) become an additional term of irreversibility. The real efficiency of any cycle is a multiplication of the irreversible term with the Carnot efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \varepsilon_{actual} = \varepsilon_{irr}\ \varepsilon_{Carnot} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A system that is operating at a Carnot efficiency of 0.8 but is only 60% efficient at doing so has an actual efficiency of 48%.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-10-11T22:24:44.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24128" PostTypeId="1" CreationDate="2018-10-12T03:48:18.403" Score="1" ViewCount="199" Body="&lt;p&gt;I am wanting to use an aluminum pole (as a Jin Pole) to lift the mast on my sailboat into place.  My mast is about 30 ft. long and weighs between 80 – 100 lbs.  The aluminum pole would be stood upright and attached to fore and lateral stays to hold it in place vertically.  It would be fitted with a pulley near the top end with a line running down and tied around the mast (near its middle) for lifting.  The other end of the line (free end) would be run downward from the pulley thus lifting upward on the mast when pulled down.  The mast would be lifted up until high enough to set it in the mast holder at its base (about 18 ft up where the lifting line is attached to the mast).  My question is how large (diameter) aluminum pipe/tube would I need to lift the 100 lb. mast vertically if I used 3/16 in. wall thickness for the aluminum pole? (Assume it will be 6061 T6 strength aluminum).  The pole will need to be an overall length of about 19 ft. if this matters.  Distance between the lifting pulley and where the line is attached to the mast to start the lift will be about 18 ft. &lt;/p&gt;&#xA;" OwnerUserId="17822" LastActivityDate="2018-10-15T21:24:13.680" Title="What diameter aluminum pole do I need to lift 100 lbs vertically?" Tags="|structural-analysis|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24129" PostTypeId="1" AcceptedAnswerId="24138" CreationDate="2018-10-12T06:48:32.800" Score="1" ViewCount="110" Body="&lt;p&gt;Consider 2 oddly shaped surfaces and pick a point on each of the surfaces. At the point considered we can always draw a normal to the surface. Now if the two surfaces are in contact at this same point will the normal of one surface always be coincident to the normal of the other surface?. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asking this because, when studying about gears, there is always talk of the common normal and I am trying to understand if the presence of common normal is due to the design of the gear tooth or there is always a common normal between 2 surfaces in contact.&lt;/p&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2018-10-12T19:03:22.440" Title="Intuitive understanding of the common normal at the point of contact of 2 gear teeth" Tags="|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24130" PostTypeId="2" ParentId="24128" CreationDate="2018-10-12T07:09:13.593" Score="1" Body="&lt;p&gt;Assuming the aluminum pole is attached to the mast at 90 degree angle with ropes attached to the end of the mast and front stay, we have a triangle of approximately 60&amp;nbsp;ft base and 18&amp;nbsp;ft height. The vertical load on the pole initially when it is upright is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P = 2\times(100/2) (30/18) = 166.66\text{ lbs}$$&lt;/span&gt;&#xA;Now say as a first guess we assume a 2.5&amp;nbsp;in diameter aluminum pole and check for Euler's critical column load:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$p_{critical}  = \frac {\pi^2EI}{L^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$L =18\text{ ft}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$E = 10000\text{ ksi}$&lt;/span&gt;, and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I = \frac {\pi(D^4- d^4)}{64} = 0.5136\text{ in}^4$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Pc = \frac{\pi^2 (10000000\times 0.5136)}{(12\times 16)^2} = 1068\text{ lbs}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an acceptable answer even though it is much more than 166.7&amp;nbsp;lbs we rough calculated, considering a factor of safety of 3 and the fact that in a windy day with a rocking boat loads can increase by several folds.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-10-15T21:24:13.680" LastActivityDate="2018-10-15T21:24:13.680" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24131" PostTypeId="1" AcceptedAnswerId="24236" CreationDate="2018-10-12T11:08:22.810" Score="1" ViewCount="48" Body="&lt;p&gt;I'm using the Eurocodes for some structural analysis and am looking for the limiting value of deflection of a plate.&#xA;EN 1993-1-7, Section 8.2 states:&#xA;&lt;em&gt;For limiting values of out of plane deflection w see application standard.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure to which application standard this is referring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone point me in the right direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="17139" LastActivityDate="2018-10-19T10:16:48.433" Title="What is the 'application standard' referred to in EN 1993-1-7" Tags="|structural-analysis|structures|standards|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24132" PostTypeId="2" ParentId="24131" CreationDate="2018-10-12T11:33:11.280" Score="3" Body="&lt;p&gt;I think it means the relevant section of EN 1993 for the specific &lt;em&gt;use&lt;/em&gt; (application) of the plated structure. A partial table of contents of EN 1993 includes&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Part 2: Steel Bridges&lt;/li&gt;&#xA;&lt;li&gt;Part 3-1: Towers, masts and chimneys&lt;/li&gt;&#xA;&lt;li&gt;Part 3-2: Towers, masts and chimneys - Chimneys&lt;/li&gt;&#xA;&lt;li&gt;Part 4-1: Silos&lt;/li&gt;&#xA;&lt;li&gt;Part 4-2: Tanks&lt;/li&gt;&#xA;&lt;li&gt;Part 4-3: Pipelines&lt;/li&gt;&#xA;&lt;li&gt;Part 5: Piling&lt;/li&gt;&#xA;&lt;li&gt;Part 6: Crane supporting structures&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-10-12T11:33:11.280" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24133" PostTypeId="2" ParentId="24109" CreationDate="2018-10-12T11:47:06.627" Score="0" Body="&lt;p&gt;Polyester film / PET / PETE, according to McMaster-Carr support, is a clear, flexible, inexpensive plastic, and is the type found around the inside of a new dress shirt collar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's minimum cold bend radius (no heating necessary), at least of certain PET materials, can be approximated from its thickness as R=150T where T is the thickness.&lt;/p&gt;&#xA;" OwnerUserId="9230" LastActivityDate="2018-10-12T11:47:06.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24134" PostTypeId="1" CreationDate="2018-10-12T14:22:47.680" Score="1" ViewCount="37" Body="&lt;p&gt;I am new to MSC NASTRAN and PATRAN.&#xA;I am running a simulation on composite laminate optimization. While creating the design variables for the study, I noticed that there is a parameter called 'Analysis Model Value'. I am unable to understand the physical significance of this term. I searched the documentation provided by MSC but I am still unable to come up with a solution. &lt;/p&gt;&#xA;" OwnerUserId="17824" LastActivityDate="2018-10-12T14:22:47.680" Title="What does 'Analysis Model Value' in NASTRAN optimization mean?" Tags="|structural-analysis|structures|aerospace-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24136" PostTypeId="1" CreationDate="2018-10-12T17:36:59.427" Score="1" ViewCount="552" Body="&lt;p&gt;Which of these poles would be straighter and stronger then the others?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are some poles that are compacted making them possible to send to space. Sending a single solid pole .5km long will not be possible.   &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;STEM ( Storable Tubular Extendible Member), a pre-stressed piece of thin metal that could be rolled up like a carpenters tape measure. On Alouette electric motors would roll out the thin pieces, which would form tubular semi-rigid antenna once they were deployed. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/rJzuh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rJzuh.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/zkIfC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zkIfC.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Seamless gutter maker modified to make a round symmetrical pole.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cGpMr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cGpMr.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://zippermast.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://zippermast.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RRbal.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RRbal.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VSQf7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VSQf7.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Each pole produced would have the same weight per measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What advantages and disadvantages does each pole made from these methods have?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other types of extendable poles I am missing that could be added to this list?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://space.stackexchange.com/questions/30972/how-would-a-long-pole-be-transported-to-space/31037#31037&quot;&gt;How would a long pole be transported to space?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastEditorDisplayName="user4139" LastEditDate="2018-10-24T19:56:17.127" LastActivityDate="2018-10-24T19:56:17.127" Title="What is the straightest, longest and strongest pole that could be launched into space?" Tags="|structural-engineering|materials|control-engineering|aerospace-engineering|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24137" PostTypeId="2" ParentId="23153" CreationDate="2018-10-12T17:45:09.767" Score="1" Body="&lt;p&gt;All real oscillators are unstable. They eventually either expand to the rails as you've demonstrated or decay to zero. So they indeed need a 'fix' to keep the poles on the imaginary axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rectifier - good to obtain a linear feedback signal of 'peak' amplitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integration of the difference between your desired amplitude and the rectified oscillation signal - necessary to reach a steady state accuracy in amplitude but you need to add a proportional component to provide damping. So basically you need a proportional-integral (PI) controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integral gain will adjust your speed of amplitude convergence, the proportional gain the damping; so overshoot and settling oscillation. Check your gain/phase margin for adequate stability.&lt;/p&gt;&#xA;" OwnerUserId="820" LastActivityDate="2018-10-12T17:45:09.767" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24138" PostTypeId="2" ParentId="24129" CreationDate="2018-10-12T17:49:46.863" Score="1" Body="&lt;p&gt;If two surfaces are tangentially in contact meaning they have the same partial derivatives at that point; then they do have a common normal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it is the cross product of the two tangent vectors, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \frac{\partial f}{\partial x} \space and\space  \frac{\partial f}{ \partial y}.$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With f being the surface function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However two surfaces can be in contact intersecting each other or touching at an edge, where they won't have a common normal!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-10-12T19:03:22.440" LastActivityDate="2018-10-12T19:03:22.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24140" PostTypeId="1" CreationDate="2018-10-12T20:05:13.413" Score="1" ViewCount="53" Body="&lt;p&gt;I'm trying to determine all forces/reactions on a disk with a hole in the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disk is loaded with a (known) uniform distributed load from A to C (all around the disk), a concentrated (known) load at D (all around the disk), is either simply supported (only force reactions, &lt;span class=&quot;math-container&quot;&gt;$M_W = 0$&lt;/span&gt;) at A and B or simply supported at A and rigidly supported (force and moment reactions) at B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bellow is a depiction of the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ktWL9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ktWL9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to determine the reactions &lt;span class=&quot;math-container&quot;&gt;$F_N$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$F_W$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$M_W$&lt;/span&gt; (if fixed and not simply supported at B).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how to do that for line problems with a sum of forces and moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this axisymmetric case however, the dum of moments does not see easy to obtain. How could I do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to get the reaction by the virtual work method. The sum of the product of force and virtual displacements needs to equal zero: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sum F \cdot \delta = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm getting however strange results with this method, and I'm probably doing some erros in the derivation of the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would the virtual work equation look for this axisymmetric case? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any resource were a similar problem is solved by this method?&lt;/p&gt;&#xA;" OwnerUserId="6364" LastActivityDate="2018-10-12T20:05:13.413" Title="Virtual work method for a disk with axisymmetric distributed and concentrated loads" Tags="|applied-mechanics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24141" PostTypeId="1" AcceptedAnswerId="24161" CreationDate="2018-10-13T00:16:18.747" Score="2" ViewCount="165" Body="&lt;p&gt;First post here, trying to get a little help with Autodesk Inventor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to make a part that is basically a wheel with little nubs on the end (think of a lego with tabs on the outer circumference). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I tried so far is making a rod and using the Constraint function to make the parts one piece. Is there a more direct way to do this? I can't seem to find a way to just add the tabs/nubs on the edge of the circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YihlA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YihlA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17829" LastEditorUserId="13311" LastEditDate="2018-10-14T19:03:13.513" LastActivityDate="2018-10-15T16:30:47.740" Title="How to add radial protrusions to a disc in Autodesk Inventor" Tags="|cad|autodesk-inventor|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24143" PostTypeId="1" CreationDate="2018-10-13T09:13:27.510" Score="4" ViewCount="2378" Body="&lt;p&gt;For example - I am installing a fold up table from Ikea, it needs to hang from the wall, as shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CEhjx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CEhjx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It explicitly says to use pan head instead of flat screw heads like shown in the picture, I would drill into hollow concrete and will use nylon anchors but my question is, would a washer on a flat screw head be a hack here to do the job of the pan head screw? &lt;/p&gt;&#xA;" OwnerUserId="14322" LastEditorUserId="7083" LastEditDate="2018-10-13T21:37:14.930" LastActivityDate="2018-10-14T11:38:34.653" Title="Can I use a wooden flat screw head with a washer to mimick a pan head screw?" Tags="|bolting|drilling|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24144" PostTypeId="2" ParentId="24143" CreationDate="2018-10-13T09:24:41.530" Score="2" Body="&lt;p&gt;It seems reasonable to believe that the flat head (countersunk) screw would apply enough force to split the wood, hence the restriction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use a flat head screw and the washer prevents the countersunk portion from exerting wedge-type force on the wood, you will have accomplished the objective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider also that if you use a fender washer (large diameter), the compression force is going to be distributed over a wider area, providing more confidence in not damaging the wood brace.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-10-13T09:24:41.530" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24145" PostTypeId="1" CreationDate="2018-10-13T11:46:29.813" Score="3" ViewCount="836" Body="&lt;p&gt;If there are secondary beams running between primary beams, and columns support the primary beams, then is it allowable to have the occasional secondary beam pinned to a column? Otherwise it seems to me that it would be impossible to connect a secondary beam to a primary beam at a location where a column is. For example, in the picture below, it appears as though the third secondary beam from the left is connected directly to the column. In this case, is the primary beam completely unaffected by this beam?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was taught that if a beam is connected to a column then it is no longer a secondary beam. Moreover, would the analysis of a single primary beam not include the concentrated loads of the secondary beams which connect only to the column? And in practice, do structural engineers often do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F7hcK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F7hcK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17832" LastEditorUserId="33" LastEditDate="2018-10-13T13:14:32.720" LastActivityDate="2018-10-13T13:35:54.563" Title="Can secondary beams be supported to columns occaisionally?" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24146" PostTypeId="2" ParentId="24145" CreationDate="2018-10-13T13:35:54.563" Score="1" Body="&lt;h2&gt;Primary Versus Secondary&lt;/h2&gt;&#xA;&lt;p&gt;Think about primary and secondary beams less about where they are connected and more about the size or amount of load that they are taking. (These mean the same thing if you think about it.)&lt;/p&gt;&#xA;&lt;p&gt;A secondary beam only takes load from the deck and a primary beam takes the multiple point loads from the ends of the secondary beams.&lt;/p&gt;&#xA;&lt;p&gt;When you look at the beams in this manner, the fact that the secondary beam needs to physically connect to a column doesn't change its type.&lt;/p&gt;&#xA;&lt;h2&gt;Connections&lt;/h2&gt;&#xA;&lt;p&gt;Beams are designed for the loads that they are expected to carry. If a member is connected to a column, then it isn't connected to a beam. In this case, there is no need to imagine connections where there aren't any.&lt;/p&gt;&#xA;&lt;p&gt;The physical constraints of the beam layout means that the end connection of every secondary beam is not the same. In this situation, there would be many typical connections and a few unique connections. This is very common.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-10-13T13:35:54.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24147" PostTypeId="2" ParentId="24136" CreationDate="2018-10-13T13:56:00.303" Score="1" Body="&lt;p&gt;I don't have direct experience with these types of contraptions (other than using a tape measure).&lt;/p&gt;&#xA;&lt;h2&gt;Straighter&lt;/h2&gt;&#xA;&lt;p&gt;For something like this, straightness would seem to depend symmetry. Any imperfections in the pole could be amplified if they occur in the weak direction (seam) of the pole. In a symmetrical configuration, these would tend to cancel out.&lt;/p&gt;&#xA;&lt;p&gt;Think about a tape measure. It can strongly support itself in one direction, but once it is flipped over, it bends easily. Also, if you have ever experienced a tape measure that has been bent out of shape, it never supports itself as well as before.&lt;/p&gt;&#xA;&lt;p&gt;The third option is the only one with symmetry. It also does not have an open slot that would be a weak point.&lt;/p&gt;&#xA;&lt;h2&gt;Stronger&lt;/h2&gt;&#xA;&lt;p&gt;We have no other information, so I am assuming that the individual size (width and thickness) of each tape is the same. In this case, the only difference between the options is the number of tapes used.&lt;/p&gt;&#xA;&lt;p&gt;Once again, the third option is the best. It uses the most individual pieces.&lt;/p&gt;&#xA;&lt;h3&gt;Result&lt;/h3&gt;&#xA;&lt;p&gt;In the end, if all you are going for is possible strength and straightness, the third option seems to be the best out of the three. There may be other reasons for choosing one over the other, but the third option seems pretty clever.&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-10-13T13:56:00.303" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24148" PostTypeId="1" CreationDate="2018-10-13T14:20:02.040" Score="0" ViewCount="243" Body="&lt;p&gt;Are there any restrictions that would not allow a steering system to cause a vehicle to be able to move sideways - at 90 degrees from the convention direction of travel? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4QmV8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4QmV8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17835" LastEditorUserId="7083" LastEditDate="2018-10-13T20:06:32.783" LastActivityDate="2018-10-14T08:07:53.283" Title="90 degree pivot steering system to allow vehicles to move sideways" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24149" PostTypeId="1" CreationDate="2018-10-13T15:00:17.030" Score="2" ViewCount="764" Body="&lt;p&gt;Helicoil from McMaster-Carr &lt;a href=&quot;https://www.mcmaster.com/catalog/124/3382&quot; rel=&quot;nofollow noreferrer&quot;&gt;18-8 Stainless Steel Helical Inserts 91732A201&lt;/a&gt;  calls out a specific tap to use, but I cannot find out what thread this tap is putting into the hole for the insert.  I want to call out the tapped thread on a drawing, and put the 2-56 insert in later.&#xA;I cannot find what the thread should be.  This would be the outside thread of the helicoil.  I know the drill size is #41 = 0.0960 but what is the thread size?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="15805" LastEditorUserId="7083" LastEditDate="2018-10-13T21:37:45.233" LastActivityDate="2018-10-14T08:21:19.910" Title="What thread to use in tapping a hole for a 2-56 Helicoil insert?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24150" PostTypeId="2" ParentId="5635" CreationDate="2018-10-13T15:18:05.407" Score="1" Body="&lt;p&gt;I think there is no unique technical term to define pressures in different situations.&#xA;As a reference states &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/pflu.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;: &lt;span class=&quot;math-container&quot;&gt;$p_\text{static}=\rho g h$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we consider the data computed from any incompressible CFD solver, the data of pressure term indeed includes the hydrostatic part when gravity &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do prefer to think the fluid pressure (incompressible) in &lt;a href=&quot;https://nptel.ac.in/courses/112104118/lecture-16/16-1a_hydro_static_pressure.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this way&lt;/a&gt; for&#xA;&lt;span class=&quot;math-container&quot;&gt;$p+\frac{\rho v^2}{2}+\rho g z=p_0$&lt;/span&gt;：&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\rho g z$&lt;/span&gt; is the potential at &lt;span class=&quot;math-container&quot;&gt;$z_A$&lt;/span&gt; not the hydrostatic pressure.&#xA;If zero is the reference position, the hydrostatic pressure or pressure due to acceleration should be &lt;span class=&quot;math-container&quot;&gt;$p_{hs}=\rho g (-z)=-\rho g z$&lt;/span&gt;, since gravity has a negative direction.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; raised in the previous equation has two components: hydrodynamic pressure &lt;span class=&quot;math-container&quot;&gt;$p_{hd}$&lt;/span&gt; and hydrostatic pressure &lt;span class=&quot;math-container&quot;&gt;$p_{hs}$&lt;/span&gt;, which is to say &lt;span class=&quot;math-container&quot;&gt;$p=p_{hd}+p_{hs}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Hence, it is quite obvious that static pressure &lt;span class=&quot;math-container&quot;&gt;$p_{s}=p_{hd}=p_0-\frac{\rho v^2}{2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Dynamic pressure &lt;span class=&quot;math-container&quot;&gt;$p_{d}=\frac{\rho v^2}{2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$p_s+p_d=p_0$&lt;/span&gt;(total pressure)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As for the common situation, pressure &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the static pressure &lt;span class=&quot;math-container&quot;&gt;$p_s$&lt;/span&gt; (hydrodynamic pressure &lt;span class=&quot;math-container&quot;&gt;$p_{hd}$&lt;/span&gt;) plus hydrostatic pressure &lt;span class=&quot;math-container&quot;&gt;$p_{hs}$&lt;/span&gt; (or pressures due to inertial force). Excluding the rest balance, &lt;span class=&quot;math-container&quot;&gt;$p_s$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$p_{hd}$&lt;/span&gt; is the real reason to make flow move (to balance dynamic pressure &lt;span class=&quot;math-container&quot;&gt;$p_d$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="17834" LastActivityDate="2018-10-13T15:18:05.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24151" PostTypeId="1" CreationDate="2018-10-13T15:18:30.063" Score="4" ViewCount="76" Body="&lt;p&gt;I am preparing a classroom demonstration about heat, including a demo of how a satellite loses heat only radiatively, since it's flying through a vacuum. I'd like to make the vacuum-production apparatus very simple and cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kind of pump used to semi-evacuate wine bottles to help preserve the flavor (supposedly) is the right kind of item in terms of cost (as cheap as $6); a slight modification of one of these improved the seal on its plunger dramatically, though I still doubt the pressure in the vacuum chamber gets much below 1/4th bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To enhance the vacuum, I'm venting steam from a kettle into the vacuum chamber (a cheap plastic jar with the wine-pump check valve attached through a hole in the lid), screwing the lid back on, pumping out as much of the hot water vapor as possible, then cooling the jar in the freezer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if this really reduces the heat conductivity of the remaining gases (mainly, water vapor) very significantly. Reducing the density of a gas doesn't really reduce the heat conductivity very dramatically, and without such a reduction, the point of the exercise (simulating vacuum) is lost. I can measure the temperature easily enough, and I will get a vacuum gauge to measure the pressure, but it would be good to work from an approximate formula based on those two measures to determine the heat conductivity. I assume there will be little but water vapor in the jar when I put on the lid. I can find tables for water vapor heat conductivity above freezing, but at the temperatures and pressures where H2O only sublimates, I've found nothing so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's a good approximate formula to use? Bear in mind that I'd also like to use not only regular water ice but also dry ice (packed around the jar) as the background for the satellite-simulating heat sensor to radiate against, so it would be best if the formula is a good approximation at dry-ice temperatures too. (Cosmic background is obviously not achievable cheaply, but the effect of even greater radiative cooling can reasonably be extrapolated from various other cooling approaches, such as burying the jar with water ice and dry ice.)&lt;/p&gt;&#xA;" OwnerUserId="17836" LastActivityDate="2018-10-15T13:40:51.053" Title="Easily compute a good approximation of heat conductivity of cooled semi-vacuum (water vapor + air)" Tags="|gas|temperature|pressure-vessel|vacuum|thermal-conduction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24152" PostTypeId="2" ParentId="24143" CreationDate="2018-10-13T18:31:07.180" Score="2" Body="&lt;p&gt;If you use a washer under the head of the flat head wood screw, get you a &lt;em&gt;cup washer&lt;/em&gt;. these are intended to let you use flat head screws as if they were pan heads. They hold the  wedge profile of the flat head screw up and out of contact with the wood so it will not split the wood as you screw it in, and they hide the sharp rim of the screw head so it will not cut your hands or get snagged on things. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-10-13T18:31:07.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24153" PostTypeId="2" ParentId="24149" CreationDate="2018-10-13T19:30:48.233" Score="4" Body="&lt;p&gt;Helicoil taps are what's known as &quot;STI&quot; or &quot;Screw Thread Insert&quot; Taps. They have the same pitch as the fastener that they're going to be used with, but a larger outer diameter, to account for the wire going into the space between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my personal experience, the hole callout on a drawing should indicate the size of the fastener to be inserted into that hole, for example: &quot;M10x1.5 STI&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This matches the way that the taps themselves are labelled and &lt;a href=&quot;https://www.cromwell.co.uk/shop/cutting-tools/thread-repair/helicoil-metric-hand-taps%3A-bottoming/f/13447?query=helicoil%20taps&quot; rel=&quot;nofollow noreferrer&quot;&gt;sold&lt;/a&gt;, with the fastener thread indicated.&#xA; &lt;a href=&quot;https://i.stack.imgur.com/YiPln.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YiPln.png&quot; alt=&quot;image of coil forming tap&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-10-14T08:21:19.910" LastActivityDate="2018-10-14T08:21:19.910" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24154" PostTypeId="2" ParentId="24151" CreationDate="2018-10-13T21:22:41.820" Score="5" Body="&lt;p&gt;In the molecular flow region of pressure, the thermal conductivity of an ideal, monatomic gas is obtained by this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;k = \frac{1}{\pi^{3/2} d^2}\sqrt{k_B^3T/m}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is the collision or molecular diameter and m is the molar mass divided by Avogadro's number. It is independent of pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximations have been derived for non-monatomic gases based on collision cross sections and collision integrals. In a nutshell, thermal conductivity is still independent of pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the mean free path of the gas reaches the diameter of the container, the gas crosses to &lt;a href=&quot;https://en.wikipedia.org/wiki/Knudsen_flow&quot; rel=&quot;nofollow noreferrer&quot;&gt;Knudsen flow&lt;/a&gt;. The mean free path is inversely proportional to pressure as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\lambda = \frac{k_B T}{\sqrt{2} \pi \sigma^2 p}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\sigma^2$&lt;/span&gt; is the collision cross sectional area (taken to first order as roughly the same as &lt;span class=&quot;math-container&quot;&gt;$d^2$&lt;/span&gt; in some cases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we reach the point where &lt;span class=&quot;math-container&quot;&gt;$\lambda &amp;gt; d_{container}$&lt;/span&gt;, we can assume the thermal conductivity of a gas in the container is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a water molecule at 0.275 nm diameter in a container with 6 cm diameter at 25 &lt;span class=&quot;math-container&quot;&gt;$^\mathrm{o}$&lt;/span&gt;C. The pressure needed to cross to Knudsen flow is 0.20 Pa. At or below this pressure should do what you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the walls of the container are below the sublimation temperature of water vapor at the given partial pressure, the walls will be coated with ice. The sublimation temperature of pure water vapor at 0.20 Pa can be &lt;a href=&quot;https://physics.stackexchange.com/questions/60170/freezing-point-of-water-with-respect-to-pressure&quot;&gt;obtained from the picture in this post&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider also a tube with radius &lt;span class=&quot;math-container&quot;&gt;$r_t$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$T_t$&lt;/span&gt; containing a probe (with zero thickness) at its center-line at &lt;span class=&quot;math-container&quot;&gt;$T_p$&lt;/span&gt;. Assuming black-body radiation for the tube and probe, the ratio of radiation to conduction in the system to first order is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;R_{rc} = \frac{\sigma\left(T_p^4 - T_t^4\right)}{k \left(T_p - T_t\right) / r_t}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A computer application (python) could be used to plot this ratio as a function of the input parameters &lt;span class=&quot;math-container&quot;&gt;$(T_t, T_p, r_t)$&lt;/span&gt; to appreciate the behavior when the tube is not evacuated (to have &lt;span class=&quot;math-container&quot;&gt;$k \rightarrow 0$&lt;/span&gt;).&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-10-15T13:40:51.053" LastActivityDate="2018-10-15T13:40:51.053" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24155" PostTypeId="2" ParentId="17803" CreationDate="2018-10-13T22:26:11.203" Score="0" Body="&lt;p&gt;As @user307640 said, if you use a mass basis, you can obtain the average molecular weight using the mass percentage composition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an alternative approach, you can do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\begin{align}&#xA;M&amp;amp;=\sum n_iMW_i\\&#xA;&amp;amp;=n_T\sum y_iMW_i&#xA;\end{align}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is total mass, &lt;span class=&quot;math-container&quot;&gt;$n_T$&lt;/span&gt; is total number of moles, &lt;span class=&quot;math-container&quot;&gt;$n_i$&lt;/span&gt; is the number of moles of species &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$y_i$&lt;/span&gt; is the mole fraction of species &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$MW_i$&lt;/span&gt; is the molecular weight of species &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$n_T = M/MW_{avg}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\begin{align}&#xA;M&amp;amp;=\frac{M}{MW_{avg}}\sum y_iMW_i\\&#xA;1&amp;amp;=\frac{1}{MW_{avg}}\sum y_iMW_i\\&#xA;MW_{avg}&amp;amp;=\sum y_iMW_i&#xA;\end{align}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is 29.2 as you pointed out previously, the average molecular weight doesn't depend on the mass basis used, which is perfectly correct.&lt;/p&gt;&#xA;" OwnerUserId="10955" LastActivityDate="2018-10-13T22:26:11.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24156" PostTypeId="2" ParentId="24148" CreationDate="2018-10-14T08:07:53.283" Score="1" Body="&lt;p&gt;Technically yes you can design it here is an &lt;a href=&quot;https://www.youtube.com/watch?v=c_1Sg1dJ7i8&quot; rel=&quot;nofollow noreferrer&quot;&gt;example&lt;/a&gt;, however designers have to develop a locking system to prevent the back wheels while the car moves in a curve otherwise the car starts spinning around itself. The only advantage here is easy parking but it makes the whole system very complexe for engineers and mechanics, think about the extra weight.  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-10-14T08:07:53.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24157" PostTypeId="1" AcceptedAnswerId="24158" CreationDate="2018-10-14T09:07:20.637" Score="1" ViewCount="903" Body="&lt;p&gt;The surface area of the head of a tunnel boring machine is usually flat. A cone-shaped head would increase the surface area. The question is if it could speed up the boring process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that with current machines the boring process is also limited by other factors. But that is not part of the question.&lt;/p&gt;&#xA;" OwnerUserId="17839" LastActivityDate="2018-10-14T10:44:38.723" Title="Why are tunnel boring machines not using cone-shaped drills?" Tags="|tunnels|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24158" PostTypeId="2" ParentId="24157" CreationDate="2018-10-14T10:44:38.723" Score="5" Body="&lt;p&gt;Because with a flat head the waste material can be (relatively easily) collected and removed from the cutting face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a cone-shaped bit that removal is not so easily accomplished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cone-shaped bits tend to be used to push or compress the material out of the way, which is fine for &quot;softer&quot; materials ...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-10-14T10:44:38.723" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24159" PostTypeId="2" ParentId="24143" CreationDate="2018-10-14T11:38:34.653" Score="2" Body="&lt;p&gt;I assume that the 'leg' is tubular steel with pre-drilled holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with using countersunk screws  is twofold, firstly  that countersunk they will only make contact with the very edge of the hole and are likely to deform it over time and effectively become less tight. In this sort of situation you want the fixings to be providing a clamping force rather than being loaded in shear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue is that the screw heads won't sit flush with the surface which doesn't look very neat and can leave a sharp edge protruding which can snag on things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using plain washers may help with the first issue, at least to a degree but not the second. You can get cup washers which are have a recess to allow countersunk screws to sit on a flat surface &lt;a href=&quot;https://i.stack.imgur.com/Ns0wn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ns0wn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; but then again anywhere which sells cup washers is also likely to have flange or pan head screws whcih is what you really want.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-10-14T11:38:34.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24160" PostTypeId="1" CreationDate="2018-10-14T15:44:42.813" Score="1" ViewCount="93" Body="&lt;p&gt;I am not sure if this question is related to this community. But nevertheless. I work on CVT engines, I had a new idea of modifying my controller during a slope or a gradient on the road so that the Vehicle speed does not take a hit.  I have a radar attached on the vehicle, which gives me information about an upcoming slope. I.e., at what distance is the slope (meters) and percentage of the slope (100% being a 90deg wall).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before I continue, I need to mention that this entire discussion is in the Cruise Control mode. That is, the controller maintains the vehicle speed irrespective of the engine RPM. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first approach I did was slowly increasing the engine RPM when the slope approaches so the vehicle speed does not take a hit. Below is the simulation I have done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ky26S.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ky26S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The red 1 is where I simulate 5% slope and the speed drops by around 10%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The red 2 is where I simulate 5% slope again but before the slope, I slowly increase the Engine RPM so the vehicle speed drops only by 4%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, I do conclude that increasing the Engine rpm is a good method. But I do it without any proper math behind it (how fast or how slowly I should do it). I want to implement it with a proper logic and math included. If anyone can help me with the proper math or guide me in a proper direction, I'd be grateful&lt;/p&gt;&#xA;" OwnerUserId="17849" LastEditorUserId="17849" LastEditDate="2018-10-14T15:52:05.733" LastActivityDate="2018-10-14T15:52:05.733" Title="Maintaining constant speed of a CVT engine during a slope" Tags="|automotive-engineering|measurements|matlab|distance-measurement|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24161" PostTypeId="2" ParentId="24141" CreationDate="2018-10-14T17:39:08.393" Score="4" Body="&lt;p&gt;The way I would model the component shown on the LHS of your image is like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://media.giphy.com/media/1yTg8TIe7D2Nfx5oze/giphy.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://media.giphy.com/media/1yTg8TIe7D2Nfx5oze/giphy.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sketch the outer diameter of the circle on one of the home planes, say XZ, with the centre of the circle at the origin.&lt;/li&gt;&#xA;&lt;li&gt;Extrude the circle an equal distance in each direction, such that the XZ plane runs through it's centre&lt;/li&gt;&#xA;&lt;li&gt;Sketch half of a 'nub' on the same plane, with a centreline along the X axis&lt;/li&gt;&#xA;&lt;li&gt;Revolve the 'master nub' by 360degrees around the centreline in the sketch (coincident with the X axis)&lt;/li&gt;&#xA;&lt;li&gt;Use the circular pattern tool to make 4 copies of this nub, equally spaced, around the Y Axis.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The reason I suggest this approach is that it ensure that the nubs are all identical to one another, and all equally spaced and pointing radially out from the centre of the circle. IF you want to edit the form of the nubs, there's a single sketch to edit, and all other nubs will auto-update.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using relations to control the position of things is dangerous, as they rely on the internal ID's of selected faces etc, and these are notorious for getting 'lost' as a result of model changes elsewhere in your workflow.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-10-15T16:30:47.740" LastActivityDate="2018-10-15T16:30:47.740" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24162" PostTypeId="1" AcceptedAnswerId="24163" CreationDate="2018-10-14T18:45:07.577" Score="0" ViewCount="273" Body="&lt;p&gt;I am looking to solve the 1D Unsteady Homentropic Flow equations using the Method of Characteristics (MOC). I am using Zucrow and Hoffman's Gas Dynamics Volume 1 as a reference and I am currently stuck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In chapter 13, the authors begin describing the use of MOC to solve the 1D unsteady flow equations. I understand their methodology till  I get to the step that states to substitute equation (13.16) back to equation (13.15).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this example:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;u = u(x,t) &#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;p = p(x,t)&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For homentropic flow, &lt;span class=&quot;math-container&quot;&gt;$$ \rho = \rho(p) $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation (13.15) is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;(\rho u \sigma_1 + \rho a^2 \sigma_2) ( u_x + \frac{\rho \sigma_1}{\rho u \sigma_1 + \rho a^2 \sigma_2}u_t) + (\sigma_1 + u\sigma_2 )(p_x + \frac{\sigma_2}{\sigma_1 + u\sigma_2} p_t) = 0&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equation (13.16) is the following:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{du}{dx} = u_x + \lambda u_t&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{dp}{dx} = p_x + \lambda p_t &#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting equation (13.18) when you substitute (13.16) into (13.15) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;(\rho u \sigma_1 + \rho a^2 \sigma_2) du + (\sigma_1 + u\sigma_2 ) dp = 0&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\lambda = \frac{\sigma_1}{u\sigma_1 + a^2\sigma_2} = \frac{\sigma_2}{\sigma_1+u\sigma_2}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have not been able to get (13.18) from back-substituting (13.16) into (13.15).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any suggestions for me to consider, look into, etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="10703" LastActivityDate="2018-10-14T19:01:45.383" Title="Solving 1D Unsteady Homentropic Flow using Method of Characteristics" Tags="|fluid-mechanics|compressible-flow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24163" PostTypeId="2" ParentId="24162" CreationDate="2018-10-14T19:01:45.383" Score="0" Body="&lt;p&gt;In (13.15) use the first equation for &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; in the first term, and the second equation in the second term. That gives&#xA;&lt;span class=&quot;math-container&quot;&gt;$$(\rho u \sigma_1 + \rho a^2 \sigma_2) \frac{du}{dx} + (\sigma_1 + u\sigma_2 )\frac{dp}{dx} = 0.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use the fact that &lt;span class=&quot;math-container&quot;&gt;$$\frac{du}{dx} \Bigm / \frac{dp}{dx} = \frac{du}{dp}.$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-10-14T19:01:45.383" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24164" PostTypeId="2" ParentId="18714" CreationDate="2018-10-14T19:22:18.967" Score="0" Body="&lt;p&gt;To elaborate on Narasimham's answer a bit, assuming a four-wheeled vehicle with fixed rear wheels, the rotation axis (which is normal to the plane on which the vehicle is moving) intersects a point (the center of rotation) on the axle line of the rear wheels. The front wheels must be rotated such that each of their rotation axes intersect the center of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an image from &lt;a href=&quot;https://en.wikipedia.org/wiki/Ackermann_steering_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia's article on Ackermann steering geometry&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xDOIp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xDOIp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another image (&lt;a href=&quot;https://mrl.nyu.edu/~dzorin/cg05/assignments/car.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zOCw2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zOCw2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the two front wheels are rotated by different angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Steering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on steering in general&lt;/a&gt; may also be of interest.&lt;/p&gt;&#xA;" OwnerUserId="11066" LastEditorUserId="11066" LastEditDate="2018-10-14T21:18:43.947" LastActivityDate="2018-10-14T21:18:43.947" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24165" PostTypeId="1" CreationDate="2018-10-14T19:54:17.900" Score="0" ViewCount="1447" Body="&lt;p&gt;Creating a drawing for a part that will be CNC milled.&#xA;Should all of my dimensions come from a baseline on the part, like an edge or is it ok to use Centerlines?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See yellow highlighted dimensions below for example.  1.775 is from the centerline of the part.  Would it be better to dimension it from the edge of the material as shown with the red arrow?  Same for the other yellow highlight, 0.100 X 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any advice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9M3cV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9M3cV.jpg&quot; alt=&quot;Example drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15805" LastActivityDate="2018-10-15T18:46:31.683" Title="Dimensions for CNC machining" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24167" PostTypeId="2" ParentId="24165" CreationDate="2018-10-14T21:00:20.710" Score="-1" Body="&lt;p&gt;It doesn't matter where you draw the dimension lines for humans to read. The software knows where everything is, otherwise it wouldn't be able to draw it on the computer screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CNC machine is actually going to move from one point to another following the order in which it machines the various features, and you shouldn't be concerned about that level of detail while you are designing and drawing the part. Whether the machine's software will work out the best (i.e. quickest) order automatically, or whether it will need a bit of human guidance to avoid trying to do impossible things, is something to think about after you have reached the final design, not before.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-10-14T21:00:20.710" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24168" PostTypeId="2" ParentId="24165" CreationDate="2018-10-14T21:35:03.027" Score="3" Body="&lt;p&gt;When you are designing you use the dimensions that make sense for designing. So if the important dimension is from the centerline then measure from the centerline. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is generally better if you measure something you can easily verify. So in many cases you wouldn't measure to the centerline but rather to the mirror entity, a bit like prefering to measure diameter rather than radius. This has some consequences for the design though, which may or may not be what you need.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be aware that your design may need to be flipped which may cause some tolerance issues. Or did you plan for the manufacturer to countersink manually? Also hard corner inner holes are a bit hard to make with the mill, so round the corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* So measuring half the distance may in some cases double up distances your general tolerance. But that may or may not come to play check your relevant standard. Remember your document is also your acceptance criteria in cases where something goes wrong.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-10-14T21:41:33.133" LastActivityDate="2018-10-14T21:41:33.133" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24169" PostTypeId="1" CreationDate="2018-10-14T22:30:24.193" Score="0" ViewCount="101" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ma7f3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ma7f3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are cross sections of identical shipping containers, except one is rotated 45 degrees. The cross sections are squares, 3 meters to a side. I drew dashed lines to show where they're suspended by a crane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also I put a cross beam inside of them, connected at the corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In general, which orientation can hold more weight? Does it matter?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you need more specifications, I can make them up. But the idea is that they're the same thing, just rotated 45 degrees, so I didn't think the exact width of the walls and beams mattered so much.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose here is to make a very heavy-duty shipping container than can hold bulk ores, which can be very dense. (Imagine lead ore, granite, etc.) That's why I put a cross beam, to help the container stay rigid. So I want to know the most efficient way of orienting the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt;, I don't know what tags are good for this question. I looked for something along the lines of &quot;abstract&quot;. Feel free to help me with tags.&lt;/p&gt;&#xA;" OwnerUserId="3510" LastActivityDate="2018-10-14T23:50:11.587" Title="Which container orientation, if any, can hold more weight?" Tags="|statics|beam|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24170" PostTypeId="2" ParentId="24169" CreationDate="2018-10-14T23:50:11.587" Score="2" Body="&lt;p&gt;Assuming you don't need to be concerned about crane operation and packing, the diamond shape is better, or ideally a geometry that would look like a tear drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tear drop container, and to some extend the diamond shape in your image, has it's bottom surface similar the shape of a plane deflected to the limit under stress, thus undergoing mostly tension, which is the most effective and economical application of steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top half of diamond container is roughly along the angle of repose of most granular materials, hence minimum stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the square cross section can be easily stacked up on a ship deck up to dozens of tiers and is much easier to handle with a crane rig in packs of two or even four. It also has lower center of mass and overall height, easier for transportation and fitting into train cars and old railroad tunnels..&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-10-14T23:50:11.587" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24171" PostTypeId="1" CreationDate="2018-10-15T01:38:29.743" Score="1" ViewCount="937" Body="&lt;p&gt;I'm trying to mesh a model of a morphing wing in Abaqus and have been running into issues. It was recommended that I partition the model into smaller parts, but it seems that the more I try to partition, the longer it takes. After partitioning into about 5 or more (small) pieces, it takes about 5 minutes or longer for each additional part. Is there a better way to go about this? I've posted a screenshot of my model so far.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XXCVQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XXCVQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17854" LastActivityDate="2024-03-18T12:04:40.717" Title="Meshing Complex models in abaqus" Tags="|structural-engineering|aerospace-engineering|finite-element-method|abaqus|meshing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24172" PostTypeId="1" CreationDate="2018-10-15T07:23:59.670" Score="1" ViewCount="44" Body="&lt;p&gt;I have the following problem: there is a copper pipe getting through a wooden wall. On one side of the wall (hot side), I have elevated temperature which is defined by radiation and convection. On the cold side I have ambient temperature, again defined by radiation and convection (picture in the attachment). It should resemble the experiment done in oven.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, between the wooden wall and the copper pipe there is a little gap which (in reality) lets some of the hot gasses to pass from one side to another. So far, the temperatures on the cold side are much lower comparing to the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to simulate this stream of hot gasses getting though the gap around the pipe and how to do it? Or if I am doing something completely wrong with defining the parameters? Is this even possible to do with Workbench only?&lt;a href=&quot;https://i.stack.imgur.com/5xQff.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5xQff.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12057" LastActivityDate="2018-10-15T07:23:59.670" Title="Flow of hot gasses around the pipe" Tags="|thermal-conduction|thermal-insulation|ansys-workbench|thermal-radiation|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24173" PostTypeId="1" CreationDate="2018-10-15T10:38:59.097" Score="0" ViewCount="44" Body="&lt;p&gt;How can I build a v shaped pathway for drones? It would be a sort of racing path and there would be multiple drones flying over the pathway at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The race would be a relay race. So there are multiple path breakpoints and each drone would have to carry a small item passed onto it while racing over the path. We have to protect the item at all cost and also be really fast at the same time in order to win the race.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of the path is &quot;V&quot; because I would like to build a v shaped inflatable structure which would protect the drones if they fall off onto the path while racing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 more point: since a drone racing envrionment would take quite a lot of space, all of this would be built in a small hilly area where it would be quite windy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what material can I use to make the drone racing pathway? What material to use for the inflatable v shape pathway? What to use so that it resists wind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other mechanism can be suggested for the pathway keeping in mind 2 important factors&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Protection of the item&lt;/li&gt;&#xA;&lt;li&gt;Speed&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="17858" LastEditorUserId="17858" LastEditDate="2018-10-16T03:05:39.300" LastActivityDate="2018-10-16T03:05:39.300" Title="Pathway for drone racing" Tags="|civil-engineering|rf-electronics|consumer-electronics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24175" PostTypeId="2" ParentId="19851" CreationDate="2018-10-15T13:43:37.540" Score="0" Body="&lt;p&gt;Not being an engineer, I'd run whatever setup fits the confines of the car and wire in a pulse width modulator.  You can tune the output of the electric motor to fit your application.  Not very pretty, but it will work and PWMs are about &lt;span class=&quot;math-container&quot;&gt;$10-$&lt;/span&gt;15 on ebay.&lt;/p&gt;&#xA;" OwnerUserId="17863" LastActivityDate="2018-10-15T13:43:37.540" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24176" PostTypeId="2" ParentId="24173" CreationDate="2018-10-15T15:58:35.767" Score="1" Body="&lt;p&gt;you can use plastic light gauge garden netting. They come in typical sizes of 50ft by 50ft at approximately under 10cents/sqft.&#xA;They can be secured by thin cords to stakes or anchors.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-10-15T15:58:35.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24177" PostTypeId="1" CreationDate="2018-10-15T17:01:09.770" Score="2" ViewCount="70" Body="&lt;p&gt;I'm working on a project concerning electricity production using biomass, and we have seen that making use of cogeneration is very attractive in order to increase the efficiency of the installation. This would be a smallish power plant, of around &lt;span class=&quot;math-container&quot;&gt;$1 \ \mathrm{MW}_e$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, we are exploring an installation that is next to an industrial plant that needs both hot water and steam. Most of the designs I've seen for a steam turbine cogeneration power plants use an open loop, where part of the turbine exit steam is feed to the installation, and feedwater has to be added in order to replenish the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems to me that using a closed loop system would be very beneficial. In this case, we were thinking of having a steam-steam condenser that heats up the steam for the industrial plant (and condenses the process fluid) followed by a water-water heat exchanger that heats up the water for the industrial system. However, I'm not very well verse in heat exchangers. Does a design like this make sense?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We read about Organic Rankine Cycles, and it seems like a good solution in order to implement closed systems; however, if it is plausible, we prefer to use water/steam as the working fluid.&lt;/p&gt;&#xA;" OwnerUserId="8973" LastActivityDate="2018-10-15T17:01:09.770" Title="Does a closed loop biomass-fired cogeneration/CHP make sense?" Tags="|thermodynamics|heat-transfer|heat-exchanger|industrial-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24178" PostTypeId="2" ParentId="24165" CreationDate="2018-10-15T18:46:31.683" Score="0" Body="&lt;p&gt;The correct answer is: it depends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to have a bolt-hole line up perfectly with some other part referenced to the same baseline, then you must reference the baseline.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, your sample part appears to be designed to mount a smaller part to a square set of four holes.  In this case, you start by referencing two holes' distance, say holes 1&amp;amp;2 along one side,   to the baseline to place the part correctly.  The distance between the two holes (in the orthogonal axis) are referenced to the first hole's centerline.  Similarly, set the distance from the orthogonal baseline to holes 1 and 4, along the next side, to the corresponding baselines in the same manner.  The other holes references all deal with  the centerlines thus produced,  so that there is the minimum tolerance error for all four holes relative to the mounted part's boltholes (which presumably are toleranced in its own drawing). &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-10-15T18:46:31.683" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24180" PostTypeId="1" AcceptedAnswerId="24205" CreationDate="2018-10-15T19:29:08.827" Score="2" ViewCount="3293" Body="&lt;p&gt;Actually, I see some papers use viscoelasticity or hyperelastic model for the same materials. It seems to me that those models were invented in different eras, and the hyperelastic is a newer model than viscoelasticity. Am I right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the difference?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the best way to decide which model to use when we want to simulate a material?&lt;/p&gt;&#xA;" OwnerUserId="17605" LastEditorUserId="17605" LastEditDate="2018-10-16T14:49:32.077" LastActivityDate="2023-09-04T09:50:42.713" Title="Viscoelasticity and Hyperelastic model, history and difference" Tags="|materials|finite-element-method|modeling|elastic-modulus|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24183" PostTypeId="2" ParentId="24171" CreationDate="2018-10-15T22:51:03.080" Score="0" Body="&lt;p&gt;An efficient way of dealing with the issue is to break the analysis into two parts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Do an analysis for the major parts of the structure without considering the details of the geometries of the attachment mechanisms, e.g., hook/hole at the left-bottom of the image you have provided.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Take the boundary conditions from the first simulation and apply them for a second substructure analysis that considers the details.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For thin structures, if the geometry is too complex for shell approximations, you should try to use as many hexahedral elements as you can with a few tetrahedral/pyramidal elements in regions where hex meshing becomes impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the partitioning of the geometry, if needed, should be carried out such that each region is simply connected (i.e., no holes).  That can be done by creating partitioning surfaces through the middle of the holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several other possibilities depending on exactly what you want out of the analysis, e.g., just static, transient/modal analysis,explicit dynamics, linear elastic-plastic fracture/X-FEM/cohesive fracture, multiphysics, etc. &lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-10-15T22:51:03.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24184" PostTypeId="2" ParentId="24180" CreationDate="2018-10-15T23:02:18.930" Score="3" Body="&lt;p&gt;Let's look at some (very rough) definitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Viscoelasticity = elastic behavior that changes if the rate of application of the load is changed or, if the load is kept fixed, the change in the elastic behavior of a material over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Hyperelasticity = rate/time-independent elastic behavior beyond linear elasticity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, elastic behavior means that once the load is removed the material returns to its original state (given enough time for viscoelastic materials or instantaneously for hyperelastic materials).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, when engineers talk about viscoelasticity they mean &quot;linear&quot; viscoelasticity, i.e, the material is linear elastic even though the elastic moduli may depend on the rate of loading or on time.  You can also have hyperelastic viscoelastic materials which are more commonly known as nonlinear viscoelastic materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyperelastic materials (such as St. Venant-Kirchhoff materials) and linear viscoelastic materials were both initially explored in the mid-1800s.  So they are approximately the same age.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of material model depends on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The amount of strain you expect: large strains -&gt; hyperelastic, small strains -&gt; linear elastic&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) The effect of loading rate: rate-dependent-&gt; viscoelastic, rate-independent -&gt; elastic (linear or hyper)&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-10-15T23:02:18.930" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24186" PostTypeId="1" AcceptedAnswerId="24189" CreationDate="2018-10-16T12:20:34.677" Score="0" ViewCount="332" Body="&lt;p&gt;I'm currently trying to run my solutions in batch mode because in the future I'll be running simulations which I want to be done in the background.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to adapt the code from here: &lt;a href=&quot;https://www.sharcnet.ca/Software/Fluent6/html/ug/node22.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sharcnet.ca/Software/Fluent6/html/ug/node22.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using some additional commands from here: &lt;a href=&quot;https://www.sharcnet.ca/Software/Ansys/16.2.3/en-us/help/flu_gs/flu_ug_sec_startup_option.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sharcnet.ca/Software/Ansys/16.2.3/en-us/help/flu_gs/flu_ug_sec_startup_option.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I tried to do was:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;fluent 3d -gr -t36 &amp;lt; inputfile &amp;gt;&amp;amp; outputfile &amp;amp;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where the inputfile has no extension and has the following content:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;; Read case file&#xA;rc CoolingChannelNoBatWall3D.cas&#xA;; Initialize the solution&#xA;/solve/initialize/initialize-flow&#xA;; Calculate 500 iterations&#xA;it 500&#xA;; Write case &amp;amp; data file&#xA;wcd CoolingChannelNoBatWall3DRes.cas&#xA;; Exit FLUENT&#xA;exit&#xA;yes&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This starts the GUI, even though I tried to suppress it. After starting the GUI it does nothing. What am I doing wrong here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;fluent 3d -t 36 -g -i &amp;lt; inputfile.jou &amp;gt;&amp;amp; outputfile.jou &amp;amp;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Does not start the gui, but it's also quite hard to see if anything's really happening.&lt;/p&gt;&#xA;" OwnerUserId="17876" LastEditorUserId="17876" LastEditDate="2018-10-16T12:46:09.970" LastActivityDate="2018-10-16T12:52:53.563" Title="Solving parallel without GUI in batch mode not working as expected (Linux)" Tags="|fluent|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24187" PostTypeId="1" CreationDate="2018-10-16T12:24:29.380" Score="-1" ViewCount="96" Body="&lt;p&gt;Consider following equations for a driving wheel in a car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ J\dot{\omega} = T - F_x R - F_r R \\&#xA;m\dot{v} = F_x + F_r $$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$F_x$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$F_r$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; are wheel inertia, angular velocity, torque input, traction force, tire radius, rolling resistance, mass and translation velocity respectively. Now assume that the vehicle (or the wheel) is in equilibrium state then both derivative should be zero. As it turns out you will get:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ 0 = T $$&lt;/span&gt;&#xA;because:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ 0 = F_x + F_r$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean that if the vehicle is in equilibrium, even though you may press the pedal gas you torque is still zero? Imagine that you are riding your car on a slippery surface and you still press the pedal, it will of course transfer torque to the wheel. But the translation velocity will remain constant. Then the second diff equation is zero but not necessarily the first one. And you can have same reasoning on the first diff equation. So despite the fact that &quot;equilibrium is when no changes are happening&quot; what does equilibrium mean in this context? Does equilibrium in first equation mean that &lt;span class=&quot;math-container&quot;&gt;$T = 0$&lt;/span&gt;? That if you don't press gas pedal it leads to equilibrium state?&lt;/p&gt;&#xA;" OwnerUserId="17875" LastEditorUserId="1832" LastEditDate="2018-10-16T12:55:09.200" LastActivityDate="2018-10-16T13:53:46.543" Title="In equilibrium is engine torque zero? Vehicle dynamics" Tags="|mechanical-engineering|dynamics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24188" PostTypeId="2" ParentId="24187" CreationDate="2018-10-16T12:51:52.583" Score="2" Body="&lt;p&gt;I'm going to present a simplification, As I think this is a case of maths blinding. In Engineering, the intuitive approach is less prone to error in many cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is the air... The air is not in equilibrium - air molecules are being smacked out of the way by the car body and doing this takes effort. If the car is traveling at a constant speed, such that it's not accelerating, and the forwards thrust from the engine equals the drag force from the air, then the car is said to be in equilibrium. The engine must produce torque with sufficient power to accelerate all of those air molecules out of the way. When you drive faster, you hit more air per second, and have to accelerate it to a higher velocity. This is why travelling at a constant 70mph burns more fuel than a constant 40mph.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-10-16T13:53:46.543" LastActivityDate="2018-10-16T13:53:46.543" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24189" PostTypeId="2" ParentId="24186" CreationDate="2018-10-16T12:52:53.563" Score="1" Body="&lt;p&gt;Solution was easier than expected, I didn't give the full file path to the fluent command: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;\ansys_inc\v192\fluent\bin\fluent 3d -t 36 -g -i &amp;lt; inputfile.jou &amp;gt;&amp;amp; outputfile.jou &amp;amp;&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="17876" LastActivityDate="2018-10-16T12:52:53.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24195" PostTypeId="1" CreationDate="2018-10-16T20:16:50.377" Score="1" ViewCount="58" Body="&lt;p&gt;Hot wire anemometry can be used to measure the temperature of a wire in fluid flow. The voltage fluctuation across the wire is used in conjunction with an energy balance on the wire to obtain the temperature. Why is it not a better idea to just measure the resistance change directly and use the material's resistivity temperature coefficient to find the temperature of the wire? That way you wouldn't be relying on empirical Nusselt correlations. What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="17888" LastActivityDate="2018-10-16T20:16:50.377" Title="hot wire anemometry" Tags="|mechanical-engineering|fluid-mechanics|heat-transfer|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24196" PostTypeId="1" CreationDate="2018-10-16T22:03:52.430" Score="2" ViewCount="121" Body="&lt;p&gt;Would pumping down on a gas cylinder regulator cause the needle of the valve to break since it’s trying to go below zero? Would this also cause a problem with the regulator itself? &#xA;I’ve included the actual picture of the regulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bp5Yg.jpg&quot; alt=&quot;Gas cylinder regulator&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17891" LastEditorUserId="1832" LastEditDate="2018-10-16T23:12:58.093" LastActivityDate="2018-10-17T00:44:01.947" Title="Would pumping down (below 0 psi) on a gas cylinder regulator cause it to break?" Tags="|gas|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24197" PostTypeId="2" ParentId="24196" CreationDate="2018-10-17T00:44:01.947" Score="4" Body="&lt;p&gt;I doubt there would be a problem.  That type of gauge has a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pressure_measurement#Bourdon_gauge&quot; rel=&quot;noreferrer&quot;&gt;bourdon tube&lt;/a&gt; inside that has a curved flattened tube.  The tube goes from flat to round as the pressure increases which straightens out the curve and moves the needle.  If you subject it to negative pressure, the tube will flatten out more and pull the needle towards the peg.  The lowest vacuum you can get is about -15 PSIG.  On the high pressure side, the needle goes up to something like 4000 PSI. 15 PSI is 0.375% of 4000 PSI.  I am not sure it would even notice if you pushed the needle the wrong way by 15 PSI.  The low pressure side is a bit more of a concern.  The peg will keep the needle from going the wrong way so the internal mechanism won't run off the end and that much pressure certainly won't bend anything.  You will be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't wish to trust your equipment to some dude on the internet, I suggest you call up the gauge manufacturer, which in this case is &lt;a href=&quot;https://store.mathesongas.com/&quot; rel=&quot;noreferrer&quot;&gt;Matheson&lt;/a&gt; and ask to talk to tech support.  They want their products to be used correctly and most vendors will be happy to answers questions. I have never called Matheson but I have asked manufacturers about equipment that has been in service for years and years (and sometimes out of service for years and years) and almost every time, they have gone out of their way to be helpful.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2018-10-17T00:44:01.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24198" PostTypeId="1" CreationDate="2018-10-17T03:43:06.577" Score="0" ViewCount="68" Body="&lt;p&gt;For shell elements, is the elasticity tensor for linear-elastic &amp;amp; isotropic materials the same in a local curve-linear (convected) system vs. a local Cartesian system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder because intuitively, for a shell element &amp;amp; linear-elastic isotropic material, the only direction that matters (material-property wise) is the z direction. And in this case, the local z axis is aligned with the curve-linear coordinate system. But how do we mathematically show it?&lt;/p&gt;&#xA;" OwnerUserId="9268" LastActivityDate="2024-03-17T05:05:25.643" Title="Curve-linear vs. Cartesian coordinate systems: linear-elastic &amp; isotropic elasticity tensor" Tags="|mechanical-engineering|applied-mechanics|finite-element-method|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24199" PostTypeId="1" AcceptedAnswerId="24200" CreationDate="2018-10-17T05:58:56.123" Score="1" ViewCount="126" Body="&lt;p&gt;Is resolution of forces similar to expressing a vector as a linear combination of some other vectors in linear algebra? &#xA;Similarly, Is the state of stress at a point in 3D similar to expressing the applied load as a linear combination of 9 other vectors(the shear and normal components along the direction of the three axes)?&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastActivityDate="2018-10-17T08:56:44.520" Title="Resolution of forces and state of stress at a point" Tags="|stresses|strength|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24200" PostTypeId="2" ParentId="24199" CreationDate="2018-10-17T07:47:24.540" Score="1" Body="&lt;p&gt;Sort of. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you resolve a vector into its component, you have to choose a coordinate system, resolving the vectors into components is the same as projecting the original vector on the coordinate axis. So yes the resolution is equivalent of linear combination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the stress tensor is symmetric then you need only six basis matrix to construct  the linear space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise you need nine component to describe the linear space. Notice, you cannot construct a &lt;strong&gt;general linear matrix space&lt;/strong&gt; with vectors, you need matrixes. So the state of stress is generally the linear combination of nine other matrixes.   &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-10-17T08:56:44.520" LastActivityDate="2018-10-17T08:56:44.520" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24201" PostTypeId="1" CreationDate="2018-10-17T11:30:41.003" Score="1" ViewCount="895" Body="&lt;p&gt;I am trying to reduce my model in ANSYS using Craig-Bampton Method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My ultimate goal is to read system matrices, stiffness, damping and mass matrix with MATLAB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of stiffness matrix that I export from ANSYS is more than 10 GB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My aim is to simulate my structure as a flexible body in Simscape Multibody using finite-elemente import method - superposition of ridig body and deformation models. Deformation model is described in space-state using system matrices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I build my flexible body with deformation model for each part (say it is a table, should I describe every leg of my table and the plate using rigid body and deformation model)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concerning model reduction I am really not sure what to do &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I use Component Mode Synthesis Method - Fixed-Interface Method (CB Method)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I generate superelements for every part selecting master degrees of freedom on the interfaces conecting to the other parts of the structure? If yes, what is the next step?&lt;/p&gt;&#xA;" OwnerUserId="17533" LastEditorUserId="17070" LastEditDate="2019-08-28T17:20:03.200" LastActivityDate="2022-03-29T20:07:20.750" Title="Model Reduction - Export matrices from ANSYS to MATLAB" Tags="|finite-element-method|ansys|matlab|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24203" PostTypeId="1" CreationDate="2018-10-17T15:29:23.027" Score="0" ViewCount="45" Body="&lt;p&gt;We have invented a solar thermal heating system with interseasonal storage, recently patented. I want to display the trees saved and carbon offset on the control display based on temperatures coming down from the solar panels and that flow rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pump station/controller has many sensors and we have added a flow meter. I can read the temperature rise (in from solar vs out to solar), &amp;amp; judge the flow rate (2-25L/Min) accurately. Typically the DeltaT is 15c but can vary wildly based on actual insolation. Flow rate is typically 15L/Min&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to  build an algorithm to show CO2e offset. I need to calculate in energy terms (kWt or Mj) the energy injected into the heating system based on temps and flow rate. From that I can calculate the CO2e and trees saved.&lt;/p&gt;&#xA;" OwnerUserId="17901" LastEditorUserId="17901" LastEditDate="2018-10-17T19:23:14.450" LastActivityDate="2018-10-19T14:42:08.150" Title="Solar Thermal heat gain formula converted to CO2e and trees saved" Tags="|heating-systems|renewable-energy|solar-energy|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24204" PostTypeId="2" ParentId="24203" CreationDate="2018-10-17T16:03:05.343" Score="1" Body="&lt;p&gt;You can get the energy gained from heating by multiplying the heat capacity with the temperature difference and the amount of medium heated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you use water as a medium with a specific heat capacity of &lt;span class=&quot;math-container&quot;&gt;$4.2~J/(g~K)$&lt;/span&gt; and mass density of &lt;span class=&quot;math-container&quot;&gt;$1000~g/l$&lt;/span&gt; your typical situation as you described (&lt;span class=&quot;math-container&quot;&gt;$\Delta T =15~K$&lt;/span&gt;, flow = &lt;span class=&quot;math-container&quot;&gt;$15~l/min$&lt;/span&gt;) would come out to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;15~K \times 15~l/min  \times 1000~g/l \times 4.2~J/(g~K) = 945000~J/min = 15750~W&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many trees per time unit that saves depends on who you talk to.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2018-10-19T14:42:08.150" LastActivityDate="2018-10-19T14:42:08.150" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="24205" PostTypeId="2" ParentId="24180" CreationDate="2018-10-17T19:28:15.203" Score="3" Body="&lt;p&gt;The hyperelastic and viscoelastic material models are both constitutive relations that relate:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Stress&lt;/strong&gt; and &lt;strong&gt;strain&lt;/strong&gt;, in the case of &lt;strong&gt;hyperelasticity&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Stress&lt;/strong&gt;, &lt;strong&gt;strain&lt;/strong&gt; and &lt;strong&gt;strain rate&lt;/strong&gt;, in the case of &lt;strong&gt;viscoelasticity&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They are both empirical models, which means that you typically need to run experiments to find the necessary parameters to fit each model (there are some ways to find elastic properties from first principles, but I'm gonna leave that discussion aside).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hyperelastic&lt;/strong&gt; models (as the name implies) deal with elastic materials - those which deform when a load is applied, and return to their original shape  after the load is released. Unlike  Hookean elastic models, which have a linear strain-stress relationship, hyperelastic models have a non-linear relationship:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zsxi6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zsxi6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One common example of simulations using hyperelastic models are about rubbers seals or gaskets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iKdYK.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iKdYK.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In &lt;strong&gt;viscoelastic&lt;/strong&gt; models the strain-stress relationship will be dependent on the strain-rate (how fast you deform the material) and time, and the loading and unloading paths are not the same:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n1xYV.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n1xYV.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your constitutive model for viscoelasticity includes creep behavior you may also have permanent deformation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5HcZq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5HcZq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The choice of one model over the other depends on what problem are you trying to solve. Are you interested in the strain-stress response when you change the rate of deformation? Are you interested in creep or stress relaxation?&#xA;Both models are not mutually exclusive, you can create an visco-hyperelastic model (or use those in the literature) if you need it.&lt;/p&gt;&#xA;" OwnerUserId="17409" LastActivityDate="2018-10-17T19:28:15.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24206" PostTypeId="2" ParentId="19411" CreationDate="2018-10-17T20:07:48.850" Score="1" Body="&lt;p&gt;You could have the parts coated with Teflon, which is one of the most chemically resistant materials known to man. During WW2, when we were processing uranium hexafluoride to make weapons grade U235, Teflon was the only material coating that could resist the corrosiveness of the uranium hexafluoride .&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2018-10-17T20:07:48.850" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="24207" PostTypeId="1" AcceptedAnswerId="24217" CreationDate="2018-10-17T20:27:39.210" Score="0" ViewCount="42" Body="&lt;p&gt;I'm building a 4' x 8' gantry-style CNC router table. The spindle motor uses pneumatics to power its draw bar (for automatic tool changes). There are three separate pneumatic supply lines that go to the spindle motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is a constant air supply of low pressure that serves to blow debris out of the tool mount during tool changes. The other two are the &quot;open&quot; and &quot;close&quot; lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can choose to place the electrically-actuated pneumatic valves and various pneumatic components on the base of the machine, on the gantry, or on the spindle mount itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, three pneumatic lines must run several meters to reach the spindle, but the bulk of the pneumatic hardware is kept off the moving parts of the machine (reducing vibration and moving mass).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the last case, the electrical control lines have to be run, but the pneumatic control lines are shortened (except for the main supply line), and a number of valves and regulators have to be placed on the gantry carriage. The second case is similar, but the equipment is mounted to the gantry itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm leaning toward just running the three pneumatic lines the full distance and keeping as much equipment as possible off the moving parts of the machine. But I don't know if the long pneumatic lines will cause other problems, perhaps sluggishness due to the elasticity of the tubing (they're typical poly or PVC pneumatic tubing). Is this even a concern?&lt;/p&gt;&#xA;" OwnerUserId="17903" LastActivityDate="2018-10-18T16:51:56.753" Title="Best balance of electrical and pneumatic control lines in CNC router table?" Tags="|pneumatic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24208" PostTypeId="2" ParentId="24201" CreationDate="2018-10-17T22:26:37.440" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So my first question would be - Should I build my flexible body with&#xA;  deformation model for each part (say it is a table, should I describe&#xA;  every leg of my table and the plate using rigid body and deformation&#xA;  model)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That would be a reasonable approach. You haven't told us much about what you want to do with your model in Simulink, but as a general principle, it's easier if you retain any positions where you want to apply things like loads or constraints to the model as master degrees of freedom, even if they are not geometrically actually on the &quot;boundary.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That might mean you need to keep some master degrees of freedom on the table top, in addition to the interface to the legs, if you want to model what happens when you apply a load at those positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative would be to model the whole table top plus legs as a single object, with the master degrees of freedom as the ends of the legs (presumably they will be constrained to be resting on the ground, in the Simulink analysis) and any other positions you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Concerning model reduction I am really not sure what to do - should I&#xA;  use Component Mode Synthesis Method - Fixed-Interface Method (CB&#xA;  Method)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would recommend that for your first attempt, definitely. It is conceptually simpler than the alternatives, and it automatically avoids some &quot;elephant traps&quot; where load paths through stiff parts of the structure (in your model, axially along the legs, for example) are &lt;em&gt;not&lt;/em&gt; captured as part of the low-frequency internal modes of the model, but they are guaranteed to be included in the &quot;rigid body&quot; part with the fixed-interface method. Leaving those stiff load paths out of the model may produce &quot;plausible looking complete nonsense&quot; - which is the worst type of error to have, when you are inexperienced in what you are doing!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Should I generate superelements for every part selecting master&#xA;  degrees of freedom on the interfaces connecting to the other parts of&#xA;  the structure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, that is implied by my answers to above. Alternatively, treat the complete table as a single superelement. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If yes, what is then my next step?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, just set up the model and run it. You should be able to find a similar tutorial example somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An important decision you need to make is how many internal modes to include in each component. One strategy to answer that is to include all the internal modes below some cutoff frequency, which depends on what frequency range you want to use the final model for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have any more problem-specific advice available, set the cutoff frequency for generating the component models to about 3 times the highest frequency you are interested in. That may mean you get many more modes in the table-top model than in the legs - but don't worry too much about that, unless the reduced model is still too big to work with when you assemble all the superelements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another issue is how to check your reduced model is correct. One fairly simple strategy is to do two vibration analyses in Ansys, on the unreduced and reduced models. Check the mode frequencies and mode shapes match up, and in particular check the reduced model contains &lt;em&gt;all&lt;/em&gt; the modes of the unreduced model, in the frequency range you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were doing this sort of modelling repeatedly, it is possible to automate some of the checking with software, but for a first attempt it's probably better to do it hand, comparing the mode shapes visually - though it may get tedious if you have a large number of modes.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-10-17T22:33:32.167" LastActivityDate="2018-10-17T22:33:32.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24210" PostTypeId="1" AcceptedAnswerId="24211" CreationDate="2018-10-18T08:40:32.807" Score="0" ViewCount="39" Body="&lt;p&gt;I want to build my own off-grid system but confused with what sizes (V, I) to get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to run a 24v system powering 4.4kWh/day.  I have 4 hours of usable sunlight a day and space for panels and equipment is not an issue (within reason).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the calculations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrics = 220v @ 60Hz&lt;br&gt;&#xA;Load = 4.4kWh/day&lt;br&gt;&#xA;Sunlight = 4 hrs  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The PV array I then calculate at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PV Array = 4400Wh / 4hrs = 1100W (10 x 100w panels)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The batteries I then work out at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4400W / 24v = 184Ah / 80% discharge = 220Ah&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A battery example [1] - 24v 220Ah deep-cycle li-ion (2000times(80%DOD))&lt;br&gt;&#xA;* or if you want to go down building your own battery pack with 18650's - 7 in series (to get 24v) x 55 packs in parallel (to get 220Ah).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation problems I cant find anyone explaining on Google is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;a) What size charge controller do I need?  I see options from 10A to 100A.  &#xA;b) What size inverter do I need? I see options up from 100W to 3000W and more.&#xA;c) Does the PV array not need to be x2 the size so it can power the house and charge the batteries at the same time during the day?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;https://www.alibaba.com/product-detail/Custom-24-Volt-Solar-24V-200Ah_60670679703.html?spm=a2700.7724857.normalList.25.15c91a82Albyfj&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.alibaba.com/product-detail/Custom-24-Volt-Solar-24V-200Ah_60670679703.html?spm=a2700.7724857.normalList.25.15c91a82Albyfj&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17908" LastActivityDate="2018-10-18T10:21:52.857" Title="Electronics confusion" Tags="|solar-energy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24211" PostTypeId="2" ParentId="24210" CreationDate="2018-10-18T10:21:52.857" Score="1" Body="&lt;p&gt;The answers to (a) and (b) depend on the &lt;em&gt;peak&lt;/em&gt; load you want to use - i.e. the highest powered devices you want to run. Note that these might be things that you only run for a few minutes (e.g. an electric kettle), so they don't eat up the whole of your daily 4.4kWh energy budget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to (c) is &quot;no&quot;. For example, suppose your appliances used 1100W for the entire four hours of daylight. The panels would be powering the appliances but not charging the battery at all. But &lt;em&gt;that doesn't matter&lt;/em&gt;, because you have used all your daily energy budget in those four hours, so &lt;em&gt;to stay within the design limits of your system&lt;/em&gt; you will not be using any electricity at all for the remaining 20 hours of the day, and it doesn't matter that the battery isn't charged!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course that is an extreme example to make the point easily - but the same thing will apply for any usage pattern over 24 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, you need to include the efficiency factors of the components in your system design. The efficiency of the battery charging system won't be 100%. The inverter should have a high efficiency, but it won't be 100% either. Its efficiency might vary depending on the load (and it might be worse for light loads than for heavy ones).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, ask yourself whether &quot;4.4kWh per day&quot; is a &lt;em&gt;long term&lt;/em&gt; average (over a week, or a month, or a year) rather than the daily limit for every day. You might have some high-power devices (e.g. a washing machine) that you don't use every day, so you need to store energy for a longer period of time - i.e. you need a bigger battery capacity.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-10-18T10:21:52.857" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24212" PostTypeId="2" ParentId="19027" CreationDate="2018-10-18T10:55:43.413" Score="0" Body="&lt;p&gt;I made some experiments long time ago, in 1990. For a bolted beams free ends, I found that:  with increasing the bolt preload, the natural frequency increased until maximum then start decreasing again. That happened in all modes. &lt;/p&gt;&#xA;" OwnerUserId="17910" LastActivityDate="2018-10-18T10:55:43.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24213" PostTypeId="1" CreationDate="2018-10-18T11:04:03.370" Score="0" ViewCount="4828" Body="&lt;p&gt;For weeks I have been searching internet , but still unable to find answer , so here it goes:&#xA;I have to calculate pressure losses in heating circuit , and am stuck at calculating losses in T fittings. Here are some crude drawings of problem &lt;a href=&quot;https://i.stack.imgur.com/5dRip.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5dRip.png&quot; alt=&quot;Flow conjunction&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/AgDCQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AgDCQ.png&quot; alt=&quot;Flow separation&quot;&gt;&lt;/a&gt;&#xA;I know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;diameters of each pipe leading to fitting (Da,Db,Dc) &lt;/li&gt;&#xA;&lt;li&gt;flow rates through these pipes (Qa , Qb , Qc)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I need to know is Pressure drops: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;∆Pc-a=Pc-Pa=?&lt;/li&gt;&#xA;&lt;li&gt;∆Pb-a=Pb-Pa=?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know of minor loss formula , but don't understand how to use if fitting has more than 2 ports. &lt;/p&gt;&#xA;" OwnerUserId="17911" LastActivityDate="2022-09-01T19:21:37.640" Title="pressure loss in a T fitting" Tags="|fluid-mechanics|pressure|hydraulics|piping|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24214" PostTypeId="1" CreationDate="2018-10-18T13:37:08.423" Score="0" ViewCount="365" Body="&lt;p&gt;There are two components given with each stress-strain curve.&#xA;Determine the component that maximise the load carrying capacity and that maximise the elastic deformation per unit applied load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it right for the maximum load carrying capacity to look for the highest yield stress and for the maximum elastic deformation per unit applied load to look for the highest strain at failure?&lt;/p&gt;&#xA;" OwnerUserId="17913" LastActivityDate="2019-08-15T19:02:15.237" Title="stress strain curve properties analyse" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24215" PostTypeId="1" CreationDate="2018-10-18T13:40:29.030" Score="0" ViewCount="168" Body="&lt;p&gt;I can't seem to figure out the proper left view for the following image, and the full and half sectional cutout for the left and front projection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DHl1y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DHl1y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17912" LastEditorUserId="17912" LastEditDate="2018-10-20T12:55:13.647" LastActivityDate="2018-10-20T12:55:13.647" Title="Can you help interpret hand-drawn technical drawing?" Tags="|mechanical-engineering|technical-drawing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24217" PostTypeId="2" ParentId="24207" CreationDate="2018-10-18T16:51:56.753" Score="1" Body="&lt;p&gt;I would keep all the control hardware/components off of the moving members as much as possible and in the same general area for ease of wiring and trouble shooting. Use nylon tubing, its more &quot;rigid&quot; such that the tubing diameter will not flex or change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In electro-pneumatic machines, I usually build a pneumatic &quot;panel/enclosure&quot; with all my air prep/sensors/valves/etc, just as you would your electrical control, and keep it close to the electrical panel/enclosure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this application, I wouldn't consider any time delays due to the tubing lengths to be significant. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-10-18T16:51:56.753" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24219" PostTypeId="2" ParentId="23795" CreationDate="2018-10-18T18:46:05.087" Score="1" Body="&lt;p&gt;I came across a project with stormwater requirements referenced to the code I was trying to identify -- the &lt;a href=&quot;https://www.iccsafe.org/codes-tech-support/topics/plumbing-mechanical-and-fuel-gas/international-plumbing-code-ipc-home-page/&quot; rel=&quot;nofollow noreferrer&quot;&gt;International Plumbing Code&lt;/a&gt;. Two chapters are relevant to what I've been trying to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://codes.iccsafe.org/content/IPC2018/chapter-11-storm-drainage&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chapter 11, Storm Drainage:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;About this chapter: Rainfall onto buildings must be removed and directed to a location that can accommodate storm water. Chapter 11 specifies the design rainfall event for the geographic area and provides sizing methods for piping and gutter systems to convey the storm water away from the building. Included in this chapter are regulations for piping materials and subsoil drainage systems.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://codes.iccsafe.org/content/IPC2018/chapter-13-nonpotable-water-systems&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chapter 13, Nonpotable Water Systems:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;About this chapter: Storm water and some liquid waste from a building can be a source of nonpotable water that can be used to reduce the volume of potable water supplied to the building. Chapter 13 provides the requirements for storage, treatment and distribution of this resource. This chapter also regulates the piping systems for reclaimed water supplied by a wastewater treatment facility.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17004" LastActivityDate="2018-10-18T18:46:05.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24220" PostTypeId="2" ParentId="24214" CreationDate="2018-10-18T18:52:09.343" Score="1" Body="&lt;p&gt;Do you know how a typical stress-strain curve looks like?&#xA;Take a look at this image:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9rakY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9rakY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load carrying capacity is the maximum load an item can carry before a failure occurs. Please note that these curves deal with stresses and not with forces. However, assuming the two components you are examining share the same geometry and are subjected to the same load pattern - you can compare their ultimate strength values. The larger the ultimate strength in this case - the larger the load carrying capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elastic deformation begins at the origin of graph and ends at the Yield Strength point (as long as the curve is linear). The larger the strain at this point the larger the elastic deformation is.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2018-10-18T18:52:09.343" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24221" PostTypeId="2" ParentId="24198" CreationDate="2018-10-18T20:07:55.297" Score="0" Body="&lt;p&gt;A linear elastic stress-strain law can be expressed in coordinate free form as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \boldsymbol{\sigma} = \mathsf{C} : \boldsymbol{\varepsilon}&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{\sigma}$&lt;/span&gt; is the Cauchy stress (a second-order tensor), &lt;span class=&quot;math-container&quot;&gt;$\mathsf{C}$&lt;/span&gt; is the stiffness tensor (a fourth-order tensor), and &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{\varepsilon}$&lt;/span&gt; is the small strain tensor (a second-order tensor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expressed in terms of the covariant components, the second-order tensors can be written in the form&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \boldsymbol{\sigma} = \sigma_{ij}\, \mathbf{g}^i \otimes \mathbf{g}^j&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\mathbf{g}^i$&lt;/span&gt; are the reciprocal basis vectors in curvilinear coordinates.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the fourth-order tensor can be expressed as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \mathsf{C} = C_{ijkl}\, \mathbf{g}^i \otimes \mathbf{g}^j \otimes \mathbf{g}^k \otimes \mathbf{g}^l&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general curvilinear coordinates, the expressions for the stiffness matrix become extremely complicated.  However, these can be simplified by using a locally orthonormal basis.  In that case, the reciprocal basis simplifies to a locally Euclidean basis, i.e.,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \mathsf{C} = C_{ijkl}\, \mathbf{e}^i \otimes \mathbf{e}^j \otimes \mathbf{e}^k \otimes \mathbf{e}^l&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most shell elements use stiffness matrices that are expressed in a local orthonormal basis at each point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the stresses are often (but not always) assumed to be zero through the thickness (the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-direction in the question?), it is sufficient to use linear elastic properties in the in-plane directions with corrections for the zero-stress assumption.  You will, of course, have to make sure that the correct coordinate system is used at each point.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-10-18T20:07:55.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24222" PostTypeId="1" CreationDate="2018-10-18T20:47:58.193" Score="2" ViewCount="195" Body="&lt;p&gt;I'm trying to describe a part -  it's an arm that holds an axle sleeve (which ultimately holds a wheel). I'm stuck right now describing it simply as a &quot;wheel arm&quot;, but was wondering if there was some better, more engineering-appropriate term.   I'd appreciate any feedback. Thanks&lt;img src=&quot;https://i.stack.imgur.com/3zpDA.png&quot; alt=&quot;![wheel arm?&quot;&gt;]&lt;a href=&quot;https://i.stack.imgur.com/3zpDA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17367" LastActivityDate="2020-09-11T02:04:30.500" Title="What is an engineering term for this type of arm" Tags="|mechanical-engineering|cad|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24223" PostTypeId="2" ParentId="23795" CreationDate="2018-10-18T20:56:54.000" Score="1" Body="&lt;p&gt;I answer this with a perspective from the United Kingdom. With an annual problem of flooding of low-lying areas especially in the Worcester area West of Birmingham the national government has written new laws that specified new housing estates and construction sites must hold 50% of water runoff and prove it before building planning can be approved. As far as I am aware there are no building codes in the eurocodes that specifically deal with stormwater runoff. More often than not stormwater runoff is dealt with by means of concrete structures that are inserted underground such as pipeworks and drainage ditches and even urban rivers lined with concrete. from this perspective, such structures are covered by Eurocode 1 and 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This document maybe of use to you in this regard. [Stormwater management][1]. An important consideration is that maybe stormwater should be dealt with in a more effective ways such as pipes that have holes to allow them to act as soakaways on route to a river and thereby allowing the ground that they are buried in to absorb a fair amount of water before actually reaches the river system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]: &lt;a href=&quot;https://ec.europa.eu/research/participants/documents/downloadPublic?documentIds=080166e5a5630bbf&amp;amp;appId=PPGMS&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ec.europa.eu/research/participants/documents/downloadPublic?documentIds=080166e5a5630bbf&amp;amp;appId=PPGMS&lt;/a&gt; &quot;&quot;&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-10-18T20:56:54.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24225" PostTypeId="2" ParentId="24215" CreationDate="2018-10-18T23:02:35.007" Score="4" Body="&lt;p&gt;I have made a 3D model of my best guess for this model from your drawing. I've then generated a crib-sheet for the computer generated standard views (first angle projection, as per your sketch), and for some section views. Hopefully one of these provides the information that you're after? - have a look below, and let me know if you have any questions!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/22JV5.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/22JV5.png&quot; alt=&quot;Best Guess 3D model&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZU9aJ.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZU9aJ.png&quot; alt=&quot;2D Drawing Cribsheet&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-10-18T23:02:35.007" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24226" PostTypeId="2" ParentId="23422" CreationDate="2018-10-18T23:41:36.177" Score="1" Body="&lt;p&gt;Usually flow orifices are very aggresive to the pressure. Everything you add has a pressure drop and doing calculations by hand of an equipment like that can be hell. So yes you should check the specs of the meter first.&lt;/p&gt;&#xA;" OwnerUserId="17922" LastEditorUserId="9" LastEditDate="2018-10-19T00:07:26.683" LastActivityDate="2018-10-19T00:07:26.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24228" PostTypeId="2" ParentId="24213" CreationDate="2018-10-19T00:21:07.180" Score="0" Body="&lt;p&gt;I have two ways you could find a solution to that problem (actually 3 but the third is using a flow simulator :P) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just assume that P2 and P3 are the same (or else you will find yourself in partial differential equation hell believe me been there and it isn't pretty) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1: Crane Flow of Fluids. Easy lazy, quick he gives you measured parameters for pressure loses and other things you just fit your numbers there &lt;a href=&quot;http://www.waterlinefountains.com/wp-content/uploads/2017/09/Crane-410.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.waterlinefountains.com/wp-content/uploads/2017/09/Crane-410.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not the best method it has a lot of error. It is the first answer that comes up in google (though they do not provide you with the tables)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2: Bird transport phenomena: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume it is plug flow, define a three pipe system in which 1 is the feed and 2 and 3 are the exits, you calculate each exit with an equalized pressure (or so I remember). Just to be certain; I checked the book and problem 7Q4 can give you the answer, look up the solution and define n as 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I think this is the answer you want, you calculate everything by hand but it isn't a problem with an easy solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend you checking if someone hasn't programmed it yet because you getting a short formula is a longshot the answer to 7Q4 is a differential equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3: Get a copy of EPAnet (free but works poorly), HYSYS (usually colleges with Chemical engineering majors have licences) or Olga, they do pressure drops though they use method 2 and 1 combined. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR: To do it by hand is possible, has been done before and that is how most simulators work, though it is so hard to do that you use equivalent lengths instead. &lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2018-10-19T00:21:07.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24229" PostTypeId="2" ParentId="17636" CreationDate="2018-10-19T00:32:17.017" Score="1" Body="&lt;p&gt;Set temperature as constant, make P change and find V solutions (if V(P,T) which according to what he is saying it is what he is doing...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You HAVE to set one constant, since you want to plot PV you need to set T as constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case it is P(V,T) you change the Volume and calculate the Pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;... I would recommend you do a loop to print out all the values then plot them in either Origin, Excel, by hand or whatever you want to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you don't know how to use loops (completely OK with that this is a place to learn)&#xA; &lt;a href=&quot;https://www.mathworks.com/help/matlab/ref/for.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mathworks.com/help/matlab/ref/for.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendation? Use Fortran because Matlab takes AGES to solve but that is just personal preferrence.&lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2018-10-19T00:32:17.017" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24231" PostTypeId="1" AcceptedAnswerId="24235" CreationDate="2018-10-19T03:46:00.880" Score="0" ViewCount="109" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bkgKJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bkgKJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I solved this problem by doing a mass balance for the mixing chamber and an energy balance for the water heater and the mixing chamber as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\dot{m}_{\text{mix}}=\dot{m}_{\text{Tank}}+\dot{m}_{\text{Cold}}\\&#xA;\\&#xA;\dot{m}_{\text{mix}}\hat{H}_{\text{mix}}=\dot{m}_{\text{Tank}}\hat{H}_{\text{tank}}+\dot{m}_{\text{Cold}}\hat{H}_{\text{cold}}\\&#xA;\\&#xA;\frac{m_{\text{water in tank}}C_{\text{v}}\left ( T_2-T_1 \right )}{\Delta t}=\dot{W}_{\text{e}}+\dot{m}_{\text{Cold}}\hat{H}_{\text{cold}}-\dot{m}_{\text{Tank}}\hat{H}_{\text{tank}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;resulting in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\dot{m}_{\text{Tank}}=0.05838 \text{ kg/s}\\&#xA;{T}_{\text{shower}}=44.67 \text{ °C}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and my results are in agreement with the solutions manual. But, if I do a mass balance in the water heater I obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\dot{m}_{\text{water heater}}=\dot{m}_{\text{Cold}}-\dot{m}_{\text{TanK}}$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if I replace the value that I have obtained previously, I do not obtain a zero mass flow for the tank. So, ¿Is my answer or mass balance wrong, or I misunderstand something in the problem statement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="17925" LastActivityDate="2018-10-19T21:20:51.570" Title="Contradictory result for mass balance" Tags="|mechanical-engineering|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24232" PostTypeId="1" AcceptedAnswerId="24234" CreationDate="2018-10-19T06:53:57.363" Score="0" ViewCount="54" Body="&lt;p&gt;I would like to know what kind of glass is used in biometric attendance machines over which we press our thumb? Is there something special about these glasses or their manufacturing?&lt;/p&gt;&#xA;" OwnerUserId="17926" LastActivityDate="2018-10-19T16:44:07.313" Title="Glass used in biometric machines" Tags="|glass|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24234" PostTypeId="2" ParentId="24232" CreationDate="2018-10-19T07:46:50.797" Score="0" Body="&lt;p&gt;Yes, it is probably the same type of glass used for smartphone screens some version of an aluminium silicate glass that resists scratches, grease, dirt etc for the very same reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also use &lt;strong&gt;oleophobic&lt;/strong&gt; coatings to avoid sticky fingeprints etc...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-10-19T16:44:07.313" LastActivityDate="2018-10-19T16:44:07.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24235" PostTypeId="2" ParentId="24231" CreationDate="2018-10-19T08:22:20.197" Score="1" Body="&lt;p&gt;The results are correct and not contradictory at all, but the way you interpret the problems is not totally correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mass or impulse balance, you have to first define a &lt;strong&gt;control volume&lt;/strong&gt;. I define it as the volume of the tank, then the continuity equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\iint_S \rho(\vec{v}_{in}.\vec{n}) \,ds+\iint_S \rho(\vec{v}_{out}.\vec{n}) \,ds+\frac{\partial }{\partial t}\iiint_V \rho \,dV = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two terms are mass flux into and out of the control volume respectively, and the third term represents the mass accumulation, here the last term is zero.&#xA;&lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the density and &lt;span class=&quot;math-container&quot;&gt;$\vec{v}_{in}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\vec{v}_{out}$&lt;/span&gt; the velocity of the fluid. Notice the terms in the above equation have the dimension of &lt;span class=&quot;math-container&quot;&gt;$\frac{kg}{s}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By defining the tank as the control volume, the mass per second that exits the tank is indeed equal to &lt;span class=&quot;math-container&quot;&gt;$0.05838\frac{kg}{s} $&lt;/span&gt; (assuming your calculations are correct). By assuming no accumulation in the tank, this mass should be replaced by the cold water, however this is not necessarily  equal to the mass of cold water mixing in the mixing chamber. You confuse the control volume of the tank with the control volume of the mixing chamber. Another way to say it, is there are two channels that provide the cold water, one goes into the tank with the same mass rate as the hot water comes out of the other side, and the one goes inside the mixing chamber. &#xA;Now if you define the control volume as the volume of the tank and the mixing chamber then you'll see, the &lt;span class=&quot;math-container&quot;&gt;$m_{mix}$&lt;/span&gt; is equal to &lt;span class=&quot;math-container&quot;&gt;$m_{tank} + m_{cold, chamber} +m_{cold, tank} $&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorUserId="10902" LastEditDate="2018-10-19T21:20:51.570" LastActivityDate="2018-10-19T21:20:51.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24236" PostTypeId="2" ParentId="24131" CreationDate="2018-10-19T10:16:48.433" Score="0" Body="&lt;p&gt;In case anyone else is ever looking for a limiting case for deflection of plates, in relation to the Eurocodes, you can find this in:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EN 1993-4-1 Section 9.8.2 (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps someone!&lt;/p&gt;&#xA;" OwnerUserId="17139" LastActivityDate="2018-10-19T10:16:48.433" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24237" PostTypeId="2" ParentId="24222" CreationDate="2018-10-19T10:22:25.553" Score="0" Body="&lt;p&gt;The circular rod? Could be something like;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Cantilever support arm/axle&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Support arm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wheel support arm/axle&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Something along those lines would probably be descriptive enough, in my opinion.&lt;/p&gt;&#xA;" OwnerUserId="17139" LastActivityDate="2018-10-19T10:22:25.553" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24238" PostTypeId="1" CreationDate="2018-10-19T12:37:52.693" Score="0" ViewCount="33" Body="&lt;p&gt;I was looking at &lt;a href=&quot;https://www.lesker.com/newweb/sample_manipulation/rotarydrives_dynamic_oring.cfm?pgid=0&amp;amp;highlight=FMH-25SS&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt; to figure out how I would install a manipulator onto my chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says 1&quot; baseplate fitting to mount to a standard 1&quot; feedthrough port&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I would have to have a tapped hole of 1 inch in diameter. I am not sure what the thread turn number is going to be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In solidworks, I see that I have 1-12, 1-14, and 1-8.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I be able to assume one of the three just based on what they call &quot;standard?&quot; &lt;/p&gt;&#xA;" OwnerUserId="17201" LastActivityDate="2018-12-16T14:01:15.267" Title="meaning of 1&quot; baseplate fitting to mount to a standard 1&quot; feedthrough port" Tags="|machining|threads|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24239" PostTypeId="2" ParentId="19801" CreationDate="2018-10-19T13:01:50.620" Score="1" Body="&lt;p&gt;There are correction factors in academic table books.&#xA;A very detailed explanation is in the &quot;FKM Richtlinie&quot; chapter 1.2 material characteristic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you mentioned that you have the book Roloff / Matek: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Look at formula 3.7: &lt;span class=&quot;math-container&quot;&gt;$R_m=K_t*R_{mN}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_p=K_t*R_{pN}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$K_t$&lt;/span&gt; is the technological size influence factor – not sure if this is the right translation&lt;/li&gt;&#xA;&lt;li&gt;And &lt;span class=&quot;math-container&quot;&gt;$K_t$&lt;/span&gt; you can find in your table book. Table 3.11 a) and b) depending what kind of metal you are using&lt;/li&gt;&#xA;&lt;li&gt;The values of &lt;span class=&quot;math-container&quot;&gt;$R_{mN}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_{pN}$&lt;/span&gt; are given in your already shown table. I guess you picked 41Cr4, then &lt;span class=&quot;math-container&quot;&gt;$R_{mN}$&lt;/span&gt; is 1000 MPa and &lt;span class=&quot;math-container&quot;&gt;$R_{pN}$&lt;/span&gt; is 800 MPa&lt;/li&gt;&#xA;&lt;li&gt;41Cr4 is a metal out of the group “heat treatable steel” – so you have to take table 3.11 a) graph 3&lt;/li&gt;&#xA;&lt;li&gt;Now you take your desired diameter, say 30 mm, and then you get &lt;span class=&quot;math-container&quot;&gt;$K_t=0.93$&lt;/span&gt; or calculated with the formula &lt;span class=&quot;math-container&quot;&gt;$K_t=1-0.26*lg(30/16)=0.929$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Put &lt;span class=&quot;math-container&quot;&gt;$K_t$&lt;/span&gt; in the formulas above you get &lt;span class=&quot;math-container&quot;&gt;$R_m=930 MPa$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_p=744 MPa$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;An additional hint also concerning the comments: the &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$d_N$&lt;/span&gt; has nothing to do with the normalization of metal (the heat treatment)! &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; stands for standard part size. In your table (nearly) all metals are tested with &lt;span class=&quot;math-container&quot;&gt;$d_N=16 mm$&lt;/span&gt; to have comparability.&lt;/p&gt;&#xA;" OwnerUserId="17780" LastEditorUserId="17780" LastEditDate="2018-10-31T12:45:04.630" LastActivityDate="2018-10-31T12:45:04.630" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24240" PostTypeId="2" ParentId="24232" CreationDate="2018-10-19T14:04:22.587" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;https://www.us.schott.com/innovation/security-at-your-fingertips/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this glass manufacturer&lt;/a&gt; it is more about the optical properties than mechanical ones. They are looking for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Homogeneous illumination of the touchsurface&lt;/li&gt;&#xA;&lt;li&gt;Transparent and durable in UV and IR light radiation&lt;/li&gt;&#xA;&lt;li&gt;Temperature-resistant to heat development in LEDs&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They use &lt;strong&gt;borosilicate glass&lt;/strong&gt; and &lt;strong&gt;aluminosilicate glass&lt;/strong&gt;, which are by themselves scratch-resistant.&lt;/p&gt;&#xA;" OwnerUserId="17409" LastEditorUserId="17409" LastEditDate="2018-10-19T15:23:59.493" LastActivityDate="2018-10-19T15:23:59.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24241" PostTypeId="1" CreationDate="2018-10-19T14:25:05.090" Score="0" ViewCount="63" Body="&lt;p&gt;I hope this is the right place to ask my question. I am conducting a feasibility study into whether or not it would be possible to tint a single region of a continuous piece of glass. Think of a single square in the centre of a much larger rectangular pane of glass. It is important that whatever process is applied to the glass that it is still a single continuous pane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen electrochromic glass, but the samples of these frames I've seen always tint the whole pane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to conduct a literature review of potential methods of tinting but this isnt my area and im not sure what to search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR;&#xA;Is it feasible to tint a small region of a much larger pane of glass via electrochromic  means or otherwise?&lt;/p&gt;&#xA;" OwnerUserId="17930" LastActivityDate="2018-10-19T18:50:24.000" Title="Automatic tinting of a region of glass" Tags="|mechanical-engineering|glass|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24242" PostTypeId="1" AcceptedAnswerId="24264" CreationDate="2018-10-19T15:20:25.460" Score="0" ViewCount="910" Body="&lt;p&gt;I am trying to bend this sheet metal concave lip inside the electrical box. I also need the top of the bent part below the top edge of the outside of the box.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pQ671.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pQ671.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4QCy2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4QCy2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17929" LastEditorUserId="13311" LastEditDate="2018-10-22T12:33:35.990" LastActivityDate="2018-10-22T12:33:35.990" Title="Bending a concave sheet metal lip in AutoDesk Inventor" Tags="|cad|autodesk-inventor|metal-folding|surface-modelling|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24243" PostTypeId="2" ParentId="24241" CreationDate="2018-10-19T16:40:12.957" Score="1" Body="&lt;p&gt;Glasses that change color with external stimulus (&lt;a href=&quot;https://en.wikipedia.org/wiki/Smart_glass&quot; rel=&quot;nofollow noreferrer&quot;&gt;smart glass&lt;/a&gt;) are usually &lt;strong&gt;laminate structures&lt;/strong&gt; in which the glass itself is not the active component that generates the change in color or transparency, but it is used to encapsulate or to serve as a substrate for the variety of coatings, films, fluids, conductive materials, etc, that make possible the color change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So one option that I see is to coat the region in which you want the change in color with a (thermally activated) &lt;strong&gt;thermochormic film&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/koqKm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/koqKm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;M. Saeli et al., Thermochromic Thin Films: Synthesis, Properties and Energy, Consumption, Modelling, 2013.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Photochromic glass&lt;/strong&gt;, on the other hand, has the ability to change color by itself. This transformation is activated by light of a particulate wavelength which depends on the composition of the glass. So you can fabricate your plate out of photochromic glass and just shine light in the area that you want tinted. You can find a quite interesting video on photochromic glass here: &lt;a href=&quot;https://www.youtube.com/watch?v=mUcUy7SqdS0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=mUcUy7SqdS0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17409" LastEditorUserId="17409" LastEditDate="2018-10-19T18:50:24.000" LastActivityDate="2018-10-19T18:50:24.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24245" PostTypeId="1" CreationDate="2018-10-19T20:08:34.167" Score="0" ViewCount="526" Body="&lt;p&gt;Is it possible to calculate the maximum load of the steel structure shown in the image using its dimensions?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vi4fY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vi4fY.png&quot; alt=&quot;steel frame&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MtEeb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MtEeb.png&quot; alt=&quot;views&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17934" LastActivityDate="2024-03-17T03:06:10.673" Title="How can I calculate the maximum load of a steel box section frame?" Tags="|steel|deflection|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24246" PostTypeId="1" AcceptedAnswerId="24268" CreationDate="2018-10-19T20:37:39.173" Score="1" ViewCount="189" Body="&lt;p&gt;I was wondering, if there is a shock load resistant electromechanical actuator that can withstand powerful impacts?&#xA;Is there an electromechanical counterpart for pneumatic cylinders that is as shockresistant as hydraulic or pneumatic actuators?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if not, is there some kind of way to make them withstand impacts via external parts?&lt;/p&gt;&#xA;" OwnerUserId="17935" LastActivityDate="2018-10-21T04:11:12.683" Title="shock-resistant electromechanical linear actuator?" Tags="|robotics|actuator|linear-motors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24247" PostTypeId="2" ParentId="17978" CreationDate="2018-10-19T21:29:50.033" Score="1" Body="&lt;p&gt;Which formula you used? Are they distilled water or tap? Remember heat transfer formulas are based upon assumptions and there is always heat loss due to transfer to the surface of the tube and turbulence generates heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The more exact you want it to be the more complex you will make the problem, and heat transfer in a tube is actually a chaotic equation I cannot remember the guy who did that experiment. Will post the name if I remember&lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2018-10-19T21:29:50.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24248" PostTypeId="2" ParentId="24245" CreationDate="2018-10-20T01:32:40.583" Score="0" Body="&lt;p&gt;No, it is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the other pieces of information needed are divided in two parts: material and structural properties of your support; and load characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Material: What alloy or steel and what specific structural properties, and section properties such as whether they are hollow sections or solid, how they are put together, what weld type or spot-welding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Load characteristics: whether loads are permeant or temporary, if they are lowered to this support gently or if they will be dropped. If they will be Yanked away or lifted. If they are solid or liquid, or a machine? this support is anchored to the ground or just sitting, or will be moveable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is some parameters to consider before one can estimate the adequacy of strength of this support.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-10-20T01:32:40.583" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24249" PostTypeId="1" CreationDate="2018-10-20T02:33:35.370" Score="0" ViewCount="276" Body="&lt;p&gt;I'm designing an electrical component into a CAD from the &lt;a href=&quot;https://media.digikey.com/pdf/Data%20Sheets/Tyco%20Electonics%20AMP%20PDFs/103166.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;manufacturer drawing&lt;/a&gt;. The version I'm interested to is the code 106166-1 (6 positions, last entry in the table).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure to understand the drawing about the following dimensions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;overall width of the connector&lt;/li&gt;&#xA;&lt;li&gt;distance between any row of pins and the case&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Width of the connector&lt;/strong&gt;&#xA;I see two different ways to calculate it:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;D * 2 + 0.240'' = 0.452''&lt;/li&gt;&#xA;&lt;li&gt;C + 0.040'' * 2 = 0.412''&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;But as you can see they are different values. And I don't understand what does &quot;3 places&quot; mean (under 0.040'', on the right of C ref).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Distance between rows and the case&lt;/strong&gt;&#xA;I can't see any actual measure of this. Only observing the drawing it &lt;em&gt;seems&lt;/em&gt; the inner row is placed adjacent to the case, hence the center of the pins should be at diameter / 2 (0.015'' / 2) from the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you please help me to understand the right approach to find those values?&lt;/p&gt;&#xA;" OwnerUserId="13452" LastActivityDate="2018-10-20T09:04:09.553" Title="Find correct dimensions from mechanical drawing" Tags="|mechanical-engineering|cad|technical-drawing|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24250" PostTypeId="1" AcceptedAnswerId="24306" CreationDate="2018-10-20T03:41:59.073" Score="0" ViewCount="45" Body="&lt;p&gt;I have read about a device called drumometer, which measures the number of strokes of hitting drum stick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is almost the same, but instead, I would like to measure the number of strokes/impacts made when I shake a very small object inside a very small container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give an example, imagine a cube where one face is removed, hence making a hole. Then, I put the end (say X) of a stick inside that cube and hold the other end  (say Y) from outside. I then start vibrating (I mean like stirring movement) the stick so that the X end hits all faces of the cube. Then I want to measure how many times the X end of that stick hits not necessarily all faces but at least one face (since I will shake it randomly anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cube could be as small as 1 cm^3 (it might not even be a cube, just something with a hole that small where a stick can be put inside and shook from outside), and the stick is just long enough for me to hold it and small enough to get through the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, in relation with that drumometer, it kinda looks like an even smaller &quot;drumometer&quot; is needed. What do you think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have good background in measurement, only have some elementary freshmen undergraduate physics background instead. This is gonna be very helpful in my experiment once I know how to measure the number of strokes.&#xA;*feel free to edit my tags if not related&lt;/p&gt;&#xA;" OwnerUserId="17338" LastActivityDate="2018-10-22T20:02:53.030" Title="How to measure number of strokes in a small space" Tags="|measurements|vibration|instrumentation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24251" PostTypeId="2" ParentId="24249" CreationDate="2018-10-20T03:55:09.813" Score="1" Body="&lt;p&gt;(Links are to a distributor website. I don't have any knowledge of this distributor, but it had pictures of the parts.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Width of the connector = 0.412&quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on pictures of the part and other models found here (&lt;a href=&quot;https://www.digikey.sg/product-detail/en/te-connectivity-amp-connectors/103166-1/A26756-ND/299058&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;), like the &lt;a href=&quot;https://www.digikey.sg/product-detail/en/te-connectivity-amp-connectors/103166-2/A26759-ND/299027&quot; rel=&quot;nofollow noreferrer&quot;&gt;103166-2&lt;/a&gt;, I can see that the slot measured by D is not centered, therefore you can't use that to get the width of the part. Instead, C + 0.040&quot; * 2 is the correct measurement. This is verified by taking the width of the edge (0.040&quot;) twice, the distance from the edge to the center of the connector (0.066&quot;) twice, and B (0.200&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&quot;3 places&quot;&lt;/strong&gt; This is confusing, but I'm pretty sure it is the width of the left, right, and bottom walls as shown on the middle left view. The top wall seems thicker in the photos and drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Distance between rows and the case.&lt;/strong&gt; Middle left view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the left: The wall is 0.040&quot; thick. Then it is 0.066&quot; to the center of the pin. Look at the top left of the view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the bottom: The wall is 0.040&quot; thick (I think). Then it is 0.090&quot; to the center of the pin. Look at the bottom right of the view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you mean the right view, the bent pins are 0.157&quot; on-center from the base of the interior of the plug.&lt;/p&gt;&#xA;" OwnerUserId="17151" LastActivityDate="2018-10-20T03:55:09.813" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24252" PostTypeId="1" CreationDate="2018-10-20T04:15:34.490" Score="0" ViewCount="187" Body="&lt;p&gt;I have looked at small engines. Once I saw Mitsubishi G350L (made around 1986), the displacement is 141cc, but it maximum power is only 3.5HP at 2200RPM. I saw another engine, Honda GX200, 196cc, 6.5HP at 3600RPM. And another amazing Bajaj (automotive maker from India) with its Qlue, the engine is 200cc but can produce 13.3HP at 5500RPM. Indeed they are different. But the difference looks to be too much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how come the engine power so different? What is the relation between the piston's displacement and its power? Those 3 engines are relatively close in the cylinder displacement.&lt;/p&gt;&#xA;" OwnerUserId="17942" LastEditorUserId="33" LastEditDate="2018-11-13T03:58:55.927" LastActivityDate="2024-03-16T21:07:26.083" Title="Relation between engine's displacement (cc) to its power" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24253" PostTypeId="2" ParentId="24252" CreationDate="2018-10-20T05:03:24.983" Score="0" Body="&lt;p&gt;Power is the product of torque times RPM. For a given amount of engine cc's displacement,  you can get more power out of those same cc's by designing the engine to spin faster. This requires changing the valve cam profiles to get the engine to spin faster and also retuning the carburetor and exhaust systems, but it is done all the time when designing engines for motorcycles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disadvantages of pulling more power out of an engine by getting it to spin faster are 1) at low speeds, an engine &quot;cammed&quot; for high speed runs very rough and will not idle at all below about 1000RPM, 2) a faster-spinning engine runs hotter and is at increased risk of overheating and failing catastrophiclly, and 3) the faster the engine turns, the faster it wears out the piston rings and the crankshaft bearings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that different engine manufacturers will design their engines to balance power output against reliability and longevity in different ways, as you have seen yourself. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-10-20T05:03:24.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24254" PostTypeId="2" ParentId="24249" CreationDate="2018-10-20T08:39:14.220" Score="1" Body="&lt;p&gt;As a general rule, there should &lt;em&gt;never&lt;/em&gt; be two different ways to find the same dimension from a correctly dimensioned drawing, so if you have two formulas, either you or the drawing have a problem somewhere!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how you got your first formula D*2 + 0.240.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the view with dimensions D and E, the length appears to be E + 0.240 + D, &lt;em&gt;assuming the two ends of the component are mirror images&lt;/em&gt; - which seems to be the way it is drawn, but there is no &lt;em&gt;guarantee&lt;/em&gt; that is the case for all the part numbers in the table, because nothing on the drawings actually &lt;em&gt;says&lt;/em&gt; it is symmetrical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since dimension E is not given for your part number, you can't use this formula anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the notation &quot;0.240 &lt;em&gt;typ[ical]&lt;/em&gt;&quot; means that the 0.240 is not an &lt;em&gt;accurate&lt;/em&gt; dimension. Compare the pins, where the pin diameter is &quot;typical&quot;, but the view of the hole pattern gives the &lt;em&gt;definitive&lt;/em&gt; size of the holes for the pins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your other formula based on a different view, C + 0.040*2, is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the distance between the end of the pins and the base of the plug: this is probably not accurately controlled, except that the pins don't protrude beyond the base. However, the length of the pins from the tip to the &lt;em&gt;middle&lt;/em&gt; of the turned part is 0.318 + 0.157 = .475. The thickness of the pin is 0.025 (from the top drawing), so the pins protrude of .475 - .0125 = .4625 into the plug. The depth of the plug itself is 0.489 so the clearance will be 0.489 - 0.4625 = 0.0265.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Wercho gives the correct answers to the other questions.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-10-20T09:04:09.553" LastActivityDate="2018-10-20T09:04:09.553" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24256" PostTypeId="2" ParentId="17978" CreationDate="2018-10-20T13:24:17.187" Score="0" Body="&lt;p&gt;The problem statement, proposed solution, and evaluation are incomplete.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The problem statement does not tell us the complete picture of the system. Are the tubes perfectly insulated or not? What kind of water is in the tubes? Is the flow laminar or turbulent? We must know all of these factors to create a proper model of the real system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The proposed solution is a simple mixing equation based on assumptions that may or may not apply to the actual mixing system. See the notes above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The evaluation quotes a number to four significant digits without reference to whether this level of precision is justifiable by all other values in the calculation. By example, when the specific heat of the water is used to only three significant digits in the calculations, the final answer is only valid as 62.9&lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The comparison is made to a measured value with four significant digits without reference to whether the measurement device has been calibrated to this level of accuracy and is precise to this level. By example, when the device has an overall &lt;span class=&quot;math-container&quot;&gt;$\pm 1$&lt;/span&gt;% calibration + measurement uncertainty, the measured value is reliable at best only as &lt;span class=&quot;math-container&quot;&gt;$63.1^o$&lt;/span&gt;C &lt;span class=&quot;math-container&quot;&gt;$\pm 0.6^o$&lt;/span&gt;C. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In summary, the answer to &quot;Why is (temperature B &lt;em&gt;slightly&lt;/em&gt; higher than temperature A)?&quot; is this: Absent further information, the best we can say is, it is higher only because you &lt;em&gt;think&lt;/em&gt; it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We cannot say anything about a proposed difference of 0.2&lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C until all of the above issues are rectified.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-10-20T13:24:17.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24260" PostTypeId="1" AcceptedAnswerId="24261" CreationDate="2018-10-20T17:52:32.217" Score="3" ViewCount="574" Body="&lt;p&gt;I'd like to use a titanium wire to make some jewelry. I ordered two kinds of titanium wire, 0.3mm thick and 0.5mm thick. Both of them are grade 1 titanium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the strength they'll have, i.e. the amount of weight they could lift before snapping? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that titanium grade 1 has an ultimate tensile strength of 240 MPa, but I don't know how to use that information to get the amount of force needed to snap the wire.&lt;/p&gt;&#xA;" OwnerUserId="11510" LastActivityDate="2018-10-20T19:10:42.590" Title="How to calculate strength for titanium wire" Tags="|materials|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24261" PostTypeId="2" ParentId="24260" CreationDate="2018-10-20T18:52:07.140" Score="0" Body="&lt;p&gt;Stress = Force / Area, and &quot;tensile strength&quot; really means &quot;tensile stress&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to be careful with units (and some people would recommend always convert everything into &quot;basic SI units&quot; of meters and Pascals), but using MPa and mm conveniently gives the force in Newtons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross section area of your thin wire is &lt;span class=&quot;math-container&quot;&gt;$3.14 \times 0.3 \times 0.3 \,/\, 4 = 0.07065$&lt;/span&gt; square mm, so the force to snap the wire is &lt;span class=&quot;math-container&quot;&gt;$0.07065 \times 240 = 16.956$&lt;/span&gt; Newtons, or a weight of &lt;span class=&quot;math-container&quot;&gt;$16.956 / 9.81 = 1.73$&lt;/span&gt; Kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the thicker wire the force is &lt;span class=&quot;math-container&quot;&gt;$4.8$&lt;/span&gt; Kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note these are only approximate. You want to have a safety factor of at least 2, and maybe as high as 5, so the &lt;em&gt;safe&lt;/em&gt; weights might be more like 350 grams and 960 grams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best thing to do would be attach something weighing that amount to a bit of wire, handle it roughly (for example jerk the wire rather than picking it up carefully) and convince yourself it isn't going to break before you start creating your jewellery.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-10-20T19:10:42.590" LastActivityDate="2018-10-20T19:10:42.590" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24263" PostTypeId="1" AcceptedAnswerId="24267" CreationDate="2018-10-20T20:22:24.083" Score="1" ViewCount="38" Body="&lt;p&gt;I am doing research into lunar regolith's hardness in relation to abrasion of lunar tools and need to choose a hardness scale (either Mohs, or Rockwell A, B, or C). Is either of these metrics more legitimate/accurate than the other?&lt;/p&gt;&#xA;" OwnerUserId="17950" LastActivityDate="2018-10-21T01:33:22.197" Title="Mineral Hardness Scales" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24264" PostTypeId="2" ParentId="24242" CreationDate="2018-10-20T20:47:11.557" Score="1" Body="&lt;p&gt;I've had a quick go - it's hard to be more precise without measurements, and it should be stressed that the actual bent shape of the metal is &lt;em&gt;very&lt;/em&gt; weird, and difficult to model. Hopefully this will give you a method to get started with, however, and you can adjust spline handle lengths and dimensions etc. as you go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AEhTY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AEhTY.png&quot; alt=&quot;First Attempt&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steps go as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Model the 'non-weird' bits, and float them in space.&lt;/li&gt;&#xA;&lt;li&gt;Create a 3D sketch with splines connecting the two bodies, set to be tangent to the edges. Right click the spline, and click &quot;activate handle&quot; to give yourself more control over its form. Also include lines connecting these splines to create a closed loop.&lt;/li&gt;&#xA;&lt;li&gt;Use &quot;Boundary Patch&quot; to generate a surface body inside the 3D sketch you just created&lt;/li&gt;&#xA;&lt;li&gt;Mirror the Surface Body to create a symmetrical copy of it.&lt;/li&gt;&#xA;&lt;li&gt;Sketch on the surface that will have the bend attached, and draw two lines. Use the split tool to split the surface accordingly.&lt;/li&gt;&#xA;&lt;li&gt;Use the Delete Face tool to remove the faces of the solid body which will join the bent part. It has now also become a surface body.&lt;/li&gt;&#xA;&lt;li&gt;Use Boundary patch again to create the outer face of the bend this time with automatic edge chain deselected, and selecting the edges of the surface bodies, rather than anything in a sketch. Set a Tangent Condition on the edges that are 'bent', and leave it as Contact for the 'cut' edges. of the tab.&lt;/li&gt;&#xA;&lt;li&gt;Repeat for the inner face of the bend&lt;/li&gt;&#xA;&lt;li&gt;Use the 'Stitch Surface' Tool, and uncheck &quot;Maintain as Surface&quot; to merge all the surface bodies and create a solid.&lt;/li&gt;&#xA;&lt;li&gt;Create a new work plane at a slight angle, and sketch a circle on this for use as an extrude cut for the cutout hole. This is an approximation, of course. You could use the draft tools to create a taper in this hole closer to real life, but without the object for reference, I've decided this is good enough - you can experiment with other methods!&lt;/li&gt;&#xA;&lt;li&gt;All Done&lt;/li&gt;&#xA;&lt;li&gt;Mirror/Rotate/Whichever to create the opposite side of the box. Also, to create a 12th point on this list and make my image nicer :) &lt;a href=&quot;https://i.stack.imgur.com/EMSrO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;High resolution version available here&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EMSrO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EMSrO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-10-20T20:47:11.557" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24265" PostTypeId="2" ParentId="24252" CreationDate="2018-10-20T21:23:23.133" Score="0" Body="&lt;p&gt;Power generation in a car has to do with many factors four of which are easily modified by manufacturer and sometimes even mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One - fuel, the more fuel you get efficiently into the cylinder and the smoother you explode it better performance you get, but of course unless you are a race care driver you want to have the engine that gives you the best MPG rather than competes with a formula one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two - compression ratio, if you can compress more air in a short cylinder head, and use a turbine to squeeze it more, you get more power. But at the same time more wear and tear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three - RPM, and that's why we used to rev-up the engine before take off when we were at high school. Same engine if you make it turn faster will give you more power, but at a cost of wear and tear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Four - designing an efficient engine, such as using lighter alloys, smaller radiators and water pumps, steeper coolant temperature gradient, minimizing moving parts, using fuel injection, designing smooth aspiration and flew of air and managing heat around the engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every manufacturer has their own design which is based on a balance between their cost of design and manufacturing, and their market.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-10-20T21:23:23.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24267" PostTypeId="2" ParentId="24263" CreationDate="2018-10-21T01:33:22.197" Score="0" Body="&lt;p&gt;Rockwell,Brinell, Knoop, Vickers, etc are for materials like metals that have at least a little ductility as the rely on plastic deformation. Mohs is for brittle minerals although it is somewhat qualitative. Abrasion of tools is not that simple anyway. For example : aluminum is softer than many metals but aluminum castings are very abrasive to cutting tools as the high silicon content combines with the aluminum to make hard abrasive particles in the soft matrix. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-10-21T01:33:22.197" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24268" PostTypeId="2" ParentId="24246" CreationDate="2018-10-21T04:11:12.683" Score="0" Body="&lt;p&gt;EDrive Corporation makes a linear actuator called the Eliminator SS rated at 65in/sec and a thrust of 4000lbf. If you want to increase its shock resistance you can take the actuator and mount it in a ruggedized housing of some sort with an opening for actuator arm to do its work. You can call the company and find out if it has a shock rating.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2018-10-21T04:11:12.683" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24272" PostTypeId="1" CreationDate="2018-10-21T10:20:51.513" Score="1" ViewCount="1432" Body="&lt;p&gt;Consider a system made of a cylindrical column (fixed on the ground) with a mass attached on top of it (B). Both the mass of the column and the mass are not negligible. Depending on the value of the mass I could have self-buckling of the column. What I would like to compare is the natural frequency at the top of the column as a function of mass and presence of self-buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If there isn't self buckling, then I can compute the natural frequency of the column+mass system as a cylindrical cantilever (still don't know how, but it seems standard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If I have self-buckling, the column bends and I expect the system to have a natural frequency around the z-axis due to system symmetry (in red). Is that true? How can I compute that frequency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LDV4S.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LDV4S.jpg&quot; alt=&quot;https://en.wikipedia.org/wiki/Self-buckling&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need help especially for &quot;2)&quot;. Any reference/help for &quot;1)&quot; is appreciated.&lt;/p&gt;&#xA;" OwnerUserId="2437" LastEditorUserId="683" LastEditDate="2018-10-21T13:47:38.607" LastActivityDate="2018-10-21T13:47:38.607" Title="Natural frequency of a column + mass" Tags="|mechanical-engineering|structural-analysis|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24273" PostTypeId="2" ParentId="24272" CreationDate="2018-10-21T10:47:34.580" Score="1" Body="&lt;p&gt;(1) What you say is correct, but you need to include the effect of the compressive stress in the column on its lateral stiffness. Names for this are the &quot;stress stiffness&quot; term (as compared with the &quot;elastic stiffness&quot; which depends on the material properties), or &quot;geometric nonlinearity&quot;. Explaining the theory would be far too long to do in an answer here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the column's mass is significant, the compressive stress will increase from the top of the column to the bottom, and you would need to make a computer (finite element) model to get a numerical answer - I doubt whether anyone has bothered to work out and publish the details of an &quot;exact&quot; solution of the equations of motion for that situation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) The frequency for &quot;orbiting around&quot; the initial configuration (which is what you seem have drawn) is the same as the frequency of side-to-side vibration. You can consider it as two side-to-side vibrations oriented 90 degrees apart, with a phase shift between them. Look up Lissajous figures for more explanation. As the mass at the top is increased until the column buckles, this frequency will decrease until it is zero.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;higher&lt;/em&gt; modes of vibration of the column and mass will still exist in theory if the mass is increased beyond that point, but in practice (and after all, engineering is a practical subject), knowing &quot;what the vibration frequency would have been if some magic had stopped the column from bucking&quot; isn't very interesting.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-10-21T10:47:34.580" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24275" PostTypeId="1" AcceptedAnswerId="24281" CreationDate="2018-10-21T12:16:01.673" Score="0" ViewCount="29" Body="&lt;p&gt;If we give a sinusoidal &lt;span class=&quot;math-container&quot;&gt;$\cos(\omega t)$&lt;/span&gt; to a 1/s, it means that you just integrate the given input. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we have the same input but Transfer Funtion with a pole, let's say 1/(s+1), what will be the output of the system ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit #1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my hand calculations, I guess there no way rather than dealing with Laplace Transform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hint: As you may know,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}&#xA;\cos(\omega t) \rightarrow s/(s^2+w^2) \\&#xA;h(s) \rightarrow 1/(s+1)&#xA;\end{gather}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;output --&gt; multiply them, and then apply partial fraction expansion, and then Laplace^-1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.(Took approx. 30 min)&lt;/p&gt;&#xA;" OwnerUserId="17954" LastEditorUserId="1832" LastEditDate="2018-10-21T18:55:28.753" LastActivityDate="2018-10-21T18:55:28.753" Title="Input-Output Relation" Tags="|transfer-function|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ClosedDate="2018-10-21T18:52:50.747" ContentLicense="CC BY-SA 4.0" />
  <row Id="24276" PostTypeId="1" CreationDate="2018-10-21T13:00:25.227" Score="1" ViewCount="959" Body="&lt;p&gt;It's easy to understand the formula for the twist angle in a section due to torsion, and all the units neatly cancel out leaving a unitless number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might sound like a very simple question that it's stupid, I mean the result is very intuitive but I still can't make out how people know that the result would be in radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I reckon it has something to do with torque in the formula since it's the only variable with something to do with circles but I can't figure out the relationship.&lt;/p&gt;&#xA;" OwnerUserId="17955" LastActivityDate="2018-10-21T19:22:44.913" Title="Why is the unit for torsional twist in radians?" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24278" PostTypeId="2" ParentId="24276" CreationDate="2018-10-21T13:16:43.463" Score="1" Body="&lt;p&gt;The results of twist angle is indeed dimensionless, but so are radians! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radian is defined  as the ratio of arc length to radius length, so both arc length and radius length have the same unit, then they cancel each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's common for many dimensionless quantities in engineering and physics to use radians. You'll get used to it.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorUserId="10902" LastEditDate="2018-10-21T14:25:22.097" LastActivityDate="2018-10-21T14:25:22.097" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24279" PostTypeId="1" CreationDate="2018-10-21T15:08:53.840" Score="1" ViewCount="22" Body="&lt;p&gt;I asked the same question in electronics sub, but since it is more general I am asking it here again. I bought the Sharp dust sensor GP2Y1010AU0F. Below are the graph of readings indoor for ~5 minutes, it seemed quite smooth to me. The peak at the 10th second is done on purpose for showing the max reading. The ug/m3 values are created using the reference graph in the data-sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hJnBO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hJnBO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to compare my readings to some references, there are some papers giving the measurements for smoke, candle, deodorant etc. However, it would be better to create and environment with known PM information.I don't have any reliable machinery for creating or measuring the environment PM density. So far, I though of creating a chemical reaction in a 10-20m3 closed room and periodically increasing the reaction parameters. I need your suggestions for testing the sensor with a known input.&lt;/p&gt;&#xA;" OwnerUserId="17957" LastActivityDate="2018-10-21T15:08:53.840" Title="Referencing for Air Quality Monitoring" Tags="|sensors|environmental-engineering|chemistry|calibration|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24281" PostTypeId="2" ParentId="24275" CreationDate="2018-10-21T17:03:18.910" Score="0" Body="&lt;p&gt;For a sinusoidal input signal at frequency &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; in rad/s you can use that the steady state response of the system would be another sinusoidal signal whose amplitude is scaled by &lt;span class=&quot;math-container&quot;&gt;$|H(j\,\omega)|$&lt;/span&gt; and its phase is shifted by &lt;span class=&quot;math-container&quot;&gt;$\angle H(j\,\omega)$&lt;/span&gt;. The total output of the system would be the sum of the steady state response and the transient response. The transient response only depends on the initial conditions and &lt;span class=&quot;math-container&quot;&gt;$H(s)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example when the input is &lt;span class=&quot;math-container&quot;&gt;$\cos (\omega\,t)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H(s)=1/(1+s)$&lt;/span&gt; you get that the steady state response is equal to &lt;span class=&quot;math-container&quot;&gt;$|H(j\,\omega)|\cos(\omega\,t+\angle H(j\,\omega))$&lt;/span&gt;. The values for &lt;span class=&quot;math-container&quot;&gt;$|H(j\,\omega)|$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\angle H(j\,\omega)$&lt;/span&gt; can be found by multiplying and dividing &lt;span class=&quot;math-container&quot;&gt;$H(j\,\omega)$&lt;/span&gt; by the complex conjugate of its denominator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;H(j\,\omega) = \frac{1}{1+j\,\omega}\frac{1-j\,\omega}{1-j\,\omega} = \frac{1-j\,\omega}{1+\omega^2}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;such that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;|H(j\,\omega)| = \frac{\sqrt{1 + \omega^2}}{1+\omega^2}&#xA;$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\angle H(j\,\omega) = atan2(-\omega, 1)&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transient response of &lt;span class=&quot;math-container&quot;&gt;$H(s)$&lt;/span&gt; are a scaled summation of &lt;span class=&quot;math-container&quot;&gt;$e^{\lambda\,t}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; are the poles of &lt;span class=&quot;math-container&quot;&gt;$H(s)$&lt;/span&gt;. So for this example the transient would be &lt;span class=&quot;math-container&quot;&gt;$C\,e^{-t}$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; depending on the initial conditions.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-10-21T17:03:18.910" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24282" PostTypeId="2" ParentId="24276" CreationDate="2018-10-21T19:16:33.233" Score="1" Body="&lt;p&gt;Get back to basics, and the definition of shear strain. (Image source: &lt;a href=&quot;http://www.materials.unsw.edu.au/tutorials/online-tutorials/5-shear-strain&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.materials.unsw.edu.au/tutorials/online-tutorials/5-shear-strain&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OljOy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OljOy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you want to clutter up your formulas with factors of &lt;span class=&quot;math-container&quot;&gt;$\pi/180$&lt;/span&gt;, measuring &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; in radians gives the shear strain &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\gamma = \frac w L = \tan \theta \approx \theta.$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-10-21T19:22:44.913" LastActivityDate="2018-10-21T19:22:44.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24283" PostTypeId="2" ParentId="4086" CreationDate="2018-10-21T22:17:03.523" Score="0" Body="&lt;p&gt;I remember doing something alike but with solids in oil pipes... As it turns out in a quick search in google I found the thesis is now published online by Arizona State University. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oscillation is given in this model by a WENO equation even though the meaning of WENO is Weighted Essentially Non Oscillatory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the idea is this, you divide the pipe system in many small cells in three levels, the last level is defined to be either one phase or the other. Here comes the interesting part; The G(x) parameter (can't remember the name but it is on the documentation for sure) is defined to solve the interphase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have any trouble compiling the software I would gladly help, I had to debug some of it as part of my social services. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://repository.asu.edu/items/34792&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://repository.asu.edu/items/34792&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2018-10-21T22:17:03.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24284" PostTypeId="1" CreationDate="2018-10-21T22:47:25.893" Score="2" ViewCount="102" Body="&lt;p&gt;When designing a concrete slab or rectangular beam, how would you compute the Mn value if there were two rows of rebars one with a yield strenght of 40 ksi and the other with the yield strenght at 60 ksi?&lt;/p&gt;&#xA;" OwnerUserId="17960" LastEditorUserId="1832" LastEditDate="2018-10-22T01:31:31.370" LastActivityDate="2018-10-22T01:48:02.000" Title="How to find resisting moment with rebars of different yield strengths?" Tags="|reinforced-concrete|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24285" PostTypeId="1" CreationDate="2018-10-21T23:57:09.720" Score="0" ViewCount="439" Body="&lt;p&gt;Are there theories that assert about the ideal shape (i.e., ignoring structural constraints) of a propeller?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just looking at the shapes that emerge in practice, it appears that both the fluid &lt;em&gt;speed relative&lt;/em&gt; to the propeller and the fluid's &lt;em&gt;viscosity&lt;/em&gt; are factors that radically affect the optimal shape.  Are there any laws that relate these factors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I have observed in practice, &lt;strong&gt;I would conclude that a ducted fan with blade angles matched to the &lt;em&gt;desired Mach number&lt;/em&gt;, and diameter matched to the &lt;em&gt;desired power&lt;/em&gt;, is a &lt;em&gt;maximally efficient propeller&lt;/em&gt;, meaning that it converts rotational energy into the maximum possible mass flow to its rear.&lt;/strong&gt;  Is this (or some statement like it) correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if so, does it apply across the full range of viscosity and flow speeds – from hypersonic air flight to trawling under water at just a few knots?&lt;/p&gt;&#xA;" OwnerUserId="77" LastActivityDate="2018-12-21T11:42:24.817" Title="Is there an ideal propeller shape?" Tags="|fluid-mechanics|aerospace-engineering|propulsion|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24286" PostTypeId="2" ParentId="24284" CreationDate="2018-10-22T01:48:02.000" Score="1" Body="&lt;p&gt;I won't bother with the actual calculations since those depend on the codes and methods you're using. I'll just focus on the philosophy of how to do it, which is the same in every case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When calculating &lt;span class=&quot;math-container&quot;&gt;$M_n$&lt;/span&gt; in the standard case, you determine a centroid which represents your rebar. If all your rows have the same number of bars and equal vertical spacing between rows, this is trivial to determine, with the centroid located at the average height of all the rows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, however, you need to calculate the adjusted centroid considering the effect of the different yield stresses for each row. You do this by finding the lowest yield stress (in this case, 40&amp;nbsp;ksi) and then, for each row, calculating a ratio between the row's yield stress and that minimum stress. In this case, the ratios would be 1 and 1.5 for the 40&amp;nbsp;ksi and 60&amp;nbsp;ksi rows, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then pretend that all the rows have 40&amp;nbsp;ksi steel, but adjust the steel area of each row by that factor. So you're basically pretending that instead of bars of equal diameter but different steels, that the bars are made of the same steel (40&amp;nbsp;ksi), but one of them uses bars with 50% more area. When calculating the rebar's centroid, this will effectively move it towards the stronger steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you then calculate the necessary steel area, do so using that minimal yield strength (40&amp;nbsp;ksi), and then adjust the necessary area for the stronger steel row by the inverse factor (divide by 1.5, in this case).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, this solution assumes similar behavior between the two steels (i.e. same Young's Modulus, etc), and that the codes don't specify anything different.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-10-22T01:48:02.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24287" PostTypeId="2" ParentId="24285" CreationDate="2018-10-22T03:48:19.990" Score="0" Body="&lt;p&gt;There are several parameters youd need to decide on first.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the viscosity of the fluid? More viscous fluid favors larger propellers with more power to surface area.&lt;/li&gt;&#xA;&lt;li&gt;How how big is the vehicle? Smaller vehicles would generate a smaller wake for a given speed and allow correspondingly propellers before turbulence takes over.&lt;/li&gt;&#xA;&lt;li&gt;What is the speed? Higher speeds result in greater cavitation and work against multi bladed or complex propellers where cavitation interactions could occur.&lt;/li&gt;&#xA;&lt;li&gt;Materials strength limits the rpm of the propeller. An infinitely strong propeller with unlimited fuel would go to lightspeed.&lt;/li&gt;&#xA;&lt;li&gt;(1), (2) and (4) determine the gear ratio and pitch.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once you answer these questions and specificy the fluid properties then one could assume an optimal propeller shaped such that the blade length is that point where material stress exceeds strength.&lt;/p&gt;&#xA;" OwnerUserId="17963" LastActivityDate="2018-10-22T03:48:19.990" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24288" PostTypeId="1" CreationDate="2018-10-22T03:55:48.773" Score="1" ViewCount="54" Body="&lt;p&gt;I need to design an active high pass filter circuit that implements the following transfer function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$G(s)= 4.15\times\dfrac{s+530}{s+2200}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, &lt;span class=&quot;math-container&quot;&gt;$K = 4.15$&lt;/span&gt;, Pole @ 2200 rad/s and Zero @ 530 rad/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already designed the non-inverting amplifier component of this circuit, using two resistors of value 1k and 4.15k. The issue I have is the passive component of the circuit and trying to control the poles and zeros of the transfer function. This is all I have attempted, I am completely stuck.&lt;/p&gt;&#xA;" OwnerUserId="8258" LastEditorUserId="8258" LastEditDate="2018-10-22T11:33:43.273" LastActivityDate="2018-11-03T00:08:01.087" Title="High Pass Filter Design from Transfer Function" Tags="|electrical-engineering|design|transfer-function|electronic-filters|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24290" PostTypeId="1" CreationDate="2018-10-22T04:49:08.893" Score="0" ViewCount="1539" Body="&lt;p&gt;I'm very new to designing shapes in multi views and drawings and got a homework problem on this. I have to draw an isometric from this multiview drawing. I have no clue where to start and am not totally sure what all the lines on the drawing mean. Any help is appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/flDRw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/flDRw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17964" LastActivityDate="2018-10-22T17:34:33.687" Title="Drawing an isometric from a multiview" Tags="|design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24291" PostTypeId="1" AcceptedAnswerId="24293" CreationDate="2018-10-22T06:09:46.147" Score="0" ViewCount="309" Body="&lt;p&gt;I'm new to solidworks, and I built my part and would like to put it into a drawing. Its a small square box-like base with one side of the base having holes in it. When I put my parts into the drawing, it shows 2 sides of the base, a bottom view, and an isometric view, but it never shows the side/plane with the holes on it. My question is this, how do I have a solidworks drawing show a different plane or side of the base?&lt;/p&gt;&#xA;" OwnerUserId="17965" LastActivityDate="2019-01-09T12:07:17.370" Title="Show all sides of a solidworks part in a solidworks drawing" Tags="|solidworks|drawings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24292" PostTypeId="1" CreationDate="2018-10-22T08:40:49.747" Score="0" ViewCount="91" Body="&lt;p&gt;When producing a set of engineering drawings for products we tend to specify the materials used for the part in the notes. For example if it's a moulded part we'd specify PA6 GF20 and then the manufacturer and the manufacturer's code. We then store the TDS for this material in the technical file. We have a new QC guy who is now insisting that we put the polymer TDS on the engineering drawing now - literally paste it into the drawing. This seems wrong to me for several reasons but I'd like to know if anyone else considers this as correct or not.&lt;/p&gt;&#xA;" OwnerUserId="17966" LastEditorUserId="1832" LastEditDate="2018-10-22T11:07:10.303" LastActivityDate="2018-10-26T11:11:02.053" Title="Technical Drawing Content" Tags="|technical-drawing|drawings|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24293" PostTypeId="2" ParentId="24291" CreationDate="2018-10-22T08:57:27.087" Score="1" Body="&lt;p&gt;You can use the &quot;Projected View&quot; button to add another view, by selecting an existing view. You should bear in mind, however, that you should avoid adding unnecessary views if you can fully define the geometry of the object with only a few. I can't comment on this without an image of your object, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sUdNJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sUdNJ.png&quot; alt=&quot;Projected View Button&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-10-22T08:57:27.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24294" PostTypeId="2" ParentId="24292" CreationDate="2018-10-22T09:03:23.723" Score="1" Body="&lt;p&gt;Consider who needs to see the drawing &lt;em&gt;outside your company&lt;/em&gt; (customers, for example) and whether you want to show them all the contents of the TDS. Are you giving away your &quot;trade secrets&quot;, for example? Could your competitors use this information against you?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you have 10,000 parts all made from the same material, do you really want 10,000 copies of the TDS, one on each drawing? What happens if you update the material specification at some future date? Somebody is going to wonder why two specifications of &quot;the same material&quot; on two drawings are different - unless you re-issue all 10,000 drawings with the updated TDS, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There might also be a legal question: if part of the TDS is actually produced by your material suppliers, not by yourselves, do you have their permission to copy and republish it?&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-10-22T09:09:31.843" LastActivityDate="2018-10-22T09:09:31.843" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24297" PostTypeId="2" ParentId="24290" CreationDate="2018-10-22T11:45:43.670" Score="1" Body="&lt;p&gt;As this is a homework question, I have placed the answers behind image links, rather than embedding the images. Think of them like spoilers - once you have a looked at them, you can no longer un-see them, and you will not learn as much from your homework. Please try to work through the problem with help from your teachers/text books before consulting the 'answer key' below!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step in any drawing problem is to fully understand the form of the object at hand. If you're not completely comfortable with what the shape should be, then you need to keep examining the source material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual process for converting those 2D drawings to ISO view, which is what your question seems to be about, primarily, is not at all easy to explain in a text-based format. I shot a quick video showing myself going through the process to create a hand-drawn ISO view from the information provided in your question sheet only. N.B. This was un-prepared, so there's a bit of erasing! I've tried to explain my thought processes as I went along. The only thing to note, is I didn't want to be obstructing the camera so opted not to use a ruler, but &lt;em&gt;you should always use a ruler&lt;/em&gt; for this kind of exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/embed/OOJT2jYBj28&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/embed/OOJT2jYBj28&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A line drawing with hidden edges, of the most useful Isometric view for showing the form of this object, in my opinion: &lt;a href=&quot;https://i.stack.imgur.com/PYmvp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/PYmvp.png&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid model views of of all possible isometric orientations. This will give you a clear understanding of the form of this object. &lt;a href=&quot;https://i.stack.imgur.com/8OgbU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/8OgbU.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-10-22T17:34:33.687" LastActivityDate="2018-10-22T17:34:33.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24300" PostTypeId="2" ParentId="24250" CreationDate="2018-10-22T12:31:52.123" Score="0" Body="&lt;p&gt;Record the sound made from the impact, zoom into and analyze the sound wave to determine what an impact &quot;looks like&quot;, count the impacts. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-10-22T12:31:52.123" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24301" PostTypeId="1" CreationDate="2018-10-22T13:36:10.540" Score="0" ViewCount="393" Body="&lt;p&gt;I have two containers of air, one at 100 kPa and rising, the other at 30 kPa and decreasing. I need a valve that lets air through when the pressure difference is &gt;70 kPa. Where would I find a valve like that? If nowhere, do you have any pointers on how to make one?&lt;/p&gt;&#xA;" OwnerUserId="16695" LastActivityDate="2018-10-22T13:56:15.853" Title="Check valve with precise cracking pressure" Tags="|pressure|valves|pressure-vessel|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24302" PostTypeId="2" ParentId="24301" CreationDate="2018-10-22T13:56:15.853" Score="3" Body="&lt;p&gt;Check Valves are available with adjustable cracking pressures, using a preload spring to control the poppet force, for example: &lt;a href=&quot;http://www.my-ssp-usa.com/products/valves/check-valves/adjustable-cpa&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.my-ssp-usa.com/products/valves/check-valves/adjustable-cpa&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to note, however, that there is a degree of histeresis with these valves, and once open, they will not close if the pressure difference drops a bit - it will have to drop lower than your application may allow. A Back Pressure Regulator may be more suitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w1JiQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w1JiQ.jpg&quot; alt=&quot;Adjustable Check Valve:&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-10-22T13:56:15.853" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24303" PostTypeId="1" AcceptedAnswerId="24311" CreationDate="2018-10-22T19:25:17.913" Score="0" ViewCount="281" Body="&lt;p&gt;I have seen anti-condensation coatings for glass, but as far as I know they're designed to cause any moisture that does form on them to sheet rather than bead. They do not, however, actually prevent condensation, and I don't see how any product could (other than extreme insulation, of course). The formation of condensation is an issue of physics; warm moist air meets cold surface and since cold air holds less moisture than warm, the moisture forms on the cold surface. On the other hand, engineers are pretty clever, so if there's a way to prevent this, it may be possible, even if I can't figure out how.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is actually prompted by some reviews of some &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B07CT98WJM&quot; rel=&quot;nofollow noreferrer&quot;&gt;ice packs&lt;/a&gt; I recently purchased from Amazon, in which reviewers complained about condensation, which seems like a nonsense complaint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry about the tag, but I tried &quot;condensation,&quot; &quot;dew,&quot; &quot;moisture,&quot; &quot;water,&quot; and maybe a couple of others and got no match.&lt;/p&gt;&#xA;" OwnerUserId="12535" LastActivityDate="2018-10-22T23:20:01.743" Title="Would it be possible to put an anti-condensation coating on, say, an ice bag such that it will not get wet in use?" Tags="|surface-preparation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24304" PostTypeId="1" CreationDate="2018-10-22T19:36:18.853" Score="1" ViewCount="394" Body="&lt;p&gt;In heat transfer, we come across Stanton number which is the ratio of heat transferred into a fluid to the thermal capacity of the fluid. But what does this physically mean and its importance in engineering applications?&lt;/p&gt;&#xA;" OwnerUserId="17075" LastEditorUserId="9" LastEditDate="2018-10-24T02:44:49.653" LastActivityDate="2018-10-24T02:44:49.653" Title="What does stanton mean physically?" Tags="|mechanical-engineering|heat-transfer|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24305" PostTypeId="2" ParentId="24303" CreationDate="2018-10-22T19:54:09.720" Score="0" Body="&lt;p&gt;Silicone coatings have been developed for these application, if I remember correctly , Exxeen Corp used to make an anti-fog silicone hard coat that might be useful to you.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2018-10-22T19:54:09.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24306" PostTypeId="2" ParentId="24250" CreationDate="2018-10-22T20:02:53.030" Score="0" Body="&lt;p&gt;Piezoelectric thin film sensors can be made as small as you want (well maybe not Planck length) :-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you have to do is pre-amp the signal from the sensor and send the output to a scope to see the waveform caused by the impact, or any other recording device you want.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2018-10-22T20:02:53.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24308" PostTypeId="1" CreationDate="2018-10-22T21:17:56.853" Score="1" ViewCount="3180" Body="&lt;p&gt;Sorry for this stupid, basic question. But I got hung up on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know from Clapeyron's theorem that the strain(internal) energy of an elastically deformed body is equal to half the work&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;U=\frac{W}{2}&#xA;$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm effectively confused about is, where does the rest of the energy from the work go? From conservation of energy the other half of the work has to be somewhere. Can you say that the other half is potential energy? Or is strain energy and potential energy effectively the same? Or am I thinking about this all wrong? &lt;/p&gt;&#xA;" OwnerDisplayName="user14808" LastActivityDate="2018-10-23T11:41:04.093" Title="Is strain energy the same as potential energy, or something differen?" Tags="|mechanical-engineering|applied-mechanics|springs|strength|elastic-modulus|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24309" PostTypeId="2" ParentId="18377" CreationDate="2018-10-22T21:19:25.707" Score="0" Body="&lt;p&gt;This web page has a beam just like you are talking about and similar to what I have previously designed for a different site.  The difference is, it right in front of the bridge as opposed to 100 m ahead.  Nice piece of sacrificial steel.  Cheaper to replace it than fix the bridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://11foot8.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://11foot8.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6993" LastActivityDate="2018-10-22T21:19:25.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24310" PostTypeId="2" ParentId="24308" CreationDate="2018-10-22T21:44:29.553" Score="4" Body="&lt;p&gt;This has been asked and answered numerous times before: &lt;a href=&quot;https://www.quora.com/Why-is-strain-energy-equal-to-1-2*force*displacement-What-about-the-remaining-half&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.quora.com/Why-is-strain-energy-equal-to-1-2*force*displacement-What-about-the-remaining-half&lt;/a&gt; basically, it's because you're calculating the area under a triangle, because the strain energy increases linearly as the displacement increases. The Work term here refers to the final position, but the energy stored is from a relaxed state, and all of the intermediate positions in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7c9Z9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7c9Z9.png&quot; alt=&quot;Load Extension Energy Graph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-10-22T21:44:29.553" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24311" PostTypeId="2" ParentId="24303" CreationDate="2018-10-22T23:20:01.743" Score="3" Body="&lt;p&gt;The condensation mechanism of water vapor onto a chilled surface requires not only a chilled surface but also the presence of nucleation seed sites in the surface to kickstart the condensation kinetics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those nucleation seeds are sites on the surface (usually microscopic pores, crevices or specks of dust) which have a little water already trapped in them. The supersaturated vapor nearby &quot;sees&quot; that water and comes out of solution right there, adding to the amount of water originally present. Growing globs of condensed water then result, which can then freeze solid if conditions are right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By coating a chilled surface with an extremely low-energy substance like a fluorosilicon oil or wax or a thin layer of a solid fluoropolymer, those surface sites are poisoned and no longer present themselves to the nearby water vapor as pre-existing bodies of condensate- and condensation is delayed or possibly eliminated altogether. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with inhibiting condensation with low-energy coatings is that they eventually degrade and lose their effectiveness by accumulating dirt and grime. So a brand-new jacket made of Gore-Tex(tm) inhibits conndensation for a few months of wear but then supports condensation like anything else after it gets dusty &amp;amp; dirty. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-10-22T23:20:01.743" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24312" PostTypeId="1" AcceptedAnswerId="24325" CreationDate="2018-10-22T23:55:12.653" Score="0" ViewCount="556" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Eh80H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Eh80H.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm an industrial engineering student taking engineering mechanics, and I'm quite confused by this problem our proffessor gave us, the chair is supposely in equilibrium, yet in order to counter the moment created by the weight pressed on the chair upward y vector vertical force on leg be would need to be eade, and cos(b) is greater than cos(a) - and she says there's no friction to insure horizntal equilibrium through resistance, could someone please show me where I'm screwing up on this, - please help! I don't want the answer I just want to see what I 'm doing wrong and fi&amp;lt; it, I just want to know how to solve this problem...  &lt;/p&gt;&#xA;" OwnerUserId="17975" LastEditorUserId="33" LastEditDate="2018-10-24T15:07:51.903" LastActivityDate="2018-10-24T15:07:51.903" Title="Statics problem ignoring horizontal reactions seems to require horizontal forces" Tags="|statics|" AnswerCount="3" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24313" PostTypeId="2" ParentId="3685" CreationDate="2018-10-23T01:58:21.237" Score="2" Body="&lt;p&gt;You can select a single body by this code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;agb.AddSelect(agc.TypeBody, bodyname);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Here is the script:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Key ideas: First, assign name &quot;GEAR&quot; to the gear body &lt;code&gt;ag.fm.Body(2).Name = &quot;GEAR&quot;;&lt;/code&gt;. Second, creat gear object by &lt;code&gt;gear = selectNode(&quot;GEAR&quot;);&lt;/code&gt; where the function &lt;code&gt;selectNode(target)&lt;/code&gt; is at the end. Third, use &lt;code&gt;agb.AddSelect(agc.TypeBody, gear);&lt;/code&gt; after &lt;code&gt;ag.bodyPick;&lt;/code&gt;)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ag.m.ClearAllErrors();&#xA;ag.gui.setUnits(ag.c.UnitMillimeter, ag.c.UnitDegree, ag.c.No);&#xA;&#xA;var p = new Object();&#xA;&#xA;//Plane&#xA;p.Plane  = agb.GetActivePlane();&#xA;p.Origin = p.Plane.GetOrigin();&#xA;p.XAxis  = p.Plane.GetXAxis();&#xA;p.YAxis  = p.Plane.GetYAxis();&#xA;&#xA;p.Sk1 = p.Plane.NewSketch();&#xA;p.Sk1.Name = &quot;Pinion&quot;;&#xA;p.Sk2 = p.Plane.NewSketch();&#xA;p.Sk2.Name = &quot;Wheel&quot;;&#xA;&#xA;//PINION&#xA;with (p.Sk1)&#xA;{&#xA;    p.Sp1 = SplineBegin();&#xA;    with(p.Sp1)&#xA;    {&#xA;        SplineFlexibility = agc.Yes;&#xA;        SplineXY(-1.76564478418, 24.6869398366);&#xA;        SplineXY(-1.51020858813, 24.7038816792);&#xA;        //......... spline coordinates of the pinion profile&#xA;        //......... removed because character limit&#xA;        SplineXY(5.26099091644, 24.1843849328);&#xA;&#xA;        SplineFitPtEnd();&#xA;    }&#xA;    p.Ln2 = Line(-1.76564478418, 24.6869398366, -1.24843570599,17.4554120057);&#xA;    p.Ln3 = Line(3.71989256718, 17.1000701545, 5.26099091644, 24.1843849328);&#xA;    p.Cr4 = ArcCtrEdge(&#xA;              0, 0,&#xA;              3.71989256718, 17.1000701545,&#xA;              -1.24843570599, 17.4554120057);&#xA;    p.Ext1 = agb.Extrude(agc.Add, p.Sk1, agc.DirNormal, agc.ExtentFixe, 34.0, agc.ExtentFixed, 0.0, agc.No, 0.0, 0.0);&#xA;}&#xA;agb.Regen();&#xA;&#xA;&#xA;var PF1 = ag.gui.CreateSpot();&#xA;ag.listview.ActivateItem(&quot;Type&quot;);&#xA;ag.listview.ItemValue = &quot;Construction Point&quot;;&#xA;ag.listview.ActivateItem(&quot;Definition&quot;);&#xA;ag.listview.ItemValue = &quot;Manual Input&quot;;&#xA;ag.listview.ActivateItem(&quot;FD8, X Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD9, Y Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD10, Z Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;agb.Regen();&#xA;&#xA;var PF2 = ag.gui.CreateSpot();&#xA;ag.listview.ActivateItem(&quot;Type&quot;);&#xA;ag.listview.ItemValue = &quot;Construction Point&quot;;&#xA;ag.listview.ActivateItem(&quot;Definition&quot;);&#xA;ag.listview.ItemValue = &quot;Manual Input&quot;;&#xA;ag.listview.ActivateItem(&quot;FD8, X Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD9, Y Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD10, Z Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;34.0&quot;;&#xA;agb.Regen();&#xA;&#xA;var LF1 = agb.LinePt();&#xA;LF1.AddSegment(PF1.GetPoint(1, 0),PF2.GetPoint(1, 0), 0);&#xA;agb.Regen();&#xA;&#xA;var Pat = ag.gui.CreatePattern();&#xA;ag.listview.ActivateItem(&quot;Pattern Type&quot;);&#xA;ag.listview.ItemValue = &quot;Circular&quot;;&#xA;&#xA;ag.listview.ActivateItem(&quot;Geometry&quot;);&#xA;ag.bodyPick;&#xA;ag.gui.PickFilter(5, true);&#xA;ag.gui.PickFilter(11, false);&#xA;ag.gui.PickFilter(11, false);&#xA;ag.gui.SelectAll();&#xA;ag.listview.ItemValue = &quot;Apply&quot;;&#xA;&#xA;ag.listview.ActivateItem(&quot;Axis&quot;);&#xA;ag.edgePick;&#xA;ag.gui.SelectAll();&#xA;ag.listview.ItemValue = &quot;Apply&quot;;&#xA;ag.listview.ActivateItem(&quot;FD2, Angle&quot;);&#xA;ag.listview.ItemValue = &quot;16.3636363636&quot;;&#xA;ag.listview.ActivateItem(&quot;FD3, Copies (&amp;gt;=0)&quot;);&#xA;ag.listview.ItemValue = &quot;21&quot;;&#xA;agb.Regen();&#xA;&#xA;//GEAR&#xA;with (p.Sk2)&#xA;{&#xA;  p.Sp1 = SplineBegin();&#xA;  with(p.Sp1)&#xA;  {&#xA;    SplineFlexibility = agc.Yes;&#xA;    SplineXY(1.9304744152, 30.2623156532);&#xA;    //......... spline coordinates of the gear profile&#xA;    //......... removed because character limit&#xA;    SplineXY(-5.79031304774, 30.3548985987);&#xA;&#xA;    SplineFitPtEnd();&#xA;  }&#xA;  p.Ln2 = Line(1.9304744152, 30.2623156532, 0.209834175566,173.751999593);&#xA;  p.Ln3 = Line(-0.629381853016, 173.762062957, -5.79031304774, 30.3548985987);&#xA;  p.Cr4 = ArcCtrEdge(&#xA;              0, 191.250741537,&#xA;              -0.629381853016, 173.762062957,&#xA;              0.209834175566, 173.751999593);&#xA;  p.Ext2 = agb.Extrude(agc.Add, p.Sk2, agc.DirNormal, agc.ExtentFixe, 34.0, agc.ExtentFixed, 34.0, agc.No, 0.0, 0.0);&#xA;}&#xA;&#xA;var PF3 = ag.gui.CreateSpot();&#xA;ag.listview.ActivateItem(&quot;Type&quot;);&#xA;ag.listview.ItemValue = &quot;Construction Point&quot;;&#xA;ag.listview.ActivateItem(&quot;Definition&quot;);&#xA;ag.listview.ItemValue = &quot;Manual Input&quot;;&#xA;ag.listview.ActivateItem(&quot;FD8, X Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD9, Y Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;77.5&quot;;&#xA;ag.listview.ActivateItem(&quot;FD10, Z Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;agb.Regen();&#xA;ag.fm.Body(2).Name = &quot;GEAR&quot;;&#xA;&#xA;var PF4 = ag.gui.CreateSpot();&#xA;ag.listview.ActivateItem(&quot;Type&quot;);&#xA;ag.listview.ItemValue = &quot;Construction Point&quot;;&#xA;ag.listview.ActivateItem(&quot;Definition&quot;);&#xA;ag.listview.ItemValue = &quot;Manual Input&quot;;&#xA;ag.listview.ActivateItem(&quot;FD8, X Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;0&quot;;&#xA;ag.listview.ActivateItem(&quot;FD9, Y Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;77.5&quot;;&#xA;ag.listview.ActivateItem(&quot;FD10, Z Coordinate&quot;);&#xA;ag.listview.ItemValue = &quot;34.0&quot;;&#xA;agb.Regen();&#xA;&#xA;var LF2 = agb.LinePt();&#xA;LF2.AddSegment(PF3.GetPoint(1, 0),PF4.GetPoint(1, 0), 0);&#xA;agb.Regen();&#xA;&#xA;var Pat = ag.gui.CreatePattern();&#xA;ag.listview.ActivateItem(&quot;Pattern Type&quot;);&#xA;ag.listview.ItemValue = &quot;Circular&quot;;&#xA;&#xA;ag.listview.ActivateItem(&quot;Geometry&quot;);&#xA;ag.gui.PickFilter(5, true);&#xA;ag.gui.PickFilter(11, false);&#xA;gear = selectNode(&quot;GEAR&quot;);&#xA;ag.bodyPick;&#xA;agb.AddSelect(agc.TypeBody, gear);&#xA;ag.listview.ItemValue = &quot;Apply&quot;;&#xA;ag.listview.ActivateItem(&quot;Axis&quot;);&#xA;ag.edgePick;&#xA;ag.gui.SelectAll();&#xA;ag.listview.ItemValue = &quot;Apply&quot;;&#xA;ag.listview.ActivateItem(&quot;FD2, Angle&quot;);&#xA;ag.listview.ItemValue = &quot;2.74809160305&quot;;&#xA;ag.listview.ActivateItem(&quot;FD3, Copies (&amp;gt;=0)&quot;);&#xA;ag.listview.ItemValue = &quot;130&quot;;&#xA;agb.Regen();&#xA;&#xA;function selectNode (target)&#xA;{&#xA;    var DM = ag.wb.AppletList.Applet( &quot;AGApplet&quot; ).App;&#xA;    var Nodes = DM.Script.ag.tree.Nodes;&#xA;    var count = Nodes.Count;&#xA;    var name, current;&#xA;    for (var i =1; i &amp;lt;= count; i++)&#xA;    {&#xA;        current = Nodes(i);&#xA;        name = current.Text.toLowerCase();&#xA;        if (name == target.toLowerCase())&#xA;        {&#xA;            DM.Script.agTree_LeftClick(current, false);&#xA;            var obj = ag.listviewSelectedObject;&#xA;            return obj;&#xA;        }&#xA;    }&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="5069" LastEditorUserId="5069" LastEditDate="2018-10-23T18:27:03.457" LastActivityDate="2018-10-23T18:27:03.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24314" PostTypeId="2" ParentId="24304" CreationDate="2018-10-23T02:20:37.017" Score="1" Body="&lt;p&gt;The physical mean of St No. How much energy the fluid can adhere. More the heat capacity more the energy accumulated by the fluid. The significance can be very well understood by the refrigerants quality criteria in the HVAC systems. &lt;/p&gt;&#xA;" OwnerUserId="17976" LastActivityDate="2018-10-23T02:20:37.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24316" PostTypeId="1" CreationDate="2018-10-23T09:49:47.583" Score="1" ViewCount="187" Body="&lt;p&gt;I want produce water filter housing (like below images)&#xA;who can tell me what are need raw material for these?&#xA;Best Regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y6nVB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y6nVB.jpg&quot; alt=&quot;water filter housing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/01iBT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/01iBT.png&quot; alt=&quot;water filter housing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17978" LastActivityDate="2018-10-23T09:49:47.583" Title="What are the raw materials for water filter housing production?" Tags="|materials|chemical-engineering|manufacturing-engineering|" AnswerCount="0" CommentCount="1" ClosedDate="2018-10-23T11:07:27.797" ContentLicense="CC BY-SA 4.0" />
  <row Id="24317" PostTypeId="1" AcceptedAnswerId="24330" CreationDate="2018-10-23T10:07:32.560" Score="0" ViewCount="64" Body="&lt;p&gt;I'm trying to wrap my head around the question: &quot;what is the optimal slip ratio and how do I go about controlling it?&quot;. I'm working on a control strategy for an electrical vehicle. The issue I'm facing is that of maximizing traction. The way I can measure slip ratio is by measuring difference between wheel velocity and vehicle velocity (both of which let's assume are measured separately).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what I need to do (I think) is to figure out a way to maximize traction by always keeping the slip ratio at some specific value. But what value? Also it seems wrong to always try to keep wheel speeds at an offset from vehicle speed. I suppose I only should be doing this during acceleration/deceleration? How can I do this in a generic way without having to implement different strategies for accel/decel? &lt;/p&gt;&#xA;" OwnerUserId="16582" LastActivityDate="2018-11-29T11:25:29.090" Title="How to actually control slip?" Tags="|automotive-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24319" PostTypeId="2" ParentId="24308" CreationDate="2018-10-23T11:28:26.583" Score="0" Body="&lt;p&gt;Say we consider a piece of flat steel  bar, length L, &#xA; And a force P pulling at the right end. So the actual force acting on that end is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total Force = P-kx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the balance state P = kx when x grows from zero to x, but at the beginning of stretching the bar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X = 0 , therefore kx = 0 and at the end of stretch, kx = P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the Total force varies linearly from &lt;span class=&quot;math-container&quot;&gt;$$ P_{initial}\space to\space P_{ final}= P\space to--&amp;gt; \space(P- kx), \space \space or\space P \space to --&amp;gt;\space 0 $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a right angle triangle starting at P on the left side and tapering down to zero at right at distance of x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the strain on the bar is the average force times the displacement which is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$U = \frac {P-0 } {2 }\space\times x $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$U = \frac {P\times x }{ 2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$U = \frac {W }{2 }  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as we see there is no 'hidden half' of work, all work gone into stretching bar has added up to its final strain.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-10-23T11:41:04.093" LastActivityDate="2018-10-23T11:41:04.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24320" PostTypeId="1" CreationDate="2018-10-23T13:33:15.647" Score="2" ViewCount="154" Body="&lt;p&gt;Consider the following circuit: &lt;a href=&quot;https://i.stack.imgur.com/kWZtg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kWZtg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let both capacitors be charged to equal voltage. The circuit acts as LC oscillator circuit with resonance at &lt;span class=&quot;math-container&quot;&gt;$f= \frac{1}{2\pi\sqrt(LC)}$&lt;/span&gt;, where C is the resultant capacitance of C1 and C2. What frequency of oscillation will be experienced by both the individual capacitors? Will it be same as f, or different?&lt;/p&gt;&#xA;" OwnerUserId="17982" LastEditorDisplayName="user14407" LastEditDate="2018-10-23T14:43:03.553" LastActivityDate="2023-12-27T10:00:20.043" Title="LC oscillator circuit" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24321" PostTypeId="2" ParentId="24320" CreationDate="2018-10-23T13:53:40.450" Score="0" Body="&lt;p&gt;Yes, the frequency remains the same, why ? Suppose you replaced the capacitors with an equivalent capacitor but the rest of circuit remains the same, so the current and voltage oscillate with frequency say &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; across the inductor, now consider the original circuit, still you will find the same current and voltage across the inductor so nothing has changed, because this system is linear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the elements in this circuit are linear, so we can define the behaviour of the system with linear differential equations,  we can rewrite the solution as a linear combination of imaginary exponential functions, the arguments of these exponential function have the common factor that we call angular frequency, is equal to &lt;span class=&quot;math-container&quot;&gt;$2\pi f$&lt;/span&gt;.  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-10-23T14:08:40.627" LastActivityDate="2018-10-23T14:08:40.627" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24322" PostTypeId="1" AcceptedAnswerId="24327" CreationDate="2018-10-23T14:45:10.660" Score="0" ViewCount="113" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thermostat#Pneumatic_thermostats&quot; rel=&quot;nofollow noreferrer&quot;&gt;According to Wikipedia&lt;/a&gt;, pneumatic thermostats were invented after electric thermostats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet &lt;a href=&quot;http://members.questline.com/Article.aspx?articleID=26707&amp;amp;accountID=4536&amp;amp;nl=15031&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; claims that &quot;at least 25 percent of U.S. commercial buildings are served by pneumatic controls, according to the U.S. Department of Energy&quot;.  I would guess that they were even more popular in the past.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems surprising.  Why set up an air compressor system and run pneumatic tubes through the walls?  Wouldn't an electric system be easier to set up and maintain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What were the factors that made pneumatic thermostats so popular?&lt;/p&gt;&#xA;" OwnerUserId="17983" LastActivityDate="2018-10-23T17:42:29.710" Title="Why were pneumatic thermostats commonly installed in buildings?" Tags="|building-design|pneumatic|engineering-history|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24325" PostTypeId="2" ParentId="24312" CreationDate="2018-10-23T16:52:02.790" Score="0" Body="&lt;p&gt;Since the weight W acts between the points A and B, the chair will not tip over. So the chair is in equilibrium. Writing the following equations will let you solve for the two support reactions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;sum of forces in Y direction = 0.&lt;/li&gt;&#xA;&lt;li&gt;sum of moments about point A = 0.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2018-10-23T16:52:02.790" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24326" PostTypeId="2" ParentId="24312" CreationDate="2018-10-23T17:03:54.300" Score="1" Body="&lt;p&gt;You should take moments around any one point, Say A. And because no data about the sections of the chair and their weight is given, you can ignore them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No horizontal component of the weight and no friction tells us no horizontal reaction. So converting to centimeters we have the moment of weight around A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}&#xA;M_{W\text{ about }A } = 32.5 \times 36\text{ cm} = 1152\text{ cmkg} \\&#xA;\sum M_A =0 \\&#xA;R_b\times 50\text{ cm} = M_{W\text{ about }A} = 1152\text{ cmkg} \\&#xA;1152/50\text{ cm} = 23.4\text{  kg} = R_B&#xA;\end{gather}$$&lt;/span&gt;&#xA;the vertical reaction of B and from there you subtract it from 32.5 to get the vertical reaction of A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there are no more reactions.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2018-10-23T20:56:04.767" LastActivityDate="2018-10-23T20:56:04.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24327" PostTypeId="2" ParentId="24322" CreationDate="2018-10-23T17:42:29.710" Score="3" Body="&lt;p&gt;In the 1940's and 50's, dynamic system controllers were invented which used air pressure to work. It was possible in fact to design and build not only analog control mechanisms but also binary logic devices that ran on compressed air and which could perform simple switching functions. These devices were less expensive and more durable than electronic logic based on vacuum tubes and relays and were popular all the way up into the early 1970's. They were called &quot;pneumatic logic&quot; or &quot;air control&quot; and they disappeared quickly when solid state logic became cheap. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-10-23T17:42:29.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24328" PostTypeId="2" ParentId="23284" CreationDate="2018-10-23T19:36:03.210" Score="4" Body="&lt;p&gt;The main issue is practicality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vast majority of car buyers have certain expectations in terms of comfort, features and interior space which are pretty much essential in their choice of vehicle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to be able to carry 4 peopel in reasonable compfort and have the ability to carry a sensible amount of luggage then that imposes certain constraints on the overall layout. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classic sports car is 2 seater, front engine, rear wheel drive with a low centre of mass and relatively short wheelbase ie it is designed for handling and performance first. These are the features whcih define its overall silhouette and body shape but also seriously impinge on interior space compared to say a fwd hatchback.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are actually a few pretty decent sports cars produced in reasonable volume, things like the MR2 and MX5. These aren't particularly expensive but they still have the basic practical issues of any sports car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the problem is not cost as such but finding a large enough market to justify volume production for something which is inherently a niche product. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally even for customers who are prepared to pay a bit of a premium many will go for a high performance hatchback or a more luxuriously equipped saloon or estate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed if anything the market is skewed towards cars which look more bulky with high seating positions. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-10-23T19:36:03.210" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24329" PostTypeId="1" CreationDate="2018-10-23T20:45:15.960" Score="1" ViewCount="39" Body="&lt;p&gt;I have a &lt;a href=&quot;https://uk.rs-online.com/web/p/pneumatic-profile-cylinders/4802282/&quot; rel=&quot;nofollow noreferrer&quot;&gt;piston made of anodized aluminium and stainless steel&lt;/a&gt; with nitrile seals which I need to connect hydraulically to our system which mainly consists of silicone rubber with polycarb and acrylic parts, but I'm having problems finding a fluid which will lubricate the pistons without damaging the seals or any of the materials in the system. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Water&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The piston and piston seals (nitrile rubber) are all pretty corrosion resistant but even so the manufacturer doesn't recommend using water in them and can't say what the effects would be, other than washing the existing lubricant out - I don't know how much of an issue water would cause? &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;As for oils:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Mineral oil attacks the silicone rubber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicone oil apparently 'hydrates' the silicone rubber, so softens and distorts it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PE-glycol based automotive brake fluid is fine for the silicone and pistons but cracks the acrylic and polycarb. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vegetable oils are fine for all parts chemically, but I was always taught not to use veg oil as a lubricant and &lt;a href=&quot;https://diy.stackexchange.com/questions/7079/what-happens-if-i-use-vegetable-oil-instead-of-machine-oil-or-grease&quot;&gt;this question&lt;/a&gt; seems to answer why - i.e. the veg oil attracts dirt and 'gums up'. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;So my questions are:&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Would water be suitable as a lubricant for the piston?&lt;/li&gt;&#xA;&lt;li&gt;If its a closed system and the oil is changed regularly, would vegetable oil (e.g. corn oil) work?&lt;/li&gt;&#xA;&lt;li&gt;Any suggestions for something to avoid all the above issues?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Many thanks all&lt;/p&gt;&#xA;" OwnerUserId="11394" LastActivityDate="2018-10-23T20:45:15.960" Title="Fluid for mechanical lubrication and hydraulic connection to plastic/rubber parts" Tags="|hydraulics|fluid|pistons|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24330" PostTypeId="2" ParentId="24317" CreationDate="2018-10-23T21:45:45.433" Score="2" Body="&lt;p&gt;Ultimately it is empirical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For absolution maximum traction rubber tyres tend to deliver maximum grip in a fairly narrow window in the transition between purely static friction and full sliding friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is very contextual and will depend on tyre construction and chemistry as well as tyre temperature and the nature and surface properties of the pavement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F1 teams expend vast resources in understanding this and need to do it again form scratch for each track they go to and it can all go out of the window if the weather changes even slightly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a road car you need to be fairly conservative. It is certainly possible that some sort of machine learning approach could provide adaptive control based on the conditions but in terms of traction the conservative approach is to aim to have zero slip. If noting else this will be safer for normal driving.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-10-23T21:45:45.433" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24331" PostTypeId="1" CreationDate="2018-10-24T04:17:22.157" Score="1" ViewCount="409" Body="&lt;p&gt;I'm asking this question as an early computer science student who is beginning to get way more interested in really low level stuff/computer/electrical engineering. I am very unfamiliar with hardware, and even after watching numerous videos and reading numerous articles there is still something I can't seem to grasp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You write a program, compile it, it is loaded into memory, and then...???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know a transistor can be made to store one bit of data. Okay, fair enough....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when you load a program into memory, what is...happening? I'm familiar with logic gates and all of that, but what does it mean to be...In memory? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a program loaded into memory, it's in its machine code computer readable format and being processed, how do those values translate from &quot;being in memory&quot; into transistors being on and off? From my understanding, the program counter is reading these pieces of data into your memory, so when you're in memory, does that automatically mean there's this on-off process going on? Is that the last step? How do physical hardware chips &quot;read&quot; this data that is in memory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think what's throwing me off is that every video I've seen representing how a transistor stores a bit of data shows a guy switching wires into different holes on a board demonstrating it. But there isn't a little man inside the computer switching wires around. How does that process take place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me out here, I am ignorant and want to learn. &lt;/p&gt;&#xA;" OwnerUserId="17994" LastActivityDate="2019-12-18T22:01:40.337" Title="Memory and Transistors" Tags="|computer-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24333" PostTypeId="2" ParentId="24317" CreationDate="2018-10-24T06:04:44.757" Score="0" Body="&lt;p&gt;If you want to design a performance tire just for best traction, just before slippage and are not at the moment concerned with endurance, heat control, and many other factors, first you need to select quality material rubber which is soft and will leave a sacrificial black track under high acceleration when it heats up. Fusing the tire to the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;like the revving drag cars do to turn the surface of tire to a highly sticky goo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you tires structure has to be layered in concentric rings and build gradually from the first adhesive layer designed for ultimate grip, to more and more tougher layers to have both strength and elasticity to transmit the power of drive shaft to the skin layer while providing adequate play to give the skin time to adjust gently to power jerks of the engine and the road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the walls of the tire have to be designed not to wrinkle or yaw at power step ups or sharp turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experienced drivers know the trick of driving on icy road without skidding, easy does it, no sudden accelerations or breaks! A good tire does this naturally, while maintaining a tight grip, it allows the engine to power up and down and the car take turns without losing grip, by letting the sanwitched layers take the impact.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-10-24T06:11:39.170" LastActivityDate="2018-10-24T06:11:39.170" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24335" PostTypeId="2" ParentId="24331" CreationDate="2018-10-24T09:00:23.170" Score="1" Body="&lt;p&gt;When somebody in a demo is &quot;switching wires&quot;, what he is really doing is applying different voltages (representing logical &quot;0&quot; and &quot;1&quot; levels) to some point in the circuit. In a typical memory circuit, there is a &quot;write enable&quot; control voltage as well as the &quot;data input&quot; voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the &quot;write enable&quot; signal is &quot;on&quot;, the memory circuit will change its internal state depending on whether &quot;0&quot; or &quot;1&quot; is applied to its input. When the &quot;write enable&quot; is &quot;off&quot;, the memory &lt;em&gt;ignores&lt;/em&gt; any changes to the input, and remains in the same state - i.e. it &quot;remembers&quot; what the last input was. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might find it easier to understand this by &lt;em&gt;not&lt;/em&gt; going directly to the level of what each transistor is doing, but to learn how digital circuits can be designed and constructed from building blocks (often called &quot;logic gates&quot;) that implement simple logical functions like &quot;and&quot;, &quot;or&quot;, and &quot;not&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In computer science terms, that is the &lt;em&gt;interface&lt;/em&gt; to a set of &quot;digital functions&quot; that can be used to design circuits, while the details of what the transistors are doing is the &lt;em&gt;implementation&lt;/em&gt; of each function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most good books and web courses on digital electronics will separate teaching about the design and implementation aspects of the problem. Many circuit designers will never be concerned with the details of the implementation - in exactly the same way that many computer programmers use the API of a library of functions, without knowing (or needing to know!) exactly how they were written.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course memory circuits are used in very large quantities, and need to be as efficient (fast, low power dissipation, physically small, etc) as possible, so the design using transistors or other semiconductor devices is often highly &quot;optimised.&quot; Just as in computer programming, this can make their internal operation hard to understand, but it doesn't change the way it is used - i.e. its interface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with other topics in computer science, an important concept here is working at different &quot;levels of abstraction&quot;. A complete, functional CPU is &quot;only&quot; a (large!) collection of transistors, which are used to create logic gates, which are used to create more complex sub-components (e.g. registers, adders, multipliers, etc) … right up the the level of &quot;the CPU&quot;. In fact there may be a course in your CS degree which takes you through the whole process in detail, for a simple CPU design. But to understand even a simple CPU (built from &quot;only&quot; thousands of transistors, not millions), trying to understand what each individual transistor is doing is &lt;em&gt;not&lt;/em&gt; the best approach - just as you wouldn't attempt to understand a computer program containing thousands of lines of code by looking at each line of code separately, and ignoring the fact that the individual lines are organized into several nested &quot;layers of abstraction&quot; like statement groups, functions, classes, design patterns, etc.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-10-24T09:20:36.533" LastActivityDate="2018-10-24T09:20:36.533" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24337" PostTypeId="1" CreationDate="2018-10-24T12:37:36.273" Score="0" ViewCount="212" Body="&lt;p&gt;I am interested in creating a nitrogen infused drink. I want to be able to make small batches of my drink (its not coffee) and testing the market before having to buy expensive equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a way to do this is to buy a strong sealable container, fill it with my beverage and add liquid nitrogen to it. I'll then seal the top and wait some time for the liquid nitrogen to infuse throughout the drink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will then decant from my stronger container into smaller sealed bottles for trials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few questions I have are these:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Will the liquid nitrogen evaporate immediately as I add it to the drink, being lost to the atmosphere and thus not infusing the drink at all.&lt;/li&gt;&#xA;&lt;li&gt;In order to counter part 1 can I just add a lot more liquid nitrogen than would be needed to fully dissolve in the drink in the knowledge that the container can handle the pressure.&lt;/li&gt;&#xA;&lt;li&gt;If a beer bottle can maintain the carbonation of a drink, can it also maintain the nitrogenation of a drink?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17998" LastActivityDate="2018-10-24T23:16:22.170" Title="Safe, cheap method to dissolve nitrogen in small batches of water (10l)" Tags="|gas|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24338" PostTypeId="2" ParentId="24337" CreationDate="2018-10-24T13:42:24.667" Score="1" Body="&lt;p&gt;The starting point to recognize the limits of your idea is to appreciate the concept and magnitude of the Henry's law constant for nitrogen dissolving in water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://webbook.nist.gov/cgi/cbook.cgi?ID=C7727379&amp;amp;Mask=10#Solubility&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://webbook.nist.gov/cgi/cbook.cgi?ID=C7727379&amp;amp;Mask=10#Solubility&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this reference, you can determine that 1 kg of water will have about 0.6 milli-moles dissolved N&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; at 25 &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C and 1 bar. By comparison, using this page&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&amp;amp;Mask=10#Solubility&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&amp;amp;Mask=10#Solubility&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can determine that CO&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; has about 34 milli-moles dissolved per kg water for the same conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially then, to obtain the same degree of &quot;fizziness&quot; for your proposed drink compared to carbonated beverages, you will have to raise the external pressure by a factor of about 50. You could also lower the temperature, but I suspect the water would freeze before the factor of 50 is reached using temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Some nitrogen will dissolve in the water according to Henry's law. The rest will not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) When this is done to increase the total pressure of nitrogen, yes. As soon as the pressure is released, the dissolved nitrogen will escape. That escaping process is the &quot;fizziness&quot; that occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Yes. But the concentration of dissolved nitrogen is about a factor of 50 times less. Hence, about a factor of 50 times less &quot;fizziness&quot; results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, a drink with dissolved-nitrogen will be flatter and be flatter faster when compared to a carbonated drink at the same pressure.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-10-24T23:16:22.170" LastActivityDate="2018-10-24T23:16:22.170" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24340" PostTypeId="2" ParentId="24331" CreationDate="2018-10-24T14:25:18.550" Score="1" Body="&lt;p&gt;You have to understand that even a stupidly simple computer is at minimum hundreds of transistors grouped into tens of separate functions. Even then the computer in question is nearly useless, since it has no pheripherals like a keyboard or a usb port.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since many of these would require some software to use. So the critical question is how do you manage the computer while every piece is not in place, and how do you program before you have any pheripherals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you are following something like &lt;a href=&quot;https://eater.net/8bit/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ben Eater's 8-bit computer&lt;/a&gt;, you have to watch each and every video untill it is a computer. And even then because it has no keyboard he has to initialize the memory by loading each memory bank by hand, after that the computer works fine. Now this may seem counter productive since in fact manual input is slower than doing the thing he programs by hand, but he has no other input devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we assume that the arcitecture is some sort of microstepping latency oriented processor, for ease of understanding. Then it works conceptually like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Control logic that activate each part of the system triggers:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Clock causes copy of program counter to instruction register. &lt;/li&gt;&#xA;&lt;li&gt;Microstep time moves one unit&lt;/li&gt;&#xA;&lt;li&gt;This causes control logic to fetch/switch the first micro step of the instruction In the control system. This turns contol signals on to modules. &lt;/li&gt;&#xA;&lt;li&gt;Increment microstep counter&lt;/li&gt;&#xA;&lt;li&gt;fetch second microstep controll pattern&lt;/li&gt;&#xA;&lt;li&gt;increment microstep counter&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;li&gt;Zero microstep counter, increment program, counter loop to beginning.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So yes the control logic is sort of the man connecting wires. But its a long way there, things need verifying a long long time before you reach this stage. Even then system needs to be programmed somehow before it can do anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And no this does not nesseserily involve a compiler, because it probably does not exist yet untill you have designed all opcodes.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-10-24T14:30:55.000" LastActivityDate="2018-10-24T14:30:55.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24341" PostTypeId="2" ParentId="24312" CreationDate="2018-10-24T14:52:24.650" Score="0" Body="&lt;p&gt;Using the principle of virtual work...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the chair is rotated infinite infinitesimally about the point A, each point will move an amount vertically that is proportional to the horizontal distance from A. The only points that we are interested are B where the reaction force acts upwards and where W acts downwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ (200+200+100) \ R_B + (-65/2) \ (60+200+100)=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$500 \ R_B = 11700$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$R_B = 117/5= 23.4$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same procedure can how be applied by rotating the chair an infinitesimal amount about the point B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$(200+200+100) \ R_A +(-65/2)\ (200-60)=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ 500 \ R_A = 4550 $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$R_A = 45.5/5 = 9.1$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2018-10-24T14:52:24.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24343" PostTypeId="1" CreationDate="2018-10-24T17:50:04.000" Score="2" ViewCount="842" Body="&lt;p&gt;I'll be the first to admit that I have always been somewhat confused by best-practice dimensioning standards surrounding pipe threads. While there is a decent amount of literature available online, (&lt;a href=&quot;https://www.engineersedge.com/hardware/taper-pipe-threads.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;such as this&lt;/a&gt;) I still find the implementation to be a bit vague -- specifically regarding thread depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://esh-docdb.fnal.gov/cgi-bin/RetrieveFile?docid=3072&amp;amp;filename=Pages%20from%20B1-20-1_2013.pdf&amp;amp;version=6&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASME B1.20.1&lt;/a&gt; (AKA &lt;em&gt;the NPT bible&lt;/em&gt;) makes me think that one should callout thread-features such that they run to the &quot;vanishing point&quot; or L4. However, L2 is designated as the &quot;maximum thread engagement distance&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience, the engagement depth on NPT fittings can vary wildly (depending on quality of machining, amount of teflon tape used, and the person wielding the wrench, etc etc). This inconsistency can cause other issues when it comes to actually assembling -- but that's another topic in itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, my questions are...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Which 'L-value' should I be using for thread depth when designing NPT fittings (i.e. in my cad model)?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;If that value is L4, do machine shops know that I'm referring to vanishing point on my drawing and not max engagement depth?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Bonus: Are there any other 'best practices' surrounding this topic which may be useful?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3t57r.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3t57r.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17105" LastEditorUserId="17105" LastEditDate="2018-10-29T16:43:56.827" LastActivityDate="2018-10-29T16:43:56.827" Title="How are NPT hole features defined?" Tags="|modeling|technical-drawing|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24344" PostTypeId="1" CreationDate="2018-10-24T18:10:01.220" Score="0" ViewCount="2892" Body="&lt;p&gt;Can somebody please explain what is ment by functional datum and manufacturing datum, with respect to dimentioning and tolerancing. &lt;/p&gt;&#xA;" OwnerUserId="16605" LastActivityDate="2019-04-24T14:02:35.473" Title="What are functional and manufacturing datums" Tags="|mechanical-engineering|design|measurements|manufacturing-engineering|tolerance|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24345" PostTypeId="1" CreationDate="2018-10-24T18:45:53.053" Score="0" ViewCount="772" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yowXU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yowXU.png&quot; alt=&quot;This is the question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have trouble solving part A of the question. I thought the resulting torque would just be 10k * L of the beam.&lt;/p&gt;&#xA;" OwnerUserId="17811" LastActivityDate="2018-10-27T18:36:20.370" Title="Resulting torque on the cross section" Tags="|torque|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24347" PostTypeId="1" CreationDate="2018-10-24T21:20:27.150" Score="4" ViewCount="2326" Body="&lt;p&gt;I'm interested in the history of the 345 kV power level for high voltage transmission in the United States. I've got two (and a half) questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;According to &lt;a href=&quot;https://books.google.de/books?id=C5W8uxwMqdUC&amp;amp;pg=PA120#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source&lt;/a&gt; 345 kV was first applied in 1953 by American Electric Power. Who has developed it and why have they chosen exactly this voltage? Because it is 3*115 kV and nearly 2*161 kV?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When you look on this &lt;a href=&quot;https://power.kreed.org/#2.91/43.18/-86.8&quot; rel=&quot;nofollow noreferrer&quot;&gt;power network map&lt;/a&gt; you will notice, that 345 kV is  still the most highest voltage for the transmission networks of the Midwest and Texas. Why does they still stick to 345 kV even for new-built power lines despite 500 kV would be much more economic (less losses at a similar right-of-way width)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;PS: I've thought that &quot;Engineering&quot; wouldn't be the right category for this question, but I was advised so &lt;a href=&quot;https://books.google.de/books?id=C5W8uxwMqdUC&amp;amp;pg=PA120#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;at Meta&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18003" LastEditorUserId="4953" LastEditDate="2018-10-25T08:14:27.067" LastActivityDate="2018-10-25T08:14:27.067" Title="What's the history of the 345 kV voltage level in power transmission?" Tags="|power|power-transmission|electrical-grid|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24348" PostTypeId="1" CreationDate="2018-10-24T23:17:18.630" Score="3" ViewCount="107" Body="&lt;p&gt;A colleague works in a closed (no public access) watershed and came across this culvert today. These smooth &quot;tracks&quot; down the center of the culvert are something we have never seen before. Any idea what they may be for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some details that may afford clues:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Culvert installation date is unknown (so 30+ years).&lt;/li&gt;&#xA;&lt;li&gt;The culvert runs underneath an access road.&lt;/li&gt;&#xA;&lt;li&gt;There are no other utilities or structures anywhere near (within a mile) of this culvert. &lt;/li&gt;&#xA;&lt;li&gt;The channel upstream and downstream of the culvert are still in their natural state.&lt;/li&gt;&#xA;&lt;li&gt;Logging use to occur in the watershed in the past, but there are no known logging zones next to this culvert.&lt;/li&gt;&#xA;&lt;li&gt;The culvert is massively over sized (serves an area of 150 acres which typically does not receive significant snow-pack).&lt;/li&gt;&#xA;&lt;li&gt;This stream empties into a main tributary which is about another 1/3 mile downstream.&lt;/li&gt;&#xA;&lt;li&gt;Surface road was originally used for access to the upper part of the watershed, and for logging vehicles/equipment.&lt;/li&gt;&#xA;&lt;li&gt;This culvert is located near Portland, Oregon&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zP8zv.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zP8zv.gif&quot; alt=&quot;Stream Culvert under Access Road&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18005" LastEditorUserId="18005" LastEditDate="2018-10-25T12:42:52.880" LastActivityDate="2018-10-26T00:46:04.027" Title="What would a track in a culvert be used for?" Tags="|hydraulics|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24349" PostTypeId="2" ParentId="24345" CreationDate="2018-10-25T00:20:31.267" Score="0" Body="&lt;p&gt;In a thin walled open tube of uniform thickness J is;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$J = \frac{1 } {3 }\times Ut^3  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t = thickness of the wall of tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U is is the length of the median boundary (perimeter of median cross section)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the rotation is &lt;span class=&quot;math-container&quot;&gt;$ \theta = \frac {TL} {GJ} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  T = 10kN \times x,\space x \space is\space 1/3 \space of\space triangle\space height. $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  Height of triangle =3x=60 \times \sqrt 3= 103.92mm \space x = 34.64mm $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T = 10kN\times 34.64mm $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque = force times distance to center of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is the answer to first part  (a).&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-10-25T00:20:31.267" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24350" PostTypeId="2" ParentId="24347" CreationDate="2018-10-25T00:37:30.040" Score="3" Body="&lt;p&gt;With the high voltage AC power lines, there are trade offs between losses due to resistance in the wires, where the higher voltages are better and losses due to capacitance to the air, etc. where lower voltages are better. That is why many big, high voltage power transmission is done with DC these days.  For example, the 500 kVDC line from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Intermountain_Power_Plant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Intermountain Power Plant&lt;/a&gt; (IPA) in Utah to Los Angeles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another factor is the higher AC voltages are considered by some to be &lt;a href=&quot;https://electrical-engineering-portal.com/environmental-impacts-hvdc-transmission&quot; rel=&quot;nofollow noreferrer&quot;&gt;bad for the environment&lt;/a&gt;. DC is ... better, not perfect but better.  Something I did not see in the article, since the higher the voltage, the lower the current, the higher voltage AC produces less of a magnetic field, 345/500 less.  In the case of IPA at full load, they are moving 1900 MW at 500 kV which works out to 3800 amps.  That much will produce a good sized magnetic field, even from the height of the transmission towers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(warstory) Years ago, I worked at power plant in a trailer under the startup lines that they said carried 10,000 amps.  The magnetic fields made the text on the CRT computer monitors almost unreadable because it jiggled up and down by almost 1/8 inch.  After a while, we got a patch from IBM that set the CRT refresh rate from 70 Hz to 60 Hz and everything was rock solid.(/warstory)&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2018-10-25T00:37:30.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24352" PostTypeId="1" CreationDate="2018-10-25T06:35:04.690" Score="0" ViewCount="434" Body="&lt;p&gt;Capacitors available in the market have a lower bound around 1 pF or 0.1 pF. Why is it difficult to manufacture capacitors in femtoFarad or lower range?&lt;/p&gt;&#xA;" OwnerUserId="17982" LastActivityDate="2018-10-25T07:25:43.183" Title="Why is it difficult to manufacture capacitors in femtoFarad or lower range?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24353" PostTypeId="2" ParentId="24352" CreationDate="2018-10-25T07:25:43.183" Score="2" Body="&lt;p&gt;It is not that such capacitors are inherently difficult to manufacture. It is that within the bounds of everyday engineering, they have few uses in actual circuits. Note also in this context that you can make an extremely low-valued capacitor simply by bringing one conductor trace on a board close to another. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-10-25T07:25:43.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24357" PostTypeId="1" AcceptedAnswerId="24462" CreationDate="2018-10-25T13:57:48.440" Score="2" ViewCount="220" Body="&lt;p&gt;Studying Fluid Mechanics, I started to notice that almost every textbook/website uses a specific point to make calculations about the pressure in a &lt;strong&gt;liquid&lt;/strong&gt; at a given depth (hydrostatic pressure): &lt;strong&gt;the geometric center&lt;/strong&gt; (&lt;em&gt;as shown in the images below&lt;/em&gt;), when presenting pressure gauges/manometers/piezometers.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; This happens regardless of the field to which the book is directed  (I looked in textbooks of Fluid Mechanics for Civil, for Electrical, for Mechanical...).&#xA;&lt;img src=&quot;https://i.stack.imgur.com/qX4KO.png&quot; alt=&quot;Pressure gauges/Manometers/Piezometers&quot; /&gt;&#xA;    &lt;sub&gt;Images sources: MATHalino/PennState College of Engineering (MNE)/The SensorsGuide/University of Sydney (MDP)/ScienceStruck/Chegg  &lt;/sub&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/V33Io.jpg&quot; alt=&quot;Pressure gauges/Manometers/Piezometers&quot; /&gt;&#xA;        &lt;sub&gt;Sources: Introduction to Fluid Mechanics - Nakayama &amp;amp; Boucher/Mecânica dos Fluidos - Noções e Aplicações - Sylvio R. Bistafa/Chegg&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;One of the textbooks I looked at even draws attention to this fact, but it doesn't explain the reason for the choice:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Note the origin of the measurement of h, in the center of the tube&lt;/p&gt;&#xA;&lt;p&gt;                                &lt;img src=&quot;https://i.stack.imgur.com/q3uxu.png&quot; width=&quot;400&quot; /&gt;&lt;br /&gt;&#xA;                                                          &lt;sub&gt;Source: Mecânica dos Fluidos - Franco Brunetti&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;A similar behavior can be identified when textbooks present liquids in motion: they use the centerline of the pipe to make calculations/measurements. Here's an example:&lt;/p&gt;&#xA;&lt;p&gt;                                    &lt;img src=&quot;https://i.stack.imgur.com/Ejj2a.jpg&quot; width=&quot;400&quot; /&gt;&lt;br /&gt;&#xA;                                                     &lt;sub&gt;Source: Fluid Mechanics for Civil Engineers - N.B. Webber&lt;/sub&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;So why is the choice of geometric center/centerline of the pipe so common when measuring/calculating pressure? Some hypotheses:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Maybe all the textbooks/websites are unconsciously copying each other?&lt;/li&gt;&#xA;&lt;li&gt;Maybe is this some kind of &amp;quot;convention&amp;quot;?&lt;/li&gt;&#xA;&lt;li&gt;Maybe it is because a point in the horizontal plane of the geometric center gives the average pressure of a tank?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4358" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-11-03T19:00:35.187" Title="Why textbooks use geometric center/centerline of the pipe when calculating/measuring pressure?" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24359" PostTypeId="2" ParentId="24357" CreationDate="2018-10-25T14:31:00.550" Score="0" Body="&lt;p&gt;Looks you could add it in any part of the vessel but only the heigth matters you put it in the side and people will be confused if it is in the center it is easy to understand.&lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2018-10-25T14:31:00.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24360" PostTypeId="1" CreationDate="2018-10-25T16:02:38.877" Score="1" ViewCount="69" Body="&lt;p&gt;Assuming the driveshaft of the rear wheeled drive RC car follows the linear DC motor theory for stall torque and no-load RPM, also voltage supplied is constant and no inductance as well. Note: the motor is running above its operating voltage 3V, the motor voltage supplied was measured 5V. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor used: product.mabuchi-motor.com/detail.html?id=9&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Forming hypothesized result:&#xA;speed are directly proportional to voltage, I directly multiply the voltage ratio with the expected speed then multiply again by the gear ratio. Gear ratio was verified by calculating its' teeth. Actual applied voltage was measured using a multimeter and measure the voltage across the motor. The hypothesized stall torque were also multiplied by the same voltage ratio and the gear ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The base of this experiment uses the  formula &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ω=Vk-Tk2*R &lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the ω is the no-load speed, v is the voltage, k is the electrical constant or the torque constant assuming they are equal to each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypothesized results of a 7.3 gear ratio&lt;br&gt;&#xA;Stall torque=0.065&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No load rpm=4200&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measured result&#xA;Stall torque=0.02&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No-load rpm=6500&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stall torque is measured using an arm and applies the force on the weight scale on both sides. The no-load speed was measured using a non-contacting tachometer. It is found out that the no-load hypothesized speed is less than the measured no-load speed, the stall torque of the hypothesized result is higher than the measured results which is expected, because of power loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is why the measured no-load speed is higher than the hypothesized speed? Is it because of inaccuracy or some unaccounted factors?&lt;/p&gt;&#xA;" OwnerUserId="18023" LastEditorUserId="18023" LastEditDate="2018-10-27T11:10:42.963" LastActivityDate="2018-10-27T11:10:42.963" Title="Performance of RC car under different gear ratio, measured data vs hypothesis" Tags="|mechanical-engineering|gears|torque|linear-motors|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24361" PostTypeId="2" ParentId="24348" CreationDate="2018-10-25T19:52:01.673" Score="2" Body="&lt;p&gt;The &quot;Smooth Tracks&quot; look like gaurd rails from the highway. Somone probably used old ones to create this in the culvert. As an engineer, I would say that this was not craeted by the state or county but an individual that need to get something from one end to the other with much less effort. Put a trail cam up and monitor it for a few months...you will probably get your answer.&lt;/p&gt;&#xA;" OwnerUserId="18027" LastActivityDate="2018-10-25T19:52:01.673" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24364" PostTypeId="2" ParentId="24348" CreationDate="2018-10-25T21:12:38.107" Score="0" Body="&lt;p&gt;Given that there are boulders near the entrance and the exit, I would say this has to do with providing a way for low flows to pass through the culvert easier. If there was a well maintained entrance and exit that showed signs of wheeled traffic, then I would have also thought someone was pushing or pulling a cart through there.&lt;/p&gt;&#xA;" OwnerUserId="18025" LastActivityDate="2018-10-25T21:12:38.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24365" PostTypeId="2" ParentId="24348" CreationDate="2018-10-26T00:46:04.027" Score="0" Body="&lt;p&gt;One possibility is that in some circumstances culverts can 'float' upwards due to forces analogous  to buoyancy so the 'rails' may just be a simple way to pin it down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally if it was made by back filling a trench this could be away to hold the pipe in place during that process. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-10-26T00:46:04.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24366" PostTypeId="2" ParentId="22894" CreationDate="2018-10-26T00:50:04.470" Score="1" Body="&lt;p&gt;In the tension-compression pattern, normal stresses are &lt;em&gt;evenly distributed&lt;/em&gt; over the cross-section of the shell, which means that the load-bearing is attributed to the whole cross-section. This is completely different from bending, when only the material fibres &lt;em&gt;adjacent to the surface of the shell&lt;/em&gt; participate in the load bearing. This feature is heavily used in so-called &quot;shallow&quot; shells (low-curvature shells), which are used instead of planar slabs. Since you've asked for a reference, you will find a good example here: &lt;a href=&quot;http://members.ozemail.com.au/~comecau/quad_shell_shallow_shell.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://members.ozemail.com.au/~comecau/quad_shell_shallow_shell.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5992" LastActivityDate="2018-10-26T00:50:04.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24367" PostTypeId="2" ParentId="23861" CreationDate="2018-10-26T01:02:14.707" Score="1" Body="&lt;p&gt;&lt;em&gt;Dynamical Systems&lt;/em&gt; are any systems that &lt;em&gt;change&lt;/em&gt; with time. Real physical systems or system models. What is changing? Any variable within the system (examples, displacement, velocity, pH, sound intensity ...). They can be linear, nonlinear time varying or not, etc. So it's a very &lt;em&gt;broad&lt;/em&gt; definition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LTI systems are a &lt;em&gt;subset&lt;/em&gt; of dynamical systems that are strictly linear and the parameters do not change with time. They are usually &lt;em&gt;models&lt;/em&gt; of real systems. Real physical systems, if you look close enough are never linear. Parameters can change with temperature for example. So LTI 'models' better suited. They approximate dynamic systems that are near linear and time invariant.&lt;/p&gt;&#xA;" OwnerUserId="820" LastActivityDate="2018-10-26T01:02:14.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24368" PostTypeId="1" CreationDate="2018-10-26T04:18:09.757" Score="2" ViewCount="1559" Body="&lt;p&gt;I'm looking at the usual UK/EU standards formula for the reinforcing steel area percentage, in a retaining wall. Essentially in &quot;usual&quot; cases it comes down to 0.13% of the cross-sectional area of Tw (wall thickness) in the upstand of the vertical wall itself, and Tb (base thickness [depth]) for the outer and inner base. If &lt;em&gt;Fy&lt;/em&gt; is below a certain amount then the percentage changes but it's still calculated as a simple percentage.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Example scenario&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I'm imagining trying to apply this to the retaining wall shown in the diagrams below. In this situation, the retaining wall is built onto one edge of a large and deep foundation. The foundation itself is very deep and very wide compared to the retaining wall built on it. Essentially when the foundation is considered as the base of the retaining wall, it continues on to a distance where the retaining wall is no longer affecting or affected by it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kind of dimensions I have in mind are - foundation/base is indefinitely wide (for these purposes), and say 900mm deep. The retaining wall upstand is say 200mm wide and 600mm high, and built at one edge of it. The foundations are deemed not to need reinforcement steel (perhaps because it's basically a large monolithic concrete block or it's on on sufficiently stable ground/rock beneath it). So the only reinforcing steel is the steel required for the retaining wall aspect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apparently, the inner and outer steel area is calculated as a percentage of the &lt;em&gt;base's&lt;/em&gt; cross-section area. But I'm looking at the three diagrams below, which are roughly to scale. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;First (left) diagram&lt;/strong&gt; - the outer steel is run down to the bottom of the foundations, so essentially the foundations act like a base that's 900mm thick and very wide. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Second diagram&lt;/strong&gt; - outer steel is only run down to 350mm deep; a more typical depth if it were on a slab not a foundation. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Third diagram&lt;/strong&gt; -  the 2nd layout is re-visualised as equivalent to two parts - essentially a 400mm base with rebar along the bottom, resting on a non-reinforced 500mm slab.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4GFYd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4GFYd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;My confusion&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;What's puzzling me is why are the steel area requirements, which are based on &lt;em&gt;Tb&lt;/em&gt;, different in these three versions? Intuitively, I expected that for any given lateral pressure, the inner/outer steel would reach a sort of limit point where there just wasn't any need to add more, because in the first two cases it's embedded with right angle bends to a point where the limiting failure mode  in the base slab is pullout failure/concrete failure rather than steel elastic limitations (ie not governed by the &lt;em&gt;cross section area&lt;/em&gt; of steel), and if the foundation itself is considered structurally adequate, then I'd have thought the second and third would be structurally equivalent and therefore need equal areas of steel too, as they are essentially the same layout. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Alternatively, the 1st layout seems &lt;em&gt;stronger&lt;/em&gt; than the 2nd, as there's even more grip and harder pullout, and the entire foundation slab is engaged rather than just the top half of it, so why does it seem to need &lt;em&gt;more&lt;/em&gt; rather than &lt;em&gt;less&lt;/em&gt; steel cross-sectional area, to achieve equal wall stability?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But instead, the standard equation is simply 0.13% of &lt;em&gt;Tb&lt;/em&gt;, so if the base is made thicker, which should &lt;em&gt;add&lt;/em&gt; not remove strength, then the area of inner/outer steel required seems to have to go &lt;em&gt;up&lt;/em&gt;, not &lt;em&gt;down&lt;/em&gt;. That feels counter-intuitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm left with a mess of confusion: Why does the steel area percentage differ in the three cases shown? Why does &lt;em&gt;adding&lt;/em&gt; strength and solidity to the bottom of the base (i.e., as extra depth) cause an &lt;em&gt;increase&lt;/em&gt; to the required steel area? Why isn't there some kind of limit based on lateral pressure rather than base dimensions beyond which more isn't needed? Why isn't a further limit reached where it doesn't matter how much deeper the base and the outer steels go, because the extra depth doesn't affect the retaining wall stability? What is the logic behind the cross section area of steel being an unqualified percentage of &lt;em&gt;Tb&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing, and what should actually happen here?&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2018-10-26T04:41:39.427" LastActivityDate="2018-10-26T18:22:54.967" Title="Percentage area of steel for a retaining wall - standard formula doesn't make sense to me?" Tags="|structural-engineering|structural-analysis|reinforced-concrete|retaining-wall|reinforcement|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24369" PostTypeId="1" CreationDate="2018-10-26T04:43:32.267" Score="0" ViewCount="340" Body="&lt;p&gt;Let’s say we have a thin band (made out of PLA, 3D printing material) with a rectangular cross section. In drawing, it would look something like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/mTj3f.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mTj3f.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s say a force of 15 N is to be applied from the top down to the band, how am I supposed to figure out the minimum thickness (and or height) that could endure the force and do not undergo deformation or failure?&lt;/p&gt;&#xA;" OwnerUserId="18024" LastActivityDate="2018-10-26T23:48:45.343" Title="Stiffness as a function of thickness" Tags="|mechanical-engineering|structural-engineering|civil-engineering|structural-analysis|structures|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24370" PostTypeId="1" CreationDate="2018-10-26T04:54:02.347" Score="0" ViewCount="35" Body="&lt;p&gt;In the design of a Rocket engines, it is understandably desirable to have a more efficient engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ensuring the gasses coming out of the engine comes out straight, and the chemistry of the propellants are all straight forward methods of improving ISP, but how can engines with identical propellants and similar nozzles have different specific impulses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the major factors besides nozzle geometry and propellant that allow for higher specific impulses, and how can one optimize an engine to improve specific impulse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Obviously I wont compromise other engine parameters for microscopic increases in isp, but I want to build an engine that gets at least 70% efficiency of theoretical Kerosene LOX combustion)&lt;/p&gt;&#xA;" OwnerUserId="14040" LastEditorUserId="14040" LastEditDate="2018-10-26T05:42:15.427" LastActivityDate="2018-10-26T05:42:15.427" Title="What factors besides nozzle geometry and propellant's affect engine ISP? How can one optimize ISP?" Tags="|thermodynamics|rocketry|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24371" PostTypeId="2" ParentId="24369" CreationDate="2018-10-26T07:06:56.483" Score="1" Body="&lt;p&gt;As a very rough first estimate we would consider this as a &lt;span class=&quot;math-container&quot;&gt;$20\pi = 62.8\text{ cm}$&lt;/span&gt; polygon made of approximately 63 equal sides of 1&amp;nbsp;cm wide by 3&amp;nbsp;cm high. After multiplying the load by a factor of safety of 2 we get 30&amp;nbsp;N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we ignore the slenderness concerns now because these columns are laterally supported and considered short, &lt;span class=&quot;math-container&quot;&gt;$L&amp;lt; 10$&lt;/span&gt; radius of gyration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now we can calculate the required thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  t= \frac{30\text{ N}}{63\text{ cm} \times \sigma} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; is the allowable compression stress of PLA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Answering some comments considering the arbitrary selection of a polygon to substitute the circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes I agree that if we add the number of sides of polygon more and more, at the limit we reach the perimeter of the circle, but that is not the point as the difference is less than 3%. I chose 20 as a simple linear approximation to circle, to give OP the kind of answer he can immediately envision. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short columns with a length smaller than 20 times radius of gyration crush ultimately rather than buckle under stresses more than yield stress . In this case with radius of gyration of 10cm, lengths of more than 200cm would be cause for buckling concerns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And local buckling happens only under concentrated loads which is not the case here.&lt;br&gt;&#xA;For deeper review &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19690013955.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; is the link to NASA research on buckling of thin wall cylinders. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-10-26T23:48:45.343" LastActivityDate="2018-10-26T23:48:45.343" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24376" PostTypeId="2" ParentId="24343" CreationDate="2018-10-26T10:33:40.663" Score="1" Body="&lt;p&gt;We are a bearing manufacturer. I design all the new special products that come in the door. We use NPT grease fittings in quite a few of our products. We use a 1/8 &amp;amp; 1/4 NPT grease fittings. As our standard practice we generally use a certified thread gauge and set our cnc to tap so the thread gauge will have a 4 to 4-1/2 turns max engagment. This has always worked for us and have never had a return in the 60+ years we have been in business. I have discussed this issue with our old school bearing designers and have learned in my 12 years with the company that this is a good way to measure. We have tested many different grease fitting brands and actual lubrication system fittings which all work great. I don't remember a time when we ever had a leak using this standard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think many times we get wrapped up in a ASME standard and forget that these guidelines are vague for a reason, they are meant to get us in the ballpark and we are suppose adjust until it works for our applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps! Good luck!&lt;/p&gt;&#xA;" OwnerUserId="18027" LastActivityDate="2018-10-26T10:33:40.663" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24378" PostTypeId="2" ParentId="24292" CreationDate="2018-10-26T11:11:02.053" Score="0" Body="&lt;p&gt;Is the drawing going to a customer, a vendor or in-house manufacturing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. If this is a customer be vague until you know for sure they are interested and &#xA;actually need a more detailed specification. Then you can discuss who besides them &#xA;will be viewing the drawings and you can make a decision on drawing content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. If this is a Vendor make sure your company has them sign a Non-Disclosure Agreement &#xA;so they do not devulgue your company secrets without the possibility of a law &#xA;suit. This is standard practice for most companies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. Also if this is a vendor, is this project for aerospace or government? If so then      be vauge, the details can be determined after you get the PO through the &#xA;procurement documentation. Don't try to tell the Vendor how to do his job. Specify&#xA;the bare minimum for him referencing a standard whether it be a company standard or&#xA;a published standard. Then you can make sure the key people get the info they need&#xA;without giving away more information than is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d. If this is in-house, then write a company standard that can be referenced on all&#xA;drawings it pertains to. Then if you need to change something you will only be &#xA;changing the one document instead of 10 thousand drawings. I actually did this in our manufacturing department. It not only saves us 100's of man hours updating component&#xA;and assembly drawings, it made a huge difference in errors due to outdated drawings. Mind you this only works well if you have a disciplined staff that tracks and controls its standard documentation properly.&lt;/p&gt;&#xA;" OwnerUserId="18027" LastActivityDate="2018-10-26T11:11:02.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24380" PostTypeId="2" ParentId="24344" CreationDate="2018-10-26T12:14:41.907" Score="1" Body="&lt;p&gt;In simplest terms, Functional Datums should have something to do with a part. E.g. a mating surface or a centerline of the bearing surface of a rotating shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Manufacturing Datum is a feature used to locate and secure a part for a manufacturing operation. A Manufacturing Datum can use the Functional Datums or would be an entirely different feature. For example, say you have a complex casting which needs additional Machining on it. In particular, it’s functional datum needs to be machined into the rough casting. In order to be consistent with locating and loading the rough casting into the machining fixture, Manufacturing Datums are defined. They are likely defined as datum target points on the casting with Basic dimensions. Your tooling will be designed to hold the part on those Manufacturing Datums in order to be able to access the Functional Datums which need machining If everything is well-designed and toleranced correctly, you should have a repeatable process for Manufacturing. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2018-10-26T12:14:41.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24381" PostTypeId="2" ParentId="24368" CreationDate="2018-10-26T18:22:54.967" Score="2" Body="&lt;p&gt;I am not sure which guidance manual you are looking at, but it must in turn be referring to EN1992-1-1 section 9.2.1.1. This section defines the minimum amount of reinforcement in a beam or slab in order for the beam or slab to be considered reinforced concrete. And it does indeed use the value of 0.13% (with adjustments for concrete and steel grade). So it is a much more generic requirement than just something for retaining walls. It is relatively easy to see why it must be a percentage of the concrete area, if you consider the different behaviour of plain concrete and reinforced concrete:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Plain concrete at low bending moments will have tension and compression in the concrete cross section. If the bending moment increases, the tension stress will reach the capacity, a crack will form and the structure will fail suddenly. This is a brittle failure.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Reinforced concrete at low bending moments will also tension and compression in the concrete. If the load increases, the tension stress in the concrete will reach the capacity and a crack will form. When the crack forms, the reinforcement will take over and transfer the tension that was previously transferred by the concrete. If there is enough reinforcement for this to be possible, the crack will be small, and the structure will not fail until the load increases enough for the reinforcement to yield, resulting in a ductile failure. (Meaning that we will get large deformations before the structure fails complete. A &quot;hey, that looks unsafe — we better fix that before someone gets hurt&quot;—thing.) The ductile failure makes the structure much safer even if it doesn't have a larger failure load.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The important bit in this context is as follows: If the amount of reinforcement is not enough to absorb the tension force that occurs in the concrete just before the first crack forms, the reinforcement will yield immediately at the creation of the first crack, and the crack will therefore become very large very fast, resulting in almost the same behaviour of plain concrete. (A brittle failure.) At a given concrete type and steel grade you will therefore want concrete area times concrete tension capacity to be less than steel area times yield stress in order to ensure a ductile failure. Which can be reformulated as steel area larger than x percent of the concrete area. (I believe the value of 0.13% also includes an empirical adjustment, so I am not going to recalculate the exact value, but hopefully the principle should be clear.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is worth noting that none of this has anything to do with the actual load on the retaining wall, so you will have to check that as well. And it is of course also perfectly possible to construct an unreinforced retaining wall which is completely safe for a given load — it just won't give a ductile failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And addition to all that, I feel I have to note that the reinforcement arrangement you have drawn on the sketches, is not a good arrangement and you should not construct it like that: When the rebar on the inner side is bend around the inner corner like that, there is a great risk of concrete cover breaking off at the bend long before the bar reaches its yield capacity. I can add a few sketches to explain this better, if you are interested, but I hesitate to add even more stuff on something tangential in case it was just a hastily drawn couple of lines? (You can also take a look at for example EN1992-1-1 annex J.2 for a couple of examples of how the reinforcement can be arranged at a corner in bending.)&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2018-10-26T18:22:54.967" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24382" PostTypeId="1" CreationDate="2018-10-26T18:41:05.717" Score="3" ViewCount="33" Body="&lt;p&gt;I'm working on a project with a company that uses sand casts during moulding processes and they provided us with samples of the sand they use before it's involved in the procedure and afterwards (Waste foundry sand). I'm wondering if there's specific parameters regarding grain size for using WFS as an aggregate for concrete mixes, my samples led me to believe that most of the WFS this company produces is about 1/2&amp;nbsp;mm to 1&amp;nbsp;mm in diameter. A brief literature review has offered no conclusive results and little information regarding specific gran diameter in WFS. Is this byproduct suitable for concrete mixing? Any help would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="18043" LastEditorUserId="1832" LastEditDate="2018-10-26T19:06:21.817" LastActivityDate="2018-10-26T19:06:21.817" Title="Waste Foundry Sand as an aggregate for concrete mixes. I have a question regarding grain diameter and it's possible applications" Tags="|concrete|industrial-engineering|construction-management|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24383" PostTypeId="1" CreationDate="2018-10-14T04:20:56.627" Score="2" ViewCount="146" Body="&lt;p&gt;Does rain cause Tarmac / asphalt pot holes? If so, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a period of rain in my city there are plenty of fresh new pot holes in the Tarmac / asphalt surfaces. The immediate cause is obviously axle overloading, due to appalling regulations, but these holes are fresh from a fourtnights heavy rain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can comprehend erosion in dirt or gravel roads as I know these substances are penetrable by water. But my impression is that Tarmac is impenetrable to water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What process am I missing?&lt;/p&gt;&#xA;" OwnerUserId="18046" OwnerDisplayName="Samuel Russell" LastEditorUserId="33" LastEditDate="2018-10-26T20:06:21.123" LastActivityDate="2018-10-26T20:33:02.470" Title="Does rain cause Tarmac / asphalt pot holes? If so, how?" Tags="|civil-engineering|roadway|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24385" PostTypeId="2" ParentId="24383" CreationDate="2018-10-26T20:33:02.470" Score="2" Body="&lt;p&gt;old tarmac has cracks in it which can admit water under the material, softening the earth underneath. when a vehicle rolls over the cracked material, it yields because the softened earth fails to support it. in yielding, the cracks open up progressively and the tarmac develops a bowl-shape which collects water, which then drains through the cracks, softening still more earth, etc. etc.  and in no time you have a pothole. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-10-26T20:33:02.470" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24387" PostTypeId="2" ParentId="8199" CreationDate="2018-10-27T06:34:25.337" Score="0" Body="&lt;p&gt;The material orientation matters. For 1D, give the  material(section) direction as &amp;lt;0 1 0&gt;&lt;/p&gt;&#xA;" OwnerUserId="16206" LastActivityDate="2018-10-27T06:34:25.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24390" PostTypeId="1" CreationDate="2018-10-27T15:30:14.297" Score="0" ViewCount="845" Body="&lt;p&gt;I am a Civil Engineering professional having around 7 years of field experience and is now working for a govt. agency. All I want is to get a Professional Engineering (PE) License. But at present, I am living outside the USA and planning to be at USA on the next year.&lt;br&gt;&#xA;I will arrive at the USA just 10 days before the exam and will leave the USA after 5 days sitting in the exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Can a non-resident USA visa holder apply to attend in the PE Exam?  &lt;/li&gt;&#xA;&lt;li&gt;What precautions/preparations I need to take to pass the PE Exam on the first attempt?  &lt;/li&gt;&#xA;&lt;li&gt;I have read many articles on PE Exam. I have heard that &lt;a href=&quot;https://bookscivil.com/pe-civil-reference-manual-16th-ed-michael-r-lindeburg-pe/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://bookscivil.com/pe-civil-reference-manual-16th-ed-michael-r-lindeburg-pe/&lt;/a&gt; this book is necessary to pass Civil Engineering PE Exam. Is that true?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18052" LastEditorDisplayName="user16" LastEditDate="2018-10-27T15:39:57.133" LastActivityDate="2018-10-27T18:53:06.543" Title="What Are The Procedures To Get a US Professional Engineer License?" Tags="|civil-engineering|licensure|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24391" PostTypeId="2" ParentId="24390" CreationDate="2018-10-27T15:58:21.660" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Can a non-resident USA visa holder apply to attend in the PE Exam?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Residency status typically isn't a requirement for most states in the US regarding applying for a professional engineer license.  That said, based upon the information you have provided, I don't believe that you meet the typical qualifications to apply for the exam.  Requirements will sometimes vary based upon which US state you are applying within.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical requirements are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Graduate of a baccalaureate engineering program accredited by the Engineering Accredita on Commission of&#xA;ABET (EAC/ABET).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Taken and passed the NCEES Fundamentals of Engineering Exam.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Completed four (4) years of satisfactory engineering experience under a licensed PE following the date you received an engineering degree.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As far as I know, ABET doesn't accredit programs outside of North America, so you likely miss the first qualification.  And I strongly suspect you haven't met the second and third qualifications regarding the FE exam and the supervised experience requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you're pretty much at &quot;No, can't happen&quot; at this point because you're ineligible to apply for the PE exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible that you can apply for licensure by comity.  But that requires you to be licensed as a professional engineer in another region, &lt;strong&gt;and&lt;/strong&gt; that the US state that you wish to receive licensure within has a comity agreement with that other region.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the sake of completeness, I'll address your other two questions although they're not as germane at this point.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What precautions/preparations I need to take to pass the PE Exam on the first attempt?&lt;br&gt;&#xA;  I have read many articles on PE Exam. I have heard that &lt;em&gt;XYZ book&lt;/em&gt;&lt;sup&gt;1&lt;/sup&gt; is necessary to pass Civil Engineering PE Exam. Is that true?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Civil PE exam has one of the lower &lt;a href=&quot;https://ncees.org/engineering/pe/pass-rates/&quot; rel=&quot;nofollow noreferrer&quot;&gt;first time pass rates&lt;/a&gt; as compared to other PE exams.  I know multiple Civil PEs, and all of them put a very significant amount of effort into studying practice exams and questions ahead of the actual exam.  Hundreds of hours of study ahead of time is &lt;em&gt;not&lt;/em&gt; unusual.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Civil PE exam also probably has one of the worst reputations for the sheer number of reference volumes required in order to successfully pass the exam.  Civil Engineers need to be able to quickly look up relevant codes and standards during the exam, and there is a wide body of material that has to be covered.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Lindeburg book you linked is one of many reference manuals that are worth considering ahead of taking the exam.  And there are a few other commercial entities that provide sample exams.  Each set of PE exams is a bit unique though, and each engineer's experience is different. I have yet to see a valid list of universally necessary reference materials that would &quot;guarantee&quot; a first time pass of the exam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, you'll need to assemble your own set of reference materials based upon your experience and your expectations regarding the exam.  Again, the practice examples and suggestions from NCEES can be invaluable here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt; Aside, I intentionally replaced the title mentioned with a more generic placeholder.  There's &lt;em&gt;lots&lt;/em&gt; of books that are supposedly &lt;em&gt;required&lt;/em&gt;. &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastEditorDisplayName="user16" LastEditDate="2018-10-27T18:53:06.543" LastActivityDate="2018-10-27T18:53:06.543" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24392" PostTypeId="2" ParentId="24345" CreationDate="2018-10-27T18:15:29.923" Score="1" Body="&lt;p&gt;Your suggested solution indicates that you seem to be confusing torque with bending moment. This is understandable since they are often used interchangeably in common parlance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, they mean very different things in engineering. To understand the difference, it's important to clarify the standard notation for a beam's local axes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;strong&gt;x-axis&lt;/strong&gt; is the beam's longitudinal axis. That is, the axis which describes the beam's length.&lt;/li&gt;&#xA;&lt;li&gt;The &lt;strong&gt;y- and z-axes&lt;/strong&gt; are the beam's transversal axes, describing it's width and height.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, given that, we can then state that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bending_moment&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bending moment&lt;/a&gt; is the internal stress state along a beam due to &lt;strong&gt;rotations around the two transversal (y and z) axes&lt;/strong&gt;. When you think of a beam or cantilever deflecting downwards due to a load, that beam is under bending moment.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Torsion_(mechanics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Torque&lt;/a&gt; is the internal stress state along a beam due to &lt;strong&gt;rotations around its longitudinal (x) axis&lt;/strong&gt;. When you think of a screw being twisted, a torque has been applied to it, causing torsion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, the calculation you expected of &lt;span class=&quot;math-container&quot;&gt;$\text{Force}\times\text{Length}$&lt;/span&gt; would give you the &lt;strong&gt;bending moment&lt;/strong&gt; of the beam, &lt;strong&gt;not the torque&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the general gist of your solution is correct: both bending moment and torque are equal to &lt;span class=&quot;math-container&quot;&gt;$\text{Force}\times\text{Distance}$&lt;/span&gt;. However, to calculate the torque you need to use another distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question states that the force is &quot;acting on the outer edge of the cross-section in the direction of the symmetry axis.&quot; From this we can conclude where on the section the force is applied and in which direction it's pointing:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I personally find &quot;outer edge&quot; to be unclear, but it's fair to assume they mean the left- or right-most point of the section (one of the top corners).&lt;/li&gt;&#xA;&lt;li&gt;Likewise, &quot;in the direction of the symmetry axis&quot; is also unclear since triangles have three axes of symmetry, but it's fair to assume they mean the vertical one, which therefore means the force is vertical.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So now we just need to calculate the torque itself. The force is vertical and on the left- or right-most edge of the section. Since the force is vertical, the torque is calculated by multiplying it by a horizontal distance. The relevant distance in this case is to the beam's centroid, which runs along that vertical axis of symmetry. Therefore the distance is equal to half the section's side, so &lt;span class=&quot;math-container&quot;&gt;$120/2=60\text{ mm}$&lt;/span&gt;. Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T = F\times d = 10 \times 60 = \pm600\text{ kNmm}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had to put that &lt;span class=&quot;math-container&quot;&gt;$\pm$&lt;/span&gt; sign because the problem as stated doesn't give us enough information to know whether the resultant torque is positive or negative. That depends on where the load is applied (left or right-most corner) and the force's direction (positive or negative, though I suppose one could argue it should be in the positive &quot;direction of the symmetry axis&quot;... which itself is unclear since we can't necessarily assume the positive direction is upwards).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2018-10-27T18:36:20.370" LastActivityDate="2018-10-27T18:36:20.370" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24393" PostTypeId="1" CreationDate="2018-10-27T18:20:48.447" Score="0" ViewCount="36" Body="&lt;p&gt;I made a pallet lift table out of 1/4&quot; SS304 and now have about 5x the deflection that beam calculations predict. I noticed that the machine was shipped supporting a 500lb pallet. &#xA;I thinking a few things: &#xA;1. Either shock loading from the 500lb pallet when the truck was going over bumps exceeding yield.&#xA;2. Creep from the pallet being supported for about a week now. Computer stress was about 6000psi, so pretty far from yield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table had two edges bent up on a press brake. One end of the table needs to be open to roll pallet lifts onto the table. I'm thing of going from 1/4&quot; SS304 with edges bent on a press brake to welded 3/8&quot; T1 A514. The yield of T1 is about 90kpsi, about 2x more than SS304.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know the creep / cold flow SS304 vs A514/T1? Is there calculations to compute? I'm also open to comments/suggestions.&lt;/p&gt;&#xA;" OwnerUserId="4311" LastActivityDate="2018-10-28T22:40:15.660" Title="Creep / cold flow SS304 vs A514/T1" Tags="|mechanical-engineering|applied-mechanics|steel|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24394" PostTypeId="1" AcceptedAnswerId="24395" CreationDate="2018-10-27T18:28:19.237" Score="-1" ViewCount="66" Body="&lt;p&gt;I am having a bar and surface (made of aluminum). I need to attach some material to the end of the bar so that, if I apply any force (25 N-sideways) to bar it has to stick with the surface.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force is applied normally to the flatter surface of the bar. I need to select a material at the end of bar so that the force applied on the bar is minimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What methodology should I choose to find an answer? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ncIZY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ncIZY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aim is lowest necessary force to keep the rod intact (with a material added to rod end)&lt;/p&gt;&#xA;" OwnerUserId="18054" LastEditorUserId="90" LastEditDate="2018-11-02T16:16:49.607" LastActivityDate="2018-11-02T16:16:49.607" Title="Approach to solve a frictional problem?" Tags="|mechanical-engineering|materials|applied-mechanics|tribology|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24395" PostTypeId="2" ParentId="24394" CreationDate="2018-10-27T21:02:02.420" Score="1" Body="&lt;p&gt;I'm trying to approach the problem at the high school level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertical force is proportional to the tangential force acting at the other end  of the rod, the ratio of these two is called the static coefficient of friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the applied forces are due to robotic arm and the weight of the rod: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$(F+mg)\mu_s=25\ [N]$$&lt;/span&gt; or: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$F\mu_s=25\ [N] - mg\mu_s$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see we have an equation with two unknowns, &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; the vertical force of robots arm and the material dependent parameter &lt;span class=&quot;math-container&quot;&gt;$\mu_s$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the plot: &lt;a href=&quot;https://i.stack.imgur.com/eR8Bs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eR8Bs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you observe if you increase the &lt;span class=&quot;math-container&quot;&gt;$\mu_s$&lt;/span&gt;, the applied force will decrease, but it makes it difficult for the rod to move freely, if you increase the force the coefficient of friction will decrease but you want to keep the applied force minimum so this is not an option. So we need to define a range for the force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; to solve the problem with bounded extrema, or simply choose a heavier rod, but make sure the force &lt;span class=&quot;math-container&quot;&gt;$25\ [N]$&lt;/span&gt; can pull it back. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-10-27T21:02:02.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24397" PostTypeId="1" CreationDate="2018-10-27T21:28:00.137" Score="2" ViewCount="116" Body="&lt;p&gt;I am wondering if there exist any species of shape memory alloy fibers with low Young's Modulus at room temperature, for an application I'm working on. So far I've played with Flexinol, but it is a bit too stiff. I know Nitinol has a Young's Modulus of ~30-40 GPa. I am wondering if there exist softer fibers - I'm interested ones that might be at around ~10 GPa, ~1GPa, and &amp;lt; 1GPa, if they exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond other materials, is there any procedure to make Nitinol naturally softer?&lt;/p&gt;&#xA;" OwnerUserId="18055" LastActivityDate="2018-11-01T21:12:57.507" Title="Shape memory alloy fibers with low Young's Modulus?" Tags="|elastic-modulus|actuator|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24398" PostTypeId="1" CreationDate="2018-10-27T20:32:22.190" Score="1" ViewCount="12" Body="&lt;p&gt;I am wondering if there exist any species of shape memory alloy fibers with low Young's Modulus at room temperature, for an application I'm working on.  So far I've played with Flexinol, but it is a bit too stiff.  I know Nitinol has a Young's Modulus of ~30-40 GPa.  I am wondering if there exist softer fibers - I'm interested ones that might be at around ~10 GPa, ~1GPa, and &amp;lt; 1GPa, if they exist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond other materials, is there any procedure to make Nitinol naturally softer?&lt;/p&gt;&#xA;" OwnerUserId="18055" OwnerDisplayName="user650261" LastActivityDate="2018-10-28T00:41:46.787" Title="Shape memory alloy fibers with low Young's Modulus?" Tags="|actuator|" AnswerCount="0" CommentCount="1" ClosedDate="2018-10-28T00:51:16.597" ContentLicense="CC BY-SA 4.0" />
  <row Id="24400" PostTypeId="2" ParentId="24393" CreationDate="2018-10-28T05:45:15.437" Score="1" Body="&lt;p&gt;Per Blacksmith37, those alloys do not creep at all at room temperature and so neither do they cold flow under that low a stress level. Suggest you check your original stress calculations. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13363" LastEditDate="2018-10-28T22:40:15.660" LastActivityDate="2018-10-28T22:40:15.660" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24401" PostTypeId="1" AcceptedAnswerId="24418" CreationDate="2018-10-28T07:51:13.117" Score="4" ViewCount="397" Body="&lt;p&gt;I've worked with &lt;a href=&quot;http://www.supraalloys.com/titanium-grades.php&quot; rel=&quot;noreferrer&quot;&gt;different grades of titanium&lt;/a&gt;. Mostly with grade 5, and also grade 1 which is ductile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that some glasses are made of titanium and are called &quot;memory titanium&quot;, I guess because they bend back to their original shape?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, which grade of titanium would that be?&lt;/p&gt;&#xA;" OwnerUserId="11510" LastActivityDate="2018-10-29T17:04:35.900" Title="What grade of titanium are &quot;memory titanium&quot; glasses made of?" Tags="|materials|metals|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24402" PostTypeId="2" ParentId="24401" CreationDate="2018-10-28T09:27:16.383" Score="1" Body="&lt;p&gt;Titanium used for frames is either 99% pure or, for memory frames 75% Titanium and 25% of nickel or copper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more here (the result of a google search) : &lt;a href=&quot;https://www.zennioptical.com/blog/whats-so-terrific-about-titanium-frames/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.zennioptical.com/blog/whats-so-terrific-about-titanium-frames/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-10-28T09:27:16.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24403" PostTypeId="1" CreationDate="2018-10-28T13:43:18.293" Score="1" ViewCount="137" Body="&lt;p&gt;Is it possible to create a mold of an aluminium motorcycle cylinder head so that you can create an exact replica with melted aluminium?&lt;/p&gt;&#xA;" OwnerUserId="18063" LastActivityDate="2018-11-06T16:47:48.723" Title="Is it possible to replicate a motorcycle engine cylinder head?" Tags="|motors|automotive-engineering|aluminum|molding|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24404" PostTypeId="1" AcceptedAnswerId="24408" CreationDate="2018-10-28T14:17:05.697" Score="0" ViewCount="126" Body="&lt;p&gt;I am wondering how the screw top of a container made of either aluminium or brass is manufactured at scale. A picture of the sort of thing I am talking about is below. Currently the only method I can think of is hydroforming. Can anyone enlighten me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oJBou.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oJBou.jpg&quot; alt=&quot;Example Image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10522" LastActivityDate="2018-10-28T18:40:43.867" Title="Manufacturing screw thread onto lid" Tags="|mechanical-engineering|manufacturing-engineering|industrial-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24405" PostTypeId="1" CreationDate="2018-10-28T14:56:33.530" Score="1" ViewCount="162" Body="&lt;p&gt;I am looking to simulate the radiation pattern of a 2.7Ghz half-wavelength dipole antenna immersed in salt water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been using Altair FEKO software to try and accomplish this. I've tried two different ideas without much success:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 1) Directly change the properties (relative permitivity, conductivity, density) of the free space medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I start changing these properties far from their defaults I start getting a lot of geometry errors. For example, to model salt water, I give the free space medium a relative permitivity of 80 and a conductivity of 4 S/m. But then trying to solve this model gives &quot;ERROR 114: Segmentation rules have been violated (wire segment is too long)&quot;. I attempt to resolve this by shortening the length of the dipole antenna to wavelength/10, but then I start getting mesh errors like &quot;WARNING 18189: The geometry is much smaller than the model extents. This could cause tolerance issues.&quot; Is this normal? Do I need to just keep playing with the antenna vs mesh dimensions until it works?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Idea 2) Create a sphere surrounding the antenna and assign the region a new medium with the correct properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I create a sphere with a width several times the length of the antenna, but no matter what medium I assign to the region inside, FEKO treats it as a lossy space. For example, giving the region the default free space medium, and the faces are all perfect electric conductor, results in all gain from the antenna being destroyed and &quot;WARNING 3397: The power loss is larger than the active power&quot;. Shouldn't a spherical region of free space be treated the same as the rest of the free space geometry? Why does FEKO treat it differently?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is possible because I've read several research papers talking about using FEKO to model antenna underwater, but so far I haven't figured out the key. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm also open to switching to Matlab if someone believes it would be easier to use that. &lt;/p&gt;&#xA;" OwnerUserId="18064" LastActivityDate="2018-10-28T14:56:33.530" Title="Method to simulate a dipole antenna in salt water?" Tags="|modeling|simulation|matlab|electromagnetism|radiation|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24406" PostTypeId="2" ParentId="24403" CreationDate="2018-10-28T17:40:00.370" Score="3" Body="&lt;p&gt;Yes it is but there a a few things you need to watch out for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to allow enough excess material for any surfaces which need to be machined subsequent to casting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All castings shrink somewhat so the pattern needs to be a some percentage larger than the finished casting needs to be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its not uncommon for castings to have internal voids which are created with cores in the mould these are a bit more difficult to reverse engineer than the exterior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally you need to add the gating (runners and risers) to the pattern. Getting this right is  very important as it not only lets the metal in and the air out of the mould, ensuring it fills properly but also controls the cooling rate, which if you don't get it right can cause shrinkage voids, hot tears and other defects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cast aluminium is also quite prone to porosity and ideally needs to be degassed if you want to make decent quality mechanical parts. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-11-06T16:47:48.723" LastActivityDate="2018-11-06T16:47:48.723" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24408" PostTypeId="2" ParentId="24404" CreationDate="2018-10-28T18:40:43.867" Score="3" Body="&lt;p&gt;These threads are rolled. The blank is automatically fed to sit around a central wheel which has the form of the thread. An external arm with the opposing half of the form then lowers to apply pressure and form the threads, before being automatically ejected from the central wheel, ready for the next part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This video &lt;a href=&quot;https://www.youtube.com/watch?v=HykVRyTsV20&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=HykVRyTsV20&lt;/a&gt; shows a machine forming the threads on a cap, but the process is identical for the container itself.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-10-28T18:40:43.867" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24409" PostTypeId="1" CreationDate="2018-10-28T19:28:32.367" Score="1" ViewCount="95" Body="&lt;p&gt;I'm fairly illiterate when it comes to Engineering, so please understand I have pretty much 0 idea what I'm talking about. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having to build a seawater distillation device for school, and we decided to go with reverse osmosis, to try and achieve this. (Reverse osmosis being the filtering of salt from water using massive applied pressure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm looking for, is a part that's a &quot;Screw Piston&quot; (I have no idea what it's actually called). So it's like a hydraulic piston, except the piston rod is threaded like a screw, so you as you tighten the screw, it applies a downward pressure. Here's a crudely made diagram of what I'm talking about &lt;a href=&quot;https://i.stack.imgur.com/4vx7k.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4vx7k.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm almost certain that something like this exists, (I mean if not, than I guess  I should probably hire a patent lawyer and get to work) but since I have no idea what it's actually called, being mechanically illiterate, I have no idea how to search it up or where I could find something like this. I'd really appreciate help telling me what it's called and where I could possibly get one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my TL;DR is, what the heck is this thing called, and where do you think I could buy one? A giant thank you in advance to anyone who bothers to read this and answers!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(btw the reason I need a screw for the piston is that I need the mechanical advantage because  reverse osmosis requires like 8000 pounds for force to filter the water.)&lt;/p&gt;&#xA;" OwnerUserId="18070" LastActivityDate="2018-10-29T15:50:48.457" Title="What is this part called, and where could I get one?" Tags="|mechanical-engineering|pressure|pressure-vessel|threads|pistons|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24411" PostTypeId="2" ParentId="24409" CreationDate="2018-10-28T21:47:45.523" Score="1" Body="&lt;p&gt;What is it called?  Your name of &quot;screw piston&quot; sounds good to me.  Where are you going to get one?  That is harder.  If that is the way you want to go, you are going to have to have it made yourself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Reverse_osmosis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reverse Osmosis&lt;/a&gt; says you will need between 600 and 1200 PSI for sea water.  At 1200 PSI, 6000 pounds means a piston 3 inches in diameter.  I would suggest a smaller piston.  I don't know where you are getting your membrane from.  That might be putting constraints on your design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were presented with an assignment to do &lt;em&gt;ANY&lt;/em&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Reverse_osmosis&quot; rel=&quot;nofollow noreferrer&quot;&gt;RO&lt;/a&gt; on some water, I would probably tap a 3/4 inch hole in one of the aluminum blocks in my basement and go to the hardware store for a 3/4 inch threaded rod.  I would then bolt the membrane across the bottom with another piece that had a hole for the filtered water to leave. If I then filled the threaded hole half way with salt water, put food grade grease (I presume you will test the result by drinking it) on the threads of the rod and cranked it down, hopefully I would get clean water out the bottom.  A rod 3/4 inch in diameter has an area of 0.44 square inches.  For 1200 PSI, you are going to need 530 pounds force. You can do that with threaded rod and aluminum, no problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastEditorUserId="4953" LastEditDate="2018-10-29T15:50:48.457" LastActivityDate="2018-10-29T15:50:48.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24412" PostTypeId="1" AcceptedAnswerId="24413" CreationDate="2018-10-29T00:19:56.063" Score="3" ViewCount="116" Body="&lt;p&gt;According to &lt;a href=&quot;https://en.wikipedia.org/wiki/Sulfur_vulcanization&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia page&lt;/a&gt;, vulcanization is an important process to make rubber harder and more durable. However, &lt;a href=&quot;https://en.wikipedia.org/wiki/Plasticizer&quot; rel=&quot;nofollow noreferrer&quot;&gt;another Wikipedia page&lt;/a&gt; says that plasticizers play a significant role in rubber manufacturing, acting as a softener.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a reason why both are needed when processing rubber? It seems to me that they exhibit an exact opposite effect on the rubber being processed and hence only one is needed.&lt;/p&gt;&#xA;" OwnerUserId="18074" LastEditorUserId="18074" LastEditDate="2018-10-29T00:27:55.290" LastActivityDate="2018-10-29T06:00:38.347" Title="Is there a reason why both plasticizer and vulcanizing agent are needed during processing of rubber?" Tags="|materials|manufacturing-engineering|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24413" PostTypeId="2" ParentId="24412" CreationDate="2018-10-29T06:00:38.347" Score="3" Body="&lt;p&gt;yes, there is, as follows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unvulcanized rubber will soften and turn into taffy when it gets warm. Vulcanization cross-links the polymer strands and prevents the rubber from turning into taffy at high temperatures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also necessarily makes the rubber stiffer and more resistant to deformation at lower temperatures, which can yield a rubber compound which is not soft enough for a particular application. This tendency can be counteracted by milling mineral oils (plasticizers) into the rubber to get the rubber molecules to slip more easily past one another (within the limits set by the cross-linking). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the process of engineering a rubber batch to have the right combination of flexibility and durability involves balancing the benefit of cross-linking against the need for rubber-like behavior, and this involves careful blending of plasticizers into the rubber resin. In so doing, the engineer can get anything from supersoft to almost rock-hard rubber, dialed-in for a broad range of special applications. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-10-29T06:00:38.347" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24414" PostTypeId="1" CreationDate="2018-10-29T11:11:26.283" Score="2" ViewCount="40" Body="&lt;p&gt;I am tasked with the design of a helipad for a remote government facility. Annex 14 Part II (ICAO Convention on International Civil Aviation) does not clearly state what materials are acceptable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common materials are concrete and asphalt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use brick pavers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of any risks or restrictions in this regard?&lt;/p&gt;&#xA;" OwnerUserId="2637" LastActivityDate="2018-10-29T11:11:26.283" Title="Wearing course for Helipad" Tags="|civil-engineering|aviation|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24415" PostTypeId="1" CreationDate="2018-10-29T14:53:48.063" Score="2" ViewCount="179" Body="&lt;p&gt;I have seen a number of products made of brass with a silver-looking coating covering the brass on the outside. These products are all quite cheap (for example a small watering can or clothes hooks). What is the cheapest silver-looking material to coat brass with and what process would be used to perform the coating operation for mass production (500,000+ per year) of these products? I'd imagine chrome plating or electro plating would be too costly for these sorts of products so I wonder how things like these are made.&lt;/p&gt;&#xA;" OwnerUserId="10522" LastEditorUserId="10522" LastEditDate="2018-10-29T15:00:40.193" LastActivityDate="2023-10-05T04:06:41.640" Title="Coating brass components cheaply" Tags="|mechanical-engineering|design|manufacturing-engineering|industrial-engineering|product-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24416" PostTypeId="1" CreationDate="2018-10-29T15:30:58.273" Score="2" ViewCount="335" Body="&lt;p&gt;According to clause 4.8.3.2 the cross-section capacity for non-slender section is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{F_c}{A_g p_y}+\frac{M_x}{M_{cx}}+\frac{M_y}{M_{cy}}\leq 1 \qquad (1)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to clause 4.8.3.3.1 simplified method of member buckling resistance, the following relationship should be satisfied:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{F_c}{P_c}+\frac{m_x M_x}{p_y Z_x}+\frac{m_y M_y}{p_y Z_y}\leq 1\quad (2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a beam-column with class 1 section (doubly-symmetric) under uniform moment (so that the equivalent moment factor &lt;span class=&quot;math-container&quot;&gt;$m_x = m_y =1$&lt;/span&gt; according to Table 26) and axial compression, and ignore lateral-torsional buckling. In this case, equation (2) is always critical than equation (1), because &lt;span class=&quot;math-container&quot;&gt;$P_c \leq A_g p_y$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$p_y Z_x \leq M_{cx}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$p_y Z_y \leq M_{cy}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the use of equation (1), or clause 4.8.3.2, if buckling resistance is always more critical? Is there any case that equation (1) is more critical?&lt;/li&gt;&#xA;&lt;li&gt;If the member is under biaxial bending but without axial force, i.e. &lt;span class=&quot;math-container&quot;&gt;$F_c=0, M_x&amp;gt;0, M_y&amp;gt;0$&lt;/span&gt;, is checking equation (2), or clause 4.8.3.3 still required?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I understand that the simplified method is at the cost of conservatism, but making one criterion obsolete seems too much to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I am aware of that the equivalent moment factor can be taken as minimum as &lt;span class=&quot;math-container&quot;&gt;$m_x = m_y =0.4$&lt;/span&gt; for double curvature member, which should make equation (2) less critical if the axial compression is not large. However, I would like to confine the discussion in the case of single curvature and uniform moment.&lt;/li&gt;&#xA;&lt;li&gt;I welcome any example that a non-doubly-symmetric class 1 section has &lt;span class=&quot;math-container&quot;&gt;$p_y Z_x \geq M_{cx}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$p_y Z_y \geq M_{cy}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="13166" LastEditorUserId="13166" LastEditDate="2018-10-30T01:04:58.847" LastActivityDate="2021-05-29T19:01:40.263" Title="Compression members with moment to BS 5950-1:2000" Tags="|structural-engineering|steel|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24418" PostTypeId="2" ParentId="24401" CreationDate="2018-10-29T15:46:59.810" Score="2" Body="&lt;p&gt;Those glasses are made out of Shape Memory Alloys (SMA), from which the most common ones include nickel, titanium, copper, zinc, or iron in their composition. I don't think the tag: &lt;strong&gt;grade of titanium&lt;/strong&gt;, is applicable for SMAs even it they contain a high percentage of titanium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a functional materials, their properties and applications are so different that they have their own system. For example, one of the most recognizable brands of SMAs, Nitinol (nickel-titanium):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...offers different &lt;strong&gt;grades of Nitinol&lt;/strong&gt;, which are distinguished by the&#xA;  austenite start temperature of the ingot: &lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;Nitinol #1 (superelastic) -35 to -10 °C&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Nitinol #2 (superelastic) -45 to -15 °C&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Nitinol #4 (superelastic) -10 to +10 °C&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Nitinol #5 (shape memory) ≥ +85 °C&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Nitinol #6 (shape memory) +35 to +85 °C&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Nitinol #8 (shape memory) +10 to + 35 °C&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Nitinol #9 (superelastic) ≥ +35 °C&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From: &lt;a href=&quot;https://cdn.thomasnet.com/ccp/00956637/229312.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://cdn.thomasnet.com/ccp/00956637/229312.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to tune the shape-memory and superelasticity properties of the alloy by modifying its composition. And speaking about the glasses (spectacles), you are more interested in the &lt;a href=&quot;https://www.youtube.com/watch?v=m4fQHgjgc3Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;superelasticity behavior&lt;/a&gt; of the material rather than in its &lt;a href=&quot;https://www.youtube.com/watch?v=XPrg8EZlD1E&quot; rel=&quot;nofollow noreferrer&quot;&gt;shape memory effect&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17409" LastEditorUserId="17409" LastEditDate="2018-10-29T17:04:35.900" LastActivityDate="2018-10-29T17:04:35.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24420" PostTypeId="2" ParentId="24415" CreationDate="2018-10-29T18:31:40.073" Score="1" Body="&lt;p&gt;a silver coating on brass is usually nickel electroplate. does your part have to be brass? if it can be made of plastic instead, a shiny coating of metal can be quickly and cheaply applied to it in mass production by either &lt;em&gt;sputtering&lt;/em&gt; or &lt;em&gt;metal evaporation&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-10-29T18:31:40.073" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24421" PostTypeId="1" AcceptedAnswerId="24424" CreationDate="2018-10-29T18:56:54.193" Score="3" ViewCount="193" Body="&lt;p&gt;I am not asking for specific numbers of any production plant. I want to know the general direct worker/employee ratio of a plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: if there are 1000 employees working usually 700 of them are direct workers etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for estimates backed up with sources or personal experience.&#xA;Also there may be multiple shifts in a production plant so I am looking for the ratio in a single shift since all the jobs may not get a night shift. The plants I am interested in are the ones that mass produce for international brands. Including the country the plant is in would also be helpful for categorizing the information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct worker means a worker directly working on the production line while employee means all the the people getting paid in that plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt; Added more information.&lt;/p&gt;&#xA;" OwnerUserId="18079" LastEditorUserId="18079" LastEditDate="2018-10-30T06:12:41.643" LastActivityDate="2018-10-30T06:12:41.643" Title="What is the percentage of direct workers in an automotive production plant compared to all the employees?" Tags="|automotive-engineering|manufacturing-engineering|car|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24422" PostTypeId="2" ParentId="24421" CreationDate="2018-10-29T22:01:33.353" Score="-3" Body="&lt;p&gt;Nowadays factories are virtually completely automated. See for example tesla.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although there may be people watching the machines they generally arent needed. Most people there would be janitors, admins, safety inspectors and so on rather than directly useful employees.&lt;/p&gt;&#xA;" OwnerUserId="18085" LastActivityDate="2018-10-29T22:01:33.353" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24424" PostTypeId="2" ParentId="24421" CreationDate="2018-10-29T23:29:01.087" Score="2" Body="&lt;p&gt;Below is a breakdown for the Fiat Chrysler Automobile Transmission plant in Kokomo, Indiana USA. (Note: Hourly = Direct, salaried = Indirect)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Employment&lt;/strong&gt;: 3,830 (3,411 hourly; 419 salaried)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Floor Space&lt;/strong&gt;: 3.1 million square feet &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Acreage&lt;/strong&gt;: 110 acre site with casting plant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transmission Plant I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Employment&lt;/strong&gt;: 2,230 (1,950 hourly, 280 salaried)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Floor Space&lt;/strong&gt;: 1.2 million square feet&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Acreage&lt;/strong&gt;: 233&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plant is fairly automated. Below are some video links of the plant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NqNWR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zO1Rz.jpg&quot; alt=&quot;Fiat Chrysler Transmission Plant 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ASwwU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/alf0B.jpg&quot; alt=&quot;Fiat Chrysler Transmission Plant 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;em&gt;Click on image for a larger version of the image.&lt;/em&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional details and video can be found &lt;a href=&quot;http://media.fcanorthamerica.com/newsrelease.do?id=322&amp;amp;mid=&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://media.fcanorthamerica.com/newsrelease.do?id=349&amp;amp;mid=&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2018-10-30T02:38:08.713" LastActivityDate="2018-10-30T02:38:08.713" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24425" PostTypeId="1" AcceptedAnswerId="24461" CreationDate="2018-10-30T03:04:51.987" Score="-2" ViewCount="381" Body="&lt;p&gt;In the saga of trying to answer the question that came into my mind while studying the basic concepts of Fluid Mechanics, &quot;Why textbooks use geometric center to calculate hydrostatic pressure when presenting pressure gauges?&quot;, after asking it to my professor of Fluid Mechanics, consulting ~20 textbooks, asking it as a part of &lt;a href=&quot;https://physics.stackexchange.com/questions/436475/&quot;&gt;this&lt;/a&gt; question on Physics.SE, &lt;a href=&quot;https://engineering.stackexchange.com/questions/24357/&quot;&gt;asking it here&lt;/a&gt; by the recommendation of a Physics.SE user, I didn't gave up and finally found an answer while reading the comments on this YouTube video:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/_VDVi_zq3nk&quot; rel=&quot;nofollow noreferrer&quot;&gt;How a Piezometer Works by Donald Elger&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zBmDJ.jpg&quot; alt=&quot;Piezometer diagram&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is it [the pressure measurement with piezometer] taken from the middle of the pipe?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Elger's answer: The pressure variation across a section of a pipe is hydrostatic; thus, the pressure will vary linearly with radius and the pressure at the center of the pipe is the average pressure. If you use this value of pressure in your calculations, this will be give you the &lt;strong&gt;most accurate results&lt;/strong&gt;. Thus, engineers nearly always apply or measure the pressure at the center of the pipe. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;With this new information, a new question arose: &lt;strong&gt;Why average pressure gives the most accurate results if used in calculations? What &quot;calculations&quot; is he referring to?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="4358" LastEditorUserId="4358" LastEditDate="2018-10-31T18:16:59.937" LastActivityDate="2018-11-02T01:35:37.513" Title="Why using average pressure in calculations gives the most accurate results?" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24427" PostTypeId="1" CreationDate="2018-10-30T09:20:44.580" Score="0" ViewCount="65" Body="&lt;p&gt;I'm mechanical engineering student who want to construct 3D printer. I have a problem choosing right and cheap material for bed. Bed will be big for a 3D printer, around 700mm x 500mm. I want to print PLA, PETG, ABS, PC. Bed MUST be heated, and i sensor for height is also necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I googled it a lot, I have been checking out forums (even here), but I still don't know what to choose. I just know i don't want to have borosilicate glass, because it is too heavy and bed will be moving up and down along Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my list of material for bed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PEI great surface, even here i think this is what everybody is suggesting. These are sheets, so I will need a base to put it on. What should be a base? Alu is great for machining but it has really big temperature coefficient. Do you think zerodur is OK? Does anybody has any experience with it? Can I buy it in EU? &lt;a href=&quot;https://en.wikipedia.org/wiki/Zerodur?fbclid=IwAR0OA479GypoKrw5Sk6sMHCHS3dhhVqct7gxVzYpka5QFVgW1hayPlXqsEI&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zerodur&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Phenolic laminated G10-FR4 this is composite that MarkForged uses it on their printers. I had an opportunity to work with Mark 2 and i like it a lot, but glue is really necessary for good adhesion. Is it because Mark2 does not have heated bed? I saw it has a lot of different variations of material and not everything is good for bed surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coropad - Tom's 3D really praised it. It is magnetic, so you can peal it of base. But what can i use for base? Regular S235 steel or C45? Can i use anything lighter and still magnetic? Does anybody uses it here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my list, I'm choosing between one of them. Do you have any other materials that you recommend? What would you choose. IMO this is is critical component of 3D printer, while we know that most of mistakes on print is because of first layer adhesion. Hopefully you'll help me choose right surface :)&lt;/p&gt;&#xA;" OwnerUserId="17673" LastEditorUserId="110" LastEditDate="2018-10-30T12:49:30.730" LastActivityDate="2018-10-31T01:43:50.933" Title="What surface to use for 3D printer?" Tags="|mechanical-engineering|3d-printing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24429" PostTypeId="1" CreationDate="2018-10-30T13:30:58.957" Score="1" ViewCount="62" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Montju%C3%AFc_Communications_Tower&quot; rel=&quot;nofollow noreferrer&quot;&gt;Montjuïc antenna&lt;/a&gt; is an interesting design for an antenna.  I presume the design is principally an antenna, but perhaps it is not?  If the ring + mast is an antenna, how does the ring work to augment the mast of the antenna?  Is there something that lets us know that the mast contains many smaller antenna, and the ring is just for aesthetics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I've heard it is also &lt;a href=&quot;https://en.wikipedia.org/wiki/Montju%C3%AFc_Communications_Tower#Sundial&quot; rel=&quot;nofollow noreferrer&quot;&gt;a sundial&lt;/a&gt;, so maybe the ring is purely to act as an horologue?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gRXmP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gRXmP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="104" LastEditorUserId="104" LastEditDate="2018-11-01T12:44:08.900" LastActivityDate="2018-11-01T12:44:08.900" Title="What are the strengths and limitations of the antenna design of the Montjuïc Communications Tower?" Tags="|electrical-engineering|telecommunication|electromagnetism|rf-electronics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24430" PostTypeId="1" AcceptedAnswerId="24435" CreationDate="2018-10-30T17:48:08.813" Score="0" ViewCount="106" Body="&lt;p&gt;Looking to add a &quot;high oil temp&quot; warning on a machine tool by adding a temperature switch into the hydraulic circuit, right after the radiator before returning to tank. I know monitoring from the tank would be better, but tapping into the tank is not possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The switch I would like to use....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mcmaster.com/7079k95&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mcmaster.com/7079k95&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;....is rated for 120/240 AC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to use the switch to trigger an input on my PLC. The switch is normally open and would close at over temperature, allowing a +24vDC to connect to the PLC input(PNP). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this switch, as only rated for AC, be used to carry the low-amperage +24vDC signal? &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-10-31T04:22:54.150" Title="Can an AC rated temperture switch be used in a DC control circuit?" Tags="|mechanical-engineering|electrical-engineering|control-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24431" PostTypeId="2" ParentId="24430" CreationDate="2018-10-30T21:08:08.200" Score="1" Body="&lt;p&gt;yes, it is OK. the rating of the switch is the MAXIMUM SAFE limit and you are nowhere near that if you are switching 24VDC. The AC and DC ratings of switches like that are sometimes different but this has to do with how the AC voltage measurement method is specified (peak, RMS, ACV, etc.). It is not a specification that the switch may not be used with DC. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-10-30T21:08:08.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24432" PostTypeId="1" CreationDate="2018-10-30T23:10:44.083" Score="0" ViewCount="443" Body="&lt;p&gt;I want a rod to freely rotate 360°, it stands vertically and the rod is gonna carry some construction below it.&#xA;The rod should be fixated to the e.g. the wall or anything on a ceiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What parts do I need to make this possible? How do I fixate the rod onto the ceiling without stopping it's rotation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked up about eccentric locking collar, is this the standard solution? &#xA;Or are there easier ways to fix the vertical rod (falling down the floor) without stopping its movement?&lt;/p&gt;&#xA;" OwnerUserId="17577" LastEditorUserId="17577" LastEditDate="2018-10-30T23:25:25.460" LastActivityDate="2020-09-23T18:01:45.197" Title="How to mount a rod (rotating)?" Tags="|mechanical-engineering|mechanisms|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24434" PostTypeId="2" ParentId="24427" CreationDate="2018-10-31T01:43:50.933" Score="1" Body="&lt;p&gt;I have used both window glass and aluminum on mine with hairspray for adhesion and have had good luck with both; so you don't need to go overkill on high cost materials. Most of the print failures I have seen (on mine and other printers) are usually print head plugging/feed issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ABS has shrink problems that can't really be addressed with just adhesion; you need to print in a heated environment to reduce the layer by layer shrink effect. You are certainly better researched on these new materials than I, but know that it is likely that some will work better in some conditions and some will work better in others. Plan on it requiring some amount of experimentation and adaptation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adhesion is just one aspect to consider:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The surface also needs to be very flat throughout a range of temperatures. Consider your mounting options.&lt;/li&gt;&#xA;&lt;li&gt;I don't know your printer design, but typically the z axis is gear reduced and experiences very little acceleration so mass is not a big issue. Mass can be an issue for high speed y (and possibly x) movements.&lt;/li&gt;&#xA;&lt;li&gt;Larger thermal mass does take longer to heat, but also provides a more stable temperature since most heat sources are controlled by an on/off relay.&lt;/li&gt;&#xA;&lt;li&gt;Higher thermal conductivity and larger thermal mass both produce a more evenly heated surface. The circuit board heating pads are not perfect especially with very thin materials.&lt;/li&gt;&#xA;&lt;li&gt;Some of the plastic bed materials have good adhesion, but only last 20 prints before needing to be replaced. Balance per-print prep time with repair time.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-10-31T01:43:50.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24435" PostTypeId="2" ParentId="24430" CreationDate="2018-10-31T04:22:54.150" Score="3" Body="&lt;p&gt;There are important differences between switches designed for AC and those designed for DC, the main distinction being &quot;breaking speed&quot;, i.e. how fast the contacts are separated when the switch is turned off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that is unlikely to matter at the current levels you describe. The problem is greatest when there is an inductive load like a motor involved, which leads to arcing across the contacts when they are separated. Briefly, energy is stored in an inductor in the form of magnetic energy and it needs to go somewhere when the circuit is broken causing the magnetic field to collapse. The voltage rises until the energy can be dissipated and this causes arcing across the contacts that can degrade them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I expect your PLC load is low inductance and low current, so it shouldn't be any problem in your specific application. Other readers might want to look further and pick a suitably rated switch for motor, transformer, or other inductive loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mouser.com/blog/which-switch-who-cares-if-its-ac-or-dc&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mouser.com/blog/which-switch-who-cares-if-its-ac-or-dc&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3937" LastActivityDate="2018-10-31T04:22:54.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24438" PostTypeId="1" AcceptedAnswerId="24439" CreationDate="2018-10-31T10:08:13.263" Score="11" ViewCount="4040" Body="&lt;p&gt;I saw this meme on Facebook showing a coiled railway overbridge and an alternative plan. I replied that straightening that overbridge will force it to start at the other side of the road, and you need to build a u-turn and there is a valuable building off the frame. Is the coil rational, and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9abjQ.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9abjQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="444" LastEditorUserId="33" LastEditDate="2018-10-31T16:35:15.467" LastActivityDate="2019-08-08T14:23:30.137" Title="Why was a coil chosen over a straight for this overbridge?" Tags="|bridges|highway-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24439" PostTypeId="2" ParentId="24438" CreationDate="2018-10-31T11:07:57.297" Score="18" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bih8n.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bih8n.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;source &lt;a href=&quot;https://www.google.be/maps/@31.2405133,74.2110404,621m/data=!3m1!1e3&quot; rel=&quot;noreferrer&quot;&gt;google maps&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between the intersection and the rail is 100m, The road would need to rise about 10m to clear the track. This would mean a average grade of 10%. However you also need to account for the vertical curve in the road so the max grade would be quite a bit higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The max grade allowed for non-local roads in the US is about 8% (the place in question is in Pakistan). This appears to be a connecting road that replaces a on-grade railway crossing a bit further east. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this design you also have a lot more span that you can cross under next to the rails for future development.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2018-10-31T12:10:10.300" LastActivityDate="2018-10-31T12:10:10.300" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="24440" PostTypeId="1" CreationDate="2018-10-31T15:46:20.620" Score="0" ViewCount="61" Body="&lt;p&gt;Our startup needs short aluminuim tubes (max. ~70 cm [~27.5 inches] long / less than 1 cm / 0.4 inch diameter) such as these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DJ2JX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DJ2JX.jpg&quot; alt=&quot;Aluminium tubes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We'd prefer to manufacture them ourselves rather than buy them from a s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there small scale solutions (machines, etc.) that would allow us to do this?&lt;/p&gt;&#xA;" OwnerUserId="18107" LastActivityDate="2020-12-22T22:29:08.767" Title="Small scale aluminium tubes manufacturing" Tags="|aluminum|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24441" PostTypeId="1" CreationDate="2018-10-31T16:16:19.187" Score="2" ViewCount="81" Body="&lt;p&gt;I am working on a vent. The vent is attached to a box and is designed to extract air out of the box using the venturi effect. Air is made to flow over the vent which creates a significant pressure difference between the face of the vent and the inside of the box making air to flow out. I have performed the Analysis using Fluent and trying to figure our the flow rate(mass flow rate) of the air escaping from the vent outlet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to CFD and want to know how to find that out from my analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MHrMe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MHrMe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;Please find attached a photo which shows the setup.&lt;/p&gt;&#xA;" OwnerUserId="17226" LastActivityDate="2018-10-31T16:16:19.187" Title="Flow rate at the vent exit" Tags="|fluid-mechanics|ansys|cfd|fluent|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24442" PostTypeId="1" AcceptedAnswerId="24449" CreationDate="2018-10-31T17:30:26.737" Score="3" ViewCount="534" Body="&lt;p&gt;I'm designing a funnel/bottle for a powder. My goal is to have all of the bottle's powder contents come out when inverted. However, the powder has a tendency to stick to itself (like flour or brown sugar) and so I will run into a chance that the powder will not fall into the bottom vessel by gravity alone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make matters more complicated, this is not an open funnel on the top. It will be more like a plastic water bottle in that when it's rotated to dumping position, the top isn't open to the air and venting doesn't currently appear to be an option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ch81f.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ch81f.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what funnel surface type will increase the flow to help dispel the powder contents? Some ideas that come to mind are smooth, vertical guides, or stepped stairs. Note my funnel outlet is fixed in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Smooth&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one makes the most sense as it's a &quot;smooth&quot; flow. However, I've been surprised by how air and water behave in relationship to fluid dynamics (ex: golf ball dimples or pickup trucks being more aerodynamic with the tailgate up).  And although fluid dynamics doesn't cover powders, I would imagine there's some similarities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nixh7.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nixh7.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Vertical Guides&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, these ridges would break up the powder clumping on the way down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vwzkP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vwzkP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stepped&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, these steps would add turbulence on the way down and help break up the powder on each &quot;step down.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XLpKf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XLpKf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Extra question:&lt;/strong&gt; Regardless of the features in the funnel (or lack thereof), what's the minimum angle from vertical that would be ideal for a funnel to dispense powders by gravity alone?&lt;/p&gt;&#xA;" OwnerUserId="15383" LastEditorUserId="15383" LastEditDate="2018-10-31T17:55:55.873" LastActivityDate="2018-11-01T02:47:44.840" Title="Ideal funnel design for powders" Tags="|fluid-mechanics|friction|flow-control|turbulence|open-channel-flow|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24443" PostTypeId="1" AcceptedAnswerId="24512" CreationDate="2018-10-31T18:30:08.960" Score="4" ViewCount="91" Body="&lt;p&gt;I am trying to select a BLE module for low power application. I have looked into different BLE modules from different chip manufacturers. For example:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.inventeksys.com/products-page__trashed/ble/ble/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ISM78G1D Bluetooth Low Energy SIP from Inventek&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.silabs.com/products/wireless/bluetooth/bluetooth-low-energy-modules/bgm13s-bluetooth-sip-module&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blue Gecko BGM13S Bluetooth® Module - SiP from Silicon Labs&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/product/CC2652R&quot; rel=&quot;nofollow noreferrer&quot;&gt;SimpleLink Multi-Standard CC2652R Wireless MCU from TI&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.cypress.com/products/ez-ble-and-ez-bt-bluetooth-modules&quot; rel=&quot;nofollow noreferrer&quot;&gt;EZ-BLE and EZ-BT Bluetooth Modules from Cypress&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.nordicsemi.com/eng/Products/nRF52840&quot; rel=&quot;nofollow noreferrer&quot;&gt;nRF52840 from Nordic Semiconductor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.artik.io/modules/artik-020/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Samsung ARTIK™ 020&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am in the process of learning strategies on selecting a BLE module given a few choices. The application will be used in a low power application which includes a sensor to detect objects. We plan to connect the BLE module to the host micro-controller controlling the sensor. The BLE module will communicate to a Gateway (currently being defined) that will transmit the data to the cloud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Appreciated if someone can provide some guidance on what criteria to use when selecting a BLE module. Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="10379" LastEditorUserId="282" LastEditDate="2018-11-02T16:18:12.577" LastActivityDate="2018-12-04T11:01:00.153" Title="What criteria to focus on when selecting a BLE module?" Tags="|electrical-engineering|wireless-communication|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24445" PostTypeId="1" CreationDate="2018-10-31T20:06:46.663" Score="5" ViewCount="1705" Body="&lt;p&gt;I'm trying to figure out how the heck some of these plastic bottle caps are injection molded. By having the thread on the inside of a cap, this obviously creates an undercut. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2OExL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2OExL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I've watched some videos on Youtube on the process, and I can see that a metal threaded core is twisted into place (automatically or manually by an operator). The mold is closed, plastic is then shot into it, the mold opens, and then the threaded core is twisted out of place to allow ejection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some caps have split threads (as seen below) and use a collapsible core. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LILJ4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LILJ4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And some caps use a long continuous thread (as seen below) which makes sense for ejection (one continuous spin).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KrnLV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KrnLV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this other style cap uses a long but non-continuous thread (as seen below). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, &lt;strong&gt;how is this achieved?&lt;/strong&gt; How does the threaded core spin out of place without colliding into the threads since the thread has an ending. In my mind, when this spins out of place, the feature on the core called out with the arrow below would collide into the freshly molded threads on its way out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gyyNs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gyyNs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15383" LastEditorUserId="33" LastEditDate="2019-04-30T12:47:52.333" LastActivityDate="2019-05-30T13:02:25.207" Title="How are non-continuous internal threads plastic injection molded?" Tags="|manufacturing-engineering|plastic|threads|molding|injection|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24446" PostTypeId="1" AcceptedAnswerId="24463" CreationDate="2018-10-31T20:20:04.153" Score="1" ViewCount="289" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gVjzl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gVjzl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I would prefer an equation. &#xA;The premise is that I'm trying to calculate how thick an iron cylinder using a screw jack would need to be to support 20kg of weight for engineering class. We haven't learned anything along these lines and I'd really like to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerDisplayName="user18111" LastEditorDisplayName="user18111" LastEditDate="2018-11-01T14:59:26.473" LastActivityDate="2018-11-05T17:23:49.907" Title="How do I calculate how much weight an iron cylinder can support?" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24448" PostTypeId="2" ParentId="24445" CreationDate="2018-11-01T01:46:49.203" Score="1" Body="&lt;p&gt;the mold contains a rotary mechanism than unscrews the part off the mold core after the plastic is solid and the mold is split open. this is called an unscrewing mold and is used to make threads like this. the giveaway is the presence of engagement features around the part which are grabbed by mating features in that part of the mold that unscrews.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-11-01T01:46:49.203" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24449" PostTypeId="2" ParentId="24442" CreationDate="2018-11-01T02:32:40.773" Score="4" Body="&lt;p&gt;The problem you have is more complicated than you think. Any industry that deals with the storage and movement of granulated/powdered materials has to deal with this and most have unique solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to look at the &lt;a href=&quot;https://www.chemengonline.com/hopper-design-principles/&quot; rel=&quot;nofollow noreferrer&quot;&gt;principles of hopper design&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the things that affect any design will be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Angle_of_repose&quot; rel=&quot;nofollow noreferrer&quot;&gt;angle of repose&lt;/a&gt;&#xA;of the material you intend to use.&lt;/li&gt;&#xA;&lt;li&gt;The moisture content of the material.&lt;/li&gt;&#xA;&lt;li&gt;The angle of the cone in the funnel/hopper.&lt;/li&gt;&#xA;&lt;li&gt;The material used in the cone and wall of the bottle/hopper/funnel&#xA;and its friction factor.&lt;/li&gt;&#xA;&lt;li&gt;The ability of the powdered material to create a near vertical wall&#xA;and stand up, despite its angle of repose.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A stepped cone for the funnel is a bad idea because it creates hang-ups for the material to get caught on. Also the steps reduce the overall angle of the cone. Generally, the steeper the angle of the cone, the better. &lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2018-11-01T02:47:44.840" LastActivityDate="2018-11-01T02:47:44.840" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24452" PostTypeId="1" CreationDate="2018-11-01T16:49:45.387" Score="1" ViewCount="68" Body="&lt;h1&gt;Time to empty a pressurize oxygen tank through a regulator.&lt;/h1&gt;&#xA;&lt;p&gt;The tank contain 12oz of oxygene at 3000psi (room temp.)&lt;/p&gt;&#xA;&lt;p&gt;The regulator is set to 120psi.&lt;/p&gt;&#xA;&lt;p&gt;The exit pipe is a 14inch pipe with 2mm inner diameter.&lt;/p&gt;&#xA;&lt;p&gt;The experiment stop when the tank reach 120psi.&lt;/p&gt;&#xA;&lt;h3&gt;I may change the tank size and the regulator output pressure.&lt;/h3&gt;&#xA;" OwnerUserId="18121" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-11-01T16:49:45.387" Title="Time to empty a pressurize oxygen tank through a regulator" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24453" PostTypeId="1" CreationDate="2018-11-01T19:24:21.090" Score="2" ViewCount="77" Body="&lt;p&gt;I'm not an engineer but a neuroscientist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do research in spinal cord injury whereby a clip is used to clamp the spinal cord to induce a bilateral compression injury model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My research looks at the differences in the injuries to the neck and lower back spinal cord. The key problem here is that the neck spinal cord is wider in diameter than the lower back cord. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our clip is 1 mm in width. So my question is, if a 35 gram clip is used to clamp a cord of a smaller diameter, and we wanted to hold the distribution of force equal in a larger cord, would the force increase or decrease? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My logic says that the force must be increased, but when we do a 35 gram injury to the neck cord, all of the animals die so it may be difficult to use force normalization in our particular case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="18123" LastActivityDate="2022-12-12T00:08:01.020" Title="Normalizing the compressive stress to a cylinder" Tags="|structural-engineering|structural-analysis|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24454" PostTypeId="1" CreationDate="2018-11-01T19:26:22.353" Score="1" ViewCount="860" Body="&lt;p&gt;Disclaimer: sadly not an engineer (I hope this is the right place for such my question. Please let me know if it isn't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm designing a tent. I know most of the dimensions of the first prototype I want to build, but I'm struggling with the poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nanzq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nanzq.png&quot; alt=&quot;Designing tent poles&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;Don't mind the short lines&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When inserting the poles into the rings at the extremities of the tent base (groundsheet), they bend to form the shape of the tent, as we can see in &lt;a href=&quot;https://www.youtube.com/watch?v=8a219dZFfpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Naturehike Cloud Up 2 pitching video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my design, I need to be able to predict &lt;em&gt;how much&lt;/em&gt; (&lt;em&gt;how far&lt;/em&gt;) the poles will be able to bend. I'd love to use &lt;a href=&quot;https://www.msrgear.com/all-season-tents/easton-syclone-poles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Easton's Syclone poles&lt;/a&gt;, but I doubt they'd sell them to me. So I'm interested in aluminium poles such as these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1FzR4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1FzR4.jpg&quot; alt=&quot;Aluminium tent poles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing this is essential for my project, because the &lt;em&gt;bendability&lt;/em&gt; of the poles will determine lots of aspects of the design (such as pole parts length for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is: how can I predict (or simulate [that'd be awesome]) the &lt;em&gt;bendability&lt;/em&gt; of my poles? Any suggestion is immensely welcomed.&lt;/p&gt;&#xA;" OwnerUserId="18122" LastActivityDate="2020-12-22T22:20:11.700" Title="How can I calculate &quot;bendability&quot; of aluminium tubes?" Tags="|aluminum|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24455" PostTypeId="1" AcceptedAnswerId="24549" CreationDate="2018-11-01T20:05:18.790" Score="1" ViewCount="564" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WVcgG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WVcgG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm an engineering student, our professor assigned us this truss problem I found &lt;span class=&quot;math-container&quot;&gt;$A_y$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$L_y$&lt;/span&gt; by using the moment about l, and found ab, now what? It seems like I have too many unknowns to solve for after that...&lt;/p&gt;&#xA;" OwnerUserId="17975" LastEditorUserId="16486" LastEditDate="2019-01-08T02:05:11.240" LastActivityDate="2019-01-08T04:26:56.700" Title="I'm taking statics, and I'm stuck on this truss problem, how do do I solve this?" Tags="|statics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24457" PostTypeId="2" ParentId="24454" CreationDate="2018-11-01T23:37:53.747" Score="0" Body="&lt;p&gt;At the most basic there are two key properties you need to look at. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yield stress of the material determines how much &lt;em&gt;stress&lt;/em&gt; the poles will take before they are permanently deformed. The elastic modulus (Young modulus) determines how far they will bend for a given load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculates stresses and deflection you will need to start with beam bending equations. &lt;a href=&quot;https://www.engineersedge.com/beam_bending/beam_bending9.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineersedge.com/beam_bending/beam_bending9.htm&lt;/a&gt; Just in terms of calculating max curvature it is probably best to treat it as a canilever with a load at one end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will also need the inner and outer diameter of the tube to calculate moment of inertia (aka second moment of area). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is complicated somewhat by the fact that the joints will have different stiffness to the rest of the pole so you may be best off calculating for one length and working with the resulting radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't take into account other loads on the poles but should at least give you an order of magnitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally it is usually a good idea to look at existing designs to get an idea of what is feasible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would also reiterate the advice in the comments that unless you really know what you're doing practical experimentation is likely to be of more practical use than theoretical analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to consider composites. Certainly GRP is readily available as prefabricated rod and tube and is not terribly expensive, &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-11-01T23:37:53.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24458" PostTypeId="2" ParentId="24453" CreationDate="2018-11-02T00:10:32.347" Score="1" Body="&lt;p&gt;Assuming you want to maintain the same pressure on the tissue on the thicker part of the cord of the animal you would need more force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say by applying 30 grams pressure to a one millimetre diameter cord you squish it to an oval with a 2 millimetre side, it means it is under 15 grams/mm stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were to apply the same stress to a two millimetre diameter cord so as to squish it to a four millimetre wide oval but with the same 15 grams per millimeter of pinched area, now you'd need 60 grams clip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is purely a mechanical outlook, as to the sensitivity of different areas of a nerve cord or spinal cord, that's your area of expertise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a personal note I owe my left thumb to a colleague of you guys, a hand surgeon at UCLA hospital who helped my chain sawed thumb get implanted back to full function by borrowing some nerve from my leg, more than twenty years ago.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-11-02T14:54:13.480" LastActivityDate="2018-11-02T14:54:13.480" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24460" PostTypeId="2" ParentId="24454" CreationDate="2018-11-02T01:15:07.583" Score="1" Body="&lt;p&gt;As has been said this question lends itself to empirical test better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But just to answer your question hypothetically:&#xA; &lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{R}= \frac {M}{EI} $$&lt;/span&gt;&#xA;Meaning the more moment or less the stiffness of the aluminum pipe, the more it bends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you need to calculate the I or second moment of area of the pipe:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ I_x = \pi (D^4-d^4)/64 $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where D is the outside radius and, d the inside one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And E is the Young modulus of the particular alum you use, which you can find it in the manufacturer data sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's just assume as an example we pick 1/2inch by 1/16 inch thickness aluminum pipes 6061-T6 with E= 10000ksi and yield at 40000psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we define. C= 1/2 diameter= 1/4inch,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;span class=&quot;math-container&quot;&gt;$ I=\pi( \frac{ 1}{2}^4-\frac{3}{8}^4)=0.0209inch^4 $&lt;/span&gt;&#xA;We know &#xA;&lt;span class=&quot;math-container&quot;&gt;$$  \sigma= \frac{M*C}{I} $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume 75% of the tube's yield strength&#xA;&lt;span class=&quot;math-container&quot;&gt;$$40000*0.75 =30000psi\quad 30000=\frac{1/4M}{0.0209} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from here we find the M and plug it into the original answer and find the R radius with a safety factor of 1.25%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for simlicity we assume the overlapped section of the pipe to be rigid and not bending.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-05-01T05:24:35.033" LastActivityDate="2019-05-01T05:24:35.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24461" PostTypeId="2" ParentId="24425" CreationDate="2018-11-02T01:35:37.513" Score="0" Body="&lt;p&gt;I also asked this question on &lt;a href=&quot;https://www.quora.com/Why-does-using-the-average-pressure-in-calculations-give-the-most-accurate-results&quot; rel=&quot;nofollow noreferrer&quot;&gt;Quora&lt;/a&gt; and started sending requests. Someone answered it. I'll post the answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On reading the context for this question, i.e the best location for measurement of pressure along a pipe and why it is the center, it helps to revisit the fundamentals of pipe flow. Essentially the center of the pipe has zero shear stress since the velocity profile is typically symmetric and almost no turbulent shear. If you trace the centerline of the pipe, you would see that the total pressure at the inlet is converted to a mix of static pressure and kinematic pressure, with almost no losses. This is not true near the wall, where there are viscous losses in the boundary layer region and there might be significant turbulence or reverse flow. So the center of the pipe is a cleaner place to read the total pressure or static pressure. Of course, the sensor will cause disturbance in the flow which needs to be accounted for.&#xA;  &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;strong&gt;Author: Roopesh Mathur&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I built an example to complement Roopesh's answer and give an example of the &quot;calculations&quot; that Elger's answer mentions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an experiment in which a Pitot tube is used and there is a flow with velocity profile given by:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v(h)=V_{max}\cdot\left(1-\frac{\left|h-R\right|}{R}\right)^{1/7}  ,\space 0\leq h\leq2R \space\space$$&lt;/span&gt;&#xA;Note that &lt;span class=&quot;math-container&quot;&gt;$v(h)=v(2R-h)$&lt;/span&gt;, so the velocity profile is symmetric, with axis of symmetry passing through &lt;span class=&quot;math-container&quot;&gt;$h=R$&lt;/span&gt;. Our goal is to &lt;strong&gt;determine &lt;span class=&quot;math-container&quot;&gt;$V_{max}$&lt;/span&gt;&lt;/strong&gt;. Below is an image that illustrates the experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;img src=&quot;https://i.stack.imgur.com/rN0qj.png&quot; width=&quot;500&quot; /&gt;&#xA;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;sub&gt;(Adapted from Fluid Mechanics - Yunus A. Çengel &amp;amp; Cimbala) &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Pitot tube can measure the stagnation pressure at a point, where &lt;span class=&quot;math-container&quot;&gt;$P_{stag}=P+\rho\frac{v^2}{2}$&lt;/span&gt;. If a piezometer is used in conjunction with a Pitot tube, it's possible to calculate the fluid velocity at a specific location, using the static pressure &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; at of this location, measured with the piezometer, and the stagnation pressure at that location, measured with the Pitot tube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v=\sqrt{\frac{2(P_{stag}-P)}{\rho}}$$&lt;/span&gt;&#xA;Since &lt;span class=&quot;math-container&quot;&gt;$v=v(h)$&lt;/span&gt; , by the velocity profile formula, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left. \begin{array}{r}&#xA;        v=v(h)\\&#xA;        P=P(h)=\gamma\cdot (h+k)\\&#xA;        P_{stag}=P_{stag}(h)=P(h)+\rho\frac{v^2(h)}{2}&#xA;  \end{array}\right\} v(h)=\sqrt{\frac{2\left[P_{stag}(h)-P(h)\right]}{\rho}}&#xA;$$&lt;/span&gt;&#xA;To determine &lt;span class=&quot;math-container&quot;&gt;$V_{max}$&lt;/span&gt; it's necessary to obtain the velocity at a specific height – using the Pitot tube, the piezometer and the Pitot velocity formula – and then replace the experimental value found in the velocity profile formula. At first, we can choose any height to do the measurements! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roopesh's answer tell us the best height to be chosen in order to get the most accurate result: the height of the pipe centerline&#xA; (&lt;span class=&quot;math-container&quot;&gt;$h=R\space$&lt;/span&gt; in my example), because there we have &quot;zero shear stress&quot; and &quot;almost no turbulent shear&quot;. Furthermore in this height there are &quot;almost no losses&quot; in the total pressure. Then, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v(R)=\sqrt{\frac{2\left[P_{stag}(R)-P(R)\right]}{\rho}}=\sqrt{\frac{2\left[P_{stag}(R)-P_{average}\right]}{\rho}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this confirms what Elger said:&quot;If you use this value [average pressure] in your calculations, this will be give you the most accurate results&quot;. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, in general, the average pressure gives the most accurate results if used in calculations because there are many applications/cases in which the locations with &lt;span class=&quot;math-container&quot;&gt;$P=P_{average}$&lt;/span&gt; are the best places for experimental data collection.&lt;/p&gt;&#xA;" OwnerUserId="4358" LastActivityDate="2018-11-02T01:35:37.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24462" PostTypeId="2" ParentId="24357" CreationDate="2018-11-02T01:43:50.303" Score="1" Body="&lt;p&gt;I'll divide my answer in two cases. First, I'll talk about &lt;strong&gt;liquids in motion&lt;/strong&gt; (assuming incompressible flow). Then, I'll talk about &lt;strong&gt;liquids at rest&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Liquid Flow:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Reading the comments of &lt;a href=&quot;https://youtu.be/_VDVi_zq3nk&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; YouTube video about piezometers made by Donald Elger, I found the answer for this case:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why is it [the pressure measurement with piezometer] taken from the middle of the pipe?&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Elger's answer: The pressure variation across a section of a pipe is hydrostatic; thus, the pressure will vary linearly with radius and the pressure at the center of the pipe is the average pressure. If you use this value of pressure in your calculations, this will be give you the &lt;strong&gt;most accurate results&lt;/strong&gt;. Thus, engineers nearly always apply or measure the pressure at the center of the pipe. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The question that came to me as soon as I read this was: &lt;a href=&quot;https://engineering.stackexchange.com/questions/24425&quot;&gt;&quot;Why using average pressure in calculations gives the most accurate results?&quot;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I recommend reading my answer to this question before proceeding)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Briefly, in general, the average pressure gives the most accurate results if used in calculations because there are many applications/cases in which the locations with &lt;span class=&quot;math-container&quot;&gt;$P=P_{average}$&lt;/span&gt; are the best places for experimental data collection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a pipe, this location is its centerline. So, I believe that this is why textbooks generally choose this location in case of liquids in motion: the centerline is associated with &lt;span class=&quot;math-container&quot;&gt;$P_{average}$&lt;/span&gt; that, in its turn, is associated with the best places for experimental data collection for many applications.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Liquids at rest:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For this case, firstly I would like to quote part of the answer written by David White to my question on Physics.SE &lt;a href=&quot;https://physics.stackexchange.com/a/436482/102936&quot;&gt;&quot;Where is the right place to put the pressure gauge to measure the pressure of a tank?&quot;&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The location depends on why you are measuring the pressure. There will be a process reason for the pressure measurement, and that will determine the location of the pressure measuring device.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When textbooks present pressure gauges/manometers/piezometers for the first time, the presentation is usually &quot;application neutral&quot; (i.e., there's no process reason), the diagrams/sketches/figures are only to illustrate the concepts/formulas. Therefore there are no best points as in the liquid flow case, for two reasons:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There is no process reason that determines the location of measurement;&lt;/li&gt;&#xA;&lt;li&gt;Since the liquid is at rest, there are no points that lead to most accurate results, they all provide the same accuracy.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But the authors need to choose a point to do the pressure-related calculations...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After everything I've researched, my hypothesis is that the &quot;point choice&quot; of hydrostatics was imported from hydrodynamics. So, instead of choosing a random point to pressure-related calculations, they choose one that at least has importance/meaning for other areas of Fluid Mechanics.&lt;/p&gt;&#xA;" OwnerUserId="4358" LastEditorUserId="4358" LastEditDate="2018-11-03T19:00:35.187" LastActivityDate="2018-11-03T19:00:35.187" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24463" PostTypeId="2" ParentId="24446" CreationDate="2018-11-02T03:41:47.880" Score="0" Body="&lt;p&gt;There are stablished criteria by codes to design columns, but we ignore these for the benefit of clarity here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically we want to design an Iron column strong enough to support a 20 kg load with a length of &lt;span class=&quot;math-container&quot;&gt;$ L= 2\times 3ft = 6ft$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We multiplied by 2 because the column is a cantilever column, it is not restrained laterally on top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we check for the slenderness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually the columns with a radius of gyration (radius of gyration is the radius at which you can assume all the mass is located), greater then L/20 are considered slender and need to be checked for buckling load.&#xA;Radius of gyration of a cylinder is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ R_{gyration} =  D/4$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we know any diameter less than 4.5 cm should be considered a slender column and checked for buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say for trial and error we pick as a first guess a 1.5 cm diameter post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ p_{critical}  = \frac {\pi^2EI}{L^2}&#xA; \\\ and \ I = \frac {pi D^4}{64}  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the Maximum buckling load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I = 0.24cm^4 and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P = \pi^2(0.24E)/ 183^2 = \text{ approx} \ 76kg$&lt;/span&gt;  Assuming Iron's E at 100 GPa. This strength is reasonable and has a factor of safety of approx. 4, I would stop here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But through the same steps picking a size smaller we gat closer to the optimal size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally we check the result Against the compression strength with a factor of safety, and check for likelihood of overturning of the base block, which is missing the width. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-11-05T17:23:49.907" LastActivityDate="2018-11-05T17:23:49.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24465" PostTypeId="1" CreationDate="2018-11-02T11:13:17.613" Score="2" ViewCount="1552" Body="&lt;p&gt;I have a concrete steel ceiling. The calculations show the ceiling is designed for 1,5 kN/m² in addition to it's own weight and traffic loads including snow. So I assume I can hang 1,5 kN/m² from that ceiling, provided I use suitable anchors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I hang any load from that ceiling, I apply discrete point loads. I probably cannot simply say: Well the room is 2x4m, so I can hang 2m x 4m x 1,5kN/m² = 12kN in the middle of the room. Which is probably wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how do I arrive at the point loads I can apply?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-01-01T17:02:30.500" Title="How to convert kN/m² into point loads?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24469" PostTypeId="2" ParentId="24465" CreationDate="2018-11-02T14:52:14.163" Score="1" Body="&lt;p&gt;Any slab or ceiling can Receive several kinds of loading which are marked on the structural plans in case that loading can be critical. e.g., a library floor is ranked both for distributed loads and concentrated loads due to stacking of the books, and they are different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or a plant floor will be rated for equipment dynamic impact loading, separately than allowable floor load. So much so that a heavy machine load has to be designed for its lateral forces impacting the lateral stability of the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However most of the ceilings can support light weight light fixture or suspended signs, provided they are attach using common sense and appropriate anchors or hardware penetrating into competent material and avoiding potential electrical or utility conduits. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-02T14:52:14.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24470" PostTypeId="1" AcceptedAnswerId="24472" CreationDate="2018-11-02T15:15:26.443" Score="0" ViewCount="476" Body="&lt;p&gt;When attempting to trap a particle in a laser beam i was able to achieve the effect.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/AvAOd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AvAOd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Pretty cool but now when i put of piece of plexiglass that is 1/8” thick i can no longer achieve the effect. The beam hits a focal lens when its diameter is around 6mm and focuses 50mm away. I put the piece if plexiglass 15mm after the focal lens. Now it is obvious that my 25mW laser will lose power but i wonder how, where, and how much? I imagine reflection and scattering accounts for most of it but there is more that plays a part. I want to know if theres a equation that takes a material, thickness, laser wavelength and power and will maybe give me my power on the other side of the material.&lt;/p&gt;&#xA;" OwnerUserId="15207" LastActivityDate="2018-11-02T15:40:01.687" Title="Calculating energy loss from laser when shining through a material" Tags="|lasers|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24471" PostTypeId="2" ParentId="24470" CreationDate="2018-11-02T15:35:36.773" Score="2" Body="&lt;p&gt;There are a number of things to deal with.  The plexi has a reflectivity of around 4% per surface (see, e.g.   &lt;a href=&quot;https://www.plexiglas.com/export/sites/plexiglas/.content/medias/downloads/sheet-docs/plexiglas-general-information-and-physical-properties.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this spec sheet&lt;/a&gt; , but minimal absorption.  So pure loss is 8% (but will increase dramatically if you tilt the plexi sheet)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you've placed the sheet between the lens and the focal point, you have not only moved the focus (see &quot;equivalent optical thickness&quot;) but almost certainly induced some coma or astigmatism, thus spoiling the power density in the focussed spot.  If you can track down a ray-trace application or even a webpage-based one, you may be able to get an estimate of these effects.&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-11-02T15:35:36.773" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24472" PostTypeId="2" ParentId="24470" CreationDate="2018-11-02T15:40:01.687" Score="2" Body="&lt;p&gt;The equation you are looking for is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I(z) = (1-R) \cdot I_0 \cdot e^{-\alpha z}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$I(z)$&lt;/span&gt;: intensity of the laser beam after the distance &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;: reflectance of material&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$I_0$&lt;/span&gt;: intensity of laser beam on the surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;: absorption coefficient (NOT the same as absorptance)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;: thickness of the material&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you plug the thickness of the plexiglas plate in as &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;, you get the intensity of the laser beam after it passes through the plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the equation considers intensity instead of power, so you need to find the beam diameter on the surface of your plexiglas plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, you need to find the absorption coefficient of the material for the given wavelength. The absorption coefficient (or &lt;a href=&quot;https://en.wikipedia.org/wiki/Attenuation_coefficient&quot; rel=&quot;nofollow noreferrer&quot;&gt;attenuation coefficient&lt;/a&gt;) is not a dimensionless number like the absorptance, but has the dimension &lt;span class=&quot;math-container&quot;&gt;$m^{-1}$&lt;/span&gt;, so they should not be mixed up.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2018-11-02T15:40:01.687" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24473" PostTypeId="1" AcceptedAnswerId="24486" CreationDate="2018-11-02T16:05:16.320" Score="3" ViewCount="291" Body="&lt;p&gt;I have a little creek in my back yard.  It's slow moving and safe (as safe as a slow moving creek can be.)  I'd like to build a little set of stone steps into the creek and am not sure of the best structural/engineering/least impactful way of do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's the main question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should the stairs be pointing up river or down river?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've attached an image (very rough) of what I'm thinking.  i'm not sure if the top version or the bottom would be better all else considered equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GLLdi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GLLdi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18135" LastActivityDate="2018-11-03T09:26:03.300" Title="Which way should I build stairs into a creek?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24476" PostTypeId="2" ParentId="17457" CreationDate="2018-11-02T17:07:11.757" Score="1" Body="&lt;p&gt;What do you do? You either use different software to solve the two equations, two unknowns (because there is definitely an answer), or you use a different technique to solve the problem. In this case, a different technique looks to be easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your equations for the sum of forces in X and Y at point O are equivalent to this force diagram (which is not to scale):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i8AHy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i8AHy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two components of the force &lt;span class=&quot;math-container&quot;&gt;$F_{OB}$&lt;/span&gt; can be determined from this diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RyNyh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RyNyh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which you can determine these answers after substituting the value of &lt;span class=&quot;math-container&quot;&gt;$F_{OA}$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F_{OB}=0.7698F=77.0N$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \gamma=6.78^o$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \beta=\gamma+\alpha = 26.78^o$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \omega=180-(90-\alpha)-\beta = 43.22^o$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2018-11-02T17:07:11.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24479" PostTypeId="1" CreationDate="2018-11-02T20:54:06.793" Score="1" ViewCount="257" Body="&lt;p&gt;I'm trying to identify the pictured fastener. It's a single piece around the arm of a gas spring, allowing it to pivot. I've had to replace the spring, but the screw is size M8 while the eyelet on the spring arm, is only big enough for an M6. Is there another way to achieve the same function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any assistance would be much appreciated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6bfOt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6bfOt.jpg&quot; alt=&quot;mystery fastener&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18142" LastActivityDate="2024-01-07T05:03:40.470" Title="Help identifying a fastener/screw" Tags="|mechanical-engineering|bolting|fasteners|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24480" PostTypeId="1" CreationDate="2018-11-02T21:35:43.190" Score="1" ViewCount="179" Body="&lt;p&gt;Are there solid materials that begin as isotropic materials (i.e. isotropic elastic parameters) in their natural (i.e. unstressed) state, but that become anisotropic under sufficient stress/strain?  What would it take to modify an existing material with isotropic elastic properties to have anisotropic elastic properties?  &lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2018-11-02T23:56:14.060" LastActivityDate="2018-11-05T21:19:13.730" Title="Can isotropic materials become anisotropic under elastic deformation?" Tags="|mechanical-engineering|materials|solid-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24481" PostTypeId="2" ParentId="24432" CreationDate="2018-11-02T22:43:19.447" Score="0" Body="&lt;p&gt;You could use an old dead electrical drill's chuck and shank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or even hang the whole drill from the ceiling. &#xA;If you need a bigger or stronger gadget, you can use an old industrial hammer drill or jack hammer tip.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-02T22:43:19.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24483" PostTypeId="2" ParentId="24479" CreationDate="2018-11-03T06:24:32.450" Score="0" Body="&lt;p&gt;There is the bolt with domed head on lhs securing the bar to the circular body. That body with flats then has a machined threaded section on the rhs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This allows the bar to rotate or be removed separately to the other part which may secure something else.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-11-03T06:24:32.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24485" PostTypeId="1" AcceptedAnswerId="24494" CreationDate="2018-11-03T06:46:42.717" Score="0" ViewCount="422" Body="&lt;p&gt;Object &lt;strong&gt;A&lt;/strong&gt; has mass &lt;span class=&quot;math-container&quot;&gt;$m_A$&lt;/span&gt;, and is travelling west with velocity &lt;span class=&quot;math-container&quot;&gt;$v_A$&lt;/span&gt;.  Object &lt;strong&gt;B&lt;/strong&gt; has mass &lt;span class=&quot;math-container&quot;&gt;$m_B$&lt;/span&gt; and is travelling east with velocity &lt;span class=&quot;math-container&quot;&gt;$-v_B$&lt;/span&gt;.  The objects collide in a perfectly inelastic collision so that the velocities of the two objects after the collision are the same.  Prior to the collision, the objects are made of uniform, identical materials and shapes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How much of the kinetic energy is absorbed by deformation of object &lt;strong&gt;A&lt;/strong&gt;, and how much of the kinetic energy is absorbed by deformation of object &lt;strong&gt;B&lt;/strong&gt;?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is not a unique solution to this question because of a lack of necessary constraints, please answer the question with some examples for the unspecified constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: when I said same shapes, I meant they are both cylinders or prisms or spheres or something.  The dimensions and masses of the shapes may be different.&lt;/p&gt;&#xA;" OwnerUserId="14562" LastEditorUserId="14562" LastEditDate="2018-11-05T07:43:58.033" LastActivityDate="2018-11-05T13:33:47.533" Title="Distribution of energy absorption in inelastic collision" Tags="|mechanical-engineering|structural-engineering|materials|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="24486" PostTypeId="2" ParentId="24473" CreationDate="2018-11-03T09:26:03.300" Score="1" Body="&lt;p&gt;From a fluid mechanics perspective, the lower the fluid interaction the better.&lt;br&gt;&#xA;As you have seen, a lot depends on how the original creek is shaped.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you face the stairs into the fluid flow, you will impact the most physical stress on the rocks. When you move water, it is going to go somewhere else. So, it will erode the banks. This approach will also result in a vortex being generated behind the stairs that will erode the ground behind your stairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pocketed entry drawn in your first example will last longer.  Just be careful to reinforce the banks.&lt;/p&gt;&#xA;" OwnerUserId="18144" LastActivityDate="2018-11-03T09:26:03.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24488" PostTypeId="1" AcceptedAnswerId="24490" CreationDate="2018-11-03T10:40:46.360" Score="1" ViewCount="210" Body="&lt;p&gt;I'm building a CNC router around a HSD ES330 spindle motor that has pneumatic tool-changing ability. The manual recommends the following pneumatic circuit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ghIvy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ghIvy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure I understand the purpose of the &quot;quick bleeder valve&quot; (circuit element #5 in the diagram). Searching for &quot;quick bleeder valve&quot; doesn't turn up anything with that symbol, or a part I'm confident buying on McMaster-Carr. I know the one-way flow restrictor (element #4) is supposed to slow the tool ejection speed; I guess during tool unlocking, air returns via #2, but I'm not sure if that exhaust air escapes through the bleeder valve, or the 5-2 valve #6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do have a one-way flow &lt;a href=&quot;https://www.mcmaster.com/6692K11&quot; rel=&quot;nofollow noreferrer&quot;&gt;restrictor&lt;/a&gt; (although I'm not entirely sure how to adjust it).&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the &quot;quick bleeder valve?&quot;&lt;/li&gt;&#xA;&lt;li&gt;Do I really need it?&lt;/li&gt;&#xA;&lt;li&gt;Where is this air exhausted?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17903" LastEditorUserId="17903" LastEditDate="2018-11-03T11:58:38.480" LastActivityDate="2018-11-03T11:58:38.480" Title="What is the &quot;quick bleeder valve&quot; in this tool changer pneumatic schematic (and other questions)?" Tags="|pneumatic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24490" PostTypeId="2" ParentId="24488" CreationDate="2018-11-03T11:52:06.713" Score="2" Body="&lt;p&gt;It's common for CNC equipment to use air pressure to manipulate tools and collets and mandrels and other &quot;non-precision&quot; mechanical components. In the case of tooling, the engagement sockets provide the precision, while the air pressure locks the collars to the mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air pressure can be rapidly applied in a closed system, but not so rapidly reversed when using a pump, for example. Considering the cost of air (as in as free as air), simply releasing the pressure to the atmosphere resolves the question of what to do with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your question, one can expect that the circuit will use air pressure to lock a tool in place, and when the sequence requires that the tool be released, the dump valve/quick bleeder valve/quick exhaust valve will be actuated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the last term, because my search resulted in better returns for various valve suppliers with &quot;quick exhaust valve&quot; as opposed to any other terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are mechanical exhaust valves (not applicable here) and solenoid exhaust valves, which become part of an electrical circuit to seal the valve and to open the valve. I found prices in the triple digit range, but also some lower cost items. &lt;a href=&quot;https://www.valvesonline.com.au/pneumatic-quick-exhaust-valve&quot; rel=&quot;nofollow noreferrer&quot;&gt;This one&lt;/a&gt; is sourced from Australia, but included as a representative example rather than a solid suggestion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/faryL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/faryL.png&quot; alt=&quot;exhaust valve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your system air pressure, it may be necessary to install an exhaust line or a deflector of some form. When the valve dumps, it will push on anything in line, which may be insignificant, or may be undesirable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The link to the inline flow restrictor shows a reference to a dial and one of the photos shows gradations along the circumference of the fitting. It's difficult to see if there is a screwdriver slot for adjustment or merely something turned with a wrench.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-11-03T11:52:06.713" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24491" PostTypeId="1" CreationDate="2018-11-03T11:55:50.880" Score="1" ViewCount="377" Body="&lt;p&gt;I see a lot of people asking why the Penny-farthing's front wheel is so big, and I think I've read enough materials to understand why (at least, at the most basic level). But even so, I feel that the rear wheel is too small — not just compared to the front one, but actually really small. &#xA;Is there a specific reason why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, why would this front-big-rear-small design not be used for other vehicles (for instance, I know some trucks have smaller front wheels and larger rear wheels, which is the exact opposite of penny farthings)? &lt;/p&gt;&#xA;" OwnerUserId="18148" LastEditorUserId="90" LastEditDate="2018-11-04T15:29:57.277" LastActivityDate="2018-11-04T21:17:17.593" Title="Why is the rear wheel of the Penny-farthing so small?" Tags="|mechanical-engineering|design|torque|wheels|bicycles|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24492" PostTypeId="2" ParentId="24491" CreationDate="2018-11-03T12:08:03.000" Score="1" Body="&lt;p&gt;The rear wheel of the penny-farthing was only necessary to stop the seat rotating down to the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Riding a monocyle is possible, but takes more skill and practice...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-11-03T12:08:03.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24494" PostTypeId="2" ParentId="24485" CreationDate="2018-11-03T16:42:19.350" Score="-1" Body="&lt;p&gt;This is one-dimensional travel (along a line). Applying conservation of energy and momentum for a perfectly elastic collision, we would find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\left(m_A + m_B\right) v_{Af}^2 - \left(m_A v_{Ao}^2 + m_B v_{Bo}^2\right) = 0&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\left(m_A v_{Af} + m_B v_{Bf}\right) - \left(m_A v_{Ao} - m_B v_{Bo}\right) = 0&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is two equations and two unknowns (final velocities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a PERFECTLY inelastic collision, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\left(m_A + m_B\right) v_{f}^2 - \left(m_A v_{Ao}^2 + m_B v_{Bo}^2\right) = 2\left(q - w - (m_A + m_B)\Delta\tilde{U}\right)&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\left(m_A + m_B\right)v_f - \left(m_A v_{Ao} - m_B v_{Bo}\right) = 0&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; is the heat loss, &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; is work done &lt;em&gt;on&lt;/em&gt; the object(s), and &lt;span class=&quot;math-container&quot;&gt;$\tilde{U}$&lt;/span&gt; is the specific internal energy (J/kg) of the combined system. Solve the momentum equation for &lt;span class=&quot;math-container&quot;&gt;$v_f$&lt;/span&gt;. For an adiabatic collision of perfectly rigid objects, find the temperature change from &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\left(m_A + m_B\right) v_{f}^2 - \left(m_A v_{Ao}^2 + m_B v_{Bo}^2\right) = -2\left(m_A + m_B\right)\tilde{C}_V\Delta T&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When energy is absorbed, &lt;span class=&quot;math-container&quot;&gt;$KE_f &amp;lt; KE_i$&lt;/span&gt;, and the temperature of the final object is greater than the initial temperature. When the solids are perfectly incompressible, &lt;span class=&quot;math-container&quot;&gt;$\tilde{C}_V = \tilde{C}_p$&lt;/span&gt;. For an isothermal collision of perfectly rigid objects,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\left(m_A + m_B\right) v_{f}^2 - \left(m_A v_{Ao}^2 + m_B v_{Bo}^2\right) = 2q&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, when energy is absorbed,&lt;span class=&quot;math-container&quot;&gt;$KE_f &amp;lt; KE_i$&lt;/span&gt;, and the process is exothermic (heat leaves). Finally, for perfectly plastic objects where all the energy is absorbed in work,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\left(m_A + m_B\right) v_{f}^2 - \left(m_A v_{Ao}^2 + m_B v_{Bo}^2\right) = -2w&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the final kinetic energy is less than the sum of the two starting objects, the final object will experience work done ON it. You can pretend this is equivalent to &lt;span class=&quot;math-container&quot;&gt;$p\Delta V$&lt;/span&gt; and say, the final object has a total volume that is less than the sum of the two starting objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The net kinetic energy change for object A is &lt;span class=&quot;math-container&quot;&gt;$(1/2)m_A (v_{f}^2 - v_{Ao}^2)$&lt;/span&gt;. Apply the same for object B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for energy &lt;em&gt;absorbed&lt;/em&gt; by the objects individually, the answer is ambiguous. The objects cease to exist as individual objects at the moment of the collision. Any energy lost is absorbed by the final combined object, not by any one individually.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-11-05T13:33:47.533" LastActivityDate="2018-11-05T13:33:47.533" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="24495" PostTypeId="2" ParentId="24093" CreationDate="2018-11-03T16:47:40.547" Score="1" Body="&lt;p&gt;I was able to order a 25mW laser in the 532nm wavelength and trap a particle using the following optics: a 50mm focal lens which after it was focused I allowed it to expand to a 1cm diameter beam width. Then I used another focusing lens of fl=25mm and that focused beam is where the trapping begins. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/VHFEx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VHFEx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;By touching a black marker to it (black because any light colors will simply be reflected and not absorbed) and when the black pieces of sharpie marker would burn off they would be trapped into the laser beam at its lowest density point. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/uROTa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uROTa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;As for how long it lasts without burning, I don't think it is burning that is the problem, its wind and air. The slightest breath can cause the particle to move away. Sometimes you can have a particle stay for minuets and some just a second.&lt;/p&gt;&#xA;" OwnerUserId="15207" LastActivityDate="2018-11-03T16:47:40.547" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24496" PostTypeId="2" ParentId="24485" CreationDate="2018-11-03T17:50:34.337" Score="0" Body="&lt;p&gt;For Simplicity we assume the objects are moving on the same path but opposite direction, however it would be only a bit complex if they were not moving colinearly if one introduces vectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the momentum of particles before the collision and after are preserved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$m_Av_A + m_Bv_B = (m_A +m_b)v_{(A+B)}   $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's consider object A. It has changed it's velocity by the amount &lt;span class=&quot;math-container&quot;&gt;$v_A -v_{(A+B)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the kinetic energy it has lost is &lt;span class=&quot;math-container&quot;&gt;$ E_k= \frac{ m \space[v_A -v_{(A+B)}]^2}{2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And same way we can calculate the loss for B.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-11-04T01:03:10.873" LastActivityDate="2018-11-04T01:03:10.873" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24497" PostTypeId="1" CreationDate="2018-11-03T18:45:34.973" Score="-1" ViewCount="1255" Body="&lt;p&gt;With the saturation pressure apparatus we did a heating and a colling process. We measured the temperature and afterwards we calculated the specifc volume of saturated vapour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question 1: What would the effect be if steam leaked out slowly during the experiment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My proposal: If steam leaks out slowly during the experiment than a higher pressure is necessary to increase the temperature. Therefore the temperature wouldn’t increase with higher pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question 2: How do the values of the specifc volume of saturated vapour obtained using the experimental data compare with the tabulated values in the steam table?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea for question 2 :(&lt;/p&gt;&#xA;" OwnerUserId="17913" LastActivityDate="2018-11-04T13:38:49.763" Title="Pressure Temperature relation of saturated steam" Tags="|thermodynamics|steam|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24498" PostTypeId="2" ParentId="24491" CreationDate="2018-11-03T18:57:06.550" Score="0" Body="&lt;p&gt;The large front wheel is basically gearing.  For every rotation of the pedals it moves about 15 feet.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other vehicles don't do this has they have gears.  Even a single speed bicycle has front and rear sprockets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rear is small as all it has to do is track.&lt;/p&gt;&#xA;" OwnerUserId="14485" LastEditorUserId="14485" LastEditDate="2018-11-03T20:41:45.610" LastActivityDate="2018-11-03T20:41:45.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24500" PostTypeId="1" AcceptedAnswerId="24507" CreationDate="2018-11-03T19:28:16.327" Score="3" ViewCount="98" Body="&lt;p&gt;I am trying to design a pipe which takes water from a river and transports it to a reservoir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not want my pipe to take all of the water from stream, because that will create all sorts of environmental problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I make sure that this does not happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flow rate in the stream is &lt;span class=&quot;math-container&quot;&gt;$0.02 m^{3}/s$&lt;/span&gt; and the stream is &lt;span class=&quot;math-container&quot;&gt;$1m$&lt;/span&gt; wide and &lt;span class=&quot;math-container&quot;&gt;$1.25m$&lt;/span&gt; deep. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I just need to make sure that the flow rate in my pipe is not greater than &lt;span class=&quot;math-container&quot;&gt;$0.02 m^{3}/s$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="17668" LastActivityDate="2019-01-01T02:06:10.147" Title="How to make sure a pipe does not take all of the water from a stream?" Tags="|hydraulics|pipelines|open-channel-flow|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24502" PostTypeId="1" CreationDate="2018-11-03T20:07:30.727" Score="1" ViewCount="91" Body="&lt;p&gt;I'm building a small pneumatic control cabinet with a couple filters, regulators, valves, and flow restrictors. I had started by using 8 mm push-to-connect fittings and nylon tube, but it seems I can make things more compact by using metal tubing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thing is, I'm having a hard time finding any kind of basics tutorial for how one does this. I've seen a few photos, and found a couple pipe-bending tutorials, but I don't know what kind of fittings to use.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It seems I can use flared fittings, is that what's usually done?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I can't find metric sizes on McMaster-Carr, so I'm going with 5/16&quot;. But while they have aluminum tubing and bending tools for 5/16&quot; OD tubing, they don't have fittings. They do have 1/4&quot; and 3/8&quot; fittings.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Can I under- or over-size the fitting and the pipe like that?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there any good resources for learning how to fabricate a control system using metal tubing?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The thing I'm controlling is the tool changer on a nice spindle motor. It has 8 mm OD push-to-connect fittings, 6 mm ID tubing. 3/8&quot; metal tubing has a 1&quot; bend radius, while 5/16&quot; tubing has an 11/16&quot; bend radius, allowing for slightly more compact layout.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17903" LastActivityDate="2019-12-30T13:14:07.470" Title="How do I connect pneumatic components with metal tubing?" Tags="|pneumatic|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24504" PostTypeId="1" AcceptedAnswerId="24505" CreationDate="2018-11-03T20:48:31.870" Score="2" ViewCount="37145" Body="&lt;p&gt;What should I choose from today. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking at washing machines, and I see that LG has this Inverter Direct Drive (6 motion cleaning) technology in their machines, I see that its not like a year old one, but I am also wondering how long do these last?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can someone tell me how does direct drive perform nowadays, especially in LG products;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If its more reliable then how cheaper is belt drive fixing?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there more stuff to go wrong with that technology?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I see some manufacturers use only inverter technology, what does that&#xA;mean to the average consumer, should I choose that over non-inverter?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14322" LastActivityDate="2021-02-10T22:54:23.133" Title="Inverter + Direct drive vs. Belt drive motor in washing machines?" Tags="|mechanical-engineering|electrical-engineering|motors|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24505" PostTypeId="2" ParentId="24504" CreationDate="2018-11-03T21:44:51.797" Score="4" Body="&lt;p&gt;There are tons of advantages and disadvantages, but I will mention only the relevant cons and pros.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indirect (belt) drive usually produces less noise in comparison to direct drive (continue power transmission vs discrete power transmission like gears and chains), but you know engineers had done a lot to reduce the noise in household equipment, unfortunately not yet in industrial sector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Belts are far more prone to relaxation and then they can transmit less torque. As you might have noticed after some years after you replace the belt. Again this is less observable for home and kitchen equipment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Belt drives dampen vibrations,  direct drives produce more vibration specially when the rpm goes up. But in washing machines, the direction of rotation changes time to time and direct drive offers more stability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maintenance cost of belt drives is lower than that for direct drive, they are easier to disengage for repairing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Belt drives apply greater load on the main shaft of the machine, it has a direct effect on the life time of bearings, shaft misalignment and as a result, oil leakage and more vibrations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct drives requires less space in comparison to belt drives, so smaller machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Belts are sensitive to the surrounding temperature, humidity, oil ..., so the probability of failure is higher plus the micro slip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inverters are a sort of speed and power control over electric motors, long story short, they make the rotation of induction motors smoother in comparison to classic motors without inverters, of course they cost the clients more, but it also regulates power consumption.  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorUserId="90" LastEditDate="2020-04-19T18:15:38.193" LastActivityDate="2020-04-19T18:15:38.193" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="24507" PostTypeId="2" ParentId="24500" CreationDate="2018-11-04T02:18:10.530" Score="3" Body="&lt;p&gt;You can put a valve on your pipe to control the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or partition the stream by means of a short concrete wall which looks like an indented grout trowel upside-down . Say two opening for the stream and one leading to a small pond downstream feeding your pipe. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-04T02:18:10.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24508" PostTypeId="1" CreationDate="2018-11-04T05:55:28.983" Score="0" ViewCount="522" Body="&lt;p&gt;I have an old classic lamp (Tolomeo).&#xA;It was purchased when we lived in Singapore, so it is designed for 220V.&#xA;Since we moved to US, I have not been able to use it as the power supply here is 110V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The adapter of the lam says:&#xA;Power Supply Type BL2UK&#xA;240V~ 50Hz TW 10ºC&#xA;For PL Lamps Type:&#xA;(then is lists some options)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I was wondering if someone can help me with these questions.&#xA;1. Will replacing this adapter solve my problem?&#xA;2. Will I also need to change the bulb? The current bulb it uses is: 10W&#xA;3. If adapter needs to be replaced, which one should I replace it with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot for any help or guidance.&lt;/p&gt;&#xA;" OwnerUserId="18156" LastActivityDate="2019-05-03T23:01:31.170" Title="Question about converting 220V lamp to 110V" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24509" PostTypeId="2" ParentId="24500" CreationDate="2018-11-04T06:04:00.210" Score="0" Body="&lt;p&gt;Some countries have Water or River authorities that decide how much water you can have from the stream and in certain seasons that can be zero...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, intakes are designed to trap debris, both floating and sediment, a coanda screen is a good option for this. Then the intake is also designed so that if there is insufficient volume in the stream then your supply stops, this can be achieved by a simple dam for example.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-11-04T06:04:00.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24510" PostTypeId="2" ParentId="24508" CreationDate="2018-11-04T06:11:23.397" Score="1" Body="&lt;p&gt;This sounds like a halogen lamp, get a new power supply that will produce the correct output in terms of power, voltage and current - you don’t mention those but they should be there stated on the original.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can get the correct bulb and use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or rewire it and change the bulb fitting for a bulb type that is convenient and locates the filament in the same position relative to the reflector.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-11-04T06:11:23.397" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24512" PostTypeId="2" ParentId="24443" CreationDate="2018-11-04T10:55:10.693" Score="1" Body="&lt;p&gt;Looking at if the device supports BLE 4.0, 4.1, 4.2 or 5.0 is a good start. Not all BLE modules are 5.0 capable. What type is distance range might be a parameter to consider some BLE module have line of sight range of 1km or more at a low data rates. Below is a short side by side comparison to help get started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jZsxQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jZsxQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2018-11-04T10:55:10.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24513" PostTypeId="2" ParentId="24502" CreationDate="2018-11-04T11:30:49.860" Score="1" Body="&lt;p&gt;As far as I know, two commonly used systems are &lt;a href=&quot;https://en.wikipedia.org/wiki/Cutting_ring_fitting&quot; rel=&quot;nofollow noreferrer&quot;&gt;cutting ring fittings&lt;/a&gt; and &lt;a href=&quot;https://www.geberit.co.uk/products/piping-systems-supply/mapress-stainless-steel-pipes-fittings/&quot; rel=&quot;nofollow noreferrer&quot;&gt;press fittings&lt;/a&gt;. Note that both need some space for installation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should talk to a pneumatic components supplier, there are systems where you simply push the pipe into a connector (no english link at hand).&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="9506" LastEditDate="2019-12-30T13:14:07.470" LastActivityDate="2019-12-30T13:14:07.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24514" PostTypeId="1" CreationDate="2018-11-04T13:23:40.963" Score="1" ViewCount="1116" Body="&lt;p&gt;I am currently working on a bridge design issue. The main goal is to correct issues of overdesign. My professor said we can do this by selecting beams with different cross sections. My question is how can I determine what is the optimal cross sectional area for the force being applied to a specific beam. So far this is what I can understand. I must first calculate the amount of stress the member is experiencing using stress = force / area of the current beam used.&lt;/p&gt;&#xA;" OwnerUserId="18162" LastActivityDate="2018-11-04T14:22:47.797" Title="How can I determine the optimal cross section of a member to correct overdesign?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24515" PostTypeId="2" ParentId="24497" CreationDate="2018-11-04T13:38:49.763" Score="0" Body="&lt;h2&gt;Question 1&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Saturation occurs at a fixed pressure and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) In a constant volume experiment, assume that you keep enough liquid to assure the system is always at the two phase point (i.e. is always saturated). Assume the leak point has no affect on the internal pressure (i.e. this is the meaning of the term &quot;slowly&quot;). When vapor &quot;leaks out (slowly)&quot;, more liquid will evaporate make up for the loss. The &lt;span class=&quot;math-container&quot;&gt;$p, T$&lt;/span&gt; relationship at the two phase line is maintained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) In a constant volume experiment, assume that you do not have enough liquid to assure the system is always at the two phase point (i.e. is always saturated). When vapor &quot;leaks out (slowly)&quot;, no liquid water is available to make up the loss. The system changes from saturation (two phase) to water vapor (single phase). The relationship between &lt;span class=&quot;math-container&quot;&gt;$p, T$&lt;/span&gt; is no longer defined by the saturation line.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question 2&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Steam tables contain values of &lt;span class=&quot;math-container&quot;&gt;$p, T$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\tilde{V}$&lt;/span&gt; (specific volume). Compare your experimental value to that from a steam table at the given &lt;span class=&quot;math-container&quot;&gt;$p, T$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-11-04T13:38:49.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24517" PostTypeId="2" ParentId="24514" CreationDate="2018-11-04T13:47:41.383" Score="0" Body="&lt;p&gt;For beam, it is often used to resist bending. Especially, when you design the bridge, the force is often in the trasverse direction, like pedestrian walking or car moving on the bridge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stress formula in the beam under bending is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sigma = -\frac{My}{EI}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; is the moment of inertia. The geometry of cross section will determine this term. For example, the rectangular cross section with &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; height and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; width have &lt;span class=&quot;math-container&quot;&gt;$I=\frac{bh^3}{12}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, there are other stresses in the beam. For instance, tension force&lt;span class=&quot;math-container&quot;&gt;$\sigma = \frac{F}{A}$&lt;/span&gt;, flexural shear force, etc. You need to determine whether or not you need to consider them as they might be much smaller than the stress introduced by bending.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to optimize your design. You need to know what is the maximum stress allowed in your bridge material and safety factor. Then, you can determine the minimum moment of inertia of the cross section. At this point, you need to choose whatever best cross section to meet the minimum moment of inertia requirement.&lt;/p&gt;&#xA;" OwnerUserId="5069" LastActivityDate="2018-11-04T13:47:41.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24518" PostTypeId="2" ParentId="24514" CreationDate="2018-11-04T14:22:47.797" Score="1" Body="&lt;p&gt;To make this easy to do by hand, start with a &lt;em&gt;statically determinate&lt;/em&gt; structural design. Then (ignoring the self-weight of the bridge) the load in each part of the structure is independent of its stiffness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you know the loads, you can choose the size of each member to create a &quot;fully stressed design&quot; where the maximum stress in each member is equal to the safe limit for the stress in the material (including a safety factor, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a bridge, you probably can't ignore its self-weight completely, but you can iterate the calculations. Start with a guess of the self weight, calculate the loads, change the size of the members, recalculate the self weight based on the new size, and repeat till nothing changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are using computer software to optimize the design, that will work for statically indeterminate structures in a similar way, but by hand it is usually too much work to recalculate how the &lt;em&gt;load paths&lt;/em&gt; through the structure change if you change the size of different parts of it by different amounts.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-11-04T14:22:47.797" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24520" PostTypeId="1" CreationDate="2018-11-04T17:19:20.963" Score="1" ViewCount="61" Body="&lt;p&gt;In a district heating system, one of the houses in the distribution net is not getting the desired heat into their house. This is because the pipes of this customer which connects them to the distribution system, are smaller than required; which leads to sudden pressure drop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solutions I can see are either to replace this customer`s supply pipe with the bigger one, or setting up a booster pump system at the customer. If I were to decide on putting a pump system, where should I put the pump; on the customer supply pipe or return pipe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LvZei.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LvZei.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18164" LastActivityDate="2018-11-04T17:19:20.963" Title="Where to put the booster pump at customer substation in a district heating system?" Tags="|mechanical-engineering|fluid-mechanics|pumps|hydraulics|heating-systems|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="24521" PostTypeId="5" CreationDate="2018-11-04T18:20:39.490" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-11-04T18:20:39.490" LastActivityDate="2018-11-04T18:20:39.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24522" PostTypeId="4" CreationDate="2018-11-04T18:20:39.490" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2018-11-04T18:20:39.490" LastActivityDate="2018-11-04T18:20:39.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24523" PostTypeId="2" ParentId="24480" CreationDate="2018-11-04T21:17:01.253" Score="1" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deformation induced anisotropy is ubiquitous in materials, whether they are metals, polymers, foams, soils, rocks etc.  A typical example is any rolled metal product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to deal only with &lt;em&gt;purely elastic deformations&lt;/em&gt;, you have to define what elastic properties mean and how you will measure these properties.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that you are interested only in small-strain elasticity in a single crystal, perfectly annealed and free of defects.  If you deform the crystal, do the elastic properties of the crystal change depending on the direction of deformation? Keep in mind that no solid material is truly isotropic except for some glasses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the speed of propagation of a wave in a particular direction of the crystal is a proxy for the elastic properties of the crystal in that direction.  When you compress the crystal, the atomic spacing decreases and the wave takes a shorter time to travel, i.e., the elastic modulus increases in that direction and the crystal anisotropy increases. However, the induced anisotropy is small and not permanent.  See, for example, &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/0041624X63900039&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sciencedirect.com/science/article/pii/0041624X63900039&lt;/a&gt; and the &lt;a href=&quot;https://en.wikipedia.org/wiki/Acoustoelastic_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Acousto-elastic effect&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any permanent anisotropy requires an increase in entropy via plastic dissipation, microcrack growth etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For induced anisotropy in bulk metallic glasses, see &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S1359646211001072&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sciencedirect.com/science/article/pii/S1359646211001072&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="3488" LastEditDate="2018-11-05T21:19:13.730" LastActivityDate="2018-11-05T21:19:13.730" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24524" PostTypeId="2" ParentId="24491" CreationDate="2018-11-04T21:17:17.593" Score="1" Body="&lt;p&gt;As stated by @SolarMike, the rear wheel is only for stability. Because the front wheel is so large, the wheelbase (distance between points of contact on the ground) is large enough to provide ample front/back stability. Making the rear wheel larger, therefore would serve only to make the wheelbase unnecessarily long, making steering harder, and the bike physically longer; neither of these traits are desirable.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-11-04T21:17:17.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24525" PostTypeId="1" CreationDate="2018-11-04T21:53:42.183" Score="2" ViewCount="6961" Body="&lt;p&gt;I'm looking at designing something using the same mechanism used in this pepper grinder. As far as my knowledge goes, it is converting vertical Linear motion (the push button) into horizontal rotary motion (the ceramic grinder).&lt;a href=&quot;https://i.stack.imgur.com/UIi4e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UIi4e.jpg&quot; alt=&quot;.&quot;&gt;&lt;/a&gt; &#xA;I would like to understand how this is achieved, especially in such a small vessel.&#xA;I'm an amateur when it comes to this stuff, and was hoping someone here could explain it to me.&#xA;I appreciate all answers.&#xA;Thanks, Beau&lt;/p&gt;&#xA;" OwnerUserId="18168" LastActivityDate="2018-11-05T07:18:05.450" Title="Converting Linear Motion into rotary motion" Tags="|linear-motion|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24526" PostTypeId="1" AcceptedAnswerId="24536" CreationDate="2018-11-04T22:14:06.183" Score="0" ViewCount="175" Body="&lt;p&gt;I'm familiar with left hand threads on bicycle pedals. The action of pedaling for that side of the bike would un-thread a right-hand threaded pedal, although some people have hacked a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a chain saw that uses a left hand threaded axle to tighten the bar which tensions the chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below shows the cover over the bar, and also the legend embossed in the plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JjeCF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JjeCF.jpg&quot; alt=&quot;chain saw cover&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right hand &quot;wing nut&quot; is an ordinary thread and serves to secure the cover and the bar, once the tension is set with the left hand knob. Clockwise rotation for the locking nut works as expected. The left hand knob turns a gear in a pinion arrangement, clockwise pushes the bar away from the motor, as per the legend.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vz2XU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vz2XU.png&quot; alt=&quot;chain saw inside with text&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand knob has unrestricted rotation (within reason) and has gear teeth as part of the construction. The pinion gear resides under the knob, engaging the teeth and rotating the axle, which resides in the groove shown in the photo. Near the end of the axle, the threads push or pull the rectangular tab (pin) which in turn pushes the chain bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O3hQM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O3hQM.png&quot; alt=&quot;chain saw with additional parts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As can be seen in this photo, the end of the axle has a severe bend, preventing movement of the tab/pin, removing the saw from service.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, all the sources I've been able to find for the replacement part have indications of &quot;no longer available.&quot; Pricing indicates a cost in the US$3.00 range, but that's not helpful without product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a Harbor Freight mini-lathe (toy lathe) that can handle replication of this part, although I'd like to avoid the left-hand thread operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see no practical reason to not use a right hand thread. There are no rotational forces as in the previous pedal example. Once the bar is properly tensioned, the locking nut secures the bar, rendering the axle and pin immobile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can easily cut a piece of steel to replace the tab/pin, cut and tap the hole to match the rod which would be cut in the lathe, threaded with an appropriately sized die and mill a flat on the gear drive end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What technical/engineering aspects would prevent me from making this work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not consider the human aspect of a reversed legend on the knob to be a factor.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-11-05T12:17:59.337" Title="What is the justification for left-hand threads?" Tags="|threads|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24527" PostTypeId="2" ParentId="24526" CreationDate="2018-11-04T22:37:06.893" Score="0" Body="&lt;p&gt;I would think the only reason was to avoid having to have an idler gear to reverse the rotation, so it was cheaper to specify a left hand thread compared to providing an extra gear or being able to extend the drive to the left side of the knob.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-11-04T22:37:06.893" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24530" PostTypeId="2" ParentId="24525" CreationDate="2018-11-04T23:29:50.473" Score="2" Body="&lt;p&gt;This is a diagram for the mechanism, &lt;a href=&quot;https://i.stack.imgur.com/0QyAv.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0QyAv.gif&quot; alt=&quot;linear actuator&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is Wikipedia link, &lt;a href=&quot;https://en.wikipedia.org/wiki/Linear_actuator&quot; rel=&quot;nofollow noreferrer&quot;&gt;click&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-04T23:29:50.473" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24531" PostTypeId="1" AcceptedAnswerId="24551" CreationDate="2018-11-04T23:51:17.083" Score="0" ViewCount="205" Body="&lt;p&gt;I am trying to heat the oil in a 1/2 inch thick rectangular steel box by using an Insulated heated blanket covered on all 6 faces of the box, the working conditions of the operation are at -50 Celsius, the temperature of the oil is required to be at 15 Celcius during the operation, so I need to determine how much heat/power is required by the blanket and how much time will it take for the temperature of the oil to change from -50 Celsius to 15 Celsius?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one face resistive diagram,&#xA;Rb = resistance by the blanket thickness of 1 inch,&#xA;Rw = rectangular steel box wall&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p7MOx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p7MOx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18170" LastActivityDate="2018-11-08T14:06:33.413" Title="Conductive Heat Transfer calculation over time" Tags="|thermodynamics|heat-transfer|energy|thermal-conduction|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24534" PostTypeId="2" ParentId="24525" CreationDate="2018-11-05T07:18:05.450" Score="-1" Body="&lt;p&gt;Universal joint and threaded rods on each end of U Joint or a ring gear and pinion with threaded rods on each&lt;/p&gt;&#xA;" OwnerUserId="18174" LastActivityDate="2018-11-05T07:18:05.450" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24535" PostTypeId="1" CreationDate="2018-11-05T11:30:45.310" Score="1" ViewCount="188" Body="&lt;p&gt;A motor is used to drive a shaft that is geared through 1:1 bevel gears to a propeller. The shafts are mounted on bearings in a hard casing that is attached to a 2nd ground. I want to know if the 2nd ground needs to react any of the torque produced by the motor, to stop the whole casing rotating? If so, how much? Any help would be greatly appreciated. I have tried sketching the system below for clarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qwrTD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qwrTD.png&quot; alt=&quot;Propeller Torque System&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18177" LastEditorUserId="90" LastEditDate="2018-11-06T16:56:37.880" LastActivityDate="2023-12-12T01:08:21.013" Title="Reacting Propeller Torque?" Tags="|mechanical-engineering|gears|torque|mechanical|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24536" PostTypeId="2" ParentId="24526" CreationDate="2018-11-05T12:17:59.337" Score="0" Body="&lt;p&gt;This is a user experience/design consideration. Users are simply conditioned to expect that rotating clockwise causes a movement to the right, and anticlockwise to the left - this is how a car steering wheel works, or bike handlebars etc. etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mark of good design is where you are able to intuitively guess what the operation of an actuator does, before you use it. If there were to be a RH thread, then you would have to rotate the knob in the un-intuitive direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Solar Mike explained, once this constraint is understood, avoiding the LH thread would cause other undesirable mechanical features such as an idler, or extended shaft. The LH thread is the simplest of these options, so it's the one that was manufactured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could replace this component with a RH threaded variant, but make sure to mark the outer cover to make the new operation direction very, very clear, particularly if it's possible you'd lend this out to a friend!&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-11-05T12:17:59.337" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24541" PostTypeId="1" CreationDate="2018-11-05T16:09:09.293" Score="0" ViewCount="321" Body="&lt;p&gt;The Poisson ratio of the material equals 0,33. If I load a cube or a cylinder that is fixed at one end with compression force in ANSYS I can confirm that the ratio is 0,33 by dividing the lateral by longitudinal strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I load a geometry that is hollow with changing cross section the lateral/longitudinal ratios across the geometry change in unreasonable intervals that make no sense to me, because I was convinced that the ratio is always constant and does not depend on cross-section or shape. My question is, what does the ratio depend on and how? If somebody could direct me to some useful literature, it would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="17533" LastActivityDate="2024-01-12T05:03:00.327" Title="Longitudinal and transversal strain" Tags="|mechanical-engineering|materials|ansys|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24542" PostTypeId="1" AcceptedAnswerId="24557" CreationDate="2018-11-05T16:47:47.563" Score="1" ViewCount="87" Body="&lt;p&gt;ICE Engines are designed to burn the fuel and air mixtures at high pressures to increase the combustion efficiency. But hydrogen combustion efficiency is already high at 1 atm and at constant Volume combustion, peak pressure may reach around 8 atm. So why compressing the hydrogen and air mixture in Hydrogen ICE engines? If we could avoid compression of fuel and air, we can design much simpler and cheaper Hydrogen engines.&lt;/p&gt;&#xA;" OwnerUserId="16175" LastActivityDate="2018-11-06T23:48:01.077" Title="Why to compress the hydrogen + air mixture in Hydrogen ICE engines when combustion efficiency of hydrogen at 1 atm is already high?" Tags="|mechanical-engineering|automotive-engineering|combustion|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24544" PostTypeId="1" CreationDate="2018-11-05T17:36:52.117" Score="1" ViewCount="306" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rvyAM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IIVdS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a statics exam I had the question in the picture attacehd and I had some trouble understanding it. The question is so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Under the effect of torque Mo rotates a wheel (radius r) with constant velocity around point A. A brake jaw presses on the wheel (height c) . The brake jaw is connected to a tilted brake handle with inclination angle α . The break handle is articulated at point B and horizontally binned. On the other side the brake handle is supported by a sliding shoe against a vertical wall ( corfficient of friction µo) . The friction coefficient between the rotating wheel and the jaw is µ. The weight of the brake handle is neglected. The system in the above described state is in equlibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I translated the question myself so execuse me if the formulation was a little bit messy. I have also attached a photo of how the answer should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What i don't understand is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) why there is no reaction at the point B  (the movable bearing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) why is the anticipated direction of brake handle motion is upwards and how the rotation of the wheel causing this motion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) which component of the reaction of the wheel on the brake handle responsible for this motion ( normal or friction )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) why is the moment at point A  Mo=Rr, isn't the opposing moment by the friction force supposed to be less than Mo because the wheel is in motion and if the opposing moment is equal to Mo doesn't that mean the wheel is at rest&lt;/p&gt;&#xA;" OwnerUserId="18183" LastActivityDate="2018-11-05T17:36:52.117" Title="statics exam question about friction" Tags="|mechanical-engineering|statics|friction|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24545" PostTypeId="1" CreationDate="2018-11-05T17:41:22.053" Score="0" ViewCount="717" Body="&lt;p&gt;I need to size bumper beams for the crumple zone of a very lightweight vehicle.  Together with the driver, the vehicle will weigh 300kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sizing the beams for a single-vehicle collision against a fixed object is simple.  To prevent brain injury, I need to keep the average deceleration under 50 G, so the average crushing force must be no more than ~ 150 kN.  Dividing the kinetic energy by the force gives me the needed crush distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the vehicle is in a collision with another (heavier) vehicle, I am unclear about when and how the two crumple zones will collapse.  Looking at NHTSA crash test data, it seems the crumple zones in most passenger cars are designed for a deceleration of about 15 G against a rigid barrier.  For a 1,700-kg car, this would mean an average crushing force of 255 kN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my vehicle with beams designed for a crushing force of 150 kN crashes into a vehicle designed with a crushing force of 255 kN, what happens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Assuming my crush zone can absorb all the excess kinetic energy in the collision, what will the deceleration of my vehicle be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Can I rely on the crush zone of the other vehicle absorbing some of the energy before the force on my vehicle over a significant period of time rises above 150 kN?  If so, how much?&lt;/p&gt;&#xA;" OwnerUserId="14562" LastActivityDate="2022-12-16T01:17:00.213" Title="Sizing crumple zone for multi-vehicle collisions" Tags="|mechanical-engineering|structural-engineering|materials|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24547" PostTypeId="2" ParentId="24545" CreationDate="2018-11-05T18:49:36.740" Score="0" Body="&lt;p&gt;As per your statement if all the crash energy is absorbed in crash zone, and the two cars fuse together and move along direction of heavy car:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\alpha = (255-150)kN / (1700 +300) \ 105000 /(2000) \times 9.8 ~ 5m/s, \  \ or \ \ g/2  $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where alpha is the excessive acceleration above your 50G. So, total G = 50.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the big car's crashing zone is going to experience a 14.5 G deceleration.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-11-06T06:58:37.867" LastActivityDate="2018-11-06T06:58:37.867" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24548" PostTypeId="2" ParentId="24541" CreationDate="2018-11-05T20:26:36.770" Score="-1" Body="&lt;p&gt;For simplicity lets assume you have a cylinder of length 20 cm which is made up of two short 10 cm cylinders, both hollow and have a 1 cm diameter hole drilled out of them, one has the cross section surface are of 2 cm^2 and the other thinner one has  a cross section surface 1 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we apply P = 2kN to the top of the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the stress on the thick part will be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \sigma =  2\, \text{kN}/ 2\, \text{cm}^2 = \ 1\, \text{kN/cm}^2$$&lt;/span&gt;&#xA;but for the thinner part&#xA;&lt;span class=&quot;math-container&quot;&gt;$$  \sigma = 2\, \text{kN}/ 1\, \text{cm}^2 = 2\,\text{kN/cm}^2. $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the thin section vertically strains twice and so horizontally its strain is two times more than the thick part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same idea applies to random changes in cross section of a bar, and even more complication arises when these changes have secondary stress concentration and distort the surface of strain plane along the length of the test sample.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="3488" LastEditDate="2018-11-06T01:26:11.107" LastActivityDate="2018-11-06T01:26:11.107" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24549" PostTypeId="2" ParentId="24455" CreationDate="2018-11-06T01:45:37.050" Score="2" Body="&lt;p&gt;You say you've already identified the zero-force members, so I'll skip part (a) of the question. The structure then becomes (deleting all the zero-force members except for DE which is needed for structural stability):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MiIlP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MiIlP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I prefer to work from the cantilever and then move in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we know DE is zero-force, we therefore know &lt;span class=&quot;math-container&quot;&gt;$EG = 20\text{ kN (compression)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the only other member with a vertical component on &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$DG$&lt;/span&gt;, we can obtain that &lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;DG &amp;amp;= 20\times\dfrac{\sqrt{7^2+DG_v^2}}{DG_v} \\&#xA;\text{where }DG_v &amp;amp;= \dfrac{5}{3\times7}\times2\times7 = 3.33\text{ m} \\&#xA;\therefore DG &amp;amp;= 46.5\text{ kN (tension)} \\&#xA;\therefore GH &amp;amp;= 46.5\times\dfrac{7}{\sqrt{7^2+3.33^2}} = 42.0\text{ kN (compression)}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;, we have the applied load, &lt;span class=&quot;math-container&quot;&gt;$CD$&lt;/span&gt; (with horizontal and diagonal components), &lt;span class=&quot;math-container&quot;&gt;$DE$&lt;/span&gt; (which we know is zero-force), &lt;span class=&quot;math-container&quot;&gt;$DG$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$DH$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$CD$&lt;/span&gt; is the only one which can absorb &lt;span class=&quot;math-container&quot;&gt;$DG$&lt;/span&gt;'s horizontal component (which is equal to the result found for &lt;span class=&quot;math-container&quot;&gt;$GH$&lt;/span&gt;, so we can find that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;CD &amp;amp;= 42.0\times\dfrac{\sqrt{7^2+CD_v^2}}{7} \\&#xA;\text{where }CD_v &amp;amp;= 5 - 3.33 = 1.67\text{ m} \\&#xA;\therefore CD &amp;amp;= 43.2\text{ kN (tension)} \\&#xA;\therefore DH &amp;amp;= 43.2\times\dfrac{1.67}{\sqrt{7^2+1.67^2}} - 46.5\times\dfrac{3.33}{\sqrt{7^2+3.33^2}} + 40 = 50\text{ kN (compression)}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, looking at &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;, only &lt;span class=&quot;math-container&quot;&gt;$CH$&lt;/span&gt; can absorb &lt;span class=&quot;math-container&quot;&gt;$DH$&lt;/span&gt;'s vertical load, so by repeating the calculations above, we get that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;CH &amp;amp;= 86.0\text{ kN (tension)} \\&#xA;HI &amp;amp;= 112.0\text{ kN (compression)}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then looking at &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;, only &lt;span class=&quot;math-container&quot;&gt;$BC$&lt;/span&gt; can absorb &lt;span class=&quot;math-container&quot;&gt;$CD$&lt;/span&gt;'s and &lt;span class=&quot;math-container&quot;&gt;$CH$&lt;/span&gt;'s horizontal components, so we then calculate that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;BC &amp;amp;= 169.3\text{ kN (tension)} \\&#xA;CI &amp;amp;= 195.3\text{ kN (compression)}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously &lt;span class=&quot;math-container&quot;&gt;$BI = HI = 112.0\text{ kN (compression)}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$IL = CI = 195.3\text{ kN (compression)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then looking at &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, we use your previously calculated reactions to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;AB &amp;amp;= 84.8\text{ kN (tension)} \\&#xA;AL &amp;amp;= 10.1\text{ kN (compression)}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And likewise with &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; (or from &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;), we obtain the last remaining &lt;span class=&quot;math-container&quot;&gt;$BL = 23.8\text{ kN (tension)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gZqIi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gZqIi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-11-06T01:45:37.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24551" PostTypeId="2" ParentId="24531" CreationDate="2018-11-06T02:51:33.717" Score="1" Body="&lt;h1&gt;Foundations&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The governing equation for the heat flow to the tank is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;k\nabla^2 T + \rho \tilde{C}_p \frac{\partial T}{\partial t} = \hat{\dot{q}}_o&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where temperature, thermal conductivity, density, and specific heat refer to the oil. The heat flow is per unit volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;--&gt; Assume the oil is well-mixed. The internal gradients disappear, giving&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\rho \tilde{C}_p V \frac{d T}{d t} = \dot{q}_o&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The governing equation for the heat flow to the air is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ (T_h - T_a) = \dot{q}_a R_a $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the temperatures of the heater and air as well as the thermal resistance to the air. Allow that &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_o$&lt;/span&gt; is also defined by a thermal resistance equation as &lt;span class=&quot;math-container&quot;&gt;$(T_h - T) = \dot{q}_o R_o$&lt;/span&gt;. The total power taken is &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_T = \dot{q}_o + \dot{q}_a$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system has three unknowns, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_T$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Solutions&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;General&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Define &lt;span class=&quot;math-container&quot;&gt;$\Theta = (T_h - T) / (T_a - T_o) = (T_h - T) / \Delta T_{ao}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$T_o$&lt;/span&gt; is the initial temperature of the oil. At time &lt;span class=&quot;math-container&quot;&gt;$t = 0$&lt;/span&gt;, we can set a boundary condition &lt;span class=&quot;math-container&quot;&gt;$\Theta = 1$&lt;/span&gt;. This means, the heater as the same temperature as the air to start. The option to set &lt;span class=&quot;math-container&quot;&gt;$\Theta(0) = 0$&lt;/span&gt; (the heater the same as the oil) is more difficult to resolve. We can combine and rewrite the first governing equations as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;-\rho \tilde{C}_p V \frac{d \Theta}{d t} = \frac{\Theta}{R_o} \Rightarrow - \frac{d \Theta}{d \tau} = \Theta&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$\tau = t / \rho \tilde{C}_p V R_o$&lt;/span&gt;. We find the solution as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Theta = \exp(-\tau)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives an equation for the temperature difference between the heater and the oil as a function of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Write the thermal resistance equations as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dot{q}_o R_o = \Theta \Delta T_{ao}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\dot{q}_a R_a = - \Phi \Delta T_{ao}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$\Phi = (T_a - T_h) / \Delta T_{ao}$&lt;/span&gt;. The boundary condition is that &lt;span class=&quot;math-container&quot;&gt;$\Phi = 0$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$t = 0$&lt;/span&gt;. We end with one equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dot{q}_T = \left(\frac{\Theta}{R_o} - \frac{\Phi}{R_a}\right)\Delta T_{ao}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Solutions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Solve the equation for &lt;span class=&quot;math-container&quot;&gt;$\Theta = \exp(-\tau)$&lt;/span&gt; to obtain &lt;span class=&quot;math-container&quot;&gt;$T_h - T$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Constant Total Power&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Solve the equation for &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_T$&lt;/span&gt; to obtain &lt;span class=&quot;math-container&quot;&gt;$\Phi(\tau)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Phi(\tau) = \frac{R_a}{R_o}\exp(-\tau) - \frac{R_a \dot{q}_T}{\Delta T_{ao}} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve this for &lt;span class=&quot;math-container&quot;&gt;$T_h(\tau)$&lt;/span&gt; from &lt;span class=&quot;math-container&quot;&gt;$T_h(\tau) = T_a - \Phi(\tau) \Delta T_{ao}$&lt;/span&gt;. Plug the equation for &lt;span class=&quot;math-container&quot;&gt;$T_h(\tau)$&lt;/span&gt; into the equation &lt;span class=&quot;math-container&quot;&gt;$T = T_h(\tau) - \exp(-\tau)\Delta T_{ao}$&lt;/span&gt; to find the temperature change of the oil with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{T_a - T}{\Delta T_{ao}} = \left(1 + \frac{R_a}{R_o}\right)\exp(-\tau) - \frac{R_a \dot{q}_T}{\Delta T_{ao}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Constant Heater Temperature&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Solve the equation &lt;span class=&quot;math-container&quot;&gt;$\Theta(\tau)$&lt;/span&gt; directly for &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;. The value of &lt;span class=&quot;math-container&quot;&gt;$\Phi$&lt;/span&gt; is a constant. Plug both &lt;span class=&quot;math-container&quot;&gt;$\Theta$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Phi$&lt;/span&gt; into the equation for &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_T$&lt;/span&gt; to find &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_T$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Translation to Picture&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The translation to the picture occurs as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$R_o = R_{CL} + R_w + R_{B1}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$R_a = R_{CA} + R_{B2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$T = T_{Lubricant} \ \ \ T_h = T_s \ \ \ T_a = T_\infty$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Example&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Here is an example of a plot of &lt;span class=&quot;math-container&quot;&gt;$\frac{T_a - T}{\Delta T_{ao}}$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; generated using a python/Jupyter code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B5Q7O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B5Q7O.png&quot; alt=&quot;temperature profile plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Take the starting point that you regulate the temperature of the heater jacket to obtain a desired heating time. Allow the power to the heater to vary. The power will start high and fall off over time (as the oil heats). Estimate the initial power using values at &lt;span class=&quot;math-container&quot;&gt;$t = 0$&lt;/span&gt;. This will give you an easy estimate to the starting point for the level power that you need when time is a constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you start with the initial power and hold it constant, you will heat the oil faster. The temperature of the jacket will increase with time. Find the time when you have to cutoff or reduce the power to avoid burning out the heater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When time is critical, you will do better to insulate (increase &lt;span class=&quot;math-container&quot;&gt;$R_a/R_o$&lt;/span&gt;) than to increase power. Notice that &lt;span class=&quot;math-container&quot;&gt;$(1 + R_a/R_o)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_a$&lt;/span&gt; are multipliers, the former especially to an exponential of time, while power is just an additive term.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-11-08T14:06:33.413" LastActivityDate="2018-11-08T14:06:33.413" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24554" PostTypeId="1" CreationDate="2018-11-06T09:12:01.913" Score="1" ViewCount="307" Body="&lt;p&gt;Unlike vertical drills where the drill bit is weighted down by gravity, in horizontal drilling, the drill string is pushed into the rocks. How does the drill pipes not snap under the immense force, especially since the pipes can end up being curved?&lt;/p&gt;&#xA;" OwnerUserId="13433" LastEditorUserId="90" LastEditDate="2018-11-06T13:24:42.087" LastActivityDate="2018-11-06T17:42:42.740" Title="In rock drilling, how does the drill pipe in horizontal directional drilling withstand the immense force?" Tags="|drilling|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24557" PostTypeId="2" ParentId="24542" CreationDate="2018-11-06T12:16:40.753" Score="1" Body="&lt;p&gt;SolarMike's comment basically answers the question, but let me flesh it out a little.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take a typical engine, say it has a compression ratio of 12:1, and weighs 500 pounds, and has a displacement of 4 liters.  Now let's reduce the compression ratio to only 2:1. To get the same amount of power output, the displacement now how to be 24 liters. (can you see why? burning one-sixth as many fuel molecules per liter, so need six times more liters).  So now the cylinders and pistons have to be about six times bigger as well, so now the car/truck has to be a lot bigger overall to fit this giant engine in there. Also those bigger cylinders are going to weigh a lot more, potentially as much as the entire car/truck, which hurts the performance of the vehicle. Overall it just doesn't make any sense to use this kind of engine.  The power density is too low.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="9275" LastEditDate="2018-11-06T23:48:01.077" LastActivityDate="2018-11-06T23:48:01.077" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24558" PostTypeId="2" ParentId="24554" CreationDate="2018-11-06T13:43:00.577" Score="1" Body="&lt;p&gt;It's called engineering design. The drill steels are designed to withstand the forces required of them. This incorporates materials used, manufacturing techniques, the design of couplings, rigidity/flexibility of the drill steels &amp;amp; drilling method: rotary, percussion or a combination of the two, rotary-percussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long holes drilled in rock, whether vertical or horizontal, are usually drilled by rotary drill rigs. Depending of the reason for drilling a hole, the type of drill bit used for such holes will be either a diamond drill bit or something like a tri-cone bit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diamond drill bit is used when rock samples need to be recovered. Tri-cone bits, or similar, are used when a hole need to be established for other reasons, such as an oil well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With rotary drilling, the drill steel experiences torsional forces from the drill head and only needs sufficient compression force to keep the drill bit engaged with the rock face at the end of the hole. If there is too much compression the drill steel with bow which can lead to rubbing against the side of the hole and excessive wear of the drill string. With too little compression there will not be enough contact with the rock face and the advance rate will be low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If drill strings do not have a good method for removal of the drill chips the drill bit and string will jam in the hole. This is why percussion and rotary-percussion drilling is not used to drill long holes, particularly horizontal holes or vertical down holes.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-11-06T13:43:00.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24561" PostTypeId="1" CreationDate="2018-11-06T16:17:16.360" Score="4" ViewCount="513" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/EwG0H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Odd ting, the table surface echo/resonance sound when speaking, especially lower lower frequencies. It helped putting on a cloth, but it did not completely remove resonance. Putting a rug beneath did not affect. Wonder whether box for roof lights may act as sound box or if it is just room size and large surfaces that are the cause? Also, would putting sound absorbing mat on wall or roof help?&#xA;Greatful for all answers. &lt;/p&gt;&#xA;" OwnerUserId="18192" LastActivityDate="2024-03-11T15:03:20.337" Title="How to reduce resonance from livingroom table" Tags="|acoustics|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="24563" PostTypeId="2" ParentId="24535" CreationDate="2018-11-06T16:39:02.883" Score="0" Body="&lt;p&gt;Yes it has to react.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say your engine can generate a maximum of 150 Lb-Ft torque. And you attach to the ground or back of a boat by a bracket which has 2 bolts on 16 (1.33 Ft) inches apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your bracket has to be able to resist the: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;150 Lb- Ft/1.33 Ft = 112.78 Lb, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;pull out tension on one bolt and push in compression on the other one. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-06T16:39:02.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24564" PostTypeId="2" ParentId="24541" CreationDate="2018-11-06T16:53:00.727" Score="0" Body="&lt;p&gt;The Poisson ratio is defined for a sufficiently uniform material (i.e., one for which the continuum assumption reasonably holds). If your material is not uniform, then what you're obtaining is some type of &lt;em&gt;effective&lt;/em&gt; Poisson ratio, which could be just about anything and doesn't hold broad meaning. This is covered in most textbooks on elasticity or mechanics of materials.&lt;/p&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2018-11-06T16:53:00.727" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24566" PostTypeId="2" ParentId="24554" CreationDate="2018-11-06T17:42:42.740" Score="0" Body="&lt;p&gt;The drill is not really pushed in the horizontal section.  The drill pipe is not very unique, mostly high strength steel like 105 Ksi yield strength. The magic is the other equipment. Mud motors turn the drill bit , the drill string itself may not even rotate ( that reduces the fatigue loads) and the &quot;measure while drilling&quot; technology that tells where the bit is and where it is going. When I retired from an oil company , turbines were sometime used the turn the bit, apparently now it is all mud motors ( look it up in Wiki, etc.).The energy to cut the rock is supplied by the very high pressure mud pumped through the pipe. ... I was on the API committee that wrote the specs for drill pipe ; we pretty much wrote what the drillers wanted. The oil company hires a drilling contractor and he rents the drill pipe from a third party , so oil company metallurgists like me, seldom sees drill pipe problems. Drill pipe never has couplings , it has &quot;rotary shouldered connections&quot;, also called integral joints. A larger diameter section with internal threads ( box) is welded to the drill pipe , male threads (pin) are cut at the other end. Often the string is &quot;hard banded&quot; where tungsten carbide is welded in bands to protect the string from wear against the steel well casing and the formation. Not a panacea as it may cut through the steel casing. The oil company must calculate the strength and size of the casing for each section of the well . The driller must calculate the curvature to be within limits for the particular drill pipe; Of coarse the driller and oil company need to work together; It would be embarrassing to find the drillers rig was not able to lift the required casing string.  This is a brief over-view of about 10 semesters hours of petroleum engineering school .  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-11-06T17:42:42.740" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24567" PostTypeId="1" CreationDate="2018-11-06T18:05:39.053" Score="0" ViewCount="189" Body="&lt;p&gt;I'm working on an Instron 5969 tester, and I'm trying to measure the residual strain in materials within their elastic zone. I'm using the test profile, which allows me to stretch the material a specific amount, hold it for a moment, release the force, and record the change in length. The problem I'm having, is when the load is removed (slowly and continuously), the machine seems to panic. It starts shaking, and the recorded values jump all over the place. I've tried multiple ramp rates (the rate at which the load is removed) to no avail. Any ideas on what else I could try?&lt;/p&gt;&#xA;" OwnerUserId="18200" LastActivityDate="2021-08-08T23:00:20.053" Title="Instron Test Profile -- Failure when steadily dropping the load" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24568" PostTypeId="2" ParentId="21555" CreationDate="2018-11-07T00:30:16.793" Score="1" Body="&lt;p&gt;&lt;strong&gt;Marine aluminium is 100 times less prone to corrosion than steel.&lt;/strong&gt; source link : &lt;a href=&quot;https://www.aluminiumleader.com/application/transport/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.aluminiumleader.com/application/transport/&lt;/a&gt;&#xA;(Not sure about Marine Aluminium vs stainless steel though) For Ships, strength to weight is ratio is not a big deal. But corrosion resistance is. Imagine a ship which doesn't get corrode even after 100 years.&lt;/p&gt;&#xA;" OwnerUserId="16175" LastActivityDate="2018-11-07T00:30:16.793" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24569" PostTypeId="2" ParentId="16952" CreationDate="2018-11-07T01:16:48.940" Score="1" Body="&lt;p&gt;Theoretically Ramjet should start working from 0.5 Mach (this is the speed where compressibility of fluids becomes significant). But it won't be much efficient until ramjet reaches around Mach 3. Because RamJet compresses incoming air by slowing down the air speed. More the speed of incoming air more compression can be achieved. Around Mack 3, diffuser can achieve upto 25 atm pressure (depending upon altitude it may vary). Nevertheless ramjet are historically used in tip-jet helicopters. Here is the link -&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Hiller_YH-32_Hornet&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Hiller_YH-32_Hornet&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16175" LastActivityDate="2018-11-07T01:16:48.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24570" PostTypeId="2" ParentId="24561" CreationDate="2018-11-07T06:01:02.077" Score="0" Body="&lt;p&gt;acoustic resonance is not uncommon in small rooms without acoustic damping. as a rule of thumb: to have an audible damping effect, you will need to entirely cover &lt;em&gt;one whole wall&lt;/em&gt; with damping material. this is most easily done by putting in a carpet that covers the whole floor of the room. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-11-07T06:01:02.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24571" PostTypeId="1" CreationDate="2018-11-07T06:04:00.867" Score="1" ViewCount="55" Body="&lt;p&gt;when I run an operation that mills a face down on my CNC, it leaves this weird texture. It's hard to describe, so I attached an image. What is it caused by and how do I fix it? &lt;img src=&quot;https://i.stack.imgur.com/WcJt3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="8440" LastActivityDate="2018-11-07T13:03:38.720" Title="Weird texture on faces when using Shapeoko CNC" Tags="|cnc|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24573" PostTypeId="1" CreationDate="2018-11-07T11:02:10.263" Score="5" ViewCount="1845" Body="&lt;p&gt;I'm watching the progress of a new office building being built over the road (essentially a manual timelapse).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the building progressed was by the pouring of a concrete core in the centre of the building to the full height (~8 storeys), before a steel beam frame was constructed around it to the full width/length of the building, obviously making a frame for the outer walls and floors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the core is substantial (not just a pillar or something), it wouldn't seem to have any structural effect on the outer parts of the floors of the building (near the outer walls).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My lay-person's guess on the purpose of the core is as a kind of protected fire space (dry-risers? wet-risers?) containing the stairs and lifts/elevators, but if I knew the answer I wouldn't be asking here.&lt;/p&gt;&#xA;" OwnerUserId="18206" LastActivityDate="2018-11-07T12:50:10.097" Title="What's the concrete core for in a building?" Tags="|civil-engineering|building-design|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24574" PostTypeId="2" ParentId="24561" CreationDate="2018-11-07T12:43:06.430" Score="0" Body="&lt;p&gt;This is meant more as a comment but my reputation is too low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to address this.  Your description seems more like a flutter echo than a true resonance, so try hanging something round over the table --a diffuser.  Maybe a basket ball or a beach ball just too see if it works.  If it does, then anything &quot;complicated&quot; like a chandelier or a ceiling fan will knock that right out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of complicated things you can do; Google &quot;architectural clouds&quot; for entertainment :-)&lt;/p&gt;&#xA;" OwnerUserId="16229" LastActivityDate="2018-11-07T12:43:06.430" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24575" PostTypeId="2" ParentId="24573" CreationDate="2018-11-07T12:50:10.097" Score="3" Body="&lt;p&gt;The concrete core at the center of the building serves in several ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A structural hard center that is fire resistant and shields emergency utilities and elevators and emergency egress means like fire and smoke resistant corridors and stairs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the main lateral load bearing component of the structure, designed to resist earthquakes and wind loads. It is basically a big shear wall cage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that it is the stiffest and heaviest part of the structure gives balance and symmetry to the massing distribution of building which is highly effective for limiting lateral torsional wobbling of the building at a big earthquake. At the same time being at the center of the building puts emergency exit corridors and stairs  equally near to all occupants at each floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason they built the core completed fully first could be they used post-stressed vertical reinforcement and high strength concrete. Post-stress bars are long continuous vertical bars tightened at the very top of the building with special jacks and make the concrete core very strong and ductile for performing under lateral and vertical loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the most effective ways of designing an earthquake resistant building is to reduce the overall weigh of the floors and by using of thin post-stressed slabs as floors that have their tendons anchored at the central concrete core this is achieved. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-07T12:50:10.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24576" PostTypeId="2" ParentId="24571" CreationDate="2018-11-07T13:03:38.720" Score="4" Body="&lt;p&gt;No milling operation will give you a perfectly flat surface. The level of finish you can achieve will depend on the capabilities of the mill itself, the tooling you use and how the tool paths are set up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also when facing. Because you are cutting with a rotating face you will inevitably have one side of the tool moving int other direction of travel and the other side moving away from it which can cause slight variation in finish from one side of the cut to the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally finishing operations use higher cutting speeds and shallower depths of cut than roughing operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often milled parts need additional finishing processes if you want a very high level of cosmetic surface finish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case probably the best and easiest option is to rub it down with a sanding block.  &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-11-07T13:03:38.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24577" PostTypeId="1" CreationDate="2018-11-07T15:34:14.243" Score="0" ViewCount="64" Body="&lt;p&gt;I am trying to write code for the buckling of composite plates using the Ritz approximation but I can't understand something. According to Reddy (Mechanics of laminated composites), this is the equation of the principle of minimum potential energy (since we are concerned with buckling, q = 0 and the frequency omega = 0):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rB2z0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rB2z0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the Ritz approximation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BvCs6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BvCs6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after substituting 6.6.5 in 6.6.4 we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wIa4o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wIa4o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that delta w_0 has been substituted with X and Y but with indexes p and q, but it seems we do not sum over them. If that is the case, how do we go around solving this? Do we iterate over them but for some reason it is not written explicitly?&lt;/p&gt;&#xA;" OwnerUserId="17981" LastActivityDate="2018-11-07T15:34:14.243" Title="Buckling of composite plates help" Tags="|materials|composite|buckling|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24578" PostTypeId="2" ParentId="3627" CreationDate="2018-11-07T15:47:31.260" Score="2" Body="&lt;p&gt;For plenty of detail on building accurate machine tools from scratch, read the excellent and detailed &quot;Foundations of Mechanical Accuracy&quot;&#xA;By Wayne R. Moore. It's an old book and there are plenty of copies around if you use your favourite web search engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic principles for increasing accuracy are lapping and comparison of surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From memory some of Moore's key steps are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Flat lapping 3 surface plates against each other to form a master surface plate&lt;/li&gt;&#xA;&lt;li&gt;Copying straight edges from the master surface plate&lt;/li&gt;&#xA;&lt;li&gt;Centreless grinding to make the first cylinders&lt;/li&gt;&#xA;&lt;li&gt;Cylindrical lapping to refine cylinders&lt;/li&gt;&#xA;&lt;li&gt;Lapping accurate screw threads from rough initial screws&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18207" LastActivityDate="2018-11-07T15:47:31.260" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24579" PostTypeId="1" CreationDate="2018-11-07T15:51:26.663" Score="2" ViewCount="452" Body="&lt;p&gt;Was the second - the &lt;em&gt;big&lt;/em&gt; - explosion of the reactor core in the Chernobyl catastrophe a &lt;em&gt;true &lt;strong&gt;nuclear&lt;/em&gt;&lt;/strong&gt; explosion? There have been those who have said that it was ... but sayings to this effect do not gain much public traction. I think that explanations that invoke hydrogen production through the reaction of steam &amp;amp; zirconium are &lt;em&gt;grossly&lt;/em&gt; implausible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any answer to this will necessarily import notions as to what &lt;em&gt;constitutes&lt;/em&gt; a &lt;em&gt;true&lt;/em&gt; nuclear explosion, and I am not &lt;em&gt;stipulating&lt;/em&gt; any &lt;em&gt;particuIar&lt;/em&gt; definition of one according to which this question 'is to be' answered. It could perhaps be said that the definition is one of the 'degrees of freedom' of it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question really hinges, I think on whether the reactor was there, fatally damaged, and &lt;em&gt;quite grossly&lt;/em&gt; overheating, with steam building up to high pressure, with free oxygen &amp;amp; hydrogen present, which at some point ignited explosively; or in a state in which essentially the same 'race' between escalation-of-nuclear-reaction &amp;amp; tendency to dissemble of the &lt;em&gt;very reacting substance&lt;/em&gt; of the core was occuring as occurs in a nuclear bomb ... but obviously terminated very much sooner by reason of the very much lesser &lt;em&gt;compaction&lt;/em&gt; of the core of a nuclear reactor compared to that of a nuclear bomb. Which latter compaction is in a bomb obviously contrived to be as extreme as is possible to be made with extant engineering resources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's quite reasonable also, I think, to consider whether the former of the two scenarios just adduced is 'worthy' of being deemed a &lt;em&gt;true nuclear explosion&lt;/em&gt; as opposed to the latter &lt;em&gt;not&lt;/em&gt; being. Certainly it is more &lt;em&gt;similar&lt;/em&gt; than the latter to what occurs in a nuclear bomb, and is only really less in force than it by reason of the lesser degree of compaction of the reacting substance; whereas in the latter scenarion the nuclear reaction is indeed a copious source of heat, but the explosion is not one &lt;em&gt;of&lt;/em&gt; the reacting substances &lt;em&gt;per se&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there &lt;em&gt;is&lt;/em&gt; some concern amongst those who are appointed to search into these matters as to which of these two kinds of event the large explosion of the Chernobyl nuclear reactor was more nearly an instance of. I think the idea that it might have been more of a 'true nuclear explosion' according to the definition I have proposed is possibly one that has been somewhat &lt;em&gt;suppressed&lt;/em&gt; ... not necessarily by force - I am not advancing a 'willful cover-up' theory - but more just through &lt;em&gt;habits&lt;/em&gt; of public discussion having become settled in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So another way of framing the question might be this: I have seen, in the course of my 'trawling' for matter about the Chernobyl incident, matter in which the concern is &lt;em&gt;with this very question&lt;/em&gt;; but not a &lt;em&gt;very&lt;/em&gt; great deal &amp;amp; quite fragmentary; and I wonder whether anyone has seen anything really substantial about it; or what &lt;em&gt;original&lt;/em&gt;  thoughts there might be out there about it amongst folks who care about these kinds of thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, I recall reading recently, in the course of the said 'trawling', that the last reading  on the meter showing the heat output of the reactor core was &lt;em&gt;33GW&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Simple Mathematical Model for Exponential Growth with Two Widely Different Timescales.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not meant to be taken as any kind of &lt;em&gt;actual engineering formula&lt;/em&gt;, however; it's merely a simple mathematical model for showing how it is &lt;em&gt;thoroughly&lt;/em&gt; intrinsic to a positive feedback-system having two different timescales to flip precipitately from one to the other as a certain parameter passes from one range to another. It is expounded, &amp;amp; its relevance examined, more thoroughly, in the &lt;em&gt;mathematics&lt;/em&gt; section. It shows just how imperative it is that the &lt;em&gt;proportion excess&lt;/em&gt; of reactivity in the core be kept to substantially less than the delayed neutron fraction. It is my understanding that one of the reasons for there not being many &lt;em&gt;plutonium&lt;/em&gt; reactors is that the &lt;em&gt;delayed neutron fraction&lt;/em&gt; is significantly smaller for plutonium than it is for uranium, making this task much more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An expression giving the factor - say &lt;span class=&quot;math-container&quot;&gt;$1+\delta$&lt;/span&gt; by which the activity of a reactor core increases over the mean duration of a neutron's being free in the core, according to a simple model, would be - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;let &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; be the &lt;em&gt;delayed neutron fraction&lt;/em&gt;;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let &lt;span class=&quot;math-container&quot;&gt;$1+\omega\alpha$&lt;/span&gt; be the mean number of neutrons yielt by a neutron upon its reacting;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt; be the &lt;em&gt;ratio&lt;/em&gt; of the delay of the neutrons to the mean duration of  a neutron's being free in the reactor core;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let &lt;span class=&quot;math-container&quot;&gt;$w()$&lt;/span&gt; be &lt;span class=&quot;math-container&quot;&gt;$\operatorname{arcf}()$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\operatorname{f}(x)=x.e^x$&lt;/span&gt; -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta = \frac{w(\zeta\alpha.\exp((1-\omega)\zeta\alpha))}{\zeta} - (1-\omega)\alpha .$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio &lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt; is a large number - a few thousand. So when &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; is substantially less that 1, as it ought to be in proper operation, the argument to the &lt;em&gt;lambert w&lt;/em&gt; function is large; and the approximation &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$w(z)≈\ln(z) - (1-\frac{1}{\ln(z)})\ln(\ln(z))$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is valid, &amp;amp; the terms &lt;em&gt;without&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt; in the denominator vanish, leaving only (terms beyond &lt;span class=&quot;math-container&quot;&gt;$\ln(z)$&lt;/span&gt; in the expansion of w(z))÷&lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;However&lt;/em&gt;, because the &lt;em&gt;lambert w&lt;/em&gt; function &lt;em&gt;tends to linearity&lt;/em&gt; as its argument becomes &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;&amp;lt;1$&lt;/span&gt;, as soon as &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; begins to &lt;em&gt;exceed&lt;/em&gt; unity, the term &lt;em&gt;without&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt; in the denominator begins precipitately to preponderate. &lt;/p&gt;&#xA;" OwnerUserId="18209" LastEditorUserId="18209" LastEditDate="2018-11-20T06:07:39.883" LastActivityDate="2023-04-16T07:04:08.613" Title="Was the explosion of the Chernobyl nuclear reactor a 'true' nuclear explosion?" Tags="|nuclear-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24580" PostTypeId="1" CreationDate="2018-11-07T16:15:29.327" Score="2" ViewCount="80" Body="&lt;p&gt;So I live in Switzerland and there is lots of funiculars in my country. Most of those were built between 1890 and 1930, and for most of those the line is on the ground most of the time, there is bridges only where necessary, just like a regular train line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Moléson - Plan-Francey funicular is very particular in that not only it was built much more recently (in the late 1990s), but also the line is basically a huge bridge, the funicular never stands on the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is very weird and should have made the line much more expensive to build for no reason at all. Why is this line built on a giant bridge rather than on the ground with bridges only where necessary? Bridges would have made sense in order to make the line straighter (and thus shorter/faster), but the line is not even straigt anywhere !&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Video of the funicular &lt;a href=&quot;https://www.youtube.com/watch?v=2w-9YD-OS9w&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Detailed photos and technical explainations about the funicular &lt;a href=&quot;https://www.remontees-mecaniques.net/bdd/reportage-funi-moleson-village-plan-francey-von-roll-1044.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3yLY3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3yLY3.jpg&quot; alt=&quot;funicular picture from www.remontees-mecaniques.net&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17069" LastEditorUserId="17069" LastEditDate="2018-11-07T19:19:46.213" LastActivityDate="2018-11-07T19:19:46.213" Title="Is there a particular reason the Moléson funicular is built as a giant bridge?" Tags="|bridges|rail|transportation|" AnswerCount="0" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="24582" PostTypeId="1" CreationDate="2018-11-07T19:35:28.293" Score="0" ViewCount="107" Body="&lt;p&gt;I'm working on a PID-controlled barn door tracker mount for astrophotography in an isosceles configuration:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KpOXz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KpOXz.png&quot; alt=&quot;Isosceles configuration barn door tracker&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those not familiar, the purpose of the barn door tracker is to compensate for the rotation of the earth during astrophotography. The axis of rotation of the tracker is aligned with Polaris, and the upper arm of the tracker rotates about that axis once in a sidereal day (approximately once per 24 hours). So the movement is very slow, but not unnoticeable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of people use a curved bolt configuration to make it so you can simply set the motor at a constant rate of rotation and get a constant angular rate, but I don't like the inability to address sources of error with that configuration. Instead, I would like to use PID control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My motivation is both for a fun project, but also to create something that is more accurate than a curved-bolt configuration and cheaper than entry-level equatorial mounts. Ignoring the cost of microcontrollers, I'm aiming to have the entire setup cost less than $120. If I can't make it for that cheap, then I might as well get a professionally-made mount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been looking at different methods of getting angular feedback, and I'm most interested in two: encoder and accelerometer. An encoder would be fairly easy to implement in code and resolution could be influenced by hardware. However, a quadrature encoder would use up almost half of my budget, and I'm skeptical of using just an incremental encoder, in case my PID tuning is off. Not to mention I would have to implement an additional way to home the tracker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After playing around with an app that lets me see the sensor outputs on my smartphone, I wondered whether an accelerometer might be a cheap alternative. There are plenty of sensors that are USB or Arduino-compatible out of the box. My biggest concern, however, is that of accuracy. The angular change is only about half a degree per minute, but in the isosceles configuration, the rate of the motor must change accurately to maintain a constant angular speed. I need something that will be able to capture that slow of a change accurately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not a controls or robotics guy, so for all I know I'm completely missing some other component that might fit the bill. Any thoughts on the advantages or disadvantages of the sensors I've mentioned? Or places I might look for more affordable types of sensors?&lt;/p&gt;&#xA;" OwnerUserId="16304" LastActivityDate="2018-11-07T22:19:38.513" Title="Cheap and accurate angle measurement for PID feedback control" Tags="|positions-measurement|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24583" PostTypeId="1" AcceptedAnswerId="24602" CreationDate="2018-11-07T21:19:42.567" Score="1" ViewCount="186" Body="&lt;p&gt;I have been searching for historical structures in the province of Van, Turkey, by examining the publicly available satellite imagery like Bing Maps. There is a historic bridge near Kaçit, Çatak. Its name is Zeril Bridge. I tried to figure out the places which are connected by the bridge. The two sides of the stream seem to be very steep and it seems to me that the bridge is situated in a strange location.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the viewpoint of civil engineering, if you build this bridge today, what would your purpose for locating the bridge at that point be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Coordinates of the Zeril Bridge: 37.915422 , 42.984053&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5SzUj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5SzUj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18213" LastActivityDate="2018-11-09T18:00:29.043" Title="From the viewpoint of a civil engineer, what was the purpose of this bridge?" Tags="|civil-engineering|bridges|architecture|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24584" PostTypeId="2" ParentId="24582" CreationDate="2018-11-07T22:19:38.513" Score="1" Body="&lt;p&gt;Since this application doesn't require continuous rotation, rather, it shuttles between limits, you can use a rotary potentiometer to provide angular feedback. Most 'standard' Pots have 270 degrees of rotation, so you would need to add some gearing - I'd recommend pulleys for this since the forces involved are very low, and you don't want backlash. Then simply set up the Pot as a divider, and use the analogue read pins to measure the voltage, and hence the rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a watch of this video for an example of this theory being implemented: &lt;a href=&quot;https://www.youtube.com/watch?v=3pYWLF8qw-g&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=3pYWLF8qw-g&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-11-07T22:19:38.513" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24585" PostTypeId="2" ParentId="15575" CreationDate="2018-11-07T22:47:59.293" Score="-2" Body="&lt;p&gt;The web have almost all the resistance to shear. I recommend to review the basic texts of mechanics of materials, such as that of James M Gere.&lt;/p&gt;&#xA;" OwnerUserId="18215" LastActivityDate="2018-11-07T22:47:59.293" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24586" PostTypeId="1" CreationDate="2018-11-08T01:08:42.680" Score="1" ViewCount="98" Body="&lt;p&gt;I have been thinking of a problem and I would like to brainstorm about it with the community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scenario:&#xA;Machined parts going through termal treatment in an oven and further abrasive treatments. Temperatures of 900 celsius in the oven. Minimal space for marking. Laser do not work. RFID neither. 10mm marking space possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can one track these parts? Any kind of paint? High temperature magnetic memory?&lt;/p&gt;&#xA;" OwnerUserId="16663" LastActivityDate="2018-11-08T15:17:11.633" Title="How to track parts through high-temperature oven and abrasive treatments?" Tags="|materials|machining|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24588" PostTypeId="2" ParentId="22377" CreationDate="2018-11-08T03:42:09.600" Score="0" Body="&lt;p&gt;If you dont use adhesives then you'll need some kind of pressure to keep things airtight.&#xA;Problem is PTFE is soft and slowly deforms (&quot;cold creep&quot;) over time , vibrations etc... Making it leak prone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to redesign and/or use a gasket made from elastic material(PTFE isnt elastic) between the PTFE and the steel&lt;/p&gt;&#xA;" OwnerUserId="5926" LastActivityDate="2018-11-08T03:42:09.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24589" PostTypeId="2" ParentId="24586" CreationDate="2018-11-08T13:58:21.133" Score="1" Body="&lt;p&gt;When they go into an oven they are on a particular sport on a rack. When they come out they will still be in the same spot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is your tracking method through the oven: track the rack the item is on and where it is on the rack. This also lets you collect data on which spots on the rack don't do as well as others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However most bulk abrasive treatments for small machined parts is putting a bunch of parts into a container along with abrasive elements and then the container is agitated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the part is larger then you would be using additional machining to run sanding/grinding disks along it (or hand sanding). Again here you can track the position of the part instead of the part itself. Up until you can add the serial number.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2018-11-08T15:17:11.633" LastActivityDate="2018-11-08T15:17:11.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24590" PostTypeId="1" CreationDate="2018-11-08T15:41:11.743" Score="1" ViewCount="48" Body="&lt;p&gt;I am currently designing a mechanical tensile testing setup to operate at different temperatures (-20 to 100 C) in a wet environment. I plan on machining two sets, one from stainless and one from PMMA. Of course, this setup will have a clamp setup similar to the below picture.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/re5R5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/re5R5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I tolerance the drawing prior to machining, I would probably want to tolerance both of the threaded shafts and all four of the smooth bolts for a sliding fit. How can I (or should I even try to) assess the need to consider the various temperatures when tolerancing these parts?&lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2018-11-08T15:41:11.743" Title="Tolerancing considerations for temperature rated mechanical assemblies" Tags="|design|machining|cad|tolerance|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24591" PostTypeId="2" ParentId="24583" CreationDate="2018-11-08T19:15:18.573" Score="0" Body="&lt;p&gt;It could be used for cattle access to grazing meadows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or may be they where moving some kind of top soil for farming, or gypsum extraction, or mining as has been mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or a mix of these and some illegal activities such as contraband trafficking with mules.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-08T19:15:18.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24592" PostTypeId="2" ParentId="21738" CreationDate="2018-11-09T01:44:54.210" Score="0" Body="&lt;p&gt;In the end, I decided to use a bottle filled with glass beads.  This provided the necessary thermal stabilization and it has the added benefit of working below -50C. &lt;/p&gt;&#xA;" OwnerUserId="16072" LastActivityDate="2018-11-09T01:44:54.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24593" PostTypeId="1" AcceptedAnswerId="24855" CreationDate="2018-11-09T02:46:40.663" Score="3" ViewCount="234" Body="&lt;p&gt;I find it difficult to understand the concept of flux and field.I have searched the internet and refered a few books.From what I understand, generally flux can be defined as &quot;some quantity per unit of some other quantity&quot; E.g: &quot;the force of attraction at a point in space per unit charge placed at that point is called electric flux&quot; , &quot;the rate of heat transfer per unit area is called heat flux in conduction&quot;. &#xA;Please suggest whether I am right or wrong.&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastActivityDate="2018-11-25T16:18:25.943" Title="The concept of flux" Tags="|electrical-engineering|heat-transfer|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24594" PostTypeId="1" AcceptedAnswerId="24595" CreationDate="2018-11-09T03:41:45.270" Score="2" ViewCount="93" Body="&lt;p&gt;I am looking for answers to the following questions concerning FreeCAD from someone who is familiar with implementation of FreeCAD :&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Which library FreeCAD uses to parse STEP file format?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Which library FreeCAD uses to convert file of STEP format to Obj format?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is it possible (or how hard is it) to use the above mentioned libraries as standalone libraries in other projects. I have checked some options, but they are all proprietary software.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If one has many files of STEP format that need to be converted to Obj format, is it possible to write a script to call FreeCAD to do it in a batch. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18226" LastActivityDate="2018-11-09T05:17:03.417" Title="The underlying library FreeCAD uses to manipulate STEP file format and format conversion" Tags="|cad|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24595" PostTypeId="2" ParentId="24594" CreationDate="2018-11-09T04:52:48.120" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;Open Cascade.&lt;/li&gt;&#xA;&lt;li&gt;Open Cascade provides triangulation. Parsing step is pretty easy, hell the  standard documents  a parser. But having a triangulation and underlying  data manipulation package makes the difference.  This is why its  called a CAD kernel.&lt;/li&gt;&#xA;&lt;li&gt;Yes. Its nontrivial to use cascade directly. That said it does include examples.&lt;/li&gt;&#xA;&lt;li&gt;Yes.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-11-09T05:17:03.417" LastActivityDate="2018-11-09T05:17:03.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24596" PostTypeId="1" CreationDate="2018-11-09T05:38:22.557" Score="0" ViewCount="1255" Body="&lt;p&gt;Can someone explain about the mechanism in this folk toy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6GStW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6GStW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Video here:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=uMqdVnP7O0A&quot; rel=&quot;nofollow noreferrer&quot;&gt;A &quot;penny toy&quot; of around 1900.&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has a spring and it converts linear motion in to rotational motion ( that drives the acrobat). Further the rotational motion is possible both anticlockwise and clockwise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What exactly is going ne? I believe the mechanism is not  uncommon and someone from this background could throw some light.&lt;br&gt;&#xA;Thanks in advance for your help.   &lt;/p&gt;&#xA;" OwnerUserId="13540" LastEditorUserId="288" LastEditDate="2018-11-09T12:20:28.670" LastActivityDate="2018-11-14T10:18:25.137" Title="How linear motion is getting converted into rotational motion in this early toy?" Tags="|mechanisms|springs|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24599" PostTypeId="2" ParentId="24596" CreationDate="2018-11-09T10:38:28.087" Score="2" Body="&lt;p&gt;&quot;The penny toy works by friction on the pin against a piece of rubber tubing.&quot; The quote is from the audio introducing the toy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the main cylinder moves in and out, likely with a compression spring at the top of the main cylinder, the pivot of the toy character, which is another cylinder, presses against the main cylinder and is forced to rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the equivalent of a rack and pinion without gear teeth, but using friction to operate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A video explaining the gear-based rack and pinion can be found on &lt;a href=&quot;https://www.youtube.com/watch?v=6R-sESemjcg&quot; rel=&quot;nofollow noreferrer&quot;&gt;The YouTube&lt;/a&gt; with the distinction that the toy uses rubber in place of the teeth of the gears. I believe that the rubber portion could be on either the linear rod or on the rotating portion, or both.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2018-11-14T10:18:25.137" LastActivityDate="2018-11-14T10:18:25.137" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24600" PostTypeId="2" ParentId="24593" CreationDate="2018-11-09T14:37:01.207" Score="3" Body="&lt;p&gt;The two disciplines use the term flux in different ways.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Electricity&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Field is the distribution of a potential (NOT potential energy) along a line. It has units V/m. It is alternatively the distribution of a force per unit charge. It has units J / C m = N m / C m = N / C. Multiply this latter by charge and you see how you obtain the force on a charge at any point in an electric field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flux is the net amount of the electrical field that is penetrating perpendicularly to a unit area. It has the units V m = N m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; / C.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Mass, Momentum, or Heat Transfer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Flux has units that can be considered from one of two perspectives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flux can be written as the amount of mass, momentum, or heat moving through a certain area in a certain time. The units are (amount / m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; s). So, specifically for each, you find&lt;/p&gt;&#xA;&#xA;&lt;p&gt;mass: kg / m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; s ...  momentum: (kg/m s) / m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; s ...  heat: J / m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, flux can be written as the concentration of mass, momentum, or heat moving at a certain velocity. The units are (amount / m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;)(m / s). The translations follow from above.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-11-09T15:28:48.993" LastActivityDate="2018-11-09T15:28:48.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24601" PostTypeId="1" AcceptedAnswerId="24603" CreationDate="2018-11-09T17:30:28.440" Score="3" ViewCount="5062" Body="&lt;p&gt;I know that a hydronic system forms a closed loop. When the boiler is turned on, the water temperature will increase and expand. Since this system forms a closed loop, there is no volume for the hot water to expand. Therefore, the pressure inside the system will increase and explosion might occur if the pressure increases significantly. To maintain a constant pressure, an expansion vessel is installed to contain the increase in the water volume as it expands as it heats up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hcdnw.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expansion vessel is divided into two compartments separated by a diaphragm. There is water on one side and air on the other side. Initially the air pressure equals the system pressure. When the water heats up its pressure increases. Eventually, the pressure exceeds that of the air causing the diaphragm to bend back in the air space, compressing the air. Since the water is allowed to expand, this means the system pressure is maintained constant. On the other hand, the air pressure has been increased because its volume decreases. Now, more increase in the water temperature will increase its pressure. The hot water will not be able to push the diaphragm more unless the pressure water becomes greater than the air pressure, which means the system pressure is not maintained constant. Am I wrong? &lt;/p&gt;&#xA;" OwnerUserId="9742" LastEditorUserId="1832" LastEditDate="2018-11-09T19:50:29.603" LastActivityDate="2020-04-05T04:13:53.360" Title="How does an expansion tank work?" Tags="|mechanical-engineering|fluid-mechanics|hvac|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24602" PostTypeId="2" ParentId="24583" CreationDate="2018-11-09T18:00:29.043" Score="3" Body="&lt;p&gt;The location makes sense from a civil engineering perspective.  The bridge has to span the river channel without being washed away during flooding events.  A narrow, steep canyon works well for this because the bridge can be built high enough over the water without it having to be too long.  The height gives it protection from floods, the steepness of the canyon walls keeps the span as short as possible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the steepness of the canyon walls gives an indication of the strength of the supporting bedrock for the abutments.  The steeper the canyon sides, the more an indication of strong, hard rock.  If the side slopes were flatter, this would indicate looser, more erodible soils under the ends of the bridge (the abutments).  Obviously, stronger, harder rock is superior to easily eroded soils for the purposes of holding up the ends of the bridge.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the bridge is built during low water times of the year, access to the site isn't too difficult.  The bridge is close enough to the main river channel to be convenient enough for travelers while still taking advantage of the favorable characteristics listed above.  &lt;/p&gt;&#xA;" OwnerUserId="9637" LastActivityDate="2018-11-09T18:00:29.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24603" PostTypeId="2" ParentId="24601" CreationDate="2018-11-09T18:06:13.697" Score="6" Body="&lt;p&gt;It is better to think of it in terms of volume than pressure. The key point is that air is compressible while water is not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you heat water is has to expand. If it is in a closed container then it can only do this by straining the container, this creates very large stresses which are the cause of the pressure increase. To put another way the pressure generated is a function of the stiffness of the container. In comparison the pressure generated by compressing air by a small proportion of its starting point is pretty small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So its not so much about keeping the pressure in the system constant as giving the water somewhere to go as it expands which doesn't involve stretching the solid parts of the system. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="90" LastEditDate="2018-11-10T16:03:55.163" LastActivityDate="2018-11-10T16:03:55.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24604" PostTypeId="2" ParentId="24601" CreationDate="2018-11-09T19:53:59.903" Score="0" Body="&lt;p&gt;For boilers, you wouldn't use an expansion tank.  You would use a steam drum.  An expansion tank is for pumps that pulse (especially piston pumps).  The difference is in the design:&lt;/p&gt;&#xA;&lt;h2&gt;Steam Drum&lt;/h2&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GiUjs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GiUjs.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Steam_drum#/media/File:Water_tube_boiler_schematic.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;)&lt;/p&gt;&#xA;&lt;p&gt;The liquid-steam interface is where expansion typically occurs.  The water is given enough thermal energy to expand up the water tubes.  However, the larger diameter down comer has different thermal heat transfer properties, and as such the water sinks inside of this tube.&lt;/p&gt;&#xA;&lt;p&gt;Any expansion then is driven towards the condenser due to this expansion.  The density differential between the heated steam and the cooled liquid is enough to overcome the pressure losses in the system and drive the steam forward.&lt;/p&gt;&#xA;&lt;h2&gt;Expansion Tank - Surge Suppressor - Pulsation Dampener&lt;/h2&gt;&#xA;&lt;p&gt;An expansion tank works how you describe, but they are mainly for pumps, such as diaphragm or piston pumps.  &lt;a href=&quot;https://i.stack.imgur.com/e7fYh.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e7fYh.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt; (&lt;a href=&quot;https://plastomatic.com/technical-article/introduction-to-pulsation-dampeners-surge-suppressors/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;The idea is to allow the pressure of the pump to stabilize against the compressed air, rather than deal with the jerky motion of the single piston compressing the liquid.  They work sort of like a pass filter in an electrical cicuit - an extra large capacitor is not needed, because the extra surge pressure (or extra electrons) are depleted when the pump cavity is refilling (or when the electrical circuit is in the negative part of the wave).&lt;/p&gt;&#xA;" OwnerUserId="2015" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-11-09T19:53:59.903" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24605" PostTypeId="1" AcceptedAnswerId="25164" CreationDate="2018-11-09T20:48:29.433" Score="1" ViewCount="81" Body="&lt;p&gt;For my 4th year dissertation in Civil Engineering I've got to investigate the difference between BS5950, Eurocode 3 and ABAQUS failures. The only thing is, in the BS5950 and Eurocode 3 the failure can only include Major OR Minor axis which is what I need. However, in ABAQUS, the software itself 'decides' in which way it will fail. It starts with Major axis but at the end it takes into account the Minor axis too which gives me an error in comparing these results with the BS and Eurocode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, is there any way you can tell ABAQUS that the only way the failure can occur is in Major axis? without taking into account the Minor axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using BBC Basic to create the ABAQUS input files and then run them through the Command. I have already tried to put 'boundary conditions' (a maximum length it can move along) but it gave me a too high of an error between with and without the boundary conditions so I can't do that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone ever do that kind of investigation? Any advice/ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="18233" LastActivityDate="2018-12-19T01:18:28.860" Title="Abaqus Failure for I-Beam Section in Major direction only" Tags="|structural-engineering|civil-engineering|beam|abaqus|software|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24606" PostTypeId="1" CreationDate="2018-11-09T21:44:11.383" Score="0" ViewCount="60" Body="&lt;p&gt;I got a question on an online quiz which leaves me clueless.&#xA;I gave my best to translate the question as accurately as possible:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a strain gauge placed upon a tension rod with a rectangular cross section in transverse direction. Which Value does it show at the onset of yielding?&#xA;The steel used is mild steel S235 with an E-Module of &lt;span class=&quot;math-container&quot;&gt;$E = 210000$&lt;/span&gt; MPa, &lt;span class=&quot;math-container&quot;&gt;$µ = 0.3$&lt;/span&gt;, yield stress &lt;span class=&quot;math-container&quot;&gt;$R_e = 235$&lt;/span&gt; MPa and UTS &lt;span class=&quot;math-container&quot;&gt;$R_m = 490$&lt;/span&gt; MPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimension of the Tension Rod: 100mm x 13,4mm x 7,4mm (Length, Width, Depth)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/gRhBH.png&quot; alt=&quot;Question&quot; title=&quot;Question&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="18236" LastEditorDisplayName="user14407" LastEditDate="2018-11-09T22:06:16.450" LastActivityDate="2018-11-09T22:06:16.450" Title="Calculating Strain gauge Values in a tensile Test" Tags="|mechanical-engineering|mechanical|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24607" PostTypeId="1" CreationDate="2018-11-09T22:15:31.617" Score="1" ViewCount="34" Body="&lt;p&gt;&lt;em&gt;First of all, I have taken a look at the Grinter's method, which is similar to Cross, but balancing angles instead of moments. My question assumes that we are using the Cross method, and &lt;strong&gt;not&lt;/strong&gt; the Grinter's method.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The literature explaining the moment distribution (i.e: Cross) method for statically indeterminate structures tends to affirm that the designer needs the moments rather than the rotation angles, and therefore all books and references I checked don't pay any attention at all at getting the rotation angles. I think that's a false affirmation, because the designer needs to check both ULS and SLS requirements (with SLS being quite often more limiting than ULS), and you need to know the rotation angles in order to find the maximum deflection and be able to check it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a bit surprising, because the &quot;we just need the moments&quot; affirmation seems to suggest that engineers didn't check the maximum deflection in structures designed with the Cross method, which allegedly was broadly used from the 1930s to the 1960s. I find it hard to believe that they didn't check the maximum deflection in their structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering what would be the most efficient way of finding the joints rotation angles after you apply the Cross method. Do you really need to apply the Grinter's method instead of Cross? Is it not possible to get the rotations starting from just the Cross solution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only solution that came to my mind is to integrate the moments diagram twice, but the conditions for solving the integration constants would depend on the structure topology, and I wouldn't call it &quot;efficient&quot;...&lt;/p&gt;&#xA;" OwnerUserId="12165" LastActivityDate="2018-11-09T22:15:31.617" Title="Most efficient way for finding the rotation angles after moment distribution (Cross method)?" Tags="|structural-analysis|moments|deflection|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24608" PostTypeId="1" AcceptedAnswerId="33250" CreationDate="2018-11-09T22:51:54.787" Score="0" ViewCount="147" Body="&lt;p&gt;I have a mass - spring - damper system with external force and I am trying to simulate it using Matlab. I want to have a linearly parameterized form and use the least squares method to find the estimators. I have reached the stage where I have the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ y = \theta^T  z $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\theta = [\theta_1 \ \ \theta_2 \ \ \theta_3]^T$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$z = [-s\frac{y}{Λ(s)} \ \ -\frac{y}{Λ(s)} \ \ \frac{u}{Λ(s)}]^T$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\ Λ(s) $&lt;/span&gt; is a stable filter &lt;span class=&quot;math-container&quot;&gt;$\ Λ(s) = s^2 + 3s + 2 $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ y $&lt;/span&gt; is the output of the system (the mass relocation) and &lt;span class=&quot;math-container&quot;&gt;$\ u $&lt;/span&gt; is the input of the system (the external force).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to calculate the elements of &lt;span class=&quot;math-container&quot;&gt;$\ z $&lt;/span&gt; vector and for &lt;span class=&quot;math-container&quot;&gt;$\ z_1 = -s\frac{y}{Λ(s)} $&lt;/span&gt; for example I have done the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ z_1 = -s\frac{y}{Λ(s)} \Rightarrow \ddot{z_1}+3\dot{z_1}+2z_1 = -\dot{y}  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$x_1 = z_1 \Rightarrow \dot{{x_1}} = x_2 - y$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$x_2 = \dot{z_1}+y \Rightarrow \dot{{x_2}} = -3x_2 - 2x_1 + 3y$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From now on I want to use the &lt;code&gt;ode45(...)&lt;/code&gt; function but I am not able to figure out how to finally calculate &lt;span class=&quot;math-container&quot;&gt;$\ z_1 $&lt;/span&gt; and then put it in the &lt;span class=&quot;math-container&quot;&gt;$\ z $&lt;/span&gt; vector.&lt;/p&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-03-12T22:51:51.290" LastActivityDate="2020-03-12T22:51:51.290" Title="Simulating a mass - spring - damper system" Tags="|control-engineering|control-theory|matlab|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24609" PostTypeId="1" CreationDate="2018-11-10T03:16:21.000" Score="0" ViewCount="778" Body="&lt;p&gt;Apologize if probably this is duplicate question. But I have tried to search if someone have asked such this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a Honda engine, GX200, 196cc, with it's gross power is 6.5HP. I would like to make its exhaust to blow something. For that purpose, I need to modify the muffler to become something like shown by the picture below. But before I modify, I need to know how strong will be the blowing force right from the  exhaust output from the piston? Is any formula to calculate the force? That modification is intended to utilize it heat and its wind blowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uOp7p.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uOp7p.jpg&quot; alt=&quot;Engine exhaust/muffler&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17943" LastActivityDate="2019-02-28T09:12:26.513" Title="How to calculate the output power of engine exhaust" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="24610" PostTypeId="1" CreationDate="2018-11-10T06:10:28.303" Score="0" ViewCount="268" Body="&lt;p&gt;If we use tethered balloons (aerostats) with wireless gateways fitted on them (let's say for Cellular or LPWAN communication), they can provide much more coverage owing to the height they can reach (e.g. while towers usually reach a maximum of &lt;a href=&quot;https://www.steelintheair.com/cell-phone-tower/&quot; rel=&quot;nofollow noreferrer&quot;&gt;2000 feet&lt;/a&gt;, tethered balloons can usually reach a maximum of &lt;a href=&quot;https://en.wikipedia.org/wiki/Tethered_Aerostat_Radar_System&quot; rel=&quot;nofollow noreferrer&quot;&gt;4.5 km&lt;/a&gt;). And increasing the height of a gateway can increase the coverage surprisingly, since at a sufficiently good height, the Fresnel Zone gets good clearance giving clear line of sight. And it has been recently demonstrated using LoRa when someone created a record of &lt;a href=&quot;https://www.thethingsnetwork.org/article/ground-breaking-world-record-lorawan-packet-received-at-702-km-436-miles-distance&quot; rel=&quot;nofollow noreferrer&quot;&gt;702 km coverage&lt;/a&gt; with a high-altitude balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder why are tethered balloons then not used for communication?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While there are many instances of high-altitude balloons (like Google's Project Loon, &lt;a href=&quot;https://www.wired.com/2006/01/lofting-balloons-for-cell-service/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Space Data SkySite&amp;reg;&lt;/a&gt; or &lt;a href=&quot;https://electronics.stackexchange.com/questions/140801/high-altitude-balloon-communication&quot;&gt;this question on this StackExchange&lt;/a&gt;), I could find scarce resource on low-altitude tethered balloons. What I could find was &lt;a href=&quot;https://ieeexplore.ieee.org/document/4798405&quot; rel=&quot;nofollow noreferrer&quot;&gt;this work done in 2008 (IEEE Conference Paper)&lt;/a&gt; by a  research lab. Another recent conference paper (2016) is &lt;a href=&quot;http://iieng.org/images/proceedings_pdf/E1216004.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; from another lab. The first paper also demonstrates that the tethered balloons turn out to be lower cost in the long run. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some reasons that directly come to mind are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Balloons would require more maintenance (and may turn out to be expensive)&lt;/li&gt;&#xA;&lt;li&gt;Cost of balloons would increase dramatically as the weight of the wireless equipments is quite high&lt;/li&gt;&#xA;&lt;li&gt;Balloons would struggle under winds (however since they'd be tethered, exactly how would wind affect them?)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I would like to know is some solid engineering answer on this, with some numbers to back the claim. &lt;strong&gt;What exactly restricts their usage from RF, Communications, Aerodynamcis or Mechanical point-of-view&lt;/strong&gt;, as compared to fixed towers? Or perhaps is it just that beyond 2000 feet (maximum tower height), you do not get any benefit in Fresnel Zone clearance?&lt;/p&gt;&#xA;" OwnerUserId="3138" LastActivityDate="2020-12-10T17:25:26.710" Title="Tethered Balloons vs Towers for Wireless Communication" Tags="|electrical-engineering|aerospace-engineering|telecommunication|wireless-communication|communication|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24611" PostTypeId="2" ParentId="17972" CreationDate="2018-11-10T07:15:39.060" Score="-3" Body="&lt;p&gt;Instant torque means that you have have &amp;lt;100 milliseconds delay from pedal/lever/throttle movement to maximum corresponding acceleration of the vehicle, which is controlled by a phase controller that sends pulses to the motor. The controller and motor are at most 100 ms time response, at least 10 ms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clutch prevents their being a direct connection in between the motor and the wheels. It also prevents continuous acceleration from low to high gear. From standstill, the delay for clutch engagement is about 1-3 seconds, compared to 50-100 milliseconds on EV's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An electric car has zero rotation difference in between the motor axle and the wheel rotation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motor-coil-pulses apply force to magnets, that's why you need a slightly slow controller to redistribute DC battery energy into multiple motor &quot;phase&quot; wires. &lt;/p&gt;&#xA;" OwnerUserId="6491" LastEditorUserId="6491" LastEditDate="2018-11-10T07:21:55.187" LastActivityDate="2018-11-10T07:21:55.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24612" PostTypeId="1" CreationDate="2018-11-10T12:22:44.493" Score="3" ViewCount="361" Body="&lt;p&gt;As part of a control systems course I have to implement control for a 3 DOF SISO mass spring damper model of a robotic cutter arm. The arm is to follow a 5Hz triangle wave reference which is positioning it. As you can see from the picture, the robotic arm requires damped and tracking control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lfx6A.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lfx6A.png&quot; alt=&quot;open loop response&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have derived the state equations and successfully simulated the system open loop in simulink to recreate the same waveform as above. I have had success implementing full state feedback with LQR control. I have also used an integrator to achieve tracking of the output waveform. The final step is to design an estimator (since in real life not all states could be accessed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I have achieved some form of control I am not entirely sure if it is correct. Does anyone have examples of 3DOF SISO with estimator, tracked and/or damped control in state space? So far all I have found is examples of active suspension:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://nopr.niscair.res.in/bitstream/123456789/7546/1/IJEMS%2013%283%29%20173-179.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Design of LQR controller for active suspension system&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, here is a link for anyone who is curious about how to derive equations of motion for MBK system:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.efunda.com/formulae/vibrations/mdof_eom.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Multiple Degree-of-Freedom Example&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18244" LastEditorUserId="7083" LastEditDate="2018-12-28T00:05:22.117" LastActivityDate="2018-12-28T00:05:22.117" Title="3 DOF SISO state space control" Tags="|control-engineering|control-theory|matlab|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24613" PostTypeId="2" ParentId="24601" CreationDate="2018-11-10T14:48:33.957" Score="2" Body="&lt;p&gt;The fundamentals are shown below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gv6Hq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gv6Hq.png&quot; alt=&quot;sealed or open compartments&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume air is an ideal gas and the process is isothermal. With a sealed compartment, the gas pressure will increase as the volume decreases (and visa-versa). With an open compartment, the gas pressure will remain constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water pressure in the tank will be a sum of the pressure from the diaphragm and the gas pressure.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-11-10T14:48:33.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24614" PostTypeId="2" ParentId="24610" CreationDate="2018-11-10T16:39:42.903" Score="2" Body="&lt;p&gt;They would be a great hazard to air traffic; The umbilical would be essentially invisible to passing aircraft. In spite of regulations not to fly here or there.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-11-10T16:39:42.903" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24615" PostTypeId="1" AcceptedAnswerId="24616" CreationDate="2018-11-10T21:33:52.277" Score="4" ViewCount="352" Body="&lt;p&gt;So, I have a situation where we are putting load on a plate from above, and the plate has a step in it, which was made by milling a bigger block of steel (i.e. there are no layers and everything is homogeneous). Conventional engineering knowledge tells us that if we increase the fillet radii, we will get a better fatigue performance for this abrupt change in EI within this plate, which is great. But a bigger fillet will also mean the item that this plate is supporting (which in this case fits into the milled area) becomes a limiting factor with how sharp their corners are. I would like to know - is there a fillet design that I can use in this situation that will allow me to improve fatigue performance without necessarily compromising the way my object fits into this milled area? I have proposed some different versions in the sketch attached, and I would like someone well versed in this area to help me understand which one is best, and for what reasons. Thanks!&lt;img src=&quot;https://i.stack.imgur.com/9nBT9.png&quot; alt=&quot;![enter image description here&quot;&gt;]&lt;a href=&quot;https://i.stack.imgur.com/9nBT9.png&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13526" LastActivityDate="2018-11-12T20:27:46.800" Title="Fatigue resistance at filleted corners - Design detail" Tags="|fatigue|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24616" PostTypeId="2" ParentId="24615" CreationDate="2018-11-10T21:45:21.623" Score="4" Body="&lt;p&gt;I have seen 'lollipop cutters' used for this purpose, just a single pass into the side (closest to option B). The thickness of the shank means it's not a perfectly square bottomed slot, but was sufficient in the instance in question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image below shows the cutter type:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/xtlZi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the profile in question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dg8EPm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dg8EPm.png&quot; alt=&quot;corner undercut radius&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method of adding an undercut radius will certainly improve the fatigue life of the part by drastically reducing the stress concentration factor at that corner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only situation in which I can imagine this having a detrimental effect is where the wall section is sufficiently thin that the removal of extra material makes it too thin for the forces involved (as illustrated below), but from the sketch in your original question, I doubt very much that this will be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hCGLSm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hCGLSm.png&quot; alt=&quot;thin wall undercut to leave not enough material&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth noting that this does leave a very small remaining interference from the 'perfectly square corner', but in my application the mating part was not so perfectly sized to cause a problem here. Also, it was fully enclosed on all sides (such that these illustrations are a section view). Lollipop cutters do not plunge well (if at all), so cutting downwards as shown on the right is only possible if you can enter the workpiece from the side (rather than above, in my case), or perhaps if a pilot slot is milled downwards first with a small endmill first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dxHYLm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dxHYLm.png&quot; alt=&quot;interference issue&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-11-12T20:27:46.800" LastActivityDate="2018-11-12T20:27:46.800" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24617" PostTypeId="1" AcceptedAnswerId="24632" CreationDate="2018-11-11T02:15:28.913" Score="0" ViewCount="604" Body="&lt;p&gt;I need to construct a table that will hold a water tank. The filled tank will weigh up to 1300 pounds. I have the table leg frames worked out to hold 2500 pounds so no worries there. (&lt;a href=&quot;https://www.benchdepot.com/cleveland-series-workbenches/1553/159/cleveland-series-frame-with-legs-industrial-work-benches.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a link to details.&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6iwCt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6iwCt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are just a 28&quot; x 34&quot; rectangle. The tank itself has a footprint of 30&quot; x 38&quot; and is made of molded polyethylene, so don't count on its rigidity for support. The table top will be 36&quot; x 48&quot;. (It is bigger than the tank solely because I can buy it off the shelf at that size without any fabrication cost.) There will be no motion to account for nor any rapid changes in the weight. So my question is how thick does the stainless steel table top need to be to safely bear 1300 pounds?&lt;/p&gt;&#xA;" OwnerUserId="18250" LastEditorUserId="8055" LastEditDate="2018-11-19T17:58:31.817" LastActivityDate="2018-11-19T17:58:31.817" Title="How thick does my stainless steel table top need to be?" Tags="|mechanical-engineering|structural-engineering|steel|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="24620" PostTypeId="1" AcceptedAnswerId="24622" CreationDate="2018-11-11T13:11:10.027" Score="0" ViewCount="172" Body="&lt;p&gt;I am simulating a mass - damper - system in Matlab and I have the following vector (1x100) as input to my system:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ut = linspace(0, 10);&#xA;u = 5 * sin(2 * ut) + 10.5; % input of our system - external force&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now, I want to take the value of the output for each value of the input vector using this differential equation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function dx = odefun_4(t,x)&#xA;    m = 15;&#xA;    b = 0.2;&#xA;    k = 2;   &#xA;    dx = [x(2); u/m - (b/m)*x(2) - (k/m)*x(1)];    &#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which is called from my main &lt;code&gt;.m&lt;/code&gt; file like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[t,X] = ode45(@odefun_4, [0 10], [0;0])&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I tried to pass the &lt;code&gt;u&lt;/code&gt; vector by making it global variable but I get an error stating: &lt;code&gt;Dimensions of arrays being concatenated not consistent.&lt;/code&gt;&#xA;I also tried to set the &lt;code&gt;Refine&lt;/code&gt; parameter using &lt;code&gt;odeset&lt;/code&gt; and the &lt;code&gt;options&lt;/code&gt; argument of &lt;code&gt;ode45&lt;/code&gt; solver but still don't get the desired results. I could really use some help.&lt;/p&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-04-27T03:17:36.827" LastActivityDate="2020-04-27T03:17:36.827" Title="Get results of output for each one value of the input - MATLAB ODE" Tags="|mathematics|matlab|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24622" PostTypeId="2" ParentId="24620" CreationDate="2018-11-11T14:08:05.020" Score="1" Body="&lt;p&gt;If you want &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; to be considered as a smooth continuous function, then you should define it as such since you also have &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; available as well. So you could use:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function dx = odefun_4(t,x)&#xA;    m = 15;&#xA;    b = 0.2;&#xA;    k = 2;&#xA;    u = 5 * sin(2 * t) + 10.5;&#xA;    dx = [x(2); u/m - (b/m)*x(2) - (k/m)*x(1)];&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I often also find it more convenient to use anonymous functions in Matlab, for example:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;m = 15;&#xA;b = 0.2;&#xA;k = 2;&#xA;u = @(t) 5 * sin(2 * t) + 10.5;&#xA;odefun_4 = @(t,x) [x(2); u(t)/m - (b/m)*x(2) - (k/m)*x(1)];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-11-11T14:08:05.020" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24623" PostTypeId="1" AcceptedAnswerId="24625" CreationDate="2018-11-11T15:36:11.660" Score="2" ViewCount="987" Body="&lt;p&gt;My professor, when discussing about Random Walk model during diffusion in metals said that &quot;Vacancies are equilibrium defects but dislocations are not&quot;. I could not understand why. Why is it so?&lt;/p&gt;&#xA;" OwnerUserId="14940" LastActivityDate="2020-02-02T22:02:56.687" Title="Why are vacancies in solids called as equilibrium defects?" Tags="|materials|metallurgy|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24625" PostTypeId="2" ParentId="24623" CreationDate="2018-11-11T17:55:18.060" Score="6" Body="&lt;p&gt;Imagine you have two boxes separated by a removable partition. Each box is filled with a different pure monatomic gas. When you remove the partition, the gases mix. Their equilibrium state tends toward an equitable mixture at every point in time. The likelihood of the gases spontaneously separating is essentially zero for any appreciable number of atoms. This is modeled by statistical mechanics, and occurs for any similar system consisting of multiple mobile components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solids are no different from gases in that they obey the principles of statistical mechanics. In the case of a crystal lattice, vacancies are lattice points with no atom, i e. empty space. All solids are constantly exposed to empty space at their surface, and tend to mix with it. This may sound strange, but remember that atoms in a lattice are mobile, and can shift lattice positions if a neighboring space is available. All of the lattice points at the surface have lots of space available outside the crystal. The same applies at grain boundaries and at dislocations. So by random atomic movement, empty space can diffuse into the lattice and create vacancies. As we established with the two-gas setup, the empty space and solid tend to be in some mixed equilibrium, i.e. the equilibrium concentration of vacancies is non-zero. The only detail remaining is that the mixture isnt equal because the atoms of the solid have strong cohesive forces binding them together. Unlike gases the crystal doesn't evenly diffuse about the room and instantly vaporize. At least, in equilibrium it wouldn't (discounting vapor pressure)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for dislocations, their equilibrium concentration is zero because they don't spontaneously form, but can spontaneously leave crystals. Recall that dislocations form in response to a net shear creating an &quot;excess&quot; half plane of atoms, or alternately an &quot;excess&quot; half plane of vacancies. This occurs by a process called &quot;slip&quot;.  Slip does not occur in equilibrium. Getting a half plane of vacancies into the crystal, all organized along a plane, and done spontaneously, is simply not going to happen by chance. In contrast, an existing dislocation can move spontaneously by absorbing or emitting dislocations in a process called &quot;climb&quot;. If a vacancy is absorbed, the half-plane of atoms decreases in number, moving the dislocation line in that direction. If a vacancy is emitted, the opposite occurs. If the dislocation line reaches the surface, it is no longer a dislocation. Thus dislocations can leave the crystal, but can't enter, resulting in a net equilibirum concentration of zero.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2018-11-11T22:02:19.527" LastActivityDate="2018-11-11T22:02:19.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24626" PostTypeId="1" CreationDate="2018-11-11T20:02:14.680" Score="0" ViewCount="2861" Body="&lt;p&gt;I've been puzzling over this for some time. In this picture, there are two steel frames, with four height-adjust screws controlling the height between them. The roller shown is applying a good amount of pressure (maybe analogous to someone rolling dough with a rolling pin). It appears that there is one screw going into a green &quot;nut&quot; fixed to the top frame and passing through to a sort of collar fixed (but allowed to rotate?) to the bottom frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking is that the screw functions as a sort of lead-screw, rotating to bring the top frame up and down, while remaining fixed on the bottom frame. Of course, it also has to rotate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing is, with the upward pressure, the green collar in the bottom frame would be pulled out, unless it snaps in on both sides. As well, there isn't anything visibly stopping the top frame's green collar/nut when upward pressure is applied on the roller (and thus the frame). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, no other information than this image of the fixture is given, and the images are evidently not great quality. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What mechanism could be used here to adjust the height? What methods are commonly used for precise height adjustment of platforms like this?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bFq8N.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bFq8N.png&quot; alt=&quot;Fusing test rig&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18257" LastActivityDate="2018-11-11T20:31:25.497" Title="How does this screw height adjustment mechanism work?" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24627" PostTypeId="1" CreationDate="2018-11-11T20:30:48.113" Score="0" ViewCount="45" Body="&lt;p&gt;I want to build some sort of SAR sensor. Since radar works high speeds and creates loads of data this should be processed faster. I am trying to find algorithms to be used for this purpose. I am at the beginning of the project so just looking around some information. I haven't had much progress yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some values in mind such as, what frequency I will use, what bandwidth I will have, what sampling frequency will I have and how many channels etc. Then I will process them. Preferably in real-time. But I am not familiar with the way how the storage capacity and processing speed needed can be calculated for this. I want to decide whether I need to use onboard (battery powered unmanned vehicle) computation or I need to send raw data to ground-station and doing all the processing in it. If so don't know if something Nvidia Xavier would be handy or standard windows installed pc would be good for this purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example scenario like this, I have a base-band signal at 10MHz and will be swept by upconverting to RF signal and received back following to downconversion back to 10MHz. Then Quadrature mixer will allow me to find the phase shift. If I am not wrong I need to sample this data at a frequency like 50MHz. This will create 50 million samples I and Q in a second for each, thus the total of 10 million data point. This is one-dimensional data. This will be scanned in a spatial direction to yield 2D radar image which should then be processed. I don't know how big will this data be and if I will be able to process it in almost real time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for some hints or some experienced advice.&lt;/p&gt;&#xA;" OwnerUserId="1287" LastEditorUserId="1287" LastEditDate="2018-11-14T19:53:56.213" LastActivityDate="2018-11-14T19:53:56.213" Title="How to determine data size created by sensor for calculation of signal processing and transmitting over wireless communication?" Tags="|robotics|signal-processing|signal|radar|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24628" PostTypeId="2" ParentId="24626" CreationDate="2018-11-11T20:31:25.497" Score="1" Body="&lt;p&gt;Without more information there's no way to know exactly how this works, but one approach would be a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Differential_screw&quot; rel=&quot;nofollow noreferrer&quot;&gt;differential screw&lt;/a&gt;.  The link is a good explanation, but essentially it's a screw with two sections of thread, with slightly different pitch.  It allows very precise movement.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2018-11-11T20:31:25.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24629" PostTypeId="1" CreationDate="2018-11-12T02:00:06.460" Score="0" ViewCount="1724" Body="&lt;p&gt;What I know, if I'm not mistaken, is that group celerity is the speed at which the energy is transferred and celerity is the speed of the waves themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since sea waves basically transfer energy and the particles won't follow along much, I don't understand why they need to be so different, specially in deep seas. And I also don't understand why the group celerity in shallow waters tends to &lt;span class=&quot;math-container&quot;&gt;$\sqrt{gd}$&lt;/span&gt; if we use Airy's linear theory (yes I know it's not very appropriate for shallow waters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any explanation will be much appreciated, thank you&lt;/p&gt;&#xA;" OwnerUserId="18258" LastActivityDate="2021-08-01T10:03:20.370" Title="Sea waves: Having trouble visualizing the difference between group celerity and celerity" Tags="|fluid-mechanics|marine-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24632" PostTypeId="2" ParentId="24617" CreationDate="2018-11-12T09:43:59.050" Score="2" Body="&lt;p&gt;The Stress formula for a rectangular plate is:&lt;a href=&quot;https://www.engineersedge.com/material_science/rectangular_plate_uniform_load_13643.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;source,&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rectangular plate, uniform load, simply supported (Empirical) equations.&#xA;Since comers tend to rise off the supports, vertical movement must be prevented without restricting rotation.&#xA;Symbols used:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; = major length of rectangular plate, (in)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; = minor length of rectangular plate, (in)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; = uniform pressure loading, ( Ibs/in')&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\nu$&lt;/span&gt; = Poisson's ratio&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; = Young's modulus, ( lbs/in2)= 29,000,000 psi for a 36 steel, with 3 as factor of safety we get 18000 psi for allowable stress.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; = plate thickness, (in)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$S_m$&lt;/span&gt; = maximum stress, (lbs/in2)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$y_m$&lt;/span&gt; = maximum deflection, (m, in)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sigma_{max} &amp;amp;= \frac {0.75 pb^2}{t^2[1.61(b/a)^3 + 1 ] } \\&#xA;y_m &amp;amp;= \frac {0.142pb^4}{Et^3[2.21(b/a )^3+1]}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for edge simply supported, but you have cantilever edges. to make the task simpler we just reduce the p by a factor of 0.75, so 0.75(1300)/ (36)(48) = 0.564 psi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;t^2 &amp;amp;= \frac{(0.564(28^2)0.75)}{ 18000[1.61(28/34)^3 +1]} \\&#xA;&amp;amp;= .0987 \\&#xA;\therefore t &amp;amp;= 0.1\text{ in}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say pick 1/8 thick sheet for practicality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check my arithmetic please.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2018-11-12T17:19:01.177" LastActivityDate="2018-11-12T17:19:01.177" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24633" PostTypeId="1" CreationDate="2018-11-12T14:55:20.577" Score="1" ViewCount="912" Body="&lt;p&gt;What is the weight bearing capacity per square yard of a 4&quot; monolithic concrete floor slab poured with 3000 psi concrete in 50' x 50' squares, reinforced with 6&quot; x 6&quot; x 10&quot; steel wire mesh, laid over 4&quot; crushed limestone slag, with 4 mil poly vapor barrier?&lt;/p&gt;&#xA;" OwnerUserId="18248" LastActivityDate="2018-11-12T14:55:20.577" Title="weight bearing capacity of a reinforced concrete floor slab" Tags="|structural-engineering|" AnswerCount="0" CommentCount="4" ClosedDate="2018-11-12T17:13:22.843" ContentLicense="CC BY-SA 4.0" />
  <row Id="24634" PostTypeId="1" CreationDate="2018-11-12T16:55:50.667" Score="0" ViewCount="153" Body="&lt;p&gt;Suppose I have a convoy composed of the following elements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A locomotive with 4 driven wheelsets (2 wheels per wheelset and therefore 8 wheels). &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; carriages with 2 wheelsets each.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Assuming the weight is uniformly distributed on the wheels, from my understanding the maximum force that can be applied by the locomotive avoiding wheels slipping is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_{max} = m_{locomotive} g \mu$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; is the adhesion coefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is that I'm not sure if I should consider also the passenger carriers and if not, then why. I think I should not consider them since the force that prevents the locomotive's wheels from slipping is just its own weight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, what if the locomotive had only 6 driven wheels out of the 8? Would that change anything? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case I think I should multiply the &lt;span class=&quot;math-container&quot;&gt;$F_{max}$&lt;/span&gt; obtained by 6/8 am I correct? &lt;/p&gt;&#xA;" OwnerUserId="17623" LastEditorUserId="1832" LastEditDate="2018-11-12T17:53:20.460" LastActivityDate="2018-11-13T00:11:50.237" Title="How to calculate the maximum starting traction force for a train with multiple carriages?" Tags="|mechanical|electric-vehicles|traction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24637" PostTypeId="2" ParentId="24629" CreationDate="2018-11-12T19:21:03.463" Score="1" Body="&lt;p&gt;Yes,  &quot;celerity&quot; is the phase velocity of the wave (waveform) itself.  &quot;Group celerity,&quot; according to &lt;a href=&quot;https://www.codecogs.com/library/engineering/fluid_mechanics/waves/group_celerity.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;codecogs&lt;/a&gt;  or  &lt;a href=&quot;https://www.essie.ufl.edu/~sheppard/OCE3016/Ocean%20Waves.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;these course notes (pdf)&lt;/a&gt;, is the group velocity.   For what it's worth, the second link there lists the following approximations, along with some derivations.   Probably more than you wanted to know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6OoIz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6OoIz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-11-12T19:21:03.463" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24639" PostTypeId="2" ParentId="24634" CreationDate="2018-11-13T00:11:50.237" Score="2" Body="&lt;p&gt;The answer to the first part of question is, you're right, you don't concern yourself with the other carriages friction or weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part, you have to consider static friction force on the wheels for traction. In steel on steel it is approximately 0.60.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the approximate traction your locomotive can provide is 60% of its weight  tributary to its powered wheels regardless of the number of the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say if you reduce the number of the axels to 3 you are puting more weight on each wheel. So the traction remains the same. Unless by changing the geometry of the axels you compromise the even distribution of the weight on the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because then the light axel will skid earlier, sinking the power.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-13T00:11:50.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24640" PostTypeId="1" CreationDate="2018-11-13T02:13:22.610" Score="1" ViewCount="2558" Body="&lt;p&gt;In my country are sold gas with different octane number (RON: Research Octane Number): 88, 90, 92, 95, 97, 98, and the latest is 100. RON-88 is still sold since still many old cars, motor bike, and agriculture machines which are using low spec gas. RON-90 is intended for transition, as alternative in case customer want to move forward using better gas but not so expensive. In our country, the higher RON, the more expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we know, every engine is designed with a specific compression ratio, and that RON is very related to the compression ratio. But many times I read explanation from automotive services claiming that higher RON will be better for car. Not specifically explained which car and what kind of car. Indeed we know that most cars are now having compression ratio 10:1, which according to the petrol producer that kind of is car suitable using petrol with RON 92-95. In my understanding, the compression ratio is related to the Top Dead Center (TDC), which it will produce specific compression that will compress the petrol inside ignition chamber, so it will increase temperature just before ignited by the spark. That compression made engine got hotter. &lt;em&gt;If an engine with specific compression ratio is fed with petrol with RON-88, the petrol will be burned before TDC&lt;/em&gt;. Vice versa, &lt;em&gt;if that kind engine is fed with RON-100 petrol, than the petrol will NOT be completely burnt&lt;/em&gt;. Hence, the automotive service claim is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question are:&#xA;1). am I right with my understanding?&#xA;2). for that 10:1 engine, if it fed with lower or higher RON, what will be the impact to the engine?&lt;/p&gt;&#xA;" OwnerUserId="17943" LastActivityDate="2023-12-18T00:05:00.550" Title="How to Understand the Relation of Compression Ratio and the Octane Number?" Tags="|automotive-engineering|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="24644" PostTypeId="2" ParentId="24640" CreationDate="2018-11-13T08:47:25.963" Score="0" Body="&lt;p&gt;Almost all gasoline-fueled engines have ignition systems which under high speed conditions will spark the charge in the cylinder before top dead center, so as to furnish enough time for the charge to burn to completion before the exhaust valve begins opening. So the issue is not whether or not the spark plug goes off before TDC, it is whether or not the fuel will combust smoothly or explode (detonate) with sudden violence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The octane number is a measure of how resistant the fuel is to detonation, which is very bad for an engine that is designed to run on gasoline and so must be avoided. High compression ratios strongly favor detonation, so engines with high compression ratios must use higher-octane fuel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In engines with lower compression ratios, detonation is still possible when the engine is running hot and under a heavy load, so there is a slight advantage to using higher octane fuels in any engine. However, most engines these days have automatic systems in them called &quot;knock detectors&quot; that detect the onset of detonation and retard the spark timing to block detonation under those conditions- so buying higher octane fuel for an engine that does not require it is mostly a waste of money.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-11-13T08:47:25.963" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24645" PostTypeId="1" AcceptedAnswerId="24646" CreationDate="2018-11-13T08:50:51.537" Score="3" ViewCount="3185" Body="&lt;p&gt;In my engineering textbook, I didnt understand the example given to explain the magnitude of the moment. It asks to calculate the magnitude of the moment about the base point O of the 600-N force in five different ways. Now, I dont understand the third way it calculated the moment. What I don't understand is why d(1) = 4 + 2 tan(40) and so the moment is 460(d(1)). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It used the principle of transmissibliity to move the vector force to point B and so we dont need one of the component as it is parallel with the distance. And to calculate the distance we have 4 + 2 tan(40). I understand upto 4 but I dont understand why we add 2tan(40). How is 2 vectored with tan(40)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use picture I have uploaded. &lt;img src=&quot;https://i.stack.imgur.com/WFbZP.jpg&quot; width=&quot;300&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="18274" LastEditorUserId="13311" LastEditDate="2018-11-13T10:28:39.513" LastActivityDate="2018-11-13T11:51:42.013" Title="A way to calculate the magnitude of the moment" Tags="|civil-engineering|applied-mechanics|moments|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24646" PostTypeId="2" ParentId="24645" CreationDate="2018-11-13T10:46:48.827" Score="4" Body="&lt;p&gt;This is a trigonometry question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Point B is a certain distance upwards from Point A. How much? You know the distance that it is to the left from Point A (&lt;span class=&quot;math-container&quot;&gt;$2\text{m}$&lt;/span&gt;) and the angle that it is from point A (&lt;span class=&quot;math-container&quot;&gt;$40°$&lt;/span&gt;), so using standard trigonometry &lt;span class=&quot;math-container&quot;&gt;$\tan(\theta)=\frac{opposite}{adjacent}$&lt;/span&gt;, you can calculate the vertical distance by rearranging to give: &lt;span class=&quot;math-container&quot;&gt;$opposite=adjacent\tan(\theta)$&lt;/span&gt;, or in this specific case, &lt;span class=&quot;math-container&quot;&gt;$x=2\tan(40)$&lt;/span&gt;, such that &lt;span class=&quot;math-container&quot;&gt;$d_1 = 4+2\tan(40)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ib8xSh.jpg&quot; width=&quot;400&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="1832" LastEditDate="2018-11-13T11:51:42.013" LastActivityDate="2018-11-13T11:51:42.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24647" PostTypeId="1" AcceptedAnswerId="24648" CreationDate="2018-11-13T12:36:28.213" Score="0" ViewCount="12724" Body="&lt;p&gt;Zinc Plating is known for its ability to prevent most corrosion, and it is done by just dipping a metal in hot pool of zinc and letting it cooldown. Meanwhile, stainless steel, an alloy that is capable of resisting corrosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc Plating Advantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cheap&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Stainless Steel Advantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is not coating (hence, scratching will not expose for corrosion)&lt;/li&gt;&#xA;&lt;li&gt;Hardnesss of Material (&lt;span class=&quot;math-container&quot;&gt;$UTS_{Stainless} \approx 860 \frac {N}{mm^2}$&lt;/span&gt;) comparatively to Mild Steel (&lt;span class=&quot;math-container&quot;&gt;$UTS_{Mild  Steel} \approx 470 \frac {N}{mm^2}$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Other than what is said, in what circumstance would you use stainless steel over galvanized steel?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="18276" LastActivityDate="2022-12-30T04:46:19.350" Title="Galvanized Steel (Zinc Plated Steel) vs Stainless Steel" Tags="|metallurgy|metals|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24648" PostTypeId="2" ParentId="24647" CreationDate="2018-11-13T13:10:38.463" Score="10" Body="&lt;p&gt;Stainless steel can be used up to temperatures of about 1000C. The corrosion resistance of zinc plating decreases rapidly above 100C, and embrittlement can occur above 500C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc plating has lower resistance to chemical corrosion from acids and alkalis than stainless steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aside from mechanical damage caused by scratching, the rate of corrosion may not be uniform over the whole structure, leading to local failures of the protective layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Zinc plating will fail if it is in contact with some common construction materials, including copper, brass, chrome plated materials, unplated iron and steel, bitumen, some species of timber (including oak, cedar, and some fir species) etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fabricating a structure from plated raw materials may damage the plating layer and reduce the corrosion resistance (e.g. assembly using bolts and screws, bending plated sheet metal, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Welding plated steel requires special precautions to make good quality joints and to protect the operator from fumes containing zinc compounds.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-11-13T13:21:14.047" LastActivityDate="2018-11-13T13:21:14.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24650" PostTypeId="1" AcceptedAnswerId="24654" CreationDate="2018-11-13T16:23:45.170" Score="0" ViewCount="371" Body="&lt;p&gt;I am actuating a small ceramic rotary valve with a cheap mg996 servo motor. The valve has a 90 degree max rotation from fully closed to fully open. Currently I have the servo shaft directly coupled to the valve shaft, and simply limit maximum rotation of the servo to 90 degrees in software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be nice to have the extra resolution afforded by utilizing the servo's full range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple way to reduce the 180 degree rotation to 90 degrees without gears? Space is tight and complexity needs to be minimized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A push/pull rod on an arm may be the best option, but I'm not aware of other ways to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I have access to a 3d printer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="1558" LastActivityDate="2018-11-14T00:38:32.910" Title="Reduce 180 degree rotation to 90 degree rotation without gears" Tags="|valves|servo|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24651" PostTypeId="1" AcceptedAnswerId="24653" CreationDate="2018-11-13T22:29:35.697" Score="-1" ViewCount="108" Body="&lt;p&gt;This is a gear that transfers force from a motorcycle kickstarter to the transmission. What method should I use to repair the missing teeth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/KIBbJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eyQP2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="18280" LastEditorUserId="18280" LastEditDate="2018-11-13T23:08:25.060" LastActivityDate="2018-11-13T23:10:16.553" Title="How to fix this gear" Tags="|motors|gears|transmission|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24652" PostTypeId="2" ParentId="24651" CreationDate="2018-11-13T22:57:23.400" Score="0" Body="&lt;p&gt;First you need to get rid of the leftover cups and cones by grinding it to have a smooth surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the best result you have to choose the same material but if you don't know the exact composition then, you can choose a similar materials but make sure the extra material has a good solderability, and it has to be compatible with the base alloy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then warm up the area, and try to distribute the smelted alloy all over the area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last step, is to machine the teeth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It won't last as long as the new one, in fact warming up the gear weaken the alloy, it means the old teeth near the missing area, are likely to fail. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-11-13T22:57:23.400" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24653" PostTypeId="2" ParentId="24651" CreationDate="2018-11-13T23:10:16.553" Score="9" Body="&lt;p&gt;Oddly enough, this problem can be fixed using a set of car keys and your wallet. Here are the steps to follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) put the keys in the ignition of your car, start it, and drive to the motorcycle scrapyard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) once there, exchange several pieces of local currency for a gear pulled out of another identical motorcycle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) place the gear in your pocket and drive your car home (optional step: purchase a six of your favorite dry-hopped IPA before arrival).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) installation procedure is the reverse of the disassembly procedure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) after washing hands, open the first bottle of IPA. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) share and enjoy.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-11-13T23:10:16.553" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24654" PostTypeId="2" ParentId="24650" CreationDate="2018-11-14T00:38:32.910" Score="1" Body="&lt;p&gt;If space is tight and complexity must be minimized, why not just buy a servo that has 90 degree travel to begin with.  There are a number of these available for cheap.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-11-14T00:38:32.910" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24656" PostTypeId="1" AcceptedAnswerId="24659" CreationDate="2018-11-14T03:46:23.920" Score="0" ViewCount="189" Body="&lt;p&gt;&lt;strong&gt;Fastest plane&lt;/strong&gt; ever built: SR-71 ,Max Speed: 3.3 Mach , Development - &lt;strong&gt;1950s&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fastest drone&lt;/strong&gt; ever built: Lockheed-D21 ,Max Speed: 3.3 Mach, Development -&lt;strong&gt;1960s&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fastest cruise missile&lt;/strong&gt; ever built: Brahmos , Max Speed: 3.3 Mach, Development - &lt;strong&gt;2000s&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why Ram/turbo-Ram jets got stuck around 3.3 Mach? Is there any technological challenge beyond 3.3 Mach like heating or combustion instability or low thrust? &lt;/p&gt;&#xA;" OwnerUserId="16175" LastActivityDate="2018-11-14T07:11:07.613" Title="Why there are no jets (missile or aircraft) faster than Mach 3.3?" Tags="|mechanical-engineering|aerospace-engineering|aerodynamics|propulsion|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24657" PostTypeId="2" ParentId="3041" CreationDate="2018-11-14T04:39:06.890" Score="2" Body="&lt;p&gt;Engineering bricks (solid) are fired at 1200 degrees in a kiln so don't worry about your complicated mathematics , your pizza oven will never get to that temp.if it did or even if it got much much higher the bricks would just melt together.(They call them clinkers)&#xA;I am a bricklayer with 47 years experience and a lot of that time was building and repairing kilns.&#xA;When you repair a kiln you only use normal clay for the crowns (internal)as once the kiln is fired up the clay turns hard just like the bricks you are cooking.&#xA;You only use fire bricks for the fire holes and they are cemented together with texacrete.&lt;/p&gt;&#xA;" OwnerUserId="18284" LastActivityDate="2018-11-14T04:39:06.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24658" PostTypeId="2" ParentId="23749" CreationDate="2018-11-14T06:07:51.633" Score="0" Body="&lt;p&gt;Simply to keep a sound and smell barrier and commonly doubles as a breezeway for the gender doors. There are plenty of bathrooms with in building code with one locking door to the common area.&lt;/p&gt;&#xA;" OwnerDisplayName="user4139" LastActivityDate="2018-11-14T06:07:51.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24659" PostTypeId="2" ParentId="24656" CreationDate="2018-11-14T07:11:07.613" Score="1" Body="&lt;p&gt;At those speeds, several effects become important. We consider air-breathing engines in what follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, aerodynamic heating of the airframe becomes important. In broad terms, to prevent structural failure, the maximum temperature of the hottest portions of the airframe must be held below red heat. Going faster than mach 3.3 pushes this limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, for an air-breathing engine that propels the airframe from sea level and zero airspeed to mach 3.3 at high altitude, the temperature at the face of the compressor stage must be below the maximum temperature limit of the first stage compressor blades, which is roughly 800F. Going faster than this means the temperature at the face of the engine will exceed the temperature rating of the blades and the engine will self-destruct. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-11-14T07:11:07.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24660" PostTypeId="1" CreationDate="2018-11-14T08:35:13.860" Score="1" ViewCount="34" Body="&lt;p&gt;If you have some experience in th field of packing and filling can you tell me which two areas should i choose to support the bottle with two starwheels ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the red area should be supported as it has a straight line and the upper area &#xA;But I'm afraid if i didn't support the bottom area the bottle would flip because the movement of the conveyor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cmEp8.png&quot; width=&quot;200&quot; /&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/5Zlvm.png&quot; width=&quot;400&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="17065" LastEditorUserId="13311" LastEditDate="2018-11-14T10:04:06.767" LastActivityDate="2018-11-14T10:04:06.767" Title="How to select the distance between two starwheeels holding a bottle?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24661" PostTypeId="2" ParentId="14867" CreationDate="2018-11-14T12:12:28.573" Score="1" Body="&lt;p&gt;I haven't seen any method of calculation to determine the cross section classification of a plate but SCI advisory note AD 391 &amp;amp; also design manual for steel frame for SAP 2000 all indicate/assume the cross section classification of a plate as class 3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See below  link for AD 391: http&#xA;//www.newsteelconstruction.com/wp/wp-content/uploads/TechPaper/AD391.pdf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this help.&lt;/p&gt;&#xA;" OwnerUserId="18288" LastActivityDate="2018-11-14T12:12:28.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24662" PostTypeId="1" CreationDate="2018-11-14T12:28:30.503" Score="0" ViewCount="538" Body="&lt;p&gt;Protecting paper documents from fire is quite straightforward with, for example, a &lt;a href=&quot;https://www.sentrysafe.com/product/CFW20201&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sentry fire chest&lt;/a&gt;. This product has been &lt;a href=&quot;https://en.wikipedia.org/wiki/Fire-resistance_rating&quot; rel=&quot;nofollow noreferrer&quot;&gt;tested&lt;/a&gt; hold the inside temperature under 177°C (350F) for 1 hour. This is fine for documents, since they will not combust or deform. However, I'm looking for an inexpensive solution for items that would take considerable damage at these temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sure, I could house the fire chest inside another larger safe. Or build myself some kind of drywall, brick and mortar box to house the fire chest in... but I'm looking for something smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea is, if I stored my valuables inside a stainless steel Thermos, could this offer sufficient additional insulation if put &lt;em&gt;inside&lt;/em&gt; the fire chest? The plastics of the Thermos might &lt;a href=&quot;https://www.quora.com/What-is-the-melting-point-of-plastic&quot; rel=&quot;nofollow noreferrer&quot;&gt;deform&lt;/a&gt; at some point, but could I get enough insulation? A Thermos works both ways, but how well...?&lt;/p&gt;&#xA;" OwnerUserId="18287" LastEditorUserId="18287" LastEditDate="2018-11-17T08:10:31.170" LastActivityDate="2021-04-09T00:03:50.453" Title="Protecting heat-sensitive media from fire: fire-safe + Thermos?" Tags="|thermal-insulation|fire|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24663" PostTypeId="2" ParentId="24662" CreationDate="2018-11-14T13:00:44.533" Score="1" Body="&lt;p&gt;Radiative heat transfer will be quite high in a house fire, so not every construction made for 100°C temperature differences will be applicable here- look for constructions that have glowing-metal temperatures to contain (ceramics ovens...), and reverse the design to work for heat applied from the outside. Also put something with high heat capacity (water is ideal) into the container to soak up the power that will inevitably make it through - you might even try a two-shell design (small thermos in fire chest?) that has the water after the first shell, so you can gain from the fluid-to-gas energy consumption of water at boiling point (remember to leave opening for steam, otherwise you built a bomb) while at the same time nailing the maximum temperature around the thermos to 100°C for as long as there is fluid water, thereby bringing the amx temperature down to what the thermos was designed for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add material beside the valuables that has a melting point slightly lower than your maximum of 52°C you can soak up more power before the temperature rises further - phase shifts usually take a lot of energy, so it might be worth a shot. Phasechange to gas would be even better, but the materials with boiling points below 52°C are probably not safe to release in a fire, and retained the pressure would rise, raising the boiling point and producing structural issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember that a house fires can trigger other violent events like falling debris, so the container should be quite sturdy to begin with, and the door-seals need thought.&lt;/p&gt;&#xA;" OwnerUserId="16248" LastEditorUserId="16248" LastEditDate="2018-11-14T13:45:10.043" LastActivityDate="2018-11-14T13:45:10.043" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24665" PostTypeId="2" ParentId="24629" CreationDate="2018-11-14T21:02:36.480" Score="1" Body="&lt;p&gt;Lets ignore wave groups for starters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are looking at a model of a steady gravitational free surface wave train of parallel waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mit.edu/2.016/www/handouts/Free-Surface-Waves.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;We have equations for that&lt;/a&gt;. See pages 16 to 39. The feature of interest is a quantity called the dispersion relation. It is discussed on page 29. It establishes the relationship between wavelength (or wave number), frequency, depth, and gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega = g k *tanh(kH)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is wave number, &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; is depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the celerity, or phase speed, is what is changing with depth all else being equal. This is a result of plugging the general solution of the wave equation back in to the kinematic boundary conditions. The physical explanation is that the particle motion implied by the velocity potential field varies with depth. For deep water, the particle orbitals are ever smaller circles as depth increases. For shallow water, the orbits are ellipses. These differently constrained motions resonate differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for wave groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Groups, or packets, can be regarded as the superposition of a continuous spectrum of waves from the above solution. The criteria is that they completely cancel out everywhere except for some bounded region. And within that region, they preserve certain features of the packet over time. So what you need to do is find the complete spectrum that does this. And of course the dispersion relation applies to each component in the spectrum. From the standpoint of understanding what is going on, the early attempts by Stokes, Rayleigh, and Osborne Reynolds considered two dispersive waves of infinitesimally different wavelength. That was in 1876-79, and it stuck for about a century, but it wasn't exactly the all singing, all dancing analysis that that was wanted, and it didn't treat the flux of the different energy components separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slowly, a more satisfying derivation took shape. It is fully consistent with the linear wave mechanics described above. See the first eight pages in the link below. This uses a bit different representation of the wave mechanics, but it is the same model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mit.edu/flowlab/NewmanBook/Tulin.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://web.mit.edu/flowlab/NewmanBook/Tulin.pdf&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;These results are consistent with the flux line diagrams, Figure 1, and quantify the net drift of kinetic&#xA;  energy relative to the wave which is apparent in the diagrams, increasing with decrease in depth.&#xA;  &lt;strong&gt;It is probably unnecessary to point out that these transport speeds do not correspond&#xA;  to the group velocity. They are in fact exactly twice the so called group velocity in the&#xA;  case of linear waves&lt;/strong&gt;  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;...&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For uniform waves, the so-called group speed does not therefore correspond to the&#xA;  actual speed of transport of any of the energy components, but to a weighted average of&#xA;  these speeds.&#xA;  It is noteworthy that dispersion does not seem to play a direct role in defining transport speeds in&#xA;  their relation to the phase speed of the wave. Rather, these speeds are finally a consequence of the&#xA;  distribution of energy within the wave structure, [1.45].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-11-15T02:47:21.380" LastActivityDate="2018-11-15T02:47:21.380" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24666" PostTypeId="1" CreationDate="2018-11-15T00:09:47.383" Score="0" ViewCount="34" Body="&lt;p&gt;I am designing a device that will measure the force of a breath through a tube. A small differential pressure sensor like this &lt;a href=&quot;https://www.digikey.ca/product-detail/en/MPX10DP/MPX10DP-ND/464046&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.digikey.ca/product-detail/en/MPX10DP/MPX10DP-ND/464046&lt;/a&gt; would be ideal but I believe that the humidity from the breath would impact the pressure reading. What kind of small, cheap pressure sensor could I use that would perform in a humid environment?&lt;/p&gt;&#xA;" OwnerUserId="18291" LastActivityDate="2018-11-15T13:39:03.760" Title="Breath Sensor: A low range pressure sensor that won't be affected by humidity" Tags="|pressure|sensors|electrical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24667" PostTypeId="1" AcceptedAnswerId="24673" CreationDate="2018-11-15T03:01:38.327" Score="0" ViewCount="629" Body="&lt;p&gt;I have both car and motorbike. Just common car and motorbike. My car is Toyota 2005, and my motorbike is Honda 2016, I bough both brand new. Both them are fuel injection. Before them, I owned car and motorbike, but both them were carburetor. With those old automotive, I have to warm the engine up until they reach their stationary. The requirement time will be around 5-10 minutes, depend on the weather and time. Morning in rainy season will be different to day time in dry season, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With my new automotive that are fuel injection, my habits still there, I still need to take that time to warm up the engine. But I read in many occasions in automotive news said that for fuel injection just required half and maximum one minute to warm up. For my car, since I start the engine up, it will go to around 1,300 RPM, and required around 7 minutes to become stationary, around 700 RPM. The stationary's RPM is clearly stated at the engine cap, with a yellow poster posted there. For the motorbike, I have no idea as no stationary indicator. But another car, Honda, I saw that it takes shorter time, and one Mazda car I saw just required around half to one minute to warm up, as said in the automotive news.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my understanding, warm up is intended to warm the engine up slowly to avoid the engine's temperature changes so quickly that of course is not good for the engine itself. So my question is, how that warming the engine could be so different? And what is actually that warming up intended? Several time I tested my car without proper warming up, the result is very bad, something like knocking in the engine.&lt;/p&gt;&#xA;" OwnerUserId="17943" LastEditorUserId="17943" LastEditDate="2018-11-15T07:14:42.993" LastActivityDate="2018-11-15T15:45:16.967" Title="What is the Meaning of To Warm Up an Engine?" Tags="|automotive-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24668" PostTypeId="2" ParentId="24667" CreationDate="2018-11-15T06:35:12.823" Score="3" Body="&lt;p&gt;Warming up an engine means making sure that the oil and water temperatures reach specified values before significant load is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are examples of engines that deliver full power from cold : emergency generators for example, but those engines are designed for that application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are engines that need to be warmed up properly before use such as engines used in racing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Car and bike engines are designed to provide power as soon as started, however most specify that gentle use is observed until it is up to temperature - gentle use for 10 minutes is better than 30 minutes sitting at idle...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the Mazda that you mention there are several possibilities, one is that the ecu is deciding the temperature that you see which has little relevance to the actual temperature of the cooling fluid or oil. This can be checked, if you wish, by fitting temperature sensors and getting the true temperature of the fluids at specific points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many cars and other motorized vehicles will have an initial high warm-up rpm for a while, this is usually now decided by the ecu and will be controlled for a smooth idle. My car, will idle at 1100 or so rpm for a minute or so when the outside temperature is -8 or -10 degrees C, then descending to its normal idle speed as the ecu decides.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-11-15T15:45:16.967" LastActivityDate="2018-11-15T15:45:16.967" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24669" PostTypeId="2" ParentId="24666" CreationDate="2018-11-15T06:53:15.233" Score="1" Body="&lt;p&gt;Put the sensor inside a membrane like a balloon so that it can detect pressure change without the humidity.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="33" LastEditDate="2018-11-15T13:39:03.760" LastActivityDate="2018-11-15T13:39:03.760" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24670" PostTypeId="1" CreationDate="2018-11-15T09:13:40.807" Score="0" ViewCount="525" Body="&lt;p&gt;How far can I drive an (average sized) electric bicycle with a 250 watt engine using three serial connected 12 V motorbike batteries?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to research the feasibility of this. I would like to serial connect three 12 V motorbike batteries and use them to run an electric bicycle. Maybe batteries similar to &lt;a href=&quot;https://www.ebay.co.uk/p/DYNAVOLT-Motorcycle-GEL-Upgrade-Mg5-3b-Battery-Mgcb5lb-Db5l-b-Yb5l-b/2118545330?iid=332402253231&quot; rel=&quot;nofollow noreferrer&quot;&gt;DYNAVOLT Motorcycle GEL Upgrade Mg5-3b Battery&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/90Xw9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/90Xw9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So how far would I be able to go with this solution? A rough estimate would be good too. Would it work in practice? I have seen a YouTube video with a guy powering his electric bicycle with one single 12 V battery so it should at least work to some extent.&lt;/p&gt;&#xA;" OwnerUserId="17642" LastEditorUserId="110" LastEditDate="2018-11-18T23:20:43.150" LastActivityDate="2019-05-19T00:02:10.387" Title="How far can I drive an electric bicycle with a 250 watt engine using three serial connected 12 V motorbike batteries?" Tags="|electric-vehicles|bicycles|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24671" PostTypeId="2" ParentId="24670" CreationDate="2018-11-15T09:35:50.363" Score="-1" Body="&lt;p&gt;How far you go depends upon your speed (assuming you're on a smooth, level road). Whitt and Wilson (page 186, 2nd ed) say 15 kcal/km at 16 km/h.  You'll travel 3.4 km per battery. (Would take about 38 min for 10.2 km)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that would more power than your 250 W motor can drive.  You'll go slower for about 45 min.&lt;/p&gt;&#xA;" OwnerUserId="16544" LastEditorUserId="16544" LastEditDate="2018-11-15T09:49:23.300" LastActivityDate="2018-11-15T09:49:23.300" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24672" PostTypeId="2" ParentId="24670" CreationDate="2018-11-15T10:21:51.183" Score="1" Body="&lt;p&gt;As a first approximation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those batteries are rated at 5Ah (more on this later).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At full power a 12V 250W motor will draw about 21A so on the face of it one battery will give you about 15 minutes of operation at full power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However standard lead acid vehicle batteries don't really like being used by this, they are designed to give short bursts of high current for starting and then be slowly recharged so a constant load doesn't do them much good nor does fully discharging them so you are never going to be able to use the full capacity and ideally you want batteries specifically deigned for deep cycle use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of estimating power requirement you are probably better off looking at data of the power output of cyclists rather than trying to estimate the actual loads on the bike. A professional cyclist can generate something like 200-300 Watts for hours on end so something like 150W+ is probably a reasonable working estimate for casual cycling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course as you add weight you will also need more power, especially in stop/start riding and climbing hills, at a constant speed on the flat weight makes much less difference.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of powered bicycles have a motor rated for the average expected load which is supplemented by pedal power for accelerating and climbing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be able to extend the range a bit with a regenerative braking system but again that comes with a cost of added mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Estimating range is more difficult but this calculator may help obviously that is aimed at competitive cycling but it is at least real world data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cyclingpowerlab.com/CyclingPowerOutput.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cyclingpowerlab.com/CyclingPowerOutput.aspx&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-11-15T21:12:51.683" LastActivityDate="2018-11-15T21:12:51.683" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24673" PostTypeId="2" ParentId="24667" CreationDate="2018-11-15T10:33:53.110" Score="3" Body="&lt;p&gt;The specific issue with carburettors are that they are purely mechanical devices and can't directly measure the mass flow rate of fuel they are delivering and as such are quite sensitive to temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carburettor engines often have a 'choke', either automatic or manual  which overrides the normal operation of the carburettor, causing the engine to run rich. This improves starting and warm up but isn't good under load so you need to wait untill the choke turns itself off before you use the engine hard. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With fuel injection the ecu has direct control over the air/fuel mix and can adjust it on the fly according to conditions so all the warm up need to do is get the oil warm enough to work and it doesn't need to worry about getting the flue system in thermal equilibrium as it can adjust air fuel ratio directly as required. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-11-15T10:33:53.110" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24675" PostTypeId="1" CreationDate="2018-11-15T11:25:32.210" Score="1" ViewCount="87" Body="&lt;p&gt;I would like to know the algorithm of switching between 4g/3g/2g. When will the phone switch from 4G to 3G or from 3G to 2G, when we select for corresponding option in the mobile phone. Is it based on dBm/RSCP etc? I am looking for some technical document. TY&lt;/p&gt;&#xA;" OwnerUserId="18294" LastEditorUserId="7083" LastEditDate="2019-08-07T20:32:35.620" LastActivityDate="2019-08-07T20:32:35.620" Title="How do mobile phones switch 4G/3G/2G?" Tags="|electrical-engineering|signal|communication|lte|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24676" PostTypeId="1" CreationDate="2018-11-15T12:45:04.283" Score="0" ViewCount="81" Body="&lt;p&gt;I'm trying to implement a new material in our processes, and that's a 2 part, highly viscous (500-1000 Pa s) liquid silicone rubber. I will also be adding a pigment, so that's 3 components that need to be mixed before entering the mold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have very little experience with injection molding, and this part has me stuck in my thought process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These will be low volume production. Maybe 5-10 runs per shot. Sometimes even 1 shot. Each shot in the range of 10-20cc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the three parts enter the static mixer, the ticker starts. at room temperature this has a pot life for about a day, but at temperatures around 100C, the cure time is in the range of 5 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I go about clearing out the injection nozzle and static mixer of the messy, thick, sticky, virtually insoluble in anything LSR so that the parts are not jammed unusable for my next use (which could be in a week)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entire process and set up is still in the concept stage, I'm open for a radical rethinking, but the material choice remains fixed. &lt;/p&gt;&#xA;" OwnerUserId="18295" LastActivityDate="2018-11-15T16:29:31.113" Title="Injection nozzles and static mixers for LSR - How do i prevent an uncleanable blockage between runs?" Tags="|molding|injection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24677" PostTypeId="2" ParentId="24676" CreationDate="2018-11-15T16:29:31.113" Score="1" Body="&lt;p&gt;Consider a &quot;one use&quot; plastic syringe / mixer nozzle with a fixed volume of material ie enough for 5 shots or 10...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then at the end of the run the plastic unit can be removed and replaced with a fresh unit of whatever mix is needed - either the same or a new mix.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-11-15T16:29:31.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24678" PostTypeId="1" CreationDate="2018-11-15T17:29:42.380" Score="1" ViewCount="295" Body="&lt;p&gt;I'm a nanoengineering student who's been stuck on the following problem for a few days. I've gone back over my professors slides, watched podcasts from an old professor, etc. etc., but don't understand it any more than I did when it first came up. All I know is that it deals with the BET Isotherm, which there isn't a tag for...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the specific surface area of the mesoporous sample described below.&#xA;Assume the effective cross-sectional area of one N&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; molecule is 0.18 nm&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;, and 1 mole of gas occupies 20 L at standard temperature and pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm then given the following graph and table:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TNfG6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TNfG6.jpg&quot; alt=&quot;A picture graphing the relative pressure to volume of adsorbed gas at STP&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not 100% sure where to start, so I just have the following equations from my notes written down:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;slope (m) = &lt;span class=&quot;math-container&quot;&gt;$\frac{c-1}{cN_m}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;y-intercept (b) = &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{cN_m}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$c-1=\frac{m}{b}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$c=\frac{m}{b}+1=e^{\frac{\delta H_{des}-\delta H_{vap}}{RT}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Area (A) = &lt;span class=&quot;math-container&quot;&gt;$N_m*N_a*A_0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A/m=$&lt;/span&gt; specific surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where Na is avogadro's number, A0 is the N2's surface area, and m is the mass of the nanoparticles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much for your help!&lt;/p&gt;&#xA;" OwnerUserId="18297" LastActivityDate="2024-03-20T01:03:10.413" Title="Calculate the specific surface area of a mesoporous sample" Tags="|chemical-engineering|experimental-physics|chemistry|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24679" PostTypeId="1" CreationDate="2018-11-15T19:20:07.380" Score="1" ViewCount="59" Body="&lt;p&gt;I have been trying to figure out how to calculate the required dimensions for liquid phase gas through a solenoid valve. Many companies (Swagelok, Bürkert, Asco) provide &quot;Flow coefficient calculators&quot; where you can enter values for the inlet pressure, outlet pressure and pick the medium flowing through the valve. There is no option to choose whether the gas flowing through the valve is in liquid phase or not though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would one go about finding the required size of a valve, if you know the pressure drop, required mass flow / volume flow and the medium flowing through it? Would you assume the flow to be just any incompressible fluid (i.e. water) and use &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Flow \ coefficient = Volume \ flow \ \times \sqrt \frac{specific \ gravity}{dP} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when would you have to be concerned about cavitation? &lt;/p&gt;&#xA;" OwnerUserId="18300" LastEditorDisplayName="user14407" LastEditDate="2018-11-16T13:49:20.520" LastActivityDate="2018-11-16T13:49:20.520" Title="Liquid phase gas flow through valve" Tags="|fluid-mechanics|gas|valves|liquid|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24680" PostTypeId="1" CreationDate="2018-11-15T19:39:45.643" Score="2" ViewCount="110" Body="&lt;p&gt;What method would be appropriate to analyze this concrete structure? &#xA;A two-way rectangular concrete slab with walls and central column. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to understand how to model it with either yield line method or equivalent frame method. Or both to be honest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm studying those methods at introductory level, but haven't found any example of a similar case, so unsure how to draw out the analytical models. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0rNYJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0rNYJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18299" LastEditorUserId="18299" LastEditDate="2018-11-15T19:47:01.557" LastActivityDate="2018-11-15T19:47:01.557" Title="Analysis method for two-way slab floor with central column" Tags="|structural-analysis|reinforced-concrete|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24681" PostTypeId="2" ParentId="3616" CreationDate="2018-11-15T20:52:34.097" Score="2" Body="&lt;p&gt;There are pretty good answers but I feel like non of them are complete answers. &#xA;99% of electric pumps are Induction motor. So for rest of the discussion I will be referring to induction motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As &lt;strong&gt;@Fred&lt;/strong&gt; said above yes, electric motor will eventually burn out. But as &lt;strong&gt;@ Olin Lathrop&lt;/strong&gt; said yes, it will take lower power while running at high speed (under no load conditions). &lt;strong&gt;Then the Question is how can an induction motor burn out while consuming low power ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Culprit is rotational friction load and properties of wire. On just starting electric motor, rotor speed keeps on increasing because there is no load/water. At rotating high speeds bearings will fail much quicker than expected leading to rapid increase in rotational friction. So load keep increases and motor speed slows down until if finally stops but power consumption keeps on increasing. Leading to higher currents. Higher the currents and higher the wire heating. Higher the wire heating, higher resistance again higher heating. At some point in time and some point on copper wire, copper might get hot enough to react with oxygen or water and we call it burn out. So its both mechanical properties of bearing and electrical properties of the wire leading eventual failure of induction motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worth to know how a system fails so that worst situations can be avoided in design level. In above induction motor failure, its an series of events happening. If we could stop at least one event, we can save the motor from burning out.&lt;/p&gt;&#xA;" OwnerUserId="16175" LastActivityDate="2018-11-15T20:52:34.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24683" PostTypeId="1" AcceptedAnswerId="24687" CreationDate="2018-11-16T12:11:30.283" Score="0" ViewCount="30" Body="&lt;p&gt;So this is very general, but... How are simulations for vehicles done? I know that computational fluid dynamic simulations are used, but why are repeated tests of actual hardware done too? Obviously this is because there are some things that are found through actual physical testing. So this leads into my question. Are simulations refined from the actual test data? how are phenomena from the actual test translated into simulation? Eyeballing things? just gathering data, making a hypothesis, and then running the simulation again from the top?&#xA;I guess what my question ultimately boils down to is: What does a simulation cycle look like?&lt;/p&gt;&#xA;" OwnerUserId="18305" LastActivityDate="2018-11-16T13:47:48.810" Title="What is the cycle of simulation editing?" Tags="|aerospace-engineering|simulation|aerodynamics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24686" PostTypeId="2" ParentId="24238" CreationDate="2018-11-16T13:08:01.017" Score="1" Body="&lt;p&gt;You can call that style of mounting similar to a &quot;bulkhead&quot; fitting, or through-the wall. The hole does not need to be tapped, just a clearance hole for the fitting to pass through, the nut then goes on the other side of the wall, holding the fitting in the wall. Per the drawing, if the wall is thicker than 1.25&quot;, then you will have to tap the wall to hold the fitting. If that's the case, you will have to contact the MFG for the thread size as its not annotated on the drawing. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-11-16T13:08:01.017" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24687" PostTypeId="2" ParentId="24683" CreationDate="2018-11-16T13:41:06.557" Score="1" Body="&lt;p&gt;That is a very broad question. It can be as &quot;simple&quot; as visually comparing the simulation and test results, and deciding on the basis of experience how the tweak the simulation if it isn't accurate enough (whatever &quot;accurate&quot; means, in a particular situation). After the model has been changed to agree with the test data, you can then use it to simulate other conditions which were not tested, with more confidence that the results will be meaningful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the other extreme, this may be treated as a mathematical optimisation problem. You set up a model with some parameters (geometry, material properties, etc) that can be varied, tell the software what the &quot;answers&quot; should be (as measured in the tests) and the software then runs the model repeatedly, changing the parameters to get the &quot;best&quot; agreement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with a fully automatic approach is that the initial model may be completely unrealistic (for example because some important feature was left out altogether, either by mistake, or because nobody thought it was important enough to be worth including!) but the software managed to make the answers line up with the test data by using unrealistic values for the parameters, and getting to a situation where several &quot;large errors&quot; cancelled out but the final results &quot;looked about right&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is fairly routine to use &quot;automatic model generation&quot; in a more limited way. For example in finite element analysis, you might define the geometry of the component, and the software automatically generates the finite element mesh, and then refines it by subdividing some of the elements in regions where the numerical solution hasn't converged, according to some predefined measure of &quot;converged&quot;. This is usually much quicker (and more accurate) than generating the complete mesh by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of work involved in validating models can cover a huge range, from one engineer working for part of a day up to several person-years of work for a large, complex, and critically important model.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-11-16T13:47:48.810" LastActivityDate="2018-11-16T13:47:48.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24688" PostTypeId="2" ParentId="24678" CreationDate="2018-11-16T14:36:01.430" Score="0" Body="&lt;h1&gt;Basic Approach&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The notes that you have are based on the first step as taking a linearization of the BET equation. This &lt;a href=&quot;https://en.wikipedia.org/wiki/BET_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; gives an explanation of the first step. You have the data plot. You can extract values of &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$p/p_o$&lt;/span&gt; from it. Fortunately, you also have a table. The table has extracted points for you. Fit the data to obtain the values of &lt;span class=&quot;math-container&quot;&gt;$v_m$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second step is to recognize the units on the term &lt;span class=&quot;math-container&quot;&gt;$v_m$&lt;/span&gt;. This is the amount (volume) of gas adsorbed as a monolayer. It is the area covered times the &quot;depth&quot; of the monolayer. The area covered is the number of gas particles times the area per particle. The depth is the diameter of the molecule.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Better Approach&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;To illustrate the better approach, I first coded the BET equation using python in Jupyter to create this plot of &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$p/p_o$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yMTLd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yMTLd.png&quot; alt=&quot;BET Isotherm Plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are really serious about doing this the right way, you will take the full data curve and fit it to a non-linearized form of the BET equation. This will give you &lt;span class=&quot;math-container&quot;&gt;$v_m \pm \Delta v_m$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$c \pm \Delta c$&lt;/span&gt;. Given just the graph, I might use a data capture application (DataThief or IgorThief) to digitize the curve and then fit the data.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-11-16T14:36:01.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24689" PostTypeId="1" AcceptedAnswerId="24692" CreationDate="2018-11-16T17:51:06.060" Score="3" ViewCount="1243" Body="&lt;p&gt;We shared &lt;strong&gt;Mechanics of materials&lt;/strong&gt; with structural engineers back in college. I learned some basic concept about structural engineering, but besides the basic knowledge i don't know anything about structural engineering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If i recall correctly, an engineer found a way to distribute the weight of buildings over a wide area to reduce the stress, i think he managed to solve the problem of sinking buildings for the first time in Chicago, and up to this moment, his solution has been popular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the area has a strong bedrock, manhattan for instance then, the buildings are much more stable, but my question targets the areas with very weak bedrock, The Hague for instance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can say, there is channel is every street, when i walk in streets i can actually hear the vibrations of water beneath my feet, if you dig the ground four or five meters, in city centre you'll reach water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder how engineers manage to build skyscrapers in this area? &#xA;The concentration of tall buildings is not as same as in manhattan, Chicago or ... . But still, how those few tall buildings are still on the ground?   &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-08-04T04:32:07.573" Title="Why skyscrapers don't sink into ground?" Tags="|structural-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24691" PostTypeId="1" AcceptedAnswerId="24693" CreationDate="2018-11-16T19:55:46.267" Score="1" ViewCount="119" Body="&lt;p&gt;My cousin is a mechanical engineer, got contacted today from someone who claimed to be working for the supreme board for international arbitrators (&lt;a href=&quot;http://sbia.org.uk&quot; rel=&quot;nofollow noreferrer&quot;&gt;sbia&lt;/a&gt;) and &lt;a href=&quot;http://www.icac-lebanon.com/lb/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ICAC&lt;/a&gt; in the Middle East and north Africa region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;She told him he would have to pay $300 in order to get a 1 day course, or workshop, afterwords he would become an international arbitrator for engineering issues related to his field. Full sized images &lt;a href=&quot;https://i.stack.imgur.com/Oa3JJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://i.stack.imgur.com/atM1d.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Oa3JJm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Oa3JJm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/atM1dm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/atM1dm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My cousin and I think this is a scam, and their website is bad and there's no mention about SBIA anywhere, no results on google, no social media presence. I would like to ask you about it though, have you ever heard about it? Is this a scam?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think they are copycatting this &lt;a href=&quot;http://www.incodir.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.incodir.org/&lt;/a&gt;, not only that, I hope you could read arabic, on the first image, it says in parentheses &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://people.bayt.com/reem-alfanek/?lang=ar&quot; rel=&quot;nofollow noreferrer&quot;&gt;ريم الفانك&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which also works in the organization I mentioned, I'm going to try to contact her somehow, either she's being impersonated in which case I'll see if we can press charges, otherwise she's part of the scam, I hope not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried contacting president@incodir.org, the email didn't get delivered to report the case, well then I must conclude that they too are a part of the scam sadly, the rabbit hole goes deep.&lt;/p&gt;&#xA;" OwnerUserId="18308" LastEditorUserId="18308" LastEditDate="2018-11-16T23:10:09.713" LastActivityDate="2018-12-16T23:58:59.053" Title="Is there such thing as supreme board for international arbitrators?" Tags="|mechanical-engineering|electrical-engineering|civil-engineering|international|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="24692" PostTypeId="2" ParentId="24689" CreationDate="2018-11-16T21:00:49.163" Score="2" Body="&lt;p&gt;As part of the process of obtaining a building permit is the requirement to investigate the soil strength and competency and possible elevated subterranean water levels by a licensed geotechnical engineer and if needed in some cases by a multitude of engineering specialists' investigation, such as geology, seismology, and any other concerns pertinent to that site investigation as determined by the building department authority.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geotechnical engineers, after studying the site and tabulating logs of test pits and doing lab test on the excavated samples, submit a report with specific recommendations for construction of that particular building on that site including modifications to the soil by additional foreign materials, borehole drainage among other things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The structural engineer following geotechnical recommendations and using their approved tables of such things as allowable bearing, shear strength, passive and active soil pressure, seismic and dynamic over loads, and if there is need of use of mat foundation or deep foundation supported on piles or need of subterranean drainage (boreholes) to drain moisture of soil to acceptable level, design the building closely following the geotechnical engineer's report. Then the structural engineers send their design to geotechnical engineer and if he approves it they submit it to the building authority to get the permits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In cases of weak soil or subterranean water, in some cases they recommend building a big subterranean concrete cube a little bigger than the entire subterranean parking and foundation of the building, like a giant dry swimming pool and water proof it and install emergency sump pumps or alternative means of drainage, then they start to build the foundation inside that giant concrete clad excavation. This is similar to the way bridges are built in rivers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some other soft incompetent soils the geotechnical engineers may recommend building a system of deep piles tied and integrated with each other, called soldier piles, and then build the building on top of it.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="16486" LastEditDate="2019-08-04T04:32:07.573" LastActivityDate="2019-08-04T04:32:07.573" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24693" PostTypeId="2" ParentId="24691" CreationDate="2018-11-16T21:17:11.797" Score="3" Body="&lt;p&gt;They can't even write English on their certificate. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... training of good stand, and is in accord with American and&#xA;  European standers...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's an obvious typo. Of course it should say&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... training of good sit, and is in accord with American and&#xA;  European sitters...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Is it worth paying $300 for a joke certificate? That depends on the buyer's sense of humour, I suppose.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-11-16T21:17:11.797" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24694" PostTypeId="2" ParentId="24689" CreationDate="2018-11-17T00:19:51.430" Score="2" Body="&lt;p&gt;You don't &lt;em&gt;need&lt;/em&gt; any modern technology to succeed. The tower of the tallest medieval building in Britain, the main church in Boston, was completed in 1520. It is 83m (280 feet) tall, and built on ground which is almost pure silt, and only 10 meters away from a tidal river. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the difficult soil conditions, digging the deep foundations was &quot;sub-contracted&quot; to coal miners from a different part of the country. The building has never needed any structural reinforcements in the 500 years it has been standing, and the top of the tower is only 5mm away from being perfectly vertical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/St_Botolph%27s_Church,_Boston#/media/File:St_Botolph%27s_Boston_ext.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/St_Botolph%27s_Church,_Boston#/media/File:St_Botolph%27s_Boston_ext.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-11-17T00:19:51.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24695" PostTypeId="2" ParentId="24004" CreationDate="2018-11-17T01:38:28.410" Score="1" Body="&lt;p&gt;There is a simple formula to determine if your planetary geartrain will assemble:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 3 planets, (R + S) / 3 must be an integer.&#xA;For n planets, (R + S) / n must be an integer.&#xA;This assumes that the planets are equally spaced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I found this in Dudley's &quot;Gear Handbook&quot; long ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The number of teeth on the planets does not enter into the formula.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;++++&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding more:&#xA;By Dudley's formula above, the gearset with R=47, P=16, S=15 should not assemble because (47+15)/3 = 20.667, which is not an integer.  But these values can be plugged into  &lt;a href=&quot;http://www.thecatalystis.com/gears/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.thecatalystis.com/gears/&lt;/a&gt;, and the picture looks like it assembles just fine:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GfId6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GfId6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you carefully check, the planets are not equally spaced.  In the picture, count the ring gear spaces from Planet A to Planet B, then B to C, then C to A.  You get 16, 16, and 15 spaces.  Not trying to bash 'thecatalystis.com'.  If you allow non-equally spaced planets, then any combination with the correct diameters will assemble.  Normally the planets are equally spaced to equalize the forces and wear on the planets.  Dudley's formula works for equally spaced planets.&lt;/p&gt;&#xA;" OwnerUserId="18312" LastEditorUserId="18312" LastEditDate="2018-11-17T22:34:22.630" LastActivityDate="2018-11-17T22:34:22.630" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24696" PostTypeId="1" CreationDate="2018-11-17T02:23:47.530" Score="1" ViewCount="156" Body="&lt;p&gt;During studying engine's exhaust as I asked &lt;a href=&quot;https://engineering.stackexchange.com/questions/24609/how-to-calculate-the-output-power-of-engine-exhaust&quot;&gt;here&lt;/a&gt;, I have to go far. The thing I though is quite simple, the fact is complicated. Thence I knew that there is water produced in combustion process. The combustion formula is: &lt;span class=&quot;math-container&quot;&gt;$C_8H_{18}+12.5O_2\rightarrow8CO_2+9H_2O$&lt;/span&gt;. Mean, every 1 of 4 strokes of 4-strokes engine will produce water (cyclic process: intake-compress-power-exhaust). This is the answer for me why the muffler always tear-dropping water especially in the morning or when the weather is cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, that happened if the car is still good enough.  Before my current car, my old car seems didn't have that kind of tear-drop water from the muffler. Not sure what had happened to the engine. The muffler was dry and clean, no smoke. So my question is, where the water have gone? Which one is better, a car with water tear-dropping from the muffler? Or a car without that water? My question is for the same ca or closed in it piston's displacement and its power.&lt;/p&gt;&#xA;" OwnerUserId="17943" LastEditorDisplayName="user14407" LastEditDate="2018-11-17T12:40:29.437" LastActivityDate="2018-11-17T19:22:51.187" Title="Where the Water Have Gone from Engine from a Combustion Process (C8H18+12.5O2==&gt;8CO2+9H2O)?" Tags="|automotive-engineering|combustion|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24698" PostTypeId="2" ParentId="24696" CreationDate="2018-11-17T06:29:09.310" Score="1" Body="&lt;p&gt;Anytime your car's engine is running, it is producing water vapor which leaves the engine through the exhaust system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the engine has just been started and the exhaust system in your car is cold, the water vapor in the exhaust condenses into liquid water and trickles out the tailpipe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the engine has been running for a while and the exhaust system is hot, the water vapor remains vapor as it flows out the pipes and exits the tailpipe as invisible vapor. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-11-17T06:29:09.310" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24702" PostTypeId="1" AcceptedAnswerId="24705" CreationDate="2018-11-17T13:44:53.293" Score="2" ViewCount="2304" Body="&lt;p&gt;I have shared the question statement, its diagram and my solution but my solution is wrong and I have no idea why is it wrong, please help me out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E5fCl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E5fCl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0dd2C.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0dd2C.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18319" LastEditorUserId="1832" LastEditDate="2018-11-18T19:21:20.793" LastActivityDate="2019-01-07T18:50:36.047" Title="Statics Question" Tags="|statics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24703" PostTypeId="1" CreationDate="2018-11-17T14:27:54.047" Score="3" ViewCount="1434" Body="&lt;p&gt;I am currently pulling a load with a wire rope that is moved through a system of sheaves, but had to replace a smaller wire rope cable (1/2&quot;) with a larger wire rope cable (3/4&quot;). The sheave/wire diameter ratio is below the recommended guidelines, which I understand affect fatigue life and strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have also noticed that the motor seems to work with a lot more effort than it did before, so I wanted to compare the frictional losses of the smaller cable with those of the larger cable to see if the diameter is making the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any formula, perhaps empirically-derived, or a table that could gauge the relative frictional losses between these two systems?&lt;/p&gt;&#xA;" OwnerUserId="12823" LastActivityDate="2018-11-17T14:27:54.047" Title="How do you calculate the frictional losses in wire ropes over sheaves?" Tags="|mechanical-engineering|pulleys|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24705" PostTypeId="2" ParentId="24702" CreationDate="2018-11-17T17:45:38.580" Score="1" Body="&lt;p&gt;you should think first about the whole structure, and then look into each joint. Apply the sum of the forces on the horizontal, the sum of the momentum and vertical forces must all be equal to zero to the structure as an object, similar to what you have been doing all the course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have found a mistake on your solution since you didn't consider the first gender support in &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;. Correct that and try to express, in the equations of ballance, expressions for the reactions relative to &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; from the equations of equilibrium of the whole structure. This will result in a problem which will be in function of &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;. In this step &lt;span class=&quot;math-container&quot;&gt;$R_a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_e$&lt;/span&gt; will be, if my calculations are correct, &lt;span class=&quot;math-container&quot;&gt;$3P$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$2P$&lt;/span&gt; in modulo, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, and only then, go to the joints and do their ballance of equilibrium. As it is all in function of &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; you then separate those suffering compression from tension. Among those suffering compression, which is taking the higher force? Equal it to &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt;kN. Among those suffering tension, which is taking the higher force? Equal it to &lt;span class=&quot;math-container&quot;&gt;$8$&lt;/span&gt;kN. The smallest of the two is the correct value of &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;, since the highest will cause a violation in the other restriction.&lt;/p&gt;&#xA;" OwnerUserId="18320" LastEditorDisplayName="user14407" LastEditDate="2018-11-17T17:56:47.100" LastActivityDate="2018-11-17T17:56:47.100" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24706" PostTypeId="1" CreationDate="2018-11-17T18:15:11.847" Score="0" ViewCount="588" Body="&lt;p&gt;I'm trying to use ANSYS to solve statically a dynamic problem.&#xA;As shown in the image below, there is an orange bar that is constraint to slide horizontally only and attached to another bar (pink) through a pin joint. When the orange bar is subjected to a momentary external force, it accelerates horizontally and as a result the pink bar behaves as a pendulum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9PrtM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9PrtM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I modeled the case in ADAMS to get all the needed kinematics/dynamics info (accelerations, pin joint reactions, angle Vs time and etc.) and now would like to use ANSYS in order to obtain the stress and deflection fields. I am familiar with the flex body option in ADAMS and actually experienced with it a little, but would like to analyze the stress with a tool I feel more comfortable with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as I understand, in cases by which the rigid body is actually moving - there is an option called Inertia relief. The general idea behind it is to &quot;remove&quot; the loads causing a motion in order to solve a static problem and compute the stress acting on the body. In this case, if I use a pin joint in ANSYS and apply horizontal acceleration on the pink bar - the all body would move. Alternatively, I can fix the bar to the pin joint in all 6 DOF but then I'll get an artificial and non-realistic bending moment (this moment generates the motion and does not bend the bar in reality).&#xA;However, when using this Inertia relief option, all the support reactions are zeroed - and this should not be the case here. I need to fix the motion in X and Y directions in the pin joint while allowing the bar to rotate along the Z axis and solve the problem statically. &#xA;I think to not use the Inertia relief option and alternately add linear acceleration (D'alembert style) but I'm not sure about it.&#xA;Any ideas or thought?&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2024-03-16T17:04:00.533" Title="Inertia relief in ANSYS - rotating model" Tags="|statics|dynamics|ansys|kinematics|inertia-matrix|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24707" PostTypeId="1" CreationDate="2018-11-17T19:02:51.080" Score="1" ViewCount="103" Body="&lt;p&gt;I am currently designing a replacement bracket for AM4 CPU coolers that will be compatible with different cooler models and allow them to be installed facing different directions, as AM4 socket is non-symmetrical, and the current design looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/23FZr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/23FZr.png&quot; alt=&quot;Mounting bracket&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This a draft and I want to use FEM to reduce the amount of metal covering, so more air could get through and also cool the VRM's and other chips in the CPU vicinity, but to do that I need to know what is the maximum force I need to design for. I think calculations using just the mass of the cooler would not be accurate, as I also have to take into account the force exerted by screws that are used to hold heatsink in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tl;dr: how much force should I account for when designing mounting bracket for CPU cooler?&lt;/p&gt;&#xA;" OwnerUserId="18323" LastActivityDate="2018-11-17T19:02:51.080" Title="Force exerted on the CPU socket by heatsink" Tags="|design|finite-element-method|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24708" PostTypeId="2" ParentId="24696" CreationDate="2018-11-17T19:22:51.187" Score="0" Body="&lt;p&gt;I think the answer to what you are asking is that, normally the water form combustion is produced as a gas as, inevitably it is at a temperature well above the boiling point of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it is a gas it is invisible and you will only see it as drops of liquid is it cools down enough to condense. whether or not this occurs depends on a variety of factors including the weather conditions, the layout of the exhaust and how hot the exhaust gases are when they enter the exhaust system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of modern cars need to run the exhausts fairly hot in order for catalytic converters and particle filters to work properly and will often adjust the air/fuel ratio to ensure that the exhaust heats up quickly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On possible difference between cars is where the catalytic converter is located. Being a fiarly hot component they need to be puts somewhere away from anything heat sensitive &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-11-17T19:22:51.187" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24709" PostTypeId="1" CreationDate="2018-11-17T21:45:39.113" Score="3" ViewCount="103" Body="&lt;p&gt;About 6 months ago I have poured concrete in a plastic mould. The mould was plastic plinth for sun umbrella. It is usually filled with water but it cracked so I filled it with concrete. Before pouring I glued the cracks with generic cyanoacrylate glue..&#xA;I prepared concrete mix with coarse and fine sand, cement and PP fibers. I was aiming for 0.3 water to cement ratio. I added a superplasticizer based on 2-octyl-2H-isothiasol-3-on. The plasticizer was dark purple fluid with distinct smell.I used quite a lot of it, somewhere near the max of recommended dose.&#xA;The concrete stayed soft for quite a while and I thought it would not harden at all. After a week it became warm and solidified. The surface at the opening where I poured it in (the top) was soft to the touch for a few weeks. You could scratch it with a fingernail. But now is rock hard.&#xA;What I have noticed after a few weeks was dark purple excretion under the glued cracks. And I think it even softened the glue. It looked purple like plasticizer, smelled like plasticizer and had a stingy taste. The concrete itself is of normal color.&#xA;Now, even after a few months a purple excretion leaches out of cracks in plastic.&#xA;Is this really plasticizer leaching out? Is this normal? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things to consider:&lt;br&gt;&#xA;Final w/c somewhere around 0.3.&lt;br&gt;&#xA;Max recommended dosage of plasticer.&lt;br&gt;&#xA;On top of the mold is an empty cavity which I left filled with water to keep the concrete moist during curing. The surface of concrete underneath is hard.&lt;/p&gt;&#xA;" OwnerUserId="18322" LastActivityDate="2018-12-18T09:01:35.297" Title="Can plasticizer leach out of cured concrete?" Tags="|concrete|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24711" PostTypeId="1" CreationDate="2018-11-17T23:50:08.650" Score="0" ViewCount="44" Body="&lt;p&gt;Recently the O-wind turbine won the James Dyson award for innovation, the item is essentially a compact wind turbine which unlike traditional wind turbines can harness wind from multiple directions, the wind is converted to kinetic energy to then generate electrical energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I am struggling to understand how this works as a concept. From what I've read the spherical shape is fitted with vents which has a large entrance hole and small exit holes. I am unsure of how this could initiate the movement of the object in the presence of wind. &lt;/p&gt;&#xA;" OwnerUserId="18324" LastActivityDate="2018-12-18T03:02:50.447" Title="Qualitatively how does the O-wind turbine function?" Tags="|renewable-energy|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24713" PostTypeId="2" ParentId="24711" CreationDate="2018-11-18T02:53:36.053" Score="1" Body="&lt;p&gt;Taken directly from the &lt;a href=&quot;https://www.jamesdysonaward.org/2018/project/o-wind-turbine/&quot; rel=&quot;nofollow noreferrer&quot;&gt;James Dyson Award&lt;/a&gt; website.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How it Works&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The turbine is of a spherical shape with a single axis of rotation going through it. Its dimensions and shape mean that it is very suitable for small-scale energy production by individual apartment dwellers e.g. by being fixed outside balconies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The turbine makes use of Bernoulli’s principle for its mechanical motion. The structure is lined up with vents which have large entrances and smaller exits for air. In the presence of wind, there is a pressure difference between the two terminals causing the turbine to move. The vents are placed all across the sphere making it receptive to wind from all directions in both the vertical and horizontal planes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The turbine will rotate in the same sense about a fixed axis regardless of wind direction. This turbine rotation is used to power a generator that can produce electricity, which can be fed into the national grid, hence providing financial incentive to users and improving the region’s sustainable energy production. &quot;&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-11-18T02:53:36.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24714" PostTypeId="1" CreationDate="2018-11-18T05:01:59.937" Score="0" ViewCount="171" Body="&lt;p&gt;I'm programming the controller for a drone and I've got a question that may be basic, but I need to understand this perfectly to continue with the most interesting part of the project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensors of my drone give me the position (x,y,z) and orientation (quaternion: x,y,z,w) of the drone, and I need to control the altitude of the drone, so I need to control the velocity in the z coordinate because that velocity is send to a topic of my drone (for those who use ROS, to the cmd_vel topic) and changing the velocity I can control the altitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, basically the input of the controller is the desired position of the drone, then I have a PID controller and the plant (unknown model, so I have to &quot;guess&quot; the parameters of the PID), and the feedback is the current position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which would be the output of the controller (I mean, the &lt;span class=&quot;math-container&quot;&gt;$u_k$&lt;/span&gt;, with the discrete model of the PID that is the input of the plant)? Is it a position or a velocity? If it is a position, then I can do something like: (position_current - position_before)/timestep. But I don't know what is theoretically correct, because maybe I can find some &lt;span class=&quot;math-container&quot;&gt;$k_p$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$k_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$k_d$&lt;/span&gt; to have the desired velocity as the output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks :D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: this is basically the model I'm using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dJa54.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dJa54.png&quot; alt=&quot;Controller model&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8089" LastEditorUserId="8089" LastEditDate="2018-11-18T18:59:37.297" LastActivityDate="2024-03-22T07:06:55.843" Title="Control: control action of a PID controller with feedback" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24715" PostTypeId="2" ParentId="24709" CreationDate="2018-11-18T05:36:22.793" Score="1" Body="&lt;p&gt;There are concrete additives to make concrete set under water. Yes, plasticiser can leach out of concrete as you have seen - one trick to help concrete “flow” was a squirt of washing up liquid but it only needs a small amount.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-11-18T05:36:22.793" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24717" PostTypeId="1" AcceptedAnswerId="24746" CreationDate="2018-11-18T11:32:43.850" Score="2" ViewCount="884" Body="&lt;p&gt;I am curious about the design considerations for a hydraulic system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you have a 3HP 1200RPM motor&#xA;you also have a hydraulic pump that's rated 4LPM and pressure of 40MPA rated for 40RPM MAX&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can put many different types of gear box on the 3HP motor to get it to operate in the N-40RPM range&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's forget about the power loss through heat when using the gearbox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand the relationship between HP and RPM of the electric motor, Pressure of the system and output Torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I am working at 200Bar, electric motor doing 40RPM and the hydraulic motor running at 4LPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would that compare to say going to 20RPM and keeping 200Bar and 4LPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do understand that raising the pressure would increase the heat and possibly require cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions on how to calculate these relationships?&lt;/p&gt;&#xA;" OwnerUserId="12586" LastActivityDate="2020-01-03T19:02:04.353" Title="Calculate hydraulic torque from motor HP, hydraulic pump and gearbox" Tags="|mechanical-engineering|fluid-mechanics|torque|hydraulics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24718" PostTypeId="1" CreationDate="2018-11-18T11:53:37.013" Score="0" ViewCount="207" Body="&lt;p&gt;I would like to mount a sensor onto a shaft which can be translated axially as well rotated independently. This arrangement permits me to capture a 360 deg scan and then translate the shaft axially to scan at a different height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should one use a lead-screw and stepper motor coupled with some other mechanism to achieve this (non-helical) motion ? Thanks for any tip.&lt;/p&gt;&#xA;" OwnerUserId="18327" LastActivityDate="2018-11-18T11:53:37.013" Title="Shaft which can independently rotate and translate axially" Tags="|mechanical-engineering|linear-motion|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24719" PostTypeId="1" CreationDate="2018-11-18T13:19:45.287" Score="2" ViewCount="1137" Body="&lt;p&gt;Attached is an example picture of a clamp that will be attached to a pipe made of Aluminium 6061-T4. The clamp should have very tight grip on the pipe and prevent it from slipping or rotating but the Aluminium pipe is fragile and so the clamp cannot be tightened too much. Thats why I am looking for a material for the inner surface of the clamp that will provide a firm grip on the Al pipe without requiring too much tightening force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clamp-pipe interface material (shown here as the black rubber-like thing) should have a very high coefficient of static friction with Al6061 in order to provive a tight grip. &lt;strong&gt;I want to know what materials are best suited for this purpose? Which material has the highest coefficient of static friction with Al6061?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My research so far:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The best material I could find is &lt;strong&gt;rubber&lt;/strong&gt; that supposedly has a coeff of static friction of &lt;strong&gt;0.8&lt;/strong&gt; with Aluminium. (source: &lt;a href=&quot;http://atc.sjf.stuba.sk/files/mechanika_vms_ADAMS/Contact_Table.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://atc.sjf.stuba.sk/files/mechanika_vms_ADAMS/Contact_Table.pdf&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Surprisingly the coeff of static friction between Aluminium-Aluminium is &lt;strong&gt;1.05&lt;/strong&gt; !! So does it mean I should use aluminium in the inner surface of the clamp instead of rubber? sounds counter intuitive. (source: &lt;a href=&quot;http://www.engineershandbook.com/Tables/frictioncoefficients.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engineershandbook.com/Tables/frictioncoefficients.htm&lt;/a&gt;)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If I use a material like rubber or silicone, should I consider any texture on the surface? Are &lt;strong&gt;treads&lt;/strong&gt; better than a &lt;strong&gt;smooth surface&lt;/strong&gt;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I am not an expert in these things and so I would greatly appreciate if someone can either suggest me a meterial or educate me on how to select the right material that will have the best grip.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SBROV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SBROV.jpg&quot; alt=&quot;Clamp with rubber-like material for grip&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 1&lt;/strong&gt;: To visualize, imagine the pipe is held vertically by the clamps and a weight is attached to the bottom of the pipe, what material at the pipe-clamp interface will allow me to hang the heaviest weight?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KiH5X.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KiH5X.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit 2:&lt;/strong&gt; The surface of the pipe is smooth to touch with matt finish, not mirror polish. Its actually anodized. Any damage to the pipe like denting or clamping marks is absolutely not acceptable. Also the pipe needs to be re-positioned occasionally, to do so, we loosen the clamp, slide the pipe to position and then tighten the clamp again&lt;/p&gt;&#xA;" OwnerUserId="17221" LastEditorUserId="17221" LastEditDate="2018-11-18T14:20:56.140" LastActivityDate="2018-11-20T17:49:18.443" Title="Material with strongest grip on Aluminium 6061" Tags="|materials|friction|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24720" PostTypeId="1" AcceptedAnswerId="24722" CreationDate="2018-11-18T14:46:47.213" Score="3" ViewCount="190" Body="&lt;p&gt;&quot;In a sling psychrometer(with dry and wet bulbs), if the moist air is not saturated(with water vapour) then it evaporates some water on the wet bulb and in turn cools the wet bulb.So the wet bulb temperature falls below the dry bulb temperature&quot; .This is the explanation given in the book Iam using(Fundamentals of thermodynamics by Moran and Shapiro).&#xA;        1)Does this mean, the wet bulb and dry bulb temperatures are initialy same(before exposing to air)?&#xA;          2)If evaporation occurs continuously as air is blown over the bulbs, why isnt the temperature reducing continuously(it stops at a point)?&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastEditorDisplayName="user17332" LastEditDate="2018-11-18T14:52:01.573" LastActivityDate="2018-11-19T05:14:33.910" Title="Wet bulb temperature in psychrometry" Tags="|thermodynamics|hvac|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24721" PostTypeId="2" ParentId="24714" CreationDate="2018-11-18T15:02:27.460" Score="0" Body="&lt;p&gt;Normally PID or any feedback control has a controller which outputs a signal that is passed to the actuators of the system. In your case the actuators would be the propeller motors of the drone. So the output would have the units volts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for systems you could also use sequential loop closing, where the output of the controller in the outer loops would have the units of the reference of the inner loop. For drones you sometimes have a separate loop for the angular speed for each propellor, such that the outer loop PID would have the controller output units of angular velocity. It is also possible that actual output of the outer loop is the square of the angular velocity, since that is proportional to the amount of thrust that a propellor generates.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2018-11-18T15:02:27.460" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="24722" PostTypeId="2" ParentId="24720" CreationDate="2018-11-18T15:03:52.363" Score="-1" Body="&lt;p&gt;1) yes,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) evaporation continues which reduces the wet bulb temperature until the temperature stays constant because the rate of evaporative heat loss is balanced by the heat absorbed due to the temperature difference between the wet bulb temperature and ambient temperature is in equilibrium. The humdity can then be obtained from a Mollier or psychrometric chart.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-11-19T05:14:33.910" LastActivityDate="2018-11-19T05:14:33.910" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24725" PostTypeId="1" CreationDate="2018-11-18T18:45:03.183" Score="3" ViewCount="639" Body="&lt;p&gt;I am trying to recreate this screw hole in 3D modeling software Siemens NX.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This drawing is from page 4-14 in &lt;a href=&quot;http://www.magnemotion.com/wp-content/uploads/2016/12/990000410_MM_LITE_Rev_E.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;MagneMover LITE User’s Manual&lt;/a&gt; document. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RMmGI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RMmGI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18330" LastEditorUserId="8055" LastEditDate="2018-11-19T16:19:37.093" LastActivityDate="2018-11-19T16:19:37.093" Title="How do I read this engineering drawing for this screw?" Tags="|cad|drafting|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24728" PostTypeId="1" CreationDate="2018-11-19T04:24:45.290" Score="0" ViewCount="61" Body="&lt;p&gt;The question is to prove the following form of helmholtz function for an ideal gas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$f=C_v (T-T_0)-C_vT \ln \frac{T}{T_0} -RT \ln \frac{V}{V_0} -s_0(T-T_0) +f_0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using &lt;span class=&quot;math-container&quot;&gt;$$df=du-Tds-sdT$$&lt;/span&gt;&#xA;Now &lt;span class=&quot;math-container&quot;&gt;$du=\frac{\partial u}{\partial T}dT +\frac{\partial u}{\partial v}dv$&lt;/span&gt; we know &lt;span class=&quot;math-container&quot;&gt;$C_v=\frac{\partial u}{\partial T}$&lt;/span&gt; and from &lt;span class=&quot;math-container&quot;&gt;$du=Tds-Pdv$&lt;/span&gt; we get &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial u}{\partial T}=T\frac{\partial s}{\partial v}_T-P$&lt;/span&gt; which from Maxwell relationship is &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial u}{\partial T}=T\frac{\partial p}{\partial T}_v-P$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unable to proceed after this.Any ideas? Thanks&lt;/p&gt;&#xA;" OwnerUserId="12580" LastActivityDate="2018-12-10T02:13:07.913" Title="Derivation of specific helmholtz function relationship for an ideal gas" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24729" PostTypeId="2" ParentId="23084" CreationDate="2018-11-19T06:31:04.060" Score="1" Body="&lt;p&gt;Nat was right that Aspen doesn't care as long as your compiler can produce a DLL from the Fortran 90 code (which most do). However, I found that the Aspen files that need to be included in most user models are written in fixed format (the formatting style used back with Fortran 77). So when using Fortran 90 code in free format you get a bunch of compiler errors because of the mix of fixed and free formatting. There is a &lt;a href=&quot;https://software.intel.com/en-us/forums/intel-fortran-compiler-for-linux-and-mac-os-x/topic/268667&quot; rel=&quot;nofollow noreferrer&quot;&gt;solution&lt;/a&gt; though: put directives around the include lines that switch to fixed format only for the included files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example I got from the Aspen support center. It is a user model that is supposed to calculate the viscosity of a compound.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;SUBROUTINE MUL2U (T, P, Z, N, IDX, &amp;amp;&#xA;      IRW, IIW, KCALC, KOP, NDS, &amp;amp;&#xA;      KDIAG, QMX, DQMX, KER)&#xA;      IMPLICIT REAL*8 (A-H,O-Z)&#xA;&#xA;#include &quot;dms_plex.cmn&quot; &#xA;#include &quot;ppexec_user.cmn&quot; &#xA;      integer b(1) &#xA;      equivalence (b(1), ib(1)) &#xA;      integer dms_ifcmnc&#xA;      integer limuusr&#xA;      limuusr=dms_ifcmnc('MUUSR') &#xA;&#xA;!     The following input language is required to define A and B&#xA;!     (just define them for the 1st component in the list)&#xA;&#xA;!     PROPERTIES RKS-MU SOLU-WATER=2 &#xA;!       PROPERTIES RK-SOAVE &#xA;&#xA;!     PROP-REPLACE RKS-MU RK-SOAVE &#xA;!       PROP MULMX MULMXUSR &#xA;&#xA;!     USER-PROPS MUUSR 1 1 2 &#xA;!&#xA;!     PROP-DATA MUUSR-1&#xA;!       PROP-LIST MUUSR &#xA;!       PVAL WATER -5 20 &#xA;&#xA;!     To check that parameters are correctly transferred, set global diagnostics:&#xA;&#xA;!     HISTORY PROP-LEVEL=5&#xA;      AA = B(limuusr + 1)&#xA;      BB = B(limuusr + 2)&#xA;      if (KDIAG .GE. 5) then&#xA;        WRITE (USER_NHSTRY, *)'A = ', AA, ' B = ', BB&#xA;      endif&#xA;      qmx = exp(AA + BB / T)&#xA;      dqmx = - BB * qmx / (T * T)&#xA;&#xA;      RETURN&#xA;      END subroutine mul2u&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The example was originally written for Fortran 77 (so it used fixed formatting), and I made the few changes necessary for it to be compiled using free formatting. However, when compiling you get a string of errors that come from the Aspen files included in the subroutine (so dms_plex.cmn and ppexec_user.cmn). Here is an example of one of the errors:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;c:\Program Files (x86)\AspenTech\AprSystem V8.8\Engine\commons/ppexec_user.cmn(7): error #5082: Syntax error, found END-OF-STATEMENT when expecting one of: %FILL &amp;lt;IDENTIFIER&amp;gt;&#xA;      INTEGER USER_IUMISS,   USER_NGBAL,    USER_IPASS,&#xA;-------------------------------------------------------^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Again, it seems that the compiler gives the error because the Aspen files are written in fixed format and we are compiling with free format. To get around this use directives that switch to fixed format and back again.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;!DEC&lt;span class=&quot;math-container&quot;&gt;$ NOFREEFORM&#xA;#include &quot;dms_plex.cmn&quot; &#xA;#include &quot;ppexec_user.cmn&quot;&#xA;!DEC$&lt;/span&gt; FREEFORM&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;When I use these directives around the include statement then the compiler works without errors and I can use a Fortran 90 user model in Aspen.&lt;/p&gt;&#xA;" OwnerUserId="6968" LastActivityDate="2018-11-19T06:31:04.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24730" PostTypeId="2" ParentId="24725" CreationDate="2018-11-19T13:36:20.050" Score="7" Body="&lt;p&gt;Let's break it down:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5x - There are 5 of these holes, with the same dimension. You should be able to spot these fairly easily, as any similar holes with no dimensions on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M4x0.7 - These holes have a Metric Thread, with 4mm outer diameter, and 0.7mm pitch. This is the &quot;M4 Coarse&quot; standard.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;↧ 6.3 - The holes are 6.3mm deep. This is the 'full diameter' depth, and not the drill depth, unless otherwise stated. See page 11 here: &lt;a href=&quot;https://shop.bsigroup.com/upload/Standards%20&amp;amp;%20Publications/Manufacturing/BIP2155-Chapter1.pdf&quot; rel=&quot;noreferrer&quot;&gt;https://shop.bsigroup.com/upload/Standards%20&amp;amp;%20Publications/Manufacturing/BIP2155-Chapter1.pdf&lt;/a&gt; for more details&lt;/p&gt;&#xA;&#xA;&lt;p&gt;MAX THD DEPTH - The thread is cut to the maximum depth possible, so will require the use of a 'bottoming tap' during manufacture - See this page for more details on different tap types: &lt;a href=&quot;https://www.cnccookbook.com/different-types-thread-taps-complete-guide-choosing/&quot; rel=&quot;noreferrer&quot;&gt;https://www.cnccookbook.com/different-types-thread-taps-complete-guide-choosing/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to model this hole inside NX, you should use the hole tool, rather than a revolved cut, for example, and then input all these details into the tool. This will ensure that the specification can be automatically added to the drawing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want an actual 3D representation of the threadform, rather than a 'symbolic thread', then you'll need to ask a separate question about modelling that, since this thread is not the right place for that information.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-11-19T13:36:20.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24731" PostTypeId="2" ParentId="24728" CreationDate="2018-11-19T13:37:46.793" Score="1" Body="&lt;p&gt;The derivative function is expanded from &lt;span class=&quot;math-container&quot;&gt;$A = U - TS $&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ dA = dU - T_odS - S_odT $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an ideal gas, &lt;span class=&quot;math-container&quot;&gt;$dU$&lt;/span&gt; depends only on temperature &lt;span class=&quot;math-container&quot;&gt;$ dU = C_V dT $&lt;/span&gt;. Expand entropy change at constant volume &lt;span class=&quot;math-container&quot;&gt;$dS_V = \delta q_V / T = C_V dT / T$&lt;/span&gt;. Assume &lt;span class=&quot;math-container&quot;&gt;$C_V$&lt;/span&gt; is constant. The final expression can be derived from this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer has a mistake in using &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; rather than &lt;span class=&quot;math-container&quot;&gt;$T_o$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-12-10T02:13:07.913" LastActivityDate="2018-12-10T02:13:07.913" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24736" PostTypeId="1" AcceptedAnswerId="24738" CreationDate="2018-11-19T20:51:51.750" Score="3" ViewCount="488" Body="&lt;p&gt;If the diameter before knurling is 80. Is the knurled diameter also 80?&lt;/p&gt;&#xA;" OwnerUserId="18345" LastActivityDate="2018-11-19T22:51:31.633" Title="Is the diameter before knurling assumed to be the same as diameter after knurling?" Tags="|technical-drawing|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24738" PostTypeId="2" ParentId="24736" CreationDate="2018-11-19T21:46:32.037" Score="3" Body="&lt;p&gt;Knurling affects the roughness of a surface, but does not affect its diameter much. No material is removed, so for every indentation, a ridge is formed. This has the effect of making the measured diameter using calipers, for example, slightly larger*. If the knurled diameter is an inspection dimension, then you may wish to place a note on the drawing, after experimental verification, to show the QA team what to look for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a manufacturer's table of expected increase in diameter for various knurls: &lt;a href=&quot;http://accu-trak.com/technicalinfo/table1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://accu-trak.com/technicalinfo/table1.html&lt;/a&gt; (thanks to @D Duck)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember the primary purpose of the drawing is to give information to ensure it will be manufactured correctly. Both the person machining the turned 'blank', and the person adding the knurl to that will be primarily concerned with the 'smooth' diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*The average diameter actually goes down, since volume is related to radius cubed, meaning the indentations are deeper than the ridges are tall, relative to the original diameter. In reality this is hard to measure, and I've never seen an engineering drawing where the knurled dimension was different to the turned dimension. &lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-11-19T22:51:31.633" LastActivityDate="2018-11-19T22:51:31.633" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24739" PostTypeId="1" CreationDate="2018-11-19T22:08:23.113" Score="0" ViewCount="495" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MUelR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if the short parallel lines at the ends mean anything?&lt;/p&gt;&#xA;" OwnerUserId="18345" LastActivityDate="2018-11-19T22:26:10.110" Title="What do the short parallel lines mean in a symmetrical value?" Tags="|technical-drawing|drawings|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24740" PostTypeId="2" ParentId="24739" CreationDate="2018-11-19T22:26:10.110" Score="1" Body="&lt;p&gt;Those short parallel lines have absolutely no meaning in terms of information relating to the part, but we put them on the drawing to identify the line of symmetry. They are always at the right angles to the symmetry line. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also another parallel lines that you may encounter later, to know more about technical drawing symbols, it's good to look at GD&amp;amp;T tables, symbols and definitions. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-11-19T22:26:10.110" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24742" PostTypeId="1" CreationDate="2018-11-20T00:31:13.847" Score="0" ViewCount="148" Body="&lt;p&gt;I recently learned about dew point and rain. From my understanding, dew point is the temperature at which the air can hold the amount of moisture present. And when the temperature decrease, the air can hold less moisture, so some of it must condenses, which comes down as rain. So with this fact, does the rain usually happens when the temperature decreases? Then how can we describe the rain which starts around noon (when it’s the hottest of the day)? Are there any other factors that causes rain?&lt;/p&gt;&#xA;" OwnerUserId="18024" LastEditorUserId="90" LastEditDate="2021-08-01T10:04:51.500" LastActivityDate="2021-08-01T10:04:51.500" Title="Dew point and raining" Tags="|water-resources|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24743" PostTypeId="1" CreationDate="2018-11-20T01:46:01.570" Score="1" ViewCount="59" Body="&lt;p&gt;I asked this question at physics but i think this suits it more. I'd like to know few things about how does one design a Thermal Power Plant? There is something fundamentally lacking in my knowledge as I think there are a lot of parameters that I’m having trouble defining or estimating. I suppose this comes from difference between classroom studying and practical knowledge. So, let's talk about a small thermal plant of 1MW capacity. I have tried my best to be as clear as possible, but if some sentence is not making sense, I apologize in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting with the steam turbine, I'm really struggling to find any specifications of that small size, while I know for sure that 1MW thermal power plants exists. And even when I do find the specifications, they mention only Temperature at which steam is to be fed and its Pressure. (Example: &lt;a href=&quot;https://www.siemens.com/global/en/home/products/energy/power-generation/steam-turbines/d-r-steam-turbines.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.siemens.com/global/en/home/products/energy/power-generation/steam-turbines/d-r-steam-turbines.html&lt;/a&gt;) While these two are important and are the only parameters needed to find the enthalpy (h1), shouldn’t mass flow rate be included too? And what about the steam that will come out of turbine, only with that I’ll know what’s enthalpy at end (h2) and will be able to find its efficiency. Sometimes in specification they mention exhaust pressure without the temperature. How does that help? By multiplying the difference in enthalpies with mass flow rate ’ll get the power that I’m pumping in Turbine in form of steam. (correct me if I’m wrong here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I understand the above, I’d like to know how the 1MW generation reacts to variable mass flow rate of constant temperature and pressure. As for the other components, I believe knowing the state of exhaust steam will help in deciding them. So, I may talk about them later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are any free source books or videos available discussing the above then I’d be happy to go through it.&lt;/p&gt;&#xA;" OwnerUserId="18347" LastActivityDate="2018-11-20T01:46:01.570" Title="How is a Steam Power Plant designed?" Tags="|turbines|power-engineering|steam|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24744" PostTypeId="2" ParentId="24579" CreationDate="2018-11-20T05:55:00.720" Score="1" Body="&lt;p&gt;We dont know. Simply we lack data beyond a certain point just before the first explosion. We therefore have to resort to mathematical modeling for deducing what happened. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this mathematical modeling depends heavily on what data we have, and the stability of the reactor was very bad at the moment of the accident. The modeling accuracy decreases from that point forward. Just like a weather forecast, it becomes more unreliable the further we go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did a nuclear explosion happen? It is possible. Atleast that possibility is coroborated by a analysis done by Yuri V. Dubasov in a 2009, this was further examined by Lars-Erik De Geer in 2017. While both suggest that probability of nuclear fizzle is there. Their conclusion is a bit different as to wether the first or second event was nuclear though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this is still very much a reasearch publication worthy material, and you need serious scientific chops to deal with this adequately. Yes its your conclusion has been suggested, and is somewhat probable. But gist of the matter is we dont know. You can not expect this forum to solve this for you.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2018-11-20T05:55:00.720" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="24746" PostTypeId="2" ParentId="24717" CreationDate="2018-11-20T06:43:46.027" Score="1" Body="&lt;p&gt;I think you need an equation to compare the hydraulic work with the mechanical work. Additionally you can add the speed to get the power, but it's not necessary...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$W_{hydr}=p*V_{geom}=M*2*\pi=W_{mech}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and now with speed &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_{hydr}=p*V_{geom}*n=p*Q=M*2*\pi*n=P_{mech}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the units:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$[bar]$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$V_{geom}$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$[cm^3/U]$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$[1/s]$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$[l/min]$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$[Nm]$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17780" LastActivityDate="2018-11-20T06:43:46.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24747" PostTypeId="1" AcceptedAnswerId="26982" CreationDate="2018-11-20T06:57:00.397" Score="0" ViewCount="244" Body="&lt;p&gt;&lt;a href=&quot;https://www.hindawi.com/journals/bmri/2017/4701481/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Research shows that applying 1 MHz non-focused ultrasound at 3 W/cm2 may reduce subcutaneous fat thickness.&lt;/a&gt; In fact, high-end and expensive (costs $5000 &gt;) &lt;a href=&quot;https://en.wikipedia.org/wiki/High-intensity_focused_ultrasound&quot; rel=&quot;nofollow noreferrer&quot;&gt;HIFU&lt;/a&gt; ultrasonic transducers are now being used in cosmetic clinics that allow the user to regulate ultrasound frequency, pulsing, power and time in order to target body fat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my surprise, I noticed that one can also buy inexpensive (costs only $50), portable 1Mhz Ultrasound cavitation device from &lt;a href=&quot;https://www.ebay.ca/itm/1MHz-Ultrasonic-Cavitation-Ultrasound-Face-Body-Massage-Skin-Tightening-Anti-Age/222369021374?hash=item33c639fdbe:g:--wAAOSwWMFa2uxK:rk:5:pf:0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ebay&lt;/a&gt; or &lt;a href=&quot;https://www.aliexpress.com/item/1MHz-Ultrasound-Cavitation-EMS-Body-Slimming-Massager-Loss-Weight-Anti-Cellulite-Fat-Burner-Galvanic-Infrared-Ultrasonic/32884572943.html?src=google&amp;amp;albslr=200431112&amp;amp;src=google&amp;amp;albch=shopping&amp;amp;acnt=494-037-6276&amp;amp;isdl=y&amp;amp;slnk=&amp;amp;plac=&amp;amp;mtctp=&amp;amp;albbt=Google_7_shopping&amp;amp;aff_platform=google&amp;amp;aff_short_key=UneMJZVf&amp;amp;&amp;amp;albcp=1633541485&amp;amp;albag=62156539197&amp;amp;trgt=296904914040&amp;amp;crea=en32884572943&amp;amp;netw=u&amp;amp;device=c&amp;amp;gclid=Cj0KCQiA28nfBRCDARIsANc5BFBp1inb-DaYexIKqC9yO8fc9QJJIJS8ERwr6Rt7CZKim8q2bmIsegsaAgDgEALw_wcB&amp;amp;gclsrc=aw.ds&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aliexpress&lt;/a&gt; that also claims to reduce body fat. Scam or not, one way to justify its claim will be to measure and verify the emitting frequency and intensity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.aliexpress.com/item/foot-massager-ultrasonic-slimming-body-firming-lifting-face-Health-Care-Products-lose-weight-burn-fat-electronic/32239480611.html?spm=a2g0s.9042311.0.0.3da24c4dN2ciMo&quot; rel=&quot;nofollow noreferrer&quot;&gt;I bought such a device from Aliexpress (costed $130)&lt;/a&gt; to play around with it. My skin can feel a definite vibration when I turn on the device, however, it is very hard to tell if ultrasound is actually exciting my tissue lesions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know how to set up an inexpensive testing setup and:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Find out if the device is really emitting ultrasound. If yes, how would I measure its frequency? Device manual says 1MHz, but I suspecting it might be a bs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How to measure the intensity and penetration depth. Finding one value should lead me to calculate another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Find out if the device's ultrasound is focused or non-focused. My gut feeling is the ultrasound is non-focused or weakly focused. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to find more clues would be to rip through the device and look at the transducer circuit. The transducer hardware might have a model number that can be used to find it's spec sheet online. However, I haven't unscrewed the device yet and I doubt I would find anything helpful on the circuit board as it probably came directly from a questionable Chinese factory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading:&#xA;There has been plenty of research conducted so far regarding effects of non-focused ultrasound on animal contouring:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pubmed/23220874&quot; rel=&quot;nofollow noreferrer&quot;&gt;The effects of nonfocused external ultrasound on tissue temperature and adipocyte morphology.&lt;/a&gt;&#xA;2) &lt;a href=&quot;http://www.electrotherapy.org/modality/ultrasound-dose-calculation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ultrasound dose calculation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18349" LastEditorUserId="18349" LastEditDate="2018-11-20T18:39:57.220" LastActivityDate="2019-04-20T00:16:14.187" Title="How to measure and verify ultrasound frequency and intensity from a cheap ultrasonic body contouring device?" Tags="|experimental-physics|biomedical-engineering|ultrasound|waves|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24748" PostTypeId="1" CreationDate="2018-11-20T07:39:48.657" Score="0" ViewCount="28" Body="&lt;p&gt;I hope this is not off-topic. I’m wondering if someone could please provide some examples of a basic mechanism (I’m thinking something like a bevel/spiral gear, but open to anything) that is attached somewhere to a vehicle, and uses the rotation of the wheels as a timer/measure to perform a task. As a fake example, maybe it would be something attached to an axle and every 10m rings a bell. I’m talking totally mechanical, not involving electricity or computers. Thank you&lt;/p&gt;&#xA;" OwnerUserId="18350" LastActivityDate="2018-11-20T07:39:48.657" Title="Mechanism timed by wheel rotations" Tags="|mechanisms|mechanical|distance-measurement|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24751" PostTypeId="2" ParentId="24742" CreationDate="2018-11-20T12:19:35.083" Score="2" Body="&lt;p&gt;The important point here is that rain tends to form at some significant height above ground level. When you get water droplets condensing in situ it is usually referred to as &lt;em&gt;mist&lt;/em&gt; or &lt;em&gt;fog&lt;/em&gt; or &lt;em&gt;dew&lt;/em&gt; when it condenses on surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rain often happens when warm, moist air is forced upwards due to large sale movements of bodies of air in particular at the boundaries (&lt;a href=&quot;https://en.wikipedia.org/wiki/Weather_front&quot; rel=&quot;nofollow noreferrer&quot;&gt;fronts&lt;/a&gt;) of warm and cold air with differnt humidity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QrnNy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QrnNy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2018-11-20T12:19:35.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24752" PostTypeId="1" CreationDate="2018-11-20T13:04:33.370" Score="1" ViewCount="226" Body="&lt;p&gt;I really do apologise if this is a stupid question but I for the life of me can't find an answer. I need to do a project where I measure the liquid of water in a tank using ultrasound and am planning on doing it with Texas Instruments TDC1000. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I'm having is that I can't find the ultrasonic transducers that they use in their examples in my country and was thinking I could maybe just use one similar to it. However I can only find ceramic benders. I would just like to know if they are the same and if not what is the difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks &lt;/p&gt;&#xA;" OwnerUserId="18355" LastActivityDate="2019-03-20T03:31:58.997" Title="Is there a difference between piezo ceramic disc transducer and a piezo ceramic bender" Tags="|sensors|ultrasound|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24754" PostTypeId="1" CreationDate="2018-11-20T14:45:47.610" Score="1" ViewCount="1824" Body="&lt;p&gt;I have come across the definition somewhere,&quot;Work hardening, also known as strain hardening, is the strengthening of a metal or polymer by plastic deformation. This strengthening occurs because of dislocation movements and dislocation generation within the crystal structure of the material.&quot;Now I have been told by my professor that strain hardening and work hardening are different.I would like to have a detailed answer on this.&lt;/p&gt;&#xA;" OwnerUserId="12558" LastActivityDate="2024-03-16T14:05:50.677" Title="What is the difference between work hardening and strain hardening?" Tags="|materials|metallurgy|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24757" PostTypeId="2" ParentId="24719" CreationDate="2018-11-20T17:41:57.563" Score="1" Body="&lt;p&gt;The anodising is for corrosion prevention, usually. In this case of have the 2 supports show free movement of the Al and clamp a collar around the bar above the top support. To increase the anoint of rich tin without damaging the surface would require 1 of 2 options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Greater force; &#xA;2) Greater surface area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you stated greater forces not possible you're left with option two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having three of those clamps that you have a picture of side by side above the collar would certainly help since the overall friction would effectively be tripled:&#xA;0.8×A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remembering that stress is force over area and force in this case needs to be limited while maintaining sufficient stress (friction) to prevent slippage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EiDQC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EiDQC.jpg&quot; alt=&quot;enterprising of clamp&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2018-11-20T17:49:18.443" LastActivityDate="2018-11-20T17:49:18.443" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24758" PostTypeId="2" ParentId="24754" CreationDate="2018-11-20T18:03:22.690" Score="0" Body="&lt;p&gt;There is stress and strain hardening. Stress hardening is done under compression. Strain hardening occurs under tension and is ultimately destructive. Strain hardening occurs due to repetitive bending that induces material fractures to increase whereas stress hardening is compression of material to increase surface contact between molecular or crystalline structures of the material to increase internal structural friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stress hardening is how steel manufacture is used to increase hardness of material after annealing.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2018-11-23T01:37:50.337" LastActivityDate="2018-11-23T01:37:50.337" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24759" PostTypeId="2" ParentId="21451" CreationDate="2018-11-20T18:28:20.123" Score="1" Body="&lt;p&gt;Horizontal forces  &lt;span class=&quot;math-container&quot;&gt;$$\sigma F_1= F_2+F_r$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_2 \sin \theta × R$$&lt;/span&gt; etc where r is radius of wheel and &lt;span class=&quot;math-container&quot;&gt;$\Theta $&lt;/span&gt; is angle between vertical impact and perpendicular to direction of travel&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-11-20T18:28:20.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24761" PostTypeId="2" ParentId="24747" CreationDate="2018-11-20T19:04:47.160" Score="0" Body="&lt;p&gt;If you have access to the right hardware with a sampling rate higher than the frequency of the transmitter, you could record the audio signal and do a fast fourier transformation on it. That will give you the amplitude and frequency of the most prominent waves in the signal.&lt;/p&gt;&#xA;" OwnerUserId="12401" LastActivityDate="2018-11-20T19:04:47.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24762" PostTypeId="1" CreationDate="2018-11-20T19:06:00.363" Score="0" ViewCount="78" Body="&lt;p&gt;In the rolling process, Why should the velocity of the strip at the entry before coming in contact with roller be lesser than the velocity of the roller?&lt;/p&gt;&#xA;" OwnerUserId="17075" LastActivityDate="2018-11-20T19:49:23.803" Title="MANUFACTURING PROCESS: ROLLING PROCESS" Tags="|mechanical-engineering|manufacturing-engineering|rolling|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24763" PostTypeId="2" ParentId="24762" CreationDate="2018-11-20T19:27:50.070" Score="1" Body="&lt;p&gt;Because the cross section of strip decreases during the milling process, its speed will increase to match the roller's speed at the point when it leaves the rollers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise the rollers my warp or leave creases on the strip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's like a flow of water in a pipe at a slower speed picking up velocity when passing through a nuzzle. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-20T19:27:50.070" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24765" PostTypeId="2" ParentId="24762" CreationDate="2018-11-20T19:44:18.243" Score="0" Body="&lt;p&gt;There are many reasons behind, and many ways to justify why it is so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to clarify it by using the conservation of the mass.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- The density of the rolling material doesn't change, so its volume. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- We assume that the width of material also stays constant (plain strain) that seems a very rough estimation but if you start with an ingot or slab after passing three or more roll machines that's doesn't change a lot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the mass conservation implies: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$v_iy_i \approx v_\omega y_\omega$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here index &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; indicates the entry speed and thickness and the index &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; indicates the thickness of slab at the moment that its linear velocity is equal to the linear velocity of the rollers. We know for a fact that &lt;span class=&quot;math-container&quot;&gt;$y_\omega$&lt;/span&gt; is smaller than &lt;span class=&quot;math-container&quot;&gt;$y_i$&lt;/span&gt; then by rewriting the first equation in this form: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{y_\omega}{y_i} \approx \frac{v_i}{v_\omega}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We may now conclude that the entry velocity should be smaller than the velocity of the roller. If &lt;span class=&quot;math-container&quot;&gt;$v_i &amp;gt; v_\omega$&lt;/span&gt; then, at the entry the slab begins to decelerate, cause minor damges to component and further energy loss. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile the Kamran's answer points out another reason. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-11-20T19:49:23.803" LastActivityDate="2018-11-20T19:49:23.803" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24768" PostTypeId="2" ParentId="24689" CreationDate="2018-11-20T20:30:16.493" Score="0" Body="&lt;p&gt;Usually using piles in multiple vertical arrays is the answer. The friction between the soil and rock against a long vertical concrete column can be substantial enough to provide more than enough reaction force to the compressive load applied by structures on it. A good example is the Shard in London which was built this way. The clay soils in London are notoriously unstable and the 97 floor tower rests on these piles. &lt;a href=&quot;https://youtu.be/PIqTIB7R5zc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Shard construction&lt;/a&gt;&#xA;&lt;a href=&quot;https://youtu.be/1DN5dTwbWUk&quot; rel=&quot;nofollow noreferrer&quot;&gt;Impossible Engineering Documentary&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-11-20T20:30:16.493" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24770" PostTypeId="1" AcceptedAnswerId="24771" CreationDate="2018-11-21T01:36:51.527" Score="1" ViewCount="1259" Body="&lt;p&gt;This is the problem I'm trying to solve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://d2vlcm61l7u1fs.cloudfront.net/media%2F4a5%2F4a5383da-7b6d-4ec7-baf6-ca99ebf309da%2FphpQwuYae.png&quot; alt=&quot;figure&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking at the worked-out solution from my professor and it shows that the sum of the forces in the x-direction is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[d/dt(momentum) in the x-direction, out] - [d/dt(momentum) in the x-direction, in]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't understand what is meant by momentum out vs. momentum in and why one is positive and the other is negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I apologize if this is a dumb question.&lt;/p&gt;&#xA;" OwnerUserId="18364" LastActivityDate="2018-11-21T01:51:16.940" Title="How to use the linear momentum equation to determine the forces required to hold a nozzle stationary?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24771" PostTypeId="2" ParentId="24770" CreationDate="2018-11-21T01:51:16.940" Score="0" Body="&lt;p&gt;If it is a solid body, the force applied on the body is the rate of change of momentum of the body in the direction of force(here we talk about the mass of the body and its velocity).In case of a fluid, we talk about the mass flow rate and velocity.So the initial momentum of the flow before the bend(and nozzle) is the momentum in and after interacting with the nozzle, the momentum of the flow leaving out is the momentum out.The difference will give the force acting on the fluid by the bend.I think you should include pressure force in x direction.&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastActivityDate="2018-11-21T01:51:16.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24772" PostTypeId="1" CreationDate="2018-11-21T02:04:24.600" Score="3" ViewCount="111" Body="&lt;p&gt;In this incident &lt;a href=&quot;https://www.bbc.com/news/world-us-canada-46274458&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.bbc.com/news/world-us-canada-46274458&lt;/a&gt; an elevator was described as falling 84 floors, but all survived apparently with little or no injury.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would seem that the car did not actually fall; a &quot;hoist rope&quot; failed, causing it to descend out of control, after which something brought it to a gentle enough stop that none of the occupants came to any notable harm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is about the way in which elevators are designed such that this event unfolded as it did. Mention of a &quot;hoist rope&quot; seems to suggest that there are separate ropes for lifting/lowering the car and for counterbalancing it, and the counterbalance ropes remained intact. But, if it was out-of-balance enough to &quot;fall&quot;, what kept it from accelerating, what terminal velocity might it have reached, and what caused it to stop without a significant impact?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to see is either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a technical account of what actually happened, or alternatively, &lt;/li&gt;&#xA;&lt;li&gt;a speculative account of what most likely happened, based on a known facts (or reasonable assumptions) about the elevator in question, and the details of the event as presented in the media.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The point of my question is to cut through the media hype and inevitable distortion of fact to better understand the mishap from a technical perspective.&lt;/p&gt;&#xA;" OwnerUserId="15169" LastEditorUserId="15169" LastEditDate="2018-11-24T00:59:58.220" LastActivityDate="2018-11-24T00:59:58.220" Title="They survived an elevator fall. How?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24773" PostTypeId="1" CreationDate="2018-11-21T02:24:15.037" Score="2" ViewCount="160" Body="&lt;p&gt;I'm not sure if this sort of question is appropriate here or not so please let me know if not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am designing a UHV vacuum system with a few interesting requirements. Inside of the vacuum chamber will be a multi-stage vibration isolation system consisting of masses sitting on springs (likely viton or other UHV compatible elastomer material). In addition, I will be using electromagnetic coils to create magnetic fields inside of the vacuum chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnetic fields will be changing in times. This means that in general eddy currents can be induced in conductors. These Eddy currents create their own magnetic fields which can be problematic for my application and thus they are to be avoided. One way to avoid Eddy currents is to avoid conductive materials in the chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vibration isolation system will operate better if its natural frequencies are low. This can be accomplished by using larger masses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I wasn't worried too much about the Eddy currents/magnetic fields I would likely make the vibration isolation out of 316 Stainless Steel (&lt;span class=&quot;math-container&quot;&gt;$8 \frac{g}{cm^3}$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am curious if there is any non-conductive material which could be machined or formed into the appropriate shapes for my vibration isolation stage which also has a high density. For example, some ceramics I find have a density of &lt;span class=&quot;math-container&quot;&gt;$~3 \frac{g}{cm^3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does any one have a recommendation for a material which could suit my needs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other requirements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;UHV compatible down to &lt;span class=&quot;math-container&quot;&gt;$10^{-11}$&lt;/span&gt; torr&lt;/li&gt;&#xA;&lt;li&gt;Non-magnetic. In addition to not supporting Eddy currents the material should not become magnetized by the changing magnetic fields which are induced (magnitude of ~300 Gauss maximum) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18365" LastActivityDate="2018-11-21T02:24:15.037" Title="Dense, non-conductive UHV compatible material?" Tags="|materials|vacuum|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24774" PostTypeId="1" CreationDate="2018-11-21T03:32:56.053" Score="1" ViewCount="28" Body="&lt;p&gt;I'm not exactly sure if this is where I'm supposed to ask, but engineering stack exchange is the closest because my question is about robotics. I'm sorry if this doesn't fit.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; Are there any intake mechanisms that are efficient and quick, but are designed to accommodate obstacles that are in the way?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let me explain my question. Essentially, let's say you have a robot with an intake that's supposed to take in different sized objects like cubes and spheres. However, to get to the cubes and spheres, there's a three inch tall wall in the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard way of retrieving the objects is to go over the wall, which might be possible, but I'm asking if there's another way without going over the wall. I've thought about some sort of extension that goes over the wall, but it's risky if your actual intake is heavy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts? I've spent some time thinking about this and I'm still haven't come up with an idea.&lt;/p&gt;&#xA;" OwnerUserId="18367" LastActivityDate="2018-11-21T03:32:56.053" Title="Intake Mechanism with Obstacle" Tags="|mechanical-engineering|robotics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24775" PostTypeId="1" CreationDate="2018-11-21T04:16:35.877" Score="-1" ViewCount="356" Body="&lt;p&gt;I'm currently working on a project but I cannot figure out how to make a 4 bit comparator (compare 2 4 bit numbers)&#xA;If cascading 1 bit comparators makes it easier then that works too&lt;/p&gt;&#xA;" OwnerUserId="18125" LastActivityDate="2018-11-21T15:37:50.183" Title="How to design a 4 bit comparator using logic gates?" Tags="|systems-design|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24778" PostTypeId="2" ParentId="24772" CreationDate="2018-11-21T08:49:41.580" Score="4" Body="&lt;p&gt;&lt;a href=&quot;https://www.washingtonpost.com/national/health-science/elevator-plunges-are-rare-because-brakes-and-cables-provide-fail-safe-protections/2013/06/07/e44227f6-cc5a-11e2-8845-d970ccb04497_story.html?noredirect=on&amp;amp;utm_term=.5fa3d07891ff&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hollywood movies&lt;/a&gt; like to show elevators plunging at high speeds to the bottom of elevator shafts for dramatic effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This used to sometimes happen in the early days of elevators. Modern elevators have &lt;a href=&quot;https://science.howstuffworks.com/transport/engines-equipment/elevator4.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;safety systems&lt;/a&gt; to prevent this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly the ropes that are used to hoist &amp;amp; lower the elevator have a factor safety designed into them so that ropes that are used are stronger than needed. This helps reduce the effects of wear and potential corrosion. Also, all elevators must undergo periodic safety inspections, which includes all aspects of the elevator system: winders, ropes sheaves, the elevator cage, counter weights, brakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, all modern elevators have &lt;a href=&quot;https://www.hydrasafebrake.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;brakes&lt;/a&gt; to slow down and safely stop elevators from an uncontrolled fall.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-11-21T08:49:41.580" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24779" PostTypeId="1" CreationDate="2018-11-21T09:16:25.533" Score="1" ViewCount="33" Body="&lt;p&gt;I’m planing to build a robot out of gears.  I was wondering what kind of (dis) engagement mechanism would be suitable for gears that are about 2.5cm (1 inch) in diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: preferably, (dis)engagement of gears accomplished without stopping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="18369" LastActivityDate="2018-11-21T09:16:25.533" Title="What are the some of the best (dis)engagement mechanism for small gears?" Tags="|gears|robotics|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24780" PostTypeId="2" ParentId="24775" CreationDate="2018-11-21T15:37:50.183" Score="0" Body="&lt;p&gt;4 XNOR gates connected to a 4-input AND. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;XNOR produces 0 on different inputs, 1 on same, so it's a 1-bit comparator. If any of 4 bits differs then the inputs are not equal, and AND will produce 0. If all are equal, you get 1.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-11-21T15:37:50.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24782" PostTypeId="1" CreationDate="2018-11-21T18:47:42.737" Score="1" ViewCount="70" Body="&lt;p&gt;The second moment of area of a rectangle is &lt;span class=&quot;math-container&quot;&gt;$I_{yy} = \dfrac{bh^2}{12}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b : base&#xA;h : height&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the second moment of area of a T beam like this one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iUXlY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iUXlY.png&quot; alt=&quot;beam section&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12478" LastEditorUserId="1832" LastEditDate="2018-11-23T12:34:14.353" LastActivityDate="2018-11-23T12:34:14.353" Title="What's the second moment of a T beam with a heel?" Tags="|mechanical-engineering|beam|" AnswerCount="1" CommentCount="4" ClosedDate="2018-11-21T19:42:44.127" ContentLicense="CC BY-SA 4.0" />
  <row Id="24783" PostTypeId="2" ParentId="24782" CreationDate="2018-11-21T20:00:32.437" Score="1" Body="&lt;p&gt;First let's find the centroid of the shape:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming each shading has its own different modulus of elasticity E1,E2,.. and just for the sake of this example let's say the ratio of E's are as follows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E1 for the top flange = 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E2 stem =1.2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E3 trapezoid bases = 1.3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We set the base as Y axis and find the centroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the trapezoid moment = &lt;span class=&quot;math-container&quot;&gt;$ 1.3 ((b_{ ta} - b_a  )(H_2/2 )+((b_{ ta} -b_a)(h_1/2)(h_1/3 + h_2/2)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stem moment + &lt;span class=&quot;math-container&quot;&gt;$1.2( b_a\times h_p/2)  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the flange moment =&lt;span class=&quot;math-container&quot;&gt;$ b_0 h_d (h_p-h_d/2)   $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the centroid is sum of these area moments divided by sum of weighted areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\overline{Y} =\frac {\sum\  area\ moments} {\sum\ weighted \ areas}  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I of the beam is the I of individual part plus parallel axis moment of inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ I=  ( \sum A_n\times E_n\times Y_n^2 ) +\sum I_n $&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-21T20:00:32.437" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24784" PostTypeId="1" AcceptedAnswerId="24833" CreationDate="2018-11-21T21:04:04.813" Score="1" ViewCount="533" Body="&lt;p&gt;I'm trying to make some code to predict the performance of a sounding rocket I want to build.  To do this, I need the atmospheric density at any point in the flight for the time-step simulation I made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got my atmospheric density in slug/ft^3 from &lt;a href=&quot;https://www.engineeringtoolbox.com/standard-atmosphere-d_604.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineeringtoolbox.com/standard-atmosphere-d_604.html&lt;/a&gt; and put all that data in this python script, under the assumption that the function that would best describe this data would be a negative exponential.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;from scipy import optimize&#xA;&#xA;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;&#xA;&#xA;&#xA;def fxn(x, a, b, c):&#xA;    return a*np.exp(-b*x) + c&#xA;&#xA;altitudes = np.array([0,5000,10000,15000,20000,25000,30000,35000,40000,45000,50000,60000,70000,80000,90000,100000,150000,200000,250000])&#xA;&#xA;density1 = np.array([23.8,20.48,17.56,14.96,12.67,10.66,8.91,7.38,5.87,4.62,3.64,2.26,1.39,.86,.56,.33,.037,.0053,.00065])&#xA;&#xA;altitudeLinSpace = np.linspace(0,500000,500000)&#xA;&#xA;#Popt returns a array containing constants a,b,c... etc for function 'fxn'&#xA;popt, pcov = optimize.curve_fit(fxn, altitudes, density1, p0=(1, 1e-6, 1))&#xA;&#xA;print(popt)&#xA;&#xA;&#xA;#Plots data&#xA;plt.grid(True)&#xA;plt.ylim((0,25))&#xA;plt.plot(altitudes, density1)&#xA;plt.plot(altitudeLinSpace, fxn(altitudeLinSpace, popt[0], popt[1], popt[2]))&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which gives me this graph when I plot the solution ( Altitude ft (x), density in slug/ft^3 * 10^-4 (y)):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/jhSkF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jhSkF.png&quot; alt=&quot;Altitude(x), density in slug/ft^3 * 10^-4 (y)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orange curve is what python found as a solution, blue is the plot of the data I used to curve fit.  Because of the poor fit, my guess is that the data is actually not a negative exponential, but I'm not really sure how to determine the kind of function this might be...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for perfect, but using a polynomial is a less than ideal solution here as the 'wiggles' give too much error at higher altitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions on how I could fit this data better would be greatly appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution from Math SE: &lt;a href=&quot;https://math.stackexchange.com/questions/3009560/curve-fitting-atmosphere&quot;&gt;https://math.stackexchange.com/questions/3009560/curve-fitting-atmosphere&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14040" LastEditorUserId="14040" LastEditDate="2018-11-25T20:16:24.553" LastActivityDate="2018-11-25T20:16:24.553" Title="Curve Fitting Atmospheric Density" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24785" PostTypeId="1" CreationDate="2018-11-21T21:13:05.097" Score="1" ViewCount="123" Body="&lt;p&gt;I make solid curved wood furniture and frequently need something like a metal plate (1/8&quot;) with a (3/4&quot; about) supporting pipe attached at 90 degrees. Welding is a little too cumbersome but I was wondering if you guys out there know about any connecting solutions similar to a concrete anchor setup?&lt;a href=&quot;https://i.stack.imgur.com/fyRgQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fyRgQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18377" LastActivityDate="2023-11-27T01:02:15.250" Title="A Steel post and base question" Tags="|steel|structures|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24786" PostTypeId="1" CreationDate="2018-11-21T22:10:58.217" Score="0" ViewCount="114" Body="&lt;p&gt;I want to couple a plate, which cannot rotate but only slide along a track, with a threaded rotating shaft. This is for a grip setup, where rotating the rod presses the plate forwards to apply pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of a type of connector for this coupling, and have searched the internet but McMaster and other sites do not seem to have one. Does anyone else know of such a connection?&lt;/p&gt;&#xA;" OwnerUserId="3419" LastActivityDate="2018-12-22T16:00:45.133" Title="Connector for sliding plate to rotating rod" Tags="|mechanical-engineering|design|connections|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24787" PostTypeId="1" CreationDate="2018-11-21T22:34:34.873" Score="0" ViewCount="431" Body="&lt;p&gt;I understand that foundations are used in structures to support and spread the weight of the bridge over a large area. However I am struggling to understand why certain materials such as concrete are used in foundations over others. My first thoughts were based on how uplift may be effected. &lt;/p&gt;&#xA;" OwnerUserId="18324" LastActivityDate="2018-11-24T20:11:49.283" Title="Is a bridge more stable on concrete or soil?" Tags="|structures|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24790" PostTypeId="2" ParentId="24787" CreationDate="2018-11-22T02:30:01.603" Score="2" Body="&lt;p&gt;Concrete is a strong, reasonably low cost, readily available, versatile construction material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It lasts forever and keeps getting even stronger over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can become into a very flexible, or very stiff part of the structure, depending on the type of reinforcement and their position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be placed in a wide range of weather conditions and will last submerge under running corroding water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be used almost all over the world from Antarctica to hottest arrid climates in the desert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be modified, drilled into, attach brackets or hardware to to receive new configuration of superstructures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is responsible for some of the wonders in modern engineering and architecture in the world. Some examples are the highways in Europe, high-rises in Dubai and Hong Kong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All are built on concrete foundations that withstand the harsh salty ocean waters, none of which would be possible without concrete foundations.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-11-24T20:11:49.283" LastActivityDate="2018-11-24T20:11:49.283" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24791" PostTypeId="2" ParentId="24786" CreationDate="2018-11-22T03:57:27.477" Score="1" Body="&lt;p&gt;Use a nut or threaded boss fixed or welded to the plate.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-11-22T03:57:27.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24792" PostTypeId="2" ParentId="24785" CreationDate="2018-11-22T03:59:57.697" Score="1" Body="&lt;p&gt;You can just thread the inside of the pipe / tube and use an ordinary bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure the pipe / tube is thick enough not to fail after threading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on a comment, a cheaper variation that is commonly used is to crimp a nut into the tube - as long as the force required is low compared to the de-crimping force...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-11-22T07:55:34.910" LastActivityDate="2018-11-22T07:55:34.910" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24794" PostTypeId="1" CreationDate="2018-11-22T10:39:19.643" Score="1" ViewCount="460" Body="&lt;p&gt;I am currently working on a project using MSC Nastran and Patran trying to model only radial stiffness between two hollow cylinders. I have been able to idealize a connection which has both radial and axial stiffness in a cylindrical coordinate system using CBUSH elements. However when only radial stiffness is required I am encountering a fatal error due to excessive pivot ratios. I am wondering if it is possible to model radial stiffness only and if so how? &lt;/p&gt;&#xA;" OwnerUserId="18380" LastActivityDate="2018-11-22T10:39:19.643" Title="Modelling radial Stiffness in Patran Usinf CBUSH Element" Tags="|finite-element-method|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24796" PostTypeId="2" ParentId="24786" CreationDate="2018-11-22T13:38:13.943" Score="0" Body="&lt;p&gt;It sounds like you're looking for a &quot;Flanged Nut&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are typically manufactured for trapezoidal lead-screw type threads, but are also available in the more standard metric/imperial threadforms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are of course all sorts of variants of these, but without more information regarding your application, there's not much I can do to provided a more targeted recommendation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/41HPqm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/41HPqm.png&quot; alt=&quot;flanged nut&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-11-22T13:38:13.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24797" PostTypeId="2" ParentId="24567" CreationDate="2018-11-22T13:53:00.353" Score="1" Body="&lt;p&gt;It sounds like you have a PID control issue which will give you an unstable machine. I wouldn't recomend playing with the settings if you don't know what you are doing, but tweaking them may fix the problem. The below may fix your problem but it would probably be good for the PID values looked at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are completely removing the load then you need to make sure your specimen is supported properly so the pull bars and the specimen can be treated as a solid bar, capable of through zero fatigue testing. As a test for this if you ramp from a positive load to a negative load, you should have a nice smooth line in both load and position/strain. If you have any slack it the load train you not have a smooth line in one or more of the transducer channels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative would be not to go all the way to zero. If you keep a small near zero load (x) on the specimen when it should be at zero load (depending on what you can control to) and ramp to and from x. Then as long as x is small enough and the specimen is elastic in that range the relative strains + the strain at x may equal the absolute strain. I have used this method with metals many times, but have not had much experience in testing plastics so cannot say or sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that helps&lt;/p&gt;&#xA;" OwnerUserId="18382" LastActivityDate="2018-11-22T13:53:00.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24798" PostTypeId="2" ParentId="16863" CreationDate="2018-11-22T14:17:12.623" Score="5" Body="&lt;p&gt;No there is no equation for it, as it is caused by the processing differences of the different sizes and has nothing to do with the size of the material.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Smaller bars are subject to more work to get them down to the required size as they start out from the same size billet as the larger bars.&lt;/li&gt;&#xA;&lt;li&gt;Smaller bars also take less time heat up during heat treatments, so the effects and/or heat treatment are slightly different. (see comment by starrise)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The values in the standards are also not the strength of the material you buy, they are miniumum (and sometimes maximum) values. It can be that a larger diameter bar exceeds the minimum strength of a smaller diameter bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the bar can have an effect for materials with a large grain structure when if the bar is only a few grains wide you can have a phenomenon called grain size effect. The other condition I know of where it can happen is with surface treated or coated parts. The same depth of a surface treatment (e.g. carburizing) will have a higher surface to bulk material ratio in smaller bars.&lt;/p&gt;&#xA;" OwnerUserId="18382" LastEditorUserId="18382" LastEditDate="2018-11-22T15:12:39.590" LastActivityDate="2018-11-22T15:12:39.590" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24799" PostTypeId="2" ParentId="13662" CreationDate="2018-11-22T14:34:19.953" Score="1" Body="&lt;p&gt;This depends on the alloy type (and specific alloy) and the quality of the material after it's been produced which can change depending on the manufacturing parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The strengths of additive manufactured material are often better than cast when it's done properly, but I have yet to see any that are better than forged/rolled material&lt;/strong&gt;. I have seen material with large amounts of porosity which had properties considerable worse than cast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The failure Criteria also depends on the application, e.g. voids in SLM material can act as initiation points for fatigue failures greatly decrease the fatigue life of a material with good strength properties. Unfortunately the only way to definitively determine the properties for a material (suitable for structural designs) is to test it.&lt;/p&gt;&#xA;" OwnerUserId="18382" LastActivityDate="2018-11-22T14:34:19.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24800" PostTypeId="2" ParentId="23099" CreationDate="2018-11-22T15:06:00.633" Score="2" Body="&lt;p&gt;The bullets won't actually 'shatter' as they are soft lead (in a copper jacket). Bullets normally deform on impact with a hard surface (such as armour) and will spread out (like plasticine). Some of the energy normally goes into deforming the armour and denting it, but if the armour was hard enough and stiff enough, it could resisit the bullet enough making (almost) all of the energy from the bullet go into deforming the bullet itself. This looks to be whats happening in the video where the armour doesn't move or deform and the energy is focused into spreading the bullet out so much it breaks into small pieces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only problem with hard materials is that they are normally also brittle (as in ceramics). While the armour works great for smaller bullets something with enough energy could potentially cause the armour to shatter (although without having details of the armour it's impossible to tell).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone is developing it as an armour then it's safe to assume they will be looking to use/sell it and will likely have some plan to do so. But there may be more development required in order to reduce costs to the point someone will pay for it, or allow it to be manufactured more easily.&lt;/p&gt;&#xA;" OwnerUserId="18382" LastActivityDate="2018-11-22T15:06:00.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24801" PostTypeId="1" CreationDate="2018-11-22T18:20:58.103" Score="1" ViewCount="25" Body="&lt;p&gt;If we would want to build a family house (ICF) regarding operational costs, should we leave the insulation on the inside, or should we remove it from the inside? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we would remove it, the walls would provide a great thermal mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we would leave it as it, as a normal ICF, then we would only need to heat/cool the AIR inside and not the walls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q:&lt;/strong&gt; which solution would have less operational cost? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;House is a very energy efficient one, almost passive house. Big windows facing south (in summer, outside shaders against the sun)&lt;/p&gt;&#xA;" OwnerUserId="18387" LastEditorUserId="18387" LastEditDate="2018-11-22T19:37:40.040" LastActivityDate="2018-11-22T19:37:40.040" Title="Heat storage mass vs. AIR in an ICF house" Tags="|civil-engineering|building-design|building-physics|architecture|home-improvement|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24804" PostTypeId="1" AcceptedAnswerId="24806" CreationDate="2018-11-22T21:14:02.670" Score="1" ViewCount="205" Body="&lt;p&gt;I have a question about the characteristic dimension. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1&#xA;The drag on an airship is known to depend on the air density, ρ , dynamic viscosity,&#xA;μ , speed relative to the air , V , and a characteristic dimension, L . An airship is to&#xA;operate at 20 m/s in air at standard conditions. A model is constructed at 1/20 scale&#xA;and tested in a wind tunnel at the same air temperature to determine drag.&#xA;(a) Perform a dimensional analysis to determine the criterion which should be&#xA;considered to obtain dynamic similarity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;like the question said for the characteristic dimension L, How can i know what dimension is L.&lt;/p&gt;&#xA;" OwnerUserId="17811" LastActivityDate="2018-11-25T23:01:28.577" Title="Dimension Analysis For fluid mechanic" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24805" PostTypeId="1" AcceptedAnswerId="24808" CreationDate="2018-11-22T22:29:55.403" Score="0" ViewCount="438" Body="&lt;p&gt;I'm trying to design a plumbing system for a school project, and I want to verify that my initial piping layout from water source to outlets is logical before I start making calculations and programs for the system.  Here is my drawing, thanks to Rhodie:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PIOxI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PIOxI.png&quot; alt=&quot;LAyout&quot;&gt;&lt;/a&gt;&#xA;The green curve represents a hill that the building is on.  My logic for this design was that by minimizing the number of fittings I have in my system I can reduce the required head I would need to generate by the pump and save some money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice or comments would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="14040" LastEditorUserId="14040" LastEditDate="2018-11-23T01:23:05.350" LastActivityDate="2018-11-23T01:23:05.350" Title="Optimal plumbing layout for 3 storey building?" Tags="|fluid-mechanics|civil-engineering|building-design|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24806" PostTypeId="2" ParentId="24804" CreationDate="2018-11-22T22:52:10.903" Score="0" Body="&lt;p&gt;The drag will actually depend on the frontal area. This can be calculated as &lt;span class=&quot;math-container&quot;&gt;$constant*L^2$&lt;/span&gt;. What the constant is depends on what you &lt;em&gt;select&lt;/em&gt; L to be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is concerned specifically with the dimensionality, i.e. how does the drag force scale as you vary each of the components in the equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example, you're looking at a 1/20th scale model. A full scale model will be 20 times as long, 20 times as wide, have wheels 20 times as big, and consequently, the frontal area will be 400 times larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesnt matter what you select as “L” since the constant is determined experimentally. I’d suggest something easy to measure, like overall length of the ship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, in this particular model, the length is 4 times the diameter, and you define L=diameter, then the constant that is measured experimentally would simply be 1/4 of the value it would have been if you defined L=length.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-11-25T23:01:28.577" LastActivityDate="2018-11-25T23:01:28.577" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24807" PostTypeId="2" ParentId="23356" CreationDate="2018-11-23T00:06:31.583" Score="0" Body="&lt;p&gt;At your level this is a major question so to keep it simple, you are asking about moments aka turning force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issues you face are calculating the specific friction coefficient of the particular rock into which you are embedding the tool. That can only be done by soil analysis by a geotechnical lab. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this specific case you can make it quite simple to calculate by pushing a sharp rod into the ground with a device to measure the force needed to push it 100mm into the ground. That would tell you the minimum amount of force required to turn the auger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation is &lt;span class=&quot;math-container&quot;&gt;$F=\frac M{A}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that the force required to penetrate the soil is going to increase proportionately to depth and add it does it will require greater force due to increasing mass of soil being sheared. Since the acceleration is actually constant velocity, this is the expected result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally you would put a strong pole through the auger eye and turn the auger with that pole. The longer the pole, the less applied effort required to force the auger deeper since we calculate moments of force perpendicular to the line of action which generates torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F*d=R_f$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$R_f$&lt;/span&gt; is the resultant force. F is applied force and &quot;d&quot; is the perpendicular distance to the radial centre of the fulcrum. In this case that fulcrum point is the auger itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this clears it up a bit. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-11-23T00:06:31.583" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24808" PostTypeId="2" ParentId="24805" CreationDate="2018-11-23T00:23:52.973" Score="0" Body="&lt;p&gt;My first question is why would you want that length of pipe? The total length could be substantially reduced if the vertical pipe was placed close to point 1. Your materials cost would be halved. The way to calculate the best vertical point would be to spur outlet 1 straight from the riser and the other 2 distant outlets from that same single riser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure head has nothing to do with number of fittings. It is calculated according to gravity×vertical volume of water above inlet: your pump. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-11-23T00:23:52.973" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="24809" PostTypeId="2" ParentId="21451" CreationDate="2018-11-23T01:35:27.577" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2ijyV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2ijyV.jpg&quot; alt=&quot;enterIng description of answer&quot;&gt;&lt;/a&gt;&#xA;The image explains some forces for consideration. Is basic to point out the obvious issues...&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-11-23T01:35:27.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24811" PostTypeId="1" CreationDate="2018-11-23T07:15:44.450" Score="1" ViewCount="36" Body="&lt;p&gt;If one press some steel part into steel hole to get unsplittable assembly, does this operation speed matter? Is it better to hit by hammer, or to slowly jam with vice? Does it relate to steel type (e.g. soft or hardened)?&lt;/p&gt;&#xA;" OwnerUserId="18392" LastActivityDate="2018-11-23T07:36:43.610" Title="Does speed matter while one steel part is pressed into another?" Tags="|mechanical-engineering|materials|steel|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24813" PostTypeId="1" CreationDate="2018-11-23T09:04:56.987" Score="1" ViewCount="63" Body="&lt;p&gt;Having the Power/Torque-RPM curves of a vehicle, how can we derive the engine power equation and the torque out of these curves with respect to time?&lt;/p&gt;&#xA;" OwnerUserId="18394" LastActivityDate="2018-11-23T09:04:56.987" Title="Engine power and torque of a vehicle" Tags="|mechanical-engineering|thermodynamics|mathematics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24814" PostTypeId="1" CreationDate="2018-11-23T09:26:03.297" Score="1" ViewCount="489" Body="&lt;p&gt;I want to create a Pneumatic cylinder, The purpose is, Inside it will will contain a Nylon piston, and one end will be filled with clay. Cylinder will be closed from either side using end caps, which also has Push-in connectors(1/4&quot; BSP). Compressed Air from a cylinder(Max. pressure of Cylinder 10 Bar) will be given to Piston side through a Non return valve and a Regulator. Which will push the Clay out through other end.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What should be the Wall thickness of Cylinder? It should safely handle Maximum pressure that might arise.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I have already provided a Safety valve calibrated at 6 Bar in my prototype that I made from a GI water Pipe, which has thickness of 3.85 mm. Is there any way the pressure inside the tank could reach a pressure higher than the maximum pressure Compressor can deliver (Compressor is rated for 10 Bar, with cutoff at 8 Bar)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm thinking about making the cylinder using a 100 x 5.75 mm Aluminium Pipe and End caps from Aluminum Rods. Would this be enough?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Should I give more thickness for End caps than cylinder?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How to find the maximum pressure rating of the cylinder if material, Wall thickness and diameter are known?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18396" LastActivityDate="2024-03-27T10:00:55.170" Title="Pneumatic cylinder: Pressure rating" Tags="|pressure|pneumatic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24819" PostTypeId="1" CreationDate="2018-11-23T13:20:19.237" Score="1" ViewCount="229" Body="&lt;p&gt;I am trying to find how do enginners build roads especially highway interchanges. &#xA;Are they relying on any topology theories such as braids and knots theory?&#xA;Is there any connection or link between topology and designing interchanges ? &#xA;I can't find any article explaining it. &#xA;Thanks a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YGZ6G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YGZ6G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18306" LastActivityDate="2018-11-24T09:58:49.710" Title="How do enginners build highway interchanges ? Are highway interchanges build based on a mathematical theory such as braids and knots theory?" Tags="|mathematics|highway-engineering|roadway|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24821" PostTypeId="2" ParentId="24784" CreationDate="2018-11-23T16:35:50.280" Score="2" Body="&lt;p&gt;Instead of a curve fit, why not just do an interpolation of the data. Cubic spline interpolation should give sufficient precision. Look at scipy.interpolate&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-11-23T16:35:50.280" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24822" PostTypeId="2" ParentId="24819" CreationDate="2018-11-23T18:00:25.417" Score="3" Body="&lt;p&gt;It's much closer to graph theory. First, traffic projections: basing on known and projected urban data - habitation, production, recreation, trade zones, transit, etc, the general transition patterns are predicted: how many cars per hour, from which origins to which destinations. Knowing how many people want to go from where to where allows projecting road profiles - numbers of lanes required to allow given throughput at given speed on given graph edge. The road layout is at this point completely absent - there are just nodes of the graph - sources/sinks for the cars - and edges - roads; given required throughput.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the general numbers you can then match your available budget against common solutions - the generalized 'generic' interchanges that fit given local conditions and are 'doable' within the budget. Usually the more efficient the solution the more expensive it is; sprawling, wide roads, long multi-level viaducts, gentle curves, the entire interchange covering a large (expensive) land area. If you tighten things, you can save up money but reduce throughput. At this point the general model is decided. Engineers don't design these. This is an academic domain. Engineers just know these, apply them, and...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...customize. Obviously one size doesn't fit all. Local conditions have their demands. So the generic model of the interchange needs to be modified - evolved to match the local conditions. Inclines, angles, matching road profiles to predicted throughputs - fitting the graph edges into actual roads and adjusting width and profile (curve radii, inclines) to the individual predicted traffic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Occasionally a design bureau with a particularly ambitious and talented (or at least reputed to be talented) engineer will create an entirely new pattern from scratch, for specific situation. If it passes the scrutiny of interested parties (one way or another) it may be implemented and either enters the canon, as a pattern for future derivatives, or becomes an eternal source of shame and mockery for that engineer.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-11-23T18:00:25.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24823" PostTypeId="1" CreationDate="2018-11-23T18:25:45.530" Score="0" ViewCount="248" Body="&lt;p&gt;Could someone please clarify the difference between flux and normal derivatives?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am studying PDE's and I came across Neumann Boundary condition (In this condition, the value of the &lt;strong&gt;normal derivative&lt;/strong&gt; is given).  I then came across the following: For a steady state thermal process, the Neumann condition describes a prescribed &lt;strong&gt;heat flux&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18402" LastEditorUserId="19906" LastEditDate="2020-12-09T15:02:41.663" LastActivityDate="2023-12-24T21:05:15.607" Title="Flux and Normal Derivatives" Tags="|modeling|mathematics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24824" PostTypeId="2" ParentId="24331" CreationDate="2018-11-23T18:59:05.747" Score="1" Body="&lt;p&gt;First: A transistor acts as a switch - amplifier. Base gets token amount of electricity - Emitter starts emitting as much as is put into the collector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connect two transistors in a row - Collector of (1) to power, emitter of (1) to collector of (2). Emitter of (2) is &quot;output&quot;. Apply voltage to bases of both, and it passes current, producing a &quot;1&quot;. But remove any, and it's &quot;0&quot;. That's an AND gate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now connect them in parallel instead. Two collectors to power, two emitters together to output. Any of them will output current into output line. You must disable both to produce a &quot;0&quot;. That's OR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now connect a resistor and a transistor between power and ground. Tap into the point between the transistor and the resistor as &quot;output&quot;. If you apply voltage to the base, the transistor passes all the current and voltage on 'output' drops to zero, connected through the open transistor to ground. Whatever comes through the resistor escapes through the path of least resistance. You have zero output on positive input. Now remove the base voltage and the transistor's resistance skyrockets. Now the resistor drives enough current from the power line to produce a clear '1' on output. That's NOT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can now combine these, connecting outputs to inputs, into a whole bunch of others. NAND = NOT A AND NOT B. XOR: A AND NOT B, OR B AND NOT A.  XNOR: (A AND B) OR (NOT A AND NOT B). And so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connect two XNOR gates in an X:  one input pin of each as input, the other connected to output of the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wHoBv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wHoBv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There, this is one bit of memory. With a rather cumbersome writing (S=1: set to 1, R=1: set to 0, S=0,R=0 - hold value, S=1,R=1 - burst in flames, or at least produce nonsensical noise) but it can be made into a much more sensible thing with a couple other gates on input and output. Attach NAND and AND on input, one input on each from 'in', the other from 'write'. Attach a 3-state buffer on output, with 3rd state (high impedance) input marked 'read'. Now you can write or read this on demand, without 'read' it won't touch the bus you connect it to, without 'write' it won't change state no matter what appears on 'in'. You may connect 'in' and 'out' to the same line. If 'write' is active, it will read and remember what's on that line. If 'read' is active, it will drive that line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's a fully usable bit of memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put 8 in a row, each output to one line. The 8 lines are a data bus. You have a byte of data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get other 8 lines (from now on known as Address Bus) and connect them to a demultiplexer. That's a construct of a mass of AND and NAND gates, that converts every possible combination of the states of the 8 lines into 256 different output lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now put repeat the row of 8 bits 256 times, every row getting an AND between the one dedicated incoming line from the demultiplexer dedicated to it, and a single global 'Write', and one AND between the same demux line and the 'Read'. And you have a working 8KB bank of memory. Set the right combination of 8 bits on the address bus, state your intent: 'read=1' or 'write=1' and you'll either store the state of the data bus in given byte, or output it onto the data bus from that byte.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add three-state buffer on all inputs and outputs (you have 18 of them so far, 8 data, 8 address, 1 read, 1 write), and activate them through 19th input: Chip Enable. Pack that into an integrated circuit and you have the 8KB RAM chip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Want more RAM? A demultiplexer controlling 'Chip Enables' of multiple chips will let you switch between multiple dies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, in the CPU, as the internal sequence moves to 'READ COMMAND', a register (just a byte, 8 RS flip-flops, set aside for a special purpose) named Program Counter is connected to the address line. Read is activated. Through the Data line the set of bits that corresponds to the new command returns to the CPU. It goes to 'command decoder' - which is just another demultiplexer. And that demultiplexer activates one line that activates a specific subsystem of a whole zoo of subsystems of the CPU - responsible for that specific instruction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, the instruction is JMP. Jump to a different place in the program, marked by the value in memory immediately after the command. The set of gates activated through the Command decoder will activate the following operations in sequence: first, advance the program counter. Then output it onto the address bus. Activate 'Read', and then stuff the output from the data bus - right into Program Counter. Command done. The control returns to reading the next command - except 'next' is now pointed by the Program Counter which just got modified. That's what the JMP command worked.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-11-23T18:59:05.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24825" PostTypeId="2" ParentId="24814" CreationDate="2018-11-23T23:29:23.080" Score="0" Body="&lt;p&gt;I recommend you to use one of the codes/standards in order to be on the safe side. For example you can calculate cylinder wall thickness according to ASME, the following formulas are taken from the pressure vessels, but they are good enough to give you a point:&#xA;for flat cover and cylinder respectively&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bBcR0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bBcR0.jpg&quot; alt=&quot;For flat covers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xZsTd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xZsTd.jpg&quot; alt=&quot;For cylinder&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;E&quot; is a joint efficiency , if you use seamless part consider it 1, if not than 0.7 for circumferential stress. For longitudinal stress in your case ,I think, you should use 0.7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;S&quot; is a permissible material stress= UTS/3.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;P&quot; is a pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;t&quot; is a wall thickness&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;C&quot; is a factor of attachment of a flat cover, usually between 0.1 to 0.3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;d&quot; inner diameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;D&quot; and &quot;R&quot; are outer diameter and radius&lt;/p&gt;&#xA;&#xA;&lt;p&gt;be consistent with proper units(SI,imperial etc.)!&lt;/p&gt;&#xA;" OwnerUserId="18401" LastEditorUserId="18401" LastEditDate="2018-11-23T23:35:08.410" LastActivityDate="2018-11-23T23:35:08.410" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24826" PostTypeId="1" AcceptedAnswerId="24836" CreationDate="2018-11-24T01:03:31.390" Score="1" ViewCount="101" Body="&lt;p&gt;It is my understanding that of all three states of matter, gases have the lowest thermal conductivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, why are &lt;em&gt;solid&lt;/em&gt; materials, such as fiberglass and polyurethane, used for this purpose? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it not make the most sense to simply fill the space with air?&lt;/p&gt;&#xA;" OwnerUserId="18404" LastActivityDate="2018-12-04T00:04:10.013" Title="Wouldn't it be most effective to insulate a house with just an air space?" Tags="|heat-transfer|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24827" PostTypeId="2" ParentId="24826" CreationDate="2018-11-24T01:13:20.677" Score="4" Body="&lt;p&gt;Insulation is used to &quot;fill the space&quot; with something other than air, because air can conduct thermal activity via convection. Convection in this application means that the air on the outside (cold in this example) will fall to the bottom of the air space, forcing the warmer air on the inside upward and outward, causing it to become colder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such movement can be observed in a simple manner. Inflate a mylar helium ballon and attach sufficient weight to create neutral buoyancy. As the balloon travels to a colder or warmer location, it will rise or fall depending on the temperature in the area where the neutral state was created. This is because it is now in the convection airflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Insulation reduces this convection by breaking up the air into smaller packets and surrounding it with fiberglass or other material. It is more difficult for sufficient convection (and conduction) to occur over a specific time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the temperature in an area does not change appreciably (inside heating), thermal equilibrium can be expected. An example would be a shed or outbuilding in a very cold environment. Even insulated, the inside temperature can be expected to match the outside temperature. This would be mitigated by some form of interior heating, but also by solar energy (sunshine) on the building walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vacuum is a much more efficient insulator, hence the dewar flasks commonly called Thermos bottles. Today's steel insulated mugs, originally manufactured by Yeti use a vacuum within a double walled sealed construction. Even though there is conduction via the steel that is common to the inside and outside, the internal vacuum removes the convection aspect. Don't drop one and break the seal, however. The cooling capacity drops from 24 hours to only a few!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I neglected to add the aspect of conduction as an additional factor in insulation. The outside material uses conduction to transfer the heat inside to the outside, otherwise there would be no problem with convection in the air. &lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2018-11-24T11:33:00.823" LastActivityDate="2018-11-24T11:33:00.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24828" PostTypeId="1" CreationDate="2018-11-24T02:07:46.380" Score="1" ViewCount="51" Body="&lt;p&gt;Does anyone know how to create the Verilog code for low pass filter in FPGA? I have no idea from where to start the project. Can anyone help?&lt;/p&gt;&#xA;" OwnerUserId="18406" LastActivityDate="2018-11-24T02:07:46.380" Title="Low Pass Filter using FPGA" Tags="|electronic-filters|" AnswerCount="0" CommentCount="2" ClosedDate="2018-11-24T12:49:01.057" ContentLicense="CC BY-SA 4.0" />
  <row Id="24831" PostTypeId="2" ParentId="24819" CreationDate="2018-11-24T09:46:06.180" Score="0" Body="&lt;p&gt;Somekind of theory like that, I reckon. I've often wondered this myself! When I was a kid I became fascinated by this &amp;amp; had a phase of 'designing' such junctions - just drawing them freehand on sheets of paper - I must've done &lt;em&gt;dozens&lt;/em&gt; of them!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing I &lt;em&gt;could&lt;/em&gt; mention &lt;em&gt;concretely&lt;/em&gt; (haha! see what I did there!?) in this connection, that your question has 'popped' out of my memory, is that the optimum curve for a road-bend is a &lt;em&gt;Cornu spiral&lt;/em&gt;, which is simply the curve on which curvature is proportional to arc-length; but the representation of which in cartesian co-ordinates broaches some fascinating mathematics - &lt;em&gt;Fresnel integrals&lt;/em&gt; which can in turn be expressed as &lt;em&gt;Bessel functions&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's distinctly possible that &lt;em&gt;knot theory&lt;/em&gt; (and that rather than &lt;em&gt;braid&lt;/em&gt; theory) &lt;em&gt;has&lt;/em&gt; been used. &lt;em&gt;Not&lt;/em&gt; (ha ... &lt;em&gt;no ... leave&lt;/em&gt; it!) ... not  so much the &lt;em&gt;profound&lt;/em&gt; depths of knot theory, but at least maybe upto the point of checking some &lt;em&gt;gallery&lt;/em&gt; of knots to find whether some configuration is equivalent to some other more-economical-to-build one ... which can be an &lt;em&gt;amazingly&lt;/em&gt; subtle question: have you read the story of the so-called 'culprit' - a knot that was only perceived to be the same as some other by a sort of &lt;em&gt;fluke&lt;/em&gt; - a guy who was actually a &lt;em&gt;lawyer&lt;/em&gt; just decided to browse a bit of knot-theory, and just somehow 'saw' that they were the same after &lt;em&gt;no-one&lt;/em&gt; else had before!&lt;/p&gt;&#xA;" OwnerUserId="18209" LastEditorUserId="18209" LastEditDate="2018-11-24T09:58:49.710" LastActivityDate="2018-11-24T09:58:49.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24832" PostTypeId="2" ParentId="24823" CreationDate="2018-11-24T13:49:57.333" Score="0" Body="&lt;p&gt;The 1-D energy balance on an object with no internal heat generation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$k \frac{\partial ^2T}{\partial z^2} = \rho \tilde{C}_p \frac{\partial T}{\partial t} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A viable boundary condition (BC) for energy flow into an object is written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$-k A\left. \frac{dT}{dz}\right|_{boundary} = \dot{q}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider an solid object with heat flow entering from the right. The BC says, the heat flow (J/s) that enters is conducted through the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This condition applies regardless of steady state or not. In steady state, &lt;span class=&quot;math-container&quot;&gt;$\partial T/\partial t = 0$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-11-24T13:49:57.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24833" PostTypeId="2" ParentId="24784" CreationDate="2018-11-24T14:13:24.103" Score="2" Body="&lt;p&gt;Here are fitting results with the equation shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZoFSm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZoFSm.png&quot; alt=&quot;Plot with fitting results&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any greater accuracy requires a model beyond this very crude, semi-empirical fit. Pay attention to the confidence limits (1&lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; values). To first order, they can be propagated through your prediction equation using linear uncertainty analysis. This can help you decide whether extra effort is needed to be more accurate, for example by splitting the pressure regions and/or by getting higher resolution data (more data points in the full or desired range).&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-11-24T14:33:04.113" LastActivityDate="2018-11-24T14:33:04.113" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24834" PostTypeId="2" ParentId="24784" CreationDate="2018-11-24T14:15:33.690" Score="2" Body="&lt;p&gt;The density varies exponentially - but at rate dependent not just on altitude but also temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z1zyz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z1zyz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://scied.ucar.edu/atmosphere-layers&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uXy5c.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uXy5c.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://acmg.seas.harvard.edu/people/faculty/djj/book/bookchap2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want a more precise fit, you'll have to split it up into atmospheric layers: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Troposphere: 0 to 12 km (0 to 7 miles)&lt;/li&gt;&#xA;&lt;li&gt;Stratosphere: 12 to 50 km (7 to 31 miles)&lt;/li&gt;&#xA;&lt;li&gt;Mesosphere: 50 to 80 km (31 to 50 miles)&lt;/li&gt;&#xA;&lt;li&gt;Thermosphere: 80 to 700 km (50 to 440 miles)&lt;/li&gt;&#xA;&lt;li&gt;Exosphere: 700 to 10,000 km (440 to 6,200 miles)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and fit the exponent into each of these - possibly introducing additional thresholds for stratosphere (at 20km and 35km) - a spline created of exponential curves. Trying to find a single simple function to fit reality with this sort of complexity is a misguided attempt.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2018-11-25T18:03:57.513" LastActivityDate="2018-11-25T18:03:57.513" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24836" PostTypeId="2" ParentId="24826" CreationDate="2018-11-24T16:54:10.583" Score="1" Body="&lt;h1&gt;Background&lt;/h1&gt;&#xA;&lt;p&gt;The trade off is between conduction and convection. Heat conduction &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_k$&lt;/span&gt; (W) is set as&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{q}_k = k A \frac{\Delta T}{w} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is thermal conductivity (W/m K), &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is area (m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;), &lt;span class=&quot;math-container&quot;&gt;$\Delta T$&lt;/span&gt; is temperature difference (&lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;K), and &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; is the width (m). Heat convection &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_h$&lt;/span&gt; is set as&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{q}_h = h A \Delta T $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is convection coefficient (W/m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; K). As the thickness of the gap &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; increases, the rate of conduction decreases. At zero thickness, the convection coefficient &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is zero. As &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; increases, &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; increases to a value indicative of a natural convection system.&lt;/p&gt;&#xA;&lt;p&gt;Armed with this information, you can look up values of &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; for air and &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; for natural convection in air. Derive a ratio as&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\dot{q}_h}{\dot{q}_k} = \frac{hw}{k}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Use this to find the value of &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; where convection dominates conduction. Any air gap thinner than this will be OK because convection will be lower than conduction, and conduction is low in any case. Any air gap thicker will be controlled by convection.&lt;/p&gt;&#xA;&lt;p&gt;This presumes that air is not &amp;quot;blowing&amp;quot; in the gap. At that point, &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; can increase by one or two orders of magnitude. By example, what was acceptable for natural convection as a 1cm air gap becomes acceptable perhaps only below 0.1mm as an air gap.&lt;/p&gt;&#xA;&lt;h1&gt;Examples&lt;/h1&gt;&#xA;&lt;p&gt;By reference with &lt;span class=&quot;math-container&quot;&gt;$k_{air} \approx$&lt;/span&gt; 0.02 W/m K as per &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_thermal_conductivities&quot; rel=&quot;nofollow noreferrer&quot;&gt;this reference&lt;/a&gt;, we can find as per &lt;a href=&quot;https://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this reference&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Free Convection (low): &lt;span class=&quot;math-container&quot;&gt;$h \approx 0.5$&lt;/span&gt; W/m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; K &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow w \approx 2$&lt;/span&gt; cm&lt;/p&gt;&#xA;&lt;p&gt;Free Convection (high): &lt;span class=&quot;math-container&quot;&gt;$h \approx 500$&lt;/span&gt; W/m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; K &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow w \approx 40~\mu$&lt;/span&gt;m&lt;/p&gt;&#xA;&lt;p&gt;Maintaining a low free convection state in a vertical gap that is 2 cm wide is difficult. Once the air at the top of the gap becomes even slightly hotter than the air at the bottom, a convection cell is established. The air rotates from bottom to top of the cell, and the convection coefficient goes up by an order of magnitude already.&lt;/p&gt;&#xA;&lt;h1&gt;A Case for an Exception?&lt;/h1&gt;&#xA;&lt;p&gt;Modern homes with brick exteriors have an air gap between the brick and the panel insulation.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.carsondunlop.com/inspection/blog/brick-houses-solid-masonry-vs-brick-veneer/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.carsondunlop.com/inspection/blog/brick-houses-solid-masonry-vs-brick-veneer/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.finehomebuilding.com/2013/03/06/best-practices-methods-for-installing-brick-or-stone-veneer&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.finehomebuilding.com/2013/03/06/best-practices-methods-for-installing-brick-or-stone-veneer&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is to avoid the case where moisture will collect at the junction of brick-touching-wall. It also allows the bricks to dry properly when they become wet.&lt;/p&gt;&#xA;&lt;p&gt;The air gap is a modest thermal insulator. However, when the sun shines on the brick, free convection cells can form in the gap. This is one reason why thermal insulation is still used between the brick+air gap and the interior house walls.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-12-04T00:04:10.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24838" PostTypeId="1" CreationDate="2018-11-24T17:20:47.810" Score="3" ViewCount="74" Body="&lt;p&gt;Is their a good method of making aluminum disks from molten aluminum?&#xA;I have a large amount of aluminum castings and I would like to make rounds that could be chucked into my lathe and used to turn parts. Typically the castings are reduced to molten and then poured into muffin tins for storage or use later.&#xA;It is difficult to hold the tapered shape in a lathe chuck. Is there a reusable method to pour longer shapes or am I stuck with sand casting the parts?&lt;/p&gt;&#xA;" OwnerUserId="18410" LastEditorUserId="19778" LastEditDate="2023-11-17T20:08:55.203" LastActivityDate="2023-11-17T20:08:55.203" Title="Pouring aluminum into disks or cylinder shapes?" Tags="|casting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24840" PostTypeId="1" AcceptedAnswerId="24887" CreationDate="2018-11-24T22:27:31.413" Score="1" ViewCount="60" Body="&lt;p&gt;Example: let's say that I have a rectangular part that I would like machined. I'd like to cut it to change the length, but the width and thickness are fine as is. The width and thickness are not nice round numbers. Is there any convention on how to communicate this through a technical drawing? Something like &quot;AS IS&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to &lt;a href=&quot;https://engineering.stackexchange.com/questions/11130/do-i-need-to-mention-every-dimension-on-a-to-scale-drawing]&quot;&gt;Do I need to mention every dimension on a to-scale drawing?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16019" LastActivityDate="2018-11-27T19:05:03.657" Title="How to communicate to keep original dimensions of a part in a technical drawing?" Tags="|drafting|technical-drawing|drawings|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24841" PostTypeId="2" ParentId="24840" CreationDate="2018-11-25T01:11:02.077" Score="1" Body="&lt;p&gt;There are standard symbols to show which surfaces should be machined, and which should not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Fig 11.3 in &lt;a href=&quot;http://www.engr.mun.ca/~dfriis/cadkey/program/textappi.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.engr.mun.ca/~dfriis/cadkey/program/textappi.html&lt;/a&gt; (or preferably, read a proper standards document!)&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-11-25T01:11:02.077" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24842" PostTypeId="2" ParentId="24838" CreationDate="2018-11-25T01:43:51.723" Score="1" Body="&lt;p&gt;A very home-spun option that should work would be to use a piece of tubing of a diameter to suit your purpose. Seamless steel tubing is available particularly on Aircraft Spruce and Specialty in many different diameters and wall thicknesses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature coefficient of steel compared to aluminum (&lt;a href=&quot;https://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;11 to 12.5 versus 8.1&lt;/a&gt;)is sufficient to allow you to pour the aluminum (ideally in one continuous pour) into the tubing and allow it to cool. As it cools, it should shrink enough to slide out of the tube. The linked page also indicates that stainless steel has an even higher differential, 16 to 17.3, but I'm sure a much higher cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An obvious advantage to this method is flexibility in cast cylinder diameters and durability of the &quot;mold,&quot; part of your requirements.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-11-25T01:43:51.723" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24843" PostTypeId="2" ParentId="7414" CreationDate="2018-11-25T01:58:00.170" Score="-1" Body="&lt;p&gt;I have no idea whether this link will work for you but EdX has been superior education to my taught degree. This video through MITx was very helpful and reminded me that engineering is not simple. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://courses.edx.org/courses/course-v1:MITx+2.01x+3T2018/courseware/W0_Intro_and_Prelim/Review_Forces_Moments/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Forces and Moments of Couples&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="1832" LastEditDate="2018-11-25T15:44:27.327" LastActivityDate="2018-11-25T15:44:27.327" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24844" PostTypeId="2" ParentId="6186" CreationDate="2018-11-25T02:06:45.730" Score="0" Body="&lt;p&gt;Higher concrete strength also has faster solidifying times. There is a trade-off with all types of concrete so you'd want to know how the resultant of higher strength is achieved and for what purpose they make it higher strength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You only really need the 90th percentile for regular use even when up to 15% variation of concrete design is above the top 5th percentile of mean production output. Usually this increased strength is a result of poor production management oversight. Check that out before staking your reputation on the final products made by another person. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2018-11-25T02:06:45.730" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24845" PostTypeId="1" CreationDate="2018-11-25T03:58:49.357" Score="1" ViewCount="33" Body="&lt;p&gt;I have been doing research on renewable energy systems and I must admit that I am a newbie in this specific area. I have noticed that stand-alone systems have their own specific monitoring system available on the market. I just wonder what functionalities and/or features would be added to these separate monitoring systems if I decided to build my own hybrid system. It's going to be like a common off-grid wind-solar system with a battery bank as backup power source.&lt;/p&gt;&#xA;" OwnerUserId="15646" LastActivityDate="2018-11-25T06:17:04.970" Title="Hybrid Energy Monitoring Systems" Tags="|renewable-energy|solar-energy|wind-power|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24846" PostTypeId="1" AcceptedAnswerId="24857" CreationDate="2018-11-25T04:31:10.227" Score="0" ViewCount="628" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UYVuT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UYVuT.png&quot; alt=&quot;question&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ca1Ot.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ca1Ot.png&quot; alt=&quot;solution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having a lot trouble understanding this problem, I don't even know where to begin or how to articulate my difficulties, but I will try my best.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the question asking to find the torque developed at the part in the picture where it says &quot;Pipe supported by wall&quot;? &lt;/li&gt;&#xA;&lt;li&gt;What is meant by the &quot;Horizontal section&quot; of the pipe? In the solution it says that it's going 24 in. into the paper. Okay, so what? Is water flowing through there? I thought it was flowing through the pipe supported by the wall. Or is it flowing through both the pipe support by the wall AND the pipe section that's going 24 in. into the paper?&lt;/li&gt;&#xA;&lt;li&gt;In the solution, I don't understand how the moment arm for the inlet flow is zero and the moment arm for the outlet flow is 24 in. Where is this being measured from? I thought it was being measured from the &quot;Pipe supported by wall&quot;? Where is the force being applied?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I apologize for my extreme confusion. I will be extremely thankful for any help that I can get. &lt;/p&gt;&#xA;" OwnerUserId="18364" LastActivityDate="2018-11-25T19:06:59.143" Title="How to use the moment-of-momentum equation for water flowing through a pipe?" Tags="|mechanical-engineering|fluid-mechanics|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24847" PostTypeId="1" CreationDate="2018-11-25T04:57:51.820" Score="1" ViewCount="43" Body="&lt;p&gt;Basically we are designing a pneumatic conveyor system for transferring 11 tons of beans per hour.&#xA;We need to decide on the specifications of the blower. What are the calculations necessary to decide on a blower wrt. static pressure, flow velocity and type of blower?&lt;/p&gt;&#xA;" OwnerUserId="18412" LastActivityDate="2019-03-06T05:31:09.540" Title="Choosing a blower for a Pneumatic Conveyor" Tags="|fluid-mechanics|pressure|pumps|airflow|pneumatic|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24849" PostTypeId="2" ParentId="11217" CreationDate="2018-11-25T07:36:20.507" Score="-1" Body="&lt;p&gt;Working on the idea that equilibrium exists in the members already, we can suggest that the vertical force is equal to the horizontal force at point A. This means we need only calculate the resultant for the truss diagonal starting at A. Take A as a point moment for 3 forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_A=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$3F_A \sin 270&#xA;-B_1 \cos 0&#xA;-R_A \sin \beta=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$3F_A \sin 270&#xA;-B_1 \cos 0&#xA;=R_A \sin \beta$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{3F_A \sin 270&#xA;-B_1 \cos 0}{ \sin \beta}&#xA;=R_A$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$R_A$&lt;/span&gt; exerts an horizontal force on all members &lt;span class=&quot;math-container&quot;&gt;$B_1, B_2, B_3$&lt;/span&gt; we should be able now to calc the rest of the truss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the horizontal reaction at B1 would be 10kN and the diagonal descent force from A would be 58.614kN compressionThis results in a compressive perpendicular force to B2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$58.614 \cos (180-\arctan \frac 3{2})=32.513kN$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That force must therefor be distributed over the 3 vertical pin joints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In spite of all this, a few examples below explain how the calculations work with statically determinate structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LG49O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LG49O.png&quot; alt=&quot;First section calculation&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UFcqC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UFcqC.png&quot; alt=&quot;Modified Truss with roller&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HiWzL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HiWzL.png&quot; alt=&quot;Modified Truss with roller calc&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2018-11-26T16:19:12.907" LastActivityDate="2018-11-26T16:19:12.907" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="24851" PostTypeId="1" CreationDate="2018-11-25T07:47:26.873" Score="0" ViewCount="104" Body="&lt;p&gt;I am studying about static pressure,dynamic pressure and the principle behind pitot tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However in pitot tubes we apply Bernoulli equation between two points which are not on the same streamline. So this should not work for rotational flows. I understand that the very application of Bernoulli equation requires there be no friction and hence no viscosity. So if the fluid was not rotating beforehand it has to be irrotational for us to apply Bernoulli equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way to do this for a rotational flow with or without viscous effects.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastEditorUserId="683" LastEditDate="2018-11-25T15:05:54.827" LastActivityDate="2018-11-25T15:05:54.827" Title="How can pitot tubes give us local velocity in a flow" Tags="|mechanical-engineering|fluid-mechanics|chemical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24852" PostTypeId="1" CreationDate="2018-11-25T12:05:58.517" Score="1" ViewCount="323" Body="&lt;p&gt;In casual conversation, and often in bill of materials, tolerances are specified without the qualifier +/- xx%, often because the specific ± character is hard to obtain (especially in legacy UNIX EDA systems) E.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100R 1% 0603 resistor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a convention as to whether this should be interpreted as +/- 1% i.e. resistance value from 99.0-101.0 ohms or as a total tolerance, i.e. resistance value from 99.5-100.5 ohms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The datasheet should specify this of course, but I was wondering if there is a convention in scientific / engineering speech whether unspecified tolerances are one or two-sided?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To all: I understand that everyone is trying to be helpful but I not asking for how to solve the problem. My question is &lt;em&gt;literally&lt;/em&gt; with regard to whether there are any &lt;em&gt;conventions in interpretation&lt;/em&gt;, in electronic engineering, of just &quot;1%&quot; as +/-1% or +/- 0.5%.&lt;/p&gt;&#xA;" OwnerUserId="7515" LastEditorUserId="7515" LastEditDate="2018-11-28T14:49:34.273" LastActivityDate="2018-11-28T14:49:34.273" Title="When tolerances are mentioned without +/-, should they be considered as a span?" Tags="|tolerance|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24853" PostTypeId="2" ParentId="24852" CreationDate="2018-11-25T13:04:20.227" Score="3" Body="&lt;p&gt;If it says +/- 1% then it means from -1% to +1%. in everyery standard i have ever seen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;+/- is simply shorthand for writing a symmetric bound one can also make unsymmetric bounds like 0 to 1% or even positive positive bounds like +0.001 to + 0.025.  Granted these are rare in electrical components  but  play a big  role in mechanical engineering where its useful to design both shaft and the hole for its  fitting  as having same nominal value but indicate the fit with tolerance entries. You see a symmetric bound hole with symmetric bound insert would not reliably go together, but would be really tedious and error prone to call a 4mm screw a 3.9 mm screw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally when people talk about percentages they mean within 1% of the target value again it means +/-1%. But one should really try to avoid doing this.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-11-26T16:35:26.100" LastActivityDate="2018-11-26T16:35:26.100" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24855" PostTypeId="2" ParentId="24593" CreationDate="2018-11-25T15:03:37.180" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;From what I understand, generally flux can be defined as &quot;some quantity per unit of some other quantity&quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's about right. The &lt;em&gt;some other quantity&lt;/em&gt; is often one of particular interest  Typically, it is a surface, or a facet of a surface, used in control volume analysis. And there is a flux operator. It returns a scalar quantity from an vector operation. In the case of fluids, the fluid advects the quantity of interest. So you have a description of the density of some thing of interest at time t (a scalar field) and a velocity field of the fluid (a vector field), and a reference surface S. The Flux operator tallies the rate at which stuff is advecting through the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, when dealing with pure field equations as in Electromagnetism, there is no need for an advective ether. Flux pertains to the vector field itself with respect to some surface of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We often have a system of equations that all involve this same control volume, and which have multiple conservation laws, which relate to a potential field. The flux idea is very useful in writing these equations in a compact form. For example, the equation below can represent a system of three conservation equations. &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; being the flux function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d}{dt} \iint_{\Omega}\mathbf q\ dx\ dy\ +  \int_{\partial \Omega} F(\mathbf q)\cdot\mathbf n\ ds = \iint_{\Omega} {\psi}\ dx\ dy $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations would be conservation of mass, conservation of x momentum, and conservation of y momentum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See page 4 of this document to see this in action. - &lt;a href=&quot;https://pdfs.semanticscholar.org/5844/0936fd380a3e0806cbb7b2507f05d5e8868b.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Numerical Approximation of the Nonlinear Shallow Water Equations&#xA;with Topography and Dry Beds:&#xA;A Godunov-Type Scheme, by&#xA;David L. George&#xA;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a nice series of three videos from the Kahn Academy that deals with the flux operator. They play in sequence. The first one is &lt;a href=&quot;https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/3d-flux/v/conceputal-understanding-of-flux-in-three-dimensions&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There should be a badge for learning mathjax. That formula took me 45 minutes.)&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-11-25T16:18:25.943" LastActivityDate="2018-11-25T16:18:25.943" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24857" PostTypeId="2" ParentId="24846" CreationDate="2018-11-25T19:06:59.143" Score="0" Body="&lt;p&gt;This pipe's  bends and the flow of water in it create two reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- momentum on the wall support created by the first 90 degree elbow, which is not asked for in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- torque created by the out flow downward which is being asked and answered by the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mswya.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mswya.jpg&quot; alt=&quot;This is a 3d sketch of the pipe.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-25T19:06:59.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24858" PostTypeId="2" ParentId="19253" CreationDate="2018-11-25T20:28:17.387" Score="0" Body="&lt;p&gt;Cornering stiffness is usually measured on Force And Moment Machine (for example:  &lt;a href=&quot;https://www.smithersrapra.com/testing-services/by-sector/automotive/tire-testing/force-and-moment&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.smithersrapra.com/testing-services/by-sector/automotive/tire-testing/force-and-moment&lt;/a&gt;).  Basically, the tire is in contact with a moving surface --either a flat track or a large drum-- which drives the tire at a particular speed under a particular load.  Then the turn angle changes and sensors measure the lateral forces as a function of turn angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is expensive, specialized equipment; probably not easily available to a university.  But you could approximate something by pulling the tire behind a vehicle and measuring the lateral forces.  Still not easy, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tires are not steel belted, there is a very old book which had formulas for these things, Mechanics of Pneumatic Tires, edited by Samuel Clark.  &lt;/p&gt;&#xA;" OwnerUserId="16229" LastActivityDate="2018-11-25T20:28:17.387" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24859" PostTypeId="1" CreationDate="2018-11-26T02:05:06.367" Score="1" ViewCount="128" Body="&lt;p&gt;If a power grid were to fully switch to renewable power generation methods how would they deal with reactive loads?&lt;/p&gt;&#xA;" OwnerUserId="18416" LastActivityDate="2018-11-26T04:13:51.203" Title="How do renewable energy sources deal with reactive loads?" Tags="|electrical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24860" PostTypeId="1" CreationDate="2018-11-26T02:41:21.413" Score="1" ViewCount="51" Body="&lt;p&gt;I am working on a project that uses a variable-speed pump to achieve a desired pressure across a medium (coffee grounds). The system includes a pressure sensor in-line between the pump and the medium. I've been reading up on water flow through soils (as this seems somewhat analogous), but I am having difficulty extracting the important information.&#xA;Does there exist a reliable way to model pressure/flow of water through saturated soils and if so, might someone be able to highlight this discussion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="5755" LastEditorUserId="61" LastEditDate="2018-11-26T09:16:25.697" LastActivityDate="2019-03-25T03:10:01.553" Title="Modeling Water Flow Through Saturated Medium" Tags="|mechanical-engineering|control-engineering|hydraulics|flow-control|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24861" PostTypeId="2" ParentId="24859" CreationDate="2018-11-26T04:13:51.203" Score="2" Body="&lt;p&gt;They would accomodate them by adding the necessary amount of either capacitance or inductance to the transmission lines to cancel the reactance. Note that this is commonly done today in order to trim the impedance of a large generator to match that of a high-voltage transmission line. It is also done to cancel the reactive component of a large electrical load in a factory that has either a significant amount of inductance or capacitance in it. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-11-26T04:13:51.203" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24862" PostTypeId="1" AcceptedAnswerId="24871" CreationDate="2018-11-26T09:33:58.443" Score="1" ViewCount="611" Body="&lt;p&gt;I'm trying to get the compressive strength (and Young's modulus) of a clay sample. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The method I found is called 'Unconfined Compressive Strength of Cohesive Soil'. According to ASTM D2166M, I have to set the loading rate to 0.5-2%/min (axial strain).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine I have though, is stress-controlled. It measures the load in Pa/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since concrete samples can be tested through a similar method; is there a method that I can use to measure the elastic modulus and/or compressive strength of the clay using something with a rate of Pa/s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I set the loading to something very low (e.g 1Pa/s) will I get acceptable results? &#xA;(Note: for concrete it is acceptable to use 0.15MPa/s)&lt;/p&gt;&#xA;" OwnerUserId="25" LastEditorUserId="25" LastEditDate="2018-11-26T21:29:36.950" LastActivityDate="2019-06-27T15:54:12.407" Title="Is it possible to calculate Young's Modulus of Clay from load-controlled tests?" Tags="|stresses|geotechnical-engineering|elastic-modulus|soil|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24863" PostTypeId="1" CreationDate="2018-11-26T10:10:06.157" Score="1" ViewCount="12" Body="&lt;p&gt;I am using 3D FEM with 8 node brick element to model a certain type of growth/expansion, which is not a plastic deformation. First, I apply a pressure/load and I get displacement and new position of nodes and calculate stress/strain. Next, I would like to change the 'edges' relaxing length (different edges by different factors) of an 8 node hexaedral element. Such change in relaxing length would be due to addition of a new material, resulting in expansion of a mesh. My question is, whether in such a model I can calculate sort of relaxing length for a brick element edges after application of pressure and consequently change it to get new nodal positions? I can imaging doing so for e.g. 1D string, but I am not sure if it makes any sense for 3D hexaedral element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be grateful for any helpful suggestion/comments&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Denis&lt;/p&gt;&#xA;" OwnerUserId="18417" LastActivityDate="2018-11-26T10:10:06.157" Title="Is it possible to change a 'relaxing length' of edges for 8 node hexaedral element in 3d FEM" Tags="|finite-element-method|solid-mechanics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24864" PostTypeId="2" ParentId="19650" CreationDate="2018-11-26T10:47:26.610" Score="0" Body="&lt;p&gt;Assuming the liquid is static, or &lt;em&gt;mostly&lt;/em&gt; static - the most simple answer: The pressure. Force exerted per unit of surface of the wall, in direction normal to the surface, is equal to pressure of the liquid in contact with that wall, and nothing else. Your liquid at the bottom is at 120kPa, it exerts 120kN per meter square of bottom surface in direction normal to the surface. This is true for any infinitesimal fragment of the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fully static liquid, that will be just density times depth times gravitational acceleration. Water, at 1000kg/m^3 times 9.8m/s^2 becomes 98kPa at 10m depth. Add ambient atmospheric pressure (~1bar) if necessary (do you consider empty tank to receive 1bar of pressure on the wall or 0?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gets trickier if the liquid is flowing at considerable speed, exerting dynamic pressure, drag due to laminar flow etc. But for &lt;em&gt;mostly static&lt;/em&gt; tank, the force is just pressure times unit of surface, and always in normal direction to that surface.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-11-26T10:47:26.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24865" PostTypeId="2" ParentId="24088" CreationDate="2018-11-26T11:19:29.357" Score="3" Body="&lt;p&gt;The simplest way to tell if a metal is ductile or brittle is to look at a fracture surface (e.g. from a tensile or impact test). To have either a ductile or brittle metal defines the failure mechanism which you can tell by having either a &lt;a href=&quot;http://home.jtan.com/~joe/bicycle/images/spoke_2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;ductile fracture&lt;/a&gt; or a &lt;a href=&quot;http://vgoinc.com/wp-content/uploads/2015/11/SEM-Image3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;brittle fracture&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elongation and reduction of area as well as notch toughness (impact test) and hardness testing can give a quantitative value to ductility, by measuring the amount of energy the material absorbs and converts to mechanical deformation. These values are normally used to allow 'design ductility' which is less interested in the failure mechanism and a lot of the time only cares if a material can absorb enough energy not to fail as soon as it is over stressed (or for deliberate deformation). But there isn't a fixed limit for defining a material as ductile or brittle as it depends on the material and is usually used to define a design criteria that must be met. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you have an elongation greater than 20% you are probably dealing with a ductile metal, but this can depend on the alloy type/condition and is not always the case for all materials. &lt;/p&gt;&#xA;" OwnerUserId="18382" LastActivityDate="2018-11-26T11:19:29.357" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24867" PostTypeId="1" CreationDate="2018-11-26T17:09:27.693" Score="0" ViewCount="1348" Body="&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; I had previously posted this question on the Space and Robotics SE sites, but received little/no response.  So I've deleted the previous posts and repasted my question here in the hope of getting more responses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to model the dynamics of a free-floating spacecraft with an single-link (revolute) robot arm attached to one end.  I'm modeling planar motion and rotation of both the spacecraft and the robot arm link.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm familiar with the process of obtaining the dynamics of a system by way of formulating the total kinetic and potential energies of the system and applying the Euler-Lagrange theorem.  The problem is that I'm not entirely sure how to formulate the energies of this system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For inspiration, I looked at modeling my spacecraft + single-link robot system as if it were a two-link robot system with a single revolute joint connecting link 1 with link 2.  For simplicity, I'm only considering planar motion in an inertial coordinate system with no gravitational potential forces acting on the spacecraft-robot arm system.  For reference, see the image below:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zbw7L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zbw7L.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Mounted Two-Link Robot Arm System&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a ordinary planar two rigid link robot system (with two revolute joints mounted to a fixed platform), the potential energies are due to gravity.  The kinetic energies for each part can be decomposed into energies from linear motion and from angular rotation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For link 1, the kinetic energy due to the linear motion of Link 1's center of gravity is constrained to circular motion because it is attached to the base at joint 1.  Its linear energy is &lt;span class=&quot;math-container&quot;&gt;$$L_1=\frac{1}{2}m_1a_{c1}^2\dot{\theta_1}^2,$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$m_1$&lt;/span&gt; is the mass of link 1 and &lt;span class=&quot;math-container&quot;&gt;$a_{c1}$&lt;/span&gt; is the half-length of link 1.  The rotational energy associated with link 1 is the energy from rotation about its center of mass.  Its rotational energy is &lt;span class=&quot;math-container&quot;&gt;$$R_1=\frac{1}{2}I_1\dot{\theta_1}^2,$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$I_1$&lt;/span&gt; is the moment of inertia of link 1 about its center of mass.  So, the total kinetic energy of link 1 is &lt;span class=&quot;math-container&quot;&gt;$L_1+R_1$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For link 2, the kinetic energy also has linear and rotational components to it.  The linear energy can be obtained geometrically by noting that the (x,y) position of link 2's center of mass can be obtained as follows:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of link 2's center of mass is &lt;span class=&quot;math-container&quot;&gt;$$P_x=a_1cos(\theta_1)+a_{c2}cos(\theta_1+\theta_2)$$&lt;/span&gt;&#xA;and&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_y=a_1sin(\theta_1)+a_{c2}sin(\theta_1+\theta_2).$$&lt;/span&gt;&lt;br&gt;&#xA;Note that &lt;span class=&quot;math-container&quot;&gt;$a_1=2*a_{c1}$&lt;/span&gt; is the full length of link 1 and &lt;span class=&quot;math-container&quot;&gt;$a_{c2}$&lt;/span&gt; is the half-length of link 2.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the chain rule, one can find the linear velocity in the x and y directions as &lt;span class=&quot;math-container&quot;&gt;$\dot{P_x}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\dot{P_y}$&lt;/span&gt;.  Then the kinetic energy due to linear motion at link 2's center of mass is &lt;span class=&quot;math-container&quot;&gt;$$L_2=\frac{1}{2}m_2v^Tv,$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$v=[\dot{P_x}, \dot{P_y}]^T$&lt;/span&gt;.&lt;br&gt;&#xA;The rotation energy of link 2's center of mass involves the sum of rotational velocities from both joint 1 and joint 2.  So, the rotational energy around link 2's center of mass is &lt;span class=&quot;math-container&quot;&gt;$$R_2=\frac{1}{2}I_2(\dot{\theta_1}+\dot{\theta_2})^2.$$&lt;/span&gt; So, the kinetic energy of link 2 is &lt;span class=&quot;math-container&quot;&gt;$K_2=L_2+R_2$&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I won't go into the details of the potential energy, but let's just assume there are expressions &lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$P_2$&lt;/span&gt; for the potential energies of link 1 and link 2 associated with the height of the center of mass of each link.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for a two-link, two joint revolute mounted planar robot arm, we can write the lagrangian as &lt;span class=&quot;math-container&quot;&gt;$$\mathcal{L}=K_1+K2-P1-P2 = L1+R1+L2+R2-P1-P2.$$&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Free-Floating Spacecraft with a Single Link Robot Arm System&lt;/em&gt;&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To obtain the equations of motion for a free floating robot with a single robot arm attached to one end, I take this model and make a few modifications:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Assume that link 1 is a free floating spacecraft.  Thus, there is no joint 1 associated with this system.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Re-define the angle &lt;span class=&quot;math-container&quot;&gt;$\theta_1$&lt;/span&gt; to be the orientation (attitude) angle of the spacecraft, with respect to an inertial reference frame.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Since the spacecraft-robot system are free-floating in space, I assume the potential energies for both the spacecraft and the robot are zero (e.g. assuming no gravity gradient effects or perturbation forces on the system).  So &lt;span class=&quot;math-container&quot;&gt;$P_1=P_2=0$&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The spacecraft has a kinetic energy associated with its linear velocity.  Denoting &lt;span class=&quot;math-container&quot;&gt;$\dot{x}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\dot{y}$&lt;/span&gt; as the linear velocity of the spacecraft (link 1), the linear energy of link 1 becomes &lt;span class=&quot;math-container&quot;&gt;$$L_1=\frac{1}{2}m_1(\dot{x}^2+\dot{y}^2).$$&lt;/span&gt;  The spacecraft also has a rotational energy associated with its angular velocity &lt;span class=&quot;math-container&quot;&gt;$\dot{\theta_1}$&lt;/span&gt; given by &lt;span class=&quot;math-container&quot;&gt;$$R_1=\frac{1}{2}I_1\dot{\theta_1}^2.$$&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;My Question&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;I'm not sure if I'm characterizing the expressions for the linear and rotational energy in link 2 correctly.  I assume link 2's (the robot arm's) center of mass has an (x,y) position given by:&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_x = x + a_{c1}cos(\theta_1)+a_{c2}cos(\theta_1+\theta_2)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_y = y + a_{c1}sin(\theta_1)+a_{c2}sin(\theta_1+\theta_2)$$&lt;/span&gt;&#xA;So, if I take time derivatives of &lt;span class=&quot;math-container&quot;&gt;$P_x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$P_y$&lt;/span&gt; as before, I should obtain an expression for the velocity &lt;span class=&quot;math-container&quot;&gt;$v=[\dot{P_x},\dot{P_y}]$&lt;/span&gt; and the kinetic energy due to linear motion should be &lt;span class=&quot;math-container&quot;&gt;$L_2=\frac{1}{2}m_2v^Tv$&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt;&lt;br&gt;&#xA;Should the kinetic energy of link 2 (the robot arm) include terms with the spacecraft (link 1's) velocity &lt;span class=&quot;math-container&quot;&gt;$\dot{x}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\dot{y}$&lt;/span&gt; in it?  My mind keeps going back and forth on whether or not to include the space craft's   velocity as part of the energy of the robot arm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt;&lt;br&gt;&#xA;Should the rotational energy involve only the sum of the angular velocities of the spacecraft's attitude (&lt;span class=&quot;math-container&quot;&gt;$\dot{\theta_1}$&lt;/span&gt;) and the robot arm's joint &lt;span class=&quot;math-container&quot;&gt;$\dot{\theta_2}$&lt;/span&gt;.  Specifically, is it correct to write out &lt;span class=&quot;math-container&quot;&gt;$R_2=\frac{1}{2}I_2(\dot{\theta_1}+\dot{\theta_2})^2$&lt;/span&gt;?  I have a doubt that the rotational energy energy of the link should be the same as that for the fixed base two-link robot arm case, but I can't seem to find any other potential  sources of rotational energy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TL;DR:&lt;/strong&gt; Are my expressions for &lt;span class=&quot;math-container&quot;&gt;$P_x$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$P_y$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_2$&lt;/span&gt; correct for the spacecraft-robot arm case?  If not, where did I go wrong or what am I missing?&lt;/p&gt;&#xA;" OwnerUserId="191" LastEditorUserId="191" LastEditDate="2018-12-18T21:06:36.980" LastActivityDate="2023-12-23T10:06:25.793" Title="Deriving the equations of motion for free-floating spacecraft with a single link robot arm" Tags="|dynamics|energy|robotics|kinematics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24868" PostTypeId="1" CreationDate="2018-11-26T20:00:46.747" Score="1" ViewCount="2243" Body="&lt;p&gt;A &quot;Carbon Phase Diagram&quot; has an upper and a lower transformation temperatures. It was my understanding that once a steel had cooled below the lower transformation temp (1333 deg F) that the austenite had changed completely to another structure (ie martensite, pearlite, cementite etc.) However when looking at the &quot;TTT&quot; diagram, it shows the transformation occurring between the two &quot;C&quot; curves. The &quot;C&quot; curves are below the 1333 deg F. The &quot;TTT&quot; diagram also shows that the transformation from austenite does not even start until the temp is below the lower transformation temp of 1333 deg F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me understand.&lt;/p&gt;&#xA;" OwnerUserId="18420" LastActivityDate="2018-11-26T23:11:38.813" Title="Carbon Phase Diagram vs. TTT diagram" Tags="|metallurgy|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24870" PostTypeId="1" AcceptedAnswerId="24885" CreationDate="2018-11-26T22:08:07.900" Score="1" ViewCount="217" Body="&lt;p&gt;Why is it that, at least in Ontario Canada, camber diagrams are required for detailed design drawings (Not shop drawings) for bridges with steel girders, but are not required for CPCI Girders?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that the camber drawing is required for the steel manufacture to put in a vertical curved to the beam so that when the dead load of the superstructure is added on top of the girders the whole thing bends to meet the road profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For concrete girder this is not done, though a similar process seems to be followed.  The screed elevation is adjusted to account for deflection of the concrete, including long term deflection from shrinkage and creep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would have thought a screed elevation table would be required for both steel and concrete girder designs, so why have to two different approaches for dealing with deflection to match the road profile?&lt;/p&gt;&#xA;" OwnerUserId="6993" LastActivityDate="2018-11-27T17:45:02.750" Title="Camber Diagram for Steel Girders but not for Concrete Girders" Tags="|structural-engineering|bridges|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24871" PostTypeId="2" ParentId="24862" CreationDate="2018-11-26T22:43:13.817" Score="1" Body="&lt;p&gt;Unconfined compressive strength test is generally a undrained soils test. The issue with a constant load machine is that you will likely load it too quickly as well close to the failure point, which will produce inaccurate readings and measurements at failure (i.e. you will end up missing the point of failure, and needing a new sample again). Otherwise, the machines are technically doing the same thing in both tests. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that said, your description of the unconfined compressive strength as &quot;young's modulus&quot; is inaccurate in the sense that the young's modulus is generally used only to describe the elastic behavior of materials and how it behaves within the elastic region (or behavior that closely mimics the elastic region), which is not really true in soils. If you are unfamiliar with soil testing, you should not perform testing on soils samples as that can cause major errors regarding designing proper foundations and prompt people to make wrong decisions. &lt;/p&gt;&#xA;" OwnerUserId="13526" LastEditorUserId="13526" LastEditDate="2018-11-27T22:18:16.287" LastActivityDate="2018-11-27T22:18:16.287" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24872" PostTypeId="2" ParentId="24868" CreationDate="2018-11-26T23:11:38.813" Score="1" Body="&lt;p&gt;Somewhere on your IRON -carbon diagram is the word &quot;equilibrium&quot; ; It describes equilibrium conditions. TTT curves are sort of the antithesis of equilibrium ; The right hand boundary is static conditions. But the route or path taken to get to the right boundary makes a substantial difference in what you have when you get there . Although they have some common features these two diagrams do not work together. About one year of metallurgical education is needed to understand them.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-11-26T23:11:38.813" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24873" PostTypeId="1" AcceptedAnswerId="24874" CreationDate="2018-11-26T23:19:01.200" Score="0" ViewCount="154" Body="&lt;p&gt;I am new to vibration analysis/simulation etc. so I may be missing something very basic here so please forgive me if I'm explaining things poorly or not making sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To familiarize myself with vibration simulation I am trying to simulate the frequency response spectrum for a tuning fork. I am using Solidworks. &#xA;I have &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;created the geometry for a tuning fork &lt;/li&gt;&#xA;&lt;li&gt;fixed the base and calculated the eigenfrequencies eigenmode shapes&lt;/li&gt;&#xA;&lt;li&gt;Applied a uniform base displacement excitation (same magnitude for all frequencies)&lt;/li&gt;&#xA;&lt;li&gt;Plotted the lateral displacement of one of the tines as a function of excitation frequency&lt;/li&gt;&#xA;&lt;li&gt;The part is made from copper (young modulus supplied by software) and I am using a modal damping ratio of 0.02 for all modes.&lt;/li&gt;&#xA;&lt;li&gt;The excitation amplitude is 1mm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is an image of the main &quot;tuning fork&quot; mode showing the fixed base (green arrows) and excitation vector (purple arrow).&lt;a href=&quot;https://i.stack.imgur.com/vDrbA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vDrbA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a plot of the magnitude lateral displacement (x direction) of a node at the tip of one of tines of the tuning force versus excitation frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wEA0b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wEA0b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak at ~500 Hz corresponds to the mode shown above. The peak at ~2.5 kHz corresponds to a higher frequency vibrational mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused about the high frequency response of the tuning fork. At low frequency I see that the response is falling off from the two resonances towards lower frequencies. This makes sense. If you slowly move the tuning fork up and down (from the base) there will be no stress or deflection since everything is slow so there will be no lateral displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I would ALSO expect the response to drop off at high frequencies as well.. I guess my intuition here is motivated by the following. Consider a mass attached to a base by a spring and a damper. If the base is shaken then at high frequencies the response of the mass falls off, it is as if there is nothing connecting the mass and the base at high frequency. I guess I expect something similar for the tuning fork. If I drive at high enough frequency the tuning forks can't even tell there is a drive..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone please explain to me:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If it is correct that this response function should be constant versus frequency above the tuning fork resonance frequencies and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Whether it is correct or incorrect can you give me any intuition for why we should expect whatever is the correct behavior?&lt;/p&gt;&#xA;" OwnerUserId="18365" LastEditorDisplayName="user16" LastEditDate="2018-11-27T02:07:23.617" LastActivityDate="2018-11-27T12:47:52.647" Title="Base Driven Tuning Fork Mechanical Response Spectrum" Tags="|vibration|simulation|frequency-response|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24874" PostTypeId="2" ParentId="24873" CreationDate="2018-11-27T00:55:18.650" Score="1" Body="&lt;p&gt;You said you applied a constant &lt;em&gt;displacement&lt;/em&gt; to the base. In that situation, if the structure doesn't have any modes that can are excited at a particular frequency, the response is going to be approximately the same as a rigid body being moved with a fixed amplitude, which is what your plot shows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you apply a fixed displacement, the &lt;em&gt;excitation force&lt;/em&gt; will increase for higher frequencies (and except where there is a resonance in the displacement response) it will be approximately proportional to the frequency squared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you applied a constant &lt;em&gt;force&lt;/em&gt;, you would see the displacement falling as the frequency increased, but unless the structure is constrained in some way, the displacement amplitude will increase to infinity as the frequency gets close to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to understand this is think abut a 2 DOF system being excited at a frequency much higher than its two resonant frequencies. In that situation the elastic forces are small compared with the inertia forces, so we can ignore the stiffness term in the equation of motion and approximate it as &lt;span class=&quot;math-container&quot;&gt;$$\begin{bmatrix} m_{11} &amp;amp; m_{12} \\ m_{21} &amp;amp; m_{22} \end{bmatrix} \begin{bmatrix} \ddot x_1 \\ \ddot x_2 \end{bmatrix} = \begin{bmatrix}F \\ 0 \end{bmatrix} $$&lt;/span&gt; where F is the excitation force applied at point 1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we apply a constant amplitude &lt;em&gt;displacement&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; at point 1, we have &lt;span class=&quot;math-container&quot;&gt;$\ddot x_1 = -\omega^2X$&lt;/span&gt; and the two unknowns in the equations are &lt;span class=&quot;math-container&quot;&gt;$\ddot x_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the second equation we have &lt;span class=&quot;math-container&quot;&gt;$$\ddot x_2 = - \frac{m_{21}}{m_{22}} \ddot x_1$$&lt;/span&gt; which means that &lt;span class=&quot;math-container&quot;&gt;$$x_2 = - \frac{m_{21}}{m_{22}}X.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, for a constant base amplitude, the tip amplitude of your tuning fork will also be approximately constant at high frequencies, except when there is a resonance that affects the motion of the system.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-11-27T12:47:52.647" LastActivityDate="2018-11-27T12:47:52.647" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24876" PostTypeId="2" ParentId="24870" CreationDate="2018-11-27T02:48:24.730" Score="1" Body="&lt;p&gt;It might have something to do with the issue of quality assurance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel beams are highly uniform, both in dimensions and material properties. Therefore, if you calculate the deflection of a steel structure and get 2&amp;nbsp;mm at midspan under dead load, you'll likely get something very close to 2&amp;nbsp;mm in the real world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meanwhile, concrete elements are far less uniform. Their real-world cross-sections tend to deviate further from what was designed than steel beams', and the material properties are obviously less accurately determined. So if you likewise calculate a 2&amp;nbsp;mm deflection at midspan, you'll probably get something... around 2&amp;nbsp;mm, more or less. Which is perfectly adequate for a serviceability assessment, but not if you're concerned about the road profile. So in the case of concrete beams, you'll probably need to adjust the screed elevation anyways.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-11-27T02:48:24.730" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24878" PostTypeId="1" CreationDate="2018-11-27T08:13:55.050" Score="1" ViewCount="261" Body="&lt;p&gt;I’m doing a “Systematic calculation of highly stressed bolted joints – Joints with cylindrical bolt“ according to VDI 2230 Part 1. In step of calculation R11 “minimum length of engagement”, when my internal thread is critical, I have to calculate the ratio of &lt;span class=&quot;math-container&quot;&gt;$s/d$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my specific application I’m using a tapped thread joint (TTJ) and I’m connecting some kind of tube (internal thread) with a threated bolt. And because my tube is thin-gauged I get a small &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; and therefore my ratio &lt;span class=&quot;math-container&quot;&gt;$s/d$&lt;/span&gt; is smaller than &lt;span class=&quot;math-container&quot;&gt;$1.4$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;VDI 2230 only considers ratios greater than &lt;span class=&quot;math-container&quot;&gt;$1.4$&lt;/span&gt; and here my questions arise:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the effect on my calculation when I’m out of the mentioned ratio-borders?&lt;/li&gt;&#xA;&lt;li&gt;Is it possible to compensate the effect?&lt;/li&gt;&#xA;&lt;li&gt;What’s the amount of error I’m introducing?&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Okay, I could reconstruct my connection – but that’s not the topic of my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help.&lt;/p&gt;&#xA;" OwnerUserId="17780" LastActivityDate="2018-11-27T08:13:55.050" Title="Calculation of critical internal thread according to VDI 2230 Part 1" Tags="|mechanical-engineering|bolting|threads|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24880" PostTypeId="1" CreationDate="2018-11-27T11:23:11.123" Score="1" ViewCount="339" Body="&lt;p&gt;Imagine a typical DC motor / planetary gear combos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rKUfl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rKUfl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my thinking correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have a 100 kg mass on a frictionless surface.  The gearbox output shaft has a 0.10m arm with which you push the mass. (ie, linearly a few inches).  You want to accelerate the mass at 0.1 m/s2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean, it would take a force of 10 N (ie 100 x 0.1), and hence a torque of 1 Nm  (10 x 0.1)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's correct, and say after the gearbox the spindle indeed delivers 1 N.  It's a 6000 rpm motor and a 100:1 gearbox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, would that mean the &lt;strong&gt;motor itself&lt;/strong&gt; indeed delivers &quot;&lt;span class=&quot;math-container&quot;&gt;$1 \text{Nm}$&lt;/span&gt;&quot; .. it &quot;doesn't matter&quot; about the gearbox?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, if the output spindle is indeed delivering &quot;&lt;span class=&quot;math-container&quot;&gt;$1 \text{Nm}$&lt;/span&gt;&quot;, would that mean the &lt;strong&gt;motor itself&lt;/strong&gt; to purchase would be rated 1/100 = &lt;span class=&quot;math-container&quot;&gt;$10 \text{mNm}$&lt;/span&gt; ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{ie, 10 milliNewton-meter ... 0.01 Newton-meter!}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure seems way too low, but I'm likely mistaken.&lt;/p&gt;&#xA;" OwnerUserId="18426" LastEditorUserId="18426" LastEditDate="2018-11-27T15:15:02.960" LastActivityDate="2018-11-27T15:15:02.960" Title="Understanding the torque of a motor/gearbox combo" Tags="|motors|gears|torque|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24881" PostTypeId="1" CreationDate="2018-11-27T11:48:16.247" Score="1" ViewCount="251" Body="&lt;p&gt;How can I find left and right directions associated with each zero for a MIMO transfer function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intia&lt;a href=&quot;https://i.stack.imgur.com/Ljoq7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ljoq7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First it can be expressed in Smith-McMillan form and can find poles and zeros. Then how to find the &quot;left and right directions associated with each zero&quot; ?&#xA;In above image poles &amp;amp; zeros are found for the first transfer function.&lt;/p&gt;&#xA;" OwnerUserId="18429" LastEditorUserId="18429" LastEditDate="2018-11-28T15:47:04.383" LastActivityDate="2018-11-28T15:47:04.383" Title="How can I find left and right directions associated with each zero for a MIMO transfer function?" Tags="|control-engineering|control-theory|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24882" PostTypeId="2" ParentId="24880" CreationDate="2018-11-27T12:16:33.470" Score="1" Body="&lt;p&gt;There are a lot of questions in here, I'll try to go through one by one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Say you have a 100 kg mass on a frictionless surface. The gearbox&#xA;  output shaft has a 0.10m arm with which you push the mass. (ie,&#xA;  linearly a few inches). You want to accelerate the mass at 0.1 m/s2.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Does that mean, it would take a force of 10 N (ie 100 x 0.1), and&#xA;  hence a torque of 1 Nm (10 x 0.1)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F=m*a$&lt;/span&gt;, so yes, it would take a force of &lt;span class=&quot;math-container&quot;&gt;$10\text{N}$&lt;/span&gt;. and &lt;span class=&quot;math-container&quot;&gt;$T=F*L$&lt;/span&gt;, so yes, that corresponds to &lt;span class=&quot;math-container&quot;&gt;$1\text{Nm}$&lt;/span&gt; in this case. So far, so good.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If that's correct [it is!], and say after the gearbox the spindle indeed&#xA;  delivers 1 N&lt;strong&gt;m&lt;/strong&gt;. It's a 6000 rpm motor and a 100:1 gearbox.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;In fact, would that mean the motor itself indeed delivers &quot;1 N&lt;strong&gt;m&lt;/strong&gt;&quot; .. it&#xA;  &quot;doesn't matter&quot; about the gearbox?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. It &lt;strong&gt;does&lt;/strong&gt; matter about the gearbox. In this example, you have a &lt;span class=&quot;math-container&quot;&gt;$6000\text{rpm}$&lt;/span&gt; motor, which corresponds to an output shaft speed of &lt;span class=&quot;math-container&quot;&gt;$60\text{rpm}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$1\text{Nm}$&lt;/span&gt; If you change the gearbox, then these two output values will vary accordingly. i.e. you could have &lt;span class=&quot;math-container&quot;&gt;$30\text{rpm}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$2\text{Nm}$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$120\text{rpm}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$0.5\text{Nm}$&lt;/span&gt;. &lt;em&gt;Theoretically&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or if the output spindle is indeed delivering &quot;1 N&quot;, would that mean&#xA;  the motor in question to purchase would be rated 1/100 = 10 mN&lt;strong&gt;m&lt;/strong&gt; ?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, this is correct. The actual motor, before the gearbox, would need to produce about &lt;span class=&quot;math-container&quot;&gt;$10\text{mNm}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The figure seems way too low, but I'm likely mistaken.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, you're mistaken, it's not that low. Here's the first datasheet that I found for a generic &lt;span class=&quot;math-container&quot;&gt;$12\text{V}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$6000\text{rpm}$&lt;/span&gt; motor, which lists the torque at maximum efficiency as &lt;span class=&quot;math-container&quot;&gt;$16.5\text{mNm}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings us onto an important point - Motor efficiency and maximum power output etc. changes as the (motor) spindle speed varies - you need to make sure that your design case is set appropriately to operate primarily near the maximum efficiency point where possible. That's the subject of another question/answer, though. You may be able to find existing questions on this topic, too - I've certainly answered at least one myself!&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-11-27T12:16:33.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24884" PostTypeId="1" CreationDate="2018-11-27T16:27:38.893" Score="2" ViewCount="7297" Body="&lt;p&gt;What is the difference between LT and T directions in MMPDS (formerly MIL-Handbook 5)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hand understood that forming metal orients its grains, destroying the isotropic nature of 'pure metal' and that the &quot;Long&quot; direction is the length of the sheet/extrusion/etc, and that there were two transverse directions. There is the &quot;Short Traverse&quot; which would be the thickness (or &quot;depth&quot;) of the sheet/extrusion/etc. There is also the &quot;Long Traverse&quot; direction which would be the width of the sheet/extrusion/etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then MMPDS throws in a &quot;T&quot;. Is this just for the times when you cannot distinguish between LT and ST (such as in a rod)? I also see datasets where specimen orientations are T &lt;em&gt;and&lt;/em&gt; ST, which furthers my confusion. MMPDS itself doesn't seem to offer much further guidance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any explanations?&lt;/p&gt;&#xA;" OwnerUserId="11631" LastEditorUserId="11631" LastEditDate="2018-11-27T19:35:56.800" LastActivityDate="2023-07-25T19:48:47.783" Title="What's the Difference Between LT and T Directions?" Tags="|metallurgy|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24885" PostTypeId="2" ParentId="24870" CreationDate="2018-11-27T17:45:02.750" Score="3" Body="&lt;p&gt;CPCI girders are prestressed precast concrete girders, right? (Every region seems to have their own acronym for these things so it's not the name I'm familiar with but the mechanics should be the same.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short version of the answer is that the prestressing will usually give you the appropriate amount of precamber even if the girder is cast in a form with a flat bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughly speaking, the amount of prestressing is chosen to give a bending moment that counters the permanent load and a small percentage of the variable load (by placing most of the prestressing in the bottom of the beam). This gives the girder an upwards precamber straight out of the form, but once it's installed and the bridge completed, the total deflection will be around zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, it will not always be possible to make it end up perfectly horizontal. And the resulting deflection is nearly impossible to calculate with millimeter accuracy due to uncertainties in creep both before and after installation and the elasticity modulus of the concrete when the prestressing is applied. But in most cases it will even out nicely enough for a precamber diagram not to be worth the trouble.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2018-11-27T17:45:02.750" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24886" PostTypeId="1" AcceptedAnswerId="24892" CreationDate="2018-11-27T18:13:16.360" Score="1" ViewCount="457" Body="&lt;p&gt;Is there any way a portal frame with pin joints at the top of the columns can be stable? Are portal frames ever done with pin joints at the bases? (Aside from those with raft foundations/combined footings)?&lt;/p&gt;&#xA;" OwnerUserId="18433" LastEditorDisplayName="user16" LastEditDate="2018-11-28T02:53:32.127" LastActivityDate="2018-11-28T04:44:20.800" Title="Portal frame stability" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24887" PostTypeId="2" ParentId="24840" CreationDate="2018-11-27T19:05:03.657" Score="0" Body="&lt;p&gt;The way to indicate dimensions that do not have to be machined is with a so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Reference_dimension&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference dimension&lt;/a&gt;. The dimension used to be followed by 'REF', but nowadays the dimension is simply placed in between parentheses, e.g. (1 mm).&lt;/p&gt;&#xA;" OwnerUserId="16019" LastActivityDate="2018-11-27T19:05:03.657" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24888" PostTypeId="2" ParentId="24884" CreationDate="2018-11-27T19:30:21.640" Score="2" Body="&lt;p&gt;I believe MMPDS is the new name for Mil-Handbook 5.  I have the 1998 version and it defines &quot;T&quot; as `&lt;strong&gt;&lt;em&gt;Applied torsional moment; transverse (grain direction); subscript “transverse”&lt;/em&gt;&lt;/strong&gt;.  Also see the figure below from Mil-Handbook 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the figure, the &quot;L&quot;, &quot;T&quot;, and &quot;S&quot; planes are marked clearly.  Note that &quot;T&quot; and &quot;L-T&quot; indicate the same plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LptZW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LptZW.png&quot; alt=&quot;MIL-Handbook 5 1998 definitions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-11-27T19:30:21.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24889" PostTypeId="1" CreationDate="2018-11-27T23:45:23.250" Score="2" ViewCount="1046" Body="&lt;p&gt;In a direct evaporative cooler, dry air is passed by a water reservoir to evaporate water into the air, causing the latent heat of vaporization to be absorbed from both the air and the water; this results in the air leaving the cooler at a lower temperature, as desired. An image for this process is below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/frkcV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/frkcV.jpg&quot; alt=&quot;direct evaporative cooler&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the type of recirculating cooling tower typically connected to the condenser of a thermoelectric power plant (image below), dry air is passed by hot water coming from the power plant's condenser. This causes water to evaporate into the  air, absorbing the latent heat of evaporation from the water and the air, resulting in a lower water temperature, as desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TkmbS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TkmbS.png&quot; alt=&quot;cooling tower&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing that is confusing me is the fact that evaporation of water into air in the direct evaporative cooler (first image) causes the air to leave the cooler at a lower temperature, but in the recirculating cooling tower, the air leaves at a higher temperature than it came in at. My questions are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) What is the main reason for the difference in air temperature leaving each of the respective cooling systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is the higher temperature of the air leaving the cooling tower due to sensible heat transfer (from warm water to cooler air) in excess of absorption of latent heat from the air?&lt;/p&gt;&#xA;" OwnerUserId="18436" LastEditorUserId="18436" LastEditDate="2018-11-28T16:42:56.033" LastActivityDate="2018-11-28T16:42:56.033" Title="Exiting air temperature in direct evaporative cooler vs cooling tower" Tags="|cooling|evaporation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24891" PostTypeId="2" ParentId="11791" CreationDate="2018-11-28T04:18:30.507" Score="0" Body="&lt;p&gt;In days of old (not sure how old) stirrups did get installed at the ends of beams in an inclined fashion.  I believe CHBDC currently still has clauses for it, but I never see it used for new design.  I have used it however in evaluating old designs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally to inclined stirrups, it was also common practice at one point to bend some of the bottom longitudinal bars up at the end of the beams where moment was decreasing and they were no longer required.   However since the shear was increasing some of this bent bar was also used for shear reinforcing.  &lt;/p&gt;&#xA;" OwnerUserId="6993" LastActivityDate="2018-11-28T04:18:30.507" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24892" PostTypeId="2" ParentId="24886" CreationDate="2018-11-28T04:44:20.800" Score="0" Body="&lt;p&gt;Yes a portal frame with at least one fixed (moment bearing) joint on top can be stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Portal frames with pin joints at the base could theoretically bear moment, provided they have fixed joints on top of the columns, but they are not advisable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fixed base and fixed top, a frame shares the moment at all joints and vibrates under lateral loads more in uniformity with the rest of the building.&#xA;Also it lends itself to smaller cross sections, hence more ductile behavior and less stiffness. Too much stiffness in a portal frame invites all the shear concentrated over its length and is detrimental to the overall lateral performance of the building. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-28T04:44:20.800" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24895" PostTypeId="1" CreationDate="2018-11-28T07:03:26.143" Score="1" ViewCount="70" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HE0Gl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HE0Gl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi, so, would you consider  the connection detailed here as fixed or pinned during your structural analysis? &lt;/p&gt;&#xA;" OwnerUserId="18433" LastActivityDate="2018-11-28T07:03:26.143" Title="STEEL CONNECTIONS" Tags="|structural-engineering|structural-analysis|steel|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24897" PostTypeId="1" CreationDate="2018-11-28T09:46:18.290" Score="0" ViewCount="169" Body="&lt;p&gt;I am building a stair from aluminium to go to my basement, the stair sides-plates are made from 10mm thick aluminium AWS 5454.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am planning to tap M10 thread into this plate. To avoid the step slipping against the stair sides a tension of 715N is required. The bolt material is A2-70.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But will this aluminium thread be able to hold this load? How do you calculate this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nKWoP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nKWoP.png&quot; alt=&quot;Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8464" LastEditorUserId="8464" LastEditDate="2018-11-28T10:18:23.977" LastActivityDate="2019-04-28T04:07:36.507" Title="Tapped thread maximum load" Tags="|aluminum|threads|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="24898" PostTypeId="1" AcceptedAnswerId="24903" CreationDate="2018-11-28T10:45:52.623" Score="1" ViewCount="157" Body="&lt;p&gt;I'm trying to design a measurement system, the primary objective of which is to reliably detect a 10% change in the the volume of fluid flowing through a pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm struggling to come up with a required accuracy/precision for the system based on something on something other than &quot;This is simply as accurate as I could make it&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note I am only really interested in relative change, i'm not sure how this effects the system requirements in terms of required precision/accuracy. For example, there two primary sources of error: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Measurement error (accuracy) &lt;/li&gt;&#xA;&lt;li&gt;Estimated internal diameter of the pipe (precision)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm assuming I can completely negate the error introduced from the estimated internal diameter of the pipe and place the requirements solely on the &quot;measurement error&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a statistical technique to determine the accuracy required to reliably detect X change in some variable?&lt;/p&gt;&#xA;" OwnerUserId="18444" LastActivityDate="2018-11-28T15:12:22.907" Title="What measurement accuracy/precision is required to detect a 10% relative change in the measured quantity" Tags="|measurements|statistics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24899" PostTypeId="2" ParentId="24889" CreationDate="2018-11-28T11:06:30.423" Score="2" Body="&lt;p&gt;Mixing two media (air and water) just averages the temperature between them (proportionally to their respective heat capacities and total volumes) - through classic conduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, evaporation absorbs a large amount of heat, reducing the resulting temperature (of both media) on exit. Rapid flow of large amounts of air speeds up evaporation considerably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air has a very low heat capacity comparing to water, so mixing the two will definitely shift the average towards the temperature of water - unless you push volume of air massive enough to make a difference; also, such volume will result in more evaporation - so even if the air is rather warm, the water temperature rise will be low - or even there will be a temperature drop, evaporative cooling outpacing heating from air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the first case a certain, moderate volume of moderately warm air is passed through cool water; the evaporative cooling is quite enough to keep the water - and the exiting air - cool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case of the cooling tower, there's a massive amount of cool air being pumped - enough to absorb (through both evaporation and plain conduction) most of the heat from the hot water. (and make no mistake, the &quot;Hot saturated discharge air&quot; is a very similar temperature as &quot;Cooled water out&quot;; maybe a little warmer as it meets hotter water later, further up the tower.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-11-28T11:06:30.423" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24900" PostTypeId="1" CreationDate="2018-11-28T11:10:00.150" Score="0" ViewCount="134" Body="&lt;p&gt;What are boundary conditions and initial conditions with respect to differential equation? Do they mean the same for partial differential equation? How to BC and IC vary for different types of PDE?&lt;/p&gt;&#xA;" OwnerUserId="18283" LastActivityDate="2018-12-28T15:00:45.163" Title="Differential equation" Tags="|mathematics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24901" PostTypeId="2" ParentId="24900" CreationDate="2018-11-28T13:46:40.837" Score="0" Body="&lt;p&gt;In general an infinite number of functions satisfy differential equations (DV). To obtain a specific solution a first order ODE must have an initial condition or constraint that specifies the value of dependent variable at a particular value of independent variable. Because the problems are typically time dependent the constraint is called an &lt;strong&gt;intial condition&lt;/strong&gt; and the problem is called &lt;strong&gt;intial value problem (IVP)&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I summarise an &lt;strong&gt;IVP&lt;/strong&gt; as: When all the constraint are specified at one value of independent variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many cases we need to solve a DV of higher order, that have constraints specified at different values of independent variable, these problems are called &lt;strong&gt;Boundary value problems (BVP)&lt;/strong&gt;  and the constraints are called &lt;strong&gt;boundary conditions&lt;/strong&gt; because the constraints are often specified at the endpoints of boundaries of domain of the solution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes those means the same for partial differential equations, if the variables behave independently or when the equation is called separable. The answer to the third question requires a deep understanding of DV's and PDE's, i suggest you to talk to your professor or aks it somewhere else.  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-11-28T13:46:40.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24902" PostTypeId="2" ParentId="24900" CreationDate="2018-11-28T14:18:56.783" Score="1" Body="&lt;p&gt;Suppose you have a parameter &lt;span class=&quot;math-container&quot;&gt;$Y$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that parameter depends only on one variable &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. You then may derive a differential equation &lt;span class=&quot;math-container&quot;&gt;$dY/dx$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose that parameter depends on more than one variable, say &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;. You then must derive a partial differential equation for each variable, in this case &lt;span class=&quot;math-container&quot;&gt;$\partial Y/\partial x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\partial Y/\partial t$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; represents spatial extent. To solve a first order differential, you will need one boundary condition. This is either the value of &lt;span class=&quot;math-container&quot;&gt;$Y$&lt;/span&gt; at a defined point &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; or the value of &lt;span class=&quot;math-container&quot;&gt;$dY/dx$&lt;/span&gt; at a defined point &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each additional differential order requires an additional boundary condition. For example, a second order differential requires two boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; represents time. Boundary conditions on time at &lt;span class=&quot;math-container&quot;&gt;$t = 0$&lt;/span&gt; are called initial conditions. Initial conditions are a subset (if you will) of boundary conditions; they apply to time. Boundary conditions on time other than &lt;span class=&quot;math-container&quot;&gt;$t = 0$&lt;/span&gt; could be called anything. Another special class are those for &lt;span class=&quot;math-container&quot;&gt;$t \rightarrow \infty$&lt;/span&gt;. These boundary conditions are called &quot;final state&quot; conditions.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-11-28T14:18:56.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24903" PostTypeId="2" ParentId="24898" CreationDate="2018-11-28T14:29:32.023" Score="1" Body="&lt;h1&gt;Direct Measurement of Volumetric Flow&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let's handle the (easy) statement of accuracy first. Validating the accuracy of the measurement of volume flow requires that you calibrate your chosen measuring device with a second approach. That second approach must be at least as precise if not better than when you put the device in to practice. A best method will use a standards test, for example as defined by NIST or other agencies, as the calibration. Alternatively, when you intend to buy a device to measure volume flow and just use it directly without in-house calibration, you should insist on getting a calibration test report.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's handle the issue of sensitivity. You want to measure a change in volumetric flow &lt;span class=&quot;math-container&quot;&gt;$\Delta \dot{V}$&lt;/span&gt; within a standard of &lt;span class=&quot;math-container&quot;&gt;$\Delta \dot{V} / \dot{V} \geq 0.1$&lt;/span&gt;. This is akin to asking to what you need in order to measure a signal &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; over a noise level &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;. A guideline to reliable measurement in this case is that &lt;span class=&quot;math-container&quot;&gt;$S/N \geq 3$&lt;/span&gt; to state that you have a signal above the noise. For a hypothetic system with no inherent flow noise (the volumetric flow is perfectly stable), the &quot;noise&quot; is the precision of the device itself. At a minimum, you must therefore assure that &lt;span class=&quot;math-container&quot;&gt;$\Delta \dot{V} / \dot{V} \leq 0.1/3 = 0.033$&lt;/span&gt;. You will want a device with a relative measurement precision of at least 3.3%, preferably (much) better. When you intend to buy a device, you should confirm that its measurement precision fits your needs.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Indirect Measurement of Volumetric Flow&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Suppose you are measuring mass flow of an incompressible fluid and you want to correlate this with volume flow in a pipe. To first order, this is an application of linear propagation of uncertainties. The starting equation is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{V} = \frac{\dot{m}}{\rho} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know that you can measure &lt;span class=&quot;math-container&quot;&gt;$\dot{m}$&lt;/span&gt; with a device that has a relative measurement precision &lt;span class=&quot;math-container&quot;&gt;$\delta \dot{m}$&lt;/span&gt;. You know that your fluid has a density that is reported with a precision of &lt;span class=&quot;math-container&quot;&gt;$\delta \rho$&lt;/span&gt;. To first order, the uncertainty propagation equation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{\Delta\dot{V}}{\dot{V}}\right)^2 = \left(\frac{\delta\dot{m}}{\dot{m}}\right)^2 + \left(\frac{\delta\rho}{\rho}\right)^2 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To &lt;em&gt;reliably&lt;/em&gt; measure a 10% change in volumetric flow, you should have a total uncertainty in your measurement system &lt;span class=&quot;math-container&quot;&gt;$\frac{\Delta\dot{V}}{\dot{V}}$&lt;/span&gt; that is &lt;em&gt;significantly&lt;/em&gt; below 10%. This is a statement of the precision of your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a second example, when you are measuring velocity to obtain volumetric flow, the equation for velocity and volumetric flow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dot{V} = v \pi d^2 / 4$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;gives this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(\frac{\Delta\dot{V}}{\dot{V}}\right)^2 = \left(\frac{\delta v}{v}\right)^2 + 4\left(\frac{\delta d}{d}\right)^2 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The precision of your measurement depends on the relative precision of the velocity flow device and the relative uncertainty of the tube diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with direct measurements, the accuracy of your system depends on how well it is calibrated. Calibration might in this case be done by using a separate device that measures volumetric flow and making a plot of volumetric flow versus mass flow. A perfect linear correlation is expected to validate the accuracy of your correlation equation.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2018-11-28T15:12:22.907" LastActivityDate="2018-11-28T15:12:22.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24904" PostTypeId="1" AcceptedAnswerId="24905" CreationDate="2018-11-28T15:52:15.353" Score="2" ViewCount="1036" Body="&lt;p&gt;I want to measure the vibrations of a solid object (a tuning fork, a bone conducting headphone, the chest during singing) and to isolate them from the vibrations of the air. So I am looking for a meter that will pick those vibrations only when it's in contact when the vibrating object (hence a meter that just reads frequency probably won't work for me)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to use a smartphone's accelerometer but it didn't pick anything. I've also to use a phone's sound meter in order to see whether it's picks more sound when in contact with such object but the reading was the same.&lt;/p&gt;&#xA;" OwnerUserId="15977" LastEditorUserId="15977" LastEditDate="2018-11-28T16:11:27.690" LastActivityDate="2018-11-29T10:58:27.293" Title="How can I measure the vibrations of a solid object?" Tags="|vibration|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="24905" PostTypeId="2" ParentId="24904" CreationDate="2018-11-28T16:33:30.277" Score="4" Body="&lt;p&gt;An accelerometer is the correct choice, but the smartphone accelerometer gets its readouts smoothed out in software before being made accessible to userspace. It won't pick up such fine vibrations. You'll need a microcontroller (for up to ~1khz frequencies) or a DSP (higher), preferably on a development board, to perform the sampling. (note you should have the sample rate at least a few times higher than the frequency you measure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is a strain gauge. This again should be read by a microcontroller/DSP, or an oscilloscope (may need some basic circuitry to provide power), although it may not perform so well on soft surfaces, like human chest. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-11-28T16:33:30.277" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24906" PostTypeId="2" ParentId="24904" CreationDate="2018-11-28T16:37:06.213" Score="3" Body="&lt;p&gt;You might want to look at contact pickups designed for acoustic musical instruments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try searching for contact microphones of contact transducers. These tend to be based on piezo electric devices and sense vibrations through direct contact as opposed to acoustic microphones whcih generally use some sort of diaphragm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The instrument pickups area available as a complete package with an analogue audio output, usually via a 1/4&quot; jack socket. Most PCs have an audio input although you should check that the impedance of the pickup matches that of the PCs input. You can also get various &lt;a href=&quot;https://en.wikipedia.org/wiki/DI_unit&quot; rel=&quot;nofollow noreferrer&quot;&gt;balancing devices&lt;/a&gt; to facilitate this although it may not be essential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned in the comments instrument pickups are designed for an audible frequency range which 200Hz is well within albeit at the low end but individual device specs should tell you  their response range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also get the sensors individually but you will at least need to fit an output connector and possibly also a pre-amp or passive filter. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2018-11-29T10:58:27.293" LastActivityDate="2018-11-29T10:58:27.293" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24907" PostTypeId="1" AcceptedAnswerId="24909" CreationDate="2018-11-28T17:32:37.327" Score="2" ViewCount="172" Body="&lt;p&gt;This stitch is composed of a single string/thread and is common on dogfood and grain sacks. It can be removed by just pulling on the end. Here is an example I found on &lt;a href=&quot;https://www.youtube.com/watch?v=Q-NHxRwds4g&quot; rel=&quot;nofollow noreferrer&quot;&gt;youtube&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2018-11-28T18:16:27.353" Title="What is the name for a removable sewn bag stitch?" Tags="|manufacturing-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24909" PostTypeId="2" ParentId="24907" CreationDate="2018-11-28T18:16:27.353" Score="6" Body="&lt;p&gt;It is called &quot;chain stitch&quot; and was the stitch used in the first designs of sewing machines, because it is easy to produce mechanically. The fact that it can be pulled out unless the end of the string is knotted (as explained in the OP's video link) is a disadvantage for most purposes. For example, if the string breaks one side of the break will often quickly unravel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard sewing machine stitch is now &quot;lock stitch&quot; which uses two threads, not one, and can't be pulled undone in the same way. Chain stitch is only used for special applications, or as decoration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://home.howstuffworks.com/sewing-machine1.htm&quot; rel=&quot;noreferrer&quot;&gt;https://home.howstuffworks.com/sewing-machine1.htm&lt;/a&gt; for an animation of how both stitches are made by machine. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-11-28T18:16:27.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24912" PostTypeId="2" ParentId="24897" CreationDate="2018-11-29T02:55:18.303" Score="-1" Body="&lt;p&gt;As a first estimate, assuming the stair is straight and runners are equally spaced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assume each step is populated with one person going down and multiply the 200kg by 3 which is not too conservative a factor of safety, 600kg per step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;say we have a ratio of 18cm riser to 31cm runner of each step we get an angle of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ alpha = arctan 18/31 = 30\ degree$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We do a first attempt at designing the stringer as a uniformly loaded horizontal beam and then multiply the depth by cos(30) = 0.86&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the horizontal projection of the length of the flight of stairs as L we have this moment, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$M = \frac {1980\times l^2} {8} \  \ $&lt;/span&gt; for a simply supported beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming allowable tensile strength of alum 2000kg-cm^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ M = \sigma\times S \ and \ S= bh^2/6   $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ M = 2000\times 1\times h^2/6 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need is plug L , the length and you have the h of your aluminum beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later you check with your local code and refine the loads, dynamic safety factor of potential several people running down the stairs scaping a fire, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just to give a sense of rough idea, otherwise this member has to be checked for longitudinal forces like column, local buckling, torsion, etc.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-11-29T03:29:45.680" LastActivityDate="2018-11-29T03:29:45.680" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24914" PostTypeId="1" CreationDate="2018-11-29T10:56:01.813" Score="1" ViewCount="214" Body="&lt;p&gt;I have adapted a refrigerator to create a temperature controlled environment for brewing beer. I have mounted an electric heater coil inside the fridge and configured a basic controller to maintain a constant temperature inside the fridge. When the temperature at the temperature probe, &lt;span class=&quot;math-container&quot;&gt;$T_p$&lt;/span&gt;, is below the target temperature, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, the heater is turned on. When &lt;span class=&quot;math-container&quot;&gt;$T_p$&lt;/span&gt; is above &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, the fridge compressor is turned on. The target temperature, &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, is in fact a small range to stabilise the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T_p \lt T-c \ \ \rightarrow \ \ Heat  $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ T_p \gt T+c \ \ \rightarrow \ \ Cool  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this:&#xA;I am trying to decide, more for interest than anything else, whether it is more effective to probe the temperature of the air inside the fridge, or the liquid (beer) inside the plastic vessel inside the fridge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My instinct says that I should probe the air temperature, because the temperature of the beer will follow the temperature of the air, subject to a time delay, and by probing the temperature of the beer I could get unstable feedback due to this delay? Would the answer change if I were to put the fridge outside, where the air temperature is subject to much more variation? I suspect in this case it might be better to probe the beer so that the controller can over-compensate with air temperature and limit temperature variation of the beer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of this application, the control system is fixed as I'm using a fixed controller ie. heating and cooling are binary on/off functions. However I'd be interested to learn of any control techniques that could be implemented with a raspberry pi, if they were suggested!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="18453" LastEditorUserId="18238" LastEditDate="2020-04-22T16:17:05.797" LastActivityDate="2024-03-03T14:02:20.417" Title="Control system for a temperature controller" Tags="|control-engineering|control-theory|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24915" PostTypeId="2" ParentId="24317" CreationDate="2018-11-29T11:25:29.090" Score="0" Body="&lt;p&gt;Ended up not worrying about slip and instead just making sure the tires don't spin out by synchronizing their velocity but without forcing them to spin at the same velocity (ie I implemented coupling). Now slip is held at optimal value by the very nature of this control strategy. &lt;/p&gt;&#xA;" OwnerUserId="16582" LastActivityDate="2018-11-29T11:25:29.090" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24917" PostTypeId="2" ParentId="24914" CreationDate="2018-11-29T12:46:37.653" Score="0" Body="&lt;p&gt;I would probe the liquid and control it with a PID + SSR relay, it'll take an initial time to stabilize, but once it figures out its equation, your good to go. You probe the liquid simply because that's what you really care about. The liquid's container will have some effect on heat transfer from the environment to the liquid, so unless the liquid is to be in the refrigerator for a very long time, you can guarantee the two temperatures will be the same. Symmetry is important here, unless everything is geometrically/ volumetrically center, temperatures will vary inside until time takes over ( may not even be possible to create complete uniform temperature in the environment, insulation,sealing,etc) So it may be possible to have temperature fluctuations throughout the liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;May I suggest...&#xA;Use the fridge as is, always on, and control the heaters, submerged in the liquid, inside its container, inside the fridge, with a PID/SSR,probe in the liquid. Stirring the liquid would ensure proper temp throughout, or multiple heater/PID/SSR/probe would help without stirring. And depending on the set point temp, you might not even need the fridge. Cold is the absence of heat, if your set temp is always above ambient, all you need to control is heat input. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-11-29T12:46:37.653" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24918" PostTypeId="1" CreationDate="2018-11-29T12:54:15.527" Score="1" ViewCount="66" Body="&lt;p&gt;I'm looking for a low-power weight sensor which can withstand impacts of up to 1000 pounds. Is there a go-to supplier or brand for this sort of product?&#xA;I want it to give its output digitally; my end goal is to have it implanted in a punching dummy and send data to a Raspberry Pi. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="18455" LastActivityDate="2024-04-02T05:03:35.530" Title="Weight Sensor module needed" Tags="|sensors|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24921" PostTypeId="1" CreationDate="2018-11-29T15:11:33.583" Score="0" ViewCount="322" Body="&lt;p&gt;How can I calculate the von Mises stress in the cross-section of a bar like this that is axially loaded?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1aptF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1aptF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18457" LastActivityDate="2018-11-30T01:51:46.980" Title="How to calculate stress in a cylindrical-uniaxial-loaded bar with sudden changes in its diameter" Tags="|mechanical-engineering|structural-analysis|stresses|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="24922" PostTypeId="2" ParentId="24921" CreationDate="2018-11-29T16:47:55.953" Score="1" Body="&lt;p&gt;Stress Concentration factors on the internal corner of a Stepped Shaft are well understood, and have some standardised calculations/equations: I'd recommend this excellent online calculator, which shows all the background equations further down the page. This  allows you to calculate the stress concentration factor in various loading conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.amesweb.info/StressConcentrationFactor/SteppedShaftWithShoulderFillet.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amesweb.info/StressConcentrationFactor/SteppedShaftWithShoulderFillet.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see for yourself how reducing the fillet radius towards zero drastically raises the stress concentration factor, and you should try to remove the requirement for a 'sharp' corner if at all possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an FEA rendering of the shaft illustrated in your question, showing how the root of the smallest diameter is the critical area (N.B. this does have a radius!)&lt;a href=&quot;https://i.stack.imgur.com/sgb3xm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sgb3xm.png&quot; alt=&quot;FEA rendering showing stress concentration at root of stepped shaft.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2018-11-30T01:51:46.980" LastActivityDate="2018-11-30T01:51:46.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24923" PostTypeId="2" ParentId="24918" CreationDate="2018-11-29T19:29:08.920" Score="0" Body="&lt;p&gt;You are actually looking for a force sensor, of which the mentioned strain gauge is one type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you are trying to measure punches, then you need to have a sensor with enough bandwidth to track the applied force on a millisecond basis (at least).  For example, the entire punching bag can tell you the total transferred energy, i.e. integral of force over time, simply by applying pendulum equations.  But that will not tell you much of anything about the momentary peak force applied.&lt;br&gt;&#xA;Next problem is that punching bags are soft, so the force is quickly dispersed from the impact area (fist cross-section) to a much wider area.  You probably will have to mount the sensing element on a small rigid plate on the surface of the bag to get an accurate force (pressure times area) reading. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-11-29T19:29:08.920" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24924" PostTypeId="2" ParentId="24921" CreationDate="2018-11-29T19:45:30.407" Score="0" Body="&lt;p&gt;You get two effects near the area of sudden changes in cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - Stress concentration and contour lines of stress distribution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 - Stress wave propogation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sharp corners are stress concentration traps. Anticipating the path of stress and desgning the part's geometry according to that helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of these  items one and two affect the performance of the member significantly, specially under repetitive or shock loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are computer simulation sofware that can analyze these up to a high level of precision.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-11-29T19:45:30.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24926" PostTypeId="2" ParentId="19633" CreationDate="2018-11-29T20:24:40.330" Score="1" Body="&lt;p&gt;I tried both ways, idlers and slide, with and without additional spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Idler with spring tends to oscillate when load torque is not constant (it is rarely constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Sliding motor and spring is less sensitive to torque fluctuations due to non-zero friction in slide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;C) Idler without spring have some minor disadvantages: accurate and not-so-compact fixture and adjustment system and reverse bending of belt (not all belts are well suited for this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D) Sliding motor withot spring is most rigid and precise system, but with drawback of no self-adjustment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus I use idler+spring when I want to track torque by measurement of idler displacement (i.e. for emergency stop in a case of stuck) because this system is most responsive.  And I use sliding motor without spring when I want precise and smooth rotation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sliding motor can be fixed in several vays. If one tighten it with screw in direction of belt stretching, he can adjust resulting tension by putting thin plates under motor. &#xA;If tightening direction is perpendicular to stretching, one adjust tension by simple pulling and fix it with screw (with extra force to avoid slipping). &#xA;Good idea is to use angular slide with one (adjustable) screw for stratching an other[s] for fixture in adjusted position.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are different increments of timing belts, but not all of then are easy to order in retail quantities. One can think about 3d-printed or even lasercut custom belts for low-loaded drives. Reinforced DIY belts of free pitch and type can be mold with 3d-printed of lasercut circular gear-like trench template, filled by steel or glassfiber cord.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pullyes can be very close, even intersect (with a slot or array of slots). Bu if they are too close, gears can be more useful.&lt;/p&gt;&#xA;" OwnerUserId="18392" LastActivityDate="2018-11-29T20:24:40.330" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24928" PostTypeId="1" CreationDate="2018-11-29T21:05:05.697" Score="1" ViewCount="207" Body="&lt;p&gt;I'm working on a project that does a finite element simulation of a mesh. I can output the stress tensor at each node, but I'm not so sure how to implement failure criteria.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my material is brittle, say concrete, what's the best way to identify failure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current approach is to compute the von mises stress and compare against the compressive yield strength of the material, however this is completely ignoring the possibility that some portions of the mesh might be under tensile stresses for which the yield strength might be much lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to resolve the stress tensor into tensile and compressive components?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&#xA;Clayton&lt;/p&gt;&#xA;" OwnerUserId="18463" LastActivityDate="2018-12-04T00:16:38.330" Title="Stress Tensor decomposition into Compression &amp; Tensile forces" Tags="|structural-engineering|structural-analysis|finite-element-method|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24930" PostTypeId="1" CreationDate="2018-11-30T02:08:21.140" Score="0" ViewCount="49" Body="&lt;p&gt;I am trying to solve the following 1-D heat equation with provided boundary conditions using explicit scheme on Matlab. I have been trying to plot the results but I realized that my temperatures are not changing. How can I solve problem and include the boundary condition?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;clc; clear all; close all; &#xA;&#xA;a=0.45*10^-5;&#xA;U_Initial= 30+273.15; %converting to kelvin&#xA;U_Wall = 20+273.15; %converting to kelvin&#xA;U_BC = 100+273.15; %converting to kelvin&#xA;dt= 0.01;&#xA;dr= 1;&#xA;Lambda= (a*dt)/(dr)^2;&#xA;t=1:dt:3600;&#xA;r= 3:dr:100;&#xA;U = zeros(length(r),length(t));&#xA;U(:,1)= U_Initial;&#xA;U(1,:)=U_BC;&#xA;U(end,:)=U_Wall;&#xA;for j= 1:1:length(t)-1&#xA;   for i=2:1:length(r)-1&#xA;     U(i,j+1)= U(i,j)+Lambda*((dr/r(i))*(U(i+1,j)-U(i-1,j))+ (U(i+1,j)-2*U(i,j)+U(i-1,j)));&#xA;   end&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18465" LastActivityDate="2018-11-30T02:37:18.900" Title="Temperature not changing in my code for 1-D heat equation: Explicit Scheme" Tags="|heat-transfer|matlab|numerical-methods|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24931" PostTypeId="2" ParentId="24930" CreationDate="2018-11-30T02:37:18.900" Score="1" Body="&lt;p&gt;This stackexchange doesn't really seek to solve these homework-like problems directly. However, I can offer some hints on how to proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you sure the scale of &lt;code&gt;a&lt;/code&gt; and &lt;code&gt;lambda&lt;/code&gt; are correct? When I assign &lt;code&gt;a=0.1&lt;/code&gt;, a reasonable profile develops. Consider non-dimensionalizing your problem so the variables all scale from 0 to 1. These kinds of errors tend to disappear when you formalize the math in this way. Once you have a non-dimensional solution, you can rescale easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be aware that MATLAB also has built in PDE solvers. You might start &lt;a href=&quot;https://www.mathworks.com/help/matlab/math/partial-differential-equations.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; to read up on the general approach and getting your problem in the right form, and then use &lt;a href=&quot;https://www.mathworks.com/help/matlab/ref/pdepe.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;pdepe&lt;/code&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note, look up &lt;a href=&quot;https://www.mathworks.com/help/matlab/ref/linspace.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;linspace()&lt;/code&gt;&lt;/a&gt; as opposed to using the colon operator &lt;code&gt;:&lt;/code&gt; for ranges with non-integer steps. The colon operator will leave out the last step in cases like &lt;code&gt;1:0.6:2&lt;/code&gt;, which produces &lt;code&gt;[ 1 1.6 ]&lt;/code&gt;. Note that &lt;code&gt;2&lt;/code&gt; is missing, which is probably not what you intended.&lt;/p&gt;&#xA;" OwnerUserId="421" LastActivityDate="2018-11-30T02:37:18.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24932" PostTypeId="2" ParentId="19205" CreationDate="2018-11-30T10:24:18.567" Score="0" Body="&lt;p&gt;If you are a bit familiar with python programming language, this can be a streamline solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Install free simple 3D engine, e.g. Panda3D or V-REP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Find appropriate STL models of helicopter, landing zone, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Import them to 3D scene, scale and set zero positions and orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) Write doing-all-the-math function in python (there is also interface to MATLAB).  Derivatives from equations now are transformed to increments, e.g.  &lt;code&gt;v += a*dt, x += v*dt&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) Call this function from 3D engine cycle loop to manipulate (and read from) scene objects. Engine loop calculates &lt;em&gt;dt&lt;/em&gt; for every frame that leads to smooth real-looking sumulation.&lt;/p&gt;&#xA;" OwnerUserId="18392" LastActivityDate="2018-11-30T10:24:18.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24934" PostTypeId="1" CreationDate="2018-11-30T17:04:53.173" Score="3" ViewCount="609" Body="&lt;p&gt;This question appeared in my midterms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pipe transports water from a storage tank A to storage tank B. Storage tank A is at an elevation of 20 m above B, the pipe length is 6.4 km.&#xA;The present discharge rate is 75% of the required discharge rate.&#xA;A new design is proposed which would lay a new 2.4 km pipe segment (alongside the older one) running from point D to B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Consider the two pipes from DB to be in parallel,  nothing complex)&#xA;Diameter of original pipe is 200mm. &#xA;Calculate the diameter for this new pipe that would allow the required discharge rate to occur.&#xA;Assume friction factor 0.02 (fanning friction factor) for all pipes.&#xA;(Assume no minor losses due to bends and other stuff. Flow friction only).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I went about the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I found the flow from the original design using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$H=\frac{fl Q^2}{3d^5}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Q= 0.0122  m^3/s  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Q = \frac {3}{4} Q_r$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Q_r = 0.01633 m^3/s $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I go to the new design and take head loss equation in the first segment of original pipe running from AB-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ H = \frac {fl_1Q_1^2}{3d^5} + \frac  {fl_2Q_2^2}{3d^5}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they're part of the same original pipe, everything is same except length and discharge, so:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ H =\frac {f}{3d^5} (4000Q_1^2 +2400Q_2^2) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ 20 =\frac {0.02}{3(0.2)^5} (4000Q_1^2 +2400Q_2^2) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac {24}{25} = 4000Q_1^2 + 2400Q_2^2 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac {24}{25} - 40000 (0.01633)^2 = 2400 Q_2^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is where I hit a little snag:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ 0.96 - 1.0667 = 2400 Q_2^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ -0.1066 = 2400 Q_2^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Negative. The square of the flow is negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I even proceed from here? The only thing that comes to mind is possibly calling the flow imaginary. But that makes no sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my solution correct? What did I do wrong? &#xA;What does one do if this happens somewhere?&#xA;Can you please look over my solution to see if I made a mistake anywhere?&lt;a href=&quot;https://i.stack.imgur.com/IQZ8I.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IQZ8I.jpg&quot; alt=&quot;diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16581" LastActivityDate="2024-04-04T13:02:50.363" Title="Negative head in Darcy's flow equation" Tags="|fluid-mechanics|pipelines|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24935" PostTypeId="1" CreationDate="2018-11-30T18:08:41.583" Score="0" ViewCount="55" Body="&lt;p&gt;So I am calculating phase velocity of a wave for Froude number scaling for model testing of a boat. I got data from Lloyd's Rules saying for the specific sea area I did &lt;span class=&quot;math-container&quot;&gt;$H_s=3.6\text{ m}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T_z=6.8\text{ s}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a regular, progressive wave (this is where I might be wrong) and using the deep water relationship of &lt;span class=&quot;math-container&quot;&gt;$\lambda = \dfrac{gT^3}{2\pi}=712.5\text{ m}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wave celerity &lt;span class=&quot;math-container&quot;&gt;$v=\dfrac{\lambda}{T}=\dfrac{712.5}{6.8}=104.7\text{  m/s} = 377\text{ km/s}$&lt;/span&gt;? Am I doing this right? Are ocean waves supposed to be that fast?&lt;/p&gt;&#xA;" OwnerUserId="18468" LastEditorUserId="1832" LastEditDate="2018-11-30T20:13:27.553" LastActivityDate="2018-12-03T07:44:37.573" Title="Finding phase velocity for froude number scaling" Tags="|fluid-mechanics|marine-engineering|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24936" PostTypeId="1" AcceptedAnswerId="24938" CreationDate="2018-11-30T21:46:55.663" Score="1" ViewCount="547" Body="&lt;p&gt;I need help figuring the distribution of a horizontal shearing force in an elastic solid trapezoid. When force &lt;span class=&quot;math-container&quot;&gt;$F_1$&lt;/span&gt; is applied to all the top surface of the trapezoid, at equilibrium, is the force in all horizontal layers of the solid the same? That is &lt;span class=&quot;math-container&quot;&gt;$F_1=F_2=F_3=F_4$&lt;/span&gt;? Or is the shear stress the same and the force varies between layers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IkUNH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IkUNH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18470" LastEditorUserId="1832" LastEditDate="2018-11-30T23:41:47.120" LastActivityDate="2018-12-01T01:34:17.647" Title="shearing force distribution in a trapezoid" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24937" PostTypeId="2" ParentId="24936" CreationDate="2018-11-30T23:40:45.920" Score="0" Body="&lt;p&gt;Newton's Second Law of Motion tells us that the force applied to an object is equal to the product of said object's mass and its acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F=ma$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, assuming that your trapezoid (both &quot;halves&quot; of it) remain static (that is, with &lt;span class=&quot;math-container&quot;&gt;$a=0$&lt;/span&gt;), then the &lt;em&gt;net force&lt;/em&gt; applied on the trapezoid must also be zero. In this case, we can therefore state that the sum of all the horizontal forces must be equal to zero:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sum F_x = F_1 - F_2 + F_3 - F_4 = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(the minus signs for &lt;span class=&quot;math-container&quot;&gt;$F_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_4$&lt;/span&gt; are because of the orientation you gave those forces in your question).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forces all being equal would satisfy that equation, but so would an infinitely many other configurations (i.e. &lt;span class=&quot;math-container&quot;&gt;$F_1 = F_2 + F_4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_3 = 0$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you've split your trapezoid into two halves. I assume this means it's actually two trapezoids, one resting upon the other, and that both are static (&lt;span class=&quot;math-container&quot;&gt;$a=0$&lt;/span&gt;). In this case, we can apply the same equation to each individual trapezoid, which gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= F_1 - F_2 = 0 \\&#xA;\sum F_x &amp;amp;= F_3 - F_4 = 0&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously solvable, with &lt;span class=&quot;math-container&quot;&gt;$F_1 = F_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_3 = F_4$&lt;/span&gt;. However, we still can't tell whether the two pairs are equal to each other (that is, whether &lt;span class=&quot;math-container&quot;&gt;$F_1 = F_3$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that, we can rely on Newton's Third Law, which states that every action produces an equal and opposite reaction. Looking at &lt;span class=&quot;math-container&quot;&gt;$F_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_3$&lt;/span&gt;, that's clearly what we're looking at: the action and reaction along the interface between the two trapezoids. So we can obviously state that &lt;span class=&quot;math-container&quot;&gt;$F_2 = F_3$&lt;/span&gt;, which combined with the previous deduction gives us&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_1 = F_2 = F_3 = F_4$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the forces are all equal. Since the areas upon which these forces are acting are different, this means the shear stresses will be different. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, as is, your trapezoid would actually rotate due to the couple formed by &lt;span class=&quot;math-container&quot;&gt;$F_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_4$&lt;/span&gt;, though that seems to be beyond the scope of this question.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2018-11-30T23:40:45.920" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24938" PostTypeId="2" ParentId="24936" CreationDate="2018-12-01T01:34:17.647" Score="0" Body="&lt;p&gt;Consider the main trapezoid, a cantilever column( assuming it is thin enough to be subject of plane shear, not a solid) with load F applied to the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that &lt;span class=&quot;math-container&quot;&gt;$ \ \Sigma F_x =0 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means if we consider the free body diagram of the column at any horizontal level including the base and the level shown in your diagram, shear, V is constant and is always:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V = -F1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the shear flow, q is not the same and shear stress increases as the length of the base of trapezoid in the free body diagram decrease. e.g. if the length of the base is half of the length on top the shear flow q is two times the shear flow of the top.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-01T01:34:17.647" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24939" PostTypeId="1" CreationDate="2018-12-01T03:00:28.483" Score="1" ViewCount="85" Body="&lt;p&gt;I have two heat pumps directly outside my house and they are noisy so I am considering the viability of using a sound barrier to block the sound. The acoustic barrier itself will definitely absorb and eliminate the sound in three directions, but the question is whether the sound will be able to go around from the open side and get into the house. Here is the basic set up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a9KAv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a9KAv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat pumps are shown in blue. The proposed acoustic barrier is shown in red.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will sound be able to go around the edges of the barrier and get back into the house or will it radiate outwards and not come back into the house?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2018-12-01T06:37:09.410" Title="Will a 3-wall acoustic barrier work or will sound go around?" Tags="|acoustics|sound-isolation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24940" PostTypeId="2" ParentId="24939" CreationDate="2018-12-01T03:43:00.167" Score="2" Body="&lt;p&gt;It will help to shield you from direct high pitch noise waves. because these waves short wavelengths will be blocked by objects in the range of your partition size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say a noise of frequency of &#xA;500-hertz has a wave length (Lambda) of 340/500 = 68cm, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but at 170-hertz lambda = 200cm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And much of the noise generated by the heat pumps will go around or be transmitted through the bases of equipment to the ground and your building walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would help if you could mount the heat pumps on rubber brackets on the concrete slab, or use noise attenuating neoprene seals . Then enclose the pumps, as you've shown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, ideally, sound proof your home from outside with acoustic foam panels, but it could be too expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can moving the pumps back a little bit can help too.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-12-01T06:10:58.070" LastActivityDate="2018-12-01T06:10:58.070" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24942" PostTypeId="2" ParentId="24939" CreationDate="2018-12-01T06:37:09.410" Score="2" Body="&lt;p&gt;For fans used to move air, their primary noise radiation frequency will be (number of blades) x (revs per second of the fan shaft) and will be emitted in all directions if the fan shaft is vertical, as in most heatpump units. A 3-blade fan running at 1750 RPM will produce a hum at 88Hz. This will have a wavelength of about 11 feet and will diffract strongly around a wall less than 11 feet in scale length (height or width). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can still get away with a &quot;small wall&quot; if it is folded into a box that completely encloses the fan unit except for an inlet baffle which must be positioned facing away from the dwelling's windows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the combination of (enclosure + inlet baffle) will have a tendency to resonate (see &quot;&lt;em&gt;helmholtz resonator&lt;/em&gt;&quot;) and if the resonance matches the fan noise, the enclosure will be ineffective in blocking the noise unless you line it with absorptive material like fiberglass. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-01T06:37:09.410" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24943" PostTypeId="1" CreationDate="2018-12-01T08:35:04.950" Score="1" ViewCount="215" Body="&lt;p&gt;Six-bar linkage is shown on the image:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/LH96m.png&quot; alt=&quot;Six-bar linkage&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;It has one degree of freedom, so if one of the variables is constrained, it's possible to calculate the positions of all the joints. In this specific case the point C is of interest. It is required to derive a function which describes trajectory of point &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; (red line on image 2), as &lt;span class=&quot;math-container&quot;&gt;$x = f(y)$&lt;/span&gt;. Constant values are positions of fixed joints &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; and links' lengths &lt;span class=&quot;math-container&quot;&gt;$L_{AB}, L_{BC}, L_{FD}, L_{FE}, L_{GE}, L_{HF}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rdnjo.png&quot; alt=&quot;Trajectory to be found&quot; /&gt;&lt;/p&gt;&#xA;&lt;p&gt;What I &lt;em&gt;can&lt;/em&gt; do, is assume the angle &lt;span class=&quot;math-container&quot;&gt;$\phi_1$&lt;/span&gt; and then use trigonometry to derive positions of points &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; and finally, &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;. The math behind that is next:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Define:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;R(\phi)&amp;amp;=[\cos\phi, \sin\phi], \arctan(A) = \arctan(A_y/A_x) \\&#xA;F&amp;amp;=H+L_{HF}R(\phi_1) \\&#xA;\phi_2&amp;amp;=\arctan(FG) \\&#xA;\phi_3&amp;amp;=\arccos\left(\frac{|FG|^2+L_{FE}^2-L_{GE}^2}{2|FG|L_{FE}}\right) \\&#xA;D&amp;amp;=F+L_{FD}R(\phi_2-\phi_3) \\&#xA;\phi_4&amp;amp;=\arctan(AD) \\&#xA;\phi_5&amp;amp;=\arccos\left(\frac{|AD|^2+L_{AB}^2-L_{BD}^2}{2|AD|L_{AB}}\right) \\&#xA;B&amp;amp;=A+L_{AB}R(\phi_4-\phi_5) \\&#xA;C&amp;amp;=B+\frac{L_{BC}}{L_{BD}}BD&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;That wasn't too hard to derive. Now, the real problem is, &lt;em&gt;angle &lt;span class=&quot;math-container&quot;&gt;$\phi_1$&lt;/span&gt; is unknown&lt;/em&gt;, instead the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; coordinate of point &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is given.&lt;/p&gt;&#xA;&lt;p&gt;What I can think of, is combing all the equations into one, expressing &lt;span class=&quot;math-container&quot;&gt;$C_y=f^1(\phi_1)$&lt;/span&gt; and trying to inverse the function into &lt;span class=&quot;math-container&quot;&gt;$\phi_1=f^2(C_y)$&lt;/span&gt;. That way I could pass &lt;span class=&quot;math-container&quot;&gt;$C_y$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$f^2$&lt;/span&gt;, get the angle &lt;span class=&quot;math-container&quot;&gt;$\phi_1$&lt;/span&gt;, pass it to the initial formulas and get &lt;span class=&quot;math-container&quot;&gt;$C_x$&lt;/span&gt; as the result. But this seems nearly impossible. I can write down the equation for &lt;span class=&quot;math-container&quot;&gt;$C_y$&lt;/span&gt; and it will have just 4 summands, but those will contain sines of &lt;span class=&quot;math-container&quot;&gt;$\phi_{2-5}$&lt;/span&gt; which in their turn are arctan or arccos of fractions which contain other angles which are again some invert function of other huge expression... Expression for &lt;span class=&quot;math-container&quot;&gt;$\phi_2$&lt;/span&gt; isn't very big, but for &lt;span class=&quot;math-container&quot;&gt;$\phi_3$&lt;/span&gt; it's enormous and &lt;span class=&quot;math-container&quot;&gt;$\phi_4$&lt;/span&gt; is getting out of my hands. Inverting all that is something I can't imagine.&lt;/p&gt;&#xA;&lt;p&gt;Are there any sane methods to calculate the six-bar linkage and get an analytic solution for the trajectory?&lt;/p&gt;&#xA;" OwnerUserId="18472" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-12-01T15:25:42.920" Title="Analytical six-bar linkage trajectory solution" Tags="|mechanical-engineering|mechanisms|linkage|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24945" PostTypeId="1" CreationDate="2018-12-01T11:36:29.253" Score="1" ViewCount="135" Body="&lt;p&gt;I have a copper tube that I'm heating (quite a lot) with a  torch, I have to connect a PVC pipe in the entrance of the copper and one in the exit (so water can flow through the copper pipe). The only problem is that the copper tube is too hot, so the PVC pipe melts, any ideas on how to connect them? Maybe a ceramic or fiberglass joint?&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="18475" LastActivityDate="2018-12-01T20:23:24.973" Title="Connect heated copper tube in PVC tube" Tags="|thermodynamics|heat-transfer|pipelines|ceramics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="24946" PostTypeId="2" ParentId="7278" CreationDate="2018-12-01T14:57:22.287" Score="1" Body="&lt;p&gt;As Karlo's comment pointed, every 2D representation of a 3D object has some loss of information. Furthermore, drawing that representation and recovering information from it (taking measures) has some level of complexity. Choosing among available representations is a trade-off between different losses of information and levels of complexity. A few pros and cons of each may be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cavalier:&#xA;- A bit easier to draw and measure in the z axis.&#xA;- Less natural perception of volume (objects seem enlarged along the z axis).&#xA;- Better perception of plan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cabinet:&#xA;- A bit harder to draw and measure in the z axis.&#xA;- More natural perception of volume.&#xA;- Plans more stretched compared with plan views.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, cavalier is likely to fit better things like drawing fortifications and cities, where plan matters the most, where cabinet is likely to be better drawings of vertical furniture (wardrobes, for example) where the elevation is more important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, if oblique projections are seen as a simple way to simulate a perspective, they simulate perspectives with different points or view (lower for cabinet, higher for cavalier). Cabinet is likely to be a better fit for objects best seen from the ground (like furniture) while cavalier is likely to be a better fit for objects best seen from air (like cities).&lt;/p&gt;&#xA;" OwnerUserId="7371" LastActivityDate="2018-12-01T14:57:22.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24947" PostTypeId="2" ParentId="24934" CreationDate="2018-12-01T17:21:14.287" Score="0" Body="&lt;p&gt;The substitution of 0.01633 for &lt;span class=&quot;math-container&quot;&gt;$Q_1$&lt;/span&gt; may be incorrect. That is the total required discharge, and not the discharge from the original pipe outlet. Did you mean to substitute the original value of &lt;span class=&quot;math-container&quot;&gt;$Q=0.0122$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$Q_1$&lt;/span&gt;? Doing that gives a positive value at the stage you are currently at. Follow that to completion and I imagine you will obtain a reasonable value.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2018-12-01T18:46:06.027" LastActivityDate="2018-12-01T18:46:06.027" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24949" PostTypeId="2" ParentId="24945" CreationDate="2018-12-01T20:23:24.973" Score="1" Body="&lt;p&gt;Any thing you add at the ends to insulate the PVC from the intense heat will heat up itself after a short time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one way of dealing with this is to add approximately 3-5 feet of the copper pipe depending on the speed of water and size of the pipes, as transition area before and after the heated part. Because the water exiting the heated pipe will cool down fast after a while and wont soften or melt the PVC. Same thing for the transition at the beginning the upstream heat conduction and convection in the length of the copper pipe will be coold off by inflow of the cold water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These transition lengths will perform better if they are narrower, because a narrow pipe offers relatively more surface to let the air or water work on heat reduction. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-01T20:23:24.973" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24952" PostTypeId="1" AcceptedAnswerId="24953" CreationDate="2018-12-02T08:35:52.440" Score="3" ViewCount="2005" Body="&lt;p&gt;All I know about welding is it is a technique in which electrode is connected to one terminal and the gripper connected to workpiece is in connection with the other terminal. As current passes and when the electrode is touched to the workpiece,an arc is generated which melts electrode and helps in the formation of a joint. As the circuit is completed with a conductor in-between,doesn't it result in short circuit?&lt;/p&gt;&#xA;" OwnerUserId="15268" LastActivityDate="2018-12-02T08:55:47.057" Title="Short circuit in welding" Tags="|welding|current|shock|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24953" PostTypeId="2" ParentId="24952" CreationDate="2018-12-02T08:55:47.057" Score="7" Body="&lt;p&gt;No, this is not classed as a &quot;short-circuit&quot;. This is a circuit with a designed current flow - ie a 60 to 150A range is common...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;short-circuit&quot; is a circuit where the current flow is not following the designed current path ie it is taking a &quot;shorter&quot; route to ground.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-12-02T08:55:47.057" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="24955" PostTypeId="1" CreationDate="2018-12-02T15:36:25.390" Score="0" ViewCount="38" Body="&lt;p&gt;I am thinking that air (if it is still) is a very good insulator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why aren't we putting the window shutter far away from the window? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would have more air between the window and the shutter and the &lt;strong&gt;Question&lt;/strong&gt;:  it would cause better insulation? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or above a few centimeters thickness, the airflow would start anyways by itself and it would just only backfire? &lt;/p&gt;&#xA;" OwnerUserId="18387" LastActivityDate="2019-12-28T05:01:50.400" Title="How much cm needed for airflow? - optimal window shutter distance" Tags="|airflow|" AnswerCount="1" CommentCount="4" ClosedDate="2019-12-28T15:30:38.967" ContentLicense="CC BY-SA 4.0" />
  <row Id="24956" PostTypeId="1" CreationDate="2018-12-02T16:19:00.953" Score="1" ViewCount="708" Body="&lt;p&gt;I am solving pipe flow problems and it came to my mind how does one find out the minor loss associated with the splitting of pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8BV12.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8BV12.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here how does one find the minor loss associated when the flow is in either direction.it is the same as contraction/expansion loss or a 90° elbow loss.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastActivityDate="2019-08-30T01:10:41.463" Title="Minor loss due to splitting of pipe" Tags="|mechanical-engineering|fluid-mechanics|chemical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24957" PostTypeId="2" ParentId="24956" CreationDate="2018-12-02T16:41:34.117" Score="0" Body="&lt;p&gt;I'm pretty certain that would qualify as a &quot;tee&quot; fitting, so the minor loss coefficient would range from 0.2 to 2.0 per &lt;a href=&quot;https://www.engineeringtoolbox.com/minor-loss-coefficients-pipes-d_626.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source&lt;/a&gt; depending upon the fitting type.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2018-12-02T16:41:34.117" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24958" PostTypeId="1" CreationDate="2018-12-02T17:25:15.483" Score="1" ViewCount="587" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MjbId.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MjbId.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;in this picture, there is a container of water and it is moving on the slope.&#xA;we want the acceleration of the container while the angle of water surface is parallel to the slope.&#xA;so the book says the answer is that we let container move by its own weight so &lt;span class=&quot;math-container&quot;&gt;$\color{red}{a_x=g\,sin\alpha}$&lt;/span&gt;. &#xA;but what i'm asking is that let us think the slop doesn't exist. so the accelerations would be like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fWGWp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fWGWp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;and the formula of the angle of surface in fluid mechanics is: &lt;span class=&quot;math-container&quot;&gt;$$\color{green}{tan\,\theta= {\Sigma a_x\over \Sigma a_y}}$$&lt;/span&gt;&#xA;so in this case &lt;span class=&quot;math-container&quot;&gt;$tan \theta $&lt;/span&gt; would be:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$tan \theta={gsin\alpha\over gcos \alpha}=tan\alpha$$&lt;/span&gt;&#xA;so &lt;span class=&quot;math-container&quot;&gt;$$\theta=\alpha$$&lt;/span&gt;&#xA;and this means:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KMKJ7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KMKJ7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is wrong because we know if we let the container move by its own weight, the &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; would be &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;&#xA;i don't know what am i missing here.&lt;/p&gt;&#xA;" OwnerUserId="18489" LastActivityDate="2021-05-21T21:02:09.523" Title="the angle of water surface in a moving car" Tags="|fluid-mechanics|fluid|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24959" PostTypeId="2" ParentId="24958" CreationDate="2018-12-02T19:46:53.870" Score="1" Body="&lt;p&gt;Let's assume the container is moving horizontally but under the forces &lt;span class=&quot;math-container&quot;&gt;$F_v =mg \cos(\alpha)$&lt;/span&gt; as vertical force and &lt;span class=&quot;math-container&quot;&gt;$F_y = mg\sin(\alpha)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water surface angle will be &lt;span class=&quot;math-container&quot;&gt;$\arctan \left(\dfrac{\sin(\alpha)}{\cos(\alpha)}\right) = \alpha$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefor &lt;span class=&quot;math-container&quot;&gt;$\theta = \alpha$&lt;/span&gt;. So if you rotate the image by &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; the water will stand parallel with the incline!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2018-12-02T23:26:52.230" LastActivityDate="2018-12-02T23:26:52.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24960" PostTypeId="1" CreationDate="2018-12-02T19:52:14.903" Score="0" ViewCount="186" Body="&lt;p&gt;I am working on analyzing a few problems, but have not really come across this type of stress concentration problem. Does anyone know where I can study the analysis of this situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressing ball through hole with smaller diameter with some force F&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W1aoh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W1aoh.jpg&quot; alt=&quot;Pressing ball through hole with smaller diameter with some force F&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem involves modelling a structure like so, and attempting to find the shear experienced at the connection E:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bodies 1, 2 and 3 connected (maybe bolts) at lines B and E, if force F_ab acts on body 3, what shear force experienced at connection E? Is it simply summation of forces? Would the geometry changes not matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oPKwd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oPKwd.jpg&quot; alt=&quot;Bodies 1, 2 and 3 connected (maybe bolts) at lines B and E&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any guidance would be much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="18490" LastActivityDate="2024-04-06T05:00:30.070" Title="stress/shear analysis of components" Tags="|mechanical-engineering|structural-analysis|stresses|shear|failure-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24961" PostTypeId="2" ParentId="24960" CreationDate="2018-12-02T20:53:50.810" Score="0" Body="&lt;p&gt;assume in the first part of your question we are measuring the stresses at the moment the ball is about to pass through the hole.&#xA;Say R1 is the radius of the hole at rest and R2 is when it is expanded to allow for the passage, and we assume no warping of the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V = F/2\pi R_2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for he second part the shear stress equation for composite beams is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \tau = VQ/It$$&lt;/span&gt;&#xA;V is the shear from the shear diagram, Q is the first moment of area of the part of the beam above or below the centroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All you need is the dimensions and if the platform is made of a different material its E and G, shear modulus of the platform.&#xA;You need to multiply the platform section area by  &lt;span class=&quot;math-container&quot;&gt;$E_{platform}/E_{base\ material}$&lt;/span&gt; to calculate the centroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first part you need some basic data like the dimension of the hole and radios of the ball, to verify if the stresses are plastic or elastic. You can model the plate around the hole as many say 60 radial cantilever beams connected to each other by concentric rings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the plate is thin enough it will allow bending around the edge of the hole or it will make the whole expand while creating a depressing. Another case that can be modeled in FEM is the bell itself will shrink into a football shape to pass through.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-12-03T17:23:28.913" LastActivityDate="2018-12-03T17:23:28.913" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24962" PostTypeId="1" AcceptedAnswerId="24976" CreationDate="2018-12-02T21:07:56.203" Score="2" ViewCount="62" Body="&lt;h1&gt;Problem&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I have two USRP antennas close to each other. I know that signal strength on the receiving end has to be at least -80&amp;nbsp;dBm. Both antennas have 2&amp;nbsp;dBi of gain. I want to compute what is maximum propagation loss so that this system still works. Signal is being transmitted with power of 10&amp;nbsp;mW.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Attempt to solve&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Since i want to know the maximum amount of power loss during transmission to one antenna to another. I can form equation as. &lt;span class=&quot;math-container&quot;&gt;$p_0=$&lt;/span&gt; power of transmitted signal, &lt;span class=&quot;math-container&quot;&gt;$p_{m}=$&lt;/span&gt; minimum acceptable signal strength, &lt;span class=&quot;math-container&quot;&gt;$x=$&lt;/span&gt; maximum amount of signal loss during transmission.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ p_0 \cdot x = p_{\text{ml}} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ x = \frac{p_0}{p_{ml}} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My solution is ratio of powers so I can use dB units, by converting them with base-10 logarithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ x_{dB}=10\cdot \log_{10}\left(\frac{p_0}{p_{ml}}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i need to change -80&amp;nbsp;dBm to millliWatts with &lt;span class=&quot;math-container&quot;&gt;$-80\text{ dBm}=10^{-80/10}\text{ mW}$&lt;/span&gt;. After this I can sum antenna gain to my answer since we are using logarithms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ x_{dB}=10\log_{10}\left(\frac{10^{-80/10}\text{ mW}}{10\text{ mW}}\right)+2\text{ dBi} +2\text{ dBi}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ x_{dB}=-86\text{ dB} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However my answer seems to be incorrect ?&lt;/p&gt;&#xA;" OwnerUserId="18491" LastEditorUserId="1832" LastEditDate="2018-12-02T23:23:13.940" LastActivityDate="2018-12-03T16:56:18.037" Title="Propagation loss in USRP antenna" Tags="|telecommunication|radio|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24964" PostTypeId="2" ParentId="17457" CreationDate="2018-12-02T23:54:35.483" Score="0" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ 100- 25cos (20) = 76.5076 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force in member ob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We divide the tributory of 75 N by this to get the cos(b-a)&#xA;75/ 76.5076 = 0.98029&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arccos (0.98029)= 11.3934&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This  added to 20 is angle beta.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beta = 20 + 11.3934 = 31.3934&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And Omega = 180- 31.3934= 148.606&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum of angle in a triangle =180°&#xA;Omega=180-(70+31.394)&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="16486" LastEditDate="2019-06-27T18:07:04.047" LastActivityDate="2019-06-27T18:07:04.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24966" PostTypeId="2" ParentId="24955" CreationDate="2018-12-03T01:09:10.957" Score="0" Body="&lt;p&gt;The distance between the window shutters and windows is mostly dependant on the thickness of the walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Air can flow through much smaller space such as the fins of amplifiers heat sinks or many electronics heatsinks that don't use a fan.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-03T01:09:10.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24967" PostTypeId="1" CreationDate="2018-12-03T03:15:06.940" Score="1" ViewCount="73" Body="&lt;p&gt;When equal and opposite torque act on  the two ends of a circular shaft,how do we know that deformations occur in a uniform fashion throughout the length of  shaft?Is it because that all cross section of shaft have same torque acting on them and hence the stress distribution on the cross section is same which further results in same strain throughout the cross section?If not,then how?&lt;/p&gt;&#xA;" OwnerUserId="19791" LastEditorUserId="19791" LastEditDate="2018-12-04T04:41:41.733" LastActivityDate="2018-12-04T19:37:55.453" Title="How do we know that deformations occur in a uniform fashion throughout the length of shaft?" Tags="|mechanical-engineering|solid-mechanics|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24968" PostTypeId="1" AcceptedAnswerId="24971" CreationDate="2018-12-03T03:59:38.890" Score="0" ViewCount="450" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/cHZIS.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the moment of inertia about X and Y axes. The area has been rotated through an angle theta. How to calculate the moment of inertia about x1 and y1 axes?&lt;/p&gt;&#xA;" OwnerUserId="18283" LastEditorUserId="1832" LastEditDate="2018-12-03T04:50:52.640" LastActivityDate="2018-12-03T08:38:08.200" Title="Moment of inertia of rotated area" Tags="|structural-engineering|mechanical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24970" PostTypeId="2" ParentId="24935" CreationDate="2018-12-03T07:44:37.573" Score="5" Body="&lt;p&gt;I'm not sure where did you brought the 'deep water relationship' in here. If you're dealing with usual water wave (gravity, surface, linear, and close to sinusoidal wave form), then the relationship you'd need to use will be dispersion relationship.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega^2 = gk\tanh(kh)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you're assuming deep water, it becomes:&#xA;&lt;span class=&quot;math-container&quot;&gt;$\omega^2 = gk$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which leads to almost exactly same equation you came up with, but one less power of T:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\lambda = \dfrac{gT^2}{2\pi} = 72m$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and phase velocity of 10.61 meter/second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Actually, your original calculation, with &lt;span class=&quot;math-container&quot;&gt;$T^3$&lt;/span&gt; was incorrect. That should have give you wave length of 491 meter, instead of 712 meter)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional comments:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;YES. Wave can be very fast, if it is extra long wave like tsunami, like 400 kph. But the wave you gave as example, (T = 6.8 s) can't be too fast.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What you did was not really worth of value for model test if you're not specifically interested in surge-direction slamming. You'll need to match the length scale first, and calculate the corresponding dimensions (in case of wave, fit the height and length first according to length ratio, and then fit time and modify your wave maker actuator) Refer page 21 of linked document. &lt;a href=&quot;http://www.ivt.ntnu.no/imt/courses/tmr7/lecture/Scaling_Laws.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://www.ivt.ntnu.no/imt/courses/tmr7/lecture/Scaling_Laws.pdf&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As you came up with zero up-crossing period for the wave, I believe you have wave spectrum instead of single wave. The input for wave maker will be pretty much different if you need to generate wave spectrum instead of regular wave.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18499" LastActivityDate="2018-12-03T07:44:37.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24971" PostTypeId="2" ParentId="24968" CreationDate="2018-12-03T08:38:08.200" Score="0" Body="&lt;p&gt;Say your Iy is the max and Ix is minimum I before rotation of the axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We build the Mohr circle plotting Ix and Iy on its x axis. And we draw a vertical along Y axis and call it xy axis, somewhere on the left outside of the circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the center of circle is at &lt;span class=&quot;math-container&quot;&gt;$ (Ix+ Iy)/2 \ $&lt;/span&gt;  and its radius is &lt;span class=&quot;math-container&quot;&gt;$ R =  (I_y - I_x)/2 \ $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;now say you want to calculate the Ix' at 30 degrees anticlockwise rotation wrt x axis. So you cut an angle 60 degrees, twice the 30, and mark the coordinates of where it intersects the circle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Y is the Ixy max of the new rotated axis and if you draw a line from this point passing through the center of the Mohr circle it intersect the circle on the other side at Ixy min. If you want you could also wright these as equations.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-03T08:38:08.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24975" PostTypeId="2" ParentId="21834" CreationDate="2018-12-03T16:32:29.800" Score="0" Body="&lt;p&gt;The valve symbols you show in your diagram are that of a simple gate valve, the symbol for the types of solenoid valves your wishing to represent are put together as such.....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q6oNB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q6oNB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are just some examples....&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-12-03T16:32:29.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24976" PostTypeId="2" ParentId="24962" CreationDate="2018-12-03T16:56:18.037" Score="0" Body="&lt;p&gt;There is simple error on second line: it should be like this&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ p_0\cdot x = p_{ml} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ x = \frac{p_{ml}}{p_0} $$&lt;/span&gt;&#xA;the fraction was wrong. &#xA;Then we have:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ x_{dB}= 10 \log_{10}(\frac{10\text{mW}}{10^{-80/10}\text{mW}})+2\text{dBi}+2 \text{dBi}= 94 \text{dB} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer seems to be correct since it corresponds to physical measurements. &lt;/p&gt;&#xA;" OwnerUserId="18491" LastActivityDate="2018-12-03T16:56:18.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24978" PostTypeId="1" CreationDate="2018-12-03T23:21:36.337" Score="0" ViewCount="122" Body="&lt;p&gt;I'm trying to figure out what size of round tube to use for a tensegrity structure that'll probably have people climbing on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rigid elements are 21' steel tubes, and in normal cases they undergo ~800 lbf of compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming it's horizontal and pinned at both ends, and is under ~4000 lbf of compression (design * safety factor 5) -- I want to know how much load it can withstand at the center, given the inner and outer diameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximate answers are fine, but I'm not sure how to approach this.&lt;/p&gt;&#xA;" OwnerUserId="18506" LastActivityDate="2018-12-04T15:56:41.410" Title="compression buckling under lateral load" Tags="|mechanical-engineering|structural-engineering|structural-analysis|structures|mechanical-failure|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24979" PostTypeId="1" CreationDate="2018-12-03T23:39:36.360" Score="0" ViewCount="376" Body="&lt;p&gt;What type of glue is best used to glue stainless steel bolt (not screw) used as self tapping to a hole (no thread) in a 3d printed nylon (polyamide) part? It will be an M2 metric bolt.&lt;/p&gt;&#xA;" OwnerUserId="414" LastActivityDate="2018-12-04T00:47:37.737" Title="Glue metal bolt to nylon hole" Tags="|metals|" AnswerCount="1" CommentCount="1" ClosedDate="2018-12-05T18:01:36.367" ContentLicense="CC BY-SA 4.0" />
  <row Id="24980" PostTypeId="2" ParentId="24928" CreationDate="2018-12-04T00:16:38.330" Score="1" Body="&lt;p&gt;A good starting point is a failure criterion such as Willam-Warnke (&lt;a href=&quot;https://en.wikipedia.org/wiki/Willam-Warnke_yield_criterion&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Willam-Warnke_yield_criterion&lt;/a&gt;).  Note that the shape of the failure surface is approximately triangular, indicating different strengths under uniaxial compression and tension.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have data on the biaxial compression of the material, you can play around with the parameters until you get a surface that approximates the material.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-12-04T00:16:38.330" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24981" PostTypeId="2" ParentId="24979" CreationDate="2018-12-04T00:47:37.737" Score="3" Body="&lt;p&gt;A quick search with &quot;adhesive for nylon to steel&quot; presents a number of results, almost all of them indicating that there is a solution. Polypropylene/polyethylene adhesive in the form of an epoxy will bond to the nylon and also to properly prepared steel surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preparation segment appears to be a critical factor. Cleanliness as well as surface roughness provides the best results. One would use an alcohol type cleaner, such as rubbing alcohol on both parts, to remove grease, skin oil, etc, but that should be performed after abrading the surfaces with 200-400 grit sandpaper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a product recommendation SE and it's impractical to suggest a specific product, as the search will also present suitable products, possibly specific to your area of the planet. For many in the USA, Loctite brand products cover a wide range of applications, including this one.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-12-04T00:47:37.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24983" PostTypeId="2" ParentId="24978" CreationDate="2018-12-04T08:59:15.703" Score="-1" Body="&lt;p&gt;Just as an illustration, without any intent for real application, we assume the following.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no restriction on deflection, however we will control for buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tube will fail under compression stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets assume R1 and R2 as outer and inner radiuses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We limit the compressive stress to 10% so that we do not need to deal with column design charts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also limit the radius of gyration to r&amp;lt;20 and r for a hollow tube is approximated to r= R2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can say &lt;span class=&quot;math-container&quot;&gt;$$ \sigma_{axial} = \dfrac{4000}{\pi(R1^2- R2^2)}&amp;lt;\dfrac{F_{yield }}{10}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we know the bending moment under concentrated load in the middle is &lt;span class=&quot;math-container&quot;&gt;$m =\dfrac{\omega \times l}{4}$&lt;/span&gt;, therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dfrac{\omega \times l}{4} = \ 0.9\times \sigma \times \dfrac{\pi\left(R_{1}^4-R_{2}^4\right)}{32R_{1}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And by plugging in R1 and R2 and the allowable sigma we find the &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt;. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2018-12-04T15:56:41.410" LastActivityDate="2018-12-04T15:56:41.410" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24984" PostTypeId="1" CreationDate="2018-12-04T10:37:37.827" Score="4" ViewCount="849" Body="&lt;p&gt;I have trouble understanding the following pneumatic circuit diagram:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/w0DvD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w0DvD.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I am new to reading circuit diagrams and i am unfamiliar with some of the symbols. I would appreciate if somebody could explain them to me. In particular I have the following questions:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What does the dotted line in bottom left of the directional control valve mean?&lt;/li&gt;&#xA;&lt;li&gt;Why is a control valve with 5 ports used? Isn't one of the port on the left of the valve redundant and a 4/2 valve could be used instead?&lt;/li&gt;&#xA;&lt;li&gt;Why are there four ports on the actuator? I am only familiar with actuators with two ports (which would correspond to the top two ports). What are the bottom port and the right port used for?&lt;/li&gt;&#xA;&lt;li&gt;What does the symbol in the top right mean?&lt;/li&gt;&#xA;&lt;li&gt;What do the triangles on top of the lines mean? Why is one white and one black?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I realize that these are a lot of questions. But as I said, I am new to this. I would also appreciate if anybody could link a free source, there such diagrams are explained systematically. So far I used the Wikipedia page, but there are a lot of symbols/subtleties which are not covered there.&lt;/p&gt;&#xA;" OwnerUserId="18510" LastEditorUserId="-1" LastEditorDisplayName="user16" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2024-01-16T13:07:50.997" Title="I want to understand this pneumatic circuit" Tags="|circuits|pneumatic|diagram|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="24986" PostTypeId="2" ParentId="24967" CreationDate="2018-12-04T19:37:55.453" Score="1" Body="&lt;p&gt;Yes you are right: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is it because that all cross section of shaft have same torque acting on them and hence the stress distribution on the cross section is same which further results in same strain throughout the cross section?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Let's cut the shaft into 100 short pieces. Each will experience the same torque and each will twist exactly 1/100 of the total shaft's deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why? because it is the same material and its crystals have identical structural properties, including identical shear modulus, G.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although in metal forging impurities are bound to enter into the alloy, by the time the quality control lets a batch through they are within the tolerance of the standards!  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-04T19:37:55.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24987" PostTypeId="1" CreationDate="2018-12-04T21:39:48.053" Score="0" ViewCount="1533" Body="&lt;p&gt;Thanks in advance, can someone explain Steady,Non steady, Uniform and Non Uniform flow and provide some example for each. &lt;/p&gt;&#xA;" OwnerUserId="17811" LastActivityDate="2018-12-05T11:17:53.897" Title="Steady Non steady and Uniform Non Uniform flow" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24988" PostTypeId="2" ParentId="24987" CreationDate="2018-12-04T21:52:08.437" Score="2" Body="&lt;p&gt;The steady state refers to any phenomenon that is independent of time. Mathematically it means: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial v}{\partial t} =0$$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a fixed point in the atmosphere and an observer on an airplane  moving with constant velocity then he sees the flow as steady the flow doesn't change as the observer approaches the point, however if the observe change his coordinate system, for instance an observer somewhere on the earth which is not moving, observes an unsteady flow pattern, or the flow pattern at the fixed point changes as the airplane approaches that point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the concept is not absolute it is coordinate dependent. Whenever an object moves through a fluid with a &lt;strong&gt;constant&lt;/strong&gt; velocity, the flow field may be transformed from an unsteady flow to a steady flow by selecting a coordinate system that is fixed with respect to the moving object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Uniform flow doend't accelerate or delay . Mathematically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial v}{\partial s}= 0 $$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about the rotational flow in your sink, it's nonuniform flow. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-12-05T11:17:53.897" LastActivityDate="2018-12-05T11:17:53.897" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="24989" PostTypeId="2" ParentId="21283" CreationDate="2018-12-04T23:21:54.707" Score="2" Body="&lt;p&gt;The keyboard contains many switches (100 in my keyboard at this moment) and the simplest solution for a keyboard assembly that contains no decoding circuitry would be for each switch to have its own line going to the motherboard, plus one common ground. This is inefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest scheme to minimize the burden of all those wires in the interconnect would be for the key wires to be arranged into a square array of vertical and horizontal wires and a switch positioned wherever any two wires cross over one another. Pressing a key then connects a specific vertical line with a specific horizontal line and when the motherboard &quot;sees&quot; that connection occurring, it interprets it as a specific key press. This is called &lt;em&gt;square matrix&lt;/em&gt; encoding and it allows a reduction in the number of address lines from N+1 to 2*(sqrt(N)) or, in this case, 20 lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a problem occurs if two keys are pressed simultaneously because then the motherboard &quot;sees&quot; four redundant possibilities instead of two and cannot tell which are real. This is called the &lt;em&gt;phantom key&lt;/em&gt; problem and requires a more complex encoding scheme than a square matrix, which means more lines are required to resolve redundancies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various schemes to eliminate redundancies like this and I do not know which was used by the manufacturer of your keyboard, but the result will be more than 20 lines. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-04T23:21:54.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24991" PostTypeId="1" CreationDate="2018-12-05T02:29:38.810" Score="1" ViewCount="142" Body="&lt;p&gt;I am using the vector cross product to calculate each surface area and thus the outward normal. However, this approach fails when the surfaces are non-planar. So what is the best approach to calculate surface area and normal for a non-planar face of a hexahedron?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xQDFp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xQDFp.png&quot; alt=&quot;Irregular Hex&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15803" LastActivityDate="2018-12-05T16:32:52.360" Title="Calculate surface area and outward normal of each face(non-planar) for a irregular hexahedron." Tags="|applied-mechanics|finite-element-method|numerical-methods|solid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="24992" PostTypeId="1" AcceptedAnswerId="24999" CreationDate="2018-12-05T03:26:58.340" Score="1" ViewCount="80" Body="&lt;p&gt;If I have a single Boolean value that I wish to transmit wirelessly over a relatively short distance (assume &amp;lt;2m) then what of the many possible options would be the most &lt;em&gt;practical&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To define what I mean my &lt;em&gt;practical&lt;/em&gt;, it is some combination of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;low cost&lt;/li&gt;&#xA;&lt;li&gt;low power&lt;/li&gt;&#xA;&lt;li&gt;environmentally robust&lt;/li&gt;&#xA;&lt;li&gt;high data integrity&lt;/li&gt;&#xA;&lt;li&gt;high reliability&lt;/li&gt;&#xA;&lt;li&gt;low bulk/weight&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All of which apply equally to both the transmitter and receiver. I assume there will be some tradeoff between these &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other concerns that might impact the answers, but are of less interest to me:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The transmission may be repeated, but not at a high frequency (assume &amp;lt;2Hz), although there is no penalty for higher frequency capability.&lt;/li&gt;&#xA;&lt;li&gt;The transmission need only be unidirectional, although bi-directionality might be a bonus.&lt;/li&gt;&#xA;&lt;li&gt;The transmission does not have any security concerns, although some level of security might be a bonus.&lt;/li&gt;&#xA;&lt;li&gt;There is only one transmitter/receiver pair in the local area, although the ability to operate with multiple unique transmissions in the same area might be a bonus.&lt;/li&gt;&#xA;&lt;li&gt;The transmission is one-to-one between transmitter and receiver, although the ability to transmit to many receivers, or from many transmitters, might be a bonus.&lt;/li&gt;&#xA;&lt;li&gt;The solution should be off-the-shelf to a certain extent, although if bespoke (or high volume) options significantly improve the solution in some way then I would be interested to hear about them.&lt;/li&gt;&#xA;&lt;li&gt;The solution should ideally be legal/unlicensed in most countries, but I'd be interested in what illegal or licensed options could be possible too.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I don't have a specific application in mind, I'm more interested as a theoretical exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that there is a related question &lt;a href=&quot;https://engineering.stackexchange.com/questions/23209/best-way-to-communicate-data-wirelessly-with-specific-conditions&quot;&gt;here&lt;/a&gt;, but currently unanswered and with slightly different criteria.&lt;/p&gt;&#xA;" OwnerUserId="157" LastActivityDate="2018-12-05T11:25:19.060" Title="Most practical way to wirelessly transmit a single bit of information?" Tags="|wireless-communication|" AnswerCount="2" CommentCount="5" ClosedDate="2018-12-05T13:10:37.537" ContentLicense="CC BY-SA 4.0" />
  <row Id="24993" PostTypeId="1" AcceptedAnswerId="24994" CreationDate="2018-12-05T04:13:08.507" Score="1" ViewCount="2457" Body="&lt;p&gt;you might find my question stupid. I am unfortunately not an engineer and didn't find the answer on google. &#xA;I want to calculate the &lt;strong&gt;braking energy&lt;/strong&gt; of a vehicle &lt;strong&gt;decelerating from v2 to v1&lt;/strong&gt; (km/h). Given are the &lt;strong&gt;Wheel inertia mass moment&lt;/strong&gt; &quot;I&quot; and the &lt;strong&gt;wheel diameter&lt;/strong&gt; &quot;d&quot; as well as the &lt;strong&gt;vehicle mass&lt;/strong&gt; &quot;M&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you give me the equation needed for that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you ! &lt;/p&gt;&#xA;" OwnerUserId="18519" LastActivityDate="2018-12-05T07:29:21.007" Title="Calculating the Braking Energy of a vehicle" Tags="|mechanical-engineering|automotive-engineering|energy|regenerative-braking|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24994" PostTypeId="2" ParentId="24993" CreationDate="2018-12-05T06:17:51.417" Score="2" Body="&lt;p&gt;there are two kinetic types of energy that are involved in a moving car: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear and rotational  kinetic energy, &lt;span class=&quot;math-container&quot;&gt;$$ K_{e \ linear} = \frac {1}{2}mv^2 \ ; K_{e \  rotatinal} = \frac {1}{2}I \omega^2  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deceleration from V1 to V2 will reduce the energy. This energy is being saved in hybrid and electrical cars as regenerative energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linear energy change is equal to: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ K_{e \ linear \ chang} = \frac {1}{2}mv_{1}^2  - \frac {1}{2}mv_{2}^2 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And rotational energy change is equal to: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ K_{e \ rotatinal \ change} = \frac {1}{2}I_{1} \omega^2 -  \frac {1}{2}I_{2} \omega^2  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;span class=&quot;math-container&quot;&gt;$ \ V = r \omega = d \omega/2 \ and\ \omega= V/r $&lt;/span&gt; with r being the radius of the wheel, r = d/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, because of small &quot;I&quot; of the wheels compared to the mass of the car, the contribution of the wheel's change of energy is not significant. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-12-05T07:29:21.007" LastActivityDate="2018-12-05T07:29:21.007" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="24995" PostTypeId="1" CreationDate="2018-12-05T06:37:07.287" Score="0" ViewCount="177" Body="&lt;p&gt;So, I just printed out &lt;a href=&quot;https://www.thingiverse.com/thing:1004996&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; battery holder off Thingaverse.  Unfortunately, it doesn't work very well, so I'm interested in redesigning it.  There's a number of issues but the most annoying one is that the batteries keep getting stuck and not gravity feeding like they are supposed to, forcing you to shake the holder or reach up the output slot to bump the batteries until they feed out again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is obviously that the holder is just a box and that the cylindrical batteries jam up against each other at the bottom where the downward motion suddenly becomes horizontal.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick perusal of things like soda vending machines and ammo feed mechanisms shows that a &lt;a href=&quot;https://images.google.com/imgres?imgurl=http%3A%2F%2Fslideplayer.com%2Fslide%2F4674141%2F15%2Fimages%2F4%2FFeeding%2BMechanisms%2BFeeding%2Blong%2Bcylinders%2BFeeding%2BU-shaped%2Bparts.jpg&amp;amp;imgrefurl=http%3A%2F%2Fslideplayer.com%2Fslide%2F4674141&amp;amp;docid=DWpSTwAlD6FRoM&amp;amp;tbnid=ORcOUHnBMnF99M%3A&amp;amp;vet=1&amp;amp;w=960&amp;amp;h=720&amp;amp;source=sh%2Fx%2Fim&quot; rel=&quot;nofollow noreferrer&quot;&gt;v-shaped feed mechanism&lt;/a&gt; seems to be the standard way of doing this correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have no idea what sort of convergence angle and other considerations need to be made for designing such a device.  I could made a tall, slowly converging angle, but that makes the storage box a lot taller and lower capacity than is idea.  Too sharp of a convergence angle and I'll have the same jamming issues found in the existing design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I assume that there is some sort of literature out there for designing feed systems for cylindrical objects.  Obviously, things like cylinder density, surface roughness and other factors will affect this, but I assume that there's at least some sort of rough rule of thumb out there for what the optimal shape should be - e.g.: 45 degree convergence angle + some factor * the cylinder W/L ratio, etc.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I can figure this out with Edisonian trial and error, but I really don't feel like burning up a bunch of PLA and time if there's a decent standard rule of thumb I can use as a starting point for experimentation.&lt;/p&gt;&#xA;" OwnerUserId="18521" LastActivityDate="2023-12-10T07:01:30.163" Title="What are rules of thumb for designing a gravity feed system for cylinders?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="24997" PostTypeId="2" ParentId="24992" CreationDate="2018-12-05T10:27:49.757" Score="2" Body="&lt;p&gt;To be totally honest there is really only one technology you should be looking for.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Infrared, all you need is a light transistor and a infrared led, thats it. Cheap as hell, well known, and robust works in wide range of situation and is not prone to noise. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Anyway,  You can also try radio waves and sound.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-12-05T10:42:25.050" LastActivityDate="2018-12-05T10:42:25.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="24999" PostTypeId="2" ParentId="24992" CreationDate="2018-12-05T11:25:19.060" Score="1" Body="&lt;p&gt;Few of the technologies that fit well are Bluetooth Low Energy and Thread. Both of them operate in the 2.4 GHZ ISM band. Also you might want take a look at Zigbee. These all offer basic level of security features. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lora and Sigfox are two other technologies that might work well. I have little knowledge on these technologies &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the other post indicate IR also can be considered. But I think you will have to create the security layer. &lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2018-12-05T11:25:19.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25000" PostTypeId="2" ParentId="24995" CreationDate="2018-12-05T16:21:03.513" Score="0" Body="&lt;p&gt;This is more a problem of 3D-printing than engineering, I think.  If you skim thru the comments on that Thingiverse page, you'll see that people have designed and posted internal baffles to ensure that AAA batteries properly dispense.  Try that first.&lt;br /&gt;&#xA;And, of course, since this is extrusion printing, make sure to clean out any support material, whiskers, etc.&lt;/p&gt;&#xA;&lt;p&gt;If you really wanted to do an engineering analysis, you'd have to determine the static coefficient of friction between battery cases and make sure the angle of inclination of the bottom level is sufficient that every battery can overcome said friction and roll to the end.  Not really worth it :-) .&lt;/p&gt;&#xA;&lt;h2&gt;ETA&lt;/h2&gt;&#xA;&lt;p&gt;Then there's always the hoary problem of batteries randomly forming a stable &amp;quot;arch&amp;quot; in this open design (similar to sand jamming in an hourglass).  If the interior were a series of ramps, the overall design would be a lot more reliable.&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2018-12-05T16:21:03.513" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25001" PostTypeId="2" ParentId="24991" CreationDate="2018-12-05T16:32:52.360" Score="1" Body="&lt;p&gt;If the surface is nonplanar then the surface normal is a (hopefully continuous and differentiable) function of location.   You will have to integrate the surface function S(x,y,z) over any two of the dimensions.    If this is an engineering problem, then of course use a finite-grid approach and numerical integration. Then treat each grid element as a planar surface. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2018-12-05T16:32:52.360" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25003" PostTypeId="1" CreationDate="2018-12-06T00:37:19.520" Score="0" ViewCount="1985" Body="&lt;p&gt;I am trying to design a peristaltic pump, in a spreadsheet where I can alter the tube and casing diameter, pump RPM and some other parameters, I'm wondering if anyone can spot anything odd in this method as in some conditions it seems to give a reasonable answer, and others not (ie: Power delivered greater than power required). My process currently is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate volume per revolution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V = \pi^2\cdot\left(\dfrac{\phi_{tube}}{2}\right)^2\cdot\phi_{case}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate discharge per occlusion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Q = V\cdot\dfrac{2\pi\cdot RPM}{60}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate flow velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v = \dfrac{Q}{A_{pipe}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate Reynolds number (where &lt;span class=&quot;math-container&quot;&gt;$\nu$&lt;/span&gt; is the dynamic viscosity):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Re = v\cdot\dfrac{\phi_{tube}}{\nu_{water}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate flow friction coefficient using Nikuradse’s experimental equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$f = 0.0008 + \dfrac{0.05525}{Re^{0.237}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate flow frictional head loss using Darcy Weisbach equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$H_l = \dfrac{f\cdot\ell_{pipe}\cdot v^2}{2 \cdot 9.81 \cdot \phi_{tube}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate torque required (where &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; is the coefficient of friction):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;T =&amp;amp; (\mu_{roller} \cdot \text{tube re-expansion force} \cdot \phi_{i,tube}) + {}\\&amp;amp;(\mu_{roller} \cdot \text{tube re-expansion force} \cdot r_{crank\ shaft})&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate power required to drive pump:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P = T \cdot \dfrac{2 \cdot \pi \cdot RPM}{60}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate total head loss assuming both ends open to atmosphere, using Bernoulli's equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$H_t = \dfrac{v^2}{2g} + H_l$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate total power out (wher &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the density):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_t = \rho_{water} \cdot Q \cdot H_t \cdot 9.81$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seems to agree with similar projects to mine, but I get some odd results. Anyone got any ideas?&lt;/p&gt;&#xA;" OwnerUserId="18529" LastEditorUserId="1832" LastEditDate="2018-12-06T02:35:31.620" LastActivityDate="2022-06-24T13:05:45.627" Title="Am I calculating the power required and power out for a peristaltic pump correctly?" Tags="|pumps|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25004" PostTypeId="2" ParentId="25003" CreationDate="2018-12-06T03:31:53.017" Score="0" Body="&lt;p&gt;Here is a simple check. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;power is always the product of a flow variable (current, flow rate, RPM) and an effort variable (voltage, pressure, torque). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can measure the voltage and current going into the motor that runs the pump. This tells you how much input power is being dissipated in the motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply this by the motor efficiency to get the net power being delivered to the pump mechanism. Ordinary electric motors are about 85% to 90% efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equate this to the mass flow rate times the pressure developed by the pump (this is the &quot;hydraulic horsepower&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between the power output of the pump and its power input tells you the efficiency of the peristaltic pump itself. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-06T03:31:53.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25005" PostTypeId="1" CreationDate="2018-12-06T04:51:44.693" Score="1" ViewCount="144" Body="&lt;p&gt;Suppose I have stress and strain relationship and it can be simplified to the equation &lt;span class=&quot;math-container&quot;&gt;$\sigma=Function(\epsilon)$&lt;/span&gt;. This is the CONSTITUTIVE equation of the material I want to do research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that I need to do the finite element analysis and it has dynamic(nonlinear) deformation, I have the model and I have the CONSTITUTIVE equation which I mentioned the 1st paragraph.&lt;br&gt;&#xA;How can we put those things into a structural equation to do the simulation and what structural equation I need to use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we put this CONSTITUTIVE equation into the following equation &lt;span class=&quot;math-container&quot;&gt;$\nabla\cdot \sigma =Force $&lt;/span&gt;? Then we can get &lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt;, we get displacement and &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; explicitly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I think the above is the linear equation, what about nonlinear equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="18530" LastEditorUserId="18530" LastEditDate="2018-12-07T02:34:25.823" LastActivityDate="2018-12-07T02:34:25.823" Title="How to introduce the CONSTITUTIVE equation into structure mechanics" Tags="|structural-engineering|finite-element-method|computer-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2018-12-07T00:56:51.733" ContentLicense="CC BY-SA 4.0" />
  <row Id="25006" PostTypeId="1" AcceptedAnswerId="25009" CreationDate="2018-12-06T13:10:10.290" Score="4" ViewCount="7750" Body="&lt;p&gt;I have been trying to understand a particular aspect of the Chernobyl accident - the role of the graphite displacers on the end of the control rods. The basic story is that the graphite displacers caused the reactivity of the core to increase when the rods moved down, a non-intuitive effect that the operators were not expecting/sufficiently aware of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All literature I can find tells basically the same story, but I shall refer here to &lt;a href=&quot;http://www.rri.kyoto-u.ac.jp/NSRG/reports/kr79/kr79pdf/Malko1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this PDF&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On page 17 of that PDF it describes the graphite displacers and shows a schematic of a control rod in the raised position and in the &quot;initial insertion&quot; position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text indicates that:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;On moving down of absorbers into the core, their displacers displace water columns from the lower part of the core. Thus, inserting of absorbers from their extreme top position introduces a positive reactivity into the core because graphite absorbs neutrons much less than water&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The graphic shows a control rod &quot;extreme top&quot; and &quot;initial insertion&quot; positions. In the second case the graphite has moved in the channel, displacing water there and increasing reactivity in the lower part of the core. The image has a &quot;+&quot; symbol showing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, there is reduced reactivity above the displacer as a result of water replacing the graphite in that position, and reduced reactivity at the very top of the core due to the presence of the boron control rod. There are &quot;-&quot; symbols showing these effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is, if the graphite displacer was already in the water column, how could moving it downwards change the overall reactivity of the core?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously the reactivity distribution changes, but it seems to me looking at the diagram that the (+) and (-) from the graphite movement would cancel out overall, leaving the (-) from the boron rod: an overall reduction in reactivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly I'm misunderstanding something.&lt;/p&gt;&#xA;" OwnerUserId="18537" LastActivityDate="2019-07-30T13:15:05.830" Title="How did the RBMK control rod design cause an increase in reactivity when moved downwards?" Tags="|nuclear-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25007" PostTypeId="1" AcceptedAnswerId="25089" CreationDate="2018-12-06T13:48:22.280" Score="1" ViewCount="126" Body="&lt;p&gt;I have a very sensitive scale used for dynamic weighing (by dynamic I mean that the scale and load cells are actually beneath a belt conveyor that actively weight product when it´s being transported along the belt).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There´s a lot of variation in the measurements (+- 8g, while the manufacturer says it should be max +-2g).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to rule out vibration as a source of enviromental influence, so I gathered some data with a pair of 3 axis accelerometer dataloggers mounted on the structure beneath the load cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I analyzed the data in Matlab and found that the maximum vibration frequency when the equipment is being static is 1.066 Hz (EDIT 3); peaks of 1.33 g in the z axis (perpendicular to the floor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It´s being a while since I´ve done anything with vibrations and I´m a little rusty on the theory and practical experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I´m not sure what I should be looking for and I don´t have (didn´t find) anything about what the normal structural vibrations are for a lab room or specifications from the scale manufacturer on how much vibrations in can withstand or filter out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I´d appreciate any suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Scale is installed on top of a mezanine 5m tall.&#xA;The Accelerometer specifications are:&#xA;Sensor Type MEMS semiconductor&#xA;Acceleration Sampling Rate (Datalogger) 200Hz&#xA;Acceleration Range ±18g&#xA;Acceleration Resolution 0.00625g&#xA;Acceleration Accuracy ±0.5g&#xA;Bandwidth 0 to 60Hz&#xA;Sampling rate (Software) 500 ms to 24 hours&#xA;Memory 4Mbit Flash; 112028 Motion Detection samples per axis,&#xA;or 168042 Normal samples&#xA;Data Format Time stamped peak acceleration, average and peak vector sum&#xA;Dimensions 3.7 x 1.1 x 0.8&quot; (95 x 28 x 21mm)&#xA;Weight 1oz (20g)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The FFT results in the X,Y and Z axis were:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kp4AB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kp4AB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/5tLZQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5tLZQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BTEV5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BTEV5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 3: Here´s both time and frequency plots:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rOgCq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rOgCq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4088" LastEditorUserId="4088" LastEditDate="2018-12-13T15:37:16.470" LastActivityDate="2018-12-14T00:49:13.253" Title="Analizing Vibrations Influence on Dynamic Weighing" Tags="|structural-analysis|vibration|matlab|industrial-engineering|modal-analysis|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="25008" PostTypeId="1" CreationDate="2018-12-06T14:05:32.027" Score="1" ViewCount="3932" Body="&lt;p&gt;What does it mean when a pump stalls? I could not find a useful explanation with the search terms pump stalling and hydraulic pump stalling. I'm trying to understand what Elon's &lt;a href=&quot;https://twitter.com/elonmusk/status/1070386062164283392&quot; rel=&quot;nofollow noreferrer&quot;&gt;tweet&lt;/a&gt; means. Maybe an animation could help but checking the videos section with these search term also didn't help me.&lt;/p&gt;&#xA;" OwnerUserId="18539" LastEditorUserId="2559" LastEditDate="2018-12-07T20:27:25.720" LastActivityDate="2018-12-07T20:27:25.720" Title="What does a stalled hydraulic pump mean?" Tags="|pumps|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25009" PostTypeId="2" ParentId="25006" CreationDate="2018-12-06T15:36:43.500" Score="10" Body="&lt;p&gt;This diagram should explain the basic problem. Prior to the accident, the control rods were in position II - that is to say, in the normal fully withdrawn position. The actual neutron absorber was sandwiched between two graphite displacer sections - so when the SCRAM button was pushed, the graphite displacer descended into the active region of the core and caused a localised spike in reactivity in the bottom of the core - the region which was most unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit for clarity:&lt;/strong&gt; the design of the control rods at the time left a 1.25m column of water above and below the graphite displacer itself when in position III. I'm unsure of the exact state of the control rods immediately prior to the accident, but, in any case: it was very likely the lower 1.25m of the reactor which suffered the power excursion when the rods were inserted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This caused water in the lower part of the reactor to flash-boil to steam, damaging the reactor and causing the rods to become stuck, and in turn fed into a positive void coefficient, causing more water to boil, and so on and so on until a steam explosion blew the reactor apart. The RBMK is what's called a &quot;loosely coupled&quot; reactor, in part due to its size. Different parts of the core can have very different power levels, and the core was particularly dangerous at low power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ai7Yp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ai7Yp.jpg&quot; alt=&quot;Diagram of RBMK control rods&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, the control rod design was the final trigger on a sequence of events which destabilised the reactor - the design of the rods themselves was deeply unsafe and caused an increase in reactor power on insertion, and then a decrease. In Chernobyl-4's case, the rods never moved far enough into the core to stem the chain reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest, as they say, is history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphic source: &lt;a href=&quot;http://consumedland.com/page_06_en.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://consumedland.com/page_06_en.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18540" LastEditorUserId="18540" LastEditDate="2018-12-07T10:23:36.673" LastActivityDate="2018-12-07T10:23:36.673" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25010" PostTypeId="1" CreationDate="2018-12-06T15:50:26.340" Score="2" ViewCount="137" Body="&lt;p&gt;Question from a validation beginner in the pharmaceutical industry.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What are the common approaches to start writing validation protocols (esp. IQ,  OQ, PQ)?&lt;/li&gt;&#xA;&lt;li&gt;What are the common approaches to start executing validation protocols?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18520" LastEditorUserId="110" LastEditDate="2018-12-09T23:02:09.510" LastActivityDate="2018-12-10T13:20:29.070" Title="Validation protocols" Tags="|manufacturing-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2018-12-22T09:22:32.753" ContentLicense="CC BY-SA 4.0" />
  <row Id="25012" PostTypeId="2" ParentId="25008" CreationDate="2018-12-06T18:45:42.243" Score="2" Body="&lt;p&gt;A stalled hydraulic pump is one the following: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;its outlet is blocked, to allow measurement of it maximum outlet pressure under &lt;em&gt;stalled conditions&lt;/em&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a pump which has stopped delivering flow because of flow separation and cavitation on the impeller blade(s)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Elon's tweet refers to the second definition.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorDisplayName="user16" LastEditDate="2018-12-07T00:56:25.610" LastActivityDate="2018-12-07T00:56:25.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25013" PostTypeId="2" ParentId="25008" CreationDate="2018-12-06T19:08:24.733" Score="0" Body="&lt;p&gt;Hydraulics stalls of compressors may have been the cause, although the term stall has a wide range of use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jet engines  at the early times could stall by disruption of flow around compressor blades exactly in the same way as an airfoil can stall, and it starts by a small pocket of air stalling around a few blades and the air turning with the compressor blades as opposed to being pushed forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was later corrected by modification in the design of jet engines and digital engine controls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you already know the more about the case it would help, otherwise a pump stalling could mean many things, not the least of which the political angle Musk may have had.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-06T19:08:24.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25014" PostTypeId="2" ParentId="25007" CreationDate="2018-12-06T19:22:00.480" Score="0" Body="&lt;p&gt;if you can compile a correct graph of sensor reading on the coordinates of time versus G, for different intervals such as noticeable changes in vibration, lapses of lesser shaky intervals, you can do a Fourier analysis and sperate the super imposed waves to clean the reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you find out the vibration frequencies you can design a dynamic insulating suspension to help minimize the undesired vibrations.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-06T19:22:00.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25015" PostTypeId="2" ParentId="25006" CreationDate="2018-12-06T20:42:32.143" Score="-3" Body="&lt;p&gt;I read Chernousenko's book in the 90s (it's one of the references in the paper).  If I remember correctly, he was one of the leads of the cleanup effort. It's a damning indictment of Gorbachov, the communist party, the people who designed the experiments while ignoring warnings of consequences, laziness on the part of the designers and deliberate inattention to safety in order to get the reactors operational as soon as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the late 80s, I attended a presentation in Chicago given to the Ukrainian community by a nuclear engineer with extensive background in the area.  His assessment was that this type of reactor was originally designed for plutonium production with power generation being an add-on.&lt;/p&gt;&#xA;" OwnerUserId="18544" LastActivityDate="2018-12-06T20:42:32.143" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25016" PostTypeId="2" ParentId="25005" CreationDate="2018-12-06T20:52:04.203" Score="3" Body="&lt;p&gt;The balance of linear momentum is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \nabla \cdot \boldsymbol{\sigma} + \rho \mathbf{b} = \rho \mathbf{a}&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{\sigma}$&lt;/span&gt; is the Cauchy stress, &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the mass density, &lt;span class=&quot;math-container&quot;&gt;$\mathbf{b}$&lt;/span&gt; is the body force, and &lt;span class=&quot;math-container&quot;&gt;$\mathbf{a}$&lt;/span&gt; is the acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a &lt;strong&gt;nonlinear&lt;/strong&gt; stress-strain relation&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \boldsymbol{\sigma} = g(\boldsymbol{F})&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{F}$&lt;/span&gt; is the deformation gradient, we can write the linear momentum equation as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \nabla \cdot [g(\boldsymbol{F})] + \rho \mathbf{b} = \rho \mathbf{a}&#xA;$$&lt;/span&gt;&#xA;Note that various strain definitions can be derived from &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{F}$&lt;/span&gt; (see, for example, &lt;a href=&quot;https://en.wikipedia.org/wiki/Finite_strain_theory#Seth%E2%80%93Hill_family_of_generalized_strain_tensors&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Finite_strain_theory#Seth%E2%80%93Hill_family_of_generalized_strain_tensors&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deformation gradient tensor (&lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{F}$&lt;/span&gt;) is related to the displacement vector (&lt;span class=&quot;math-container&quot;&gt;$\mathbf{u}$&lt;/span&gt;) as explained in great detail in &lt;a href=&quot;https://en.wikipedia.org/wiki/Finite_strain_theory#Material_coordinates_(Lagrangian_description)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Finite_strain_theory#Material_coordinates_(Lagrangian_description)&lt;/a&gt;.  For some problems (but not all) we can write&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \nabla \cdot [g(\nabla \mathbf{u} + \boldsymbol{I})] + \rho \mathbf{b} = \rho \ddot{\mathbf{u}}&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{I}$&lt;/span&gt; is the second-order identity tensor.&#xA;This is clearly a highly nonlinear problem that can be solved for the displacement &lt;span class=&quot;math-container&quot;&gt;$\mathbf{u}$&lt;/span&gt; given the appropriate boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the answer to your question is that the balance of linear momentum holds whether the problem is linear or not. &lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2018-12-06T20:52:04.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25018" PostTypeId="2" ParentId="15040" CreationDate="2018-12-07T00:46:20.393" Score="1" Body="&lt;p&gt;The fixed end moment is the moment at the joint if it were held to not be rotated, or if it were fixed. This is why the moment is 3PL/16, because B is &quot;fixed&quot; and C is pinned. &lt;/p&gt;&#xA;" OwnerUserId="18547" LastActivityDate="2018-12-07T00:46:20.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25020" PostTypeId="1" AcceptedAnswerId="25022" CreationDate="2018-12-07T09:50:14.303" Score="0" ViewCount="48" Body="&lt;p&gt;In a grid-connected (utility-interactive) battery-based PV system, what is the purpose of the 'Isolated Subpanel' terminal of the inverter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Found this descriptive image and hence answered the question myelf.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3PDbQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3PDbQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15765" LastEditorUserId="15765" LastEditDate="2018-12-07T11:39:33.930" LastActivityDate="2018-12-07T11:39:33.930" Title="What is the purpose of the isolated subpanel terminal of a solar inverter?" Tags="|electrical-engineering|solar-energy|photovoltaics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25021" PostTypeId="1" CreationDate="2018-12-07T10:29:32.050" Score="0" ViewCount="122" Body="&lt;p&gt;I’m not sure if my procedure to simulate the lifting force &lt;span class=&quot;math-container&quot;&gt;$F_{Kab}$&lt;/span&gt; is correct. So I illustrate my way how to get the lifting force &lt;span class=&quot;math-container&quot;&gt;$F_{Kab}$&lt;/span&gt; using SolidWorks and the simulation tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you can give me a hint how to solve my issue correctly or perhaps you can also confirm my procedure. Many thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My procedure:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to VDI 2230 I have to do two variants of calculation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;applying the minimum assembly preload (see step of calculation R1)&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;in R0 I get &lt;span class=&quot;math-container&quot;&gt;$F_{Mmax}$&lt;/span&gt; using the table A7&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;table A7 you can find via a google search. Enter &quot;vdi 2230 pdf&quot; and I think then you get a pdf and look for table A7 (see annex).&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;with &lt;span class=&quot;math-container&quot;&gt;$\alpha_A$&lt;/span&gt; the tightening factor I get the &lt;span class=&quot;math-container&quot;&gt;$F_{Mmin,variant1}=F_{Mmax}/\alpha_A$&lt;/span&gt; which is applied to every bolt&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;applying the maximum reduction in preloading force (see step of calculation R4)&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;here my &lt;span class=&quot;math-container&quot;&gt;$F_{Mmin,variant2}=F_{Mmin,variant1}-F_z$&lt;/span&gt; which is also applied to every bolt&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$F_z$&lt;/span&gt; is the loss of preload as a result of embedding during operation&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the simulation all possible loads are active and are acting on the housing. The bolts are tighten in the first simulation with &lt;span class=&quot;math-container&quot;&gt;$F_{Mmin,variant1}$&lt;/span&gt; and in the second simulation with &lt;span class=&quot;math-container&quot;&gt;$F_{Mmin,variant2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perform the two simulations…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the lifting force &lt;span class=&quot;math-container&quot;&gt;$F_{Kab}$&lt;/span&gt; I right click the simulation results and pick “list result forces”. And now I choose “free body force”, klick the appropriate housing surface and klick the update-button. Now SolidWorks shows me the (total) lifting force &lt;span class=&quot;math-container&quot;&gt;$F_{Kab}$&lt;/span&gt;. This force divided by quantity of bolts gives me the &lt;span class=&quot;math-container&quot;&gt;$F_{Kab}$&lt;/span&gt; per bolt which I’m interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my further calculations according to VDI 2230 I have to take the smallest &lt;span class=&quot;math-container&quot;&gt;$F_{Kab}$&lt;/span&gt; per bolt.&lt;/p&gt;&#xA;" OwnerUserId="17780" LastEditorUserId="17780" LastEditDate="2018-12-10T07:39:48.723" LastActivityDate="2018-12-10T07:39:48.723" Title="Calculating the lifting force $F_{Kab}$ with SolidWorks – according to VDI 2230" Tags="|solidworks|bolting|threads|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25022" PostTypeId="2" ParentId="25020" CreationDate="2018-12-07T11:14:39.910" Score="1" Body="&lt;p&gt;I found out the answer. The isolated subpanel serves a load-center or distribution panel for backup loads (incase of power outage, etc).&lt;/p&gt;&#xA;" OwnerUserId="15765" LastActivityDate="2018-12-07T11:14:39.910" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25025" PostTypeId="2" ParentId="24535" CreationDate="2018-12-07T12:47:50.017" Score="0" Body="&lt;p&gt;Yes, it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine you &lt;em&gt;block&lt;/em&gt; the propeller entirely, say, fuse it to the casing, or tangle it in a huge wad of fishing net that causes huge drag (or instead, just fuse the two gears together, so they can't turn - oh, say, put a wrench between them.) In this case the motor, through the shaft and its bevel gear will try to spin the entire assembly of the propeller, its shaft, its two bearings, the lower bevel gear and the entire casing, making the whole thing turn around the motor shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ground2 is what prevents that and causes the torque to be transferred to spinning the propeller as opposed to spinning the entire assembly. It doesn't 'feel' much torque&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the gears and the propeller. The upper two bearings are negligible, as they only transfer as much torque as their internal friction. But any torque put by environment on the propeller will be applied to the gear, and transferred to ground2 through the two lower bearings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be some extra torque from friction of two upper bearings and the bevel gears, but that's just losses, as opposed to torque from the work done. I believe the torque will distribute equally between the two grounds, assuming ground 2 surrounds the casing instead of just being attached to one edge. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-12-07T12:47:50.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25026" PostTypeId="1" AcceptedAnswerId="25033" CreationDate="2018-12-08T05:16:49.007" Score="1" ViewCount="32" Body="&lt;p&gt;I am interested in discovering a method to relate air flow to temperature reduction efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose of this is in application to atmospheric water generation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The apparatus will involve a fan (with a CFM rating), a long coil that will be externally cooled, and the temperature it is cooled to with perhaps another post processing stage. That would involve the narrow diameter pipe opening into a larger diameter &quot;dew point chamber&quot; or a way to cool the air further by narrow to wider expansion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of the condensed water will be collected into a collection reservoir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question is: How can I relate the CFM (cubic feet per minute) of air flow, the diameter of the pipe, the length of the pipe, the temperature of the pipe, the diameter of the expansion chamber --&gt; the amount of water collected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not an engineer. I am a simple person with aspirations to design a viable atmospheric water generator. I do not want to discuss energy requirements. I only want to discuss the variables I have outlined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I understand the idea of relative humidity vs. water available to &quot;mine&quot; from the air via condensation. My question pertains to the apparatus itself and how the variables relate. My concern is a high air flow will only have a certain rate of cooling based upon the length, diameter, and temperature of the pipe. My goal is to develop, with someone's kind help, a sort of formula for determining the temperature of the starting air (ambient) and the output temperature based on the variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you, my fellow smart humans!&lt;/p&gt;&#xA;" OwnerUserId="18555" LastEditorUserId="18555" LastEditDate="2018-12-08T05:38:22.670" LastActivityDate="2018-12-08T23:23:04.107" Title="Variables Relating to Atmospheric Water Generation" Tags="|airflow|cooling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25029" PostTypeId="1" CreationDate="2018-12-08T16:48:57.687" Score="2" ViewCount="954" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mSwKI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mSwKI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown above, two identical rectangular beams are placed and NO pinned together. I know each beam has its own neutral surface but why they carry half of the bending moments? I thought the bending moments they carry are same because each of the beam is subjected same force P at the end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="17209" LastActivityDate="2018-12-14T03:57:17.360" Title="Why two identical beams without being pinned together will share half of the bending moments?" Tags="|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25031" PostTypeId="1" AcceptedAnswerId="25032" CreationDate="2018-12-08T17:18:13.297" Score="0" ViewCount="80" Body="&lt;p&gt;Recently I've been scratching my head over an assignment, and I have trouble understanding a couple of the parts.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fHTpU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fHTpU.png&quot; alt=&quot;Sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue I have is with the overall calculations, since we've just started on mechanical issues and I have no experience except easy transmissions. Therefor, I am wondering how I calculate the ratio of the compound gear between A and B, and how many turns the worm gear have to turn to make the C spur gear travel a specified distance of 100mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question goes as follows;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How many revolutions does the worm gear have to complete, for the&#xA;  spring to compress entirely?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The C gear is connected to a rack that then compresses a spring, that'll have to travel 100mm. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="18560" LastActivityDate="2018-12-08T19:27:33.883" Title="Calculations of compound gears?" Tags="|mechanical-engineering|transmission|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25032" PostTypeId="2" ParentId="25031" CreationDate="2018-12-08T19:27:33.883" Score="1" Body="&lt;p&gt;In order for a tooth on C to travel 100mm, the teeth on B also need to travel 100mm. Let's not worry about how many revolutions that is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A has 2.6 times as many teeth as B,and travels the same number of rotations, so, that means that a tooth on A must travel 260mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pitch of the worm is 4mm, so, it needs to rotate 260/4=65 times in order to cause the teeth if A to move 260mm, and consequently the teeth of the rack meshed with C 100mm&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-12-08T19:27:33.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25033" PostTypeId="2" ParentId="25026" CreationDate="2018-12-08T23:23:04.107" Score="0" Body="&lt;p&gt;This is a straightforward calculation for an engineer trained in the art of what is called HVAC (heating/ventilation/air conditioning) engineering but a full outlining of the solution process would require you to grasp many new physics concepts (most of which are not simple), an unfamiliar dictionary of engineering jargon, and the mastery of coupled algebraic equations. As such it is beyond the scope of what can be dealt with here. I would recommend that you partner up with an HVAC engineer to help you with this project.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-08T23:23:04.107" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25034" PostTypeId="1" AcceptedAnswerId="25041" CreationDate="2018-12-08T23:29:06.387" Score="0" ViewCount="2077" Body="&lt;p&gt;In an experiment used to find the Young's modulus of a steel bar, the formula for change in height is given as follows:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\delta_x=\frac{Fx^2}{6EI}(3L-x)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$I=\frac{bh^3}{12}$$&lt;/span&gt;&#xA;where:&#xA;    &lt;span class=&quot;math-container&quot;&gt;$δ$&lt;/span&gt;- the change in height of the steel bar under load at a distance L&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$δ_x$&lt;/span&gt;- the change in height of the steel bar under load at the point where deflection is measured&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;- the distance between the point of application of the load and the end support&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;- the distance between the point at which deflection is measured and the end support&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;- Young’s modulus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;- applied load&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt;- second moment of area of the cross- section&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;- the width of the beam&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;- the height of the beam&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt;- the weight of each value in the weighted mean&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;- the mass hung from the bar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is outlined below:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LsnEF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LsnEF.png&quot; alt=&quot;formula background&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I do not understand how they have derived this top formula. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="18450" LastActivityDate="2018-12-09T20:34:08.930" Title="derivation of Young's modulus experiment formula" Tags="|materials|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25036" PostTypeId="2" ParentId="25029" CreationDate="2018-12-09T00:52:03.250" Score="3" Body="&lt;p&gt;Let's consider the two beams as two springs which will deflect under the load &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; the amount:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ D = \frac {P*L^3}{3EI}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So K or stiffness of these two springs is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ k = \frac{P}{d} = \frac{3EI}{L^3} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two springs are working in parallel, each sharing half of the force, therefore half of the momentum (if they were to act in series they should have been attached lengthwise).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they were pinned together the combined composite &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; of the new beam would have been &lt;span class=&quot;math-container&quot;&gt;$8I$&lt;/span&gt; because the height of the beam would have doubled and we know &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; is related to &lt;span class=&quot;math-container&quot;&gt;$H^3$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or consider initially the top beam is taking all the load and deflects fully under the load, then we add the second beam right next to it and the new additional support decreases the deflection by half. Now if we slide the second beam under the first, it is the exact same situation, each beam shares half of the force.&#xA;This is similar to old car leaf springs.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-12-14T03:57:17.360" LastActivityDate="2018-12-14T03:57:17.360" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25037" PostTypeId="2" ParentId="25029" CreationDate="2018-12-09T07:57:06.643" Score="1" Body="&lt;p&gt;That they share half the force or half the momentum isn't immediately obvious; there is no physical law saying so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is through geometric constraints (under the assumption of zero vertical compression of the beams!) that we know they must have the same deflection and thus stress-state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But imagine having a big stack of short stubby rubbery beams (say pencil erasers). If you make the stack big enough the lowest beam won't feel anything. And if your two beams are of any realistic material (not Eulerium) the first beam will carry ever slightly so much more load than the second.&lt;/p&gt;&#xA;" OwnerUserId="14899" LastEditorUserId="14899" LastEditDate="2018-12-09T14:53:41.497" LastActivityDate="2018-12-09T14:53:41.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25038" PostTypeId="2" ParentId="10326" CreationDate="2018-12-09T13:00:08.647" Score="1" Body="&lt;p&gt;Take &lt;a href=&quot;https://en.wikipedia.org/wiki/Tesla_valve&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tesla airflow check valve&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vYnOP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vYnOP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 'pass' direction it's very much linear per  Hagen-Poiseuille equation. In the reverse direction for tiny &lt;span class=&quot;math-container&quot;&gt;$\Delta P$&lt;/span&gt; it's linear, but as the pressure increases, the flow increase is vanishingly low. I'm not sure what the exact relation is in the inverse direction, but it's way far from linear - and the device is entirely passive, no moving parts whatsoever.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-12-09T13:00:08.647" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25040" PostTypeId="1" AcceptedAnswerId="25042" CreationDate="2018-12-09T18:56:47.760" Score="0" ViewCount="103" Body="&lt;p&gt;Referring to this diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CZju9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CZju9.png&quot; alt=&quot;Peltier Element&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the height of the P and N components affect heat or cold generated on this thermoelectric element?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea I keep having is that it seems like the hot and cold sides are too close to each other. Would increasing the distance, via lengthening the height of the P and N components help to segregate the differential more or would this decrease the differential produced?&lt;/p&gt;&#xA;" OwnerUserId="18555" LastActivityDate="2018-12-09T20:00:44.760" Title="Peltier Element - Segregation of the Hot/Cold" Tags="|heat-transfer|cooling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25041" PostTypeId="2" ParentId="25034" CreationDate="2018-12-09T19:21:18.920" Score="2" Body="&lt;p&gt;I try to tackel the problem in two different ways, as suggested in the comments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first approach is based on Bernoulli beam theory, a set of differential equations, (if you want to know how drive them, then you can ask it here or in physics SE):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$q(x)=-\frac{dv}{dx}=\frac{d^2M}{dx^2}=-EI_{yy}\frac{d^3\alpha}{dx^3}=EI_{yy}\frac{d^4u}{dx^4}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I respect the &lt;strong&gt;right handed coordinate system&lt;/strong&gt;, the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; axis lies along the beam, &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; represent the vertical force, shear force, moment, curvature and displacement respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you know how to drive the moment, if you don't, then take a look at the second method: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M=F(L-x)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Integrating the equation twice results in: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$EI_{yy}u(x)=-F(L\frac{x^2}{2}-\frac{x^3}{6})-C_1x+C_2$$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary conditions in this case are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$EI_{yy}\alpha(x=0)=0 \rightarrow C_1=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$EI_{yy}u(x=0)=0 \rightarrow C_2=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$EI_{yy}u(x)=-F(L\frac{x^2}{2}-\frac{x^3}{6})$$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second way is a bit more lengthy but it helps you to develop your insight:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gMiWZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gMiWZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image: Mechanics of materials Prof. Wim VAN PAEPEGEM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we find the reaction forces: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$R_A+F=0 \rightarrow R_A=F$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$RM_A=-F.L$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we wish to find the shear forces and the moment, you can choose between the right hand side of the beam or the left hand side, as you can see in the photo, i find it easier to write the equilibrium equations for the right hand side (BC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$-V+F=0 \rightarrow V=-F $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M+F(L-x)=0 \rightarrow M=F(L-x)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can now integrate the moment as i did before. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-12-09T20:34:08.930" LastActivityDate="2018-12-09T20:34:08.930" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25042" PostTypeId="2" ParentId="25040" CreationDate="2018-12-09T20:00:44.760" Score="1" Body="&lt;p&gt;Here is the problem: the electrical current has to flow through the chunks of junction material. That material has electrical resistance and will therefore heat up as the current flows, negating the cooling effect. So you do not want the P &amp;amp; N chunks to be too thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if you make then too thin in response, then it's too easy for heat to flow across the chunks from the hot side to the cold side, short-circuiting the cooling effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means two things: 1) the thickness of the chunks is a compromise between these two competing influences, and 2) the holy grail of this business is a material that simultaneously possesses high electrical conductivity and low thermal conductivity, which  are to a great extent contradictory properties. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-09T20:00:44.760" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25045" PostTypeId="1" CreationDate="2018-12-09T23:55:18.147" Score="1" ViewCount="76" Body="&lt;p&gt;I'm building a motorized standing desk and am looking into what to use for the linear actuators. I'm targeting a stroke of 450mm and a lifting force of 100kg. I found some &lt;a href=&quot;https://www.ebay.com/itm/2-18-Inch-Stroke-Linear-Actuator-1500N-330lbs-Pound-Max-Lift-12V-Volt-DC-Motor/112887896341?hash=item1a48a49915:m:mdfXAjchkXMgJEmUIpXpOCQ:rk:3:pf:0&quot; rel=&quot;nofollow noreferrer&quot;&gt;Linear Actuator&lt;/a&gt; on ebay that would work but I was wondering if I could go cheaper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I'm using this &lt;a href=&quot;https://www.daycounter.com/Calculators/Lead-Screw-Force-Torque-Calculator.phtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;calculator&lt;/a&gt; right then a for lead screw with 2mm pitch and 8mm diameter to lift 100kg I need a torque of ~114 N*cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this &lt;a href=&quot;https://www.aliexpress.com/item/RepRap-3D-Printer-THSL-300-8D-Lead-Screw-Dia-8MM-Thread-8mm-Length-300mm-with-Copper/32435304660.html?spm=2114.search0104.3.9.359b61c3w8Ukqd&amp;amp;ws_ab_test=searchweb0_0,searchweb201602_5_10065_10068_10130_10890_5730315_10547_319_10546_317_10548_10545_10696_453_10084_454_10083_10618_5729215_10307_538_537_536_10059_10884_10887_100031_321_322_10103,searchweb201603_51,ppcSwitch_0&amp;amp;algo_expid=db65bc82-a987-45fb-84de-8a7e6f34e441-1&amp;amp;algo_pvid=db65bc82-a987-45fb-84de-8a7e6f34e441&quot; rel=&quot;nofollow noreferrer&quot;&gt;lead screw&lt;/a&gt; and a &lt;a href=&quot;https://www.aliexpress.com/item/NEMA23-stepper-motor-57X56mm-3-0A1-8-Degree-stepping-motor-175Oz-in-Nema-23-CNC-for/32951270990.html?spm=2114.search0104.3.7.e1ac68c00GHjot&amp;amp;ws_ab_test=searchweb0_0,searchweb201602_5_10065_10068_10130_10890_5730315_10547_319_10546_317_10548_10545_10696_453_10084_454_10083_5729211_10618_10307_538_537_536_10059_10884_10887_100031_321_322_10103,searchweb201603_51,ppcSwitch_0&amp;amp;algo_expid=7ca8b519-e947-4831-bdfb-a8e78c4859f1-4&amp;amp;algo_pvid=7ca8b519-e947-4831-bdfb-a8e78c4859f1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nema 23&lt;/a&gt; should work. Together that's half the price of the linear actuator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that right? I don't have much experience and could easily be missing something obvious.&lt;/p&gt;&#xA;" OwnerUserId="18571" LastEditorUserId="18571" LastEditDate="2018-12-10T18:06:59.113" LastActivityDate="2018-12-10T18:06:59.113" Title="Linear actuator calculations" Tags="|actuator|linear-motion|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25046" PostTypeId="1" AcceptedAnswerId="25053" CreationDate="2018-12-10T03:36:19.297" Score="-2" ViewCount="49" Body="&lt;p&gt;&lt;strong&gt;Answer briefly how the object-oriented methodology transforms an information system.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I really do not know how to answer that question. I know what is an information system (&quot;System capable of collecting, storing, processing and distributing information in time and form for decision making within an organization&quot;) and also what is object-oriented methodology (&quot;discipline related to programming&quot;) but I do not know how the latter models the first one.&lt;/p&gt;&#xA;" OwnerUserId="18572" LastEditorDisplayName="user16" LastEditDate="2018-12-10T12:57:49.293" LastActivityDate="2018-12-10T12:57:49.293" Title="In what way the object-oriented methodology model an information system?" Tags="|systems-engineering|" AnswerCount="1" CommentCount="8" ClosedDate="2018-12-10T12:58:25.733" ContentLicense="CC BY-SA 4.0" />
  <row Id="25047" PostTypeId="1" AcceptedAnswerId="25054" CreationDate="2018-12-10T05:20:43.253" Score="1" ViewCount="106" Body="&lt;p&gt;This is from Stephen Hawking's latest book; &lt;em&gt;Brief Answers to the Big Questions&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The speed at which we can send a rocket is governed by two things, the speed of the exhaust and the fraction of its mass that the rocket loses as it accelerates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="18574" LastEditorUserId="18574" LastEditDate="2019-01-20T12:32:23.710" LastActivityDate="2019-01-20T12:32:23.710" Title="the speed of the exhaust" Tags="|aerospace-engineering|rocketry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25048" PostTypeId="1" AcceptedAnswerId="25049" CreationDate="2018-12-10T05:37:21.947" Score="3" ViewCount="135" Body="&lt;p&gt;Are solar panels placed horizontally at the equators since the solar altitude (solar window) changes direction from north to south?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qq7AU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qq7AU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15765" LastActivityDate="2018-12-10T05:44:02.477" Title="How are solar panels at equator placed?" Tags="|solar-energy|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25049" PostTypeId="2" ParentId="25048" CreationDate="2018-12-10T05:44:02.477" Score="5" Body="&lt;p&gt;Basically, yes, but any shading caused by buildings / other obstructions may make one direction preferential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find sites that will do simulations online...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-12-10T05:44:02.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25050" PostTypeId="2" ParentId="25047" CreationDate="2018-12-10T06:16:42.930" Score="2" Body="&lt;p&gt;&quot;the fraction of its mass that the rocket loses as it accelerates&quot; means that assuming the power delivered is constant, the total mass of the rocket decreases as the fuel is spent, therefore altering the thrust (power) to mass ratio.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2018-12-11T09:52:07.293" LastActivityDate="2018-12-11T09:52:07.293" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25051" PostTypeId="1" AcceptedAnswerId="25052" CreationDate="2018-12-10T08:09:08.770" Score="0" ViewCount="285" Body="&lt;p&gt;What is the solar altitude at Bangalore at solar noon during summer solstice, March equinox, Winter solstice and September equinox? In other words, What is the solar window (solar altitude angle range) of Bangalore?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: Consider latitude of Bangalore to be 11 deg, North.&lt;/p&gt;&#xA;" OwnerUserId="15765" LastActivityDate="2018-12-10T10:40:54.263" Title="What is the solar window (solar altitude angle range) of Bangalore?" Tags="|solar-energy|" AnswerCount="1" CommentCount="2" ClosedDate="2018-12-11T13:52:36.043" ContentLicense="CC BY-SA 4.0" />
  <row Id="25052" PostTypeId="2" ParentId="25051" CreationDate="2018-12-10T10:40:54.263" Score="2" Body="&lt;p&gt;Earth axial tilt is 23.5°. For 11°N the solar altitude at noon of the equinoxes will be 90-11=79°; 90-11-23.5=55.5° at winter solstice and - it would be 90-11+23.5=102.5° at the summer solstice, but we normalize it to angle between nearest horizon and the Sun, so for values &gt;90 we take 180-(value): 180-102.5=77.5° at the summer solstice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual solar window though will be the range between minimum and maximum altitude: 55.5°-90° as the Sun passes through Zenith sometime in the late spring and early summer.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-12-10T10:40:54.263" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25053" PostTypeId="2" ParentId="25046" CreationDate="2018-12-10T11:02:23.440" Score="1" Body="&lt;p&gt;Object-oriented methodology allows representing the system as a hierarchical structure of self-contained subsystems and entities comprising it. This allows separate development and testing of the subsystems apart from each other, good code reuse (through inheritance - similar objects; and through use of object libraries - use of pre-made objects by 3rd parties) and abstracting internals of subsystems apart from the systems using them, creating consistent interfaces between them, easy to extend, debug and modify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an easy analogy: You build the big system out of big bricks, each of which you build of smaller bricks, and these - of yet smaller, each with own quirks and special purposes, some being just special variants of others, others being pre-packaged ready-to-use bigger bricks. They all need to fit just right, but if you make a small brick of the tiny bricks and it fits just right with the bigger brick, you can just reuse it everywhere it's needed (and if it's faulty, you can pinpoint the fault and fix it easily.) If you started building the entire system out of the tiniest bricks from start, you'd end up with a mess where every segment, no matter how common, needs to be made from scratch all over again, and if something's broken, you don't follow the simple chain big brick; smaller brick within it, even smaller within, the bad tiny one - you'd need to track all the hundreds of tiny bricks from top to the faulty one through that mess.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2018-12-10T11:09:50.130" LastActivityDate="2018-12-10T11:09:50.130" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25054" PostTypeId="2" ParentId="25047" CreationDate="2018-12-10T11:33:20.973" Score="5" Body="&lt;p&gt;He refers to &lt;a href=&quot;https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation&quot; rel=&quot;noreferrer&quot;&gt;Tsiolkovsky's Rocket Equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta v=v_e \ln {\frac {m_0}{m_f}} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$v_e$&lt;/span&gt; is the exhaust velocity;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;${\frac {m_0}{m_f}}$&lt;/span&gt; is the fraction of mass; &lt;span class=&quot;math-container&quot;&gt;$m_f$&lt;/span&gt; - the &quot;dry mass&quot;/&quot;final mass&quot; (rocket without fuel) and &lt;span class=&quot;math-container&quot;&gt;$m_0$&lt;/span&gt; - &quot;wet mass&quot;/&quot;launch mass&quot; (rocket fully fueled up.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is one of the most important in rocket science - describing the change of velocity a rocket can achieve. The implications are that the larger the difference between the mass of fuel and the mass of the craft, the larger the velocity achievable (but the &lt;span class=&quot;math-container&quot;&gt;$\ln$&lt;/span&gt; results in diminishing returns as mass of fuel is increased) and that engines that impart the exhaust with most velocity provide most performance - linearly, without that pesky &lt;span class=&quot;math-container&quot;&gt;$\ln$&lt;/span&gt; - but then... with square root of energy needed; &lt;span class=&quot;math-container&quot;&gt;$E={1\over 2} mv^2; v = \sqrt{2E \over m}$&lt;/span&gt;. So, increasing power of the engine - chemical energy of fuel, amount of electrical energy imparted by ion drive - results in diminishing returns again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of Hawking's last ideas - &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Breakthrough_Starshot&quot; rel=&quot;noreferrer&quot;&gt;Breakthrough Starshot&lt;/a&gt;&quot; nicely sidesteps both problems. The propellant is photons, moving at speed of light, and the craft doesn't carry any fuel - the propellant is beamed from a ground-based station through a powerful laser. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2018-12-10T11:33:20.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25056" PostTypeId="2" ParentId="25010" CreationDate="2018-12-10T13:20:29.070" Score="0" Body="&lt;p&gt;Commonly validation protocols are derived from Requirements documents. Therefore review the requirements document to create the validation protocol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Installation Qualification (IQ)&lt;/strong&gt;:  Procedure to evaluate the means of the equipment and the testing of it material. Below is an example of IQ document. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MfqWC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MfqWC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Operational Qualification (OQ)&lt;/strong&gt;: Procedure to validate the equipment or product parameters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BWgxh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BWgxh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Performance Qualification (PQ)&lt;/strong&gt;: This processes mostly like a stress test, that will validate the boundary conditions. For example if the Motor is specified to run at a minimum speed of 25 RPM and a maximum speed of 150 RPM, this process will test and document the results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Em3ou.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Em3ou.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometime some of these steps are combined, such as OQ and PQ. &lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2018-12-10T13:20:29.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25057" PostTypeId="1" CreationDate="2018-12-10T15:40:35.540" Score="0" ViewCount="132" Body="&lt;p&gt;Can somebody please explain what relief amount at the end of cut in a grooving operation on lathe is?&lt;/p&gt;&#xA;" OwnerUserId="16605" LastActivityDate="2018-12-10T17:37:57.190" Title="What is relief amount in grooving operation" Tags="|mechanical-engineering|machining|cnc|lathe|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25059" PostTypeId="2" ParentId="25057" CreationDate="2018-12-10T17:37:57.190" Score="1" Body="&lt;p&gt;A groove will have a very slight internal radius in its corner, but if this is undesirable, then a relief grooving tool is used to remove this material. The relief grooving tool comes in at 45 Deg to cut as shown: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/20RLq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-12-10T17:37:57.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25062" PostTypeId="1" CreationDate="2018-12-11T00:03:45.017" Score="0" ViewCount="54" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LDRXG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LDRXG.png&quot; alt=&quot;addition of shearing forces and stresses&quot;&gt;&lt;/a&gt;&#xA;I would like to ask for help regarding the summation of non parallel shearing forces acting on an elastic body.&#xA;In the image force F (Left to right direction,in yellow) acts on the top horizontal surface &quot;A&quot; of an elastic cuboid body that is attached to the table only at its bottom surface.  I know how to calculate the stress and strain. But what happen if a second force &quot;F_vertical&quot; (in black) acts on the top portion of the vertical surface facing the reader, again direction left to right. how do I add the stress and strain imposed by both forces.&#xA;(I am a physician not an an engineer or physicist ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thank you&#xA;Yehuda Raveh &lt;/p&gt;&#xA;" OwnerUserId="18470" LastActivityDate="2018-12-11T03:28:30.283" Title="addition of non paralle shearing forces and stresses" Tags="|stresses|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25065" PostTypeId="2" ParentId="25062" CreationDate="2018-12-11T03:28:30.283" Score="1" Body="&lt;p&gt;There's some ambiguity in your diagram and description because you have one force acting on a point and another force acting over a portion of a surface. Also, although you say, &quot;I know how to calculate the stress and strain [for the first loading configuration],&quot; it's not clear which idealizations you're applying to get these answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, one possible idealization is that the force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; acts over the entire top surface, thus producing a state of pure shear throughout the body. (An equal and opposite force would act over the bottom surface due to its attachment to the table.) Assuming that the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-axis points in the upward vertical direction, the original state of stress would be&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\sigma =&#xA;\begin{pmatrix}&#xA;0 &amp;amp; 0 &amp;amp; F/A_\mathrm{top} \\&#xA;0 &amp;amp; 0 &amp;amp; 0 \\&#xA;F/A_\mathrm{top} &amp;amp; 0 &amp;amp; 0&#xA;\end{pmatrix},$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$A_\mathrm{top}$&lt;/span&gt; is the cross-sectional area of the top surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you don't apply this idealization of &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; acting across the entire top, then you'll get a stress variation around the point where &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; is drawn as acting. Think about pushing at a single location on a very compliant material such as a sponge; most likely, you'll see a lot more deformation around the point your finger touches than elsewhere.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the material, any small element would be subjected to the following stress state:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/luUz0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/luUz0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These four forces are all necessary to apply a state of pure shear upon an internal element; any fewer would result in translation and/or rotation. This stress state becomes poorly defined at the surface because you're not applying any upward or downward forces; the material is effectively generating these forces internally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that's just the simple case of a single force! The revised state of stress is more complex. Just slightly inside the material where the force &lt;span class=&quot;math-container&quot;&gt;$F_\mathrm{vertical}$&lt;/span&gt; is applied, the revised stress state is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma^\prime =&#xA;\begin{pmatrix}&#xA;0 &amp;amp; F_\mathrm{vertical}/A_\mathrm{side} &amp;amp; F/A_\mathrm{top} \\&#xA;F_\mathrm{vertical}/A_\mathrm{side} &amp;amp; 0 &amp;amp; 0 \\&#xA;F/A_\mathrm{top} &amp;amp; 0 &amp;amp; 0&#xA;\end{pmatrix},$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$A_\mathrm{side}$&lt;/span&gt; includes only the area upon which &lt;span class=&quot;math-container&quot;&gt;$F_\mathrm{vertical}$&lt;/span&gt; acts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elsewhere in the block, however, the stress will vary. The force on the bottom, which originally needed to resist only &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;, now must resist a torque generated from the off-center application of &lt;span class=&quot;math-container&quot;&gt;$F_\mathrm{vertical}$&lt;/span&gt;. I don't know of a solution for the general state of stress; you may need to use finite element analysis for that. &lt;/p&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2018-12-11T03:28:30.283" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25066" PostTypeId="1" CreationDate="2018-12-11T04:05:30.393" Score="-1" ViewCount="383" Body="&lt;p&gt;It is will known that increasing cooling of a metal casting will lead to finer micro-structure which mean higher mechanical properties.&#xA; If I have 2 molds one is thick copper die ,&amp;amp; second is just thin-walled copper cup or mold cooled by water. which one of them excepted to cool the molten metal i.e. aluminium faster ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the calculated heat diffusivity of copper is : thermal conductivity x specific heat x density = 385 x 0.385 x 8960 = 1328096 m²/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;while for water it is just : 0.6 x 4.2 x 1000 = 2520 m²/s only !!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, the copper can transfere heat 527 times faster than water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by calculations it seems that copper die have higher cooling power as it has higher thermal diffusivity , but in real life it seems water have more cooling power. I am so confused.&lt;/p&gt;&#xA;" OwnerUserId="18585" LastActivityDate="2019-02-04T00:38:36.177" Title="which type of casting will give higher cooling rate, casting in thin water-cooled copper mold or casting in thick copper die?" Tags="|materials|thermodynamics|manufacturing-engineering|metallurgy|casting|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25067" PostTypeId="1" CreationDate="2018-12-11T06:21:06.673" Score="0" ViewCount="38" Body="&lt;p&gt;I am designing a savonius type wind pump for low wind speeds(avg speed = 2.7&amp;nbsp;m/s), with these speeds and rotor area of about 2&amp;nbsp;m&lt;sup&gt;2&lt;/sup&gt;. Acc. to my calculations, I am getting a volume flow rate of 50&amp;nbsp;ml/s at 10&amp;nbsp;m lift head, for this flow rate, is it feasible to build a piston pump of diameter 1.5&amp;nbsp;inch and stroke of about 4&amp;nbsp;cm. Is there a minimum limit on stroke length of a piston pump which is feasible?&lt;/p&gt;&#xA;" OwnerUserId="18589" LastEditorUserId="18589" LastEditDate="2018-12-12T19:49:47.153" LastActivityDate="2018-12-12T19:49:47.153" Title="Designing a piston pump for windpump" Tags="|mechanical-engineering|structural-engineering|fluid-mechanics|design|renewable-energy|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25070" PostTypeId="1" AcceptedAnswerId="25072" CreationDate="2018-12-11T18:23:06.213" Score="0" ViewCount="168" Body="&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$\omega_0 \in \mathbb{R}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\omega_0 &amp;gt; 0$&lt;/span&gt;. Be &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt; a transfer function defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$G(s)=\frac{1-\frac{s}{\omega_0}}{1+\frac{s}{\omega_0}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're interested into evaluating its phase response in a continuous LTI system. First, we split &lt;span class=&quot;math-container&quot;&gt;$G(j\omega)$&lt;/span&gt; into its real and imaginary parts (assuming &lt;span class=&quot;math-container&quot;&gt;$\tau=\frac{\omega}{\omega_0}$&lt;/span&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$G(j\omega) = \frac{1}{1+\tau^2}\left[1-\tau^2+j\left(-2\tau\right)\right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then using the definition of &lt;em&gt;phase&lt;/em&gt; of a transfer function we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\angle G(j\omega)=\arctan\left(\frac{\mathbb{Im}\{G(j\omega)\}}&#xA;{\mathbb{Re}\{G(j\omega)\}}\right)=\arctan\left(\frac{-2\tau}{1-\tau^2}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for any &lt;span class=&quot;math-container&quot;&gt;$\tau \gg 1$&lt;/span&gt; (which implies that &lt;span class=&quot;math-container&quot;&gt;$\omega \gg 1$&lt;/span&gt;) then the following should hold for Taylor's series:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\angle G(j\omega) \simeq \arctan\left(\frac{2}{\tau}\right) \simeq \frac{2}{\tau}$$&lt;/span&gt;&#xA;which approaches &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; gets bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here comes the discrepancy. Let &lt;span class=&quot;math-container&quot;&gt;$\omega_0 = 1$&lt;/span&gt; (for plotting reasons), why does &lt;a href=&quot;https://www.wolframalpha.com/input/?i=bode%20(1-s)%2F(1%2Bs)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wolfram Alpha&lt;/a&gt; give me the following Bode plot?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Mxxzu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mxxzu.png&quot; alt=&quot;Wolfram Alpha Bode plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I expected one discontinuity point at &lt;span class=&quot;math-container&quot;&gt;$\omega = \omega_0 = 1$&lt;/span&gt; (and another one for &lt;span class=&quot;math-container&quot;&gt;$\omega = -\omega_0$&lt;/span&gt; not shown in the plot), but I guess I messed up something since the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2018-12-12&lt;/strong&gt; (in response to Sam F.) Here's how I've separated the real from the imaginary part of &lt;span class=&quot;math-container&quot;&gt;$G(j\omega)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;G(j\omega)&#xA;&amp;amp;=\frac{1-j\frac{\omega}{\omega_0}}{1+j\frac{\omega}{\omega_0}} \\&#xA;&amp;amp;=\frac{1-j\tau}{1+j\tau} \\&#xA;&amp;amp;=\frac{1-j\tau}{1+j\tau}\cdot\frac{1-j\tau}{1-j\tau} \\&#xA;&amp;amp;=\frac{(1-j\tau)^2}{1^2-(j\tau)^2} \\&#xA;&amp;amp;=\frac{1^2+(j\tau)^2-2\cdot 1 \cdot j\tau }{1+\tau^2} \\&#xA;&amp;amp;=\frac{1- \tau^2+j(-2\tau)}{1+\tau^2} \\&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="16932" LastEditorUserId="16932" LastEditDate="2018-12-12T10:16:17.003" LastActivityDate="2019-01-11T11:00:24.317" Title="Phase response of a continuous transfer function" Tags="|transfer-function|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25072" PostTypeId="2" ParentId="25070" CreationDate="2018-12-11T20:54:57.657" Score="0" Body="&lt;p&gt;The plots are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how we do it on our heads in about a minute after years of practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the numerator and denominator separately. Numerator rise with a slope of &lt;span class=&quot;math-container&quot;&gt;$20$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$dB/dec$&lt;/span&gt;, but at the same time denominator falls with the exact same slope, now it's clear that these two tend to cancel each other out, the results is a constant number. As you know the &lt;span class=&quot;math-container&quot;&gt;$20log1 = 0$&lt;/span&gt; so it explains the magnitude plot. Notice if you look at the magnitude of the transfer function you'll see no real pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The phase &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$arctan(*)$&lt;/span&gt; is a continue function, at the break frequency it is equal to &lt;span class=&quot;math-container&quot;&gt;$-\frac{\pi}{2}$&lt;/span&gt;, as we plot the phase in logarithmic scale we have to add the phases. Both denominator and numerator drop &lt;span class=&quot;math-container&quot;&gt;$-\frac{\pi}{2}$&lt;/span&gt; over the whole frequency band. So &lt;span class=&quot;math-container&quot;&gt;$-\pi$&lt;/span&gt; or 180°.   &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2018-12-12T10:08:18.373" LastActivityDate="2018-12-12T10:08:18.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25074" PostTypeId="1" CreationDate="2018-12-12T04:27:45.387" Score="-1" ViewCount="174" Body="&lt;p&gt;For example, initially this reactor is having a 2nd order reaction of &lt;span class=&quot;math-container&quot;&gt;$-ra=kaCa^2$&lt;/span&gt; and it has a fractional conversion of 50% in a CSTR. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I replace it with something 1/6 as large. What happens to the conversion assuming everything else stays the same&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through theory, I know the conversion will decrease as its a smaller CSTR, but is there a way to calculate it using the formula - &lt;span class=&quot;math-container&quot;&gt;$fa = Fao - Fa1 / Fao$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="18600" LastEditorUserId="1832" LastEditDate="2018-12-12T15:18:09.070" LastActivityDate="2019-01-13T16:02:05.417" Title="CSTR properties - What happens to conversion of products if I change the reactor size?" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25075" PostTypeId="1" AcceptedAnswerId="25084" CreationDate="2018-12-12T09:39:52.277" Score="0" ViewCount="182" Body="&lt;p&gt;What is the degree of cooling caused by air flowing from smaller diameter into a large diameter? How much does air cool as it expands?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YhlXF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YhlXF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the air inlet was the biggest hole on the right and the outlets were the full smaller holes, would the air cool as it left each hole? Let's say that the surface area of the inlet equals the outlets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could these types of apparatus be linked in series to produce a staged cooling effect? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does cooling only occur in change of pressure?&lt;/p&gt;&#xA;" OwnerUserId="18555" LastActivityDate="2018-12-13T02:57:25.423" Title="Air Cools As It Expands" Tags="|heat-transfer|airflow|cooling|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25078" PostTypeId="1" CreationDate="2018-12-12T19:03:41.630" Score="0" ViewCount="63" Body="&lt;p&gt;I'm an EE, but I'm working on a project at work where I need to venture into materials science &amp;amp; chemistry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, does thermoplastic, Polybutylene terephthalate (PBT), fall into any of these material categories?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/paEnw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/paEnw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18607" LastActivityDate="2022-08-24T08:04:15.363" Title="Generic category of PBT thermoplastic" Tags="|materials|plastic|chemistry|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25079" PostTypeId="1" AcceptedAnswerId="25082" CreationDate="2018-12-12T20:35:34.760" Score="2" ViewCount="154" Body="&lt;p&gt;Referring to the pasted in ductwork diagram, if an inline ventilation fan with a high CFM airflow was used to circulate air within a sealed ductwork system, would the fast moving airflow result in a significant pressure drop within the ductwork and cause a section of it to implode?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or since it is a sealed system and air can neither leave nor enter it, would the result be that the fan would essentially just be circulating pressurized air around the ductwork and there would thus be no pressure drop within the ductwork? Referring to the ductwork diagram again, would static and dynamic pressure (in PSI) at points A, B, C, and D be approximately the same if measured first with the fan turned off and then measured again with the fan turned on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this appears to be a homework question, it is not. I am contemplating running ductwork from my house to my garage and I want to make sure that what I construct will not implode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1CJFG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1CJFG.png&quot; alt=&quot;Ductwork diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user18610" LastEditorUserId="33" LastEditDate="2018-12-13T02:44:54.700" LastActivityDate="2018-12-13T02:44:54.700" Title="Will circulating air within a sealed ductwork system cause it to implode?" Tags="|fluid-mechanics|airflow|heating-systems|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25080" PostTypeId="2" ParentId="25079" CreationDate="2018-12-12T22:42:47.697" Score="2" Body="&lt;p&gt;Would the pressures be the same with the fan On/Off? &lt;strong&gt;No.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure in front of the fan would be higher than that behind it - that is to say, the pressure would be highest at A, and then drop gradually around the loop B, D, C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it implode? &lt;strong&gt;No.&lt;/strong&gt; - The pressure at C &lt;em&gt;would&lt;/em&gt; be lower with the fan on than off, but in order to implode, the pressure would have to drop to below atmospheric pressure, so unless the system is already at a very low pressure, and on the point of failure, this would not push it over the edge. This is not a concern for ductwork between house/garage!&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-12-12T22:42:47.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25081" PostTypeId="2" ParentId="25078" CreationDate="2018-12-13T02:22:46.203" Score="1" Body="&lt;p&gt;No . A more common polymer is polyethylene terepthalate  ( PET) from which many bottles are made . I guess they would be the same group if it has a name. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-12-13T02:22:46.203" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25082" PostTypeId="2" ParentId="25079" CreationDate="2018-12-13T02:24:08.297" Score="2" Body="&lt;p&gt;Closed or not, a fan will always raise the pressure in front of it, and lower the pressure behind it.  In order for the ducting to implode you would need a relatively strong vacuum.  What is needed would be based on the rigidity of the ducting.  However, a couple points that are important to remember:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Its a closed system, so mass is conserved.  Regardless of where the air density is at any given point, the air is available.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;While a fan will slightly raise the pressure on the outlet side, even a powerful fan, is not a compressor.  It may move the air but it will not compress the air significantly, so the &quot;make up air&quot; will be readily available on the intake side.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Natural convection dictates that an increase pressure gradient will increase air flow.  Even if there was compression, in a closed system, the rate of compression would be compensated by the rate of airflow.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16931" LastActivityDate="2018-12-13T02:24:08.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25084" PostTypeId="2" ParentId="25075" CreationDate="2018-12-13T02:57:25.423" Score="0" Body="&lt;p&gt;The principal you are referring to is a fairly fundamental law.  The Charles law.  V/T = k. This assumes that there is no compression, and constant ambient temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer your question without measurements is difficult, however it would appear from inspection that the unit surface area of the output is greater than the unit surface area of the input, so I would imagine temperature would drop proportionately. &lt;/p&gt;&#xA;" OwnerUserId="16931" LastActivityDate="2018-12-13T02:57:25.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25085" PostTypeId="1" CreationDate="2018-12-13T09:21:39.467" Score="2" ViewCount="855" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UiB55.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UiB55.jpg&quot; alt=&quot;https://mega.nz/#!ddxACSoK!fVdumMALh_Mq-zWNwBArQ5xdTWsa2Wg8OEPrq8YL8m0&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't understand why &lt;span class=&quot;math-container&quot;&gt;$Z_{th}$&lt;/span&gt; doesn't include &lt;span class=&quot;math-container&quot;&gt;$3\ \Omega$&lt;/span&gt; resistance.&lt;/p&gt;&#xA;" OwnerUserId="18614" LastEditorUserId="1832" LastEditDate="2018-12-14T01:27:32.430" LastActivityDate="2019-09-10T03:01:43.843" Title="Thevenin equivalent impedance in laplace domain" Tags="|electrical-engineering|circuits|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25086" PostTypeId="2" ParentId="25085" CreationDate="2018-12-13T20:32:56.617" Score="1" Body="&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\frac{4}{s}$&lt;/span&gt; is a current source so it's open when you calculate the Thévenin equivalent, so no current flows thorough the 3 Ohm resistor however &lt;span class=&quot;math-container&quot;&gt;$\frac{12}{s}$&lt;/span&gt; is a voltage source so it's a short circuit while you calculate the equivalent impedance.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-12-13T20:32:56.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25087" PostTypeId="1" CreationDate="2018-12-13T23:45:17.283" Score="1" ViewCount="105" Body="&lt;p&gt;Vanadium is present in all high speed steel grades. Unfortunately vanadium is not only very expensive but is also a well known cytotoxic metal. Some workers get a severe persistent cough lasting for months for from contact with dust or vapors containing vanadium which are present in a workshop!&lt;/p&gt;&#xA;&lt;p&gt;I searched a lot but can not find any high speed steels without vanadium,  except some research grades which replaced vanadium by niobium but unfortunately niobium forms very coarse carbides which are very detrimental to steel toughness.&lt;/p&gt;&#xA;&lt;p&gt;My question is do vanadium-free high speed steels or any temper-resisting steels with similar hardness exist?&lt;/p&gt;&#xA;" OwnerUserId="18586" LastEditorUserId="90" LastEditDate="2023-11-17T19:11:59.420" LastActivityDate="2023-12-17T20:02:25.247" Title="Are there is vanadium-free high speed steels?" Tags="|mechanical-engineering|structural-engineering|materials|steel|metallurgy|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25089" PostTypeId="2" ParentId="25007" CreationDate="2018-12-14T00:49:13.253" Score="2" Body="&lt;p&gt;So, look at the time history that you added in edit 3.  Notice how it is never ever ever below 1.  That's very telling.  Let us assume that you have a MEMS accelerometer is able to read down to 0 Hertz.  So in the vertical direction, we expect to see 1g due to gravity PLUS the vibration.  In other words, we expect to see &lt;span class=&quot;math-container&quot;&gt;$y(t) = 1 + A cos(\omega t)$&lt;/span&gt;, where A is the amplitude of vibration and &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; is the frequency. Now A may be a function of time, and there might be multiple frequencies, but this is the general idea.  So, we expect to see something that is symmetric about 1.  Sometimes greater than 1 and sometimes less than 1.  E.g. if A = 0.3g, then you'd expect to see between 0.7 and 1.3g. Given that signal, we could do a fourier transform or other analysis to determine the values of A and &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that is NOT what we see here. The value is always positive.  So, based on this, and the fact that this is billed as &lt;em&gt;datalogger&lt;/em&gt;, and not just a raw accelerometer, I very strongly suspect that what you are getting is not actually the variable y(t).  Instead, I think it is an already processed value.  i.e. the datalogger is probably sampling a block of time, perhaps 1 second, perhaps 100 ms, could be anything, and then doing some kind of processing, perhaps RMS over that time period.  i.e. you aren't getting y(t), but instead something more akin to &lt;span class=&quot;math-container&quot;&gt;$\sqrt{\frac{1}{N} \sum y_i ^2} $&lt;/span&gt;.  Because this is an already processed value, taking a fourier transform of it does not give any meaningful result.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you work out the math, you'll see that if we take that &lt;span class=&quot;math-container&quot;&gt;$1 + A cos(\omega t)$&lt;/span&gt;, the RMS value will never be less than 1, which is consistent with the data you posted.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put another way, when you datalogger spits out a number like &quot;1.3g&quot; at a given time, it is not telling you that the instaneous value of acceleration is 1.3g.  It is saying that the average value, over a range of frequencies, and over a certain time range, is 1.3g.  There may be various settings to your datalogger, and it might be doing a different type of processing, perhaps it's a peak over a frequency range, or something else.  But I don't think you are getting raw acceleration.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2018-12-14T00:49:13.253" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25090" PostTypeId="2" ParentId="25087" CreationDate="2018-12-14T01:44:52.277" Score="0" Body="&lt;p&gt;No. Surprisingly ( to me) it is in every high speed steel. It forms a very hard ,very stable carbide , VC. This carbide does not even dissolve during the normal very high temperature austenize heat.  So people must actually allergic to VC , as there is no available V. It has been used in essentially all HSS since 1910 . Makes me wonder why this problem was unknown until now . ( ASM Tool Steel handbook , 1962.)&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2018-12-14T01:44:52.277" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25091" PostTypeId="1" CreationDate="2018-12-14T05:25:01.863" Score="2" ViewCount="89" Body="&lt;p&gt;In one of my application, there is a requirement to dispense the concrete paste at the required rate continuously. some technical details of the application given below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W1fmE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W1fmE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Dispense rate 5-50 LPM&lt;/li&gt;&#xA;&lt;li&gt;Dispense should be continuous&lt;/li&gt;&#xA;&lt;li&gt;The material is concrete paste&lt;/li&gt;&#xA;&lt;li&gt;Solid content is up to 4 mm&lt;/li&gt;&#xA;&lt;li&gt;delivery length is 5-6 meters&lt;/li&gt;&#xA;&lt;li&gt;bends - 3&lt;/li&gt;&#xA;&lt;li&gt;Hose dia 38 mm&lt;/li&gt;&#xA;&lt;li&gt;density 1.9 to 2.2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Currently, I am using a screw pump, which can pump up to the full length of 5-6m. But with some issues. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The flow often stops, for which we had to reverse the motor few rotations and then run forward. But this operation is not acceptable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The materials flown out of the pump was heating up by 8-10 deg.C.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;We have also tried the piston pumps that are used in construction sites for concrete spraying operation. It works continuously, but there are no linear flow rate adjustments, and it is bigger in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The screw pump works fine, but it often stops. The reversing and restarting pumping operation is not desirable for the process. &#xA;We noticed that when reversing the motor for few rotations, the material in the hopper is raising. So we are guessing that the motor is stopping due to air trap. But supplier commenting it is because of vacuum created in the pump. because the concrete paste is not perfect fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are any screw pump users came across such situations? How did you solve it? Any suggestions to fix this issue I can get to rectify it??&lt;/p&gt;&#xA;" OwnerUserId="15386" LastEditorUserId="15386" LastEditDate="2018-12-14T06:56:45.093" LastActivityDate="2019-05-13T08:02:40.590" Title="How to resolve the interruptions in the screw pump operation due to air trap?" Tags="|pumps|concrete|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25094" PostTypeId="2" ParentId="25074" CreationDate="2018-12-14T15:54:15.127" Score="1" Body="&lt;p&gt;The mole balance on component A through the reactor balances flow rates out and in with rate of change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{n}_{Ao} - \dot{n}_{Ai} = \frac{dn_A}{dt} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left side can be written in terms of overall conversion for a consumption reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ f_A \equiv \frac{\dot{n}_{Ai} - \dot{n}_{Ao}}{\dot{n}_{Ai}} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right side can be rewritten as a rate expression on concentration &lt;span class=&quot;math-container&quot;&gt;$dn_A / dt = V dC_A/dt$&lt;/span&gt;. Use an empirical rate law expression for the rate. Assume the reaction consumes component A. In a CSTR, the outlet concentration is the reactor concentration. Combined we obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ f_A\ \dot{n}_{Ai} = V\ k\ C_{Ao}^n $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you want to decrease the volume of the reactor &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, you must change these factors with these consequences:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Decrease the volumetric flow rate of A in to keep the overall conversion &lt;span class=&quot;math-container&quot;&gt;$f_A$&lt;/span&gt; constant.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Decrease the overall conversion &lt;span class=&quot;math-container&quot;&gt;$f_A$&lt;/span&gt; to keep the same volumetric flow rate of A in.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Increase the concentration of A in the outlet for the same overall conversion and volumetric flow rate of A in.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2018-12-14T15:54:15.127" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25096" PostTypeId="1" AcceptedAnswerId="25101" CreationDate="2018-12-14T20:11:40.530" Score="2" ViewCount="432" Body="&lt;p&gt;I'm currently researching the strength of domes underwater and I went down the rabithole of material science. The basics of this research (or though experiment) is to bring a dome filled with air to the seafloor and see if it implodes or not. The specifics are as follows. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; The dome is 3D-printed&#xA; Outer Diameter: 10 cm&#xA; Inner Diameter: 8 cm (so the thickness of the dome is 1 cm) (this can vary)&#xA; Material: PLA&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;QUESTION: What is the maximum amount of pressure this dome can take before it breaks?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already found the following data from this &lt;a href=&quot;https://downloads.makerbot.com/legal/MakerBot_R__PLA_and_ABS_Strength_Data.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Properties PLA:&#xA; - Impact strength:&#xA;   - STD: 1.8 ft-lb/in&#xA;   - MAX: 4.1 ft-lb/in&#xA; - Compressive Strength&#xA;   - STD: 2600 PSI&#xA;   - MAX: 13600 PSI&#xA; - Tensile Strength&#xA;   - STD: 6783 PSI&#xA;   - MAX: 9531 PSI&#xA; - Flexural Strength&#xA;   - STD: 8970 PSI&#xA;   - MAX: 13731 PSI&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It seems likely that I'll be needing the compressive strength here, because the theoretical dome is being put at the bottom of the theoretical sea. It's going to be compressed. But these numbers have very little context (at least for me, I don't have an engineering degree). I mainly worry about thickness. In my mind, when a wall is thicker, it means it can hold more weight, can withstand heavier punches and generally breaks less easily. I don't see this reflected in this number. It also doesn't take the shape of the object in mind. A sphere is supposedly  stronger then any other shape (I found this &lt;a href=&quot;https://www.monolithic.org/benefits/benefits-strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;cool site&lt;/a&gt; selling dome bunkers). But this too doesn't seem reflected in the compressive strength value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might also have to do with &lt;a href=&quot;https://en.wikipedia.org/wiki/Size_effect_on_structural_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;the size effect&lt;/a&gt;, but that seems to weaken - not strengthen - structures and would only be applicable if I made the outer diameter variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can help me any further it would be very much appreciated. I really want to make clear that I'm not asking you to do my homework (this isn't really homework, but still). Pointing me vaguely in the right direction is also extremely helpful. Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="18634" LastActivityDate="2018-12-15T14:26:24.727" Title="Strength of Plastic Underwater Domes with various thicknesses" Tags="|materials|pressure|strength|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25097" PostTypeId="1" CreationDate="2018-12-14T21:07:52.957" Score="0" ViewCount="48" Body="&lt;p&gt;I am working on an assembly where I have various screws, mainly metal on metal fastening.&#xA;Where can I find a realiable source for recommended torque settings based on screw size, screw material and fastening surfaces materials?&#xA;I tried reading on the Machniery's Handbook but I can't seem to find anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;JL&lt;/p&gt;&#xA;" OwnerUserId="18635" LastActivityDate="2018-12-14T21:07:52.957" Title="Recommended torque" Tags="|mechanical-engineering|fasteners|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25098" PostTypeId="1" CreationDate="2018-12-15T08:21:02.350" Score="0" ViewCount="317" Body="&lt;p&gt;I'm looking to control a commercial deadbolt using a small servo, motor, or gear motor. The deadbolt has a common Schlage cylinder such as &lt;a href=&quot;https://images.homedepot-static.com/productImages/39eda453-f5f0-4c06-bf4c-d32bc4a8c01f/svn/prime-line-door-kits-locks-trims-e-2103-c3_1000.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; that has a flat shaft that runs through a slotted hole in the deadbolt. In &lt;a href=&quot;http://www.mrlock.com/image/catalog/mfg/ilc/lori_deadbolt/4500-parts-breakdown.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this diagram&lt;/a&gt; the shaft is labeled 'tailpiece' and is connected to the thumb-turn on the inside of the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most servos and electric motors come with short, round shafts on the end of them. Presumably in common usage a gear or cam is mounted to that shaft with a screw. For this situation, it would be convenient to couple the two shafts directly together inline, such that the motor would replace the thumb-turn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not in any way a mechanical engineer, so I'm really grasping at straws but I think what I need is a &lt;a href=&quot;https://www.mcmaster.com/shaft-couplings&quot; rel=&quot;nofollow noreferrer&quot;&gt;shaft adapter/coupling&lt;/a&gt;? Something that would encase the round shaft of the servo on one side and the flat shaft of the lock cylinder on the other and clamp down on both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively the server / motor could be mounted perpendicular to the flat shaft and two gears could mesh to make a 90 degree angle. I don't have anything to measure torque properly, but I know the thumb-turn is not quite an inch in radius and it takes more force to turn it then it takes to pull a stiff trigger, so probably 5-10 lb inches? I feel like the gear-based option might not stand up to that. Then again, the whole lock set could do with some grease. It's an old house.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another thought I just had was a through-hole motor. Or servo. Do they make through-hole servos?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;This is a one-off solution for home application, but I'll document it and share that documentation so it'd be nice if the solution was broadly applicable to common lockset configurations. I'm leaving the standard keyway on the outside of the door for power outages. It may be hard to turn due to the servo on the other end, but it would be infrequently used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that I've disassembled it and refreshed my memory of what it looks like on the inside, I realize there's a second option. The thumb-turn engages the lock cylinder via a component that I don't know the name for. On the thumb-turn side it's a washer with an oblong inner diameter. On the door side it's a collar with a matching outer diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's some photos of the actual parts: &lt;a href=&quot;https://i.stack.imgur.com/Io0Tz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/Io0Tz.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could potentially make a wood part that fits over the door-side collar and could screw into the arms of a servo motor. Might be easier than a shaft-coupler.&lt;/p&gt;&#xA;" OwnerUserId="18639" LastEditorUserId="18639" LastEditDate="2018-12-15T22:18:34.103" LastActivityDate="2018-12-15T22:18:34.103" Title="Question about how to couple round servo shaft to flat shaft" Tags="|mechanical-engineering|design|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25099" PostTypeId="2" ParentId="25096" CreationDate="2018-12-15T12:39:53.333" Score="1" Body="&lt;p&gt;For simplicity let's first consider the case of a sphere with thickness much smaller than the radius ('Thin-Walled' pressure vessel). Say &lt;span class=&quot;math-container&quot;&gt;$\frac{r}{t} &amp;lt; 10$&lt;/span&gt;, with &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; the radius of the sphere and &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; the thickness. It is straightforward to show (see &lt;a href=&quot;http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) that the stress in two orthogonal circumferential directions is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sigma = \frac{Pr}{2t} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where P is the difference between the external and internal pressures. If &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is compressive (exterior pressure is larger than internal) then &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; is compressive. You will then need to apply a failure model appropriate to your material (not sure about PLA). But a simple criteria could be if this stress is greater than the maximum stress your material can handle then it will fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your pressure vessel is not 'thin' then you need to consider variation of stress through the thickness which complicates matters significantly ('Thick-Walled' pressure vessel). A discussion of this case can be found &lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Mechanics/Cylinders.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A dome is much more complicated since the flat face on the bottom will also be under the same pressure and will therefore be subject to bending. The interface between the hemisphere and the base also complicates the stress state. This is why you virtually never see pressure vessels with flat faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality you will need to consider imperfections in the fabrication, variation in material properties, buckling, etc. &lt;/p&gt;&#xA;" OwnerUserId="3010" LastEditorUserId="3010" LastEditDate="2018-12-15T12:47:44.087" LastActivityDate="2018-12-15T12:47:44.087" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25101" PostTypeId="2" ParentId="25096" CreationDate="2018-12-15T14:05:07.690" Score="3" Body="&lt;p&gt;A shell structure under external pressure is likely to fail in buckling before it reaches the compressive strength of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buckling of curved shells is a complex problem in general, but for a complete sphere a simple formula (source &lt;a href=&quot;http://eprints.keele.ac.uk/1563/1/Stability.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) is &lt;span class=&quot;math-container&quot;&gt;$$P = \frac{2E}{\sqrt{3(1-\nu^2)}}\frac{t^2}{r^2}$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\nu$&lt;/span&gt; are Young's modulus and Poisson's ratio, &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; is the shell thickness and &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the mean radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; = 510,000 psi and &lt;span class=&quot;math-container&quot;&gt;$\nu$&lt;/span&gt; = 0.33 (sources, &lt;a href=&quot;https://www.makeitfrom.com/material-properties/Polylactic-Acid-PLA-Polylactide&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20160010284.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) This gives about &lt;span class=&quot;math-container&quot;&gt;$31,000$&lt;/span&gt; psi for your dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These buckling formulas are very imprecise and I would take a safety factor of at least 10 times, which would give a safe buckling load of about 3000 psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the maximum pressure using the &quot;STD&quot; compressive stress is about 1200 psi (using the formula in mg4w's answer), in this case buckling is not the limiting factor, but the buckling load is proportional to &lt;span class=&quot;math-container&quot;&gt;$t^2$&lt;/span&gt; while the compressive load is proportional to &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;, so for a thinner shell the buckling load decreases faster than the compressive load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So taking the pressure gradient in water as about 0.5 psi per foot of depth, (there's no point using more than one significant figure here!) a sphere with your dimensions should be good for a depth of about 2,500 feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a dome with a flat base would fail before that, since the base would break at the edges where it joined the sphere. Assuming the base is rigidly supported by the dome around its edges, the maximum stress (source &lt;a href=&quot;http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) is &lt;span class=&quot;math-container&quot;&gt;$$\frac{3Pr^2}{4t^2}$$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the maximum tensile strength as 6783 psi, that gives a pressure of about 450 psi at failure - much smaller than the other failure modes considered above, and limiting the depth to about 900 feet (with no safety factor applied!)&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-12-15T14:26:24.727" LastActivityDate="2018-12-15T14:26:24.727" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25103" PostTypeId="1" CreationDate="2018-12-15T18:31:04.323" Score="1" ViewCount="1957" Body="&lt;p&gt;The moisture-carrying ability of air goes up as the temperature rises. So if I take a volume of air and heat it or cool it (allowing it to expand/contract so the pressure remains constant) how can I calculate the (approximate) relative humidity of the warmer/cooler air that contains the same amount of water vapor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I blow 30°c air at 20% relative humidity over a cool object (but not cool enough to condense water out of the air.) The air is cooled to 20°. What would the relative humidity of the cooled air be if it contained the same amount of water vapor? How do I calculate that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am building an Arduino-based control for a warm air dryer for ski boots and bike gear. I will measure the temperature/humidity of the heated air as it comes out of the dryer, and then measure it again as it escapes the gear that I am drying. While the gear is still damp, the waste air should pick up additional moisture, and so its relative humidity should be higher that it would be if I simply cooled it down. When my gear is completely dry, the cooler waste air should contain the same amount of water vapor that it started with, so the relative humidity should be the amount calculated for the measured change in temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want a &quot;good enough&quot; solution, not one that's mathematically pure. If it gets me answers that are within a percent but involves less computation, that's better than an ideal solution that involves lots of time-consuming calculations. The Arduino microcontroller is fairly slow and cannot do high speed/high precision floating point calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Would this question be better on an engineering forum? I could see arguments either way.)&lt;/p&gt;&#xA;" OwnerUserId="18643" LastActivityDate="2022-02-02T12:24:01.570" Title="What is a formula to approximate the change in relative humidity cause by a change in temp?" Tags="|dehumidification|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25104" PostTypeId="1" CreationDate="2018-12-15T19:57:13.893" Score="2" ViewCount="99" Body="&lt;p&gt;I'm trying to understand the output of clockwork coils in joules or watts for different coils. You know the pullback car or windup pocket watch mechanism. I have heard it also called power in the barrel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this below image as is the best approximation of the power output, but I don't know the coil size, or how to determine the size of a coil barrel for a particular out requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question, how to determine output of a clockwork motor of given sizes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6kIqo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6kIqo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18644" LastActivityDate="2018-12-15T19:57:13.893" Title="clockwork motor output in joules formula" Tags="|motors|power|energy-storage|engineering-general|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25105" PostTypeId="1" CreationDate="2018-12-15T20:02:19.773" Score="1" ViewCount="31" Body="&lt;p&gt;I will be a freshman in University next year. After graduating, I would like to get a job in the R&amp;amp;D sector of drinking water treatment technologies. I'm unsure if I should get a degree in ME or CE. ME seems to better prepare students for the development of new technologies, but CE deals a lot more with drinking water management.&lt;/p&gt;&#xA;" OwnerUserId="14574" LastActivityDate="2018-12-15T20:02:19.773" Title="Drinking Water Treatment R&amp;D: Mechanical or Civil Engineering Degree?" Tags="|mechanical-engineering|civil-engineering|water-resources|research|" AnswerCount="0" CommentCount="2" ClosedDate="2018-12-15T20:55:02.823" ContentLicense="CC BY-SA 4.0" />
  <row Id="25106" PostTypeId="1" CreationDate="2018-12-15T20:35:35.543" Score="2" ViewCount="95" Body="&lt;p&gt;I am doing finite element analysis of a donut shape for my school work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LTn2V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LTn2V.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm increasing the thickness of the object with a 1&amp;nbsp;mm increment and run FEA each time, and then comparing the maximum displacement of the object. According to the mathematical relationship between the stiffness and thickness of this object (&lt;span class=&quot;math-container&quot;&gt;$k = 4Ebh^3/L^3$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is the thickness), I should be getting a &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; that increases with the cube of the thickness. Although it roughly follows the trend, it doesn't match the mathematical model perfectly... why is that? Is anything happening in that donut shape that I haven't caught, like buckling? And how would that affect the displacement or stiffness? &#xA;&lt;a href=&quot;https://i.stack.imgur.com/9ucei.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ucei.png&quot; alt=&quot;displacement vs thickness&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ko5kO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ko5kO.png&quot; alt=&quot;displacement vs height&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18024" LastEditorUserId="18024" LastEditDate="2018-12-15T22:47:34.660" LastActivityDate="2018-12-15T22:47:34.660" Title="FEA using abaqus, and it does not follow the trend" Tags="|mechanical-engineering|structural-engineering|civil-engineering|" AnswerCount="0" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="25107" PostTypeId="2" ParentId="25103" CreationDate="2018-12-15T20:44:30.877" Score="0" Body="&lt;p&gt;You need to know the vapor pressure of water at different temperature. No simple formula to calculate this exists, however you can hunt down a table - &lt;a href=&quot;https://en.wikipedia.org/wiki/Vapour_pressure_of_water&quot; rel=&quot;nofollow noreferrer&quot;&gt;like the one on wikipedia&lt;/a&gt; - and then use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clausius-Clapeyron equation&lt;/a&gt;. I've learned to use this form:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ln\frac{P_1}{P_2}=-\frac{L}{R}(\frac{1}{T_1}-\frac{1}{T_2})$$&lt;/span&gt;&#xA;With latent heat &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; and gas cosntand &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;. Use thge closest known value pair auf T and P from your table to calculate the desired values.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2018-12-15T20:44:30.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25108" PostTypeId="2" ParentId="25103" CreationDate="2018-12-15T21:00:50.080" Score="1" Body="&lt;p&gt;The formula for relative humidity is the saturation vapor pressure at the dew point divided by the saturation vapor pressure at the ambient temperature. The formula for saturation vapor pressure is:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;=610.78e&lt;sup&gt;(t/(t + 238.3 ) * 17.2694)/100&lt;/sup&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I don't remember where I found this formula, but I use it in an Excel spreadsheet to calculate water loss in my pond. Since I'm already using the formula, I also calculate relative and absolute humidity. The formulas seem to be reasonably accurate. Of course the temperature is in degrees Celsius. &lt;br&gt;&lt;br&gt;I have the spreadsheet on my Dropbox account and can post the URL if you're interested.&lt;br&gt;&lt;br&gt;I found the link for the paper where I found the formula; it's &lt;a href=&quot;https://www.conservationphysics.org/atmcalc/atmoclc2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12535" LastEditorUserId="90" LastEditDate="2022-02-02T12:03:29.450" LastActivityDate="2022-02-02T12:03:29.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25110" PostTypeId="1" CreationDate="2018-12-15T18:26:34.520" Score="0" ViewCount="72" Body="&lt;p&gt;For my rube golberg project I have to twist a lightbulb in and have it turn on. I have seen different videos on how they do it but they never explain it and most of the videos are blurry. We can't have the lightbulb already in partway to screw it in.   Does anyone know what the best way to have something screw the lightbulb in and a great power source for it to turn on? &lt;/p&gt;&#xA;" OwnerDisplayName="ScienceDaily" LastEditorUserId="33" LastEditDate="2018-12-16T01:15:19.437" LastActivityDate="2018-12-16T01:15:19.437" Title="Rube Goldberg, twisting in a lightbulb" Tags="|mechanical-engineering|lighting|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25111" PostTypeId="1" CreationDate="2018-12-15T22:04:16.410" Score="1" ViewCount="34" Body="&lt;p&gt;I've done some rough calculations on how much water could be produced by using the waste heat of a powerplant to boil seawater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in a ballpark figure and would like to know if I have a major flaw in the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 1MW plant with 1/3 efficiency can boil 17 moles of water per second assuming that the water is at the BP and it takes 40KJ/mol to boil it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;17 moles of water is 300g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one day the plant boils 26 tons of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One acre foot is around 1200 tons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;California uses 110,000 acre feet per day.  This is 132,000,000 tons per day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be produced by 5,100,000MW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems like this potential source would be a drop in the bucket in terms of meeting water needs.&lt;/p&gt;&#xA;" OwnerUserId="398" LastActivityDate="2018-12-15T22:04:16.410" Title="How much seawater could be desalinated using the waste heat from a powerplant?" Tags="|civil-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25112" PostTypeId="1" CreationDate="2018-12-15T23:49:13.470" Score="1" ViewCount="23" Body="&lt;p&gt;What would be the best way to design a sealed computer containing a gas or liquid circulated by a fan/pump to cool the components? ( With a metal case ) Then the computer would not be susceptible to liquids or dust, while avoiding the hassle of heat pipes or complex heat spreaders for all components.&lt;/p&gt;&#xA;" OwnerUserId="18646" LastEditorUserId="18646" LastEditDate="2018-12-16T01:18:24.090" LastActivityDate="2018-12-16T01:18:24.090" Title="DIY Thermal Solution" Tags="|heat-transfer|thermal-conduction|computer-hardware|" AnswerCount="0" CommentCount="2" ClosedDate="2018-12-16T01:10:17.223" ContentLicense="CC BY-SA 4.0" />
  <row Id="25113" PostTypeId="1" AcceptedAnswerId="25181" CreationDate="2018-12-16T09:21:26.677" Score="1" ViewCount="56" Body="&lt;p&gt;Please bear with me - I'm not an engineer, I'm a software developer so lack the language / knowledge to describe what I want precisely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm controlling a servo to rotate /pan a platform to hold a camera (think pan/tilt). This is fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'd like to house the servo/electronics in a waterproof box, and have a platform on top of the box that Pans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I transfer the rotation through a waterproof enclosure? Is there such a thing as a waterproof bearing, that the servo can rotate one side within the enclosure, and have the rotation exposed on the outside, so I can attach the Panning platform to it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does such a thing exist? If so, what is it called?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, surely this is a problems that has already been solved&#xA; Can anyone direct me to a solution for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="18649" LastActivityDate="2018-12-19T18:11:19.130" Title="Waterproof Pan control/bearing/mechanism" Tags="|bearings|servo|waterproofing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25119" PostTypeId="1" CreationDate="2018-12-16T21:32:14.917" Score="0" ViewCount="575" Body="&lt;p&gt;Reduction gearboxes made of machine steel gear have a maximum limit to prevent changing their physical properties (softening). They change their properties as a result of the temperature increase which is subject of many variables: friction between gear teeth, ambient temperature, quality of bearings are among the main dependencies. The most typical engineering solutions are such as making gear of different stronger materials which withstands higher temperatures, oil bath or liquid cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know what are possible ways to increase the durability of the same gear in a given gearbox. I can heat the gear pieces to a high temperature using torch that makes it &quot;tempered&quot; or &quot;annealed&quot; making it stronger. Which one of those known processes intended specifically for this purpose and how do I find the procedure (heating, followed by cooling or not...) and temperatures needed to alter the structure of the steel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also are there other ways to increase the rotational speed limit beside gear tempering and gearbox cooling? Assume for this topic that bearings do not limit the speed.&lt;/p&gt;&#xA;" OwnerUserId="18655" LastActivityDate="2020-06-14T15:17:08.470" Title="How to increase maximum rotational speed limit of a reduction gearbox?" Tags="|gears|metals|" AnswerCount="3" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="25120" PostTypeId="1" CreationDate="2018-12-16T21:52:26.913" Score="0" ViewCount="19" Body="&lt;p&gt;How would I go by getting around 2KV ac rms able to support an average current of 15amp and max of 20amp out of 2 lines of 220? And what kind of relay could take that? Looking for a winding ratio.&lt;/p&gt;&#xA;" OwnerUserId="18656" LastEditorUserId="18656" LastEditDate="2018-12-17T20:57:11.297" LastActivityDate="2018-12-17T21:00:39.203" Title="Get 2KV AC 15AMP draw from 2 lines of 220" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25121" PostTypeId="1" CreationDate="2018-12-16T22:29:42.040" Score="1" ViewCount="79" Body="&lt;p&gt;I am using GRBL and UGS with an arduino uno and its CNC shield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I send the $H command, the z motor is moving a little bit and then it goes crazy. I mean it's starts to hum and doesn't move anymore. After that, the machine is in ALARM state. ALARM:8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I send $X command, the machine goes back to idle. So i am now able to use my cnc machine again. But why am I not able to set it to home position ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PpMRD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PpMRD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;config:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$0 = 10    (Step pulse time, microseconds)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$1 = 25    (Step idle delay, milliseconds)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$2 = 0    (Step pulse invert, mask)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$3 = 0    (Step direction invert, mask)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$4 = 0    (Invert step enable pin, boolean)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$5 = 0    (Invert limit pins, boolean)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$6 = 0    (Invert probe pin, boolean)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$10 = 1    (Status report options, mask)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$11 = 0.010    (Junction deviation, millimeters)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$12 = 0.002    (Arc tolerance, millimeters)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$13 = 0    (Report in inches, boolean)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$20 = 0    (Soft limits enable, boolean)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$21 = 1    (Hard limits enable, boolean)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$22 = 1    (Homing cycle enable, boolean)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$23 = 0    (Homing direction invert, mask)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$24 = 25.000    (Homing locate feed rate, mm/min)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$25 = 500.000    (Homing search seek rate, mm/min)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$26 = 250    (Homing switch debounce delay, milliseconds)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$27 = 1.000    (Homing switch pull-off distance, millimeters)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$30 = 1000    (Maximum spindle speed, RPM)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$31 = 0    (Minimum spindle speed, RPM)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$32 = 0    (Laser-mode enable, boolean)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$100 = 250.000    (X-axis travel resolution, step/mm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$101 = 250.000    (Y-axis travel resolution, step/mm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$102 = 250.000    (Z-axis travel resolution, step/mm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$110 = 500.000    (X-axis maximum rate, mm/min)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$111 = 500.000    (Y-axis maximum rate, mm/min)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$112 = 500.000    (Z-axis maximum rate, mm/min)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$120 = 10.000    (X-axis acceleration, mm/sec^2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$121 = 10.000    (Y-axis acceleration, mm/sec^2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$122 = 10.000    (Z-axis acceleration, mm/sec^2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$130 = 200.000    (X-axis maximum travel, millimeters)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$131 = 200.000    (Y-axis maximum travel, millimeters)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;$132 = 200.000    (Z-axis maximum travel, millimeters)&lt;/p&gt;&#xA;" OwnerUserId="18657" LastActivityDate="2018-12-16T22:29:42.040" Title="cnc z axis motor huming when homing" Tags="|cnc|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25123" PostTypeId="1" CreationDate="2018-12-17T01:41:31.057" Score="1" ViewCount="38" Body="&lt;p&gt;I have been working on a product that involves the reflection of low wavelengths of UV light (200-250nm). However, I have not come across many materials that are able to do so and the high-reflectivity mirrors I come across are extremely expensive. What mirrors would be able to reflect UV light at such wavelengths at around a 50-75% reflectivity rate?&lt;/p&gt;&#xA;" OwnerUserId="18659" LastActivityDate="2019-08-16T12:21:14.193" Title="Looking for something to reflect ultraviolet light at extremely low wavelengths" Tags="|materials|thermodynamics|building-physics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25124" PostTypeId="1" CreationDate="2018-12-17T02:11:00.480" Score="2" ViewCount="167" Body="&lt;p&gt;I've been trying to solve the dynamics of this system for quite a while, but cannot seem to get the right answer. I've tried solving it with Lagrangian method but I am not so confident with the equations I'm getting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, looking at the picture below,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cR8lk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cR8lk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there are three components in the system where 1 is the link between &lt;a href=&quot;https://i.stack.imgur.com/88Yzu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt; and &lt;a href=&quot;https://i.stack.imgur.com/x2iAs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;. &lt;a href=&quot;https://i.stack.imgur.com/x2iAs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt; moves linearly along the y axis and is driven by a linear drive. 1 is actually a linear drive that moves &lt;a href=&quot;https://i.stack.imgur.com/88Yzu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt; up and down, but because there is a surface when x=0, the friction between the surface and &lt;a href=&quot;https://i.stack.imgur.com/88Yzu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt; cause 1 and &lt;a href=&quot;https://i.stack.imgur.com/88Yzu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt; to rotate just as in the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to solve the dynamics of this system in terms of x,y,theta, and velocity and acceleration (dot(x),dot(y),dot(theta),twodot(x), etc) from the lagrangian method. But my initial attempt had l to be constant when it actually moves due to ux. I also did not consider the surface to be deformable; the surface in which the linear drive 1 exerts a force of ux deforms as it is not a rigid body. For simplicity's sake, I considered the reaction force from the surface to be f and tried to solve the dynamics again, but got totally confused as I couldn't figure out how much work was being done by f and as there were new variable of l ldot l two dot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, it would be really helpful if anyone could define the work done by the external force, (the u's and friction and f)as well as kinetic and potential energy of each parts so that I could use them in the lagrangian method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you think that there are other methods other than lagrangian to solve for the acceleration and velocity of x y and theta of this particular system, that would also be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the problem to solve is the acceleration of x y and theta at point (x,y) when the linear drives on m3 and m2 are given a force of uy and ux, respectively. I am trying to solve the accelerations in terms of uy ux Ffr theta and f (the springy reaction force of the surface at x=0). m's are all masses, s is the thickness of &lt;a href=&quot;https://i.stack.imgur.com/88Yzu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt; r is the vector of (x,y) with respect to the the point of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial attempt at this problem doesn't consider l to be changing (but it changes since its a linear drive that gives a force of ux).&lt;a href=&quot;https://i.stack.imgur.com/88Yzu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/88Yzu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i first found the acceleration of x y relative to the point of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then i used the lagrangian work theorem to derive the following equation:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/x2iAs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x2iAs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the equation above i found theta two dot which i plugged into the acceleration equation of x and y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically how I attempted to solve the problem and it would be great if anyone could also following this logic to derive a correct answer as opposed to my worthless one.&lt;/p&gt;&#xA;" OwnerUserId="18660" LastEditorUserId="18660" LastEditDate="2018-12-19T09:48:51.607" LastActivityDate="2023-12-23T21:08:15.930" Title="Dynamics of a rigid body in a system of two linear drives" Tags="|dynamics|kinematics|acceleration|linear-motion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25125" PostTypeId="1" CreationDate="2018-12-17T03:05:07.997" Score="1" ViewCount="540" Body="&lt;p&gt;I would like to get some opinions on using a low flow hydraulic MOTOR to drive a high flow pump ? Would the high flow pump be able to create a useable pressure to drive a piece of equipment that requires the greater flow ? Or will the system loss to friction , restriction and mechanical drag make it impossible to realize any useful gains ? &lt;/p&gt;&#xA;" OwnerUserId="18661" LastActivityDate="2020-02-02T16:01:00.103" Title="Low flow hydraulic motor to run high flow pump?" Tags="|hydraulics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25127" PostTypeId="1" CreationDate="2018-12-17T15:25:59.160" Score="1" ViewCount="61" Body="&lt;p&gt;I have a pretty basic question but I would like to be sure of what I think. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a setup like the one in the figure. I have two motors which can generate two forces &lt;span class=&quot;math-container&quot;&gt;$F_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_2$&lt;/span&gt; perpendicular to the plane of the drawing (coming out of the plane of the drawing). The black box is a rigid body of mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; which has the Center of Gravity (CoG) in the middle (the white circle). We can think about two forces coming from two propellers as on a quadrotor UAV or a helicopter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LpMuR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LpMuR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&#xA;regarding &lt;span class=&quot;math-container&quot;&gt;$F_1$&lt;/span&gt;, since the corresponding arm is aligned with one of the axes of symmetry of the body (specifically y), if it will be activated it will contribute in torque terms only to a rotation &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; with respect to the x-axis and therefore the acceleration around the x-axis will have this shape:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$I_x\ddot{\phi}=...+L \cdot F_1$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$I_x$&lt;/span&gt; is the moment of inertia with respect to the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, if I activate the motor 2, corresponding to the force &lt;span class=&quot;math-container&quot;&gt;$F_2$&lt;/span&gt;, since it is displaced from the axes of symmetry, but still parallel to one of them, I should have: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$I_x\ddot{\phi}=...+L \cdot F_2 - \underbrace{f(I_y,d_2,a,mg)}_{M_d}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$I_y$&lt;/span&gt; is the moment of inertia with respect to the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; axis, &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; the total mass of the rigid body and &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; the acceleration due to gravity. &#xA;The term &lt;span class=&quot;math-container&quot;&gt;$M_d$&lt;/span&gt; should take into account the torque coming from the motor and the counter-torque coming from the fact that the point of attachment is not aligned with the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; axis. &#xA;&lt;strong&gt;I would like to know how to write the expression of &lt;span class=&quot;math-container&quot;&gt;$f(\cdot)$&lt;/span&gt;&lt;/strong&gt; which should be quite easy to do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help!&lt;/p&gt;&#xA;" OwnerUserId="18666" LastActivityDate="2018-12-18T07:12:41.667" Title="Torque computation for different attachment point" Tags="|torque|kinematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25128" PostTypeId="1" AcceptedAnswerId="25139" CreationDate="2018-12-17T17:02:11.403" Score="1" ViewCount="90" Body="&lt;p&gt;I am trying to construct a lift handle in the CAD Sketchup. The circled part of the image shows the arch I'm referring to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sUgRv.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The program, Sketchup, has a 'tool' called push/pull, and another tool called rotate. I have a video uploaded on streamable demonstrating this, but I'm not sure if it's within the site's rule to do so. Here's a brief clip partially demonstrating it in action: &lt;a href=&quot;https://i.imgur.com/pgVRw9d.mp4&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.imgur.com/pgVRw9d.mp4&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use the push/pull tool, to raise the face of the object by 250&amp;nbsp;mm, and the rotate tool, to rotate it by 15 degrees. &lt;span class=&quot;math-container&quot;&gt;$90/15 = 6$&lt;/span&gt;. The arch ends up with an overall height of 1074&amp;nbsp;mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I repeat the task 2 more times, but with the angle of 7.5 and 30 degrees.&#xA;With 7.5 degrees, the arch looks much curvier, but the height is 2032&amp;nbsp;mm. With 30 degrees, the arc sides are noticeably pointy, but its height is more acceptable, 591&amp;nbsp;mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image of all 3 arcs together: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vw9DT.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimension of each side on the inner part of the arc, are all 250&amp;nbsp;mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what equation may I use, so I end up with an arch that looks curvy enough, in this case 15 degrees or lower, and has a height of 508&amp;nbsp;mm from the 'ground' (or x-axis)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried the following, with no success:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Circumference of a circle &lt;span class=&quot;math-container&quot;&gt;$C=2\pi r$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The arc in this case represents 1/4th of a circle; so &lt;span class=&quot;math-container&quot;&gt;$C_{arc}=2\pi r/4$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;2\pi508/4 &amp;amp;= 797.96 \\&#xA;\dfrac{797.96}{90/15} &amp;amp;= 132.99&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I push/pull the face of the object by 132.99, and rotate its top face by 15 degrees. I repeat this process 5 more times. The 6th arch rotated, is perfectly perpendicular to the x-axis. Unfortunately, the height is 571&amp;nbsp;mm, instead of 508&amp;nbsp;mm.&lt;/p&gt;&#xA;" OwnerUserId="18668" LastEditorUserId="1832" LastEditDate="2018-12-17T18:47:33.557" LastActivityDate="2018-12-20T00:30:57.120" Title="Building a half-arch of a specific height, while consistently using an angle divisible by 90" Tags="|cad|geometry|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25129" PostTypeId="1" AcceptedAnswerId="25144" CreationDate="2018-12-17T17:11:48.860" Score="3" ViewCount="466" Body="&lt;p&gt;Does notching or cutting out the corner of an angle brace distribute stress better along the flange of a member with shear lag or reduce the stressed edge of the flange's ability to resist deformation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Previously, user CableStay &lt;a href=&quot;https://engineering.stackexchange.com/a/4196/33&quot;&gt;had illustrated Shear Lag&lt;/a&gt; concept in a bolted angle iron member very well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/niLm3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/niLm3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question then asks if it is possible to better distribute the stress of the uniform stress closer to the end of the member by notching or cutting the corner of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aVMdj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aVMdj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="18669" LastEditorUserId="1832" LastEditDate="2018-12-17T18:40:20.937" LastActivityDate="2018-12-18T19:09:48.543" Title="Does notching the corner of an angle brace member help or hurt when shear lag is present?" Tags="|structural-engineering|shear|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25130" PostTypeId="1" AcceptedAnswerId="25131" CreationDate="2018-12-17T18:28:02.557" Score="3" ViewCount="7780" Body="&lt;p&gt;I am trying to find out if the 12&amp;nbsp;V DC electric motors that I have are powerful enough to launch a 100&amp;nbsp;g tennis ball 30&amp;nbsp;m away. The tennis ball is put down a pipe and ends up wedged in between 2 wheels (radius = 0.05&amp;nbsp;m) driven by the two 12&amp;nbsp;V DC electric motors. The picture below should hopefully clarify the situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6aljX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6aljX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The electric motors are spinning at ~7000&amp;nbsp;RPM (no load), the motor specs are here: &lt;a href=&quot;http://www.banebots.com/product/M5-RS555-12.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.banebots.com/product/M5-RS555-12.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;A wheel (with mass of 0.05&amp;nbsp;kg and radius of 50&amp;nbsp;mm) is connected to each motor&lt;/li&gt;&#xA;&lt;li&gt;The tennis ball is dropped down the pipe and ends up in between the two rotating disks&lt;/li&gt;&#xA;&lt;li&gt;The launching angle is 20&amp;nbsp;degrees&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I calculated how far the ball flies if I keep the RPM constant based on the horizontal component of the linear velocity in combination with the flying time and launching angle. However, this does not account for the mass of the tennis ball. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the ball passes through the two wheels, the electric motors will see an opposite moment. I guess I can calculate the moment by multiplying the mass * g * radius. However, it is not clear to me how I can then accurately calculate the effect on the RPM of the motors. I can optimize the location of wheels so that the ball isn't squeezed too much, so I was hoping to ignore this in the calculation and build in some margin instead. In the end, I would like the ball to fly at least 30&amp;nbsp;m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have an idea on how I can calculate if the motors are strong enough to launch the ball 30&amp;nbsp;m away?&lt;/p&gt;&#xA;" OwnerUserId="18670" LastEditorUserId="1832" LastEditDate="2018-12-17T18:51:02.323" LastActivityDate="2021-11-02T15:13:14.043" Title="Torque required to launch a tennis ball" Tags="|mechanical-engineering|torque|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25131" PostTypeId="2" ParentId="25130" CreationDate="2018-12-17T19:06:16.783" Score="1" Body="&lt;p&gt;Well, color me amazed.  The work's been done for you, sort of.&#xA;Here's a &lt;a href=&quot;http://www.actawm.pb.edu.pl/volume/vol5no4/WOJCICKI_KULESZA_PUCILOWSKI_EN_2010_085.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt;{dead link}  &lt;a href=&quot;http://yadda.icm.edu.pl/baztech/element/bwmeta1.element.baztech-article-BPB2-0062-0016/c/httpwww_actawm_pb_edu_plvol5no4wojcickikuleszapucilowskien2010085.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;new link that works&lt;/a&gt; that analyzes exactly your machine in excruciatingly precise mathematical and engineering detail... except I didn't see estimates of the achieved range, as they were more concerned with exit speed and covering an actual tennis court.&lt;/p&gt;&#xA;&lt;p&gt;Now, I found a hands-on review of a tennis ball &lt;a href=&quot;https://woofwhiskers.com/max-launcher-dog-ball-launcher-review/&quot; rel=&quot;nofollow noreferrer&quot;&gt;launcher&lt;/a&gt; which successfully does send a ball up to 90 feet, close to your 30 m goal.  Granted, this is a dramatically different mechanism -- basically a large spring with a launch platform to propel the ball -- but it does suggest that your design can fire a ball that far if you can achieve similar exit velocities.&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="615" LastEditDate="2021-11-02T15:13:14.043" LastActivityDate="2021-11-02T15:13:14.043" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25134" PostTypeId="2" ParentId="25120" CreationDate="2018-12-17T21:00:39.203" Score="1" Body="&lt;p&gt;You cannot get 2kV from a relay unless you have 2kV available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transformer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Turns\ ratio = \frac {V_{High}} {V_{Low}}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2018-12-17T21:00:39.203" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25135" PostTypeId="2" ParentId="25127" CreationDate="2018-12-17T21:24:52.447" Score="1" Body="&lt;p&gt;You would need to calculate &lt;span class=&quot;math-container&quot;&gt;$ \ I_{xy} \ for \ the \ angle\ \alpha $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can apply the torque &lt;span class=&quot;math-container&quot;&gt;$F2*(d2^2+L^2)^{1/2}$&lt;/span&gt;, which is F2 working around the new axis for &lt;span class=&quot;math-container&quot;&gt;$I_{xy}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One easy way of calculating&#xA;&lt;span class=&quot;math-container&quot;&gt;$I_{xy} \  $&lt;/span&gt;is by constructing the Mohr circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the x axis you draw the circle at the centre of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ Ix - (Ix - Iy)/2 \ and\ radius\ of: \ (Ix- Iy)/2  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you draw an angle equal to &lt;span class=&quot;math-container&quot;&gt;$ 2*\alpha \ $&lt;/span&gt; counter clockwise from Iy on x axis of the Mohr circle and draw the diameter by continuing this angle to intersect the other side of the circle. The projections of these two points on y and x axis are your new &lt;span class=&quot;math-container&quot;&gt;$I_{xy}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-12-18T07:12:41.667" LastActivityDate="2018-12-18T07:12:41.667" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25136" PostTypeId="1" CreationDate="2018-12-17T21:47:23.273" Score="4" ViewCount="8553" Body="&lt;p&gt;My question may sound silly as it’s mainly good sense but I may be mixing things up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s consider a flexible cantilever beam of a given length and material is fixed at one end and free at the other end. &#xA;Let’s suppose a dead weight is put at the free end.&#xA;This results in a bending moment at the fixed end of say 10 Nm and a given deflection at the free end of say 1 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let’s suppose the beam thickness gets increased and as a result stiffer. The deflection at the tip will be reduced for sure, but what about the bending moment at the fixed end? Is making the beam more rigid going to reduce the bending moment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking forward to your thoughts!&#xA;Many thanks &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thanks for you answers, much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get that the bending moment caused by a static load applied at the free-end does not depend on the elasticity of the beam. Only the deflection does. I would have thought that the greater the deflection the higher the bending moment... Can you confirm this isn't true? Imagine a 100 m long vertical monopile support structure looking like a slender cylinder anchored in the seabed and subjected to dynamic wave loadings exciting all the cylinder modes. The thinner the cylinder wall is, the more elastic the cylinder gets and the greater the deflections at the top get. I would think that the maximum bending moment would be reached when the deflection is the most significant and therefore making the cylinder more rigid with thicker wall would both reduce the deflections and the max bending moment, wouldn't it?&lt;/p&gt;&#xA;" OwnerUserId="18672" LastEditorUserId="1832" LastEditDate="2018-12-18T21:43:52.870" LastActivityDate="2018-12-19T02:37:30.477" Title="How does stiffness/rigidity affect the bending moment of a beam" Tags="|mechanical-engineering|stresses|beam|stiffness|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25137" PostTypeId="2" ParentId="25136" CreationDate="2018-12-17T22:15:19.403" Score="4" Body="&lt;p&gt;There are two basic types of structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statically determinate structures&lt;/strong&gt; are those where you can calculate the forces at the restraints without knowing anything about the flexibility of the structure itself (you assume it is strong enough to carry the loads without breaking, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Statically indeterminate&lt;/strong&gt; or &lt;strong&gt;redundant structures&lt;/strong&gt; are those where the way the loads are transferred to the restraints depend on the relative stiffness (or flexibility) of the different parts of the structure. A simple example of a statically indeterminate structure would be a heavy rectangular box resting on a table. The &lt;em&gt;total&lt;/em&gt; force exerted on the table top is simply the weight of the box, but the way the force is &lt;em&gt;distributed over the area the box is resting on&lt;/em&gt; depends on the flexibility of both the box and the table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cantilever beam fixed at one end and loaded at the other end is statically determinate. For most beam structures you can construct a &lt;a href=&quot;https://en.wikipedia.org/wiki/Shear_and_moment_diagram&quot; rel=&quot;noreferrer&quot;&gt;shear force and bending moment diagram&lt;/a&gt; which shows the forces and moments everywhere along the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the &lt;em&gt;deflection&lt;/em&gt; of the beam depends on its flexibility, but the distribution of the internal forces on it does not, and the forces don't change for different &quot;shapes&quot; of the beam, e.g. changes in cross section along it lengths, or sections made from different materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Statically determinate structures are very convenient for making engineering designs, because the design process splits into two parts: first you calculate the internal forces everywhere in the structure, and the you can select the &lt;em&gt;size&lt;/em&gt; of the structure to support those loads with acceptable stress levels (to avoid failure) and sufficiently small deformations. For example structures like the trusses which support the roof of buildings are not &quot;perfectly&quot; statically determinate, but they are close enough that they can be designed assuming they are.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-12-17T22:15:19.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25138" PostTypeId="1" AcceptedAnswerId="25142" CreationDate="2018-12-17T23:01:15.587" Score="3" ViewCount="1527" Body="&lt;p&gt;For building something like a big building or a parking garage, there is a need for a large flat surface of concrete. Wondering how they make the dirt and the concrete flat. It's hard to make a tiny flat surface at the beach in sand let along thousands of square feet. Wondering what sorts of machines tools or techniques are used at a high level.&lt;/p&gt;&#xA;" OwnerUserId="18674" LastActivityDate="2018-12-18T10:47:56.100" Title="How land is made flat at scale" Tags="|surface-preparation|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25139" PostTypeId="2" ParentId="25128" CreationDate="2018-12-17T23:45:08.887" Score="0" Body="&lt;p&gt;It looks like you need a = 118.1978.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came up with this using the formulas on &lt;a href=&quot;https://www.calculatorsoup.com/calculators/geometry-plane/polygon.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;.  A section through your handle looks like part of a regular polygon with 12, 24, or 48 sides.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use the calculator on that site, but you have to make a minor adjustment because your first segment is perpendicular to the surface (so if you placed a mirror image below it you would have two adjacent sides with no angle in between).  You can account for this by rotating their illustration by 1/2 the angle of each segment, and using r instead of R in their computation, and then adding half the side length to the answer.  In other words, their calculated r + a/2 = height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this approach, I perfectly reproduced all four of your results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate a, given the desired height is a little more complicated because of the modification I described above.  In general, you can do it by solving two equations simultaneously: a = 2 r tan(pi/n) and r + a/2 = height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging in your desired height = 508, and n = 24, I get the answer a = 118.1978.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here's a sketch that explains why r + a/2 = 508 mm:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/p6nzN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p6nzN.png&quot; alt=&quot;sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6284" LastEditorUserId="6284" LastEditDate="2018-12-20T00:30:57.120" LastActivityDate="2018-12-20T00:30:57.120" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25140" PostTypeId="2" ParentId="25138" CreationDate="2018-12-18T00:28:03.027" Score="3" Body="&lt;p&gt;&lt;a href=&quot;https://www.cat.com/en_US/products/new/equipment/dozers/large-dozers/18500099.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tear it up:&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kx7Tw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kx7Twm.jpg&quot; alt=&quot;dozer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cat.com/en_US/products/new/equipment/motor-graders/motor-graders/1000034758.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Smooth it out:&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/s7Xyq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s7Xyqm.jpg&quot; alt=&quot;grader&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;John Deere makes similar equipment, too.  See &lt;a href=&quot;https://www.deere.com/en/dozers/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://www.deere.com/en/motor-graders/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concrete is either poured in sections or pre-cast and delivered on-site.  So the traditional techniques still work at scale.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2018-12-18T00:28:03.027" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25142" PostTypeId="2" ParentId="25138" CreationDate="2018-12-18T02:20:54.807" Score="5" Body="&lt;p&gt;For big constructions, there is a survey plan and a few bench marks marked and cemented by the surveyor. Then at the time of excavation there are auto rotating laser levels that cast a level moving beam which is easy to see and follow. there are reflecting mirrors and elevation readers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excavation is done in stages, from rough to fine contours using different heavy machinery depending on the type and configuration of the site. Many jobs call for presence of soils engineer at intervals to test the soils and determine their properties and if they match the project's soils report.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concrete slabs can be leveled with vibrating, moving, or, rotating screeds or operator driven motorized finishers or manually. There are plastic stakes that mark the finish level and the have fittings for long aluminum levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structural drawings usually call for 1/8&quot; to 1/16&quot; tolerance in slabs level.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-12-18T07:19:53.703" LastActivityDate="2018-12-18T07:19:53.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25143" PostTypeId="2" ParentId="25136" CreationDate="2018-12-18T02:42:14.133" Score="2" Body="&lt;p&gt;In your example the change in the cross section of the beam doesn't have any effect on the end moment even if the beam is a hollow section such as a pipe as long as the ratio of length to depth is greater than 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the beam is very deep and this ratio is less than 10, shear deformation and web warping effect could change the picture. We follow Euler- Bernoulli beam theory which is a great simplification of linear elasticity albeit a genius one. And that theory is very effective predictive of stresses when the beam is slender.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See a brief history of the theory &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;here.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-18T02:42:14.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25144" PostTypeId="2" ParentId="25129" CreationDate="2018-12-18T03:05:04.950" Score="1" Body="&lt;p&gt;It makes sense to stream line the shape of connection at the end so as to eliminate the part that is not in the path of stress for two reasons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A - In a situation of dynamic loading such as a truss in a bridge under impact of vibration of passing traffic the part which is not sharing the stress, does not strain with the rest of the structure either, and in the long run this sea-saw effect causes fatigue in the connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B - In a large structure with many complex joints removing excessive material leads to easier inspection and maintenance of the structure and lighter weight, think about power towers, bridges, stadium roofs, etc.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-12-18T19:09:48.543" LastActivityDate="2018-12-18T19:09:48.543" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25145" PostTypeId="2" ParentId="25138" CreationDate="2018-12-18T05:39:21.810" Score="2" Body="&lt;p&gt;Regarding Kamran's citation of laser leveling, note that this technique is also used to produce rice fields covering tens to hundreds of acres with very precise drainage slopes. In this case, the earth moving machine receives the light flashes from the rotating laser mirror and lifts or lowers the grading blade in response. The machine operator then simply drags the grader blade back and forth across the field to be leveled and as long as the laser can be seen by the grader, the required slope is produced. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-18T05:39:21.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25146" PostTypeId="1" CreationDate="2018-12-18T07:30:23.400" Score="-1" ViewCount="67" Body="&lt;p&gt;Is there any method or formula that can be used to find the operating voltage of a coil using the resistance of the coil? If so, is it also applicable to AC coils or any other technique?&lt;/p&gt;&#xA;" OwnerUserId="18678" LastEditorUserId="90" LastEditDate="2018-12-18T17:12:10.810" LastActivityDate="2022-06-01T04:07:25.507" Title="DC relay relation with resistance" Tags="|electromagnetism|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25147" PostTypeId="2" ParentId="17106" CreationDate="2018-12-18T08:07:52.000" Score="2" Body="&lt;p&gt;An excellent question.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The short answer is - it wouldn't know what materials to bind to the objects - and that's a headache for many.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Revit has a set of textures it uses to render - but there's no clear existing path on how to neatly export those textures into a non-autodesk program (or there wasn't when I worked on the issue a year+ ago).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even exporting into another autodesk program, like Maya for example - wasn't a smooth process - there was a provided way to link up the programs so textures would go across correctly into the maya environment )  but I recall it being error-prone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there are some more obscure ifc standards for specifying specific files - but Revit doesn't (or didn't) use them when exporting to ifc.&lt;/p&gt;&#xA;" OwnerUserId="18679" LastActivityDate="2018-12-18T08:07:52.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25148" PostTypeId="1" AcceptedAnswerId="25159" CreationDate="2018-12-18T08:29:00.177" Score="1" ViewCount="29" Body="&lt;p&gt;Anyone tried mounting thermal imaging camera on ordinary optical microscope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it is unlikely to work. Because of different refraction index and possible absorption of IR radiation by the lens.&lt;/p&gt;&#xA;" OwnerUserId="18680" LastEditorUserId="18680" LastEditDate="2018-12-18T09:49:09.653" LastActivityDate="2018-12-18T23:21:53.820" Title="Mount thermal imaging camera on ordinary optical microscope?" Tags="|thermal-radiation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25150" PostTypeId="2" ParentId="25138" CreationDate="2018-12-18T10:47:56.100" Score="0" Body="&lt;p&gt;Easier done than thought.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Grading&quot; and &quot;levelling&quot; are carried out in stages of approximation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A grading machine or a common dozer can work up an even surface (horizontal or uniform gradient).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the good question, &quot;It's hard to make a tiny flat surface at the beach in sand let alone thousands of square feet. Wondering what sorts of machines tools or techniques are used at a high level.&quot;: Remember that the surface of a liquid is &lt;strong&gt;always&lt;/strong&gt; horizontal. A layer of concrete usually evens out the minor variations (bumps and pits), giving you a horizontal &quot;flat&quot; surface.  &lt;/p&gt;&#xA;" OwnerUserId="18682" LastActivityDate="2018-12-18T10:47:56.100" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25153" PostTypeId="2" ParentId="25124" CreationDate="2018-12-18T16:42:47.943" Score="0" Body="&lt;p&gt;I am having a really hard time following your setup. First, you have the x-axis as the vertical axis, and the y-axis as the horizontal axis. Is this correct? And then there's a... surface?... when x = 0, and that surface has friction, so that's what's causing your [m2] to rotate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically you're dragging a block on the ground at x=0, is that correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if your &quot;hoist&quot; drive, [1], is actually at zero, which we'll say means that [ux] is actually the length from the surface (x=0) to the trolleying platform [3], then you can't actually have the load [2] drag the surface because that would imply that you have something supporting the load [2] that isn't your hoist drive [1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your picture, your load [2] has some width, and you appear to be dragging it about the corner. If you're actually doing this - your load [2] is actually pivoting about that forward corner - then you now need to take into account the rotation of the load [2], meaning also its moment of inertia, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe it's not actually pivoting, but it's continuing to drag because your hoist drive [1] is &quot;springy&quot; and stretches to allow the load [2] to drag? But the especially confusing thing here for me is that you're asking how you solve for the dynamics of ux and uy, despite the fact that you have &lt;em&gt;actuators&lt;/em&gt; on both of those axes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have linear drives on ux and uy, then those dynamics are whatever you make them to be. The only thing you have left to solve for is how &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is a function of everything else. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2018-12-18T16:42:47.943" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25154" PostTypeId="2" ParentId="25146" CreationDate="2018-12-18T18:16:33.783" Score="0" Body="&lt;p&gt;Most relay coils carry a specification on how much current must be applied in order to pull in the contacts. Knowing this and the coil resistance, ohm's law then tells you what voltage will be necessary. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-18T18:16:33.783" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25155" PostTypeId="2" ParentId="25136" CreationDate="2018-12-18T19:22:00.037" Score="3" Body="&lt;p&gt;As mentioned by other answers, when dealing with a statically determinate structure, the stiffness of each element is irrelevant when calculating the bending moment, but a key variable when calculating the deflection. Meanwhile, for statically indeterminate structures, even the calculation of bending moment requires the stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simplified terms, this is because in statically determinate structures one can determine how the load is transferred through the structure to the supports without caring about the stiffness itself. In indeterminate structures, however, the stiffness directly impacts how the load is shared by the supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's try a variation on your cantilever example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WmO8Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WmO8Z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's just a cantilever with a force applied at the free end. There's also a dangling vertical beam at the end (which is unsupported at the bottom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I don't need to know anything about the stiffness of each of those beam segments to determine how the load will be divided between them. We trivially know that dangling beam won't do anything, and all the load will be resisted by the cantilever. After all, that dangling beam might be infinitely stiff, but it's a dead end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine you get the thinnest possible steel wire and use it to connect the bottom of the dangling beam to the support (with fixed connections, not pinned). Your structure diagram then becomes something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vRtQU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vRtQU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, given how weak that new diagonal &quot;beam&quot; is (just a tiny wire), we can probably approximate its stiffness as equal to zero (especially when compared to the stiffness of an actual steel beam). Which basically means we can pretend the diagonal doesn't exist. Therefore, the structure will actually behave exactly as it did in the original case: the entire load will be resisted by the horizontal beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, that was easy enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now imagine the diagonal and horizontal members were reversed: the diagonal is the steel beam and the horizontal is just a pitiful wire. In this case, we can obviously state that the load will travel down the vertical (previously dangling) beam and then up the diagonal towards the support, with the tiny wire not really doing much of anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if both the horizontal and diagonal beams are steel beams, and therefore both contribute to resisting the load? Well, then we can't trivially figure this out anymore.&lt;sup&gt;1&lt;/sup&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's why calculating the shear force and bending moment of statically indeterminate structures depends on the elements' stiffness: each element's stiffness defines how much load it must support. And the proportion of the load going to each element is directly proportional to its stiffness: the stiffer the beam (as compared to others sharing the load), the more load it must support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your follow-up question, the answer is simply &quot;no&quot;. Just think of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;fundamental beam equation&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dfrac{\partial^2}{\partial x^2}\left(EI\dfrac{\partial^2 w}{\partial x^2}\right) = q$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This tells us that the first integral of the loading is the shear force, the second integral is the bending moment, the third integral is the angle of rotation times the stiffness, and the fourth integral is the deflection times the stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the greater the load applied on a beam, the greater the shear force and bending moment will be and therefore, the greater the deflection. But if you change the beam's stiffness (and the applied loading remains the same), the bending moment will remain the same, but the deflection will change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you double a beam's stiffness, how do you expect its deflection to behave?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is that if you double the beam's stiffness, then the deflection will be halved. This is consistent with what I've said: the loading remained the same, therefore so did the shear force (first integral of loading) and bending moment (second integral of loading) diagrams. However, since the third and fourth integrals of loading are equal to the product of stiffness and rotation or deflection, the resulting rotation/deflection is halved if the stiffness is doubled (basically, &lt;span class=&quot;math-container&quot;&gt;$EIw = (2EI)(w/2)$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your idea were correct, then doubling the stiffness would lead to another answer: the bending moment diagram itself would change (say, decrease by half), and then the deflection due to that reduced moment would itself be reduced by half due to the stiffness coefficient when integrating from moment to rotation and deflection.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;This is a bit off-topic, so putting this as a footnote. To actually calculate this case, you need to use compatibility equations which effectively guarantee that the diagonal's deflection at the free end is equal to the horizontal's deflection at its free end plus the vertical's compressive deflection. Basically equations which guarantee that everyone agrees where (and at which angle of rotation) the nodes end up in the structure's deflected shape.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2018-12-19T02:37:30.477" LastActivityDate="2018-12-19T02:37:30.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25157" PostTypeId="2" ParentId="24867" CreationDate="2018-12-18T20:44:58.737" Score="0" Body="&lt;p&gt;Sorry I missed this over at &lt;a href=&quot;https://robotics.stackexchange.com&quot;&gt;Robotics&lt;/a&gt;. Your question is really unclear and you don't have a good diagram that helps it out at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You say there is no gravitational field, there are no springs, there is no body flexure, there is no friction - &lt;em&gt;it's not clear what dynamics you're expecting to get out of this problem!&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Even beyond that, the remainder of the premise isn't clear. You mention in &lt;code&gt;Question 1&lt;/code&gt; that the spacecraft is link 1, but then what is pinning link 1? Earlier in the text of your question, you said: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For inspiration, I looked at modeling my spacecraft + single-link robot system as if it were a two-link robot system with a single revolute joint connecting link 1 with link 2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but then you go on to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For link 1, the kinetic energy due to the linear motion of Link 1's center of gravity is constrained to circular motion because it is attached to the base at joint 1.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Again, you start out saying there's only one joint, then it's two joints and a fixed platform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the Question 1 question itself: if link 1 has a velocity, and link 2 is connected to link 1, then link 2 also has that velocity. The only difference in velocities would be the result of any motion in the joint between links 1 and 2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding question 2, there is no &quot;rotational energy&quot; exactly, there's &lt;em&gt;kinetic&lt;/em&gt; energy and &lt;em&gt;potential&lt;/em&gt; energy. Any rotational kinetic energy gets grouped with the other kinetic energy terms. That is, link 1 has the sums of its kinetic energies and link 2 has the sums of its kinetic energies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you don't actually have a fixed platform (or do you?), your link 1 isn't going to rotate about the first joint. Instead, the entire system rotates about the system's center of mass. The location of that center of mass then depends on the ratio of masses and the joint angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe, as I mention at the start, you would be better served to clarify what you're trying to study and use that as a tool to guide the rest of your approach. I think you should be looking to apply conservation of linear momentum and conservation of angular momentum, for the system as a whole, and then try to define your reference datum and other terms in a manner that best supports that approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't help much more than this because again I'm not sure that I follow the exact setup for the question. I will say, though, that I think continuously re-defining your reference datum (relative to m1 or m2) is a mistake. The &lt;strong&gt;system&lt;/strong&gt; has a center of mass. The &lt;strong&gt;system&lt;/strong&gt; has its angular and linear momentum conserved, so you should probably look to define everything else relative to the center of mass of the system. Then you should start finding things like translational velocities are just ratios of masses, etc., because you have to keep the vertical, horizontal, and angular components of momentum the same. These constraints (along with the joint kinematic constraint) are how you're going to solve the problem. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2018-12-18T20:44:58.737" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="25159" PostTypeId="2" ParentId="25148" CreationDate="2018-12-18T23:21:53.820" Score="0" Body="&lt;p&gt;The glass used in visible light microscopes does not pass much IR. For strong signals you need calcium fluorite lenses, which are expensive and rare. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-18T23:21:53.820" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25160" PostTypeId="1" AcceptedAnswerId="25196" CreationDate="2018-12-18T23:34:52.410" Score="1" ViewCount="1492" Body="&lt;p&gt;I'm looking at googles mocktails maker. And it puts 2 holes in a container. 1 hole is connected to a pump that will take water out and the other connects to a backflow prevention valve that only allows air in and not liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the point of having that valve? It doesn't connect to anything. Also when I test it with and without having that valve the water does the exact same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference:&#xA;&lt;a href=&quot;https://github.com/deeplocal/mocktailsmixer#step-19-attach-tubing-and-check-valve&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://github.com/deeplocal/mocktailsmixer#step-19-attach-tubing-and-check-valve&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/reknM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/reknM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Does it have to do with keeping the container pressurized? Why would I need to have a constant pressure if the pump doesn't require it.&lt;/p&gt;&#xA;" OwnerUserId="18690" LastEditorUserId="18690" LastEditDate="2018-12-19T00:54:00.010" LastActivityDate="2022-10-03T08:58:10.793" Title="What is the point of this valve that only allows air in and not liquid?" Tags="|mechanical-engineering|pumps|airflow|valves|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25161" PostTypeId="1" CreationDate="2018-12-19T00:12:59.840" Score="1" ViewCount="203" Body="&lt;p&gt;I'm asking if I'm doing this correctly.&#xA;I've estimated the thrust of a ducted-fan model aircraft neglecting pressure changes across the duct as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F = Q * \rho * (V_e - V_0)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$= Q * rho * (Q/A_e - Q/A_0)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  = 131 [cu. ft. / sec] * 0.00237 [slug/cu. ft] * (131/0.74 -131/0.92)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$  = 0.31 * (177.03 - 142.4)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$= 0.31 * 34.6$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$= 10.7 lbf$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;F = Thrust, &lt;/li&gt;&#xA;&lt;li&gt;Q = volume flow rate of air = 131 [cu. ft. / sec], &lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; = density of air = 0.00237 [slug / cu. ft], &lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$A_0$&lt;/span&gt; = an circular inlet area = 0.92 [sq. ft], &lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$A_e$&lt;/span&gt; = an exit area = 0.0.74 [sq. ft]. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;That seems an awfully high value for thrust to me. Am I missing something?&lt;/p&gt;&#xA;" OwnerUserId="18691" LastEditorUserId="18691" LastEditDate="2018-12-24T16:30:43.027" LastActivityDate="2018-12-24T16:30:43.027" Title="Am I estimating thrust of a ducted fan correctly?" Tags="|fluid-mechanics|modeling|thrust|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25162" PostTypeId="1" AcceptedAnswerId="25168" CreationDate="2018-12-19T00:21:14.080" Score="1" ViewCount="48" Body="&lt;p&gt;I'm trying to make a quick release male and female connector similar to the one shown here.&lt;a href=&quot;https://i.stack.imgur.com/JjbiS.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JjbiS.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One connector is attached to the bottom of a container filled with liquid. When it is not connected it is closed and none of the liquid can flow out of it. However when it is in contact with another connector it then opens and allows the liquid to flow out of it. Any ideas where these kind of connectors can be found or made?&lt;/p&gt;&#xA;" OwnerUserId="18690" LastActivityDate="2018-12-19T02:05:44.653" Title="Is there a male and female valve that is closed until connected together?" Tags="|mechanical-engineering|valves|liquid|connections|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25163" PostTypeId="2" ParentId="25136" CreationDate="2018-12-19T00:39:09.360" Score="3" Body="&lt;p&gt;About the 2nd question after you read the answers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually it's the other way around. Imagine a load on your beam. The integration of the loading is the shear force. The integration of the shear force is the moment. The angle of deflection is the integration of the moment divided by E*I (this is where the material kicks in, E is Young's modulus and I is the second moment of area of beam's section). Then you integrate the angle of deflection to find the vertical deflection. (This may be helpful: &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Direct_integration_of_a_beam&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Direct_integration_of_a_beam&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a statically determined system, axial &amp;amp; shear forces along with bending and torsional moments are all unrelated to your material or your section (considering been in the elastic region, small deflections etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest moment is not at the point where the biggest displacement is, in a double fixed beam, the supports have the biggest moment but the deflections there are 0.&lt;/p&gt;&#xA;" OwnerUserId="18692" LastActivityDate="2018-12-19T00:39:09.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25164" PostTypeId="2" ParentId="24605" CreationDate="2018-12-19T01:18:28.860" Score="0" Body="&lt;p&gt;Standards use simplified method of defining failures most times based on experimental data. On that other hand, FEA Software try to simulate exactly what happens in your material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not familiar with abacus but I am familiar with FE and other programs. In order to trigger failure, you have to reach the stress of your yield criterion. I am not really sure if you can change your criterion for each direction thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An other way is to cheat by assuming an anisotropic material, rasing enough your material's stiffness of one of the axis will eventually trigger failure in the other axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were you, I would first try to connect the failure in the other axis with a different part of the standard (that failure should be checked somewhere). Then setup the problem in a way that your main axis will be the crucial.&lt;/p&gt;&#xA;" OwnerUserId="18692" LastActivityDate="2018-12-19T01:18:28.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25165" PostTypeId="2" ParentId="17099" CreationDate="2018-12-19T01:26:52.653" Score="0" Body="&lt;p&gt;In the real world this is the &quot;real&quot; stiffness because nothing is perfect and homogeneous. We just simplify things assuming plane stress or stain because we don't have the whole picture or it's not easy to get it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example for you to think about is wood. Wood has different stiffness in each direction (anisotropic material).&lt;/p&gt;&#xA;" OwnerUserId="18692" LastActivityDate="2018-12-19T01:26:52.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25166" PostTypeId="2" ParentId="17369" CreationDate="2018-12-19T01:41:42.690" Score="0" Body="&lt;p&gt;Since you only have the ratio, and you are programming it, you have to assume an &lt;span class=&quot;math-container&quot;&gt;$E \times l$&lt;/span&gt; for your column or beam and then apply the ratio for the other one (using 1 will do).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, to define your stiffness matrix you will also need a ratio for your section areas. This is why I would suggest picking 2 realistic sections and approximately find that ratio for each stiffness ratio.&lt;/p&gt;&#xA;" OwnerUserId="18692" LastEditorUserId="16486" LastEditDate="2019-01-20T04:16:00.350" LastActivityDate="2019-01-20T04:16:00.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25167" PostTypeId="2" ParentId="16311" CreationDate="2018-12-19T01:49:33.023" Score="0" Body="&lt;p&gt;If I remember correctly, first, you do the static condensation and then you rotate the degrees of the roller appling the transformation only on those. So it's like going from your roller local axis to the local axis of you element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, getting rid of degrees in the global system will be the same with doing it locally and then transforming it.&lt;/p&gt;&#xA;" OwnerUserId="18692" LastActivityDate="2018-12-19T01:49:33.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25168" PostTypeId="2" ParentId="25162" CreationDate="2018-12-19T02:05:44.653" Score="2" Body="&lt;p&gt;There are many, many different commercially-available designs of fluid connectors that shut off automatically when disconnected and open up when connected. They are available in plastic and metal, permanent and disposable, in all sorts of sizes. Try a search on &quot;make-and-break fluid connectors&quot;.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-19T02:05:44.653" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25171" PostTypeId="1" CreationDate="2018-12-19T07:07:44.447" Score="0" ViewCount="283" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hP9lU.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand why the pressure fluctuations during a hydraulic hammer phenomenon would go below the line pressure (Pmin &amp;lt; Pi) I thought there will just be a rise in pressure and then the pressure would came back to the settling pressure but thats not the case as the pressure seems to go well below the settling pressure too (which could cause cavitation if the pressure drops below vapor pressure). I know there are multiple peaks because of the pressure wake being reflected back and fourth in the hydraulic line (and that the pressure at any point in the pipe at any given time is a superposition of all the wave, incident and reflected, at the point). &#xA;I am not able to understand how the pressure can go below the line pressure as surge is suppose to be a compression wave. The Pmin suggest that a expansion wave is travelling behind the compression wave. Any material to understand this better would also be helpful.&lt;/p&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="16275" LastEditDate="2018-12-19T16:40:46.350" LastActivityDate="2018-12-19T19:31:31.073" Title="Why does pressure during a hydraulic transient phenomenon like fluid hammer goes below the line pressure?" Tags="|hydraulics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25172" PostTypeId="1" CreationDate="2018-12-19T08:19:08.403" Score="0" ViewCount="805" Body="&lt;p&gt;I'm working in an automotive company that must follow &lt;a href=&quot;https://www.iatfglobaloversight.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;IATF&lt;/a&gt; (International Automotive Task Force) requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I administrate a test-bench that performs RF (radio frequency) tests using an appropriate measure instrument.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As IATF requires, the test-bench has been successfully validated using MSA (Measurement System Analysis) procedures for the production process where the test is performed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After one year without changes both in production process and test-bench measure chain the instrument must be calibrated (and adjusted accordingly if necessary) from the producer (calibration lasts one year).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IATF requires that every change in the process must be verified with the MSA. The problem is that &quot;change&quot; is too much generic, and I can't find the proper IATF documentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I to re-perform MSA after the calibration/adjustment of the instrument?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Facts are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the instrument is the same as before (same serial number) or is the same model (different serial number)&lt;/li&gt;&#xA;&lt;li&gt;the instrument is fresh calibrated (like when I performed the MSA first validation)&lt;/li&gt;&#xA;&lt;li&gt;the instrument is plugged out, calibrated or replaced with a calibrated one (same model) and than plugged in. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;so, my opinion is that the measurement test chain is the same as before so I don't need to provide MSA documentation to auditors. The calibration + optional adjustment process don't modify the instrument, it restores the instrument to the conditions of the previous calibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not able to find official IATF documentation or others sources about that: does calibration + optional adjustment procedure modify the measure process?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the calibration definition from the MSA &lt;a href=&quot;https://www.aiag.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;AIAG&lt;/a&gt; manual:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A calibration system is a set of operations that establish, under specified&#xA;  conditions, the relationship between a measuring device and a traceable&#xA;  standard of known reference value and uncertainty. Calibration may also&#xA;  include steps to detect, correlate, report, or eliminate by adjustment any&#xA;  discrepancy in accuracy of the measuring device being compared.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;so, by definition, a calibration process is a check between instrument actual characteristics and nominal characteristics that comprises adjustments to align real and nominal characteristics. In any case after this process real characteristics are compliant to nominal.&lt;/p&gt;&#xA;" OwnerUserId="18695" LastEditorUserId="18695" LastEditDate="2018-12-20T15:03:43.750" LastActivityDate="2024-02-01T17:01:05.110" Title="Must MSA (Measurement System Analysis) be re-performed after annual instrument calibration?" Tags="|automotive-engineering|measurements|statistics|product-testing|calibration|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25174" PostTypeId="1" CreationDate="2018-12-19T11:31:34.247" Score="3" ViewCount="87" Body="&lt;p&gt;In the semiconductor industry and research, silicon wafers are transported in round plastic sample containers with a screw lid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WstiXm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WstiXm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than any other screw top though, the lid always opens (releases) when turning in the &lt;em&gt;clockwise&lt;/em&gt; direction, as opposed to e.g. a bottle cap that opens counterclockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a practical reason for this? Why are they made like this even by different brands?&lt;/p&gt;&#xA;" OwnerUserId="844" LastActivityDate="2021-01-08T06:10:00.760" Title="Why do transport containers for wafers open clockwise?" Tags="|design|semiconductor|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25179" PostTypeId="1" CreationDate="2018-12-19T16:31:52.710" Score="1" ViewCount="66" Body="&lt;p&gt;I'm trying to create a mechanical setup to transfer motion from a motor to two shafts. I need a method to do it that fits the following criteria to the greatest extent possible:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does not slip&lt;/li&gt;&#xA;&lt;li&gt;Does not introduce vibration &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It's not a strenuous job with a lot of torque behind it. None the less I did quite like the idea of the chain or timing belt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is twofold&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) Would a chain or timing belt introduce extra vibration to the system compared with a v belt?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B) Is there a way to get V belts slip free?&lt;/p&gt;&#xA;" OwnerUserId="18703" LastActivityDate="2018-12-19T21:19:40.880" Title="Chain, V pulley, Timing Belt" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25180" PostTypeId="2" ParentId="25066" CreationDate="2018-12-19T16:46:41.623" Score="0" Body="&lt;p&gt;It depends on what you're casting. For a smaller castings with thin sections, the amount of heat that needs to be absorbed is low and the surface area and casting to mold volume ratio is high. So a chunky solid copper die may absorb the heat quickly enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the heat transfer slows as the die heats up (temperature difference decreases). If the casting is a larger block, a lot more heat has to diffuse through a smaller surface area and as the surface of the die heats up, the cooling rate reduces. In this case having water that can take the heat and move it away from the die surface (by pump or convection), allows the heat to be distributed throughout the cooling system more evenly, keeping the temperature difference high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also use chillers or cool the mould to increase the temperature difference and speed up the cooling, which will give you more nucleation points when solidifying and a smaller grains. But an increased cooling rate often actually decreases the strength due to increased porosity and other casting defects and may not even fill the die properly (I have seen people pre-heat moulds to slow down the cooling rate and get a better casting). Increasing the strength is really best done by alloying with other elements.&lt;/p&gt;&#xA;" OwnerUserId="18382" LastActivityDate="2018-12-19T16:46:41.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25181" PostTypeId="2" ParentId="25113" CreationDate="2018-12-19T18:11:19.130" Score="0" Body="&lt;p&gt;I needed a &quot;shaft gland&quot; / &quot;gland seal&quot; used in boats (and for my application, radio controlled boats&quot; to solve the problem.&lt;/p&gt;&#xA;" OwnerUserId="18649" LastActivityDate="2018-12-19T18:11:19.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25182" PostTypeId="1" CreationDate="2018-12-19T18:11:28.230" Score="1" ViewCount="344" Body="&lt;p&gt;I'm having trouble understanding the plot attached. The bode plot here shows the bode plot of a plant to be controlled with PID. So, if I want a phase margin of 30 degrees at the bandwidth frequency, then as shown on the picture, based on the formula (PM=180 + ...) my bandwidth after PID controller is applied is 30 hz, with a phase margin of 30 degrees at that bandwidth frequency (30 hz). Say now I want a phase margin of 50 degrees. From calculation this will then occur at the frequency corresponding to phase of -130 degrees, in the phase plot of the plant. But this is weird then; I will obtain the same frequency if I want, say, 60 degrees of phase margin (i.e. phase in plant of ~-110 degrees)! So the same frequency for different possible phase margins? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused.... Is it then the bandwidth that can give this phase margin is 0 i.e. not realisable? So there is a maximum phase margin that can be realised?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p9FR1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p9FR1.jpg&quot; alt=&quot;Bode plot of plant that wants to be PID controlled&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16854" LastActivityDate="2018-12-19T18:11:28.230" Title="Is there a maximum phase margin a PID controller can give?" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25183" PostTypeId="2" ParentId="5750" CreationDate="2018-12-19T18:16:30.573" Score="3" Body="&lt;p&gt;There are several options for doing this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If there are only a few values to be saved per design point, you could use 'output parameters'.  &lt;/li&gt;&#xA;&lt;li&gt;If there is a lot of data to be saved, the report generator may be useful. Some information can be found here: &lt;a href=&quot;https://www.sharcnet.ca/Software/Ansys/17.2/en-us/help/wb2_help/wb2h_projectreports.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Working with Project Reports&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;An APDL snippet can be used in the results* tree. Some information here: &lt;a href=&quot;http://www.padtinc.com/blog/tag/design-points&quot; rel=&quot;nofollow noreferrer&quot;&gt;Saving Mechanical APDL Plots in a Design Study&lt;/a&gt;    &lt;/li&gt;&#xA;&lt;li&gt;IronPython may be used. This requires you are an expert in Ansys and know programming well.  &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;See also &lt;a href=&quot;https://caeai.com/sites/default/files/automated_postprocessing.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Path Plots and Automated Post Processing in MADPL and Workbench&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Be careful to place the snippet in the results section of the model, otherwise it will run before in stead of after solving. &lt;/p&gt;&#xA;" OwnerUserId="17524" LastEditorUserId="17524" LastEditDate="2018-12-19T20:56:49.717" LastActivityDate="2018-12-19T20:56:49.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25184" PostTypeId="2" ParentId="25171" CreationDate="2018-12-19T19:31:31.073" Score="1" Body="&lt;p&gt;The negative pressure excursions occur because the moving fluid in the system possesses inertia. This means it wants to keep moving beyond its equilibrium position even after the driving force has gone to zero. Here is an easy way to visualize how this creates negative excursions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine we have a very, very long gun barrel loaded with an explosive charge and a tight-fitting projectile. We set off the explosive and the pressure in the barrel behind the projectile goes positive, and the projectile starts to move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it moves, the hot gas behind it expands and its pressure diminishes. Now we have set the projectile into motion and even though the positive driving pressure will eventually fall to atmospheric, the projectile remains in motion at that point and begins pulling a vacuum behind it where there once was positive pressure. This vacuum now acts to slow down the projectile. Eventually the projectile is brought to a halt, but as there is now a hard vacuum behind it, its direction of travel reverses and it slides back towards its starting position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the projectile causes it to overshoot its equilibrium point on each cycle and it exhibits harmonic oscillation back and forth until friction bleeds off the kinetic energy and the projectile stops. This is how a transient impulse can create negative pressure excursions.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-19T19:31:31.073" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25185" PostTypeId="1" CreationDate="2018-12-19T21:02:13.323" Score="0" ViewCount="56" Body="&lt;p&gt;With two gimbals, the angles of the gimbals can effectively move to allow the flywheel to point in any direction with respect to the frame. This much is clear to me, since the flywheel direction is only one vector (as opposed to a coordinate system) and you only need two angles, like yaw and pitch or spherical coordinates, to rotate a vector to any direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the third gimbal has to do with the fact that, because the flywheel's direction is 'just' a vector, the flywheel's direction with respect to the frame orientation is not unique. You could rotate a given local coordinate system around the flywheel direction and every rotation would be a possible orientation with respect to the global coordinates (for a given flywheel direction in local coordinates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not clear in how a third gimbal would fix that, though. Although in my head I can understand that yaw-pitch-roll rotation steps each occur by rotating one of the gimbals, and that together they give a well-defined orientation, I can't seem to reconcile that with the aforementioned fact that the frame's orientation should not be uniquely determined by the flywheel's direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that issue, as I suspect, have to do with the concept of gimbal lock? (which I am also having a bit of trouble fully understanding) At first I was under the impression that the main practical issue with gimbal lock was requiring very fast, frictionless movements (or even flips). I wasn't sure about the implications of &quot;losing one degree of freedom&quot; (I could see one angle was made redundant, but I thought without friction everything would still be normal after moving away from 90° pitch*). Now, my intuition says even under in an ideal, frictionless environment, going in and out of gimbal lock will mean that we may suddenly have switched to one the wrong orientations (as in, even if the flywheel direction is correct in local coordinates, the yaw-pitch-roll reconstruction of our orientation based on gimbal angles would be wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I've said so far may be somewhere between a little and &lt;em&gt;completely&lt;/em&gt; off, but if it's close to correct, then my last question is: what prevents us from going into an incorrect orientation when not gimbal locked. For instance, if initially the flywheel is pointing along the roll axis**, then what prevents angles of, say, 0° yaw, 45° pitch, 0° roll from drifting to 45° yaw, 0° pitch, 90° roll (in both cases the flywheel direction would be the same in an aircraft's local coordinates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. It's hard to google for gyroscope because a lot of the results are about the sort of spinning top device that is also called gyroscope or about the gyroscopic effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt; * In the case of using angles yaw-pitch-roll, applied in that order, to describe positions. &lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;** If my understanding is correct, angles, starting from the initial/zero-angle orientation, are applied from inner gimbal angles first to outer gimbal angles last. Assuming the flywheel is in the inner gimbal, and considering the inner gimbal axis must be the yaw axis, it should be okay to have the flywheel pointing along the (initial) roll axis (and what would not be okay would have the flywheel pointing along the yaw axis, since it must be perpendicular to that). &lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="18698" LastActivityDate="2018-12-19T22:37:00.003" Title="Aren't there infinitely many orientations some given gyroscope angles could represent? (Generally understanding gyroscopes)" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25186" PostTypeId="2" ParentId="25179" CreationDate="2018-12-19T21:19:40.880" Score="1" Body="&lt;p&gt;Vibration is everywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a principal frequency and its harmonics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since most of the energy is in the principal frequency we concern ourselves with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say we call the distance between the two pullies L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ L = \lambda /2  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy of vibration is inversely related to the tension in the belt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to judge which belt lend itself to your setting better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The chain belts are heavy and call for more substantial pullies and tension free wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rubber indented belts are versatile and won't slip either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can google for standing vawes on cords and belts and gather detailed information.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-19T21:19:40.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25187" PostTypeId="2" ParentId="25185" CreationDate="2018-12-19T22:30:25.437" Score="1" Body="&lt;p&gt;You see this is a real system with real world constraints not a mathematical model. If it were just magic of mathematics then its true only 2 axes would be ok. You could just jump from one solution to the next, but that is not how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the gyroscope to actually work it needs to be free to move unopposed. That means it needs to have well formed 3 degrees of freedom at all times. Or otherwise the structure starts causing forces on the gyro. As it can not turn into the direction it wants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a real physical object needs to have a path to the orientation it needs to be in that also makes sense. You cant just unwind it to a different direction then back to where it was it would break.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this same reason even the third gimbal is not enough. At the poles of the 3 gimbals you have a singularity that makes one or two of the gimbals to become meaningless in allowing rotation in all directions. This is called gimbal lock. But atleast the gyro can work on all shallow angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR 2 angles is enough to point a vector in all directions. But it can not reach all directions from all positions via the shortest path.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2018-12-19T22:37:00.003" LastActivityDate="2018-12-19T22:37:00.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25189" PostTypeId="1" AcceptedAnswerId="25197" CreationDate="2018-12-20T04:36:25.170" Score="4" ViewCount="144" Body="&lt;p&gt;Suppose you were to fasten a string to a ping pong ball, then fasten the string to a stationary object, and then use a vacuum cleaner with a strong vacuum to suck the ball partially into the vacuum hose. Due to the strong vacuum's pull on the ball, the string should be very taut and the ball should also be centered inside the vacuum hose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this setup and referring to the two drawings below, if you were to then push down on the string with your finger at the point indicated by the arrow in Figure A, would the tautness of the string move the ball from a centered position in the hose down until it comes to rest on the bottom of the vacuum hose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, would pushing down on the string at that point cause a bend in the string, as shown in Figure B, with the result being that the ball would be pulled back towards the opening of the vacuum hose with the ball remaining centered inside the hose due to the strong airflow flowing around it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I'm asking this question is because I have been unsuccessful with keeping the string fastened to the ping pong ball, even with super glue, so I am unable to see what the ball will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z450S.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z450S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user18610" LastEditorDisplayName="user18610" LastEditDate="2018-12-20T12:51:59.123" LastActivityDate="2018-12-20T18:53:26.910" Title="Will the airflow inside a vacuum hose keep a sphere centered inside of it?" Tags="|fluid-mechanics|airflow|aerodynamics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25192" PostTypeId="1" AcceptedAnswerId="25195" CreationDate="2018-12-20T12:53:45.567" Score="2" ViewCount="220" Body="&lt;p&gt;I have checked the definition of tonal masking on Wikipedia and google but I saw the definition of sound masking and auditory masking.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Auditory_masking&quot; rel=&quot;nofollow noreferrer&quot;&gt;Auditory masking&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Sound_masking&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sound masking&lt;/a&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I want to know what exactly is tonal masking?&lt;/p&gt;&#xA;" OwnerUserId="12397" LastEditorUserId="8055" LastEditDate="2018-12-26T19:38:07.857" LastActivityDate="2018-12-26T19:38:07.857" Title="What is tonal masking?" Tags="|electrical-engineering|telecommunication|acoustics|sound-isolation|electronic-filters|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25194" PostTypeId="2" ParentId="19624" CreationDate="2018-12-20T14:48:10.277" Score="1" Body="&lt;p&gt;As for your need, as mentioned by Solar Mike, I believe OpenFOAM is the best option as there wouldn't be any issues for licence cost as well ,  since it is an opensource CFD platform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can simulate almost anything in OpenFOAM as per your needs. You can quite very easily program your equations to create a solver of your own, and for doing that you don't have to fall much deep into the CFD aspects or programming. You can familiarize yourself with proramming the equation by having a look at the already written solver which come with the package for tutorial cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Programming the equations isn't hard at all in OpenFOAM, as while initially writing the solver you only have to look if you are to perform your calculation in implicit or explicit ways,(yes you do need to have a considerable grasp over concepts, syntax, available libraries) while you can specify your discretization schemes and matrix solvers during setting up your case. You can get a basic idea by having a look at pre-compiled solvers and tutorial cases, and if needed can modify tose already existing solvers to work for your case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per your case I believe there are already many 1D solvers in OpenFOAM dealing with fluid flow(incompressible and compressible ), all you would need is to specify your fluid properties in transport properties file of a case(you should be able modify a tutorial case to run your own case).&#xA;Many tutorials, forums and youtube channel(&lt;a href=&quot;https://www.youtube.com/channel/UCjdgpuxuAxH9BqheyE82Vvw&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/channel/UCjdgpuxuAxH9BqheyE82Vvw&lt;/a&gt;) are there where you can learn basics of it.&#xA;I hope this helps!!!&lt;/p&gt;&#xA;" OwnerUserId="18714" LastEditorUserId="18714" LastEditDate="2018-12-20T18:41:23.577" LastActivityDate="2018-12-20T18:41:23.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25195" PostTypeId="2" ParentId="25192" CreationDate="2018-12-20T15:15:08.280" Score="1" Body="&lt;p&gt;Tonal masking is using tones at specific frequencies for sound masking, as compared with using noise containing a continuous band of frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most practical situations noise is more effective for masking than tones, but experiments with pure tones are a way to try to understand how masking actually &quot;works&quot; in human hearing.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-12-20T15:15:08.280" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25196" PostTypeId="2" ParentId="25160" CreationDate="2018-12-20T16:37:39.467" Score="1" Body="&lt;p&gt;yes it is just to maintain the pressure in the container at the same level as atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many small pumps malfunction when the source reservoir has vacuum built up in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even simple bottle water dispensers have check valves that lets air replace used water and break the vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially unable to handle the negative head are peristaltic pumps, which function by a rotary triangle of wheels squeezing the water in a tube forward, because of a random bubble of air trapped in the tube between the wheels. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="13311" LastEditDate="2018-12-21T13:04:54.437" LastActivityDate="2018-12-21T13:04:54.437" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25197" PostTypeId="2" ParentId="25189" CreationDate="2018-12-20T18:53:26.910" Score="3" Body="&lt;p&gt;No - the sphere would likely stick to one wall, or bounce between them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A sphere is not a very aerodynamic shape, having both a relatively blunt front, and sudden cut off at the back. This causes a lot of turbulence, such as the familiar vortex shedding shown here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/Images/mix.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://www.grc.nasa.gov/www/k-12/airplane/Images/mix.gif&quot; alt=&quot;Vortex Shedding around a sphere&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows how the air flows around a fixed sphere in free space - in your hose, the vortex forming on one side of the sphere will push it towards the wall, whereupon the flow will change as a result of the wall being there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key thing to note, is that the highly turbulent flow is not a stable equilibrium state - think of trying to balance a pencil vertically on your finger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you replaced the sphere with a different shape, imagine a Nerf football-type situation, that is designed to create a stable flow, then you would be able to get the behaviour that you're after - more like trying to keep the pencil vertically by holding the top, and dangling it downwards.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-12-20T18:53:26.910" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25198" PostTypeId="2" ParentId="25172" CreationDate="2018-12-20T20:55:23.597" Score="0" Body="&lt;p&gt;A &lt;em&gt;change&lt;/em&gt; may be that the target measurement is different, or that the hardware you're testing is different, or you move the machine into a different location in the shop and there may be different interference level in that area(if some type of electronic testing). Use your judgement or ask someone who knows more about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless there is a specification or customer requirement for MSA, or something defined in your quality manual, you should determine your MSA frequency. I don't know how long the MSA takes or if there's some cost etc, so take with a grain of salt.  You could perform an MSA at the start of each shift, or each lot of parts, or once a week, or once a month, every 3 months, whatever. If the MSA fails, your containment lot size of possible affected parts is all of the parts made since the last MSA. Larger frequency, larger containment lot and more risk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're unsure of the reliability of the inspection equipment, start at a higher frequency initially and then gradually reduce frequency to where you're comfortable. Once a day, for a 1 week. Once a week for 3 weeks, once a month, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Prove and document&lt;/strong&gt; that your equipment is reliable. If it fails, this way you have data to back your decision on MSA frequency.&lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2018-12-20T20:55:23.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25199" PostTypeId="1" AcceptedAnswerId="25300" CreationDate="2018-12-20T21:01:09.910" Score="1" ViewCount="2500" Body="&lt;p&gt;My in-view sketch toolbar has disappeared recently -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Q8Jr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Q8Jr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can you make this reappear? So confused, don't even know what it's name is!&lt;/p&gt;&#xA;" OwnerUserId="18716" LastActivityDate="2019-01-05T05:44:19.443" Title="How to restore In-view Sketch Toolbar?" Tags="|solidworks|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25201" PostTypeId="2" ParentId="24285" CreationDate="2018-12-21T07:15:38.487" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Propeller Diameter&#xA;  It is observed that the speed of the shaft and the propeller diameter are very closely related. A low shaft speed (given diameter) is very beneficial from efficiency point of view. However, it leads to high shaft torque subsequently large shafts and gearboxes. Therefore, while designing the propeller a balance must be found to ensure its performance.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;Generally, propulsive efficiency can be increased by installing propellers with large diameters. However, the diameter behind the ship is limited by the draught of the vessel. Optimization of propeller design is done to meet the balance.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;marine propeller&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;&lt;p&gt;RPM&#xA;  Selection of RPM of the ship plays a very important role in the propeller design. The rotational speed chosen for the vessel must be different from the resonant frequencies of the shaft, hull and other propulsion machinery. It is often seen that low RPM design increases the propulsive efficiency by 10 to 15 percent.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Number of blades&#xA;  The number of blades chosen has an effect on the level of unsteady forces acting on them. Considering the efficiency point of view, optimum open water efficiency increases with increase in the number of blades.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Blade Outline&#xA;  Blade outline plays a very important role in propulsive efficiency. Research and experiments using propellers with various blade areas have shown that efficiency increases by decreasing the blade area. This is because the frictional drag increases with increased blade area. However, it should be kept in mind that the strength of the propeller cannot be compromised by decreasing the area to large extent. Tip of the propeller has to be kept narrower in order to ensure efficient flow of water through the blade.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Angle of Attack &amp;amp; Camber&#xA;  The design of angle of attack of the propeller and its corresponding chamber depends on the design lift which has to be determined by the naval architect. If a larger angle of attack is chosen then the section designed would be less susceptible to pressure side cavitation and more susceptible to suction side cavitation. The reverse also holds true if the angle of attack is decreased.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Pitch/Diameter ratio&#xA;  To achieve the best propulsive efficiency for a given propeller diameter, an optimum Pitch/Diameter ratio is to be found, which again corresponds to a particular design rate of revolution.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;Apart from the conventional changes in the parameters of the propeller in order to achieve higher efficiency, there are certain external components which help in improving the flow around the propeller, hence, improving propulsive efficiency. They are as follows –&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;7&quot;&gt;&#xA;  &lt;li&gt;&lt;p&gt;Stern Tunnels&#xA;  These devices help in reducing the wake peak effect for ships having V-shaped sterns, thus reducing the effect of vibration. They are horizontal hull appendages placed above the propeller, they deflect the water towards the propeller ensuring even flow through the propeller.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Schneekluth ducts&#xA;  These type of devices redirect the flow to the upper portion of the propeller disk thus equalising the effect of wake and improving hull efficiency. These are also capable of accelerating the flow by means of the aerofoil shape cross section. This is done by creating a low pressure area in front of the duct which is achieved through its design.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;design of propeller&#xA;  Image Credits: schneekluth.com&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;9&quot;&gt;&#xA;  &lt;li&gt;Grouches Spoilers&#xA;  Their function is to redirect the flow horizontally in towards the propeller. They are small curved triangular plates welded in front, top and side of the propeller. These were designed to prevent the formation of keel vortices for U-shaped stern ships.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;  &#xA;  &lt;p&gt;propeller design for u shaped stern ships&#xA;  Image Credits: schneekluth.com&lt;/p&gt;&#xA;  &#xA;  &lt;ol start=&quot;10&quot;&gt;&#xA;  &lt;li&gt;Rudder Bulb System&#xA;  Kappel propeller comprises of smoothly curved extended blade tip towards the suction side of the blade which reduces the energy loss from the tip vortex flow. Integrating a rudder bulb and flaring the Kappel propeller hub cap will provide additional reduction in the energy loss from hub vortex and drag from the hub-rudder profile. The rudder bulb system also reduces cavitation increasing the total efficiency of the propeller assembly. A Kappel propeller and a rudder bulb system is designed for a target ship and accordingly engine is selected as per the power/ rpm requirement. It is important to properly optimize these two parts to achieve improvement in fuel saving, efficiency and reduction in emission.&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;http://seq.vindicatrix.com/newsletters/17-%20Aug%202016.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://seq.vindicatrix.com/newsletters/17-%20Aug%202016.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18718" LastEditorUserId="1832" LastEditDate="2018-12-21T11:42:24.817" LastActivityDate="2018-12-21T11:42:24.817" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25204" PostTypeId="2" ParentId="5469" CreationDate="2018-12-21T13:18:53.223" Score="0" Body="&lt;p&gt;My thinking is that it can be done.&#xA;I look at the dripping front window glass on the RV and it is pouring down sometimes. old boat portholes even have drip guides to collect the run off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you had a black steel plate outside with a strong thermal conductor to a heatsink inside, then as the temp drops in the evening and you get to the dew point, it will condense on the coldest surface. Perhaps a little pc fan could help here with negligable current.&#xA;I even made a steel 'door' for my yacht, as I would often see the roof dripping in water when stored over winter, the idea being that every temperature fall would condense water on the door and run outside via a simple little guide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Peltier mini dehumidifiers work pretty well, but they draw about 3amps @12v which is not passive enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently playing with these ideas, can't see any reason they won't work, again, it does depend on temp ranges on inside vs outside.&lt;/p&gt;&#xA;" OwnerUserId="18722" LastActivityDate="2018-12-21T13:18:53.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25206" PostTypeId="2" ParentId="25119" CreationDate="2018-12-21T13:46:23.350" Score="0" Body="&lt;p&gt;If you wish to increase the speed of your gearbox, you must alter the gear ratios. Properly cooling the mechanical components may increase its efficiency, but that will not increase its speed above what it is designed for, only bring its actual speed closer to its designed speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To increase durability, you can run it lubricated with oil. That will reduce friction/temperatures reducing wear and improving efficiency, thus leading to longer service life&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to harden the gears via heat treatment, you must know what material they are made out of. Not all steel can be hardened by heat treatment. Now given a steel that can be hardened by heat treatment, you must know the proper pre-heat(if necessary) and hardening temperatures, proper soak/hold times, and proper method of quenching(water,oil,air,etc). Maximum hardness is retained in minimal quenching time. After hardening, the part needs to be tempered to reduce brittleness and re-introduce toughness. Without knowing the gear material, its not possible to do this properly*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Id bet there are some old-time toolmakers/blacksmiths that can properly heat treat without much information or process controls, that's due to skill, practice, knowledge, and experience.&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2018-12-21T13:46:23.350" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25208" PostTypeId="1" AcceptedAnswerId="25210" CreationDate="2018-12-22T05:03:42.363" Score="2" ViewCount="81" Body="&lt;p&gt;Geothermal systems are used in regions across the U.S. for home use, but industrial geothermal plants are only used in areas where subteranian temperatures are very high. Why is this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Geothermal_energy_in_the_United_States&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Geothermal_energy_in_the_United_States&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.energy.gov/eere/geothermal/geothermal-maps&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.energy.gov/eere/geothermal/geothermal-maps&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2018-12-26T18:33:02.240" Title="Why are home geothermal systems usable anywhere in the U.S. but industrial geothermal plants limited in geographical use?" Tags="|heat-transfer|geotechnical-engineering|heating-systems|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25209" PostTypeId="2" ParentId="25208" CreationDate="2018-12-22T08:32:05.313" Score="0" Body="&lt;p&gt;For a home system, it makes sense if you can generate energy for less than you'd pay an energy company. If this is the case, the system will pay for itself after a number of years. This works, because the cost per unit of energy that is charged to the consumer by the companies is relatively the high. If it takes an extra year to pay off because you've got a cooler subterranean temperature, that's not a huge reason not to install it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an industrial system, the cost per unit needs to be &lt;em&gt;much&lt;/em&gt; lower. A difference in temperature gradient of just a few degrees could affect operational efficiency significantly, and drastically alter the potential profit margin. Building an enormous generating plant on site A will cost much the same as site B, (compare this to a consumer who is already tied to their house - it wouldn't make sense to move!), so as a result, geothermal plants are concentrated in the areas where they will provide the greatest financial benefit to their owners.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2018-12-22T08:32:05.313" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25210" PostTypeId="2" ParentId="25208" CreationDate="2018-12-22T09:20:27.107" Score="7" Body="&lt;p&gt;Because they're two completely different systems that have, in some circles, accidentally been given the same names.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The things that are sometimes (particularly in the USA) called home geothermal systems are ground-source heat pumps. The energy for these comes from sunlight on the ground over the year. Electricity is used to move heat from a source temperature that is broadly equal to the average ambient air temperature over the year (let's say 10-20 degrees C), to 30-90 degrees C to provide space- and water- heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Industrial geothermal systems are heat exchangers. The energy for these comes from the much hotter temperatures underground, that arises from the heat of magma, heating rocks above it. So in this case, the source temperature is 300 degrees celsius or higher. That's high enough to provide space and water heating without a heat-pump. It's also high enough (relative to ambient temperatures) to drive a heat-engine, which is pretty much the opposite of a heat pump. This heat engine is then used to generate electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A heat pump uses work to generate a temperature difference. A heat engine is the inverse of this: it uses a temperature difference to generate work.)&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2018-12-23T12:19:06.463" LastActivityDate="2018-12-23T12:19:06.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25216" PostTypeId="1" CreationDate="2018-12-22T16:25:40.053" Score="0" ViewCount="1249" Body="&lt;p&gt;Is there a reliable online reference for head losses of PEX tubing at various nominal sizes and flow rates?&lt;/p&gt;&#xA;" OwnerUserId="6489" LastActivityDate="2018-12-22T16:36:23.613" Title="Head loss curves for PEX tubing?" Tags="|pressure|hydraulics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25217" PostTypeId="2" ParentId="25216" CreationDate="2018-12-22T16:25:40.053" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://www.pmengineer.com/articles/86617-minitube-distribution-systems-for-hydronic-radiant-floor-heating&quot; rel=&quot;nofollow noreferrer&quot;&gt;This article&lt;/a&gt; (&quot;Minitube Distribution Systems for Hydronic Radiant Floor Heating&quot;, by John Siegehthaler) has a somewhat small / low-quality chart which doesn't seem to cite a refererence / source:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OXKpr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OXKpr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is useful at least to get you &quot;in the ballpark&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Uponor &lt;a href=&quot;https://www.uponorpro.com/~/media/Extranet/Files/manuals/PressureLossTables_H191_1210.aspx?sc_lang=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Pressure Loss Tables&quot;&lt;/a&gt; could be useful; example:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dZpeN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dZpeN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although not a graph I believe you could generate a curve from this data; and it it is usefully broken down by tube size, fluid type, and temperature.&lt;/p&gt;&#xA;" OwnerUserId="6489" LastEditorUserId="6489" LastEditDate="2018-12-22T16:36:23.613" LastActivityDate="2018-12-22T16:36:23.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25218" PostTypeId="2" ParentId="1794" CreationDate="2018-12-22T17:16:36.967" Score="0" Body="&lt;p&gt;All material exchanges hardness for toughness.&#xA;The harder it is, the more brittle.&#xA;Some material can be treated or combined (alloyed) to make something with the desired traits, but a pure material is subject to the limitations of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brittle knife is not a functional knife.  It is a knife shaped object.  This is perfectly usable of a one-time (or short time) use until the fractures make it unusable.  It is also perfectly fine for a wall hanging... would make an interesting art piece especially if lit well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shivs/spikes (long shards) made of diamonds could be used as throwing &quot;knives&quot;.  That would be a single-use weapon from diamonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just not a suitable material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience, the heat treating of an alloy (such as steel) is the best way to get a balance between toughness and hardness for a long-term use functional knife.&lt;/p&gt;&#xA;" OwnerUserId="18732" LastActivityDate="2018-12-22T17:16:36.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25219" PostTypeId="1" CreationDate="2018-12-22T20:53:43.870" Score="0" ViewCount="679" Body="&lt;p&gt;I searched about this, but couldn't find any helpful explanations. All it says is that magnetic field is generated by either permanent magnets or electromagnets. In both cases, how does the exciter generator start, so that it creates magnetic field according to physics, (e.g. Lorentz law, or another) ?&lt;/p&gt;&#xA;" OwnerUserId="18731" LastActivityDate="2023-12-27T11:05:50.707" Title="Difference between permanent magnet and electromagnet exciter generator in principles wise?" Tags="|electrical-engineering|electromagnetism|magnets|generator|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25220" PostTypeId="1" CreationDate="2018-12-22T21:37:58.257" Score="2" ViewCount="1337" Body="&lt;p&gt;Relative to the &lt;a href=&quot;https://en.wikipedia.org/wiki/CIE_1931_color_space#CIE_xy_chromaticity_diagram_and_the_CIE_xyY_color_space&quot; rel=&quot;nofollow noreferrer&quot;&gt;CIE chromaticity diagram&lt;/a&gt;, it is usually said that rgb monitors can't actually reproduce the entirety of colors in that diagram. The best they can do is cover a triangular region inside the &quot;horseshoe&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fZIBb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fZIBb.png&quot; alt=&quot;CIE chromaticity diagram and typical gamut for an rgb monitor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the subject I read this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;An additive colour space defined by three primary colors has a chromaticity &lt;a href=&quot;https://en.wikipedia.org/wiki/Gamut&quot; rel=&quot;nofollow noreferrer&quot;&gt;gamut&lt;/a&gt; that is a color triangle, when the amounts of the primaries are constrained to be nonnegative.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: why is that the amounts of the primaries being constrained to be nonnegative leads to a triangle? Is the triangle a projection of a slanted slice in another color space? (I'm rusty on my math, just trying to understand this point).&lt;/p&gt;&#xA;" OwnerUserId="18733" LastActivityDate="2020-05-18T21:03:30.320" Title="Why is the color gamut of an RGB monitor triangle-shaped?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25221" PostTypeId="2" ParentId="25219" CreationDate="2018-12-22T23:57:49.477" Score="0" Body="&lt;p&gt;There are two ways that exciter generators start, so they can then spin up the excitation current to start a larger generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easy way is to use permanent magnets in the exciter generator. As soon as it begins to turn, it will generate the current to get the main generator going. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky way is to rely on the fact that the iron in the armature and rotor of a generator usually retains a small amount of magnetic field even when the generator is shut down. When you start spinning this kind of generator, that residual field allows the generator to make its own excitation current and the generator will then start itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The self-starting principle using residual magnetism is commonly used in small gasoline engine powered AC generators. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-22T23:57:49.477" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25222" PostTypeId="2" ParentId="25220" CreationDate="2018-12-23T00:27:36.317" Score="0" Body="&lt;p&gt;Thats simply because of the nature of the XY diagram. Yes you are just seeing one slice of color intensity, and no its not the same triangle in all slices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram is showing pure rainbow colors on the outer curved part (the numers are wavelengths). Now the interior is just interpolation of colors on the edge (this is why it has the sail like shape), mixtures of pure edge colors. Therefore if you have 3 colors near any of the rainbow edges and mix and match them then you get a triangle since that is a interpolation between 2 points (well is not quite this simple, but the diagram tries to make it this simple really really hard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the magenta end is a bit special, see it does times not exist in nature. Its purely a human interpretation. Thats why it's straight. As a result any primary in magenta space suddenly veers off to a different direction so CMY is a square or pentacle, even if it has 3 primaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now since the pure single wavelength rainbow is the sail shaped curve there is nothing outside this. No negative color. There is something below the magenta line that we dont see so it can not mix. Also would be pointless since each animal has a different shape of the curve (arguably every human).&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2018-12-23T00:27:36.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25223" PostTypeId="1" CreationDate="2018-12-23T02:22:21.157" Score="7" ViewCount="239" Body="&lt;p&gt;It is will known that most of tool steels are quenched in oil to prevent cracking and distortion which appears when quenching in water due to non-uniform cooling caused by formation of insulating vapor blanket &quot;film boiling&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but according to some  literature &lt;a href=&quot;http://www.gmferd.com/journalcra.com/sites/default/files/31683_0.pdf&quot; rel=&quot;noreferrer&quot;&gt;1&lt;/a&gt; using 14% brine solution or 1% polymr solution eliminate this vapor blanket so we can expecting more uniform cooling compared to water quenching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;according to another literature &lt;a href=&quot;https://www.researchgate.net/profile/Nikolai_Kobasko/publication/260124962_Steel_Superstrengthening_Phenomenon/links/5692bca808ae0f920dcd8fb1/Steel-Superstrengthening-Phenomenon.pdf&quot; rel=&quot;noreferrer&quot;&gt;2&lt;/a&gt; the probability of quench crack formation&#xA;at first increases to the maximum value, and then decreases to zero. &lt;a href=&quot;https://i.stack.imgur.com/kdAYI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kdAYI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;increasing cooling rate is possible by using strongly circulated brine quenching&#xA;bath, and using brine eliminate vapor blanket which ensure uniform cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, can we excepting maximum hardening depth, minimal cracking and distortion and eliminated fire hazard by quenching high alloy tool steels by quenching in circulated brine instead of oil ?&lt;/p&gt;&#xA;" OwnerUserId="18586" LastEditorUserId="16486" LastEditDate="2020-01-31T17:46:39.657" LastActivityDate="2020-06-29T18:04:45.570" Title="Can high alloy tool steels be quenched in brine?" Tags="|mechanical-engineering|materials|steel|metallurgy|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25227" PostTypeId="2" ParentId="23284" CreationDate="2018-12-23T19:35:57.260" Score="0" Body="&lt;p&gt;Your question is why don't most automotive companies manufacture cars that look as good as a sports car? Specifically what is the major cost driver for the production of a sports car's body?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well to answer this we have to look at certain aspects in fact many aspects, like:&#xA;1) Material that is being used, sports car have a huge engine at the very least and are very well aerodynamically built. Plus generally major automotive companies prefer Aluminium or other blends of steel with plastic because they need to think about the cost. Sports cars presently prefer carbon fiber blends which itself is very costly when compared to normal chassis building materials. So the material is one very major constraint.&#xA;2) Design, when designing a normal passenger car you have priorities which restrict the flexibility of the design. For instance in passenger car you need to think about is space, road clearance and boot space where as in sports car you are more concerned about aerodynamic behaviour, decreasing the weight of the chassis so their priorities are different hence the design and look is different. Also sports cars are more sleek and have low ride height, having a low height is a passenger vehicle will make them use a higher blend of plastic which when used with steel weakens the material and will be difficult to obtain with their current manufacturing tools. So this is one of the major reasons why design differs.&#xA;3) If you build a passenger vehicle with sports car features like low ride height, low ground clearance, lesser boot space and better aerodynamic structure. The down force on the tires will increase and eventually make its users go for high end Michelin or Pirelli tires and will be an added cost. Less boot space might act as obstructing factor for the customers as passengers vehicles are meant to have a certain boot space which makes it convenient for the passengers.And decrease in ground clearance might become problematic on normal streets on certain countries and could cause damage to the vehicle.&#xA;4) Lastly the time and effort put into to build a sports car is far more than any normal passenger vehicle which is also one of the main reason there is a cost difference between them. So manufacturing a passenger car with sports car design and features will have high price tag and the company would incur losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So above mentioned can be one of the many reasons automotive company do not manufacture normal or passenger cars who look as good as sports car. Hope I was able to answer your doubt.&lt;/p&gt;&#xA;" OwnerUserId="18736" LastActivityDate="2018-12-23T19:35:57.260" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25228" PostTypeId="2" ParentId="23174" CreationDate="2018-12-24T01:08:37.660" Score="3" Body="&lt;p&gt;Let’s consider gears as being represented by a pitch circle on which lie several equally spaced dots with a number equal to the number of teeth. These dots can either represent the teeth or the gaps between teeth, and they follow the rule that, as a pair of meshing gears rotate, the dots of one gear will coincide with the dots of the other gear as the dots pass the point of tangency of the pitch circles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Klw0A.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Klw0A.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between neighbouring points measured along the pitch circle is the circular pitch &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; for both gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we know the location of a dot on one gear, we can infer the location of a corresponding point on a meshing gear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tnGhu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tnGhu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, if a dot on gear 1 is an arclength &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; away from the pitch circle tangency point, there will exist a corresponding dot on gear 2 that is also an arclength &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; away. We know this to be true since rotating the gears result in moving the dots by the same distance along the pitch circles; corresponding dots will coincide at the tangency point, and a rotation of the gears will move both dots by &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt;. Care is to be taken in the directions of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use this information to consider tooth spacing for a closed loop of four arbitrary gears:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QeXWO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QeXWO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, for &lt;span class=&quot;math-container&quot;&gt;$i=1, 2, 3, 4$&lt;/span&gt;, each gear has &lt;span class=&quot;math-container&quot;&gt;$N_i$&lt;/span&gt; teeth, and the centres of the pitch circles connect to form a closed polygon with internal angles &lt;span class=&quot;math-container&quot;&gt;$\theta_i$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let’s consider a dot on gear 1 at the point of tangency of pitch circles 1 and 2. A corresponding dot will occur for gear 2 coincidently. From the above information, we can use the dot on gear 2 to infer a corresponding dot on gear 3, and then use the dot on gear 3 to infer a corresponding dot on gear 4, and finally use gear 4’s dot to obtain another dot on gear 1. This should appear as follows, with arclengths of interest indicated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mm1t9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mm1t9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how the consideration of a single dot on gear 1 led to the discovery of a second dot on the same gear. If these are to be two dots on the same gear, they must be separated by an integer number of tooth spacings &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;, i.e. they must be separated by &lt;span class=&quot;math-container&quot;&gt;$zp$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is an integer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note how the arclengths are related to the radii of the pitch circles &lt;span class=&quot;math-container&quot;&gt;$r_i$&lt;/span&gt; and the internal angles &lt;span class=&quot;math-container&quot;&gt;$\theta_i$&lt;/span&gt;. This yields the equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$s_A = r_1 \theta_1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$s_A + s_B = r_2 \theta_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$s_B + s_C = r_3 \theta_3$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$s_C + zp = r_4 \theta_4$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elimination of the unknown arclengths leads to the following condition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_4 \theta_4 - r_3 \theta_3 + r_2 \theta_2 - r_1 \theta_1 = zp$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By noting that the radius of each gear &lt;span class=&quot;math-container&quot;&gt;$r_i = N_i p/2\pi$&lt;/span&gt;, this condition simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$N_4 \theta_4 - N_3 \theta_3 + N_2 \theta_2 - N_1 \theta_1 = 2\pi z$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, for gears of given numbers of teeth, you need to alter the interior angles of the closed polygon such that the above condition holds for some integer &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;, which may be positive, negative or zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your particular case where &lt;span class=&quot;math-container&quot;&gt;$N_1 = 12$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$N_2 = N_3 = N_4 = 44$&lt;/span&gt;, this becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$44( \theta_4 - \theta_3 + \theta_2) - 12 \theta_1 = 2\pi z$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find a geometrically valid set of internal angles that obey this rule, and the teeth should all mesh. Once you know the internal angles, you can determine the distance between the top and bottom gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rule can also be generalised to any closed loop of an even number of gears, where the rule can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sum_{i=\textrm{even}} N_i \theta_i - \sum_{i=\textrm{odd}} N_i \theta_i  = 2\pi z$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2018-12-24T01:08:37.660" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25233" PostTypeId="2" ParentId="23987" CreationDate="2018-12-24T21:52:41.317" Score="2" Body="&lt;p&gt;This is a very common scheme to obtain tilt/pan action. Lots of variations of it are used in drum set hardware as adjustable mounts for cymbals and in microphone stands as adjustable mounts for microphones. In these contexts, this sort of mechanism is called a &lt;em&gt;tilt mount&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The facing cuts are usually knurled with a series of radial ridges that engage and interlock when the screw connecting them is tightened, so the assembly will not slip out of position during use. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, the plane containing the tilt mount faces is parallel to the axis of the parts A and B rather than being at 45 degrees.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-24T21:52:41.317" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25235" PostTypeId="1" CreationDate="2018-12-24T23:11:11.240" Score="2" ViewCount="139" Body="&lt;p&gt;One of the many challenges facing PV power generation is how to store surplus energy for night time demand. My question is this: since the sun is always shining somewhere on the planet, would it be possible to design an international smart energy grid that can transmit DC current for long distances? Is any research related to this question being conducted?&lt;/p&gt;&#xA;" OwnerUserId="18743" LastActivityDate="2024-02-29T13:07:55.990" Title="Transferring energy generated by photovoltaic methods" Tags="|photovoltaics|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="25240" PostTypeId="2" ParentId="23911" CreationDate="2018-12-25T08:29:53.177" Score="0" Body="&lt;p&gt;Magnetic compasses for serious work will contain two small permanent magnets with their N/S axies perpendicular to each other built into the compass housing; these are carefully rotated one way or the other to null out the influence of other magnetic materials nearby on the compass readings. I do not know if this technique would be effective in your case, where the magnet nearby is large and powerful. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are better off devising another attachment method that does not rely on magnets! &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2018-12-25T08:29:53.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25242" PostTypeId="2" ParentId="25220" CreationDate="2018-12-25T17:48:38.420" Score="1" Body="&lt;p&gt;Summary: human vision has four physiological components giving rise to three color components in opponent color theory. The CIE chromaticity diagram was constructed by realizing all colors can be created by varying intensity of two appropriate monochromatic sources. Reaching a color by two different combinations is meramerism, and the colors are metamers, and indistinguishable. Computer displays have three monochromatic components interfering with a backlight. The nature of component masking reducing intensity with three components results in a triangular gamut strictly inside the CIE chromaticity diagram, enclosing a white point. All of the colors in the gamut are reachable by continuous variation of monochromatic intensity. The gamut has straight sides by definition of the CIE chromaticity diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical human vision involves four components: one type of rod which excels at accumulating the effects of sequentially arriving photons, and three types of cones: short, medium, and long. All four are sensitive to light across a range of wavelengths in the general shape of a bell curve. However, peak wavelength sensitivity differs among the four. Short cones are most sensitive to violet wavelengths, medium cones to green, long cones to yellow, and rods to blue-green. The following image shows the relative response profules for each, with cones color coded. The black line profile is for rods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.unm.edu/~toolson/human_cone_action_spectra.gif&quot; alt=&quot;Graph of rod and cone response profiles.&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A theory of human vision is the trichromatic model. This model of human vision, which leads to the CIE chromaticity diagram, is essentially that the rods are responsible for controlling the overall perceived intensity, while each cone is responsible for varying the position in the chromaticity diagram. One can determine the position by integrating the input wavelength profile against the three bell-curve-like responses of the cones. Repeat the integral for the rods and multiply that by each of the cone integrals to obtain a vector with three components. The result is a position in the overall gamut of human vision. Note that the chromaticity diagram is two dimensional, while the vector is three dimensional. The chromaticity diagram is a slice through the gamut of human vision, which has the shape of a cone in the trichromatic model. When intensity is very low, rods and cones receive no photons, so all colors tend to lose saturation and become what we call black, reflected by low values of the rod integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CIE chromaticity diagram was constructed by experiment. Participants were asked to match a displayed color by adjusting the intensities of various light sources. The adjustable light sources were comprised of a variety of near-monochromatic profiles. The lights were controlled by knobs whose positions (and thus intensities) could be recorded. It turns out different participants would match the colors using different combinations. The same color produced by different combinations of light are called metamers. The diagram was constructed by realizing that interpolation of relative variation of different monochromatic sources led to some of the same colors. Put another way, there was overlap if different monochromatic sources were used. To sum up, if you can reach any two positions on the chromaticity diagram, you can reach any position on a straight line between them by varying the relative intensities of your sources. If you have three non-collinear sources, you have a triangle, and can reach every point in the triangle by the same method of varying intensities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shifting gears, consider the properties of consumer computer displays. They have some form of color control, almost always with the three components red, green, and blue. The control generally involves a mask which can be varied continuously for each component at each pixel. They also have some form of backlighting which, by design, is as close to standard white points as is commercially feasible (D50 or D65, depending on requirements). Maximum intensity is only reached when all three components are fully engaged. When any component is disengaged, even partly, the overall intensity decreases. When only one component is engaged, much of the backlight is blocked, intensity is reduced, and the extremes of the chromaticity diagram can't be reached. So any display gamut will be strictly within the bounds of the chromaticity diagram. We also only have three monochromatic components to work with, so we expect the gamut of the display to be a triangle, by definition of the chromaticity diagram. Furthermore, every point in the triangular gamut is reachable by varying the intensities. Finally, because the backlight is a standard white point, we expect white to be somewhere inside the triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this answers your questions!&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2018-12-25T18:35:20.143" LastActivityDate="2018-12-25T18:35:20.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25246" PostTypeId="1" CreationDate="2018-12-26T00:44:55.793" Score="1" ViewCount="38" Body="&lt;p&gt;first time on this board even if I've been lurking for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a tennis racquet technician and, for the sake of saving some money and just for the fun of it, I'm trying to build a machine that could measure the inertial properties of a racquet in a more efficient way than the typical pendulum setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a picture of the typical commercially available (and overly expensive) machines:&#xA;&lt;img src=&quot;https://i.stack.imgur.com/X6ZlG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figure that an accelerometer ans gyroscope would do ths trick measuring the angular acceleration of the baseline setup (shaft and clamping device) and a motor that could provide data of torque variation due to increased inertia of the added racquet for a given angle should do the trick. Im having trouble finding such a motor or another sensor that could perform thia task. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions or pointers here? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Claudio &lt;/p&gt;&#xA;" OwnerUserId="18753" LastActivityDate="2018-12-26T00:44:55.793" Title="How to measure the inertia of an object using a motor" Tags="|mechanical-engineering|measurements|prototyping|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25250" PostTypeId="1" CreationDate="2018-12-26T10:44:09.873" Score="4" ViewCount="1783" Body="&lt;p&gt;I have a small endoscopic camera (custom made). I wanted to push the camera inside a horizontal pipe of dimensions varying from (25 mm to 65 mm ID). But the wire attached to the camera is not stiff enough to push the camera. After a certain length, camera stops moving (cz the wire coils inside the pipe). But I have seen Companies like Olympus etc., push their camera even up to 25 meters. I know that they have a special cover over the wire which is stiff and flexible for pushing the camera. How can I make a cover similar to that ? I have enquired many places but could not find a solution. Thanks in advance for your help.&lt;/p&gt;&#xA;" OwnerUserId="18754" LastActivityDate="2018-12-26T22:38:45.467" Title="How to push an endoscopic camera upto 20 meters in a horizontal pipe?" Tags="|materials|springs|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25251" PostTypeId="2" ParentId="25250" CreationDate="2018-12-26T12:46:47.687" Score="3" Body="&lt;p&gt;You can buy endoscope or borescope cables which should &quot;just work,&quot; though 20m is longer than the typical cable you will find on Amazon or Ebay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to try making your own, look at the design of Bowden cables (used for brake cables on bikes, etc). The outer cover is made like a coil spring, so it can bend around a large radius but won't &quot;buckle&quot; into a small radius at one point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also consider flexible chimney sweeping rods, which are made in 1 or 2 meter lengths which you can join together as you move the camera down the pipe. They would have the disadvantage that the camera cable would be separate from the rods, unlike a Bowden cable where it would be inside the cable and protected from dirt and damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact chimney sweeping suppliers sell inspection cameras that fit onto standard sweeping rods, and are self contained with battery power and a wireless data transmission link, so you don't need a separate cable to the camera.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2018-12-26T12:52:04.107" LastActivityDate="2018-12-26T12:52:04.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25253" PostTypeId="2" ParentId="25235" CreationDate="2018-12-26T15:46:43.727" Score="0" Body="&lt;p&gt;As stated before in the comments, natural and political challenges let you down to follow this project but electrical engineers and providers face another challenge, look the graph here onder: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QoXhv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QoXhv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the horizontal axis represent time in hour and vertical axis represent power in megawatt. As you can see the irregularities implies having bigger capacity groups than we usually need to provide enough in peak hours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cost of bigger capacity made us to find a solution that nowadays is called interconnection electrical networks. &#xA;The surplus somewhere can compensate for smaller capacity groups somewhere else in peak hours and within some hours later the groups will exchange their role.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete example is France and Belgium in west Europe and Czech and Hungry in east Europe. All the transmission lines are HVDC &lt;span class=&quot;math-container&quot;&gt;$400 kV$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interconnection follows of course by costs, but it reduce the capital for installation or maybe it prevent the urge for building new power plants, the economical effect of interconnection is strong enough to overpower political problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photovoltaic energy is DC in nature so it means no rectifier or under-station is needed and DC transmission lines can transport 100 % and 50 % more power in respect to 1-phase and 3-phase AC transmission lines. Spare more material and simple transmission towers, no mutual induction and capacitance between wires and ground, all of these advantages push us towards more interconnections. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do we need it now? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No at this moment we don't. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- The production capacity of solar or wind power plants are way too small in compare to traditional groups, and this small fraction does't compensate for the start and the maintenance costs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- The possible surplus can be transmitted within a small region, for instance San francisco and New York or even closer distance Belgium to Nederland. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- The most important reason is short circuit detection, grounding and the cost of safety instruments at this moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But do we need it later? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes and that's why thousands are working together hours and hours each day to overcome all those challenges above.   &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2018-12-26T15:46:43.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25254" PostTypeId="2" ParentId="25250" CreationDate="2018-12-26T16:23:43.580" Score="3" Body="&lt;p&gt;Most electrical engineers use an electrician Conduit Ducting wire for that (which you could use in your cable setup), but in your case it probably easier to use a piece of string and a vacuum cleaner. Put the vacuum cleaner at one end and suck the piece of string through. Then pull the camera through using the string like explained in &lt;a href=&quot;https://www.youtube.com/watch?v=0HUWhb4D4B4&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to get cables in a long pipe - with vacuum cleaner&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="18756" LastEditorUserId="8055" LastEditDate="2018-12-26T22:38:45.467" LastActivityDate="2018-12-26T22:38:45.467" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25257" PostTypeId="2" ParentId="25250" CreationDate="2018-12-26T18:33:41.250" Score="2" Body="&lt;p&gt;I think I have a very similar USB camera. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tape it to an electrician's tool called, in trade jargon, fish tape, this is a narrow strong, springy tape, which is uses to pull the wires through conduits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My sewer line is prone to clogging at the point when it connects to old City sewer main, which even though is public, has never been serviced by the city because it runs in my property, (easement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have taped a keychain led flash light to this fish tape and it helps a lot.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-26T18:33:41.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25258" PostTypeId="1" AcceptedAnswerId="35894" CreationDate="2018-12-26T22:08:17.980" Score="2" ViewCount="172" Body="&lt;p&gt;This is a screenshot of a lab where we were experimenting Vapor-Compression Refrig. Systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to understand what is happening between points 1-2 because I thought that because of max isothermal efficiency the gradient of the line shouldn't be more than the constant entropy line between the two points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I am not sure how to explain the dip between points 5-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YwAJE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YwAJE.png&quot; alt=&quot;Refrigerant: R134a&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VyyaJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VyyaJ.png&quot; alt=&quot;Schematics&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Lb3pf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lb3pf.png&quot; alt=&quot;System schematics&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18758" LastActivityDate="2020-05-22T08:14:20.280" Title="Is this refrigeration cycle thermodynamically possible?" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25259" PostTypeId="2" ParentId="23987" CreationDate="2018-12-26T23:19:13.967" Score="3" Body="&lt;p&gt;I think this &lt;a href=&quot;https://www.scuba.com/scuba-gear-37/034061/Omni-Swivel-Regulator-Omniswivel.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;omniswivel&lt;/a&gt; comes pretty close to what you describe. The only difference is that the surface between the two parts is circular instead of elliptical. This allows for 360° of rotation instead of only two fixed positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image: &lt;img src=&quot;https://decadiving.com/wp-content/uploads/2016/02/OM-SW-A.jpg&quot; alt=&quot;omniswivel&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="18759" LastEditorUserId="18759" LastEditDate="2018-12-26T23:49:19.660" LastActivityDate="2018-12-26T23:49:19.660" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25260" PostTypeId="2" ParentId="22707" CreationDate="2018-12-27T06:04:01.837" Score="2" Body="&lt;p&gt;When you hold the long arm (rifle or shotgun), there are three places where the gun is held:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The forearm is held with your off hand,&lt;/li&gt;&#xA;&lt;li&gt;near the action or on a pistol grip with your strong hand,&lt;/li&gt;&#xA;&lt;li&gt;and the butt of the stock up against your shoulder, which is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lever&quot; rel=&quot;nofollow noreferrer&quot;&gt;fulcrum&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When you work a pump action on a long arm, &#xA;&lt;a href=&quot;https://i.stack.imgur.com/W2abx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W2abx.jpg&quot; alt=&quot;Pump action rifle, showing fulcrum and where force applied by working action&quot;&gt;&lt;/a&gt;&#xA;you are introducing much more deviation from the original point of aim, at multiple points along the path of the pump foreend; even though it seems that a fore-and-aft action would result in minimal deviation minimal, the force is exerted out, farther from the fulcrum of stock-butt-on-shoulder, at multiple positions along the path, so any deviation is magnified by that distance. Also, the force of working the action is applied with your off-hand, presumably weaker, and since the off-hand typically is less dexterous (no pun intended), it will take longer for the sight picture to settle down and give you another good alignment to the target.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hv9Io.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hv9Io.jpg&quot; alt=&quot;Lever action rifle, showing fulcrum and where force applied by working action&quot;&gt;&lt;/a&gt;&#xA;When you work a lever action or bolt action, the force which disturbs your aim is much closer to the fulcrum, so any deviation caused by working the action is much less, so reacquiring the sight picture is much quicker. You're also using your strong hand, which gives you greater dexterity and therefore quicker recovery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rifle photographs courtesy Hmaag's Own work, CC BY-SA 3.0, from &lt;a href=&quot;https://commons.wikimedia.org/wiki/Main_Page&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikimedia&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18762" LastEditorUserId="18762" LastEditDate="2018-12-28T19:07:24.000" LastActivityDate="2018-12-28T19:07:24.000" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25263" PostTypeId="1" AcceptedAnswerId="25266" CreationDate="2018-12-27T13:17:30.937" Score="1" ViewCount="2976" Body="&lt;p&gt;I need help calculating required torque and power for an elevator to select appropriate motor and gearbox. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass (weight of elevator plus max load) is 2000kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of the elevator is 10 m/s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The radius of the pulley is 0,5 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use a counterweight - mass of counterweight is not specified, what is appropriate here? &#xA;I dont get it how to count when I have a counterweight in the system. &lt;/p&gt;&#xA;" OwnerUserId="18767" LastActivityDate="2018-12-27T17:57:16.823" Title="Torque and power for elevator with counterweight" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25264" PostTypeId="1" CreationDate="2018-12-27T14:55:07.020" Score="1" ViewCount="28" Body="&lt;p&gt;In the midst of trying to redesign the pneumatic subsystem of a combined system, I'm at the point of validating a new valve would serve as a direct replacement. The reason for the redesign is an inconsistency of operation over many cycles. The delay between the shifting of the solenoid and the pressurization of the Retract side of the Cylinder is inconsistent, causing errors throughout the rest of the system. The goal with the redesign is to return the system to a closer approximation of the original configuration (shorting piping), while using modern valve technology to retain remote control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current Pneumatic Circuit:&lt;br&gt;&#xA;The system pressure is 90-110 psi, then Filter, Regulated to 60-80 psi, and Lubricated.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yzTZo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yzTZo.png&quot; alt=&quot;Nope, I can&amp;#39;t draw&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An interest in operator safety moved flow control valves outside of machine guarding at a midpoint in the life of the machine, though there isn't documentation available demonstrating that artificial volume increase in the cylinder. This added roughly 236 inches of 3/4 piping to the system.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For pneumatic reasons, the flow controls have been installed to allow free flow to the cylinder from the source, but metered flow from the cylinder to the exhaust. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Flow coefficient of the flow control valve is 8.3.&lt;br&gt;&#xA;The Proportional Electro-pneumatic valve supports pressure control of the (A) port, aligned to the retract side of the pneumatic cylinder. The maximum effective area of the valve is 52 mm^2, with a flow coefficient of 2.9 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Issues investigated beyond the valves:&#xA;Lubrication is regularly checked throughout operation.&#xA;The cylinder is new and properly configured&#xA;Header pressure doesn't vary&#xA;regulated pressure doesn't vary.&#xA;The solenoid is in good working order.&#xA;There isn't an issue with the electronic control circuitry.&lt;/p&gt;&#xA;" OwnerUserId="18686" LastActivityDate="2018-12-27T14:55:07.020" Title="Is this Proportional Electro-pneumatic valve a direct replacement for other valves listed?" Tags="|valves|piping|pneumatic|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25266" PostTypeId="2" ParentId="25263" CreationDate="2018-12-27T17:13:41.757" Score="1" Body="&lt;p&gt;Let's say we ignore the friction for simplicity now, and assume mass of counter weight 600kg, this should be the weight of elevator plus cable and depending on the traffic pattern some extra weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More important than the speed is what acceleration you need. Say we choose 4m/s otherwise at constant speed of 10m/s we just need much smaller tension of holding the difference weight of 2000- 600=1400kg. So the tension would be only 13729N and torque would be 6860N.m, but we will see the torque needed to accelerate the elevator is much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tension force in the cable to accelerate the elevator and its counter weight is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F_{net} = T_{cable} - mg =  m\alpha $$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$ and \ F_{net}= \ m\alpha = 2600* \alpha = 2600*4 = 10400N   $$&lt;/span&gt;&#xA;So:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T_{cable}= 10400 + mg =10400 + 1400 *9.8 = 24120N   $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now we can calculate the torque&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \tau = f*r = 24120 * 0.5 = 12060 N.m  $$&lt;/span&gt;&#xA;So we see the torque needed to accelerate the elevator is nearly two times that which is needed to move it at 10m/s. The real scenario is each manufacturer uses their own proprietary pattern of relays and controls that adjust the acceleration in a curve , ramping up, for comfort and minimizing wear and tear.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-12-27T17:57:16.823" LastActivityDate="2018-12-27T17:57:16.823" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25267" PostTypeId="1" CreationDate="2018-12-27T18:09:58.757" Score="3" ViewCount="1286" Body="&lt;p&gt;I have a W-shape steel beam that is fixed-fixed and the top flange is fully braced for lateral torsional buckling. The negative moment at the end controls the design and there is no lateral bracing for the bottom flange. What will be the unbraced length (&lt;span class=&quot;math-container&quot;&gt;$L_b$&lt;/span&gt;) for the bottom flange? Also, if I provide brace points at the points of moment inflection then what will &lt;span class=&quot;math-container&quot;&gt;$L_b$&lt;/span&gt; be? In general, what is the rule for measuring &lt;span class=&quot;math-container&quot;&gt;$L_b$&lt;/span&gt; for bottom flange under negative moment? If possible, please provide proof from AISC code.&lt;/p&gt;&#xA;" OwnerUserId="18770" LastEditorUserId="9" LastEditDate="2019-04-05T00:44:56.523" LastActivityDate="2019-04-05T00:44:56.523" Title="How to measure the unbraced length of lateral torsional buckling for negative moment in a W-shape steel beam?" Tags="|structural-engineering|steel|beam|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25268" PostTypeId="1" AcceptedAnswerId="25275" CreationDate="2018-12-27T19:00:49.153" Score="2" ViewCount="163" Body="&lt;p&gt;I have a Delta ASB0912L DC Brushless fan.  PDF &lt;a href=&quot;https://www.tedss.com/DataSheets/2031/2031000026.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the datasheet it says, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RPM 3800&#xA;CFM 67.80 &#xA;IN H2O 0.302&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;On the blower, squirrel cage fan, Dayton from Grainger, found &lt;a href=&quot;https://www.grainger.com/product/DAYTON-Rectangular-Permanent-Split-1TDP1&quot; rel=&quot;nofollow noreferrer&quot;&gt;here (Rectangular Permanent Split Capacitor OEM Specialty Blower, Flange: Yes, Wheel Dia: 3&quot;, 115VAC)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says,&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RPM 3010&#xA;CFM @ 0.200-In. SP78&#xA;CFM @ 0.300-In. SP74&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I want to exhaust fumes from a spray booth.  I know there are many articles and plans online.  I am trying to understand the math here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems intuitive to me the Grainger fan is more powerful and will remove more air than the CPU fan.  But the math makes no sense to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the CPU fan it has .302 for Static Pressure and .3 for the Grainger and the CFM is pretty close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am moving air through 6' of ductwork, 4&quot; in diameter, and neither tell me what CFM means in this context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a 2'Lx2'Wx2'H spray box.  Again, I'm not asking for plans.  I can get that already.  I only want to understand what these numbers mean and the spray box application gives the question context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why doesn't the CPU fan move almost as much as the Blower based on the specifications?  Also, is the Delta IN H20 given static pressure or is that something else?&lt;/p&gt;&#xA;" OwnerUserId="7865" LastEditorUserId="7083" LastEditDate="2018-12-28T00:05:47.897" LastActivityDate="2018-12-28T06:53:42.360" Title="What is the difference in CFM, at a given static pressure, between a CPU fan and a Squirrel Cage Blower fan when moving air through ductwork?" Tags="|pressure|statics|airflow|hvac|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25270" PostTypeId="1" CreationDate="2018-12-27T19:05:31.760" Score="2" ViewCount="21" Body="&lt;p&gt;I'm looking at a spec sheet for a high density polyethylene woven fabric and I'm seeing the following parameters on it&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tensile Grab Strength: Warp 155 / Weft 170 - tested ASTM D-5034-09&lt;br&gt;&#xA;  Grab Tear: Warp 30 / Weft 25 - tested ASTM D-751&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I understand how warp and weft are the horizontal and vertical so clearly they're running these tests twice, once for each dimension, but I tried reading the papers on the tests that were run and they were quite dense (no pun intended) and I quickly got confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I translate this into something I can use? Is there a standard unit these are in? What is the difference between grab strength and grab tear?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm putting my products in these bags and I need to understand what these terms mean!&lt;/p&gt;&#xA;" OwnerUserId="707" LastActivityDate="2018-12-27T19:05:31.760" Title="How do I relate Grab Strength and Grab Tear to the real world?" Tags="|terminology|strength|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25271" PostTypeId="1" CreationDate="2018-12-27T21:32:38.427" Score="3" ViewCount="1627" Body="&lt;p&gt;Suppose, you put a point load on a plate that is completely supported. The pressure under this plate is Force divided by the area of the plate.&#xA;If I make this base very large, then pressure will be negligible.&#xA;There has to be a limit to the size of this base right, after which it becomes useless to increase the area.&#xA;Also, how does thickness of that plate come into play. Like at the top layer of the plate , the point load does engage the area completely , as we move vertically downwards does the area increases. The closest to reference I got was the force transfer under wheel surfaces in bridge design. How is this assumption made that the contact area increases with 45 deg. angle.&lt;/p&gt;&#xA;" OwnerUserId="18772" LastActivityDate="2020-03-16T18:57:34.420" Title="How is Load Transferred in a fully supported base" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25272" PostTypeId="1" CreationDate="2018-12-27T23:41:36.843" Score="1" ViewCount="629" Body="&lt;p&gt;I have heard titanium is the strongest and the lightest material. It is used for my glasses' frame. Is it true that it is the strongest and the lightest as well?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Strongest, the biggest forces can it bear without damage or breaking &lt;span class=&quot;math-container&quot;&gt;$\frac{Newton}{m^2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Lightest, the least weight of a certain volume of material &lt;span class=&quot;math-container&quot;&gt;$\frac{kg}{m^3}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;i.e.: a material is quite strong to bear some load, but also heavy at the same time. Another material is light enough, but can not bear any load. So, what I am asking is what is the strongest material but lightest as well.&lt;/p&gt;&#xA;" OwnerUserId="17943" LastEditorUserId="6767" LastEditorDisplayName="user16" LastEditDate="2022-02-22T05:57:29.940" LastActivityDate="2023-09-01T10:29:32.583" Title="What is the strongest and lightest material?" Tags="|materials|metallurgy|" AnswerCount="3" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25273" PostTypeId="2" ParentId="25271" CreationDate="2018-12-28T02:09:13.557" Score="1" Body="&lt;p&gt;It depends on how the plate is supported. For example, if it is supported on two edges like a one-way slab or four edges, and these edges are simply supported or can take moment, or the plate supported by a concrete pad.&#xA;AISC has some equations for calculating base plates that are based on assumption roughly that base plate distributes the load by acting as a two-way Cantilever beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Westrgaard equation assigns an equivalent radius to concentrated load on a very small radius circle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$r_\text{equivalent} = \sqrt{1.6r_0 +t^2}  - 0.675t $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$r_0$&lt;/span&gt; is the initial radius, and &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; is the plate thickness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only concern for a very large concentrated load on a very small area on a steel plate is punching shear, otherwise given enough thickness a point load will penetrate into the plate and quickly crush a rather large area of the plate and then can be dealt with as a larger radius of bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel is very tough and resilient and methods like the angle of projection are not a good model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much research has been done in the area of ballistics, armor, etc. but I guess it is not easily accessible. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="20995" LastEditDate="2020-03-16T18:57:34.420" LastActivityDate="2020-03-16T18:57:34.420" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25274" PostTypeId="1" AcceptedAnswerId="25295" CreationDate="2018-12-28T05:52:29.033" Score="2" ViewCount="3405" Body="&lt;p&gt;When we increase the temperature of water in an open container, the vapour pressure increases.But boiling (phase transition at constant temperature and pressure) occurs only when the vapour pressure equals atmospheric pressure.My understanding is, till that point the atmosphere can hold the water vapour like a cover or lid but when equilibrium partial pressure of water vapour(vapour pressure) tries to increase beyond atmospheric pressure, the atmosphere cant hold the vapour and boiling starts.Is this right?&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastActivityDate="2021-06-04T13:01:45.403" Title="Why boiling occurs when vapour pressure equals the local atmospheric pressure" Tags="|fluid-mechanics|thermodynamics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25275" PostTypeId="2" ParentId="25268" CreationDate="2018-12-28T06:53:42.360" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Why doesn't the CPU fan move almost as much as the Blower based on the specifications? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is because of the working principle of fan and blower. The fan takes the air axially(axis of rotation) and pushing it in the axial direction. But the Blower works like a centrifugal compressor in which axial inlet and radial/tangential outlet are present. Fundamentally, due to the pressure reactions, the centrifugal pump/turbines are more efficient than the axial pump/turbines. That is what you are seeing as the data in the sheets.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Also, is the Delta IN H20 given static pressure or is that something else? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Both are in the units of inches of water column.&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2018-12-28T06:53:42.360" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25276" PostTypeId="1" AcceptedAnswerId="25277" CreationDate="2018-12-28T11:58:17.907" Score="2" ViewCount="1107" Body="&lt;p&gt;Please could someone explain the reason for using negative superscripts in SI units, and more precisely when to use negative versus positive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not understand, for example, why the volume of a fuel would be given in m&lt;sup&gt;3&lt;/sup&gt; but heat of combustion would be given in J m&lt;sup&gt;-3&lt;/sup&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18779" LastActivityDate="2018-12-28T12:25:14.783" Title="What is the meaning of negative exponents in SI units?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25277" PostTypeId="2" ParentId="25276" CreationDate="2018-12-28T12:25:14.783" Score="4" Body="&lt;p&gt;Negative exponents means &lt;strong&gt;per&lt;/strong&gt;. In countries that were heavily influenced by the British, units such as meters per second would most likely be written &lt;span class=&quot;math-container&quot;&gt;$m/s$&lt;/span&gt;. In Europe and elsewhere it could be written with negative exponents as &lt;span class=&quot;math-container&quot;&gt;$m.s^{-1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example &lt;span class=&quot;math-container&quot;&gt;$J m^{-3}$&lt;/span&gt; means joules per cubic meter.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2018-12-28T12:25:14.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25279" PostTypeId="1" AcceptedAnswerId="25285" CreationDate="2018-12-28T16:14:51.590" Score="0" ViewCount="58" Body="&lt;p&gt;During a lecture my teacher told me that the UPSTREAM in ADSL is very much affected from near end crosstalk due to DOWNSTREAM, while DOWNSTREAM is not affected by this  phenomena. This lead to the decision to dedicate lower frequency to the UPSTREAM and higher frequency to the DOWNSTREAM. Is this correct? In case it is, how is the DOWNSTREAM not affected by near end crosstalk effect?&lt;/p&gt;&#xA;" OwnerUserId="18781" LastActivityDate="2018-12-28T20:31:51.047" Title="ADSL near end crosstalk (NEXT)" Tags="|telecommunication|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25282" PostTypeId="2" ParentId="23943" CreationDate="2018-12-28T19:14:41.230" Score="0" Body="&lt;p&gt;lets try not to complex out the answer, so consider a volume of liquid freely falling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets talk in a frame of reference attached to that liquid.&#xA;Effective weight of every infintesimally small element would be zero during a free fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also since we are in frame of the liquid a=0 , this a hydrostatic condition. And how does pressure difference in z direction arise in hydrostatic condition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;due to the fluids own weight , which is effectively zero in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no pressure difference should arise in this case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes Euler's equation yields the same(in ground frame), and you have done it correctly.&lt;/p&gt;&#xA;" OwnerUserId="18714" LastActivityDate="2018-12-28T19:14:41.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25283" PostTypeId="2" ParentId="25258" CreationDate="2018-12-28T19:30:32.270" Score="0" Body="&lt;p&gt;Are you sure that the diagram is correct because I believe that point 5 should be at the saturation point on the bell shape curve, and the dip should start after that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You see 4 to 1 is meant to be constant pressure process in evaporator(that absorbs heat and cause cooling) and after the saturation point constant pressure lines have a negative slope, thus causing a dip in the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A standard process that is generally demonstrated in the books has point 5(and point 1 overlapping) at the saturation point and thus no dip is shown in such a case.&lt;/p&gt;&#xA;" OwnerUserId="18714" LastActivityDate="2018-12-28T19:30:32.270" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25284" PostTypeId="1" AcceptedAnswerId="25286" CreationDate="2018-12-28T19:38:19.627" Score="1" ViewCount="142" Body="&lt;p&gt;I'm building a motorized standing desk and am trying to calculate what linear actuators I need. If I have 4 linear actuators (one on each corner) each able to lift 25kg does that mean that the desk would be able to lift a full 100kg?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, when linear actuators are sharing a load is the total drive force equal to the sum of the actuators individual drive force? Or is it more complicated than that?&lt;/p&gt;&#xA;" OwnerUserId="18571" LastActivityDate="2018-12-29T01:05:22.763" Title="Sharing load between linear actuators" Tags="|linear-motion|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25285" PostTypeId="2" ParentId="25279" CreationDate="2018-12-28T20:31:51.047" Score="0" Body="&lt;p&gt;Most of the noise is physically generated where the ISP's multiplexors split up the signal onto individual wires to each consumer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upstream data signal is generated by the consumer and has been attenuated by the cable when it reaches that point in the network. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For downstream data, the cable attenuates both the downstream signal and the noise which reaches the consumer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the signal to noise ratio is better for downstream data and the bandwidth can be higher.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-12-28T20:31:51.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25286" PostTypeId="2" ParentId="25284" CreationDate="2018-12-28T21:27:14.703" Score="2" Body="&lt;p&gt;Yes, the four will share the 100kg, each carrying 25kg. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only thing is you want to connect them to same rocker switch and install the linear actuators symmetrically with respect to the center of your desk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some comments I modify my answer this way. Lets say there is a heavy vise you need to have on the desk and it throws the center of gravity of the desk from the center to x and y centimeters off the center. Lets call the distance between the base of the legs X on the long side and Y on the short side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the loads distributed to the four legs are going to change as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load to the right hand side pair of legs &lt;span class=&quot;math-container&quot;&gt;$$ is = l00*x/X = R_{load}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ R_{load}*y/Y =load\ on\ the\ upper\ right\ leg.  $$&lt;/span&gt;&#xA;We repeat this process and find the individual loads for each leg.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="18571" LastEditDate="2018-12-29T01:05:22.763" LastActivityDate="2018-12-29T01:05:22.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25287" PostTypeId="1" CreationDate="2018-12-28T22:07:39.913" Score="1" ViewCount="28" Body="&lt;p&gt;Consider a flexible square of uniform density with vertices (0,0), (0,1), (1,0), (1,1). I need the 4 coordinates for rolling contacts along the diagonals that will stably support it. Someone could just test it. Open to all thoughts and ideas. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be similar to the Bessel Points problem, but now in 2D. Guessing the answer is around (.22, .22), (.22, .78), (.78, .22), (.78, .78). How can I find this answer?&lt;/p&gt;&#xA;" OwnerUserId="18785" LastActivityDate="2018-12-28T23:04:59.150" Title="Supporting flexible square plate of uniform density in equilibrium with only 4 rolling contacts placed along the square's diagonals" Tags="|structural-analysis|statics|beam|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25288" PostTypeId="2" ParentId="14731" CreationDate="2018-12-28T22:59:40.793" Score="1" Body="&lt;p&gt;A thick paper and treated the same way you treat a catalyst bead to become porous so the surface area can get more light. From this if you use a thick clear uncoated paper and leave it a few seconds in a ultrasonic cleaner then leave it to dry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea is a non-toxic dye that changes color when placed in a water environment like a compound molecule that only associates when water is present, after it dries the compound breaks and the color fades. You could try with copper but be careful how you handle it because some ions are toxic. &lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2018-12-28T22:59:40.793" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25289" PostTypeId="2" ParentId="25287" CreationDate="2018-12-28T23:04:59.150" Score="1" Body="&lt;p&gt;I recommend you check Roark's hand book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I remember it is around page 425.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically the idea is to have the slope of the plate after deflection to be the same on inside and outside of the supports rollers.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-28T23:04:59.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25290" PostTypeId="1" CreationDate="2018-12-28T05:58:42.493" Score="1" ViewCount="1016" Body="&lt;p&gt;I am having trouble determining the maximum weight capacity a motor can handle. I provided the specs below. The calculations are for the top row or the 580KV model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to start off,I'm only taking maximum values, I took the Watts and converted it to HP. About 10.06HP. Then I took this and converted it to Newtons. Around 7480N and my weight of about 725N + the vehicle at 88N so 813N. So would this motor be able to move 813N from a stand still on 27in tires? Assuming it's flat ground and no wind. If not would it be able to if already traveling at 10mph? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side not how fast would it be traveling if the wheel diameter is 27in, the rpm is 30,000, it's producing 7500W of power, with 82Kg of weight onboard, and the gear ratio is 1.25:1? Also acceleration? thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xuFMK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xuFMK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user6596282" LastActivityDate="2024-04-01T15:04:00.467" Title="Calculating max weight capacity of brushless motor" Tags="|power|mechanical|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25291" PostTypeId="2" ParentId="25290" CreationDate="2018-12-28T07:16:44.380" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;27&quot; diam= 13.5&quot; radius &lt;/li&gt;&#xA;&lt;li&gt;circumference  = 84.8&quot; = 7.0686 '&lt;/li&gt;&#xA;&lt;li&gt;10 MPH = 14.6667 fps ['/s]  &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;27&quot;D @ 10 MPH = 2.075 revs per sec = 124.5 RPM or 12.45 RPM/MPH&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Max RPM = 30 kRPM @ 580 RPM/V &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;yet max mechanical HP may be at 80~85% no load RPM, depending on gear or pulley losses and range.&lt;/li&gt;&#xA;&lt;li&gt;Max Power = 7500 W&lt;/li&gt;&#xA;&lt;li&gt;Gear ratio, G needed for max voltage,power @ X MPH&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;assume 82% of no load RPM&lt;/li&gt;&#xA;&lt;li&gt;G = 0.82*20,000/(12.45 X) = 1317:1 pulley ratio ( cascaded) e.g. 131.7 for 10 MPH, 65.9 for 20 MPH&lt;/li&gt;&#xA;&lt;li&gt;or G= 66 at 20 MPH @ 54V and 10 MPH @ 27Vdc&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8847" OwnerDisplayName="Tony EE rocketscientist" LastActivityDate="2018-12-28T07:16:44.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25292" PostTypeId="1" AcceptedAnswerId="25293" CreationDate="2018-12-29T05:41:42.103" Score="4" ViewCount="552" Body="&lt;p&gt;I am interested in building a very basic water jet propulsion system for a toy boat. Before I build it, I would like to know from a conceptual standpoint if this design of a water jet propulsion system will actually produce forward thrust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please refer to the conceptual drawing below of this water jet design. This drawing shows a top-down view of the toy boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An embedded motor-propeller will pull water into the pipe section coming from the stern of the toy boat and will at the same time be forcing water out of the pipe section going back to the stern of the toy boat. The force which should propel the toy boat forward should come from the dynamic pressure of the rushing water pushing against the inner walls of the two 90 degree elbow sections of the pipe. Will this particular design of a water jet produce forward thrust as expected?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I know that a traditional inline water jet system would be the more ideal thing to construct, I am very interested in finding out if a boat can be propelled using just the dynamic pressure(s) generated within a pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RhbnN.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RhbnN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user18610" LastActivityDate="2018-12-30T16:58:26.230" Title="Will this water jet propulsion system design produce any forward thrust?" Tags="|mechanical-engineering|fluid-mechanics|propulsion|marine-engineering|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25293" PostTypeId="2" ParentId="25292" CreationDate="2018-12-29T08:24:57.630" Score="5" Body="&lt;p&gt;No, it would just create a torque couple and rotate the boat anticlockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the small propeller has an output volume, q grams/s, and the distance between inlet pipe and outlet is 5cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thrust and suction of each end of pipe.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F = ρ q (v2 - v1) $$&lt;/span&gt; &#xA;say density of water is =1, and V1 is initially zero, for simplicity, even though it wouldn't affect the outcome either way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F = qV2   \ and\ V2 = q/a\ :a\ is\ pipe's\ area$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you have a torque,&#xA;&lt;span class=&quot;math-container&quot;&gt;$T = 5*q^2/a. $&lt;/span&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;anticlockwise\ direction   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will turn the boat in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some comments I added a bit more detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The OP's sketch has 6 nodes or bends that we calculate. Note we analyse the sketch as it is, not introducing any changes or recommendations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entrance and exit momentums and thrusts have been done above and remain the same, including the torque they cause.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let's call the four bends from left to right c1, c2, c3 , c4. these corners each experience reaction force &lt;span class=&quot;math-container&quot;&gt;$ \ F= ρ q (v2 - v1) = ρ q (v1sin(45) - v1) + ρ q (v1 cos(45) - v1) $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if we project these two vector component on the x and y axis at C1 we have &lt;span class=&quot;math-container&quot;&gt;$$ \ F_{c1} = q^2/a$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and its direction is 135 degrees at c1. And it's reaction is pushing the boat back at 135 degrees out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At c2 we end up with the same reaction as Fc1 and same 135 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At c3 we have same reaction but pushing the ship at 45 dgrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At c4 we have the same reaction again pushing the boat out at 45 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal components of these vectors cancel out and the vertical components add up to &lt;span class=&quot;math-container&quot;&gt;$$ q^2/a*4* \sqrt{2}/2 = 4* 0.707 = 2.82 q^2/a $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is forward thrust pushing the boat forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the boat is still rotating under the combined torque and thrust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is actually one of the methods cruise ships use as a docking maneuver to turn in place in ports with limited space.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2018-12-30T00:24:55.713" LastActivityDate="2018-12-30T00:24:55.713" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25295" PostTypeId="2" ParentId="25274" CreationDate="2018-12-29T10:38:03.683" Score="2" Body="&lt;p&gt;The molecules in the liquid phase are more confined than in the vapor phase.Temperature and molecular repulsion tend to separate the molecules while the pressure and molecular attraction tend to confine the molecules. When any substance in the liquid phase, at atmospheric pressure, it means that this pressure is sufficient to confine the molecules together to form liquid. &#xA;The molecules should possess equal energy to overcome this pressure for the liquid to boil. When this energy is supplied as heat, the repulsion increases so that the molecules are able to overcome the atmospheric pressure which is holding the molecules together. Hence the molecular confinement can be broken only when the vapor pressure becomes equal to or greater than the atmospheric pressure which is referred to as boiling.&lt;/p&gt;&#xA;" OwnerUserId="18783" LastActivityDate="2018-12-29T10:38:03.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25296" PostTypeId="2" ParentId="5865" CreationDate="2018-12-29T17:06:40.973" Score="0" Body="&lt;p&gt;You are conflating two different ideas here, probably by accident. The spline curve achieved with a spline and ducks is a minimum energy curve. The defining points have support over the entire length of the spline, meaning moving one affects the entire length of the curve. That doesn't seem to be the thrust of you question, but it plays a major roll in the history of surface representation. By contrast, with nurbs and subdivision surfaces, control points exercise local control, beyond which the surface is unaffected. When designing a surface, it is often very useful to be able to say with certainty that some variation is localized. Imagine wanting to run a series of different ship bow designs through a performance model, all affixed the same aft hull.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Support_(mathematics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Local Support: Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In mathematics, the support of a real-valued function f is the subset of the domain containing those elements which are not mapped to zero. If the domain of f is a topological space, the support of f is instead defined as the smallest closed set containing all points not mapped to zero. This concept is used very widely in mathematical analysis.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Non-uniform_rational_B-spline&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nurbs: Wikipedia&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In many applications the fact that a single control point only influences those intervals where it is active is a highly desirable property, known as local support. In modeling, it allows the changing of one part of a surface while keeping other parts unchanged.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now for what you did ask about - interpolation vs approximation schemes. These can apply to either case - local or non-local, but the effort required increases as the span of the variation increases (ie. using higher order nurbs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a comment, alephzero said &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There is no reason why you can't define B-splines using only points on&#xA;  the curve. If your CAD system won't let you do that, it's a limitation&#xA;  of your CAD software, not of the spline formulations. It doesn't&#xA;  require any math beyond high-school algebra.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is a bit disingenuous with respect to the algorithms needed to pull off interpolating schemes. It requires an optimization step not found in approximation schemes. These have to be &lt;a href=&quot;https://www.hindawi.com/journals/ijmms/2013/893414/&quot; rel=&quot;nofollow noreferrer&quot;&gt;implemented with some care&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Subdivision_surface&quot; rel=&quot;nofollow noreferrer&quot;&gt;Subdivision Surfaces: Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Subdivision surface refinement schemes can be broadly classified into two categories: interpolating and approximating. Interpolating schemes are required to match the original position of vertices in the original mesh. Approximating schemes are not; they can and will adjust these positions as needed. In general, approximating schemes have greater smoothness, but editing applications that allow users to set exact surface constraints require an optimization step.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;[...]  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;After subdivision, the control points of the original mesh and the new generated control points are interpolated on the limit surface. The earliest work was the butterfly scheme by Dyn, Levin and Gregory (1990), who extended the four-point interpolatory subdivision scheme for curves to a subdivision scheme for surface. Zorin, Schröder and Sweldens (1996) noticed that the butterfly scheme cannot generate smooth surfaces for irregular triangle meshes and thus modified this scheme. Kobbelt (1996) further generalized the four-point interpolatory subdivision scheme for curves to the tensor product subdivision scheme for surfaces. Deng and Ma (2013) further generalized the four-point interpolatory subdivision scheme to arbitrary degree.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For free-form design, the approximation technique is easier to implement. The control points need to be chosen with care by the designer in both cases, but it is easier to understand &lt;em&gt;why&lt;/em&gt; you have to exercise care with the approximation techniques. For importing a surface from another format, interpolation techniques may be desirable, at least to form an initial coarse mesh, with an operator putting these in the right place for the job.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-12-29T17:42:12.723" LastActivityDate="2018-12-29T17:42:12.723" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25297" PostTypeId="2" ParentId="25292" CreationDate="2018-12-29T18:02:08.750" Score="1" Body="&lt;p&gt;The water starts out stationary. It ends up being jetted out the stern. Yep, that makes thrust. There will be some pressure interaction over the entire hull of the ship. That is geometry dependent. It can be engineered to minimize the parasitic drag from collecting the inlet from the stern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically a &lt;a href=&quot;https://en.wikipedia.org/wiki/Thrust_reversal&quot; rel=&quot;nofollow noreferrer&quot;&gt;thrust reverser&lt;/a&gt; in reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amrican Airlines jet powerback video - &lt;a href=&quot;https://www.youtube.com/watch?v=-Zkxh903s_w&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=-Zkxh903s_w&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-12-29T18:08:22.497" LastActivityDate="2018-12-29T18:08:22.497" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25298" PostTypeId="1" CreationDate="2018-12-29T18:22:55.677" Score="0" ViewCount="236" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ssOxo.jpg&quot; alt=&quot;the radiator&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently designing a Hottub for my scouting. This isn't the first time we are doing this. In the past, we build a wood stove around a radiator as shown above. Now I'm wondering if this is efficient. I've read that a radiator uses convection where warm air goes up and warm water cools down. We use the radiator to warm up water my applying warm air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any great thoughts about this system are more than welcome.&lt;/p&gt;&#xA;" OwnerUserId="18793" LastEditorUserId="1832" LastEditDate="2018-12-29T22:12:30.863" LastActivityDate="2018-12-30T09:54:39.517" Title="Is using a radiator to warm water inefficient and if so why" Tags="|thermodynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25299" PostTypeId="1" AcceptedAnswerId="25304" CreationDate="2018-12-29T18:53:18.250" Score="1" ViewCount="13750" Body="&lt;p&gt;Here is a picture of a caliper that I found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mo0uR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mo0uR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calipers are such basic tools, but I have trouble understanding how they work. Here the main scale says &quot;mm&quot; at the end, and the Vernier scale says &quot;0.05mm&quot;. I've read some explanations saying that if the Vernier constant is, say, 0.1, on the Vernier scale the &quot;1&quot; is at a distance of 9/10 from 0, so that if an object that is 0.1mm thick is measured, the &quot;1&quot; on the Vernier scale is moved 1/10 to the right and the &quot;1&quot; is aligned with the &quot;1&quot; on the main scale, and therefore the measurement is taken to be 0.1mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But here we have a Vernier scale with 0.05mm as the constant. Shouldn't this mean that the &quot;1&quot; of the Vernier scale should be 0.05mm short of the &quot;1&quot; on the main scale? But this is clearly not the case, the &quot;1&quot; spacing is clearly much larger than the &quot;1&quot; spacing on the main scale. &lt;strong&gt;So how does this caliper function?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2018-12-30T19:09:58.457" Title="Confused as to how to read this caliper" Tags="|measurements|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25300" PostTypeId="2" ParentId="25199" CreationDate="2018-12-29T19:25:05.343" Score="1" Body="&lt;p&gt;Solution was simple, select '&lt;em&gt;View-&gt;Toolbars-&gt;View(Heads-Up)&lt;/em&gt;' from the toolbar -&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YM3K3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YM3K3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lol &lt;/p&gt;&#xA;" OwnerUserId="18716" LastActivityDate="2018-12-29T19:25:05.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25301" PostTypeId="1" CreationDate="2018-12-29T19:37:22.737" Score="1" ViewCount="32" Body="&lt;p&gt;I'm trying to find state-space representation of following &lt;code&gt;2DOF&lt;/code&gt; systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ispYk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ispYk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried classic &lt;code&gt;Double pendulum equations&lt;/code&gt; for it but I'm confused how to handle it. now I'm looking for any help and tip to solve it. &lt;/p&gt;&#xA;" OwnerUserId="18795" LastEditorUserId="18795" LastEditDate="2018-12-29T19:42:51.267" LastActivityDate="2018-12-29T19:42:51.267" Title="Differential equations strange 2 DOF systems" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25302" PostTypeId="2" ParentId="25298" CreationDate="2018-12-29T19:47:44.677" Score="1" Body="&lt;p&gt;So ideally, you would want to counterflow the hot air and water. But the problem here is capacity. The efficiency gains from counterflow designs are only significant when you have gobs of capacity and not much in the way of temperature differences. Since you don't want to boil the water, and you really just need to maximize the amount of heat transfer, and the wood stove is capable of boiling the water, you would apply parallel flow here. The hot air can't boil the cool feedwater, and you design it so the air cools to the point it can't boil the hot side either. This is a matter of controlling the water flow and airflow at the design stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot tub heaters typically are several times the size of a house furnace. So trying to get that kind of heat to transfer though a single radiator is going to be tough. We used 210,000 btu heaters for the tubs at a resort in Florida where I worked. Hot water cast iron radiators work at about 170 btu per segment. You can bump this up quite a bit using forced hot air and greater water flow, but you still have a capacity problem. You are short by a factor of 100. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have to be extremely careful that the heater materials are compatible with the water treatment chemicals for whatever type of spa you have. The spa may need a chemistry that can't be tolerated by the metals in the radiator at the temps you need to run it at as a heater. Plastered tile and concrete spas are the toughest and require SS plumbing throughout to handle the chemicals. Plastic and fiberglass spas are a bit easier.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2018-12-29T19:57:32.447" LastActivityDate="2018-12-29T19:57:32.447" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25303" PostTypeId="2" ParentId="25299" CreationDate="2018-12-29T20:05:48.503" Score="1" Body="&lt;p&gt;The 0.05mm is the readability  of the caliper and thereby, the tolerance of your measurement.&lt;br&gt;&#xA;Reading the caliper is the same as any other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Vernier_scale&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Vernier_scale&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18796" LastActivityDate="2018-12-29T20:05:48.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25304" PostTypeId="2" ParentId="25299" CreationDate="2018-12-29T20:19:19.423" Score="1" Body="&lt;p&gt;The spacings are larger just to make it easier to read and make more room for the numbers 0 … 10 on the Vernier scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The marks on the Vernier scale correspond to 0.00, 0.05, 0.10 … 1.00 mm. For example, when the mark labeled &quot;2&quot; lines up with a tick mark on the main scale, you add 0.20mm to the main scale reading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't matter &lt;em&gt;which&lt;/em&gt; tick mark on the main scale the &quot;2&quot; lines up with. Obviously the &quot;2&quot; mark will line up with different marks if you measure objects with size 0.20 mm, 1.20 mm, 2.20 mm, etc. The essential point is that &lt;em&gt;only&lt;/em&gt; the Vernier scale &quot;2&quot; mark will line up with &lt;em&gt;any mark&lt;/em&gt; on the main scale for an increment of 0.20mm.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2018-12-29T20:19:19.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25305" PostTypeId="1" AcceptedAnswerId="25307" CreationDate="2018-12-30T00:56:48.123" Score="1" ViewCount="2019" Body="&lt;p&gt;So I’m driving down the highway in my sedan on a speed of lets say 60 mph. Why is it that my car sways just before it passes or it passes me? Almost always where the semis first tire is (closest to the hood)? What forces are acting upon it? What could be a possible explanation to why it happens? Why is it only at that spot that the car sways? &lt;/p&gt;&#xA;" OwnerUserId="18797" LastEditorUserId="110" LastEditDate="2018-12-30T01:37:52.670" LastActivityDate="2018-12-30T01:59:40.250" Title="Passing by a semi truck on a highway" Tags="|automotive-engineering|airflow|car|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25306" PostTypeId="2" ParentId="2076" CreationDate="2018-12-30T01:43:42.447" Score="0" Body="&lt;p&gt;Nikola Tesla stated in his patent that in order to achieve the highest economy,the &lt;a href=&quot;https://teslauniverse.com/nikola-tesla/patents/us-patent-1061206-turbine&quot; rel=&quot;nofollow noreferrer&quot;&gt;changes in the velocity and direction of movement of the fluid should be as gradual as possible.&lt;/a&gt; This observation led to the turbine's &lt;strong&gt;cylindrical design&lt;/strong&gt; where, by the sheer shape of the turbine, the incoming fluid is forced to move in a spiral. In this way the fluid moves in the longest path and changes in the direction of movement are minimized. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The object of my invention is to overcome these deficiencies and to&#xA;  effect the transmission and transformation of mechanical energy&#xA;  through the agency of fluids in a more perfect manner and by means&#xA;  simpler and more economical than those heretofore employed. I&#xA;  accomplish this by causing the propelling fluid to move in &lt;strong&gt;natural&#xA;  paths or stream lines of least resistance,&lt;/strong&gt; free from constraint and&#xA;  disturbance such as occasioned by vanes or kindred devices, and &lt;strong&gt;to&#xA;  change its velocity and direction of movement by imperceptible&#xA;  degrees,&lt;/strong&gt; thus avoiding the losses due to sudden variations while the&#xA;  fluid is imparting energy.&quot; — &lt;a href=&quot;https://teslauniverse.com/nikola-tesla/patents/us-patent-1061206-turbine&quot; rel=&quot;nofollow noreferrer&quot;&gt;TESLA PATENT 1,061,206 TURBINE&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Rotor is designed to provide the incoming fluid least resistance and to use the fluid's &lt;a href=&quot;https://en.wikipedia.org/wiki/Boundary_layer&quot; rel=&quot;nofollow noreferrer&quot;&gt;boundary-layer&lt;/a&gt; property in running the turbine. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;It is well known that a fluid possesses, among others, two salient&#xA;  properties, adhesion and viscosity. Owing to these a solid body&#xA;  propelled through such a medium encounters a peculiar impediment known&#xA;  as “lateral” or “skin resistance,” which is twofold, one arising from&#xA;  the shock of the fluid against the asperities of the solid substance,&#xA;  the other from internal forces opposing molecular separation. As an&#xA;  inevitable consequence a certain amount of the fluid is dragged along&#xA;  by the moving body. Conversely, if the body be placed in a fluid in&#xA;  motion, for the same reasons, it is impelled in the direction of&#xA;  movement. These effects, in themselves, are of daily observation, but&#xA;  I believe that I am the first to apply them in a practical and&#xA;  economical manner in the propulsion of fluids or in their use as&#xA;  motive agents.&quot; — &lt;a href=&quot;https://teslauniverse.com/nikola-tesla/patents/us-patent-1061206-turbine&quot; rel=&quot;nofollow noreferrer&quot;&gt;TESLA PATENT 1,061,206 TURBINE&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The Tesla's turbine should be able to pick up the energy of propelling fluid very efficiently. This is the unique design of Tesla's turbine.&lt;/p&gt;&#xA;" OwnerUserId="18760" LastActivityDate="2018-12-30T01:43:42.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25307" PostTypeId="2" ParentId="25305" CreationDate="2018-12-30T01:58:24.327" Score="4" Body="&lt;p&gt;A truck speeding down a highway creates an envelope stream of high pressure air surrounding it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is basically a wake, composed of layers of high pressure shockwaves, created when the front of the truck, hood and cabin penetrate the still air and push it open. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shockwave moves with the truck and after turning into a small turbulent tail at the backside dissipate slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you drive near a truck in passing, or it passes by you, your car cuts into the boundary of this wake. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your car's aerodynamics and your driving habits, the impact of incursion may become more pronounced at a certain point and angle. Some times this impact may be strong enough to steer the car closer to the truck. Or you my lose control by trying to avoid the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I keep away from the big trucks as much as I can, or else try to anticipate the shock and be prepared.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-30T01:58:24.327" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25308" PostTypeId="2" ParentId="25305" CreationDate="2018-12-30T01:59:40.250" Score="1" Body="&lt;p&gt;As far as I understand a semi track travelling on highway has at least three different zones of interest. They are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Front &lt;/li&gt;&#xA;&lt;li&gt;Middle  &lt;/li&gt;&#xA;&lt;li&gt;Back &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Below are the three regions pictorially.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3pZAj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3pZAj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kOyZH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kOyZH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lKyrK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lKyrK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n5c3w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5c3w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order for the truck to pass through it diverts outwards air in front of the semi truck. The empty area  middle (between the tractor and rear wheels) it creates vacuum which is reason sometime small car like the sedan get dragged toward the semi. The back of the semi also creates air void thus and air gets pulled in thus a small sedan could swayed into the lane the semi is travelling. &lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2018-12-30T01:59:40.250" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="25309" PostTypeId="2" ParentId="25298" CreationDate="2018-12-30T09:54:39.517" Score="0" Body="&lt;p&gt;The simplest I have seen is where people have made a hot tub in the garden and built a small woodstove below the level of the tub that has a small water jacket or coil which is directly connected to the tub water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the woodstove is lit it heats the water using thermosyphon flow (all pipes slope so the flow is not impeded), once the tub is hot enough - stop adding wood ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No controls needed and have the drain at the lowest point...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2018-12-30T09:54:39.517" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25311" PostTypeId="2" ParentId="25292" CreationDate="2018-12-30T11:24:17.693" Score="0" Body="&lt;p&gt;Based on the other answers to this question, this simple water jet design should produce forward thrust. Yet for the toy boat to travel forward in a straight line it needs an additional part added to it. To cancel out the anticlockwise rotation which will be caused by the combined thrust and torque, I believe adding a small rudder into the inlet pipe section will provide the means of generating a clockwise torque to cancel out the anticlockwise torque. See modified picture below. I will probably need to keep readjusting the rudder towards the Port side and fixing it in place until I find the ideal position to make the toy boat travel in a straight line. I will then use the toy boat's main rudder to steer the boat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/POHaB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/POHaB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user18610" LastEditorDisplayName="user18610" LastEditDate="2018-12-30T16:58:26.230" LastActivityDate="2018-12-30T16:58:26.230" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25312" PostTypeId="2" ParentId="21002" CreationDate="2018-12-30T12:06:01.723" Score="4" Body="&lt;h1&gt;If it worked he would not need the engine to maintain speed&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Ask your friend: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If this is &quot;perfect&quot; — as you say — with no friction, no losses... then why do you need the engine to maintain cruising speed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once you reach cruising (i.e. constant) speed, the job of an engine is to give energy to the rest of the car for all the energy it loses to air friction and road friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your friend did not have air or road friction — say by being in a &lt;a href=&quot;https://en.wikipedia.org/wiki/Hyperloop&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hyperloop&lt;/a&gt; — then all they would need is a shove to get up to speed, and &lt;strong&gt;then they would turn the engine off&lt;/strong&gt; and just keep coasting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So their own premise proves why it will not work, because if they &lt;strong&gt;need&lt;/strong&gt; an engine to maintain speed, then they are having losses, and the system is not &quot;perfect&quot; as postulated.&lt;/p&gt;&#xA;" OwnerUserId="4716" LastEditorUserId="4716" LastEditDate="2018-12-30T17:25:15.553" LastActivityDate="2018-12-30T17:25:15.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25313" PostTypeId="1" AcceptedAnswerId="25338" CreationDate="2018-12-30T12:31:42.407" Score="3" ViewCount="267" Body="&lt;p&gt;I had an appliance (wireless audio receiver), which has operational range 662-698 mhz (and had same ranged antenna exactly from factory), like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U9Jtw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U9Jtw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, that antenna was broken accidentally and I want to replace the antenna, but can't find exactly that model. However, I've found a bit wider ranged antenna:  632-698 mhz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can that work without problems? may i replace with that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: exact model is: shure &lt;code&gt;ULXP4 (662-698 mhz)&lt;/code&gt;: &lt;a href=&quot;https://web.archive.org/web/20140113201654/http://cdn.shure.com/brochure/upload/38/pdf_2010_ulx_brochure.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sophisticated, scalable&#xA;wireless solutions&lt;/a&gt; (also here specs: &lt;a href=&quot;http://alss1.com/wp-content/uploads/2013/07/ULX4PK.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ULX Wireless System Specification Sheet&lt;/a&gt;)  and replacement antenna I want is: &lt;a href=&quot;http://bit.do/fgergsadf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://bit.do/fgergsadf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12812" LastEditorUserId="282" LastEditDate="2019-05-30T23:35:42.460" LastActivityDate="2019-05-30T23:35:42.460" Title="Can wider range (i.e. 600-700 MHz) antenna replace lower range (650-700 MHz) antenna?" Tags="|wireless-communication|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25315" PostTypeId="2" ParentId="25299" CreationDate="2018-12-30T19:09:58.457" Score="4" Body="&lt;p&gt;The problem with most analog meters and measuring devices is you have no feedback on your reading to confirm you are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.stefanelli.eng.br/en/virtual-vernier-caliper-simulator-05-millimeter/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Metric vernier caliper, (calliper or pachymeter), read in millimeter and vernier scale 0.05 mm&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This site explains how the vernier caliper works in detail, but more importantly has an interactive applet at the bottom.  You can move the caliper and it provides the reading.  The magnifying glass allows you to zoom in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 0 marker gives you the first two digits.  Where the movable arm aligns exactly with the fixed caliper gives you the digits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cSICw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cSICw.png&quot; alt=&quot;Vernier Calipers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image shown, the 0 on the movable arm is after 27 and checking the 0 to 10 scale on the movable arm aligns with the fixed arm at 7.5.  This makes the reading 27.75mm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice there are 20 divisions on the movable arm, which corresponds to the nearest 0.5mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make a few measurements and this will get you over the learning curve.  Put paper over the bottom of the screen to help you learn.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2018-12-30T19:09:58.457" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25316" PostTypeId="2" ParentId="16976" CreationDate="2018-12-30T23:24:16.400" Score="0" Body="&lt;p&gt;Rough estimate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the car is not moving fast- Let's call the flat contact surface of tire A, and forget about the bending flexure resistance of the side-walls and the leading and trailing edges of this surface.&#xA;&lt;span class=&quot;math-container&quot;&gt;$ A*air\ pressure \ = tire\ load  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;span class=&quot;math-container&quot;&gt;$ A = \frac{tire\ load }{air\ pressure} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from there having the section dimensions of the tire we can calculate the effective radius, which is the distance from a circle section A to its center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this can hardly be a good estimate when the car is accelerating and the torque from the wheel is warping the walls of the tire, pinching and wrinkling the attack edge of the flat surface, or the tire is hot after lengthy trip. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-30T23:24:16.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25318" PostTypeId="1" AcceptedAnswerId="25319" CreationDate="2018-12-31T11:49:09.797" Score="0" ViewCount="925" Body="&lt;p&gt;I am trying to calculate some rather long and complex drive trains, this is a project that I have worked up to, but its the first  problem of its kind I have come across, to try and explain the position I am in please follow my example of how I am trying to calculate the final ratios and if I am going wrong, please can you point out where I have gone wrong and if you have the time, why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&#xA;In the image below is a simple drive train, it consists of three shafts with four gears turning.( I know it doesn't look it, but please assume all teeth are of the correct size and mesh with no problems)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Shaft 1&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;Input Gear 1 (Purple) has 24 teeth&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Shaft 2&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;Driven 2 (Red) has 12 teeth&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;Driven 3 (Blue) has 15 teeth&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Shaft 3 &lt;/li&gt;&#xA;&lt;li&gt;&lt;ul&gt;&#xA;&lt;li&gt;Output Gear(Green) has 10 teeth.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the power travels through Purple,Red, Blue Green.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for my Math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;though a 2:1 is connected to a 3:2 I have been calculating  the ratio like this, but I am sometimes getting different answers with long (up 12 gears) trains in my project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would calculate the final ratio like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(12/24)x(15/12)x(10/15) = 0.416(RECUR.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I am going wrong&#xA;Is the Final ratio 100:416 (1:0.416)&#xA;or is it &#xA; 1/0.416 = 1.666 = ~16:10  or is it something else?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/P57ZR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P57ZR.png&quot; alt=&quot;Image shows example drive train as described in the example above&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is greatly appreciated, im stuck at the moment and need to get on!&lt;/p&gt;&#xA;" OwnerUserId="18806" LastActivityDate="2018-12-31T12:11:14.447" Title="How can I calculate final gear ratio when the drive train contains multiple gears with some reductions on the same shaft?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25319" PostTypeId="2" ParentId="25318" CreationDate="2018-12-31T12:11:14.447" Score="1" Body="&lt;p&gt;According to the &lt;a href=&quot;https://woodgears.ca/gear/ratio.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Woodgears&lt;/a&gt; web site, one takes the gear ratios independently and &quot;adds them together&quot; as shown in the video within the linked site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, the progression is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;24:12, 15:10 which reduces to 2:1 and 3:2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two (first ratio) times one and one-half (second ratio) is 3, which provides for a 3:1 increase or a 1:3 reduction if the reverse is created. There is no math between the red and blue gears as one rotation of the red is one rotation of the blue.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2018-12-31T12:11:14.447" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25320" PostTypeId="1" AcceptedAnswerId="25325" CreationDate="2018-12-31T12:40:27.913" Score="0" ViewCount="105" Body="&lt;p&gt;For example we have an inverted pendulum attached to a cart, with the cart having a speed limitation. Having derived the linearized model, how can I calculate the maximum possible deviation on the angle, and angular rate from the linearized model and the speed limitation ?&lt;/p&gt;&#xA;" OwnerUserId="18808" LastActivityDate="2018-12-31T19:05:57.897" Title="How can I calculate the maximum possible deviation from equilibrium, that can still be compensated, given a limitation on the control input?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25321" PostTypeId="1" AcceptedAnswerId="25324" CreationDate="2018-12-31T14:00:58.020" Score="0" ViewCount="64" Body="&lt;p&gt;I can roughly understand the working of an axial flow gas turbine i.e. the fluid moves over the blade and gets redirected, its whirl velocity component changes and the blade is pushed in the opposite direction.But I couldnt understand the working of a table fan or an aircraft propeller.I know that its the opposite of an axial flow gas turbine but couldnt really imagine it.Please explain the principle?&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastEditorDisplayName="user17332" LastEditDate="2018-12-31T16:01:05.140" LastActivityDate="2018-12-31T18:45:05.813" Title="Working of a fan or propeller" Tags="|thermodynamics|turbomachinery|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25322" PostTypeId="2" ParentId="25313" CreationDate="2018-12-31T15:46:27.240" Score="2" Body="&lt;p&gt;As mentioned in the comments, it should work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only concern that I have is &lt;a href=&quot;http://alss1.com/wp-content/uploads/2013/07/ULX4PK.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;per the spec sheet&lt;/a&gt;, you're moving from a 1/2 wave antenna to a full wave antenna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the other 1/2 wave antennas though, it appears that Shure has picked frequencies that don't overlap when looking at the full wave.  So I don't think that you'll see any cross channel issues when using the replacement full wave antenna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The replacement full wave antenna is not likely tuned to the same frequency as the old 1/2 wave antenna.  So that could (will) impact the efficiency in transmit and receive, which translates into a decreased range for the device.&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2018-12-31T15:46:27.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25323" PostTypeId="1" AcceptedAnswerId="25329" CreationDate="2018-12-31T18:36:14.217" Score="0" ViewCount="95" Body="&lt;p&gt;According to Elon in &lt;a href=&quot;https://youtu.be/nSIzsMlwMUY?t=786&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video (@13:06)&lt;/a&gt;, he states that the reinforced segments are 70% just dirt, but in normal tunneling, they actually truck the dirt out and bring new dirt in which seems very inefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it really that simple or is Elon oversimplifying the challenge?&lt;/p&gt;&#xA;" OwnerUserId="18812" LastActivityDate="2019-01-01T08:25:14.803" Title="Why do normal tunnel boring operations not use the dirt (from the boring itself) for reinforcement rather than bringing it from other sources?" Tags="|civil-engineering|tunnels|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25324" PostTypeId="2" ParentId="25321" CreationDate="2018-12-31T18:45:05.813" Score="1" Body="&lt;p&gt;A table fan or a plane propeller have blades, pitched into an angle, which cuts  into stationary air and acelerates it forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are basically rotating wings, but working in reverse sense. Wings move up and create lift when wind goes through them. Fans and proppellers go through the air and cause it to move. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a moving car if you hold your hand out of the window and try to keep your open hand like a row but gently angled you feel the pressure of the air trying to lift your arm up, and if you resist it you feel you are pushing the air down. And you may wish you had bigger hands to fly with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fan does the exact thing, by moving through the air in a screwdriver motion, it pushes it forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could imagine tiny stationary air bubbles being plowed through by the blade and proppeled by the slant on the surface of it.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-31T18:45:05.813" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25325" PostTypeId="2" ParentId="25320" CreationDate="2018-12-31T19:05:57.897" Score="1" Body="&lt;p&gt;Lets set the following variables,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Length of the pendulum = r&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- Mass of pendulum = m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Small angle from vertical = a&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the horizontal acceleration of you pendulum, &quot;A&quot; is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A = \frac{mg *a}{r}  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your control system has to be able to compensate for at least this acceleration A just to keep the tilted pendulum in check, a bit more to bring it back to stable, vertical position.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2018-12-31T19:05:57.897" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25326" PostTypeId="2" ParentId="25323" CreationDate="2019-01-01T00:07:16.910" Score="2" Body="&lt;p&gt;With present technology dirt is one of the worst possible concrete additives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It brings the compressive strength of concrete down to unacceptable levels even at fractions of less than 2%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elon Musk is one of the few people who has earned the license to say unbelievable things. If his team has come up with a paradigm changing concrete making method,  many desert countries, with unlimited dirt supplies would love to hear about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for making brick out of dirt, cost of brick making is not in  the polluted raw material, it is manufacturing, fuel, removing pollutants, etc. That's why clean, quality clay and other ingredients are essential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brick manufacturing also has a large carbon footprint.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-01T00:07:16.910" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25327" PostTypeId="2" ParentId="24500" CreationDate="2019-01-01T02:06:10.147" Score="0" Body="&lt;p&gt;As long as the pipe intake is higher than the stream bed, it cannot drain it dry.&#xA;The bottom is sized to maintain a minimum safe flow rate for the stream ecology.&#xA;The middle is sized for the maximum flow rate of your application.  A dry pipe signals that the stream ecology is in danger.&#xA;The top is for overflow when your intake pipe is full.&lt;/p&gt;&#xA;" OwnerUserId="9532" LastActivityDate="2019-01-01T02:06:10.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25329" PostTypeId="2" ParentId="25323" CreationDate="2019-01-01T08:25:14.803" Score="1" Body="&lt;p&gt;Regarding the word &lt;strong&gt;dirt&lt;/strong&gt;. It depends on what a person means by dirt. Generally, it is understood that dirt means loose material such as soil or sand on or near the surface of the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In mining however, and this may apply to tunneling, the word dirt can sometimes be used used to describe the broken rock, or mined material, that is excavated from the mine or tunnel in the process of mining or tunneling. In this situation it's a form of jargon or colloquialism used by some in the English speaking mining community and possibly the tunneling community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One reason why locally excavated &quot;dirt&quot; may not be suitable as a ground reinforcing or ground supporting material is that it's strength properties are inadequate for it to be used. Another reason is that it could be a material that expands and contracts when subject to changes in moisture content. This could cause the supporting medium to buckle and decouple from the ground mass. Another reason is the material being removed might be soft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Materials that are unsuitable include: clay; chalk; talc; limestone; ultramafic rocks such as komatiites ; most sedimentary rocks such a sandstone and mud stone; platey material such as mica; and oxidized or weathered rocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Materials that are better suited to be used for ground support purposes are hard, competent, non-friable materials with high strength characteristics, such as basalt and dolerite. Which is why these types or rock are used as aggregate materials for concrete. Generally, these materials need to be brought in sites.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In countries like Bangladesh, situated in a massive river delta, where there are no hard rock sources, broken baked clay is sometimes used as aggregate for road construction purposes.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2019-01-01T08:25:14.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25331" PostTypeId="1" CreationDate="2019-01-01T18:23:55.190" Score="1" ViewCount="570" Body="&lt;p&gt;I am running a DSLR to take time lapse photos for long periods of time, often overnight, often in extreme cold (-20F).  If at any point the camera does not get enough power, it shuts down - time lapse over.  The stock lithium batteries I believe are rated 7.2 or 7.4 volts but measure 8.4v when fully charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been using a 12v 12AH SLA battery connected to a voltage regulator (DROK DC-DC 6-32V to 0.8-28V 150W 15A) to output 8.4v.  This works well but fails in the coldest of temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I am losing camera power, is it likely because battery voltage is dropping in the cold?  Meaning I should wire two batteries in series increasing voltage before the regulator and allowing the voltage to be maintained at 8.4v?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is current dropping because of the cold and I should put them in parallel?  Or both?&lt;/p&gt;&#xA;" OwnerUserId="18815" LastActivityDate="2024-01-06T14:02:50.387" Title="Maintain voltage and current supply to device in extreme cold" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25332" PostTypeId="2" ParentId="25331" CreationDate="2019-01-01T20:42:42.687" Score="0" Body="&lt;p&gt;Based on the extremely low temperature you've described, it's likely the temperature is causing the drop-out of your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://all-about-lead-acid-batteries.capnfatz.com/all-about-lead-acid-batteries/lead-acid-battery-maintenance/battery-voltmeter-soc-temperature-correction/&quot; rel=&quot;nofollow noreferrer&quot;&gt;All About Lead Acid Batteries&lt;/a&gt;, the voltage you read at lower temperatures is not accurate. The voltage shown on the meter will by higher than reality. The chart shows that at zero degrees F, a battery that reads 12.082 volts is effectively fully discharged. Extrapolating from the chart, at minus twenty degrees F, your battery is dead even though it might display 12.14 volts. For high current draws, it is inadvisable to discharge a lead acid battery below fifty percent depth of discharge, as it results in reduced battery life. At lower current levels (your application) going below twenty five percent state of charge will also reduce life span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://www.thoughtco.com/why-batteries-discharge-quickly-cold-weather-607889&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thoughtco.com&lt;/a&gt;, as the temperature drops, the current available also is reduced. Especially in lead-acid batteries, this turns the Peukert effect into a fairly detrimental situation. Your device requires more current, the lower temperature reduces the current, causing the device to &quot;demand&quot; more, which likely drops the voltage, increasing the current demand. It's a downward spiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could consider alternative options. Dedicate a battery to a low wattage heater that will keep the primary battery warm. Especially insulate the enclosures for the battery or batteries. With the low temperatures involved, a fairly large expanded polystyrene cooler with thick walls (more than one!) nested together could give you the thermal stability required to complete the tasks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connecting batteries in parallel will provide more current to the device, but will still suffer from the cold temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium chemistry based batteries are not to be charged at temperatures below freezing. If your camera contains such a battery and the supply voltage is at a charging level, you may be damaging the internal battery as well. These batteries also suffer from reduced output and capacity in extremely low temperatures. Our EVs present a dash panel warning at low temperatures indicating that regenerative braking may be reduced. The onboard computers prevent the system from charging the pack during extreme cold, reducing the chance of damage to the pack.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-01-01T20:42:42.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25333" PostTypeId="1" CreationDate="2019-01-01T21:38:38.087" Score="1" ViewCount="745" Body="&lt;p&gt;In a horizontal channel that feeds off a river, does the head/pressure loss due to friction, turns etc in the channel translate to a drop in the water level between its inlet and outler? If so, could someone provide a basic insight how? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="18817" LastActivityDate="2019-06-01T06:03:05.343" Title="Open Channel flow with pressure losses - flow depth?" Tags="|fluid-mechanics|hydraulics|hydrology|open-channel-flow|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25334" PostTypeId="2" ParentId="25333" CreationDate="2019-01-01T22:51:49.030" Score="0" Body="&lt;p&gt;The flow velocity is not uniform over a cross section through the channel because of the viscosity of the water. It is zero at the walls and a maximum in the center of the free surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The viscous forces tend to reduce the flow velocity. The work done by the viscous forces is counterbalanced by work done by gravity, as the depth of the stream decreases and the water moves vertically down as well as flowing horizontally.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-01-01T22:51:49.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25335" PostTypeId="2" ParentId="9026" CreationDate="2019-01-02T00:17:44.037" Score="0" Body="&lt;p&gt;Note that there are multiple physical constraints in the design. Unlike in a turboshaft engine, in a turbofan, the torque of the turbine spool has to match the torque of it's associated compressor spool. Also, the axial load on the two spools must largely cancel, because thrust bearings are a pain in this environment. Balancing these requirement puts a severe crimp on the relative size of the two spools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, you have the problem of managing tip losses overall. Tip losses account for as much as 1/3 the aerodynamic losses in the machine. Low hub ratio designs minimize tip losses. Better tip design has allowed more efficient cascades and bigger hub ratios, but you still need a clearance that works for the entire spool. This tends to limit how much the hub ratio can vary over the stages of a single spool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you also want to minimize the surface area of anything that requires cooling. Cooling stuff robs efficiency.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-01-02T00:17:44.037" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25336" PostTypeId="2" ParentId="25331" CreationDate="2019-01-02T00:32:05.583" Score="0" Body="&lt;p&gt;Batteries depend on chemistry, and chemical reactions slow down in the cold, so any kind of batteries are always going to have poor performance in extreme cold weather. The other suggestions given so far are good.  Here is an another option (depending on your budget and general familiarity with electronics).  You could just ditch batteries altogether and go with supercapacitors.  Supercapacitors have lower power density than batteries, but since there is no chemistry involved, performance is pretty good, even down to -40C. You do lose a little bit of performance at low temps, but nothing close to what a battery loses. It has been said that the Russian military uses supercapacitors to start their tanks in the winter in Siberia (see Ref 1) Here is one example of a commercial available supercap]&lt;a href=&quot;https://www.mouser.com/ProductDetail/Maxwell-Technologies/BCAP0350-E270-T13?qs=WBQIFPjBBZvo1y4AnDEsTA%3D%3D&amp;amp;gclid=Cj0KCQiAvKzhBRC1ARIsANEXdgy7OAagqOHSxJt6RYh-BG1rY8TEdsNhztRHI-2S1HGfMGYXbqI8jNYaAvgvEALw_wcB&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;.  It's fairly cheap, but only goes to about 3V so you'd have to stack a few of them together to get up to 7-8V. 3000 farads means it'll store quite a bit of energy. You'd need to do the calculations on current draw to see if that's enough to get you through the night.  It won't be as much as your 12 Ah battery at room temp, but it's probably better than the battery at extreme cold.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Encyclopedia of Electrochemical Power Sources edited by Jürgen Garche, Chris K. Dyer, Patrick T. Moseley, Zempachi Ogumi, David A. J. Rand, Bruno Scrosati&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-01-02T00:32:05.583" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25337" PostTypeId="1" CreationDate="2019-01-02T01:21:56.367" Score="0" ViewCount="260" Body="&lt;p&gt;I am currently building a pour over coffee maker(personal use), but I am having trouble sourcing a water pump to use. I am hoping to create something that performs similar to the &lt;a href=&quot;http://poursteady.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Poursteady&lt;/a&gt; commercial machine. My current plan is to boil the water to a specific temperature in a kettle, then use a water pump to pump the water from the kettle into the coffee funnel containing the grounds. I'm doing all the automation using an arduino. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Requirements for the pump:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Food grade &lt;/li&gt;&#xA;&lt;li&gt;Safe to pump ~100c water&lt;/li&gt;&#xA;&lt;li&gt;A flow rate of &gt;=500ml/minute&lt;/li&gt;&#xA;&lt;li&gt;Either AC or DC is fine&lt;/li&gt;&#xA;&lt;li&gt;The more affordable the better.&#xA;Self priming&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I believe peristaltic pumps, magnetic pumps, and vibration pumps all seem to be okay options, but having zero experience with pumps, I am wondering if any of them are recommended over the other given the specific requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="18820" LastEditorUserId="18820" LastEditDate="2019-01-16T02:13:12.157" LastActivityDate="2019-03-16T23:40:07.303" Title="Best affordable food grade hot water pump" Tags="|pumps|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25338" PostTypeId="2" ParentId="25313" CreationDate="2019-01-02T03:40:19.580" Score="2" Body="&lt;p&gt;The operational frequency operational range of the proposed antenna is within the frequency range of the old antenna. It should not be an issue from frequency standpoint. If the gains are similar chances are that the replacement antenna might work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also noticed the following statement for your alternative antenna. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bcrem.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bcrem.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the alternative antenna is not design your UXLP4 wireless receiver. Wanted to make you aware of the differences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did get chance to review &lt;a href=&quot;https://www.shureasia.com/productdocumentsfiles/default/discontinued/wireless/ulxwirelessuserguide-36fc66950413b32b410df0a3e73af99c.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ULX Wireless System User Guide&lt;/a&gt; Looks like you might have other antenna options for your receiver. Some those options are listed below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9eHRC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9eHRC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on region you might have to select the appropriate regulatory transmit and receive frequency / channel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UYKm9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UYKm9.png&quot; alt=&quot;transmit and receive Frequency / Channel &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a side note United States of America, Federal Communication Commission (FCC) has reassigned the 600MHz frequency spectrum for cellular communication as licenses frequency band. Therefore as of October 2018 it is illegal to use unless authorized by the FCC. That most likely explains the absence of 600 MHz frequency spectrum antenna for wireless receivers.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://nofilmschool.com/2017/11/how-600-mhz-wireless-auction-will-impact-filmmaking&quot; rel=&quot;nofollow noreferrer&quot;&gt;Everything You Need to Know About How the 600 MHz Wireless Auction Will Impact Filmmaking&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en-us.sennheiser.com/spectrum&quot; rel=&quot;nofollow noreferrer&quot;&gt;Guide to Wireless Microphone Operation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-01-02T03:40:19.580" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25339" PostTypeId="1" CreationDate="2019-01-02T05:46:12.513" Score="0" ViewCount="67" Body="&lt;p&gt;I want to build a battery for an electric bicycle using 18650 batteries. But I'm not into spot welding so much. Would it be possible to build a large case for the 18650 batteries with these types of black plastic cases serial connected together and then just put the batteries in place when I want to use them instead?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tybzw.jpg&quot; alt=&quot;batteries&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="17642" LastEditorUserId="17642" LastEditDate="2019-01-02T15:08:56.710" LastActivityDate="2019-01-02T16:39:53.410" Title="Larger battery from 18650 batteries - using a case?" Tags="|battery|bicycles|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25340" PostTypeId="1" CreationDate="2019-01-02T05:46:31.630" Score="0" ViewCount="1968" Body="&lt;p&gt;From &lt;a href=&quot;https://community.plm.automation.siemens.com/t5/Testing-Knowledge-Base/Stress-and-Strain/ta-p/355027&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; page it can be seen that young's modulus is given by, &#xA;E = (Δstress) / (Δstrain)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means E is a point function,  that changes as the strain in member is increased. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am feeling skeptical about this definition, as I haven't read E as derivative anywhere before. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nU9QV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nU9QV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2116" LastActivityDate="2024-01-05T12:20:25.660" Title="Is youngs modulus a derivative of stress and strain?" Tags="|stresses|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25341" PostTypeId="2" ParentId="25333" CreationDate="2019-01-02T05:46:52.937" Score="1" Body="&lt;p&gt;If the channel is horizontal, the flow will set a slope to use the head lost as a potential energy to turn it to kinetic energy that is required to overcome the head loss at the bends and also the viscosity losses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can guess a head loss and use it as slope and write the Manning formula for discharge and calculate the flow and from there the height of the water at exit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the height is more than say two feet we change our guess by reducing the height to one foot and adding the head as slope, S, recalculating. Until the results become closer and closer to our tolerance.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-02T05:46:52.937" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25342" PostTypeId="2" ParentId="25339" CreationDate="2019-01-02T10:20:28.403" Score="1" Body="&lt;p&gt;From a strictly electronic point of view, one can construct a larger battery using the cases you've posted. In your example, an electric assist bicycle, you will have relatively high current draw requirements. The wiring and spring loaded contacts will fare poorly under such high currents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's possible that the wires will melt the insulation due to the heat generated, or the spring contacts will get hot enough to cause the case around them to melt. Once the insulation is melted, short circuits can occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's common for batteries for e-bikes to be rated to 40 amperes current draw and those wires and contacts will not tolerate the level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With soldering of this type of battery, the heat from the tool will likely damage the battery and might cause explosion. Multiple batteries composed of this type of cell use spot welders specifically made for this purpose. You can find a DIY resource by searching &quot;DIY battery spot welder&quot; for options to complete your project.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-01-02T10:20:28.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25344" PostTypeId="2" ParentId="25340" CreationDate="2019-01-02T12:33:13.320" Score="-1" Body="&lt;p&gt;I dont see any problem with that definition.&#xA;You can check ie. ISO 527. Where E is calculated from a tensile test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To further understand the meaning or defining E I'd recommend you to read something about anisotropic materials.&#xA;You can check for instance long fiber composites - which is orthotropic material, look at the E matrix/tensor.&lt;/p&gt;&#xA;" OwnerUserId="18819" LastActivityDate="2019-01-02T12:33:13.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25346" PostTypeId="1" CreationDate="2019-01-02T15:13:30.513" Score="1" ViewCount="54" Body="&lt;p&gt;I have a Statics assignment which I don't seem to fully understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the full assignment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1vNQI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1vNQI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0EBTj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0EBTj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ez8H6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ez8H6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x4v8z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x4v8z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first question I have tried making several tables with Areas of each parts and the centre of gravity for each recognizable shape (triangle, square, circle, quarter circle, half circle). The problem is I don't understand how to piece it together. The figure doesn't have an xy axis (should it have one?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have made these calculations for the areas so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HD7vP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HD7vP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what I don't fully understand, how should one make use of this table. I have stumbled upon several tables which some contain the area A and coordinates x and y and others with only the area A and coordinate z. Which one to use? Here is an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Yi5lE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yi5lE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18829" LastActivityDate="2019-01-02T15:13:30.513" Title="HELP WITH ASSIGNMENT. CENTER OF GRAVITY" Tags="|statics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25348" PostTypeId="2" ParentId="25339" CreationDate="2019-01-02T16:39:53.410" Score="0" Body="&lt;p&gt;Any type of welding is not a good idea.  But more to the point - if you don't include load-balancing electronics both for charging and discharging, you will  be VerySorry(TM) .&lt;br&gt;&#xA;To start out, you should read up on  topics such as &quot;Charge in parallel, discharge in series&quot; depending on the anticipated charging voltage and output/drive voltage intended.  each cell can only put out its own rated current (without voltage sag), and can only charge up to certain voltages and current rates; those current rates vary dramatically with the charge level of the cell.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a reason that Tesla Corp's vehicles have gigantic amounts of hardware and software control of the battery packs. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2019-01-02T16:39:53.410" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25349" PostTypeId="1" AcceptedAnswerId="25354" CreationDate="2019-01-02T19:31:17.567" Score="1" ViewCount="44" Body="&lt;p&gt;Is there any kind of board or mat that when we place it on the floor and wire it, a self-made connected device can read these data when it is being pushed and holding onto it, at multiple points?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Location where the force is applied onto. Something working similar to a touchscreen, or a trackpad, or at least a keyboard. Except it's not any of those :-D&lt;/li&gt;&#xA;&lt;li&gt;Magnitude of the force&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I need to build a footstep detecting device consisting a board that can be wired and detect point of pressure and how much amount it is. It is something working like a capacitive sensing device. But able to handle greater pressure.&lt;/p&gt;&#xA;" OwnerUserId="18809" LastEditorUserId="18809" LastEditDate="2019-01-02T21:11:15.627" LastActivityDate="2019-01-03T10:36:38.090" Title="A board or mat that we can detect a push at an exact point" Tags="|electrical-engineering|power-electronics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25350" PostTypeId="1" AcceptedAnswerId="25351" CreationDate="2019-01-03T00:54:19.653" Score="0" ViewCount="156" Body="&lt;p&gt;I have an RV garage but no RV.  I'm considering building a platform 10ft above the floor.  The garage is 14 feet tall and the area that I want the platform is 16ft a 16ft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The details are stopping me though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that I'll use a 2x6 every 2 feet as the joists.  (see my picture attached.)  Would a,b,c,or d be better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;a:&lt;/strong&gt; each joist runs the full length and is attached to a 2x6 that has already been attached to the wall&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;b:&lt;/strong&gt; i split the joists in half and have a middle joist running perpendicular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;c:&lt;/strong&gt; similar to a but the joists are connected directly to the wall studs rather than an intermediary 2x6&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;d:&lt;/strong&gt; combination of b and c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe there's a better way, so I'm open to suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are 2x6s strong even strong enough for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Photo attached to show what I mean.  The perspective view is to show how it fits in my garage.  There are 3 solid walls and 1 open.  I'm thinking that the open side will be where the joists are parallel to, (so each end of the main 8 joists will be connected to a solid wall)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tickP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tickP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18135" LastEditorUserId="18135" LastEditDate="2019-01-03T19:58:20.930" LastActivityDate="2019-01-03T19:58:20.930" Title="Structurally sound option for overhead storage platform in garage" Tags="|structural-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25351" PostTypeId="2" ParentId="25350" CreationDate="2019-01-03T02:24:32.267" Score="1" Body="&lt;p&gt;Depends on the load, but if you guess your load per square feet is going to be near 25lbs, 2x6 is not enough, consider 2x8 @ 16&quot;, Doug Fur Larch, grade 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to use a 2x8 or other means of ledger beam at two ends which is properly attached to your garage framing, so that you nail the joist hangers to it. You should consult a type5 framing sheet for nailing and connectors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to consider earthquake loads and brace the existing garage walls for that. Finally you need to block your rafters at least at midspan.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-03T02:24:32.267" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25352" PostTypeId="2" ParentId="25340" CreationDate="2019-01-03T07:52:04.967" Score="1" Body="&lt;p&gt;In the theory of elasticity, the constitutive relation in three dimensions is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sigma_{ij} = C_{ijkl}\epsilon_{kl}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives the &lt;span class=&quot;math-container&quot;&gt;${\bf linear}$&lt;/span&gt; relationship between stress and strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fourth order tensor - &lt;span class=&quot;math-container&quot;&gt;$C_{ijkl}$&lt;/span&gt; takes 81 components in its own form and reduces to 21 if it is a symmetric stress, and 9 for planar stress. Finally it takes only one component for one dimensional problem such as uni-axial tensile problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e., &lt;span class=&quot;math-container&quot;&gt;$\sigma = C\epsilon$&lt;/span&gt; or, &lt;span class=&quot;math-container&quot;&gt;$\sigma = E\epsilon$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the relation between stress &amp;amp; strain is linear (with in elastic limit) it is sufficient having just &lt;span class=&quot;math-container&quot;&gt;$\sigma $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$ \epsilon$&lt;/span&gt;  values to get &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then when why we need &lt;span class=&quot;math-container&quot;&gt;$\delta\sigma $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$ \delta\epsilon$&lt;/span&gt; to calculate &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are testing the specimen in UTM, will it show zero deformation for zero load applied? at least for me during my undergraduate studies the load was non-zero when the displacement/deformation was zero. This is due to the friction between the model and fixtures/Jigs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also within the elastic region, the material tends to show &lt;span class=&quot;math-container&quot;&gt;$hysteresis$&lt;/span&gt; effects if you are using the same sample again and again. This leads to non-zero deflection for zero load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, the calculation of &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; at any specific &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt; is prone to error if the &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; - &lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt; curve have local error due to the instrumentation/observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid these errors it is better to take the infinitesimal changes in stress and strain or slope of the &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; - &lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt; curve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2019-01-03T07:52:04.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25353" PostTypeId="2" ParentId="25340" CreationDate="2019-01-03T10:22:09.283" Score="1" Body="&lt;p&gt;The Young's modulus (sometimes refered to as elastic modulus) is only relevant for the elastic region of the material and doesn't apply after plastic deformation begins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rather than a derivative, the Young's modulus is better defined as the gradient of the linear (elastic) section at the begining of the graph. Or in real life terms it is the gradient of a (linear) line of best fit applied to the data in the elastic (linear) region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This aproach works fine for most solid materials. But things can start to get complicated for some materials like rubbers.&lt;/p&gt;&#xA;" OwnerUserId="18382" LastActivityDate="2019-01-03T10:22:09.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25354" PostTypeId="2" ParentId="25349" CreationDate="2019-01-03T10:36:38.090" Score="0" Body="&lt;p&gt;You are looking for Pressure sensing mats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are used for exactly what you want to use them for, getting stepped on for footprint and gait analysis. &lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2019-01-03T10:36:38.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25355" PostTypeId="1" AcceptedAnswerId="25373" CreationDate="2019-01-03T12:01:12.097" Score="1" ViewCount="80" Body="&lt;p&gt;I have a cold body of constant temperature A and a hot surface several times it's area. What is the best way I can absorb the heat from that surface using A. Will sheets of metal like alluminium or copper help?&lt;/p&gt;&#xA;" OwnerUserId="18840" LastActivityDate="2021-01-23T19:01:20.117" Title="Absorbing heat from a surface" Tags="|heat-transfer|cooling|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25356" PostTypeId="1" CreationDate="2019-01-03T13:04:47.417" Score="4" ViewCount="896" Body="&lt;p&gt;It’s well known that quenching most alloyed steels starting from AISI 4140 to high speed steels in water, brine , or sometimes oil will cause cracking and distortions.&lt;/p&gt;&#xA;&lt;p&gt;In industry, oil quenching is sometimes replaced by molten salt, which is considered better than oil in decreasing crack probability and this advantageous, attributed to uniform cooling provided by salt baths, compared to other quenching mediums.&lt;/p&gt;&#xA;&lt;p&gt;Cracks and distortions caused by water or brine quenching are usually attributed to rapid cooling, but sometimes they are attributed to non-uniform cooling caused by vapor blanket. That’s so confusing!!&lt;/p&gt;&#xA;&lt;p&gt;It is expected to obtain uniform cooling by quenching steel in high concentration salt solution &amp;quot;nearly saturated solution&amp;quot; thanks to large amounts of ion particles which disturb vapor blanket, but these type of solutions will also have high cooling rates compared to oil. Such rapid cooling is usually attributed to be the of quench cracks in water quenching!&lt;/p&gt;&#xA;&lt;p&gt;O1 steel can’t fully hardened in parts larger than 1 inch!&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i4eD9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i4eD9.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;so my problem if it is safe to quench thick parts of such low hardenability steels in very high concentration salt solution which are expected to combining uniform cooling with relatively high speed cooling to fully harden thick parts of such low hardenability steel ?&lt;/p&gt;&#xA;" OwnerUserId="18830" LastEditorUserId="90" LastEditDate="2023-05-14T07:57:23.633" LastActivityDate="2023-05-14T07:57:23.633" Title="What actually cause quench cracking, non-uniform cooling or rapid cooling?" Tags="|structural-engineering|materials|steel|manufacturing-engineering|metallurgy|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="25357" PostTypeId="2" ParentId="6531" CreationDate="2019-01-03T14:38:44.410" Score="0" Body="&lt;p&gt;This is a little late to help you with your 2015 project, but hopefully it can shed some light on what's going on. I'm a PE in FL/GA, that specializes in bridge design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Why are the piles stopped short of the cap? Most codes do not have a requirement to locate pile heads close to the bearing points. Embedment is generally detailed based on the required fixity of the design, but constructability or other project-specific design constraints may also be necessary. For driven piles in non-seismic areas, an embedment of 1-ft into a concrete element is considered adequate for a pinned connection, and 4-ft is considered adequate for a fully-fixed connection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this specific example, it appears the designer is likely using the longer piles to act as flexural resisting elements in the wall pier's weak axis to reduce the need for flexural reinforcement.  This is why the (c) bars appear close to the minimum reinforcing required for temp/shrinkage. The two shorter piles are most likely to avoid pile-driving conflicts with the battered piles during construction, but could also simply be a cost-saving measure since the piles were not needed to resist flexural loads. The embedment of the short piles appears to be ~4-ft, which would be a fixed connection, allowing these piles to resist flexure at the base of the wall and lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) What is the load path to the piles?  This would be up to the Engineer to determine based on their own judgement and comfort level, but would be based on standard beam or deep-beam theory. Personally, I would design this as follows: &#xA;    1) The cap as a continuous beam between the top three piles, ignoring contribution of the wall portion, as this is conservative. Typical reinforcing sizes for cap bars are #6 thru #10 for (a) bars, and #4 or #5 for (b) bars. I nearly always use #8 or #10 for (a) and #5 for (b) because steel is relatively cheap and it's very common sizes for highway bridges.&#xA;    2) Distribute vertical loads between all 5 piles based on lever rule for bearing loads, and tributary width for self-weight of the pier. Remember that battered piles will have additional axial forces induced from lateral loads, and all load combinations should be checked to determine what controls the design. All piles should be the same size, and driven to the same minimum embedment and nominal bearing capacity, regardless of which pile controls the design.&#xA;    3) Flexural design of the wall-pier (weak axis) would assume no contribution from the concrete/rebar, and all flexure resisted by the 3 long steel h-piles. Wall rebar would be minimum for temp/shrinkage/crack control. Lateral rebar in wall may need to be slightly larger to resist horizontal forces by &quot;spanning&quot; between the 3 piles, but it is still probably controlled by min rebar reqs.  Bar sizes for (c) and (d) bars are commonly #4 or #5 to achieve min reqs in wall piers, but sometimes #6 or #7 may be necessary for larger concrete elements. I almost always specify #5 bars @ 12&quot; max as a minimum in foundations, regardless of whether #4s can be used.&#xA;    4) Flexure at bottom of wall pier and lower resisted by all 5 piles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="18842" LastActivityDate="2019-01-03T14:38:44.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25358" PostTypeId="2" ParentId="25272" CreationDate="2019-01-03T15:06:48.207" Score="3" Body="&lt;p&gt;Your question does not have a definitive answer, as the lightest material known is far from the strongest. On the contrary, there is a correlation between density and strength, meaning that in general denser materials are stronger than those that are less dense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might want to look in to Ashby diagrams, a plot where two properties are plotted against each other for many materials. In this case a Strength-Density-Diagram would be advisable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On &lt;a href=&quot;http://teachingresources.grantadesign.com/Charts-overview&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;, you can find a variety of these plots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second more &quot;numeric&quot; might be to look up the &lt;em&gt;breaking length&lt;/em&gt; of materials, which is a theoretical calculation for how long of a bar you could make out of a certain material until it fails under its own weight. There are some astonishing numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation is pretty easy,&#xA;You get the force exerted by the bar itself &lt;span class=&quot;math-container&quot;&gt;$F_T=\rho\cdot A\cdot g\cdot L$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the density, &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the cross-sectional area, &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is the gravitational acceleration and &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the length of the bar.&#xA;On the other hand you get the maximum force the bar can support &lt;span class=&quot;math-container&quot;&gt;$F_R=R\cdot A$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the maximum bearable stress (i.e. ultimate tensile strength, UTS).&#xA;If you set the forces equal you get&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F_T=\rho\cdot A\cdot g\cdot L = R\cdot A = F_R$$&lt;/span&gt;&#xA;As you can see the area cancels out and if you solve for &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; you get&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ L=\frac{R}{\rho\cdot g} $$&lt;/span&gt;&#xA;which means the breaking length of a material is only defined by its UTS and density (assumed &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is constant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a neat table on &lt;a href=&quot;https://en.wikipedia.org/wiki/Specific_strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; on breaking length / specific strength (which is just &lt;span class=&quot;math-container&quot;&gt;$=\frac{R}{\rho}$&lt;/span&gt;) but there surely are other sources as well.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2019-01-03T15:06:48.207" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25359" PostTypeId="2" ParentId="25350" CreationDate="2019-01-03T16:41:34.597" Score="0" Body="&lt;p&gt;Whatever joist plan you use , you can make it substantially stronger (and neater appearance) by attaching plywood to the bottom. I used glue and screws because the connection must be strong . This more or less makes an &quot;H&quot; beam , the plywood becomes the primary tension member; chose 1/2 , 3/4 , etc. depending on how stiff and strong you want. With your relatively long span , you could consider steel strapping on the bottom of the joists , they would be covered by the plywood . A little more hassle drilling screw holes but serious increase in strength.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-01-03T16:41:34.597" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25360" PostTypeId="1" AcceptedAnswerId="25365" CreationDate="2019-01-03T17:17:34.500" Score="2" ViewCount="71" Body="&lt;p&gt;&lt;a href=&quot;https://www.engineeringtoolbox.com/thermal-conductivity-d_429.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Engineering Toolbox&lt;/a&gt; presents a chart of &quot;common&quot; materials and their thermal conductivity coefficients. I'd read or learned some time ago that diamond was superior in this respect. The chart shows diamond at 1000 W/m K compared to the next closest, silver at 429 W/m K. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My objective is to reduce the grip of a thermal pad between a heated bed and a glass surface. The heat transfer from the bed to the glass was improved by the addition of this thermal transfer pad, but the pad is very sticky, one millimeter thick and relatively fragile. I've only now discovered that &lt;a href=&quot;https://www.sparkfun.com/products/retired/14283&quot; rel=&quot;nofollow noreferrer&quot;&gt;this product&lt;/a&gt; is discontinued!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pursuant to this objective is the ability to lift the glass, with the thermal pad (dusted?) remaining on the heater, without the pad adhering to the glass. It's not adhesive, but it sticks with the power of gekko.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering to powder one surface of the pad, but do not wish to reduce the thermal conductivity. I'm aware that diamond powder exists for lapping/grinding purposes, but have also found that most of the product listings provide the expression &quot;synthetic diamond&quot; in the description.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though SE discourages more than one question per post, these are all related to my objective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a better choice than diamond powder to provide a thermally conductive release mechanism between the thermal gap filler pad and the glass?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related:&#xA;Will synthetic diamond powder display the same thermal transfer characteristics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit Added from comments:&#xA;The bed is a flat electrical heater (3D printer) with a range from ambient to about 100°C. There is normally a 1mm air gap to the glass bed. As such, the air gap has to be heated before the glass reaches the desired temperature. Having installed the 1mm transfer pad reduced the elapsed time involved.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2019-01-04T21:48:12.873" LastActivityDate="2019-01-05T15:04:06.437" Title="Powder form of material with high thermal transfer" Tags="|thermal-conduction|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="25361" PostTypeId="1" AcceptedAnswerId="25829" CreationDate="2019-01-03T18:15:36.543" Score="0" ViewCount="288" Body="&lt;p&gt;There is an interesting and potentially useful property to do with the procession of Gyroscopes, as well as devices like the &quot;Powerball&quot; toy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NXH2h.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NXH2h.png&quot; alt=&quot;Gyroscope&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the gyroscope is rotating (W&lt;sub&gt;s&lt;/sub&gt;) and then force is applied to increase the procession (W&lt;sub&gt;p&lt;/sub&gt;) the W&lt;sub&gt;s&lt;/sub&gt; rotation increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Experiments with the Powerball show that a small increase in W&lt;sub&gt;p&lt;/sub&gt; results in a very large increase in both W&lt;sub&gt;s&lt;/sub&gt; and the resistance to an increase of rotational velocity in W&lt;sub&gt;p&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More intuitively, imagine the dynamics here as being like a variable gear that increases in gear ratio at an accelerating rate as the speed of the input increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be practical to design a vehicle such as a human powered bike or a motorised car that used this property of a gyroscope instead of a conventional gearing system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/me9RW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/me9RW.png&quot; alt=&quot;Bike Gears and Chain&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12002" LastEditorUserId="12002" LastEditDate="2019-01-03T18:30:33.157" LastActivityDate="2019-02-04T19:08:22.243" Title="Gyroscope/Powerball gearing system for bikes?" Tags="|mechanical-engineering|mathematics|mechanisms|mechanical|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25365" PostTypeId="2" ParentId="25360" CreationDate="2019-01-03T23:37:42.877" Score="1" Body="&lt;p&gt;Here's the answer to one of your questions: Synthetic diamond is identical to naturally- occurring diamond. It's made in a laboratory instead of deep underground and has the same properties as the &quot;underground&quot; variety.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-01-03T23:37:42.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25367" PostTypeId="2" ParentId="13473" CreationDate="2019-01-04T01:47:41.477" Score="0" Body="&lt;p&gt;The idea is you don't want the shocks to dissipate usable energy in the freestream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\displaystyle \frac{{\mit\Delta}{\cal S}}{{\cal R}}	\displaystyle \simeq \frac{(\gamma+1)}{12\,\gamma^{\,2}}\left( \frac{{\mit\Delta}p}{p_1}\right)^{\,3}$$&lt;/span&gt;&#xA;This the formula for the change in entropy, where &lt;span class=&quot;math-container&quot;&gt;$\Delta p$&lt;/span&gt; is the change in pressure across the shock. Since the change in entropy is proportional to the shock strength cubed, Several weak shocks fritter away less energy than a single strong one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See here for a detailed explanation. You may need to back up a few pages to unpack some of the equations.&lt;a href=&quot;https://farside.ph.utexas.edu/teaching/336L/Fluidhtml/node208.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Supersonic Compression by Turning&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-01-04T01:47:41.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25370" PostTypeId="1" AcceptedAnswerId="25380" CreationDate="2019-01-04T13:24:47.050" Score="0" ViewCount="926" Body="&lt;p&gt;I need a long arm which will transmit motion from a motor to a pen. It needs to be about 500mm long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to try and do is reduce the bend (i.e make it stiff) but also make it fairly light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question A)&#xA;In this case, am I better off going for a tube rather than a solid rod?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question B) &#xA;Our of the materials available to me (which is carbon fibre, aluminium, steel or wood) - is there any of this which would have an obvious benefit?&lt;/p&gt;&#xA;" OwnerUserId="18703" LastActivityDate="2019-01-05T07:22:06.367" Title="Stiff rod / tube - What material and diameter" Tags="|materials|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25373" PostTypeId="2" ParentId="25355" CreationDate="2019-01-04T16:41:07.043" Score="0" Body="&lt;p&gt;A shape such as the one below will help. Make the piece from metal (copper). Insulate the edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z81Wq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z81Wq.png&quot; alt=&quot;metal adapter piece&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the differences in area get larger, this will eventually have diminishing returns in improvement. When the goal is high heat transfer, thinner is better. Of course, at some point, the thin piece will be the same as just putting the smaller cold piece directly on the larger hot piece. The trade off is between the heat transfer perpendicular versus lateral in the piece. When the goal is uniform temperature on the (larger) hot surface, thicker may be better. In the latter case, an optimal shape (curvature of the sidewalls) may be possible instead to minimize lateral temperature differences versus the perpendicular temperature difference.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-01-04T16:41:07.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25374" PostTypeId="1" CreationDate="2019-01-04T18:53:06.187" Score="1" ViewCount="41" Body="&lt;p&gt;I have an air supply grate that is making a roaring noise. It has rectangular slats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the sound is coming from the grate, not the duct, because when I remove the grate, the duct is silent. It is only when the grate is in place that it is noisy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a shape I can make a grate that will generate little or no noise when air passes through it?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2019-01-04T21:14:11.297" Title="Grate that does not cause air rush noise?" Tags="|acoustics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25375" PostTypeId="2" ParentId="25374" CreationDate="2019-01-04T19:06:59.880" Score="0" Body="&lt;p&gt;You can try variable gravity adjusting horizontal slats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are hinged at the top and swing down, so as they can open more with larger air flow and hang shut when there is no flow.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-04T19:06:59.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25376" PostTypeId="2" ParentId="25374" CreationDate="2019-01-04T20:19:21.973" Score="0" Body="&lt;p&gt;If the grate is designed to only prevent larger things getting through then using a mesh 1cm or 1/2 inch squares may be appropriate.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-01-04T20:19:21.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25377" PostTypeId="2" ParentId="25340" CreationDate="2019-01-04T21:14:00.907" Score="-1" Body="&lt;p&gt;As has been mentioned in other answers, the Young's modulus, E, is the ratio of stress to strain in the elastic range of stress/strain curve and is a constant, not a derivative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a constant as opposed to what a derivative implies which is a variable depending on the local curvature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysis and design of structures and members is based heavily on the assumption that E is a constant and will remain constant for the periods of loading and unloading a beam, column, or a whole structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that E is a slope of straight line and a constant as such, at least assumed, allows all our methods of analysis of the behavior of a structural member work. Otherwise we had to go to unlimited number of trial and errors, assuming a different E for a different stress configuration.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;** Edit**&#xA;If we don't handle Youngs modulus as a constant, we can not assume E constant  the entire system of analysing beam stresses and bending collapses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We calculate I, second area moment which is fundamental in structural analysis based on the assumption that E is constantan and cross section of beams and columns remain planar after bending. If the E is not constant the strain is not linear and I will be meaning less.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-02-03T20:33:58.903" LastActivityDate="2019-02-03T20:33:58.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25378" PostTypeId="2" ParentId="25374" CreationDate="2019-01-04T21:14:11.297" Score="1" Body="&lt;p&gt;If it is a metal grate, the sound may be amplified by the grate itself vibrating, and possibly transferring the vibrations into the duct and then the structure of the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A plastic grate (if such a thing is available) may be quieter because the material has more internal damping than metal.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-01-04T21:14:11.297" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25379" PostTypeId="2" ParentId="25356" CreationDate="2019-01-04T21:23:07.213" Score="1" Body="&lt;p&gt;To answer your specific question ; maybe.  O-1 tool steel has good hardenability ; the standard Jominy test showing HRC 60 two inches from the end ( austenitized at 1525 F). From experience with O-1 , cracking occurred in small parts ( less than 1/2&quot; thick) when oil quenched from 1475 F , but when quenched into oil from 1600 F no cracking occurred. This was because the higher temperature produced retained austinite ; being relatively soft and ductile it prevented cracking. The same affect would occur with water quenching ( retained austenite). One would need to experiment with the best austenitizing temperature for each part configuration using a water quench. Double tempering will be needed as the retained austenite will mostly transform during the first temper. Heat-treated steels tend to crack and distort because of the stresses caused by the contraction and expansion: As the steel cools ( contraction) and transforms to martensite ( expansion), then continues cooling (contraction).   &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-01-04T21:23:07.213" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25380" PostTypeId="2" ParentId="25370" CreationDate="2019-01-04T22:03:23.633" Score="3" Body="&lt;p&gt;When maximizing the shaft lateral stiffness vs. weight, one generally wants to optimize both the geometry and material, as you yourself have wisely decoupled in your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's very common to optimize a system by constructing a ratio (termed a figure of merit) of geometric or material parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to assume that the rod is loaded as a cantilever beam:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OCjLk.png&quot; width=&quot;300&quot;/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tube is stiffer than a solid rod per weight, so it always benefits you (in terms of stiffness vs. weight alone) to use a tube, and in fact as wide a tube as possible. You can verify this by taking the ratio of the polar &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_second_moments_of_area&quot; rel=&quot;nofollow noreferrer&quot;&gt;area moment of inertia &lt;/a&gt; (which mediates the stiffness) to the area (which mediates the weight): &lt;span class=&quot;math-container&quot;&gt;$$\frac{I}{A}\propto\frac{r_2^4-r_1^4}{r_2^2-r_1^2}=r_2^2+r_1^2.$$&lt;/span&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/wjup4.png&quot; width=&quot;300&quot;/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit is the strength of the tube; thin-walled tubes easily buckle under a lateral load, unlike solid rods. And of course space limitations may constrain the use of a very wide tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For selecting the material, a useful technique is to use an Ashby chart:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RB4lB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RB4lB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, in terms of material properties, the stiffness is mediated by the elastic modulus, and the weight is mediated by the density. You're looking for materials that lie to the left and to the top. (Here, we're implicitly taking the ratio of the elastic modulus to the density.) If low weight is very important (at the expense of not maximizing stiffness), then wood is appropriate. If stiffness is very important (at the expense of not minimizing weight, then a ceramic would be suitable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the constraints that come into play: neither wood nor a ceramic are as easy to machine into a thin-walled tube as metal, and metal can generally resist weathering and high temperature better than wood and can generally resist brittle fracture better than a &#xA;ceramic. Consequently, geometry and material selection is often going to involve tradeoffs. The figure-of-merit approach allows to quantitatively identify the best solution by suitably weighting the parameters.&lt;/p&gt;&#xA;" OwnerUserId="10123" LastEditorUserId="10123" LastEditDate="2019-01-04T22:46:48.930" LastActivityDate="2019-01-04T22:46:48.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25381" PostTypeId="1" CreationDate="2019-01-05T01:56:44.663" Score="0" ViewCount="61" Body="&lt;p&gt;Hello everyone this project is for my aquarium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a big bucket beside it on the floor and I was planning on using it for the water changes. Will be a neocaridina shrimp breeding aquarium so my goal is to maintain water parameters stable with very slowly changing between water changes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the main goal is to pour the new water into the bucket and slowly geting it released to the aquarium at a constant and even dripping rate so water never overflows on either side as well. Basicly the water will be forever flowing from one to the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving the water from aquarium to bucket is easy thanks to gravity but uphill.... how can I do it with a dripping system at the end at 2+ meters high? I guessed If I just use a tiny water pump with a airline tube leading into the aquarium could work but water pressure worries me caused by the dripping output... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other materials must be used? Is this idea viable or just a waste of time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much.&lt;/p&gt;&#xA;" OwnerUserId="18859" LastActivityDate="2020-09-27T02:01:40.130" Title="Water loop with a drop output?" Tags="|water-resources|flow-control|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25382" PostTypeId="2" ParentId="25199" CreationDate="2019-01-05T05:44:19.443" Score="1" Body="&lt;p&gt;To reset the default toolbar to its original tool set, select Toolbar Options &gt; Add or Remove Buttons &gt; Main Toolbar &gt; Reset Toolbar, or on the Customize dialog box'sToolbars tab, select Main Toolbar and click the Reset mbutton. From the menu bar, select View &gt; Customize Toolbar.&lt;/p&gt;&#xA;" OwnerUserId="18852" LastActivityDate="2019-01-05T05:44:19.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25384" PostTypeId="1" CreationDate="2019-01-05T07:21:55.560" Score="1" ViewCount="300" Body="&lt;p&gt;For water steam cycle in a &quot;combined cycle power plant&quot; containing 3 boiler drums and dosing of 5% ammonia solution in the condenser. How can I calculate the pH of the boiler water in each of the drums? (LP- 4.6 bar, IP- 36 bar, HP- 162 bar)&lt;/p&gt;&#xA;" OwnerUserId="18856" LastActivityDate="2019-02-15T20:03:15.750" Title="pH calculations for boiler water" Tags="|chemical-engineering|steam|chemistry|boilers|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25385" PostTypeId="2" ParentId="25370" CreationDate="2019-01-05T07:22:06.367" Score="1" Body="&lt;p&gt;A compromise between efficiency and practicality:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a 1/4&quot; type L copper tubing with brass fitting adaptor (1/4 to 1/2) or similar as base connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then solder 2 gauge 16 electrical copper wires from the sides of brass fitting to the tip of tubing at the 50cm end so it forms a rigid horizontal long triangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Readily available materials, total cost under $30.00 including the soldering gun, stiff enough vertically to support its own weight. And reinforced laterally by the two wires to take the load of pen.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-05T07:22:06.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25388" PostTypeId="2" ParentId="25360" CreationDate="2019-01-05T15:04:06.437" Score="0" Body="&lt;p&gt;Based on the question and comments, it seems you want something akin to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jEoEC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jEoEC.png&quot; alt=&quot;powder tray&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a tray that holds a conductive powder. The powder is used to allow the glass bed to &quot;self-level&quot; independently of the hot plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the conductive powder, any metal powder would do. The thermal conductivity increase by using a powder in the gap versus having an air gap is significant enough, let alone that you want to use the most conductive material possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would have two further issues. First, the powder should not stick to the glass bed. The choice may be have to be made through experiments. Second, the gap of 1 mm is small, and this may make such a tray design nearly impossible from a mechanical perspective. Indeed, even when you can create a 1 mm tray, the depth of the powder may be too thin to allow the self-leveling of the glass independent of the hot plate. In other words, when the powder is too thin, any thermal movement of the hot plate will be transmitted to the glass because the powder is essentially compacted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method may avoid the second problem. Increase the gap between the hot plate and the glass. This will allow the (thicker) powder to act both as a thermal conductor as well as an expansion/compression &quot;blanket&quot; between the hot plate and the glass. The downside is, as the gap increases, the temperature difference between the hot plate and the glass will increase proportionally. The end point will be that what you gain in mechanical stability by increasing the gap will eventually put you back to the same situation as having a 1 mm air gap. This will happen when &lt;span class=&quot;math-container&quot;&gt;$(k/d) \approx h$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is conductivity, &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is gap width, and &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is the convection coefficient of air).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A second approach to this problem is to replace the air gap with a liquid gap. Flow water or oil between the hot plate and the glass. This will increase the convection coefficient in the gap by a factor of 10 - 100.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-01-05T15:04:06.437" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25389" PostTypeId="1" AcceptedAnswerId="25390" CreationDate="2019-01-05T16:14:35.803" Score="1" ViewCount="61" Body="&lt;p&gt;I think I'm missing something important here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a fluid with a velocity &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; in a pipe with cross section area &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, the volumetric flow rate is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Q_v = uS$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we now consider vectors, the volumetric flow rate is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Qv = \int \vec{u} ·\vec{n} dS$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\vec{n}$&lt;/span&gt; is the normal vector to the cross sectional area. For example putting numbers, considering only one axis &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \vec{u} = 10 \vec{i} \qquad \vec{n} = \vec{i} \qquad S = 10 \\ Qv = 100$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However from the divergence theorem we know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \int \vec{u} · \vec{n} dS = \int \nabla · \vec{u} dV$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for incompressible fluids &lt;span class=&quot;math-container&quot;&gt;$\nabla · \vec{u} = 0$&lt;/span&gt;, so it would give &lt;span class=&quot;math-container&quot;&gt;$Q_v = 0$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously i'm doing something very wrong here, but I cannot see what. Am I considering something wrong in these equations? Is the normal vector in the divergence theorem refering to another surface in this case, or to both surfaces, these being, both &lt;span class=&quot;math-container&quot;&gt;$\vec{n} = \vec{i}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\vec{n} = -\vec{i}$&lt;/span&gt;? &lt;/p&gt;&#xA;" OwnerUserId="13681" LastActivityDate="2019-01-05T16:22:10.713" Title="Volumetric flow in a pipe from velocity" Tags="|pipelines|fluid|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25390" PostTypeId="2" ParentId="25389" CreationDate="2019-01-05T16:22:10.713" Score="1" Body="&lt;p&gt;The divergence theorem only applies to a &lt;strong&gt;closed&lt;/strong&gt; surface that encloses the volume &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you construct a closed surface by &quot;shrink wrapping&quot; a section of pipe and closing off &lt;em&gt;both&lt;/em&gt; ends, enclosing a volume &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, the volumetric flow out the &lt;em&gt;complete&lt;/em&gt; surface is zero, because the flow into one end is equal and opposite to the flow into of the other end - i.e. whatever goes into one end must come out of the other end.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-01-05T16:22:10.713" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25391" PostTypeId="1" CreationDate="2019-01-05T17:35:07.510" Score="0" ViewCount="43" Body="&lt;p&gt;I know this is a silly question, but shouldn't the high speed rotating blades (in case of an axial air compressor or an axial air fan for example) block the air from flowing? What makes air flow between blades?&lt;/p&gt;&#xA;" OwnerUserId="1293" LastActivityDate="2019-01-05T21:51:30.067" Title="Why doesn't axial turbomachinery blades block fluid from flowing?" Tags="|mechanical-engineering|fluid-mechanics|turbomachinery|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25392" PostTypeId="2" ParentId="25391" CreationDate="2019-01-05T18:41:29.007" Score="2" Body="&lt;p&gt;The blades are set at an angle which causes them to &quot;dig in&quot; to the air and propel it through the fan disc as the disc rotates. In this manner, the blades add kinetic energy to the air by performing work on it.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-01-05T18:41:29.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25394" PostTypeId="2" ParentId="25391" CreationDate="2019-01-05T21:51:30.067" Score="1" Body="&lt;p&gt;The blades are basically small helicopter wings. They are set to a small angle to the shaft so they create &quot;lift&quot;, but because they can't move up with the lift they move the air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However they can block the air in some cases. When they stall. Stall is when the smooth flow of air around a blade gets turbulent and the blade/wing loses its aerodynamics properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In early era of jet engines stalling inside a jet engine was a serious problem. It has been taken care of by digital automatic control of thrust and fuel regimen independent of pilot input.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-05T21:51:30.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25395" PostTypeId="2" ParentId="25381" CreationDate="2019-01-05T22:04:32.513" Score="1" Body="&lt;p&gt;Yes, you definitely can do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are small filler pumps that will automatically kick into power when the water falls below the present level and the floater activates the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can have the discharge drip continuously but the infill intermittently, automatically. And by putting a call control valve on the drain path, you control the whole circulation.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-05T22:04:32.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25396" PostTypeId="2" ParentId="25381" CreationDate="2019-01-05T23:11:36.267" Score="1" Body="&lt;p&gt;Drip irrigation parts cost about five cents a piece. They come in fractional gallons per hour at 25 psi. So pump the water into the tank from the bucket via a drip head, and use a simple float valve to regulate the outflow. First thought is a carb float. Toilet float might work as well.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-01-05T23:11:36.267" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25399" PostTypeId="1" CreationDate="2019-01-06T04:42:43.153" Score="0" ViewCount="364" Body="&lt;p&gt;I know you can apply displacement to a rigid body (cylindrical rod) to perform a 3 point bending test in ABAQUS. Is it possible to apply force to the rod (e.g. rod with super high young's modulus so it wont deform). I am trying to apply topology optimisation on a structure, so the displacement will change when material is removed at every iteration.&lt;/p&gt;&#xA;" OwnerUserId="18873" LastActivityDate="2020-10-27T18:06:40.670" Title="How to perform 3 point bending test by using force instead of displacement in ABAQUS?" Tags="|mechanical-engineering|materials|finite-element-method|abaqus|solid-mechanics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25401" PostTypeId="2" ParentId="25399" CreationDate="2019-01-06T12:17:49.970" Score="1" Body="&lt;p&gt;If the model is a three dimensional object, restrain exactly 6 degrees of freedom, so it can not translate or rotate, but it can deform in any way without creating reaction forces at the restraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the elements in the model have rotation degrees of freedom, this is easy: you just restrain all 6 degrees of freedom at one grid point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.digitalengineering247.com/article/free-floating-fea-models/&quot; rel=&quot;nofollow noreferrer&quot;&gt;This link&lt;/a&gt; shows some examples of how to do this for models that only have translation degrees of freedom - i.e. almost all solid elements in FE modelling software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you then apply self-equilibrating loads, there will be no reactions at the constrained degrees of freedom. You should check the reactions are very small compared with the applied loads (they probably won't be exactly zero, because of numerical rounding errors) to make sure the model is working correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a two-dimensional model use the same basic idea, but you only need to restrain 3 degrees of freedom to prevent translation and rotation in the plane of the model.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-01-06T12:17:49.970" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25405" PostTypeId="2" ParentId="13663" CreationDate="2019-01-07T00:29:32.617" Score="2" Body="&lt;p&gt;The modulus of elasticity is not going to significantly change for concrete in ambient temperatures but there are provisos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assumption is that you are talking about set concrete that has achieved 28 day strength. If you are talking about fluid concrete then that is down to the slump test and results are highly variable between strength types.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High strength concrete hardens rapidly and more so in higher temperatures. Obviously low temperatures will affect crystal lattice formation and ice would interfere with the rate of formation which is why ice is added to high strength concrete on sites in hot countries while being poured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to find out the effects on elastic range then an extensometer test would be the best way to find out the thermal effects. That said, concrete is not terribly good under tension so the results will likely be quite inconclusive. A Charpy test would demonstrate energy absorption and therefore the compressive variation but again the issues you face are that concrete is highly variable in composition and quality control is very limited, unlike steel which is manufactured to a much higher quality and far lower variability of structural properties during manufacture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing about tons modulus and concrete is that 5% of specimens will achieve maximum strength, 10% expected to fail strength and the rest will be median strength within allowed safety limits and thus modulus of elasticity is accounted safe within legislated limits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find more on the subject through Eurocode specifics in various text books on the subject.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-07T00:29:32.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25406" PostTypeId="2" ParentId="11804" CreationDate="2019-01-07T00:45:21.330" Score="1" Body="&lt;p&gt;Thermal expansion axially in pipes will be far less than you expect and the calculations for it demonstrate it. Steel is quite strong and elastic in both tensile and compressive states compared to concrete or stone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these pipes I shouldn't expect them to expand much beyond 1mm in every 200m and steel can take enormous compressive stress loads and generally its capacity is described by Young's modulus as &lt;span class=&quot;math-container&quot;&gt;$210kN/mm^2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is only above 800°C that steel becomes more ductile and faces structural creep issues of any concern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buried S bend joints can absorb expansion stresses every 500m and remain within the elastic range of the material so, no need to worry much beyond the completion date of the project and traffic woes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is far more interesting to me is the design lifespan of the project as there is less than 50 years of oil left and systems like this will become white elephants of a bygone energy age.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2019-01-11T10:37:41.833" LastActivityDate="2019-01-11T10:37:41.833" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25407" PostTypeId="2" ParentId="11316" CreationDate="2019-01-07T00:54:43.570" Score="1" Body="&lt;p&gt;As an honours civil engineering student I have been taught by three different doctors of civil and structural engineering that there is no difference between the two scenarios in design practice. What matters is how you prevent actual punching of slabs using collars, rebar design and other methods.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-07T00:54:43.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25409" PostTypeId="2" ParentId="12442" CreationDate="2019-01-07T05:15:02.220" Score="0" Body="&lt;p&gt;The particles ( atoms or molecules) of any fluid(liquid or gas) in a container will be in continuous random motion.During that motion they collide on the container walls and their velocity (direction or magnitude or both)changes.By momentum conservation the container wall experiences pressure.So if you move the container walls, the particles travel straight untill they  hit the walls in the new position thus they fill the additional volume or at least this how my teacher explained pressure.&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastActivityDate="2019-01-07T05:15:02.220" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25410" PostTypeId="2" ParentId="18815" CreationDate="2019-01-07T06:49:12.817" Score="0" Body="&lt;p&gt;If you reduce the number of layers of bearings to just one large diameter, and try to add all their mass to exterior layer, you almost double the angular moment of inertia of the system, which is going to be more efficient in conserving the momentum, but also will save on the leak and waste of energy due to wobbling of intermediary layers and friction due to out of plane vibration of those layers and fluttering of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I better alternative would be just one thin disk or even some radial spokes leading to the ball bearing with the total mas of your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets compare two cases, first one you have many many layers of rings, ilmost like a solid disk, and second one with just one large disk same size as the first case and mass concentrated on the perimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ I_{many\ rings} = \frac{1}{2} MR^2  $$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$ I_{one \ ring} = MR^2  $$&lt;/span&gt; &#xA;M= mass , R = radius  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we see I of one ring is twice as much as I of many rings. And also we don't have all the extra friction of many rings.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-07T06:49:12.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25411" PostTypeId="2" ParentId="21128" CreationDate="2019-01-07T07:20:29.090" Score="1" Body="&lt;p&gt;here are two alternatives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;first, you can &lt;em&gt;conformally coat&lt;/em&gt; the assembly with a sprayed-on and heat-cured waterproofing compound, commonly know as &quot;paint&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;second, you can immerse the circuit assembly in a liquid which will cure into a solid. substances like this are known as &quot;potting compounds&quot; and there are dozens of different varieties which are not particularly expensive. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-01-07T07:20:29.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25412" PostTypeId="2" ParentId="6987" CreationDate="2019-01-07T11:14:51.007" Score="0" Body="&lt;p&gt;As pointed out, non-conductive plastics can build up a charge. In certain applications industry code call for all piping in ex-hazard areas (or where the inside is an ex-hazard area!) to be conductive material, and grounded. To my knowledge,there's no conducting PVC. Hoever both PE and PP can be bought in a coducting variety (they mix carbon into the plastic matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beaty of PVC ist that the pipes can be glued, with PP or PE you need to weld the pipes or use plug sleeve. I don't know the prices of conductive PP, PE-el is 3-4x as pricey as ordinary PE (and I think with PVC the difference is even bigger).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would seriously consider using (stainless) steel pipes, maybe with press fittings. &lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-01-07T11:14:51.007" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25414" PostTypeId="1" AcceptedAnswerId="25417" CreationDate="2019-01-07T15:23:19.850" Score="0" ViewCount="321" Body="&lt;p&gt;In the term of &lt;strong&gt;machine design&lt;/strong&gt; and &lt;strong&gt;mechanical engineering&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If same spring is to be used in both smooth and rough road what will be the effect in suspension of the vehicle and if instead of using same spring for smooth and rough road how about using spring with &lt;strong&gt;variable spring rate&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to this and sorry for my bad English.&#xA;But i am stocked in finding the comparative answer of this question.&#xA;Can anyone help me out here????&#xA;Please....&lt;/p&gt;&#xA;" OwnerUserId="13703" LastEditorUserId="13703" LastEditDate="2019-01-17T08:47:53.987" LastActivityDate="2019-01-17T08:47:53.987" Title="Racing cars have spring with constant rate whereas Off road vehicles have variable spring rate" Tags="|mechanical-engineering|design|" AnswerCount="2" CommentCount="0" ClosedDate="2019-01-09T03:34:34.863" ContentLicense="CC BY-SA 4.0" />
  <row Id="25415" PostTypeId="1" AcceptedAnswerId="25473" CreationDate="2019-01-07T16:00:16.703" Score="0" ViewCount="51" Body="&lt;p&gt;I designed a LQR controller and want to implement it on a microcontroller. I don't know about digital control and don't know when to apply it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a brief research, I found out that I need to apply the z-Transform, instead of Laplace-Transform. However the gain matrix is derived from time-domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The microcontroller I use is a OpenCR1.0.&lt;/p&gt;&#xA;" OwnerUserId="18808" LastActivityDate="2019-01-11T00:43:21.453" Title="When is it necessary to consider digial control rather than directly coding in the gain matrix?" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25416" PostTypeId="2" ParentId="25414" CreationDate="2019-01-07T16:02:30.113" Score="0" Body="&lt;p&gt;Road Race cars only need a relatively small suspension movement - about +-3 inches or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Off road vehicles have a much larger suspension travel - often +-10 inches or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variable rate springs allow for supple movements over a short range of movement and and &quot;tighten up&quot; for larger movements as necessary.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-01-07T16:02:30.113" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25417" PostTypeId="2" ParentId="25414" CreationDate="2019-01-07T18:12:41.207" Score="2" Body="&lt;p&gt;It is a compromise between comfort and practicality. Yes the off-road cars would have a smother ride if they have constant rate springs, but then they would need a much larger suspension to accommodate all the play.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say two cars with similar mass on two surfaces with same speed: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1-a well leveled, smooth track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- a coarse bumpy off-road trail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first car is expected to hit obstacles of say 1 inch height and the second 3 inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a spring with constant ratio k, F= kx and the jerk or acceleration the bump will give to the car will be F = m*a therefore a = f/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but if the spring in the second car is to absorb a 3inch obstacle with the same constant k it needs to recoil 3-5 times x or even more depending on the speed, and this will call for a very deep spring box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to optimise the performance of the spring they design the spring with a gradually increasing stiffness, e.g. &lt;span class=&quot;math-container&quot;&gt;$F = k*x^2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way the suspension can take a much larger range of bumps and at the same time is comfortable when meeting small obstacles and tough enough to take the big ones and still not bottom out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the accomplish this is to build the spring with gradually thicker coil diameter, or use spring leaves, or design the geometry in a way that the play of spring becomes stiff as it recoils.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="10902" LastEditDate="2019-01-08T07:37:40.790" LastActivityDate="2019-01-08T07:37:40.790" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25419" PostTypeId="2" ParentId="21083" CreationDate="2019-01-07T18:31:56.730" Score="0" Body="&lt;p&gt;Hello we are currently in the process of working on a True Horizontal Centrifugal casting machine at work as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am no expert but from my understanding the required value range of N can be calculated via the formula driven by optimal G-Factor values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zsQu0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zsQu0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18888" LastActivityDate="2019-01-07T18:31:56.730" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25420" PostTypeId="2" ParentId="24702" CreationDate="2019-01-07T18:50:36.047" Score="0" Body="&lt;p&gt;Using method of joints we can calculate the maximum compression according to the values given being 6kN. As that strain is greater than 8kN we should assume 6kN to be achieved first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force P should not exceed the maximum compression value of 6kN at the vertical component of the truss CD.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$6 \sin 60=3 \sqrt {3}=5.196kN$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no way that tension can even come close to 8kN in the truss without exceeding the 6kN compression.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-07T18:50:36.047" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25421" PostTypeId="2" ParentId="2704" CreationDate="2019-01-07T20:29:31.900" Score="0" Body="&lt;p&gt;As there are 3 feet below a radial surface we need to calculate the maximum moment of area between the feet induced by table weight. This requires calculation of the midpoint of triangular line between two feet which describe an arc of 292mm with an angle between them of 120°.&#xA;The radius calculates to &lt;span class=&quot;math-container&quot;&gt;$292 / 2=146mm$&lt;/span&gt;&#xA;Using cosine rule we can determine the length of the chord midpoint to be &lt;span class=&quot;math-container&quot;&gt;$$L=\frac{\sqrt{ 2(146 ^2)-2(146^2)&#xA; \cos 120}}{2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$L=126.44$$&lt;/span&gt;&#xA;Now we calculate the inscribed radial distance as the perpendicular distance to L calculates by Pythagoras&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \sqrt{146^2-126.44^2}=r_2$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$r_2=73mm$$&lt;/span&gt;&#xA;Comparing the table to radius to the effective minimum support cantilever radius between the feet to be &lt;span class=&quot;math-container&quot;&gt;$\frac {800}{2}-73=327mm$&lt;/span&gt; from the table edge making that centroid quite high up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my opinion, this table is not going to be terribly safe to use considering that the support radius is just 73mm against a top heavy design of 400mm &lt;span class=&quot;math-container&quot;&gt;$\frac{3}{4}$&lt;/span&gt;m above the ground giving an axial load of 13.3kg which is 130.8N self weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the tipping point, we really only need to treat this as a simple beam equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result is that 8.32kg at the most precarious edge would be the limit of tipping. For safety I would factor 25% minimum which would reduce this to 6.24kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the weight of the table falling on a small child would likely cause serious injury and probably break a person's foot if tipped over accidentally, I'd add lateral stabilizers to the table feet parallel to the table radius in the shape of triangles pointing outwards. The base of them being 140mm long as isosceles triangles with 2x80mm sides&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-07T20:29:31.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25422" PostTypeId="2" ParentId="13457" CreationDate="2019-01-07T21:01:26.840" Score="0" Body="&lt;p&gt;In a rigid system, the roller transfers lateral forces to the pinned fixings and can only resist perpendicular forces.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-07T21:01:26.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25423" PostTypeId="2" ParentId="21982" CreationDate="2019-01-08T00:14:41.897" Score="1" Body="&lt;p&gt;The best way is to create a rectangle by cutting the triangular portion from the left to the right side where it is missing and calculate from the original zero coordinate which is the base width of the triangular portion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logic is this the centroid of both triangles is calculated from &#xA;&lt;span class=&quot;math-container&quot;&gt;$$[\frac {bc}{2}\times \frac {2}{3}b]_{left} + [\frac {bc}{2} \times \frac {1}{3}b]_{right}$$&lt;/span&gt;&#xA;By this statement we realise that the 2 triangular portions are equivalent to a single rectangle added to the rectangular portion of the p'gram but critically we cannot change the origin point to begin at the origin+b or the answer will be wrong.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-08T00:14:41.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25424" PostTypeId="1" CreationDate="2019-01-08T03:57:41.560" Score="7" ViewCount="240" Body="&lt;p&gt;I am using an LED as a point source for a school project. I have two questions in regards to the LED(s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that light follows the inverse square law when measuring perceived brightness, but all sources I can find show examples of a round bulb light. Is there an altered version of the formula for LED’s that incorporates factors such as the fan angles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I will most likely be adding more LED’s to my project to increase the total lumens/lux. I’ve seen many people ask the question whether lumens will add in such a scenario. In theory they would but it wouldn’t be perceived that way. My question is; is there a formula/equation/calculation that can help me predict the lumens perceived by x number of LED’s, d inches apart from one another, measured at y feet away, with regard to joining and isolated parts of the fan angles of each LED.&lt;/p&gt;&#xA;" OwnerUserId="18889" LastEditorUserId="18889" LastEditDate="2019-01-08T04:40:38.013" LastActivityDate="2022-11-21T13:22:19.543" Title="LED Lumen Calculations" Tags="|mathematics|lighting|photonics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25426" PostTypeId="2" ParentId="6213" CreationDate="2019-01-08T04:34:20.963" Score="1" Body="&lt;p&gt;Why not use parallel arms which not only maintain level foot but also lift the entire apparatus as a unit.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-08T04:34:20.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25427" PostTypeId="2" ParentId="19910" CreationDate="2019-01-08T04:44:54.663" Score="1" Body="&lt;p&gt;A foil coated plastic would be ideal as a metallic surface prevents light bleed and is easily manufactured. The higher the density of any material the less transparent they tend to be except in the case of transparent ones, obviously.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-08T04:44:54.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25428" PostTypeId="1" CreationDate="2019-01-08T05:22:08.103" Score="2" ViewCount="210" Body="&lt;p&gt;I'm using thermoelectric module to get the voltage. With the help of it, I'm getting 3 volts Open circuit voltage. I've attached DC-DC converter which regulates this voltage to 5 volts. I've attached USB at the output of the converter to connect my phone for charging. The thing is a basic cell phone is getting charged some 20% and then it doesn't progress even after 2 to 3 hours of connecting to USB. And whenever I connect smartphone, for a second charging symbol pops up and then charging stops.&lt;/p&gt;&#xA;" OwnerUserId="18890" LastEditorUserId="18890" LastEditDate="2019-01-08T05:41:06.953" LastActivityDate="2024-01-12T14:05:50.680" Title="Why are phones not getting charged when I supply 5 V and 0.3 A?" Tags="|electrical-engineering|power-electronics|battery|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25429" PostTypeId="2" ParentId="23014" CreationDate="2019-01-08T05:39:51.290" Score="0" Body="&lt;p&gt;Use of something with a massive surface to volume ratio would be hugely advantageous and this is where biological designs are useful. A red blood platelet has the greatest surface area in relation to its volumetric size. Use of that shape with a thermal sink inside it would encourage condensation but humidity is a vital factor next to ambient temperatures.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-08T05:39:51.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25430" PostTypeId="1" AcceptedAnswerId="25433" CreationDate="2019-01-08T05:41:14.463" Score="2" ViewCount="119" Body="&lt;p&gt;I read an article about Alberta, Canada losing millions per day from air barrels and am now doing some research, but having a hard time finding information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am curious to know more about air barrels and why they occur. Are there better systems out there that prevent air barrels or is it a constant, inevitable issue for pipelines? In particular, I am interested in the economics that result in air barrels, as well as the infrastructural issues or limitations that companies use to justify the purchase of air barrels. If you happen to know anything else about them, please do tell.&lt;/p&gt;&#xA;" OwnerUserId="18891" LastEditorUserId="33" LastEditDate="2019-01-08T13:56:03.080" LastActivityDate="2019-01-08T13:56:03.080" Title="What are air barrels (with regards to oil pipelines) and why are they a problem?" Tags="|petroleum-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25432" PostTypeId="2" ParentId="19357" CreationDate="2019-01-08T05:48:15.773" Score="0" Body="&lt;p&gt;A radial solution seems the best idea to step 2.8 times in every degree. An epicyclic gear agreement could pivot the lense through the required range by creating an offset centre fulcrum. The best part of this idea is that the lense itself could be a gear in the system.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-08T05:48:15.773" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25433" PostTypeId="2" ParentId="25430" CreationDate="2019-01-08T06:45:14.030" Score="1" Body="&lt;p&gt;Air barrel means oil pipeline capacity which is booked or contracted by an oil company while they already know they can not deliver that volume of oil. Basically they over book the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alberta pipeline in order to maintain the line or repair accidents has to cut down the flow of oil, so the oil companies overbook to be able to have guaranteed delivery of what they can produce. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However some companies game this known loophole to leverage the markets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://thetyee.ca/Opinion/2018/11/28/Air-Barrels-Pipeline-System/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is an article.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-08T06:45:14.030" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25434" PostTypeId="2" ParentId="25428" CreationDate="2019-01-08T06:51:34.517" Score="0" Body="&lt;p&gt;Then the phone charge circuit is not measuring enough capacity of either voltage or current to stay engaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solution is to make sure that the voltage does not drop under load and that sufficient current is available. Some phones need to see 2A available, but will only use what they need of that.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-01-08T07:21:40.183" LastActivityDate="2019-01-08T07:21:40.183" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25435" PostTypeId="1" CreationDate="2019-01-08T07:51:57.010" Score="0" ViewCount="303" Body="&lt;p&gt;I recently became curious why some lights flicker in some videos, and I read some explanation about it, and I still have some questions. &#xA;1. Does every light have to have a frequency? can it have an infinite frequency? (= can it possibly light continuously without &quot;flickering&quot;)&#xA;2. From what I understand, the lights in the video flicker when the camera frequency rate is higher than that of the lights. So, does that mean the lights won't flicker when the camera frequency rate is higher than light frequency rate? (but they still don't align perfectly) is there any visible difference between the camera frequency rate being higher than the light frequency rate and the two frequency rate being exactly the same?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much!&lt;/p&gt;&#xA;" OwnerUserId="18024" LastActivityDate="2019-01-11T16:00:54.777" Title="Camera and light frequency rate" Tags="|mechanical-engineering|electrical-engineering|control-engineering|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25436" PostTypeId="1" AcceptedAnswerId="25451" CreationDate="2019-01-08T08:10:48.060" Score="0" ViewCount="186" Body="&lt;p&gt;I'm currently trying to do an initial design for a propeller. In order to do this I'm trying to use &lt;a href=&quot;http://web.mit.edu/drela/Public/web/xrotor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Xrotor&lt;/a&gt;. Xrotor allows the user to enter certain information about both the propeller geometry, flight conditions and the airfoil lift and drag data after which it performs its calculations. Xrotor requires the following values for the lift and drag data (example values are given):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;========================================================================&#xA;1) Zero-lift alpha (deg):   0.00       7) Minimum Cd           : 0.0070&#xA;2) d(Cl)/d(alpha)       :  6.280       8) Cl at minimum Cd     : 0.150&#xA;3) d(Cl)/d(alpha)@stall :  0.100       9) d(Cd)/d(Cl**2)       : 0.0040&#xA;4) Maximum Cl           :  2.00       10) Reference Re number  : 2000000.&#xA;5) Minimum Cl           : -1.50       11) Re scaling exponent  : -0.2000&#xA;6) Cl increment to stall:  0.200      12) Cm                   : -0.100&#xA;                                      13) Mcrit                :  0.620&#xA;========================================================================&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This information can be obtained from the drag polars of the airfoil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HS92z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HS92z.png&quot; alt=&quot;&lt;span class=&quot;math-container&quot;&gt;$\alpha,C_l$&lt;/span&gt; plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Utyzq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Utyzq.png&quot; alt=&quot;&lt;span class=&quot;math-container&quot;&gt;$C_d,C_l$&lt;/span&gt; plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[&lt;img src=&quot;https://i.stack.imgur.com/MV3lx.png&quot; alt=&quot;360 degree extrapolated polar using JBlade&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 360 degree polar was made using &lt;a href=&quot;https://sites.google.com/site/joaomorgado23/Home&quot; rel=&quot;nofollow noreferrer&quot;&gt;JBlade&lt;/a&gt;. The only value that Xrotor requests that I am not sure on how to calculate is the &lt;code&gt;d(Cd)/d(Cl^2)&lt;/code&gt; figure. I'm fairly confident that this can be obtained using the &lt;span class=&quot;math-container&quot;&gt;$C_d,C_l$&lt;/span&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Drag_polar&quot; rel=&quot;nofollow noreferrer&quot;&gt;drag polar&lt;/a&gt;, but I'm not completely confident on how I should go about it. What is the correct way of determining this value?&lt;/p&gt;&#xA;" OwnerUserId="17876" LastEditorUserId="17876" LastEditDate="2019-01-08T08:24:25.590" LastActivityDate="2020-01-09T07:34:09.840" Title="How to obtain the $dC_d/dC_l^2$ value from the drag polar of an airfoil for Xrotor?" Tags="|mechanical-engineering|fluid-mechanics|drag|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25437" PostTypeId="1" AcceptedAnswerId="25450" CreationDate="2019-01-08T10:07:31.500" Score="0" ViewCount="198" Body="&lt;p&gt;We are heating up our houses in winter to ex.: 21C to then have an insulated box inside the house, which we cold down to 5 and -15C, and we call it a refrigerator.. :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;we would buy a new refrigerator and put it in a house having 21C constantly&lt;/li&gt;&#xA;&lt;li&gt;we would also have a &quot;heat-pump boiler&quot; for heating water. &lt;/li&gt;&#xA;&lt;li&gt;the &quot;heat-pump boiler&quot; outputs colder air than what is there in the house normally, that is how &quot;heat-pump boilers&quot; work, ok. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Like if I would put out the whole refrigerator to outside, where there is ex.: day: 5C night: -5C, I would be sure that the electricity consumption (operation cost) would go down, because it wouldn't have to battle with the indoor 21C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question&lt;/strong&gt;: would blowing cold air from the &quot;heat-pump boiler&quot; to the back (to the bottom or top of the back?) of a refrigerator help reduce the operational cost for the refrigerator? &lt;/p&gt;&#xA;" OwnerUserId="18894" LastActivityDate="2019-01-16T20:23:27.433" Title="Save money with using a refrigerator correctly - blowing cold air to the back of it?" Tags="|thermodynamics|heat-transfer|temperature|refrigeration|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25438" PostTypeId="2" ParentId="23455" CreationDate="2019-01-08T10:41:46.720" Score="0" Body="&lt;p&gt;To copy a sketch you already must have a sketch so I created one.Now I am gonna close this sketch.&#xA;    1. Now I am going to select the sketch on the feature tree at the left and I will press Right Ctrl+C, you need to hold Ctrl and then press C.&#xA;       Now I have made a plane as a reference and moved it a little bit to make it more clear, you do not need to create one, you can copy sketch to whatever plane you want.&#xA;       I am going to start a sketch on this plane, and then I will just press ight Ctrl+V, hen the sketch should appear.&#xA;       As you can see the same sketch was copied to a different plane, it is an easy way to save your time, if you want the same sketch on parallel planes you can use Convert Entities.&lt;/p&gt;&#xA;" OwnerUserId="18852" LastActivityDate="2019-01-08T10:41:46.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25441" PostTypeId="1" CreationDate="2019-01-08T16:05:07.343" Score="0" ViewCount="119" Body="&lt;p&gt;So I want to tap a super small hole on grade 5 titanium alloy with M1.8x0.2 (which should be possible according to &lt;a href=&quot;https://www.quora.com/Can-you-add-a-thread-inside-a-2mm-diameter-hole-of-Ti-6Al-4V-How&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Quora's answer&lt;/a&gt;. But is there any tap available for me to do this job? I mean M1.8 with grade 5 titanium, the normal tap should break very easily. Since it's a super small hole with super tough material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, can it be done by normal CNC, like the one you'll find in a normal company? Since the document looks like some scientific research which might be using some advanced machines. And can it be done on CNC engraving and milling machine or it needs a CNC lathe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last, why you should &lt;a href=&quot;https://www.mmsonline.com/articles/tips-for-tapping-titanium-alloys&quot; rel=&quot;nofollow noreferrer&quot;&gt;tap titanium with CNC&lt;/a&gt; but not by hand?&lt;/p&gt;&#xA;" OwnerUserId="18854" LastEditorUserId="6790" LastEditDate="2020-05-02T20:17:16.130" LastActivityDate="2020-05-02T20:17:16.130" Title="Tapping small hole on titanium" Tags="|machining|metals|cnc|alloys|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25442" PostTypeId="1" CreationDate="2019-01-08T16:31:52.637" Score="0" ViewCount="28" Body="&lt;p&gt;In an average family house, where should we let the colder air from a &quot;heat-pump boiler&quot; out? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) - to the inside of the family house (ex.: heated in winter, so letting cold air in it..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) - to the outside of the family house? &lt;/p&gt;&#xA;" OwnerUserId="18894" LastEditorUserId="18382" LastEditDate="2019-01-08T19:12:13.797" LastActivityDate="2019-01-08T19:12:13.797" Title="Where to output the colder air from a &quot;heat-pump boiler&quot;?" Tags="|heating-systems|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25443" PostTypeId="2" ParentId="10495" CreationDate="2019-01-08T16:45:06.107" Score="2" Body="&lt;p&gt;A 8' wide gate should be good enough. Compare the scenario with a 8' wide car parking space, between two parked car. standard Driveway is 15' wide and its a common practice all around the world, specifically in Asia, however, the driver needs to maneuver the car couple of times to get in. As the road is not so busy; it should not be a problem. &lt;/p&gt;&#xA;" OwnerUserId="18901" LastActivityDate="2019-01-08T16:45:06.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25444" PostTypeId="2" ParentId="25441" CreationDate="2019-01-08T17:41:47.050" Score="3" Body="&lt;p&gt;It should be doable by hand; the Quora answer missed what I found to be the most important factor for cutting Ti; use chlorinated cutting fluid like carbon tetra chloride. I once drilled small holes ( 1/8 &quot;) in grade 2 or 3 ( lower strength more ductile than 5 ) ; I broke a bit ( high speed steel, not Home Depot) before making the first hole . I used the carbon tet and then drilled several holes . Not easy but not as tough as drilling 18-8 stainless.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-01-08T17:41:47.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25445" PostTypeId="2" ParentId="13922" CreationDate="2019-01-08T17:59:56.767" Score="0" Body="&lt;p&gt;As you point out, the underlying elements on the surface will take the applied surface load and the FE element formulation allocates, consistently, to the nodes as nodal forces. Hope this explanation helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;CR&lt;/p&gt;&#xA;" OwnerUserId="18902" LastActivityDate="2019-01-08T17:59:56.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25446" PostTypeId="2" ParentId="25442" CreationDate="2019-01-08T18:00:07.343" Score="2" Body="&lt;p&gt;At least in the UK, domestic air source heat pumps are normally used to take heat from outside the house (cooling the air outside) and use it to heat water and/or the air inside the house. So the cooled air would go outside the house. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use a heat pump the other way round to take heat from inside the house and put it outside then we call it air conditioning. I have heard of reversible heat pumps that cool in the summer (heat inside to outside) and heat  in the winter (heat outside to inside), but it always goes one way or the other and never heats and cools inside the building at the same time.&lt;/p&gt;&#xA;" OwnerUserId="18382" LastActivityDate="2019-01-08T18:00:07.343" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25448" PostTypeId="1" AcceptedAnswerId="25460" CreationDate="2019-01-08T18:40:25.797" Score="1" ViewCount="338" Body="&lt;p&gt;The setup: I'm trying to mount a leadscrew powered by this motor &lt;a href=&quot;https://www.pitsco.com/TETRIX-MAX-TorqueNADO-Motor-with-Encoder&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tetrix Max Torquendado Motor&lt;/a&gt; on a 40ish pound robot, in order to lift it 4&quot; off the ground. I already have the linear motion, latching mechanism, etc. I'd like it to go as fast as possible, but for now, let's just say that it can take 4 seconds to lift the robot. There will be 40ish pounds of tension acting on the screw when the robot is hanging, and minimal compression. Also, resolution is unimportant. It seems like there's a simple way to calculate the pitch/diameter leadscrew needed based on the torque, motor, and time requirements. What leadscrew should be used, and generally how could I calculate this sort of thing?&lt;/p&gt;&#xA;" OwnerUserId="18903" LastEditorUserId="110" LastEditDate="2019-01-08T23:02:11.763" LastActivityDate="2019-01-09T22:24:30.513" Title="What lead screw to use for specific torque and speed requirements?" Tags="|robotics|linear-motion|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25450" PostTypeId="2" ParentId="25437" CreationDate="2019-01-09T00:18:55.007" Score="2" Body="&lt;h2&gt;Foundations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The CoP of a refrigerator uses the cold heat flow and the work input, expressed as &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_c / \dot{w}$&lt;/span&gt;. The Carnot expression based on temperature is &lt;span class=&quot;math-container&quot;&gt;$T_c / (T_h - T_c)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$T_c$&lt;/span&gt; is the internal cold temperature and &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt; is the external hot temperature. Decreasing &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt; will increase the CoP, meaning less work flow &lt;span class=&quot;math-container&quot;&gt;$\dot{w}$&lt;/span&gt; will be needed to remove the same amount of heat &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_c$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Application to System at Hand&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In a picture of a control volume for the system at hand, &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt; can be taken as the temperature of the coils used to remove the heat. The heat flow to the air that surrounds the coils happens outside of the control volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt; is controlled by two factors. The first is the amount of heat that is being pumped out of the refrigerator in to the cooling fluid. The second is the amount of heat that is being dumped from the coil to the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the first factor is constant, anything that can be done to increase the second (heat flow from coil to air) will decrease &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt;. Two approaches are to decrease the temperature of the air &lt;span class=&quot;math-container&quot;&gt;$T_{air}$&lt;/span&gt; and to increase the flow rate of air. The first approach increases the temperature difference in the convection coefficient &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; in the equation &lt;span class=&quot;math-container&quot;&gt;$\dot{q} = hA(T_h - T_{air})$&lt;/span&gt;, thereby increasing the heat flow from the coils, thereby decreasing &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt;. For the same &lt;span class=&quot;math-container&quot;&gt;$T_{air}$&lt;/span&gt;, the latter approach increases &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; of the air, thereby increasing the heat that is removed from the coil, thereby decreasing &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Proposed Approach&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Based on your overview of the system, you might try a passive convection approach to move the cool air from the heat-pump boiler to the back of the refrigerator. Basically, the passive convection systsem is a channel where the hot air rising from the back of the refrigerator &quot;sucks in&quot; the cooler air from the heat-pump boiler. The better that you seal the channels, the better will be the operation. No forced convection fan needed, therefore no cost is lost by balancing the power drain to run the fan versus improving the cooling efficiency at the back of the refrigerator. The only investment is the cost in materials and labor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qhPGw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qhPGw.png&quot; alt=&quot;passive cooling&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aesthetics are a different story.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-01-13T16:55:48.803" LastActivityDate="2019-01-13T16:55:48.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25451" PostTypeId="2" ParentId="25436" CreationDate="2019-01-09T03:18:50.847" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$d C_d/dC_l^2 = dC_d/(2 \cdot C_l \cdot dC_l) = (1/(2\cdot C_l)) \cdot(1/$&lt;/span&gt;Slope of &lt;span class=&quot;math-container&quot;&gt;$C_d/C_l$&lt;/span&gt; curve&lt;span class=&quot;math-container&quot;&gt;$)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a few &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;s of interest and read the &lt;span class=&quot;math-container&quot;&gt;$C_l$&lt;/span&gt;s. Find the slope at those &lt;span class=&quot;math-container&quot;&gt;$C_l$&lt;/span&gt;s. The ratio should be consistant.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-01-09T03:18:50.847" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25454" PostTypeId="2" ParentId="25435" CreationDate="2019-01-09T10:01:31.650" Score="3" Body="&lt;p&gt;There are a number of reasons why lights might flicker. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incandescent bulbs running on AC will, of course have a cyclical input although the fact that the filament stays hot tends to smooth the actual light output out a lot. For similar reasons fluorescent tubes will also flicker at the mains frequency and this tends to be rather more pronounced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;LED lights of course run on DC power but most are controlled by pulse width modulation as this tends to be the most efficient way to provide a regulated current supply and dimming. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How this interacts with cameras can be a bit complex and has to do not just with the frequency of the lights but the sampling rate of the camera and the phase angle between them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if you have a situation where they camera and lights have the same frequency the light level can seem (to the camera) brighter or darker depending on the phase, ie if the camera is recording at 40 frames per second and the lights are flickering at 40Hz then depending on exactly when you start sampling you could be consistently in the dark or light stage of the cycle or somewhere in between. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously it is very difficult to measure or control this phase difference whcih can make consistent exposure difficult. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equally if the frequencies are different you can get a variation in light levels very similar to the wagon wheel effect. &lt;a href=&quot;https://en.wikipedia.org/wiki/Wagon-wheel_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Wagon-wheel_effect&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also get problems with still cameras on short exposures.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly lights designed for use with film and video tend to be of a type with little or no flicker and for this reason LED lights are much more common in theatre than film. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2019-01-11T16:00:54.777" LastActivityDate="2019-01-11T16:00:54.777" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25456" PostTypeId="2" ParentId="6101" CreationDate="2019-01-09T11:56:25.373" Score="1" Body="&lt;p&gt;2024 is precipitation hardened, meaning it is first solution treated to remove any precipitates already present and then quenched to stop more forming during cooling. Afterwards it is aged to reform the precipitates in a controlled manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the solution treatment effectively gives you a fresh start repeating the whole cycle will give you the correct temper. But you will get some grain growth every time you do the solution treatment, which normally reduces the tensile strength slightly (not sure how much with 2024).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you were thinking of just doing the aging stage (similar to tempering hardened steels), then the precipitates will already be present and it won't have much of an effect.&lt;/p&gt;&#xA;" OwnerUserId="18382" LastActivityDate="2019-01-09T11:56:25.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25458" PostTypeId="1" CreationDate="2019-01-09T20:39:30.777" Score="0" ViewCount="71" Body="&lt;p&gt;I am reading &lt;a href=&quot;https://web.stanford.edu/group/GGG/pelliccione-rsi-2013.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;, which talks about dealing with mechanical noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When characterizing their existing noise, it says the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The integrated RMS vibrations up to 1 kHz on the mixing chamber plate&#xA;  are approximately 2 μm in the vertical z direction and 6 μm in the&#xA;  lateral directions x and y.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It seems they integrated noise spectral density.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In implementing some of the mechanical noise isolation technique, what I care about is what the actual, physical size of the total vibration I will have, rather than the magnitude of individual components, which I believe Power Spectral Density represents.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I think the integrated RMS is what I am interested. (let me know if I am wrong on this point)  But does this figure actually equal the actual magnitude of oscillation?  For example, if I bolt an object onto a plate which has an integrated RMS vibrations of 2 μm in the vertical z direction, will it actually oscillate vertically by about 2μm? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am having a hard time understanding is their 1kHz integration bound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't one be interested in the integrated RMS vibrations up to the infinity? why stop at 1kHz?  &lt;/p&gt;&#xA;" OwnerUserId="17201" LastActivityDate="2019-01-09T22:00:16.220" Title="Amplitude RMS vs Amplitude in understanding vibration" Tags="|vibration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25459" PostTypeId="2" ParentId="25458" CreationDate="2019-01-09T22:00:16.220" Score="1" Body="&lt;p&gt;The paper says the main source of excitation is the &quot;pulse tube&quot; which has a fundamental frequency of about 1 Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you look at the plots in section IV, it is pretty clear that the large responses are at low frequency (well below 100 Hz) so there is no point in measuring the response at higher frequencies if there is nothing that is going to excite them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the plots, and assuming the authors haven't deliberately omitted something that is happening in the 100Hz - 1kHz frequency range, there probably wouldn't be any significant difference in the measurements from a design point of view if the cutoff was reduced to 100Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if a different mechanical system is subject to noise input that &lt;em&gt;does&lt;/em&gt; contain high frequencies (e.g a small internal combustion engine running at 12,000 RPM where the fundamental excitation frequency would be 200Hz not 1Hz) an arbitrary cutoff point of 1kHz would be much too low.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-01-09T22:00:16.220" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25460" PostTypeId="2" ParentId="25448" CreationDate="2019-01-09T22:24:30.513" Score="1" Body="&lt;p&gt;The relationship between torque and force for a frictionless single-start leadscrew is [1]:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{F}{\tau} = \frac{2 \pi}{p}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; is linear lifting force, &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; is torque, and &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can choose a suitable torque which that motor is best operated at, you know the force is the weight of the robot, so you can calculate pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed of lifting is determined by the power delivered by the motor. So you can get maximum speed by operating the motor at a torque that maximizes its power output. That's typically half its stall torque (700 oz-in) so choose &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; = 350 oz-in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1] &lt;a href=&quot;https://en.wikipedia.org/wiki/Screw_(simple_machine)#Torque_form&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Screw_(simple_machine)#Torque_form&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2019-01-09T22:24:30.513" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25461" PostTypeId="2" ParentId="12201" CreationDate="2019-01-09T22:34:14.430" Score="-1" Body="&lt;p&gt;If you want to approximately find it, you probably need the dashed line part too, else it's like having being on the edge where shear is 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any way, this is not the proper way of doing so. Take a look at this which is similar: &lt;a href=&quot;https://engineering.stackexchange.com/questions/6601/interpretation-of-shear-flow-shear-center-of-l-profile&quot;&gt;Interpretation of shear flow/shear center of L-profile&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18692" LastActivityDate="2019-01-09T22:34:14.430" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25462" PostTypeId="2" ParentId="24416" CreationDate="2019-01-09T23:01:08.543" Score="1" Body="&lt;p&gt;For stocky columns, material yielding and ultimate failure will come before even reaching the buckling load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just think about Euler's equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_{cr} = \dfrac{\pi^2 EI}{L_{cr}^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$L_{cr} \rightarrow 0$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$P_{cr} \rightarrow \infty$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, equation one can be more critical and &lt;span class=&quot;math-container&quot;&gt;$P_{cr}$&lt;/span&gt; can be greater than &lt;span class=&quot;math-container&quot;&gt;$Af_y$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18692" LastEditorUserId="1832" LastEditDate="2019-01-10T14:57:09.010" LastActivityDate="2019-01-10T14:57:09.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25463" PostTypeId="1" CreationDate="2019-01-10T00:45:36.010" Score="0" ViewCount="180" Body="&lt;p&gt;I have questions about a well pump application I am working on. It consists of a well that with a submersible pump that fills a reservoir tank. As far as I know, the tank is not pressurized. Here are some values associated with the system.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Static well level = 39 ft&lt;/li&gt;&#xA;&lt;li&gt;Tank Discharge level = 28 ft&lt;/li&gt;&#xA;&lt;li&gt;Well drawdown = 67 ft&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A couple of questions I have are listed below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in assuming that the static head requirement for the system curve is 67 ft? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawdown is  a function of flow, so it's considered dynamic head. I keep on getting tripped up on the difference in suction vessel and discharge vessel pressure, though. The well is open to the atmosphere so there is no suction &quot;vessel&quot; pressure, but the tank is closed and so I keep on thinking there might be discharge vessel pressure, but if the water is releasing into open space and then dropping to fill the tank, wouldn't this also be atmospheric pressure, unless the tank is pressurized? Appreciate the assistance.&lt;/p&gt;&#xA;" OwnerUserId="18916" LastActivityDate="2024-02-14T05:04:45.523" Title="Static Head Assumptions" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25464" PostTypeId="2" ParentId="25463" CreationDate="2019-01-10T01:27:10.667" Score="0" Body="&lt;p&gt;In the absence of a digram, I assume the well drawdown is 67 ft. And if the tank discharge level is at 28 then the total head for pump should be 95 ft if the tank is at grade level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the tank head if it's closed then as it gets filled with water its air volume shrinks so the pressure increases and depending on the ratio of volume of air lost to water you need to add the increased pressure of trapped air to the head required for the pump. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that you need to stablish a relationship between the q of pump WRT the volume of the tank. Regardless of where is the pump outlet in the tank, above water level or below. If you do that then you need to modify the well drawdown, because the pump will be pumping less water due to more head it needs to overcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the numbers that you have in you question are suggesting that the tank is configured to have a constant water height, may be a floater switch device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you edit your question and add a diagram, it will clarify the situation.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-10T01:27:10.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25466" PostTypeId="2" ParentId="21472" CreationDate="2019-01-10T14:01:57.303" Score="0" Body="&lt;p&gt;you should use include file&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#include &amp;lt;SMAAspUserSubroutines.hdr&amp;gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;initialize Mutex #1 at begin of the subrotine after include vaba.....&#xA;call MutexInit( 1 ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;use &quot; call MutexLock( 1 ) &quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your program documentation to use for 4 cpus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at the end of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;use &quot; call MutexUnlock( 1 ) &quot;&lt;/p&gt;&#xA;" OwnerUserId="18924" LastEditorUserId="9" LastEditDate="2019-01-10T14:17:58.923" LastActivityDate="2019-01-10T14:17:58.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25467" PostTypeId="2" ParentId="25337" CreationDate="2019-01-10T14:47:10.740" Score="1" Body="&lt;p&gt;The simple answer begins with this:  You will be pumping water that is at the boiling point, and such conditions will cause cavitation&lt;sup&gt;&lt;a href=&quot;http://www.physics.org/article-questions.asp?id=134&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;,&lt;a href=&quot;https://blog.craneengineering.net/what-is-pump-cavitation&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;&lt;/sup&gt; in many pump designs.&lt;br&gt;&#xA;Cavitation in a pump occurs when the liquid passing through the pump begins to vaporize in the low–pressure regions upstream of the impeller or similar mechanism which causes the liquid to flow through the pump casing.  Such bubbles of vaporized liquid cause various forms of damage to the pump; see to the links for more information.&lt;br&gt;&#xA;Every pump is susceptible to cavitation, so you need to choose a pump which is tested to operate at the conditions of the liquid which you will be pumping with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any pump which has been designed and certified for use in coffee makers like that which you are describing would be suitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following that, your second concern will be the size and physical configuration of the pump.  Since you have not provided any information as to the physical dimensions and configuration in which you plan to build your coffee maker, I looked at some of the schematics&lt;sup&gt;&lt;a href=&quot;http://poursteady.com/wp-content/themes/poursteady-theme/images/20160415_Poursteady_Specs_PS1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;,&lt;a href=&quot;http://poursteady.com/wp-content/themes/poursteady-theme/images/20160415_Poursteady_Specs_PS13c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;4&lt;/a&gt;&lt;/sup&gt; available on the Poursteady website which you referenced.  Going off those, it appears as though you will are expecting to build a device which will have ample space in which to install your pump. &lt;/p&gt;&#xA;" OwnerUserId="10707" LastActivityDate="2019-01-10T14:47:10.740" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25468" PostTypeId="2" ParentId="25424" CreationDate="2019-01-10T16:05:52.840" Score="0" Body="&lt;p&gt;If you want to look at an area that light is illuminating then you are probably best working with &lt;strong&gt;Illuminance&lt;/strong&gt; which is measured in lux. The inverse square law looks how the area reflecting the light increases as you get further away, but if you can calculate the area reflecting the light from a known angle, distance and luminous flux (lumens).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to your second question it would appear there is a roughly square relationship between the amount of light and what the eye perceives so... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perceived light ≈ √Actual light&lt;/p&gt;&#xA;" OwnerUserId="18382" LastActivityDate="2019-01-10T16:05:52.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25470" PostTypeId="1" AcceptedAnswerId="25477" CreationDate="2019-01-10T16:46:50.570" Score="-1" ViewCount="178" Body="&lt;p&gt;I am wondering if a spinning permanent magnet's interaction with the Earth's magnetic field will produce a propelling force strong enough to propel a toy boat through the water. This propelling force would be most effective when the toy boat travels due North and stays on that course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please reference the drawing below. A permanent magnet would be fastened to a non-metallic bar that is perpendicular to an electric motor's shaft and on the other end of this bar would be a non-metallic counterweight. Say that the toy boat is 1 foot long, the magnet is a .5 inch cube N52 Neodymium magnet, the motor is spinning at 3000 rpm, and boat is in an indoor swimming pool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic concept is that as the magnet is rotated from the position of facing due North to facing due South (180 degree rotation), the motor will have to push the magnet through the Earth's magnetic field and the reaction to this is that the toy boat will be pushed through the water. As the magnet is rotated from the position of facing due South to facing due North (180 degree rotation), the motor will not have to push the magnet through the Earth's magnetic field. So, during each full/360 degree rotation of the motor's shaft, the magnet's interaction with the Earth's magnetic field should create a paddle wheel effect that should propel the toy boat forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this magnetic paddle wheel effect propel this boat through the water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I am wondering if the rotational speed of the motor has a direct effect on the amount of propelling force that is generated. For example, will the motor have a harder time pushing the magnet through the Earth's magnetic field when it is spinning at 3000 rpm as compared to spinning at only 500 rpm? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had to insert a revised drawing because the original drawing had the magnet positioned in the wrong orientation. Also added text to help convey concept.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GvUhS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GvUhS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user18610" LastEditorDisplayName="user18610" LastEditDate="2019-01-11T13:14:27.227" LastActivityDate="2019-01-11T13:56:47.837" Title="Could a spinning permanent magnet's interaction with the Earth's magnetic field propel a boat through the water?" Tags="|mechanical-engineering|design|applied-mechanics|experimental-physics|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="25471" PostTypeId="1" CreationDate="2019-01-10T19:18:37.333" Score="1" ViewCount="199" Body="&lt;p&gt;My problem is a bridge with a horizontal cord of span 12&amp;nbsp;m between two piers of arbitrary height. It is required to use a pin-connected truss, consisting of steel members bolted together to steel gusset plates. Assume the end supports at the piers to be a pin and a roller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A vertical loading of a concentrated force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; [19.5&amp;nbsp;kN] acting 4&amp;nbsp;m from the left-hand support, and linearly distributed load &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; [1250&amp;nbsp;N/m] acting on the central 6&amp;nbsp;m of the span. This load can be applied in part to several joints on the top cord within the given region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force of the wind and the weight of all members are to be neglected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume the maximum tensile force in each member cannot exceed 4.25&amp;nbsp;kN, and the maximum compressive force cannot exceed 3.5&amp;nbsp;kN regardless of the length of the member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MGic3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MGic3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I knew that I can reduce the concentrated force by adding a pin on that force and join the pin to multiple members, but doing that leaves me with too much unknowns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also, am I able to reduce the 19.5k&amp;nbsp;N load by replacing the single pin, where the load acts and adding two pins between the load so the load will be 9.75&amp;nbsp;kN in each pin? This approach is used &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BS9ca.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BS9ca.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no restriction about how the bridge should look like, but I designed mine so to have in each member tension not more then 4.25KN and compression less than 3.5 KN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EReVm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EReVm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18927" LastEditorUserId="1832" LastEditDate="2019-01-12T18:28:40.960" LastActivityDate="2019-01-12T18:32:26.040" Title="designing bridge problem" Tags="|structural-engineering|" AnswerCount="0" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25473" PostTypeId="2" ParentId="25415" CreationDate="2019-01-11T00:43:21.453" Score="0" Body="&lt;p&gt;You can also solve the LQR problem for discrete systems instead of continuous. One does needs to solve the &lt;a href=&quot;https://en.wikipedia.org/wiki/Algebraic_Riccati_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;DARE instead of CARE&lt;/a&gt;. But this does require having a discretized state space model. A simple but bad discetization method would be &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;forward Euler&lt;/a&gt;. But there are a lot of other methods out there, such as zero-order-hold, first-order-hold and bilinear-transform/Tustin. In certain situations one method would be preferable over another and a decent overview is given in &lt;a href=&quot;https://www.youtube.com/playlist?list=PLUMWjy5jgHK0MLv6Ksf-NHi7Ur8NRNU4Z&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; YouTube series by Brian Douglas. But at low frequencies they all yield the same response, so if the sample rate is significantly faster than the dynamics of the system then it won't matter much which method you pick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be noted that it is very likely that the continuous LQR gain you calculated for the continuous system will also work on a digital controller if the absolute values of the closed loop poles lie significantly below the sample frequency. Namely most often zero-order-hold would be a good representation of what the digital controller does. Namely it calculates the controller output and holds it for one sample time. This roughly adds half a sample time delay, which might make the system unstable if the closed loop poles lie close to the sample frequency. But a delay only reduces the phase margin of the controller, but LQR has a guaranteed phase margin of 60°, so should always be able to handle some decent amount of delay. For example when use would use poles placement you would not have this 60° phase margin guarantee. But with LQR the performance might not be as optimal as the LQR designed for the discretized system.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2019-01-11T00:43:21.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25474" PostTypeId="1" AcceptedAnswerId="25483" CreationDate="2019-01-11T10:03:36.823" Score="3" ViewCount="916" Body="&lt;p&gt;I have a physics background so apologize my ignorance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confused by the constants &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; when it comes to torsion of beams. From what I understand &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; are the same thing, &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; is (usually) for round beams with isotropic materials while &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt; is for beams with unusual shapes and/or material properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I really don't understand is how you get from &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt; though. In a &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S0263822309004693&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper&lt;/a&gt;, call it paper 1, I found, in the case of an orthotropic beam with rectangular cross section, they derive equations for torsional rigidity &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; then mention equations such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\kappa=\frac{D}{2ba^3G^0_x}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; are the height and width of the beam while &lt;span class=&quot;math-container&quot;&gt;$G^0_x$&lt;/span&gt; is the shear modulus. Where did this equation come from? Is it a standard result? Did they derive the equation for &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt; from FEM simulations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then in &lt;a href=&quot;https://link.springer.com/article/10.1007%2Fs00707-017-2067-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; paper (call it 2), which deals with torsion of arbitrarily shaped orthotropic composites, they talk about the torsional rigidity &lt;span class=&quot;math-container&quot;&gt;$GJ$&lt;/span&gt;, which makes me doubt my previous statement about &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt;. One of the equations that they give is that the torsional rigidity factor &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; (which is &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt;, right?) is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\beta=\frac{GJ}{ab^3G_{zx}}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I can tell &lt;span class=&quot;math-container&quot;&gt;$GJ=D$&lt;/span&gt; from paper 1 and like I said &lt;span class=&quot;math-container&quot;&gt;$\beta=\kappa$&lt;/span&gt;. So once again I wonder how &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; (or &lt;span class=&quot;math-container&quot;&gt;$GJ$&lt;/span&gt;) and &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt; (or &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;) are related.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should explain that the reason why I'm looking to understand better these relationships is because I'm 3D printing springs. I'm looking to find a relationship between the shear moduli of an orthotropic material and the spring constant of a coil spring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One method that I found is to relate the energy stored in a spring to the energy stored by a beam under torsion &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}kx^2=\frac{1}{2}\kappa \theta^2$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; are the spring constant and deflection, and &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; are the torsional constant and angle of twist of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two big problems that I have are that my springs have square cross sections (because they're easier to print) and are orthotropic. Hence why I'm having so much trouble.&lt;/p&gt;&#xA;" OwnerUserId="18933" LastEditorUserId="18933" LastEditDate="2019-01-14T03:26:38.463" LastActivityDate="2019-06-13T04:07:50.663" Title="Torsional rigidity D vs torsional rigidity factor k vs torsion constant J" Tags="|materials|beam|modeling|springs|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25477" PostTypeId="2" ParentId="25470" CreationDate="2019-01-11T13:56:47.837" Score="3" Body="&lt;p&gt;Like anyone who has used a compass knows, earth's magnetic field does exert a force on magnetic objects.  In theory this force could be harnessed to do work, but in practice it is too small to be a concern.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a 1 meter long wire carrying 1 amp of current sitting perpendicular to earth's magnetic field will be subject to about 0.0001 Newtons of force.  That's not going to be enough to overcome friction let alone build up any noticeable amount of speed.  For comparison, dropping a paper clip off the side of a table exerts 100 times that force.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your &quot;paddle wheel&quot;:  The magnet is always in earth's field and is the equivalent of a paddle wheel that is fully submerged in the water.  That means the force exerted on your spinning magnet will follow a typical sinusoid pattern, resulting in net zero work.  You would be better off just leaving the magnet stationary, but don't expect your boat to move a noticeable distance unless you see all the iron objects in the world flying towards the poles.  &lt;/p&gt;&#xA;" OwnerUserId="18936" LastActivityDate="2019-01-11T13:56:47.837" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25478" PostTypeId="1" AcceptedAnswerId="25481" CreationDate="2019-01-11T16:02:41.070" Score="0" ViewCount="909" Body="&lt;p&gt;Does surface finish affect steels strength properties? If so, how does strength increase or decrease with the type of surface finish applied?&lt;/p&gt;&#xA;" OwnerUserId="18938" LastActivityDate="2019-05-09T23:07:08.923" Title="Surface Finish and Strength" Tags="|strength|surface-preparation|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25480" PostTypeId="2" ParentId="25478" CreationDate="2019-01-11T17:16:23.030" Score="3" Body="&lt;p&gt;Shot peening improves the fatigue life , mostly used for steel springs. It does not change the bulk strength much . Shot peening is also used for aluminum. A very smooth ,polished surface can also improve &quot;strength&quot; in VERY highly stressed high strength steels ( this is likely of little use in ordinary applications ; I measured the effect in hydrogen stress cracking coupons stressed nearly to yield.). Generally steels are so tough that surface finish does not have a large affect.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-01-11T17:16:23.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25481" PostTypeId="2" ParentId="25478" CreationDate="2019-01-11T18:20:15.390" Score="1" Body="&lt;p&gt;Finishing operations on steel will not affect the hardness significantly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roughing or forging operations on steel, however, do increase its hardness (by work hardening). Generally, the more a machining process deforms the surface of steel, the harder it becomes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a research article on, &lt;a href=&quot;http://www.scielo.br/pdf/jbsmse/v32n1/v32n1a06.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Effects of Milling Condition on the Surface Integrity of Hot Forged Steel &lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="13526" LastEditDate="2019-05-09T23:07:08.923" LastActivityDate="2019-05-09T23:07:08.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25482" PostTypeId="1" CreationDate="2019-01-12T00:57:25.337" Score="1" ViewCount="1782" Body="&lt;p&gt;On a technical drawing for resistors, the lead diameter was given as &lt;span class=&quot;math-container&quot;&gt;$\varphi 0.55\text{mm}$&lt;/span&gt;. Since they &lt;em&gt;also&lt;/em&gt; specify the wire is 24 SWG, and 24 SWG is a diameter of 0.559mm, did they simply use φ (phi) when they should have used Ø, or is this actually accepted notation? And if it &lt;em&gt;is&lt;/em&gt; accepted usage, does the usage of φ differ from Ø?&lt;/p&gt;&#xA;" OwnerUserId="18942" LastActivityDate="2019-01-12T05:09:32.777" Title="Lowercase phi (φ) seemingly used for diameter in diagram instead of Ø?" Tags="|drafting|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25483" PostTypeId="2" ParentId="25474" CreationDate="2019-01-12T01:59:45.227" Score="0" Body="&lt;p&gt;Different texts use different notation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; is sometimes called &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt;, torsional stiffness constant, units of &lt;span class=&quot;math-container&quot;&gt;$L^4$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; is shear modulus of rigidity which is sometimes assumed to be 40% of &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$GJ$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$JK$&lt;/span&gt; is called section modulus of torsion rigidity as &lt;span class=&quot;math-container&quot;&gt;$EI$&lt;/span&gt; is in a beam bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The frequently used equations can be found in Roark's Formulas for Stress and Strain, 7th edition, chapter 10, pp 404.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using my phone app, but if you need more detail I responded on my computer using &lt;span class=&quot;math-container&quot;&gt;$\LaTeX$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations for torsion in a rectangular solid section:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\theta =\frac { TL }{ KG }$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$${ \tau }_{ max }=\frac { 3T }{ 8a{ b }^{ 2 } } \left[ 1+0.6095\left(\frac { b }{ a }\right) + 0.8865{ \left(\frac { b }{ a } \right) }^{ 2 }-1.8023{ \left(\frac { b }{ a } \right) }^{ 3 }+0.9100{ \left(\frac { b }{ a } \right) }^{ 4 } \right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$K=a{ b }^{ 3 }\left[ \frac { 16 }{ 3 } -3.36\frac { b }{ a } \left(1-\frac { { b &#xA;                }^{ 4 } }{ { 12a }^{ 4 } } \right) \right] \text{ for } a\ge b $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\theta = \text{angle of rotation [rad]} \\&#xA;\tau = \text{torsion stress} \\&#xA;K = \text{torsional stiffness constant}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2019-01-12T15:35:58.020" LastActivityDate="2019-01-12T15:35:58.020" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25484" PostTypeId="2" ParentId="25482" CreationDate="2019-01-12T05:09:32.777" Score="2" Body="&lt;p&gt;As engineers, symbols get re-used due to the fact that there are so many things we need to give a label to, so if someone uses a lower case does it change the thing described?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as it is clear what is referred to, then that is what matters. If someone heard verbally &quot; phi &quot; and used lower instead of uppercase...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table in this link gives the diameter of 24SWG as 0.558mm, see &lt;a href=&quot;https://www.esr.co.uk/electronics/cable-copper2.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.esr.co.uk/electronics/cable-copper2.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-01-12T05:09:32.777" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25485" PostTypeId="1" AcceptedAnswerId="25492" CreationDate="2019-01-12T05:40:43.470" Score="1" ViewCount="192" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qMspd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qMspd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I would be appreciate if anyone can solve problems below for this beam.  &lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;・Find the all sectional forces &lt;span class=&quot;math-container&quot;&gt;$F_x,F_y,M_x,M_z$&lt;/span&gt; for both elements① and element②.  &lt;br&gt; &#xA;・Find elastic strain energy of element① and element②. &lt;br&gt;&#xA;Flexural rigidities  of two elements are both &lt;span class=&quot;math-container&quot;&gt;$EI$&lt;/span&gt; and torsional rigidities of them are &lt;span class=&quot;math-container&quot;&gt;$GJ$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18943" LastActivityDate="2019-01-12T22:17:07.553" Title="A problem on mechanics of materials" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25486" PostTypeId="1" CreationDate="2019-01-12T05:57:20.040" Score="1" ViewCount="68" Body="&lt;p&gt;The current capacity of screw press is only 100 kg/hr. Motor is 22kw, shaft RPM is 16, applied torque is 13,120 Nm, theoretical volume is 0.000967 cu m. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Am I correct to say the press is going = 13,120 Nm divided by 0.000967 cu m will get the pressure Nm/cu m?&lt;/li&gt;&#xA;&lt;li&gt;If it is ok how can I increase the capacity to meet the same applied pressure equal to above answer?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Regards,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Romy&lt;/p&gt;&#xA;" OwnerUserId="18944" LastActivityDate="2020-10-04T01:08:55.670" Title="How to increase kg per hr capacity of screw press to achieve the same applied pressure per unit volume of raw material?" Tags="|torque|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25487" PostTypeId="1" CreationDate="2019-01-12T06:39:48.157" Score="-1" ViewCount="96" Body="&lt;p&gt;I was wondering if this portable power supply will be sufficient for running 4 UVC lamps all at once. If so, for approximately how long?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power Supply: &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B07GFRDM5W&quot; rel=&quot;nofollow noreferrer&quot;&gt;SUAOKI G100 27000mAh /3.7V Power Bank Station 100W AC, Type-C Quick Charge 3.0 USB, Solar Rechargeable Portable Power Generator for Travel Outdoor Use for MacBook Smartphone Camera&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;UVC Lamps: &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B079SZ8NZR&quot; rel=&quot;nofollow noreferrer&quot;&gt;UV Linear Clean &amp;amp; Germicidal Lamp Kit for up to 100 sq. ft. Room with 5ft Cord and Plug (Ozone Free)&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="18945" LastEditorUserId="110" LastEditDate="2019-01-12T14:12:40.443" LastActivityDate="2021-09-28T20:06:50.660" Title="How much power will 4 germicidal lamps use?" Tags="|electrical-engineering|power-electronics|electrical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25490" PostTypeId="2" ParentId="25487" CreationDate="2019-01-12T07:23:18.573" Score="0" Body="&lt;p&gt;Rough idea without any losses is 4 * 6 = 24 W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so 100Wh / 24W is just over 4 hours, but no other losses are considered and assumes the full capacity of 100Wh can be used which may not be the case... Also the stated 6W per lamp may not be fully accurate ie start up demand may be different.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-01-12T07:31:46.097" LastActivityDate="2019-01-12T07:31:46.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25492" PostTypeId="2" ParentId="25485" CreationDate="2019-01-12T17:27:33.027" Score="0" Body="&lt;p&gt;Lets start from the point A and name the other end of the beam B and the end at support, where element 1 is connected  c, and radius of the beams, R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loads are P concentrated at A and q distributed between points A and B over L= a. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the circular beam ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$I = \frac{\pi R^4}{4} ,\ \quad J = \frac{π R^4}{  2} ,\quad Tmax = (π / 2) τ_{max} R^3 \quad τ = T r / J                                $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;τ = shear stress (Pa, psi)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T = twisting moment (Nm, in lb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;r = distance from center to stressed surface in the given position (m, in)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J = Polar Moment of Inertia of Area (m4, in4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deflection at the point A is sum of two deflections due to loads P and q,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \delta = P a^3 / (3 E I) \quad + q a^4 / (8 E I)   $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deflection on point B is rotation under torsion due to beam AB torque and deflection under P an q, which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(P + q*a)a^3/3EI $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \theta_{rotation} = L T / (J G)= a(P*a + q*a^2/2  )/GJ $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$and\quad \delta = (P +qa )a^3 / (3 E I) $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elastic, strain, energy of beam AB , U is the work done by the forces which caused deflection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$U_{AB} =  m^2L/2EI= (a*P+a^2q/2)^2/2EI$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ U_{C B}= \theta *T/2 = T^2*a/2GJ = (P*a + q*a^2/2)^2/2GJ&#xA; $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F_{x, ab} = 0 \quad F_{y, ab} = P +qa\quad \\ M_{z,ab} =aP +qa^2/2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F{x,bc }=0\quad F{y,bc} =P+aq \\ M{z,bc}= a( P+aq )$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: all coordinate are assumed local WRT the beam.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-01-12T22:17:07.553" LastActivityDate="2019-01-12T22:17:07.553" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25493" PostTypeId="2" ParentId="25486" CreationDate="2019-01-12T17:59:22.293" Score="1" Body="&lt;p&gt;If you need to calculate the pressure,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A first estimate, ignoring friction, which can be substantial is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F_{screw}= T/(r*tan (\alpha)) \quad$&lt;/span&gt; r is screw radius , a is pitch angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P= F/area of press.&#xA;by increasing the motor rpm, or increasing the screw pitch you can increase the output, if your motor has enough power.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-12T17:59:22.293" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25494" PostTypeId="1" AcceptedAnswerId="25500" CreationDate="2019-01-12T20:08:00.660" Score="1" ViewCount="69" Body="&lt;p&gt;I am trying to construct a torsion pendulum/balance for a physics experiment but the research papers are always missing specifics on how the certain connections are made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for recommendations on rigid connections for:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How to attach a glass mirror to tungsten wire.&lt;/li&gt;&#xA;&lt;li&gt;How to attach tungsten wire to torsion pendulum/balance beam.&lt;/li&gt;&#xA;&lt;li&gt;How to attach PLA to glass.&lt;/li&gt;&#xA;&lt;li&gt;How to attach glass to glass (I do not have the capability to melt them).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.leydenscience.org/physics/gravitation/cavend.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a link&lt;/a&gt; to an example experimental setup. The balance beam I am using is actually quartz glass tube. The wire is tungsten. The plastic components are 3d printed and are PLA. The total weight is about 22lbs suspended from the tungsten wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I posted this in the Physics exchange and they closed it claiming it is an engineering problem)&lt;/p&gt;&#xA;" OwnerUserId="18950" LastEditorUserId="1832" LastEditDate="2019-01-13T02:59:52.083" LastActivityDate="2019-01-13T13:51:29.880" Title="Recommendations on a rigid connection between materials" Tags="|mechanical-engineering|structural-engineering|materials|welding|joining|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25496" PostTypeId="1" CreationDate="2019-01-12T22:52:28.080" Score="3" ViewCount="77" Body="&lt;p&gt;I build an awesome skating rink for the kids. Now I have to flood it. I have the rink built 450 feet from the lake. The rink is 7 feet higher than the water. I have a 1.5-inch polyurethane rigid hose running from lake to rink (100% sure no air leak). I have that hose connected to a brand new, Flotec FP5172 1-1/2 HP Self-Priming High Capacity Pump running on 115V, and then from the pump a 15-foot hose to irrigate the rink. With the show about to start, I plug in the pump and it’s not siphoning water. The guys that sold the pump to me did it was ample strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I have the pump at the lake pushing the water uphill versus my current setup of pump beside rink and pulling it through 450 feet of hose? There’s barely any suction at the lake end when I pull out the hose from the lake.&lt;/p&gt;&#xA;" OwnerUserId="18952" LastEditorUserId="1832" LastEditDate="2019-01-13T02:57:59.653" LastActivityDate="2019-01-13T18:51:49.183" Title="Pump Placement Question" Tags="|pumps|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25497" PostTypeId="2" ParentId="25496" CreationDate="2019-01-12T23:58:36.443" Score="4" Body="&lt;p&gt;Yep, it would probably take half a day for that pump to fill a suction hose that long. Also, there is pressure loss once the flow rate picks up. This means the pump will cavitate before it reaches rated flow due to friction losses. So put the pump close to the water. 50' max on the suction, and even at 50', you should use a 2&quot; diameter suction hose and reduce it at the pump. Now all you need is a 450' extension cord. Which won't work either because of the voltage loss in the wire. The cost in wire that would work would be several times the cost of the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the pump back. Get a 240 volt model and run a 240 volt 12ga circuit down to the lake. Or buy a gas powered pump. Very handy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some good news - you can take 400' of rigidwall back and replace it with flat hose that costs 1/4 as much. That should pay for the electrical mtls.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2019-01-13T00:05:30.243" LastActivityDate="2019-01-13T00:05:30.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25499" PostTypeId="1" CreationDate="2019-01-13T02:28:35.447" Score="0" ViewCount="166" Body="&lt;p&gt;I was wondering how I would configure some sort of setup to get lights like &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/B079SZ8NZR&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; turn on a timer once activated by a motion sensor. I want them to turn on for about 15 seconds and then turn off once tripped by a motion sensor but I have no engineering experience and I don't know where to start. One of my friends told me I would need to configure an Arduino board setup causing it to shut off power to the lamps after 15s but after extensive research, I'm not even sure that's possible. Is there any way to pull this task off?&lt;/p&gt;&#xA;" OwnerUserId="18659" LastEditorUserId="1832" LastEditDate="2019-01-13T03:01:37.577" LastActivityDate="2022-05-29T04:08:15.840" Title="Activating lights using a motion sensor?" Tags="|electrical-engineering|control-engineering|sensors|building-design|lighting|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="25500" PostTypeId="2" ParentId="25494" CreationDate="2019-01-13T06:09:14.173" Score="0" Body="&lt;p&gt;Keeping things simple...&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;epoxy&lt;/li&gt;&#xA;&lt;li&gt;design a fitting suitable for 3D printing to mechanically attach the parts &lt;/li&gt;&#xA;&lt;li&gt;if the part needs to detach, design a fitting, otherwise epoxy&lt;/li&gt;&#xA;&lt;li&gt;epoxy&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7021" LastEditorUserId="1832" LastEditDate="2019-01-13T13:51:29.880" LastActivityDate="2019-01-13T13:51:29.880" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25502" PostTypeId="1" CreationDate="2019-01-13T14:53:00.283" Score="1" ViewCount="293" Body="&lt;p&gt;I'm currently designing my own centrifugal fan for one of my projects, everything excluding the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5TAWb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5TAWb.jpg&quot; alt=&quot;A diagram of a generic centrifugal fan&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My design is similar to the above just that the outlet is much shorter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I'm facing is that my fan produces a high pitch noise. Assuming I keep the blades design and the power parameters the same, how can I reduce the high pitch frequency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my understanding, the volute casing has a tiny gap at the start of the volute tongue (indicated as Rc), could this be the cause of the high frequency noise?&lt;/p&gt;&#xA;" OwnerUserId="16469" LastActivityDate="2019-01-13T23:51:20.683" Title="Noise in centrifugal fan" Tags="|airflow|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25503" PostTypeId="1" CreationDate="2019-01-13T14:55:13.200" Score="1" ViewCount="211" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2ZVDY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2ZVDY.png&quot; alt=&quot;Exam Question on Equations of Motion and Free Body Diagrams&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;Assumptions from this:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Sum of forces on the beam to be calculated&lt;/li&gt;&#xA;&lt;li&gt;Sum of moments on the beam to be calculated&lt;/li&gt;&#xA;&lt;li&gt;Sum of forces on the motor to be calculated&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F=-kx = kθL^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Equations of Motion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Beam&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Msq4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Msq4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Would anyone be able to explain why the + and - L/2theta are the way they are?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I understand why k1 is positive and k2 are negative as taking moments about the centre, assuming clockwise is +ve, means that k2 opposes and k1 follows rotation but I don't understand how to assign L/2 a + or -.&lt;/p&gt;&#xA;" OwnerUserId="18956" LastEditorUserId="18956" LastEditDate="2019-01-14T15:37:43.483" LastActivityDate="2019-01-14T16:47:09.383" Title="Equations of motion for a motor supported by a spring resting at the centre of a beam" Tags="|mechanical-engineering|springs|linear-motion|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25504" PostTypeId="2" ParentId="25496" CreationDate="2019-01-13T18:51:49.183" Score="1" Body="&lt;p&gt;There are pumps with a pressure priming tank attached that have enough suction to pump up to 18 feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EsjXs.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EsjXs.gif&quot; alt=&quot;pressure tank assembly&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is done routinely in Canada for farming.&#xA;You can try to install an open basic tank with a check valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is the info sheet &lt;a href=&quot;https://www1.agric.gov.ab.ca/$department/deptdocs.nsf/all/agdex1143&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-13T18:51:49.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25505" PostTypeId="2" ParentId="25502" CreationDate="2019-01-13T19:15:40.757" Score="1" Body="&lt;p&gt;Yes, you are probably hearing the noise generated when the blades on the fan pass by the start of the volute tongue. Modeling this process will be hard, so experiments are the best way to proceed. Try making that intersection more gradual by, for example, radiusing that lip or cutting shallow slots in it that run in the same direction as the fan movement. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-01-13T19:15:40.757" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25506" PostTypeId="2" ParentId="25502" CreationDate="2019-01-13T23:51:20.683" Score="0" Body="&lt;p&gt;Spin it faster or add more blades. The threshold of human hearing is about 22.5 kHz for 90% of people. You want to get the primary frequency over 25 kHz to be on the safe side, and it will still drive dogs, cats, and birds crazy. Also, some people have &quot;poodle ears&quot;. I met a lady who could hear 26 kHz coming from a grease trap vent fan located across the street in at a restaurant. I shut their fan off for two days rather than comp her a hotel room.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-01-13T23:51:20.683" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25508" PostTypeId="1" CreationDate="2019-01-14T01:01:38.480" Score="0" ViewCount="40" Body="&lt;p&gt;Given a system describing a 2-vehicles platoons that can be described as such :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$v_k$&lt;/span&gt; the speed of each vehicle&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; the gap between the vehicles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can describe the state of the system as :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\dot{s} = v_0 - v_1$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\dot{v_k} = u_k$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Hence, writing the state of the system as &lt;span class=&quot;math-container&quot;&gt;$x = (v_0, s, v_1)^T$&lt;/span&gt; yields the following state equation :&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\dot{x} = Ax + Bu$$&lt;/span&gt;&#xA;with :&#xA;&lt;span class=&quot;math-container&quot;&gt;$A = \begin{pmatrix}&#xA;0 &amp;amp; 0 &amp;amp; 0 \\&#xA;1 &amp;amp; 0 &amp;amp; -1 \\&#xA;0 &amp;amp; 0 &amp;amp; 0&#xA;\end{pmatrix}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B = \mathbb{1}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eigenvalues of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; are all &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;, so the system should be marginally stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it seems to me that taking any &lt;span class=&quot;math-container&quot;&gt;$v_1 \neq v_0$&lt;/span&gt; as initial conditions gives &lt;span class=&quot;math-container&quot;&gt;$s \to \infty$&lt;/span&gt; so &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; isn't bounded, so it isn't marginally stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where am I wrong ?&lt;/p&gt;&#xA;" OwnerUserId="18959" LastActivityDate="2019-01-14T16:30:24.897" Title="2-Vehicles platoon stability" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25510" PostTypeId="1" CreationDate="2019-01-14T08:53:51.937" Score="1" ViewCount="27" Body="&lt;p&gt;I am working on a project of linear positioning where a bed is coupled with a motor driven high pitch multi-start back driving lead screw. I need the bed to be freely moving when the motor is not energized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I am using a NEMA 23 hybrid stepper motor of 1.8 Nm torque with gears since it has significant detent torque. But i need to eliminate the gears.&#xA;What kind of motor will be suitable for me? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approx specs needed are as follows:&lt;br&gt;&#xA;Torque: 0.7-1 Nm&lt;br&gt;&#xA;max speed: 200 rpm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I read about variable reluctance stepper motors but was unable to find any manufacturers or distributors for it. Can someone please direct me to one, also I am open to consider any other types of electric motors.&lt;/p&gt;&#xA;" OwnerUserId="18966" LastEditorUserId="18966" LastEditDate="2019-01-14T14:09:10.540" LastActivityDate="2019-01-14T14:09:10.540" Title="Which type of motor can I use for my project?" Tags="|motors|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25511" PostTypeId="1" CreationDate="2019-01-14T10:50:53.070" Score="0" ViewCount="282" Body="&lt;p&gt;I hope you are well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am writing to you because I am working on a sports project and I have a little trouble interpreting my results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we placed the accelerometer on the legof a runner and we have the acceleration data of the player according to the 3 axes (x,y,z) according to the points of measurement (1 point = 2.6ms).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have for the moment considered that the acceleration (m/ms) in y and I get this for 200 points:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Zgu7M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Zgu7M.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I subtract 9.81 from my data and then I calculate the integral of this function by the method of the strangles on the left and i get this :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VdNzz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VdNzz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have several questions to ask you: - Knowing that the accelerometer is placed on one leg of the runner, How can I determine the speed of the runner according to the result of my integral? what conversion do I have to do? - Finally, should I also consider the speed in x and z?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot for your help.&lt;/p&gt;&#xA;" OwnerUserId="18970" LastActivityDate="2019-01-14T21:45:47.790" Title="Calcultate speed from accelerometer for runner" Tags="|mechanical-engineering|acceleration|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25512" PostTypeId="2" ParentId="25511" CreationDate="2019-01-14T11:24:23.997" Score="3" Body="&lt;p&gt;You are unlikely to get any useful results from this, for several reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accelerometer &quot;x,y,z&quot; directions are relative to the accelerometer. Since the runner's leg is rotating, they are constantly changing orientation relative to the ground. You don't know how to split the &quot;1g&quot; gravitational acceleration into the correct components at each measured point because you don't know the orientation of the accelerometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you solved that issue, the measured acceleration will include a lot of &quot;noise&quot; - for example a sharp peak each time the runner's foot hits the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the worst (and unsolveable) problem is the fact that integrating the acceleration for a &quot;long&quot; time period is a very ill-conditioned process. If the runners &lt;em&gt;average velocity&lt;/em&gt; is constant (ignoring the changes in the leg acceleration during each stride) the &lt;em&gt;average acceleration&lt;/em&gt; will be zero at any running speed (including standing still). You are trying to do that calculation in reverse, and get the correct &quot;average velocity&quot; from measuring an acceleration with an average value of zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be much better to measure the &lt;em&gt;position&lt;/em&gt; of the runner, for example using a GPS device for long distances, or a camera system for short distances.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-01-14T11:24:23.997" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25513" PostTypeId="2" ParentId="25499" CreationDate="2019-01-14T12:13:31.957" Score="1" Body="&lt;p&gt;Use standard PIR sensor for common motion-activated spotlights, like these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vC3WP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vC3WP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Open it, and  find the relay inside (it will usually be the bulkiest part, likely colored blue). Disconnect the output from the &quot;NO&quot; (Normally Open) pin and connect it to &quot;NC&quot; (Normally Closed) (you may need to cut a path and draw a wire to the pin). That reverses the PIR sensor operation; it switches the light off when it would switch it on normally. Then just use the 'timer' dial on the sensor to make it deactivate (and switch the light on) 15 seconds after motion stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's quite possible you may find PIR sensors on the market that expose the NC contact alongside with the NO one, which would make the task even easier. Or if you don't want to modify the device, use its output to power another relay, and use the NO contact of that relay to control the light. &lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2019-01-14T12:19:38.350" LastActivityDate="2019-01-14T12:19:38.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25514" PostTypeId="1" CreationDate="2019-01-14T12:53:53.817" Score="1" ViewCount="120" Body="&lt;p&gt;I'd like to make an oven for curing adhesives that require an elevated temperature for a proper cure. I'd like for the temperature to be settable from 60 degrees C to 75 C plus or minus 5 degrees over time, and over the volume of the container. I'd like for the volume to be about 6&quot; x 6&quot; x 6&quot; or 15 cm x 15 cm x 15 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Primary objectives are to keep the cost down and to make the oven out of easily available materials from home depot or mcmaster carr or similar. Ideally would prefer to repurpose some home appliance. The ideal for me might be just repurposing a toaster and adding a temperature controller to it, but I worry about a toaster having consistent temperature through its volume.&lt;/p&gt;&#xA;" OwnerUserId="18971" LastActivityDate="2019-01-15T10:20:46.067" Title="Cheap low-temperature oven" Tags="|design|heating-systems|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25515" PostTypeId="2" ParentId="25514" CreationDate="2019-01-14T14:35:23.257" Score="0" Body="&lt;p&gt;you can just put a pot full of water with a thermometer on an oven and adjust the heat to bring the water to a steady 65 c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then submerge the adhesive container in it. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-14T14:35:23.257" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25516" PostTypeId="1" CreationDate="2019-01-14T15:13:41.750" Score="0" ViewCount="897" Body="&lt;p&gt;So I need to calculate how big geothermal heat pumps I need for a building. Heat pump is used for heating and to heat used water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I know how many kW's are needed for heating the building and the calculated flow rate of used water, what is the simplest and quickest way to calculate how much power is required from the pump(s)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say AC heating requires 20 kW, underfloor heating requires 5 kW and flow rate of used warm water 0,3 l/s. And I know dT for AC heating is for example 35 and underfloor heating is 10 and used warm water leaves the pump at 58 degrees celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I calculate it? And how do I convert this all to annual kWh used? &lt;/p&gt;&#xA;" OwnerUserId="18972" LastActivityDate="2022-08-03T02:40:31.510" Title="Calculating geothermal heat pump" Tags="|thermodynamics|heat-transfer|heating-systems|heat-exchanger|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25517" PostTypeId="1" CreationDate="2019-01-14T15:15:55.947" Score="1" ViewCount="154" Body="&lt;p&gt;Question from my Fluid Dynamics book (White 7e - 5.15)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thrust F of a propeller is generally thought to be a function of its diameter D and angular velocity Ω, the forward speed V, and the density ρ and visocity μ of the fluid. Rewrite this relationship as a dimensionless function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to solve this question using the Buckingham Pi Theorem. The solutions provided by my professor only use the Ipsen's method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't understand what repeating variables to choose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the Buckingham Pi theorem each pi must contain the repeating variables. But in this example that is not possible??&lt;/p&gt;&#xA;" OwnerUserId="18973" LastActivityDate="2019-01-14T15:15:55.947" Title="Buckingham Pi Theorem doesnt work for this case?" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25518" PostTypeId="2" ParentId="25508" CreationDate="2019-01-14T16:30:24.897" Score="1" Body="&lt;p&gt;Your system is in fact not marginally stable. It can be verified with a condition stated &lt;a href=&quot;https://en.wikipedia.org/wiki/Marginal_stability#Continuous_time&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; on wikipedia&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the system is in state space representation, marginal stability can&#xA;  be analyzed by deriving the Jordan normal form: if and only if the&#xA;  Jordan blocks corresponding to poles with zero real part are scalar is&#xA;  the system marginally stable.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can find the Jordan normal form with the MATLAB function &lt;span class=&quot;math-container&quot;&gt;$\texttt{jordan()}$&lt;/span&gt;, which yields&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{bmatrix} 0&amp;amp;1&amp;amp;0\\0&amp;amp;0&amp;amp;0\\0&amp;amp;0&amp;amp;0 \end{bmatrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding Jordan blocks are &lt;span class=&quot;math-container&quot;&gt;$\begin{bmatrix} 0&amp;amp;1\\0&amp;amp;0 \end{bmatrix}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;. Clearly, the first Jordan block is not a scalar but a 2x2 matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Therefore, the system is not marginally stable.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also see it by finding the eigenvalues and corresponding eigenvectors, it turns out that one of the eigenvalues has multiplicity 2, which violates the condition under the link above.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;and all poles with zero real part are simple roots &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2019-01-14T16:30:24.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25519" PostTypeId="1" AcceptedAnswerId="25528" CreationDate="2019-01-14T17:21:35.630" Score="1" ViewCount="347" Body="&lt;p&gt;I'm modifying a lawn tractor to pull single axle trailers in a woodland environment.    They are rear wheel drive, have enough power, but traction is our biggest issue even when using tractor/ATV tyres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is - what height (in relation to the rear axle) should the tow hitch be at and why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at tractor pulling for comparisons and &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;they seem to go as high as they can within the rules but I understand this is due to benefits of lifting the front of the sled off the ground.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;they extend the hitch to influence front wheel lift - I guess this is to put more weight on the back axle.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Front wheels lifting on woodland slopes isn't ideal so I have always opted to go low (below axle line) as this pulls the front down and I don't see how it could reduce the weight on the back wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a physics explanation would be really helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;  I also found these two links which undertake a complex analysis of sled pulling with trucks under the title 'Physics of a truck pull, using a 2WD pickup truck'.             &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://jimhawley.ca/downloads/Truck_Pull/Single_equation_model.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Part I - A simplified physical model captured by a single equation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://jimhawley.ca/downloads/Truck_Pull/Enhancements_to_the_dynamics.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Part II - Enhancements to the dynamics of the simplified model&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12340" LastEditorUserId="12340" LastEditDate="2019-01-15T08:11:53.217" LastActivityDate="2019-01-15T08:11:53.217" Title="Physics of Trailer Hitch Height" Tags="|car|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25520" PostTypeId="2" ParentId="25514" CreationDate="2019-01-14T18:31:49.753" Score="1" Body="&lt;p&gt;I like the toaster + PID controller idea. The big challenge to insuring even temperature distribution in the oven is properly insulating the chamber. If the chamber is insulated to minimize temperature change from design discrepancies and thermal &quot;leaks&quot;, the PID will eventually figure out the switching. Similar to heat treating ovens/kilns. Just a PID, thermocouple, and heating elements. The 6&quot; or so thick, overlapped, fire brick chamber is what makes it work right. More PID's/thermocouples/heating elements to create more &quot;zones&quot; will increase your level of control too. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2019-01-14T18:31:49.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25521" PostTypeId="2" ParentId="25499" CreationDate="2019-01-14T21:27:59.823" Score="1" Body="&lt;p&gt;Here's a product that does what you want:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Socket outlet to plug your light into &lt;/li&gt;&#xA;&lt;li&gt;Timer settable to 10s - 4 minutes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm sure they'll exist with other country's plugs on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QUp0R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QUp0R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.aliexpress.com/item/ALLSOME-Infrared-PIR-Sensor-Socket-Automatic-Wall-Light-Lamp-Holder-Socket-PIR-Motion-Switch-Detector-EU/32860755459.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.aliexpress.com/item/ALLSOME-Infrared-PIR-Sensor-Socket-Automatic-Wall-Light-Lamp-Holder-Socket-PIR-Motion-Switch-Detector-EU/32860755459.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2019-01-14T21:27:59.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25522" PostTypeId="1" AcceptedAnswerId="25537" CreationDate="2019-01-14T21:37:32.373" Score="1" ViewCount="365" Body="&lt;p&gt;If you draw a diagram with a spring, is there any convention about how it should behave with lateral loads? I guess it might be either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rigid &lt;/li&gt;&#xA;&lt;li&gt;Free&lt;/li&gt;&#xA;&lt;li&gt;Elastic with bending and/or shear stiffness&lt;/li&gt;&#xA;&lt;li&gt;Pin jointed - Initially free but as the spring rotates, it increases&#xA;stiffness due to its orientation changing.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BIECP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BIECP.png&quot; alt=&quot;spring with lateral force diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example where it looks like we're supposed to neglect any bending or shearing of the springs. For instance, if the object moves in the y direction, only the 3 bottom springs would deflect and we wouldn't consider the bending stiffness of the other 3 springs. Is that right and common?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WJaf9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WJaf9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;From &lt;a href=&quot;http://ecoursesonline.iasri.res.in/mod/page/view.php?id=126222&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ecoursesonline.iasri.res.in/mod/page/view.php?id=126222&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="4437" LastEditDate="2019-01-15T07:55:00.030" LastActivityDate="2019-01-16T06:03:10.620" Title="Convention for spring symbols with lateral force" Tags="|springs|diagram|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25523" PostTypeId="2" ParentId="25511" CreationDate="2019-01-14T21:40:09.547" Score="0" Body="&lt;p&gt;I run religously. 72 years, been running all my life and will run my 28th marathon this March in Los Angeles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I check my performance such as speed, length, heart rate using a GPS app on my phone.&#xA;To confirm other answers that acceleration would not be a good tool to measure speed. I share my experience which is along the way many endurance runners do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I change my gait, stride, landing on my heel or toe, in a sequence of algorithms according to the temperature, slope, my heart rate, my muscle condition, to repair a small tore, on the go, thousands of variables would affect the acceleration of my ankle and would not have anything to do with my speed. And that's the beauty of it, you optimise the output of limited system (your body) by taking advantage of allocating the burden to alternative part in a rotation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accelerometer if held in position well and calibrated is good for measuring your errors in your ankle movement, landing, twisting, etc. of course combined with other tool such as fast video.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-01-14T21:45:47.790" LastActivityDate="2019-01-14T21:45:47.790" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25525" PostTypeId="1" CreationDate="2019-01-14T23:51:43.390" Score="-1" ViewCount="450" Body="&lt;p&gt;I have a 337 watt motor, and want to lift ten pounds. What kind of gear ratio will I need for a gearbox I attach to this in order to lift the ten pounds at a constant speed?&lt;/p&gt;&#xA;" OwnerUserId="18975" LastActivityDate="2019-10-12T12:03:10.487" Title="Calculating Minimum Gear Ratio To Lift Weight" Tags="|motors|gears|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25526" PostTypeId="2" ParentId="25522" CreationDate="2019-01-15T00:26:57.557" Score="0" Body="&lt;p&gt;Initially it will be constrained by torsional resistance of the cross section of the spring and will behave like a column laterally loaded, or a cantilever beam with a concentrated load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at large deflections the compound reaction of torque and linear expansion of the spring will interact to the applied force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sketch of different degrees of freedom of a joint as you asked.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JjUUV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JjUUV.jpg&quot; alt=&quot; here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some other degrees of freedom such as guided, semi fixed with spring constrains. Or a joint free to move laterally but not rotationally.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-01-15T04:38:15.973" LastActivityDate="2019-01-15T04:38:15.973" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25527" PostTypeId="2" ParentId="25516" CreationDate="2019-01-15T05:02:11.153" Score="0" Body="&lt;p&gt;Usually supply is the limiting factor not the pump.&#xA;However say you have a pump that has an output of q liters of water per second, with a usable Delta temperature of 10 Celsius degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this pump can deliver q*10 kcalories energy per second, which is q*11.622 Watt hour or q*41.80 kilojoules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can either calculate how much available geothermal energy your system has by calculating liters per second times dt available. Or you calculate your demand and concert it to flow of the pump.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-15T05:02:11.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25528" PostTypeId="2" ParentId="25519" CreationDate="2019-01-15T06:01:21.067" Score="0" Body="&lt;p&gt;Let's say you need to tow 1000kg weight. And you have your hitch at 10 cm, 01meters above the axel of your tractor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pull it with an acceleration of 2 meters per second or 1/5 g. You have an overturning torque trying to lift up the front wheels of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1000*2*0.1= 200 *9.8 Nm.$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is enough torque to lift the front wheel of a small garden tractor. If you put the connection lower than axel you loose part of traction due to the redistribution of part of the mass to front wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So may be best is to hook the trailer hitch lined up with axel or barely above it.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-01-15T06:34:27.143" LastActivityDate="2019-01-15T06:34:27.143" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25529" PostTypeId="2" ParentId="25514" CreationDate="2019-01-15T09:32:16.170" Score="1" Body="&lt;p&gt;Even distribution of temperature is the worst part of the challenge, and a toaster oven, being a radiative heat source, will irradiate the sides exposed to the heating elements to way higher temperature than the remaining sides. Consider a convection oven or an air fryer instead; it keeps the temperature quite constant throughout the volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PID is overkill for this kind of application and risky because it can easily overshoot the control and overheat your adhesive; you should use plain hysteresis control (lower 'on' threshold, higher 'off' threshold) - you might even be able to tweak the bimetallic switch built into the oven/fryer, by tuning it way down (they normally rarely go below 100C.) And you can get after-market air fryers quite cheap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on time of curing, another good alternative may be a slow-cooker aka crock-pot. Seal your item in a zip-loc bag and immerse in water. It may again need tweaking of the bimetallic switch (they usually operate around 80-90C so you should be able to adjust it easier), and it will provide good several hours of 'cooking' at far lower electricity cost and without the noise convection ovens or air fryers produce. While convection ovens and air fryers come with timers that allow setting up up to about 30 minutes of operation, crock-pots often allow values in range of 8-10 hours. &lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2019-01-15T10:20:46.067" LastActivityDate="2019-01-15T10:20:46.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25530" PostTypeId="2" ParentId="25525" CreationDate="2019-01-15T10:07:43.387" Score="0" Body="&lt;p&gt;Without RPM answering this with a simple number is impossible. 337 watt is 337 Joules/second. Potential energy of 1kg lifted 1 meter grows by 9.8 Joules; for 10 pounds (4.5 kg) it's 44.1J/m, so the payload without resistance needs to move upwards at 7.64m/s to maintain constant speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what gear ratio and rotary to linear conversion (a pulley?) will convert the rotary motion of your motor to linear motion of the payload at that speed will depend eon RPM of the motor which is completely independent from its power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given motor RPM &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; (in RPM) and assuming radius of gear coupled to its shaft ('input') &lt;span class=&quot;math-container&quot;&gt;$r_1 = 1$&lt;/span&gt;, you have &lt;span class=&quot;math-container&quot;&gt;$2 \pi \omega_1$&lt;/span&gt; of linear speed of the gear's teeth. For the desired output's linear speed to be 7.64 m/s, or 458.4 m/minute, &lt;span class=&quot;math-container&quot;&gt;$v = {458.4 \over {2 \pi \omega}}$&lt;/span&gt; or about 73/&lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; - where &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; is your motor's RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is obviously assuming no friction, which in case of such mechanisms will in practice be quite significant.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-01-15T10:07:43.387" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25531" PostTypeId="1" AcceptedAnswerId="25532" CreationDate="2019-01-15T13:57:22.197" Score="0" ViewCount="206" Body="&lt;p&gt;Some sources point to difficulties which may destroy the bridge, so the Wikipedia article on the Crimean Bridge e.g. has: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The geology of the Kerch Strait is difficult: it has a tectonic fault, and the bedrock is covered by a 60 m (197 ft) layer of silt.[44] About 70 mud volcanoes have been found in the area of the strait.[44] More than 7,000 piles support the bridges; these piles have been driven up to 91 m (300 ft) beneath the water surface.[44] Some of the piles are at an angle to make the structure more stable during earthquakes.[44]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some experts have expressed doubts that the construction is durable, given the tectonic and sea current conditions in the strait.[44][45]&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;44&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Pollock, Emily (6 July 2018), &lt;a href=&quot;http://web.archive.org/web/20181013090024/https://www.engineering.com/BIM/ArticleID/17007/Europes-Longest-Bridge-Spans-Troubled-Waters.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;Europe’s Longest Bridge Spans Troubled Waters&lt;/a&gt;, Engineering.com, archived from the original on 13 October 2018 &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://economics.unian.info/10296951-kerch-strait-bridge-may-collapse-at-any-time-expert.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kerch Strait Bridge may collapse at any time – expert&lt;/a&gt;, UNIAN, 12 October 2018&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18977" LastEditorUserId="18977" LastEditDate="2019-01-16T00:36:16.633" LastActivityDate="2019-01-16T00:36:16.633" Title="Is the Crimean Bridge in danger of collapsing?" Tags="|bridges|" AnswerCount="1" CommentCount="9" FavoriteCount="0" ClosedDate="2019-01-15T18:40:03.523" ContentLicense="CC BY-SA 4.0" />
  <row Id="25532" PostTypeId="2" ParentId="25531" CreationDate="2019-01-15T17:43:01.537" Score="2" Body="&lt;p&gt;Evaluating safety and risks involved in a project of this magnitude and complexity need a multifaceted scientific research, composed of teams representing different areas of technology, with experts to sample the soils, water current; chemical engineers, meteorologist, seismologists, etc, expensive imaging, testing machinery. It could cost millions of dollars and many years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had two classmates in college many years ago who spent their entire Ph.D. time doing an assessment of the safety of a small single span 110-foot bridge in Massachusets, with unlimited help from undergrad students at the lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question is way too ambitious for a site like this.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-01-15T23:46:12.947" LastActivityDate="2019-01-15T23:46:12.947" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25534" PostTypeId="2" ParentId="15765" CreationDate="2019-01-15T22:53:30.200" Score="1" Body="&lt;p&gt;Disclaimer: This only applies to one state, but it is likely that other states have similar language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Illinois, the law that defines what a &lt;a href=&quot;http://www.ilga.gov/legislation/ilcs/ilcs3.asp?ActID=1344&amp;amp;ChapterID=24&quot; rel=&quot;nofollow noreferrer&quot;&gt;Professional Engineer&lt;/a&gt; is has this clause:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;No officer, board, commission, or other public entity charged with the enforcement of codes and ordinances involving a professional engineering project shall accept for filing or approval any technical submissions that do not bear the seal and signature of a professional engineer licensed under this Act. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So in this case, the law that creates the Professional Engineer also forces public entities to require technical submissions to be sealed. It appears that private companies are on their own as to what they require.  The term &quot;technical submissions&quot; is also pretty broad.&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2019-01-15T22:53:30.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25535" PostTypeId="1" CreationDate="2019-01-16T03:50:46.253" Score="1" ViewCount="184" Body="&lt;p&gt;Physics has never been my strong point... I have a task to size a motor. The question is:  Based on information below, how do I calculate required motor torque to accelerate a mass with the information below? Please explain the formulas and where they originate from. I understand how to calculate the force required. I don't understand how I compute the required torque to achieve this force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Variables:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Motion Direction:  Horizontal - Perpendicular to gravity&#xA;Mechanics:  Ballscrew &amp;amp; Linear Slide mechanism.&#xA;&#xA;(W) Weight of Load to move:           100KG&#xA;(D) Distance load should move:        1 m&#xA;(A) Speed load should move:          .04 m/s&#xA;(C) Coef of Friction of Mechanism:    .003&#xA;(p) all Screw Pitch:                 2.75 mm&#xA;(d) Ball Screw Dia:                   20mm&#xA;(T) Min motor torque required:         ?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;*Please arbitrarily choose variables for any missing information and declare them in your post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="18987" LastActivityDate="2019-10-13T09:01:30.340" Title="Electric Motor Sizing : Understanding how much torque I need given Force" Tags="|motors|mechanical|electrical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25536" PostTypeId="1" CreationDate="2019-01-16T05:44:37.483" Score="1" ViewCount="316" Body="&lt;p&gt;&lt;strong&gt;Problem description:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to know the liquid-liquid eqilibrium behavior of a THF-water-toluene system. I need to use NRTL because UNIFAC does not predict THF-water systems accurately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have NRTL parameters only for THF and water. UNIFAC is predicts THF-toluene, so I could simulate that system with UNIFAC and fit NRTL parameters to the UNIFAC results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leaves toluene-water parameters. Since toluene and water are immiscible, VLE data does not exist, only VLLE data or LLE data. I have heard that it requires some special expertise to fit NRTL parameters to LLE data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Main question:&lt;/strong&gt; How can I fit toluene-water NRTL parameters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Related side question:&lt;/strong&gt; According to Wikipedia, there can be two sets of NRTL parameters: 1.) for LLE prediction and 2.) for VLE prediction. My thermodynamics textbook doesn't discuss this. To what extent can parameters fit to VLE data be used for LLE? Please link to a reference that discusses this, if you know of one.&lt;/p&gt;&#xA;" OwnerUserId="11933" LastActivityDate="2019-07-30T05:02:45.373" Title="Fit NRTL parameters for an immiscible system" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25537" PostTypeId="2" ParentId="25522" CreationDate="2019-01-16T06:03:10.620" Score="1" Body="&lt;p&gt;The convention for an idealized spring in such a schematic is that the lateral stiffness is zero but also that it does not move out of its original orientation with respect to its base (i.e., it does not tilt). The idealized spring is characterized by only one parameter: the axial stiffness. (It is thus assumed that the axial deflections are sufficiently small that the stiffness is constant and displacement independent. This assumption holds for small perturbations of any stable solid.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the lateral stiffness is zero, other constraints must limit a body's motion in the directions orthogonal to the spring axis. This can be seen in your second schematic, in which each degree of freedom is constrained by the axial stiffness of one or more springs. In contrast, your first diagram is not well defined: the block should accelerate to the right for any positive &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;, which then results in a discrepancy because a real spring would naturally resist such acceleration. To indicate this resistance, you'd either add a pseudospring with horizontal orientation or move away from lumped components altogether, perhaps modeling the spring using the method of compliant joints or by finite element analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good source of sample schematics for springs and rigid bodies is vibration textbooks, e.g., Tongue's &lt;em&gt;Principles of Vibrations&lt;/em&gt;. Thorough and rigorous textbooks will emphasize that the block in your first diagram must be constrained (e.g., by walls) from moving left or right or in or out of the screen/page.&lt;/p&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2019-01-16T06:03:10.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25538" PostTypeId="2" ParentId="15394" CreationDate="2019-01-16T06:08:21.463" Score="0" Body="&lt;p&gt;Usually, the frequency of the ultrasonic anemometer is mainly 200 KHz probe, and there are 4 detectors. Through the description in this paper, it is uncertain whether the wind speed of a large range can be monitored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer to this article:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.weatherstation1.com/what-is-wind-sensor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.weatherstation1.com/what-is-wind-sensor/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This covers almost all the principles of wind monitoring, and I hope it will be helpful to you.&lt;/p&gt;&#xA;" OwnerUserId="18988" LastEditorUserId="1832" LastEditDate="2019-01-16T13:05:06.867" LastActivityDate="2019-01-16T13:05:06.867" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25539" PostTypeId="2" ParentId="25535" CreationDate="2019-01-16T07:33:31.977" Score="1" Body="&lt;p&gt;assuming constant acceleration, a, we need to cover 1 meter with an average speed of &lt;span class=&quot;math-container&quot;&gt;$$04m/s =\frac{0 +V_{final}}{2} \\  V_{final}= 0.8m/s $$&lt;/span&gt;&#xA;Force, F, needed for this acceleration is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V_{final}= \sqrt{2a*x}: 0.8 = \sqrt{2a*1}\\ a=0.8^2/2= 0.32 m/s^2 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F = ma =100*0.32m/s^2 =32kgm/s^2 \\ and\ 32* 1.003 = 32.09kgm/s^2$$&lt;/span&gt;&#xA;This is force demand increased by friction coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the torque is &lt;span class=&quot;math-container&quot;&gt;$$\tau = 32.09kgm/s^2/(mechanical\ advantage\ of\ screw)=\frac{32.19}{(\pi*0.02 )/0.00275)}\\=  1.4kgm*m = 1.4* 9.8 =13.76Nm$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have assumed &lt;span class=&quot;math-container&quot;&gt;$sin( small\ angle\ \alpha)=\alpha$&lt;/span&gt; on the screw pitch.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-16T07:33:31.977" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25540" PostTypeId="1" CreationDate="2019-01-16T08:54:24.800" Score="0" ViewCount="115" Body="&lt;p&gt;I am currently wondering is there any way or device that can transform a picture into 3D model automatically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or a machine learning thing to learn from picture (2D) into 3D model?&lt;/p&gt;&#xA;" OwnerUserId="18989" LastActivityDate="2021-06-05T05:01:40.160" Title="Is there any device that can capture a picture into 3D model?" Tags="|modeling|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25541" PostTypeId="1" AcceptedAnswerId="25543" CreationDate="2019-01-16T08:56:41.913" Score="3" ViewCount="1210" Body="&lt;p&gt;I have been doing a bunch of research online into how to build a basement that is below the water table.  The common wisdom is to &quot;not waste the money doing that&quot; but as someone who doesn't like to be told no and always likes to fully understand their options, I am curious how I could go about building a basement that is below the water table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it is &lt;em&gt;possible&lt;/em&gt; to build a waterproof basement because submarines exist, marinas often build underwater sections of buildings, pools are a thing, etc.  The ideal would be to have the primary structure for the basement be poured concrete with no windows or other holes in the walls (plumbing/electricity/air would come in through the ceiling from the higher floors).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, lets assume that I don't need sewer out of the basement (perhaps by using an up-flush system to pump sewage/water out). Lets also imagine that the basement is on a flood plain so I can build it while it is dry, but it &lt;em&gt;will&lt;/em&gt; be completely submersed in water later in the year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, again for simplicity lets assume that I am building in a location that has no building codes I need to abide by (so any option is on the table), but I do care about the structure being sound for a long time (e.g., 100 years).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the techniques I could use to water &lt;strong&gt;PROOF&lt;/strong&gt; the basement so that it can be fully submerged under water for extended periods of time without leaking?&lt;/p&gt;&#xA;" OwnerUserId="18990" LastEditorUserId="110" LastEditDate="2019-01-16T11:06:26.527" LastActivityDate="2019-01-16T19:04:38.870" Title="How does one build a waterproof basement?" Tags="|structural-engineering|civil-engineering|structures|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="25543" PostTypeId="2" ParentId="25541" CreationDate="2019-01-16T10:38:24.380" Score="3" Body="&lt;p&gt;IT's certainly possible to build a cellar below the GW table. You need to ensure two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;water proof&lt;/li&gt;&#xA;&lt;li&gt;countering buoancy due to the groundwater&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As for point one, water proof concretes exist, it's a matter of concrete recipe and care in execution. A bit of information on this &lt;a href=&quot;http://www.waterproof-concrete.co.uk/&quot; rel=&quot;nofollow noreferrer&quot;&gt;can be found here&lt;/a&gt;. Coating, e.g. water proof PE liners that can be applied to the concrete also exist. Again, this requires care in execution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for point 2, the buoancy can be countered by the weight of the building, the weight of the contents (water tanks don't tend to swim when full, oil tanks may be an issue!), by building a larger baseplate (so a bit of ground would have to be lifted as well) or even by friction with the ground. This depends on size, dpeth, and shape of your cellar and needs to be designed together with a structural engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During construction you need to manage the groundwater some way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buildings below ground water level are built all the time when unavoidable, usually when you need the depth for some reason.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2019-01-16T10:55:37.357" LastActivityDate="2019-01-16T10:55:37.357" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25544" PostTypeId="1" CreationDate="2019-01-16T11:53:42.930" Score="0" ViewCount="233" Body="&lt;p&gt;Here goes a recently opened underground railway station in Moscow, Russian Federation (image is from &lt;a href=&quot;https://pikabu.ru/story/_6392489&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BT7ON.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BT7ON.jpg&quot; alt=&quot;Cast Iron Tubing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the left is an undecorated (just painted and it will likely remain this way) tunnel wall. The tunnel wall is lined with cast iron tubings - specially crafted pieces of cast iron connected with thick steel bolts and having the seams sealed with cement. This lining forms a huge pipe which holds the ground from collapsing and the shot is made from inside the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cast iron is something like 30 millimeters thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there actual ground right behind that (not very thick) cast iron? What if the ground underneath is not hard enough and the loaded pipe has to bend downwards?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the cast iron hold all the possible mechanical loads alone or is there anything else in the structure?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="374" LastEditDate="2019-01-16T12:03:35.103" LastActivityDate="2019-06-15T18:03:40.490" Title="Does tunnel lining made of cast iron tubings actually accept all the mechanical loads?" Tags="|structural-engineering|materials|stresses|tunnels|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25545" PostTypeId="1" CreationDate="2019-01-16T12:07:44.837" Score="0" ViewCount="78" Body="&lt;p&gt;So I am working with COMSOL Multiphysics and I have a very complex simulation. I have a boundary condition that evaluates to 1 if the value of a variable is 20 and 0 otherwise. I could implement my code just fine in COMSOL. However, I am facing a problem (see a draft that I made below please). Basically, for the first iterations whenever the variable reaches 20 it turns the boundary condition to 0, but as I have more iterations the limit slightly increases (thus reaching values above 20). After some time the limit comes back to 20 and the same problem as before occurs again. Any thoughts?&#xA;I have already tried to increase the solver tolerance but the some problem occurs&#xA;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="18993" LastActivityDate="2020-02-11T17:05:25.903" Title="Finite element method, question about the solver" Tags="|finite-element-method|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25546" PostTypeId="2" ParentId="25540" CreationDate="2019-01-16T12:24:37.030" Score="1" Body="&lt;p&gt;Look up '3D scanners'. There are plenty of such devices, professional and amateur, for small items, microscopic, and for architecture, from complex and precise to as simple as &lt;a href=&quot;https://all3dp.com/2/5-best-3d-scanner-apps-for-your-smartphone/&quot; rel=&quot;nofollow noreferrer&quot;&gt;apps for a smartphone&lt;/a&gt;. There are also programs that can transform a series of photos into a model, and ones that aid manual 3D modelling with use of a single photo (where you manually add depth). &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-01-16T12:24:37.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25548" PostTypeId="1" CreationDate="2019-01-16T13:07:40.960" Score="0" ViewCount="110" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dq054.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dq054.jpg&quot; alt=&quot;Example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like the picture shown I want to have the length from the front end to the windshield in horizon, and want to have the length from the lowest point to the highest point of the hood in vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would like to know what tools I need to measure stuff like this, its mostly with in 1-2 feet in dimension&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: If this is in wrong site please help to relocate thank you&lt;/p&gt;&#xA;" OwnerUserId="18994" LastActivityDate="2021-08-04T10:02:25.350" Title="How can I get the horizon or vertical length of the uneven object?" Tags="|measurements|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25549" PostTypeId="2" ParentId="25545" CreationDate="2019-01-16T13:18:59.770" Score="1" Body="&lt;p&gt;In a nonlinear problem where you change some parameter, there is no guarantee that solutions for all values of the parameter will be stable. It is quite likely that for &quot;small&quot; values of the applied &quot;load&quot; (whatever that might be - not necessarily a physical force) there are stable solutions with your variable &lt;span class=&quot;math-container&quot;&gt;$x &amp;lt; 20$&lt;/span&gt; and the boundary condition = 0, and for &quot;large&quot; loads there are stable solutions with &lt;span class=&quot;math-container&quot;&gt;$x &amp;gt; 20$&lt;/span&gt; and the boundary condition = 1, but there is an intermediate range of loads with no stable solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure out what is going on, I suggest you make three simulations with different conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;fix the &quot;0 or 1&quot; boundary condition to 0 for all values of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;fix it to 1 for all values of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Change the boundary conditions in the model so that you fix &lt;span class=&quot;math-container&quot;&gt;$x = 20$&lt;/span&gt; and let the &quot;0 or 1&quot; quantity vary. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You will probably find there are some situations where there are stable solutions with &lt;span class=&quot;math-container&quot;&gt;$x = 20$&lt;/span&gt; but your &quot;0 or 1&quot; variable has values &lt;em&gt;between&lt;/em&gt; 0 and 1. Most likely, there are no stable solutions of your original model in that region. For example you may find there is a stable solution with &lt;span class=&quot;math-container&quot;&gt;$x &amp;gt; 20$&lt;/span&gt; and the boundary value = 0, and another stable solution with &lt;span class=&quot;math-container&quot;&gt;$x &amp;lt; 20$&lt;/span&gt; and the boundary value = 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we don't know anything about your model, it's impossible to say whether the fix is to change the model to get rid of the instabilities, or use a more sophisticated &quot;path following&quot; solution method (for example the Riks algorithm, or the &quot;arc length&quot; method - two names for pretty much the same thing) which can find the unstable solutions without failing numerically.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-01-16T13:18:59.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25550" PostTypeId="2" ParentId="25548" CreationDate="2019-01-16T13:29:06.323" Score="0" Body="&lt;p&gt;Park the nose against a wall and measure, or have a measuring stick (long ruler 6ft or 2m) from the base of the windscreen with a drop line to find the bumper edge.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-01-16T13:29:06.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25551" PostTypeId="2" ParentId="25548" CreationDate="2019-01-16T15:54:35.963" Score="0" Body="&lt;p&gt;Use a masons' 6ft aluminum level with a plumb bob.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drop the plumb line to any edge you want to measure from a point on horizontal level while the end of the level is at the end of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;measure the level from plumb line to the end.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-16T15:54:35.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25552" PostTypeId="2" ParentId="25541" CreationDate="2019-01-16T16:56:08.620" Score="2" Body="&lt;p&gt;you need to build your basement with concrete, on deep foundation that penetrates into competent layer of soil and is below the level of under ground water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming your structure is built strong enough to withstand all the hydrostatic and soil pressure, we need to insulate and water proof it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are waterproof plastic sheets that are applied over a waterproof rolled over polyurethane film.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This insulation covers all around and from top to the bottom of the exterior of concrete wall, leading on the bottom to a trench sloped to carry any water to a sump pump. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wall is then covered with 12 inches of gravel built over a 4 inch diameter perforated PVC pipe running on the trench behind the wall, to collect the water and take it to the said sump pump. then backfilled with the dirt from excavation and compacted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is standard practice in areas with aquifer. The packages are readily available at the big box building material stores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They come with manuals and have link to instruction youtube videos. here is a sketch,&lt;a href=&quot;https://i.stack.imgur.com/W9doa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W9doa.jpg&quot; alt=&quot;wall detail&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-01-16T19:04:38.870" LastActivityDate="2019-01-16T19:04:38.870" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25553" PostTypeId="1" CreationDate="2019-01-16T17:15:14.963" Score="0" ViewCount="1075" Body="&lt;p&gt;I want to make a cheap thermoforming fixture out of melamine particle board so I can form some PETG plastic sheets. The forming temperature ( hence oven temp) for thermoforming PETG is around 300 deg. F. So the melamine particle board will be exposed to this temp for about 3 minutes. Are there any dangers to doing this? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="18995" LastActivityDate="2019-01-16T17:15:14.963" Title="Heat Resistance of Melamine Particle Board" Tags="|thermal-radiation|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25554" PostTypeId="2" ParentId="25540" CreationDate="2019-01-16T17:17:12.897" Score="0" Body="&lt;p&gt;There are many apps with variety of precision for the cellphones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had one that required a series of photos taken from around a real object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will then turn it into a mesh frame and render it as you like, with warped texted over the faces for example. I deleted it due to memory consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google without too much fanfare is stitching all the photos from defferent angle taken at deferent scales in social media and is creating 3d cities. If you check on Google map or Google earth, you see some cities are now populated with real 3d objects such as ships on ports or airplanes, aside from the buildings which were already there. I guess their goal is to digitise the whole world eventually. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-16T17:17:12.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25555" PostTypeId="2" ParentId="25384" CreationDate="2019-01-16T17:29:44.310" Score="1" Body="&lt;p&gt;Do they provide you with the temperature of the system? In this case what they want you to do is solve first the boiler mass balance, in case there are heat/temperature specs you could make a energy balance (taking into account the Keq of the dissociation of ammonia along with the Keq of the disociation of water that does change at higher temperatures) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To solve the energy balance I would recommend you using Bird but given the nature of your problem I would recommend things closer to &lt;a href=&quot;http://faculty.poly.edu/~rlevicky/Handout6.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2019-01-16T17:29:44.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25556" PostTypeId="2" ParentId="25544" CreationDate="2019-01-16T17:32:59.597" Score="1" Body="&lt;p&gt;They have done their homework, one hopes.&#xA;Many competent dense soils are basically self supporting, like all the caves. Yes there is compact or dens soil behind the cast iron panels to take advantage of soils structural strength. The cast iron is sufficient to take all the loads, even if it rarely is called in to take the full load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have tested the soil and it chemical and structural content and have chosen the cast iron alloy suitable for that condition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are methods to protect cast iron from corrosion. Here is what they have done in London Road Shunt Tunnel. &lt;a href=&quot;https://flexcrete.com/corrosion-protection-london-road-shunt-tunnel/&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-16T17:32:59.597" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25557" PostTypeId="1" CreationDate="2019-01-16T17:49:18.050" Score="1" ViewCount="360" Body="&lt;p&gt;The following image shows a CAD model in its folded and unfolded form.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MF5Xs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MF5Xs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to successfully give it the ability for it to be adjusted between a 25 and 45 degree form, at 5 degree increments; but would still like to know a mathematical equation, for future projects.&lt;br&gt;&#xA;The following images displays the arms at various angles:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sGTjS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sGTjS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the pivot bracket that moves to hold the arm at each adjusted angle, it is rotated at the center of the small sky-blue circle; please note the green circle has a triangle within it as well: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/jWNv7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jWNv7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One end of the pivot bracket slides along 5mm deep notch/track/mortice, that is sloped at 2.5 degree angle:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/coXHC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/coXHC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This the approximate distance of each holes:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pHmvm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pHmvm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve gone ahead and measured each internal angles and the distance between the arm &amp;amp; pivot bracket at each adjusted angle; they each add up to 180 degrees:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0HFWB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0HFWB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m interested in the mathematical formula that would be used to obtain the values: 29.1, 17.0, 15.0, 14.1  &lt;/p&gt;&#xA;" OwnerUserId="18668" LastActivityDate="2020-10-07T22:02:15.220" Title="Equation used to obtain the adjustment angle of a pivoting lid support mechanism?" Tags="|geometry|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25558" PostTypeId="2" ParentId="25557" CreationDate="2019-01-16T18:40:41.467" Score="1" Body="&lt;p&gt;This is an impressive amount of work but unfortunately the CAD drawings don't seem to match the assumption in drawing the triangles. From the side views, The locking wheel moves along a line that does not pass through the pivot point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But ignoring that, and taking the 45 degree case as &quot;exact&quot; with the support vertical, Pythagoras' theorem says that the length of the sloping arm is &lt;span class=&quot;math-container&quot;&gt;$\sqrt{2}$&lt;/span&gt; times the length of the support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the other angles, you can use the &lt;a href=&quot;https://www.bbc.com/bitesize/guides/zsgjxfr/revision/7&quot; rel=&quot;nofollow noreferrer&quot;&gt;sine rule&lt;/a&gt; to find the other angle at the base of the triangle. If the right-hand angle is &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, the left hand one is &lt;span class=&quot;math-container&quot;&gt;$\sin^{-1}(\sqrt 2 \sin x)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That gives the angles as &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;25 36.7&#xA;30 45 (exactly) &#xA;35 54.2&#xA;40 65.3&#xA;45 90 (exactly)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;These aren't quote the same as yours, probably for the reason I gave at the start of the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then calculate the third angle by subtracting the others from 180 degrees:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;25 36.7 118.3&#xA;30 45   105&#xA;35 54.2  90.8&#xA;40 65.3  74.7&#xA;45 90    45&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I don't know what your angles of 29.1 etc are supposed to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can calculate the base of the triangles using the sine rule again. The calculated value for the 25 degree case would be &lt;span class=&quot;math-container&quot;&gt;$54.45\, \sin(118.3)\, / \sin(25) = 111.35$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;angle base-length&#xA;45     53.45&#xA;40     80.09&#xA;35     93.18&#xA;30    103.26&#xA;25    111.35&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I don't know what your lengths are measuring. They are obviously not the base of the triangles, because they don't consistently get longer as the angle reduces from 45 to 25.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-01-16T18:53:10.437" LastActivityDate="2019-01-16T18:53:10.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25559" PostTypeId="1" AcceptedAnswerId="25566" CreationDate="2019-01-16T19:31:34.313" Score="1" ViewCount="120" Body="&lt;p&gt;Background:&#xA;My professor for system dynamics at my university gave us an engineering dynamics concept quiz. A particular question I am interested in presented two figures I have re-drawn below. The pulleys are stated to have no mass or friction. No gravitational accelerations, or mass was given for the left figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor said (without proof) that the figure with the idealized force (right figure) will accelerate faster. I would like to know why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am posting here because my professor could not understand where my confusion was and thus could not answer my question to my satisfaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that regardless of the source of force, the accelerations should be the same. How is this possible? &lt;a href=&quot;https://i.stack.imgur.com/kKuWO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kKuWO.png&quot; alt=&quot;Left figure has a mass weighting 50N. Right figure is an idealized force of the same magnitude of 50N&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18998" LastEditorUserId="18998" LastEditDate="2019-01-16T20:59:18.377" LastActivityDate="2020-08-27T22:45:14.740" Title="Pulley question: Which 10N block will accelerate faster?" Tags="|mechanical-engineering|dynamics|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="25560" PostTypeId="1" CreationDate="2019-01-16T19:54:34.023" Score="1" ViewCount="515" Body="&lt;p&gt;Back when railways were being massively build, in the 1900-1910 decade for example, most new lines were made with narrow gauge, instead of the standard gauge which is 1435mm for most of the world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in Switzerland, most new train lines built between 1890 and 1914 used a gauge of 1m. Urban tramways used 1m gauge too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that this was much less expensive. I wonder however what factors lead to the construction of such smaller railways to be less expensive. Understandably the earthworks are lighter, but not so much as the width of the vehicles dictates this and not the width of the tracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quotes as for where I learned that narrow gauge was supposedly cheaper : From &lt;a href=&quot;https://en.wikipedia.org/wiki/Narrow-gauge_railway&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page on narrow gauge railways&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since narrow-gauge railways are usually built with smaller radius curves, smaller structure gauges and lighter rails, they can be less-costly to build, equip and operate than standard- or broad-gauge railways (particularly in mountainous or difficult terrain).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Also quoting the &lt;a href=&quot;https://fr.wikipedia.org/wiki/Ligne_La_Chaux-de-Fonds_%E2%80%93_Le_Noirmont_%E2%80%93_Glovelier&quot; rel=&quot;nofollow noreferrer&quot;&gt;page for the railways line between Glovelier and La Chaux de Fonds, Switzerland&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;La voie normale permet de mettre en service des trains directs depuis&#xA;  Bâle jusqu'à Saignelégier pour les jours de Marché-concours. Mais la&#xA;  construction a coûté cher. La compagnie est rapidement en difficultés&#xA;  financières et n'a plus d'autres choix que demander de l'aide à sa&#xA;  rivale. Les deux compagnies font direction commune dès 1906.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;(My translation (emphasis mine) )&#xA;  The normal gauge allowed to have trains directly from Basel to Saignelégier on &#xA;  days of Market. &lt;strong&gt;But the construction costs were prohibitive. The&#xA;  company was quickly in financial difficulties&lt;/strong&gt; and had no other choice&#xA;  than asking the rival company [translation note:The narrow gauge&#xA;  railwy Saignelégier-La Chaux de Fonds] for help. Both companies had&#xA;  common leadership since 1906&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If a new line were to be built today, now that massive industrial construction machinery is available, would it still be significantly less expensive to make it narrow gauge ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="17069" LastEditorUserId="17069" LastEditDate="2019-01-16T20:22:37.407" LastActivityDate="2023-08-04T18:00:40.780" Title="What are the factors that made narrow-gauge railways cheaper in the early 1900s, and are they still applicable today?" Tags="|civil-engineering|rail|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25561" PostTypeId="1" CreationDate="2019-01-16T19:58:29.597" Score="0" ViewCount="520" Body="&lt;p&gt;I bought a Everlast 205 AC/DC and I am new to welding. I would like to experiment with aluminum welding but my electrode doesn't last very long. The arc-cone is pretty wide and I cannot establish a nice weld pot and I don't understand why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My welder is set on AC with 40% cleaning and 140 A, 80 Hz. If I set the cleaning above 60% my electrode last only 2 seconds, but below 50% it lasts about 10-15 seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here a picture of my experiment aluminum which is a 4 mm thick plate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AqAud.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AqAud.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's wrong with my setting?&lt;/p&gt;&#xA;" OwnerUserId="2070" LastEditorUserId="2070" LastEditDate="2019-01-16T20:31:51.353" LastActivityDate="2019-01-16T20:31:51.353" Title="What settings to set to weld aluminum with TIG?" Tags="|welding|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25562" PostTypeId="1" AcceptedAnswerId="25583" CreationDate="2019-01-16T20:15:40.710" Score="1" ViewCount="1795" Body="&lt;p&gt;My question is about a tank filled with some liquid (properties of the liquid are known and can be considered constant. Specific heat equals 1.5 kJ/(kg.K)). The initial temperature of the liquid in the tank is 90 ºC. Now imagine that 2000 L/h of this liquid are recirculating, but before returning to the tank they are heated somehow (doesn't matter how) until they reach 240 ºC (assume it remains a liquid at this temperature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I estimate &lt;strong&gt;how much time&lt;/strong&gt; it would take before the tank temperature reaches 230 ºC. The liquid volume on the tank is constant and equal to 10000 L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's kind of an open problem (the values are only hypothetical so that the problem can be implement on Excel). How should I start analyzing it? What equations should I be looking for?&lt;/p&gt;&#xA;" OwnerUserId="18999" LastEditorUserId="18999" LastEditDate="2019-01-17T08:01:06.727" LastActivityDate="2019-01-17T14:16:18.143" Title="How to calculate the time to heat up a tank using recirculating liquid" Tags="|heat-transfer|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25563" PostTypeId="2" ParentId="25437" CreationDate="2019-01-16T20:23:27.433" Score="1" Body="&lt;p&gt;I would design the fridge's heat exchanger in such a way that the exothermic side is kept wetted and exposed to a natural air stream. The resulting evaporation will have a greater cooling effect than just cool air. &lt;/p&gt;&#xA;" OwnerUserId="13133" LastActivityDate="2019-01-16T20:23:27.433" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25564" PostTypeId="2" ParentId="25561" CreationDate="2019-01-16T20:30:04.550" Score="1" Body="&lt;p&gt;On AC the electrode is supposed to form a ball end slightly larger than the diameter of the electrode, you can use the +ve/-ve balance (aka cleaning) to fine tune the formation of this ball. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to make sure you are using the right electrodes, ceriated, lanthanated and zirconated are the ones generally used for AC, pure  tungsten will also work but has a lower current capacity and thoriated won't work properly at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That also looks like a very thick chunk of aluminium. Because it has such a high thermal conductivity if you don't have enough current then you just won't get a molten puddle to form at all as heat is conducted away as fast as you can put it in,m unlike steel where you can usually at least form a puddle regardless of material thickness, even if you don't get much penetration. For a 200 amp machine I would suggest that material around 4mm thick will be a good place to start in terms of getting a feel for the process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fairly wide arc for AC compared to DC is normal for TIG. If your machine has frequency control you can increase the frequency to tighten the arc. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2019-01-16T20:30:04.550" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25565" PostTypeId="1" CreationDate="2019-01-16T20:58:04.160" Score="0" ViewCount="94" Body="&lt;p&gt;A steel parallelepiped with a square cross-section fits exactly into the cavity.&#xA;it's loaded from the top surface(see picture).&#xA;how can I calculate the stress in X- and Y-direction if the strain in X- and Y-direction is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jhrDC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jhrDC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/juFLz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/juFLz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11393" LastActivityDate="2019-01-17T08:10:30.643" Title="stress in X- and Y-direction" Tags="|mechanical-engineering|structural-engineering|stresses|ansys-workbench|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25566" PostTypeId="2" ParentId="25559" CreationDate="2019-01-16T21:28:50.497" Score="1" Body="&lt;p&gt;The accelerations will not be same. What you do is you reduce the masses as rotational inertia then apply the force as moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the fïrst drawing has more rotational inerta it will accelerate slower.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2019-01-16T21:28:50.497" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25567" PostTypeId="1" CreationDate="2019-01-16T21:51:46.760" Score="1" ViewCount="1158" Body="&lt;p&gt;I am trying to understand some analytic expressions given in the handbook Roark's Formulas for Stress and Strain for bimetallic strip stresses. Namely, Roark's Formulas give the formulas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{Equivalent } EI = {w t_b^3 t_a E_b E_a \over 12(t_a E_a + t_b E_b)}K_1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where&#xA;&lt;span class=&quot;math-container&quot;&gt;$$K_1 = 4 + 6{t_a \over t_b} + 4\left({t_a \over t_b}\right)^2 + {E_a \over E_b}\left({t_a \over t_b}\right)^3 + {E_b \over E_a}{t_b \over t_a}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and stresses of the top and bottom layers of the bi-metal strip as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{alignat}{4}&#xA;\sigma &amp;amp;= -&amp;amp;&amp;amp;{(\gamma_b - \gamma_a)\Delta T E_a \over K_1}\left[{3{t_a \over t_b} + 2\left({t_a \over t_b}\right)^2 - {E_b t_b \over E_a t_a}}\right] \\&#xA;\sigma &amp;amp;= &amp;amp;&amp;amp;{(\gamma_b - \gamma_a)\Delta T E_b \over K_1}\left[{3{t_a \over t_b} + 2 -{E_a \over E_b}\left({t_a \over t_b}\right)^3}\right]&#xA;\end{alignat}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where subscripts &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; indicate the two different metal layers, &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; are the Young's moduli, &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; are the layer thicknesses, &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; is the beam width, &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; is the coefficient of thermal expansion, and &lt;span class=&quot;math-container&quot;&gt;$\Delta T$&lt;/span&gt; is the elevation in temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I understand the Timoshenko bi-metallic strip curvature derivation (S. Timoshenko, &quot;Analysis of Bi-Metal Thermostats,&quot; J. Opt. Soc. Am. 11, 233-255 (1925)), which also appears on Wikipedia (&lt;a href=&quot;https://en.wikipedia.org/wiki/Bimetallic_strip&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Bimetallic_strip&lt;/a&gt;). I cannot understand how the formulas above can be derived from the Timoshenko derivation or what additional assumptions were made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roark's Formulas is completely unhelpful in not providing a clear reference in which this derivation is done. If someone could point out a reference I can use to obtain the result shown or can show how the derivation is done, I would be immensely grateful.&lt;/p&gt;&#xA;" OwnerUserId="19001" LastEditorUserId="19001" LastEditDate="2019-01-19T18:50:35.513" LastActivityDate="2019-01-19T18:50:35.513" Title="Derivation of Bimetallic Strip Stress in Roark's Formulas" Tags="|mechanical-engineering|stresses|beam|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25569" PostTypeId="1" CreationDate="2019-01-16T23:36:54.213" Score="1" ViewCount="539" Body="&lt;p&gt;I’m not an engineer, however I think this place is the best one to find an answer to my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my problem is as follows: I need to create a mechanism, preferably driven by a dc motor, that will hit the surface above it as hard as possible for a small DC motor. Like a sort of hammer, that is supposed to hit the surface above it (above is the keyword, I found DaVinci’s hammer mechanism, but it works based on gravity, which is not really helpful in the opposite direction). I’ve tried to research rotary to linear mechanisms, but all of them look like they will stop when the linear part hits the surface above and cannot go further. Can you suggest something?&lt;/p&gt;&#xA;" OwnerUserId="19003" LastEditorUserId="90" LastEditDate="2019-01-17T02:58:05.857" LastActivityDate="2019-01-17T06:33:54.180" Title="Rotary to linear motion - how to make upside down “hammer” mechanism?" Tags="|mechanical-engineering|applied-mechanics|mechanisms|linear-motion|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25570" PostTypeId="1" CreationDate="2019-01-16T23:47:09.667" Score="0" ViewCount="68" Body="&lt;p&gt;A bit vague but a question for you working engineers out there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a couple of years now I've been designing fluid systems, mostly using Crane's TP 410 as the guiding resource for pressure drop calculation and flow rate calculations. One thing that has bothered me is that I've never really gotten to validate my models against real life systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, &lt;strong&gt;in the field how well does this method hold up ? Are the modeled results fairly accurate?&lt;/strong&gt; Anecdotal evidence is welcome as well as something more concrete like a trade study. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand this isn't a traditional question you'd get on this site, so if this is the wrong forum just let me know and I'll take down my question.&lt;/p&gt;&#xA;" OwnerUserId="14961" LastActivityDate="2019-01-16T23:47:09.667" Title="Method of Hydraulic Resistances (Crane TP 410), how accurate is it really?" Tags="|fluid-mechanics|fluid|systems-design|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25572" PostTypeId="2" ParentId="25565" CreationDate="2019-01-17T00:53:02.387" Score="1" Body="&lt;p&gt;Depending on the structural properties of the surrounding material, you will have transfer of stress and strain between the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But let's assume there is no side strain for now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you have to superpose two cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- the vertical loading&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- assuming 1/3 strain along X and Z axis, calculate the pressure that would deflect them back to zero and apply that as lateral loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do this by a couple of iterations, and I guess some FEM programs allow for setting this up as a support constrain.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-17T00:53:02.387" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25573" PostTypeId="2" ParentId="25569" CreationDate="2019-01-17T01:41:41.357" Score="0" Body="&lt;p&gt;A consideration of an upward-striking hammer is that you have to have some method of preventing damage to the mechanism as the hammer reaches the surface being struck. This could be the absorption of energy or the displacement of the item/surface being struck. Recoil is the method used in the gravity-fed DaVinci design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of methods come to mind, but I'll focus on one. Consider DaVinci's design, but place the cam under the lever end containing the hammer. As the cam rotates, the hammer falls to a conveniently placed platform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other end of the lever is extended and also has a hammer head attached. As the cam rotates and the first hammer falls, the second hammer rises. It should be configured to strike the desired surface before the first hammer contacts the platform, for optimum results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above described configuration, the weight on the first hammer end has to be greater than that on the second hammer end. The placement and size of the cam allows for flexibility. One could place the cam outboard of the first hammer head and have a substantial size and displacement, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You've noted that the mechanisms you've found appear to stop the hammer at the upper surface. This is to be expected if you are striking this surface. Will you expect your upward-striking hammer to move the surface it strikes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, your design will have to allow for greater travel via cam diameter, displacement and location along the lever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this does not appropriately address your design considerations, please provide additional constraining criteria.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-01-17T01:41:41.357" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25575" PostTypeId="2" ParentId="25569" CreationDate="2019-01-17T06:33:54.180" Score="1" Body="&lt;p&gt;check this graph,&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cyzOv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cyzOv.png&quot; alt=&quot;cam hammer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you attach a spring loaded vertical ramrod released upwards when the little roller cam, bearing inside this rotating gear falls into the step, you have your hammer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is based on Leonardo De Vinci's hammer, except like a negative or reverse mechanism. the cam is inside the gear not outside.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-17T06:33:54.180" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25576" PostTypeId="1" CreationDate="2019-01-17T07:40:14.270" Score="0" ViewCount="40" Body="&lt;p&gt;I am currently researching the feasibility of setting up a coilgun that launches Cubesats for a paper.&lt;/p&gt;&#xA;" OwnerUserId="19007" LastActivityDate="2019-01-17T07:40:14.270" Title="Is it possible to protect electronics on a cubesat from 100-1000g of acceleration" Tags="|electromagnetism|magnets|propulsion|acceleration|rocketry|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25578" PostTypeId="2" ParentId="24706" CreationDate="2019-01-17T09:14:27.833" Score="0" Body="&lt;p&gt;Well, I'm going to answer myself here - just for sharing the information with others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured out that what I was missing is adding the effect of the angular acceleration.&#xA;I avoided using the Inertia relief option, and alternatively added both the linear and angular accelerations in opposite directions (D'alembert principle). I got those accelerations from the ADAMS dynamic analysis as stated in the question.&#xA;The pin joint is constraint in all 6 DOF artificially. However, since there was no net moment about this axis (the effect of both accelerations cancelled each other) - there was no bending moment and the stress distribution seems realistic now.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2019-01-17T09:14:27.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25580" PostTypeId="1" CreationDate="2019-01-17T10:32:16.667" Score="2" ViewCount="1193" Body="&lt;p&gt;I am trying to assemble the parts of Plummer block and draw the half sectional views from the front as given &lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;url=https://www.ut.edu.sa/documents/10156/2181404/CW7%2B(Plummer%2BBlock)&amp;amp;ved=2ahUKEwi2tYnuy_TfAhVQWysKHZdFBjIQFjAIegQIARAB&amp;amp;usg=AOvVaw1Twfvx_Etvk0tGL6yqR9L8&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I am facing some trouble understanding a few sectional views. In this view &lt;a href=&quot;https://i.stack.imgur.com/09gP7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/09gP7.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I am unable to understand why there are two different radii R32 and R34 marked in the front view of base.&lt;/p&gt;&#xA;&lt;p&gt;In this view &lt;a href=&quot;https://i.stack.imgur.com/djOUZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/djOUZ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;I am unable to get why there are two different radii for brass shown R32 and R30. Also which component R38&#xA;represent in the front view.&lt;/p&gt;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-01-17T10:45:52.527" Title="Half sectional views of Plummer block" Tags="|mechanical-engineering|autocad|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25581" PostTypeId="2" ParentId="25580" CreationDate="2019-01-17T10:45:52.527" Score="3" Body="&lt;p&gt;The semicircle in that front view steps to a larger diameter for part of its length. See the top view for reference:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zyg81.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zyg81.png&quot; alt=&quot;front/top views showing double dimensions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, looking at the alternate view for the second component shows where those radii come from:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yVMoP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yVMoP.png&quot; alt=&quot;front/top views showing double dimensions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-01-17T10:45:52.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25583" PostTypeId="2" ParentId="25562" CreationDate="2019-01-17T14:11:11.603" Score="1" Body="&lt;p&gt;The mass balance gives mass flow in = mass flow out = &lt;span class=&quot;math-container&quot;&gt;$\dot{m}$&lt;/span&gt;. The energy balance on an insulated, constant volume tank gives&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dot{m}\tilde{C}_p(T - T_h) + m_T\tilde{C}_V\frac{dT}{dt} = 0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the variables are tank temperature &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, hot temperature &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt;, specific heat capacities &lt;span class=&quot;math-container&quot;&gt;$\tilde{C}_p$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\tilde{C}_V$&lt;/span&gt;, and tank fluid mass &lt;span class=&quot;math-container&quot;&gt;$m_T$&lt;/span&gt;. The ratio &lt;span class=&quot;math-container&quot;&gt;$m_T/\dot{m}$&lt;/span&gt; is the residence time in the tank &lt;span class=&quot;math-container&quot;&gt;$t_R$&lt;/span&gt;. The residence time is also a ratio of tank volume to volumetric flow. Allow &lt;span class=&quot;math-container&quot;&gt;$\tau = t/t_R$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\theta = (T_h - T)/(T_h - T_o)$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$T_o$&lt;/span&gt; is the initial temperature in the tank. The differential equation becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\theta = \gamma \frac{d\theta}{d\tau} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$\gamma = \tilde{C}_V / \tilde{C}_p$&lt;/span&gt;. When &lt;span class=&quot;math-container&quot;&gt;$\gamma = 1$&lt;/span&gt;, the answer follows with the boundary condition that &lt;span class=&quot;math-container&quot;&gt;$\theta = 1$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\tau = 0$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\theta = \exp(-\tau) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example plot is below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j9Eqe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j9Eqe.png&quot; alt=&quot;plot of dimensionless temperature profile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modifications for &lt;span class=&quot;math-container&quot;&gt;$\gamma \neq 1$&lt;/span&gt; are left as an exercise.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-01-17T14:16:18.143" LastActivityDate="2019-01-17T14:16:18.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25585" PostTypeId="1" CreationDate="2019-01-17T16:21:17.037" Score="1" ViewCount="156" Body="&lt;p&gt;I am trying to use Autodesk Fusion 360's generative design feature to produce more efficient turbine blades for hydroelectric power generators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I simulate the load of water-fall on a turbine and use it to optimize the shape of the turbine? Also where can i obtain a starting shape model of a turbine?&lt;/p&gt;&#xA;" OwnerUserId="19013" LastActivityDate="2019-01-17T16:21:17.037" Title="Simulating fluid dynamics for Generative Design of water turbine in Autodesk Fusion 360" Tags="|fluid-mechanics|simulation|computer-aided-design|autodesk-inventor|" AnswerCount="0" CommentCount="2" ClosedDate="2019-01-18T13:39:58.057" ContentLicense="CC BY-SA 4.0" />
  <row Id="25586" PostTypeId="1" AcceptedAnswerId="25594" CreationDate="2019-01-17T19:36:15.963" Score="0" ViewCount="50" Body="&lt;p&gt;I'm trying to do an experiment on high pressure's effect on the electrical impedance of a really corrosive solution. I'm aiming for at least 2,000 psi with a pressure chamber but the only material I found that's chemically inert is PTFE and it's really limiting. One of my design is to use a PTFE diaphragm at the opening of the tube and press it inside the chamber to create pressure, and the other one is to push a PTFE piston into a tube directly but I don't know if it will hold much pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I can't really find a pre-made diaphragm that's under $500 that suits my chamber. Also, I can't find a way to fit a pressure gauge into the chamber and seal it properly. If you have an idea where to find these parts or any suggestions on how I should improve my design to better accommodate these challenges, that would be a great help. Thanks in advance!    &lt;/p&gt;&#xA;" OwnerUserId="19016" LastActivityDate="2019-01-18T17:45:26.383" Title="Issues creating a high pressure chamber for corrosive solution" Tags="|design|pressure|chemical-engineering|experimental-physics|seals|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25589" PostTypeId="2" ParentId="17099" CreationDate="2019-01-18T04:38:50.200" Score="1" Body="&lt;p&gt;As mentioned by Apostolos above, the real world is complicated. The stiffness tensor that you wrote correct but for most applications assuming plane stress or strain is enough, especially when dealing with isotropic materials. If you don't make assumptions, you have to find the 6 constants, which is not necessarily easy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you extend the stiffness tensor to 3D you get the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{bmatrix}&#xA;\sigma_{1} \\&#xA;\sigma_{2} \\&#xA;\sigma_{3} \\&#xA;\sigma_{4} \\&#xA;\sigma_{5} \\&#xA;\sigma_{6}&#xA;\end{bmatrix}&#xA;=&#xA;\begin{bmatrix}&#xA;C_{11} &amp;amp; C_{12} &amp;amp; C_{13} &amp;amp; C_{14} &amp;amp; C_{15} &amp;amp; C_{16} \\&#xA;C_{21} &amp;amp; C_{22} &amp;amp; C_{23} &amp;amp; C_{24} &amp;amp; C_{25} &amp;amp; C_{26} \\&#xA;C_{31} &amp;amp; C_{32} &amp;amp; C_{33} &amp;amp; C_{34} &amp;amp; C_{35} &amp;amp; C_{36} \\&#xA;C_{41} &amp;amp; C_{42} &amp;amp; C_{43} &amp;amp; C_{44} &amp;amp; C_{45} &amp;amp; C_{46} \\&#xA;C_{51} &amp;amp; C_{52} &amp;amp; C_{53} &amp;amp; C_{54} &amp;amp; C_{55} &amp;amp; C_{56} \\&#xA;C_{61} &amp;amp; C_{62} &amp;amp; C_{63} &amp;amp; C_{64} &amp;amp; C_{65} &amp;amp; C_{66}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\varepsilon_{1} \\&#xA;\varepsilon_{2} \\&#xA;\varepsilon_{3} \\&#xA;\varepsilon_{4} \\&#xA;\varepsilon_{5} \\&#xA;\varepsilon_{6}&#xA;\end{bmatrix} &#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here it's much worse than the 2D case because you have to find 36 constants, assuming that the matrix is not symmetric. This is the most complete characterization of a material. The problem is that that's a lot of testing which is very hard! Therefore assumptions are made in order to simplify and reduce the number of constants required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most real world applications, the worse case scenario is assuming the material is orthotropic. The stiffness matrix for an orthotropic material is simpler than that for fully anisotropic so once inverted and having introduced &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\nu$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; the result is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{bmatrix}&#xA;\varepsilon_{1} \\&#xA;\varepsilon_{2} \\&#xA;\varepsilon_{3} \\&#xA;\varepsilon_{4} \\&#xA;\varepsilon_{5} \\&#xA;\varepsilon_{6}&#xA;\end{bmatrix}&#xA;=&#xA;\begin{bmatrix}&#xA;\frac{1}{E_1} &amp;amp; -\frac{\nu_{21}}{E_2} &amp;amp; -\frac{\nu_{31}}{E_3} &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xA;-\frac{\nu_{12}}{E_1} &amp;amp; \frac{1}{E_2} &amp;amp; -\frac{\nu_{32}}{E_3} &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xA;-\frac{\nu_{13}}{E_1} &amp;amp; -\frac{\nu_{23}}{E_2} &amp;amp; \frac{1}{E_3} &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \frac{1}{2G_{23}} &amp;amp; 0 &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \frac{1}{2G_{13}} &amp;amp; 0 \\&#xA;0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; \frac{1}{2G_{12}}&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;\sigma_{1} \\&#xA;\sigma_{2} \\&#xA;\sigma_{3} \\&#xA;\sigma_{4} \\&#xA;\sigma_{5} \\&#xA;\sigma_{6}&#xA;\end{bmatrix}&#xA;$$&lt;/span&gt;&#xA;This means you have to measure &quot;only&quot; nine material constants, assuming the matrix is symmetrical, making life a whole load easier. Examples of situations where the material is assumed to be orthotropic are 3D printing and composites.&lt;/p&gt;&#xA;" OwnerUserId="18933" LastActivityDate="2019-01-18T04:38:50.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25593" PostTypeId="2" ParentId="1910" CreationDate="2019-01-18T15:23:43.513" Score="4" Body="&lt;p&gt;There are a number of factors that have limited wider adoption of Turbo-Compounds and it all comes down to how they actually work and where they receive their greatest benefits.  In the early studies of the concept by NACA in the 1940's they determined that the most critical factor is the Exhaust Exit Pressure (Pe) to Intake Manifold Absolute Pressure (MAP, or Pm) ratio as this determines the Effective Mean Exhaust Jet Velocity, Ve(eff), available to the turbine power recovery stage.  This means that a Turbo-Compound produces the largest benefits at low values of the ratio (low Pe, high Pm), making them ideal for use in conditions with either exceptionally high MAP or exceptionally low Pe.  Excluding other exhaust recovery stages (such as a following turbo-supercharger), the Pe experienced by the Turbo-Compound turbine--the Power Recovery Turbine--is normally the same as the static atmospheric pressure (Pa).  This means they make a lot of sense in Aircraft which operate at high altitudes with low Pa thereby naturally lowering the Pe:Pm ratio, and therefore increasing the Ve(eff) and the power available to the Power Recovery Turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this in mind, the limited efficacy of the Turbo-Compound in most road applications should be apparent.  Since cars generally operate at or near sea level pressures they have a higher Pe, which increases the value of the Pm:Pe ratio and reduces the amount of recovery possible with the turbine.  In order for a car to receive more than marginal benefit from a Turbo-Compound it would need to operate with a continuous boost in the MAP, and fairly significant one at that.  This means it first must use Forced Induction (Supercharger or Turbocharger) and second, that the blower maintains a steady MAP of 30% or greater boost over Pa at normal operating conditions.  Since most road cars operate at partial throttle under almost all conditions this means that it needs to have positive boost under those partial throttle conditions, which isn't a very common arrangement due to strain on the engine and the heat produced.  This is why it is really only seen in Compression Ignition engines, Diesels, which operate within a limited set RPM range the vast majority of time at speeds which produce sufficient airflow for effective cooling (both charge-air and coolant).  As others have pointed out, they could theoretically be of use in racing as well, especially long-track/oval racing, where the car is operated at continuous RPM and high-speed for extended periods.  Another place where I could see good benefit to Turbo-Compounding is in Pike's Peak Hill Climb racing--short, full-throttle, high-boost sprints at high elevation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these factors:  the requirements to get the sufficient benefit from the system; the added cost, weight and complexity; the potential reduced reliability; and other available technologies; have relegated the traditional Turbo-Compound to a largely historical curiosity of &quot;what could have been.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, we are seeing variations of the idea.  In the current F1 era they use something resembling Turbo-Compound which stores the recovered energy as electricity instead of applying directly back to the crank.  Ferrari is developing a similar system for use in their next generation of sportscars as well but will use the recovered power to drive an electrical supercharger as a sort of uncoupled turbocharger set up.  Either way, the power recovery is used more for electrical generation that anything and the electricity is then used to produce work through independent motors.&lt;/p&gt;&#xA;" OwnerUserId="19027" LastActivityDate="2019-01-18T15:23:43.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25594" PostTypeId="2" ParentId="25586" CreationDate="2019-01-18T17:45:26.383" Score="0" Body="&lt;p&gt;High Pressure corrosion testing is usually done in alloys like  Hastelloy C ( or similar). Autoclave Engineers corp makes a lot of this equipment. Of course in this area it is usually done on a &quot;money is no factor&quot; basis.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-01-18T17:45:26.383" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25598" PostTypeId="1" CreationDate="2019-01-19T07:28:53.320" Score="2" ViewCount="179" Body="&lt;p&gt;Comparing the cross-section of modern high-speed trains to old train waggons, which had straight walls and were limited to speeds around 200 km/h, I notice that most of them have curved outer walls. The width of the cabin decreases with height (and at heights occupied by passengers) and this obviously reduces space and comfort inside the cabin. So my question is, what are the main benefits of this? I am thinking of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;lower cross section of the train, i.e. reduced aerodynamic drag.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;lower center of gravity. This is not bad with strong side winds!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;lower moment of inertia for rolling around the longitudinal axis between the wheels for driving with high speed trough curves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So what are the main benefits of curved outer walls? Is a circular shape the best?&lt;/p&gt;&#xA;" OwnerUserId="19034" LastActivityDate="2019-01-19T22:18:40.767" Title="Why do (some) high-speed trains have curved (convex) walls?" Tags="|mechanical-engineering|fluid-mechanics|aerodynamics|transportation|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25599" PostTypeId="2" ParentId="25598" CreationDate="2019-01-19T08:30:53.543" Score="2" Body="&lt;p&gt;All the three items you said are correct. And what Solar Mike said is correct too.&#xA;Here are the list of the other items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- with manufacturing an design advances they can build ideal shapes more easily than the old times when a box was default structure for trains and cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- cylindrical shapes are ideal for longitudinal torsion caused by the vibration due to suspension and turbulence created by the wake of train itself. Also the entire car acts as an integrated structural pipe and is inherently stronger. While the old train cars would warp and crumble easily.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- economy in material, for the same volume a cylinder has less surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4- structural efficiency, round surfaces share and transfer stresses much better than the rectangular ones which have stress concentration on the corners and rattle and suffer fatigue damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5- cross winds and stormy gusts go around the round cross sections more easily.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-19T08:30:53.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25600" PostTypeId="1" CreationDate="2019-01-19T13:13:34.610" Score="0" ViewCount="24" Body="&lt;p&gt;I was wondering if I could restrict the rotation of a part over another in one direction alone using an electromagnet but allowing me to have a continuous rotation if forced (maybe by using some leverage?) to rotate in that direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concern with using a ratcheted device is intermittent restriction- what I am looking for is a continuous restriction - at any angle of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to add an image to the question, but not sure how. thus attaching a google drive image. url is &lt;a href=&quot;https://drive.google.com/file/d/1ugcji9pDeVQiaCE9BgmFnaj6gt8gPMcf/view&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://drive.google.com/file/d/1ugcji9pDeVQiaCE9BgmFnaj6gt8gPMcf/view&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, 1 will rotate clockwise or counter cw based on an input (not shown) and 2 rotates w.r.t 1, but need an electromagnet that can stop its rotation in counter clockwise direction unless otherwise turned using a 20 kg force (that is leveraged from elsewhere)&lt;/p&gt;&#xA;" OwnerUserId="19039" LastActivityDate="2019-01-19T13:13:34.610" Title="Restricted Rotation using electromagnets" Tags="|mechanical-engineering|electromagnetism|magnets|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25601" PostTypeId="1" CreationDate="2019-01-19T15:10:22.433" Score="-3" ViewCount="626" Body="&lt;p&gt;There is something wrong with the way we describe how a cornering vehicle wheel creates a cornering force. I think a rolling wheel must first rotate on a vertical axis to then create the cornering force. This is fundamentally different than what we think about cornering wheels. Can anyone else see this?&lt;/p&gt;&#xA;" OwnerUserId="16865" LastEditorUserId="16865" LastEditDate="2021-12-17T12:29:14.927" LastActivityDate="2021-12-17T12:29:14.927" Title="Can a rolling wheel create a side force without first rotating on a vertical axis at its contact patch?" Tags="|mechanical-engineering|dynamics|wheels|car|rolling|" AnswerCount="1" CommentCount="1" ClosedDate="2019-01-21T23:49:55.000" ContentLicense="CC BY-SA 4.0" />
  <row Id="25603" PostTypeId="1" CreationDate="2019-01-19T18:16:34.190" Score="1" ViewCount="519" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MEaZo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MEaZo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that for a system to be BIBO stable the eigenvalues need to be stable so real(lambda)&amp;lt;0&#xA;so get that is only valid for 4&#xA;but how can i find out the other systems which are BIBO stable?&lt;/p&gt;&#xA;" OwnerUserId="19026" LastActivityDate="2019-01-20T15:42:57.687" Title="How to find BIBO stability fast?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25604" PostTypeId="1" CreationDate="2019-01-19T20:54:47.517" Score="1" ViewCount="192" Body="&lt;p&gt;I need to settle a discussion on the following question: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part A:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Find the equivalent spring constant for springs in series&lt;/li&gt;&#xA;&lt;li&gt;Use equations of motion, sum of torques to find natural frequency of the system&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;where: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}&#xA;\dfrac{1}{k_{eq}} = \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dfrac{1}{k_3} \\&#xA;I_p\ddot\Theta = -(k_{eq} + k_4 + k_5 r^2 + k_6 r^2)\Theta&#xA;\end{gather}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Part B:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;System in torsion therefore: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$w_{nf}=\sqrt{\dfrac{k}{J}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$J = \dfrac{mr^2}{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Issues&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solutions from my lecturer suggest adding the three strings in series and not using an equivalent &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$w_{nf}^2=\dfrac{k_1 + k_2 + k_3 + k_4 + k_5 r^2 + k_6 r^2}{J}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe you have to use the following, taking in to account the equivalent spring constant for the springs in series:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$w_{nf}^2 = \dfrac{k_eq{} + k_4 + k_5 r^2 + k_6 r^2}{J}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please clarify whether the use of K123 being the equivalent is the right method. All literature suggests equating spring constants in series yet my lecturer appears to think differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vKKI4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vKKI4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18956" LastEditorUserId="1832" LastEditDate="2019-01-21T02:01:45.420" LastActivityDate="2019-01-21T02:01:45.420" Title="Moment of Inertia of a Rotor System" Tags="|mechanical-engineering|applied-mechanics|springs|kinematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25605" PostTypeId="2" ParentId="25603" CreationDate="2019-01-19T21:55:52.363" Score="1" Body="&lt;p&gt;In order for a linear time invariant system to be BIBO all modes who are observable and controllable need to have a negative eigenvalue. A quick way to check the observability and controllability is with the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hautus_lemma&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hautus lemma&lt;/a&gt;. Calculating the eigenvalues of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; should be straightforward when it is diagonal or upper/lower triangular. For the matrices that are not of that form you could use the determinant and trace of the matrix and from that deduce the sign of the real part of the eigenvalues (at least for two by two matrices). Namely the determinant and trace of a matrix are equal to the product and sum of all its eigenvalues respectively.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastEditorUserId="1765" LastEditDate="2019-01-20T15:42:57.687" LastActivityDate="2019-01-20T15:42:57.687" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25606" PostTypeId="2" ParentId="25598" CreationDate="2019-01-19T22:18:40.767" Score="0" Body="&lt;p&gt;Old carriage design relied upon 4 surfaces which meant that velocity was restricted. With an increase in velocity there is a fourfold increase in structural stresses, generally. Just as early tunnels were built with rectangular cross sections it was only with my great great grandfather's invention that tunnels became cylindrical tubes under London. This was possibly due to Alfred Ely Beach's suggestion in 1840 that a circular section would be more efficient.&#xA;See James Henry Greathead for further info. Early airframe design was based on box but now is elliptoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rectangular design has 4 corners that have to resist the combined stresses transmitted by 2 concurrent surfaces in maintaining structural integrity. This flaw is eliminated with elliptiod shapes and removes potential pinion points which are weak and heavy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Centroids of trains are taken axially through their length and due to lightweight materials in the carriage, the CoG is usually &lt;span class=&quot;math-container&quot;&gt;$\frac {1}{3}$&lt;/span&gt;rd from the bottom of the carriage. The onion points removed allow increased velocities due to better structural strain management in fabric of the train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A key issue also is removing opportunity for strain hardening to occur and thereby increase mechanical safety of materials used.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-19T22:18:40.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25608" PostTypeId="2" ParentId="25604" CreationDate="2019-01-20T02:02:41.043" Score="1" Body="&lt;p&gt;You are correct. The schematic shows that k1,k2,k3 work in series with k4 and k5 and k6 being parallel at the upper side of the disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your lecturer has made an error, and next time most likely will add an errata to his notes.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-20T02:02:41.043" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25611" PostTypeId="2" ParentId="25601" CreationDate="2019-01-20T02:47:46.650" Score="1" Body="&lt;p&gt;Your understanding is missing substantial information regarding statics and equilibrium. Think for a minute:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The instantaneous forces acting in the wheel system are radial velocity, momentum, inertia, kinetic energy and friction. Above all there is also gravity. When the wheel changes direction, its centroid remains constant but its centre of gravity changes. It is possible for a body in motion to have its centre of gravity outside itself. This occurs because of a constant resistance force applied to the control arm which exceeds the losing radial velocity gyroscopic forces. In addition there is the moment of inertia imbalance that is corrected by asked force through the control arm. The axle or more correctly, constant velocity joint (cvj), transmits force to the issuing wheel which also provides resistance to direction change observing by steering control arm. This is why wheel bearings are often the conical thrust type or specially designed ball race type instead of rod type bearings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lateral forces are the result of radial inertia change with an equal and opposite force in the directing of steering. Newton's third law of motion... &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-20T02:47:46.650" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="25612" PostTypeId="2" ParentId="18545" CreationDate="2019-01-20T03:11:16.570" Score="1" Body="&lt;p&gt;To answer your question requires simple calculations of equilibrium. In order to close the valve, the external force must exceed internal force by a fraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking logically, &lt;span class=&quot;math-container&quot;&gt;$\rho = \frac {Force} {Area} $&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is applied stress. This means that the volumetric value of the float directly affects the force it exerts against incoming water pressure. To reduce the volume of the float, it is placed on an arm because the fulcrum effect substantially multiplies the force exerted against incoming water pressure. This is why the ballcock is used in toilet cysterns so effectively. Were the ballcock arm half as long, the ball would be double in size... and therein less your design problem: float size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lever arm system is the most effective at keeping float size to a minimum. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-20T03:11:16.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25613" PostTypeId="2" ParentId="6546" CreationDate="2019-01-20T03:26:28.220" Score="1" Body="&lt;p&gt;You have two moments which means you need to find the neutral axis which usually coincides with the system centroid along &lt;span class=&quot;math-container&quot;&gt;$I_{xx}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty is overcome by applying simple rules of statics and equilibrium.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Sigma M_A + \Sigma M_B=\Sigma M_{W}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that your 2 arms could actually behave as a single pivot if their apex pinion point was cosymmetrical above the neutral axis in the x-x plane. Remember that force through a single point removed the resonance?!&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2019-07-29T12:33:10.503" LastActivityDate="2019-07-29T12:33:10.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25614" PostTypeId="2" ParentId="23663" CreationDate="2019-01-20T03:44:12.073" Score="1" Body="&lt;p&gt;I would use a collar lock with a slide action locking gate. Solves the durability issues and can be used thousands of times. The collar can be square and slide into a dual keyway to support the blue item while a cylindrical gate (spring loaded or not) shuts behind it. Collar thickness determines load bearing capacity.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-20T03:44:12.073" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25615" PostTypeId="2" ParentId="21828" CreationDate="2019-01-20T04:05:29.577" Score="0" Body="&lt;p&gt;I am not yet a qualified civil engineer which means this is not certifiable as correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems the cut is in the end of the beam which rests on a load bearing wall. Since punching shear is at 45° the cut is likely outside the critical zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that beam extends towards the camera and the hole is through the maximum critical shear point there may be structural issues to think about. I would put a thick steel pipe through and ensure it is sealed in the beam properly with good grade cement before putting the electrical cables back through. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-01-20T04:05:29.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25616" PostTypeId="1" CreationDate="2019-01-20T04:14:25.547" Score="-1" ViewCount="104" Body="&lt;p&gt;If you had a flywheel acting as a mechanical battery in an all mechanical system, such that you could pump energy into the flywheel up to a desired speed and then slowly drain its kinetic energy - then how could you &lt;strong&gt;mechanically&lt;/strong&gt; transmit this power at a constant output speed, no matter how fast or slow the flywheel moves? Obviously there will be a minimum speed threshold to drive the output, but we want a constant speed from its upper rpm limit down to this threshold.&lt;/p&gt;&#xA;" OwnerUserId="19044" LastActivityDate="2021-02-08T09:00:40.080" Title="Transmission for variable-speed flywheel to output a steady output speed?" Tags="|mechanical|transmission|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25617" PostTypeId="2" ParentId="25616" CreationDate="2019-01-20T06:25:30.833" Score="0" Body="&lt;p&gt;The flywheel can pump air into a sealed container with a check valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the container will act as a regulator to keep the outlet air at say a steady 100psi, even though the container pressure could vary between say 5kpsi and 200 psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a wide number of pneumatic storage and distribution of energy at work in many contexts and applications, which can be modified for this purpose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OP asks for no electrical systems, otherwise there is already working magnetically controlled gyros for the exact same purpose&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-01-20T23:42:43.300" LastActivityDate="2019-01-20T23:42:43.300" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25618" PostTypeId="2" ParentId="25616" CreationDate="2019-01-20T07:47:52.773" Score="0" Body="&lt;p&gt;The DAF Variomatic transmission is/was a transmission based on a vee-belt and two pulleys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The space between the pulley faces was controlled to give the &quot;gear ratio&quot; ie driven pulley effective diameter compared to the driver pulley effective diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque capability was a limiting factor so worked well for smaller engine size cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is but one link for moreinfo : &lt;a href=&quot;https://www.uniquecarsandparts.com.au/how_it_works_variomatic_transmission&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.uniquecarsandparts.com.au/how_it_works_variomatic_transmission&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-01-20T08:09:13.340" LastActivityDate="2019-01-20T08:09:13.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25620" PostTypeId="1" AcceptedAnswerId="25622" CreationDate="2019-01-20T11:51:09.253" Score="0" ViewCount="438" Body="&lt;p&gt;I have shared a question from mechanics of solids and my solution with it, as you can see I am trying to take the derivative of the stress equations to find the min, max points but its going wrong. I need to know why my solution is wrong.&lt;a href=&quot;https://i.stack.imgur.com/B0SWq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B0SWq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fVygq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fVygq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tYRob.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tYRob.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18319" LastActivityDate="2019-01-20T20:00:29.057" Title="Mechanics of Solids / Strength of Materials" Tags="|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25621" PostTypeId="1" CreationDate="2019-01-20T19:03:23.020" Score="0" ViewCount="305" Body="&lt;p&gt;Does anyone know how i can read the angle on the Nyquist plot&#xA;without knowing the open loop transfer function?&#xA;So by only having the plot.&lt;/p&gt;&#xA;" OwnerUserId="19026" LastActivityDate="2019-01-20T19:03:23.020" Title="how to calculate the angle in Nyquist plot" Tags="|control-engineering|control-theory|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25622" PostTypeId="2" ParentId="25620" CreationDate="2019-01-20T20:00:29.057" Score="1" Body="&lt;p&gt;Let us assume that the greatest stress in any segment is at its top. (You can verify this on your own.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us further assume that if the maximum stresses in the segments are not all equal, then we can adjust the geometry (here, &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;) to reduce the larger value. (You can verify this on your own.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, we've optimized the geometry when the maximum stresses in all segments are equal. (The limit for a continuously changing cross section, i.e., an infinite number of segments, would be a constant stress throughout the hanging rod. This problem comes up in the context of &lt;a href=&quot;https://en.wikipedia.org/wiki/Space_elevator#Cable_section&quot; rel=&quot;nofollow noreferrer&quot;&gt;space elevator tethers&lt;/a&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum stresses in the upper and lower section are therefore &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_\mathrm{max,\,upper}=\frac{\mathrm{load}}{\mathrm{cross}\mbox{-}\mathrm{sectional\,area}}=\rho\left(\frac{A_bb+A_a(100\,\mathrm{m}-b)}{A_a}\right)=\rho\left[\left(\frac{A_b}{A_a}-1\right)b+100\,\mathrm{m}\right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_\mathrm{max,\,lower}=\frac{\mathrm{load}}{\mathrm{cross}\mbox{-}\mathrm{sectional\,area}}=\frac{\rho A_bb}{A_b}=\rho b,$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively, where &lt;span class=&quot;math-container&quot;&gt;$A_i$&lt;/span&gt; is the cross-sectional area of segment &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;, and we wish to equate these stresses to determine the optimum geometry. The result is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$b_\mathrm{optimum}=\frac{100\,\mathrm{m}}{2-A_b/A_a},$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the minimum maximum stress is &lt;span class=&quot;math-container&quot;&gt;$\rho_\mathrm{optimum} b$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this result in the answer you were looking for?&lt;/p&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2019-01-20T20:00:29.057" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25623" PostTypeId="1" AcceptedAnswerId="25626" CreationDate="2019-01-21T04:25:16.773" Score="2" ViewCount="649" Body="&lt;p&gt;I am using strut-and-tie method to design a corbel. I noticed that Eurocode mentions strut-and-tie only in section 6, which is ultimate limit state design. How do I perform serviceability limit state design for the corbel using strut-and-tie method?&lt;/p&gt;&#xA;" OwnerUserId="19049" LastActivityDate="2019-01-21T20:23:06.370" Title="How to perform SLS check with strut-and-tie method?" Tags="|structural-engineering|reinforced-concrete|eurocodes|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25624" PostTypeId="1" CreationDate="2019-01-21T04:35:05.680" Score="1" ViewCount="183" Body="&lt;p&gt;I want to make a DIY dough press, so need to know how much pressure I would need to apply. I don't have mechanical engg background, so based on some googling, I think this is what I need to do, but wanted to validate here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took a small dough ball, and placed a very light flat surface on its top, and then put a 16.9 fl oz water bottle (weighing 1.1 lbs) on top of it. It squished the dough ball to a thick circle of 2.09 inch diameter. (3.145 sq inch area).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I put 7 more water bottles on top of it (total 8.8 lbs approx), and the dough squished flatter to a circle of 2.91 inch (6.647 sq inch).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my questions are, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Is this correct way to calculate compressive strength of this dough ball by dividing force / area, where it came to rest. So with 1 bottle on top, when it squished somewhat flatter, the PSI = 1.1/3.145 = 0.349 lbs/sq inch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) If so, then shouldn't the compressive strength be constant as it is tied to the material. So dividing the final area after adding more weight should yield the same value? The second value, 8.8/6.647 =  1.323 lbs/sq inch.&lt;/p&gt;&#xA;" OwnerUserId="9580" LastActivityDate="2021-06-10T19:52:13.227" Title="Manually Attempting Empirical Measurement of Flour Dough Compressive Strength" Tags="|pressure|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25625" PostTypeId="1" CreationDate="2019-01-21T06:56:18.013" Score="-1" ViewCount="121" Body="&lt;p&gt;A rock is falling from an embankment of mass &lt;span class=&quot;math-container&quot;&gt;$m_a$&lt;/span&gt; at a height of &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;. Below, a fall protection fence catches the rock, dissipating the energy through tension cables to steel posts and then into concrete piles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fyBOS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fyBOS.png&quot; alt=&quot;An example from a RTA standard drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you calculate the &quot;impact&quot; capacity of the tension cables, steel posts and concrete foundations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: At this height, the rock initially has a potential energy of &lt;span class=&quot;math-container&quot;&gt;$PE=mgh$&lt;/span&gt;. Assuming there are no losses due to friction, bouncing, heat, sound, etc (a very conservative assumption), the potential energy will be mostly converted to kinetic energy, &lt;span class=&quot;math-container&quot;&gt;$KE=0.5mv^2$&lt;/span&gt;. The fence will absorb this complete energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say the rock hits in the middle of the bay between the two posts. The cables will deflect with the impact of the rock. How do you calculate how much energy the cables absorb and whether the cables have sufficient capacity? How is this different from the applied force?&lt;/p&gt;&#xA;" OwnerUserId="19050" LastEditorUserId="19050" LastEditDate="2019-01-21T21:53:08.700" LastActivityDate="2019-01-24T11:17:04.650" Title="Impact capacity of tension cables, columns and foundations" Tags="|mechanical-engineering|structural-engineering|structural-analysis|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25626" PostTypeId="2" ParentId="25623" CreationDate="2019-01-21T09:09:20.113" Score="2" Body="&lt;p&gt;One important thing to remember about the strut-and-tie method is that as a lower-bound method it is based on the plasticity theory. That is, the principle is to find a safe and statically admissible stress distribution, and if any such distribution is possible the structure will not collapse at this load. This principle only makes sense for the ultimate limit state. However, if the assumed load path is close to what linear elasticity theory would give, the stresses will be a decent approximation to a detailed finite element calculation and may very well be sufficiently accurate. Therefore, the difficult part of the calculation is knowing how well your strut-and-tie model aligns with elasticity theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a bit more on this in EN 1992-1-1 section 5.6.4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you do your SLS check with a strut-and-tie model, the actual calculations are quite straightforward. Calculate the stress in concrete and reinforcement and check them against the SLS limits in section 7.2. Use the reinforcement stress to calculate a crack width and check that against the limit from section 7.3. And you're done. The strut-and-tie approach will not give you a way to calculate a deflection but for a typical corbel that wouldn't be very interesting anyway.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastEditorUserId="1832" LastEditDate="2019-01-21T20:23:06.370" LastActivityDate="2019-01-21T20:23:06.370" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25627" PostTypeId="2" ParentId="25625" CreationDate="2019-01-21T09:12:23.807" Score="0" Body="&lt;p&gt;Say speed of the rock when it hits the fence is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V = \sqrt {2gh}\quad$&lt;/span&gt; and its kinetic energy is &lt;span class=&quot;math-container&quot;&gt;$E = 1/2 mV^2  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This energy will be trapped by the fence and transferred to the posts. Assume  inelastic collision with the post, or for simplicity ignore the mass of the post'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elastic energy in a deflected cantilever post should be equated with the kinetic energy of the falling rock. &#xA;From there you can calculate all other reactions and moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Commercial chain-link usually is strong enough for heights of up to 5 feet at spans of 6-8 feet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you show your effort we can help you.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-21T09:12:23.807" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25629" PostTypeId="1" CreationDate="2019-01-21T11:41:09.837" Score="1" ViewCount="29" Body="&lt;p&gt;What design responses must be given in the TOSCA gui. I chose volume reduction as an objective function with the constraints of displacement. Is this approach correct?&lt;/p&gt;&#xA;" OwnerUserId="19051" LastActivityDate="2019-01-21T11:41:09.837" Title="How to run a steady state dynamics, Modal step based optimisation using TOSCA?" Tags="|dynamics|abaqus|modal-analysis|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25630" PostTypeId="1" AcceptedAnswerId="26281" CreationDate="2019-01-21T14:14:58.943" Score="0" ViewCount="230" Body="&lt;p&gt;I apologize for the vague title. I know my problem, but I am having problems describing it in concise words, which is making it difficult for me to google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am building a CNC machine that is similar to a pick and place machine. I used extruded aluminum for the rails, and I am certain that everything is squared up. I had a local machine shop create a plate that has a grid of holes, so that I can attach &quot;parts&quot; that can be used by the CNC machine. However, this grid of holes is not machined at the tolerance that I need (budget constraints). I need a system for making sure that the &quot;parts&quot; are square to the cnc frame. One of my ideas was to use leveling springs similar to how you adjust a the bed on a 3D printer to be square with the frame. However, I am unsure if this is the best idea, and I would like a second opinion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is important that the machine has a resolution of 200um (if you were to move in a straight line across the machine ~500mm all parts that should be on that line should not be more than 200um above or below it). I know that some solutions maybe more expensive than having the plate machined to lower tolerance, but if I were to mass produce this machine I would like a way to adjust the alignment of parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for taking the time to read and/or answer my question!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NVsonl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NVsonl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19052" LastActivityDate="2019-04-06T18:02:10.427" Title="System for aligning/truing objects to each other in a plane for CNC machine" Tags="|machining|cnc|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25631" PostTypeId="1" CreationDate="2019-01-21T14:39:09.020" Score="1" ViewCount="44" Body="&lt;p&gt;Are there any satellite thrusters that mimic the &quot;thrust reversal&quot; feature of airplanes i.e. (A320...).&#xA;I already know of procedures to decelerate a satellite such as using two separate thrusters firing in opposite directions however a single  thruster would be far more economical -encorporating the thrust reversal.   &lt;/p&gt;&#xA;" OwnerUserId="19053" LastActivityDate="2019-01-21T19:41:50.003" Title="Satellite Thruters with reversers" Tags="|mechanical-engineering|electrical-engineering|aerospace-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25632" PostTypeId="1" CreationDate="2019-01-21T19:29:03.807" Score="1" ViewCount="117" Body="&lt;p&gt;I have a scenario where we have encountered a cracked steel plate where the crack has propagated within the plate longitudinally for an extensive length. The plate thickness is 1&quot;, and the crack length is approximately 12&quot; with somewhat of a bend horizontally, yet staying within maybe 3/8&quot; max variation propagating right between the top and bottom of the plate. Stresses are complex but the highest load is likely a vertical load, and may be close to 10-20 tons. Impact factor typically ranges between 3.5-5.5 for this load case, but it is complex and not precisely known for the location for the crack, which is suspected to have received only perhaps loads upward to an impact factor of 2-3. Precise stresses for failure are unknown. Steels in question is 4140. Growth marks have been observed throughout most of the entire crack, with somewhat of a &amp;lt; shape, which seems to indicate that it cracked both top and bottom and then finally through perhaps? But that Seems strange that it would crack just right near the middle of the plate thickness for such an extensive length when loads are not applied linearly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to review the theories how the crack propagated based on the described observations. What can a &amp;lt; crack surface tell me? Can someone list that out and provide appropriate literature backup? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZTBkD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZTBkD.png&quot; alt=&quot;Picture showing grow patterns&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/E0Y9i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/E0Y9i.png&quot; alt=&quot;Overall details&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D7tzC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D7tzC.png&quot; alt=&quot;Overall details&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13526" LastEditorUserId="13526" LastEditDate="2019-01-21T21:59:28.407" LastActivityDate="2020-06-17T03:07:55.840" Title="Factors favoring crack propagation in the middle of a plate for an extensive length relative to plate thickness (Reasons for &lt; marks on crack surface)" Tags="|structural-engineering|failure-analysis|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25633" PostTypeId="2" ParentId="25631" CreationDate="2019-01-21T19:41:50.003" Score="1" Body="&lt;p&gt;Satellites use a variety of systems to control their exact position in orbit. Some are tiny rockets, others take advantage of earth's magnetic field. Some are passive control, some active.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Vernier_thruster&quot; rel=&quot;nofollow noreferrer&quot;&gt;Vernier thrusters&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All the methods are obviously extremely lightweight,( solar panels are thin as paper), to the extent they will fail in earth atmosphere conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Airplane thrust reversers function relaying on the atmosphere pressure and the fact that the exhaust of the jet engine when redirected will create a large cone of wake acting like parachute with an impact zone much greater than just the discharge flow of jet engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the vacuum of space this advantage is not available and the additional weigh of the heavy structures will be an extra burden&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-21T19:41:50.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25634" PostTypeId="2" ParentId="25624" CreationDate="2019-01-21T19:46:49.070" Score="1" Body="&lt;p&gt;It sounds like you're assuming the dough follows a &lt;em&gt;perfectly-plastic&lt;/em&gt; [2] material model, which might be reasonable but it depends on the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) For a general material, strength is the stress at which failure occurs. So you would &#xA;find the smallest load which causes any permanent deformation. In practice, that's difficult to measure so you have tricks like what you're doing which depend on assumptions about the material. For a perfectly plastic material, your calculation would be correct but since the stress seems to increase with strain, it looks like the dough is strain-hardening and a different material model may be appropriate, such as &lt;em&gt;linear hardening&lt;/em&gt; &lt;a href=&quot;https://i.stack.imgur.com/cr6E1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;. If it's a &lt;em&gt;rigid/linear-hardening&lt;/em&gt; material, you can plot the pressure and strain values on a curve like in the picture below and extrapolate to zero strain to see what the yield stress is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) Yes, pressure at equilibrium should be constant for a perfectly-plastic material, but it should be a function of strain for a more general material. However, there may be a mistake in your experiment which is that if the dough grips onto the surfaces then you're measuring a more complicated quantity that also depends on the geometry of the blob and it will become stiffer the flatter it is because it's constrained from spreading outwards by friction with the surfaces. To properly measure the relationship between stress and strain, the dough should be free to slide frictionlessly on the two surfaces. See the picture of different deformation shapes here [3]. Case (e) is the correct one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cr6E1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cr6E1.jpg&quot; alt=&quot;Some plastic material models&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[2] &lt;a href=&quot;https://nptel.ac.in/courses/105108072/mod04/lec4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://nptel.ac.in/courses/105108072/mod04/lec4.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[3] &lt;a href=&quot;http://www.instron.co.uk/en-gb/our-company/library/test-types/compression-test&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.instron.co.uk/en-gb/our-company/library/test-types/compression-test&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2019-01-21T19:46:49.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25635" PostTypeId="1" CreationDate="2019-01-21T20:59:54.967" Score="2" ViewCount="81" Body="&lt;p&gt;Why do aluminum clad steel wires have high corrosion resistance despite being two dissimilar metals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does the aluminum act as a protective &amp;amp; sacrificial layer similar to zinc in galvanization or is there some other mechanism going on?&lt;/p&gt;&#xA;" OwnerUserId="10570" LastActivityDate="2019-01-21T23:12:23.527" Title="Why does aluminum clad steel wire have high corrosion resistance?" Tags="|electrical-engineering|materials|chemistry|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25636" PostTypeId="2" ParentId="25635" CreationDate="2019-01-21T21:38:00.323" Score="0" Body="&lt;p&gt;It has good corrosion resistance in dry air ( same as pure aluminum ).In an ordinary location with occasional rain it works well. Dissimilar metals is not a problem if there is no electrolyte ( eg. water). In wet conditions, especially along a sea coast (salt) corrosion will be significant. The aluminum is sacrificial protecting the steel , with a little more voltage than zinc. Offshore steel structures are usually protected with aluminum anodes. Although life is longer in sea water because a &quot;calcarous&quot; deposit builds up and stiffels corrosion ( not always, that is why there are galvanic protection engineers and contractors).  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-01-21T21:38:00.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25637" PostTypeId="2" ParentId="25635" CreationDate="2019-01-21T23:12:23.527" Score="1" Body="&lt;p&gt;The dominant protection mechanism is that the aluminum excludes seawater from contact with the steel underneath, and if passivated with a layer of oxide, the corrosion rate of the aluminum via direct chemical attack will be low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aluminum does not protect steel the way a coating of zinc will. If the aluminum gets pierced and the steel underneath is exposed, the aluminum tends to get undercut via crevice corrosion and come off the steel. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-01-21T23:12:23.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25638" PostTypeId="1" CreationDate="2019-01-21T23:46:09.763" Score="1" ViewCount="245" Body="&lt;p&gt;&lt;strong&gt;Objective:&lt;/strong&gt; I want to connect multiple sensors to an item of equipment and visualize the data real-time on an iPad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mock of Desired Output:&lt;/strong&gt;&lt;a href=&quot;https://i.stack.imgur.com/rH717.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rH717.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt; I need advice on the best methods to use to connect multiple sensor data and transfer the data through either wifi or Bluetooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I use / Is it possible to use Arduino (I have experience with this)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there existing wifi modules that allow for such a connection between multiple sensors and devices?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Considerations&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I have experience with Reactjs and react native to build the application&lt;/li&gt;&#xA;&lt;li&gt;Sensors are (RPM, flow rate, weight, and PSI)&lt;/li&gt;&#xA;&lt;li&gt;This is pretty vague but I'm new to the IoT and acknowledge it will be a challenge.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19058" LastActivityDate="2019-01-21T23:46:09.763" Title="Multiple sensor data to WiFi visualized on React Native app: Guidance" Tags="|mechanical-engineering|computer-engineering|wifi|iot|" AnswerCount="0" CommentCount="2" ClosedDate="2019-01-22T13:05:17.740" ContentLicense="CC BY-SA 4.0" />
  <row Id="25639" PostTypeId="1" CreationDate="2019-01-22T02:26:40.697" Score="0" ViewCount="51" Body="&lt;p&gt;What is the general practice for selecting time interval for RMS calculation for vibration data (mm/s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should it be linked to the the frequency of the motor or the sampling rate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="19059" LastActivityDate="2019-04-09T20:24:15.043" Title="time interval selection for RMS of vibration" Tags="|mechanical-engineering|motors|vibration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25640" PostTypeId="1" CreationDate="2019-01-22T03:34:19.730" Score="0" ViewCount="3237" Body="&lt;p&gt;For the transfer function &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt;, I tried to design a lead compensator for the function to have a response to the step with the following specifications:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\%OS = 10\% \ \rightarrow \ Overshoot $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ T_s = 5sec \ \rightarrow \ 2\% \ Criterion$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I simulate the function already with the lead compensator by the &lt;code&gt;rlocus()&lt;/code&gt; command the system shows the desired behavior. However when I simulate with the &lt;code&gt;step()&lt;/code&gt; function the parameters do not match.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9EcqX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9EcqX.png&quot; alt=&quot;Behavior of the compensated function with rlocus() and step() functions&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4UtM5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4UtM5.png&quot; alt=&quot;Matlab code used&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TR5Vi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TR5Vi.png&quot; alt=&quot;Steps used to design the lead compensator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19060" LastEditorUserId="18238" LastEditDate="2020-04-27T14:44:12.187" LastActivityDate="2022-01-17T20:04:25.443" Title="Design of a lead compensator with root locus in matlab" Tags="|control-engineering|control-theory|matlab|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25641" PostTypeId="1" CreationDate="2019-01-22T11:12:02.953" Score="3" ViewCount="529" Body="&lt;p&gt;I would like to know what does the &quot;D&quot; means in steel grade 16MND5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the standard EN 10027:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;16 means 0.16 % C&#xA;MN 5 means 1.25 % Mn (5 must be divided by 4 to get the Mn mass fraction)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But nothing is specified for D.&lt;/p&gt;&#xA;" OwnerUserId="19062" LastActivityDate="2024-02-16T19:05:25.627" Title="What does the D stand for in the steel grade 16MND5" Tags="|steel|standards|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25642" PostTypeId="2" ParentId="25641" CreationDate="2019-01-22T11:17:00.900" Score="0" Body="&lt;p&gt;Having looked at the link,&lt;/p&gt;&#xA;&lt;p&gt;D is for Flat products for cold forming, and is usually followed by two numbers characterising steel&lt;/p&gt;&#xA;&lt;p&gt;See &lt;a href=&quot;https://en.wikipedia.org/wiki/Steel_grades#Category_1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Steel_grades#Category_1&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Based on the diagram in the same link:&lt;br /&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sBlAx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sBlAx.png&quot; alt=&quot;&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="28917" LastEditDate="2020-10-04T13:42:18.723" LastActivityDate="2020-10-04T13:42:18.723" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25643" PostTypeId="1" CreationDate="2019-01-22T13:10:46.613" Score="0" ViewCount="720" Body="&lt;p&gt;Im seeking an alternative to thermal grease/ pastes. I was researching graphite pads when I stumbled across the soft PGS line from Panasonic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to choose a suitable thermal interface material to connect the CPU die to a heat sink. I do not wish to use thermal paste ( it is greasy). I was wondering which material would be most suitable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PGS :Excellent thermal conductivity : 700 to 1950 W/(m·K) ( no data for z direction; I found some sources stating that the z direction was about 15W/m-K but I can not verify this).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“GraphiteTIM(Compressible Type) PGS with thermal conductivity : X-Y direction 400W/m∙K, Z direction (28W/m∙K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat sink will be made of aluminium ( roughly 200-250W/m-k) or copper  (roughly 350-386W/m-k).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is more important? XY thermal conductivity or Z direction thermal conductivity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the heat-sink has a defined thermal conductivity, is there any advantage in choosing a conductivity greater than the heat sink's value? (Example, two materials A and B with a thermal conductivity of 700 and 1000W/m-k  respectively for xy plane; assume both have similar z direction conductivity. Does choosing the 1000W/m-k over the 700W/m-k material provide any benefit to the overall system, considering that the heat sink is limited to 386W/m-k? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="19063" LastActivityDate="2019-01-22T20:58:53.620" Title="Which is more important XY plane thermal conductivity or z direction thermal conductivity?" Tags="|mechanical-engineering|materials|heat-transfer|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25644" PostTypeId="1" CreationDate="2019-01-22T14:20:43.843" Score="2" ViewCount="67" Body="&lt;p&gt;I have an instrument that is technically a humidity controlled chamber. It measures how much moisture is picked up by the sample through a microbalance(Zone 1). Nitrogen gas flows from zone 3 and split into 2 chambers (by a union Tee ) in Zone 2. Finally the gas is vented in Zone 1, each chamber with its own exhaust vent (tubings that vent to the back).&lt;/p&gt;&#xA;&lt;p&gt;The Problem is that the flow is uneven. or let me rephrase, the flow is even as long as the resistance against the flow is equal for both chamber. with the slightest change in resistance at one of the vents, the flow is completely diverted to the other chamber. This happens if one of the vent tubings got slightly bent, one is slightly longer than the other, etc.. It is ridiculous sometimes&lt;/p&gt;&#xA;&lt;p&gt;My question is: if I joined the exhaust tubings together at the end (by a Y-connection), would that equalize the pressures and gas flow in the chambers?&#xA;I'm no engineer but I can't think of a better place to ask&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MpIdS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MpIdS.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Jy0tx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jy0tx.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19064" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-01-22T14:20:43.843" Title="How to balance flow and pressure in nitrogen purged chambers?" Tags="|airflow|compressed-air|compressible-flow|compressed-gases|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25646" PostTypeId="2" ParentId="25641" CreationDate="2019-01-22T18:07:18.653" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://hal.archives-ouvertes.fr/hal-01213899/document&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; (Table 1, page 4) gives the composition of 16MND5-A508 as &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt; C     S     P     Mn   Si   Ni  Cr   Mo   V Cu Co Al N&#xA;&amp;gt; 0.159 0.008 0.005 1.37 0.24 0.7 0.17 0.5  &amp;lt;0.1&#xA;&amp;gt; Weight % - iron balance.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A508 is the relevant &lt;a href=&quot;https://www.astm.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASTM&lt;/a&gt; standards document (not a free publication.)&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-01-22T18:07:18.653" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25648" PostTypeId="2" ParentId="25643" CreationDate="2019-01-22T20:58:53.620" Score="2" Body="&lt;p&gt;Since the interface material is very thin, its thermal resistance in the z direction will be very low compared to the much thicker heatsink as long as its thermal conductivity is similar to or higher than the heatsink's. Thermal resistance = thickness / thermal conductivity and thermal resistances in series add to give the overall thermal resistance. So I think small differences in z direction thermal conductivity are not important since they're dwarfed by the much higher thermal resistance of the heatsink in series with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There would be value in a higher xy thermal conductivity than the heatsink material because that would spread the heat over wider area of the heatsink which would increase the heat flow rate overall as well as reducing the temperature at the interface. I think a small increase in this property would be significant as long as it's already similar to or higher than the heatsink's thermal conductivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how I visualize heat flow with a high xy thermal conductivity interface. Without that interface, it would be more concentrated directly over the die, leaving the outer bottom edges of the heatsink relatively unused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZNlkU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZNlkU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, please only use capital K for kelvin and dot for multiplication in SI units since those are standard. The hyphen is a minus sign, not a multiplication sign! W/(m·K), not W/m-k.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2019-01-22T20:58:53.620" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25649" PostTypeId="1" CreationDate="2019-01-22T20:59:46.057" Score="0" ViewCount="405" Body="&lt;p&gt;can someone tell me how to calculate the bearing forces of the two bearings?&#xA;I can't get any further because I don't have the angle for the second bearing.&#xA;I don't have an idea anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F =   20KN&lt;br&gt;&#xA;q0= 2000N/m&lt;br&gt;&#xA;a =   1000mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the bottom under Lösungen are the results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QFRe9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QFRe9.png&quot; alt=&quot;Test&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19069" LastActivityDate="2021-02-11T23:16:17.333" Title="Calculate bearing forces" Tags="|mechanical-engineering|statics|" AnswerCount="3" CommentCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="25650" PostTypeId="1" CreationDate="2019-01-22T21:22:31.123" Score="0" ViewCount="188" Body="&lt;p&gt;I understand that for magnetic lines of the rotating magnetic field to cut the windings of the rotor (and induce current), the field has to rotate at a slightly higher rate than the rotor. But if we assume the motor starts from zero RPM, at what speed does the field rotate at this moment in time? &lt;strong&gt;Do we have to somehow sense the rotating speed of the rotor and adjust the frequency of the 3-phase input voltage somehow?&lt;/strong&gt; I think this is not what happens, since such a requirement seems somewhat complicated to achieve, especially given that induction motors have been around for quite a long time. So what am I missing?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2024-03-25T21:08:41.113" Title="Does the frequency of input voltage in induction motors have to be continuously altered?" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25651" PostTypeId="2" ParentId="25649" CreationDate="2019-01-22T21:38:45.403" Score="0" Body="&lt;p&gt;Lets take the moments about point a.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ Ma = 5a*2000(4+5/2) - 20000a = 45000*a\ Nm \\  F_{vertical S}= 45000*a/5a = 9000N$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_{vertical}A = 10000-9000= 1000N  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F_{S horizontal} = 9000*-1/2 = -4500N$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F_{horizonatal A}= 4500N  $&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2019-01-22T21:44:52.557" LastActivityDate="2019-01-22T21:44:52.557" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25653" PostTypeId="1" AcceptedAnswerId="25670" CreationDate="2019-01-23T01:20:41.303" Score="1" ViewCount="119" Body="&lt;p&gt;I am trying to figure out what the efficiency loss of a leaking hydraulic cylinder is.  This &lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;rct=j&amp;amp;q=&amp;amp;esrc=s&amp;amp;source=web&amp;amp;cd=1&amp;amp;cad=rja&amp;amp;uact=8&amp;amp;ved=2ahUKEwjm5pPi3YLgAhUQCKwKHfM1BX8QFjAAegQICRAC&amp;amp;url=http%3A%2F%2Fwww.mate.tue.nl%2Fmate%2Fpdfs%2F10299.pdf&amp;amp;usg=AOvVaw0UI7OlgfDT9rIm3kF0_Dc8&quot; rel=&quot;nofollow noreferrer&quot;&gt;master's thesis&lt;/a&gt; found that a controllable pitch propeller (essentially a hydraulic cylinder inside a rotating shaft to control the pitch of the propeller) had external leakage of  0.5 ltr/min at 40 bar and internal leakage of 3 ltr/min at 100 bar.  I was wondering if it would be possible to convert these numbers into a rough efficiency loss?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been google searching for days to try to find a reference to indicate how to make this calculation.  I also went all out and went to a local university library to use Engineering Village -- hoping to find some paper that did an analogous study but couldn't find much to help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems relevant but not sure how to use it: &lt;a href=&quot;https://engineering.stackexchange.com/questions/19228/losses-through-holes-drilled-in-a-pipe&quot;&gt;Losses through holes drilled in a pipe&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17228" LastActivityDate="2019-01-24T15:51:55.897" Title="Calculate Efficiency Loss of a Leak in a Hydraulic Cylinder" Tags="|hydraulics|energy-efficiency|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25654" PostTypeId="2" ParentId="25632" CreationDate="2019-01-23T06:54:45.970" Score="1" Body="&lt;p&gt;At the time of 1994 Northridge earthquake in California, I along with  many other engineers inspected several four and five story damaged buildings to assess the urgent safety situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We observed many torn base plates of various thickness and sizes up to 2&quot; thickness. The most probable cause was shear rupture under large pendulum motion of the columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those observations are collected in a document available online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google, 1994 Northridge earthquake- gpo.gov.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This failure of shear plates is a complex area and propagation of cracks has to do with dynamic repetitive loading, fatigue, exposure to chemicals, incorrect welding, defective product, structural design problems. It may have to be investigated by teams of different expertise.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-23T06:54:45.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25655" PostTypeId="1" CreationDate="2019-01-23T09:10:58.923" Score="1" ViewCount="42" Body="&lt;p&gt;I have a very simple little T8 CNC without endstops that I use for PCB's. After a failed drill job, the position of the machine is lost and I would like to resume the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine is running GRBL 0.9 and I'm using UGS for sending the gcode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would like to be able to jog the machine to match a known hole that's already been drilled and then say &quot;hey, you're now at position x, y&quot;, but I can't seem to find either a gcode or function in UGS to do this. Is there any such gcode?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My workaround is currently to jog to a known hole, reset to zero, then send gcode to move to the inversion of that known hole and set zero again. Is there any smarter way?&lt;/p&gt;&#xA;" OwnerUserId="19072" LastActivityDate="2019-01-23T09:10:58.923" Title="Best way to resume aborted cnc job without endstops by zeroing in on known hole…?" Tags="|cnc|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25658" PostTypeId="2" ParentId="25632" CreationDate="2019-01-23T15:44:36.157" Score="0" Body="&lt;p&gt;Your diagram shows a typical brittle fracture propagation , likely caused by an impact load. The chevrons point to the origin of the fracture. The impact strength of 4140 depends on the heat-treatment; it can be brittle at room temperature. In failure analysis, the chevron area ( purple ) is of no particular interest other than it points to the origin. Refer to LEFM ( linear elastic fracture mechanics) in Wikipedia; a relatively good summary.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-01-24T01:03:01.257" LastActivityDate="2019-01-24T01:03:01.257" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25659" PostTypeId="1" AcceptedAnswerId="25660" CreationDate="2019-01-23T20:05:12.633" Score="1" ViewCount="3223" Body="&lt;p&gt;After reading about submarines in World War II, I was curious about their battery capacity, specifically in comparison to modern Battery Electric Vehicles (ie Tesla, Bolt, Leaf, i3, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to find a source that either answers the question in kWh or gives me enough information to calculate it myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My current guess is about 30 kWh&lt;/strong&gt;, based on a figure of 12000 Ah in a 120 cell system, and a voltage of 2.75 V - 1.05 V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would accept an answer for any class of submarine in WWII, but I was looking at the US Balao class.&lt;/p&gt;&#xA;" OwnerUserId="8538" LastEditorUserId="90" LastEditDate="2020-04-23T19:37:16.777" LastActivityDate="2020-04-28T15:27:41.383" Title="What is the capacity of a WWII Submarine battery in kWh?" Tags="|electrical-engineering|battery|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25660" PostTypeId="2" ParentId="25659" CreationDate="2019-01-23T20:38:04.117" Score="3" Body="&lt;p&gt;Collecting bits from sites,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://fleetsubmarine.com/battery.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;fleetsubmarines&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;World War II American fleet submarines had two batteries, each&#xA;  composed of 126 cells. By comparison, a 12-volt car battery contains&#xA;  only 6 cells, each producing about 2.25 volts when fully charged, with&#xA;  a maximum power output of about 45-50 amps. Each cell in a submarine&#xA;  battery produces from 1.06 volts when fully discharged, to 2.75 volts&#xA;  at the optimum output, so connecting the 126 cells in each battery in&#xA;  series gives a usable output of from about 210 to 350 volts, and a&#xA;  power output of as much as 15,000 amps with both batteries connected&#xA;  in parallel.&#xA;  (no mention of total amp-hrs) &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.quora.com/How-long-does-a-diesel-submarine-have-to-run-its-engines-to-fully-charge-its-batteries&quot; rel=&quot;nofollow noreferrer&quot;&gt;quora&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My submarines (Oberon class of the 1960s-1990s) had two lead acid&#xA;  batteries containing 224 cells each with a nominal voltage of 440&#xA;  volts.. The cells were rated 74.20 ampere-hours at a 5 hour rate&#xA;  (nominal voltage of each cell was 2.2 V)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;448*74*2.2 = 73kWh &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://uboat.net/articles/id/54&quot; rel=&quot;nofollow noreferrer&quot;&gt;uboat.net&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The US Navy &quot;Balao&quot; type submarine (1944/45) was fitted with 4 four&#xA;  Elliot Main (Electric Motors) two on each shaft, with a total&#xA;  horsepower of 2,740. While submerged, these motors were powered by two&#xA;  massive (each cell weighing 1650#) 126-cell batteries (in series)&#xA;  capable of delivering 5,320 Amp/Hrs each.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Assuming they meant Amp-hrs, and guessing 2.2 V per cell, 2.2 * 2*5320 = 23.4 kWh&lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2019-01-23T20:38:04.117" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25661" PostTypeId="1" AcceptedAnswerId="25686" CreationDate="2019-01-23T23:24:22.030" Score="0" ViewCount="247" Body="&lt;p&gt;Let's say I have a 300 rpm motor and a roller connected to it's shaft and I would like to use it for pulling fabric from a printer-kind machine. But think that printer's printing speed is changing according to printed surface and I want this motor's shaft to spin freely when opposite load occurs on this roller by fabric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically, if printer prints 100rpm long fabric, 200rpm of motor will spin freely as printer pull other side of fabric. When printer prints another length of fabric, motor will roll again. Roller (or idler?) should be able to spin asynchronously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know much about bearings and mechanical parts so my question might be weird, but I heard some kind of bearing can solve my problem. I tried to explain but sorry if it's nonsense. Thank you for your help in advance.&lt;/p&gt;&#xA;" OwnerUserId="19082" LastEditorUserId="19082" LastEditDate="2019-01-23T23:31:01.207" LastActivityDate="2019-01-26T10:20:45.837" Title="What kind of bearing should be used for free rotation under load?" Tags="|mechanical-engineering|motors|bearings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25662" PostTypeId="1" CreationDate="2019-01-24T00:36:47.160" Score="1" ViewCount="38" Body="&lt;p&gt;Long story short I'm looking to build an ultrasonic sensor that can also be used to communicate with a seperate sensor. (So it must be able to send out a pitch around ~42kHz, but also alter the pitch so that a second sensor could listen to the output and determine an FSK signal, so I could change the output signal to 43kHz if I wanted). Optimally it'd have an operating range of around 10kHz or more (so 38kHz to 48kHz), but I'm not especially set on an particular range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I was thinking I'd just buy a piezo electric speaker such as &lt;a href=&quot;http://122.55.220.155/piezo/files/KSN-1076A-Datasheet.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;, but I'm unable to find a microphone that can listen to a 42kHz signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next I was thinking I could just buy a prebuilt ultrasonic sensor, but the only ones I can find don't allow you to modulate the output frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know things like this already exist as the ones on cars can modulate the output frequency, however I'm unsure what the best route for me would be. (I haven't had tremendous success finding an OEM sensor that I could modulate from a car but I'm open to that possibility).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="19083" LastActivityDate="2019-01-24T00:36:47.160" Title="What's the best way to get an ultrasonic sensor that can vary its operating frequency" Tags="|wireless-communication|audio-engineering|frequency-response|ultrasound|communication|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25663" PostTypeId="1" CreationDate="2019-01-24T03:31:43.550" Score="0" ViewCount="76" Body="&lt;p&gt;Knit command on SW 2017 results blue sketch lines even with all surfaces selected &lt;a href=&quot;https://i.stack.imgur.com/a0Pqm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a0Pqm.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I was following the steps in a tutorial and I am sure i got them all 100% correct up to that point.&lt;/p&gt;&#xA;&lt;p&gt;what could be the reason?&lt;/p&gt;&#xA;" OwnerUserId="13823" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2021-02-13T00:06:05.613" Title="After I knit the surface some blue sketch lines presist" Tags="|cad|solidworks|surface-modelling|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25665" PostTypeId="1" CreationDate="2019-01-24T11:09:20.320" Score="3" ViewCount="86" Body="&lt;p&gt;I am trying to relate the degree of corrosion with the pressure exerted in surrounding concrete. My common sense tells me that higher the degree of corrosion, higher should be the pressure exerted, but my equation tells me it is constant. Obviously, I am making a calculation mistake but could not figure it out. What mistake am I making (figure attached)?&#xA;Consider unit length along the rebar:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S3uHb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S3uHb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Area after corrosion is &lt;span class=&quot;math-container&quot;&gt;$A_1=A_0(1-p)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the percentage of corrosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The volume occupied by rust if there was sufficient space is given by converting the volume of steel to equivalent volume or rust: &lt;span class=&quot;math-container&quot;&gt;$A_2=A_1+pA_0\dfrac{\rho_s}{\rho_c}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Initial volume: &lt;span class=&quot;math-container&quot;&gt;$V_1=A_2-A_1$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Final volume: &lt;span class=&quot;math-container&quot;&gt;$V_2=A_0-A_1$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Change in volume: &lt;span class=&quot;math-container&quot;&gt;$\Delta V=V_1-V_2$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;strain: &lt;span class=&quot;math-container&quot;&gt;$\epsilon=\dfrac{\Delta V}{V_1}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Solving, we get &lt;span class=&quot;math-container&quot;&gt;$\epsilon=\dfrac{-\rho_c + \rho_s}{\rho_s}$&lt;/span&gt;. There is no &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="19087" LastEditorUserId="1832" LastEditDate="2019-01-24T17:25:48.963" LastActivityDate="2019-04-15T02:21:33.103" Title="Does the pressure due to corrosion in concrete depends on degree of corrosion?" Tags="|civil-engineering|corrosion|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25666" PostTypeId="2" ParentId="25625" CreationDate="2019-01-24T11:17:04.650" Score="0" Body="&lt;p&gt;Theoretically, impact energy will partially be absorbed by deformation of the rock and partially by deformation of the structure. However, the deformation of the rock will be very small, so just assume all energy will be absorbed by the structure. And since you don't know the exact shape of the rock, it is not practical to do any calculations on it anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will want to calculate the contact force as a function of the deformation of the structure, &lt;span class=&quot;math-container&quot;&gt;$F(u)$&lt;/span&gt;. I recommend for an initial approximation to assume either linear-elastic behaviour or perfectly plastic behaviour depending on the stiffness of the component the rock hits. The energy absorbed by the structure is the work of the contact force, so you solve &lt;span class=&quot;math-container&quot;&gt;$\int_0^{u_{max}} F(u)du = mgh$&lt;/span&gt; for the deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give you an equivalent static impact force and a deformation to check the structure for. Just remember that the load on the foundation varies a lot depending on whether the rock hits a post or a cable midspan between posts.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2019-01-24T11:17:04.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25667" PostTypeId="2" ParentId="25665" CreationDate="2019-01-24T13:40:31.967" Score="2" Body="&lt;p&gt;I'd say the mistake is assuming infinite stiffness for concrete and steel so only the rust has a strain. In reality, rust is probably the stiffest material involved, and you will get a better approximation by assuming infinite stiffness for rust and steel and calculating the strain and stress in concrete.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2019-01-24T13:40:31.967" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25669" PostTypeId="1" AcceptedAnswerId="25674" CreationDate="2019-01-24T15:45:22.050" Score="1" ViewCount="7614" Body="&lt;p&gt;I'm really confused. I've only ever used metric in my life. Recently I'm trying to build a IOT device and there is plumbing and water involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking for a 19mm fitting, but on Amazon and other sites it's always advertised in inches. &lt;a href=&quot;https://www.amazon.co.uk/gp/product/B0087OSUY4/ref=ppx_yo_dt_b_asin_title_o00__o00_s00?ie=UTF8&amp;amp;psc=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;John Guest 3/4&quot; BSP x 1/4&quot; Push Fit Tap Adapter&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did the conversion and 3/4 inch, according to my calculation, should be around 19mm. But whenever I order these fittings they always send me a larger size. The resulting size is always around 25mm when I measure it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me it seems like 3/4 inch doesn't actually mean 3 quarters of an inch?&lt;/p&gt;&#xA;" OwnerUserId="19091" LastEditorUserId="33" LastEditDate="2019-01-25T13:57:40.380" LastActivityDate="2019-01-25T13:57:40.380" Title="Plumbing Imperial units confusion - 3/4&quot; is supposed to be around 19mm, right?" Tags="|measurements|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25670" PostTypeId="2" ParentId="25653" CreationDate="2019-01-24T15:51:55.897" Score="1" Body="&lt;p&gt;The efficiency is a dimensionless quantity and is calculated by power output devided by power input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope you have the value of your power input? But you can also calculate it with the following formula. The left one is for your (electro-)motor which is convertible in your hydraulic pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_{mechanical}=M[Nm]*2*n[1/sec]*\pi=p[bar]*Q[l/min]=P_{hydraulic}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three (main-)possibilities to get some energy loss in your system:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the mechanical one caused by friction in your pump, motor, valves … - &lt;span class=&quot;math-container&quot;&gt;$P_{loss-mechanical} [W]$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;the pressure one caused by perfusion-resistance of the elements in your hydraulic system, e.g. in your piping – &lt;span class=&quot;math-container&quot;&gt;$P_{loss-\Delta p} [W]$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;the volumetric one caused by leakage in your hydraulic system – &lt;span class=&quot;math-container&quot;&gt;$P_{loss-volumetric} [W]$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The volumetric loss is your issue - mentioned by the two leakage (internal and external). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_{l- vol} [W]= \Sigma (p[bar]*Q[l/min])=10^5*40bar*1/60000*0,5l/min + 10^5*100bar*1/60000*3l/min=533,33 W$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the &lt;span class=&quot;math-container&quot;&gt;$10^5$&lt;/span&gt; and the &lt;span class=&quot;math-container&quot;&gt;$1/60000$&lt;/span&gt; are values because of the different units…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the equation for efficiency:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$n= P_{output}/P_{input}=(P_{input}-\Sigma(P_{loss}))/P_{input}=(P_{input}-P_{l- vol})/P_{input}=(P_{input}-533,33 W)/P_{input}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="17780" LastActivityDate="2019-01-24T15:51:55.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25671" PostTypeId="2" ParentId="25669" CreationDate="2019-01-24T15:58:02.310" Score="0" Body="&lt;p&gt;you are right 3/4&quot; means &lt;span class=&quot;math-container&quot;&gt;$ 25.4mm/inch* \frac{3}{4}= 19.05mm $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is any mismatch, it is there fault.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-24T15:58:02.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25672" PostTypeId="2" ParentId="25669" CreationDate="2019-01-24T16:18:17.733" Score="0" Body="&lt;p&gt;Plumbing fittings from good plumbing suppliers are available in metric and imperial sizes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to use either imperial or metric all the way through, but there are some copper fittings that are available that are conversion fittings that will convert from imperial to metric...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These fittings are usually compression or solder fittings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, yes, I have faced the same issue on my parent's house with old imperial sized pipes and I now keep the conversion fittings to convert to the metric sizes ie 1/2&quot; pipe is now 15mm etc.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-01-24T16:38:56.783" LastActivityDate="2019-01-24T16:38:56.783" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25673" PostTypeId="2" ParentId="25669" CreationDate="2019-01-24T17:37:40.753" Score="3" Body="&lt;p&gt;Yes it is BUT BSP (and pipe fittings in general)  fittings are specified by the nominal bore rather than the thread diameter, which can be confusing if you are used to dealing with metric fasteners. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2019-01-24T19:21:08.357" LastActivityDate="2019-01-24T19:21:08.357" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25674" PostTypeId="2" ParentId="25669" CreationDate="2019-01-24T18:17:12.157" Score="3" Body="&lt;p&gt;Pipes and pipe fittings in the US are mostly NPT (National Pipe Thread (or Tapered, depending upon who you ask). (Check out the &lt;a href=&quot;https://en.wikipedia.org/wiki/National_pipe_thread&quot; rel=&quot;nofollow noreferrer&quot;&gt;NPT Wikipedia article&lt;/a&gt;.)  A 3/4 inch NPT fitting is not 3/4 inch anywhere, except in name. The inside diameter of a 3/4 inch nominal pipe is slightly more than 3/4 inch but if you want to drill to tap a 3/4 inch NPT, you will need to drill a hole 59/64 in diameter, which sounds like the one inch you were seeing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that is not what you were asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you are looking at Amazon, UK at a BSP, British Standard Pipe, link, I expect you are not US based. BSP is another standard, incompatible both with NTP and Metric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I suggest is going down to your local hardware store and seeing what they have most of and going with that.  If you need anything subtle, you will need to go to a plumbing supply house or Amazon.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2019-01-24T18:17:12.157" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25676" PostTypeId="2" ParentId="25669" CreationDate="2019-01-24T20:24:52.540" Score="0" Body="&lt;p&gt;No, as noted by@ user 1683793 , 3/4 inch is the NAME of a size of ASTM pipe ( not the size) , very close to API pipe. Any 3/4 pipe will thread to any other whether is is Schedule 10 or Sch 120 , stainless or copper, etc. ; only the wall thickness and inner diameter change. Most often Sch 40 is used for domestic service. And tubing has nothing to do with pipe other than it may look similar. I am not familiar with pipe threaded to ISO,BS, DIN, JIS, GOST, etc. Although I suspect CSA ( Canadian) is very close to ASTM. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-01-25T00:56:52.480" LastActivityDate="2019-01-25T00:56:52.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25677" PostTypeId="2" ParentId="25663" CreationDate="2019-01-24T21:16:33.270" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zDmsF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zDmsF.png&quot; alt=&quot;gaps not auto knitted&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your gaps are larger than your knitting tolerance, then they will not automatically be closed. Pull the sliders to the minimum and maximum respectively to ensure that all gaps are shown, and either increase your tolerance or manually select the larger gaps that you would like to knit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This screenshot is not taken from the same model as you - I just spent a long time trying to force some significant gaps - not easy to have an error when you're trying to get one!!&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-01-24T21:16:33.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25679" PostTypeId="1" CreationDate="2019-01-25T08:59:55.260" Score="0" ViewCount="47" Body="&lt;p&gt;Consider a simple example:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;there's a vessel with water and a continuos level sensor (4-20 mW signal)&lt;/li&gt;&#xA;&lt;li&gt;there's a water line with a magnetic valve, feeding the vessel&lt;/li&gt;&#xA;&lt;li&gt;When the water level is below a certain treshold, the valve opens - this is realized via a PLC unit somewhere&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now, consider the same scenario, the water level is critical for human safety so the whole refilling function needs to be SIL II. Now I - not an electrical or control systems engineer - need to understand how much more complications, cost, etc. will arise due to the SIL II requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;p.s. To give you an idea how much or little I know: My understanding is that I either need a safety PLC or a SIL II certified measurement amplifier with an additional potential free contact set to the correct level, hardwired to the relais controlling the valve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since SIL are describesd in an international IEC norm, I assume the answer will not be highly dependent on local codes.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-07-04T02:18:02.350" Title="What is the actual difference between a control function with Securtiy Integritiy Level (SIL) II and one without?" Tags="|control-engineering|safety|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25680" PostTypeId="1" CreationDate="2019-01-25T14:48:56.660" Score="1" ViewCount="114" Body="&lt;p&gt;I've a Rigol DS2072A. It works great except for an offset problem that I noticed the past few days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Channel 2 always shows a DC offset. This is independent of any settings such as channel coupling, internal termination resistance or bandwidth limiting. The input connector only has a 50 Ohm terminator connected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a constant height above 'zero' on the screen, independent of vertical scale. This means that for higher ranges, the scope calculates that there is more offset. (e.g. 80mV@1V/div vs. 8V@100V/div).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I noticed that the offset lowers when the scope warms up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only for the lowest scales (in the millivolts range) the offset changes with the scale setting, but there is no sound of relays for these scales, so it's probably software scaling what we're looking at in these regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attached a picture where you can clearly see the offset:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OFMfl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OFMfl.png&quot; alt=&quot;Channel 2 offset&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running any calibration commands doesn't change anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way I can fix this? Or does this indicate that the scope is broken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a case of 'you get what you pay for' or just bad luck?&lt;/p&gt;&#xA;" OwnerUserId="17095" LastActivityDate="2019-01-25T14:48:56.660" Title="Strange offset on oscilloscope channel" Tags="|measurements|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25681" PostTypeId="1" AcceptedAnswerId="25682" CreationDate="2019-01-25T14:53:14.510" Score="0" ViewCount="130" Body="&lt;p&gt;In this flow control simple experiment in the lab, i am made to control the flow through the pipe and at the same time record down the output given by the sensor in voltage. I am told to record down values for increasing output and decreasing output. &#xA;The lowest flow is 0cc/min, and highest flow is 3000cc/min. Just wondering why must i take datas for both increasing and decreasing output. Is this so that i can determine the recommended operating range of the sensor? which will be the min. value to the max value of the both sets of outputs (increasing and decreasing) ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Objective of this experiment is by the way to investigate the linearity and hysteresis of the flow sensor. &lt;/p&gt;&#xA;" OwnerUserId="15037" LastActivityDate="2019-01-26T12:07:45.907" Title="Flow control experiment" Tags="|flow-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25682" PostTypeId="2" ParentId="25681" CreationDate="2019-01-25T15:42:35.723" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The Objective of this experiment is by the way to investigate the&#xA;  linearity and hysteresis of the flow sensor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The key word here is &lt;strong&gt;hysteresis&lt;/strong&gt;: the output of the sensor at the same flow rate might not the same depending on whether you are in the increasing or decreasing output direction. This is why you are being asked to test in both directions. Check out &lt;a href=&quot;https://en.wikipedia.org/wiki/Hysteresis&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Hysteresis&lt;/a&gt; for more details on hysteresis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a typical characteristic curve with hysteresis (albeit from a different domain):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GnKvh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GnKvh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also ask about &lt;strong&gt;linearity&lt;/strong&gt;: it's pretty basic but it means that if you plot your sensor output vs. flow rate, you basically get a straight line if the sensor is linear, or a curved line if not:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z5eEa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z5eEa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sensor is linear but has hysteresis, then you'll essentially get 2 parallel lines. If it's linear but without any hysteresis, both lines will be on top of each other and you'll get a unique characteristic regardless of which direction you go.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2019-01-26T12:07:45.907" LastActivityDate="2019-01-26T12:07:45.907" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25686" PostTypeId="2" ParentId="25661" CreationDate="2019-01-26T10:20:45.837" Score="1" Body="&lt;p&gt;Ok, so what you are searching for is a torque limiter (slip type) which only transmits torques below a certain threshold (be careful here, since you need to allow some torques to accelerate your shaft). Slip type means that it automatically continues transmitting torque if the torque is below a certain threshold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most simple type are so called friction plates which are literally just two plates which are preloaded to each other and attached to the shaft on both sides. If too much torque is acting on it, frictional forces will be lower than transmitted force and the rotate relatively to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beside, keep in mind that the mass of the roller is quite important!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Torque_limiter&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.motioncontroltips.com/torque-limiter/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Motioncontroltips&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13505" LastActivityDate="2019-01-26T10:20:45.837" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25688" PostTypeId="2" ParentId="25560" CreationDate="2019-01-26T13:59:52.463" Score="1" Body="&lt;p&gt;I think that the main reason is indeed for the weights of the trains, locomotives and carriages alike. Of course, one could also make standard gauge rolling stock lighter, but there is some savings already for everything being smaller (not just narrower, but scaled down to keep its proportions) and so it goes from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a way, &quot;narrow gauge&quot; at that time was also synonymous with smaller, lighter, slower, and cheaper, which lead to rolling stock being adapted to this. So while one could of course build luxurious long-distance carriages in that size, &quot;narrow-gauge&quot; implied &quot;local&quot; and therefore attracted those who were making things more affordable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today, we see a similar trend with so-called &quot;light-rail&quot; vehicles. The main difference to &quot;heavy rail&quot; is that light rail doesn't need physical collision protection with heavy trains, since they don't share track with anything other than light-rail itself. (I am not going into exceptions from this.) But going from there, light-rail means using trams that are not just lighter, but also cheaper to build and operate, need much less in terms of stations and platforms, and are of course slower. No point paying and maintaining a fast vehicle when it stops at every village on the way. &lt;/p&gt;&#xA;" OwnerUserId="19115" LastActivityDate="2019-01-26T13:59:52.463" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25689" PostTypeId="2" ParentId="17292" CreationDate="2019-01-26T14:32:25.030" Score="0" Body="&lt;p&gt;I don't have time right now to research examples, but I have heard of many subway systems that have much better brakes than the cheap ones on freight trains. Since subways brake every few minutes to a full stop, that's a few hundred times per day and means that there is also a lot more maintenance to do. So separate disk brake disks are often used instead of brake shoes on the wheels like on freight trains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can also testify that acceleration and deceleration, for example on the Frankfurt U-Bahn is stronger than on other systems I have used. (Unfortunately, there are a few years between me in Frankfurt/Germany and me in San Francisco/USA, so I can't compare with DART.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for DART's headways, besides signalling and brakes, there might also changes to tracks be needed to withstand higher lateral forces. And as @agentp points out in his answers, braking of train cars has to be coordinated, so the train needs the required electronics to do that. Add in some other small things and it becomes so complex that you better get new trains. So smaller headways on subways (which unlike freight cars should be build with this in mind) are definitely possible, but adding them to an existing system is a major engineering project that requires long-term planning. (Munich S-Bahn did something like that successfully.)&lt;/p&gt;&#xA;" OwnerUserId="19115" LastActivityDate="2019-01-26T14:32:25.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25690" PostTypeId="1" CreationDate="2019-01-26T23:17:14.077" Score="0" ViewCount="1628" Body="&lt;p&gt;I got a terrible &lt;strong&gt;error&lt;/strong&gt; in the &lt;strong&gt;Ansys&lt;/strong&gt; when I try to create a &lt;strong&gt;new project&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't be able to open the geometry. Every time i open it i get the error &quot;&lt;strong&gt;Geometry editor was closed abnormally&lt;/strong&gt;&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked for the solution but didn't get the proper answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I useing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Windows 10;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ansys 19.2;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;notebook Specifications:&#xA;Lenovo ideapad 320&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intel Pentium 4415U (2.3 ГГц)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RAM 4 ГБ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;nVidia GeForce 920МХ &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xu3yX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xu3yX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19119" LastActivityDate="2019-01-26T23:17:14.077" Title="Geometry editor was closed abnormally" Tags="|ansys|ansys-workbench|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25691" PostTypeId="2" ParentId="17292" CreationDate="2019-01-27T00:56:32.663" Score="0" Body="&lt;p&gt;The complexity of the system is directly related to the length of the train. Any differential in the applied rate of deceleration among the cars accumulates in the couplers. The cars don't just have to apply the same braking effort, they have to actually each decelerate identically. This would require very intense load monitoring at the couplers and very fine and rapid response to a controller that can handle all varieties of cars and loads. And the system would have to be maintainable by the existing maintenance infrastucture. Have you ever been to a transit system maintenance yard? Do you think they will take a train out of service if one car has a brake fault?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An analogy would be water hammer. Try stopping the flow in a mile long pipe by shutting a valve quickly and see what happens.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-01-27T00:56:32.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25692" PostTypeId="1" AcceptedAnswerId="26230" CreationDate="2019-01-27T01:20:35.037" Score="2" ViewCount="119" Body="&lt;p&gt;Would a passive and controlled amount of electricity fed through post-tension cables, heat up a slab, without ruining the dependability and health of the post-tension cables, and the concrete itself?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering because I coat concrete slabs with different types of paint, and in the winter, it is super slow to add layers of coats on top of one another because of the waiting time for drying and curing of the paint. If I could heat up the slab through the post tension, then re-grout the post tension ends, it would totally change the game for me. I am just afraid of possible snapping the cables. In the winter I can get away with only one coat for the whole day. In the summer I can put down 5 coats of paint over a 7,200 square foot area with ease.&lt;/p&gt;&#xA;" OwnerUserId="10348" LastEditorUserId="1832" LastEditDate="2019-02-26T15:51:31.570" LastActivityDate="2019-03-03T20:06:50.027" Title="Using electricity to heat up a slab through post-tension cables?" Tags="|reinforced-concrete|prestressed-concrete|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25694" PostTypeId="2" ParentId="25692" CreationDate="2019-01-27T06:44:00.593" Score="1" Body="&lt;p&gt;Yes, theoretically...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But electric underfloor heating is done with thin wire to limit the current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concrete rebar is 1/2&quot; or 3/4&quot; in diameter so will need a lot of current - have you considered a source for that... Also the connections with the rebar were not designed for electricity and you may find bars that you only get to one end.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-01-27T06:44:00.593" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25696" PostTypeId="2" ParentId="25536" CreationDate="2019-01-27T12:26:48.900" Score="1" Body="&lt;p&gt;Parameter fitting is parameter fitting. What you need is experimental data (often hard to come by), a model (NRTL), an algorithm (method to calculate LLE) and a fitting routine (I recommend something like the Nelder-Mead). If you are using something like &lt;code&gt;Matlab&lt;/code&gt;, you can just use &lt;code&gt;fmincon&lt;/code&gt; or one of its many solvers to minimize the error between predicted and experimental by changing the parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You guess initial parameters, use your algorithm to predict LLE, compare the compositions in both phases with the corresponding experimental data, calculate the sum of the errors squared, and then use your fitting routine to calculate new parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process simulator &lt;code&gt;VMGSim&lt;/code&gt; is capable of doing this, but it was not openly documented at least a few years ago. I am sure &lt;code&gt;Aspen/HYSIS&lt;/code&gt; can do this as well. You can always use &lt;code&gt;Matlab&lt;/code&gt; or another program too, the solvers in &lt;code&gt;Matlab&lt;/code&gt; or any other package will be good enough.&lt;/p&gt;&#xA;" OwnerUserId="19124" LastEditorUserId="19124" LastEditDate="2019-03-02T03:41:39.457" LastActivityDate="2019-03-02T03:41:39.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25697" PostTypeId="2" ParentId="25274" CreationDate="2019-01-27T12:39:25.783" Score="1" Body="&lt;p&gt;Molecules favor the phase that requires the least amount of work to be in. When I say work, I mean the chemical potential. At equilibrium the chemical potential for a species is the same in both phases. As you add heat to the liquid you change the energy of the system, sure, but the real phenomena is that you change how much work is required to keep a molecule in the liquid phase. The vapor phase in this example is staying constant. If your heat addition/removal increases the amount of work required to insert a molecule into the liquid, meaning it becomes easier to insert a molecule into the vapor, then that is where the molecules will shift, to the easier phase to exist in. They will shift until the chemical potentials are the same in both phases. Boiling is simply a point where molecules rush from one phase to another rather than slowly migrating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At all times, there are molecules going back and forth between both phases,at equilibrium they are the same rates, at boiling there is more going to the vapor because the work required for a molecule to exist is the vapor is lower (constant in this case) and the work required for a molecule to exist in the steadily heating (boiling) liquid is increasing (high).&lt;/p&gt;&#xA;" OwnerUserId="19124" LastActivityDate="2019-01-27T12:39:25.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25698" PostTypeId="1" CreationDate="2019-01-27T17:51:36.733" Score="0" ViewCount="66" Body="&lt;p&gt;Is there a relatively inexpensive (let's say under $500US) crack-detection device and related mobile app that would be easy enough for a non-engineer/non-metallurgist to use to determine whether a bicycle frame was forming cracks that would be undetectable to the human eye?&lt;/p&gt;&#xA;" OwnerUserId="21526" LastActivityDate="2020-12-22T22:12:36.470" Title="small relatively inexpensive crack detection apparatus for bike shops" Tags="|aluminum|welding|heat-treatment|ultrasound|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25699" PostTypeId="1" CreationDate="2019-01-27T17:58:29.823" Score="1" ViewCount="406" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/m1qFj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m1qFj.png&quot; alt=&quot;The question and Image are shown below:&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I tried&lt;/strong&gt;: Using the conjugate beam method to find deflection and hence stiffness of the beam, but I am unable to find reactions in the conjugate beam.&lt;/p&gt;&#xA;" OwnerUserId="19130" LastActivityDate="2019-01-29T16:58:38.547" Title="Calculation of natural frequency in a hinged beam" Tags="|structural-analysis|beam|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25701" PostTypeId="1" CreationDate="2019-01-27T19:01:23.510" Score="0" ViewCount="85" Body="&lt;p&gt;The component I am interested in is the blue &quot;head&quot; in the attached image. I call it the sensor head but will like to know if it has a specific industry name&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9lBuQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9lBuQ.jpg&quot; alt=&quot;enter image desciption here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19131" LastEditorUserId="110" LastEditDate="2019-01-28T00:52:23.733" LastActivityDate="2019-05-08T10:15:33.880" Title="What is the name and purpose of this Instrumentation / Mechanical fitting" Tags="|sensors|mechanical|electrical|instrumentation|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25702" PostTypeId="2" ParentId="25698" CreationDate="2019-01-27T20:16:15.970" Score="2" Body="&lt;p&gt;Solar Mike is exactly right, the method he cites is called &lt;em&gt;dye penetrant&lt;/em&gt; and it works for ferrous metals, aluminum alloys, and composites as well. all you need is a bottle of the fluorescent dye, the spray-on developer powder, and a UV light. This will cost less than $100. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-01-27T20:16:15.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25703" PostTypeId="2" ParentId="25699" CreationDate="2019-01-27T20:48:18.190" Score="1" Body="&lt;p&gt;Let us ignore the mass of the beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then by inspection of symmetry and the rigidity of the two beams, we note that the deflection of beam under mass m, Yb is equal to deflection of the point D, Yd, because when the beam vibrates it will form an isosceles triangle whit the point c vibrating from maximum to minimum and points A and E stationary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we stablish &lt;span class=&quot;math-container&quot;&gt;$ \quad Y_{B} = Y{_ d} \quad $&lt;/span&gt; meaning the deflection of spring is exactly as if it were under mass m, so the natural frequency is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F = m*\alpha = k*y $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \omega_{ n} = \sqrt{\frac{ k}{m}} $&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-01-29T16:58:38.547" LastActivityDate="2019-01-29T16:58:38.547" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25704" PostTypeId="2" ParentId="25698" CreationDate="2019-01-27T21:58:59.890" Score="2" Body="&lt;p&gt;There are three dye-pene types. The cheapest one is likely good enough ;1- A dye is applied ( usually red), 2- allow to soak ,3 - clean off surface dye( May use a cleaner for this ), 4- spray on a white developer: Depending on experience ,very small cracks can be found in any material. The second type of dye-pene uses a fluorescent dye and UV light ; faster than the red dye but requires a dark room and UV light. The third type uses magnetic fluorescent dye , a magnet and UV light but only works with steel . Cracks revealed by the red dye viewed days later. Also , I think red dye is the best method to reveal porosity or other non-linear imperfections . Also best if results need to be viewed by several people to make a judgement. Not likely to be your problem , but I remember well several people tying to squeeze under the curtain to view the UV light findings.  The red dye only requires three aerosol cans ( cleaner, dye, and developer) but I have no idea what the price is today....PS - The method I call &quot;magnetic fluorescent dye&quot; is commonly called &quot;wet mag&quot;.   &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-01-28T16:16:47.600" LastActivityDate="2019-01-28T16:16:47.600" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25706" PostTypeId="1" CreationDate="2019-01-28T00:18:42.103" Score="0" ViewCount="47" Body="&lt;p&gt;using the offset tool on surfaces can result a shape different from the one offset from, as in broken lines or edges instead of curves like this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gABmn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gABmn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is there a way to know the maximum distance i can offset to without resulting such deformation other than trial and error especially if i dont have designe constrains? &lt;/p&gt;&#xA;" OwnerUserId="13823" LastActivityDate="2019-01-28T00:18:42.103" Title="How to know the maximum offset distance in solidworks that doesn't result a defromed shape" Tags="|cad|solidworks|autocad|surface-modelling|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25708" PostTypeId="1" CreationDate="2019-01-28T01:35:43.710" Score="1" ViewCount="55" Body="&lt;p&gt;Can boundary layer theory be applied in finite element methods as it is applied in finite volume method? i.e The flow away from the boundary is considered inviscid and solved first, then the flow inside the boundary layer is solved? &lt;/p&gt;&#xA;" OwnerUserId="19137" LastActivityDate="2019-06-27T17:01:45.193" Title="Can boundary layer theory be applied in finite element methods as it is applied in finite volume method?" Tags="|finite-element-method|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25709" PostTypeId="1" CreationDate="2019-01-28T04:31:13.107" Score="0" ViewCount="1157" Body="&lt;p&gt;Consider I have two type of cooking kettles, which the shape is just like a pure tube with the volume is 1000 cc (1 liter) with the radius and the height as below (volume=pi*r^2*h):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The first kettle: cross-section radius=5.6419cm and the height=10cm.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The second kettle: cross-section radius=3.9894cm and the height=20cm.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Those two kettles filled up with the same kind water from a same source. Consider that I have a small stove that that its flame will be enough to boil the water inside those kettles. Also consider that those kettles has enough additional height so the water will not get spilled when it start boiled. Then my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Which one will be get boiled faster (shorter time)?&lt;/li&gt;&#xA;&lt;li&gt;When they are get boiled, which one is healthier (due to it distribute the heat uniformly to that water inside the kettle)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;:&#xA;Water will be filled up from top. Say that there is a small cap to fill water into it. There is no other hole.&#xA;Flame on the stove are put the same distance for both the kettles, close enough so the flame will be effective to heat the water. &lt;/p&gt;&#xA;" OwnerUserId="17943" LastEditorUserId="33" LastEditDate="2019-01-28T20:59:03.930" LastActivityDate="2021-11-19T17:08:01.880" Title="Does kettle shape affect the speed at which water boils?" Tags="|heat-transfer|boilers|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25713" PostTypeId="1" CreationDate="2019-01-28T10:09:15.110" Score="0" ViewCount="1389" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DB7Ul.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DB7Ul.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi all. I am building a machine to deform sheet metal into corrugations. I have attached an image that I got off a presentation from MIT's website. In the example, they provide a formula for the creation of a single indentation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I edited the image to give you an idea of the shape I am going for. It is essentially a &quot;W&quot; shape after pressing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know how to apply the formula to my design. Is it the same amount of force that I would require? Double the force or even triple the force. My gut feel is that it should be somewhere between the same and double the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards&#xA;Stefan&lt;/p&gt;&#xA;" OwnerUserId="19143" LastActivityDate="2024-03-04T05:05:20.640" Title="Force required to deform sheet metal" Tags="|mechanical-engineering|materials|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25714" PostTypeId="2" ParentId="25709" CreationDate="2019-01-28T10:24:52.673" Score="1" Body="&lt;p&gt;The difference between the two containers may be miniscule, but a reference located in the &lt;a href=&quot;https://physics.stackexchange.com/questions/10470/how-does-water-evaporate-if-it-doesnt-boil&quot;&gt;Physics StackExchange&lt;/a&gt; presents the statement that evaporation depends on the surface area, local humidity and air pressure. The latter two factors will be presumed to be identical for your question.&lt;/p&gt;&#xA;&lt;p&gt;Evaporation cools the water, requiring more heat to maintain the water temperature. Allowing that you have sufficient heat to boil the water, this factor can be considered unimportant regarding time-to-boil.&lt;/p&gt;&#xA;&lt;p&gt;The surface area of the larger diameter container will present faster evaporation than that of the smaller one. Evaporation does not require boiling temperature, which means it will begin in the larger container sooner than the smaller container.&lt;/p&gt;&#xA;&lt;p&gt;Your second question may not have a solution, as there will be convection of the water within both containers and heat distribution for either would be about the same and would likely be uniform throughout.&lt;/p&gt;&#xA;&lt;p&gt;The smaller diameter container should boil sooner than the larger one.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="-1" LastEditDate="2021-11-19T17:08:01.880" LastActivityDate="2021-11-19T17:08:01.880" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25716" PostTypeId="2" ParentId="25709" CreationDate="2019-01-28T11:43:37.633" Score="1" Body="&lt;p&gt;Backpacker here. I have some practical experience getting the most out of a bottle of fuel. This really is a matter for experiment. But wide pans do better than tall ones in general. If you have enough of a source to actually boil the water, then maximizing heat transfer into the water is your first concern. So check the temp of the air as it rises around the edge of the pan. You want it protected from drafts and just above the boiling point. Even a fairly small flame will favor the wider pan. it would have to be something tiny like a candle to favor the taller one. Also, I'm assuming a lid is available. Doing this without a lid is basically silly. Lids come with pots, and have done for quite a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the real point of the question may be to challenge you to think of all the factors that could effect the result.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;lid or no lid&lt;/li&gt;&#xA;&lt;li&gt;material and thickness of pans. Note: you have to heat the pan up too.&lt;/li&gt;&#xA;&lt;li&gt;fuel type and combustion gas temps.&lt;/li&gt;&#xA;&lt;li&gt;hob design&lt;/li&gt;&#xA;&lt;li&gt;fuel rate.&lt;/li&gt;&#xA;&lt;li&gt;rate of heat loss from lid or possibly sides.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-01-28T11:43:37.633" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25719" PostTypeId="2" ParentId="25708" CreationDate="2019-01-28T15:41:48.497" Score="-1" Body="&lt;p&gt;It can but you are wasting the full potential of the method when doing it so. I recommend you look into &quot;Applied partial differential equations&quot; by Trim. You can also check a few articles in perturbation theory which solves part of the boundary layer. Doing finite element is just useful when you are programming it. &lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2019-01-28T15:41:48.497" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25720" PostTypeId="1" CreationDate="2019-01-28T19:11:10.403" Score="-1" ViewCount="89" Body="&lt;p&gt;Usually I see composite as a laminate wrapped around the core of a sandwich structure. Usually in loading such as bending the bending stress is the highest furthest away from the neutral axis, while usually the maximum transverse shear appears at the outer surface of a structure. I believe this is why laminate is orientated differently as it's orthotropic  + wrapped around the outer surface of the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also seen people making sandwich core e.g. 2d extruded lattice, by using composite. I suppose they are the same as the conventional laminate in terms of manufacturing and material properties (layers of lamina), the only difference is probably that in order to take the shape of a 2d extruded lattice, it's cut from layers of rectangular lamina? Why would people do this though, are they better than traditional material? Is it because people make them quasi-isotropic?    &lt;/p&gt;&#xA;" OwnerUserId="18873" LastEditorUserId="12827" LastEditDate="2019-07-26T18:24:31.690" LastActivityDate="2019-08-25T19:02:15.283" Title="Why do people build the core of a sandwich structure using composite" Tags="|materials|composite|solid-mechanics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25722" PostTypeId="2" ParentId="25713" CreationDate="2019-01-28T19:26:41.367" Score="0" Body="&lt;p&gt;The plastic section modulus for the plate is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$S = \frac {bh^2}{6} = \frac {LT^2}{6} \  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;W defines the narrowness of the grove, the narrower W, the bigger F is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I your case because you have two groves next to each other with no flat band in between there is tension in the two groves by the lateral dragging of the sheet into the jig.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were to estimate force needed to bend only the two indentations it would be just 2F. But because there is some extra  force spent on dragging the sheet  up the inclined walls of the center ridge, it's sate to allow 2.5 F.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-01-29T06:14:43.277" LastActivityDate="2019-01-29T06:14:43.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25723" PostTypeId="1" AcceptedAnswerId="25726" CreationDate="2019-01-28T19:45:39.140" Score="0" ViewCount="87" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Don't try this at home. I am taking personal responsibility for what happens in my car.&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi all,&#xA;I recently purchased a fire extinguisher for my car (it weighs 1.5 lbs or 0.68 kg). I browsed the internet for a decent bracket to mount it inside my car, under the front passenger seat. Everything was fairly expensive which made me want to make a DIY version that would be cheaper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been trying to find what sort of materials I can/should use. The last thing I want is the bracket to break and have a 1.5 lbs metallic projectile flying inside the car in the event of a crash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bracket should be 16&quot; or ~40 cm wide so that I can use the bolts on which the passenger seat is mounted to attach the bracket. (See picture below)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rPUZ6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rPUZ6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously I could use some thick steel and be safe but I don't really have easy access to tools to cut metals so I would like to try something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My requirement for the bracket is that it should be able to hold on to the fire extinguisher in the event of a frontal collision on the freeway. Which would be an impact of, if my memory serves me right, of &lt;/p&gt;&#xA;&lt;code&gt;65 mph + 65 mph = 130 mph or ~210 kmh&lt;/code&gt;&lt;/p&gt;&#xA;If my math is correct this would be equal to &lt;/p&gt;&#xA;&lt;code&gt;210000 * 0.68 = 3966 N&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that I'm not sure where to go...&#xA;If I was considering using a PVC pipe, would I be able to relate those Newtons to the tensile strength of the pipe to figure out if it would hold up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had also thought about 3D printing the bracket but a friend pointed out that it's hard to make measurements with PLA because the printer layers the material so the &quot;resistance&quot; or &quot;strength&quot; is different in different directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also open to other material suggestions!&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerDisplayName="user19148" LastEditorDisplayName="user19148" LastEditDate="2019-01-28T19:55:26.633" LastActivityDate="2019-01-28T21:40:57.590" Title="Material for car mounted fire extinguisher bracket" Tags="|structural-engineering|materials|automotive-engineering|structural-analysis|strength|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25724" PostTypeId="2" ParentId="25723" CreationDate="2019-01-28T20:11:44.027" Score="0" Body="&lt;p&gt;A couple thoughts here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost nothing will hold the bracket (and extinguisher) in case of a hard crash unless the fire extinguisher &amp;amp; bracket are mounted flat to a surface parallel to the firewall, i.e. perpendicular to front-back axis.  The last thing you want is a projectile, so start with this requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, a reasonably thick (say 8 mm)  3D-printed bracket made from ABS (or nylon if you have a printer that can handle that temperature) is the quickest and easiest way to go. Yes, the strength depends on layer orientation, but if you print it 90 degrees from  the installation orientation, so that the layers are in the X-Z plane when installed,  that should suffice.  IANAME (mech Eng) so do not treat this as a guarantee. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that I think a bit, I would recommend nylon strapping with quick-release buckles. That's cheap, incredibly strong, and takes no time at all to install. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2019-01-28T20:11:44.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25725" PostTypeId="2" ParentId="25709" CreationDate="2019-01-28T20:16:38.040" Score="0" Body="&lt;p&gt;You have left out several things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the flame as wide as the larger diameter kettle?  Are there covers?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the flame is as wide as the larger, then you will lose a lot of its heat when using the smaller kettle, because it'll just create a hot air convection path outside the kettle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so on.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you need boiling everywhere (&quot;rolling boil&quot;) or just at the bottom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you mean by &quot;healthy&quot; ?   Just bringing to onset of boiling will suffice if you are worried about things that die instantly at 90 - 100 \degree C , but not if you need to kill things (which exist in nature) that require a few minutes in that temperature range. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2019-01-28T20:16:38.040" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25726" PostTypeId="2" ParentId="25723" CreationDate="2019-01-28T21:12:07.823" Score="2" Body="&lt;p&gt;I don't know what you thought you were doing multiplying &lt;span class=&quot;math-container&quot;&gt;$210000 \times 0.68$&lt;/span&gt; but it certainly doesn't give a force in Newtons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3503406/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Measured crash data&lt;/a&gt; shows the maximum deceleration is of the order of &lt;span class=&quot;math-container&quot;&gt;$50g$&lt;/span&gt;. In other words your 1.5lb object has a force of &lt;span class=&quot;math-container&quot;&gt;$1.5\times50 = 75$&lt;/span&gt; lb acting on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to restrain it is to realise that in a frontal crash it will be trying to accelerate forwards, relative to the vehicle. Therefore if you mount it on something where it can't move forwards, such as the front of the footwell, the rear of a seat, or even inside the glove box, you don't have much to worry about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: the reference was based on crashes at lower speeds, but that is not a critical issue. If the crash is enough to crumple the front of the car by a significant amount, the maximum &lt;em&gt;force and acceleration&lt;/em&gt; will be more or less independent of the speed, but the &lt;em&gt;duration&lt;/em&gt; of the impact and the &lt;em&gt;total amount of deformation&lt;/em&gt; will increase. There is not much point worrying about the state of the fire extinguisher, if a high speed frontal crash deforms the car so much that you are trapped in the seat with the steering wheel pushed against your chest, and you can't reach the extinguisher to operate it even if you are still conscious!&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-01-28T21:40:57.590" LastActivityDate="2019-01-28T21:40:57.590" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25733" PostTypeId="1" CreationDate="2019-01-29T13:19:10.023" Score="0" ViewCount="46" Body="&lt;p&gt;I want to be able to precisely rotate a item. For this I have positioned the item on top of a very large gear. It is driven by a smaller gear. When the item is in the correct position, I will lock the smaller gear and the item is fixed in the correct position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I achieve a more precise system without making the large gear larger and the small one smaller(without using motors)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X2slX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X2slX.png&quot; alt=&quot;Large gear has item on top of it and is driven by small gear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19157" LastActivityDate="2019-01-29T13:44:31.557" Title="How to get precision with a small gear?" Tags="|design|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25734" PostTypeId="2" ParentId="25733" CreationDate="2019-01-29T13:44:31.557" Score="1" Body="&lt;p&gt;add intermediate gears, or consider an epicyclic reduction...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or use a worm and gear, these can do higher ratios compared to gears in a similar volume... see &lt;a href=&quot;https://www.motioncontroltips.com/worm-gears-what-are-they-and-where-are-they-used/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.motioncontroltips.com/worm-gears-what-are-they-and-where-are-they-used/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-01-29T13:44:31.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25735" PostTypeId="1" CreationDate="2019-01-29T15:19:35.820" Score="1" ViewCount="129" Body="&lt;p&gt;Consider an isothermal expansion of an ideal gas (always at ambient temperature &lt;span class=&quot;math-container&quot;&gt;$T_0$&lt;/span&gt;), from some initial pressure &lt;span class=&quot;math-container&quot;&gt;$p_i$&lt;/span&gt; to the ambient pressure &lt;span class=&quot;math-container&quot;&gt;$p_0$&lt;/span&gt;, such that the maximum work available is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$W = mRT_0\ln(\frac{p_i}{p_0}) = p_iV_i\ln(\frac{p_i}{p_0})$&lt;/span&gt;. (1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pressure is throttled to some intermediate pressure, &lt;span class=&quot;math-container&quot;&gt;$p_1&amp;lt;p_i$&lt;/span&gt;, then presumably the expansion work can be be split into two phases, one with constant pressure &lt;span class=&quot;math-container&quot;&gt;$p_1$&lt;/span&gt; and one with variable pressure, i.e. &lt;span class=&quot;math-container&quot;&gt;$p&amp;lt;p_1$&lt;/span&gt;. the maximum work available should then be the sum of these two phases, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$p_1V_1\ln(\frac{p_1}{p_0})$&lt;/span&gt; (variable) + &lt;span class=&quot;math-container&quot;&gt;$V_i(P_i-P_1)$&lt;/span&gt; (constant) (2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an ideal gas, the exergy is then just &lt;span class=&quot;math-container&quot;&gt;$mRT\ln(p_i/p_0)$&lt;/span&gt;, so without the throttling the exergy efficiency is 100%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, firstly, is this correct? Secondly, is the exergy destruction in the valve (1)-(2)? Thirdly, can you throttle (regulate pressure) without destroying a significant amount of exergy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, how do you draw ideal gas throttling on a PV diagram?&lt;/p&gt;&#xA;" OwnerUserId="19158" LastEditorUserId="19158" LastEditDate="2019-01-29T16:30:14.410" LastActivityDate="2019-01-29T16:30:14.410" Title="What happens when throttling in an isothermal process" Tags="|compressed-air|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25736" PostTypeId="2" ParentId="15789" CreationDate="2019-01-29T22:18:42.063" Score="1" Body="&lt;p&gt;2015 International Plumbing Code, and prior editions, definition: TEMPERED WATER. Water having a temperature range between 85°F (29°C) and 110°F (43°C). This agrees with other codes and previously mentioned standards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since water does not usually occur naturally in those temperature ranges, and I'm talking only about water used for human consumption and in buildings (obviously we see water, steam, &amp;amp; ice in a huge range of temps naturally; but we aren't pulling our water supply from geysers or ice pipes), common usage of tempered means water that has been heated or cooled for a specific use or equipment (and it doesn't have to be between 85-110, it just commonly is), as it usually is supplied by utility companies or wells at about 55-degrees Fahrenheit, give or take 5 degrees maybe. This can be domestic water used for drinking or touching humans (including toilets, sinks, showers, faucets, etc.) or hydronic water used within systems for equipment or comfort (heated water system, cooled water system, chilled water).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heated water hydronics can go as high as 180-200 degrees (steam is obviously higher/much higher, but that's a change of state), kitchens/cleaning usually use 140 degree - up to 168 degrees (my regular 'ole home dishwasher even heats water to 163-168), domestic systems by code cannot exceed 140 degrees, though we use mixing valves to get that down further to 110-120 degrees, schools for young kids temper down to 95-110 degrees in the hot pipe (kids are less careful with stuff), hot tubs at 104 degrees, give or take, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a hot day you probably want your water at 34 degrees (many beer brands are best at 34, but some better or different quality beers should be served at temps. a bit higher) but your stomach likes water right at 98.6 degrees... And 60 degree water can feel very cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chilled, cold hydronic water can be anywhere from 33 degrees to much higher although we usually see systems are designed for water 45-65 degrees to do their job of pulling heat out of the air or off equipment. For colder temps, either in the environment or for cooling equipment, we use water/refrigerant mixes. And remember: there's no such thing as cold, the definition of cold is the absence of heat, to some degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So chilled air does not equal 85 degrees!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's all the details:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So Apple is correct that they are modifying the temperature (tempering) of air/water to do a job. And they are correct the methods they are using are naturally occurring, very energy efficient (think geothermal fields), and effective (btw, the Hoover Dam's concrete was cooled the very same way - otherwise the Dam would have taken hundreds of years to set due to the amount of heat concrete gives off when installed; and with a concrete dam that large, it wouldn't have been usable for many lifetimes!). But they are not using 85 degree water or air to cool off people! Unless those people are just trying to be cooler than 86 degrees or higher. That's where the ad is misleading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how it works: the Second Law of Thermodynamics says heat will naturally move from hotter areas to areas with less heat (colder); and the bigger the difference in temperature between the two, the faster heat will transfer!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using chemicals in/or alone from the water at certain pressures improves performance of heat transfer and protects pipes from freezing if outdoors. And since hydronic systems are usually 99.9% closed, it's cheap to operate as you don't have to add more water or chems very often.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to finally answer. Apple is introducing cool water through material, pipes, and/or equipment which is naturally picking up the heat being transferred, then that water moves through the loop to a cooler area, and the heat continues to transfer out. The same with air, they take cool air from some area, blow it through the working space, it naturally picks up heat, then that air is taken through a cooler area where the heat leaves again, and then they can use that same air to blow through a space again as it has lost a lot of the heat. Everything is relative!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Men like their offices at 68 degrees, women like theirs at 74. So as long as the water or air is colder or hotter (depending on if it's winter or summer) than the thing/area that needs comfort/tempering/temperature, it will feel good to us and give or take heat naturally. This is exactly how air conditioners, radiators, heaters, humidifiers, chillers, heating coils, heat exchangers, etc. work! It's all the same basic, simple process/physics but just engineered perfectly. But then some owners want the results but at a cheaper price, so then you get spaces and people that can't get comfortable, warm or cold enough.&#xA;You just need somewhere to dump the heat (like a rooftop outside), and somewhere to get a cooling effect (through high pressure chemicals, underground, or normally cool water or coils), and then the equipment to supply and return the medium used.&lt;/p&gt;&#xA;" OwnerUserId="19162" LastEditorUserId="9770" LastEditDate="2020-04-09T17:51:19.510" LastActivityDate="2020-04-09T17:51:19.510" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25738" PostTypeId="1" CreationDate="2019-01-30T06:06:54.870" Score="0" ViewCount="69" Body="&lt;p&gt;I am trying to develop an equation to determine the cooling rate of a steel tube in air. I'm using Fourier's Law, Stefan-Boltzman Law, Newton's Law as well as the specific heat capacity equation. The tube transfers heat by radiation and convection to the air and it transfers heat through conduction to the steel cooling bed it is laying on. The equation I have determined from these is &lt;span class=&quot;math-container&quot;&gt;$$T_{tube, final} = T_{tube, intial}- \frac{(\dot Q_{radiation} + \dot Q_{convection} + \dot Q_{conduction})\times\Delta t}{\rho VC_{p}}$$&lt;/span&gt; where, &lt;span class=&quot;math-container&quot;&gt;$$\dot Q_{radiation}=\epsilon\sigma A(T_{tube, initial}^4-T_{air}^4),$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$\dot Q_{convection}=hA(T_{tube, initial}-T_{air}),$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$\dot Q_{conduction}=kA\frac{T_{tube, intial}-T_{cooling bed}}{dx}.$$&lt;/span&gt; The thermodynamic properties of the steel are: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$C_{p}=416\frac{J}{kgK}$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$\rho=7667 \frac{kg}{m^3}$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$k=24.2 \frac{W}{mK}$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$\epsilon=0.86$$&lt;/span&gt; My assumptions making this equation are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Air temperature and cooling bed temperature is at a constant 300 K.&lt;/li&gt;&#xA;&lt;li&gt;Temperature dependent properties do not vary much and should be considered constant.&lt;/li&gt;&#xA;&lt;li&gt;The heat transfer coefficient is 5 tW/(m^2K).&lt;/li&gt;&#xA;&lt;li&gt;The temperature throughout the tube is consistent&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I know the tube starts at about 1300 K and drops to about 1150 K in about 40 seconds, when I use this equation though I get really low final temperatures, sometimes even negative ones. Can you help point me in the right direction? I am not looking to develop a 100% precise equation. Just one that will give me a close estimate with only a few percent error.&lt;/p&gt;&#xA;" OwnerUserId="19165" LastEditorUserId="19165" LastEditDate="2019-01-30T06:13:00.263" LastActivityDate="2019-01-30T06:13:00.263" Title="Developing a model of rate of cooling of steel tube in air" Tags="|thermodynamics|heat-transfer|modeling|cooling|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25739" PostTypeId="1" CreationDate="2019-01-30T13:39:17.750" Score="1" ViewCount="474" Body="&lt;p&gt;I learnt that, there are two types of weight plates in strength training, cast iron plates and bumper plates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bumper plates are coated with rubber material so that if dropped comparatively lesser damage is caused. I know this has something to do with the rubber absorbing energy but can't understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take one in each type with same mass and drop it from the same height (neglecting air resistance and assume same contact area during impact), what is the difference between the two impacts? &lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastEditorUserId="90" LastEditDate="2019-01-31T13:23:44.543" LastActivityDate="2019-07-08T13:04:41.237" Title="How rubber and other ductile materials absorb impact energy" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25740" PostTypeId="1" CreationDate="2019-01-30T13:19:01.450" Score="0" ViewCount="18" Body="&lt;p&gt;Is it safe to use a 1/2&quot; NPT connection for thermowells when the process  pressure is  70 bar and temperature is about 85°C  ? &lt;/p&gt;&#xA;" OwnerUserId="19169" OwnerDisplayName="user96729" LastEditorUserId="19169" LastEditDate="2019-01-30T13:53:35.553" LastActivityDate="2019-01-30T13:53:35.553" Title="Thermowell NPT at 70bar" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="25741" PostTypeId="2" ParentId="19206" CreationDate="2019-01-30T14:03:45.817" Score="0" Body="&lt;p&gt;Point of inflection is the point where a curve changes its curvature, that is f''(x) = 0 and f&quot;'(x) not equal to 0. Contraflexure is a point in Bending Moment diagram where Bending Moment changes its sign and since it changes sign, local value is taken as zero (Bending Moment=0 only will not give Contraflexure point, its sign should also be different before and after the point). Now let k be the second derivative of the elastic curve ( curve followed by beam under Bending ) wrt beam length. We know the second derivative gives curvature. We know that E&lt;em&gt;I&lt;/em&gt;k = M (E - Elastic Modulus,  I = Area moment of cross-section of Beam, M - Bending Moment Applied), Now as we know that Bending Moment is equal to the second derivative of the elastic curve. So, at the point of contraflexure (M=0), the second derivative of elastic curve = 0. Also when bending moment changes sign at contraflexure curvature of the elastic curve also changes. That is, the point of contraflexure is for Bending Moment Diagram while Point of inflection is the corresponding point in Elastic curve of the beam and here beam changes curvature. &lt;/p&gt;&#xA;" OwnerUserId="19170" LastEditorUserId="1832" LastEditDate="2019-01-30T14:42:00.877" LastActivityDate="2019-01-30T14:42:00.877" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25742" PostTypeId="1" CreationDate="2019-01-30T16:33:03.210" Score="0" ViewCount="107" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ayumj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ayumj.jpg&quot; alt=&quot;hydrostatics image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can deduce the Forces for F1 and F2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;F_1 &amp;amp;= \dfrac{1}{2}\rho g d_1^2 \\&#xA;&amp;amp;= \dfrac{1}{2} \cdot (1000 \cdot 9.81) \cdot d_1^2 \cdot 8 \\&#xA;\dfrac{F_1}{d_1^2} &amp;amp;= 39240\text{ N}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've moved the &lt;span class=&quot;math-container&quot;&gt;$d_1^2$&lt;/span&gt; over to left and worked out the rest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;F_2 &amp;amp;= \dfrac{1}{2}\rho g \dfrac{d_2^2}{2} \cdot 8 \\&#xA;&amp;amp;= \dfrac{1}{2} \cdot (1000 \cdot 9.81) \cdot \dfrac{d_2^2}{2} \cdot 8 \\&#xA;2\dfrac{F_2}{d_2^2} &amp;amp;= 39240\text{ N}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on solving b and c?&lt;/p&gt;&#xA;" OwnerUserId="19174" LastEditorUserId="6790" LastEditDate="2020-05-04T11:28:44.950" LastActivityDate="2020-10-01T12:06:20.650" Title="Hydrostatics pressure forces" Tags="|fluid-mechanics|hydrostatics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25743" PostTypeId="1" AcceptedAnswerId="25748" CreationDate="2019-01-30T18:08:58.927" Score="0" ViewCount="135" Body="&lt;p&gt;A uniform block is in the form of a square ABCD of sides 2 m and a unit thickness. E is a point on AD such that ED = xm. The portion EDC is then removed from the block. What is the maximum value of x such that the block will not topple?&lt;/p&gt;&#xA;" OwnerUserId="13632" LastActivityDate="2019-01-31T02:17:16.013" Title="Centroid and toppling" Tags="|mechanical-engineering|civil-engineering|statics|dynamics|" AnswerCount="1" CommentCount="1" ClosedDate="2019-01-31T02:15:52.263" ContentLicense="CC BY-SA 4.0" />
  <row Id="25744" PostTypeId="1" CreationDate="2019-01-30T18:18:06.923" Score="0" ViewCount="83" Body="&lt;p&gt;I am designing a rocket engine as my final project at college, and I tried using a &lt;a href=&quot;https://github.com/jte0419/Rocket_Nozzle_Design&quot; rel=&quot;nofollow noreferrer&quot;&gt;MatLab code developed by youtuber JoshTheEngineer.&lt;/a&gt; However, after running the code, it seems that the output points and plot show the variation of area ratio (A/A*) along the nozzle length. Is this how the method is supposed to work? My understanding was that the method should provide the nozzle contour, hence the code must output the coordinates of points along the nozzle contour. Is there another manipulation I have to do to obtain the coordinates?&lt;/p&gt;&#xA;" OwnerUserId="19175" LastActivityDate="2019-01-30T18:18:06.923" Title="Method of Characteristics nozzle - outputs area ratio?" Tags="|aerospace-engineering|propulsion|rocketry|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25745" PostTypeId="2" ParentId="19206" CreationDate="2019-01-30T18:28:09.153" Score="1" Body="&lt;p&gt;From what my teacher taught me, both are the same point.It is called contraflexure in the bending moment diagram (where the bending moment changes sign, i.e hoging to saging or the other way) .Here bending moment neednot be zero, it just changes sign(it neednot change gradually, it can also make a sudden jump).It is called the point of inflection in the deflection curve also called elastic curve (the point where the deformation changes from concave to convex or the other way)&#xA;&lt;img src=&quot;https://i.stack.imgur.com/6pF1p.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refer this question I asked in Math stack exchange&#xA; &lt;a href=&quot;https://math.stackexchange.com/questions/3095917/can-point-of-inflection-occur-at-a-point-where-second-derivative-doesnot-exist&quot;&gt;https://math.stackexchange.com/questions/3095917/can-point-of-inflection-occur-at-a-point-where-second-derivative-doesnot-exist&lt;/a&gt; ,  for better understanding of point of inflection. For furthur reference use these links&#xA;&lt;a href=&quot;https://math.stackexchange.com/questions/1021582/definition-of-point-of-inflection&quot;&gt;https://math.stackexchange.com/questions/1021582/definition-of-point-of-inflection&lt;/a&gt;&#xA;&lt;a href=&quot;https://math.stackexchange.com/questions/402459/an-inflection-point-where-the-second-derivative-doesnt-exist&quot;&gt;https://math.stackexchange.com/questions/402459/an-inflection-point-where-the-second-derivative-doesnt-exist&lt;/a&gt;&#xA;&lt;a href=&quot;https://math.stackexchange.com/questions/688503/what-is-inflection-point&quot;&gt;https://math.stackexchange.com/questions/688503/what-is-inflection-point&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastEditorDisplayName="user17332" LastEditDate="2019-02-01T08:13:35.703" LastActivityDate="2019-02-01T08:13:35.703" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25746" PostTypeId="2" ParentId="25739" CreationDate="2019-01-30T18:58:11.607" Score="1" Body="&lt;p&gt;There is no difference in the rebound energy unless the coefficient of restitution is different.  In brief - a perfectly elastic collision, whether with cold steel or sponge rubber, will return all the energy to the object which collided.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rubber bumpers or other devices designed to reduce damage or rebound do so by being inelastic - either the energy is used to deform (permanently) the material, or it's dispersed in the form of random energy (heat, sonic waves, etc), or in some designs redirected to safe locations.  An example of the last is an airbag, which responds to a collision (with a person...) by forcing the compressed air out numerous exit holes in all directions. &lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="615" LastEditDate="2019-07-08T13:04:41.237" LastActivityDate="2019-07-08T13:04:41.237" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25748" PostTypeId="2" ParentId="25743" CreationDate="2019-01-30T19:10:01.080" Score="0" Body="&lt;p&gt;The CG of the cut shape should fall inside the base footprint which is, 2- xm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The CG is &lt;span class=&quot;math-container&quot;&gt;$ \dfrac{ \sum A_n* d\bar{x}}{ \sum A} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Area of square is 2*2= 4 and its dx=2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Area of the triangle cut off is xm*2/2= xm and its dx = 2-xm/2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ CG_x = \dfrac {4 *2- ( 2-xm/2 ))*xm }{4-xm } \quad$&lt;/span&gt;, note we factored the E, thickness, out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this cg must be smaller than the base, &lt;span class=&quot;math-container&quot;&gt;$2-xm$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2019-01-31T02:17:16.013" LastActivityDate="2019-01-31T02:17:16.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25750" PostTypeId="2" ParentId="25742" CreationDate="2019-01-30T19:58:07.480" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$   F_1= 4*F_2 $&lt;/span&gt; because the thrust is related to h squared.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forces acting on the gate  are -8*F1 at the height of D1/3 and  (+8/4)*F1 = 2*F1 acting at 1/6*D1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So they create a resultant acting at the height of (D1/3 - D1/(6*4) = 7/24D1.&#xA;The resultant is &lt;span class=&quot;math-container&quot;&gt;$ F_{final}= 8*(D_1^2/2-D_{1}^2/8)= 4*D_1^2(3/4)*9.8 $&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-30T19:58:07.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25752" PostTypeId="2" ParentId="25739" CreationDate="2019-01-30T20:19:26.217" Score="0" Body="&lt;p&gt;Its question of reducing impact acceleration, or in some texts G force.&#xA;If you have two elastic collision with same mass but different elasticity, one though one flexible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both of them rebound reducing their velocity to zero and reverse, but the softer one over a longer time and smaller acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \alpha = m\frac{ V_1 - V_2}{\delta t} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller acceleration means smaller force, &lt;span class=&quot;math-container&quot;&gt;$ F=m*\alpha$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-01-31T05:40:12.323" LastActivityDate="2019-01-31T05:40:12.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25755" PostTypeId="1" CreationDate="2019-01-30T23:10:57.600" Score="1" ViewCount="63" Body="&lt;p&gt;I'm trying to build this circuit for a force transducer such that the data is logged by an Arduino Uno. I have the op-amp recommended in the schematic, but I haven't been able to get stable readings from the transducer. I've attached the schematics for the force transducer and op-amp, as well a picture of my attempt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qiyqJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qiyqJ.jpg&quot; alt=&quot;Force Transducer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jWSLN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jWSLN.jpg&quot; alt=&quot;MCP6004&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J2Czz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J2Czz.jpg&quot; alt=&quot;Schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19179" LastEditorUserId="1832" LastEditDate="2019-02-04T21:26:00.103" LastActivityDate="2019-02-04T21:26:00.103" Title="How to prototype this circuit using an Arduino and breadboard?" Tags="|electrical-engineering|circuits|electrical|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25757" PostTypeId="1" AcceptedAnswerId="25785" CreationDate="2019-01-31T08:46:48.003" Score="0" ViewCount="37" Body="&lt;p&gt;After testing a power transformer(pt) and closing its wiring terminals I've encountered that phase S 63 kV correct voltage and phases R and T shows incorrect voltage 32 kV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any body tell why?&lt;/p&gt;&#xA;" OwnerUserId="4411" LastEditorUserId="90" LastEditDate="2019-01-31T13:21:16.570" LastActivityDate="2019-02-02T04:31:29.313" Title="Unbalance voltage mesurements from 63 kV power transformer" Tags="|electrical-engineering|control-engineering|circuits|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="25758" PostTypeId="1" AcceptedAnswerId="25761" CreationDate="2019-01-31T10:37:39.183" Score="1" ViewCount="1620" Body="&lt;p&gt;This question is about a technical aspect of naval warfare, and I think that this site is the closest fit. &lt;em&gt;History&lt;/em&gt; would have been my next guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the TV show “The Last Ship” the protagonists have an Arleigh-Burke class guided missile destroyer. It seems to engage targets mainly with rockets of different sizes. Incoming rockets are shot down either with the point defense turrets or with other rockets. It is probably skewed for the plot of the show, but they seem to be able to shoot down most incoming missiles, making the ship appear near invincible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the fifth season they are up against a battleship and are struck with incoming shells from canons in two different episodes. The first time the destroyer gets rendered disabled, the second time it starts to sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reading up on this I noticed that modern fleets do not have battleships with huge cannons any more, guided missile destroyers are the largest warships that are built now. Given that in the show the rockets appear somewhat ineffective and the shells seem to be unblockable, I wonder why modern ships use mostly rockets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Judging by their presentation of cyber warfare, the military accuracy is probably not good enough to extrapolate this. But if guided missiles are supposed to be so much better, did the destroyer have a serious problem with the battleship just for plot reasons or is that somewhat realistic?&lt;/p&gt;&#xA;" OwnerUserId="19181" LastEditorUserId="19181" LastEditDate="2019-01-31T19:35:14.423" LastActivityDate="2019-01-31T19:35:14.423" Title="Why don't modern warships still use cannons?" Tags="|ships|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25759" PostTypeId="1" CreationDate="2019-01-31T11:58:09.670" Score="2" ViewCount="102" Body="&lt;p&gt;I was reading through the  ITTC – Recommended&#xA;Procedures and Guidelines - Practical Guidelines for Ship CFD Applications (&lt;a href=&quot;https://ittc.info/media/4196/75-03-02-03.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ittc.info/media/4196/75-03-02-03.pdf&lt;/a&gt;) document and this little thing kept me thinking. In page 9 the distance y of the first point from the wall can be computed is said to be computed as &lt;span class=&quot;math-container&quot;&gt;$y = \frac { y ^ { + } L } { \operatorname { Re } _ { L } \sqrt { \frac { C _ { f } } { 2 } } }$&lt;/span&gt;. And right below &lt;span class=&quot;math-container&quot;&gt;$C _ { f } = \frac { 1.328 } { \sqrt { R e _ { L } } }$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's try to deduce the formula for &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;. &#xA; &lt;span class=&quot;math-container&quot;&gt;\begin{cases}&#xA;	  \tau  _ {  w  } = \frac { 1 } { 2 }  \rho   U  _ { e } ^ { 2 }  C  _ {  f  } \\&#xA;	u ^ { * } = \sqrt { \frac { \tau _ { w } } { \rho } }&#xA;\end{cases}&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\Rightarrow u ^ { * } = \sqrt { \frac { \frac { 1 } { 2 }  \rho   U  _ { e } ^ { 2 }  C  _ {  f  } } { \rho } } \Rightarrow u ^ { * } = U_e\sqrt{\frac{C_f}{2}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\begin{cases}&#xA; y ^ { + } = \frac { yu ^ { * } } { \nu } \Rightarrow y=y ^ { + }\frac{\nu}{u^{*}}= y^{+}\frac{\nu}{U_e\sqrt{\frac{C_f}{2}}}\\ Re_L= \frac{U_e L}{\nu} \Rightarrow \frac{Re_L}{L} = \frac{U_e}{\nu} \Rightarrow \frac{\nu}{U_e}=\frac{L}{Re_L}&#xA; \end{cases}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Rightarrow \boxed{y =\frac{y ^ { + }L}{Re_L\sqrt{\frac{C_f}{2}}} }$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the deduction &lt;span class=&quot;math-container&quot;&gt;$C_f$&lt;/span&gt; is the &lt;strong&gt;local&lt;/strong&gt; friction coefficient, but in the guidelines, &lt;span class=&quot;math-container&quot;&gt;$C_f$&lt;/span&gt; is obtained using a formula for an average friction coefficient. So in my opinion &lt;span class=&quot;math-container&quot;&gt;$C _ { f }$&lt;/span&gt; should be calculated by &lt;span class=&quot;math-container&quot;&gt;$\frac { 0.6642 } { \sqrt { Re _ { x } } }$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think?&lt;/p&gt;&#xA;" OwnerUserId="5528" LastActivityDate="2019-01-31T11:58:09.670" Title="Regarding the ITTC Guidelines for Ship CFD applications" Tags="|fluid-mechanics|cfd|friction|shear|ships|" AnswerCount="0" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25760" PostTypeId="1" CreationDate="2019-01-31T12:01:05.190" Score="1" ViewCount="164" Body="&lt;p&gt;[Summary: If two forces are such that the component of one of these forces forms a couple with the other, than which of the two cases (obtaining resultant via adding forces or obtaining moment and a resultant force via the method shown below) are correct? Read further for clear and detailed explanation]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let two forces, F1 and F2, act on a rigid body and have a common point of application at Point A. Then their resultant force, R, can be calculated by the normal use of parallelogram as shown in the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B5YOb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B5YOb.png&quot; alt=&quot;Adding Forces F1 and F2 to get Resultant Force R&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, since forces can be moved along their action lines in rigid bodies (statics), let the same forces be moved along their action lines as shown in the image below (Img.2):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6C9eE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6C9eE.png&quot; alt=&quot;Adding forces F1 and F2 to get moment along with resultant force&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that the y-component of F2 is equal but opposite to F1. This means that the y-component of F2 and the force F1 act as a couple, and cause Moment in the rigid body, which is equal to F1 multiplied by the distance between them. In addition to this moment, we also have the x-component of F2 left over, which is the resultant force on the rigid body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does this mean that, given the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) If any two forces Fa and Fb have intersecting action lines,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A line perpendicular to the action line of Fa at the point of intersection divides the plane (or rigid body) in two parts,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) And Fb points towards a different plane (or different part of the rigid body) than Fa,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have a resultant force (found by the parallelogram law) and moment at the same time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then what is the point of application, O, of the resultant force found in the second Image (Img. 5)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if this is so, then shouldn't the parallelogram be only defined in a way where the two forces are directed towards the same plane?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is this entire thing true? Can the forces actually be moved and resolved in a way they are here? Why or why not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Sorry if the post is messy and incomprehensible. I tried, as much as I can, to explain the question in the most organized fashion...]&lt;/p&gt;&#xA;" OwnerUserId="19182" LastEditorUserId="1832" LastEditDate="2019-02-01T01:51:42.400" LastActivityDate="2019-02-01T01:51:42.400" Title="Can a simple application of two forces have two different outcomes on a rigid body?" Tags="|mechanical-engineering|materials|statics|mathematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25761" PostTypeId="2" ParentId="25758" CreationDate="2019-01-31T12:20:38.253" Score="3" Body="&lt;p&gt;Ship-to-ship combat is now restricted to the internet's various fandoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are close enough to launch a massive projectile to a nearby ship, you are close enough to get hit by one of it's lighter guided missiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those guided missiles tend to be more accurate over distance and much lighter to carry. Missiles can also do the penetrate and then explode for more damage where the big-hunk-o-metal can only do penetration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A well designed ship can still sail with a few holes in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shooting a massive shell requires a lot more propellant than a self propelled projectile. And the launching rig needs to be very sturdy to catch the recoil.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2019-01-31T12:20:38.253" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25764" PostTypeId="2" ParentId="25742" CreationDate="2019-01-31T14:47:22.490" Score="0" Body="&lt;p&gt;Magnitude of resultant is the algebraic sum.There are expressions to find the depth of the points F1 and F2 (centre of pressures).You have the magnitude of the three forces (F1,F2 and resultant), assume right direction for resultant.Then apply the concept that &quot;the sum of moments due to individual forces equal to the moment due to the resultant force&quot;.&lt;/p&gt;&#xA;" OwnerDisplayName="user17332" LastActivityDate="2019-01-31T14:47:22.490" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25766" PostTypeId="2" ParentId="25760" CreationDate="2019-01-31T16:57:52.447" Score="1" Body="&lt;p&gt;Any two force regardless of their graphic representation if they have a point mass destination have no torque unless the mass is at a fixed distance from a center fixed as a rotation arm. Or if they act on two separate points on a real object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case it is ok to break the forces into their components, however you have to start from the intersection point, A, add all x components together, y components together and then apply the resultants which will give you the exact outcome of resolving the forces by parallelogram method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have not yet added the components in your second sketch, if you do the F2y will be cancelled by F1.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-01-31T16:57:52.447" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25767" PostTypeId="2" ParentId="25760" CreationDate="2019-01-31T18:32:38.787" Score="2" Body="&lt;p&gt;There is only one possible resultant for two forces acting through a single point, and it is not possible for those forces to produce a moment about said point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is valid to move a force along it’s line of action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is valid to break a force into x and y components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the situation you describe, if you moved the force along its line of action and then broke it into x and y components, you would still solve the three equilibrium equations. Sum of forces in the y direction. Sum of forces in the x direction. And sum of the moments about a point. The moments about the point produced by the x and y components would cancel. (In your writeup, you seemed to be ignoring the moment produced by the y component, which would be incorrect)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wD6Xgl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wD6Xgl.jpg&quot; alt=&quot;Free Body Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2019-01-31T18:47:13.253" LastActivityDate="2019-01-31T18:47:13.253" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25769" PostTypeId="1" AcceptedAnswerId="29425" CreationDate="2019-01-31T20:47:18.667" Score="5" ViewCount="90" Body="&lt;p&gt;From my office on the 21st floor of a building in Vancouver, BC two parking garages in the process of being demolished are visible. A third is just out of view a few more blocks away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of these structures is being taken down by a team consisting of an excavator with a jackhammer, an excavator with a claw and a man holding a hose. The jackhammer punches holes in the cement, and the claw breaks it up and allows it to fall to the floor below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple of blocks away a different approach has been taken where it seems they are cutting the concrete into chunks and then lowering it down with a crane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We've all seen the news clips and YouTube videos of demolition via pyrotechnics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there other approaches?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably, demolition by explosion requires a large amount of upfront time, effort and engineering skill as &lt;a href=&quot;https://engineering.stackexchange.com/questions/16490/controlled-demolition-of-a-sky-scraper/16495#16495&quot;&gt;detailed in this answer&lt;/a&gt;. All of this culminates in one significant event, the detonation, and then substantial cleanup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The jackhammer and claw approach seems like it is at the opposite end of the spectrum and requires relatively little skill and lots of labor to proceed at an excruciatingly slow and noisy pace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cutting and lowering seems to be in the middle with more expensive machinery to cut and skilled operators required. It also seems to move very slowly but anecdotally seems to have none of the noise associated with jackhammering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does the decision tree/matrix look like for picking an approach? Surely cost plays heavily into it but watching these sites move forward day by day I'm curious as to how two very similar sites in the same city (so permitting is presumably similar) chose different approaches and neither chose to use explosives.&lt;/p&gt;&#xA;" OwnerUserId="19177" LastActivityDate="2021-04-19T22:47:39.653" Title="What approaches are there for demolishing a structure and how is an approach chosen?" Tags="|structural-engineering|civil-engineering|building-physics|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="25771" PostTypeId="1" AcceptedAnswerId="25773" CreationDate="2019-02-01T01:02:58.973" Score="1" ViewCount="147" Body="&lt;p&gt;If I am carrying out a reaction by flowing 200 SCCM of a gas in a 20mm diameter pipe at 100mm length, if I change the pipe diameter to 10mm:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do I adjust the mass flow rate of the gases?&lt;/li&gt;&#xA;&lt;li&gt;If so, how do you calculate this?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thank you for helping. Looking to learn.&#xA;Cheers&lt;/p&gt;&#xA;" OwnerUserId="19189" LastActivityDate="2019-02-01T04:24:38.643" Title="Do you adjust gas flow rate (SCCM) to volume of chamber?" Tags="|flow-control|compressed-gases|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25773" PostTypeId="2" ParentId="25771" CreationDate="2019-02-01T04:24:38.643" Score="0" Body="&lt;p&gt;It depends on if you have a smooth laminar flow with little friction and steady state flow, or transient, and or turbulent flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you pipe lining is smooth, by changing the inside diameter of the pipe to 10mm, you get the same  mass flow rate but at four times the speed.&#xA;Bernoulli equation applies, conservation of kineticand potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if your pipe lining is rough, then you may need to use Poiseuille's Law if the flow is laminar:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ Flow = \frac {P_1 -P_2}{R} \quad, \ R = \frac{8\eta*L}{\pi*r^4} \quad \eta = viscosity $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many practical cases both rules apply, meaning you have pressure loss due to viscosity and you have pressure loss to constrictions, however for a short length as 100mm the friction loss is just nominal and the flow rate is constant.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-01T04:24:38.643" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25775" PostTypeId="1" CreationDate="2019-02-01T09:11:17.113" Score="0" ViewCount="116" Body="&lt;p&gt;If I know the tip and hub diameter of a centrifugal compressor ,the inlet velocity and outlet velocity, how do I determine the number of blades on a centrifugal compressor?&lt;/p&gt;&#xA;" OwnerUserId="19194" LastEditorUserId="19194" LastEditDate="2019-02-01T09:20:15.160" LastActivityDate="2019-02-01T09:20:15.160" Title="How is the number of blades on a centrifugal compressor determined?" Tags="|mechanical|turbomachinery|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25777" PostTypeId="1" AcceptedAnswerId="25780" CreationDate="2019-02-01T11:10:22.763" Score="0" ViewCount="84" Body="&lt;p&gt;In Mechanisms and Machines Theory we were introduced to Assur groups and structural analysis where a mechanism is subdivised into elementary Assur groups of different classes. &#xA;But what is really the practical use of it? &lt;/p&gt;&#xA;" OwnerUserId="19196" LastActivityDate="2019-02-01T17:14:02.057" Title="What is the practical use of Assur Groups and structural analysis?" Tags="|mechanisms|robotics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25778" PostTypeId="1" AcceptedAnswerId="25779" CreationDate="2019-02-01T15:01:29.627" Score="1" ViewCount="111" Body="&lt;p&gt;When performing a pavement design using AASHTOWare Pavement ME for AC on AC Overlay, the pavement distress prediction for bottom-up (alligator) cracking and AC thermal cracking seem to give incorrect results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the Reliability for both these distresses seems fixed at 50%. It does not allow me to increase the value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, if I go ahead and run the program for 10 years the two distresses almost always remain zero (or close to zero).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on this?&lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-02-01T15:01:29.627" Title="Why do the distress values for Pavement ME Design for alligator cracking and thermal cracking for AC pavements seem to give incorrect results?" Tags="|pavement|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25779" PostTypeId="2" ParentId="25778" CreationDate="2019-02-01T15:01:29.627" Score="1" Body="&lt;p&gt;Note that you can change the reliability for total fatigue cracking and total transverse cracking which includes the cracking in the existing AC layer that reflects through the AC overlay and new cracks that develop in the overlay. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reliability is set at 50 percent for new fatigue cracking in the AC overlay and new transverse cracks in the AC overlay. The reason the reliability is set to 50 percent is because it is impossible to separate new crack development in the AC overlay from the reflective fatigue and transverse cracks in the existing AC layer that reflect through the overlay. As such, the reliability is applied to the total fatigue and total transverse cracks which combines new crack development with reflective cracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To provide an answer to the second part of your question, more information is required. AC overlay thickness and the location of the site in terms of climate is important relative to version 2.3.1. If the site is located in the south or in a mild climate, version 2.3.1 will almost never calculate transverse cracks &amp;ndash; they will be close to &amp;ldquo;0&amp;rdquo; when using the global calibration coefficient for transverse cracks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the AC overlay thickness is relatively thick, then it will take a long time for additional fatigue cracks to develop in the existing AC layer because the tensile strains are significantly reduced. The version to be released in July of this year has a coefficient for transverse cracks that is climate dependent.&lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-02-01T15:01:29.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25780" PostTypeId="2" ParentId="25777" CreationDate="2019-02-01T17:01:10.157" Score="0" Body="&lt;p&gt;Things have to be assembled. Assur group gives you a catalog of possible ways to make a rigid assembly out of joints. Each joint in this case represents a connection that will be assembled later, or locked down in case of foldable structures meant for reuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being 0 degrees of freedom means that analysis of these is straight forward. Recognizing them will also make kinematic and dynamic analysis easier since you can replace them with a single rigid body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since assur groups can be predicted in a organized manner nothing stops you from generating a very big catalogue of these. Allowing you to systematically search for new solutions for the task at hand.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2019-02-01T17:14:02.057" LastActivityDate="2019-02-01T17:14:02.057" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25782" PostTypeId="1" AcceptedAnswerId="25786" CreationDate="2019-02-01T18:47:03.187" Score="0" ViewCount="81" Body="&lt;p&gt;Which is more economical to improve or change the soil that is used to build a structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my answer is that it depends on each case, but I would like to get more information, or where to search about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="6397" LastEditorUserId="90" LastEditDate="2022-08-03T00:53:23.303" LastActivityDate="2022-08-03T00:53:23.303" Title="Soil mechanics-Geotechnical engineering" Tags="|fluid-mechanics|civil-engineering|geotechnical-engineering|soil|soil-mechanics|" AnswerCount="1" CommentCount="3" ClosedDate="2019-02-02T17:01:50.587" ContentLicense="CC BY-SA 4.0" />
  <row Id="25784" PostTypeId="1" CreationDate="2019-02-02T01:58:46.933" Score="0" ViewCount="179" Body="&lt;p&gt;A common problem with ultra-high vacuum viewport windows is that often the window material deforms, into a saddle or Pringles chip shape. . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An engineer from a viewport  manufacturer explained to me that this can happen when the hoop stress around the window is not fully symmetric. If it were symmetric, the window would simply bulge or cup in the middle. That seems intuitive, but why would an uneven stress cause a saddle-like strain in particular? What is a simple physical explanation for this? So far, the fact that uneven stress in x/y directions cause different curvatures seems intuitive. What is also important is that glass is a very stiff material and cannot stretch (so volume/thickness is assumed to be conserved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a rough illustration of a stressed window in a metal sleeve typically found in experiments:&lt;a href=&quot;https://i.stack.imgur.com/hJH4I.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hJH4I.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: the brazing pad (thick black) is where the metal sleeve (thin black lines) is brazed to the braze adapter (red lines representing another cylinder). The glass is hermetically sealed to the bottom of the metal sleeve.&lt;/p&gt;&#xA;" OwnerUserId="19201" LastEditorUserId="19201" LastEditDate="2019-02-03T22:45:11.220" LastActivityDate="2021-07-23T16:09:20.907" Title="hoop stress on a thin glass window" Tags="|stresses|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25785" PostTypeId="2" ParentId="25757" CreationDate="2019-02-02T04:31:29.313" Score="0" Body="&lt;p&gt;As I Know when you close the breaker you have seen the  voltages.&#xA;You should check the secondary circuits fuses.&#xA;This voltages shows open circuits of one poles.&#xA;If the polarity of poles were incorrect you shoud have seen 37Kv on other poles.&#xA;If fuses were healthy you shoud check the open circuits of PT secondery circuits.&lt;/p&gt;&#xA;" OwnerUserId="4412" LastActivityDate="2019-02-02T04:31:29.313" CommentCount="0" CommunityOwnedDate="2019-02-02T04:31:29.313" ContentLicense="CC BY-SA 4.0" />
  <row Id="25786" PostTypeId="2" ParentId="25782" CreationDate="2019-02-02T04:59:31.827" Score="0" Body="&lt;p&gt;There is many factors leading to the decision of, either improving the existing soil by adding the missing aggregate and mixing and compacting, bringing it to optimal moisture, density and strength. Or it is better to remove it entirely and import good soil and compact it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are decisions that can only be answered by a licenced soils engineer, or if the remedy is known to your local building department they usually do a jobsite observation and give you their recommendation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they request a soils report then a soils engineer will drill exploratory holes into the ground and take enough samples as he deems needed to take to his lab and run the whole battery of tests on the samples. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then he will compile a report and recommends on what needs to be done. Sometimes they recommend stabilizing the soils by adding the certain sizes of sand and gravel, sometimes they recommend replacing the bad strata. And sometimes they recommend a mix of both and also drilling deep pylons and building certain retaining walls and concrete or steel soldier piles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They give detailed specific instructions to modify the soil and design the foundation. and how to maintain the site in the future as far as drainage and surface water control.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-02T04:59:31.827" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25787" PostTypeId="2" ParentId="18074" CreationDate="2019-02-02T05:38:56.450" Score="4" Body="&lt;p&gt;I am a welder, and here to tell you, you can definitely get shocked. True, if everything is in good condition, proper PPE you are golden. This means your ground has good connection (the bolt holding the wire to the clamp is tight, strong spring for clamp, clean surface on base metal in contact with the ground clamp, not going through any moving parts i.e. bearings, swivels), wearing gloves and good thick soled shoes, the more you sweat the higher chance you have, being engineers you understand salt water is a great conductor I am sure. High Frequency increases your chance to get a pants wetting experience (never happened to me but have seen guys pee themselves while TIGing). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is really why I am scouring the web to understand more about why HF will reach out and get ya. I get that's what it's for in welding, the electrical arc will reach out to the base metal instead of touching the electrode (tungsten) resulting in a contamination, but &lt;em&gt;how&lt;/em&gt; does it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back to it though, some reasons for getting shocked. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;bad ground of course&lt;/li&gt;&#xA;&lt;li&gt;filings inside the welder from grinding (make sure you blow it out every 6 months &#xA;or so) Best is to always unplug and take off the housing.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;being wet (sweat, standing in water, early morning dew soaked gloves, ect.) making &#xA;yourself a better conductor than the copper wire ground. This goes for other people &#xA;around you. Early in my career I almost killed a guy (stick welding) on the other &#xA;end of a project (around 20' away) because he was standing in water with soaked &#xA;shoes touching it with bare hands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My advice is to never count on perfect conditions and always act like the gun is loaded, so to speak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many other hazards to welding but this should answer your question on getting shocked and how to best avoid it.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;              (source for the lower portion)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.brighthubengineering.com/power-plants/89792-ac-and-dc-shock-comparison/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.brighthubengineering.com/power-plants/89792-ac-and-dc-shock-comparison/&lt;/a&gt;               &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AC current is alternating in nature and follows a sine curve. It is continuously changing direction and passing through zero to a maximum positive value and then to a maximum negative value. The voltage of an AC current is a RMS or root mean square value, and the peak or maximum value is 1.4 times the RMS value. However DC current will make a single continuous contraction of the muscles compared to AC current, which will make a series of contractions depending on the frequency it is supplied at. In terms of fatalities, both kill but more milliamps are required of DC current than AC current at the same voltage.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19204" LastEditorUserId="19204" LastEditDate="2019-02-03T23:53:31.430" LastActivityDate="2019-02-03T23:53:31.430" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25788" PostTypeId="2" ParentId="25784" CreationDate="2019-02-02T06:00:58.577" Score="0" Body="&lt;p&gt;Your sketch is not clear, But any collapse and wrinkling is a process of least energy configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you had very fast video camera and reviewed the recording at normal speed, you would see every step of deformation follows the path of least strain energy.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-02T06:00:58.577" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25789" PostTypeId="2" ParentId="6907" CreationDate="2019-02-02T08:06:36.217" Score="1" Body="&lt;p&gt;You would have to have a raised edge to catch the excess material which will cause some turbulence, the amount of material would dictate the height of the raised edge. Unless there is a material that will naturally attract the media being used, really don't see a way around it. I would try a blade with either side angled upwards towards the middle with a flat bottom and a channel on the top side in the middle that is just above the liquid to prevent turbulence on the trailing end that leads to a runoff area to the sides of the tank. The rate of travel will dictate angle as much as the depth and width of the channel. It can also be cammed in the middle so the trailing edge either way will raise some. Or looking from the top of the rectangle blade the raised portion could be a diamond in shape leading to the runoff area, more turbulence but easier to make.&lt;/p&gt;&#xA;" OwnerUserId="19204" LastEditorUserId="1832" LastEditDate="2019-02-02T17:05:53.917" LastActivityDate="2019-02-02T17:05:53.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25790" PostTypeId="2" ParentId="25698" CreationDate="2019-02-02T08:40:53.280" Score="0" Body="&lt;p&gt;Dye penetrant is a good solution but it's finite, once gone you'll need to buy more. A long term solution is an Ultrasonic Flaw Detection device. Unfortunately your looking at around 2k for a cheap one.&lt;/p&gt;&#xA;" OwnerUserId="19204" LastActivityDate="2019-02-02T08:40:53.280" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25792" PostTypeId="1" AcceptedAnswerId="25793" CreationDate="2019-02-02T18:14:59.667" Score="1" ViewCount="44" Body="&lt;p&gt;I'm a neuroscientist interested in building a device to replicate the following study.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nature.com/articles/s41598-018-24924-9&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nature.com/articles/s41598-018-24924-9&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially, they sent air through a nose catheter to stimulate mechanically sensitive cells of the nose. They had 8 seconds of continuous air, followed by 12 seconds off (0.05 Hz). This was to mimic the breathing associated with deep meditation, and they report it induced alterations to consciousness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in building a device that can replicate their paradigm, as well as send air bursts at an arbitrary frequency of my choosing. Ideally the range should be between very slow (~0.05 Hz) and higher frequencies (~100 Hz). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand pulsing most standard pumps won't work for those higher frequencies. My mechE friend proposed some two way valve that would be able to pulse between two directions, the nose catheter, or just out to the air. I could then just use a standard continuous pump or fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unfamiliar with the terminology and haven't been able to find such an electrical valve. Does anyone know of any such valves, or perhaps has a better design in mind?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks everyone!&lt;/p&gt;&#xA;" OwnerUserId="19209" LastActivityDate="2019-02-02T21:17:21.980" Title="Build Device for Emitting Weak Air Bursts at Specified Frequency" Tags="|mechanical-engineering|electrical-engineering|valves|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25793" PostTypeId="2" ParentId="25792" CreationDate="2019-02-02T21:17:21.980" Score="1" Body="&lt;p&gt;The RDWorks forum has a discussion which covers adding an &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B019PTHS7Y&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;air diverter valve&lt;/a&gt; that can be operated by the Ruida class of laser cutter controllers. It is a simple matter of directing a relay to the line that trips the solenoid of this valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Priced at US$22.49 currently, it won't break the bank. Combined with an arduino and a MOSFET to handle the 24v load current of the coil, you should be able to regulate the airflow in the manner you describe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/62AJW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/62AJW.jpg&quot; alt=&quot;air valve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-02-02T21:17:21.980" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25804" PostTypeId="1" CreationDate="2019-02-03T15:50:59.743" Score="0" ViewCount="590" Body="&lt;p&gt;I can't figure out this huge difference in values between these two mechanical material properties. When its refer to flexural strength and modulus there's a difference in value (if the material is not isotropic or homogeneous), but not quite significant as the first two.&lt;/p&gt;&#xA;" OwnerUserId="14265" LastActivityDate="2019-02-03T15:50:59.743" Title="Why the Tensile modulus is always larger than Tensile strength?" Tags="|mechanical-engineering|materials|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25805" PostTypeId="1" CreationDate="2019-02-03T17:36:15.393" Score="3" ViewCount="501" Body="&lt;p&gt;I'm trying to make a swing rotor centrifuge and find a proper BLDC motor to reach 6000G with capacity of 4x50 ml Falcon tubes. I did the calculations roughly but I'm not so sure about them and also couldn't find a proper motor for my calculations.&lt;/p&gt;&#xA;&lt;p&gt;My rotor will be something like this;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HPCVM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HPCVM.jpg&quot; alt=&quot;Rotor Design&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Well for my calculations;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mass of the system : 250 g (50 g for body and 4x50 g for tubes)&lt;/li&gt;&#xA;&lt;li&gt;Radius of the body : 6.5 cm (I'm trying to minimise it to minimise the inertia)&lt;/li&gt;&#xA;&lt;li&gt;RCF : 6000&lt;/li&gt;&#xA;&lt;li&gt;Ramp-up from 0 to 6000G : 10 s&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;And probably my supply voltage will be 12 V or 15 V and my custom ESC can deliver up to 20 A.&lt;/p&gt;&#xA;&lt;p&gt;And from those inputs I get;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Inertia : 0.000528125 kgm&lt;sup&gt;2&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;Angular Acc. : 95.1056763541775 rad/s&lt;sup&gt;2&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;RPM : 9100&lt;/li&gt;&#xA;&lt;li&gt;Torque (with 20% safety margin) : 60.27 mNm&lt;/li&gt;&#xA;&lt;li&gt;Power : 57.4 W&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;But for example when I go and search for a motor such as &lt;a href=&quot;https://www.alibaba.com/product-detail/500W-High-Power-Brushless-Outrunner-DC_60283824747.html?spm=a2700.7724857.normalList.33.4ab2c274DKm1Vs&quot; rel=&quot;nofollow noreferrer&quot;&gt;200 W High Power Brushless Outrunner DC Motor Bldc Motor Price&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;It seems the torque is well within specs. But to get 9100 RPM I need to supply 10.1 V which may cause decrease in torque (?).&lt;/p&gt;&#xA;&lt;p&gt;Also taking Torque Constant into account I get; 5.7 A and it seems too low to me. I mean can I really make this thing spin from 0 to 9100RPM  in just 10 seconds with just 10 V and 5 A?&lt;/p&gt;&#xA;&lt;p&gt;Additionally, there is also Back EMF which I still didn't quiet understand. For the motor in the link, BEMF constant is 0.0011 V/rpm and when I calculate torque producing voltage I get 0.6 V, does this mean that I actually drive this with 0.6V? If so, than I need 95 A to spin it with 9100 RPM? But with terminal resistance involved, my available current is 0.6/0.1 = 6 A, so I can't drive this with 95 A, because of terminal resistance?&lt;/p&gt;&#xA;&lt;p&gt;As you can see, I'm really confused and I'm really new at this things, so any help would be great.&lt;/p&gt;&#xA;" OwnerUserId="19220" LastEditorUserId="90" LastEditDate="2022-02-07T10:38:35.963" LastActivityDate="2022-02-07T10:38:35.963" Title="Choosing a proper BLDC motor for lab centrifuge" Tags="|mechanical-engineering|electrical-engineering|motors|torque|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="25807" PostTypeId="2" ParentId="25066" CreationDate="2019-02-04T00:38:36.177" Score="1" Body="&lt;p&gt;A company I worked at used tool steel molds and channels in the mold with flowing water to increase cooling. Heed the advice of others, never let ANY water come into contact with molten aluminum. I have seen it hit a 20' ceiling (would have gone farther) and as far as 15' in circumference.  &lt;/p&gt;&#xA;" OwnerUserId="19204" LastActivityDate="2019-02-04T00:38:36.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25809" PostTypeId="1" CreationDate="2019-02-04T01:02:54.410" Score="1" ViewCount="103" Body="&lt;p&gt;I have 5/8&quot; PEX piping (I.D. .475&quot;) throughout house.  Am connected to city water supply (@ 40# psi)&#xA;all new PEX pipe, all new valves....no leaks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In April, 57,000 gallons reportedly flowed thru the house.&#xA;In May, 37,000 gallons reportedly flowed thru the house.&#xA;In June, 11,00 gallons reportedly flowed through the house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;History, In January, February, and March...only 2000-3000 gallons flowed each month.  From July to Dec, the flow rate was back to 2000-3000 each month.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the 5/8&quot; PEX pipe diameter, with the 40 psi pressure, how many gallons is possible to flow in one hour?   With a 30 psi, whats the flow rate per hour?  With a 50 psi, whats the flow rate...in gallons per hour for the 5/8&quot; PEX?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I figure the flow rate, (gallons per hour) coming to/through the 5/8&quot; PEX depending only, on knowing the cities psi at the time?  The psi could be a variable.&lt;/p&gt;&#xA;" OwnerUserId="19224" LastActivityDate="2020-08-16T14:38:36.903" Title="need to know flow rate" Tags="|fluid-mechanics|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25810" PostTypeId="1" CreationDate="2019-02-04T01:43:46.467" Score="2" ViewCount="197" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KFJ9l.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KFJ9l.png&quot; alt=&quot;setup&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume we have a beam with density &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; that is subject to three-point bending according to the image above. The beam cross section is rectangular with the variable height &lt;span class=&quot;math-container&quot;&gt;$H(x)$&lt;/span&gt; and constant width &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What function describing the non-constant height, &lt;span class=&quot;math-container&quot;&gt;$H(x)$&lt;/span&gt; will yield a minimal total mass of the beam assuming a maximum allowed deflection of &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the beam can easily be described as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$m(H) = \int_0^L \rho BH(x)\text{d}x$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started off trying to solve this using Euler-Bernoulli beam theory but later realized that this formula is only valid for constant second moments of area. Any ideas how to solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="19226" LastEditorUserId="1832" LastEditDate="2019-02-04T20:43:43.980" LastActivityDate="2019-02-04T20:43:43.980" Title="What is the optimal shape of a beam subject to three point bending?" Tags="|mechanical-engineering|beam|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25812" PostTypeId="2" ParentId="25809" CreationDate="2019-02-04T02:47:43.273" Score="0" Body="&lt;p&gt;Instead of coming up with values that can be explained away, you should have concrete evidence on your side. Most pipes coming from the city water supply meter to your house are 3/4&quot; (check yours first). Install or have a flow meter installed directly off the water meter supplied by the water company for numbers that can't be easily refuted by the water company or you, should there be an underground leak (since the values went back down isn't the case here). Everything from the water companies meter to your house is the responsibility of the home owner so as long as you follow your areas building code there wont be a problem installing it. A lot cheaper than an additional 34k gallon water bill. You should also be able to check the water companies meter yourself by looking at it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FLOW METER,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B06XJZSWQ2&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/DAE-AS200U-75-Couplings-Measuring-Gallons/dp/B06XJZSWQ2/ref=sr_1_7?s=industrial&amp;amp;ie=UTF8&amp;amp;qid=1549247199&amp;amp;sr=1-7&amp;amp;keywords=flow+meters&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19204" LastActivityDate="2019-02-04T02:47:43.273" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25813" PostTypeId="1" CreationDate="2019-02-04T03:53:00.337" Score="1" ViewCount="309" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F0MBs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F0MBs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have a centrifugal compressor as shown in the figure and I know the following parameters----1&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Thrust&lt;/li&gt;&#xA;&lt;li&gt;Inlet and outlet pressure&lt;/li&gt;&#xA;&lt;li&gt;Power required to operate the compressor&lt;/li&gt;&#xA;&lt;li&gt;Inlet and outlet velocity&lt;/li&gt;&#xA;&lt;li&gt;Static pressure and pressure&lt;/li&gt;&#xA;&lt;li&gt;The blade angle&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now I want to know how do I calculate the force exerted by the exhaust coming out?&lt;/p&gt;&#xA;" OwnerUserId="19194" LastEditorUserId="8055" LastEditDate="2019-02-13T19:38:16.970" LastActivityDate="2024-02-18T05:03:45.417" Title="How do I calculate the force coming from the centrifugal compressor exhaust?" Tags="|fluid-mechanics|mechanical|turbomachinery|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="25815" PostTypeId="2" ParentId="25813" CreationDate="2019-02-04T05:33:07.090" Score="0" Body="&lt;p&gt;measure the pressure inside the discharge diffuser with a miniature pitot tube, and the ambient air pressure. the difference will be (delta p). Now measure the area of the discharge diffuser outlet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the reaction force generated when the blower is running will be approximately &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(delta p) x (outlet area). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also simply instrument the blower with a force gauge that measures the reaction from the exhaust flow. This has several advantages, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, it gives you the actual answer you seek.&#xA;Second, if you modeled the thrust-generation process mathematically and used that model to predict the behavior of the system, due diligence would require you to verify the correctness of that model by measuring the thrust reaction force in real life. This means you'll be doing this experiment in the future, regardless of the form your model takes. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13363" LastEditDate="2019-02-04T07:01:42.397" LastActivityDate="2019-02-04T07:01:42.397" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25816" PostTypeId="2" ParentId="25809" CreationDate="2019-02-04T05:42:11.470" Score="1" Body="&lt;p&gt;The easy way is to put a big bucket under each water outlet in the house and turn every one of them on at the same time, and let them run for exactly one minute. then shut them all off and measure how many gallons of water you got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this, you turn on all the faucets, shut the main supply off, position your buckets, and turn the main on for one minute, then shut it off again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you do this experiment, you should also have a pressure gauge near the main valve so you can repeat the test for different values of the source pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to convince someone else (especially an engineer ;-)) that your water billing is in error, you'll have a much easier time using experimental data than by theoretical calculations!&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-02-04T05:42:11.470" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25817" PostTypeId="2" ParentId="25809" CreationDate="2019-02-04T06:54:23.627" Score="1" Body="&lt;p&gt;Instead of going to all the hassle of using all the taps and 1 minute of flow etc.&lt;/p&gt;&#xA;&lt;p&gt;Consider closing the main isolating valve to the property overnight for 8 hours or 6 and see if the main meter records a reading - will at least confirm any leak upstream of the property.&lt;/p&gt;&#xA;&lt;p&gt;Did exactly that on a house I had and there was a leak. Then the water people replaced the supply pipe for &lt;strong&gt;free&lt;/strong&gt; ! Special offer to reduce leaks...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2020-06-30T07:54:21.460" LastActivityDate="2020-06-30T07:54:21.460" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25819" PostTypeId="2" ParentId="270" CreationDate="2019-02-04T08:06:52.730" Score="4" Body="&lt;p&gt;In terms of historical interest, water was used in the first hydraulic brakes and very soon the obvious problems of freezing (due to ambient temperature) and boiling (due to lots of brake use) became evident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To begin avoiding freezing, alcohols and glycerol/water mixtures were used, but issues were still evident.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaking was one, but seals were soon developed to cure that, back in 1921.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This led to the progression to the current options of mineral oils (Citroen still uses these) as well as glycol/ether and silicone products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See:&#xA;&lt;a href=&quot;https://www.total.co.in/advice/brake-fluid.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.total.co.in/advice/brake-fluid.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://didyouknowcars.com/the-history-of-brakes/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://didyouknowcars.com/the-history-of-brakes/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-02-04T08:06:52.730" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25820" PostTypeId="1" CreationDate="2019-02-04T10:47:38.957" Score="0" ViewCount="50" Body="&lt;p&gt;When dimensioning bolts I have to decide if the applied load is concentrically or eccentrically to the bolt axis (according to VDI 2230 but I think in other approaches too). But how do I decide and justify my decision for concentrically or eccentrically load?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet I made my decision if the axial load of a specific bolt (the working load is proportionally related to the bolts in the bolted joint) is concentrically or eccentrically to the bolt axis. My worry is that this decision criterion is to rigorous and my dimensioning of the bolt leads (always) to an overdesigned bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there more decision criterion to consider?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I give some parameters which might be relevant – but mind the following list is not complete:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the material of the housing and cover (plastic or metal)&lt;/li&gt;&#xA;&lt;li&gt;the thickness of the housing and cover&lt;/li&gt;&#xA;&lt;li&gt;may be some parameters of a FEM simulation&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;bend of the considered bolt (&lt;span class=&quot;math-container&quot;&gt;$[N/mm^2]$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$[°]$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;surfaces of the used materials&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the coating&lt;/li&gt;&#xA;&lt;li&gt;the roughness of material (caused by kind of production)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;and more parameters…&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;And what are the thresholds of these parameters to say the load is concentrically or eccentrically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m expecting a statement like: When the parameter A fulfils the following, then you can say that the applied load on the bolt is concentrically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m glad to hear from your experiences how you decide if a bolt is loaded concentrically or eccentrically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="17780" LastActivityDate="2019-02-04T10:47:38.957" Title="Decision criterion if bolted joints are loaded concentrically or eccentrically" Tags="|mechanical-engineering|threads|bolting|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25823" PostTypeId="1" AcceptedAnswerId="25842" CreationDate="2019-02-04T12:25:10.307" Score="1" ViewCount="97" Body="&lt;p&gt;I am working on a telemetry system where a liquid level sensor (Hydrostatic pressure sensor) is used to monitor the volume of fuel (gasoline / diesel) in an underground storage tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensor has to immersed in the fuel and has a cable that is to exit the tank and then be connected to an acquisition unit housed in a panel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem i will like ideas on is that, the sensor cable needs to form a liquid-tight seal at the point it exits the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come across a component called a CordGrip / Cable gland but will like to know if any other methods exist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final solution should take into account that the exit point of the cable is exposed to the outside environment of sunshine and rain&lt;/p&gt;&#xA;" OwnerUserId="19131" LastActivityDate="2019-03-07T13:02:40.573" Title="How to create a Liquid tight cable seal in a fuel storage tank" Tags="|sensors|process-engineering|instrumentation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25826" PostTypeId="1" AcceptedAnswerId="25867" CreationDate="2019-02-04T16:16:59.107" Score="3" ViewCount="1272" Body="&lt;p&gt;I'm designing a precision adjustment bracket, and I'd like to have a component that allows the field tech to make small adjustments to push and pull on each axis of adjustment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went looking to see how others have solved this problem and found an assembly that seems common to several precision adjustment assemblies. I'm a bit embarrassed to say that, as a degreed mechanical engineer.... I don't know what these are called. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9ZZn2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ZZn2.png&quot; alt=&quot;Adjustment assembly 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would say that this is a captive screw that threads into a barrel nut, but that doesn't seem to be it. At least, I can't turn up anything suitable on &lt;a href=&quot;https://www.mcmaster.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;McMaster's site&lt;/a&gt; when I try it. Pro tip - Probably don't Google &quot;captive &lt;em&gt;bolt&lt;/em&gt;.&quot; Apparently &lt;em&gt;that&lt;/em&gt; name is more commonly used to refer to the device used to slaughter animals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also can't seem to find barrel nuts that have a threaded end as pictured in the assembly; &lt;a href=&quot;https://en.wikipedia.org/wiki/Barrel_nut&quot; rel=&quot;nofollow noreferrer&quot;&gt;barrel nut&lt;/a&gt; seems to be exclusively the drop-in fastener used for a lot of flat-pack furniture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's another view of the same item from above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eS51O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eS51O.png&quot; alt=&quot;Adjustment assembly 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then what I believe is the same style assembly in a different product:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jl60g.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jl60g.png&quot; alt=&quot;Adjustment assembly 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all the brackets where it's not just &quot;loosen the nut and tap with a wrench until it's generally correct,&quot; the assemblies I've shown seem to be the standard solution. I'd like to incorporate it into my solution, but &lt;em&gt;I need to know what it's called&lt;/em&gt; to be able to source the parts. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2021-08-05T19:19:37.397" Title="What is this captive screw and barrel nut adjustment assembly called? Picture attached" Tags="|mechanisms|terminology|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25827" PostTypeId="2" ParentId="7245" CreationDate="2019-02-04T17:03:43.733" Score="0" Body="&lt;p&gt;Other recycled materials such as &lt;a href=&quot;https://millscale.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;mill scale&lt;/a&gt; can be used in refractories as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refractories&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refractory material is made by crushing dolomite and mixing it with a flux suspension liquid or paint. Mill scale can be used as flux material that is combined with the liquid binder and ultimate used to produce the refractory material.  &lt;a href=&quot;https://en.wikipedia.org/wiki/Mill_scale&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Mill_scale&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19237" LastActivityDate="2019-02-04T17:03:43.733" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25828" PostTypeId="1" CreationDate="2019-02-04T18:39:10.003" Score="0" ViewCount="67" Body="&lt;p&gt;So I am in the process of working on a rather strange and specific project. I am looking to capture as much energy as possible from a small steam turbine using a DC motor and then use that to power another DC motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially my idea was to charge a capacitor using the first dc motor until a specific voltage and then discharge it as fast as possible into the other dc motor. The only part I am not sure of is creating the circuit that will allow me to flip the switch when a certain voltage is reached.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also as a secondary question what would be the advantages and disadvantages of using a battery vs a capacitor in this situation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I apologies if this is a simple question, I am still new to circuit design so any recommendations would be super appreciated. &lt;/p&gt;&#xA;" OwnerUserId="19239" LastActivityDate="2024-03-16T06:03:30.453" Title="Voltage controlled switch?" Tags="|electrical-engineering|circuits|electrical|circuit-design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25829" PostTypeId="2" ParentId="25361" CreationDate="2019-02-04T19:08:22.243" Score="1" Body="&lt;p&gt;Naively, this looks as if it would be rather efficient, as there are few obvious mechanisms for heat loss.  However, I'm unsure as to how make such a mechanism reliable.  What if the wheel starts turning in the wrong direction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gearing would probably still be required, given the rotational speeds required for the exploited gyroscopic effects, so heat would be lost there.  But the lack of chain makes for one fewer mode of failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one make the gyroscopic effect reliable, such an engine might be reliable, perhaps more so than the traditional cog &amp;amp; chain, but given the need to step down the rotation speed to drive the wheel, there's not likely to be a noticeable energy saving compared to your traditional bike.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, a nice idea.  It would be worth building such a thing to investigate further.&lt;/p&gt;&#xA;" OwnerUserId="19240" LastActivityDate="2019-02-04T19:08:22.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25830" PostTypeId="1" CreationDate="2019-02-04T19:11:01.347" Score="3" ViewCount="124" Body="&lt;p&gt;What is the best way to test the weld strength of a cross wire weld? For example, a wire grid with .212 dia wire x .212 dia wire. &#xA;I cannot find any solid information online for testing cross wire welding (projection welding). I have seen a  picture of someone using a torque wrench to test individual intersections. &lt;/p&gt;&#xA;" OwnerUserId="19241" LastActivityDate="2019-03-04T13:25:57.897" Title="How to test Cross Wire Welds" Tags="|mechanical-engineering|welding|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25832" PostTypeId="2" ParentId="25826" CreationDate="2019-02-04T19:26:41.207" Score="-1" Body="&lt;p&gt;those are used to hold together do-it-yourself furniture, like from IKEA. hardware stores sell replacement assemblies. Take your photo to a well-equipped hardware store and ask the help what they are called. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-02-04T19:26:41.207" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25834" PostTypeId="2" ParentId="25828" CreationDate="2019-02-04T20:38:00.090" Score="0" Body="&lt;p&gt;I believe you are describing a &lt;a href=&quot;https://en.wikipedia.org/wiki/Charge_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;charge pump&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't give any specifics on what you're trying to do, though, so I can't really give meaningful feedback on anything else. Generally speaking, though:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am looking to capture as much energy as possible from a small steam turbine using a DC motor and then use that to power another DC motor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you're driving it, it's a generator. Not all DC motors are the same, so you can't just grab any DC motor and spin the output shaft and expect to get voltage at the terminals. Shunt motors, brushless motors, etc. will require extra inputs to get a useful output. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Essentially my idea was to charge a capacitor using the first dc motor until a specific voltage and then discharge it as fast as possible into the other dc motor.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You don't get to pick how fast a capacitor discharges unless you can control the load resistance/impedance. The motor will determine how quickly the capacitor discharges. And again, you haven't stated anything about the drive cycle, supply/consumption voltages, etc., so I'm not clear why you're not just driving the motor directly with the steam generator. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also as a secondary question what would be the advantages and disadvantages of using a battery vs a capacitor in this situation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Capacity/energy density, but also inherent voltage regulation. Battery voltages are determined by their chemistries; a fully discharged lead-acid battery may still read about 10.8 volts. A capacitor still has a charge at 1 V, but that may not be useful energy because the &lt;em&gt;voltage&lt;/em&gt; isn't in a useful band. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Don't re-invent the wheel. Don't re-invent the switching mode converter. If you need to change the bus voltage, use a buck/boost converter to a voltage that is suitable for your output, and then use a battery float if you need surge capacity (where the instantaneous power demand might be greater than the source can generate, even if your source is capable of delivering &gt;= the average power demands.)  &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2019-02-04T20:38:00.090" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25835" PostTypeId="1" CreationDate="2019-02-04T21:07:25.927" Score="1" ViewCount="204" Body="&lt;p&gt;I'm trying to make a DIY tennis ball machine. However, I'm not quite sure how to attach the wheels to the steel rods, which will be attached to the motors. In other words, the wheel and the rod should become like one single object, so that rotating the rod will rotate the wheel too. The design is something like the picture below, but the wheels I'll be using can be found &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B07CVJNLDG&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. What are your suggestions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3hAum.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3hAum.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19244" LastActivityDate="2019-02-04T21:07:25.927" Title="How to attach a wheel to a turning axle for a ball machine" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25836" PostTypeId="2" ParentId="25130" CreationDate="2019-02-04T22:56:29.300" Score="3" Body="&lt;p&gt;I'm using another type of 12 V motor (here's the &lt;a href=&quot;https://www.ebay.com/itm/DIY-DC-12V-24V-3000-6000-RPM-High-Speed-Adjustable-Marshmallows-Machine-Motor/183249689117?_trkparms=aid%3D111001%26algo%3DREC.SEED%26ao%3D1%26asc%3D20160908105057%26meid%3D596a41df3df24f59ba9119e416bcd7ca%26pid%3D100675%26rk%3D5%26rkt%3D15%26sd%3D302910615730%26itm%3D183249689117&amp;amp;_trksid=p2481888.c100675.m4236&amp;amp;_trkparms=pageci%3Aa57edf9d-28ce-11e9-bbe4-74dbd1809234%7Cparentrq%3Abab096751680aa16af9741a9ffd48612%7Ciid%3A1&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; to it on eBay) that has 1.2 Kg of torque and 3000 rpm speed. I used two 5 inch diameter plastic wheels and tested how far the two motors can launch the ball. At about 45 degrees of angle, it can shoot the ball ~22-23 meters. I hope this info helps! &lt;/p&gt;&#xA;" OwnerUserId="19244" LastActivityDate="2019-02-04T22:56:29.300" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25840" PostTypeId="1" CreationDate="2019-02-05T08:15:04.293" Score="0" ViewCount="133" Body="&lt;p&gt;I have a State Space model and I need to create an Internal delay. Therefore I want to take a certain Padé approximation and convert it to ABCD matrices and extend my current model. However I do not no how to implement an Internal Delay to the State Space representation. I want to construct this by myself because my model is nonlinear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="19249" LastActivityDate="2019-02-05T08:15:04.293" Title="How to implement Pade Appromixation into State Space form" Tags="|control-engineering|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="25842" PostTypeId="2" ParentId="25823" CreationDate="2019-02-05T12:39:07.123" Score="0" Body="&lt;p&gt;As this is a fuel tank, you need to consider explosion hazards. If I understood you correctly, you plan to place a pressure sensor low in the (underground) tank and have the cable enter the tank near the top. This means the seal will not only need to be tight against fuels, but also against fuel vapor. I don't know where you are, so I don't know which ex-protection codes apply if any. Make sure YOU know which codes are relevant and follow them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you use cable ducts, make sure they are compliant with the relevant codes and stand up to a the fuel-ladden atmosphere (organic solvents attack some plastics).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other ways to solve your problem:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;screw the pressure sensor directly into a socket on the tank (&lt;a href=&quot;https://www.vega.com/en/home_us/products/product-catalog/pressure/process-pressure/vegabar-14&quot; rel=&quot;nofollow noreferrer&quot;&gt;possible product&lt;/a&gt;), this is likely impossible in your case as the tank is underground.&lt;/li&gt;&#xA;&lt;li&gt;screw a pressure transmitter into a socket at the top of the tank, have the actual sensor hanging at a steel rope into the fluid (&lt;a href=&quot;https://www.vega.com/en/home_us/products/product-catalog/level/hydrostatic/vegabar-87&quot; rel=&quot;nofollow noreferrer&quot;&gt;possible product&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Use a different level mesurement technology like radar or ultrasonic that measuers from the top (though that is likely more expensive, but you can maintain the sensor more easily). The beauty of radar is that with some tank materials (not metals obvs.), you can measure through the tank so the whole sensor does not contact your medium, there's one less seal you need to worry about re. explosive vapors and so on.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would go with option two as a compromise between ease of service and cost. Find a reputable vendor of measuring devices near you and talk to their application engineer. &lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-02-05T12:39:07.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25845" PostTypeId="1" CreationDate="2019-02-05T13:59:38.083" Score="0" ViewCount="28" Body="&lt;p&gt;I wish to find a bending and torsion test standard for a surfboard (non-prismatic) that is roughly 2000mm long. The closest I've found for bending test is probably those ones that are designed for concrete where the specimen is really long and thick.&lt;/p&gt;&#xA;" OwnerUserId="18873" LastActivityDate="2019-02-05T13:59:38.083" Title="What bending test and torsion test standard are recommend for a surfboard?" Tags="|mechanical-engineering|materials|solid-mechanics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25847" PostTypeId="2" ParentId="25805" CreationDate="2019-02-05T15:19:53.940" Score="1" Body="&lt;p&gt;This is not an answer to your question, per se, but I suggest you consider purchasing a centrifuge rather than trying to make your own. Centrifuges are highly engineered products. Rotors especially are designed carefully because a rotor failure can be catastrophic. A decent centrifuge will have a containment housing designed to keep a rotor failure from injuring users. There will be programmed ramps up to and down from speed and safety interlocks. All of these features are necessary. Commercially available centrifuges are also tested thoroughly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While a new centrifuge may seem expensive, it is probably cheaper than designing and building an equivalent device yourself, especially if you value your time at all. As of the writing of this answer, there are currently 9,729 listings for centrifuges on eBay, 4,247 of which are used.&lt;/p&gt;&#xA;" OwnerUserId="9506" LastActivityDate="2019-02-05T15:19:53.940" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25848" PostTypeId="1" CreationDate="2019-02-05T16:22:19.307" Score="0" ViewCount="35" Body="&lt;p&gt;Can anyone share the equations involved, or some relevant theory material for frquency domain analysis of steering rack and pinion using bode plot?&lt;/p&gt;&#xA;" OwnerUserId="18714" LastActivityDate="2019-02-05T16:22:19.307" Title="Frequency domain analysis of steering system of a car" Tags="|dynamics|vibration|electric-vehicles|car|frequency-response|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25849" PostTypeId="1" CreationDate="2019-02-05T17:08:55.247" Score="0" ViewCount="109" Body="&lt;p&gt;I know the derivation of phase equilibrium criteria derivation for 2 phases. I am not getting how to extend the derivation to more than 2 phases. Please help me out on that. Thank you&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DbRRA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DbRRA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19256" LastEditorUserId="19256" LastEditDate="2019-02-10T09:25:02.707" LastActivityDate="2020-11-10T02:06:00.600" Title="How to derive phase equillibrium criteria for more than 2 phases mathematically?" Tags="|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25850" PostTypeId="1" AcceptedAnswerId="25852" CreationDate="2019-02-05T17:25:15.373" Score="0" ViewCount="123" Body="&lt;p&gt;I'm thinking of making a RC tank that should be simple as possible. But I'm not sure how to make proper steering for the tank. I googled about tank steering and found (for me) complex solutions as dual differential steering.&#xA;Now, my question is - Is it possible to steere the tank with momentum?&#xA;Idea for project is - Left motor will power left Wheele(tracks are too complicated to design), right motor will power right Wheele. Powered Wheele are at the back. So if I want to steere the tank to the left side, right motor will be on, left will be off. Does this method Works? Should right motor go forward but left go backward?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the sketch. This is sketch for rotating/steering from place.&#xA;I guess it will work fine, but what if tank moves forward/backward and wants to steere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FKrPo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FKrPo.png&quot; alt=&quot;The sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18346" LastEditorUserId="18346" LastEditDate="2019-02-05T17:46:33.950" LastActivityDate="2022-02-20T03:03:40.340" Title="Steering with momentum" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="25851" PostTypeId="2" ParentId="11729" CreationDate="2019-02-05T18:27:40.127" Score="1" Body="&lt;p&gt;I think you can take this all the way back to photography, film was always shot square because its more economical, you waste less film, and your developing paper is also cut square (Think about cutting cookies out of cookie dough, theres a lot of waste)&#xA;To enjoy a round display you'd need round format media. &#xA;There are quite a few reasons you wouldn't want to film in a round format&#xA;You'd need larger film reels, the projected image would be smaller to fit on the same size film, and it would be harder to frame your shots and exclude extra stuff from the shot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The eventual driver of screen development is media consumption, and since all the preexisting equipment is set up to use square or rectangular formats the need for a square display becomes more obvious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, while we may have developed and could have continued to develop round screens, there isn't much that needs to be displayed on them in the mainstream. &lt;/p&gt;&#xA;" OwnerUserId="19257" LastActivityDate="2019-02-05T18:27:40.127" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25852" PostTypeId="2" ParentId="25850" CreationDate="2019-02-05T19:30:51.993" Score="1" Body="&lt;p&gt;You want to either use swivel pivot on the front wheels like shopping cart front wheels, or place the CG of your toy near the rear end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or else the steering will not be very effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do that by applying different power to left or right rear wheels you get turned to right or left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you watch delivery fork lifts of say Home Depot, you see they can maneuver precisely into tight corner. Their front wheels are free to pivot. And back wheels are attached to a joy stick.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-05T19:30:51.993" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25853" PostTypeId="1" CreationDate="2019-02-05T21:30:12.740" Score="-1" ViewCount="157" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OG14R.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is in the attached picture. I know how to solve the first part of the problem, but the second part confused me a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when the oven (system) is unsealed, doesn’t make that an open system where ideal gas equation is not applicable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, then would the new pressure and temperature be STD since it’s open system now? Which parameter is constant in this case and which is variable? I’m totally confused.&lt;/p&gt;&#xA;" OwnerUserId="19258" LastEditorUserId="1832" LastEditDate="2019-02-06T18:46:44.030" LastActivityDate="2021-01-05T23:02:10.297" Title="Classic Thermodynamics: Ideal gas question" Tags="|thermodynamics|gas|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25854" PostTypeId="1" CreationDate="2019-02-06T01:23:20.543" Score="0" ViewCount="140" Body="&lt;p&gt;I'm trying to make a gear to mesh with a motor shaft that I bought online, but it doesn't give any information about the pitch or diameter. When I measured the diameter and pitch of the motor shaft, it doesn't seem to conform to anything metric or imperial. So I just guesstimated how big the teeth should be and how spaced apart they should be and designed a gear from scratch. I 3D printed some test gears to see how they mesh, fine tuned it, then printed a full size gear. The teeth appear to mesh very well, but its very very loud, which I am predicting is due to the two gear pitches being different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a bit of research, I figured using a gear generator would be much more accurate. I generated the gear based off of the circular pitch (tooth size). Now I've narrowed the correct tooth size down to 3.2mm &lt;span class=&quot;math-container&quot;&gt;$\pm$&lt;/span&gt; 0.2mm. But I don't know if this will work any smoother than my original. I don't want to waste a huge 12+ hour print on a gear that wont mesh well. What kind of tolerances do I need for proper meshing? Will 3D printing even provide enough resolution for a quiet and smooth fit? I am using an Ultimaker 2 by the way. I appreciate any insight I can get. &lt;/p&gt;&#xA;" OwnerUserId="8974" LastActivityDate="2019-02-06T01:23:20.543" Title="What is the tolerance on gear pitch?" Tags="|gears|3d-printing|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25855" PostTypeId="1" CreationDate="2019-02-06T03:10:52.953" Score="-2" ViewCount="65" Body="&lt;p&gt;what should be the size of gate?&#xA;if street 8 ft&#xA;according to house map Car entrance gate 11 ft&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can corolla grande car enter into the gate easily with some moves? &lt;/p&gt;&#xA;" OwnerUserId="19261" LastActivityDate="2019-02-06T07:11:16.850" Title="what should be the size of gate?" Tags="|mechanical-engineering|fluid-mechanics|autocad|" AnswerCount="1" CommentCount="1" ClosedDate="2019-02-06T12:08:17.123" ContentLicense="CC BY-SA 4.0" />
  <row Id="25856" PostTypeId="1" CreationDate="2019-02-06T04:12:21.323" Score="0" ViewCount="26" Body="&lt;p&gt;How would I go about detecting a very small amount time dilation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to accelerate something and measure the time dilation even if I reach 1km/s time dilation is very small but it looks just barely within reach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm currently thinking a spinning disc, but the numbers are all just a bit beyond what I know how to build. For the timer I obviously want maximum resolution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could spin a 0.6m diameter disc at 1000hz(60,000rpm *fixed conversion error).. I suppose... in a vacuum...errrrmmm... I'm trying not to kill myself.....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to test a hypothesis...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cosmic inflation redshift destroys energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cosmic inflation create new space&lt;/p&gt;&#xA;&#xA;&lt;p&gt;new space has vacuum energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HYPOTHESIS 1: expansion red-shift of certain energies(eg photons) has space-energy conservation. space to energy / energy to space conversion is occurring, and if you factor in space as energy there is conservation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I roughly calculated the energy the numbers and they fall within reasonable range, but I cant think of a way to test it except correlating expansion rate with radiation energy traversing various regions of spacetime,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I while I'm working on that I moved on to hypothesis 2, which is much more of a long shot, because relative velocity redshift doesn't have lost energy, the energy is conserved as momentum. But I figure is one form of redshift involves space-energy conversion then perhaps so do the other types of redshift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypothesis 2 : All red/blue shifting includes an energy to space conversion. While traveling at high speed if light is emitted forward it is blue shifted and converts some space into energy, emit light backwards and it creates space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my plan is to arrange a time dilation test by accelerating a time counter. Sandwich it between 2 powerful LED arrays, one facing forward and one facing backwards. check if time dilation is affected when light is emitted forward and/or backwards. (try both pointing at the counter and both pointing, always aligned so lights are pointed parallel to motion vector)&lt;/p&gt;&#xA;" OwnerUserId="19262" LastEditorUserId="19262" LastEditDate="2019-02-06T10:15:06.753" LastActivityDate="2019-02-06T10:15:06.753" Title="giga/tera hertz timer? super fast spinning disc?" Tags="|motors|vacuum|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25857" PostTypeId="1" CreationDate="2019-02-06T04:37:16.817" Score="1" ViewCount="284" Body="&lt;p&gt;I want to build a &lt;a href=&quot;https://www.youtube.com/watch?v=EZ5bowfjV7o&quot; rel=&quot;nofollow noreferrer&quot;&gt;rock and roll bed&lt;/a&gt; out of 8020 aluminium extrusion, in order to avoid the bother of welding steel pipe. However, I don't have much of a 'feel' for how strong this stuff is. Will 20x20mm T-Slot extrusion be strong enough to support the weight of two people (in a simple frame like the one in the link) or would I need to go for a thicker strength?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="19263" LastActivityDate="2019-02-06T04:37:16.817" Title="Building bed out of 8020 aluminium extrusion" Tags="|aluminum|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25859" PostTypeId="2" ParentId="25855" CreationDate="2019-02-06T07:11:16.850" Score="2" Body="&lt;p&gt;For a 90 degree entrance of a compact car you need minimum 20 feet clear radius per city of Los Angeles parking code. This means if your alley is 8 foot wide you need a space leading to the gate of 12 feet by 12bfeet clear. But the gate itself can be 8'6&quot; wide provided it is located at the far side of that clear space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can try measuring the space you need by finding a vacant parking lot, tree orange traffic cones. Use one to Mark the inside turn corner and the other two to Mark star and end of turn for the front bumper. Allow at least 6 inches clearance for the mirrors and the sweeping inside mid column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pay extra attention to backing out, you be surprised how much the front suspension drags out before a turn starts. If you watch the semi trucks backing into a truck dock you see how much extra space is needed.&#xA;I have attached a screenshot of the city of LA chart, they basically require a 37 feet square clear. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/DJBX2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DJBX2.jpg&quot; alt=&quot;City of Los Angeles digram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-06T07:11:16.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25862" PostTypeId="1" CreationDate="2019-02-06T16:57:22.997" Score="0" ViewCount="23" Body="&lt;p&gt;I need to do the simulation with a specified material. It is known as a thermoplastic.&#xA;During the simulation, I need to define the material property.&#xA;I need to choose whether it is viscoplastic or viscoelasticity model.&#xA;I'm wondering whether the problem I want to model is truly visco-plastic (as it is thermo&lt;strong&gt;plastic&lt;/strong&gt; material)&#xA;Or I can choose a viscoelastic model with considering the permanent deformation after sufficient relaxation time?&lt;/p&gt;&#xA;" OwnerUserId="17605" LastActivityDate="2019-02-06T16:57:22.997" Title="Must I use viscoplastic model to fit the thermoplastic material" Tags="|simulation|plastic|abaqus|elastic-modulus|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25864" PostTypeId="1" CreationDate="2019-02-06T21:26:58.353" Score="2" ViewCount="1816" Body="&lt;p&gt;Suppose we have a tank containing compressed air.  If a hole is made in the tank the air will start flowing out.  The mass flow rate of exiting air will be heavily dependent on the diameter of the orifice (ie the narrowest opening on the system) - the bigger the orifice the greater the initial mass flow and vice versa.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's say we want a certain mass flow rate (x) coming out of the tank, whilst maintaining a constant pressure in the exiting gas.   We can achieve the desired mass flow rate &#xA;by making an appropriately sized opening in the tank (I'll call this opening O[i]) and adjusting the pressure using a pressure regulator.  The issue here is that the pressure regulator &lt;em&gt;could&lt;/em&gt; have a smaller opening (O[pr]) on its inside, ie O[i]&gt;O[pr].  This would consequently reduce the mass flow rate coming from the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's obvious that a pressure regulator can't be used above a specific desired mass flow rate.  What other devices could be used?  A wider pressure regulator maybe? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have included a diagram of a pressure regulator, which clearly shows the small opening inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/dtj6N.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19275" LastActivityDate="2022-03-24T20:01:20.150" Title="Do pressure regulators restrict the mass flow rate?" Tags="|fluid-mechanics|flow-control|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25865" PostTypeId="1" AcceptedAnswerId="25868" CreationDate="2019-02-07T09:42:21.687" Score="0" ViewCount="121" Body="&lt;p&gt;Please suggest some effective repair solution for spalling overhead slab due to high corrosion in rebars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/mEp6V.jpg&quot; alt=&quot;Spalling Slab with corroded rebars&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19280" LastEditorUserId="1832" LastEditDate="2019-02-07T21:54:35.260" LastActivityDate="2019-02-07T21:54:35.260" Title="Finding good and effective solution for overhead RCC slab spalling due to corrosion in Rebars" Tags="|structural-engineering|civil-engineering|reinforced-concrete|" AnswerCount="1" CommentCount="1" ClosedDate="2019-02-07T21:56:59.690" ContentLicense="CC BY-SA 4.0" />
  <row Id="25867" PostTypeId="2" ParentId="25826" CreationDate="2019-02-07T17:06:19.273" Score="6" Body="&lt;p&gt;I never quite found exactly what I'm looking for, but I did come close, so I'll post what I've got. If anyone ever does have a better answer, please post it and I'll accept that one as correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parts seem to go by a lot of names, but &lt;a href=&quot;https://engineering.stackexchange.com/questions/25826/what-is-this-captive-screw-and-barrel-nut-adjustment-assembly-called-picture-at?noredirect=1#comment47559_25826&quot;&gt;Phil Sweet's comment&lt;/a&gt; led me down the track to get where I am. An IPB, &quot;illustrated parts breakdown,&quot; of a reel mower showed a similar assembly where they called one part a &lt;strong&gt;pivot pin&lt;/strong&gt;. I had tried &quot;swivel pin&quot; to no avail, but &lt;strong&gt;pivot pin&lt;/strong&gt; started to turn up some results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not affiliated with the following site, but they seemed to be the only people I could find that had a McMaster-style site that let me configure the parts by material grade, thread spec, length, etc., so I'll mention them here for that reason: &lt;a href=&quot;https://us.misumi-ec.com&quot; rel=&quot;noreferrer&quot;&gt;https://us.misumi-ec.com&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They sell the components for an assembly that they depict in their catalog as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/llDS9.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/llDS9.png&quot; alt=&quot;Adjustment Assy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The assembly is made up of an &quot;adjusting bolt&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9R6WP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9R6WP.jpg&quot; alt=&quot;Adjusting bolt1&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZtW2T.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZtW2T.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;neck&quot; on the adjusting bolt, noted with dimension &quot;T&quot; above, sits in a &quot;block for adjusting bolts&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wvzd4.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wvzd4.jpg&quot; alt=&quot;Blocks for adjusting bolt&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The block wraps around the &quot;neck&quot; of the adjusting bolt, such that the bottom of the hex head cap is on one side of it and the top of the shoulder is on the other side. Once you put the adjusting bolt into the slot in the block it can't move axially. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you thread the adjusting bolt into the &quot;Brackets for Adjustment Screws Bolt Type&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KGQYS.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KGQYS.jpg&quot; alt=&quot;Bracket for adjustment screws bolt type&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My issue with this setup is that the block can't swivel/pivot, so it wasn't of any use for me. I continued to look for a while and found a relatively common part for tractors called a &lt;strong&gt;stabilizer stub pin.&lt;/strong&gt; This is used in the three-point hitch that attaches towed accessories to the tractor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ybhre.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ybhre.jpg&quot; alt=&quot;Stabilizer Stub Pin&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These pins have a smooth bore, so the adjusting screw could slide into it, &lt;strong&gt;BUT&lt;/strong&gt; there isn't any way to get it to work with the neck/shoulder on the adjusting bolt. I thought that I might be able to use just a standard bolt if it had a machined groove in it; my idea was that I could drop the bolt into the stabilizer stub pin and then use a circlip/snap ring/retaining ring to act as the shoulder on the Misumi adjusting bolt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling around led me back to Misumi's site, actually, for their &quot;bearing shaft screws&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5sEd6.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5sEd6.jpg&quot; alt=&quot;Bearing shaft screws&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately I'm backing out and am going to take a different approach entirely to the problem. I need relatively low-cost assemblies, but they need to be ALL stainless steel because of the installation environment. The only way to get the pivot on the bolt is to use the stabilizer stub pin and the bearing shaft screws, but the stub pin is not stainless and the bearing shaft screw is prohibitively expensive because it's machined to actually provide a bearing surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record, too, you can find these assemblies pretty easily if you look for &quot;chainsaw tensioners,&quot; but again I &lt;em&gt;need&lt;/em&gt; the stainless steel material so they were out of the question for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ed3sU.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ed3sU.jpg&quot; alt=&quot;Chainsaw tensioner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2019-02-07T17:06:19.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25868" PostTypeId="2" ParentId="25865" CreationDate="2019-02-07T20:43:49.113" Score="0" Body="&lt;p&gt;It's eminently dangerous. I would immediately inform the authorities of unsafe condition of the structure and request immediate evacuation of the building, had I observed something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next step would be shoring up by a qualified contractor and adding beams or demolition and reconstruction of the slab per instructions of an engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would try to find out the reason for this defect to correct it for reconstruction.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-07T20:43:49.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25870" PostTypeId="1" CreationDate="2019-02-07T22:21:51.763" Score="1" ViewCount="530" Body="&lt;p&gt;So we have this homework problem, it is a rectangular filleted bar and we have to find the minimum thickness. I go and calculated the ratios but here's the problem: (D/d) = 1.375. What may the problem be? Well for this intro to Civil Engr. Course we are taught to find the stress concentration factors using a graph. This graph only has the lines plotted for (D/d) = [ 1.1 1.2 1.3 1.5 ]. How do you guys deal with this when the (D/d) ratio is not listed on a table? Is there some formula to derive the function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a Mechanical Engineering Major taking an intro Civil Engineering Course (Mechanics Materials) , so I have limited knowledge in Civil Engineering terms.&lt;/p&gt;&#xA;" OwnerUserId="19286" LastActivityDate="2019-02-07T23:46:59.393" Title="Finding Stress Concentration Factor when (D/d) ratio is not on a table?" Tags="|materials|civil-engineering|stresses|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25872" PostTypeId="2" ParentId="25870" CreationDate="2019-02-07T23:46:59.393" Score="2" Body="&lt;p&gt;In a comment under your question, you've said you don't think interpolation is allowed. You're wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interpolating between such graphs is standard fare in engineering. You'll do it many, many times for multiple different types of graphs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case, get the stress concentration factor for &lt;span class=&quot;math-container&quot;&gt;$D/d = 1.3$&lt;/span&gt; and for &lt;span class=&quot;math-container&quot;&gt;$D/d = 1.5$&lt;/span&gt;, and then just interpolate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$f_{1.375} = f_{1.3}+\dfrac{f_{1.5}-f_{1.3}}{1.5-1.3}(1.375-1.3)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The only other method would be to find out the underlying equations used to plot those charts and then just calculate the actual result yourself. Problem is, you were likely given those charts instead of the equations because the equations are ugly and complicated.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-02-07T23:46:59.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25873" PostTypeId="1" CreationDate="2019-02-08T00:58:11.320" Score="0" ViewCount="76" Body="&lt;p&gt;I need help knowing if I am selecting the correct wood beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load will be all near the center and a 250 pound man will be doing pull ups on it. I would estimate a 1,000 pound dynamic load to be safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking at a 3.5in x 12 in x 20ft Big Beam (LP makes it, it's an LVL). It will only be supported at the ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the correct beam size? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="19287" LastActivityDate="2021-10-25T17:07:32.073" Title="Big Beam LVL or Glu Lam load capacity to support a gymnast" Tags="|structural-engineering|beam|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25875" PostTypeId="1" CreationDate="2019-02-08T02:46:28.437" Score="1" ViewCount="111" Body="&lt;p&gt;Yeah.. I dont know if this belongs here, I coulnd´t find a proper stack exhange site for this but it sort of involves physics I guess so... yeah.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I have a ventilator, that after a few mins of functioning starts making this weird noise, like if the air is somehow trapped, and indeed the air flow is reduced considerably.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feels like if something like a curtain got glued behind the fan, however there is nothing there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No idea how to even approach this problem, all I know is that this sound comes along with an air flow reduction of about 75%, which sucks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestion is very welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the ventilator model: &lt;a href=&quot;https://http2.mlstatic.com/D_NP_627075-MLA29187536462_012019-Q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://http2.mlstatic.com/D_NP_627075-MLA29187536462_012019-Q.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if I turn it off and wait until the blades stop spinning, then turn it on again the problem goes away for a few seconds/minutes. It´s like it resets.&lt;/p&gt;&#xA;" OwnerUserId="19289" LastEditorUserId="19289" LastEditDate="2019-02-08T13:50:11.260" LastActivityDate="2021-10-26T08:02:55.280" Title="Fan starts making noise and air flow is reduced considerably" Tags="|aerodynamics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25876" PostTypeId="2" ParentId="25873" CreationDate="2019-02-08T03:49:03.110" Score="1" Body="&lt;p&gt;If you can approximate it as isotropic material, then you can use Euler beam theory to find the stress and compare it to the yield stress:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \sigma = My/I$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \sigma $&lt;/span&gt; = normal stress&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ y $&lt;/span&gt; = distance from neutral plane (midplane of symmetric section&#xA;beam). Maximum stress occurs at the outside edge so use &lt;span class=&quot;math-container&quot;&gt;$y=h/2$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is the height of the beam (12 in).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ I $&lt;/span&gt; = 2nd moment of area of the section about the axis perpendicular to y and also in the plane of the section. For a rectangular section, &lt;span class=&quot;math-container&quot;&gt;$I=b h^3/12$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; is the width of the beam. For other shapes like I-beam, see Google.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ M $&lt;/span&gt; is the bending moment. For a simply-supported beam (is it simply supported?) with point load at the center, the bending moment at the center (where it's maximum), is &lt;span class=&quot;math-container&quot;&gt;$ M = FL/4 $&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;=load (1000 lb_f) and &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; = length (20 ft).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Beware that there might be other failure modes too.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2019-02-08T03:49:03.110" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25877" PostTypeId="1" CreationDate="2019-02-08T04:38:21.113" Score="0" ViewCount="74" Body="&lt;p&gt;how to define the shape of a heavy duty chain link when the pitch of the chain and the traction load that the chain is subjected to is known?&#xA;lets say i consider the traction load as the maximum load the chain can support, how can i do the back calculation from that to find the required dimensions?&lt;/p&gt;&#xA;" OwnerUserId="16489" LastEditorUserId="16489" LastEditDate="2019-02-08T17:00:53.190" LastActivityDate="2020-03-04T22:01:30.233" Title="defining the dimensions of a bucket elevator chain link" Tags="|mechanical-engineering|structural-engineering|structural-analysis|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25878" PostTypeId="2" ParentId="25875" CreationDate="2019-02-08T05:02:03.947" Score="0" Body="&lt;p&gt;Possibly it stalls, an aviation term. It means the angle of attack of the fan blades is too much and they creates turbulence as opposed to blowing the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try to gently reduce the blades' pitch by bending the attack edge of the blades slightly flatter. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A prominent symptom of stall is delayed vibration. All the signs points to the fan stalling and because of worn bushing going into a wild vibration. If you can open the safety cage, try to reduce the angle of attack of the blades as in my answer.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-02-08T17:33:51.040" LastActivityDate="2019-02-08T17:33:51.040" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25879" PostTypeId="1" CreationDate="2019-02-08T05:16:10.193" Score="0" ViewCount="1076" Body="&lt;p&gt;I have produced a part drawing which calls up a machined mark on a flat surface, and the supplier has manufactured the part with this surface painted. I haven't added a note to the drawing that says that machined surfaces are not to be painted. Can I ask for the part to be re-worked by the supplier based on a generic 'drawing to BS 8888' note in the drawing border?&lt;/p&gt;&#xA;" OwnerUserId="19291" LastActivityDate="2019-02-08T05:16:10.193" Title="What standards apply when specifying coating/painting in a mechanical part drawing?" Tags="|drafting|standards|technical-drawing|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25883" PostTypeId="2" ParentId="25864" CreationDate="2019-02-08T15:43:17.330" Score="0" Body="&lt;p&gt;I do that kind of thing regularly. My advice would be as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Work out what kind of pressure you need in the tank for the maximum flow rate&lt;/li&gt;&#xA;&lt;li&gt;Use a pressure regulator, as you mention in your question, rated to the pressure you worked out in #1.&lt;/li&gt;&#xA;&lt;li&gt;Make sure the supply pressure to the tank is sufficiently above the maximum pressure you worked out in #1 and within range of the pressure regulator chosen in #2.&lt;/li&gt;&#xA;&lt;li&gt;To control the mass flow rate, rather than using an orifice, use a mass flow controller on the outlet of the tank. This will ensure that you get the desired mass flow rate, while maintaining the correct pressure inside the tank. They can be expensive, but are well worth the money. I have had good experience with &lt;a href=&quot;https://www.bronkhorst.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bronkhorst&lt;/a&gt; ones. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt;: I do not work nor do I have any affiliation whatsoever with Bronkhorst, it's just from experience of using their products.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-02-08T15:43:17.330" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25886" PostTypeId="2" ParentId="19910" CreationDate="2019-02-08T18:07:12.627" Score="1" Body="&lt;p&gt;You want a polymer that is strong and easy to injection mold in thin sections (.7mm). Black nylon would be perfect.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2019-02-08T18:07:12.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25887" PostTypeId="2" ParentId="25877" CreationDate="2019-02-08T18:16:20.280" Score="1" Body="&lt;p&gt;Assuming you have the load on the Conveyor belt and dynamic load factor, usually from 3 to 10, let's call that load Ft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pitch of a chain is the distance between two bushings in mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ft is supported by shear stress in the pin and is 0.6 times the tensile strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You decide what steel alloy you use and calculate the surface area of the pin that is adequate to take the Ft, from there you have the diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many types of chains and depending on the task same are better. the following is a quote from a handbook of design based on ISO standard. I added the link to the handbook at the end of my answer.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Block chains have a high tensile strength for a reduced width. They are used to convey heavy loads, abrasive loads&#xA;  or when violent shocks are expected (e.g.: draw benches).&#xA;  To increase the service life of the chain (better wear resistance), the blocks are sometimes bushed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sedis.com/fichiers/uploads/sedis-white-paper-technical-advice-conveyor-chains-1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;design conveyor belt handbook.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-08T18:16:20.280" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25888" PostTypeId="1" CreationDate="2019-02-08T18:30:57.147" Score="0" ViewCount="34" Body="&lt;p&gt;When we did the simulation, we would get deformation and got each nodes displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In physics, it shows the trajectory of a system of particles of the objects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it the same as the displacement of the Gaussian point? Does displacement of the Gaussian point (obtained from the Subtraction from two phases gaussian point) also mean the trajectory of a system of particles of the objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much.&lt;/p&gt;&#xA;" OwnerUserId="17605" LastActivityDate="2019-02-08T18:30:57.147" Title="Displacement of Gaussian point after structure simulation" Tags="|structural-analysis|simulation|deformation|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25890" PostTypeId="1" CreationDate="2019-02-08T21:20:28.967" Score="0" ViewCount="82" Body="&lt;p&gt;I have a 200 mile pipeline with a few pump stations in between.  I am trying to find ways to optimize the electricity and DRA consumption, and was wondering if anyone has ever done something similar before?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I am building models of the outlet flow rate as a function of energy and DRA. I am also building models of the flow &lt;em&gt;normalized by energy&lt;/em&gt; as a function of DRA usage, vice versa.  Is there any other more proper ways to do this optimization?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formal problem would be: How to achieve optimal flow rate with least DRA / electricity usage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Data&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow rate at the beginning and end of the pipeline &lt;br&gt;&#xA;Current of the pumps at the intermediate pump housing &lt;br&gt;&#xA;DRA ppm  at several pump stations &lt;br&gt;&lt;/p&gt;&#xA;" OwnerUserId="19302" LastActivityDate="2019-02-08T21:20:28.967" Title="Ideas on optimization of DRA and energy consumption of a pipeline" Tags="|pipelines|energy|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25891" PostTypeId="2" ParentId="25875" CreationDate="2019-02-09T01:39:04.940" Score="2" Body="&lt;p&gt;When a fan with a small AC motor in it like this one develops a &quot;squeal&quot; and slows down after running for a short time, it indicates that the motor bearings (usually porous bronze bushings saturated with oil) are worn out or that they need to be cleaned out and relubricated with oil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find a piece of felt surrounding each motor bearing which is soaked with light machine oil (SAE 10 typically) at the factory. The felt feeds oil into the porous structure of the bronze bushing to maintain the space between the motor shaft and the bushing wet with oil. When the oil is all used up, the bearing goes dry and begins to squeal or &quot;chatter&quot;. If you stop the fan, it gives the remaining oil a chance to seep into the bearing which lubricates it for a short time; the fan then runs normally for a short time and then starts squealing again, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a small oil can full of SAE 10 motor oil, carefully drip some oil onto the exposed portion of the felt block or (if you are lucky) into the oil fill pipe sticking out of the bushing (on more expensive types of motor). Continue adding oil until the felt wick no longer soaks it up, and test the fan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the squeal stops, you are good. If it does not, the bushings are worn out and the motor must be replaced.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-02-09T01:39:04.940" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25893" PostTypeId="1" CreationDate="2019-02-09T02:57:53.317" Score="2" ViewCount="160" Body="&lt;p&gt;Mold for Injection Molding is very expensive and is justified for mass production.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, at Prototype stage, how we make the same part with less expensive method. I know 3D Printing is a way but it doesn't have Isotropic properties as its a layered structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3D Printing is usually meant for visualization purposes and doesn't fulfill needs of Prototype testing. What are methods can be used to produce one-off injection mold type items during prototype development?&lt;/p&gt;&#xA;" OwnerUserId="18407" LastEditorUserId="90" LastEditDate="2019-02-09T13:18:00.243" LastActivityDate="2019-02-09T13:18:00.243" Title="Prototype for Injection Molding" Tags="|3d-printing|prototyping|injection|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25895" PostTypeId="2" ParentId="25893" CreationDate="2019-02-09T06:31:06.963" Score="0" Body="&lt;p&gt;Some would make it out of clay or similar- but that does not mean it is cheap...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clay is cheap but the time and skills necessary to create these models are  expensive - expensive is a characteristic of unique pre-production models.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-02-09T06:31:06.963" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25896" PostTypeId="1" CreationDate="2019-02-09T07:02:35.177" Score="1" ViewCount="961" Body="&lt;p&gt;I'm trying to control the speed of DC motor but unsure of the relationship between speed and PWM. I'm aware that the bigger the duty cycle, the larger the speed, but how exactly would this relationship play out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would double the duty cycle make the speed double too?&lt;/p&gt;&#xA;" OwnerUserId="19306" LastActivityDate="2019-03-14T13:02:30.463" Title="Is the speed of a DC motor proportional to PWM?" Tags="|control-engineering|motors|pid-control|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25897" PostTypeId="2" ParentId="25893" CreationDate="2019-02-09T07:05:42.343" Score="2" Body="&lt;p&gt;There are 3D printers that are meant for quick prototyping of injection molds. They print the molding parts, and remove most of the cost of machining the mold. They arent suitable for mass manufacturing since the resulting mold wont last for very long, but is good enough for a small production runs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you dont have access to something like this. Then we have used 3D printed moulds and injected them with big syringes or vacum casted the parts. This works well for urethanes and silicones. Just as long as your parts are small enough for your 3D prints and you have acess to a suitable 3D printer (DLP or SLA printer, FDM wors very badly). Or if your shape can be machined faser out of some softer material and the costs dont exceed your budget, use that instead.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also make a positives (this is often cheaper). And cast a mould out of that thenn use this mould the same way as outlined.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2019-02-09T07:11:34.410" LastActivityDate="2019-02-09T07:11:34.410" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25899" PostTypeId="2" ParentId="25893" CreationDate="2019-02-09T09:46:58.567" Score="0" Body="&lt;p&gt;You must focus on 2d-fying your objects if possible, so they can slot and fit together, and place multiple parts in the same injection mould... in that way you can reduce the cost of your final product by many times, by thinking of the product from a production-simplicity viewpoint rather than topology performance, and end up with a crazy cheap product. it's sometimes 10 times cheaper to make 2 half moulds that are ultrasound-welded or latched than a full mould.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I highly recommend that you get involved with a 3d printing shop. 1 kilo of plastic costs 1500 rupees and the electric cost is about 50 rupees. You can refine a design over many generations of testing and consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It forces you to learn about plastic melting points, adhesion, topology optimization, using i.e. ABS which is the most common plastic in toys like lego and car-plastics (strong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you 3d print with ABS plastic, you can print at 100 percent fill rate. It glues the filament completely-glued to undernearth and neighboring plastic, thereby 70-80% as strong as ABS moulding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can study ABS print versus mould strength comparison and apply them to your 3d shape...&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bXE1P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bXE1P.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can strengthen an ABS prototype 2x strength by treating it with Acetone Vapor, and painting it with acetone with a brush, where liquid acetone percolates into the part and fuses any voids together. It can make perfectly glassy surfaces. It can acheive strength very near to an injection moulded part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on the complexity of your part also. Some moulds are 3-4-5 pieces coming in at different angles. Some have only one CNC milled piece of metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bought a RepRap Printer for 22000 rupees. It was 8 hours of great fun to build following a video guide. The resale value two years later is 17000 rupees after printing 7-8 kilos of plastic. It was a very valuable asset for designing and learning plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also made silicone-detergent cheap silicone moulds for my parts and poured epoxy resine /polyurethane resin in them for fun, but that was a waste of time. &lt;/p&gt;&#xA;" OwnerUserId="6491" LastActivityDate="2019-02-09T09:46:58.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25900" PostTypeId="1" AcceptedAnswerId="25904" CreationDate="2019-02-09T16:19:03.940" Score="0" ViewCount="39" Body="&lt;p&gt;requirements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a panel of wood 4 ft long by 2 ft wide laying flat&lt;/li&gt;&#xA;&lt;li&gt;this panel should be able to tilt side to side (2 way)&lt;/li&gt;&#xA;&lt;li&gt;the weight on the panel will be about 250 lbs up to 400 lbs.&lt;/li&gt;&#xA;&lt;li&gt;controlled lift of each side should be max 3 inches&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Solutions investigated:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;pneumatic cylinders (price unknown)&lt;/li&gt;&#xA;&lt;li&gt;air suspension (for cars) (roughly $250)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any ideas how this can be done without breaking the bank?&lt;/p&gt;&#xA;" OwnerUserId="19308" LastEditorUserId="1832" LastEditDate="2019-02-09T21:23:37.220" LastActivityDate="2019-02-10T02:00:29.703" Title="Ideas needed: how to tilt a panel side to side" Tags="|mechanical-engineering|control-engineering|pneumatic|" AnswerCount="2" CommentCount="2" ClosedDate="2019-02-10T03:42:41.083" ContentLicense="CC BY-SA 4.0" />
  <row Id="25901" PostTypeId="1" CreationDate="2019-02-09T19:31:14.760" Score="0" ViewCount="109" Body="&lt;p&gt;I was doing a case study on the plastic used in cars for controlling the AC in a manual type system which actually snapped in mine it just rotates according to the knob position in the dashboard, pushing and pulling with a stiff cable/rod. It finally separated as you can see in the picture. So rotating the knob effectively does nothing. &lt;strong&gt;I am not considering electronically actuated (Automatic type) AC in cars&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8ifo6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8ifo6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note, the notch is indeed a design flaw as stresses are more concentrated there. I did ANSYS on a model of this however this is only a Material Selection Course that I am studying. So I need to limit myself to just that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went over the Ashby textbook and saw a (similar?) example in the 4th Edition.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;6.10 DEFLECTION-LIMITED DESIGN WITH BRITTLE POLYMERS&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{K1c}{E}$&lt;/span&gt; was derived here by substituting the hooke's law and ignoring all material independent variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything else except Young's modulus and Fracture Toughness is independent of material so we exclude that according to Ashby. And arrived at ABS, NYLON PLA as the best materials like me in the textbook too. I reckoned it could apply here too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three limits for the same objective, displacement limited, load limited and energy limited. I went with deflection limited since I imagined the part would require some deflection to snugly fit during assembly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also put an arbitrary limit of 10% strain since any significant cant change in geometry will result in obsolete part, prices not more expensive than ABS, not weaker than ABS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got cardboard and paper, ABS, PLA in level 2 on CES Edupack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part I don't understand &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Clearly, steel for example is cheaper and stronger than ABS OR any polymer, it wouldn't fail. But my problem is why do automakers around the world use plastic in all road cars when metal does the job better? Is it for lower production costs? My professor doesn't buy into the argument of &quot;Planned Obsolesence&quot; here.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is a better material index here? The Young's modulus in the denominator looks suspicious, so I should just be maximizing &lt;strong&gt;K1c&lt;/strong&gt;?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I should consider &quot;Toughness&quot; and not &quot;Fracture Toughness&quot; since clearly there are notches/imperfections and a better Fracture Toughness only makes things worse?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Yes, I know top trim cars have electronic actuators but I meant only on manual type AC controls&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=lLGyuur92Ho&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is the video of this mechanism&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a mirror image to my picture because it's in an Australian car where they drive on the left.&lt;/p&gt;&#xA;" OwnerUserId="19309" LastEditorUserId="3505" LastEditDate="2022-06-24T07:03:08.460" LastActivityDate="2022-06-27T23:18:27.030" Title="Which material index should be maximized for a component in a car's manual type HVAC?" Tags="|mechanical-engineering|materials|automotive-engineering|equipment-selection|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25902" PostTypeId="1" CreationDate="2019-02-09T19:44:21.630" Score="0" ViewCount="183" Body="&lt;p&gt;I did the math and figured out that according to the laws of physics, for any specific type of wire between poles a given distance apart, the more slack the wire has, the less its weight between those two poles gets magnified into tension force. When there's a buildup of ice on the wire and the wire has very little slack, the weight of the ice is already a lot heavier than the wire but the weight of the ice also gets greatly magnified into a force of tension that's much bigger than the weight of the ice on the part between those two poles. My question is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did most of the places where part of a power line that fell down during a freezing rain storm fell down only from the weight of the ice on the power lines and not from a tree branch that fell down from a buildup of ice on it bringing down a power line with it? If so, did most of the spots where part of a power line fell down only from the weight of the ice on it have very little slack?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so, this fact might be so useful for research. Maybe in the future, power lines could be built with more slack to account for the possibility that so much ice might build up on them during a freezing rain storm.&lt;/p&gt;&#xA;" OwnerUserId="6419" LastActivityDate="2019-02-11T05:28:32.483" Title="When freezing rain brings down power lines, is it usually because the power lines have too little slack?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25903" PostTypeId="2" ParentId="25901" CreationDate="2019-02-09T19:45:49.663" Score="0" Body="&lt;p&gt;You forgot one one parameter : mass... if all the plastic in a car was replaced with metal, then the total mass of a car would go up by 30% (wag) and the subsequent fuel consumption down....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why lots of bits are plastic... also cost of course...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-02-09T19:45:49.663" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25904" PostTypeId="2" ParentId="25900" CreationDate="2019-02-09T22:55:50.777" Score="0" Body="&lt;p&gt;Within the restrictions you've provided, one could establish rotating shafts (axles) under the panel in the appropriate locations. Cams added to these shafts with properly ramped profiles would provide a controlled lift at relatively low cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on materials, some mitigation of friction would be needed. Bearing assemblies on which the cams make contacts (cam followers) would add some expense in exchange for extended life.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-02-09T22:55:50.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25905" PostTypeId="2" ParentId="25900" CreationDate="2019-02-10T02:00:29.703" Score="0" Body="&lt;p&gt;Fix a lever to the panel and lift or lower one side by hand. Hook it into stops at all the positions you want to hold it at.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2019-02-10T02:00:29.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25906" PostTypeId="1" CreationDate="2019-02-10T02:57:22.517" Score="2" ViewCount="49" Body="&lt;p&gt;Trying to design a circuit to accomplish the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;5V PWM signal is converted to 0-2.5V analog voltage&lt;/li&gt;&#xA;&lt;li&gt;The analog voltage should never drop below a certain minimum voltage level, when PWM is 0 or filtered PWM signal results in a voltage level less than this minimum.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know I can do this with a microcontroller and other IC's but looking for a simple solution. I've come up with the following. A voltage regulator produces 2.5V. An n-channel FET switched by the PWM signal produces a 2.5V PWM signal. This 2.5V PWM signal is converted via RC filter to 0 - 2.5V analog voltage.&#xA;I also have a voltage divider to feed the minimum voltage level to the capacitor through a diode. I am aware I will need to compensate for the voltage drop over the diode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to keep the 2.5V voltage regulator to guarantee a 2.5V voltage level, since the 5V PWM signal may be 3.3V PWM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u9M0y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u9M0y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19310" LastEditorUserId="8055" LastEditDate="2019-02-11T14:52:36.090" LastActivityDate="2019-02-11T14:52:36.090" Title="PWM to analog RC based filter with minimum analog voltage level" Tags="|electrical-engineering|pwm|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25907" PostTypeId="2" ParentId="25902" CreationDate="2019-02-10T05:54:07.403" Score="1" Body="&lt;p&gt;Slack means length of wire is longer than the distance between the towers, means more ice accumulation: more weight. I don't know what kind of calculation you did. I'd say you managed to get the exact wrong answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geometry of a wire or cable suspended between the supports under self weight, meaning not supporting a bridge or deck, is a catenary. the span and slack and tension in the wires are highly standard and optimised for many factors, only a sub section of it is elements, wind, icing, earthquake, thunder, dynamic vibration and resonance in gusty storms, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is always room for improvement, but that shouldn't start with wrong assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some comments by OP, I think this will help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call the sag in the wire S and the distance at post or tower between the insulators X. Under certain wind condition the wires will vibrate and swing around the insulators possibly into a paraboloidal kind of surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At no moment these individual surfaces shouldn't get closer than is likely to touch or even close enough to create an electrical arc and start a fire. This mean one restriction for the sag, 2*S &amp;lt; X+ arcing distance. This is just one of concerns&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-02-11T05:28:32.483" LastActivityDate="2019-02-11T05:28:32.483" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="25908" PostTypeId="1" CreationDate="2019-02-10T09:47:03.153" Score="0" ViewCount="110" Body="&lt;p&gt;The formula for change in enthalpy for ideal gas is derived using ideal-gas law which cannot be extended to ideal solutions (my intuition). Then why can we extend the formula of delta S mix to ideal solutions from ideal gas.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9weeR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9weeR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19256" LastActivityDate="2019-02-10T09:47:03.153" Title="Why can we extend the derivation of delta(S mix) to ideal solution from ideal gas behavior?" Tags="|thermodynamics|chemical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25909" PostTypeId="1" AcceptedAnswerId="25912" CreationDate="2019-02-10T13:35:36.860" Score="2" ViewCount="3808" Body="&lt;p&gt;For a two node beam element there are four shape functions for four degree of freedom:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7TqiA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7TqiA.png&quot; alt=&quot;Hermitian Shape function for two node beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a straight three node beam element how shape functions are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note that beam is straight and not curved.&lt;/p&gt;&#xA;" OwnerUserId="8108" LastEditorUserId="8108" LastEditDate="2019-02-10T14:16:23.760" LastActivityDate="2021-04-13T11:54:24.197" Title="Shape Functions of beam element with 3 nodes (quadratic element)" Tags="|structural-engineering|beam|finite-element-method|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25910" PostTypeId="1" CreationDate="2019-02-10T13:41:51.367" Score="-1" ViewCount="61" Body="&lt;p&gt;I've been given the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;CL (lift coefficient)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\eta_p$&lt;/span&gt; (efficiency of the propeller)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$n_p$&lt;/span&gt; (number of revolutions of the propeller)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$n_e$&lt;/span&gt; (number of revolutions of the engine)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\eta_g$&lt;/span&gt; (efficiency of the gearbox)&lt;/li&gt;&#xA;&lt;li&gt;T (which is equal to R)&lt;/li&gt;&#xA;&lt;li&gt;Z (which is needed to calculate &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So far I've considered that &lt;span class=&quot;math-container&quot;&gt;$\eta_p = \dfrac{T\cdot V}{c_p \omega_p}$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$c_p$&lt;/span&gt; is the torque of the propeller whereas &lt;span class=&quot;math-container&quot;&gt;$\omega_p$&lt;/span&gt; is the acceleration of the propeller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore &lt;span class=&quot;math-container&quot;&gt;$c_p \cdot \omega_p = \dfrac{T \cdot V}{\eta_p}$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$c_p \omega_p = \eta_g c_e \omega_e$&lt;/span&gt; (torque and acceleration of the engine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there we know that &lt;span class=&quot;math-container&quot;&gt;$c_e \omega_e = \dfrac{c_p \omega_p}{\eta_g}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally &lt;span class=&quot;math-container&quot;&gt;$P_e = c_e \omega_e$&lt;/span&gt; which is equal to &lt;span class=&quot;math-container&quot;&gt;$P_e = \dfrac{mep \cdot V \cdot n_{cyl}}{2\pi\alpha}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$P_e$&lt;/span&gt; is the power of the engine, &lt;span class=&quot;math-container&quot;&gt;$mep$&lt;/span&gt; is the mean effective pressure, &lt;span class=&quot;math-container&quot;&gt;$n_{cyl}$&lt;/span&gt; is the number of cylinders, and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\alpha = \dfrac{\text{mass of air}}{\text{mass of combustible}}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="19312" LastEditorUserId="1832" LastEditDate="2019-02-10T19:28:30.677" LastActivityDate="2020-07-04T22:05:05.553" Title="calculate number of cylinders and volume in a turboprop" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25911" PostTypeId="2" ParentId="25910" CreationDate="2019-02-10T13:55:18.370" Score="0" Body="&lt;p&gt;You can find valid solutions for 2, 3, 4, 5, 6, 8, 10 and 12 (or more) cylinders - 360 just happens to be divisible by all of those and so is 720... 7 cylinders is also possible&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the important point is that &lt;strong&gt;nothing&lt;/strong&gt; in your question limits or controls the &lt;strong&gt;number&lt;/strong&gt; of cylinders.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-02-10T15:28:36.523" LastActivityDate="2019-02-10T15:28:36.523" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25912" PostTypeId="2" ParentId="25909" CreationDate="2019-02-10T16:16:50.620" Score="1" Body="&lt;p&gt;If the nodes are at &lt;span class=&quot;math-container&quot;&gt;$\xi = -1, 0, +1$&lt;/span&gt; you can find the shape functions using &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrange_polynomial&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lagrangian polynomial interpolation&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact you don't need to work through the general procedure, since you can write down the general form the shape functions must take with only a few unknown parameters, and then solve for the unknown values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the shape functions that are non-zero, and have non-zero derivative, at &lt;span class=&quot;math-container&quot;&gt;$\xi = 1$&lt;/span&gt;. They must be zero, and have zero derivatives, at &lt;span class=&quot;math-container&quot;&gt;$\xi = -1, 0$&lt;/span&gt;. Therefore they must have the general form&#xA;&lt;span class=&quot;math-container&quot;&gt;$$N(\xi) = (\xi+1)^2 \xi^2 (a\xi + b)$$&lt;/span&gt;&#xA;for some values of &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;.&#xA;Using the product rule, the derivative is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$N'(\xi) = 2(\xi+1)\xi(2\xi + 1)(a\xi + b) + (\xi+1)^2 \xi^2 a.$$&lt;/span&gt;&#xA;At &lt;span class=&quot;math-container&quot;&gt;$\xi = 1$&lt;/span&gt; we therefore have &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}N(1) &amp;amp;= 4a + 4b \\&#xA;N'(1) &amp;amp;= 16a + 12b\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one shape function we want &lt;span class=&quot;math-container&quot;&gt;$N(1) = 1, N'(1) = 0$&lt;/span&gt; and for the other, &lt;span class=&quot;math-container&quot;&gt;$N(1) = 0, N'(1) = 1$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving the simultaneous equations for &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; gives the two shape functions as &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}(\xi+1)^2 \xi^2 (-3\xi+4)/4 \\&#xA;(\xi+1)^2 \xi^2 (\xi-1)/4 \end{gather}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nv75Q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nv75Q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get the shape functions that are nonzero at &lt;span class=&quot;math-container&quot;&gt;$\xi = -1$&lt;/span&gt; simply by changing &lt;span class=&quot;math-container&quot;&gt;$\xi$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$-\xi$&lt;/span&gt; (be careful with the sign of the non-zero slope shape function!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape functions for the middle node can be found in the same way, but if you see what is going on you can just write them down by inspection:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}(\xi-1)^2(\xi+1)^2 \\&#xA;(\xi-1)^2 \xi (\xi+1)^2 \end{gather}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-02-10T16:36:34.963" LastActivityDate="2019-02-10T16:36:34.963" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25913" PostTypeId="1" AcceptedAnswerId="25914" CreationDate="2019-02-10T17:46:32.620" Score="0" ViewCount="79" Body="&lt;p&gt;The Washington DC water treatment plant appears to use sedimentation tanks for both treating water before consumption and for treating sewage that is then discharged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: What happens to the sedimentation after it is collects? None of the articles that I have read discuss this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to me that the sediment in the water source won't necessarily be toxic, but I suspect that the sediment in the sewage contains a brew of toxic metals and organics. So does that need to be treated as toxic waste? Is the source sediment used for fertilizer, or is it just sent to a landfill? Is there an issue with dust?&lt;/p&gt;&#xA;" OwnerUserId="19317" LastActivityDate="2019-02-10T17:52:33.460" Title="What happens to the sedimentation collected in water treatment plants?" Tags="|water-resources|waste-water-treatment|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25914" PostTypeId="2" ParentId="25913" CreationDate="2019-02-10T17:52:33.460" Score="4" Body="&lt;p&gt;I was part of a team of contractors to remove and spread the dried sediment from a water treatment facility - the sediment was a red cake/ powder and was spread on the fields as fertiliser. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-02-10T17:52:33.460" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25915" PostTypeId="2" ParentId="18991" CreationDate="2019-02-10T20:27:35.817" Score="1" Body="&lt;p&gt;In Piper Cherokee airplane, the power throttle assembly is just a bushing and shaft with an adjustable pressure handle to increase the grip.&#xA;It has worked on my airplane for 41 years, no problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a really simple mechanism.&#xA;Here are some photos from Ebay. they sell it for $85. &lt;a href=&quot;https://www.ebay.ie/itm/Piper-Cherokee-PA28-151-161-Control-Quadrant-Assy-P-N-65850-00-/112844567379?hash=item1a460f7353&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cherokee quadrant throttle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-10T20:27:35.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25918" PostTypeId="1" CreationDate="2019-02-11T06:46:29.240" Score="0" ViewCount="249" Body="&lt;p&gt;If in both cases if there is an application of Shear Stress then why do we have Screw dislocation in one case and Edge dislocation on the other?&lt;/p&gt;&#xA;" OwnerUserId="19123" LastActivityDate="2019-07-12T01:01:55.483" Title="Screw Dislocation and Edge dislocation" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25920" PostTypeId="1" AcceptedAnswerId="25921" CreationDate="2019-02-11T11:03:40.087" Score="1" ViewCount="169" Body="&lt;p&gt;I have a project which involving several mating parts and also shaft guided through bearings, my issue here is that I don't have much experience with tolerances and manufacturing processes and I cannot determine the perfect balance between cost and accuracy. I am using Aluminium 6061 for raw material. For example I have some guidance &quot;shafts&quot; which are &quot;made&quot; by removing a large material from the part, I made some research and I saw that a +- 0.025 mm is a typical tolerance for a decent CNC machine. My question is, could I go down with that value, in terms of cost or is already a tight tolerance ?&lt;/p&gt;&#xA;" OwnerUserId="3807" LastEditorUserId="203" LastEditDate="2019-02-11T11:55:30.910" LastActivityDate="2019-02-11T11:55:30.910" Title="Mechanical tolerances on a CNC machine" Tags="|mechanical-engineering|tolerance|cnc|prototyping|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25921" PostTypeId="2" ParentId="25920" CreationDate="2019-02-11T11:37:20.230" Score="1" Body="&lt;p&gt;It's going to be an uphill battle from that point on. While you can get every part of the machine and the process to tighter tolerances, their errors may compound to the sort of tolerance of 0.025; you may manufacture relatively simple parts with lower error but whether you succeed or not is more up to luck than your efforts - whether the tolerances of the tool, the table drive, the head drives, the frame rigidity, the sliders rigidity, and the material itself add up and ruin the accuracy or cancel each other resulting in near-perfect workpiece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One nasty property to consider is the internal stresses of the material, resulting from the manufacturing process. If you remove large parts of it, the stresses that were canceled out by say, two outer surfaces pulling with equal strength against the central bulk pushing, will result in the workpiece ending as soon as you unlatch it. I had worked with 7mm brass sheets, and upon removing most of material from one side, I't observe good 2cm of deflection on a 30cm bar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, if you aim at sub-0.01mm accuracy in all domains you can control (tool shape, material alignment, table flatness and machine being level), and the machine is good quality, you can expect better than 0.03mm accuracy in the result. If you want better, you'll need to get into the &quot;guts&quot; of the process - drives of the head and table, bearings, vibration dampening etc.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-02-11T11:37:20.230" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25924" PostTypeId="1" CreationDate="2019-02-11T13:42:26.047" Score="0" ViewCount="52" Body="&lt;p&gt;I'm putting a steel frame in, to &quot;open plan&quot; the rear of a brick cavity-wall terrace house in the UK, which will then have a single storey extension added behind it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already retained a structural engineer for earlier work on the project, and he seems quite sharp and pretty good. But I've also had considerable conflicting advice and have concerns on 2 key issues, so I want to ask for other people's suggested approaches in this situation, to inform me and let me think a bit before the fees clock starts ticking:)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The house is a pretty ordinary 1920 - 1940s brick London terrace house. It's got a 100mm cavity rear wall, pointing is good (or will be by the time we're done), and the foundations are quite wide (500mm front/800mm rear) and the subsoil is a mix of clay and large-ish pebbles, probably good for any likely increase in load despite the location (100kN+ bearing?). The oversite slab (about 350mm below finished floor level and 150mm above foundations) is solid with no cracks, and very dense - I killed  a concrete saw blade just sampling through it, and had to use a breaker on the subsoil just to get 400mm down. The adjoining houses have a 200mm+ solid brick party wall on one side, and single leaf only (100mm) on the other. The adjoining house masonry is very poor and weak indeed and missing in many places.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plan is to open plan the entire rear ground floor, with a symmetrical &quot;figure 8&quot; moment resisting 203x203x46 frame resting on the foundations, the bottom member then encased in C40 self-consolidating concrete for load spreading and protection. The &quot;figure 8&quot; would be about 3m high, and the upstairs floor is being replaced with dense block&amp;amp;beam running front-to-back &lt;em&gt;(this is a settled point: despite the extra load and that the house wasn't designed for it, it's something we want to do for various reasons that don't matter here)&lt;/em&gt;. The block and beam upper floor is being supported by load-spreading C40 on the oversite slab (calculated already) and so the rear concrete beams resting on the frame will only be ~ 2 - 2.5m long. The extension roof when complete will slope, and meet the external wall quite a bit above the top of the frame, so the top of the frame can be exposed outside the current external wall face - it won't affect water integrity or corrosion issues when completed. The engineer has indicated his default choice is to propose calcs based on the concrete beams sitting in the flange of the upper UC with stiffeners, and the splice in the upper UC joined by moment resisting join to the central UC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 2 questions I keep having, that just keep coming to mind, and I want to see what suggestions exist before incurring fees or expressing any ideas to the engineer.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;6m of 203 UC is heavy - just under 300kg. I'd like to get it as 2 parts with end plates and splice them, just to make it practical/safer to manage and place, but I'm not sure what sort of end plate/splice options I have. It'll be encased, so there will be uniform support along the underside of the beam, as well as the underside of the top flange too, as well as along the web, so torsion won't be an issue. But the load is at 3 points - ends and middle - and the middle is where any splice is likely to be.&lt;br/&gt;&lt;br/&gt;What kind of splicing approach should I expect, and how easy will it be to make it rigid enough, if spliced close to the middle load?&lt;/li&gt;&#xA;&lt;li&gt;The torsion and eccentric load on the upper beam worry me. It just feels like weak designing, even if the calcss did stand up, because it doesn't nullify those issues like other similar-cost solutions probably could. The UC is 203mm wide, the upper masonry it supports is 300mm wide (2x100 brick + 100 cavity), so it'll need some kind of 320mm wide plate or channel welded/bolted onto it to provide full-width support to the upper walls. Yes both upper leaves would indeed be 100% supported, but they'd also be COG considerably further from the UC web, creating extra torsion.&lt;br/&gt;&lt;br/&gt;Given those, I'm tempted to ask if he can design the upper frame from say 300wx200H RHS instead of 203x203 UC - same height, 2 webs one each edge, broader hence no additional top plate, much less of a torsion concern, and enough extra stiffness that I can cut 125Wx150H holes in one side every 520mm c/c, and insert the floor beams into the interior of the RHS. The concrete beams can then rest centrally spanning the entire inner bottom width of the RHS rather than resting eccentrically on a UC flange. It also avoids stiffeners that a UC would need, compensating for the extra material cost and there's a wider range of thicknesses than UC would have, for more or less rigidity, to avoid wasted over-speccing.&lt;br/&gt;&lt;br/&gt;Does this make sense to propose, or are there issues I'm not seeing?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="6366" LastActivityDate="2019-02-11T13:42:26.047" Title="Possible improved approaches to 6mx3m moment-resisting frame with eccentricity+torsion, in advance of detailed engineer discussion" Tags="|structural-engineering|structural-analysis|steel|beam|building-design|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25925" PostTypeId="1" CreationDate="2019-02-11T14:10:29.890" Score="0" ViewCount="249" Body="&lt;p&gt;I'm designing a machine that will seal plastic tubes. The idea is to clamp the tube with two beams. These two beams have to move away to let a piece of the tube through, then those beams should move to each other so that the plastic tube gets folded and hold between the two beams. One beam moves exactly the same as the other, only in the reverse direction. &#xA;Currently, the design uses timing belts to accomplish this, but I'm searching for other ways to accomplish this type of movement, but I'm having a hard time finding mechanisms like this on the internet because I don't know how to call it. Is there a name for this kind of mechanism/movement? Or do you know any other ways to accomplish this movement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some illustrations that might clarify my explanation:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QPZfI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QPZfI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BizbN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BizbN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way I came up with is using a ball screw mechanism where one half of the axle has it's screw tapped the other way around, but that part is probably expansive to built. An rack and pinion system with two racks would also work. Other alternatives I came up with all use two motors.&lt;/p&gt;&#xA;" OwnerUserId="14151" LastEditorUserId="14151" LastEditDate="2019-02-11T14:49:47.273" LastActivityDate="2019-02-11T14:49:47.273" Title="Mechanism that moves two parts linear to each other" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="25927" PostTypeId="2" ParentId="25918" CreationDate="2019-02-11T23:02:56.633" Score="1" Body="&lt;p&gt;For edge dislocations, the defect line appears to move parallel to the direction of application of shear.  That can happen if there is an extra plane of atoms in the lattice that creates the dislocation core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, the defect line in screw dislocations moves perpendicular to the direction of shear.  An extra plane of atoms is not necessary for screw dislocations to form.  Rather, the process is very similar to ripping a strip of paper in two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are numerous resources on the web that discuss the issue, e.g., &lt;a href=&quot;https://web.iit.edu/sites/web/files/departments/academic-affairs/academic-resource-center/pdfs/Material_Deformations_Workshop.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://web.iit.edu/sites/web/files/departments/academic-affairs/academic-resource-center/pdfs/Material_Deformations_Workshop.pdf&lt;/a&gt; and &lt;a href=&quot;http://www.ndhu.edu.tw/ezfiles/29/1029/img/3034/Chapter4-1_1041020.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ndhu.edu.tw/ezfiles/29/1029/img/3034/Chapter4-1_1041020.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2019-02-11T23:02:56.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25928" PostTypeId="1" CreationDate="2019-02-12T01:48:59.400" Score="2" ViewCount="104" Body="&lt;p&gt;Suppose there's a 10 liter perfectly-insulated sealed tank of saturated liquid nitrogen at 4 bar. Therefore the fluid has properties,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\rho=740 \,\frac{\text{kg}}{\text{m}^3}$&lt;/span&gt;,&#xA;&lt;span class=&quot;math-container&quot;&gt;$m=7.4 \,\text{kg}$&lt;/span&gt;,&#xA;&lt;span class=&quot;math-container&quot;&gt;$\hat{u}=-93\,\frac{\text{kJ}}{\text{kg}}$&lt;/span&gt;,&#xA;and &lt;span class=&quot;math-container&quot;&gt;$U=-690\,\text{kJ}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you drain all the liquid from the tank, then eventually in the tank you'll have 10 liters of saturated nitrogen vapor at atmospheric pressure with properties,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\rho=4.6 \,\frac{\text{kg}}{\text{m}^3}$&lt;/span&gt;,&#xA;&lt;span class=&quot;math-container&quot;&gt;$m=0.046 \,\text{kg}$&lt;/span&gt;,&#xA;&lt;span class=&quot;math-container&quot;&gt;$\hat{u}=55\,\frac{\text{kJ}}{\text{kg}}$&lt;/span&gt;,&#xA;and &lt;span class=&quot;math-container&quot;&gt;$U=2.5\,\text{kJ}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total internal energy has increased, despite removing most of the mass! Have I made an error in my calculations, or is this correct? Is there an intuitive way to think about this phenomenon?&lt;/p&gt;&#xA;" OwnerUserId="19330" LastActivityDate="2019-02-13T01:04:28.297" Title="Does a tank of liquid nitrogen increase in total internal energy after emptying?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|cryogenics|mass-transfer|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25929" PostTypeId="1" CreationDate="2019-02-12T02:37:05.900" Score="0" ViewCount="42" Body="&lt;p&gt;Hi i would like to ask if there's any ways i can shift weights and lock it in place at different angles (180 degree to 90 degree)  without using electronic devices- preferably using wood as the main material. I have tried using a slider to guide the weights and by pushing the rod that is holding the weights to lock it in place. However, it is not feasible as it is too complicated and the weights does not move smoothly. &lt;/p&gt;&#xA;" OwnerUserId="19227" LastActivityDate="2019-02-12T05:56:00.237" Title="shifting weights and locking in place at different angles" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25930" PostTypeId="2" ParentId="25928" CreationDate="2019-02-12T02:49:07.843" Score="2" Body="&lt;p&gt;Internal energy is a measure of the distribution of translation, rotation, and vibration in the molecules. A liquid has far less in translational and rotational energy per molecule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As proof, consider that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\tilde{U}_{vap} - \tilde{U}_{liq} = \Delta_{vap}\tilde{U} = \Delta_{vap}\tilde{H} - \Delta_{vap}(p\tilde{V})$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an ideal gas, &lt;span class=&quot;math-container&quot;&gt;$\Delta_{vap}(p\tilde{V}) = RT_{vap}/M$&lt;/span&gt; (SI units J/kg). Since vaporization is endothermic, the first term is positive. The term &lt;span class=&quot;math-container&quot;&gt;$RT_{vap}$&lt;/span&gt; is typically smaller. For nitrogen with &lt;span class=&quot;math-container&quot;&gt;$T_{vap} \approx 77$&lt;/span&gt; K and &lt;span class=&quot;math-container&quot;&gt;$\Delta_{vap}\tilde{H} \approx 6$&lt;/span&gt; kJ/mol (&lt;a href=&quot;https://webbook.nist.gov/cgi/cbook.cgi?ID=C7727379&amp;amp;Mask=4#Thermo-Phase&quot; rel=&quot;nofollow noreferrer&quot;&gt;NIST Webook&lt;/a&gt;), one has this in units of J/mol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta_{vap}\bar{U} \approx 6000 - 8.3(77) \approx 5400$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to provide this energy to vaporize the liquid. You loose this internal energy when you condense the vapor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compare with what you have, take only the gas and condense it to liquid. The internal energy will be less by 9.3 kJ just for the 46 g as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta_{cond}{U} \approx (46/28)(-5.4) = -9.3$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a transformation entirely at 1 bar pressure. Changing pressure on an ideal gas does not change its internal energy. It does change however change enthalpy, so the &lt;span class=&quot;math-container&quot;&gt;$\Delta(pV)$&lt;/span&gt; term has to be done differently. Basically, you have to do the equivalent of vaporization at 4 bar and then an expansion to 1 bar. The specific difference in your case between liquid at 4 bar and gas at 1 bar is about -4.1 kJ/mol. Without deeper consultation, I might suspect the difference is from the loss at the expansion of the gas since, for real gases or liquids, &lt;span class=&quot;math-container&quot;&gt;$U(V,T)$&lt;/span&gt;, meaning ultimately also that &lt;span class=&quot;math-container&quot;&gt;$\tilde{U}(p,T)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the end, we have no surprise to find that a liquid has a lower internal energy than a gas even though the liquid has a higher mass than the gas. The difference has no reference to how the tank was emptied, since internal energy is a state property/function.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-02-13T01:04:28.297" LastActivityDate="2019-02-13T01:04:28.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25932" PostTypeId="2" ParentId="25929" CreationDate="2019-02-12T05:56:00.237" Score="-1" Body="&lt;p&gt;yes, we can shift weights and lock it in place at different angles with or without using electronic devices. we can use a screw jack mechanism to lift and place weights. we have to select screw jack as per the weight &amp;amp; height.&lt;/p&gt;&#xA;" OwnerUserId="19332" LastActivityDate="2019-02-12T05:56:00.237" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25933" PostTypeId="1" CreationDate="2019-02-12T06:04:54.340" Score="1" ViewCount="25" Body="&lt;p&gt;What happens when you over torque a joint comprising of two different materials ? Will the corresponding material wear make the joint leak proof ?&lt;/p&gt;&#xA;" OwnerUserId="19333" LastActivityDate="2019-02-12T09:41:02.600" Title="Over torquing two different materials" Tags="|automotive-engineering|manufacturing-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25934" PostTypeId="2" ParentId="25933" CreationDate="2019-02-12T06:40:31.523" Score="1" Body="&lt;p&gt;No, not necessarily - one material could deform sufficiently to make the joint leak as the clamping force may be reduced or non-existent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better answer may be possible if you gave more detail - materials, shape of joint, clamping method etc&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-02-12T06:40:31.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25936" PostTypeId="2" ParentId="25933" CreationDate="2019-02-12T09:41:02.600" Score="0" Body="&lt;p&gt;Depends on the materials. The materials may bind, sealing the joint tightly, or not - one deforming elastically in some places, plastically in others, and as soon as the torque is removed, the elastic deformation areas push back, producing extra gaps in the plastic deformation zones. You would produce a tight seal in most cases if you never remove the torque, keeping the joint stressed... on the other hand, the long-term stress may cause deformations or even fractures in weaker materials completely independently from the seal.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-02-12T09:41:02.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25937" PostTypeId="1" CreationDate="2019-02-12T11:11:31.437" Score="1" ViewCount="442" Body="&lt;p&gt;My small (1.5hp) air compressor is connected to a water trap made of 8mm copper of about 4 metres length. The idea is that you push the compressed air through the piping which cools it, condensing the water out (which you drain away) and thus sending dry air to your air tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the system works, but i would like to understand the maths behind it. I started trying to work it out with help of the below graph from engineeringtoolbox.com but didn't know where to go next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FQMmc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FQMmc.png&quot; alt=&quot;heat dissipation of uninsulated copper tube - lines represent temperature difference.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I mathematically determine the temperature drop of the air in the piping? We can assume an average ambient temperature of 25 degrees Celcius with air coming out of the compressor at 100 degrees Celcius.&lt;/p&gt;&#xA;" OwnerUserId="19335" LastActivityDate="2022-03-31T21:03:50.347" Title="How can I calculate the amount of heat my copper water trap can dissipate?" Tags="|thermodynamics|hvac|piping|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25938" PostTypeId="2" ParentId="25896" CreationDate="2019-02-12T12:34:20.703" Score="0" Body="&lt;p&gt;You can look at that through voltage at terminals of DC motor, because with PMW, You are controlling input voltage. That is connection, and how input voltage affects its speed, well it depends on the parameters of motor. For separately excited motor, use this equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$U = RaI + k\Phi n$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where U is input voltage, Ra is armature resistance, and I is armature current, n is speed of motor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If You are using PMW to control excitation voltage, then you only affect the excitation, so it will change speed in different way&lt;/p&gt;&#xA;" OwnerUserId="19267" LastEditorUserId="19267" LastEditDate="2019-02-12T12:40:51.240" LastActivityDate="2019-02-12T12:40:51.240" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25939" PostTypeId="1" CreationDate="2019-02-12T12:42:34.383" Score="0" ViewCount="34" Body="&lt;p&gt;I was seeing how an electromagnet could attract a car when it's destroyed, and I was wondering, because I have to find the water tubes in my house (20/40 cm depeer), if I can put a wire inside the tubes and move the magnet until I find the force of the attraction of the two elements.&#xA;Unknown evaluation: any electromagnet/wire is good enough? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expected result: the electromagnet will stick to the wire path.&lt;/p&gt;&#xA;" OwnerUserId="19337" LastActivityDate="2019-02-12T15:09:37.413" Title="Can a electromagnet find a tube underground?" Tags="|electromagnetism|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25940" PostTypeId="2" ParentId="25896" CreationDate="2019-02-12T12:59:00.770" Score="1" Body="&lt;p&gt;In a complex way that is only approximately linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you have the static friction - at lowest duty cycles the motor won't budge until it overcomes the static friction, and once it does move, the dynamic friction is lower, so it will immediately get up to a somewhat higher RPM which you could then dial down lowering the duty cycle (so, a hysteresis effect, zero on the up-ramp, decreasing RPM on the down-ramp).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then losses and efficiency start playing a role; the dynamic friction offsets the curve from y=ax by a certain value, magnetic field losses will make the curve bend, as both very low and very high field strength create more magnetic losses than &quot;nominal&quot; (the field of the rotor extending outside the stator and short-circuiting through structural elements and the air), and at higher RPM viscous friction of lubricant in bearings and air drag will apply a quadratic RPM drop-off factor. Add to that response to rapid changes of input delayed by inertia of the rotor, add induction reactions due to rapid voltage changes of PWM, and you're getting a model complex enough there's no point trying to write exact equations - either apply the linear approximation accepting the resulting error, or measure the response, and create a look-up table for input:output values.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-02-12T12:59:00.770" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25942" PostTypeId="2" ParentId="25939" CreationDate="2019-02-12T15:09:37.413" Score="1" Body="&lt;p&gt;Actually if you can run the wire through the tube there is a simpler method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Send a radio signal through the wire and use a detector tuned to that signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This requires some specialized equipment though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tube was buried recently there may be a tracer wire already next to it you can use for this.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2019-02-12T15:09:37.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25943" PostTypeId="1" AcceptedAnswerId="25944" CreationDate="2019-02-12T18:22:14.380" Score="3" ViewCount="31928" Body="&lt;p&gt;for this question: &quot;What is the difference between strength, hardness and toughness in materials?&quot; i have searched and have found these following definitions&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Strength refers to resistance to deformation, and also to a large&#xA;  elastic range. In the Elastic region of the stress-strain&#xA;  relationship, the relationship is described by a linear function, such&#xA;  that σ = E ϵ, where σ is the stress, E is the Elastic modulus, and ϵ&#xA;  is the strain.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Toughness is the resistance to failure or crack propagation. It is&#xA;  somewhat related to strength. Very strong materials will have low&#xA;  toughness, i.e. low tolerance for flaws or defects, i.e. incipient&#xA;  cracks.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Reference&#xA;  &lt;a href=&quot;https://www.physicsforums.com/threads/difference-between-toughness-and-strength.67220/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.physicsforums.com/threads/difference-between-toughness-and-strength.67220/&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i don't understand those definitions. Aren't deformation and failure one? Are toughness, strength and hardness both the ability to resist external forces?&lt;/p&gt;&#xA;" OwnerUserId="19339" LastActivityDate="2021-08-22T08:30:55.583" Title="Strength vs. Hardness vs. Toughness" Tags="|materials|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25944" PostTypeId="2" ParentId="25943" CreationDate="2019-02-12T19:58:28.650" Score="6" Body="&lt;p&gt;Strength = Ability of a material to withstand an applied load. There are several different measures of strength. Two common measures are the ability to withstand a load without plastic deformation (yield strength) or without failure (ultimate strength). In the sketch below, Material 1 has higher strength than Material 2. It can carry more load both before deforming plastically and before failing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Toughness = Ability of a material to absorb energy without fracture. In a stress-strain curve, the area under the curve is often considered a measure of toughness. In the sketch below, Material 2 has higher toughness than Material 1. (I should have drawn them to be more obviously different, but let’s say the area under the Material 2 curve is greater than the area under the Material 1 curve.) So, Material 2 may have a lower strength than Material 1 but it is able to absorb more total energy before failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hardness = ability of a material to resist plastic deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the sketch reflects two hypothetical (imaginary) material curves. Many other stress-strain curve profiles are possible. Also note that the description of toughness presented here is based on general material toughness, which is only one way to assess toughness. There is also impact toughness, notch toughness, and fracture toughness (the &lt;a href=&quot;https://www.nde-ed.org/EducationResources/CommunityCollege/Materials/Mechanical/Toughness.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;NDT resource center&lt;/a&gt; offers an introductory discussion of each) but these quantitative toughness measures are outside my personal experience so I won’t attempt to comment on them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dx1dXl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dx1dXl.png&quot; alt=&quot;stress strain curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2019-02-13T18:15:29.793" LastActivityDate="2019-02-13T18:15:29.793" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="25945" PostTypeId="2" ParentId="25943" CreationDate="2019-02-12T20:12:02.463" Score="1" Body="&lt;p&gt;Hardness is the measure of a material resistance to scratching, like it's hard to drill a hole into, or hard to sand. Or diamond that can cut many surfaces but is hard to cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Toughness is the ability of material to resist cracking or breaking under stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strength is the ability of material to withstand great tension or compression or other forces. Like a steel cable that can support great tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This properties overlap at certain functionalities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have added a diagram that clarifies this more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i0bPu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i0bPu.jpg&quot; alt=&quot;digram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is &lt;a href=&quot;https://www.polymersolutions.com/blog/wp-content/uploads/2015/09/Tough-Strong-Hard-Infographic.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;link to the site&lt;/a&gt; to the site.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-12T20:12:02.463" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25947" PostTypeId="1" AcceptedAnswerId="25951" CreationDate="2019-02-12T21:24:57.607" Score="0" ViewCount="350" Body="&lt;p&gt;I have a question that I can't solve.&#xA;Coefficient of drag= x,&#xA;Coefficient of Lift= y,&#xA;it is given that x= 0.015 + 0.05*(y^2), FInd Max Lift to drag ratio&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I know that Lift to drag ratio is L/D= Coefficient of lift/Coefficient of drag. Can anyone tell me how to solve this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="17811" LastActivityDate="2019-02-14T00:58:58.087" Title="Max Lift to drag ratio" Tags="|aerodynamics|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25948" PostTypeId="1" CreationDate="2019-02-12T21:45:11.717" Score="0" ViewCount="34" Body="&lt;p&gt;Maybe you have heard about Sonomotors and its futuristic electric car, the Sion.&#xA;If not, here is the link to its website: &lt;a href=&quot;https://sonomotors.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://sonomotors.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S6MBC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S6MBC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, the manufacturer claims that this car can recharge up to 30km on one day using its own solar-cells. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come to Physics Stackexchange, because I want to know: &lt;strong&gt;Can this be true ?&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I am looking for well-researched, fact based, answers without any bias for or against this car.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: got some good answers... but let's really crack it down... Can we chose Germany for instance, look at the solar radiation per month and compute an expected range per month? &lt;/p&gt;&#xA;" OwnerUserId="9039" LastActivityDate="2019-02-13T12:14:17.793" Title="Sonomotors, Sion: Can this electric car really charge up to 30km only from its solar cells?" Tags="|electrical-engineering|solar-energy|car|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25950" PostTypeId="2" ParentId="25928" CreationDate="2019-02-12T21:49:33.087" Score="0" Body="&lt;p&gt;Starting with Liquid Nitrogen at:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \rho = 740 \frac{kg}{m^3}, m = 7.4 kg, \hat{u}= -93 \frac{kJ}{kg}, P = 4 bar, U=-690 kJ $$&lt;/span&gt;&#xA;Two things:&#xA;1) You can't drain all the liquid from the tank because you've said that &lt;span class=&quot;math-container&quot;&gt;$Q = \frac{m_{liquid}}{m_{vapor}}=0 $&lt;/span&gt; (saturated liquid)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) If the specific internal energy of the nitrogen is &lt;span class=&quot;math-container&quot;&gt;$\hat{u} = -93 \frac{kJ}{kg}$&lt;/span&gt;, then the specific internal energy of the liquid that you drain from the tank will be &lt;span class=&quot;math-container&quot;&gt;$\hat{u} = -93 \frac{kJ}{kg}$&lt;/span&gt;. If you drain 7.54 kg of liquid from the tank, as you did before, you'll reach this thermodynamic state:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \rho = 4.6 \frac{kg}{m^3}, m = 0.046 kg, \hat{u}= -93 \frac{kJ}{kg}, P = 0.23 bar, U=-4.27 kJ $$&lt;/span&gt;&#xA;So the pressure is actually lower, but the specific internal energy is the same.  I looked this up using density and specific internal energy.  The problem is that this still looks like you've added internal energy to the tank, because it's a smaller negative number.  Internal energy is given on a relative scale.  As Jeffrey J Weimer pointed out, the absolute internal energy of the fluid is the sum of kinetic and potential energies of all the fluid particles, which can never be negative.  Maybe the simplest answer is that you certainly removed energy from the tank by removing mass, but engineering thermodynamics was mostly invented before the atomic model, or the idea of absolute zero temperature, or so many important models in physics. Really, engineers don't use internal energy, they use the change in internal energy:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \delta U = \delta W + \delta Q$$&lt;/span&gt;&#xA;Which is saying that you can increase the internal energy of the fluid by adding heat to the fluid, or doing work on the fluid, but those are your only choices.&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Internal_energy&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Internal_energy&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4058" LastActivityDate="2019-02-12T21:49:33.087" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25951" PostTypeId="2" ParentId="25947" CreationDate="2019-02-12T21:52:53.657" Score="-2" Body="&lt;p&gt;The lift(y) to drag(x) ratio can be expressed as:&#xA;&lt;span class=&quot;math-container&quot;&gt;$LD_{ratio} = y/x = \frac{y}{0.015 + 0.05*(y^2)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will get the maximum when the denominator is 0, so when:&#xA;&lt;span class=&quot;math-container&quot;&gt;$0.015 + 0.05*(y^2)=0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, only when &#xA;&lt;span class=&quot;math-container&quot;&gt;$$y = 0. - 0.547723i$$&lt;/span&gt;&#xA;or&#xA;&lt;span class=&quot;math-container&quot;&gt;$$y = 0. + 0.547723i$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...are you sure about the formula ? &lt;/p&gt;&#xA;" OwnerUserId="9039" LastEditorUserId="9039" LastEditDate="2019-02-13T06:51:07.613" LastActivityDate="2019-02-13T06:51:07.613" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25952" PostTypeId="2" ParentId="25947" CreationDate="2019-02-12T23:03:30.543" Score="1" Body="&lt;p&gt;As an alternative to solving with calculus, if you are provided with a graph of the L/D ratio, you can find the line which passes through the origin (0 lift and 0 drag), and is tangential to the L/D line on the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point at which this straight line touches the L/D graph is the maximum L/D.&lt;/p&gt;&#xA;" OwnerUserId="19260" LastActivityDate="2019-02-12T23:03:30.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25953" PostTypeId="2" ParentId="25937" CreationDate="2019-02-13T00:36:35.780" Score="0" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T_{out} = \frac{\dot{Q}}{\dot{m}C_p}+T_{in}$$&lt;/span&gt;&#xA;Where T is temperature, &lt;span class=&quot;math-container&quot;&gt;$\dot{Q}$&lt;/span&gt; is heat rate, &lt;span class=&quot;math-container&quot;&gt;$\dot{m}$&lt;/span&gt; is mass flowrate and &lt;span class=&quot;math-container&quot;&gt;$C_p$&lt;/span&gt; is the specific heat of air at constant pressure.  To check the units:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$T_{out}[C] = \frac{\dot{Q} \left[\frac{J}{s}\right]}{\dot{m}\left[\frac{kg}{s}\right]C_p\left[\frac{J}{kgC}\right]} + T_{in} [C]$$&lt;/span&gt;&#xA;To clarify, &lt;span class=&quot;math-container&quot;&gt;$\frac{J}{s} = W$&lt;/span&gt;, and these are SI units (Joule, kilogram, Celsius).&lt;/p&gt;&#xA;" OwnerUserId="4058" LastEditorUserId="4058" LastEditDate="2019-02-13T15:35:57.400" LastActivityDate="2019-02-13T15:35:57.400" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25954" PostTypeId="1" CreationDate="2019-02-13T01:52:52.630" Score="0" ViewCount="40" Body="&lt;p&gt;When portholes are put into a cylindrical hull, such as of a submarine or an airplane fuselage, is the glass window curved to match the curvature of the cylinder?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if so, is the porthole wrapped around the cylinder so that orthographically it would appear wider than it is tall, but it retains the area of a circle,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OR&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is the porthole projected onto the side of the cylinder, so that it looks like a circle when viewed orthographically.&lt;/p&gt;&#xA;" OwnerUserId="19347" LastActivityDate="2019-02-13T18:44:20.430" Title="How are circular portholes mapped onto cylindrical hulls?" Tags="|naval-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25956" PostTypeId="2" ParentId="25943" CreationDate="2019-02-13T02:40:12.267" Score="-1" Body="&lt;ul&gt;&#xA;&lt;li&gt;Strength = Stress required to produce failure. It is often used to determine how much force an object can take without breaking, but there are many measures of strength.&lt;/li&gt;&#xA;&lt;li&gt;Hardness = Force / Area. The larger the area produced by a given force, the lower the hardness. It can be correlated to, but is not the same as strength.&lt;/li&gt;&#xA;&lt;li&gt;Toughness = Energy required to form a crack. Ceramics have low toughness because cracks form easily.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="6446" LastActivityDate="2019-02-13T02:40:12.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25957" PostTypeId="2" ParentId="25947" CreationDate="2019-02-13T07:40:13.293" Score="3" Body="&lt;p&gt;As has been mentioned above by Eric Shain the correct answer is to take the partial with respect to y= CL and equate it to zero. Let's Call the max L/D, M.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M = \frac{C_L}{C_D} = \frac{C_L}{C_{D_0} +K*C_L^2}, \quad C_{D_0}=0.015\ and\ K=0.05 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{\partial M}{\partial C_L} =\ \frac{\partial}{\partial C_L}&#xA;[ \frac{C_L}{0.015 +0.05*C_L^2}]\ =\quad \frac{[0.015 + 0.05*C_L^2&#xA;]-C_L [0 + 2(0.05)C_L^2]}{[0.015 + 0.05*C_L^2&#xA;]^2} =0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Rightarrow 0.015- 0.05C_L^2=0\\ C_L^2=0.015/0.05=0.3\\C_{L_M}=\sqrt{0.3} \quad and\ C_{D_M}= 2*C_{D_0}=0.03$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And &lt;span class=&quot;math-container&quot;&gt;$$M=\frac{C_{L_M}}{C_{D_M}}= \frac{\sqrt{0.3}}{0.03}=18.257$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum lift to drag ratio is 18.257. please double check my arithmetic, but it looks correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&#xA;The following is a quote from Wikipedia on lift to drag ratio. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A House sparrow has a 4:1 L/D ratio, a Herring gull a 10:1 one, a Common tern 12:1 and an Albatross 20:1, to be compared to 8.3:1 for the Wright Flyer to 17.7:1 for a Boeing 747 in cruise.[4] A cruising Airbus A380 reaches 20:1.[5] The Concorde at takeoff and landing had a 4:1 L/D ratio, increasing to 12:1 at Mach 0.95 and 7.5:1 at Mach 2.[6] A Helicopter at 100 kn (190 km/h) has a 4.5:1 L/D ratio.[7] A Cessna 172 glides at a 10.9:1 ratio.[8] A cruising Lockheed U-2 has a 25.6 L/D ratio.[9] The Rutan Voyager had a 27:1 ratio and the Virgin Atlantic GlobalFlyer 37:1&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-02-14T00:58:58.087" LastActivityDate="2019-02-14T00:58:58.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25958" PostTypeId="2" ParentId="25948" CreationDate="2019-02-13T12:14:17.793" Score="1" Body="&lt;p&gt;It could well be true, but that doesn't mean it is useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pick a day in the middle of summer, with a clear sky. Go to a flat test track. Spend 6 hours (three hours each side of midday) driving round it at walking pace, say 5km/h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oops, I forgot to tell you to inflate the tires to 50% higher pressure than normal, to reduce rolling resistance. They won't blow if you are only doing 5km/h on a perfectly maintained surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Congratulations, you have now traveled 30km in a day using only solar cells. But so what?&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-02-13T12:14:17.793" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25959" PostTypeId="1" CreationDate="2019-02-13T14:41:00.823" Score="0" ViewCount="42" Body="&lt;p&gt;I'm trying to build a keyboard addon for my smartphone (&lt;a href=&quot;https://gitlab.com/xengi/mkbd65&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://gitlab.com/xengi/mkbd65&lt;/a&gt;). The last challenge I'm facing is how the two pieces, namely the smartphone and the keyboard should slide back and forth.&#xA;I know I need some kind of linear slider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like that: &lt;a href=&quot;http://en.smoothgroup.cc/goods/detail/58.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.smoothgroup.cc/goods/detail/58.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I construct or implement something like that? Is there a better way to do it? What do I need to consider which I maybe haven't thought about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think I'm able to do something like that: &lt;a href=&quot;http://phonesdata.com/files/models/Nokia-N950-482.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://phonesdata.com/files/models/Nokia-N950-482.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my goal is to build something like that: &lt;a href=&quot;http://phonesdata.com/files/models/Nokia-N900-684.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://phonesdata.com/files/models/Nokia-N900-684.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help appreciated!&lt;/p&gt;&#xA;" OwnerUserId="19353" LastActivityDate="2019-02-13T14:41:00.823" Title="How to slide 2 small pieces of metal into each other" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25960" PostTypeId="1" CreationDate="2019-02-13T15:15:41.953" Score="1" ViewCount="1230" Body="&lt;p&gt;I was trying to drive up an icy hill with a front-wheel-drive car (2006 Honda Civic Hybrid) when the front tires lost traction and the car started to slide backwards. The car then easily spun sideways as it slid backwards making it veer left/right and I nearly hit a telephone pole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I repeated this a few times hoping to get up the hill. It seemed like I had much better control of the car during the backwards slide if I shifted the car into reverse (instead of drive) the moment I lost traction. In reverse, it didn't seem like the car would spin sideways. In these &quot;Reverse vs Drive Experiments&quot;, I think I was sliding backwards at about the same speed (just &quot;riding the brake&quot; and not touching the accelerator pedal), but the reverse mode still seemed much more effective at keeping my car straight (while drive mode seemed to cause a spin-out with more random left/right motion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My conclusion is that I am much safer changing to reverse mode if I start sliding backwards. Is this expected?&lt;/p&gt;&#xA;" OwnerUserId="2132" LastActivityDate="2019-02-13T16:48:07.313" Title="Reverse vs Drive When Slipping Backwards" Tags="|automotive-engineering|safety|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25962" PostTypeId="2" ParentId="25960" CreationDate="2019-02-13T16:48:07.313" Score="2" Body="&lt;p&gt;The control of a typical motor vehicle is based on the traction of the steering wheels in relation to the surface on which it operates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you describe, your slipping is happening when you are attempting to travel forward, but gravity is  interfering with your plans. In a forward gear selection, the wheels will attempt to continue to operate in that direction, constantly breaking traction with the road. This results in the described veering, as the vehicle no longer has appropriate guidance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you shift to reverse, the tires will rotate and gain some traction while the vehicle travels downward. This provides for steering control, just as you've described.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even though you are &quot;riding the brake,&quot; you are modulating the braking force from full-on to zero, but at the lower braking levels, the wheels are allowed to rotate in the direction of travel, providing steering control. In the forward gears, your braking force may also be full-on to zero, but at the lower braking levels, the engine is continuing to remove the traction desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also note that you can shift to neutral to accomplish the same (improved steering) results.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-02-13T16:48:07.313" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25963" PostTypeId="2" ParentId="25954" CreationDate="2019-02-13T18:44:20.430" Score="1" Body="&lt;p&gt;Commercial aircraft windows (for the passengers) are cylindrical to match the cylindrical body of the plane.  This is both stronger -- matching shape -- and more or less maintains a smooth, even drag surface.  Their boundaries are near-rectangular just to match internal view angles from the seats.&lt;br&gt;&#xA; OTOH,  the cockpit windows are close to planar so as to minimize any view distortions that could make navigation difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to submarines -- there aren't any on military (in general) subs;  research subs designed for deep investigations are as close to spherical as possible, and the glass is shaped that way as well. This is because strength is the main constraint, and as you probably know a sphere is the strongest configuration for pressure resistance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Portholes on surface ships may be round for various historical reasons, but since they are usually above the water surface, they can be manufactured planar, saving cost and still meeting any strength needs. &lt;/p&gt;&#xA;" OwnerUserId="615" LastActivityDate="2019-02-13T18:44:20.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25964" PostTypeId="1" AcceptedAnswerId="25965" CreationDate="2019-02-14T00:57:04.287" Score="2" ViewCount="86" Body="&lt;p&gt;I work in a pharmaceutical plant where saline solution is produced and bottled. The environment has to be clean (by pharma standards), which rules out many greases and other protective coatings.&#xA;As expected from a saline environment (from my experience it can be as agressive as a marine one), the corrosion over all parts in the room is heavy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have mostly controlled the corrosion of structures (all of them have to be made of stainless steel: 316L for those parts in contact with the product, 316 for some others and some structural parts of the equipment are made of 304 SS) by carefully selecting materials, specially material pairs (when I have to), introducing electrical discontinuities in structural joints (if I have a bronze piece, for example, in contact with a SS one, I introduce teflon washers/separators to reduce galvanic corrosion), and at least in one case by using a sacrifice anode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the corrosion of electrical components and wiring is something I haven't been able to control, to the point that I've had to replace all the wiring in one of the machines in as little as 5 months. Wires (not only power lines, but also control ones, such as termocouple wiring) get rusted not only in the terminals (where it's evident), but also under their plastic sleeves. In the case of control wiring, this causes noisy signals that result in machine stoppages and downtime (my most recent case is a termocouple that sent a correct signal of 180°C but showed (unreal) peaks of up to 270°C that, of course, triggered an alarm in the PLC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only blame it on corrosion, as the problem goes away by replacing all the wiring and terminals. The wires I took off show corroded under the sleeves, and some of them were brittle. (If you think it's not corrosion, I'd like to hear your opinion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't hide all the wiring from the product (in particular some of the termocouples my get some product spray) and my electrical boxes are as separated from product and tigth as possible (even they don't hold to any IP standard).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there anything I can do to prevent this sort of corrossion in electrical components and wiring?&lt;/li&gt;&#xA;&lt;li&gt;Is there anything akin to a sacrifice anode that I can use in electrical parts / wiring?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17733" LastActivityDate="2019-02-20T17:00:28.673" Title="corrosion in electrical parts and wiring" Tags="|electrical|corrosion|maintenance|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25965" PostTypeId="2" ParentId="25964" CreationDate="2019-02-14T01:44:25.923" Score="2" Body="&lt;p&gt;I guess you are using iron/Constantan thermocouples for the relatively low temperatures; so iron is going to corrode badly . Can you change to chromel / alumel couples ? It would require some instrumentation change as they have lower emf voltage. Chromel / alumel is a very common corrosion resistant pair generally used for higher temperatures, but I do remember using them for ambient temperature measurements. Of course an anode connected to a thermocouple would defeat the purpose of the couple. Possibly a coating like a liquid epoxy that the thermocouples could be coated with after they are installed.   Can non contact infra-red temperature sensors be used ?&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-02-14T01:44:25.923" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25966" PostTypeId="2" ParentId="25964" CreationDate="2019-02-14T02:33:10.023" Score="0" Body="&lt;p&gt;can you reduce the humidity of your production environment? If moisture can be reduced, the corrosion rates can in general be minimized. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, there are commercially available corrosion inhibitors which you can spray onto the stripped ends of your wires, which will seep into them and shut off access to oxygen and moisture- and thereby shut down the major corrosion mechanisms you are faced with. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The classic one is WD-40 spray, but in the aerospace industry there are others that are used to corrosion-proof the insides of aircraft; you can probably find references to these on the web. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;there are also aerosol lacquers which are used to corrosion-proof electronic assemblies that are used in humid tropical environments; these were first developed during WWII and are still in use today. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-02-14T02:33:10.023" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25967" PostTypeId="1" CreationDate="2019-02-14T05:17:25.753" Score="0" ViewCount="688" Body="&lt;p&gt;I am trying to perform structural analysis of composite pressure vessel in APDL/ANSYS. I have completed the solution and now I want to extract ABD matrix.Do anyone have an idea about this? &lt;/p&gt;&#xA;" OwnerUserId="19325" LastEditorUserId="9" LastEditDate="2019-09-06T18:17:45.520" LastActivityDate="2019-09-06T18:17:45.520" Title="How to get laminate stiffness matrices (ABD) in APDL ANSYS?" Tags="|mechanical-engineering|finite-element-method|ansys|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25968" PostTypeId="1" CreationDate="2019-02-14T10:14:14.947" Score="0" ViewCount="33" Body="&lt;p&gt;My field isn't dealing with motors and such at all, however, I now need to build something that controls two motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have two guiding reels with tape between them, similar to old tape recorders. Now the important thing I want to control is the speed that the tape moves and the tension. Another important factor is that I want to be able to switch direction relatively fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I went on the journey of figuring out what type of motors I need and it's quite overwhelming, there are so many types and when I read stuff online it's always &quot;it depends&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am looking at now is buy two brushless dc motors with sensors that can sense the position, an Arduino and two motor drivers like the L6234 (&lt;a href=&quot;https://www.st.com/en/motor-drivers/l6234.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.st.com/en/motor-drivers/l6234.html&lt;/a&gt;). Then use the Arduino to control both motors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot find online if it is possible to change the tension in this way. My idea would be to use one motor to give a virtual load on the other motor to increase tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible using brushless dc motors and would my general setup up? Or do I need something completely different?&lt;/p&gt;&#xA;" OwnerUserId="15945" LastEditorUserId="14244" LastEditDate="2019-02-14T10:29:01.540" LastActivityDate="2019-02-14T10:29:01.540" Title="What motors to use to guide tape with high speed" Tags="|electrical-engineering|control-engineering|motors|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="25969" PostTypeId="1" AcceptedAnswerId="26767" CreationDate="2019-02-14T11:15:57.127" Score="0" ViewCount="77" Body="&lt;p&gt;Below are two cropped views of &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Aerial_telescope&quot; rel=&quot;nofollow noreferrer&quot;&gt;Johannes Hevelius's 8 inch telescope with an open work wood and wire &quot;tube&quot; that had a focal length of 150 feet to limit chromatic aberration.&lt;/a&gt;&quot; from &lt;a href=&quot;https://iiif.lib.harvard.edu/manifests/view/ext:aHR0cHM6Ly9paWlmLmxpYi5oYXJ2YXJkLmVkdS9tYW5pZmVzdHMvaWRzOjE2NDcwMzE1$1i&quot; rel=&quot;nofollow noreferrer&quot;&gt;Harvard University, Houghton Library, pga_typ_620_73_451_fig_aa&lt;/a&gt; (found &lt;a href=&quot;https://hollis.harvard.edu/primo-explore/fulldisplay?context=L&amp;amp;vid=HVD2&amp;amp;search_scope=everything&amp;amp;tab=everything&amp;amp;lang=en_US&amp;amp;docid=01HVD_ALMA211764767520003941&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;) the first of which I've zoomed and enhanced the contrast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've used these images in two other questions in Astronomy SE:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/29317/7982&quot;&gt;How did Johannes Hevelius' long telescope work? Why all the round holes?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/q/29597/7982&quot;&gt;How does making a refracting telescope very long reduce the chromatic aberration of an uncorrected lens?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://astronomy.stackexchange.com/a/29324/7982&quot;&gt;This answer&lt;/a&gt; provides a &lt;a href=&quot;https://archive.org/stream/Machinacoelesti1Heve#page/410/mode/2up&quot; rel=&quot;nofollow noreferrer&quot;&gt;link to the original text and image&lt;/a&gt; but the discussion was published in 1673 and may be difficult to understand or paste into google translate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the structure itself may be familliar to engineers. I'm getting the feeling that the series of square blocks with round holes and string or lines connecting them (presumably under tension?) have no specific optical function, and are purley part of a system to keep the long &quot;telescope tube&quot; stiff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this combination of blocks and lines a recognizable implementation of some named structure or technique? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iFTbX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iFTbX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1I0eh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1I0eh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6264" LastActivityDate="2019-04-06T22:19:01.583" Title="Is the combination of blocks and lines in this old, very long telescope an implementation of some named structure or technique?" Tags="|structural-engineering|terminology|engineering-history|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25971" PostTypeId="1" CreationDate="2019-02-14T12:42:50.767" Score="-1" ViewCount="54" Body="&lt;p&gt;I am interested in mathematically describing as well as formalising a manufacturing plant so that other people in the engineering field understand how the system works. I am not from the engineering field, and so I don't know what the typical standards for this type of task are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the manufacturing plant consists of three work units: A warehouse, a transport system and a processing station. The materials are transported from the warehouse to the processing station, where they are processed and then stored again. This process repeats itself indefinitely. What is the best and most professional way how an engineer would describe this system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far, I have described the process textually, but I shall only explain it by mathematical formulas or figures. Are there various specifications or guidelines in the engineering environment? I don't know what a good starting point is.&lt;/p&gt;&#xA;" OwnerUserId="14138" LastActivityDate="2019-02-15T06:21:56.043" Title="What is the engineering way to describe a factory / manufacturing plant system in a formel and mathematical way" Tags="|modeling|manufacturing-engineering|mathematics|systems-engineering|diagram|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25973" PostTypeId="2" ParentId="8825" CreationDate="2019-02-14T13:27:30.697" Score="2" Body="&lt;p&gt;Here the small bubbles are formed due to low pressure occurred at the discharge port of the pump. Generally cavitation is formed due to the sudden fluctuations in the fluid i.e. sudden decrease and increase or vice versa in the pressure of the fluid. When the pressure of the fluid is above vapor pressure there will not be any cavitation effect but as it decreases below vapor pressure it started forming cavitation bubbles. When the fluid pressure starts decreasing below vapor (saturation) pressure and at a particular pressure below vapour pressure i.e. at negative pressure (negative pressure is also known as tensile strength of the fluid), the fluid starts forming small cavitation bubbles (cavitation nuclei) which are in the size of microns or lesser. As the pressure decreases more, bubble size starts to increase and it increases up to certain limit. The maximum bubble size depend on pressure fluctuations and fluid properties. Now, when the pressure increases above vapor pressure then formed bubbles starts collapsing. The growth and the collapse of the bubble produce the sound which is also known as the cavitation noise. At the time of collapse of the bubble, the pressure generated is very high which will erode the material of the pump very fastly and it will also affect performance and the life of the pump. Small impurities present in the fluid such as in condensed gases, solid particles, etc. also increases cavitation effect.&#xA;Cavitation bubbles are also formed due to the high flow rate. Cavitation can be avoided by maintaining pressure in the system above vapor pressure or above atmospheric pressure. High flow rate and velocity of the fluid is also reason of the formation small cavitation bubbles. Use standards for maintaining the flow rate and velocity of the fluid.&lt;/p&gt;&#xA;" OwnerUserId="19364" LastEditorUserId="19364" LastEditDate="2019-02-15T08:07:40.473" LastActivityDate="2019-02-15T08:07:40.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25974" PostTypeId="2" ParentId="13025" CreationDate="2019-02-14T13:46:13.120" Score="1" Body="&lt;p&gt;1) I usually use online voronoizing tool (&lt;strong&gt;&lt;a href=&quot;https://www.voronator.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.voronator.com/&lt;/a&gt;&lt;/strong&gt;) for generating voronoization in 3D file. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The website asks you to upload your 3D file for generating the&#xA;voronoi.&lt;/li&gt;&#xA;&lt;li&gt;It supports formats like &lt;strong&gt;stl, stla, stlb, ply, off, obj, dae, 3ds&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;but the file has to be less than &lt;strong&gt;20mb&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;li&gt;You can even import a file which is already meshed.&lt;/li&gt;&#xA;&lt;li&gt;There is also an advance settings option, where you can select the&#xA;output format, number of holes and thickness of the layer.&lt;/li&gt;&#xA;&lt;li&gt;You can get output in two formats, either an &lt;strong&gt;STL&lt;/strong&gt; or &lt;strong&gt;PLY&lt;/strong&gt; file.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;2) The second solution would be &lt;strong&gt;&quot;Voronoi Sketch Generator&quot;&lt;/strong&gt; add-in for fusion 360. The add-in asks you for some reference values to generate a sketch. You have options to select the number holes and also you can choose whether the holes have to be straight or curved. &#xA;There are also plenty of youtube tutorial videos on how to use this tool which can come in handy for you. I'll post the links of videos which I found useful for me.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/9YEEd81MeCU&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/9YEEd81MeCU&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/6q51guUtB4I&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/6q51guUtB4I&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What ever may be the option, but you still have to export and import across the softwares to have job done.&lt;/p&gt;&#xA;" OwnerUserId="19362" LastActivityDate="2019-02-14T13:46:13.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25977" PostTypeId="2" ParentId="25937" CreationDate="2019-02-14T14:19:18.033" Score="0" Body="&lt;h1&gt;Pipe (Dry) Air Flow&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Theory&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Start from the air flow through the pipe. The energy balance equation below considers only convection from the outer air and takes a lumped analysis of the fluid in the pipe. In this, P is the perimeter length of the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{m}\tilde{C}_p\frac{dT}{dx} = - h_a P (T - T_a) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the tube length &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; and the entering hot temperature &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt;, we recast this as a dimensionless expression using &lt;span class=&quot;math-container&quot;&gt;$\Theta = (T - T_a)/(T_h - T_a)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$X = x/L$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{d\Theta}{dX} = -\beta \Theta $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dimensionless factor &lt;span class=&quot;math-container&quot;&gt;$\beta = h_a P L / \dot{m}\tilde{C}_p$&lt;/span&gt; is the ratio of heat leaving the pipe by convection to the heat entering the pipe with the fluid flow. The factor &lt;span class=&quot;math-container&quot;&gt;$PL$&lt;/span&gt; is the total external tube area &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;. With the boundary condition &lt;span class=&quot;math-container&quot;&gt;$\Theta = 1$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$X = 0$&lt;/span&gt;, we have the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln(\Theta) = -\beta X $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a plot of &lt;span class=&quot;math-container&quot;&gt;$\Theta$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$\beta = 2$&lt;/span&gt; (black upper curve) and &lt;span class=&quot;math-container&quot;&gt;$\beta = 3$&lt;/span&gt; (blue lower curve).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fb032.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fb032.png&quot; alt=&quot;Temperature Profile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Typical Values&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The pipe in your problem has a DIAMETER of 8 mm and a length of 4 m. This gives &lt;span class=&quot;math-container&quot;&gt;$A = (4)(2)\pi(0.004) = 0.1$&lt;/span&gt; m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stagnate air outside the tube has a value &lt;span class=&quot;math-container&quot;&gt;$h_a \approx 5$&lt;/span&gt; W/m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C [&lt;a href=&quot;https://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;engineering toolbox&lt;/a&gt;]. Air has a specific heat of &lt;span class=&quot;math-container&quot;&gt;$\tilde{C}_p \approx 0.7$&lt;/span&gt; J/g &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C [&lt;a href=&quot;https://www.ohio.edu/mechanical/thermo/property_tables/air/air_cp_cv.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ohio MAE Website&lt;/a&gt;].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass flow of air as an ideal gas is &lt;span class=&quot;math-container&quot;&gt;$\dot{m} = M(p\dot{V}/RT)$&lt;/span&gt;. At 25 &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C and 1 bar, this gives a relationship for mass flow (g/s) to volumetric flow (m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;/s) as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dot{m} = 28(101325\dot{V})/((8.314)(298)) = 1150 \dot{V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, we have the relationship for &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; to mass flow (g/s) or STP volumetric flow (m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;/s) as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \beta = 0.71/\dot{m} = 6.2 \times 10^{-4}/ \dot{V} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we take a mass flow of 1 g/s of air through the tube. This gives &lt;span class=&quot;math-container&quot;&gt;$\beta \approx 0.7$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt; To proceed further, we need the mass or volumetric flow through the tube.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Condensation of Water Vapor&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Water vapor will condense when its partial pressure is equal to (or above) its vapor pressure. We can consider &lt;a href=&quot;https://en.wikipedia.org/wiki/Antoine_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Antoine equation&lt;/a&gt; to track the vapor pressure as a function of temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \log( p_{vap} ) = A - \frac{B}{C + T} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first relationship we need to make is to translate the curve of &lt;span class=&quot;math-container&quot;&gt;$\Theta$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; to a corresponding curve of &lt;span class=&quot;math-container&quot;&gt;$p_{vap}$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt;. For ease in what follows, define that relationship as &lt;span class=&quot;math-container&quot;&gt;$AE \equiv 10^{\log (p_{vap})} = f(\Theta)$&lt;/span&gt;. We can then use the &lt;a href=&quot;https://en.wikipedia.org/wiki/Relative_humidity&quot; rel=&quot;nofollow noreferrer&quot;&gt;relative humidity&lt;/a&gt; of the incoming air to obtain the &lt;span class=&quot;math-container&quot;&gt;$p_{H_2O}$&lt;/span&gt; in the incoming stream based on it &lt;span class=&quot;math-container&quot;&gt;$p_{vap}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ p_{H_2O}(X = 0) = RH_o\ p_{vap}(X = 0) = RH_o\ AE(\Theta(X = 0)) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The incoming stream has an initial value of &lt;span class=&quot;math-container&quot;&gt;$p_{H_2O}(X = 0)$&lt;/span&gt;. We have a plot of &lt;span class=&quot;math-container&quot;&gt;$p_{vap}(X)$&lt;/span&gt; from &lt;span class=&quot;math-container&quot;&gt;$AE(\Theta(X))$&lt;/span&gt;. When &lt;span class=&quot;math-container&quot;&gt;$p_{vap}(X)$&lt;/span&gt; falls below the value &lt;span class=&quot;math-container&quot;&gt;$p_{H_2O}(X = 0)$&lt;/span&gt;, water vapor will condense. The analysis needed to find the point &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; when condensation starts is left as an exercise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Starting at the point where water vapor condenses, the latent enthalpy of the exothermic condensation will act as an additional heat source in the energy balance. The temperature of the air will fall less rapidly. This analysis is also left as an exercise.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-02-15T03:09:54.273" LastActivityDate="2019-02-15T03:09:54.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25978" PostTypeId="2" ParentId="21866" CreationDate="2019-02-14T15:17:14.000" Score="0" Body="&lt;p&gt;I was going to suggest a Streacom case, I've had one for years and it works excellently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at your original design, if you could get the triangular block made as a vapour chamber instead of a solid block it might actually work.&#xA;You'd need to consult with a company that makes heat pipes to see if the triangular shape would work or not.&#xA;Given the diagrams I've seen of how they normally work I'm guessing not due to the varying and long distances between input and output surfaces, but would be an interesting engineering exercise that could have unexpected results.&#xA;Who knows, you might have stumbled on a great design and they could manufacture it for you...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your build, these high density heatpipes might be of interest:&#xA;&lt;a href=&quot;https://www.fanlesstech.com/2018/09/flat-cool-pipes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.fanlesstech.com/2018/09/flat-cool-pipes.html&lt;/a&gt;&#xA;&lt;/p&gt;&#xA;" OwnerUserId="19368" LastActivityDate="2019-02-14T15:17:14.000" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="25979" PostTypeId="1" CreationDate="2019-02-14T16:03:50.823" Score="2" ViewCount="59" Body="&lt;p&gt;From: &lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4033560/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4033560/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that concrete can really go bad after a few hundred freeze and unfreeze cycles: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j3Roa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j3Roa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If an ICF house has insulation from the inside of the house, then the concrete will be cold, if the outside temperature is cold too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More detail on the layers: insulation for the ICF is grey EPS, 15cm outside, 15cm concrete, inside: 5cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question&lt;/strong&gt;: for how many years can concrete survive without any issue in an ICF house? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, because of the 5cm grey EPS insulation from inside, the concrete will receive many freeze/unfreeze states and will go bad earlier/deform/crack?. Living in ~eastern EU, winters are max -10C. But it goes about ~60 days under 0C yearly. &lt;/p&gt;&#xA;" OwnerUserId="19367" LastEditorUserId="90" LastEditDate="2019-02-15T13:52:13.410" LastActivityDate="2019-02-15T13:52:13.410" Title="How does concrete perform in an Insulated Concrete Form (ICF) house?" Tags="|concrete|thermal-insulation|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25980" PostTypeId="1" AcceptedAnswerId="25983" CreationDate="2019-02-14T17:10:12.113" Score="1" ViewCount="60" Body="&lt;p&gt;I have data from some ships, basically the base speed for a list of RPM. But it is know that this speed is not constant and it will change mainly based on whether and load of the ship. Is it any rule of thumb to stime the effect of this varialbes into a ship speed without going to heavy on the math.&lt;/p&gt;&#xA;" OwnerUserId="6590" LastActivityDate="2019-02-14T19:12:39.987" Title="How to estimate a ship speed based on weather conditions?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="25983" PostTypeId="2" ParentId="25980" CreationDate="2019-02-14T19:12:39.987" Score="2" Body="&lt;p&gt;Speed of a ship and power demand and fuel consumption have to do with many factors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are roughly enumerated as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ship's engine power and draft versus speed curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seas' conditions: draft, roughness, undercurrents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wind and meteoritical conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These information are sett into many navigation equipment on board of even small yachts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are taught as part of marine engineering courses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one article covering some of the general concepts and charts, &lt;a href=&quot;https://ac.els-cdn.com/S2468013315300127/1-s2.0-S2468013315300127-main.pdf?_tid=a6cb3bd2-99a9-48b1-a97b-b00096995428&amp;amp;acdnat=1550170555_633ee92829157e4dd3a6b28a73d9d197&quot; rel=&quot;nofollow noreferrer&quot;&gt;On the estimation of ship’s fuel consumption and speed curve: A statistical approach&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-14T19:12:39.987" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="25984" PostTypeId="2" ParentId="25971" CreationDate="2019-02-14T19:37:16.813" Score="1" Body="&lt;p&gt;Solar Mike is correct (as usual); the &lt;em&gt;block diagram approach&lt;/em&gt; is the classic method for &quot;formalizing&quot; the dynamics inside a large and complicated system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it is important to realize that this paradigm is of limited utility in analyzing the actual behavior of a big factory in that it fails to take into account the behavior of the humans that manage the flow of material and information inside that factory, in particular when the factory enters an out-of-control state. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In that case the system can enter a death spiral called &quot;ISOC&quot; or &lt;em&gt;in specification but out of control&lt;/em&gt;, about which great volumes of literature could be written... by me, a recovering ex-engineer, for example.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-02-14T19:37:16.813" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25985" PostTypeId="1" AcceptedAnswerId="25996" CreationDate="2019-02-14T22:13:59.673" Score="2" ViewCount="445" Body="&lt;p&gt;I'm designing a cooling system to cool nitrogen gas stream at 300°C with a mass flow rate of 15.6 g/s moving through an undulating channel in a cylindrical aluminum ring block with external fins. How much mass of aluminum or what special feature is needed to cool 15.6 g of nitrogen moving through and exiting to 25°C EVERY SECOND. It's to be noted that the block is immersed in a large pool of water, and that should be factored in as a cooling advantage. I'll love a method of calculation that allows flexibility with variables like mass flow rate, temperature, and channel diameter.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastEditorUserId="19370" LastEditDate="2019-02-20T11:10:18.063" LastActivityDate="2019-02-23T15:37:44.923" Title="Cooling Hot Nitrogen with an aluminum block" Tags="|thermodynamics|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="25989" PostTypeId="2" ParentId="25971" CreationDate="2019-02-15T06:21:56.043" Score="0" Body="&lt;p&gt;To represent it, like already mentioned, a diagram would be a good tool to use. If you wanted to go a little deeper, some type of simulation software would be of value. For example, Arena is a program that allows you to simulate a process. You could watch a few videos on YouTube about Arena to see if the program would be helpful to you.&lt;/p&gt;&#xA;" OwnerUserId="17829" LastActivityDate="2019-02-15T06:21:56.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="25991" PostTypeId="2" ParentId="25985" CreationDate="2019-02-15T06:58:13.077" Score="-1" Body="&lt;p&gt;Lets say you have Q per second calories of nitrogen heat to loose to the Aluminum for every 1 cm length of your pipe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we have &lt;span class=&quot;math-container&quot;&gt;$$Q_{1\ cm\ length} = \frac{K*A(T_{hot}-T_{cold}  )}{d(R^2_{aluminium}/R^2_{pipe})}\   $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-K= thermal conductivity of the aluminum = &lt;span class=&quot;math-container&quot;&gt;$o.5 (cal/sec)/(cm^2  C/cm)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-t = time, second&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-R = radius&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-T = temperature Celsius&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-A = area of contact per one cm length of pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-d = thickness of aluminum around pipe, approximate 0.8* one side, if the aluminum section is square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So If we have the diameter of the pipe as D, we have A=pi*D and we can calculate d the thickness of aluminum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we have all the dimensions by plugging in the equation we can calculate the Q heat loss by nitrogen per second per each centimeter length of the pipe and if we multiply that by t we get total heat lss after some time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some comments I modify my answer as follows to include the length of the pipe and the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Q_{nitrogen} = \frac{K*A*L(T_{hot}-T_{cold}  )}{d(R^2_{aluminium}/R^2_{pipe})}t\   $$&lt;/span&gt;&#xA;-L=length of pipe&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation gives the Q loss by nitrogen after certain time t passing through the block. So if you need to loose more heat you have to make the pipe length L longer, or change other variables.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-02-15T16:26:27.220" LastActivityDate="2019-02-15T16:26:27.220" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="25994" PostTypeId="2" ParentId="25979" CreationDate="2019-02-15T10:08:22.587" Score="2" Body="&lt;p&gt;I don't have experience with ICF houses specifically, but I do have experience with specifying concretes for outdoor use with significant frost exposure, so that is what I am basing my answer on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you choose the concrete recipe well, 50 years should not be a problem. And with small repairs now and then, neither should 100 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is no worse than using concrete outdoors, where it is directly exposed to the environment. And designing concrete for outdoors exposure for a lifetime of 100 years (with a bit of maintenance) is routine nowadays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also remember that the problem is the number of times, the concrete freezes — it can stay frozen for long periods of time without any problems. And the outer layer of insulation will reduce the number of freeze/thaw cycles significantly. (The concrete may freeze during a winter where the outer temperature reaches -10 degrees, but it will not freeze/thaw repeatedly on days where the temperature cycles between, say, +2 and -2 degrees.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comment on the article, you are linking to: It may interest you to know that where I live (Denmark), all but possibly one of the concrete recipes, they are testing, would be banned from use, if there is significant frost exposure. I believe they deliberately included concrete recipes that they knew would fail.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2019-02-15T10:08:22.587" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25995" PostTypeId="2" ParentId="8825" CreationDate="2019-02-15T15:38:00.460" Score="0" Body="&lt;p&gt;Air leak in the suction. No cavitation ; cavitation voids ( almost vacuum ) do not leave the immediate area of the pump.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-02-15T15:38:00.460" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25996" PostTypeId="2" ParentId="25985" CreationDate="2019-02-15T16:19:00.553" Score="3" Body="&lt;h1&gt;Cooling a Gas in a Channel (Tube)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Assume we have a cylindrical channel that has gas flowing in it. The channel is in a material that itself is cylindrical. We have a single pass tube heat exchanger or alternatively the equivalent of a 1-D extended system but with gas flow. The outside of the tube is exposed to air or water at a temperature &lt;span class=&quot;math-container&quot;&gt;$T_\infty$&lt;/span&gt;. The energy balance at a given point &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; for the flow of the gas through this heat exchanger is basically as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{m}\tilde{C}_p \frac{dT}{dz} = - U P (T - T_\infty) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is the perimeter area of the inside of the tube, &lt;span class=&quot;math-container&quot;&gt;$P = 2\pi r_{i}$&lt;/span&gt;. The factor &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; is the &lt;a href=&quot;https://www.engineeringtoolbox.com/overall-heat-transfer-coefficient-d_434.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;overall heat transfer coefficient&lt;/a&gt; (based from the inside of the tube). It is a combination of the convection of the gas in the tube, the conduction through the tube, and the convection outside the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recast this in a dimensionless form with &lt;span class=&quot;math-container&quot;&gt;$\Theta = (T - T_\infty)/(T_h - T_\infty)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$Z = z/L$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt; is the entering hot temperature and &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the tube length. Allow &lt;span class=&quot;math-container&quot;&gt;$\beta = UPL/\dot{m}\tilde{C}_p$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ d\ln(\Theta) = - \beta\ dZ $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the boundary condition &lt;span class=&quot;math-container&quot;&gt;$\Theta(Z=0) = 1$&lt;/span&gt;, the answer is &lt;span class=&quot;math-container&quot;&gt;$\ln(\Theta) = -\beta Z$&lt;/span&gt;. This gives the temperature profile of the gas in the tube. See &lt;a href=&quot;https://engineering.stackexchange.com/questions/25937/how-can-i-calculate-the-amount-of-heat-my-copper-water-trap-can-dissipate/25977#25977&quot;&gt;this post&lt;/a&gt; for a comparable question and answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the system proposed, &lt;span class=&quot;math-container&quot;&gt;$T_h = 300$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C. The value of &lt;span class=&quot;math-container&quot;&gt;$T_\infty$&lt;/span&gt; should be BELOW 25 &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C. Otherwise, cooling could eventually require an infinite area heat exchanger (in essence an infinite length tube). Assume &lt;span class=&quot;math-container&quot;&gt;$T_\infty = 0$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C, and the goal is &lt;span class=&quot;math-container&quot;&gt;$\Theta = (25 - 0)/(300 - 0) = 0.083$&lt;/span&gt;. Use a 1 mm diameter tube with values of &lt;span class=&quot;math-container&quot;&gt;$\dot{m} =$&lt;/span&gt; 16 g/s, &lt;span class=&quot;math-container&quot;&gt;$\tilde{C}_p = 1$&lt;/span&gt; J/g &lt;a href=&quot;https://www.engineeringtoolbox.com/nitrogen-d_977.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;for nitrogen&lt;/a&gt;, and &lt;span class=&quot;math-container&quot;&gt;$U = h_a = 50$&lt;/span&gt; W/m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C. This latter assumption says that gas flow in the channel is the limiting value. The result is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\beta = UPL/\dot{m}\tilde{C}_p = (50)\pi(0.001)L/((16)(1)) = 0.010 L$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ln(0.083) = -0.010 x \Rightarrow x = 250$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need a 1 mm diameter tube that is 250 m long with walls that are cooled at 0 &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every 10x increase in heat transfer coefficient cuts the required length by 10x. Every 10x decrease in flow rate cuts the required length by 10x. Decreasing the external temperature &lt;span class=&quot;math-container&quot;&gt;$T_\infty$&lt;/span&gt; gives a modest decrease in required length.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Flowing Gas Thru a Tube&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The velocity of the gas through the tube is &lt;span class=&quot;math-container&quot;&gt;$v = \dot{V}/A$&lt;/span&gt;. Use &lt;span class=&quot;math-container&quot;&gt;$A = \pi D$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is the diameter. The volumetric flow of an ideal gas is related to its mass flow as below with &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; as temperature, &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; as molar mass, and &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; as pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{V} = \dot{m} \frac{RT}{Mp} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider 1 bar at 300 &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C with nitrogen to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{V} = \dot{m} \frac{(8.314)(573)}{(28)(101325)} = 0.00170 \dot{m}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the value 16 g/s to obtain &lt;span class=&quot;math-container&quot;&gt;$\dot{V} = 0.027$&lt;/span&gt; m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;/s. In a 1 mm diameter tube, the required gas velocity becomes 8.6 m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ v = 0.027 / ((\pi)(0.001)) = 8.6$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, we are left to find the required pressure drop to sustain 8.6 m/s through a 250 m long tube with a 1 mm internal diameter. This involves a &lt;a href=&quot;https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;review of the friction factor&lt;/a&gt; for fluids in pipe flow. The gas contracts as it cools. This demands further investigation too, because the basic calculations with incompressible fluids will fail with such cases.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Other Options&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The analysis used a 1 mm diameter tube. Increasing the tube diameter will decrease the required length and decrease the required velocity. The issue at some point will be that a gas does not sustain a uniform temperature profile. It will cool along the wall of the tube but less at the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analysis used only one tube. An alternative is to flow the same mass (volume) of gas through a multi-channel system. With &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; tubes, the contact area increases by &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;. This means, the required tube length decreases by &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;. This alternative says, rather having the gas flow through one longer channel at higher speed, you have the gas flow through N channels at 1/N speed because area increases by N. The contact time calculated as area per volumetric flow is the same. The appreciation of why we flow gases through mesh grids to heat or cool them rapidly strikes home. As for the pressure drop, this issue should be addressed in a separate discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gas can be cooled by &lt;a href=&quot;https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;adiabatic expansion&lt;/a&gt;. An effective design may just be to allow the hot gas at high pressure to expand rapidly to lower pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, analysis &lt;a href=&quot;https://engineering.stackexchange.com/questions/26086/calculating-heath-exchange-for-custom-boiler/26096#26096&quot;&gt;at this problem&lt;/a&gt; shows an example of how a heat exchanger design equation is used for fluid flow in a pipe surrounded by a constant temperature fluid. This may be an easier approach to take from a practical standpoint.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Cooling a Gas Flowing in a Channel through a Block&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The first part of the analysis above gives you the LENGTH OF A TUBE that is required to cool a gas to a given end temperature when the tube walls are held at a constant (lower) temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second part of the analysis above gives you the GAS VELOCITY that is required to pass the gas through the tube at a specific mass flow rate (g/s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume that you will make your channel with a diameter &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;. Assume the block is cooler than your desired end temperature. Determine the length of channel that you need using the first part (the equation for &lt;span class=&quot;math-container&quot;&gt;$\Theta$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$Z$&lt;/span&gt;). Determine the gas velocity you need to support the desired mass flow rate using the second step. Finally, follow up with pressure drop calculations for gas flow in pipes. I suspect you will find out that you likely will not be able to get the velocity that you need in a single channel because the pressure drop that you will need across such a long channel is so high as to be physically unreachable. Also, any curves in the channel only act to increase the required pressure drop. So, a &quot;curvy&quot; channel will be far longer but require far higher pressure to force the gas through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the end, you may have to realize that a better option to cool the amount of gas you want as rapidly as you want is to run the gas through a multi-channel (porous) solid that is held at a lower temperature than you need.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Heating a Mass of Solid in a Given Time (Optional)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Admittedly, the first part of the question is likely not addressed in the above. Specifically, what time is needed for a mass of a solid to absorb a specific amount of heat (from whatever source).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The approach to answer this first question depends on making one or another of many different assumptions. When the gas at constant temperature is in full contact with the solid, the problem is an unsteady state convection + conduction problem. It may require analysis comparable to the use of &lt;a href=&quot;https://en.wikipedia.org/wiki/Heisler_chart&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heisler charts&lt;/a&gt;. When the gas is in full contact with the solid but also changing its temperature, the Heisler chart analysis is replaced by an incremental or numerical integration approach (solve the Heisler chart at t, increment by dt and solve again). Finally, when the gas is flowing through the solid and the solid is not maintained at a constant temperature, one goes to the need for a Heisler chart or its foundations plus time integration plus position analysis.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-02-23T15:37:44.923" LastActivityDate="2019-02-23T15:37:44.923" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="25997" PostTypeId="1" CreationDate="2019-02-15T16:20:29.047" Score="1" ViewCount="48" Body="&lt;p&gt;A little bit of context. My friends and I were watching the &quot;Dark Knight Rises&quot; and a discussion about the Batmobiles ensued. I, being a former ME student, said it was stupid that batman has a turbine in his car at all because they aren't responsive enough for the very Fast and Furious way Batman tends to drive. Batman seems to need a car with an engine that can produce a lot of power and adjust its power output very quickly; Batman needs a car without much acceleration lag. Turbines can produce a lot of power, but they simply can't be responsive. I suggested Batman use a rocket instead, though he should probably just be using some turbo and super chargers or be driving an electric vehicle at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point my friend, with an English degree, said that Batman did in fact have a rocket on the back of his car, and pointed out that the jet engine in the Batmobile has flames coming out of the back of it, which means it has an afterburner. He then argued that would make the Batmobile responsive enough for the driving style to be fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I disagree, but I can't find any sources on how an afterburner actually effects the responsiveness of a turbine. After looking I learned that an afterburner produces extra thrust. My thought is that the turbine would already need to be working in order for an afterburner to do anything. In a car that would still lead to a responsiveness problem right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can an afterburner make the turbine in Batman's Batmobile responsive enough satisfy his insane driving requirements?&lt;/p&gt;&#xA;" OwnerUserId="19377" LastEditorUserId="19377" LastEditDate="2019-02-15T21:46:16.510" LastActivityDate="2019-02-15T21:46:16.510" Title="Would an afterburner on a turbine drastically improve acceleration lag?" Tags="|thermodynamics|automotive-engineering|turbines|car|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="25998" PostTypeId="1" CreationDate="2019-02-15T17:25:57.223" Score="0" ViewCount="982" Body="&lt;p&gt;Context: introductory mechanical vibrations, 300-level engineering. ISBN13: 978-0-13-287169-3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See problem solution below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vNcL5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vNcL5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As seen in the &quot;Potential Energy&quot; section, the author writes that the distance the springs compress is &lt;span class=&quot;math-container&quot;&gt;$$s=(a+r)\theta$$&lt;/span&gt;When I derive  &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; I get &lt;span class=&quot;math-container&quot;&gt;$\tan(\theta)=s/a$&lt;/span&gt; , &lt;span class=&quot;math-container&quot;&gt;$s=atan(\theta)$&lt;/span&gt; , and by small angle approximation &lt;span class=&quot;math-container&quot;&gt;$s=a\theta$&lt;/span&gt;. How is he getting &lt;span class=&quot;math-container&quot;&gt;$(a+r)$&lt;/span&gt;? &lt;/p&gt;&#xA;" OwnerUserId="19383" LastEditorUserId="8055" LastEditDate="2019-02-15T20:41:04.877" LastActivityDate="2019-03-18T07:25:33.057" Title="Mechanical vibration: 2 springs attached horizontally to a wheel at a vertical distance a above the center, find the natural frequency of the system" Tags="|mechanical-engineering|vibration|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="25999" PostTypeId="1" AcceptedAnswerId="26001" CreationDate="2019-02-15T17:39:55.580" Score="1" ViewCount="121" Body="&lt;p&gt;I am working on a 1.5-meter satellite dish with a custom receiver cantenna-like feed. The dish didn't come with feed arms, so I need a way to hold the feed into place. What I tried was using a 1-meter threaded rod. My threaded rod was 1m (length) x D=14mm (thickness). I used nuts to hold the feed in place and held the threaded rod from the little hole in the center of the dish (again, using nuts). This has the advantage of allowing me to adjust the position of the feed (focal length) so I can play around and experiment to see where my signal strength is at its peak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage is that I don't need to use metallic arms (like many large dishes do) to keep it in place, as they can block a (small) portion of the radio waves from hitting the reflector and reduce its gain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's a problem which leads me to come here and ask for help: even slightly touching the feed (or the edge of the rod) will have it wobble around for quite a while. This means that if there's a small wind, the feed will be wobbling around, which is not good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: what can I do about this? Can I use another type of rod, or make any adjustments to my current threaded rod in order to stop it from wobbling back and forth by the slightest touch? Could using a thicker threaded rod (the hole is 1 inch = 2.54 cm, but I could always drill a larger hole if needed), or would that not fix the problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few pictures of the dish:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jqfkJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jqfkJ.jpg&quot; alt=&quot;https://cdn.discordapp.com/attachments/417338610191302656/530112919938269194/image0.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CwSft.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CwSft.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kwWqG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kwWqG.jpg&quot; alt=&quot;https://cdn.discordapp.com/attachments/417338610191302656/530112852120567829/thumbnail_20181202_164746_resized.jpg&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas are appreciated!&lt;/p&gt;&#xA;" OwnerUserId="12819" LastActivityDate="2019-02-15T19:41:42.653" Title="Stopping/reducing wobbliness in threaded rod" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26000" PostTypeId="1" CreationDate="2019-02-15T19:11:35.680" Score="1" ViewCount="791" Body="&lt;p&gt;I am doing calculations on costs for alternatives for manufacturing this children's toy I invented, a puzzle where letters in plastic fit into wood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been using a CNC router to carve subtract the letter shapes from wood and a laser cutter to cut those same shapes in plastic (with some clearance). I have a good idea of the costs involved in manufacturing it myself, hiring someone to supervise the machines, or hiring a service bureau. My understanding of manufacturing at scale is much more limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it would be possible to get similar and cheaper results by moulding Medium Density Fiber or Perspex. Is that possible? Can you think of other scalable alternatives?&lt;/p&gt;&#xA;" OwnerUserId="19384" LastEditorUserId="19384" LastEditDate="2019-12-07T18:20:34.270" LastActivityDate="2019-12-07T19:57:05.317" Title="Scalable alternatives to CNC router and laser cutter" Tags="|lasers|cnc|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26001" PostTypeId="2" ParentId="25999" CreationDate="2019-02-15T19:27:10.800" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Increasing the rod thickness would be very effective. If I've visualizing the configuration correctly, the rod is acting as a cantilevered beam (i.e., a beam rigidly attached at one end): &#xA;&lt;br&gt;&lt;img src=&quot;https://i.stack.imgur.com/oxlSk.png&quot; width=&quot;300&quot;&gt;&lt;br&gt; The lateral stiffness (which governs the vibration amplitude of such a cantilevered beam) scales up with the area moment of inertia, which remarkably has a &lt;em&gt;fourth-power&lt;/em&gt; dependence on the diameter. This means that you could double the stiffness with only a 19% increase in rod diameter, and doubling the diameter would increase the stiffness (and suppress the vibration amplitude) by 16×.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What about stabilizing the end of the rod by attaching nonconductive tensioned cord or string stretching back to a support?&lt;br&gt;&lt;img src=&quot;https://s3.amazonaws.com/answer-board-image/0d9258c3-11a4-4b3a-a062-77f7f3792f7f.png&quot;&gt;&lt;br&gt; This is how tall and narrow towers (which would otherwise be unacceptably laterally compliant) are stabilized. The limit of the tensioning load would be the buckling threshold of the rod (if the tension points back to the base; therefore, try to orient the tension loads so that they tend to elongate the rod).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(Whimsical) If you have substantially excess time and money, you might add a set of strain gauges and actuators to the length of the rod to detect vibrations and suppress them, respectively. One example of such a &quot;smart structure&quot; is &lt;a href=&quot;https://patents.google.com/patent/US5631437&quot; rel=&quot;nofollow noreferrer&quot;&gt;a cannon barrel that can stabilize itself against vibration&lt;/a&gt; as the shell moves through it, thus increasing accuracy, for example. Overkill?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="10123" LastEditorUserId="10123" LastEditDate="2019-02-15T19:41:42.653" LastActivityDate="2019-02-15T19:41:42.653" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26002" PostTypeId="2" ParentId="25998" CreationDate="2019-02-15T21:21:12.880" Score="0" Body="&lt;p&gt;If the wheel were rotating around its center, then the small-displacement approximation of motion at point &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; would be &lt;span class=&quot;math-container&quot;&gt;$a\theta$&lt;/span&gt; in the horizontal direction (and the corresponding movement at the bottom of the wheel would be &lt;span class=&quot;math-container&quot;&gt;$r\theta$&lt;/span&gt; in the opposite direction). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, however, the wheel is rotating around its point of contact with the underlying base. The small-displacement motion is therefore identical to that of a wheel centered at the point of contact, which is &lt;span class=&quot;math-container&quot;&gt;$(r+a)\theta$&lt;/span&gt; horizontally at point &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; (and &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; at the bottom of the wheel, which is consistent with the figure). Does this make sense?&lt;/p&gt;&#xA;" OwnerUserId="10123" LastEditorUserId="10123" LastEditDate="2019-02-15T21:26:41.280" LastActivityDate="2019-02-15T21:26:41.280" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26007" PostTypeId="1" AcceptedAnswerId="26008" CreationDate="2019-02-16T08:28:12.957" Score="0" ViewCount="4029" Body="&lt;p&gt;I have one tile that came off. I got a replacement tile, and some grout. But the online guides say you need mortar to glue the tile to the ground. Can I use the grout as mortar? I want to save money so I only want to buy one thing.&lt;/p&gt;&#xA;" OwnerUserId="19388" LastActivityDate="2022-06-03T20:35:25.973" Title="Can I use grout as mortar?" Tags="|civil-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26008" PostTypeId="2" ParentId="26007" CreationDate="2019-02-16T09:33:09.797" Score="1" Body="&lt;p&gt;Yes you can do it, is simple answer, if what you have is sanded grout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can add a little acrylic glue available in 4oz tubes at building material centers for a couple of dollars it will be even better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you may be able to ask to take a torn bag of ThinSet for free, such as I have done. They actually encourage this to keep the shelves clean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mortar or what is known by many in trade as Readymix, is for applications where you may need to level more than one inch thickness. or your tiles or marble is bigger than 12x12 inches for it offers structural  volum integrity and easier trawling.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-16T09:33:09.797" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26009" PostTypeId="1" CreationDate="2019-02-16T09:57:24.263" Score="0" ViewCount="65" Body="&lt;p&gt;How many gallons of water (70 degrees F) can flow thru a pipe diameter (I.D. 0.475&quot;) at 50 psi over a 30 foot length in a 24 hour time frame?&lt;/p&gt;&#xA;" OwnerUserId="19224" LastActivityDate="2019-02-16T09:57:24.263" Title="Can you help me with a flow rate question?" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26010" PostTypeId="1" CreationDate="2019-02-16T10:19:31.437" Score="0" ViewCount="284" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/6of0z.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To model a swivel arm on column that rotates by hand &#xA;What type of bearing should i use &#xA;Should it support radial and axial load or just supporting axial load is enough ? &lt;/p&gt;&#xA;" OwnerUserId="17065" LastActivityDate="2021-01-30T12:11:34.697" Title="What type of bearing used in swivel arm?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26011" PostTypeId="1" AcceptedAnswerId="26016" CreationDate="2019-02-16T11:29:18.870" Score="2" ViewCount="52" Body="&lt;p&gt;I am currently working on desalination project for my final year. The design is based on flashing of hot liquid in low pressure vessel and then condensing the extracted the steam to produce pure water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I am facing is that there doesn't seem to be any thermodynamic data (under saturation dome) available for saline water. The need for saturation data is because I need to calculate the required mass flow rate based on daily desalination goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the goal is to produce 100 L/day with average sunlight of 8 hrs, then desalination rate must be 0.0035 kg/s. If I know the quality (percentage of steam produced) of the liquid gas mixture, I can easily calculate the mass flow rate of saline water corresponding to desalination goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only data I could find related to this problem was from a MIT resource &lt;a href=&quot;http://web.mit.edu/seawater/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thermophysical properties of seawater&lt;/a&gt;, but unfortunately it only provides a single value for enthalpy at any given saturation temperature or pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I can see only three options:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use steam table, and consider the values an approximation and incorporate the error produced in my design by slightly overdesigning&lt;/li&gt;&#xA;&lt;li&gt;Consider the enthalpy of seawater (from MIT resource) as enthalpy of saturated liquid (i.e. &lt;span class=&quot;math-container&quot;&gt;$h_f$&lt;/span&gt;), and enthalpy of vapor (from steam table) as &lt;span class=&quot;math-container&quot;&gt;$h_g$&lt;/span&gt; and then perform my calculations&lt;/li&gt;&#xA;&lt;li&gt;Thirdly, I could take the enthalpy of seawater (from MIT resource) as the value that is halfway between &lt;span class=&quot;math-container&quot;&gt;$h_f$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$h_g$&lt;/span&gt; (from steam table), and then calculate the approximated value of &lt;span class=&quot;math-container&quot;&gt;$h_f$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$h_f=$&lt;/span&gt; Enthalpy at saturated liquid line; &#xA;&lt;span class=&quot;math-container&quot;&gt;$h_g=$&lt;/span&gt; Enthalpy at saturated vapor line&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help/guidance in this regard will be highly appreciated&lt;/p&gt;&#xA;" OwnerUserId="19390" LastActivityDate="2019-02-16T21:50:06.970" Title="Is thermodynamic data of fresh water a good approximation for saline water?" Tags="|thermodynamics|water-resources|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26013" PostTypeId="1" CreationDate="2019-02-16T17:49:04.650" Score="0" ViewCount="185" Body="&lt;p&gt;After receiving help on my previous post (&lt;a href=&quot;https://engineering.stackexchange.com/questions/25999/stopping-reducing-wobbliness-in-threaded-rod&quot;&gt;Stopping/reducing wobbliness in threaded rod&lt;/a&gt;), I've decided to come back and ask another question: how can I actually hold the satellite dish into place? I don't wish to have the dish steerable (as this will probably require too much work considering the size of the dish), but I want it to &lt;em&gt;at least&lt;/em&gt; be able to point (&quot;look&quot;) directly upwards (to local zenith). The dish weights 16kg and has a diameter of 1.5 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I plan on installing the dish:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/J8zc0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J8zc0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it seems, there's nothing the dish has to offer for this kind of things (I got it used from someone who no longer needed it):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/jqfkJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jqfkJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CwSft.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CwSft.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/kwWqG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kwWqG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on how to hold the dish in place to point to local zenith would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="12819" LastActivityDate="2024-02-21T08:01:20.180" Title="Installing satellite dish to point to local zenith" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="26014" PostTypeId="2" ParentId="26010" CreationDate="2019-02-16T18:40:27.490" Score="1" Body="&lt;p&gt;you have both:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ M= 9kg*L+ lever\ weight*L/2$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;L = length of the lever arm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ P = 9 + Lever\ weight $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need a bearing that can carry both radial and axial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not any less important is to make sure the post and its attachment at base are strong enough to support sudden jerks and vibration of the start and stopping action of the lever.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-16T18:40:27.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26015" PostTypeId="2" ParentId="26013" CreationDate="2019-02-16T18:50:15.793" Score="0" Body="&lt;p&gt;You can always use cinderblocks and place plastic anchors to attach to your antenna.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically you make three short pedestals with cinderblocks and cement. No need to penetrate the roofing and worry about leaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is provided your antenna is out of view and wont be a like a sore eye.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or you can search eBay for brackets.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-16T18:50:15.793" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26016" PostTypeId="2" ParentId="26011" CreationDate="2019-02-16T21:28:56.390" Score="1" Body="&lt;p&gt;The EES software library files provided on MIT website is not recognized (there seems to be some error, atleast on my PC). The MATLAB files works just fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$h_f$&lt;/span&gt; can be approximated from enthalpy of sub-cooled liquid near saturation temperature using &lt;code&gt;SW_Enthalpy&lt;/code&gt; function and the value of &lt;span class=&quot;math-container&quot;&gt;$h_{fg}$&lt;/span&gt; can be easily obtained by using the function &lt;code&gt;SW_LatentHeat&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. I was searching for the required data under the keyword of 'enthalpy' and that's probably the reason I was unable to find the required data.&lt;/p&gt;&#xA;" OwnerUserId="19390" LastEditorUserId="19390" LastEditDate="2019-02-16T21:50:06.970" LastActivityDate="2019-02-16T21:50:06.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26017" PostTypeId="1" AcceptedAnswerId="26027" CreationDate="2019-02-16T22:00:53.707" Score="0" ViewCount="623" Body="&lt;p&gt;Say a stepper motor has fixed value of &quot;&lt;code&gt;units&lt;/code&gt;&quot;, given as a holding torque related to distance of the center of its shaft. Like &lt;strong&gt;1kg/cm&lt;/strong&gt; or &lt;strong&gt;14oz/inch&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a GT2 16-teeth pulley on the motor shaft. (I'm using a 5mm GT2 belt as a load transfer if that matters) Knowing that the motor can hold 1kg weight, 1cm away of the center of its shaft, and the 16-teeth pulley &quot;outher-teeth&quot; diameter is 9.7mm, I can easily assume that it can hold about 1kg of load safely (at least trusting manufacturer's datasheet). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, what if I add a GT2 48-teeth pulley on the load part (I don't know the outer diameter)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any algorithm I could use (to atleast &lt;em&gt;roughly&lt;/em&gt;) calculate load capabilites of the motor, based only on the pulley teeth count?&lt;/p&gt;&#xA;" OwnerUserId="19393" LastEditorUserId="19393" LastEditDate="2019-02-16T22:11:18.367" LastActivityDate="2019-02-17T16:22:03.313" Title="Stepper motor torque calculation based on pulleys ratio" Tags="|stepper-motor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26020" PostTypeId="2" ParentId="1910" CreationDate="2019-02-17T08:46:25.577" Score="2" Body="&lt;p&gt;Modern compound engines use electric motors to translate the speed of the turbine to the speed of the wheel or propeller. For this, modern electronics is needed. It works in F1 and in the trucks and would work in hybrid cars with a small internal combustion engine. Turbines have a narrower working range than a piston engine. Cars most of the time idle or need full power for acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if it is possible to use a small compression ratio piston engine and a burning chamber (covered with a catalyst material, padded in insulation), where temperatures are low enough to reduce NOx, but still high enough to burn soot and CO. These temperatures aneal and thus flatten the catalyst's surface (so called pre-cat) and thus are avoided in usual catalyst converters. The high temperature, high pressure exhaust can be used to drive a two stage turbine. Note that this turbine housing costs more than a housing for relatively cold diesel exhaust. Also the first stage may need Nickel blades and only the second stage can have variable guide planes. A Cyclon (or even a filter) in the chamber may help to prolonge the duration of soot particles in the chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All in all this is stuff for a premium hybrid car or truck. Agilty is due to the electric motor driven by a super-capacitor. FADEC enforces high temperatures at the filter in order to be never congested by soot. Excess energy is dumped into the batteries.&lt;/p&gt;&#xA;" OwnerUserId="19398" LastActivityDate="2019-02-17T08:46:25.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26023" PostTypeId="1" CreationDate="2019-02-17T09:39:11.383" Score="0" ViewCount="147" Body="&lt;p&gt;I am working on a robotics project, I got a SolidWorks sldprt/step model of the robot(cca. 40 solids). However, I have only Autocad/and Inventor available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to measure a distance from n ellipse center (endpoint) and a circle center - joint axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the ellipse and the circle lay in different z levels, are hollow and belong to different solids. No matter what I try, I can get neither the inventor nor the autocad to measure the distance using annotate or measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help.&lt;/p&gt;&#xA;" OwnerUserId="19378" LastActivityDate="2019-02-17T09:39:11.383" Title="Measure distance between a circle and an ellipse center in autocad/inventor" Tags="|robotics|autocad|autodesk-inventor|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26026" PostTypeId="1" AcceptedAnswerId="29142" CreationDate="2019-02-17T14:06:06.297" Score="1" ViewCount="1663" Body="&lt;p&gt;Is it possible to ( how do you ) export a series of points along a spline.&#xA;A spline has been created in 3d from various segments.&#xA;I would like to export a table of 3d points along the spline starting at one end and every mm or so along the spline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The result would be a file with Z,Y,Z points that are every mm along the spline.&lt;br&gt;&#xA;I want to end up with lots of little line segments that approximate the spine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen a few tutorials on how to export intersection points but that in not what I am looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="19402" LastActivityDate="2019-07-11T15:28:39.483" Title="Solidworks export the 3d points of a spline" Tags="|solidworks|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26027" PostTypeId="2" ParentId="26017" CreationDate="2019-02-17T16:22:03.313" Score="0" Body="&lt;p&gt;Circumference is a fixed ratio (&amp;pi;) times the diameter. Therefore the number of teeth is in proportion to the diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If everything is 100% efficient then the torque will step up by the pulley ratio. 1 kg·cm with 16 teeth on the motor and 48 on the load will give 3 kg·cm available on the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that torque is force &lt;strong&gt;times&lt;/strong&gt; distance so it's kg·cm, not kg/cm.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2019-02-17T16:22:03.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26029" PostTypeId="2" ParentId="8619" CreationDate="2019-02-17T20:48:41.707" Score="1" Body="&lt;p&gt;The best way to stop the flow of a siphon is to have a shutoff valve inline at the crest. The valve can have a mechanical timer.&#xA;This will work if an anti backflow foot valve is inline at the inlet and another check valve is inline at the outlet to allow flow in one direction while preventing air from entering and migrating up the long leg.                          This allows complete control of the siphon and when it is shutoff before air is allowed to enter the line, it stays primed for further use.&#xA;US Patent 5,358,000 was the first to teach this method of siphon control.&lt;/p&gt;&#xA;" OwnerUserId="19404" LastEditorUserId="-1" LastEditDate="2023-03-15T14:23:37.277" LastActivityDate="2023-03-15T14:23:37.277" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26030" PostTypeId="1" CreationDate="2019-02-17T21:40:31.770" Score="1" ViewCount="771" Body="&lt;p&gt;So I need to design a cylindrical cam (aka barrel cam I think). I'm an aerospace engineer so I've never had a course on cams/gear design (wish I had though). Is there a general equation for the curve (path) on a cylindrical cam of known radius? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture of my application; it's basically like a fishing reel. I need to use it to wind thread from a bigger spool to a smaller spool that will work on my sewing machine. The top rod with the purple thing is the input and the black thing is a rubber belt that turns the green drum. The green drum is the part I need to have a cam path on so that it moves the red arm. Any help or pointing to references is greatly appreciated! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L7kee.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L7kee.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19405" LastEditorUserId="1832" LastEditDate="2019-02-18T03:22:14.323" LastActivityDate="2019-02-18T03:22:14.323" Title="Designing cylindrical cam" Tags="|design|kinematics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26031" PostTypeId="2" ParentId="26030" CreationDate="2019-02-17T23:22:58.323" Score="1" Body="&lt;p&gt;You have half the answer in your question. The device is known as a level wind mechanism and is a groove cut into the drum for the follower/rod in your drawing. A segment of video on the &lt;a href=&quot;https://youtu.be/Zf_2RJqduoo?t=173&quot; rel=&quot;nofollow noreferrer&quot;&gt;YouTube will&lt;/a&gt; show you a general shape for this mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's effectively a helix that reaches from end to end with a gentle curve to reverse direction (and cross the previous groove).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More information can be found at &lt;a href=&quot;https://patents.google.com/patent/US20100187346A1/en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google Patents&lt;/a&gt; including a hard-to-find diagram of the rod:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7zNYC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7zNYC.png&quot; alt=&quot;cam winder rod&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2019-02-18T00:21:16.667" LastActivityDate="2019-02-18T00:21:16.667" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26034" PostTypeId="1" AcceptedAnswerId="26044" CreationDate="2019-02-18T14:27:44.487" Score="0" ViewCount="388" Body="&lt;p&gt;One end of a small insulated rod is attached to a support which is maintained at a constant temperature of 473 K,and once the rod has reached a uniform temperature in its entire length the insulation is removed. Now the surrounding air is at 298 K with convective heat transfer coefficient h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am required to find the time elasped when the temperatures at some length of rod reaches a given temperature.The radius,thermal conductivity and thermal diffusivity are all given&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a unsteady state heat conduction and convection problem.I got the following pde &lt;span class=&quot;math-container&quot;&gt;$$k \frac{\partial^2 T }{\partial z^2} - \frac k\alpha \frac{\partial T}{\partial t}=\frac{2h}{R} (T-T_0)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is thermal conductivity &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is thermal diffusivity and &lt;span class=&quot;math-container&quot;&gt;$T_0$&lt;/span&gt; is surrounding ambient temperature.I am facing trouble finding out the two boundary conditions.The initial condition is that at &lt;span class=&quot;math-container&quot;&gt;$t=0$&lt;/span&gt; the rod is at a uniform temperature of 473K.The first boundary condition is at &lt;span class=&quot;math-container&quot;&gt;$z=0$&lt;/span&gt; the temperature is 473K.However i am unable to find out the second boundary condition.The tip of the fin is not insulated and is neither maintained at a constant temperature.&lt;/p&gt;&#xA;" OwnerUserId="12580" LastActivityDate="2019-02-20T15:55:04.863" Title="Boundary conditions in a heat transfer problem" Tags="|mechanical-engineering|heat-transfer|chemical-engineering|finite-element-method|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26035" PostTypeId="2" ParentId="23016" CreationDate="2019-02-18T16:32:36.820" Score="0" Body="&lt;p&gt;To close the loop on this, and in case it's helpful for anyone else in the future... Here's the solution we ended up picking, and a few of the reasons and resources that moved us in that direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me preface by saying, as several have pointed out in the comments and answers, that &lt;strong&gt;the key to implementing version control is the mindset change and consistency of following a new process&lt;/strong&gt;. We &lt;em&gt;had&lt;/em&gt; a process that emulated what we do now, but it was complicated, time consuming, and prone to errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's what we ended up doing.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;git&lt;/strong&gt; &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;While looking into SVN I found a helpful article, but even more helpful was this comment on the article: &lt;a href=&quot;http://disq.us/p/ru9xu7&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;5 reasons, coming from someone who has moved from subversion to git twice voluntarily&quot;&lt;/a&gt;. The ability to seamlessly work off-line or outside of our internal network were big selling points. &lt;/li&gt;&#xA;&lt;li&gt;A comment I came across a few times (see &lt;a href=&quot;https://softwareengineering.stackexchange.com/a/35080&quot;&gt;this answer on softwareengineering.se&lt;/a&gt;, for example) is that merging in SVN is complicated, and for those less comfortable with SVN can lead to a reluctance to commit. This defeats the purpose.&lt;/li&gt;&#xA;&lt;li&gt;In contrast, I found &lt;a href=&quot;https://www.atlassian.com/git/tutorials/saving-changes/git-commit&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article from Atlassian&lt;/a&gt; explaining that in git, &quot;commits are cheap.&quot; Version control is most useful when frequent commits are made, so I wanted to lower the bar as much as possible for the team.&lt;/li&gt;&#xA;&lt;li&gt;Because I knew I needed to get a certain level of buy-in before devoting significant resources to the project, setting up a server (such as would be needed for SVN) was off the table. I don't have the skills to do this, and convincing our small team to put time/money into something that they aren't convinced they need (when there are other things we all agree we need but that we can't afford yet) was a non-starter. Of course, this is a Catch-22: If we had a server with SVN set up, we likely could have gotten a satisfactory SVN process going.&lt;/li&gt;&#xA;&lt;li&gt;I set up TortoiseSVN on my PC and used it for a couple of weeks. It was functional, but not exceptional.&lt;/li&gt;&#xA;&lt;li&gt;Once I decided to move away from SVN to DCVC, the choice between Mercurial and git was decided by the fact that I had used git in the past. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;GitLab&lt;/strong&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Free and open source.&lt;/li&gt;&#xA;&lt;li&gt;We did consider GitHub, but it is limited to three collaborators on private repos.&lt;/li&gt;&#xA;&lt;li&gt;Bitbucket is limited to five users for free teams.&lt;/li&gt;&#xA;&lt;li&gt;As we (and our resources) grow, we may revisit hosting. This part of the process would be relatively easy to change.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Sourcetree&lt;/strong&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Based on &lt;a href=&quot;https://unmethours.com/question/38/revision-control-for-text-based-modeling-programs/&quot; rel=&quot;nofollow noreferrer&quot;&gt;comments on a forum for energy modeling&lt;/a&gt; (which is our use case), I decided to give Sourcetree a try. I found the interface helpful and intuitive.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some other helpful resources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Oliver Steele's &lt;a href=&quot;https://blog.osteele.com/2008/05/my-git-workflow/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;My git workflow&quot;&lt;/a&gt;. The graphics were particularly useful for explaining git to our team.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://mindfulmodeller.ca/version-control-for-energy-models/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Version control for energy models&quot;&lt;/a&gt;. Helped us think through developing/implementing a workflow for our particular application.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17004" LastActivityDate="2019-02-18T16:32:36.820" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26037" PostTypeId="2" ParentId="191" CreationDate="2019-02-18T16:42:46.543" Score="4" Body="&lt;p&gt;First the scrap is separated at the source ; for example cast iron generally only contains Si and Mn. High vapor pressure elements boil off or collected in the flux/slag : eg, Zn, Pb, Sn, Bi, An ,,,,Aluminum oxidizes and goes into the slag. Steels do pick up Cr, Ni, Mo, and Cu residuals , generally these are advantageous ; they all add to hardenability except Cu. ( Cu is important in atmospheric corrosion resistance). V and Nb and W are present in very small amounts so insignificant. , And Co , expensive and has specialized applications so it is also separated at the scrape source ; Co scrape is easy to identify; medical prosthesis and jet engine hot section blades and vanes ,also in some high speed tools -again separated at the scrape source. Ni alloys and austenitic stainless are separated at the source as they are not ferromagnetic. Magnetic Martensitic and ferritic stainless ( typically 13 % Cr) can be separated at the scrape source . The separation of steels at the sources is done because all the alloy elements are worth more than carbon steel . There must be books available on this ; it is a major factor in the steel industry.  An example of what happens in the real world ; grade A 516 carbon steel plate is the workhorse of industry but when a thick section with high strength is ordered , &quot;somehow&quot; the Cr, Mo, Ni residuals are high enabling acceptable heat-treatment results.      &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-02-18T20:43:13.337" LastActivityDate="2019-02-18T20:43:13.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26039" PostTypeId="2" ParentId="26000" CreationDate="2019-02-18T21:37:06.870" Score="2" Body="&lt;p&gt;You can certainly drive the cost down by injection moulding the plastic parts. It has  a relatively high setup  cost though. Acrylic can also be extruded. In both cases laser cutting is probably also much faster than you think. Just be sure to choose the right laser: what takes forever in a fablab can take seconds in a real high power laser setup. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway you can die cut acrylic. This is pretty cheap to set up and cheap to operate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the wood well there are a few alternatives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might also be possible to die cut the veneer but I don't actually know much about that. Actually it seems possible to die cut plywood and softer &lt;a href=&quot;https://m.youtube.com/watch?v=BFU-LEyQ0y8&quot; rel=&quot;nofollow noreferrer&quot;&gt;woods&lt;/a&gt;. Which is how puzzle pieces are made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then there exist mills that have multiple spindles. So that they would mill several of the characters at once. By choosing right millhead you should be able to do this quite economically.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="19384" LastEditDate="2019-12-07T19:57:05.317" LastActivityDate="2019-12-07T19:57:05.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26041" PostTypeId="2" ParentId="25849" CreationDate="2019-02-18T23:55:14.600" Score="1" Body="&lt;p&gt;The starting point at constant &lt;span class=&quot;math-container&quot;&gt;$T, p$&lt;/span&gt; is the Gibb's energy in any phase &lt;span class=&quot;math-container&quot;&gt;$\Pi$&lt;/span&gt; with components &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ dG^\Pi = \sum_j \mu_j^\Pi\ dz_j^\Pi $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At phase equilibrium, the total Gibb's energy change is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ dG = 0 = \sum_\Pi \sum_j \mu_j^\Pi\ dz_j^\Pi $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any given phase, we have the mass balance equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sum_j dz_j^\Pi = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have as many of these as we have phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remaining analysis is an exercise in applying the proper mathematics to combine the phase equilibrium equation with the &lt;span class=&quot;math-container&quot;&gt;$\Pi$&lt;/span&gt; mass balance equations.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-02-18T23:55:14.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26044" PostTypeId="2" ParentId="26034" CreationDate="2019-02-19T12:54:19.237" Score="2" Body="&lt;p&gt;The BC at the far end is the flux condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ -k \left. \frac{dT}{dx}\right|_{x=L} = h(T - T_o) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See Chapter 17 in the book by &lt;a href=&quot;https://www.wiley.com/en-us/Fundamentals+of+Momentum%2C+Heat%2C+and+Mass+Transfer%2C+Revised+6th+Edition-p-9781118947463&quot; rel=&quot;nofollow noreferrer&quot;&gt;Welty etal&lt;/a&gt; for examples of extended surfaces and their solutions at steady state. Finally, just for fun, here is an example temperature profile of an extended beam at steady state with the heat transfer BC. Your final answer will end at a comparable profile when &lt;span class=&quot;math-container&quot;&gt;$m^2 = 10$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mjlHZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mjlHZ.png&quot; alt=&quot;temperature profile in extend beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-02-20T15:55:04.863" LastActivityDate="2019-02-20T15:55:04.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26045" PostTypeId="1" CreationDate="2019-02-19T17:11:47.793" Score="1" ViewCount="56" Body="&lt;p&gt;My task is to write a code from scratch for 3D elastoplasticity, namely a DP model (see for example DeSouza Neto, &lt;em&gt;et al&lt;/em&gt;., 2008, Chapter 8) having yield surface&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ f(\mathbf{\sigma},c) = \sqrt{J_2(\mathbf{s(\mathbf{\sigma})})} + \eta\,p(\mathbf{\sigma}) - \xi \, c(\bar{\epsilon}^p) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;being &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the first term the square root of the second invariant of the stress tensor;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\eta$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\xi$&lt;/span&gt; are chosen according to the Mohr-Coulomb surface that one approximates, which in turn depend upon&lt;/li&gt;&#xA;&lt;li&gt;the friction angle &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; and dilatancy angle &lt;span class=&quot;math-container&quot;&gt;$\psi$&lt;/span&gt;, this latter used since the flow rule is not assumed to be associative, so another parameter &lt;span class=&quot;math-container&quot;&gt;$\bar{\eta}$&lt;/span&gt; is computed&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is the cohesion, which in my linear assumed model depends on the cumulated plastic strain &lt;span class=&quot;math-container&quot;&gt;$\bar{\epsilon}^p$&lt;/span&gt;. The linear functional form this one takes is &lt;span class=&quot;math-container&quot;&gt;$c(\bar{\epsilon}^p) = c_0 + H \, \bar{\epsilon}^p$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is&lt;/strong&gt;: I know how to implement the numerical integration algorithm to solve the (rate independent) rate equations, but I couldn't find a set of parameters &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; (shear and bulk modulus respectively, or in alternative &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\nu$&lt;/span&gt;, Young and Poisson moduli), &lt;span class=&quot;math-container&quot;&gt;$\xi$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\eta$&lt;/span&gt; or equivalently &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\psi$&lt;/span&gt; and lastly &lt;span class=&quot;math-container&quot;&gt;$c_0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt;, initial cohesion and plastic modulus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to ask for some references where such parameters are available, say for granular media. This task is developing oriented, it is just intended to implement an elastoplastic constitutive model, I only need some physically plausible parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference&lt;/strong&gt;: E. A. de Souza Neto D. Perić D. R. J. Owen, &lt;em&gt;Computational Methods for Plasticity: Theory and Applications&lt;/em&gt;, 2008 John Wiley &amp;amp; Sons, Ltd.&lt;/p&gt;&#xA;" OwnerUserId="19425" LastEditorUserId="1832" LastEditDate="2019-02-20T03:11:53.427" LastActivityDate="2019-02-20T03:11:53.427" Title="Drucker-Prager Model Parameters, a Proper Choice" Tags="|civil-engineering|structural-analysis|modeling|geotechnical-engineering|elastic-modulus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26046" PostTypeId="1" CreationDate="2019-02-19T17:31:10.163" Score="1" ViewCount="264" Body="&lt;p&gt;I think it would be more efficient for a vertical wind turbine to have a covered funnel about half of it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like this figure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x1tcd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x1tcd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would protected the turbine from opposite wind and also redirect another half wind to the correct side. It should increase overall efficiency&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any problem of this model so it not being used?&lt;/p&gt;&#xA;" OwnerUserId="1606" LastActivityDate="2019-03-21T22:01:10.563" Title="Why vertical wind turbine don't be half covered" Tags="|fluid-mechanics|renewable-energy|wind-power|generator|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26047" PostTypeId="1" CreationDate="2019-02-19T17:42:53.727" Score="1" ViewCount="358" Body="&lt;ol&gt;&#xA;&lt;li&gt;I know that magnetic stripes used in credit cards and audio tapes are typically read by a magnetic reader head. But is it theoretically possible for the stripe to be read by a laser in a similar fashion to a barcode scanner?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I also heard of technologies like magnetic secure transmission by Samsung that emulate the magnetic stripe within 3 inches of a card reader. I'm guessing the smartphone or device that does the MSE is using more power to ensure the magnetic field reaches the card reader's reader head from a distance of 3 inches. &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;I'm curious about the opposite. Could a magnetic reader head theoretically read a magnetic stripe with a gap distance (even a very small gap say 0.01mm) or without physically touching the magnetic stripe? Or is the magnetic field generated by the magnetic reader head too weak to handle any gap between the reader head and the magnetic stripe?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19426" LastActivityDate="2019-11-16T23:00:25.043" Title="Can magnetic stripes be read wirelessly with lasers or other technologies?" Tags="|electrical-engineering|electromagnetism|magnets|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26048" PostTypeId="2" ParentId="26046" CreationDate="2019-02-19T18:21:41.687" Score="0" Body="&lt;p&gt;That &quot;shield&quot; would create a rotational force that needs to be countered - given vertical turbines use an aerofoil section the &quot;counter&quot; force produced by the aerofoil is probably much less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also consider that keeping all that mass of &quot;funnel&quot; pointing into the wind as well will need to be considered.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-02-19T18:21:41.687" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26049" PostTypeId="2" ParentId="26047" CreationDate="2019-02-19T20:06:23.483" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Reading a magnetic field with a laser or other light source will be &quot;difficult&quot; because light is not charged and it does not have its own magnetic field.... except it is electromagnetic radiation.  I will speculate that it is &quot;science fictionally possible&quot; to affect the phase of a laser with a rapidly (~10^14 Hz) changing magnetic field but that is not going to happen with a magnetic strip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1a. &lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_secure_transmission&quot; rel=&quot;nofollow noreferrer&quot;&gt;Magnetic secure transmission (MST)&lt;/a&gt; appears to produce a magnetic field that looks to a card reader as if it had a card being scanned. Pretty neat, I never heard of it before but I can see how it would work.  If an external coil produced a magnetic field that looked like a magnetic card was being swiped, the reader would not know the difference.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Magnetic card stripes are effectively the same media as magnetic disks, in a different shape.  The heads of old style (and maybe new style) magnetic disks never touched the surface; they rode along on a cushion of air about 0.1 micron thick.  I can't find a good picture but &lt;a href=&quot;http://www.datarecovery.com.sg/data_recovery/hard_disk_head_crash.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this web site&lt;/a&gt; may give you the idea.  The magnetic field is in the magnetic media (disk or stripe), not in the reading head.  As the media moves near the coil in the head, it induces a very small current which is amplified to be used as ones and zeros.  As a card moves in the reader, different parts move under the read head and the associated computer can put the pieces together to form a card number.  One of the tracks on a payment card is recorded at 210 bits per inch. Each bit is about the size of the thickness of a thin plastic bag. To tell the bits apart, you need to be close, &lt;em&gt;VERY&lt;/em&gt; close.  From a distance, the best you can tell is there is a magnetic field on the card, but you knew that already.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2019-02-19T20:06:23.483" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26050" PostTypeId="2" ParentId="26045" CreationDate="2019-02-19T20:12:41.583" Score="1" Body="&lt;p&gt;Granular materials come in an infinite number of varieties and the best place to look for model parameters is the soil/rock mechanics literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, in some sands (but definitely not most sands), &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; = 35 GPa, &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; = 44 GPa, &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; = 35&lt;span class=&quot;math-container&quot;&gt;$^{\circ}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\psi$&lt;/span&gt; = 15&lt;span class=&quot;math-container&quot;&gt;$^{\circ}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$c_0$&lt;/span&gt; = 0 Pa.  The cohesion of clays is, of course, higher; anywhere from 10 - 1000 KPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hardening modulus &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; is typically not used in soil/rock mechanics because perfect plasticity is assumed. If your model has it, you will either have to estimate it from experimental data or look for papers that use your particular model and see what numbers they have used.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2019-02-19T20:12:41.583" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26051" PostTypeId="1" CreationDate="2019-02-19T20:59:36.250" Score="1" ViewCount="669" Body="&lt;p&gt;Can anybody tell me how to make a similar welding job - that is to spot weld a thin stainless steel wire mesh to a stainless steel frame. What machine and technique  should be used to accomplish this job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The time has come to replace the torn wire mesh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/amMs0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/amMs0.jpg&quot; alt=&quot;spot welded stainless steel wire mesh&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19430" LastActivityDate="2019-02-20T15:53:45.053" Title="Stainless steel wire mesh spot welded to a frame" Tags="|mechanical-engineering|welding|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26052" PostTypeId="2" ParentId="26046" CreationDate="2019-02-19T20:59:36.847" Score="1" Body="&lt;p&gt;I think you may be confusing vertical axis wind turbine based on lift mechanism as in an airplane wing, with the low efficiency models that are used to show the wind speed on rooftops as part of weather vane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in a sail ship as opposed to intuition, the best wind is not the back wind which pushes the ship forward. it at best will push the ship with the speed of the wind minus frictions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is side wind at 90 degrees to the direction of the ship, because it gives lift to the sails, then the ship will go forward up to 8 times the speed of the wind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These vertical axis wind turbines are designed to work based on the lift and they are designed not to create too much drag on the back swing. They are designed to turn in a way that they face the wind deploying in best angle of attack to gain lift and turn around and feather out to make the least drag, basically this aeronautic bias makes them embrace the wind regardless its direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also any vertical shield would be an engineering headache to build and keep solid against 80-100 miles/hr winds.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-19T20:59:36.847" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26053" PostTypeId="1" CreationDate="2019-02-19T21:31:56.523" Score="0" ViewCount="88" Body="&lt;p&gt;Does it matter whether you pressurize a vessel from zero to 4000 psi in 40 millisecond or 10 hours? Will that affect the response of the tank wall in terms of the tensile stress in the wall during or after full pressurization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given I have a scenario where I'm making a large amount of gas in a closed chamber by reacting some chemical compounds. If the reaction takes place so fast that the vessel went from zero to 4500 psi in less than 1 second, how can I calculate (or at worst estimate) the thickness of steel wall of the vessel that can take the pressure surge and still remain intact with the gas contained. If you need further details, I'll be happy to provide.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastEditorUserId="19370" LastEditDate="2019-03-01T02:12:41.580" LastActivityDate="2019-03-01T02:12:41.580" Title="Does pressurization speed affect the tensile stress in the vessel wall" Tags="|structural-engineering|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26054" PostTypeId="1" CreationDate="2019-02-19T22:33:29.743" Score="1" ViewCount="22" Body="&lt;p&gt;I have to model through software a wireless (my choice: 3g,4g, 802.11...) channel with given Bit Error Rate. I'm searching some distributions, I'm reading tons of papers but I can't find how to simulate in realistic way the Channel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model should take as input a bit sequence and BER and produce the bit sequence as it could be received with noise.&lt;/p&gt;&#xA;" OwnerUserId="19431" LastEditorUserId="33" LastEditDate="2019-02-20T15:54:30.230" LastActivityDate="2019-02-20T15:54:30.230" Title="Modeling of a Wireless Channel with known BER" Tags="|modeling|telecommunication|wireless-communication|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26055" PostTypeId="1" AcceptedAnswerId="26082" CreationDate="2019-02-20T00:13:14.293" Score="1" ViewCount="841" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/4OhNj.jpg&quot; alt=&quot;The truss&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the reactions in this truss?I would usually cut in the joint where beams 5,6,7 and 8 are and calculate reactions in the two trusses that are created.How do I do that when there is a force acting on the joint?&lt;/p&gt;&#xA;" OwnerUserId="19433" LastActivityDate="2019-02-21T21:33:04.747" Title="How do I calculate the reactions in this truss?" Tags="|civil-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26056" PostTypeId="1" CreationDate="2019-02-20T05:17:18.577" Score="1" ViewCount="113" Body="&lt;p&gt;I need to buy a level switch for a quench tower water drainage system. I intend to use its output directly to open a solenoid valve to drain the tank when the switch detects water and close it afterward when the water recedes. I have identified make and model of a level switch I think is good. But the manufacturer provides multiple types of electrical outputs. It would be great if someone could advise on which of the following type is best for a simple level switch - solenoid valve circuit.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Relay&lt;/li&gt;&#xA;&lt;li&gt;NAMUR&lt;/li&gt;&#xA;&lt;li&gt;Transistor&lt;/li&gt;&#xA;&lt;li&gt;2-wire &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;P.S: I am a newbie regarding electrical stuff. I'll consult a friend who will help me to design the circuit, but I just need to place the order for the level switch.&lt;/p&gt;&#xA;" OwnerUserId="19228" LastEditorUserId="110" LastEditDate="2019-02-21T01:54:57.467" LastActivityDate="2024-03-19T21:01:05.193" Title="How to choose type of output from a water level switch?" Tags="|electrical-engineering|circuits|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26057" PostTypeId="1" CreationDate="2019-02-20T09:53:26.173" Score="6" ViewCount="147" Body="&lt;p&gt;How are dangerous to build things studied?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such as, nuclear reactors, jet engines, sawing machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do engineers discover, how they ought to be built, while avoiding getting or getting others hurt?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there perhaps a component of &quot;anticipating to avoid risk based on earlier experience&quot;? But how does this allow designing potentially dangerous things?&lt;/p&gt;&#xA;" OwnerUserId="2643" LastActivityDate="2019-02-21T16:30:18.460" Title="How are dangerous to build things studied?" Tags="|design|safety|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26058" PostTypeId="2" ParentId="26051" CreationDate="2019-02-20T13:10:06.700" Score="1" Body="&lt;p&gt;Not exactly a spot welder but I would consider something like this: &lt;a href=&quot;https://sunstonewelders.com/product/pistol-grip-single-probe-hand-piece-pg2-with-roll-spot/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pistol Grip Single Probe Hand Piece – PG2 with Roll Spot&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But judging from your picture you might want to consider using a different method then welding. There is quite a bit of corrosion which can result from improper welding. If you need support, just let me know.&lt;/p&gt;&#xA;" OwnerUserId="19438" LastEditorUserId="14629" LastEditDate="2019-02-20T15:53:45.053" LastActivityDate="2019-02-20T15:53:45.053" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26059" PostTypeId="1" CreationDate="2019-02-20T15:07:31.543" Score="0" ViewCount="294" Body="&lt;p&gt;This may be a very simple question and I think I know the answer but I would like it to be confirmed by somebody else.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My questions regard the zeros of a systems (the roots of the numerator of a Transfer Function).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1)&lt;/strong&gt; Is it possible to move zeros around (kind of how one can do with poles)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current answer to this (correct me if I'm wrong) is no. The reason is because the zeros of the closed loop transfer function are at least the same with those of the plant (of course more zeros may be presented by a controller). As a results, regardless of the controller method, the zeros of the plant/system will not move. If this is correct, this leads me to my second questions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2)&lt;/strong&gt; If the Plant/system has an unstable zeros, doesn't this automatically mean that, as far as loop gain is concerned, Robust Stability is never achieved? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I think this is because if one plots the Root Locus, a pole will always move to that zero (assuming at least proper closed loop transfer function). So from a control engineering point of view, the only thing one can do, as far as stability is concerned, is to move the poles as far away from the imaginary axis as possible. The further away the poles, the more gain can be presented in the system that would not lead to instability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct to think the above? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;P.S. as far as Robust Stability, this is what I currently understand it to be. That being said, I am not very sure if it is correct.&lt;/p&gt;&#xA;" OwnerUserId="19439" LastActivityDate="2019-02-20T15:07:31.543" Title="Can one move the zeros of a transfer function?" Tags="|control-engineering|control-theory|transfer-function|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26060" PostTypeId="1" CreationDate="2019-02-20T15:13:38.057" Score="0" ViewCount="32" Body="&lt;p&gt;I'm trying to design a cooling system for cooling nitrogen at 300°C. My original design (which is very open to modifications) is an aluminum ring block with exterior fins and vertical holes through the block connected end to end to form a continuously undulating gas line. An amazing engineer proposed flowing the gas through multiple holes simultaneously to address the pressure drop problem along the original single gas line, but any better approach is encouraged.&#xA;My design operates by submerging the block into a water the size an Olympic swimming pool at 20°C. 468 g of nitrogen driven out of a vessel originally pressurized to ~5000 psi goes through the cooling system in roughly 10 seconds. The aim is to be able to cool the nitrogen down to 25°C as it exits the cooling system at that staggering flow rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What mass, geometry, and feature of the aluminum block, and/or modifications to the original design is required to achieve the stated aim.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastActivityDate="2019-02-20T15:13:38.057" Title="Designing a submersible cooling system for rapid flow hot nitrogen gas" Tags="|thermodynamics|design|experimental-physics|" AnswerCount="0" CommentCount="12" ClosedDate="2019-02-20T17:27:29.907" ContentLicense="CC BY-SA 4.0" />
  <row Id="26062" PostTypeId="2" ParentId="26057" CreationDate="2019-02-20T15:47:18.680" Score="1" Body="&lt;p&gt;That depends on the degree and extent of the hazard involved. Analysis of all the individual components with precision and specificity is highly imperative, but no amount of numericals can exactly potray or define the real world outcome of an event. Destructive testing is one way engineers simulate a catastrophic outcome of a system, and it done in heavy protection/containment. If you're talking about something comparable to a nuclear meltdown, engineers prefer making conservative estimate which in the end is worth the risk. But in the case of cutting tools, I believe dummies tend to suffer the blow until it's tweaked to an allowable safety level.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastActivityDate="2019-02-20T15:47:18.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26063" PostTypeId="2" ParentId="26057" CreationDate="2019-02-20T16:35:36.897" Score="1" Body="&lt;p&gt;Building models, computer simulations, and making the real thing and going away helps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes they screw up despite that, either by not knowing everything about the subject, or simply not thinking about rare/unusual events.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most well known example is probably Castle Bravo, where a nuclear test went wrong at the Bikini Atoll. They evacuated everyone close to the blast, but they didn't now the bomb's outer shell would react and add 9 megatons to the original 6 MT yield. They didn't know everything about the subject (U238), so they didn't bring everyone far enough.&#xA;Marie Curie suffered from radiation poisoning, because she didn't know enough at the time about the heath effects. Lake Karachay became polluted because soviet scientists didn't know enough (or simply ignored) about handling radioactive materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rockets are tested before the first real flight without people/expensive equipment. Using a cheap concrete block (or an outdated electric car) is better if something goes wrong. Because things tend to go wrong the first time. New airplanes are tested by simulations, then without engines (like a towed glider), possibly with remote control.&#xA;Rocket or jet engines are first fixed to something, and only cameras and sensors are close to be safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you need to know what you're dealing with. When something goes wrong with a sawing machine, you might want to be behind a thick wall. If something goes wrong with a nuclear reactor, you want to be on a different planet. :)&lt;/p&gt;&#xA;" OwnerUserId="19441" LastActivityDate="2019-02-20T16:35:36.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26064" PostTypeId="2" ParentId="26057" CreationDate="2019-02-20T16:37:26.387" Score="2" Body="&lt;p&gt;It is hard to give a general answer, a lot will be highly domain specific. It's also hard to answer as engineers don't work in a vacuum. Depending on wether one designs a product to be mass-produced or a one off industrial plant, safety relevant specifications may arrive at the engineers desk from the client, from an official ordinance, their professional conscience or even from their marketing department. The designing engineers may then use a few tools to arrive at a reasonably safe design, these include industry codes and norms, simulations, testing and risk assessments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Codes:&lt;/strong&gt;&lt;br&gt;&#xA;For a lot of machines and components, you have codes that specify how the device is to be constructed. For example &lt;a href=&quot;https://www.engineeringtoolbox.com/flanges-pn-pressure-ratings-d_46.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;pipe class ratings&lt;/a&gt; tell you what wall thicknesses and flange types to use for many temperature and pressure ranges, a comparativly simple example. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tests&lt;/strong&gt;&lt;br&gt;&#xA;Certain safety critical behaviors can be tested, a classical example would be a crash test dummy in a vehicle. This is not always viable or economical, AFAIK testing is common for complex, mass produced goods like cars where the costs of the test are ameliorated over many produced units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Simulations&lt;/strong&gt;&lt;br&gt;&#xA;Tests can sometimes be replaced or aided by complex simulations, this is again common in car design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Risk assessments&lt;/strong&gt;&lt;br&gt;&#xA;Depending where you live, every machine sold will have a CE marking, which (among other things) means the manufacturers did a risk assessment. Industrial plants, too, need one to operate legally (in theory). This means someone or many someones made a long list of all liekly hazards like burning on open flames, on hot surfaces, electrocution, cutting on sharp edges, injuries due to fast moving parts etc. For each risk, they will evaluate how often it will ocur, how many persons will be in danger, how serious the danger is, and if - in sum - the risk is deemd unacceptable, the design goes back to the drawing board to install further safeguards or remove the risk some other way.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficulty in answering this uestion is that the answer depends highly on the field you work in. You can crash a car into a wall and examine the test dummy, you can't test a nuclear plant the same way (too heavy) and to do it with a sawing machine would be plain dumb. Every domain has it's own standards.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there perhaps a component of &quot;anticipating to avoid risk based on earlier experience&quot;?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Previous experience is (hopefully) baked into the codes and norms the engineers work with (often very implicitly so, it must be said).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last not least, there's many overlapping concerns regarding safety. The area I know best, plant design, you have the safety of the individual machines, you need to consider hazardous incidents (release of dangerous substances) as well as workplace safety, explosion protection, fire protection.  The last three usually have their own dedicated expert who, at several points, looks over the design and it's execution.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2019-02-21T16:30:18.460" LastActivityDate="2019-02-21T16:30:18.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26065" PostTypeId="2" ParentId="25964" CreationDate="2019-02-20T17:00:28.673" Score="1" Body="&lt;p&gt;&lt;strong&gt;Conformal Coatings are your Friend!&lt;/strong&gt;&lt;br&gt;&#xA;I suggest you look into &quot;conformal coatings&quot; used to protect electrical and/or electronic components from corrosion. These coatings are commonly used on Naval electronics -- especially for gear used aboard submarines. Use as directed (well ventillated, away from potential spark/ignition sources until fully dry, etc.) on pretty much everything except electrical contacts and moving parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So long as you have different metals in contact, you've gotta keep moisture away or else you'll get electro-chemical corrosion (AKA ad-hoc battery.) FYI, such coatings do a pretty good job, but they make probing circuitry harder. I've had to grind sharp points on a set of DVM leads to pierce these coatings; they dry pretty tough, by design.&lt;/p&gt;&#xA;" OwnerUserId="13014" LastActivityDate="2019-02-20T17:00:28.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26068" PostTypeId="2" ParentId="26055" CreationDate="2019-02-20T22:19:06.403" Score="0" Body="&lt;p&gt;All the forces to the left of point at joint between member 6 and 8 add up to 0.25P down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming we ignore the weight of truss members, we get P/2 +P = 1.5P acting at the left of member 4. This will cause -0.75P at the joint between members 6 and 8. And adding this to 1P already shown will give 0.25P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking moment about 1st support from right,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ M= \frac{(0.25P*75 + P*25)}{50 } = 0.875P \ reaction\ at\ that\ support$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Sigma F_y =0, \quad 0.8725P-(0.25P +P)= 0.375P \ reaction \ at \ rightmost\ post.  $$&lt;/span&gt;&#xA;Note, the reaction at rightmost post is directed up. so I ignored minus sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the rest of it I leave to you.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-02-21T04:22:20.970" LastActivityDate="2019-02-21T04:22:20.970" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26069" PostTypeId="2" ParentId="26056" CreationDate="2019-02-21T02:57:45.263" Score="0" Body="&lt;p&gt;How about using a latching solenoid valve like the following?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/te9Hj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/te9Hj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ussolid.com/1-2-magnetically-latching-solenoid-valve-dc-6-v.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;1/2&quot; Magnetically Latching Solenoid Valve DC 6 V&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a h-bridge driver to switch on and off the valve. Below is one that I have used most recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.pololu.com/product/2130&quot; rel=&quot;nofollow noreferrer&quot;&gt;DRV8833 Dual Motor Driver Carrier&lt;/a&gt;&#xA;\&#xA;&lt;a href=&quot;https://i.stack.imgur.com/PPJOA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PPJOA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HTH&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-02-21T02:57:45.263" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26070" PostTypeId="1" CreationDate="2019-02-21T06:21:19.117" Score="0" ViewCount="36" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RfswV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RfswV.jpg&quot; alt=&quot;Kitchen&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignore the ugly look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes I cooked food with very strong smell. That smell sometimes is not expected by many neighbors, they protested. Rather than to use exhaust fan, I considered to install funnel on top of the stove, then I will pipe it to the roof. I expected it will send the air-smell high enough. But, learned from another neighbor, that way is still not solve the problem. The air-smell sometimes is still there, especially when wind blows, the smell went anywhere. It considered become pollutant. Then, I was thinking how to solve. What is the best way to treat that air-smell?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is any method how to treat that smell so it become neutral (or &quot;clean&quot;) before it &quot;delivered&quot; the the free air? Just consider like water-waste treatment of any industry, it must treated before released to the nature/river. I expected that there is one device which such the smell, and proceed until it clean, and then deliver to the air.&lt;/p&gt;&#xA;" OwnerUserId="17943" LastEditorUserId="615" LastEditDate="2019-02-21T20:02:05.483" LastActivityDate="2019-02-21T20:02:05.483" Title="How to purify/treat air smell produced by a kitchen?" Tags="|air-quality|drain-waste-vent|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26071" PostTypeId="1" CreationDate="2019-02-21T07:19:49.237" Score="2" ViewCount="2064" Body="&lt;p&gt;I am studying mechanics of materials while I came across these terms called the displacement fields and the deformation fields. I am not able to understand the difference between them. Can someone explain it to me. &lt;/p&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2019-02-21T19:19:36.907" Title="What are displacement field and Deformation field?" Tags="|mechanical-engineering|kinematics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26072" PostTypeId="1" AcceptedAnswerId="26081" CreationDate="2019-02-21T07:24:38.443" Score="1" ViewCount="1097" Body="&lt;p&gt;In CFD we focus on the NS equations as the governing equations (together with other equations) and solved them using Some numerical method (usually FVM). What are the governing equations (counter parts of the NS equation) in solid mechanics?. &lt;/p&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2022-09-23T15:06:15.537" Title="What are the governing equations solved in FEA for structural mechanics?" Tags="|finite-element-method|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26075" PostTypeId="1" CreationDate="2019-02-21T09:56:17.553" Score="0" ViewCount="755" Body="&lt;p&gt;The concept of Adiabatic is basically the isolation of a system in a realm where energy doesn't flow either way. Adiabatic compression and expansion is a common phenomenon harnessed in equipments and machines to heat or otherwise cool. The practical application of adiabatic can vary in design and technique, hence my question. If I have a tank of nitrogen gas at 300°C and 4500 psi and I want to cool the gas to 25°C and fill the nitrogen into a large balloon to a pressure of 20 psi, how can I harness adiabatics (especially adiabatic expansion) to cool the gas to my desired temperature between the tank and balloon.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastActivityDate="2019-02-21T15:35:51.970" Title="Practical Application of Adiabatic processes" Tags="|thermodynamics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26076" PostTypeId="2" ParentId="26075" CreationDate="2019-02-21T15:35:51.970" Score="2" Body="&lt;p&gt;Adiabatic expansion of nitrogen at room temperature will produce &lt;a href=&quot;https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Joule-Thomson&lt;/a&gt; cooling.  If your tank were only slightly above room temperature, the resulting balloon would be quite cold.  The bad news is all real gases have an inversion temperature where the J-T coefficient becomes negative and throttling the gas to a lower pressure makes the temperature go &lt;em&gt;UP&lt;/em&gt;. Looking at the chart on Wikipedia, I have to wonder if you (or your professor) chose that temperature because it is close to the inversion temperature of nitrogen (348 °C).  Adiabatically, taking 300 °C nitrogen from 310 bar (no mixed units, please) to 1 bar, you will get a little cooling, exactly how much requires an Mollier (Pressure-Enthalpy) diagram for nitrogen.  Above 350 °C, you are not going to cool it by any adiabatic process.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2019-02-21T15:35:51.970" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26077" PostTypeId="1" AcceptedAnswerId="26169" CreationDate="2019-02-21T16:50:37.047" Score="2" ViewCount="1455" Body="&lt;p&gt;I apologize in advance if this is off topic - maybe it is more of an English language query... Please feel free to migrate or close if needed :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen mine cages/service lifts referred to as Mary-Ann (or Maryann/Mary Anne etc.) for example &lt;a href=&quot;https://www.sudbury.com/letters-to-the-editor/mine-cage-technology-flawed-jodi-blasutti-243919&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, &lt;a href=&quot;http://www.canadianminingjournal.com/features/red-lake-just-hitting-its-stride/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;http://www.meglab.ca/en/portfolio/instrumentation-and-control/automation-of-the-mine-cage-maryann-underground-mine-canada-s-mid-north-2013/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Does anyone know where this comes from? And is it a common term - I have heard it used in South Africa and the sources linked are Canadian, which suggests this is the case. &lt;/p&gt;&#xA;" OwnerUserId="19460" LastEditorUserId="90" LastEditDate="2022-08-03T00:10:06.973" LastActivityDate="2022-08-03T00:10:06.973" Title="Why is a mining cage called a Mary-Ann?" Tags="|mining-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26078" PostTypeId="2" ParentId="26077" CreationDate="2019-02-21T17:12:35.257" Score="0" Body="&lt;p&gt;I would guess the name derives from the fact that St Anne, the apocryphal mother of Mary and grandmother of Jesus (in both Christian and Islamic traditions) was the patron saint of miners, among many other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Saint_Anne&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Saint_Anne&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;As the mother of Mary, this devotion to Saint Anne as the patron of&#xA;  miners arises from the medieval comparison between Mary and Christ and&#xA;  the precious metals silver and gold. Anne's womb was considered the&#xA;  source from which these precious metals were mined.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Mining certainly has a long history of its own traditions, and even its own laws. For example the &quot;barmote courts,&quot; created in the 13th century to regulate lead mining in the UK, still exist today. Their function is now almost entirely ceremonial, but they still have the power to make legally binding judgments over mining disputes - the most recent judgment AFAIK was made in 2013.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-02-21T17:26:22.400" LastActivityDate="2019-02-21T17:26:22.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26079" PostTypeId="2" ParentId="26071" CreationDate="2019-02-21T19:19:36.907" Score="1" Body="&lt;p&gt;Conventions vary. As defined &lt;a href=&quot;https://nptel.ac.in/courses/105106049/lecnotes/mainch3.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;on that page&lt;/strong&gt;&lt;/a&gt;, the deformation field &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{\chi}_\kappa$&lt;/span&gt; represents the positions of the current points/elements/particles in an object right now as a function of a reference configuration (designated configuration &lt;span class=&quot;math-container&quot;&gt;$\kappa$&lt;/span&gt;) that typically corresponds to where the points were earlier: &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{x}=\boldsymbol{\chi}_\kappa(\boldsymbol{X})$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xo3VQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xo3VQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In contrast, the displacement field &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{u}$&lt;/span&gt;  is a comparison of where the material particles are now relative to where they were earlier: &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{u}=\boldsymbol{x}-\boldsymbol{X}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Again, conventions vary. &lt;a href=&quot;https://en.wikipedia.org/wiki/Deformation_(mechanics)#Displacement&quot; rel=&quot;nofollow noreferrer&quot;&gt;In another context&lt;/a&gt;, deformation may be defined as motion other than the rigid-body motion consisting of translation and rotation. Care is required to ensure that the definitions are being applied consistently.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this help clarify things?&lt;/p&gt;&#xA;" OwnerUserId="10123" LastActivityDate="2019-02-21T19:19:36.907" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26081" PostTypeId="2" ParentId="26072" CreationDate="2019-02-21T20:37:22.880" Score="1" Body="&lt;p&gt;&lt;strong&gt;Momentum equation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Navier-Stokes equations represent the equations for the conservation of linear momentum.  In convective form they are written as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\rho\frac{D\mathbf{v}}{Dt} = - \nabla p + \nabla \cdot \boldsymbol \tau + \rho\,\mathbf{g}&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\mathbf{v}$&lt;/span&gt; is the velocity and the stress is  &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{\sigma} = -p\,\mathbf{I} + \boldsymbol{\tau}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The linear momentum equation in solids is identical&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\rho~\frac{D\mathbf{v}}{Dt} = \boldsymbol{\nabla} \cdot \boldsymbol{\sigma} +\rho~\mathbf{b} &#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Constitutive relation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For compressible Newtonian fluids, the relationship between stress and velocity is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \boldsymbol \sigma = \lambda (\nabla\cdot\mathbf{v}) \mathbf I + 2 \mu \dot{\boldsymbol \varepsilon}&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is the bulk viscosity, &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; is the dynamic viscosity, and&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{\boldsymbol \varepsilon} = \tfrac{1}{2} \left(\nabla\mathbf{v} + ( \nabla\mathbf{v})^\mathrm{T}\right)&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For linear elastic solids, the stress-strain relation is &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \boldsymbol \sigma = \lambda (\nabla\cdot\mathbf{u}) \mathbf I + 2 \mu \boldsymbol \varepsilon&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; are Lame parameters, &lt;span class=&quot;math-container&quot;&gt;$\mathbf{u}$&lt;/span&gt; is the displacement, and&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\boldsymbol \varepsilon = \tfrac{1}{2} \left(\nabla\mathbf{u} + ( \nabla\mathbf{u})^\mathrm{T}\right)&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Eulerian vs Lagrangian&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because it typically does not make sense to solve CFD equations from the point of view of individual particles, an Eulerian FVM approach is often used.  In contrast, Lagrangian methods are essential for solids because we typically need to know where each point in a structure moves to under the action of loads.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2019-02-21T20:37:22.880" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26082" PostTypeId="2" ParentId="26055" CreationDate="2019-02-21T21:33:04.747" Score="1" Body="&lt;p&gt;You need to realize that this isn't one large truss. In fact, it's three separate trusses, each supporting the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's your structure with some noteworthy points added and the three &quot;sub-trusses&quot; shown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gPJZz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gPJZz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start by looking at the left-most truss which ends at node B. If you deleted the vertical member at B, that truss would no longer be stable. This means this truss is basically supported by the rest of the structure at A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means we can treat this truss as having a vertical support at B and solving it independently from the rest of the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is trivial, and we get that &lt;span class=&quot;math-container&quot;&gt;$R_A = R_B = 50\text{ kN}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, when looking at the rest of the structure, we can pretend that left-most truss doesn't exist and replace it with a concentrated downwards force of 50&amp;nbsp;kN at B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iy8QU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iy8QU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, looking at the central truss (which ends at node D), we can see the same thing applies: if it weren't for member 8 (the right-most diagonal), this truss would also be unstable. Therefore, we can once again consider node D to be a vertical support for this truss and solve for the reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you can choose whether to include the load applied at node D on this truss or on the next one. Since the load would be applied directly on the fictional support at D, that support will absorb the entirety of that load, which will then be applied when calculating the last truss. Or you can choose not to consider the load now, and then add the reaction you find for D to that load when calculating the last truss. It's exactly the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one is still pretty straightforward though, and you get &lt;span class=&quot;math-container&quot;&gt;$R_C = 225\text{ kN}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_D = 25\text{ kN}$&lt;/span&gt; (assuming you added the applied load now, otherwise it'd be -75&amp;nbsp;kN).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you move onto the last truss, replacing everything else with a downwards concentrated load of 25&amp;nbsp;kN at D (or with an upwards 75&amp;nbsp;kN but then adding in the applied load at the same location).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4t8Jq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4t8Jq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike the other trusses, this one is self-standing, so you can just solve it as-is. This is also simple, and you get &lt;span class=&quot;math-container&quot;&gt;$R_E = 87.5\text{ kN}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_F = 37.5\text{ kN}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In image form, this is what we did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TfPiU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TfPiU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-02-21T21:33:04.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26083" PostTypeId="1" CreationDate="2019-02-22T06:23:51.297" Score="1" ViewCount="934" Body="&lt;p&gt;Sorry if this is elementary but here's what I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to power a relay via 12v DC (vehicle) to open a closed switch while the 12v power is sent to the relay. Essentially, when the blinkers blink, it will open the switch each time they blink. The switch must remain closed when the blinkers are not switched on (blinking).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to that, I need to open the resting closed circuit without feeding power to it. I've attached an image as an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mggFg.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mggFg.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize the above example may not even exist, but I hope it helps you understand what we're trying to accomplish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="19463" LastActivityDate="2019-02-22T06:23:51.297" Title="12v relay to open and an close switch without sending power to switch" Tags="|electrical-engineering|automotive-engineering|electrical|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26084" PostTypeId="1" CreationDate="2019-02-22T10:14:48.177" Score="0" ViewCount="52" Body="&lt;p&gt;Gear A and B will have the same speed and tangential component of acceleration at the point where they mesh.why? From plane motion of rigid bodies energy and momentum methods.&lt;/p&gt;&#xA;" OwnerUserId="19466" LastActivityDate="2019-02-27T18:06:45.493" Title="Motion of gears" Tags="|mechanical-engineering|dynamics|" AnswerCount="1" CommentCount="3" ClosedDate="2019-03-24T19:51:34.057" ContentLicense="CC BY-SA 4.0" />
  <row Id="26086" PostTypeId="1" CreationDate="2019-02-22T14:31:15.777" Score="1" ViewCount="102" Body="&lt;p&gt;I am playing with the Idea that I should make a my own heath exchanger to generate warm water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as a hobbyist I would like to calculate the potential of my design and here is the question for me, how do I do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to calculate the heat transfer of a pipe passing trough liquid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: I have a boiler containing 100liters of water, in this boiler there is a spiral with length of 10 meters with liquid of 50°C. How do you calculate the heath transfer from the liquid in the pipe to the water inside the boiler ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets forget the fact that my boiler water is loosing heat to it's surrounding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional info: the spiral pipe will be made out 1/2&quot; (Ø21.3mm) x sched. 5 (1.65mm)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the liquid speed is 1m³/h&lt;/p&gt;&#xA;" OwnerUserId="8464" LastActivityDate="2019-02-25T14:59:30.963" Title="Calculating heath exchange for custom boiler" Tags="|heat-transfer|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26087" PostTypeId="1" CreationDate="2019-02-22T18:22:51.393" Score="1" ViewCount="54" Body="&lt;p&gt;I am currently in the process of creating a heat budget model to compute stigma surface temperature. Stigmas are part of the reproductive organ of a flower (which in my case are flowers of orchards). I've attached a picture to show how it looks like. The stigmas are the white shaped bits that are popping out of the flower center. (picture not taken by me, &lt;a href=&quot;https://thumbs.dreamstime.com/z/white-fruit-flowers-spring-beautiful-cherry-petals-anthers-stigma-branch-orchard-early-sunny-day-ready-109463924.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://thumbs.dreamstime.com/z/white-fruit-flowers-spring-beautiful-cherry-petals-anthers-stigma-branch-orchard-early-sunny-day-ready-109463924.jpg&lt;/a&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These things are hyper small, like in the fractions of the mms. Still, I've decided to simply consider the entire organ as if it only had 1 stigma and compute a budget for that one specifically. I consider this stigma as a cylinder and perform the necessary calculations there. My question is hence the following: would neglecting the other stigmas and only considering one would have a significant impact on the evaluation of drag and convection resistance? I am asking this because for objects of such small size, the turbulence they must generate must be so small, that I feel like the turbulence created by one stigma would not effect the others, but I'd like to know the thoughts of other people who have more knowledge than me on this matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a meteorologist by formation so I still have lots to learn in terms of understanding the mechanics of turbulence, but I have a fundamental grasp at it .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="19472" LastEditorUserId="19472" LastEditDate="2019-02-22T18:31:11.237" LastActivityDate="2020-08-16T17:03:55.310" Title="Evaluating drag and convection for a flower heat budget model" Tags="|mechanical-engineering|fluid-mechanics|heat-transfer|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26088" PostTypeId="2" ParentId="26087" CreationDate="2019-02-22T19:18:56.863" Score="0" Body="&lt;p&gt;Admittedly fascinating subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If heat model scope scale is smaller than say a 5mm cube, then you will be reasonably safe to ignore other Stigmas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However at that fine level, you should at least acknowledge the curvature of the stem and the textured spherical top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curve of the stem will encourage spirals of upward draft and the top will seem to current bigger than its actual size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you where to consider the entire bunch of stigmas then it will behave more like a heat sink. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-22T19:18:56.863" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26089" PostTypeId="1" CreationDate="2019-02-22T20:30:31.207" Score="0" ViewCount="49" Body="&lt;p&gt;If I want to expand a rubber cylinder by a certain internal volume - with all else held constant (type of material, length and radius (without constraining it), etc.) will a thinner or thicker walled cylinder cope with greater volumetric expansion before rupture? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The thick and thin versions have the same ID, it is the OD which changes in each case.) &lt;/p&gt;&#xA;" OwnerUserId="11394" LastEditorUserId="11394" LastEditDate="2019-02-23T16:45:59.980" LastActivityDate="2019-02-23T16:45:59.980" Title="Thin or thick wall for expanding rubber pressure cylinder" Tags="|stresses|strength|pressure-vessel|elastic-modulus|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26094" PostTypeId="1" CreationDate="2019-02-23T10:56:32.213" Score="0" ViewCount="429" Body="&lt;p&gt;We've been given a DWG file from a topographic survey of our garden. It looks great but when I try to measure dimensions of objects in Autodesk I know the real world size of, they're off by a factor of about 6 which means I cannot actually use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here in the UK I would expect everything to be in metres but how can I check, because something seems screwy?&lt;/p&gt;&#xA;" OwnerUserId="19481" LastActivityDate="2020-12-17T00:07:50.833" Title="How can I see the scale of a DWG file?" Tags="|cad|autocad|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26096" PostTypeId="2" ParentId="26086" CreationDate="2019-02-23T15:28:29.820" Score="3" Body="&lt;h1&gt;Foundations&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The starting point is the general heat exchanger equation below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dot{q} = U A \Delta T_{LM} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;span class=&quot;math-container&quot;&gt;$\dot{q}$&lt;/span&gt; is heat flow (J/s = W), &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_transfer_coefficient&quot; rel=&quot;nofollow noreferrer&quot;&gt;overall heat transfer coefficient&lt;/a&gt; of the system (W/m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C), &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the tube area (m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;), and &lt;span class=&quot;math-container&quot;&gt;$\Delta T_{LM}$&lt;/span&gt; is the log mean temperature through the exchanger (&lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next equation is the energy balance on the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{q} = \dot{m}\tilde{C}_p \Delta T $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;span class=&quot;math-container&quot;&gt;$\dot{m}$&lt;/span&gt; is the mass flow (kg/s) and &lt;span class=&quot;math-container&quot;&gt;$\tilde{C}_p$&lt;/span&gt; is the specific heat of the fluid in the pipe (J/kg &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Approach&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Reasonable values for the overall heat transfer coefficient &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; can be obtained using charts or correlation tables in a range of textbooks on heat transfer equipment or from on-line sources. Exact values are only obtained by measurements on the final working system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is obtained by geometry along the tube. For a tube of length &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; and external radius &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;, the external tube area is &lt;span class=&quot;math-container&quot;&gt;$2\pi r L$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The specific heat and flow rate of the fluid in the tube are supposedly known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assign the temperature of the external fluid as &lt;span class=&quot;math-container&quot;&gt;$T_\infty$&lt;/span&gt; and the inlet temperature as &lt;span class=&quot;math-container&quot;&gt;$T_i$&lt;/span&gt;. You end with one equation and one unknown, the outlet temperature &lt;span class=&quot;math-container&quot;&gt;$T_o$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ U A \frac{\left(T_i - T_o\right)}{\ln\left(\frac{T_i - T_\infty}{T_o - T_\infty} \right)} = \dot{m}\tilde{C}_p \left(T_o - T_i\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use this to estimate the output temperature of the fluid through the tube. Alternatively, use this to determine what tube length you need to get a desired temperature.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Example&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Recast the above as a dimensionless equation with &lt;span class=&quot;math-container&quot;&gt;$\Theta = (T_o - T_\infty)/(T_i - T_\infty)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\beta = U/\dot{m}\tilde{C}_p$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Theta = \exp(\beta\ A)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a plot of &lt;span class=&quot;math-container&quot;&gt;$\Theta$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; for values of &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; of 1 (lower right black), 3, and 10 (upper left red).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K6aui.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K6aui.png&quot; alt=&quot;Heat Exchanger Temperature Profile Plot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Estimates with Variations (Uncertainties)&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Variations (uncertainties) in &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\Theta$&lt;/span&gt; can related through linear uncertainty propagation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \left(\frac{\Delta \Theta}{\Theta}\right)^2 = \beta^2 \Delta^2 A + A^2 \Delta^2 \beta $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By example, when you know the area precisely, &lt;span class=&quot;math-container&quot;&gt;$\Delta A = 0$&lt;/span&gt;. When you have a 10% uncertainty in &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;, this translates to an estimate of the relative uncertainty in &lt;span class=&quot;math-container&quot;&gt;$\Theta$&lt;/span&gt; as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \left(\frac{\Delta \Theta}{\Theta}\right) = 0.10\ \beta A $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In plain English, when you know the area of the exchanger, the relative uncertainty in the output temperature increases linearly with the area and effective heat transfer function. Larger heat exchangers will have greater uncertainty in their end temperature, as will exchangers with low flow or high heat transfer coefficient.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-02-25T14:59:30.963" LastActivityDate="2019-02-25T14:59:30.963" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26097" PostTypeId="1" AcceptedAnswerId="26573" CreationDate="2019-02-23T16:19:12.617" Score="0" ViewCount="5030" Body="&lt;p&gt;I'm having trouble welding the inner corners of my material. Welding material is sometimes deposited on either one side or both sides, but never really connecting the two parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side/flat welds work fine: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/C8x6g.jpg&quot; alt=&quot;photo&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But those pesky inner corners don't: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/du8MX.jpg&quot; alt=&quot;photo&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material I'm using is 1mm thick 16x16mm square steel tubing.&#xA;Welding settings are 20 Amps DC electrode positive and welding electrode is 1.6mm steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I doing something wrong? Is there a trick to this I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="16583" LastEditorUserId="1832" LastEditDate="2019-02-24T00:04:49.250" LastActivityDate="2020-12-05T11:51:46.770" Title="Stick welding inner corner" Tags="|welding|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26098" PostTypeId="2" ParentId="26097" CreationDate="2019-02-23T16:23:42.203" Score="0" Body="&lt;p&gt;Practice...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not the best welder, and there are a couple on here...&#xA;But I think you just need a bit more practice and try to get the arc down at the bottom of the joint - it may be that you are just a &quot;bit&quot; too far away sometimes only a mm makes all the difference...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take the arc closer and when it sounds like you're about to extinguish it stay about there... Also &quot;pushing&quot; the arc from side to side may help...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I can't remember the times my instructor would mention &quot;birds crap smoother than that&quot;... :)  No offence meant.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-02-23T16:31:03.847" LastActivityDate="2019-02-23T16:31:03.847" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26100" PostTypeId="1" CreationDate="2019-02-23T20:39:14.803" Score="-1" ViewCount="218" Body="&lt;p&gt;Rotary ramjet is an engine that uses ramjets attached to a shaft to allow them to operate at the supersonic speeds they require without the engine itself being in translational motion. The huge advantege of such an engine is that it essentially contains a single moving part instead of thousands. I've seen patents for using deflectors at the nozzle to direct some of the exhaust along the shaft for propulsion. Why is that they are not currently replacing turbojets?&lt;/p&gt;&#xA;" OwnerUserId="18744" LastActivityDate="2019-02-24T06:29:57.217" Title="Why are rotary ramjets not replacing turbojets yet?" Tags="|mechanical-engineering|aviation|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26101" PostTypeId="1" AcceptedAnswerId="26110" CreationDate="2019-02-23T20:47:14.787" Score="0" ViewCount="73" Body="&lt;p&gt;Just began studying automata. I've already seen several examples of the applications of finite automata: pattern matching, automatic doors, etc. And so far, I understand the logical connection between those mechanisms and state diagrams and formal definitions. But I'm curious, what &lt;em&gt;thing&lt;/em&gt; is implementing time steps—if that's what they're called—allowing states to transition. I know a tiny bit about computer architecture so I'm thinking it's like a processor clock, you know, something just ticking on and off very rapidly. Then the automata takes input on every &quot;tick&quot; of said clock. But with different devices using different current types (AC/DC) how is this actually being accomplished in an arbitrary physical device?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that I'm writing this out, it seems clear there are multiple ways to accomplish that ticking depending on what we're talking about, but I'm just looking for the most common examples found in real-life. For example what is allowing the sequence of input signals to occur in an automatic door?&lt;/p&gt;&#xA;" OwnerUserId="19485" LastActivityDate="2019-02-24T11:55:35.533" Title="What is the mechanism of transition in applications of finite state automata?" Tags="|electrical-engineering|computer-engineering|consumer-electronics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26103" PostTypeId="1" CreationDate="2019-02-23T23:31:16.203" Score="0" ViewCount="1391" Body="&lt;p&gt;If I need to calculate the head coefficient using &lt;span class=&quot;math-container&quot;&gt;$c_h=\dfrac{gH}{N^2D^2}$&lt;/span&gt; and the volume coefficient using &lt;span class=&quot;math-container&quot;&gt;$c_q= \dfrac{v}{ND^3}$&lt;/span&gt; while keeping &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; (impeller rotational speed in rpm), how can I make the units cancel out and have a dimensionless answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am measuring &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; (impeller diameter in meters), &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; (gravity m/s^2), &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; (total pressure head in meters), &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; (volumetric flow rate in m/s).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made it dimensionless by changing RPM to 1/s, however I have been asked to keep RPM and change the other units, can someone plz help.&lt;/p&gt;&#xA;" OwnerUserId="19489" LastEditorUserId="1832" LastEditDate="2019-02-23T23:57:46.770" LastActivityDate="2020-08-17T15:05:10.613" Title="head coefficient and volume coefficient for centrifugal pump" Tags="|chemical-engineering|pumps|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26105" PostTypeId="2" ParentId="26100" CreationDate="2019-02-24T01:46:37.687" Score="0" Body="&lt;p&gt;The reason you haven't seen one yet is because any &lt;em&gt;mechanical&lt;/em&gt; engineer who has worked on rotating structures would take one look at this design and fall about laughing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Werner J.A. Dahm, Andrew P. Lapsa and Peter E. Hamlington, &quot;Inside-Out Rotary Ramjet Turbogenerator,&quot; &lt;a href=&quot;http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.894.5367&amp;amp;rep=rep1&amp;amp;type=pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;AIAA 2006-4169&lt;/a&gt;, claims&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The inside-out design has the advantage of placing the ramjet flowpath&#xA;  material on the inside rim of the rotor under purely compressive&#xA;  centrifugal loading.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Sorry, but that claim is pure fantasy, and the paper doesn't even attempt to explain why the authors think it is correct. If the authors don't understand the most elementary ideas about stress fields in rotating structures, it's not too surprising if their ideas aren't taken very seriously as a practical design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If somebody wants to construct an &quot;inside-out&quot; rotor with a supersonic rim speed, well, good luck with that. Maybe they could try making the entire machine out of a single artificial diamond, or something similar. Even then, they would have to run the rotor inside a vacuum tank, or the external windage would absorb most of the generated power producing a vortex trail extending many miles behind the aircraft, but zero thrust. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any conventional structural material would simply explode before it got up to its working rotation speed. Even if it did work, any structural failure would be catastrophic, not only for the engine but for the entire plane as well. It would be impossible to design a containment system for such a rotor that was light enough to be useful in practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be fair to the authors, after a few decades of working on aircraft engine design, this sort of thing doesn't really surprise me, though. I've seen dozens of &quot;fantasy engines&quot; invented by aerodynamicists which were no more impractical than this one.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-02-24T01:46:37.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26106" PostTypeId="1" CreationDate="2019-02-24T05:48:23.823" Score="0" ViewCount="31" Body="&lt;p&gt;{Equilibrium proportion P, this is the proportion which is reached at a constant temperature after an infinitely long time:P is also the austenite proportion in the case of extremely slow heating. It is determined by the fact that the free enthalpy of the pearlite-austenite mixture is a minimum at the temperature T for this particular proportion P. It can be read from an iron-carbon diagram (incorporating effects of alloying elements).}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;above {Equilibrium ....} read from journal paper.&#xA;My question is how to find P from the iron-carbon diagram??&#xA;Explain in detail if possible with example.&#xA;Thanks &lt;/p&gt;&#xA;" OwnerUserId="17778" LastEditorUserId="17778" LastEditDate="2019-02-26T12:04:15.033" LastActivityDate="2019-02-26T12:04:15.033" Title="How to find phase proportion at equilibrium - from iron-carbon diagram - Leblond" Tags="|mechanical-engineering|materials|metallurgy|welding|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26107" PostTypeId="2" ParentId="26103" CreationDate="2019-02-24T06:05:13.773" Score="0" Body="&lt;p&gt;Volumetric flow rate is not m/s, it is m^3/s...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you want rpm in minutes then have the volumtric flow rate in m^3/minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just balancing units.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-02-24T06:05:13.773" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26108" PostTypeId="2" ParentId="26100" CreationDate="2019-02-24T06:29:57.217" Score="0" Body="&lt;p&gt;In addition to alephzero's points, here is another:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We take the case of a fan with ramjets on its blade tips. By some clever contrivance we feed fuel and ignition out to each ramjet, spin it up to speed and ignite the ramjets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A brief flow perturbation induces a flameout on one of the ramjets. The system is now severely out of dynamic balance and we need a method of restarting the dead jet before the entire device tears itself to pieces. This is a nontrivial problem.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-02-24T06:29:57.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26109" PostTypeId="1" CreationDate="2019-02-24T08:11:00.770" Score="0" ViewCount="104" Body="&lt;p&gt;I have a system involving a tank of nitrogen gas inflating a balloon, the tank pressure drops from about 4500 psi to 120 psi, inflating the balloon to 20 psi. How can I calculate the average flow rate (both mass and volumetric), and the time it takes the balloon to reach 20 psi. From my amateur standpoint, Bernoulli's equation seem to involve the pressure drop rather than the driving pressure relative to flow velocity. Can anyone help with that?&lt;/p&gt;&#xA;" OwnerUserId="19370" LastActivityDate="2019-02-24T08:11:00.770" Title="Flow rate and Pressure relation" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26110" PostTypeId="2" ParentId="26101" CreationDate="2019-02-24T11:55:35.533" Score="2" Body="&lt;p&gt;A synchronous system with a clock is not necessary. The transition between states can be triggered asynchronously by inputs to the system (e.g. someone pushes a button to operate an automatic door). The finite state machine needs to have some logic elements to &quot;remember&quot; of what state it is currently in, but that does not have to be anything like the addressable memory used in digital computer system. It could simply be a set of mechanical switches and electrical relays. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example a mechanical combination lock is an example of an asynchronous finite state machine where the &quot;memory&quot; is just the physical positions of the different parts of the lock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To interface the real world inputs to a synchronous control system such as a digital computer, there are two basic techniques: either the computer program regularly &lt;em&gt;polls&lt;/em&gt; the input devices to check what state they are in, or the input devices generate &lt;em&gt;interrupts&lt;/em&gt; which force the program to jump to a different point and continue executing from there. Textbooks on computer system design explain both methods in detail.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-02-24T11:55:35.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26112" PostTypeId="1" CreationDate="2019-02-24T12:37:21.273" Score="0" ViewCount="116" Body="&lt;p&gt;Taking into account the fact that what matters in hydroelectric power generation is the potential energy of water, one would be led to the conclusion that we do not need the rest of the reservoir on the picture below, i.e. everything what is to the left of the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why one has to flood big areas of land if even a reservoir with small surface gives us what we need, i.e. equal pressure at the turbine (given by the height of the water column above it, the hydrostatic pressure is not affected by the area of the dam) and the same flow rate of water, given by river's flow rate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one wouldn't be restricted by the need to regulate the flow rate to the turbine, one could in principle build a power plant at a waterfall's edge, e.g. by digging a hole in the rock from the of the waterfall to its foot and putting a turbine there. Much smaller cost, no land flooded. Correct, or are there any other restrictions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NU7pa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NU7pa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;[&lt;a href=&quot;https://en.wikipedia.org/wiki/File:Hydroelectric_dam.svg]&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/File:Hydroelectric_dam.svg]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19494" LastEditorUserId="19494" LastEditDate="2019-02-25T01:55:11.723" LastActivityDate="2019-02-25T01:55:11.723" Title="Why hydroelectricity needs to flood so much land surface if only the height difference matters (hydrostatic pressure is not affected)?" Tags="|electrical-engineering|" AnswerCount="4" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="26114" PostTypeId="1" CreationDate="2019-02-24T14:49:34.620" Score="0" ViewCount="95" Body="&lt;p&gt;I have been tasked to mathematically model and simulate a 2 propeller, 2 motor UAV. But I am struggling with determining the thrust equations, of the propeller, for my design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I know that:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ft = Ct * rho * (w/(2*pi))^2 * D^4&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where &lt;strong&gt;rho&lt;/strong&gt; is the density of air ;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;D&lt;/strong&gt;, the diameter in meters of the propeller;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Ct&lt;/strong&gt;, the thrust coefficient;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;And &lt;strong&gt;w&lt;/strong&gt; is the angular velocity of the propeller.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However, I was not given Ct although it is necessary for this equation.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been told I can use something called the  &quot;experimental propeller force-thrust constant&quot; or &lt;strong&gt;Kf&lt;/strong&gt; (units: N/V) to calculate &lt;strong&gt;Ct&lt;/strong&gt; however, after multiplying &lt;strong&gt;Kf&lt;/strong&gt; by the voltage of the motor driving the propeller, I am at a total loss as to how to do so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I used dimensional analysis in an attempt to receive a clue to the equation:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ct = Ft/(rho*(w/(2*pi))^2*D^4)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Therefore &lt;strong&gt;Ct&lt;/strong&gt;'s units are&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;N/((kg/(m^3))*(rev^2/sec^2)*m^4)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Which is equivalent to:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(N * sec^2)/(kg * rev^2*m)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Using the definition N = Kgm/s^2 the units of Ct are rev^-2.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Using the last two unit expressions I have arrived at this equation:&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Ct = Kf * V / ((w/(2*pi))^2 * D * m)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Where V is voltage and m is mass.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The units work, but I know that doesn't mean its correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone kindly verify if this equation is actually correct? Otherwise can someone please explain how to achieve the equation for &lt;strong&gt;Ct&lt;/strong&gt; using &lt;strong&gt;Kf&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any constructive comments are much appreciated. Please forgive my visually unfriendly web math notation.&lt;/p&gt;&#xA;" OwnerUserId="11039" LastActivityDate="2019-02-24T14:49:34.620" Title="Determining thrust equations for a custom UAV" Tags="|mechanical-engineering|electrical-engineering|aerospace-engineering|aerodynamics|thrust|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26116" PostTypeId="1" CreationDate="2019-02-24T15:26:50.883" Score="0" ViewCount="79" Body="&lt;p&gt;I need to design a vibration table (approx. 1000mx1000mm) to test products between 7-200Hz. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if it is recommended to make the table structure from standardised, extruded aluminum profiles bolted together to reduce the cost (material + time) or does one have to go with a welded structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I am aware of the results of the Junker vibration test with regards to bolted connections and it seems like a bad idea because all the fasteners would loosen. However, maybe there is a workaround to use these aluminum profiles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sincerely appreciate any advice and/or previous experiences with these.&lt;/p&gt;&#xA;" OwnerUserId="19498" LastEditorUserId="1832" LastEditDate="2019-02-24T20:44:16.043" LastActivityDate="2019-07-26T17:02:25.240" Title="Vibration table for tests (7-200Hz) from extruded aluminium profiles bolted together - good idea (?)" Tags="|structures|vibration|fasteners|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26117" PostTypeId="1" CreationDate="2019-02-24T16:26:35.190" Score="2" ViewCount="355" Body="&lt;p&gt;The upper floor of my house has a split A/C, &lt;code&gt;three-ton&lt;/code&gt;. When the temperature gets up above about &lt;code&gt;95 Fahrenheit&lt;/code&gt;, the unit can't keep the upstairs cool. That means that for a few weeks a year, my kids' bedrooms are still almost &lt;code&gt;80 F&lt;/code&gt; when it's time to put them to bed. That's not really acceptable. Multiple technicians say there's nothing wrong with the unit. I understand that A/C units are sized to keep you comfortable 99% of the year, so it looks like I'm just struggling with that 1%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attic has an approximate &lt;code&gt;R value of 32&lt;/code&gt;. Not super, but the calculators I've seen indicate that upgrading to 49 won't pay for itself in less than twenty years. So I don't think insulation is the main issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The house is about eleven years old, so I expect the compressor to require replacement any minute. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I could upgrade to a larger unit, but then you run into&#xA;short-cycling issues.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What if I got a larger unit with a two-stage compressor?&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I could get &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;a &lt;code&gt;Goodman GSX14 three-ton&lt;/code&gt; as a direct replacement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;or I could also get a &lt;code&gt;GSXC16 four-ton&lt;/code&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Would the ability to run at half-speed compensate for the short-cycling concerns, and allow me to have a higher peak rating without the downside of lost dehumidification?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;An alternate solution would be to get some portable 12,000 BTU units&#xA;and just use those during the peak temperatures.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I'm not sure if that might have other side-effects.&lt;/p&gt;&#xA;" OwnerUserId="18" LastEditorUserId="19124" LastEditDate="2019-02-28T14:50:10.100" LastActivityDate="2020-02-05T16:02:40.357" Title="Can I oversize my air conditioner if I use a two speed compressor?" Tags="|hvac|cooling|home-improvement|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26118" PostTypeId="2" ParentId="26112" CreationDate="2019-02-24T16:50:35.463" Score="3" Body="&lt;p&gt;If the reservoir were deep and small, it would be possible to get the required power for a short period of time, but, it would not take long for the water level to drop, and the system would no longer be able to keep going at the same rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power generated is a function of the drop in height, and the flow rate out of the upper reservoir.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a very large reservoir, you are able to sustain the flow rate out of the upper reservoir for a much longer period of time before the level drops too low to be useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydro plants are often used to smooth supply/demand fluctuations due to their ability to rapidly adjust their output by opening/closing valves - it doesn't make sense to run them at a constant rate based on the flow rate of the river, which will change seasonally in any case. There are plenty of other technologies better suited to a constant load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, there isn't an easy way to get the drop in height without flooding the area behind the dam - you'd have to build walls around the river on all sides, and this would be such a monumental engineering challenge that the benefit of reduced land use would be completely eclipsed by the cost and risk. (see @joojaa's answer, too)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Analogy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q) An electric wheelchair runs on 24V. You could get the same voltage from putting together 8x AA batteries, so why do they need to have two very large Lead Acid Batteries?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) The AA batteries could make the motor turn, but they wouldn't get you all the way to the shops and back before running flat.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2019-02-24T22:41:03.760" LastActivityDate="2019-02-24T22:41:03.760" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26122" PostTypeId="2" ParentId="26112" CreationDate="2019-02-24T17:20:04.193" Score="1" Body="&lt;p&gt;One of the things you're missing is that for constant power you need a constant head. Power will be proportional to &lt;em&gt;mgh&lt;/em&gt; where &lt;em&gt;m&lt;/em&gt; is the mass/unit time through the turbine, &lt;em&gt;g&lt;/em&gt; is the acceleration due to gravity and &lt;em&gt;h&lt;/em&gt; is the head. Since the power will decrease linearly with head height it is important to keep the head constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be done by flooding a large elevated plain and putting the turbine at the bottom of a valley leading away from the plain. I suspect that there aren't many suitable geographical locations.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2019-02-24T17:20:04.193" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26124" PostTypeId="2" ParentId="26112" CreationDate="2019-02-24T17:37:57.443" Score="1" Body="&lt;p&gt;I would like to add geography into the discussion. While building the dam is one thing, and having enough water is another. The  reality is that  most  basin areas that  are exactly of right  size are for most part already dammed. So the  area is also partly a function of the size and shape of the area  on the other side of the dam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It  would be economically very expensive to build the entire area surronded by  a dam. The aim is to use the natural features as well as possible. This encourages the slight exageration of the  area inorder to maximize power output. The area is softof &quot;free&quot; anyway. Having more area afford you to have more turbines in parallel before you start to affect the water level of your dam.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2019-02-24T20:15:01.387" LastActivityDate="2019-02-24T20:15:01.387" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26125" PostTypeId="1" AcceptedAnswerId="36852" CreationDate="2019-02-24T17:49:52.810" Score="0" ViewCount="96" Body="&lt;p&gt;I want to use a round-over bit on a CNC router after cutting a profile with a 3mm (1/8&quot;) bit. So the round-over bit needs a shank thinner than that. I could only find shanks with 6mm (1/4&quot;) diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do round-over bits with shanks thinner than 3mm (1/8&quot;) exist? If not, what is the physical limitation that prevents it?&lt;/p&gt;&#xA;" OwnerUserId="19384" LastActivityDate="2020-07-24T15:53:20.797" Title="CNC round-over bits with thin shank" Tags="|cnc|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26126" PostTypeId="2" ParentId="26112" CreationDate="2019-02-24T19:49:54.090" Score="0" Body="&lt;p&gt;The simple explanation is that the dam storage is more of a capacitor than a water mountain. Try to picture a capacitor in a rectification circuit, it's filled as a dam so when the voltage drops a bit in the case of full rectification or totally in the case of half, the capacitor sustains the supply of current in the circuit until the periodic electricity flow comes back up and recharges the depleted capacitor. So the large space occupied is more for storage and steady/uniform water supply over time than gaining pressure. And if you consider how much it'll cost to build a concrete channel along the river bank to raise the water linear upward, it's just not practical. Your idea might thrive in an apocalyptic situation where there's heavy shortage of flowing water, else, it's not really a great idea.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastActivityDate="2019-02-24T19:49:54.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26128" PostTypeId="1" CreationDate="2019-02-25T02:29:53.947" Score="0" ViewCount="51" Body="&lt;p&gt;I'm working on a design problem, and I'm having trouble figuring out how to model the torque needed to drive a bike tire at a set acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have a tire with normal force between tire and ground &lt;em&gt;Fn&lt;/em&gt;, coefficient of static friction between tire and ground &lt;em&gt;u&lt;/em&gt;, and diameter &lt;em&gt;d&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the necessary axle torque to drive the tire at desired linear acceleration &lt;em&gt;a&lt;/em&gt; from standstill without slipping? How great can &lt;em&gt;a&lt;/em&gt; be if you want to avoid slip?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the problem doesn't give me a tire mass or moment of inertia. Is this problem possible without those?&lt;/p&gt;&#xA;" OwnerUserId="19502" LastActivityDate="2019-02-25T16:00:06.410" Title="How to model a driven wheel with no slip" Tags="|mechanical-engineering|design|dynamics|mechanical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26129" PostTypeId="1" CreationDate="2019-02-25T02:33:41.747" Score="0" ViewCount="125" Body="&lt;p&gt;I am building a gearbox that needs to lift around 50lbs very rapidly. I already have a single planetary gearbox that does the job in around 7-10 seconds, but I would like to cut this time down significantly. In order to do this, I have heard that I should be using a gearbox with multiple input motors as this will be stronger. However, I don't know how I would calculate the torque provided or the speed with which the output shaft will be turning.  &lt;/p&gt;&#xA;" OwnerUserId="19503" LastActivityDate="2019-11-22T18:01:25.170" Title="How can you calculate the torque and speed of a single output gear being driven by two different input gears?" Tags="|mechanical-engineering|motors|gears|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26131" PostTypeId="2" ParentId="26116" CreationDate="2019-02-25T06:29:14.933" Score="1" Body="&lt;p&gt;Most vibration fixtures are made of magnesium, at least they are for spacecraft launch vibration simulation.&lt;/p&gt;&#xA;" OwnerUserId="19507" LastActivityDate="2019-02-25T06:29:14.933" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26132" PostTypeId="1" CreationDate="2019-02-25T07:22:18.410" Score="0" ViewCount="43" Body="&lt;p&gt;In my efforts to characterise a 0.67 NA microscope objective (working distance of about 15mm, effective focal length of 25mm), I have placed a 20 micron precision pinhole at the focal plane of the objective, and back illuminated the pinhole with a 767nm laser. The light coming out of the objective is focused onto the imaging plane (camera plane) with a F=1 metre lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By my understanding, the image field &lt;span class=&quot;math-container&quot;&gt;$U_i$&lt;/span&gt; should be a convolution of the Point Spread Function (PSF) &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; and the object field &lt;span class=&quot;math-container&quot;&gt;$U_o$&lt;/span&gt;, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;U_i(x_i,y_i) \propto \int_{-\infty}^\infty\int_{-\infty}^\infty h(x_i-\xi, y_i-\eta) U_o(\xi,\eta)\, d\xi d\eta\, ,&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the proportional sign is for good measure as there are some coefficients in front, but that does not affect the profile. The image intensity profile is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;I_i(x_i, y_i) = |U_i(x_i, y_i)|^2\,.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case the microscope objective has a circular aperture, and that means that the PSF &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; is given by the airy function, i.e., &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;h(x,y) = \frac{J_1 \left( \frac{2\pi}{\lambda} NA \sqrt{x^2 + y^2}\right)}{\sqrt{x^2 + y^2}} \, ,&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$J_1$&lt;/span&gt; is the Bessel function of the first kind, order one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have plotted out the intensity profile (gauss quad integral) that I expect for the current pinhole and wavelength, taking into account the magnification of the system, and I got the following plot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W7jBa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W7jBa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this differs from my measurements taken in experiment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SlVVs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SlVVs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the image is approximately 20 microns (pixel size = 3.75/40 microns). &#xA;The number of peaks somewhat represent what is shown from the plot above. What concerns me more is how the intensity profiles goes to zero between peaks, unlike what the theory predicted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my understanding of the current theory incorrect?&lt;/p&gt;&#xA;" OwnerUserId="16399" LastActivityDate="2019-03-28T19:26:56.573" Title="What is the theory behind imaging a precision pinhole?" Tags="|optics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26133" PostTypeId="1" CreationDate="2019-02-25T08:46:12.693" Score="0" ViewCount="3239" Body="&lt;p&gt;I want to understand the meaning of &lt;strong&gt;headwater&lt;/strong&gt;, and I stumble across &lt;a href=&quot;http://epg.modot.org/index.php/748.4_Headwater_and_Backwater&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;, and this diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/chaRM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/chaRM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where it got me. Headwater is defined as &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;measured from the flow line (invert) of the culvert inlet, to the&#xA;  &lt;strong&gt;water surface elevation&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I have no idea what does &lt;strong&gt;water surface elevation&lt;/strong&gt; mean. Is it the actual water surface ( doesn't look like it as there is another &lt;strong&gt;normal water surface elevation&lt;/strong&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if it's not, what it is, and how it is being defined?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="9504" LastEditDate="2019-05-17T15:48:13.673" LastActivityDate="2019-05-17T15:55:07.663" Title="What is &quot;water surface elevation&quot; and &quot;headwater&quot;?" Tags="|hydraulics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26134" PostTypeId="2" ParentId="26128" CreationDate="2019-02-25T09:32:53.210" Score="1" Body="&lt;p&gt;Assuming the road level, because no information on slope is given, the maximum force before slipping is &lt;span class=&quot;math-container&quot;&gt;$F_{max}= F_n* \mu =mg\mu  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the torque will be &lt;span class=&quot;math-container&quot;&gt;$  \tau = mg\mu D/2=I \alpha$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;assuming the mass of the wheel is concentrated at the perimeter its &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ I= mD^2/4 $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\alpha = \frac{mD^2/4}{mg \mu D/2}$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$ \alpha=&amp;lt;\frac{D}{2g\mu} $$&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-02-25T16:00:06.410" LastActivityDate="2019-02-25T16:00:06.410" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26135" PostTypeId="2" ParentId="26125" CreationDate="2019-02-25T10:56:24.237" Score="1" Body="&lt;p&gt;Depending on the material, likely strength doesn't allow smaller ones. Regardless, depending on shape and gap width, you can do this with a larger tool, by driving it multiple times over the edge, with the tip in consecutive passes going through paths lying on a quarter-circle of radius equal to tool radius minus desired fillet radius. (if the tool has a flat tip, you need to add that flat tip radius to the path offset.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say, the upper-right shows the bit cross-section (well, half of it). With consecutive passes with tip offset at positions on the red line you achieve a fillet of a much narrower radius (blue overlaid by yellow). The only problem is if there are narrow gaps or the depth doesn't allow to lower the tool deep enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AMTPa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AMTPa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-02-25T10:56:24.237" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26136" PostTypeId="2" ParentId="26129" CreationDate="2019-02-25T11:17:21.430" Score="2" Body="&lt;p&gt;The first order approximation, likely sufficient for your needs, would be to just sum all the &quot;motor-side&quot; factors and treat two motors and their axis-attached gears as a single motor and a single gear of twice the torque, twice the friction, twice the durability, twice the moment of inertia, same diameter, and just calculate it as for a single such motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It becomes trickier if the motors or their gears are different, because different speeds and torques don't add up that nicely, but if you just double the input, simple doubling of the parameters and treating it as a single input will cover most of of your bases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effect on durability of the output gear is changed, as two motors distribute the load, vs one gear of twice the torque being able to snap the teeth or cause skipping, but if the output gear can take input from the single, double-torque input, it will take it from two single-torque ones easy.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-02-25T11:17:21.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26138" PostTypeId="1" AcceptedAnswerId="26145" CreationDate="2019-02-25T14:45:47.863" Score="0" ViewCount="73" Body="&lt;p&gt;If we wanted to open gate with maximum force that man can apply at distance of 1m from pivoted joint.&#xA;So I want torque to determine motor and gearbox that can be used to automate gate.&lt;/p&gt;&#xA;" OwnerUserId="19509" LastActivityDate="2019-02-25T22:20:10.977" Title="How much torque is needed?" Tags="|torque|" AnswerCount="1" CommentCount="3" ClosedDate="2019-03-22T11:49:22.173" ContentLicense="CC BY-SA 4.0" />
  <row Id="26139" PostTypeId="1" AcceptedAnswerId="26194" CreationDate="2019-02-25T15:08:56.310" Score="3" ViewCount="1695" Body="&lt;p&gt;I am trying to understand what is the difference between the wave equation ad the advection equation? Bot the equation seems to move a quantity. While in literature it is said that the advection equation is the simpler of the two I don't see how the wave equation is that much different than the advection equation except moving the quantity in both directions when talking in terms of 2d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ref :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.rose-hulman.edu/~bryan/lottamath/wave.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.rose-hulman.edu/~bryan/lottamath/wave.pdf&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://farside.ph.utexas.edu/teaching/329/lectures/node90.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://farside.ph.utexas.edu/teaching/329/lectures/node90.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="1832" LastEditDate="2019-02-25T22:35:45.510" LastActivityDate="2019-03-02T08:17:41.390" Title="What is the difference between the wave equation and the advection equation?" Tags="|convection|waves|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26140" PostTypeId="1" AcceptedAnswerId="26141" CreationDate="2019-02-25T15:25:26.203" Score="1" ViewCount="79" Body="&lt;p&gt;Got a Tech Drawing today with measurements and all that. I understand everything except for one part where they've specified a bolt with M4 P=0.35. The M part I get but what does the P specify?&lt;/p&gt;&#xA;" OwnerUserId="19510" LastActivityDate="2019-02-25T15:33:49.700" Title="What does P0.35 stand for in a Tech drawing for a Bolt" Tags="|bolting|technical-drawing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26141" PostTypeId="2" ParentId="26140" CreationDate="2019-02-25T15:33:49.700" Score="3" Body="&lt;p&gt;I would go with Pitch of 0.35mm ie 1 turn advances by 0.35mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A view of the drawing may clarify this.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-02-25T15:33:49.700" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26142" PostTypeId="2" ParentId="26133" CreationDate="2019-02-25T16:25:28.520" Score="-1" Body="&lt;p&gt;It means normal water surface elevation across the flood plain in the absence of roadway or bridge or culvert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is a quote from the same site document.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://epg.modot.org/index.php/750.3_Bridges#750.3.2.4.1_Normal_Water_Surface_Elevation&quot; rel=&quot;nofollow noreferrer&quot;&gt;normal water elev.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-25T16:25:28.520" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26143" PostTypeId="2" ParentId="26116" CreationDate="2019-02-25T19:16:14.920" Score="0" Body="&lt;p&gt;In addition to &lt;a href=&quot;https://www.eng-tips.com/viewthread.cfm?qid=449698&quot; rel=&quot;nofollow noreferrer&quot;&gt;the replies you got on Eng-Tips site&lt;/a&gt;; at normal stress levels, steel has an unlimited fatigue life while the fatigue life of the aluminum will depend on the stress level and number of cycles. And the greater stiffness of welded steel may be an advantage.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="1832" LastEditDate="2019-02-26T15:55:39.523" LastActivityDate="2019-02-26T15:55:39.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26144" PostTypeId="1" CreationDate="2019-02-25T21:13:06.100" Score="0" ViewCount="1678" Body="&lt;p&gt;I have faced a problem with &lt;a href=&quot;https://www.socomec.com/range-ups-single-phase_en.html?product=/ups-itys-es_en.html&amp;amp;view=technical#technical&quot; rel=&quot;nofollow noreferrer&quot;&gt;ITYS E-3KVA&lt;/a&gt; UPS device. While operating in the in-line mode the measured neutral-ground voltage is almost equal to zero. However, during the battery mode the neutral-ground voltage is equal to 102 V. Knowing that the UPS is properly grounded and the wiring network was checked many times and everything was ok. This neutral-ground voltage caused a serious damage to the medical devices we use. Could you please provide me with a thorough analysis, from your point of view, and tell me how to overcome this problem. Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="19474" LastEditorUserId="110" LastEditDate="2020-11-21T23:40:19.610" LastActivityDate="2024-04-05T08:04:00.547" Title="Hot neutral in UPS devices in battery mode" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26145" PostTypeId="2" ParentId="26138" CreationDate="2019-02-25T22:20:10.977" Score="0" Body="&lt;p&gt;The maximum push force for an adult male, is 818N according to NASA - &lt;a href=&quot;https://msis.jsc.nasa.gov/sections/section04.htm#Figure%204.9.3-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://msis.jsc.nasa.gov/sections/section04.htm#Figure%204.9.3-6&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying this at a distance of 1m from the pivot, corresponds to a torque of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T=F*d=818\text{ N} * 1\text{ m}=818\text{ Nm}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is likely much more torque than is required to automate the gate, however - your calculations should be based on the mass and shape of the gate, and any speed requirements for opening time and/or acceleration, rather than on the force of a human.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-02-25T22:20:10.977" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26146" PostTypeId="1" CreationDate="2019-02-26T00:41:10.117" Score="0" ViewCount="271" Body="&lt;p&gt;I'm working on a team to build a liquid sounding rocket as a school project.  We need to present our plumbing and instrumentation diagram, but we are butting heads on what program and standards we should utilize to make the PID more readable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any strategies we can use to improve readability and represent different 'flow states' that the plumbing system can provide?  Are there any good examples of P&amp;amp;ID diagrams that we should reference for rocketry specifically?  Is there an industry standard tool we can buy for relativity cheap?&lt;/p&gt;&#xA;" OwnerUserId="14040" LastActivityDate="2019-02-26T07:46:47.843" Title="Liquid Rocket Plumbing and Instrumentation Diagram: Standards and Presentation" Tags="|fluid-mechanics|valves|rocketry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26147" PostTypeId="1" CreationDate="2019-02-26T05:53:34.157" Score="0" ViewCount="39" Body="&lt;p&gt;Can the two columns and triangle thing be taken off so I  can add a porch.Is it a major job?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hKSmq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hKSmq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19514" LastEditorUserId="1324" LastEditDate="2019-02-26T06:06:23.777" LastActivityDate="2019-02-26T09:50:27.280" Title="Can the two columns and little triangle thing be removed?" Tags="|columns|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26149" PostTypeId="2" ParentId="26146" CreationDate="2019-02-26T07:46:47.843" Score="1" Body="&lt;p&gt;The term is piping and instrumentation diagramm, though plumbing sounds cute. I've never done rocket science so maybe there's conventions I'm not aware of here, but I often work with P&amp;amp;ID in plant design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;MS Visio has been used for P&amp;amp;ID drafting (Maybe the 90 day free trial period is enough for your project), there's also the free tool Diag from the openoffice or libreoffice family. Both should have a symbol library for ASME (US American codes) P&amp;amp;ID symbols, maybe also for ISO (rest of the world) ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To represent different flow states, you can print the P&amp;amp;ID several times, mark the valves and gates as open or closed and highlight the lines that carry medium. You see this done in professional contexts too.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-02-26T07:46:47.843" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26150" PostTypeId="2" ParentId="26147" CreationDate="2019-02-26T09:50:27.280" Score="1" Body="&lt;p&gt;That triangle marquee is called a gable. And the house is built in Colonial style.&#xA;And of course it can be removed, provided you get the required permit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One alternative that may be less costly and is in line with this style, and may be even a bit more classical, is to keep the columns and gable roof but add a balcony on the first level to do the job of both a porch and create an open air entertainment space. here is roughly an example. again you can build it even bigger than this so that your columns penetrate through the balcony.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0PJY2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0PJY2.jpg&quot; alt=&quot;circular portico.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-26T09:50:27.280" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26151" PostTypeId="1" CreationDate="2019-02-26T09:54:55.053" Score="0" ViewCount="83" Body="&lt;p&gt;I need to make a flat pattern for a cylinder made from a steel plate 44mm thick laminated with explosion cladding stainless steel plate of 8mm total plate thickness is 52mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How I usually work is take the mean diameter of the cylinder (Inside diameter + thickness ) * PI. To get the developed length of the cylinder. resulting in use of Kfactor 0.5&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now there are materials bonded to each other I fear the Kfactor of 0.5 might not be correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional the laminated plate 8mm is on the inside diameter side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plate will be rolled into shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What Kfactor should I use in this case or is 0.5 a good Kfactor?&lt;/p&gt;&#xA;" OwnerUserId="8464" LastEditorUserId="8464" LastEditDate="2019-02-26T10:04:23.100" LastActivityDate="2024-03-02T09:07:00.923" Title="Kfactor for laminated steel plates rolling" Tags="|metallurgy|metal-folding|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26152" PostTypeId="2" ParentId="25692" CreationDate="2019-02-26T13:23:49.040" Score="1" Body="&lt;p&gt;Based on your most recent comment, you seem to be looking for a DIY improvised way of drying your painting. I'm thinking of something, but it depends on the scale of the painting you do in the winter. I think it's way better, more efficient, and less risky to actually dry the paint from the outside given it'll take quite a lot of energy and temperature gradient to get the slab surface to temperature which increases the risk of damaging something. Why not section the slab surface and then use a heat gun to preheat each section before coating or you add the paint then shower with some hot air.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastActivityDate="2019-02-26T13:23:49.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26153" PostTypeId="2" ParentId="25720" CreationDate="2019-02-26T13:27:42.513" Score="0" Body="&lt;p&gt;Structured composites are often designed with multiple loading directions/conditions, and something being &quot;better&quot; has to be defined by the application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honeycomb structures may be used in a panel that needs to resist crushing as well as bending. With material being at 90° to the surface it can provide more protection against forces normal to the panel surface than say a soft foam core with the same density. &lt;/p&gt;&#xA;" OwnerUserId="18382" LastActivityDate="2019-02-26T13:27:42.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26155" PostTypeId="1" CreationDate="2019-02-26T17:01:36.480" Score="2" ViewCount="343" Body="&lt;p&gt;I am working with CR-P2 batteries. I have a few questions that I am trying to answer. I would appreciate if anyone kindly help me with these questions. Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I have found in google two different types of CR-P2 batteries from different vendors. One group has 1500 mAh capacity and another group has 1400 mAh capacity. Most of them claims 10 years battery life. Why these two battery capacity numbers are significant? How significant is the capacity difference of 100 mAh?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;When we draw large pulse current from the battery (let's say 3500 mA) it generates heat. I am assuming it is because of the resistor inside the battery that heats up. Is there any other reason? It causes battery faster degradation. Why does that happen?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks again for your help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&#xA;Niloy&lt;/p&gt;&#xA;" OwnerUserId="10379" LastEditorUserId="8055" LastEditDate="2019-07-22T18:29:04.653" LastActivityDate="2019-07-23T10:44:33.260" Title="Questions regarding CR-P2 battery specification" Tags="|electrical-engineering|battery|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26156" PostTypeId="2" ParentId="26155" CreationDate="2019-02-26T17:53:51.287" Score="2" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The mAh is a unit for measure of energy dished out by 1 ampere flowing continuously for 1 hour. It's a measure of the capacity and strength of battery over time. So 100 J of energy should count when used for a low resistance load, but for high resistance load that seems to be your case, it doesn't matter too much in a practical sense. For the battery life expectancy, that obviously depends on the quality of materials and manufacturing process used, so if you wish to investigate, you'll have to cut open one and do some chemical analysis. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If there is a resistor along one of the terminals, it means the resistivity in ohms might be too high for what you're loading the battery with. My advice, try doing a circuit analysis of the battery, the resistor, and your load, that way you can vary the resistance of the resistor and optimize how much power is consumed by the resistor as the current increases. You might decide to take out the resistor if you wish, but doing so will increase the voltage of the battery relative to the voltage with resistor in it. Also ordinarily, batteries tend to heat up due to cumulative energy losses over time as large current is dished out, so you can do nothing to that, unless you'll tradeoff reducing the resistance of your load. As for the degradation issue, there might be numerous causes, like your load, or battery quality, but I don't think the heating up has anything to do with that. Anybody else might have a better answer to that question.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19370" LastEditorUserId="19370" LastEditDate="2019-02-26T18:13:25.090" LastActivityDate="2019-02-26T18:13:25.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26157" PostTypeId="2" ParentId="26132" CreationDate="2019-02-26T18:28:25.417" Score="1" Body="&lt;p&gt;The fact that you have a dark center in your image indicates that you are not in focus.   That's one problem. The diffraction pattern of a circular aperture is the well-known &lt;a href=&quot;https://en.wikipedia.org/wiki/Airy_disk&quot; rel=&quot;nofollow noreferrer&quot;&gt;Airy disk&lt;/a&gt;, which has a bright center.   I'm not sure how you came up with the intensity profile in your graph.&lt;/p&gt;&#xA;&lt;p&gt;Another issues is that you don't know the spatial quality of your source laser. I would recommend expanding that beam with a lens and then refocussing to a spot at least twice the diameter of your pinhole. This will provide a light source (from the perspective of the pinhole output) that is as close to a pure spherical wavefront as possible. Oops, and as close to uniform spatial intensity as possible.&lt;/p&gt;&#xA;&lt;h2&gt;addendum&lt;/h2&gt;&#xA;&lt;p&gt;Somewhere in there you need to convolve the Optical Transfer Function of every lens in the system. It's certainly possible that one or another lens (and its NA as well) is degrading the pure Airy pattern.  But I would certainly start by adjusting focus until the center is bright.&lt;/p&gt;&#xA;" OwnerUserId="615" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-02-26T18:28:25.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26158" PostTypeId="1" CreationDate="2019-02-26T20:41:53.047" Score="0" ViewCount="761" Body="&lt;p&gt;I am developing a toy for 2-6-year old children with parts in acrylic ( Poly(methyl methacrylate)). I searched &lt;a href=&quot;http://ec.europa.eu/DocsRoom/documents/4208/attachments/1/translations/en/renditions/native&quot; rel=&quot;nofollow noreferrer&quot;&gt;Toy Safety Directive 2009/48/EC&lt;/a&gt; and &lt;a href=&quot;https://osha.europa.eu/en/legislation/directives/regulation-ec-no-1907-2006-of-the-european-parliament-and-of-the-council&quot; rel=&quot;nofollow noreferrer&quot;&gt;Regulation 1907/2006 REACH&lt;/a&gt; for mentions of perspex or acrylic and found nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is acrylic OK for children's toys under EU regulations?&lt;/p&gt;&#xA;" OwnerUserId="19384" LastEditorUserId="19384" LastEditDate="2019-02-27T10:55:59.720" LastActivityDate="2019-02-27T10:55:59.720" Title="Is acrylic OK for children's toys under EU regulations?" Tags="|materials|regulations|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26159" PostTypeId="2" ParentId="26151" CreationDate="2019-02-26T21:54:32.877" Score="0" Body="&lt;p&gt;Stainless still's modulus of elasticity is 28*10^6 psi, compared to 29*10^6 of steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order to calculated the neutral axis, centroid of bend you can the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ A_{stainless} * 28/29 =A_{convert\ to\ stell} $$&lt;/span&gt; &#xA;&lt;span class=&quot;math-container&quot;&gt;$$So\quad  8*28/29 =7.724mm$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Y_{Neutraal} = \frac{(44+7.724/2) * 7.724 + 44*22)}{44+7.724}= 25.862$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{ 25.862}{26} *0.05    =0.4973 \ Kfactor $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we see the difference is lees than 0.003 so you may even assume Kfactor of 0.5.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-26T21:54:32.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26162" PostTypeId="1" AcceptedAnswerId="26166" CreationDate="2019-02-27T04:50:36.950" Score="1" ViewCount="162" Body="&lt;p&gt;According to uniaxial stress strain curve, specimen fails at a certain stress value since necking begins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that, same stress value appears before necking especially around yield point however specimen keeps its integrity and continue up to uts point and then rupture. Why ?&lt;/p&gt;&#xA;" OwnerUserId="19522" LastActivityDate="2019-02-27T07:58:18.887" Title="Stress strain curve" Tags="|mechanical-engineering|structural-engineering|materials|solid-mechanics|" AnswerCount="1" CommentCount="1" ClosedDate="2019-02-27T16:02:25.443" ContentLicense="CC BY-SA 4.0" />
  <row Id="26163" PostTypeId="2" ParentId="26053" CreationDate="2019-02-27T05:25:39.323" Score="-1" Body="&lt;p&gt;My answer is yes it affects if material of the vessel is rate dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some ductile materials behave brittle at high strain rates. &lt;/p&gt;&#xA;" OwnerUserId="19522" LastActivityDate="2019-02-27T05:25:39.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26165" PostTypeId="1" CreationDate="2019-02-27T07:49:35.620" Score="1" ViewCount="86" Body="&lt;p&gt;How can I design a planetary gearbox with 1:20 ratio that the planets do not interfere with each other?&lt;/p&gt;&#xA;" OwnerUserId="19523" LastEditorUserId="110" LastEditDate="2019-02-28T02:00:39.100" LastActivityDate="2019-02-28T02:00:39.100" Title="how can I design a planetary gearbox?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26166" PostTypeId="2" ParentId="26162" CreationDate="2019-02-27T07:58:18.887" Score="0" Body="&lt;p&gt;It just seems the way you mention because most of the books use the so called engineering graph. Which assumes the specimen surface area as constant throughout the loading which as we know is not right. Under tension the specimen extends axially and becomes narrower by the Poison's contraction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we just plot stress versus strain curve without modifying the stress due to reduction in cross section of the specimen we get the known curve with straight elastic part and the large extension under higher stress due to hardening and the brittle rupture. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because at the first point on the graph stress level the specimen is still ductile and has flexibility to accept even a little bit higher stress and keep stretching while it becomes harder and more brittle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after a large expansion it has lost almost all its ductility and it starts a sort of free fall elongation without increasing the stress and breaks at that point. &#xA; In reality there is never negative slope on the curve and it keeps rising albeit with different shapes at distinct stress levels.&#xA;The Wikipedia graph shows this, the blue curve,B, is the real and thered,A, is the engineering curve..&lt;a href=&quot;https://i.stack.imgur.com/AVhKj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AVhKj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They explain this clearly here.&lt;a href=&quot;https://en.wikipedia.org/wiki/Stress%E2%80%93strain_curve&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia Stress/strain&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-27T07:58:18.887" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26167" PostTypeId="2" ParentId="26165" CreationDate="2019-02-27T09:06:38.400" Score="1" Body="&lt;p&gt;Offset the planet gears longitudinally from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However this can create a twisting force on the sun gear or the ring gear as torque is transferred.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2019-02-27T09:06:38.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26168" PostTypeId="1" CreationDate="2019-02-27T10:49:41.017" Score="0" ViewCount="744" Body="&lt;p&gt;I want to convert rotational motion to linear motion. Since available space is limitation in my project, I was planning to use screw jack mechanism as a lift. But screw hack has very low efficiency - about 15% to 20% only. The rest of energy is lost to friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was thinking about using worm gear to convert rotational motion to linear motion of worm by not allowing the worm to rotate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible and will it have better efficiencies as the contact area between worm gear and worm is much less than screw jack mechanism ?  I can use any other gear mechanism also as long as efficiency is high....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I need a quick release mechanism to dis-engage the worm gear from the worm. Can some one please guide me on how to disengage the worm gear ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&#xA;Deval&lt;/p&gt;&#xA;" OwnerUserId="19525" LastActivityDate="2020-03-24T12:01:20.153" Title="worm gear linear motion with release mechanism" Tags="|mechanical|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26169" PostTypeId="2" ParentId="26077" CreationDate="2019-02-27T11:51:31.160" Score="1" Body="&lt;p&gt;Someone has solved this for me offline. According to Anglo American's &lt;a href=&quot;https://southafrica.angloamerican.com/~/media/Files/A/Anglo-American-South-Africa-V2/Attachments/media/a-magazine-issue-9.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;The A Magazine&lt;/a&gt; (page 25), the origin of the name is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;‘Mary-Ann’ is a name that is often given to service cages. It originates from&#xA;  the United Kingdom and it is said that the aristocratic mine owner who&#xA;  installed the first device named it after his mistress.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19460" LastActivityDate="2019-02-27T11:51:31.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26170" PostTypeId="1" CreationDate="2019-02-27T12:03:57.727" Score="4" ViewCount="1908" Body="&lt;p&gt;Is there any difference between resonance and flutter. Both of them are related to overlap of external excitation frequency and Natural frequency then increase in amplitude even inifite amplitude theorytically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference between them?&lt;/p&gt;&#xA;" OwnerUserId="19522" LastEditorUserId="19522" LastEditDate="2019-02-27T12:09:32.480" LastActivityDate="2022-01-28T18:03:28.197" Title="Resonance vs flutter" Tags="|mechanical-engineering|dynamics|aerospace-engineering|vibration|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26171" PostTypeId="2" ParentId="26170" CreationDate="2019-02-27T12:21:22.983" Score="4" Body="&lt;p&gt;In flutter, the force applied to the structure depends on the deformation and velocity of the structure. Therefore, &lt;em&gt;the force is necessarily at exactly the same frequency as the motion&lt;/em&gt;, though it may not be at the correct &lt;em&gt;phase&lt;/em&gt; to cause the motion to increase in amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often, flutter is caused by the structure having &lt;em&gt;two&lt;/em&gt; different vibration modes with similar (but not identical) frequencies, and the flutter occurs at a frequency which is &lt;em&gt;not&lt;/em&gt; equal to either of the mode frequencies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In resonance, in frequency of the force is not necessarily the same as the natural frequency of the structure, and the amplitude of the response depends how close the two frequencies are to each other (and also on the amount of damping the system, of course).&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-02-27T12:21:22.983" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26172" PostTypeId="1" CreationDate="2019-02-27T12:57:16.983" Score="1" ViewCount="604" Body="&lt;p&gt;Honeycomb structures are made of top and bottom shells and honeycomb geometry at core. In terms of bending load, maximun normal strains take place on top and bottom hence, sheet parts can support them as expected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However maximun shear strain take place at the center of the cross section which coincide honeycomb structure. As I know, stiffness of the honeycomb very low even negligible. So, what is the idea behind shear stiffness of a honeycomb structure?&lt;/p&gt;&#xA;" OwnerUserId="19522" LastActivityDate="2024-04-02T23:05:50.697" Title="Honeycomb structures vs shear stress" Tags="|mechanical-engineering|structural-analysis|aerospace-engineering|solid-mechanics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26173" PostTypeId="2" ParentId="26172" CreationDate="2019-02-27T13:33:49.800" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;As I know, stiffness of the honeycomb very low even negligible.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That is only correct if all the edges of the honeycomb are free.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the top and bottom of the honeycomb are constrained by bonding to the face plates, the honeycomb can &lt;em&gt;not&lt;/em&gt; &quot;squash flat&quot; with negligible stiffness, and it uses the material very efficiently to resist shear, because applying a &quot;global&quot; shear strain means that the walls of the honeycomb are in either tension or compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honeycomb structures can be much stiffer than solid metal with the same weight. I have used 2m x 1m sheets of aluminium honeycomb (with top and bottom sheets) which are light enough for one person to pick up and carry around easily, but stiff enough and strong enough to stand on without damaging then, if they are supported at each end and used as a 2m long beam.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-02-27T13:33:49.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26174" PostTypeId="2" ParentId="1903" CreationDate="2019-02-27T15:36:46.363" Score="2" Body="&lt;p&gt;IC engines work in cycles, in a four-stroke IC engine there are four cycles&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Intake stroke&lt;/li&gt;&#xA;&lt;li&gt;Compression stroke&lt;/li&gt;&#xA;&lt;li&gt;Combustion stroke&lt;/li&gt;&#xA;&lt;li&gt;Exhaust stroke&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;During intake stroke the air-fuel mixture is sent into the engine cylinder, later in the compression stroke, it is compressed and then in the ignition stroke fuel gets ignited to take out power. In the exhaust stroke, the residue is sent out of the chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two types of IC Engines based on their method of fuel ignition&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;SI (spark ignition: typical gasoline or petrol engines)&lt;/li&gt;&#xA;&lt;li&gt;CI (Compression Ignition: typical diesel engines) &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The way they are named like that is because in SI engines the fuel is ignited using a spark plug and in CI engines fuel is compressed to such high pressure that it reaches a temperature near to its autoignition temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*This auto-ignition temperature is an important thing to notice&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The autoignition temperature or kindling point of a substance is the lowest temperature at which it spontaneously ignites in normal atmosphere without an external source of ignition, such as a flame or spark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GovS3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GovS3.png&quot; alt=&quot;table with different fuels&amp;#39; auto-ignition temperatures&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown in the figure, it is clear that diesel has a lower auto-ignition temperature than gasoline. For diesel, it is 210°C; for gasoline, it is 280°C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In CI engines the air is taken during the intake stroke and it is compressed to a high pressure where the air will reach a temperature of more than 210°C, then diesel is injected using a fuel injector. As soon as the diesel comes in contact with the air it gets ignited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when you put gasoline inside the diesel engine, the air during compression still will be at a temperature ranging 210°C - 220°C and the Gasoline entered has an auto-ignition temperature of 280°C. This makes fuel impossible to ignite, hence the engine will not start. If it starts it will stop within seconds due to the accumulation of fuel in the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whereas SI Engines use a different method to ignite the fuel, first the fuel is mixed with air in the carburetor. This mixing can take place only when the fuel is in vapour state. For that, the flash point of the fuel must be very low.&#xA;(flash point is the lowest temperature at which a liquid can form an ignitable mixture in air near the surface of the liquid. The lower the flash point, the easier it is to ignite the material.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now coming to flash point of gasoline it is -44°C at normal atmosphere temperatures it can easily form vapours. But diesel has a flash point of 55°C. When you put diesel into the petrol engine (SI) it will not form an ignitable mixture with air, even when a spark is generated inside the cylinder the diesel inside will be in liquid form which is not the ideal condition for flame generation. Thus the engine will not get started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the reason why the vehicles are not compatible with other fuels than they are designed for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*NOTE:- This is most raw explanation for this question, more in-depth explanations may be there please refer to them as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fun To Know:- Due to higher compression ratios diesel engines are usually used for high torque applications like load carrying and power generation. Petrol or gasoline engines are used in commercial cars and bikes due to their light weight and faster response. &lt;/p&gt;&#xA;" OwnerUserId="19528" LastEditorUserId="1832" LastEditDate="2019-02-27T16:16:25.043" LastActivityDate="2019-02-27T16:16:25.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26176" PostTypeId="2" ParentId="26170" CreationDate="2019-02-27T17:59:26.787" Score="2" Body="&lt;p&gt;I think flutter is when self excitation happens and there may not be any external vibration. Flutter mostly happens in aviation on airfoils such as wing-tips, ailerons , empennage, and because of self excitation it can grow in intensity to dangerous catastrophic levels even in smooth air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Airplane manufacturers design their planes in the wind tunnel to eliminate flutter under design operational envelop, even at failure of some sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resonance is when the external excitation frequency is close to that of the natural frequency of the object and causes the vibration amplitude and energy to increase and even become very high if excitation frequency matches that of natural frequency. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-27T17:59:26.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26177" PostTypeId="2" ParentId="26084" CreationDate="2019-02-27T18:06:45.493" Score="1" Body="&lt;p&gt;when the two gears mesh with each other,there is a point contact in between the two gears.&#xA;Since a single point in 3D space can not have different velocity or acceleration at a given point in time,thus both the gears have same tangential acceleration. &lt;/p&gt;&#xA;" OwnerUserId="19532" LastActivityDate="2019-02-27T18:06:45.493" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26178" PostTypeId="2" ParentId="16277" CreationDate="2019-02-27T18:56:35.290" Score="0" Body="&lt;p&gt;The thing that accelerates your electric vehicle is not torque- it is &lt;em&gt;power&lt;/em&gt;, which is the product of torque times RPM. If the RPM is very small (as it is when you are starting off from a standstill) then the &lt;em&gt;power&lt;/em&gt; generated by the motor is similarly small and your acceleration will be small. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you introduce a gearset between the wheel and the motor, you can gear down the motor so it spins faster at a low wheel speed and thereby accelerate faster. Then, as you speed up, you select different gears to match the peak &lt;em&gt;power&lt;/em&gt; point of the motor to the wheel speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that the efficiency of any electric motor is a function of its rotating speed, and there is a point on that efficiency curve where it is a maximum. Prudent design then requires that you design the motor so when operating at its rated speed, it is also operating at its peak efficiency point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since vehicles have to operate at a range of different speeds, optimum design dictates a variable gearset or &lt;em&gt;transmission&lt;/em&gt;  to both extract peak power from the motor and also achieve peak efficiency.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-02-27T18:56:35.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26180" PostTypeId="2" ParentId="26172" CreationDate="2019-02-27T20:38:36.303" Score="0" Body="&lt;p&gt;A honeycomb member has usually hexagonal pattern shear members. Because it is closest to an isotropic solid member within the constrains of manufacturing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets imagine we put a honeycomb Formica door flat and use it as a beam supporting 100lbs at the center, which will have 50lbs shear at support. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the support end cuts the hexagons exposing the diagonal blades at abutment, they are under out of plane shear and torque, which they can poorly take and will rotate into a mechanism, starting thee chain of collapse in the entire door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However as has been said by alephzero these 60 degree lose blades are contained in a solid frame such as a 2in solid wood and they won't develop the rotation mechanism and the door will act very stiff, supporting the load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do we use the honeycomb slabs like concrete ceilings, parking slabs, shopping mall floors, is because it is the most efficient use of material for large span slabs. For a slab to have a large 'I', it needs most of material on top and bottom and the shear can be taken care of hollow, lightweight honeycomb.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-02-27T20:38:36.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26181" PostTypeId="2" ParentId="26053" CreationDate="2019-02-27T21:30:02.793" Score="2" Body="&lt;p&gt;No, a gun barrel pressurizes rapidly yet the designs ( strength and thickness ) are based on the peak pressure independent of stress rate. Impact loading is different story and requires a blow from a solid object , not pressurization with a fluid. Hundreds of years ago cannon barrels were commonly made with cast irons for which a very strong case could be made that had low impact toughness , yet on average they performed well as gun barrels.  The same can said of cast steel of that era; I promise you that Bessemer steels were not impact tough because of high nitrogen , phosphorus, and relatively high carbon ( and low manganese).I looked in an old reference for impact values but found none, apparently it was not worth even trying to measure on these old irons and steels.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-02-27T21:30:02.793" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26182" PostTypeId="1" AcceptedAnswerId="26185" CreationDate="2019-02-28T00:12:58.860" Score="-1" ViewCount="30" Body="&lt;p&gt;Is it common to take equal values for the proportional and derivative gains of a PID controller? if so what does it mean?&lt;/p&gt;&#xA;" OwnerUserId="16692" LastActivityDate="2019-02-28T06:19:34.753" Title="Equal proportional and derivative gains of PID" Tags="|pid-control|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26184" PostTypeId="1" CreationDate="2019-02-28T02:06:38.297" Score="2" ViewCount="79" Body="&lt;p&gt;&lt;a href=&quot;https://simplicable.com/new/tensile-strength&quot; rel=&quot;nofollow noreferrer&quot;&gt;16 Examples of Tensile Strength&lt;/a&gt; has a list of items and their tensile strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;bamboo has these tensile strength ratings&#xA;Bamboo  MPA 350 PSI 50,763&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I needed to have a hydraulic motor to shred bamboo, the hydraulic motor ratings are in Nm, example&#xA;cc/rev 492&#xA;Nm/bar 6.86&lt;/p&gt;&#xA;&#xA;&lt;p&gt;cc/rev 197&#xA;Nm/bar 2.51&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the second hydraulic motor rated at 197 cc/rev and 2.51 Nm/bar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at 20bar system pressure the motor should output 50.2Nm of torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the required Nm needed to crush bamboo.&#xA;Newton Meters and MPA are different.&lt;/p&gt;&#xA;" OwnerUserId="19536" LastEditorUserId="8055" LastEditDate="2019-02-28T17:51:24.013" LastActivityDate="2019-02-28T17:51:24.013" Title="Hydraulic motor sizing" Tags="|pressure|hydraulics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26185" PostTypeId="2" ParentId="26182" CreationDate="2019-02-28T06:19:34.753" Score="1" Body="&lt;p&gt;The transfer function of a PID controller is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;C(s) = K_p + \frac{K_i}{s} K_d\,s = \frac{K_d\,s^2 + K_p\,s + K_i}{s}.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &lt;span class=&quot;math-container&quot;&gt;$K=K_p=K_d$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K_i$&lt;/span&gt; is relatively small compared to &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$K_i=\varepsilon\,K$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$|\varepsilon|&amp;lt;1$&lt;/span&gt; this can also be written as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;C(s) = K\frac{s^2 + s + \varepsilon}{s},&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which has a zero close to &lt;span class=&quot;math-container&quot;&gt;$s=-1$&lt;/span&gt;. This means that &lt;span class=&quot;math-container&quot;&gt;$C(s)$&lt;/span&gt; has an asymptote of slope zero below a frequency of one rad/s (0.16 Hertz) and a slope of plus one above that frequency. With slope I am referring how many times the magnitude in the Bode of &lt;span class=&quot;math-container&quot;&gt;$C(s)$&lt;/span&gt; increases by 20 dB per decade. Usually this transition frequency is placed close to or below the bandwidth of your closed loop system (assuming your plant acts like a second order system at that frequency range).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the assumption that &lt;span class=&quot;math-container&quot;&gt;$K=K_p=K_d$&lt;/span&gt; might imply that the closed loop bandwidth is roughly one rad/s or higher.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2019-02-28T06:19:34.753" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26186" PostTypeId="2" ParentId="16277" CreationDate="2019-02-28T08:39:03.213" Score="1" Body="&lt;p&gt;In general cases, a gear system is usually linked with an electric motor for extreme grade requirements like to carry heavy load or to attain very high speeds. The reason we need a gear system when it comes to internal combustion engines is cause they produce useful torque and power in a narrow band of engine speeds and the gear system helps the engine to be within this useful range and hence lets the vehicle attain high speeds.&#xA;For example, a hero splendor gives more mileage than a honda activa although both of them have almost the same power output. Both have an internal combustion engine but the honda activa lacks a gear system.&#xA;But when it comes to electric motors and electric vehicles the scenario seems to be completly different as electric motors give useful torque and power through a wide band of engine speeds hence a gear system isn't necessary at all.&#xA;Coming to the question, I think the efficiency of the two wheeler will reduce if we add a gear system as weight and friction  will increase unnecessarily hence more load on the electric motor and inturn reducing efficiency and increasing the complexity of a rather simple two wheeler.&#xA;Whilst on the other hand for an electric car, a gear system  isn't that bad an idea at all as the wide useful range in electric motors is found at lower rpms and this is where gear systems can be of vital importance for higher velocities.&lt;/p&gt;&#xA;" OwnerUserId="19539" LastActivityDate="2019-02-28T08:39:03.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26187" PostTypeId="2" ParentId="24609" CreationDate="2019-02-28T09:12:26.513" Score="1" Body="&lt;p&gt;To calculate force at exhaust port specifically is of little use, but nevertheless, it is possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honda GX200 specs - Max power 6.5HP@3600 rpm or 60 rps. We will calculate thrust at this rps. &#xA;196 cc engine. The exhaust port diameter from diagrams is estimated to be around 2.8cm. &#xA;Assuming complete combustion of exhaust gases we will use the density of air at 1000k as a substitute as the density is subject to large variation due to active air-fuel ratio, load conditions etc. Density = .3716kg/m^3&#xA;Engine is pumping .196L of exhaust gases every 2 revolutions as it is a 4 stroke engine.&#xA;Therefore at 60 rps, it is pumping .196L 30 times every second. That is equal to 5.88L/s or .00588m^3/s. This is the volumetric flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross sectional area of the exhaust port is pi*(d/2)^2 = .0006157m^2&#xA;Therefore the Exhaust gas velocity is .00588/.0006157 = 9.54m/s&#xA;We can now calculate thrust or force by using the following equation&#xA;F=density* Me* Ve =.3716*.00588*9.54 = .020 N of force. This is negligible and not enough to force the engine forward r in any particular direction, but can be used as an effective blower or pump of sorts with appropriate tubing. The only way to increase thrust at this rpm would be to restrict the exhaust manifold by reducing diameter and increasing length. This would cause the exhaust gases to be forced through a smaller space and hence will develop the tendency to escape quicker, ultimately increasing exhaust velocity. Hope this answer helped.&lt;/p&gt;&#xA;" OwnerUserId="19540" LastActivityDate="2019-02-28T09:12:26.513" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26188" PostTypeId="2" ParentId="26168" CreationDate="2019-02-28T10:26:11.260" Score="1" Body="&lt;p&gt;Rightly, the screw jack is not efficient for the job. For rotation to linear conversion usually Rack and Pinion arrangement is used. (Restricting the rotation of worm can also do the almost same thing.) The efficiency of rack and pinion is also high as the linear motion is raised because of normal forces at contact points unlike sliding in Screw jacks. The efficiency can go as high as 80-90%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the disengagement, you can instead disengage the pinion from it's shaft as is done in gearboxes. Attach a sleeve to the shaft and pinion for engaging and disengage and a synchronizing ring for smooth operation. For this, you will need to lose the shaft from its power source also. Consider a clutch plate in case of a continuous power supply. Refer to &lt;a href=&quot;https://www.youtube.com/watch?v=wCu9W9xNwtI&quot; rel=&quot;nofollow noreferrer&quot;&gt;manual transmission mechanisms&lt;/a&gt; used in vehicles. &lt;/p&gt;&#xA;" OwnerUserId="19541" LastActivityDate="2019-02-28T10:26:11.260" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26189" PostTypeId="1" CreationDate="2019-02-28T11:14:09.670" Score="1" ViewCount="121" Body="&lt;p&gt;Let me preface my question by saying I am a recent chemical engineering graduate and apologize if my inexperience shows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have designed a batch process which results in a water slurry stream containing an insoluble precipitate (biurea) with impurities such as chloride, complex salts and DOC which leaves the reactor at high temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to separate the precipitate, remove impurities and produce a dry product (with no particular size distribution requirements).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To achieve this I plan to use a plate filter press to produce an output stream of about 50-70 wt% solids. Is this a good choice? it seems to have low capital and operating costs (higher labour intensity) than a centrifuge and more suited to a batch process compared to a belt filter press.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am unsure how cleaning is typically implemented in an industrial process. I would like to wash the precipitate thoroughly to remove as much chloride, salt, DOC etc. impurities as possible.Is washing supposed to be performed INSIDE the plate filter press? or is the solids washed in a separate tank to be recycled to the filter press until the desired level of washing is achieved? Should I heat the washing liquid to improve solubility?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For drying I would like to use a simple tray vacuum dryer but read that this does not scale well - I need to produce about 600kg/day dry product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you and please let me know if I should clarify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;stream leaving the reactor is at 120 degC before entering the filter press and contains 5-15 wt% solids of precipitate. I am designing that the stream leaving the filter via conveyor is about 50-70% solids&lt;/p&gt;&#xA;" OwnerUserId="10923" LastEditorUserId="10923" LastEditDate="2019-03-03T16:05:36.500" LastActivityDate="2019-03-03T16:05:36.500" Title="Solid-liquid separation of a crystal slurry" Tags="|chemical-engineering|vacuum|fluid-filtration|drying|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26191" PostTypeId="1" CreationDate="2019-02-28T12:08:09.977" Score="2" ViewCount="271" Body="&lt;p&gt;I am solving an optics exercise that requires to design a Keplerian telescope with a given magnification M (negative), a certain object size L and the restriction of not having lenses faster than F/1 (wavelength of the signal and resolution are also given, but they are used for solve successive questions). In particular, the exercise requires to retrieve the NA of the objective lens from the optical invariant of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given these information, I would say that, using the fastest lenses possible, I would need a first lens with radius R1=L and a focal length of F1=2L (so that the F/1 requirement is satisfied), while the second lens should have a focal length of F2=|M|F1, a distance from the first of d=F1+F2 and a radius of R2=|M|L. However, I am not sure how to retrieve the &lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrange_invariant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lagrange invariant&lt;/a&gt;  from this theoretical design: for once, since the system is afocal, the object should be at infinity, thus how can the marginal ray be defined? Moreover, in this case both the first and the second lens limit the marginal rays of the system, and being the object at infinity I am not sure how should it be tilted (shouldn't it be parallel to the optical axis? But then wouldn't it coincide with the chief ray?).&lt;/p&gt;&#xA;" OwnerUserId="19544" LastEditorUserId="8055" LastEditDate="2019-02-28T14:57:58.427" LastActivityDate="2024-01-14T08:06:45.840" Title="What is the optical Lagrange invariant of an afocal system?" Tags="|optics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26193" PostTypeId="1" AcceptedAnswerId="26195" CreationDate="2019-02-28T13:10:45.570" Score="2" ViewCount="147" Body="&lt;p&gt;I am releasing CO2 from a cylinder into a ventilated room and want to mimic a human breathing (let's ignore the in/out breath). Problem is that the gas that comes out is at room temperature (or lower due to decompression) not body temperature. How can I heat the gas that comes out of the cylinder to 37ºC?&lt;/p&gt;&#xA;" OwnerUserId="434" LastActivityDate="2019-03-01T19:51:26.267" Title="How to heat CO2 coming out of a gas canister to human breath temperature?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26194" PostTypeId="2" ParentId="26139" CreationDate="2019-02-28T13:13:34.533" Score="3" Body="&lt;p&gt;As alephzero mentioned the difference is huge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will explain mathematically and then physically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any given partial differential equation of the form, &lt;span class=&quot;math-container&quot;&gt;$$A \frac {\partial^2\phi}{ \partial x^2} + B \frac {\partial^2\phi}{ \partial x \partial y} + C \frac {\partial^2\phi}{ \partial y^2} + \rm first\ derivative \ terms\ =0 $$&lt;/span&gt;&#xA;We get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\rm B^2 - 4AC = 0 , - Parabolic, $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\rm B^2 - 4AC &amp;lt; 0 , - Elliptic  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\rm B^2 - 4AC &amp;gt; 0 , - Hyperbolic. $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the above conditions the wave equation is a hyperbolic equation and the diffusion equation is a parabolic equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These conical conditions decides the zone of influence and zone of dependance in your domain of interest.  i.e., To be get disturbed at any point of interest in the domain, you should know where and when you should pinch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically, &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Consider a 2D square plane, apply heat on the west side edge, then&#xA;the heat will flow from west to east. if you are running from north&#xA;to south you will feel the same temerature.  &lt;span class=&quot;math-container&quot;&gt;$-&amp;gt;$&lt;/span&gt; This process is&#xA;goverend by unsteady heat diffusion equation mathematically&#xA;parabolic and zone of dependance is all east side to where apply&#xA;heat.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Throw a stone in the water pool which is still. after some time you will find, the wave reach all the corners of the pool. So you disturb anywhere to get disturbed anywhere. These waves are goverend by Laplace equation which is elliptic. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Now consider a moving channel, throw a stone, then see the waves now. Do the disturbances reach upstream of the channel? Nope. The disturbances created by the stone are carried downstream by the moving stream. This is in the 2D sense. Imagine same thing for 1D also. The created disturbances are carried by the material elasticity (noted by the wave/sound speed). when and where (the disturbed region) is decided by the wave speed. These inrteresting problems are governed by hyperbolic equations. So disturb only certain location to get disturbed somewhere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can google more to get pictorial explanations for these. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this answer helps!&lt;/p&gt;&#xA;" OwnerUserId="13721" LastEditorUserId="19124" LastEditDate="2019-03-02T08:17:41.390" LastActivityDate="2019-03-02T08:17:41.390" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26195" PostTypeId="2" ParentId="26193" CreationDate="2019-02-28T14:18:15.343" Score="2" Body="&lt;h2&gt;Options&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;You could purchase something equivalent to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://farnam-custom.com/standard-products/process-air-heaters&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://farnam-custom.com/standard-products/process-air-heaters&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could make something equivalent to this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=mEExlOX458Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=mEExlOX458Y&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Principles&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The principles for the heat exchanger are comparable to this post, only using air inside and water outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/26086/calculating-heath-exchange-for-custom-boiler/26096#26096&quot;&gt;Calculating heath exchange for custom boiler&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are interested in the heat flow needed to change a the temperature of a flowing fluid, consider the equation below with &lt;span class=&quot;math-container&quot;&gt;$\dot{q}$&lt;/span&gt; as heat flow (W), &lt;span class=&quot;math-container&quot;&gt;$\dot{m}$&lt;/span&gt; as mass flow (kg/s), &lt;span class=&quot;math-container&quot;&gt;$\tilde{C}_p$&lt;/span&gt; as specific heat (J/kg &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C), and &lt;span class=&quot;math-container&quot;&gt;$\Delta T$&lt;/span&gt; as the desired temperature change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dot{q} = \dot{m}\tilde{C}_p \Delta T $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an ideal gas, mass flow and volume flow &lt;span class=&quot;math-container&quot;&gt;$\dot{V}$&lt;/span&gt; are related as below with &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; as molar mass (g/mol), &lt;span class=&quot;math-container&quot;&gt;$\dot{n}$&lt;/span&gt; as molar flow (mols/s), &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; as pressure (Pa), and &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; as gas law constant (J/mol K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{m} = M\dot{n} = M p \dot{V} / R T $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is not constant, the volume flow rate will change for the same mass flow rate. Basically, as an ideal gas is heated in a tube, it expands and moves faster to keep the same mass flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total energy that you need to supply the heat is calculated from the efficiency of the power to heat conversion.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-03-01T19:51:26.267" LastActivityDate="2019-03-01T19:51:26.267" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26196" PostTypeId="1" CreationDate="2019-02-28T14:25:40.230" Score="1" ViewCount="115" Body="&lt;p&gt;I am looking into using this &lt;a href=&quot;http://www.htc-sensor.com/products/179.html&quot; rel=&quot;nofollow noreferrer&quot;&gt; TAL230B miniature type load cell&lt;/a&gt; to measure forces in x and y direction. The normal sensor loading is displayed in image 1. What is the effect of loading them as in image 2? And what is the effect of putting a rotational force on the end?&lt;img src=&quot;https://i.stack.imgur.com/rOX6b.jpg&quot; alt=&quot;&quot;&gt; &lt;img src=&quot;https://i.stack.imgur.com/oIEUT.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19546" LastEditorUserId="8055" LastEditDate="2019-02-28T17:51:15.577" LastActivityDate="2019-02-28T17:51:15.577" Title="How much does loading a load cell in other directions than the direction it was designed for effect the readings?" Tags="|measurements|sensors|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26198" PostTypeId="1" CreationDate="2019-02-28T20:17:44.100" Score="0" ViewCount="210" Body="&lt;p&gt;So, I'm doing a group project for a materials and manufacturing engineering class, where our job is to make a product using injection molding. For our project, we chose a sphere (keeping it simple)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in charge of calculating cooling time for the part while its in the mold. Herein lies the issue:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to figure out how heat (power) flows through our mold, Which can be represented by a hollow sphere in a perfect cube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equation for thermal conductivity is (K)(A)(T2-T1)/(L), Where L is the parallel distance of hard object in which the heat has to flow. This equation is set up for a solid, uniform plate, The problem is that the object in which the heat must flow is NOT a uniform object!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what do I set L as? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have Schaum's outline for college physics, and it allows me to define the rest of the parameters. I just need L.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, another question: Our mold is being 3-D Printed, to where the mold isn't truly a solid object; a certain percent is plastic (Given by % infill) and the rest as air. I believe I got the K-value correct for this part but I would like to backcheck -&gt; ((%infill)(K-PLA)) + (100-%infill)(K-air) = K-incomplete_infill&lt;/p&gt;&#xA;" OwnerUserId="19548" LastActivityDate="2024-03-04T18:07:35.967" Title="Sphere in a cube - Thermal conductivity" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26199" PostTypeId="2" ParentId="26198" CreationDate="2019-02-28T20:49:32.460" Score="0" Body="&lt;p&gt;You probably don't need an &quot;accurate&quot; answer here - I guess you really want to know whether to wait 5 minutes or an hour before the molding is cool enough to handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A practical way to get an approximate answer would be to take the outside of the mold as another sphere, with an &quot;average&quot; thickness corresponding to the real shape. You know the maximum and minimum thickness of the mold from the size of the sphere and the cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-050-thermal-energy-fall-2002/lecture-notes/10_part3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt; gives an equation for heat flow through a spherical shell (section 2.10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if you want it to cool down quickly, maybe you should change the shape of the mold so it is closer to a sphere, to reduce the thickness at the corners of the cube.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-02-28T20:49:32.460" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26201" PostTypeId="2" ParentId="26193" CreationDate="2019-03-01T01:32:08.083" Score="1" Body="&lt;p&gt;take a cheap hair dryer apart so you can control the heater element separate from the fan blower. Connect the heater element to 120VAC through a variac (variable AC transformer) so you can control the voltage it sees. Put a temperature sensor in the outlet stream of the hair dryer and build a duct that carries CO2 from the tank to the inlet of the hair dryer. turn on the CO2 so it blows through the hair dryer with the dryer's fan off. adjust the variac to get the desired discharge temperature. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-03-01T01:32:08.083" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26202" PostTypeId="1" CreationDate="2019-03-01T02:21:22.763" Score="0" ViewCount="97" Body="&lt;p&gt;I'm designing a pressure vessel with a relatively large cap that provides access into the vessel, what recommended thread specifications is needed to maximized locking and safety against faliure or popout.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastActivityDate="2019-03-01T02:21:22.763" Title="Screw Cap design for high pressure surge vessel" Tags="|design|threads|" AnswerCount="0" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="26204" PostTypeId="2" ParentId="26198" CreationDate="2019-03-01T04:37:42.470" Score="0" Body="&lt;p&gt;You can find an upper bound to the cooling time by approximating the mold with a simpler shape that you know will take more time than the actual shape. That could mean treating the cube as a sphere whose surface touches the corners of the cube. Then you can confidently decide to open the mold or whatever after that maximum time has elapsed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also find a lower bound on the cooling time by using a sphere the touches the thinnest parts of the cube.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2019-03-01T04:37:42.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26205" PostTypeId="1" CreationDate="2019-03-01T04:42:31.173" Score="0" ViewCount="5532" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FOscq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FOscq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's given that L1 = 5m, L2 and L3= 7.5m, w = 8.9kN/m , I am asked to find the bending moment using these coefficient.. Prior to this , i need to have the value of shear force first.Shear force = coefficient x force.I having problem getting the force for support 2 and 3 .... I think that the 'bottom'shear force for support 2 should be = F x Cs = w x L1 x Cs = 8.9 x 5 &#xA;x 0.66 = 29.4kN, while 'top'shear force for support 2 should be = F x Cs = w x L1 x Cs = 8.9 x 7.5 x 0.484 = 32.3kN, am i right ? Correct me if i am wrong. &lt;/p&gt;&#xA;" OwnerUserId="9254" LastActivityDate="2019-03-01T12:37:40.867" Title="Shear force and bending moment for continuous span" Tags="|structural-engineering|beam|moments|shear|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26206" PostTypeId="1" CreationDate="2019-03-01T04:45:16.847" Score="0" ViewCount="97" Body="&lt;p&gt;&lt;strong&gt;Assumptions&lt;/strong&gt;&lt;br&gt;&#xA;Building Process exists&lt;br&gt;&#xA;Waterproofing is able to be replaced&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Notes&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Please answer yes or no with the science backing.  &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This question is only about the strength and lifespan. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This question is &lt;strong&gt;NOT&lt;/strong&gt; about cost, or process. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Additional material may be required. Please include in your answer if it can overcome an obstacle. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Potentially Relevant Values&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compressive Strengths:&lt;br&gt;&#xA;Granite: ~200MPa&lt;br&gt;&#xA;Concrete: ~70MPa   &lt;/p&gt;&#xA;" OwnerUserId="19551" LastEditorUserId="19551" LastEditDate="2019-03-03T05:51:54.470" LastActivityDate="2019-03-03T05:51:54.470" Title="Is a granite block tunnel longer lasting than reinforced concrete?" Tags="|structural-engineering|civil-engineering|tunnels|masonry|" AnswerCount="1" CommentCount="9" FavoriteCount="0" ClosedDate="2019-03-01T13:16:45.633" ContentLicense="CC BY-SA 4.0" />
  <row Id="26207" PostTypeId="2" ParentId="26205" CreationDate="2019-03-01T05:45:07.557" Score="1" Body="&lt;p&gt;Right, or just as a double check, we say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction at leftmost support plus the lower shear at the 2nd support must add up to total load on the span, meaning (1)*5*8.9kN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$0.34 + 0.66 = 1  \quad ,Check. $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,it is 29.37kN at the bottom of the 2nd support and the top shear is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vtop= 8.9*7.5*0.484=32.307.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-03-01T06:42:49.320" LastActivityDate="2019-03-01T06:42:49.320" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26208" PostTypeId="2" ParentId="26206" CreationDate="2019-03-01T05:55:48.230" Score="1" Body="&lt;p&gt;It is possible to use granite instead of concrete in construction (remember that buildings were made out of stone long before concrete was invented), but only in those instances where the structure is guaranteed to never experience tensile stresses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By reinforcing the concrete with prestressed steel rods, it is possible to render the resulting structure resistant to tensile loads as well- something which cannot be done with granite blocks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Careful preparation of concrete results in uniform properties. Blocks of granite, being natural products mined out of the earth's crust, can possess faults, cracks and inhomogenieties which compromise their structural integrity in a way that is out of the designer's control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of no way to reinforce granite blocks with any sort of nanomaterials. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-03-01T05:55:48.230" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="26209" PostTypeId="2" ParentId="26205" CreationDate="2019-03-01T10:14:21.473" Score="2" Body="&lt;p&gt;You're on the right track, though there looks to be a typo in the last sentence of your question (you wrote &quot;L1&quot; and then typed the length for L2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the &quot;bottom&quot; shear coefficient refers to the span extending toward the left. The &quot;top&quot; shear coefficient refers to the span extending toward the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a brief discussion of how to use coefficient tables/diagrams of this sort might be useful to you. Below is a sketch (not to scale) of a beam configuration, shear diagram, and moment diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lLp3N.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lLp3N.jpg&quot; alt=&quot;example multi-span beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For Shear&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conceptually:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$V_{span,\ end} = C_{shear,\ end} * L_{span} * w$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking your figure as an example:&#xA;&lt;span class=&quot;math-container&quot;&gt;$V_{2,\ right} = 0.516 * L_2 * w$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the reaction at your Support 3 will equal &lt;span class=&quot;math-container&quot;&gt;$V_{2,\ right} + V_{3,\ left}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For Moment&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate moment within a span:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$M_{span} = C_{moment} * (L_{span})^2 * w$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate moment at a support:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$M_{support} = C_{moment} * (L_{average})^2 * w$$&lt;/span&gt;&#xA;Where &lt;span class=&quot;math-container&quot;&gt;$L_{average}$&lt;/span&gt; is the average of the spans directly to the left and right of the support under consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Caveat&lt;/strong&gt;&#xA;I’m operating on the assumption that these coefficients were provided as part of a homework assignment, along with the beam geometry and loading. It’s important to note that in cases with unequal spans, the exact coefficients will vary with the span ratio(s).&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2019-03-01T12:37:40.867" LastActivityDate="2019-03-01T12:37:40.867" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26210" PostTypeId="1" AcceptedAnswerId="26219" CreationDate="2019-03-01T15:19:50.127" Score="1" ViewCount="580" Body="&lt;p&gt;I'm new to plasma cutting with a hand-held plasma cutter (as opposed to CNC). Its straightforward to cut with a fence/guide and I am getting good results that way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I want to freehand a rough shape without setting up a complicated jig, I can't figure out a way to see a line I want to cut. I wear a welding mask on a low setting for eye safety and once it dims I can't see any lines I've drawn on the part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a good way to safely see a cut line when plasma cutting with a dimming mask?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2019-03-02T21:51:01.440" Title="Visible markings for plasma cutting?" Tags="|machining|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26212" PostTypeId="2" ParentId="26144" CreationDate="2019-03-02T00:20:16.463" Score="0" Body="&lt;p&gt;This was more complicated than I thought.  The APC &lt;a href=&quot;https://forums.apc.com/spaces/5/smart-ups-symmetra-lx-rm/forums/general/583/high-earth-voltage-at-smart-ups-output-on-battery-operation&quot; rel=&quot;nofollow noreferrer&quot;&gt;web site&lt;/a&gt; says&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While &quot;On Battery,&quot; safety agencies ... require the UPS input&#xA;  connections to be disconnected from the utility input to prevent&#xA;  &quot;backfeed&quot; power/voltage from the UPS batteries being applied to the&#xA;  UPS input terminals. Because the Neutral must also include this&#xA;  &quot;backfeed&quot; protection ..., the output Neutral to Ground voltage will&#xA;  &quot;float&quot; to approx half the voltage between output Line &amp;amp; Neutral. For&#xA;  a 230Vac system, the N-GND voltage can be 115Vac.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They don't want you to backfeed so you don't electrocute the workers trying to fix things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;APC goes on to say:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the UPS output must have a N-GND connection for zero volts while&#xA;  the UPS is &quot;On Battery,&quot; an output isolation transformer must be&#xA;  attached to the UPS output so that a new N-GND connection can be made&#xA;  without creating a ground loop.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If it were my equipment, I would open it up and connect neutral and ground, with an interlock so you have to be physically disconnected from the mains to make the connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea is to have a transfer switch such as is used with a generator.  When the circuit is to be powered from the generator, you are physically disconnected from the mains. If you wire it correctly you could have the N-GND connection with a transfer switch.  This is not a UPS but it will keep you from blowing up your equipment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your best bet is to get the isolation transformer.  It would not be as much fun as reworking the equipment but it also won't void your warranty.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2019-03-02T00:20:16.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26213" PostTypeId="2" ParentId="16277" CreationDate="2019-03-02T03:17:57.423" Score="0" Body="&lt;p&gt;Motors for electric bikes may be brushed, or they may be brushless DC motors (BLDC motors). The industry default has become brushless motors, because generally they are quieter, smaller, and lighter, and they don't need to serviced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, what happens is thick copper coils of wire electric power from the battery into the movement that pushes you along. Instead of having one motor powering all the wheels using gears or chains, they build a motor directly into the hub of each wheel—so the motors and wheels are one and the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now due to the availability of high torque over wide range of motor speed directly coupling the motor with the motor is not a great idea. Thus, generally the electric motor is always connected to the drive wheel through fixed gear reduction ratio. for this purpose, we generally use a combination of spur gears or helical gears in order to incorporate the advantages of regenerative breaking (charging of battery during breaking or when no throttle is applied / dynamo) which is an essential requirement for all electrical vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the other part of your question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From basic physics, power is defined as rate of energy consumption. In electrical engineering load is defined as power but also at particular voltage &amp;amp; frequency. You often find the heaters are rated as 1.2 kW at 250 Volt that means it will consume 1200 watt if supplied from 250 volt, and if the voltage is less than 250 volts then its power won’t be 1200 watt. similarly, motor is rated to give full load output when connected the source of specified voltage and frequency.&lt;/p&gt;&#xA;" OwnerUserId="19532" LastEditorUserId="19532" LastEditDate="2019-03-02T06:00:07.067" LastActivityDate="2019-03-02T06:00:07.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26214" PostTypeId="1" CreationDate="2019-03-02T04:23:03.030" Score="0" ViewCount="151" Body="&lt;p&gt;I am curious whether this will become an option, especially for longer lasting structures. Could pre-stressed graphene or another high tensile strength material by used to reinforce granite block construction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Potentially relevant values:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermal Expansion Coeffiecients:&lt;br&gt;&#xA;granite  7.9−8.4⋅10−6 m/mK,&lt;br&gt;&#xA;graphene: it has a negative thermal expansion value &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compressive Strengths:&lt;br&gt;&#xA;concrete: ~40 MPa&lt;br&gt;&#xA;granite: ~200 MPa  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tensile Strengths:&#xA;Rebar: 420 MPa&lt;br&gt;&#xA;Graphene: 130.5 GPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note I am not certain of these values. &lt;/p&gt;&#xA;" OwnerUserId="19551" LastEditorUserId="19551" LastEditDate="2019-03-02T04:37:16.430" LastActivityDate="2023-08-11T12:05:00.700" Title="Could pre-stressed graphene be used to reinforce granite block construction?" Tags="|materials|civil-engineering|building-design|masonry|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26215" PostTypeId="1" CreationDate="2019-03-02T04:37:03.530" Score="1" ViewCount="1206" Body="&lt;p&gt;I have done a basic calculation of the force at an angle on the speed bump when the car makes a contact on it (please check the attached image), now I wanted to calculate the kinetic energy that I can get from this considering the vehicle is 1850&amp;nbsp;kg and travelling at 15&amp;nbsp;mph from this force. I have also calculated the potential energy that I can get from the car just using gravity and the mass. Once I obtain that how do I add both the energies? Can I just add them or is there an equation for it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main aim here is to calculate the energy that I can get when a car at 15&amp;nbsp;mph hits a speed bump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5A5I4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5A5I4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19560" LastEditorUserId="1832" LastEditDate="2019-03-02T15:43:18.287" LastActivityDate="2019-03-02T15:43:18.287" Title="Force and Kinetic Energy on a Speed Bump" Tags="|energy|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26216" PostTypeId="2" ParentId="26214" CreationDate="2019-03-02T07:33:30.890" Score="-2" Body="&lt;p&gt;I think you're better off using a carbon fiber wrapped into a rod and impregnated with epoxy. That's more practical and cost effective than graphene.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastActivityDate="2019-03-02T07:33:30.890" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26217" PostTypeId="2" ParentId="26215" CreationDate="2019-03-02T07:48:33.923" Score="1" Body="&lt;p&gt;The kinetic energy in this case is horizontal and it doesn't have any interaction with the potential. The potential isn't really much to consider since a speed bump is so near to ground level and to harness that energy, the bump has to sink to ground level as each wheel pairs stands on it. The amount of energy absorbed by the bump equal to the difference between the kinetic energy for the initial velocity just before the car hits the bump and final velocity just after leaving it. I'm not sure what you intend to use the energy for, but it doesn't work like you're approaching it. Your attached math is right by the way.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastActivityDate="2019-03-02T07:48:33.923" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26218" PostTypeId="2" ParentId="11796" CreationDate="2019-03-02T10:45:59.453" Score="1" Body="&lt;p&gt;When designing a gear or gear system, there are two important angles to consider: The pressure angle and the helix angle (in worms also called the lead angle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure angle determines the shape of the tooth and must be the same for two gears to mesh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The helix angle determines the angle of the tooth on a gear. Spur gears have this angle set to 0 (and then the tooth becomes straight).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a worm drive system the helix or lead angle cannot be 0. If you do that there would be no worm: You would end up with just a single (in case of a 1-start worm) concentric tooth around a hub. However, you &lt;em&gt;can&lt;/em&gt; give it a very low helix angle (&amp;lt;6). In that case, the helix angle of the straight spur gear and the low lead angle worm would be close enough, so that, with a little extra clearance, it would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Toy worm drives use this all the time, as demonstrated beautifully in your LEGO example. Just don't use it in a mission critical production environment.&lt;/p&gt;&#xA;" OwnerUserId="19561" LastActivityDate="2019-03-02T10:45:59.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26219" PostTypeId="2" ParentId="26210" CreationDate="2019-03-02T17:11:53.287" Score="2" Body="&lt;p&gt;I use the liquid correction pens, the sort of thing used for whiting out marks on paper, Tippex etc. &lt;a href=&quot;https://uniball.in/shop/img/p/4/3/1/431-thickbox_default.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://uniball.in/shop/img/p/4/3/1/431-thickbox_default.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They produce a reasonably thin line about the same as the kerf of a plasma cutter and work pretty well on most metals also because they are quite bright white the marks are very visible when cutting with the sort of shade of eye protection you are likely to be using for plasma cutting.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is lightly scoring the surface with a thin slitting disk in an angle grinder which can work well for straight lines and shallow curves and has the advantage of not being burned off by the cutting process so you keep a good reference for any final grinding to shape you might need to do. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastEditorUserId="4357" LastEditDate="2019-03-02T21:51:01.440" LastActivityDate="2019-03-02T21:51:01.440" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26222" PostTypeId="1" AcceptedAnswerId="26235" CreationDate="2019-03-03T00:09:53.427" Score="-1" ViewCount="48" Body="&lt;p&gt;Which is more reliable. For example, in setting steel posts in the ground, the pole should go under ground 1/3 of its height from the surface up. In IBC, steel post depth is made rather on a storey foundation, calling for 18&quot; depth for 1 storey/floor, and 24&quot; depth for 2 storey/floors. Which should be followed?&lt;/p&gt;&#xA;" OwnerUserId="10348" LastActivityDate="2019-03-04T03:00:28.070" Title="International Building Code vs Industry Standards" Tags="|concrete|foundations|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26224" PostTypeId="1" AcceptedAnswerId="26250" CreationDate="2019-03-03T13:47:36.740" Score="0" ViewCount="129" Body="&lt;p&gt;My understanding of a typical jet engine is that when running, the turbine extracts energy from the hot exhaust gases. This drives the turbine, which is coupled to the compressor at the front of the engine. As a result, when the turbine accelerates, the compressor accelerates. This increases the mass flow rate of air into the engine, which in turn increases the mass flow rate out of the engine, producing more thrust. It also means more energy is extracted by the turbine, which drives the compressor even faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems as though it is a cycle that should continue to increase without limit. Therefore, my question is what stops the engine spin speed from increasing to infinity (ignoring material limits)? Why does the spin speed of the engine stabilise if the compressor and the turbine are continually driving each other at increasing rates?&lt;/p&gt;&#xA;" OwnerUserId="7802" LastActivityDate="2019-03-05T10:56:17.593" Title="If the turbine drives the compressor, then what limits the engine speed?" Tags="|aerospace-engineering|turbines|compressors|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26227" PostTypeId="2" ParentId="26224" CreationDate="2019-03-03T17:10:08.747" Score="1" Body="&lt;p&gt;Simply put, fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fuel is the source of energy by burning and expanding the flow of air in combustion chamber, and the flow of fuel is controlled by the pilot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pilot or auto pilot, have a chart and an algorithm for fuel demand and push fuel levers to adjust the fuel being injected into combustion chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there are other applications for jet engines such as marine or power generation for electrical plants, on those the fuel control could be semi or fully automatic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is Wikipedia's diagram of a jet engine, &lt;a href=&quot;https://i.stack.imgur.com/LEalp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LEalp.png&quot; alt=&quot;turbo fan jet&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-03T17:10:08.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26229" PostTypeId="1" AcceptedAnswerId="26233" CreationDate="2019-03-03T19:49:17.497" Score="0" ViewCount="46" Body="&lt;p&gt;I'm trying to get an answer for something potentially quite simple, but I can't visualize it (nor do I have the parts yet to build it). It's for a little hobby project, but I'm slightly concerned I have made an assumption about what will happen, which may not be correct. I've drawn the problem below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OA3y7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OA3y7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence you have two pulleys connected by a timing belt.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Pulley A = 80 Teeth&#xA;Pulley B = 120 Teeth&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Pulley A is static, it does not rotate at all. Pulley B is free to spin around its own axis. It also then rotates around the axis of A, keeping the timing belt taught. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one revolution of pulley B around the axis of A (dotted line), how many times would pulley B have rotated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption would be that it would be 2/3 rotations of pulley B for every time it does an entire revolution around pulley A.&lt;/p&gt;&#xA;" OwnerUserId="19578" LastActivityDate="2019-03-03T20:51:43.830" Title="Trying to visualize the resulting interaction between two pulleys" Tags="|pulleys|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26230" PostTypeId="2" ParentId="25692" CreationDate="2019-03-03T20:06:50.027" Score="0" Body="&lt;p&gt;Nobody else applied any numbers to this so here is my take.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smallest post tension strand I found on the web was 1/2 inch in diameter or (0.25^2*PI) about 0.2 square inches.&#xA;We want to heat up a 7,200 square foot area slightly to assist in drying paint.&#xA;A square that size would be about 85 feet on a side or 1020 inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So assume we have an iron conductor 1020 inches long with an area of 0.2 square inches.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resistance of iron is about 4E-6 Ohm-Inch so doing the math:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ohms=(4e-6 ohm inch)*(1020 inch)/(0.2 inch^2)=0.02 ohms&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Remembering P=IE and E=IR we have P=I^2R  so if you put your welder on the strand and pump some current through, I get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;20A      8W&lt;/li&gt;&#xA;&lt;li&gt;40A      32W&lt;/li&gt;&#xA;&lt;li&gt;60A      72W&lt;/li&gt;&#xA;&lt;li&gt;100A     200W&lt;/li&gt;&#xA;&lt;li&gt;200A     800W&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now,this is inside the concrete. If you put in 800W without being sure what is happening inside, you are ... not going to be pleased with the results.  A little reading shows that if you keep the temperature of the strand below 200°C, you won't compromise the strength... of the strand but I would not want to bet on the concrete if it was expected to set at that temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I would do is connect a welder at some minimum current and feel around and feel around to see if the strand is warm before entering the slab.  I expect that the heat transfer inside with contact with the concrete will be greater than outside to air so if it is not too hot to touch, you are OK.  Let it sit for a while and turn up the current a little, keeping in mind that the power goes up as square of the current.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2019-03-03T20:06:50.027" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26233" PostTypeId="2" ParentId="26229" CreationDate="2019-03-03T20:51:43.830" Score="1" Body="&lt;p&gt;yes, you are right, but only partially. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one clockwise revolution, the belt is wound around, pulled, by A at a rate of 80 teeth. So pully B will have turned&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$80/120 =\frac{2}{3}\quad$&lt;/span&gt; counterclockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we have to add to this one clockwise rotation that we initiated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total B rotation is 1-2/3 = 1/3 clockwise.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-03T20:51:43.830" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26235" PostTypeId="2" ParentId="26222" CreationDate="2019-03-04T03:00:28.070" Score="1" Body="&lt;p&gt;You don't get to pick what code or standard to use for a project, the local governing authority has established a code and all applicants for a building permit have to follow it, even state when it has to construct roads and bridges, or other public projects has to follow these codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the examples you used don't sound correct. In the case of steel columns, there are specified load combination factors(such as LFRD) and design criteria, building types, fire protection, the utility of the column as axial or lateral load bearing member, etc. But in all the cases the columns are supported by a concrete foundation through a base plate and anchor bolts all designed according to governing codes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the state of California has its building code. The city of Los Angeles has adopted this code with some modifications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building industry institutes, such as AISC, American Steel Institute, or ACI  work closely with current codes and upgrade to comply with the latest codes. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-04T03:00:28.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26236" PostTypeId="2" ParentId="26214" CreationDate="2019-03-04T08:23:06.653" Score="-1" Body="&lt;p&gt;I think so. The simplest example I can think of is this: You take a beam of granite and glue (potentially pre-stressed) graphene to the side that will be under tension in your building. If the beam spans between two coloumns, you would reinforce the underside. If it cantilevers out of a wall, the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now there's several questions here that I can't answer:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;will you need to pre-stress the graphene? &lt;/li&gt;&#xA;&lt;li&gt;Can you actually securely glue graphene to granite?&lt;/li&gt;&#xA;&lt;li&gt;Is there any case where such a construction is superior to reinforced concrete?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-03-04T08:23:06.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26239" PostTypeId="2" ParentId="25830" CreationDate="2019-03-04T13:25:57.897" Score="3" Body="&lt;p&gt;The only test method I've seen is in the RWMA Resistance Welding Manual. They recommend the use of a pusher block and support to test the pull strength of cross wire welds. It is shown on page 6-5 of the revised 4th edition manual. There is also a reference table with weld strengths based on setdown and various weld parameters on the same page.&lt;/p&gt;&#xA;" OwnerUserId="19585" LastActivityDate="2019-03-04T13:25:57.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26240" PostTypeId="1" CreationDate="2019-03-04T17:10:49.700" Score="0" ViewCount="82" Body="&lt;p&gt;I would like to know if there is any theory behind designing the fixation for any part. To optimise the number of mounting holes and fasteners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If so what is it called and where can I get it?&#xA; Please help me.&lt;/p&gt;&#xA;" OwnerUserId="19588" LastActivityDate="2019-04-04T03:02:45.440" Title="Regarding design of fixation of a part" Tags="|mechanical-engineering|structural-engineering|design|computer-aided-design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26242" PostTypeId="1" CreationDate="2019-03-04T18:01:15.733" Score="2" ViewCount="1956" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nucXZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nucXZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my lecture material on different types of structural supports, there is the above picture of a hinge. It is explained that a hinge is a type of support that exerts no moment on the rod, but prevents the end of the rod from moving. On the right, the supporting force is shown not to be necessarily parallel to the rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, often in solving problems we assume that the force in this type of rod is parallel to the rod itself, such as here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gktKo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gktKo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it assumed that the force exerted by the hinge against the rod is parallel to the rod, if the resultant force in a hinge is not always parallel to the rod it supports?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2019-03-05T03:37:18.867" Title="Direction of the resultant force in hinge supports" Tags="|structural-engineering|civil-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26243" PostTypeId="2" ParentId="26242" CreationDate="2019-03-04T18:31:48.890" Score="2" Body="&lt;p&gt;If the rod is pinned at both ends, the pin joints can transmit forces, but not bending moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the equilibrium of the rod by taking moments about one end, the line of action of the force at the other end must be along the rod, otherwise there would be a non-zero moment acting on the rod and it would rotate.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-03-04T18:31:48.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26244" PostTypeId="2" ParentId="26240" CreationDate="2019-03-04T21:29:17.253" Score="2" Body="&lt;p&gt;It depends on the context. In structural engineering it means a connection that would be strong enough to develop a plastic hinge after the applied stresses surpass the design stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In trusses and spaceframes a joint that is not pinned can be designed to be fixed up to a certain limit and then fail in a controlled, energy consuming, ductile way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Machinery a fixation usually means minimum strength that can resist all the forces, vibrations and torques, so it is important the node be bolted in a balanced and symmetrical way so that the center of rigidity of the part coincides with the axis and point of action of the load, it was designed to support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are rules on how to place the bolts and fasteners so that they contribute evenly to rigidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is AISC handbook of steel connections. &lt;a href=&quot;https://www.aisc.org/globalassets/continuing-education/education-archives/connections/connections_viii_2017.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Connections in steel structures&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For lumber connections and fasteners Google Lumber or wood Type5 construction connectors and fasteners.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-03-05T02:39:21.753" LastActivityDate="2019-03-05T02:39:21.753" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26247" PostTypeId="2" ParentId="26242" CreationDate="2019-03-05T03:37:18.867" Score="0" Body="&lt;p&gt;You've made a small misunderstanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difference between pinned and fixed supports is that pinned supports do not resist rotation, and therefore do not apply &lt;strong&gt;&lt;em&gt;concentrated&lt;/em&gt;&lt;/strong&gt; bending moments on the beams. Meanwhile, fixed supports resist rotations by applying said concentrated bending moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your observation is correct, though: the forces applied on the beam might not be parallel to the beam, which implies in a bending moment developing along the beam. However, the fact remains: &lt;strong&gt;&lt;em&gt;the pinned support is only applying forces&lt;/em&gt;&lt;/strong&gt;. What happens to the beam is another matter entirely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, &lt;strong&gt;&lt;em&gt;fixed supports apply not only forces but also concentrated bending moments&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The consequence of this is that the beam's bending moment at the support due to the support's reactions is zero in pinned supports (since the lever arm between the force and the studied point is obviously zero) and (usually) non-zero in fixed supports (since it's equal to the support's concentrated bending moment reaction).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-03-05T03:37:18.867" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26249" PostTypeId="1" CreationDate="2019-03-05T06:47:07.113" Score="1" ViewCount="233" Body="&lt;p&gt;I have a tipping bucket rain gauge whose accuracy I feel could be improved by redesigning the shape of the bucket lip. In particular, after tipping, there tends to be a bead of water (approx 0.1 - 0.2ml) attached to the lip of the bucket. The bucket tips with only 4.2mls of water and that is collected close to the pivot point so that a 0.1ml drop at the lip of the bucket (as far from the  pivot as you can get) is not insignificant and adds quite a bit of random noise to the measurements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As seen in the image the current bucket lip is a smooth scallop shape and (although difficult to see in the image) the lip profile is a flat downward sloping bevel. Essentially the bucket has a similar shape and lip profile as a curved woodworking chisel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thoughts are that there must be some modification that I could make to the lip to encourage that drop to fall off and follow its brethren down the drain? I've had many different ideas to try: cutting a valley in the centre of the lip, having a thin protrusion sticking out from the lip, or maybe a fine sawtooth edge would work to draw the drop off? I have also thought that having a semi-circular rather than bevelled lip profile would work better to draw the drop over the edge?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I have is that I have no easy way of testing any of these hypotheses. I only have one tipping bucket and can't afford to buy more just to experiment on so I don't want to go modifying it unless I am reasonably sure that it will be an improvement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Google searches have brought up plenty of results on hydrophobic surfaces and the like (which would be fantastic but are a little beyond my reach) but nothing on surface shape or edge geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously I'm looking for ideas for modifications that I can easily make at home but any links to theoretical info on the subject would be welcome also.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://i.stack.imgur.com/vvHs1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/vvHs1.jpg&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="19593" LastEditorUserId="9504" LastEditDate="2019-05-17T15:48:03.447" LastActivityDate="2019-05-17T15:48:03.447" Title="Best spout shape to ensure minimal water sticks after pouring?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="0" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="26250" PostTypeId="2" ParentId="26224" CreationDate="2019-03-05T10:56:17.593" Score="0" Body="&lt;p&gt;This an oversimplification, but it should get the idea across&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the turbine extract energy from the air?  The turbine stator (nozzle) turns the air from axial to tangential (converting internal energy to kinetic energy), and then the turbine rotor removes the tangential velocity from the air and converts it into rotational energy of the rotor.  So, the amount of energy extracted by the rotor depends on the relative tangential velocity between the air and the turbine.  Put another way, &lt;strong&gt;as the turbine speeds up, it extracts less work from the air.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact opposite is true for a compressor.  The compressor rotor add tangential velocity to the air, and then the compressor stator turns the air from tangential back to axial.  So &lt;strong&gt;as the compressor speeds up, it is doing more work on the air.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So start at a low speed, and then dump a bunch of fuel in the combustion.  Now the turbine produces a lot of torque, and the compressor is not consuming much.  The excess torque accelerates the rotor.  Now at the high speed, the compressor is consuming more torque, but the turbine is producing less.  There's still some excess torque, but not as much.  The rotor keeps accelerating.  As it goes the compressor is consuming more and more and more torque, and the turbine is producing less and less and less, until some point it balances out.  The turbine is producing exactly as much as the compressor is consuming.  No excess torque.  Therefore it stops accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now what this simplified description has assumed is that the amount of fuel burned (i.e. amount of heat added) is constant with speed.  That's not really true.  But again this should get you the idea.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-03-05T10:56:17.593" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26251" PostTypeId="1" CreationDate="2019-03-05T14:50:47.920" Score="0" ViewCount="158" Body="&lt;p&gt;In terms of calculated stress, strain, displacement etc which method yields more conservative results. Analytical calculation or numeric calculation?&lt;/p&gt;&#xA;" OwnerUserId="19522" LastActivityDate="2019-03-05T14:50:47.920" Title="Analytical calculation vs FEA" Tags="|mechanical-engineering|structural-analysis|finite-element-method|aerospace-engineering|solid-mechanics|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26253" PostTypeId="1" CreationDate="2019-03-05T16:02:02.997" Score="2" ViewCount="136" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LDh1K.jpg&quot; alt=&quot;Beam loading condition&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been battling with a statics question (as attached) and can't for the life of me figure it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've attempted removing the left side guide joint so then there are only 3 reactions and it should be statically determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't know how to reapply the loading conditions for the removed guide as it is axial not transverse to which the boundary condition applies. (X direction in Cartesian coordinates).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone point me in the right direction on this matter? &lt;/p&gt;&#xA;" OwnerUserId="19600" LastEditorUserId="1832" LastEditDate="2019-03-05T17:08:30.933" LastActivityDate="2019-03-12T16:46:20.463" Title="Indeterminate beam with a guide at each end" Tags="|civil-engineering|statics|beam|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26254" PostTypeId="1" AcceptedAnswerId="26265" CreationDate="2019-03-05T17:55:28.843" Score="1" ViewCount="762" Body="&lt;p&gt;I am given these sets of equations for a gantry crane:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;span class=&quot;math-container&quot;&gt;$M\frac{d^2x}{dt^2}+D\frac{dx}{dt}-mg\theta =u$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;span class=&quot;math-container&quot;&gt;$ml\frac{d^2\theta }{dt^2}+mg\theta +m\frac{d^2x\:}{dt^2}=0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;span class=&quot;math-container&quot;&gt;$y=x+l\theta $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is the position of the trolley; &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the mass of the trolley; &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is the mass of the payload; &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the angular position of the payload; &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; is the damping coefficient; &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; is the payload position; &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; is the force applied to the trolley; &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; is the cable length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8fZt3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8fZt3.png&quot; alt=&quot;Gantry Crane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the equations above, I have found the transfer function of this system to be &lt;span class=&quot;math-container&quot;&gt;$G\left(s\right)=\frac{y\left(s\right)}{u\left(s\right)}=\frac{s^2\left(l-1\right)+g}{Mls^4+Dls^3+\left(M+m\right)gs^2+Dgs}$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I have to find the system response if &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; is zero and the payload is shifted by an angle of &lt;span class=&quot;math-container&quot;&gt;$10^{\circ }$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that we must now consider the initial condition of a &lt;span class=&quot;math-container&quot;&gt;$10^{\circ }$&lt;/span&gt; angle when we do the Laplace transform; however I can't figure out how to incorporate the initial condition in my equations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; is zero, how is it mathematically possible to find &lt;span class=&quot;math-container&quot;&gt;$y(s)$&lt;/span&gt; to be not zero, given that &lt;span class=&quot;math-container&quot;&gt;$y(s)=u(s)G(s)$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="18341" LastActivityDate="2019-03-07T16:53:56.150" Title="Find the system response given the initial condition" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26255" PostTypeId="2" ParentId="26253" CreationDate="2019-03-05T19:51:46.057" Score="0" Body="&lt;p&gt;Let's call the supports A and B from left to right and the mid support C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load P creates a moment in the beam, &lt;span class=&quot;math-container&quot;&gt;$M= P \cdot b$&lt;/span&gt; which will cause the beam to rotate clockwise by an angle &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the moments generated at the two supports counter this moment to settle at equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can write&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum M &amp;amp;= 0 \\&#xA;P \cdot b &amp;amp;= M_{A} +M_{B } \tag 1 \\&#xA;M_{A} &amp;amp;= \left(\frac{  \theta}{AC}\right) \frac {6EI}{L_{AC}} \tag 2 \\&#xA;M_B &amp;amp;= \left(\frac{ \theta}{BC}\right)\frac {6EI}{L_{BC}} \tag 3&#xA;\end{align}$$&lt;/span&gt;&#xA;These are the fixed end moments created by support settlement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say &lt;span class=&quot;math-container&quot;&gt;$BC= K \cdot AC$&lt;/span&gt;, so we substitute in the above equation (1) and we get,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P \cdot b = M_{A} = \left(\frac{  \theta}{AC}\right) \frac {6EI}{L_{AC}} + M_{B}=\left(\frac{  \theta}{K \cdot AC}\right) \frac {6EI}{K \cdot L_{AC}} =\frac{\theta(1+K^2)6EI}{K^2 \cdot L_{AC}^2} $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And from here we calculated the &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \theta = \frac{(P \cdot b)K^2L_{AC}^2}{(1+K^2)6EI} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added a sketch of the beam, after deflection, and of course, highly exaggerated and inaccurate. &lt;span class=&quot;math-container&quot;&gt;$ \theta$&lt;/span&gt; is the angle the line AB makes with the level original beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I have used &lt;span class=&quot;math-container&quot;&gt;$AC, L_{AC}$&lt;/span&gt; and similar notations for the same lengths just to emphasize where it has been used to delineate an angle and where to show the length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2WlPV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2WlPV.png&quot; alt=&quot;sketch of the beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="19600" LastEditDate="2019-03-11T15:00:06.213" LastActivityDate="2019-03-11T15:00:06.213" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26264" PostTypeId="2" ParentId="26254" CreationDate="2019-03-06T12:33:04.503" Score="0" Body="&lt;p&gt;Laplace transform works in the &lt;strong&gt;frequency&lt;/strong&gt; domain, whereas initial conditions are for transient simulations in the &lt;strong&gt;time&lt;/strong&gt; domain, so it does not make sense to apply an initial condition to a Laplace transform. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does make sense is to do a time-domain simulation with the initial condition you mention. The fact that &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; is equal to zero simply means that there are no external horizontal forces applied on the trolley. However, with an initial condition of 10° for &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;, the angular acceleration of mass will result in a non-zero &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; as a function of time, which in turn will result in a non-zero &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; as a function of time (c/f your first and second equations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are various tools out there for time-domain simulations, you may not need to use the Laplace transform at all for this, it's simply solving coupled differential equations. Here is an example with GNU Octave. I am using a &lt;a href=&quot;https://en.wikipedia.org/wiki/State-space_representation&quot; rel=&quot;nofollow noreferrer&quot;&gt;state-space representation&lt;/a&gt; rather than a Laplace transform because it allows me better control of the states, which makes it easier to specify the initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I define my state vector as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ &#xA;X = \begin{bmatrix} x \\ \dot{x} \\ \theta \\ \dot{\theta} \end{bmatrix}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{X} = \begin{bmatrix} \dot{x} \\ \ddot{x} \\ \dot{\theta} \\ \ddot{\theta} \end{bmatrix}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The input is &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; and output is &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;. The state-space representation is written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{X} = AX + Bu \\&#xA;y = CX + Du&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is a 4x4 matrix, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; a 4x1 matrix, &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; a 1x4 matrix and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; a 1x1 matrix, i.e. a scalar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After re-arranging the differential equations to have &lt;span class=&quot;math-container&quot;&gt;$\dot{X}$&lt;/span&gt; as a function of &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; only, I get the following matrices:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;A = \begin{bmatrix} 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; -D/M &amp;amp; mg/M &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 \\ 0 &amp;amp; D/(Ml) &amp;amp; -(g/l)*(1+m/M) &amp;amp; 0 \end{bmatrix} \\&#xA;B = \begin{bmatrix} 0 \\ 1/M \\ 0 \\ -1/(Ml) \end{bmatrix} \\&#xA;C = \begin{bmatrix} 1 &amp;amp; 0 &amp;amp; l &amp;amp; 0 \end{bmatrix} \\&#xA;D = 0&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Octave, with some randomly chosen values, the code looks something like this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;% Parameters&#xA;M=1;&#xA;m=0.15;&#xA;D=0.5;&#xA;g=9.81;&#xA;l=1;&#xA;theta_0=10*pi/180;&#xA;&#xA;% State-space matrices&#xA;A = [0 1 0 0; 0 -D/M m*g/M 0; 0 0 0 1; 0 D/(M*l) -(g/l)*(1+m/M) 0];&#xA;B = [0 1/M 0 -1/(M*l)]';&#xA;C = [1 0 l 0];&#xA;DD = 0; % using DD to avoid confusion with parameter D&#xA;G = ss(A,B,C,DD);&#xA;&#xA;% Time-domain simulation&#xA;t = linspace(0,10,500); % time vector&#xA;u = zeros(size(t));     % input vector u&#xA;x0 = [0 0 theta_0 0];   % initial conditions&#xA;lsim(G,u,t,x0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which gives the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4EURn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4EURn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other method is to solve the differential directly with an ode solver, which yields the same results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you need to define a function to implement your differential equations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function dx = gantry_crane(t,x)&#xA;&#xA;  M = 1;&#xA;  m = 0.15;&#xA;  D = 0.5;&#xA;  l = 1;&#xA;  g = 9.81;&#xA;&#xA;  dx(1) = x(2); % dx/dt&#xA;  dx(2) = (m*g*x(3) - D*dx(1))/M; %d^2x/dt^2&#xA;  dx(3) = x(4); % dtheta/dt&#xA;  dx(4) = -(m*g*x(3)+m*dx(2))/(m*l); %d^2theta/dt^2&#xA;&#xA; end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is saved under the file name &lt;code&gt;gantry_crane.m&lt;/code&gt;. You then call the &lt;code&gt;ode&lt;/code&gt; solver as follows:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;theta_0=10*pi/180;&#xA;x0 = [0 0 theta_0 0]&#xA;[t,x] = ode45(@gantry_crane,[0 10],x0);&#xA;y = x(:,1) + x(:,3); % y = x + l*theta and l=1&#xA;plot(t,y)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which gives the following result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/47iGj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/47iGj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2019-03-07T16:53:56.150" LastActivityDate="2019-03-07T16:53:56.150" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26265" PostTypeId="2" ParentId="26254" CreationDate="2019-03-06T14:03:15.857" Score="0" Body="&lt;p&gt;When deriving the transfer function it is usually implicitly assumed that the initial conditions are zero. However the &lt;a href=&quot;https://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems&quot; rel=&quot;nofollow noreferrer&quot;&gt;Laplace transform&lt;/a&gt; of the derivative of a variable with respect to time with initial conditions is defined by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\mathcal{L}\left\{\frac{d^n y}{dt^n}\right\}(s) = s\,\mathcal{L}\left\{\frac{d^{n-1} y}{dt^{n-1}}\right\}(s) - \left.\frac{d^{n-1} y}{dt^{n-1}}\right|_{t=0^+}.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example &lt;span class=&quot;math-container&quot;&gt;$\mathcal{L}\{\tfrac{d}{dt}y\}(s)=s\,\mathcal{L}\{y\}(s)-y(0^+)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mathcal{L}\{\tfrac{d^2}{dt^2}y\}(s) = s\,\mathcal{L}\{\tfrac{d}{dt}y\}(s) - \tfrac{d}{dt}y(0^+) = s^2\,\mathcal{L}\{y\}(s) - s\,y(0^+) - \tfrac{d}{dt}y(0^+)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unforced system response of the system with initial conditions can be calculated by taking the inverse Laplace transform of the resulting expression. However it might be easier to not switch to and from the frequency domain and just solve it directly in the time domain.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2019-03-06T14:03:15.857" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26267" PostTypeId="1" CreationDate="2019-03-06T18:30:30.563" Score="0" ViewCount="686" Body="&lt;p&gt;I'm wondering if it's correct to write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \alpha = \theta_{lg} - \theta_{th} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x6GYS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x6GYS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;EDIT: the theta angles are defined positive in counter-clock direcction, instead alpha in clockwise direction. My goal is to find a relationship between the three angles I wrote above (the relationship should be valid at any time istant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="7006" LastEditorUserId="7006" LastEditDate="2019-03-06T19:35:36.187" LastActivityDate="2022-09-27T05:00:30.057" Title="Polarity of a relative angle - Knee joint angle" Tags="|applied-mechanics|mechanisms|mechanical|biomedical-engineering|biomechanics|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26269" PostTypeId="2" ParentId="26267" CreationDate="2019-03-06T20:14:58.407" Score="0" Body="&lt;p&gt;As shown in your diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\alpha = suplement(\theta_K )$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \alpha = 180- (\theta_K)  = 180-(\theta_{th}- \theta{lg} )=180 +\theta_{lg} -\theta_{th}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you need to add 180 to your answer.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-06T20:14:58.407" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26270" PostTypeId="1" AcceptedAnswerId="26271" CreationDate="2019-03-06T20:55:57.440" Score="0" ViewCount="350" Body="&lt;p&gt;I'm trying to build a cablecam, a picture that serves only for illustration is:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9NMGa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9NMGa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The DC motor is&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;48V&lt;/li&gt;&#xA;&lt;li&gt;5000W&lt;/li&gt;&#xA;&lt;li&gt;6000rpm&lt;/li&gt;&#xA;&lt;li&gt;6000 oz-in stall torque&lt;/li&gt;&#xA;&lt;li&gt;85% effiencity&lt;/li&gt;&#xA;&lt;li&gt;0.066 Ohms Resistance&lt;/li&gt;&#xA;&lt;li&gt;135Kv (RPM / Volt)&lt;/li&gt;&#xA;&lt;li&gt;Reduction ratio with timing belt is 1:2, so speed on 60mm output pulley is 3000rpm.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Battery is Li-Ion&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;48V&lt;/li&gt;&#xA;&lt;li&gt;15Ah&lt;/li&gt;&#xA;&lt;li&gt;70A constant/120A peak for 20sec.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Mechanical properties  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Max load 40 kg (including motor, wires, frame, electronics, pulleys...) &lt;/li&gt;&#xA;&lt;li&gt;Friction coefficient is 0.7&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I want to know if this 48V DC motor can even start this mass on the rope without burn itself. I would like to know if it could reach speed of 50km/h within 10 seconds or 50 meters of travel on rope.&lt;/p&gt;&#xA;" OwnerUserId="19614" LastEditorUserId="19614" LastEditDate="2019-03-09T10:10:46.843" LastActivityDate="2019-03-09T10:10:46.843" Title="Calculating proper specs of DC motor for rig driven on the rope" Tags="|motors|torque|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26271" PostTypeId="2" ParentId="26270" CreationDate="2019-03-06T21:47:53.510" Score="2" Body="&lt;p&gt;&lt;strong&gt;MOTOR REQUIREMENTS&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Start-up&lt;/strong&gt;&lt;br&gt;&#xA;The force to overcome the friction equals &lt;span class=&quot;math-container&quot;&gt;$$ F_s^{max} = \mu_s m g=0.7 \cdot 40\text{ kg} \cdot 9.8\text{ m/s}^2=274.4\text{ N}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Required torque of the output pulley to start up equals &lt;span class=&quot;math-container&quot;&gt;$$ \tau_{pulley} = F_s^{max}  \cdot \dfrac{d_{pulley}}{2} = 274.4\text{ N} \cdot 30\text{ mm} = 8.2\text{ Nm}  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reduction ratio is 1:2, assuming a efficiency of 90% means the motor should provide a torque that equals &#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \tau_{motor} = \tau_{pulley} \frac{1}{i}  \frac{1}{\eta} = 8.2\text{ Nm} \cdot \dfrac{1}{2} \cdot \dfrac{1}{0.9}= 4.6\text{ Nm}$$&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6000&amp;nbsp;oz-in stall torque (which is the torque at 0&amp;nbsp;rad/sec) equals 42.3&amp;nbsp;Nm which is more than enough to overcome the required 4.6&amp;nbsp;Nm.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Acceleration&lt;/strong&gt;&lt;br&gt;&#xA;The force to accelerate the load equals &lt;span class=&quot;math-container&quot;&gt;$$ F_{acc} = ma = 40\text{ kg} \cdot \dfrac{ 50\text{ km/h}}{10\text{ sec}} = 55.6\text{ N}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the friction still needs to be overcome, the required force equals &lt;span class=&quot;math-container&quot;&gt;$$ F_{total}=274.4\text{ N} +55.6\text{ N}= 330\text{ N}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the same equations leads to&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \tau_{motor} = 330\text{ N} \cdot 30\text{ mm}  \dfrac{1}{2} \dfrac{1}{0.9} = 5.5\text{ Nm}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In motion&lt;/strong&gt;&lt;br&gt;&#xA;When the rig is at 50&amp;nbsp;km/h, the friction force still needs to be overcome.&lt;br&gt;&#xA;The required angular velocity of the motor (using the reduction ratio) equals&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \omega_{motor} = \frac{v_{rig}}{d_{pulley}/2} \cdot i= \frac{13.9\text{ m/s}}{30 \text{ mm}} \cdot 2 = 926\text{ rad/s}=8842 \text{ rpm}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is exceeding the motor speed&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When changing the reduction/gear ratio to 1:1, the motor torque to overcome friction becomes &lt;span class=&quot;math-container&quot;&gt;$$ \tau_{motor} = 9.1\text{ Nm} $$&lt;/span&gt; and the torque to accelerate &lt;span class=&quot;math-container&quot;&gt;$$ \tau_{motor} = 11\text{ Nm} $$&lt;/span&gt;.&lt;br&gt;&#xA;The required velocity in motion becomes &lt;span class=&quot;math-container&quot;&gt;$$ \omega_{motor} = 4421 \text{ rpm}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE&lt;/strong&gt;&lt;br&gt;&#xA;I used a load of 40&amp;nbsp;kg, but such a big motor has quite some weight as well. This should be taken into account and recalculation needs to be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;br&gt;&#xA;&lt;strong&gt;GRAPH WITH MOTOR CURVES&lt;/strong&gt;&lt;br&gt;&#xA;Some motor manufacturers provide curves with speed, power and efficiency with respect to torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nuXAc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nuXAc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Mechanical working points&lt;/strong&gt;&lt;br&gt;&#xA;In this graph the blue curve is relevant for the (mechanical) equations done above.  This line represents what speed the motor can achieve for given torque at the rated voltage of 48V.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;At start up, the motor speed is 0&amp;nbsp;rpm and the required motor torque is 9.1&amp;nbsp;Nm, so we can draw a point at (9.1&amp;nbsp;Nm, 0&amp;nbsp;rpm).&lt;/li&gt;&#xA;&lt;li&gt;In motion, the motor speed is 4421&amp;nbsp;rpm and the required motor torque is 9.1&amp;nbsp;Nm, so we can draw a point at (9.1&amp;nbsp;Nm, 4421&amp;nbsp;rpm).&lt;/li&gt;&#xA;&lt;li&gt;During acceleration the required motor torque is 11&amp;nbsp;Nm and the motor speed increases from 0&amp;nbsp;rpm to 4421&amp;nbsp;rpm, so we all these points (forming a line) from (11 Nm, 0 rpm) to (11&amp;nbsp;Nm, 4421&amp;nbsp;rpm).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All points are drawn left and below the blue line, which means the motor is able to start up, accelerate and keep the rig in motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems dull to add the start up and in-motion points, as the acceleration points cover these already. However, some systems need to overcome more start-up forces and/or apply a different acceleration profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For dc motors applies:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the motor torque is proportional to the motor current&lt;/li&gt;&#xA;&lt;li&gt;the motor speed is proportional to the motor voltage&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Reducing the applied motor voltage shifts the blue line. Controlling a dc motor is &quot;shifting the blue line downwards/to the left&quot; by lowering the motor voltage till it intersects with the points drawn above. (It is possible to overdrive the motor with a higher voltage than 48V, but I wouldn't recommend it. Moreover, as your battery voltage is 48V, it would require addition means)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the battery voltage may lower during use, so, changing the reduction/gear ratio to 4:3, gives more headroom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motor current&lt;/strong&gt;&#xA;The red curve in the graph is the current for given torque.&#xA;The current ratings for the motor have been given as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;no load current = 4.5&amp;nbsp;A (at zero torque = 0&amp;nbsp;Nm)&lt;/li&gt;&#xA;&lt;li&gt;stall current = 650&amp;nbsp;A (at stall torque = 42.3&amp;nbsp;Nm) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, the relation between torque and current is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ I_{motor} = \tau_{motor} \frac{650\text{ A}-4.5\text{ A}}{42.3\text{ Nm}} + 4.5A $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Using  the reduction/gear ratio to 1:1) the required torque to accelerate eqauls 11&amp;nbsp;Nm, so the required current becomes  172&amp;nbsp;A.&lt;br&gt;&#xA;To keep the wig in motion, a torque of 11&amp;nbsp;Nm is required, requiring a current of 143&amp;nbsp;A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**This exceeds the current ratings of the motor controller (and battery) **  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the former reduction ratio to 2:1 requires 4.6&amp;nbsp;Nm to keep the wig in motion, so a contineous current of 74&amp;nbsp;A. This almost satisfies the current rating of the motor controller, but the required speed cannot be achieved as shown above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SUMMARY&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The motor is able to comply the requirements, especially using a gear ration of about 4:3.&lt;/li&gt;&#xA;&lt;li&gt;The motor controller fails to comply as it either limits the maximum torque or the maximum speed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADVISE&lt;/strong&gt;&lt;br&gt;&#xA;Note that the friction force is 5 times as big as the acceleration force. So, down-scaling the requirement for acceleration has hardly no use.&lt;br&gt;&#xA;The friction force is proportional to the payload as well as the friction coefficent. Improving the latter may yield interesting results.&lt;/p&gt;&#xA;" OwnerUserId="19613" LastEditorUserId="19613" LastEditDate="2019-03-07T23:11:46.080" LastActivityDate="2019-03-07T23:11:46.080" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="26272" PostTypeId="1" CreationDate="2019-03-07T00:21:39.117" Score="0" ViewCount="161" Body="&lt;p&gt;I am learning CFD and I have come across an example in which the thermal conductivity needs to be interpolated at a face between two rectangular cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The face weighting factor is given as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$g = \frac{V_1}{V_1+V_2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt; are cell volumes.  I expected the interpolated value, &lt;span class=&quot;math-container&quot;&gt;$k_{1-2}$&lt;/span&gt;, would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$k_{1-2} = gk_2+(1-g)k_1$$&lt;/span&gt;   (linear interpolation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but instead the author of the example gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$k_{1-2} = \frac{k_1k_2}{[1-g]k_1+gk_2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the context of finite volume method CFD, what is this method of interpolation called and why would it be preferred over linear interpolation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The example I am referring to above is Example 1, Chapter 8, of &lt;strong&gt;&lt;em&gt;The Finite Volume Method in Computational Fluid Dynamics&lt;/em&gt;&lt;/strong&gt; by Moukalled et al.&lt;/p&gt;&#xA;" OwnerUserId="19615" LastEditorUserId="1832" LastEditDate="2019-03-07T00:42:28.933" LastActivityDate="2019-03-07T00:42:28.933" Title="Interpolation at a face, Finite Volume Method in Computational Fluid Dynamics" Tags="|finite-element-method|cfd|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26275" PostTypeId="2" ParentId="270" CreationDate="2019-03-07T04:57:16.540" Score="3" Body="&lt;p&gt;Even the presence of &lt;a href=&quot;https://whyps.com/water-in-hydraulic-oil&quot; rel=&quot;nofollow noreferrer&quot;&gt;water in hydraulic oil&lt;/a&gt; will damage the system components. Then you can think what will happen when you choose water as a hydraulic fluid. A hydraulic fluid is selected depending on the application the system will perform. For systems operating under high temperature, fire resistant fluids can be used and for systems operating under low temperature petroleum based fluids can be used.&#xA;Water is considered as a wrong selection because of many reasons. Hydraulic fluid has many functions like providing lubrication and sealing, heat transfer, and power generation. But, while using water most of these functions can’t be achieved. The water will reduce the strength of lubricating film and hence internal or external leakage will occur. Also, the water molecules will result in oxidation/corrosion of metal surfaces. The low boiling point and freezing point of water is another drawback of using water as a hydraulic fluid.&lt;/p&gt;&#xA;" OwnerUserId="19619" LastActivityDate="2019-03-07T04:57:16.540" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26276" PostTypeId="1" CreationDate="2019-03-07T08:12:24.243" Score="1" ViewCount="28" Body="&lt;p&gt;For an electrochemical flow cell stack employed in electrosynthesis, I require an proton exchange membrane (PEM) in order to enable &quot;divided cell&quot; electrolysis mode. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Nafion proton exchange membrane that is currently used exhibits massive dimensional changes due to &lt;strong&gt;swelling upon contact with organic solvents&lt;/strong&gt;, for example a length increase of close to 100% in Methanol, or 50% in Acetonitrile. These solvents are crucial for many experiments and thus cannot be avoided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This dimensional change lead to failing seals between the gaskets and the membrane and thus to crossover of material between the anodic and cathodic compartments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thus looking for &lt;strong&gt;alternatives to Nafion&lt;/strong&gt;, the requirements are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-swelling in the mentioned solvents &amp;lt;20%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-high conductivity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-commercially available (or otherwise sourceable)&lt;/p&gt;&#xA;" OwnerUserId="19621" LastActivityDate="2019-03-07T08:12:24.243" Title="Ion exchange membrane that is dimensionally stable in organic solvents" Tags="|materials|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26277" PostTypeId="1" CreationDate="2019-03-07T09:41:14.857" Score="2" ViewCount="52" Body="&lt;p&gt;So I am designing a MEMS structure consisting of cantilever arrays.&#xA;For my problem, I want the cantilevers to vibrate only at their 1st mode of resonance and reject all the other modes of resonance passively. &#xA;Is there anyway I can achieve this mechanically, that is by modulating the cantilever structure? &lt;/p&gt;&#xA;" OwnerUserId="19622" LastActivityDate="2019-04-06T20:02:15.473" Title="How to allow only one mode of resonance in a cantilever beam" Tags="|mechanical-engineering|structural-analysis|vibration|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26280" PostTypeId="1" AcceptedAnswerId="26285" CreationDate="2019-03-07T16:20:45.583" Score="0" ViewCount="361" Body="&lt;p&gt;The device have to climb a pipe vertically. I have a DC motor with power rating 21.2 w, output speed 13360 rpm and maximum output torque 154.4 gcm.&#xA;My device is 1 kg and I have wheel with 0.012 m radius. I did calculations and I found that my gear box should be 1:72 but I am not sure about that. I need the device to go upwards as fast as possible. With gear ratio 1:72 is enough for the DC motor to lift the device? And what I have to do to make it go faster? Make the gear ratio smaller or bigger?&lt;/p&gt;&#xA;" OwnerUserId="19627" LastEditorUserId="8055" LastEditDate="2019-03-07T17:01:28.160" LastActivityDate="2019-03-07T21:00:19.417" Title="The torque I need to lift the device and correct gear ratio for the device to be as fast as possible" Tags="|motors|gears|torque|power|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26281" PostTypeId="2" ParentId="25630" CreationDate="2019-03-07T17:25:57.073" Score="0" Body="&lt;p&gt;At a bare minimum,I would accurately mount a &quot;fence&quot; in one of the directions, X or Y. pretty much a piece of tool steel stock, or anything precision ground say, 1/4&quot; x 1&quot; wide x however long you can fit, mounted to the working table at one of its ends. The fence should be indicated to within .001&quot; or so, alignment with whatever axis you pick. Now when you go to put parts on the work table, you can use squares and parallels, referenced off your fence which you know is whiten .001&quot; alignment of a particular axis(square to the other), to accurately align whatever feature on the part your using as the parts reference. Then your parts will be accurately located withe respect to the fence, which is accurately located with respect to the axis' of travel. This would be the most &quot;universal&quot; way to set up your machine. If your working on the same parts constantly, I would design fixtures for holding them. All depends how fast you want to work. Dowel pin the fence to the table so it can be removed and put back without indicating again.&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2019-03-07T17:25:57.073" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26283" PostTypeId="2" ParentId="26277" CreationDate="2019-03-07T19:24:07.863" Score="1" Body="&lt;p&gt;It is physically impossible to make a real-world structure with only &lt;em&gt;one&lt;/em&gt; mode of vibration, though it may be possible to design something with only one mode &lt;em&gt;in the frequency range that you are interested in.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you are designing a MEMS system, this is probably irrelevant, because you can easily filter out the response from the unwanted modes. Any MEMS system is likely to have enough computing power to do some real-time digital filtering of the signal from the MEMS transducer.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-03-07T19:24:07.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26285" PostTypeId="2" ParentId="26280" CreationDate="2019-03-07T21:00:19.417" Score="0" Body="&lt;p&gt;lifting a 1kg weight requires 9.8 WATT, we need to add to that the power needed to accelerate your mass from zero to the final speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we assume an acceleration of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a =1.8m/s   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power needed &lt;span class=&quot;math-container&quot;&gt;$ \ P= 9.8 + 1.8*1kg =11.8N.s =11.8W$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say your system efficiency is 80% then you need 11.8W* 1.25 = 14.75 W which is less than your DC motor output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you need a torque of &lt;span class=&quot;math-container&quot;&gt;$14.75* (1.2cm/2*100cm/m) = 0.0885Nm \ &amp;lt; $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;your DC motor torque of &lt;span class=&quot;math-container&quot;&gt;$0.1513Nm = 154.4*9.8/100cm/m*1000g/kg$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means no gear needed, your motor with the 1.2cm wheel directly attached to its shaft will lift the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the question of benefit of increasing gear ratio, you need to have the motor's torque and performance ratio diagrams, but adding gear will add to friction.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-07T21:00:19.417" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26286" PostTypeId="1" AcceptedAnswerId="26292" CreationDate="2019-03-07T21:32:59.593" Score="0" ViewCount="1226" Body="&lt;p&gt;My middle schooler is building a catapult that’s supposed to hurl a ping-pong ball 5 meters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t6CiT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t6CiT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the pressure on the latch is so high that the kid just cannot trigger it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4KiQH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4KiQH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The door latch used as a trigger is being under so much pressure that it just wouldn’t slide out. I doubt that adding lubricant would help enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the question: what latch design could you suggest so that it would operate smoothly under lateral pressure?&lt;/p&gt;&#xA;" OwnerUserId="3548" LastActivityDate="2019-03-08T15:03:07.160" Title="A latch that’s easy to open under lateral pressure" Tags="|pressure|mechanisms|mechanical|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26287" PostTypeId="2" ParentId="26286" CreationDate="2019-03-07T21:43:19.340" Score="0" Body="&lt;p&gt;With caution and your supervision:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He can use a mouse trap spring and release mechanism. They are efficient and easily release, but require caution handling. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-07T21:43:19.340" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26288" PostTypeId="2" ParentId="26286" CreationDate="2019-03-07T22:33:09.093" Score="1" Body="&lt;p&gt;Consider to replace the latch socket with a pin. The pin extends from the side of the moving part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pin A is held down by a smaller version of the catapult platform, with a hinge at the opposite end. It also has a pin (B) which is held down by another small hinged lever. When the device is unloaded all the hinged portions are held upright out of the way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The catapult platform is pushed into position and the first lever drops over the pin. The second lever drops over the first lever's pin. The latch holds the second lever down, with a substantial reduction of force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a derivation of the three-ring release designed, patented and produced by Bill Booth and is created in various ways for similar purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7cV63.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7cV63.png&quot; alt=&quot;force reduction image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keeping the hinges close to the pin improves the force reduction for each stage.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-03-07T22:33:09.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26292" PostTypeId="2" ParentId="26286" CreationDate="2019-03-08T05:48:01.093" Score="5" Body="&lt;p&gt;How about a levered latch like this (very rough sketch) showing the catching tooth above the self latching spring and the pivot point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ugoy6.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ugoy6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-03-08T05:48:01.093" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26293" PostTypeId="2" ParentId="26270" CreationDate="2019-03-08T09:46:59.717" Score="0" Body="&lt;p&gt;I think it goes something like this - the energy from the motor is just it's rated power &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt; time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;, which produces kinetic energy (mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;, velocity &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt;) in the system: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$P t = \frac{1}{2} m v^2$$&lt;/span&gt;  solve for &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;, knowing &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; (5000 W), &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; (40 kg) and final &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; (50 km/h) results in the time to reach 50 km/h is about  0.8 s. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a simplification so I'm disregarding&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;slipping of the drive wheel on the rope - likely to be some initial slippage with a constant power input because the initial acceleration is infinite.&lt;/li&gt;&#xA;&lt;li&gt;additional mass of the motor raising the total accelerated mass over OP's 40 kg &lt;/li&gt;&#xA;&lt;li&gt;any motor/battery/efficiency related power limitations&lt;/li&gt;&#xA;&lt;li&gt;might need a different gearing ratio over what OP has specified (3000 rpm on 60 mm diameter pulley is 33 km/h)&lt;/li&gt;&#xA;&lt;li&gt;friction in pulleys, inertia of pulleys&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The frictional between the drive pulley and the rope isn't just related to the mass of the cart, it's also related to the tension in the rope. &lt;/p&gt;&#xA;" OwnerUserId="16544" LastActivityDate="2019-03-08T09:46:59.717" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26295" PostTypeId="1" CreationDate="2019-03-08T11:12:48.173" Score="1" ViewCount="329" Body="&lt;p&gt;I want to use steel (or other material) rods to support weight over a 900mm cavity. The weight that needs to be supported is technically unknown, but I'd specify some rough guidelines as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;100KG (max) - some give, but no permanent deformity&lt;/li&gt;&#xA;&lt;li&gt;&amp;lt;50KG - no real visible bend&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm not hugely attached to the weights listed above as they could be dropped with max weight advisories etc but I'd say those are probably ideal. Exceeding them would be fantastic, reducing them isn't a huge issue. Ideally I'd be able to calculate myself and adjust the design to use a bigger rod if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rod would be used to back a slat of wood which presumably would add to the strength of the unit, and there would be multiple slats (laid horizontally) covering an overall length of about 1.6m. I'd probably add an overlap between the slats also which I would assume would add strength to the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to find my feet with CAD design and engineering principals so I'm very much new to the (seemingly) advanced maths I've seen glimpses of for this sort of calculation, so a resource I can read through or a breakdown of a calculation would be much appreciated!&lt;/p&gt;&#xA;" OwnerUserId="19638" LastEditorDisplayName="user16" LastEditDate="2019-03-08T12:56:48.893" LastActivityDate="2022-02-03T23:05:19.127" Title="Calculating diameter of steel rod required to not bend over a given length" Tags="|structures|metals|wood|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26296" PostTypeId="1" CreationDate="2019-03-08T12:27:00.937" Score="0" ViewCount="45" Body="&lt;p&gt;Is it possible to thermally decompose an indicator like wood in a induction heating container made of stainless steel if the induction frequency is low enough to heat the container about 1000°c.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the radiant heat from the steel cause the wood to heat to a point where they could break down and combust? &lt;/p&gt;&#xA;" OwnerUserId="19640" LastActivityDate="2019-03-08T12:27:00.937" Title="Can radiant induction heat thermally decompose wood" Tags="|heat-transfer|heating-systems|thermal-conduction|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26297" PostTypeId="2" ParentId="26286" CreationDate="2019-03-08T15:03:07.160" Score="2" Body="&lt;p&gt;My son made a model trebuchet which is a sort of medieval catapult. The trigger mechanism is quite simple using three screw eyes and a pin. In case the picture below isn't clear, the there are two screw eyes attached to the frame and a single screw eye attached to the arm with a pin through all three screw eyes. Attaching a string to the pin keeps hands away from things when launching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/djuQp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/djuQp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9506" LastActivityDate="2019-03-08T15:03:07.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26298" PostTypeId="1" CreationDate="2019-03-08T15:32:01.433" Score="0" ViewCount="43" Body="&lt;p&gt;I am working on a 1.5m satellite dish with a custom feed. The problem is that the feed is being very wobbly once slightly pushed, as shown here: &lt;a href=&quot;https://www.youtube.com/watch?v=rAFGbEhhDZ8&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=rAFGbEhhDZ8&lt;/a&gt; (it swings back and forth, which would not be a good thing on a windy day).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would adding a second thick plate to the back of the feed (where it's held on by the threaded rod (24M)) make the feed any stiffer and reduce/stop wobbliness? I had this problem with my previous thin rod (see Stopping/reducing wobbliness in threaded rod), but I fixed it by increasing the diameter of the rod. What could I do with the feed?&lt;/p&gt;&#xA;" OwnerUserId="12819" LastActivityDate="2019-03-08T15:32:01.433" Title="Stopping metallic construction held by a threaded rod from wobbling" Tags="|mechanical-engineering|applied-mechanics|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26303" PostTypeId="1" AcceptedAnswerId="26306" CreationDate="2019-03-08T19:58:13.017" Score="2" ViewCount="116" Body="&lt;p&gt;I have been trying to calculate the analytical solution for the following problem but have not succeeded to reach an exact solution. Can anyone propose a way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A concrete beam is to be lifted by two cranes. The beam is 20&amp;nbsp;m long and weights 900&amp;nbsp;kN. The cranes have lifting capacities of 500&amp;nbsp;kN and 400&amp;nbsp;kN. What are distances &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; in the scheme below so the cranes are not overloaded? (position of cranes are marked with | |)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;            &amp;lt;---- a----&amp;gt; &amp;lt;----- b -------&amp;gt; &amp;lt;--- c ---&amp;gt;            &#xA;            ===========================================  &amp;lt;-- the beam&#xA;                        ||                ||    &#xA;                     A = 500 kN        B = 400 kN  &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A &amp;amp; B: reaction forces of cranes&lt;/li&gt;&#xA;&lt;li&gt;distributed self weight q = 900 kN / 20 m = 45 kN/m&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a + b + c = L = 20\text{ m}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The beam is statically determinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My attempted solutions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started from the equilibrium equations of the sum of bending moments about the left and right ends which are equal to zero:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum M_{left} &amp;amp;= 0 = Aa + B(a + b) = \dfrac{qL^2}{2} \\ &#xA;\sum M_{right} &amp;amp;= 0 = Bc + A(b + c) = \dfrac{qL^2}{2}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A third equation is needed; I have tried &lt;span class=&quot;math-container&quot;&gt;$a + b + c = L$&lt;/span&gt;, but I arrive at the &quot;apple = apple&quot; situation, which implies that the three equations are not independent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried to add an additional constrained by putting any of the &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; values equal to a constant value. The issue then is that each of the equilibrium equations become dependent on only one of the two remaining unknowns, as in the following case, where I have put &lt;span class=&quot;math-container&quot;&gt;$b = L/3$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum M_{left} &amp;amp;= 0 = \dfrac{qL^2}{2} = Aa + B(a + L/3) \\&#xA;\sum M_{right} &amp;amp;= 0 = \dfrac{qL^2}{2} = Bc + A(c + L/3)&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And thus, each solved unknown will not satisfy the other equilibrium equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have run out of tricks in my sleeve ...&lt;/p&gt;&#xA;" OwnerUserId="19642" LastEditorUserId="1832" LastEditDate="2019-03-09T01:55:03.333" LastActivityDate="2019-03-09T02:03:06.820" Title="Exact anlytical solution for beam-lifting problem constrained by crane capacity" Tags="|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26304" PostTypeId="2" ParentId="26303" CreationDate="2019-03-08T20:26:57.970" Score="2" Body="&lt;p&gt;There is not a unique solution to the problem, which explains why you can't find one!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see why, think about a different problem: two people who weigh 50kg and 40kg want to balance each other on a see-saw. How far should they sit from the middle of the see-saw, to make it balance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be obvious that the &lt;em&gt;ratio&lt;/em&gt; of their distances from the center is the only thing that  matters.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-03-08T20:26:57.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26306" PostTypeId="2" ParentId="26303" CreationDate="2019-03-08T23:07:20.697" Score="2" Body="&lt;p&gt;Let's call the distance from the anchor to 500&amp;nbsp;kN crane from the middle of the beam &lt;span class=&quot;math-container&quot;&gt;$D_1$&lt;/span&gt; and the distance to 400&amp;nbsp;kN, &lt;span class=&quot;math-container&quot;&gt;$D_2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do exact analysis we get the moment of cranes' hook forces about the midpoint of the beam at equilibrium configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_1 D_1= P_2 D_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;denoting &lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$P_2$&lt;/span&gt; as the tributary weight of the beam on 500&amp;nbsp;kN crane and 400&amp;nbsp;kN, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting &lt;span class=&quot;math-container&quot;&gt;$P_2$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$4/5 P_1$&lt;/span&gt; in above equation we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P_1 D_1 = 4/5 P_1 D_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing both sides by &lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$D_1= \dfrac{4}{5}D_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the safest ways to lift the beam is to attach the 400&amp;nbsp;kN crane to one end of concrete beam and 500&amp;nbsp;kN to a point at the other end 90% the length of the beam.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2019-03-09T02:03:06.820" LastActivityDate="2019-03-09T02:03:06.820" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26307" PostTypeId="1" CreationDate="2019-03-09T00:43:59.627" Score="0" ViewCount="40" Body="&lt;p&gt;I have a rod mounted inside of a 2x4 that is allowed to freely spin inside of its hole. Attached to that rod's head is a range finder. I want the rod to reliably point down at the ground at all times to accurately read the 2x4's distance from the ground. The rod cannot be fixed in place. The 2x4 will be hoisted into the air and lowered by hand to about 8 ft max. My first idea was to weight the range finder so that it spins around and points down, however playing around with it, it's not as consistent as I would like. The 2x4 allows for free spin but isn't frictionless and I have to tap it a bit by hand to get it to fully position itself downward. I thought about adding a vibrator or something to the rod. Are there any other suggestions I may not be thinking of?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8RWx9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8RWx9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19644" LastActivityDate="2019-03-09T18:20:15.053" Title="How can I keep a free spinning range finder pointing down?" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26308" PostTypeId="2" ParentId="26307" CreationDate="2019-03-09T00:54:14.700" Score="2" Body="&lt;p&gt;You mean it is NOT frictionless? You should try to get it frictionless and then use a weight. Probably one on each side of the wood to reduce friction from tilting inside the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way would be to add some ball bearing between the wood and the rod, you could ask at the local bike store or get some hightech thingy. Also depends how long you are going to use it, you will need to oil them or get a version that works without oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to go with just the wood and rod, maybe you can polish both and use some lubricant, but that would need constant maintenance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you need the wood at all? you could just use a thread and a wellbalanced weight at the end. Or attach a thread to the end of the rod?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to edit confusing wording, english is not my native language.&lt;/p&gt;&#xA;" OwnerUserId="19645" LastActivityDate="2019-03-09T00:54:14.700" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26309" PostTypeId="1" CreationDate="2019-03-09T01:29:01.293" Score="0" ViewCount="269" Body="&lt;p&gt;Can a telsa powerwall 2 be rate limited (discharge/charge)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have access to realtime electricity rates (30 min update interval) and also access to 12 hour predicted look ahead. So I wish to limit the charging (if the buyback rate is high) and limit the discharging (if the buyback rate is low).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can see that I can basically just use an on/off controller by literally shutting the powerwall off entirely (the api supports this). But it would be nicer to rate limit it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any public (or private) api that could allow me to achieve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: I found this API however it doesn't appear to have figured out any rate limiting. &lt;a href=&quot;https://github.com/vloschiavo/powerwall2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tesla Powerwall 2 - Local Gateway API documentation&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19646" LastEditorUserId="110" LastEditDate="2019-03-10T14:31:22.557" LastActivityDate="2019-03-10T14:31:22.557" Title="Can a Tesla Powerwall 2 be rate limited?" Tags="|electrical-engineering|battery|renewable-energy|solar-energy|energy-storage|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26310" PostTypeId="1" AcceptedAnswerId="26311" CreationDate="2019-03-09T01:52:41.777" Score="0" ViewCount="29" Body="&lt;p&gt;I'm pressing glass and I want to know if the pressure that the walls of my mold feel is the same through all the internal surfaces. I'm willing to assume that glass will behave purely as a fluid (very viscous) in the temperature range I'm working it.&lt;/p&gt;&#xA;" OwnerUserId="18457" LastActivityDate="2019-03-09T02:38:38.273" Title="Will the pressure inside a pressing mold will be constant if I'm working with a fluid inside?" Tags="|fluid-mechanics|pressure|glass|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26311" PostTypeId="2" ParentId="26310" CreationDate="2019-03-09T02:38:38.273" Score="0" Body="&lt;p&gt;Yes, in an enclosed container, with fluid at rest, the pressure is equal on all the walls and at any point inside the container. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluid at rest indicates the Kinetic energy of the fluid,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \ E =\frac{1}{2}\dot{m}v^2 $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;is insignificant, there is no rapid flow of matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is Pascal’s law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This law is used in hydraulic presses as well.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-09T02:38:38.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26313" PostTypeId="2" ParentId="26307" CreationDate="2019-03-09T11:54:27.153" Score="0" Body="&lt;p&gt;A great solution is to increase the moment of inertia of the rod by attaching another rod pointing downward and perpendicular to the original. At the end of that rod attach a mass with considerable weight so it will require a lot of torque to spin the original rod off your desired position, and every time it spins by maybe human contact or strong wind, it quickly find it way back by oscillating around the position.&lt;/p&gt;&#xA;" OwnerUserId="19370" LastEditorUserId="19370" LastEditDate="2019-03-09T18:20:15.053" LastActivityDate="2019-03-09T18:20:15.053" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26314" PostTypeId="1" AcceptedAnswerId="26316" CreationDate="2019-03-09T13:25:54.950" Score="0" ViewCount="213" Body="&lt;p&gt;The 40 kg weight accelerates and runs on a rope from 0 to 50 km/h in 10 seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the minimum required rolling friction that I need (acceleration without slipping) to achieve acceleration in a time of 10 seconds from 0-50km / h. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friction adjustment (later for different speeds and accelerations) I would change by increasing /decreasing the angle between the drive pulley and the pulley 2 as shown in the figure. Currently, the angle is 15 '.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&#xA;How to calculate the minimum required rolling friction (or better say, minimum required angle which is 15' now) between the drive pulley and the rope of 40kg of weight moving along the rope, as shown in the drawing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XC7fz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XC7fz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All pulleys are covered with rubber, and the rope is synthetic, Dyneema. What aditional information do I have to apply for the correct calculation?&lt;/p&gt;&#xA;" OwnerUserId="19614" LastEditorUserId="19614" LastEditDate="2019-03-10T10:06:27.397" LastActivityDate="2019-03-10T10:06:27.397" Title="Minimum required rolling friction between pulley and the rope" Tags="|friction|pulleys|acceleration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26316" PostTypeId="2" ParentId="26314" CreationDate="2019-03-09T16:09:50.137" Score="0" Body="&lt;p&gt;let's find out acceleration first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v= at = 50000M/h * 1h/3600s *1000M/kM =13.888M/s \\ a= \frac{v}{t} =13.88/10 =1.388 m/s^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the tributary weight on the two pullies is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;weight on the right pully is &lt;span class=&quot;math-container&quot;&gt;$$40 *(1147.5-154.6)/1147.5= 34.6kg \ and \ on \ the \ left =40-34.6 = 5.4 kg $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the angle of rope to left is &lt;span class=&quot;math-container&quot;&gt;$15 * 154.6/1147.5 = 2.02$&lt;/span&gt; degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;say negligible, we assume the entire friction from the right wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friction force is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F_f = \mu*40*9.8/ sin(15) \quad = 40*1.388$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-03-09T16:17:10.340" LastActivityDate="2019-03-09T16:17:10.340" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26318" PostTypeId="1" CreationDate="2019-03-10T04:15:05.303" Score="1" ViewCount="16" Body="&lt;p&gt;I know that channel bandwidth is the range of frequencies it can transmit . Let's say channel bandwidth is 100 Hz . Does that mean it can transmit any signal whose bandwidth is 100Hz regardless of actual values of those frequencies .&#xA;For example signal 1 has  frequencies 200Hz- 300Hz .&#xA;Signal 2 has frequencies 800Hz-900Hz.&#xA;Both signals have bandwidth 100Hz&#xA;Can the channel transmit both of these signals?&lt;/p&gt;&#xA;" OwnerUserId="19655" LastEditorUserId="19655" LastEditDate="2023-08-28T04:05:55.087" LastActivityDate="2023-08-28T04:05:55.087" Title="Can channel bandwidth transmit any VALUES of frequencies as long as they are within range" Tags="|communication|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26320" PostTypeId="2" ParentId="23614" CreationDate="2019-03-10T18:07:40.177" Score="1" Body="&lt;p&gt;I just bought the same tensioner. It took 20 mins (well, maybe longer!) to sus out but I reached the same conclusion.  The back-tensioner is just an adjustable overtension alarm where the microswitch  wires in to the coilwinder to stop it if the tension gets too high.&#xA;The wire feeds in from below, normally off the wire drum standing on its end and through an upturned funnel guiding the wire. The wire enters between a couple of felt pads to stabilise it from its rotation off the wire drum,  then around the magnetic tension spool to a free-wheeling pulley and thence over the alarm arm pulley to the coilwinder.&#xA;The Magnetic tensioner works a bit like a car speedo, where a rotating magnetic drags a cup around against a spring the faster it goes. I guess by using a magnetic feild to add drag to the tensioning pulley, set by the big knob on the other side,  it is a more consistant tension compared to using a slipping clutch based on friction materials. &lt;/p&gt;&#xA;" OwnerUserId="19658" LastActivityDate="2019-03-10T18:07:40.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26321" PostTypeId="1" AcceptedAnswerId="26324" CreationDate="2019-03-10T19:32:30.883" Score="1" ViewCount="1849" Body="&lt;p&gt;This problem is about bending moment of a simple beam subject to a mixture of 4 or 5 UDLs and point loads. The beam is assumed to be rigid enough that deflection of the combined load is &quot;small&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this analytically for all load combinations simultaneously, in order to graph the result, is a tricky formula. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering if, in this kind of simple case without significant distortion, one can simply use standard formulae for bending moments for single UDLs and point loads on a simple beam, and sum the results, which would be a much easier approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, it should work. But is this actually a correct intuition? Can bending moments be summed in this way for multiple loads acting on a beam, to generate a graph of total bending moment acting at different positions on the beam?&lt;/p&gt;&#xA;" OwnerUserId="6366" LastActivityDate="2019-03-10T21:38:48.150" Title="Can bending moments for a combination of loads on a simple beam, be calculated by adding the individual B.M. of each load?" Tags="|beam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26324" PostTypeId="2" ParentId="26321" CreationDate="2019-03-10T21:38:48.150" Score="3" Body="&lt;h1&gt;YES&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This is known as the &lt;a href=&quot;https://en.wikipedia.org/wiki/Superposition_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;superposition principle&lt;/a&gt;. For beams composed of linear-elastic materials where all loads are constant and independent from changes in beam geometry (i.e. not rain or snow loads, which increase as the beam deflects, creating an ever-deeper &quot;pool&quot;), the loads can all be considered independently. This means you can model each load alone, obtain the relevant results and then add them all up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just note that determining which results are &quot;relevant&quot; is not necessarily easy to do without modeling everything simultaneously. For example, a beam with complicated loading won't &lt;strong&gt;&lt;em&gt;necessarily&lt;/em&gt;&lt;/strong&gt; have maximum bending moment at midspan (unless all the loads are maximized at midspan, of course). In these cases, it may be hard to determine &lt;em&gt;a priori&lt;/em&gt; which points to sample the bending moment, so finding the maximum bending moment may require solving all the loads simultaneously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cTgCR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cTgCR.png&quot; alt=&quot;simply-supported beam with asymmetric loading and resulting bending moment diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This beam has a maximum bending moment 4.10&amp;nbsp;m from the left support, which couldn't be determined &lt;em&gt;a priori&lt;/em&gt; since the loading can be split into three &quot;sub-loads&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;UDL of 10&amp;nbsp;kN/m along the entire span, which we trivially know will cause maximum bending at midpsan;&lt;/li&gt;&#xA;&lt;li&gt;UDL of 10&amp;nbsp;kN/m on the middle 2&amp;nbsp;m, which we trivially know will cause maximum bending at midpsan;&lt;/li&gt;&#xA;&lt;li&gt;UDL of 30&amp;nbsp;kN/m applied 2nbsp;m to 4&amp;nbsp;m from the left support. This one is harder to determine, but has maximum bending 3.40&amp;nbsp;m from the left support.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Given this information, you couldn't possibly know that the most important point along the span is actually at 4.10&amp;nbsp;m. The only way to know this is by considering all the loads simultaneously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the loading is more trivial or if you already know which points to sample, then the superposition principle is an excellent friend to have.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-03-10T21:38:48.150" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26326" PostTypeId="1" CreationDate="2019-03-11T02:51:19.223" Score="0" ViewCount="86" Body="&lt;p&gt;I need to cool 3 galons of water from 77 to 32°F to form an ice bank that I will use to cool beer (it is a noble effort, you see).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figured I need to install a cooling capacity of 0.34 kw to lower the temperature and form the ice in four hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I should use R134a as my refrigerant, and think I need a 1/2 hp compressor (0.34 kw plus 20%).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use a single 42 feet,  5/16&quot; copper pipe as my evaporator (26°F), and a similarly sized condenser (86°F) hopefully without forced air (ambiente temperature is 68°F), but are most uncertain about these calcs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guidance would be very much appreciated...&lt;/p&gt;&#xA;" OwnerUserId="19664" LastActivityDate="2019-03-11T02:51:19.223" Title="Dimension of an evaporator coil" Tags="|refrigeration|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26328" PostTypeId="1" CreationDate="2019-03-11T08:37:48.187" Score="1" ViewCount="283" Body="&lt;p&gt;I'm trying to get informative resource material through this.&lt;/p&gt;&#xA;" OwnerUserId="19619" LastActivityDate="2019-03-12T10:29:28.607" Title="What are the difference between open loop and closed hydraulic system?" Tags="|mechanical-engineering|fluid-mechanics|hydraulics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26329" PostTypeId="2" ParentId="26328" CreationDate="2019-03-11T09:09:01.073" Score="1" Body="&lt;p&gt;One discards the depressurized hydraulic fluid, the other recirculates it back to the pump. That's all there is to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the Falcon 9 rocket first stage uses open loop, using pressurized nitrogen to pressurize the hydraulic fluid to control the grid fins, then directs the depressurized fluid that leaves the actuators into the fuel tank and uses it as fuel during landing. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-03-11T09:09:01.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26331" PostTypeId="1" CreationDate="2019-03-11T12:23:27.257" Score="0" ViewCount="75" Body="&lt;p&gt;Hello all I need to glue some ABS plastic parts together some of them are transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used AA glue before but it will make my transparent parts foggy and after sticking them together its nearly impossible to disassembly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there a better option than the AA 502 glue for easier removal and not fog my transparent parts?&#xA;Thank you &lt;/p&gt;&#xA;" OwnerUserId="18994" LastActivityDate="2019-03-12T22:38:13.853" Title="Which glue is good for transparent ABS plastic and easy removal?" Tags="|adhesive|abs|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26332" PostTypeId="1" CreationDate="2019-03-11T15:40:27.767" Score="3" ViewCount="715" Body="&lt;p&gt;I just modded my &lt;strong&gt;Basetech BT-305&lt;/strong&gt; laboratory power supply to become an IoT device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything works as planned, except the volt display in the multimeter-part of the supply, which is directly shown below. I disassembled the display part and saw a missing resistor, but am unable to find out how many ohms it had, as I am not an electrical engineer, so I hoped you guys could help me.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the problematic part:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Yk9vd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yk9vd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.produktinfo.conrad.com/datenblaetter/500000-524999/513812-sp-01-en-BASETECH_BT_305_LABOR_NETZTEIL.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;I also found the circuit diagram PDF on the page I bought this device&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z6RH9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z6RH9.jpg&quot; alt=&quot;Basetech BT-305&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bought the above from &lt;a href=&quot;https://www.conrad.de/de/basetech-bt-305-labornetzgeraet-einstellbar-0-30-vdc-0-5-a-150-w-anzahl-ausgaenge-1-x-513812.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Basetech BT-305 Labornetzgerät, einstellbar 0 - 30 V/DC 0 - 5 A 150 W Anzahl Ausgänge 1 x&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case anyone has the same device and has a picture of this part, that would solve the problem of searching for the right resistor immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A high-enough resistor that would fit in there is 0.5 kilo Ohms. This was measured by using a very high potentiometer and slowly decreasing the resistance until it worked as it should. Thanks to William for the idea in the comments below.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lSjPu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lSjPu.jpg&quot; alt=&quot;Picture of the part that was missing:&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As there seems to be interest in how i modded this device for iot in the comments. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;the goal is to switch the output on/off over the internet (app/http requests)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short: disassembled the device, added a second AC power plug (female) into the bt-305, so that a 5v wallcharger could be powered in the case. The 5v charger is used to power a &quot;Sonoff SV&quot;  which is an esp-based smart switch from itead. The sonoff sv supports a so called &quot;isolated mode&quot;, which means it input is 5v, and the 2 output pins just short each other when getting the command to switch, acting like a normal switch on 2 cables. Those output pins are soldered to the pins of the &quot;output&quot; button of bt-305, so sonoff is just simulating a button press. Sonoff has an app to communicate with the switch, or you can connect it to IFTTT and send http requests to a rest-url via web-hooks to set the state of the switch. I generally use the app, as it supports &quot;inching mode&quot; which means the switch-output gets shorted for just 0.5 seconds, which is necessary to simulate a button press. &lt;/p&gt;&#xA;" OwnerUserId="19671" LastEditorUserId="19671" LastEditDate="2019-06-08T19:46:49.237" LastActivityDate="2024-02-13T06:06:35.830" Title="Basetech BT-305 missing resistor" Tags="|electrical-engineering|power|" AnswerCount="1" CommentCount="8" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26333" PostTypeId="2" ParentId="26331" CreationDate="2019-03-11T15:48:05.597" Score="1" Body="&lt;p&gt;You can try ARclear optically transparent adhesives made by Adhesives Research Inc No guarantees though, you will have to experiment to get what you want, of course. :-)&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2019-03-11T15:48:05.597" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26336" PostTypeId="1" CreationDate="2019-03-11T18:51:08.137" Score="0" ViewCount="35" Body="&lt;p&gt;I am trying to figure out the actual (Range of ) communication protocols used to communicate with Fuel Dispensers at fuel retail outlets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see them write 2-wire protocol which is not specific as there are a bunch of communication protocols that communicate over 2 wires. I am suspecting it to be CAN but not sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be glad if anyone can assist with even further information on communication with fuel station devices&lt;/p&gt;&#xA;" OwnerUserId="19131" LastActivityDate="2019-03-11T18:51:08.137" Title="Name of protocol used to communicate with Fuelling Station Dispensors" Tags="|electrical-engineering|control-engineering|automotive-engineering|microelectronics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26337" PostTypeId="1" AcceptedAnswerId="26352" CreationDate="2019-03-11T21:42:47.480" Score="2" ViewCount="119" Body="&lt;p&gt;I'm searching for an approach on how to calculate the adiabatic temperature loss of air when increasing its humidity by a humidifier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumed we have the following setup:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Air measuring position 1 (before humidifier):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T = 30{ °C}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x = 10\text{ g/kg (absolute humidity)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Air measuring position 2 (after humidifier):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T = \text{? °C}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x = 20\text{ g/kg (absolute humidity)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The question now is:&lt;/strong&gt; How can I calculate the air temperature at measuring position 2? With a &lt;code&gt;Mollier&lt;/code&gt;-diagram it would be easy to figure out, but I want to solve it the analytical way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess there is something possible with the enthalpy of air:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$h_1=h_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Q_1\cdot x_1\cdot T_1=Q_2\cdot x_2\cdot T_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{Q_1\cdot x_1\cdot T_1}{Q_2\cdot x_2}=T_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is missing? :-)&lt;/p&gt;&#xA;" OwnerUserId="19676" LastEditorUserId="19676" LastEditDate="2019-03-11T21:50:22.713" LastActivityDate="2019-03-12T22:53:36.697" Title="How to calculate adiabatic air temperature loss caused by increase of humidity?" Tags="|thermodynamics|cooling|temperature|vapor-pressure|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26338" PostTypeId="2" ParentId="11525" CreationDate="2019-03-11T22:09:00.060" Score="1" Body="&lt;p&gt;They don't give a complete description, hence why there's no accurate conversion between the various hardness scales.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hardness and toughness&lt;/strong&gt; have arbitrary and specific test conditions because they represent a combination of more fundamental properties that we either can't measure or perhaps don't have a model of. They're useful for specifying materials that have consistent performance to each other in typical applications, not for putting into any theoretical model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Young's modulus and Poisson's ratio&lt;/strong&gt; are less empirical because we have a model that can use them to make predictions (linear elasticity). That gives a complete description of how a material will behave as long as it stays linearly elastic. In that regime, there's no need for hardness or toughness because those describe non-elastic behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you're willing to neglect time dependence, and I guess any kind of history dependence, then I think you're limiting yourself to simple linear elasticity which is well defined with its 21 constants. You don't need to consider hardness and toughness because they describe permanent deformation as well as load rate dependence. Large elastic strains (like in elastomers) can also have history dependence if stretching it makes it temporarily stiffer, for example. I'm not sure if there's room for a nonlinear elasticity model that excludes any time dependence. But if there is, then I guess that's what you're looking for.&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="4437" LastEditDate="2019-03-11T22:14:27.387" LastActivityDate="2019-03-11T22:14:27.387" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26339" PostTypeId="2" ParentId="26253" CreationDate="2019-03-12T03:01:26.170" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\newcommand{\a}[1]{\langle#1\rangle}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I've attempted removing the left side guide joint so then there are only 3 reactions and it should be statically determined.&lt;/p&gt;&#xA;&lt;p&gt;However, I don't know how to reapply the loading conditions for the removed guide as it is axial not transverse to which the boundary condition applies. (X direction in Cartesian coordinates).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I think this is what you are suggesting. By removing the left guide, you get the beam shown in the left-half of the following figure. If you know the correct formula, you can calculate the slope of the beam at end A. Then what you need to do is add a moment to the ends of the beam (like in the right-half of the following figure) and calculate what moment &lt;span class=&quot;math-container&quot;&gt;$M_A$&lt;/span&gt; is required to bring the slope of point A to zero. The sum of those two are your beam. (Note that you can visualize your original diagram as a half symmetry model about the left guide. My figure below shows a &amp;quot;whole&amp;quot; beam because finding the beam formulas for such an arrangement may be easier. Half of this beam is identical to your original beam.) That is a clever approach if you have the two formulas for the slope at end A!&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vzo0Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vzo0Y.png&quot; alt=&quot;superposition of loads&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;I do not have the two formulas for those arrangements. Instead of deriving two equations from the two diagrams, I find it easier to derive one equation from the diagram that includes all of the loads. I will use singularity functions which makes it easy to write one equation for the entire half-beam. (A singularity function such as &lt;span class=&quot;math-container&quot;&gt;$\a{x-a}^n$&lt;/span&gt; is zero when x is less than a, and &lt;span class=&quot;math-container&quot;&gt;$(x-a)^n$&lt;/span&gt; for x greater than a.) The load diagram for your beam with all of the symbols is as follows:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tcjYQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tcjYQ.png&quot; alt=&quot;beam diagram&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;From &lt;span class=&quot;math-container&quot;&gt;$\sum F_{vertical} = 0$&lt;/span&gt;, we know that the reaction at the pin constraint is &lt;span class=&quot;math-container&quot;&gt;$R_B=P$&lt;/span&gt;. The shear &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, moment &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;, and slope &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; equations as a function of position (x, measured from end A) can be written as follows:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;V(x) &amp;amp;= P\a{x-a}^0-P\a{x-L}^0 \tag 1 \\&#xA;M(x) &amp;amp;= -M_A\a{x}^0+P\a{x-a}^1-M_C\a{x-L}^0-P\a{x-L}^1 \tag 2 \\&#xA;EI \cdot \theta (x) &amp;amp;= -M_A\a{x}^1+ \frac12P\a{x-a}^2-M_C\a{x-L}^1-\frac12P\a{x-L}^2+c_1\tag 3 &#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where L is the total length (L=a+b). For example, equation 2 indicates that the moment is equal to &lt;span class=&quot;math-container&quot;&gt;$-M_A$&lt;/span&gt; between x=0 and x=a, then it begins to increase by &lt;span class=&quot;math-container&quot;&gt;$P(x-a)$&lt;/span&gt; when x&amp;gt;a, and decreases by &lt;span class=&quot;math-container&quot;&gt;$M_C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$P(x-L)$&lt;/span&gt; when x&amp;gt;L. If you were to plot this, you would see the expected moment diagram for the beam. The constant of integration &lt;span class=&quot;math-container&quot;&gt;$c_1$&lt;/span&gt; in equation 3 can be found from knowing that the slope is 0 at x=0. From equation 3 (and remembering that &amp;lt;0-a&amp;gt; is 0),&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;EI \cdot 0 &amp;amp;= -M_A \cdot 0+\frac12*P(0)^2-M_C \cdot 0 -\frac12P(0)^2+c_1 \tag 4\\&#xA;\therefore c_1 &amp;amp;=0 \tag 5 \\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;From knowing that the slope is 0 at x=L, the moment can be found as follows using equation 3 again:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;EI \cdot 0 &amp;amp;= -M_A\a{L}^1+ \frac12P\a{L-a}^2-M_C\a{L-L}^1-\frac12P\a{L-L}^2 \tag 6 \\&#xA;0 &amp;amp;= -M_A \cdot L+ \frac12P(b)^2-M_C(0)-\frac12P(0)^2 \tag 7 \\&#xA;\therefore M_A &amp;amp;= \frac12 \frac PL b^2 \tag 8 \\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The moment at end C can be found by writing the sum of the moments about point B, as follows:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum M_{about\; B} &amp;amp;= M_A - P\cdot b + M_C=0 \tag 9 \\&#xA;\therefore M_C &amp;amp;= P\cdot b -M_A \tag {10} \\&#xA; &amp;amp;= P\cdot b - \frac12 \frac PL b^2 \tag {11} \\&#xA; &amp;amp;= P \cdot b\left(1-\frac{b}{2L}\right) \tag {12} \\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="12984" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-03-12T16:46:20.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26342" PostTypeId="2" ParentId="26328" CreationDate="2019-03-12T10:29:28.607" Score="1" Body="&lt;p&gt;In closed loop hydraulic system motor return is connected directly to the hydraulic pump inlet and in open loop  the hydraulic pump inlet and the motor are connected to the reservoir. If you are interested to find more details on this, check: &lt;a href=&quot;https://whyps.com/types-of-hydraulic-system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Types of hydraulic system&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="19682" LastActivityDate="2019-03-12T10:29:28.607" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26343" PostTypeId="1" CreationDate="2019-03-12T10:50:12.153" Score="0" ViewCount="47" Body="&lt;p&gt;I have a set of little flywheels that have a magnet to break the motion of the system. I have one aluminum and several copper ones all water jet cut roughly to size. The copper and aluminum disks I have are all the same dimensions but bought from three different sources.&#xA;Yesterday I machined the Aluminum disc and three of the copper discs to the correct specs on the lathe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I checked the amount of drag caused by these discs some time ago and found the copper was far more effective than the aluminum. Today I needed to check the operating distance of the magnet so repeated this test on the newly machined parts and was shocked to find the three copper discs provided so little drag it bearly registered. The aluminum and the other copper ones worked very well. The drag the magnet courses on the other copper wheels is considerable so the difference is huge, one copper disc brings the whole system to a halt immediately but with the other discs I machined yesterday I can barely feel any drag at all, at first I was unsure there was any until I spun it up faster then I could just about feel some slight resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the strength of the eddy currents produced was dictated by the conductivity of the metal used so I am now confused as to what happened here. As I said before the copper discs are from three different sources so maybe different coppers. Does anyone know what has gone wrong and if it's the copper which coppers are best for use in this job?&lt;/p&gt;&#xA;" OwnerUserId="8597" LastActivityDate="2019-03-12T10:50:12.153" Title="Missing eddy currents in copper?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26344" PostTypeId="1" CreationDate="2019-03-12T10:52:43.840" Score="2" ViewCount="2910" Body="&lt;p&gt;How do I determine the internal normal force, shear force and moment at the point C?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IvrQ8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IvrQ8.jpg&quot; alt=&quot;Figure1&quot;&gt;&lt;/a&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$V_c = 0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$N_c = 0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$M_c=3.75 kN/m$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have already done this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dxDCC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dxDCC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4DtcR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4DtcR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where is the problem?&lt;/p&gt;&#xA;" OwnerUserId="19683" LastActivityDate="2019-04-11T20:01:30.253" Title="The internal normal force, shear force and moment" Tags="|mechanical-engineering|structural-engineering|statics|moments|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26345" PostTypeId="1" CreationDate="2019-03-12T14:44:15.253" Score="2" ViewCount="108" Body="&lt;p&gt;Bulk material is conveyed over a slat conveyor, and the material is leveled with overhead rollers while being transported in the conveyor to maintain an even material height. Most of these &quot;levelers&quot; operate at a surface speed that is much faster than the speed of the conveyor. However, one particular overhead roll serves to compress the material slightly and feed a machine downstream, and it operates with a surface speed close to the conveyor speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This particular roller operates at a constant speed, but the conveyor does not. As such, there are times when the roller has a faster surface speed than the conveyor, but other times when the roller operates more slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to calculate the effect of these external devices on the force or energy required to operate the conveyor? I am particularly worried about this last roller, since it seems to cause the most resistance to motion. I have been unable to find even empirical information on the effect of these rollers in standard chain pull calculation guides.&lt;/p&gt;&#xA;" OwnerUserId="12823" LastActivityDate="2023-12-17T19:04:40.520" Title="How do you calculate the chain tension required when compressing bulk material over a conveyor?" Tags="|mechanical-engineering|friction|transportation|conveyor|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26347" PostTypeId="2" ParentId="26344" CreationDate="2019-03-12T16:39:57.983" Score="2" Body="&lt;p&gt;As stated by &lt;a href=&quot;https://engineering.stackexchange.com/questions/26344/the-internal-normal-force-shear-force-and-moment#comment48643_26344&quot;&gt;@CableStay in a comment&lt;/a&gt;, you just made a small mistake when determining the equivalent concentrated load for the distributed load between A and B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Triangular loads are equivalent to a concentrated load applied 1/3 of the way from the maximum load. You know this since you did it correctly every other time in this exercise, but when you did it for the load between A and B, you placed the concentrated load at C, halfway between A and B, when it should actually be 2&amp;nbsp;m away from A and 1&amp;nbsp;m away from B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeating your work fixing this one mistake, you get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum M_B &amp;amp;= \dfrac{1}{2} \cdot 10 \cdot 3 \cdot 1 - \dfrac{1}{2} \cdot 10 \cdot 1.5 \cdot 0.5 - 3A_y = 0 \\&#xA;\therefore A_y &amp;amp;= 3.75\text{ kN} \\&#xA;w_C &amp;amp;= 10\dfrac{1.5}{3} = 5\text{ kN/m} \\&#xA;V_C &amp;amp;= 3.75 - \dfrac{1}{2} \cdot 5 \cdot 1.5 = 0\text{ kN} \\&#xA;M_C &amp;amp;= -3.75 \cdot 1.5 + \dfrac{1}{2} \cdot 5 \cdot 1.5 \cdot 0.5 = -3.75\text{ kNm}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-03-12T16:39:57.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26349" PostTypeId="2" ParentId="26344" CreationDate="2019-03-12T19:54:02.563" Score="0" Body="&lt;p&gt;Let's just handle this from the other end of the beam, of course the result will be the same as in Wasabi's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;lets get the moment about A&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \Sigma M_a =  1/2\cdot 10\cdot 3 \cdot2 + 1/2\cdot10\cdot 1.5 \cdot 3.5 - Y_b =0 \\ \therefore Y_B = 18.75kN  \quad and;  \quad Y_A = 15+7.5- 18.75= 3.75kN     $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: slope of the load on the left side of B is 3.333 kN/m, so the load increases by 5kN at each 1.5 meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v_c = 3.75 - 1.5\cdot 5/2 =0 kN  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ M_C = -3.75\cdot1.5 + (1.5\cdot5/2) 0.5= -3.75 $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \Sigma F_x = 0, \quad N_C =0 $$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-12T19:54:02.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26351" PostTypeId="2" ParentId="26331" CreationDate="2019-03-12T22:38:13.853" Score="0" Body="&lt;p&gt;I can't say this will work for sure, but you should try clear silicone RTV. There are any number of companies that make the stuff, like &lt;a href=&quot;https://www.homedepot.com/p/Permatex-3-oz-Clear-Silicone-Adhesive-Sealant-75151/302775054&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; for example. Silicone RTV is rubber like and fairly think viscosity. As caulking, it is possible to peel off once cured. You should be able to buy clear silicone RTV at any hardware or home improvement store.&lt;/p&gt;&#xA;" OwnerUserId="9506" LastActivityDate="2019-03-12T22:38:13.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26352" PostTypeId="2" ParentId="26337" CreationDate="2019-03-12T22:53:36.697" Score="1" Body="&lt;p&gt;You are missing an equation of state that can be solved analytical or preprogram SRK or PR in a TI calculator when make it solve the enthalpy for you (enthalpy 1=enthalpy 2 but the phases are different and so is the volume), because what you want will have changes in the molar Volume and it has to be solved, that is why you solve it with a Mollier Diagram or Steam tables; Someone already solved/measured those properties so any engineer (without access to a simulator) can solve it with a pen and a napkin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try solving it yourself (and it is quite fun) but please by all means use a computer because you will be wasting time for a simple problem otherwise. &lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2019-03-12T22:53:36.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26354" PostTypeId="2" ParentId="26345" CreationDate="2019-03-13T01:24:17.627" Score="0" Body="&lt;p&gt;A very crude way of estimate the part of tension needed to compact the bulky material is get an average mass per second if material which is taller than the clearance between the roller and the conveyor belt and assume that part looses it's kinetic energy to inelastic collision with the roller and add that to other tension loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$E= 1/2 \dot {m}v^2= T_{belt}\cdot v$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-13T01:24:17.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26356" PostTypeId="1" AcceptedAnswerId="26358" CreationDate="2019-03-13T04:15:24.060" Score="0" ViewCount="78" Body="&lt;p&gt;I built/assembled a small drill press recently and used a part from a Junk Drill because I couldn't figure out what a particular component I needed was called. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I was looking for was a way to step up in chuck size so I could accommodate a bigger bit on a rod that was a smaller size (see: 3.17mm motor shaft driving 1/4&quot;+ bits).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now know the function of what I was looking for, it's a chamber of gears driven by a central D shaft gear the motor shaft can be inserted into. It's the internal component used to drive keyless chucks available in most retail drills. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what this would be called so I could look for it properly next time? I resorted to stripping an existing finished product for a part just because I couldn't figure it out, I was originally wanting to mount a keyed chuck&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt; &lt;a href=&quot;https://i.stack.imgur.com/8pBSj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8pBSj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19695" OwnerDisplayName="Sani Evans" LastEditorUserId="6767" LastEditDate="2019-03-13T07:49:17.000" LastActivityDate="2019-03-13T07:49:17.000" Title="Terminology for drill component?" Tags="|tools|drilling|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26357" PostTypeId="2" ParentId="26356" CreationDate="2019-03-13T04:18:33.790" Score="0" Body="&lt;p&gt;Chuck &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Some filler text because SE won't allow me to post fewer than 30 characters!&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="JayEye" LastActivityDate="2019-03-13T04:18:33.790" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26358" PostTypeId="2" ParentId="26356" CreationDate="2019-03-13T04:33:03.340" Score="0" Body="&lt;p&gt;I think it is not the chuck you want, but you've taken the &lt;em&gt;planetery gearbox&lt;/em&gt;(full of planetary gears, may be called orbital gears instead) and the &lt;em&gt;torque limiter&lt;/em&gt; from a battery operated drill, so it seems likely one of those words will help you.  If I may ask, is there a reason you've gone this route instead of buying a cheap drill press?  Mine isn't industrial grade, but it's more robust than the one in your picture and only cost $40 CAD on sale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also be advised that if you want the torque limiting function, broken battery operated hand tools can often be found cheaply on second hand sources, so you could take the gearbox from a higher quality hand drill and put yours back in service.  If you go shopping for broken drills I'd go for Milwaukee(Extremely well documented, high quality) if you can find it.  &lt;/p&gt;&#xA;" OwnerUserId="17734" LastActivityDate="2019-03-13T04:33:03.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26363" PostTypeId="1" CreationDate="2019-03-13T12:36:01.183" Score="1" ViewCount="895" Body="&lt;p&gt;So I'm given a question in which a Truck drives over a pratt truss bridge. In the question the truss is in 3D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm given:&#xA;1)The loads from each axle of the truck&#xA;2)The weight of each member in the truss&#xA;3)The weight of the concrete and asphalt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question also states that &quot;The forces acting upon the bridge shown in Figure 3.1 can be decomposed into forces acting in the x-y plane and forces acting in the x-z plane (i.e. the forces in the lateral bracing members). Show how this observation permits a very simple analysis of the structure whereby the cross-bracing members can be simply ignored and where the truss to be analysed is a planar truss, i.e. a truss in which all members lie in one plane.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few questions here, (3) being the most important:&#xA;1) How can I &quot;show that the observation permits a very simple analysis....&quot;&#xA;2) If the cross bracing members can be simply ignored are their weights ignored as well?&#xA;3) Considering the truss would be mirrored on either side but I am now only analysing one side (one plane), are the loads from the lorry, and the weights from the concrete and asphalt divided in half?&lt;/p&gt;&#xA;" OwnerUserId="19701" LastActivityDate="2019-03-13T16:30:16.457" Title="Member weight in a planar truss?" Tags="|structural-engineering|structural-analysis|bridges|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26364" PostTypeId="1" AcceptedAnswerId="26383" CreationDate="2019-03-13T12:58:21.060" Score="0" ViewCount="567" Body="&lt;p&gt;Part of my project is to compare linear elements and quadratic elements with a Quad mesh within Ansys workbench - static structural. The button that my classmates have to change between element orders is missing on my Ansys. Has anyone had ths issue and knows a solution or knows something I need to check to get the button to show? the option should appear under relevance in the Defualts section of the mesh.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RdPqm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RdPqm.png&quot; alt=&quot;Mesh details&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards,&lt;br&gt;&#xA;George Packham&lt;/p&gt;&#xA;" OwnerUserId="19702" LastActivityDate="2020-07-04T09:18:08.810" Title="Element order option missing" Tags="|ansys|ansys-workbench|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26365" PostTypeId="1" CreationDate="2019-03-13T14:02:42.170" Score="0" ViewCount="51" Body="&lt;p&gt;I cant exactly give you the  complete relevant details of the variables but assume the following (bold means vector):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B(x)&lt;/strong&gt; ~ &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{8\pi}*(\nabla\frac{1}{r})\times\int \mathbf X'\times \mathbf \Omega' dV(\mathbf X')$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where r=|&lt;strong&gt;X-X'&lt;/strong&gt;|, prime denotes evaluation at point &lt;strong&gt;X'&lt;/strong&gt; and dV is the volume throughout the domain and it holds as &lt;span class=&quot;math-container&quot;&gt;$r \rightarrow \infty$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let this integral be evaluated for a large value of r (say R). My question is why is the following possible if the above integral holds on a spherical surface, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$8\pi R^2(\mathbf n \times \mathbf B)_{r=R} = \int \mathbf x \times \mathbf \Omega dV - \mathbf n \mathbf n. \int \mathbf x \times \mathbf \Omega dV$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For exact reference see pg 519 of the book in the link &lt;a href=&quot;http://users.df.uba.ar/cobelli/Libros/Batchelor.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fluid Dynamics by G.K.Batchelor&lt;/a&gt;. What I am asking about is the derivation between equation 7.2.3 and 7.2.4. Though this book is considered a classic, but damn its hard to keep up with these technicalities.&lt;/p&gt;&#xA;" OwnerUserId="19703" LastActivityDate="2019-03-14T07:59:00.157" Title="How does this transformation actually happen?" Tags="|mechanical-engineering|fluid-mechanics|aerospace-engineering|fluid|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26367" PostTypeId="2" ParentId="26363" CreationDate="2019-03-13T16:30:16.457" Score="2" Body="&lt;p&gt;First, let's summarize the assumptions we generally make when analyzing trusses.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Truss members are connected by frictionless pins&lt;/li&gt;&#xA;&lt;li&gt;Truss members carry axial loads only&lt;/li&gt;&#xA;&lt;li&gt;Loads are applied only at the joints&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Now, for your questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(1) Show the system can be analyzed as two independent, parallel plane trusses.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before jumping to 3D space, let's see what we can understand from 2D space. Consider a truss arrangement on an x-y plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1ctRJm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ctRJm.jpg&quot; alt=&quot;2D Truss Joint&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a vertical load applied to a system of orthogonal vert/horiz members. With knowledge of the equations of equilibrium and Assumption #2, we can observe that Load P will be carried entirely by the vertical member. The horizontal members will not carry any of Load P because no component of Load P is along the axis of the horizontal members. If we had diagonal members, then we could imagine calculating a component of Load P that acts along the diagonal member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this gets you thinking about how vertical loads would be carried by out-of-plane members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(2) Are the weights of the cross-bracing ignored?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In structural analysis courses, the self-weight of the truss is often completely ignored, on the premise that the effect of the self-weight is small compared to the applied loads (and likely also because representing self-weight requires a lot of nodal loads, which is inconvenient for introductory analysis). In your more real-world case, it's still likely the lateral bracing self-weight is negligible, but since this is about the learning process, you might consider analyzing it just to convince yourself of that fact. Remember that loads (including self-weight) can only be applied at the joints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(3) Are loads from the truck and bridge deck divided in half for plane-truss analysis?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, and no. For purposes of a homework assignment, it's likely that they do just want you to split the load in half (though you still have to consider how to turn that into point loads at truss joints.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gets into considering the framing system for the deck/stringers/floor beams. A not-to-scale sketch of one possible framing system is below, along with a picture of an actual truss bridge that used a similar framing plan. (Note that it may actually be more common to have stringers frame directly into the floor beam rather than running continuously over the top of the floor beams as I sketched.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As I sketched it, the point loads from the truck axles are applied to the deck. The deck acts to distribute those loads to the stringers. The stringers then carry the load to the floor beams. The floor beams transfer the loads to the truss joints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pcgYI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pcgYI.png&quot; alt=&quot;Framing System&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turning the deck weight into loads at the truss joints can be done by considering the length of deck that is tributary to each floor beam. For example, if the floor beams are spaced at 30 feet, then the load from 30 feet of deck goes to each interior floor beam, and from 15 feet of deck to the first/last floor beam. Then the floor beams will transfer that load equally to the two trusses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying the truck weight to the truss can get a little thornier, particularly considering that trucks can move both longitudinally and transversely along the bridge and you don't state whether the assignment specifies truck position. If the floorbeam spacing is more than twice the distance between the truck's first and last axles, it's reasonable to assume the truck load is all applied to one floorbeam. If the floorbeam spacing is tighter than that, you might consider how much truck load goes to each floorbeam. Assuming it all goes to one floorbeam is conservative though. If the truck can be placed somewhere other than perfectly in-between the two trusses (ie. closer to one truss than the other) you might want to account for one truss receiving more than 50% of the truck load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how the homework assignment is written, the professor may just want you to experience the process of making simplifying assumptions. However you choose to represent the truck/deck loads, document and justify your choices.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2019-03-13T16:30:16.457" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26370" PostTypeId="1" CreationDate="2019-03-14T00:47:26.223" Score="0" ViewCount="148" Body="&lt;p&gt;Hello all I'm designing a joint that needs a bearing to fit central but I'm wondering whether I have to protrude a bit on both parts so there will be some gaps between the bearing and the parts which allowing both components rotate freely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to describe this exactly or accurately so I attached some pictures for explanation The first picture is without protruding the second one is with protrude and the final picture is the location where I want to put the bearing(represented as red rectangle)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I am right how much should I protrude? The bearing is 6mm outer diameter 3mm inner hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8Zj80.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Zj80.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5p4cW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5p4cW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Kj0D.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Kj0D.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18994" LastActivityDate="2019-08-13T17:29:51.850" Title="Designing a joint with bearing, how much do I need to protrude for the bearing to stand? With pictures" Tags="|bearings|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26372" PostTypeId="1" CreationDate="2019-03-14T07:23:39.013" Score="1" ViewCount="149" Body="&lt;p&gt;I need a 3D CAD software that permits to create parts and assembly.&#xA;Right now, I try FreeCad, Salome and BRL-CAD.&#xA;The first one can create only single parts, not assembly.&#xA;Salome have the same problem.&#xA;BRL-CAD can create parts and assembly, but it's use is not intuitive and the graphical module right now is not mature.&#xA;Any other?&lt;/p&gt;&#xA;" OwnerUserId="19707" LastActivityDate="2019-03-14T08:24:34.777" Title="which open source CAD?" Tags="|mechanical-engineering|cad|" AnswerCount="1" CommentCount="2" ClosedDate="2019-03-14T19:31:32.440" ContentLicense="CC BY-SA 4.0" />
  <row Id="26374" PostTypeId="2" ParentId="26372" CreationDate="2019-03-14T08:24:34.777" Score="1" Body="&lt;p&gt;Almost all the big Companies offer free version of their products for students. If you are a student, you could use Solidworks (12 month) or Inventor. &#xA;For non students (or something comparable) I would recommend using trial versions of reputable software. From what I have experienced, free CAD software isn't all that great (at least for my purpose).&lt;/p&gt;&#xA;" OwnerUserId="19438" LastActivityDate="2019-03-14T08:24:34.777" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26375" PostTypeId="2" ParentId="26370" CreationDate="2019-03-14T08:33:40.587" Score="0" Body="&lt;p&gt;It depends on the type of bearing (and the direction of force). &#xA;But generally I would build the part with a little bit of spacing so that the bearing is no dragging on the housing. If you are using a taper roller bearing for example, you might not need the spacing because the housing might already have a &quot;spacer&quot;.&lt;/p&gt;&#xA;" OwnerUserId="19438" LastActivityDate="2019-03-14T08:33:40.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26376" PostTypeId="1" CreationDate="2019-03-14T08:36:21.370" Score="2" ViewCount="1032" Body="&lt;p&gt;I'm driving a DC motor (actually a linear actuator) with a Polulu motor controller from a Raspberry Pi. The motor drives if you ask for anything between 20 and 100% duty cycle, so there is a dead zone between -20% and 20% duty cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using a laser distance sensor to measure position of the actuator, and I'd like to make the actuator track a desired position using a PID controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I'm facing is that as the position crosses over from being too low to too high, the PID controller takes a load of time to cross through this deadband, after which the position is completely wrong!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried creating a lookup table for the motor speed based on the requested speed, but this doesn't help a huge amount. I was thinking of creating two lookup tables, one for when the position is above setpoint and one for when it is below which cross over instantly from -20 to 20 and vice-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on this problem (or just what it is called so I can Google it!) would be much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="19711" LastActivityDate="2020-09-08T05:06:15.667" Title="PID control of a DC motor with PWM from Raspberry Pi - how to cross dead area?" Tags="|control-engineering|motors|pid-control|pwm|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26378" PostTypeId="2" ParentId="26364" CreationDate="2019-03-14T08:58:25.307" Score="3" Body="&lt;p&gt;I dont have a direct solution for your problem but you might want to check your software version. If you are using the student version of ANSYS there might be a difference to the version you are using in class. And keep in mind that the student version is very limited regarding the number of nodes and elements you can use!&lt;/p&gt;&#xA;" OwnerUserId="19438" LastActivityDate="2019-03-14T08:58:25.307" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26379" PostTypeId="1" AcceptedAnswerId="26414" CreationDate="2019-03-14T12:03:45.540" Score="0" ViewCount="37" Body="&lt;p&gt;I'm in the process of building a hovercraft. In order to lift it, I'll need a motor directing the flow of air into a skirt. The specs for the motor I'm looking at say that it can turn a 15x5 dual-blade propeller at 6,000 RPM with 262W of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the motor is designed to be used by a large drone. In that case, the motor only has to work against atmospheric pressure. My hovercraft will have about .17 psi of pressure within the skirt. How do I adjust the required power for a given RPM based on the pressure below the propeller?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mass is 87kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.6m x .45m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Airgrap: 12.7mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cushion Pressure: 1.185 kPa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inward flow during static flight is 3,000 CFM (I believe that means flow out is also 3,000 CFM?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the motors I'm looking at: &#xA;&lt;a href=&quot;https://www.getfpv.com/tiger-motor-mn4014-330kv-antigravity-2-motors.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.getfpv.com/tiger-motor-mn4014-330kv-antigravity-2-motors.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16660" LastEditorUserId="16660" LastEditDate="2019-03-14T12:13:55.340" LastActivityDate="2019-03-17T01:25:27.070" Title="How to adjust power requirements for a propeller working against a pressure head" Tags="|motors|pressure|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26380" PostTypeId="1" CreationDate="2019-03-14T15:11:30.740" Score="0" ViewCount="458" Body="&lt;p&gt;when the sides of a pontoon are verticle, BM(distance between center of buoyancy and metacenter) is given by I/V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as for a small angle that expression is true. But when the heel angle increases what heppens to the BM? &lt;/p&gt;&#xA;" OwnerUserId="19714" LastEditorDisplayName="user16" LastEditDate="2019-03-24T19:44:55.580" LastActivityDate="2024-01-10T10:04:35.697" Title="Is the metacentric height independent of heel angle?" Tags="|fluid-mechanics|marine-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26381" PostTypeId="1" CreationDate="2019-03-14T15:52:33.927" Score="1" ViewCount="2384" Body="&lt;p&gt;I am trying to determine the approximate CFM of several axial flow fans being used in agricultural storage in the developing world. The only information the storers have is the power, RPM, and diameter of their fans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to estimate the CFM at various levels of Static Pressure, and within a range of likely blade pitches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to to do this? I've looked at the fan laws, but it seems like these are meant to calculate &lt;em&gt;changes&lt;/em&gt; in variables, not the absolute values of the variables themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="19716" LastActivityDate="2024-03-18T07:06:20.780" Title="Estimating fan CFM from RPM, HP, and diameter" Tags="|airflow|hvac|airfoils|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26383" PostTypeId="2" ParentId="26364" CreationDate="2019-03-14T17:06:45.700" Score="1" Body="&lt;p&gt;Solution is based on the fact that the user has an older version of Ansys, in this case user has 16.2. There is the option for lines and quadratic nodes but it is found under automatic method as Element midside nodes. The option are either dropped (linear) or kept(quadratic). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue arose as my colleagues are using the new version whilst my licence is an older model &lt;/p&gt;&#xA;" OwnerUserId="19702" LastActivityDate="2019-03-14T17:06:45.700" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26384" PostTypeId="1" AcceptedAnswerId="26385" CreationDate="2019-03-14T17:33:07.537" Score="0" ViewCount="49" Body="&lt;p&gt;I'm hoping some kind person can tell me if I have this right or not.  I swear this isn't a homework problem.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be using a heated pneumatic press to heat seal lidding to plastic containers (imagine I'm sealing yogurt containers).  I need to verify that the machine I'm looking at can deliver a certain psi at the interface of the plastic and the lidding.  That interface has much less area than the total area of the heated platen applying the force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand I take the pressure delivered to the machine, subtract the ambient air pressure, multiply by the surface area of the cylinder and then add the total weight of the platen to get the full downforce the machine is capable of.  From there I divide that by the surface area of the contact between the object under the press and the platen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please enjoy my wonderful art below (these are just random numbers):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/06uqT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/06uqT.png&quot; alt=&quot;psi diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in this case the total downforce of the platen is:&#xA;(100psi - 14.7psi) * 1 sq in + 10 lbs = 95.3 lbs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the force experienced by the red object under the press is:&#xA;95.3lbs / 2 sq in = 47.65psi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like that's right but all my attempts to google for this answer only lead me to finding the total downforce of the press and not the force experienced by a smaller object &lt;em&gt;under&lt;/em&gt; the press.  Do I have it right?  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="19717" LastActivityDate="2019-03-14T18:05:43.190" Title="Pressure delivered to an object under a pneumatic press" Tags="|pressure|hydraulics|compressed-air|compressors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26385" PostTypeId="2" ParentId="26384" CreationDate="2019-03-14T18:05:43.190" Score="1" Body="&lt;p&gt;This looks right- but note the units of &lt;em&gt;pressure&lt;/em&gt; (not &lt;em&gt;force&lt;/em&gt;) experienced by the red object are pounds per square inch.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-03-14T18:05:43.190" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26386" PostTypeId="1" AcceptedAnswerId="26389" CreationDate="2019-03-14T19:57:40.993" Score="2" ViewCount="212" Body="&lt;p&gt;The following image is  an example that will help to demonstrate what i am saying and this below is a blender image version of an object with a slant surface at the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EaNI2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EaNI2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to draw the multi view projection of this view all the necessary dimensions are given, so the top and the front will be as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;top:&lt;a href=&quot;https://i.stack.imgur.com/IEfXx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IEfXx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;front:&lt;a href=&quot;https://i.stack.imgur.com/mGSNw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mGSNw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i very well understand how the front and the top projections worked and here is where my question arises, as seen both from the pictorial view and the two side projection there are two thoroughly subtracted cylinders in the slant surface which means when i try to do the right side projection i will be seeing two ellipses rather than the original circles i have seen by the top view as shown below, so how could i figure out the dimensions of the major and minor axis of the ellipses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;right view:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fAynY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fAynY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19718" LastEditorUserId="19718" LastEditDate="2019-03-14T20:19:09.897" LastActivityDate="2019-03-14T22:06:03.203" Title="How to calculate the dimensions of an ellipse in a multi-view projection" Tags="|mechanical-engineering|technical-drawing|architecture|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26387" PostTypeId="2" ParentId="26381" CreationDate="2019-03-14T20:52:19.153" Score="0" Body="&lt;p&gt;For a particlar fan blade or impeller type an empirical relationship could be developed. I am not familiar with an empirical  correlation of this type for fan or blowers. I have seen empirical correlations for liquid mixing impellers with fitting parameters for specific impeller types to estimate power. To know for a specific geometry fan blade without an empirical correlation I believe you would need a sort of finite element analysis.&lt;/p&gt;&#xA;" OwnerUserId="14901" LastActivityDate="2019-03-14T20:52:19.153" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26389" PostTypeId="2" ParentId="26386" CreationDate="2019-03-14T22:06:03.203" Score="2" Body="&lt;p&gt;This is a simple trigonometric exercise. You really only need the &quot;front&quot; view to solve this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the cleaned up version of that view (I have omitted the holes for now, and I assumed the right-hand &quot;vertical&quot; is 1.50&amp;nbsp;m thick, which isn't explicitly stated in your diagrams):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cCfLG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cCfLG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, really, the only things we need to observe here is the slope of the incline, which can be trivially found to be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dfrac{\text{d}y}{\text{d}} = \dfrac{0.99}{9.00} = 0.11$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for every meter traveled horizontally, the incline has traveled 11 centimeters vertically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, one of the cylindrical holes is shown with a diameter of 1&amp;nbsp;m&lt;sup&gt;[1]&lt;/sup&gt;. I'm assuming this is valid for both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the cylinder's diameter and the slope of the incline, it's easy to see that the vertical drop will be of 0.11&amp;nbsp;meters. After all, if you're at the top-most lip of the hole and move horizontally to the other end, you'll need to drop 0.11&amp;nbsp;meters to get to the incline once again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you're drawing the holes in the &quot;right&quot; view, your ellipses need to have a &lt;strong&gt;major diameter of 1&amp;nbsp;meter&lt;/strong&gt; (just like the actual circle, since there's no &quot;warping&quot; at this angle) and a &lt;strong&gt;minor diameter of 0.11&amp;nbsp;meters&lt;/strong&gt;, since that's the vertical drop between the top and bottom lips of the circle.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;[1]&lt;/sup&gt; There's a point of uncertainty here as to whether the cylinder's cross-section is circular or whether the hole on the inclined surface is circular. Only one of these can be true (if the cylinder is circular, then the inclined hole will be an ellipse; if the inclined hole is circular, then the cylinder is elliptical). I have further assumed here that the cylinder is circular since, well, that seems much more likely.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;For completeness, in the unusual case where the inclined hole is circular (and the cylinder and hole at the bottom face are therefore elliptical), the solution would require a tiny bit more math: we'd need to use the Pythagorean Theorem to find the hypotenuse associated with the given slope, and then just use proportions to find the vertical step equivalent to a 1&amp;nbsp;m hyotenuse with the same slope.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\small\begin{align}&#xA;h &amp;amp;= \sqrt{0.11^2 + 1^2} = 1.00603\text{ m} \\&#xA;\dfrac{x}{1} &amp;amp;= \dfrac{0.11}{1.00603} \\&#xA;\therefore x &amp;amp;= 0.10934\text{ m}&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-03-14T22:06:03.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26390" PostTypeId="2" ParentId="11525" CreationDate="2019-03-15T10:00:53.267" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Put another way, what is the argument that the mechanical properties covered by the usual texts (Young's, Poisson's, hardness, toughness, etc.) can give me a complete mechanical description of a system under arbitrary geometries and loads?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There is no such argument.  All models are wrong.  Some models are useful.  In other words, there is absolutely nothing that can give you a &lt;em&gt;complete&lt;/em&gt; mechanical description of any system.  But the usual properties are quite useful in making predictions under many (but not all) practical situations.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-03-15T10:00:53.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26391" PostTypeId="1" CreationDate="2019-03-15T12:37:05.283" Score="0" ViewCount="125" Body="&lt;p&gt;Admittedly, this question is related to an earlier one on the same subject.  I would like to mount a heavy bag using a system detailed in the attached picture (FirstLaw Fitness Shock Mount).  The system is designed to be mounted between and directly to 2 joists that are 16&quot; on center with included lag bolts. My garage ceiling is covered with drywall with a bedroom above.  I assume (hope) that the gas strut on the mounting plate will somewhat help with noise and vibration, but my largest concern is mitigating unsafe loading conditions on the ceiling.  I saw an earlier post where a structural engineer suggested not attaching the mount directly to the joist. Was this to provide mitigation for vibration/noise? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone provide a &quot;safer&quot; installation idea or are the instructions adequate? Based on an earlier answer, could I further distribute the load if I attached a 2x6 or 2x4 on the ceiling so that it runs under the floor joist for a couple of feet, do the same to the next joist over and then attach the mount perpendicular to and on the 2x6's?  If this is acceptable, what size and length of lag bolts would be required since they would need to go through the 2x6 and into the joist? When attaching the mount to the 2x6's, would the lag bolts also need to go into the joist?  And finally, if this idea provides for better load distribution/safer install, is there concern that it does require more holes being drilled into the joist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has a better/easier/safer idea, I would be most interested.  A diagram with dimensions would be great. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;. &lt;a href=&quot;https://i.stack.imgur.com/AsQDl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AsQDl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19722" LastActivityDate="2024-03-19T03:02:05.227" Title="Optimal method of attaching commercially-available heavy bag mount to finished ceiling with least risk of structural damage" Tags="|structures|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26392" PostTypeId="1" AcceptedAnswerId="26395" CreationDate="2019-03-15T16:11:20.737" Score="2" ViewCount="6238" Body="&lt;p&gt;I am working on calculating the stiffness of a cantilever beam by applying a distributed load. I wanted to ask if the distribution of the load on the beam effects its stiffness.&lt;br&gt;&#xA;For eg., if I apply a point load at the free end vs a constant distributed load throughout the beam. Will the stiffness I get be different? How do I calculate the stiffness in later case?&lt;br&gt;&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="3345" LastActivityDate="2019-03-15T19:34:17.633" Title="Stiffness of a cantilever beam" Tags="|mechanical-engineering|beam|vibration|stiffness|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26393" PostTypeId="2" ParentId="26391" CreationDate="2019-03-15T16:15:24.633" Score="0" Body="&lt;p&gt;It would help a lot if you knew the actual joist construction in your garage ceiling. Assuming it is some kind of truss system about 12&quot; deep, I would cut an access where you want to mount the thing (small one at first to check for services above) and then scab about a 3' length of decent 5/8 ply full height to the sides of the trusses. Then I'd nail a sill and hang blocking at least 2 x 8 and lagbolt to that. You can use storm hangers in leiu of a sill if desired (I'd probably double up on the plywood in this case). Stuff with as much insulation as you can get in there, patch the sheetrock, and mount the bag. If height is an issue, just box in the cutout. If you have solid wood joists, you could get away with lagging to them. But you should have some heavy, full height blocking between the two joists right where the plate is going. The old-style X bracing, if present, is probably not up to snuff for this job. &lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-03-15T16:15:24.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26394" PostTypeId="2" ParentId="26392" CreationDate="2019-03-15T18:11:54.623" Score="2" Body="&lt;p&gt;The stiffness of a beam does not change with the loading if the equivalent loads and their points of action on the beam are equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First lets do the stiffness of the beam under q uniform load.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\delta = \frac{qL^4}{8EI}  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's load a cantilever beam with a point load equivalent to uniform load.&#xA;in the distribuited load we have total load &lt;span class=&quot;math-container&quot;&gt;$ \ P=qL \ $&lt;/span&gt; acting at the center witch is L/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta= \frac{P(L/2)^2}{6EI}\cdot(3L-a_{distance\ from\ end}) \\ \therefore \delta=\frac{qL^3}{24EI}(3L-0)= \frac {qL^4}{8EI}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as we see the stiffness doesn't change under different loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to remind us a simple point load at the end of a cantilever beam causes deflection &lt;span class=&quot;math-container&quot;&gt;$ \delta=PL^3/3EI.$&lt;/span&gt; If we compare this to the above result we see placing the same load on the beam but uniformly distributed causes 3/8 of point load deflection, even though the stiffness remains the same.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-03-15T18:24:20.527" LastActivityDate="2019-03-15T18:24:20.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26395" PostTypeId="2" ParentId="26392" CreationDate="2019-03-15T19:34:17.633" Score="8" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Stiffness&lt;/em&gt;&lt;/strong&gt; is a murky term frequently used ambiguously in engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the most common definition of stiffness is the product of a beam's Young's Modulus &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; (which is a function of its material) and its moment of inertia &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; (which is a function of its cross-section). So &lt;span class=&quot;math-container&quot;&gt;$\text{Stiffness} = EI$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Loading has nothing to do with stiffness according to this definition, which you could say describes an isolated beam's stiffness. This value allows you to say whether one beam would be stiffer than another in identical circumstances (whatever those may be).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, another possible definition is stiffness as the deflection a beam or structure suffers under load. This would be an analogy with a spring's stiffness (which is literally measured in force needed to move the spring a unit distance).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This could be described as the whole structure's stiffness (even if that structure is a single cantilever beam). Obviously, a structure with pinned supports will be less stiff than one with fixed supports. Likewise, a cantilever beam with a concentrated load at midspan will deflect less (be stiffer) than one where the load is at the free end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as mentioned in &lt;a href=&quot;https://engineering.stackexchange.com/a/26394/1832&quot;&gt;@kamran's answer&lt;/a&gt;, different loading patterns will lead to different deflections even if the total applied load is equal:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A cantilever beam with a uniformly distributed load will have a deflection at the free end of &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\delta = \dfrac{qL^4}{8EI}$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;The same beam with a concentrated force (equal to &lt;span class=&quot;math-container&quot;&gt;$P=qL$&lt;/span&gt; to keep the same total force) at the free end will have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\delta = \dfrac{PL^3}{3EI} = \dfrac{qL^4}{3EI}$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;The same beam where the concentrated force is applied at the midspan will have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\delta = \dfrac{5PL^3}{48EI} = \dfrac{5qL^4}{48EI}$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that all these solutions have the constant &lt;span class=&quot;math-container&quot;&gt;$EI$&lt;/span&gt; given as the more common definition of stiffness above. So a stiffer beam (one with greater &lt;span class=&quot;math-container&quot;&gt;$EI$&lt;/span&gt;) will deflect less than a more flexible one in any and all cases. However, how much any given beam will deflect depends on the loading and support conditions.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Given this, the answer to your question depends on what you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for the isolated beam's stiffness, then the result will be the same no matter your loading pattern (since you'll use different equations depending on the loading to get that beam's stiffness).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, you're looking to get the stiffness as &quot;force per deflection&quot;, then the loading will change your result. Any loading is valid, but your result will only be valid for that type of loading (i.e. this beam needs 10&amp;nbsp;kN/mm at midspan, but only 3.2&amp;nbsp;kN/mm at the free end).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-03-15T19:34:17.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26398" PostTypeId="1" CreationDate="2019-03-15T22:20:59.950" Score="1" ViewCount="54" Body="&lt;p&gt;Is it possible to increase an air column's diameter over a short distance? Let me explain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a small 10 cm electric computer cooling fan that's pushing air into a cone-shaped duct 20 cm long. If the exit side of the cone-shaped duct has a diameter of 5 cm, I would think the diameter of the air &quot;column&quot; at the exit of the duct will be halved, and the pressure to be double of what it is at the beginning of the duct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the diameter of the cone-shaped duct is instead increased to 20 cm on the exit side, &lt;strong&gt;will the air column have double the diameter and half the air pressure than at the beginning of the cone-shaped duct?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="2080" LastActivityDate="2019-03-16T05:42:13.643" Title="Is it possible to &quot;expand&quot; air flow area in a duct over a short distance?" Tags="|fluid-mechanics|airflow|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26399" PostTypeId="2" ParentId="26398" CreationDate="2019-03-15T23:53:24.107" Score="2" Body="&lt;p&gt;probably not until the flow has traveled several pipe diameters downstream of the transition. To speed up the establishment of uniform flow you'll have to put in several  fine mesh screens (like a window screen) right at the end of the transition section, spaced about 2&quot; apart; this is a common trick used to smooth out flow fields in old-school wind tunnels. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-03-15T23:53:24.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26400" PostTypeId="2" ParentId="26398" CreationDate="2019-03-16T00:50:36.880" Score="1" Body="&lt;p&gt;Make the changes of diameter gradual to minimize the losses : about 6 degrees is best... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do the analysis to prove it... It was an analysis done in class by our professor... Of course if space is an issue then 6 degrees may not be appropriate.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-03-16T00:50:36.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26401" PostTypeId="2" ParentId="26398" CreationDate="2019-03-16T04:43:28.847" Score="1" Body="&lt;p&gt;Actually Bernuoulli law says that when you decrease the flow diameter the pressure goes down, while speed goes up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{v^2}{2}+gz+\frac{P}{\rho}=\text{constant}$$&lt;/span&gt;&#xA;Assuming the fluid incomprehensible, which at these low Mach numbers it applies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when the fan outlet is decreased the fluid is accelerated, by leaching energy from the pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your assumption is not right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And inversely when you increase the duct diameter the flow speed goes down giving back some energy to pressure, causing it to go up.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-03-16T05:42:13.643" LastActivityDate="2019-03-16T05:42:13.643" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26403" PostTypeId="1" CreationDate="2019-03-16T12:50:52.527" Score="0" ViewCount="561" Body="&lt;p&gt;Would it be possible to install solidworks on Microsofts cloud platform (azure) and access solidworks remotely. Actually, my computer is quite old and cannot install the newer versions, so i thought i could access it on cloud for the time being. &lt;/p&gt;&#xA;" OwnerUserId="9537" LastActivityDate="2019-03-16T12:50:52.527" Title="Solidworks on Azure" Tags="|mechanical-engineering|solidworks|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26404" PostTypeId="2" ParentId="26332" CreationDate="2019-03-16T13:10:07.100" Score="0" Body="&lt;p&gt;The resistor value can be determined fairly accurately by process of observation, comparison and elimination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Observation&lt;/strong&gt;&lt;br&gt;&#xA;There are two 7 segment LED displays and supporting driver circuits. These include CS7107GN driver IC, a block of identical resistor blocks and similar resistor and capacitor blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Comparison&lt;/strong&gt;&lt;br&gt;&#xA;Within the Similar resistor and capacitor Circuit block (highlighted in red) has approximately 5 capacitors and about 3 to 4 resistors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Elimination&lt;/strong&gt;&lt;br&gt;&#xA;Measure the resistor values in each block. Most likely the values are similar, close or some combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use a potentiometer to determine the appropriate value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an image pictorially describing the process. Blocks are segment in colours&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2gayn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2gayn.jpg&quot; alt=&quot;7 Segment Display PCB&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-03-16T14:00:04.997" LastActivityDate="2019-03-16T14:00:04.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26405" PostTypeId="2" ParentId="26370" CreationDate="2019-03-16T14:24:23.337" Score="0" Body="&lt;p&gt;The upper part should protrude a little bit so that it isn't in contact with anything but the bearing face. This will introduce risk of debris which you should take into account if the item will be in an unclean environment (but only if the bearing is under great dynamic load). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the &quot;arm&quot; is in contact with the &quot;cylinder&quot; (that houses the bearing), it seems to me like that would defeat the function of the bearing. Thus the protrusion should only be in contact with the inner circle of the bearing, so as to not collide with the outer that should only be in contact with the cylinder.&lt;/p&gt;&#xA;" OwnerUserId="19731" LastActivityDate="2019-03-16T14:24:23.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26406" PostTypeId="1" CreationDate="2019-03-16T15:17:57.447" Score="1" ViewCount="51" Body="&lt;p&gt;I've read (1) that the efficiency (COP) of a heat pump does not change much with size. If that is true, why don't power companies employ them to produce yet more power from fossil fuels or renewables? I'm aware that air source heat pump efficiency drops in cold wet climates, but aren't there other more stable sources that could be tapped by an industrial scale plant? Ocean currents, for example.&#xA;(1) &lt;a href=&quot;https://engineering.stackexchange.com/questions/6203/are-smaller-heat-pumps-more-efficient&quot;&gt;Are smaller heat-pumps more efficient?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19732" LastActivityDate="2019-03-16T15:17:57.447" Title="Industrial scale heat pumps" Tags="|thermodynamics|heat-transfer|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26407" PostTypeId="1" CreationDate="2019-03-16T16:31:24.703" Score="2" ViewCount="91" Body="&lt;p&gt;I have to build a box to extend the length of a projector bracket:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u3RDE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u3RDE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projector bracket weighs 7kg and has an arm of about 500mm. The projector weighs another 5kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The box has a depth of 140mm. That is the length by which I need to extend the length of the projector arm because the wall on which the bracket is mounted is recessed by 140mm behind the image surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does one of the following two ways of joining offer a stronger structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YwDLj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YwDLj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have tools to make complex joints. I can only screw ½&quot; plywood at right angles.&lt;/p&gt;&#xA;" OwnerUserId="19734" LastActivityDate="2019-03-17T00:00:03.367" Title="Which way of joining is stronger?" Tags="|structural-engineering|materials|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26408" PostTypeId="1" CreationDate="2019-03-16T17:33:35.253" Score="0" ViewCount="571" Body="&lt;p&gt;In the following problem of heat transfer, what does characteristic length actually mean? In this problem, they have found the heat transfer coefficient (h) separately for the vertical and horizontal surface with the characteristic length being the height and width, respectively. Is it okay to take the length of the duct as characteristic length? If yes how should I proceed to solve this problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/evxeV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/evxeV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19735" LastEditorUserId="1832" LastEditDate="2019-03-16T21:42:29.820" LastActivityDate="2019-03-16T21:42:29.820" Title="Characteristic length for the given problem of heat transfer" Tags="|mechanical-engineering|heat-transfer|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="26409" PostTypeId="1" CreationDate="2019-03-16T21:13:44.003" Score="0" ViewCount="30" Body="&lt;p&gt;I am changing career paths and will be working on ECUs/VCUs and associated downstream electronic control modules. I am interested to find a research starting point in terms of Real-Time-Operating-Systems as related to Automotive applications and I figured that the highest priority task in Automotive applications might be braking or fire-related. Is this assumption correct and if not, what (in modern automotive design) is considered the highest priority task to deal with?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your responses.&lt;/p&gt;&#xA;" OwnerUserId="19738" LastActivityDate="2019-03-16T21:13:44.003" Title="Engine Control Unit theory regarding highest priority task?" Tags="|control-theory|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26410" PostTypeId="1" CreationDate="2019-03-16T22:40:22.967" Score="1" ViewCount="291" Body="&lt;p&gt;The Transfer Function of a Closed-Loop system is: &lt;span class=&quot;math-container&quot;&gt;$T(s)=\frac{Z(s)}{1+G(s)Z(s)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The BIBO stability of &lt;span class=&quot;math-container&quot;&gt;$T(s)$&lt;/span&gt; is guaranteed if all the poles of &lt;span class=&quot;math-container&quot;&gt;$T(s)$&lt;/span&gt; have negative real value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many textbooks I've read, it's written without further explanation that the zeros of the characteristic polynomial &lt;span class=&quot;math-container&quot;&gt;$1+G(s)Z(s)=0$&lt;/span&gt; are the poles of &lt;span class=&quot;math-container&quot;&gt;$T(s)$&lt;/span&gt;, &lt;strong&gt;without any conditions on &lt;span class=&quot;math-container&quot;&gt;$Z(s)$&lt;/span&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one textbook, it was written that &lt;span class=&quot;math-container&quot;&gt;$Z(s)$&lt;/span&gt; should be bounded. If &lt;span class=&quot;math-container&quot;&gt;$Z(s)$&lt;/span&gt; is bounded, i.e. has no poles, it's obvious that the characteristic polynomial contains all the poles of a closed-loop transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in many physical systems, &lt;span class=&quot;math-container&quot;&gt;$Z(s)$&lt;/span&gt; has poles. I have a theoretical example in which &lt;span class=&quot;math-container&quot;&gt;$Z(s)$&lt;/span&gt; has a pole at &lt;span class=&quot;math-container&quot;&gt;$s_0$&lt;/span&gt;, while &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt; has a zero at &lt;span class=&quot;math-container&quot;&gt;$s_0$&lt;/span&gt;, such that their product is finite and different from 1 &lt;span class=&quot;math-container&quot;&gt;$Z(s_0)G(s_0)\rightarrow A_0 \neq -1$&lt;/span&gt;. Therefore, &lt;span class=&quot;math-container&quot;&gt;$T(s)$&lt;/span&gt; will have a pole at &lt;span class=&quot;math-container&quot;&gt;$s_0$&lt;/span&gt;, while the characteristic polynomial will not have a root at &lt;span class=&quot;math-container&quot;&gt;$s_0$&lt;/span&gt;. Therefore, &lt;strong&gt;we found an example for characteristic polynomial that &lt;em&gt;doesn't&lt;/em&gt; contains all the poles of a closed-loop transfer function.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, it seems as if a necessary condition for working only with the characteristic polynomial is that the &lt;span class=&quot;math-container&quot;&gt;$Z(s)$&lt;/span&gt;'s poles won't coincide with the &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt;'s zeros. But is there a necessary and sufficient condition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this question not addressed in control-theory textbooks? If it does, please refer me to those textbooks.&lt;/p&gt;&#xA;" OwnerUserId="19739" LastEditorUserId="19739" LastEditDate="2019-03-17T20:27:31.653" LastActivityDate="2019-03-17T20:27:31.653" Title="Does the characteristic polynomial contains all the poles of a closed-loop transfer function?" Tags="|control-theory|transfer-function|stability|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26413" PostTypeId="2" ParentId="26407" CreationDate="2019-03-17T00:00:03.367" Score="4" Body="&lt;p&gt;In addition to the excellent suggestion of glue with the screws, consider to add a batten or bracing at the corners, as per the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6PFac.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6PFac.png&quot; alt=&quot;box support&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red squares would run the width of your box and allow you to drive the screws through the thickness of the plywood and into long grain of the battens, rather than into edge grain of the other plywood. This will give you the best possible strength in the simplest manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ends of the boxes should also have battens and be screwed in place. The box ends will provide torsion strength which is going to be a substantial portion of the design success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure why, but one red square is missing from the drawing. The four box end battens are omitted for clarity.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-03-17T00:00:03.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26414" PostTypeId="2" ParentId="26379" CreationDate="2019-03-17T01:25:27.070" Score="0" Body="&lt;p&gt;Alright, I think I figured out the answer. Bernoulli's equation tells us about the energy density of air. Each pascal of pressure is one joule of energy per cubic meter. If we multiply the in-skirt pressure of the hovercraft by the volume of flow per second, we should get the cost in additional power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1,185 Pa * 1.42 m^3/s = 1,683W&lt;/p&gt;&#xA;" OwnerUserId="16660" LastActivityDate="2019-03-17T01:25:27.070" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26415" PostTypeId="1" AcceptedAnswerId="26488" CreationDate="2019-03-17T03:55:01.420" Score="1" ViewCount="105" Body="&lt;p&gt;I recently started as designer at a making custom signage (small signs viewed up close, so there are lots of sizing limitations) and none of the other designers have any solid rules to rely on for building these signs when they are light through light diffusing acrylic. Often signs will look patchy, with individual LED's being distinguishable through the acrylic, due to the variation in lighting, but not the consistency of the acrylic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any established equations/relationships between the light diffusing properties of a material and the resulting scattering and intensity of the light? How can I measure or estimate the light diffusing properties?&#xA;The variables I can currently estimate or measure are: light intensity of the LED's, their spacing, thickness of material its shining through and the resulting light intensity, as well as, grading the resulting light pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the issues we face is that our supplier does not send a consistent opacity of light diffusing acrylic, so a thickness that would normally look fine, will now either not let enough light through, or it will look patchy (because of point source LED's.) To be clear, the variation is batch to batch, not on the same sheet.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ideally, these relations/equations would help me understand the&#xA;  difference in thicknesses required. Any starting point would be helpful.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="8539" LastEditorUserId="8539" LastEditDate="2019-03-23T19:19:58.630" LastActivityDate="2019-03-27T21:38:16.273" Title="Relations/Equations to do with light diffusing materials" Tags="|optics|lighting|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26419" PostTypeId="2" ParentId="26376" CreationDate="2019-03-17T16:46:13.787" Score="0" Body="&lt;p&gt;You just need a two loops control strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An inner speed loop will simply set PWM needed to achieve requested rotational speed .&#xA;This unfortunately needs a speed transducer on the motor spindle but it's definitely worth.&#xA;The input of this sub-system is requested speed coming from outer position loop &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer position loop processes requested position set point and position feedback and feeds inner loop. This is basically what you've already implemented .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This approach to position control roots back to the very beginning of control theory, it shouldn't be hard to Google much more details &lt;/p&gt;&#xA;" OwnerUserId="7668" LastActivityDate="2019-03-17T16:46:13.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26421" PostTypeId="1" AcceptedAnswerId="26427" CreationDate="2019-03-17T18:19:22.263" Score="3" ViewCount="238" Body="&lt;p&gt;I have been reading &lt;a href=&quot;https://www.mdpi.com/1996-1073/12/2/272/htm#B22-energies-12-00272&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; paper, which addresses the advantages of diffusers in cross flow turbines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The abstract says (emphasis is mine):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A CP’s extraordinary improving resulted when yaw increased up to 22.5° for the hydrofoil shaped and up to 30° for the &lt;strong&gt;symmetrical diffuser&lt;/strong&gt;. Similar behaviour in yawed flows also occurred in case of a &lt;strong&gt;ducted&lt;/strong&gt; single rotor, demonstrating that it is a characteristic of CFTs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am confused as to the difference between a duct and a diffuser in terms of turbines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, the diffuser sounds similar to a duct.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;a wind turbine modified with a cone-shaped wind diffuser that is used to increase the efficiency of converting wind power to electrical power. The increased efficiency is possible due to the increased wind speeds the diffuser can provide. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is, what is the difference between the two? Do they have different functions?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ask because it seems like the author is implying that the two are different.&lt;/p&gt;&#xA;" OwnerUserId="19486" LastActivityDate="2019-11-12T05:59:23.737" Title="What is the difference between a diffuser and a duct in turbines?" Tags="|power|turbines|energy|energy-efficiency|diffusion|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26422" PostTypeId="1" AcceptedAnswerId="26423" CreationDate="2019-03-17T21:02:23.833" Score="0" ViewCount="32" Body="&lt;p&gt;I was looking in google, but I did not find any information on how the drilling was done, I would appreciate it if someone who knew how it was answered. As typical it could be an axis of 200 [mm] in diameter and a hole of 24 [mm]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="19749" LastActivityDate="2019-03-17T21:35:00.007" Title="How is the drilling done along the axis for a typical synchronous motor shaft?" Tags="|manufacturing-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26423" PostTypeId="2" ParentId="26422" CreationDate="2019-03-17T21:35:00.007" Score="2" Body="&lt;p&gt;Holes or threaded holes can be done on a lathe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are bored or machined using drill bits to start then a cutting tool or boring bar to reach the final internal diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a video of a boring bar being made to give you an idea : &lt;a href=&quot;https://www.youtube.com/watch?v=P8kmZ-9SjRA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=P8kmZ-9SjRA&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-03-17T21:35:00.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26424" PostTypeId="2" ParentId="23021" CreationDate="2019-03-17T22:00:30.107" Score="1" Body="&lt;p&gt;Use a filter as a noise gate. Your signal or data with it phase reversed will drive the filter the filter in turn drives a vca. The original signal is being passed through the filter proper. Then you will balance the difference in amplitude or and phase of either or both signals to remove the noise. This has the least buffer and lag time I think. Also is this in hardware anolog, digital or programming? Should be no noise in the first place? IO streams or is this a type of tap? Maybe to look into something without cross talk.Fortran? Go back to your hardware to see how it is working with your ...data?&lt;/p&gt;&#xA;" OwnerUserId="19750" LastEditorUserId="19750" LastEditDate="2019-03-17T22:17:02.787" LastActivityDate="2019-03-17T22:17:02.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26425" PostTypeId="2" ParentId="25998" CreationDate="2019-03-18T05:06:17.543" Score="1" Body="&lt;p&gt;For this kind of system we can assume that pure rolling happens at the point of contact between wheel and the surface. It means that there is no slip between the wheel and the surface so we can take the system to be at zero velocity at the point of contact. For calculation purpose we take this to be a reference point and calculate kinetic energy and potential energy about this point. &lt;/p&gt;&#xA;" OwnerUserId="19753" LastActivityDate="2019-03-18T05:06:17.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26426" PostTypeId="2" ParentId="25998" CreationDate="2019-03-18T07:13:17.507" Score="0" Body="&lt;p&gt;The disk is rotating without slippage around its center. Ans its angular moment of inertia is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ I\  =1/2 mr^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But potential energy of springs, is 1/2 square their displacement times K, which for small angle is horizontal and moves at the point of connection to the disk at height of a+r. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for the strain energy of the springs the statement&#xA;&lt;span class=&quot;math-container&quot;&gt;$$s= (a+r)\theta   $$&lt;/span&gt;&#xA;And &lt;span class=&quot;math-container&quot;&gt;$ \ U=2[\frac{1}{2}kx^2]=\ k(a+r)^2\theta^2 $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine the springs were connected to a point even outside of the disk 3 times its radius via a weightless lever. still the S would be &lt;span class=&quot;math-container&quot;&gt;$ \ 4r\cdot\theta \ $&lt;/span&gt; and would not effect the &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; of the disk and its kinetic energy.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-03-18T07:25:33.057" LastActivityDate="2019-03-18T07:25:33.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26427" PostTypeId="2" ParentId="26421" CreationDate="2019-03-18T08:57:44.180" Score="2" Body="&lt;p&gt;A definition of a diffuser is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The  (divergent) diffuser is a duct so shaped that the fluid flowing through it decelerates, the pressure increasing from inlet to outlet.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from Engineering Thermodynamics Work &amp;amp; Heat Transfer, Rogers &amp;amp; Mayhew, page 38.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A duct is assumed to have neither a convergent or divergent shape.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-03-18T08:57:44.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26428" PostTypeId="1" CreationDate="2019-03-18T16:16:41.933" Score="0" ViewCount="27" Body="&lt;p&gt;What thickness of stainless steel and grade will I need for a step cover that is 30.25&quot; by 14&quot;.  One long side will have a1&quot; bend lip and the three other sides will rest on a solid flat surface.&lt;/p&gt;&#xA;" OwnerUserId="19756" LastActivityDate="2019-03-18T16:16:41.933" Title="Product engineer" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26429" PostTypeId="1" AcceptedAnswerId="26464" CreationDate="2019-03-18T17:22:43.917" Score="1" ViewCount="396" Body="&lt;p&gt;I have a standard non-linear static problem to solve with FEA. I am applying displacement boundary conditions only, no external force. The equation to solve  look like these:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K(u)u=R(u)$&lt;/span&gt;  Eq(1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms &lt;span class=&quot;math-container&quot;&gt;$K, R$&lt;/span&gt; are the stiffness matrix and residual vector(in this case internal force vector), in any standard FEA formulation, &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; is the field variable(displacement). The relationship between &lt;span class=&quot;math-container&quot;&gt;$K, R $&lt;/span&gt; can be given as per standard FEM definitions, &lt;span class=&quot;math-container&quot;&gt;$\partial F/ \partial U=R$&lt;/span&gt;(obtained from Taylor series expansion).  Due to the non-linear nature, the equations cannot be solved directly, an incremental iterative approach is needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused on how to set up the NR algorithm. All examples I see are breaking up the external load into steps, and run iterations in each step to check for the convergence &lt;span class=&quot;math-container&quot;&gt;$Residual=External Force-Internal Force= 0$&lt;/span&gt;. But I don't have an external force vector, then how should I proceed with the incremental steps? Moreover, in this case, &lt;span class=&quot;math-container&quot;&gt;$External Force=0$&lt;/span&gt; so how should the convergence be checked in each iteration step?  &lt;/p&gt;&#xA;" OwnerUserId="15803" LastEditorUserId="15803" LastEditDate="2019-03-18T17:30:08.630" LastActivityDate="2019-03-20T23:03:55.130" Title="FEA: Newton-Raphson algorithm for Dirichlet BC non-linear static analysis" Tags="|mechanical-engineering|applied-mechanics|finite-element-method|modeling|numerical-methods|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="26430" PostTypeId="1" CreationDate="2019-03-18T17:30:57.673" Score="0" ViewCount="29" Body="&lt;p&gt;I’m designing a machine which has an actuator on a carriage sliding on two round linear guide shafts. I got an idea of transmitting electricity to the actuator through the shafts. There would be carbon brushes on the carriage against the shafts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is the selection of the linear bearings. The carbon brushes will wear and some graphite residue will stick on to the shafts. I wonder which bearing type would be best here. There are ball circulating bearings, plastic sliding bearings and self lubricating bronze bushings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some self lubricating bushings use graphite as lubricant so that seems like an obvious choice but I’m not sure how they would work here. On the other hand plastic sliding bearings like Igus claim to work even with sand on the guides. The ball circulating bearings seem like the worst choice in my opinion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So which bearing would be best in here?&lt;/p&gt;&#xA;" OwnerUserId="19757" LastActivityDate="2019-03-18T17:30:57.673" Title="Using carbon brushes on linear guides" Tags="|linear-motion|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26431" PostTypeId="1" CreationDate="2019-03-18T17:37:09.223" Score="1" ViewCount="90" Body="&lt;p&gt;first of all i am must say that i don't know much about cad systems .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i am a software developer and i was given a task to make a webgl application that show some products in 3d .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the product is composed from multiple parts and i was given each part in a separate &quot;STEP&quot; file .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;converting these file in other format like obj, fbx etc and showing them in 3d is not a problem&#xA;but how to position / rotate each part(file) in 3d space accordingly to other parts ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is there any option in this cad system specification that can help on that ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks&lt;/p&gt;&#xA;" OwnerUserId="19758" LastActivityDate="2023-12-24T20:07:00.900" Title="multi parts step files" Tags="|automotive-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26432" PostTypeId="1" CreationDate="2019-03-18T20:07:06.333" Score="2" ViewCount="163" Body="&lt;p&gt;I have spanned ~15' with 2x 10mm tempered glass panels.  They're connected in the middle with a 1.5&quot; x 1.5&quot; 16-gauge, 316 stainless steel post filled with concrete.  This post has a 316 base welded on and this is securely bolted to structure below the floor.  My spans of 88&quot; are well within the specs of both the glass and the glass clamps, per their respective manufacturers.  You can see the setup in this photo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CwRz5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CwRz5.jpg&quot; alt=&quot;Balcony Post Detail&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;base&quot; footer on the post is ornamental only (shown with the green arrow).  I did not want to introduce any more holes into my waterproofing, so I epoxied some screw heads on it to appear functional.  But, in reality, it's just sitting there.  Either way, the base of the post is solid and cannot be moved.  A 9-layer waterproofing was then installed on top of the post's base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other ends of these glass panels are into structural elements of the house.  The one end is into wood studs, and the other end is into an 11-gauge 316 post, again filled with concrete, that is attached to structure at the top and the bottom.  Both ends are holding solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uKc9B.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uKc9B.jpg&quot; alt=&quot;Balcony in full&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The glass panels have already survived a 60mph wind gust storm without any issues.  I was watching it through the window and I couldn't see it move at all.  So, I'm happy about that.  Also, if I lean on the pole away from the house, it does not budge.  Same thing if I pull on it towards the house.  It's completely sturdy and does not move.  The problem starts if I start moving the top of the pole back-and-forth (shown with the blue arrow in the first picture), away and then towards the house.  Just moderate pressure from my hand forward and backwards will start the post and both glass panels to sway (swing, vibrate, undulate) back-and-forth.  This is what I'd like to reduce and/or eliminate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've come up with some ideas on how to tackle this problem, and I've even discounted one idea.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Since the bottom of the post is stable and does not sway, I have to assume the sway is from the deflection of the tube steel.  So, I considered a simple, straight piece of 1/4&quot; or 3/8&quot; flat bar steel (piece A in red below), sistered to the post at the top and bottom.  My rationale here is that since the bottom is sturdy and the flat bar steel will provide rigidity and strength and will not deform, that this flat bar would provide strength to the post.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Again in thinking this sway or vibration is caused by the metal bending away from the base, I wanted to add additional support from the structure of the house.  Since this is a view-facing balcony, I want to block the view as little as possible, so I came up with the following cane-like design that would be made of flat bar steel and bolt into the post near the top, and then bolt into the side of the house (piece B in orange below).  The strength of the flat bar should then remove the motion of the post, like a flying buttress.  This allows me to add structure without puncturing the floor's waterproofing.  The post and glass is about 1-1/2&quot; from the edge of the balcony.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tear this all up and use an 11-gauge post.  Obviously this is my last resort.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do nothing and live with the sway.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do something else.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cdf0L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cdf0L.png&quot; alt=&quot;Solution option designs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TQ0kP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TQ0kP.jpg&quot; alt=&quot;Solution #1 and Solution #2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also considered that the post was creating a wave effect between the glass panels.  I then considered a stabilizer between the glass panels, but I tested this with a piece of flat bar steel that I clamped onto both pieces of glass.  This did nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Can you please let me know which option, 1-5, they think I should choose and why?&lt;/strong&gt;  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="18528" LastEditorUserId="18528" LastEditDate="2019-03-18T23:45:50.973" LastActivityDate="2019-03-19T04:27:43.453" Title="Need Help to Remove Sway from Steel Post + Glass Panels" Tags="|structural-engineering|steel|structures|vibration|glass|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26433" PostTypeId="2" ParentId="26431" CreationDate="2019-03-18T21:12:37.500" Score="0" Body="&lt;p&gt;A STEP file's orientation depends on two things:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;How it was modeled in its originating CAD software. The designer is the one who decides how to orient the model. When exported to STEP, the model should keep the original orientation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Whether said CAD software considers the Z axis as &quot;up.&quot; While AFAIK a great majority of CAD programs do consider Z as &quot;up&quot;, and XY as the floor, SolidWorks does not. In SolidWorks, XY is the front plane, so the Y axis is &quot;up&quot;.  That's a &quot;feature&quot; in their eyes.  SolidWorks is the most used parametric solid modeler in the world, so many STEP files out there were made in it. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In short, they should already be oriented in the same 3D space.  There is nothing more to do except perhaps rotate the parts' axes if it came from SolidWorks.  Good luck!&lt;/p&gt;&#xA;" OwnerUserId="18528" LastActivityDate="2019-03-18T21:12:37.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26434" PostTypeId="2" ParentId="26432" CreationDate="2019-03-18T23:09:49.927" Score="1" Body="&lt;p&gt;You may not see it but the vibration is slowly loosening the base support, compromising waterproofing, cracking the slab and causing many other damages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you may have seen in glass guard rails which have been engineered and permitted, they usually either have a wedge, or fin, of same thickness glass, say 6&quot; wide perpendicular to the post bolted to the post or a chrome ornate buttress, like a 3/4&quot; pipe, angled from the top of the post to the deck anchored to the deck at 6&quot; set-back from the glass or at least a 4&quot;x 4&quot; post. 1.5&quot; support is really too springy for this kind of application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your suggestion of adding an extra bar, or post will not be very helpful; it will add to the strength of the post but not to its rigidness. it will be like the leaf spring in car suspension, soft against bumps, but resilient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would think to add the chrome pipe as buttress is the easiest. the only problem is you need to drill a hole in the slab and caulk it with epoxy waterproofing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I say should not be considered professional advice by any means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the US any work done on a project that can affect the safety of occupants needs a permit from the pertinent governing agency.&#xA;e.g, in California the railing must meet minimum required lateral uniform and concentrated load; be 42 inches high. Any tempered glass is required to be approved and show the manufacturers name and info printed permanently, etc.... You need to check with your building official for your particular case. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-03-19T04:27:43.453" LastActivityDate="2019-03-19T04:27:43.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26435" PostTypeId="2" ParentId="26432" CreationDate="2019-03-19T00:05:38.787" Score="1" Body="&lt;p&gt;Make the post go from floor to ceiling just like the one on the corner. It won't sway.&lt;/p&gt;&#xA;" OwnerUserId="16544" LastActivityDate="2019-03-19T00:05:38.787" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26441" PostTypeId="1" CreationDate="2019-03-19T12:12:11.627" Score="0" ViewCount="16" Body="&lt;p&gt;I'm running a linear elastic solver generating the tetrahedron's using tetGen. It's not clear to me though how one defines what's an acceptable tetra density that provides sufficient accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that in the picture below the maximum displacement drastically decreases alongside an increase in tetra count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a much as my machine is able to process but I assume it would converge around 90k tetrahedron. Is it at that point that we can consider the results to be accurate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone provide a guideline that can be translated programmatically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DmCDi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DmCDi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19769" LastEditorUserId="19769" LastEditDate="2019-03-19T12:17:12.530" LastActivityDate="2019-03-19T12:17:12.530" Title="What's the methodology for adequate tetrahedralisation?" Tags="|materials|finite-element-method|simulation|elastic-modulus|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26442" PostTypeId="1" AcceptedAnswerId="26444" CreationDate="2019-03-19T13:19:50.497" Score="1" ViewCount="130" Body="&lt;p&gt;I am currently working on developing a simulation for a restricted UAV, and I’m having some issues with Simulink while doing this. I have my mathematical equations which have been approved by my professor, but while I’m trying to simulate the system I am running into strange issues. Moreover, neither my professor nor my TA can figure out what is causing this issue, despite multiple private sessions to troubleshoot. If anyone can shed some light on the issue it would be highly appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Firstly, the equation I’ve modeled on Simulink is this:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Fxxg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Fxxg.png&quot; alt=&quot;equation 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9ZtJh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ZtJh.png&quot; alt=&quot;diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My Simulink block diagram is above. Input 1 Is Fb+Ff which, in this scenario = 6.8. Input 2 is theta which = 0.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have annotated the values that should be and are at each of the nodes before the summing junction. In this section of the block diagram the theoretical value and the simulation agree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem comes in the more complex half of the diagram. The graph I am validating should look like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UPdzV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UPdzV.png&quot; alt=&quot;graph1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an initial rise and then oscillations about a non-zero number. Obviously, my graph will be ideal so oscillations will not persist like they do in the graph above.&#xA;My final graph is this…&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0U7j1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0U7j1.png&quot; alt=&quot;graph2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where the y-axis is elevation angle in degrees and the x-axis is in seconds. Obviously these two graphs are in no way similar. In an attempt to analyze where I could have made an error in my block diagram, I analyzed the behavior at each node. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Immediately after the cos block to the lower right I get oscillating behavior&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DEKuo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DEKuo.png&quot; alt=&quot;graph 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Immediately after the summing junction and gain (1/0.91) I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2KcID.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2KcID.png&quot; alt=&quot;graph 4&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above bears a resemblance to the graph I’m validating against.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the first integrator, however, I get this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mWhEX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mWhEX.png&quot; alt=&quot;graph 5&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding the two graphs together, should I not be getting a graph that rises and then oscillates about a non-zero number? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have I made a mistake in my block diagram somewhere? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The only value I’m allowed to change is the gain marked &lt;em&gt;De&lt;/em&gt;.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I am starting to hit a wall and really need some insight on this. Surely this cant be something wrong with the Simulink source code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nothing I do gets me anywhere near the validation graph. Can someone please point me in the right direction or let me know if you see any errors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any insight is highly appreciated, thank you for reading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Should you want to view the Simulink file for further analysis, it is available for download here: &lt;a href=&quot;https://ufile.io/08aqv&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ufile.io/08aqv&lt;/a&gt;&lt;/strong&gt; However please note that the attached file is my entire Simulink project. The circuit reference in the question above is contained in the subsystem &quot;elevation&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11039" LastActivityDate="2019-03-19T14:43:28.823" Title="Strange simulink error. Professor &amp; TA don't know. Any Ideas?" Tags="|mechanical-engineering|electrical-engineering|control-engineering|aerospace-engineering|simulink|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26443" PostTypeId="1" AcceptedAnswerId="26447" CreationDate="2019-03-19T13:32:40.697" Score="1" ViewCount="240" Body="&lt;p&gt;I'm a computational person, and I don't quite understand how rotational DoFs are manifested in the linear elasticity equations. I understand the translational DoFs are simply the displacement components. When specifying boundary conditions for linear elasticity simulations, we typically specify the displacement and/or traction boundary conditions at the boundaries. Where do the rotational boundary conditions come in? Are they manifested in the traction boundary conditions? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not very intuitive to me and I was hoping someone can explain.&lt;/p&gt;&#xA;" OwnerUserId="19770" LastActivityDate="2019-03-21T02:53:23.217" Title="Rotational and translation DoFs in structural mechanics" Tags="|structural-engineering|structural-analysis|solid-mechanics|traction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26444" PostTypeId="2" ParentId="26442" CreationDate="2019-03-19T14:43:28.823" Score="4" Body="&lt;p&gt;Your results are correct. I have solved the differential equations in Octave using the &lt;code&gt;ode&lt;/code&gt; solver and I get the same results:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function dX = elevation(t,X)&#xA;  g = 9.81;&#xA;  m_cw = 1.87;&#xA;  l_w = 0.47;&#xA;  m_uav = 1.17;&#xA;  l_a = 0.66;&#xA;  J_e = 0.91;&#xA;  D_e = 0.04;&#xA;  F_in = 0.68;&#xA;  theta = 0;&#xA;&#xA;  dX(1) = X(2);&#xA;  dX(2) = (F_in * l_a * cos(theta) + (m_cw*l_w - m_uav*l_a)*g*cos(X(1)) - D_e * X(2)) / J_e;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which gets saved as &lt;code&gt;elevation.m&lt;/code&gt; and solved by:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;tspan = [0 60];&#xA;X0=[0,0];&#xA;opts = odeset('MaxStep',0.1,'InitialStep',0.1);&#xA;[t,X] = ode45(@elevation,tspan,X0,opts);&#xA;plot (t,X(:,1))&#xA;grid&#xA;xlabel('time')&#xA;ylabel('elevation [rad]')&#xA;title ('elevation vs. time')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and which gives the following results:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cZUkc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cZUkc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the only possible explanations are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The equations you use are incorrect (unlikely because you say they have been checked)&lt;/li&gt;&#xA;&lt;li&gt;The initial conditions of the differential equations are incorrect or do not match those used to get the validated results (assumed to be zero for both elevation and elevation rate at t=0). For example, it looks like the initial condition for the elevation in the validated graph is about -30 deg or so.&lt;/li&gt;&#xA;&lt;li&gt;The numerical values for the various parameters are incorrect or do not match those used to get the validated results&lt;/li&gt;&#xA;&lt;li&gt;The inputs used (&lt;code&gt;Fb+Ff&lt;/code&gt; and &lt;code&gt;theta&lt;/code&gt;) do not match those used to get the validated results.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;To me, #4 seems like the most plausible explanation. You have assumed both inputs to be constant. Is that really the case? Are they not a function of time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried changing the value of &lt;code&gt;D_e&lt;/code&gt; as you suggest and you do get some quite different results. For example, this is with &lt;code&gt;D_e = 0.2&lt;/code&gt; instead of &lt;code&gt;0.04&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hvYwA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hvYwA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-03-19T14:43:28.823" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26447" PostTypeId="2" ParentId="26443" CreationDate="2019-03-19T15:41:11.073" Score="1" Body="&lt;p&gt;In the general theory of 3D elasticity, there is no need for rotation variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However many engineering structures such as shells and beams have one or two dimensions that are small compared with the others. Representing the behaviour through the &quot;thin&quot; dimension as the &lt;em&gt;difference&lt;/em&gt; between the translational displacements at the &quot;top&quot; and &quot;bottom&quot; surfaces is very badly conditioned numerically. To overcome that problem, the behaviour can represented &lt;em&gt;approximately&lt;/em&gt; as the translation of the mid-plane (or more generally, the &quot;neutral axis&quot; of the object), and the slope (or rotation) of the cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example of this type of approximation is &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler-Bernouilli beam theory&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formulation for shells and plates are similar in principle, but more complicated because the shell itself may be curved in one or both directions (e.g. part of a cylinder or a sphere). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction &quot;force&quot; corresponding to a constrained rotation variable is a &lt;em&gt;moment&lt;/em&gt; - i.e. the finite value of the product &lt;span class=&quot;math-container&quot;&gt;$Fd$&lt;/span&gt; when two equal and opposite forces &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; are separated by a distance &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;, in the limit as &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; increases to infinity and &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; decreases to zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These approximations are useful in most engineering situations because of &lt;a href=&quot;https://engineering.stackexchange.com/questions/16666/what-is-saint-venant-principle&quot;&gt;St. Venant's principle&lt;/a&gt;, which says that the &lt;em&gt;overall&lt;/em&gt; response of a structure does not depend on the exact &lt;em&gt;details&lt;/em&gt; of how the loads are applied, so long as the different applied loads are statically equivalent.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-03-19T15:41:11.073" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26449" PostTypeId="2" ParentId="23398" CreationDate="2019-03-19T17:26:33.497" Score="1" Body="&lt;p&gt;Street lights are installed in order to provide a minimum lighting level along the roadway. Some municipalities or states or countries may have minimum lighting standards, but these often vary depending on the context (i.e. rural, urban, intersections).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For certain situations, such as lighting for mid-block or pedestrian crossings, there may be additional requirements beyond merely lighting level. Putting the lights so they illuminate the front of a pedestrian in a crosswalk provides better visibility than placing the lights overhead or behind the pedestrians, so some newer standards may require lighting be placed accordingly so they are in front of the crosswalk on the approach legs to a crosswalk or intersection.&lt;/p&gt;&#xA;" OwnerUserId="2606" LastActivityDate="2019-03-19T17:26:33.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26450" PostTypeId="2" ParentId="22345" CreationDate="2019-03-19T17:36:13.477" Score="3" Body="&lt;p&gt;In the US this is commonly referred to as the Right-of-Way (property line to property line on either side of the roadway). Or, if the roadway right-of-way is wider than needed for construction, it may be referred to as the limits of construction.&lt;/p&gt;&#xA;" OwnerUserId="2606" LastActivityDate="2019-03-19T17:36:13.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26451" PostTypeId="1" CreationDate="2019-03-19T17:51:21.163" Score="0" ViewCount="45" Body="&lt;p&gt;I'm writing a discussion paper on the potential energy efficiency benefits of integrating different manufacturing processes by co-location on the same site. One possible field for integration is in the coupling of motors to loads. Motors are not my field, but I know that motor efficiency (within a range of speeds) increases up to about 150 kW. Would it be more efficient for some processes (loads) to be driven by one powerful motor, operating at low torque but differentially connected to loads via efficient transmissions, than using multiple lower rated (and therefore less efficient) motors?  Obviously the efficiency gains would need to be greater than total transmission losses!&lt;/p&gt;&#xA;" OwnerUserId="19775" LastActivityDate="2019-08-16T21:51:10.880" Title="Would some manufacturing processes be more efficient if driven differentially by a single motor, up to ~150 kW, than multiple lower-rated motors?" Tags="|motors|torque|manufacturing-engineering|energy-efficiency|transmission|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26452" PostTypeId="1" AcceptedAnswerId="26453" CreationDate="2019-03-19T18:08:50.470" Score="0" ViewCount="2562" Body="&lt;p&gt;So, I have a 1,000 pounds load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the required torque in inch/pounds to lift that load for a motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the motor is installed at the center of the radius generating that torque, does it matter how long the radius is if the end requirement still 1,000 pounds load?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the answer be 1,000 in-lbs? &lt;/p&gt;&#xA;" OwnerUserId="19777" LastActivityDate="2019-03-19T19:00:31.557" Title="How much torque do I need to lift 1,000 pounds?" Tags="|torque|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26453" PostTypeId="2" ParentId="26452" CreationDate="2019-03-19T19:00:31.557" Score="1" Body="&lt;p&gt;Any torque and power demand greater than your motor rating will cause it to stall.&#xA;for a 1000lbs.inch motor the pully's radius r must be, &lt;span class=&quot;math-container&quot;&gt;$$r &amp;lt; 1 \ \text{inch}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise you need a gear. Also the motor must have at least the power of,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P= 9.8  \cdot 1000\cdot 0.45 \ lbs/kg \cdot0.305 \ ft/meters=1345 \ \text{watts}$$&lt;/span&gt;&#xA;This is if you want to lift it in 1 second. If you want to lift it in 10 seconds then the power will be 134.5 watts, but the required torque  will still be the same.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-19T19:00:31.557" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26455" PostTypeId="2" ParentId="26451" CreationDate="2019-03-19T19:43:58.763" Score="0" Body="&lt;p&gt;Some pros and cons of distributing power mechanically:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Firms switching to electric power showed significantly less employee sick time, and, using the same equipment, showed significant increases in production.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Line_shaft&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Line_shaft&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-03-19T19:43:58.763" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26461" PostTypeId="1" CreationDate="2019-03-20T02:28:30.410" Score="0" ViewCount="194" Body="&lt;p&gt;Blocks A and B weigh 30lb and 15lb respectively, and they are both at a height 8ft above the ground when the system is released from rest. Assume the pulley is frictionless. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Determine the speed of the block A just before hitting the ground. &lt;/li&gt;&#xA;&lt;li&gt;Determine the tension of the cord during the motion.  &lt;/li&gt;&#xA;&lt;li&gt;Determine the maximum height reached by block B&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UznPc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UznPc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far for part 1 I've tried to apply the conservation of energy.&#xA;Using mgh only for the initial because velocity is 0 and the kinetic energy of both blocks, plus the potential energy of block B at height 16ft for final.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This creates:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(30+45)8=\frac{1}{2}(\frac{30}{32.2})v^{2}+\frac{1}{2}\frac{15}{32.2}v^{2}+15(16)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which simplifies to &lt;span class=&quot;math-container&quot;&gt;$v=13.10 ft/s$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For part 2,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using &lt;span class=&quot;math-container&quot;&gt;$T_{1}+U_{1-2}=T_{2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Rightarrow T_{1}+(W_{A}-F_{A})h=T_{2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$= F_{A}= W_{A}-\frac{T_{2}}{h}=30-\frac{(.5)(.932)(13.10)^2}{8}=20.003lb$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would use the same process for the tension on B.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to my actual question, is this the correct and easiest way to approach this problem? I don't have access the the solution and am looking for some feedback.&lt;/p&gt;&#xA;" OwnerUserId="18587" LastEditorUserId="110" LastEditDate="2019-03-20T10:31:42.307" LastActivityDate="2019-03-20T17:52:53.653" Title="Dynamics work and energy problem" Tags="|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26463" PostTypeId="2" ParentId="24752" CreationDate="2019-03-20T03:31:58.997" Score="1" Body="&lt;p&gt;So, the difference between both is their working and the mode of excitation. the bender piezo shall be loaded in cantilever (better for application with higher loads) less accuracy. While the disk transducer is better accuracy with lower stroke.&#xA;I feel like this is correct. Look at the product specifications to be sure. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="19783" LastActivityDate="2019-03-20T03:31:58.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26464" PostTypeId="2" ParentId="26429" CreationDate="2019-03-20T03:51:06.440" Score="2" Body="&lt;p&gt;The discretized problem that you are trying to solve is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \text{find}\,\, \mathbf{u}^{n+1} \,\, \text{such that} \,\, \mathbf{r}(\mathbf{u}^{n+1}) = \mathbf{0} \,\, \text{subject to the constraints} \,\, \mathbf{g}(\mathbf{u}^{n+1}) = \mathbf{0}&#xA;$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$\mathbf{u}^{n+1}$&lt;/span&gt; is the displacement at the &lt;strong&gt;end of load step&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; and the residual is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \mathbf{r}(\mathbf{u}^{n+1}) := \mathbf{f}^{\text{int}}(\mathbf{u}^{n+1}) - \mathbf{f}^{\text{ext}}(\mathbf{u}^{n+1}) \,.&#xA;$$&lt;/span&gt;&#xA;In the &lt;strong&gt;absence of external forces&lt;/strong&gt;, we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \mathbf{r}(\mathbf{u}^{n+1}) = \mathbf{f}^{\text{int}}(\mathbf{u}^{n+1}) \,.&#xA;$$&lt;/span&gt;&#xA;Therefore, at the end of the load step, the displacements should be such that the internal forces are zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's method can be used to find the values of &lt;span class=&quot;math-container&quot;&gt;$\mathbf{u}^{n+1}$&lt;/span&gt; at which &lt;span class=&quot;math-container&quot;&gt;$\mathbf{r} = \mathbf{0}$&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We start with the solution at the &lt;strong&gt;beginning&lt;/strong&gt; of load step &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \mathbf{u}^{n+1}_0 = \mathbf{u}^n&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;strong&gt;Caveat&lt;/strong&gt;: For implicit dynamic computations with the Newmark-&lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; method, a better initial guess is &lt;span class=&quot;math-container&quot;&gt;$\mathbf{u}^{n+1}$&lt;/span&gt;.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the iterations proceed, we will get better and better estimates of the quantity &lt;span class=&quot;math-container&quot;&gt;$\mathbf{u}^{n+1}_k$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is the &lt;strong&gt;iteration number&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Taylor series expansion of the residual about the current value of the displacement leads to&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;  \mathbf{0} = \mathbf{r}(\mathbf{u}^{n+1}_k) = \mathbf{r}(\mathbf{u}^{n+1}_{k-1}) + \frac{\partial \mathbf{r}(\mathbf{u}^{n+1}_{k-1})}{\partial \mathbf{u}} \cdot (\mathbf{u}^{n+1}_{k} - \mathbf{u}^{n+1}_{k-1}) + \dots&#xA;$$&lt;/span&gt;&#xA;Reorganizing, we get&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \mathbf{u}^{n+1}_{k} = \mathbf{u}^{n+1}_{k-1} - \left[\frac{\partial \mathbf{r}(\mathbf{u}^{n+1}_{k-1})}{\partial \mathbf{u}}\right]^{-1} \cdot \mathbf{r}(\mathbf{u}^{n+1}_{k-1})&#xA;$$&lt;/span&gt;&#xA;In the absence of external forces, we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA; \begin{align}&#xA; \mathbf{u}^{n+1}_{k} &amp;amp;= \mathbf{u}^{n+1}_{k-1} - \left[\frac{\partial \mathbf{f}^{\text{int}}(\mathbf{u}^{n+1}_{k-1})}{\partial \mathbf{u}}\right]^{-1} \cdot \mathbf{f}^{\text{int}}(\mathbf{u}^{n+1}_{k-1}) \\&#xA;  &amp;amp; = \mathbf{u}^{n+1}_{k-1} - \left[\mathbf{K}^{\text{int}}\right]^{-1} \cdot \mathbf{f}^{\text{int}}(\mathbf{u}^{n+1}_{k-1})&#xA; \end{align}&#xA;$$&lt;/span&gt;&#xA;The quantity &lt;span class=&quot;math-container&quot;&gt;$\mathbf{K}^{\text{int}}$&lt;/span&gt; is called the &lt;strong&gt;tangent stiffness matrix&lt;/strong&gt;.  We continue iterating until a convergence criterion is satisfied - typically some sort of the energy norm computed from force and displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Constraints&lt;/strong&gt;:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I haven't talked about the process of applying the displacement constraints in the above procedure.  There are several ways of applying these constraints.  The most easily understood approach is to use Lagrange multipliers.  A more common approach is to use a &lt;a href=&quot;https://en.wikipedia.org/wiki/Penalty_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;penalty method&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;strong&gt;Computing the tangent stiffness&lt;/strong&gt;:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The tangent stiffness matrix can be derived in several ways.  For example, in the Belytchko et al. book on nonlinear FE, they proceed using the convected rate of the Kirchhoff stress.  Simo's book on computation inelasticity tends to focus on algorithmically consistent tangent stiffness matrices.  Deriving the tangent stiffness for complex material models and implementing it without errors is one of the major difficulties experience by authors of FE codes.  Very few commercial codes get everything right.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="3488" LastEditDate="2019-03-20T23:03:55.130" LastActivityDate="2019-03-20T23:03:55.130" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26465" PostTypeId="1" CreationDate="2019-03-20T04:05:08.510" Score="0" ViewCount="242" Body="&lt;p&gt;What is the best approach to fix the pockets seen below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q2G4L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q2G4L.jpg&quot; alt=&quot;wall_air_pocket&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/g2IFI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g2IFI.jpg&quot; alt=&quot;hole&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/JHLeQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JHLeQ.jpg&quot; alt=&quot;some_other_pocket_in_same_spot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pockets were due to the stones in place which did not allow to pass the concrete material, and later on gravity made it's way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Options on my head:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fill with liquid cement, although it won't be as tough.&lt;/li&gt;&#xA;&lt;li&gt;Fill with concrete, can't guarantee it will make it into every spot of the air pocket, nor sure how to do it well without failing miserably.&lt;/li&gt;&#xA;&lt;li&gt;Fill with liquid cement and little pieces of rock (one of the 3 components that concrete uses) so it can fill as much as possible and the pieces of rock can fill the big part.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cutting down the affected parts are not an option. &lt;/p&gt;&#xA;" OwnerUserId="19784" LastEditorUserId="1832" LastEditDate="2019-03-20T16:06:04.423" LastActivityDate="2019-03-20T16:06:04.423" Title="What is the best approach to fix empty pockets in a concrete slab/wall" Tags="|civil-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26466" PostTypeId="1" CreationDate="2019-03-20T06:01:07.870" Score="0" ViewCount="55" Body="&lt;p&gt;Can someone please help in finding out the phase proportion of each phase w.r.t time , cooling rate and temperature by using (mathematical model to predict phases from given cct diagram)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Leblond method&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="17778" LastActivityDate="2019-03-20T06:01:07.870" Title="Phase prediction by cct diagram" Tags="|materials|manufacturing-engineering|metallurgy|welding|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26467" PostTypeId="1" AcceptedAnswerId="26468" CreationDate="2019-03-20T08:33:57.367" Score="1" ViewCount="142" Body="&lt;p&gt;The datasheet of the AD630 mentions that the device can be used as a Precision Phase Comparator or as a Lock-in Amplifier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the difference? They both seem to map the amplitude of a modulated signal to around DC. That is, the output of both circuits is the demodulated signal. (Assuming the reference signal is the modulated signal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would I want to pick one circuit over the other?&lt;/p&gt;&#xA;" OwnerUserId="17095" LastActivityDate="2019-03-20T12:24:16.937" Title="Difference between AD630 'Phase Comparator' and 'Lock-in Amplifier' modes" Tags="|electrical-engineering|modulation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26468" PostTypeId="2" ParentId="26467" CreationDate="2019-03-20T12:24:16.937" Score="0" Body="&lt;p&gt;The same circuit may be used for different purposes depending on what inputs you apply. And you pick depending on what you want to achieve. Amplifiers are very versatile components and you can achieve quite different goals with the same component.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;em&gt;Assuming the reference signal is the modulated signal&lt;/em&gt; is one of possible inputs and then you use the circuit as a demodulator - you get the demodulated signal on output - a simple receiver of modulated signal.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you pass two identical, or very similar signals (modulated or not) which are phase-shifted, you can treat the output as measurement of the phase shift - Phase Comparator - measuring system delay, e.g. round-trip time of echo vs reference in distance measurements using ultrasound sensors.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you apply a noisy signal and the reference input gets attenuated noise plus carrier, you're getting a demodulator that can deal with the noise - just like the first case but filtering high-frequency noise away.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-03-20T12:24:16.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26470" PostTypeId="2" ParentId="26461" CreationDate="2019-03-20T17:14:54.030" Score="1" Body="&lt;p&gt;The acceleration is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F =m\alpha, \quad \text{F is the difference between the 2 masses}, \  (m_1-m_2)g= (m_1+m_2)\alpha \\ \alpha = \frac{(m_1-m_2)g}{(m_1+m_2)}=32(30-15)/(30+15)=32\cdot15/45=32/3\text {ft.s^2}= 10.66$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T_{cord }= m_1g+\alpha =m_2\alpha =30\cdot 10.66=320lbs $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v_{final}=\sqrt{2\alpha H}= 13.05 ft.s$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum height reached by block B is &lt;span class=&quot;math-container&quot;&gt;$16+ 13.05^2/ 2g=16+13.05^2/2*32=18.66ft$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did not explain every step but if you need clarification ask, please.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-03-20T17:52:53.653" LastActivityDate="2019-03-20T17:52:53.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26471" PostTypeId="1" CreationDate="2019-03-20T19:17:31.127" Score="1" ViewCount="32" Body="&lt;p&gt;I need to create a variable 3-7 psi / 0.2-0.5 bar in a low density gas running through some piping. I can calculate the required Cv factor to do this with a valve would be in the 30-200 range. Looking at some valve spec sheets, &lt;a href=&quot;https://www.mcmaster.com/47865k45&quot; rel=&quot;nofollow noreferrer&quot;&gt;a typical valve&lt;/a&gt; to get to the lower end of that range would have an effective orifice size of about 1&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is that the gas is supersonic going through that size orifice. Similarly, if I use a 4&quot; valve (such that the velocity would be reasonable in the connecting pipe) the valve needs to be over 90% closed, making the effective orifice size only an inch anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've thought about just running through a long length of pipe, but its difficult and expensive to make the pressure drop variable. I've also thought about running the gas through a vertical liquid column to create a hydrostatic head on the gas, but if the density is near 1 kg/L, I'd need a 16' column, which would be both expensive and impractical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Has anyone every solved this problem before? Is there a good way to create substantial pressure drops in gases without creating high velocities?&lt;/p&gt;&#xA;" OwnerUserId="2781" LastActivityDate="2020-01-12T00:18:51.390" Title="Create low velocity pressure drop?" Tags="|pressure|valves|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26474" PostTypeId="2" ParentId="26443" CreationDate="2019-03-21T02:53:23.217" Score="0" Body="&lt;p&gt;Just answering the traction part of your question. A traction on the surface of a shell or beam that has a component tangent to the surface is equivalent to forces on the translational DOFs of the nodes plus moments on their rotational DOFs.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/mVIXY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mVIXY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4437" LastActivityDate="2019-03-21T02:53:23.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26477" PostTypeId="1" CreationDate="2019-03-21T06:18:49.550" Score="1" ViewCount="94" Body="&lt;p&gt;I am reading about pure bending from Beer Johnston's mechanics of materials. There, a prismatic member with a plane of symmetry was subjected to equal and opposite couples M and M' acting in that plane. It was said about this member that since the bending moment M is same in any cross section, the member will bend uniformly. I could not understand this.&lt;/p&gt;&#xA;&lt;p&gt;Is it because that stress will be same on each cross section and therefore since same stresses cause same deformation, it will bend uniformly?&lt;/p&gt;&#xA;" OwnerUserId="19791" LastEditorUserId="19906" LastEditDate="2021-08-06T05:51:56.587" LastActivityDate="2021-08-06T05:51:56.587" Title="Does same stress always produce same strain?" Tags="|mechanical-engineering|stresses|solid-mechanics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26479" PostTypeId="2" ParentId="26477" CreationDate="2019-03-21T08:06:47.467" Score="4" Body="&lt;p&gt;Since the bending moment is constant across any section, by the bending equation, the beam will deform as a pure circular arc, i.e. with constant radius of curvature across any small element length, in other words, uniform deformation. This compares directly to the case when &quot;M&quot; varies along the length, then the radius of curvature, in turn the deformation would have been continuously varying.&lt;/p&gt;&#xA;" OwnerUserId="19792" LastActivityDate="2019-03-21T08:06:47.467" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26481" PostTypeId="1" AcceptedAnswerId="26482" CreationDate="2019-03-21T08:52:35.597" Score="0" ViewCount="155" Body="&lt;p&gt;I suspect I know the answer to this, but hoping someone may know something I don't here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way I can determine the specific heat capacity at constant pressure (Cp) of an oil if I know the density, temperature and viscosity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oil I am interested in: &lt;a href=&quot;http://www.bakerprecision.com/neo/75w90rhd.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.bakerprecision.com/neo/75w90rhd.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The company is only small and hasn't got a value of Cp as far as I can tell, I have contacted them. They are unlikely to have tested it in my experience of dealing with them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I know this could only be done by experimental correlation, and there is no universal equation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Is there any theory you know that can estimate the value of Cp of an oil from basic properties without experimentation?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know: &#xA;Dentity, viscosity and surface tension at room temperature. Plus viscosity at various other temperatures. &lt;/p&gt;&#xA;" OwnerUserId="14498" LastActivityDate="2019-03-21T09:57:55.710" Title="Determine Cp of an oil from density and temperature?" Tags="|fluid-mechanics|heat-transfer|fluid|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26482" PostTypeId="2" ParentId="26481" CreationDate="2019-03-21T09:51:38.083" Score="1" Body="&lt;p&gt;There is no practical way to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to estimate specific heats from first principles using computer simulation and quantum mechanics. There are also some semi-empirical &quot;laws&quot; which apply to particular situations, such as materials whose elements have large atomic numbers, (bigger than iron, and therefore irrelevant for lubricating oil) or at very low (cryogenic) temperatures. None of those is applicable to gear oil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measuring specific heat is not difficult - you could do it yourself with a &quot;kitchen table&quot; experiment. Put a known mass (or volume) of oil in a thermally insulated container (e.g. a thermos flask), apply a known amount of heat (for example using an electrical resistor immersed in the oil), and measure the temperature change.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-03-21T09:57:55.710" LastActivityDate="2019-03-21T09:57:55.710" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26484" PostTypeId="1" CreationDate="2019-03-21T11:47:56.370" Score="0" ViewCount="29" Body="&lt;p&gt;Under what conditions would steel be able to be modelled as a 3D SMD model, and under what conditions would the model break down?&lt;/p&gt;&#xA;" OwnerUserId="19796" LastEditorUserId="19796" LastEditDate="2019-03-21T12:10:04.597" LastActivityDate="2019-03-21T12:10:04.597" Title="Why would a SMD (spring mass damper) model of steel break down?" Tags="|structural-engineering|structural-analysis|modeling|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26485" PostTypeId="1" CreationDate="2019-03-21T11:49:02.363" Score="0" ViewCount="48" Body="&lt;p&gt;I'm fascinated by how robot arm nowadays can combine precision with operating speed (e.g. KIRA by Motorized Precision).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can imagine building a robot arm with stepper motors would be pretty precise but (from my experiences with stepper motors) wouldn't be able to deliver the speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Servo motors (from my experience using them in combination with a raspberry pi) would be a bit faster but not fast enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DC Brushed motors are lack position awareness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have experience with brushless motors but I've mainly seen them being used for drones which is quite a different use case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I want to know what kind of motor might in those kinds of industrial robot arms that are able to do the split between speed and precision.&lt;/p&gt;&#xA;" OwnerUserId="19797" LastEditorUserId="9" LastEditDate="2019-03-21T13:42:05.977" LastActivityDate="2019-03-21T17:48:22.090" Title="What kind of electric motors are used to build industrial robot arms?" Tags="|motors|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26487" PostTypeId="1" CreationDate="2019-03-21T13:22:14.157" Score="1" ViewCount="166" Body="&lt;p&gt;In a conventional nuclear reactor the amount of neutrons 'flying around' is maintained with control rods and water is usually used as a moderator. From a certain point of view the control rods keep the nuclear fuel just critical enough. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a molten salt reactor where the fuel is dissolved in the salt irrelevant of this is thorium, plutonium or uranium. And in a reactor where the nuclear fuel is solved into the salt :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is it possible to maintain the mass critical enough in a molten salt reactor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand that they use control rods in the reactor itself, but if the salt is dropped into the safety reservoir below the reactor isn't the mass getting uncontrolled supercritical? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="19798" LastEditorUserId="19798" LastEditDate="2019-03-23T08:22:28.937" LastActivityDate="2021-01-11T15:03:05.343" Title="Critical mass in a molten salt reactor" Tags="|nuclear-technology|nuclear-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26488" PostTypeId="2" ParentId="26415" CreationDate="2019-03-21T14:42:54.307" Score="2" Body="&lt;p&gt;You seem to face a few issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First is to account for the potential to have patchiness in any one your sheets. Theoretical equations for light transmission through a system will be easiest when applied for isotropic materials, not for &quot;patchy&quot; systems. This problem has to be solved at the distributor end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second is to account for the transmission (opacity) for a given thickness in an isotropic sheet. For an isotropic material that is uniformly absorbing light at a constant amount per unit volume, the transmission falls off exponentially with thickness. For a given thickness, the total amount of light that is absorbed is a function of such factors as the concentration of absorber and its absorbing or scattering efficiency. The theoretical formulations are found under themes such as Beer's law for absorption of light. In essence, the light transmission equation is as below, with &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; as the transmitted intensity, &lt;span class=&quot;math-container&quot;&gt;$I_o$&lt;/span&gt; as the incoming intensity, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; as an absorption factor, and &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; as thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{I}{I_o} = \exp(-\alpha t) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation neglects reflection at the interfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third is to account for scattering to cause opacity. In polymers, light scatters from crystalline region. A perfectly glassy polymer is transparent. Uniform scattering comes from uniformly sized, uniformly distributed regions of crystallinity. A rough &quot;off the top of my head&quot; equation is to model the opacity as a rule of mixtures approach using the volume fraction of uniformly-sized, uniformly distributed crystalline regions &lt;span class=&quot;math-container&quot;&gt;$f_c$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ O \equiv \frac{I}{I_o} = 1 - f_c $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This models a polymer with a &quot;full volume distribution&quot; of crystalline regions &lt;span class=&quot;math-container&quot;&gt;$f_c = 1$&lt;/span&gt; will be fully opaque. You are free to set &lt;span class=&quot;math-container&quot;&gt;$f_c$&lt;/span&gt; as some other metric for linearity between scattering and opacity. Patchiness in any given sheet of polymer indicates uneven distribution of sizes or number density of crystalline regions. This problem has to be solved at the distributor end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourth is to account for inconsistent stocks from sheet to sheet. When for example the concentration of absorber, type of absorber, or size/number density of scattering regions varies from sheet to sheet, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; or the scattering potential will vary from sheet to sheet. The light transmission will therefore vary from sheet to sheet even when the sheet has the same thickness &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;. This problem has to be solved at the distributor end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally is account for the LEDs as point sources. Even in the best cases, LEDs are point sources not uniform sources of light when compared to tubes or bulbs. One suggestion to get around this is to invert your LEDs. Point them toward a diffuse scattering surface. Have that surface project the scattered light outward.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-03-27T21:38:16.273" LastActivityDate="2019-03-27T21:38:16.273" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26489" PostTypeId="2" ParentId="26487" CreationDate="2019-03-21T16:13:10.680" Score="0" Body="&lt;p&gt;If the fuel concentration drops you can extract the salt from the solution as well perhaps using a chemical reaction that converts one of the components of the salt to gas or solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That safety reservoir isn't going to be a single big hollow tank. Instead it's going to be several smaller tanks lined with neutron absorbent material and designed to quench the reaction as fast as feasible.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2019-03-21T16:13:10.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26492" PostTypeId="2" ParentId="26477" CreationDate="2019-03-22T00:27:12.603" Score="1" Body="&lt;p&gt;Radius of a beam curvature at any cross section is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{1}{\rho}= \frac{d\phi}{d_x}=\frac{d^2w}{d_x^2}= - \frac{M}{EI} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \therefore \ \text{for a constant moment we have a constant radius.} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically in any interval of the span of a beam wher there is only constant moment, the beam bends into a section of a circle.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-22T00:27:12.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26493" PostTypeId="1" CreationDate="2019-03-22T00:36:18.617" Score="1" ViewCount="136" Body="&lt;p&gt;I need a motor for my project. My aim is to move a 10 kg mass using a belt pulley system, with a velocity of 15km/hr to a distance of 3.6m. The maximum time for moving the same is 1.5 sec.&#xA;Can somebody please help me with the calculation and finding out the right motor, that is the max power and max rpm of the motor required?&lt;/p&gt;&#xA;" OwnerUserId="19804" LastActivityDate="2023-11-27T04:10:51.397" Title="how to find the motor specification" Tags="|mechanical-engineering|motors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26494" PostTypeId="1" CreationDate="2019-03-22T03:00:19.543" Score="0" ViewCount="153" Body="&lt;p&gt;I am trying to model the dynamics of some system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the system consists of one horizontal linear actuator that drives an object along a planar surface and one vertical actuator that brings the object in contact with the planar surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planar surface (plate) deforms as the vertical motion of the object comes into contact with it. After the plate deforms, the object is dragged along the surface, giving arise to frictional force that opposed the horizontal motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, does the normal reaction force from the elastic plate depend on the frictional force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initially I thought that because the two forces (normal and frictional/tangential force) should not interfere with each other as they are described by unit vectors perpendicular to each other. But placing an encoder on the vertical actuator, I found that moving the object tangentially along the deformed surface (along x-axis) affects the vertical motion (displacement/velocity along y-axis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="18660" LastActivityDate="2024-03-26T17:00:40.053" Title="Effect of friction (or horizontal motion) on vertical surface deformation" Tags="|mechanical-engineering|dynamics|friction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26495" PostTypeId="1" CreationDate="2019-03-22T05:45:17.843" Score="1" ViewCount="1027" Body="&lt;p&gt;I ran an two aerodynamic simulations, one for a spoiler angle of 0 deg and the other for the same spoiler declined at 30 deg. The results showed a considerable increase in downforce, but only a slight change in drag. Is this the expected behavior of a spoiler?&lt;/p&gt;&#xA;" OwnerUserId="19807" LastActivityDate="2021-10-09T02:37:14.787" Title="How do drag and downforce in a rear spoiler of a car vary with angle of attack?" Tags="|mechanical-engineering|fluid-mechanics|automotive-engineering|aerodynamics|cfd|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26496" PostTypeId="1" CreationDate="2019-03-22T06:58:44.793" Score="0" ViewCount="36" Body="&lt;p&gt;I have a 50,500 kgs semi-trailer to modify to be electric truck.&#xA;I will mount a motor direct to differntial axis ( without gearbox ).&#xA;Differential ratio is 841 ( 5.125 ).&#xA;If I want this 50,500 kgs semi-trailer to climb a 26 % slope , what is the suitable motor power for this truck.&lt;/p&gt;&#xA;" OwnerUserId="19809" LastActivityDate="2019-03-22T06:58:44.793" Title="how to calculate motor power for electric truck" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26499" PostTypeId="2" ParentId="26493" CreationDate="2019-03-22T09:51:00.367" Score="1" Body="&lt;p&gt;There are two masses here to accelerate, the target &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; kg and the mass of the belt which is yet unknown(usually it is given as weight per meter), the total mass is &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;. &#xA;I make some assumptions here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- There is no relative motion between the belt and the mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- There is no air resistance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- The mechanical losses are also negligible (This is very rough assumption). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4- Constant acceleration from the beginning to the terminal speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective pull force (on the belt) can be calculated as: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F = \mu_sMg$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here &lt;span class=&quot;math-container&quot;&gt;$\mu_s$&lt;/span&gt; is the static coefficient of friction, you can find it by looking at the brochure, &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is about &lt;span class=&quot;math-container&quot;&gt;$9,81$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\frac{m}{s^2}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiplying the force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; by safety factor &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; and of course not to forget the gearing effect &lt;span class=&quot;math-container&quot;&gt;$\eta$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Phi = \frac{sF}{\eta}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the required power is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$p = v\Phi$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need at least one another parameter, the diameter of the conveyor disc &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; to calculate the speed of the motor: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\omega = \frac{v}{d/2}$$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have the power and the speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, there remain some design consideration here, for example, you have to specify the dimensions of the machine, before calculating the speed of the motor, and countless other small things, but here i tried to demonstrate the very basic steps. I hope the other engineers here, also contribute. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-03-22T09:51:00.367" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26500" PostTypeId="2" ParentId="15473" CreationDate="2019-03-22T09:55:54.630" Score="2" Body="&lt;p&gt;For your question, 1 m2 = 10^12 Darcy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the next question, I think your question is units involved in Darcy's Law, i.e., the Darcy's units and SI units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, we define the permeability of porous media as 1 Darcy, it means the porous media can transmit 1 cm3/s of water with viscosity of 1 cP (1 mPas) under pressure gradient of 1 atm/cm cross an area of 1 cm2. The units involved here are called Darcy's units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For SI unit, we use flow rate of m3/s, viscosity of Pas, pressure of Pa, length of m, cross section area of m2, and we can get the permeability unit with m2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use the units system for both cases, Darcy's equation is the same without any conversion factor.&lt;/p&gt;&#xA;" OwnerUserId="19812" LastActivityDate="2019-03-22T09:55:54.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26502" PostTypeId="1" CreationDate="2019-03-22T12:43:54.587" Score="0" ViewCount="23" Body="&lt;p&gt;we currently need some help in analyzing the real (!) content of our design files of Solidworks 2018. Recently we are faced with ghost-parts within our assembly files. Unfortunately, support is not able to solve the issue by now. Since version 2015 the files are no more based on MS-CFB (Compound File Binary File Format). Does anyone have an idea how to explore the file contend? Greetings&lt;/p&gt;&#xA;" OwnerUserId="19816" LastActivityDate="2019-03-22T12:43:54.587" Title="Analyse Soldiworks 2018 Files" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26503" PostTypeId="2" ParentId="26495" CreationDate="2019-03-22T13:03:26.400" Score="0" Body="&lt;p&gt;To answer your question, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;only a slight change in drag. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The spoilers cross sections are made by the aerofoils/airfoils. &#xA;These are called &quot;Streamlined bodies&quot; in which you tend to get more lift than drag. (In the other way, they produce less drag than bluff bodies like sphere/cylinder etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To explain a bit more, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume your car moves at 100 km/hr ~ 28 m/s, and the spoiler (chord) size is 0.3m and also made up of NACA 0012 aerofoil (Google for this profile!). &#xA;Then the &lt;span class=&quot;math-container&quot;&gt;$Re$&lt;/span&gt; number of the flow will be roughly ~ 5&lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt;10&lt;span class=&quot;math-container&quot;&gt;$^5$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this Reynold's number, the NACA 0012 aerofoil produces the lift of (&lt;span class=&quot;math-container&quot;&gt;$C_l$&lt;/span&gt; - coefficient of lift)  1.5 for ~15&lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt; of angle of attack (remeber you have mentioned angle of spoiler not the angle of attack. They are very different. pls. check this!) at the same time, the drag coefficient (&lt;span class=&quot;math-container&quot;&gt;$C_d$&lt;/span&gt;) is 0.03!. So the ratio between these two is called efficiency of the aerofoil &lt;span class=&quot;math-container&quot;&gt;$C_l / C_d = $&lt;/span&gt; 45.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Drag force is less for streamlined bodies.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: Please check your angle, the effieciency is a strong function of angle of attack. At very high angle of attack  the flow tends to seperate and can create adverse effects. &lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2019-03-22T13:03:26.400" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26504" PostTypeId="1" CreationDate="2019-03-22T12:47:44.613" Score="2" ViewCount="49" Body="&lt;p&gt;I am not an engineer, so apologies if this one is a bit of an ill-thought out...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone show me an example of something like the following scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small stepping motor driving rotation of a ringed encasement, via a gearhead on the rotation shaft interlocking with a 'teethed track' which is inlayed on the wall of the encasement (I know, see the diagram).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, why is this a stupid idea? The stepping motor has to be small to fit in the encasement (for a 'wearable') - are the forces involved here unreasonable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ringed encasement is supposed to rotate to a precise position - I imagine that in these situations, rotation would continue after the power to the stepping motor was killed, necessitating some kind of braking mechanism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate your help&lt;img src=&quot;https://i.stack.imgur.com/qK4Eb.jpg&quot; alt=&quot;![enter image description here&quot;&gt;]&lt;a href=&quot;https://i.stack.imgur.com/qK4Eb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="OzzWal" LastEditorUserId="86" LastEditDate="2019-03-22T13:31:11.117" LastActivityDate="2023-02-11T14:51:13.867" Title="Gearhead &amp; 'teeth-track' to drive rotation of ringed encasement?" Tags="|gears|stepper-motor|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26507" PostTypeId="1" AcceptedAnswerId="26508" CreationDate="2019-03-22T14:56:01.293" Score="1" ViewCount="1101" Body="&lt;p&gt;I need to make PTFE caps on my test tubes. Why PTFE? The test tubes contain industrial solvent mixtures &amp;amp; the rubber caps that came with the test tubes are very, very slowly being dissolved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need about a dozen caps. If I were to make it out of solid PTFE, turned on a lathe creating the cap, then it would be expensive. So I'm thinking of having thick cups of PTFE. Then it would be filled by a soft elastomer, creating the cap. Seeing that most PTFE pieces are created by taking PTFE powder &amp;amp; thermoforming it in molds, the production method is expensive even if the cups are lightweight. So I'm thinking &quot;what about vacuum forming?&quot;, so I'm here. Can Vacuum forming be used on PTFE?&lt;/p&gt;&#xA;" OwnerUserId="8098" LastActivityDate="2019-07-10T03:22:09.720" Title="PTFE sheet vacuum forming" Tags="|plastic|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26508" PostTypeId="2" ParentId="26507" CreationDate="2019-03-22T17:48:38.163" Score="1" Body="&lt;p&gt;Normally no, after all the point of PTFE is that its has not the problems of thermoplastics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there exist products like moldflon that enables you to process them in this way. OTOH i would machine them as its cheaper on my time.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2019-03-23T12:58:30.763" LastActivityDate="2019-03-23T12:58:30.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26509" PostTypeId="2" ParentId="26507" CreationDate="2019-03-22T18:04:28.653" Score="1" Body="&lt;p&gt;the cost to make the tooling required to sheet-mold the plastic will exceed the cost to machine the caps on a lathe, for quantities of caps in the hundreds. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-03-22T18:04:28.653" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26511" PostTypeId="1" CreationDate="2019-03-22T22:00:17.690" Score="1" ViewCount="133" Body="&lt;p&gt;I am trying to design and construct a working replica aircraft instrument (a Horizontal Situation Indicator to be exact) but am facing a couple of issues. The first one can best be demonstrated with the below picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hXIf2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hXIf2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inner disk (brown-ish, including the white markings) can rotate freely around a central axis. The orange bar is fixed to that disk, so it rotates with it. At the same time, the bar should be movable along the slot on the lower part, so with the disk orientation depicted in the picture it would be able to move left and right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought the easiest way to move that bar is using a rack that moves left and right. However, I don't see a way to keep that rack coupled with its driving mechanics throughout the rotation. The goal is for it to be moved completely independently of the rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first solution that comes to mind is fixing the motor that moves the rack to the disk so it rotates with it and the coupling between the rack and its pinion remains stationary, but that comes with other drawbacks as I'd have to incorporate a slip ring for the wiring and have to deal with the extra torque that is required to rotate the disk to begin with.&lt;/p&gt;&#xA;" OwnerUserId="19821" LastActivityDate="2023-12-29T08:05:10.613" Title="Rotate rack while keeping it engaged with gears" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26512" PostTypeId="1" AcceptedAnswerId="26549" CreationDate="2019-03-22T22:33:42.440" Score="1" ViewCount="2905" Body="&lt;h1&gt;The context&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I’ve taken an interest in the design of pressure vessels for use in submarine applications. From a cursory survey of what is used in industry, it seems the &lt;em&gt;de facto&lt;/em&gt; standard housing is a thick-walled cylinder (rather than a thin-walled cylinder which would be prone to buckling when loaded in hydrostatic compression). While the equations for stress and deformation in a thick-walled cylinder subject to internal and external pressures are well established (&lt;a href=&quot;https://www.amazon.ca/Roarks-Formulas-Stress-Strain-8th/dp/0071742476&quot; rel=&quot;nofollow noreferrer&quot;&gt;Young&lt;/a&gt;, &lt;a href=&quot;https://www.amazon.ca/Machine-Design-5th-Robert-Norton/dp/013335671X&quot; rel=&quot;nofollow noreferrer&quot;&gt;Norton&lt;/a&gt;), I’ve also read that prestressed compound cylinders can result in a housing with a greater strength to weight ratio (&lt;a href=&quot;https://nptel.ac.in/courses/112105125/pdf/module-9%20lesson-2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kharagpur&lt;/a&gt;, &lt;a href=&quot;https://www.mae.ust.hk/~meqpsun/Notes/Chapter2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sun&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/l7W3C.png&quot; alt=&quot;Stress distributions in compound pressure cylinders&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a result, I’ve been trying to understand the derivation of the thick-walled cylinder equations to better understand how they can be applied and optimized.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;The question&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;With the help of multiple online sources (primarily &lt;a href=&quot;http://engweb.swan.ac.uk/~c.kadapa/teaching/2017-2018/EGF316/week2/EGF316%20Thin%20and%20Thick%20Cylinders%20-%20notes.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kadapa&lt;/a&gt; and &lt;a href=&quot;http://www.engr.mun.ca/~katna/5931/Thick%20Walled%20Cylinders(corrected).pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Katna&lt;/a&gt;), I’m confident that I understand the derivation of Lamé’s equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_r = C_1 + \frac{C_2}{r^2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_\theta = C_1 - \frac{C_2}{r^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where things get interesting. In both sources mentioned above, and indeed everywhere else I can find it, the equation for the change in radius of a differential shell is calculated from the hoop strain, rather than the radial strain as I would expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\varepsilon_\theta = \frac{\sigma_\theta - \nu \left( \sigma_r + \sigma_z \right)}{E}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$u = \varepsilon_\theta r$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I’m satisfied this is correct, I don’t understand why. Why doesn’t the radial deformation result from the radial strain as I expect it should, according to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\varepsilon_r = \frac{\sigma_r - \nu \left( \sigma_\theta + \sigma_z \right)}{E}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$u = \varepsilon_r r$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;What I've tried so far&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;I’ve tried evaluating both equations for &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; for the simple case of a single cylinder subject to internal and external pressure (the boundary conditions given in every textbook).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/B4DSt.png&quot; alt=&quot;Differential cylinder element showing stresses and deformation&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I haven’t compared the values to a more reliable source, the equation with hoop strain always produces deformations in the direction of the pressure differential (an external pressure causes the cylinder to contract). The equation using radial strain has so far always produced deformations of differing magnitude and in the direction opposite the pressure differential (an external pressure causes the cylinder to expand).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both Kadapa and Katna, it is offhandedly mentioned that  and &lt;span class=&quot;math-container&quot;&gt;$\varepsilon_r$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\varepsilon_\theta$&lt;/span&gt; are equal on the bases that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\varepsilon_r = \frac{u}{r} = \frac{2 \pi u}{2 \pi r} = \frac{\delta\ circumference}{circumference} = \varepsilon_\theta$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While I can’t find fault with this statement, I think I have shown that it is only true for thin-walled cylinders where the hoop stress is assumed to be constant throughout the wall as below:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Based on the methods presented in Kadapa, Katna, and others,&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\varepsilon_r = \frac{du}{dr}$$&lt;/span&gt;&#xA;  &lt;span class=&quot;math-container&quot;&gt;$$\varepsilon_\theta = \frac{u}{r}$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Assume that&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\varepsilon_r = \varepsilon_\theta$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;If so, &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{du}{dr} = \frac{u}{r}$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;This first order linear ODE has the solution&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u = kr$$&lt;/span&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;As a result, &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\varepsilon_\theta = \frac{u}{r} = k$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="19822" LastActivityDate="2020-08-16T20:04:05.420" Title="When calculating the deformed radius of a pressurized thick-walled cylinder, why is the hoop strain used rather than the radial strain?" Tags="|mechanical-engineering|pressure-vessel|solid-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26513" PostTypeId="2" ParentId="26511" CreationDate="2019-03-22T22:34:27.243" Score="0" Body="&lt;p&gt;Assuming you want to use this as a teaching model, because I don't know if they are certified, you can have the lateral deviation indicator's motor connected trough something like these 2-3 wire rotary &lt;a href=&quot;http://www.mercotac.com/html/205.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;connectors&lt;/a&gt; mounted on your vertical axis.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-22T22:34:27.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26514" PostTypeId="1" AcceptedAnswerId="26526" CreationDate="2019-03-22T23:34:25.323" Score="7" ViewCount="4952" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0lfOo.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0lfOo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From doing some research I've learned that internal stresses inside an object can &quot;concentrate&quot; on sharp edges, as shown in the above picture. For that reason we often make fillets so that the change in diameter is smoother and we get a smaller stress concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why does the stress concentrate on sharp edges? From Googling I've found numerous inadequate answers, such as imagining &quot;stress flow lines&quot; getting packed at the sharp edge. &lt;strong&gt;Here I can see an analogy to electromagnetism, where we talk about flow lines of an electric field, but what is &quot;flowing&quot; here? What do the lines represent?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively, I can imagine this: If we press the object on the left in the above picture, there is a higher stress after the edge, since there is now a smaller area enduring the same force. Hence the stress is higher. But this does not explain why the stress concentrates &lt;em&gt;at&lt;/em&gt; the edge. I mean, even with the fillet, the radius decreases anyway, so why is there not a similarly high stress on the right of the fillet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar effect also occurs when we have a continuous piece of material with a hole in it. The stress is the highest around the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cCo0b.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cCo0b.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once again, force lines, what do they represent? Again I can see the similarity to electromagnetism. But what kind of a force field are we talking about here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One source I found simply explained that we can use methods like finite element analysis to confirm this concentrating of stress does happen. But this is not an explanation either. &lt;strong&gt;Use of FEA implies there is some kind of differential equation to be solved, so what kind of theory is there behind it?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="13372" LastEditorUserId="13372" LastEditDate="2019-03-26T13:00:26.350" LastActivityDate="2019-03-29T10:28:09.810" Title="Why do stress concentrations really occur?" Tags="|structural-analysis|stresses|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26515" PostTypeId="2" ParentId="26514" CreationDate="2019-03-23T01:36:29.847" Score="0" Body="&lt;p&gt;Those lines in your diagrams are normal to the equal stress contour planes in the section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let's say in your top left diagram the area to the left is A and to the right is A/2, and the section is under compression load P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now assume we cut the section from the notch in two parts, A the wide and A/2 the narrow. Then we stress them under axial load P. part A/2 is stressed twice and coiled under a lot of pressure. Par A is more relaxed; it is under half the stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we bring them beck together, while still stressed. Part A/2 will try to push the contact boundary surface back into part A. But due elastic behavior of the material, its edges bend into a curve and there is stress concentration around those edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exactly as when you shoot a  stubbed cylindrical slug into a steel plate target. There is so much stress concentration on the slug's perimeter that it crumbles and tears off around the edges.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-23T01:36:29.847" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26516" PostTypeId="1" CreationDate="2019-03-23T04:42:22.050" Score="0" ViewCount="411" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/f2ENb.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F=32KN; Mf1=60KN-m; Mf2=-58KN-m; g=30KN/m I need to solve this using Force method. Find the reaction at the pinned end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/kv97Y.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sAcIg.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19825" LastEditorUserId="19825" LastEditDate="2019-04-03T13:08:03.963" LastActivityDate="2019-04-05T00:54:40.887" Title="Statically indeterminate fixed-pinned beam problem" Tags="|mechanical-engineering|applied-mechanics|beam|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26517" PostTypeId="1" CreationDate="2019-03-23T04:54:06.480" Score="1" ViewCount="168" Body="&lt;p&gt;I am trying to find a cheap power supply that will produce the high voltage needed to create lichtenberg figures on wood.  I ruled against using a microwave oven transformer due to the high current, so only the neon sign power source sold on ebay for ~$10 would be in my price range.  Wikipedia's article on neon sign transformers suggests just those with an iron core are used by hobbyists to make lichtenberg figures, and that manufacturers are now producing switch mode power supplies to power the neon signs.  I can't be certain of the specs of the ebay transformers since they are generic units produced in China, but assuming that they are switch mode power supplies, will they be able to fit my needs for a power supply?&lt;/p&gt;&#xA;" OwnerUserId="19824" LastActivityDate="2019-12-18T08:03:05.477" Title="Can a switched mode power supply for a neon sign be used for making lichtenberg figures?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26518" PostTypeId="2" ParentId="26494" CreationDate="2019-03-23T05:01:20.777" Score="0" Body="&lt;p&gt;If your object has to move through a depression created by vertical force, then the force F is the sum of two forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Friction force required to pull it horizontally and the viscosity, and plastic deformation of the pit on the support platform.&#xA;The first part &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F= \mu F_y $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky part is force needed to deform the plate and keep rolling this depression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need material properties on the support.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-23T05:01:20.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26519" PostTypeId="2" ParentId="26517" CreationDate="2019-03-23T06:20:11.350" Score="1" Body="&lt;p&gt;I have tried this, with negative results. My suspicion is that the power supply unit has an internal current limiting circuit built-in which prevents it from delivering enough current to do the job. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more powerful supply might work but must be treated with respect, since the voltage and current could potentially kill you.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-03-23T06:20:11.350" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26522" PostTypeId="1" AcceptedAnswerId="26530" CreationDate="2019-03-23T18:59:48.207" Score="5" ViewCount="770" Body="&lt;p&gt;In educational construction material as well as engineering publications, the corbel angle for significant excess loads, and also for substantial openings, in an ordinary brick wall is usually stipulated as being taken as 60 degrees to the horizontal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hyypO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hyypO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;In the first illustration, the reduced load is spread out, in the second a distributed load is spread out, and in the third, a point load is spread out. There are no buttresses or other &quot;out-of-plane&quot; support. The wall can be imagined as an indefinite length ordinary 100mm thick brick wall.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to find the source of that 60 degree value, to check when it's applicable and how it's calculated, but I can find nothing whatsoever on the topic. The few mentions online of &quot;corbel angle&quot; are all of a form that just state/assume 60 degrees (with no reasoning). I can't find anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Related to this, when dealing with an opening, which of these is the correct way to use the corbel angle, and why? They can't both be correct, as they contradict, but there doesn't seem to be an obvious reason why the answer should be one rather than the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3S2vF.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3S2vF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;In the first diagram, the reduced weight is treated as causing a reduced load in nearby masonry at 60 degrees outward; in the second the heavier surrounding weight is treated as causing an increased load at 60 degrees inward.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the corbel angle actually work in these typical situations, where there is an excess loads + openings at some point in an ordinary brick wall, and how universally appropriate is it to always assume it's 60 degrees in ordinary brick or concrete construction?&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2019-03-23T19:07:31.380" LastActivityDate="2023-09-17T02:19:14.403" Title="Should the corbel angle always be estimated as 60 degrees in masonry construction, or is there a formula to use instead?" Tags="|structural-engineering|design|structural-analysis|masonry|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26523" PostTypeId="2" ParentId="26507" CreationDate="2019-03-23T19:05:20.833" Score="6" Body="&lt;p&gt;Don't make them, just google &quot;ptfe stoppers&quot;, and buy them. (You've asked an XY problem.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mKtGS.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mKtGS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LWfTi.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LWfTi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/93eiI.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/93eiI.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16544" LastEditorUserId="16544" LastEditDate="2019-03-24T09:03:47.597" LastActivityDate="2019-03-24T09:03:47.597" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26524" PostTypeId="1" CreationDate="2019-03-23T20:31:55.150" Score="0" ViewCount="30" Body="&lt;p&gt;What have you found to be some effective entrainers to distill MTBE from Methanol, given the azeotropic conditions at 67%. I have tried to simulate 1-butanol, as well as ethanol, and numerous other components although have come up short in distilling with high purity. &lt;/p&gt;&#xA;" OwnerUserId="19832" LastEditorUserId="-1" LastEditDate="2019-06-18T12:10:14.150" LastActivityDate="2019-06-20T09:00:28.600" Title="Entrainers to distill MTBE from Methanol? - Chemical Engineering Question" Tags="|mechanical-engineering|chemical-engineering|process-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26525" PostTypeId="1" CreationDate="2019-03-23T22:24:34.247" Score="1" ViewCount="64" Body="&lt;p&gt;If we have a closed cylinder filled with a gas and we heat it, by &lt;span class=&quot;math-container&quot;&gt;$1^{st}$&lt;/span&gt; law we have: &lt;span class=&quot;math-container&quot;&gt;$\Delta q=\Delta U$&lt;/span&gt; (?) since &lt;span class=&quot;math-container&quot;&gt;$\Delta V$&lt;/span&gt; = 0. But &lt;span class=&quot;math-container&quot;&gt;$\Delta H= \Delta U+ V. \Delta P$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What does that difference between the enthalpy change and heat supplied physically signify in this case?&lt;/p&gt;&#xA;" OwnerUserId="19833" LastEditorUserId="19833" LastEditDate="2019-03-23T22:39:56.593" LastActivityDate="2019-03-23T22:39:56.593" Title="Enthalpy vs Heat" Tags="|mechanical-engineering|thermodynamics|chemical-engineering|energy|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26526" PostTypeId="2" ParentId="26514" CreationDate="2019-03-24T04:59:31.400" Score="3" Body="&lt;h2&gt;What do the lines represent?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Stress has units of force per area, so stress can be thought of as analogous to force flux. The lines represent force, and the density of lines represents force through an area: stress. Curvature of the lines represents changing force directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In equilibrium the force lines cannot intersect the geometry boundary except where an external force is applied. If they did intersect the boundary somewhere else then there would be a net force on the geometry and it would not be in equilibrium. The force lines must instead flow around the boundaries. The disruption of flow is stronger closer to boundaries perpendicular to the direction of applied stress, so the density of flow lines changes there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the boundary is re-entrant---like with a hole or inside corner or crack---then the density of lines increases. If the boundary protrudes---like with the external corners of the first example provided by the asker---then the density of lines decreases. If there are no perpendicular boundaries (i.e. a consistent cross-section) then the density of lines is the same as the density of the applied force.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Why does stress concentrate near perpendicular geometry changes?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Consider the bar shown below. It is made from a pure, single-element, single-crystal, closest-packed (i.e. hexagonal) isotropic material one atom thick (i.e. 2D) with a symmetrical crack one atom wide. The bar is under uniaxial tension perpendicular to the crack. We will also assume deformation is negligible so that the force vectors don't change direction, simplifying the analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t6esz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t6esz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Atoms far from the crack tip have six neighboring atoms. An example is shown in the upper-left of the image below. Each neighbor applies a force to the central atom. The neighbors above and below apply a tensile force which arises from the global tensile stress. The neighbors left and right each apply a compression force which arises from the Poisson effect. The atoms are color-coded by position. Those on the left are red, those on the right are blue. Their applied forces match colors. The resultant tension on the central atom is shown by black arrows in the lower-left of the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t6epO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t6epO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The upper-right part of the image shows the same analysis repeated for the atom at the left crack tip. There are only five neighbor atoms, as shown. Because the atom to the right is missing, there is no compression force to the left. To balance the left-right forces requires the leftward two tensile forces to increase in magnitude. The result is an increase in the resultant tension on the atom parallel to the global stress direction. Thus the atom at the crack tip experiences a concentration of stress.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2019-03-27T21:53:30.267" LastActivityDate="2019-03-27T21:53:30.267" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26527" PostTypeId="1" CreationDate="2019-03-24T05:39:45.707" Score="0" ViewCount="368" Body="&lt;p&gt;Why do I see rocket nozzle CFD like the one &lt;a href=&quot;https://www.youtube.com/watch?v=txk-VO1hzBY&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; in this SpaceX video done in 2D?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real Fluid flow is three dimensional and involves an x, y and z axis. Is fluid flow done in 2D only because computers are not powerful enough to run 3d fluid flow simulation? Or is CFD done in 2D because that gives an accurate enough real world approximation? Or is it a bit of both.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Freshman majoring in Aerospace engineering.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19835" LastActivityDate="2019-11-27T22:36:41.910" Title="Why is Rocket Nozzle CFD done in 2D instead of 3D?" Tags="|cfd|rocketry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26530" PostTypeId="2" ParentId="26522" CreationDate="2019-03-24T08:07:16.323" Score="4" Body="&lt;p&gt;I recommended reading EN1996-1-1, and especially section 6.1.3, at least if you are in an area where eurocodes are applicable. This section contains the load distribution angle of 60 degrees. It has a few figures showing applications of the angle, but no formulas for calculating an alternative value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not expect to be able to find a formula to calculate a more exact value of the angle. To me, it appears very much to be an empirical value, which is backed by experiments rather than theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the question on window openings: The angle of 60 degrees is for distributing concentrated over a higher length of wall. Therefore the figure on the right in your diagram is the correct one. Strictly speaking, you could use the figure on the left for distributing the weight of the window sill and the figure on the right for distributing the weight of the wall between and above the windows, but that will rarely be worth the bother. Just think of it as distributing a load, not distributing the absence of a load.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastEditorUserId="6366" LastEditDate="2019-03-27T09:47:39.730" LastActivityDate="2019-03-27T09:47:39.730" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26532" PostTypeId="2" ParentId="26527" CreationDate="2019-03-24T14:15:58.627" Score="1" Body="&lt;p&gt;It isn't really done in 2D. It is done in pseudo 3D. The actual annular areas are factored in. It is primarily based on a uniformity assumption that the gas dynamics are the same all around the nozzle for any constant axial and radial coordinate value. But all the physics are correct for 3D. In the analytic kernel, the uniformity assumption allows you to extract the azmuthal terms because they are constant, and typically very small in this case compared to the axial and radial terms. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that gas injection thrust vectoring destroys this symmetry and is used to steer a fixed nozzle rocket.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-03-24T14:15:58.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26533" PostTypeId="1" CreationDate="2019-03-24T17:23:23.280" Score="0" ViewCount="2110" Body="&lt;p&gt;If I want to calculate the efficiency of a microwave oven, I have the input Energy (= power (watts) x time (s)) and the output which I measured using temperature difference (q=mct). &#xA;So do I divide output/input x 100 or do I find the difference between them (=work) then divide it by the input energy?&lt;/p&gt;&#xA;" OwnerUserId="19489" LastEditorUserId="19489" LastEditDate="2019-03-24T18:41:55.563" LastActivityDate="2019-11-18T15:39:11.297" Title="calculating the efficiency of a microwave oven" Tags="|thermodynamics|heat-transfer|energy-efficiency|radiation|thermal-radiation|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26534" PostTypeId="1" CreationDate="2019-03-24T18:23:59.163" Score="0" ViewCount="1834" Body="&lt;p&gt;I have been trying to understand if compressor-based systems in air conditioners can be replaced to make something more economical and energy efficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand, the compressors primary job is to pressurize the refrigerant to bring it back into liquid state. Can't the same job be done via a heat exchanger?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fig. 3 in this resource is an example of what I am talking about: &lt;a href=&quot;https://www.aiche.org/resources/publications/cep/2015/september/cool-down-liquid-nitrogen#fig1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.aiche.org/resources/publications/cep/2015/september/cool-down-liquid-nitrogen#fig1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the existing compressor in an air conditioner be replaced by a cryogenic heat exchanger that uses liquid nitrogen to condense the liquid and pump it with a much smaller compressor? If so, how would I go about calculating the amount of liquid nitrogen needed to cool a room ~200 sq ft in size?&lt;/p&gt;&#xA;" OwnerUserId="19838" LastActivityDate="2020-12-22T08:52:39.947" Title="Air conditioner using liquid nitrogen and a cryogenic heat exchanger" Tags="|mechanical-engineering|thermodynamics|hvac|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26535" PostTypeId="2" ParentId="26533" CreationDate="2019-03-24T18:25:52.290" Score="1" Body="&lt;p&gt;The equation for efficiency in thermodynamics is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\eta = W_{out}/W_{in}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and to get this as a percentage you just multiply &lt;span class=&quot;math-container&quot;&gt;$\eta \;\mathrm{x}100=\%$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the link below!&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thermal_ef&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thermal_ef&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19839" LastActivityDate="2019-03-24T18:25:52.290" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26537" PostTypeId="1" CreationDate="2019-03-24T21:29:22.860" Score="1" ViewCount="5126" Body="&lt;p&gt;I need the equation of a discrete PID controller and I find different answers from different websites. For example, &lt;a href=&quot;http://portal.ku.edu.tr/~cbasdogan/Courses/Robotics/projects/Discrete_PID.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; the answer is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u[k] = u[k-1] + (K_p + K_i \frac{T_s}{2} + \frac{K_d}{T_s}) e[k] + (-K_p + K_i \frac{T_s}{2} - 2 \frac{K_d}{T_s}) e[k-1] + \frac{K_d}{T_s} e[k-2]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that's the equation that is used in &lt;a href=&quot;http://dspace.ucuenca.edu.ec/bitstream/123456789/21401/1/IEE_17_Romero%20et%20al.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; paper, for example. If I write that equation in function of &lt;span class=&quot;math-container&quot;&gt;$T_i$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$T_i = K_p/K_i$&lt;/span&gt;) and &lt;span class=&quot;math-container&quot;&gt;$T_d$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$T_d = K_d/K_p)$&lt;/span&gt;, I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u[k] = u[k-1] + K_p \left[ (1+\frac{T_s}{2 T_i} + \frac{T_d}{T_s}) e[k] + (-1 + \frac{T_s}{2 T_i} - 2 \frac{T_d}{T_s}) e[k-1] + \frac{T_d}{T_s} e[k-2]\right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in &lt;a href=&quot;https://en.wikipedia.org/wiki/PID_controller&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia&lt;/a&gt; and in &lt;a href=&quot;http://dspace.ucuenca.edu.ec/bitstream/123456789/21401/1/IEE_17_Romero%20et%20al.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; paper (although they say they're using the written equation above) they use the following equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u[k] = u[k-1] + (K_p + \frac{K_d}{T_s}) e[k] + (-K_p - 2 \frac{K_d}{T_s}) e[k-1] + \frac{K_d}{T_s} e[k-2]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I think that &lt;a href=&quot;http://ipco-co.com/PET_Journal/Papers%20CEIT%2714/248.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; they use the same equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what can I do? Which do I implement? Any demonstration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="8089" LastEditorUserId="17070" LastEditDate="2019-09-28T02:29:58.240" LastActivityDate="2020-06-24T04:07:45.877" Title="What is a definitive discrete PID controller equation?" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26539" PostTypeId="1" CreationDate="2019-03-24T23:56:14.730" Score="1" ViewCount="143" Body="&lt;h2&gt;Problem outline:&lt;/h2&gt;&#xA;&lt;p&gt;I'm trying to figure out the distance travelled at the articulation point of a turning vehicle with 4 driven wheels (each wheel houses a motor and can all rotate at different RPMs). If this isn't a suitable place for such a question, please let me know if there's a better exchange/forum for it.&lt;/p&gt;&#xA;&lt;h2&gt;Motivation:&lt;/h2&gt;&#xA;&lt;p&gt;This is a work-related odometry calculation problem used in a larger analytic, not something &amp;quot;just for the sake of doing it&amp;quot; (I wish!), so I unfortunately don't have anyone academic to turn to for assistance/grading etc.&lt;/p&gt;&#xA;&lt;h2&gt;Known:&lt;/h2&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The distance from the articulation point to the front axle, &lt;span class=&quot;math-container&quot;&gt;$l_1$&lt;/span&gt;,&lt;/li&gt;&#xA;&lt;li&gt;The distance from the articulation point to the rear axle, &lt;span class=&quot;math-container&quot;&gt;$l_2$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$l_1$&lt;/span&gt; != &lt;span class=&quot;math-container&quot;&gt;$l_2$&lt;/span&gt;),&lt;/li&gt;&#xA;&lt;li&gt;The length of each axle, &lt;span class=&quot;math-container&quot;&gt;$l_a$&lt;/span&gt;,&lt;/li&gt;&#xA;&lt;li&gt;The articulation angle, γ,&lt;/li&gt;&#xA;&lt;li&gt;The RPM of each wheel, and&lt;/li&gt;&#xA;&lt;li&gt;The diameter of each wheel.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I have a timeseries of RPM/articulation angle data with a 2Hz sampling rate, but if I can get some help with the general equations, I'm more than comfortable doing this bit.&lt;/p&gt;&#xA;&lt;p&gt;I've done similar instant centre problems with non-articulated vehicles in the past, but my uni years are now a fair bit behind me and these days I spend most of my time programming, so my engineering skills are a bit rusty. Unfortunately, &amp;quot;reasonable&amp;quot; accuracy is important here (still making allowable assumptions such as disregarding slip, etc) so I can't use the assumption of linear distance. Any help would be very much appreciated!&lt;/p&gt;&#xA;&lt;p&gt;Rough idea:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/FyUCa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FyUCa.png&quot; alt=&quot;Turning articulated vehicle&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19842" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-03-25T00:06:03.217" Title="Distance travelled at articulation point of turning vehicle" Tags="|mechanical-engineering|applied-mechanics|modeling|mathematics|distance-measurement|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26540" PostTypeId="2" ParentId="26534" CreationDate="2019-03-25T00:32:50.313" Score="2" Body="&lt;p&gt;Air conditioning, or cooling of the sort you are describing, uses a compressor to compress the gas. In so doing, it gains thermal energy (heat) which has to be removed from the system via a heat exchanger. This means that your description of a compressor or a heat exchanger is close, but not close enough. You have to have both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat of compression is removed by a heat exchanger. The most common design uses forced air to cool the compressed gas. Heat is removed from the gas, the air around the heat exchanger absorbs it, so to speak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the process of exchanging heat to the environment, the gas condenses and becomes a liquid (or a more dense gas) and is passed on to the evaporator, which is the cooler part of the system. A pressure reducing device is also incorporated into the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The liquid evaporates (hence the name), absorbing thermal energy from the environment, cooling it. In the process it becomes a gas again and the cycle continues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reference to figure three of your link points to a liquid to gaseous nitrogen heat exchanger. This is a net-loss cooling system. The LIN evaporates and cools the intermediate heat exchanger in a manner similar to the above description, but the nitrogen is lost to the atmosphere. You'd have to have a recovery system in place to capture and re-compress/re-condense the nitrogen. That stuff is pretty expensive compared to a conventional refrigeration system and conventional refrigerant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your last question regarding the amount of LN2 would have to have a specific duration, as the nitrogen isn't going to stick around for long and the heat would return as soon as the supply tank was empty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider this experiment. Ordinary rubbing alcohol (or any alcohol!) will evaporate, creating a cooling action. While the bottle is full and dripping onto a container, the container is causing the liquid to evaporate (just as nitrogen would). Eventually, the container would reach the evaporation temperature of the alcohol and it would not get cooler. That temperature is higher than nitrogen's, but the concept is valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the alcohol bottle is empty, the container that was cooling will begin to warm again. If the entire assembly was encased in another container and the evaporated alcohol was recovered, compressed and cooled, it would work longer, but you're back to the original refrigerator and compressor system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A less complex example is a typical shop air compressor. As the compressor is running, the storage tank will become warm. The heat exchange portion is the tank mass and airflow around it. Eventually the tank temperature becomes that of the surrounding air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the air is released, the reduction of pressure results in cooling and can be felt on the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is simpler and less expensive to stick with conventional refrigeration unless you need unusually low temperatures. Also consider to research ammonia cooling systems, used for some winter sport venues such as bobsled and luge. (can be dangerous)&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2019-03-25T21:55:26.250" LastActivityDate="2019-03-25T21:55:26.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26541" PostTypeId="2" ParentId="24860" CreationDate="2019-03-25T03:10:01.553" Score="1" Body="&lt;p&gt;Saturated flow through sediments is governed by &lt;a href=&quot;https://www.e-education.psu.edu/earth111/node/926&quot; rel=&quot;nofollow noreferrer&quot;&gt;Darcy's Law&lt;/a&gt;, which relates the pressure drop across the porous medium to the permeability of the medium to calculate the fluid flow rate. The fluid viscosity affects the relationship but is usually assumed to be constant in groundwater studies. That may or may not be the case for your coffee (e.g. if the temperature changes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you know the gauge pressure on the pump side of the coffee grounds and if the pressure on the downstream side is atmospheric, then you know the pressure drop. You need to know/estimate/calibrate the permeability of the media. You don't say what kind of coffee maker, but it will depend on the grind and tamping for an espresso machine. You need to decide if you can assume the value is constant. If it is not, you need to determine how it varies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Groundwater studies use analytical or digital model solutions to determine how flow changes with differences in boundary conditions and variation in permeability. Your system sounds similar to Darcy's original experiments so you can easily calculate flow if you know the other parameters.&lt;/p&gt;&#xA;" OwnerUserId="19841" LastActivityDate="2019-03-25T03:10:01.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26543" PostTypeId="1" AcceptedAnswerId="26544" CreationDate="2019-03-25T05:05:55.667" Score="2" ViewCount="18387" Body="&lt;p&gt;These holes are caused by bars that are put through the frame before the concrete is poured and then clamped, so the mold can't swell outward. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for these holes though that is widely recognized?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5GpL9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5GpL9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19347" LastEditorUserId="33" LastEditDate="2019-03-25T14:23:04.040" LastActivityDate="2023-05-11T01:47:49.677" Title="What are the regular holes in poured concrete walls called?" Tags="|concrete|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26544" PostTypeId="2" ParentId="26543" CreationDate="2019-03-25T07:02:49.853" Score="5" Body="&lt;p&gt;These are left over from concrete ties which are used to hold the formwork together ready for pouring concrete. They are usually called tie holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the concrete is cured and forms need to be removed their ends protruding out are bent and snapped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6JBRN.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6JBRN.jpg&quot; alt=&quot;snapped.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-03-25T07:15:39.940" LastActivityDate="2019-03-25T07:15:39.940" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26545" PostTypeId="1" CreationDate="2019-03-25T10:28:34.423" Score="0" ViewCount="304" Body="&lt;p&gt;Outlet nozzle - 1 1/4 inch&lt;br&gt;&#xA;Cut off pressure - 8 bar&lt;br&gt;&#xA;Tank capacity - 35 ltrs   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the discharge flow rate?&#xA;Based on the outlet nozzle, discharge rate will vary. &lt;/p&gt;&#xA;" OwnerUserId="19847" LastEditorUserId="110" LastEditDate="2019-03-25T10:50:05.550" LastActivityDate="2024-03-29T03:03:25.363" Title="How can I calculate the discharge flow rate of compressor of 1 HP?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26547" PostTypeId="1" AcceptedAnswerId="26548" CreationDate="2019-03-25T14:16:27.620" Score="0" ViewCount="250" Body="&lt;p&gt;Hello everyone I am designing a parts that needs to rotate so I am planning to  insert a round bushing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am curious of how much space I need to leave between the round bushing and the parts that need to rotate? (Assuming the round bushing will be fixed by tighten the screw)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer diameter of the round bushing is 3.5mm so how many mm hole should I design for that 3.5mm round bushing? If produce by a CNC machine how accurate can it mill? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much&lt;a href=&quot;https://i.stack.imgur.com/B32Al.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B32Al.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18994" LastActivityDate="2019-03-25T15:20:33.370" Title="How much space should I leave between round bushing and the rotating parts to minimize the play?" Tags="|cnc|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26548" PostTypeId="2" ParentId="26547" CreationDate="2019-03-25T15:20:33.370" Score="1" Body="&lt;p&gt;This is one of those questions that can be answered by &quot;you get what you pay for.&quot; Ignoring the poor grammar of that statement, one does have to spend money for quality performance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 3D printer world, you'd have to have as much as 0.2 mm spacing to ensure the rotation you seek. In the CNC world, it's not that large. A friend who owned a machine shop created parts with his CNC with tolerances of as small as 0.0005&quot;. That's five ten-thousandths of an inch. In these circumstances, one must ensure temperatures are uniform for both parts. If the parts were machined in 65°F conditions, those parts might not fit together in colder or warmer conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can get a good sliding fit with room for lubrication by having a clearance of two or three thousandths of an inch. When I want a clean sliding fit for tubing, I'll purchase a &quot;standard&quot; size from Aircraft Spruce and Specialty, say, 0.375 outside diameter. The matching tubing, next size up would be 7/16&quot; in the list, and I'd have to pick a wall thickness to match the 0.375, which in this case is 0.035, giving me an inside diameter of 0.368, a difference of 0.007&quot;. The next size down would cause too much slop and the next size up would result in an impossible-to-fit set of tubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seven thousandths of an inch is tight enough that a small burr will prevent a fit. One last consideration is your bushing size. Typically a manufacturer will specify that part's tolerance as well. Is the 3.5 mm listed as +/- 0.1 or something similar? &lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-03-25T15:20:33.370" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26549" PostTypeId="2" ParentId="26512" CreationDate="2019-03-25T16:41:28.187" Score="0" Body="&lt;p&gt;I think I've identified my misconception. As alephzero commented on my question, I was conflating the local strain with the overall radial deformation. The strain is defined as the normalized change in length of an infinitesimal line segment, and not as the change in length of a macroscopic dimension such as radius. As such, the results &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\varepsilon_r = \frac{du}{dr}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\varepsilon_\theta = \frac{u}{r}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;give me immediate access to the deformations as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u = \int \varepsilon_r dr = \varepsilon_\theta r$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for pointing me in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="19822" LastActivityDate="2019-03-25T16:41:28.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26550" PostTypeId="1" AcceptedAnswerId="26560" CreationDate="2019-03-25T17:32:47.160" Score="1" ViewCount="84" Body="&lt;p&gt;I have a device 4cm X 5cm X 15cm that stands on asphalt/concrete/road. So, the device is taller than its base which makes contact with the asphalt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to maximize the friction so that the device stands as well as possible without any attachment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering a thin rubber add-on attached at the bottom of the device and I'm wondering what is the best design for such rubber to maximize the friction with asphalt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I selected rubber because it has the highest friction coefficient with asphalt - there is a reason why tires are made of rubber...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how should I design the bottom side of this rubber part that will be in contact with the asphalt? Here is an idea which I got inspiration from a bathtub mat. What is the best pattern possible? And what should be the pattern size (small or big relative to the 5cm X 4cm area)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Dq7D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Dq7D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5984" LastActivityDate="2019-03-26T02:08:48.913" Title="Which design to maximize friction for a device standing on concrete?" Tags="|mechanical-engineering|friction|concrete|industrial-engineering|product-engineering|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26551" PostTypeId="2" ParentId="26545" CreationDate="2019-03-25T18:10:26.630" Score="0" Body="&lt;p&gt;Horsepower is the product of pressure times mass flow rate. Mind the units, and solve for the flow rate. This assumes 100% compressor efficiency, so it will be an upper limit to the true answer. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-03-25T18:10:26.630" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26552" PostTypeId="2" ParentId="26550" CreationDate="2019-03-25T18:14:50.557" Score="2" Body="&lt;p&gt;If you have a rubber that you believe has the best coefficient of friction possible then it wont matter too much how it is shaped unless you need to deal with water or mud like a car tire. The best way to increase the force of friction is to add weight. As you probably know the friction equation is &lt;span class=&quot;math-container&quot;&gt;$F=\mu N$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$(F=$&lt;/span&gt;Friction Force, &lt;span class=&quot;math-container&quot;&gt;$\mu=$&lt;/span&gt;friction coefficient, &lt;span class=&quot;math-container&quot;&gt;$N=$&lt;/span&gt;normal force) &lt;a href=&quot;https://www.school-for-champions.com/science/friction_equation.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friction force&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what this device does, but if it sees side load and is top heavy it is more likely to tip with a gripy base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the issue is that you don't want it to fall over when you set it down... then I'd guess you need this base to help counteract the variation in how level the ground is? In this case, my shot in the dark is three bumps so you just get the minimum three points of contact. Three-legged tables don't wobble.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you take nothing else away from this and are simply looking for rubber pattern advice I'd say lighting bolts... Lightning bolts are badass. &lt;/p&gt;&#xA;" OwnerUserId="18926" LastEditorUserId="1832" LastEditDate="2019-03-26T02:08:48.913" LastActivityDate="2019-03-26T02:08:48.913" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26553" PostTypeId="1" CreationDate="2019-03-25T20:02:33.953" Score="0" ViewCount="28" Body="&lt;p&gt;We need to determine the physical damage in some area of overhead lines that can't be inspected visually. The damage's in general caused by short circuit and air pollution (mostly sulphur compound). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our idea is to attach strain gauges to the area and measure the change in strain which should be increase as wire's damaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Was wondering if I could get some opinions on whether this could potentially work and are there any issues we should considering? &lt;/p&gt;&#xA;" OwnerUserId="14738" LastActivityDate="2019-03-25T20:16:28.953" Title="Measuring wire damage using strain gauges" Tags="|mechanical-engineering|electrical-engineering|measurements|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26554" PostTypeId="1" AcceptedAnswerId="26591" CreationDate="2019-03-25T20:11:20.350" Score="0" ViewCount="329" Body="&lt;p&gt;I am building a roll stabilized model rocket. I am beginner at controls systems. How do I find out the transfer function of the rocket. Although, I am aware of definitions and stuff of control systems. I just a need a nudge in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what your thought on using State Space Analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks so much! :D&lt;/p&gt;&#xA;" OwnerUserId="19852" LastActivityDate="2019-03-27T21:11:27.563" Title="How to design a transfer function for a Model Rocket?" Tags="|control-engineering|pid-control|rocketry|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26555" PostTypeId="2" ParentId="26553" CreationDate="2019-03-25T20:16:28.953" Score="1" Body="&lt;p&gt;It isn't obvious that the strain &lt;em&gt;will&lt;/em&gt; change when the wire is damaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To take a simple case, consider a wire hanging vertically. The strain depends on the material properties and the weight of the wire. If the dimensions of the wire change globally, the weight and cross section area change in proportion, so the stress and strain remain the same. If there is some local damage to the cable, there will be no significant change in strain except at that local position, and you can't put strain gauges everywhere along the length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a horizontal cable hung from poles, the tension will depend on the shape of the cable between the poles, and will change with the thermal expansion of the cable in different weather conditions. That would probably be a bigger effect than what you are trying to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspecting the cable &quot;visually&quot; with a camera carried on a drone might be a more practical idea.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-03-25T20:16:28.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26556" PostTypeId="1" CreationDate="2019-03-25T20:33:56.287" Score="1" ViewCount="102" Body="&lt;p&gt;I want to build an aluminum tube frame strong enough to support a 2300 lbs water tank. The dimensions of the frame would be 28 &quot; wide, 72&quot; long and 16 inches tall.&#xA;My intention is to use 2&quot; x 2&quot;,  1/8 wall tube for the sides, back, and vertical supports.&#xA;However, on the front face i would like it open for a 'drawer step'  spanning the 75 inches. &#xA;My question is what kind of aluminum tube could i use for a header across that face to support its share of the load.&#xA;I was considering 4&quot; x 2&quot; tube with 1/4 wall, installed on end, making a 4&quot; header.  Would this suffice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there more information i need to provide?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you to anyone looking at my inquiry&lt;/p&gt;&#xA;" OwnerUserId="19853" LastEditorUserId="19853" LastEditDate="2019-03-25T20:39:53.967" LastActivityDate="2021-09-11T05:02:05.177" Title="Aluminum tube frame to support 2300 lbs water tank" Tags="|structural-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26558" PostTypeId="2" ParentId="26556" CreationDate="2019-03-25T21:56:47.563" Score="1" Body="&lt;p&gt;What you're planning to do has lateral instability. It will collapse by rotation of one or both of the side walls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You either need to brace all four sides with possibly 1.5 inch tube or on front use some short walls, say 12 inches wide by 16 tall, on each side to provide lateral stability, and possibly use 2.5 inch tube for the front top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best idea is to design and calculate the support for loads and apply 20% of that as lateral load and calculate for lateral support.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-03-25T21:56:47.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26559" PostTypeId="1" CreationDate="2019-03-25T23:48:25.240" Score="0" ViewCount="93" Body="&lt;p&gt;I'm trying to solve this&#xA; &lt;a href=&quot;https://i.stack.imgur.com/WIQnq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WIQnq.png&quot; alt=&quot;problem&quot;&gt;&lt;/a&gt; &#xA;which is to find the total deformation of the bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to implement the finite element approach, but I am getting way too large of numbers. Here is what I did in Excel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l8xuS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l8xuS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought my units were wrong, but they seem to be fine. The only difference from the other problem I solved was that the second boundary condition was supported; this one has a free end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="16935" LastActivityDate="2019-03-25T23:48:25.240" Title="Need help with FEA bar analysis problem" Tags="|structural-engineering|structural-analysis|applied-mechanics|structures|finite-element-method|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26560" PostTypeId="2" ParentId="26550" CreationDate="2019-03-26T00:26:28.630" Score="1" Body="&lt;p&gt;I would use three(triangulated) very sharp tip toes, stainless steel conical points so that they dig into the asphalt a little bit with a little pressure from the installer :-)&#xA; I don't know if they have them anymore, but track and field shoes use to have threaded conical spikes with the shoe soles having threaded inserts to receive the spikes. You could use something similar. Me thinks this will grip the ground better than a textured rubber pad.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2019-03-26T00:26:28.630" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26564" PostTypeId="1" CreationDate="2019-03-26T12:21:56.133" Score="0" ViewCount="50" Body="&lt;p&gt;I have a question about modelling a nonlinear 2nd order ODE for an impact of a sphere on a thin plate in Simulink. Depending on the solver and solver options, my results in Matlab vary a lot. Some are physically reasonable, some not. Strange enough, refining the time step makes the solution less plausible from a physical point of view. Now I thought about modelling the ODE in Simulink an see which results I can get there.The problem is the dimensionless ODE for the indentation of a sphere in a plate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d^2 \sigma}{d \tau^2}+\left( 1 + \lambda \frac{d}{d\tau}\right)\sigma^\frac32 = 0$$&lt;/span&gt;&#xA;Where &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; is a dimensionless displacement and &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; a dimensionless time. &lt;span class=&quot;math-container&quot;&gt;$\lambda = 110 = constant$&lt;/span&gt;. The initial conditions are: &lt;span class=&quot;math-container&quot;&gt;$\sigma\left(\tau = 0\right)=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\frac{d\sigma}{d\tau}|_{\tau=0}=1$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I do hardly have any experience with Simulink. Does anyone have any experience/tips for modeling this kind of problem in Simulink?&lt;/p&gt;&#xA;" OwnerUserId="19862" LastEditorUserId="1832" LastEditDate="2019-03-26T12:48:24.230" LastActivityDate="2019-03-26T12:48:24.230" Title="Solving Hertz Impact ODE with Simulink" Tags="|applied-mechanics|mathematics|matlab|simulink|solid-mechanics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26571" PostTypeId="1" CreationDate="2019-03-26T19:47:54.840" Score="0" ViewCount="90" Body="&lt;p&gt;Could you please tell me why the singularity function is one when x=0&#xA;I mean &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;0&amp;gt; ^0=1$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks you in advance&lt;/p&gt;&#xA;" OwnerUserId="19870" LastEditorUserId="19870" LastEditDate="2019-03-26T22:24:51.300" LastActivityDate="2020-04-21T12:05:50.633" Title="Singularity function" Tags="|mechanical-engineering|mechanical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26572" PostTypeId="2" ParentId="26514" CreationDate="2019-03-26T19:59:59.677" Score="2" Body="&lt;p&gt;The first picture of the sharp corner(or notch) is an example of stress intensity. I would like to point out that stress intensity is different from that of stress concentration. Stress intensity occurs in a crack tip, similar sharp tips like re-entrant corners, whereas stress concentration is what happens in the plate with a hole(or with a fillet), where the stress is the highest near the hole. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both are different in the sense that you can never obtain convergence in the case of stress intensity, whereas in the other case of stress concentration once the mesh is fine enough, the stress value will be higher than the surrounding, but will be converged( meaning value won't increase with mesh refinement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for stress intensity can be explained very simply on the basis of linear elastic mechanics, given a radial co-ordinate system based on the sharp corner, the analytical displacement solution in the surrounding has the form &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$u \propto r^{\alpha}$&lt;/span&gt;&lt;/strong&gt;. Thus the corresponding strain/stress will be given as &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$\epsilon / \sigma \propto r^{-1+\alpha}$&lt;/span&gt;&lt;/strong&gt; (where &lt;strong&gt;&lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;&lt;/strong&gt; is the order of singularity and is always less than one). Thus you can see, as &lt;strong&gt;&lt;em&gt;&lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;&lt;/em&gt;&lt;/strong&gt; approaches &lt;strong&gt;zero&lt;/strong&gt;, the stress becomes infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Greater details can be found in any Fracture mechanics textbook. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now coming to stress concentrations, they occur due to abrupt changes in geometry. I think the figures you presented are the best explanations for them. The lines of force represented in the picture are basically the internal force that develops due to the application of an external load. Due to a constant external load, a certain internal force develops in response( and is uniform given a case of plate with no hole), now if the geometry is abruptly changing( a hole is drilled in the middle), the responsive internal force, still being the same in magnitude because the external load hasn't changed, has to be concentrated in some regions( in this case around the hole).  This will be clearer if you just look at the force lines for the bar without the hole in it. &lt;/p&gt;&#xA;" OwnerUserId="15803" LastActivityDate="2019-03-26T19:59:59.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26573" PostTypeId="2" ParentId="26097" CreationDate="2019-03-27T01:41:32.403" Score="3" Body="&lt;p&gt;It's hard to give a definitive answer for welding questions like this because there are so many variables, but here are some possible things to try:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Technique&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;From the photos it looks like you aren't able to maintain an arc on&#xA;the inside corner. Likely you need to decrease your arc length (if using stick/SMAW) or wire feed speed (if using MIG/GMAW)&#xA;and/or increase your 'heat' (current or voltage depending on the type&#xA;of power supply you are using.) Until you have a stable arc, nothing&#xA;else will matter much.&lt;/li&gt;&#xA;&lt;li&gt;20 amps is a very low amount of current for welding. Your concern about melting through is obviously valid on such thin material, but there are technique options to help. One option is to try weaving (aka transverse oscillation) from side to side as you proceed up the weld. This gives the heat a little time to distribute in each side while also ensuring that you get good fusion on both sides. For many types of weld this technique is frowned upon, but for this type of joint it will be fine. The other major technique change you can try here is traveling much faster across the weld. The photo of your side weld shows an excessive amount of weld metal deposited. While this isn't necessarily a problem, it means you are putting in more heat than you need to, making the risk of burn-through bigger. It's also correlated with a lack of good fusion. Running hotter and faster will likely produce a better weld all around this joint.&lt;/li&gt;&#xA;&lt;li&gt;Make sure if you don't weave that your electrode is running straight down the center of the joint. If you are having trouble seeing, you may need a bright light or an adjustable shade welding hood.&lt;/li&gt;&#xA;&lt;li&gt;Position the joint to be as favorable to a good weld as you can. If you were welding the inside corner in the orientation photographed, gravity will pull your weld pool from the vertical wall down on to the horizontal wall. If at all possible, rotate the assembly 45 degrees so that both tubes are at an equal 45 degree angle from the earth. This way gravity will help equally on both faces. The orientation of the joint makes such a big difference that in America structural welders aren't allowed to perform welds in a harder orientation than they have taken a qualification test on.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Joint Design&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On a tube joint like this, consider if you can simply skip welding the harder faces. If you don't need the joint to be totally sealed, and can weld the other faces strong enough, then you save time and effort by skipping the harder welds. Typically the inside corner would be considered the easier face in a t joint, but if the faces are easier for you then do them.&lt;/li&gt;&#xA;&lt;li&gt;If it's possible to add a steel backing inside of the joint, that will allow you to burn much hotter without worrying about burning through. A short slug of square bar or slightly smaller tube would be good candidates.&lt;/li&gt;&#xA;&lt;li&gt;If you could add a T shaped gusset plate to the outside and weld to that instead, your welding situation would be much easier.&lt;/li&gt;&#xA;&lt;li&gt;Tacking additional metal (scrap tube or plate) at the beginning and end of the joint gives you a place to start and stop your weld, which will help you move faster through the joint. These are called 'run-on' and 'run-off' tabs and are common in structural joints.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Equipment&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Thin metals like this are easiest to weld with the TIG (GTAW) process. If you can get access to TIG equipment, try the weld with that and very little filler metal.&lt;/li&gt;&#xA;&lt;li&gt;MIG will be the next easiest process. I can't tell for sure from the photo if you're attempting MIG or stick, but if it's stick you could try switching to MIG. Avoid self-shielded (flux core) wires as they are specifically designed for deeper penetration.&lt;/li&gt;&#xA;&lt;li&gt;If you're using MIG, short circuit transfer is generally preferred for thin metals as it has limited penetration (and thus risk of burn-through.) 75% argon 25% CO2 will be best for that.&lt;/li&gt;&#xA;&lt;li&gt;If you have access to a pulsed spray MIG machine, it will have settings to limit the amount of penetration which will control burn-through.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="122" LastActivityDate="2019-03-27T01:41:32.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26574" PostTypeId="1" CreationDate="2019-03-27T01:56:02.233" Score="1" ViewCount="1362" Body="&lt;p&gt;I am designing an end plate connection. My lecture notes show its bending capacity is calculated using elastic section modulus. For class 1 I-beams, we can use its plastic section modulus. I wonder whether plates could be designed this way since it's capacity would be 50% higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For info, I'm using Eurocode.&lt;/p&gt;&#xA;" OwnerUserId="19871" LastEditorUserId="19871" LastEditDate="2019-03-27T04:42:21.993" LastActivityDate="2020-05-15T12:25:22.290" Title="Can a plate under bending be designed using its plastic section modulus?" Tags="|structural-engineering|steel|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26575" PostTypeId="1" CreationDate="2019-03-27T06:42:35.550" Score="-1" ViewCount="75" Body="&lt;p&gt;We all know that manpower hydrofoils can fly over calm water. here &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/WszbRFF_WLk&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydrofoil Surface Flight&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is the manpower aircraft in the link below feasible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/uS-JLuLIQ6Q&quot; rel=&quot;nofollow noreferrer&quot;&gt;Manpower flight&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this kind of manned aircraft is feasible, as long as the area and lift center are well controlled.&lt;/p&gt;&#xA;" OwnerUserId="19873" LastEditorUserId="19873" LastEditDate="2020-03-22T21:27:26.060" LastActivityDate="2020-03-22T21:27:26.060" Title="Is such manned flight feasible?" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="26576" PostTypeId="1" CreationDate="2019-03-27T08:07:04.850" Score="4" ViewCount="122" Body="&lt;p&gt;In Eulerian description of fluid flow, we observe how the properties of the fluid are changing at a point. Since a point has zero dimension hence a zero area, how can we make sense of the 'finite pressure at the point' phrase?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since fluids are discrete in nature, a collection of molecules, the way I think about it is that Imagine that we are measuring the pressure with a measurement device [assume] which is cylindrical in shape having diameter enough so that Knudsen number is quite low [continuum hypothesis is valid]. At the molecular scale, Pressure is collisions of the fluid molecules with the measurement device. We position the device in the fluid with direction of fluid flow in such a way that diameter is the characteristic length. Let's say we keep on decreasing the diameter of the device hence the Knudsen number will keep on increasing and it will reach a point when the continuum hypothesis is not a good approximation anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't it be that the diameter should approach a finite threshold value [no matter how small it is] rather than zero for continuum hypothesis to hold true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[On the same note, when a pitot tube measures local velocity, it does not measure velocity at a point 'strictly' because it has a finite area of opening for the fluid to enter and I guess surface tension effects will also come in to play if we decrease the area of the opening. There is only so much area we can decrease theoretically for the continuum hypothesis to hold true.]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence the way I visualize a point in Eulerian description of fluid flow is that it is a collection of the minimum number of molecules for which the continuum hypothesis must hold true. Since molecules occupy space hence the point in this context does not have a zero area but a finite positive area [no matter how small it is].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please tell me if this interpretation is correct or not and flaw in reasoning.&lt;/p&gt;&#xA;" OwnerUserId="19874" LastEditorUserId="1832" LastEditDate="2019-03-27T14:30:02.367" LastActivityDate="2021-02-21T23:50:11.563" Title="Eulerian description of fluid flow" Tags="|mechanical-engineering|fluid-mechanics|civil-engineering|chemical-engineering|aerospace-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26578" PostTypeId="1" CreationDate="2019-03-27T12:27:50.863" Score="0" ViewCount="267" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uDQ4H.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uDQ4H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I tried to solve it but I don't know how to bring the Area A in this equation. I used the area of the small vessel, but this one shouldn't be in the equation. Thanks a lot&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S53FA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S53FA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19877" LastEditorUserId="110" LastEditDate="2019-03-28T14:31:05.950" LastActivityDate="2019-12-28T15:48:39.453" Title="Fluid Mechanic Buoyancy problem. How to determine the height?" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|pressure|statics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26580" PostTypeId="2" ParentId="26575" CreationDate="2019-03-27T13:45:02.927" Score="2" Body="&lt;p&gt;The video shows a human bean on a hydrofoil wake board or small surfboard using inertia to propel the board and passenger forward. The mass of the board is small compared to the human and the viscosity of the water is high compared to that of air. The foils are substantially smaller in proportion to those of an aircraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aircraft mass will be substantially greater than the human if it is capable of flight with such a payload.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider also the plight of the wind-up flying toy bird. It's much lighter than a person and has a greater power-to-weight ratio (rubber band versus muscle) than a person. It does fly, but only barely.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2019-03-27T20:08:01.767" LastActivityDate="2019-03-27T20:08:01.767" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26581" PostTypeId="1" AcceptedAnswerId="26586" CreationDate="2019-03-27T14:11:49.427" Score="0" ViewCount="4511" Body="&lt;p&gt;I'm trying to find the maximum bending moment and deflection for this symmetrical simple beam. The formulae I can find, deal with either cantilever beams with one fixed end, or continuous beams without cantilever ends, not this kind of situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0UhVh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0UhVh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to model the beam as two fixed-end cantilevers, on the basis that symmetry demands the middle of the beam is horizontal, and then modelled the loads as separate loads superimposed. But I'm not getting anywhere and not sure how I'd tell whether my answer was correct, if I did. I don't know if the fact that the beam can rise or fall in the middle (depending how F compares to W) is the problem, or if it's something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beam won't be exposed to non-linear behaviour. It could bend either way, as I don't know whether the 2 point forces will be larger or smaller than the UDL, or the exact lengths involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an easy way to work out the maximum deflection and B.M. of this beam, and if so, what are they?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note, this isn't homework or coursework (I haven't had those for many years!). It's part of a real-world practical design I'm trying to work on.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="6366" LastActivityDate="2022-06-15T19:18:23.373" Title="Simple beam with cantilever ends" Tags="|beam|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26582" PostTypeId="2" ParentId="26575" CreationDate="2019-03-27T14:49:10.250" Score="1" Body="&lt;p&gt;Unpowered heavier-than-air machines for human flight have been around for a long time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Official endurance records for gliders (sailplanes) were abandoned long ago, since the only thing being measured was the ability of the pilot to stay awake and still be in a fit mental state to make a controlled landing at the end of the flight. The last official record, a flight of just over 71 hours (almost 3 days), was made over 60 years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just throw away the &quot;surfboard,&quot; and redesign the rest of the structure to make it more aerodynamic!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-03-27T14:49:10.250" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26583" PostTypeId="1" AcceptedAnswerId="26623" CreationDate="2019-03-27T15:31:30.517" Score="0" ViewCount="936" Body="&lt;p&gt;I have a cable which dissipates &lt;span class=&quot;math-container&quot;&gt;$\dot Q_{total}$&lt;/span&gt; and I need to determine the surface temperature  of the insulation &lt;span class=&quot;math-container&quot;&gt;$T_{ins}$&lt;/span&gt; and the surface temperature of the conductor &lt;span class=&quot;math-container&quot;&gt;$T_{cond}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an insulating layer with the thermal resistance &lt;span class=&quot;math-container&quot;&gt;$R_{cond}$&lt;/span&gt; and the convective thermal resistance &lt;span class=&quot;math-container&quot;&gt;$R_{conv}$&lt;/span&gt;. The radiation component is neglected. From the geometry of the insulating layer I can obtain &lt;span class=&quot;math-container&quot;&gt;$R_{cond}$&lt;/span&gt;, but for &lt;span class=&quot;math-container&quot;&gt;$R_{conv}$&lt;/span&gt; I need to calculate the convection coefficient via the Nusselt-number. I have an empirical equation for the Nusselt-number, but the Rayleigh-number calculation requires the value of &lt;span class=&quot;math-container&quot;&gt;$T_{ins} - T_{ambient,air}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have the following equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\dot Q_{total} = (1/R_{total}) \cdot (T_{cond} - T_{ambient,air})$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$R_{total} = R_{cond} + R_{conv} = R_{cond} + Nu \cdot \lambda/L$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is the thermal conductivity and &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; the length of the cable. There are two unknown values in this equation &lt;span class=&quot;math-container&quot;&gt;$T_{cond}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T_{ins}$&lt;/span&gt;, because &lt;span class=&quot;math-container&quot;&gt;$Nu$&lt;/span&gt; depends on &lt;span class=&quot;math-container&quot;&gt;$T_{ins}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\dot Q_{total}$&lt;/span&gt; depends on &lt;span class=&quot;math-container&quot;&gt;$T_{cond}$&lt;/span&gt;. Is there a way to calculate the temperatures or do I have to define a temperature? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore I cannot determine the film temperature because  &lt;span class=&quot;math-container&quot;&gt;$T_{ins}$&lt;/span&gt; is unknown. At which temperature should I determine the thermodynamic properties for the Rayleigh- and Prandtl-number calculation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Summary:&#xA;Known: heat flow &lt;span class=&quot;math-container&quot;&gt;$\dot Q_{total}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$R_{conv}$&lt;/span&gt; (because geometry is known), &lt;span class=&quot;math-container&quot;&gt;$T_{ambient,air}$&lt;/span&gt; Unknown: &lt;span class=&quot;math-container&quot;&gt;$T_{ins}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$T_{cond}$&lt;/span&gt;, film temperature &lt;span class=&quot;math-container&quot;&gt;$T_{f}$&lt;/span&gt; and therefore also the exact values of Rayleigh, Prandtl and Nusselt-number.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want the thermal resistance of convection to be as accurate as possible, that is why I wanted to calculate it with these dimensionless quantities &lt;/p&gt;&#xA;" OwnerUserId="16313" LastEditorUserId="16313" LastEditDate="2019-03-29T10:05:06.947" LastActivityDate="2019-03-29T17:32:28.570" Title="Calculating the Surface Temperature of a Cable with known heat flow" Tags="|mechanical-engineering|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26584" PostTypeId="2" ParentId="26581" CreationDate="2019-03-27T15:39:31.570" Score="1" Body="&lt;p&gt;I don't think we can have an exact solution or even a proper estimate without having some numbers. If you look at the BM diagram of the structure, it looks something like this: &lt;a href=&quot;https://i.stack.imgur.com/tKOrY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tKOrY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it is quite obvious that without knowing the lengths and the load values involved we will never be able to estimate which peak will be higher. Similarly, for deflection, the magnitudes of the load affect the deflection pattern.  &lt;/p&gt;&#xA;" OwnerUserId="15803" LastActivityDate="2019-03-27T15:39:31.570" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26585" PostTypeId="1" CreationDate="2019-03-27T16:20:59.913" Score="1" ViewCount="478" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZkeEf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZkeEf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;im monitoring an overhung pump 19kw . and it shows a high peak at 7x rpm . on Thrust Vertical DE of pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what does it mean ? &lt;/p&gt;&#xA;" OwnerUserId="19879" LastEditorUserId="19879" LastEditDate="2019-03-27T16:33:51.550" LastActivityDate="2021-07-16T15:50:29.197" Title="Vibrations analysis 7x rpm" Tags="|vibration|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26586" PostTypeId="2" ParentId="26581" CreationDate="2019-03-27T16:43:52.737" Score="3" Body="&lt;p&gt;You are correct that symmetry demands that rotation at midspan be zero. However, as you also mentioned, there will obviously be deflection at midspan as well. You therefore cannot simplify the structure by placing a fixed support at midspan (not entirely sure what you meant by &quot;two fixed-end cantilevers&quot;, since you'd obviously still need to consider the pinned support).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, you need to simplify the structure by putting a &quot;fixed rotation&quot; support. That is, one which only resists rotation, presenting no force reactions, just a bending moment reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, you seem to be trying to do this from tables, without going into the math. In that spirit, let's see what we can do. For the record, I don't have a table handy, so I did some googling and will be using &lt;a href=&quot;https://i.stack.imgur.com/RWlOb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For starters, let's make use of the superposition principle and look at the loads individually (or rather, look at the distributed load first and then the pair of concentrated forces later).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Let's start with the distributed load:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the results for the midspan is straightforward. Just pretend this is a simply supported beam with no cantilevers (obviously this only works because the distributed load is only on the span and not on the cantilevers).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at tables, we know that the bending moment and deflection at midspan are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;M_{midspan} &amp;amp;= \dfrac{wL^2}{8} \\&#xA;\delta_{midspan} &amp;amp;= -\dfrac{5wL^4}{384EI}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note I'm using &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; and later &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; positive in the downward direction, feel free to invert the signs if you prefer another convention)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about the cantilevers? Well, we trivially know that &lt;span class=&quot;math-container&quot;&gt;$M_{support} = 0$&lt;/span&gt; since the entirety of the distributed load is absorbed by the supports, so there's nothing on the cantilevers to generate any moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the cantilevers will present a deflection due to the rotation at the support. Though the lack of loading means that the cantilevers will act as undeformed straight beams, they will suffer a rigid-body rotation which will lead to a deflection at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic trigonometry tells us that the vertical deflection of the cantilever will be equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta_{cant} = d\cdot\tan\theta_{support}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, civil engineers are lazy, so we just take the first-order Taylor series of &lt;span class=&quot;math-container&quot;&gt;$\tan\theta$&lt;/span&gt;, and simplify it to merely &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; (this is only valid for small deflections and rotations, but that's the usual space inhabited by civil engineering, so it's fine for our purposes). So we can simplify the above to merely&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta_{cant} = d\theta_{support}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, &lt;span class=&quot;math-container&quot;&gt;$\theta_{support}$&lt;/span&gt; can also be found in tables, and is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\theta_{support} = \dfrac{wL^3}{24EI} \\&#xA;\therefore \delta_{cant} = d\dfrac{wL^3}{24EI}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now let's move onto the concentrated forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that the bending moment diagram for these forces will look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/whhtt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/whhtt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can trivially observe that &lt;span class=&quot;math-container&quot;&gt;$M_{support} = -Fd$&lt;/span&gt; (negative because it generates tension on the top fibers). The constant bending moment diagram along the span tells us that the totality of each concentrated load is entirely absorbed by the nearest support. This means that, as far as the span is concerned, there are no concentrated forces at all! It's only being acted upon by a constant bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, we can change the model (when looking at the midspan) into this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lZbUL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lZbUL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankfully, this case is present in many tabled results. We already know the bending moment, but we're still looking for the deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My table only has the result for a single concentrated bending moment at one end, so we'll just have to double the deflection it gives us for the midspan (which we know is where the maximum deflection actually happens in our case):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta_{midspan} = 2\times\dfrac{ML^2}{16EI} = \dfrac{FdL^2}{8EI}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with that we have the bending moment diagram due to the concentrated loads and the maximum deflection at midspan due to those loads at the cantilever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we don't yet have, however, is the deflection at the cantilevers themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For that, we'll need to get a bit creative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deflection at the cantilevers is a function of two things: the rotation at the support and the additional deformation which happens between the support and the cantilever's end. Thankfully, both of those things are easy to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotation at the support can be found using the same concentrated-bending-moments model we used to find the deflection at midspan, and then looking at the same table. In this case, the table gives us the rotation at the support with the bending moment and at the other. It also gives them different signs since the beam has a positive slope on one end and a negative slope on the other. In our case, we're just going to add both of those values ignoring the signs, since our double-bending-moment case will increase the rotation at both supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\theta_{support} = \dfrac{ML}{6EI} + \dfrac{ML}{3EI} = \dfrac{ML}{2EI}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, just as we did with the distributed load, the deflection at the cantilever due to this rotation at the support can be easily found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta_{cant,rotation} = d\theta_{support}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the additional deflection due to the cantilever's deformation after the support, we can simply model the cantilever alone as if it had a fixed support instead of a pinned one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The table I used before doesn't actually have cantilever results, but I remember that a cantilever's maximum deflection is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta_{cant,deform} = \dfrac{Fd^3}{3EI}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can therefore conclude that the total deformation at the cantilever due to the concentrated loads is equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\delta_{cant} &amp;amp;= d\theta_{support} + \dfrac{Fd^3}{3EI} \\&#xA;&amp;amp;= d\dfrac{ML}{2EI} + \dfrac{Fd^3}{3EI} \\&#xA;&amp;amp;= d\dfrac{FdL}{2EI} + \dfrac{Fd^3}{3EI} \\&#xA;&amp;amp;= \dfrac{F}{EI}\left(\dfrac{d^2L}{2} + \dfrac{d^3}{3}\right) \\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After all that, we now bring everything together by adding the results from the distributed load and concentrated forces:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;M_{midspan} &amp;amp;= \dfrac{wL^2}{8} - Fd \\&#xA;\delta_{midspan} &amp;amp;= -\dfrac{5wL^4}{384EI} + \dfrac{FdL^2}{8EI} \\&#xA;M_{support} &amp;amp;= 0 - Fd \\&#xA;\delta_{cant} &amp;amp;= \dfrac{wdL^3}{24EI} - \dfrac{F}{EI}\left(\dfrac{d^2L}{2} + \dfrac{d^3}{3}\right)&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-03-27T16:51:27.907" LastActivityDate="2019-03-27T16:51:27.907" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26587" PostTypeId="2" ParentId="26571" CreationDate="2019-03-27T16:53:59.857" Score="1" Body="&lt;p&gt;The definition of the singularity function &lt;span class=&quot;math-container&quot;&gt;$ &amp;lt;x-a&amp;gt;^0 $&lt;/span&gt; is 0 if &lt;span class=&quot;math-container&quot;&gt;$x&amp;lt;a$&lt;/span&gt; and equal to 1 if &lt;span class=&quot;math-container&quot;&gt;$x\ge a$&lt;/span&gt;. It does not matter if x-a is 0 or infinity; the value is still 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;0&amp;gt;^0 = 1$&lt;/span&gt; by definition.&lt;/p&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2019-03-27T16:53:59.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26589" PostTypeId="1" CreationDate="2019-03-27T19:11:57.570" Score="0" ViewCount="130" Body="&lt;p&gt;I work in a sleep research lab. We have a couple of Masimo pulse oximeters that report pulse oximetry values in terms of integers (e.g. 90, 91, 92, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now our principle investigators used to use an Ohmeda Biox 3700 that reported data via analog output in a nice (noisy) curve (e.g. after filtering, 90, 90.1, 90.3, 90.4, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's some discussion that because of the lack of decimal, the Masimo is useless for data reporting. The Ohmeda though is from the 90s and I can't find any documentation that indicates if taking the decimal is even acceptable for the device's resolution. Any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="19880" LastEditorUserId="33" LastEditDate="2019-03-28T14:11:05.310" LastActivityDate="2020-08-20T11:06:30.597" Title="Pulse Oximetry, does the decimal matter?" Tags="|biomedical-engineering|medical-devices|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26590" PostTypeId="1" CreationDate="2019-03-27T19:29:20.647" Score="0" ViewCount="204" Body="&lt;p&gt;In order to get the bending and the shear function using the singularity functions we integrate loading. could you please tell me why the two constants are zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know that  the load is related by the bending M(x) and the shear equation using these equations: q(x)=dV(x)/dx and V=dM/dx. In order to find the M(x) and V(x) Using the singularity method I found that the constants of integration are null:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ V(x) = \int q(x)dx\,.$$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$$ M(x) = \int V(x)dx\,.$$&lt;/span&gt; without constants.&#xA;Could you please clarify why these constants are always null?&lt;/p&gt;&#xA;" OwnerUserId="19870" LastEditorUserId="33" LastEditDate="2019-03-28T12:35:01.103" LastActivityDate="2019-03-28T22:35:07.240" Title="why the constants of the shear and bending are zero" Tags="|statics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26591" PostTypeId="2" ParentId="26554" CreationDate="2019-03-27T21:11:27.563" Score="0" Body="&lt;ul&gt;&#xA;&lt;li&gt;Define your inputs (i.e., control surfaces on the fins, steerable motors, etc.)&lt;/li&gt;&#xA;&lt;li&gt;Define your output (roll position of the rocket)&lt;/li&gt;&#xA;&lt;li&gt;Find the differential equation that describes the roll in terms of the input.&lt;/li&gt;&#xA;&lt;li&gt;Linearize (note that if you're using fins then the linearized equation will be dependent on airspeed, and that if you're using steerable motors then the linearized equation will be dependent on thrust -- this will complicate your life, but get over the first hurdle first!)&lt;/li&gt;&#xA;&lt;li&gt;Extract the transfer function&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;State space is not a bad way to go, but if you're only concerned with roll control, and if you're confident that you're not going to be affecting the behavior in yaw or pitch, then this is a classic single-input, single-output system that can be handled with transfer functions.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-03-27T21:11:27.563" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26593" PostTypeId="2" ParentId="26537" CreationDate="2019-03-27T21:30:51.670" Score="0" Body="&lt;p&gt;I prefer none of the above.  This works better, in my experience.  If you work out the math, you'll find that it's equivalent if you set &lt;span class=&quot;math-container&quot;&gt;$d_d = 0$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u_d[k] = (1-d_d)(e[k]-e[k-1]) + d_d u_d[k-1] \\&#xA;  u_i[k] = e[k] + u_i[k] \\&#xA;  u[k] = k_i u_i[k] + k_p u[k] + k_d u_d[k]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I've introduced the &lt;span class=&quot;math-container&quot;&gt;$d_d$&lt;/span&gt; term -- this has the effect of band-limiting the derivative term (the closer &lt;span class=&quot;math-container&quot;&gt;$d_d$&lt;/span&gt; is to 1, the slower the derivative), which will save you a world of grief from high-frequency oscillations.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-03-27T21:30:51.670" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26594" PostTypeId="1" CreationDate="2019-03-27T23:54:52.100" Score="1" ViewCount="80" Body="&lt;p&gt;I will try to lift up 25kg (50 pounds)(electrical child car) on a linear actuator and I know that there would be a 30cm long lever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cihmZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cihmZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my question is if there will be some lever. I think it won´t because the linear actuator will be in the middle above the load. Or am I wrong and there will be a 20cm or 30cm or 50cm lever?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wycOp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wycOp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your advice.&lt;/p&gt;&#xA;" OwnerUserId="19883" LastActivityDate="2019-03-29T17:17:58.793" Title="Is it a Lever Or Not?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26595" PostTypeId="1" AcceptedAnswerId="42636" CreationDate="2019-03-28T00:21:27.540" Score="1" ViewCount="552" Body="&lt;p&gt;I have a shaft inside a bushing, the shaft has a gear on one end that is supposed to provide resistance to other gears it is driven with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a hard time finding the correct term for the sort of grease or fluid, or in more lay-man's terms a &quot;goo&quot;, that will provide &lt;em&gt;increased&lt;/em&gt; friction between the shaft and the bushing, rather than reducing friction, as it is with regular oil or grease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially a tar-like substance, that isn't as messy as actual tar. Perhaps also similar to tree-sap, except more manageable in terms of putting it into a mechanical device. Also tree-sap gets hard over time, I'd like to keep its mechanical properties, like viscosity, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you're wondering, this is going to be part of a soft-open gearing, such that a small door doesn't fling open when the lock is released, instead it provides some resistance.&lt;/p&gt;&#xA;" OwnerUserId="19884" LastActivityDate="2021-04-16T17:50:31.407" Title="What are fluids or &quot;goos&quot; called, that increase friction?" Tags="|mechanical|friction|fluid|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26596" PostTypeId="2" ParentId="26595" CreationDate="2019-03-28T00:55:41.093" Score="0" Body="&lt;p&gt;That made for some interesting links... You may be looking for shear thickening lubricant. According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Dilatant&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiki page&lt;/a&gt;, some of these products can also be called non-Newtonian fluids, but I would not suggest cornstarch and water for your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional research resulted in a question posted &quot;elsewhere&quot; that suggests the product you seek is rare, uncommon and difficult to find. One application listed is as projectile defense in body armor!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen slow opening devices which use a cylinder linkage in which the cylinder allows &quot;normal&quot; fluid to travel through a circuit with a restricting opening. Think about a syringe in that you have a large diameter plunger and a small diameter needle. Plenty of force does not result in rapid travel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can create a closed system similar to a syringe using air as the working fluid, if one can drill a small enough hole in the restrictor plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, I have seen speed limiters that use a high gear ratio assembly to spin a simple squirrel-cage impeller fan. Similarly, there is arresting gear for exiting a building that also uses a spinning fan of paddle wheel design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you used &quot;small&quot; in your question, I suggest paddle wheel or squirrel cage fans may not work well in tiny scale, but a cylinder and orifice can be reduced and still be effective. You can get away with a clean wall cylinder and a piston with such tight tolerances that the entire perimeter is the orifice. If properly machined, no seals (o-rings) would be needed, only a light lubricant and possibly periodic lubing.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-03-28T00:55:41.093" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26597" PostTypeId="1" AcceptedAnswerId="26598" CreationDate="2019-03-28T03:05:08.820" Score="0" ViewCount="1040" Body="&lt;p&gt;Hello all I am drawing a hole for M3 countersunk screw(flathead) I have checked the specs online the diameter of the head is largest 5.5mm thick 1.65 90degree&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I draw a 5.5 hole 90degree it's only 1.3 thick as I remember, so I measure the actual product that comes with M3 countersunk it's roughly 6.5mm diameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is which size should I draw for a M3 countersunk screw? they're on 3mm abs plastic and 2mm aluminum, thank you very much. &lt;/p&gt;&#xA;" OwnerUserId="18994" LastActivityDate="2019-03-28T08:39:33.323" Title="Curious of drawing a hole for M3 countersunk screw, 5.5mm or 6.5mm?" Tags="|cnc|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26598" PostTypeId="2" ParentId="26597" CreationDate="2019-03-28T08:39:33.323" Score="1" Body="&lt;p&gt;This is what Inventor recommends for a clearance hole:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4nQPb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4nQPb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for a tapped hole:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o5OFD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o5OFD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-03-28T08:39:33.323" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26602" PostTypeId="1" CreationDate="2019-03-28T14:33:37.170" Score="0" ViewCount="40" Body="&lt;p&gt;How can we know the evolution of the oulet concentration of a perfectly well-mixed continuous flow reactor taking into account the residence time distribution function called &lt;span class=&quot;math-container&quot;&gt;$E(t)$&lt;/span&gt;? I have heard about this model&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$C_{RP} = \int_0^\infty C_{RF}(t)E(t)\text{d}t$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;E(t) &amp;amp;= \dfrac{1}{t}\exp\left(-\dfrac{t}{\tau}\right) \\&#xA;\tau &amp;amp;= \dfrac{V}{q}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation only gives the outlet concentration at the steady-state  called &lt;span class=&quot;math-container&quot;&gt;$C_{SS,RP}$&lt;/span&gt; (so one value). I am looking for the evolution as a function of time &lt;span class=&quot;math-container&quot;&gt;$C_{RP}(t)$&lt;/span&gt; for the concentration in a perfectly well-mixed batch reactor.&lt;/p&gt;&#xA;" OwnerUserId="19892" LastEditorUserId="1832" LastEditDate="2019-03-28T22:56:09.610" LastActivityDate="2019-03-28T22:56:09.610" Title="performances of a continuous flow reactor" Tags="|chemical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26603" PostTypeId="1" AcceptedAnswerId="26608" CreationDate="2019-03-28T17:48:51.347" Score="0" ViewCount="29" Body="&lt;p&gt;After introducing a quantization block in the simulink model, the output of the controlled system starts to oscillate around the zero fixpoint.&#xA;I am using a lqr controller to regulate the system, which was designed based on the continuous system. &#xA;Can I tune the parameters of the lqr design to fix this problem ? Which method is recommended here ?&lt;/p&gt;&#xA;" OwnerUserId="18808" LastActivityDate="2019-03-28T22:25:17.540" Title="How can I damp oscillating behaviour caused by quantization in a digital control system?" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26605" PostTypeId="1" CreationDate="2019-03-28T18:54:01.517" Score="0" ViewCount="128" Body="&lt;p&gt;Does anybody know the name of this type of bearing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ScM0i.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ScM0i.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19897" LastEditorUserId="1832" LastEditDate="2019-03-28T22:57:16.710" LastActivityDate="2019-04-27T23:01:35.280" Title="Name that bearing!" Tags="|bearings|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26606" PostTypeId="2" ParentId="26605" CreationDate="2019-03-28T19:07:04.803" Score="2" Body="&lt;p&gt;&quot;&lt;a href=&quot;https://www.rollon.com/IN/en/products/linear-line/1-compact-rail/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Compact rail bearing&lt;/a&gt;&quot; looks like a Timken subsidiary makes them. &lt;/p&gt;&#xA;" OwnerUserId="18926" LastActivityDate="2019-03-28T19:07:04.803" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26608" PostTypeId="2" ParentId="26603" CreationDate="2019-03-28T22:25:17.540" Score="0" Body="&lt;p&gt;The typical method is to choose hardware whose quantization is lower than the system's ambient noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can't damp it out -- it's a nonlinear phenomenon, which, when linearized around a switch in the output, has infinite incremental gain.  If for some reason you can't just make the quantization small enough, you can add some dither to it -- either add a sawtooth right before the worst quantization step, or implement a delta-sigma modulator (a 1st-order is enough in my experience).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-03-28T22:25:17.540" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26609" PostTypeId="2" ParentId="26590" CreationDate="2019-03-28T22:35:07.240" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\newcommand{\a}[1]{\langle#1\rangle}$&lt;/span&gt;The reason is simple: the singularity-function form of the shear and bending moment equations already perfectly describes the results without the need for integration constants.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because the constants serve to reflect the impact of the beam's support reactions on the shear and bending moment, but the singularity-function form already includes these reactions in its basic form.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, imagine we have a free-floating beam (so no supports) with a uniformly distributed load &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; throughout its span. This would lead to a rigid-body movement in the direction of the load (since the beam has no supports to resist that external force) and no internal stresses, but just pretend that's not the case: the beam magically resists that motion, generating shear and bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using simple integration, we can find these shear and bending moment equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;q(x) &amp;amp;= q \\&#xA;V(x) &amp;amp;= \int q(x)\text{d}x = qx + C_1 \\&#xA;M(x) &amp;amp;= \int V(x)\text{d}x = \dfrac{qx^2}{2} + C_1 x + C_2&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the specific values of &lt;span class=&quot;math-container&quot;&gt;$C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_2$&lt;/span&gt;, those are defined by the boundary conditions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the beam is simply-supported, then the constants are (&lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the support reaction)&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;V(0) &amp;amp;= C_1 = R \\&#xA;M(0) &amp;amp;= C_2 = 0 \\&#xA;V(x) &amp;amp;= qx + R \\&#xA;M(x) &amp;amp;= \dfrac{qx^2}{2} + Rx&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the beam is a cantilever fixed at &lt;span class=&quot;math-container&quot;&gt;$x=L$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;V(0) &amp;amp;= C_1 = 0 \\&#xA;M(0) &amp;amp;= C_2 = 0 \\&#xA;V(x) &amp;amp;= qx \\&#xA;M(x) &amp;amp;= \dfrac{qx^2}{2}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the beam is a cantilever fixed at &lt;span class=&quot;math-container&quot;&gt;$x=0$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$R_y$&lt;/span&gt; is the vertical reaction, and &lt;span class=&quot;math-container&quot;&gt;$R_M$&lt;/span&gt; is the moment reaction)&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;V(0) &amp;amp;= C_1 = R_y \\&#xA;M(0) &amp;amp;= C_2 = R_M \\&#xA;V(x) &amp;amp;= qx + R_y \\&#xA;M(x) &amp;amp;= \dfrac{qx^2}{2} + R_y x + R_M&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Even if the supports are &quot;within&quot; the span (as in, not at the extremities), the same basic equations will remain valid. The only difference will be that the reactions will lead to discontinuities in the beam results, so you'll need to break the beam down into its component parts and use the basic equations for each part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the basic equation (i.e. &lt;span class=&quot;math-container&quot;&gt;$V = qx + C_1$&lt;/span&gt;) is defined purely by the loading, since that's where it comes from. The support reactions then set the value of the integration constants &lt;span class=&quot;math-container&quot;&gt;$C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_2$&lt;/span&gt; to make that basic equation represent the specific case being studied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With singularity functions, this happens naturally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's because singularity functions aren't described only by the loading, but also by the supports. After all, the support reactions are placed directly into the &quot;loading&quot; function itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, while the loading equation in the simple-integration case was just &lt;span class=&quot;math-container&quot;&gt;$q(x) = q$&lt;/span&gt;, it now depends on the boundary conditions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the beam is simply-supported, then we get&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;q(x) &amp;amp;= q + R\a{x-0}^{-1} \\&#xA;V(x) &amp;amp;= \int q(x)\text{d}x = qx + R\a{x-0}^0 \\&#xA;&amp;amp;= qx + R\\&#xA;M(x) &amp;amp;= \int V(x)\text{d}x = \dfrac{qx^2}{2} + R\a{x-0}^1 \\&#xA;&amp;amp;= \dfrac{qx^2}{2} + Rx&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the beam is a cantilever fixed at &lt;span class=&quot;math-container&quot;&gt;$x=L$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;q(x) &amp;amp;= q \\&#xA;V(x) &amp;amp;= \int q(x)\text{d}x = qx \\&#xA;M(x) &amp;amp;= \int V(x)\text{d}x = \dfrac{qx^2}{2}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the beam is a cantilever fixed at &lt;span class=&quot;math-container&quot;&gt;$x=0$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$R_y$&lt;/span&gt; is the vertical reaction, and &lt;span class=&quot;math-container&quot;&gt;$R_M$&lt;/span&gt; is the moment reaction)&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;q(x) &amp;amp;= q + R_y\a{x-0}^{-1} + R_M\a{x-0}^{-2} \\&#xA;V(x) &amp;amp;= \int q(x)\text{d}x = qx + R_y\a{x-0}^0 + R_M\a{x-0}^{-1} \\&#xA;&amp;amp;= qx + R_y \\&#xA;M(x) &amp;amp;= \int V(x)\text{d}x = \dfrac{qx^2}{2} + R_y\a{x-0}^1 + R_M\a{x-0}^0 \\&#xA;&amp;amp;= \dfrac{qx^2}{2} + R_y x + R_M&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Comparing these results to those found by traditional integration, they are the same. And that's precisely because of the difference in how the support reactions are treated in each case: traditional integration ignores them and then later adds their effect via the integration constants. Meanwhile, singularity functions include the boundary conditions at the very start in how they define &lt;span class=&quot;math-container&quot;&gt;$q(x)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you continue integrating past the bending moment into the slope and deflection equations, you then have another set of conditions which haven't been taken into account yet: the physical constraints such as &lt;span class=&quot;math-container&quot;&gt;$\theta=\delta=0$&lt;/span&gt; at fixed supports. Since this effect isn't incorporated into the singularity equations, you'll need to add integration constants for each of these additional integrations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to take the example of the cantilever with fixed end at &lt;span class=&quot;math-container&quot;&gt;$x=0$&lt;/span&gt; above, the integration up to the deflection is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;q(x) &amp;amp;= q + R_y\a{x-0}^{-1} + R_M\a{x-0}^{-2} \\&#xA;V(x) &amp;amp;= \int q(x)\text{d}x = qx + R_y\a{x-0}^0 + R_M\a{x-0}^{-1} \\&#xA;M(x) &amp;amp;= \int V(x)\text{d}x = \dfrac{qx^2}{2} + R_y\a{x-0}^1 + R_M\a{x-0}^0 \\&#xA;\theta(x) &amp;amp;= \int M(x)\text{d}x = \dfrac{qx^3}{6} + \dfrac{R_y}{2}\a{x-0}^2 + R_M\a{x-0}^1 + C_1\\&#xA;\delta(x) &amp;amp;= \int \theta(x)\text{d}x = \dfrac{qx^4}{24} + \dfrac{R_y}{6}\a{x-0}^3 + \dfrac{R_M}{2}\a{x-0}^2 + C_1 x + C_2\\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, &lt;span class=&quot;math-container&quot;&gt;$C_1 = C_2 = 0$&lt;/span&gt;, since the fixed support at &lt;span class=&quot;math-container&quot;&gt;$x=0$&lt;/span&gt; has &lt;span class=&quot;math-container&quot;&gt;$\theta(0)=\delta(0)=0$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-03-28T22:35:07.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26610" PostTypeId="2" ParentId="26585" CreationDate="2019-03-28T23:17:37.223" Score="1" Body="&lt;p&gt;According to the OP's comment, the pump has a 7 bladed impeller. Each blade will cause some disturbance to the pressure and/or velocity in the flow. Those disturbances will cause a force on the mounting of the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The text on the OP's picture says the pump RPM was 2940, or approximately 50 revolutions per second. The frequency of the disturbances in the flow is therefore about 350 Hz. Maybe the pump and the structure it is mounted on has a natural vibration frequency close to 350 Hz moving in the vertical direction, and that is why you are getting a large amplitude on your plot.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-03-28T23:17:37.223" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26611" PostTypeId="1" AcceptedAnswerId="26612" CreationDate="2019-03-29T02:45:19.180" Score="0" ViewCount="55" Body="&lt;p&gt;I want to secure a box (roughly 20kg in total) that contains electronic equipment to the underside of a flat steel structure. We can weld onto the underside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steel structure has a high chance of receiving mechanical impacts from above (for example, large rocks falling onto it), so I'd like to find ways to dampen the mechanical shock and vibration reaching the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen shock mounts/isolation mounts before (like moulded rubber mounts), but they seem to be designed for supporting objects that rest on top of the mount -- in other words, the mount is designed to be under compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What options do I have if the object is &quot;hanging&quot; from the mount?&lt;/p&gt;&#xA;" OwnerUserId="19904" LastEditorUserId="19904" LastEditDate="2019-03-29T03:18:17.270" LastActivityDate="2019-03-29T06:26:28.277" Title="Shock/Isolation mounting under tension" Tags="|structural-engineering|structures|vibration|shock|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26612" PostTypeId="2" ParentId="26611" CreationDate="2019-03-29T06:26:28.277" Score="1" Body="&lt;p&gt;Use U brackets with the tops of the U welded to the table underside and flatten the bottom of the U to provide the support - then the vibration mounts can be in compression also side supports can be used to minimize the sidways movement.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-03-29T06:26:28.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26615" PostTypeId="2" ParentId="26589" CreationDate="2019-03-29T09:31:47.210" Score="1" Body="&lt;p&gt;The accepted error in pulse oximtery isn't that good.  It's about &lt;span class=&quot;math-container&quot;&gt;$\pm$&lt;/span&gt;2 % so your additional significant figures are useless.&#xA;A good review is &lt;em&gt;&lt;strong&gt;Aoyagi, T. J Anesth (2003) 17: 259.&lt;/strong&gt;&lt;/em&gt; &lt;a href=&quot;https://doi.org/10.1007/s00540-003-0192-6&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://doi.org/10.1007/s00540-003-0192-6&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="16544" LastActivityDate="2019-03-29T09:31:47.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26616" PostTypeId="2" ParentId="26514" CreationDate="2019-03-29T10:28:09.810" Score="1" Body="&lt;p&gt;I would not overthink the lines of force explanation too much; its a concept that really only makes sense in the context of vanishing shear forces, like masonry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, the simplest explanation of why stress concentrations occur is as follows: consider a rod with a step function in its radius, like in your initial picture. Given a fixed axial load, the most optimal distribution for carrying that load is a uniform distribution over the cross section. Every element carries the same load, and no one area will have a reason to fail before any others do. A uniform rod might reach that distribution if the force is led-in sufficiently gently. But can a rod with a sudden change in cross-section do the same thing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best case scenario would be if the thin end of the rod would still maintain its optimal uniform stress distribution. But resistance to deformation in the thick end will tend to favor the same flat stress distribution profile in the thick end, and it will settle into that at least far away from the cross sectional change. But the uniform distribution in the thick and thin end will have a different stress magnitude; they cant just meet as two uniform distributions of different magnitudes and be in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by that argument alone, we have proven that the stress distribution near the sectional change, can and will not be the optimal uniform profile; that is it will have some local maxima; concentrations if you will.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now that still leaves open the possibility that at least the thin end would have the good taste to maintain its optimal uniform distribution, gradually spreading to a wider distribution in the thin end, without unnecessarily weakening our construction. And you can engineer a situation like that, with a nicely rounded transition. But if you consider the effect of the 'extra' material that exists, from a sharp geometry relative to a rounded geometry, you realise that relative to the deformation in the rounded case, that extra material would have to experience a deformation to match the deformation of the rod. You can think of the stress concentration as the reaction force required to deform that 'extra' material along with the rest of the rod. And since the extra material is located out there near the sectional change on the perimeter, thats where you will find a stress peak. Even though the thin end of the rod would love to stay in a nice flat stress distribution, if you put some extra material squeezing it into another shape away from a flat stress profile; well, ask and you shall receive.&lt;/p&gt;&#xA;" OwnerUserId="14899" LastActivityDate="2019-03-29T10:28:09.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26617" PostTypeId="1" CreationDate="2019-03-29T12:29:05.587" Score="1" ViewCount="295" Body="&lt;p&gt;It's a known fact that you cannot place a curtain near a radiator because it will trap hot air behind the curtain and also block cold air from going through the radiator from bellow.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But obviously there must be a minimum distance you can place a curtain that goes from ceiling to floor in front of the radiator that would allow it to do its normal work. What is this distance in centimeters(hopefully) ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NNr0v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NNr0v.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NG2eW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NG2eW.jpg&quot; alt=&quot;As you can see the air can go up and to the room from the sides&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JX7jU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JX7jU.jpg&quot; alt=&quot;I placed the curtains 10 cm in front of the radiator so that it doens&amp;#39;t stop airflow, I hope this is good enough&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1885" LastEditorUserId="1885" LastEditDate="2019-03-29T21:04:45.013" LastActivityDate="2019-03-29T21:04:45.013" Title="Minimum distance between long curtains and radiator for good air flow?" Tags="|thermodynamics|convection|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26619" PostTypeId="2" ParentId="26594" CreationDate="2019-03-29T12:43:47.733" Score="2" Body="&lt;p&gt;If the actuator is linear, it would be best if you followed the second option where there is no moment with respect to the axis of movement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you would need to make sufficiently rigid the platform that raises the load. &lt;/p&gt;&#xA;" OwnerUserId="19906" LastEditorUserId="1832" LastEditDate="2019-03-29T17:17:58.793" LastActivityDate="2019-03-29T17:17:58.793" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26620" PostTypeId="1" CreationDate="2019-03-29T13:48:18.400" Score="1" ViewCount="268" Body="&lt;p&gt;The wheel is made from a 10-kg thin ring and four 0,5-kg slender rods. If the torsinal spring attached to the wheel's center has a stiffnes &lt;span class=&quot;math-container&quot;&gt;$ k = 2 N*m/rad$&lt;/span&gt;, so that the torque on the center of the wheel is &lt;span class=&quot;math-container&quot;&gt;$ M = (2* \theta)*N*m$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is in radians. The wheel is rotated two full turns, &lt;span class=&quot;math-container&quot;&gt;$\theta = 4*\pi$&lt;/span&gt; Determine the angular velocity when the &lt;span class=&quot;math-container&quot;&gt;$\theta = 0$&lt;/span&gt; and determine the angular acceleration immediately after release, just before the movement begins. Determine the angular acceleration at the moment when the wheel is back to its original position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have calculated the angular velocity but I have problems calculating the angular acceleration. Where to start?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$w = 10.9 rad/s$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\alpha_1 = 9.5 rad/s^2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\alpha_2 = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GJaqj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GJaqj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wdzkR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wdzkR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19908" LastActivityDate="2024-01-05T19:03:05.340" Title="Angular acceleration and the moment of inertia" Tags="|mechanical-engineering|dynamics|acceleration|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26621" PostTypeId="2" ParentId="26578" CreationDate="2019-03-29T16:41:11.980" Score="0" Body="&lt;p&gt;Since the mass of the fluid and its density are constant, the volume of the fluid is constant. Use that to find a formula that relates the change in height of the fluid on the side of the small cylinder to the change in height of the fluid on the side of the large cylinder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the small cylinder lowers by 1 unit, the fluid must rise on the big cylinder by (Aouter-Ainner)/A inner. So the level rises on the inner cylinder by 1 + (Aouter - Ainnier)/Ainner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus the change in the level of the outer cylinder is [(Aouter-Ainner)/Ainner]/[1 + (Aouter-Ainner)/Ainner]. This can be further simplified with algebra.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-03-29T16:41:11.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26623" PostTypeId="2" ParentId="26583" CreationDate="2019-03-29T17:32:28.570" Score="0" Body="&lt;h1&gt;Problem Statement&lt;/h1&gt;&#xA;&lt;p&gt;You have three unknowns: the temperature at the conductor &lt;span class=&quot;math-container&quot;&gt;$T_c$&lt;/span&gt;, the temperature at the insulation &lt;span class=&quot;math-container&quot;&gt;$T_i$&lt;/span&gt;, and the air heat transfer coefficient &lt;span class=&quot;math-container&quot;&gt;$h_a$&lt;/span&gt;. You know the air temperature &lt;span class=&quot;math-container&quot;&gt;$T_a$&lt;/span&gt;, heat flow per length &lt;span class=&quot;math-container&quot;&gt;$\dot{q}/L \equiv q_L$&lt;/span&gt;, the geometry of the cable + insulation (radii &lt;span class=&quot;math-container&quot;&gt;$r_c$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$r_i$&lt;/span&gt;), and the insulation material (thermal conductivity &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;).&lt;/p&gt;&#xA;&lt;p&gt;You have two energy balance equations.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ q_L =  \frac{2\pi\ k}{\ln(r_i/r_c)}\ (T_c - T_i)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ q_L = 2\pi\ r_i\ h_a\ (T_i - T_a) $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In the set of unknowns, you also know that &lt;span class=&quot;math-container&quot;&gt;$h_a = f(T_i, T_a)$&lt;/span&gt;. The relation is solved through a correlation equation, chart, or table using the Nusselt number &lt;span class=&quot;math-container&quot;&gt;$Nu$&lt;/span&gt; (and Reynolds number and Prandtl number no doubt). The flow is (ostensibly) laminar (because you mention film temperature, meaning boundary layer theory, meaning laminar flow).&lt;/p&gt;&#xA;&lt;h1&gt;Proposed Approach&lt;/h1&gt;&#xA;&lt;p&gt;When we know &lt;span class=&quot;math-container&quot;&gt;$h_a$&lt;/span&gt;, we can solve for both unknown temperatures.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt; Make a plot of &lt;span class=&quot;math-container&quot;&gt;$T_c$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T_i$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$h_a$&lt;/span&gt; over a range of values for &lt;span class=&quot;math-container&quot;&gt;$h_a$&lt;/span&gt; that are expected to be relevant for the physical system.&lt;/p&gt;&#xA;&lt;p&gt;We do not know &lt;span class=&quot;math-container&quot;&gt;$h_a$&lt;/span&gt; until we know &lt;span class=&quot;math-container&quot;&gt;$T_i$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt; Calculate &lt;span class=&quot;math-container&quot;&gt;$h_a$&lt;/span&gt; as a function of &lt;span class=&quot;math-container&quot;&gt;$T_i$&lt;/span&gt;. Superimpose the values correspondingly on the above graph.&lt;/p&gt;&#xA;&lt;p&gt;The answer for the three unknown values will be at the point where the line for &lt;span class=&quot;math-container&quot;&gt;$T_i$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$h_a$&lt;/span&gt; made using the energy balance equations intersects with the line for &lt;span class=&quot;math-container&quot;&gt;$h_a$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$T_i$&lt;/span&gt; made using the correlation expressions.&lt;/p&gt;&#xA;&lt;p&gt;Here is an example of the plot that I believe you might have at the end.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WqkRK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WqkRK.png&quot; alt=&quot;profile plot&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The circles show how to obtain the three unknown values for your system.&lt;/p&gt;&#xA;&lt;p&gt;You &lt;em&gt;could&lt;/em&gt; also iterate through the equations. I prefer a graphical approach.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-03-29T17:32:28.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26625" PostTypeId="2" ParentId="26574" CreationDate="2019-03-29T18:31:32.637" Score="1" Body="&lt;p&gt;Consider the failure mode, and compare the type of loading you expect. In the states, we often design to (ultimate load) or the maximum load expected. At this load level we accept damage in the form of ductile deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You must consider what level loading you are comparing to strength, and how the damage of the element at this level will occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is additional strength when using the plastic section modulus, but it also implies permanent deformation.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastEditorUserId="19912" LastEditDate="2019-03-29T18:58:09.783" LastActivityDate="2019-03-29T18:58:09.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26627" PostTypeId="2" ParentId="26620" CreationDate="2019-03-29T21:34:58.107" Score="0" Body="&lt;p&gt;For the acceleration after releasing,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;\sum M &amp;amp;= I\alpha \\&#xA;2*4\pi &amp;amp;= 2.667 kg m^2*\alpha_1 \\&#xA;\alpha_1 &amp;amp;= 9.42 rad/s^2 \\&#xA;\end{align}&#xA;$$&lt;/span&gt; (Not sure how they rounded up to 9.5.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the original position (no torsion in spring) of a simple harmonic oscillator, the velocity is a maximum (magnitude) and the acceleration is a minimum (magnitude). Therefore, &lt;span class=&quot;math-container&quot;&gt;$\alpha_2=0$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2019-03-29T21:34:58.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26628" PostTypeId="1" AcceptedAnswerId="32120" CreationDate="2019-03-29T23:22:40.037" Score="0" ViewCount="1391" Body="&lt;p&gt;I want to connect this to wires , I know where the negative terminal is on this ( the spring) but I can’t figure out where the positive terminal is, I can’t find and answer online anywhere else either.  It’s a blue laser. Please help&lt;img src=&quot;https://i.stack.imgur.com/7LcbF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="19916" LastActivityDate="2020-01-07T20:52:44.873" Title="Where is the positive terminal on the laser pointer" Tags="|battery|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26629" PostTypeId="1" CreationDate="2019-03-30T03:39:41.013" Score="2" ViewCount="161" Body="&lt;p&gt;I'm sure there is a simple answer here, so I apologize in advance for the unchallanging &amp;amp; low level question, but here it is:&lt;/p&gt;&#xA;&lt;p&gt; Why does the water at water oulets, through out a building, stop their water flow near instantly when the Rising Stem Gate valve at the out side water main is shut closed, But will trickle and trickle with little flow for almost an hour when the indoor building main ball valve, downstream of RSG valve, is shut closed.&lt;/p&gt;&#xA;&lt;p&gt;Building has 3 floors.&lt;/p&gt;&#xA;&lt;p&gt;Anything helps.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2moHi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2moHi.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19919" LastEditorUserId="19778" LastEditDate="2021-01-21T12:42:24.333" LastActivityDate="2023-10-08T19:03:10.317" Title="Water &amp; plumbing pipes vs gravity or air pressure" Tags="|fluid|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26630" PostTypeId="1" CreationDate="2019-03-30T04:34:44.683" Score="1" ViewCount="866" Body="&lt;p&gt;This may sound crazy but bear with me. I was thinking about why cargo ships are built from such heavy material as steel, while smaller boats are often made of plastics. I know cargo ships carry a lot of weight and the sea can be rough. Those ships of course need to maintain structural integrity and such. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wouldn't it be possible to build a structure from steel, supported by the strength of the containers that fill the ship. And then build a thick hull around it made from a combination of plastics and compartments of some kind of foam (polyurethaan?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantages are that plastics are flexible (a bump doesn't result in damage as easily), easy to work with, easy to repair, cheap, and really light compared to steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is kinda like how tall buildings are built from a strong structure, but have an outside &quot;hull&quot; of glass (see image below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't do any calculations on it (I'm a software engineer, unfortunately not a hardware engineer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody know the reasons why big (cargo) ships are not made from plastics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/V5fJA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V5fJA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11975" LastActivityDate="2022-03-17T16:34:51.227" Title="Why not build container ships from plastic?" Tags="|structural-engineering|structural-analysis|metals|plastic|ships|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26631" PostTypeId="1" CreationDate="2019-03-30T04:41:21.013" Score="2" ViewCount="2416" Body="&lt;p&gt;There a few neat mechanisms that allows only one way motion without the use of a pawl or ratchet.  This is one of them:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.instructables.com/id/LEGO-Technic-One-Way-Gear-Mechanism/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.instructables.com/id/LEGO-Technic-One-Way-Gear-Mechanism/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm looking for is a mechanism that can't be undone once it has gone from one position to another (without taking it apart; i.e., removing a pin) after it is rotated.  This could be useful in certain applications, but my search turns up nothing like I remember.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can think there might be a four or five-bar mechanism that has a critical point where the mechanism can't undone, but can't remember (or where to search for it)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do remember seeing one many years ago at one of my first engineering jobs, but can't recall the details other than &quot;that was neat!&quot;  The mechanism I saw had I believe 3 to 4 moving parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some other mechanisms that don't use springs or latches; a search for this turned up something also nifty, but not quite what I was looking for:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=E4cMyPqE3lg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=E4cMyPqE3lg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any thoughts on this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No magnets, no springs; this was a strictly mechanical thing (nothing hidden; I could take it apart, and the mechanical lock was fascinating).  The closest analogy I can think of is to those Sokoban games (and perhaps this is a way to think about it) where if you perform a certain order of operations, you lock up the puzzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one solution I came up with this morning, but I'm wondering if there is a list of them, a search term I'm missing, or other examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this example, a plunger enters from the left; I'm not too sure of the geometry, but I think this is an example of how it could work.  When the plunger brings the two locking elements behind it, it can't go back, because of a stop at the end of the plunger, prevents the linkage from reversing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mTUXa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mTUXa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even cooler - I believe the mechanism I saw had no linkages - just a series of Sokoban-like moves where the system went from one state to another, but couldn't be reversed without taking a part out of the 2D plane it was in.&lt;/p&gt;&#xA;" OwnerUserId="16011" LastEditorUserId="16011" LastEditDate="2019-03-30T15:44:24.877" LastActivityDate="2020-06-24T08:16:29.393" Title="One way/locking mechanism without springs or flexible members" Tags="|mechanisms|mechanical|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26632" PostTypeId="2" ParentId="26630" CreationDate="2019-03-30T04:46:33.280" Score="8" Body="&lt;p&gt;In short, because steel is less expensive on a strength per dollar basis of anything else, and for large structures, it's less expensive to work, again on a strength per dollar basis.  That's why steel construction predominates in cars, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small boats, the cost equation leans more toward the ease of manufacture of &lt;em&gt;that size&lt;/em&gt; of structure out of fiberglass (which isn't &lt;em&gt;just&lt;/em&gt; plastic) or aluminum compared, plus a consideration for the corrosion resistance of fiberglass.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-03-30T04:46:33.280" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26634" PostTypeId="1" CreationDate="2019-03-30T06:10:21.060" Score="1" ViewCount="139" Body="&lt;p&gt;I was designing a whit-worth quick return mechanism, I have provided the contact sets but they fail when I try to animate. The contact set is defined for the crank and slotted link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Abspj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Abspj.png&quot; alt=&quot;the crank just rotates passing through the slotted link.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19921" LastEditorUserId="7083" LastEditDate="2019-03-30T21:42:49.310" LastActivityDate="2020-09-24T19:06:29.160" Title="Fusion 360 animation , contact sets" Tags="|mechanical|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26636" PostTypeId="1" AcceptedAnswerId="26637" CreationDate="2019-03-30T10:39:17.623" Score="1" ViewCount="1016" Body="&lt;p&gt;Please explain how the normal shear stress (shear stress normal to the cuve, C, defining the cross-section of a thin-walled beam) is zero. I have attached a screenshot from a book which says that the outer surfaces of the beam are stress free. This cannot be true except in very special cases. Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o4Mtr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o4Mtr.jpg&quot; alt=&quot;Structural Analysis with Applications to Aerospace Structures by Craig&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider the following image. Here axis &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; is the axial direction of the beam, and axes &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$3$&lt;/span&gt; define the cross-section of the beam. For the normal shear stress to be zero in the member &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; of the cross-section, the shear stress &lt;span class=&quot;math-container&quot;&gt;$\sigma_{21}$&lt;/span&gt; should be zero on the lateral face &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; which is possible only if there is no axial force on the lateral face &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;. Am I right?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GnYlE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GnYlE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17712" LastEditorUserId="17712" LastEditDate="2019-03-30T11:11:01.520" LastActivityDate="2019-03-30T11:11:01.520" Title="Why/how normal shear stress is zero in thin-walled beams?" Tags="|structural-analysis|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26637" PostTypeId="2" ParentId="26636" CreationDate="2019-03-30T10:50:20.393" Score="0" Body="&lt;p&gt;We don't know what equation 1.5 says in your book, but the statement &quot;the outer surfaces are stress free&quot; is nonsense, unless the book is only talking about some particular &lt;em&gt;component&lt;/em&gt; of the stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stress distribution on the outer surface of any component has to be in equilibrium with the external forces on that surface. That doesn't mean the stress must be zero. The simplest example I can think of is a rod under tension. The axial stress at the surface is the same as everywhere else in the rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram in the book, the stress component &lt;span class=&quot;math-container&quot;&gt;$\tau_n$&lt;/span&gt; will be zero at the surface &lt;em&gt;if there is no pressure load on the surface&lt;/em&gt; at that position along the beam. Otherwise, it will not be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The book says it is &lt;em&gt;assumed&lt;/em&gt; that &quot;&lt;span class=&quot;math-container&quot;&gt;$\tau_n$&lt;/span&gt; is zero through the wall thickness.&quot; That is an &lt;em&gt;assumption,&lt;/em&gt; and it is only an approximation. In fact, for a uniform beam there is a parabolic variation of &lt;span class=&quot;math-container&quot;&gt;$\tau_n$&lt;/span&gt;  through the thickness. This can be &lt;em&gt;ignored&lt;/em&gt; for a slender beam (which is thin relative to its length) because it is small compared with the other stress components, but it is not &lt;em&gt;zero&lt;/em&gt;. For a thick beam, this shear stress component &lt;em&gt;is&lt;/em&gt; included in the beam formulation.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-03-30T11:02:01.047" LastActivityDate="2019-03-30T11:02:01.047" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26638" PostTypeId="1" CreationDate="2019-03-30T13:09:31.047" Score="0" ViewCount="93" Body="&lt;p&gt;in my project I am required to find how much two surfaces slide against each other (to see how much it'll will rub against skin). Does anyone know how I can find these results?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="19923" LastActivityDate="2023-12-12T05:04:35.533" Title="Find amount of sliding between 2 surfaces" Tags="|finite-element-method|abaqus|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26639" PostTypeId="1" AcceptedAnswerId="26698" CreationDate="2019-03-30T15:07:44.373" Score="1" ViewCount="136" Body="&lt;p&gt;&quot;Work (as in Thermodynamics) is said to be done by a system if sole effect on the surroundings could be the raising of a weight.&quot; So, how is non mechanical work like : Chemical work, Magnetization work or even electric current flow work be reduced to the same?&lt;/p&gt;&#xA;" OwnerDisplayName="user19926" LastActivityDate="2019-04-02T23:11:02.220" Title="equivalent deduction of Non-mechanical thermodynamic work" Tags="|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26640" PostTypeId="2" ParentId="26631" CreationDate="2019-03-30T15:10:18.470" Score="0" Body="&lt;p&gt;In order for a mechanical movement to be &quot;one way&quot;, something has to change state as the elements in the mechanism move. The expansion and contraction of a spring and the appearance or disappearance of a magnetic field constitute such a change of state. I don't think what you are looking for exists , I would be amazed if there was a device that was &quot;one way&quot; without having any springs or electro-magnetic phenomena. You claim you saw such a device early in your career, but did you actually inspect it to see if it had any hidden springs or magnets?&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2019-03-30T15:10:18.470" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26641" PostTypeId="1" CreationDate="2019-03-30T16:09:26.743" Score="0" ViewCount="206" Body="&lt;p&gt;I am trying to code a simple FEM problem. A bar is fixed at the bottom edge, and a displacement is applied at the top. I want to apply the displacement gradually , so I have divided it into a number of steps. A rough outline of the code looks like this: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; %********************************************************* %          &#xA;&amp;gt; Newton-Raphson Control parameters&#xA;&amp;gt; %*********************************************************&#xA;&amp;gt; &#xA;&amp;gt; uload=5;         % the total displacement  &#xA;  numu=10;         % number of steps &#xA;  du=uload/numu;   % step size &#xA;  tu=0;            %initial value&#xA;&amp;gt; tol=1e-2; &#xA;&#xA;&#xA;&amp;gt;   for ii=1:numu  %Loop over total number of steps&#xA;&amp;gt; &#xA;&amp;gt;       iter=0;&#xA;&amp;gt;       tu=tu+du;  %increment displacement at each step&#xA;&amp;gt; &#xA;&#xA;&amp;gt;          while(error&amp;gt;=tol)  %Iteration&#xA;&amp;gt;            iter=iter+1&#xA;&amp;gt;            Ku=sparse(sdof,sdof);    %Initialize stiffness matrix&#xA;&amp;gt;            [Ku] = Kmat(Ku);  %Create stiffness matrix by user defined function&#xA;&amp;gt;               Ru = sparse(sdof,1);  %Initialize residual force vector&#xA;&amp;gt;            [Ru] = Rvec(Ru,u);  %Calculate residual force vector by user defined function(consists of internal force only)&#xA;&amp;gt;            [Ku,Ru] = applyBC(Ku,Ru); %Apply BCs by user defined function&#xA;&#xA;&amp;gt;            delu = sparse(sdof,1) ;  %Initialize solution vector       &#xA;&amp;gt;            delu = Ku\Ru ;   % Calculate displacement&#xA;             u=u+delu; %   %Increase displacement value&#xA;&amp;gt;            error=norm(Ru)  %Calculate Error (This step is probably causing the divergence due to selection of improper criterion)&#xA;&#xA;&#xA;&#xA;&amp;gt;          end %iteraion  &#xA;&amp;gt; end %step&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This code won't compile in MATLAB because all the functions are user defined and the input data is not provided. I just wanted to portray a rough outline of what I am trying to do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This algorithm, works fine if I only have a few elements(suppose 100) defined. If I decrease the mesh size, the solutions tends to diverge, the error keeps on increasing. I figure that is probably because the way I calculate the error is not useful for this simple case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is what can be the best convergence criterion for this problem? Since the problem is simple linear elastic, no non-linearity involved, I think the values calculated in the first iteration is correct, so no need for further iterations. But later I would like to extend this code to perform non-linear static analysis, so is there a convergence criterion that can be used for both? If not, then what can be used for each case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: I have modeled a similar problem in a commercial code, where obviously it works, but when I check the details of the solution, it says that it has used one iteration per step(which is expected). In my code with more elements, the iterations keep increasing after the 2nd step. Thus my doubt regarding the choice of convergence criterion. &lt;/p&gt;&#xA;" OwnerUserId="15803" LastEditorUserId="15803" LastEditDate="2019-03-30T17:00:12.527" LastActivityDate="2019-10-07T12:11:43.230" Title="Finite Element Code: Convergence issues" Tags="|mechanical-engineering|applied-mechanics|finite-element-method|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26645" PostTypeId="1" AcceptedAnswerId="26661" CreationDate="2019-03-30T20:25:27.773" Score="16" ViewCount="3915" Body="&lt;p&gt;Surge protectors are banned on cruise ships, and so are confiscated during boarding.  Extension cords and cube taps etc. are discouraged, but not banned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm asking specifically about surge protector incompatibility, and would like an answer that includes the technical details of the ship's cabin power supply.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I've asked in various places (not StackExchange) and have never been able to get a technical answer as to how a surge protector can cause damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most responses are along the lines of &quot;&lt;em&gt;they cause fires&lt;/em&gt;&quot;, &quot;&lt;em&gt;fire is really bad on a ship&lt;/em&gt;&quot;, and &quot;&lt;em&gt;they ban all extension cords&lt;/em&gt;&quot;, so please don't give a similar answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To discourage &quot;&lt;em&gt;they ban all devices&lt;/em&gt;&quot; responses, please look at this ad from Amazon and note the &quot;Non Surge Protection &amp;amp; Ship Approved&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rdzls.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rdzls.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;(Note that I wrote this last night before the answer from @StainlessSteelRat)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Learning that ships use an isolated-neutral system was a great help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a summary (many details omitted) of where I am so far.&#xA;In particular, I run into what seems like dead ends (in bold)&#xA;where the bad effects aren't nearly as bad as they at first sound.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Wiring Systems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Normal house wiring (North American standards) is asymmetric, having one line at 120V,&#xA;and the other shorted to ground at the main breaker panel.&#xA;The neutral side (larger slot in outlet) is, &lt;em&gt;in theory&lt;/em&gt;, safe to insert a fork without shock.&#xA;The live side is, in practice, dangerous to insert a fork.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Ship wiring is symmetric, having two lines, each at 60V and 180 degrees out of phase.&#xA;Both sides &lt;em&gt;in theory&lt;/em&gt; pass the fork test (similar to birds safely sitting on a distribution wire).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Surge Protection:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;With normal house wiring, when a surge protector notices a voltage spike between ground and the live wire,&#xA;it diverts the excess current.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;a surge protector has a continuous but small current (for the detection), and occasionally a brief but large current (when clamping).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Surge Protectors on Isolated-neutral systems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The connection between one of the live wires and ground causes the system to behave like a grounded system.&#xA;The current is normally small, but with many cabins on the same circuit, it might become significant.&#xA;This makes the system more dangerous to users, &lt;strong&gt;but no more dangerous than home wiring&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The grounding occurs on the side of the circuit that connects to the live socket.&#xA;Devices that rely on the difference between the two sides will be compromised.&#xA;For instance, the on/off switch is normally on the live side, but would now be on the neutral side, leaving voltage on the device&#xA;(e.g. even when switched off, the threaded part of a light-bulb socket would be live).&#xA;This makes the system more dangerous to users, &lt;strong&gt;but no more dangerous than home wiring that has live and neutral reversed&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There will be multiple connections to ground in the circuit, which can create ground-loop currents.&#xA;This can be annoying (e.g. hum in sound systems), &lt;strong&gt;but doesn't create any danger&lt;/strong&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Speculation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the two wires aren't different colours, or if the installer knows that it doesn't make any difference,&#xA;one line could end up connected to the neutral outlet in one cabin and to the live outlet in another.&#xA;That could increase the continuous current leakage through ground.&#xA;In the case of an actual spike, both sides would be briefly clamped to ground creating a large current and possibly tripping the circuit breaker.&#xA;This would be annoying, &lt;strong&gt;but not dangerous&lt;/strong&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm sure that the ship's electrical system has devices to monitor and regulate the power supply.&#xA;The current leak to ground caused by surge protectors might cause this mechanism to get false readings.&#xA;This could be dangerous to users.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is an effect on metals known as &lt;em&gt;stray current corrosion&lt;/em&gt;.&#xA;Even without any added electricity, random currents occur in metals, and, particularly in the presence of salt,&#xA;will cause the metal to corrode at points that act as anodes.&#xA;Many marine vessels have magnesium-anode devices attached to their hulls to mitigate this effect.&#xA;I imagine this is the reason that ships use isolated-neutral wiring, to avoid introducing any electrical currents into their hulls.&#xA;This isn't immediately dangerous, but would be immensely expensive in the long run.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19930" LastEditorUserId="19930" LastEditDate="2019-04-01T14:21:10.333" LastActivityDate="2023-08-30T05:53:14.947" Title="What is it about the cabin electrical supply on cruise ships that makes surge protectors dangerous?" Tags="|electrical-engineering|power-electronics|marine-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26646" PostTypeId="2" ParentId="17625" CreationDate="2019-03-30T23:55:46.710" Score="1" Body="&lt;p&gt;Short version: EM pulse weapons are nowhere near as small and convenient as they look in the movies, and they are also much easier to protect against than the movies would have you believe.  Military hardware is protected, and even a lot of civilian hardware has a reasonable level of protection, as small EM pulses are a fact of everyday life (coming from lightning, faulty lightswitches, cheap DC motors and many other things). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, what is an EM pulse, and why is it a problem? It is a pulse of electromagnetic radiation, usually spread out over a very wide range of frequencies, but lasting a short time.  It will cause currents to flow in metal objects it reaches, and if those metal objects are part of an electronic circuit the unexpected current will damage the circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, how can one protect against it? A sheet of metal will block most of the electromagnetic pulse. It will travel through small holes and round corners though. Military vehicles and ships tend to be made of metal, and tend to not have many holes which might let blast waves or toxic gas in, so they are pretty good shielding anyway.&#xA;The second way to protect against an EM pulse is to harden the actual electronics by providing somewhere for the big current pulse caused by the EMP to go where it won't do any damage. For example, you can build in special diodes to short out the unwanted current.&#xA;Neither of these methods is perfect. With a strong enough EMP, the pulse will overwhelm the protective elements on the circuit board, and the metal sheets only reduce the pulse, not block it entirely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, is it practical to build an EMP weapon? You can produce a pulse with something which works a lot like a big radio. You can produce a bigger pulse by connecting a powerful electromagnet to an antenna, then crushing the magnet flat with explosives. But neither produces that big of a pulse, and because the pulse spreads out very quickly the range is miserable.&#xA;The second way to make an EM pulse is to detonate a nuke. Preferably a specially adapted nuke, but any nuke will do if it is detonated in the right part of the atmosphere.  The hardening on military equipment, is really designed to deal with this, so they can keep fighting after a nuke near miss, but it has the side effect of making non-nuclear EMPs pretty pointless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, is a railgun more susceptible to damage than e.g. a modernised 16&quot; gun? Both probably have similar fire control systems, which will be equally vulnerable. The railgun has an electric system for accelerating the projectile, but that system is designed to operate with truly enormous currents, so the currents induced by an EMP won't do any damage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for examples of EMP use - I have never come across any examples of non-nuclear EMP devices having any real effect except at blank point range.  At which point a regular bomb would be much more destructive.  Nuclear EMPs are another story however, successful tests have been made, and in one case (starfish prime) they accidentally caused significant damage to satellites ins orbit and electrical installations in Hawaii.&lt;/p&gt;&#xA;" OwnerUserId="12022" LastActivityDate="2019-03-30T23:55:46.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26648" PostTypeId="1" CreationDate="2019-03-31T05:56:02.360" Score="3" ViewCount="451" Body="&lt;p&gt;I'm building a loft ladder with alternating-tread steps. The stringers are 2x6 White Spruce, and I'm hoping to build the steps out of the same material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TvWJV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TvWJV.jpg&quot; alt=&quot;picture of alternating-step ladder&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The span for the steps is 21&quot;. Code requires handling a maximum point load of 300 pounds at the step center, or a live load of 40psf, whichever is higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I've done the math right, steps made from 2x6 white spruce laid flat will  meet that requirement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm unable to find reliable information about the connection between the steps and the stringer. I'm hoping to attach the steps with three 16d 3.5&quot; nails on the wide side, and two 16d 3.5&quot; nails on the narrow side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My main concern being the limited amount of wood above the fastener in the step member causing failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I determine the maximum load the connection can bear in this configuration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5yBho.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5yBho.png&quot; alt=&quot;CAD drawing of member connection and nail&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18240" LastActivityDate="2019-03-31T15:02:45.237" Title="Strength of face-nailed connection for stair steps" Tags="|structural-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26649" PostTypeId="2" ParentId="26648" CreationDate="2019-03-31T06:56:06.610" Score="4" Body="&lt;p&gt;I suggest cutting slots in the stringers to hold the steps putting the load on wood and not 3 nails which will work loose or split the wood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AwLrz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AwLrz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;image from finehomebuilding.com&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-03-31T07:54:07.817" LastActivityDate="2019-03-31T07:54:07.817" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26653" PostTypeId="2" ParentId="26648" CreationDate="2019-03-31T13:28:16.537" Score="4" Body="&lt;p&gt;The item shown would be ripped out on the spot by our local inspectors. (Technically, they'd make you do it while they were writing out the fine.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twenty-one inch treads aren't allowed on stairs (OP says it's okay where he is, see comment below). That's probably not a staircase according to code. It looks like it would be classed as a ladder (I reckon the tread spacing to be greater than that allowed since it appears to cover about 9 feet with 11 treads). Permanently installed ladders are prohibited. Folding or temporary ladders can be used to access attics and such (unheated and unfurnished spaces). All stairs within the structure must have identical rise and run — are there any other grade breaks in the building? Code prohibits nailing stair treads as proposed. The banisters have to be on both sides and equal in height, spaced to code, and have approved banister profiles and mountings (maybe they are, hard to tell from wide angle shot). Handrails can't be bang-ended like that. Open treads are usually restricted. They have to be really wide treads so you cant get anything trapped in there and must pass the same tennis-ball-can't-pass-through rule. Talk to your building codes dept. and buy a book on stair construction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend of mine's father died when he fell on an open staircase while changing a light bulb. His leg went through the open step and he got a compound fracture in the leg. The resulting blood clot killed him two days later. I got the call and had to go find my friend and tell him. I remember that day quite well.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2019-03-31T15:02:45.237" LastActivityDate="2019-03-31T15:02:45.237" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26654" PostTypeId="1" CreationDate="2019-03-31T14:57:01.190" Score="0" ViewCount="52" Body="&lt;p&gt;So I'm trying to get to the condition as shown in the picture below, where it says that if we have zero free length springs and k1r2=k2r1, then any configuration (pose) of the link is in equilibrium. So this means I need to construct the potential energy of this system and show that for the potential energy to be constant, I need to satisfy the condition below. So far no luck. Here the springs are assumed to have 0 natural length (as said in the text). Anyone could help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NNd9O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NNd9O.png&quot; alt=&quot;picture&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16854" LastActivityDate="2019-07-31T19:06:26.400" Title="Static balancing of link mechanism" Tags="|mechanical-engineering|structural-engineering|civil-engineering|mechanisms|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26655" PostTypeId="2" ParentId="26629" CreationDate="2019-03-31T19:22:52.027" Score="1" Body="&lt;p&gt;Valves tend to leak, the amount depends on the design, size , seat materials, condition/age of the valve, pressure , and probably things I missed. So it would be difficult to guess the specific problem . Because of these factors , refineries use two valves in series for critical lines even though they get &quot;good&quot; valves and ( sometimes) verify construction quality. And well head valves are leak tight when new; however, a 3 -9/16 &quot; ( standard size ) well head valve can weigh 5,000 lb  ( for a 15,000 psi rating), ( I have no idea what the price is now). &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-04-01T15:14:37.980" LastActivityDate="2019-04-01T15:14:37.980" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26656" PostTypeId="1" AcceptedAnswerId="26731" CreationDate="2019-03-31T23:14:57.093" Score="1" ViewCount="305" Body="&lt;p&gt;I saw these specifications listed for a compressed air dryer (desiccant type). I also wonder, can we extract moisture without make it liquid (dew) first when in it is in a vapor state? How would we do this?&lt;/p&gt;&#xA;" OwnerUserId="4630" LastEditorUserId="16981" LastEditDate="2019-04-12T12:55:05.857" LastActivityDate="2019-04-12T12:55:05.857" Title="What is the meaning of dew point -20 or -40 or -70 deg.C? Since below zero centigrade, water cannot dew (already freeze)?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26658" PostTypeId="2" ParentId="26630" CreationDate="2019-04-01T00:44:36.050" Score="7" Body="&lt;p&gt;Anything is possible and lowering the lightship weight (LS) of ships will save fuel, but the key issue would be any change must not reduce the seaworthiness of ships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is very broad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regulations have been established for commercial steel ships (DNV GL, Lloyds, ABS, individual countries).  These regulations take into account the harsh marine environment.  In many cases, they represent the hard learned lessons from accidents like the Titanic.  Regulatory agencies, would have to be overcome.  It is on thing to do it for small craft, another for commercial ships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Safety (which probably has more to do with regulations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fire: Steel melts at around 1370°C.  Plastics melt at temperatures below 300°C.  A candle burns at 1000°C.  A typical fire in a steel compartment can be extinguished by closing off the air supply.  On a plastic hulled vessel, the hull would burn (which would probably extinguish the fire, but cause stability problems).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dapzI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dapzI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strength:  A ship is built from the keel up.  The hull steel plates and stiffeners must displace the displacement weight of the ship.  The hull must be able to withstand hurricane force wind and waves.  Any plastic would not have the strength of steel, so more structure would have to be used, so any loss in hull weight would probably be offset in structure weight.  A steel ship already flexes in heavy seas.  The longer the ship, the greater the flex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fex3J.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fex3J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impact:  There are a number of estimations for steel weight estimation.  They are summarized in &lt;a href=&quot;https://link.springer.com/article/10.1007/s00773-015-0345-y&quot; rel=&quot;noreferrer&quot;&gt;Preliminary parametric estimation of steel weight for new ships&lt;/a&gt;.  They report a 6.25% Error for a 8500TEU container ship when using the following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Steel Weight = 0.034 \times L^{1.7}\times B^{0.7}\times D^{0.4}\times C_B^{1.7} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ Steel Weight = 0.034 \times {317m}^{1.7}\times {45.8m}^{0.7}\times {25}^{0.4}\times {0.6}^{0.5} = 24, 780mt$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume half is structural and half hull or 12,390mt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://edoc.pub/queue/calculation-of-displacement-lwt-and-dwt-4-pdf-free.html&quot; rel=&quot;noreferrer&quot;&gt;Electrical Systems and Engine Room: Displacement, LWT and DWT&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Displacement = C_B \times L \times B\times D\times \rho_{SW}$$&lt;/span&gt;&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Displacement = 0.6 \times 317m \times 45.8m\times 25\times 1.025 tonnes/m^3 = 223,224mt$$&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are just estimates, but at best with no increase in structure and if plastic had no weight, you'd save &amp;lt; 5.5%.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2019-04-01T00:44:36.050" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26659" PostTypeId="1" CreationDate="2019-04-01T02:43:31.610" Score="1" ViewCount="14" Body="&lt;p&gt;I'm pretty interested in space exploration. Here I'm interested in a moon of Jupiter (Encealadus) which is covered by a thick ice crust, but apparently has at some depth a liquid water ocean which contains interesting molecules/salts/etc. Therefore has been some talks about how/if a probe could be sent there to investigate whether there might be life in that ocean. The most obvious way would seem to melt your way through the ice crust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I've decided to try to make a napkin calculation about the feasibility of this idea, from an energy perspective. I'd be curious to know if you see wholes in that reasoning, if some constants are wrong, etc. It's a napkin calculation - I just wrote it down &amp;amp; calculated mentally. So obviously not meant to be an accurate answer, most about a ballpark estimate. It's rounded to whatever integer makes calculations easier to do with just pencil/paper.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;Energy needed to melt the ice&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We assume @ stp (probably not right), use common values for water (there are salts in it so probably not right)&lt;/li&gt;&#xA;&lt;li&gt;To estimate that, we must know the mass of ice to be melted. &lt;/li&gt;&#xA;&lt;li&gt;To get the mass of ice, we need the volume of ice our probe would displace&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Volume &amp;amp; mass of ice&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assuming a radius of 1m for our probe for 1km of ice crust. This is essentially a huge cylinder, so V = d*(2pi*r²), where d = thickness of crust (1km) and r = radius of our probe (1m). You get:&#xA; - V = pi*10⁴ m³ &#xA; - density of ice = 10³kg/m³&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that, we can get the mass of ice to melt:&#xA; - pi*10^7 kg of ice to melt through&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider specific heat for ice of ~2kJ/kg, it seems we would need:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;~10^8 kJ to melt all that ice.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Energy available from radioactive source&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;For P-239, I found ~200 meV as value for energy of a &lt;a href=&quot;http://www.world-nuclear.org/information-library/nuclear-fuel-cycle/introduction/physics-of-nuclear-energy.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;fission event&lt;/a&gt;.&#xA;Molar mass = 239*10⁻² kg/moles or ~50 moles/kg&#xA;Avocadro number ~6*10²³ atomes/mole&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we bring 1kg. P-239 has as half-life of ~24 000 years or 600*10³ months. We'll use the half-life to calculate a rate of energy emitted per month, which we can then compare to the energy needed to melt the ice to see if have any chances of melting through it in a reasonable amount of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, taking avocadro's number with the molar mass &amp;amp; considering 1kg of P239 we have ~4*10¹¹kJ of energy. Since we're using the half-life to get a rate of emission, we need to divide that by two (e.g. only half the Pu decays during the half-life).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We find that we emit ~10⁶kJ/month/kg of Pu we bring.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It seems we would need quite a few months to melt through just 1km of ice.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Questions&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What might be wrong about that calculations, implicit assumptions that may not be right, etc.&lt;/li&gt;&#xA;&lt;li&gt;What other considerations we omitted that may have significant impact? E.g. our probe might be much smaller, will also have things other than Pu (affects heat dissipations?), etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19941" LastActivityDate="2019-04-01T02:43:31.610" Title="Feasability of melting through ice crust with radiocative heater" Tags="|thermodynamics|aerospace-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26661" PostTypeId="2" ParentId="26645" CreationDate="2019-04-01T05:57:16.250" Score="12" Body="&lt;p&gt;From &lt;a href=&quot;https://www.dco.uscg.mil/Portals/9/DCO%20Documents/5p/CSNCOE/Safety%20Alerts/USCG%20Marine%20Safety%20Alert%2003-13%20Surge%20Protective%20Devices%20Onboard%20Vessels.pdf?ver=2017-08-08-082206-293&quot; rel=&quot;nofollow noreferrer&quot;&gt; USCG Surge Protective Devices Onboard Vessels&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A marine casualty investigation of two separate stateroom fires onboard a U.S. Flag Container ship revealed that the sources of the fires were attributed to the use of SPDs plugged into a lighting circuit. It was discovered that a ground had developed on another circuit that was connected to the same distribution panel providing power to the staterooms. This ground created an imbalance of voltage between the two power conductors supplying the SPDs which caused excessive currents, overheating, and subsequently, a fire. In this instance, even if the SPDs automatically tripped as designed, only one power conductor would have been secured while the other would continue to provide power, possibly shorting to the device’s ground wire and the structure of the vessel.&lt;/p&gt;&#xA;&lt;p&gt;This safety alert only applies to vessels with alternating current power systems and may be most likely applicable to larger industrial and commercial vessels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;There are two main methods to supply power on ships.  Insulated neutral and high impedance grounded neutral.&lt;/p&gt;&#xA;&lt;p&gt;The insulated neutral system is used for low voltage ships (&amp;lt;1000V, typically 440V @ &amp;lt; 5MW).  An insulated neutral is used to maintain the integrity of the power supply system.&lt;/p&gt;&#xA;&lt;p&gt;On land, a grounded neutral is used. Any ground fault (hard or instantaneous) will cause breakers to trip or fuses to open. At worst, you are in the dark.&lt;/p&gt;&#xA;&lt;p&gt;At sea, loss of power coming through narrows or channel may cause a dangerous situation.  So an insulated neutral is used. A single ground fault (hard or instantaneous) will energize the hull, but much like birds sitting on uninsulated wires, the crew is not at an immediate risk, but the ground fault must be cleared because:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;if a crew member makes contact with a second power line, they may be shocked.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;and&lt;/strong&gt; a second ground fault will interrupt the flow of power, possibly blacking out the ship.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;and&lt;/strong&gt; cathodic corrosion protection which protects the hull from rusting may be impacted.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;and specifically&lt;/strong&gt; may cause a fire because current greater than wires ampacity flows, I2R losses create heat, melt/burn insulation and causing a fire.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Circuit protection is designed to protect the insulation of wires to prevent fires.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5dSC8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5dSC8.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;So ground faults are &lt;strong&gt;dangerous&lt;/strong&gt; and must be cleared as soon as possible.  Circuits must have circuit protection (fuses or breakers) on all power lines because ground faults can develop on different power circuits.&lt;/p&gt;&#xA;&lt;p&gt;Ground faults occur in ships in all spaces.  In panels, in wiring, in devices, dirt collecting on surfaces, maintenance, failures, etc.  All metal surfaces are bonded to the ship hull so that ground faults can be detected and cleared.&lt;/p&gt;&#xA;&lt;p&gt;As ships get larger and more electrical power (&amp;gt;6MW) is required, the voltage must increases to decrease current to reduce copper usage.  But higher voltage (&amp;gt;1000V) means higher short circuit currents.  To limit short circuit current, 6.6kV systems use a high impedance or resistance (1.32kΩ) connected to ground (ship hull) to limit short circuit to 5A.  The integrity of maintaining power if a single ground fault occurs is given up because it is to dangerous.  Ground faults still occur and may black out the ship, but they must be detected and cleared.  The dangers to crew or the ship power or cathodic corrosion protection are more complex.&lt;/p&gt;&#xA;&lt;p&gt;For the question this means the ship hull is at 0V, but without hull return (no current is meant to flow in the hull).&lt;/p&gt;&#xA;&lt;p&gt;From IEEE 45.1-2017 - IEEE Recommended Practice for Electrical Installations on Shipboard - Design.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q2Sew.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q2Sew.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Lower voltages (440V, 120V) are achieved by transformers.  At 120V, the loads can be connected as 3-wire delta or 4-wire wye.  Individual consumers get one phase and every third light or cabin is a different phase to keep the load balanced as much as possible.&lt;/p&gt;&#xA;&lt;p&gt;At LV Bus, Neutral is solidly grounded to ship's hull at the transformer.  If 4-wire wye (120V/208V) (neutral, 3 phases) is used, neutral is not grounded at receptacles, but power cables may have 3 prongs, Line, Neutral and Ground.  So the Receptacle Ground is connected to the hull, so single ground faults in devices plugged in can be detected and cleared.&lt;/p&gt;&#xA;&lt;p&gt;If 3-wire delta (120V) is used, there is no ground connection at the transformer, but Receptacle Ground is connected to the hull.  This is probably where the problem occurs.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EbO6X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EbO6X.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;USCG show 60V at each terminal, so the system cannot be an insulated neutral.  If LV bus is a delta, 60V at each terminal makes sense for a high impedance neutral with hull set at 0V.&lt;/p&gt;&#xA;&lt;p&gt;120V power strips or surge protectors are designed to interrupt the live L, but there is no protection on neutral N.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Here lies the problem with inexpensive and older SPDs that only disconnect one “hot” terminal lead. The other “hot” terminal remains hot if the circuit breaker supplying the receptacle and SPD does not trip.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;All circuits have breakers/fuses on all power lines.  The problem occurs when the current is enough to activate surge protectors or power bars, but not enough to trigger built in circuit protection devices.&lt;/p&gt;&#xA;&lt;p&gt;This fails for either wye or delta, but easier on delta.  Delta has three phases, A, B &amp;amp; C.  Each outlet gets 2 phases at a voltage of 120V.  Cabin 1 has phase AB, 2 has BC, 3 has CA, repeat.&lt;/p&gt;&#xA;&lt;p&gt;Cabin 1 overloads circuit, power strip breaker trips, but only interrupts A.  B is live with 60V (possibly 69V) and because of the plugs ground plug, possibly connects B to ship hull.&lt;/p&gt;&#xA;&lt;p&gt;Now any other ground fault anywhere on ship occurs and higher than designed currents flow, melting the wire insulation, and causing a fire.&lt;/p&gt;&#xA;&lt;p&gt;Most likely, fault will occur in another cabin.  Hair dryer in Cabin 2 trips power bar circuit breaker, interrupting phase B, but leaving phase C connected and possibly to Ground.&lt;/p&gt;&#xA;&lt;p&gt;Phase B in Cabin 1 is connected to Phase C in Cabin 2.  Short circuit with no circuit protection.  Current greater than wires ampacity flows, I2R losses create heat, melt/burn insulation, creating a fire.&lt;/p&gt;&#xA;&lt;p&gt;The outlet has dual pole breakers connected to a surge protector with a breaker on one side.  The problem occurs when a surge occurs activating the surge protector but not meeting the outlet breakers, killing power to one half of the circuit with a potential 60V on the neutral of devices not designed to have a potential on the neutral, causing funky behavior in the wiring, device, surge protector to cause a fire in something probably a device connected to the surge protector.&lt;/p&gt;&#xA;&lt;p&gt;Properly designed surge suppressors and power strips for ships must have dual pole breakers to disconnect both lines in event of a fault.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-04-01T19:19:22.727" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26665" PostTypeId="1" CreationDate="2019-04-01T08:58:40.640" Score="1" ViewCount="530" Body="&lt;p&gt;Is there generally any cheaper way to get a bunch of gear wheels. I've been to some local companies who specialize in gear production and the price is always somewhere in the range 100-150 USD for a single gear. This is well beyond what I'm willing to pay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side question, why are the prices so high? I understand that the machine has to be set such that it must cut my desired dimensions but $100 is 3 days wage in my area and surely it wont take 3 days to set it up, nor does it consume anything?&lt;/p&gt;&#xA;" OwnerUserId="19947" LastActivityDate="2019-04-12T19:42:58.890" Title="Is there a cheap way to get gears" Tags="|gears|manufacturing-engineering|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26667" PostTypeId="2" ParentId="26665" CreationDate="2019-04-01T11:27:12.303" Score="1" Body="&lt;p&gt;It depends on the specs of the gears you need. If you need small high-torque, high-precision gears you will need to pay the price. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can deal with larger lower-torque gears then you can cut them out of the sheet of plywood following a template from a &lt;a href=&quot;https://woodgears.ca/gear_cutting/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;gear generator&lt;/a&gt; (not affiliated to that site, just mentioning a good one) or 3D print one.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2019-04-01T11:27:12.303" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26669" PostTypeId="1" AcceptedAnswerId="26687" CreationDate="2019-04-01T18:45:05.730" Score="1" ViewCount="195" Body="&lt;p&gt;At first, I had a particular goal, but I'm starting to doubt feasibility which has led me here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Original Goal:&lt;/strong&gt; Create a small &lt;em&gt;(3in²)&lt;/em&gt; electromagnet with 100% duty cycle capable of holding at least 50lbs for 1-2 hours at a time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't fully killed this idea off, but I'm starting to think heat may be a significant issue here; so I'm considering a different approach.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;New Question&lt;/strong&gt;: Can I use a permanent magnet with holding force of +50lbs and use an electromagnet to disrupt the holding ability, which would then release it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea here being that I could possibly use a magnet capable of holding hundreds of pounds, but it'd always be &quot;on&quot; so to speak. If my original electromagnet idea is plagued by heat problems; maybe I could leverage a powerful electromagnet at &amp;lt; 5s intervals to separate a metal object from a permanent magnet.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Does any of this make sense? Am I speaking jibberish? I'm not sure exactly which direction I can go, but my end goal is that I need to hold ~50lbs for 1-2 hours and then release it... with a magnet that won't become overwhelmingly hot.&lt;/p&gt;&#xA;" OwnerUserId="19954" LastActivityDate="2019-04-03T15:40:08.767" Title="Can you use an electromagnet to break the connection of a permanent magnet?" Tags="|electrical-engineering|electromagnetism|magnets|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26670" PostTypeId="1" AcceptedAnswerId="26671" CreationDate="2019-04-01T20:52:49.733" Score="0" ViewCount="67" Body="&lt;p&gt;I'm stuck on this question. I got the first part right which is to work out the dimension change in the x direction but I got the y and z directions wrong. I had to submit this online and it doesn't show what you did wrong so I'm really frustrated on why I got this wrong I was hoping someone could correct me. Since I put my answers online I left the negatives in both my y and z values which might be why I got it wrong I just wanted someone to help to see if I have actually done something wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sy7El.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sy7El.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I forgot to add in my diagram the length of the x which is 200mm sorry. It's in my working though. &lt;/p&gt;&#xA;" OwnerUserId="19956" LastEditorUserId="1832" LastEditDate="2019-04-02T04:22:27.560" LastActivityDate="2019-04-02T04:22:27.560" Title="Help with determining the dimension change in the xyz direction of a component under a load?" Tags="|mechanical-engineering|materials|mathematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26671" PostTypeId="2" ParentId="26670" CreationDate="2019-04-01T21:12:28.713" Score="0" Body="&lt;p&gt;Did you include the stress applied to the Y direction (your drawing shows Py = 10 kN) this contributes to the strain in the Y direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also If you take the Y direction as &quot;up&quot; (as you have with the load 10 kN) then your Z dimension is out of the page, when you calculate the displacement in those directions it looks like you got the signs mixed up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Remember the load in the Y direction will cause expansion in Y and contraction in X and Z. You need to add up all these effects in each direction.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-01T21:12:28.713" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="26672" PostTypeId="2" ParentId="26665" CreationDate="2019-04-01T22:31:31.880" Score="1" Body="&lt;p&gt;If you are getting custom gears made, I am surprised if you can get them that cheaply.  I did a search for Boston Gear (a company I dealt with years ago) and found them for sale on Amazon ranging from not bad to waaaay too much money. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have metal working equipment, making a toothed wheel that will function as something like a gear is easy.  Making a for-real involute gear that transfers power smoothly and does not self-destruct is hard but not as hard as it would be if you could not use Ratchet Freak's gear generator.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending upon what you are trying to accomplish, sprockets and roller chain, something like what is used on a bicycle, or a belt and pulley, such as some cars have to drive the fan, maybe a much more economical solution.  &lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2019-04-01T22:31:31.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26676" PostTypeId="1" CreationDate="2019-04-02T07:44:21.650" Score="0" ViewCount="43" Body="&lt;p&gt;What is the definition of BIPV?&#xA;Building integrated photovoltaics emerged as a new way of reducing energy consumption. However, the question is how do these systems work?&lt;/p&gt;&#xA;" OwnerUserId="19945" LastEditorUserId="19912" LastEditDate="2019-04-02T22:02:39.023" LastActivityDate="2024-03-17T14:05:55.680" Title="What is the definition of Building integrated photovoltaics?" Tags="|photovoltaics|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26677" PostTypeId="2" ParentId="26676" CreationDate="2019-04-02T07:48:50.560" Score="-1" Body="&lt;p&gt;One of the most recent energy consumption methods is using PV panels in front of the buildings. In abbreviation, the system calls BIPV (Building Integrated PhotoVoltaic). The generated electricity in this type of system strongly depends on the outdoor conditions which could affect the temperature of the panel and consequently increase the efficiency. &lt;/p&gt;&#xA;" OwnerUserId="19945" LastActivityDate="2019-04-02T07:48:50.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26678" PostTypeId="1" AcceptedAnswerId="26688" CreationDate="2019-04-02T08:28:00.073" Score="0" ViewCount="708" Body="&lt;p&gt;The primary purpose of a vehicle (car, locomotive, etc.) horn is to &lt;em&gt;warn&lt;/em&gt; other traffic units that something potentially dangerous is about to happen. And so, others must pay attention to the signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why not to tune the horns to dissonant intervals or chords (for example, several notes semitone or even quarter-tone apart, e.g. C natural, C half-sharp and C sharp sounding simultaneously), which are much more unpleasant to listen and so more &quot;attention-payable&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe this is not a proper place for the question, but I've unfortunately not found a better one, since is more related to engineering and design (than to, for example, music).&lt;/p&gt;&#xA;" OwnerUserId="19960" LastActivityDate="2019-04-02T16:50:35.493" Title="Why vehicle horns are tuned to consonant intervals?" Tags="|design|car|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26679" PostTypeId="2" ParentId="26676" CreationDate="2019-04-02T09:21:56.090" Score="0" Body="&lt;p&gt;&quot;Building Integrated&quot; can mean that the pv panels are the waterproofing roof layer in place of the roof tiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is compared to pv panels mounted on the roof supported some small distance (10cm) above the existing roof tiles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pv panels work the same in either case as long as the roof integrated panels have sufficient cooling so that the performance is not degraded with the temperature (pv panels suffer from reduced output as their temperature increases).&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-04-02T09:21:56.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26682" PostTypeId="1" CreationDate="2019-04-02T14:09:49.770" Score="0" ViewCount="65" Body="&lt;p&gt;So I have to design a vibrating feeder of 55kg which moves the material (light objects eg. olives) on it in forward direction (linear motion). I am stucked on the positon of the spring. I mean which arrangement of spring will be better:&#xA;1. Spring can be placed at 4 corner of the vibrating feeder.&#xA;2. 2 in back corners and 2 in middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any calculation or relation regarding this topic...?&#xA;I have attached the simple arrangment of what I have explained above.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/V4VL0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/V4VL0.png&quot; alt=&quot;with 4 springs in corners&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ocB64.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ocB64.png&quot; alt=&quot;2 springs in middle and 2 at the back&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19966" LastActivityDate="2019-04-02T14:09:49.770" Title="Position of Spring to support Vibrating feeder" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26683" PostTypeId="2" ParentId="26676" CreationDate="2019-04-02T14:24:17.943" Score="0" Body="&lt;p&gt;Photovoltaics are build-integrated when they are part of the fabric of the building envelope itself, rather than modules which are mounted onto the external envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BIPV can be &lt;a href=&quot;https://www.tesla.com/en_GB/solarroof&quot; rel=&quot;nofollow noreferrer&quot;&gt;roof tiles&lt;/a&gt;, &lt;a href=&quot;https://www.pilkington.com/en/global/products/product-categories/solar-energy/pilkington-sunplus-bipv&quot; rel=&quot;nofollow noreferrer&quot;&gt;windows&lt;/a&gt; or &lt;a href=&quot;http://en.hanergythinfilmpower.com/index.php?m=content&amp;amp;c=index&amp;amp;a=lists&amp;amp;catid=77&quot; rel=&quot;nofollow noreferrer&quot;&gt;curtain-wall&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While they come in and out of fashion, they have been completely eclipsed by the growth in rooftop PV and utility-scale standalone PV, which now completely dominate the market, leaving BIPV as a small niche market. While it is conceivable that this could become a signicant growth market, it is more likely to remain little more than a rounding error in global PV figures.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2019-04-02T14:24:17.943" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26684" PostTypeId="2" ParentId="26516" CreationDate="2019-04-02T15:04:00.483" Score="1" Body="&lt;p&gt;The steps to find the reaction at the pin support using conventional methods:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Remove the redundant support at the pin to create a statically determinate &lt;em&gt;virtual structure&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Solve for the cantilever tip support deflection under the provided loading. Use principle of linear superposition to add up the deflection from each load source separately.&lt;/li&gt;&#xA;&lt;li&gt;In order to make the &lt;em&gt;virtual structure&lt;/em&gt; compatible with the real structure, a point load at the tip is required to counter act the deflection of the &lt;em&gt;virtual structure.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It looks like you went through these steps. My guess is there is a problem with the deflection computation. I would try breaking apart each load individually and finding the deflection for each. Then add them up. Then enforce compatibility.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastEditorUserId="19912" LastEditDate="2019-04-02T15:32:25.440" LastActivityDate="2019-04-02T15:32:25.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26686" PostTypeId="1" AcceptedAnswerId="26694" CreationDate="2019-04-02T16:20:44.677" Score="0" ViewCount="544" Body="&lt;p&gt;I'm trying to figure out how to model projectile motion which can be defined using a simple [quadratic] equation which defines projectile parabola:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y = y_0 + v_{0y}t - \dfrac{1}{2} gt^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across some &lt;a href=&quot;http://mlrg.cs.brown.edu/dynamics-slides.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;slides&lt;/a&gt; from Brown University which model the motion as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\text{Matrix form: }&amp;amp; \begin{pmatrix} 0 &amp;amp; 100 &amp;amp; -10\end{pmatrix} \\ \\&#xA;\text{state: }&amp;amp; \begin{matrix}y &amp;amp; y' &amp;amp; y'' \\ p &amp;amp; v &amp;amp; a\end{matrix} \\ \\&#xA;&amp;amp;\begin{pmatrix}&#xA;1 &amp;amp; 0 &amp;amp; 0 \\&#xA;1 &amp;amp; 1 &amp;amp; 0 \\&#xA;0 &amp;amp; 2 &amp;amp; 1 \\&#xA;\end{pmatrix}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm not quite sure how they came up with that transition matrix. Can anyone shed some light how to pick the state variables and how to model the motion?&lt;/p&gt;&#xA;" OwnerUserId="19968" LastEditorUserId="1832" LastEditDate="2019-04-02T22:10:06.747" LastActivityDate="2019-04-02T22:10:06.747" Title="How to model projectile motion using dynamical system matrices" Tags="|dynamics|systems-engineering|systems-design|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26687" PostTypeId="2" ParentId="26669" CreationDate="2019-04-02T16:50:17.247" Score="3" Body="&lt;p&gt;These are called electropermanent magnets. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q964U.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q964U.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/g00S6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g00S6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Electropermanent_magnet&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Electropermanent_magnet&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control can either come from rotating one of the magnets, or by using a control coil. If the magnet is rotated this can change the state of the coil to whether current is needed for the magnet to be on with current or off with current. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can I use a permanent magnet with holding force of +50lbs and use an&#xA;  electromagnet to disrupt the holding ability, which would then release&#xA;  it?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, this would be possible, however, the current or coil size needed to generate an opposing field may be more than is acceptable for a design. Check the &lt;a href=&quot;https://sciencing.com/calculate-force-electromagnet-5969962.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;force for the coil&lt;/a&gt; that may be used in a design.&lt;/p&gt;&#xA;" OwnerUserId="4921" LastActivityDate="2019-04-02T16:50:17.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26688" PostTypeId="2" ParentId="26678" CreationDate="2019-04-02T16:50:35.493" Score="2" Body="&lt;p&gt;Tuning car horn pairs to a major third seems to be a tradition, better suited for discussion on the automotive stack exchange than here. But it is definitely not the only option, and it boils down to cost versus benefit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the horn in my 1965 Cadillac hearse had three trumpets, two of which were tuned to a minor interval, which created dissonance. The sales literature described that horn as being &quot;oddly compelling&quot;, and it was, and the extra expense of furnishing a three-trumpet horn was small in comparison to the cost of the car, which was about $12000 in 1965 money- or about twice the price of a base model Caddy at the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In comparison, my 1960 VW Bug had a single-tone horn, because it was a cheap car. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-04-02T16:50:35.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26689" PostTypeId="1" CreationDate="2019-04-02T17:12:03.723" Score="0" ViewCount="143" Body="&lt;p&gt;I have a fundamental misunderstanding about the role of gimbals and gyroscopes in determination of angular rotation rates (in inertial measurement units). A set of three gimbals enables an object mounted on it's innermost gimbal to assume an arbitart orientation in space. Viewed differently, a three axis gimbals set enables full freedom (3 degrees of freedom) of relative motion between the object mounted on them and it's support, what paves the way for the design of stabilized platforms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, one can mount three orthogonal gyroscopes on a gimballed stabilized platform, so that the gimbals isolate the gyros from any external torques and the three gyros will maintain a fixed orientation relative to the inertial space. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Up to this point, i agree with the chains of reasoning. What i don't understand is why one can't simply incorporate angle sensors in order to measure the angles between the axes of the gyros and spacecraft axes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I now quote the wikipedia article on &quot;gimbal&quot;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In inertial navigation, as applied to ships and submarines, a minimum of three gimbals are needed to allow an inertial navigation system (stable table) to remain fixed in inertial space, compensating for changes in the ship's yaw, pitch, and roll. In this application, the inertial measurement unit (IMU) is equipped with three orthogonally mounted gyros to sense rotation about all axes in three-dimensional space. The gyro outputs are kept to a null through drive motors on each gimbal axis, to maintain the orientation of the IMU. To accomplish this, the gyro error signals are passed through &quot;resolvers&quot; mounted on the three gimbals, roll, pitch and yaw. These resolvers perform an automatic matrix transformation according to each gimbal angle, so that the required torques are delivered to the appropriate gimbal axis. The yaw torques must be resolved by roll and pitch transformations. The gimbal angle is never measured. Similar sensing platforms are used on aircraft.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if the initial purpose of the stabilizes gimballed platform was to isolate the gyros from any external torques, why do we now need a complicated drive motors system and a matrix calculator in order to apply torques to the components of this platform?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question might result from a lack of knowledge on practical engineering issues, or maybe from a basic misunderstanding of the mechanics involved here. Anyway, my attempts at reading other sources than wikipedia didn't help me a lot, so i'll be glad to get an exhaustive answer.&lt;/p&gt;&#xA;" OwnerUserId="3284" LastActivityDate="2022-04-18T09:39:44.010" Title="Why do gimballed gyroscopes platforms need drive motors in order to maintain it's fixed orientation?" Tags="|instrumentation|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26690" PostTypeId="2" ParentId="26689" CreationDate="2019-04-02T18:12:05.380" Score="0" Body="&lt;p&gt;The matrix calculator sounds fancy, untill you realize that this is one of the earliest things engineers learn in uni and is realy just a way of avoid using sinus and cosinus ad nauseum*. A control system needs to be present  ayway and the computations needed arent really all that hard, they just sound fancy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the gimbal system has  a nasty problem. If you turn it to a certain orientation it  ceases  to work. Since the disks need to be fed with energy anyway or they would cease to eventually rotating. You would need motors anyway to counteract  friction. But you can actually read the energy changes and keep the gyro in base position. This then will never end up having to deal diminished degrees of freedom leading eventually  to gimbval lock. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a way this is technically simpler the letting the gimbal free float and reading  angles. Even if you were to read angles directly im not sure you could or even would want  to skip the matrix calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* Although, most students dont realize it at this point that engineering math is not about doing computations like in earlier school stages. So freshmen tend to learn abhor matrices since they are superficially labour intensive. Not realizing yet that labour intensive is way better than hard or complicated (which is what they equate labour intensive with). Labour intensive just simply means can be done by a machine. This frees you to think on higher order things. But then freshmen arent engineers, yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh and also its numerically more stable.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2019-04-02T18:22:38.057" LastActivityDate="2019-04-02T18:22:38.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26691" PostTypeId="2" ParentId="22562" CreationDate="2019-04-02T18:33:09.053" Score="0" Body="&lt;p&gt;According to BS EN ISO 17892-12:2018 there are three methods to determine liquid limit and plasticity index of soils. Previously, BS 1377 PART 2:1990 used the definitive method of a cone penetrometer. This method superceded the Cassagrande method after 1975 (Smith's Elements of Soil Mechanics, 9th Ed, 2014, pp11-13)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/AKgpu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AKgpu.jpg&quot; alt=&quot;Liquid limit apparatus diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with the Cassagrande method is that it is highly dependant on the person doing the test and therfore far less reliable than the cone penetrometer test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the variable results of the Cassagrande method, I would hazard a statement of &quot;yes&quot; because there are soils that cannot release their water content completely due to their molecular alignment and ionic bonds which are also affect by Van der Waal's (pronounced fun-dur-Varl's) forces too.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-04-02T18:33:09.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26694" PostTypeId="2" ParentId="26686" CreationDate="2019-04-02T19:28:41.163" Score="0" Body="&lt;p&gt;He started from &lt;span class=&quot;math-container&quot;&gt;$$ y = y_0 + v_0 t + \frac 1 2 a t^2$$&lt;/span&gt; but then for some reason best known to himself decided to take &lt;span class=&quot;math-container&quot;&gt;$\frac 1 2 a$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$y''$&lt;/span&gt; rather than a, (and ignored the fact that his &lt;span class=&quot;math-container&quot;&gt;$y''(t)$&lt;/span&gt; is no longer equal to &lt;span class=&quot;math-container&quot;&gt;$d y'(t) / dt$&lt;/span&gt;, but hey, Lewis Carroll was also a mathematician, and this guy, like Humpy Dumpty, can make words mean whatever he feels like they ought to mean).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So he gets the approximate equations&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{matrix} y(t) \\ y'(t) \\ y''(t) \end{matrix} \approx&#xA;\begin{matrix} y(t-1) + y'(t-1) \\ y'(t-1) + 2y''(y-1) \\ y''(t-1)&#xA;\end{matrix}$$&lt;/span&gt;&#xA;or&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{bmatrix} y(t) &amp;amp; y'(t) &amp;amp; y''(t) \end{bmatrix} \approx&#xA;\begin{bmatrix} y(t-1) &amp;amp; y'(t-1) &amp;amp;y''(t-1) \end{bmatrix}&#xA;\begin{bmatrix} 1 &amp;amp; 0 &amp;amp; 0 \\ 1 &amp;amp; 1 &amp;amp; 0 \\ 0 &amp;amp; 2 &amp;amp; 1\\ &#xA;\end{bmatrix}&#xA;$$&lt;/span&gt;&#xA;He can't seem to make up his mind whether &lt;span class=&quot;math-container&quot;&gt;$y''$&lt;/span&gt; is really a variable, or just the arbitrary constant &lt;span class=&quot;math-container&quot;&gt;$-10$&lt;/span&gt;, (or was that &lt;span class=&quot;math-container&quot;&gt;$+10$&lt;/span&gt;, or &lt;span class=&quot;math-container&quot;&gt;$\pm 20$&lt;/span&gt; - I've lost the will to care...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be more obvious what's going on if you replace the 2's with 1's in his equations, and let &lt;span class=&quot;math-container&quot;&gt;$y''(t)$&lt;/span&gt; actually BE the second derivative of &lt;span class=&quot;math-container&quot;&gt;$y(t)$&lt;/span&gt;, not something close.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do people actually keep their jobs as university lecturers for teaching like this nowadays? It's 50 years since I was an undergrad, so I don't know....&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-04-02T19:28:41.163" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26695" PostTypeId="2" ParentId="26669" CreationDate="2019-04-02T19:44:19.583" Score="0" Body="&lt;p&gt;An electropermanent magnet exists in theory, but in practice hard to find.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To meet your goal of a 50 lb load without overheating if attached to some thermally conductive angle iron. It is cheap and easy to find, which may be better than a theoretical answer that has no cheap commercial solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Does any of this make sense?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;example of random web electromagnetic solenoid   110lb/50kg @ 8W $21 (1pc)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3ztKo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3ztKo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.amazon.com/Electric-Lifting-Magnet-Electromagnet-Solenoid/product-reviews/B01GJWLGZU&quot; rel=&quot;nofollow noreferrer&quot;&gt;YXQ 50Kg 110LB 24V 50mm Electric Lifting Magnet Electromagnet Solenoid&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;8Watts   Thermal rise depends on heatsink to base and applied voltage Pd=V^2/R&lt;/li&gt;&#xA;&lt;li&gt;Holding power depends on no air gap to steel surface.&lt;/li&gt;&#xA;&lt;li&gt;if object is suspended object jarred loose to cause any separation, it falls then a swivel connection is needed.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8847" LastEditorUserId="8847" LastEditDate="2019-04-03T15:40:08.767" LastActivityDate="2019-04-03T15:40:08.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26697" PostTypeId="2" ParentId="26516" CreationDate="2019-04-02T22:52:48.317" Score="3" Body="&lt;p&gt;Your bending moment diagram is very wrong. It is continuous even though there are concentrated bending moments (which create discontinuities), and you have &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; generating positive bending moment when it should be negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using your notation, where subscript:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 represents the pinned end;&lt;/li&gt;&#xA;&lt;li&gt;2, where &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$M_{f1}$&lt;/span&gt; are applied;&lt;/li&gt;&#xA;&lt;li&gt;3, where &lt;span class=&quot;math-container&quot;&gt;$M_{f2}$&lt;/span&gt; is applied;&lt;/li&gt;&#xA;&lt;li&gt;4, the fixed end&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and adding that superscript - means the result on the left side of the point and +, on the right side of the point, we can calculate that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{alignat}{4}&#xA;M_1 &amp;amp;= &amp;amp;&amp;amp;=0\text{ kNm} \\&#xA;M_2^- &amp;amp;= &amp;amp;&amp;amp;=0\text{ kNm} \\&#xA;M_2^+ &amp;amp;= M_2^- - 60 &amp;amp;&amp;amp;= -60\text{ kNm} \\&#xA;M_3^- &amp;amp;= -60 - 32\cdot3.6 &amp;amp;&amp;amp;= -175.2\text{ kNm} \\&#xA;M_3^+ &amp;amp;= M_3^- + 58 &amp;amp;&amp;amp;= -117.2\text{ kNm} \\&#xA;M_4 &amp;amp;= -60 - 32\cdot6.4 + 58 - 30\cdot2.8\cdot\dfrac{2.8}{2} &amp;amp;&amp;amp;= -324.4\text{ kNm} \\&#xA;\end{alignat}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running this on &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt; to confirm:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/idsnB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/idsnB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to be using the moment-area method to calculate deflections. As far as I can tell, you've done so correctly, just using the wrong moment diagram. Just repeat your work and it should be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To compare your results, here are the ones given by Ftool &lt;span class=&quot;math-container&quot;&gt;$(EI = 1\text{ kNm}^2)$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Total deflection at the &quot;free end&quot;: 5812&amp;nbsp;m&lt;/li&gt;&#xA;&lt;li&gt;Deflection due to moment-area 1: 820.8&amp;nbsp;m&lt;/li&gt;&#xA;&lt;li&gt;Deflection due to moment-area 2: 912.4&amp;nbsp;m&lt;/li&gt;&#xA;&lt;li&gt;Deflection due to moment-area 3: 2297&amp;nbsp;m&lt;/li&gt;&#xA;&lt;li&gt;Deflection due to moment-area 4: 1782&amp;nbsp;m&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;where moment-area 1 is the rectangular component of the 3.6&amp;nbsp;m stretch, area 2 is the inclined component of the 3.6&amp;nbsp;m stretch, area 3 is the rectangular component of the 2.8&amp;nbsp;m stretch, and area 4 is the parabolic component of the 2.8&amp;nbsp;m stretch.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-04-05T00:54:40.887" LastActivityDate="2019-04-05T00:54:40.887" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26698" PostTypeId="2" ParentId="26639" CreationDate="2019-04-02T23:11:02.220" Score="0" Body="&lt;p&gt;Other work is also a flow of energy between system and surroundings (across a boundary). Heat is flow of energy too. Consider these three sayings as analogous:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Mechanical work is said to be done BY a system if the sole effect on the surroundings could be the raising of a weight (in the surroundings).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat flow is said to be EXOTHERMIC if the sole effect on the surroundings could be an increase in the temperature of the surroundings.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other work is said to be done BY a system if the sole effect on the surroundings could be either the raising of a weight (in the surroundings) or an increase in the temperature of the surroundings.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-04-02T23:11:02.220" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26699" PostTypeId="2" ParentId="26170" CreationDate="2019-04-02T23:44:53.440" Score="1" Body="&lt;p&gt;Ok, you have a system of some kind. Normally, (not resonance, and not flutter),  you apply a force, you get a displacement. e.g. for sake of argument let's just say application of 10 N force gets 0.1 m displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, at &lt;em&gt;resonance&lt;/em&gt;, you can apply a small force and get a large displacement.  e.g. 10 N force gets 10 m displacment.  That's 100X the normal relationship, a big amplification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, at &lt;em&gt;flutter&lt;/em&gt;, what happens is that the displacement itself causes the size of the force to change.  e.g. there's an intial 0.01m displacement.  that displacement, in and of itself, causes a force of 0.1N.  The force then causes a larger displacement, 0.1m, which then causes a larger force 1N, when then causes a larger displacement 1m, etc, etc, etc.  Both the force and displacement grow exponentially until something happens to stop it.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="9275" LastEditDate="2019-04-03T00:47:06.280" LastActivityDate="2019-04-03T00:47:06.280" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26701" PostTypeId="1" AcceptedAnswerId="33079" CreationDate="2019-04-03T05:26:38.783" Score="0" ViewCount="720" Body="&lt;p&gt;Refer to &lt;a href=&quot;https://www.concretepipe.org/wp-content/uploads/2014/09/DD_17.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this document&lt;/a&gt;, and particular to Figure 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bvgWf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bvgWf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder what is the formula that governs the box culvert partial flow with respect to full flow? In other words, base on what formula the above graph is generated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.cedengineering.com/userfiles/Partially%20Full%20Pipe%20Flow%20Calculations.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;formula for pipe partial flow&lt;/a&gt; is as thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eprDU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eprDU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So essentially I'm trying to find the analogous formula for pipe culvert.&lt;/p&gt;&#xA;" OwnerUserId="3353" LastEditorUserId="3353" LastEditDate="2019-04-03T05:34:54.047" LastActivityDate="2020-02-05T09:14:29.270" Title="The formula for partial flow of Box Culvert" Tags="|civil-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26703" PostTypeId="1" CreationDate="2019-04-03T08:11:51.633" Score="0" ViewCount="29" Body="&lt;p&gt;One typical way to heat sewage sludge digesters is to pump the sludge through heat exchangers, raw sludge is injected before or after the HX. What kind of injectors are typically used for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not loking to brainstorm ideas what &lt;em&gt;could&lt;/em&gt; be used, I'm looking for solutions that are common and proven in the field.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-04-03T08:11:51.633" Title="What injectors are typically used for mixing sewage sludge?" Tags="|civil-engineering|waste-water-treatment|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26704" PostTypeId="1" AcceptedAnswerId="26706" CreationDate="2019-04-03T09:00:41.503" Score="4" ViewCount="1048" Body="&lt;p&gt;I have heard many times that four-wheel drive is less efficient than two-wheel drive. Many car producers have a normal drive mode where only two wheels are powered, and turn the four-wheel drive on only when needed, to &quot;save energy&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never understood why it should be so. You need to transfer only half the energy to each wheel (which must also save the material wear), so the total energy consumption must be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please answer the question also for the case of electric cars with four separate small engines each attached to a single wheel.&lt;/p&gt;&#xA;" OwnerUserId="19978" LastActivityDate="2019-04-03T21:03:17.780" Title="Is it true that four-wheel drive is less efficient and if so, why?" Tags="|automotive-engineering|energy-efficiency|car|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26705" PostTypeId="1" CreationDate="2019-04-03T09:24:59.263" Score="2" ViewCount="128" Body="&lt;p&gt;Strain in the x and y directions are defined by the following equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ε_x=\frac{du}{dx}+\frac{1}{2}[(\frac{du}{dx})^2 + (\frac{du}{dy})^2]$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ε_y=\frac{dv}{dx}+\frac{1}{2}[(\frac{dv}{dx})^2 + (\frac{dv}{dy})^2]$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is when the strain is negative, the first term will be negative and the terms in the square brackets will always be positive.  However, this does not occur if the strain is positive, does this mean that the terms in the square brackets need to be directional?&lt;/p&gt;&#xA;" OwnerUserId="16825" LastActivityDate="2019-04-03T12:39:32.633" Title="Compressional Strain Definition" Tags="|materials|solid-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26706" PostTypeId="2" ParentId="26704" CreationDate="2019-04-03T09:38:07.283" Score="9" Body="&lt;p&gt;Simply put, power transmission is not free.  Every time you have some energy over here, and you want to move it to over there, you are going to lose some of it in the process.  In this case, if your engine is in the front and you want to drive the back wheels, you'll need some combination of gears, bearings, and shafts to transmit that power.  Gears are pretty efficient, but it's not 100% efficient.  Maybe 97-98% efficient. Look at the schematic of a 4 wheel drive train (good one here: &lt;a href=&quot;https://www.artofmanliness.com/articles/gearhead-101-how-part-time-four-wheel-drive-works/&quot; rel=&quot;noreferrer&quot;&gt;https://www.artofmanliness.com/articles/gearhead-101-how-part-time-four-wheel-drive-works/&lt;/a&gt;)  There are 2 differentials and a transfer case. All of those contain gears, and every single gear costs you a little bit of efficiency.  Each one is not a large loss, but it does add up. With 2 wheel drive, you'd just have 1 differential and no transfer case.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-04-03T09:38:07.283" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26709" PostTypeId="2" ParentId="26704" CreationDate="2019-04-03T08:44:10.573" Score="3" Body="&lt;p&gt;It is an engineering question. Weight, space, and friction are a problem when you have a four-wheel drive. There are more mechanical (rotating) parts, which produces lots of friction. If I remember correctly, there is about 1/4 energy loss in a typical four-wheel driven car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engineering problematics hold true for electric cars, despite the fact that there are fewer moving parts. However, I don't think that there is a basic physical principle underlying.&lt;/p&gt;&#xA;" OwnerDisplayName="Userman" LastActivityDate="2019-04-03T08:44:10.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26710" PostTypeId="1" AcceptedAnswerId="26720" CreationDate="2019-04-03T11:44:37.393" Score="3" ViewCount="12853" Body="&lt;p&gt;I've always wondered if this is just a marketing thing, but all across the electronic supply spectrum I see &quot;gold plated&quot; listed as the best conductors for various contacts and connectors. However, silver is considered the most conductive element (6.2×107 S/m), followed by copper (5.9×107 S/m), with gold (4.5×107 S/m) being third. I get that any of those are better conductors than the most common (tin and steel) but I would think silver, being less expensive, more abundant, and a better conductor, would be preferred over gold. What am I missing here?&lt;/p&gt;&#xA;" OwnerUserId="19980" LastActivityDate="2024-03-24T08:01:43.167" Title="If gold is a worse electrical conductor than silver and copper, why are gold plated contacts considered &quot;better&quot; by the market?" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26711" PostTypeId="1" CreationDate="2019-04-03T11:54:22.940" Score="0" ViewCount="44" Body="&lt;p&gt;I am currently designing a device that can pick up 20 kid's plastic balls from one box and transfer them all into another box. I have decided to use a vacuuming method to vacuum up the balls as my way of picking up the balls. Right now I'm just testing if my idea can just pick up one ball. But anyway, I'm planning to use a 12V DC motor that comes with 18,800 RPM and 4.9kg/cm torque connected to a 120mm diameter centrifugal impeller made of plastic to provide suction. This is all then attached to the top of an L shape pvc pipe of 120mm diameter at the top to fit the impeller and motor, and 80mm diameter at the bottom for the ball fit through. I am also powering this device with 8 1.5V AA batteries and won't be able to use other sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point I've been doubting if this potential vacuum would even have the sucking power to even vacuum up just one of these 65mm diameter balls. If this device fails which it likely will, would it be that my 12V DC motor isn't powerful enough, or my pvc pipe is too wide or something else. I've read online that it's something to do with a good CFM volume or airflow volume for good suction power, but I'm not too sure what component of the vacuum cleaner would provide this. Any suggestions would be really appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture of ball: (&lt;a href=&quot;https://hiphiphooray.com.sg/product/colourful-air-filled-balls/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://hiphiphooray.com.sg/product/colourful-air-filled-balls/&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basic prototype of vacuum: &lt;a href=&quot;https://ibb.co/q183PCM&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ibb.co/q183PCM&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19979" LastActivityDate="2019-04-03T14:35:54.067" Title="Is it possible design a portable high powered vacuum cleaner from a 12V DC motor?" Tags="|airflow|vacuum|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26712" PostTypeId="2" ParentId="26710" CreationDate="2019-04-03T12:01:16.020" Score="4" Body="&lt;p&gt;While gold has a slightly higher resistance, it has the major advantage of not corroding or at least corroding very very slowly in normal conditions which is why it is used for so many contacts on our day to day devices...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silver and copper tarnish and corrode easily in the atmosphere leading to poor connections.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-04-03T12:01:16.020" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26713" PostTypeId="2" ParentId="26705" CreationDate="2019-04-03T12:33:58.740" Score="2" Body="&lt;p&gt;I presume you got these equations from a derivation similar to &lt;a href=&quot;https://en.wikipedia.org/wiki/Deformation_(mechanics)#Normal_strain&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Deformation_(mechanics)#Normal_strain&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small (infinitesimal) strains, the second order (squared) terms are negligible compared with &lt;span class=&quot;math-container&quot;&gt;$\partial u/\partial x$&lt;/span&gt;, and are simply ignored. This gives what is commonly called &quot;engineering strain&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For large strains, you need a more careful definition of &quot;strain&quot; that gives zero strain for arbitrarily large rigid body rotations of a small element of material. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Finite_strain_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Finite_strain_theory&lt;/a&gt; for details. The Cauchy-Green strain tensor is commonly used to model situations where small elastic deformations of the material are superimposed on large rigid body translations and rotations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general comment, there is no single &quot;theoretically correct&quot; definition of strain. If you assume the material properties (Young's modulus, Poisson's ratio, etc) are constant in the stress-strain equations, you will get different results depending on which strain measure you take - but in real life, the material properties are strain-dependent in any case, so you can make an accurate (but nonlinear) model of the material with any reasonable definition of strain.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-04-03T12:39:32.633" LastActivityDate="2019-04-03T12:39:32.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26716" PostTypeId="1" CreationDate="2019-04-03T15:33:11.607" Score="0" ViewCount="68" Body="&lt;p&gt;Why is Flaring done in industries? Why can't the same gases be used for production of energy (Thermal energy)? Or why can't the excess gas be recycled in the previous stream from which it came?&lt;/p&gt;&#xA;" OwnerUserId="19256" LastActivityDate="2019-04-04T11:14:05.043" Title="Flaring System in Industries" Tags="|chemical-engineering|petroleum-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26719" PostTypeId="2" ParentId="26716" CreationDate="2019-04-03T17:30:56.257" Score="0" Body="&lt;p&gt;The usual answer is that nobody in the vicinity of the flared gas wants to buy it, at least not in the quantities in which it is available, and so to not release a flammable gas mixture into the air around the plant, it is ignited and burned off instead. This is a wasteful process, but in the case of a petroleum refinery the energy content of the flared gas is tiny compared to the refinery's total output of useful fuels. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-04-03T17:30:56.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26720" PostTypeId="2" ParentId="26710" CreationDate="2019-04-03T20:00:31.210" Score="6" Body="&lt;p&gt;Gold is used for connectors not so much for its conductivity as it's chemical inertness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even ostensibly corrosion resistant materials like aluminium and stainless steel form a thin metal oxide layer on the surface, indeed this is what gives them their corrosion resistance and metal oxides have poor conductivity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So gold plating is potentially useful for improving the electrical connection between two contacts it tends to be most useful in signal or data connections where you are usually dealing with low voltages and maintaining a consistent resistance across connections is important. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The downside of gold is that it is soft and so prone to wear and so you tend to see it most often in connectors which need to be physically small and aren't subject to a lot of wear eg things like SIM cards, flash cards, graphics cards, phone batteries etc etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where connections are changed frequently eg in music, sound and lighting equipment chrome plating or nickel plating is usually preferred as it is much harder and you can mitigate any small increase in resistance by having a larger contact area. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2019-04-03T20:00:31.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26721" PostTypeId="2" ParentId="26704" CreationDate="2019-04-03T21:03:17.780" Score="2" Body="&lt;p&gt;As others have pointed out, this isn't really a matter of physics so much as the choice of how to implement the 4WD system. With modern ECU governed AWD systems, the mileage penalty is much less than it used to be with the hydraulically controlled AWD units. A common system was to hang a small hydraulic pump on the front and rear shaft, piped in a loop. If the machine was FrontWD normally, and everything was hooked up, no pressure would develop in the system. If the front axle's tires started to slip, it's pump would pump more fluid than the rear and transfer some torque. If the imbalance was large, the pressure would engage a clutch and transmit power directly though the shafts. These systems had appreciable parasitic losses.&#xA;The two pumps can be eliminated now and the entire thing controlled with sensors and an ECU. They can also talk to other ECU's and as a coordinated group comprise a dynamic stability control system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this short video on the Honda CRV comparing the pre 2012 system with the post 2012 system.&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=d2fnBxjAWso&quot; rel=&quot;nofollow noreferrer&quot;&gt;Honda CRV AWD systems&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-04-03T21:03:17.780" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26722" PostTypeId="1" CreationDate="2019-04-03T21:31:59.513" Score="1" ViewCount="2711" Body="&lt;p&gt;Determine:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;the distributed load &lt;span class=&quot;math-container&quot;&gt;$w_0$&lt;/span&gt; at the end of D of the beam for which the reaction at B is zero&lt;/li&gt;&#xA;&lt;li&gt;the corresponding reaction at C. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7ulX9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7ulX9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please show me how to get the right answer to this question.: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}&#xA;R_1 = 1/2 \cdot w_0 \cdot 9 = 0/2 w_0 \\&#xA;R_2 = 1/2 \cdot 3.5 \cdot 9 = 63/4\text{ kN} \\&#xA;\text{For }B = 0 \\&#xA;\sum M_c = 0 \\ &#xA;9/2 w_0 \cdot 8 - 63/4 * something \\&#xA;\end{gather}$$&lt;/span&gt;&#xA;Thats where I stopped I don't know what to do &#xA;May someone please show me the solution and explain to me his/her steps? &lt;/p&gt;&#xA;" OwnerUserId="19987" LastEditorUserId="1832" LastEditDate="2019-04-18T02:48:19.830" LastActivityDate="2019-04-18T22:17:26.473" Title="Statics Distributed load question" Tags="|mechanical-engineering|statics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26724" PostTypeId="1" CreationDate="2019-04-03T23:00:01.603" Score="0" ViewCount="51" Body="&lt;p&gt;I bought a CNC linear rail guide slide from ebay(&lt;a href=&quot;https://www.ebay.com/itm/202481581216&quot; rel=&quot;nofollow noreferrer&quot;&gt;seen here&lt;/a&gt;). I'm trying to find a proper worm gear to mate with the wormscrew. There is no technical specification that came with the kit however on the wormscrew. I've manually measured the pitch of the wormscrew to be 2mm(12.7 threads per inch).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking at &lt;a href=&quot;http://shop.sdp-si.com/catalog/?cid=p390&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site's&lt;/a&gt; catalog, but I'm just not sure what minimum measurements of the wormscrew I need in order to find a proper gear that fits. I just need to keep track of the number of turns the gear makes. I don't think I really care about the gear's quality class, gear ratio, bore/hub config. I just need it to fit the worm. Am I correct to assume the module(pitch) of the gear also needs to be 2mm? Do the pressure angle, lead angle, and number of teeth affect the fitting of the gear onto the worm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've seen suggestions online that I should just take the kit to grainger or mcmaster and ask for a proper gear that fits, but I'm not near either of those places.&lt;/p&gt;&#xA;" OwnerUserId="19989" LastActivityDate="2019-04-03T23:00:01.603" Title="Finding the right worm gear for a wormscrew that didn't come with technical specs" Tags="|mechanical-engineering|gears|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26725" PostTypeId="2" ParentId="26722" CreationDate="2019-04-03T23:17:19.537" Score="1" Body="&lt;p&gt;In summary the steps are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Write equilibrium equations in terms of unknown load magnitude Wo.&lt;/li&gt;&#xA;&lt;li&gt;Set the vertical reaction  RB = 0.&lt;/li&gt;&#xA;&lt;li&gt;Solve equation for Wo.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It helps me to break the trapezoidal distribution into a rectangular distribution magnitude Wo, and a triangular distribution with peak magnitude = (3.5 - Wo).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the steps you have shown. There is already a problem when you calculate the resultant of the triangular distribution. Look closely at the diagram at the top of the solution below to see the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bzwde.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bzwde.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-03T23:17:19.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26726" PostTypeId="2" ParentId="26716" CreationDate="2019-04-03T23:40:36.713" Score="2" Body="&lt;p&gt;There are 2 general reasons for flaring; In a refinery or chemical plant for safety and at well site for economics - subject to Dept of Interior rules. Refineries and chemical plants have a large number of safety valves , most are automatic . Just like the safety valve you have on your hot water heater : the hot water heater vents hot water outside , no problem . But refinery/chem units release very hot hydrocarbons ( generally) , they must be contained in a piping system and taken to a safe location and ignited/ flared ( during an upset- over-pressure or over-temperature condition). Also, during a start-up or shut-down there may be flaring.  This piping system must not have air in it or burning may occur anywhere in the system during an upset: So , the piping is flushed with gas to keep air out. Also there are ignition burners at the flare tip to ignite any emergency release , these are the small flares one routinely sees around the plants...  Flares at well sites are to permit oil production . There is essentially always some gas produced with oil , this is separated at the well head . Usually the oil goes into a pipeline ( gathering line) . In a larger field there will be gas lines that take the gas to a compressor . From the compressor, the gas is reinjected into the oil field or it may be combined with other gas and piped to a gas plant where LPG, water and sulfur, etc. are removed and some portion becomes pipeline gas.If there is limited gas from a few wells it is flared according to rules of the Interior Dept; There are time limits and volume limits. The time allows for more wells to be drilled, or pipelines to be built or gas plants to be built. There are various situations.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-04-03T23:40:36.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26727" PostTypeId="2" ParentId="26701" CreationDate="2019-04-03T23:51:12.383" Score="1" Body="&lt;p&gt;Page 2 of the DD_17.pdf document that you linked tells you the answer in the first paragraph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manning's equation is what is used. There are different formulas used to solve for the hydraulic radius R for different shape culverts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a fundamental topic for hydraulic engineering can I suggest that you find a text book on this topic? It will offer much more in depth background information than can be offered in this setting.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-03T23:51:12.383" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26728" PostTypeId="2" ParentId="26716" CreationDate="2019-04-04T11:14:05.043" Score="0" Body="&lt;p&gt;I think, it's because sometimes the flowrate can be very high (at fire blowdown case of light hydrocarbon liquids). At such high flowrate, consumer may not able to absorb thus plant safety is not guaranteed.&#xA;I think plants and governments are trying to recover the gas for economic and environmental issue, so it has been taken care of.&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2019-04-04T11:14:05.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26731" PostTypeId="2" ParentId="26656" CreationDate="2019-04-04T13:13:13.800" Score="4" Body="&lt;p&gt;Start with the phase diagram of water found &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Phase_diagram_of_water.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;at this link&lt;/a&gt;. The solid &amp;lt;-&gt; vapor transformation occurs at pressures below 1 bar. Using a temperature of -40 &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C puts the pressure around 7 Pa. This is vacuum conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike all five other transformations (e.g. fusion is solid -&gt; liquid), the vapor -&gt; solid transformation has no standard definition. In some cases, it is called solid-state condensation. In others, it is simply called vapor deposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The better term for the transition vapor -&gt; solid is the frost point. However, the use of the word dew point is a hypothetical approach. It says that, when you would try to cool the air to saturate it with water, you must cool to that temperature. Yes, in this case, when you would cool an infinitesimal amount just below that temperature, you would create solid ice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, you can use a desiccant to extract water vapor from air without going through a liquid state. The water vapor adsorbs on to the solid.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-04-04T13:13:13.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26732" PostTypeId="1" CreationDate="2019-04-04T13:39:26.903" Score="3" ViewCount="3366" Body="&lt;p&gt;I am suffering from a strange hum in my home. I have just bought my home and living in 3 months. A month ago I've started to hear a humming noise. Then, I recorded the sound and looked at its frequency response. The peak was at 50 Hz. After a while, I started to use an application named &quot;SignalScopeX&quot; for iPhone. It gave me the same results, in this app you can see the response in real time and there is a peak at exactly 49 Hz. I measured several times (in night and day) and the sound exists every time. Is there a possibility that in pure silent mics response 50&amp;nbsp;Hz, or is it real sound? I hear it but still I want to be sure about that. I suspected an electrical machine vibrating my walls (I am living in Turkey, main electricity is 50 Hz). Me (living in 6th floor) and my neighbours (living in 5th and 7th floors) cut of our home electricity but still, the sound is exists. What can be the reason of the sound?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of my thoughts are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There may be a constant working motor like an aquarium on the 10th floor. Is it possible?&lt;/li&gt;&#xA;&lt;li&gt;There is a general electricity line of building in one of my room. If there is any leakage may it cause the sound? How can I measure it? Are there any more possibilities?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19995" LastEditorUserId="1832" LastEditDate="2019-04-04T14:51:42.783" LastActivityDate="2023-11-07T22:30:40.437" Title="What can be the reason of the constant 50 Hz sound in my home?" Tags="|frequency-response|sound-isolation|waves|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26733" PostTypeId="1" CreationDate="2019-04-04T14:27:21.963" Score="0" ViewCount="29" Body="&lt;p&gt;If using a laminar flow element, in conjunction with Poiseuille's equation, to calculate mass and volumetric flow rate, what would the benefit be to using two absolute pressure transducers to measure dP instead a single differential transducer? I am currently using a single differential sensor and a single absolute (for STP &amp;amp; mass flow calculations), but have become aware of an instrument that uses 3 absolute sensors to accomplish the same calculations. Why would this be done?&lt;/p&gt;&#xA;" OwnerUserId="19866" LastActivityDate="2019-04-04T14:27:21.963" Title="Use of Absolute vs. Differential Pressure Transducers to Calculate Flow Through Annular Restriction" Tags="|mechanical-engineering|fluid-mechanics|pressure|airflow|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26736" PostTypeId="2" ParentId="26732" CreationDate="2019-04-04T17:20:06.413" Score="0" Body="&lt;p&gt;Indeed, if the spectral content of the noise is peaked at the mains frequency, it is evidence of a &lt;em&gt;motor&lt;/em&gt; somewhere in the building that runs all the time. This could be a ventilator or exhaust fan for a kitchen or bath, an aquarium pump, a refrigerator, or a component of the plumbing system that either boosts the water pressure for the upper floors or scavenges sewage from them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that motor is well-coupled to the frame of the building, you will hear it almost everywhere, so I would recommend walking through the building with your sound meter app running and check the basement, attic, hallways, lobby, etc. Somewhere you will find that motor!&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-04-04T17:20:06.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26739" PostTypeId="1" CreationDate="2019-04-04T20:07:08.913" Score="0" ViewCount="81" Body="&lt;p&gt;The related &lt;a href=&quot;https://en.wikipedia.org/wiki/Breeder_reactor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; article says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A breeder reactor is a nuclear reactor that generates more fissile&#xA;  material than it consumes. Breeder reactors achieve this because&#xA;  their neutron economy is high enough to create more fissile fuel than&#xA;  they use, by irradiation of a fertile material, such as uranium-238 or&#xA;  thorium-232 that is loaded into the reactor along with fissile fuel.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thus, a breeder reactor somehow solves, that the thermal neutrons in it create equal or more &lt;span class=&quot;math-container&quot;&gt;$^{239}\rm Pu$&lt;/span&gt; (by the neutron capture of &lt;span class=&quot;math-container&quot;&gt;$^{238}\rm U$&lt;/span&gt;, which decays to &lt;span class=&quot;math-container&quot;&gt;$^{239}\rm Pu$&lt;/span&gt; in two steps), than the fission (of &lt;span class=&quot;math-container&quot;&gt;$^{239}\rm Pu$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$^{235}\rm U$&lt;/span&gt;) wastes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is, how do they reach this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a thermal reactor, the neutrons are thermal (i.e. their kinetical energy is roughly the same, than the medium around them). Thus, also the cross sections for neutron capture are roughly fixed. And these are far smaller, which would be required for breeding (as far I know, roughly half of it).&lt;/p&gt;&#xA;" OwnerUserId="9" LastActivityDate="2021-04-04T03:29:13.203" Title="How is the neutron capture cross-section increased in thermal breeder reactors?" Tags="|nuclear-technology|nuclear-engineering|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26741" PostTypeId="1" AcceptedAnswerId="26748" CreationDate="2019-04-04T23:00:56.343" Score="0" ViewCount="626" Body="&lt;p&gt;When we use Abaqus to solve the static problem, one step is to set:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Time period, the number of increments and increments size&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since the static problem is not time-dependent, there is no time, which means the above time is virtual time. Then I was wondering how we look at those parameters in the static case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we think that way:&#xA;The Time period is the real/physical total deformation time for the static problem and is scaled to 1. And its substep which from Time period/number of increments comes from the sub deformation which is also scaled.&lt;/p&gt;&#xA;" OwnerUserId="17605" LastActivityDate="2019-04-05T18:01:08.707" Title="Physical meaning of increment in static problem in abaqus" Tags="|statics|abaqus|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26742" PostTypeId="2" ParentId="25267" CreationDate="2019-04-04T23:08:09.697" Score="2" Body="&lt;p&gt;On page 16.1-306 of AISC 360-10... &quot;The unbraced length is defined as the spacing between locations where twist is restrained.&quot; Without the addition of any bottom flange bracing, this would imply that the unbraced length is the distance between the supports.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-04T23:08:09.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26743" PostTypeId="2" ParentId="26739" CreationDate="2019-04-04T23:21:05.030" Score="0" Body="&lt;p&gt;Thermal neutrons get lots of attempts. What you need, which the wiki article doesn't cover, is a picture of the overall &quot;neutron budget&quot; inside the  reactor at a given instant. Some reactors run in the 98% neutron efficiency range. With fission averaging about 2.5 neutrons produced per fission event, there is plenty of headroom (at least on paper). You need good moderators and perhaps a hotter fuel load. Plus an external neutron accelerator to help manage both power output and breed ratio simultaneously doesn't hurt. The accelerator gives you a second knob. It's hard to manage two separate variables with just one knob.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-04-04T23:21:05.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26744" PostTypeId="1" CreationDate="2019-04-05T03:00:38.623" Score="0" ViewCount="684" Body="&lt;p&gt;Is there any equation for minimum area ratio &lt;em&gt;(Throat area (At)/ Combustion Chamber area (Ac) )&lt;/em&gt; required for choked flow in &lt;strong&gt;convergent nozzle&lt;/strong&gt; &lt;em&gt;(just like equations for minimum Pressure ratio and Temperature ratios)&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While searching online for minimum Area ratio required, most of them are related to CD nozzle. But in my case I'm looking for Convergent nozzle only so that I can get an estimate how wide combustion chamber should be to the exhaust area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to derive equation from mass flow continuity equation but it is dependent on flow velocity inside combustion chamber which is generally assumed to be near zero. &lt;em&gt;(and I got stuck here.)&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="16175" LastActivityDate="2023-12-20T04:10:26.300" Title="What's the minimum area ratio required to achieve choking in Converging nozzle?" Tags="|fluid-mechanics|thermodynamics|propulsion|rocketry|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26745" PostTypeId="2" ParentId="26741" CreationDate="2019-04-05T04:24:12.763" Score="2" Body="&lt;p&gt;Static analysis is an ideal scenario, no real event is actually static- either quasi-static or dynamic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is based on the assumption that all loads are applied slowly and gradually until they reach their full magnitudes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So when you have some (pseudo) time steps, what you are basically doing is discretizing the steps of load application. &lt;/p&gt;&#xA;" OwnerUserId="15803" LastActivityDate="2019-04-05T04:24:12.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26746" PostTypeId="1" AcceptedAnswerId="26758" CreationDate="2019-04-05T06:27:18.590" Score="0" ViewCount="101" Body="&lt;p&gt;i have Question that Mechatronics Engineering is all about Precision or intelligence , from precision i mean that to follow pattern precisely with precise time and from intelligence i mean making intelligent decisions based on input using image processing etc. &#xA;for example i google these Mechatronics keyword and i got these images &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) auto industry robotic arms manufacturing cars : &lt;a href=&quot;https://i.stack.imgur.com/5NIJe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/5NIJe.jpg&lt;/a&gt;&#xA;so my Question is these auto industry robotic arms just follow precise pattern or they have cameras in them and they assemble or cut parts of cars based on what they see in camera added to them ( image processing ).&#xA;if no camera then it mean if car assembly is bit off from its position they could make mistake and could mess up things ?&lt;/p&gt;&#xA;" OwnerUserId="20004" LastActivityDate="2019-04-05T19:52:17.223" Title="Mechatronics Engineering is about Precision or intelligence?" Tags="|robotics|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26748" PostTypeId="2" ParentId="26741" CreationDate="2019-04-05T09:30:45.010" Score="3" Body="&lt;p&gt;&quot;Statics&quot; usually is defined as meaning that the &lt;em&gt;acceleration&lt;/em&gt; of the structure is too small to cause significant &lt;em&gt;inertia&lt;/em&gt; forces, i.e. the &quot;mass &lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt; acceleration&quot; terms in the equations of motion can be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the &lt;em&gt;response&lt;/em&gt; of the structure may still change with time when the acceleration is negligible, if the material behaviour is nonlinear and time-dependent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in a creep analysis, the amount of creep strain is time-dependent, and may eventually cause structural failure even when there is no change in the external loads. In this type of analysis, time is a real physical quantity, not an arbitrary scale factor between 0 and 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the material behaviour is linear, in a coupled thermal - structural analysis the changes in the temperature distribution depending on the thermal conductivity and specific heat of the material are modelled in &quot;real time&quot;, and produce corresponding real-time changes in the thermal stress in the structure.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-04-05T09:37:38.513" LastActivityDate="2019-04-05T09:37:38.513" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26750" PostTypeId="2" ParentId="26746" CreationDate="2019-04-05T09:48:56.200" Score="1" Body="&lt;p&gt;It is neither, one or both. Think of Mechatronics as just modern machine building. The key idea is that there is no single reason for mechatronical machines.&#xA;The aim can be any of, but not limited to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Faster&lt;/li&gt;&#xA;&lt;li&gt;Cheaper&lt;/li&gt;&#xA;&lt;li&gt;Better&lt;/li&gt;&#xA;&lt;li&gt;With less parts&lt;/li&gt;&#xA;&lt;li&gt;more flexibility&lt;/li&gt;&#xA;&lt;li&gt;Higher precision&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now since your main method is to include control systems into all this intelligence follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most robots are static. They just dumbly do things like a CNC. But since they are programmable, you end up bolting methods to aim, or pick things that dont come with a superior control. Having a digital intrface enables you to do that. And it will happen over time that you add more inteligence. Its a side effect, and then a goal.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2019-04-05T09:48:56.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26751" PostTypeId="2" ParentId="26741" CreationDate="2019-04-05T09:51:17.027" Score="3" Body="&lt;p&gt;In linear static formulation of FE problem the equations do not include time, so the time you enter is not used. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If, however, you are doing an implicit time dependent analysis, the time and time-steps are used to advance the solution in time. Note that each step is still a static solution of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a material non-linearity then the solver will have to converge in small steps towards the solution (using for example the Newton method or line search), but it is not related to the time you are asked to enter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope it helps!&lt;/p&gt;&#xA;" OwnerUserId="11712" LastEditorUserId="11712" LastEditDate="2019-04-05T12:23:51.250" LastActivityDate="2019-04-05T12:23:51.250" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26752" PostTypeId="1" CreationDate="2019-04-05T13:26:25.670" Score="9" ViewCount="422" Body="&lt;p&gt;Recently this water tank was constructed (poured concrete).  (I presume it is a water tank, as they laid blue plastic/PVC water pipe, perhaps 24 in. dia., in the vicinity at the same time.)  After the outside was finished, they erected this dark (almost black in the picture) inverted umbrella-like screen around the base.  I have never seen such a thing, either during construction, or around a finished tank.  What is its purpose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tOizW.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tOizW.jpg&quot; alt=&quot;water tank surrounded by what?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20010" LastActivityDate="2019-07-01T20:50:24.003" Title="What is the umbrella-like thing surrounding this new structure?" Tags="|structural-engineering|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26754" PostTypeId="1" AcceptedAnswerId="28799" CreationDate="2019-04-05T17:46:22.100" Score="1" ViewCount="1341" Body="&lt;p&gt;We can get local coordinate system information (location and rotation angle (in degrees) relative to the global Cartesian coordinate system.) by using &lt;code&gt;*GET, Par, CDSY, N, LOC, X (Y or Z)&lt;/code&gt; or &lt;code&gt;*GET, Par, CDSY, N, ANG, XY (YZ or ZX)&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to get the element coordinate system information for each element?&lt;/p&gt;&#xA;" OwnerUserId="5069" LastActivityDate="2019-07-14T16:00:50.380" Title="How to get element coordinate system information (location, angle) for each element using ANSYS Mechanical APDL" Tags="|ansys|ansys-workbench|ansys-apdl|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26755" PostTypeId="2" ParentId="26741" CreationDate="2019-04-05T18:01:08.707" Score="0" Body="&lt;p&gt;While the other answers make some good points the only one really addressing the issue is Schneider's response.&#xA;The time increments in a static problem in abaqus means that the total load, say for example 10 N, is divided into chunks of say 1 N each (if you have 10 substeps), and the software adds 1 N each time the previous substep has converged. This is done because it is easier for the solver to converge with small loads.&lt;/p&gt;&#xA;" OwnerUserId="18457" LastActivityDate="2019-04-05T18:01:08.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26756" PostTypeId="1" CreationDate="2019-04-05T19:10:14.140" Score="1" ViewCount="251" Body="&lt;p&gt;Recently I had need to prove the beam deflection equation for a simple cantilever beam with a point load at the end. The equation would be in reference to the Euler-Bernoulli theory taught in most basic engineering undergraduate courses. The only caveat is to not use &quot;I&quot; at any point in the derivation, but rather use basic concepts of force, moment, and some conceptual ingenuity to build the equation. The derivation would be specifically to find the angle (&lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;) that is the result of the intersection of lines projecting from the two beam faces post-deflection. Another starting assumption would be a rectangular beam face.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My problem is that every time I derive the equation, I come out 2x too large. My initial equation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$2\int_0^{h/2} \frac{(r_i + h/2 + y)\theta - l_o}{l_o} (E) (y) (z) dy = Wl_o$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{(2)(E) (z)(\theta)}{l_o}\int_0^{h/2} y^2 dy = Wl_o$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately leading to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{(E) (z)(h^3)(\theta)}{12l_o}= Wl_o$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \theta = \frac{12W{l_o}^2}{Ezh^3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, I come up with an answer that is twice as large as (what I believe to be) the correct answer, and I can't seem to reconcile where I lost a 2.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My thought process behind the initial equation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The initial fraction represents the strain&lt;/li&gt;&#xA;&lt;li&gt;E represents the Elastic Modulus (which needs to be multiplied against the strain to get a force/area (pressure point))&lt;/li&gt;&#xA;&lt;li&gt;For each pressure point you need to multiply by a distance to get the moment (I chose the center axis)&lt;/li&gt;&#xA;&lt;li&gt;To account for the depth of the beam, multiply by &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Integration from &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$h/2$&lt;/span&gt; is necessary to sum the moment from center axis to top of beam&lt;/li&gt;&#xA;&lt;li&gt;The &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; represents the bottom half of the face of the beam that is equal in moment to the top half (so double the top half)&lt;/li&gt;&#xA;&lt;li&gt;The &lt;span class=&quot;math-container&quot;&gt;$W*l_o$&lt;/span&gt; represents the total moment being accounted for&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With all that said, I've racked my brain and can't seem to find the discrepancy. Hopefully it's not a stupid mistake.. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This is a X-Post from physics SE.&lt;a href=&quot;https://i.stack.imgur.com/nhwEq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nhwEq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20011" LastEditorUserId="20011" LastEditDate="2019-04-13T00:53:44.530" LastActivityDate="2019-04-13T00:53:44.530" Title="Simple beam deflection of cantilever beam" Tags="|statics|beam|deflection|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="26757" PostTypeId="1" AcceptedAnswerId="26770" CreationDate="2019-04-05T19:27:21.353" Score="1" ViewCount="112" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KFnDw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KFnDw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a)Z=(B+C')A+B(C+D')+BD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z=AB+AC'+BC+BD'+BD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z=AB+A'C+B&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z=B+AC'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) A B C|Z&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0 0 0|0&#xA;0 0 1|0&#xA;0 1 0|1&#xA;0 1 1|1&#xA;1 0 0|1&#xA;1 0 1|0&#xA;1 1 0|1&#xA;1 1 1|1&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;(c)       BC  00 01 11 10&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; A 0   |0  |1 |1 |1|&#xA;&#xA;   1   |1  |0 |1 |1|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I was able to complete the 1st 3 Q's i.e.(a),(b),(c).But i didn't understand the last 2 Q's i.e. (d),(e)&lt;/strong&gt;.Can someone give me a hint or push me in the right direction? Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="19969" LastActivityDate="2019-04-07T07:04:29.193" Title="Implementing logic expression and the truth table of logic function" Tags="|electrical-engineering|electrical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26758" PostTypeId="2" ParentId="26746" CreationDate="2019-04-05T19:52:17.223" Score="1" Body="&lt;p&gt;Modern robots are precise. given the same inputs they produce the same outputs. A signal turns on and the arm moves to a position. Every time the same signal turns on, the robot arm moves back to the same position, often with in 10's of microns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The intelligence is to some degree illusionary. A computer program running on the robot controller is simply executing a program responding to various inputs such as single sensors detecting the presence of part or values from a vision camera providing an X and Y coordinate of part center in its field of view. Based on the inputs in accordance with program code it produces outputs, including moving the robot arm to various locations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some applications the programming is fairly simplistic, it other cases it maybe vary complicated. Besides the application code there will be code to actually control robot functions such as motion planing, handling emergency stop signals, and controlling and monitoring its own power supplies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mechatroincs combines knowledge of mechanical systems such as the robot arms transmission systems, electrical / electronic such as the electrical motors and electronic circuits for things like inputs and outputs, with the controls and software. it is a large skill set to master, but the career field is rewarding.&lt;/p&gt;&#xA;" OwnerUserId="20012" LastActivityDate="2019-04-05T19:52:17.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26759" PostTypeId="2" ParentId="26732" CreationDate="2019-04-05T19:55:03.900" Score="0" Body="&lt;p&gt;It does &quot;sound&quot; like it could be from an electrical device assuming you are in region that used 50 Hz power. If you are able, try turn off the power at the circuit breaker panel one circuit at a time and see if the sound goes away. That would help limit your search for the cause. &lt;/p&gt;&#xA;" OwnerUserId="20012" LastActivityDate="2019-04-05T19:55:03.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26761" PostTypeId="1" AcceptedAnswerId="26771" CreationDate="2019-04-05T21:01:36.230" Score="0" ViewCount="150" Body="&lt;p&gt;I have rope on a spool that is on a cart that is powered and moving at a constant velocity of 0.5 m/s. The end of the rope is passed thru a break and fixed to the wall. one break has a compression spring with an unknown force to create friction to maintain a constant tension on the rope. The spool has a friction of .1 and is .5 m in diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the force the spring needs to put on to maintain a constant tension as a result of the movement of the cart but do not know where to start&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6ykC9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6ykC9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20014" LastEditorUserId="20014" LastEditDate="2019-04-08T14:28:33.873" LastActivityDate="2019-04-09T11:23:25.680" Title="Dynamic spring tension" Tags="|statics|dynamics|springs|" AnswerCount="2" CommentCount="1" FavoriteCount="0" ClosedDate="2019-04-07T15:11:37.023" ContentLicense="CC BY-SA 4.0" />
  <row Id="26762" PostTypeId="2" ParentId="26761" CreationDate="2019-04-06T00:05:01.993" Score="2" Body="&lt;p&gt;The constraints set out in your question don't make sense. You have described an object which is moving at a constant velocity - this means that there must be no resultant force on it. As illustrated, the tension in the rope would be slowing the cart down, unless there is some driving force inside the cart which you have not illustrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess here is that the cart is powered, and you're asking what the force should be on the roller to keep the tension at the wall anchor at 38N?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, you've set this out as a 'physics type' problem, with 2D weightless string, constant velocities etc. whereas the resistive force generated by squashing a 3D rope between some rollers is an 'engineering type' problem, which is best answered by physical testing, as the force required will vary over the lifetime of the system as the rope and rollers wear...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the above goes some way to helping you out - feel free to edit your question to provide more specific details if you can.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-04-06T00:05:01.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26766" PostTypeId="2" ParentId="26752" CreationDate="2019-04-06T19:00:50.680" Score="0" Body="&lt;p&gt;A very similar one has just been built in this area. The steel &quot;umbrella&quot; may be to retain any water leak from the central tank. I had never seen one previously. Update -June 28. It is a conventional water tank, they do most construction close to the ground then lift it to the top and finish it.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-06-29T15:18:00.643" LastActivityDate="2019-06-29T15:18:00.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26767" PostTypeId="2" ParentId="25969" CreationDate="2019-04-06T22:19:01.583" Score="1" Body="&lt;p&gt;Lateral stiffness is from the diamond stays and notched spreaders. The weight is distributed along the beam using a running &quot;crow's foot&quot; arrangement. The square blocks and holes are presumably there to aid in telling the rigger what they need to do to keep it all straight and true.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-04-06T22:19:01.583" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26768" PostTypeId="2" ParentId="26732" CreationDate="2019-04-06T22:22:35.807" Score="0" Body="&lt;p&gt;In addition to a motor, a transformer carrying a relatively high current at 50 Hz can also be a source of mechanical vibration.&lt;/p&gt;&#xA;" OwnerUserId="19568" LastActivityDate="2019-04-06T22:22:35.807" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26770" PostTypeId="2" ParentId="26757" CreationDate="2019-04-07T07:04:29.193" Score="0" Body="&lt;p&gt;Question D refers to the fact you can use NAND gates in different combinations to replicate different types of gates. IE to replicate a NOT gate, simply tie the 2 inputs of the NAND gate together, and there you go. This &lt;a href=&quot;https://www.instructables.com/id/NOT-AND-OR-Gates-From-NAND-Gates/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt; details the different connections of NAND gates to replicate your gates. &#xA;Question E refers to your schematic you'll have made in question D once you've removed any redundant NAND gates, if there are any. Every gate has a propogation delay, or the time it takes for the signal to go from the input pins the the output pin. This is usually in nanoseconds. As the question states it'll take 20ns to go from the input of a gate to the output, simply add up the gates in series to come to the answer.&lt;/p&gt;&#xA;" OwnerUserId="12187" LastActivityDate="2019-04-07T07:04:29.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26771" PostTypeId="2" ParentId="26761" CreationDate="2019-04-07T10:24:40.043" Score="0" Body="&lt;p&gt;You have not specified the diameter of the spool nor the friction constant. Neither have you specified the friction constant in the pulleys. I’m guessing you’re trying to find the required spring force in the pulleys so that an appropriate force is subtracted from 38 N such that the remaining force cancels out the rotation friction in the spool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;I am unable to comment so I will write here. You want to calculate the required spring force to maintain a constant tension in the rope. However, since the rope is fixed to a wall it cannot move between the pulleys and therefore the tension results purely from radius, friction and moment of inertia (if there is angular acceleration) of the spool. The friction in the spool causes a moment &lt;span class=&quot;math-container&quot;&gt;$M_{friction}$&lt;/span&gt; (opposing uncoiling of the spool). This is balanced by tension in the rope so&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ M_{total} = M_{friction}+M_{inertia} = Tr \:\: =&amp;gt; \:\:T=\frac{M_{total}}{r}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, if you wish to maintain a constant tension in the rope, you must maintain a constant &lt;span class=&quot;math-container&quot;&gt;$\frac{M}{r}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="20027" LastEditorUserId="20027" LastEditDate="2019-04-09T11:23:25.680" LastActivityDate="2019-04-09T11:23:25.680" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26773" PostTypeId="1" CreationDate="2019-04-07T17:49:50.860" Score="4" ViewCount="424" Body="&lt;p&gt;Internet search results don't give much results of real life examples of bladeless turbine. Has anyone ever seen a working real example of &lt;a href=&quot;https://teslauniverse.com/nikola-tesla/patents/us-patent-1061206-turbine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tesla's bladeless turbine,&lt;/a&gt; in some industrial or other application? How did it perform? What was the application?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Marino&lt;/p&gt;&#xA;" OwnerUserId="18760" LastEditorUserId="18760" LastEditDate="2019-04-07T18:37:56.910" LastActivityDate="2021-03-02T13:05:15.610" Title="Tesla's Turbine in Real Life" Tags="|turbines|turbomachinery|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26774" PostTypeId="2" ParentId="26773" CreationDate="2019-04-07T18:12:04.590" Score="2" Body="&lt;p&gt;An inspection of his patent reveals that his pump relies on the viscosity of the fluid to perform work- that is, the fluid is coupled to the impelling cylinder by shear forces alone. This means the fluid being pumped is being strongly sheared whenever the pump is running, and this inevitably produces significant frictional losses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are  many other ways of pumping fluid in which the pumping forces are not generated by viscosity, and hence will not suffer the efficiency losses of Tesla's scheme. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall efficiency limit of this thermodynamic process can be accurately determined by calculating the &lt;em&gt;carnot efficiency&lt;/em&gt; for the process, which limit holds regardless of the details of the process steps. This is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carnot efficiency = &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(source temperature - sink temperature)/(source temperature * sink temperature)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the source temperature is the temperature of the &quot;hot side&quot; of the cycle and the sink temperature is the temperature at which the cycle dumps its waste heat. All temperatures are in degrees &lt;em&gt;absolute&lt;/em&gt; (Kelvin). &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="16486" LastEditDate="2020-02-06T11:18:20.337" LastActivityDate="2020-02-06T11:18:20.337" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26775" PostTypeId="1" CreationDate="2019-04-07T18:25:32.107" Score="1" ViewCount="141" Body="&lt;p&gt;In the past few weeks I've seen several European steam locomotives in museums. I noticed several of them have a box in the driver's cabin, similar to the one in the picture I've attached. Several cables or tubes leave this box in a circular arrangement and I wonder what this box is doing. So, what is it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pS1Q7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pS1Q7.jpg&quot; alt=&quot;Box in steam locomotive&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Engineering.SE seems to be the best fit for this question; if there's a better one, please let me know.)&lt;/p&gt;&#xA;" OwnerUserId="20031" LastActivityDate="2019-04-17T01:13:50.373" Title="What's this box in steam locomotives?" Tags="|rail|steam|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26776" PostTypeId="1" CreationDate="2019-04-07T20:38:46.400" Score="0" ViewCount="102" Body="&lt;p&gt;How can I ignore the thread in standard parts in Solidworks (for Example nut with thread) in the collision simulation without deactivating the component? I can disable all the 180 nuts in the list of components, but there has to be a way to solve this problem more elegantly. On the other hand, the collision detection notifies me about a possible collision with components (those that include thread) even before the movement simulation.&lt;/p&gt;&#xA;" OwnerUserId="20034" LastEditorUserId="20034" LastEditDate="2019-04-07T20:55:34.837" LastActivityDate="2019-04-07T20:55:34.837" Title="Ignore the thread in Collision Detection in Solidworks" Tags="|solidworks|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26778" PostTypeId="2" ParentId="26732" CreationDate="2019-04-07T21:26:17.553" Score="0" Body="&lt;p&gt;Probably it is a small transformers or even an abandoned electrical door chime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try both closing the electrical breakers one by one and or using a stethoscope or directional mic to pick up the location of it. It may well have been buried under some cabinetry or tile work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is actually shorted by connection with a pipe or similar conductor, you pay the utility bill for it. Use magnetic Ohmmeter to find it.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-07T21:26:17.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26780" PostTypeId="2" ParentId="26739" CreationDate="2019-04-08T02:45:12.437" Score="0" Body="&lt;p&gt;There are two questions asked here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; How neutron cross section increased for Thermal neutrons? (As per the title)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nuclear reaction is just like any other chemical reaction. Just the presence the reactant does not cause reaction. There should be enough time and collisions between reactants. By slowing down the neutrons, other atoms get enough time to capture neutron. In thermal reactor, fast neutrons are slowed down by thousands of times. (When fission happens, released neutrons have around 80% total energy emitted. So Fast neutrons are incredibly fast). This is all just my theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; Few of the neutrons are absorbed by some other atoms which are either don't undergo fission or doesn't emit further neutrons. And even then how more fissile atoms are created in breeder reactors? (As per second paragraph in the content.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple answer - by lot of engineering work. A nuclear reactor is designed around the solution for this problem. For example, Xe is formed as part of fission product. Xe eats lots of neutrons if present in reactor. But in LFTR, Xe comes out of reactor as it forms because Xe is a gas and reactants are in molten state.&lt;/p&gt;&#xA;" OwnerUserId="16175" LastActivityDate="2019-04-08T02:45:12.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26781" PostTypeId="1" AcceptedAnswerId="26783" CreationDate="2019-04-08T04:30:26.057" Score="1" ViewCount="197" Body="&lt;p&gt;I've read that desert sand is too round for concrete mixes. But, I'm not building a dam, I'm building a 3 foot landscaping wall. I'm interested in using desert sand for the color, so this is an aesthetic choice. How will using desert sand affect the strength of my mix? Are there ways to counteract the effect? I've been reading everything. Is desert sand ok to use, just uneconomical for large projects?&lt;/p&gt;&#xA;" OwnerUserId="20037" LastEditorUserId="20037" LastEditDate="2019-04-08T04:41:30.860" LastActivityDate="2019-04-08T12:37:56.487" Title="How will using desert sand instead of concrete sand affect my concrete recipe?" Tags="|materials|civil-engineering|concrete|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26782" PostTypeId="1" CreationDate="2019-04-08T04:55:28.897" Score="0" ViewCount="86" Body="&lt;p&gt;I am trying to find warping function &lt;span class=&quot;math-container&quot;&gt;$\mathcal{T}$&lt;/span&gt; for a thin-walled multi-cellular beam. There are four rectangular cells. The governing differential equation for &lt;span class=&quot;math-container&quot;&gt;$\mathcal{T}$&lt;/span&gt; is Eq. 8.98 as shown in attached figure. But I don't know the boundary condition.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nVP4q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nVP4q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2019-04-08T04:55:28.897" Title="Determination of Warping Function for Thin-walled Multi-cellular Beams" Tags="|structural-engineering|structural-analysis|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26783" PostTypeId="2" ParentId="26781" CreationDate="2019-04-08T05:27:16.907" Score="3" Body="&lt;p&gt;You can use desert sand ( or use the approved concrete coloring admixtures which will give the tint you need) with the following considerations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Wash the sand thoroughly to remove any possible dirt or clay mixed with it, roundness is not a big issue.&lt;/li&gt;&#xA;&lt;li&gt;Add more cement to compensate for possible reduction in strength.&lt;/li&gt;&#xA;&lt;li&gt;Add concrete acrylic glue as per manufacturer's recommendation.&lt;/li&gt;&#xA;&lt;li&gt;Generally the following is recommended proportion of concrete materials:  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Add 6 bags of 94lbs of cement, water not more than 40% of cement weight, to a cubic yard of sand and gravel. The more water the weaker your concrete. so consider the moisture of your sand as part of water ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aggregate should be roughly half less than 3/8&quot; and half between 3/8 to 1.25&quot; for your application and lack of slump tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget curing the wall, watering it for at least a week.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could seek advice from local trades-people as to the ways to make your concrete stronger and as far as plasticity workable. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2019-04-08T12:37:56.487" LastActivityDate="2019-04-08T12:37:56.487" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26784" PostTypeId="1" CreationDate="2019-04-08T05:35:40.997" Score="0" ViewCount="43" Body="&lt;p&gt;I have implemented a PI compensator for an integrator plant through floating point variables and everything works fine. The design intent is tracking filter and it does not track well the input signal. In simulation, everything works well. What could be the problem and what is the workaround?&lt;/p&gt;&#xA;" OwnerUserId="17773" LastActivityDate="2019-04-08T05:35:40.997" Title="Digital control implementation in MCU" Tags="|control-engineering|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="26788" PostTypeId="1" CreationDate="2019-04-08T21:03:09.180" Score="0" ViewCount="2474" Body="&lt;p&gt;Of course, this is not true for all engines (after all the pre-2014 F1 engines were 4-stroke and were rpm limited to 18,000 rpm). However, from my experience, 2-stroke engines tend to run at a higher rpm than a 4-stroke engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a 4-stroke engine only has a power stroke on every other revolution of the crankshaft, surely it would make more sense for it to in fact be 4-stroke engines that run at higher rpm, given it could theoretically run at twice the rpm for the same fuel consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess I must be missing something here, because I fail to understand why 2-stroke engines tend to run at higher rpm to a 4-stroke engine?&lt;/p&gt;&#xA;" OwnerUserId="20043" LastActivityDate="2019-04-09T01:51:25.927" Title="4-stroke vs. 2-stroke rpm" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26789" PostTypeId="1" CreationDate="2019-04-08T21:32:55.473" Score="1" ViewCount="149" Body="&lt;p&gt;I'm trying to find an answer to a question in one of my thermodynamics textbooks. (Undergraduate chemical engineering is the context).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a conceptual question, asking, &quot;Do &lt;em&gt;all&lt;/em&gt; systems in equilibrium continue to be in equilibrium if they are instantly isolated from their surroundings?&quot; What does everyone think?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've consulted a few authoritative sources, and the answer becomes confusing and conceptual and even philosophical. I figured there might be a very straightforward answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="20044" LastActivityDate="2019-04-10T03:17:21.007" Title="Systems In &quot;Thermodynamic Equilibrium&quot;" Tags="|chemical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26790" PostTypeId="2" ParentId="26788" CreationDate="2019-04-09T00:23:12.700" Score="2" Body="&lt;p&gt;Valve train design gets more difficult as RPM goes up, because those valves need to be started and stopped faster, without beating the entire valve train apart.  So while there may not be any theoretical reasons that a 4-stroke engine can't turn super-fast, there's just fewer things you have to worry about when you're trying to make a 2-stroke engine turn super-fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And big marine diesels turn at 100RPM.  And they're 2-strokes.  So there's a counter-example for you).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-04-09T00:23:12.700" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26791" PostTypeId="2" ParentId="26789" CreationDate="2019-04-09T01:48:34.283" Score="3" Body="&lt;p&gt;Certainly, an isolated system at equilibrium will remain so when isolated. This example is a bit of a tautology though. But the implications are important, because we next have to consider non-isolated systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An open system allows mass flow but may have an insulated boundary. A closed system allows no mass flow but may have a diathermal boundary. Finally, a system may allow mass and heat flow (as well as flow of other work).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While these systems can be in steady state, can they be in true equilibrium? See the discussion &lt;a href=&quot;http://press-files.anu.edu.au/downloads/press/p128271/mobile/ch11s03.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. The answer is no. They can be at steady state, a quasi-equilibrium condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The instant a non-isolated system under steady state conditions is isolated, its steady state is lost. The once non-isolated system that is now isolated can then move to its true equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, though a non-isolated system that is suddenly isolated can move its state, &lt;em&gt;will&lt;/em&gt; it move its state? Think about a refrigerator that is losing heat and being cooled to make up. Allow a bit of air flow leaking in and out at various places around the doorway. The inside is at a steady state temperature and pressure (and chemical potential and ...). Now, isolate that system. No heat or mass flow is allowed at all. The result is, the inside of the refrigerator will remain at the temperature (and pressure and chemical potential) that it had at the instant it is isolated. Why would it change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, we have taken out chemical reactions from this consideration (oh no, we forgot that we left both yogurt and mold spores inside). When a chemical reaction is occurring in a steady state (non-isolated) system, it may not be at its true chemical equilibrium within the confines of the system. We only need to realize that we purge mold spores (and yogurt) in/out of the refrigerator over a steady period to see the difference. Once the system is isolated, the mold can grow freely to a new equilibrium state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, systems that are not isolated to start and are at steady state but include chemical reactions may not remain at the state they have when they are instantly isolated. They can change internally because the chemical reaction itself was only at a quasi-equilibrium state, not a true (isolated case) equilibrium state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about non-isolated systems that act as isolated systems because they have no flows across their boundaries. If they are at equilibrium, they remain so when isolated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, the connection that is to be made is that equilibrium is a state only of isolated systems or systems that behave exactly as such to start. The question itself is therefore somewhat a tautological one.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-04-10T03:17:21.007" LastActivityDate="2019-04-10T03:17:21.007" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26792" PostTypeId="2" ParentId="26788" CreationDate="2019-04-09T01:51:25.927" Score="1" Body="&lt;p&gt;Most 2-stroke engines are small (think: weed-eater, model airplane, etc.), which allows them to spin at absurd speeds because the reciprocating masses are small. But once you scale 2-stroke engines up into the ~100HP range, their wide-open throttle RPM is in the 4500-5500RPM range, which is not that different from the redline limit of a four-stroke engine of similar power rating. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-04-09T01:51:25.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26794" PostTypeId="2" ParentId="3467" CreationDate="2019-04-09T18:26:04.330" Score="2" Body="&lt;p&gt;Bulk metallic glasses can often be even stronger than that. These are materials that can take 2% strain (steels are typically around 0.5% (this is elastic strain)) and have UTS north of 1800 MPa. Some of the iron-based ones are supposed to be north of 2200MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The commercially used ones are Zr-based (Zr-Ti-Cu-Ni-Al (in the exact right proportions) is most common).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are also some Zr-based versions containing Be that are ductile and have absolutely monstrous stress-strain curves. With a lower density.&lt;/p&gt;&#xA;" OwnerUserId="2955" LastActivityDate="2019-04-09T18:26:04.330" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26795" PostTypeId="2" ParentId="24120" CreationDate="2019-04-09T18:46:49.140" Score="3" Body="&lt;p&gt;The stresses from bending and axial should be added up by superposition. This moves the location of zero stress (neutral axis). I have included a figure to demonstrate the principle, and provided the calculations again, please note the axial stress is 40 kN/mm^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sigma_{bending}= m/S =\frac{1kN*1m}{(50\times 100^2)/6} =6000N/500mm^2 = 12N/mm^2$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \sigma_{axial}= 200kN/(50*100)mm^2 = 40Nmm^2  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sigma_{max} = \sigma_{bending} + \sigma_{axial} = 12N/mm^2 + 40N/mm^2 = 52 N/mm^2$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \sigma_{min} = -\sigma_{bending} + \sigma_{axial} = -12N/mm^2 + 40N/mm^2 = 28 N/mm^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can see from the scaled image, as well as the math, that the axial stress overcomes the compression stress due to bending. There is no location on the cross section where the stress is zero. Therefore the neutral axis is below the beam. We can determine this distance by similar triangles (or equation of a line, or whatever you prefer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{52-28}{100} =\frac{52}{D_{NA}+D} -&amp;gt; (D_{NA} + D) = \frac{52*100}{52-28} = 216.67mm $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the total depth from the bottom surface of the beam to the neutral axis is the difference of the depth and the calculated dimension. &#xA;&lt;span class=&quot;math-container&quot;&gt;$D_{NA} = (216.67 - 100)mm = 116.67mm$&lt;/span&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/xFXm2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xFXm2.png&quot; alt=&quot;scaled view showing the stress diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-09T18:46:49.140" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26796" PostTypeId="2" ParentId="24118" CreationDate="2019-04-09T19:50:57.030" Score="3" Body="&lt;p&gt;We really want to avoid putting wood members into torsional stress states. The previous answers are valid and applicable mostly for isotropic materials. Wood is an-isotropic and its material properties are the result of the complex structures naturally formed in wood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to rely on planks in torsion is particularly problematic because the wood is very weak when tension is applied perpendicular to the grain. Wood material has very little strength between the grain lamina.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Significant levels of torsion on a wood plank would almost certainly split the plys and have a non-ductile failure. This is a failure state the you would probably not capture by simply taking the allowable shear stress into the 'stress due to torsion' equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My recommendation is to add more framing to eliminate the reliance on torsion in wood.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-09T19:50:57.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26797" PostTypeId="1" CreationDate="2019-04-09T20:10:29.557" Score="0" ViewCount="109" Body="&lt;p&gt;I'm currently designing an air rifle and am looking for ways to increase projectile velocity without resorting to combustion, modifying the projectile, barrel length or changing max pressure within the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've considered using a De Laval nozzle to increase the gas's velocity, but from what I've researched it would actually perform worse due to the decreased pressure on the projectile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, I've been looking for some sort of nozzle or other device with no moving parts to aid in increasing the gas's pressure, but haven't found any (poor wording when searching or does it not exist?). Is there such a thing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking forward for answers.&lt;/p&gt;&#xA;" OwnerUserId="20051" LastEditorUserId="20051" LastEditDate="2019-04-10T00:00:59.320" LastActivityDate="2024-03-16T13:03:10.493" Title="Device with no moving parts to increase fluid pressure?" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26798" PostTypeId="2" ParentId="26789" CreationDate="2019-04-09T20:11:58.673" Score="0" Body="&lt;p&gt;Equilibrium is not the same as either maximal entropy or constant temperature. All it really means is you can't set up a heat engine &lt;em&gt;inside&lt;/em&gt; the region and extract work from it. It is possible for the system to be interacting with the environment in such a way that the environment is determining the equilibrium point, and equilibrium is then lost when the environment is changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thermal equilibrium just means the temperature is uniform throughout a region at a given instant. But it does not mean the temperature is constant. An insulated body that is being slowly heated uniformly throughout can be in thermal equilibrium during the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Black body radiation provides a simple example of how the environment can change an equilibrium. In a vacuum, a body subject to uniform surface radiation will eventually reach thermal equilibrium. Suddenly remove the radiation source and the equilibrium is destroyed. The surface will begin to cool through black body radiation.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-04-09T20:11:58.673" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26799" PostTypeId="2" ParentId="25639" CreationDate="2019-04-09T20:24:15.043" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;Your sampling frequency &lt;span class=&quot;math-container&quot;&gt;$f_{samp}$&lt;/span&gt; should be at least 2 - Times higher the highest expected relevant frequency &lt;span class=&quot;math-container&quot;&gt;$f_{max}$&lt;/span&gt; because of the Nyquist - Shannon Theorem. However you minimize error by using a higher sampling frequency (Rule of thumb &lt;span class=&quot;math-container&quot;&gt;$f_{samp} = 5  f_{max}$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;li&gt;If you want to get the RMS of smaller frequency bands (for example by performing a FFT), your measuring time &lt;span class=&quot;math-container&quot;&gt;$t_M$&lt;/span&gt; should be long enough to get a high enough frequency resolution &lt;span class=&quot;math-container&quot;&gt;$\Delta f $&lt;/span&gt; (difference between two interesting frequency bands, formula see next point). &lt;/li&gt;&#xA;&lt;li&gt;Also if you are not aiming for a fft, you should measure long enough to measure at least one full cycle of the expected minimal frequency of interest &lt;span class=&quot;math-container&quot;&gt;$f_{min}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$f_{min} = \Delta f = 1/t_M$&lt;/span&gt;)&lt;/li&gt;&#xA;&lt;li&gt;Longer measurements provide kind of a averaging. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You want to measure a fan with 9 blades rotating with 10 Hz. &lt;/li&gt;&#xA;&lt;li&gt;In one measurement, you want to have at least 12 revolutions. &lt;/li&gt;&#xA;&lt;li&gt;You definitly want to measure the 90 Hz blade pass frequency.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You also want to measure 5 harmonics  -&gt; 450 Hz&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To make sure you get the 450 Hz you need to sample with at least 900 Hz, &lt;span class=&quot;math-container&quot;&gt;$f_{samp} = 2250 Hz$&lt;/span&gt; would be better though.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;To measure one cycle you have to measure at least 0.1 s. If you want to measure over 12 revolutions you need to measure at least &lt;span class=&quot;math-container&quot;&gt;$t_M = 1.2 s$&lt;/span&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;--&gt; You end up with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$f_{samp} = 2250 Hz$&lt;/span&gt;  -&gt; &lt;span class=&quot;math-container&quot;&gt;$f_{max} = 1225 Hz$&lt;/span&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$T_{m} = 1.2 s$&lt;/span&gt; -&gt; &lt;span class=&quot;math-container&quot;&gt;$f_{min} = \Delta f = 0.83 Hz$&lt;/span&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="20050" LastActivityDate="2019-04-09T20:24:15.043" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26801" PostTypeId="1" AcceptedAnswerId="26811" CreationDate="2019-04-10T01:18:26.863" Score="0" ViewCount="292" Body="&lt;p&gt;This is an off-level rigging problem. When solving for the tension forces in the red slings (lifting from point A), can the assumption be made that the y force vector at points B and C are each half of the weight of the object? Or must one calculate the force vectors at point A? In which case the y vectors are not symmetrical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AWCMW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AWCMW.jpg&quot; alt=&quot;Off-level pick points&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18926" LastEditorUserId="18926" LastEditDate="2019-04-10T20:26:27.577" LastActivityDate="2019-04-11T07:25:12.127" Title="Can you assume that the forces in the vertical direction are evenly distributed for an off level rigging problem?" Tags="|statics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26802" PostTypeId="2" ParentId="26801" CreationDate="2019-04-10T02:17:26.420" Score="-1" Body="&lt;p&gt;Yes, we know the Y components of the cables are equal and each half of the weight because the axis of A passes through CG of the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The X components are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \text{For B, }\quad  X_B = \frac{10 \cdot W}{12\cdot 2} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \text{For C,} \quad X_C =\frac{10 \cdot W}{6 \cdot 2}   $$&lt;/span&gt;&#xA;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;After comments indicating my errors by @PhilSweet and @ingen0rd I thought best way is to calculate the tension in the cables by taking moment of forces about point B.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$  \Sigma M _B =0 \quad \therefore -W\cdot10 + F\ C_{vertical} \cdot 20 + F\ C_{ horizontal}\cdot6 =0 $$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$\\ F \ C_{vertical} =6/10 \cdot F \ C_{horizontal} $$&lt;/span&gt;&#xA;So we have two equations and two unknowns which we can solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F\ C_{horiz}\cdot6 +6/10*F\ C_{horiz}\cdot 20 -10W =0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ 18 F\ C_{horiz}= 10W ,\quad F\ C_{horiz}= W/1.8 $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$F\ C_{vert}= 6W/18 = W/3$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so &lt;span class=&quot;math-container&quot;&gt;$F\ B_{ vert}=2W/3  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please check my numbers as always.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-04-11T07:25:12.127" LastActivityDate="2019-04-11T07:25:12.127" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26804" PostTypeId="1" CreationDate="2019-04-10T13:20:22.007" Score="1" ViewCount="346" Body="&lt;p&gt;I understand that there are quite a few fairly simple and cheap methods to measure the flow of gas in terms of volume, allowing you to get a flow rate in terms of litres per minute for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what would be the cheapest method to measure the mass of gas flowing along a pipe, so that differences in pressure etc. don't impact the results taken?&lt;/p&gt;&#xA;" OwnerUserId="20042" LastActivityDate="2019-05-14T09:27:10.097" Title="Cheapest way to measure gas flow (mass)?" Tags="|gas|flow-control|compressed-gases|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26805" PostTypeId="2" ParentId="10769" CreationDate="2019-04-10T15:15:22.137" Score="1" Body="&lt;p&gt;I'm running into a similar issue at the original poster. It seems like a workable mechanical solution would be to mount your float valve upside down, so as the water level drops, the valve would close.  In my setup, I have several source tanks for irrigation. Because of the different height of the tanks, I need to automate the shutoff of the two higher tanks as the water is pumped out so the pump will start to draw from a lower level tank as flow is shut off from the higher level tank.&lt;/p&gt;&#xA;" OwnerUserId="20061" LastActivityDate="2019-04-10T15:15:22.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26806" PostTypeId="1" CreationDate="2019-04-10T15:16:33.193" Score="0" ViewCount="72" Body="&lt;p&gt;Would anyone be able to run an air flow simulation through the following geometry at 100 cc/min? I'm curious if any dead spots occur and if the geometry of the annular gap will produce laminar flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R6f7d.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R6f7d.png&quot; alt=&quot;e&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimensions are in inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reynold's number analysis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydraulic Diameter* = 0.003 in = 7.62E-5 m&#xA;Absolute Viscosity of Air at 20.2C = 1.983E-5 N-s/m^2&#xA;Density of Air at 20C = 1.293 kg/m^3&#xA;Mean Velocity = &#xA;   =&gt; A_cs = 0.004089 cm^2&#xA;   =&gt; Q = 100 cc/min&#xA;   =&gt; V = 24455.9 cm/min = 4.076 m/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re = (rho&lt;em&gt;V&lt;/em&gt;D)/nu&#xA;   = (1.293 kg/m^3 * 4.076 m/s * 7.62E-5 m) / 1.983E-5 N-s/m^2&#xA;Re = 20.25&lt;/p&gt;&#xA;" OwnerUserId="19866" LastEditorUserId="19866" LastEditDate="2019-04-11T18:15:47.583" LastActivityDate="2019-04-11T18:15:47.583" Title="Flow Simulation for Laminar Flow in Annular Gap?" Tags="|mechanical-engineering|fluid-mechanics|airflow|simulation|ansys|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26807" PostTypeId="2" ParentId="26804" CreationDate="2019-04-10T15:38:32.673" Score="0" Body="&lt;p&gt;It depends what you call &quot;cheap&quot; but a mass flow meter/controller, such as &lt;a href=&quot;https://www.bronkhorst.com/products/gas-flow/meters/&quot; rel=&quot;nofollow noreferrer&quot;&gt;these ones&lt;/a&gt;, will measure the flow rate, and also control it to a set point regardless of pressure variations (for the mass flow controller). They rely on a &lt;a href=&quot;https://www.bronkhorst.com/service-support/technologies/thermal-mass-flow-sensor-for-gases-bypass-principl/&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermal bypass&lt;/a&gt; to measure the flow rate:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LEUNq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LEUNq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no control of the flow rate, and you're just measuring it, then any pressure change &lt;em&gt;will&lt;/em&gt; result in a change of flow rate, it's the underlying physics and you can't avoid that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want a rough idea, putting the end of the pipe into an upturned cylinder full of water, and measuring how much volume/level the water has come down over a certain period of time, does work pretty well and gives you a good idea. Obviously, this isn't an &quot;online&quot; measurement, it's just a check, and it doesn't have the same level of accuracy as a mass flow meter. Again, as with any measurement technique, this will depend on the pressure.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2019-04-11T11:42:36.833" LastActivityDate="2019-04-11T11:42:36.833" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26808" PostTypeId="1" CreationDate="2019-04-10T16:02:19.097" Score="2" ViewCount="4345" Body="&lt;p&gt;I migrated to Linux and don't want to use solidworks , so i need one software which could opening my already solidworks file like *.sldprt and *.sldasm .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So i have installed FREECAD but it could not opening those files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question i have is this:&#xA;  could i do my modeling designing which only need to fit dimension and shape without any characteristic simulation and ... in solidworks alternative, and which one is your suggestion?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank A lot.&lt;/p&gt;&#xA;" OwnerUserId="20064" LastActivityDate="2020-08-23T18:41:05.860" Title="Free alternative to openinig solidwork defualt files in linux( *.sldprt &amp; *.sldasm)?" Tags="|cad|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26809" PostTypeId="1" CreationDate="2019-04-10T16:25:00.403" Score="2" ViewCount="109" Body="&lt;p&gt;I always wonder why nuclear power plants aren't designed to handle a loss of the heat sink. At the point at which hydrogen is being made from the water, can't you just remove the water and catch the molten corium into a concrete container until it cools off?&lt;/p&gt;&#xA;" OwnerUserId="20065" LastActivityDate="2019-04-10T20:49:24.247" Title="Why aren't nuclear reactors designed to handle the loss of the heat sink?" Tags="|nuclear-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26811" PostTypeId="2" ParentId="26801" CreationDate="2019-04-10T18:48:26.060" Score="1" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To convince yourself, why an equal distribution of the vertical component is wrong, try looking at the corresponding horizontal components (Kamran has calculated these values in his answer): The horizontal component at C is twice as large as the horizontal component at B, which results in a net horizontal load on the object, giving it a constant, horizontal acceleration off to the side. (And there is no moment equilibrium, so it would also rotate.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, the easiest way to calculate the correct distribution of the vertical component is to extend the line A-C until it reaches the horizontal level of B. I call this point C'. The horizontal distance from C' to the center of gravity is &lt;span class=&quot;math-container&quot;&gt;$10'\cdot\frac{12'}{6'}=20'$&lt;/span&gt;, which is twice the distance from B to the center of gravity, so the vertical component of the reaction at B is twice the corresponding value at C or &lt;span class=&quot;math-container&quot;&gt;$\frac{2}{3}$&lt;/span&gt; of the total weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drawing the force vectors at point A and using horizontal and vertical equilibrium as you suggest in your question is also a good way to do it.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2019-04-10T18:48:26.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26812" PostTypeId="1" CreationDate="2019-04-10T19:04:05.843" Score="0" ViewCount="6367" Body="&lt;p&gt;Given pressure is directly proportional to density, my understanding is that you can use the pressure of gas in a container to calculate the mass of the gas in the container. My reasoning being;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$p=k\rho$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; = pressure and &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; = density)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\rho = \frac{m}{V}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; = mass and &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; = Volume)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{p}{k} = \frac{m}{V} \rightarrow m = \frac{pV}{k}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that assuming &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is kept constant (i.e. constant temperature etc.) the pressure in the container should be directly proportional to the mass of the gas in the container, and therefore you can calculate the mass of gas in the container by using the pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is whether this is correct? And my second question is, assuming this is correct, would this still be accurate if the container is in use i.e. gas is being added/removed from the container, or would this only work when the container is sealed with no pipes in or out?&lt;/p&gt;&#xA;" OwnerUserId="20042" LastActivityDate="2019-04-12T14:52:41.037" Title="Calculating mass of gas in container" Tags="|gas|compressed-gases|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26813" PostTypeId="2" ParentId="26812" CreationDate="2019-04-10T20:16:28.210" Score="0" Body="&lt;p&gt;Yes for a closed system you can calculate the mass in a container if you know the volume, pressure and temperature just as you did. In terms of solving this if it is not a closed system you can do it but you will need more information.&lt;/p&gt;&#xA;" OwnerUserId="19424" LastActivityDate="2019-04-10T20:16:28.210" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26815" PostTypeId="2" ParentId="26809" CreationDate="2019-04-10T20:31:59.773" Score="3" Body="&lt;p&gt;This is a really good question and your question is answered directly here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Corium_(nuclear_reactor)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Corium_(nuclear_reactor)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically the corium is so hot that it will react with concrete producing water vapor and carbon dioxide which may further react into just hydrogen gas and carbon monoxide and that the corium can melt the concrete to about a depth of one meter in one hour.&lt;/p&gt;&#xA;" OwnerUserId="19424" LastActivityDate="2019-04-10T20:31:59.773" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26816" PostTypeId="2" ParentId="26801" CreationDate="2019-04-10T20:35:57.683" Score="-1" Body="&lt;p&gt;No. The issue here is that the problem appears to be overconstrained at first glance. So you can't even assume that there is a physically valid solution at all. How many equations do you have? How many unknowns? Has it been constructed so that two of the equations are linear multiples of each other and one can be dropped?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to take shortcuts like this &lt;em&gt;will&lt;/em&gt; bite you in the butt someday.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-04-10T20:35:57.683" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26817" PostTypeId="2" ParentId="26809" CreationDate="2019-04-10T20:49:24.247" Score="0" Body="&lt;p&gt;They are. The heat sink is also the neutron moderator. If you take some of the water away, reactors are designed to slow down by more than the loss of coolant. This is what &lt;em&gt;negative void coefficient&lt;/em&gt; means. The only reactors that didn't do this were the Chernobyl type designs (RBMKs). They were known to be a danger by 1955. A similar design was built near Arco, Idaho and cycled to destruction to prove the point.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-04-10T20:49:24.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26818" PostTypeId="1" CreationDate="2019-04-10T20:56:34.857" Score="2" ViewCount="122" Body="&lt;p&gt;I have read the Wikipedia article on &lt;a href=&quot;https://en.wikipedia.org/wiki/Generation_IV_reactor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Generation IV nuclear reactors&lt;/a&gt;, but from that article it is not clear to me which of the main reactor types that use nuclear waste to fuel the process. As far as I have been able to understand, Molten Salt Reactors (MSR) and some Sodium-Cooled Fast Reactors (SFR) are capable of doing this. Are there any other other reactor types that also can be fueled by nuclear waste?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, when using nuclear waste as fuel, how large a percentage of the waste is actually consumed?&lt;/p&gt;&#xA;" OwnerUserId="20068" LastEditorUserId="20068" LastEditDate="2019-04-11T08:11:12.023" LastActivityDate="2021-04-04T03:05:39.033" Title="Which Gen IV reactor types use nuclear waste as fuel?" Tags="|nuclear-engineering|nuclear-technology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26819" PostTypeId="1" CreationDate="2019-04-10T21:54:55.040" Score="0" ViewCount="66" Body="&lt;p&gt;I was wondering if the force generated from the engines of a jet plane would lift it up from the ground if in a free treadmill.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it take off?&lt;/p&gt;&#xA;" OwnerUserId="1853" LastEditorUserId="13297" LastEditDate="2019-04-10T22:27:40.897" LastActivityDate="2019-04-12T19:54:40.903" Title="Would a plane lift from ground if in a free treadmill?" Tags="|dynamics|airflow|aerodynamics|" AnswerCount="2" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26820" PostTypeId="2" ParentId="26819" CreationDate="2019-04-10T22:30:47.397" Score="-3" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aircraft fly by the action of air over the wings. That's a simplification, but all that is needed to answer your question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the aircraft is not moving with respect to the ground (as would be the case on a treadmill), the only airflow over the wings is local wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accordingly, aircraft will take off if there is wind that meets or exceeds the aircraft's safe speed for conditions. I have flown an ultralight in very smooth, very strong wind and had a zero distance ground roll for take off and landing. No treadmill required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many small aircraft pilots will attest to the same circumstances.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-04-10T22:30:47.397" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26821" PostTypeId="1" AcceptedAnswerId="26822" CreationDate="2019-04-10T22:54:01.307" Score="0" ViewCount="145" Body="&lt;p&gt;Why there are different dimensions for the same measurements on this drawing?&#xA;For example 0.127mils and 0.097 mils pin to pin. Or highlighted 405 and 385 mils. &#xA;I can't find anything on the datasheet indicating the difference?? Is this typical and max? Or typical and min? (some actually say max)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Need to know highlighted distance so I know where to place this part on the PCB. I could just make it 405mils to be safe, but I still want to know why they did this on the drawing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete datasheet link below drawing. Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4twdp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4twdp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.mouser.com/datasheet/2/418/NG_CD_1218434_D1-1234416.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mouser.com/datasheet/2/418/NG_CD_1218434_D1-1234416.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20069" LastActivityDate="2019-04-10T23:45:33.030" Title="Help interpreting dimensions from this drawing" Tags="|cad|drafting|technical-drawing|drawings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26822" PostTypeId="2" ParentId="26821" CreationDate="2019-04-10T23:45:33.030" Score="1" Body="&lt;p&gt;The link you've provided has a pretty comprehensive set of dimensions. One bit of text that jumped out to me is worded &quot;See sheet 2 for recommended PC board layout.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would read the reference of 0.127 / 0.097 as a difference of 0.030, which would then translate to 0.112 +/- 0.015 which appears consistent on many of the measurements. A couple appear a bit tighter at 0.020 or +/1 0.010 here and there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reference to Max is not so much to determine hole placement as it is to provide you with a dimension that the part will not exceed. If you allocate 0.225 for the shell, you'd have 0.001 clearance for the housing/components/accessories placed adjacent to that portion of the connector. It's not impossible to consider that the 0.224 Max could be measured at 0.220 which would still not cause problems for a design that allocates 0.225 in the housing/case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another location on the document specifies that the tolerance is +/- 0.010 unless otherwise specified. That means that the diagram on sheet two which has no tolerance listed individually for the PCB hole layout would be assigned that figure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above is a layman's interpretation mixed with a bit of mechanical drawing instruction and a good bit of tradesman's osmotic absorption from a career machinist (60+ years). It could be incorrect.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-04-10T23:45:33.030" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26823" PostTypeId="2" ParentId="26819" CreationDate="2019-04-11T01:27:15.720" Score="3" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aircrafts, other than a few which could lift vertically, create lift by moving forward in the air and bending the stream of air flow around their wings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Airplanes create the traction forward by the thrust of their engines not by traction of their wheels, so they will start to move even on water like a sea plane or on a free treadmill, ice, even from a wood carriage on set of rails like the Wright brothers &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypothetically there is no issues, other than control and breaking in case of aborting a take off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a private pilot I have had to take off from glazed, icy tarmacs, or sllippery muddy dirt runways, albeit while  following proper pilot's instructions by the manufacturer.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-04-12T19:54:40.903" LastActivityDate="2019-04-12T19:54:40.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26824" PostTypeId="1" AcceptedAnswerId="26829" CreationDate="2019-04-11T02:15:44.207" Score="0" ViewCount="73" Body="&lt;p&gt;I am creating a technical drawing of a system I created. Let's call my system, system X. System X is a mounting mechanism for already invented system Y. System Y is customizable, however, system X is a general mechanism that can be fitted to any system Y. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, when labelling my technical drawing, which includes a physical outline of system Y (otherwise the paper would have a lot of blank space), how can I articulate that my general depiction of system Y is replaceable with any customized version of system Y?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am creating other drawings that detail the mechanisms of X up close without Y, however, in the case of my question, I am trying to create an overview drawing of what system X would look like when applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="19486" LastEditorUserId="19486" LastEditDate="2019-04-11T03:09:32.917" LastActivityDate="2019-04-11T12:35:25.383" Title="Technical Drawing: Labelling a component that's not part of my design but integral to my system" Tags="|technical-drawing|drawings|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26825" PostTypeId="1" CreationDate="2019-04-11T06:21:37.667" Score="0" ViewCount="200" Body="&lt;p&gt;I am given this transfer function for a system:&#xA;&lt;span class=&quot;math-container&quot;&gt;$T\left(s\right)=\frac{9.81}{s^4+0.5s^3+11.2815s^2+4.905s}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I want to design a controller that will make the closed loop system have an overshoot due to a step input equal to &lt;span class=&quot;math-container&quot;&gt;$5\%$&lt;/span&gt;. A controller with a proportional gain &lt;span class=&quot;math-container&quot;&gt;$K_p=1$&lt;/span&gt; gives me this step response of the closed-loop system on MATLAB:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N5QXa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N5QXa.png&quot; alt=&quot;pcontrol&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, the overshoot with this proportional gain is way bigger than &lt;span class=&quot;math-container&quot;&gt;$5\%$&lt;/span&gt;. As K increases, the response becomes weirder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, I have tried &lt;span class=&quot;math-container&quot;&gt;$K_p=6$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p3PxS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p3PxS.png&quot; alt=&quot;pcontrol&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PI, PD or PID controllers give a similar response to the figure above. So, it seems the best I can do is a controller with &lt;span class=&quot;math-container&quot;&gt;$K_p=1$&lt;/span&gt;, which is not acceptable in my case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the appropriate type of controller for this system to get a small overshoot?&lt;/p&gt;&#xA;" OwnerUserId="18341" LastEditorUserId="18341" LastEditDate="2019-04-11T06:58:21.207" LastActivityDate="2019-09-08T14:02:15.287" Title="What is the appropriate type of controller for this system?" Tags="|mechanical-engineering|electrical-engineering|control-engineering|transfer-function|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26826" PostTypeId="2" ParentId="26824" CreationDate="2019-04-11T09:44:35.950" Score="2" Body="&lt;p&gt;It is possible to use dashed lines for things that are in the image for reference, or not directly part of the item in the drawing. But in general adding something because you have blank space seems redundant. Add something if it makes things clearer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also perfectly ok to write instructions and explanations. Drawings are human readable after all. So you can do prettymuch what you want as long as your clear about it. But then being clear is about the hardest thing on earth... Which is why deviating too much from convention is problematic.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2019-04-11T09:44:35.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26827" PostTypeId="1" CreationDate="2019-04-11T10:02:33.440" Score="1" ViewCount="722" Body="&lt;p&gt;From my internet research, it seems there's no systematic way of tuning the design parameters for the LQR controller. I have found Bryson’s rule as an initialization. However I don't know how to proceed from this point besides trial and error or grid search.&#xA;Ideally I want to tune the parameters such that the settling time for specific state variables are below a threshold. Additionally I want to make constraints on the sensitivity function, such that disturbances up to a specific frequency are attenuated.&#xA;Is there a recommended way to tune my parameters such that requirements can be fulfilled ?&lt;/p&gt;&#xA;" OwnerUserId="18808" LastActivityDate="2023-12-24T14:07:20.867" Title="How can I tune the parameters of a lqr controller?" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26828" PostTypeId="2" ParentId="26825" CreationDate="2019-04-11T10:47:22.367" Score="1" Body="&lt;p&gt;I used &lt;a href=&quot;https://octave.sourceforge.io/control/function/optiPID.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;optiPID&lt;/code&gt;&lt;/a&gt; in Octave where the plant P(s) is controlled by a PID controller with second-order roll-off:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                 1                1&#xA;C(s) = Kp (1 + ---- + Td s) -------------&#xA;               Ti s         (tau s + 1)^2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;in the usual negative feedback structure&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;         L(s)       P(s) C(s)&#xA;T(s) = -------- = -------------&#xA;       1 + L(s)   1 + P(s) C(s)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With your transfer function, and by and tweaking the weighing factors to put more weight on minimising overshoot, I got the following values (the ones of interest here are the optimised values):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;optiPID: Astrom/Hagglund PID controller parameters:&#xA;kp_AH =  0.032753&#xA;Ti_AH =  0.63360&#xA;Td_AH =  0.16028&#xA;warning: optiPID: optimization starts, please be patient ...&#xA;warning: called from&#xA;    optiPID at line 215 column 1&#xA;Max no. of function evaluations exceeded...quitting&#xA;optiPID: optimized PID controller parameters:&#xA;kp_opt =  0.017183&#xA;Ti_opt =  0.14305&#xA;Td_opt =  0.50410&#xA;optiPID: closed-loop stability check:&#xA;st_AH = 1&#xA;st_opt = 1&#xA;optiPID: gain margin gamma [-] and phase margin phi [deg]:&#xA;gamma_AH =  2.2796&#xA;phi_AH =  80.408&#xA;gamma_opt =  4.3438&#xA;phi_opt =  64.139&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The step response looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/laZnR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/laZnR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives and overshoot of approx. 5.7%. The amount of overshoot is actually governed by the integral gain, not the proportional one (which only influences the speed of response) so I have played around with it to reduce the overshoot further and using &lt;code&gt;Ti_opt =  0.2&lt;/code&gt;, I get the following step response:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/suVRM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/suVRM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives only about 1.6% overshoot. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roll-off parameter &lt;code&gt;tau&lt;/code&gt; is calculated as &lt;code&gt;Td/10&lt;/code&gt;, which gives the following transfer function for the PID controller:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;&amp;gt;&amp;gt; C_opt&#xA;&#xA;Transfer function 'C_opt' from input 'u1' to output ...&#xA;&#xA;      0.001732 s^2 + 0.003437 s + 0.01718&#xA; y1:  -----------------------------------&#xA;      0.0005082 s^3 + 0.02016 s^2 + 0.2 s&#xA;&#xA;Continuous-time model.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;If you're not bothered about the response time, the Astroem/Haegglund PID (with &lt;code&gt;kp_AH&lt;/code&gt;, &lt;code&gt;Ti_AH&lt;/code&gt; and &lt;code&gt;Td_AH&lt;/code&gt;) has essentially zero overshoot:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; &amp;gt;&amp;gt; C_AH&#xA;&#xA;Transfer function 'C_AH' from input 'u1' to output ...&#xA;&#xA;        0.003326 s^2 + 0.02075 s + 0.03275&#xA; y1:  --------------------------------------&#xA;      0.0001628 s^3 + 0.02031 s^2 + 0.6336 s&#xA;&#xA;Continuous-time model.&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2019-04-11T13:01:10.457" LastActivityDate="2019-04-11T13:01:10.457" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26829" PostTypeId="2" ParentId="26824" CreationDate="2019-04-11T12:35:25.383" Score="1" Body="&lt;p&gt;If the paper has a lot of blank space without drawing Y, then you can draw X at a bigger scale to fill it up, or use smaller paper!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Y is physically bigger than X (which seems possible, if X is just a mounting system) you might want to add a reduced scale &quot;sketch&quot; drawing showing how X and Y are assembled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If Y could be any component with a standard type of mount (e.g. any make of TV with a VESA standard mounting) explain that in a text comment, and just draw a &quot;generic&quot; version of Y in your assembly sketch drawing.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-04-11T12:35:25.383" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26835" PostTypeId="1" CreationDate="2019-04-11T15:30:28.107" Score="0" ViewCount="51" Body="&lt;p&gt;Consider a generic second order state space SISO system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{equation}&#xA;\left[ \begin{array}{c}{\dot{\tilde{x}}_{1}} \\ {\dot{x}_{2}}\end{array}\right]=A \left[ \begin{array}{c}{\tilde{x}_{1}} \\ {\tilde{x}_{2}}\end{array}\right]+B \tilde{u}&#xA;\end{equation}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{equation}&#xA;\mathbf{\tilde{y}}=C \left[ \begin{array}{c}{\tilde{x}_{1}} \\ {\tilde{x}_{2}}\end{array}\right]&#xA;\end{equation}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can interpret &lt;span class=&quot;math-container&quot;&gt;$\tilde{x}_1$&lt;/span&gt; as the position, &lt;span class=&quot;math-container&quot;&gt;$\tilde{x}_2$&lt;/span&gt; as the velocity, and &lt;span class=&quot;math-container&quot;&gt;$\dot{\tilde{x}}_2$&lt;/span&gt; as the acceleration. If I were to measure the state of my system using an accelerometer, how would I include it in this model? Is there a way of choosing matrix &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; such that &lt;span class=&quot;math-container&quot;&gt;$\tilde{y}=\dot{\tilde{x}}_2$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first intuition would be to copy the equation for &lt;span class=&quot;math-container&quot;&gt;$\dot{\tilde{x}}_2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{equation}&#xA;\mathbf{\tilde{y}}=a_{21}\tilde{x}_{1} + a_{22}\tilde{x}_{2} + b_{2}\tilde{u}&#xA;\end{equation}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;though I am not sure about that. Having a direct input-to-output term would mean that the system had infinite bandwidth (and therefore an instantaneous response).&lt;/p&gt;&#xA;" OwnerUserId="4659" LastEditorUserId="4659" LastEditDate="2019-04-11T21:38:36.537" LastActivityDate="2019-04-11T21:38:36.537" Title="Output matrix for system with accelerometer" Tags="|control-engineering|instrumentation|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26838" PostTypeId="1" CreationDate="2019-04-12T07:51:00.847" Score="1" ViewCount="208" Body="&lt;p&gt;It is an aluminium section for a mullion and transom and the autocad drawing I have is very complex and I tried using autocad mechanical but it does not work as I am not able to select the area as it is very complex. The properties say that the selection is a closed area however on going to content&gt;calculations&gt;moment of inertia, or using aminertia, I cannot select the enclosed area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give me an application where I can find the moment of inertia and corresponding values. &lt;/p&gt;&#xA;" OwnerUserId="20081" LastActivityDate="2019-05-19T10:02:45.550" Title="Which program should I use to find moment of inertia of complex sections?" Tags="|autocad|cross-section|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26839" PostTypeId="1" AcceptedAnswerId="26840" CreationDate="2019-04-12T10:19:03.953" Score="2" ViewCount="247" Body="&lt;p&gt;I've just started to study reinforced concrete design and I came across this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PjxWU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PjxWU.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt; in the &quot;Design of Concrete Structures&quot; by Dr.Nilson.&#xA;and I was wondering why do we increase the minimum thickness if we are using stell grade more than 60 ksi instead of decreasing it?&lt;/p&gt;&#xA;" OwnerUserId="13708" LastActivityDate="2019-04-12T11:41:29.867" Title="Why do we increase the minimum beam/slab depth if we increase the grade of the reinforcement to control the deflection" Tags="|deflection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26840" PostTypeId="2" ParentId="26839" CreationDate="2019-04-12T11:41:29.867" Score="2" Body="&lt;p&gt;The principle behind that table is that you will be doing a strength calculation in any case and need the table for a guideline of whether deflections or bending strength will be the design driver. High strength reinforcement does not contribute any more to the stiffness than low strength reinforcement, as the elasticity modulus of steel is the same for reinforcement of different strength grades. (If you used a more accurate approach and actually calculated the deflection, the yield strength of the reinforcement would not change the deflection.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you are using low strength reinforcement you will need relatively more or larger rebars to satisfy the strength requirement, resulting in a stiffer beam or slab, which in turn will make it less likely that deflections become the design driver.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2019-04-12T11:41:29.867" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26841" PostTypeId="2" ParentId="26812" CreationDate="2019-04-12T13:25:40.270" Score="2" Body="&lt;h1&gt;Isolated System&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;An isolated system allows no mass or heat flow. The equation for an ideal gas is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ p V = n R T$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is pressure, &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; is volume, &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is number of moles, &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the gas law constant, and &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is temperature. Use the molar mass &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; and the relationship that density &lt;span class=&quot;math-container&quot;&gt;$\rho = m/V = n/MV$&lt;/span&gt; to obtain your equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ m = \frac{V}{MRT}\ p $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real gases must use a modification of the above. The pressure may not be a linear function of moles (mass) of gas. To the extent that you are operating in a region where the gas is ideal or where the equation of state keeps a linear relation between pressure and moles, you can continue with the ideal gas law expression.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Non-isolated System&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;For a steady state process, mass flow in equals mass flow out. The mass inside the container is constant with time. Also, the temperature of the system is constant with time. Under this condition, you can use the equation for an isolated system with no reservations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the system has flow, the modification of the ideal gas law becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{dm}{dt} =  \frac{V}{MRT}  \frac{dp}{dt} - \frac{pV}{MRT^2}  \frac{dT}{dt}  - \frac{pV}{M^2RT}  \frac{dM}{dt} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When temperature and gas composition in the system are not changing with time, you can use pressure change with time as a measure of mass change with time.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-04-12T14:52:41.037" LastActivityDate="2019-04-12T14:52:41.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26842" PostTypeId="1" CreationDate="2019-04-12T13:55:49.077" Score="1" ViewCount="115" Body="&lt;p&gt;I ran across this company at their booth at a Tradeshow (2018). &lt;a href=&quot;https://www.superiorwaterconditioners.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;They&lt;/a&gt; claim to have a chemical free (and as far as I can tell - a power free) treatment method for hard water to minimize scale and corrosion effects downstream. My intuition tells me that this can't be for real, partly because it's &quot;power free&quot;, but I also don't understand their explanation for how it works:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;When water is subjected to temperature change (delta T), pressure change (delta P), friction, turbulence and/or evaporation, dissolved minerals, primarily calcium carbonate (CaCO3) and magnesium carbonate (MgCO3), will precipitate out of solution and deposit on the heat transfer surfaces of plumbing systems and equipment in the form of a rock-like build-up, commonly called lime/scale. At the molecular level, the negatively and positively charged ions of these minerals are attracted to, and bond tightly together with, one another, thus forming the lime/scale deposit known as calcite. These mineral deposits are a great insulator and require much more energy to heat or cool water in many different types of residential, commercial, institutional and industrial applications.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They've been around for since 1964, so they are either:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;very good salesmen, with a critical threshold of gullible people, or&lt;/li&gt;&#xA;&lt;li&gt;they've discovered some niche effect (unlikely)...&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thoughts?&lt;/p&gt;&#xA;" OwnerUserId="7622" LastActivityDate="2022-08-25T17:21:21.010" Title="Magnetic treatment of hard water to prevent scale and corrosion? How does this work?" Tags="|mechanical-engineering|chemical-engineering|electromagnetism|water-resources|waste-water-treatment|" AnswerCount="5" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26844" PostTypeId="1" CreationDate="2019-04-12T14:19:08.400" Score="2" ViewCount="409" Body="&lt;p&gt;I have a question regarding gasturbine energy balance,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the total energy that flows into the gasturbine is given 12 MW and the power needed for the compressor is 4.4 MW and the turbine power is 9.3 MW. So the net output is 9.3-4.4 = 4.9 MW which is around 40% efficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what happened to the 12-9.3 = 2.7 MW, where it is lost ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also says, 60% losses but actually 4.4 MW of the turbine power is needed to run the Compressor which I dont think is a loss, how would you explain this ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At last, how do I create a Sankey Diagram of this energy flow ?&lt;/p&gt;&#xA;" OwnerUserId="20086" LastEditorUserId="20086" LastEditDate="2019-04-12T14:44:04.423" LastActivityDate="2020-06-06T05:07:30.767" Title="Gasturbine energy balance" Tags="|mechanical-engineering|thermodynamics|turbines|energy|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26846" PostTypeId="1" CreationDate="2019-04-12T15:11:21.540" Score="0" ViewCount="30" Body="&lt;p&gt;I'm trying to make a cooling device to used in hot areas for cooling down newborns. I'm planning to cool down the mattress to a temperature of 37C, how effective will such a system be to keep the body core temp to 37C? &#xA;I understand that the baby can not be analysed as a lumped body.&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="19792" LastActivityDate="2019-04-12T16:24:45.720" Title="How effective will a cooled mattress be for cooling down a baby?" Tags="|cooling|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26847" PostTypeId="2" ParentId="26842" CreationDate="2019-04-12T15:25:57.450" Score="2" Body="&lt;p&gt;There's very little data available to evaluate these claims.  The manufacturers tend to rely on anecdotal information and testimonials rather than hard data, and no-one else really has any incentive to spend money to test these devices.  I did find an &lt;a href=&quot;https://extension.psu.edu/magnetic-water-treatment-devices&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; from Penn State University that cites several studies which reported no effect from the devices, and recommends that consumers use caution and have water independently tested both upstream and downstream of the device if they choose that route.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2019-04-12T15:25:57.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26848" PostTypeId="2" ParentId="26846" CreationDate="2019-04-12T16:05:46.523" Score="1" Body="&lt;p&gt;I've had a waterbed since 1981.  It is very effective at both cooling and heating one's body.  At 37C, a mattress would make you feel like you're being cooked.  At 27C, you'll feel like you're freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I keep mine at around 30C, a little more in winter, a little less in summer, but either way, not much different.  The thermostat might not be accurate, so don't rely on those numbers.  The important point is that there is a very small Goldilocks zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It also helps to have a mattress pad, to provide some insulation to slow down the rate of heat transfer.  As I learned when I first got the bed, without it, the temperature has to be very precisely set, and even a slight variation will be noticeable, while anything more will be uncomfortable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might start with a comfortable swimming pool temperature and make it a few degrees warmer since heat generation from muscular activity won't be happening.&lt;/p&gt;&#xA;" OwnerUserId="19930" LastActivityDate="2019-04-12T16:05:46.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26849" PostTypeId="2" ParentId="26846" CreationDate="2019-04-12T16:24:45.720" Score="0" Body="&lt;p&gt;Answering just the engineering side of the question, not the medical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skin temperature is usually 3 to 6 degrees lower than the body's core, more like 32-33 centigrade, and not uniform over all areas of the body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A cooled mattress has to be frequently replaced after it warms up, or be connected to a cooling system, like a water bed with cold water circulating in it as CableStay's comment mentions, and has to have a thermometer to stabilise its temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions of safety and hygiene and maintenance make the design more involved.    &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-12T16:24:45.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26850" PostTypeId="1" CreationDate="2019-04-12T18:18:34.520" Score="2" ViewCount="80" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jui8G.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jui8G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I have two alum plates (45cm x 20cm x 1mm) that are hinged on the long side with two 30 x 80 hinges (30 x 40 each tang) in the middle, and &quot;L&quot; bracket / 20mm brace at the ends secured with a M4 loc nut  and bolt. As you might imagine, however much you tighen the lock nut on the M4 mushroom bolt, it cannot keep the flap up due to the &quot;mechanical advantage&quot; of the plate weight. Or, if tighten to the nth degree to hold plate ... the latter cannot be repositioned!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked at 'stays', but they will only 'lock' in the fully open state, (I need to hold the flap between -90 (closed) to +20 from horizontal)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea at present is to construct a stay using two flat alum bars constucted as a &quot;V&quot; with the outer points at around 4&quot; on the unit and barn door, and the pivot at about 7&quot; (ie pointing forwards) The pivot will have a fixed bolt / thumbwheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's this latter part I seek advice on. Even though the surface area will be larger - and so I could use a M6 bolt - is there any action I can take to make the friction grip between the approx 1&quot; diameter area on each bar stronger? (A fibre washer will reduce the contact area of metal, so that might not be an option)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an ideal World, I'd opt for something like a plastic 'crown' gear fixed on each component. When loosened, you can turn the item to a new angle before re-engaging teeth again. (A system used on some cheap Chinese photo light fittings)&lt;/p&gt;&#xA;" OwnerUserId="20089" LastEditorUserId="20089" LastEditDate="2019-04-13T18:26:16.347" LastActivityDate="2019-04-29T15:09:03.157" Title="How to keep lighting &quot;barn doors&quot; open" Tags="|mechanical-engineering|friction|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26851" PostTypeId="2" ParentId="26665" CreationDate="2019-04-12T19:42:58.890" Score="0" Body="&lt;p&gt;Plenty of cheap gear options these days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I don't need precise rotational accuracy or high torque capabilities, like when testing prototypes, I just 3d print the gears. It will be fairly inexpensive when done this way, and about the crappiest quality I can tolerate for gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For moderately higher strength, get it laser cut from a sheet of plastic or composite. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For even higher strength, and if the gear is large enough, get it CNC plasma/torch cut from steel plate. Water jet cut it if better tolerances are needed.&lt;/p&gt;&#xA;" OwnerUserId="1459" LastActivityDate="2019-04-12T19:42:58.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26852" PostTypeId="2" ParentId="26844" CreationDate="2019-04-12T21:52:56.560" Score="1" Body="&lt;p&gt;The energy to drive the compressor is Work ie not useful output. Remember Q - W is change in internal energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any engine « burning » or combusting fuel is not 100% efficient - internal combustion engines are between 20 and 40% efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, other losses such as  friction, unburnt fuel, losses due to clearances etc The first and second laws of thermodynamics always apply: you can change energy from one form to another, but you never get all of it...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-04-12T21:52:56.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26853" PostTypeId="2" ParentId="26844" CreationDate="2019-04-12T23:15:26.990" Score="2" Body="&lt;p&gt;energy flow rate towards turbine after compression and combustion is 12MW (chemical energy of fuel + compression energy of compressor). Energy extracted by turbine is 9.3MW. &#xA;Efficiency of turbine 9.3(output) /  12(input)  * 100  = 77.5%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;rest of the energy flow 12MW - 9.3MW = 2.7 MW is in outgoing hot gases / rejected heat to environment. &lt;/p&gt;&#xA;" OwnerUserId="16175" LastActivityDate="2019-04-12T23:15:26.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26854" PostTypeId="2" ParentId="17552" CreationDate="2019-04-12T23:25:38.513" Score="0" Body="&lt;p&gt;If you try to be too smart, you'll outsmart yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complicated designs have some things in common:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;They have more parts to break.&lt;/li&gt;&#xA;&lt;li&gt;They are heavier.&lt;/li&gt;&#xA;&lt;li&gt;They produce a lot of friction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For a device that has a very simple goal and is very underpowered, those are not good design attributes.  Consider this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Six wheels will produce twice as much friction as three wheels.&lt;/li&gt;&#xA;&lt;li&gt;Two strings will produce twice as much friction as one string.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The best version of one of these cars I've seen used bearings taken from broken CD players (very low friction) and 3 CDs as wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it also had an extension on the trap's arm and simply used reaction mass to launch the car.&lt;/p&gt;&#xA;" OwnerUserId="19930" LastActivityDate="2019-04-12T23:25:38.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26855" PostTypeId="2" ParentId="22121" CreationDate="2019-04-13T03:57:52.443" Score="1" Body="&lt;p&gt;This is done in a jet engine, and partly, by the cowl around the front fan on a jet airliner engine, but on a large propeller, you could only do it slightly without slowing the airflow through the propeller. The extra weight would outweigh the benefits. &lt;/p&gt;&#xA;" OwnerUserId="20261" LastActivityDate="2019-04-13T03:57:52.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26856" PostTypeId="2" ParentId="11013" CreationDate="2019-04-13T04:08:49.467" Score="1" Body="&lt;p&gt;The short answer is yes, but they will be much more effective if they turn it opposite directions, or have straightening vanes between to take the spin out of the air.&lt;/p&gt;&#xA;" OwnerUserId="20261" LastActivityDate="2019-04-13T04:08:49.467" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26857" PostTypeId="2" ParentId="25091" CreationDate="2019-04-13T07:46:08.210" Score="1" Body="&lt;p&gt;Your pump is having difficulties to suck from hopper. That's because the material in hopper is not fluid enough, not enough weight to push material down the inlet throat, so it depends on suction effect - but eventually suction effect is not functioning when a hole created down to screw impeller. My suggestion is to have better screw pump design (Hopper facing open top, screw to down, with extended screw sticking out from inlet throat to hopper, hopper taper should steep more than material slide angle).&lt;/p&gt;&#xA;" OwnerUserId="4630" LastActivityDate="2019-04-13T07:46:08.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26858" PostTypeId="1" AcceptedAnswerId="26868" CreationDate="2019-04-13T12:27:40.907" Score="3" ViewCount="1980" Body="&lt;p&gt;Of course, most 2-stroke engines require oil to be mixed with the fuel (I appreciate there are 2-stroke engines that don't require this). Therefore, it makes sense to me that oil will be burnt during the combustion process, given it is mixed with the fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what doesn't make sense to me is why 4-stroke engines don't burn just as much oil. In both engines, my understanding is that the oil has to be used within the cylinders to keep them from seizing, so surely a 4-stroke engine would also be burning oil too, with the oil just being added at a different stage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to understand why 2-stroke engines burn more oil than 4-stroke engines?&lt;/p&gt;&#xA;" OwnerUserId="20042" LastEditorUserId="9262" LastEditDate="2019-04-15T20:41:56.290" LastActivityDate="2019-04-16T01:25:05.863" Title="Why do 2-stroke engines burn more oil?" Tags="|mechanical-engineering|automotive-engineering|mechanical|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26860" PostTypeId="2" ParentId="26858" CreationDate="2019-04-13T13:41:23.990" Score="2" Body="&lt;p&gt;Simply because many small 2 strokes use the oil / fuel mix as the lubrication system as the lubrication occurs as the mix passes through the crankcase on its way to the combustion chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A four stroke, and some 2 strokes, have a dedicated oil lubrication system with an oil reservoir, often in the sump, so the oil is not consumed with the fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4 strokes do have a small amount of oil left on the bore as the piston descends because the scraper ring is not perfect while it does remove most of the oil film... But this is a really tiny amount.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-04-13T13:41:23.990" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26861" PostTypeId="1" AcceptedAnswerId="26866" CreationDate="2019-04-13T14:24:12.007" Score="1" ViewCount="109" Body="&lt;p&gt;I need to build a steel frame with a steel web in it, as illustrated below.&#xA;Flanges of the frame will be welded, as well as the web, along its whole perimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would adding a diagonal fold to the web allow this structure to whistand a bigger load? Or, for an equivalent loading, would the fold allow me to use a thinner web?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lXcHH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lXcHH.jpg&quot; alt=&quot;no fold&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8Ug8u.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8Ug8u.jpg&quot; alt=&quot;fold&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Addendum:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the doubly folded strap of the web material welded on the ends to the frame, is there a critical angle for this double fold, where, at some point it is stronger to have no folds at all, rather than two folds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively and I may be wrong about this, concentrating matter close to the diagonal tends to make the web behave as a truss. However shear flow must be disrupted as well with two folds of very obtuse angle. See illustration below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this case can it be assumed that one single fold, whatever the angle of the fold, weakens the structure, but two diagonal folds, whatever the angle, makes it stronger?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jD5LI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jD5LI.jpg&quot; alt=&quot;two fold&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user12750" LastEditorDisplayName="user12750" LastEditDate="2019-04-15T10:15:25.727" LastActivityDate="2019-04-15T16:41:24.183" Title="Influence of a diagonal fold on a web + double fold situation" Tags="|structural-analysis|steel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26862" PostTypeId="1" CreationDate="2019-04-13T14:57:35.313" Score="2" ViewCount="228" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CIhQY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CIhQY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be a silly question but can someone explain how this shear force on the right side of the beam causes Qy throught the whole beam? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand it mathematically but not conceptually so I would be grateful is someone could help me understand this via an analogy or something, because when I solve mechanics I am often stuck on such simple problems because intuitively it feels incorrect to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware there is also the M force on the beam but I left it out of the picture because that one I can visualize and I understand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, why does the whole beam &quot;feel&quot; this force (if we ignore bending for a moment)? if you press down on my hand I don't feel that pressure anywhere else on my arm.&lt;/p&gt;&#xA;" OwnerUserId="8851" LastActivityDate="2019-04-14T07:49:39.127" Title="Why does concentrated shear force on a beam influence the whole beam?" Tags="|beam|shear|solid-mechanics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26864" PostTypeId="2" ParentId="26862" CreationDate="2019-04-13T16:13:51.773" Score="5" Body="&lt;p&gt;So, we apply a point load to the end of the cantilever beam and can see that for &lt;em&gt;external&lt;/em&gt; equilibrium, there must be shear reaction at the support. (Leaving aside for now the moment reaction that also exists.) We have an intuitive understanding of this Newtonian action-reaction from our daily lives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ClO8J.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ClO8J.jpg&quot; alt=&quot;cantilever beam with shear reaction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the next question is, how is the load 'getting' from the point of application to the support? We can assess that by looking at &lt;em&gt;internal&lt;/em&gt; equilibrium. If we start from the right end of the beam (where the load is applied) and cut the beam at any section, there must be a shear reaction for the beam segment to be in equilibrium. We can see this using the same logic as for the external equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vWEhL.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vWEhL.jpg&quot; alt=&quot;shear reaction at various section cuts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This illustrates that the applied load creates a shear force at every point along the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example you gave from the human body is perhaps a difficult one for developing this intuition, given the complexity of muscle, tendon, and bone. However, you mention that if you you press down on your hand you don't feel the &quot;pressure&quot; anywhere else, so this may be about distinguishing between internal and external equilibrium. Pressure is external. The beam (or your hand) only feels that external pressure at the point of load application and at the support. However, if you hold out your forearm and hand rigidly and push down on the tips of your fingers, you will feel muscles and tendons reacting to the load to maintain equilibrium.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2019-04-13T16:13:51.773" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26865" PostTypeId="2" ParentId="26862" CreationDate="2019-04-13T16:14:43.790" Score="0" Body="&lt;p&gt;The shear force has to continue along the beam until a point where it meets a force of equal magnitude and opposite direction to set it off, which in you case is the left hand support reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively if you cut the beam at 1 cm from the load, at the left you have P1 force which has not been countered by any reaction yet, pushing this section down and you need to apply a balancing vertical force of P1 directed up, to keep it in equilibrium, otherwise it will fall down. Same thing if you cut the beam at 2,3,.. cm. So throughout the beam the shear continues until it has encountered the opposing force at the left support.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-13T16:14:43.790" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26866" PostTypeId="2" ParentId="26861" CreationDate="2019-04-13T16:27:46.840" Score="1" Body="&lt;p&gt;if you add a fold you disrupt the flow of shear force which is acting in a rotating pattern all around the web, and make it weaker actually. because you are encouraging it to warp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you just put a doubly folded strap of the web material with strong welds on the ends to the frame and leave the web undisturbed it would act as a truss, and be stronger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attached a sketch of shear flow below. as you see you create two triangles of shear panel with much less shear resistance, usually shear strength of a section is related to its side length , a^3 , so we have here roughly 1/2a^3=1/8a multiplied by two, 1/4a or times less shear strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6UUBt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6UUBt.jpg&quot; alt=&quot;here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-04-14T19:02:40.277" LastActivityDate="2019-04-14T19:02:40.277" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26867" PostTypeId="1" CreationDate="2019-04-13T17:55:47.453" Score="0" ViewCount="55" Body="&lt;p&gt;I'm trying to calculate the energy per unit length the vacuum pumps require to vacuum a cylinder with a diameter D. Which formulas should I use? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used a typical formula for compressors and the result I got was that the air exited at 12,000 K at the end of the process! Also 500 kWh/km to vacuum a 3.3 m diameter cylinder (the initial pressure is 101,325 Pa and the end pressure is 99 Pa).&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-13T17:55:47.453" Title="Energy required to vacuum a cylinder" Tags="|vacuum-pumps|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26868" PostTypeId="2" ParentId="26858" CreationDate="2019-04-13T19:29:05.493" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;In both engines, my understanding is that the oil has to be used within the cylinders to keep them from seizing, so surely a 4-stroke engine would also be burning oil too, with the oil just being added at a different stage?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oil is not added because piston rings separate it from fuel. Rings keep the oil at one side of the piston (the crankcase side) and the burning fuel-air mixture at the piston head side. Ideally, the oil is not supposed to get to the head side at all: no mixing and no burning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, some is oil is left on the inner cylinder wall in the form of a film, some of it get washed away by fuel and burned by the combusting mix. But this amount is much, much smaller than the amount of oil that has to be mixed into fuel in cheap 2-stroke engines. While the mixed-in oil passes through the crankcase just once yet it still has to provide similar level of lubrication as the oil that's (almost) permanently contained in the oil pan of a 4-stroke engine (or a proper2-stroke engine with a dedicated scavenging blower, as you've mentioned). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically: 4-stroke keeps reusing oil and loses only as much oil as slips past piston rings. 2-stroke uses oil once and throws it overboard. That's why so much more oil is necessary when it's mixed into fuel.&lt;/p&gt;&#xA;" OwnerUserId="1931" LastEditorUserId="20042" LastEditDate="2019-04-16T01:25:05.863" LastActivityDate="2019-04-16T01:25:05.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26870" PostTypeId="1" AcceptedAnswerId="26872" CreationDate="2019-04-13T23:43:18.920" Score="0" ViewCount="383" Body="&lt;p&gt;For example, the step response of &lt;span class=&quot;math-container&quot;&gt;$\frac{s+1}{s+2}$&lt;/span&gt; is decreasing although its gain is +1, and this system has higher gain at high frequencies than low frequencies based on its Bode plot. But if I make the transfer function strictly proper, no matter how transient the pole I added is, such as &lt;span class=&quot;math-container&quot;&gt;$\frac{s+1}{(s+2)(s+1000000000000)}$&lt;/span&gt;, the system tends to respond &quot;nicely&quot; like those responses shown in textbooks and professor's examples.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Mathematically, &lt;span class=&quot;math-container&quot;&gt;$\frac{s+1}{s+2}=1-\frac{1}{s+2}$&lt;/span&gt;, so its response  to a step input is an exponential decay, is there a more physical/intuitive way to explain this? &lt;/li&gt;&#xA;&lt;li&gt;Why systems with higher gain at high frequencies have decreasing response to step inputs?&lt;/li&gt;&#xA;&lt;li&gt;Is there a reason that we prefer increasing(nice) step response (or do we prefer it at all)?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;By nice, I mean something like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ime0l.png&quot; width=&quot;400&quot; /&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By not nice, I mean something like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/yuRkz.png&quot; width=&quot;400&quot; /&gt;&lt;/p&gt;&#xA;" OwnerUserId="20099" LastEditorUserId="20099" LastEditDate="2019-04-13T23:48:24.043" LastActivityDate="2019-04-14T03:17:31.927" Title="Why does a lead-compensator type system decrease to a step response" Tags="|mechanical-engineering|electrical-engineering|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26871" PostTypeId="2" ParentId="26870" CreationDate="2019-04-14T01:41:28.430" Score="0" Body="&lt;p&gt;There's something wrong with the simulation program you're using. When you add the transient pole the program realizes the main pole is -2 and makes the calculations right. Try to simulate the step responses after eliminating s+1 (the zero). Maybe the problem is that the zero is confusing the program. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-14T01:41:28.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26872" PostTypeId="2" ParentId="26870" CreationDate="2019-04-14T03:17:31.927" Score="2" Body="&lt;p&gt;Normally a lead compensator is used to increase the phase of the open loop transfer function in a certain frequency range, while trying to keep the magnitude close to constant. So one would normally not encounter a lead compensator all by itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to also answer the question regarding the shape of the step response you can use that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\lim_{s\to0}&#xA;\frac{s+1}{s+2} = \frac{1}{2}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\lim_{s\to\infty}&#xA;\frac{s+1}{s+2} = 1&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So low frequencies get halved, while high frequencies pass straight through. The moment the step starts the input has an infinite slope which one could also view as a &quot;high frequency&quot;, so passes straight through the lead compensator. But the steady state response to the step would be a &quot;low frequency&quot;, so that magnitude would get halved. So initially the step would pass straight through the lead compensator, but as time progresses its magnitude would drop to a half.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2019-04-14T03:17:31.927" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26873" PostTypeId="2" ParentId="26862" CreationDate="2019-04-14T07:49:39.127" Score="3" Body="&lt;p&gt;If we cut a section at any point of the beam, an equal and opposite force will be there (shear reaction) to maintain the equilibrium. So if we think it as a differential section, it will be easy to understand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine this differential section extended to length 'l', each fiber of the beam will have a shear resistance until it becomes equal to the force magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An understandable analogy (probable explanation) to this would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine this cantilever beam is cut at any point away from the end points, the section not connected to rigid support will end up in a circular motion along the trajectory of the tangent. This is because the force applied and shear resistance at the cut end will form a couple which will make the beam rotate. Hence this proves that if we apply force at one end of the cantilever beam, it's resisting force is felt at every point of the beam. Hope this clears you confusion.&lt;/p&gt;&#xA;" OwnerUserId="20102" LastActivityDate="2019-04-14T07:49:39.127" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26875" PostTypeId="2" ParentId="8935" CreationDate="2019-04-14T11:56:11.547" Score="3" Body="&lt;p&gt;I am an ex rigger used to slinging heavy loads. You cannot spread the load equally over 4 cables. Three of the cables will support the load - the fourth will provide stability. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this from practical experience. I was involved in the construction of the Westgate Bridge. The rig we used to lift the box girded which weighed anywhere from 70 to 120 tonnes was a 3 cable arrangement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was you I would assume the load is carried over two cables.&lt;/p&gt;&#xA;" OwnerUserId="20105" LastActivityDate="2019-04-14T11:56:11.547" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26876" PostTypeId="1" CreationDate="2019-04-14T12:21:53.540" Score="0" ViewCount="423" Body="&lt;p&gt;I am bit confused about the curve: &lt;a href=&quot;https://i.stack.imgur.com/MSXoZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MSXoZ.jpg&quot; alt=&quot;Head Vs flow rate Curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could observe from the [Pump curve] that Total Head decreases with increase in flow rate, but for [System Curve], it is opposite, i.e, Total Head increases with flow rate. Why is that so? I need a detailed explanation or a link where I could understand this curve.&lt;/p&gt;&#xA;" OwnerUserId="20106" LastEditorUserId="13297" LastEditDate="2019-04-14T22:17:43.543" LastActivityDate="2019-04-14T22:17:43.543" Title="Pump Characteristic Curve" Tags="|mechanical-engineering|fluid-mechanics|motors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26877" PostTypeId="2" ParentId="26876" CreationDate="2019-04-14T13:50:27.340" Score="1" Body="&lt;p&gt;It's quite simple:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you're pumping a fluid, as the flow rate increases, head losses also increase. That's why the pump head decreases. As there are more internal losses, the pump provides the fluid with less energy. Pump curve: A-B*Q^2 (A and B being positive coefficients).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the system, the increase of the flow rate increases losses as well, so the resistive head goes up. System curve: C+D*Q^2 (C and D being positive coefficients).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operation point is the point where the pump head equates the resistive head of the system and that is true for a determinate flow rate. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorDisplayName="user20096" LastEditDate="2019-04-14T18:33:32.067" LastActivityDate="2019-04-14T18:33:32.067" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26880" PostTypeId="2" ParentId="26744" CreationDate="2019-04-14T18:55:23.610" Score="0" Body="&lt;p&gt;Of course there is, you can find it on the NASA webpage:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/astar.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.grc.nasa.gov/www/k-12/airplane/astar.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can demonstrate the equation with mass flow conservation equation (continuity), ideal gas law and isentropic relations. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-14T18:55:23.610" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26881" PostTypeId="2" ParentId="3183" CreationDate="2019-04-14T19:03:59.860" Score="3" Body="&lt;p&gt;There is a list of some widely used devices and their transmit power (and many other properties) in an easily digestible table: &lt;a href=&quot;https://clients.mikealbano.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://clients.mikealbano.com/&lt;/a&gt;&#xA;The most smartphones seem to be just around 20 dBm and the maximum lies around 30 dBm.&lt;/p&gt;&#xA;" OwnerUserId="20109" LastActivityDate="2019-04-14T19:03:59.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26882" PostTypeId="2" ParentId="25665" CreationDate="2019-04-15T02:16:49.580" Score="1" Body="&lt;p&gt;I found the error. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strain in concrete should be &lt;span class=&quot;math-container&quot;&gt;$\dfrac{A_2-A_0}{A_0}$&lt;/span&gt; which will result in strain equal to &lt;span class=&quot;math-container&quot;&gt;$p\left(\dfrac{\rho_s - \rho_c}{\rho_c}\right)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strain calculated previously would only account for strain in corrosion particles.&lt;/p&gt;&#xA;" OwnerUserId="19087" LastEditorUserId="1832" LastEditDate="2019-04-15T02:21:33.103" LastActivityDate="2019-04-15T02:21:33.103" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26883" PostTypeId="1" CreationDate="2019-04-15T04:14:49.477" Score="0" ViewCount="106" Body="&lt;p&gt;I looked through the stack exchange list of sites, hoping to find one more suitable for my question, but decided that this was the most likely place to get the answer I'm looking for. I'm not an engineer, but I'll try to explain myself as clearly as possible. Here's the set up: Two fifty-five gallon fish tanks, each on a metal shelf stand. Each tank has two holes (drilled or cut out, idk, and wouldn't mind having the neccessary steps for this explained) in their glass floors. There are corresponding holes drilled through the metal shelves that they sit on. Two U shaped glass tubes seem to hang under it, one bigger than the other, so the smaller one is above it. One end of each is set securely through the holes of one shelf and the holes of one tank, same on the other side (picture to better explain). The idea is that, when both tanks are filled with water, the tubes would fill, too, and the fish would be able to use them as highways, so they could swim freely from one tank to the other. I'm not sure how the tubes could be securely connecting, though I was imagining in the same way that the walls of the tanks are connected (I think it's glue, not sure, clarification welcome). Also, a friend proposed the problem of pressure from the water above being too much on the tubes. I really just want to know if this is doable. And, if doable, but there are some glaring obstacles, to have them explained. Thanks in advance and sorry for disrupting the well-versed engineer-y air of the site. (Side note -just because I'm intruding: This is actually an idea I came up with for a tank set up in a novel I'm writing. I've lived around fish tanks, but have never actually tried doing anything remotely like this with them- not sure how to put that. Anyway, thanks again.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z1MkM.jpg&quot; alt=&quot;ignore the upside down fish&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="20111" LastEditorUserId="1832" LastEditDate="2019-04-15T16:42:24.337" LastActivityDate="2019-04-15T18:24:32.670" Title="Unusual idea for fish tank design. Unsure if plausible or not" Tags="|pressure|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26884" PostTypeId="2" ParentId="26883" CreationDate="2019-04-15T04:20:14.757" Score="0" Body="&lt;p&gt;This is possible, but the stress concentration around the holes in the tank bottoms will probably cause them to crack and fail catastrophically! Here is an easier way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of connecting the bottoms of the tanks, simply invert your bent tubes, fill them with water, and place one end of each in each tank to form a siphon. Fish can swim up into the water-filled tube, traverse the gap between tanks, and then swim down and out of the tube in the next tank over. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-04-15T04:20:14.757" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26885" PostTypeId="1" AcceptedAnswerId="26886" CreationDate="2019-04-15T04:59:00.140" Score="0" ViewCount="82" Body="&lt;p&gt;Is it true that computer chips &quot;float&quot; inside of the computers ? Can someone give me references(books, articles, papers...) about that fact ?&lt;/p&gt;&#xA;" OwnerUserId="7129" LastEditorUserId="19930" LastEditDate="2019-04-15T15:22:04.397" LastActivityDate="2019-04-15T15:22:04.397" Title="A naïve question about computer chips" Tags="|fluid-mechanics|rf-electronics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26886" PostTypeId="2" ParentId="26885" CreationDate="2019-04-15T05:05:07.023" Score="2" Body="&lt;p&gt;No, they are securely fixed (soldered) to the circuit boards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think you have heard the term &quot;float&quot; with reference to a chip, then you are probably hearing about &quot;floating point arithmetic&quot; which is the process used inside the chip to solve the mathematical equations and calculations we want.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-04-15T05:05:07.023" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26887" PostTypeId="2" ParentId="26883" CreationDate="2019-04-15T06:09:06.367" Score="0" Body="&lt;p&gt;Yes, you can do it by either using flex silicon sealants or some rubber compression washers like those commonly used in copper tubing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your supports have to be solid and offer easy access to the bottom for maintenance of the pipes. The pipes on the suspended part have to be supported by strong but flexible brackets.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-15T06:09:06.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26888" PostTypeId="1" CreationDate="2019-04-15T12:18:44.000" Score="5" ViewCount="4702" Body="&lt;p&gt;I know that in places where there are convenient rivers and streams the tendency is to dump the communities sewage into those waterways so that it is carried away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what about places where there is no river nearby? So, for example in Phoenix, or other midwestern cities where rivers are either very small or non-existent, what do they do?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2019-04-15T20:41:46.853" Title="What happens to sewage if there is no river near by?" Tags="|waste-disposal|" AnswerCount="5" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26889" PostTypeId="2" ParentId="26888" CreationDate="2019-04-15T12:24:23.797" Score="3" Body="&lt;p&gt;Three options come to mind : septic tanks and sewage systems and reed bed treatment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some properties far away from others ie in rural areas have septic tanks, which get emptied every so often by &quot;sludge gulpers&quot; - tankers designed to suck up and remove the contents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being on mains drainage is much easier and the sewage ends up at Water Treatment Plants or Sewage Farms (either name is used). The sewage is treated and the residue can be spread on fields as fertilizer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reed beds can treat sewage but needs the ground area and climate for it to work well. One book that explains them is &quot;Lifting the Lid&quot; By Halestrap &amp;amp; Harper.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="-1" LastEditDate="2019-04-15T20:41:46.853" LastActivityDate="2019-04-15T20:41:46.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26890" PostTypeId="2" ParentId="26888" CreationDate="2019-04-15T12:58:04.390" Score="1" Body="&lt;p&gt;Dumping your untreated poop directly into the river is actually a bad practice. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of municipalities (at least in my neck of the EU) have a main line running through it which collects the sewage from the buildings and brings it to a water treatment plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is not an option then a septic system can be installed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical septic system works with a large holding tank which overflows out into the drainage field. Which is a large area where the sewage gets slowly percolated into the groundwater where natural processes take care of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building a septic system required a permit in most areas. This permit is to ensure the field and tank are sized properly, its location is recorded and that the ground can percolate the waste well enough.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2019-04-15T12:58:04.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26891" PostTypeId="1" CreationDate="2019-04-15T14:00:24.407" Score="0" ViewCount="461" Body="&lt;p&gt;In geodetic surveying, we consider the spherical geometry of the surface though the surface is not perfectly spherical.&lt;/p&gt;&#xA;&lt;p&gt;In-plane surveying, we consider a small surface area which is considered as a plane or flat surface whose results upon surveying do not get affected.&#xA;But can't totally neglect the curvature of the earth's surface so the assumption of flatness must affect results to some extent.&lt;/p&gt;&#xA;&lt;p&gt;I know that geodetic surveying has great precision than plane surveying. However, both will be accurate though.&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If both survey methods were performed for an area then what would be the &amp;quot;precision difference&amp;quot;(in terms of significant figures/numbers) between these two surveying methods for the same area?&lt;/p&gt;&#xA;&lt;p&gt;I am questioning about the &amp;quot;closeness of the results&amp;quot; of plane surveying to the actual/real parameters. So what is reality then? This is what we are trying to achieve with well equipped, periodically improved and advanced technology for great accuracy and precision that hadn't been achieved before.&#xA;I want to compare their precision because I think results from geodetic surveying are very close to reality so &lt;strong&gt;&amp;quot;comparing these readings with plane surveying allows us to find out the extent to which the assumption of flatness affects results?&amp;quot;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And generally what is the precision (in numbers) for these(individual) methods when performed in their standard confined areas? I think this question can be answered with some experimental readings and experiences.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="20115" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-09-22T04:57:12.000" Title="Comparing the precision work of plane surveying and geodetic surveying" Tags="|civil-engineering|surveying|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26892" PostTypeId="2" ParentId="26883" CreationDate="2019-04-15T14:18:36.930" Score="1" Body="&lt;p&gt;I see three immediate problems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The tubes, and their joints, will experience high pressure.&lt;/li&gt;&#xA;&lt;li&gt;If there is even a slight movement in one tank but not the other, unless the tubes are flexible, or perhaps even if they are, the tubes or the joints will break.&lt;/li&gt;&#xA;&lt;li&gt;Gravel, dirt, etc. will accumulate in the tubes, which will be difficult to clear.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The siphon proposal sounds better, but the tube would soon accumulate air and have to be frequently refilled with water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you really want to put holes in your tanks (and risk destroying them), I'd suggest putting the holes an inch or so below the water line.  That would eliminate the problems of high pressure, dirt accumulation, and air in the siphon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your biggest concern will be how to ensure that the seals are strong and flexible.  Eventually someone (even you) will accidentally bump into a tube.  You really don't want the tube to crack or the seal to leak.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I was about to say it sounds like an interesting idea, like a Habitrail for fish, and then I googled it and found an &lt;a href=&quot;https://www.youtube.com/watch?v=WpajelnO6oM&quot; rel=&quot;nofollow noreferrer&quot;&gt;interesting fish habitrail video&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/S6fzT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S6fzT.png&quot; alt=&quot;three tanks interconnected by siphon tubes&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It uses the siphon suggestion.  If you look at the video you can see the air accumulation problem starting, but I think that could be mostly eliminated if you could arrange for continuous water circulation between the tanks (e.g. a common external filtration system taking water from one tank and returning it to the other).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could make the tubes longer and have the tanks separated farther apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It certainly seems like a better idea than potentially destroying two tanks.&lt;/p&gt;&#xA;" OwnerUserId="19930" LastActivityDate="2019-04-15T14:18:36.930" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26893" PostTypeId="1" AcceptedAnswerId="26899" CreationDate="2019-04-15T14:35:34.867" Score="-1" ViewCount="129" Body="&lt;p&gt;I am trying to calculate the &lt;span class=&quot;math-container&quot;&gt;$I_{sp}$&lt;/span&gt; of a rocket with a thrust of 500&amp;nbsp;kN. I know that you can calculate &lt;span class=&quot;math-container&quot;&gt;$I_{sp}$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$I_{sp} = \dfrac{v_e}{g_0}$&lt;/span&gt;. However, to calculate &lt;span class=&quot;math-container&quot;&gt;$v_e$&lt;/span&gt;, you need the &lt;span class=&quot;math-container&quot;&gt;$I_{sp}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$v_e = g_0 \cdot I_{sp}$&lt;/span&gt;. How do I calculate both without having to set one value (Other than &lt;span class=&quot;math-container&quot;&gt;$g_0$&lt;/span&gt;)?&lt;/p&gt;&#xA;" OwnerUserId="20095" LastEditorUserId="1832" LastEditDate="2019-04-15T16:48:21.487" LastActivityDate="2019-04-15T17:44:09.663" Title="Is there an easier way to calculate rocket engine Isp?" Tags="|aerospace-engineering|rocketry|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26894" PostTypeId="2" ParentId="26888" CreationDate="2019-04-15T14:42:33.297" Score="8" Body="&lt;p&gt;As all the other answers have said, nobody in the developed world dumps sewage in nearby rivers.  What &lt;em&gt;is&lt;/em&gt; discharged into the river is the treated water that comes out of the Sewage Plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your city is in a very arid area like Phoenix without a river your choices are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Discharge the treated water into the soil&lt;/li&gt;&#xA;&lt;li&gt;Use the treated water for irrigation&lt;/li&gt;&#xA;&lt;li&gt;Treat the water a bit more and use it as drinking water&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Most&lt;/em&gt; of the solids will be digested by the treatment process.  The rest can be incinerated, used as fertilizer, or trucked to landfill.&lt;/p&gt;&#xA;" OwnerUserId="20116" LastActivityDate="2019-04-15T14:42:33.297" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26895" PostTypeId="2" ParentId="26888" CreationDate="2019-04-15T15:06:22.207" Score="10" Body="&lt;p&gt;Phoenix has two monster &quot;wastewater&quot; treatment plants that produce &quot;highly treated effluent&quot; as output. This water is used to water golf courses, etc. before it soaks into the ground water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the treated water is piped out to the Palo Verde Nuclear Generating Station which uses it in the cooling towers.  When I worked there, if the wind was in the right direction, the mist from the cooling towers would land on your car and you  had trouble seeing through the windshield on the way home. PVNGS is a zero (liquid) effluent site.  Anything they take from Phoenix that they don't use and all onsite sewage goes to the water reclamation plant where they extract any water they can and the rest goes to the evaporation pond where desert the sun and air do their thing, leaving nothing behind but minerals.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2019-04-15T15:06:22.207" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26896" PostTypeId="2" ParentId="26888" CreationDate="2019-04-15T15:24:04.227" Score="6" Body="&lt;h1&gt;Re-use and recharging.&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Arizona is rather proud of this, it seems.  &amp;#9786;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After filtration and treatment, the waste water is supplied for various re-use purposes.  These can range from various categories of irrigation through snowmaking (sic!) to industrial coolants, and there is a system of categories of waste water and what each category may be used for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be run off into dry streams and rivers, or dumped into infiltration ponds from which it passes into the groundwater table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The treatment plant at 91st Avenue in Phoenix supplies water to the Palo Verde nuclear plant as coolant, for example.  It is claimed that 100% of the water treated at that and at the 23rd Avenue plant in Phoenix is re-used.  According to a 2013 report, as of 2010 over the Phoenix Active Management Area, 82% of all waste water from treatment plants is reused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were &lt;em&gt;plans&lt;/em&gt; back in 2010 for the 91st Avenue plant to do more recharging into the Agua Fria River.  I do not know whether they ever came to fruition.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Further reading&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A. Middel, R. Quay, and D. D. White (2013).  &lt;em&gt;&lt;a href=&quot;https://sustainability.asu.edu/docs/dcdc/website/documents/DCDC_WaterReuse_Final.pdf&quot; rel=&quot;noreferrer&quot;&gt;Water reuse in central Arizona&lt;/a&gt;&lt;/em&gt;. Decision Center for a Desert City Technical Report 13&amp;ndash;01.  Arizona State University.&lt;/li&gt;&#xA;&lt;li&gt;Mike Fulton and Chuck Graf (2011-11-05). &lt;em&gt;&lt;a href=&quot;https://www.flagstaff.az.gov/DocumentCenter/View/15091/ADEQ_Flagstaff-Reclaimed-Water-Presentation_Final&quot; rel=&quot;noreferrer&quot;&gt;Wastewater Treatment and Reclaimed Water Reuse in Arizona: Past and Present&lt;/a&gt;&lt;/em&gt;.  Arizona Department of Environmental Quality.&lt;/li&gt;&#xA;&lt;li&gt;Mike Fulton (2014-10-10). &lt;em&gt;&lt;a href=&quot;https://legacy.azdeq.gov/environ/water/apec/download/101014_fulton_wwreuse.pdf&quot; rel=&quot;noreferrer&quot;&gt;Arizona’s Legacy and Leadership in Wastewater Reuse&lt;/a&gt;&lt;/em&gt;.  Arizona Department of Environmental Quality.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://cap-az.com/public/blog/75-arizona-a-national-leader-in-wate-reuse&quot; rel=&quot;noreferrer&quot;&gt;Arizona: A National Leader in Water Reuse&lt;/a&gt;.  2014-10-04.  Central Arizona Project.&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;&lt;a href=&quot;https://www.usbr.gov/lc/phoenix/AZ100/2010/phoenix_water_reclamation.html&quot; rel=&quot;noreferrer&quot;&gt;Phoenix Metropolitan Water Reclamation and Reuse Project&lt;/a&gt;&lt;/em&gt;.  2010.  U.S.  Bureau of Reclamation.&lt;/li&gt;&#xA;&lt;li&gt;Steve Rohrer, Tim Francis, and Andrew Brown (2012). &quot;&lt;a href=&quot;http://www.reclaimedwater.net/data/files/149.pdf&quot; rel=&quot;noreferrer&quot;&gt;91st Avenue Unified Wastewater Treatment Plant Targets 100 Percent Reuse&lt;/a&gt;&quot;.  &lt;em&gt;Guidelines for Water Reuse&lt;/em&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15164" LastEditorUserId="15164" LastEditDate="2019-04-15T15:31:16.290" LastActivityDate="2019-04-15T15:31:16.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26897" PostTypeId="2" ParentId="26861" CreationDate="2019-04-15T16:41:24.183" Score="0" Body="&lt;p&gt;There is potential value to adding stiffener plates that are transverse to the plane of the web. For tall welded plate wide flange girders, it is very common to provide a stiffener element in order to reduce the web plate thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned, the shear wants to flow through the plate analogous to a truss, this same analogy is used by AISC in Chapter G they call it Tension Field Action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not this is the most economical solution depends on your actual configuration, loads, and fabricator capabilities. &lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-15T16:41:24.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26899" PostTypeId="2" ParentId="26893" CreationDate="2019-04-15T17:11:19.180" Score="0" Body="&lt;p&gt;Easiest: from thrust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T = v_e {dm\over dt}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both fuel mass flow and thrust are simply, directly measurable quantities - measure the force exerted by the engine on a test stand and fuel flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That way you're getting the exhaust speed, and as result, the specific impulse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative is using the rocket equation - &lt;span class=&quot;math-container&quot;&gt;$\Delta v = {I_{sp} \over g_0} \ln {m_0 \over m_f}$&lt;/span&gt; - measure rocket mass change (fuel expenditure) over time, and change in velocity, find the specific impulse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But since &lt;span class=&quot;math-container&quot;&gt;$I_{sp}$&lt;/span&gt; is directly proportional to &lt;span class=&quot;math-container&quot;&gt;$v_e$&lt;/span&gt;, and nothing else (&lt;span class=&quot;math-container&quot;&gt;$g_0$&lt;/span&gt; is a constant), you can't change one without changing the other.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2019-04-15T17:44:09.663" LastActivityDate="2019-04-15T17:44:09.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26901" PostTypeId="1" AcceptedAnswerId="26903" CreationDate="2019-04-15T17:49:29.620" Score="1" ViewCount="38" Body="&lt;p&gt;The type of hose involved is the thinnest one on the right in this &lt;a href=&quot;http://www.flexiblerubberhoses.com/photo/pl6863074-id_8mm_to_22mm_epdm_air_conditioning_hose_for_conveying_r134a_r12_etc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;picture&lt;/a&gt;. This hose contains an external rubber layer, a textile reinforcement layer, an inner rubber layer and a nylon lining.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hose has been ziptied to a plastic electrical conduit and it serves as a discharge hose of an automotive air conditioner, so it gets warm during the system operation, and internal pressures are high (the reason for the reinforcement layer).&#xA;Also, it's located not very far from the exhaust plumbing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the zip tie and the conduit possibly restrict the hose internally due to thermal expansion as the hose warms up, or would the zip tie expand together with the hose? Can the reinforcement layer prevent the hose from developing a restriction except in the case of a lining delamination?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an exemplificative picture of the arrangement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KUiYo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KUiYo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15424" LastActivityDate="2019-04-15T18:25:49.487" Title="Can a reinforced hose develop an internal restriction due to thermal expansion?" Tags="|reinforcement|thermal-expansion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26902" PostTypeId="2" ParentId="26883" CreationDate="2019-04-15T18:24:32.670" Score="0" Body="&lt;p&gt;It is common for tanks intended for salt water to have holes with bulkhead fittings. The bulkheads normally come with NPS threads in the US. It would be just a matter of connecting them the way you want ; clear flexible hoses would be best. Tanks would need to be ordered with the size and location of the holes specified as it is not reasonable for an amateur to cut the standard tempered glass bottoms. Apparently one would not have gravel in these tanks.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-04-15T18:24:32.670" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26903" PostTypeId="2" ParentId="26901" CreationDate="2019-04-15T18:25:49.487" Score="1" Body="&lt;p&gt;No, the zip tie and conduit won't cause the pipe to develop an internal restriction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even squeezing the outer cover will not compress the inner sleeve... There are metal pipe clamps that will compress the inner but they have serious leverage that you won't get from a nylon zip tie.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-04-15T18:25:49.487" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26904" PostTypeId="1" CreationDate="2019-04-15T19:33:28.543" Score="0" ViewCount="74" Body="&lt;p&gt;Artificial gills aren't used for diving, not because they can't extract oxygen out of water, but because they won't provide enough oxygen for someone to breathe (only arround 1% oxygen in water instead of 20% in air). So they stick with oxygen tanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why don't they use them both at the same time? The gills will on give a little bit of oxygen, true, but over time it will allow the diver to stay underwater longer with the same oxygen tank. If you add a rebreather to that, so only arround 5% of your air goes away every breathe, it will be even longer.&lt;/p&gt;&#xA;" OwnerUserId="20123" LastActivityDate="2019-04-15T19:33:28.543" Title="Why aren't artificial gills used together with oxygen tanks for diving?" Tags="|biomedical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26905" PostTypeId="2" ParentId="26891" CreationDate="2019-04-15T20:31:32.240" Score="1" Body="&lt;p&gt;While I do not discount your question's validity, I think it is quite hard, if not impossible for you to answer the first question as there are &lt;strong&gt;too many fine details that affect precision&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, even within plane surveying, the proximity of your survey points, how often you do look back checks, and all those can play a major role in how much error that your survey will have. Equipment can play a role, weather can play a role, and obviously, as you have already recognized, the size of the area surveyed can play a role as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the number of satellites that you can &quot;see&quot;, the number of epochs that you take for each determination of your actual coordinate results, atmospheric conditions, latency and other factors can affect the actual accuracy of your geodetic survey results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what I am essentially pointing at here is that accuracy generally is less so affected by your method of data collection, but more so by the surrounding conditions of your data collection efforts. There are no necessarily any inherent errors within each method that will produce errors that will sufficiently invalidate the respective methodologies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For question two, it depends on what you mean by &quot;standard confined areas&quot; mean. I recommend you reaching out to specific equipment manufacturers or survey equipment dealers for comparison of actual equipment that you may use.&lt;/p&gt;&#xA;" OwnerUserId="13526" LastActivityDate="2019-04-15T20:31:32.240" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26906" PostTypeId="1" AcceptedAnswerId="26957" CreationDate="2019-04-15T21:36:12.807" Score="1" ViewCount="50" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XSKMh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XSKMh.png&quot; alt=&quot;Parshall Flume Outflow&quot;&gt;&lt;/a&gt;&#xA;How do I calculate the discharge from a Parshall flume that flow over a bulkhead (XY) into three partitions (A, B and C) divided 25%/50%/25%.  There is no restriction of flow when the pipe gates are open (the incoming volume does not exceed the capacity in any of the partitions).  The partitions A &amp;amp; C feed 8&quot; pipe, while the partition B feeds a 10&quot; pipe. When either A or B partitions are closed (using gates A1 and C1), a downstream gate (A2 and C2) can be opened to dump that partition's portion into the 10&quot; pipe (Section C).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If gate C1 is closed (effectively shutting off the water to that pipe), and the downstream gate C2 is closed, what percentage of the water goes into Pipe A and B? (My guess is A 33% and B 66%)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If gate C1 is closed (effectively shutting off the water to that pipe), and the downstream gate C2 is open, what percentage of the water goes into Pipe A and B? (My guess is A 25% and B 75%)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Thanks, Chris&lt;/p&gt;&#xA;" OwnerUserId="20126" LastEditorUserId="20126" LastEditDate="2019-04-16T17:49:21.303" LastActivityDate="2019-04-18T13:33:51.800" Title="How to calculate divided irrigation discharge from a Parshall flume" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26908" PostTypeId="1" CreationDate="2019-04-16T07:18:40.473" Score="3" ViewCount="624" Body="&lt;p&gt;Assume I have a honecomb(sandwich) panel under distributed normal load. Each edge of this panel is simply supported(pinned) as on the picture below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lxR6s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lxR6s.png&quot; alt=&quot;panel under normal load P&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is how to find the core stresses σ-z (not on sides of the panel) as on the picture below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SxsDU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SxsDU.png&quot; alt=&quot;core stress σ-z&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also it will be very useful if you present some equations that describe σ-z in an ordinary (not honeycomb) panel. I'll be very pleased if you recommend some literature.&lt;/p&gt;&#xA;" OwnerUserId="11713" LastActivityDate="2023-01-26T06:00:40.077" Title="How to compute core stress in honeycomb structure?" Tags="|structural-analysis|stresses|strength|composite|buckling|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26909" PostTypeId="1" CreationDate="2019-04-16T07:48:47.457" Score="0" ViewCount="2877" Body="&lt;p&gt;As shown in the figure provided, a Uniform load acts upon a rectangular plate with edges at Y = 0 and Y = b which are simply supported. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YGxJr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YGxJr.jpg&quot; alt=&quot;Rectangular Plate Simply supported on edges with Uniform Acting Load&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this form of the Plate Equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{\partial^4 \omega}{\partial x^4} + 2\frac{\partial^4 \omega}{\partial x^2\partial y^2}+\frac{\partial^4\omega }{\partial y^4} = \frac{P}{D}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where P is the Uniform Load and, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ D = \frac{Et^3}{12(1-v^2)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you determine an equation that describes the Deflection of the plate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how is load  'P' put into the equation?&lt;/p&gt;&#xA;" OwnerUserId="20131" LastActivityDate="2023-06-26T20:03:20.450" Title="Deflection of Rectangular Plate with 2 Simply Supported Edges" Tags="|mechanical-engineering|stresses|deflection|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26911" PostTypeId="2" ParentId="26908" CreationDate="2019-04-16T09:41:57.393" Score="1" Body="&lt;p&gt;This is just &quot;stress = force / area&quot; where the area is the cross section area of the honeycomb walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data sheet for the honeycomb will give the wall thickness. To understand the data sheet, you need to know how honeycomb is manufactured: flat strips of material are bonded together in the appropriate places, and the pack of strips is them pulled apart to create the hexagonal spaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the wall thickness in one of the three directions along the sides of the cells is double that in the other two directions, because that is where the strips were bonded together. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If all the cell walls were thickness &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;, each individual cell would be surrounded by a wall of thickness &lt;span class=&quot;math-container&quot;&gt;$t/2$&lt;/span&gt; (since each wall is the boundary of two cells), but in fact two parallel walls of each cell have thickness &lt;span class=&quot;math-container&quot;&gt;$2t$&lt;/span&gt; not &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; so the &quot;average&quot; wall thickness surrounding an individual cell is &lt;span class=&quot;math-container&quot;&gt;$2t/3$&lt;/span&gt; not &lt;span class=&quot;math-container&quot;&gt;$t/2$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-04-16T09:41:57.393" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26912" PostTypeId="1" CreationDate="2019-04-16T13:20:29.807" Score="0" ViewCount="159" Body="&lt;p&gt;Capacitive touchscreens do not rely on touch as the name suggests but rather measure the distortion of the electrostatic field. To my understanding a touchscreen might be able depending on the build to detect touches from a distance of a couple of tenths of a millimeter above the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to influence the electric field of the touchscreen from a greater distance such as a centimeter above the screen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I am thinking is that creating an electrostatic charge might perhaps help, for example similar to the way an ionizer works. Unfortunately I lack the electrical knowledge to understand the implications of 'the other half of the capacitor' which is needed to excite the touchscreen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to create a 'hover stylus' similar to the wacom digitizer (which of course works differently based on electromagnetic resonance).&lt;/p&gt;&#xA;" OwnerUserId="20133" LastActivityDate="2023-12-24T11:08:05.937" Title="Hover Stylus for capacitive touchscreen" Tags="|electrical-engineering|microelectronics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26913" PostTypeId="1" AcceptedAnswerId="26924" CreationDate="2019-04-16T15:44:53.420" Score="3" ViewCount="132" Body="&lt;p&gt;Yesterday, Notre Dame cathedral caught fire. The roof and the central spire were destroyed and some of the stonework was subjected to high temperatures for an extended period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the likely implications for the strength of the individual stones, the walls and other structures?&lt;/p&gt;&#xA;" OwnerUserId="20135" LastEditorUserId="1832" LastEditDate="2019-04-16T15:50:12.793" LastActivityDate="2019-04-18T02:50:47.630" Title="What are the possible implications of the heating of the structural stonework of Notre-Dame cathedral during the fire?" Tags="|structural-engineering|structural-analysis|fire|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26914" PostTypeId="1" AcceptedAnswerId="26915" CreationDate="2019-04-16T16:06:20.717" Score="0" ViewCount="40" Body="&lt;p&gt;Is it possible to power an AC motor, using alternating positive and negative DC current?  I would assume toggling between each at 60hz would work.&lt;/p&gt;&#xA;" OwnerUserId="20137" LastActivityDate="2019-04-16T18:06:38.640" Title="Simulating AC using DC" Tags="|power-electronics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26915" PostTypeId="2" ParentId="26914" CreationDate="2019-04-16T18:00:02.453" Score="0" Body="&lt;p&gt;This is possible! The AC motors are designed to run with AC currents so this is not a very good idea. A very popular way to do it is PWM, it basically mimics a sin wave by switching between two states, you need a filter and proper timing to run a single phase AC motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some spacial applications require to run at  desired frequencies, so we design a rectifier to gain a DC signal and then turn the rectified signal to an AC signal with desired frequenties(this is not possible without the presence of an AC grid). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A multiphase AC motor needs a rotating field, three sin waves make a very smooth rotating field, however a modulated DC signal can't generate a rotating field as smooth as sin wave no matter how well it mimics the sin wave (think about the harmonics). The results are extra vibrations, arising of recoils between mechanical parts of the motor and introducing harmonics to the AC grid. The motor won't work efficient at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen many researches and papers coming out each day, DC drive is yet in development stage.  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2019-04-16T18:06:38.640" LastActivityDate="2019-04-16T18:06:38.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26916" PostTypeId="2" ParentId="26909" CreationDate="2019-04-16T18:21:02.250" Score="0" Body="&lt;p&gt;Analyses of one way slabs deflections are complicated.  Boundary conditions and even the radius of curvature affect the behavior of the slab and contribution of twisting of hypothetical strip sections to carrying moment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many codes adopt using effective width for design of strength and deflection of short, wide slabs. for example for a ratio of support b to span a of 1.6 &lt;span class=&quot;math-container&quot;&gt;$(effective\ width)/a\quad is\ 1.519\ and\ E= E/(1-v^2)$&lt;/span&gt;,  (in pp:170 Roark’s Formulas&#xA;for Stress and Strain by YOUNG and BUDYNAS 7th ed).&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-16T18:21:02.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26917" PostTypeId="2" ParentId="26913" CreationDate="2019-04-16T18:40:09.353" Score="0" Body="&lt;p&gt;Two factors that are positive.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Masonry, grout, concrete are some of the best fire resistant materials.&lt;/li&gt;&#xA;&lt;li&gt;Exposure to the hottest flames has been avoided by the fact that geometry of the building and its long hallways mostly acted as a ventilation manifold to feed the flames with fresh oxygen, and kept the walls awash with cool air. &lt;/li&gt;&#xA;&lt;li&gt;They have some of the best experts in preservation and rehab of these world heritage structures.&lt;/li&gt;&#xA;&lt;li&gt;It's incredible and offensive even why safety measures, fire sprinklers, alarms were not installed especially during the construction or failed to work.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2019-04-16T19:20:20.360" LastActivityDate="2019-04-16T19:20:20.360" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26919" PostTypeId="1" CreationDate="2019-04-16T20:44:40.297" Score="2" ViewCount="851" Body="&lt;p&gt;I can find many figures online for the extra power generated when a turbo is fitted to an engine. However, I can't find any figures on how much energy the turbocharger turbine actually takes out of the exhaust stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do know however that in Formula 1 they have an &quot;MGU-H&quot; system, which take rotational energy from the turbocharger turbine and converts it to electrical energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how much power could be produced by a generator fitted to a turbocharger turbine and could it be more effective to use this power to drive the wheels, rather than use it to spin a compresser as with a conventional turbocharger system?&lt;/p&gt;&#xA;" OwnerUserId="20042" LastEditorUserId="20042" LastEditDate="2019-04-16T20:53:20.387" LastActivityDate="2022-06-06T15:39:45.643" Title="How much power could a turbocharger generate?" Tags="|automotive-engineering|turbines|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26920" PostTypeId="1" CreationDate="2019-04-16T22:10:32.043" Score="1" ViewCount="380" Body="&lt;p&gt;I would like to plan a small road network for an imaginary city, and I am looking for a way to create curved road segments. I am familiar with cubic Bezier curves, but I am not sure if it's the best fit for this problem, and it's not trivial to offset. Circle arcs are easy to calculate and offset, but they are not as flexible and hard to chain. What kind of splines do enginieers use, when they plan roads?&lt;/p&gt;&#xA;" OwnerUserId="20144" LastEditorUserId="16486" LastEditDate="2019-04-17T04:05:27.570" LastActivityDate="2019-05-11T01:43:22.400" Title="What kind of splines should be used when planning roads?" Tags="|highway-engineering|roadway|traffic-intersections|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26921" PostTypeId="2" ParentId="26919" CreationDate="2019-04-16T23:53:01.663" Score="2" Body="&lt;p&gt;That actually exists. It's called turbocompounding (&lt;a href=&quot;https://www.dieselnet.com/tech/engine_whr_turbocompound.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.dieselnet.com/tech/engine_whr_turbocompound.php&lt;/a&gt;). Sometimes the turbine is coupled directly to the crankshaft and there's no compressor. Sometimes there's no turbine (&lt;a href=&quot;http://mechstuff.com/differences-between-superchargers-vs-turbochargers/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mechstuff.com/differences-between-superchargers-vs-turbochargers/&lt;/a&gt;).&lt;/p&gt;&#xA;&lt;p&gt;Here's the power a turbine can produce:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P = \eta* \frac{dm}{dt} * c_{p,mean} * T_{inlet}*[(\frac{p_{outlet}}{p_{inlet}})^{((k-1)/k)}-1]$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T_{inlet}$&lt;/span&gt;: Inlet temperature [K]. Usually 400-500 ºC for Otto engines and a around 600 ºC for Diesel engines. Let's take 873 K.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$c_{p,mean}$&lt;/span&gt;: Mean heat capacity at constant pressure [J/kg-K]. Mean between T_inlet and T_outlet. For air it would probably be around 1050 J/kg-K, but consult tables.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\eta$&lt;/span&gt;: Efficiency of the turbine. 0.80 is a typical value.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$p_{outlet}$&lt;/span&gt;: Outlet pressure [Pa]. Normally atmospheric pressure (101,325 Pa).&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$p_{inlet}$&lt;/span&gt;: Inlet pressure [Pa]. Let's assume 200,000 Pa (2 bar), because the expansion isn't full.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;: Adiabatic index, usually 1.4 for air and similar for exhaust gases.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;: Mass, in [kg]. Its temporal derivative is mass flow [kg/s]. The same mass that enters must exit (ignoring leaks through the gaps). Let's say 4*10^-3 kg (I can't remember the figures I knew now). And 4 kg/s as the exhaust process takes place at a few miliseconds.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;: Power, in [W].&lt;/p&gt;&#xA;&lt;h4&gt;Worked example&lt;/h4&gt;&#xA;&lt;p&gt;Using the &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; = &lt;code&gt;0.8 * 4e-3 * 1050 * 873 * ((101325/200000)^((1.4 - 1)/1.4) - 1)&lt;/code&gt; = -514W, which is in the right range. It could be from 1 to 10 kW, for example, depending chiefly on the mass flow and the expansion ratio (p_outlet/p_inlet).&lt;/p&gt;&#xA;&lt;p&gt;Sorry for the terrible format, but I'm new to this forum and I don't know how to format yet. If you are going to convert the power to electrical power, multiply to 0.90 (usual efficiency of a electric generator/motor).&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorUserId="27170" LastEditorDisplayName="user20096" LastEditDate="2022-06-06T15:39:45.643" LastActivityDate="2022-06-06T15:39:45.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26922" PostTypeId="2" ParentId="26920" CreationDate="2019-04-16T23:58:30.077" Score="5" Body="&lt;p&gt;Circular arcs are not used except in special situations, because the sudden change in curvature from a straight road to a circular arc would mean drivers had to quickly turn the steering wheel to the correct position to follow the circular arc, hold the wheel steady, and quickly turn back to straight at the end of the arc. There would also be problems if the road was banked around the curve (which is standard practice for high speed roads) since the bank angle can't suddenly jump from zero to the correct value for the circular arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard shape is a spiral curve, where the driver follows the curve by turning the wheel gradually to enter and leave the curve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could probably make a reasonable approximation to the standard spiral from Bezier curves, once you realize what you are trying to achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are lots of web references to the design process for  spiral curves, including designing the road camber around the curve (which may be irrelevant for your project). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;http://cc4w.net/spiral/SpiralTraining.pdf&quot; rel=&quot;noreferrer&quot;&gt;http://cc4w.net/spiral/SpiralTraining.pdf&lt;/a&gt; for example.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-04-16T23:58:30.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26923" PostTypeId="2" ParentId="26913" CreationDate="2019-04-17T00:19:10.733" Score="1" Body="&lt;p&gt;Concrete can be significantly damaged at 600F but specifics are highly variable according to the Portland Cement association. It depends on the ratio of cement to aggregate , sizes of aggregate, composition of the aggregate, heating/cooling rate, &#xA;duration, etc. Specific information on sandstone and limestone is even more difficult to find. Rock can even crack during the fire depending on moisture content and how fast it is heated.  I expect it will be many weeks or months before objective information is available. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-04-17T00:19:10.733" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26924" PostTypeId="2" ParentId="26913" CreationDate="2019-04-17T00:23:07.927" Score="1" Body="&lt;p&gt;There is a specific temperature range that causes calcination of limestone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{CaCO}_3 + \text{Heat} = \text{CaO} + \text{CO}_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature range of reaction is 840°C-900°C and is used for creating cement in a kiln. Adding oxygen and sulphur dioxide causes limestone to become gypsum... which is structurally significantly weaker than limestone from which the old cathedral is built. Above 900°C the carbon dioxide emissions accelerate beyond permissible emissions levels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Effectively, superheating limestone causes chemical reactions that result in physical changes of the structural material blocks. Exfoliation is a likely result with degradation of surfaces reaction to other chemical compounds in the fire itself. If sulphur compounds were present then there is a risk that the stone may have been changed into another type of rock to varying depths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add water to carbon dioxide and it creates carbolic acid which may attack limestone.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Carbonic acid, hydrochloric acid and acetic acid are some acids that react with limestone, causing it to dissolve. Each of these acids reacts with limestone in different ways. Limestone is made up mainly of calcite that is the chemical compound calcium carbonate.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Carbonic acid, which is a weak acid, forms when rainwater and carbon dioxide in the air react with one another. When carbonic acid comes into contact with limestone or calcium carbonate, it can cause it to dissolve over a long period of time. In nature, this reaction leads to the formation of caves.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://www.reference.com/science/acids-dissolve-limestone-8485a52922d6ebe1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.reference.com/science/acids-dissolve-limestone-8485a52922d6ebe1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="1832" LastEditDate="2019-04-18T02:50:47.630" LastActivityDate="2019-04-18T02:50:47.630" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26925" PostTypeId="1" AcceptedAnswerId="26947" CreationDate="2019-04-17T00:29:55.287" Score="0" ViewCount="152" Body="&lt;p&gt;If two surfaces rub together with a normal force between them, friction is generated.  You could add lubrication between those two surfaces and reduce that friction.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bearings essentially consist of two surfaces rubbing together and, as before, under load, friction will occur.  As before, you could lubricate the bearings and reduce that friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So absent a better understanding, I can't see how bearings add much of anything to efficiency.  This must be wrong or all of industry would be spending money frivolously on unnecessary parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is, in what way is my argument wrong?&lt;/p&gt;&#xA;" OwnerUserId="17228" LastActivityDate="2020-07-09T06:43:12.680" Title="Refute my assumption: Bearings Don't Improve Efficiency" Tags="|friction|bearings|energy-efficiency|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26926" PostTypeId="2" ParentId="26676" CreationDate="2019-04-17T00:48:04.220" Score="0" Body="&lt;p&gt;BIPV works by converting light energy into electrical energy. The voltage output is usually connected to batteries and possibly by inverter to the national grid. Heat is an enemy of energy conversion in PV cells which means surface cooling is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wind turbines similarly use voltage  inversion to convert energy to national grid power. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-04-17T00:48:04.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26927" PostTypeId="2" ParentId="26925" CreationDate="2019-04-17T00:56:35.753" Score="3" Body="&lt;p&gt;Bearing have rolling elements specifically designed to insulate the two moving surfaces from coming into contact with each other and let the move freely (almost).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many types of bearings, ball bearings, cylinder bearing, cone bearing, etc. Each designed to best serve its purpose with least amount of stress, friction, wear and tear. Some resist normal stresses, some lateral and some angular stresses. &lt;a href=&quot;https://en.wikipedia.org/wiki/Bearing_(mechanical)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia bearings link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all of them share one important feature and that is intermediary rolling elements rolling in races or tracks to rotate between the two surfaces and change the direct rubbing into ideally friction less rolling. here is Wikipedia diagram, by I, Solaris2006, CC BY-SA 3.0,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/w/index.php?curid=2496983&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://commons.wikimedia.org/w/index.php?curid=2496983&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/5Bj2h.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Bj2h.jpg&quot; alt=&quot;diagram, &quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-17T00:56:35.753" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="26928" PostTypeId="2" ParentId="26920" CreationDate="2019-04-17T00:59:10.680" Score="5" Body="&lt;p&gt;Roads are usually designed using catenary curves. &lt;a href=&quot;https://sketchup.engineeringtoolbox.com/catenery-curve-c_169.html&quot; rel=&quot;noreferrer&quot;&gt;https://sketchup.engineeringtoolbox.com/catenery-curve-c_169.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Road curves are designed around vehicle velocities and they are not uniform. As vehicles approach a curve they tend to slow down and then accelerate out of the bend which means a non uniform curve is best suited to the purpose. This can be parabolic or more accurately, catenary. The curves are usually calculated using differential calculus.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-04-17T00:59:10.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26929" PostTypeId="2" ParentId="26797" CreationDate="2019-04-17T01:09:53.990" Score="0" Body="&lt;p&gt;One way is &quot;dieseling&quot; but is considered a criminal practice in many countries. The idea is to place a dose of petroleum gel into the pellet base prior to ejection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not recommend modification of air rifles because the muzzle velocity is what determines lethality and legal classification of armament type. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-04-17T01:09:53.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26930" PostTypeId="2" ParentId="26775" CreationDate="2019-04-17T01:13:50.373" Score="1" Body="&lt;p&gt;I may be wrong but that looks like a pressure feed for the boiler. Considering it is a machine that builds water pressure, this is probably the distribution point into the burner of the water to be converted into steam. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-04-17T01:13:50.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26931" PostTypeId="2" ParentId="26732" CreationDate="2019-04-17T02:04:53.277" Score="1" Body="&lt;p&gt;Some large buildings have a transformer attached to them for the mains supply for all apartments. The vibrations are probably being transmitted via harmonic resonance due to acoustic amplification. This may have been why previous owners sold the property.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Human body operates at 25Hz which means harmonic octave frequencies are more noticeable to us. It may be possible to use noise cancelling sound to negate the impact and an audio specialist may be able to help out with creation of an inverse sine wave oscillator.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-04-17T02:04:53.277" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26932" PostTypeId="2" ParentId="26722" CreationDate="2019-04-17T02:16:42.803" Score="-1" Body="&lt;p&gt;The distance from the nearest support is 1m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distributed load is 3.5kN/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the load is not uniform, you need to calculate the rectangular area load and halve it to obtain triangular load which is half base by height.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$W_0 = 0.5×3.5=1.75kN$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is effectively a point load half-way between D and the nearest support. While I've not calculated the reaction moment, I think you might be able to figure it out.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2019-04-18T22:17:26.473" LastActivityDate="2019-04-18T22:17:26.473" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26933" PostTypeId="1" CreationDate="2019-04-17T08:27:45.040" Score="0" ViewCount="42" Body="&lt;p&gt;Which &lt;strong&gt;90 degree bracket&lt;/strong&gt; would you use for this one? Do you have any suggestions for what i should be looking for, I find it very hard to find something. It is 5cm between all the outer holes.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iOsPW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iOsPW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Would love to use &lt;strong&gt;something like this&lt;/strong&gt;, but that doesn't works for my uneven, off-centre, geared dc motor. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/KaXj7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KaXj7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20143" LastActivityDate="2019-04-17T22:47:29.760" Title="Which mounting bracket to use with a geared box?" Tags="|mechanical-engineering|motors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26934" PostTypeId="1" CreationDate="2019-04-17T08:41:45.440" Score="0" ViewCount="71" Body="&lt;p&gt;I have a tiny reactor, and the volume I need is 425ml. If there is no/few other reactants, then the volume in &lt;code&gt;hydraulic retention time&lt;/code&gt; is clear: reactor/container's volume. and hrt is &lt;code&gt;V/u&lt;/code&gt;(V is 425ml, u is flow rate) But if I fill 80% of container with something like lime mud, the real volume of water is only 20% left -- 85ml. How is my &lt;code&gt;V&lt;/code&gt; should be? 425 or 85ml? IOW, the volume in hydraulic retention time is of reactor/container or liquid?&lt;/p&gt;&#xA;" OwnerUserId="16508" LastActivityDate="2023-12-23T09:06:30.813" Title="the volume in hydraulic retention time is of reactor/container or liquid" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26936" PostTypeId="1" CreationDate="2019-04-17T15:19:31.877" Score="0" ViewCount="166" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D0OuE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D0OuE.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;So hypothetically, I'd like a to build a latching system to prevent a steel block experiencing a horizontal force from moving  until another upwards force releases a latch, at which point it would allowed to move. Not shown, downward force is applied to the latch by a spring.&lt;/p&gt;&#xA;&lt;p&gt;So obviously, this design will not work because releasing the latch would require pushing the steel block in the opposite direction of the force, but there is another object in the way preventing that motion.&lt;/p&gt;&#xA;&lt;p&gt;Perhaps the latch could flex? But that does not seem durable given a large horizontal force?&lt;/p&gt;&#xA;&lt;h3&gt;Is there a &lt;em&gt;simple&lt;/em&gt; soltuion to allow this basic design to work as intended without requiring the moving object to move counter to the resisted horizontal force when the latch is released?&lt;/h3&gt;&#xA;&lt;p&gt;I am not a professional mechanical engineer, so I apologize if I have not worded this in an easily understandable way; please feel free to suggest edits to make things more clear.&lt;/p&gt;&#xA;" OwnerUserId="20150" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-04-17T15:27:06.200" Title="Using a rotational latch to restrain an object under a horizontal force" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26937" PostTypeId="1" AcceptedAnswerId="26946" CreationDate="2019-04-17T15:33:40.040" Score="0" ViewCount="213" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/hP4rg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers often use Finite Element Analysis software to solve for stresses inside complicated structures to generate pictures like the one above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am familiar with FEA as a method to solve differential equations. So what (differential) equation are we using to describe forces/stresses inside objects?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to analyzing forces, I know only Newton's laws. But how can we use the laws in here, when the internal stresses are dependent on the geometry of the object?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2019-04-17T21:14:06.690" Title="What equation are we using to describe forces in FEA?" Tags="|structural-analysis|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26938" PostTypeId="1" CreationDate="2019-04-17T15:47:53.340" Score="1" ViewCount="145" Body="&lt;p&gt;I am trying to design a simple drop / shock testing machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We want to rate our products to be shock rated to 5000g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to test our products to 5000g I plan on dropping a weight onto the product imparting a shock. (See drawing)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sq128.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sq128.png&quot; alt=&quot;Drop Test Rig&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is; How high would I need to drop my 72 kg&#xA;weight from the test piece in order to rate by product to survive a 5000g shock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the impulse time and distance is quite critical in the final numbers regarding the calucations. Lets say for calucation sake, that both materials were to be steel.&lt;/p&gt;&#xA;" OwnerUserId="20151" LastActivityDate="2020-01-12T19:24:40.037" Title="What height is needed to drop a 72 kg mass to impart a 5000g shock onto a test piece" Tags="|product-testing|shock|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26939" PostTypeId="2" ParentId="26938" CreationDate="2019-04-17T16:53:01.803" Score="2" Body="&lt;p&gt;it depends on how far your steal drop table travels after hitting the sample, the less travel the greater g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kinetic energy of the drop table when it hits the sample is &lt;span class=&quot;math-container&quot;&gt;$$KE=1/2mv^2 \quad and\ v=\sqrt{2gh} $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But depending on the softness or hardness of your sample and the apparatus and your weight the force of collision changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_{impact}X_{distance\ traveled}= 1/2mv^2  $$&lt;/span&gt;&#xA;For example if your steel drop weight fall 20cm and travels 2mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v=1.979m/s\ and\ F=(1/2)72*1.979*1000/1mm/m = 71276.3kg\ force $$&lt;/span&gt;&#xA;And if your sample is one gram it will feel&lt;span class=&quot;math-container&quot;&gt;$ \ 71276.3*1000gr-kg/9.8 \ g\ $&lt;/span&gt; acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the 72kg jerks up by the impact you need to add that impulse too.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-01-12T19:24:40.037" LastActivityDate="2020-01-12T19:24:40.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26940" PostTypeId="2" ParentId="26934" CreationDate="2019-04-17T17:26:34.993" Score="0" Body="&lt;p&gt;Water volume is 85 ml now, so the hrt is 85 ml/u. The liquid retention time because the mud doesn't react, right?&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-17T17:26:34.993" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26941" PostTypeId="1" CreationDate="2019-04-17T17:29:00.177" Score="0" ViewCount="11" Body="&lt;p&gt;I know the inlet, throat and outlet areas for a nozzle expander. Is there a formula to know its axial length? Something like L= f(A) = f(D).&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-17T17:29:00.177" Title="Is there a formula to relate the axial length of a nozzle expander with its radius?" Tags="|thermodynamics|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26942" PostTypeId="1" CreationDate="2019-04-17T17:32:19.587" Score="0" ViewCount="170" Body="&lt;p&gt;I was wondering if there is a motor/generator design where the rotor moves gradually in and out of the stator along the axis of rotation, or the stator moves over the rotor.  If so, I'd like some resource(s) to read more about it.  The application I have in mind is for a wildly variable load like a wind turbine. If this is a bad design, please briefly explain why.  Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="20152" LastEditorUserId="9770" LastEditDate="2020-01-13T14:39:46.127" LastActivityDate="2023-12-20T11:11:59.117" Title="Is there a motor design where a rotor gradually moves into a stator?" Tags="|mechanical-engineering|electrical-engineering|motors|electromagnetism|generator|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26943" PostTypeId="2" ParentId="26937" CreationDate="2019-04-17T17:33:21.027" Score="0" Body="&lt;p&gt;The laws implied there are complex equations hard to resolve by hand. The computer divides the realm into several little pieces (elements) and applies the equations for those elements, resolving them by iterating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Finite_element_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Finite_element_method&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the equations for the weak formulation. There's also a strong formulation: &lt;a href=&quot;https://www.researchgate.net/post/What_is_the_difference_between_strong_form_and_weak_form&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.researchgate.net/post/What_is_the_difference_between_strong_form_and_weak_form&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weak one is preferred because integrating is generally easier than derivating for computers.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-17T17:33:21.027" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26944" PostTypeId="2" ParentId="26942" CreationDate="2019-04-17T17:38:39.117" Score="1" Body="&lt;p&gt;That actually exists and it's applied in magnetic levitation (maglev) trains. It's called linear induction motor (LIM). LSM if it's synchronous. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Linear_induction_motor&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Linear_induction_motor&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotor is addressed as &quot;slider&quot; here because it doesn't rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More info here: &lt;a href=&quot;https://e-archivo.uc3m.es/bitstream/handle/10016/13575/PFC_CARLOS_PEREZ_RODRIGUEZ.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://e-archivo.uc3m.es/bitstream/handle/10016/13575/PFC_CARLOS_PEREZ_RODRIGUEZ.pdf&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's in Spanish, but use Google Chrome translator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Download these scientific articles as well: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Study of Non-Symmetric Double-Sided Linear Induction Motor for Hyperloop All-In-One System (Propulsion, Levitation, and Guidance) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Woo-Young Ji 1, Geochul Jeong 2, Chan-Bae Park 1, Ik-Hyun Jo 1, and Hyung-Woo Lee 1  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And some formulas here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyperloop Transportation System: Control, and Drive System Design &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jawwad M. Sayeed  Deperatment of Electrical, Computer, and Software engineering University of Ontario Institute of Technology, Oshawa, Ontario  jawwad.sayeed@uoit.ca &#xA;Ahmed AbdelRahman  Deperatment of Electrical, Computer, and Software engineering, University of Ontario Institute of Technology, Oshawa, Ontario  ahmed.abdelrahman@uoit.ca &#xA;Mohamed Z. Youssef  Deperatment of Electrical, Computer, and Software engineering, University of Ontario Institute of Technology, Oshawa, Ontario  mohamed.youssef@uoit.ca&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the Hallbach Array:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hyperloop Transportation System: Control, and Drive System Design &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jawwad M. Sayeed  Deperatment of Electrical, Computer, and Software engineering University of Ontario Institute of Technology, Oshawa, Ontario  jawwad.sayeed@uoit.ca &#xA;Ahmed AbdelRahman  Deperatment of Electrical, Computer, and Software engineering, University of Ontario Institute of Technology, Oshawa, Ontario  ahmed.abdelrahman@uoit.ca &#xA;Mohamed Z. Youssef  Deperatment of Electrical, Computer, and Software engineering, University of Ontario Institute of Technology, Oshawa, Ontario  mohamed.youssef@uoit.ca&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-17T17:38:39.117" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26945" PostTypeId="1" CreationDate="2019-04-17T18:03:42.060" Score="-1" ViewCount="234" Body="&lt;p&gt;I just work on the engineering company that specialize in water piping. I am confuse about the term of nipple. What is the different between Single Nipple, Double Nipple, and Pipe Nipple? &#xA;Hope somebody can help.&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="12341" LastActivityDate="2021-10-04T07:05:55.570" Title="What Is the Different Between Single Nipple, Double Nipple, and Pipe Nipple" Tags="|mechanical-engineering|piping|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26946" PostTypeId="2" ParentId="26937" CreationDate="2019-04-17T18:44:39.903" Score="1" Body="&lt;p&gt;For structural FEA, there are a few approaches. Most commonly differential equations are developed that relate the deformation state and the stress state of each element in the structure in order to satisfy equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good resource for the details of structural FEA is a book by Ed Wilson Called &#xA;Three Dimensional Static and Dynamic Analysis Of Structures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a chapter that shows the basis of equilibrium and compatibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.edwilson.org/Book/02-equi.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.edwilson.org/Book/02-equi.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**To be more descriptive: For linear elasticity problems that use FEA to find internal forces and deformations, most FEA solvers do not directly solve the the partial differential equations. Instead FEA is used to approximate the solutions to the PDEs. I would like to note that using this method, solutions can be found to problems which are nearly impossible (if not impossible) to write closed formed elastic solutions.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastEditorUserId="19912" LastEditDate="2019-04-17T21:14:06.690" LastActivityDate="2019-04-17T21:14:06.690" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="26947" PostTypeId="2" ParentId="26925" CreationDate="2019-04-17T19:11:06.257" Score="6" Body="&lt;p&gt;Your initial assumption is a bit  flawed. You can not just  draw the conclusion that all friction is the same. It is like saying all birds i know of can fly thus all birds can fly. Which is obviously untrue. Moreover, continuing this similie, you also imply that there is only one kind of flight.&lt;/p&gt;&#xA;&lt;p&gt;In reality friction is a highly complicated phenomena that includes a lot of the things we don't deem worth thinking about in our day to day life. In fact if you have ever seen the formula:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;F_f=  \mu F_n&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;You have seen a physicist saying i have no idea of how it  works but  seen a simple property. Everything in your question is hidden in the symbol &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; and the formula is approximate. Also by  the  way the formula is  actually &lt;span class=&quot;math-container&quot;&gt;$F_f &amp;lt;= \mu F_n$&lt;/span&gt; which  is even less useful to answer the question asked.&lt;/p&gt;&#xA;&lt;p&gt;See, the friction coefficient  is highly dependent on not just the material pair, but also the quality of surface finish, surfaces capability  to carry  the load, how the materials react on the interface, temperature, lubrication and geometry. But it gets better, rolling is way different it has much lower friction so a roller bearing  has  just the fraction of the friction a conventional sliding contact pair. Even better you can have no solid contact  at  all. Typically  this is achieved by lubrication, but there are even bearings that do this naturally like airbearings or magnetic bearings that don't even have touch anything.&lt;/p&gt;&#xA;&lt;p&gt;It is not always even about the efficiency increase. Rather about controlling the wear, or some other  characteristic. See by having  specialized pieces you can change them if damaged, you can spread the load so they last longer, and you can get these cheaper than manufacturing them yourself. The bearing can even have  a function in aligning the shaft properly and thereby reduce vibration.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2020-07-09T06:43:12.680" LastActivityDate="2020-07-09T06:43:12.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26948" PostTypeId="1" CreationDate="2019-04-17T21:54:55.477" Score="1" ViewCount="355" Body="&lt;p&gt;I am trying to design a gear that has a 1:2 speed ratio (input:output).  This design is very limited in space and for simplicity and space savings requires the input and output shafts to be coaxial.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads me to think of using a planetary gearset, but in order to achieve a 1:2 gear ratio, we must use a fixed planet carrier gear, which reverses the direction of the output, which is not desired.  Using a fixed ring or sun is not possible when targeting a 1:2 ratio because the ring and sun would have to be the same size.  Am I missing something, or is there no simple way to make a 1:2 gear ratio with coaxial shafts, or do I require a reverse gear attached to the output of a planetary gear with fixed planet carrier?&lt;/p&gt;&#xA;" OwnerUserId="20154" LastActivityDate="2021-04-01T09:30:04.817" Title="How to design a coaxial 1:2 gear without reversing direction?" Tags="|mechanical-engineering|gears|mechanisms|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26949" PostTypeId="2" ParentId="26945" CreationDate="2019-04-17T22:45:22.227" Score="1" Body="&lt;p&gt;It's really easy:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Single nipple: One end with threads (grooves): &lt;a href=&quot;https://www.google.com/search?q=single+nipple&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google Image Search&lt;/a&gt; (NSFW, due to unintended results)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Double nipple: Both ends with threads (grooves): &lt;a href=&quot;https://www.google.com/search?q=double+nipple&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google Image Search&lt;/a&gt; (NSFW, due to unintended results)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And pipe nipples are like double nipples: &lt;a href=&quot;https://www.google.com/search?q=pipe+nipple&amp;amp;oq=pipe+nipple&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google Image Search&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorUserId="472" LastEditDate="2019-09-15T02:32:40.990" LastActivityDate="2019-09-15T02:32:40.990" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26950" PostTypeId="2" ParentId="26933" CreationDate="2019-04-17T22:47:29.760" Score="1" Body="&lt;p&gt;Couldn't find anything so i recommend you creating a custom bracket. Created this one instead&lt;a href=&quot;https://i.stack.imgur.com/i5M01.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i5M01.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20143" LastActivityDate="2019-04-17T22:47:29.760" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26951" PostTypeId="1" AcceptedAnswerId="26953" CreationDate="2019-04-17T23:23:41.450" Score="0" ViewCount="1735" Body="&lt;p&gt;I have a Ford F350 cab &amp;amp; chassis and a gooseneck trailer. For the gooseneck hitch I want to use a steel plate.&#xA;The steel plate will be BOLTED to the frame, above the rear axle And the gooseneck ball will be welded to the steel plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between the frames is about 34&quot;. So the plate Is going to be 34&quot;long x 8&quot; wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trailer is 16,000 lbs GVWR. And a maximum tongue weight of 4,000 lbs (at 25%)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A regular A36 steel plate, 3/8&quot; thick, will bend under load. (A36 mechanical properties: Brinell=112, Tensile=58,000 psi, Yield=36,000 psi)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm thinking to use a 1/8&quot; thick, high strength steel ASTM 514 / AR400F (Tensile Strength=184,000 psi, Yield Point=150,000 psi, Elongation = 23% in 2&quot;,Brinell Hardness = 360/444)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a 1/8&quot; AR400F steel plate strong enough, for maximum 4,000 lbs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, what do you think about AR400F 3/16&quot; thick ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HvqLm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HvqLm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MT76b.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MT76b.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/R7mzS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R7mzS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/J8Wrc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J8Wrc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20156" LastActivityDate="2019-04-18T23:29:16.147" Title="Is a 1/8&quot; AR400F steel plate strong enough, for 4,000 lbs?" Tags="|structural-engineering|materials|steel|strength|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26952" PostTypeId="2" ParentId="26951" CreationDate="2019-04-18T01:40:31.003" Score="0" Body="&lt;p&gt;What do you mean &quot;bend&quot;? Permanent plastic deformation or: deflect under load and spring back when the load is removed? The elastic deflection is the same for both steels (at the same thickness). The high strength steel will resist plastic/permanent deformation better but I can't calculate the thickness. However, when the specification says the A 514 is weldable, they mean when a welding engineer develops a procedure (low hydrogen, etc.) for a welder experienced with the material. So, if you are welding it may be tricky and even torch-cutting should be done with caution. Some middle ground is ASTM A 516, or A 515 very common, weldable plates that come at a few strength levels between A 36 and A 514. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="1832" LastEditDate="2019-04-18T02:31:44.593" LastActivityDate="2019-04-18T02:31:44.593" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26953" PostTypeId="2" ParentId="26951" CreationDate="2019-04-18T05:16:02.117" Score="0" Body="&lt;p&gt;I could see a moment when an overloaded trailer with defective brakes is rushing into the cabin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to consult the trailer's hitch manual, or a professional engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My estimate is you need a minimum 5/8&quot; thick plate with manufacturer recommended geometry and anchored at a the correct distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a safety critical motion-dynamics problem which needs due deliberation as to the possible forces which may be encountered during acceleration, turning, hitting a pothole, braking, jackknife possiblity, limping due to unbalanced load, and many more issues. It's not just a motorcycle trailer.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="16486" LastEditDate="2019-04-18T23:29:16.147" LastActivityDate="2019-04-18T23:29:16.147" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26954" PostTypeId="1" CreationDate="2019-04-18T10:18:37.307" Score="1" ViewCount="110" Body="&lt;p&gt;If I were to look at &quot;normal&quot; batteries online, they often have charging C-rates of less than 1. However, with a car that uses regenerative braking, obviously such a low charging rate would not be acceptable, as this would mean very little energy can be recovered by this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, F1 cars can deploy 160 horsepower for approximately 33 seconds per lap from the battery system (4MJ). Given Lithium-Ion batteries have a specific energy of up to 0.875 MJ/kg, this means the 20kg battery can only have a maximum possible capacity of 17.5 MJ. Therefore, a deployment of 4MJ in 33 seconds would mean a discharge C-rate of around 25. This is without even considering the extreme charging C-rate that would be required to recharge the battery during such heavy braking (little time) as in F1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, my question is how electric/hybrid car batteries can handle the charging/discharging demands of regenerative braking?&lt;/p&gt;&#xA;" OwnerUserId="20042" LastEditorUserId="20042" LastEditDate="2019-04-22T18:07:57.800" LastActivityDate="2019-04-22T18:07:57.800" Title="How can batteries in electric/hybrid cars charge at such a high rate?" Tags="|automotive-engineering|battery|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26955" PostTypeId="2" ParentId="26954" CreationDate="2019-04-18T11:53:48.773" Score="2" Body="&lt;p&gt;The &quot;batteries&quot; you find online are probably lithium cells.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The batteries in electric cars are not single cells. Instead they combine thousands of cells in a configuration with some of them parallel to each other (for higher current) and others in series (for higher voltage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that each single cell is only providing a fraction of the total power and each cell only gets a fraction of the total power input when charging (either by being plugged in of regenerative braking).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note: to keep the batteries viable for the expected lifetime of the car the charge controller only uses a small part of the total capacity of the batteries because fully charging or fully discharging the battery will decrease the lifetime of the battery. That longevity range is around 50% charged. &lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2019-04-18T11:53:48.773" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="26956" PostTypeId="1" AcceptedAnswerId="27035" CreationDate="2019-04-18T13:17:03.173" Score="2" ViewCount="1707" Body="&lt;p&gt;Risers are added reservoirs used to feed liquid metal to solidifying casting. In the case of alloys that shrink during solidification (this is not the case of all alloys), one uses such reservoirs to make liquid metal available throughout solidification. Risers should be kept liquid throughout the solidification, which means the solidification must be directed toward the risers, which must solidify last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What are the principles used to design such risers?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I already know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Their volume should be sufficient to compensate the solidification shrinkage&lt;/li&gt;&#xA;&lt;li&gt;They should remain liquid last, so maximizing the volume/surface ratio seems appropriate&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I am looking for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How to ensure the riser can efficiently feed the rest of the cast part?&lt;/li&gt;&#xA;&lt;li&gt;Are there other rules (some kind of general algorithm maybe) for efficiently designing risers?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="20155" LastActivityDate="2019-04-22T21:06:41.760" Title="What are the design principles for risers in sand casting?" Tags="|materials|manufacturing-engineering|casting|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26957" PostTypeId="2" ParentId="26906" CreationDate="2019-04-18T13:33:51.800" Score="0" Body="&lt;p&gt;Hydraulics get complicated fast, the beaty of weirs (that's the function of your bulkhead hear) is that you simplify the hydraulics, as you separate the flow upstream of the weir from the flow conditions downstream the weir: As long as water level downstream is a bit below the crest, flow conditions upstream can be calculated backwards from the weir withour regarding downstream conditions. This is why you can build flow distributing structures as per your drawing: Flow into one compartment depends only on weir length (again, only so long as the water level downstream is below the weir crest). You &quot;pay&quot; for this with hydraulic losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your guesses is correct, provided two things:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the wall partinioning B &amp;amp; C is higher than the resulting water level in those compartments.&lt;/li&gt;&#xA;&lt;li&gt;There's enough distance between upstream of the bulkhead for the flow to evenly distribute, or the flow velocity is low (how much distance? How low? I don't know. You want even distribution along the bulkhead). Given the shape of your flume upstream, I'd assume that you are good.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If gate C1 is closed (effectively shutting off the water to that pipe), and the downstream gate C2 is closed, what percentage of the water goes into Pipe A and B? (My guess is A 33% and B 66%)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reservoir C will fill up to the water level upstream of the bulkhead, then flow over the bulkhead into the compartments A and B will be proportional to bulkhead lengths. IF the bulkhead between B and C is lower than that level, you will have an additional flow into compartment B, to be calculated by the &lt;a href=&quot;https://engineering.stackexchange.com/a/8039/61&quot;&gt;weir formula&lt;/a&gt;. You need to check the hydraulic losses for the pipes from reservoir A and B and calculate the water levels in A and B at the relevant flow rates.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If gate C1 is closed (effectively shutting off the water to that pipe), and the downstream gate C2 is open, what percentage of the water goes into Pipe A and B? (My guess is A 25% and B 75%)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;IF the hydraulic losses from the pipes A or B are so high that the water level rises above weir crest, you'd need to account for a lower discharge coefficient and the math gets hariy fast.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-04-18T13:33:51.800" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26961" PostTypeId="2" ParentId="26912" CreationDate="2019-04-18T22:42:26.217" Score="0" Body="&lt;p&gt;The Samsung Galaxy note 8 and later used a stylus with a rare earth magnet... which interferes with electrostatic fields.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-04-18T22:42:26.217" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26962" PostTypeId="2" ParentId="26850" CreationDate="2019-04-18T22:46:35.983" Score="0" Body="&lt;p&gt;Caravans use friction stays to hold their windows open. You might be able to obtain them and use a couple to keep the barn doors open with slight modification. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-04-18T22:46:35.983" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26964" PostTypeId="1" CreationDate="2019-04-18T23:16:19.867" Score="1" ViewCount="418" Body="&lt;p&gt;I’m working on a project for my home gym. I want to be able to drop a loaded barbell without bothering my downstairs neighbors. My main approach is to use 2 blocks of dense foam on which to drop the barbell (when loaded with weights, barbells have 2 contact points to the ground). Ideally the foam absorbs the full impact of the weights, letting no vibration travel through the the floor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’ve learned that dense foam is expensive! A 4” tall block of decent foam &lt;a href=&quot;http://www.usafoam.com/mattress/memorytopper.html#Memory4&quot; rel=&quot;nofollow noreferrer&quot;&gt;costs $106&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To help me determine what material to buy, is there an equation that calculates how much force foam of a given density and thickness can absorb, without any force passing through to the floor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some initial values:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Assume barbell weighs 250 lbs, so 125 lbs dropping on each foam block &lt;/li&gt;&#xA;&lt;li&gt;Barbell is dropped from 72 inches above foam. &lt;/li&gt;&#xA;&lt;li&gt;Foam comes in densities of 0-8 lbs per cubic ft, escalating in price substantially. &lt;/li&gt;&#xA;&lt;li&gt;The contact point between barbell and foam pad is the width of the weight plates, an average of 7 square inches. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I’m looking for a way to find the optimal trade off between foam density and thickness. For example, a dense foam costs more, but requires less volume to absorb impact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any assistance!&lt;/p&gt;&#xA;" OwnerUserId="20163" LastActivityDate="2019-05-19T13:51:47.180" Title="How much foam is required to fully absorb the impact of dropping a barbell?" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26965" PostTypeId="2" ParentId="24670" CreationDate="2019-04-18T23:21:42.503" Score="0" Body="&lt;p&gt;In my opinion, your motor is 250W/12V as almost no 36V motor. If your motor is 250W/12V, then your battery connection will be parallel, and that will give you 12V/15Ah. By theory, your motor will finish your battery is 0.72hour or in 43.2 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind that 12V/5Ah is mean, if you have device with 12V and 1A voltage and current requirement, then the such device will finish the power source (battery) exactly one hour). Then, the label 250W/12V in a device, mean that it needs 12V power source and it will finish 250W power in one hour. By so, the current it required is 250W/12V=20.8333A. So, if now your batteries are connected parallel, then they will have 12V/15Ah. With so, then the 250W/12V motor will finish the 12V/15Ah batteried (with parallel connection) in (12V*15A)/250W=0.72 hour or 43.2 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your motor really 250W/36V, then your batteries need to be serial connection, and it will give you 12V/5Ah=60Watt. Then your 36V/5Ah will be finish (12V*5A)/250W=0.24H or 14.4 minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But all of this just a theory. The fact, you have to consider loses of the motor. You have to consider loses of the battery. You have to consider drag coefficient of the the bicycle to the road. Also, the weigh of the the bicycle itself.&lt;/p&gt;&#xA;" OwnerUserId="17943" LastActivityDate="2019-04-18T23:21:42.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26966" PostTypeId="2" ParentId="26956" CreationDate="2019-04-19T01:07:36.857" Score="2" Body="&lt;p&gt;Casting terminology can vary a bit. My experience is in art casting in cast iron. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience with sand casting &lt;em&gt;risers&lt;/em&gt; specifically tend to be used as vents to allow gas to escape from the mould and tend to be small diameter and placed at high points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm more used to seeing the term &lt;em&gt;well&lt;/em&gt; used in the context of feeding metal. There is no reason not to use risers as feeders of course although this sometimes conflicts with the fact that the can be in awkward places where you want to keep them as thin as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general terms a casting will cool from the thinnest sections first so its the thicker bits which need feeding. In many cases it is also convenient to have a well which feds the mold indirectly form the side or below which helps to control the flow in the mould as a whole as it is filling from head pressure rather directly from a falling  stream of metal straight from the ladle this also helps to settle out any air and solid debris which gets mixed up in the initial stream . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in general &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;You want to be feeding the thickest parts of the mould&lt;/li&gt;&#xA;&lt;li&gt;Transitions between feeders and the mould (gates) should be thinner than both but short, a 'fishtail' shape is often desirable with the flare on the pattern side. &lt;/li&gt;&#xA;&lt;li&gt;The typical section size of the feeder should be larger than the section it feeds. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In terms of an intuitive approach try to imagine the mould cooling from thin to thick and look for islands of thick sections of metal, it is these which will need feeding if they aren't already fed from the main well. You can think of this as the thinner sections 'sucking' metal from the rest of the mould as they solidify. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in pattern design try to avoid abrupt changes of section &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2019-04-19T01:07:36.857" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26967" PostTypeId="2" ParentId="26964" CreationDate="2019-04-19T01:08:04.877" Score="0" Body="&lt;p&gt;Calculate the kinetic energy of the barbell before falling and get more info on the foam. When it deforms, how much energy does it absorb? That's a paremeter called resilience [J/mm3]. Search that parameter first or get data from similar or generic foams.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-19T01:08:04.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26968" PostTypeId="2" ParentId="26827" CreationDate="2019-04-19T03:47:24.580" Score="0" Body="&lt;p&gt;There are many ways. But, all of them are either trivial or subjective.&#xA;The most fair way I have found so far is to use Interactive Genetic Algorithm (IGA). IGA uses the human subjectivity to lead the GA optimization. &lt;a href=&quot;https://ieeexplore.ieee.org/abstract/document/8396849/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is a research performing all you need for Model Predictive Control (MPC) [the same as LQR but respecting constraints too]. In your case, the constraints are offline. In this work, a multi-objective GA is used and applied constraints on the results. Then I got a huge pareto-front and applied an IGA to get the best result out of them which looks appealing to a human. You can directly apply IGA without a multi-objective GA too. But, a GA before IGA does a great filtering. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have access to IEEE, follow the chapter 4 of my &lt;a href=&quot;http://dro.deakin.edu.au/eserv/DU:30112367/mohammadi-enhancinghuman-2018.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;thesis&lt;/a&gt; (sorry for the goatic font. I ran out of the symbols).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can apply constraints either during optimization (at a GA before IGA) or in real-time (MPC instead of LQR). In my case, both of the constraints are applied.&lt;/p&gt;&#xA;" OwnerUserId="14288" LastEditorUserId="14288" LastEditDate="2019-04-19T03:56:22.453" LastActivityDate="2019-04-19T03:56:22.453" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26969" PostTypeId="1" CreationDate="2019-04-19T05:03:49.857" Score="0" ViewCount="65" Body="&lt;p&gt;I need to get a piece of stiff material cut and drilled for a project I'm working on.  It has the following requirements:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1.6mm thick&lt;/li&gt;&#xA;&lt;li&gt;should be stiff (similar to PCB material, aluminum, fiberglass)&lt;/li&gt;&#xA;&lt;li&gt;about 12&quot; x 3&quot;&lt;/li&gt;&#xA;&lt;li&gt;6 holes need to be drilled (all around 5mm in diameter)&lt;/li&gt;&#xA;&lt;li&gt;will be making ~500 of these plates&lt;/li&gt;&#xA;&lt;li&gt;material type is not important - just needs to be about as stiff as PCB material/fiberglass&lt;/li&gt;&#xA;&lt;li&gt;conductivity doesn't matter - metal or non-metal is fine - my PCB will have plastic board supports spacing it off of the mounting surface.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out the best material that will yield the best pricing were I to outsource this to a machine shop or manufacturer of some kind.  In other words, what combination of material and machining process will cost the lowest.  I was looking at having a PCB manufacturer use PCB material (because this will be the mounting plate for my PCB and they'd already have all the tooling), but I'm open to use fiberglass, metal, plastic, or any other material.  I was hoping that someone here might have done projects with various machine shops or manufacturers and be able to provide some recommendations as to the best materials and processes to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to provide a few more details - I am manufacturing a PCB and I need a mounting plate on which to mount DIN rail clips.  So, the PCB will have board supports (which allow for a loose-ish fit) which hold it to a mounting plate (of some material type), and the mounting plate will have DIN rail clips attached to it.  The reason for the plate is to provide something to which the DIN rail clips can mount to.  I do not want the PCB directly attached to DIN rail clips because I don't want twisting forces applied to the PCB during DIN rail mounting/removal.&lt;/p&gt;&#xA;" OwnerUserId="20164" LastEditorUserId="20164" LastEditDate="2019-04-19T15:05:38.723" LastActivityDate="2019-04-20T02:00:25.923" Title="What's the best material to use for my project" Tags="|manufacturing-engineering|metals|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26970" PostTypeId="1" CreationDate="2019-04-19T05:36:00.340" Score="1" ViewCount="664" Body="&lt;p&gt;You are comparing different A/D converters that will have 3.2 V and -3.2 V as their positive and negative reference voltages, respectively. Out of the following, converter &lt;strong&gt;_ can resolve the smallest change in voltage, which is approximately_&lt;/strong&gt;  . (m = milli and u = micro)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A) 16-bit, 100 kHz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B) 8-bit, 1 MHz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C) 15-bit, 512 kHz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(D) 19-bit, 256 kHz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(E) 17-bit, 330 kHz&lt;/p&gt;&#xA;" OwnerUserId="20165" LastActivityDate="2020-05-18T06:03:35.343" Title="How to find the smallest change in voltage" Tags="|electrical-engineering|computer-engineering|computer-hardware|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26971" PostTypeId="2" ParentId="26964" CreationDate="2019-04-19T06:20:25.747" Score="0" Body="&lt;p&gt;No foam or any rubber gym mat will be able to help you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Too much weight to drop 6 feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just have to lower your weight gently with the help of a friend spotting you on a steel frame with as many grab steps as your workout possibly requires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That frame should be supported on a gym hard rubber sheet. like everyone who works out at home. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could even be held liable for structural or property damage.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-19T06:20:25.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26972" PostTypeId="1" CreationDate="2019-04-19T08:21:16.553" Score="0" ViewCount="120" Body="&lt;p&gt;I tried proving the condition dG&amp;lt;=0 where G is Gibb's free energy and instead ended up proving dB&amp;lt;=0 where B is exergy. Can someone please explain what went wrong with my derivation. I have attached the photo of derivation below. Thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit 1: I looked up on the net and it said exergy and Gibb's free energy are same at NPT ensemble. I cannot understand how when both the type of energies use different Temperatures in there formula. Can someone also give an explanation for this??&#xA;&lt;a href=&quot;https://i.stack.imgur.com/CLF0F.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CLF0F.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19256" LastEditorUserId="19256" LastEditDate="2019-04-20T05:48:39.583" LastActivityDate="2019-04-20T05:48:39.583" Title="Doubt in proof of condition dG&lt;=0" Tags="|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26973" PostTypeId="2" ParentId="26838" CreationDate="2019-04-19T09:25:39.697" Score="1" Body="&lt;p&gt;Okay, one program I found that can help calculate the different properties of a section is &quot;ShapeDesigner SaaS&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the reason that I am not doing it manually is that I would have to break it into at least 35+ parts. This program calculates it in a few moments. Hope this helps some people. &lt;/p&gt;&#xA;" OwnerUserId="20081" LastActivityDate="2019-04-19T09:25:39.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26974" PostTypeId="2" ParentId="26954" CreationDate="2019-04-19T10:25:33.353" Score="0" Body="&lt;p&gt;What matters is the stability and longevity of the electrolyte system in the batteries. Even wet lead batts can be dump charged at a fantastic rate (c 100 or so) if their state of charge is low and they are manufactured to high tolerances. But the rate of charge acceptance that doesn't cause damage drops of sharply starting around 80% charge. From a design point of view, Charge acceptance is controlled by surface area of the elecrtolyte/anode/cathode interface. It is also limited by internal heat generation and heat dissipation. Most electric vehicles use fan-cooled batteries with sophisticate cooling. So the first item is physical design - surface area, heat dissipation, and eletrolyte stability enhancements such as gels and glass mats in the case of lead batts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second item is the controller or BMS. A sophisticated BMS will monitor each cell individually and regulate the charge rate of each individually. This can be several hundred cells in an electric car. This greatly enhances the overall battery lifespan when pushing it through hard charge/discharge cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third item is the choice of battery chemistry. There are dozens in production, and hundreds that have been tried. Some are just better at taking a high charge rate or work better in the configurations used in high charge rate batteries. Lion is well known for taking a high charge rate up to 99% of fully charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd also like to comment of the regenerative braking ability of cars. Most can't and don't need to regenerate a very large proportion of their capacity. The average stop will store about 1/2 mile's worth of cruise power. Some cars (and trains) use a special section of the battery just to handle small amounts of high rate regeneration. These supercapacitors then bleed the power into the rest of the battery at a slower rate. These tend to be found in lead-acid traction batteries, but are being researched for other chemistries as well.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-04-19T10:25:33.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26975" PostTypeId="1" CreationDate="2019-04-19T12:31:48.673" Score="0" ViewCount="60" Body="&lt;p&gt;How will we find the overall drag coefficient for a steady flow which is laminar initially and then becomes turbulent after a critical distance?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7C1gV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7C1gV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19256" LastEditorUserId="19256" LastEditDate="2019-04-20T06:00:16.607" LastActivityDate="2019-04-20T06:00:16.607" Title="Finding Drag coefficient for flows with both Laminar and Turbulent part" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26976" PostTypeId="1" AcceptedAnswerId="26977" CreationDate="2019-04-19T13:18:19.163" Score="0" ViewCount="76" Body="&lt;p&gt;I am writing my first 3D cad-file and I want to place into it a &lt;a href=&quot;https://en.wikipedia.org/wiki/Turnstile&quot; rel=&quot;nofollow noreferrer&quot;&gt;turnstile mechanism&lt;/a&gt; (a mechanism that allows a fixed axis to rotate only one way, and not in the opposite). Are there cad-templates for devices I can find in some online-platform for engineers? I found &lt;a href=&quot;https://grabcad.com/library/turnstile-mechanism/details?folder_id=130515&quot; rel=&quot;nofollow noreferrer&quot;&gt;this cad-file&lt;/a&gt;, though it looks too large and complicated, I want to place it in a device that is 20cm*20cm*20cm. As an engineer, where would you look for a template of a device you want to create physically? Or any suggestions how to create my own simple turnstile cad-file?&lt;/p&gt;&#xA;" OwnerUserId="20173" LastEditorUserId="20173" LastEditDate="2019-04-19T14:23:20.823" LastActivityDate="2019-04-19T14:23:20.823" Title="Cad file for turnstile mechanism/ one-way clutch/ one-way rotation?" Tags="|mechanical-engineering|cad|bearings|3d-printing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26977" PostTypeId="2" ParentId="26976" CreationDate="2019-04-19T14:22:33.757" Score="0" Body="&lt;p&gt;I found what I was looking for in this &lt;a href=&quot;https://www.youtube.com/watch?v=UIhCPl8eb7s&quot; rel=&quot;nofollow noreferrer&quot;&gt;video&lt;/a&gt; - apparently what I need is a Sprag Clutch or a Roller Clutch. The authour of the video provides his 3d-cad files to a Roller Clutch &lt;a href=&quot;https://makersmuse.podia.com/roller-clutch&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, although they cost all together 4$. Anyways, I purchased them, because it would save me a lot of time and it looks like a decently working clutch.&lt;/p&gt;&#xA;" OwnerUserId="20173" LastActivityDate="2019-04-19T14:22:33.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26978" PostTypeId="1" CreationDate="2019-04-19T15:35:52.033" Score="1" ViewCount="104" Body="&lt;p&gt;I am trying to calculate the exact rolling resistance (&lt;span class=&quot;math-container&quot;&gt;$F_{rr}$&lt;/span&gt;) of my car. The approach I am going with is a coastdown test at a low speed so no drag slows down the car. From there I am going to measure the speed of the car several times and find the acceleration and therefore &lt;span class=&quot;math-container&quot;&gt;$F_{rr}$&lt;/span&gt;. With the help of an online polynomial regression, I am going to find the function of &lt;span class=&quot;math-container&quot;&gt;$F_{rr}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am not sure if my approach is correct and if the function that I am going to find is indeed capable to give me the &lt;span class=&quot;math-container&quot;&gt;$F_{rr}$&lt;/span&gt; at higher speeds.&lt;/p&gt;&#xA;" OwnerUserId="20175" LastEditorUserId="1832" LastEditDate="2019-04-19T21:55:38.027" LastActivityDate="2023-12-27T09:04:55.540" Title="Calculate Rolling Resistance Experimentally" Tags="|mechanical-engineering|automotive-engineering|applied-mechanics|wheels|deformation|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26979" PostTypeId="1" CreationDate="2019-04-19T18:45:04.657" Score="0" ViewCount="29" Body="&lt;p&gt;What is the range of compressive strength of carbon nanotube reinforced polymers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although CNT's themselves have phenomenal (oriented) compressive strength, and one can find references to that, CNRP's incorporate CNTs in the polymer matrix and polymers are not known for their compressive strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot of materials data on various carbon nanotube reinforced polymers but it all seems focused on applications where compressive strength isn't limiting. &lt;/p&gt;&#xA;" OwnerUserId="19896" LastActivityDate="2019-04-19T18:45:04.657" Title="What is the compressive strength of carbon nanotube reinforced polymer?" Tags="|materials|carbon-fiber|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26980" PostTypeId="1" CreationDate="2019-04-19T19:42:37.843" Score="1" ViewCount="102" Body="&lt;p&gt;I come from a fairly remote island where we need a 150 min ferry ride to reach the mainland.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While the island has few resources that would encourage a government to investigate the possibility of a large bridge such as the Denmark-Sweden one, there is one resource that we have in abundance - wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to combine a mega-bridge with an offshore wind-farm? With turbines in the bridge towers. Possibly also wave-generators underneath? Or is this just fantasy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is a realistic possibility, then maybe, just maybe, we could at least get someone to consider this rather than laugh-out any suggestion of connecting us to the mainland.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shortest distance is 23-24 km. An alternative would be a total distance of 29Km, but hitting a small island after about 20Km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect it's just fantasy, but it'd be good to know.&lt;/p&gt;&#xA;" OwnerUserId="20176" LastActivityDate="2019-08-09T21:20:47.953" Title="Can a bridge's towers be used as wind turbines?" Tags="|bridges|renewable-energy|wind-power|offshore|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26981" PostTypeId="2" ParentId="26980" CreationDate="2019-04-19T20:06:51.313" Score="1" Body="&lt;p&gt;One huge safety issue would be protection against turbine blade failure. There is no realistic way to catch and restrain a broken piece of blade, and apart from any loss of life during the incident itself, there would be a risk of the bridge being unusable for a long time while repairs to the damage were carried out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.telegraph.co.uk/news/uknews/8948363/1500-accidents-and-incidents-on-UK-wind-farms.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Accident statistics&lt;/a&gt; for UK wind farms published in 2011 showed a total of 1500 reported incidents, four deaths, and 300 injuries, over a 5 year period. That may be &quot;acceptable&quot; for people who are voluntarily employed in a high risk industry, but it certainly isn't an acceptable risk for the general public, especially if they are in effect &lt;em&gt;required&lt;/em&gt; to use the bridge regularly for transportation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a more general regulatory issue here, since in the UK wind farms are classed for accident reporting purposes as &quot;machines&quot; not as &quot;buildings,&quot; and therefore any design regulations relating to the general public's access to such structures may be non-existent.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-04-19T20:06:51.313" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26982" PostTypeId="2" ParentId="24747" CreationDate="2019-04-20T00:16:14.187" Score="0" Body="&lt;p&gt;You can measure this with a piezoelectric transducer and an oscilliscope. Several companies sell thin film piezo transducers that you can just stick onto your device under test. The transducer has leads that you would connect to a piezo pre-amplifier which will amplify and buffer the signal coming from your DUT. Then attach a scope lead to the output of the pre-amp and you can see the signal on your scope.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2019-04-20T00:16:14.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26983" PostTypeId="1" CreationDate="2019-04-20T00:52:01.700" Score="0" ViewCount="24" Body="&lt;p&gt;Equilibrium concentration of point defects in solids is defined as &lt;span class=&quot;math-container&quot;&gt;$\chi=\exp(\frac{\Delta s}{k_B})\exp(\frac{-\Delta h}{k_BT})$&lt;/span&gt;. At high temperatures, may be even higher than melting temperature of the solid, does the entropy become undefined meaning there will be no equilibrium concentration for point defects? &lt;/p&gt;&#xA;" OwnerUserId="20178" LastEditorUserId="20178" LastEditDate="2019-04-20T01:51:16.553" LastActivityDate="2019-04-20T01:51:16.553" Title="Equilibrium concentration of point defects in solids at sufficiently elevated temperature" Tags="|materials|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26984" PostTypeId="2" ParentId="26972" CreationDate="2019-04-20T01:08:13.963" Score="1" Body="&lt;p&gt;You have assumed constant temperature and pressure conditions for your system. Also it is implied from your derivation that there is no mass transfer or diffusion. This is an NPT ensemble which achieves equilibrium by minimizing Gibbs Free Energy, &lt;span class=&quot;math-container&quot;&gt;$\Delta G=\Delta H-T \Delta S$&lt;/span&gt; and looks like that is what you derived. Note that thermodynamic potentials are defined such that they minimize the total energy content of the system under various physical conditions (constant N, P &amp;amp; T for this case gives rise to &lt;span class=&quot;math-container&quot;&gt;$\Delta G$&lt;/span&gt; while &lt;span class=&quot;math-container&quot;&gt;$\Delta$&lt;/span&gt; H is defined for a system with constant N,P &amp;amp; S).&lt;/p&gt;&#xA;" OwnerUserId="20178" LastActivityDate="2019-04-20T01:08:13.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26985" PostTypeId="2" ParentId="26969" CreationDate="2019-04-20T02:00:25.923" Score="2" Body="&lt;p&gt;If you wanted to do 50,000 units, the answer is almost certainly injection molding of some kind.  You could make what you are describing easily for probably pennies per piece... but it would cost quite a bit to set up the mold.  That's probably not worth the setup cost just for 500 pieces.  I think I would just go to mcmaster.com and search their plastics section to see what is lowest cost. Buy some big sheets and have a machine shop slice them into 3&quot; sections with a band saw. Looks like you could get the raw material in acrylic for less than $1 per piece.  That should be easily enough to drill holes in and should be stiff enough. There might be cheaper options, I didn't spend too much time looking. Not sure how much the machine shop would charge for the manufacturing, but it shouldn't be too much.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-04-20T02:00:25.923" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="26986" PostTypeId="1" CreationDate="2019-04-20T05:39:37.757" Score="0" ViewCount="59" Body="&lt;p&gt;I’ve seen folks submerge computers in mineral oil to cool them- but would doing the same to a video projector profoundly throw off its internal optical aspects? I’m considering one of those small “hand held” projectors and having it project from inside an aquarium. I expect it to look cool but will it work?&lt;/p&gt;&#xA;" OwnerUserId="20181" LastActivityDate="2019-04-20T05:39:37.757" Title="Will a video projector submerged in mineral oil still work?" Tags="|fluid-mechanics|materials|design|heat-transfer|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26987" PostTypeId="1" CreationDate="2019-04-20T06:55:14.593" Score="2" ViewCount="98" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xdfFl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xdfFl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;A typical radial piston rotor can have simple holes machined from the outside through the inner diameter to create flow channels for hydraulic fluid to actuate the pistons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across this model and I was wondering, how would something like this get machined?&lt;/p&gt;&#xA;" OwnerUserId="20182" LastEditorUserId="20182" LastEditDate="2019-04-20T12:26:53.060" LastActivityDate="2020-01-22T08:00:36.757" Title="how are these radial piston rotors machined" Tags="|mechanical-engineering|hydraulics|pistons|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26989" PostTypeId="1" CreationDate="2019-04-20T16:16:37.883" Score="1" ViewCount="578" Body="&lt;p&gt;I have a rectangular metal plate with a hole in it (with diameter of 300 mm). The plate has a temperature gradient going from one of its short sides to the other, I can measure the temperature anywhere in the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to compute how much the hole is deforming from its round shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that I can calculate the expansion of the hole at uniform temperature with &lt;span class=&quot;math-container&quot;&gt;$\frac{ΔL}{L_0}=αΔT$&lt;/span&gt; . So I was wondering if it is correct to measure a bunch temperature points at the edge of the hole and just apply that formula to each of them independently. But since there's a temperature gradient I'm assuming that there will be mechanical stresses between hot and cold zones working against the expansion, is this the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to compute this by hand?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture &lt;span class=&quot;math-container&quot;&gt;$T1&amp;gt;T2$&lt;/span&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/HEPvd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HEPvd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18413" LastEditorUserId="18413" LastEditDate="2019-04-20T22:12:32.310" LastActivityDate="2019-05-21T01:02:00.527" Title="Thermal expansion of a hole in a plate with a temperature gradient" Tags="|materials|stresses|thermal-expansion|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="26990" PostTypeId="2" ParentId="26987" CreationDate="2019-04-20T17:04:08.977" Score="0" Body="&lt;p&gt;They have lathes with special tools which can hold the odd parts firmly to be worked on. Watch around minute 1:25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=4mhT1a28qO0&quot; rel=&quot;nofollow noreferrer&quot;&gt;you tube drilling a hole in a shaft.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-20T17:04:08.977" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26991" PostTypeId="1" CreationDate="2019-04-20T18:16:38.417" Score="1" ViewCount="58" Body="&lt;p&gt;What is meant by thermodynamic efficiency of an engine?&#xA;and what are the factors that affect the thermodynamic efficiency of a diesel engine?&lt;/p&gt;&#xA;" OwnerDisplayName="user19926" LastActivityDate="2019-04-21T01:12:36.520" Title="Efficiency of an engine" Tags="|energy-efficiency|engines|diesel|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26993" PostTypeId="2" ParentId="26991" CreationDate="2019-04-20T19:08:30.453" Score="0" Body="&lt;p&gt;The thermodynamic efficiency of an engine is the energy output divided by the energy input. For a car engine for example,  the output is the shaft horsepower (torque times RPM) and the input is the heat added by burning fuel inside the engine. A low efficiency engine burns a lot of fuel and produces a little horsepower (it wastes the heat input); a high efficiency engine burns a little fuel and produces a lot of horsepower (it makes good use of the heat input). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different operating cycles (otto, brayton, stirling, etc.) exhibit different thermodynamic efficiencies depending on how they are designed and operated, but all of these are limited to being less than the theoretical maximum possible thermodynamic efficiency of any heat engine, which is called the &lt;em&gt;Carnot efficiency&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-04-20T19:08:30.453" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="26994" PostTypeId="2" ParentId="26991" CreationDate="2019-04-20T22:00:20.817" Score="1" Body="&lt;p&gt;It's basically: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \eta = \frac {Brake power (kW)}{\frac{dm_f}{dt} \cdot  LCV}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where m_f is the fuel mass that enters the engine and its temporal derivative the mass flow [kg/s]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LCV is the Low Calorific Value [kJ/kg], it depends on the fuel used. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorUserId="4816" LastEditDate="2019-04-21T00:44:28.500" LastActivityDate="2019-04-21T00:44:28.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26995" PostTypeId="1" CreationDate="2019-04-20T22:26:40.490" Score="0" ViewCount="53" Body="&lt;p&gt;Air bearings are a cheap solution for Hyperloop, unlike the costly magnetic levitation system. I'm consulting some sources and it seems there's enough air in the tube at 100 Pa to feed the air bearings. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-20T22:26:40.490" Title="Why have some Hyperloop researchers abandoned the air bearings initial idea?" Tags="|bearings|air|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="26996" PostTypeId="2" ParentId="26978" CreationDate="2019-04-20T22:56:28.247" Score="0" Body="&lt;p&gt;After obtaining a mean value, compare it with &lt;span class=&quot;math-container&quot;&gt;$F_{rr} = \mu W$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt; is the weight of the car and &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; the static friction coefficient. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorUserId="1832" LastEditDate="2019-04-21T22:09:19.023" LastActivityDate="2019-04-21T22:09:19.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="26997" PostTypeId="2" ParentId="26989" CreationDate="2019-04-20T23:03:42.823" Score="-1" Body="&lt;p&gt;Assuming the plate is insulated on the sides and around the hole, only open at two ends, and assuming is settled into a stationary heat profile, yes you can corollate your temperature contour line with the expansion in the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It takes some time for the expansion of the plate and deformation of the circle into a kind of egg shape to settle. Then stresses redistribute and relax.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reminds me of a jobsite I had where the stock pile of steel beams heaped on top of each other would groan at sunrise or when the clouds opened up because of the squeezing stresses due to expansion, but shortly thereafter things became quiet down again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the stationary state, if we assume the warping of surface limited to small angles, &lt;span class=&quot;math-container&quot;&gt;$$\frac{\Delta L_{I,j,k}}{L_{i,j,k}} =\alpha\cdot\Delta T  $$&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-04-21T00:24:00.503" LastActivityDate="2019-04-21T00:24:00.503" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="26998" PostTypeId="1" CreationDate="2019-04-21T00:34:55.380" Score="0" ViewCount="31" Body="&lt;p&gt;This may be difficult to answer but, theoretically, if you had a car with a sterling engine that produces say approximately 300 horsepower how much of any specific heat source would need to be used if the car had a mass of 1.5 tonnes in order for the car to travel 100 km. Furthermore, what would a combustion engine that produces the same power in a car that has the same body and mass use in terms of fuel compared to the sterling engine? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the requirements for a sterling engine to perform at the same level as a combustion engine in an automobile?&lt;/p&gt;&#xA;" OwnerUserId="20177" LastEditorUserId="20177" LastEditDate="2019-04-21T01:05:41.907" LastActivityDate="2019-04-21T01:05:41.907" Title="PLEASE ANSWER Sterling vs Combustion Engine" Tags="|mechanical-engineering|motors|automotive-engineering|engines|fuel-economy|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="26999" PostTypeId="2" ParentId="26991" CreationDate="2019-04-21T01:12:36.520" Score="0" Body="&lt;p&gt;Thermal engines take in Q amount of heat energy out of a fuel and convert part of it as mechanical work and part of it as wasted heat back to the environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  Q_{in}=W_{out} +Q_{out}$$&lt;/span&gt;&#xA;The thermal efficiency is the percentage of heat energy taken in to work put out.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$  \eta= \frac{W_{out}}{Q_{in}}=\frac{Q_{in}-Q_{out}}{Q_{in}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \eta=1-\frac{Q_{out}}{Q_{in}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The efficiency of even the best heat engines is low; usually below 50% and often far below. So the energy lost to the environment by heat engines is a major waste of energy resources, Since a large fraction of the fuels produced worldwide go to powering heat engines, perhaps up to half of the useful energy produced worldwide is wasted in engine inefficiency, ..&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quoted from Wikipedia page engine efficiency.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-21T01:12:36.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27000" PostTypeId="2" ParentId="63" CreationDate="2019-04-21T04:11:16.163" Score="2" Body="&lt;p&gt;Roundabouts let traffic flow more smoothly and there is also less reliance on technology that can malfunction due to a series of factors (i.e. power outage, etc).&lt;/p&gt;&#xA;" OwnerUserId="20177" LastActivityDate="2019-04-21T04:11:16.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27001" PostTypeId="2" ParentId="23749" CreationDate="2019-04-21T04:14:06.630" Score="1" Body="&lt;p&gt;In order to limit noise from the appliances that are being used (i.e. shower) as well as to stop any steam from showers escaping and causing fire alarms to be set off. Furthermore, the use of double-layered doors limits any odour produced in the bathroom from escaping into other rooms.&lt;/p&gt;&#xA;" OwnerUserId="20177" LastActivityDate="2019-04-21T04:14:06.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27003" PostTypeId="1" CreationDate="2019-04-21T07:06:13.663" Score="0" ViewCount="846" Body="&lt;p&gt;I recently bought a 2020 profile and after I got the product I realised that I needed a 2040 profile instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I was thinking of buying one more 2020 profile and connecting it with the 2020 profile parallel to each other to make it a 2040 profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find a way I can do this using a joining plate so that I can easily assemble and disassemble it whenever required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone can guide how? and where? I can buy the parts it would be great.&lt;/p&gt;&#xA;" OwnerUserId="20184" LastActivityDate="2019-04-23T23:26:35.857" Title="Joining two 2020 aluminium profiles to make a 2040profile" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27005" PostTypeId="1" CreationDate="2019-04-21T08:55:34.037" Score="0" ViewCount="31" Body="&lt;p&gt;I am working on a 500 kg roller, which I want to pull using electric motor. I am considering hub motor, since it is easier to frame a basic ev with it. Now I need to know the capacity of the hub motor, I need to choose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roller weight = 500 kg&#xA;Frame weight + driver weight = 150 kg&#xA;speed required = 10 km/hr&#xA;Surface on which the roller is to run is predominantly flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things I need to know,&#xA; 1. Torque requirement for my problem.&#xA; 2. motor wattage &#xA; 3. battery requirement, to run for 2 hours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being new to such design projects, I may have missed details. Please ask me for them. Help me with the calculation.&lt;/p&gt;&#xA;" OwnerUserId="20186" LastActivityDate="2019-04-21T08:55:34.037" Title="ev hub motor spec, for pulling roller" Tags="|electric-vehicles|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27006" PostTypeId="1" CreationDate="2019-04-21T12:42:48.297" Score="-1" ViewCount="203" Body="&lt;p&gt;If I am having a fifth order system,  it is said that the dominant two poles nearer to the jw axis  determine the nature of the output, is there any criterion to neglect the other three poles , should we neglect them even if they are just to the left of the dominant poles... &lt;/p&gt;&#xA;" OwnerUserId="20185" LastActivityDate="2023-01-01T10:02:55.287" Title="Criterion to neglect poles of higher order" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27007" PostTypeId="1" CreationDate="2019-04-21T12:43:10.230" Score="0" ViewCount="38" Body="&lt;p&gt;In modeling a wing with the finite element method, I am asked to consider an engine (approximated as a lumped force and moment) and it's mass, inertia, prestress and gravity contributions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the engine contribution (in terms of mass and inertia) to the Virtual Work is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta \mathcal{W}_{engine} = - \delta s^{T} M \ddot{s} - \delta \theta^{T} J \ddot{\theta}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; is the displacement of the engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I add the prestress and the pendulum-like contribution?&lt;/p&gt;&#xA;" OwnerUserId="20189" LastEditorUserId="4753" LastEditDate="2019-06-13T23:51:17.813" LastActivityDate="2019-06-13T23:51:17.813" Title="Pendulum-like effect" Tags="|applied-mechanics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27008" PostTypeId="1" CreationDate="2019-04-21T17:29:10.727" Score="2" ViewCount="264" Body="&lt;p&gt;What is the significance of asymptotes in the root locus of a linear time invariant system. Does it say that for that particular pole on which it is starting doesn't have a practical pole in the closed loop if the open loop gain is taken to infinity? &lt;/p&gt;&#xA;" OwnerUserId="20185" LastEditorUserId="19776" LastEditDate="2019-04-22T03:47:34.357" LastActivityDate="2023-07-05T01:06:35.750" Title="Significance of asymptotes in root locus" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27009" PostTypeId="1" CreationDate="2019-04-21T18:14:56.247" Score="0" ViewCount="2325" Body="&lt;p&gt;I've built &lt;a href=&quot;https://www.appropedia.org/Open-source_syringe_pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; syringe pump following those instructions. Now I would need how do I calculate the output pressure given the step rate of the stepper motor, and vice-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if it is helpful anyway but I know that the relationship between the torque exerted by the motor and the linear force upon the syringe plunger is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M = F r_a \tan(\alpha), \; \tan(\alpha) = \frac{p}{2 \pi r_a}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$r_a$&lt;/span&gt; is the radius of the threaded rod, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; the thread angle and &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; the pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M = \frac{F p}{2 \pi} = \frac{P r^2 p}{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the syringe radius. Is it right until now? Moreover, is it any useful? I found myself unwillingly stuck in this project but I never studied such cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that the exerted torque depends on the load so I wouldn't know how to link output liquid pressure with motor step rate.&lt;/p&gt;&#xA;" OwnerUserId="20190" LastEditorUserId="1832" LastEditDate="2019-04-21T18:20:50.737" LastActivityDate="2019-04-23T12:32:06.417" Title="Calculate output pressure of a syringe pump" Tags="|fluid-mechanics|stepper-motor|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27010" PostTypeId="1" CreationDate="2019-04-21T20:58:25.360" Score="0" ViewCount="17" Body="&lt;p&gt;I'm trying to find a source for this type of plastic rack gear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mniJF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mniJF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image came from &lt;a href=&quot;https://www.youtube.com/watch?v=9ErxoOQGvm4&quot; rel=&quot;nofollow noreferrer&quot;&gt;this youtube video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can find short pieces of straight rack gear on ebay, but nothing like this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to be a vinyl/plastic type material, has 90-degree mounting flange, and is toothed on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any information would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="1558" LastActivityDate="2019-04-21T20:58:25.360" Title="Source for plastic rack gear strips" Tags="|gears|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27011" PostTypeId="1" AcceptedAnswerId="27016" CreationDate="2019-04-22T00:26:31.603" Score="2" ViewCount="116" Body="&lt;p&gt;In the engineering material book. It shows the following figure&#xA;&lt;a href=&quot;https://i.stack.imgur.com/B5ht0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B5ht0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is the text to depict the picture&#xA;&quot;The origin of rubber elasticity is more difficult to picture than that of a crystal or glass. The long molecules, intertwined like a jar of exceptionally long worms, from entanglements—points where molecules, because of their length and flexibility, become knotted together (Figure 25.6).&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I still don't know the point and line.&lt;/p&gt;&#xA;" OwnerUserId="17605" LastActivityDate="2019-04-22T01:56:23.043" Title="What is the meaning of line and points in the schematic of a polymer" Tags="|structural-engineering|materials|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27012" PostTypeId="2" ParentId="27006" CreationDate="2019-04-22T00:38:55.060" Score="0" Body="&lt;p&gt;The dominant pole approximation makes analysis much easier. Basically, you choose the poles closest to the j&lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; axis because they have small real parts, so they will be slower to decay and therefore dominate the response (Analogous to a rate limiting step in chemistry). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of any hard and fast rule for which poles to neglect, it depends on the application. If possible it's probably best to experiment numerically. If you're a student then in the case of a class-work, if you're dealing only with 2nd order systems then it makes sense to drop the other 3, even if they're not too far from the two dominant poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can read more here:&#xA;&lt;a href=&quot;https://lpsa.swarthmore.edu/PZXferStepBode/DomPole.html##section8&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://lpsa.swarthmore.edu/PZXferStepBode/DomPole.html##section8&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20193" LastActivityDate="2019-04-22T00:38:55.060" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27013" PostTypeId="2" ParentId="27011" CreationDate="2019-04-22T00:44:04.543" Score="2" Body="&lt;p&gt;The lines are long polymer chains. They contain C, H, O, sometimes F, Cl, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The points are the knots that tie the chains together. They are normally carbon or hydrogen atoms shared by two chains. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-22T00:44:04.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27014" PostTypeId="1" CreationDate="2019-04-22T00:52:43.930" Score="0" ViewCount="34" Body="&lt;p&gt;In Hyperloop air bearings, the air escapes through the rear cross-sectional area of each ski, you know? Yes, there is a pressure differential, but the air comes out of the skis, crashes into the tube and then is accelerated to a certain speed below 760 mph/1,220 kph and escape. The air cannot escape through the front cross-sectional area because of the ski high speed and for the same reason they cannot escape through the lateral areas (well, there will be some minor leaks). That's Couette-Poiseuille flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the mass flow is rho&lt;em&gt;A&lt;/em&gt;c&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where c is the speed the air is accelerated to, rho is the density, and A the rear cross-sectional area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think? &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-22T00:52:43.930" Title="Is this is a valid estimation for the mass flow required for Hyperloop air-bearings?" Tags="|airflow|bearings|compressed-air|air|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="27015" PostTypeId="2" ParentId="27008" CreationDate="2019-04-22T00:55:10.003" Score="0" Body="&lt;p&gt;The aymptotes show the tendency of the pole movements in open loop. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-22T00:55:10.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27016" PostTypeId="2" ParentId="27011" CreationDate="2019-04-22T01:56:23.043" Score="2" Body="&lt;p&gt;The &lt;strong&gt;dots&lt;/strong&gt; represent &lt;strong&gt;atoms&lt;/strong&gt; (as Dibujo de Croquis pointed out, they can be C, H, O, F, Cl, etc.), and the &lt;strong&gt;lines&lt;/strong&gt; represent that there's a &lt;strong&gt;chemical bond&lt;/strong&gt; between two atoms, as exemplified by this picture: &lt;a href=&quot;https://i.stack.imgur.com/X2c90.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X2c90.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Your picture can be particularly confusing because:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;It just draws dots for the atoms for the branches of each monomer, and skips the atoms that make the chain itself, this is standard practice (but you may imagine an atom the end of each line segment, usually carbon).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It decides to suddenly stop drawing those atoms at the branches near the center of the picture. They do that because they want to show how it is tangling in the middle, so in order to get a cleaner picture, they just decided to spot drawing those atoms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It includes some dash lines between the atoms in the branches. This is to show that, in spite of not having a strong chemical bond, there's some weak interaction between this atoms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18413" LastActivityDate="2019-04-22T01:56:23.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27017" PostTypeId="2" ParentId="7965" CreationDate="2019-04-22T03:23:06.563" Score="0" Body="&lt;p&gt;To put it simply the use of camera's will work for 99% of the time but they can still fail in certain scenarios (i.e. extreme weather, malfunction, object blocking/obstructing camera, etc) whereas mirrors can be used in any situation furthermore they reduce the level of distraction. &lt;/p&gt;&#xA;" OwnerUserId="20177" LastActivityDate="2019-04-22T03:23:06.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27018" PostTypeId="2" ParentId="7965" CreationDate="2019-04-22T04:51:45.240" Score="0" Body="&lt;p&gt;Aside from other good reasons already mentionned :&#xA;Because the eye &amp;amp; brain have a better dynamic range than most mainstream cameras &lt;/p&gt;&#xA;" OwnerUserId="5926" LastActivityDate="2019-04-22T04:51:45.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27019" PostTypeId="1" CreationDate="2019-04-22T08:00:31.800" Score="0" ViewCount="213" Body="&lt;p&gt;I am very much confused finding the time constant and DC gain for the below step response. The idea is to find PI gains for speed control and I am following this tutorial online &lt;a href=&quot;http://ctms.engin.umich.edu/CTMS/index.php?aux=Activities_DCmotorB&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tutorial Link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motor specifications&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KP9Oz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KP9Oz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is the open loop step response when a 10 V is applied across the DC motor. The battery is of 26 V, which means that a PWM duty cycle of 38% or 0.38 (10/26). The X-axis is time(seconds) and the y-axis is Gear box output shaft RPM. Initially it stayed at 100 RPM and then it fell down to 50RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x3BoH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x3BoH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As explained in the tutorial link provided above,I am trying to find a first order transfer function between PWM Duty cycle and the output shaft angular velocity. What should be my DC gain ? Currently i am applying only 0.38(38%) duty does the time constant change if applied 1 (100 %) duty cycle ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new to this topic let me know if any further information is needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="20194" LastActivityDate="2024-03-26T10:03:40.387" Title="How to tune PID for a step response" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27020" PostTypeId="1" CreationDate="2019-04-22T11:02:42.633" Score="0" ViewCount="165" Body="&lt;p&gt;We have a home, which built before 50 years, now we are planning to built one more home on top [ terrace] , Is it right to go ahead &amp;amp; built new home on terrace ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or do we need to check any other things before building new home ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the current home is 50 years old, will it &lt;strong&gt;survive&lt;/strong&gt; if we build new home on top ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our Main intention is to keep the terrace &lt;strong&gt;secure&lt;/strong&gt;.... &lt;/p&gt;&#xA;" OwnerUserId="20196" LastActivityDate="2019-04-24T20:45:42.893" Title="Build new home on top of 50 years old house" Tags="|home-improvement|" AnswerCount="3" CommentCount="4" ClosedDate="2019-04-24T21:46:07.780" ContentLicense="CC BY-SA 4.0" />
  <row Id="27021" PostTypeId="2" ParentId="27020" CreationDate="2019-04-22T12:08:02.973" Score="2" Body="&lt;p&gt;Yes, you should check that the foundation and intermediate floors will carry the extra load.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-04-22T12:08:02.973" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="27022" PostTypeId="1" CreationDate="2019-04-22T13:30:59.493" Score="-1" ViewCount="38" Body="&lt;p&gt;My front gate is 13 feet  wide and length is 20 feet  and the road is 11 feet wide I have a Scorpio SUV  can it be park in my house&lt;/p&gt;&#xA;" OwnerUserId="20198" LastActivityDate="2019-04-22T15:48:32.137" Title="Help me to park my car" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ClosedDate="2019-04-22T17:55:59.093" ContentLicense="CC BY-SA 4.0" />
  <row Id="27023" PostTypeId="2" ParentId="27022" CreationDate="2019-04-22T14:54:00.600" Score="0" Body="&lt;p&gt;It can be a big challenge. Definitely is much smaller than standard, 9 meters which is approximately 30 feet, square turning area by our city code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Backing into gate may be a bit easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try on a open area with blocks used as your width and corners of the gate.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-04-22T15:03:54.643" LastActivityDate="2019-04-22T15:03:54.643" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27024" PostTypeId="2" ParentId="27020" CreationDate="2019-04-22T15:02:04.447" Score="1" Body="&lt;p&gt;You have to investigate your existing building and compare it to the local codes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may turn out that the expenses of upgrading the existing foundation and walls to the new code to make them safe for a second story, are too high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the cities have free or reasonably priced initial engineering consultation in their building permit department. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the comments by the OP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can only tell you the steps needed to be taken in California, but it may be similar at your place.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;you go to your building department with a very rough sketch of what you plan to build.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;They send a geologist or technician to pick up available documents and data about your project and visit your home.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;he may give you a report of allowable soil bearing and drainage codes, etc, or he may decide that you need to provide a geology- soils engineering report.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Soil engineer asks you for a more refined plan of your second floor reviews those and digs trenches to collect samples and give a report addressed to the city as to competency of your soil and details of modification to the foundations and bearing walls and seismic provisions, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You hire an architect and engineer to design the addition. Or you can prepare the plans and specifications by certain home designers if the existing foundation and first story deemed to be strong enough.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In many cases the modifications of the existing building to make it strong for additional support for second floor prove to be more expensive than demolition and new construction of a two story building. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-04-24T20:45:42.893" LastActivityDate="2019-04-24T20:45:42.893" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27025" PostTypeId="1" CreationDate="2019-04-22T15:16:42.600" Score="0" ViewCount="1414" Body="&lt;p&gt;I'm confused as to what is the difference between the above said stresses and their corresponding strains. If possible can anybody explain in detail what is engineering stress&lt;/p&gt;&#xA;" OwnerUserId="20200" LastActivityDate="2020-04-29T10:41:00.193" Title="Engineering stress and true stress" Tags="|mechanical-engineering|stresses|solid-mechanics|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="27026" PostTypeId="2" ParentId="27022" CreationDate="2019-04-22T15:48:32.137" Score="0" Body="&lt;p&gt;If SUV has a turning radius of 5.6m =18.5 ft then the road is not wide enough or you need a curved driveway across property to reach gate 20' from road.&lt;/p&gt;&#xA;" OwnerUserId="8847" LastActivityDate="2019-04-22T15:48:32.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27027" PostTypeId="2" ParentId="27019" CreationDate="2019-04-22T16:03:25.653" Score="0" Body="&lt;p&gt;You cannot control speed by PWM without RPM feedback since speed varies with load.&lt;/p&gt;&#xA;" OwnerUserId="8847" LastActivityDate="2019-04-22T16:03:25.653" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27028" PostTypeId="2" ParentId="27025" CreationDate="2019-04-22T16:11:15.447" Score="2" Body="&lt;p&gt;A tensile test of a steel rod is typically used to demonstrate the difference. Necking occurs due to the lateral contraction according to poissons ratio. Because of this, the cross section changes in area. Because the load is constant, the true stress is increases in the area where necking occurs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers base calculations off of the original cross section to keep things uniform.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-22T16:11:15.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27029" PostTypeId="1" CreationDate="2019-04-22T17:01:37.513" Score="0" ViewCount="118" Body="&lt;p&gt;I am looking for feedback on my camera rig design. I am going to run a track along the ceiling of my studio to move my camera back in forth. I also want the camera to be able to pan, so I want to incorporate one of my spare nema 17 stepper motors. Below is a rough sketch of my proposed design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A couple concerns I have:&#xA;1. Is this too much axial load on the stepper motor shaft? The camera weighs about 2.5 lbs including the housing that will attach to the flange.&#xA;2. The flange is attached to the stepper motor by a set screw. I do not want the safety of my camera to rely on a set screw as this is mounted from the ceiling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any design suggestions would be very helpful. Thank you!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ASYjt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ASYjt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19052" LastActivityDate="2023-12-28T03:01:50.203" Title="Ceiling Mounted Camera Slider with Pan" Tags="|mechanical-engineering|stepper-motor|linear-motion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27030" PostTypeId="2" ParentId="27009" CreationDate="2019-04-22T17:08:00.257" Score="1" Body="&lt;p&gt;The question asks how to derive output pressure from step-rate.&lt;br&gt;&#xA;The answer does not perform all the calculations but attempts to define the variables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Minimum required specs:   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;s = Step incr. [m] (after gear reduction)&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;(°/step * mm/rev(wormgear) * 360°/1000[mm/m])   &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;F = minimum stepper force [kg] ( &lt;a href=&quot;https://www.orientalmotor.com/stepper-motors/technology/speed-torque-curves-for-stepper-motors.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;that must exceed load to prevent skipping at speed&lt;/a&gt;)   &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Torque * gear Radius, r [m] = F [kg]  (converted from N-m)&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;The minimum stepper force, F must exceed syringe input load force &#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;F=Pi/Area to avoid skipping&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Pi  = inlet pressure&lt;br&gt;&#xA;Po = outlet pressure&lt;br&gt;&#xA;L = length of tube&lt;br&gt;&#xA;η  = viscosity&lt;br&gt;&#xA;R = radius&lt;br&gt;&#xA;V = volume of the fluid at outlet pressure&lt;br&gt;&#xA;v = velocity of the fluid at outlet pressure  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;repeat for each concentric stage&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Flowrate = \dfrac{π\cdot r^4 \cdot (P-Po) }{ 8\cdot η\cdot L}  [m^3/s]&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reference &lt;a href=&quot;https://www.physicsforums.com/threads/flow-rate-of-a-syringe.854294/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.physicsforums.com/threads/flow-rate-of-a-syringe.854294/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If below output is into air, can we assume P1=P2?  NO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be better to measure Flow rate with Force input for target syringe then define desired flow rate it is capable of then determine gear ratio and motor torque and step rate needed to perform task..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nfEfh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nfEfh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet considering the optimization of stepper speed is a controlled acceleration to prevent skipping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;If you have the open source GRBL Panel software (S/W) with these variables in an Arduino CNC shield, you can maximize velocity then control a ratio of this for flow rate, with this S/W better than anything else, that I have come across.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="8847" LastEditorUserId="8847" LastEditDate="2019-04-23T12:32:06.417" LastActivityDate="2019-04-23T12:32:06.417" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="27031" PostTypeId="2" ParentId="27025" CreationDate="2019-04-22T17:51:00.180" Score="1" Body="&lt;p&gt;In many ductile alloys like A36 steel, the real stress versus strain curve is different by the fact that as gradually the test machine increases the tension they measure the new reduced area, due to the Poisson effect, of the sample and calculate the stress based on this reduced area. &lt;span class=&quot;math-container&quot;&gt;$$\sigma= \frac{Tension}{A_{reduced}}$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in engineering stress strain curve they assume the area of the sample stays constant as the original area and graph the apparent stress versus strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore the engineering, apparent, plot bends down after a stage called hardening while the real graph keeps rising until the rupture point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Her is a comparison graph from Wikipedia page on Stress versus strain of A36 steel which is similar to many other ductile alloys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W97jm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W97jm.jpg&quot; alt=&quot;stress strain by Slashme (david Richfield) Wikipedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graph by, Wikipedia: [User:Slashme] (David Richfield) - Own work&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-22T17:51:00.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27032" PostTypeId="2" ParentId="27020" CreationDate="2019-04-22T17:56:18.930" Score="1" Body="&lt;p&gt;Please consider this as a major modification to your building structure. In my area, there are laws that govern this type of modification and require that the design be performed by a qualified professional engineer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very unlikely that your house will tolerate this modification as is. An engineer will do an in depth review of the building, and work with you and your design team to determine what will be required to make the change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many possibilities to make it work, but it will be constrained by your budget and the ability of the construction team.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*The civil / structural engineers will likely want a full assesment of the existing structure. Types and sizes of materials used, method of construction. For the soil they may not need to perform any test, but a geotechnical assesment of your site is highly advised (this will make sure there are no problems with the site and provide for a more optimized design of any new foundations that need to be built.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend having a contractor (or someone very knowledgeable about construction) on your design team as early as possible. Much of what can be done will hinge on the ability of the constructors.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-22T17:56:18.930" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27033" PostTypeId="2" ParentId="27029" CreationDate="2019-04-22T17:56:32.423" Score="0" Body="&lt;p&gt;Have the stepper motor turn a worm and a gear with the shaft (with a decent fixing) supported on a bearing to hang the camera from. The worm and gear will also give more control over the rotation of the motor relative to the camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IQsee.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IQsee.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-04-22T18:01:45.417" LastActivityDate="2019-04-22T18:01:45.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27034" PostTypeId="1" CreationDate="2019-04-22T19:09:55.503" Score="0" ViewCount="38" Body="&lt;p&gt;So we are trying to build an AC generator, an alternator, if you will. We are using a 5&quot; PVC pipe for the body with a hole drilled diametrically, through which a bolt passes. Inside of this pipe, on the bolt, we have a six pole rotor magnet, which rotates as we rotate the bolt. We then have 18 gauge, plastic coated wire wrapped around the pipe. We used a drill to spin the magnet at respectable speeds, but were not able to get any electrical outputs. Are there any glaring issues in this design? If so, how can we address them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first image is of the inside of the generator, showing the rotor magnet (apologies for the image). The second is of the outside of the generator, showing the wire wrappings. The bare copper wire is shown there to test an unprotected wire and is not used in the actual design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yn4gpm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yn4gpm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I4XIjm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I4XIjm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14362" LastEditorDisplayName="user14362" LastEditDate="2019-04-24T14:24:29.247" LastActivityDate="2019-04-24T14:24:29.247" Title="Issues with Building AC Generator" Tags="|generator|ac|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27035" PostTypeId="2" ParentId="26956" CreationDate="2019-04-22T20:53:46.203" Score="2" Body="&lt;p&gt;I am a metal casting process researcher who has several past and ongoing collaborations with commercial foundry and tooling experts. In commercial steel sand casting, I have observed the terms &quot;riser&quot; and &quot;feeder&quot; to be used more-or-less interchangeably. In the literature, British commonwealth authors tend to prefer feeder, while American authors tend to prefer riser. I will use feeder because it is a more descriptive term. When I use the term &quot;section&quot; I mean a contiguous region of the casting. The term is left vague because it is not a particularly well-defined term in practice. When I use the term &quot;hotspot&quot; I mean the last part of a section or feeder to solidify.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The Seven Feeding Rules&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;John Campbell defines seven rules for feeder design in &lt;em&gt;Complete Casting Handbook&lt;/em&gt; (located on p. 661-688 of 2004 edition).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;strong&gt;Do not feed unless necessary&lt;/strong&gt; &amp;mdash; only attach feeders to locations that have demonstrated shrinkage problems. Demonstration could take the form of simulation results. Prefer to use thin sections which taper away from the gates so feeding is done by the filling system. Ignoring this requirement increases the cost of production by consuming more liquid metal per casting. It is worth noting that this is not always possible nor the most economical route.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Heat transfer requirement&lt;/strong&gt; &amp;mdash; The feeder must solidify later than the section it is intended to feed. Influenced by feeder size and shape, and feeder-mold interfacial heat transfer properties. Ignoring this requirement can result in a feeder freezing before the section, leaving the section without feed metal.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Mass transfer requirement&lt;/strong&gt; &amp;mdash; The feeder must have sufficient &lt;em&gt;useful&lt;/em&gt; volume to meet or exceed the contraction of the section it is intended to feed. Useful volume means the amount of liquid available to feed the section that isn't consumed by direct solidification of the feeder itself, or by thermal contraction. Influenced by all of the same factors as (2.). Ignoring this section can result in the feeder freezing before it has supplied sufficient feed metal to the section. The geometry of the section determines which of rules (2.) and (3.) is a stricter requirement. Heat transfer (2.) tends to be more important for thick sections, and mass transfer (3.) for thin sections. The reason is that thin sections will rapidly solidify, well before attached risers, so that useful feeder volume becomes the dominant factor.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Junction requirement&lt;/strong&gt; &amp;mdash; The point of attachment between feeder and section can become a hotspot. If this occurs, porosity will form at the attachment point. What results is referred to as &quot;under-riser shrinkage porosity&quot;. Increasing the size of the feeder, or changing its location relative to the section, can help resolve this issue. A common rule is to use a feeder whose diameter (via Heuvers' circles or spheres) or section modulus (care of Chvorinov, Wlodawer, et al.) is 1.0 to 1.2 times that of the section. For sufficiently thick sections as much as 2.0 times may be necessary to avoid junction issues.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Feed path requirement&lt;/strong&gt; &amp;mdash; There must be a clear feeding path available from the feeder to every point in the section intended to be fed. If there is a narrow feature between the feeder and some part of the overall section, then that narrow feature will solidify before the section, cutting off feeding. The rest of the section will be isolated. To achieve this, ensure there is a monotonically increasing gradient of modulus or section thickness from most extreme points back to the feeder. Attaching excess metal in the form of feedpads can assist with this requirement. Feeding distance is limited by both feeder geometry and section geometry. Tapering sections can increase the feeding distance. Attempting to feed too far can result in microscopic shrinkage called center-line shrink or microshrink.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pressure gradient requirement&lt;/strong&gt; The feeder must have higher metallostatic pressure than the section intended to be fed, or the porosity will form in the section by the action of gravity. Ensuring the feeder is located at a higher point than the attached section can help, but is not sufficient. Vented feeders exert more pressure because they are not working against atmospheric pressure &amp;ndash; atmosphere supports up to 1.5 meters of steel. The metallostatic pressure gradients change over the course of solidification as sections become cut off from one another. All of these changes must be accounted for.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Pressure requirement&lt;/strong&gt; The feeder must have sufficient metallostatic pressure to ensure that points in the attached section are above points in the feeder. If the liquid level in the feeder drops below an un-solidified part of the attached section, that part will experience shrinkage porosity. Whereas the pressure gradient requirement considers the entire course of solidification, the pressure requirement is concerned with a single feeder-section system once it has become isolated from the rest of the casting.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h2&gt;Additional considerations&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Other literature sources give additional criteria. Some of these are obvious, some follow from the seven rules above, and a couple are less obvious.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Locate feeders near thick sections&lt;/li&gt;&#xA;&lt;li&gt;Place side feeders on gates&lt;/li&gt;&#xA;&lt;li&gt;Place top feeders away from gates&lt;/li&gt;&#xA;&lt;li&gt;Do not cluster feeders (minimum distance)&lt;/li&gt;&#xA;&lt;li&gt;Blind feeders height:diameter in range of 1:1 and 3:1&lt;/li&gt;&#xA;&lt;li&gt;Feeders located on flat, accessible surfaces to ease removal and minimize risk of damage to the casting&lt;/li&gt;&#xA;&lt;li&gt;Select feeder shapes from a list of standards&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are sourced from a Master's Thesis by Jensen, 1998, who cited:&#xA;- S Guleypoglu and J L Hill AFS Transactions, 1995&#xA;- J L Hill, et al., AFS Transactions, 1991&#xA;- J L Hill, et al., AFS Transactions, 1993&#xA;- B Ravi and M N Srinivasan, Computer-Aided Design 22(1), 1990.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Numerical models&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A few mathematical and geometric models exist that can guide engineers toward better solutions. These models tend to center around requirements (2.) and (3.), with a little bit into (4.) and (5.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two models that are still widely used 70 years after their invention are &lt;a href=&quot;https://www.giessereilexikon.com/en/foundry-lexicon/Encyclopedia/show/heuvers-circle-method-4649/?cHash=8b1a894ab6cde93d18d8ce7070c9b2c7&quot; rel=&quot;nofollow noreferrer&quot;&gt;Heuvers' circles&lt;/a&gt; and &lt;a href=&quot;https://www.giessereilexikon.com/en/foundry-lexicon/Encyclopedia/show/chvorinovs-rule-4627/?cHash=2d0751fb7828d4549443fd324577ef4f&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chvorinov's rule&lt;/a&gt;. Heuvers' circles are used to ensure that the diameter of the largest inscribed circle or sphere in a feeder is larger than that of the attached section. Chvorinov's rule is based on experimental work that shows solidification time is proportional to &lt;span class=&quot;math-container&quot;&gt;$\left(\frac{V}{A}\right)^{n}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is often taken to be &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;. If the modulus of a feeder is larger than its attached section, then it should solidify later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wlodawer and Heine independently published work extending Chvorinov's rule to individual parallelpiped sections of castings. They each noticed that castings can be broken up into sections with shared boundary segments. The value of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; for each section can then be discounted by shared area since very little heat is transferred across shared boundaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorelmetal's Ductile Iron book &lt;a href=&quot;http://www.sorelmetal.com/en/publi/Gating-risering/Gating-Risering.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;(pdf link)&lt;/a&gt; and the Redbook Rules from Steel Founders' Society of America &lt;a href=&quot;http://user.engineering.uiowa.edu/~becker/documents.dir/redbook.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;(pdf link)&lt;/a&gt; both deal directly with sizing feeders based on geometric considerations. The models are too involved to share here, but are based on practical and empirical considerations from experts in the field.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What about computer models?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I am unaware of any commercially available computational models that can accurately guide an engineer through every aspect of feeder design. For what it's worth, &lt;a href=&quot;https://www.magmasoft.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;MAGMASoft&lt;/a&gt; is a casting process simulation software capable of both solidification simulation and parametric geometry optimization based on simulation results. So if you have a &quot;close-enough&quot; design, it can optimize the geometry to minimize defects and maximize process efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting that &quot;close-enough&quot; design is the part that still largely relies on human judgement and experiential knowledge. While there are a number of mathematical and geometric models (described above), none of them are perfect or cover all of the above requirements. Furthermore, manual calculations associated with these methods are time-consuming. They can also be error-prone if used by engineers unfamiliar with the effects of geometry on feedpaths. There is at least one open-source program for automatically placing feeder geometries based on (5.) &lt;a href=&quot;https://github.com/wwarriner/casting_geometric_toolsuite&quot; rel=&quot;nofollow noreferrer&quot;&gt;(GitHub link)&lt;/a&gt;. Full disclosure: I am the author of that program. The original intent with the software is to divide the casting into feed-able sections using a watershed approach and then determine feeder size based on section geometry. It works reasonably well, and when we have an open source solidification solver it will be that much more accurate.&lt;/p&gt;&#xA;" OwnerUserId="421" LastEditorUserId="421" LastEditDate="2019-04-22T21:06:41.760" LastActivityDate="2019-04-22T21:06:41.760" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27036" PostTypeId="1" CreationDate="2019-04-22T20:54:54.117" Score="0" ViewCount="1663" Body="&lt;p&gt;If a pipe (of unknown/arbitrary length) with cross section area &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and with a continuous supply of a simple fluid with a known pressure &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; at the opening to the atmosphere, which has a pressure of &lt;span class=&quot;math-container&quot;&gt;$p_a$&lt;/span&gt;, what would be the rate of flow out of the pipe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked but not found an answer anywhere, but apologies if this seems trivial or seems to be a duplicate of another question. &lt;/p&gt;&#xA;" OwnerUserId="20202" LastEditorUserId="20202" LastEditDate="2019-04-22T23:27:38.510" LastActivityDate="2023-12-29T07:08:25.967" Title="Open pipe and relation between pressure and flow rate" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27037" PostTypeId="1" AcceptedAnswerId="27038" CreationDate="2019-04-22T21:40:13.983" Score="1" ViewCount="619" Body="&lt;p&gt;So I'm given this question with the following diagram:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two sections of a pipe are joined in a flanged joint as shown in the Figure below. The difference in pressure from interior to exterior is 2 MPa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The joint is connected with 12 off bolts, equally pitched around a pitch circle diameter of 350 mm. If the maximum allowable tensile stress on the bolts is 300 MPa, what diameter do the bolts need to be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lIh8H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lIh8H.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My working out&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qdYTc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qdYTc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I find the area of the section that the bolt's diameter takes up:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A = \pi (0.175+r)^2 - \pi(0.175-r)^2 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A = 0.7\pi r\ $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I find out the force acting on that area of section and divide by twelve to obtain the force for a section that has one bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_{12} = (2\times 10^6)(0.7\pi r) = 14 \times 10^5 \pi r$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_1 = 366519.1429r$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then sub this into the stress equations since I know the maximum tensile stress for 1 bolt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A = {F_1 \over \sigma_{max}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pi r^2 = {366519.1429r \over 300 \times 10^6}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Diameter = 0.039 m$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the answer I'm given says it is supposed to be 5.9mm. Is there anything wrong in my thinking?&lt;/p&gt;&#xA;" OwnerUserId="20204" LastEditorUserId="20204" LastEditDate="2019-04-22T21:52:13.427" LastActivityDate="2019-04-22T21:58:36.650" Title="How to find the diameter of bolts in a flanged joint?" Tags="|structural-engineering|structural-analysis|stresses|bolting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27038" PostTypeId="2" ParentId="27037" CreationDate="2019-04-22T21:44:07.983" Score="2" Body="&lt;p&gt;Your calculation for the maximum forces to the bolt appears flawed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forces resisted by the bolts in tension is equal to the total force of the area on either side of the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have the tube diameter and thickness, you can calculate the interior area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple the pressure by this area to find the total force trying to pull these two sides apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide that by 12 and you have your force in one bolt. Then divide by the allowable bolt stress to get the minimum required area and then diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I do this I get 5.8mm this is the root area (doesn't include the threads).&lt;/p&gt;&#xA;" OwnerUserId="19912" LastEditorUserId="19912" LastEditDate="2019-04-22T21:58:36.650" LastActivityDate="2019-04-22T21:58:36.650" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="27039" PostTypeId="2" ParentId="27036" CreationDate="2019-04-22T22:23:24.140" Score="0" Body="&lt;p&gt;Poiseuille's Law,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  Flow=\dfrac{π\cdot r^4 \cdot (P-Po) }{ 8\cdot η\cdot L}  cm^3/s  $$&lt;/span&gt; &#xA;And in your case assuming you have a Pipe with the area &lt;span class=&quot;math-container&quot;&gt;$Acm^2$&lt;/span&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Flow= A\dfrac{r^2 \cdot (P-Po) }{ 8\cdot η\cdot L}  cm^3/s $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-P = pressure at the entrance, Bar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-P0 = atmosphere pressure, Bar&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-eta = viscosity at dyne second/cm2 for water at 20c it is 0.01&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-L = length cm.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-22T22:23:24.140" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27040" PostTypeId="1" CreationDate="2019-04-22T22:59:35.653" Score="0" ViewCount="46" Body="&lt;p&gt;The air is at 13 kPa and 400 K. I can raise both the supply temperature and the pressure if necessary. No problem with that. I could get 1.7 kg/s for the air bearings, but I have no more air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought levitation this way was possible, but I'm doubtful now because there's an Australian engineer who's skeptical. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Elon Musk said it was possible, but Hyperloop critics deny it.  &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-22T22:59:35.653" Title="Can I lift 15 tons on air bearings with 0.2 kg air/s?" Tags="|airflow|bearings|air|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27041" PostTypeId="1" AcceptedAnswerId="27043" CreationDate="2019-04-23T01:57:18.333" Score="0" ViewCount="134" Body="&lt;p&gt;I am trained as a seismic geophysicist but really enjoy getting to translate what I know into what other people can understand and use for their applications, if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those with working in and/or retain practical knowledge of Oil/Gas E&amp;amp;P in terms of the engineering side, what is a usable or comprehensive list of terms to know in order to &lt;em&gt;read&lt;/em&gt; (and understand) data from my engineer colleagues? For example, I know terms such as Water Cut, GOR, etc. I want to learn more and expand this list.&lt;/p&gt;&#xA;" OwnerUserId="19365" LastActivityDate="2019-04-23T03:34:20.340" Title="Basic Drilling/Production/Reservoir Engineering Terms" Tags="|mechanical-engineering|petroleum-engineering|drilling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27042" PostTypeId="2" ParentId="27036" CreationDate="2019-04-23T02:34:46.383" Score="0" Body="&lt;p&gt;If you don't know the length L, then you'll have to disregard head losses. Then you can simply apply the Bernoulli equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ p + \rho *(c^2/2) + \rho*g*z = constant $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the pressure, &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; the density, &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; the speed, &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; the gravity, &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; the height. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apply this equation in the entry and in the exit and clear &lt;span class=&quot;math-container&quot;&gt;$c^2$&lt;/span&gt;. After that, multiply &lt;span class=&quot;math-container&quot;&gt;$c_2$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$A_2$&lt;/span&gt; and you'll get the flow rate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry for the wrong formatting, I still don't know how to format equations properly. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorDisplayName="user20096" LastEditDate="2019-04-23T19:01:10.183" LastActivityDate="2019-04-23T19:01:10.183" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27043" PostTypeId="2" ParentId="27041" CreationDate="2019-04-23T03:34:20.340" Score="2" Body="&lt;p&gt;There are a number of online reference. I entered &quot;dictionary oil terms&quot; into a search engine &amp;amp; came up with these as an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.glossary.oilfield.slb.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Schlumberger Oilfield Glossary&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://cogcc.state.co.us/COGIS_Help/glossary.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Glossary of Oil and Gas Terms - &lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://bruinep.com/glossary-of-oil-and-natural-gas-terms/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bruin Glossary of Oil and Natural Gas Terms&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://bfs-usa.com/oil-gas-glossary-of-terms/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oil and Gas Glossary of Terms for Production Financing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.opisnet.com/resources/glossary-of-terms/&quot; rel=&quot;nofollow noreferrer&quot;&gt;OPIS Glossary of Terms&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.oil150.com/about-oil/oil-gas-dictionary/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Oil &amp;amp; Gas Dictionary of Historical Terminology&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many more references.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2019-04-23T03:34:20.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27044" PostTypeId="1" CreationDate="2019-04-23T06:01:25.263" Score="1" ViewCount="38" Body="&lt;p&gt;I would like to calculate the vertical velocity of a drone by integrating accelerometer data, problem is that the accelerometer measures in the frame of reference of the drone and not the the earth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know how much the frame of reference is rotated along the x and y axes and I know the position of the vector in this frame of reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to rotate the vector back to get actual vertical acceleration but the order of operations matter for matrix rotations.&#xA;The drone rotated on both axes simultaneously so I don't know which rotation to calculate first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't necessarily believe that this is the right place for this question but other communities suggested I try here.&lt;/p&gt;&#xA;" OwnerUserId="20201" LastActivityDate="2019-05-23T23:18:46.933" Title="Order of matrices when rotating a vector" Tags="|acceleration|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27046" PostTypeId="1" CreationDate="2019-04-23T06:55:54.157" Score="1" ViewCount="116" Body="&lt;p&gt;I have a book with the formulas to calculate the length of engine exhaust pipe, but these formulas don't take into account the muffler.&#xA;If I add an expansion chamber, how the calculated length change?&lt;/p&gt;&#xA;" OwnerUserId="19708" LastActivityDate="2023-12-29T06:04:45.540" Title="Calculate exhaust with expansion chamber muffler" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27047" PostTypeId="1" CreationDate="2019-04-23T07:04:31.490" Score="0" ViewCount="196" Body="&lt;p&gt;I'd like to use a worm drive to achieve the same results as pinion gear with a rack on either side. namely that the rotation of the worm gear would cause one rack to move left and the other to move right.&#xA;is it possible to have one thread on the worm gear and create right hand thread on one rack and left hand thread on the other rack? &lt;/p&gt;&#xA;" OwnerUserId="20209" LastActivityDate="2019-09-20T15:03:05.427" Title="convert rotational movement to linear in opposing directions using worm gear" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27048" PostTypeId="2" ParentId="27047" CreationDate="2019-04-23T11:50:13.257" Score="1" Body="&lt;p&gt;No, this is not possible. As a worm screw rotates, the 'tooth front' moves along its axis, this is the same on the top or the bottom of the worm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Worm Wheel is often not cut 'handed' at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BTnVa.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BTnVa.gif&quot; alt=&quot;Worm Gear Animation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While it would be possible to 'straighten out' the wheel into a rack, it would not be possible to get opposing linear motion from a single worm screw.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-04-23T11:50:13.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27050" PostTypeId="1" CreationDate="2019-04-23T13:34:57.783" Score="1" ViewCount="259" Body="&lt;p&gt;I am not sure what the difference is between using a spider of RBE2s. and multiple SPCs across a face. I've heard that the spider is the more accurate method as multiple SPCs make the model overly stiff but I am not sure why. In this particular instance I am trying to model the shown face being thru-bolted to a much larger plate. I've included 2 images for further clarification of my definitions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spider of RBE2s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1VG5L.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1VG5L.png&quot; alt=&quot;Spider of RBE2s with a SPC in the center&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiple SPCs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JXkqB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JXkqB.png&quot; alt=&quot;Multiple SPCs across a face&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20211" LastEditorUserId="20211" LastEditDate="2019-04-23T20:24:41.010" LastActivityDate="2019-04-23T20:24:41.010" Title="FEA: What is the difference between a spider of RBE2s and multiple SPCs" Tags="|structural-analysis|finite-element-method|modeling|deflection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27051" PostTypeId="1" CreationDate="2019-04-23T14:37:36.530" Score="3" ViewCount="208" Body="&lt;p&gt;Your typical Foucault's current brakes operate on a disk, which is placed in an electromagnetic field in order to transform kinetic energy into heat without friction. The problem is that the disk still overheats, just like an usual disk brake. In the case of railways this also needs that the wheels have adherence to rail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Theoretically, it would be as simple as to place two huge magnets around both rails in order to brake the train. A simple mechanical design could move the magnets up and down whether we want braking or not. The magnetic field would turn kinetic energy into heat, but this would not overheat as the heat would advance along the train itself. Also this would not require any friction, and as thus, no mechanical parts have to be changed due to use. (Friction braking is still required for stopping the train once a low speed would has been attained with Foucault's current braking).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would have the following advantages :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Braking doesn't rely on rail adherence&lt;/li&gt;&#xA;&lt;li&gt;Since permanent magnets can be used, no electricity needed to operate the braking device (work on non-electrified railways, or in emergency when there's damage to the overhead lines)&lt;/li&gt;&#xA;&lt;li&gt;No risk of overheating in the long run since the dissipated heat moves along the rail.&lt;/li&gt;&#xA;&lt;li&gt;Don't damage brake pads, and don't damage rails as does the usual emergency brake which applies friction to the rails.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The only major problem I'd see with such a system is that the magnets would also attract the rail itself (made from steel) - but if magnets on both rails are in reverse directions each force will cancel the other, so the magnets just have to be very solidly attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is such a system possible or have I missed something ? If it's possible, why isn't it more widely used ? Or would the magnets needs to be too strong for this idea to be any effective, and this would have effects on passengers or their belonging ?&lt;/p&gt;&#xA;" OwnerUserId="17069" LastEditorUserId="13297" LastEditDate="2019-04-23T15:04:19.617" LastActivityDate="2019-05-21T14:19:36.287" Title="Can a Foucault's current brake be used on the rails themselves to brake trains?" Tags="|electromagnetism|magnets|rail|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="27053" PostTypeId="2" ParentId="27050" CreationDate="2019-04-23T16:25:51.400" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;... the spider is the more accurate method as multiple SPCs make the model&#xA;  overly stiff ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That depends what you are trying to do. Remember that (1) the RBE2 does not have to connect to all 6 DOFs at the end of each &quot;leg&quot;, and (2) not all 6 DOFs at the &quot;independent&quot; grid (GN in the documentation) have to be constrained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If either of those conditions is relevant, the quote is likely to be true in the general situation, but there are many special cases where it makes no difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even when it makes no difference, the RBE2 has the advantage that it automatically sums the reaction forces and moments over the face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we don't know the physical situation you are trying to model, it's impossible to say which method will be &quot;more accurate&quot; for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the elements in the model have rotation variables, what do to with them requires thought - for example if the face of the model is &quot;rigidly constrained,&quot; and the surface is in the XY plane (for example,) do you want to constrain the XZ and YZ shear strains at the surface of the model, or not? If the answer is no, those shear components will give non-zero rotations in the .5 and .6 directions at the grid points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Local coordinate systems in the model are another consideration. Note that an RBE2 can apply &quot;constraints&quot; at the ends of the leg &lt;em&gt;which are not in the same direction as the local coordinates at the corresponding grid point&lt;/em&gt;. As before, that may be either a good or a bad feature, depending on what you want to do. (Be careful if your &quot;constrained surface&quot; is curved not flat, for example...)&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-04-23T16:46:47.683" LastActivityDate="2019-04-23T16:46:47.683" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27054" PostTypeId="2" ParentId="27044" CreationDate="2019-04-23T16:29:11.270" Score="0" Body="&lt;p&gt;You can apply a matrix coordinate transformation. Basic dynamics class has the development of these matrices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This document provides a review of the matrices and also contains some practical ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec03.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec03.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the end, you can make a T matrix for each rotation and then multiply them in together.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-23T16:29:11.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27056" PostTypeId="2" ParentId="27046" CreationDate="2019-04-23T21:42:24.097" Score="0" Body="&lt;p&gt;Use typical relations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$mass flow= density*area*speed$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And bear in mind the waves synchronizing. The speed of the wave and its reflections: It reflects when it reaches the muffler (open end). That way you can calculate the length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Time_(exhaust)*speed_(wave)=length$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many times do you want the wave to reflect? Bear that in mind too.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-23T21:42:24.097" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="27057" PostTypeId="2" ParentId="27003" CreationDate="2019-04-23T21:46:05.557" Score="1" Body="&lt;p&gt;Weld them together, that's the best solution for metal profiles.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorDisplayName="user20096" LastEditDate="2019-04-23T23:26:35.857" LastActivityDate="2019-04-23T23:26:35.857" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="27058" PostTypeId="1" CreationDate="2019-04-23T23:03:31.107" Score="1" ViewCount="22" Body="&lt;p&gt;I would like to create an experiment where I would hold ionized air molecules. Most likely negative O2-1 ions. I am looking for the most appropriate enclosure so that the ions do lose as little of charge to the surroundings as possible. I couldn't find any suggestions... Any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2019-04-23T23:29:22.937" Title="How to contain air ions inside a chamber without loosing their charge?" Tags="|electrical-engineering|gas|electromagnetism|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="27059" PostTypeId="2" ParentId="27058" CreationDate="2019-04-23T23:29:22.937" Score="1" Body="&lt;p&gt;The chamber should be made from a dielectric material so it doesn't interact with your ions. Use plastic, for example. PVC or methacrylate if the pressure is high. The enclosure should be tight, so melt down the plastic sheet borders and join them properly.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-23T23:29:22.937" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28058" PostTypeId="1" CreationDate="2019-04-24T02:24:58.287" Score="0" ViewCount="27" Body="&lt;p&gt;I post it because I want to know if anyone of you see any error in this circuit made in proteus 8.6. I accept any observation, this circuit provide 12v of output and 3 modes of charging fast(15 A), medium(10 A) and slow(5 A).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3TstG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3TstG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20216" LastEditorUserId="110" LastEditDate="2019-04-24T03:14:01.850" LastActivityDate="2019-04-24T03:14:01.850" Title="Charger for lead-calcium battery circuit" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28059" PostTypeId="2" ParentId="26970" CreationDate="2019-04-24T03:36:02.253" Score="2" Body="&lt;p&gt;I order to solve this problem there one needs to understand the following&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ADC &lt;strong&gt;range&lt;/strong&gt; which is the maximum and minimum ADC input. For this problem the range is 3.2V, -3.2V.&lt;/li&gt;&#xA;&lt;li&gt;ADC &lt;strong&gt;resolution&lt;/strong&gt; which is the smallest distinguishable change in input. For a 12 bit converter this would be 6.4V/4096 = 0.0015625V &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Below is an example for range of 3.3V based on a 12 bit converter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LSQ2w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LSQ2w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base on the above detail it should be possible to find the solution to above problem. In order to learn details I suggest checking out the following references.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference&lt;/strong&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://users.ece.utexas.edu/~valvano/Volume1/E-Book/C14_ADCdataAcquisition.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Analog to Digital Conversion, Data Acquisition and Control&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.edx.org/course/embedded-systems-shape-the-world-microcontroller-inputoutput&quot; rel=&quot;nofollow noreferrer&quot;&gt;Embedded Systems - Shape The World: Microcontroller Input/Output&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-04-24T03:36:02.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28061" PostTypeId="1" CreationDate="2019-04-24T06:17:04.710" Score="0" ViewCount="24" Body="&lt;p&gt;If I am using a 9&quot; long by 1&quot; diameter coil for my generator, which way will be better to generate the most power? Slice the 9&quot; coil into nine coils and connect them in series, or leave it as one coil?&lt;/p&gt;&#xA;" OwnerUserId="20219" LastActivityDate="2019-04-24T06:17:04.710" Title="Generator coils" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28063" PostTypeId="1" CreationDate="2019-04-24T10:42:16.203" Score="0" ViewCount="268" Body="&lt;p&gt;I want to derive a transfer function of my system so I can plug it into MATLAB and use the PID tuner tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My system is a balancing beam with a motor on each end of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first principles calculations for this system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking only the clockwise moment (at 0 degrees, equilibrium):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HJqMG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HJqMG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the clockwise moment goes from being &lt;span class=&quot;math-container&quot;&gt;$\text{Thrust Left} + Mg$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\text{Thrust Left} + Mg\cos\theta$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the tilt of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the overall moment is decreased. In order to hold the beam at equilibrium (at any value of tilt), the ThrustLeft needs to be multiplied by &lt;span class=&quot;math-container&quot;&gt;$\dfrac{1}{\cos\theta}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because thrust is proportional to the square of motor speed, the motor speed needs to be multiplied by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$RPM_{\text{left thrust}} = k \dfrac{1}{\sqrt{\cos \theta \ }}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Subsequently, the force required by ThrustLeft to maintain equilibrium at any angle is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7BYuH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7BYuH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the thrust required increases at an increasing rate as the angle of tilt increases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're still with me, thanks very much. Subsequently, the tilt against time is the same as the graph above (albeit on a slightly different scale). The angle of tilt increases at an increasing rate over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to represent this as a transfer function! But my theory is poor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming it would be something like &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$f(t) = k \dfrac{1}{\sqrt{\cos t}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$f(t)$&lt;/span&gt; is a function of tilt with respect to time.&lt;/p&gt;&#xA;" OwnerUserId="20226" LastEditorUserId="1832" LastEditDate="2019-04-24T15:49:20.887" LastActivityDate="2019-04-24T15:49:20.887" Title="Derive a Transfer Function from first principles (PID, system identification, statics)" Tags="|mechanical-engineering|control-engineering|applied-mechanics|statics|pid-control|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28065" PostTypeId="1" CreationDate="2019-04-24T15:11:55.373" Score="-1" ViewCount="1490" Body="&lt;p&gt;I don't know how to decide between the &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$O$&lt;/span&gt; arrangement, i don't have any specific example at this moment, but i want to know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Catalogues offer only some basic calculations. I'm not quite sure if there any guidelines or instruction exist, i have already consulted many books, but so far my researches failed to help me justify the choice between the two arrangements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/drbDO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/drbDO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://debearings.se/wp-content/uploads/2016/08/12-Tapered_roller_bearings.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://debearings.se/wp-content/uploads/2016/08/12-Tapered_roller_bearings.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2023-08-19T09:05:15.640" Title="How to choose between two possible arrangements of tapered roller bearings?" Tags="|bearings|machine-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28066" PostTypeId="2" ParentId="28065" CreationDate="2019-04-24T15:26:47.907" Score="0" Body="&lt;p&gt;The arrangement on the left is often used for fully floating hubs on vehicle axles, as the end float is easily set and controlled.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-04-24T15:26:47.907" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28067" PostTypeId="2" ParentId="27051" CreationDate="2019-04-24T15:34:18.833" Score="2" Body="&lt;p&gt;Yes they can and they are in use.  Details on &lt;a href=&quot;https://en.wikipedia.org/wiki/Eddy_current_brake&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiki&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wwmB5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wwmB5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/w/index.php?curid=3599846&quot; rel=&quot;nofollow noreferrer&quot;&gt;By Sebastian Terfloth User:Sese_Ingolstadt - Own work, CC BY-SA 3.0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A linear eddy current brake in a German ICE 3 high-speed train in action.&lt;/p&gt;&#xA;" OwnerUserId="8847" LastEditorUserId="6767" LastEditDate="2019-05-21T14:19:36.287" LastActivityDate="2019-05-21T14:19:36.287" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28068" PostTypeId="1" CreationDate="2019-04-24T15:54:29.517" Score="0" ViewCount="22" Body="&lt;p&gt;I'd like to do a DIY project to reuse my home AC heat output to heat my pool that is located very nearby the AC radiator. I planned on fitting a second radiator directly onto the heat output and hook the pool pump onto it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I know or have an idea of the amount of heat that the AC can output, how do I size the second radiator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For optimal heat transfer, would the 2 radiator needs to be identical?&#xA;What happen if it is oversized, will it reduces the performances or just be too expensive for nothing? What parameter should I look for, only thermal resistance?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="19321" LastActivityDate="2019-04-24T15:54:29.517" Title="Sizing radiator" Tags="|heat-transfer|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28069" PostTypeId="1" CreationDate="2019-04-24T18:59:29.403" Score="1" ViewCount="459" Body="&lt;p&gt;I have been looking at papers describing how to do non linear control with neural networks. A lot of these papers have block diagrams with a symbol I don't understand. There is a block with an angled arrow drawn through it. The tail of this arrow is an output of another part of the block diagram.&lt;a href=&quot;https://i.stack.imgur.com/WKwLd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WKwLd.png&quot; alt=&quot;Sample Block Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sample is from a paper called &quot;Nonlinear System Control Using Neural Networks&quot; by authors Jaroslava Žilková, Jaroslav Timko, and Peter Girovský.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A link to their paper is here:&lt;a href=&quot;https://uni-obuda.hu/journal/Zilkova_Timko_Girovsky_8.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nonlinear System Control Using Neural Networks&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20232" LastActivityDate="2019-04-24T21:37:35.530" Title="In a control block diagram what does a block with an angled arrow through it mean?" Tags="|control-theory|diagram|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28070" PostTypeId="1" CreationDate="2019-04-24T19:20:44.983" Score="0" ViewCount="40" Body="&lt;p&gt;so i was planning to make a project were a system of computers( in my case raspberry pi's) need to communicate to do a certain job. And i am having a trouble choosing a way of communications. so i would like it if any one has some thing to offer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;here are my needs, and thoughts,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Needs&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;i want a data to be transferred between multiple raspberry pi boards. The data includes videos frame by frame, and some sensor data.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;i want it to use the smallest amount of time possible for transferring data. since the boards are going to be attached on moving objects i need to have a fast communication in order for each of them to make decision depending on the others data.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;i want each board not to be dependent on the other during network fails&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thoughts&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Use a computer as server or the man in the middle to handle data transfers and also do some computations.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;use WiFi or Bluetooth modules to create connections.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;those are my thoughts and needs so, suggest me any design principles you now, their advantages, and disadvantages.&lt;/p&gt;&#xA;" OwnerUserId="19718" LastEditorUserId="19718" LastEditDate="2019-04-25T06:02:33.693" LastActivityDate="2019-04-25T06:02:33.693" Title="what is the best way to communicate among multiple raspberry pi boards?" Tags="|wireless-communication|circuit-design|networking|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28071" PostTypeId="2" ParentId="28069" CreationDate="2019-04-24T20:20:37.003" Score="1" Body="&lt;p&gt;Given the title of the figure, I'm inferring that the slanted line means &quot;this block adapts slowly based on the given information&quot;.  The author is probably trying to convey the notion that you could disconnect that slanted input and things would still work -- for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Block diagrams are of interest to me because they're very much a living language.  I work, or have worked, in both control systems and RF signal processing, where there's a lot of overlap between the language, and yet a lot of differences.  Watching a controls engineer and an RF engineer show each other block diagrams is like watching someone from London try to talk to someone from Glasgow -- it's officially the same language, but there are significant differences.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastEditorUserId="19776" LastEditDate="2019-04-24T21:37:35.530" LastActivityDate="2019-04-24T21:37:35.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28072" PostTypeId="1" CreationDate="2019-04-25T00:41:19.477" Score="3" ViewCount="92" Body="&lt;p&gt;I would like to make a Van der Graff generator for my science fair project out of aluminized mylar. Do you think this might be feasable or do I need to use thick stainless steel ball? Is there some sort of charge or conductivity limit that I need to consider?&lt;/p&gt;&#xA;" OwnerUserId="1630" LastActivityDate="2023-01-04T16:00:05.343" Title="What is the material requirement for Van der Graff generator sphere?" Tags="|electrical-engineering|electromagnetism|electrical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28073" PostTypeId="2" ParentId="28072" CreationDate="2019-04-25T02:06:12.410" Score="1" Body="&lt;p&gt;You want to deposit charge inside the high voltage terminal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll need some mechanical strength for that and because the charge is deposited near the middle you'll need an conducting path to the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wants to be very smooth and every radii as big as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the 4 MV one I worked on had a dome of about 1/16&quot; stainless polished to a mirror finish. After 34 years it had quite a few dings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a photo of the dome. The rings to the left go around the structure to control the potential gradient to the base. The top deck is supported by layers of glass and metal (square columns) and the beamline in contained in the round glass and metal structure. The long acrylic rods passed from the base to the deck to control variacs at the high voltage terminal. The terminal could either go to +4 MV or -4 MV. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TQqpa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TQqpa.jpg&quot; alt=&quot;Dome of VdG, 4 MV&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pfurp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pfurp.jpg&quot; alt=&quot;Top deck and support structure of 4 MV VdG&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16544" LastEditorUserId="16544" LastEditDate="2019-04-25T08:17:17.257" LastActivityDate="2019-04-25T08:17:17.257" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28074" PostTypeId="2" ParentId="23934" CreationDate="2019-04-25T04:45:23.477" Score="0" Body="&lt;p&gt;There are some points and some assumptions we need to make.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If the ball floats the water will drain from the hole.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If we assume there is a fill valve or somehow we close he hole just after the ball starts to float, your assumption is correct, the ball has to be lighter than water..&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If these are taken care of then by doing a bit of geometry we see by your sketch the hole is very small compared to the sphere, we can assume small angles, 9 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$sin(2/20)\pi/2 = (1/10)\cdot \pi/2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and subtract the surface area of the hole multiplied by the water pressure, as opposed to calculating the sphere sectional surface and pertinent water pressure on it, at the bottom of the tank &lt;span class=&quot;math-container&quot;&gt;$$ F_{floating}= \frac{4}{3}\pi R^3\rho_{water}&#xA;- \pi*4\cdot(o.8*o.1kg/m height) \\ = \frac{4}{3}\pi (1/5)^3*1kg/liter \\ -\pi*4mm*(1)liter/(1000)milL&#xA;\cdot(o.8*o.1kg/m height)  $$&lt;/span&gt; assuming the atmosphere pressure constant for the top and bottom of the tank. I did a rough calculation and the effect of the hole is minimal as with your dimensions less than 3%. This is within 4 decimal point rounding.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-04-25T05:10:27.877" LastActivityDate="2019-04-25T05:10:27.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28075" PostTypeId="1" CreationDate="2019-04-25T06:55:09.403" Score="1" ViewCount="146" Body="&lt;p&gt;I am currently trying to redesign a pipe with 2 dents from the inside.&lt;br&gt;&#xA;What is the best way to manufacture something like this, also how would i go about describing this process in a drawing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the pictures you can see the pipe before and after the dents are placed. The sides on the pipes are welded on, so it is possible to acces the inside before doing this.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cutout is present on both pipe, but this is also an extra process that could be done after the denting process.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Material is 3mm AISI 304&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: The design is meant to be used for a prototype, demanding production solutions will not be optimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vCcDB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vCcDB.png&quot; alt=&quot;No dents&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6AnW9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6AnW9.png&quot; alt=&quot;Dents&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20237" LastEditorUserId="20237" LastEditDate="2019-04-29T12:12:12.187" LastActivityDate="2019-04-29T12:12:12.187" Title="Best way to dent a pipe from the inside?" Tags="|manufacturing-engineering|metals|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28076" PostTypeId="2" ParentId="28075" CreationDate="2019-04-25T07:35:07.983" Score="0" Body="&lt;p&gt;Fit it into a mold which has the indents correctly placed and then pressurize the inside - hydraulic pressure with oil or water may be used. A pulse may be effective for this as opposed to a gradual increase in pressure.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-04-25T07:35:07.983" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28077" PostTypeId="2" ParentId="28075" CreationDate="2019-04-25T08:01:20.907" Score="1" Body="&lt;p&gt;This is a three part precise jig that can repeat the process many times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there are left and right solid pipes with a round notch to support the  moving quarter circle hammer shown in black.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you put the part in the vise you secure the indentation spot with stainless steel belts upside and downside of the spot to be dented, not to let it bulge outside the desired zone, then tighten the vise sliding the round hammer exactly into its spot. then remove the jig and repeat the next piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6lf5c.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6lf5c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-25T08:01:20.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28078" PostTypeId="1" AcceptedAnswerId="28082" CreationDate="2019-04-25T08:24:35.203" Score="1" ViewCount="2120" Body="&lt;p&gt;I have a top level project directory with about 40 Assembly files and several hundred SLDPRT files and they are all highly dependent on each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original version of this works fine and has no errors that we know of.  Now I wish to &lt;strong&gt;safely&lt;/strong&gt; clone this entire set of design files to form the basis of the next iteration of our design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(The originals are of course backed up off-site and are entirely safe from any accidental corruption by stray references in the new cloned version.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, given the complexity of the project and the huge number of cross-references in the design, how can I make a cloned copy that has absolutely no residual links to the old design files, so that I can be 100% confident that the new cloned directory is going to be self-contained and free from broken reference errors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a single assembly or part it would be a simple case of using the &quot;Pack and Go&quot; feature of solidworks, but how can I reliably do this for multiple assemblies that are themselves cross-linked?  I'm worried that I might make a mistake somewhere and not realise it until a year later by which time it would be almost impossible to fix.&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastActivityDate="2019-04-25T10:25:15.853" Title="Solidworks - How to safely clone a large multi-assembly project?" Tags="|solidworks|computer-aided-design|project-management|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28079" PostTypeId="2" ParentId="28078" CreationDate="2019-04-25T08:29:19.940" Score="1" Body="&lt;p&gt;I would make a copy onto a USB stick then, at best, check that on a separate computer - any broken links should show up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have separate computer then a new account may do the same - as long as it does not get access to your user file directory.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-04-25T08:29:19.940" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28080" PostTypeId="2" ParentId="28078" CreationDate="2019-04-25T09:43:11.580" Score="1" Body="&lt;p&gt;Most operating systems can substitute the path so that after the operation another path looks like another. So from the softwares perspective it is the same path even if it is not. This is a very easy way out as you dont need to change anything at all. How you do this depends a bit how your computer/network is set up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This said solidworks includes a tool called solidworks explorer that will handle updating file level structure changes.  Use that for the moving of files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally, instead of copying disk level items, you would have let your PDM handle this, as this leaves a link between versions for traceability and other operations. But it sounds like you dont have one and or dont want to use one. In this kind of cases i might be tempted to just set the stff up for a vrsion control system.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2019-04-25T09:43:11.580" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28082" PostTypeId="2" ParentId="28078" CreationDate="2019-04-25T10:25:15.853" Score="1" Body="&lt;p&gt;You hinted at the answer in your original question - Pack and Go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this were my project, I would create a 'Package Master' top level assembly that contains your 40 assembly files as sub-assemblies. No need to mate them to each other, just have them there contained within the file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then use the Pack and Go feature that you're familiar with. I always use the Select/Replace tool to remove the previous 'version number', and update with the latest one, so that you never get in trouble with two different parts having the same filename, and either windows or a user getting confused between these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure you tick the 'Include Drawings', 'Include Suppressed Components' etc. boxes at the top left of the Pack and Go dialogue according to your needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can delete the 'Package Master' from the target directory once everything has gone through, if you don't need it any more.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-04-25T10:25:15.853" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28083" PostTypeId="1" CreationDate="2019-04-25T10:57:46.780" Score="0" ViewCount="2207" Body="&lt;p&gt;The buckling stress, using the Rankine and Gordon Equation or the Perry-Robertson Equation, is always less than the yield stress of the material?&lt;/p&gt;&#xA;" OwnerUserId="20241" LastActivityDate="2019-04-29T23:09:45.090" Title="The buckling stress and yield stress" Tags="|structural-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28084" PostTypeId="1" CreationDate="2019-04-25T12:03:29.623" Score="1" ViewCount="245" Body="&lt;p&gt;I want to build a monitor supporting arm for three monitors besides each other. The outer monitors shall be able to swivel forward to reduce size of the whole thing when not in use. My problem's that I have no idea how to design the hinge - the outer monitors should rotate around their inner forward vertical edge, but I cannot put a hinge there, because the monitors have to be as close to each other as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for something I can put into the supporting arm behind the monitors that acts like a hinge at the front (outside the arm itself)... how to do this?&lt;/p&gt;&#xA;" OwnerUserId="20243" LastActivityDate="2021-10-13T13:06:50.543" Title="Off-center hinge construction" Tags="|mechanical-engineering|mechanisms|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28086" PostTypeId="1" CreationDate="2019-04-25T12:45:36.140" Score="0" ViewCount="51" Body="&lt;p&gt;I have a sample of 600 milligrams which is supported by a rope. So the force is in the vertical direction. Now I want to change the direction to the horizontal direction while keeping the force equal(1 microgram accuracy). The first idea would be to use bearings. But if you calculate the resolution which is caused by friction, it would correspond 0.6 milligrams accuracy while I want a resolution of 0.001 milligrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have any ideas how to change the direction of the force while retaining a high resolution?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kind regards&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/50sFd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/50sFd.png&quot; alt=&quot;Bearing used to change direction of force&quot;&gt;&lt;/a&gt;&#xA;Bearing converting force to horizontal plane.&lt;/p&gt;&#xA;" OwnerUserId="19157" LastActivityDate="2019-04-25T12:45:36.140" Title="How to change the direction of the force with a high resolution" Tags="|mechanical-engineering|design|" AnswerCount="0" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28087" PostTypeId="1" CreationDate="2019-04-25T12:53:47.380" Score="1" ViewCount="504" Body="&lt;p&gt;How is the bore diameter of a flange Type 21 - acc. EN 1092-1 determinated?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the standard:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;NOTE 5 The bore diameters of flanges type 21 are not specified in this standard, the effective bore diameters are usually&#xA;  given in the relevant component standard(s).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are the relevant component standard(s)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mP9i7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mP9i7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8464" LastActivityDate="2021-07-06T23:00:25.060" Title="Flange EN1092-1 Type 21 Bore Diameter" Tags="|pressure-vessel|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28088" PostTypeId="1" CreationDate="2019-04-25T13:39:05.757" Score="0" ViewCount="40" Body="&lt;p&gt;I need a clear material that will allow sun in but won't insulate heat. Sun sun rays get in but doesn't trap heat. Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="20246" LastActivityDate="2019-04-27T01:24:54.600" Title="Thermal Non-Insulators?" Tags="|materials|heat-transfer|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28089" PostTypeId="1" CreationDate="2019-04-25T14:27:42.783" Score="2" ViewCount="477" Body="&lt;p&gt;I'm looking to manufacture some steel pins and plates, with a view to large scale production in the future. One of the pins of the design has a constant diameter. This pin passes through holes in two different plates. One of the plates has an interference fit with the pin and the other has a clearance fit with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, this sounds like it should use a shaft-based system, where the shaft diameter is constant and the hole sizes vary. Given, however, that the difference between hole diameters is only a matter of tens of microns and I've heard that it is typically harder to make non-standard holes than non-standard shafts, I am wondering whether I should use a hole base system, where the hole sizes in both plates are the same and the pin has a small step (tens of microns) to accommodate the different fits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the hole-base system could work but the question really comes down to whether it is easier to accommodate the tiny difference in fit through the manufacture of the shaft or the hole.&lt;/p&gt;&#xA;" OwnerUserId="20247" LastActivityDate="2024-01-30T10:02:55.430" Title="Should I use shaft or hole basis on a constant diameter shaft with both clearance and interference fits?" Tags="|mechanical-engineering|manufacturing-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28090" PostTypeId="2" ParentId="18536" CreationDate="2019-04-25T15:15:19.523" Score="0" Body="&lt;p&gt;The plastic section modulus can be calculated for a given cross section by:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Locating the plastic neutral axis. PNA&lt;/li&gt;&#xA;&lt;li&gt;Bringing the stress distribution to full plastic stress (constant) above and below the PNA.&lt;/li&gt;&#xA;&lt;li&gt;Summing the moments on the cross section.&lt;/li&gt;&#xA;&lt;li&gt;Finally divide the resulting moment by the yield stress to find the plastic section modulus.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Designing structural steel to perform in it's plastic deformation region is one of the pillars of good seismic design. It is very common to consider the strength of lateral elements and connections post elastic behavior.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-25T15:15:19.523" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28091" PostTypeId="2" ParentId="28089" CreationDate="2019-04-25T15:44:33.190" Score="0" Body="&lt;p&gt;I guess it depends on the respective materials of each part. Normally for an interference fit, you'd the shaft to be a lot stiffer than the plate in which the hole is, to ensure a good interference. Also, some materials are a lot easier to machine than others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having said that, and given that the difference in diameter is very small (0.01-0.02m), I would have thought that it easier/more reliable to drill/ream a hole in a plate of the right size with the right tolerance, than to turn down a shaft to the right diameter, even if this means buying a specialised reamer. So my vote is for the hole-based system.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-04-25T15:44:33.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28092" PostTypeId="2" ParentId="28088" CreationDate="2019-04-25T15:44:51.097" Score="2" Body="&lt;p&gt;What you seem to be asking for is a transparent material that is a good thermal conductor. As pointed out by others here, ordinary glass conducts heat (but not as well as a metal). It isn't possible in practice to make a transparent material that conducts really well because the properties that make for good thermal conductivity (delocalized valence electrons) also make the material a poor transmitter of light. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-04-25T15:44:51.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28093" PostTypeId="1" AcceptedAnswerId="28099" CreationDate="2019-04-25T15:57:31.020" Score="0" ViewCount="207" Body="&lt;p&gt;Suppose we have a water reservoir at some elevation say z and the flow is frictionless then by Bernoulli's equation we have the exit velocity (2gz)^0.5 . This means that the static pressure in the middle of horizontal pipe would always be atmospheric.&#xA;Now what if we split the pipe into two at the exit, would the velocity at each exit be the same (2gz)^0.5 ? If so, then this implies that the pressure in the middle of the same horizontal pipe is negative (gauge). Does this make sense ?&lt;/p&gt;&#xA;" OwnerUserId="20249" LastActivityDate="2019-04-25T23:37:48.410" Title="Frictionless Flow" Tags="|fluid-mechanics|fluid|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28094" PostTypeId="1" CreationDate="2019-04-25T16:43:01.427" Score="0" ViewCount="36" Body="&lt;p&gt;I'm using a TEC12706 Peltier module for making a cooling setup. The hot side rejects heat to a Cooler Master Hyper 410R cooler and the cold side fin gets cold enough to form water condensate on itself.&#xA;I then tried to connect a small 12V CPU fan to the cold fin and measured the temperature of the air coming from the fan, the fan basically throws air onto the cooled fin and the air then escapes from the sides, but I'm only able to achieve a temperature drop of 0.2C in the air. I think it may be due to the less contact time of air with the fin, how can I increase the contact time of air with the cool surface or is there something else going on?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fVGp0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;A similar setUp&lt;/a&gt;!&lt;/p&gt;&#xA;" OwnerUserId="19792" LastEditorUserId="19792" LastEditDate="2019-04-25T16:45:14.137" LastActivityDate="2019-04-25T16:45:14.137" Title="How to achieve successful Peltier cooling" Tags="|heat-transfer|cooling|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28096" PostTypeId="1" CreationDate="2019-04-25T17:40:01.013" Score="0" ViewCount="57" Body="&lt;p&gt;For one of my courses the teachers asked me to find two samples of 30 or queries of data, and I was thinking on how I could begin to search it?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the scientific articles that i looked for just show the data into graphs or simplified, but that's not really what I'm looking for, any help into how i could begin to start searching for it? &lt;/p&gt;&#xA;" OwnerUserId="20250" LastActivityDate="2019-04-26T07:55:27.887" Title="How to find raw data sets for engineering?" Tags="|mechanical-engineering|statistics|data|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28098" PostTypeId="2" ParentId="28096" CreationDate="2019-04-25T20:23:56.550" Score="2" Body="&lt;p&gt;Search for &lt;em&gt;daily maximum and minimum temperatures&lt;/em&gt; for your local region. If there is a river nearby, search for &lt;em&gt;river level&lt;/em&gt; or &lt;em&gt;river gauge height&lt;/em&gt;. If it rains, search for &lt;em&gt;annual rainfall&lt;/em&gt; in your region. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-04-25T20:23:56.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28099" PostTypeId="2" ParentId="28093" CreationDate="2019-04-25T23:37:48.410" Score="1" Body="&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conservation of mass dictates that the velocity at an exit with double the area of the upstream pipe is 1/2 the velocity in the upstream pipe. For an ideal fluid in an ideal (horizontal) expansion transition, there is no change in entropy. So Bernoulli's equation gives you the relationship for calculating the pipe's static pressure given that the exit is at atmospheric pressure. This pressure will be the same throughout the pipe, and will be less than atmospheric pressure. We are also assuming an ideal bell transition from the reservoir to the pipe.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-04-25T23:37:48.410" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28100" PostTypeId="1" AcceptedAnswerId="28102" CreationDate="2019-04-26T01:28:02.830" Score="0" ViewCount="78" Body="&lt;p&gt;What are the non-practical features of a sterling engine that cause the internal combustion engine to be used instead (in a car)? (e.g. price, size, the power to weight ratio, etc.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you name any that aren't as obvious as those stated above?&#xA;Please provide answers that aren't obvious but are still relevant.&lt;/p&gt;&#xA;" OwnerUserId="20177" LastEditorUserId="20177" LastEditDate="2019-04-26T02:04:58.287" LastActivityDate="2019-05-26T03:02:55.313" Title="What are the non-practical features of a sterling engine that cause the internal combustion engine to be used instead?" Tags="|mechanical-engineering|motors|automotive-engineering|combustion|engines|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28102" PostTypeId="2" ParentId="28100" CreationDate="2019-04-26T01:41:11.493" Score="0" Body="&lt;p&gt;I found out this for one of the other features but would still like more. &#xA;City driving where there is stop-start activity is not beneficial for the usage of sterling engines as well as the fact that you are not able to link the cylinders of multiple sterling engines to one another like you can with a combustion engine.&lt;/p&gt;&#xA;" OwnerUserId="20177" LastActivityDate="2019-04-26T01:41:11.493" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28103" PostTypeId="2" ParentId="1910" CreationDate="2019-04-26T02:04:08.450" Score="0" Body="&lt;p&gt;I’ve been doing a lot of reading on this and what we haven’t seen is a form of turbo compounding but instead using a small turbine engine capable of running itself when a steady torque is required. If married to a reciprocating engine with variable valve timing, the valves could be held open; the turbine generating the power. Under acceleration the turbine could operate as a turbocharger using bleed air intercooled as boost. &lt;/p&gt;&#xA;" OwnerUserId="20252" LastActivityDate="2019-04-26T02:04:08.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28105" PostTypeId="2" ParentId="28096" CreationDate="2019-04-26T07:55:27.887" Score="0" Body="&lt;p&gt;If you have a graph that you want to get data from - to plot it differently for example, then DataThief is a handy option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Works on many systems, and no I am not associated, but found it years ago.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can be found : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://datathief.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://datathief.org/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-04-26T07:55:27.887" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28106" PostTypeId="2" ParentId="28084" CreationDate="2019-04-26T09:52:04.927" Score="0" Body="&lt;p&gt;You can use offset hinges at the top and bottom of the screen. That way you can still have the hing axis where you want it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other option is a multibar linkage. Though designing that requires a bit more intuition.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2019-04-26T09:52:04.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28108" PostTypeId="1" CreationDate="2019-04-26T12:49:53.157" Score="1" ViewCount="37" Body="&lt;p&gt;I'm searching for a mathematical approach to well describe the creep （or stress relaxation）of helical-coil compression spring.I've got the spring parts and can Experiment with these parts .I need extra theoretical analysis which my experiment is trying to support.&lt;/p&gt;&#xA;" OwnerUserId="20238" LastActivityDate="2019-04-26T12:49:53.157" Title="A creep mathematical model of alloy steel for making springs" Tags="|springs|creep|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28109" PostTypeId="1" CreationDate="2019-04-26T13:07:18.530" Score="1" ViewCount="81" Body="&lt;p&gt;In random vibration analysis, the load is given as a PSD (power spectral density). The values are often &lt;span class=&quot;math-container&quot;&gt;$G^2/Hz$&lt;/span&gt;, but I have an example where the PSD is a pressure &lt;span class=&quot;math-container&quot;&gt;$Pa^2/Hz$&lt;/span&gt;. The value is written as &lt;span class=&quot;math-container&quot;&gt;$(100 N/m^2)^2/Hz$&lt;/span&gt; which brings me to these questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is that value &lt;span class=&quot;math-container&quot;&gt;$(100 \;N/m^2)^2/Hz$&lt;/span&gt; equal to &lt;span class=&quot;math-container&quot;&gt;$10000 \;(N/m^2)^2/Hz$&lt;/span&gt;? (The results of the analysis imply that the answer is Yes, but I want to confirm my understanding.)&lt;/li&gt;&#xA;&lt;li&gt;Are both styles of writing the value acceptable? Is one style more common or acceptable than the other? For example, I have never seen an acceleration PSD written as &lt;span class=&quot;math-container&quot;&gt;$(5 \;G)^2/Hz$&lt;/span&gt; to represent a value of &lt;span class=&quot;math-container&quot;&gt;$25 \;G^2/Hz$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2019-04-26T13:07:18.530" Title="How to state the PSD value (random vibration)" Tags="|structural-analysis|vibration|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28111" PostTypeId="2" ParentId="28088" CreationDate="2019-04-26T19:19:58.707" Score="0" Body="&lt;p&gt;The reason most of the energy of light gets trapped in an enclosed space is not the glass or the transparent material. Glass can let radiant heat filter out as it does let it in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also there are glasses or films to cover the glasses, designed to absorb ultraviolet part of the spectrum of sun light and they help reducing the pass through of some of the heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem is the sun rays passing though heat up the objects such as floors, furniture, and air. These objects are opaque and do not reflect most of the light back as radiation, only a small part of it as infrared which glass will let pass. They transmit the heat by convection and conduction and heating up the air and walls and everything (greenhouse effect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The smart way of controlling the heat infiltered by the glass is to install the window where they face a side of the building which is not exposed to direct sun light, such as north side.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-04-26T19:26:01.860" LastActivityDate="2019-04-26T19:26:01.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28112" PostTypeId="2" ParentId="19442" CreationDate="2019-04-26T21:38:08.537" Score="1" Body="&lt;p&gt;As Mahendra Gunawardena said, there's not one universally used method of pricing manufacturing projects, but there are some pretty common facts that result from the economics of the situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, if you're talking about small runs at a large company and the design doesn't include anything exotic, the pricing will probably have less to do with the specifics of your design, and more to do with the overhead costs and benefits of dealing with you as a client. In the kinds of jobs shops that I've worked for and with, the algorithm for bidding small projects goes something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1: Should we quote this work at all or decline to bid?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This evaluation is usually done very quickly by someone senior enough to understand the business relationship with the particular client and the overall workload of the company. They need very few details about the actual design, often one image and a brief description would be enough if the client is familiar with the jargon of the industry and thus able to explain the task. They'll be considering questions like:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does the company have the spare capacity to complete the work on a schedule the client will accept?&lt;/li&gt;&#xA;&lt;li&gt;Is this the kind of project that the company can produce to an adequate quality level?&lt;/li&gt;&#xA;&lt;li&gt;Is this the kind of project that the company is efficient at doing, so they can do the work for a competitive rate.&lt;/li&gt;&#xA;&lt;li&gt;Is the client competent and clear enough about what their needs are that the product is going to work, or are they likely to confuse a design flaw on their part for a manufacturing defect on our part?&lt;/li&gt;&#xA;&lt;li&gt;Is the client likely to chose us to do this work, or are they likely to use another vendor meaning that our labor to bid the project is wasted.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;If any of the above items are 'maybe's or 'no's then the person will take into account if there's another compelling reason to do the work. For instance even it's s stretch on their quality abilities, if it's a loyal client who brings them other work they may decide to take it on and accept the costs of improved tooling, attention to management, procedures, etc. Similarly, even if it's a kind of work where the company has to give up most of its profits to be competitive, they may still decide to bid on the work if they are very low on work and at risk of having to lay off workers. If the company doesn't need the work and it seems like 'bad' work according to the criteria above, they will likely politely decline to save the cost of investing any further management labor.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Step 2: Assuming the person in step 1 decides that the work is worth going after, the next step is to figure out the pricing they should offer.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This step may be done by the same person who handled step 1, but in most companies, especially larger ones, most of the heavy lifting of this process will be handled by someone else.&lt;/li&gt;&#xA;&lt;li&gt;Especially for small projects, this person won't want to spend more time than they have to developing the pricing, so before they do any detailed analysis of tooling or design they will look at the project qualitatively and decide what needs to be penciled out and what can be 'eyeballed.' For smaller projects, lots more can and should be eyeballed. The reason for this is that a small loss (say 10% on a small project represents an amount of money that the company can probably tolerate losing occasionally when there's a mistake. Conversely, a few too many hours of manager or engineer time in the bidding process can quickly become a big enough cost as a percentage of the contract sum that either the company still loses money on it, or their bid is non-competitive. If a company doesn't win the work based on a non-competitive bid, all of the money invested in the bidding process is loss.&lt;/li&gt;&#xA;&lt;li&gt;In this evaluation, these kinds of questions will be considered:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Have we done a very similar project in the past? If so, often reviewing that project and the actual profit or loss can form a solid basis to bid the new work scaled by quantity and complexity.&lt;/li&gt;&#xA;&lt;li&gt;Do I know what the client is expecting this to cost? If the expectation is comfortably more than actual costs then they can pick a 'safe' number without doing any more work.&lt;/li&gt;&#xA;&lt;li&gt;Are the materials, manufacturing labor, tooling, or management labor are obviously the majority of the costs? If it's clear that one or more of these cost categories was pretty small then they can focus efforts on pricing the ones that matter more and just apply a percentage to cover the others.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;After that, for items where they feel like the the bid warrants more thorough evaluation, they will proceed to estimate the costs:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;For materials, they would start by checking if the project can be done from a  material they ordinarily stock. Even if the yield isn't great, for a small run using common materials the waste will probably be worth the reduced cost of not having to order and receive another stock size. If the material isn't something they keep around, then they will look for prices they have recently paid for similar materials. Failing that, they will need to actually contact a material supplier and get pricing and lead times.&lt;/li&gt;&#xA;&lt;li&gt;For tooling, they would probably start by talking to either an engineer or the folks who actually work on the shop floor. They would start by seeing if any existing tooling can be used or adapted for the project. If not, they would check if there's a tool that can be purchased that accommodates the design. If not, then they would start the process with an engineer or determining what custom tooling (dies, stamps, workholding, etc) would be required and what the design and production effort would take.&lt;/li&gt;&#xA;&lt;li&gt;For fabrication labor they would discuss the project with the folks who run the shop floor and estimate what time it would take to set up tooling, produce, and inspect the whole run.&lt;/li&gt;&#xA;&lt;li&gt;For management labor, they would estimate the amount of time to support the project, This would include things like material procurement, submittal drawings, first article production, functional tests, shipping, etc.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="122" LastEditorDisplayName="user6335" LastEditDate="2019-04-28T03:51:57.253" LastActivityDate="2019-04-28T03:51:57.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28113" PostTypeId="1" CreationDate="2019-04-27T00:31:23.413" Score="0" ViewCount="43" Body="&lt;p&gt;Why isn't the usage of air bearings for HL feasible?&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-27T00:31:23.413" Title="Why can't we use air bearings for Hyperloop? Why isn't it feasible?" Tags="|bearings|air|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28114" PostTypeId="2" ParentId="28088" CreationDate="2019-04-27T01:24:54.600" Score="0" Body="&lt;p&gt;The question is far too vague.  But based on what it actually says, the best clear material that I know of that meets all the requirements is not only readily available, it's trivial to install.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That material is called &quot;air&quot;.  Simply cut a hole of appropriate size in whatever it is you want the sunlight to pass through.  The sunlight will enter unobstructed, and heat will not be trapped.&lt;/p&gt;&#xA;" OwnerUserId="19930" LastActivityDate="2019-04-27T01:24:54.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28115" PostTypeId="1" CreationDate="2019-04-27T03:07:23.637" Score="0" ViewCount="45" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yATbz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yATbz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi everyone,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I approach questions like part (ii)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From what I understand now, I believe the shortest method is that as long as the &lt;strong&gt;ANGLE of the relative velocity of B with respect to velocity of A is = Angle of thetha which in this case is arctan(50/150&lt;/strong&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another method I tried was to find the time taken for projectile and intercepter to reach the same x coordinate and use this to obtain the time in terms of thetha. Then I equated the y coordinate using this time in terms of thetha but I ended up being unable to solve them with 10sin - 10/3cos sin + cos = sqrt(3) + 3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other ways are possible as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="19499" LastActivityDate="2019-04-27T17:52:27.687" Title="Approaches to whether a projectile gets intercepted" Tags="|mechanical-engineering|dynamics|projectiles|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28116" PostTypeId="2" ParentId="11963" CreationDate="2019-04-27T03:08:47.097" Score="2" Body="&lt;p&gt;I see, like me, you want a simple answer. In the kettle, the water is heated to boiling point around the element, but the surrounding water is not. Steam cannot exist at less than 100° C. at atmospheric pressure, so when the steam contacts the cooler water, it immediately condenses, leaving no cushioning effect, so it's like metal against metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cavitation doesn't have to involve vaporization. A liquid like hydraulic fluid, if the pump inlet is restricted, vacuum bubbles form. Because there is no air in them to cushion the impact, again, it's like metal against metal. The pump sounds like it's crunching metal shavings. Even though it's a liquid, it impacts like metal &amp;amp; will fatigue metal parts. It also occurs if the flow is over a surface like the side of a ball &amp;amp; there is not enough pressure to keep it following the surface, or it flows off the edge of a surface like a propeller. The liquid is thrown off the surface and vacuum bubbles form, then collapse with no cushioning, giving a metallic crackling sound and eroding the propeller edges.&#xA;It is even worse for submarines. It says &quot;Here I am!&quot;  to the enemy. &#xA;You can demonstrate the effect with a garden hose &amp;amp; a bucket of water. Take it up a flight of stairs. remove the fittings, and drape the hose over the handrail or hold it deep in the bucket and start syphoning the water, then slam your finger over the inlet. You'll here a faint metallic crack from inside the highest part of the hose as the water keeps going, then slams back against itself. &lt;/p&gt;&#xA;" OwnerUserId="20261" LastEditorUserId="20261" LastEditDate="2019-04-27T04:08:27.340" LastActivityDate="2019-04-27T04:08:27.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28117" PostTypeId="1" AcceptedAnswerId="35899" CreationDate="2019-04-27T03:21:54.803" Score="0" ViewCount="135" Body="&lt;p&gt;Say I am trying to create a metal composite and have a mixture of 3 different metal powders each with varying weight %'s. I'm going to compact them into the shape of a ring by applying a force P. The inner and outer radius of the ring are given and I am asked to determine the thickness of this ring after compaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is, each of these metals have a true density (ie. 8900 kg/m3 for Al). Looking at this powder alone, if density is reached, can it be further compressed or does adding any more force P do nothing because it physically cannot obtain a greater density (due to atoms themselves etc.). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question, assuming you cannot compress more than the true density, is there a formula to determine how much force corresponds to the density achieved?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="16036" LastActivityDate="2020-06-21T14:04:45.460" Title="What are equations related to the compaction of composite metal powders?" Tags="|materials|metallurgy|composite|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28118" PostTypeId="2" ParentId="28115" CreationDate="2019-04-27T03:52:44.700" Score="0" Body="&lt;p&gt;You can write the parametric equation of trajectories:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y= -1/2gt^2+v_{y_{0}}t+y_0  \quad and\quad x= v{x_{0}}t+x_{0}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the A it is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y_A=-1/2gt^2+1/2*30*t+50 $$&lt;/span&gt;&#xA;And for B,we just use sin(theta)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ y_B=-1/2gt^2+50*sin(\theta)t+0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We eliminate t and find the tetha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we plug in the below equations to find t and x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ x_B=150- 50 cos(\theta)t$$&lt;/span&gt;&#xA;And &#xA;&lt;span class=&quot;math-container&quot;&gt;$$  x_A = 30 cos(30)t+0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$150-20cos(\theta)t  =0 \quad cos(\theta)t=7.5$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-04-27T17:52:27.687" LastActivityDate="2019-04-27T17:52:27.687" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28119" PostTypeId="1" CreationDate="2019-04-27T04:07:31.567" Score="0" ViewCount="156" Body="&lt;p&gt;Most people are probably familiar with &lt;a href=&quot;https://gfycat.com/boldwellmadeiberianmidwifetoad&quot; rel=&quot;nofollow noreferrer&quot;&gt;Leonardo hammer cams&lt;/a&gt;. But I'd like to make a motor-powered hammer that swings more like this (apologies for the crude diagram):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hjwdk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hjwdk.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hammer would fall due to gravity, then get dragged along the ground on its way back up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this diagram does not work, because the hammer on the way up needs to make a &gt;= 90 degree angle with the wheel radius in order to fall (step 2), but needs to collapse to &amp;lt; 90 degrees in step 4 in order to drag along the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this doable with simple, hardware store components? &lt;/p&gt;&#xA;" OwnerUserId="20262" LastActivityDate="2019-04-27T17:39:27.057" Title="How can I make a motor powered hammer swing more naturally than a Leonardo cam?" Tags="|mechanical-engineering|motors|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28120" PostTypeId="2" ParentId="28119" CreationDate="2019-04-27T05:46:58.330" Score="1" Body="&lt;p&gt;It may work if you modify the legs of the handles properly and possibly allow a bit of vertical play in the axis of the wheel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just as an alternative you can put the hammer passing straight through a spring loaded sleeved hole in the disk.&#xA;here is a sketch.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BqFfJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BqFfJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-04-27T17:39:27.057" LastActivityDate="2019-04-27T17:39:27.057" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28121" PostTypeId="2" ParentId="28117" CreationDate="2019-04-27T06:00:46.223" Score="0" Body="&lt;p&gt;Many factors come into play, such as the size of the grains, their roundness, and most important the granular aggregate distribution, so that in microscopic scale they are of not similar size but of a range of sizes which leave the least amount of cavity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how much force needed to achieve maximum compactness again depends on the shape of the grains. If the grains have sharp corners, they require less forces to flatten and compact. One would estimate that after  compression beyond just shaking  and initial compression, the force required to bring the grains past the yield point would increase fast.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-27T06:00:46.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28122" PostTypeId="2" ParentId="28115" CreationDate="2019-04-27T17:26:04.917" Score="1" Body="&lt;p&gt;First, use consistent angles. Define &lt;span class=&quot;math-container&quot;&gt;$\theta^*$&lt;/span&gt; to be the supplementary angle to the one illustrated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an intercept to occur, &lt;span class=&quot;math-container&quot;&gt;$x_A(t) = x_B(\theta^*,t)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y_A(t) = y_B(\theta^*,t)$&lt;/span&gt; for some &lt;span class=&quot;math-container&quot;&gt;$\theta^*$&lt;/span&gt; and time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; as the origin - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$x_A = 30 cos(30) t \qquad\qquad\&#xA;       y_A=-1/2gt^2 +30sin(30)t\qquad$$&lt;/span&gt; &#xA;&lt;span class=&quot;math-container&quot;&gt;$$x_B = 50 cos(\theta^*) t + 150 \qquad y_B = -1/2gt^2 + 50sin(\theta^*)t - 50$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve for the angle and time, and don't forget to change the angle back to it's supplement.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-04-27T17:26:04.917" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28123" PostTypeId="1" CreationDate="2019-04-27T21:32:41.913" Score="0" ViewCount="126" Body="&lt;p&gt;Im interested in creating a 3d printed pendulum clock, &#xA;all the diagrams and files I can find are mounting the pallet at the top of the escapement wheel which causes more complex design to avoid pendulum rod with the rest of the clock, &#xA;why cant the pallet be mounted downwards from the escapement wheel? is there any particular reason? &lt;/p&gt;&#xA;" OwnerUserId="20266" LastActivityDate="2023-02-06T11:08:00.967" Title="Pendulum clock, escapement pallet mounting position" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28124" PostTypeId="2" ParentId="28123" CreationDate="2019-04-27T22:34:22.430" Score="1" Body="&lt;p&gt;If pallet was to be on the bottom it would tend to fall into the escape gear at the left and right of each tick tack, so it needed to have ballasts balancing it, which adds more inertia than the escape gear can handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While when it is positioned on top the connection to pendulum is naturally balanced with lighter assembly.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-27T22:34:22.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28126" PostTypeId="1" CreationDate="2019-04-28T11:36:22.007" Score="1" ViewCount="124" Body="&lt;p&gt;I understand that most turboshaft engines tend to run at (or close to) their most efficient rpm all of the time, and it is the load that is changed e.g. the pitch of the blade pitch on helicopters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If everything else is kept the same and the load is increased, the rpm would obviously drop. Therefore, I assume that to maintain the constant rpm, the fuel flow must be increased as load is increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How exactly does the engine set the correct fuel flow to keep the engine at its most efficient rpm? I guess that on modern engines, this is probably done by a computer system, but is there also a mechanical way to do this, as would have been used in older engines?&lt;/p&gt;&#xA;" OwnerUserId="20042" LastActivityDate="2019-05-28T18:04:10.457" Title="How do turboshaft engines maintain a constant rpm regardless of load?" Tags="|turbines|engines|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28127" PostTypeId="2" ParentId="3369" CreationDate="2019-04-28T12:58:22.730" Score="-1" Body="&lt;p&gt;I wish our fan would expel more air? &#xA;I have considered changing the pitch of the blades by bending them but each blade, the pitch must be exact or else the blades will not rotate properly and could shake itself apart. &#xA;I'm going to try adding washers to one side of the blade which holds the blades in place to the rotating motor.&#xA;In theory, this should work as the weight and tilt of each blade should be exact at minimal effort. &lt;/p&gt;&#xA;" OwnerUserId="20272" LastActivityDate="2019-04-28T12:58:22.730" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28128" PostTypeId="2" ParentId="28126" CreationDate="2019-04-28T12:58:39.803" Score="2" Body="&lt;p&gt;In old steam machines a centrifugal governor was used: &lt;a href=&quot;https://en.wikipedia.org/wiki/Centrifugal_governor&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Centrifugal_governor&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later they used other governors with cables and mechanical devices like that.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-04-28T12:58:39.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28129" PostTypeId="1" CreationDate="2019-04-28T16:11:48.377" Score="3" ViewCount="1134" Body="&lt;p&gt;As an example of my question, imagine I have a small generator attached to a bicycle pedals. If this generator is connected to one light bulb, how would it compare with the generator being connected to multiple bulbs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My assumption is that, as you add more bulbs, either;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The resistance on the pedals would remain the same, but you would now have to pedal faster in order to light the bulbs to the same brightness, or;&lt;/li&gt;&#xA;&lt;li&gt;It would become harder to pedal when the extra bulbs are added, but the same rpm would be needed to produce the same brightness.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Which (if either) of my assumptions is the correct one?&lt;/p&gt;&#xA;" OwnerUserId="20042" LastActivityDate="2019-04-28T17:42:07.493" Title="Do generators produce a fixed load?" Tags="|generator|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28130" PostTypeId="1" CreationDate="2019-04-28T16:18:18.203" Score="0" ViewCount="516" Body="&lt;h1&gt;Description&lt;/h1&gt;&#xA;&lt;p&gt;We had been through several issues with K-factor handling of two teams due to the same name assigned by different standards.&lt;/p&gt;&#xA;&lt;p&gt;ANSI and DIN has defined the same term for the same purpose. However they defined different reference points, so there is a serious ambiguation: &lt;a href=&quot;https://en.wikipedia.org/wiki/Bending_(metalworking)#K-factor&quot; rel=&quot;nofollow noreferrer&quot;&gt;K-factor (ANSI)&lt;/a&gt;, &lt;a href=&quot;https://de.wikipedia.org/wiki/Biegeverk%C3%BCrzung&quot; rel=&quot;nofollow noreferrer&quot;&gt;K-factor (DIN)&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Expected behavior from the latter institute would be it should have chosen some different name for the &amp;quot;their&amp;quot; K-factor; like L-factor or M-factor (who cares?).&lt;/p&gt;&#xA;&lt;h1&gt;Question&lt;/h1&gt;&#xA;&lt;p&gt;I'm trying to find out the responsible institute that defined the same term without doing a research and caused the confusion. I couldn't get any helpful documents from the internet. Where can I find which K-factor is declared first?&lt;/p&gt;&#xA;" OwnerUserId="20275" LastEditorUserId="20275" LastEditDate="2021-01-18T03:26:13.810" LastActivityDate="2021-01-18T03:26:13.810" Title="Which K-factor (metal bending) is defined first, ANSI or DIN?" Tags="|standards|metal-folding|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28131" PostTypeId="2" ParentId="28126" CreationDate="2019-04-28T17:36:28.060" Score="1" Body="&lt;p&gt;Setting aside newer turboshaft engines that because of advances in technology do use variable rpm, they control the rotational speed is governed by a Full Authority Digital Engine Control (FADEC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the quote below is from &lt;a href=&quot;https://pdfs.semanticscholar.org/49a8/64f6083715c9879e1c231e48a45e8a9af765.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gianluigi Alberto Misté and Ernesto Benini&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;fuel&#xA;  control system which, by adjusting the amount of fuel injected,&#xA;  ensures that the rotor speed is as constant as possible for each&#xA;  type of flight maneuver. The main reason for choosing a&#xA;  constant rotational speed is linked to the supposed decrease in&#xA;  engine efficiency at part load (typical variations in speed do not&#xA;  exceed 15% &lt;a href=&quot;https://pdfs.semanticscholar.org/49a8/64f6083715c9879e1c231e48a45e8a9af765.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;), as well as because variable speed drives trains&#xA;  of resonant frequencies into the airframe&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This in conjunction with the pilot control on adjusting the rotors angle of attack and speed ensure optimal operation..&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-28T17:36:28.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28132" PostTypeId="2" ParentId="28129" CreationDate="2019-04-28T17:42:07.493" Score="5" Body="&lt;p&gt;As you add bulbs in parallel, the load impedance drops and to light all the bulbs you have to pedal harder. When the load impedance becomes less than the output impedance of the generator, then more work is expended in heating the generator coils than in lighting the bulbs. In the limit of near-zero load impedance (short circuit), all the pedal work is lost as I^2 x R losses in the generator windings. The pedal effort then does not change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add bulbs in series, then to light all the bulbs you again have to pedal harder. However, in this case the load impedance grows bigger and less current is allowed to flow through the load. The pedal effort declines and the bulb string grows dimmer. In the limit of infinite load impedance, the current goes to zero and the pedal effort goes to zero too. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-04-28T17:42:07.493" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28133" PostTypeId="1" CreationDate="2019-04-28T18:26:53.230" Score="0" ViewCount="185" Body="&lt;p&gt;I often read that turbine engines can run on almost any fuel source. I understand that turbine engines are less &quot;picky&quot; in terms of the fuel source that they can use, but surely there would still be some limitations as to which fuel source can be used e.g. a certain range for flash point, auto-ignition temperature etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, could a turbine engine be built from scratch to run on a fuel not normally suitable for turbine engines? And if so, would this engine be less efficient?&lt;/p&gt;&#xA;" OwnerUserId="20042" LastActivityDate="2019-05-28T21:02:35.373" Title="Can turbine engines really run on &quot;just about anything&quot;?" Tags="|turbines|fuel-economy|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28134" PostTypeId="2" ParentId="25853" CreationDate="2019-04-28T19:36:17.493" Score="0" Body="&lt;p&gt;For the first part, with an ideal gas in a closed system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ V = \mathrm{constant} \ \Rightarrow \ \frac{p_i}{T_i} =  \frac{p_f}{T_f} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second part, with an ideal gas in an open system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ p = \mathrm{constant} \ \Rightarrow \ \frac{V_i}{T_i} =  \frac{V_f}{T_f} = \frac{V_i + \Delta V}{T_f} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both cases are one equation with one unknown.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-04-28T19:36:17.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28135" PostTypeId="2" ParentId="28133" CreationDate="2019-04-28T19:46:57.643" Score="1" Body="&lt;p&gt;Your range of &quot;gasoline to sunflower oil&quot; (in a comment) is too small. Turbine engines run very well on natural gas, when used for gas pumping applications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combustion is continuous, so auto-ignition and flash point are not very relevant except for light-up, and you can use a different fuel for that if need be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only real limitation is clean burning. A burnt-on deposit of &quot;crud&quot; covering everything can seriously screw up the heat transfer of the hot end cooling system, and even the blade aerodynamics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The efficiency is mainly driven by the combustion temperature, and therefore by the chemical reactions involved. Hydrocarbons are all pretty similar in that respect, but (as a theoretical concept only!) a turbine fueled by say carbon monoxide, or by acetylene gas, or more practically by hydrogen, would be somewhat different.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-04-28T19:55:24.753" LastActivityDate="2019-04-28T19:55:24.753" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28136" PostTypeId="1" AcceptedAnswerId="28141" CreationDate="2019-04-28T20:08:08.697" Score="0" ViewCount="781" Body="&lt;p&gt;I often find these metal cylinders in the streets of London: a silver-colored tube with a round tapering at one end and a hole with no thread at the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J65Ea.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J65Ea.jpg&quot; alt=&quot;Top view of the metal cylinder&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8KQMx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8KQMx.jpg&quot; alt=&quot;Front view of the metal cylinder with hole&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are they?&lt;/p&gt;&#xA;" OwnerUserId="19384" LastEditorUserId="19384" LastEditDate="2019-12-07T18:12:46.613" LastActivityDate="2019-12-07T18:12:46.613" Title="What are these metal cylinders often found in the streets of London?" Tags="|metals|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28137" PostTypeId="1" AcceptedAnswerId="28146" CreationDate="2019-04-28T20:13:28.620" Score="4" ViewCount="139" Body="&lt;p&gt;I have this circuit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tIgot.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tIgot.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If D2 was a resistance, I could just combine it with R2 to get some composite resistance that I could put in series with R1. That way, I could calculate the resistance of the entire circuit and, thus, the total current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this is not the case here but I know a few things here:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;whatever voltage drops at D2, also drops at R2&lt;/li&gt;&#xA;&lt;li&gt;the current through R1 is the same as through R2+D2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But I do not know the current from the beginning, so I cannot calculate the voltage drop at R1. And the diode is non-ohmic, so I cannot find out anything about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, I do not see how I could apply Kirchhoff's laws or any other methods because I seem to be missing information... where do I start here?&lt;/p&gt;&#xA;" OwnerUserId="20277" LastActivityDate="2019-04-29T11:39:44.940" Title="How do I calculate voltage at an LED in parallel?" Tags="|electrical-engineering|circuits|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28139" PostTypeId="1" CreationDate="2019-04-28T20:38:43.317" Score="-1" ViewCount="95" Body="&lt;p&gt;I have been assigned a hands-on project with an adequate budget. The client has given a minimal but adequate description of what they want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, it seems impossible, especially at certain (or all) elevations. Even with a perfect pump (but I must use an actual pump from a manufacturer). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've linked a rudimentary illustration &lt;a href=&quot;https://i.stack.imgur.com/UzP4V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the necessary Net Positive Suction Head cannot exceed that which is required. Is this the case? (Due to max elevation) If so, I may recommend to the client a re-design - any recommendations?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please ask if any additional details are needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="20044" LastActivityDate="2021-01-06T00:03:00.293" Title="Is this pump design possible? (NPSH Question)" Tags="|fluid-mechanics|pumps|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28140" PostTypeId="2" ParentId="28139" CreationDate="2019-04-28T20:48:24.323" Score="0" Body="&lt;p&gt;Just mount the pump in the vertical section, either just above the tank or less than 2/3 the way up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the height you show the pump will probably not be able to self-prime.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can put the outlet on the side or even from the bottom of the tank that would be an advantage.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-04-28T21:59:37.990" LastActivityDate="2019-04-28T21:59:37.990" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28141" PostTypeId="2" ParentId="28136" CreationDate="2019-04-28T21:15:01.210" Score="4" Body="&lt;p&gt;These containers are &lt;em&gt;nitrous oxide canisters&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.talktofrank.com/drug/nitrous-oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.talktofrank.com/drug/nitrous-oxide&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Recreational_use_of_nitrous_oxide&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Recreational_use_of_nitrous_oxide&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16544" LastEditorUserId="16544" LastEditDate="2019-04-28T21:20:01.170" LastActivityDate="2019-04-28T21:20:01.170" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28142" PostTypeId="1" AcceptedAnswerId="28144" CreationDate="2019-04-28T21:34:24.237" Score="0" ViewCount="139" Body="&lt;p&gt;Is there a mathematical analysis of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydraulic_jumps_in_rectangular_channels#Jump_location&quot; rel=&quot;nofollow noreferrer&quot;&gt;submerged hydraulic jump&lt;/a&gt;, such as under what condition it is formed, the speed and trajectory of the circular flow therein, etc.? Is there a good mathematically sophisticated reference on this topic?&lt;/p&gt;&#xA;" OwnerUserId="20279" LastActivityDate="2019-04-29T01:19:05.960" Title="The mechanism of submerged hydraulic jump" Tags="|fluid-mechanics|hydraulics|open-channel-flow|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28143" PostTypeId="2" ParentId="28139" CreationDate="2019-04-28T22:34:36.947" Score="0" Body="&lt;p&gt;As mentioned in Solar Mike's answer the lower you have the pump the more efficient operation, and easier priming.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can even install it submerged under the water, as a sump pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the second part of your question, &#xA;Simply no unless you have a huge specially built pump available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a very rough estimate one kilowatt rated pump's uotput is 50- 80 Gallons per minutes, depending on the head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anything bigger better be 240 or 480 volts and you would need a dedicated electrical panel with an array of the pumps and proper electrical grounding such as red copper stick and bonding.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-04-28T22:34:36.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28144" PostTypeId="2" ParentId="28142" CreationDate="2019-04-29T00:46:42.760" Score="1" Body="&lt;p&gt;In open channels when the location of supercritical flow and hydraulic jump are very close, or restriction in down stream encourages higher elevation, submerged or partially submerged hydraulic jump happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, when they use flumes to measure the flow rate of an irrigation open channel because of the downstream conditions or obstacles the hydraulic jump may build up and partially or totally submerge the supercritical flow up-stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In closed pipe flow such as sewer system, the incidental blockage before an elbow or other constrain will cause submerged jump leading to deposit of more solids, that's why they design those locations with access openings, clean-outs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because of wide range of conditions leading to this situation, I think the only practical way of detailing and calculation the submerged jump is using FEM and simulation programs.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-04-29T01:19:05.960" LastActivityDate="2019-04-29T01:19:05.960" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28146" PostTypeId="2" ParentId="28137" CreationDate="2019-04-29T00:59:33.283" Score="2" Body="&lt;p&gt;One good place is to start is &lt;a href=&quot;https://media.digikey.com/pdf/Data%20Sheets/Lite-On%20PDFs/LTL-307EE.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;LTL-307EE&lt;/a&gt; datasheet. From the datasheet take look at current profile for the device. Below is the forward current and voltage characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/51RzZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/51RzZ.png&quot; alt=&quot;LED forward current and voltage graphs&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also take look at absolute maximum rating for the device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iIL9b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iIL9b.png&quot; alt=&quot;Absolute Max ratings&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have to size &lt;em&gt;R1&lt;/em&gt; resistor such that no more than in this case 30mA of current flows through the LED. &lt;em&gt;R1&lt;/em&gt; is called a current limiting resistor. I would size the R1 to make ensure 25mA of current, with will provide Relative Luminous Intensity of around 3.0. Like wise &lt;em&gt;R2&lt;/em&gt; can be used to manage current LED current flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this help you get started. &lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-04-29T11:39:44.940" LastActivityDate="2019-04-29T11:39:44.940" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28148" PostTypeId="1" CreationDate="2019-04-29T05:57:39.027" Score="0" ViewCount="207" Body="&lt;p&gt;How to get the bending moment for tie member?  For the tie member, there is no uniformly distributed load and concentrated load, right? How to get the bending moment? The tension force was found to be 18.2kN btw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/U6K3F.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="9254" LastEditorUserId="1832" LastEditDate="2019-04-29T23:13:27.397" LastActivityDate="2019-09-27T19:02:40.380" Title="How to get the bending moment for tie member?" Tags="|beam|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28149" PostTypeId="1" CreationDate="2019-04-29T13:23:22.073" Score="0" ViewCount="281" Body="&lt;p&gt;I understand what an implicit and explicit form of finite-difference (FD) discretization for the transient heat conduction equation means. But I am not able to understand if it is possible to categorize the discretization of steady state heat conduction equation (without a source term, i.e Laplace equation) as implicit or explicit scheme as well?(or is it just implicit form of discretization as we use iterative solvers?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Whenever I try to understand the meaning of explicit an implicit solvers, the notes and other resources always explain them using a (transient) time-varying equation. According to such resources For a transient equation if the solution for the next time step is given in terms of current time step values (which we already know, either by initial condition or by previous calculation) it is called as an Explicit solution, and it could be solved for directly without forming a system of equation. If the solution for next time step value is given in terms of current time step values (which we know) and next time step values (which we don't know) it leads to an implicit solution and this gives us a system of equations which is solved for by iterative solvers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My confusion is, if we take a steady state equation there are no time steps involved (so there is no current and next time step) so what would be an explicit and implicit solution? and why would an explicit solution require an iterative solver as they can be solved for directly?&lt;/p&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="16275" LastEditDate="2019-04-30T02:12:45.990" LastActivityDate="2019-04-30T06:29:00.087" Title="Implicit and Explicit finite difference discretization of steady state heat conduction equation" Tags="|numerical-methods|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28150" PostTypeId="1" CreationDate="2019-04-29T13:24:55.573" Score="0" ViewCount="610" Body="&lt;p&gt;I am new to SolidWorks and am designing an enclosure. My model looks like this - &lt;a href=&quot;https://i.stack.imgur.com/N5f0j.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N5f0j.png&quot; alt=&quot;Enclosure&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to add a projection onto the sides of the surface with the slotted hole, so that the front view vaguely looks like this (just for representation,I have omitted the slotted rectangle) -&lt;a href=&quot;https://www.google.com/search?q=rectangle%20with%20semicircular%20ends&amp;amp;rlz=1C1CHBF_hiIN791IN791&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwislp6usvXhAhXaxMQBHRnsAtsQ_AUIDigB&amp;amp;biw=1366&amp;amp;bih=626#imgrc=2DW79XoDUXZJtM:&quot; rel=&quot;nofollow noreferrer&quot;&gt;Front View&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I do this?&lt;/p&gt;&#xA;" OwnerUserId="20285" LastActivityDate="2019-05-29T20:03:30.490" Title="How to extrude an edge in SolidWorks?" Tags="|mechanical-engineering|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28151" PostTypeId="2" ParentId="28149" CreationDate="2019-04-29T14:16:59.977" Score="1" Body="&lt;p&gt;A non-iterative solution for steady-state heat flow must be implicit. You are solving for the temperatures &quot;everywhere&quot; in the structure in a single &quot;step&quot;, so every boundary condition must affect the entire temperature distribution in the equations you are solving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Iterative solutions for steady-state conditions can be either implicit or explicit. You can think of each iteration as defining some sort of time increment, and calculating how the temperature distribution changes from the initial guess to the final steady state conditions. Implicit methods are likely to converge faster, but explicit methods will work and have some advantages - for example, it is much easier to program an implicit method on a massively parallel computer system like a GPU with hundreds or thousands of cores.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="9" LastEditDate="2019-04-30T06:29:00.087" LastActivityDate="2019-04-30T06:29:00.087" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28152" PostTypeId="2" ParentId="26850" CreationDate="2019-04-29T15:09:03.157" Score="2" Body="&lt;p&gt;As stated above, I went for 'parallelograms' As it happens,the wing nut clamping against a square M6 nut between bars has proven sufficient to hold the doors in any fixed position&lt;a href=&quot;https://i.stack.imgur.com/ekswI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ekswI.jpg&quot; alt=&quot;The bars holding the flap open&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9Au0f.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9Au0f.jpg&quot; alt=&quot;The bars in the closed position along case edge&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20089" LastActivityDate="2019-04-29T15:09:03.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28153" PostTypeId="1" CreationDate="2019-04-29T19:10:27.483" Score="0" ViewCount="32" Body="&lt;p&gt;What is the principle of a fogging machine ?&#xA;on what process a fogging machine depends ? &#xA;distillation or vapourization ?&#xA;what is the harmful effect of fogging machine ?&lt;/p&gt;&#xA;" OwnerUserId="20288" LastActivityDate="2019-04-29T19:10:27.483" Title="What is the smoke made by the fogging machine? Is it harmful to humans?" Tags="|chemistry|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28154" PostTypeId="2" ParentId="28150" CreationDate="2019-04-29T19:45:00.123" Score="1" Body="&lt;p&gt;Disclaimer - Your question wasn't all that clear, so I'm guessing the shape you want...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are so, so many ways to model the shape you're after, and which one is best will depend on your design intent. What dimensions are the critical ones? You want to be able to change those in one sketch and then have the rest of the model update if you need to change things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to assume the rectangular cutout is the critical item, and that the wall thickness should follow this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully the screen capture below is useful, and shows some tricks you can use. Let me know if you have any more specific questions - remember that drawings/images/etc showing what you're after will &lt;em&gt;always&lt;/em&gt; be more helpful than words.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://media.giphy.com/media/U5UsdAszu2DB2aNb6V/giphy.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://media.giphy.com/media/U5UsdAszu2DB2aNb6V/giphy.gif&quot; alt=&quot;SW modelling a basic enclosure as per request&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-04-29T19:45:00.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28155" PostTypeId="1" CreationDate="2019-04-29T20:01:42.643" Score="1" ViewCount="61" Body="&lt;p&gt;Most home reverse osmosis water systems use a low pressure bladder type tank that has 5-10 PSI. That way the back pressure on the RO system is not too high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use a glass-lined tank of the type used for people with well water and have it pressurized to a faucet-type pressure of around 30 PSI. The only way I can figure out how to do this is to have two tanks, one unpressurized and the other pressurized. So, the RO system would drip water into the unpressurized tank. Then a pump would draw water out of the unpressurized tank and force it into the pressure tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the only way to do this, or is there a simpler solution?&lt;/p&gt;&#xA;" OwnerUserId="1672" LastEditorUserId="6767" LastEditDate="2019-12-01T21:21:08.140" LastActivityDate="2019-12-01T21:21:08.140" Title="How can I set up a reverse osmosis water system to use a pressure tank?" Tags="|water-resources|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28156" PostTypeId="1" AcceptedAnswerId="29621" CreationDate="2019-04-29T21:42:05.840" Score="0" ViewCount="46" Body="&lt;p&gt;So, I have a problem with fatigue cracking originating from hot punched holes. From failure forensic reviews, it seems like surface conditions contributed to some of the failures, especially where there are leftover shear marks and laps from punching operations. I did a quick research but it doesn't seem like there are well established standards regarding the finishes of holes generally in terms of hole surface quality. Am I over thinking and that nobody had the same problem previously? Or are there some standard that I am not already aware of that exists? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of my thoughts is to establish Ra Rp Rz acceptable ranges that can be applied for the holes. But I have doubts about measurement consistency as those measurements are usually more geared towards flat surfaces. What are your thoughts on this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for all your help!!&lt;/p&gt;&#xA;" OwnerUserId="13526" LastActivityDate="2019-08-12T23:37:56.703" Title="Engineering standards or testing standards for hot punched metal holes for rolled steel" Tags="|metallurgy|surface-preparation|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28157" PostTypeId="2" ParentId="28083" CreationDate="2019-04-29T23:09:45.090" Score="2" Body="&lt;h1&gt;Both formulae cap the column's maximum stress to its yield stress&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Both the Rankine-Gordon and Perry-Robertson formulae are means of determining the maximum average stress (&lt;span class=&quot;math-container&quot;&gt;$\sigma = P/A$&lt;/span&gt;) a given column can support. Due to the fact that short columns will collapse under compression rather than buckling, it's important that these equations cap the result to the material's compressive yield stress. As we will see, this is precisely what both equations do.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Rankine-Gordon Formula&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The formula is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma = \dfrac{\sigma_c}{1+a\left(\dfrac{L}{k}\right)^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\sigma_c$&lt;/span&gt; is the material's compressive yield stress;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a = \dfrac{\sigma_c}{\pi^2 E}$&lt;/span&gt; (where &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is the material's Young's modulus), but is usually determined experimentally;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the column's length;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$k = \sqrt{\dfrac{I}{A}}$&lt;/span&gt;, the column's least radius of gyration.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, in the case where the column is infinitely long, we know the column should buckle immediately. This is precisely what happens, since &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; is inversely proportional to &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if instead we have a very short beam (&lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; tending to zero), the equation simplifies to &lt;span class=&quot;math-container&quot;&gt;$\sigma = \dfrac{\sigma_c}{1} = \sigma_c$&lt;/span&gt;. Since you can't have &lt;span class=&quot;math-container&quot;&gt;$L &amp;lt; 0$&lt;/span&gt;, we get that the compressive yield stress is the maximum value possible according to the Rankine-Gordon Formula.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Perry-Robertson Formula&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This formula is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma = \dfrac{1}{2}\left(\sigma_c + \sigma_e(1+\theta) - \sqrt{(\sigma_c + \sigma_e(1+\theta))^2 - 4\sigma_c\sigma_e}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\sigma_e$&lt;/span&gt; is the average stress (&lt;span class=&quot;math-container&quot;&gt;$P/A$&lt;/span&gt;) associated with the beam's Euler buckling load&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\theta = \dfrac{w_{0,i}c}{k}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$w_{0,i}$&lt;/span&gt; is the initial imperfection&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is the maximum distance to the section's centroid (in the relevant direction)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Once again, let's look at the case of an infinitely long column. In this case, we know that &lt;span class=&quot;math-container&quot;&gt;$\sigma_e = 0$&lt;/span&gt;, so the equation simplifies to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sigma &amp;amp;= \dfrac{1}{2}\left(\sigma_c - \sqrt{\sigma_c^2}\right) \\&#xA;&amp;amp;= \dfrac{1}{2}\left(\sigma_c - \sigma_c\right) \\&#xA;\sigma &amp;amp;= 0&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This formula also has infinitely long columns buckling immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the column with a length approaching zero. This is a more complicated case. Such a beam would have &lt;span class=&quot;math-container&quot;&gt;$\sigma_e = \infty$&lt;/span&gt;, but let's pretend we can actually play with such numbers. Let's also adopt the simplification suggested by Robertson of assuming &lt;span class=&quot;math-container&quot;&gt;$\theta \approx 0.003\lambda = 0.003\dfrac{L}{k}$&lt;/span&gt;, which when combined with our zero-length column gives us &lt;span class=&quot;math-container&quot;&gt;$\theta = 0$&lt;/span&gt;. The equation then simplifies to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sigma &amp;amp;= \dfrac{1}{2}\left(\sigma_c + \infty - \sqrt{(\sigma_c + \infty)^2 - 4\sigma_c\infty}\right) \\&#xA;&amp;amp;= \dfrac{1}{2}\left(\sigma_c + \infty - \sqrt{\sigma_c^2 + 2\sigma_c\infty + \infty^2 - 4\sigma_c\infty}\right) \\&#xA;&amp;amp;= \dfrac{1}{2}\left(\sigma_c + \infty - \sqrt{\sigma_c^2 - 2\sigma_c\infty + \infty^2}\right) \\&#xA;&amp;amp;= \dfrac{1}{2}\left(\sigma_c + \infty - \sqrt{(\sigma_c - \infty)^2}\right) \\&#xA;&amp;amp;= \dfrac{1}{2}\left(\sigma_c + \infty - |\sigma_c - \infty|\right) \\&#xA;&amp;amp;= \dfrac{1}{2}\left(\sigma_c + \infty - (\infty - \sigma_c)\right) \\&#xA;&amp;amp;= \dfrac{1}{2}\left(\sigma_c + \sigma_c\right) \\&#xA;\sigma &amp;amp;= \sigma_c&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We had to play a bit loose with our mathematical concepts, but this demonstrates that even for very short columns, the stress is never greater than the compressive yield stress using the Perry-Robertson Formula.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Given this, we can see the answer to the question is &lt;strong&gt;&lt;em&gt;YES&lt;/em&gt;&lt;/strong&gt;, both methods always give results which are lower than the material's yield stress.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd mention, however, that the proper term is just &quot;ultimate stress&quot;, not &quot;buckling stress&quot; as in the question. Short columns collapse under compression at stresses far below the stress required for them to buckle. Basically: for short columns, the ultimate stress is roughly equal to the yield stress &lt;span class=&quot;math-container&quot;&gt;$\sigma_c$&lt;/span&gt;. For slender columns, it's equal to the buckling stress since that's lower.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;1&lt;/sup&gt; If you want to be a real pedant, you could argue the answer is NO, since both methods give zero-length columns ultimate stresses which are equal to (not less than, as asked by the question) the yield strength. But that would require you to actually consider zero-length columns as a real thing which they obviously aren't, and that you don't accept an implicit &quot;or equal to&quot; after the question's &quot;less than&quot;.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-04-29T23:09:45.090" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28158" PostTypeId="2" ParentId="26891" CreationDate="2019-04-30T00:40:43.833" Score="3" Body="&lt;p&gt;I'm not sure I totally understand your question. I believe you are saying in your third paragraph that &quot;geodetic surveying has greater accuracy than plane surveying&quot;? This is not really true. Every survey is as accurate as the methods employed to perform the survey irrespective of if the work is done in a geodetic coordinate system or a rectangular plane coordinate system.&#xA;Your coordinate system is made up of several parts, an ellipsoid which approximates the shape of the earth, a datum which positions this ellipsoid relative to the earth, and for what you are calling &quot;plane surveying&quot;, a map projection which &quot;projects&quot; a section of this ellipsoid onto a flat plane so we can use &quot;grid&quot; coordinates instead of &quot;spherical&quot; coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your two questions it appears you are mixing the terms accuracy and precision. These are very different things. Accuracy is closeness to the &quot;true&quot; value and is never known. Precision is closeness between multiple measurements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In designing a survey, we look at the precision of the instruments and procedures we will use to determine an expected precision for the observations. We then look at the observations we will make (including redundant observations) and from all of this we build a stochastic model which we can use to predict the precision of our end results. We adjust our observation model until we are achieving the precision we are seeking for the survey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until now we have not mentioned a geoid, which throws another whole level of complexity into the discussion. The &quot;z&quot; component above is based on ellipsoidal elevations. Typically we work in orthometric elevations (elevations relative to some form of mean sea level). The geoid model is the mathematical model which attempts to model this &quot;mean sea level&quot; surface, and allows use to convert our ellipsoidal elevations to orthometric elevations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you can clarify what it is you are trying to ask so we can give a better answer.&lt;/p&gt;&#xA;" OwnerUserId="20290" LastActivityDate="2019-04-30T00:40:43.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28159" PostTypeId="1" CreationDate="2019-04-30T02:12:15.730" Score="1" ViewCount="58" Body="&lt;p&gt;Pardon me if this is a silly question, because I don't quite study RF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have a set of transmitter (radio + antenna) and a set of receiver (radio + antenna), at location A and B respectively. Then I transmit a signal at a certain power, for example 5W, and I measured the signal at the receiver, say 3.4W, due to losses incurred when it travels from point A to B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, I swapped the locations of the transmitter and receiver. I have also rotated the antennas by 180 degrees, this is to ensure consistent gain at the angle towards the other antenna, if directional antennas are used (I'm not sure if I'm making sense here).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will this result in the same received power (i.e. same loss)? What are some things that would affect the results and that I have overlooked?&lt;/p&gt;&#xA;" OwnerUserId="20291" LastActivityDate="2019-04-30T02:12:15.730" Title="Swapping transmitter and receiver locations" Tags="|electrical-engineering|rf-electronics|radio|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28160" PostTypeId="2" ParentId="26752" CreationDate="2019-04-30T04:57:42.330" Score="0" Body="&lt;p&gt;I think that the umbrella-shaped object may be some sort of structure built to retain any water leak as well as collect any possible rain water and add to the water stored in the pipe. Another reason could be to stop any people from climbing the water tank. &lt;/p&gt;&#xA;" OwnerUserId="20177" LastActivityDate="2019-04-30T04:57:42.330" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28161" PostTypeId="1" AcceptedAnswerId="28162" CreationDate="2019-04-30T05:04:32.567" Score="2" ViewCount="1388" Body="&lt;p&gt;I have designed a system using a minimum realization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the A, B, C matrices and I have assigned closed loop poles for the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the place command to find my K matrix. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if it is possible to use the lsim matlab command to simulate my closed loop system (A-B*K)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$x' = Ax+Bu$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$u=-Kx$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$x' = (A-BK)x$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$y=Cx$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All examples I have found for lsim use the system transfer function or the open loop system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen examples such as this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;t = 0:0.01:10;         % simulation time = 10 seconds&#xA;u = zeros(size(T));    % no input&#xA;x0 = [0.1 0.1 0.1];    % initial conditions of the three states&#xA;sys= ss(A,B,C,D);&#xA;lsim(sys,u,t,x0)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I am not sure how to use this with the closed loop (A-B*K) system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated! Thank you.&lt;/p&gt;&#xA;" OwnerUserId="20292" LastEditorUserId="20292" LastEditDate="2019-04-30T23:14:19.673" LastActivityDate="2019-05-01T08:28:05.623" Title="How to use lsim in matlab for a closed loop system" Tags="|control-engineering|control-theory|matlab|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28162" PostTypeId="2" ParentId="28161" CreationDate="2019-04-30T09:00:07.213" Score="0" Body="&lt;p&gt;Yes, you can using the &lt;a href=&quot;https://uk.mathworks.com/help/control/ref/feedback.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;feedback&lt;/code&gt;&lt;/a&gt; command. However, if your open-loop system is defined by your state-space system &lt;code&gt;sys= ss(A,B,C,D)&lt;/code&gt;, then your closed-loop system would be defined as follows (and not by &lt;code&gt;A-B*K&lt;/code&gt; as you suggest):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CL_loop = feedback(sys,K); % K is your controller/compensator, this needs to be an LTI object&#xA;lsim(CL_loop,u,t,x0); % you might have additional states due to the feedback loop, and so would need additional IC&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Note that you may also need to change your input vector, as it's now representing your desired trajectory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update after comment &amp;amp; question edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's just a matter of creating a new state-space representation with the appropriate matrices, e.g.:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;CL_loop = sys(A-B*K,zeros(size(B)),C,zeros(size(D))); % implement x'=(A-B*K)*x &amp;amp; y = C*x&#xA;lsim(CL_loop,u,t,x0);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2019-05-01T08:28:05.623" LastActivityDate="2019-05-01T08:28:05.623" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28163" PostTypeId="1" CreationDate="2019-04-30T09:31:07.640" Score="1" ViewCount="196" Body="&lt;p&gt;All values of metric units are expressed in decimal or scientific notation. Some imperial units, use fraction numbers as primary notation (3/4&quot; for 0.75 inch).&#xA;As far as i know, only inch and foot values are using this notation. Are there any other units using this notation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;My question is about the &lt;em&gt;written&lt;/em&gt; form&lt;/li&gt;&#xA;&lt;li&gt;IMO it is not about language but more engineering related. To be more specific: the background of my question is about how to store those notations in digital form (so maybe ask this at stackoverflow?). In other words: what data type should be used for number values of specific units of measurement. Millimeters can always be stored as decimal. For inches this is weird, and a fraction number is expected.&lt;/li&gt;&#xA;&lt;li&gt;We have a system, used to store a lot of technical data (characteristics of technical components). We are wondering if the data type should be dependent on the System of Measurement used, imperial or metric:&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;metric: data type = decimal&lt;/li&gt;&#xA;&lt;li&gt;imperial: data type = fraction &lt;strong&gt;BIG question&lt;/strong&gt;: for inch and ???&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="20293" LastEditorUserId="20293" LastEditDate="2019-04-30T18:32:56.540" LastActivityDate="2019-04-30T18:32:56.540" Title="Which imperial units of measurement use vulgar (simple) fraction numbers?" Tags="|measurements|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28164" PostTypeId="1" CreationDate="2019-04-30T11:48:00.007" Score="0" ViewCount="220" Body="&lt;p&gt;Steam at 1 MPa, 300°C flows through a 30 cm diameter pipe with an average velocity of 10 m/s. The&#xA;mass flow rate of this steam is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A) 0.731 kg/s B) 2.74 kg/s&#xA;C) 3.18 kg/s D) 3.78 kg/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me out please? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought I had to use m=PVA &#xA;P=1000 pa&#xA;V=0.2579 (property table)&#xA;A= pi*15^2&#xA;So m=1000*0.2579*((pi*15^2)/1000)= 183 kg/s which is not correct.&lt;/p&gt;&#xA;" OwnerUserId="20294" LastActivityDate="2019-04-30T13:09:03.063" Title="Steam at 1 MPa, 300°C flows through a 30 cm diameter pipe with an average velocity of 10 m/s. The mass flow rate of this steam is:" Tags="|thermodynamics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28166" PostTypeId="2" ParentId="28148" CreationDate="2019-04-30T17:18:52.340" Score="0" Body="&lt;p&gt;While the term 'tie-member' is typically used to describe an axial only link member, you must consider any loading effects. As you have shown it in the image it looks like a simple link. But just because the terminology suggests that the member is axial only does not necessarily mean that it is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other connecting elements or applied loading could bending of the member. So you must identify any other loading that may be applied to the element. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also a bending moment can be generated by axial forces that have eccentricity.&#xA;If your connections are off center, then there would be some bending moment due to only the axial load.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-04-30T17:18:52.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28167" PostTypeId="2" ParentId="28148" CreationDate="2019-04-30T18:19:59.817" Score="1" Body="&lt;p&gt;You say &quot;[t]he tension force was found to be 18.2kN&quot;. I assume you were the one who got this result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you did, you will have noticed that this force is perfectly parallel to the tie's longitudinal axis and that there was no perpendicular (transversal) component applied to the tie on either end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, we can look at the tie in isolation and rotate it to make it horizontal. What we then have looks exactly like a simply-supported beam (since the tie's pinned connections allow rotation but not displacement). It just so happens that beam has no transversal external loads (concentrated forces or distributed loads).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the bending moment for a beam with no transversal external loads? Well, it's just zero. The beam suffers no bending moment. You're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only potential wrinkle is the fact that everything in the world has at least one load applied to it: its weight. Though it's not shown in your image, that tie does have to support its own weight. That weight will be pointing straight down, so it can be decomposed into a parallel and a transversal component. The parallel component will alter the tie's axial load (it will actually vary along the tie's length), while the transversal component will generate bending moment. Once again, the comparison to the simply-supported beam applies, and the bending moment diagram will be identical to the one for such a beam, just remember to calculate it using only the weight's transversal component, not its totality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, since ties are theoretically meant to be axial-only and since their self-weight is usually much lower than the loads applied to the entire structure, it is very common to calculate the ties' weight and apply them as concentrated forces at the endpoints (half to each node, of course). This will give an accurate description of the axial force but will eliminate the bending moment on the assumption that it's negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your question, if the weight (or density, through which you could calculate the weight) wasn't given, then it's probably fair to assume the teacher wants you to disconsider the weight, in which case the tie's bending moment is simply zero.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-04-30T18:19:59.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28170" PostTypeId="1" AcceptedAnswerId="28171" CreationDate="2019-05-01T08:01:17.103" Score="1" ViewCount="669" Body="&lt;p&gt;I am looking for non-destructible way which can be performed in home conditions to tell if the material is titanium (in my case it is a mug). Advices like &quot;grind it&quot; or &quot;use electronic microscope&quot; could be useful but I cannot apply them as stated above. I am targeting primarily aluminium and steel alloy because of their competitive prices and availability. In pure form detection would be easy (weight) but what about the alloy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am also asking for non-expert level ways -- I can weight something, boil water, measure temperature or resistance, those kinds of stuff. I am not a chemist for sure.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="20299" LastActivityDate="2019-05-01T13:07:26.277" Title="How to make sure the material is titanium?" Tags="|materials|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28171" PostTypeId="2" ParentId="28170" CreationDate="2019-05-01T08:47:53.490" Score="4" Body="&lt;p&gt;Do what &lt;a href=&quot;https://en.wikipedia.org/wiki/Archimedes%27_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Archimedes&lt;/a&gt; did in the 3rd century BCE, when the king asked him to find a nondestructive way to ascertain whether this crown was made of pure gold, as request, or whether it was made of gold alloy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accurately find the mass of the object and also accurately find its volume. Divide the mass by the volume to obtain the density of the object and compare that to published values of density for the material the object is supposedly made from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An accurate set of scales will give you the mass and by collecting all the water displaced by placing the object in a container of water and accurately measuring the volume using a finely &lt;a href=&quot;https://en.wikipedia.org/wiki/Graduated_cylinder&quot; rel=&quot;nofollow noreferrer&quot;&gt;graduated measuring cylinder&lt;/a&gt; will give you the volume you need.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2019-05-01T08:47:53.490" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28172" PostTypeId="1" CreationDate="2019-05-01T10:19:29.157" Score="0" ViewCount="65" Body="&lt;p&gt;I have an application where a person drives a bush into its housing using a hammer of about 3kgs.&#xA;Suppose the same application is to be done with a hydraulic press,is it possible to calculate the press tonnage(probably using an approximated calculation of the kinetic energy of the hammer).&#xA;TIA&#xA;Thomas&lt;/p&gt;&#xA;" OwnerUserId="16939" LastActivityDate="2020-05-25T16:07:10.643" Title="Calculate press tonnage from impact momentum" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28173" PostTypeId="1" CreationDate="2019-05-01T10:28:01.783" Score="0" ViewCount="33" Body="&lt;p&gt;I want to design a control system with microcontroller to control position or orientation of antenna with BLDC motor with high precision like 0.1 deg and I don't have idea which motor position controller IC I can use or if there is an other idea please share with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="20300" LastEditorUserId="110" LastEditDate="2019-05-01T15:33:57.320" LastActivityDate="2019-05-01T15:33:57.320" Title="Control systems to control position of antenna with motor?" Tags="|electrical-engineering|control-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28176" PostTypeId="2" ParentId="28170" CreationDate="2019-05-01T12:54:50.007" Score="2" Body="&lt;p&gt;In adition to measuring the density of your object as described by Fred there is a few other tests you can easily make. You can try to measure:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Heat capacity of your object. By heating the object in hot water measure tempertaure. Then move it over to another cooler water container where you know the volume and measure the temperture of the water after this.Then compare to table values&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Specific electrical resistance is also very quick to do if you happen to have a multimeter. This would very quickly rule out aluminium. Requires very minimal setup time to do this measurement.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now density is best but adding these to the mix can give you a better confidence in your results. Resistance measurement also works if you suspect there is a hollow cavity in your object.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2019-05-01T13:07:26.277" LastActivityDate="2019-05-01T13:07:26.277" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28177" PostTypeId="2" ParentId="28172" CreationDate="2019-05-01T14:02:09.693" Score="1" Body="&lt;p&gt;Let's say that your hammer strike, moving at a speed of 4 meters/s, drives the bushing into its housing by 1 centimeter .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the kinetic energy of hammer &lt;span class=&quot;math-container&quot;&gt;$ K_e=\frac{1}{2}m*v^2 $&lt;/span&gt; has Been used to move the bushing. Annotating &quot;d&quot; as the distance the bushing travels and ignoring its weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F_{average}d=- \frac{1}{2}m*v^2 \\ F_{average}= \frac{ \frac{1}{2}m*v^2}{d} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_{press}= \frac{\frac{1}{2}3kg*4^2}{1cm}*\frac{100cm}{1meter}=2400kg $$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-01T14:02:09.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28179" PostTypeId="1" CreationDate="2019-05-01T18:18:58.570" Score="-1" ViewCount="155" Body="&lt;p&gt;How do I calculate the width of a slot if the overall width of the part is .310 and from the edge of the part to the center of the slot is .155?&lt;/p&gt;&#xA;" OwnerUserId="20305" LastActivityDate="2019-05-01T23:47:41.233" Title="Calculating slot width or opening" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28180" PostTypeId="1" CreationDate="2019-05-01T19:30:01.763" Score="0" ViewCount="66" Body="&lt;p&gt;In a 4-bar linkage, all the joints are regarded as hinge joints. However, when I searched &quot;spherical 4-bar linkage system&quot;, I didn't get a 4-bar system with spherical joints but instead a 4 bar system curved over the surface of a sphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone show me an example of a 4-bar linkage system with straight rods that uses spherical joints? I want to understand the mathematics of the kinematic equations for one and I need a standard example. Is there a name for this? &lt;/p&gt;&#xA;" OwnerUserId="648" LastActivityDate="2019-05-01T19:30:01.763" Title="Is there a spherical joint equivalent to a 4 bar linkage?" Tags="|mechanical-engineering|linkage|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28181" PostTypeId="1" CreationDate="2019-05-01T19:33:16.770" Score="4" ViewCount="195" Body="&lt;p&gt;In an Uncle Scrooge (or Scrooge McDuck, not sure of the translation here) comic named &lt;a href=&quot;https://en.wikipedia.org/wiki/Treasure_Under_Glass&quot; rel=&quot;nofollow noreferrer&quot;&gt;&quot;Treasure Under Glass&quot;&lt;/a&gt;, Scrooge orders a giant glass dome to be manufactured in order to aid his endeavours exploring a 17th shipwreck containing a valuable treasure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The giant dome, which is Donald's idea is described in the following image by a fine engineering drawing &lt;s&gt;since I couldn't find a pdf of the comic&lt;/s&gt; :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cIhaO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cIhaO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary objective is to clear the area of sharks so it can be explored on foot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the comic, I made the following rough estimates : &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The spanish galeon is 60m long (based on &lt;a href=&quot;https://en.wikipedia.org/wiki/Neptune_(galleon)&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Neptune&lt;/a&gt;), which would make the dome bottom diameter approximately equal to 100 m&lt;/li&gt;&#xA;&lt;li&gt;We will assume the dome is a perfect half-sphere (assuming from the drawings in the comic)&lt;/li&gt;&#xA;&lt;li&gt;Dome height estimated at 50 m according to the previous statement&lt;/li&gt;&#xA;&lt;li&gt;The dome is tightened to the bottom of the ocean with several ropes as in the drawing, each rope being fixed into the wet sand at its end.&#xA;According to the comic, we see that there are 5 ropes extending from one side to another, distributed in an even fashion.  &lt;/li&gt;&#xA;&lt;li&gt;Following 3., the estimated depth at the bottom of the wreck is 100 m&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Following these assumptions, would such a thing be feasible from an engineering point of view, assuming close to infinite funds and that Scrooge agrees to pay the bill ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular :&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What kind of tension would each rope have to withstand ? Would&#xA;regular ropes be enough or should they have used steel cables ?&lt;/li&gt;&#xA;&lt;li&gt;How would those be tighten into the sand in a reliable way ?&lt;/li&gt;&#xA;&lt;li&gt;What kind of tremendous forces would the dome have to withstand ?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4946" LastEditorUserId="4946" LastEditDate="2019-05-02T08:20:19.183" LastActivityDate="2019-09-30T23:01:45.207" Title="Is Scrooge's giant glass dome feasible in real life?" Tags="|structural-engineering|civil-engineering|pressure|compressors|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28183" PostTypeId="1" CreationDate="2019-05-01T20:57:13.267" Score="0" ViewCount="142" Body="&lt;p&gt;I am taking a nonlinear control class. As we were discussing stability, we came across 'weaker' and 'stronger' stability conditions. For example asymptotic stability (AS) is stronger than stability in the sense of lyapunov (SISL), and exponential stability (ES) is stronger than AS. Then we across local and global definitions of stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question then arises, if for some system we have a global form of some stability, say AS, and a local form of another 'stronger' stability for the same system, say ES, then which statement is actually stronger. Or does it so happen that global AS + local ES =&gt; global ES?&lt;/p&gt;&#xA;" OwnerUserId="20310" LastActivityDate="2019-05-02T01:32:11.527" Title="If a system is globally asymptotically stable AND locally exponentially stable, can we say claim that it is globally exponentially stable?" Tags="|control-engineering|stability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28184" PostTypeId="1" CreationDate="2019-05-01T21:06:28.957" Score="2" ViewCount="36" Body="&lt;p&gt;I'm currently a sophomore Mechanical engineering student and need some help with an equation for my project. I am coding a simulation in Matlab and have been given a very short time frame to do it. I chose to model the thermal efficiency of a house with the main output being cost to run heating/cooling system per day. Due to time constraints I tried to simplify the analysis a bit. I am neglecting heat transfer through siding, framing, the roof/ceiling, the floor, people inside, and appliances. Basically just four walls of insulation. I am looking at windows and solar heat gain from them. Since I am ignoring siding I am also ignoring solar heat gain transfer to it. Currently I have the program written to produce an output on a graph, but my equation to calculate the new inside temperature doesn't seem correct. I am using conduction through the insulation, convection from the ambient temperature inside, and solar heat gain due to the windows. I have restricted the size of the house to a 30'X40' single story with 8ft walls. The walls were also 1ft thick. So I ended up with the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q_(conduction) + Q_(solar H.G. from windows) = Q_(convection due to ambient inside temp)&#xA;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q_(conduction) = -k&lt;em&gt;A&lt;/em&gt;(T_wall-T_outside)/(Thickness of wall)&lt;br&gt;&#xA;[A = total surface area of walls on house]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q_(solar H.G. from windows) = (Area of window)&lt;em&gt;(Shading Coefficient)&lt;/em&gt;(Solar Cooling Load Factor)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q_(convection) = h&lt;em&gt;A&lt;/em&gt;(T_wall-T_ambient)&lt;br&gt;&#xA;[A = total surface area of walls on house]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From there I solved for the new temperature of the wall inside (T_wall). T_wall was used to show how the house's inside temperature changed from the desired inside temperature due to the outside temperature and solar heat gain. I'm currently finishing up my first thermodynamics class so this information is still relatively new to me. What information I could find for solar heat gain was from this website: &lt;a href=&quot;http://energy-models.com/heat-gains-and-losses-windows-and-skylights-glass&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solar Heat Gain&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ten years of outside temperature data was pulled from the usclimatedata website. Does this equation look correct? Thanks in advance for any helpful advice/help! &lt;/p&gt;&#xA;" OwnerUserId="20309" LastActivityDate="2019-05-01T21:06:28.957" Title="Solar Heat Gain/Outside Temperature Effects on a House" Tags="|mechanical-engineering|thermodynamics|hvac|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28185" PostTypeId="2" ParentId="28179" CreationDate="2019-05-01T23:47:41.233" Score="2" Body="&lt;p&gt;You do not have sufficient information to solve the question. If you have a part of 0.310 width, the center line of the part is always going to be at 0.155 from either edge. The slot width can be any dimension smaller than 0.310 within limits of the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The information you have provided can only determine that the slot is centered.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-05-01T23:47:41.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28186" PostTypeId="2" ParentId="28183" CreationDate="2019-05-02T01:32:11.527" Score="1" Body="&lt;p&gt;Nope.  A system can have better local stability than global.  As an extreme example, a system can be stable within some boundary in the state space, and unstable beyond it.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-05-02T01:32:11.527" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28188" PostTypeId="1" AcceptedAnswerId="28190" CreationDate="2019-05-02T10:12:19.463" Score="1" ViewCount="102" Body="&lt;p&gt;I have a self balancing robot control loop that uses the PID controller. I feed the robot's angle (rad) to sum block that generates the error between the set point and the angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The error then goes to the PID and goes to DC motor block that needs the input to be in volts and generate an output in N.m.  Then the signal goes to the self balancing robot block that need the input to be in N.m and outputs a rad angle.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oS7OU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oS7OU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;As you see I don't know how to make the PID output a voltage (V). How i can do that?&lt;/p&gt;&#xA;" OwnerUserId="29384" LastEditorUserId="272" LastEditDate="2019-05-02T12:26:07.827" LastActivityDate="2019-05-02T12:26:07.827" Title="Do variables dimensions matter in PID simulation + I want the PID output to be volts?" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28189" PostTypeId="1" AcceptedAnswerId="28191" CreationDate="2019-05-02T10:48:11.617" Score="0" ViewCount="368" Body="&lt;p&gt;Given the following situation:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/m5cgk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m5cgk.png&quot; alt=&quot;in&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pipe A and B have a known and identical length and diameter, and are in parallel. I'm assuming the flow trough pipes A and B are also identical, as is the roughness of the pipes. &#xA;Pipe C also has a known length and diameter (which is different from the diameter of pipe A/B).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested in the total head loss at the end of Pipe C. I am aware of the following Darcy-Weisbach equations that are relevant in this case (parallel and series):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zZ8D4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zZ8D4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the total headloss is the sum of the headloss per pipe, placed in series. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/udlw6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/udlw6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here the total headloss over set of 2 parallel pipes is equal to the headloss of one of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, how do I combine these two equations to retrieve the total headloss over pipes A, B and C?&lt;/p&gt;&#xA;" OwnerUserId="20318" LastEditorUserId="20318" LastEditDate="2019-05-02T10:53:56.453" LastActivityDate="2019-05-02T16:21:31.600" Title="Determining the headloss according to Darcy-Weisbach with pipes in parallel and series" Tags="|fluid-mechanics|hydraulics|pipelines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28190" PostTypeId="2" ParentId="28188" CreationDate="2019-05-02T12:23:25.990" Score="2" Body="&lt;p&gt;The PID simulation you've built does not care about the units of numbers being passed around your control loop.  However, you may want to ensure that the units are correct at different points in the loop to make it easier for you to interpret and debug.  To do that, you can simply add gain blocks (no frequency dependence) with the appropriate conversion factors.  If you do not add separate gain blocks, then the unit conversion will be implicitly included in the gain of each transfer function. &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2019-05-02T12:23:25.990" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28191" PostTypeId="2" ParentId="28189" CreationDate="2019-05-02T16:21:31.600" Score="0" Body="&lt;p&gt;The total headloss through pipes A+C is equal to the headloss through pipes B+C. Therefore, you only need to calculate one of the paths, and your first equation gives you the desired total headloss (where 1 and Q in the first term are for pipe A, and 2 and Q in the second term are for pipe C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, the total headloss is equal to the pressure difference that is driving the flow. For example, if the tank is filled with a liquid to depth H, and if the end of the pipe C is opened to the atmosphere, the pressure driving the flow due to the elevation difference is known. In other words, you usually know the total headloss because you know the pressure difference, and the calculation is to find the flow rate. You can determine that from the first equation because you are assuming that &lt;span class=&quot;math-container&quot;&gt;$Q_A=0.5Q_C$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to the Darcy-Weisbach wall losses, there are also losses due to the inlets into pipes A+B, the tee where A+B combine into C, and the exit loss at the end of C. These may be significant depending on the flow rate and length of pipes.&lt;/p&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2019-05-02T16:21:31.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28192" PostTypeId="1" CreationDate="2019-05-02T19:05:57.190" Score="1" ViewCount="38" Body="&lt;p&gt;I have a question regarding worm drives, specifically double-enveloping types:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Do steel worm/steel wheel combinations last longer than steel worm/bronze wheel or steel worm/brass wheel combinations? If one were to use hardened tool steel for both the worm and wheel?&lt;/li&gt;&#xA;&lt;li&gt;Are they also stronger than steel/bronze or steel/brass drives? &lt;/li&gt;&#xA;&lt;li&gt;How much stronger are they? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note: I know little about the underlying science behind gear design, just curious.&lt;/p&gt;&#xA;" OwnerDisplayName="user10249" LastEditorDisplayName="user10249" LastEditDate="2019-05-02T19:41:51.197" LastActivityDate="2019-05-02T19:41:51.197" Title="Steel Worm/Steel Wheel vs Steel Worm/Bronze Wheel" Tags="|mechanical-engineering|gears|metallurgy|metals|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28195" PostTypeId="1" CreationDate="2019-05-02T21:26:59.317" Score="1" ViewCount="117" Body="&lt;p&gt;I have created this simulation ( see below) of a transistor and zener diode voltage regulator. The output is supposed to be a steady 5 V. But it is like the top bit s cut, to an approx 4.4 V value, then it plunges to the minimum voltage value. I really dont understand why and how to fix it&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uiIf5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uiIf5.png&quot; alt=&quot;my simulation of transistor zener diode circuit&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20326" LastActivityDate="2019-05-03T04:03:15.357" Title="My output is much lower than I expect it to be for my transistor-zener diode voltage regulator" Tags="|simulation|power-electronics|circuits|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28197" PostTypeId="2" ParentId="28181" CreationDate="2019-05-03T01:04:41.850" Score="1" Body="&lt;p&gt;Spheres buckle under external pressure.  This is a stability problem.  Stability problems are not as clear cut as we might prefer. Everything looks fine and then surprise, it fails, apparently for no reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two sites I found that discuss this (which my Marks' Mechanical Engineering Handbook did not) are:&#xA;&lt;a href=&quot;https://engineering.stackexchange.com/questions/15717/how-to-calculate-external-pressure-resistance-in-spheres-and-cylinders/15725&quot;&gt;How to calculate external pressure resistance in spheres and cylinders?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.physicsforums.com/threads/hypothetical-hollow-steel-sphere-collapse-from-outside-pressure.730230/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.physicsforums.com/threads/hypothetical-hollow-steel-sphere-collapse-from-outside-pressure.730230/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They give the same formula and the link from this site says &quot;When tested, however, this formula was found to be completely inaccurate.&quot;  I don't think we really care about the exact values, right now, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But that was not your question.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The buoyancy of the dome minus its weight determines how much stress will be on the ropes.  First, let's find how much the dome wants to float.  A hemispherical dome 100M in diameter, using the V=4/3 pi R^3 formula for a sphere and knowing that a cubic meter of water has a mass of 1000 kg, the dome wants to float with a force equal to 262,000 metric tons (1000kg).  Wow, that's a lot for five ropes to hold.  But wait how heavy is the dome?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water at 100 M exerts a pressure of about 1000 kPa. Waving my hands and saying the pressure will be the same for the whole dome (which it is not), we can use the magic formula from the other web site with the values &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Modulus of elasticity 50 gPa&lt;/li&gt;&#xA;&lt;li&gt;Poisson's Ratio 0.22&lt;/li&gt;&#xA;&lt;li&gt;Radius 50 M&lt;/li&gt;&#xA;&lt;li&gt;Pressure 1000 kPa&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and get a required thickness of glass of 0.2 M. Given that glass weighs around 2500 Kg/M^3, and the surface of the hemisphere would be (A=0.5*(4*PI*R^2) ) about 16,000 M^2, we can calculate the weight of the dome as 7,859,000 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks to me as if the weight of the dome exceeds the weight of the displaced water so it will sink.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Tension? Zero since no ropes are required.&lt;/li&gt;&#xA;&lt;li&gt;Tighten into the sand? Not necessary&lt;/li&gt;&#xA;&lt;li&gt;Tremendous Pressure? 10 bar is roughly ten times atmospheric.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For comparison, my car tires have about 2 bar of (gauge) pressure. My home air compressor can put out about 6 bar. Mind you, 10 bar over an area of 16,000 m^2 is a force of about 1.6 billion kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After typing this up, it occurs to me that the density of glass I gave above is only 2.5 times that of water and the volume inside the dome is a LOT greater than the volume of the material that makes up the dome so I expect I lost a decimal somewhere.  I encourage readers to comment and correct as appropriate.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastEditorUserId="4953" LastEditDate="2019-05-03T20:43:19.053" LastActivityDate="2019-05-03T20:43:19.053" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28198" PostTypeId="2" ParentId="28195" CreationDate="2019-05-03T03:13:54.337" Score="1" Body="&lt;p&gt;The simulation has an AC source feeding the Zener Diode - Series pass transistor circuit. Replacing the AC source with the DC source should eliminate voltage plung situation. Alternatively inserting a rectification circuit between AC source and the Zener regulation circuit will also solve this issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 4.4V output voltage situation can be solved by selecting a 5.6V Zener diode such as a &lt;a href=&quot;https://www.vishay.com/docs/85816/1n4728a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1N4734A&lt;/a&gt; from Vishay &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a example zener diode  - series pass transistor 5V output circuit &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EZ05h.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EZ05h.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a list of parts from other vendors&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M6lCd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M6lCd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively below is list of 1N47 series zener diodes from Vishay&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RBzog.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RBzog.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.vishay.com/docs/85816/1n4728a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;1N4728A to 1N4764A Zener Diodes from Vishay Semiconductors&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.bristolwatch.com/ele/zener_power_supply.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Transistor-Zener Diode Regulator Circuits&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.eleccircuit.com/simple-fixed-dc-regulator-using-transistor-zener-diode-with-pcb/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Small Zener diode voltage regulator circuit with PCB&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.circuitstoday.com/voltage-regulators&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voltage Regulators&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-05-03T04:03:15.357" LastActivityDate="2019-05-03T04:03:15.357" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28199" PostTypeId="1" AcceptedAnswerId="28212" CreationDate="2019-05-03T08:00:18.267" Score="0" ViewCount="87" Body="&lt;p&gt;I'm having some work done in the UK (Eurocode) which involves a steel column to support a beam at 1st floor (US: 2nd floor?) level. The column bears a very-close-to-vertical total load of 47.3kN  (including any relevant safety factors I guess, as that's the final figure in his calcs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engineer has specified a 152x152 23kg UC to be used for the column, with suitable end+base plates and fixings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My issue is that the engineer has offered a low rate as its for a &quot;charity job&quot; (house belongs to an elderly frail relative). I've gone back to him with a set of 3 queries and got replies that are professional but curt and I don't want to go back again to ask if anything else is possible, unless I'm already privately very sure it is and can back it with calcs, as I don't want to lose his goodwill. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only time he's said yes to any change, was when I also attached calcs to justify why I thought it should be OK. Unhelpful and not my job, I know!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I really need a slimmer column, to run flush with the 100mm partition wall.  Unfortunately I don't understand column slenderness calculations well enough to check if there's scope to replace the 3m 152x152 23kg UC by, say, a 100x100x10 25kg RHS, or even a 100x180x5 21kg RHS.  My tentative figures suggest that because of the RHS vs UC profile, both of these would be stiffer in any direction, than the UC and they'd fit nicely flush to the wall, so maybe they'd work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't need an absolute calculation &lt;em&gt;(&quot;Does this profile work for that length and load&quot;)&lt;/em&gt;, but only a relative one &lt;em&gt;(&quot;Would this profile be at least as satisfactory as the one he's already said is good enough&quot;)&lt;/em&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What calculations are needed, to confirm if a more slender (in one dimension at least!) RHS profile would be at least as satisfactory/rigid in performance as a column, compared to the currently-proposed UC?&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2019-05-03T08:19:36.997" LastActivityDate="2019-05-03T21:09:26.030" Title="RHS column profile able to replace 3m UC 152x23 column?" Tags="|steel|columns|eurocodes|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28200" PostTypeId="1" CreationDate="2019-05-03T11:48:22.973" Score="1" ViewCount="104" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HADTV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HADTV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am new in engineering, could anyone explain the approach to calculate the max heat generation, the heat transfer from the cylinder to the atmosphere and from cylinder to the piston.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am also wondering about how does changing L3 affect the max heat generation which is allowed and why&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the picture is the cross-section of the engine.&lt;/p&gt;&#xA;" OwnerUserId="20330" LastActivityDate="2019-05-03T11:48:22.973" Title="Heat Transfer in A Cylinder of an Internal Combustion Engine" Tags="|mechanical-engineering|heat-transfer|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28201" PostTypeId="2" ParentId="22598" CreationDate="2019-05-03T12:13:54.503" Score="2" Body="&lt;p&gt;Assuming you mean a &lt;a href=&quot;https://en.wikipedia.org/wiki/Ground-effect_vehicle&quot; rel=&quot;nofollow noreferrer&quot;&gt;wing-in-ground-effect vehicle&lt;/a&gt;, which is in flight during its cruise phase, the ground effect is relatively easy to calculate, at least for potential flows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boundary condition imposed on potential airflow by the ground is zero transversal flow, which can be recreated by using an imaginary wing flipped around the ground line respective to the one you are calculating. In other words, if your problem wing flies 10m above the ground, you can simulate the ground contribution by calculating the lift of that wing combined with another flying 10m under the ground and inverted around the ground axis:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yhXZx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yhXZx.jpg&quot; alt=&quot;potential flow streamlines around a 2D airfoil in ground effect&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2019-05-03T12:13:54.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28202" PostTypeId="1" AcceptedAnswerId="28205" CreationDate="2019-05-03T12:29:52.543" Score="2" ViewCount="860" Body="&lt;p&gt;What causes the spider gear to spin about it's own axis while the vehicle is taking a turn?&lt;/p&gt;&#xA;" OwnerUserId="15268" LastEditorUserId="10902" LastEditDate="2019-05-03T17:42:17.580" LastActivityDate="2019-05-03T17:42:17.580" Title="Working of spider gear in differential axle wheel mechanism" Tags="|automotive-engineering|power-transmission|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28203" PostTypeId="1" CreationDate="2019-05-03T14:25:22.820" Score="1" ViewCount="637" Body="&lt;p&gt;I am currently trying to design a rear gate to retrofit to an existing caged trailer (similar to image &lt;a href=&quot;https://i.stack.imgur.com/bArej.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;3&lt;/a&gt;), with the goal of having a system where the rear gate can pivot down (on its bottom edge) to form a ramp, or from either side so that it can swing out of the way to the side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently I have the following design mocked up (not to scale): images &lt;a href=&quot;https://i.stack.imgur.com/ONh4s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt; &amp;amp; &lt;a href=&quot;https://i.stack.imgur.com/sdyIK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;. In these images, parts are coloured as below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Black: Existing trailer body&lt;/li&gt;&#xA;&lt;li&gt;Grey: Hinges with removable locking pins, as seen in image &lt;a href=&quot;https://i.stack.imgur.com/cZqpP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;4&lt;/a&gt;. Horizontal pins would have some kind of extra or spring loaded locking to avoid loss during vibration. This part should be straightforward.&lt;/li&gt;&#xA;&lt;li&gt;Blue/light blue: Ramp and attachment points&lt;/li&gt;&#xA;&lt;li&gt;Yellow: Gas strut/damper or hydraulic ram (this is where the issue arises).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The current issues I am facing at this point:&#xA; - How can I attach and detatch the strut from the bottom of the rear gate to allow the gate to swing open sideways, while reducing the risk of sudden release? I am thinking of some kind of mechanism where the strut slots in to the rear gate section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strut appears necessary as a method to dampen the rear gate movement, as it will likely have significant weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My end objective here is to be able to have the rear gate function as a ramp for loads up to approximately 500kg, as well as being able to swing out of the way for loads that are simply pushed off the back of the tray (e.g. gravel, rubbish dump runs).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ground clearance ideally should not be reduced (below axle level), and a dependence on electricity should also be avoided.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is, how can I safely and efficiently detatch this strut to swing the gate open sideways with the above design, or should I consider an alternative design to achieve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ONh4s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ONh4s.png&quot; alt=&quot;Side view of design&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sdyIK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sdyIK.png&quot; alt=&quot;Rear quarter view&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bArej.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bArej.jpg&quot; alt=&quot;trailer example&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cZqpP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cZqpP.png&quot; alt=&quot;latch pin example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20332" LastEditorUserId="20332" LastEditDate="2019-05-03T15:56:46.013" LastActivityDate="2024-02-07T07:08:35.953" Title="Pivoting trailer gate on multiple axes" Tags="|automotive-engineering|mechanical|safety|car|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28204" PostTypeId="2" ParentId="28203" CreationDate="2019-05-03T15:27:29.300" Score="0" Body="&lt;p&gt;So why not use the pin and socket you show in the last image, two on each side but instead of the handle at the top put it at the bottom with a lock position. That can take care of the sides with either hand opening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then two more ( or 3 depending on load) along the bottom, again with a lock position which will allow drop down use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are clips you can get to make sure the pins stay in the secured position.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-05-03T15:27:29.300" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28205" PostTypeId="2" ParentId="28202" CreationDate="2019-05-03T15:43:04.770" Score="2" Body="&lt;p&gt;The reason it turns about its center is the difference in rotation of the two wheels, which are connected through the sun-gears to planet-pinion gears. By rotating it allows the two sun-gears and consequently, the two wheels go at deferent speeds without either losing power, or having to skid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the reason for the difference in the wheels speed is because in a turn the inside wheel has less speed then the outside wheel, having to go around a smaller radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bE5fE.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bE5fE.gif&quot; alt=&quot;diagram of the differential&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-03T15:43:04.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28206" PostTypeId="2" ParentId="28203" CreationDate="2019-05-03T16:40:59.130" Score="0" Body="&lt;p&gt;It looks like I might be able to do this with damping swings. In this figure, the springs are welded to the ramp, with the other end of the spring coil welded to a (high tensile steel) axle across the length. The axle is then welded at the middle to a plate, which contacts another plate fixed to the bottom of the trailer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two plates in contact tension the springs as the ramp lowers, damping its movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll probably swap the rear plate (attached to the spring axle) for smaller sections to make it a bit more manageable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uTrTv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uTrTv.png&quot; alt=&quot;Spring design&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20332" LastActivityDate="2019-05-03T16:40:59.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28207" PostTypeId="1" CreationDate="2019-05-03T17:14:05.493" Score="2" ViewCount="19" Body="&lt;p&gt;I'm trying to assemble a simple setup where a line which is typically under vacuum can be pressurized by an external nitrogen supply, with this pressure regulated to less than 1 bar absolute (let's say 700 mbar). This nitrogen supply comes from the building and is at ~6 bar. What kind of regulator would let me do this? Would a conventional regulator which just reads gauge pressure work? As an additional note, there would not be any flow to or from a process occurring in this line while it is being pressurized or at the nominal 700 mbar regulated pressure.&lt;/p&gt;&#xA;" OwnerUserId="20323" LastEditorUserId="20323" LastEditDate="2019-05-03T17:57:36.040" LastActivityDate="2019-05-03T17:57:36.040" Title="Parts needed for controlling vacuum line pressure with pressurized nitrogen" Tags="|pressure|gas|vacuum|compressed-gases|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28208" PostTypeId="1" CreationDate="2019-05-03T17:35:50.583" Score="-1" ViewCount="954" Body="&lt;p&gt;For example, I have an axial force and a torsion force applied on a rock socket installed in bedrock. If I am considering the side friction between socket and rock as a resistance, the socket surface will have two shear stress generated, one in axial direction, and the other in radial direction. If I want to design this, what is the maximum I should use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I combine the two stresses by triangle rule, or simply adding them up, or only take the larger one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a conceptual sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RB6WPm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RB6WPm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20334" LastEditorUserId="33" LastEditDate="2019-05-03T19:23:14.977" LastActivityDate="2019-09-30T21:02:15.400" Title="Can I combine two shear-stresses acting on one plane?" Tags="|mechanical-engineering|stresses|" AnswerCount="1" CommentCount="18" ContentLicense="CC BY-SA 4.0" />
  <row Id="28209" PostTypeId="2" ParentId="28208" CreationDate="2019-05-03T18:51:59.577" Score="1" Body="&lt;p&gt;You are right, the stresses in this case to are at the same plane, at any given point, so simply by triangle method or vector addition just add them, to verify that you do not surpass the allowable stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However keep in mind their projection axially and horizontally for later when you need to calculate Say the torque as opposed to pull out force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the stresses where not on the same surface then you could calculate maximum stress and its direction using Mohr circle.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-03T18:51:59.577" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="28211" PostTypeId="1" CreationDate="2019-05-03T21:03:12.310" Score="1" ViewCount="625" Body="&lt;p&gt;I am trying to find simulate an op amp based linear regulator negative voltage regulator circuit. &#xA;I have done the circuit for positive voltages, but does not know how to do it for negative voltage. Anyone could help me?  I dont know how to connect the op amp. &#xA;I have included a screenshot of my simulation&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9uFQ8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9uFQ8.png&quot; alt=&quot;my simulation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20326" LastActivityDate="2020-04-13T18:25:39.350" Title="I am trying to create an op amp based negative voltage regulator" Tags="|simulation|circuits|circuit-design|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28212" PostTypeId="2" ParentId="28199" CreationDate="2019-05-03T21:09:26.030" Score="1" Body="&lt;p&gt;The root problem of your question is that you're asking us to make some engineering judgement calls and calculations for a building where we don't know all the relevant details. You could however calculate the slenderness of the existing column and then propose a section that has this slenderness or less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slenderness parameter that is commonly used here in the US is &lt;span class=&quot;math-container&quot;&gt;$ \sqrt{\frac{KL}{r}} $&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; is the radius of gyration &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the column length, and &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; is the effective length factor you can assume for comparison that this is &lt;span class=&quot;math-container&quot;&gt;$K=1$&lt;/span&gt;.  Be sure and use consistent length units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find tabulated section properties that should give you the &lt;span class=&quot;math-container&quot;&gt;$r_{x}$&lt;/span&gt; and the &lt;span class=&quot;math-container&quot;&gt;$r_{y}$&lt;/span&gt; value (radius of gyration) for different shapes (including the one originally specified). In general you will want to take the smaller of the &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; values as that direction buckling will control.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-05-03T21:09:26.030" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28213" PostTypeId="1" CreationDate="2019-05-03T22:23:21.390" Score="2" ViewCount="1773" Body="&lt;p&gt;I have seen some theories saying stress is a scalar. I don't understand why. Anyone can explain.&lt;/p&gt;&#xA;" OwnerUserId="20334" LastEditorUserId="33" LastEditDate="2019-05-07T12:53:32.693" LastActivityDate="2023-08-30T03:45:18.163" Title="Is stress a vector or scalar?" Tags="|mechanical-engineering|stresses|" AnswerCount="2" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28214" PostTypeId="2" ParentId="24508" CreationDate="2019-05-03T23:01:31.170" Score="0" Body="&lt;p&gt;Search for &quot;120 to 240 transformer adaptor usa to uk step up&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll need to choose one with a power rating higher than your lamp. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something suitable is about £10 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You shouldn't need to even change the plug.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QTetk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QTetk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16544" LastActivityDate="2019-05-03T23:01:31.170" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28215" PostTypeId="2" ParentId="28211" CreationDate="2019-05-03T23:38:43.397" Score="0" Body="&lt;p&gt;Below is a good example of configuring the op-amp based voltage regulator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N44dr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N44dr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take look at &lt;a href=&quot;https://www.circuitlib.com/index.php/lessons/96-op-amp-voltage-regulator-concept&quot; rel=&quot;nofollow noreferrer&quot;&gt;Op-amp voltage regulator concept&lt;/a&gt; post, this might help answer some of your questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.mouser.com/pdfdocs/linear_reg_basic_appli-e.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Basics of Linear Regulators&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-05-03T23:38:43.397" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28218" PostTypeId="1" CreationDate="2019-05-04T01:26:16.413" Score="2" ViewCount="459" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0hl6M.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0hl6M.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reference to this example problem (only partially shown), I need to understand why for &lt;span class=&quot;math-container&quot;&gt;$\Delta h$&lt;/span&gt; the change in enthalpy, which is &lt;span class=&quot;math-container&quot;&gt;$\Delta P/\rho + \Delta u$&lt;/span&gt;, the change in pressure is ignored in the solution even though it is considerably large. Could somebody please explain this to me?&lt;/p&gt;&#xA;" OwnerUserId="19383" LastActivityDate="2019-05-07T14:32:21.413" Title="When To Ignore Changes In Pressure When Computing Flow Energy For Gases?" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28219" PostTypeId="1" AcceptedAnswerId="28221" CreationDate="2019-05-04T05:56:39.310" Score="1" ViewCount="266" Body="&lt;p&gt;It might be a little off topic but I'm trying to find source for why is that model called magic formula? I read that it's a simple function of commonplace functions, a data fitting. Is it simply an empirical model with an unusual name? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model (or rather series of models) I'm referring to:&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Hans_B._Pacejka&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Hans_B._Pacejka&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far only this lecture mentions the history of the name:&#xA;&lt;a href=&quot;http://textofvideo.nptel.ac.in/107106080/lec15.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://textofvideo.nptel.ac.in/107106080/lec15.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20336" LastEditorUserId="20336" LastEditDate="2019-05-04T07:20:22.547" LastActivityDate="2019-05-04T14:24:57.477" Title="Pacejka &quot;Magic Formula&quot; tire models origin of the name" Tags="|automotive-engineering|simulation|friction|wheels|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28220" PostTypeId="1" CreationDate="2019-05-04T10:30:30.063" Score="1" ViewCount="122" Body="&lt;p&gt;I'm a bit confused by torque motors datasheets as it never mentions any nominal voltage. I would like to run a torque motor on a 24&amp;nbsp;V battery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required max torque is 150&amp;nbsp;Nm (=21000&amp;nbsp;oz-in) and max speed 15&amp;nbsp;RPM. (=1.57&amp;nbsp;rad/s). I have found a motor that shows max torque, max current at this torque but I don't find any nominal voltage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I multiply &lt;span class=&quot;math-container&quot;&gt;$K_e$&lt;/span&gt; (V/rad.s^-1) by max speed and add the voltage drop in the resistor (&lt;span class=&quot;math-container&quot;&gt;$R*I_{max}$&lt;/span&gt;), I can find 132&amp;nbsp;V.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Why does the torque motor not have a nominal voltage? (compared to the classical BLDC motors)&lt;/li&gt;&#xA;&lt;li&gt;Are the supply voltage calculations (132&amp;nbsp;V) correct?&lt;/li&gt;&#xA;&lt;li&gt;In this case, should I choose a boost converter (24&amp;nbsp;VDC -&gt; +132&amp;nbsp;VDC?) followed by a motor drive that can handle this voltage?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can find the motor datasheet below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ufWCK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ufWCK.png&quot; alt=&quot;Datasheet of a motor that may fit my requirements&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20337" LastEditorUserId="1832" LastEditDate="2019-05-04T14:29:42.437" LastActivityDate="2019-05-04T14:29:42.437" Title="Finding Supply/Nominal voltage for Torque Motor BLDC" Tags="|electrical-engineering|motors|electrical|electric-vehicles|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28221" PostTypeId="2" ParentId="28219" CreationDate="2019-05-04T14:24:57.477" Score="0" Body="&lt;p&gt;&quot;Magic X&quot; is a common expression for things which aren't adequately explained but that &quot;just work&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, in programming, it is considered poor form to use &quot;magic numbers&quot;. These are numbers which simply appear in a given code without any explanation of what they are. For example, if code has &lt;code&gt;x * 0.707&lt;/code&gt;, you immediately start wondering what that 0.707 comes from. You'll need to look at the code and think about it and remember that the cosine of 45 degrees is approximately equal to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, one should probably define a named variable (for example &lt;code&gt;var cos45 = 0.707&lt;/code&gt;) with that number as its value. Then you just do &lt;code&gt;x * cos45&lt;/code&gt; and it's clear as day what the code does, no thinking required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, if you do come across a program which does &lt;code&gt;x * 0.707&lt;/code&gt;, you might not really understand where that number comes from, but you can tell it &quot;just works&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a similar vein, &quot;magic formula&quot; simply means it's a formula which, for reasons unknown, &quot;just works&quot;. That much is stated in the &lt;a href=&quot;https://en.wikipedia.org/wiki/Hans_B._Pacejka#The_Pacejka_%22Magic_Formula%22_tire_models&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article&lt;/a&gt; you linked to (emphasis added):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pacejka has developed a series of tire design models over the last 20 years. &lt;strong&gt;They were named the 'magic formula' because there is no particular physical basis for the structure of the equations chosen, but they fit a wide variety of tire constructions and operating conditions.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One of the &lt;a href=&quot;http://phors.locost7.info/phors21.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;references&lt;/a&gt; to the Wikipedia article even keeps the symbolism going (emphasis added):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The next step is to compute the product of a new helper, &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, times &lt;span class=&quot;math-container&quot;&gt;$b_0$&lt;/span&gt; and the aforecomputed &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;. The &lt;strong&gt;magicians&lt;/strong&gt; who created the formula tell us that &lt;span class=&quot;math-container&quot;&gt;$B b_0 D = (b_3 F_z^2 + b_4 F_z)\exp(-b_5F_z)$&lt;/span&gt;. This slurps up a few more of the &lt;strong&gt;magical&lt;/strong&gt; eleven empirical numbers, and a pattern emerges [...]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, the &quot;magic formula&quot; is so-called because it is totally empirical. There are many other empirical formulas in life, though, so you might be asking why &quot;only&quot; this one is called thus. The answer is that this one is &lt;strong&gt;&lt;em&gt;really&lt;/em&gt;&lt;/strong&gt; empirical. Other empirical equations are mostly intuitive: they may have empirically-determined coefficients, but the overall form of the equation makes some sense. This one is just a bunch of coefficients added, multiplied and exponentiated and somehow it all works out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, as that same reference I linked to above puts it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Once again, don't try to find any physics in here: it's just a convenient formula that fits the data reasonably well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-05-04T14:24:57.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28223" PostTypeId="1" CreationDate="2019-05-04T16:43:43.233" Score="1" ViewCount="158" Body="&lt;p&gt;I have created a bent right angled tube in solid works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has a constant inner diameter of 1&quot; and thickness of 1/8&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tube is comprised of a straight portion 1 ft in length, followed by a 90 deg bend with a 10 cm radius of curvature, then another straight portion 10 cm in length. I apologize for mixed units, I am just trying to make a proof of concept at this time so I will change the dimensions later.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've fixed the first flat face of 1 ft section at the top, and I have set the material to natural rubber for now but I would probably want it to be neoprene or polyethylene in the finished product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to deflect the second shorter straight portion by an angle of 45 deg in the plane and measure the deflection in the curved section. &lt;strong&gt;Is there a way that I can directly apply a deflection (angular preferably) to the end of the 10 cm length tube and measure the change of the arc in the curved section?&lt;/strong&gt; I have had to apply a load, but I don't really have any idea of knowing what that should be to correspond to the desired deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an image using alloy steel because the simulation kept failing if I applied a load to rubber?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vetxb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vetxb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18234" LastEditorUserId="18234" LastEditDate="2019-05-04T17:15:13.467" LastActivityDate="2019-05-04T17:15:13.467" Title="Trying to simulate the bending of a right angled pipe in solidworks" Tags="|pipelines|solidworks|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28224" PostTypeId="2" ParentId="28218" CreationDate="2019-05-04T18:48:54.750" Score="0" Body="&lt;p&gt;When calculating flow energy, you should always take pressure or temperature changes into account as they are linked by ideal gas law (&lt;span class=&quot;math-container&quot;&gt;$PV = nRT$&lt;/span&gt;). In this case, calculating change in enthalpy due to temperature change is equivalent to calculating it using pressure change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dXlj3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dXlj3.png&quot; alt=&quot;Table of thermodynamic quantities for specific cases from https://en.wikipedia.org/wiki/Table_of_thermodynamic_equations&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20337" LastEditorUserId="20337" LastEditDate="2019-05-07T08:51:26.150" LastActivityDate="2019-05-07T08:51:26.150" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28225" PostTypeId="1" CreationDate="2019-05-05T15:20:36.667" Score="1" ViewCount="306" Body="&lt;p&gt;As an Aerospace Engineering Student, I have been trying to make a project of a PID Controller to control the attitude response of a satellite using the parameters from a project called Kufasat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parameters are described below (moments of inertia, mass, initial angular velocity, etc):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Aeknm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Aeknm.png&quot; alt=&quot;Attitude Dynamics parameters&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attitude Kinematics equations are also below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vxjTU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vxjTU.png&quot; alt=&quot;Attitude Kinematics equations and its results&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to design a PID controller to stabilize the angular motion of the satellite, reducing it to zero values. I have been trying to create some models in Matlab to insert the PID parameters, but all I got is divergences only. I feel like my model is unstable, or that I have been missing something here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also tried to use the Aerospace Toolbox, but got no success at all. I also do not know if this toolbox would help me with this problem.&#xA;Furthermore, I have been thinking of State Space Formulation, but I have no idea of how I would organize all these variables in order to formulate the four matrices (A, B C and D), to make me able to apply a PID Control technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear that I am lost, and I am even willing to pay someone to help me, as my advisor professor do not have enough time to help me build this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please, someone help me. I am desperate and need to finish this by the end of July.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I have written a code in Matlab with all the calculations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;clear all&#xA;&#xA;omega = [0.06 0.03 0.09];&#xA;I = [0.0547 0.0519 0.0574];&#xA;Kp = 0.008;&#xA;Kd = 0.08;&#xA;Ki = 10^-6;&#xA;phi_EAngle = [70; 40; 50]&#xA;&#xA;Theta(1) = phi_EAngle(1)&#xA;Theta(2) = phi_EAngle(2)&#xA;Theta(3) = phi_EAngle(3)&#xA;&#xA;R1 = [1 0 0; 0 cosd(Theta(1)) sind(Theta(1));0 -sind(Theta(1)) cosd(Theta(1))];&#xA;R2 = [cosd(Theta(2)) 0 -sind(Theta(2));0 1 0;sind(Theta(2)) 0 cosd(Theta(2))];&#xA;R3 = [cosd(Theta(3)) sind(Theta(3)) 0; -sind(Theta(3)) cosd(Theta(3)) 0;0 0 1];&#xA;&#xA;R = R3*R2*R1&#xA;&#xA;phi = acos((1/2)*(trace(R)-1))&#xA;a_x = (1/(2*sin(phi)))*(R'-R)&#xA;&#xA;a(1) = a_x(3,2);&#xA;a(2) = a_x(1,3);&#xA;a(3) = a_x(2,1);&#xA;a = a'&#xA;&#xA;q = a*sin(phi/2);&#xA;q_x = q'*q&#xA;q(4) = cos(phi/2)&#xA;&#xA;phi_p = a'*omega';&#xA;a_p = 1/2*(a_x-cot((phi/2)*a_x*a_x))*omega'&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="20340" LastActivityDate="2019-05-05T15:20:36.667" Title="Designing a PID Controller for Satellite Control" Tags="|control-engineering|aerospace-engineering|pid-control|matlab|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28226" PostTypeId="1" CreationDate="2019-05-05T16:00:33.690" Score="0" ViewCount="73" Body="&lt;p&gt;There is this problem I can't figure out. One measurements wheel needs to be attached to a slowly moving but wildly shaking and twitching construction machinery. I am trying to greatly absorb the shocks by means of a spring pictured below (and a damper not pictured). The wheel also needs to move in parallel with the machinery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this design I have coupled the wheel to the machinery by a guided spring. The spring stretches and compresses between two metal plates with the upper plate sliding over &lt;em&gt;rod1&lt;/em&gt; and &lt;em&gt;rod2&lt;/em&gt;. I am not a mechanical engineer, but I think there will be problems due to the way the wheel is dragged alongside the machinery, i.e. there will be some friction between the two rods and the upper metal plate, causing the spring guide not to work very well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is the case, what is a standardized known to work manner to approach this problem - shock reduction / isolation between the machinery and such a wheel ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gRhff.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gRhff.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3412" LastActivityDate="2019-05-05T20:40:55.237" Title="shock absorbing rigid joint" Tags="|springs|shock|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28227" PostTypeId="2" ParentId="18806" CreationDate="2019-05-05T17:42:08.373" Score="2" Body="&lt;p&gt;To find the purely theoretical limit depending totally on limits of the rolling resistance of the tire and air drag. So here is the work.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Formulas&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$F_d = \dfrac{1}{2}pA\mu_d v^2$&lt;/span&gt;: this is the formula for drag force.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$f_s = \mu_s mg$&lt;/span&gt;: this is the formula for static friction force.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;NOTE: I am using values of a Tesla model 3.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Known values&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A = 0.23&amp;nbsp;m&lt;sup&gt;2&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;g = 9.80665&amp;nbsp;m/s&lt;sup&gt;2&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;li&gt;m = 1611&amp;nbsp;kg&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\mu_s$&lt;/span&gt; = 0.7&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\mu_d$&lt;/span&gt; = 0.23&lt;/li&gt;&#xA;&lt;li&gt;p = 1.2754&amp;nbsp;kg/m&lt;sup&gt;3&lt;/sup&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Assumptions&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Car is perfectly perpendicular to the force of gravity.&lt;/li&gt;&#xA;&lt;li&gt;Car is at sea level.&lt;/li&gt;&#xA;&lt;li&gt;Conditions are ideal.&lt;/li&gt;&#xA;&lt;li&gt;The power of the engine is infinite.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Results&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V_{theo} = 1280.9\text{ mph}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="20341" LastEditorUserId="1832" LastEditDate="2019-05-05T18:23:33.097" LastActivityDate="2019-05-05T18:23:33.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28228" PostTypeId="2" ParentId="28226" CreationDate="2019-05-05T19:46:18.977" Score="2" Body="&lt;p&gt;If all three things labeled &quot;joint&quot; in your drawing are hinges, I don't think the spring will do anything at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't say what you are trying to measure here, but the design doesn't seem to support the wheel properly. For example what happens it the wheel &quot;gets stuck&quot; in a depression? It looks like the linkages will just twist, and leave the wheel &quot;stuck&quot; while the machine moves forwards. Whether the wheel unsticks itself, of something gets permanently bent or breaks, seems a matter of luck rather than good design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the first thing you need is a linkage that constrains the axle of the wheel properly so it can only move up and down, and not twist in any direction. In two dimensions a four-bar linkage would do that, but you also need something so the wheel can't move forwards or backwards out of step with the machine it is attached to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about adding a damping mechanism after you have got that sorted out - though you might not need any damping unless the measuring wheel is very light.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-05-05T19:46:18.977" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28229" PostTypeId="2" ParentId="28226" CreationDate="2019-05-05T20:40:55.237" Score="1" Body="&lt;p&gt;Here is a simple design as a starting point for you to modify according to what are your constraints and what are the data that you need to collect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the track of the wheel, there is a graduated scale that could be fit with acceleration sensors or what not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HFZum.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HFZum.jpg&quot; alt=&quot;diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-05T20:40:55.237" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28230" PostTypeId="1" CreationDate="2019-05-05T22:14:20.237" Score="1" ViewCount="90" Body="&lt;p&gt;I tried to find the answer in past posts, but couldn't locate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a pump (100% efficient, for simplicity) that runs on 100kW and follows the equation: Power(pump) = (Specific Weight)&lt;em&gt;(Flow)&lt;/em&gt;(Pump Head). The flow is steady, but not inviscid. We are given info for losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My specific question is this: Where should points A and B be located for a Bernoulli equation designed to account for the energy imparted to the fluid by the pump? &lt;em&gt;I must calculate flow imparted by pump&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I place point &quot;A&quot; at the pipe entrance? Or, perhaps, right before it? I'm confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An &lt;a href=&quot;https://i.stack.imgur.com/Od1ky.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;illustration&lt;/a&gt; is attached and, as this question is quite specific, please let me know if any information is needed. I have a solution, but want to grasp the concepts enough to understand what is happening, physically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know if more info is needed, or if I've overlooked a post from the past. Thanks, everyone.&lt;/p&gt;&#xA;" OwnerUserId="20044" LastActivityDate="2021-05-25T05:06:55.647" Title="When finding flow generated by a pump, where do points A and B go?" Tags="|fluid-mechanics|pumps|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28231" PostTypeId="2" ParentId="28230" CreationDate="2019-05-06T01:26:44.947" Score="1" Body="&lt;p&gt;We put A and B just before and after the pump.&#xA;IN Bernoulli equation the pump pressure goes to the left side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_{p}= W{in}\cdot\gamma /Q \quad and\ \gamma =1$&lt;/span&gt;&#xA;in your pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_1+1/2 \rho v_1^2+ \rho gh_1 +P_p= P_2+1/2 \rho v_2^2+ \rho gh_2 + f_h\rho g $&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-06T01:26:44.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28232" PostTypeId="1" CreationDate="2019-05-06T08:28:32.543" Score="1" ViewCount="174" Body="&lt;p&gt;Can you please explain the difference between high-energy pulsed lasers and frequency combs?&lt;/p&gt;&#xA;" OwnerUserId="20212" LastEditorUserId="272" LastEditDate="2019-05-06T11:51:47.617" LastActivityDate="2021-09-22T17:07:05.753" Title="What's the difference between pulsed lasers and frequency combs?" Tags="|electrical-engineering|optics|lasers|fiber-optics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28233" PostTypeId="2" ParentId="28232" CreationDate="2019-05-06T11:50:12.790" Score="1" Body="&lt;p&gt;There is not necessarily any relationship between the two.  In theory, a pulsed laser could exist without exhibiting a frequency comb spectrum, and a frequency comb could exist without operating in pulsed mode.  Your confusion probably arises because one has a discrete set of pulses in the &lt;em&gt;time domain&lt;/em&gt; while the other has a discrete set of 'pulses' in the &lt;em&gt;frequency domain&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pulsed Lasers:&lt;/strong&gt; Pulsed lasers emit very short bursts of laser energy.  The width of the pulses can be as short as ~10s of femtoseconds with no upper limit on the pulse length.  Common pulsed lasers used in industrial processing applications operate with pulse lengths of ~10s of nanoseconds.  Most pulsed lasers operate in a repetitive pulse mode with pulse repetition frequencies (PRF) ranging from Hertz up to Megahertz.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Frequency Combs:&lt;/strong&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Frequency_comb&quot; rel=&quot;nofollow noreferrer&quot;&gt;Frequency combs&lt;/a&gt; may or may not be pulsed (although the most common type is a pulsed laser).  The requirement to be a frequency comb is that the frequency spectrum (or equivalently wavelength spectrum) exhibit discrete and evenly-spaced frequencies.  This may come as a surprise because most lasers are designed to operate at a single well-defined frequency, but that is not a requirement to be a laser. &lt;/p&gt;&#xA;" OwnerUserId="272" LastActivityDate="2019-05-06T11:50:12.790" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28235" PostTypeId="2" ParentId="28232" CreationDate="2019-05-06T12:34:40.277" Score="1" Body="&lt;p&gt;A pulsed laser is defined via the difference between phase and group velocity - the frequency &lt;span class=&quot;math-container&quot;&gt;$f_{ceo}$&lt;/span&gt; - and the repetition frequency of the pulses - &lt;span class=&quot;math-container&quot;&gt;$f_{rep}$&lt;/span&gt;. In the frequency domain (because the pulses are following each other in time with frequency &lt;span class=&quot;math-container&quot;&gt;$f_{rep}$&lt;/span&gt;), the spectrum is a line spectrum since it is the Fourier transform of&#xA;arbitrarily thin pulses in the time domain AND vice versa. &lt;span class=&quot;math-container&quot;&gt;$f_{ceo}$&lt;/span&gt; is the offset of&#xA;the line spectrum against frequency &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;. This is always a comb spectrum/frequency comb.&lt;/p&gt;&#xA;" OwnerUserId="14377" LastActivityDate="2019-05-06T12:34:40.277" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28236" PostTypeId="1" AcceptedAnswerId="28239" CreationDate="2019-05-06T13:03:46.977" Score="1" ViewCount="166" Body="&lt;p&gt;I'm recently interested in print-heads, especially thermal inkjet print heads.&#xA;I would like to design a simple print head with nozzles in a row with no chambers, but only a straight linear pipe-like structure, and the heating resistors stuck on the side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm quite lost here. I don't know what kind of material I should use, or what should be the diameter of the nozzle, and etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to look at the earliest version of thermal inkjet print heads but all that I can get in google patent is the latest, and high tech version of print heads that I can barely understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone tell me what to do, or where to look at?&lt;/p&gt;&#xA;" OwnerUserId="20348" LastActivityDate="2019-05-06T16:42:13.283" Title="How to design and make thermal inkjet print head / nozzles" Tags="|mechanical-engineering|electrical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28237" PostTypeId="1" AcceptedAnswerId="28238" CreationDate="2019-05-06T14:50:37.853" Score="1" ViewCount="124" Body="&lt;p&gt;ME Design uses hourly climate data files (HCD) as input into the pavement design algorithm. As a user I can enter an arbitrary number of years of climate data. If I enter say 30 years of climate data and then set an analysis period of 20 years, what data does the program use? Does it use the first value in the list or the last 20 years of the file? What if my analysis period is greater than the amount of climate data entered?&lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-06-05T15:02:00.200" Title="How does Pavement ME Design handle climatic data?" Tags="|pavement|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28238" PostTypeId="2" ParentId="28237" CreationDate="2019-05-06T14:50:37.853" Score="1" Body="&lt;p&gt;Pavement ME Design checks to see if the pavement construction date is within the entered climatic data range. If it is, the analysis uses that date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pavement construction date is not included in the entered climate data range, the analysis always uses the earliest month/day in the list that matches the pavement construction date. It does this so that the answer is the same at years 1,2,5,10 whether an analysis is run for 10, 20 or 30 years.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the analysis period is longer than the available climate data, the data is reused starting from the earliest applicable date.&lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-05-06T14:50:37.853" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28239" PostTypeId="2" ParentId="28236" CreationDate="2019-05-06T16:42:13.283" Score="2" Body="&lt;p&gt;For starters, read the May 1985 issue of the &lt;em&gt;HP Journal&lt;/em&gt; magazine; it explains the construction of the very first thermal inkjet printhead that HP brought to market in March of 1984, which I helped design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That printhead was called the &lt;em&gt;Thinkjet&lt;/em&gt;; you can try googling on that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wrote extensively about printhead device physics and design during my 34 years in the inkjet field and would be happy to discuss your project with you. You can contact me directly through my website, www.nielsenkillowatt.com.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-05-06T16:42:13.283" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28242" PostTypeId="1" CreationDate="2019-05-07T11:06:37.513" Score="2" ViewCount="62" Body="&lt;p&gt;I am preparing a basic course on Separation Processes (5th semester, BSc Chemical Engineering). I can see that many well-known textbooks (e.g. Seader &amp;amp; Henley, Wankat) start by teaching in detail distillation and only then, the chapters on absorption/stripping are presented. As a student, I think I was taught exactly the opposite: first absorption and then distillation, which I believe it is a natural way as absorption is significantly simpler than distillation. My question is, what are the pros and cons of teaching first absorption and then distillation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My own list would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Pros&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Absorption/stripping equilibria are often described by Henry's law (or similar). It is an easier first step than complex VLE expressions&lt;/li&gt;&#xA;&lt;li&gt;It's a good way to introduce McCabe-Thiele algorithms&lt;/li&gt;&#xA;&lt;li&gt;No condenser or reboiler. Therefore, no need to do energy balances at first (only when vaporization heat is very significant) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cons&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Distillation is the most relevant separation process and it is important to allocate a lot of time to its thorough study&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3991" LastEditorUserId="3991" LastEditDate="2019-05-07T14:12:10.937" LastActivityDate="2019-05-07T14:12:10.937" Title="When teaching separation processes, which one should be introduced first? Distillation or absorption?" Tags="|chemical-engineering|process-engineering|mass-transfer|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28243" PostTypeId="2" ParentId="28218" CreationDate="2019-05-07T13:30:38.447" Score="1" Body="&lt;h1&gt;Starting Point&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The starting point of the difference form of the steady state energy balance equation is below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{q} - \dot{w} = \dot{m}\Delta\left(\tilde{U} + \frac{p}{\rho} + \frac{v^2}{2} + gz\right) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, &lt;span class=&quot;math-container&quot;&gt;$\dot{q}$&lt;/span&gt; is the heat flow (J/s - negative leaving the system), &lt;span class=&quot;math-container&quot;&gt;$\dot{w}$&lt;/span&gt; is the Clausius form of work (J/s - positive when done by the system), &lt;span class=&quot;math-container&quot;&gt;$\dot{m}$&lt;/span&gt; is the mass flow (kg/s), &lt;span class=&quot;math-container&quot;&gt;$\tilde{U}$&lt;/span&gt; is the specific internal energy (J/kg), &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is pressure (Pa), &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is density (kg/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;), &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; is velocity (m/s), &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt; is gravity (m&lt;span class=&quot;math-container&quot;&gt;$^2$&lt;/span&gt;/s), and &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is height (m).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any system, the definition of specific enthalpy is &lt;span class=&quot;math-container&quot;&gt;$\tilde{H} \equiv \tilde{U} + p\tilde{V} = \tilde{U} + \frac{p}{\rho}$&lt;/span&gt;. The expression becomes as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{q} - \dot{w} = \dot{m}\Delta\left(\tilde{H} + \frac{v^2}{2} + gz\right) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;System&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Take the control volume as the fluid inside the compression flow. Allow adiabatic so that &lt;span class=&quot;math-container&quot;&gt;$\dot{q} = 0$&lt;/span&gt;. Redefine &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; as purely shaft work because the system (fluid) does not expand or contract &lt;em&gt;as a whole&lt;/em&gt; (see below about incompressible fluids or ideal gases).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is horizontal in &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;. This is identical to stating categorically that &lt;span class=&quot;math-container&quot;&gt;$\Delta z \equiv 0$&lt;/span&gt;. The somewhat questionable approach to propose to neglect change in potential energy misses the clarity that zero change in height is a fact of the choice of control volume and not a (randomly applied) assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity of the fluid at any point is found from volumetric flow &lt;span class=&quot;math-container&quot;&gt;$\dot{V}$&lt;/span&gt; and area &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dot{m} = \dot{V}\rho = vA\rho$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Incompressible Fluid&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;For an incompressible fluid AND ONLY FOR THIS CASE, we can use the conservation of energy expression to obtain this expression at any point in the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ v = v_i \frac{A_i}{A} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The statement that we &quot;neglect&quot; changes in kinetic energy can therefore only mean that we correspondingly assume that the area changes are negligible. Otherwise, the full representation of the energy balance is as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ - \dot{w}_s = \dot{m}\Delta\tilde{H} +\frac{\dot{m}v_i^2}{2} \left(\frac{A_i^2}{A_f^2} - 1 \right) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a system with constant specific heat capacity &lt;span class=&quot;math-container&quot;&gt;$\tilde{C}_p$&lt;/span&gt; (J/kg K), this can be written as below using change in temperature &lt;span class=&quot;math-container&quot;&gt;$\Delta T = T_f - T_i$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;K or &lt;span class=&quot;math-container&quot;&gt;$^o$&lt;/span&gt;C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ - \dot{w}_s = \dot{m}\tilde{C}_p\Delta T + \frac{\dot{m}v_i^2}{2} \left(\frac{A_i^2}{A_f^2} - 1 \right) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Ideal Gas&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The opposing case to an incompressible fluid is the ideal gas. In this case, we obtain the expression below using molar mass &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; (kg/mol) and gas law constant &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; (J/mol K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{m} = \frac{\dot{n}}{M} = \frac{p}{MRT}vA $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to a modification of the velocity equation as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ v = v_i \frac{\beta_i A_i}{\beta A} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$\beta = p/MRT$&lt;/span&gt;. For a system with negligible pressure drop from inlet to outlet, the energy balance becomes as below for an ideal gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ - \dot{w}_s = \dot{m}\tilde{C}_p\Delta T + \frac{\dot{m}v_i^2}{2} \left(\frac{\beta_i^2 A_i^2}{\beta_f^2 A_f^2} - 1 \right) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The solution states that potential energy change is negligible. In fact, &lt;span class=&quot;math-container&quot;&gt;$\Delta z = 0$&lt;/span&gt; is an absolute truth, not a required assumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution states that kinetic energy change is negligible. This can be a gross mistake depending on the type of fluid (incompressible or ideal gas), the extent of the throttling ratio &lt;span class=&quot;math-container&quot;&gt;$A_f/A_i$&lt;/span&gt;, and for an ideal gas the ratio of temperature and pressure drops through the system.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-05-07T14:32:21.413" LastActivityDate="2019-05-07T14:32:21.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28244" PostTypeId="2" ParentId="28213" CreationDate="2019-05-07T16:01:06.127" Score="3" Body="&lt;p&gt;As a mechanical engineer, I've come across the stress being represented mainly as a second order tensor, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Cauchy_stress_tensor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cauchy stress tensor&lt;/a&gt;, however vector representations are also common. As for scalar representations? This might be common in scenarios of uniaxial stress, such as a bar under uniform tensile stress, where there is only one component of stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cauchy stress tensor&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, for a given coordinate system &lt;span class=&quot;math-container&quot;&gt;$O(x,y,z)$&lt;/span&gt;, there are 6 components of stress:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_{xx}\quad \sigma_{yy}\quad \sigma_{zz}\quad \sigma_{xy}\quad \sigma_{xz}\quad \sigma_{yz}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a solid object of arbitrary shape. It can be a beam, a brick, a gear, etc. In general, the stress will vary throughout the solid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To consider the stresses at a point in the solid body, let's surgicaly remove an infinitesimally small cube from within the solid at the point of interest, to observe the stress acting on the faces of the cube. The faces of the cube is aligned with the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-, &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;- and &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-axes of our coordinate system. On each face of the six faces of the cube, three components of stress act: two components lie tangential to the face (these components are called shear stresses), and one component is perpendicular to the face (this component is called the normal stress). The diagram below shows the stresses acting on three of the faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/op2Oh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/op2Oh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Six faces, each with three stress components, might suggest a total of 18 components. However, balance of forces (stress &lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt; infinitesimal face area) onto the cube means pairs of stress components on opposite faces are equal and opposite (&lt;span class=&quot;math-container&quot;&gt;$\rightarrow$&lt;/span&gt; 9 unique components).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The convention for defining stress components is to use the form &lt;span class=&quot;math-container&quot;&gt;$\sigma_{ij}$&lt;/span&gt;, where subscript &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt; indicates the face on the cube which the stress component acts, and &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt; indicates the direction of the stress. (&lt;span class=&quot;math-container&quot;&gt;$\sigma_{xy}$&lt;/span&gt; acts in the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; direction on the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;-face of the cube).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By balance of moments, it can be shown that &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_{xy} = \sigma_{yx}, \qquad \sigma_{xz} = \sigma_{zx}, \qquad \sigma_{yz} = \sigma_{zy}$$&lt;/span&gt;&#xA;reducing the number of unique stress components to six.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, at any point in a 3D solid, the stress state is fully described by six numbers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sigma_{xx}\quad \sigma_{yy}\quad \sigma_{zz}$&lt;/span&gt; are the normal stress components&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sigma_{xy}\quad \sigma_{xz}\quad \sigma_{yz}$&lt;/span&gt; are the shear stress components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Frequently, shear stresses (but not normal stresses) are alternatively expressed as &lt;span class=&quot;math-container&quot;&gt;$\tau_{ij}$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$\sigma_{ij}$&lt;/span&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A natural means of expressing this stress state is as a Cauchy stress tensor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\sigma] = \begin{bmatrix}&#xA;\sigma_{xx} &amp;amp; \sigma_{xy} &amp;amp; \sigma_{xz} \\&#xA;\sigma_{yx} &amp;amp; \sigma_{yy} &amp;amp; \sigma_{yz} \\&#xA;\sigma_{zx} &amp;amp; \sigma_{zy} &amp;amp; \sigma_{zz} \\&#xA;\end{bmatrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or, equivalently, to highlight the symmetric nature of the stress tensor as well as using only the 6 unique components:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\sigma] = \begin{bmatrix}&#xA;\sigma_{xx} &amp;amp; \sigma_{xy} &amp;amp; \sigma_{xz} \\&#xA;\sigma_{xy} &amp;amp; \sigma_{yy} &amp;amp; \sigma_{yz} \\&#xA;\sigma_{xz} &amp;amp; \sigma_{yz} &amp;amp; \sigma_{zz} \\&#xA;\end{bmatrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In 2D stress problems where, say, stresses in the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;-direction are not of concern, this tensor is often reduced to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\sigma] = \begin{bmatrix}&#xA;\sigma_{xx} &amp;amp; \sigma_{xy} \\&#xA;\sigma_{xy} &amp;amp; \sigma_{yy} \\&#xA;\end{bmatrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of the Cauchy tensor is that it is possible to re-express the stress components in a different frame of reference &lt;span class=&quot;math-container&quot;&gt;$O(x',y',z')$&lt;/span&gt; by performing a change of basis on the tensor. This is done by performing a matrix transformation on the array of elements &lt;span class=&quot;math-container&quot;&gt;$[\sigma]$&lt;/span&gt; to obtain the array of stress components in the new coordinate system &lt;span class=&quot;math-container&quot;&gt;$[\sigma']$&lt;/span&gt;. That is, if a vector with components &lt;span class=&quot;math-container&quot;&gt;$[a]$&lt;/span&gt; expressed in &lt;span class=&quot;math-container&quot;&gt;$O(x,y,z)$&lt;/span&gt; is related to the components &lt;span class=&quot;math-container&quot;&gt;$[a']$&lt;/span&gt; re-expressed in &lt;span class=&quot;math-container&quot;&gt;$O(x',y',z')$&lt;/span&gt; via the equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[a'] = [Q][a]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$[Q]$&lt;/span&gt; is an orthogonal matrix, then the Cauncy stress tensor elements in both systems are interrelated by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\sigma'] = [Q][\sigma][Q]^T$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is nothing stopping you from re-arranging the 6 stress components into a 6-dimension column vector, such as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$[\sigma] = \begin{bmatrix}&#xA;\sigma_{xx} \\&#xA;\sigma_{yy} \\&#xA;\sigma_{zz} \\&#xA;\sigma_{xy} \\&#xA;\sigma_{xz} \\&#xA;\sigma_{yz} \\&#xA;\end{bmatrix}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is convenient for expressing material laws in linear elasticity problems, as in this form, the stress vector is related to a similarly defined strain vector via multiplication with a &lt;span class=&quot;math-container&quot;&gt;$6\times 6$&lt;/span&gt; constitution matrix (as opposed to the stress tensor being related with the strain tensor via appropriate multiplication of a fourth order tensor!). However, re-expressing this vector in different coordinate systems is not a readily available action. A transformation matrix &lt;span class=&quot;math-container&quot;&gt;$[Q]$&lt;/span&gt; cannot be used directly here, least to say that &lt;span class=&quot;math-container&quot;&gt;$[Q]$&lt;/span&gt; is a &lt;span class=&quot;math-container&quot;&gt;$3\times 3$&lt;/span&gt; matrix, and the stress vector here is &lt;span class=&quot;math-container&quot;&gt;$6 \times 1$&lt;/span&gt;!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Cauchy stress vector&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, we used the unit cube to describe the stress state at a point by observing the faces of an infinitesimal cube. An alternative way to examine stress is to analyse how the stress act on an imaginary surface at that point. Imagine now that we cut the solid body about a plane that passes through the point, and the plane has a normal vector&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[n] = \begin{bmatrix}n_x \\ n_y \\ n_z\end{bmatrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/s82Sv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s82Sv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's analyse the stress that act on this plane at the point of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like for the case of the cube, three perpendicular components of stress act on the cutting plane, and these are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$t_x \quad t_y \quad t_z$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the total stress acting on this surface can be represented as a vector, in the form&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[t] = \begin{bmatrix}t_x \\ t_y \\ t_z\end{bmatrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This vector is a true vector is the sense that it can be transformed to re-expressed its elements in a new coordinate system. Considering the transformation matrix &lt;span class=&quot;math-container&quot;&gt;$[Q]$&lt;/span&gt; from earlier, the stress vector representations in the two coordinate systems are interrelated by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[t'] = [Q][t]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stress vector is referred to as the Cauchy stress vector. Note that expression of stress as a vector requires a reference surface of a given normal &lt;span class=&quot;math-container&quot;&gt;$[n]$&lt;/span&gt; (The Cauchy stress tensor can be defined without providing a reference surface).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cauchy stress vector is related to the Cauchy stress tensor by the Cauchy stress equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$[\sigma][n] = [t]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In full:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{bmatrix}&#xA;\sigma_{xx} &amp;amp; \sigma_{xy} &amp;amp; \sigma_{xz} \\&#xA;\sigma_{xy} &amp;amp; \sigma_{yy} &amp;amp; \sigma_{yz} \\&#xA;\sigma_{xz} &amp;amp; \sigma_{yz} &amp;amp; \sigma_{zz} \\&#xA;\end{bmatrix}&#xA;\begin{bmatrix}n_x \\ n_y \\ n_z\end{bmatrix}&#xA;=&#xA;\begin{bmatrix}t_x \\ t_y \\ t_z\end{bmatrix}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Stress vs traction&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To define our stress vector, we had to cut at a plane to obtain a surface on which to describe stress. However, what happens if we consider a point on the outer surface of our solid? A similar vector can be computed by considering the force per unit acting on the outer surface of the solid, but, as a mechanical engineer, I would argue that this vector is called a traction vector instead of a stress vector at this point, since now this vector arises due to external forces rather that internal. Stresses are internal forces per unit area, whereas tractions are external forces per unit area. However, this is just a matter of terminology, and other disciplines may use different terminologies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;To summarise&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stress can be expressed as a vector or a tensor. Stress can only really be described by a scalar for simple uniaxial stress problems.&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2019-05-07T16:01:06.127" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28245" PostTypeId="1" CreationDate="2019-05-07T19:20:01.017" Score="1" ViewCount="296" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dppEM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dppEM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bsb4s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bsb4s.png&quot; alt=&quot;ucy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please help me find the answer to the above question, I am unsure how to work with F2 as I can't find the distance from Moment A or Moment J to this. Thank you&lt;/p&gt;&#xA;" OwnerUserId="20365" LastActivityDate="2024-03-17T12:06:30.747" Title="Help with finding support reactions of a truss" Tags="|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28246" PostTypeId="2" ParentId="28245" CreationDate="2019-05-07T20:32:56.213" Score="0" Body="&lt;p&gt;Welcome to Engineering! Have you considered that the force &lt;span class=&quot;math-container&quot;&gt;$F2$&lt;/span&gt; has a component in the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; direction?&#xA;If you break &lt;span class=&quot;math-container&quot;&gt;$F2$&lt;/span&gt; down this way, you can observe that you now have a perpendicular distance to each of these components that you can use so sum the moments about one of the hinges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you sum the moments about the hinge &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; you should only have one variable which is the horizontal support reaction at &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then take that reaction into the sum of the forces to maintain equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a conceptual guide to solving your problem, if you have any more specific issues I think it would be appropriate for you to show your work, and your issue here.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-05-07T20:32:56.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28247" PostTypeId="1" AcceptedAnswerId="28248" CreationDate="2019-05-07T21:08:06.780" Score="0" ViewCount="1783" Body="&lt;p&gt;The following images are taken from Mechanics of Materials, Ferdinard Beer et al. 6th ed (2012)&#xA;Just wondering that why the effective length of the column is taken 0.7 L (highlighted in the sample problem) ? Souldn't it be 0.5 L since in the xy plane we have two plates supporting the column with a reaction force in the y direction and a moment (which is the same as if the column were fixed at both ends).&#xA;Any thoughts would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/MuIhF.png&quot; alt=&quot;enter image description here&quot;&gt;&#xA;&lt;img src=&quot;https://i.stack.imgur.com/h1D8J.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="20249" LastActivityDate="2019-05-08T02:27:03.753" Title="Buckling in Columns" Tags="|solid-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28248" PostTypeId="2" ParentId="28247" CreationDate="2019-05-07T21:20:50.823" Score="0" Body="&lt;p&gt;The text says... &quot;smooth and rounded fixed plates&quot;. So although the plates are fixed, they are not rigidly attached to the column. They will restrain translation in the y and that is all. So for all intents and purposes you can consider the top support to be &quot;pinned&quot;. And the effective length can be determined from The first figure, condition c).&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-05-07T21:20:50.823" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28249" PostTypeId="2" ParentId="28245" CreationDate="2019-05-07T21:22:26.277" Score="0" Body="&lt;p&gt;Whenever confused start from the basics:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum M &amp;amp;=0 \\&#xA;\sum F_x &amp;amp;=0 \\&#xA;\sum F_y &amp;amp;=0&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore &#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F_{a_y} = F_1 + F_2 \cos30= 1.5 + 2.598 = 4.09\text{ kN (up)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J has no vertical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the moment sum we have:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\sum M =0 \therefore F_{j,x}\cdot4 - (1.5\cdot2 + 2.598\cdot6 -3\cdot1/2\cdot2) =0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last part in the parenthesis is the horizontal component of &lt;span class=&quot;math-container&quot;&gt;$F_2$&lt;/span&gt; multiplied by its moment arm to get its contributory moment about A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here you can calculate horizontal reaction of J and then the horizontal reaction of A.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2019-05-11T15:21:04.430" LastActivityDate="2019-05-11T15:21:04.430" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28251" PostTypeId="2" ParentId="28247" CreationDate="2019-05-08T02:27:03.753" Score="0" Body="&lt;p&gt;In real life cases these diagrams are all highly exaggerated to convey the idea that effective length has to do with the inflection point and or, it starts at a pin support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise for all intents and purposes the smooth rounded edge of the two plates are close enough to a pin support at that joint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we want to be too fussy,  even if the two plates are too tight and fit to the column, under the load they yield just a smidge to make it a pin connection. And we would see the yield line as shiny horizontal score lines after. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-08T02:27:03.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28252" PostTypeId="1" AcceptedAnswerId="28273" CreationDate="2019-05-08T04:06:53.857" Score="0" ViewCount="90" Body="&lt;p&gt;So I am working on a project. I need a drone (quadcopter) to ascend for 5 seconds, then cut off power to the drone, then let it stay in the air for as long as possible without power. My idea for allowing it to glide is to have some rubber bands attached to some parachute like material, and the rubber bands will be stretched and held in place by some pins. After 4 to 5 seconds, the pins will disengage, allowing the rubber bands to compress, expanding the parachute-like material. An image is included to illustrate what I mean. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The red lines illustrate the rubber bands. The black are the quadcopter arms. The blue circles are where the pins would be located. After the timer stops, the image on the right demonstrates the rubber band straightened, and the purple is the parachute area. There will be one parachute per rubber band, and each rubber band will have its own locking mechanism. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ApvpG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ApvpG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I want to figure out is how to design a mechanism that will retract a piston/pin after 4 to 5 seconds. I did some research on pneumatic cylinders, and I figured I could have pressurized air entering the cylinders with air leaving the system slowly until the piston disengages the rubber band. There would be some measuring to get the required air pressure needed to have the pistons last 5 seconds, but these cylinders are too expensive and too long for this application. Then, I read the post linked at the bottom of this post, and I think the mag-locks would be great, but I have only found them in rectangular forms; none of them have the geometry needed for this, so I figured I could have the mag-lock attached to the bottom of the drone, and the mag-lock would pull pins that were attached to springs that are attached to the top of the drone; when the mag-lock deactivates, the pins would retract, allowing the rubber band to disengage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l4s1m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l4s1m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am having issues with finding timers that are suitable for my task. Kitchen timers don't work well when it comes down to setting them for 5-seconds. I want to know if you guys have any ideas regarding getting a timer that would be well suited for this task and if you guys have any suggestions as to how I could improve the design of the parachute mechanism I have planned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Post mentioning mag-locks: &lt;a href=&quot;https://engineering.stackexchange.com/questions/10288/is-there-a-device-like-this&quot;&gt;Is there a device like this?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20368" LastActivityDate="2019-05-08T19:23:06.080" Title="Use 4 to 5-second timer to disengage/engage mechanical system" Tags="|design|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28253" PostTypeId="2" ParentId="25701" CreationDate="2019-05-08T09:24:55.367" Score="3" Body="&lt;p&gt;I'd call the blue part the &lt;code&gt;transmitter housing&lt;/code&gt;.&#xA;This usually contains electronics for sensor signal conversion, which converts the sensor (or probe) low level signal to a standard 4-20 mA signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensor (probe) is the metallic cylinder. And the &lt;code&gt;sensor head&lt;/code&gt; (containing the sensing element, membrane/coil) is usually located at the end surface of the sensor. &lt;/p&gt;&#xA;" OwnerUserId="20228" LastActivityDate="2019-05-08T09:24:55.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28254" PostTypeId="1" AcceptedAnswerId="28258" CreationDate="2019-05-08T10:11:53.450" Score="25" ViewCount="14177" Body="&lt;p&gt;As I understand, a steam machine needs a pressurised gas to work. This can be &lt;a href=&quot;https://en.wikipedia.org/wiki/Pneumatic_motor&quot; rel=&quot;noreferrer&quot;&gt;compressed air&lt;/a&gt; but it used to be &lt;a href=&quot;https://en.wikipedia.org/wiki/Steam_engine&quot; rel=&quot;noreferrer&quot;&gt;steam&lt;/a&gt;. Energy was provided to the steam engine by heating up water. I don't understand why it used water as the pressurised gas could have been generated only by heating air in a closed tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the design of heating air in a tank instead of heating water to generate steam feasible? If so, why do steam engine work with steam instead of hot air?&lt;/p&gt;&#xA;" OwnerUserId="2817" LastEditorUserId="90" LastEditDate="2019-05-08T16:43:52.397" LastActivityDate="2023-10-22T05:18:01.957" Title="Why use steam instead of just hot air?" Tags="|steam|" AnswerCount="9" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28255" PostTypeId="2" ParentId="25701" CreationDate="2019-05-08T10:15:33.880" Score="1" Body="&lt;p&gt;The part in your image is a Level Transmitter, &lt;a href=&quot;https://www.aliexpress.com/i/32711941471.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.aliexpress.com/i/32711941471.html&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The silver bit is the transducer, and the blue bit is the transmitter/&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-05-08T10:15:33.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28256" PostTypeId="2" ParentId="28254" CreationDate="2019-05-08T11:08:09.933" Score="43" Body="&lt;p&gt;Because boiling a volume of water creates a &lt;em&gt;much bigger&lt;/em&gt; volume of steam this volume increase is about 1700 times. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the gas laws doubling the absolute temperature of a volume of gas only doubles the volume at the same pressure, or doubles the pressure at the same volume. Doubling the absolute temperature means going from 300 K (27°C) to 600 K (327°C). So to get working pressures you need to heat to extremely high temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an additional benefit the pressure of steam is related to the temperature of the boiler. This means that if you can maintain temperature there will be no pressure drop until you run out of water to boil. With pressurized gas the pressure will drop as you use gas unless you increase the temperature even more.&lt;/p&gt;&#xA;" OwnerUserId="288" LastEditorUserId="288" LastEditDate="2019-05-10T12:16:23.947" LastActivityDate="2019-05-10T12:16:23.947" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28258" PostTypeId="2" ParentId="28254" CreationDate="2019-05-08T12:41:00.423" Score="27" Body="&lt;p&gt;I would say that using a warm pressurized gas is not very feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ratchet freak already mentioned how you can get much more volume out of heating water into steam than just heating up air until it's warmer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This touches on, but doesn't completely address an important factor about steam as power.  Converting to steam includes a &lt;em&gt;phase change&lt;/em&gt; from liquid to gas.  This phase change actually acts as an additional storage of energy.  You can draw this energy out of the steam later in the system (through a heat exchanger for example), converting it back into a liquid, which can then be fed back into a boiler to produce more steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, it does come down to what Solar Mike said in his comment.  Steam contains more energy than air, and it especially contains more &lt;em&gt;available&lt;/em&gt; energy compared to air. This is due to the additional energy stored in the form of phase change, along with the greater specific heat capacity of steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  Another factor I forgot about is that it is more effective to have a liquid phase when it comes time to re-pressurize the system.  It's easier to pressurize the water because it doesn't really compress.&lt;/p&gt;&#xA;" OwnerUserId="7736" LastEditorUserId="7736" LastEditDate="2019-05-08T18:48:11.163" LastActivityDate="2019-05-08T18:48:11.163" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="28259" PostTypeId="1" AcceptedAnswerId="28262" CreationDate="2019-05-08T14:05:43.033" Score="0" ViewCount="57" Body="&lt;p&gt;I have designed a flow resistance element and I would like to be able to keep the total pressure drop across the element to &amp;lt; 5.0 inches of water (from 5cc/min to 5,000 cc/min). I have run some bench tests on the current design and the pressure drops are a bit too high. I would like to avoid changing the diameter of the bore and would prefer to shorten the length of the flow path. What would be the best way to calculate the ideal flow path length to achieve the pressure drops I am looking for?&lt;/p&gt;&#xA;" OwnerUserId="19866" LastActivityDate="2019-05-08T14:29:32.010" Title="How to determine ideal flow path length for specified pressure drop" Tags="|mechanical-engineering|fluid-mechanics|pressure|airflow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28260" PostTypeId="1" CreationDate="2019-05-08T14:23:51.970" Score="1" ViewCount="551" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JOrUw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JOrUw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm new to beams and I'm struggling to find the correct vertical reactions forces at support C and E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct answer is 8.58&amp;nbsp;kN for &lt;span class=&quot;math-container&quot;&gt;$V_c$&lt;/span&gt; and -3.58&amp;nbsp;kN for &lt;span class=&quot;math-container&quot;&gt;$V_e$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I calculate the sum of the moments and set equal to 0 I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{gather}&#xA;1.5 + 2V_c - (3\cdot3) + (2\cdot(2+\frac{2}{3}))= 0 \\&#xA;V_c = 2.4\text{ kN} \therefore V_e= 2.6\text{ kN}&#xA;\end{gather}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is it that I'm doing wrong in my calculation?&lt;/p&gt;&#xA;" OwnerUserId="20365" LastEditorUserId="1832" LastEditDate="2019-05-11T15:07:13.427" LastActivityDate="2019-05-11T15:07:13.427" Title="Reaction Forces on a Beam" Tags="|statics|beam|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28261" PostTypeId="1" CreationDate="2019-05-08T14:24:32.690" Score="0" ViewCount="426" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/beJow.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While trying to solve this question and using the flexure formulas of bending for composite beams  I obtained a lower value of moment for steel than for aluminum.How is that possible?&lt;/p&gt;&#xA;" OwnerUserId="20374" LastActivityDate="2019-05-12T03:35:04.107" Title="Composite beam- increase in bending moment due to reinforced steel" Tags="|beam|composite|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28262" PostTypeId="2" ParentId="28259" CreationDate="2019-05-08T14:29:32.010" Score="1" Body="&lt;p&gt;Use the Darcy-Weisbach equation and the Swanee-Jain correlation in order to calculate the pressure drop and the friction factor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$pressure drop = rho*f*(L/D)*v^2/2 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this formula (the first one): &lt;a href=&quot;https://es.m.wikipedia.org/wiki/Ecuaci%C3%B3n_de_Swamee-Jain&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://es.m.wikipedia.org/wiki/Ecuación_de_Swamee-Jain&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where rho is the density, f the friction factor, L the ideal length, D the diameter and v the speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, K is the absolute roughness and are the Reynolds number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your element isn't exactly a pipe, you can model it as a pipe by using the equivalent length: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cedengineering.com/userfiles/Pneumatic%20Conveying%20Systems.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cedengineering.com/userfiles/Pneumatic%20Conveying%20Systems.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Go to page 57.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-05-08T14:29:32.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28263" PostTypeId="2" ParentId="28254" CreationDate="2019-05-08T14:31:18.917" Score="4" Body="&lt;p&gt;&lt;strong&gt;For Air:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the ideal gas law PV = nRT rearranged to be V = (nRT)/P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n and R are considered constant, and for our purposes, P is also constant. Thus, to increase the volume, (i.e. create a flow of volume across the turbine to power it), T has to be increased. From this equation, V is proportional to T. so for every multiplicative increase of T you get the same increase in V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an increase of 1000x in volume, T has to be increased by 1000x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For Steam:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now when we consider steam, it is a much different story. This equation is no longer valid. I think it would be best to look at the density of water (liquid) vs water (steam). liquid water has a density of 1000 kg/m^3 and we will say steam is 1 kg/m^3. There are variations possible here based on pressure and temperature etc, but we will go with what I listed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since density is mass/volume, we can see from these numbers that when water is boiled off, its volume increases 1000X!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since water can be liquid &amp;lt; 100 °C and Steam &gt; 100 °C, the temperature would only have to be increased by a few degrees to get 1000x expansion! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional note here too, is that the steam loses a lot of energy when it is running a turbine, and is condensed back to a liquid state. Not all of it is, but a lot of it is (Conservation of energy). This condensed steam may now be used again and be boiled off to run the turbine more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For air, this phenomena does not exist and the air has to be compressed/cooled down all over again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conclusion:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a lot more going on here, but this is the main principle here. Air &lt;em&gt;can&lt;/em&gt; be used, it just &lt;em&gt;shouldn't&lt;/em&gt; be used. Water is just more efficient due to the phase change -&gt; density/volume change.&lt;/p&gt;&#xA;" OwnerUserId="20373" LastActivityDate="2019-05-08T14:31:18.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28265" PostTypeId="2" ParentId="28260" CreationDate="2019-05-08T14:51:19.420" Score="0" Body="&lt;p&gt;Here are the mistakes:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The triangle load acts on the point 2 + 4/3 = 10/3 or 3 + 1/3. That's the load center, the baricenter of the triangle. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The sign of this term should be equal to the sign of 3*3 because both loads tend to bend the beam counter-clowisely. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerDisplayName="user20096" LastEditorUserId="7736" LastEditDate="2019-05-08T14:58:12.870" LastActivityDate="2019-05-08T14:58:12.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28267" PostTypeId="2" ParentId="28261" CreationDate="2019-05-08T16:10:04.360" Score="1" Body="&lt;p&gt;The difference in stiffness is important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First let's consider the aluminum W8x40 section without any steel plates: It can be loaded until the aluminum reaches a stress of 12 ksi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then let's consider what happens if we bolt some very thin steel plates unto the flanges of the aluminum section. Then we can increase the load until the steel plates reach a stress of 18 ksi. As the steel is 3 times stiffer than aluminum, the aluminum will have a stress that is 3 times lower than the steel plates. (I'm assuming we can disregard the slip in the bolts as well as a linear-elastic stress distribution — a plastic stress distribution would be a different calculation.) That results in a maximum stress in aluminum of 6ksi, so we get a total capacity of only half of what we had before we added the steel plates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the steel plates aren't extremely thin, they may increase the capacity, but the point is that the steel plates may fail well before we reach the capacity of the aluminum.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2019-05-08T16:10:04.360" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28268" PostTypeId="1" AcceptedAnswerId="28283" CreationDate="2019-05-08T17:27:02.043" Score="0" ViewCount="207" Body="&lt;p&gt;&lt;strong&gt;Disclaimer&lt;/strong&gt;: I'm not a (device) engineer. I don't know the terminology. So I suppose some of my statements, questions or ideas could sound either naive or vague.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;As far as I undestand, the most advanced (and publicly known) omnidirectional treadmill is the Infinadeck. But with current state of the progress on this treadmill, it's clearly not ready for a mass market. One of my biggest concerns is the shape of the treadmill: why don't the Infinadeck's engineers use a disk instead of a square? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking about possible implementation. I'm not sure how easy it would be to implement. Because, of course, the biggest issues are often related to  implementation details. I would like to hear your thoughts about it. As well as some comments on Infinadeck's shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This description isn't touching the vector generation or movement initiation, it's just about directionality of the treadmill.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iu5LA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iu5LA.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 1.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z3JEc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z3JEc.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure 2. Single engine: the sphere and the conveyor parts. This figure doesn't show the spikes, though.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So the shape of the device is cylindrical. There are three major elements: shell, engines and platform. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Shell is just a container for the engines. &lt;/li&gt;&#xA;&lt;li&gt;Platform is a thick metal disk wrapped in a thin elastic and seamless case. So it's like in regular one-directional threadmills, but radial. The case, of course, is smooth on the inside to lower the friction. But on the outside it has numerous tiny sockets that are used for pulling the case in the desired direction. And of course it's not fixed by some kind of joints or anything. It lies on top of the engines. There also may be some supporting columns beside the engines. On the other hand it needs to be restricred by the shell, so that it stays inside.&lt;/li&gt;&#xA;&lt;li&gt;There are multiple engines, spread inside the shell. Each one of them consists of the two major parts:&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The sphere that translates its movement into movement of the case above it. The sphere isn't a smooth ball. It's covered with numerous spikes all around it which precisely fit the sockets of the platform's case. So when the sphere rotates, it pulls the case in the same direction.&lt;/li&gt;&#xA;&lt;li&gt;The same spikes are used to rotate the sphere. There is a pair of conveyor belts(see Figure 2) that translate their movement into the sphere's. I suppose the easiest way is to make the spikes penetrate the belts. And this pair of conveyors can move around the sphere horizontally depending on the required direction but independently of the vertical movement of the belts itself.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPD&lt;/strong&gt;: The platform should be able to move in &lt;em&gt;any&lt;/em&gt; direction at &lt;em&gt;any&lt;/em&gt; moment (with reasonable latency, of course). And the vector is generated in real time by Neural Network watching the user of the treadmill. So the person just walks in random direction and the neural network instantly generates the vector, which is then translated into engines movement.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How far this concept from reality?&lt;/p&gt;&#xA;" OwnerUserId="20378" LastEditorUserId="20378" LastEditDate="2019-05-09T15:06:06.783" LastActivityDate="2019-05-09T15:06:06.783" Title="Why is there still no truly omnidirectional treadmill? (technical reasons)" Tags="|mechanical-engineering|mechanisms|mechanical|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28269" PostTypeId="2" ParentId="28261" CreationDate="2019-05-08T18:14:46.613" Score="1" Body="&lt;p&gt;Let's approximately calculate the S of composite beam, knowing that aluminum section I is 146in^4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ I_{COMP}= I_{alum}+ 2*A_{steel} *4.25^2 = 146+(8*0.25*2)4.25^2=146+72.25=218.25in^4  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The S of aluminum section is 35.5in^3 and the S of steel 72.25/4.5= 16.05in^3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At yield point the steel plates can take a moment of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ M_s= \sigma*S=18*16.05=288.9 k.in $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the aluminum beam has not fully loaded at this stage, it has strained the same as steel but stressed only 1/3 of  18ksi so it can take only 18/3=6ksi&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ M_a= 6*35.5=213k.in $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we compare the difference adding the steel plates made with the original aluminum section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ M_{alumalone}= 12*35.5=426k.in \\ which\ is\ less\ than \ 288.9+213=501.9$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After a comment by @Wassabi I corrected the stress of Aluminum and recalculated the beam. The concept is the same but the numerical answer is changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we gain roughly 20%  of additional strength. But from here on out the advantage of adding steel plates will increase rapidly by adding to the thickness of the plates.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-05-12T03:35:04.107" LastActivityDate="2019-05-12T03:35:04.107" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28270" PostTypeId="1" CreationDate="2019-05-08T18:15:15.963" Score="0" ViewCount="110" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/TiMsi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am having a hard time taking care of signs. Whether we’ve made the cut at right or at left. Whether should we make the rotation according to axis x, y or z.&#xA;In this case, they assumed that P was applied at the left of the axis of rotation and obtained M as negative. Why couldn’t we assumed that P is at right as well and with the calculations shown be positive?&lt;/p&gt;&#xA;" OwnerUserId="20374" LastActivityDate="2019-05-08T18:37:51.377" Title="Sign convention-moment" Tags="|moments|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28271" PostTypeId="2" ParentId="28270" CreationDate="2019-05-08T18:37:51.377" Score="1" Body="&lt;p&gt;Both moments tend to turn the plate in the same direction (counter-clockwise). So both should have the same sign. If you use convention used in the book, then both are negative: &lt;a href=&quot;https://classes.mst.edu/civeng2210/concepts/07/internal/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://classes.mst.edu/civeng2210/concepts/07/internal/index.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you use the screw rule both are positive: &lt;a href=&quot;http://www2.eng.cam.ac.uk/~hemh1/gyroscopes/screwrule.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www2.eng.cam.ac.uk/~hemh1/gyroscopes/screwrule.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So both must have the same sign and the result is then the same. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-05-08T18:37:51.377" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28272" PostTypeId="2" ParentId="28254" CreationDate="2019-05-08T18:48:51.293" Score="4" Body="&lt;p&gt;Do a web search on &quot;Rankine cycle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The real key is that in a practical heat engine, you need to pump in cold working fluid, heat it, let it do work (and cool) as it expands, and then repeat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you ignore losses, the energy to pump a working fluid up by a given pressure difference is proportional to the volume times that pressure difference.  A mole of liquid water is &lt;em&gt;much smaller&lt;/em&gt; than a mole of gas.  So a heat engine that pumps liquid water (or any other liquid that will subsequently be boiled) has a significant advantage over a heat engine that does the job with a gas.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-05-08T18:48:51.293" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28273" PostTypeId="2" ParentId="28252" CreationDate="2019-05-08T19:23:06.080" Score="0" Body="&lt;p&gt;Have you worked with Arduino at all? A single servo motor looped with an Arduino code could do that at a fraction of the weight. Have an input that determines current to the motors and when that input is zero, output a half-twist on the servo.&lt;/p&gt;&#xA;" OwnerUserId="20380" LastActivityDate="2019-05-08T19:23:06.080" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28274" PostTypeId="2" ParentId="28260" CreationDate="2019-05-08T19:39:13.490" Score="2" Body="&lt;p&gt;Step one: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choose a right handed coordinate system. As you can see in my schematic: &lt;a href=&quot;https://i.stack.imgur.com/7blPp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7blPp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt; axis lies along the beam: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/sZWsE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sZWsE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in &lt;span class=&quot;math-container&quot;&gt;$ZX$&lt;/span&gt; plane, the counterclockwise moment is positive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Horizontal and vertical equilibrium: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal equilibrium is evident so i skip it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertical equilibrium: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$R_E +R_C + -2 [\frac{kN}{m}]\times 2 \times \frac{1}{2}-3 [kN] = 0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R_C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_E$&lt;/span&gt; are the reaction forces,  i demonstrate those as they point upward but you can choose those pointing down. The third term represents the resultant of the distributed load, (the area under curve), notice its point of application.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment equilibrium:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I choose point &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; as the reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$1.5 [kN.m]+3[kN] \times 1[m]+2[kN] \times \frac{4}{3}[kN.m]+V_E \times 2 [m]=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can eventually find the value of &lt;span class=&quot;math-container&quot;&gt;$R_E$&lt;/span&gt; and substituting it in the first equation gives us the value of &lt;span class=&quot;math-container&quot;&gt;$R_C$&lt;/span&gt;. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-05-08T19:39:13.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28275" PostTypeId="2" ParentId="28254" CreationDate="2019-05-08T21:31:07.233" Score="15" Body="&lt;p&gt;Hot air engines are feasible, and have a 200 year history behind them, s&lt;a href=&quot;https://en.wikipedia.org/wiki/Stirling_engine&quot; rel=&quot;noreferrer&quot;&gt;tarting in 1816 with the Reverend Robert Stirling&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other answers are largely correct : steam offers more energy per unit volume, but theoretically lower achievable efficiency while the Stirling Cycle can (in theory) match the ideal efficiency of the Carnot Cycle, using a regenerator to store teh bulk of the heat removed in the cooling phase, and return it to the air in the heating phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there are practical difficulties transferring heat into and out of the working fluid fast enough, while minimising flow losses; good heat exchangers keep gas in close contact with the heat exchange surfaces, maximising heat exchange rates, but also maximising frictional losses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steam wins here; the bulk of the heat exchange is in the liquid phase, which is much more effective (compare water cooling with air cooling).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot air engines surface every few decades; &lt;a href=&quot;https://en.wikipedia.org/wiki/Stirling_engine#Twentieth_century_revival&quot; rel=&quot;noreferrer&quot;&gt;Philips did some impressive work&lt;/a&gt; achieving close to 40% thermal efficiency between the 1950s and 1970s, and this work lives on in some &lt;a href=&quot;https://en.wikipedia.org/wiki/Gotland-class_submarine&quot; rel=&quot;noreferrer&quot;&gt;Swedish submarines&lt;/a&gt; where its relative silence is an advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also make cute toys capable of running off m&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B00BXVBNBU&quot; rel=&quot;noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;odest heat sources - like the warmth of your hand.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="435" LastActivityDate="2019-05-08T21:31:07.233" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28276" PostTypeId="1" AcceptedAnswerId="28277" CreationDate="2019-05-08T21:49:45.923" Score="0" ViewCount="46" Body="&lt;p&gt;I am currently for the past few years creating a Brushless Motor on and off. The stator I have built has 8 Stators and I cannot seem to find out how to wire this. I see most Brushless Motors with 9 Stators. Is this possible and if so how would I be able to get this working? If you do know how to wire this can you please provide a small diagram on wiring this? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="20382" LastActivityDate="2019-05-08T23:01:36.390" Title="How to Wire a Brushless Motor with 8 stators?" Tags="|motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28277" PostTypeId="2" ParentId="28276" CreationDate="2019-05-08T23:01:36.390" Score="1" Body="&lt;p&gt;I believe that you mean you have built a motor with eight &lt;strong&gt;poles&lt;/strong&gt; on one stator.  The reason that brushless motors generally have some multiple of three poles on the stator is because they are three phase.  If you want to carry on with your 8-pole stator, you need to wind it as a 2- 4- or 8-phase motor -- and then you need to design and build a driver that can handle that.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-05-08T23:01:36.390" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28278" PostTypeId="1" AcceptedAnswerId="28286" CreationDate="2019-05-09T00:35:20.313" Score="2" ViewCount="344" Body="&lt;p&gt;I have a motor with a 4mm shaft. I've purchased a propeller that's designed to fit a 1/2 inch shaft. How can I securely couple the two?&lt;/p&gt;&#xA;" OwnerUserId="16660" LastEditorUserId="334" LastEditDate="2019-05-12T09:00:08.403" LastActivityDate="2019-05-12T09:00:08.403" Title="How to connect a motor with a 4mm shaft to a propeller that takes 1/2 inch shaft?" Tags="|mechanical-engineering|motors|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28280" PostTypeId="2" ParentId="28260" CreationDate="2019-05-09T01:36:49.257" Score="3" Body="&lt;p&gt;You have your signs wrong and the CG of distributed load is at 3.33m from E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming counter clockwise moment positive, and the reaction direction up positive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ M_E=0 \rightarrow \ 1.5 +2*3.33 + 3*3 -2 V_c =0 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ V_c=\frac{1.5+6.66+9}{2}= \frac{17.16}{2}=8.58kN $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I leave the rest to you.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-05-09T16:38:36.127" LastActivityDate="2019-05-09T16:38:36.127" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28281" PostTypeId="2" ParentId="28278" CreationDate="2019-05-09T02:43:33.653" Score="0" Body="&lt;p&gt;Use gears, mechanical losses will only be 1 % and that's negligible. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-05-09T02:43:33.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28283" PostTypeId="2" ParentId="28268" CreationDate="2019-05-09T04:30:01.253" Score="2" Body="&lt;p&gt;A disk shaped threadmill seems somewhat problematic from a technical perspective. Placement of actuators and moving of the surface material seems much more problematic in a disk than other shapes. Speed of actuators need to be nonuniform for one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tension on a disk seems like a unknown problem. On a disk the corners make up more of the area while in a square its uniform. Wether you can even roll such a surface at all without elevating the user quite much is a question i cant answer. In anycase a disk shaped structure needs to withstand much more two directional bending and nonuniform stretching.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But really many of the technical problems could be solved by adequate funding.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2019-05-09T04:35:22.163" LastActivityDate="2019-05-09T04:35:22.163" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28284" PostTypeId="2" ParentId="28254" CreationDate="2019-05-09T04:32:33.400" Score="10" Body="&lt;p&gt;There are really countless possible answers to this.  I'd like to add one that hasn't been made clear yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are doing a closed cycle, where you aren't letting steam/gas out into the atmosphere, at some point you need to pump the fluid back into the boiler to gain energy again.  Pumping gas is very difficult to do.  Pumping liquid water is &lt;em&gt;very&lt;/em&gt; easy.  So in the case of closed systems, it's terribly convenient that we can condense the steam back into water, and then move it mechanically back into the boiler to create new steam.&lt;/p&gt;&#xA;" OwnerUserId="678" LastActivityDate="2019-05-09T04:32:33.400" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28286" PostTypeId="2" ParentId="28278" CreationDate="2019-05-09T07:13:29.973" Score="4" Body="&lt;p&gt;If the motor speed matches the speed range needed for the prop, then make an adapter with relevant locking screws or equivalent, balanced as necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;something like :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q7eGj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q7eGj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source &lt;a href=&quot;https://shaft-collars-couplings.staffordmfg.com/viewitems/shaft-adapter-couplings/shaft-adapter-couplings---step-up-type-with-keyway&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://shaft-collars-couplings.staffordmfg.com/viewitems/shaft-adapter-couplings/shaft-adapter-couplings---step-up-type-with-keyway&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-05-09T12:24:09.270" LastActivityDate="2019-05-09T12:24:09.270" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28287" PostTypeId="1" AcceptedAnswerId="28295" CreationDate="2019-05-09T07:19:14.920" Score="-1" ViewCount="202" Body="&lt;p&gt;Is it possible to build a brushed DC motor with a ring magnet stator, instead of a stator with gaps as shown in the GIF below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0LzJx.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0LzJx.gif&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20392" LastEditorUserId="20392" LastEditDate="2019-05-09T10:04:10.787" LastActivityDate="2023-10-02T05:31:27.170" Title="Can two-pole brushed DC motors have a ring magnet stator?" Tags="|electromagnetism|" AnswerCount="2" CommentCount="18" ContentLicense="CC BY-SA 4.0" />
  <row Id="28290" PostTypeId="2" ParentId="28254" CreationDate="2019-05-09T12:08:52.240" Score="7" Body="&lt;p&gt;While toy steam engines often require that one pre-fill the boiler, and then only operate until the boiler is nearly empty, most practical steam engines require that every gram of steam which leaves the boiler be replaced by a gram of water during operation.  The amount of work required to pump a gram of cold air into an engine is slightly less than the amount of work that can be obtained from a gram of hot air that leaves, but not by a whole lot.  The amount of work required to pump in a grab of water, however, is far lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, if a steam engine includes a condenser, that can be used to reduce the pressure downstream of the pistons to below atmospheric.  This partial vacuum contributed substantially to the power output and efficiency of early steam engines, but becomes less significant as working pressures increase.  Steam locomotives don't bother with a condenser because getting high power outputs requires using higher working pressures, and the relative usefulness of the condenser for power generation goes down as working pressures go up.  The general lack of condensers is perhaps somewhat curious given that maintenance costs are strongly related to the total quantity of contaminants brought into an engine, and recirculating water would reduce the quantity of new water (and thus contaminants) it would be necessary to add.&lt;/p&gt;&#xA;" OwnerUserId="397" LastActivityDate="2019-05-09T12:08:52.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28291" PostTypeId="1" AcceptedAnswerId="28294" CreationDate="2019-05-09T13:57:21.200" Score="0" ViewCount="232" Body="&lt;p&gt;In an electric motor like this one, are the stator magnets radially magnetised, as in, each of the two magnets has opposite poles on the inside and outside diameter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Uvr4M.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uvr4M.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20392" LastActivityDate="2019-05-09T16:01:49.223" Title="Are stator magnets in brushed DC motors radially magnetised?" Tags="|motors|electromagnetism|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28292" PostTypeId="1" CreationDate="2019-05-09T15:03:12.610" Score="0" ViewCount="66" Body="&lt;p&gt;Support structures like lattice supports, tree supports or wire supports are being widely used in metal printing or FDM as they minimise the wastage of material and power consumption. In addition, they are easy to remove.&#xA;But compared to volume support, are these structures stable enough to stand firm against the build tray during the printing process?&#xA;How these structures avoid buckling depending on the height of supported zones?&lt;/p&gt;&#xA;" OwnerUserId="20401" LastActivityDate="2024-01-14T00:06:50.970" Title="Stability of lattice or mesh supports for printing" Tags="|mechanical-engineering|3d-printing|stability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28293" PostTypeId="2" ParentId="28278" CreationDate="2019-05-09T15:51:34.337" Score="4" Body="&lt;p&gt;Following on from Solar Mike's answer, I'm not sure you'll be able to find something off-the-shelf with the diameters you have (mix of imperial and metric units), so you may need to get something custom made instead. I have sketched the following (not knowing what shaft lengths you had, I made a few assumptions - adjust to suit your needs):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/63brf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/63brf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer diameter is 1/2 inch, while the ID is 4mm. I have toleranced it as H7 hole so that it's a snug fit on the 4mm shaft. The adapter is secured onto the 4mm shaft with 2 M3 grub screws at 90 degrees, you can also use Loctite 638 if you want to be sure. I have added a 1mm chamfer on the end of the shaft adapter.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-05-09T15:51:34.337" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28294" PostTypeId="2" ParentId="28291" CreationDate="2019-05-09T16:01:49.223" Score="1" Body="&lt;p&gt;You should be able to find a picture on the web, but -- yes.  One is north on the inside and south on the outside, the other is south on the inside and north on the outside (so pay attention if you have a pile of taken-apart motors that you're reassembling!).  The steel can of the motor helps to convey the flux between the outside poles of the magnets, which increases the motor's torque -- this is why some of the larger &quot;can&quot; motors have an extra ring around the outer case right where the magnets are located.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-05-09T16:01:49.223" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28295" PostTypeId="2" ParentId="28287" CreationDate="2019-05-09T16:11:07.430" Score="0" Body="&lt;p&gt;See my answer to &lt;a href=&quot;https://engineering.stackexchange.com/questions/28291/are-stator-magnets-in-brushed-dc-motors-radially-magnetised&quot;&gt;your other question&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it is possible, and since permanent magnet DC motors have been around for well over 100 years, it's undoubtedly been done.  There are many downsides:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It would be difficult to magnetize such a magnet to be as powerful as a pair of correctly-magnetized magnets&lt;/li&gt;&#xA;&lt;li&gt;Magnetic materials -- even steels -- are brittle, hard to machine, and won't have thermal expansion characteristics that match whatever you want to use for a case.&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;So getting one to match the size of the case would be difficult.&lt;/li&gt;&#xA;&lt;li&gt;And Once you got your magnet machined to match the case at room temperature, chances are low that it would &lt;em&gt;stay&lt;/em&gt; matched, because of differences in thermal expansion coefficients.  &lt;/li&gt;&#xA;&lt;li&gt;This, in turn, means that you can't press the magnet and case together -- you'd need to assemble them with glue or clips.&lt;/li&gt;&#xA;&lt;li&gt;And all of that means that except in very specific circumstances, a full-ring magnet would be &lt;em&gt;more&lt;/em&gt; expensive, &lt;em&gt;more&lt;/em&gt; trouble, &lt;em&gt;less&lt;/em&gt; reliable, and very likely lower performance than a &quot;traditional&quot; arrangement.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-05-09T16:11:07.430" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28296" PostTypeId="2" ParentId="28292" CreationDate="2019-05-09T16:35:51.727" Score="0" Body="&lt;p&gt;It is a fast evolving industry.&#xA;I have been looking into some printers and it seems at the present time the limit of the size is 30cm cube roughly, but it is changing every day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the shapes don't even need support, some printers can bridge spans of up to 25mm. Vertical angles less than 45degree don't need support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are new ways of printing that don't need any support at all, such as laser printing, however some those are not publicly available.&#xA;All hardware manufacturers have manuals and recommend what to do as for support and filler. considering the fact that filaments get cold and hard fast after printing, the structural strength of support is not of primary concern. Also digital model makers are trying to design the model with least need for lattice or any support and make it easy to release the support with least post processing.&#xA;Here is an article about the methods and manufacturers of both software and hardware.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://all3dp.com/1/3d-metal-3d-printer-metal-3d-printing/#metal-3d-printing-filament&quot; rel=&quot;nofollow noreferrer&quot;&gt;3d metal fuse deposition modeling&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-09T16:35:51.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28297" PostTypeId="1" AcceptedAnswerId="28299" CreationDate="2019-05-09T18:05:33.553" Score="2" ViewCount="2838" Body="&lt;p&gt;I've been going down the rabbit hole as far for pipe vs tube and have ran into the same statement from co-workers and on the internet. That statement is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;TUBE IS STRONGER THAN PIPE&quot;&#xA;&lt;a href=&quot;https://www.metalsupermarkets.com/difference-between-pipe-and-tube/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.metalsupermarkets.com/difference-between-pipe-and-tube/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which makes no sense to me, shouldn't how strong they are be a function of wall thickness and pipe ID? It seems like it's just comparing apples and oranges. Anybody got some insight on this ?&lt;/p&gt;&#xA;" OwnerUserId="14961" LastActivityDate="2023-01-24T18:34:41.863" Title="Tube Strength vs Pipe Strength" Tags="|materials|piping|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28299" PostTypeId="2" ParentId="28297" CreationDate="2019-05-09T18:14:03.080" Score="3" Body="&lt;p&gt;Firstly, I assume that when you say &quot;tube&quot; what you mean is Hollow Structural Sections, and when you say &quot;pipe&quot; I assume that you mean typical scheduled pipe sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the US, the AISC Code specifies material properties for structural shapes. From the manual Tables 2-4, we can see that the Pipe material is ASTM A53 Gr. B, and that for round HSS we have ASTM A500 Gr. B as the most common shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two materials are characterized by their yield and tensile stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For A53 Gr. B, Fy = 35 ksi and Fu = 60 ksi.&#xA;For A500 Gr. B, Fy = 42 ksi and Fu = 68 ksi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only is the pipe wall thickness and pipe ID/OD part of the strength of a given element, but also the Material Strength plays a large roll.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your intuition that the wall thickness and ID/OD effect the strength is correct. But if you have a pipe with identical dimensions as an HSS tube. The HSS(tube) will be stronger because it has higher Fy and higher Fu.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastEditorUserId="19912" LastEditDate="2019-05-09T18:52:17.627" LastActivityDate="2019-05-09T18:52:17.627" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28300" PostTypeId="1" AcceptedAnswerId="28301" CreationDate="2019-05-09T19:51:27.083" Score="1" ViewCount="147" Body="&lt;p&gt;I'm currently working on a project which involves a small fridge encased in a wooden box.  It's designed for hotels and the boxes vary slightly, but one common trait is that the box has an open vent towards the back which is meant for ventilation.  Unfortunately, the hotels like to jam the box up against a wall, which prevents the vents from working properly.  We have no control over how the wooden box is manufactured (in fact, the fridge was made to fit inside this box) and the vent hole varies in height from hotel to hotel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The point is, the hotels demand that we be able to cool down the temperature inside the box, without any added electricity.  I have already looked at Thermoelectric Couples and the temperature difference is about 18°C between the ambient interior of the box and the compressor which isn't enough to run a small fan.  I have attached some passive heat sinks to the compressor itself, which did reduce the ambient temperature, but increased the temperature inside the fridge.  I'm stuck, and I am thinking of using aluminum foil extending from the fridge to outside of the box to help conduct the heat away.  Can anyone suggest any other ideas?&lt;/p&gt;&#xA;" OwnerUserId="20406" LastEditorUserId="3445" LastEditDate="2019-05-10T19:34:18.707" LastActivityDate="2019-05-11T13:13:32.523" Title="Cooling Down Fridge Encased in wooden box" Tags="|thermodynamics|heat-transfer|cooling|temperature|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28301" PostTypeId="2" ParentId="28300" CreationDate="2019-05-09T20:47:10.880" Score="2" Body="&lt;h1&gt;Theoretical Refridgeration&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The coefficient of performance metric for the fridge is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$CoP_R = \dot{q}_c/\dot{w}_R = T_c/(T_h - T_c)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_c$&lt;/span&gt; is removed from the fridge with power &lt;span class=&quot;math-container&quot;&gt;$\dot{w}_R$&lt;/span&gt;. For an ideal case, this is defined by the temperature in the fridge &lt;span class=&quot;math-container&quot;&gt;$T_c$&lt;/span&gt; and the temperature at the external coils &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt;. You want to keep &lt;span class=&quot;math-container&quot;&gt;$T_c$&lt;/span&gt; as low as possible. You control &lt;span class=&quot;math-container&quot;&gt;$\dot{w}_R$&lt;/span&gt; through the electrical power to the fridge &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; and the (irreversible) efficiency of the motor &lt;span class=&quot;math-container&quot;&gt;$\epsilon_M$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$\dot{w}_R = \epsilon P$&lt;/span&gt;. The system defines &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_c$&lt;/span&gt; as the heat flow that enters the fridge from the outside. You must remove at least that to sustain the temperature &lt;span class=&quot;math-container&quot;&gt;$T_c$&lt;/span&gt;. The temperature &lt;span class=&quot;math-container&quot;&gt;$T_h$&lt;/span&gt; is a design setting for the compressor and fluid that are used in the cooling coils themselves. Finally, &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_c$&lt;/span&gt; is related to the heat that must be pulled from the coils &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_h$&lt;/span&gt; by the expression &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_c + \dot{w}_R = \dot{q}_h$&lt;/span&gt;. Heat transfer into the fridge and away from the coils are addressed as theoretical problems in heat transfer.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Theoretical Heat Transfer&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The heat flow into the fridge &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_i$&lt;/span&gt; has to be removed as &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_c$&lt;/span&gt; by the heat pump (cooling cycle). How can it be decreased? Consider it from heat transfer as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{q}_i = U A (T_a - T_c) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_transfer_coefficient&quot; rel=&quot;nofollow noreferrer&quot;&gt;overall heat transfer coefficient&lt;/a&gt; for the system, &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the essentially external area of the fridge, and &lt;span class=&quot;math-container&quot;&gt;$T_a$&lt;/span&gt; is the surrounding air temperature. We can expand &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; as a sum of resistors for external convection &lt;span class=&quot;math-container&quot;&gt;$R_a = 1/h_a A$&lt;/span&gt;, wall conduction &lt;span class=&quot;math-container&quot;&gt;$R_w = w/kA$&lt;/span&gt;, and internal convection &lt;span class=&quot;math-container&quot;&gt;$R_c = 1/h_cA$&lt;/span&gt;. In these expressions, &lt;span class=&quot;math-container&quot;&gt;$h_j$&lt;/span&gt; is the air or internal convection coefficient, &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; is the wall thickness, and &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is the thermal conduction of the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the back side, convection around the coil to the surrounding air pulls out the heat that is dumped to the air. This affects the temperature of the coil as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{q}_h = h_a A_c (T_h - T_a) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, &lt;span class=&quot;math-container&quot;&gt;$A_c$&lt;/span&gt; is the area of the coils.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Practical Insights&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Let us assume that you have no control on &lt;span class=&quot;math-container&quot;&gt;$T_a$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As two first steps to lower &lt;span class=&quot;math-container&quot;&gt;$T_c$&lt;/span&gt; with all else the same, you can improve the efficiency of your motor to increase &lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt; or you can increase the power to the motor to increase &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the heat transfer realm, you can make changes to decrease the heat transfer into the fridge to lower &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_i$&lt;/span&gt;. Ideally, you would want to DECREASE the convection around and inside the fridge to lower &lt;span class=&quot;math-container&quot;&gt;$h_a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$h_i$&lt;/span&gt;. This means that a fridge sitting in stagnate air is the best case. Alternatively, you can increase the thickness of the wall to the fridge &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; or decrease the thermal conductivity of the insulation in the walls &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The opposing case is to make changes that will increase &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_h$&lt;/span&gt;. Here, the immediate inverse recommendation arises to INCREASE &lt;span class=&quot;math-container&quot;&gt;$h_a$&lt;/span&gt; as convection around the coils or to increase the area of the coils &lt;span class=&quot;math-container&quot;&gt;$A_c$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Practical Suggestions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;One suggestion is to change the insulation in your fridge to a material with a higher &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; value. This will decrease &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; and lower the heat flow into the fridge from the outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another suggestions is to work on increasing the convection flow around the coils. Ask the hotels to have holes drilled in the bottom back of the box. This will improve the natural convective flow of air along the coils at the back of the fridge. The wider the holes or the opening is for air flow, the better. The best option is when the entire bottom of the box is a mesh that is mechanically strong enough to hold the weight of the fridge yet open enough to allow full air flow from below the fridge. This will only work when the box also includes a wide enough hole at the top to vent the hot air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Absent the option to change the box design, design a channel system attachment to the fridge. The channel should allow natural convection to pull cooler air from the front along the bottom, run it up the coils in the back, and vent it out the back of the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative answer is to redesign the fridge to include cooling fans to blow the air. The power could be pulled from a part of the power to the fridge itself. This will be a better cooling system than the passive one above. The downside is, this may generate more noise than a hotel will tolerate.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Other Thoughts&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;When the hotel has their specifications on cooling demand, you should set your specifications on ventilation to meet that cooling demand. Otherwise it seems that you put yourself in a box that you allow them to dictate unilaterally that your system must meet their criteria regardless of where they put it. I know of no heat transfer system that functions the same always independently of its surroundings.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-05-11T13:13:32.523" LastActivityDate="2019-05-11T13:13:32.523" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28302" PostTypeId="2" ParentId="25478" CreationDate="2019-05-09T20:50:32.273" Score="0" Body="&lt;p&gt;There are a lot of different types of strength to describe steel (Compressive, tensile, yield, ultimate, fatigue, hardness, toughness, etc). Surface finish affects some of these strength parameters, while some others are independent of surface finish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface finish generally cannot affect the intrinsic property of the underlying steel, which dictates strength parameters that are inherent to the steel material in question. That will include strength parameters such as compressive strength, tensile strength, yield strength, ultimate strength, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surface finish has significant impacts however to fatigue strength, since different surface finishes will generally have a large impact on how the S-N curve will look like for that particular piece of steel. Generally, smoother surfaces that has been machined or work hardened will have an improved fatigue strength over rougher surfaces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hardness, as indicated by @kamran, will generally be impacted by compressive surface finishing techniques applied, making it harder in the process. Stated in @kamran's answer as well, roughening will increase hardness since most roughening methods generally apply some compressive force to steel surfaces. I do want to note that other finishing operations can affect hardness indirectly as well. For example, while milling does not really change hardness at all of the underlying material, you can change the hardness of the surface that you are testing by milling, because hardness of steel do not necessarily stay consistent throughout the entire piece.&lt;/p&gt;&#xA;" OwnerUserId="13526" LastActivityDate="2019-05-09T20:50:32.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28303" PostTypeId="2" ParentId="28300" CreationDate="2019-05-09T20:51:15.197" Score="0" Body="&lt;p&gt;All the refrigerators have specific clearance behind the fridge radiators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many hotels have louvered doors for the cabinet enclosing the fridge, for the exact reasons you mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure wrapping the aluminum foil around the radiator will help.&#xA;It will impede the convection circulation around the fridge.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-09T20:51:15.197" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28304" PostTypeId="1" CreationDate="2019-05-09T21:14:56.990" Score="2" ViewCount="79" Body="&lt;p&gt;I have a mean time to failure (MTTF) recorded from 100 systems which is 6.5 years. The shape parameter of my component is given with k = 1.3 (based on a list the from literature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to sample (generate) further failure times based on a Weibull distribution. Can I use 6.5 years (=MTTF) as a scale parameter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to do this in python code, if anybody knows a solution for this.&lt;/p&gt;&#xA;" OwnerUserId="14138" LastEditorUserId="3445" LastEditDate="2019-05-10T09:16:19.043" LastActivityDate="2019-05-10T09:16:19.043" Title="How to sample Mean-Time-To Failure values from a Weibull Distribution with shape 1.5?" Tags="|mathematics|statistics|failure-analysis|maintenance|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28305" PostTypeId="2" ParentId="28297" CreationDate="2019-05-09T22:20:27.417" Score="3" Body="&lt;p&gt;The advertisement referenced in the question was written by a liberal arts major without much of a clue what he was writing about.  Pipe comes in certain diameters and wall thicknesses and threads defined by ANSI OR API ( in the US, code bodies in other countries usually follow these same sizes - such as CSA in Canada). Pipe and tube may be made of the same material so to say one is stronger is nonsense. Tubes may be square and other shapes. ASTM writes the specifications for pipe and tube ( in the US ). Although there are 2 separate committees ( 10 and 11 if I remember correctly)  about 80% of the people are on both committees. For clarification : API has specifications for pipe, casing, and tubing . API pipe is essentially the same sizes as ANSI pipe but is likely different steels. API casing and tubing is very specific to oil/gas wells and rarely used elsewhere.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-05-09T22:40:06.277" LastActivityDate="2019-05-09T22:40:06.277" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28306" PostTypeId="2" ParentId="11806" CreationDate="2019-05-09T23:09:06.490" Score="1" Body="&lt;p&gt;As a very crude analogy, it takes more air to blow a feather across a table than a pen. The friction force translates to the energy converted from the air pressure to motion. To move a more stubborn object, higher pressure is needed. The higher pressure means more air will leak around the object resulting in a higher volume of air needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the mechanism behind the turbine speed, the rotation of the turbine is what generates electric potential to drive the load. The interaction of the generator's magnets and coils effectively couple the turbine shaft to whatever is using the power. As more load machines connect to the grid (couple to the turbine), the turbine will become harder to turn. It thus takes more work (steam) to keep the turbine spinning the same speed. Power is torque times speed, so if you must maintain speed, the torque must vary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important that the turbine remain the same speed because that controls the output voltage and frequency, and the load machines are designed to operate in a specific window.&lt;/p&gt;&#xA;" OwnerUserId="20411" LastActivityDate="2019-05-09T23:09:06.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28308" PostTypeId="1" CreationDate="2019-05-10T00:27:42.457" Score="1" ViewCount="437" Body="&lt;p&gt;I am wondering about the magnetic field on the inside of hollow magnets. As an example, will use a ring magnet. In a diametrically magnetised ring magnet, does the magnet have the same pole on the inside and outside diameter, like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5nzL2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5nzL2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, does it have opposite poles, like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xmmok.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xmmok.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20392" LastEditorUserId="20392" LastEditDate="2019-05-10T01:39:35.220" LastActivityDate="2021-11-17T22:03:00.227" Title="Diametrically magnetised ring magnet, opposite poles on the inside?" Tags="|electromagnetism|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28309" PostTypeId="1" CreationDate="2019-05-10T01:31:53.667" Score="1" ViewCount="278" Body="&lt;p&gt;I'm trying to apply the associated flow rule to elastic-perfectly plastic material, &#xA;i.e.,&lt;span class=&quot;math-container&quot;&gt;$$d\epsilon_{ij}^p=\lambda\dfrac{\partial f}{\partial \sigma_{ij}}$$&lt;/span&gt;,&#xA;where &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; is a yield function with no hardening variable, &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is a plastic multiplier. I understand that &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; can be found through consistency condition (&lt;span class=&quot;math-container&quot;&gt;$df=0$&lt;/span&gt;), which gives &lt;span class=&quot;math-container&quot;&gt;$$\lambda=\dfrac{\dfrac{\partial f}{\partial \sigma_{ij}}C_{ijkl}d\epsilon_{kl}}{\dfrac{\partial f}{\partial \sigma_{ij}}C_{ijkl}\dfrac{\partial f}{\partial \sigma_{kl}}}$$&lt;/span&gt;&#xA;This seems confusing to me since after yielding is activated for perfectly plastic deformation &lt;span class=&quot;math-container&quot;&gt;$d\bf\sigma=0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$d\bf\epsilon^e=0$&lt;/span&gt; when &lt;span class=&quot;math-container&quot;&gt;$d\bf\epsilon^p&amp;gt;0$&lt;/span&gt;. Therefore &lt;span class=&quot;math-container&quot;&gt;$d\epsilon_{kl}=d\epsilon^p_{kl}$&lt;/span&gt;, substituting this into &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt;'s expression renders &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; indeterminate. &#xA;Maybe there's something I understand wrong, but if this is the case, how would the plastic deformation be predicted for such the material? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="20413" LastActivityDate="2019-05-10T01:31:53.667" Title="Plastic Multiplier for Perfectly Plastic Materials" Tags="|solid-mechanics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28310" PostTypeId="1" CreationDate="2019-05-10T02:26:35.517" Score="1" ViewCount="64" Body="&lt;p&gt;As I understand from reading about air conditioners is that the way they work is that air sucked in from the room is passed by the cold evaporator pipes and then the air is released again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is how does the air sucked in from the room change temperature so quickly since it passes very quickly by the cold evaporator pipes and released again? &lt;/p&gt;&#xA;" OwnerUserId="20414" LastActivityDate="2019-06-09T06:02:40.310" Title="In an air conditioner, how does the released air get cold so quick?" Tags="|mechanical-engineering|airflow|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28312" PostTypeId="2" ParentId="28254" CreationDate="2019-05-10T03:17:01.313" Score="4" Body="&lt;p&gt;As said, boiling water creates much more gas. To make a useful engine for the kind of applications that steam engines are used in, you want to build &lt;em&gt;lots&lt;/em&gt; of pressure to be able to move heavy stuff, e.g. a train for which the locomotive alone may be 10 Mg (megagrams, &quot;tonnes&quot;) or more in mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at the ideal gas law, written to emphasize pressure, i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P = \frac{nRT}{V}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can see there are three ways to increase the pressure: increase the temperature &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;, which is what you're suggesting, &lt;em&gt;de&lt;/em&gt; crease the volume &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt;, which you can't do here because it's a fixed-size chamber, &lt;em&gt;or&lt;/em&gt; you can increase the &lt;em&gt;number&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; of gas particles (molecules) (technically, bulk number in moles rather than direct number, for that, replace &lt;span class=&quot;math-container&quot;&gt;$nR$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$N k_B$&lt;/span&gt; above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Another, hypothetical way, as you can see, would be to somehow change &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; ... but that's called &quot;magic&quot; and &quot;changing the laws of physics&quot;. Sadly, we don't have that kind of power )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid water contains about 1000 times the &lt;em&gt;number&lt;/em&gt; of particles per unit volume than gaseous water does and - more importantly - air. And it generates all that gas simply by boiling it. Thus, with a volume of material that easily fits in the chamber, you can create from it up to 1000x more gas than a similar volume of air would give you and thus 1000x the pressure at any given temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the gas law certainly also works just as well for &lt;em&gt;air&lt;/em&gt;, too, and thus you could also, with air, increase &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; and thus also &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; in the same way. The problem with that is &lt;em&gt;what it takes to do that&lt;/em&gt;. Air is, under ordinary conditions, &lt;em&gt;already&lt;/em&gt; a gas: and thus to get more &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; into the chamber, you have to pump in &lt;em&gt;gas&lt;/em&gt; and increase the pressure. That means you are now &lt;em&gt;compressing&lt;/em&gt; air ... and congrats, you have just re-invented the compressed-air engine, only now with an added heater. Better to just burn all that fuel to run the compressor and compress the air from the get-go, mount the tank, and leave out the pyrotechnics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative would be to use air that is condensed into a liquid or solid form. You can do that - it's called liquid nitrogen (well, okay, that's 75% of air, not 100%). The problem with that is you have to expend energy in cooling it down, now, to make it liquid, since room temp - 295 K (or 300 K if you're in China or many other parts of the world) - is way above the boiling point of nitrogen, 77 K. Thus, it's much more efficient to just heat up and boil something that is already liquid at room temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now on that last note, water isn't the only option in &lt;em&gt;theory&lt;/em&gt; - another would be alcohol, but alcohol is flammable: you heat that stuff up in your engine and unless it were a completely inert atmosphere therein, it would catch on fire and burn &lt;em&gt;all at once&lt;/em&gt; and you'd have a bomb, not an engine. Moreover, even if there is an inert atmosphere, the alcohol may pyrolyze (decompose, i.e. the molecules fall apart) before it reaches a good working temperature, and while in theory you could actually say this is good because it produces more gas by breaking big gas molecules into little ones and hence rising &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; even further, decomposing the material consumes even more energy. Moreover, the breakdown products of this include gases that you cannot re-liquefy at room temperature and hence we're back to the compressed air engine. Even moreso, they include water (&lt;span class=&quot;math-container&quot;&gt;$\mathrm{H_2O}$&lt;/span&gt;) and thus we're &lt;em&gt;also&lt;/em&gt; back to the steam engine. Finally, the &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; of alcohol yielded per unit volume is smaller than water owing to the larger sizes of its molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So at the end of the day, you might as well tank it up with water. It is liquid at room temp, boils at a relatively low temp, has lots of molecules to give up, is &lt;em&gt;extremely&lt;/em&gt; common and cheap (esp. if you get it from the ocean and not use precious freshwater reserves since you don't need to drink any of this, moreover, distillation is trivial since effectively your engine &quot;distills&quot; it many times and thus doing a pre-run outside is trivial added expense), and anything else you might use will stand a strong likelihood of turning into it anyways, and insofar as it is concerned, it itself does not pyrolyze until around 3000 K or so, well above the melting point of iron, 1811 K and thus above the point where your engine turns into a lava puddle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is where it's at. In fact, I might hazard that it's the optimal material for this purpose: generally speaking, more complex molecules fall apart more easily (more complex &lt;em&gt;things&lt;/em&gt;, &lt;em&gt;break&lt;/em&gt; more easily in general, as a universal principle), and of simple molecules (e.g. 3 atoms or so) all are either gases at room temp and/or dangerous reactive substances (e.g. hydrogen chloride).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you were building an engine to operate on an extraterrestrial body like Titan or even Pluto, you'd have the option of using solid or liquid nitrogen or methane/ethane as it's cold enough these are now available &quot;for free&quot; like how water is, without any special chilling steps. While the problem of decomposition temp for the latter still technically remains, you have to remember engines, and the gas laws with regard to temperature, operate on &lt;em&gt;ratios&lt;/em&gt; of expansion and with liquid methane being boiled you're starting at 111 K instead of 373 K, thus going from just over 111 K to 333 K, still below the decomp. point of methane, gives you roughly the same expansion as going from 373 K to 1119 K, as may be done in a real steam engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is, now you're going to be more worried about fuel/oxidizer combos, and you'll probably thus want to just use them more directly in an internal combustion-like setup. The whole point of steam engines was that you could use felled wood and/or coal, together with &lt;em&gt;atmospheric&lt;/em&gt; oxidizer. Nothing like that on these worlds.&lt;/p&gt;&#xA;" OwnerUserId="20415" LastEditorUserId="20415" LastEditDate="2019-05-10T03:37:59.870" LastActivityDate="2019-05-10T03:37:59.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28313" PostTypeId="2" ParentId="28310" CreationDate="2019-05-10T03:56:38.480" Score="1" Body="&lt;p&gt;The air gets cold because the cold evaporator is designed to have a large surace area and be good at transferring energy from the air to the refrigerant quickly.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-05-10T03:56:38.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28314" PostTypeId="1" CreationDate="2019-05-10T04:31:42.987" Score="1" ViewCount="50" Body="&lt;p&gt;I have a very specific and weird task I need to accomplish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a concrete wall that I need to get samples of the concrete dust to grind and get them tested, i plan on using a plunge router with a vacuum collar type setup with a grinding head of some sort ... problems are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;needs to be in-situ ( i can't do a standard core then chop it up )&lt;/li&gt;&#xA;&lt;li&gt;I can only find one grinding head that appears to do what i need &lt;a href=&quot;https://i.stack.imgur.com/52mVo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/52mVo.png&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;needs to be about 50-70mm diameter sample area and not miss any of each &quot;layer&quot;&lt;/li&gt;&#xA;&lt;li&gt;i need to do this about 50 times and i'm not sure a vacuum bezel head will last long?&lt;/li&gt;&#xA;&lt;li&gt;i need to drill in 0.5mm increments each sample of the dust&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;i thought id check and see if anyone else has an idea of an easier way I could accomplish this! Surely i'm not the only one who has encountered something like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="20417" LastActivityDate="2019-05-10T04:31:42.987" Title="How to grind into concrete ( not core drill )" Tags="|mechanical-engineering|control-engineering|concrete|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28315" PostTypeId="2" ParentId="28310" CreationDate="2019-05-10T04:43:13.507" Score="1" Body="&lt;p&gt;The cold refrigerant is circulated into winding tubes passing through two partitions with hundreds of tightly spaced thin vanes to absorb the heat of the air coming from the room very quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numerous blades create Maxim contact for the warm room air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, after the room air has given much of its heat to the evaporative coil it gets recycled back to the room. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optimally the temperature of the air returning to the room should be around 5-degrees Celsius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the phote the insulated copper line brings the cold refrigerant from the condenser and cycles is through the thin aluminum pipes. The Evaporative coil is installed in the air-conditioning ducting in a position that all the room air is pumped through it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o8suq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o8suq.jpg&quot; alt=&quot;evaparaative coil.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-10T04:43:13.507" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28317" PostTypeId="1" CreationDate="2019-05-10T06:09:15.280" Score="0" ViewCount="135" Body="&lt;p&gt;If a nozzle is projected upwards with water flowing with a certain flow rate&#xA;then in how many ways can you find the diameter of the jet of water at a certain &#xA;height from the nozzle exit?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One approach- We can find velocity at that height and we know the discharge so we can find diameter from here&#xA;Do you have some other approach&lt;/p&gt;&#xA;" OwnerUserId="20389" LastEditorUserId="20389" LastEditDate="2019-05-10T06:56:16.273" LastActivityDate="2019-05-14T09:32:37.010" Title="Fluid is flowing from nozzle in upward direction" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28318" PostTypeId="2" ParentId="28297" CreationDate="2019-05-10T07:00:41.767" Score="0" Body="&lt;p&gt;The toughness of a material only has to do with some of its properties.If the tube and the pipe are made from the same material and then they would have the same toughness.&lt;/p&gt;&#xA;" OwnerUserId="20420" LastActivityDate="2019-05-10T07:00:41.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28320" PostTypeId="1" CreationDate="2019-05-10T09:20:15.933" Score="0" ViewCount="27" Body="&lt;p&gt;I' m trying to understand attacks on Quantum key distribution protocols. In the COW QKD protocol, pulses from Alice are passed through an asymmetric&#xA;coupler and a fraction 'T' of the photons are sent to generate key, and the remaining fraction '1 − T' is sent into an interferometer to monitor coherence of consecutive pulses. In one of the &lt;a href=&quot;https://arxiv.org/abs/quant-ph/0609090&quot; rel=&quot;nofollow noreferrer&quot;&gt;references&lt;/a&gt; &quot;Zero-Error Attacks and Detection Statistics in the Coherent One-Way Protocol for Quantum Cryptography&quot;by Gisin et al. I came across the sentence: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If any two non-empty pulses are coherent and all pulses are equidistant, any attempt of measuring the number of photons on a finite number of pulses breaks some coherence and contributes to errors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;which I'm unable to understand. &lt;/p&gt;&#xA;" OwnerUserId="20422" LastEditorUserId="5724" LastEditDate="2019-12-13T18:15:48.997" LastActivityDate="2019-12-13T18:15:48.997" Title="How would measuring the number of photons on a finite number of pulses break coherence of a weak coherent laser source in COW QKD protocol" Tags="|lasers|photonics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28321" PostTypeId="2" ParentId="12953" CreationDate="2019-05-10T11:58:39.590" Score="4" Body="&lt;p&gt;Sailboat 2-speed winches do exactly this. Winding the handle clockwise directly engages the drum via one ratchet, and when you wind it anti-clockwise a different ratchet engages an epicyclic gear. Both directions move the drum in the same direction, and both ratchets prevent the drum from moving backwards. The key feature is two ratchets in opposing directions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another directional mechanism is Bendix gears, which slide axially along a coarse spiral shaft until they engage or disengage the output gear. The direction of the torque controls which way they slide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another is a one-way bearing or &lt;a href=&quot;https://en.wikipedia.org/wiki/Sprag_clutch&quot; rel=&quot;nofollow noreferrer&quot;&gt;sprag clutch&lt;/a&gt;; a roller bearing where one (usually the inner) race is not circular but a series of ramps. In one direction the rollers jam, in the other they are free to rotate. It acts like a ratchet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;KalleMP mentions that pull-back toy cars have a intermediate gear that switches engagement from one position to another with different ratios, one for winding up the spring fast and the other for unwinding it slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine that there are hydraulic solutions too.&lt;/p&gt;&#xA;" OwnerUserId="20423" LastEditorUserId="20423" LastEditDate="2019-05-11T22:09:15.863" LastActivityDate="2019-05-11T22:09:15.863" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28322" PostTypeId="1" CreationDate="2019-05-10T12:34:59.573" Score="2" ViewCount="523" Body="&lt;p&gt;Would it be possible to install a dead-mans brake on a wheelchair? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been calling around and wheelchairs do not seem to come with a dead-mans brake. With that i mean the brake found on airport trolleys: you let go the trolley stops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem that needs to be solved is that we are often pushing a grown man on a steep hill. If the person pushing the wheelchair were to stumble and release the chair, it would probably end in death as the road is not only steep, it is also full of bends and has irresponsible, speeding cars and buses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to install one made from off-the-shelf parts? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen this: &lt;a href=&quot;https://casterconnection.com/dead-man-brake-system.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://casterconnection.com/dead-man-brake-system.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is part of a set of small wheels, i'm not sure if that would be the right solution. If i replace the front wheels with this braking pair, would the chair not flip over? &lt;/p&gt;&#xA;" OwnerUserId="20426" LastActivityDate="2022-03-11T05:06:35.973" Title="Dead mans brake on a hand-pushed wheelchair" Tags="|design|friction|wheels|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28323" PostTypeId="2" ParentId="28322" CreationDate="2019-05-10T12:43:37.093" Score="0" Body="&lt;p&gt;If you have the brake on the front wheels then even if they lock - it will probably slide backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a dead-man (dead-person for the politically correct...) brake, then if it comes on suddenly - which is what they tend to do, then the chair will probably go over backwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would need to make sure the centre of gravity with the person in the chair is sufficiently far forward of the rear axle to prevent this tipping, BUT this may be more of a disadvantage as the person in the chair may balance on the rear wheels to avoid obstacles in tight spaces...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interesting issue and I have pushed a friend many many miles in a wheelchair... His party trick is to balance on the rear wheels only...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-05-10T12:43:37.093" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28324" PostTypeId="2" ParentId="28322" CreationDate="2019-05-10T15:04:57.797" Score="2" Body="&lt;p&gt;You could have a spring loaded brake acting on the rear wheels, released only when the push handles are pushed, and bent down, like many airport carts. But in this case with light cables transferring the controls from the brakes to the push handles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One would need to attach a pair of balance legs to stabilize the chair from flipping back, to the back of the chair like this photo. This chair already has attendant breaks with lock, but they engage when pulled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RWjc4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RWjc4.jpg&quot; alt=&quot;safety back legs.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-10T15:04:57.797" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28325" PostTypeId="1" CreationDate="2019-05-10T16:43:35.253" Score="0" ViewCount="361" Body="&lt;p&gt;It seems based on the magnetic field of ring magnets like the  magnetic field lines of a bar magnet actually runs from N to S also &lt;em&gt;inside&lt;/em&gt; the magnet, yet schematic images tend to show the field lines running from N to S outside the magnet, then through south pole up to north pole inside magnet. That seems wrong, like it contradicts the behaviour of for example this axially magnetised ring magnet --&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RnVYR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RnVYR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20392" LastActivityDate="2023-10-02T06:39:12.970" Title="Are the magnetic field lines flowing from north to south _inside_ a bar magnet as well?" Tags="|electromagnetism|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28326" PostTypeId="1" AcceptedAnswerId="28328" CreationDate="2019-05-10T17:42:50.200" Score="0" ViewCount="110" Body="&lt;p&gt;I am wondering if a pneumatic-based, paddle wheel-type lifting apparatus would have any practical application within the aeronautical or aerospace fields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please reference the conceptual drawing below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is showing radial bladed impellers ('paddle wheels') embedded halfway within an enclosure. The impellers are rotated by electric motors (although they can be any kind of motor). This enclosure has lower ambient pressure within it due to the suction of the intake pipe of the attached centrifugal blower. The centrifugal blower is being rotated by its own electric motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the impeller blades that are outside of the enclosure encounter higher air resistance/drag as compared to the blades within the enclosure, the paddle wheels should generate a net force in the direction indicated by the arrows on the impellers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I do not believe that this lifting apparatus would be a replacement for aircraft propellers, I believe that it may have a practical application somewhere within the aeronautical/aerospace field, yet at this time I don't know what that would be. Perhaps one application would be that this lifting apparatus could be used on a STOL/VTOL aircraft or a drone to provide vertical lift/hovering capabilities. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this pneumatic-based, paddle wheel-type lifting apparatus have any practical application within the aeronautical or aerospace fields?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited the Title text and made wording edits to the Body text. Also, revised the original drawing and reposted it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7VYF5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7VYF5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user18610" LastEditorDisplayName="user18610" LastEditDate="2019-05-13T12:36:06.060" LastActivityDate="2019-05-13T12:36:06.060" Title="Pneumatic-based, paddle wheel-type lifting apparatus for aircrafts" Tags="|mechanical-engineering|fluid-mechanics|design|aerospace-engineering|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="28328" PostTypeId="2" ParentId="28326" CreationDate="2019-05-10T21:19:16.997" Score="3" Body="&lt;p&gt;Simply put it will not work. And the reason is very simple. It's the fact that lift is created when the wind is almost tangent to the wing or the blade of a windmill, or else it will stall.&#xA;Not when it is pushing it. This is what made the Wright brothers machine fly as opposed to those who made flapping wings and crashed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many people who start learning sailing make the same mistake. Best wind is when it is blowing almost parallel to the sail, not when is blowing into the sail and stagnates. Best you get from 10 miles back wind is probably 7 miles boat speed, because of the friction. But same wind when it blows 90 degrees will give the sail lift at 50 miles speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Impeller has to let the air wash around it to give it lift, not to shovel the air.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-10T21:19:16.997" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28329" PostTypeId="2" ParentId="28325" CreationDate="2019-05-10T22:15:16.097" Score="0" Body="&lt;p&gt;The field lines run south to north.  The magnet lends magneto-motive force, which drives the field to increase as it goes through the magnet.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-05-10T22:15:16.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28330" PostTypeId="1" AcceptedAnswerId="28332" CreationDate="2019-05-10T22:30:23.227" Score="1" ViewCount="53" Body="&lt;p&gt;I am an undergraduate researcher and I am working at enhancing a calibrating system for a hydrographic surveying system that my supervisor uses on his hydrography course. The rope is used to lower a bar check system, and it comes out of a coil. I want to find a reliable way to measure how much rope has come out of the coil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently the rope has markings on it at every meter so that the person deploying it can know how much rope has been used by counting the number of markings that passed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering if there is a more reliable way to measure how much rope has come out of the coil, preferable by some sort of measuring device. The most obvious idea would be some sort of pulley system that records the number of rotations of the pulley, but that is prone to slipping and the error could add up over 30 something meters. Do you know of such a device or idea that could help with that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(p.s.: this is a very minor aspect of the whole design, and doesn't affect at all my work, it would just be a &quot;nice to have&quot; thing in the design)&lt;/p&gt;&#xA;" OwnerUserId="11240" LastEditorUserId="11240" LastEditDate="2019-05-10T22:37:29.633" LastActivityDate="2019-05-10T23:46:45.887" Title="Whats a reliable way to measure the amount of rope going out of a coil?" Tags="|mechanical-engineering|measurements|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28331" PostTypeId="1" CreationDate="2019-05-10T23:27:29.567" Score="1" ViewCount="31" Body="&lt;p&gt;Ok so I am a machinist, not an engineer, but figured someone here might be able to provide some insight into my situation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part is 2.75&quot; diameter and is on a 3.3* angle in the X axis on my mill. I have milled the surface flat at 3.3*, and now I need to indicate center to drill a bolt circle, accurately. tolerance is +-.005&quot;&lt;/p&gt;&#xA;" OwnerUserId="20434" LastActivityDate="2019-05-11T01:16:20.413" Title="Bolt circle indication on angled ellipse" Tags="|measurements|manufacturing-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28332" PostTypeId="2" ParentId="28330" CreationDate="2019-05-10T23:46:45.887" Score="2" Body="&lt;p&gt;Depends on how much accuracy and at what cost you're looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several manufacturers which make devices like what you are asking.&#xA;Here is one.  &lt;a href=&quot;https://www.safewareinc.com/itemdetail/CMC%20267030?ads_cmpid=1774051606&amp;amp;ads_adid=75849742504&amp;amp;ads_matchtype=&amp;amp;ads_network=g&amp;amp;ads_creative=342995678301&amp;amp;utm_term=&amp;amp;ads_targetid=pla-737801560713&amp;amp;utm_campaign=&amp;amp;utm_source=adwords&amp;amp;utm_medium=ppc&amp;amp;ttv=2&amp;amp;gclid=CjwKCAjwwtTmBRBqEiwA-b6c_zisxVy5lBK-HOOeMktWypeKYksA5AintRAr3syNy4q8DthBPzd8JhoC3-gQAvD_BwE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cord meter&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you can use a basic computer optical mouse and an LED counter attached to an Arduino sensor for around $20-30. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-10T23:46:45.887" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28333" PostTypeId="2" ParentId="28331" CreationDate="2019-05-11T01:16:20.413" Score="1" Body="&lt;p&gt;The center of an ellipse inscribed in a rectangle is the center of the rectangle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above property, you can randomly set two perpendicular axes by marking the middle of any line using precision parallels and the intersection of those two axis will be the center of the ellipse and the center of your bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assumed your part is a cylinder; correct me if not!&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-11T01:16:20.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28334" PostTypeId="2" ParentId="26920" CreationDate="2019-05-11T01:43:22.400" Score="1" Body="&lt;p&gt;&lt;strong&gt;Clothoid?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recall hearing that some roadway curves are designed using the Clothoid Spiral AKA the Spiral of Cornu AKA &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler_spiral&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Euler_spiral&lt;/a&gt; &#xA;in which the (radius of) curvature varies linearly with arclength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Caveat: this is from ages ago (AKA grad school); YMMV.&lt;/p&gt;&#xA;" OwnerUserId="13014" LastActivityDate="2019-05-11T01:43:22.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28335" PostTypeId="1" CreationDate="2019-05-11T08:55:06.897" Score="0" ViewCount="27" Body="&lt;p&gt;I am working with a Physical Vapoor deposition (PVD) machine. I have 3*3cm Transistor Samples that I need to attach to the Sample holder of the PVD. I want to fabricate a &quot;mesh&quot; that will hold my sample holder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sample holder is a metal plate with holes for screws. The structure needs to be made of metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a picture I drew in pain of what I have in mind, with the screw holes in the corners.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lPzUa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lPzUa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't seem to be that hard but I have no idea how to start or where to look. How do I go about doing this? are there companies that can do this ? &lt;/p&gt;&#xA;" OwnerUserId="20437" LastActivityDate="2019-05-11T08:55:06.897" Title="how to fabricate an aluminum square mesh?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28336" PostTypeId="1" CreationDate="2019-05-11T16:37:12.967" Score="0" ViewCount="51" Body="&lt;p&gt;I purchased a &lt;a href=&quot;https://newenglandarbors.com/products/12-x-16-barcelona/&quot; rel=&quot;nofollow noreferrer&quot;&gt;pergola kit&lt;/a&gt; and realized afterwords I'm not happy with the placement of the center post.  If possible I'd like to eliminate it.  After searching for a while I haven't been able to find any information regarding span lengths for such a small load.  The inside dimensions of the sleeve surrounding the  horizontal 16' span is 1 3/4 x 5 7/8.  Can anyone help point me in the right direction or recommend a solution?  &lt;/p&gt;&#xA;" OwnerUserId="20443" LastActivityDate="2023-03-11T11:35:52.127" Title="Pergola Span and Support Modification" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28337" PostTypeId="2" ParentId="28336" CreationDate="2019-05-11T17:22:42.153" Score="1" Body="&lt;p&gt;You can remove the two 8ft beams on front and replace them with a 3-1/8 by 6 inch Glulam by 16ft long.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is close enough in size after paint and strong enough. if your posts are vinyl you need to change them to Doug Fir 4x4 posts.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-11T17:22:42.153" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28339" PostTypeId="1" CreationDate="2019-05-11T19:05:20.977" Score="1" ViewCount="467" Body="&lt;p&gt;I am studying for the FE exam and have come across a question regarding calculating cancer risk.  I'm fairly certain that the answer provided has typos, but I'm not 100% sure.  Here is the question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A carcinogen is in drinking water at a concentration of 0.01 mg/L.  The cancer risk of 30 years of adult (75 kg) exposure, given a cancer slope factor of 0.80 [mg/(kg*day)]^-1, is most nearly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the chronic daily intake equation is CDI = [(Concentration)(Ingestion Rate)(Exposure Frequency)(Exposure Duration)]/[(Body Weight)(Averaging Time)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are my issues with solution provided:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The ingestion rate is 2.3 L/day.  That wasn't given in the question, and the standard ingestion rate anyways given by the EPA is 2 L/day.  How did they come up with this number?  Is it likely just a typo?&lt;/li&gt;&#xA;&lt;li&gt;For &quot;Averaging Time&quot;, they have multiplied 75*365...shouldn't it be 30*365 since the question states that it is 30 years of exposure??&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19369" LastActivityDate="2023-11-13T07:05:25.623" Title="Calculating Chronic Daily Intake" Tags="|environmental-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28341" PostTypeId="2" ParentId="28261" CreationDate="2019-05-12T01:20:51.390" Score="0" Body="&lt;p&gt;The big problem here is that the beam and plates are of different materials. More specifically, that they have different moduli of elasticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are accustomed to seeing the following bending stress profile:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eqUO8m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eqUO8m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This obviously doesn't work for composite sections, though. Since the different parts are meant to work monolithically, the different materials must have the same elongation at the interface. However, since the materials have different elastic moduli (and that &lt;span class=&quot;math-container&quot;&gt;$\sigma = E\epsilon$&lt;/span&gt;), equal elongations mean different stresses at the interface. So we know there'll be a stress discontinuity at the interface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the standard bending elongation equation (trivially obtained from the more common bending stress equation)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\epsilon = \dfrac{My}{EI}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we also know that the slopes of the composite-beam elongation and stress diagrams will be different in each material since they have different elastic moduli.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, we can't yet determine what the slopes will actually be. It may seem clear that they'll just be &lt;span class=&quot;math-container&quot;&gt;$\dfrac{M}{EI}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\dfrac{M}{I}$&lt;/span&gt;, but that won't work: using those equations with each part's &lt;span class=&quot;math-container&quot;&gt;$EI$&lt;/span&gt; will find the results if it were working alone (well, if the beam were alone or if the plates were isolated from the beam). Instead, we'd need to find how much moment &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; each part will absorb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Personally, I find that a hassle, though. Why not replace the steel plates with equivalent aluminum plates, finding the new &quot;all-aluminum beam&quot;'s moment of inertia and do all the math assuming a single elastic modulus? That's way easier.&lt;sup&gt;1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's do that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://engr.bd.psu.edu/rxm61/MET210/CourseSupplements/Steel%20W-Shape%20Section%20Data.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;steel shape tables&lt;/a&gt;, we know that the moment of inertia of a W8x40 beam is equal to 146&amp;nbsp;in&lt;sup&gt;4&lt;/sup&gt;. We now need to calculate the change in moment of inertia due to the steel plates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since the beam and plates have different materials, we also need to take their different elastic moduli into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be done by considering the fact that bending moment doesn't actually care about moment of inertia &lt;span class=&quot;math-container&quot;&gt;$(I)$&lt;/span&gt;, nor about elastic modulus &lt;span class=&quot;math-container&quot;&gt;$(E)$&lt;/span&gt;. Bending moment cares about stiffness (&lt;span class=&quot;math-container&quot;&gt;$EI$&lt;/span&gt;). If two beams have the same &lt;span class=&quot;math-container&quot;&gt;$EI$&lt;/span&gt;, they'll behave the same under bending (assuming elastic behavior), even if one is tiny but made of hard material and the other is large but made of plastic. Same &lt;span class=&quot;math-container&quot;&gt;$EI$&lt;/span&gt; means same behavior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use this knowledge to our advantage in this case and transform the steel plates into &quot;equivalent aluminum&quot; ones. We do this by setting the steel plates' elastic modulus as equal to aluminum's and then adjusting the plates' moment of inertia so that their final &lt;span class=&quot;math-container&quot;&gt;$EI$&lt;/span&gt; is unchanged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question then becomes how to change the plate's geometry (and therefore their moment of inertia) to satisfy that condition. Thankfully, we know that moment of inertia is directly proportional to width and to the cube of height. Obviously, the easiest dimension to change is therefore the width: if you double the width, you double the moment of inertia; changing the height would require taking cubed-roots, and ain't nobody got time for that.&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the question tells us to assume that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dfrac{E_{steel}}{E_{alum}} = 3$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means that the steel plates are three times stiffer than similarly-sized aluminum plates. So, for aluminum plates to be as stiff as the steel ones, they'd have to be three times wider. After all&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;(EI)_{plate} &amp;amp;= E_{steel}\cdot I_{plate\ on\ beam} \\&#xA;&amp;amp;= E_{steel}\cdot (I_{plate} + A_{plate}d^2) \\&#xA;&amp;amp;= 3E_{alum}\cdot \left(\dfrac{bh^3}{12} + bhd^2\right) \\&#xA;&amp;amp;= E_{alum} \cdot 3\left(\dfrac{bh^3}{12} + bhd^2\right) \\&#xA;&amp;amp;= E_{alum} \cdot \left(\dfrac{(3b)h^3}{12} + (3b)hd^2\right)&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we replace our 8x0.25&quot; steel plates with 24x0.25&quot; aluminum ones, the beam's behavior will be identical (we'll get to the difference in allowable bending stress further ahead).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now let's calculate this &quot;equivalent beam&quot;'s moment of inertia. Using the &lt;a href=&quot;https://en.wikipedia.org/wiki/Parallel_axis_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;parallel axis theorem&lt;/a&gt; and being thankful that the symmetric reinforcement won't change the neutral axis, we know it'll be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;I_{equiv\ beam} &amp;amp;= I_{beam} + 2\left(I_{equiv\ plate} + A_{equiv\ plate}\cdot d^2\right) \\&#xA;&amp;amp;= I_{beam} + 2\left(\dfrac{(3b)h^3}{12} + (3b)h\cdot d^2\right) \\&#xA;&amp;amp;= 146 + 2\left(\dfrac{24\cdot0.25^3}{12} + 24\cdot0.25\cdot\left(\dfrac{8.25 + 0.25}{2}\right)^2\right) \\&#xA;&amp;amp;= 362.8\text{ in}^4&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can use the aluminum's elastic modulus with this equivalent moment of inertia to get the bending stress throughout the beam with the standard equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma = \dfrac{My}{I_{equiv}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there's a hitch: the stress on the plates is actually three times what we calculate with that equation. After all, we're simulating the plates with an area three times greater than it actually is. The actual area of steel will need to absorb that same internal force but with a smaller area and therefore greater stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this moment of inertia, the allowable stress in the plates is therefore just &lt;span class=&quot;math-container&quot;&gt;$f_{y,steel}/3 = 18/3 = 6\text{ ksi}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The allowable bending moment of the reinforced beam is therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M = \dfrac{\sigma\cdot I}{y} = \dfrac{6\cdot 362.8}{8.25/2+0.25} = 497.6\text{ k.in}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original, unreinforced beam, however, could resist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M = \dfrac{\sigma\cdot I}{y} = \dfrac{12\cdot 146}{8.25/2} = 424.7\text{ k.in}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reinforcement therefore increased the beam's bearing capacity by ~17%.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;1&lt;/sup&gt; All I wrote might make it seem complicated, but this method really is simple. It's basically: (1) multiply the plate's width by &lt;span class=&quot;math-container&quot;&gt;$E_{plate}/E_{beam}$&lt;/span&gt;; (2) calculate the equivalent beam's moment of inertia; (3) divide the plate's allowable bending stress by &lt;span class=&quot;math-container&quot;&gt;$E_{plate}/E_{beam}$&lt;/span&gt;; (4) calculate the allowable bending moment. It's effectively the same as the traditional method under the hood, but this one just seems more intuitive for me.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;2&lt;/sup&gt; Changing the height would also complicate the stress calculation later on, since it depends on the distance to the fiber. Changing the width doesn't affect this at all.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-05-12T01:27:44.677" LastActivityDate="2019-05-12T01:27:44.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28342" PostTypeId="1" CreationDate="2019-05-12T02:07:25.673" Score="1" ViewCount="25" Body="&lt;p&gt;I am wondering at a high level what are the &lt;em&gt;material components&lt;/em&gt; that might be used in building several large scale fish tanks that are &lt;em&gt;temperature regulated&lt;/em&gt; like on a fishing farm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, what I am imagining is something like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; __________________________________________________&#xA;|                |                |                |&#xA;|     Tank 1     |    Tank 2      |       3        |&#xA;|                |                |                |&#xA;|________________|________________|________________|&#xA;|                |                |                |&#xA;|                |                |                |&#xA;|       4        |       5        |      6         |&#xA;|                |                |                |&#xA;|________________|________________|________________|&#xA;|                |                |                |&#xA;|                |                |                |&#xA;|       7        |       8        |      ...       |&#xA;|                |                |                |&#xA;|                |                |                |&#xA;|________________|________________|________________|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where each tank is let's say 10 meters by 10 meters. There could be potentially a hundred or more of these tanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for the many tanks are twofold:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;To separate certain species of fish from one another.&lt;/li&gt;&#xA;&lt;li&gt;To create different environments (salinity, nutrients, and most relevant, &lt;strong&gt;temperature&lt;/strong&gt;).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The question is &lt;em&gt;how you would keep these tanks at a certain temperature&lt;/em&gt;. Perhaps, let's say, they are indoor. Or if it can be answered both ways, then including outdoor would be interesting to know but I imagine there are too many additional constraints in that case. So let's say they are indoor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am not an engineer, I do software development, but I am  interested in learning how to build real stuff too.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine that these are built with concrete, then somehow perhaps the concrete is heated or cooled with some technology. Or perhaps instead of regulating the temperature of the concrete, the water is cooled somehow directly. I would like to know how this cooling process would work, so I can understand how something like this could be built. Right now I can imagine the structural components for the most part (reinforced concrete, for example, and digging the dirt, etc.), but I don't know how the actual &lt;em&gt;temperature regulation&lt;/em&gt; would work at this scale. Also would be interested to know (if there are multiple ways), what the best vs. the most cost effective way would be (or just if this type of thing is considered expensive or not). But if that's too much to ask, just knowing generally how you would implement this sort of thing would be the main question.&lt;/p&gt;&#xA;" OwnerUserId="20446" LastActivityDate="2019-05-12T02:07:25.673" Title="How to build multiple large scale fish tanks with water at different temperatures" Tags="|thermodynamics|concrete|heating-systems|water-resources|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28343" PostTypeId="1" AcceptedAnswerId="28347" CreationDate="2019-05-12T08:55:46.720" Score="1" ViewCount="4173" Body="&lt;p&gt;I am learning the method of joints, but there's no explanation on how to figure out the direction of internal forces in my course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you know that BA is tension and BC is compression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aCKFg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aCKFg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user20448" LastActivityDate="2019-05-12T18:20:57.640" Title="Determining the direction of internal forces in a truss" Tags="|mechanical-engineering|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28344" PostTypeId="1" AcceptedAnswerId="28346" CreationDate="2019-05-12T11:20:24.540" Score="0" ViewCount="1906" Body="&lt;p&gt;Given &lt;a href=&quot;https://engineering.stackexchange.com/a/28258/2817&quot;&gt;this answer&lt;/a&gt;, I understand the importance of phase change (liquid --&gt; gas) for the fluid used in a steam engine. Thus I understand it is important to use a fluid that is liquid at pressure and temperature this engine may work in (between 600 and 1000hPa, between -20°C and 50°C (I may take large margin for extreme climate condition including high altitude mountains)).&#xA;Thus, we may use any fluid that is liquid in this condition and easy to boil (I'm thinking of ethanol that has a lower boiling point but other fluids may be suited).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: why water and not other fluids such as ethanol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;: I cite ethanol but I don't want the answer to be focused on this fluid in particular. It can be any fluid that is liquid in normal temperature and pressure condition and easy to boil. I have no idea what fluid can meet this requirement.&lt;/p&gt;&#xA;" OwnerUserId="2817" LastEditorUserId="2817" LastEditDate="2019-05-12T17:45:33.967" LastActivityDate="2021-05-18T13:23:58.313" Title="Why use steam water and not other fuild in steam engines?" Tags="|steam|" AnswerCount="3" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28345" PostTypeId="2" ParentId="28344" CreationDate="2019-05-12T14:59:53.227" Score="2" Body="&lt;p&gt;I can imagine using other liquids than water in a steam engine, but it would have to be for a very specific reason and purpose.  In general, water is in so many ways superior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steam engines designs are very inefficient in terms of preserving the boiled liquid.  In most cases, the steam is just released into the air, resulting in a very simple and inexpensive design.  Capturing, cooling and condensing, pumping, etc. would be a far more complicated process that is not only much more expensive, but far more likely to experience mechanical breakdowns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a steam engine (as in what pulls a train).  Its water continuously boils away and it has to stop to &lt;a href=&quot;https://en.wikipedia.org/wiki/Water_stop&quot; rel=&quot;nofollow noreferrer&quot;&gt;refill&lt;/a&gt; from a trackside storage tank.  Early trains needed to do this every 10 km or so, and later, onboard storage tanks extended the range to 200 km.  Those trackside tanks can be automatically kept full using a windmill powered well pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine the same process using ethanol:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Who is going to keep the trackside tanks full of ethanol, and how?&lt;/li&gt;&#xA;&lt;li&gt;Who is going to keep people from stealing and drinking the alcohol?&lt;/li&gt;&#xA;&lt;li&gt;What is going to happen when the ethanol &quot;steam&quot; released from the engine gets near the sparks pouring out of the chimney?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So yes, in theory any liquid with a reasonable boiling point could be used, but in practice there really isn't even the slightest justification for it.&lt;/p&gt;&#xA;" OwnerUserId="19930" LastActivityDate="2019-05-12T14:59:53.227" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28346" PostTypeId="2" ParentId="28344" CreationDate="2019-05-12T16:00:34.737" Score="9" Body="&lt;p&gt;There's some interesting information on a number of alternative working fluids &lt;a href=&quot;http://www.douglas-self.com/MUSEUM/POWER/oddfluid/oddfluid.htm&quot; rel=&quot;noreferrer&quot;&gt;here&lt;/a&gt;, including ammonia, alcohol, mercury, sulpur dioxide, freon, and many others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Refrigerants are used today in &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Organic_Rankine_cycle&quot; rel=&quot;noreferrer&quot;&gt;organic rankine cycle&lt;/a&gt; applications which operate at lower temperatures than steam systems and are used to generate electricity from waste heat sources.  These are small systems which are not practical for large power generation applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these materials were considered in the past because high temperatures could be achieved at lower pressures than with water, but modern metallurgy has given us metals that easily contain the necessary pressures in water boilers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water has great benefits over these other materials because is has very low cost, is non-toxic, is ready available, and is non-combustible. As a result, there's no compelling reason to seek other fluids.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2019-05-12T16:00:34.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28347" PostTypeId="2" ParentId="28343" CreationDate="2019-05-12T18:20:57.640" Score="1" Body="&lt;p&gt;There are a few ways of doing it.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Just math it out&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This basically means doing all the calculations paying attention to your sign convention and seeing what the final result is. In complex cases, with a large truss dealing with many loads in multiple directions, this is the only real option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a simple truss such as this one, though, there are two easier ways of doing it via inspection:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;So flexible!&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This method is basically just looking at the deformed configuration under load. Feel free to exaggerate as much as you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you have a horizontal load at node B. This force is going to try to move B rightward. Let it. In fact, lets pretend that force moves B a full meter to the right (remember, exaggerate!), such that the truss becomes an &lt;a href=&quot;https://en.wikipedia.org/wiki/Isosceles_triangle&quot; rel=&quot;nofollow noreferrer&quot;&gt;isosceles triangle&lt;/a&gt;.&lt;sup&gt;1&lt;/sup&gt; Now consider what that does to beams AB and BC:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;AB went from being 2&amp;nbsp;m long to &lt;span class=&quot;math-container&quot;&gt;$\sqrt{2^2 + 1^2} = 2.2\text{ m}$&lt;/span&gt; long.&lt;/li&gt;&#xA;&lt;li&gt;BC went from being &lt;span class=&quot;math-container&quot;&gt;$\sqrt{2^2 + 2^2} = 2.8\text{ m}$&lt;/span&gt; long to &lt;span class=&quot;math-container&quot;&gt;$\sqrt{2^2 + 1^2} = 2.2\text{ m}$&lt;/span&gt; long.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The numbers themselves are irrelevant (after all, they're based on a made-up displacement of B). All that matters is that AB got longer and BC got shorter. No matter what the real displacement of B is, this will always be the case: AB gets slightly longer, BC gets shorter. Which tells us that AB is under tension and BC, under compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the basic gist of this method: look at the expected deformed configuration and see which elements get stretched (tension) and which get squeezed (compression).&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Node equilibrium&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The last way is to look at the individual nodes. The equilibrium diagram in your question (just the arrows, ignoring the values) is a cheat-sheet: all arrows pointing towards the node (such as the one representing BC) are compression, those pointing away (such as AB) are tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way to see this is to think in terms of &lt;a href=&quot;https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton&amp;#39;s_third_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;Newton's Third Law&lt;/a&gt;: the node is applying a force on the beam, and the beam is applying an equal and opposite one on the node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a beam is under compression, the external force it feels is pointing inwards (think of a column with a downwards force above and an upwards reaction force below). Therefore, the column's reaction points outwards from the column... which is &quot;towards&quot; the node. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, a beam under tension feels outward-facing external forces, so its reaction points towards the column, and away from the node.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore you just need to figure out the direction for each of the arrows in the node diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, you have a rightwards external force at B. Only BC has a horizontal component, so it has to be solely responsible for countering that force. That happens by making BC's horizontal component point towards the left. That being so, BC's vertical component must point up, since BC's resultant force must be purely axial and therefore parallel to BC's inclination. Since that resultant will point left-and-up, it's pointing towards the node and is therefore compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, the node now has an upwards force which must be countered by AB, which must point down. Since that makes it point away from the node, it must be tension.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;&lt;sup&gt;1&lt;/sup&gt; In reality, there'd also be some vertical movement, but it's much smaller than the horizontal and can be safely ignored for this thought experiment.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-05-12T18:20:57.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28348" PostTypeId="1" CreationDate="2019-05-12T18:36:35.423" Score="1" ViewCount="533" Body="&lt;p&gt;I feel a little lost. I think my problem is quite simple, but unfortunately, my knowledge of math is also simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two rigid bodies that move relative to each other. I know two points of each body (see picture). &lt;a href=&quot;https://i.stack.imgur.com/FkJNR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FkJNR.png&quot; alt=&quot;Bodies with points&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I find the place or point where the lower body rotates against the upper body? My idea was to just connect the points of each body and the intersection is the point I am looking for. &lt;a href=&quot;https://i.stack.imgur.com/tM5vr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tM5vr.png&quot; alt=&quot;lines on bodies&quot;&gt;&lt;/a&gt; but this doesnt seem right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for your help, I really apreciate it.&lt;/p&gt;&#xA;" OwnerUserId="20450" LastActivityDate="2019-10-11T17:02:15.293" Title="finding the center of rotation between 2 rigid bodies" Tags="|mathematics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28349" PostTypeId="2" ParentId="28348" CreationDate="2019-05-12T19:00:46.703" Score="0" Body="&lt;p&gt;It depends on how those two rigid bodies are moving. Are they turning around a common center, or are they orbiting around in an ellipse? if it is an ellipse, is this ellipse wobbling, or may be they are being attracted by the force of a field which changes by coordinate and time.&#xA;What is the cause of this rotation, magnetic field, or they are floating in a whirlpool, or gravity or tied by a wire?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming they are just simply rotating around a common center are the facing each other continually or they have their different rotation period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in the most basic case they could be rotating around a common point but like Earth and Mars they have different orbits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need some basic information before you can start to draw lines.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-05-12T20:44:56.647" LastActivityDate="2019-05-12T20:44:56.647" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28350" PostTypeId="1" CreationDate="2019-05-13T03:23:45.877" Score="3" ViewCount="696" Body="&lt;p&gt;I am in the process of completing a design project for my fluids course in which Buckingham-Pi /Dimensional Analysis is needed to find the optimal parameters needed to achieve a rotational rate of 120 rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;So the question at hand is that the rotational rate of an anemometer is a function of density, viscosity, wind speed, rod length, sphere diameter, rod diameter, and number of spheres.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Omega = f(rho,mu,Ds,Dr,Lr,N,U)] and from that we need to non-dimensionalize the parameters and get Pi groups in which we analyze and determine the optimal dimensions required to achieve a rotational rate of 120 rpm.*&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QmzRV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QmzRV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have met with the professor regarding the project and I was told that the &lt;em&gt;dependent variable (group with the omega) is to be plotted on the y axis against all other groups (independent Pi groups on the x axis)&lt;/em&gt;. He also mentioned that &lt;em&gt;rho, and mu are parameters because the same results should occur regardless of the fluid used in testing.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;**I have worked out the problem and have achieved the Pi Groups and even plotted them; but I do not know how to account for the the non-dimensional value N( # of spheres); or what needs to be done next. **&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;All work for Pi-Groups&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sDJ3L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sDJ3L.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nXLcF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nXLcF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Graphs/Excel Stuff&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I chose random ranges that fit within the spec so is that even correct to get some data points to graph so can you do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(All pi groups which use rho and mu, are plugged in as  the values at the top since I'm not sure what range to use for them since it can be any fluid, so not sure if my approach is even remotely correct?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NOTE :&lt;/strong&gt; &lt;em&gt;Pi (4) when plotted was taken to be the Reynolds's # which is the inverse of the computed Pi group (I mention this on excel spreadsheet).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*First graph is dependent pi group (5) against all other pi groups.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Second is with the avg rotational rate of 120 rpm as the parameter for the dependent pi group (5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Third is just wind speed Pi group (4) against omega dependent pi group (5).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wpucx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wpucx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0daTM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0daTM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look and let me know if I'm doing something wrong/right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Any and all help is appreciated :)&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="20453" LastActivityDate="2020-10-05T21:03:35.373" Title="Buckingham-Pi Dimensional Analysis when there is a non-dimensional parameter: How is it accounted for?" Tags="|mechanical-engineering|fluid-mechanics|design|fluid|prototyping|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28351" PostTypeId="2" ParentId="13583" CreationDate="2019-05-13T04:44:39.850" Score="0" Body="&lt;p&gt;Such an explanation of the voltage drop at the beginning of the yielding plateau is generally accepted. However, the same voltage drops occur in other materials without solid carbon solution (doi:10.1016/j.ijsolstr.2009.09.018). In addition, the same short voltage drops are found at the end of the yielding plateau (doi:10.1016/j.ijsolstr.2011.02.019).&lt;/p&gt;&#xA;" OwnerUserId="20454" LastEditorUserId="20454" LastEditDate="2019-05-13T08:08:15.743" LastActivityDate="2019-05-13T08:08:15.743" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28353" PostTypeId="2" ParentId="1977" CreationDate="2019-05-13T11:07:27.057" Score="2" Body="&lt;p&gt;In addition to the other answers, some soils (e.g. clay) expand and contract as their moisture content changes, and wet soil expands as it freezes. This movement is rarely even across site, and differential movement causes a building to crack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Foundations are dug down to below an expansive soil and below the frost line to avoid movement. They are then built up with crushed rock or concrete, which doesn't expand with moisture or freezing temperatures.&lt;/p&gt;&#xA;" OwnerUserId="20423" LastActivityDate="2019-05-13T11:07:27.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28355" PostTypeId="1" CreationDate="2019-05-13T14:15:54.073" Score="0" ViewCount="80" Body="&lt;p&gt;Can anyone help me to identify this electrical connector set?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IkSiem.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IkSiem.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Eqmqgm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Eqmqgm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8P0E0m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8P0E0m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I7LQrm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I7LQrm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vcUuOm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vcUuOm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vPsdGm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vPsdGm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oi5IRm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oi5IRm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x0EMWm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x0EMWm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zcOPnm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zcOPnm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SBuBum.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SBuBum.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DllNJm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DllNJm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Hint: One male plug style connector has some text saying &quot;HC 0617 CANNON&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="20459" LastEditorUserId="33" LastEditDate="2019-05-13T15:28:37.550" LastActivityDate="2019-05-13T17:33:49.787" Title="What kind of plug is this?" Tags="|electrical-engineering|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28356" PostTypeId="2" ParentId="28355" CreationDate="2019-05-13T15:13:31.130" Score="0" Body="&lt;p&gt;The look like professional-grade DIN or XLR connectors, with screw locking threads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cannon is a specialist connector manufacturer - now part of the ITT conglomerate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They were probably originally used for professional audio cabling.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-05-13T15:13:31.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28357" PostTypeId="2" ParentId="28355" CreationDate="2019-05-13T15:18:21.200" Score="1" Body="&lt;p&gt;Cannon is a manufacturer of electrical connectors used in the audio engineering and scientific instrumentation fields. This Cannon connector looks like an instrumentation cable. Contact CANNON on the web for more info.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-05-13T15:18:21.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28358" PostTypeId="2" ParentId="28355" CreationDate="2019-05-13T17:33:49.787" Score="0" Body="&lt;p&gt;It seems balanced XLR stereo connection with screw lock. In pro audio cabling balanced means each side (left or right of the stereo) will be carried on three wires, positive, negative, ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In receiver positive or negative which carry the same signal with different polarity will be flipped and coupled with the other one, chocking noise picked by the cable on its way, because once flipped the noise wave will be the mirror image of self. &#xA;This has 5 connectors. one seems to be ground two for each right and left.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-13T17:33:49.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28359" PostTypeId="1" AcceptedAnswerId="28361" CreationDate="2019-05-13T18:33:49.793" Score="2" ViewCount="5656" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fk3zH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fk3zH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where does the T-beam come from? There's just a straight beam resting on two supports. And where do 10 and 80 come from? I am so confused.&lt;/p&gt;&#xA;" OwnerDisplayName="user20448" LastActivityDate="2019-10-24T16:48:33.143" Title="Maximum tensile and compressive stress in a T-beam" Tags="|mechanical-engineering|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28360" PostTypeId="1" CreationDate="2019-05-13T19:17:58.110" Score="3" ViewCount="594" Body="&lt;p&gt;I went through all of the work and experience at a previous job to get my Professional Engineer's license, however, my new firm is not licensed to provide design services.  Is it still allowable to place &quot;P.E.&quot; after my name since &lt;em&gt;I&lt;/em&gt; am a licensed professional engineer even though I will not be providing sealing and stamping of documents?&#xA;I know some engineers will get their licenses even if it is not required, but I wasn't sure how this lined up with presenting your credentials on a business card.  &lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2019-05-13T19:46:29.323" Title="PE Designation on business card if firm doesn't offer engineering services" Tags="|licensure|" AnswerCount="2" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28361" PostTypeId="2" ParentId="28359" CreationDate="2019-05-13T19:19:32.653" Score="1" Body="&lt;p&gt;To find the neutral axis of any section including the T section in your question, ( called T beam because its cross section is a T):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;one picks a horizontal axis parallel to x axis (or the axis they need to find the beam's neutral axis).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;break the section into simple geometric shapes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;find the area moment of each section by multiplying its area by  the distance from its CG to the axis picked at the first step. (in your case they assumed the top surface's, Line Z, as that random reference axis, and 10 is the distances of the horizontal 80 by 20 flange and 60+20 is the vertical web CG from the top axis)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;sum all the area moments and dived by total area of all the part, you get the neutral axis, which they did and got 52 mm from the top surface as the neutral axis of the T beam.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="23239" LastEditDate="2019-10-24T16:48:33.143" LastActivityDate="2019-10-24T16:48:33.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28362" PostTypeId="1" CreationDate="2019-05-13T19:21:41.943" Score="2" ViewCount="186" Body="&lt;p&gt;In induction heaters metal seems to levitate in center of radial axis, as well as longitudinal axis (see video). Is by any chance the reason for that like in this schematic, that magnetic field lines of a metal rod, forming closed loops, move in opposite direction to the coil when going around the sides of the metal? So, repulsion at the sides, and attraction along the poles. Or, is it other reason?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fSNKF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fSNKF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=WRPoOaaJCIA&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xy95G.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20456" LastActivityDate="2021-07-03T15:04:25.373" Title="In induction heaters why do metal pieces levitate in center of radial axis?" Tags="|electromagnetism|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28363" PostTypeId="2" ParentId="28360" CreationDate="2019-05-13T19:45:42.177" Score="1" Body="&lt;p&gt;If you put PE on your card, it suggests that you &lt;em&gt;are&lt;/em&gt; able to sign off documents right now. (I assume that even if you wanted to do that outside your employment, you aren't paying for professional indemnity insurance since you don't need it for your job, so that would NOT be a good idea...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just make sure you don't mislead somebody about what you (or your employer) can do, and they file you (or your employer) under &quot;P&quot; for &quot;pretentious time-waster&quot; when they find out it's not true :)&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-05-13T19:45:42.177" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28364" PostTypeId="2" ParentId="28360" CreationDate="2019-05-13T19:46:29.323" Score="1" Body="&lt;p&gt;You worked for it, it represents a level of skill relevant to your &quot;trade&quot; so it belongs on your card.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to not mention it, that is your choice and you may have good reason for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a colleague who has a PhD and does not mention it... But the work does tend to give the game away....&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-05-13T19:46:29.323" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28365" PostTypeId="2" ParentId="28362" CreationDate="2019-05-13T20:11:45.680" Score="1" Body="&lt;p&gt;In the clip the induced magnet into the metal creates a magnetic field opposing the coil's magnetic field and make it levitate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your sketch,there will be acceleration from left to right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could be used as a gun, hypothetically.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-13T20:11:45.680" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28367" PostTypeId="1" CreationDate="2019-05-14T02:27:54.847" Score="0" ViewCount="46" Body="&lt;p&gt;With a magnet inside another hollow magnet, for example an iron rod in a hollow cylinder, the iron bar placed inside the cylinder with poles attracted to opposite poles of cylinder (likes repel opposites attract), is there also a force along the radius of the cylinder? Per the closed loop rule, the magnetic flux along the sides of the iron rod flows in opposite direction to that of the cylinder. It would seem like the magnetic flux on the inside of the cylinder &quot;squeezes&quot; its way past opposite flux from the iron rod (the iron rod being fixed in place along the longitudinal axis by attraction at poles), compressing it from all directions at once, repelling the iron rod with a vector towards the centre of the radial axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6joGd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6joGd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20456" LastActivityDate="2019-05-14T02:27:54.847" Title="Is there an inward (radial) force acting on a magnet inside a magnet?" Tags="|electromagnetism|" AnswerCount="0" CommentCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="28368" PostTypeId="2" ParentId="26804" CreationDate="2019-05-14T09:27:10.097" Score="1" Body="&lt;p&gt;By knowing the volume flow rate of a gas , you can also find the mass passing through a cross section area for certain thermodynamic conditions.&lt;/p&gt;&#xA;" OwnerUserId="20420" LastActivityDate="2019-05-14T09:27:10.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28371" PostTypeId="2" ParentId="28348" CreationDate="2019-05-14T15:40:33.193" Score="1" Body="&lt;p&gt;What you might be looking for is the &lt;em&gt;relative instantaneous centre of rotation&lt;/em&gt; (ICR for short) of the two bodies. This is the point in space that, at a given point in time, one body will rotate about the other body, as though there is a pivot connecting the rigid bodies. The reason why this is called &quot;instantaneous&quot; is because, for a general arbitrary motion of two rigid bodies, this apparent pivot may change its location on either body as time progresses (note: the relative centre of rotation need not be located physically on either body).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(this answer only considers the 2D case, i.e. planar motion. This answer doesn't apply to the more general case of spatial motion, where not all velocity vectors lie within a single plane)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An illustrative example that shows how an instantaneous centre of rotation might change location on either body is the rolling of a ruler over the curved part of a fixed semi-circle without slip:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3fZZj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3fZZj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Three different points in time are shown, and it can be seen that the instantaneous pivot , i.e. the ICR, occurs at the point of contact. However, at different times, the pivot has changed location, with the coloured segments showing how much the ICR has moved on each body since the first point of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if there is a physical pivot directly linking the two rigid bodies, then the relative centre of rotation does not change location on either body, and so the term &quot;instantaneous&quot; is not necessary. The presence of a such a pivot restricts the possible relative motions of the rigid bodies!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how is the relative ICR defined mathematically? In the general case of two rigid bodies moving, there is one and only one point at which the velocity of body 1 and the velocity of body 2 are the same: this point is the relative ICR. (If body 2 is not rotating with respect to body 1, then the ICR cannot be evaluated: how can there be a centre of rotation without any rotation, after all?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the &lt;strong&gt;relative&lt;/strong&gt; ICR of the rigid body pair, we need to find the &lt;strong&gt;absolute&lt;/strong&gt; instantaneous centres of rotation for the individual bodies. The (absolute) ICR of body 1 is defined as the point on body 1 where there is no velocity. Similarly, the (absolute) ICR of body 2 is defined as the point on body 2 where there is no velocity. The absolute ICRs are easy to find, since it is the point to which all velocity vectors lie perpendicular to. Therefore, to find the (absolute) ICR for a body, we need to know the velocity vectors at two points on that body. Then, for each of those two points, the (absolute) ICR will lie on the line passing through the point perpendicular to the direction of velocity. We can therefore draw two lines and the (absolute) ICR must lie on the intersection of those lines. If not, the ICR would not lie perpendicular to one of the velocities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following diagram shows the case in the question, with velocities that I random assigned to each of the four points &lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$P_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$P_3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$P_4$&lt;/span&gt;, such that the magnitudes of their velocities are &lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$V_3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$V_4$&lt;/span&gt; respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0ZSAS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0ZSAS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the diagram, lines are drawn perpendicular to the velocity vectors, and the intersections of the lines reveal the absolute ICRs, labelled as points &lt;span class=&quot;math-container&quot;&gt;$I_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$I_2$&lt;/span&gt; for bodies 1 and 2 respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the positions of the absolute ICRs determined, we can now measure the distances between the points at which velocities are known, and the absolute ICRs: these distances are represented in the diagram by &lt;span class=&quot;math-container&quot;&gt;$a_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$a_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$a_3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$a_4$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We know from circular motion that the tangential velocity is equal to the angular velocity multiplied by the distance from the pivot, and so the &lt;strong&gt;absolute value&lt;/strong&gt; of the angular velocities of each body can be determined with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$|\omega_1| = \frac{V_1}{a_1} = \frac{V_2}{a_2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$|\omega_2| = \frac{V_3}{a_3} = \frac{V_4}{a_4}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that, for each body, both fractions must give the same angular velocity magnitude. Otherwise, the velocities of the body are not valid for rigid body motion. (Consider the case of two points on the same body moving in opposite directions: this would suggest the body is deforming, which contradicts the assumption of a rigid body)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign of the angular velocities can be established by using a sign convention. Here, I'll use the convention where anticlockwise angular velocities are postive, so that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\omega_1 = |\omega_1|$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\omega_2 = -|\omega_2|$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to determine our &lt;strong&gt;relative&lt;/strong&gt; ICR, the point at which the velocities of both bodies are equal, it is important to note that the relative ICR must lie on the line connecting &lt;span class=&quot;math-container&quot;&gt;$I_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I_2$&lt;/span&gt;. This is because the velocities of both bodies considered at the same point can only be parallel to one another along this line, due to the velocities being perpendicular to &lt;span class=&quot;math-container&quot;&gt;$I_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I_2$&lt;/span&gt;. The following diagram shows the velocities at a point on the line and another point off the line to illustrate this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j6WZj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j6WZj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the relative ICR must lie on this line, but where on this line? Note that as the point gets further from &lt;span class=&quot;math-container&quot;&gt;$I_1$&lt;/span&gt;, the velocity of body 1 at that point increases proportionally. Similarly, as the point moves away from &lt;span class=&quot;math-container&quot;&gt;$I_2$&lt;/span&gt;, the velocity of body 2 at that increases proportionally. Therefore, there must be a point at which the two velocities become equal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$I_{21}$&lt;/span&gt; represent the point of relative ICR. We can determine the length of line connecting &lt;span class=&quot;math-container&quot;&gt;$I_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I_2$&lt;/span&gt;, which we will represent as &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;. Then, let's define &lt;span class=&quot;math-container&quot;&gt;$l_1$&lt;/span&gt; as the distance between &lt;span class=&quot;math-container&quot;&gt;$I_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I_{21}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$l_2$&lt;/span&gt; the distance between &lt;span class=&quot;math-container&quot;&gt;$I_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I_{21}$&lt;/span&gt;. Note that &lt;span class=&quot;math-container&quot;&gt;$l_1 + l_2 = L$&lt;/span&gt;. Calculating the value for &lt;span class=&quot;math-container&quot;&gt;$l_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$l_2$&lt;/span&gt; will then determine the position of &lt;span class=&quot;math-container&quot;&gt;$I_{21}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uk79y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uk79y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine &lt;span class=&quot;math-container&quot;&gt;$l_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$l_2$&lt;/span&gt;, we use the fact the velocity &lt;span class=&quot;math-container&quot;&gt;$V_{21}$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$I_{21}$&lt;/span&gt; must be the same:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_{21} = l_1 \omega_1 = -l_2 \omega_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(This expression holds provided a sign convention is used where both angular velocities have the same direction if they have the same sign.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this, it can be seen that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$l_1 = \frac{\omega_2}{\omega_2 - \omega_1}L$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$l_2 = \frac{\omega_1}{\omega_1 - \omega_2}L$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For those familiar with the radii of the pitch circles of gears, these expressions might be familiar!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It can be seen that both &lt;span class=&quot;math-container&quot;&gt;$l_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$l_2$&lt;/span&gt; lie in the range &lt;span class=&quot;math-container&quot;&gt;$0 &amp;lt; l_{1,2} &amp;lt; L$&lt;/span&gt; if and only if &lt;span class=&quot;math-container&quot;&gt;$\omega_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\omega_2$&lt;/span&gt; has different signs.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the position of &lt;span class=&quot;math-container&quot;&gt;$I_{21}$&lt;/span&gt; is revealed. Note however that the relative ICR has been calculated only for a single instance in time. To calculate for other points in time, this above must be repeated using the velocities of points &lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$P_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$P_3$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$P_4$&lt;/span&gt; at that time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In vector notation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the above can be adapted into vector notation. This way, it is useful to consider a point &lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt; on body 1 with position &lt;span class=&quot;math-container&quot;&gt;$\mathbf{R}_1$&lt;/span&gt; and known velocity &lt;span class=&quot;math-container&quot;&gt;$\mathbf{V}_1$&lt;/span&gt;, and body 1 has known angular velocity &lt;span class=&quot;math-container&quot;&gt;$\omega_1\mathbf{k}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathbf{R}_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\mathbf{V}_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\omega_2\mathbf{k}$&lt;/span&gt; are defined similarly for body 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The velocity &lt;span class=&quot;math-container&quot;&gt;$\mathbf{v}_1$&lt;/span&gt; of body 1 at an arbitrary point &lt;span class=&quot;math-container&quot;&gt;$\mathbf{r}$&lt;/span&gt; can be expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{v}_1 = \mathbf{V}_1 + \omega_1 \mathbf{k} \times \left(\mathbf{r} - \mathbf{R_1} \right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mathbf{v}_2$&lt;/span&gt; of body 2 at position &lt;span class=&quot;math-container&quot;&gt;$\mathbf{r}$&lt;/span&gt; is similarly expressed as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{v}_2 = \mathbf{V}_2 + \omega_2 \mathbf{k} \times \left(\mathbf{r} - \mathbf{R_2} \right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By cross-multiplying these expressions with unit vector &lt;span class=&quot;math-container&quot;&gt;$\mathbf{k}$&lt;/span&gt;, the positive vectors of the &lt;strong&gt;absolute&lt;/strong&gt; ICRs are seen to be equal to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{I}_1 = \mathbf{R}_1 + \frac{1}{\omega_1} (\mathbf{k} \times \mathbf{V}_1)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{I}_2 = \mathbf{R}_2 + \frac{1}{\omega_2} (\mathbf{k} \times \mathbf{V}_2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the &lt;strong&gt;relative&lt;/strong&gt; ICR is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{I}_{21} = \frac{\mathbf{k} \times (\mathbf{V}_2 - \mathbf{V}_1) + \omega_2 \mathbf{R}_2 - \omega_1 \mathbf{R}_1}{\omega_2 - \omega_1}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="276" LastActivityDate="2019-05-14T15:40:33.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28372" PostTypeId="1" CreationDate="2019-05-14T15:50:58.147" Score="1" ViewCount="208" Body="&lt;p&gt;Fellow engineers, I need your help in improving my design for a pneumatic system for maintaining a constant supply of at least 90 PSI and 20 CFM for a particle analyzer with stringent air requirements. Its expected to draw air every 2 seconds. Here's the breakdown of the components used in the setup (in this order). Picture attached. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A: Bulk water separator. This should help with removing moisture from the air because our plant air is heavily moisture-laden at times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: Particulate filter: 0.01 micron air filter to ensure clean air, free from particles.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C: Desiccant air dryer: Heat-less desiccant dryer for dry air. -40 C pressure dew point to ensure that there is no further condensation beyond this point and the air stays dry till it makes its way all the way to the analyzer. The dryer is rated for an inlet flow of 45 SCFM. I up-sized it because this is a heat-less dryer so a certain percentage of the air going in to the dryer will be used to regenerate the desiccants as well. The dryer has an in-built filter that will prevent any of the desiccants from leaving the dryer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;D: Globe valve: This is to contain the flow rate coming out of the dryer. This was suggested to me by another engineer. Since we are filling up a large tank, the air coming out of the dryer could potentially exceed the rated flow rate of the dryer which could damage the desiccants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;E &amp;amp; F: Check valve and booster pump in parallel. The booster pump is meant to charge the receiver tank to a pressure of 95 - 100 PSI. The check valve is there to reduce the charge time by allowing air to pass through up to the inlet pressure. This could help take some of the burden off of the booster pump so that the pump only cycles when the tank pressure needs to be raised to setpoint. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;G: Air receiver tank: This will help create a reservoir of air that will ensure that the analyzer never has to wait for air. The tank is rated for 200 PSI but I won't be charging it more than 100 PSI. Still debating if I should go with a smaller tank because 80 gallons is an awful lot of volume for the booster to fill.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;H: Filter: To get rid of any of the remaining particles and oil vapors left in the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I: Pressure regulator: To maintain the desired pressure for the equipment&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J: Equipment - Particle analyzer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions on improving this design?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MNc1F.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MNc1F.jpg&quot; alt=&quot;Pneumatic system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20451" LastActivityDate="2019-06-20T04:07:30.173" Title="Designing a pneumatic system for consistent pressure/flow" Tags="|pneumatic|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28373" PostTypeId="2" ParentId="28350" CreationDate="2019-05-14T17:37:28.313" Score="1" Body="&lt;p&gt;Let's assume that you will work out the right combinations of dimensionless terms on your own. Let's consider only the question of what to plot once you have those terms. By example, in heat transfer, the Nusselt number &lt;span class=&quot;math-container&quot;&gt;$Nu$&lt;/span&gt; is for forced convection is a function of the Reynolds number &lt;span class=&quot;math-container&quot;&gt;$Re$&lt;/span&gt; and the Prandtl number &lt;span class=&quot;math-container&quot;&gt;$Pr$&lt;/span&gt;. In this case we can make one of two general plots.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Plot &lt;span class=&quot;math-container&quot;&gt;$Nu$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$Re$&lt;/span&gt; for fixed values of &lt;span class=&quot;math-container&quot;&gt;$Pr$&lt;/span&gt;. This is equivalent to tracking the effects of fluid velocity for a fixed type of fluid.&lt;/li&gt;&#xA;&lt;li&gt;Plot &lt;span class=&quot;math-container&quot;&gt;$Nu$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$Pr$&lt;/span&gt; for fixed values of &lt;span class=&quot;math-container&quot;&gt;$Re$&lt;/span&gt;. This is equivalent to tracking he effects of different types of fluids at the same flow velocity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Suppose later that, using further analogies in laminar flow, we predict that &lt;span class=&quot;math-container&quot;&gt;$Nu = C Pr^{1/3}Re^{1/2}$&lt;/span&gt;. We can revise our plots to be &quot;linearized&quot;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Plot &lt;span class=&quot;math-container&quot;&gt;$Nu$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$Re^{1/2}$&lt;/span&gt; for fixed values of &lt;span class=&quot;math-container&quot;&gt;$Pr$&lt;/span&gt;. This is equivalent to tracking the effects of fluid velocity for a fixed type of fluid.&lt;/li&gt;&#xA;&lt;li&gt;Plot &lt;span class=&quot;math-container&quot;&gt;$Nu$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$Pr^{1/3}$&lt;/span&gt; for fixed values of &lt;span class=&quot;math-container&quot;&gt;$Re$&lt;/span&gt;. This is equivalent to tracking he effects of different types of fluids at the same flow velocity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Finally, with the assumption that only two parameters influence this system, we can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ d Nu = \left(\frac{\partial Nu}{\partial Re}\right)_{Pr} d Re + \left(\frac{\partial Nu}{\partial Pr}\right)_{Re} d Pr$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partial derivatives are the slopes of the curves on the respective plots. Absent a theoretical equation, we can use the above equation and the graphs to predict which term has the greater influence on &lt;span class=&quot;math-container&quot;&gt;$Nu$&lt;/span&gt;. By example, when we ant to increase &lt;span class=&quot;math-container&quot;&gt;$Nu$&lt;/span&gt; for the case when &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \left(\frac{\partial Nu}{\partial Re}\right)_{Pr} &amp;gt; \left(\frac{\partial Nu}{\partial Pr}\right)_{Re}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we predict that we will do better to increase &lt;span class=&quot;math-container&quot;&gt;$Re$&lt;/span&gt; (e.g. increase the flow velocity for the given fluid) than to increase &lt;span class=&quot;math-container&quot;&gt;$Pr$&lt;/span&gt; (e.g. change the fluid at the same flow).&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-05-14T17:37:28.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28375" PostTypeId="1" CreationDate="2019-05-14T18:49:20.940" Score="0" ViewCount="100" Body="&lt;p&gt;Recently we put 500 feet borewell in our area, it's hot summer season now. Person who drilled the borewell told me that we can get 1.5 inch of water now. During winter season we can expect at least 2 inch of water. Now we are planning to install the motor at 410 feet and buy 5hp motor with 2 inch mouth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is&#xA;Can I put 2 inch pipe inside borewell and control water outlet at the end of borewell with 1.5 inch? Whenever water level increase can I increase outlet with 2 inch? Can I control water outlet at the end?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question rephrased:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we adjust water outlet of the submersible motor at the ground surface or edge of the pipe? Usually people adjust water outlet at the end of motor mouth section which is inside the borewell not at the ground. Is there any signigficance where the reducer kept?&lt;/p&gt;&#xA;" OwnerUserId="20466" LastEditorUserId="20466" LastEditDate="2019-05-16T04:07:03.293" LastActivityDate="2023-02-17T11:53:06.983" Title="Borewell motor water outlet adjustment" Tags="|motors|pumps|water-resources|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28376" PostTypeId="1" CreationDate="2019-05-14T19:01:13.823" Score="0" ViewCount="121" Body="&lt;p&gt;Does a car starter motor use a starting resistor to limit the high starting current? What is the type of connection between field coil and armature? Series, parallel or mixed?&lt;/p&gt;&#xA;" OwnerUserId="20467" LastActivityDate="2019-05-15T04:07:45.500" Title="Car dc starter motor-starting resistor" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28379" PostTypeId="1" AcceptedAnswerId="28388" CreationDate="2019-05-14T19:27:41.663" Score="1" ViewCount="415" Body="&lt;p&gt;So there is a lot of information out there about designing an external Geneva drive, but I can't seem to find anything about designing an internal one. I'm trying to design something similar to this (but with a different number of positions):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WQu5K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WQu5K.png&quot; alt=&quot;Internal Geneva Drive&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the internal drive is based off a calculations from a triangle, and I'd assume the the internal one is too, but it's just not clicking for me. Can anyone point me in the right direction? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="20468" LastActivityDate="2019-05-15T15:42:41.570" Title="Internal Geneva drive calulations/design" Tags="|gears|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28380" PostTypeId="2" ParentId="26752" CreationDate="2019-05-14T19:33:03.063" Score="2" Body="&lt;p&gt;Turns out this was the beginnings of a larger tank (shown in this image) that will encompass the inner concrete tank.  As work has progressed, they are welding plates onto the edges of the umbrella-like structure, vertically, to form a cylinder.  Having exceeded the height of the inner concrete tank, they are now rounding off the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That poses a better question:  Why are they building a steel tank around an inner concrete tank?  (It's not rainwater collection; that isn't done in this part of the country.)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/AAgjl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AAgjl.jpg&quot; alt=&quot;concrete tank with beginning of encompassing steel tank&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20010" LastEditorUserId="20010" LastEditDate="2019-06-28T15:50:34.323" LastActivityDate="2019-06-28T15:50:34.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28381" PostTypeId="2" ParentId="28376" CreationDate="2019-05-14T20:18:12.733" Score="1" Body="&lt;p&gt;Having stripped and rebuilt many starter motors from those on lawn mowers to those for 10 litre diesels - there were no extra resistors fitted except, of course, the resistance of the field coils or armature windings themselves. Not ignoring the resistance of the cables and internal resistance of the battery though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are series or parallel connected depending on application. Designed for maximum power at low speed to get the stationary engine rotating.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-05-14T20:18:12.733" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28382" PostTypeId="1" CreationDate="2019-05-14T21:19:18.063" Score="1" ViewCount="538" Body="&lt;p&gt;I am currently trying to derive a model to help estimate as precisely as possible how much time can be saved by insulating the jacket of a reactor used for eating and cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT: I edited my post so that it is clearer for everybody, I understand it could have be quite confusing. I also corrected a few typos.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Presentation of the system&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider the following reactor:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SjMhJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SjMhJ.png&quot; alt=&quot;Jacket agitated vessel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, &lt;span class=&quot;math-container&quot;&gt;$T_{iJ}$&lt;/span&gt; denotes the temperature at the jacket's inlet, &lt;span class=&quot;math-container&quot;&gt;$T_{oJ}$&lt;/span&gt; denotes the temperature at the jacket outlet, &lt;span class=&quot;math-container&quot;&gt;$T_{surf}$&lt;/span&gt; is the temperature at the surface of the jacket in contact with the ambient air and $T_{film} is the temperature very close to the jacket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature of the air far away from the jacket (compare to the width of the film at the surface of the jacket) will be denoted by &lt;span class=&quot;math-container&quot;&gt;$T_{\infty}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Energy balance with insulation of the jacket&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reactor is supposed to be perfectly adiabatic and operating as batch, therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(\sum_i m_i c_{p_{i}}\right)\frac{dT_r}{dt} = UA \Delta T_{ln} = UA \frac{T_{iJ}-T_{oJ}}{\ln \left(\frac{T_{iJ}-T_r}{T_{oJ}-T_r}\right)} \tag 1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature in the jacket can be averaged knowing the temperature at both inlet and outlet. If the temperature at the outlet is not know on the process and if the jacket is insulated, thus:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_j c_{p_j}\left(T_{iJ} - T_{oJ}\right) = UA \Delta T_{ln} \tag 2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; is the mass flow rate of water into the jacket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which leads to,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T_{oJ} = T_r + \left(T_{iJ}-T_r\right)\exp\left(-\frac{UA}{F_j c_{p_j}}\right) \tag 3$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt; can be directly solved. It can also be solved as equation &lt;span class=&quot;math-container&quot;&gt;$(4)$&lt;/span&gt; using &lt;span class=&quot;math-container&quot;&gt;$T_j = (T_{iJ} + T_{oJ})/2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\left(\sum_i m_i c_{p_{i}}\right)\frac{dT_r}{dt} = UA (T_j - T_r) \tag 4$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Energy balance without insulation of the jacket&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the jacket is not insulated, therefore &lt;span class=&quot;math-container&quot;&gt;$(2)$&lt;/span&gt; becomes the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_j c_{p_j}\left(T_{iJ} - T_{oJ}\right) = \underbrace{UA \Delta T_{ln}}_{\text{exchange jacket/reactor}} + \underbrace{h_c A_c \left(T_{film}-T_{\infty}\right) + h_r A_r \left(T_{surf}-T_{\infty}\right)}_{\text{convective and radiative exchange between the jacket and the room}} \tag 5$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging the terms it leads to equation &lt;span class=&quot;math-container&quot;&gt;$(6)$&lt;/span&gt; as follow,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\ln \left(\frac{T_{iJ}-T_r}{T_{oJ}-T_r}\right) = \frac{UA}{F_j c_{p_j}} + \frac{h_c A_c}{F_j c_{p_j}}\cdot\frac{T_{film}-T_{\infty}}{T_e - T_s} + \frac{h_r A_r}{F_j c_{p_j}}\cdot\frac{T_{surf}-T_{\infty}}{T_e - T_s} \tag 6$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$T_{film} = (T_{surf} + T_{\infty})/2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T_{surf} = f(T_j)$&lt;/span&gt; but can be set for this problem as &lt;span class=&quot;math-container&quot;&gt;$T_j - 3$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here comes the tricky part as the temperature we need to solve, &lt;span class=&quot;math-container&quot;&gt;$T_{oJ}$&lt;/span&gt;, depends on partially itself, &lt;span class=&quot;math-container&quot;&gt;$T_j$&lt;/span&gt;, and must be used during the numerical solving to calculate so solve &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea on how to proceed to solve the whole system of equations &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(6)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I would be glad if you could bring some hints on either how to proceed to solve this messy equations or another method much easier and sufficiently precise to forecast the savings on the process.&lt;/p&gt;&#xA;" OwnerUserId="19799" LastEditorUserId="19799" LastEditDate="2019-05-15T17:52:37.703" LastActivityDate="2019-05-16T20:13:35.757" Title="Modelling reactor jacket with or without insulation, how to proceed?" Tags="|heat-transfer|process-engineering|industrial-engineering|heat-exchanger|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28384" PostTypeId="1" CreationDate="2019-05-15T05:34:15.400" Score="1" ViewCount="155" Body="&lt;p&gt;We are using Cold rolled Carbon steel(SPC 270)grade. We have observed rust formation due to exposure to air(8-12hrs). The rust analysis resulted 1.6% chlorine whereas allowable limin is &amp;lt;0.01%. Air ppm shows 4ppm for 24hrs. We would like to know the allowable chlorine ppm in air which will cause rust and reasons for it.&lt;/p&gt;&#xA;" OwnerUserId="20472" LastActivityDate="2019-05-15T05:34:15.400" Title="Rust formation on Cold rolled carbon steel(SPC 270D)" Tags="|materials|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28387" PostTypeId="1" CreationDate="2019-05-15T11:41:33.630" Score="2" ViewCount="146" Body="&lt;p&gt;Background: I'm designing a machine that uses an actuator to push an object. The connecting component may not take certain bending forces, therefore I'm trying to evaluate the system with a free body diagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moving system is support by preloaded linear rails. The red component is a load cell that may only take limited bending forces (it is however rated to handle the 50N in tension or compression)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both sliders are preloaded and can be assumed perfectly parallel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition tells me that there will be some bending forces, but I'm not able to be clear about this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XtUC3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XtUC3.png&quot; alt=&quot;partial free body diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18295" LastEditorUserId="18295" LastEditDate="2019-05-15T11:56:50.770" LastActivityDate="2019-05-15T16:18:57.640" Title="Is there a moment at a point between two connected sliders, when one is subjected to a force? (schematic shown)" Tags="|mechanical-engineering|statics|moments|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28388" PostTypeId="2" ParentId="28379" CreationDate="2019-05-15T14:31:07.303" Score="3" Body="&lt;p&gt;So i finally stumbled upon something last night that had had some good tables and figures. See the link below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://uotechnology.edu.iq/tec_magaz/2016/volum342016/No.09.A.2016/[11Text.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kinematics Characteristics of The Internal &amp;amp; External Geneva&#xA;Mechanism Under The Effect of Multi Design Parameters&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically it had this image which helped to visualize things better. Turns out it is using the same triangle that the external gear uses for figuring out the center point of the driver and driven gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Tm6kR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tm6kR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20468" LastEditorUserId="20468" LastEditDate="2019-05-15T15:42:41.570" LastActivityDate="2019-05-15T15:42:41.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28389" PostTypeId="1" AcceptedAnswerId="28402" CreationDate="2019-05-15T15:19:44.337" Score="1" ViewCount="63" Body="&lt;p&gt;I have an application wherein I need a hermetic seal at cable exit points from a plastic housing. The cables are not removable. I'd like to achieve this seal using the plastic housing itself by creating a friction fit between the hole and PVC cable jacket (since there's only about 5mm of vertical space to work with). A couple questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Is this possible? How can one prevent the seal from breaking as the cable is moved into different positions?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What are the best practices for such designs? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help!&lt;/p&gt;&#xA;" OwnerUserId="20477" LastActivityDate="2019-05-16T18:22:56.370" Title="Hermetic Cable Seal With Space Limitations" Tags="|product-engineering|waterproofing|seals|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28390" PostTypeId="2" ParentId="28387" CreationDate="2019-05-15T16:18:57.640" Score="2" Body="&lt;p&gt;Let's call the length of the handles L and the distance between the two vertical rollers D. This is for now ignoring the details of rails being preloaded. Add details to your question if you feel it may be consequential on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let's start off by noting that the red load sensor is not going to receive any moment because it is a pin support. It will just receive the 50N normal force from the left handle and will transfer it up to the end of right side handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's look at the forces and reactions of the two parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the left side we have a clockwise moment of 50L.N at the connection point of the handle to the square slider and this moment is balanced by two horizontal forces of 50L/D on the top right and bottom left rollers. The one on top right roller pointing to the right and the one on the bottom left pointing to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the connection point of the two handles we have a 50 N force directed up acting on the right handle. This causes another 50L N clockwise moment at the right slider which again will be balanced by two horizontal forces on the vertical rollers by the amount of 50L/D.N. One is acting on the top right roller to the right and the other on the bottom left roller to the left.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-15T16:18:57.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28391" PostTypeId="1" AcceptedAnswerId="28394" CreationDate="2019-05-15T17:14:36.203" Score="2" ViewCount="356" Body="&lt;p&gt;Imagine a cantilevered beam, fixed on one end and with a tension line connecting the other end to ground (e.g. perpendicular to the length of the beam).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I increase the tension, does the modal frequencies of the beam change?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;On the one hand, tension affects the frequency of a string.  So, would pulling on the beam (in tension) cause it to have a higher frequency?&lt;/li&gt;&#xA;&lt;li&gt;On the other hand, the bending stiffness of a beam is not affected by a lateral force.  So, if the stiffness is not affected, then neither should the frequency.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What can help me reconcile these two comments?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bonus aspects of the question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If the stress is high enough to affect the slope of the stress-strain curve, then the stiffness changes because the modulus changes.&lt;/li&gt;&#xA;&lt;li&gt;In Nastran, you cannot apply a load while performing an eigenmode / eigenvalue normal modes analysis.  Does that mean it is not a linear phenomena?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="7569" LastEditorUserId="33" LastEditDate="2019-05-15T19:50:17.990" LastActivityDate="2019-05-19T19:36:38.393" Title="Do internal / external stresses affect modal frequency?" Tags="|beam|vibration|frequency-response|modal-analysis|nastran|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28392" PostTypeId="2" ParentId="28391" CreationDate="2019-05-15T17:56:18.740" Score="0" Body="&lt;p&gt;You can look at it as springs in series and find K effective of the two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \frac{1}{K_e}= \frac{1}{K_1}+\frac{1}{K_2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{1}{K_e}= \frac{L^3}{3EI}+ \frac{1}{K_{wire}} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then it gets a bit complicated. you need to find the mode shapes of the beam and calculate the beam's participatory mass for each mode and apply that to the wire as a spring mass system.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-15T17:56:18.740" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28394" PostTypeId="2" ParentId="28391" CreationDate="2019-05-15T19:08:20.323" Score="5" Body="&lt;p&gt;Consider a small piece cut out of a structure that has non-zero internal stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To maintain equilibrium, there must be some forces applied to the boundary of the piece. (Of course when it was part of the complete structure, those forces came from the stress in the adjacent parts of the structure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you deform the piece further, &lt;em&gt;those external forces can do work,&lt;/em&gt; which is simply &quot;force &lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt; distance&quot; applied to the geometry of the deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can describe this work for &lt;em&gt;all possible&lt;/em&gt; deformations as being equivalent to an additional stiffness matrix, called the &lt;em&gt;stress stiffness.&lt;/em&gt; (Historically it was sometimes called the &quot;geometric stiffness&quot;, but that name is not very self-explanatory!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when a structure with internal stresses vibrates, the relevant stiffness matrix is not the elastic stiffness &lt;span class=&quot;math-container&quot;&gt;$K_e$&lt;/span&gt;, but the sum of the elastic and stress stiffness &lt;span class=&quot;math-container&quot;&gt;$K_e + K_\sigma$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K_\sigma$&lt;/span&gt; can either increase or decrease the vibration frequencies. In fact &lt;span class=&quot;math-container&quot;&gt;$K_\sigma$&lt;/span&gt; is the same stress stiffness that is used in finite element elastic buckling models, and the buckling load is the magnitude of the stress when the lowest natural frequency of the structure is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a &quot;linear&quot; phenomenon in the sense that &lt;span class=&quot;math-container&quot;&gt;$K_\sigma$&lt;/span&gt; depends on the particular stress distribution in the whole structure, and that  (obviously) depends on the applied loads, including inertia loads if the structure is rotating, etc). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can model this in Nastran, for example using SOL 106. You first do a static analysis to find the internal stresses (that step can be either a linear or nonlinear stress analysis) and then a vibration analysis of the stressed structure. If the stress analysis is nonlinear, you may want to do several vibration analyses with different levels of applied load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The natural frequencies of a stretched string is a simple special case of the general situation, and the usual &quot;Dynamics 101&quot; derivation of the frequencies just ignores all the terms in the general equations which are 0 in that special case, without really explaining what is going on.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-05-15T19:15:13.507" LastActivityDate="2019-05-15T19:15:13.507" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28395" PostTypeId="1" CreationDate="2019-05-16T06:58:00.970" Score="0" ViewCount="48" Body="&lt;p&gt;I'm learning about the basics of converging/diverging nozzles in my thermodynamics class, and it has me curious. Could you theoretically blow into a tiny converging/diverging nozzle and break the sound barrier, or do other effects prevent that from happening?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming one could blow 10MPH, dividing by the speed of sound in Wolfram gives me: &#xA;M=0.013&#xA;Using an online calculator I get &#xA;A/A*=45&#xA;Using the width of a straw being 7.5mm (sorry for mixing units, but M is unitless so no harm done) I get &#xA;A*=1.25mm, or ~&#xA;D*=1.25mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which seems like something you could realistically 3D print. I saw somewhere that Lulzbot claims an xy precision of .05mm, so I would imagine the tolerance would be a manageable issue. However, I assume there's something I'm missing. I've heard somewhere that there's a fourth power relation between the width of arteries and the ability to push blood through. Is that an incompressible substance relation, or can I apply that here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also notice that my textbook uses a few assumptions, such as ideal gas relations and isentropic flow. I believe air should behave ideally in this range, but isentropic implies frictionless (right?), and I'm not sure this would apply with such a small 3D printed part, or such a large change in speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TL;DR can you reach choke flow through a converging/diverging nozzle with Ai = Ae = the cross sectional area of a straw?&lt;/p&gt;&#xA;" OwnerUserId="20482" LastActivityDate="2019-05-16T06:58:00.970" Title="Difficulty to reach Mach 1 by blowing into a small converging-diverging nozzle?" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|" AnswerCount="0" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="28396" PostTypeId="1" AcceptedAnswerId="28410" CreationDate="2019-05-16T08:54:41.140" Score="1" ViewCount="157" Body="&lt;p&gt;I'm designing a control system for a balloon-borne telescope experiment. As the balloon is constantly moving and the gondola is rotating, I want to stabilize the telescope which can be controlled in 3 axis of rotation. I have encoders for each axis motors and gyroscope located on the electronics box around 20cm apart from the telescope's center of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a possibility of performing a sensor fusion, like Kalman filter, for these two groups of sensors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a prepared model of the control loop, but have then realized that the sensors are in different places and while gyroscopes measure gondola movement, the encoders measure the telescope rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e9zNg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e9zNg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20483" LastActivityDate="2019-05-17T04:38:02.383" Title="Stabilization of telescope on gimbal using data from encoders and gyroscope" Tags="|control-engineering|control-theory|matlab|" AnswerCount="1" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28397" PostTypeId="1" CreationDate="2019-05-16T09:23:26.510" Score="1" ViewCount="123" Body="&lt;p&gt;I am looking for a away I can convert the horizontal rotation into vertical, typically we use bevel gear for such thing. &#xA;But I am using a SG90 servo motor for which I am struggling to get the correct gears to achieve my requirement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure is there is any other way I can do this using the SG90 servo motor. I am looking for something which I can 3D print if nothing is readily available.&lt;/p&gt;&#xA;" OwnerUserId="20184" LastEditorUserId="33" LastEditDate="2019-05-16T12:48:01.190" LastActivityDate="2019-05-16T12:48:01.190" Title="Need bevel gear for SG90 servo motor" Tags="|gears|servo|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28400" PostTypeId="2" ParentId="28382" CreationDate="2019-05-16T17:13:45.223" Score="2" Body="&lt;h1&gt;Full Analysis&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Perfectly Insulated System&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The best case is to analyze this system in both the time and position domains. The time domain considers the variation of the reactor temperature &lt;span class=&quot;math-container&quot;&gt;$T_r(t)$&lt;/span&gt; and the outlet temperature of the jacket system &lt;span class=&quot;math-container&quot;&gt;$T_{jL}(t)$&lt;/span&gt;. The position domain considers the variation of the temperature profile through the jacket &lt;span class=&quot;math-container&quot;&gt;$T_{j}(z)$&lt;/span&gt;. In this approach, you might hope for a set of differential equations that lend themselves to separation of variables for the jacket temperature &lt;span class=&quot;math-container&quot;&gt;$T_{j}(z,t) = T_{jz}(z) T_{jt}(t)$&lt;/span&gt;. The position domain analysis through the jacket at any instant in time then follows an approach similar to what is &lt;a href=&quot;https://engineering.stackexchange.com/questions/25562/how-to-calculate-the-time-to-heat-up-a-tank-using-recirculating-liquid/25583#25583&quot;&gt;posted here for time&lt;/a&gt; or &lt;a href=&quot;https://engineering.stackexchange.com/questions/25985/cooling-hot-nitrogen-with-an-aluminum-block/25996#25996&quot;&gt;posted here for position&lt;/a&gt;. The temperature drops exponentially as a function of position along the jacket. The time domain presents a more difficult analysis. In the end, the position domain result is modified by an exponential of time with a respective time constant that contains a ratio of the mass flow and heat capacity of the fluid in the jacket &lt;span class=&quot;math-container&quot;&gt;$\dot{m}\tilde{C}_p$&lt;/span&gt; to the overall coefficient and area of the reactor to jacket system &lt;span class=&quot;math-container&quot;&gt;$U_r A_r$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Open to Air&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The complexity of the analysis for the above system is increased when the jacket is not perfectly insulated.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Estimate Analysis&lt;/h1&gt;&#xA;&#xA;&lt;h2&gt;Perfectly Insulated System&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;We can estimate the system with one assumption. Assume the temperature profile of the jacket does not change with time. This is a gross assumption that has serious limitations on the accuracy. We can use it to first approximation to determine the orders of magnitude that parameters have to influence the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To first approximation, the energy balance on the well-insulated system can be written as below to solve for the temperature in the reactor &lt;span class=&quot;math-container&quot;&gt;$T_r$&lt;/span&gt; as a function of time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sum m \tilde{C}_p \frac{dT_r}{dt} = U_r A_r (T_r - &amp;lt;T&amp;gt;_j) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\tilde{C}_p$&lt;/span&gt; are the component masses and specific heat capacities in the reactor, &lt;span class=&quot;math-container&quot;&gt;$U_r$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$A_r$&lt;/span&gt; are the overall transfer coefficient and area of the reactor, and &lt;span class=&quot;math-container&quot;&gt;$&amp;lt;T&amp;gt;_j$&lt;/span&gt; is a constant, average temperature for the jacket over its length and over a representative time frame. This equation is immediately solvable with the appropriate boundary condition &lt;span class=&quot;math-container&quot;&gt;$T_r(t=0)$&lt;/span&gt;. Be aware that this answer may predict that &lt;span class=&quot;math-container&quot;&gt;$T_r$&lt;/span&gt; goes to a nonsensical result as &lt;span class=&quot;math-container&quot;&gt;$t \rightarrow \infty$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Open to Air&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;To first approximation, the energy balance on the un-insulated system can be written as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sum m \tilde{C}_p \frac{dT_r}{dt} = U_r A_r (T_r - &amp;lt;T&amp;gt;_j) + U_j A_j (&amp;lt;T&amp;gt;_j - T_a)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, &lt;span class=&quot;math-container&quot;&gt;$U_j$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$A_j$&lt;/span&gt; are the jacket overall coefficient and area exposed to air while &lt;span class=&quot;math-container&quot;&gt;$T_a$&lt;/span&gt; is the (constant) air temperature. Here again, the equation can be solved immediately with the appropriate boundary condition &lt;span class=&quot;math-container&quot;&gt;$T_r(t=0)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Making Predictions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Both solutions to the above are exponentials with time. They differ by a factor on the order of &lt;span class=&quot;math-container&quot;&gt;$\alpha \equiv U_jA_j(&amp;lt;T&amp;gt;_j - T_a)/\sum m \tilde{C}_p(T_r(t=0) - &amp;lt;T&amp;gt;_j)$&lt;/span&gt;. The perfectly insulated system has this factor as ZERO. Increasing values of &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; go away from perfect insulation. The system open to air has a defined value for this factor. An insightful approach is to plot the dimensionless temperature &lt;span class=&quot;math-container&quot;&gt;$\Theta \equiv (T_r - &amp;lt;T&amp;gt;_j)/(T_r(t=0) - &amp;lt;T&amp;gt;_j)$&lt;/span&gt; as a function of time &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; for variations in the ratio &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;. You can then make slices horizontally (at constant &lt;span class=&quot;math-container&quot;&gt;$\Theta$&lt;/span&gt;) for given values of &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; to estimate the effect of improving the insulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best approach might be akin to either of these:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I &lt;strong&gt;need&lt;/strong&gt; at least a 20% improvement on time to heat or cool to make the use of insulation worthwhile.&lt;/li&gt;&#xA;&lt;li&gt;The equations predict that I will gain only a 10% improvement on time.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt; I will make no changes&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I &lt;strong&gt;need&lt;/strong&gt; at least a 20% improvement on time to heat or cool to make the use of insulation worthwhile.&lt;/li&gt;&#xA;&lt;li&gt;The equations predict that I may gain a 25% improvement on time.&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\Rightarrow$&lt;/span&gt; I will plan further experimental bench top studies to validate the approximations&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I cannot immediately see a shortcut to estimate whether this first estimate will be consistently too low or too high compared to a &quot;real&quot; model. At that level, the project is better handed to a (chemical) engineer to tackle in a dedicated manner (rather than as a forum discussion).&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-05-16T20:13:35.757" LastActivityDate="2019-05-16T20:13:35.757" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28401" PostTypeId="1" AcceptedAnswerId="28406" CreationDate="2019-05-16T17:17:40.910" Score="2" ViewCount="368" Body="&lt;p&gt;I'm looking for a lightweight material that will hold its shape when stretched out. Something like a rubber band that doesn't snap back. So it starts out small, but once it's stretched it stays large. (Apologies if there's a name for such a property that I don't know.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="20488" LastActivityDate="2019-05-18T13:18:35.443" Title="Material that holds shape when stretched" Tags="|materials|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28402" PostTypeId="2" ParentId="28389" CreationDate="2019-05-16T18:22:56.370" Score="1" Body="&lt;p&gt;The solution you have suggested will not be waterproof for long. The reason is that, if you pull on the cable it will stretch and thin very slightly. This is enough to break the seal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can make the space, a cable gland is the best option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't, and O-ring seal is probably the next best thing. Make sure to use a thick o-ring with some compression so the seal will be maintained even if the cable stretches slightly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is to use a rubber &quot;plug&quot; where the cable passes through the center. This can be water tight, but you will have to take great care to make sure the plug is retained and remains under compression with regards to the seal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do a similar thing using potting compound, but it's not trivial. A rigid epoxy will have the same issues as the press fit you have suggested, so you need something semi-flexible. I don't know enough to recommend a product unfortunately.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-05-16T18:22:56.370" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28403" PostTypeId="1" CreationDate="2019-05-16T18:37:26.873" Score="1" ViewCount="94" Body="&lt;p&gt;I have a reaction chamber where gasses are injected to maintain a certain mixture. The container is sort of like a bell, where the bottom is submerged in water to make a seal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue I have is that when injecting new gas into the chamber the excess has nowhere to go, so the water level is forced downward. I would like to maintain the water level difference inside and outside the chamber of &amp;lt;0.1in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This rules out most kinds of check valves, because the pressure differential is too low. What I'm currently doing, is intentionally leaving a small hole open in the reaction chamber to allow pressure to equalize. Gas is injected at a very slow rate, so this works ok, but unwanted diffusion is always occurring through the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of a mechanism that will allow the pressure to equalize, but which will not allow significant gas exchange once the pressure is equal? I'm thinking something similar to a water trap, but with even lower pressure drop.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-05-17T08:38:53.263" Title="What's a good way to equalize pressure between 2 spaces without allowing gas exchange under normal circumstances?" Tags="|fluid-mechanics|pressure|chemistry|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28404" PostTypeId="1" AcceptedAnswerId="28418" CreationDate="2019-05-16T22:06:10.890" Score="4" ViewCount="825" Body="&lt;p&gt;As I've progress through my undergraduate engineering program, I've noticed a shift away from physical intuition - understanding why certain phenomena or equations make sense because of basic patterns we are already familiar with - to mathematical intuition - showing why equations and their phenomena should occur by mathematical derivation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this I came across today in aerodynamics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Isentropic Flow: &lt;span class=&quot;math-container&quot;&gt;$\frac{p_0}{p} = (1 + \frac{\gamma-1}{2}M^2)^{\gamma/(\gamma-1)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can usually follow the derivation of the equation, identify the variables in it, and apply it to appropriate cases. But, I cannot describe why this equation is justified by simply looking at it - I must follow the mathematical derivation. There is no way I'd be able to describe this equation to a layperson in a way that could satisfy them. I can't come up with an explanation that satisfies myself; I want to know why &lt;span class=&quot;math-container&quot;&gt;$1+$&lt;/span&gt; is there, why &lt;span class=&quot;math-container&quot;&gt;$\gamma- 1$&lt;/span&gt; should be divided by &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;, and why the Mach number should be squared, without solely relying on the mathematical derivation. I have this problem with many  formulas I have learned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know a fundamental part of engineering is understanding the equations and phenomena we use to to describe physical systems. Now, it seems like I should just trust the equations and make sure I'm not using them improperly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I, as an engineer-in-training hoping to complete research, focus on trying to understand equations to my satisfaction, or should I instead just become well acquainted the equations, their use cases, and their general behavior?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I acknowledge questions on this site should have answers and not just be discussed. I hope this question has an answer that has some consensus among experienced engineers. &lt;/p&gt;&#xA;" OwnerUserId="13453" LastActivityDate="2020-12-10T18:08:49.640" Title="Should Engineers Understand Equations without Relying on Derivations?" Tags="|aerospace-engineering|mathematics|aerodynamics|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28405" PostTypeId="1" CreationDate="2019-05-16T22:19:51.087" Score="2" ViewCount="74" Body="&lt;p&gt;In this photo, what does &lt;code&gt;'LBLMY&lt;/code&gt; signify? Seeing it directly under the capacity, and 5000 pounds lower, I'm guessing it's a &quot;limited capacity&quot; of some sort. But I can't come up with anything that would abbreviate to &lt;code&gt;'LBLMY&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only things I've found while searching for &quot;lblmy&quot; look like OCR errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QeRdJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QeRdJ.jpg&quot; alt=&quot;Railcar label&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20492" LastActivityDate="2019-05-16T22:19:51.087" Title="What does &quot;'LBLMY&quot; mean on the side of a railcar?" Tags="|rail|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28406" PostTypeId="2" ParentId="28401" CreationDate="2019-05-16T22:41:55.667" Score="4" Body="&lt;p&gt;You're looking for a material that has &lt;a href=&quot;https://en.wikipedia.org/wiki/Deformation_(engineering)#Plastic_deformation&quot; rel=&quot;nofollow noreferrer&quot;&gt;plastic deformation&lt;/a&gt; properties and is highly &lt;a href=&quot;https://en.wikipedia.org/wiki/Ductility&quot; rel=&quot;nofollow noreferrer&quot;&gt;ductile&lt;/a&gt;. To achieve your desired material behavior, you'll want a polymer like &lt;a href=&quot;https://en.wikipedia.org/wiki/Parafilm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Parafilm&lt;/a&gt;, which you can &lt;a href=&quot;https://smile.amazon.com/Parafilm-PM-996-Purpose-Laboratory-Film/dp/B004VQP6CQ/ref=sr_1_1?keywords=parafilm&amp;amp;qid=1558046487&amp;amp;s=gateway&amp;amp;sr=8-1&quot; rel=&quot;nofollow noreferrer&quot;&gt;buy online&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: More generally, you should look at different thermoplastics and find one that meets your needs. Available material properties at room temperature range from extreme malleability/ductility (“floppy” plastic like Parafilm) to relative stiffness and structural integrity (you need to heat these materials, although this can be easily achieved with readily available hot water in examples like &lt;a href=&quot;https://www.youtube.com/watch?v=Kp8tduAkbus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ingo&lt;/a&gt;). &lt;/p&gt;&#xA;" OwnerUserId="16783" LastEditorUserId="16783" LastEditDate="2019-05-18T02:35:36.177" LastActivityDate="2019-05-18T02:35:36.177" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28407" PostTypeId="2" ParentId="28404" CreationDate="2019-05-17T00:15:00.983" Score="2" Body="&lt;p&gt;As with a lot of things in engineering (and life in general), you have to weight the cost vs benefit associated with the application. Sometimes it’s enough to treat the equation like a black box, needing only to know the input output behavior.  Other times, knowing the how to derive equations enables you to derive new things from new assumptions.  You won’t memorize how to derive every equation that you’ll end up using only because there just too much to remember. But having and actively practicing analytical derivations will help you to read research papers in case you need to.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How deep you need to go in understanding each equation is really a function of how much you’ll need it in your application. &lt;/p&gt;&#xA;" OwnerUserId="191" LastActivityDate="2019-05-17T00:15:00.983" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28408" PostTypeId="2" ParentId="28404" CreationDate="2019-05-17T04:26:01.770" Score="5" Body="&lt;p&gt;Knowing the derivation is important because it usually tells you what initial assumptions were made in the derivation and what the limits of applicability of the resulting equation are. Understanding both of these things is an essential skill all engineers must master. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-05-17T04:26:01.770" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28409" PostTypeId="2" ParentId="28372" CreationDate="2019-05-17T04:26:01.880" Score="1" Body="&lt;p&gt;100 psi in the buffer tank doesn't give you a lot of headroom. The closer that is to your 90psi output pressure, the larger the tank will have to be. You'll have to calculate how low the pressure will drop during the output burst. Make sure the minimum pressure is higher than output + regulator pressure drop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also just at a glance, placing the pump in the middle of all this seems a little suspect. I would be worried about debris from the pump, although to be fair they would probably be caught by the expansion tank and oil filter. Why not place it at the beginning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that the check valve will not do anything after start up. And you want to limit the inrush airflow anyway right? In steady state the buffer tank will be fluctuating between 100 and 90 something PSI, so why include an extra part that can fail?&lt;/p&gt;&#xA;" OwnerUserId="20490" LastEditorUserId="20490" LastEditDate="2019-06-20T04:07:30.173" LastActivityDate="2019-06-20T04:07:30.173" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28410" PostTypeId="2" ParentId="28396" CreationDate="2019-05-17T04:38:02.383" Score="0" Body="&lt;p&gt;I think you're getting confused because you're trying to combine the gyroscopes and encoders into one data source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easier way would be to calculate an angular deflection in each axis using your gyros. Now lets call wherever the telescope is pointing &quot;zero&quot;. The gyroscope provides a target angle that is necessary to maintain &quot;zero&quot;, and the motors and encoders will use PID to try to stay at that angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the stars are far away, so you can probably just pretend your gyro and telescope exist at a single point.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-05-17T04:38:02.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28411" PostTypeId="2" ParentId="17552" CreationDate="2019-05-17T05:11:50.623" Score="1" Body="&lt;p&gt;It will be hard to ensure that all the wheels have equal traction. There will be no suspension, so your boost wheels may simply slip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds like what you're trying to do is have 2 different gear ratios. This is a cool idea, but if they are both active simultaneously some weird things are going to happen. Probably, the high gear will limit the speed at which the low gear can unwind so low gear will not do anything. You include some slack in the high gear to begin with, but this will get complicated an finicky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An easier way would be modify the arm, so it &quot;gets longer&quot; during the run due to a curved profile or other design feature. See terrible drawing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gyMQT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gyMQT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-05-17T05:11:50.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28413" PostTypeId="2" ParentId="15802" CreationDate="2019-05-17T06:23:10.410" Score="1" Body="&lt;p&gt;The main material property that determines the bursting strength of a pressure tank is the tensile strength (the stress at the yield point) of that material. Knowing the tensile strength allows you to specify the required wall thickness to withstand the maximum design pressure inside the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the amount of energy stored in a pressure tank is large, the amount of damage that can be done by a tank when it bursts is very great. For this reason, pressure tanks are designed to have large factors of safety and in commercial settings their design is highly regulated by safety codes that are established by (for example) the American Society For Testing And Materials, or ASTM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This in turn means that pressure tank design is not a DIY enterprise. Your legal exposure is severe if a pressure tank you designed from scratch blew up and was not ASTM compliant; in fact, your insurance company will not write you a liability policy unless your pressure vessels are provably built &quot;to code&quot; or purchased from an ASTM-compliant manufacturer.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-05-17T06:23:10.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28414" PostTypeId="2" ParentId="28403" CreationDate="2019-05-17T08:38:53.263" Score="2" Body="&lt;p&gt;There are specialty sensors and relays that are sensitive enough for the range of pressure you need, but they may be expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;one way is to hang the container in balance with a ballast weight to a gas valve like a flush mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a sketch which is basically you container hung in balance with a submerged&#xA;Wider flange to take in more volume with little change in pressure and higher leverage, to control a small gas valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d2HvQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d2HvQ.jpg&quot; alt=&quot;gas flush valve mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-17T08:38:53.263" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28415" PostTypeId="2" ParentId="28391" CreationDate="2019-05-17T12:54:03.487" Score="0" Body="&lt;p&gt;The statement of the question, using &quot;internal/external stresses&quot; is confusing. I don't understand what &quot;external stress&quot; can be here. You can have external forces but not external stresses.  From the question elaboration, it appears to ask whether the internal state of stress affects vibration modes, yet the questioner's example of a &quot;tension line&quot; cannot change the internal state of static stress without altering the boundary conditions on the beam.  Altering the boundary conditions will certainly affect vibration modes.  In addition, an &quot;internal state of stress&quot; always exists in a material, whether undergoing vibration or not.  It seems that the questioner is asking whether the &quot;static&quot; state of stress affects the modes, and static stress are internal stresses of motionless material.  I suggest the questioner clear up these issues, and I'll give the answer to the simple question which I believe he really wants to ask: &quot;Does the static internal state of stress affect vibration modes of a beam&quot;? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to that question is No, as long as the elastic limit of stress in the material is not exceeded.  A different static state of stress can result in total stresses during vibration that exceed the elastic limit, which may affect the vibration modes.  But the answer to that would be too complicated to discuss here.  For purposes here, the simple answer is &quot;No.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Explanation is that, first of all, vibration of a taught string is a different phenomenon, where the restoring (spring) force is the tension itself.  With the beam, called a &quot;bar&quot; in vibration science, the restoring force is the elastic force, and even more, it's the change of elastic force, not the total (change + static), caused by the vibration. Static forces don't enter into the equations of motion.  If you change the static situation, the dynamic changes of internal stress only occur about a changed mean value, and as long as the response is elastic (obeys Hook's law), the dynamic solution will not change. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, we know very well that the static state of internal stress normally has no effect on vibration modes  Most all hardened metals have some nonzero internal static stresses, and we are not concerned with them when we calculate vibration modes in such materials. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, with 1095 Swedish spring steel, there can be residual internal stresses in high temperature quenching during the hardening process that are not eliminated by sufficient uniform annealing (tempering), and that doesn't affect the vibration of a cantilever made of such material.  Such a device is used as sound source in free reed musical instruments, and makers aren't directly concerned with the internal state of stress in the reeds they make.  Now, there are complications when you consider other practical effects, such as metal fatigue and breakage.  Fatigue can occur over many cycles when stress levels are even below the elastic limit, a static concept.  There is also an &quot;endurance limit,&quot; which is below the elastic limit, above which fatigue lifetime is greatly reduced.&lt;/p&gt;&#xA;" OwnerUserId="14623" LastEditorUserId="14623" LastEditDate="2019-05-19T19:36:38.393" LastActivityDate="2019-05-19T19:36:38.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28416" PostTypeId="5" CreationDate="2019-05-17T14:50:17.257" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-05-17T14:50:17.257" LastActivityDate="2019-05-17T14:50:17.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28417" PostTypeId="4" CreationDate="2019-05-17T14:50:17.257" Score="0" Body="The study of the occurrence, distribution, movement, and properties of the waters of the earth an their relationship with the environment within each phase of the hydrologic cycle." OwnerUserId="9504" LastEditorUserId="9504" LastEditDate="2019-05-17T15:47:20.857" LastActivityDate="2019-05-17T15:47:20.857" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28418" PostTypeId="2" ParentId="28404" CreationDate="2019-05-17T15:06:14.173" Score="4" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Should I, as an engineer-in-training hoping to complete research, focus on trying to understand equations to my satisfaction, or should I instead just become well acquainted the equations, their use cases, and their general behavior?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Engineers apply equations to predict behavior. The critical requirement is to know the constraints of the equation. The constraints are either from first principles or from practical applications. The application demand is to appreciate the trends that arise from the equation. This arises in cases where we need &quot;seat of the pants&quot; estimates or &quot;back of the envelope&quot; calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the ideal gas law. From practice, it has general constraints to be at low pressure and high temperature relative to the critical point of the gas. From first principles, it has specific constraints for molecules to have little to no molecular interactions and to be small spheres. The trends it predicts are encapsulated in such statements as the Boyles or Charles laws.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider Bernoulli's equation. From its fundamental derivation, it has constraints to be applied only for incompressible fluids under inviscid flow. The trends it predicts are encapsulated in statements that say as the height of the discharge increases, the pressure at the outlet decreases in proportion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What then should you do when you confront an equation for the first time? As an engineer who wants to make accurate predictions, you must not fail to learn all of the constraints of that equation. Correspondingly, as an engineer who wants to move efficiently through a practice of problem solving, you must not fail to deconstruct the various trends of that equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you go about the first requirement? You study the roots of the equation. It may arise from empirical analysis. A case in point is the derivation of the ideal gas law from the Avagadro, Boyles, and Charles laws. Alternatively, it may arise from first principles. A case in point is the derivation of the ideal gas law from kinetic theory of gases. Each approach will provide insights to the practical limits or conceptual assumptions behind the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you go about the second requirement? You take the equation and do a thorough unit analysis. You also take the equation and (literally if not figuratively) &lt;em&gt;plot the trends&lt;/em&gt; it should predict. In essence, as an engineer in training, you have tools that transcend the hand-held calculator (that was the revolution from the slide-rule of my days). Go plot the left side of the equation as a function of the various parameters it contains!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, when your goal is to be know for accuracy and efficiency in your engineering practice, you cannot just focus on trying to understand an equation &lt;em&gt;to your satisfaction&lt;/em&gt;. This nebulous, subjective statement will fail you in the end. You must become well-acquainted with the equation and its trends. This goal, and the habit it demands of you, will stand you well through your coming career.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-05-17T15:06:14.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28419" PostTypeId="2" ParentId="26133" CreationDate="2019-05-17T15:08:34.820" Score="1" Body="&lt;p&gt;The picture you've provided includes the answer that you require, the water surface elevation is the elevation of the water as depicted on the upstream portion of the culvert.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zJBN8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zJBN8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, the headwater elevation above the culvert pipe is providing a certain amount of pressure head that is contributing to pushing water through the pipe (in addition to gravity and velocity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, the situation depicted in that image is a pipe in a pressure flow situation (as opposed to an open channel flow situation). The Normal Water Surface Elevation is what calculations would indicate the surface elevation should be when the system is analyzed without considering the downstream impacts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's not depicted is what those downstream impacts are. For example, they could be a blockage or other constriction in the pipe or perhaps a tailwater elevation, which is when the elevation of the water surface at the point of discharge. Were the pipe to discharge partially or fully below the surface of a wet pond, that tailwater induces a back pressure on the system which affects the hydraulic grade line profile across the pipe network.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are dealing with a pipe network, it is extremely important to be cognizant about when and what causes your system to go into pressure flow conditions. Generally, though, if your observations or calculations indicate a headwater at some point, you've entered a pressure flow scenario for that pipe and it is very important to review the hydraulic grade line profile through the network to ensure you don't have water coming out of your inlets.&lt;/p&gt;&#xA;" OwnerUserId="9504" LastEditorUserId="9504" LastEditDate="2019-05-17T15:55:07.663" LastActivityDate="2019-05-17T15:55:07.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28420" PostTypeId="2" ParentId="21297" CreationDate="2019-05-17T16:31:03.413" Score="0" Body="&lt;p&gt;As stated in the comments, based on your pictures, you don't need a siphon to start that flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your influent line should have a cutoff valve to stop flow and the lower tank will also need an air bleed off valve so that the air can exit the tank as it's filling, but also so that you can introduce air back into the system when emptying the tank. If you don't permit air to exit and enter the system will stall either due to a vacuum or excessive air pressure. Vacuums and excessive air pressure are bad for the longevity of tanks not designed specifically for them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Controls for the air bleed off valve can be tied to floats in the tank. As water enters the tank from the upper reservoir, the float will rise and eventually close the air release valve; once this closes, water from the upper reservoir will stop flowing down because the air pressure won't permit it to continue into the tank. At that point, manual valves can be utilized to close the influent line. With the influent line closed, you can manually open the air release valve which will permit air back into the tank and then permit the lower tank to drain. When it's time to refill the tank, simply open the valve on the influent line and let the tank fill until the float closes the air valve again.&lt;/p&gt;&#xA;" OwnerUserId="9504" LastActivityDate="2019-05-17T16:31:03.413" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28421" PostTypeId="1" AcceptedAnswerId="29600" CreationDate="2019-05-17T17:13:47.563" Score="2" ViewCount="2915" Body="&lt;p&gt;I have a second order system and I need to control it using a PI controller. I need to tune the gains of the controller in order for the system to satisfy the below specifications:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ OS\% &amp;lt; 10\% \rightarrow Overshoot $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ t_r &amp;lt; 1.2 \ seconds \rightarrow Rise \ Time $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have derived the third order transfer function of the closed loop system with the controller and I am not able to understand which characteristic polynomial I have to use in order to achieve the specified requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transfer function of the plant is:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ G(s) = \frac{10}{(s+1)(s+9)} $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transfer function of PI controller is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ C(s)=\frac{K_p(s+c)}{s} $$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\ c = \frac{K_i}{K_p} $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed loop transfer function (linear unity feedback) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T(s) = \frac{10K_p(s+c)}{s(s+1)(s+9) + 10K_p(s+c)} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now how should I proceed in order to compute the gains &lt;span class=&quot;math-container&quot;&gt;$\ K_p $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ K_i $&lt;/span&gt; of the controller in order to satisfy the given specifications ? &lt;/p&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-03-08T01:53:39.420" LastActivityDate="2020-03-08T01:53:39.420" Title="PI controller for second order system" Tags="|control-engineering|control-theory|pid-control|transfer-function|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28423" PostTypeId="1" CreationDate="2019-05-17T19:26:20.797" Score="1" ViewCount="137" Body="&lt;p&gt;I have a Couette-Poiseuille flow (both flows combined) where the plates are not parallel. So h, the thickness of the flow, changes along the x axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I use a mean h to calculate the viscid stress? Is this procedure correct? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$τ_v=-μ_a  v/h+|∆p_L |h/2L$&lt;/span&gt; (I know the Poiseuille equation for parallel plates includes the y coordinate, but I substituted for h, the height I'm interested in to calculate the stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$τ_v=-1/(h_1-h_2 ) ∫〖μ_a  v/h〗 dh+1/(h_1-h_2 ) ∫|∆p_L |h/2Ldh$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The integration limits are &lt;span class=&quot;math-container&quot;&gt;$h_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$h_2$&lt;/span&gt; and the former is greater than the latter. Both are the thickness limits and I'm still too new to this forum, so I can't put them where they should be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the result I got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$τ_v=-μ_a  v/(h_1-h_2 )  ln⁡(h_1/h_2 )+|∆p_L |(h_1+h_2)/4L$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right or wrong?&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-05-17T19:26:20.797" Title="Couette-Poiseuille flow between non-parallel plates" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28424" PostTypeId="1" CreationDate="2019-05-17T19:34:13.337" Score="1" ViewCount="1181" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k3hNx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k3hNx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can provide me information about this part right here, please tell me.&lt;/p&gt;&#xA;" OwnerUserId="20506" LastEditorUserId="33" LastEditDate="2019-05-22T13:22:41.367" LastActivityDate="2021-03-10T19:23:48.903" Title="Jet Boat Engine Part Identification" Tags="|materials|bearings|machine-elements|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28425" PostTypeId="2" ParentId="28424" CreationDate="2019-05-17T20:22:31.550" Score="7" Body="&lt;p&gt;It is a high pressure seal and bearing unit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The metal assembly to the right with the yellow parts is the drive coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After Sam’s comment and looking at that part again, there are 3 distinct metal parts - the middle one with the holes is aluminium, while the outer two are the outer races of the two bearings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two bearings are often used due to either the load, controlling end float or controlling play or vibration in the shaft.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-05-17T20:44:55.980" LastActivityDate="2019-05-17T20:44:55.980" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28426" PostTypeId="2" ParentId="28424" CreationDate="2019-05-17T21:18:04.150" Score="-1" Body="&lt;p&gt;It's a boat jet engine. And as Solar Mike's answer says, it is a bearing and seal. I guess it is a brush less electrical motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to an eBay ad. &lt;a href=&quot;https://www.ebay.com/i/282171897630?chn=ps&amp;amp;mkevt=1&amp;amp;mkrid=711-117182-37290-0&amp;amp;mkcid=2&amp;amp;itemid=282171897630&amp;amp;targetid=541454106452&amp;amp;device=m&amp;amp;adtype=pla&amp;amp;googleloc=9031022&amp;amp;poi=&amp;amp;campaignid=1669934873&amp;amp;adgroupid=65058351339&amp;amp;rlsatarget=pla-541454106452&amp;amp;abcId=1139336&amp;amp;merchantid=113679796&amp;amp;gclid=Cj0KCQjwt_nmBRD0ARIsAJYs6o3EJLz2Ff_hVHIE0sz2awrof8O1eLN__qlMJ4CyMAfESd_X6F5OoUoaAstQEALw_wcB&quot; rel=&quot;nofollow noreferrer&quot;&gt;ElectricalJet motor &lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-17T21:18:04.150" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28427" PostTypeId="1" CreationDate="2019-05-17T22:01:01.563" Score="2" ViewCount="276" Body="&lt;p&gt;At work, we cut a 6&quot; hole in the wall for a duct. Later we decided we now want a 10 inch duct in the same place. what tricks are there to center the larger hole over the center of the smaller hole if it is already cut? I'm using a hand saw to cut into dry wall.&lt;/p&gt;&#xA;" OwnerUserId="8974" LastActivityDate="2019-05-19T17:58:53.237" Title="Cutting a larger circle on center of a smaller circle" Tags="|measurements|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28428" PostTypeId="2" ParentId="28427" CreationDate="2019-05-17T22:10:35.203" Score="1" Body="&lt;p&gt;Use a couple of steel straps or bars with a hole at the centre clamped or screwed and re-mark and drill...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-05-17T22:10:35.203" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28429" PostTypeId="2" ParentId="28427" CreationDate="2019-05-17T22:18:59.997" Score="3" Body="&lt;p&gt;You can take a scrap piece of wood (maybe a scrap piece of drywall might work) and use the 10&quot; hole saw to make a hole in the scrap piece (this will be your guide). Then center the guide onto your wall and use the 10&quot; hole as a guide for the hole saw to keep it centered. It might be easier with two people, one holding the guide and the other using the drill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can see what I'm trying to say at around the 4:30 mark of&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=EXhDlDBzXjY&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="20468" LastEditorUserId="17612" LastEditDate="2019-05-18T22:41:39.163" LastActivityDate="2019-05-18T22:41:39.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28430" PostTypeId="1" CreationDate="2019-05-17T22:59:04.140" Score="1" ViewCount="164" Body="&lt;p&gt;I'm sorry if this is not the right site but I really don't see another more suitable on Stack Exchange. Some Soviet tanks have these canister-looking things that hang down off the sides of the hull, very near the wheels. Some of them appear to be tilted. Here's a photo of them ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fGUeZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fGUeZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, here is a 3D model of an IS-3 so you can get a really good view of these. The ones that are tilted on this one are different. They appear to be different on each one:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://sketchfab.com/3d-models/tank-is-3-5e1df63a21074eee9343618686a971bd&quot; rel=&quot;nofollow noreferrer&quot;&gt;3D Model&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another 3D model where you can see these. Again these are different ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://3dwarehouse.sketchup.com/model/ebd0485d-ba6f-41f1-8c07-a3438c0b37fc/IS-3&quot; rel=&quot;nofollow noreferrer&quot;&gt;3D Model&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these things just become loose and end up twisting? What do you think they could be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if you look at the first photo, on many of these tanks the left side of the suspension and wheels are set back slightly from the right side. It's quite subtle, and strange. Does anyone know why this is? This offset I'm speaking of is really subtle. The first model has it, the second model doesn't. The first one being obviously more accurate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Someone has suggested it looks like a shock absorber, here is a screenshot from a good position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qCXmB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qCXmB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, someone pointed out that in the first model I linked some of these are missing above some wheels, about 3 or four. In this picture you can also see some missing. I'm really confused about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6C5Mf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6C5Mf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20508" LastEditorUserId="20508" LastEditDate="2019-05-18T17:06:59.410" LastActivityDate="2019-12-22T14:54:13.250" Title="What are these parts on these Soviet tanks?" Tags="|mechanical-engineering|mechanisms|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28431" PostTypeId="1" CreationDate="2019-05-17T23:09:02.287" Score="1" ViewCount="89" Body="&lt;p&gt;I have modelled a multi-tubular reactor performing a Fischer-Tropsch synthesis of alkanes and I was wondering about the stability of such a reactor. Even a very tiny deviation of the wall temperature can lead to a potential destruction of the catalyst if not more... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Fischer-Tropsch synthesis I am modelling as an adiabatic temperature around 1200 K. This is the result I get when the wall temperature varies from 201.000 °C to 201.470 °C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/m9cg3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m9cg3.png&quot; alt=&quot;Wall temperature Fischer Tropsch synthesis&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So theoretically a variation of less than 0.5 K can lead to a reactor temperature variation of more than 150 K.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;How to control the temperature along the reactor (it can be more than 10 meter long) as precisely as let 0.1 K say of difference to avoid such unpleasant drift leading to a potentially bad situation? &lt;/p&gt;&#xA;" OwnerUserId="19799" LastActivityDate="2020-02-13T21:02:20.237" Title="How to control precisely the wall temperature of a tubular reactor?" Tags="|heat-transfer|chemical-engineering|heat-exchanger|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28432" PostTypeId="2" ParentId="28430" CreationDate="2019-05-17T23:50:08.203" Score="0" Body="&lt;p&gt;Those things appear to be shock absorbers, to cushion the up-and-down action of the idler rollers. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-05-17T23:50:08.203" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28434" PostTypeId="2" ParentId="28430" CreationDate="2019-05-18T04:30:22.247" Score="3" Body="&lt;p&gt;I don't know if there's a tank-specific term, but they look like travel limiters: the idler wheels would hit them when they bottom out.  In your picture, the wider piece that would make contact with the swing arm is probably something softer that would give just a bit.  Not a full-fledged shock absorber, but it would absorb some of the shock.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first 3D model only shows them on some of the wheels, but your picture and the second 3D model have them on all of the wheels. &lt;/p&gt;&#xA;" OwnerUserId="3445" LastActivityDate="2019-05-18T04:30:22.247" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28435" PostTypeId="1" CreationDate="2019-05-18T07:26:04.763" Score="1" ViewCount="204" Body="&lt;p&gt;Goodmorning,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to understand how 2D solid elements work. I want to study the local buckling of a simple aluminium pipe with thin walls.&#xA;From what I understood, I can use 2D solid elements and model just a section of this pipe .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've drawn a surface on the XZ plane (Z axis of simmetry). This surface is just a long but very thin rectangle that rapresents the pipe's wall. The pipe is open so no caps etc.&#xA;I applied the property 2d solid axisimmetric, my costrains and my loads. I then created my mesh with quadx elements.&#xA;In analysis, I selected SOL105 for bucking analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I run Nastran, I get some errors, the first one being:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;USER FATAL MESSAGE 2011 (TA1)&#xA;NO PROPERTY ENTRY FOR ELEMENT TYPE QUADX&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Indeed, checking the bdf file, there's no property to be found, nor the material definition!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to group elements with no properties in Patran, but it found none.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for helping me out.&lt;/p&gt;&#xA;" OwnerUserId="20510" LastEditorUserId="20510" LastEditDate="2019-05-18T07:31:09.433" LastActivityDate="2019-05-18T11:12:12.987" Title="2D Solid - Axisimmetric - Missing properties and material" Tags="|structural-engineering|structural-analysis|buckling|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28436" PostTypeId="2" ParentId="28435" CreationDate="2019-05-18T11:12:12.987" Score="1" Body="&lt;p&gt;You almost certainly selected the wrong element type, and then didn't set up the special material data it needs. To quote from &lt;a href=&quot;https://docs.plm.automation.siemens.com/data_services/resources/nxnastran/10/help/en_US/tdocExt/pdf/element.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this online Nastran manual&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The CTRIAX and CQUADX elements ... are fully nonlinear elements supported in solution types&#xA;  SOL 106 and 601 &lt;strong&gt;but only support hyper-elastic material&lt;/strong&gt; defined on the&#xA;  MATHP bulk data entry.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(My emphasis). Aluminium is not a hyperelastic material. For a linear buckling analysis, the element type you probably wanted was the CQUADX4.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-05-18T11:12:12.987" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28437" PostTypeId="2" ParentId="28401" CreationDate="2019-05-18T13:18:35.443" Score="0" Body="&lt;p&gt;&lt;strong&gt;Ductile and Plastic Material for Making Parts? Suggest you try Polycaprolactone.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're looking for a highly ductile/purely plastic material to form into a shape -- one that's easy to work with at home, you might try:  &lt;a href=&quot;https://en.wikipedia.org/wiki/Polycaprolactone#Hobbyist_and_prototyping&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Polycaprolactone#Hobbyist_and_prototyping&lt;/a&gt; &#xA;granules, after warming in hot water. There are a host of brand-names all selling essentially the same granules: Polymorph, Thermomorph, and Friendly Plastic are just a few. Quite safe; medicos use this stuff to splint fingers sometimes.&lt;/p&gt;&#xA;" OwnerUserId="13014" LastActivityDate="2019-05-18T13:18:35.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28438" PostTypeId="1" AcceptedAnswerId="28441" CreationDate="2019-05-18T14:38:13.463" Score="1" ViewCount="50" Body="&lt;p&gt;I'm wondering how to compare the efficiencies of vertical and horizontal ground loops for heating. [Assumptions follow, please correct if any are wrong!]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest differences seem to be the higher cost of vertical loops vs the land required for horizontal loops, so therefore I'm having trouble finding any information on how to compare the energy input/output of similarly sized systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm interested comparing the two systems, so at the risk of getting off the question: in my area, average winter temp is a low of -6F/-20C, and a high of 12F/-11C, with cold snaps of -40deg. Frost gets to 70in/ 1.7m in the colder winters. If comparisons want to use that info, I would be very curious. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="20514" LastActivityDate="2019-05-18T18:56:04.613" Title="Vertical vs Horizontal Ground Loop Heating" Tags="|mechanical-engineering|heat-transfer|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28439" PostTypeId="1" CreationDate="2019-05-18T14:43:26.417" Score="0" ViewCount="139" Body="&lt;p&gt;I have some samples of readings from an accelerometer (unit: g) and I did a FFT (on MATLAB) on the vector generated from these samples to see its frequency spectrum. Then I have isolated the peaks and located its frequency.&#xA;OK, now the control theory question.&#xA;Let's say, to simplify the example, the &lt;strong&gt;only&lt;/strong&gt; peak is 2 g at the frequency of 10 Hz. The accelerometer acquired the samples over a vibration system, so I assume it is undamped oscillating endelessly at a natural frequency of 10 Hz. My transfer function would be:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F(s) = \dfrac{(10\cdot2\pi)^2}{s^2 + (10\cdot2\pi)^2} = \dfrac{3948}{s^2 + 3948}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;I multiplied by &lt;span class=&quot;math-container&quot;&gt;$2\pi$&lt;/span&gt; before raising because it's given in Hz and I have to convert it into rad/s (right?)&lt;/p&gt;&#xA;&lt;p&gt;I intend to make a PID control (on Simulink) to analyze some parameters, as overshoot, settling time, etc.&lt;/p&gt;&#xA;&lt;p&gt;The &lt;strong&gt;REAL&lt;/strong&gt; question is: how do I &amp;quot;tell&amp;quot; the system that the amplitude of it is 2 g?&#xA;Is it like a gain that I insert before the transfer function block?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZvgWv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZvgWv.png&quot; alt=&quot;PID control&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20513" LastEditorUserId="1832" LastEditDate="2021-01-03T16:38:57.517" LastActivityDate="2021-01-03T16:38:57.517" Title="How to input amplitude on a second order transfer function?" Tags="|pid-control|matlab|transfer-function|simulink|" AnswerCount="1" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28440" PostTypeId="2" ParentId="26919" CreationDate="2019-05-18T15:49:27.640" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;... could it be more effective to use this power to drive the wheels,&#xA;  rather than use it to spin a compresser as with a conventional&#xA;  turbocharger system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. The purpose of the e-machine on a turbocarger system is to make up any differences between the instantaneous torque demand of the compressor and the instananeous torque produced by the turbine. In order to have acceptable driveability and maximize performance, most turbochargers have a waste gate to manage torque mismatches which are designed to err on the side of excess turbine torque. With an e-motor adding or removing torque as needed, the turbocharger can be completely redesigned for better efficiency using simpler mechanicals and providing a wider operating envelope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the e-motor is there to service the turbocharger's need, not the other way around. You can build in a bias if you want, so that there is a net generator output, but this is extremely time variable, so all you can due is dump it into a battery storage system where it can be consumed as needed. This arrangement can't compete with the efficiency of crankshaft/tranny/shafting/wheels. A crankshaft generator would be a better way to generate electrical traction power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, the biggest problem with producing these systems is cooling them. Electrical motors don't like the sort of temps found next to the turbo. I expect the first couple generations of these that we will see will not be doing much extra generator-wise. That only adds to the heat problem.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-05-18T15:49:27.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28441" PostTypeId="2" ParentId="28438" CreationDate="2019-05-18T18:47:13.943" Score="1" Body="&lt;p&gt;Geothermal heating requires detailed gathering of several kind of data for planning and then you can judge which system is better for your particular case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of Those factors are&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Outdoor design temperature (coldest night of the winter usually published by the building authority), &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Soils and geology data (frost line and aggregate of the layers of soil at 10-20- 50 feet) &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tonnage of the heat pump, related to the size of your home and its thermal leakiness, etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The availability of land.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Vertical ground loops cost more but don't need a large area of land so they lend themselves more to urban areas, they may need high pressure plumbing and bigger pumps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are schematic diagrams of , horizontal loop&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vE1Rk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vE1Rk.jpg&quot; alt=&quot;horizontal loop&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and vertical loop, --- &lt;a href=&quot;https://i.stack.imgur.com/RRvPO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RRvPO.jpg&quot; alt=&quot;vertical loop&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are courses that teach how to estimate and design an efficient systems, International Ground Source Heat Pump Association has a course called, IGSHPA training. Here is the &lt;a href=&quot;https://igshpa.org/training/&quot; rel=&quot;nofollow noreferrer&quot;&gt;course&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are people who have installed their system after taking similar courses, but one needs to check with the manufacturer of the pumps to see what kind of a warranty restriction the may have on installer qualification.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-05-18T18:56:04.613" LastActivityDate="2019-05-18T18:56:04.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28442" PostTypeId="1" CreationDate="2019-05-19T01:40:02.463" Score="0" ViewCount="147" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mBQJu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mBQJu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to draw this in orthograhpic views. I'm a bit confused about the weight of lines I should use to draw. Like ,for the pat (I pointed the arrow at) you can see it's sticking out a bit, would I use just a regular weight line or what? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also. We've been given some dimensions like 30mm and 70mm. And the thing also says the drawings need to be done on a 1:1 scale. What does this mean? DOes this mean I need to draw exactly at 30mm and 70mm, not make it bigger ration? &lt;/p&gt;&#xA;" OwnerUserId="20518" LastActivityDate="2023-02-02T07:01:30.157" Title="Drawing in orthograhphic" Tags="|technical-drawing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28443" PostTypeId="2" ParentId="28424" CreationDate="2019-05-19T12:35:46.660" Score="1" Body="&lt;p&gt;Combined thrust bearing and shaft seal / stuffing box. Yes, it's really called that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nipple on top is to inject lubricant which has dual function of preventing water ingress to engine compartment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thrust unit forward loads cannot be transferred to drive shaft / engine without:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;damage&lt;/li&gt;&#xA;&lt;li&gt;or serious design compromise&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The thrust bearing transfers these loads to the housing which is of course secured to the hull &amp;amp; framing. This also improves thrust vector geometry.&lt;/p&gt;&#xA;" OwnerUserId="10415" LastActivityDate="2019-05-19T12:35:46.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28444" PostTypeId="2" ParentId="26964" CreationDate="2019-05-19T13:51:47.180" Score="1" Body="&lt;p&gt;Even if the foam &lt;em&gt;does&lt;/em&gt; absorb most of the impact, it's going to get damaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try a sheet of plywood on top of the foam.  Not only would it take the damage, it would distribute the energy over more of the foam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could even try multiple layers:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;— barbell&#xA;— plywood&#xA;— foam&#xA;— plywood&#xA;— foam&#xA;— floor&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You might be able to get away with relatively thin layers of foam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Though I suspect that eventually you'll end up damaging either the floor or yourself.)&lt;/p&gt;&#xA;" OwnerUserId="19930" LastActivityDate="2019-05-19T13:51:47.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28445" PostTypeId="1" CreationDate="2019-05-19T16:32:12.847" Score="1" ViewCount="80" Body="&lt;p&gt;I found this switchable coupling in Rollof/Matek machine elements, i have never seen this kind of coupling before. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zr0Ky.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zr0Ky.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the dutch edition its name is: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Tweevlakskoppeling&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Which literally means &lt;em&gt;Two-surface coupling&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some research on the web i couldn't find any reliable information, i thought the translation is wrong, then I checked the original edition in german, apparently, the name is correct. The original edition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8cgIA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8cgIA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone identify this coupling?   &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-06-18T18:02:45.270" Title="I need help identifying a special kind of coupling" Tags="|machine-design|machine-elements|coupling|" AnswerCount="1" CommentCount="3" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28446" PostTypeId="1" AcceptedAnswerId="28453" CreationDate="2019-05-19T16:35:29.067" Score="-1" ViewCount="76" Body="&lt;p&gt;how to know if a deflection in a supersonic isentropic flow causes compression (oblique shock) or expansion (Prandtl-Meyer expansion)&#xA;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.chegg.com/homework-help/air-flows-supersonically-surface-changes-direction-twice-fig-chapter-9-problem-146p-solution-9780073398273-exc&quot; rel=&quot;nofollow noreferrer&quot;&gt;an example of a problem that is related to my problem&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wR1my.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wR1my.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3134" LastEditorUserId="3134" LastEditDate="2019-05-19T19:53:06.743" LastActivityDate="2020-04-13T23:14:53.067" Title="Supersonic isentropic flow" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28447" PostTypeId="2" ParentId="28445" CreationDate="2019-05-19T17:38:48.813" Score="1" Body="&lt;p&gt;Look for overrunning clutch or power transmission elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HkSn9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HkSn9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EimNw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EimNw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find a lot of info in Ringspann site:&#xA;&lt;a href=&quot;https://www.ringspann.com/en/products/freewheels/complete-freewheels/for-bolting-to-the-face/complete-freewheels-fkh&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ringspann power transmission section&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or here:&#xA;&lt;a href=&quot;https://www.yumpu.com/en/document/view/25781671/power-transmission-ringspann&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ringspann catalogue&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2019-05-19T17:38:48.813" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28448" PostTypeId="2" ParentId="28427" CreationDate="2019-05-19T17:58:53.237" Score="1" Body="&lt;p&gt;If the first hole is neat, you can just take some calipers and set them to 2in, then scratch them around in a circle following the edge of the first hole and extending out radially. Not the most accurate method, but fast and will give decent results. &lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-05-19T17:58:53.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28449" PostTypeId="2" ParentId="28431" CreationDate="2019-05-19T18:07:23.830" Score="2" Body="&lt;p&gt;I've had this problem before, albeit not with such a tight temperature range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend immersing the reactor in a fluid with high thermal mass, then I would circulate and temperature control the fluid. The circulation will minimize the effects of temperature differences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is ideal, but dangerous since it would have to be under pressure, and it would explode into steam if there was a breach. Oil might work fine.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-05-19T18:07:23.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28451" PostTypeId="1" CreationDate="2019-05-19T23:21:15.740" Score="0" ViewCount="3420" Body="&lt;p&gt;The Integral Absolute value Error (IAE) and the Integral Squared Error (ISE) is to be analysed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A PID Controller is given as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{gather}&#xA;C(s)=K_p+K_i\frac{1}{s}+K_ds&#xA;\end{gather}&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$K_p=0.07847$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$K_i=0.03587$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K_d=0.04291$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transfer function of the controlled system is:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{gather}&#xA;G(s) = \frac{1}{s^2+s+1}&#xA;\end{gather}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as a feedback loop I have following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WosRm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WosRm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculating the overall Transfer function &lt;span class=&quot;math-container&quot;&gt;$F(s)$&lt;/span&gt; we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F(s) = \frac{0.04291s^2+0.07847s+0.03587}{s^3+1.043s^2+1.078s+0.03587} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The step response looks like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/05G6o.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/05G6o.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I wanted to find the Integral Absolute value Error (IAE) and the Integral Squared Error (ISE). By researching google I came to know the principle behind it and I also saw the formulas but how really to mathmetically solve it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hg9dR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hg9dR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example to solve for ISE for &lt;span class=&quot;math-container&quot;&gt;$W(s)$&lt;/span&gt; as step response in frequency domain:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{gather}&#xA;e(s)=w(s)-y(s) \\&#xA;\\&#xA;e(s) = w(s)-(F(s)w(s)) \\&#xA;\\&#xA;e(s) = \frac{1}{s} - \left(\frac{F(s)}{s}\right)&#xA;\end{gather}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I replace &lt;span class=&quot;math-container&quot;&gt;$F(s)$&lt;/span&gt; and solve for &lt;span class=&quot;math-container&quot;&gt;$e(s)$&lt;/span&gt; ??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that integrate from &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; to infinity to the &lt;span class=&quot;math-container&quot;&gt;$e(s)^2$&lt;/span&gt;?? &#xA;I don't get it. How am I supposed to get the tuned value for &lt;span class=&quot;math-container&quot;&gt;$K_p$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$K_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K_d$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Matlab code is &lt;a href=&quot;https://pastebin.com/LjhtWGh6&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="20525" LastEditorUserId="1832" LastEditDate="2020-05-10T02:34:08.853" LastActivityDate="2020-05-10T03:24:41.053" Title="How to find Performance assessment (IAE and ISE) of a PID Controlled system?" Tags="|control-engineering|control-theory|pid-control|matlab|transfer-function|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28452" PostTypeId="1" AcceptedAnswerId="28456" CreationDate="2019-05-19T23:59:55.497" Score="-1" ViewCount="7065" Body="&lt;p&gt;This &lt;a href=&quot;http://wentec.com/unipower/calculators/power_torque.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; shows a motor torque, power, and speed relationship, as shown below: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FbMrJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FbMrJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regards to its units, how does kW and RPM become equivalent to N.m?&lt;/p&gt;&#xA;" OwnerUserId="15712" LastActivityDate="2023-04-13T09:41:00.343" Title="Motor torque calculation, about the units" Tags="|motors|torque|power|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28453" PostTypeId="2" ParentId="28446" CreationDate="2019-05-20T00:22:56.753" Score="2" Body="&lt;p&gt;Really easy: Calculate the deflection angle beta. If the angle is positive, then it's an oblique shockwave, but if it's negative, then you have an expansion. If the angle beta is zero, then there's neither compression nor compression and if beta is 90º (π/2) you have a normalmente shockwave. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An expression that may help you (I did a report on this and demonstrated the formulæ):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3O8G2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3O8G2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorDisplayName="user20096" LastEditDate="2019-05-20T00:31:09.577" LastActivityDate="2019-05-20T00:31:09.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28455" PostTypeId="2" ParentId="28451" CreationDate="2019-05-20T00:40:49.077" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$e(s)=w(s)(1-F(s))$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know both F and w (w is the reference). Then calculate e(s), later e(t) (use the Laplace anti-transform) and integrate. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-05-20T00:40:49.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28456" PostTypeId="2" ParentId="28452" CreationDate="2019-05-20T00:57:36.450" Score="5" Body="&lt;p&gt;It's quite a simple calculation, especially if you're used to dealing with metric units:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Power=Torque \times angular Speed$&lt;/span&gt; (rotational equivalent of &lt;span class=&quot;math-container&quot;&gt;$force \times speed$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Power [kW]= Torque [Nm] \times Speed [\frac{rev}{min}] \times \frac{1}{60} [\frac{min}{seconds}] \times 2π [\frac{rad}{rev}] \times 10^{-3} [\frac{kW}{W}]$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;RPM= rev/min&lt;/p&gt;&#xA;&lt;p&gt;Finally you clear torque and get the result:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Torque [Nm]=\frac{(60 \times \frac{1000}{2π})\times Power[kW]}{ Speed [rpm]}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;And 60,000/2*π is actually 9548.8, not 9.5488, the equation online is wrong.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorUserId="-1" LastEditorDisplayName="user20096" LastEditDate="2023-04-13T09:41:00.343" LastActivityDate="2023-04-13T09:41:00.343" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28457" PostTypeId="1" AcceptedAnswerId="28470" CreationDate="2019-05-20T02:22:28.450" Score="1" ViewCount="1982" Body="&lt;p&gt;What's an authoritative definition of whether the idealized pinned support and pin joint can transfer moments in 3D?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most sources say both of them cannot transfer moments but are apparently only considering 2D. A real pin joint obviously can transfer moments about all axes normal to the pin's axis. So is an implementation of the ideal pin joint really more like a ball joint than a pin joint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume a pinned support is the same as a pin joint but with one end fixed to ground. But maybe they have different numbers of degrees of rotational freedom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if this support can carry moments about two perpendicular axes (red arrows), would it be correctly called a pinned support?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UlygO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UlygO.png&quot; alt=&quot;Pinned support&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In comparison, this one has all 3 rotational degrees of freedom, so is it still a pinned support?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jrNHn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jrNHn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4437" LastEditorUserId="4437" LastEditDate="2019-05-20T08:19:14.470" LastActivityDate="2019-05-20T18:20:28.570" Title="Can pinned support / pin joint transfer moments in 3D?" Tags="|mechanical-engineering|structures|connections|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="28460" PostTypeId="1" AcceptedAnswerId="60177" CreationDate="2019-05-20T07:10:31.147" Score="0" ViewCount="2451" Body="&lt;h3&gt;TL;DR&lt;/h3&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is it possible to calculate the mean diameter of a collar?&lt;/li&gt;&#xA;&lt;li&gt;Can I ignore the collar's component when calculating the raising torque?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;I have been trying to &lt;a href=&quot;https://3dprinting.stackexchange.com/questions/7695/z-axis-stepper-motor-and-lead-screw-torque-calculation/9827#9827&quot;&gt;answer&lt;/a&gt; this question, &lt;a href=&quot;https://3dprinting.stackexchange.com/questions/7695/z-axis-stepper-motor-and-lead-screw-torque-calculation&quot;&gt;Z axis: Stepper motor and lead screw torque calculation&lt;/a&gt;, over on the &lt;a href=&quot;https://3dprinting.stackexchange.com&quot;&gt;3D Printing&lt;/a&gt; site.&lt;/p&gt;&#xA;&lt;p&gt;I obtained an equation for the raising torque from&#xA;&lt;a href=&quot;https://ncalculators.com/mechanical/torque-required-raise-load-calculator.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Torque Required to Raise Load (TR) Calculation&lt;/a&gt; and the &lt;em&gt;full&lt;/em&gt; &lt;a href=&quot;https://i.stack.imgur.com/3G23E.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;equation&lt;/a&gt;, for &lt;em&gt;both&lt;/em&gt; thread and collar, is given by:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T_R = \frac{Fd_m}{2}\left(\frac{l+\pi\mu{d_m}}{\pi{d_m}-\mu{l}}\right)+\left(\frac{F\mu_cd_c}{2}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Neither of the mean diameters for the square thread, nor the collar were given.&lt;/p&gt;&#xA;&lt;p&gt;However, I found that the mean diameter of the square thread can be  determined from the pitch, &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt;, and nominal diameter, &lt;span class=&quot;math-container&quot;&gt;$d_n$&lt;/span&gt;, by&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d_m = d_n-\frac{l}{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, I do not know how to calculate the mean diameter of the collar. I was hoping that it would simply be&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d_c = d_n+\frac{l}{2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;but, I think that might be incorrect, although I am not sure - as this would seem to be the formula for calculating the major diameter of the screw, when given its mean diameter.&lt;/p&gt;&#xA;&lt;p&gt;Looking at a &lt;a href=&quot;http://users.encs.concordia.ca/%7Enrskumar/Index_files/Mech344/Solutions/Problem%20Set%206-Chapter%2010_Power%20Screws%20and%20Fasteners-Selected%20Problems.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;series of tutorials&lt;/a&gt;, I see that the mean diameter of the collar may need to be provided. For example, from Solution 10.3&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uS0Wy.png&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Diagram of load, screw and collar&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uS0Wy.png&quot; alt=&quot;Diagram of load, screw and collar&quot; title=&quot;Diagram of load, screw and collar&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;After reading around the topic, I see that often the component relating to the collar's friction is discarded (or, as the tutorials state, considered to be eliminated). Why is that? Is that a common simplification?  The OP had this simplified version of the above formula in their question,&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T_R = \frac{Fd_m}{2}\left(\frac{l+\pi\mu{d_m}}{\pi{d_m}-\mu{l}}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So, do I actually need to worry about the collar's mean diameter?&lt;/p&gt;&#xA;" OwnerUserId="5899" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2024-03-13T21:28:10.857" Title="Mean diameter of the collar when calculating the raising torque" Tags="|torque|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28462" PostTypeId="1" CreationDate="2019-05-20T10:51:24.553" Score="1" ViewCount="41" Body="&lt;p&gt;Please, I have to solve a problem and I really need some help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to give an efficient method to consider a maximum amplitude as a starting point to consider as noise the whole spectrum below itself. Like on the example of the FFT below. Let's consider only on X axis, the red plot, the maxium peak is at a magnitude of 0.4, between 7 and 8 Hz. The next peak is at a magnitude of ~0.3, and the next after that is at ~0.25, and so on (in a decreasing order).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8f2n6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8f2n6.png&quot; alt=&quot;FFT of vibration sample&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I give a consistent method to find the threshold between noise and real signal in terms of magnitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The perfect answer would be transforming it in a closed-loop control where I could go adding transfer functions with natural frequencies corresponding to those peaks until the response are no longer affected (for a unitary step). But for that, I would have to tell the system what is the magnitude of each peak. That's where I lose myself...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much in advance.&lt;/p&gt;&#xA;" OwnerUserId="20513" LastActivityDate="2019-05-20T10:51:24.553" Title="How can I measure the impact of noise on a FFT?" Tags="|matlab|signal|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28463" PostTypeId="2" ParentId="28439" CreationDate="2019-05-20T11:45:06.970" Score="1" Body="&lt;p&gt;It sounds like you have some experimental data, and from that you would like to be able to estimate the transfer function of your system so that you could use it for control system design purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's the case, and given that you are using MATLAB, I would suggest looking at the &lt;a href=&quot;https://uk.mathworks.com/help/ident/ref/tfest.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;tfest&lt;/code&gt;&lt;/a&gt; function, which is part of the System Identification Toolbox:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Description&lt;/strong&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;sys = tfest(data,np)&lt;/code&gt; estimates a continuous-time transfer function, &lt;code&gt;sys&lt;/code&gt;, using time- or frequency-domain data, &lt;code&gt;data&lt;/code&gt;,&#xA;  and contains &lt;code&gt;np&lt;/code&gt; poles. The number of zeros in the &lt;code&gt;sys&lt;/code&gt; is &lt;code&gt;max(np-1,0)&lt;/code&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;sys = tfest(data,np,nz)&lt;/code&gt; estimates a transfer function containing &lt;code&gt;nz&lt;/code&gt;&#xA;  zeros.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;sys = tfest(data,np,nz,iodelay)&lt;/code&gt; estimates a transfer function with&#xA;  transport delay for input/output pairs &lt;code&gt;iodelay&lt;/code&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;sys = tfest(___,Name,Value)&lt;/code&gt; uses additional options specified by one&#xA;  or more &lt;code&gt;Name&lt;/code&gt;,&lt;code&gt;Value&lt;/code&gt; pair arguments. All input arguments described for&#xA;  previous syntaxes also apply here.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;sys = tfest(data,init_sys)&lt;/code&gt; uses the linear system &lt;code&gt;init_sys&lt;/code&gt; to&#xA;  configure the initial parameterization of &lt;code&gt;sys&lt;/code&gt;.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;code&gt;sys = tfest(___,opt)&lt;/code&gt; specifies the estimation behavior using the&#xA;  option set &lt;code&gt;opt&lt;/code&gt;. All input arguments described for previous syntaxes&#xA;  also apply here.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Once you have your transfer function as an &lt;a href=&quot;https://uk.mathworks.com/help/ident/ref/idtf.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;code&gt;idtf&lt;/code&gt;&lt;/a&gt; model, you should be able to use it in your Simulink. It may be necessary to do some pre-processing on your data before fitting.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-05-20T11:45:06.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28465" PostTypeId="1" CreationDate="2019-05-20T14:12:00.873" Score="2" ViewCount="183" Body="&lt;p&gt;I am having hard time understanding how does the axial component stay constant through the rotor. Every resource i look just says it is due to mass conservation but never gives the actual conservation equation and i can't write it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For clarity, i am talking about wake rotation theory and not the simplified 1D momentum theory.&lt;/p&gt;&#xA;" OwnerUserId="20535" LastActivityDate="2024-02-25T10:05:15.633" Title="In horizontal axis wind turbines, why does axial component of absolute velocity stay constant (wake rotation theory)" Tags="|aerodynamics|turbines|wind-power|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28467" PostTypeId="2" ParentId="28452" CreationDate="2019-05-20T17:05:56.593" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;With regards to its units, how does kW and RPM become equivalent to N.m?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Power = force * speed.  When you're dealing with rotary motion, power = torque * rotation rate.  Their calculation of torque from HP and RPM is just a matter of getting the units consistent, and solving for torque (noting that for the power as a function of torque and rotation rate, the rotation rate must be in radians/second).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-05-20T17:05:56.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28468" PostTypeId="2" ParentId="25119" CreationDate="2019-05-20T17:15:58.520" Score="1" Body="&lt;p&gt;The easiest way to accomplish this is to install a lubricating pump that circulates the oil in the gearbox through an external cooling radiator. You may also need to force-feed the pressurized oil into the bearings through passageways machined into the bearing mounts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oil cooling and pressurized oil feed is what allows a sport bike engine to run at 16,000 RPM while developing more than 150HP, and why the gearbox on the weedcutting deck (which did not have pressurized lubrication or an oil cooler) on my tractor was limited to 3200 RPM and 15 HP. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-05-20T17:15:58.520" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28469" PostTypeId="1" CreationDate="2019-05-20T17:18:35.483" Score="0" ViewCount="90" Body="&lt;p&gt;I'm modelling a pipe in Patran with CQUAD8 elements, but I get this error, like the structure is not costrained. I do have costrains on both ends of the pipe. What can the problem be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I use automatic costrains it works, but I don't know why and where these constrains are put. I don't have any control over the solution. I'd like to know the actual problem here. With CQUAD4 works fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for helping me out.&lt;/p&gt;&#xA;" OwnerUserId="20510" LastActivityDate="2019-05-20T17:18:35.483" Title="MSC Nastran CQUAD8 on a pipe - excessive pivot ratio" Tags="|buckling|nastran|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28470" PostTypeId="2" ParentId="28457" CreationDate="2019-05-20T18:20:28.570" Score="2" Body="&lt;p&gt;Yes it can, and in many designs it should be a pin support along one axis and fixed support along other. In many trusses and bridges that is the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design software when defining a joint have the joint degrees of freedom choices. Among those are the option to define the joint a hinge along one axis but fixed along the others.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can even have joints or supports that can have, rigid, pin, or spring or even predefined varying stiffness and ductility restriction on rotation, settlement, lateral displacement, moment transfer. This is becoming more advantageous in seismic design. Where one expects different behavior of a structure under different spectra of earthquake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between the two extremes of a pin joint and fixed join one can define the semi fixed joint or fixed but after a certain stress behaving like a hinge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real construction  of a joint there is always joints that are legally considered pin or fixed but they behave a certain mix of both, but you design them depending on what the code defines that type ofjoint.   &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-20T18:20:28.570" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28471" PostTypeId="1" CreationDate="2019-05-20T18:26:30.977" Score="0" ViewCount="470" Body="&lt;p&gt;I'm modeling a pipe in Patran, and I'd like to try bar3 and bar4 elements. &#xA;I assing the properties CBEAM to my geometry, but when I create the mesh, Patran tell me bar3 and bar4 don't support that property, and I need to use bar2. &#xA;Doesn't work even with CBAR, CROD or CPIPE. How can I use bar3 then? Seems like no properties work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for helping me out.&lt;/p&gt;&#xA;" OwnerUserId="20510" LastActivityDate="2024-01-27T10:04:15.487" Title="MSC Nastran - BAR3/4 elements - can't assign any properties" Tags="|beam|buckling|nastran|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28472" PostTypeId="2" ParentId="3374" CreationDate="2019-05-21T01:04:19.917" Score="5" Body="&lt;p&gt;The fundamental problem is that hot air must leave the room and cold air must enter it, both at the same time through the same single opening.  The conflicting forces will mostly cancel each other out, resulting in very little actual exchange of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you really need is old technology.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Double hung&lt;/em&gt; windows are what our ancestors used.  The bottom sash lifts, just as in most windows, but the top sash, rather than being fixed, can be lowered to create an opening at the top.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With both a high and a low opening in the window, hot air tends to be forced out the top as cooler air enters at the bottom.&lt;/p&gt;&#xA;" OwnerUserId="19930" LastEditorUserId="19930" LastEditDate="2019-05-22T16:34:13.387" LastActivityDate="2019-05-22T16:34:13.387" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28473" PostTypeId="2" ParentId="28465" CreationDate="2019-05-21T04:34:00.670" Score="0" Body="&lt;p&gt;The windmills based on lift action don't change the speed of air passing through them same as an airplane's wing flying through the air, the air stream washing around the wing's top and bottom doesn't pick up any speed. It Just gets bent down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blades of the windmill in this case work like a wing they let the air stream through but they bend it by twisting it as a screw driver, albeit that twisting will calm down rather fast by the viscosity of the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A crude example of the flux of the air before and after the windmill would be like those striped colored toothpaste, going straight till it hits the blades then twists a quarter of circle then keeps roping out straight behind not turning anymore, and getting mixed with the rest of the wind. this is kind of cleaned up version of the affair, other wise there is some turbulence and a bit of drag, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the axial speed of the air remains the same but the tangential speed changes for the transit time starting from the encountering the tip of the blade till some time after leaving the trailing edge of the blade.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-21T04:34:00.670" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28475" PostTypeId="1" CreationDate="2019-05-21T09:52:01.310" Score="1" ViewCount="49" Body="&lt;p&gt;Iron ore and few compounds added will result in steel which are casted to steel billers form. this billet is again reheated to melting point of 1200c and then sent to roughing mill for desired shape. can we avoid reheat furnance and directly cast and then roughing mill?&lt;/p&gt;&#xA;" OwnerUserId="17042" LastActivityDate="2019-05-21T21:38:35.883" Title="Is there a way to avoid reheating of steel billets in Reheat furnace and directly go from casting to roll milling?" Tags="|steel|manufacturing-engineering|casting|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28477" PostTypeId="1" CreationDate="2019-05-21T15:08:48.230" Score="1" ViewCount="156" Body="&lt;p&gt;The problem : I need to create a mechanism with 2 parallel shafts, at some distance apart, rotating in the same direction, either clock- or anticlock-wise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance (&gt; 1m) makes it impractical to set a chain of gears between the shafts, so my approach is to have a third shaft at 90 degrees with bevel gears; the driving motor could attach to this third shaft. It has been pointed out to me that a pulley-belt approach could also be used, but the distance between the parallel shafts makes me prefer the 90-degree shaft approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not being a mechanical engineer, I tried to understand a bit more about bevel gears, and found out there are straight, spiral and other types. The 3D library I'm currently using for generating the gears creates angled teeth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lrMux.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lrMux.png&quot; alt=&quot;bevel gears with angled teeth&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's the 1st question : for any type of bevel gear except the straight ones, does the geometry of the teeth have any impact on the direction of rotation ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I don't want different rotation speeds on the parallel shafts, and there is no constraint on the rotation speed of the third, 90-degree shaft, so far my plan is to create gears with identical sizes (number of teeth).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here comes the 2nd question : Should I use larger or smaller gears ? What's the impact of the number of teeth on the gear operation for the bevel gears ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note, I plan to 3d-print these gears for the prototype, so any comment that can factor this in is highly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you all in advance.&lt;/p&gt;&#xA;" OwnerUserId="20543" LastActivityDate="2019-05-21T22:22:13.423" Title="Bevel gear design &amp; usage" Tags="|mechanical-engineering|gears|mechanisms|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28481" PostTypeId="2" ParentId="28475" CreationDate="2019-05-21T18:51:51.480" Score="1" Body="&lt;p&gt;Historically production was : ingot &gt; reheat&gt; bloom &gt; reheat &gt; billet. For about the last 40 years production has been  : continuous casting directly produces billets. An ingot was roughly 2 ' X 4' X  8 ' . Continuous strands can be down to about 6 &quot; X very long. There are a great number of variations depending on final product , alloy, the individual mill , etc. A billet form is inventoried until needed , then heated and shaped. The homogenization and grain refinement provided by reheat also need to be considered.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-05-21T21:38:35.883" LastActivityDate="2019-05-21T21:38:35.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28482" PostTypeId="2" ParentId="28477" CreationDate="2019-05-21T22:22:13.423" Score="1" Body="&lt;p&gt;For your selection of gear models to test via 3D printer, consider a convenient search for &quot;openscad bevel gears&quot; as your starting point. You will find quite a few creators of the code that can be customized to meet your specific requirements. The creators will often provide &quot;canned&quot; or &quot;out-of-the-box&quot; models that may also fit your objective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You ask about the number of teeth for the gears. From a layman's point of view, more teeth means less wear, longer life and smaller loads per tooth. A bevel gear will have substantial mesh surface which, with a larger diameter, means the forces are better spread along the radius.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-05-21T22:22:13.423" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28483" PostTypeId="1" CreationDate="2019-05-22T05:33:18.897" Score="1" ViewCount="39" Body="&lt;p&gt;I have an odour control system where foul air is being extracted from multiple sources and then pushed into a scrubber vessel. Normally, all my designs look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ibb.co/58YSdpc&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ibb.co/58YSdpc&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With this arrangement, I'm able to maintain negative pressure at all sources which ensures that the foul air is being sucked into the treatment system, and that it doesn’t escape into the air through any openings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I currently have a project where one of the sources is a sludge buffer tank that is being mixed through pressurised  air. The system looks like this: (Fans A/B are duty/standby arrangement with rating: 4650m3/hr @1500Pa)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ibb.co/bj04pBB&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ibb.co/bj04pBB&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compressor blows air into the tank from the bottom through the water, and the water level in the tank varies from 1 m to 4 m. This means the pressure in the headspace could range from 10,000 to 40,000 Pa (The way I did this is that  I subtracted the pressure of a water column 1m to 4m deep from the 50 kPa). I would like to know how the air flow rate directions and relative magnitudes  would be like in this system, and whether there will still be negative pressure in the sources. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="20558" LastActivityDate="2019-05-22T05:33:18.897" Title="How to calculate flow rate and direction in system with multiple blowers" Tags="|fluid-mechanics|chemical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28486" PostTypeId="2" ParentId="25223" CreationDate="2019-05-22T12:06:52.273" Score="0" Body="&lt;p&gt;I would say it really depends on the tool steel being used. I read that its recommended for water hardening steels to be quenched in a brine solution opposed to straight water, so perhaps there's some validity to the statement. Whether or not you can quench an oil hardening steel in water/brine, id imagine water quenching an O type steel would &quot;shock&quot; the part too much, promoting cracking of thin sections and brittleness. I also wouldn't try and quench an air hardening material in a liquid for the same reasons. &lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2019-05-22T12:06:52.273" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28490" PostTypeId="2" ParentId="14192" CreationDate="2019-05-22T20:05:49.673" Score="0" Body="&lt;p&gt;First I would like to clear the doubt that the difference between FEA and CFD is that FEA is for structural applications and CFD for fluid dynamical applications, is wrong. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computational Fluid Dynamics (CFD) refers to the use of the numerical techniques to solve fluid dynamical problems. When I say 'numerical techniques', I am referring to a very broad range of techniques, including, but not limited to, Finite Difference Methods, Finite Element Methods, Finite Volume Methods, Polynomial Fitting, Spectral Methods, Boundary Element Methods and so on. Even though the basic philosophy is the same i.e. discretize a system with infinite degrees of freedom into a finite system, these are all different techniques with different mathematical foundations. For instance, Finite Difference Methods are based upon discretization of the differential form of governing equations and work by approximating derivative via truncation of Taylor Series expansions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The order of accuracy of FDM depends on the highest order of the Taylor series expansion terms that are eliminated. FDM is the most intuitive of these methods to understand. There are compact schemes such as Pade Schemes which allows one to improve the accuracy of Finite Differences for the same numerical stencil. The FVM and FEM method on the other hand involve discretization of the integral form of the PDE. Spectral Methods involve discretization of the grid into a finite set of points and representing the solution as a linear combination of periodic functions (for instance, a Fourier Series). In CFD the governing PDEs are of course the Navier-Stokes equations (Navier–Stokes equations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finite Element Method(FEM), also known as Finite Element Analysis(FEA) is a specific numerical technique that, of course, solves a continuous problem stated in the form of a PDE, by discretizing the problem into a finite number of nodal points but it does so by first multiplying the differential form of the governing equation(PDE) with an arbitrary weighting function and using Integration by parts and the Divergence Theorem to obtain, what is known as the 'Weak Form' of the governing equation, and then formulating a system of linear equations by approximating the solution field as a linear combination of a finite number of basic functions, each of which are pairwise orthogonal and altogether satisfy partition of unity. The beauty of the Finite Element Method and the weak form is that it allows us to approximate our quantity of interest using functions that have 'weaker' continuity requirements while at the same time maintaining compatibility at element interfaces.&lt;/p&gt;&#xA;" OwnerUserId="20566" LastActivityDate="2019-05-22T20:05:49.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28491" PostTypeId="1" CreationDate="2019-05-22T20:33:18.500" Score="2" ViewCount="146" Body="&lt;p&gt;I need to create a simple machine to assist in rotating a cylindrical load, akin to a roll of carpet, from a horizontal position to vertical. The load weighs 450 pounds, which is distributed evenly along a 35 foot length. The machine would use a geared motor to rotate the load from the bottom of the cylinder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need two questions answered: 1 how much torque would the motor need to rotate the load, and 2 how heavy would the machine need to be to keep it from tipping over?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine would be assisted by human power, but I'd like to begin with the values needed for the machine to do all the work, then work my way backwards from there. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ldtYN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ldtYN.jpg&quot; alt=&quot;Lift Machine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20567" LastActivityDate="2019-05-23T00:16:30.307" Title="What is the torque required to rotate a 450 lb., 35' long cylindrical load 90 degrees?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28493" PostTypeId="2" ParentId="28491" CreationDate="2019-05-22T22:29:01.980" Score="2" Body="&lt;p&gt;The CG of the load is at 17.5 feet.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ Torque= 450*17.5=7875 \ lbs.ft $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the connection of the base plate to concrete should be able to support this torque (moment) with a say 50% safty factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example if your baseplate connecting the lift to concrete is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$18\ inches *18\ inches.$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_{safety.factor }=7875*1.5= 11812.5    $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you have 4 anchor bolts attaching the base plate to the concrete slab at 1.25 ft OC,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F_{AB }= 11812.5/1.25= 9450\ lbs$$&lt;/span&gt;&#xA;The plate needs to be engineered bot as a first estimate you can start at a 18 inch square plate with 3/4 thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could have the bottom of you lift attached to a horizontal steel beam leg of at least longer than 18 ft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**Edit **&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After OP' comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the lifter needs to be portable, then you need a base wide and heavy to resist overturning moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say you have a bas of 6ft SQ with the crane mounted near the left side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weight needed to keep the balance under the crane is approximately,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12000/6 = 2000 lbs. But you need to have say 500 lbs right above each wheel as well, just for dynamic loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would add grab bars to the crane for operators to hold on to.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-05-23T00:16:30.307" LastActivityDate="2019-05-23T00:16:30.307" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28494" PostTypeId="1" CreationDate="2019-05-22T22:43:51.503" Score="2" ViewCount="84" Body="&lt;p&gt;Without using FEA, is there a computer algorithm or numerical method available to calculate the shear center of any 2D geometry if the following are known:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;geometry of the individual components of the section mesh;  centroid;&#xA;   principal axes angle of rotation;&#xA; area;&#xA; perimeter;&#xA; moments of inertia&#xA;   x and y;&#xA; radius of gyration;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="20565" LastEditorUserId="20565" LastEditDate="2019-05-23T06:13:06.517" LastActivityDate="2019-05-23T06:13:06.517" Title="computer algorithm to calculate shear center" Tags="|structural-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28495" PostTypeId="2" ParentId="28471" CreationDate="2019-05-22T23:40:13.063" Score="0" Body="&lt;p&gt;It's been forever since I've used patran, but in Nastran what you are looking for is CBEAM3. that's a 3 noded beam. CBEAM is strictly 2 nodes. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-05-22T23:40:13.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28497" PostTypeId="1" CreationDate="2019-05-23T00:32:16.280" Score="0" ViewCount="1136" Body="&lt;p&gt;I have been wondering about the different methods of finding the center of mass of a robot? Mine is a balancing robot, so it has two wheels and a chassis on top. I need to find the center of mass so that I can tune the robot to balance. I have done some research but I have only been able to find the following 2 ways:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Use the conventional method of finding the center of mass an irregular object with weighted lines and then the junction of all these lines would give the center of mass. But I don't know how reliable this method is? Since the robot is a 3D object, so its center of mass will lie somewhere at a depth inside the robot chassis?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Design the whole robot, part by part in a CAD software, like Inventor and then use the software's features to let it automatically find the center of mass. Unfortunately, my CAD skills are not good enough.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Any help and suggestions are appreciated.&lt;/p&gt;&#xA;" OwnerUserId="21627" OwnerDisplayName="WiredMaker" LastEditorUserId="6767" LastEditDate="2019-05-26T10:20:17.620" LastActivityDate="2019-05-26T10:20:17.620" Title="How to find the center of mass of a robot?" Tags="|robotics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28498" PostTypeId="2" ParentId="28497" CreationDate="2019-05-23T01:18:06.327" Score="0" Body="&lt;p&gt;Multiply the mass of each object (M) by the distance between the object and some reference point (D) or datum.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;centre of mass = &lt;span class=&quot;math-container&quot;&gt;$$\dfrac{\sum M*D}{\sum M}$$&lt;/span&gt;  for each axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, if you have the robot, is to tilt it to the tipping angle on each axis and measure this. Then if you can control acceleration repeat back and forth to measure the triangular velocity and slack (hysteresis). Repeat until you can measure the tipping angle . If inherently unstable support the top with a string to keep quasi stable while finding the max angle vs acceleration at low f that can be recovered without falling . Constant acceleration means constant current limiter to the driver. This measures the maximum g vs tipping angle that the motors can control without loss of torque. As the frequency of the triangle velocity is reduced the angle increases and peak momentum of it can control before a recovery error.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;balance approach.&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;weight the robot with an accurate scale.&lt;/li&gt;&#xA;&lt;li&gt;balance it on each axis . this is your CoM.  verify with a TILT angle for sin 30 deg = 50 % support force.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8847" OwnerDisplayName="Sunnyskyguy EE75" LastEditorUserId="8847" LastEditDate="2019-05-24T19:29:40.530" LastActivityDate="2019-05-24T19:29:40.530" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28500" PostTypeId="1" CreationDate="2019-05-23T03:42:33.467" Score="2" ViewCount="63" Body="&lt;p&gt;So I made a design for the Autodesk Competition but for some reason, I have never had this issue, but when I export to .obj and open in S3D, it doesn't convert the measurements. My meaning: I have a 3.5x3.5x10 inch design but in S3D it reads as 3.5x3.5x10mm and I do not know how to fix this. Any help would be great!&lt;/p&gt;&#xA;" OwnerUserId="20571" LastActivityDate="2019-05-23T03:42:33.467" Title="Fusion 360: Measurement issues to Slicer Software" Tags="|mechanical-engineering|cad|autocad|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28503" PostTypeId="2" ParentId="28497" CreationDate="2019-05-23T06:48:16.437" Score="1" Body="&lt;p&gt;If the robot is made up of simple geometrical solids like cubes, cylinders, spheres, ellipsoids, etc. it is easy to find the x,y,z of the CG of each part. If not, try and imagine cutting it with partitions in the 3 Cartesian planes into cubes with easily identifiable CGs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now pick an arbitrary point, P ( x,y,z  ) as reference and measure the distance components x,y,z of all the CGs of the parts' you defined above,  from that refrence point P. Then the CG of the robot in 3d space will be: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ X_{  cg}= \frac{\Sigma X_{i}m_{i}}{\Sigma_{m}} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Y_{  cg}= \frac{\Sigma Y_{j}m_{j}}{\Sigma_{m}} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Z_{  cg}= \frac{\Sigma Z_{k}m_{k}}{\Sigma_{m}} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And repeat this procedure for when the limbs of the robot, or any part articulated, is extended to its limit, like if the arms are extended out. And if the lims move fast you need to consider dynamic forces as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will give you an area under the wheels, which the sensors should be programed to read and send signals to balancing mechanism. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-05-23T08:25:15.250" LastActivityDate="2019-05-23T08:25:15.250" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28504" PostTypeId="1" CreationDate="2019-05-23T07:00:01.057" Score="1" ViewCount="244" Body="&lt;p&gt;what happens if starting arm of starter does not return to &quot;OFF&quot; position after a temporary supply failure?&lt;/p&gt;&#xA;" OwnerDisplayName="user19926" LastActivityDate="2020-08-20T02:55:24.367" Title="Starting of direct current Motor" Tags="|electrical-engineering|motors|power|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28505" PostTypeId="1" AcceptedAnswerId="28591" CreationDate="2019-05-23T08:06:33.983" Score="2" ViewCount="49" Body="&lt;p&gt;I made a simulation of this cylinder that is being suspended mid air and held up by 8 rods, 4 on top and 4 on the bottom. The entire model is cooled to 15K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I managed to find the forces required for 6 of the rods to have a safety factor of above 1 as pictured below (green being above 1):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gWVGL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gWVGL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two rods in red have a safety factor of less than 1. All rods have a parametrised force so I would like to know how to create a design study to get all of the rods to be above 1 or at least all have the same safety factor under 1. Basically, I am only interested in changing the forces that are applied to the rods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried an Adaptive Multiple Objective with the goals of all rods having a safety factor above 1 but it didn't produce any meaningful results.&lt;/p&gt;&#xA;" OwnerUserId="1434" LastEditorUserId="1434" LastEditDate="2019-05-23T08:17:39.280" LastActivityDate="2019-05-30T05:40:42.073" Title="Direct Optimisation - How to Create Time Efficient Study" Tags="|mechanical-engineering|ansys|optimal-control|thermal-expansion|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28506" PostTypeId="1" CreationDate="2019-05-23T12:14:26.577" Score="1" ViewCount="51" Body="&lt;p&gt;I would like to measure viscosity, but do not have access to a viscometer. If possible, I would like to avoid having to purchase one. I have access to many other  flow measuring devices, including flowmeters, pressure sensors, and a turbidimeter, and was wondering if taking measurements with these would somehow allow me to make, at the very least a proportional measurement to viscosity (I only need relative measurements between different fluids, exact specifications are not necessary). I dont have reason to believe these liquids would behave significantly non-newtonian, and they would likely all be water with additives such as salt and detergents. &lt;/p&gt;&#xA;" OwnerUserId="20576" LastActivityDate="2019-05-23T17:23:23.703" Title="Measuring viscosity with tools other than viscometer" Tags="|fluid-mechanics|measurements|fluid|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28507" PostTypeId="2" ParentId="28506" CreationDate="2019-05-23T12:38:55.070" Score="2" Body="&lt;p&gt;A long glass tube and a steel ball which matches the internal diameter closely - then measure the different times compared to some known liquids for the ball to drop a given distance.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-05-23T12:38:55.070" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28509" PostTypeId="2" ParentId="28506" CreationDate="2019-05-23T17:23:23.703" Score="2" Body="&lt;p&gt;punch a small hole in the bottom of a plastic container. measure the time it takes for the container to empty through the hole. experiment with different hole diameters to &quot;scale&quot; the effect appropriately for your expected viscosity values, and calibrate it against a standard.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-05-23T17:23:23.703" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28510" PostTypeId="2" ParentId="28442" CreationDate="2019-05-23T19:25:10.837" Score="1" Body="&lt;p&gt;You Should use a regular line with HB pencil to draw that pointed part (it's important its the part that popping out can be seen in front view but you can't see that in side view, but you can see the line).&#xA;1:1 Scale shows that you have to make the diagram in 30mm and 70mm only you cant make it 60mm and 140mm as that would be changed to 2:1. Usually 1:1 implies that use the dimensions exactly given in the question you can't make it bigger or smaller.&lt;/p&gt;&#xA;" OwnerUserId="20582" LastActivityDate="2019-05-23T19:25:10.837" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28512" PostTypeId="2" ParentId="27044" CreationDate="2019-05-23T23:05:48.213" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know how much the frame of reference is rotated along the x and y axes and I know the position of the vector in this frame of reference.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you know how much, then you must know the order. How much is dependent on the order. The problem is, there are several ways to get from one reference frame to another via rotations. You need to look at what you mean above and use the system of transforms that makes the most sense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;TLDR&lt;/strong&gt; If what you have is azimuth, elevation, and roll from Earth frame to Body frame; then the matrix operations are performed in that order, meaning the matrices are written in the opposite order followed by a column vector and evaluated last first. To go from Body to Earth, the operations are performed in the opposite order, or you use the transpose of the combined &lt;em&gt;direction cosine transformation matrix&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you start by considering a level flight scenario. You have a RHR body frame and a RHR local earth frame. Conventionally, the Earth frame is NED - positive x,y,z is North, East, Down. The body frame is usually FSU - positive x,y,z is Forward, Starboard, Up. But Aft, Starboard, Down; and Forward, Port, Down are also pretty common. Pick one - it doesn't matter as far as the construction of the transform is concerned, the difference is only in the angles that you feed the transform and whether positive pitch is nose up or nose down, positive roll is port side up or port side down, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is assumed that the earth frame is an inertial frame here. If you need to treat it as a non-inertial frame, things can quickly get tedious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we habitually write x,y,z, that is the order conventionally used to construct the &lt;em&gt;direction cosine transformation matrix&lt;/em&gt; from the three square simple rotation matrices based on the &lt;em&gt;Euler angles&lt;/em&gt;. The Euler angles are the appropriate angle set for this arrangement of operations. Starting in the Earth frame, Phi is the angle in the xy plane from earth x axis to the projection of xBody axis, yielding x',y',z (this is the azimuth angle). Theta is the angle in the new rotated x',z plane from x' to xBody, yielding xBody, y', z' (this is the elevation angle). Phi is the roll angle about xBody from y' to yBody, yielding xBody, yBody, zBody (this is the roll angle).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, you want to go the opposite way. No problem. The inverse transformation matrix from body frame to earth frame is just the Matrix Transpose of the original matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very good reference for this is Mark Drela's &lt;em&gt;Flight Vehicle Aerodynamics&lt;/em&gt;. It is well worth acquiring, relatively cheap in paperback, and can be viewed online via the EDX website, in the course archives for Flight Vehicle Aerodynamics.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2019-05-23T23:18:46.933" LastActivityDate="2019-05-23T23:18:46.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28513" PostTypeId="1" AcceptedAnswerId="28515" CreationDate="2019-05-24T04:53:05.207" Score="0" ViewCount="180" Body="&lt;p&gt;I have a flat roof (dead load = 20 psf, live load = 30 psf) spanning 11 feet with a 2 ft. overhang. I'm trying to calculate the total load in plf on the header marked as X. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/FFKdT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FFKdT.png&quot; alt=&quot;building profile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this calculation correct:&#xA;TL = (DL + LL) × (11/2 + 2)&#xA;= (20 + 30) × 7.5&#xA;= 375 plf&lt;/p&gt;&#xA;" OwnerUserId="20584" LastActivityDate="2019-05-24T06:23:15.990" Title="Window header sizing: total load on header" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28515" PostTypeId="2" ParentId="28513" CreationDate="2019-05-24T06:23:15.990" Score="0" Body="&lt;p&gt;No it is not correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you just want to calculate the load on the header with service loads and no load factors, you find the load on the header by getting the moments about an arbitrary point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case the easy choice is the point at the left end of the rafters. let's denote it A and the point right on top of the wall with the header B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Sigma M_{A}=0 \quad M_{A} = w*l*l/2 -R_{B}*11\ ft=0 \\ (20+30)*(11+2)*13/2 =R_{B}*11  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ R_{B}= 4225\ lbs/11 = 384,09\ lbs.ft $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So each lineal foot of the header is supporting 384.09 lbs.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-24T06:23:15.990" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28519" PostTypeId="2" ParentId="17271" CreationDate="2019-05-25T00:27:29.123" Score="1" Body="&lt;p&gt;Assuming by performance you mean the total power or torque output of the engine, the displacement is certainly a major factor, but it cant be analysed individually. Other factors such as number of cylinders, number of strokes, Compression ratio, mode of aspiration, bore/stroke ratio, cam profiles/ timing, head geometry, redline,etc etc etc... when considered for a given displacement reveal the overall picture and are equally important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single cylinder 1000cc engine will have a entirely different performance figure than a 4 cyl, and a 1000 cc 2 stroke will behave highly unlike a 4 stroke, and with a different powerband.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if the engine size, weight and NVH is also considered towards overall performance, a single cylinder will require a much heavier flywheel and thus a heavier overall engine. Also a multicylinder engine of the same capacity allows for reciprocating load balance and better vibration characteristics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Along the same lines, a litre class bike engine has a horsepower figure of around 190 HP, whereas a 1.2 L car engine could peak at sub 100 HP. What makes the difference in this case is a higher redline and a different bore/stroke ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Displacement also doesnt reveal a correct picture when forced induction comes into the picture. Small displacement turbocharged engines make much more power than NA higher capacity engines; current 1.6L F1 engines are the smallest ever displacement wise, but generate the highest power ever, all thanks to turbocharging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then race spec versions of road vehicles having the same engine with the same displacement churn out entirely different numbers. ECU fuel and ignition maps matter too. A lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In all, displacement alone isnt enough to predict the engine performance, knowing at least the 3 other factors described above as well can help get a clearer idea. And in the end it always comes down to what application the engine is designed for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="20598" LastEditorUserId="20598" LastEditDate="2019-05-25T10:23:09.343" LastActivityDate="2019-05-25T10:23:09.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28520" PostTypeId="1" CreationDate="2019-05-25T05:46:56.197" Score="1" ViewCount="128" Body="&lt;p&gt;I'm thinking about a system that would replace water in a swimming pool with clean water from an Artesian well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The water in the well is about 1C and the swimming pool is around 20C. Reheating the whole pool is obviously very costly, so I'm looking to use a counter-current heat exchanger. What kind of efficiency can I expect from it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is just a concept idea for now, just looking at possible angles.&lt;/p&gt;&#xA;" OwnerUserId="20599" LastActivityDate="2019-06-02T20:18:26.157" Title="Heat exchanger efficienccy" Tags="|heat-transfer|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="28522" PostTypeId="1" CreationDate="2019-05-25T10:34:13.193" Score="-1" ViewCount="212" Body="&lt;p&gt;I've researched in the Internet but can't find any conclusive answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When will your phone emit more radiation, downloading 1 Mb of data from a cell network (3G, 4G) or downloading the same amount of data using WiFi? What is likely to consume more energy?&lt;/p&gt;&#xA;" OwnerUserId="20601" LastActivityDate="2019-05-27T14:00:22.200" Title="What emits more radiation, cell network or wifi?" Tags="|electromagnetism|radiation|waves|wifi|networking|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28523" PostTypeId="1" CreationDate="2019-05-25T13:23:13.317" Score="2" ViewCount="62" Body="&lt;p&gt;When running stepper motors with a chopper driver, the motor voltage applied is much higher than what the motor is rated for, because the chopper is a current source.&#xA;The question however is, what voltage to use in order to get the maximum performance out of the motor (for example, to get the holding torque the motor claims to have)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came across the following formula in some forum (without explanation provided)&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Vmax = SQRT(inductance) * 32&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;So that would mean that if I'm using some stepstick stepper drivers that support a maximum of 36V I should look for stepper motors with inductance below (36/32)^2=1.26mH, otherwise I won't be able to run them at their rated performance (of course they also should be rated for current less than the maximum RMS current of the stepper).&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1. Is this correct?&#xA;2. Where does this formula come from? Why 32?&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="20603" LastActivityDate="2019-05-25T13:23:13.317" Title="Stepper motor - optimal voltage when using a chopper driver" Tags="|stepper-motor|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28524" PostTypeId="1" CreationDate="2019-05-25T14:46:57.550" Score="1" ViewCount="50" Body="&lt;p&gt;My master thesis deals with systems with several operating states that generate heat and store it in a heat storage. Thereby I analyze combined heat and power plants with combustion engines, but also heat pumps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am looking for a comprehensive term for such systems in my thesis and in the title. The term &quot;heat generators&quot; seems a bit too simplified to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any more specific alternatives for a comprehensive term that might also point to the heat storage tank and the limited number of operating states?&lt;/p&gt;&#xA;" OwnerUserId="20604" LastActivityDate="2019-05-25T14:46:57.550" Title="Search for comprehensive term for plants with heat generation with heat storage and operating states" Tags="|thermodynamics|heat-transfer|heating-systems|combustion|heat-exchanger|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28525" PostTypeId="2" ParentId="88" CreationDate="2019-05-25T16:00:53.007" Score="0" Body="&lt;p&gt;The answer by Frédéric Bourgeon is great. You may also find useful this &lt;a href=&quot;https://www.eurocodeapplied.com/design/en1993/elastic-critical-moment&quot; rel=&quot;nofollow noreferrer&quot;&gt;free online calculation tool for elastic critical moment&lt;/a&gt; in eurocodeapplied.com. It is based on the aforementioned NCCI but it is much simpler to apply as compared to LTBeam, at least for the typical cases encountered in design practice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It covers typical I- or H- profiles (IPE, HEA, HEB, HEM), three end support conditions (both ends simply supported, both ends fixed, one end simply supported and one end fixed), and three loading arrangements (linear bending moment, parabolic bending moment, triangular bending moment).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="20607" LastActivityDate="2019-05-25T16:00:53.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28526" PostTypeId="2" ParentId="28089" CreationDate="2019-05-25T21:46:13.783" Score="0" Body="&lt;p&gt;In a common workshop you will probably find H6/H7 reamers for different hole diameters.&#xA;If you are not going to fabricate special tools for your project, you will be limited to standard hole sizes - meaning you will easily find a 8 mm H7 reamer but would never find a 7.38 mm H7 reamer on the shelves.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ggJIQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ggJIQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The manufacturing capability of a standard lath machine (turning machine) should provide you a greater flexibility, since it offers a great accuracy for a wide range of shaft sizes. &#xA;If, for instance, the machine gives you a +/- 0.001 mm tolerance up to a 15mm shaft diameter, you will get the same accuracy for both  8 mm and 7.38 mm shafts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/brwc9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/brwc9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope It helps...&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2019-05-25T21:46:13.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28528" PostTypeId="1" CreationDate="2019-05-26T11:11:03.277" Score="0" ViewCount="23" Body="&lt;p&gt;I have two no of motors connected in parallel of 7.5 kw max current capacity 30Amps.&#xA;The load is about 18Ton is the motor sufficient to lift this load.&#xA;The mechanical arrangement of screw jack system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current now taken by motor is as such,&#xA;Up movement: 37amps initial current and normal 33Amps.&#xA;Down movemenf: 20amps initial current and normal 18amps.&lt;/p&gt;&#xA;" OwnerUserId="20616" LastActivityDate="2019-05-26T11:11:03.277" Title="Tundish Lift Lower mechanism" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28530" PostTypeId="1" CreationDate="2019-05-26T15:47:41.630" Score="0" ViewCount="33" Body="&lt;p&gt;I am installing shades on a section of deck. The shade will be made of sections of fabric (which will let some wind through), but I assume a solid section.&#xA;Dimensions are 3.5ft*4ft = 14ft^2 = 1.3m^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will put the shades on and off, but I want to make sure it won't break if there is a sudden gust of wind. Suppose a 60 mph wind gust = 26.8 m/s and 1.2kg/m^3 air density. This leads to 561.4N of force. (I know the 4X4 red cedar post will hold. I know the fabric will hold. The seam may break, but that's an easy fix.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shades will be supported by 2 tubes (beams), one on top, one at bottom. So, I assume an even load. I also assume the load is even on the whole length on the tube. So, 1 tube of 1.066 meter should support roughly 281 N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was able to find how to compute the maximum moment of 37.43 Nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where I can't finish my calculations. Will a (generic) 1in aluminum tube (1/16in thickness) I find at the store enough?&#xA;I found the tensile strength of aluminum is between 33000 to 76000. So I assume the generic tube I find at the store will be the lowest one at 33000.&lt;/p&gt;&#xA;" OwnerUserId="20617" LastActivityDate="2019-06-25T18:02:30.253" Title="Selection of Aluminum (or steel) tubing dimension based on uniform load on supported beam" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28531" PostTypeId="1" AcceptedAnswerId="28533" CreationDate="2019-05-26T17:18:26.667" Score="2" ViewCount="108" Body="&lt;p&gt;I wonder if there exist a formal guideline to help engineers choosing the right type of bearing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except in some few cases, this is not very evident to me. &#xA;As far as i know, the plain bearing systems are really simple, so why bothering to design the rolling element bearings? I guess there is a reason, but i don't see it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've done some research on the web, besides the economic factors and some other vague statements i couldn't find anything useful.  &lt;/p&gt;&#xA;" OwnerUserId="17181" LastActivityDate="2019-05-26T18:41:43.733" Title="How to make a choice between rolling element bearing and journal (plain) bearing?" Tags="|mechanical-engineering|bearings|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28532" PostTypeId="2" ParentId="28530" CreationDate="2019-05-26T17:28:29.627" Score="1" Body="&lt;p&gt;I just assume your calculations are correct up to just the bending moment strength of aluminum tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sigma_x= \frac{M}{S} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ S_{alum\ tube}= \frac{\pi*(D^4-d^4)}{32D}= \frac{\pi*(1-0.7724)}{32} =0.02233\ in^3&#xA; =365mm^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sigma = \frac{37430}{365}= 102.5 $$&lt;/span&gt;&#xA;please check my numbers.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-26T17:28:29.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28533" PostTypeId="2" ParentId="28531" CreationDate="2019-05-26T17:37:09.020" Score="3" Body="&lt;p&gt;rolling element bearings produce less friction, last longer, can position a rotating shaft with greater accuracy, and do not require pressurized lubrication in order to carry large loads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;sliding contact bearings are cheap, easy to install, and are adequate for light loads without pressure lubrication, and are fine for cost-sensitive applications. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-05-26T17:37:09.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28534" PostTypeId="2" ParentId="28531" CreationDate="2019-05-26T17:53:44.270" Score="1" Body="&lt;p&gt;It’s difficult to say when we should select the rolling element bearing or the other one, sometimes we have to consider the upsides and downsides of each to decide. Also the working condition and the load tell us which is the best option. Here I’m going to mention the most important advantages and disadvantages of both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantages of plain bearing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- This type usually have a large bearing surface which reduces the sensitivity to impact forces &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- Noiseless&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Suitable for high and rough rpm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4- Easy to assemble and disassemble thanks to split joints &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5- Very long lifespan &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disadvantages &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- High starting torque &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- High consumption of oil &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Continue control  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4- Low rendement thanks to friction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantages of rolling element bearings &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Low friction &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- The starting torque is not larger than the service torque &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Slight consumption of lubricant &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4- These types don’t need continue inspection and maintenance &lt;/p&gt;&#xA;&#xA;&lt;p&gt;5- Low response time &lt;/p&gt;&#xA;&#xA;&lt;p&gt;6- Thanks to normalisation the interchangeability is often possible &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The disadvantages &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Vibrations and impacts lead to pitting and fretting so eventually the failure of rings and rolling elements &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- Limited lifespan (usually measured in hours)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Thanks to centrifugal force and its impact on the rolling elements these types aren’t suitable for high rpm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4- Excessive sealing &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So according to upsides and downsides we can conclude the combination of high rpm, long lifespan and large loads push us toward journal bearings, for instance gas turbines, or hydraulic turbines, generators, huge ships &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To damp the strong impact in low rpm we can also use plain bearings, hamers, press machines &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some industries require low maintenance, like oil and gas, here the rolling element bearing is a good option. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-05-26T17:53:44.270" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28535" PostTypeId="2" ParentId="28531" CreationDate="2019-05-26T18:41:43.733" Score="2" Body="&lt;p&gt;To add to the list of criteria in other answers:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cooling. If a bearing supporting a rotor transmitting say 10MW of power dissipates only 0.01% of that power in friction, that is still 1kW of heat being generated in the bearing. Rolling element bearings are relatively easy to cool via the lubricating oil since there an unobstructed flow path between the rolling elements. Sliding bearings are not.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;At the other end of the design spectrum:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Size. Manufacturing a 2mm diameter journal bearing can be as trivial as drilling a 2mm diameter hole in something. Manufacturing a 2mm diameter rolling element bearing is slightly more difficult - though it has been done: &lt;a href=&quot;https://www.nskmicro.co.jp/english/products/products04.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nskmicro.co.jp/english/products/products04.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-05-26T18:41:43.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28536" PostTypeId="1" CreationDate="2019-05-26T20:02:53.117" Score="2" ViewCount="1002" Body="&lt;p&gt;I have expanded steel mesh that I use to make garden cages:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Llvrd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Llvrd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how much sunlight the mesh blocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I need to calculate the area that the mesh takes up (per square inch) and then from there I could calculate a &lt;em&gt;percentage of sunlight blocked&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How can I calculate the area of the mesh?&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="10536" LastEditorUserId="10536" LastEditDate="2019-05-31T02:33:48.770" LastActivityDate="2019-05-31T12:32:04.607" Title="Calculate area of mesh?" Tags="|materials|mathematics|geometry|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28538" PostTypeId="2" ParentId="28536" CreationDate="2019-05-26T23:28:13.833" Score="1" Body="&lt;p&gt;Every 1/2 of an inch you have two slats.&#xA;You  need to measure the horizontal thickness of an slat. by eye balling the tape it looks like 1/8th of inch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area the mesh masks is 12*2*2*height* thickness of the slats per each foot of the length of the mesh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This needs to be subtracted by the 1/2 area of intersection to eliminate the overlap, if you want to be exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area lighted per lineal foot length of the mesh is 1* height of the mesh subtracted by the masked area. Or if you divide the masked area by the area of each lineal foot of the mesh you get the percentage of the light masked.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-05-27T00:30:57.410" LastActivityDate="2019-05-27T00:30:57.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28539" PostTypeId="1" CreationDate="2019-05-27T04:11:00.120" Score="0" ViewCount="51" Body="&lt;p&gt;I have an image of lines (they're actually crystal defects viewed through an SEM)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IUl1P.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IUl1P.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm really after is a histogram of line lengths, but as can be seen from the above micrograph, most lines exit through the image area, making measuring a histogram of their lengths impossible. Also these lines are very long, and taking a much larger picture would reduce the ability to resolve closely spaced lines, and stitching together many pictures of this resolution is generally prohibitively time-consuming. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be measured easily is the spacing between lines (say, the black lines), and a histogram made from these. One could draw many perpendicular lines randomly through the lines, then measure the length between each intersection, for example. I think there should be a relationship between this line spacing distribution and the line length distribution, given that the lines are randomly placed and parallel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas on how I could extract some kind of distribution of line lengths from a line separation histogram?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I essentially believe that there is a measurement or set of measurements that I can do to extract a length distribution. For example, the measurement offered above. Now, this measurement may contain information on not just the length distribution, but the position distribution of the lines (vertical and horizontal), and their orientation distribution (these are distributions based on the set of info of a line: length, x and y positions, and orientation). Thus, if we assume random positions and parallel orientation, there is one distribution left, that being length distribution, which may be extractable given the measured distribution. To clarify, I think the measurement offered would give the set of 4 distributions just by a kind of &quot;solution by extremes&quot;: Imagine changing lines position so that they are all in a row or a column, changing the lengths so that they are either very short or very long, or changing the orientations so that some are nearly parallel to the measurement direction. This all intuitively changed the shape of the measured distribution for me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is all my intuition, so I am asking if anyone has relevant know-how, or can point me in the right direction on how I may be able to, first show that this is possible, and second extract a line length distribution from some measurement(s). I may just post this in the math stack exchange&lt;/p&gt;&#xA;" OwnerUserId="20620" LastEditorUserId="20620" LastEditDate="2019-05-29T20:53:17.533" LastActivityDate="2019-05-29T20:53:17.533" Title="Can you convert a histogram of linear separation of lines into a distribution (histogram, PDF, PMF) of their lengths?" Tags="|statistics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28541" PostTypeId="1" AcceptedAnswerId="28552" CreationDate="2019-05-27T11:34:01.003" Score="1" ViewCount="86" Body="&lt;p&gt;I was passing by a row of motorbikes and mopeds yesterday, and noticed they all have really small radius wheels compared to bicycles. Not all motorbikes are equal (some motorbikes are literally bicycles with engines), but generally it seems that motorbikes have smaller wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume both vehicle types have been optimised and it's not just a fashion choice on the part of the manufacturers. What are they optimising for which makes them different? And why do mopeds/scooters have the broadest range of wheel radiuses when they're the lowest power type of motorbike?&lt;/p&gt;&#xA;" OwnerUserId="20626" LastActivityDate="2019-05-28T15:06:05.683" Title="Bike wheel sizes" Tags="|automotive-engineering|bicycles|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28547" PostTypeId="2" ParentId="28522" CreationDate="2019-05-27T14:00:22.200" Score="1" Body="&lt;p&gt;So, after a little googling around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cellphones typically transmit on 3/4 G at between 0.3 and 3 W - subject to tower location &amp;amp; interference. The out put power is managed by the cell tower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mobile phone wifi is typically 15mW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume min output for 3/4 G at 300 mW (0.3 W) this is, at best, 20 times greater than wifi at 15 mW.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On that basis I would expect wifi to be the more energy efficient medium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However protocol efficiency also plays a part. The simplest crude measure of this is probably to time how long each channel takes to recieve the same block of data and scale the power usage accordingly. Note that subject to your data plan upload / download speeds will usually differ &amp;amp; so there may be different results for each direction.&lt;/p&gt;&#xA;" OwnerUserId="10415" LastActivityDate="2019-05-27T14:00:22.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28548" PostTypeId="1" AcceptedAnswerId="28550" CreationDate="2019-05-27T14:19:56.370" Score="0" ViewCount="625" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sHebp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sHebp.png&quot; alt=&quot;Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, L, theta and P are given, there will be 4 reactions at the wall? (2 horizontal and 2 vertical). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum F_v &amp;amp;= 0 \therefore F_{v,B} + F_{v,A} = P \\&#xA;\sum F_h &amp;amp;= 0 \therefore F_{h,B} = F_{h,A}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What will be the tensile stresses along AB and AC?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not know how to obtain the other 2 equations to solve all the reaction forces at the wall and the tensile stresses? Can anyone please help? One of those equations are from the moment?&lt;/p&gt;&#xA;" OwnerUserId="20631" LastEditorUserId="1832" LastEditDate="2019-05-30T01:34:51.603" LastActivityDate="2019-05-30T01:39:22.313" Title="Truss Analysis Problem" Tags="|mechanical-engineering|civil-engineering|solid-mechanics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28549" PostTypeId="1" CreationDate="2019-05-27T14:22:20.110" Score="6" ViewCount="208" Body="&lt;p&gt;A recent news &lt;a href=&quot;https://www.dailymail.co.uk/news/article-7073489/Anger-controlled-explosion-World-War-II-bomb-London-street-wrecks-cars-smashes-windows.html&quot; rel=&quot;noreferrer&quot;&gt;item&lt;/a&gt; featured this photo of a WW2 bomb being prepared for detonation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone identify the purpose of the large widget sitting on the bomb? it's got a lot of cooling fins on it, and there appears to be fan on the right hand side. There's also what looks like an earthing cable in the foreground and what might be a power cable in the background, so my guess is that its either a heater, or a cooler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that some explosives are inert when cooled sufficiently, so maybe it's an attempt to freeze the bomb to make it safe. I'm also aware that some early bombs could be made safe by injecting steam into them which would melt the explosive and allow it to be poured out, so maybe it's some form so steam injector?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm just guessing. Does anyone know what it really is and what it's supposed to do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EohZG.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EohZG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20630" LastActivityDate="2020-02-06T15:40:49.377" Title="Identify this bomb disposal equipment" Tags="|mechanical|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28550" PostTypeId="2" ParentId="28548" CreationDate="2019-05-27T15:23:23.780" Score="2" Body="&lt;p&gt;As member AB is orthogonal to Point C it cannot impart any vertical reaction, hence Point B provides the only vertical support, and hence has to have a vertical reaction equal to P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to find point B horizontal reaction, we set the sum of moments about point C equal to zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum M_C &amp;amp;=0 \\&#xA;-PL +F_{h_{B}}H &amp;amp;=0 \\&#xA;F_{h_{B}} &amp;amp;= PL/H\text{ pointing left} \\&#xA;\therefore F_{h_{C}}&amp;amp;=PL/H\text{ pointing right}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tensile stress along AB is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;T&amp;amp;=\dfrac{\sqrt{F_{h_{B}}^2+ F_{v_{b}}^2}}{A} \\&#xA;&amp;amp;= \dfrac{\sqrt{(PL/H)^2+P^2}}{A}&#xA;\end{align}$$&lt;/span&gt;&#xA;And I let you do the stress on AC. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2019-05-30T01:39:22.313" LastActivityDate="2019-05-30T01:39:22.313" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28552" PostTypeId="2" ParentId="28541" CreationDate="2019-05-27T17:25:36.963" Score="1" Body="&lt;p&gt;Kick scooters' small wheels have set the inspiration for the battery powered scooters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Small wheels make it easy to mount and dismount the scooter. They have little angular momentum, so speed up faster and slow down to a quick stop faster, which is ideal for urban busy traffic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They turn direction and weave through the traffic or sometimes sidewalks easier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the trade off is, they hardly can take the road bumps or puddles. They are wobbly and unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;New designs are trying to remedy theses defects by using a larger wheel, suspension, setting the caster angle more leaned back to add stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In scooters design, fashion plays a big roll in among other things, sizing the wheels, mostly because of the young consumer in this industry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In bicycles the engineering of efficiency, low drag, balance call for large, narrow size wheels up to a 29-inch diameter. The frame should be designed to be comfortable, give the rider flexibility of a more vertical gait for power pedaling, at the same time let them lean into the wind when cruising. They have wind tunnels and test drag in different configurations. Another advantage of a large wheel aside from control and a smooth ride is large angular momentum gives the wheel stability when rolling fast if they hit an obstacle. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-05-28T15:06:05.683" LastActivityDate="2019-05-28T15:06:05.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28555" PostTypeId="1" CreationDate="2019-05-27T20:19:39.677" Score="1" ViewCount="31" Body="&lt;p&gt;We are in desperate search of suitable pumps for one of our applications.&#xA;We need to be able to pump Diesel fuel in a range going from 5bars to 125bars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do such thing, I've been looking at &lt;a href=&quot;http://www.bosch.co.jp/tms2015/en/products/pdf/GS-Datenblatt_HDP5_EN_low.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bosch HDP5&lt;/a&gt; gasoline direct injection pump (and it's prop valve conveniently attached).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok now... two problems : &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I have doubts about it's behaviour at low pressure (5bars)&lt;/li&gt;&#xA;&lt;li&gt;It's a gasoline pump and we plan to pump diesel&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Structurally, it's a piston pump and my limited understanding of it don't see where the problem could be in pumping lower pressure than it's max.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as far as diesel fuel go, it's got a higher lubricity so I don't see it being a problem either...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But being far from an expert, I'd like to have your feedback on it.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The other option would be to use a Bosch CP1 or CP3 which is a high&#xA;pressure diesel pump.&#xA;Now the fuel is right but the pressures involved are way higher than we need. The pressure compatibility question still stands...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And the last option would be to use a simple External gear pump such&#xA;as the &lt;a href=&quot;https://www.boschrexroth.com/en/xc/products/product-groups/industrial-hydraulics/pumps/external-gear-pumps&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bosch Rexroth&lt;/a&gt; ones but also derating them a lot in terms&#xA;of the pressure and putting diesel fuel in it...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;They are also given for a minimum viscosity of 5mm²/s which is pretty much the viscosity of diesel fuel at ambient temperature, which is a problem since it is usually heating up with time (return/bypass system)&#xA;I can understand the limitation of pumping a liquid with too high a viscosity but I have trouble seeing where it could fail pumping a liquid with too little viscosity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your inputs!&lt;/p&gt;&#xA;" OwnerUserId="20636" LastEditorUserId="20636" LastEditDate="2019-05-27T20:42:51.363" LastActivityDate="2019-05-27T20:42:51.363" Title="Low pressure Diesel pumping with High pressure Gasoline pump?" Tags="|fluid-mechanics|pressure|pumps|hydraulics|diesel|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28557" PostTypeId="2" ParentId="23943" CreationDate="2019-05-28T00:06:18.640" Score="0" Body="&lt;p&gt;Tanklike objects freefall towards a system barycenter. We don't know the spacial distribution of masses that comprise the system, so we can't say much about the gravitational field in the area of the tank. But if the barycenter lies outside the tank, then there is some gradient to the gravitational field, and there will be a corresponding pressure gradient in the fluid in the tank. The rigid tank will accelerate at some average value of the field bounded by the tank, with the fluid at the &quot;front&quot; end at higher pressure, the fluid in the middle at lower pressure, and the fluid at the &quot;back&quot; at higher pressure. These pressures cause tides on a free surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inside the tank, a fluid also self-gravitates. This turns out to be a surprisingly thorny problem to solve for a general self-gravitating fluid mass. The unbounded, self-gravitating, incompressible, rotating fluid solution was finally put to bed by Chandrasekhar in 1967 after the world's best mathematicians having worked on it for about 400 years (Newton started the whole business trying to explain the ocean's tides). The bounded case still appears to be an open problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So no more hand waving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a fun paper - &lt;a href=&quot;http://faculty.georgetown.edu/tl48/lsfinal.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://faculty.georgetown.edu/tl48/lsfinal.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you might want to warm up with Chandra - &lt;a href=&quot;https://onlinelibrary.wiley.com/doi/10.1002/cpa.3160200203&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://onlinelibrary.wiley.com/doi/10.1002/cpa.3160200203&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... and a model for compressible fluids based on Chandra's work - &lt;a href=&quot;https://www.researchgate.net/publication/23881101_Ellipsoidal_figures_of_equilibrium_-_Compressible_models&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.researchgate.net/publication/23881101_Ellipsoidal_figures_of_equilibrium_-_Compressible_models&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-05-28T00:06:18.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28558" PostTypeId="2" ParentId="28541" CreationDate="2019-05-28T01:14:18.887" Score="2" Body="&lt;p&gt;It's all to do with the torque output and rpm of the &quot;motor&quot;. A fit cyclist puts out about the same torque as a schoolbus diesel engine. But they do so at much lower rpm and therefor much less power. So a bicycle needs to gear the 80 rpm source up by a factor of 5 or so to hit 40 mph. Five is about as high as you want to go with sprocket ratios. The rest has to be done with wheel diameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My old Katana spun 9,600 rpm, and needed to convert that to about 160mph. That would be 2400 rpm for 40 mph. That's 30 times the gearing, so instead of gearing up by a factor of 5 as on the bicycle, it would have to gear down by a factor of 6 if it had the same sized tires as a bicycle. It's better to make the wheel diameter smaller and go with less gearing.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2019-05-28T09:29:58.523" LastActivityDate="2019-05-28T09:29:58.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28560" PostTypeId="1" AcceptedAnswerId="28561" CreationDate="2019-05-28T05:49:47.407" Score="0" ViewCount="2547" Body="&lt;p&gt;Is it in the same direction of the applied load or in the co-planar perpendicular direction to the load applied?&lt;/p&gt;&#xA;" OwnerUserId="20638" LastActivityDate="2019-05-28T06:32:46.590" Title="Along which direction is the area moment of inertia taken in bending moment equation?" Tags="|mechanical-engineering|civil-engineering|beam|stresses|strength|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28561" PostTypeId="2" ParentId="28560" CreationDate="2019-05-28T06:32:46.590" Score="1" Body="&lt;p&gt;Bending moment of inertia or second moment of inertia is taken on the surface of the beam parallel to applied loads' vertical component and perpendicular to the long axis of the beam, about a horizontal line on that surface, called neutral axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When loads are applied to a beam they want to bend the beam into a deflected down curve. They create bending moment in the beam which is countered by tension stresses on the lower part and compression stresses on the upper part of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment of inertia is a characteristic property of the beam section which defines how that cross section bends and how much stress both tension and compression it takes to bend, and how this stress is distributed on that section of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a engineering representation of the stiffness of the beam at that point.&#xA;In prismatic beams, extruded shapes, moment of inertia or I is constant along the length of the beam. But if the Beam's cross section varies along its length, then the I varies along the length of the beam.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-28T06:32:46.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28565" PostTypeId="2" ParentId="28548" CreationDate="2019-05-28T15:30:41.887" Score="0" Body="&lt;p&gt;Resolve at point A, remembering that in a truss the members only carry axial loads:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Vertically P = F&lt;sub&gt;AB&lt;/sub&gt; * sin theta&lt;/li&gt;&#xA;&lt;li&gt;Horizontally F&lt;sub&gt;AB&lt;/sub&gt; * cos theta + F&lt;sub&gt;AC&lt;/sub&gt; = 0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This gives you the member forces without even needing to worry about the reactions (though they're pretty trivial once you've worked out the member forces).&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2019-05-28T15:51:59.517" LastActivityDate="2019-05-28T15:51:59.517" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28566" PostTypeId="2" ParentId="18536" CreationDate="2019-05-28T15:50:20.097" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://archive.org/details/en.1993.1.5.2006&quot; rel=&quot;nofollow noreferrer&quot;&gt;Eurocode 1993-1-5 Design of Steel Structures&lt;/a&gt; - Plated structural elements, &lt;em&gt;Annex C&lt;/em&gt; Clause &lt;em&gt;C.6 Material properties&lt;/em&gt; allows the following assumptions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) elastic-plastic without strain hardening &lt;em&gt;(i.e. elastic-perfectly plastic)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) elastic-plastic with a nominal plateau slope &lt;em&gt;(this uses a very small young's modulus post-yield, E/10000, to help with convergence difficulties experienced in finite element analysis)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) elastic-plastic with linear strain hardening &lt;em&gt;(this uses E/100 to model the post-yield young's modulus)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) true stress-strain curve modified from test results&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nRWy3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nRWy3.png&quot; alt=&quot;Figure C.2 showing options a to d as indicative stress-strain curves&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option c would be applicable for your situation: beyond the elastic region you can approximate the Young's Modulus as 1/100th of the elastic Young's Modulus.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2019-05-28T15:50:20.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28568" PostTypeId="1" CreationDate="2019-05-28T16:09:14.057" Score="0" ViewCount="416" Body="&lt;p&gt;I have this projection of two intersecting cones, and I am asked to draw the intersection curve of the two cones on the projection. What is the right way to do it? &lt;a href=&quot;https://i.stack.imgur.com/2wkgj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2wkgj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20644" LastActivityDate="2019-05-29T09:30:55.250" Title="Engineering Drawing-Projection of intersection curve of two standing cones" Tags="|mechanical-engineering|technical-drawing|drawings|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28569" PostTypeId="2" ParentId="28548" CreationDate="2019-05-28T18:02:30.810" Score="1" Body="&lt;p&gt;@AndyT and @kamran explained how to technically solve this question.&#xA;I would like to add a short comment about the thing seemed to confuse you the most.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, there are 4 reactions in this problem. If the triangle was made of a solid body, you would deal with a statically indeterminate problem and find it quite hard to resolve those reactions. However, since we are dealing with a truss, you should examine carefully whether its statically determinate or not:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The number of unknowns is 6 - the above mentioned 4 reactions + axial&#xA;forces along the truss members&lt;/li&gt;&#xA;&lt;li&gt;The number of equation is also 6 - 3 pin joints, each has 2&#xA;equilibrium equations (in X and Y directions)&lt;/li&gt;&#xA;&lt;li&gt;Since the number of unknown equals the number of equations - you deal&#xA;with statically determinate problem and can resolve all the&#xA;reactions without dealing with the truss stiffness&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2019-05-28T18:02:30.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28572" PostTypeId="1" CreationDate="2019-05-28T23:59:13.170" Score="0" ViewCount="387" Body="&lt;p&gt;so i was given a problem by my mechanical principles teacher to work out the max vertical deflection of a simply supported beam with two point loads one a distance of 5.3 away from A and another 5.9 away from a on a 6m beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the point loads are  both 15.7KN and the UDL is 1.2Kn/M &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now ive worked out most of the constants using the double integration method but the main problem im getting is that c1=c3=c5 = a positive number when an online calculator shows me that its meant to be negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0uFDE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0uFDE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kdok2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kdok2.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fWJew.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fWJew.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JnvB5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JnvB5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20646" LastActivityDate="2020-10-20T12:04:15.293" Title="simple supported Beam deflection question involving two point loads and a UDL" Tags="|beam|deflection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28573" PostTypeId="1" AcceptedAnswerId="28588" CreationDate="2019-05-29T01:57:14.743" Score="1" ViewCount="57" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ P_m=P_n*{G\frac1{G_{STC}}}*[1-\gamma*(T_c-25)]$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This equation is applied on sollar cells. &lt;span class=&quot;math-container&quot;&gt;$ P_m$&lt;/span&gt; is the mean power produced in 15 minutes, &lt;span class=&quot;math-container&quot;&gt;$P_n$&lt;/span&gt; is the nominal power [W], G is the insolation [W/m2], &lt;span class=&quot;math-container&quot;&gt;$G_{STC}$&lt;/span&gt; (the subscript still in the wrong place, I don't know why) is the radiation [W/m2] in standard testing conditions &lt;a href=&quot;https://sinovoltaics.com/learning-center/quality/normal-operating-cell-temperature-noct-definition-noct-definition/&quot; rel=&quot;nofollow noreferrer&quot;&gt;reference here&lt;/a&gt;. Last, but nost least, &lt;span class=&quot;math-container&quot;&gt;$T_c$&lt;/span&gt; is the cell temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the parameter gamma (&lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt;)? I have a value for gamma, 0.005 [adimensional], but I don't know what it means or its definition. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorDisplayName="user20096" LastEditDate="2019-05-29T11:11:54.393" LastActivityDate="2019-05-29T15:38:47.957" Title="What's the name of this parameter?" Tags="|solar-energy|photovoltaics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28574" PostTypeId="2" ParentId="28520" CreationDate="2019-05-29T02:49:05.503" Score="1" Body="&lt;p&gt;Unless you are going to pump water removed, true counter flow will be difficult due to low pressure drop to support flow velocity on discharge. Perhaps you could coil up thin gauge tubing in an overflow bucket or drum. Run the supply water through the tubing. For heat transfer you can get good heat transfer on tube side but on return side it would be low due to low velocity unless you are going to pump the return as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For overall heat transfer coefficient, it's the inverse of the sum of the inner and outer convective resistances conduction resistance and any fouling factor. For convective losses there are correlations called Nusselt number correlations. It is a non-trivial effort to figure out but can be done with some assumptions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The big obstacle is return flow is not the same as supply flow due to evaporation losses so this may make heat recovery impractical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without having done the calculation of the overall heat transfer coefficient for this problem I wonder whether there is much savings to be had&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for efficiency, the commenter above is correct: it depends on surface area, and I think also on ambient heat losses to environment for hot/returned fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &lt;span class=&quot;math-container&quot;&gt;$1 \frac{\mathrm{l}}{\mathrm{s}}$&lt;/span&gt; flow, l from the author comment above, neglecting any heat recovery the heat load would be&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;Q &amp;amp;= \dot{m} \, C_{\text{p}} \, \Delta T&#xA;  \\[10px]&#xA;  &amp;amp;= 1 \frac{\mathrm{l}}{\mathrm{s}}&#xA;     \times \left(20\sideset{^{\circ}}{}{\mathrm{C}}-1 \sideset{^{\circ}}{}{\mathrm{C}}\right)&#xA;     \times 1 \frac{\mathrm{kg}}{\mathrm{l}}&#xA;     \times 4.18 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}&#xA;  \\[10px]&#xA;  &amp;amp;= 79 \, \mathrm{Watts}&#xA;\end{align}&#xA;\,.$$&lt;/span&gt;&#xA;This is the heat requirement before any heat recovery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you are saving electric heat in the United States, this is worth about&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;0.13 \frac{\mathrm{USD}}{\mathrm{kW} \cdot \mathrm{hr}} \times .079 \, \mathrm{kW}&#xA;~~=~~ .103 \frac{\mathrm{USD}}{\mathrm{hr}}&#xA;~~\approx~~ 900 \frac{\mathrm{USD}}{\mathrm{year}}&#xA;\,,$$&lt;/span&gt;if you could recover 100 percent.&lt;/p&gt;&#xA;" OwnerUserId="14901" LastEditorUserId="10678" LastEditDate="2019-06-02T20:18:26.157" LastActivityDate="2019-06-02T20:18:26.157" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28575" PostTypeId="1" CreationDate="2019-05-29T04:11:18.020" Score="1" ViewCount="24" Body="&lt;p&gt;What is the smallest grain size achievable (both theoretical and practical) using &lt;em&gt;mechanical&lt;/em&gt; methods to crush a fracturable mineral composed primarily of a single molecular species? Example minerals include: calcium carbonate minerals such as limestone or chalk; silicon dioxide minerals (quartz, sand); diamond (carbon); sapphire (aluminum oxide).&lt;/p&gt;&#xA;" OwnerUserId="20609" LastEditorUserId="20609" LastEditDate="2019-05-29T05:12:52.113" LastActivityDate="2019-05-29T05:12:52.113" Title="What is the smallest grain size achievable using mechanical methods?" Tags="|materials|crushing|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28577" PostTypeId="1" CreationDate="2019-05-29T05:12:59.247" Score="0" ViewCount="394" Body="&lt;p&gt;There are examples of using SOL 108 and SOL 111 to create a frequency response.  In Siemens Simcenter (NX Nastran), the examples focus on a tower with a seismic mass.  The model drives the mass with an acceleration, attaching the mass to the structure using rigid elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S3eUP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S3eUP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This creates a model that adds (?) stiffness at the mass attachment points.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if I want to simulate the frequency response function (FRF) of a structure that is unconstrained?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The test I conducted was to hang my part on a wire and tap (laterally) it with an impulse hammer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PlFPf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PlFPf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might be equivalent to constraining the support point as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fixed in DOF's 1, 2 (Tx and Ty)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I could apply a lateral acceleration (in the direction of the hammer tap).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the conventional way to produce the FRF?  If not, what is?&lt;/p&gt;&#xA;" OwnerUserId="7569" LastActivityDate="2019-05-29T13:12:13.143" Title="Nastran / Frequency Response / Free-Free" Tags="|frequency-response|nastran|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28579" PostTypeId="2" ParentId="28572" CreationDate="2019-05-29T05:43:22.283" Score="1" Body="&lt;p&gt;I calculated the same but I got a different value from yours.&#xA;The maximum deflection that I got is .31818 mm at x = 3.25 m&#xA;The constant from the double integration method is -23.189.&#xA;I checked it with an online calculator and its perfect...&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1QbHm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1QbHm.jpg&quot; alt=&quot;Picture 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gdehn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gdehn.jpg&quot; alt=&quot;Picture 2&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YYvIz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YYvIz.jpg&quot; alt=&quot;Picture 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20638" LastEditorUserId="20638" LastEditDate="2019-05-29T09:44:36.620" LastActivityDate="2019-05-29T09:44:36.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28582" PostTypeId="2" ParentId="28568" CreationDate="2019-05-29T07:52:11.410" Score="2" Body="&lt;p&gt;Have look at this and then see which intersections give you the points for the outline (this is not exact - just lines drawn on a simple graphics package...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C2CmJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C2CmJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-05-29T07:52:11.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28585" PostTypeId="1" CreationDate="2019-05-29T09:23:25.337" Score="1" ViewCount="638" Body="&lt;p&gt;I am designing an automated hot wire foam cutter for my project, and I am stuck on the computation for the most viable cutting speed of my design. The cutting tool used is a nichrome wire and the medium is a polystyrene board. How do I solve for the best cutting speed of a heated nichrome wire given the melting point of styrofoam boards have a melting point of 240 degrees Celsius, and the Styro board has a thickness of about half an inch.&lt;/p&gt;&#xA;" OwnerDisplayName="Shan Jearren Latasa" LastEditorUserId="6767" LastEditDate="2019-05-29T15:30:08.937" LastActivityDate="2019-05-29T22:23:02.047" Title="How do you determine the apt cutting speed of a hot wire foam cutter?" Tags="|tools|cutting|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28586" PostTypeId="2" ParentId="28577" CreationDate="2019-05-29T13:12:13.143" Score="2" Body="&lt;p&gt;If your structure was literally hung on a piece of (plastic covered) wire then it &lt;em&gt;was&lt;/em&gt; constrained vertically, at that point, by the wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better method is to hang it from a soft bungee cord, so its natural frequency of vibration as a &quot;mass on a spring&quot; moving vertically is a factor of 10 (or more) lower than the frequencies you are interested in measuring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to answer the question, to find a FRF for an unconstrained structure in Nastran, you just run the analysis with no constraints and no other &quot;tweaks&quot;. So long as you don't try to calculate the response at a frequency of 0 Hz (which obviously doesn't mean anything physically), the analysis should &quot;just work&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modelling your &quot;stiff&quot; wire support correctly is actually quite difficult, because if you think about it, the structure can behave like a compound pendulum consisting of a rigid wire and a rigid mass, with different vibration modes in different planes because of the different rotational inertia properties of the mass in different planes. Those responses are &lt;em&gt;not&lt;/em&gt; &quot;linear&quot; behaviour, because they depend on the stress distribution in the model caused by its self weight - the most significant effect being the tension in the string.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-05-29T13:12:13.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28587" PostTypeId="1" CreationDate="2019-05-28T08:32:27.030" Score="2" ViewCount="4410" Body="&lt;p&gt;I have an exercise bike that works by a (I'm assuming solid iron) flywheel and a 'horseshoe' of magnets that are moved in and out to set the difficulty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attached an image of the same design found on Google Images so you can see what I have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RfsCh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RfsCh.jpg&quot; alt=&quot;view of exercise bike mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to do is make the bike harder to pedal. For this, I have bought a bunch of neodymium magnets online, after reading that they were the strongest magnets you can easily get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I bought a bunch of circular ones and smaller flat ones. I've routed an 'arm' out of MDF to screw them onto - so far I've only done the circular ones but I would have gone on to try the flat ones as well. I've illustrated what I'm doing here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jk0Ti.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jk0Ti.png&quot; alt=&quot;demo of lever arm&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm using the MDF arm as a lever to move the magnets in and out in the same way as the horseshoe would. The idea is to test the concept before I go messing around with the real horseshoe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And I'm glad I tested it because so far, nothing is working as I would intuitively expect. The magnets have no effect at all on the flywheel!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's going on?&lt;/p&gt;&#xA;" OwnerUserId="20657" OwnerDisplayName="Tim" LastEditorUserId="6767" LastEditDate="2019-05-29T15:30:03.757" LastActivityDate="2021-02-25T22:04:15.140" Title="How do I add magnets to the mechanism of my exercise bike?" Tags="|magnets|bicycles|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="28588" PostTypeId="2" ParentId="28573" CreationDate="2019-05-29T15:27:24.773" Score="2" Body="&lt;p&gt;It's the &lt;a href=&quot;https://sinovoltaics.com/solar-basics/measuring-the-temperature-coefficients-of-a-pv-module/&quot; rel=&quot;nofollow noreferrer&quot;&gt;temperature coefficient&lt;/a&gt; of the panel. It varies by panel, but 0.005 (0.5%) is a fairly normal value for it. Its unit is 1/degree C. It's the rate at which cell efficiency drops off, as the cell temperature rises above 25 degrees C.&lt;/p&gt;&#xA;" OwnerUserId="163" LastEditorUserId="163" LastEditDate="2019-05-29T15:38:47.957" LastActivityDate="2019-05-29T15:38:47.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28590" PostTypeId="2" ParentId="28585" CreationDate="2019-05-29T22:23:02.047" Score="1" Body="&lt;p&gt;Here is what I know, from my slight experience hand-cutting foam for model airplanes, and from things I've heard from actual experts:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;It varies with wire temperature, which, in turn, varies with applied voltage and cutting speed.  So there's no one right speed, and things may change dramatically if you actively control for wire temperature.&lt;/li&gt;&#xA;&lt;li&gt;It varies with foam density (although usually the model airplane guys are using the lightest foam they can lay their hands on).&lt;/li&gt;&#xA;&lt;li&gt;You want to look for a range, because if you're cutting something tapered then the big end needs to go faster than the little end.&lt;/li&gt;&#xA;&lt;li&gt;To some small extent you can compensate for slow cutting by allowing for a bigger kerf -- a hot wire going slowly through foam melts more material, and you can sorta-kinda fix that.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I think you'll just need to experiment.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-05-29T22:23:02.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28591" PostTypeId="2" ParentId="28505" CreationDate="2019-05-30T05:40:42.073" Score="1" Body="&lt;p&gt;Found out about Response Surfaces and using all the points that I already created, created a Response Surface Optimization which let me easily find the answer without a significant amount more computation.&lt;/p&gt;&#xA;" OwnerUserId="1434" LastActivityDate="2019-05-30T05:40:42.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28593" PostTypeId="1" CreationDate="2019-05-30T07:35:40.140" Score="3" ViewCount="109" Body="&lt;p&gt;We are using a load cell as shown below, the data sheet for which can be found on this &lt;a href=&quot;https://www.te.com/commerce/DocumentDelivery/DDEController?Action=srchrtrv&amp;amp;DocNm=FS20&amp;amp;DocType=DS&amp;amp;DocLang=English&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/xcj4N.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xcj4N.jpg&quot; alt=&quot;Load Cell&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vAErF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vAErF.jpg&quot; alt=&quot;Drawing of the Load cell from Data sheet&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load cell needs to coupled with a plunger as shown below.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/y2qcg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y2qcg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plunger is supposed to measure the axial force acting on it. Initially, we had just stuck the plunger with the help of glue on top of the load cell. But with this setup, &lt;strong&gt;if the force acting is not completely axial&lt;/strong&gt;, it tends to cause a moment which leads to &lt;strong&gt;incorrect measurement of the force&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;We are interested in knowing a reliable way of coupling to solve the issue.&lt;/strong&gt;&#xA;Also, we would like to know if there is some reading reference available regarding this.&lt;/p&gt;&#xA;" OwnerUserId="20664" LastActivityDate="2019-05-31T07:01:07.593" Title="Coupling of load cell with a plunger" Tags="|mechanical-engineering|coupling|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28594" PostTypeId="2" ParentId="28593" CreationDate="2019-05-30T13:22:15.813" Score="2" Body="&lt;p&gt;Could you not make a guide of sort to constrain the plunger action on the load cell to be along the right axis. Something like that (probably can do something better, but that's all I had time to do):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yG429.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yG429.png&quot; alt=&quot;perspective&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hEoZ3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hEoZ3.png&quot; alt=&quot;side view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UEa3s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UEa3s.png&quot; alt=&quot;top view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so fairly tight around the diameter of the pin (say 5.5mm hole), and resting on the surfaces of the load cell to ensure perpendicularity. Can the plunger move up and down or does it have to be stuck to the load cell?&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-05-30T13:22:15.813" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28596" PostTypeId="1" AcceptedAnswerId="28630" CreationDate="2019-05-30T15:22:16.273" Score="3" ViewCount="459" Body="&lt;p&gt;After doing some research, I have come to the realization that 3 blade design horizontal turbines have the greatest maximum coefficient of performance. Most sources seem to agree a Cp of ~0.45 at TSR=7 is optimal for 3 blade designs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power output of a turbine is P=0.5*Cp&lt;em&gt;rho&lt;/em&gt;A*V^3. So what I cannot wrap my head around is why, all other things being equal, more blades (of course up to a limit) does not equal a greater coefficient of performance and thus more power. How is only 3 blades this limit on something 200m in diameter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems there would be a significant portion of the air passing freely undisturbed through the rotor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I welcome any answers and explanations, but preferably backed by formulas.&lt;/p&gt;&#xA;" OwnerUserId="20668" LastActivityDate="2019-06-01T11:57:13.490" Title="Do 3 blade wind turbines generate more power than a 9 blade, all else being equal?" Tags="|power|turbines|energy-efficiency|wind-power|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28597" PostTypeId="1" AcceptedAnswerId="28601" CreationDate="2019-05-30T16:34:10.963" Score="2" ViewCount="316" Body="&lt;p&gt;Does anyone know what these three symbols are? I am working to identify them but they do not seem to appear in the site: &lt;a href=&quot;https://www.edrawsoft.com/pid-legend.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.edrawsoft.com/pid-legend.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know a better reference for P&amp;amp;ID diagram symbols?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xwgJi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xwgJi.png&quot; alt=&quot;Sample from a project I am working on&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20670" LastActivityDate="2019-08-27T00:41:49.260" Title="P&amp;ID Symbol Identification Question" Tags="|piping|technical-drawing|instrumentation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28599" PostTypeId="2" ParentId="28596" CreationDate="2019-05-30T17:55:10.623" Score="2" Body="&lt;p&gt;Very basically in modern wind turbines which act as a wing and get lift, if you add too many blades because of the wake interference you will get kind of a stall in lift because you will develope a big ellipsoid of air rotating with the blades as opposed to going through and imparting lift on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In jet engines they counter this by adding stator blades and adjustable vanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the number of blades is limited to a few, in contrast with old windmills which worked on the wind drag effect, but were not nearly as effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have a choice of 4 or 3 or 2 blades engineering concerns such as cost, size, ease of maintenance, transportation of the parts, erection, noise, lead to the optimal choice as a three blade.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-05-30T17:55:10.623" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28600" PostTypeId="2" ParentId="5344" CreationDate="2019-05-30T21:21:33.423" Score="5" Body="&lt;p&gt;I think answer to the question &lt;em&gt;would the bearing stress differ&lt;/em&gt;   is ==&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;YES!&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;bearing stress &lt;em&gt;would&lt;/em&gt; vary along the bolt in different sections in plates...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and you &lt;strong&gt;cannot&lt;/strong&gt; look at the entire bolt at once to calculate the bearing stress &lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is a pictorial explanation why do we need to consider it in different sections &amp;amp;  how to calculate bearing stresses (if you want to).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let us assume that we have a 3 plate arrangement like shown in pic.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/W3kr1.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W3kr1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where this blue thing on the right is called &lt;strong&gt;CLEVIS&lt;/strong&gt; and on the LHS we have a plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the cross sectional view let us say:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q8TbY.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q8TbY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This connection consists of a flat bar A (which i will refer here as just plate), a clevis C, and a bolt B that passes through holes in the plate and clevis. Under the action of the tensile loads P, the plate and clevis will press against the bolt in bearing, and contact stresses, called bearing stresses, will be developed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bearing stresses being contact stresses depend upon the contact between two surfaces so the answer to * Do I have to look at the entire bolt at once to calculate the bearing stress * is answered here&lt;/strong&gt; and the answer is  &lt;strong&gt;NO&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;you have to look at each contact area not just because you want to calculate but more because looking at the entire bolt at once to calculate the bearing stress is&lt;/em&gt; CONCEPTUALLY WRONG!!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;amp; &#xA;NOW, assume that clevis has upper and lower plate of different thickness... even if it doesn't look like... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but let's just say it for a moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;these two plates of CLEVIS (upper and lower) are under forces say: &lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$P_3$&lt;/span&gt; (respectively).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and using equations of static equilibrium:     &lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt;  +   &lt;span class=&quot;math-container&quot;&gt;$P_3$&lt;/span&gt;  =  &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Bearing&lt;/strong&gt; stresses exerted by the clevis against the bolt appear on the left-hand side of the free-body diagram (see 1 and 3) . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;amp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Bearing&lt;/strong&gt; stresses from the plate appear on the right-hand side (see 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, we make an assumption i.e.:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, The actual distribution of the bearing stresses is difficult to determine, so we customarily assume that the stresses are uniformly distributed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based upon the assumption of uniform distribution, we can calculate an average bearing stress &lt;span class=&quot;math-container&quot;&gt;$\sigma_{b}$&lt;/span&gt; by dividing the total bearing force &lt;span class=&quot;math-container&quot;&gt;$F_b$&lt;/span&gt; by the bearing area &lt;span class=&quot;math-container&quot;&gt;$A_b$&lt;/span&gt; , since stress is Force divided by acting area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The projected area &lt;span class=&quot;math-container&quot;&gt;$A_b$&lt;/span&gt; on which they act is a rectangle having a height equal to the thickness of chosen plate (upper or lower) of the clevis and a width equal to the diameter of the bolt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bwEKl.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bwEKl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SO,following would be the EXPRESSIONS for bearing stresses at different contacts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$plate_1$&lt;/span&gt; (of clevis) &lt;span class=&quot;math-container&quot;&gt;\begin{align} \begin{pmatrix} \sigma_1 = \dfrac{P_1}{d * t_1}  \end{pmatrix} \end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$plate A$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;\begin{align} \begin{pmatrix} \sigma_2 = \dfrac{P}{d * t_2}  \end{pmatrix} \end{align}&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$plate_3$&lt;/span&gt; (of clevis) &lt;span class=&quot;math-container&quot;&gt;\begin{align} \begin{pmatrix} \sigma_3 = \dfrac{P_3}{d * t_3}  \end{pmatrix} \end{align}&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and until and unless &lt;span class=&quot;math-container&quot;&gt;$t_1$&lt;/span&gt; == &lt;span class=&quot;math-container&quot;&gt;$t_3$&lt;/span&gt;  bearing stresses would differ &lt;/p&gt;&#xA;" OwnerDisplayName="user19926" LastActivityDate="2019-05-30T21:21:33.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28601" PostTypeId="2" ParentId="28597" CreationDate="2019-05-30T21:53:36.757" Score="2" Body="&lt;p&gt;Two of those are not standard usage symbols in that way.  You may to have to use context and/or a physical inspection to determine what they are. &lt;strong&gt;Where is that diagram from?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Various organizations define symbols for piping and/or control.  The &lt;em&gt;International Society of Automation&lt;/em&gt; uses &lt;em&gt;ANSI/ISA-5.1-2009&lt;/em&gt;, which you can officially &lt;a href=&quot;https://www.isa.org/store/products/product-detail/?productId=116630&quot; rel=&quot;nofollow noreferrer&quot;&gt;find on their website&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The &lt;strong&gt;top symbol&lt;/strong&gt; is probably meant to be a shut-off valve, probably ball type.  The usual symbol is something like:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tOJRs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tOJRs.png&quot; alt=&quot;ball valve&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;br&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;middle symbol&lt;/strong&gt; is a fairly standard &lt;strong&gt;Solenoid-controlled valve&lt;/strong&gt;.&lt;br&gt;&#xA;&lt;br&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;bottom symbol&lt;/strong&gt; could be meant to be a butterfly valve (for flow control), or a &quot;blowout&quot; valve / rupture disk (for over-pressure safety).&lt;br&gt;&#xA;The standard symbols for those are like:&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3Q3p7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Q3p7.png&quot; alt=&quot;butterfly valve&quot;&gt;&lt;/a&gt;  &lt;strong&gt;and&lt;/strong&gt;  &lt;a href=&quot;https://i.stack.imgur.com/zLMDy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zLMDy.jpg&quot; alt=&quot;pressure&quot;&gt;&lt;/a&gt;  &lt;strong&gt;or&lt;/strong&gt;  &lt;a href=&quot;https://i.stack.imgur.com/ZqlBF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZqlBF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3339" LastEditorUserId="3339" LastEditDate="2019-08-27T00:41:49.260" LastActivityDate="2019-08-27T00:41:49.260" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28603" PostTypeId="1" CreationDate="2019-05-31T01:40:59.407" Score="-1" ViewCount="528" Body="&lt;p&gt;I have a difficult and interesting design project to work on and I am trying to get my hands on some ballpark figures from people who know engineering a little better. Help is appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying design a lightweight pressure hull for a hybrid between a submarine and sailing vessel. The real question here is how thin does the hull have to be and of what possible materials. (All of the below figures are only approximate because their final form is affected so much by all of the other figures final forms.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a large number of constraints in this problem. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The pressure hull needs to be ~40' long with ~8' diameter. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It needs to hopefully weigh under 20,000 pounds&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It needs to dive to 40' with a safety margin of 60' (built for 100' or ~44psi).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It's only internal structure would ideally be shape keeping rings to keep a little of the force in tension.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It needs to be manufacturable under 1,000,000 USD (foregoing labor costs)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;It needs to allow for a hatch, portholes, and other imperfections without completely eliminating the structural integrity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Possible materials I have done minimal research in are steel, carbon fiber, fiber glass, and some impracticably expensive exotic alloys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option I have looked into is taking an existing &lt;em&gt;internal&lt;/em&gt; pressure steel tank, like those below, and converting it. The issue is I cannot figure out how to calculate the conversion from internal to external pressure. &lt;a href=&quot;http://www.arcenergyequipment.com/index.php/inventory/h-separators-vessels-128.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.arcenergyequipment.com/index.php/inventory/h-separators-vessels-128.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Light, cheap, big, strong... something has got to go. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the constraints are impossible, any numbers and equations you could point me to would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a great day!&#xA;-Ben&lt;/p&gt;&#xA;" OwnerUserId="20673" LastEditorUserId="6790" LastEditDate="2020-05-04T08:02:46.023" LastActivityDate="2020-05-04T08:02:46.023" Title="What is the optimal building material for a submarine pressure hull?" Tags="|materials|pressure|naval-engineering|submarines|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28604" PostTypeId="2" ParentId="28603" CreationDate="2019-05-31T02:04:02.443" Score="0" Body="&lt;p&gt;Let me first iterate that I’m not a professional engineer but if you can settle for a harder material, a composite like Kevlar or fiberglass could do. However, if you are willing to use a more flexible material, you’ll want more than rings inside the hull just to be safe.&lt;/p&gt;&#xA;" OwnerUserId="18656" LastActivityDate="2019-05-31T02:04:02.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28606" PostTypeId="2" ParentId="28593" CreationDate="2019-05-31T07:01:07.593" Score="0" Body="&lt;p&gt;Responding to @am304&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually making a guide would have worked perfectly if the application had involved measuring vertical axial force. In our case, the whole assembly is supposed to be horizontal with the purpose of measuring the horizontal axial force. The image shown below will explain it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7LmLo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7LmLo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that although the distributive load acting on the plunger due to the guide causes a moment, it will cause an error in the measurement of the axial force to some extent since we plan on having the least count of 100 mgs (milligrams). I am not sure about how true this is. But it feels like it will have an impact in the magnitude of the force calculated by the plunger.&#xA;It was also seen while doing a check on the load cell that the same weight placed at the centre of the load cell and at the periphery was giving a &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you feel about it? Also, is there any way I can eliminate this moment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. - The plunger can have a relative movement w.r.t. load cell and need not be stuck to it.&lt;/p&gt;&#xA;" OwnerUserId="20664" LastActivityDate="2019-05-31T07:01:07.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28607" PostTypeId="1" AcceptedAnswerId="28609" CreationDate="2019-05-31T08:07:57.110" Score="0" ViewCount="295" Body="&lt;p&gt;I'm a programmer working on a project where we will be controlling a crane's boom for motion control. We are currently a year out from having the actual crane in house, but are doing the software ahead of time as we can 3d simulate a lot of it. I thought it would be a fun idea to make a little miniature version to play with (and it would be a fun toy for the office). Our crane boom consists of 4 segments and to scale it down somewhat proportionally each square tube will be 1.5 feet long and the first extension will be 2 inch outer diameter, with each inner tube getting 1/4 smaller. Making a little mechanism to pan and tilt the boom is simple, but I'm not sure how to go about controlling the extension with a motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pco3T.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pco3T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ideally I'd like to control the extension with just one motor I can mount at the base of the boom. My question is, is there a mechanism that can create this collapsible column with just one motor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize the easy solution is probably 4 separate motors that use a rack and pinion to extend each segment, but I don't think Ill have the space for the motors at this scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option I've pondered is using a spring and then having the motor just wind up a string that pulls to compress the spring but I feel acquiring a spring of the right diameter, length and strength would prove challenging as at full extension this will be about 4 feet, though I'd like to make it as small as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydraulics and pneumatics are completely out of my ability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, each segment should move the same amount as the boom extends. I feel like there has to be some sort of mechanism for this but I can't think of anything else. I think our actual crane uses a bunch of pulleys but I'm not sure how that works. My boss is designing the crane, but I'd like to do this as a surprise for him so I don't want to ask him for his input.&lt;/p&gt;&#xA;" OwnerUserId="20678" LastActivityDate="2019-05-31T10:07:05.830" Title="Motorized 4 part extension column mechanism?" Tags="|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28608" PostTypeId="1" CreationDate="2019-05-31T09:38:07.547" Score="0" ViewCount="17" Body="&lt;p&gt;I would like to know if submersible pumps can be used to extract leachate from leachate wells. The leachate is likely to contain solids of size 15-20mm. Further the pump should have diameter less than 200mm. The head required is 65-70m and a min flow of 500 L/hour. If not then please suggest an alternative. &lt;/p&gt;&#xA;" OwnerUserId="20679" LastActivityDate="2019-05-31T09:38:07.547" Title="can submersible pumps be used for leachate from landfill that may have some solid particles of size 15-20mm?" Tags="|pumps|environmental-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28609" PostTypeId="2" ParentId="28607" CreationDate="2019-05-31T10:07:05.830" Score="0" Body="&lt;p&gt;Many moons ago, I was involved in a process which required a transportable antenna tower. The assembly was similar to your requirements. Tilt the tower module vertical and crank a winch handle to cause each of the &quot;telescoping&quot; segments to elevate. One crank, many pulleys and a strong wire cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T9tTE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T9tTE.png&quot; alt=&quot;extensible crank assembly&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The drawing above is not to scale and placement of the pulleys are not critical. The individual segments to be extended would require guiding mechanisms as the segments are offset from the centerline. In the case of the antenna tower, the guiding mechanisms were a combination of low-friction plastic guide blocks and boat trailer rollers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A series of nylon or teflon buttons would likely serve your requirements for guidance.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-05-31T10:07:05.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28610" PostTypeId="1" AcceptedAnswerId="28615" CreationDate="2019-05-31T10:24:05.113" Score="19" ViewCount="4245" Body="&lt;p&gt;According to many sources, one of the purposes of building's foundations is &quot;To distribute the weight of the structure over large area so as to avoid over-loading of the soil beneath.&quot; &lt;a href=&quot;https://en.wikipedia.org/wiki/Foundation_(engineering)&quot; rel=&quot;noreferrer&quot;&gt;(Wikipedia)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the Wikipedia page on foundations, there is the following picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FnVb1.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FnVb1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the foundation seems to consist of some type of &quot;legs&quot;, with a larger surface area on the bottom. But if the point is to distribute the weight of the building over an area, why not put the bottom of the building in contact with ground? Isn't the bottom of the building itself a much larger contact area than the area the foundation can provide? In the picture, the bottom of the building does seem to rest at ground, so why do we require the foundation to have these large surface area rectangles at the bottom if the whole weight of the building is already resting on ground?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is another picture from the same page:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9QElx.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9QElx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems to be some sort of lodge or a cabin, with stones used as foundation. Here it is evident that the building itself is raised above ground with the stones used as a contact to ground. But now the small stones suffer all the weight of the structure. Wouldn't they be under quite a heavy load over small contact area and possibly sink into the ground? Why not simply have the building's floor, a much larger contact area, rest on the ground?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastEditorUserId="13372" LastEditDate="2019-05-31T10:42:40.210" LastActivityDate="2019-05-31T21:32:29.297" Title="What is the purpose of building foundations?" Tags="|structural-engineering|foundations|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28611" PostTypeId="2" ParentId="28610" CreationDate="2019-05-31T10:33:28.557" Score="6" Body="&lt;p&gt;Those stones have larger stones underneath them in the ground so that they will not sink - as the load is spread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The purpose is to prevent rats getting into the storage - this was a technique used in many countries for barns etc &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This was also done to stop snow sitting against the lower part of the wall in many cases as well.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-05-31T13:01:01.593" LastActivityDate="2019-05-31T13:01:01.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28613" PostTypeId="2" ParentId="28536" CreationDate="2019-05-31T11:59:26.257" Score="2" Body="&lt;p&gt;The occlusion of your mesh is the ratio between the area occluded by a solid plate of the same mass (and thickness) and the area covered by your mesh (which you can measure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Occlusion = \dfrac{m/(t \cdot p)}{(l \cdot h)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; = mass of your mesh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; = thickness of your mesh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; = density of your mesh (steel being 8050kg/m^3)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; = length and height of your mesh&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously only works on mesh manufactured by slitting and expanding, it doesn't hold for punched mesh.&lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2019-05-31T11:59:26.257" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28614" PostTypeId="2" ParentId="28536" CreationDate="2019-05-31T12:32:04.607" Score="1" Body="&lt;p&gt;Why not calc the area of one (2 triangles) of the open areas and knowing the triangle count per screen total area you have it!&lt;/p&gt;&#xA;" OwnerUserId="1322" LastActivityDate="2019-05-31T12:32:04.607" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28615" PostTypeId="2" ParentId="28610" CreationDate="2019-05-31T12:41:26.777" Score="18" Body="&lt;h3&gt;Down, Not Out&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Building foundations don't always have to spread the load out to a larger area. Sometimes the load only needs to be transferred to a stronger (harder) layer. This layer may be deep in the earth and have a softer layer on top of it. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Layers&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Say you want to build a building in an area that has a lot of soft clay at the surface. This clay will not support much of anything (especially when it is wet). Underneath this layer is hard bedrock. You now have two choices:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Add a basement to your building so that you can rest on bedrock.&lt;/li&gt;&#xA;&lt;li&gt;Install long steel or concrete piles from the bottom of your building to the bedrock.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;There isn't always budget to do #1, so #2 becomes the cheaper option.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Something Smaller&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Sometimes you are just building a house or a shed. Piles and deep basements are out of the picture. You could put the entire house on the ground to use the entire area, but that causes problems with water infiltration, pest access, heat lost, etc. Those are all reasons why even smaller building may not want to use the entire area of the building as a foundation, but some thought still needs to go into what the building actually rests on.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Related Question&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/q/1977/33&quot;&gt;Why dig out and then fill in before building a large structure?&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="33" LastEditorUserId="33" LastEditDate="2019-05-31T21:32:29.297" LastActivityDate="2019-05-31T21:32:29.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28616" PostTypeId="2" ParentId="2046" CreationDate="2019-05-31T16:32:17.177" Score="0" Body="&lt;p&gt;You can use Bernoulli, for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dfrac{P_1}{\gamma} + \dfrac{V_1^2}{2g} + z_1 = \dfrac{P_2}{\gamma} + \dfrac{v_2^2}{2g} + z_2 + h_f$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt; = pressure after the plunge&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_2$&lt;/span&gt; = pressure at the exit of the needle (atmospheric)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$z_1 = z_2$&lt;/span&gt; ...for a horizontal setup&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\gamma = $&lt;/span&gt; density times gravitational acceleration &lt;span class=&quot;math-container&quot;&gt;$=pg$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$v_1 =$&lt;/span&gt; the speed of the plunger is pretty much negligible when compared to the speed of the fluid expelled through the needle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$h_f =$&lt;/span&gt; all the frictions due to diameter reduction etc. For this case, I will consider &lt;span class=&quot;math-container&quot;&gt;$h_f =0$&lt;/span&gt; (ideal case)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence if we solve for &lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dfrac{P_1}{pg} = \dfrac{P_2}{pg} + \dfrac{1}{2}\dfrac{V_2^2}{g}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$(P_1-P_2) = \delta P = \dfrac{1}{2}pV_2^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the minimum force will be calculated as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{Force} = \text{Area of plunger} \cdot \dfrac{1}{2}pV_2^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the calculation of the velocity &lt;span class=&quot;math-container&quot;&gt;$v_2$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$v_2 =$&lt;/span&gt; (volume of liquid in the syringe)/(time to empty the syringe)/(Area of the needle inner diameter)&lt;/p&gt;&#xA;" OwnerUserId="20683" LastEditorUserId="1832" LastEditDate="2019-06-01T00:36:56.297" LastActivityDate="2019-06-01T00:36:56.297" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28617" PostTypeId="2" ParentId="28610" CreationDate="2019-05-31T17:40:20.770" Score="8" Body="&lt;p&gt;Building foundations perform many tasks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They provide a strong base to take the weight of the building, penetrate through incompetent layers of soil to reach a layer strong enough, they resist the heaving forces of expansive clay soils, they anchor the building down against toppling in earthquakes and getting blown by the wind or washed away by the floods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add to this many other tasks such as insulating the floors from moisture, freezing, animals, enemies, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many parts of the word they act as a passive thermal climate control. Because the soil strata under ground have more moderate temperature and do not change much during winter and summer, and because the foundations are massive they have large capacity to act as a thermal battery and moderate the temperature of the building, which has been done in hot arid areas of middle east, or in basement and cellars in many urban areas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some areas of my old home country, Iran, specially in rugged mountainous villages the foundation of the entire village is integrated to resist the strong winds, heavy snows and bridge the soft patches of ground soils, span melted snow streams, and provide the support to the covered protected alleys.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-05-31T18:16:21.760" LastActivityDate="2019-05-31T18:16:21.760" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28618" PostTypeId="1" AcceptedAnswerId="28619" CreationDate="2019-05-31T19:39:47.157" Score="-1" ViewCount="122" Body="&lt;p&gt;Suppose that the engine cranking has been successful but I  keep holding the ignition switch at the START position instead of releasing the switch to go at the RUN position. Does this mean that the starter motor is still spinning or has it somehow disengaged (probably through the starter solenoid ) from the battery?&lt;/p&gt;&#xA;" OwnerUserId="20686" LastActivityDate="2019-05-31T19:56:22.650" Title="Is starter motor running if I hold the ignition switch at the START position?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28619" PostTypeId="2" ParentId="28618" CreationDate="2019-05-31T19:49:30.297" Score="1" Body="&lt;p&gt;The flywheel will accelerate faster than the starter motor and the ring gear will accelerate the starter dog since there is a bendix drive or over-run clutch to allow this to happen so the starter motor itself will not overspeed. This does create noise which most notice...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-05-31T19:56:22.650" LastActivityDate="2019-05-31T19:56:22.650" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28622" PostTypeId="1" CreationDate="2019-05-31T21:30:40.270" Score="0" ViewCount="328" Body="&lt;p&gt;[&lt;img src=&quot;https://i.stack.imgur.com/YCJp4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UfmIW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UfmIW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hello, I try to extrude a lot of circle sketches but I can't select all of them and when I try, there is so many lags. Does someone has a solution ? I am sorry I'm a noob in CAD. Thank you so much ! :)&lt;/p&gt;&#xA;" OwnerUserId="14830" LastEditorUserId="14830" LastEditDate="2019-05-31T22:32:48.423" LastActivityDate="2020-02-26T16:01:25.407" Title="(CAD) Inventor - Extrude rectangular pattern of circle sketches" Tags="|cad|autodesk-inventor|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28626" PostTypeId="1" AcceptedAnswerId="28627" CreationDate="2019-06-01T09:20:39.607" Score="1" ViewCount="1282" Body="&lt;p&gt;Considering the fact that the starter solenoid needs a maximum of 50 Amps initially (in order for the pinion gear to engage to the flywheel through the plunger and lever) and then some 8 Amps during cranking, do we need a relay to supply power to the solenoid? Is this relay of electromechanical type or a solid state one?&lt;/p&gt;&#xA;" OwnerUserId="20686" LastActivityDate="2019-06-01T15:14:22.560" Title="Is a starter relay needed to supply the current to the starter motor solenoid?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28627" PostTypeId="2" ParentId="28626" CreationDate="2019-06-01T09:29:05.583" Score="2" Body="&lt;p&gt;Earlier cars did not use a relay, as the ignition switches were built sufficiently well to carry the 50A current for the pull-in winding current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newer cars have components like the ignition switch that are lighter and have more functions and need a relay to drive the starter solenoid. This relay is usually in the fuse box in the engine compartment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We had a Commer van (petrol) that the ignition switch had the following positions : off, ignition, start. Another van (diesel) had off, on, heat, start...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most relays are the electro-mechanical type as they are cheap and easy to get hold of, but as cars get more computer well more micro-processor control then some things are being replaced with solid-state components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: here is a link kindly provided by a comment :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://autosystempro.com/starter-control-circuit-components/&quot; rel=&quot;nofollow noreferrer&quot;&gt;example starting circuits&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not include all possibilities, early Landrovers just had a big spring-loaded push button that directly operated the starter - no relays, solenoids at all...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-06-01T15:14:22.560" LastActivityDate="2019-06-01T15:14:22.560" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28629" PostTypeId="2" ParentId="28587" CreationDate="2019-06-01T10:33:10.853" Score="1" Body="&lt;p&gt;if you fix/glue the magnets to the flywheel spaced equally apart and have a coil of wire wound around an iron core close to it you have in effect a generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short circuiting the ends of the 2 coil wires will cause a resistance to the pedal effort and in effect a brake that is variable if you also have a variable resistance in the coil wires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 2mm gap between the magnets and the coil face will be sufficient to generate a reasonable amount of electricity which can be controlled by the variable resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You &quot;could&quot; have a fair bit of heat generated in the coil of wire when the electricity is almost a dead short so I would advise having a load in the circuit in the form of a 240 volt 60 to 100 watt light bulb (by experiment)  to safely burn off the power as a light source......no cooling fan needed. &#xA;Ian. &lt;/p&gt;&#xA;" OwnerUserId="20697" LastActivityDate="2019-06-01T10:33:10.853" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28630" PostTypeId="2" ParentId="28596" CreationDate="2019-06-01T11:57:13.490" Score="3" Body="&lt;p&gt;Because in the real world, all else is never equal, but let's start there analytically. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What does &quot;all else being equal&quot; actually mean?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;With respect to a flat swept disk perpendicular to an ideal fluid with a uniform free stream flow field, from Momentum Theory, the induced flow field associated with optimum performance has slowly been specified in ever more realistic and convenient forms. Prandl, Glauert, Betz, and a host of others have have developed the theory to the point where we now have a closed solution describing the induced flow that is associated with maximum power extraction. (see, for example, section 3.5 in &lt;a href=&quot;https://pdfs.semanticscholar.org/9ecb/d17ae8d276d0395927e6604d94ca9d97ffde.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://pdfs.semanticscholar.org/9ecb/d17ae8d276d0395927e6604d94ca9d97ffde.pdf&lt;/a&gt; )&#xA;The relevant parameter is the ratio of the blade speed at a given radius to the free stream wind velocity. This is r/R * TSR. This determines the magnitude and direction of the ideal induced flow at this radial position. The actual rotor lift is proportional to, and opposed to, the momentum in the induced flow. Now all you have to do is design a blade set that creates this precise induced flow across it's entire span. It doesn't matter how you do it - whether with one blade or a thousand, you just need to match the theoretical induced flow field. One note - Differing numbers of blade will have significantly different properties in terms of &lt;em&gt;individual&lt;/em&gt; span loading distributions in order for the overall rotor assembly to have the required span loading distribution. This is due to the interaction of the shed vorticity on the trailing blades. But according to momentum theory &lt;strong&gt;there exists a blade design for each differing number of blades which will produce the same optimum performance given the same TSR&lt;/strong&gt;. So all else being equal means the TSR is the same, and the blades are optimized for the actual blade count so as to create the optimal induced flow vector at every radial distance. Blade count does affect the design of the individual blades, but not the theoretical performance limit of the rotor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we add in an actual vortex model, the &lt;em&gt;potential jumps&lt;/em&gt; in the flow change from a single jump at the disk to spiral jumps in the wake trailing behind each blade. These two models are highly compatible. Using a complex Fourier expansion of the vortex flow, you can separate the induced flow into a constant induced flow and a cyclic induced flow. Interestingly, the proportion of these two are constant radially, and they always have the same vector direction. This is why the actuator disk model appears to overperform it's rather pathetic assumptions, as does lifting line theory applied to modest AR wings, and Michell's integral applied to the wakes of not-so-thin ships. &lt;a href=&quot;https://apps.dtic.mil/dtic/tr/fulltext/u2/433976.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;DTIC vortex rotor model evaluated with complex Fourier Analysis, 1964&lt;/a&gt; The takeaway is that TSR needs to be matched to the blade number, and it has to be reduced for greater blade numbers. &lt;strong&gt;This combined with the result from the above paragraph puts a downward trend on best possible performance as the blade number increases.&lt;/strong&gt; This performance limit vs TSR curve is called the Schmitz whirlpool power coefficient. See page 7 here - &lt;a href=&quot;https://mragheb.com/NPRE%20475%20Wind%20Power%20Systems/Optimal%20Rotor%20Tip%20Speed%20Ratio.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://mragheb.com/NPRE%20475%20Wind%20Power%20Systems/Optimal%20Rotor%20Tip%20Speed%20Ratio.pdf&lt;/a&gt; (This result probably isn't the same as the power coefficient formula in the first link, but I'm not certain.)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;So what is missing from Momentum theory?&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Aerodynamic drag&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Tip losses due to the real vorticies not behaving like the highly&#xA;concentrated ideal tip vortices.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Structural constraints related to performance.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Economic constraints and &quot;rest of system&quot; interfaces.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Real wind behaves badly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;These all influence blade count.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aerodynamic drag limits HAWTs to TSRs best suited to at least three blades. The best two-blade designs don't quite match the best 3 blade designs because they can't quite run at their optimum speed due to excessive drag. Here I'm taking about skin friction and profile drag. These are the components of basic 2D airfoil L/D ratios. Vortex drag is handled next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very early on researchers realized that wing tips didn't work as well as the simple models predicted they should. Glauert and Prandl each had tip correction factor formulae. These favor more blades, not fewer, in an all else being equal scenario. But for high aspect ratio designs, the differences are quite small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biggest problem is structural. If you change from a three bladed to a four bladed rotor, you have to run a bit lower TSR to maximize performance. This means a higher torque at the same power. So the rotor ends up using more material with more blades.  Also, the profile drag total is higher because you have to support that extra torque with thickness. And the blade area total has to be larger because with lower TSR, the apparent wind speed is less. So you don't save much on aero drag even though it is running slower (but this part can get tricky due to the complicated details of the geometry differences).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rest of system issues - for rotors of any respectable size, you don't want to have to deal with slower. You already have gearboxes with 30 ton gears in two stage gearboxes. There are huge losses in gearing the rotor rpm up to a suitable generator rpm. This is one of the reasons why three-blade rotors are designed to run a bit quicker than the calculated aerodynamic optimum. It's a bigger problem for the large machines, but still an issue for kilowatt sized units. To save on gearbox cost and losses, you want to run at the fast end of the flattish Cp vs TSR area. So you can trade a bit of aero Cp off for a more manageable and efficient gearbox. Gearbox losses are typically proportional to the maximum torque rating of the gearbox, so looking at this a different way, you'd need a lower cutout speed with the higher blade count and the same gearbox torque rating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And lastly, real wind behaves badly. It isn't the uniform flow that the models assume. There is usually a substantial vertical wind gradient, and some transverse wind gradient, wind speed variations (gusts), and wind direction variations. These each produce excursions from the ideal induced flow field that we are chasing. And these excursions are not symmetric around the rotor, so imbalances occur that produce large fluctuating forces and moments on the rotor shaft. The largest imbalance is normally the flap moment of the blade. The load imbalance isn't a statics problem. The shed vorticity from the blades &quot;remembers&quot; the load at the time it was created and carries that information into the swirling wake. Here, the wake influences the induced flow of the following blades based on earlier conditions seen at a different location on the disk. The result is often described as a form of hysteresis, but that isn't the right term. Hysteresis is statically stable, what's going on here is strictly dynamical outside of actual blade stall conditions. The physical response of the individual blades and the net imbalances seen at the hub are a massive engineering concern. We just know more about these issues in the three-bladed rotor than any of the others. The two-bladed rotor has a particular imbalance problem that can be molified with the three bladed design. And more than three blades make all but the smallest machines heavier and more costly to produce without any performance improvements.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-06-01T11:57:13.490" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28631" PostTypeId="2" ParentId="28622" CreationDate="2019-06-01T14:39:15.923" Score="1" Body="&lt;p&gt;Thank you for the help, actually, I was trying to extrude one of the circles and then pattern the extrusion, it didn't work because I used rectangular pattern in the tab &quot;sketch&quot; instead of the retangular pattern in the &quot;3D model&quot; tab.&lt;/p&gt;&#xA;" OwnerUserId="14830" LastActivityDate="2019-06-01T14:39:15.923" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28632" PostTypeId="1" CreationDate="2019-06-01T17:22:19.293" Score="2" ViewCount="180" Body="&lt;p&gt;I have this question and I can't seem to find any sources which I can used to determine a potential answer. &lt;/p&gt;&#xA;" OwnerUserId="20700" LastActivityDate="2019-06-01T20:05:30.640" Title="Why are moulds sometimes dimensionally larger than the metal components they make?" Tags="|mechanical-engineering|materials|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28633" PostTypeId="2" ParentId="28632" CreationDate="2019-06-01T17:24:26.220" Score="1" Body="&lt;p&gt;Moulds have to be fractionally larger and have a release angle so that the parts will come out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the parts are larger than the mould then that is an interference fit and it can be very difficult to remove parts unless one is heated...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-01T17:24:26.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28634" PostTypeId="1" CreationDate="2019-06-01T18:16:58.727" Score="-1" ViewCount="59" Body="&lt;p&gt;What is the heat transfer difference between parallel flow and counter-flow heat exchanger?&#xA;Are different equations used to calculate q?&lt;/p&gt;&#xA;" OwnerUserId="20330" LastActivityDate="2019-06-03T00:17:53.827" Title="Heat Exchanger- the calculation of q" Tags="|heat-transfer|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28635" PostTypeId="2" ParentId="121" CreationDate="2019-06-01T18:30:35.453" Score="0" Body="&lt;p&gt;They say a picture is worth a thousand words, so here's a picture of the &quot;Sweet Spot&quot; for various machining operations including drilling, milling, and turning:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cnccookbook.com/feeds-speeds/&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://www.cnccookbook.com/feeds-speeds/&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you can see what going a little faster or slower on feedrate or spindle rpm is doing for you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;RPM's are all about heat.  If the cutter gets too hot, it softens and rapidly dulls.  Feedrate is all about the ability to clear the chips.  If they pack in the flutes too much and jam, the cutter breaks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are the basics.  There's a lot more too it than that, and you can learn a lot more from this &lt;a href=&quot;https://www.cnccookbook.com/feeds-speeds/&quot; rel=&quot;nofollow noreferrer&quot;&gt;free feeds and speeds course&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="20702" LastActivityDate="2019-06-01T18:30:35.453" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28637" PostTypeId="2" ParentId="28632" CreationDate="2019-06-01T20:05:30.640" Score="4" Body="&lt;p&gt;For a traditional foundry sand mold ; the metal freezes as essentially the size of the cavity but then shrinks as it cools to room temperature. So the cavity is larger to accommodate this shrinkage. For a permanent mold or die casting it is similar except the metal die is warm&gt;hot so when it is machined at room temperature , the expansion of the warm die and the shrinkage or the part must be considered. Then there are more complicated situations like investment /shell casting. A plastic pattern is made by injecting into a warm die &gt; the plastic shrinks to room temp . The plastic is embedded in ceramic &gt; the ceramic is heated to high temp ( eg. 1600 F) and expands &gt; the metal is poured in and freezes at about 1600 F then shrinks down to room temp. Experience helps in determining what size the die cavity for the plastic part should be at room temp.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-06-01T20:05:30.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28640" PostTypeId="1" AcceptedAnswerId="28650" CreationDate="2019-06-02T11:57:44.513" Score="1" ViewCount="443" Body="&lt;p&gt;I'm designing a steering mechanism following the Ackerman principle. I've met the condition that my steering arms' normal projections must cross in the middle of the rear axle however when I do a turn my wheels' axles do not cross in a single point on the rear axle. I'm applying some visuals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aU4dB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aU4dB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the bottom most line is the x axis of my drawing, so the wheel base is 1210mm. I've set the length of my steering arms at 150mm for demonstration purposes, where 756.82mm seems to be the distance between their joint points. The white lines that go off to the right are the projections of my wheels' axles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a visual of when turning at 15 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jW6Hi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jW6Hi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lines should cross, I can't figure out what I'm doing wrong.&lt;/p&gt;&#xA;" OwnerUserId="20333" LastActivityDate="2019-06-03T07:46:32.667" Title="Ackerman steering" Tags="|mechanical-engineering|automotive-engineering|mechanisms|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28641" PostTypeId="1" AcceptedAnswerId="28643" CreationDate="2019-06-02T15:51:47.973" Score="1" ViewCount="1016" Body="&lt;p&gt;Sorry, it seems like the answer to this question is just a quick google search away; but I have not found a page defining what the axis is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess was that the axis of a piple is the imaginary line running down through the middle of the pipe. However, this doesn't make sense based on a question I am looking at in a calculus book. The question is &quot;write the vector field describing the flow if the axis of the pipe is the unit circle in the &lt;span class=&quot;math-container&quot;&gt;$xy$&lt;/span&gt; plane.&quot; This makes me think that the axis of a pipe is cross section of a pipe, but that doesn't really  make sense either because usually an axis is a line, not a circle.&lt;/p&gt;&#xA;" OwnerUserId="20708" LastActivityDate="2019-06-02T16:42:26.970" Title="What is the axis of a pipe?" Tags="|mechanical-engineering|fluid-mechanics|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28643" PostTypeId="2" ParentId="28641" CreationDate="2019-06-02T16:42:26.970" Score="4" Body="&lt;p&gt;I'm guessing that the pipe in the problem is a torus.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if &quot;centerline&quot; would be a better word than &quot;axis&quot; in this case?&lt;/p&gt;&#xA;" OwnerUserId="63" LastActivityDate="2019-06-02T16:42:26.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28644" PostTypeId="2" ParentId="28520" CreationDate="2019-06-02T19:36:43.233" Score="0" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;tl;dr&lt;/em&gt;-&lt;/strong&gt;  Sounds like you'll need to spend about &lt;span class=&quot;math-container&quot;&gt;$84,000 \frac{\mathrm{USD}}{\mathrm{year}}$&lt;/span&gt; to heat the water for your pool.  The proposed heat exchanger might realistically give you about &lt;span class=&quot;math-container&quot;&gt;$50 \%$&lt;/span&gt; savings, or &lt;span class=&quot;math-container&quot;&gt;$42,000 \frac{\mathrm{USD}}{\mathrm{year}},$&lt;/span&gt; if we neglect its own operational expenses, capital cost, depreciation, etc..&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;If you've raising &lt;span class=&quot;math-container&quot;&gt;$1 \frac{\mathrm{L}}{\mathrm{s}}$&lt;/span&gt; of water from &lt;span class=&quot;math-container&quot;&gt;$1\sideset{^{\circ}}{}{\mathrm{C}}$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$20\sideset{^{\circ}}{}{\mathrm{C}} ,$&lt;/span&gt; then that requires a heat duty of about&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;Q&#xA;&amp;amp;~~=~~&#xA;\frac{\mathrm{d}m}{\mathrm{d}t}&#xA;~~&#xA;C_{\text{p}}&#xA;~~&#xA;\Delta T&#xA;\tag{1}&#xA;\\[10px]&#xA;&amp;amp;~~\approx~~&#xA;\left(1 \frac{\mathrm{L}}{\mathrm{s}} \times \frac{1000\,\mathrm{g}}{1 \, \mathrm{L}}\right)&#xA;\left(4.2 \frac{\mathrm{J}}{\mathrm{g} \cdot \mathrm{K}}\right)&#xA;\left(20\sideset{^{\circ}}{}{\mathrm{C}} - 1\sideset{^{\circ}}{}{\mathrm{C}}\right)&#xA;\\[10px]&#xA;&amp;amp;~~=~~ 79.8 \, \mathrm{kW}&#xA;\,.&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;h3&gt;Math check&lt;/h3&gt;&#xA;&lt;p&gt;If you want to check my math, then you can &lt;a href=&quot;https://www.wolframalpha.com/input/?i=(heat+capacity+of+water)+*+((20+-+1)+Kelvin)+*+(molar+density+of+water)+*+(1+liter)+%2F+(1+second)&quot; rel=&quot;nofollow noreferrer&quot;&gt;query WolframAlpha with the following&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;pre&gt;&lt;code&gt;(heat capacity of water) * ((20 - 1) Kelvin) * (molar density of water) * (1 liter) / (1 second)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&lt;p&gt;, which, checking just now, yields a result of &lt;span class=&quot;math-container&quot;&gt;$78.84 \, \mathrm{kW} .$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Then, if we assume that your cost of electricity is &lt;span class=&quot;math-container&quot;&gt;$0.12 \frac{\mathrm{USD}}{\mathrm{W} \cdot \mathrm{hr}},$&lt;/span&gt; that's about &lt;span class=&quot;math-container&quot;&gt;$9.58 \frac{\mathrm{USD}}{\mathrm{hr}}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$83,943 \frac{\mathrm{USD}}{\mathrm{year}}$&lt;/span&gt; worth of electricity to heat the water.&lt;/p&gt;&#xA;&lt;p&gt;Since you're looking at burning more than &lt;span class=&quot;math-container&quot;&gt;$80,000 \frac{\mathrm{USD}}{\mathrm{year}}$&lt;/span&gt; on just heating a pool, it does seem reasonable to try to find a way to reduce that cost.&lt;/p&gt;&#xA;&lt;p&gt;Next, as you've noted, we can't generally get ideal heat exchange.  The closer we try to approach it, the more costly the heat exchanger, plus the greater other operational expenses become.&lt;/p&gt;&#xA;&lt;p&gt;While it's really hard to give an exact figure, especially we don't have many details, the heuristic that folks in industry have previously recommended to me is that a decent heat exchanger can have a temperature approach of about&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\left.\Delta T\right|_{\text{approach}}&#xA;~=~ T_\text{hot-in} - T_\text{cold-out}&#xA;~\approx~ 10\,\mathrm{K}&#xA;\tag{2}&#xA;\,,$$&lt;/span&gt;&#xA;while a more robust heat exchanger might try for a &lt;span class=&quot;math-container&quot;&gt;$\left.\Delta T\right|_{\text{approach}}$&lt;/span&gt; of about &lt;span class=&quot;math-container&quot;&gt;$5 \, \mathrm{K} .$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Since the numbers above were all pretty linear (this is a nice, simple system!), we can basically say that you'll have a savings-factor of&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\xi_{\text{savings}}&#xA;~\approx~&#xA;\frac{19\,\mathrm{K} - \left.\Delta T\right|_{\text{approach}}}{19\,\mathrm{K}}&#xA;~=~ 1 - \frac{\left.\Delta T\right|_{\text{approach}}}{19\,\mathrm{K}}&#xA;\tag{3}&#xA;\,.$$&lt;/span&gt;&#xA;So,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\xi_{\text{savings}}&#xA;~\approx~&#xA;\left\{&#xA;\begin{array}{cl}&#xA;47.4 \% &amp;amp; \text{if decent heat exchanger with}~\left.\Delta T\right|_{\text{approach}} \approx 10\,\mathrm{K} \\&#xA;73.9 \% &amp;amp; \text{if good heat exchanger with}~\left.\Delta T\right|_{\text{approach}} \approx 5\,\mathrm{K} &#xA;\end{array}&#xA;\right.&#xA;\tag{4}&#xA;\,.&#xA;$$&lt;/span&gt;&#xA;We're definitely roughly guesstimating here, but it looks like the heat exchanger might realistically save you &lt;span class=&quot;math-container&quot;&gt;$50 \%$&lt;/span&gt; on heating costs – though, to be clear, it may offset some of those savings with additional maintenance costs, depreciation, etc..&lt;/p&gt;&#xA;&lt;p&gt;So, in short, I'm guessing maybe &lt;span class=&quot;math-container&quot;&gt;$50 \%$&lt;/span&gt; efficiency.  Given the prior estimate of &lt;span class=&quot;math-container&quot;&gt;$83,943 \frac{\mathrm{USD}}{\mathrm{year}}$&lt;/span&gt; for heating, that'd seem to be about &lt;span class=&quot;math-container&quot;&gt;$42,000 \frac{\mathrm{USD}}{\mathrm{year}}$&lt;/span&gt; in savings on your electric bill, before factoring in stuff like potential increases in electrical costs to run pumps, maintenance fees, capital costs, etc..&lt;/p&gt;&#xA;" OwnerUserId="10678" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-06-02T19:53:01.587" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28645" PostTypeId="2" ParentId="28634" CreationDate="2019-06-03T00:17:53.827" Score="0" Body="&lt;p&gt;There are two methods: The logarithmic mean and the NUT efficiency methods. Get info about those methods and you'll be able to calculate Q.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-06-03T00:17:53.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28648" PostTypeId="2" ParentId="17271" CreationDate="2019-06-03T06:21:45.783" Score="1" Body="&lt;p&gt;If you take &quot;performance&quot; as a mix of how fast the bike will accelerate and travel along its intended terrain, you can roughly approximate it as a ratio of thrust to mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by others, thrust can be increased by having a larger capacity engine with a corresponding mass penalty. Depending on the terrain, this penalty may be more or less important - there are plenty of 1000cc street bikes but very few off road bikes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other reasons for numbers in the name though, one being marketing (I don't remember Mazda being big on publishing the engine capacity of their rotary cars) the other being licensing. In the UK, I can ride a 125cc bike on my current license. Having numbers in the name makes it incredibly clear what I need to be looking at. &lt;/p&gt;&#xA;" OwnerUserId="512" LastActivityDate="2019-06-03T06:21:45.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28650" PostTypeId="2" ParentId="28640" CreationDate="2019-06-03T07:46:32.667" Score="2" Body="&lt;p&gt;If you just first turn the king pin on the interior wheel by say 25 degrees and draw a line perpendicular to the same wheel going up till it intersect the rear axel continuation, then from that point draw a line to the center of the other wheel and make that wheel perpendicular to the line, there is only one straight tie-bar that will connect the ends of wheels' bars so that they mach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now try for 15 degrees, 30 degrees and compare your length of rods and tie-bar. This gives you a pretty good idea of the geometry.&#xA;Here is a link with more info. &lt;a href=&quot;http://datagenetics.com/blog/december12016/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Steering geometry&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;just so you know the car manufacturers take advantage of three dimensional toe, caster, camber, drop arc, and lateral expansion of suspension and the lateral fricshion in road puddles as to give the car its road handling and comfort characteristics and sharing the handling of bumps with the shocks. basically the suspension twists and deform in a way that breaks the sudden shakes or jerks and vertical moves. Also they make the car surefooted in sudden jerks and even they design them to self destroy the whole suspension and wheels as sacrifice mass in an accident to break the crash and absorb a lot of energy. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-03T07:46:32.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28652" PostTypeId="1" CreationDate="2019-06-03T10:58:07.917" Score="2" ViewCount="558" Body="&lt;p&gt;I have been looking online and found there should be two points of contact, one in the blades and one in the pin. In order to achieve this, I assume there's a slight angle of the blades into the opposite blade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I couldn't find any information about specific numbers. How would I go about designing scissors that ensure the aforementioned contact while not being too hard to actuate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="20718" LastActivityDate="2021-02-23T15:06:30.800" Title="When designing scissors, how do you design the blade geometry to ensure good cutting capabilities?" Tags="|mechanical-engineering|mechanisms|product-engineering|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28653" PostTypeId="2" ParentId="23114" CreationDate="2019-06-03T11:46:43.893" Score="0" Body="&lt;p&gt;&lt;strong&gt;I suspect that by 'pre-load' they mean putting a flow &lt;em&gt;restriction&lt;/em&gt; between the fluid pressure supply and the bearing.&lt;/strong&gt; Doing this is essentially adding a (large) fluidic resistor between the pressure supply (analog of voltage source.) This may seem wasteful, dropping all that pressure through that restriction, but the result is a flow rate that's basically LargePressure/LargeFluidResistance; this gives you almost constant flow, -- because there's almost nothing you can do at the bearing that would be nearly as much flow resistance as that (seemingly wasteful) upstream restriction, the flow stays quite  constant (with respect to motions at the bearing), making bearing act mechanically stiff.&lt;/p&gt;&#xA;" OwnerUserId="13014" LastActivityDate="2019-06-03T11:46:43.893" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28654" PostTypeId="1" CreationDate="2019-06-03T14:14:01.173" Score="6" ViewCount="229" Body="&lt;p&gt;I am currently an engineering student and a member of the aerospace club of my engineering school (ESILV) which is called LéoFly. We’re a team of 12&amp;nbsp;students and are building a Fusex (Experimental Rocket) for the launch campaign of the CNES in partnership with Planète Sciences association. The goal of our experiment is to counter the “natural” roll of the rocket after the boost-phase. I am the one who must program the algorithm that counters the roll using a gyroscope and four solenoid valves (2 for the clockwise direction and 2 others for the counter-clockwise direction). I decided to use Simulink to develop the closed loop control using the following image, equations and data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image is the cross-section of the rocket with the four fins. This is an example of a “natural” roll in the counter-clockwise direction, so the program gives the order to the two solenoid valves of the clockwise direction to eject gases and cancel the roll of the rocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R0fG9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R0fG9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the equations we have started with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum C = J \dot\omega &amp;amp;= J \ddot\theta \\&#xA;2F_a R_1 + 4F_b R_2 &amp;amp;= J\ddot\theta \\&#xA;\dfrac{1}{J}(2F_a R_1 + 4F_b R_2) &amp;amp;= \ddot\theta \\&#xA;\omega(t) = \ddot\theta(t)t + \omega_0 &amp;amp;= \int \ddot\theta(t)\text{d}t&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;’s are the torques considered in the experimentation (Thrust torque and Roll torque), &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; is the inertia, &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the angle and &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; equals the derivative of &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; in the time domain. After that, &lt;span class=&quot;math-container&quot;&gt;$F_a$&lt;/span&gt; is the force of thrust, &lt;span class=&quot;math-container&quot;&gt;$F_b$&lt;/span&gt; is the force of roll, &lt;span class=&quot;math-container&quot;&gt;$R_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_2$&lt;/span&gt; are two different radii of the rocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the data, taken from our experiments and our rocket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OzMxf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OzMxf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, in Simulink, I’ve crafted the closed loop thanks to the schematic representation, the equations and the data. We don’t want roll, so the input is 0 (in radians per second) and the output is the roll of the rocket (in radians per second). This is a closed loop because of the gyroscope, and the dynamic subsystem is composed of the thrust torque and the roll torque which is the disturbance in this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cgRR7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cgRR7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thêta-lim is the saturation block that represents the max roll that the rocket can have physically. The White-Noise block is here to add randomness in the representation of the roll torque. The roll torque can be clockwise or counter-clockwise and is never always at the maximum of him (here the Roll Torque block is approximately 10&amp;nbsp;N.m so the disturbance which is added to the thrust torque is between 10&amp;nbsp;N.m and -10&amp;nbsp;N.m). Thrust Torque is approximately equal to 0.29&amp;nbsp;N.m and the inverse of Inertia is equals to 90&amp;nbsp;m&lt;sup&gt;2&lt;/sup&gt;.kg&lt;sup&gt;-1&lt;/sup&gt;. The switch subsystem is here just to permit a better implementation in the Arduino to pin the solenoid valves in the program (which will be generated from Simulink). The PID has the next values: &lt;span class=&quot;math-container&quot;&gt;$P = 114$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I = 28$&lt;/span&gt;. I didn’t use the auto-tuner because it gives the values for a step of 1 while we don’t want a step of 1 but an input of 0. Thus, here is what the scope gives me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Clvma.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Clvma.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is what we wanted for our roll control system: counter the “natural roll” and converge to an angular velocity of 0 radians per second. Of course, it is a student Fusex so we don’t want to have a perfect curve that converges to 0, we don’t have the time and the money to do that!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the problem is at &lt;span class=&quot;math-container&quot;&gt;$t = 0^+$&lt;/span&gt;: I put in Simulink (in the integrator block) a “natural” roll (angular velocity) of 2 radians per second. How is it possible to go from 19&amp;nbsp;rpm to approximately 0&amp;nbsp;rpm in 0&amp;nbsp;seconds? It does not seem physically possible, and we don’t know if it is a problem due to Simulink or if it is a problem in our blocks, if our closed-loop is poorly conceived or if the starting equations are not good. One of our professors told us that it can certainly be the problem of Simulink (because of the side effects apparently).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, what are your impressions about this and what do you recommend me to resolve this (these) problem(s)?&lt;/p&gt;&#xA;" OwnerUserId="20719" LastEditorUserId="1832" LastEditDate="2019-06-22T23:19:52.940" LastActivityDate="2024-01-28T11:02:40.287" Title="I have a problem with my representation of roll in Simulink" Tags="|control-engineering|aerospace-engineering|aerodynamics|simulink|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28657" PostTypeId="2" ParentId="28652" CreationDate="2019-06-03T16:38:14.773" Score="1" Body="&lt;p&gt;Yes, there is a gentle curve (not angle ) built into the blades, designed to counter the small play in the hinge. This curve holds the cutting edges close together but at the same time it should not be too much as to interfere with the smooth mechanism of the scissors, my estimate is the curve trace angle is under one degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has to do with the stiffness of the alloy of the blades and what kind of heat profile they have been hardened with too.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-03T16:38:14.773" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28658" PostTypeId="1" CreationDate="2019-06-03T17:56:38.610" Score="1" ViewCount="2073" Body="&lt;p&gt;In a text I'm reading on Euler-Bernoulli beam theory it is said that &lt;strong&gt;as the beam is assumed to be thin, the effect of transverse shear stress is ignored&lt;/strong&gt;. If the beam is thick, we need to use Timoshenko beam theory which accounts for transverse shear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's consider a simple loading case:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/rgZBF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same load F, if the beam is thin, the normal stress distribution inside the cross section of the beam is bigger since the smaller cross sectional area has to be in higher state of stress to counter the moment caused by the load. If the beam is thicker, the same moment is spread across a larger area and the stresses across the cross section are smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these same normal forces also cause the transverse shear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/jiUje.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Transverse shear is caused by the fact that since the moment felt by the beam is higher when we move to the left, the normal forces inside the beam when considering a small box inside the  beam (as in the above picture) are at imbalance and thus a transverse shear has to be present to maintain equilibrium. But if the beam is thick and the normal forces are smaller, &lt;strong&gt;why does the transverse shear now become relevant? Shouldn't the situation be the other way around: Thicker the beam, smaller the transverse shear?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2019-06-03T17:56:38.610" Title="Why can we ignore transverse shear stress in thin beams?" Tags="|structural-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28659" PostTypeId="1" AcceptedAnswerId="28661" CreationDate="2019-06-03T20:25:04.613" Score="1" ViewCount="408" Body="&lt;p&gt;Since copper is excellent at dissipating heat and does not rust, I am surprised that brake manufacturers do not electroplate cast iron/steel brake rotors with a thin layer of copper (~.5mm). There are brake manufacturers who offer brake rotors with cadmium or zinc-plated coatings at an additional cost but these coatings usually do not last long especially in areas that use a lot of rock salt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think copper plating would be the most ideal coating for ventilated brake rotors since it should keep the cooling vanes from rusting and becoming clogged with rust.  Clogged cooling vanes reduce the rotor's cooling capacity resulting in higher temperatures, faster rotor wear, premature warping, and a shortened lifespan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moreover, since copper is very good at dissipating heat, air flowing through copper-plated cooling vanes should draw away more heat from the brake rotor, resulting in a cooler ventilataed brake rotor, maximizing or perhaps even extending its intended lifespan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since there should be almost no rust on a copper-plated brake rotor, removing an worn out, copper-plated brake rotor should be no problem. Anyone who has worked on brakes knows how time-consuming it is to remove a rusted-on brake rotor or drum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brake manufacturer should not electroplate the friction areas (as shown in the picture below) since the high pressure and friction from contact with the brake pads would most likely warp the copper coating, but all the other surfaces would have a copper coating, most importantly the surfaces of the cooling vanes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will electroplating a ventilated brake rotor with copper improve its cooling capacity and maximize its lifespan?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zBI28.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zBI28.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user18610" LastEditorDisplayName="user18610" LastEditDate="2019-06-06T12:21:39.453" LastActivityDate="2019-06-06T12:21:39.453" Title="Will electroplating a ventilated brake rotor with copper improve its cooling capacity and maximum its lifespan?" Tags="|mechanical-engineering|materials|automotive-engineering|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28660" PostTypeId="1" CreationDate="2019-06-03T22:01:50.047" Score="1" ViewCount="173" Body="&lt;p&gt;In Etabs, if I model a building with semi-rigid diaphragms assigned to the floors, the floors will now transfer gravity loads when default meshing is assigned (as compared to rigid diaphragms transferring only mass and lateral loads). If i'm working with a simple model with only walls and floors modeled (no columns), the weight distribution will be incorrect since all the weight from the entire floor is going to the walls. One way to correct that would be to model the columns. Can another way to correct it be to assign a weight modifier of 0.01 to the floors and use dummy beams on top of the walls with assigned gravity loads based on tributary area?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd appreciate some insight. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="20727" LastActivityDate="2019-06-03T22:01:50.047" Title="Etabs Semi Rigid Diaphragms" Tags="|structural-engineering|structural-analysis|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28661" PostTypeId="2" ParentId="28659" CreationDate="2019-06-03T23:24:18.050" Score="4" Body="&lt;p&gt;Copper is better than cast iron at &lt;em&gt;conducting&lt;/em&gt; heat, and is used as a heat spreader in some high power density electronics products.  However, I don't believe that it is better at dissipating heat from a surface.  You might get marginally better heat dissipation if you made the whole rotor out of copper, but it'd be weaker, heavier, more expensive, and you'd have to reengineer your brake pads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any time you've got a hot thing that needs to be cooled by convection, the limitation is the interface between the air and the object to be cooled.  A thin copper plating -- or even making the whole rotor out of copper -- isn't going to help this.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-06-03T23:24:18.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28662" PostTypeId="2" ParentId="28659" CreationDate="2019-06-04T00:22:16.490" Score="2" Body="&lt;p&gt;It would be too thin to make a significant difference for heat dissipation. It may give better oxidation resistance , especially in locations where the rotors are exposed to salt water from deicing. I believe some aircraft rotors are made of beryllium copper for good heat conductivity but that is completely different. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-06-04T00:22:16.490" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28663" PostTypeId="1" AcceptedAnswerId="28667" CreationDate="2019-06-04T02:17:50.630" Score="0" ViewCount="34" Body="&lt;p&gt;This question is a bit general but hopefully someone can point me in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a metal container, welded shut that's fitted into a hole that is say 10 meters wide and 30 meters deep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has the volume of about 21,205&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you pump pure oxygen into the tank until all the water was pushed out of that cylinder, what would be the pressure of the oxygen in the tank. How much oxygen would be in the tank?&lt;/p&gt;&#xA;" OwnerUserId="20730" LastEditorUserId="33" LastEditDate="2019-06-04T03:13:24.887" LastActivityDate="2019-06-04T07:22:03.147" Title="hydrostatic storage of gasses" Tags="|mechanical-engineering|pressure|pressure-vessel|compressed-gases|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28664" PostTypeId="1" CreationDate="2019-06-04T03:21:59.727" Score="2" ViewCount="594" Body="&lt;p&gt;How you would calculate engine oil pressure of an engine? I've noticed on my car I just got, it uses a pressure switch in the oil system to measure engine oil pressure it seems in my case the switch has a threshold of 6psi to turn the pressure switch on. The dash will actually display a psi value that does change with engine speed,etc for oil pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am assuming the ecu is using a algorithm to determine the oil pressure I see on my display. This is interesting to me where exactly would you start to create a algorithm to determine engine oil pressure.&lt;/p&gt;&#xA;" OwnerUserId="20731" LastEditorUserId="9" LastEditDate="2019-07-25T07:59:43.113" LastActivityDate="2019-07-25T07:59:43.113" Title="Calculate engine oil pressure" Tags="|automotive-engineering|pressure|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28665" PostTypeId="2" ParentId="28664" CreationDate="2019-06-04T03:50:10.857" Score="3" Body="&lt;p&gt;There are pressure sensors fitted to determine the pressure - similar to temperature sensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Been available for years and cars used to have pressure gauges either Bourdon or electric but they went out of fashion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all the ecu fitted on cars and the multilevel screens they may include them again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using odb readers can show a lot of instantaneous readings oil, fuel, turbo pressure etc There are pressure sensors for all of them...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-04T03:50:10.857" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28666" PostTypeId="2" ParentId="28659" CreationDate="2019-06-04T06:07:02.753" Score="2" Body="&lt;p&gt;A thin layer of copper plating will have no effect on &lt;em&gt;convective&lt;/em&gt; heat transfer to rapidly flowing air, which is what gets the heat out of a cored/ventilated steel brake disc. In addition, a few taps of the brakes would scrape most of the plating off the disc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To have any effect on heat rejection rate in a predominantly steel disc, the copper cladding would need to be thick enough to conduct heat laterally outwards from under the brake puck's contact patch against the disc before the heat has a chance to soak into the steel underneath it. This requires a significantly thick copper layer relative to the thickness of the steel disc, as in the case of a copper-clad stainless steel frying pan bottom.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-06-04T06:07:02.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28667" PostTypeId="2" ParentId="28663" CreationDate="2019-06-04T07:22:03.147" Score="0" Body="&lt;p&gt;If the hole that the water drains out from is a hole with a check valve located at the height H from the top of the tank the pressure of the gas is equal to the pressure of water at the depth H. And the bottom part below that level will be water only. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However if you want to drain the entire tank and replace the water with the gas you should have the hole on the bottom of the tank, otherwise there is always water remaining below the discharge level.&#xA;In that case the pressure of the gas will be 3 atmosphere. Therefore you can use ideal gas low to calculate the weight of the oxygen:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ PV=nRT $$&lt;/span&gt;&#xA;-P is pressure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-V is volume, liters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-n is number of gas molecules&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-R is the ideal gas constant= 0.08206 L atm mol–1 K–1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-T is temperature in kelvin.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-04T07:22:03.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28668" PostTypeId="1" CreationDate="2019-06-04T08:01:59.740" Score="2" ViewCount="972" Body="&lt;p&gt;Given a three member truss structure shown here:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7tvOP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7tvOP.png&quot; alt=&quot;Truss structure&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I find the vertical and horizontal displacements at M?&#xA;Assume all members are elastic and E,A, I and constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I start with finding the reaction forces with &#xA;&lt;span class=&quot;math-container&quot;&gt;$F_x=2P-H_B=0, H_B=2P$&lt;/span&gt; (rightwards) and&#xA;&lt;span class=&quot;math-container&quot;&gt;$M_B=2l*V_A-2Pl=0, V_A=P$&lt;/span&gt; (upwards)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence &lt;span class=&quot;math-container&quot;&gt;$N_{AB}=2P$&lt;/span&gt; and here is where I get stuck. I cannot find the forces that put joint A into equilibrium. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Additionally, show all member forces&lt;/p&gt;&#xA;" OwnerUserId="20734" LastEditorUserId="20734" LastEditDate="2019-06-05T20:20:18.773" LastActivityDate="2019-06-06T09:37:30.437" Title="How do I calculate the member forces of a truss when external force acts on the member (not joint)?" Tags="|structural-engineering|structural-analysis|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28669" PostTypeId="2" ParentId="28668" CreationDate="2019-06-04T10:46:39.880" Score="3" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;TL;DR&lt;/em&gt;&lt;/strong&gt; This analysis is long-winded and I realise at the end that it's incorrect.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your reactions are right. You are incorrect to state that &lt;span class=&quot;math-container&quot;&gt;$N_{AB}=2P$&lt;/span&gt;, as you are ignoring the shear in member BC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After getting the reactions, you should calculate the member forces at A:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Resolving vertically: &lt;span class=&quot;math-container&quot;&gt;$N_{AC}*cos(45)=P$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Resolving horizontally: &lt;span class=&quot;math-container&quot;&gt;$N_{AC}*sin(45) + N_{AB}=0$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Given that &lt;span class=&quot;math-container&quot;&gt;$cos(45)=sin (45)$&lt;/span&gt;, we find that &lt;span class=&quot;math-container&quot;&gt;$N_{AB}=P$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly resolving at C, we'll find that the shear in member BC is &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;, and the horizontal component of member AC is also &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So basically member BC will act as a simply supported member spanning from B to C, with a point load at midspan of &lt;span class=&quot;math-container&quot;&gt;$2P$&lt;/span&gt;, and an axial compression of &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ignoring second order effects (which is probably oversimplistic), the horizontal displacement from &lt;span class=&quot;math-container&quot;&gt;$2P$&lt;/span&gt; on a simply supported beam is given by the standard solution &lt;span class=&quot;math-container&quot;&gt;$WL^3 / (48EI)$&lt;/span&gt;, or in this case &lt;span class=&quot;math-container&quot;&gt;$2P(2l)^3 / (48EI)$&lt;/span&gt;; and the vertical displacement from the standard solution &lt;span class=&quot;math-container&quot;&gt;$FL/EA$&lt;/span&gt;, or in this case &lt;span class=&quot;math-container&quot;&gt;$Pl/EA$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Horizontal displacement &lt;span class=&quot;math-container&quot;&gt;$2P(2l)^3 / (48EI)$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Vertical displacement &lt;span class=&quot;math-container&quot;&gt;$Pl/EA$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I've ignored shear in members AB and AC; I think these may be zero but I've not proved that. I've also ignored that the axial tension in AB will cause position A to move, dragging member AC and hence C with it, plus the axial compression in AC will move point C even further.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;On second thoughts this analysis presented here is inadequate, certainly for the displacements and possibly for the forces. Perhaps building a stiffness matrix is required.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="255" LastEditorUserId="255" LastEditDate="2019-06-04T14:09:20.673" LastActivityDate="2019-06-04T14:09:20.673" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="28670" PostTypeId="1" CreationDate="2019-06-04T13:12:26.500" Score="1" ViewCount="27" Body="&lt;p&gt;Using two cases studies: I. When the windows are wound up and air conditioner is turned on. II. When the windows are wound down and air conditioner is turned off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case studies are evaluated for driving in city and highway but more importantly how can the fuel consumption rate of the AC be measured so as to suggest when it is more economical to use it and reduce the effects of drag on the vehicle?&lt;/p&gt;&#xA;" OwnerUserId="20737" LastActivityDate="2019-06-04T13:12:26.500" Title="How to evaluate the fuel consumption rate of an air conditioner in a vehicle?" Tags="|aerodynamics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28672" PostTypeId="1" AcceptedAnswerId="28673" CreationDate="2019-06-04T15:47:15.823" Score="0" ViewCount="271" Body="&lt;p&gt;I am interested in knowing how many newton's of force is needed to keep 1 kg of mass hovering above the moon's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And whether the current most powerful ion thrusters would be up to the task? I believe ESA tested an ion thruster recently which achieved 100 kilo newton's. (I massively mis-remembered this part. I was thinking of the X3 thruster, and the figure is 5.4 Newtons.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance! &lt;/p&gt;&#xA;" OwnerUserId="20740" LastEditorUserId="20740" LastEditDate="2019-06-04T18:28:43.523" LastActivityDate="2019-06-04T18:43:11.717" Title="Amount of thrust needed to Hover on the moon?" Tags="|aerospace-engineering|propulsion|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28673" PostTypeId="2" ParentId="28672" CreationDate="2019-06-04T17:10:34.503" Score="2" Body="&lt;ul&gt;&#xA;&lt;li&gt;Look up the surface gravity of the moon.  It'll be in units of meters per second squared.&lt;/li&gt;&#xA;&lt;li&gt;Look at the definition of the Newton.&lt;/li&gt;&#xA;&lt;li&gt;Use those two things to determine the thrust needed to keep 1kg of mass aloft&lt;/li&gt;&#xA;&lt;li&gt;Look up the specifications for current ion thrusters.  Pay attention to mass vs. thrust ratio, or maximum acceleration.&lt;/li&gt;&#xA;&lt;li&gt;Do some math.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think you'll find that even current ion thrusters have significantly less acceleration available than the surface acceleration of the moon.  Ion thrusters are mass efficient, and they're a good candidate for interplanetary travel because small continuous acceleration will get the job done as well as short bursts of big acceleration.  But takeoff &amp;amp; landing will still require chemical rockets.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-06-04T17:10:34.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28674" PostTypeId="2" ParentId="28668" CreationDate="2019-06-04T18:10:04.447" Score="1" Body="&lt;p&gt;As the OP and others Have done we set the moment about B = 0 to get reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Sigma M_B=0,\ 2lA_v-2Pl=0\ A_v=P $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Sigma F_v=0\ A_v+B_v=0\ B_v=-P\ and\ C_v=P\ BC\ under\ tension$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F_{AC}=\sqrt(P^2+P^2=\sqrt2P^2 =P\sqrt2\\ and\ F_{AB}=P \ tension $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Sigma F_x=0,\quad B_h=2P$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Horizontal displacements of M as has been mentioned in @AndyT, but we need to add to that the horizontal component of movement of point C multiplied by 1/2, which is &lt;span class=&quot;math-container&quot;&gt;$1/2 \sqrt2*(P*l/AI  )\sqrt2/2= Pl/2AI \ \text{and add to that the horizontal expansion of AB/2 which is }=lP/EA $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \therefore H_{displacement\ of\ M}= 2P(2l)^3 / (48EI)&#xA;+ Pl/2EA+Pl/EA $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On vertical displacement we note that under the tension point M located at hlf height of BC is moving up by &lt;span class=&quot;math-container&quot;&gt;$Pl/EA$&lt;/span&gt; . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;please check my arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-06-04T23:29:34.367" LastActivityDate="2019-06-04T23:29:34.367" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28676" PostTypeId="2" ParentId="28672" CreationDate="2019-06-04T18:43:11.717" Score="0" Body="&lt;p&gt;You don't need any energy or thrust, if your 1kg mass is already in orbit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically an object orbiting the moon relatively near the surface, say one hundred miles, stays on orbit with no energy needed and rotates with a speed of v = SQRT(G*M/R). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case approximately 1.022 km/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lifting up some mass and placing it on mons orbit is a different story!&#xA;you need to calculate the difference between the potential energy of the mass on the moon and in orbit altitude&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$E1-E2 = -GMm*((1/R)-(1/(R+d))$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R= moon radius and d is the orbital altitude.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-04T18:43:11.717" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28677" PostTypeId="1" CreationDate="2019-06-04T20:05:23.010" Score="1" ViewCount="156" Body="&lt;p&gt;I am attempting to model a steel sample subjected to combined mechanical and thermal loading. The thermal loading can be quite elevated, where reduction in the elastic modulus will occur.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know if ANSYS's static structural module automatically accounts for the reduction of stiffness? &lt;/p&gt;&#xA;" OwnerUserId="20742" LastEditorUserId="20742" LastEditDate="2019-06-05T00:46:46.817" LastActivityDate="2019-11-06T20:20:35.197" Title="Structural analysis + thermal expansion using ANSYS static module" Tags="|structural-engineering|structural-analysis|thermal-expansion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28681" PostTypeId="1" CreationDate="2019-06-05T08:50:41.057" Score="1" ViewCount="63" Body="&lt;p&gt;The &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Timoshenko_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article on Timoshenko beam theory&lt;/a&gt; says:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The model takes into account shear deformation and rotational bending effects,&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I am familiar with shear deformation, but what is meant by &quot;rotational bending&quot;? Google search yields no good explanations of the term, so what do we mean by that? &lt;strong&gt;What kind of rotational bending does the Timoshenko model account for that the Euler beam model does not?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13372" LastEditorUserId="1832" LastEditDate="2019-06-08T21:20:01.247" LastActivityDate="2019-06-08T21:20:01.247" Title="Rotational bending in Timoshenko beam theory" Tags="|structural-engineering|beam|" AnswerCount="0" CommentCount="6" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28684" PostTypeId="1" CreationDate="2019-06-05T15:53:12.083" Score="3" ViewCount="264" Body="&lt;p&gt;I am trying to make a pen-like product that moves out fluid from its tip for some special chores, but I am having issues about the motion of the clutch. The picture you see is a &lt;strong&gt;sectioned&lt;/strong&gt; concept because my thought isn't fully fleshed out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gtHvl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gtHvl.png&quot; alt=&quot;full view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Legend: Black stripes - fluid, Yellow - Button, Orange - Body, Pink - Spring, Teal - Clutch, Green - Clutch ring, Blue - Plunger, Red - Cap, Grey - Cartridge&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goal: When I push the button once, the plunger pushes up the red cap, which pushes up the fluid in the cartridge. As more liquid is expended, the red cap moves up until it reaches the top, when there's no liquid. &lt;strong&gt;Each time the button is pushed&lt;/strong&gt;, I want the plunger to move and stop &lt;strong&gt;in accordance to the level of fluid left in the cartridge&lt;/strong&gt;. For example, if the liquid is halfway, the plunger will be more upward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dfP94.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dfP94.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a clutch system where the clutch ring is press fitted inside and a clutch has a tight fit against the clutch ring, to force the clutch to hold on to the plunger as it moves upward. This way, the plunger can move up then stay still without any external force acted on it (the button being pressed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eYKWz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eYKWz.png&quot; alt=&quot;clutch mechanism motion&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Main problem for me: After a certain level of fluid is gone but fluid still remains, the button won't be able to touch the plunger no matter how much I press the button. I tried introducing a small connection between the button and the clutch, so the plunger can still move even if the button doesn't have contact with it, but that introduces more problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jLnMy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jLnMy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may notice more problems but the bottom line question is: &lt;strong&gt;will this mechanism work? If so, how can I resolve this main issue and if not, what other mechanism(s) will work? I'm not looking for specific numbers&lt;/strong&gt;, but just needed to be point in the right direction (mechanism choice) to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate any and all responses!&lt;/p&gt;&#xA;" OwnerUserId="20752" LastEditorUserId="20752" LastEditDate="2019-06-05T16:01:45.423" LastActivityDate="2019-07-29T12:37:53.147" Title="What mechanism do I use for this pen-like design?" Tags="|mechanical-engineering|design|mechanisms|machining|machine-design|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28685" PostTypeId="1" CreationDate="2019-06-05T18:13:23.653" Score="3" ViewCount="69" Body="&lt;p&gt;I am having a very difficult time identifying this thread. This part is a plug that I am trying to replace because the threads were stripped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best fit for pitch is 11 threads per inch (imperial), but the major thread diameter is 32mm. It does not appear to be tapered and the rubber seal would support this observation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have looked through many standard thread size charts and have been unable to locate a match. I thought BSPP-16 was close enough (at 33mm) but when I tried, the BSPP plug was too large to fit the associated female thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5xoYS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5xoYS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QqAww.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QqAww.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9WaEC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9WaEC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20755" LastEditorUserId="20755" LastEditDate="2019-06-05T18:30:32.180" LastActivityDate="2019-06-05T18:30:32.180" Title="Thread identification" Tags="|mechanical-engineering|threads|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28686" PostTypeId="1" CreationDate="2019-06-05T20:47:37.300" Score="-1" ViewCount="92" Body="&lt;p&gt;What are some examples of motors, preferably less than $300, that would be able to function in a vacuum?&lt;/p&gt;&#xA;" OwnerUserId="20750" LastActivityDate="2019-06-07T04:16:45.363" Title="Motors that would work in a vacuum" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28690" PostTypeId="2" ParentId="28668" CreationDate="2019-06-06T02:18:42.133" Score="0" Body="&lt;p&gt;After some discussion with my classmates, I analyzed the &lt;span class=&quot;math-container&quot;&gt;$BC$&lt;/span&gt; member as a simple beam. and horizontal displacement &lt;span class=&quot;math-container&quot;&gt;$\delta_{mh} = \delta_{bending} + \delta_{truss}$&lt;/span&gt; &#xA;Considering load &lt;span class=&quot;math-container&quot;&gt;$2P$&lt;/span&gt; on &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt;, we have bending load &lt;span class=&quot;math-container&quot;&gt;$Pl$&lt;/span&gt; on point &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;. The displacement from this displacement load can be calculated via whichever method of choice. Here I used the principle of virtual work:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\delta_{bending} = \frac{2}{EI}\int_0^l Px \cdot \frac{x}{2} dx = \frac{Pl^3}{3EI}$$&lt;/span&gt;&#xA;Next, the displacement of the truss structure is also determined through virtual work. Here is where I am unsure of the member forces. Positive: right, upwards, tension&#xA;We have reaction forces &lt;span class=&quot;math-container&quot;&gt;$V_A=-P,V_B=P,H_A=-2P$&lt;/span&gt; from equilibrium conditions. Point load &lt;span class=&quot;math-container&quot;&gt;$2P$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is treated as two loads &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; on &lt;span class=&quot;math-container&quot;&gt;$C,B$&lt;/span&gt;. Hence we have member forces &lt;span class=&quot;math-container&quot;&gt;$N_{AB}=-\sqrt2P , N_{BC}=P, N_{CA}=P$&lt;/span&gt;. Placing virtual horizontal load at &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;, we compute &lt;span class=&quot;math-container&quot;&gt;$$ 2\delta_{truss}=\frac{Pl}{EA}(2+2+4\sqrt2) \\ \delta_{truss}=\frac{Pl(2+2\sqrt2)}{EA}$$&lt;/span&gt;&#xA;and hopefully I have the correct displacement here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I believe vertical displacement is a simple virtual load on C. i.e. &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\delta_{CV} = \delta_{MV} = \frac{2Pl}{EA}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="20734" LastEditorUserId="20734" LastEditDate="2019-06-06T09:37:30.437" LastActivityDate="2019-06-06T09:37:30.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28692" PostTypeId="1" CreationDate="2019-06-06T08:22:54.033" Score="-1" ViewCount="60" Body="&lt;p&gt;An apartment yield analysis is a set of draft drawings in which the optimum residential apartment layout and resultant yield is calculated for a given building site, taking into account development constraints such as apartment area sizes, floor-to-space ratios, building setbacks, solar exposure, internal corridor widths etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This bulk of this is done by an architect / draftsperson manually; the problem is not trivial and it can take man-hours of trial and error to approach an optimal layout.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My view is that once most of the constraints are established, it falls into the 'packing problem' domain that computers are useful at attacking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any service or CAD software component that wholly or partly automates the packing of such shapes and produces the drawings for a building or individual floorplans?&lt;/p&gt;&#xA;" OwnerUserId="20707" LastEditorUserId="20707" LastEditDate="2019-06-06T22:50:41.253" LastActivityDate="2019-07-07T05:01:55.447" Title="Software for automatic building floorplan calculation / drawing (packing problem)" Tags="|structural-engineering|cad|building-design|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28693" PostTypeId="1" CreationDate="2019-06-06T08:45:52.150" Score="2" ViewCount="32" Body="&lt;p&gt;Is there some temperature at which all &quot;5w&quot; motor oils should have about the same viscosity?  (5w-20, 5w-30, 5w-40, 5w-50, etc.) &#xA;I am looking for SAE or ASTM wording or chart that says it straight out, by definition.  But it's not exactly jumping out at me.&lt;br&gt;&#xA;Maybe it is not that simple?  Maybe they don't define it that way? &#xA;Could I find the answer by using ASTM D341 graph paper to chart out a few oils with known mfr specs for viscosity at 40 deg C and 100 deg C ? (which they all have)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One little problem there is I would need the chart that covers from 100 deg C all the way down to, I don't know where, but some pretty low temps, and I can't see the limits on any of their 7 charts.  And I don't know if those charts are accurate for modern motor oil either.  &lt;/p&gt;&#xA;" OwnerUserId="20758" LastActivityDate="2019-06-06T08:45:52.150" Title="Is there some temperature at which all &quot;5w&quot; oils should have about the same viscosity?" Tags="|mechanical-engineering|fluid-mechanics|automotive-engineering|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="28694" PostTypeId="1" CreationDate="2019-06-06T11:44:59.133" Score="1" ViewCount="2436" Body="&lt;p&gt;In a number of designs, a beam will have an underslung welded plate attached, as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B6Cvy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B6Cvy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this sketch, a symmetrical standard steel beam/column of dimensions H&lt;sub&gt;1&lt;/sub&gt; x B&lt;sub&gt;1&lt;/sub&gt;, with web and flange thickness w and t, has a wider plate welded under it of dimensions H&lt;sub&gt;2&lt;/sub&gt; x B&lt;sub&gt;2&lt;/sub&gt; (with B&lt;sub&gt;2&lt;/sub&gt; &gt;&gt; H&lt;sub&gt;2&lt;/sub&gt;, ratio around 10:1 ~ 30:1). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For simplicity, I'm only considering stiffness in the vertical plane, not torsion/eccentric/lateral loads. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example might be, if symmetrical static vertical point loads/UDLs were applied on the plate of such a beam, at both sides, at various points along the beam's length, and one wanted to evaluate the beam's bending moments or deflection along its length, or something like that. The usual methods for loads on a simple beam would be used, but you'd need to know its area moment of inertia first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming one doesn't just add the area M.o.I. (or 2nd moment of inertia - same thing) of the two together, as the value of the composite depends on the positions of the two. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the easiest way to calculate the value for this setup?&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2019-06-06T20:48:44.543" LastActivityDate="2020-07-02T05:07:40.793" Title="Calculating area moment of inertia for a beam with underslung plate?" Tags="|mechanical-engineering|structural-engineering|beam|geometry|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28697" PostTypeId="1" CreationDate="2019-06-06T14:27:05.433" Score="1" ViewCount="66" Body="&lt;p&gt;Biochar is a major by-product from pyrolysis along with bio-oil and it is considered an important chemical feed-stock because of many light oxygenates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read some papers on using bio-char use as an adsorbent to remove furfural from bio-oil and use the acid rich liquid for anaerobic digestion. But I was wondering if someone has tried desorption of furfural from bio-char? This could help recover it as a pure chemical co-product. &lt;/p&gt;&#xA;" OwnerUserId="18316" LastActivityDate="2019-11-17T10:01:40.200" Title="What is the mechanism of adsorption-desorption on biochar obtained from slow pyrolysis/ torrefaction?" Tags="|chemical-engineering|chemistry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28698" PostTypeId="2" ParentId="18763" CreationDate="2019-06-06T14:43:08.280" Score="1" Body="&lt;p&gt;Simple answer is this, the Heat exchange equation accounts for the amount of time the fluids are in contact, while the conservation of energy equation assumes the fluids are in equilibrium.  So if the temperature differences between the 2 fluids is smaller you need the fluids to be in contact longer to get the amount of heat exchange you are calculating for the higher temperature difference.  Since the conservation of energy equations have flow rates in the terms, this is the equivalent to saying the fluids are traveling across the heat exchanger surface at a constant rate, the only way to get the  same amount of heat transfer with a smaller temperature differential is to increase the surface of the heat exchanger, so the fluids stay in contact for a longer period of time. &lt;/p&gt;&#xA;" OwnerUserId="20766" LastActivityDate="2019-06-06T14:43:08.280" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28700" PostTypeId="2" ParentId="28694" CreationDate="2019-06-06T17:33:24.837" Score="1" Body="&lt;p&gt;To calculate the second moment of area of a composite beam, we need to find its neutral axis and then use the parallel axis theorem to add individual sections moments of area. I am assuming you made a mistake and meant to say &lt;span class=&quot;math-container&quot;&gt;$B_2\gg H_2$&lt;/span&gt;, if we go by the scale of your diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First the H beam. Because of symmetry, we subtract the I of the rectangular left between the 2 flanges and then add the I of the web.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ I_{Hbeam}= \dfrac{B_1H_1^3}{12} +\dfrac{(- B_1+W)(H_1-2T)^3}{12}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the neutral axis again by symmetry is at &lt;span class=&quot;math-container&quot;&gt;$ H1/2$&lt;/span&gt; and the area is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ A_{Hbeam}= B_1H_1+(-B_1+W)(H_1-2T) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we are set to find the neutral axis of the composite beam.&#xA;setting the bottom of the red flange as reference axis we get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Y_{neutral\ axis}=\frac{ \sum A_n \bar{Y}_n}{\Sigma A_n}  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We denote the distance from each section's neutral axis to the reference line on the bottom of the red plate, &lt;span class=&quot;math-container&quot;&gt;$\bar{Y} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the composite I:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ I_{composite}=\sum (A_n d_y^2 + I_n) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$d_y$&lt;/span&gt; is the distance between the section's and composite section's neutral axes.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2019-06-08T02:15:59.450" LastActivityDate="2019-06-08T02:15:59.450" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28701" PostTypeId="1" AcceptedAnswerId="28702" CreationDate="2019-06-06T18:08:03.937" Score="0" ViewCount="378" Body="&lt;p&gt;Stress = Force / Area &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets see with static force... what happens to stress output when decrease the area. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# Stress = Force / Area &#xA;F = 50000&#xA;A = [15.0,10.0,5.0]&#xA;&#xA;S = zero(A) # Initialize the output &#xA;for i in 1:length(A)&#xA;    S[i] = F / A[i]&#xA;end&#xA;&#xA;m = DataFrame(hcat(A,S))&#xA;names = [&quot;area&quot;,&quot;stress&quot;]&#xA;names!(m, Symbol.(names))&#xA;print(m)&#xA;&#xA;3×2 DataFrame&#xA;│ Row │ area    │ stress  │&#xA;│     │ Float64 │ Float64 │&#xA;├─────┼─────────┼─────────┤&#xA;│ 1   │ 15.0    │ 3333.33 │&#xA;│ 2   │ 10.0    │ 5000.0  │&#xA;│ 3   │ 5.0     │ 10000.0 │&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With decreasing area we increase the stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in thinking about hoop stress then how the minimum required thickness is obtained. Using the ASME B31.3 pressure tmin to find t &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;304.1.2 Straight Pipe Under Internal Pressure&#xA;t = PD / 2(SEW + PY)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The numerator has P * D so anything in the denominator will dilute the numerator...  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to see how P and D affect the t output we can iterate through a range of static pressure and changing Diameter... so lets see what this looks like: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;# Define numerator inputs &#xA;P = 200 # static P&#xA;D = collect(1.0:1.0:24.0) # varying D&#xA;&#xA;# define the denominator static inputs &#xA;S = 20000&#xA;E = 1.0&#xA;W = 1.0 &#xA;Y = 0.4 # temp coefficient &#xA;&#xA;# initialize output &#xA;out = zero(D)&#xA;for i in 1:length(D)&#xA;    out[i] = P * D[i] / (2*(S*E*W) + (P*Y))&#xA;end&#xA;&#xA;# plot the output&#xA;plot(D,out,seriestype = :bar, title=&quot;t = static P in relation to Diameter&quot;)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GjK34.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GjK34.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a big factor in the minimum required thickness for pressure containment is the P vs diameter.. at say static P increase the D and we need a thicker wall for pressure containment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in thinking about S = F / A &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does this relate to the pressure design thickness calculation.... S in the tmin calculation is in the denominator and S is static in this equation governed by the 3:1 ASME safety factor in relation to materials tensile stress. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Intuitively i can see how at a static F over varying Area - how the stress INCREASES... (as we saw top of the post) because there is less area to distribute the stress over... so stress would increase pounds per sq inch... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But doesn't that mean for smaller D in a pipe, we have less A so that the stress would increase... but running pressure design thickness calculation.. at smaller D we need a thinner wall for pressure containment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For sure im missing something here! (then again i see no area in the pressure tmin calculation) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone can help me connect the dots? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="15762" LastActivityDate="2019-06-06T19:14:43.060" Title="Pressure design thickness and how relates to hoop stress" Tags="|stresses|piping|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28702" PostTypeId="2" ParentId="28701" CreationDate="2019-06-06T19:14:43.060" Score="2" Body="&lt;p&gt;The &quot;area&quot; terms in the equations all cancel out. Think about a section of pipe of internal diameter &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;, length &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt;, thickness &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; with internal pressure &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine you slice the section of pipe (and the fluid causing the internal pressure) into two semicircular parts, and think about the forces acting one of the parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;area&quot; of the cut through the fluid is d wide by &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; long, and the fluid pressure of &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;. So the force exerted by the fluid is pressure &lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt; area = &lt;span class=&quot;math-container&quot;&gt;$P\,d\,l$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is balanced by the hoop stress in the pipe wall. The area of the cut through the walls is &lt;span class=&quot;math-container&quot;&gt;$2\,t\,l$&lt;/span&gt; (2, because there are two cuts, on either side of the pipe).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the average hoop stress is &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt;, the force on the cut surfaces of the pipe is stress &lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt; area = &lt;span class=&quot;math-container&quot;&gt;$2S\,t\,l$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;span class=&quot;math-container&quot;&gt;$2Stl = Pdl$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$$S = \frac{Pd}{2t}.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual hoop stress varies between the outside and inside of the pipe wall, and this formula doesn't include any safety factors, so in the ASME formula the &lt;span class=&quot;math-container&quot;&gt;$2t$&lt;/span&gt; term is replaced by something more realistic for design purposes.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-06-06T19:14:43.060" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28704" PostTypeId="1" CreationDate="2019-06-07T01:46:42.590" Score="0" ViewCount="189" Body="&lt;p&gt;There's an adiabatic efficiency because real processes can't be adiabatic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want an isothermal compression (yes, I know an adiabatic compression consumes less work, but my pressure differential is too high and a high temperature rise is undesired).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know real compressions can't be isothermal because there will always be a higher or lower temperature rise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I define it? Something like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\eta_{iso}=1-\Delta T/Factor$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how could apply this to my compressor? I can't apply it directly as if it were the adiabatic efficiency, which is based on theorical vs. real enthalpy jump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just came up with this idea: What about multiplying the isothermal work by &lt;span class=&quot;math-container&quot;&gt;$1+\Delta T/Factor$&lt;/span&gt; to get the real compression work?&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorDisplayName="user20096" LastEditDate="2019-06-07T01:54:15.723" LastActivityDate="2019-06-07T01:54:15.723" Title="Could I define an isothermal efficiency?" Tags="|thermodynamics|compressors|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="28705" PostTypeId="2" ParentId="28686" CreationDate="2019-06-07T04:07:20.350" Score="1" Body="&lt;p&gt;You will not be able to afford a real vacuum rated motor on a budget. BUT, as others have said almost any motor will work in a vacuum, so you will need to &quot;rate&quot; it yourself. Consider the factors that may cause the motor to fail, for example lack of cooling, tin whisker growth, insulation failure, lubricant issues etc. Then ultimately you will need to run some tests. This is what you would be paying for if you bought a motor that was rated for vacuum from the get-go.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-06-07T04:07:20.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28706" PostTypeId="2" ParentId="28686" CreationDate="2019-06-07T04:16:45.363" Score="0" Body="&lt;p&gt;First, consider that your motor does not nesseserily need to be in the vacuum (unless you are in space). There exist several of the shelf components that allow you to connect a shaft through the wall your vacuum container without leakage. This will make your design much simpler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second many electric motors will work in vaccum, but they will have hard time getting rid of heat since many systems are air cooled. However there are structural methods you can use to conduct the heat away from the system. This also applies to all your other components.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2019-06-07T04:16:45.363" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28707" PostTypeId="2" ParentId="28692" CreationDate="2019-06-07T04:18:41.453" Score="0" Body="&lt;p&gt;That's called architecture and no there is no software to replace the job of an architect. There are many tools to make this process easier from conception to preliminary modeling through finished working drawings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the same way you can not have a word processor write a novel, there is no software to design architectural projects. That actually leaves the process free to skills and experience and artistic flare of the designer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't mean that highly technical or specialty function spaces like hospital ORs or say refinery control rooms or some military buildings don't have prescribed standard plans and specs.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-07T04:18:41.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28709" PostTypeId="1" CreationDate="2019-06-07T08:29:33.807" Score="1" ViewCount="156" Body="&lt;p&gt;We need to add ventilation holes to an existing building. The holes will be core-hole drilled through a limestone brick wall. The bricks have crevices. The contractor is acting stupid and obstructive or I would ask them. How would one mount a short (24 cm lond, D= 300mm) piece of round air duct into the core hole? Simply put in? Sikaflex all around? How about risk of moisture creeping in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first idea is to let the duct sick out 1 cm on both sides, and to grout the wedge with sikaflex. Criticisms? Better ideas?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2024-02-12T00:08:26.117" Title="How to professionally mount an air duct through a core hole in limestone brickwork" Tags="|civil-engineering|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28710" PostTypeId="1" CreationDate="2019-06-07T14:31:21.470" Score="3" ViewCount="56" Body="&lt;p&gt;So I have a client that is asking about if they should switch from one lubricant to another in a planetary gearbox they have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main advantage that this new lubricant has is that it has a much lower viscosity (about 1/2 as much) and it is more thermodynamically stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all things being equal, I'm looking to try to get a rough order of accuracy estimate for the efficiency increase in switching to this new lubricant. I've been looking through some hydrodynamic lubrication theory, but haven't found anything that is really what I'm looking for. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any known references or established theories related to this would be extremely helpful! :)&lt;/p&gt;&#xA;" OwnerUserId="20776" LastActivityDate="2019-06-07T14:31:21.470" Title="Efficiency of a planetary gearbox as a function of the lubricants viscosity" Tags="|mechanical-engineering|gears|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28711" PostTypeId="2" ParentId="28709" CreationDate="2019-06-07T16:17:31.887" Score="0" Body="&lt;p&gt;In a project which was a mixed use 5 story building including a movie theater in LA in 1988 we needed similar set up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For under-floor ventilation in the theater we created concrete subfloor channels with service access hatches and sloped to drainage with flex insulated double wall ducts running through opening to vertical registers on the riser of steps on the isles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been there a couple of times since for minor structural changes required by other anchor tenants, but have not heard of any issues with floor ventilation system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think you need anything around your duct, makes it easy to replace if needed. Just make sure you have drainage for potential condensation.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-07T16:17:31.887" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28714" PostTypeId="1" AcceptedAnswerId="28720" CreationDate="2019-06-07T18:20:00.987" Score="9" ViewCount="11302" Body="&lt;p&gt;Small deformation shear strain &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; in engineering &lt;a href=&quot;https://en.wikipedia.org/wiki/Deformation_(mechanics)#Shear_strain&quot; rel=&quot;noreferrer&quot;&gt;can be expressed&lt;/a&gt; as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\gamma_{xy}=\alpha +\beta ={\frac {\partial u_{y}}{\partial x}}+{\frac {\partial u_{x}}{\partial y}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; represents the displacement of the edges of the infinetesimal element. But in the strain tensor we have defined&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\epsilon_{xy} = 1/2 \gamma_{xy}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear strain represents the change in the angle between the lines of the infinitesimal element. So why do we define the tensorial component as half of this angle? Why not just use the angle itself?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2022-12-17T23:04:00.900" Title="Why is tensorial shear strain half of engineering shear strain?" Tags="|structural-engineering|deformation|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28715" PostTypeId="1" AcceptedAnswerId="28717" CreationDate="2019-06-07T20:51:04.020" Score="2" ViewCount="529" Body="&lt;p&gt;At any point inside a structural element under deformation, we can always find three planes such that when we define a coordinate system with basis vectors lying on those planes, the shear stresses at that point are zero and the stress tensor is diagonal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principal stresses are the eigenvalues of the stress tensor, and are &lt;strong&gt;invariant of the rotation of the coordinate system.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This last feature on bold is recited everywhere.But my question is, so what? Why is it so useful to find these stresses whose components don't change with rotation of the coordinate system? And why would I even want to rotate my coordinate system?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2019-06-08T01:05:55.687" Title="What really is so special about principal stresses or stress invariants?" Tags="|structural-engineering|stresses|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28717" PostTypeId="2" ParentId="28715" CreationDate="2019-06-08T01:05:55.687" Score="5" Body="&lt;p&gt;I'll take your word for it that your quote is &quot;recited everywhere&quot;. The point of it is not that you can &quot;rotate your coordinate system&quot; just for fun, but that the principal stresses are &lt;em&gt;independent&lt;/em&gt; of whatever coordinate system you choose to work in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The physical structure doesn't know or care what coordinate systems you are using, so any conclusions you draw from your analysis about failure criteria, crack propagation directions, the onset of plasticity, etc, etc, &lt;em&gt;must be independent of your choice of coordinate system&lt;/em&gt; - otherwise they are guaranteed to be wrong!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The principal stress values, and the direction of the coordinate system corresponding to them, at every point in the structure, &lt;em&gt;are&lt;/em&gt; a complete description of the state of stress which is independent of the analyst's coordinate system(s), and also very easy to work with since the stress tensor is diagonal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a couple of examples, in brittle materials cracks propagate in the direction of the largest (tensile) principal stress. In ductile materials, a useful failure criterion is the maximum shear stress acting in any direction at a point, and that is simply the difference between the maximum and minimum principal stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it is possible to devise formulas that reproduce those results in a different coordinate system, but they are much more complicated, and don't give much &lt;em&gt;insight&lt;/em&gt; into what is going on in the structure.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-06-08T01:05:55.687" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28720" PostTypeId="2" ParentId="28714" CreationDate="2019-06-08T14:01:40.407" Score="7" Body="&lt;p&gt;A tensor is a mathematical object which has to obey certain rules about how to transform it from one coordinate system to another. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineers started using and measuring strains a century or more before tensors were invented (by Ricci, in around 1900, and not in the context of continuum mechanics). They just happened to make a bad choice of how to measure shear strains (though they got it right for shear stresses).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can create an &quot;engineering formulation&quot; of continuum mechanics that doesn't use tensors at all, by pretending the 6 stress components are a &quot;vector,&quot; using matrix arithmetic in an ad hoc way, and inventing complicated-looking formulas for transforming stresses and strains into different coordinate systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, you can change the definition of shear strain by a factor of two, and use mathematics that doesn't need any &quot;special&quot; definitions, just standard vector calculus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second way has obvious advantages if you want to combine continuum mechanics with other phenomena such as fluid dynamics, or with special or general relativity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Really, this is no more of an issue than measuring angles in degrees in real life, or in radians for mathematical purposes.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-06-08T14:01:40.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28721" PostTypeId="2" ParentId="28322" CreationDate="2019-06-08T14:16:25.607" Score="0" Body="&lt;p&gt;I would use an active baking system that disengages while actively used. Think of a bicycle brake being applied to release deadbolt / ratchet and allow free rotation of wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems with such a system are that you need to consider the friction required to slow and then stop the wheelchair &lt;em&gt;safely&lt;/em&gt;. A dead stop engaging mechanism would be horribly dangerous whereas a spring loaded energy absorption collar would certainly be requisite in the design. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2019-06-08T15:25:06.287" LastActivityDate="2019-06-08T15:25:06.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28722" PostTypeId="1" AcceptedAnswerId="28725" CreationDate="2019-06-08T15:33:23.283" Score="1" ViewCount="57" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0mtA3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0mtA3.jpg&quot; alt=&quot;opamp&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, can I say that &lt;span class=&quot;math-container&quot;&gt;$V_E$&lt;/span&gt; correspond to -0.7? if yes, why can this be?&lt;/p&gt;&#xA;" OwnerUserId="20789" LastEditorUserId="110" LastEditDate="2019-06-09T10:29:22.953" LastActivityDate="2019-06-09T10:29:22.953" Title="Operational Amplifier" Tags="|electrical-engineering|electrical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28723" PostTypeId="1" AcceptedAnswerId="28726" CreationDate="2019-06-08T16:45:05.210" Score="-1" ViewCount="1513" Body="&lt;p&gt;I need to find the effective length of a column (kl) from this chart which can be found in the &quot;design of concrete structure&quot; By Dr. Arthur Nilson.&lt;a href=&quot;https://i.stack.imgur.com/uwFOm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uwFOm.png&quot; alt=&quot;chart&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the question states &quot;if Psi top is 1.5 and the bottom joint is a firm (fixed) joint what is k? if the column is braced&quot; what I don't know is if the joint is fixed should I take Psi as 0 or infinity? this is a diagram of the column &lt;a href=&quot;https://i.stack.imgur.com/oYw4b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oYw4b.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13708" LastActivityDate="2019-06-09T00:14:36.617" Title="How do i find the effective length factor (k) for a column from this chart?" Tags="|civil-engineering|design|reinforced-concrete|columns|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28725" PostTypeId="2" ParentId="28722" CreationDate="2019-06-08T20:46:41.353" Score="0" Body="&lt;p&gt;In simplistic explanation Bipolar Junction Transistor is made of two diodes back to back. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BSh1f.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BSh1f.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The minimum voltage required switch on the diode between Base and Emitter is 0.7V. Since  &lt;span class=&quot;math-container&quot;&gt;$V_B$&lt;/span&gt; = 0V, in order for the diode to be on the voltage across &lt;span class=&quot;math-container&quot;&gt;$V_{BE}$&lt;/span&gt; has to be -0.7V. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.electronics-tutorials.ws/transistor/tran_1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bipolar Transistor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-06-08T20:46:41.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28726" PostTypeId="2" ParentId="28723" CreationDate="2019-06-09T00:14:36.617" Score="1" Body="&lt;p&gt;You should take &lt;span class=&quot;math-container&quot;&gt;$\psi=0$&lt;/span&gt;, typically for fixed braced concrete column, design aids set the bottom as zero flexibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Common sense then tells us that if we have a column with top and bottom fixed; it has 2 inflection points at 1/4 length of the column from the bottom and top. Hence we get K= 1/2, which is as shown on this chart.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-09T00:14:36.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28727" PostTypeId="1" AcceptedAnswerId="40656" CreationDate="2019-06-09T19:32:10.643" Score="3" ViewCount="286" Body="&lt;p&gt;Cooling towers for nuclear power plants often have the shape of a hyperboloid. At first glance, I thought that an obvious reason for this would be the smaller sectional area, meaning the tower would be less exposed to wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I did a quick calculation, based on a cooling tower nearby. I modeled the cooling tower as a hyperboloid and calculated its volume and its sectional area. I then calculated the radius needed for a cylindrical tower with the same volume. Its sectional area was just 1.5% larger. By reducing its height from 150 to 145 meters (and increasing the radius, of course), the sectional areas would have been the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that hyperboloids can be built using only straight pieces of reinforcement. However, those towers are quite big and the circular shape could be very well approximated by some sort of polygon with e.g. 1000 vertices making circularly shaped reinforcement unnecessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can imagine other reasons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The waisted shape could probably lead to a certain compression of the hot vapor and thus help/accelerate the condensation. Is this true? How big could that effect be?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Some documents I have read say one needs less material to build a hyperboloid cooling tower. Is there really a big difference in terms of concrete? After a rough calculation, assuming a thickness of 1 meter of concrete all around, I'd estimate that a cylindrical shape would only need approximately 1.5% more concrete.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, why do almost all cooling towers have this particular shape?&lt;/p&gt;&#xA;" OwnerUserId="20793" LastActivityDate="2021-03-02T12:06:45.997" Title="Cooling tower for nuclear power plants" Tags="|structural-engineering|civil-engineering|cooling|building-physics|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28728" PostTypeId="2" ParentId="28727" CreationDate="2019-06-09T20:49:34.580" Score="3" Body="&lt;p&gt;The shape is to take advantage of the Bernoulli law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The narrower throat of the tower creates reduced pressure encouraging faster convection discharge of warm, moist vapors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot water from the turbines is sprayed in at roughly 1/3 height. that compared with cool water on the floor, usually from a near river creates uplift and convective ventilation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That hyperbolic shape of the tower accelerates this vertical flow.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="512" LastEditDate="2019-06-10T15:53:00.073" LastActivityDate="2019-06-10T15:53:00.073" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28729" PostTypeId="1" CreationDate="2019-06-09T22:47:54.867" Score="-1" ViewCount="52" Body="&lt;p&gt;The work done by a turbocharger increases the inlet charge without much of a burden to the exhaust flow.  This increases the amount of air (and fuel) that flows into the engine. How does this affect work done for a given amount of fuel consumed?&lt;/p&gt;&#xA;" OwnerUserId="20797" LastActivityDate="2019-07-10T00:02:10.390" Title="Does utilizing exhaust energy to supercharge (&quot;turbocharging&quot;) increase the thermal efficiency of a gasoline internal combustion engine?" Tags="|mechanical-engineering|thermodynamics|energy-efficiency|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28730" PostTypeId="2" ParentId="28729" CreationDate="2019-06-09T23:36:55.233" Score="0" Body="&lt;p&gt;Yes, turbocharging improves both the volumetric and thermal efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as most drivers tend to use the power that does not always translate to an improved fuel consumption. &lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-09T23:36:55.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28731" PostTypeId="2" ParentId="28729" CreationDate="2019-06-09T23:37:58.680" Score="0" Body="&lt;p&gt;It's complicated and depends on application and demands such as emissions restrictions and required throttle response, but generally, figure on about a 2 percentile improvement with a turbo. An example would be going from 34% to 36% thermal efficiency by adding a turbo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see this for yourself if you look at the BSFC fuel maps of a family of engines that include turbo and normally aspirated versions.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-06-09T23:37:58.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28732" PostTypeId="1" AcceptedAnswerId="28740" CreationDate="2019-06-09T23:43:51.520" Score="2" ViewCount="891" Body="&lt;p&gt;My understanding of swash plates led me to believe that the pistons in a swash plate pump rotate around with the driving shaft along with the chambers in which they aspirate (e.g. &lt;a href=&quot;https://www.youtube.com/watch?v=1FpT9Z4at-w&quot; rel=&quot;nofollow noreferrer&quot;&gt;M&amp;amp;S Hydraulic Youtube video on swash plate plumps&lt;/a&gt;).  As the axial position of the pistons change, the piston chambers draw and expel fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I just watched a training video from &quot;Our Virtual Academy&quot; (which costs money to view unfortunately) where the pistons do not rotate with the shaft; instead, the swash plate rotates causing the pistons to move axially.  The angle of the swash plate is varied by forcing pressurized fluid into the control chamber (here is an &lt;a href=&quot;https://www.toyota-industries.com/company/business/automobile/compressor/kind_4/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;example&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't imagine why one would use this latter form of swash plate pump.  What are the reasons for the different approaches? (What is the correct terminology to distinguish the two so I can carry on my own research)&lt;/p&gt;&#xA;" OwnerUserId="17228" LastEditorUserId="17228" LastEditDate="2019-06-10T16:56:23.283" LastActivityDate="2020-01-07T16:58:29.863" Title="Swash Plate Pump Design" Tags="|pumps|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28734" PostTypeId="1" CreationDate="2019-06-10T10:20:43.187" Score="1" ViewCount="37" Body="&lt;p&gt;I'm trying to optimize the weight of a truss with the constraint that the trusses don't face more stress than the material resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a sketch of my truss:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/okef3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/okef3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a Python function that given the list of X,Y coordinates for my fours nodes, gives the weight of the truss. Let's call it Weight(X).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also a function that calculates the stress in each truss: Stress(X).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a classical unconstrained gradient approach I've minimized the weight of my truss. I only act on Node 2's x and y coordinates (lower right node). Here is the configuration I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n5mWS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n5mWS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like now to be sure that the stresses in my trusses are not higher than the material resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read some things on the Lagrangian dual approach. Here is an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is to minimize &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}(x_1-1)^2+\frac{1}{2}(x_2-1)^2$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$x_1+x_2 &amp;lt; 1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x_1&amp;gt;0$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First here is the Lagrangian of the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$L(x_1,x_2,\lambda_1,\lambda_2)=\frac{1}{2}(x_1-1)²+\frac{1}{2}(x_2-1)^2 + \lambda_1(x_1+x_2-1)-\lambda_2 x_1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to find the root of derivate of &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; regarding the prime-variables &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x_2$&lt;/span&gt;. The following are called the &quot;primal-dual relations&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\frac{\partial L}{\partial x_1} &amp;amp;= (x_1-1)+\lambda_1 x_1 - \lambda_2 \\&#xA;\frac{\partial L}{\partial x_2} &amp;amp;= (x_2-1)+\lambda_1 x_2 \\&#xA;\therefore x_1 &amp;amp;= 1 - \frac{\lambda_1 - \lambda_2}{2} \\&#xA;x_2 &amp;amp;= 1 - \frac{\lambda_1}{2}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whith these relations, the dual fonction can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$l(\lambda_1,\lambda_2)= -\frac{\lambda_1^2}{2}-\frac{\lambda_2^2}{4} + \frac{\lambda_1 \lambda_2}{2} + \lambda_1 - \lambda_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have then to maximize the function in terms of &lt;span class=&quot;math-container&quot;&gt;$\lambda_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\lambda_2$&lt;/span&gt;. When this &lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; couple is found and with the primal-dual relations, we get back to the optimal &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x_2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I solve my optimization problem following this methodology?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how to do it because my cost and constraints functions are implicit (I cannot analytically derivate them toward the X-Y coordinates of my moving node).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've writting my Lagrangian function as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;L(X,\lambda_1, \lambda_2, \lambda_3, \lambda_4) = &amp;amp;\mathrm{weight}(X) - \\&#xA;&amp;amp;\lambda_1(400-\mathrm{abs}(\mathrm{stress1}(X))) -\\&#xA;&amp;amp;\lambda_2(400-\mathrm{abs}(\mathrm{stress2}(X))) -\\&#xA;&amp;amp;\lambda_3(400-\mathrm{abs}(\mathrm{stress3}(X))) -\\&#xA;&amp;amp;\lambda_4(400-\mathrm{abs}(\mathrm{stress4}(X)))&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a Lagrangian multiplier per truss and the stress() functions give the nominal stress in each truss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think this is properly defined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've set &lt;span class=&quot;math-container&quot;&gt;$l_1\ldots l_4=1$&lt;/span&gt; at the beginning, I've found the &lt;span class=&quot;math-container&quot;&gt;$x\ast$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y\ast$&lt;/span&gt; that root the derivative toward &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x_2$&lt;/span&gt;. But then, I don't know what to do with these values because these are implicit relations, I don't have my primal-dual relations, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, do I have to do something when my Lagrangian multipliers are negative or something?&lt;/p&gt;&#xA;" OwnerUserId="20803" LastEditorUserId="20803" LastEditDate="2019-06-11T09:51:22.690" LastActivityDate="2019-06-11T09:51:22.690" Title="Constrained optimization using Lagrangian" Tags="|structural-analysis|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28736" PostTypeId="1" CreationDate="2019-06-06T18:12:58.340" Score="0" ViewCount="120" Body="&lt;p&gt;Following image is something I want to design. The uniform stainless steel round plate of diameter 1000mm and thickness 10mm is rotated by a stepper motor. But after every turn, a pressure of 3000 kg/cm^2 is applied at he edge of the plate by a hydraulic piston. Will my disc deflect? Does it need a support at bottom where pressure is applied? How should I couple my motor to the disc so that the coupling can withstand the load? Can my motor withstand such pressure? &#xA;Please help me with the design.After ever turn the plate stops and pressure is applied. Which stepper can withstand such load?&lt;a href=&quot;https://i.stack.imgur.com/lsURl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lsURl.jpg&quot; alt=&quot;System Design&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="Lalit Chavan" LastActivityDate="2019-11-07T13:03:05.523" Title="Selection Of coupling and determining size of motor to rotate heavy disc" Tags="|design|stepper-motor|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28738" PostTypeId="2" ParentId="28736" CreationDate="2019-06-10T11:41:59.973" Score="1" Body="&lt;p&gt;There are a lot of questions here, and not really enough information to answer them...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Will my disc deflect? Yes. How &lt;em&gt;far&lt;/em&gt; it will deflect and whether or not that is a problem depends on the force applied (you have not said - the pressure in the piston is not enough information on its own).&lt;/li&gt;&#xA;&lt;li&gt;Does it need a support at bottom where pressure is applied? Maybe? This depends on how much deflection is acceptable, see above.&lt;/li&gt;&#xA;&lt;li&gt;How should I couple my motor to the disc so that the coupling can withstand the load? I would recommend &lt;em&gt;decoupling&lt;/em&gt; the motor from the disc. You don't want to use the motor shaft to support it. Something like a soft jaw coupling will allow the motor to keep driving even as the disk deflects. You then need to support the disk on a separate shaft, which can be much beefier and less sensitive than the motor.&lt;/li&gt;&#xA;&lt;li&gt;Can my motor withstand such pressure? As mentioned above, it's not recommended to take any force on a motor shaft.&lt;/li&gt;&#xA;&lt;li&gt;Which stepper can withstand such load? Once you've successfully decoupled the load, you can pick any stepper with suitable performance re: spinning your disc.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-06-10T11:41:59.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28739" PostTypeId="1" CreationDate="2019-06-10T12:34:26.727" Score="0" ViewCount="708" Body="&lt;p&gt;I'm CADing a tooth profile for a bicycle chainring model and I'm referring my measurements and formulas to these sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.gearseds.com/files/design_draw_sprocket_5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.gearseds.com/files/design_draw_sprocket_5.pdf&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://data.epo.org/publication-server/image?imageName=imgf0003&amp;amp;docId=2977669&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://data.epo.org/publication-server/image?imageName=imgf0003&amp;amp;docId=2977669&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;One of them is the main instructional document while the other just presents a drawing of the profile with the same variables and equations as the first one. (though they do use different units)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They ask me to calculate some variables which will provide a graphical Point A and they make me calculate for the Angle B between a Line C and the x-axis with Point A as vertex, and it always ends up that Line C just connects the center of the roller circle and Point A. They also instructed me calculate for a length of Line D which is the same length of Line C from the circle's edge to Point A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why can't they just tell me to do that instead? I'm just wasting my time calculating for these practically redundant variables. It's not like they're teaching this for manual sketching, as they specifically use terms that are present in 2D drawing/CAD programs.&lt;/p&gt;&#xA;" OwnerUserId="20804" LastActivityDate="2019-06-10T12:34:26.727" Title="Why do instructions on creation of roller chain sprocket tooth profile require redundant measurements?" Tags="|mechanical-engineering|bicycles|transmission|chain|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28740" PostTypeId="2" ParentId="28732" CreationDate="2019-06-10T13:57:58.447" Score="3" Body="&lt;p&gt;There exist three types of axial plunger pumps. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- Rotating swash plate axiale plunger piston pumps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually for low-power systems, as the temperature rises above 120° C, it can degrade the fluid properties and cause some damages to the sensitive components. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the advantage of the first type is it simplicity and efficiency, driving a light swash plate requires less power than the heavy block of barrel and pistons&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- Fixed swash plate axiale plunger piston pumps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this type, the cylinder barrel rotates with the shaft, of course, designing this type needs more consideration and have more complexities than the previous type, but the clearance between the cylinder barrel and the housing block, allows leakage of the fresh fluid so it serves a lubricant of the rotating body and &lt;strong&gt;cooling&lt;/strong&gt; of the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3- Bent axis axial plunger piston pump: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cylinder barrel makes an angle with the rotating shaft, by changing the angle, you can acquire a variable displacement pump.     &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2019-06-10T14:12:55.660" LastActivityDate="2019-06-10T14:12:55.660" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28742" PostTypeId="1" CreationDate="2019-06-10T15:31:46.087" Score="2" ViewCount="854" Body="&lt;p&gt;For a small unit like a drill: Is it better to charge after any use or wait until the lithium battery is exhausted ?&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2022-12-23T06:18:30.393" Title="Should lithium batteries be recharged after every use , or when they are exhausted?" Tags="|battery|tools|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28743" PostTypeId="1" CreationDate="2019-06-10T16:08:35.867" Score="1" ViewCount="47" Body="&lt;p&gt;Tell me how can the 's' in first equation described with the 'integration' in second equation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hyu4B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hyu4B.png&quot; alt=&quot;pic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20789" LastEditorUserId="20789" LastEditDate="2019-06-10T16:18:45.543" LastActivityDate="2019-06-10T17:26:08.220" Title="Integrator OpAmp" Tags="|electrical-engineering|electrical|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28744" PostTypeId="2" ParentId="28743" CreationDate="2019-06-10T16:54:56.923" Score="0" Body="&lt;p&gt;The first equation is the Laplace transform of the second.  This is a subject that's touched on in a differential equations class (usually the second semester of a 2-semester Junior-level college class in the US), and is covered extensively in signal processing course work in Electrical Engineering.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-06-10T16:54:56.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28745" PostTypeId="2" ParentId="28742" CreationDate="2019-06-10T17:16:03.627" Score="0" Body="&lt;p&gt;It depends on the battery. You can discharge some batteries until 0-10 % and battery life won't be reduced. Examples: NCA (Nickel-cobalt-aluminum) and LTA (Lithium titanate oxide) lithium-ion batteries. The final state of charge (SOC) is 0-10 % and the depth of discharge (DOD) is 100-90 %. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other batteries in which is better to charge them after any use because their life gets shortened when the DOD it too high. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Check this papers: El-Sayed recommends a 20 % DOD:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;El-Sayed, M. A. H. “Lithium-ion energy storage battery in PV-smart building application”, Renewable Energy and Power Quality Journal, no. 19, April 2019.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Abdon, A., Zhang, X., Parra, D., Patel, M. K., Bauer, C., Worlitschek, J. “Techno-economic and environmental assessment of stationary electricity for different time scales”, ELSEVIER Energy, vol. 139, pp. 1173-1187, November 2017.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorDisplayName="user20096" LastEditDate="2019-06-10T20:18:43.637" LastActivityDate="2019-06-10T20:18:43.637" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28746" PostTypeId="2" ParentId="28743" CreationDate="2019-06-10T17:26:08.220" Score="2" Body="&lt;p&gt;This is a Laplace transform. 1/s is an integrator, is the Laplace transform of an integral.&#xA;s is the Laplace transform of a derivative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$v_1(t)=-R*C*d(v_0(t))/dt ------&amp;gt;V_1(s)=-R*C*s*V_0(s)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; is dimensional, its dimension is &lt;span class=&quot;math-container&quot;&gt;$time^{-1}$&lt;/span&gt;. It's basically a complex variable associated with frequency. The Laplace realm is the frequency realm, as opposed to the temporal realm.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-06-10T17:26:08.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28747" PostTypeId="1" AcceptedAnswerId="28807" CreationDate="2019-06-10T19:03:28.043" Score="1" ViewCount="59" Body="&lt;p&gt;I'm doing an experiment where 2 electrodes are inserted into 2 openings in a cylinder holding electrolytes to measure its impedance. We want the openings to be tightly sealed because the cylinder holding the electrolyte will be pressurized. Our idea is to insert the electrodes through the holes of tube fittings. The tube fitting will screw onto the threaded cylinder and it will lock into the electrodes. Will this work, and what type and size of tube fitting should I use for this? Please give any other suggestions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K4X8U.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K4X8U.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a picture of what it looks like. The cylinder will be remade with only 2 holes. That's the electrode with a rod attached. We're hoping the tube fitting will dig into the rod. The diameter of the rod is about 1/4 in.  &lt;/p&gt;&#xA;" OwnerUserId="19016" LastEditorUserId="19016" LastEditDate="2019-06-12T18:41:02.870" LastActivityDate="2019-06-14T20:47:27.893" Title="Are tube fittings a feasible way to create tight seal?" Tags="|pressure|pressure-vessel|seals|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28749" PostTypeId="1" AcceptedAnswerId="28761" CreationDate="2019-06-11T01:08:58.003" Score="0" ViewCount="43" Body="&lt;p&gt;I am working on a personal project and wanted to use a solid oxide fuel cell to power up a load (I have not decided what yet), so I am trying to find a way to estimate the power I would need to heat up my fuel cell to a working temperature, so I was considering calculating the heat capacity of the fuel cell. But as it turns out the fuel cell is composed of different materials. Should it be enough to obtain the heat capacity from each material and then add each value to compute the overall energy needed? &#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="14978" LastActivityDate="2019-06-11T16:39:19.863" Title="Is there a way to calculate the heat capacity of a solid or volumetric heat capacity of a solid oxide fuel cell?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28750" PostTypeId="1" CreationDate="2019-06-11T01:32:09.113" Score="1" ViewCount="82" Body="&lt;p&gt;Because adiabatic processes would lead to a high temperature rise due to the high reduction in pressure (from 1 atm to roughly 0 atm). &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-06-11T01:32:09.113" Title="Are most vaccum pumps isothermal?" Tags="|vacuum|vacuum-pumps|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28751" PostTypeId="1" CreationDate="2019-06-11T03:48:44.373" Score="0" ViewCount="153" Body="&lt;p&gt;I am trying to write a python code to solve the neutron diffusion equation to model neutron flux distribution in a one-dimensional two-group setting. The governing equations of the system are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$-D_1\nabla^2\phi_1+(\mathcal{E}_{a_1}+\mathcal{E}_{s12})\phi_1=\frac{1}{k}[\nu_1\mathcal{E}_{f_1}\phi_1+\nu_2\mathcal{E}_{f_2}\phi_2]+\mathcal{E}_{s21}\phi_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$--D_2\nabla^2\phi_2+(\mathcal{E}_{a_2}+\mathcal{E}_{s21})\phi_2=\mathcal{E}_{s12}\phi_1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;=Diffusion Co-efficient, &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt;=neutron flux, &lt;span class=&quot;math-container&quot;&gt;$\mathcal{E}_{a}$&lt;/span&gt;=absorption cross-section, &lt;span class=&quot;math-container&quot;&gt;$\mathcal{E}_{s}$&lt;/span&gt;=scattering cross-section, &lt;span class=&quot;math-container&quot;&gt;$\nu$&lt;/span&gt;=neutrons per fission,  &lt;span class=&quot;math-container&quot;&gt;$\mathcal{E}_{f}$&lt;/span&gt;=fission cross-section and &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;=multiplication factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I divided the geometry into 420 mesh elements and discretized the equations using Forward Difference Method (FDM). Putting the whole system into matrix form yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Bigl[C_1\Bigr]\Bigl[\phi_1\Bigr]=\frac{1}{k}\Bigl[h\Bigr]\Bigl[\nu_1\mathcal{E}_{f_1}\Bigr]\Bigl[\phi_1\Bigr]+\frac{1}{k}\Bigl[h\Bigr]\Bigl[\nu_2\mathcal{E}_{f_2}\Bigr]\Bigl[\phi_2\Bigr]+\Bigl[h\Bigr]\Bigl[\mathcal{E}_{s21}\Bigr]\Bigl[\phi_2\Bigr]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Bigl[C_2\Bigr]\Bigl[\phi_2\Bigr]=\Bigl[h\Bigr]\Bigl[\mathcal{E}_{s12}\Bigr]\Bigl[\phi_1\Bigr]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\Bigl[C\Bigr]$&lt;/span&gt;=Co-efficient Matrix and &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;=mesh element length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;th iteration, I am supposed to use &lt;span class=&quot;math-container&quot;&gt;$\phi_{1}^{n-1}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\phi_{2}^{n-1}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$k^{n-1}$&lt;/span&gt; to solve the two above equations and calculate  &lt;span class=&quot;math-container&quot;&gt;$\phi_{1}^{n}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\phi_{2}^{n}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$k^{n}$&lt;/span&gt; using the iterative scheme:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Bigl[C_1\Bigr]\Bigl[\phi_1\Bigr]^n=\frac{1}{k}\Bigl[\nu_1\mathcal{E}_{f_1}\Bigr]\Bigl[\phi_1\Bigr]^{n-1}+\frac{1}{k}\Bigl[\nu_2\mathcal{E}_{f_2}\Bigr]\Bigl[\phi_2\Bigr]^{n-1}+\Bigl[\mathcal{E}_{s21}\Bigr]\Bigl[\phi_2\Bigr]^{n-1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Bigl[C_2\Bigr]\Bigl[\phi_2\Bigr]^n=\Bigl[\mathcal{E}_{s12}\Bigr]\Bigl[\phi_1\Bigr]^n$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$k^n=k^{n-1}\frac{\int dr(\nu_1\mathcal{E}_{f_1}\phi_1^n+\nu_2\mathcal{E}_{f_2}\phi_2^n)}{\int dr(\nu_1\mathcal{E}_{f_1}\phi_1^{n-1}+\nu_2\mathcal{E}_{f_2}\phi_2^{n-1})}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused about how to code the last equation. Do I take the differential length &lt;span class=&quot;math-container&quot;&gt;$dr$&lt;/span&gt; as mesh element length &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; and turn the integration into a summation over the geometry? Or do I actually try some kind of integration? I cannot imagine how an integration can be introduced here given that &lt;span class=&quot;math-container&quot;&gt;$\phi_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\phi_2$&lt;/span&gt; are not known as functions but as matrices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help! And thanks for reading through such a long question!&lt;/p&gt;&#xA;" OwnerUserId="20809" LastActivityDate="2021-10-19T04:15:00.720" Title="How to code this iterative scheme for solving 2-group diffusion equation?" Tags="|numerical-methods|nuclear-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28752" PostTypeId="1" CreationDate="2019-06-11T05:34:59.617" Score="0" ViewCount="250" Body="&lt;p&gt;I would like to make an ice cream machine for home use. That is a mixing bowl which can be stored in the freezer over night and then keeps cold long enough to freeze the ingredients while mixing them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I thought I would just wrap a mixing bowl in commercial cooling packs, but apparently, in the hard shell cooling packs, they often simply use water. Now I would like to come up with a more fancy design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea would be to get a long copper pipe and bend it around my cooling bowl. After bending, it would be like a long spiral around the bowl. At one end, I would plug it shut, then it's filled with water and with a bit of air left, the other end is plugged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would the pressure from freezing water be enough to unbend the copper?&#xA;Do you have ideas on fastening the copper spiral?&lt;/p&gt;&#xA;" OwnerUserId="20810" LastActivityDate="2019-06-11T15:19:46.730" Title="Freezing water in bent copper pipe" Tags="|thermal-insulation|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28753" PostTypeId="2" ParentId="28752" CreationDate="2019-06-11T05:42:34.137" Score="0" Body="&lt;p&gt;Water, on freezing can easily split copper pipe - have had to repair several. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air-gap idea, while good, may not work in all situations... If the water freezes in the middle of the copper pipe then the blocked end can then split.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you should consider a solution that will not freeze until it gets down to -20 degrees C or so... A temperature chosen on the grounds that is what most household freezers get to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have liquid nitrogen about for your cooling, then you will need a different fluid...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-11T05:42:34.137" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28755" PostTypeId="1" CreationDate="2019-06-11T09:40:32.493" Score="0" ViewCount="81" Body="&lt;p&gt;I'm looking to design or purchase an evaporative cooling system for a small 70ft^2 California dorm room without air conditioning. This room has a single window and door to a large common space with a freezer. The common space and room currentl reach temperatures of 80+, while outside the room reaches temperatures of 90+. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An evaporative cooling system cools air by using heat to vaporize water. This can only go so far, as a closed room will eventually become too humid and water will cease to evaporate. For this to work in steady-state, then, moisture or moist air must continuously be sucked away from the room. Practically, I can only think of doing this by having the moist air leave the room. Mass balance implies that new (unfortunately hot) air must enter the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So our system is now dry hot air entering through one window, passing through a fan and over a basin of water to cool it, and leaving through my door towards warm common space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Would simply having an ultrasonic humidifier blow water into a continuously flowing air stream be sufficient to cool the air, or otherwise be more efficient than just running the air over a water bucket? If the heat is truly just going into enthalpy of vaporization, this seems like it'd be more efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) This design relies on the air cooling process being sufficiently &quot;fast&quot; relative to the air's flow rate. Otherwise the hot air will enter but only be slightly cooled by the system before leaving our cooling system, entering the main room and discomforting the occupants. Everyday experience suggests that opening a window, even with an ice bucket in front of the window, won't absorb heat quickly enough when it's 90 degrees outside and 80 inside. At the same time, evaporative coolers have found commercial success, so is this simply not true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) How can I effectively have moist air &quot;leave&quot; a room? Is simply opening a door enough such that the cool, moist air will diffuse out on its own? If I leave the door closed and turn on a fan and ice water in front of an open window, can I expect warm dry air to come in through the bottom of the window and by mass balance necessarily leave as cool, moist air (unintuitively) through the top?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) I'm hoping that my moisture sink will be the common space outside my door, but given that this is almost, if not just-as warm as my room, it seems that would let heat back in from that end as well. Does this backflow doom our system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that, in more industrial contexts, these kinds of questions require simulation tools to answer. I just wonder if there's back-of-the-hand heuristics (e.g., convection is faster than conduction) that I can use to do this without needing to break out Ansys or fuss around too much with system configurations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If cooling my room like this is, indeed, hopeless, how does a $40 evaporative cooling system manage to circumvent these problems from a heat/ mass balance perspective? Where does it dump the moisture, and how does it prevent heat backflow? Is there a better system design I'm missing?&lt;/p&gt;&#xA;" OwnerUserId="8066" LastActivityDate="2019-06-12T00:55:01.150" Title="How to dump moist air without heat backflow in evaporative cooling system?" Tags="|cooling|hvac|evaporation|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28756" PostTypeId="1" CreationDate="2019-06-11T11:18:25.237" Score="0" ViewCount="44" Body="&lt;p&gt;For example, i have this differential equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dot{y} =  1.8 + 0.58u \sqrt{y}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to model this system in open loop?? My understanding is that since the term &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; is in the output as well as one of the input terms, then it is only possible to model with a feedback loop.&lt;/p&gt;&#xA;" OwnerUserId="21627" LastEditorUserId="21627" LastEditDate="2019-06-11T21:50:45.010" LastActivityDate="2019-06-11T21:50:45.010" Title="Is it possible to have open loop with same input term as the output?" Tags="|modeling|matlab|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28757" PostTypeId="2" ParentId="28752" CreationDate="2019-06-11T12:48:26.390" Score="3" Body="&lt;p&gt;If you really want to use the copper pipe, insert a thin strip of closed-cell foam into the length of the pipe.  The ice expansion will crush it rather than the pipe, and unlike leaving an air gap, it will remain positioned along the entire interior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this might actually be an instance of &quot;&lt;a href=&quot;https://meta.stackexchange.com/questions/66377/what-is-the-xy-problem&quot;&gt;the X Y Problem&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;One possible alternative (I'm sure there are better) is to use a larger bowl that fits under your metal mixing bowl.  Put the water into the larger bowl, insert the metal bowl, and using a weight across the rim of the outer bowl, force the inner bowl to remain sunken almost to the rim while the liquid is being frozen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you start with too much liquid, when it freezes it will simply spill out through the gap between the two bowls.  Next time use less.  When you have the correct amount, you can permanently keep it in a large and tough plastic bag and avoid all the mess.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The other problem is the use of water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are going to chill the water in a freezer that provides some unknown temperature.  Experiment with different concentrations of salt until you find the strongest concentration that will freeze solid in your freezer.  This will ensure that as the ice melts, the liquid will be as cold as possible.&lt;/p&gt;&#xA;" OwnerUserId="19930" LastEditorUserId="19930" LastEditDate="2019-06-11T15:19:46.730" LastActivityDate="2019-06-11T15:19:46.730" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28758" PostTypeId="1" CreationDate="2019-06-11T13:40:20.177" Score="0" ViewCount="1959" Body="&lt;p&gt;Two gear wheels A and B are touching each other and they can rotate about fixed points P and Q respectively. how to calculate the degree of freedom of the system by using gubbler's equation?&lt;/p&gt;&#xA;" OwnerUserId="19714" LastActivityDate="2019-06-11T20:38:44.227" Title="what is the degree of freedom of a pair of gear wheels?" Tags="|mechanical-engineering|gears|mechanisms|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28759" PostTypeId="2" ParentId="28752" CreationDate="2019-06-11T14:20:55.697" Score="1" Body="&lt;p&gt;The best ice cream is made with the outside bucket temperature below 0 oC. See this link &lt;a href=&quot;https://www.icecreamnation.org/science-of-ice-cream/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.icecreamnation.org/science-of-ice-cream/&lt;/a&gt;. The common approach is the tried and true method to use ice and rock salt in the outer bucket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you want to make a bucket-in-a-bucket, use a liquid with a freezing point below 0 oC. You could use tips from the links below to make your own liquid ice pack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.spine-health.com/blog/how-make-your-own-gel-ice-pack-or-moist-heat-pack&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.spine-health.com/blog/how-make-your-own-gel-ice-pack-or-moist-heat-pack&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.themakeyourownzone.com/5-ways-to-make-homemade-ice-packs/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.themakeyourownzone.com/5-ways-to-make-homemade-ice-packs/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, to your direct questions ...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Freezing water will split the copper tube.&lt;/li&gt;&#xA;&lt;li&gt;Copper can be brazed to copper. See this link &lt;a href=&quot;https://www.copper.org/applications/plumbing/techcorner/soldering_brazing_explained.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.copper.org/applications/plumbing/techcorner/soldering_brazing_explained.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-06-11T14:20:55.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28760" PostTypeId="2" ParentId="28755" CreationDate="2019-06-11T16:17:56.010" Score="0" Body="&lt;p&gt;Evaporative cooling system's efficiency is greatly influenced by the ambient moisture, and is not even advisable for high humidity zones. And it needs to be installed in a shaded area or have a vented enclosure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way they operate is by blowing dry air through filters kept wet by a small pump circulating water from a reservoir on the floor of the cooler box refilled by a flush ball valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The optimal setting for releasing the moist air is to maintain a small opening somewhere in the room, small enough to keep the air pressure in the room fractionally higher than outside preventing the hot air to circulate back to the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The methods you suggest are not going to be practical.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-11T16:17:56.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28761" PostTypeId="2" ParentId="28749" CreationDate="2019-06-11T16:39:19.863" Score="0" Body="&lt;p&gt;This is a case where you want to determine a property in a composite using some form of a rule of mixtures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ P^n = \sum f_j P^n $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is the composite property, &lt;span class=&quot;math-container&quot;&gt;$f_j$&lt;/span&gt; is typically the volume fraction, and &lt;span class=&quot;math-container&quot;&gt;$P_j$&lt;/span&gt; is the property of the component &lt;span class=&quot;math-container&quot;&gt;$j$&lt;/span&gt; in the mixture. When &lt;span class=&quot;math-container&quot;&gt;$n = 1$&lt;/span&gt;, this is the simple rule of mixtures. When &lt;span class=&quot;math-container&quot;&gt;$n = -1$&lt;/span&gt;, this is the inverse rule of mixtures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example of this rule applied to fiber composites is given at the link below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Rule_of_mixtures.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Rule_of_mixtures.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For heat capacity, the formulation is written using mass specific heat capacity and mass fraction. The link below gives an explanation and a calculator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://thermtest.com/rule-of-mixtures-calculator&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://thermtest.com/rule-of-mixtures-calculator&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-06-11T16:39:19.863" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28762" PostTypeId="2" ParentId="28756" CreationDate="2019-06-11T17:38:17.663" Score="1" Body="&lt;p&gt;The answer really boils down to how you define your system.  If you say &quot;this is my plant&quot; and give that equation, then as long as &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; does not depend on &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; then the system is operating in open loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, there is some inherent feedback &lt;em&gt;internal&lt;/em&gt; to your plant, but that's the case with just about anything that might be described with a differential equation.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-06-11T17:38:17.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28763" PostTypeId="2" ParentId="28755" CreationDate="2019-06-11T19:38:43.170" Score="1" Body="&lt;p&gt;I'm not an expert on swamp coolers, but I would try to keep all the evaporation &lt;em&gt;outside&lt;/em&gt; of your dorm room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example: You build a heat exchanger between inside and outside, then you mist the exterior with water. You'd put fans on both sides preferably. If you mist enough, the exterior will reach the dew point temperature of the exterior environment. There are charts you can use to calculate this temperature. Assuming it's dry outside, the temperature will probably be quite low. Heat leaves your room through the heat exchanger (as long as your room is warmer than the outside dew point). And the humidity in your room does not increase.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-06-11T19:38:43.170" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28764" PostTypeId="2" ParentId="28758" CreationDate="2019-06-11T20:38:44.227" Score="1" Body="&lt;p&gt;The gear is just  two otherwise independent items linked together. What  you do is calculate the freedom as you would without  the gear and then eliminate one from the other of the gearpairs to account for the gear forcing the other object. So if your gears can just  rotate and  have only one degree of fredom then the system of 2 gears  has freedom of 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not generally a good idea to think this way  unless you idealize the gear quite much. A real gear sytem does have backlash.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2019-06-11T20:38:44.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28770" PostTypeId="2" ParentId="28755" CreationDate="2019-06-12T00:55:01.150" Score="0" Body="&lt;p&gt;You are right, evaporative cooling systems require a continuous flow of large amounts of air through the space being cooled in order to be effective. These work well in places where the ambient humidity is very low (Arizona deserts or the central valley in California). Bowls of water and fans are not going to make a difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will find useful rules of thumb for evaporative air conditioning in heating/ventilating/air conditioning (HVAC) handbooks published by the makers of air conditioning equipment, like Trane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But... if you are in a dorm, this means you are not paying for electricity, yes? Then the smart thing to do is buy a small window-mount refrigeration-style air conditioning unit that plugs into the wall, and build a wooden mount for it to fit the window you have. These sell for &lt;span class=&quot;math-container&quot;&gt;$100 new or $&lt;/span&gt;50 used. As a recovering ex-engineer, I would do this and declare VICTORY!&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-06-12T00:55:01.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28774" PostTypeId="1" CreationDate="2019-06-12T18:05:25.040" Score="-1" ViewCount="66" Body="&lt;p&gt;Good morning. I have been reading the threads about &lt;em&gt;&quot;Understanding required torque for a motor lifting a weight&quot;&lt;/em&gt; in hopes it would apply to my situation. I would like to utilize an R/C servo to rotate a mass about 100 degrees. the specifics are for my 1959 Impala, I would like to automate the license plate bracket to rotate down when the transmission is put into reverse so the back up camera will be able to view behind the car.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The license plate holder and license plate weight approximately 20oz&#xA;(0.57kg) Dim: 10&quot;L x 6&quot;H x 0.075&quot;W)&lt;/li&gt;&#xA;&lt;li&gt;The rotating shaft is 0.25&quot; (0.00635m) stainless steel&lt;/li&gt;&#xA;&lt;li&gt;The license plate holder (and license plate) are offset from the shaft by 0.5&quot;&#xA;(0.0127m)&lt;/li&gt;&#xA;&lt;li&gt;The total rotational distance for the holder and license plate to travel is 0.27m in 3 seconds&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Solving F=ma, I come up with 5.59N (gravity) plus .5N (up) =&gt; F = 6.1N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T=Fr solves to be 0.02Nm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am not sure how the offset plays into the calculations?? or does it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PzZif.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PzZif.jpg&quot; alt=&quot;drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/69b41.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/69b41.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20825" LastEditorUserId="20825" LastEditDate="2019-06-13T18:21:20.347" LastActivityDate="2019-06-14T09:20:13.880" Title="understanding required torque for a servo" Tags="|mechanical-engineering|torque|servo|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28775" PostTypeId="1" CreationDate="2019-06-12T22:30:06.440" Score="0" ViewCount="1710" Body="&lt;p&gt;I am trying to understand the fundamental difference between the following terms in the context of control systems: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Hysteresis &lt;/li&gt;&#xA;&lt;li&gt;Low pass filter&lt;/li&gt;&#xA;&lt;li&gt;Deadband&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Are these just fancy names that can be &lt;em&gt;tweaked&lt;/em&gt; to provide &lt;strong&gt;same&lt;/strong&gt; result? By tweaking I mean some intelligent way of formulating the variables, such as using the absolute value of the output variable, instead of delta between current and previous value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can conceptually think of examples where the same behavior can be achieved by how we tweak any of the above three techniques. However, I am concerned that such tweaking is essentially changing the fundamental meaning of the term itself. &lt;/p&gt;&#xA;" OwnerUserId="20826" LastEditorUserId="8055" LastEditDate="2019-06-13T23:51:25.373" LastActivityDate="2019-06-13T23:51:25.373" Title="Difference between Hysteresis, low pass filter and deadband in control theory?" Tags="|control-engineering|control-theory|signal-processing|electronic-filters|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28776" PostTypeId="2" ParentId="28775" CreationDate="2019-06-12T23:15:21.590" Score="3" Body="&lt;p&gt;No, or at least not as they are generally used and understood.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Hysteresis is a nonlinear phenomenon where a variable trending in one direction tends to &quot;pull&quot; another variable along with it, and that other variable &quot;sticks&quot; until it's pulled the other way.  Think of backlash in gears, or the B-H curve of a hard magnetic material.&lt;/li&gt;&#xA;&lt;li&gt;A low-pass filter is usually linear, usually constructed.  It passes low-frequency signals, and blocks high frequency signals.&lt;/li&gt;&#xA;&lt;li&gt;Deadband is a memoryless nonlinear action where inputs within some band (the dead band) are not passed to the output.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-06-12T23:15:21.590" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28777" PostTypeId="1" CreationDate="2019-06-12T23:32:51.667" Score="1" ViewCount="118" Body="&lt;p&gt;I think i understand the mechanics in a steam locomotive:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fill boiler with water&lt;/li&gt;&#xA;&lt;li&gt;Close off boiler&lt;/li&gt;&#xA;&lt;li&gt;Heat boiler until pressure is sufficient&lt;/li&gt;&#xA;&lt;li&gt;Keep boiler hot and use steam pressure until no more water left&lt;/li&gt;&#xA;&lt;li&gt;repeat&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Boiler is heated and energy is given to water in form of heat. Water gets hot and vaporizes. Water vapor tries to use a lot more space, this expansion is used in pistons to create mechanical force. Since the boiler is closed off while in use, the water/steam pressure has nowhere to go but to the pistons. This seems clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, in power plants (all types that heat water: nuclear, coal, gas and so on) water is circulated. This means that they do not shut down the plant to refill it with water. Here's what i have trouble understanding: if you have to use X amount of energy to pump in the water to a cold boiler to keep the pressure up, you would also have the same amount on energy available to extract on the other end. But if you heat it, you add energy to the water while it is in the boiler, you increase the pressure. Logic would say you have more energy now available on the other end. But wouldn't the pressure try to press out form the pump side also? The pump would have to fight more to keep the water in and waste more energy. So i think the pressure is not increased in the boiler. How is the heat energy then extracted by the turbines? Is the water converted to steam and the speed of the steam is increased using heat while keeping the pressure the same and then the energy is extracted from the very fast speed of the flowing steam(kinetic energy essentially)? Much like a jet engine, the last compressor stage is the highest pressure point and downstream only speed is added while pressure is lowered? If so then it would seem the pump would have to maintain pressure while the steam slowly leaves via turbines. Or is there some other mechanism that eludes me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If somebody says that turbines work by using the pressure difference then please explain where does the energy come from if the pressure is generated by the high pressure feedwater pump before the boiler. All you could extract would be what the pump puts in and boiler should increase pressure both ways making it harder for the pump to put anything in at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="20827" LastActivityDate="2019-06-13T00:39:24.903" Title="Why doesn't steam/water flow backwards from expansion in power plant feedwater pump/boiler?" Tags="|thermodynamics|power|steam|flow-control|" AnswerCount="1" CommentCount="1" ClosedDate="2019-06-13T08:53:00.090" ContentLicense="CC BY-SA 4.0" />
  <row Id="28778" PostTypeId="1" CreationDate="2019-06-13T00:15:57.130" Score="2" ViewCount="877" Body="&lt;p&gt;Been referencing this website for some time but now need to seek specific help.  I am not an engineering student but rather a DIY hack who loves engineering and machining neither of which I am particularly good at yet!! :(&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently looking at a project to design an aluminium cylinder head for an engine.  The engine block is an existing cast iron one with a 92mm cylinder bore.  The strength of the block is not in question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am specifically trying to understand how to calculate the thickness of the aluminium combustion chamber walls in order to minimise deflection (else the head gasket will fail) and to ensure sufficient strength so as not to fail.  To simplify lets assume the cylinder head is a flat plate sitting on top of the block.  It is bolted down with a bolt spacing of 95mm across the block and 102mm along the length (not sure if this is relevant or not).  I realise that a real cylinder head will have braces and other features that will provide strength but that is too complicated and this simple model will give me a starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure within the combustion chamber is expected to get as high as 1500psi (race engine) at which point lets say the volume of the combustion chamber will be 90 cubic centimeters (the piston will already be travelling back down the cylinder).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to use the formula I found on this site below;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;S = PD / (2 x Tw)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tw = PD / (2 x S)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipidea shows the yield strength of aluminium (6061) to be 35,000psi so lets be safe and use 25,000psi as a working strength.  The formula becomes;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tw = (1500 x 3.622) / (2 x 25000)&#xA;Tw = 0.10866in or 2.76mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, I am not sure I have the thickness correct in terms of being strong enough not to fail as it seems very thin.  Secondly, this does not take into account the bending of the head.  I really do not know how to calculate for that.  I would say 0.001 - 0.002&quot; for deflection is the max.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to research and learn for myself but this one has beaten me.  Any help would be hugely appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;* UPDATE - EDIT *&lt;/strong&gt;&#xA;Not sure the best way to address concerns in the answers so shall do it here.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Not sure what in-cylinder temperatures will be but understand that it should be included.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I didn't make it clear as to which thickness I am looking at.  The cylinder walls are cast iron and are from an existing engine so I do not have any concerns about their ability to withstand the forces.  I am only concerned about the aluminium plate on the top.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What I am designing will be almost a flat plate on top of the cylinder block so this calculation should provide indicative results.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I agree that 356 is a common alloy for engine use.  6061 is almost identical and available to me here in Australia.  Maybe I need to use something like SimScale to analyse stresses.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="20828" LastEditorUserId="20828" LastEditDate="2019-06-14T07:04:06.620" LastActivityDate="2019-06-17T01:59:08.570" Title="Engine Combustion Cylinder Wall Thickness" Tags="|structural-analysis|combustion|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28779" PostTypeId="2" ParentId="28777" CreationDate="2019-06-13T00:39:24.903" Score="2" Body="&lt;p&gt;Google &quot;Rankine cycle&quot;.  The boiler operates at an elevated pressure (sometimes several atmospheres).  But because steam at a given pressure is many times bigger than liquid water at the same pressure, the energy required to pump water into the boiler is considerably less than the energy obtained from using that water (as steam) to power the extracting mechanism (piston engine, turbine, whatever).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-06-13T00:39:24.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28783" PostTypeId="2" ParentId="22087" CreationDate="2019-06-13T15:23:56.247" Score="2" Body="&lt;p&gt;The possible air flow restriction caused by a shade or awning will more than offset the advantage of slightly cooler air ( provided by shade) entering the compressor/condenser. I see nothing on my outdoor unit that could be exposed to solar UV except paint. My previous unit ( 15 years old) showed no damage from UV except a decal with a name that was somewhat curled on an edge.I had my new unit installed a little further from the house to avoid the roof/eave over hang which gave some shade but at a cost of some interference with the upward hot air flow from the unit. Update : Just had a new unit put in and I checked the installation instructions ( American Standard unit , in the US) : There is no mention of shading the unit. However, it does say that there must not be anything above it for at least 5 ft. so that air flow is not impaired.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-06-14T02:35:06.453" LastActivityDate="2019-06-14T02:35:06.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28785" PostTypeId="1" CreationDate="2019-06-13T18:50:55.323" Score="0" ViewCount="1941" Body="&lt;p&gt;Trying to find examples of structures, components etc. in plane strain (under normal loading conditions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is a thin-walled-pressure-vessel in plane strain?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I thought it was plane stress but could be wrong)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards&lt;/p&gt;&#xA;" OwnerUserId="20841" LastActivityDate="2019-06-14T02:19:41.923" Title="Plane Strain examples" Tags="|pressure-vessel|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28786" PostTypeId="2" ParentId="28785" CreationDate="2019-06-13T19:23:18.330" Score="1" Body="&lt;p&gt;A section of a long straight pipeline can be considered as plane strain. When the internal pressure creates radial strain in the pipe, the Poisson's ratio effect can not reduce the &lt;em&gt;length&lt;/em&gt; of the pipe so the strain in the axial direction is always zero. (Think about a pipeline that is say 100mm in diameter and 100km long - it doesn't make sense to say that if the diameter increased by 1 mm because of internal pressure or thermal expansion, Poisson's ratio would make the length of the pipeline decrease by a few hundred meters!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A thin walled pressure vessel is &lt;em&gt;approximately&lt;/em&gt; a plane stress situation, if you ignore the radial stress through the thickness of the vessel. To be consistent with the boundary conditions, the radial stress must be equal to the internal pressure on the inside of the wall and zero on the outside, but that is a small stress compared with the circumferential and axial stresses, which are both of the order of &lt;span class=&quot;math-container&quot;&gt;$R/t$&lt;/span&gt; times bigger, where &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the pipe radius and &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; the wall thickness. (The previous sentence ignored a few factors of 2, but they don't affect the point I'm making).&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-06-13T19:23:18.330" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28787" PostTypeId="2" ParentId="28785" CreationDate="2019-06-13T20:15:39.427" Score="0" Body="&lt;p&gt;If you confine a column laterally by rigid walls so it can not laterally expand under the load it is under plane strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In structural members  under plane strains the  principal plane strains can be found by rotating the coordinate system very similar to Mohr circle to find the principal max and min strain and the rotation angle is.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ tan2\theta=\frac{2 \epsilon_{xy}}{\epsilon_{xx}-\epsilon_{yy}} $$&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-13T20:15:39.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28788" PostTypeId="2" ParentId="28778" CreationDate="2019-06-13T20:19:57.293" Score="2" Body="&lt;p&gt;You text says cylinder head, not cylinder. A cylinder would be easier as it has a cast iron liner (usually) in the aluminum casting. I can promise you the head is not 6061 or any other wrought alloy. You are looking for 3 digit casting numbers. I found 356 and 319 are ( or were) very common alloys in automotive castings. Strength will be lower, Like all aluminum, they are age hardenable to some degree; both would heat-treat to about 24,000 psi yield strength. Stresses would be seriously difficult to calculate, I think finite element techniques would be needed. Even those few approximately hemispherical combustion chambers would need to allow for 6 openings for valves (4), spark plug, and fuel injection.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="1832" LastEditDate="2019-06-17T01:59:08.570" LastActivityDate="2019-06-17T01:59:08.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28790" PostTypeId="1" CreationDate="2019-06-13T23:43:18.790" Score="1" ViewCount="120" Body="&lt;p&gt;The steel series called kigami/shirogami (yellow/white paper steel, high purity carbon steels in different grades), aogami (blue paper steel, a tungsten alloyed tool steel series), aogami super (a refinement of the aogami series), gingami (silver paper steel, a rudimentary but pure and apparently precise stainless series) are commonly used for quality japanese hairdressing, kitchen, gardening, woodworking and shaving tools. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such tools are, in the west, usually marketed as very traditional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nowadays, it seems these steel types are proprietary to the hitachi subsidiary &quot;Yasugi Special Steel Works&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When were these steel types introduced? English language internet research yields no clues except that YSS was founded in 1899 and merged multiple times, and that some rather old YSS-steel based razors exist with S2 and A2 as part of their inscriptions, and that these steels use some kind of electric crucible process and are made from iron sands, and that the highest carbon grade of &quot;white&quot; steel (shirogami #1) is considered similar to traditionally made &quot;tamahagane&quot; sword/knife steel in material properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were these steel types an invention of YSS; if not, who introduced them when; if yes, when did they introduce them? What preceded them in the relevant markets?&lt;/p&gt;&#xA;" OwnerUserId="15221" LastActivityDate="2019-06-13T23:43:18.790" Title="When were the japanese yellow/white/blue/silver paper steels introduced?" Tags="|steel|metallurgy|engineering-history|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28791" PostTypeId="2" ParentId="28785" CreationDate="2019-06-14T02:19:41.923" Score="1" Body="&lt;p&gt;Thin wall vessels are not in plain strain. The lower the strength the thicker a vessel must be to reach plain strain. A reference you maybe able to find is &quot;Fracture of Structural Materials &quot; by Tetelman and Mcevily , Pub. Wiley and sons 1967.A note on pg 21 :&quot; In general , plain -strain conditions exist in thick bodies and plain-stress conditions exist in thin ones&quot;. For petrochemical vessels we generally would not do a fracture mechanics analysis /brittle fracture evaluation unless wall thickness was over 1 in. : that is, there was a small chance of plain strain conditions. ( Low temperature brittle fracture aside). For line pipe and casing; toughness tests ( eg. Charpy) are often required not because of plain strain ,but because the high strengths and high stresses; casing couplings can easily reach yield stress.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-06-14T02:19:41.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28792" PostTypeId="1" AcceptedAnswerId="28794" CreationDate="2019-06-14T03:53:00.317" Score="0" ViewCount="67" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CdcaR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CdcaR.jpg&quot; alt=&quot;image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the minimum value of &lt;span class=&quot;math-container&quot;&gt;$R_2$&lt;/span&gt; (in &lt;span class=&quot;math-container&quot;&gt;$R_1$&lt;/span&gt;) to minimize loading effect?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please explain the concept.&lt;/p&gt;&#xA;" OwnerUserId="20789" LastEditorUserId="110" LastEditDate="2019-06-14T10:26:33.347" LastActivityDate="2019-06-14T10:26:33.347" Title="LPF Second Order use cascade" Tags="|electrical-engineering|electrical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28793" PostTypeId="2" ParentId="28774" CreationDate="2019-06-14T09:20:13.880" Score="0" Body="&lt;p&gt;This is a pitfall I see all the time. You will not be able to calculate the required force analytically in a situation like this. Friction, and exterior forces will dominate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Honestly the best approach short of turning this into a scientific experiment would be to eyeball it, and test the results. Hobbyking makes a &lt;a href=&quot;https://hobbyking.com/en_us/hobbykingtm-hk15338-giant-digital-servo-mg-25kg-0-21sec-175g.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;large robotics servo&lt;/a&gt; that may be about the right size.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-06-14T09:20:13.880" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28794" PostTypeId="2" ParentId="28792" CreationDate="2019-06-14T10:25:42.793" Score="0" Body="&lt;p&gt;As described in &lt;a href=&quot;https://www.electronics-tutorials.ws/filter/filter_2.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Passive Low Pass Filter&lt;/a&gt; &quot;&lt;em&gt;to reduce the loading effect, 10x from the previous stage implement the impedance such that R2 = 10 x R1 and C2 = 1/10th C1&lt;/em&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conceptually the target 2nd-Order Filter Corner Frequency and 2nd-Order Low Pass Filter -3dB Frequency will dictate the R and C component values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice the ideal R and C valued don't exist in the open market, and engineers make compromises to design the best circuit. The calculation may require a resistor value of 11.6 Kohm and capacitor value of 157 micro farads. These components don't physically exists in the market. Therefore designers are required to make compromises. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below are some equations and graphs to complete this discussion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd-Order Filter Corner Frequency&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GUOsS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GUOsS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2nd-Order Low Pass Filter -3dB Frequency&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o4WEe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/o4WEe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n is the order number, which 2 for this discussion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequency Response of a 2nd-order Low Pass Filter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t2xj6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t2xj6.png&quot; alt=&quot;*enter image description here*&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-06-14T10:25:42.793" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28795" PostTypeId="1" AcceptedAnswerId="28842" CreationDate="2019-06-14T10:46:06.467" Score="3" ViewCount="67" Body="&lt;p&gt;I have in mind a device somewhere in between a firearms trigger and a clock escapement. One way it could be constructed is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A crank would displace a linearly moving slider backwards, so that it compresses a spring.&lt;/li&gt;&#xA;&lt;li&gt;Eventually the crank would let the slider slip, so that the spring shoots it forward, where it could again be caught by the crank on its next turn.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;However, this description is not essential to my question. I simply wish to deliver the idea to the reader, all because I do not have a name for it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is this name that I am seeking, so that I can further look it up, to hopefully find, study and re-use previously developed constructions of this kind.&lt;/p&gt;&#xA;" OwnerUserId="20844" LastActivityDate="2019-06-17T22:17:12.990" Title="How do I call a kinematic assembly that converts rotary motion to sawtooth linear?" Tags="|mechanical-engineering|kinematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28797" PostTypeId="1" CreationDate="2019-06-14T13:16:07.377" Score="1" ViewCount="452" Body="&lt;p&gt;Is the term &amp;quot;English units&amp;quot; still in current everyday use? I do hear this term among other engineers I work with but I'm unsure if this is a localized usage or is consistent with the wider community.&lt;/p&gt;&#xA;&lt;p&gt;Mainly I'm focused in the HVAC field but I don't know if terminology would be different in other areas. I noticed that &lt;a href=&quot;https://www.ashrae.org/technical-resources/ashrae-handbook&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASHRAE uses&lt;/a&gt; the term &amp;quot;IP&amp;quot; or &amp;quot;inch-pound&amp;quot; to refer to what could also be called &amp;quot;English&amp;quot; units, e.g.:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&amp;quot;The ASHRAE Handbook ... is available in I-P (Inch-Pound) or SI&#xA;(System International) units of measurement.&amp;quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Formerly I had thought &amp;quot;US customary&amp;quot; was the appropriate term to use.&lt;/p&gt;&#xA;&lt;p&gt;There might be some subtle technical differences among how &amp;quot;English&amp;quot;, &amp;quot;IP&amp;quot;, &amp;quot;US customary&amp;quot;, or any other scheme is defined. But what I'm interested in is understanding which term mechanical engineers are likely to use &lt;em&gt;in everyday discussion or writing&lt;/em&gt; where those technical differences are generally not relevant.&lt;/p&gt;&#xA;&lt;p&gt;I don't have the impression these units are widely used outside the USA, but I do have an international &amp;quot;audience&amp;quot; in mind so international differences would be interesting, if any.&lt;/p&gt;&#xA;" OwnerUserId="6489" LastEditorUserId="6489" LastEditDate="2022-03-18T12:15:12.150" LastActivityDate="2022-03-18T12:15:12.150" Title="What's the current nomenclature for &quot;English&quot; system of units?" Tags="|hvac|unit|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28798" PostTypeId="2" ParentId="28797" CreationDate="2019-06-14T13:54:23.793" Score="2" Body="&lt;p&gt;The term &quot;English Units&quot; might be confusing, especially with the English and American...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the term is &quot;Imperial Units&quot; for those in the UK, but not sure how other countries refer to them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gallons are defined as US or Imperial and have different volumes. But a gallon does have different volumes in history anyway...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inch-pounds is a &quot;standard&quot; name, as well as foot-pounds, both used on torque wrenches smaller and larger.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-14T13:54:23.793" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28799" PostTypeId="2" ParentId="26754" CreationDate="2019-06-14T15:02:25.270" Score="1" Body="&lt;p&gt;Just like any other &lt;code&gt;*get&lt;/code&gt; function you can use in APDL, you can loop on each element (or on each element of a group) and fill an array:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;allsel&#xA;*get,nbe,elem,all,count&#xA;wanted_SYS= 11&#xA;*del,myarray&#xA;*dim,myarray,array,nbe,6&#xA;*do,i,1,nbe,1&#xA;    esel,s,,,i&#xA;    *get,myarray(i,1),cdsy,wanted_SYS,loc,x&#xA;    *get,myarray(i,2),cdsy,wanted_SYS,loc,y&#xA;    *get,myarray(i,3),cdsy,wanted_SYS,loc,z&#xA;    *get,myarray(i,4),cdsy,wanted_SYS,ang,xy&#xA;    *get,myarray(i,5),cdsy,wanted_SYS,ang,yz&#xA;    *get,myarray(i,6),cdsy,wanted_SYS,ang,zx&#xA;*enddo&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="12198" LastActivityDate="2019-06-14T15:02:25.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28800" PostTypeId="1" AcceptedAnswerId="31910" CreationDate="2019-06-14T15:24:14.323" Score="0" ViewCount="83" Body="&lt;p&gt;I have thin, variable diameter PVC wires that run along a metal housing. Problem being, they can get pulled on and disconnect. I want to bond them down to the metal housing as a sort of strain relief but have not had success. There are holes in the metal so a glue would have to be pretty viscous. Taping over (tried VHB, teflon) still allows them to move around under the tape and instant adhesives (tried hot melt, acrylic) peel. Accepting all suggestions, and thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="20478" LastActivityDate="2019-11-13T06:59:21.247" Title="Bonding PVC Cables To Metal Housing" Tags="|mechanical-engineering|design|adhesive|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28802" PostTypeId="1" AcceptedAnswerId="28839" CreationDate="2019-06-14T18:32:36.300" Score="2" ViewCount="9770" Body="&lt;p&gt;I have a quick question related to the topic of mechanics of materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were to have a beam that is fixed at both ends and has a central point force, do moments develop at the fixed supports?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tVPCm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tVPCm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking, wouldn't there be an absence of moments at the fixed supports since the balanced vertical reaction forces at both ends would prevent the development of a moment?&lt;/p&gt;&#xA;" OwnerUserId="20847" LastActivityDate="2019-06-17T20:54:54.110" Title="Moments of Fixed-Fixed Beam" Tags="|mechanical-engineering|applied-mechanics|mechanical|solid-mechanics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28803" PostTypeId="2" ParentId="28802" CreationDate="2019-06-14T18:46:23.233" Score="4" Body="&lt;p&gt;This is a statically indeterminate problem. I don't know what exactly you mean by&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;moments develop at the fixed supports&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But I guess you mean if there exist any non-zero components of moment at the fixed points. The answer is yes. A simple and lengthy calculations yields: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_{\text{Fixed points}} = \left|\frac{FL}{8}\right|$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sign of the moments at the endpoints are different, so the total moment is zero, otherwise the beam would rotate. &lt;strong&gt;But the moment itself at the endpoints is not necessarily zero.&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the moment is not zero there. The graph of the moment vs the length of the beam is a parabola. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2019-06-17T20:54:54.110" LastActivityDate="2019-06-17T20:54:54.110" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28804" PostTypeId="1" CreationDate="2019-06-14T18:54:58.127" Score="2" ViewCount="80" Body="&lt;p&gt;When implementing the control for my application I got this Nyquist plot which made me flabbergasted.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/elHdw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/elHdw.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;If I zoom on the point &lt;span class=&quot;math-container&quot;&gt;$-1 + 0j$&lt;/span&gt; I get the following image&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gydX3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gydX3.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;How do I interpret this contour going to infinity? Do I have 1 or 2 counter-clockwise turns around &lt;span class=&quot;math-container&quot;&gt;$-1 + 0j$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="20842" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-06-18T08:41:05.050" Title="Interpreting Nyquist plot" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28805" PostTypeId="1" CreationDate="2019-06-14T19:30:32.657" Score="1" ViewCount="655" Body="&lt;p&gt;If a torque and axial force is applied to a beam with a circular cross section, where along the cross section of the circular beam is the location of the max shear stress, max bending moment, max torque, and max axial stress? Correct me if I am wrong, but from class, I learned that max bending moments occur away at the neutral axis at the top and bottom of the beam (i.e. 90 and 270 degrees). Secondly, I think that the max shear stress occurs at the neutral axis (i.e. 0 and 180 degrees) since that is the apex of the parabolic shear profile but I am unconfident in that answer because I read that fracture points due to shear occur at 45 degrees. Lastly, I am not sure where along the beam the maximum torque and axial stress would occur. Thank you. &lt;/p&gt;&#xA;" OwnerUserId="20848" LastActivityDate="2020-03-11T07:02:55.467" Title="Locations of max stresses in a circular beam?" Tags="|mechanical-engineering|applied-mechanics|stresses|mechanical|strength|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28807" PostTypeId="2" ParentId="28747" CreationDate="2019-06-14T20:47:27.893" Score="0" Body="&lt;p&gt;Further to my comment about connectors or fittings with olives, consider something like this as the olive will compress around the electrode and usually seal nicely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These can be had with tapered or metric threads, you will need to tap the cylinder as needed. PTFE tape can be used on the thread into the cylinder and can also be put around the olive to aid the seal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bDWSX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bDWSX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility is to use chamfered holes and a fixing plate so that those rubber seals you show get compressed onto the electrode.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-14T20:47:27.893" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28808" PostTypeId="1" AcceptedAnswerId="28809" CreationDate="2019-06-14T22:06:19.037" Score="1" ViewCount="499" Body="&lt;p&gt;Would it be considered safe to use super glue, which contains ethyl cyanoacrylate, to attach a PVDF film to a metal alloy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says super glue is not recommended for polyethylene or polypropylene. However, I am not sure about the hazards for polyvinylidene fluoride. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="20849" LastActivityDate="2019-06-15T20:40:32.943" Title="Adhesion of Super Glue (containing ethyl cyanoacrylate) and PVDF (polyvinylidene fluoride) film" Tags="|mechanical-engineering|materials|plastic|building-design|adhesive|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28809" PostTypeId="2" ParentId="28808" CreationDate="2019-06-15T01:40:36.757" Score="1" Body="&lt;p&gt;The issue is not one of safety; it is just that PE, PP and PVDF are almost impossible to get anything to stick to, including glue. It is doubtful that crazy glue would work on it. you can try latex-based glue, this is what's used in stickytape. There are industrial-strength versions which fall under the category of pressure-sensitive adhesives. 3M makes these. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13363" LastEditDate="2019-06-15T20:40:32.943" LastActivityDate="2019-06-15T20:40:32.943" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28810" PostTypeId="2" ParentId="28800" CreationDate="2019-06-15T01:47:27.853" Score="1" Body="&lt;p&gt;the way this is usually accomplished in practice is to use a &lt;em&gt;zip tie anchor&lt;/em&gt; which consists of a plastic square with a loop in it through which a zip tie can be threaded. the bottom surface of the square has a piece of doublesticky foam tape (3M brand) on it. When you have the anchor positioned where you need it, you peel the backing paper off the doublesticky and smush it down onto the cabinet. Then you affix the wire bundle in place by running a zip tie around the bundle and through the loop on the top of the anchor, and zip it down tight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These anchors are usually spaced 6 to 12 inches apart and if stuck onto a clean surface, they will adhere so tightly that a screwdriver is needed to pry them free. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-06-15T01:47:27.853" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28811" PostTypeId="2" ParentId="28805" CreationDate="2019-06-15T06:39:31.003" Score="1" Body="&lt;p&gt;Axial load on a beam does not make any moment or shear. Just normal stress sigma= F/A, stress is evenly distributed over total area of the cross section of the rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum shear due to the torque is at the surface of the rod along it's entire length and minimum shear stress is zero at the neutral axis.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-15T06:39:31.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28812" PostTypeId="2" ParentId="15931" CreationDate="2019-06-15T07:02:06.080" Score="2" Body="&lt;p&gt;Pneumatic tires are a good option when you want to drive around the cart on sand. A pair will set you back by roughly $30. Also, they are easily available on amazon. The DIameter of 10 inches is a good choice to go with. There are numerous such tires available on amazon with different dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PVumo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PVumo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the &lt;a href=&quot;https://amusingoutdoors.com/best-beach-cart/&quot; rel=&quot;nofollow noreferrer&quot;&gt;top beach carts&lt;/a&gt; also use such tires. As for driving around on sand, unless you get larger wheels, wet sand will require more effort. Unfortunately, the larger ones also cost more. However, on dry sand, these pneumatic tires work great. Before you make your final decision, it is a good idea to take into account a few factors:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Diameter&lt;/li&gt;&#xA;&lt;li&gt;Load carrying capacity &lt;/li&gt;&#xA;&lt;li&gt;Construction material&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your budget is more than enough in a nutshell and also, pneumatic ones can provide you with proper traction on dry sand as well.&lt;/p&gt;&#xA;" OwnerUserId="20853" LastActivityDate="2019-06-15T07:02:06.080" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28814" PostTypeId="1" AcceptedAnswerId="28816" CreationDate="2019-06-15T07:43:24.957" Score="0" ViewCount="71" Body="&lt;p&gt;I am a language editor of academic papers in civil engineering (typically written in English by Poles).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I frequently encounter the term 'assumption' being used to describe a value that is adopted for the sake of calculations.  For example, an author might not precisely know the original thickness of a corroded steel shell of a silo, but makes a decision about what it might have been in order to perform calculations/modelling.  This would typically be written as something like: &quot;assumption for thickness of steel shell = 6 mm&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really feel uncomfortable with words like 'assumed' and 'assumption' on the basis that they are, almost by definition, unscientific.  What would be the correct way to describe the above example?&lt;/p&gt;&#xA;" OwnerUserId="20854" LastEditorDisplayName="user16" LastEditDate="2019-06-15T14:09:14.930" LastActivityDate="2019-06-15T14:09:14.930" Title="Describing values as being 'assumptions' or 'assumed' - what is the correct approach in academic papers for civil engineering?" Tags="|terminology|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28815" PostTypeId="2" ParentId="28814" CreationDate="2019-06-15T07:48:43.090" Score="1" Body="&lt;p&gt;Why are &quot;assumptions&quot; unscientific?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many many problems are solved by stating things like :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density is assumed constant,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure is assumed constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, there is nothing wrong with &quot;the original thickness of the steel shell is assumed to be 6mm&quot; or variation thereof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, perhaps guessing the direction of the paper(s), assuming an original thickness to predict the rate of corrosion so as to be able to estimate the lifetime or the strength of the steel is very relevant to some engineers.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-06-15T07:56:37.623" LastActivityDate="2019-06-15T07:56:37.623" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28816" PostTypeId="2" ParentId="28814" CreationDate="2019-06-15T08:57:23.387" Score="1" Body="&lt;p&gt;Engineering has been defined as &quot;the art of reaching the correct conclusions from insufficient data&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Everything an engineer does is based on assumptions and approximations. Most of them are never explicitly written down, otherwise papers would be ten times longer than they currently are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would make no difference to the conclusions of the paper if the engineer took some expensive 3D measuring equipment to the silo, and measured the profile of the entire corroded shape to an accuracy of say 0.1mm, and then made a 500,000 degrees of freedom structural model of every tiny bump. And even if that was done, the material properties used would be &quot;assumptions&quot;, unless a sample of the material from the silo itself was tested. And of course the effect of corrosion on the material properties of every bump would be different. And so on, to infinity...&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-06-15T08:57:23.387" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28817" PostTypeId="1" CreationDate="2019-06-15T09:20:01.247" Score="0" ViewCount="396" Body="&lt;p&gt;How to design gating and runner system in permanent dies so that material flows smoothly throughout the design without spilling back from the gating channel.&#xA;What should be the shape of gating system and how the volume of gating system be related to the volume of the piece that is required to be casted through gravity die casting?&lt;/p&gt;&#xA;" OwnerUserId="20856" LastActivityDate="2023-12-25T20:08:01.090" Title="Design of Gating &amp; Runner system in Gravity Die Casting" Tags="|mechanical-engineering|fluid-mechanics|design|cad|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28818" PostTypeId="2" ParentId="28817" CreationDate="2019-06-15T09:30:30.617" Score="0" Body="&lt;p&gt;Provide air vents so that it reduces or prevents the &quot;spitting back&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These air vents need to be placed so that any excess can be trimmed of the final part without affecting the final part.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-15T09:30:30.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28820" PostTypeId="2" ParentId="28814" CreationDate="2019-06-15T11:56:28.350" Score="0" Body="&lt;p&gt;There are a lot of assumptions in science. Thing is most of us dont notice these assumptions because people leave them out. Does not mean assumptions are not there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can guarantee that a lot of the physics formulas you were given at any level of school had some simplifications associated with them, you just missed that section in the text.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you introduce a new formulation of something it is prudent to tell what simplifications of reality you made. So using a sentence with a assumption is a perfectly normal way to do this. This way the reader knows wether or not the simplification is suitable for them. In fact leaving this out, except for cases where the assumptions are grandfathered from known formulas, is not very engineringly or scientifical for that matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both engineering and science is to a great extent about honesty to yourself and others. Not documenting it as such is dishonest. There is nothing unscientific about simplifying reality, in fact that is what science is.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2019-06-15T11:56:28.350" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28821" PostTypeId="1" CreationDate="2019-06-15T14:49:34.650" Score="1" ViewCount="39" Body="&lt;p&gt;Im new to rotary solenoids and I had like to understand the definition of stroke.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, some rotary solenoid has in its technical information a parameter called stroke options and it says 25-45 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HCjOj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HCjOj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to some example - &lt;a href=&quot;https://www.testco-inc.com/ledex#rotary&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it mean that the solenoid has a maximum rotation of 25 or 45 degrees but I can also use limits of +-10 for example it will have to be +- 25/45 degrees?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="12940" LastEditorUserId="12940" LastEditDate="2019-06-21T16:57:19.523" LastActivityDate="2019-06-21T16:57:19.523" Title="Solenoid stroke limits" Tags="|mechanical-engineering|actuator|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28823" PostTypeId="1" CreationDate="2019-06-15T19:53:49.233" Score="1" ViewCount="171" Body="&lt;p&gt;I have a 4 wheel skid steer robot. My motor can deliver up to 6 Nm with a gearbox efficiency of 70% up to 170 rpm. I want to find out whether I will be able to make a pivot turn and move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I assumed that pivot turn force = force to move the blocked wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force required to move a 100 kg weight is F = c * m*a = 0.02 (friction, not a rolling friction as I am taking worst case scenario) * 100 * 9.81 = 20 N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wheel will be placed 10 cm below the weight so the torque required is 0.1 * 20 = 2 Nm. Is that correct?&lt;/p&gt;&#xA;" OwnerUserId="20860" LastActivityDate="2019-06-15T19:53:49.233" Title="Torque required to move the skid steer robot" Tags="|torque|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28824" PostTypeId="1" CreationDate="2019-06-15T20:19:17.977" Score="0" ViewCount="64" Body="&lt;p&gt;my team is working on a project where fluid (mostly water) need to be pushed through small channels (like small pipes) in order to reach larger reservoir. We were considering using air to push the water but we fear that the water will simply  bubble through the water. Does anyone have any advice on how to work around this to move the water through the channels? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/90TLm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/90TLm.jpg&quot; alt=&quot;*Very* Rough design schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17950" LastEditorUserId="17950" LastEditDate="2019-06-15T20:35:51.293" LastActivityDate="2021-01-04T23:34:27.957" Title="Pushing fluid through microfluidics" Tags="|fluid-mechanics|pumps|microfluidics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28825" PostTypeId="2" ParentId="28817" CreationDate="2019-06-15T21:19:15.717" Score="0" Body="&lt;p&gt;Without pressure on the liquid metal ,it is not &amp;quot;die casting&amp;quot; ; it is permanent mold casting. The answers depend on the alloys used and the size if the parts. For smaller parts in zinc or lead , no problem ; the various turns in the runners and sprue  will prevent spit back; Provided there some mechanism to vent air and gas from the mold. For steels , copper and other alloys you will need risers to supply liquid metal to compensate for shrinkage as the parts freeze. Riser design is at least as important as gate design. The consideration of relative volumes is critical in riser design, not much in gate and runner design. For investment casting smaller parts ( such as 1 in. typical dimension) steels , the runners and sprue  provide the riser : ( Very similar to permanent mold casting).&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2021-04-09T14:16:03.470" LastActivityDate="2021-04-09T14:16:03.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28826" PostTypeId="2" ParentId="28802" CreationDate="2019-06-15T22:46:38.227" Score="-1" Body="&lt;p&gt;The moments are Pl/8 at support and at the middle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After comments by @Wasabi, and the fact that my answer is correct, I add this explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This beam if it was not fixed end, supports would rotate at the ends over the supports by an angle, &lt;span class=&quot;math-container&quot;&gt;$ \theta= Pl^2/16EI ,$&lt;/span&gt; but we know it's a fixed end, so we know the support has created a moment which has bend the beam back from &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; into horizontal angle of zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;section on fixed end support from NDS, 2005 edition on fixed end supported beams with a concentrated load on the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b0yqf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b0yqf.jpg&quot; alt=&quot;fixed end beams&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-06-17T07:00:38.450" LastActivityDate="2019-06-17T07:00:38.450" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28827" PostTypeId="1" CreationDate="2019-06-16T03:10:48.467" Score="1" ViewCount="39" Body="&lt;p&gt;I am having problems locating references of the typical thickness and material of the coating along the hypotenuse side of 50/50 polarizing beam splitter cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what these coatings usually are? Have references or published paper references? etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="18950" LastActivityDate="2019-06-16T03:10:48.467" Title="What is a common polarizing 50/50 beam splitter cube coating" Tags="|materials|optics|thin-films|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28828" PostTypeId="1" AcceptedAnswerId="28835" CreationDate="2019-06-16T12:42:42.713" Score="1" ViewCount="3793" Body="&lt;p&gt;I am trying to understand this problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AaHM5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AaHM5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we break down the 10T, they behave like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/R4TRl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/R4TRl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The right end of the beam can move as much as it wants, but the left end is pinned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is there no horizontal reaction at A, if there is a horizontal part of the forces being applied on the beam axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine, the force will push one section, and each section will push the next one, until the end, where it will create a reaction, just like the vertical forces are doing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you a lot.&lt;/p&gt;&#xA;" OwnerUserId="20865" LastActivityDate="2019-06-17T01:09:26.053" Title="Pin support, why is there no horizontal reaction force?" Tags="|structures|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28829" PostTypeId="2" ParentId="28828" CreationDate="2019-06-16T14:51:41.747" Score="2" Body="&lt;p&gt;Think about the three members that form the triangle EBC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no horizontal load applied at E.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know there is no horizontal load applied at C by beam CD, because there is a roller bearing at D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore by equilibrium, there is no horizontal load applied at B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore there is no horizontal tension in beam AB and no horizontal reaction at A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course there will be horizontal tension in BC, but that will be balanced at each end by the horizontal components of tension in EB and EC.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-06-16T14:51:41.747" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28832" PostTypeId="2" ParentId="28824" CreationDate="2019-06-16T16:53:23.333" Score="1" Body="&lt;p&gt;yes. knowing the surface tension of the fluid and its contact angle with the channel walls, you can select a channel width that does not support &lt;em&gt;two-phase flow&lt;/em&gt;, where bubbles and fluid can pass by one another and get mixed up in the channel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the tube diameter on a turkey baster's tip is made small enough so that air cannot flow up in the tube while juice flows down, which would cause the baster to spill out its contents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tradeoff between the effects of capillarity and inertial or gravitational effects is what sets the break point between the baster holding its contents and spilling them. The nondimensional similitude parameter that establishes this is called either the &lt;em&gt;Froude number&lt;/em&gt; or the &lt;em&gt;Weber number&lt;/em&gt;, I cannot remember which.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-06-16T16:53:23.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28833" PostTypeId="1" CreationDate="2019-06-16T17:14:40.260" Score="-1" ViewCount="1333" Body="&lt;p&gt;I am looking for a cheap (preferably under 2000 USD) electrical motor that can spin freely in either direction when the motor is unpowered. The motor will be used as an alternative source of power to a hand-cranked ferry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three cranks: one on the ferry + one at each docking station. All three cranks are connected - when one turns, all turn the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, we're just looking for a quick proof-of-concept hack that interferes minimally with the current operation of the ferry. What we have in mind is a small electrical motor mounted near each of the two cranks on shore, directly connected to the crank axles using belts or bike chains. To keep it as simple and cheap as possible for this quick proof-of-concept, we were hoping to find a motor that spins with minimal resistance in either direction when unpowered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our first idea was to use a direct drive brushless motor designed for bikes, since they presumably do not need a clutch to spin freely, but from what I have read, such motors don't like high load at low RPMs, so they don't seem ideal for our use-case. Our cranks will spin at about 30 RPM. The load varies depending on the weather, but rarely more than an adult person can crank with one arm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of an off-the-shelf product that meets our requirements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have googled for a geared motor with an internal magnetic clutch that automatically releases when the motor is unpowered, but I've found no good candidates so far.&lt;/p&gt;&#xA;" OwnerUserId="20868" LastEditorUserId="8055" LastEditDate="2019-06-18T16:20:40.867" LastActivityDate="2019-06-20T04:27:52.830" Title="Cheap electrical motor that can spin freely in both directions when unpowered" Tags="|electrical-engineering|motors|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28834" PostTypeId="2" ParentId="28833" CreationDate="2019-06-16T17:28:37.430" Score="2" Body="&lt;p&gt;Any motor that does not have permanent magnets should work fine, given the low rpm.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-16T17:28:37.430" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28835" PostTypeId="2" ParentId="28828" CreationDate="2019-06-16T18:36:48.857" Score="5" Body="&lt;p&gt;Assuming I'm understanding the problem statement correctly and we have a vertical load transferred to a beam via a sort of angled frame, then here's how I'd think about the beam reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Quick Way...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The quickest approach is to solve for the support reactions by remembering the structure must be globally stable. In this case, we can consider the beam and the frame as a single structure for purposes of calculating the support reactions. Our equations of statics say the sum of the forces in the horizontal direction, the sum of the force in the vertical direction, and sum of the moments, must each be zero. Because there is no externally-applied horizontal load, there is only one possible horizontal force (the support reaction). For the sum of horizontal forces to be zero, the horizontal support reaction must therefore also be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Longer Way...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longer way round is perhaps to convince ourselves that the frame applies equal and opposite horizontal loads to the beam and that this, again, means the horizontal support reaction must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, imagine we have a simply supported beam with equal and opposite horizontal load applied at a point.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/avl9A.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/avl9A.jpg&quot; alt=&quot;beam with coincident equal-opposite loads&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no net load at the point, so it's reasonably easy to convince ourselves that there's no support reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What if we applied equal and opposite loads at separate points along the beam?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sytAP.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sytAP.jpg&quot; alt=&quot;Beam with separate equal-opposite loads&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's perhaps less clear that there's no support reaction. But we can convince ourselves easily using the equations of statics. The beam reactions are about global stability of the structure. As is always the case in statics, the sum of horizontal forces must be zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Sigma F_x = (+P) + (-P) + H_A = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by some quick math, we prove that the horizontal reaction at Point A must be zero. In fact, for this beam configuration, the only segment of the beam &quot;feeling&quot; axial load is the segment in between the applied loads. It can be tempted to picture the left load &quot;pushing&quot; on the beam segment to the left of it, but that's not what's happening. Imagine a game of tug-of-war where the two innermost opponents perfectly oppose each other's forces...the players farther down the rope wouldn't feel anything, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now, all we have to do is convince ourselves the horizontal components of the applied loads are equal and opposite.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/AgBtD.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AgBtD.jpg&quot; alt=&quot;components of applied load&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we zoomed in and just looked at the bit of structure right were the vertical load is applied...well, that bit of structure must also be in equilibrium. The applied load is going to distribute between the two angled members, which means there will be horizontal components. However, again, by our lovely equations of statics, for this bit of structure to be in equilibrium, the sum of horizontal forces must be zero. Here, the only possible horizontal forces are the horizontal components of the loads in the angled members -- and thus, these components must be equal and opposite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ta, da! Sometimes our initial intuition can lead us a little astray but the equations of statics never lie. Forces and moments must always sum to zero in order for the structure to remain...static.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2019-06-16T18:36:48.857" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28836" PostTypeId="2" ParentId="28797" CreationDate="2019-06-16T19:24:45.557" Score="1" Body="&lt;p&gt;I teach the American Engineering system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.learnthermo.com/T1-tutorial/ch01/lesson-B/pg04.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.learnthermo.com/T1-tutorial/ch01/lesson-B/pg04.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I see references also to it being called the English Engineering system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/English_Engineering_units&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/English_Engineering_units&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At one point, I remember being taught the inch-pound and ft-slug systems. I imagine the former could still be common in systems where pressure is expressed in pounds per square inch (psia or psig). An analogy is the use of the cgs (centimeter gram second) system by chemical physicists to avoid large exponential factors that would arise otherwise with the mks (meter kilogram second) system&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-06-16T19:24:45.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28837" PostTypeId="1" CreationDate="2019-06-16T20:19:13.540" Score="0" ViewCount="166" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/0pRzl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to calculate the hydraulic efficiency for this turbine by using the formula:&#xA;E(hyd) = (Vw&lt;em&gt;U)/g&lt;/em&gt;H&lt;br&gt;&#xA;where  Vw= Velocity of whirl at inlet, U=velocity of runner at inlet, H=Given Head, and g=acceleration due to gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But every time I get the efficiency more than 1 (which of course isn't possible). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried this multiple but everytime it is more than 1. So I'd really appreciate if someone would properly guide me on how to make velocity triangles for this specific problem. Thanks.&lt;img src=&quot;https://i.stack.imgur.com/G9vPv.jpg&quot; alt=&quot; I have added my own calulations here so please let me know where I&amp;#39;m wrong.&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="20869" LastEditorUserId="20869" LastEditDate="2019-06-17T04:17:26.653" LastActivityDate="2019-06-17T11:46:31.337" Title="Can someone please give me a hint on how to draw velocity triangles for this specific inward flow reaction turbine problem?" Tags="|fluid-mechanics|turbines|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28838" PostTypeId="2" ParentId="28828" CreationDate="2019-06-17T01:09:26.053" Score="0" Body="&lt;p&gt;You are more than overthinking this.  Keep it simple. For an object at rest:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ΣF(x) = 0 [sum of vertical forces = zero]&lt;/li&gt;&#xA;&lt;li&gt;ΣF(y) = 0 [sum of horizontal forces = zero]&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;ΣMoments() = 0 [sum of moments about any point = zero]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Free body diagram of beam.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2CmYj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2CmYj.jpg&quot; alt=&quot;Free Body Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, ΣF(x) = 0.  With the diagram drawn this way, what do you thing Force A (in the x direction) is?&lt;/p&gt;&#xA;" OwnerUserId="1567" LastActivityDate="2019-06-17T01:09:26.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28839" PostTypeId="2" ParentId="28802" CreationDate="2019-06-17T02:22:02.450" Score="4" Body="&lt;p&gt;When trying to figure out whether or not a given reaction will exist at a given support, it's worth remembering what a reaction actually is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reaction is the means by which the support resists the movement of the beam at that point. Force reactions resist the beam's attempts to deflect up-down or left-right at the support. Moment reactions resist the beam's attempts to rotate at the support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, let's assume there's no bending reaction at the beam's extremities. In that case, we'll be simply dealing with a simply-supported beam with pinned supports at the ends. These pinned supports will generate vertical forces which resist the beam's deflection at those points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how would the beam deflect in this case? Well, it'd obviously be shaped like a parabola (well, it'd actually be a cubic function, but it looks parabola-ish).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/P3Dwv.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P3Dwv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But that means there was rotation at the extremities: the beam was previously horizontal at the supports, but now it's tilted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the definition of a fixed support is that it doesn't allow for rotations at that point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what does the fixed support do to stop that rotation? Well, it applies a concentrated bending moment to the beam, rotating it back to horizontal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HgWlb.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HgWlb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Figures created with &lt;a href=&quot;https://www.ftool.com.br&quot; rel=&quot;noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free structural analysis program.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-06-17T02:22:02.450" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28841" PostTypeId="1" CreationDate="2019-06-17T18:16:45.973" Score="2" ViewCount="165" Body="&lt;p&gt;Can somebody explain relation between combustion efficiency and pressure in lean mixtures using any kind of chart or mathematical equations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance I thought, combustion efficiency increases with pressure initially and then gets constant. But not sure by how much amount it varies based upon pressure for different fuels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I tried to derive a formula to calculate or estimate Combustion Efficiency using Collision theory. As per my understanding, more collisions need to happen between reactants for better combustion efficiency. But collision theory alone is not sufficient to calculate or estimate Combustion Efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also does Combustion efficiency varies between simple fuels like Hydrogen and methane under similar conditions? If so, why and at what margin?&lt;/p&gt;&#xA;" OwnerUserId="16175" LastEditorUserId="16175" LastEditDate="2019-06-20T19:57:44.703" LastActivityDate="2019-06-20T19:57:44.703" Title="Effect of Pressure on Combustion efficiency (of hydrogen) in excess air" Tags="|thermodynamics|automotive-engineering|energy-efficiency|combustion|engines|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28842" PostTypeId="2" ParentId="28795" CreationDate="2019-06-17T22:17:12.990" Score="1" Body="&lt;p&gt;I remembered the concept but not the name so I searched a few terms and found that Wikipedia calls it a &lt;a href=&quot;https://en.wikipedia.org/wiki/Quick_return_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;quick return mechanism&lt;/a&gt;. Many such mechanisms are &lt;a href=&quot;https://en.wikipedia.org/wiki/Linkage_(mechanical)&quot; rel=&quot;nofollow noreferrer&quot;&gt;linkages&lt;/a&gt; (requiring no springs and catches, unlike your idea) and therefore may be called more specifically quick return linkages.&lt;/p&gt;&#xA;" OwnerUserId="9262" LastActivityDate="2019-06-17T22:17:12.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28843" PostTypeId="1" CreationDate="2019-06-18T05:52:20.740" Score="2" ViewCount="73" Body="&lt;p&gt;In modern times bubble levels and water levels are used to check if a surface if strictly horizontal/vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did people ensure vertical/horizontal surfaces in older times?&lt;/p&gt;&#xA;" OwnerUserId="20875" LastActivityDate="2019-06-18T15:55:11.067" Title="How were horizontal and vertical structures ensured in old times?" Tags="|civil-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28844" PostTypeId="2" ParentId="28843" CreationDate="2019-06-18T07:03:50.343" Score="4" Body="&lt;p&gt;Well, the Romans did lots of surveying and built some excellent roads, using the &lt;a href=&quot;https://legioneromana.altervista.org/content/how-use-roman-groma?language=en&quot; rel=&quot;noreferrer&quot;&gt;Groma&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plumb line is still handy...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, water find its own level so a crude level is also available and before lasers came along, many large sites were set up with flexible pipes with transparent sections to provide a level across a longer distance, 20 or 30m or more.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-18T07:03:50.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28845" PostTypeId="2" ParentId="28841" CreationDate="2019-06-18T08:34:36.460" Score="2" Body="&lt;p&gt;In an engine, higher pressures lead to a higher brake power. Why? Check the Otto and Diesel diagrams:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mechanicalbooster.com/2017/10/diesel-cycle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mechanicalbooster.com/2017/10/diesel-cycle.html&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.mechanicalbooster.com/2017/10/otto-cycle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mechanicalbooster.com/2017/10/otto-cycle.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase the combustion pressure, then you will maximize the area contained among the lines and the work output will increase. As you can see in the pictures, the expansion is considered to be isentropic, so these are the formulae you can use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T_3/T_4=(p_3/p_4)^{(\gamma-1)/\gamma}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; is the adiabatic constant for the combustion gases, roughly 1.4. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you increase the pressure ratio you increase the temperature ratio too and the output work of this process can be calculated this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sum_{p_3}pv^{\gamma}dv$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where v is the specific volume or the volume if you prefer. This is an integral between &lt;span class=&quot;math-container&quot;&gt;$p_3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$p_4$&lt;/span&gt; (sorry for the wrong format). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a politropic isentropic process, that's why the politropy index equals &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt;. And remember that &lt;span class=&quot;math-container&quot;&gt;$pv^{\gamma}=constant$&lt;/span&gt;, that's the curve you must integrate. So &lt;span class=&quot;math-container&quot;&gt;$p_3 v_3 ^{\gamma}=p_4v_4^{\gamma}$&lt;/span&gt;. You can use &lt;span class=&quot;math-container&quot;&gt;$v=R*T/p$&lt;/span&gt; in both occasions, rembering that the gas can be modeled as ideal and R has an approximate value of 287 J/kgK. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, as that pressure ratio increases, the integral will get bigger. That's why efficiency will increase, because the output work is increasing higher than the input (compression) energy (heat+work) and the losses. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\eta=W_o/E_i$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\eta=1-Losses (heat+work)/E_i$&lt;/span&gt;. &lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-06-18T08:34:36.460" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="28846" PostTypeId="2" ParentId="28804" CreationDate="2019-06-18T08:41:05.050" Score="1" Body="&lt;p&gt;It seems you have 1 counter-clockwise turn around. One trajectory comes from the left, then turns around and then moves to the right as you change the parameters. I think this behavior is normal as you have 1 integrator (PID controller) and you have one pole in s=0. As you change the parameters you annulate some function chunks and you get at s=0. Then you continue changing your parameters and get backwards to the right, though this time the trajectory is different due to the different values your function is taking.&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-06-18T08:41:05.050" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28847" PostTypeId="1" AcceptedAnswerId="28854" CreationDate="2019-06-18T08:51:55.563" Score="1" ViewCount="1759" Body="&lt;p&gt;When rubber (I don't know the specifics, but I assume it's butyl rubber) gets old, it turns into one of two states usually:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;It becomes sticky and gummy.&lt;/li&gt;&#xA;&lt;li&gt;It becomes hard and brittle.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'd like to know if there's a way to change hardened or sticky rubber back into its original condition. The only thing I tried so far, is heating it up gently with a hot air gun to ~60°C which makes hard rubber more malleable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand there are different kinds of synthetic rubbers, made of different polymers, and henceforth, there might be a range of chemicals that might be used to treat those various kinds. I couldn't find any information on such &quot;pairings&quot;, for instance I was searching specifically to treat butyl rubber, but I couldn't find anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air ducting in cars is made of some sort of hard rubber (I'm not sure what kind of polymer), and I've heard that Ballistol can be used to soften it, but it's hearsay, essentially, I'm not sure about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not looking for a way to completely &quot;rejuvenate&quot; rubber parts like that, just to have them restored a bit, so they are easier to work with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same thing with sticky rubber. Is there a way to make it less sticky?&lt;/p&gt;&#xA;" OwnerUserId="19884" LastActivityDate="2019-06-18T17:28:17.477" Title="What chemical can be used to restore synthetic rubber?" Tags="|chemistry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28849" PostTypeId="1" AcceptedAnswerId="28852" CreationDate="2019-06-18T12:49:14.670" Score="0" ViewCount="58" Body="&lt;p&gt;Following situation: I need to mount a console holding a load to the edge of a 25cm concrete plate. I want you to check my reasoning. This is the situation:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oZ364.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oZ364.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking is: The load is 170kg, which amounts to 1.67 kN. The length of the cantilever is 600 mm. So the tension load on the upper two fasterners (of four, installed in square patterns) is a total of 1.67 kN * 600 mm / 150 mm = 6.69 kN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking the &lt;a href=&quot;https://www.hilti.de/medias/sys_master/documents/h89/h84/9342591827998/Technical-data-sheet-for-Hilti-HIT-1-and-Hilti-HIT-1-CE-Technical-information-ASSET-DOC-7567931.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;specifications of a likely system&lt;/a&gt;, I see that 2 M10 fasteners at embedment depth 100mm offer me 2 * 8.7 kN tensile load, which contains decent margin of safety. The shear load per fastener is in the same range, there's 4 fasteners, so even considering 10 or 20 kg for the console (likely too much) this is ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm out of school too long. Is the above  correct?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-06-18T15:25:37.290" Title="Correct sizing of fastener for console" Tags="|structural-engineering|fasteners|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28850" PostTypeId="1" CreationDate="2019-06-18T14:52:17.960" Score="0" ViewCount="39" Body="&lt;p&gt;I have a basement which floods, the water comes through a drain in the floor. I am therefore looking to use an &lt;a href=&quot;https://www.huntingdonfusion.com/index.php/en/products-list/101-mechanical-pipe-plugs-and-stoppers/steel-plugs/683-ordering-steel&quot; rel=&quot;nofollow noreferrer&quot;&gt;HFT Pipestoppers&lt;/a&gt; to stop the water coming in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the water comes from an old tank (B) (which I can't find).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what pressure I would expect in the pipe to the basement (x)&#xA;and advise if this is a good option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YgCby.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YgCby.jpg&quot; alt=&quot;]&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20880" LastEditorUserId="1832" LastEditDate="2019-06-18T23:01:36.267" LastActivityDate="2019-06-20T14:10:26.903" Title="Water in basement plug install" Tags="|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28851" PostTypeId="2" ParentId="28850" CreationDate="2019-06-18T14:56:10.767" Score="2" Body="&lt;p&gt;The pressure will be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P = \text{density} \times \text{gravity} \times \text{height}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Density is 1000&amp;nbsp;kg/m&lt;sup&gt;3&lt;/sup&gt;, and gravity is 9.81&amp;nbsp;m/s&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But your problem is defining the height above the measuring point as you say you don’t know where the tank is.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="1832" LastEditDate="2019-06-18T23:03:32.537" LastActivityDate="2019-06-18T23:03:32.537" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28852" PostTypeId="2" ParentId="28849" CreationDate="2019-06-18T15:25:37.290" Score="1" Body="&lt;p&gt;You'd have to check for these factors.:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The fastener must penetrate into and fully develop its strength inside the steel cage tie bar or L shaped rebar inside the concrete which is a continuous part of its reinforcement. Concrete alone is susceptible to cracking under tension and is not reliable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Dynamic load factor. A load of P, even if lowered gently onto a support has 2 to 3 times its weight impact factor. Because it needs to deforms the member to its deflected shape and beyond and back to equilibrium. So the load factor of 3 is advisable.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-18T15:25:37.290" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28853" PostTypeId="2" ParentId="28843" CreationDate="2019-06-18T15:55:11.067" Score="3" Body="&lt;p&gt;Egyptians used shallow water trenches as level complemented with &quot;&lt;a href=&quot;http://www.surveyhistory.org/egyptian_surveying_tools1.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Merchet and Groma&lt;/a&gt;&quot; to build the great pyramids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In old Persia they had plumbed levels and used ropes and triangulation.&#xA;They used those tools to great accuracy to build those Qanats, (subterranean irrigation canals)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the old Roman empire they recognized surveying as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Surveying#Ancient_history&quot; rel=&quot;nofollow noreferrer&quot;&gt;profession&lt;/a&gt; and had skilled masons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-18T15:55:11.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28854" PostTypeId="2" ParentId="28847" CreationDate="2019-06-18T17:28:17.477" Score="2" Body="&lt;p&gt;The embrittlement and goop-forming reactions are fundamentally different, and different measures would be needed to deal with each. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rubber embrittlement occurs when the rubber oxidizes, and stops being rubber anymore because the rubber molecule chains get cut and the chemical bonds between adjacent molecules get broken. This degradation process is made worse by exposure to heat, ozone, and ultraviolet. Once it happens, you cannot repair the rubber part simply because it's no longer rubber. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goop reaction occurs because to make rubber soft and flexible, it has mineral oil or other oils called &lt;em&gt;plasticizers&lt;/em&gt; milled into it during its manufacture. The oil molecules form tiny globs of oil within the rubber matrix and interfere with the bonding between adjacent rubber molecules, letting them slip past one another instead of getting caught and pulling back against each other. With the passage of time, the oil sometimes tends to diffuse out of the bulk and accumulate at the surface, which renders the surface sticky. Sometimes the oil globs slowly dissolve into the bulk rubber and disconnect most of the molecules from one another, and the once-solid rubber object turns into something like tar and dribbles apart. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a substance called &lt;em&gt;ARMOR ALL&lt;/em&gt; which can be used to rejuvenate decaying rubber. It contains a water emulsion of silicone oil that soaks into the rubber surface, but the improvement is just cosmetic since it can't re-connect broken chemical bonds.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-06-18T17:28:17.477" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28855" PostTypeId="2" ParentId="28833" CreationDate="2019-06-18T19:36:19.097" Score="1" Body="&lt;p&gt;I sounds like what you might want is an electrostatic motor. if there is current running through the motor then there will be a magnetic field which will help the motor lock in place. if there is no current, then the motor should spin freely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: this is just for the motor, if there is a gearbox on the output shaft of the motor, it will still take some force to spin the output of the gearbox. If you need a gear box, but you still want the shaft to free spin when it's not in use, then you might need to have a clutch type system which engages and disengages the output of your gearbox. &lt;/p&gt;&#xA;" OwnerUserId="20468" LastActivityDate="2019-06-18T19:36:19.097" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28856" PostTypeId="1" AcceptedAnswerId="28863" CreationDate="2019-06-18T22:07:38.527" Score="5" ViewCount="1602" Body="&lt;p&gt;Why do some guyed broadcast towers come to a point at the base. It seems that it would be more stable if each vertical bar was anchored separately rather than being balanced at one point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xBs0Y.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xBs0Y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20882" LastEditorUserId="20882" LastEditDate="2019-06-18T22:14:43.387" LastActivityDate="2019-06-19T10:05:44.507" Title="Why do some guyed towers come to a point at the bottom?" Tags="|structural-engineering|structures|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28858" PostTypeId="2" ParentId="28856" CreationDate="2019-06-18T23:35:23.473" Score="2" Body="&lt;p&gt;The tower can be in a location of extreme winds or temperature change. Or it may be located at a point where there is interruption in direction of cables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wherever the reason, there is anticipation of excessive stress on the guys and demand for flexibility at the bottom to allow the tower freely deflect with the expansion and shortening of the guy cables and not experience undue stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In such cases if the tower were to be self standing on its own legs, it would be necessary for the guy cables to counter undesirable resistance of a self supporting tower to tilt it into optimal angle. Here is Wikipedia link&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Additionally_guyed_tower&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;guyed tower advantage s&quot;&gt; guyed tower&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-18T23:35:23.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28859" PostTypeId="1" CreationDate="2019-06-19T00:02:19.473" Score="2" ViewCount="85" Body="&lt;p&gt;I am currently prototyping a part for a personal project. My CAD design was contracted out and there are some off the shelf parts that I need to order per the design (springs, screws, etc.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My designer included two types of springs in the BOM:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3ID x 15mm (possibly 3LD??)&lt;/li&gt;&#xA;&lt;li&gt;5ID x 5mm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I am not sure what the ID (or LD) naming convention means. Could use some help.&lt;/p&gt;&#xA;" OwnerUserId="20883" LastEditorUserId="1832" LastEditDate="2019-06-19T01:45:48.873" LastActivityDate="2019-06-19T01:45:48.873" Title="Spring Industry Naming Convention" Tags="|mechanical-engineering|design|springs|prototyping|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28860" PostTypeId="2" ParentId="28859" CreationDate="2019-06-19T00:13:24.003" Score="1" Body="&lt;p&gt;Usually spring's geometry is denoted as : OD* ID* Length. Outside diameter* Inside diameter* Length.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-19T00:13:24.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28862" PostTypeId="1" CreationDate="2019-06-19T09:01:21.280" Score="0" ViewCount="358" Body="&lt;p&gt;How do I select the cross section for a machine frame? Do I need to refer to the AISC steel construction manual to find the capacity of beams, columns, etc?&lt;/p&gt;&#xA;" OwnerUserId="20822" LastEditorUserId="20822" LastEditDate="2019-06-20T06:25:43.183" LastActivityDate="2019-06-20T06:25:43.183" Title="Machine Frame Design" Tags="|mechanical-engineering|structural-engineering|design|structural-analysis|machine-design|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28863" PostTypeId="2" ParentId="28856" CreationDate="2019-06-19T10:05:44.507" Score="6" Body="&lt;p&gt;If the base of the tower is &quot;rigidly&quot; fixed to the ground, the foundations have to withstand the bending moments from the side loads (e.g. wind) on the tower. Otherwise, the foundations will work loose and not support the tower at all. Depending on the local geology, that may mean deep foundations to support a tall tower. Foundations are expensive, because unlike the tower itself they can't be prefabricated to a standard design off site and delivered by truck!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By using guy ropes and a tower base as shown, the only force on the foundation is downwards, and if the foundation gradually sinks into soft ground over time, that can be compensated for cheaply by adjusting the guy ropes.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-06-19T10:05:44.507" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28864" PostTypeId="2" ParentId="28372" CreationDate="2019-06-19T10:13:36.600" Score="1" Body="&lt;p&gt;I see a few possible issues:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;do the amounts of air add up? you need 20 CFM @ 90 psi, that's about 120 CFM @ 15 psi (close to normal conditions) but the dryer is rated for 45 CFM&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In all applications I've seen (that I talked through with sales engineers for the air supply) the dryer was installed downstream of the compressor&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The globe valve seems suspect, wherre should the excess airflow you fear com from? Size the compressor so that airflow does not exceed the rated wirflow of the dryer&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't understand the check valve paralell to the compressor. When will outside pressure ever be significantly higher than pressure in tank?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;One compressor? Did you test the actual runtime it will experience, to supply the air? Is the compressor designent for that? This depends on actual air demand and size of tank.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You did not tell us how long the ai demand is every two seconds. Just make sure you size your tank accordingly&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;95 psi tank pressure for a process that needs 90 seems low. Check your components from back to tank for actual pressure loss&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A lot will depend on the qulitiy of the pressure regulator. Get a chart with the pressure loss at the flow you expect and talk your application through with an application engineer. Often, pressure regulators have a minimum delta p of about 15 psi.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Alternativly, you can get a sales engineer (many compressor manufacturers will deliver complete stations) to design the system for you, and learn as much as possilbe from them.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-06-19T10:13:36.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28867" PostTypeId="1" AcceptedAnswerId="28905" CreationDate="2019-06-19T15:21:18.067" Score="2" ViewCount="414" Body="&lt;p&gt;As per Bryton Cycles, most of TurboJet engines' Thermodynamic efficiency should be around 40%. Rest of energy gets rejected as waste heat in exit gases of after the turbine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if these hot gases from the turbine are passed through Nozzle of 95% efficiency (either Convergent or CD), then overall efficiency should be 95% Right? Because Nozzle is scavenging all the waste heat and pressure, converting it into Kinetic energy of gases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per my knowledge &lt;strong&gt;nozzles are extremely efficient devices with efficiencies above 95%&lt;/strong&gt; (Please do comment if that's not the case). And for the sake of discussion please assume Combustion efficiency as 100% and no mechanical loss. And here I'm considering whole purpose of turbojet engine it propell the gases as fast as possible. So overall efficiency here is kinetic energy of exit gases vs chemical energy released by fuel.&lt;/p&gt;&#xA;" OwnerUserId="16175" LastEditorUserId="16175" LastEditDate="2019-06-22T02:30:28.683" LastActivityDate="2019-06-22T04:51:48.357" Title="Efficiency of TurboJet engines with nozzles" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|jet|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28868" PostTypeId="1" CreationDate="2019-06-19T16:01:32.717" Score="-1" ViewCount="64" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GFY7B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GFY7B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in question A2 part (ii) why P2 is considered to be zero&lt;/p&gt;&#xA;" OwnerUserId="20890" LastActivityDate="2019-06-19T16:24:48.380" Title="fluid mechanics" Tags="|fluid|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28869" PostTypeId="2" ParentId="28868" CreationDate="2019-06-19T16:09:40.593" Score="3" Body="&lt;p&gt;P2 is given as atmospheric and P1 is given as a gauge pressure of 80kN/m^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gauge is assumed to measure the pressure against atmospheric ie it could be a Bourdon tube type gauge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means if atmospheric is taken as 1 then the inlet pressure would be 80 + 1 kN/M^2 absolute. Then, one would subtract 1 from the 81 to get the pressure difference...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-06-19T16:24:48.380" LastActivityDate="2019-06-19T16:24:48.380" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28872" PostTypeId="1" CreationDate="2019-06-19T23:46:20.283" Score="0" ViewCount="42" Body="&lt;p&gt;Is a regular roller train drive used in most car engine's timing chain drive? If so, wouldn't the polygonal effect affect the accuracy of strict angular synchronization between the camshafts and the crankshaft?&lt;/p&gt;&#xA;" OwnerUserId="20893" LastActivityDate="2019-06-20T04:27:00.733" Title="How does polygonal effect affect the effect of angular synchronization in car engines' timing train drive?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28873" PostTypeId="1" CreationDate="2019-06-20T00:58:27.963" Score="2" ViewCount="68" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QppfR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QppfR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a part like the one shown, let's say I wanted to find the dimension labeled 3&quot;, if it wasn't already labeled. Since there are tolerances already on the part, how would I determine the 3&quot; value along with its tolerances? Can I just add and subject the tolerances? For example, 4.00 +- 0.3 - 1.00 +-0.1 = 3.00 +- 0.2 ? What if I wanted to find the distance from the right end of the hole to the right face of the part, assuming diameter D = 1&quot; +- 0.01? Could I simply do 3.00 +- 0.2 - (1/2) +-(0.01/2) = 3.00 +- 0.2 - 0.5 +- 0.005 = 2.5 +- 0.195? Thanks for the help!&lt;/p&gt;&#xA;" OwnerUserId="20894" LastActivityDate="2019-06-20T00:58:27.963" Title="How to find a dimension that isn't explicit if there are tolerances" Tags="|mechanical-engineering|tolerance|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28874" PostTypeId="2" ParentId="17994" CreationDate="2019-06-20T03:15:43.460" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;https://www.pumpsandsystems.com/topics/understanding-npsh-npsh-definitions&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pump &amp;amp; Systems&lt;/a&gt;,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Note that suction pressure is stagnation pressure (total pressure); it includes velocity head. Adding velocity head puts all pumps on the same basis, otherwise a pump would require different amounts of NPSH when tested with different sizes of suction lines (assuming that suction pressure is measured in the suction line, which is normally the case).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20895" LastActivityDate="2019-06-20T03:15:43.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28875" PostTypeId="2" ParentId="28872" CreationDate="2019-06-20T04:27:00.733" Score="1" Body="&lt;p&gt;One checks if the crank and cams are in time after fitting a new chain by completing two full revolutions of the cranshaft, they should, and normally do if all is correct, come back to the correct positions.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-20T04:27:00.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28876" PostTypeId="2" ParentId="28833" CreationDate="2019-06-20T04:27:52.830" Score="0" Body="&lt;p&gt;There are plenty of motors that spin freely when unpowered. BUT, given the low RPMs you're going to need a gear reduction, and &lt;em&gt;that&lt;/em&gt; is not going to spin freely when unpowered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend using a &lt;a href=&quot;https://en.wikipedia.org/wiki/Freewheel&quot; rel=&quot;nofollow noreferrer&quot;&gt;freewheel&lt;/a&gt; gear or sprocket, that way the gearbox only turns when the motor is running. One designed for a &lt;a href=&quot;https://www.amazon.com/freewheel-sprocket/s?k=freewheel%20sprocket&quot; rel=&quot;nofollow noreferrer&quot;&gt;bicycle&lt;/a&gt; might work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't make this work with only one direction of rotation, look into centripetal clutches (or electronic ones).&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-06-20T04:27:52.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28877" PostTypeId="2" ParentId="26524" CreationDate="2019-06-20T09:00:28.600" Score="2" Body="&lt;p&gt;Few points:&#xA;First off, see if you can figure out a way to NOT use MTBE in the first place since it is not an environmentally friendly chemical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, in dealing with separating azeotropic mixtures (assume 2 components), as chemical engineers we have a few tools available:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Adsorption (such as with molecular sieves)&lt;/li&gt;&#xA;&lt;li&gt;Reaction (i.e. reactive distillation)&lt;/li&gt;&#xA;&lt;li&gt;Using entrainers to adjust the relative volatility of one of the components &lt;/li&gt;&#xA;&lt;li&gt;Adjusting the pressure to shift the vapor liquid equilibrium (VLE). (i.e. distillation under pressure or vacuum)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My hunch is that in practice this particular separation is done using (4). In fact a quick lookup I found a reference stating that distillation occurs in two columns. The first having P = 6 bar, and the second P=12 bar claiming 99% purity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try re-running your VLE calculations with these pressures. If we've already solved it without entrainers, then no need to reinvent the wheel. From a cost perspective, you can save on the cost of the entrainer (and future separation) at the cost of operating under pressure.&lt;/p&gt;&#xA;" OwnerUserId="20896" LastActivityDate="2019-06-20T09:00:28.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28878" PostTypeId="2" ParentId="28697" CreationDate="2019-06-20T09:42:24.043" Score="1" Body="&lt;p&gt;I haven't done furfural desorption from biochar,   but I have done furfural desorption from metals. Here are some thoughts off the top of my head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your question of furfural adsorption. I would be concerned about the adsorption selectivity of furfural (a fairly &quot;mid sized&quot; molecule) from bio-oil. My expectation is that it's not that good since bio-oil has so many different components. Then, you may have to deal with recovery losses from the biochar adsorbent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The biochar also presents two &lt;em&gt;other&lt;/em&gt; issues, namely:&#xA;1. Properties (pore size etc) may not be very consistent between biochar from different feedstocks (switchgrass etc)&#xA;2. Biochar particle attrition if it occurs during the adsorption may hinder (i.e. clog) further downstream bio-oil refining processes if not dealt with.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the &lt;em&gt;mechanism&lt;/em&gt; oh boy. I don't think anyone really knows. Here's a guess. Diffusion into pores, then furfural is likely physisorbed (VdW interactions) on carbon substrate (the biochar). I'd really really really (did I say really!) doubt chemisorption, but I could be wrong. Then, desorption should also be easy followed by diffusion back out the pore (assuming pore size and shape were large enough and right shape). Pretty standard. Check out some kinetics textbook like Fogler or read up on how molecular sieves work if you aren't familiar.&lt;/p&gt;&#xA;" OwnerUserId="20896" LastEditorUserId="20896" LastEditDate="2019-06-20T09:50:44.193" LastActivityDate="2019-06-20T09:50:44.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28879" PostTypeId="1" CreationDate="2019-06-20T10:02:15.780" Score="4" ViewCount="173" Body="&lt;p&gt;I hope I can get some help with this problem as I've been struggling with it for a long time. I have a helical tension spring with radius &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; and a square wire cross section and made of orthotropic material. I'm trying to derive an equation for the spring constant &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; by equating the energy stored by the spring &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}kx^2$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; is the distance stretched, with the energy stored &lt;span class=&quot;math-container&quot;&gt;$U_t$&lt;/span&gt; by a shaft of equivalent length so&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{2}kx^2=U_t$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gHU2K.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gHU2K.png&quot; alt=&quot;Cross section of the wire&quot;&gt;&lt;/a&gt;&#xA;As explained in the title, I have a particular situation because the material is orthotropic and the cross-section of the shaft is a rectangle as shown in the image above. I already have an expression for the torsional rigidity &lt;span class=&quot;math-container&quot;&gt;$GJ$&lt;/span&gt; of an orthotropic rectangular shaft with shear moduli &lt;span class=&quot;math-container&quot;&gt;$G_{zx}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$G_{zy}$&lt;/span&gt;:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$GJ=a b^3  G_{zx} \beta$$&lt;/span&gt;&#xA;where&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\beta=\sum_{n=1,3,5...}^{\infty}\frac{32 c^2}{\pi ^4 n^4} \left(1-\frac{2 c }{\pi  n}\tanh \left(\frac{\pi  n}{2 c}\right) \right)$$&lt;/span&gt;&#xA;and &lt;span class=&quot;math-container&quot;&gt;$c=\frac{a}{b}\sqrt{\frac{G_{zy}}{G_{zy}}}$&lt;/span&gt;. When I equate&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{2}kx^2=U_t=\frac{GJ}{2 L}\theta^2$$&lt;/span&gt;&#xA;I can plug in &lt;span class=&quot;math-container&quot;&gt;$\theta=\frac{x}{R}$&lt;/span&gt; (the angle the shaft has been twisted by is roughly equal to the angle each coil of the spring gets stretched by) so&#xA;&lt;span class=&quot;math-container&quot;&gt;$$k=\frac{GJ}{L R^2}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$L=2\pi R N_a$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$N_a$&lt;/span&gt; are the number of active coils in the spring. To me the equation looks complete but the problem is that when I enter values like &lt;span class=&quot;math-container&quot;&gt;$a=0.005$&lt;/span&gt;m, &lt;span class=&quot;math-container&quot;&gt;$b=0.0025$&lt;/span&gt;m, &lt;span class=&quot;math-container&quot;&gt;$G_{zx}=1159.1$&lt;/span&gt;MPa, &lt;span class=&quot;math-container&quot;&gt;$G_{zy}=974.3$&lt;/span&gt;MPa, &lt;span class=&quot;math-container&quot;&gt;$N_a=5$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$R=0.01$&lt;/span&gt; I get &lt;span class=&quot;math-container&quot;&gt;$k=632.642$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$k=2780$&lt;/span&gt;, which is the right answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I have that the torque on the shaft is &lt;span class=&quot;math-container&quot;&gt;$T=ab^3G_{zx}\theta\beta$&lt;/span&gt; I tried other equations for &lt;span class=&quot;math-container&quot;&gt;$U_t$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$U_t=\frac{1}{2}T\theta$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$U_t=\frac{1}{2}\frac{T^2 L}{GJ}$&lt;/span&gt; but none of the answers come even close, &lt;span class=&quot;math-container&quot;&gt;$198.75$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$62.4393$&lt;/span&gt;. Am I doing something wrong when equating the two energies?&lt;/p&gt;&#xA;" OwnerUserId="18933" LastActivityDate="2019-06-20T10:02:15.780" Title="Equation for the spring constant of an orthotropic, rectangular wire spring" Tags="|applied-mechanics|torque|springs|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28880" PostTypeId="1" CreationDate="2019-06-20T14:52:04.233" Score="2" ViewCount="189" Body="&lt;p&gt;I'm reading &lt;a href=&quot;http://www.geosci.usyd.edu.au/users/prey/Teaching/Geol-3101/Strain/stress.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this text on stress.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is mentioned that the Cauchy stress tensor can be split into a sum of two other tensors: hydrostatic pressure &lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; and deviatoric stress. Hydrostatic pressure is defined as the mean of the normal stresses. Deviatoric stress tensor is what we get when we subtract a tensor with the pressure on diagonal from the original Cauchy stress tensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align} \ \,\\ \left[{\begin{matrix} s_{11} &amp;amp; s_{12} &amp;amp; s_{13} \\ s_{21} &amp;amp; s_{22} &amp;amp; s_{23} \\ s_{31} &amp;amp; s_{32} &amp;amp; s_{33} \end{matrix}}\right] &amp;amp;=\left[{\begin{matrix} \sigma_{11} &amp;amp; \sigma_{12} &amp;amp; \sigma_{13} \\ \sigma_{21} &amp;amp; \sigma_{22} &amp;amp; \sigma_{23} \\ \sigma_{31} &amp;amp; \sigma_{32} &amp;amp; \sigma_{33} \end{matrix}}\right]-\left[{\begin{matrix} \pi &amp;amp; 0 &amp;amp; 0 \\ 0 &amp;amp; \pi &amp;amp; 0 \\ 0 &amp;amp; 0 &amp;amp; \pi \end{matrix}}\right] \\ &amp;amp;=\left[{\begin{matrix} \sigma_{11}-\pi &amp;amp; \sigma_{12} &amp;amp; \sigma_{13} \\ \sigma_{21} &amp;amp; \sigma_{22}-\pi &amp;amp; \sigma_{23} \\ \sigma_{31} &amp;amp; \sigma_{32} &amp;amp; \sigma_{33}-\pi \end{matrix}}\right]. \end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\pi=\frac{\sigma_{11}+\sigma_{22}+\sigma_{33}}{3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is then said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The second component is the Deviatoric stress and is what actually causes distortion of the body.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;Why do the diagonal stresses in the deviatoric stress tensor cause distortions, considering they are still normal stresses?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been taught that normal stresses cause only volumetric changes, and don't cause distortions, that is, don't change the shape of the element. It is the shear stresses, which lay off-diagonal, that cause shape changes to the element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shear stresses are present unchanged on the deviator tensor, but the normal stresses are now differences between the normal stresses on the element and their average. But they are still &lt;em&gt;normal&lt;/em&gt; stresses, so why are they present in the deviator tensor, if the purpose of the deviator tensor is to include only stresses that cause distortions? &lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2019-06-20T16:03:51.953" Title="Why is deviatoric stress the one causing distortions?" Tags="|structural-engineering|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28881" PostTypeId="2" ParentId="28880" CreationDate="2019-06-20T16:03:51.953" Score="1" Body="&lt;p&gt;The components in the stress tensor depend on the coordinate system being used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about a simple situation like pure shear in two dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one coordinate system, the only non-zero entries in the stress tensor are the &lt;span class=&quot;math-container&quot;&gt;$\tau_{xy}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\tau_{yx}$&lt;/span&gt; components (which are equal, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you rotate the coordinate axes by 45 degrees, the &lt;span class=&quot;math-container&quot;&gt;$\tau_{xy}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\tau_{yx}$&lt;/span&gt; components become zero and you have equal and opposite values for &lt;span class=&quot;math-container&quot;&gt;$\sigma_{xx}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\sigma_{yy}$&lt;/span&gt; components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But note that the &lt;em&gt;sum&lt;/em&gt; of the two direct stress components is zero, so there is no hydrostatic pressure in the stress field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a three dimensional stress field, you can &lt;em&gt;always&lt;/em&gt; find a set of coordinate directions where &lt;em&gt;all&lt;/em&gt; the shear stress terms are zero - namely the &quot;principal stress directions&quot; for the state of stress.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-06-20T16:03:51.953" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28882" PostTypeId="2" ParentId="28797" CreationDate="2019-06-20T16:20:06.543" Score="0" Body="&lt;p&gt;English units of measure are no longer used in England. Understanding the history of American independence and British colonial rule will clarify their origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to American independence the imperial units were used. After independence the use of American  imperial units changed according to American standards independent of British imperial units. This explains why US gains are not the same as British gallons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When UK adopted ISO standards officially, all English measures in engineering became metric standard however, US Imperial measures still remained in use on American soil while rest of the world uses metric based measures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why US imperial units are called English units when in fact they are actually American.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explanation was given to me by my Civil Engineering head of department at University of Wolverhampton, England where I am a student of civil engineering.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-06-20T16:20:06.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28884" PostTypeId="1" CreationDate="2019-06-20T16:49:48.687" Score="1" ViewCount="112" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/za3HA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/za3HA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is known that brittle materials tend to fail under normal stresses, that is, tensile or compressive stresses. Brittle materials often are bound together into crystals by ionic bonds between atoms, with alternating positive and negative charged atoms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I remember also that brittle materials fail when ionic bonds between layers are broken and the layers shift, causing like-charges to move close and therefore repel each other. The above picture I found illustrates this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But wouldn't this situation happen more likely in &lt;em&gt;shear&lt;/em&gt;, instead of compression or tension? If we had a cube of brittle material, wouldn't a shearing stress effectively shift those atomic layers in opposite directions, bringing like charges together, just like in the above picture? But apparently brittle materials fail more likely in tension or compression, and that's why maximum principal stress failure theory is used for brittle materials. This seems contradicting to me. &lt;strong&gt;Wouldn't a compressive stress, for example, simply squeeze the sample of material together instead of moving its layers?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2019-06-25T21:20:05.030" Title="Shouldn't brittle materials be the ones failing under shear deformations?" Tags="|stresses|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28885" PostTypeId="1" CreationDate="2019-06-20T17:05:08.987" Score="6" ViewCount="355" Body="&lt;p&gt;I am analyzing a beam, that has a bend in it. Intuitively and inspecting FEA results using a linear solver, there is a stress concentration at the bend of this I-beam.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/51QHg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/51QHg.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The part was not a straight I-beam that was later bent. Instead it was made with this bend by welding a curved web onto two flanges constructed of formed pieces of sheet metal. While there was some plastic deformation in the forming, the I-beam has mostly not gone through any yielding. My question is more related to, what stress concentrations arise from this modified geometry. Most analysis of beams pertains only to straight beams.&lt;/p&gt;&#xA;&lt;p&gt;When I was learning about stress concentrations in Shigley's, they talked about notches and holes, but not bends. Is there a factor one should apply for an x degree bend?&lt;/p&gt;&#xA;&lt;p&gt;Any thoughts?&lt;/p&gt;&#xA;" OwnerUserId="20900" LastEditorUserId="1832" LastEditDate="2020-07-14T21:25:21.657" LastActivityDate="2023-12-21T14:05:05.597" Title="How to analyze a bent beam: Stress concentration?" Tags="|mechanical-engineering|beam|finite-element-method|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28886" PostTypeId="2" ParentId="28885" CreationDate="2019-06-20T17:47:36.347" Score="0" Body="&lt;p&gt;If the bend has passed the elastic limit ( assuming material is steel) it have caused strain hardening in that spot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hardening or stiffness both attracts stress and keeps residual stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We take advantage of this hardening when we want to break metal wires by bending and unbending them repeatedly until stress concentrated at the bend will break it.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-06-20T18:12:36.597" LastActivityDate="2019-06-20T18:12:36.597" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28888" PostTypeId="1" CreationDate="2019-06-21T02:44:23.087" Score="1" ViewCount="498" Body="&lt;p&gt;I have two related questions&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is a thicker tempered glass stronger than thinner?&#xA;For example, I have to hit much harder at 3/4” &#xA;tempered glass panel surface than I hit 1/4” &#xA;panel in order for glass to shatter. My understanding &#xA;is that tempering pre stresses and creates a &#xA;tension in a surface layer. Why does the &#xA;glass thickness matter if the “shield” is about &#xA;the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did anybody study possibility of tempering the glass panel &#xA;all the way through as opposed to just the surface?&lt;/p&gt;&#xA;" OwnerUserId="20904" LastActivityDate="2019-06-21T13:57:57.947" Title="Tempered glass physics" Tags="|glass|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28889" PostTypeId="2" ParentId="28888" CreationDate="2019-06-21T03:27:33.220" Score="3" Body="&lt;p&gt;Tempered glass resists fracture because it is manufactured in such a way as to place its surfaces into a state of residual compressive stress. To cause it to experience brittle failure in tension (which is normally how glass breaks) requires overcoming those frozen-in surface stresses, which normally act to squeeze shut any surface pits or scratches at which brittle failure would begin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's going to take more work to shatter a thick piece of tempered glass simply because the thicker a piece of material is, the better it is at distributing applied stresses so that they do not cause it to fail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the greatest stresses in a material being bent by a load are at its surfaces, a strengthening treatment which is applied to those surfaces will yield significant improvement in its performance even though the interior of the part hasn't been strengthened. This approach has been satisfactory for things like sliding glass doors and car windshields, but the invention of smart phone touchscreens has required the development of highly-strengthened glass called &quot;gorilla glass&quot;, which I recommend you run a search on for the details of its manufacture.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-06-21T03:27:33.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28891" PostTypeId="1" AcceptedAnswerId="28892" CreationDate="2019-06-21T08:38:02.440" Score="2" ViewCount="2130" Body="&lt;p&gt;I'm struggling to identify which equation is used within the first red square. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please help identify which equation was used as well as why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SXlm7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SXlm7.png&quot; alt=&quot;Reference&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankyou!&lt;/p&gt;&#xA;" OwnerUserId="20873" LastActivityDate="2019-06-21T23:47:21.837" Title="Max Tensile &amp; Compressive Stress Question - Identify the equation" Tags="|mechanical-engineering|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28892" PostTypeId="2" ParentId="28891" CreationDate="2019-06-21T10:35:21.443" Score="5" Body="&lt;p&gt;The calculation for &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; that you've boxed is essentially arrived at through geometry. We simply want to find the distance from the centroid where total stress is zero. Drawing the axial, bending, and total stress along the cross-section makes this clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gd9Co.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gd9Co.jpg&quot; alt=&quot;stress distribution&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sigma_{axial} &amp;amp;= \frac{P}{A} \\&#xA;\sigma_{flexure} &amp;amp;= \frac{My}{I} \\&#xA;\sigma_{total} &amp;amp;= \sigma_{axial} + \sigma_{flexure}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we can readily calculate &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; based on the slope of the flexural stress. I tend to phrase it in my head as &quot;at what distance &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; from the centroid does the flexural stress perfectly counteract the axial stress?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; such that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_{axial} = \frac{2\sigma_{flexure}}{d}e$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is the depth of the section and &lt;span class=&quot;math-container&quot;&gt;$\sigma_{flexure}$&lt;/span&gt; refers to the stress at the extreme fiber (distance &lt;span class=&quot;math-container&quot;&gt;$d/2$&lt;/span&gt; from the centroid). Rearranging that expression, and substituting, we see the equation used in the given solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$e = \frac{\sigma_{axial} \cdot \frac{d}{2}}{\sigma_{flexure}} = \frac{\sigma_{axial} \cdot I}{M}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth emphasizing, that from the standpoint of developing understanding, memorizing their particular expression for &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; is not all that useful. It's more important to understand what's happening - that we're simply adding axial and flexural stress, and finding the location where total stress is zero.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="1832" LastEditDate="2019-06-21T23:47:21.837" LastActivityDate="2019-06-21T23:47:21.837" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28893" PostTypeId="2" ParentId="28888" CreationDate="2019-06-21T13:57:57.947" Score="2" Body="&lt;p&gt;By definition, &quot;tempering the glass panel all the way through&quot; is not possible.&#xA;The interior and exterior parts of the glass &lt;em&gt;must&lt;/em&gt; have different properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Tempered_glass&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tempered glass&lt;/a&gt; is made by cooling the exterior and interior at different times:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The glass panel is heated to a uniform temperature hot enough to soften it.&lt;/li&gt;&#xA;&lt;li&gt;The exterior surface is cooled by air until it becomes hard.&lt;/li&gt;&#xA;&lt;li&gt;The interior part remains soft enough to flow.&lt;/li&gt;&#xA;&lt;li&gt;The entire panel is allowed to slowly cool back to room temperature.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The essential part happens during the final cooling period.  As it cools, the glass shrinks, but the outer surface is already cooler, so shrinks less than the interior.  By the time the entire panel has cooled, the outer surface is pulled tightly together by the interior, which has shrunk more than it has.  At the same time, the interior is pulled outward by the exterior layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stress (compression of the surface layer and tension of the interior) is what gives the glass its strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, while the compressed outer layer is extremely resistant to damage, when it does experience damage, it is catastrophic.&#xA;A hard blow by a sharp object can create a crack in the outer layer.&#xA;This will relieve some of the internal stress and cause the crack to spread rapidly.&#xA;This will distort that part of the glass, which will create extra stress on the surrounding areas, which will themselves crack and propagate the destruction throughout the entire panel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explosive shattering into small pieces is what makes tempered glass much safer, as it does not leave any large jagged pieces that could cause severe injuries.&#xA;Side windows on cars are made of tempered glass for this reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More interestingly, it is possible to break the interior of tempered glass &lt;em&gt;without&lt;/em&gt; breaking the exterior layers, such as by a single hammer blow to a nail directed into the edge of a glass table top.  Though it can look beautiful, the result should not be used, as it will no longer have its original strength nor its safety features.  It should be discarded or safely framed as &quot;art&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;Time Warp&quot; TV show featured an &lt;a href=&quot;https://youtu.be/0QeLz_iUF6c?t=73&quot; rel=&quot;nofollow noreferrer&quot;&gt;example of interior shattering&lt;img src=&quot;https://i.stack.imgur.com/YSWAV.png&quot; alt=&quot;example of interior shattering&quot;&gt;&lt;/a&gt;:&lt;/p&gt;&#xA;" OwnerUserId="19930" LastActivityDate="2019-06-21T13:57:57.947" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28894" PostTypeId="1" AcceptedAnswerId="28895" CreationDate="2019-06-21T15:10:34.193" Score="0" ViewCount="125" Body="&lt;p&gt;I am trying to get an idea of the amount of hydrogen that I can store in some pressurized steel tanks underground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the ideal gas law PV=nRT&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a simple calculation&#xA;pressure = 200bar&#xA;volume   = 2000Liters&#xA;temp     = 25C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My results were about&#xA;16136.7 moles of hydrogen&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 mole of any gas is about 22.4 liters but that's at standard temperature and pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At these elevated pressure it should be about 361,671 liters?&lt;/p&gt;&#xA;" OwnerUserId="20730" LastActivityDate="2019-06-22T02:14:46.193" Title="Ideal gas law, theory vs reality" Tags="|thermodynamics|pressure|gas|pressure-vessel|compressed-gases|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28895" PostTypeId="2" ParentId="28894" CreationDate="2019-06-21T15:51:26.030" Score="1" Body="&lt;p&gt;Have a look at the following:&#xA;&lt;a href=&quot;https://industry.airliquide.us/volume-compressed-gas-cylinder&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://industry.airliquide.us/volume-compressed-gas-cylinder&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on this, if you assume a gas cylinder of 47 litres (standard size) at 200 bar fill pressure, you have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_1 V_1 = P_2 V_2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$P_1$&lt;/span&gt; = 200 bar (above atmospheric pressure) = 201.325 bar (absolute pressure)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt; = 47 &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$P_2 = P_{atm}$&lt;/span&gt; = 1.01325 bar (absolute pressure)&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt; is what you're trying to determine&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You end up with &lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt; = 9338.6 &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; approximately.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-06-21T15:51:26.030" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28897" PostTypeId="1" AcceptedAnswerId="28899" CreationDate="2019-06-21T17:32:21.510" Score="0" ViewCount="760" Body="&lt;p&gt;I'm seeking an explanation in reference to the photo below. I'm unsure how the solution within the red box calculates the max tension of the Aluminium beam and the max compression off the Aluminium beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How are you able to differentiate whether the Aluminium is in tension or compression within this scenario?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as how they are able to make the statement that &lt;em&gt;&quot;the steel is entirely compressive bending stress and the max bending stress is at the top of the flange&quot;&lt;/em&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rahAI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rahAI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankyou!&lt;/p&gt;&#xA;" OwnerUserId="20873" LastActivityDate="2019-06-21T22:12:52.213" Title="Max compression &amp; Tension" Tags="|mechanical-engineering|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28899" PostTypeId="2" ParentId="28897" CreationDate="2019-06-21T18:26:05.250" Score="3" Body="&lt;p&gt;It has to deal with the location of the Neutral Axis (NA).  That whole calculation for Ῡ =37 mm is key here.  Bending occurs about the NA.  That big yellow arrow represents the moment in the beam, or in other words which direction it is bending.  Based on the moment arrow everything above the NA will be compression, and everything below it will be tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculations show the NA at a distance of 37 mm from the bottom and from the cross section dimensions we see the aluminum section is 40 mm deep.  This means all the steel is above the NA which in turns means all the steel for this sections is in compression.  The portion of the aluminum between the NA and the steel is also in compression.  The maximum stress occurs at the furthest distance from the neutral axis.&lt;/p&gt;&#xA;" OwnerUserId="6993" LastEditorUserId="6993" LastEditDate="2019-06-21T22:12:52.213" LastActivityDate="2019-06-21T22:12:52.213" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28900" PostTypeId="1" CreationDate="2019-06-21T22:44:09.417" Score="1" ViewCount="195" Body="&lt;p&gt;I have written my own finite element code for a linear elastic (plane stress) problem consisting of a plate with a circular hole in the middle, fixed on one side, having a tensile force acting on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have obtained displacements and calculated the stresses and strains at the nodes. I want to calculate the stress concentration factor numerically and compare it with theoretical values for the same case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the stress concentration factor &lt;span class=&quot;math-container&quot;&gt;$K_{t}$&lt;/span&gt; is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;K_t = \frac{\sigma_{max}}{\sigma_{nom}}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sigma_{max}$&lt;/span&gt; is very easy to find but I have trouble understanding how to calculate &lt;span class=&quot;math-container&quot;&gt;$\sigma_{nom}$&lt;/span&gt; for my problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="20914" LastActivityDate="2019-06-22T00:18:28.070" Title="How is the stress concentration factor calculated in FEM?" Tags="|mechanical-engineering|structural-engineering|structural-analysis|stresses|finite-element-method|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28902" PostTypeId="2" ParentId="28900" CreationDate="2019-06-22T00:18:28.070" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\sigma_{\text{nom}}$&lt;/span&gt; is the stress in the component without a hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to calculate that by hand, without needing to create another FE model ;)&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-06-22T00:18:28.070" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28903" PostTypeId="2" ParentId="28894" CreationDate="2019-06-22T00:36:23.453" Score="0" Body="&lt;p&gt;Theory: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 mole at those conditions takes up 0.123 l. Why? If 16,136.7 moles of hydrogen take up 2,000 l, then 1 mole will take up 0,123 l. You can also apply the ideal gas equation to check it: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V=n*R*T/p=1 [mole]*0.082 [atm*l/mol*K]*298.15 [K]/[(200*10^5/101,325) [atm]]=0.123 [l]$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what about reality? Well, ideal gas equation fails at such high pressures because it doesn't regard the space occupied by the mollecules of the gas (important at those high pressures becuase collisions are more frequent and repulsion forces more intense, therefore there's an increase in volume). Note: This is true for hydrogen, but not for all real gases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's correct the ideal gas equation with the compressibility factor &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we'll have &lt;span class=&quot;math-container&quot;&gt;$p*V=z*n*R*T$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At 20 MPa (200 bar) and 25 ºC we have &lt;span class=&quot;math-container&quot;&gt;$z=1.124$&lt;/span&gt; (mean value between 1.0601 and 1.1879, highlighted down here):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EdQNr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EdQNr.jpg&quot; alt=&quot;z of hydrogen for different temperatures and pressures. Source: https://h2tools.org/hyarc/hydrogen-properties&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source:&lt;a href=&quot;https://h2tools.org/hyarc/hydrogen-properties&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydrogen compressibility at different temperatures and pressures&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's do the math:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V=z*n*R*T/p=1.124*1 [mole]*0.082 [atm*l/mol*K]*298.15 [K]/[(200*10^5/101,325) [atm]]=0.139 [l]$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And 2,246.7 l if you consider the 16,137.7 moles. So prepare a bigger tank for your hydrogen!&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastEditorDisplayName="user20096" LastEditDate="2019-06-22T02:14:46.193" LastActivityDate="2019-06-22T02:14:46.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28904" PostTypeId="1" AcceptedAnswerId="28929" CreationDate="2019-06-22T01:02:43.343" Score="2" ViewCount="1142" Body="&lt;p&gt;Im looking for clarification on how the solution identified the moment to be hogging, as well as the axial stress to be in compression?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also I'm confused as to how the solution came to a conclusion that the bottom of the cross section is compressive stress, whilst the top of the cross section is tensile stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please reference the photo below with the red box's&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RwzG1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RwzG1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thankyou!&lt;/p&gt;&#xA;" OwnerUserId="20873" LastActivityDate="2019-06-23T20:14:47.537" Title="Identifying Compression and Tension as well as Hogging" Tags="|mechanical-engineering|structural-engineering|civil-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28905" PostTypeId="2" ParentId="28867" CreationDate="2019-06-22T02:31:23.147" Score="1" Body="&lt;p&gt;Thruth to be told, you cannot have an engine 95 % efficient. Why? You have a combustion process and combustion is irreversible (entropy rises). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And no, you cannot calculate overall efficiency like that. In a turbojet you calculate it this way: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node85.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node85.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Read this too: &lt;a href=&quot;https://www.quora.com/How-efficient-is-a-turbojet-engine&quot; rel=&quot;nofollow noreferrer&quot;&gt;How efficient is a turbojet engine&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you intend to recover the waste heat? It's WASTE, its exergy is too low!&lt;/p&gt;&#xA;" OwnerDisplayName="user20096" LastActivityDate="2019-06-22T02:31:23.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28906" PostTypeId="2" ParentId="28867" CreationDate="2019-06-22T04:51:48.357" Score="0" Body="&lt;p&gt;The complete Brayton cycle of an aircraft gas turbine engine has a couple of sources of energy loss - loss from the propulsive thrust power that is. The nozzle you mention is only part of the whole cycle, and yes can be very efficient if shaped properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/opHF3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/opHF3.png&quot; alt=&quot;from an old uni book, English terms are my own translations&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Above image quantifies the energy streams for a bypass engine with a high bypass ratio. Every stage in energy transformation comes with a loss, quantified in the stage efficiency. The stages in the energy transformation are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Chemical energy =&gt; heat. Has a very high efficiency, heat radiation to the environment is minimal, as is CO and soot formation.&lt;/li&gt;&#xA;&lt;li&gt;Heat energy =&gt; gas power. We're converting heat into useful power, which can never be fully accomplished: after full expansion, gas temperature has increased.&lt;/li&gt;&#xA;&lt;li&gt;Gas power =&gt; propulsive power. Either into shaft power for turboprops/turboshafts, or into kinetic power for jet engines. Cannot occur loss free (isentropically) due to heat transfer, friction losses etc.&lt;/li&gt;&#xA;&lt;li&gt;Propulsive power (of the medium = air, combustion gasses) into thrust power (of the aircraft). The medium is propelled with a higher velocity than that of the aircraft, which is an efficiency loss. A necessary one of course.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;End efficiency of the whole cycle is about 34%, probably a bit higher nowadays since the depiction is several decades old. Notice that the jet nozzle (exhaust) mentioned in the OP is only one aspect, with a high efficiency indeed. &lt;/p&gt;&#xA;" OwnerUserId="12193" LastActivityDate="2019-06-22T04:51:48.357" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28907" PostTypeId="2" ParentId="28904" CreationDate="2019-06-22T04:55:38.933" Score="1" Body="&lt;p&gt;Your beam will bend down as opposed to what you think, under the moment of 100kN*((200/2)-30 =70 mm)and will at the same time compress under the stress of  P/A= 100kN/ (20*200).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the force of 100kN is applied below the neutral axis of the beam it causes counterclockwise moment hence compression below the neutral axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb you always get compression on the part of the beam's section which is under axial load. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-22T04:55:38.933" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28908" PostTypeId="1" CreationDate="2019-06-22T08:15:49.727" Score="2" ViewCount="3015" Body="&lt;p&gt;I need to build special equipment for a microwave oven ( 2.45 (GHz)).&#xA;But even glass and plastics do heat up in microwave ovens. Where can I find a list of the materials that are the most transparent to this frequency ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="5926" LastActivityDate="2020-07-06T22:44:14.570" Title="What are the materials most transparent to microwaves?" Tags="|microwaves|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28909" PostTypeId="1" CreationDate="2019-06-22T09:05:22.633" Score="2" ViewCount="92" Body="&lt;p&gt;Looking for the correct terminology to for the cable/filament feeding mechanism used in 3d printer filament feeder where the filament is gripped between a bearing and a gnarled gear tooth driven by motor - also seen in MIG-welder wire feeders. Would like to be able to do some searching to find out more on integrating similar mechanism to retract either stainless steel cable or Dyneema rope in something I am working on.&lt;/p&gt;&#xA;" OwnerUserId="2201" LastActivityDate="2019-06-22T10:29:13.317" Title="Whats the correct name for the filament driving mechanism used in 3d printer extruders and MIG wire feeders?" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28910" PostTypeId="2" ParentId="28909" CreationDate="2019-06-22T10:29:13.317" Score="1" Body="&lt;p&gt;I believe the term you seek is &quot;hobbed gear.&quot; It's a cylinder typically attached to the extruder motor or drivetrain with a series of cuts about the circumference. A pressure roller is positioned parallel to the hobbed gear, or in the case of Bondtech drives, a second hobbed gear is located similarly and geared to the drive side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hF7VR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hF7VR.jpg&quot; alt=&quot;hobbed gear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-06-22T10:29:13.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28911" PostTypeId="2" ParentId="28654" CreationDate="2019-06-22T14:36:53.317" Score="0" Body="&lt;p&gt;Isn't there something off about your equations already?&#xA;If I look at the sketch that you provided looks to me as if the forces Fa and Fb should act against each other which would result in opposite signs in the balance equation. And then of course you would also have to take into account the angle of the forces relative to the axis of rotation.&#xA;But if I see the equation to calculate Fb might also be that they are just put wrong in the sketch.&lt;/p&gt;&#xA;" OwnerUserId="20870" LastActivityDate="2019-06-22T14:36:53.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28917" PostTypeId="2" ParentId="14822" CreationDate="2019-06-22T20:34:35.357" Score="0" Body="&lt;p&gt;Try to use ECEF data in perturbation calculation then execute integration part with ECI data after making the proper conversion from ECEF to ECI reference frame. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ref: D. A. Vallado, Fundamentals of Astrodynamics and Applications, 4th Editon, Microcosm Press, 2013, pp.591-595     &lt;/p&gt;&#xA;" OwnerUserId="20924" LastActivityDate="2019-06-22T20:34:35.357" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28921" PostTypeId="1" AcceptedAnswerId="28946" CreationDate="2019-06-23T01:23:06.900" Score="-1" ViewCount="1258" Body="&lt;p&gt;I am trying to calculate &lt;em&gt;mass&lt;/em&gt; fractions of a Hydrogen Oxygen gas mixture as a function of equivalence ratio before ignition. My work is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equivalence Ratio: &lt;span class=&quot;math-container&quot;&gt;$$\phi=\frac{FA}{FA_s}=\frac{\frac{m_f}{m_{air}}}{\frac{m_{fs}}{m_{air,s}}}=\frac{\frac{n_f}{n_{air}}}{\frac{n_{fs}}{n_{air,s}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$FA$&lt;/span&gt; is the fuel to air ratio and a subscript &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; stands for the stoichiometric reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The balanced combustion reaction equation (assuming a perfectly stoichiometric mixture) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$2H_2+1O_2\rightarrow2H_2O$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, &lt;span class=&quot;math-container&quot;&gt;$n_{fs} = 2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n_{air,s}=1$&lt;/span&gt;. Now, when I write the mass fractions as a function of equivalence ratio, my work is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y_{O_2}=\frac{m_{air}}{m_f+m_{air}}=\frac{1}{FA+1}=\frac{1}{\phi\frac{n_{fs}}{n_{air,s}}+1}=\frac{n_{air,s}}{\phi n_{fs}+n_{air,s}}=\frac{1}{2\phi+1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By a similar calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y_{H_2}=\frac{n_{fs}\phi}{n_{fs}\phi+n_{air,s}}=\frac{2\phi}{2\phi+1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quick check that should be true reveals that &lt;span class=&quot;math-container&quot;&gt;$y_{H_2}+y_{O_2}=1$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I am asking about this calculation is that I did this calculation for a previous project and the results the class said was correct was the reverse (i.e. the expression with &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; in the numerator was for &lt;span class=&quot;math-container&quot;&gt;$y_{O_2}$&lt;/span&gt; not &lt;span class=&quot;math-container&quot;&gt;$y_{H_2}$&lt;/span&gt;. Is there something I did wrong/did I calculate them the wrong way round? The code I am basing my previous work off of (written in MATLAB) is: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;        yO =  32/(4*phi+32);   &#xA;        yH =  4*phi/(4*phi+32);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The 32 crops up because the molar mass of &lt;span class=&quot;math-container&quot;&gt;$O_2$&lt;/span&gt; was in the expression for some reason and I have not been able to figure out exactly how the above code expressions were derived. I just know they gave results that were marked correct for the class I did this for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I have figured out the source of my confusion. The statement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\phi=\frac{FA}{FA_s}=\frac{\frac{m_f}{m_{air}}}{\frac{m_{fs}}{m_{air,s}}}=\frac{\frac{n_f}{n_{air}}}{\frac{n_{fs}}{n_{air,s}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is true by itself, but this does not mean that &lt;span class=&quot;math-container&quot;&gt;$FA_s=\frac{n_{fs}}{n_{air,s}}$&lt;/span&gt;! I made this assumption in my head after I canceled out the molar masses used to transform the masses into moles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The correct expressions are: &lt;span class=&quot;math-container&quot;&gt;$$y_{O_2} = \frac{n_{o,s}M_O}{\phi n_{f,s}{M_f}+n_{o,s}M_O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y_{H_2} = \frac{\phi n_{f,s}M_f}{\phi n_{f,s}{M_f}+n_{O,s}M_O}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="9226" LastEditorUserId="9226" LastEditDate="2019-06-26T06:24:42.327" LastActivityDate="2019-06-26T12:03:01.603" Title="Calculating Mass Fractions of a Hydrogen and Oxygen Gas Mixture Pre Ignition" Tags="|combustion|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28922" PostTypeId="1" CreationDate="2019-06-23T02:35:59.563" Score="0" ViewCount="60" Body="&lt;p&gt;You can see it in the video here for the Barracuda: &lt;a href=&quot;https://youtu.be/_NZKyvQVRd0?t=124&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/_NZKyvQVRd0?t=124&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here's a pic of the SMX-31: &lt;a href=&quot;http://www.hisutton.com/images/Fr_SMX-31_cutaway.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.hisutton.com/images/Fr_SMX-31_cutaway.jpg&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the advantages of this rudder layout over a more conventional one without the gap? &lt;/p&gt;&#xA;" OwnerUserId="20925" LastActivityDate="2019-06-23T14:28:17.827" Title="What is up with the spaced rudders on the SMX-31/Barracuda?" Tags="|fluid-mechanics|naval-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28923" PostTypeId="2" ParentId="28922" CreationDate="2019-06-23T05:57:44.910" Score="0" Body="&lt;p&gt;Because of the lack of any longitudinal fins or a keel, it will give authority and stability in control and prevent the sub from undue yaw and oversteering when turning or diving and rising (pitching up, down).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be also structurally supporting the rudder when the sub is doing accelerated maneuvers or dodging an obstacle or torpedo. It changes the rudder structure from a hinged flagpole to a simply supported beam.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-23T05:57:44.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28924" PostTypeId="1" CreationDate="2019-06-23T08:11:33.530" Score="0" ViewCount="101" Body="&lt;p&gt;Used to connect the chain for a window actuator to a U-shaped window bracket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The threaded end screws into a bolt hole (M5.5 female thread) on one side of the U-bracket. The tip of the fastener fits into a 3mm diameter slot on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The end of the chain pushes and pulls perpendicular to the axis of the fastener on the unthreaded section between the two connections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dimensions:&lt;/strong&gt; Total length: &lt;strong&gt;34mm&lt;/strong&gt;.&#xA;Threaded section length: &lt;strong&gt;6mm&lt;/strong&gt;.&#xA;Unthreaded section: &lt;strong&gt;3mm&lt;/strong&gt; diameter.&#xA;Threaded section: &lt;strong&gt;5.5mm&lt;/strong&gt; diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gMxYO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gMxYO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What type of fastener is this, and how do I go about procuring a replacement, short of resorting to a bespoke service?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The unthreaded section can be thicker to some degree, or be threaded, as long as there is a 3mm diameter tip of at least 4mm length to connect to that side of the bracket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a M5.5 bolt with a 3mm diameter dogpoint would do the job. However, as far as I understand, there are no bolts standardized to a M5.5 diameter.&lt;/p&gt;&#xA;" OwnerUserId="20926" LastEditorUserId="20926" LastEditDate="2019-07-06T08:29:15.443" LastActivityDate="2019-07-06T08:29:15.443" Title="What type of fastener is this?" Tags="|fasteners|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28925" PostTypeId="2" ParentId="28924" CreationDate="2019-06-23T08:24:47.003" Score="2" Body="&lt;p&gt;It is a pin with a threaded head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A normal bolt would not work in many chains that go through restricted passages, so the head in this case is designed to sit flush with the surface.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-23T08:24:47.003" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28928" PostTypeId="2" ParentId="28922" CreationDate="2019-06-23T14:28:17.827" Score="2" Body="&lt;p&gt;I assume you mean the added flap arrangement, as opposed to a simple spade rudder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR 99% of the time, these are doing nearly nothing, so you want the smallest, lowest drag arrangement when doing nothing. That translates to a very high max lift coefficient arrangement, even if the L/D isn't all that great at high lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is these have a significantly higher maximum coefficient of lift, making them smaller for a given amount of force. This is very desirable since 99% of the time they aren't doing much and being smaller, these have less drag then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the best sources detailing the performance of this type mechanical arrangement (which is different from aircraft flap actuation) is probably the BoatDesign.net/hydrodynamics thread on the wing-sail designs used Americas Cup's boats and a few other ultra high performance sailboats.  Tom Speer is a retired Boeing engineer who has been working on the last three or four generations of these boat's wings, and has posted analyses of their geometry and performance. The images below are from one of Tom's post's via a bunch of cut and pastes. The entire tread is here - &lt;a href=&quot;https://www.boatdesign.net/threads/understanding-wing-technology.34697/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Understanding Wing Technology&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also Tom's homepage, under rigid wingsail rigs. - &lt;a href=&quot;http://www.tspeer.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.tspeer.com/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Huhkk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Huhkk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/h2dGB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h2dGB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-06-23T14:28:17.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28929" PostTypeId="2" ParentId="28904" CreationDate="2019-06-23T19:37:01.283" Score="3" Body="&lt;p&gt;Once again, the &lt;a href=&quot;https://en.wikipedia.org/wiki/Superposition_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;superposition principle&lt;/a&gt; is our friend. It basically means we can look at each load in isolation, find the individual results, and then add them all together to get the final result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we are actually dealing with two loads:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;There's obviously the force &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;;&lt;/li&gt;&#xA;&lt;li&gt;And then there's the bending moment &lt;span class=&quot;math-container&quot;&gt;$P\cdot e$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$e$&lt;/span&gt; is the eccentricity of the load &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; in relation to the beam's neutral axis. Since the beam is rectangular with height 200&amp;nbsp;mm, we trivially know the neutral axis is at 100&amp;nbsp;mm. And since the load is only 30&amp;nbsp;mm from the bottom, we know &lt;span class=&quot;math-container&quot;&gt;$e=100-30=70\text{ mm}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now let's look at these loads in isolation:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The force &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This one is pretty obvious. The force alone (remember, we're not yet considering the bending moment effect, so you can think of the force as acting at the neutral axis) obviously causes uniform compression throughout the beam. This means the beam remains perfectly straight, just a bit shorter. Not much to discuss here, I believe.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;The bending moment &lt;span class=&quot;math-container&quot;&gt;$P\cdot e$&lt;/span&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Herein lies the rub.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bending moment is counter-clockwise in this case. This seems pretty intuitive to me, so I won't spend any time demonstrating this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is how the beam deflects given this bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the free end, it can rotate in one of two ways: clockwise or counter-clockwise. It once again seems pretty intuitive to me that the beam will rotate counter-clockwise, but here's a demonstration, just in case it isn't to you:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's transform the bending moment into a &lt;a href=&quot;https://en.wikipedia.org/wiki/Couple_(mechanics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;force couple&lt;/a&gt;: two opposing forces which cancel each other out other than generating a bending moment. A classic example is a &lt;a href=&quot;https://i.stack.imgur.com/jEAOz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;crank valve&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lc04I.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lc04I.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you had a crank valve at the end of a beam, and you applied the forces shown above, how would you expect it to rotate? It seems clear the beam would rotate counter-clockwise. By this I mean that the free end (and indeed any other slice along the beam) would rotate counter-clockwise, such that the top fiber moves a bit to the left and the bottom fiber moves to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that tells us all we need to know: if the top fiber moves to the left, then the total length of that fiber has increased (from 700&amp;nbsp;mm to... 701&amp;nbsp;mm or whatever). And we know that if an element is lengthened, it must be under tension&lt;sup&gt;1&lt;/sup&gt;. Likewise, if the bottom fiber moved to the right, that fiber's length shortened and it is therefore under compression.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Total result&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;So, summing both loads' results, we get that the top fiber is under compression due to the force and under tension due to the bending moment. Whether the final result is compression or tension will depend on which stress is greater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom fiber is under compression due to both loads, so that's simple enough.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Hogging&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;As for why the moment is considered hogging, well, that's just the technical definition of hogging moment.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When the beam bends in such a way that it forms concavity downwards (cup-shaped) it is called as sagging. Whereas the bending which results in convexity upwards (like a hump); is known as hogging. &lt;a href=&quot;https://civilengineer.webinfolist.com/mech/bm.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;[source]&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Basically, positive moment is sagging, negative moment is hogging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And if you remember your sign conventions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OXLrT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OXLrT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see that counter-clockwise moments on the left side of the beam are negative, and therefore hogging.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; Or possibly under a positive thermal load, if you want to be a pedant.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-06-23T20:14:47.537" LastActivityDate="2019-06-23T20:14:47.537" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28930" PostTypeId="1" CreationDate="2019-06-23T23:21:51.217" Score="5" ViewCount="3676" Body="&lt;p&gt;If one looks up the meaning of shear strain, most sources talk about this situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a7rys.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a7rys.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, a force acts &lt;em&gt;along&lt;/em&gt; a surface of an element at it is elongated, turning a rectangle into a rhombus. Apparently this is called &quot;simple shear&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I found out there is something called &quot;pure shear&quot;, which is illustrated in this picture:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IV4DX.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IV4DX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So pure shear, on the other hand, is a &quot;flattening&quot; of a body; the element is squeezed flatter on one axis and elongated on another. There is no change in angle between the lines, as there was in the simple shear case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;Why are these two things both called &quot;shear strains&quot;? What do they have in common, besides that neither change the area of the element? What is &quot;pure&quot; about &quot;pure shear&quot; and &quot;simple&quot; about &quot;simple shear&quot;?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple shear is caused when the force is acting &lt;em&gt;along&lt;/em&gt; the surface of the element, without normal forces. But what about pure shear? It involves squeezing in one axis and elongating on other, so it must be caused by normal forces, right? But if this is right, then it wouldn't make sense to put this in the same category (&quot;shear&quot;) as simple shear, as one would be caused by normal forces that act directly on the surfaces of the element, and the other by forces &lt;em&gt;along&lt;/em&gt; the surfaces of the element.&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2021-08-18T20:20:54.507" Title="Why is pure shear a &quot;shear&quot;?" Tags="|structural-analysis|stresses|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28931" PostTypeId="1" CreationDate="2019-06-24T00:32:41.627" Score="2" ViewCount="87" Body="&lt;p&gt;On &lt;a href=&quot;https://en.wikipedia.org/wiki/Timoshenko_beam_theory#Quasistatic_Timoshenko_beam&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia derivation on Timoshenko beam equation&lt;/a&gt; we arrive at this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \delta U = \int_L \left[-M_{xx}\frac{\partial (\delta\varphi)}{\partial x} + Q_{x}\left(-\delta\varphi + \frac{\partial (\delta w)}{\partial x}\right)\right]~\mathrm{d}L $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Integration by parts, and noting that because of the boundary conditions the variations are zero at the ends of the beam, leads to &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \delta U = \int_L \left[\left(\frac{\partial M_{xx}}{\partial x} - Q_x\right)~\delta\varphi - \frac{\partial Q_{x}}{\partial x}~\delta w\right]~\mathrm{d}L $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But why are the variations on the angle &lt;span class=&quot;math-container&quot;&gt;$\delta\varphi$&lt;/span&gt; and displacement &lt;span class=&quot;math-container&quot;&gt;$\delta w$&lt;/span&gt; zero at the ends?&lt;/strong&gt; If we have a beam resting on two supports at the ends, then I can understand why the displacement variation would be zero since the beam is bound there, but I don't see how the angle variation could be zero, since more the beam is displaced along its length, bigger the change in angle at the ends. And if we are dealing with a beam, say, attached to a wall with one end free, the one of the ends experiences a difference in both the angle and the vertical displacement.&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2019-07-24T18:02:50.373" Title="Why are variations zero on the endpoints of a beam?" Tags="|structural-engineering|beam|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28933" PostTypeId="1" AcceptedAnswerId="28939" CreationDate="2019-06-24T09:25:43.303" Score="1" ViewCount="5947" Body="&lt;p&gt;sorry for my english. I was looking to buy a house. I saw some in my hometown, which is very affordable.&#xA;but on a picture of the house they are selling I found on the net troubled me.&#xA;no pillar/column, just hollowblocks and some steel bars.&lt;a href=&quot;https://i.stack.imgur.com/Rl8eC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rl8eC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;kindly look at the pic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found it in this site&#xA;&lt;a href=&quot;https://fiestacommunitiesblog.wordpress.com/tag/fiesta-communities/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://fiestacommunitiesblog.wordpress.com/tag/fiesta-communities/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20932" LastActivityDate="2019-06-24T15:29:45.827" Title="Is it OK for a 2 story house not have a column or pillar" Tags="|civil-engineering|building-design|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28934" PostTypeId="1" AcceptedAnswerId="28938" CreationDate="2019-06-24T09:47:22.757" Score="1" ViewCount="1461" Body="&lt;p&gt;I'm looking at designing a small gadget (think something like a Fitbit in size) where the case has a couple of buttons on it.  For various reasons the buttons have to be around the narrow side (edge) of the case.  I am fairly lost as to what the standard button mechanical design for this type of application would be.  I mean, I know about (much bigger) buttons with a cantilever or a linear spring or spring and pivot, or rubber dome switches, but here... the whole thing it tiny!  The largest buttons that can fit are roughly 12x5mm, and the &quot;mechanism&quot; has to fit in a few mm of thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The prototype has oval plastic buttons which are ~12x5mm and 4mm deep, and C&amp;amp;K PTS840 tactile switches on a PCB behind them.  These switches have a tiny little plunger 1.7x1mm with 0.2mm of travel that the buttons can hit.  In general the buttons can push the plunger just fine.  However there isn't really anything keeping the buttons in place or restricting them to move in a straight line, they kinda rattle around in holes in the case and can both tilt / jam at an angle and fall out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is probably a solved problem for cell phones / smartwatches / whatever, but I don't know what the standard solutions are.  Could you please let me know what the major different of approaches are and what their pros and cons are?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This would be designed for injection molding in the future; PC/ABS material.  Wall thickness is around 1.2-1.4mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;  I took apart some sports watches; they all have buttons with a metal spring loaded plunger that goes through a O-ring seal and hits a tactile switch.  The total height of the mechanism is about 5mm (from the outside surface of the button to the switch on the PCB).  That's a good mechanism (and waterproof) but I don't have room to fit something like that; and it probably can't be easily made out of just plastic parts.  I'm still looking for a good solution.&lt;/p&gt;&#xA;" OwnerUserId="2335" LastEditorUserId="2335" LastEditDate="2019-08-09T01:36:05.960" LastActivityDate="2019-08-09T23:08:36.633" Title="Push button design for small gadget" Tags="|mechanical-engineering|mechanisms|plastic|molding|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28935" PostTypeId="1" AcceptedAnswerId="28937" CreationDate="2019-06-24T12:11:55.853" Score="-1" ViewCount="49" Body="&lt;p&gt;So I am learning for thermodynamics now and there is one thing that is bugging me.&#xA;In the book it shows a laval nozzle (very simplified) and a diffuser. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OljO2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OljO2.jpg&quot; alt=&quot;laval nozzle&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aOp0T.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aOp0T.jpg&quot; alt=&quot;diffuser&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the formulas:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TkWUi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TkWUi.jpg&quot; alt=&quot;formulas&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(a is the speed of sound)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to those images and its explanation that would mean, if I make a paper cone with a hole in it and blow into it, at the hole speed of sound would be reached. That is obviously not the case. So what am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the beginning it says &quot;Isentropic flows&quot; but how does one actually know that the flow is isentropic? Or is it because of turbulence?&lt;/p&gt;&#xA;" OwnerUserId="10919" LastActivityDate="2019-06-24T15:10:40.587" Title="What happens in a cone-shaped nozzle?" Tags="|thermodynamics|airflow|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28936" PostTypeId="1" CreationDate="2019-06-24T14:03:43.983" Score="0" ViewCount="44" Body="&lt;p&gt;I need to design an electrically powered gas heater, but i need to do so in detail. I'm looking for a guide similar to the shell and tube heat exchanger design guide in &lt;em&gt;coulson and richardson's chemical engineering design volume 6&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="20934" LastActivityDate="2019-06-24T14:03:43.983" Title="Are there any books which show how to design electrical coil heat exchangers?" Tags="|electrical-engineering|chemical-engineering|heating-systems|heat-exchanger|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28937" PostTypeId="2" ParentId="28935" CreationDate="2019-06-24T15:10:40.587" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;According to those images and its explanation that would mean, if I make a paper cone with a hole in it and blow into it, at the hole speed of sound would be reached. That is obviously not the case. So what am I missing?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are not blowing hard enough.  There needs to be enough pressure drop from the big part of the cone to atmospheric to accelerate the air to the speed of sound.  Do the math to figure out how hard to blow, and blow that hard.  (You may want to figure out the bursting strength of your cone.  And your lungs).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-06-24T15:10:40.587" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28938" PostTypeId="2" ParentId="28934" CreationDate="2019-06-24T15:21:31.213" Score="3" Body="&lt;p&gt;You can get switches designed to be soldered onto the edge of a PCB and actuated from the side.  It's your responsibility to make the buttons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get some cheap electronic watches and tear them down, and maybe some broken cell phones, ditto.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learn about this sort of stuff by getting onto a distributor website (DigiKey and Mouser in the US) and looking at what they have available.  If I'm seriously looking, I'll go from there to manufacturer's websites.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-06-24T15:21:31.213" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28939" PostTypeId="2" ParentId="28933" CreationDate="2019-06-24T15:29:45.827" Score="5" Body="&lt;p&gt;As @SamFarjamirad states it's best to consult a knowledgeable construction engineer where you live, or in your region, because such people would know the building codes &amp;amp; construction practices for your region. A column or pillar for a two storey house may not be necessary if the load bearing walls are designed &amp;amp; built properly. This may include internal load bearing wall, inside the house. The other thing to be wary of is the quality of the foundations for the load bearing walls. Also, if the house to be renovated later &lt;strong&gt;all load bearing walls must not be removed or altered&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like the steel is being threaded through the hollow block in the external walls. This would be done to increase the flexibility of the house during earthquakes and should a typhoon/cyclone/hurricane affect the house. Internal load bearing walls should also have such steel reinforcement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If designed and built correctly such steel reinforced walls &lt;strong&gt;should not&lt;/strong&gt; collapse. They might crack during extreme natural events but they &lt;strong&gt;should not&lt;/strong&gt; fall and collapse the house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the picture you included, some things that I found strange were:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Looking at the bricks surrounding the doorway, the bricks on the left&#xA;side of the doorway vary in width and they are not aligned vertically&#xA;very well, so there are gaps of varying sizes between the bricks and&#xA;the door frame. It would be better if the bricks were uniform in size&#xA;and vertically aligned.&lt;/li&gt;&#xA;&lt;li&gt;It might be the way the picture has been taken but the bricks at the&#xA;top of the window holes in the left of picture look like they may be&#xA;sagging. They should be horizontal and supported by a solid lintel.&lt;/li&gt;&#xA;&lt;li&gt;I'm used to seeing bricks laid in an interlocking pattern where the&#xA;bricks that are laid above one layer are slightly offset. This may&#xA;not be possible if the walls are threaded with steel reinforcing. Ask&#xA;a local engineer about this. In the picture, the bricks are laid vertically on top of one another.&lt;/li&gt;&#xA;&lt;li&gt;I looked at the website you provided a link to and the second picture&#xA;showed what appear to be how the roofing iron was attached to the&#xA;house. Just laying the roofing iron on the walls as shown I consider&#xA;to be a primitive form of construction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="90" LastActivityDate="2019-06-24T15:29:45.827" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28940" PostTypeId="2" ParentId="28931" CreationDate="2019-06-24T17:10:06.143" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;but I don't see how the angle variation could be zero&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The angle variation would be zero if the beam is simply supported at each end, because angle variations come from bending, bending comes from moment, and a simple support does not exert a moment on the beam.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't see that called out in the problem statement on the Wikipedia page, but that pretty much has to be it.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-06-24T17:10:06.143" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28941" PostTypeId="1" CreationDate="2019-06-24T18:32:52.637" Score="1" ViewCount="90" Body="&lt;p&gt;I know it's possible for a PVC pipe to develop a staic charge due to fluids flowing through them. This is a twist on that idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a sealed length of pvc (say 3.5 feet in length) - tethered below the surface in a flow of water (river or stream), develop an ESD charge on the inside of the PVC pipe surface? Would the conductivity of the water disipate any potential charge? Can the inside surface develop a different potential than the outside surface?&lt;/p&gt;&#xA;" OwnerUserId="20935" LastActivityDate="2019-06-24T18:32:52.637" Title="Static charge, PVC pipe and water" Tags="|electrical-engineering|fluid-mechanics|materials|hydraulics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28946" PostTypeId="2" ParentId="28921" CreationDate="2019-06-25T13:18:47.550" Score="2" Body="&lt;p&gt;The derivation has a mistake in setting &lt;span class=&quot;math-container&quot;&gt;$FA = \phi n_{fs}/n_{O2s}$&lt;/span&gt;. Here is the correction. For simplicity, I use &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; rather than &lt;span class=&quot;math-container&quot;&gt;$FA$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$m_O$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n_O$&lt;/span&gt; as the mass and moles, respectively, for O&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt;, and a &lt;span class=&quot;math-container&quot;&gt;$\star$&lt;/span&gt; for stoichiometric. I use the term &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; for mass fraction and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; (capital M) for molar mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \phi \equiv \frac{F}{F_s} = \frac{m_f/m_O}{m^\star_f/m^\star_O} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \omega_O \equiv \frac{m_O}{m_f + m_O} = \frac{1}{F + 1} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F = \phi\ F_s = \phi\ \frac{m^\star_f}{m^\star_O} = \phi\ \frac{n^\star_f\ M_f}{n^\star_O\ M_O}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the given reaction, &lt;span class=&quot;math-container&quot;&gt;$n^\star_f = 2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n^\star_O = 1$&lt;/span&gt;. Therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \omega_O = \frac{1}{2\phi M_{f/O} + 1} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$M_{f/O}$&lt;/span&gt; is the ratio of molar mass of fuel (H&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt;) to oxygen O&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expression for &lt;span class=&quot;math-container&quot;&gt;$\omega_f$&lt;/span&gt; is found from a mass balance &lt;span class=&quot;math-container&quot;&gt;$\omega_O + \omega_f = 1$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-06-26T12:03:01.603" LastActivityDate="2019-06-26T12:03:01.603" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28948" PostTypeId="1" CreationDate="2019-06-25T14:08:51.460" Score="0" ViewCount="129" Body="&lt;p&gt;I'm interested in using a pump to inject liquid into a mains water pipe (3-6 bar pressure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know what pump to use as they all seem too big for the task they are carrying out. I want to use a variable speed 12V/24V pump that only needs to inject a max of 2L per hour but is this going to be too small to deal with injecting into a pressurised pipe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could somebody show me an example of the smallest pump you could use for this?&lt;/p&gt;&#xA;" OwnerUserId="20946" LastActivityDate="2019-06-25T14:33:23.600" Title="Choosing a pump to inject liquid into a pressurised water pipe" Tags="|fluid|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28949" PostTypeId="2" ParentId="28948" CreationDate="2019-06-25T14:33:23.600" Score="1" Body="&lt;p&gt;IMO, the wide range of pressures you are talking about rules out a centrifugal pump. 2l/h is really little flow. I would look for a membrane pump, maybe even a dosing pump (here: a membrane pump with variable stroke) as these are often available or adjustable for small volume flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using these search terms (&quot;membrane dosing pump 2l/h 6 bar&quot;) I found &lt;a href=&quot;https://www.sera-web.com/pdf/dosierpumpen/magnetmembranpumpen/produktinformationen/pi-2001_01_magnetmembranpumpen_204_1__en.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this by Sera&lt;/a&gt;. So &lt;em&gt;something&lt;/em&gt; is available, depending on your project maybe not within your budget. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;AFAIK, peristaltic pumps are seldom rated for 6bar but I could be wrong and a search in this direction might yield a cheaper solution.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-06-25T14:33:23.600" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28952" PostTypeId="1" CreationDate="2019-06-26T05:45:24.123" Score="2" ViewCount="173" Body="&lt;p&gt;I am calculating the transmit power for &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; single antenna IoT devices in an Indoor scenario. The receiver has 64 antennas. I am trying to minimize the transmit sum-power, i.e., &lt;span class=&quot;math-container&quot;&gt;$\sum_k p_k$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$p_k$&lt;/span&gt; is the transmit power of &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;-th IoT device. The noise power is -168dBm/Hz. The BW is 180kHz and the transmit frequency is 2.4GHz. I have a minimum requirement for the SNR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upon solving I am getting a transmit power in range of -60 dBm to -70dBm. Is the transmit power feasible considering the noise is too less and the path loss is minimal considering a room size of 10m, i.e, the max distance between transmitter and receiver is 10m?&lt;/p&gt;&#xA;" OwnerUserId="20951" LastEditorUserId="20302" LastEditDate="2019-09-28T02:30:28.683" LastActivityDate="2019-09-28T02:30:28.683" Title="Negative transmit power in dBm" Tags="|electrical-engineering|power|telecommunication|wireless-communication|transmission|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28953" PostTypeId="1" CreationDate="2019-06-26T07:10:56.583" Score="0" ViewCount="123" Body="&lt;p&gt;Using a motor as a generator but  looking to test something &#xA;Why would or would they not be good generators &#xA;thanks &lt;/p&gt;&#xA;" OwnerUserId="20936" LastActivityDate="2019-06-26T07:16:25.627" Title="How effective are pancake motors as generators and why" Tags="|electrical-engineering|motors|electrical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28954" PostTypeId="2" ParentId="28953" CreationDate="2019-06-26T07:16:25.627" Score="3" Body="&lt;p&gt;DC motors make fine generators. You can test them by connecting the shaft of the motor to the chuck of a variable-speed electric drill and connecting the motor wires to a load resistor; you then measure output current and voltage as functions of drill speed. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-06-26T07:16:25.627" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28958" PostTypeId="1" AcceptedAnswerId="29018" CreationDate="2019-06-26T17:14:35.287" Score="1" ViewCount="250" Body="&lt;p&gt;I am trying to create a program which estimates the final temperature of a magnet made of copper wire. I have different configurations in mind, but I will start off with a straight square wire with a circular cooling channel inside of it. Inside runs water which cools it off. The coil is wrapped in some type of electrical insulator. The wire is exposed to air. I am trying to model this using unsteady-state conditions and run it up to a steady state position to calculate the final temperature of the inside (middle if it makes it easy) of the wire, the temperature of the outside of the wire as well as the final temperature of the water coolant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been at this, and coming up with some equations, but I have not taken heat transfer before and am struggling somewhat on this problem. I know that for unsteady state heat transfer &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&amp;part;T/&amp;part;t = &amp;alpha; &amp;nabla;^2(T)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;where the partial of temperature versus time is the laplacian of temperature times a factor alpha. There are some numerical methods to go about doing this and I am able to include a convective heat transfer coefficient, but I am not sure how to add in the rest of the geometries nor how to include a heat generation term. It just seems like there re so many things that have to be simultaneously solved and I am not sure where to go. I am not sure if the textbook I am reading is not sufficient or if I am not seeing where to go in general. Thanks in advance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/wJSoT.jpg&quot; alt=&quot;general sketch&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wire is square with an inner cooling channel. A layer of insulation wraps around the wire which may be neglected for now. I'm trying to reconcile both the convection from inside the wire, the convection outside the wire and the conduction inside the wire. No temperature is fixed which makes this very difficult for me to try to design around.&lt;/p&gt;&#xA;" OwnerUserId="20953" LastEditorUserId="20953" LastEditDate="2019-06-27T18:28:33.600" LastActivityDate="2019-07-02T20:15:09.683" Title="Heat Transfer in Different Wire Configurations" Tags="|mechanical-engineering|fluid-mechanics|heat-transfer|thermal-conduction|convection|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="28961" PostTypeId="1" CreationDate="2019-06-26T21:18:04.530" Score="-1" ViewCount="191" Body="&lt;p&gt;I have seen videos or heard about 3D printed buildings, 3D printed space shuttle parts and more. There seems to be a lot of craze developing around 3D printed manufacturing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have never heard anyone discuss stress analysis of 3D printed parts. I have seen in the past that 3D printed plastic parts were rather fragile. However, it is possible that they were manufactured using the most coarse and cheap settings since they were just models of certain concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since we are moving to manufacture more and more things through 3D printers, has anyone done research into stress analysis on 3D printed mechanical parts?&lt;/p&gt;&#xA;" OwnerUserId="20958" LastActivityDate="2019-06-27T10:05:18.970" Title="What does stress analysis of 3D printed parts reveal vs their normal counterparts?" Tags="|stresses|finite-element-method|manufacturing-engineering|3d-printing|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="28964" PostTypeId="2" ParentId="28961" CreationDate="2019-06-27T04:17:12.777" Score="0" Body="&lt;p&gt;They 3d print metal, concrete, alloys, and even bone from your own bone tissues. &#xA;All have structural constraints that need to be worked at. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example printing metal has temperature issues. Or 3d printing concrete can hardly have reinforcement in it. The try to use microfiber reinforcement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They constantly test these materials under different circumstances and improve the systems. Obviously it's a very promising emerging field.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering centuries of experience that has gone into conventional methods of building things, like forging metals, casting, rolling, or huge construction projects like bridges or multi story buildings, any practical application of 3D printing for complex projects is far into the future. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more realistic goal would be a mix of 3D printing and robotics post processing and assembly.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-06-27T04:17:12.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28966" PostTypeId="1" CreationDate="2019-06-27T06:50:53.840" Score="1" ViewCount="42" Body="&lt;p&gt;&lt;strong&gt;why can't remove the Image frequency using a band-pass filter?&lt;/strong&gt;&lt;br&gt;&#xA;I've been asked to search for the reason why we can't use BPF (band-pass filter) in order to remove the effects of the Image frequency?&#xA;&lt;br&gt;I've searched but couldn't find anything useful about it.&#xA;I would be glad if you could help me.&lt;/p&gt;&#xA;" OwnerUserId="20965" LastActivityDate="2019-06-27T10:16:16.730" Title="why can't remove the Image frequency using a band-pass filter?" Tags="|telecommunication|communication|digital-communication|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28967" PostTypeId="2" ParentId="28961" CreationDate="2019-06-27T10:05:18.970" Score="0" Body="&lt;p&gt;The analysis itself is exactly the same as in case of any other elements.&#xA;Another point can be made on if you want to perform analysis of 3D printed element to see its residual stresses right after the 3D print or if you want to get the information how the element will perform once it is in-service. The first analysis will be highly dependent on the technology of 3D print and settings you have used in addition to the material used. Important part in determining in service behaviour is what are input and output data. Usually, in the input data one needs to consider the way the element has been built and the way the material is oriented to account for highly non-isotropic material properties dependent on the factors I have just mentioned.&lt;/p&gt;&#xA;" OwnerUserId="20967" LastActivityDate="2019-06-27T10:05:18.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28968" PostTypeId="2" ParentId="28966" CreationDate="2019-06-27T10:16:16.730" Score="2" Body="&lt;p&gt;You only know something is an &quot;image frequency&quot; because you know what the signal was like &lt;em&gt;before&lt;/em&gt; you sampled it and reduced its bandwidth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do an experiment, either with real hardware or a computer simulation. Use a sampling rate of say 1000/sec and sample analog sine wave signals at 200 Hz and 800 Hz. The sampled data will be &lt;em&gt;identical&lt;/em&gt; even though you would describe the sampled 800 Hz as an &quot;image frequency&quot;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-06-27T10:16:16.730" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28971" PostTypeId="2" ParentId="16830" CreationDate="2019-06-27T14:22:24.883" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;em&gt;Does the pore size limit, in any way, the surface area available for hydrostatic loading on the back of a retaining wall (I assume it does not)?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Pore size has no bearing on hydrostatic surface area load on a retaining wall. What does change is the structural load-bearing strength of the soil in the water saturated soil. The more water the less strength in the soil. Imagine that where there is saturated soil it behaves exactly like a fluid (water causes mudslides because the water content exceeds the liquid limit of that soil under gravity and explosively collapses like a river down slopes.)&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;em&gt;Soil particles become suspended in water while a hydrostatic load is present: does this (or any other water related factor) affect the overall soil load or does it remain the same?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Soil particles don't become suspended in an hydrostatic load but become inert structurally proportionally to the liquid limit state of that soil relative to the depth of soil. Think of doing a rock in a bucket of water. The surface of water rises volumetricaly. The problems arise when the rock particles become very small and are classed as clay. It is impossible to remove all the water from clay because it is bonded electronically to the soil itself. The capacity for clay to &lt;strong&gt;adsorb&lt;/strong&gt; water is well understood and documented and causes immense damage to buildings worldwide but it also means that the mass of clay behind a retaining wall will be directly influenced by the water content. This water content will also determine the pressure against which a retaining wall must theoretically be designed to withstand in addition to the safety factor relevant to the local building regulations for such structures. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-06-27T14:22:24.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28973" PostTypeId="1" CreationDate="2019-06-27T15:03:00.753" Score="0" ViewCount="32" Body="&lt;p&gt;I derived a non linear differential equation system for a mechanical system. However I doubt the correctness of it.&#xA;So I compared the model's dynamics with a VREP simulation. I did this by&#xA;by Euler integrating the model with the same step size as in the simulation, and then comparing the state variable from the simulation and the  model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if this is a valid approach, and if there are other ways of doing it &lt;/p&gt;&#xA;" OwnerUserId="18808" LastActivityDate="2019-06-27T15:09:27.657" Title="How can I validate the validity of a model of a dynamical system?" Tags="|dynamics|modeling|simulation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28974" PostTypeId="2" ParentId="28973" CreationDate="2019-06-27T15:09:27.657" Score="2" Body="&lt;p&gt;Verify your model against actual physical hardware.  If you're sure of parts of the model, and unsure of others, then just verify the parts you're worried about.  If the physical hardware is big and expensive, then if possible build a simplified physical model that still captures the behavior you're concerned about.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-06-27T15:09:27.657" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28975" PostTypeId="2" ParentId="16877" CreationDate="2019-06-27T15:24:41.877" Score="-1" Body="&lt;p&gt;All soil is rock. Looking at this from a geotechnical perspective rather than agrarian or geologically will clarify why.&lt;/p&gt;&#xA;&lt;p&gt;Rock consists of up to 50% metal and that is mostly aluminium, calcium, silicon and a few other light metals in the earth's lithosphere.  Rock weathers either mechanically or chemically. The processes that cause weathering are wind, rain, thermal variation and kinetic erosion due to gravity and solar energy.&lt;/p&gt;&#xA;&lt;p&gt;Some rocks are almost devoid of metal in their composition because they are mostly carbon, hydrogen and oxygen compounds with a metal atom bond. As rock decomposes due to oxidation its density changes.&lt;/p&gt;&#xA;&lt;p&gt;Igneous rocks create very dense, dark soils and often are highly metallic. Black granite, basalt and gabbro are good examples of such rock. These rocks decompose to metal rich souls which can be very heavy and highly compacted to produce surfaces &amp;quot;as hard as rock&amp;quot; because they are made from rock. The finer the soil particles, the greater the compaction and if treated with chemicals, they may bond together like rock and behave like bedrock structurally. To say compaction is as strong as rock is accurate but soil integrity to resist penetrative pressure requires lateral strength only available from extreme heat that made the parent material to begin with: in excess of 1750°C&lt;/p&gt;&#xA;&lt;p&gt;In summation, so can only achieve the strength of the rock from which it consists less the pore content and frictional factors of the grains. These some may reduce structural load-bearing strength by a much as 60%. When water is added to the mix, then it may reach liquefaction and behave as a fluid instead of a solid at ambient temperatures. However, as permafrost it is another animal. Technically it is theoretically possible but in practice not. 95% compaction is considered safe in engineering applications but any more than that raises economic issues without the mechanical properties being greatly improved.&lt;/p&gt;&#xA;&lt;p&gt;Hope this helps&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2021-03-20T05:39:50.603" LastActivityDate="2021-03-20T05:39:50.603" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28976" PostTypeId="2" ParentId="24862" CreationDate="2019-06-27T15:54:12.407" Score="1" Body="&lt;p&gt;I would use a Shear Box test for this. Stress testing is for maximum load-bearing capacity...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shearing is what gives you the modulated graphic results you seek. Have a look at Smith's Soil Mechanics 9th ed for more details. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-06-27T15:54:12.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28977" PostTypeId="2" ParentId="15761" CreationDate="2019-06-27T16:00:10.480" Score="0" Body="&lt;p&gt;&lt;em&gt;One more question I was thinking about is How much does the strength of the concrete get affected in a pile (due to the presence of water and some unwanted soil in it) ?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strength of concrete improves with water. The crystalline lattice formation rate in concrete is a result of the chemical reaction of cement and water. I would not be too concerned with groundwater water while putting concrete in a pile as it will not harm the overall  concrete ultimate strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get advice from a geotechnical engineer for proper treatment. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-06-27T16:00:10.480" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28978" PostTypeId="2" ParentId="10199" CreationDate="2019-06-27T16:06:36.900" Score="0" Body="&lt;p&gt;Water is incompressible. Remember this fact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the water content is exceeded you reduce soil structural strength and therefore its load-bearing capacity. As you try compact a drenched soil you are trying to compress the incompressible. Once the liquid limit is reached you cannot compress the soil any further. To compact soil you need to effectively remove whatever is in the soil void spaces. When water eventually drains there will never be a time when there is no water in the compacted soil. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-06-27T16:06:36.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28979" PostTypeId="1" CreationDate="2019-06-27T16:25:45.417" Score="1" ViewCount="24" Body="&lt;p&gt;I am in the process of designing a chiller test rig, of which a very simplified diagram would be (the quick coupling might just be replaced by a non-return valve):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qXkFs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qXkFs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am struggling to find the type of valve I need where the &quot;??&quot; is. The purpose of this valve is to create some load on the chiller and simulate having a real system connected to it. In other words, I (think I) want some resistance (flow restriction/increased pressure?) at that point in the circuit. Can you advise what kind of valve I would need for this (something manual ideally, not electrically actuated).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fluid is a mix of ethylene glycol and water at 0°C. The chiller is set up to run at ~25psi at the outlet and the liquid flow rate is ~6 l/min.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-06-27T16:25:45.417" Title="Valve advice for chiller test rig" Tags="|pressure|valves|flow-control|liquid|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28980" PostTypeId="1" CreationDate="2019-06-27T18:26:35.137" Score="0" ViewCount="33" Body="&lt;p&gt;I am trying to collect signals from preamplifiers (Aquarian Audio preamps, PA-4) and have them sent to a laptop nearby wirelessly, I have not found a dedicated device for this process, and the preamplifiers have BNC connector. I need suggestions and this is the specs for the pre-amplifiers that I am trying to collect the data from &lt;a href=&quot;http://www.aquarianaudio.com/pa4.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.aquarianaudio.com/pa4.html&lt;/a&gt;. If anyone has been into this similar situation, I need your advice. &lt;/p&gt;&#xA;" OwnerUserId="20972" LastEditorUserId="110" LastEditDate="2019-06-27T21:37:07.607" LastActivityDate="2019-06-27T21:37:28.227" Title="How can I transmit signals from pre-amplifiers to a laptop nearby using bluetooth?" Tags="|electrical-engineering|computer-engineering|computer-hardware|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28981" PostTypeId="2" ParentId="28980" CreationDate="2019-06-27T21:37:28.227" Score="1" Body="&lt;p&gt;The output of the preamp is analog.  The computer accepts digital. You will need to use an analog to digital converter before trying to send the data via blue tooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without knowing anything about it besides what it says on the web site, NewEgg has an &lt;a href=&quot;https://www.newegg.com/p/N82E16812117304?Description=adc&amp;amp;cm_re=adc-_-12-117-304-_-Product&quot; rel=&quot;nofollow noreferrer&quot;&gt;ADC&lt;/a&gt; that might be what you need.  The output of the ADC is not something you can plug into your computer so you will need something like &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B01KO1JNCA&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; from Amazon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get from the BNC output of your preamp to the ADC, you will need a BNC to RCA adapter, something like &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B078481ZDJ&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; from Amazon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I think you would do better with something like &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B000KW2YEI&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; device from Amazon that (the web page says) will accept analog input and run it to your computer via USB.  Sorry, you will be tripping over cables but it's a much simpler solution.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2019-06-27T21:37:28.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28982" PostTypeId="1" CreationDate="2019-06-28T08:37:42.943" Score="3" ViewCount="64" Body="&lt;p&gt;I've recently visited an old (about 60 years old) micro hydroelectric standalone installation. Total electric power is about 3 kW A/C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A local told me between generator + fly-wheel and turbine there is a &lt;strong&gt;clutch&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question is: what is the role of a such mechanical part in this system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;NB: local was not an expert.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="2157" LastActivityDate="2019-06-28T13:48:01.990" Title="Clutch role in a micro hydropower" Tags="|mechanical|energy|renewable-energy|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28983" PostTypeId="1" CreationDate="2019-06-28T09:23:28.847" Score="1" ViewCount="46" Body="&lt;p&gt;This is a project regarding linear actuators for a 4 Wheeled Robot, with spider like hinged legs. The first question on my mind was to figure out is if a hydraulic system has any weight to Force(strength) advantage on this load category of max 6 kN to an electric-mechanical system with a brush-less motor. Since its a mobile unit and has to be water and dirt resistant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also most important question is what Force class does the hydraulics start to outshine the feather weight electric actuators?&#xA;&lt;strong&gt;Spider Robot needs to operate within the parameter:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;dynamic load of max 6 kN&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;operated with 12 V battery pack.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;quick actuation speeds at least 6 mm/s&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;brush-less DC motor 1000 W (&lt;strong&gt;with Feedback&lt;/strong&gt;) preferred in electric or hydraulic systems for efficiency.&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thank you for any advice or for sharing your experiences.&lt;/p&gt;&#xA;" OwnerUserId="8677" LastEditorUserId="8677" LastEditDate="2019-06-29T09:03:04.803" LastActivityDate="2019-06-29T09:03:04.803" Title="Deciding between Electric or Hydraulic Actuators for a Dynamic Load of around 4 kN to 6 kN" Tags="|mechanical-engineering|electrical-engineering|hydraulics|robotics|actuator|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28984" PostTypeId="1" CreationDate="2019-06-28T12:04:43.930" Score="5" ViewCount="9432" Body="&lt;p&gt;I opened a new sheet in Solidworks drawing, in the title block, there was a field written &quot;Do not scale drawing&quot;. What does that mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a scale already being chosen over there. It was just a new sheet and the default title box appeared. I searched the internet but could not find any simple straight description about that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1dUPM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1dUPM.png&quot; alt=&quot;do not scale drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14893" LastEditorUserId="14893" LastEditDate="2019-07-02T20:41:05.243" LastActivityDate="2019-07-06T02:04:23.367" Title="What does &quot;do not scale drawing&quot; in a drawing sheet mean?" Tags="|solidworks|technical-drawing|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="28985" PostTypeId="2" ParentId="28984" CreationDate="2019-06-28T12:34:41.470" Score="6" Body="&lt;p&gt;This is information to be shown on the printed copy - it is telling the reader to please not print this A1 drawing sheet on A4 paper, or else the information you have highlighted &quot;Scale 1:8&quot; would not be true any more.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-06-28T12:34:41.470" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="28986" PostTypeId="2" ParentId="28982" CreationDate="2019-06-28T13:48:01.990" Score="4" Body="&lt;p&gt;The clutch provides an instant disconnect either for scheduled maintenance or in an emergency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the water flow is stopped instantly, then the energy of that moving pipeful of water has to be dissipated and can cause a lot of damage. Some systems are fitted with surge pipes for exactly that reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On one system I worked with, slowing the flow from max to zero without causing the surge pipe to overflow took 30 seconds and at 30 litres a second that adds up...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-28T13:48:01.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28987" PostTypeId="1" CreationDate="2019-06-28T16:25:49.530" Score="1" ViewCount="259" Body="&lt;p&gt;first post&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I designed a 2000 watts 1000 degrees Celsius furnace with following chamber size: 11 cm length, 12 cm width and 15 cm depth. The refractory that I use is 1880 C high temp tabular alumina refractory with 94 percent al2o3, 5 percent sio2, 0.2 percent fe2o3 and 0.6 percent cao. The heating element is 1.2mm in diameter with 2000 watts and is planted inside the refractory material with 3mm offset from inside of the chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;everything is OK so far but something is confusing me, here are the questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- The furnace heat-rate is so slow (about 10 degrees in minutes). Is this problem is related to refractory? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- What kind of refractory should I use for best performance and longer life cycle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;any kind of help would be very useful.&lt;/p&gt;&#xA;" OwnerUserId="20973" LastActivityDate="2019-06-28T16:25:49.530" Title="best refractory cement for embedded heating element furnace" Tags="|heating-systems|electrical|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28988" PostTypeId="2" ParentId="22087" CreationDate="2019-06-28T18:08:20.900" Score="2" Body="&lt;p&gt;In this type, the rated power is determined in such a way that the outdoor unit is not in any way protected by shadows or shields; That is, the outer plate casing essentially casts a shadow on the condenser, i.e. the heat sink. If the system is completely enclosed, ie well-installed and well-fitted, it will not leak for many decades. This, of course, depends on the quality of the copper tube used, the quality of the flange on the tubes. Of course, it increases the lifetime of the climate if it is in a place where it is more protected. Generally speaking, if the outdoor unit is in a sheltered place, it will dust sooner. With a hand-held sprayer, it is advisable to wash more frequently with clean water from the lamellae of the outdoor unit. This keeps the factory performance. If you are in shade and in a sheltered place, less energy is invested in the operation, but of course this is very small overall. Regular cleaning is more important. I noticed that the pipes running outdoors do not have any protection. In Europe, it is standard that these pipes should be routed in an outdoor plastic air channel, because their outdoor insulation will eventually dissipate from the sun's UV radiation. If this happens, it is advisable to put in a new tube. Furthermore, the pipe is more aesthetic when running in a plastic channel. Be careful to drop the pipeline at all points towards the outdoor unit. Many people forget, but the refrigerants in them are no longer as perfectly blended as the banned ozone damaging R22. (Mixing of oil and refrigerant). For this reason, not every point of the system is lubricated (low oil consumption) or even an oil plug may develop. Therefore, they are instructed by thermal engineering engineers to make the piping slightly sloping (ie not a rough slope). Of course there is also a horizontal section if there is no other solution. However, very little climb in the tube is forbidden, especially for inverters with low power. If you pay attention to this, you will be in good friendship with this device for long years. Durable, high quality device. The indoor unit, however, often needs to be cleaned with a fungicidal composition, such as alcohol-containing hospital and solarium disinfectants. Use of chlorine-containing agents is strictly prohibited! Especially, the blower of the indoor unit needs to be cleaned with this alcoholic substance, because in this type of black fungus is highly prone to settling.Greetings from Hungary.&lt;/p&gt;&#xA;" OwnerUserId="20974" LastActivityDate="2019-06-28T18:08:20.900" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28989" PostTypeId="2" ParentId="26752" CreationDate="2019-06-28T19:43:01.913" Score="1" Body="&lt;p&gt;Looks like you are looking at a water tower under construction.  There tends to be three main types of water tanks for water distribution systems:  Volume, Pressure and Combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first is for mainly for volume and it basically the equivalent of a reservoir.  These tend to be built at ground level, and have a large uniform radius from bottom to top.  This gives a large volume without much pressure due to ground proximity.  Though if built on a nearby hill could provide a decent pressure to the system below.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sE42W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sE42W.png&quot; alt=&quot;High Volume&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second tends to be for pressure.  The main matter for supplying pressure pressure is simply height above ground.  These type of water towers tend to have narrow central shaft and a wide short top.  The pipe running up the centre provides a connection between the water distribution system and a small reservoir at the top of the tower.  The reservoir at the top of the tower allows for slight draw downs due to water use without dramatically affecting the height of the water column and thus the pressure in the water system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OppxG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OppxG.png&quot; alt=&quot;Pressure Tower&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Third type is a combination pressure and storage water tower.  These tend to have a a cylindrical shaft smaller than a pure storage water tower and larger than a pure pressure water tower.  The tower tends to maintain the radius for the full height of the tower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IEYRz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IEYRz.png&quot; alt=&quot;Combination Tower&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suspect that what you are witnessing is the construction of a pressure water tower.  I am guessing they are building the reservoir at ground level where its easier to work on.  Also less damaging should something go wrong and it falls.  They will jack/hoist the reservoir up as as the central shaft is built under neath is to support it.  In the end it might look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rw8x3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rw8x3.png&quot; alt=&quot;Water tower final&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6993" LastEditorUserId="6993" LastEditDate="2019-06-29T16:40:05.127" LastActivityDate="2019-06-29T16:40:05.127" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28990" PostTypeId="1" AcceptedAnswerId="28994" CreationDate="2019-06-29T04:55:39.220" Score="1" ViewCount="91" Body="&lt;p&gt;Is there any material that will absorb very low frequency radio light? I know that stealth aircraft use radar absorbant materials in their design that absorb radio light best at certain frequencies, but I cannot find any information on materials that are explicitly stated to absorb very low frequency radio light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not looking for any aspect of geometry such as faraday cages - only materials.&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2019-07-30T01:00:55.117" Title="Material to absorb very low frequency radio" Tags="|materials|telecommunication|radio|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="28992" PostTypeId="1" CreationDate="2019-06-29T20:46:01.983" Score="-1" ViewCount="6952" Body="&lt;p&gt;Here is my attempt but I am not getting the right answer. Can someone tell me what I am doing wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/OkucW.jpg&quot; alt=&quot;my attempt at derivation&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="20980" LastEditorUserId="1832" LastEditDate="2019-06-29T21:33:31.077" LastActivityDate="2019-08-01T09:02:20.297" Title="How to derive fixed-end moment of a beam?" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28993" PostTypeId="1" AcceptedAnswerId="28997" CreationDate="2019-06-29T23:44:12.790" Score="1" ViewCount="786" Body="&lt;p&gt;I'm working on a balancing robot. I'm familiar with the simple PID control loop but i can't understand this PID control loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pIB0b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pIB0b.png&quot; alt=&quot;pdpi&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is this actually a cascade control loop? The programming states that the angle control PD loop is executed every 5ms, the speed control PI loop is executed every 40ms and the steering PD loop every 20ms. From my understanding, based on the loop/function execution timing, the innermost loop is the angle control loop, then the steering loop and the outermost loop is the speed control loop. Correct?&lt;/li&gt;&#xA;&lt;li&gt;Why is this complex control loop better than a classic control loop with P, I and D in one single controller?&lt;/li&gt;&#xA;&lt;li&gt;How many control loops can be nested within each other? I have only seen one inner loop and one outer loop. Is it possible to have 2 inner loops and one outer loop? Or 3 loops, each nested within the other?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21627" LastEditorUserId="21627" LastEditDate="2019-06-30T00:04:51.487" LastActivityDate="2019-07-05T08:14:19.407" Title="Is this a PID cascade control loop?" Tags="|control-engineering|pid-control|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28994" PostTypeId="2" ParentId="28990" CreationDate="2019-06-30T00:57:17.057" Score="1" Body="&lt;p&gt;a chunk of material will absorb electromagnetic radiation if 1) its characteristic impedance is close to that of a vacuum (to minimize reflections) and 2) if its impedance contains the right amount of ohmic resistance, so as to dissipate the radiation into heat and not re-radiate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that said, to interact significantly with EM radiation of any wavelength requires (ideally) that the size of the interacting object be a significant fraction of the wavelength of that radiation. This means that to absorb centimeter-range radar requires something like a centimeter of thickness to the absorbing layer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different rules apply if you are interested in just reflecting away the incident EM waves. In this case, a faraday cage will work.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-06-30T00:57:17.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28995" PostTypeId="1" AcceptedAnswerId="29016" CreationDate="2019-06-30T03:46:09.383" Score="-1" ViewCount="534" Body="&lt;p&gt;So, I was trying to make something like this &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HJpst.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HJpst.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, While I have the motor, the motor has a vertical shaft and I need something to connect a rod to it in a T shape&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this even possible? And if yes, then what should I get?&lt;/p&gt;&#xA;" OwnerUserId="20983" LastEditorUserId="110" LastEditDate="2019-07-02T01:07:29.650" LastActivityDate="2019-07-02T12:18:47.130" Title="Can a rod be attached perpendicular to the motor shaft" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28997" PostTypeId="2" ParentId="28993" CreationDate="2019-06-30T05:52:51.733" Score="5" Body="&lt;p&gt;I would not call this a cascaded control loop but rather a MIMO control loop. Typically a cascaded control loop has an inner and an outer loop and the output of the outer loop is used as the setpoint for the inner loop. Here you have multiple loops, for some reason running at different update rates, but each with their own setpoint and (presumably) sensors. Therefore I would call this a MIMO control loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your other questions; the above configuration is not necessarily better than a single PID controller. For example, due to the possible interaction between the loops, it is typically more difficult to evaluate robust stability in practice (i.e. based on measurements). On the other hand, multiple controllers allow for more flexibility and/or might even be needed in order to be able to control all performance variables of interest. Controlling angle, speed and turn likely requires multiple sensors, hence multiple inputs and controllers are required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can of course nest as many loops as you want and/or deem necessary. The only real limitation is the ability/time required to analyze the resulting complex interaction. &lt;/p&gt;&#xA;" OwnerUserId="11869" LastActivityDate="2019-06-30T05:52:51.733" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="28998" PostTypeId="1" CreationDate="2019-06-30T07:22:35.317" Score="1" ViewCount="732" Body="&lt;p&gt;I came across with &lt;a href=&quot;https://groups.google.com/forum/#!topic/rec.autos.tech/XUhXh0SxXGI&quot; rel=&quot;nofollow noreferrer&quot;&gt;this discussion&lt;/a&gt; about this topic, however it doesn't address the answer clearly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the safety role of center hole (&lt;code&gt;A&lt;/code&gt;) on the vehicle wheels? Is it supposed to carry the weight on the wheel and take the stress over from the wheel lugs?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, is it there only to help on assembly of the wheel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://user-images.githubusercontent.com/6639874/60393560-0bc17f80-9b20-11e9-8a30-705bda8e9eae.png&quot; alt=&quot;image&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="20275" LastActivityDate="2019-06-30T07:28:50.177" Title="Is the center hole important on car wheels?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="28999" PostTypeId="2" ParentId="28998" CreationDate="2019-06-30T07:28:50.177" Score="1" Body="&lt;p&gt;It depends on the design of the hub and wheel assembly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some wheel / hub assemblies use a controlled fit to locate the wheel central as the nuts do not have cones. This is usual for aluminium or alloy wheels as the cones of the lugs or nuts will compress and damage the soft metal of the wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steel wheels may not use the central register as the cones on the lug nuts and the wheels are designed to centralize the wheel on the hub.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-06-30T07:28:50.177" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29000" PostTypeId="1" AcceptedAnswerId="29001" CreationDate="2019-06-30T20:54:41.780" Score="1" ViewCount="82" Body="&lt;p&gt;Some tight-coiled springs are made so that they are already under tension - how are they manufactured?&lt;/p&gt;&#xA;" OwnerUserId="20990" LastEditorUserId="-1" LastEditDate="2020-04-25T12:31:54.923" LastActivityDate="2020-04-25T12:31:54.923" Title="How are tight coiled springs made" Tags="|springs|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29001" PostTypeId="2" ParentId="29000" CreationDate="2019-07-01T08:17:44.497" Score="2" Body="&lt;p&gt;I believe you are referring to Extension Springs, which have a certain degree of pre-tension that holds the coils against each other until a certain force has been applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the spring is wound, the wire is under tension. Imagine wrapping a some elastic around your finger a few times while it is stretched, there will be an inwards force, the diameter wants to be smaller, it wants to crush your finger. So too with the spring - the wire wants to relax, and become shorter, but it can't - the other spring coils are in the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a spring, hold it at opposite ends, and twist - you will see that it naturally wants to get longer and fatter, or shorter and narrower depending on which way you twist. With an extension spring, it can't get any shorter, so the initial tension from winding remains.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-07-01T08:17:44.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29002" PostTypeId="1" AcceptedAnswerId="29003" CreationDate="2019-07-01T10:44:12.113" Score="0" ViewCount="166" Body="&lt;p&gt;In the case of a static calculation of a structural model, one can verify the model by checking the applied and reacting forces to be equal zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to adapt the same principle for a structural model with a response spectrum as an input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it correct to do the following:&#xA;1) perform a frequency extraction procedure to obtain 90% of the eigenmodes contributing to the mass and rotational inertia.&#xA;2) calculate the mass of structure being excited and multiply it by the peak acceleration amplitude given in the response spectrum, eg 20g. &#xA;3) its not clear what this step should be... is there something I can do with the modal participation factors to help me calculate the total reaction load?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions will be welcome.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards&lt;/p&gt;&#xA;" OwnerUserId="14216" LastActivityDate="2019-07-01T14:46:02.807" Title="How do I verify the reaction forces of an FE model due to a response spectrum" Tags="|frequency-response|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29003" PostTypeId="2" ParentId="29002" CreationDate="2019-07-01T14:12:42.357" Score="1" Body="&lt;p&gt;In response spectrum analysis of MDOF structures, the most common methods of modal combination are: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;absolute sum (assume the peak response of all modes occurs simultaneously)&lt;/li&gt;&#xA;&lt;li&gt;SRSS (square root of the sum of the squares)&lt;/li&gt;&#xA;&lt;li&gt;CQC (complete quadratic combination) &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If what you're aiming for is a pencil-and-paper check of the software output, the first two methods are most appropriate, and the absolute sum method should be conservative (sometimes overly conservative). CQC is more rigorous, and can give better solutions for structures with closely spaced modes, but it's best suited to a computer. It's worth noting that many off-the-shelf analysis packages will allow the user to choose a modal combination method when setting up the response spectrum analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if eigenanalysis has supplied the frequency and mass participation for each mode, we can pull the peak acceleration for each mode from the acceleration response spectrum curve. Then, mass * acceleration gives the the base shear for each mode, and we find total base shear with our chosen modal combination method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for how many modes to consider, the standard of practice I'm familiar with (United States - bridges) is to include sufficient modes to capture at least 90% mass participation (not 90% of the modes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that for seismic design purposes, you may need to take one final step and consider some combination of the longitudinal and transverse responses. The pertinent design code should specify.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2019-07-01T14:46:02.807" LastActivityDate="2019-07-01T14:46:02.807" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29004" PostTypeId="1" CreationDate="2019-07-01T14:56:40.940" Score="0" ViewCount="40" Body="&lt;p&gt;My project is a DIY talkbox using Monacor KU-516 compression driver. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a hard time finding the right connector to fit the hose to the driver, M35 is a rare size it seems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's one option of M35 x 1.5 looking through the tables for metric-thread sizes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no information on TPI threads per inch on the datasheet for the driver. I read a tutorial from another builder, he 3D printed a part with 18TPI and didn't fit perfectly.      &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone knows any item I could buy to solve without 3D printing a custom please let me know. (I've searched tap/hose adapters with no luck) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I have to 3D print a part how should I measure the TPI accurately to make the correct thread?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming the 18 TPI fitting didn't work as 25.4 (1 inch) / 18 = 1.41 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the flange is M35 x 1.5, at 18TPI would the fitting not snug enough? &lt;/p&gt;&#xA;" OwnerUserId="20992" LastActivityDate="2019-07-01T14:56:40.940" Title="Help finding connector to fit 1 3/8in / 35mm mounting flange on horn driver" Tags="|design|computer-aided-design|connections|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29005" PostTypeId="1" AcceptedAnswerId="29047" CreationDate="2019-07-01T17:38:58.450" Score="4" ViewCount="9951" Body="&lt;p&gt;Consider a four-wheeled cart that has swiveling casters for the two front wheels, and has wheels that don't swivel in back.  For example, a common shopping cart, or trolley if you prefer.  If such a cart is pushed and let go, it will continue to roll in the same general direction, and so I'll call it stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, if the same cart is turned around, pushed, and let go, then the cart won't continue in the same direction for long; it has a tendency to spin around, so I'll call it unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is a cart with swiveling casters in front stable, but is unstable when the same cart is turned around?&lt;/p&gt;&#xA;" OwnerUserId="20993" LastActivityDate="2019-07-13T17:32:26.457" Title="Why is a cart with swiveling casters in front stable, but not if the swiveling casters are in back?" Tags="|mechanical-engineering|wheels|stability|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29007" PostTypeId="1" CreationDate="2019-07-01T23:24:05.460" Score="0" ViewCount="428" Body="&lt;p&gt;I am learning Control Theory and i have a confusion about whether the following PID controllers are interchangeable or similar. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Are they equivalent? Why?&lt;/li&gt;&#xA;&lt;li&gt;Is there any difference in how i would implement both of them in code? I have to program this in Arduino.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here is the first control loop which is the normal definition of a PID controller:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sjk9O.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sjk9O.png&quot; alt=&quot;pid1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is the changed one with the negative feedback changed to positive and the setpoint changed from positive to negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i6kIN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i6kIN.png&quot; alt=&quot;pid2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21627" LastActivityDate="2019-07-02T01:56:23.823" Title="Can i switch negative feedback to positive feedback in PID control?" Tags="|control-engineering|pid-control|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29009" PostTypeId="1" CreationDate="2019-07-02T01:37:41.847" Score="1" ViewCount="173" Body="&lt;p&gt;I'm a noob if it comes to electronic components but I'm trying to achieve the following and I would really appreciate some suggestions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following requirement:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When a circuit is closed the first time (via some sort of pulse switch), an LED will receive power.&lt;/li&gt;&#xA;&lt;li&gt;After the circuit is closed the first time, it will be open again for a long duration of time.&lt;/li&gt;&#xA;&lt;li&gt;When a circuit is closed the second time, LED should not receive any power.&lt;/li&gt;&#xA;&lt;li&gt;Loop continues 1-3&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Basically I want a pulse of electricity to toggle circuit on every other time. &#xA;Note that during the off state, the circuit will have no power at all. So any solutions involving 555 timers etc won't work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have bee reading about Dual Coil Bistable relays and I was thinking to use it in conjunction with some logic gates to make this happen, but I feel like I'm going a little bit too far with this and maybe this is simpler than I make it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="20996" LastActivityDate="2024-03-09T11:00:40.237" Title="Persistent Bistate Toggle Switch" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29010" PostTypeId="2" ParentId="29007" CreationDate="2019-07-02T01:56:23.823" Score="1" Body="&lt;p&gt;No, you can't change that willy-nilly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second diagram is perfectly valid, however, &lt;em&gt;if&lt;/em&gt; the plant tends to operate in reverse from normal.  If you have a plant whose output gets more negative when its input goes positive then you'd use the second one.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-07-02T01:56:23.823" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29012" PostTypeId="2" ParentId="28992" CreationDate="2019-07-02T08:14:21.397" Score="1" Body="&lt;p&gt;your calculations of reaction forces &lt;span class=&quot;math-container&quot;&gt;$F_A\ and\ F_B \  $&lt;/span&gt; are wrong. This beam is indeterminate and you are handling it as a simply supported determinate beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way of finding the fixed end reactions is to select the two fixed end moments as two degrees of redundancy and solve the beam. we will refer to this diagram at the bottom of the page.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$R_A = \frac{Pb}{L}+ \frac{M_a}{L} -\frac{M_b}{L}\\\ RB = \frac{Pa}{L}-\frac{M_A}{L}+\frac{M_S}{L}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force displacement relations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ (\theta A)_1 = \frac{Pab(L + b) }{6LEI}\quad$&lt;/span&gt;and &lt;span class=&quot;math-container&quot;&gt;$\ (\theta B)_1=\frac{Pab(L + a) }{6LEI}  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ (\theta A)_2=\frac{M_AL}{3EI}( \theta B )_2=\frac{M_AL }{6EI}  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$( \theta A )_3=\frac{M_BL}{6EI}(\theta B  )_3=\frac{M_BL }{3EI}  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compatibilty equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\theta A = (\theta A)_1 - (\theta A)_2 - (\theta A)_3 = 0  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \theta B = (\theta B)_1 - (\theta B)_2 - (\theta B)_3 = 0 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \therefore \frac{M_AL}{3EI} +\frac{M_BL}{6EI}= \frac{Pab(L + b) }{6LEI}\\ \frac{M_AL}{6EI }+\frac{M_BL}{3EI }= \frac{Pab(L + a) }{6LEI} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving these equations, we obtain &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ M_A =\frac{Pab^2}{L^2} \quad M_B= \frac{Pa^2&#xA;b}{L^2} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the reactions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ R_A =\frac{Pb^2 }{L^3}(L + 2a ) \quad R_B=\frac{Pa^2}{L^3}(L + 2b ) $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/29s9w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/29s9w.jpg&quot; alt=&quot;diagram&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-02T08:14:21.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29014" PostTypeId="1" CreationDate="2019-07-02T12:00:15.800" Score="1" ViewCount="24" Body="&lt;p&gt;I am fabricating &lt;strong&gt;Bottom gate Thin-film-Transistors&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;Gate&lt;/strong&gt; is 60nm Molibdenum. The &lt;strong&gt;dielectric&lt;/strong&gt; structure changes with the table below. The temperature represent the annealing temperature after having deposited the IGZO &lt;strong&gt;semiconductor&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;sample  Structure temperature&#xA;1       SiTa      300&#xA;2       SiTa      180&#xA;3       SiTaSi    300&#xA;4       SiTaSi    180&#xA;5       TaSi      300&#xA;6       TaSi      180&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;after heating the samples, I am getting these bubble-like formation on the gate structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;what could be causing this issue? &#xA;Prebake SiTa-300&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ifLP2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ifLP2.png&quot; alt=&quot;prebake&quot;&gt;&lt;/a&gt;&#xA;Postbake SiTa-300&#xA;&lt;a href=&quot;https://i.stack.imgur.com/D1XDS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D1XDS.png&quot; alt=&quot;Postbake&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prebake SiTa-180&#xA;&lt;a href=&quot;https://i.stack.imgur.com/B7cyX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B7cyX.png&quot; alt=&quot;Prebake&quot;&gt;&lt;/a&gt;&#xA;Postbake SiTa-180&#xA;&lt;a href=&quot;https://i.stack.imgur.com/xuYRo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xuYRo.png&quot; alt=&quot;Postbake&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;prebake SiTaSi-300&#xA;&lt;a href=&quot;https://i.stack.imgur.com/63B6q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/63B6q.png&quot; alt=&quot;prebake&quot;&gt;&lt;/a&gt;&#xA;postbake SiTaSi-300&#xA;&lt;a href=&quot;https://i.stack.imgur.com/el3Ov.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/el3Ov.png&quot; alt=&quot;Postbake&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prebake SiTaSi-180&#xA;&lt;a href=&quot;https://i.stack.imgur.com/52Jlo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/52Jlo.png&quot; alt=&quot;Prebake&quot;&gt;&lt;/a&gt;&#xA;Postbake SiTaSi-180&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tZb6w.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tZb6w.png&quot; alt=&quot;Posbake&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20437" LastEditorUserId="20437" LastEditDate="2019-07-02T12:07:13.237" LastActivityDate="2019-07-02T12:07:13.237" Title="&quot;bubbles&quot; forming after heating my Thin film transistors" Tags="|materials|thin-films|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29015" PostTypeId="2" ParentId="29009" CreationDate="2019-07-02T12:01:18.970" Score="0" Body="&lt;p&gt;Good news, this is &lt;em&gt;much&lt;/em&gt; simpler than you're making it out to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You simply want an On-Off-On-Off alternating Toggle Switch, like this one: &lt;a href=&quot;https://www.adafruit.com/product/1684&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.adafruit.com/product/1684&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connect your LED to only one of the outputs, and away you go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I just spotted that you want to trigger this with a pulse of electricity, not a button push (I misread 'pulse switch' as 'push switch').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that you want this to be reliable when there is no power at all in the system, I'd still use a mechanical toggle system like this - simply set up a small solenoid actuator to push the switch when you send the appropriate pulse of electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are using a power-down of the whole system as the 'off' state, you may be able to go with a simple On-On Alternating toggle, or SPDT switch, although this will flash the LED for a short time at the start of point 3. while the solenoid switches the LED off.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2019-07-02T12:08:48.267" LastActivityDate="2019-07-02T12:08:48.267" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29016" PostTypeId="2" ParentId="28995" CreationDate="2019-07-02T12:18:47.130" Score="1" Body="&lt;p&gt;Yes, you could design a &quot;top hat&quot; that sits on the motor shaft and is secured with a couple of grub screws at 90°, with a hole for the rod to go through (the rod can also be secured with a couple of grub screws or if the rod is threaded, use a threaded hole instead).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example with a 6mm diameter motor shaft and a 6mm diameter rod:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8vXV0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8vXV0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A detailed drawing of the part in question would look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iA0Gl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iA0Gl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-07-02T12:18:47.130" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29017" PostTypeId="1" CreationDate="2019-07-02T16:20:41.643" Score="1" ViewCount="209" Body="&lt;p&gt;I’m doing research on the sliding of 2 fracture surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do that, I’m trying to understand the Small-sliding deformable interaction in the Abaqus (2018), and I use the Node-to-surface discretisation method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, I’m interested in how exactly the ‘slip’ amount is calculated and imposed to the Abaqus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understood the derivation, that is presented in the Abaqus Theory Manual (6.11),  where the calculation of an ‘overclosure’ and a ‘slip’ is shown:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iejav.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iejav.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To check how ‘slip’ is calculated I created in Abaqus a very simple example with the slip upwards of the LHS body and downwards of the RHS body: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ycp4h.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ycp4h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the problem I have is the slip in Abaqus is quite different from the one I understand. If we analyse the lower contacting cells with nodes 13 and 168 (see picture above), then using the logic in the derivation we can decompose it to this schematics: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PL9mP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PL9mP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in the derivation, we have slave node13 penetrating RHS body which becomes node Xnplus1. Then we draw the anchor node ‘xo’ that coincides with the normal direction from the surface of RHS body to node13. Consequently, xi node is drawn to the normal from the Xnplus1 to the face. &#xA;Also, unplus1 = xnplus1 as in (5.1.1-4) as in the manual. I understand that the slip is then the distance from xi to xo in the vertical ‘y’ direction, i.e. slip_y = 0.352. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, Abaqus gives me slip 0.28 all various configurations e.g. small or finite sliding, instantaneous load variation, fixed 1 increment size, surface-to-surface, and all other options are similar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’d be glad if you please could help me to understand this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sincerely, &#xA;AMAN&lt;/p&gt;&#xA;" OwnerUserId="21003" LastEditorUserId="17070" LastEditDate="2019-08-25T06:22:52.163" LastActivityDate="2019-08-25T06:22:52.163" Title="How is the slip calculated in Abaqus small-sliding interaction?" Tags="|applied-mechanics|simulation|solid-mechanics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29018" PostTypeId="2" ParentId="28958" CreationDate="2019-07-02T17:10:58.117" Score="0" Body="&lt;h1&gt;Foundations&lt;/h1&gt;&#xA;&lt;p&gt;The heat transfer problem has three considerations. First is steady state versus unsteady state. The former case is far easier and is all that I will consider here. Second is the temperature profiles axial in &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; and radial in &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;. The best approach is to presume that separation of variables will work. Finally, the last consideration is the influence of heat transfer internally and externally to the cylinder while generating uniform power in the shell &lt;span class=&quot;math-container&quot;&gt;$P = i^2R$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Also, for simplicity, neglect the (electrical) insulation layer on the outside of the wire.&lt;/p&gt;&#xA;&lt;p&gt;The starting point is to build each of the cases one step at a time.&lt;/p&gt;&#xA;&lt;h1&gt;Axial versus Radial Gradients - Biot Number Estimates&lt;/h1&gt;&#xA;&lt;p&gt;In the ideal case, we allow for separation of variables to make &lt;span class=&quot;math-container&quot;&gt;$T(r,z) = Tr(r)Tz(z)$&lt;/span&gt;. A starting point for this is to determine the effective Biot numbers for radial heat transfer.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Bi_{rw} = h_w \Delta R/k \hspace{1cm} Bi_{ra} = h_a \Delta R/k $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;In these, &lt;span class=&quot;math-container&quot;&gt;$h_w, h_a$&lt;/span&gt; are the convection coefficients of the flowing water or surrounding air, &lt;span class=&quot;math-container&quot;&gt;$\Delta R$&lt;/span&gt; is the outer minus inner radius &lt;span class=&quot;math-container&quot;&gt;$\Delta R = r_o - r_i$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is the thermal conductivity of the wire. When &lt;span class=&quot;math-container&quot;&gt;$Bi &amp;lt; 0.1$&lt;/span&gt;, the wire functions in a lumped analysis. The radial gradients can be neglected.&lt;/p&gt;&#xA;&lt;h1&gt;Axial Gradients Only&lt;/h1&gt;&#xA;&lt;h2&gt;Water Cooling Only&lt;/h2&gt;&#xA;&lt;p&gt;Consider the system for the case that &lt;span class=&quot;math-container&quot;&gt;$Bi_{rw} &amp;lt; 0.1$&lt;/span&gt; with perfect thermal insulation around the wire. This will give the maximum temperature that can be expected in the wire. The energy balance along the flow says that total power input &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; must be taken out by the flowing fluid (water) with mass flow &lt;span class=&quot;math-container&quot;&gt;$\dot{m}$&lt;/span&gt; and specific heat &lt;span class=&quot;math-container&quot;&gt;$\tilde{C}_p$&lt;/span&gt;. The fluid temperature changes from input &lt;span class=&quot;math-container&quot;&gt;$T_{wi}$&lt;/span&gt; to output &lt;span class=&quot;math-container&quot;&gt;$T_{wf}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{m}\tilde{C}_p(T_{wf,max} - T_{wi}) = P$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The subscript &amp;quot;max&amp;quot; indicates the case of maximum output temperature on the fluid. You can also use this to establish that, in this case, the temperature profile in the fluid is linear along the axis.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{m}\tilde{C}_p(T_w(z) - T_{wi}) = P \frac{z}{L}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T_w(z) - T_{wi} = P\left(\frac{z}{L}\right)\left(\frac{1}{\dot{m}\tilde{C}_p}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The differential energy balance on the wire with no radial gradients and with the external wall as a perfect thermal insulator is as below with &lt;span class=&quot;math-container&quot;&gt;$r_i$&lt;/span&gt; as the radius of the channel.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ h_w 2\pi r_i\left(T_z(z) - T_w(z)\right) dz = P \frac{dz}{L}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This gives the temperature in the wire as below with &lt;span class=&quot;math-container&quot;&gt;$A_i$&lt;/span&gt; as the area of the water channel.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T_w(z) - T_{wi} = P \left(\frac{z}{L}\right)\left(\frac{1}{\dot{m}\tilde{C}_p} + \frac{1}{h_w A_i}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;h2&gt;Including External Convection to Air&lt;/h2&gt;&#xA;&lt;p&gt;When we add convection to air, the energy balance to the internal fluid (water) written in differential form is as below with &lt;span class=&quot;math-container&quot;&gt;$r_o$&lt;/span&gt; as the external radius of the wire and &lt;span class=&quot;math-container&quot;&gt;$T_a$&lt;/span&gt; as the air temperature.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{m}\tilde{C}_p dT_w + h_a 2\pi r_o (T_z(z) - T_a) dz = P \frac{dz}{L}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The differential energy balance on the wire at any axial position &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; is the equation below with &lt;span class=&quot;math-container&quot;&gt;$r_i$&lt;/span&gt; as the channel radius.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ h_w 2\pi r_i(T_z(z) - T_w(z))dz + h_a 2\pi r_o(T_z(z) - T_a)dz = P \frac{dz}{L}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;These are two coupled differential equations to solve for &lt;span class=&quot;math-container&quot;&gt;$T_w(z)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T_z(z)$&lt;/span&gt;. In all cases, the water and wire temperatures will be lower than the case for water cooling only because a portion of the power generation is now removed by the air.&lt;/p&gt;&#xA;&lt;h1&gt;Radial Gradients&lt;/h1&gt;&#xA;&lt;p&gt;The radial gradient problem is easily solved for a hollow cylinder with fixed internal and external wall temperatures &lt;span class=&quot;math-container&quot;&gt;$T_{ci}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T_{co}$&lt;/span&gt;. The steady state profiles are well-documented in textbooks on heat transfer (see Chapter 17 in &lt;a href=&quot;https://www.wiley.com/en-us/Fundamentals+of+Momentum%2C+Heat%2C+and+Mass+Transfer%2C+Revised+6th+Edition-p-9781118947463&quot; rel=&quot;nofollow noreferrer&quot;&gt;Welty, et.al&lt;/a&gt;). At any position &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;, the radial temperature profile will have the form below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{T_r - T_{co}}{T_{ci} - T_{co}} = \frac{\ln(r/r_o)}{\ln(r_i/r_o)} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Following the format of separation of variables, the radial gradient is superimposed on the axial profile from above.&lt;/p&gt;&#xA;&lt;p&gt;The link between the wall temperatures &lt;span class=&quot;math-container&quot;&gt;$T_{ci}, T_{co}$&lt;/span&gt; and the water or air temperatures occurs as convection. This is an entirely separate issue to tackle. It can only be done once a proof is established that separation of variables is valid.&lt;/p&gt;&#xA;&lt;h1&gt;Unsteady State&lt;/h1&gt;&#xA;&lt;p&gt;This level of analysis is best done step-wise. Include time on the axial profile with insulated boundary, then include time on the axial profile including air convection, and finally consider time only on the radial boundary.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-07-02T20:15:09.683" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29019" PostTypeId="1" CreationDate="2019-07-02T18:57:25.430" Score="0" ViewCount="44" Body="&lt;p&gt;I am trying to apply a compression force to samples using air. I do not want to directly apply my loading since my samples are irregular. I do not wish to process them with a flattening step. I hope to use fluid and pressure to get a single compression force with even distribution across the surface. In the left example I have a spacer that has a negligible poisson's ratio so i will neglect expansion orthogonal to the applied pressure. In the right hand example I would have pressure on the side correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the cartoon below I am hopping to have the pressurization of the air transfer to the samples. Assume the vessel is air tight with the exception of the flow restriction to control pressure levels.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I make this assumption?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/stKOi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/stKOi.png&quot; alt=&quot;re&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11634" LastActivityDate="2019-07-03T17:07:10.587" Title="How to ensure uniaxal and uniform force application on uneven samples using pressurized fluids" Tags="|mechanical-engineering|fluid-mechanics|pressure|pneumatic|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29021" PostTypeId="1" CreationDate="2019-07-03T02:53:36.463" Score="1" ViewCount="118" Body="&lt;p&gt;I saw a machine that fed a sheet of laminate through some rollers.  The motor had a pulley attached to its shaft, and a spring connected to the pulley.  Why not use a belt?  Wouldnt the spring make it too difficult to know the exact position of the load on the other side of the spring?&lt;/p&gt;&#xA;" OwnerUserId="4403" LastActivityDate="2019-07-03T02:53:36.463" Title="Springs on a pulley... When to use?" Tags="|springs|pulleys|machine-design|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29023" PostTypeId="1" CreationDate="2019-07-03T08:01:27.923" Score="1" ViewCount="1048" Body="&lt;p&gt;I am a placement student and was given a project asking to measure the time taken for the laser to cut a metal and send it to Excel into a spreadsheet automatically. I am a computer science student and am unaware of how I should begin this project or how to do it. How do I measure the time taken for the laser to cut each metal? &lt;/p&gt;&#xA;" OwnerUserId="21012" LastEditorUserId="110" LastEditDate="2019-11-30T12:19:20.003" LastActivityDate="2023-09-11T01:02:20.333" Title="How do I measure the time taken for a laser to cut a metal?" Tags="|mechanical-engineering|electrical-engineering|lasers|software|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29024" PostTypeId="2" ParentId="29023" CreationDate="2019-07-03T09:14:54.087" Score="1" Body="&lt;p&gt;Cut various metals of various thicknesses, across various lengths, with the laser in question. Measure the time taken, using a digital stopwatch. Create a model that reproduces these observations, using the fewest assumptions and the simplest form, and accounts for the different type of metal, and different thicknesses. Apply the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll find such a model built into most laser-cutting software interfaces: the software will typically give a fairly good estimate of how long a given cut will take, once it has the pattern and has been told what type of material is to be cut.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2019-07-03T09:14:54.087" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29025" PostTypeId="1" CreationDate="2019-07-03T10:00:30.663" Score="0" ViewCount="70" Body="&lt;p&gt;I have a thick-walled pressure cylinder, that transports fluid and experiences only internal pressure.  During the process, the outer wall temperature is measured.  And I have no clue about the inner wall temperature.  The only fact, that I know is, the temperature on the outer wall is due to inner wall temperature that is conducted outside.  So, the inner wall temperature should be always greater than that of outer wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I would like to reproduce the same situation in a simulation software by providing the outer wall temperature as input and read out the inner wall temperature data on the corresponding nodes.  To do this, I am applying surface loads on the outer wall, which is the measured value as the boundary condition.  But with this I see the inner wall temperature are always lesser than the outer wall temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I tried to give the internal pressure as boundary condition for the inner wall, the software says, pressure data for the given nodes are not required.  Any idea about, where am i going wrong, or what am i missing? thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="21013" LastActivityDate="2020-06-10T01:04:50.393" Title="Wall temperature of pressure cylinder - Boundary condition" Tags="|pressure-vessel|thermal-conduction|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29027" PostTypeId="1" AcceptedAnswerId="29028" CreationDate="2019-07-03T12:10:21.420" Score="5" ViewCount="689" Body="&lt;p&gt;I am rebuilding a car engine. With some of the fasteners I am unscrewing, I notice that a lot of torque is required to undo the fastener. It feels like a snap when undoing a fastener, rather than a smooth twist. If I tighten the fastener to a set torque again and loosen it, it requires the same torque as it took to fasten it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why would this be? My guess would be something to do with heat cycling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: There is no threadlocker on these bolts.&lt;/p&gt;&#xA;" OwnerUserId="21017" LastActivityDate="2019-07-03T17:15:24.393" Title="What causes a fastener to lock?" Tags="|bolting|fasteners|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29028" PostTypeId="2" ParentId="29027" CreationDate="2019-07-03T12:19:38.490" Score="5" Body="&lt;p&gt;Oxidation, dried lubricant, contaminants, rust - all that acts as a layer of glue binding the elements. As you apply a firm torque, you snap that bond. It takes a considerable time to re-form. &lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-07-03T12:19:38.490" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29029" PostTypeId="2" ParentId="29027" CreationDate="2019-07-03T13:06:27.300" Score="3" Body="&lt;p&gt;Under high pressure and high heat metal bonds by a process called diffusion. Where the atoms of the two parts intersperse over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This and oxidation, slow chemical burning of unintentional matrial and debris between the two parts, cause a strong bond that many times cuase the fastener to break before it can be removed.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-03T13:06:27.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29032" PostTypeId="2" ParentId="29019" CreationDate="2019-07-03T17:07:10.587" Score="1" Body="&lt;p&gt;Both water and air are fluid. And they transfer the pressure equally every where. On the right hand case if your samples are submerged in water they feel the pressure everywhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In material testing labs when they need to test a sample for its compressive strength they prepare the sample with capping it smooth with a special paste which is chemically neutral and  and solidifies to a smooth level finish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then they can test it under uniaxial pressure. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-03T17:07:10.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29033" PostTypeId="2" ParentId="29027" CreationDate="2019-07-03T17:15:24.393" Score="2" Body="&lt;p&gt;A standard bolt will stretch when tightened.  This stretching increases force on the threads of both the bolt and hole and thus increases the force required to overcome static friction and loosen the bolt.  If we exclude other materials and chemical changes to the surfaces then that snap that you experience is the moment you overcome that static friction between the two parts.&lt;/p&gt;&#xA;" OwnerUserId="20957" LastActivityDate="2019-07-03T17:15:24.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29034" PostTypeId="2" ParentId="29005" CreationDate="2019-07-03T17:24:38.180" Score="0" Body="&lt;p&gt;Your release doesn't need to be perfectly inline with motion with swivel casters on the front.  The single direction casters in the rear make any left/right force upon release irrelevant because they are constrained.  Flipped around with swivels on the rear and your release has to be perfectly inline for straight movement.  Any left/right force upon release and it'll turn the casters.&lt;/p&gt;&#xA;" OwnerUserId="20957" LastActivityDate="2019-07-03T17:24:38.180" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29035" PostTypeId="1" AcceptedAnswerId="29039" CreationDate="2019-07-03T18:28:23.030" Score="1" ViewCount="5252" Body="&lt;p&gt;Given an Archimedes screw pump (shown below) how would you go about calculating the torque required to operate it?  I'm assuming a situation where you have to design for a given flow rate, Q (m^3/s) and the diameter, angle of incline and head are known.  The fluid to be pumped is water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the power required to lift water in such a pump is just the hydraulic power, which we have all the variables for:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  P(hyd) = rho * g * Q * H&#xA;&#xA; (Power = density * gravity * flow rate * head )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;But I'm struggling to conceptualise how this translates into a value for torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g7fRx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g7fRx.jpg&quot; alt=&quot;Archimedes screw pump&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21023" LastActivityDate="2019-07-08T07:14:41.317" Title="Calculating the torque required by an Archimedes screw pump" Tags="|torque|pumps|power|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29036" PostTypeId="1" CreationDate="2019-07-03T22:41:34.883" Score="1" ViewCount="38" Body="&lt;p&gt;I am learning pure substances in thermodynamics.I read a sentence &quot;Consider a piston cylinder containing liquid at 20 degree celsius and 1 atm pressure.&quot;My question is that the pressure of liquid  inside the cylinder will increase as we will go downward .So,is 1 atm pressure at the top of the piston or the average pressure of liquid in the piston?&lt;/p&gt;&#xA;" OwnerUserId="20605" LastActivityDate="2019-07-04T00:37:01.767" Title="What pressure is taken as the pressure of liquid" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29037" PostTypeId="2" ParentId="29036" CreationDate="2019-07-03T23:32:20.510" Score="2" Body="&lt;p&gt;For a gas, the difference is so small it can be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The density of air at 20C and 1 atm pressure is about 1.2 kg/m^3, so the pressure difference in a 1m tall cylinder is about 1.2 x 9.8 = about 12 Pa. Compare that with the 1 atm pressure which is about 100,000 Pa. The pressure difference is about 0.01% of the total pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, this pressure gradient is often &lt;em&gt;not&lt;/em&gt; negligible in liquids, where the density is about 1,000 times greater than for gases.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-07-03T23:32:20.510" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29038" PostTypeId="1" CreationDate="2019-07-04T00:07:41.553" Score="1" ViewCount="319" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KgsLy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KgsLy.png&quot; alt=&quot;Strange upside u shaped piping&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is this tall upside down shaped U pipe near a sugar refinery on Maui and I can't figure out what it's for ? I've also see these in other industrial areas near salt flats.&lt;/p&gt;&#xA;" OwnerUserId="21027" LastEditorUserId="21027" LastEditDate="2019-07-04T00:26:47.613" LastActivityDate="2019-07-04T00:26:47.613" Title="What is this weird upside down U shaped pipline for?" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29039" PostTypeId="2" ParentId="29035" CreationDate="2019-07-04T00:35:48.757" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P = 2 \pi T \omega$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is torque and &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; is the angular velocity in radians per second.  Assuming that you know the flow and RPM, you can solve for torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must add a caveat -- the above assumes no losses.  Any friction, or water leaking backward in the assembly after it's been pumped part way up is going to reduce the efficiency, and thus increase the torque for a given flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Edit, after the fact -- sorry for not thinking to include this before):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know almost nothing about these pumps.  I could come up with the above equation purely by using conservation of energy.  If you're adding potential energy to something at a given rate, then conservation of energy says that if the process is 100% efficient then you need to be expending that same amount of energy, be it electrical, mechanical, or whatever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So all I needed to do was to (A) know the power vs. torque and speed relation, and (B) assert 100% efficiency.  You can use this sort of thing everywhere -- motors, pumps, switching power supplies (that's kind of a specific example, but I'm an EE), cars, trains, whatever.  If you know the power out, and you can reasonably guess at the efficiency (or look it up in a book), then you know the power in.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastEditorUserId="19776" LastEditDate="2019-07-07T20:04:27.107" LastActivityDate="2019-07-07T20:04:27.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29040" PostTypeId="2" ParentId="29036" CreationDate="2019-07-04T00:37:01.767" Score="0" Body="&lt;p&gt;At any point at the depth of D from the surface of liquid the pressure of liquid is the sum of overbearing one atm pressure and the hydrostatic liquid pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ P= \rho *g * D + 1atm  $&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-04T00:37:01.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29042" PostTypeId="2" ParentId="25679" CreationDate="2019-07-04T02:18:02.350" Score="1" Body="&lt;p&gt;Within the U.S. the standard is ISA 84.00.01 and internationally it is IEC 61511 which is the process industry standard for functional safety. &#xA;Typically some type of risk analysis is driving the SIL requirement. Often this is a risk analysis of some type providing a quantitative failure rate of a safety function.  This helps determine the level of SIL protection based on some risk tolerance criteria which is often company specific. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A SIL 2 function will require devices which have a SIL rating which includes the safety relay or PLC or DCS. A calculation should be done according to the requirements of the standard to determine the SIL reliability based on the probability of failure on demand for the input devices and control elements. Often a more reliable function requires multiple input or control devices. &#xA;If you truly have a SIL requirement, I suggest to hire a qualified engineering firm to help design it. &lt;/p&gt;&#xA;" OwnerUserId="14901" LastActivityDate="2019-07-04T02:18:02.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29043" PostTypeId="2" ParentId="29009" CreationDate="2019-07-04T04:01:18.800" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/toNAw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/toNAw.png&quot; alt=&quot;Falstad&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H would have to be your switch.  You may have to use switch debounce.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\bar Q$&lt;/span&gt; as required to get triggering as required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adjust 1kΩ and 10μF to get LED to function as you want.  Little unclear how it works between your 1. &amp;amp; 2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;out goes to LED  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Falsted simulation &lt;a href=&quot;http://www.falstad.com/circuit/circuitjs.html?cct=$%201%200.000005%2010.20027730826997%2056%205%2050%0A165%20208%20144%20320%20144%204%200%0Aw%20208%20176%20208%20272%200%0Aw%20208%20176%20176%20176%200%0Aw%20208%20240%20128%20240%200%0Aw%20176%20176%20176%20272%200%0Ac%20176%20272%20176%20320%200%200.000009999999999999999%200.04950495049504958%0Ag%20176%20320%20176%20336%200%0Ar%20176%20176%20176%20112%200%201000%0Aw%20176%20112%20272%20112%200%0AR%20128%20112%2048%20112%200%200%2040%205%200%200%200.5%0AO%20336%20208%20400%20208%200%0Ar%20128%20112%20128%20240%200%20100%0Aw%20128%20112%20176%20112%200%0Ac%20128%20240%2080%20240%200%200.00001%203.5438318946034997e-13%0AL%20-64%20272%20-96%20272%200%201%20true%205%200%0Ag%20304%20304%20304%20320%200%0Aw%20272%20112%20336%20112%200%0Aw%20336%20112%20336%20176%200%0A193%20-32%20240%2080%20240%200%205%0Aw%2064%20240%2080%20240%200%0Aw%2080%20304%2064%20304%200%0Aw%20-32%20272%20-64%20272%200%0AR%20-32%20240%20-64%20240%200%200%2040%205%200%200%200.5%0Ao%2014%2064%200%204102%205%200.00009765625%200%202%2014%203%0Ao%2010%2064%200%204102%200.0000762939453125%200.00009765625%200%201%0Ah%202%207%205%0A&quot; rel=&quot;nofollow noreferrer&quot;&gt;T flip-flop triggering 555 one-shot&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2019-07-04T04:01:18.800" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29046" PostTypeId="1" CreationDate="2019-07-04T13:30:50.783" Score="0" ViewCount="29" Body="&lt;p&gt;I was reading about reaction turbines on Wikipedia, it says: &quot;&lt;em&gt;Reaction turbines develop torque by reacting to the gas or fluid's pressure or mass. The pressure of the gas or fluid changes as it passes through the turbine rotor blades.&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand the working of impulse turbine by utilisation of kinetic energy, but this confuses me. Can anybody explain?&lt;/p&gt;&#xA;" OwnerUserId="19792" LastActivityDate="2019-07-04T13:30:50.783" Title="What exactly makes reaction turbines work?" Tags="|turbines|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29047" PostTypeId="2" ParentId="29005" CreationDate="2019-07-04T14:25:18.773" Score="4" Body="&lt;p&gt;Consider what happens in &lt;strong&gt;front wheel steering&lt;/strong&gt;, such as an automobile:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Turn the steering wheel slightly clockwise.&lt;/li&gt;&#xA;&lt;li&gt;The front wheels each turn slightly clockwise.&lt;/li&gt;&#xA;&lt;li&gt;Their rolling action causes the wheels to pull the front end of the vehicle slightly to the right.&lt;/li&gt;&#xA;&lt;li&gt;The back wheels follow the track of the front wheels, but in a slightly tighter turn.&lt;/li&gt;&#xA;&lt;li&gt;Holding the steering wheel in that position will cause the vehicle to continuously turn in a circle.&lt;/li&gt;&#xA;&lt;li&gt;The wheels' caster makes them try to return to a neutral position.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The wheels are mounted with &quot;caster&quot;, meaning that the almost vertical axis about which the wheels rotate when steered doesn't meet the ground at the same position as where the wheels meet the ground.&#xA;This causes sideways friction when the wheels are not in a straight-forward position (negative feedback).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a very stable arrangement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Turns are circular.&lt;/li&gt;&#xA;&lt;li&gt;The sharper the desired turn, the more the steering wheel must be turned.&lt;/li&gt;&#xA;&lt;li&gt;Releasing control causes the steering to automatically return to a neutral position.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now consider what happens in &lt;strong&gt;rear wheel steering&lt;/strong&gt;, such as a fork-lift (or when driving a car in reverse):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Turn the steering wheel slightly clockwise.&lt;/li&gt;&#xA;&lt;li&gt;The back wheels each turn slightly counter-clockwise.&lt;/li&gt;&#xA;&lt;li&gt;Their rolling action causes the wheels to push the back end of the vehicle slightly to the left.&lt;/li&gt;&#xA;&lt;li&gt;The front wheels continue forward, turning to the right, but in a slightly tighter turn.&lt;/li&gt;&#xA;&lt;li&gt;Holding the steering wheel in that position will cause the back wheels to steer more and more widely around the front wheels, which make an increasingly tighter turn.&lt;/li&gt;&#xA;&lt;li&gt;Depending on the caster, the wheels might have a tendency to steer even more when released.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is a very unstable arrangement:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;To drive a circular path, the driver must initiate the turn by swinging the back end out to the left, and then must continuously reduce the amount of steering to avoid following a spiral path.&lt;/li&gt;&#xA;&lt;li&gt;Releasing control does not return the steering to neutral, and will likely result in a very tight spiral.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Rear wheel steering is very handy at slow speed.&#xA;A fork-lift for example can position the front end, push it into place, and adjust the back end to line up square.&#xA;Similarly, it is easier to back a car into a parking spot than to drive in forward, and much easier to drive out forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at higher speed, rear wheel steering is inherently unsafe.  With front wheel steering, if something goes wrong, the vehicle ends up going straight.  With rear wheel steering, if something goes wrong, the vehicle ends up spiraling out of control.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Edit: a simpler explanation:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The OP commented: &quot;&lt;em&gt;Shopping carts have no caster, but they do have trail. But I don't see how trail can be responsible, seeing as the same swiveling casters are stable if in front, but unstable if in the back. There must be another explanation.&lt;/em&gt;&quot;, and yes, there &lt;em&gt;is&lt;/em&gt; another explanation.  I &lt;em&gt;mentioned&lt;/em&gt; negative feedback, but failed to expand on the idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine what happens when a vehicle turns to the right.&#xA;As a passenger, you feel yourself pulled to the left by a force (some physicists say a virtual force) known as centrifugal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This same force acts on the vehicle itself, though because it is a rigid object you don't notice it except perhaps for a slight tip to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider how the vehicle's wheels react to this force, looking at three different types of vehicles:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;On a railroad car, the wheels are not steerable and are attached to a fixed axle.  This means that both wheels must roll at the same rate, even though when going around a curve one wheel has to travel a greater distance.  Centrifugal force behaves as if it acts on the car's center of gravity and pushes all the wheels to the left.&#xA;The track, and the surface of the wheels themselves are not, as one might think, perfectly level.  Instead they slope to the inside so that the part of the wheel that hangs over the inside of the rail has a larger diameter than the part of the wheel that hangs over the outside of the rail.&#xA;When the car turns, and is pushed to the left, the left wheels end up rolling on their larger diameter section while the right wheels run on their smaller diameter section.  In effect, the car has larger wheels on the left and smaller wheels on the right, and it naturally turns to the right, following the track.  (At slow speed this effect doesn't happen, hence the loud squealing of slipping wheels in shunting yards.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On a vehicle with front-wheel steering, the centrifugal force also acts on the center of gravity and pushes all the wheels to the left.  The rear wheels are fixed and so continue following the vehicle.  The front wheels can turn, so as they push to the left against the ground, the ground pushes them to the right.  With caster, or trail, this force provides negative feedback that makes the wheels return to their normal straight-forward position unless a steering wheel is forcing them to stay in a turned position.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On a vehicle with rear-wheel steering, it is the front wheels that are relatively unaffected by the sideways force, and the rear wheels that react to it.  The force of the road on the wheels makes them want to turn to the left.  But the vehicle is turning because the rear wheels were already turned to the left, so in this case the centrifugal force provides positive feedback.  The more the vehicle turns, the more the wheels tend to make it turn.  Rather than forcing the vehicle to turn, the steering wheel must be used to &lt;em&gt;prevent&lt;/em&gt; the vehicle from turning.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The rail car is symmetric front to back, asymmetric side to side, and steers by a very different mechanism using both front and rear wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The front-steering vehicle's front wheels are affected by negative feedback, which forces the steering back to a neutral position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rear-steering vehicle's rear wheels are affected by positive feedback, which forces the steering into ever increasing turn positions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Negative feedback is generally a good thing, enforcing stability (e.g. thermostats).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Positive feedback is generally a bad thing, causing rapid unrestricted growth (e.g. explosions, PA system howling).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Historical side notes:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Since they don't use rolling wheels, ships and planes steer from the rear because in that situation, rear steering is more stable.&#xA;It's no longer done this way, but the traditional ship steering wheel used to be turned the opposite way to how it is turned in a car.  (Notice in the description above how steering clockwise caused the rear wheels to turn counter-clockwise.  Having the steering wheel turn the same way as the rudder made sense.)&#xA;One would move the top of the steering wheel to the left in order to turn the ship to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Planes on the other hand &lt;em&gt;do&lt;/em&gt; still use this &quot;backward&quot; steering.&#xA;When steering, the pilot pushes on the right rudder pedal in order to yaw to the right.&#xA;This is the exact opposite of how anyone that has used a soap-box car with single axle front steering would expect it to work.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Bureaucrats and social idealists are often ignorant of the laws of physics, which tend to conflict with their world-view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 1970s Joan Claybrook, as head of the US National Highway Transportation and Safety Administration, somehow discovered that motorcycles were more dangerous to their drivers than were cars.&#xA;As a result, she commissioned, at great expense to the taxpayers, the design of a &quot;safe&quot; motorcycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Naturally it had a steel-pipe roll-cage to protect the driver.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And since her friend Ralph Nader had already killed the Chevrolet Corvair because he incorrectly believed that rear-engine vehicles were &quot;unsafe at any speed&quot;, she had the motorcycle's engine mounted at the front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This of course meant that the steering had to go at the back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone that has ever walked along, pushing a bicycle without holding the handlebars, knows that it's easy to go forward and impossible to go backward.&#xA;The engineers and other professionals working on the project pointed out that her design wouldn't work, but she dismissed negative &quot;opinions&quot; of the naysayers, her ideals of highway safety taking precedence over reality.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually it was given a pair of outrigger wheels (effectively the same as training wheels on a child's first bicycle), because no one was able to drive it without it almost immediately tipping over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an image of the prototype (the driver would sit facing the right):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sqSkV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sqSkV.jpg&quot; alt=&quot;rear-steering motorcycle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.profbobsfunwithhistoricalstuff.com/2016/07/joan-claybrook-and-safe-motorycle.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.profbobsfunwithhistoricalstuff.com/2016/07/joan-claybrook-and-safe-motorycle.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19930" LastEditorUserId="19930" LastEditDate="2019-07-11T18:23:51.470" LastActivityDate="2019-07-11T18:23:51.470" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29048" PostTypeId="1" CreationDate="2019-07-04T17:21:21.823" Score="1" ViewCount="63" Body="&lt;p&gt;We have to detect vibration/sound produced at rail track for upcoming train at 1000 m distances . So which sensor is appropriate here? thanks.&lt;/p&gt;&#xA;" OwnerUserId="21033" LastActivityDate="2020-07-30T04:08:50.923" Title="Which kind/model of sensor is appropriate for detecting vibration at one end of 1000 m steel block?" Tags="|vibration|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29052" PostTypeId="1" CreationDate="2019-07-04T20:31:36.550" Score="0" ViewCount="61" Body="&lt;p&gt;I have read in a paper that the for formation of new crystals in a solution the energy released due to formation of solid from solution should be more than the energy needed to overcome solid-liquid interactions. As in the meta-stable this is not favorable, no new crystals form in the meta-stable region i.e. low supersaturation state. My doubt is, if it is thermodynamically unfavorable to form new crystals in meta-stable region then why does crystals grow in meta-stable region. In a way, it is also the formation of new crystals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exact lines from the text is: &quot;The formation of stable nuclei only happens if the increase in interfacial energy resulting from the formation of a solid{liquid interface is offset by a decrease in energy released by the formation of a (thermodynamically favored) solid. At low supersaturations, the energy required to form a solid{liquid interface is bigger than the energy released upon formation of a cluster which are consequently unstable resulting in a very low nucleation rate. Only when the supersaturation is high enough, i.e. when we enter the labile region, stable solid particles will form.&quot;&lt;/p&gt;&#xA;" OwnerUserId="19256" LastActivityDate="2024-03-11T13:03:05.330" Title="Doubt regarding meta-stable and labile region of a supersaturated solution" Tags="|thermodynamics|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29053" PostTypeId="2" ParentId="28587" CreationDate="2019-07-05T07:57:34.010" Score="2" Body="&lt;p&gt;I've designed magnetic brakes like this and have had no issues with a setup similar to yours.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the system will work even without a special arrangement of magnets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess as to why it's not working, is that the magnet bracket may be made from iron, and may be concentrating the flux on the old system, improving it's power compared to your new one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would gather more data by having someone spin the flywheel, then (carefully) holding a magnet near it. You should be able to feel the drag force and get an idea of what's happening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gap will need to be very small, is your new gap as small as the old one?&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-07-05T07:57:34.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29054" PostTypeId="2" ParentId="28993" CreationDate="2019-07-05T08:14:19.407" Score="1" Body="&lt;p&gt;To address your second and 3rd question it's important to remember the purpose of cascading PID loops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say you want to control a ships heading, so you create a rudder position signal using a PID loop.  But the rudder doesn't take a position signal, it's just got actuators that take a &quot;force&quot; signal. So you run &lt;em&gt;those&lt;/em&gt; on a PID loop to match the position setpoint. And now you've got a cascaded loop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be possible to combine these into one loop, but you can see that logically it wouldn't make sense, they're each separate control problems with their own loops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now lets go deeper. Lets say the actuators are self contained devices that take a force signal, but internally they're just a dumb cylinder and valve. So you add a PID loop to that: it takes the force signal, and adjusts the valve using PID to match the force setpoint. That's 3 cascaded loops, and you can see how this could go on and on.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-07-05T08:14:19.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29055" PostTypeId="1" CreationDate="2019-07-05T21:12:15.960" Score="0" ViewCount="61" Body="&lt;p&gt;I am MAthematics and MEchanical Engineering student and this is my senior year. For reserch and personal interest, I need to learn about contact mechanics. From where I should start and what is the prerequisites for this subject. &lt;/p&gt;&#xA;" OwnerUserId="21042" LastActivityDate="2019-07-05T23:43:06.943" Title="Where should I study contact mechanics (for introductory level)" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29056" PostTypeId="1" CreationDate="2019-07-05T22:58:51.200" Score="1" ViewCount="173" Body="&lt;p&gt;I'm looking for good software to design control systems, and more specifically to neatly design &quot;block diagrams&quot; for control systems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something like that?&lt;/p&gt;&#xA;" OwnerDisplayName="user21045" LastActivityDate="2019-12-28T12:01:50.237" Title="Designing Block Diagrams For Control Systems" Tags="|control-engineering|design|control-theory|systems-design|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29058" PostTypeId="2" ParentId="29055" CreationDate="2019-07-05T23:43:06.943" Score="1" Body="&lt;p&gt;Start with either Slaughter's or Barber's book on linear elasticity where they deal with Hertzian non-adhesive contact and the theory that leads to that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you  know the basics, move on to K L Johnson's book on contact mechanics.  That will give you a good feel for adhesive contact too along with various solution methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond a point you will have to start finding numerical solutions.  A good starting point is the book by Peter Wriggers (though I don't like the pedagogy in the book).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you know these three well, you can go to the scientific literature and start exploring the details of other solutions and approaches.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2019-07-05T23:43:06.943" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29059" PostTypeId="1" CreationDate="2019-07-06T01:13:48.393" Score="0" ViewCount="245" Body="&lt;p&gt;I am designing a functional replica of the Apollo FDAI (aka 8 ball).&#xA;See pics &lt;a href=&quot;https://i.stack.imgur.com/bVYYi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Real FDAI&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/F4tx3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Another view&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;From the Apollo 8 Flight Journal:&#xA;  The FD/AI (Flight Director/Attitude Indicator) or &quot;8-ball&quot; is one of the most important instruments in the spacecraft. Designers had originally intended to give the crew three separate displays to show their attitude; one each for roll, pitch and yaw. Being pilots, the crews quickly threw out the three displays for a development of the artificial horizon familiar from aircraft instrument panels.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Here's what I have so far (I know, those aren't steppers, it's only V1)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3VYpO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;V1_ball&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/Rguiz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;V1_view2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V2 will rely on three stepper motors to position the ball. Previously I put the motors more or less in line with their respective axes. Unfortunately, this introduced the challenges of dealing with rotating electrical connections, balance of the ball, motors turning the weight of other motors... Etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, why not eliminate these issues at the cost of increased mechanical complexity!&#xA;&lt;a href=&quot;https://i.stack.imgur.com/b9JXS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;V2 sketch&lt;/a&gt; (please excuse the humanCAD)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My plan is to use gears (and belts potentially) to move the control axes to be colinear with the roll axis. All control axes will be concentric and will have its own shaft (two of which will be hollow). Basically like the three hands on a clock. This will allow all the motors to be outside the moving ball mechanism.&#xA;The problem with this is it demands more performance of the motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: if I wish for the ball to pitch, I have to move the yaw motor in tandem with pitch to prevent the yaw bevel gear from moving relative to the pitching platform. Furthermore, if I want to move pitch and yaw at the same rate, the yaw motor must move twice as fast as the pitch motor.&#xA;This issue is even worse if I want to roll, requiring compensation in both pitch and yaw axes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't expect this to be a problem at low rates, but it severely limits the maximum rate of the ball. The pitch motor must be able to move at least twice as fast as the roll, even worse, the yaw must rotate 3x as fast as the roll motor. I'm not willing to sacrifice the accuracy of the instrument, so simply gearing the motors &quot;faster&quot; is not a solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that an elegant mechanical solution exists, but I'm stumped. I think if I had a mechanism that behaves in a &quot;I can turn you, but you can't turn me&quot; manner might go a long way. &lt;/p&gt;&#xA;" OwnerUserId="21044" LastEditorUserId="21044" LastEditDate="2019-07-06T01:31:34.600" LastActivityDate="2020-05-07T15:02:30.323" Title="Master/Slave Gear Mechanism for Navball" Tags="|mechanical-engineering|design|gears|stepper-motor|computer-aided-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29060" PostTypeId="2" ParentId="29048" CreationDate="2019-07-06T01:36:11.590" Score="1" Body="&lt;p&gt;If you want to detect sound, use a microphone.  If you want to detect vibration, there are a few options, but probably the default option would be a piezoelectric accelerometer.  For both microphones and accelerometers, there are thousands of models available depending on the frequency and amplitude range you care about. Cost will be in the hundreds of US dollars range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can provide more detail about what you need to do, we can give you a better answer.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-07-06T01:36:11.590" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29061" PostTypeId="2" ParentId="28984" CreationDate="2019-07-06T02:04:23.367" Score="1" Body="&lt;p&gt;In mechanical drafting and machine shop school, it meant &lt;em&gt;not to measure the drawing using a 6&quot; machinist scale, and use those measurements to fabricate the part&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(You can replace 6&quot; scale with any type of measuring device.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my career I have worked with hundreds of mechanical engineers, machinists, toolmakers, tool-and-die makers, etc. To a man, I'm confident that they would all agree. &lt;/p&gt;&#xA;" OwnerUserId="20985" LastActivityDate="2019-07-06T02:04:23.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29062" PostTypeId="1" AcceptedAnswerId="29137" CreationDate="2019-07-06T02:46:27.793" Score="0" ViewCount="74" Body="&lt;p&gt;A question about performance of a turbojet.&#xA;Why are compression ratio and thrust are not in linear but increase exponentially with RPM?&lt;/p&gt;&#xA;" OwnerUserId="18943" LastEditorUserId="18943" LastEditDate="2019-07-06T06:27:57.720" LastActivityDate="2019-07-11T08:52:25.863" Title="Relationships among turbo jet rpm and compression ratio and thrust" Tags="|aerospace-engineering|turbomachinery|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29063" PostTypeId="2" ParentId="29052" CreationDate="2019-07-06T05:57:01.843" Score="0" Body="&lt;p&gt;What the text is saying is that there are two different processes at work. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;creation&lt;/strong&gt; of a solid-liquid interface. In other words, going from not having a crystal to having a crystal at all. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The &lt;strong&gt;growth&lt;/strong&gt; of a crystal when there is already something solid for it to grow on.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It is possible (the book says) for &lt;strong&gt;2&lt;/strong&gt; to be thermodynamically favoured while &lt;strong&gt;1&lt;/strong&gt; is not. In this régime new crystals will not spontaneously form out of the liquid but if a solid surface or particle is present then a crystal will grow on it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words: nucleation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are further distinctions to be made based on the nature of the solid-liquid interface. A speck of dust is better than the smooth wall of a vessel, and a tiny crystal &lt;em&gt;of the substance in question&lt;/em&gt; is better than a random speck of dust. The principle is the same, though: it is easier to add molecules to something that is already the right shape than to create “the right shape” out of nothing. &lt;/p&gt;&#xA;" OwnerUserId="21047" LastActivityDate="2019-07-06T05:57:01.843" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29064" PostTypeId="1" AcceptedAnswerId="29066" CreationDate="2019-07-06T10:22:15.880" Score="2" ViewCount="82" Body="&lt;p&gt;I'm trying to understand how telephone wires work.&lt;br&gt;&#xA;I've stumbled upon this kind of identification method with colours.&lt;br&gt;&#xA;I can't manage to understand the reasons why is used and if it's still common to use it.&lt;/p&gt;&#xA;" OwnerUserId="14109" LastEditorUserId="10902" LastEditDate="2019-07-06T15:52:04.783" LastActivityDate="2019-07-06T15:52:04.783" Title="What is the purpose of colour code for telephone wires?" Tags="|telecommunication|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29065" PostTypeId="1" AcceptedAnswerId="29067" CreationDate="2019-07-06T13:33:20.623" Score="2" ViewCount="347" Body="&lt;p&gt;Before I describe the problem I would just like to say that I'm no engineer this is my hobby, all I know is from google or if someone told me or from books. Something obvious to you may not be obvious to me. Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to move a gear with the shaft like they were one in one moment and in the other I want to stop the gear using clutch brake while continuing to rotate the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to achieve this is using air bearings which would be the perfect solution if they weren't so impractical because of the required compressor and other parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second solution would be to use two clutches. One to connect the gear to shaft and another clutch to brake the gear when the first one is not connected but I don't want to use more than one clutch pack because they are expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third solution is to use sync cone but that is much slower than clutch and that is why I don't want to use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fourth solution which I found is using a shaft mounted brake to lock the gear with the shaft but I'm not sure if this would count as a clutch brake and if it would lower the efficiency of the system as whole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question would be is there a way to achieve the problem described without using more breaks or clutches (if shaft mounted brake wouldn't count as a break then that is also OK)? The perfect solution would be a bearing which I could &quot;turn off&quot; and let it rotate with the shaft but I'm not sure if something like that exists.&lt;/p&gt;&#xA;" OwnerUserId="21049" LastActivityDate="2019-07-06T13:53:03.657" Title="Is it possible to move gear on a shaft in one moment and in the other allow it to rotate freely?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="29066" PostTypeId="2" ParentId="29064" CreationDate="2019-07-06T13:51:14.483" Score="3" Body="&lt;p&gt;At the most basic level, wires are color coded to simplify things during installation.  And that's all wiring, not just telephone wires.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of the challenges involved if you're running a new telephone jack into a room that's a reasonable distance away from the phone block.  If the wires weren't color coded, you would have to set up some sort of ringer on each of the lines in order to determine which of the wires to use and where.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a basic phone outlet (two wire) it's not horribly complicated.  On a standard outlet (four wire) it starts to get annoying.  With standard telephone wire (8 wire) then it gets really annoying.  Color coding simplifies all that.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case you haven't wired any of those types of connections, &lt;a href=&quot;http://www.westernet.net/Help/RJ45.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;this webpage&lt;/a&gt; shows the standard color coding for telephone and ethernet wiring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional aspect of color coding is standardized use of the wires.  When someone new comes in to perform maintenance on an existing installation, they'll have a reasonable degree of assurance that specific color wires are used for certain tasks.  Not following the standard can actually impact the data transfer rate on a cat-5 ethernet cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to higher voltages, there is a safety element as well.  Incorrectly wiring a 240V outlet's multiple wires can potentially cause a fire or damage the equipment. &lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2019-07-06T13:51:14.483" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29067" PostTypeId="2" ParentId="29065" CreationDate="2019-07-06T13:53:03.657" Score="0" Body="&lt;p&gt;Yes use a electromagnetic clutch, often used on car a/c compressors to disconnect them when not needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When engaged they drive and  freewheel when not energised.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-07-06T13:53:03.657" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29070" PostTypeId="1" CreationDate="2019-07-06T22:00:28.253" Score="1" ViewCount="50" Body="&lt;p&gt;Often beeswax is used to seal small air leaks on some musical instruments. One of the disadvantages of beeswax is that it has a melting point of around 140 degrees Fahrenheit and can therefore flow to other parts of instruments with deleterious effects. Do you know of a nontoxic substitute for beeswax with a higher melting point?&lt;/p&gt;&#xA;" OwnerUserId="21056" LastActivityDate="2020-01-04T05:02:05.333" Title="Material for filling small air leaks" Tags="|materials|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29071" PostTypeId="2" ParentId="10341" CreationDate="2019-07-06T23:25:54.933" Score="1" Body="&lt;p&gt;this is incredibly late, but for people still googling this question: I understand it as Hg being the vapour state with almost 0 liquid state. The quality x of this would 1. And the saturated liquid state is where the liquid is on the cusp of changing state, thus an infinitesimal amount of the liquid has vaporised, with a subsequent quality x would be 0. Denote saturated liquid state as Hf. &lt;/p&gt;&#xA;" OwnerUserId="21058" LastActivityDate="2019-07-06T23:25:54.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29072" PostTypeId="1" CreationDate="2019-07-07T05:17:08.263" Score="0" ViewCount="290" Body="&lt;p&gt;How do I quantify how much kinetic energy would be released from popping a balloon?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that any additional kinetic energy that results from the balloon will come from it being pressurized, but how is that calculated?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2023-12-24T13:05:00.677" Title="Kinetic energy of a balloon" Tags="|pressure|pressure-vessel|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29073" PostTypeId="1" CreationDate="2019-07-07T13:14:47.977" Score="0" ViewCount="73" Body="&lt;p&gt;When -on a hot day- the sun is incident on a window covered from the outside by a closed window shutter, the resulting heat is absorbed by the layers of air surrounding the shutter, subsequently circulating upward by convective motion thus drawing in new cooler air from below to cool down the shutter. If instead a curtain was shut on the inside of the window (window shutter absent in this case), the same heating of the surrounding air occurs, but this time around this air is tapped from the interior atmosphere of the house and thus it screws up the overall insulation of the house's interior conditions from the outside. To (partially) offset this effect, one is forced to pick light-tainted curtains. Even then, it seems to me that in a sunnier climate, curtains are going to shed more particulate matter into the surrounding room and thus adversely affect the air quality?&lt;/p&gt;&#xA;&lt;p&gt;Q: &lt;strong&gt;Is this a legitimate/complete assessment? Have I overlooked something?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6692" LastEditorUserId="90" LastEditDate="2021-07-28T05:23:02.440" LastActivityDate="2021-07-28T05:29:26.943" Title="In a warmer/sunnier climate, window (or roller shutters) superior to curtains?" Tags="|thermodynamics|thermal-conduction|thermal-insulation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29074" PostTypeId="1" CreationDate="2019-07-07T13:38:02.803" Score="0" ViewCount="160" Body="&lt;p&gt;Let's say we have one simple box and its bottom has been cut in a circular shape. We want to attach another box to this box that can rotate 180deg around the first one. What mechanism can be used to join these two boxes? &lt;/p&gt;&#xA;" OwnerUserId="21060" LastActivityDate="2019-07-07T17:39:49.977" Title="What are the different methods to join 2 parts so that one can rotate 180 degrees around the other?" Tags="|mechanical-engineering|machine-design|machine-elements|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29075" PostTypeId="2" ParentId="29070" CreationDate="2019-07-07T16:54:09.083" Score="0" Body="&lt;p&gt;You might try &lt;a href=&quot;https://en.wikipedia.org/wiki/Polycaprolactone&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Polycaprolactone&lt;/a&gt; &#xA;It softens at hot-water temperatures and is safe to handle. (You'll probably need to heat it a bit above that to get to fully liquid.) FDA approved for some applications, so IMHO pretty safe. This material is usually sold in granular form; some names I've seen are Polymorph, ThermoMorph and &quot;friendly plastic.&quot;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But especially if this is a wooden instrument, be careful and if possible test on similar materials (to your instrument) &lt;em&gt;before&lt;/em&gt; using it on your instrument. My concern is that it'll soak into wood and change the stiffness and thus tone.&lt;/p&gt;&#xA;" OwnerUserId="13014" LastActivityDate="2019-07-07T16:54:09.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29076" PostTypeId="2" ParentId="29074" CreationDate="2019-07-07T17:39:49.977" Score="1" Body="&lt;p&gt;Perhaps a &quot;turntable bearing&quot; also called a &quot;lazy susan&quot; bearing. Something like this one: &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B00ZSQSWTM&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/Square-Inch-Susan-Turntable-Bearing/dp/B00ZSQSWTM&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="13014" LastActivityDate="2019-07-07T17:39:49.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29077" PostTypeId="2" ParentId="29072" CreationDate="2019-07-07T18:01:12.733" Score="0" Body="&lt;p&gt;There are two components to the total energy in the balloon:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The P*V work done on the air (or whatever fluid you've filled it with) and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) the stress*strain work done on the rubber as the balloon expands. The latter is tricky to calculate (due to rubber's nonlinearity, large-strain modeling of the membrane, etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one starts from an uninflated balloon, all the work is done by forcing the fluid into the balloon. So you might measure the pressure and flow rate (dv/dt) -- ideally sampled simultaneously) -- from uninflated to the bursting point, and numerically integrate P * delta_Volume, from (negligible initial volume) up to the bursting volume (and pressure) of the balloon. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that consumer balloons have somewhat uneven wall thicknesses, so expect a fairly wide range of results.&lt;/p&gt;&#xA;" OwnerUserId="13014" LastEditorUserId="13014" LastEditDate="2019-07-09T10:25:37.987" LastActivityDate="2019-07-09T10:25:37.987" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29078" PostTypeId="2" ParentId="6388" CreationDate="2019-07-07T18:17:31.260" Score="0" Body="&lt;p&gt;1) I've used several brands of FSRs, including FlexiForce and you're very unlikely to get results usable in trade; their responses to pressure vary with temperature (because, I suspect, the stiffness of the polymers in the sensor itself vary w/temp.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2) The application guides I've read strongly suggest NOT using extremely stiff materials or sharp radii in direct contact with the sensor that can cause stress concentrations.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) The stiffness and shape of both what's on top of the sensor and what's underneath both matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) all of the materials you're considering are orders of magnitude stiffer than the sensor, so they'll likely give you equally wrong and expensive results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) I suggest you first try either thin disks or very-slightly rounded domes made with something like Nylon or HDPE (High-density polyethylene, AKA cutting-board plastic.) I suspect you can back those with disks of a more rigid material; steel would be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) The disk has to be same-size or slightly smaller than the sensor's active area, or else some of your load will bypass the active area, leading to systemic errors. &lt;strong&gt;And you'd better radius the edges, lest you create a stress concentration right there.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13014" LastEditorUserId="13014" LastEditDate="2019-07-07T18:25:40.863" LastActivityDate="2019-07-07T18:25:40.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29079" PostTypeId="1" CreationDate="2019-07-07T18:18:45.990" Score="1" ViewCount="824" Body="&lt;p&gt;I previously posted a different question that was very vague.  I have kept working on the problem, and now I have a different question.  Please let me know if I should have just added it to the old one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find the drag that a go-kart sees using a coast-down test, in which you accelerate then allow the cart to decelerate on its own (without using the brakes.)  I have data on the speed of the cart vs. the time.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a cart that is decelerating due to drag, my data is saying that the velocity is proportional to the square of time (a parabolic curve of velocity vs. time).  I have looked at the data of other people's coast-down tests (I found them online), and their velocity appears to follow the same trend that mine does.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above would mean that the derivative of velocity, the acceleration, would be linear if plotted against time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given these two relationships:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v \propto t^2$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$a \propto t$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it would follow that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v \propto a^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that you can calculate drag using &lt;span class=&quot;math-container&quot;&gt;$F=ma$&lt;/span&gt;, where  is the drag force &lt;span class=&quot;math-container&quot;&gt;$F=D$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would mean that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$D \propto \sqrt{v} \quad (?)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have read that drag is proportional to &lt;span class=&quot;math-container&quot;&gt;$v^2$&lt;/span&gt;.  The data that I have looked at from other people's tests also shows the relationship that drag is proportional to &lt;span class=&quot;math-container&quot;&gt;$v^2$&lt;/span&gt;, but I can't figure out how they got that relationship when their velocity curves are parabolic with respect to time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?  I can't figure out how these two equations can both be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I guess my main question at this point is&lt;/strong&gt;: should the velocity vs. time plot for a decelerating vehicle have a curve fit of &lt;span class=&quot;math-container&quot;&gt;$$t \propto v^2$$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$$t \propto \sqrt{v}$$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've attached an image of my data to show what I'm seeing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5HQUu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5HQUu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21053" LastEditorUserId="10400" LastEditDate="2019-07-11T14:11:46.870" LastActivityDate="2024-02-17T13:07:35.897" Title="How can I connect F=ma to Drag=1/2*Cd*(rho)*V^2*A?" Tags="|mechanical-engineering|fluid-mechanics|automotive-engineering|fluid|" AnswerCount="4" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29080" PostTypeId="1" CreationDate="2019-07-07T18:21:07.577" Score="0" ViewCount="822" Body="&lt;p&gt;Let's say we have a flat object. We are asked to create some mechanism so that the part can rotate in the shown angles above. What mechanisms can be used?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nwrAN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nwrAN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21060" LastActivityDate="2020-12-02T09:00:44.320" Title="What kind of mechanism can be used to rotate a part in 180 degrees (Ψ (yaw)) and 30degrees in  (pitch) direction?" Tags="|mechanical-engineering|control-theory|microelectronics|machine-design|machine-elements|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29081" PostTypeId="2" ParentId="29080" CreationDate="2019-07-07T18:29:27.883" Score="1" Body="&lt;p&gt;Gimbals are the classic solution for this: &lt;a href=&quot;https://en.wikipedia.org/wiki/Gimbal&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Gimbal&lt;/a&gt; &#xA;And if you don't need any roll, you can omit one of the gimbal rings, in which case a &lt;strong&gt;pan/tilt&lt;/strong&gt; mechanism is sufficient (often sold for photography.) You might look for a gimballed gyroscope toy and start with that. (Flight quality gimbals are &lt;em&gt;expensive&lt;/em&gt;.)&lt;/p&gt;&#xA;" OwnerUserId="13014" LastEditorUserId="13014" LastEditDate="2019-07-07T21:08:50.583" LastActivityDate="2019-07-07T21:08:50.583" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29082" PostTypeId="2" ParentId="29079" CreationDate="2019-07-07T18:41:29.997" Score="0" Body="&lt;p&gt;There are two parts to drag for a vehicle, aerodynamic and rolling friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aerodynamic drag is the larger part above about 40mph, while that due to rolling friction is larger below 40mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total drag is both combined, obviously rolling friction is there at any speed, so the curve you plot is representing both, unless, your results are below 40mph and are mostly due to rolling friction...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drag is usually calculated as a function of several things: Drag coefficient, density, velocity squared and area.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;D = Cd * rho * (V^2 / 2) * A&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;rho is density, Cd is drag coefficient, V is velocity and A is area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as you put in your question the drag is basically proportional to the velocity squared plus some other bits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to the drag formula, but inclination will be relevant to an aeroplane, but not really relevant for a vehicle...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/drageq.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Drag formula&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another link you may find interesting is this one as it expands on those things you are looking at:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sciencedirect.com/topics/engineering/aerodynamic-drag&quot; rel=&quot;nofollow noreferrer&quot;&gt;overview of drag on a vehicle&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-07-07T19:29:27.813" LastActivityDate="2019-07-07T19:29:27.813" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29083" PostTypeId="1" AcceptedAnswerId="29088" CreationDate="2019-07-07T18:43:34.067" Score="1" ViewCount="3041" Body="&lt;p&gt;Determine the resultant internal loadings acting on the cross section at B of &#xA;the pipe shown in Fig. 1–7a. End A is subjected to a vertical force of 50 N, a &#xA;horizontal force of 30 N, and a couple moment of 70 N m. Neglect the &#xA;pipe’s mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BOCvQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BOCvQ.png&quot; alt=&quot;fig 1-7a&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/me0iw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/me0iw.png&quot; alt=&quot;solution of problem in book&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, I can't feel the solution. I would add upward reaction force (70/0.5=140 N) at B due to moment. But I wouldn't add 70 N m to moment equation at point B. can you explain the reason?&lt;/p&gt;&#xA;" OwnerUserId="21063" LastActivityDate="2019-07-08T18:42:17.013" Title="In static equilibrium body, how can an applied moment about an axis contribute to reaction moment to another point away from the axis?" Tags="|mechanical-engineering|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29084" PostTypeId="1" CreationDate="2019-07-07T20:53:25.047" Score="0" ViewCount="22" Body="&lt;p&gt;Are there impulse turbines that operate efficiently on hot exhaust gases rather than water as the working fluid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im thinking about something analogous to a water wheel, since it seems that bucket/blade design in such a turbine would be much easier to do for an amateur since it wouldn't depend so closely on the intended operating conditions as for turbines that rotate in a perpendicular plane to the flow&lt;/p&gt;&#xA;" OwnerUserId="18744" LastActivityDate="2019-07-07T20:53:25.047" Title="Are there impulse turbines operating on hot exhaust gas?" Tags="|turbines|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29086" PostTypeId="1" CreationDate="2019-07-08T04:42:36.047" Score="2" ViewCount="234" Body="&lt;p&gt;I'm trying to design a closed-loop guidance system for rockets - something that can be implemented in Kerbal Space Program, for example. To do this, I'm following the design process described in Ashish Tewari's &lt;em&gt;Advanced Control of Aircraft, Spacecraft and Rockets&lt;/em&gt;. His approach breaks the system up into a &lt;em&gt;guidance controller&lt;/em&gt; for trajectory tracking and 3 &lt;em&gt;attitude controllers&lt;/em&gt; for pitch/roll/yaw control:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RAJcu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RAJcu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The controllers are designed with a state space approach. The attitude controllers send gimbal angle commands &lt;a href=&quot;https://i.stack.imgur.com/t0TZ6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t0TZ6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; to the engine to maintain attitude control (zero angle-of-attack), while the guidance controller also has a feedthrough term &lt;a href=&quot;https://i.stack.imgur.com/dPmka.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dPmka.png&quot; alt=&quot;\frac{\Deltau}{T}&quot;&gt;&lt;/a&gt; to make trajectory corrections in order to follow a preset nominal trajectory &lt;a href=&quot;https://i.stack.imgur.com/eiSIO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eiSIO.png&quot; alt=&quot;{r_n(t),\phi_n(t),v_n(t)}&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He then demonstrates how to tune each of the controllers individually using LQR, but leaves integrating them as an exercise to the reader.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a first step, I tried implementing the guidance and pitch controllers individually in Simulink, then combining them as prescribed. However, while the individual performance is very good (as expected), their combined performance is abysmal (see below image). The pitch and guidance controllers' commanded gimbal angles start 'fighting' each other immediately, saturating the gimbal actuator and resulting in zero control over either the trajectory or angle of attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZttMa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZttMa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all that said, my questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is there a fundamental error with this controller architecture, or is the poor performance I'm getting resolvable with a better controller tuning?** My theory is that the main issue here is that both the guidance and pitch controllers directly feed commands to the gimbal actuator; in other examples of hierarchical control systems, the outer 'guidance loop' only seems to set the setpoint for the inner 'actuator loop' to follow. &lt;/li&gt;&#xA;&lt;li&gt;Could this also be a consequence of tuning the controllers individually, then combining them, as opposed to the other way round?** In a controls problem where the linearized system comprises multiple sets of weakly coupled dynamics, can the controllers really be tuned individually with LQR, or do we still need to consider the full system when performing LQR?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21067" LastEditorUserId="110" LastEditDate="2019-07-08T12:48:10.447" LastActivityDate="2019-07-08T16:42:56.163" Title="Closed loop guidance and control: how to tune a system with multiple feedback controllers?" Tags="|control-engineering|control-theory|aerospace-engineering|aerodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29087" PostTypeId="2" ParentId="29035" CreationDate="2019-07-08T07:14:41.317" Score="0" Body="&lt;p&gt;Hydraulic power of a pump is &lt;span class=&quot;math-container&quot;&gt;$P_h=QgH\rho$&lt;/span&gt;, with flow &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt;, gravitational constant &lt;span class=&quot;math-container&quot;&gt;$g$&lt;/span&gt;, hydraulic height &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; and fluid density &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt;. Note that &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; is a bit higher than the actual useful hydraulic height:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/os1PB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/os1PB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shaft power (The power delivered at the shaft of the pump) is &lt;span class=&quot;math-container&quot;&gt;$P_s=P_h/\eta$&lt;/span&gt;, with hydraulic effiency &lt;span class=&quot;math-container&quot;&gt;$\eta$&lt;/span&gt; - for screw pumps, this is usually around 75%. The strength of screw pumps is that this effiency is achieved over a wide range of flow rates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you know the shaft power an can plug this into the formula provided by Tim: &lt;span class=&quot;math-container&quot;&gt;$T=\frac{P_s}{2\pi\omega}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you need to know &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt;. RPM is &lt;span class=&quot;math-container&quot;&gt;$n=\frac{50}{\sqrt[3]{D^2}}$&lt;/span&gt; with outer Diameter &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; given in meters, then &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; is given in rpm. This is the formula of Muysken, quoted after Nagel and Radlik. The actual rpm may be lower, at higher speeds the effiency suffers because more water flows back over the &quot;shovels&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wether your sizing a motor and gear or building the pump house, you should work closely with the manufacturer of the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: Nagel / Radlik: Wasserförderschnecken - Planung, Bau und Betrieb von Wasserhebeanlagen. &lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-07-08T07:14:41.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29088" PostTypeId="2" ParentId="29083" CreationDate="2019-07-08T11:34:45.007" Score="0" Body="&lt;p&gt;You cannot replace a couple moment with a single force. That would introduce an erroneous term to the vertical force balance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a reason why it is called a &lt;strong&gt;couple&lt;/strong&gt; moment. If you want to replace it with forces you can do the following. Add 2 forces that are equal in magnitude, opposite in direction and do not share the same line of action such that their net moment along the axis of interest gives you the same value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, as long as you do not change the direction of axis, you can move the point of application of moments and even combine them all with one along each orthogonal direction. Therefore, you can imagine moving the moment at point A to point B and performing the moment balance to find your reaction forces. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be careful, however, when you try to do the same with forces. In that case, you might have to add additional moments to your equations in order to include the effect of forces on moment balance. &lt;/p&gt;&#xA;" OwnerUserId="12979" LastActivityDate="2019-07-08T11:34:45.007" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29090" PostTypeId="1" CreationDate="2019-07-08T12:41:56.740" Score="0" ViewCount="107" Body="&lt;p&gt;When your first idea is drafted you calculate which forces/pressures and so on are working on your system. And then you break the forces down into the single screws which are built in your device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exemplary let’s say, that screw is loaded axially with &lt;span class=&quot;math-container&quot;&gt;$8000 \ N$&lt;/span&gt; – the transverse load could be unattended.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you decide to take the right screw dimension?&#xA;Things that should in my opinion be also considered are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;How is the load working on the screw?&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;concentrically vs. eccentrically&lt;/li&gt;&#xA;&lt;li&gt;static vs. dynamic&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;How is the montage of the screw realized?&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Simple tightening spindle vs. tightening by angle monitoring in the plastic range&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;What degree of utilization is allowed?&lt;/li&gt;&#xA;&lt;li&gt;What is the strength grade of the screw?&lt;/li&gt;&#xA;&lt;li&gt;And perhaps many more points…&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The aim that I’m expecting is, something like: &lt;em&gt;With these forces, this montage, this strength grade and … (simple calculation) … you have to take a screw M16x2  10.9&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m curious to hear your experiences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help.&lt;/p&gt;&#xA;" OwnerUserId="17780" LastActivityDate="2019-07-08T17:52:19.317" Title="Is there a rule of thumb to construe a screw" Tags="|mechanical-engineering|bolting|threads|fasteners|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29091" PostTypeId="1" CreationDate="2019-07-08T15:24:16.240" Score="0" ViewCount="41" Body="&lt;p&gt;Can someone tell me what the type of connector is??&lt;a href=&quot;https://i.stack.imgur.com/5G4jl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5G4jl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20972" LastActivityDate="2019-07-08T16:24:57.217" Title="What is the type of this connector?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29092" PostTypeId="2" ParentId="25864" CreationDate="2019-07-08T16:13:14.867" Score="1" Body="&lt;p&gt;Your assessment of the situation is correct - the solution is of course to specify a pressure regulator with a large enough orifice that it does not cause you a problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure regulators are specified with a Maximum Flow Rate, in the same way they are specified with Maximum Operating Pressure. You simply need to pick one that is suitable for your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yMnQB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yMnQB.png&quot; alt=&quot;Screenshot of RS Pressure Regulators&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-07-08T16:13:14.867" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29093" PostTypeId="2" ParentId="29091" CreationDate="2019-07-08T16:24:57.217" Score="2" Body="&lt;p&gt;Based on the limited information in the photo, I used &quot;female breadboard jumper wire&quot; as a preliminary search. This resulted in many returns, all of which appear as in the image. Additionally, the term &quot;Dupont&quot; arose. More searching results in returns for single pin Dupont connectors as well as multiple pin connectors, often found on computer mainboards for USB connections and similar.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-07-08T16:24:57.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29094" PostTypeId="2" ParentId="29086" CreationDate="2019-07-08T16:42:56.163" Score="1" Body="&lt;p&gt;I cannot really comment on the LQR part of your question as I'm only marginally familiar with it (I've never seen a practical/industrial implementation but that's probably due to the specific application area I'm working in). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general it is not possible to tune multiple SISO controllers for a MIMO system without considering the interaction between the different loops. Otherwise the combination of SISO controllers does not necessarily result in optimal performance or can even result in an unstable system. Look up relative gain array (RGA) for one approach towards investigating this interaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As your example shows tuning these controllers requires a MIMO approach. The simplest approach would be to apply sequential loop closing in combination with your favorite controller tuning algorithm. &lt;/p&gt;&#xA;" OwnerUserId="11869" LastActivityDate="2019-07-08T16:42:56.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29095" PostTypeId="1" AcceptedAnswerId="29096" CreationDate="2019-07-08T16:46:00.300" Score="4" ViewCount="1265" Body="&lt;p&gt;I am curious as to how pump salad spinners work: not in terms of how they spin and use centripetal forces to get rid of water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean in terms of how the pump itself works. How does the 2D motion of pushing the pump get converted into the rotation and spinning of the salad spinner itself? Are there gears involved? I would like to know the underlying mechanism of it.&lt;/p&gt;&#xA;" OwnerUserId="21073" LastActivityDate="2019-07-09T11:23:15.550" Title="How do Salad Spinners work?" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29096" PostTypeId="2" ParentId="29095" CreationDate="2019-07-08T17:42:55.813" Score="4" Body="&lt;p&gt;Inside the pump-like mechanism, there is a vertical shaft with helical grooves cut into its length, which wrap around the shaft. the bottom of this shaft is connected to the salad basket and supported by a bearing at its base, so the shaft and the basket can revolve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;pump&quot; handle is a hollow tube which slips over the basket shaft. Inside it are a couple of protrusions which fit into the spiral groove. When you press down on the handle, those protrusions engage the groove and as the protrusions descend, they urge the shaft to revolve, which sets the basket into rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the bottom of the handle's stroke, the sprial groove blends into a circumferential cut or &quot;runout groove&quot; in the shaft so the shaft and basket can continue to revolve without any further motion on the part of the handle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you release pressure on the handle, a spring urges the handle back up again while the protrusions retract. At the top of the handle's travel, there is another runout groove that the protrusions fall into, and when you push down a second time on the handle, the protrusions find the spiral groove and the downward force applied by your hand on the handle forces the shaft to turn faster, and the process repeats. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13311" LastEditDate="2019-07-09T11:23:15.550" LastActivityDate="2019-07-09T11:23:15.550" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29097" PostTypeId="2" ParentId="29090" CreationDate="2019-07-08T17:52:19.317" Score="2" Body="&lt;p&gt;In a handbook of mechanical engineering facts you will find tables containing maximum load ratings for different diameters and types of screw fasteners, and required torque values to hold those loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will also find this data on the website of a company that manufactures screw fasteners. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most mechanical engineering design textbooks will contain a chapter on &lt;em&gt;threaded fasteners&lt;/em&gt; which explain the connection between tensile load and required torque, thread friction, thread pitch, screw diameter, screw grade, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;American Society For Testing And Materials&lt;/em&gt; (ASTM) specification book will contain a chapter on how screw strength is measured, and how different screws are ranked for strength. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-07-08T17:52:19.317" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29098" PostTypeId="2" ParentId="29083" CreationDate="2019-07-08T18:18:34.960" Score="0" Body="&lt;p&gt;The counterclockwise moment applied at A will be transferred as torque to the 90 elbow and from there it will impart a moment of 70N.m all the way along the length of that section of the pipe, in similar way as when you apply this moment to the free end of a cantilever beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the moment of 70N.m is the constant along the length of that section of the pipe starting from the elbow to and including the point B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is correct and you must include 70N.m in the sum of the moments.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-07-08T18:42:17.013" LastActivityDate="2019-07-08T18:42:17.013" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29100" PostTypeId="1" CreationDate="2019-07-08T21:53:42.780" Score="1" ViewCount="79" Body="&lt;ol&gt;&#xA;&lt;li&gt;Is there a kind of &quot;handbook&quot; for general engineering ideas and possible applications?  I vaguely remember a machinist friend mentioning some standard book to glean ideas from...&lt;/li&gt;&#xA;&lt;li&gt;I'm in need of an extendable arm.  I wish to raise something into the air ~10-12 feet (with no anchor above), hold ~3-5 lbs, and be as compact as possible when retracted.  Telescoping mechanisms are the obvious choice, but I think their collapsed size makes them already too long (especially to reach 12 feet, with their sections being &gt;2 feet).  I'd like to be able to carry this on a belt or small harness without being cumbersome.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;My mind keeps turning to something like a retractable tape measure but I'm sure there's a better solution, and that's not stable enough to hold the weight steady.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance, I appreciate you reading this!&lt;/p&gt;&#xA;" OwnerUserId="21077" LastEditorUserId="8055" LastEditDate="2019-07-09T17:46:33.283" LastActivityDate="2021-04-30T02:10:10.950" Title="Extendable apparatus" Tags="|mechanical-engineering|statics|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29101" PostTypeId="1" CreationDate="2019-07-09T00:30:48.907" Score="0" ViewCount="2997" Body="&lt;p&gt;I am reading thermodynamics from Cengel and Boles. I am currently studying flow work. The following image contains text of the book:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/eCGXR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eCGXR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work done is calculated by considering the pressure of upstream fluid but the downstream fluid also applies pressure on it,but that is not considered why? According to me,the work done should be (P2-P1)*V where P2 is pressure of fluid upstream and P1 is pressure  downstream at the boundary of control volume.&lt;/p&gt;&#xA;" OwnerUserId="20605" LastEditorUserId="33" LastEditDate="2019-08-12T04:51:37.617" LastActivityDate="2024-01-18T21:05:35.780" Title="How is flow work calculated?" Tags="|mechanical-engineering|thermodynamics|fluid|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29103" PostTypeId="2" ParentId="29080" CreationDate="2019-07-09T01:57:50.537" Score="0" Body="&lt;p&gt;You can mount your head part on a neck globe with two pins protruding out and  screw that into a  sphere socket like a kids doll neck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This spherical socket is threaded in a pitch that spirals around the sphere with the slope of 30 degrees over a half a turn and the two pins slide back and forth in this thread. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-09T01:57:50.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29104" PostTypeId="2" ParentId="29073" CreationDate="2019-07-09T02:17:54.960" Score="1" Body="&lt;p&gt;There are two advantages for having a curtain versus not.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Even though the curtains absorb and convert sun's energy into infrared heat, but because of their relative small mass they can not absorb much energy radiated by the sun light, as opposed to sun heating up the entire room and its contents. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The  little heat generated by the convection around the shades goes up directly without interface with the space used by people.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In designing the interior heating and air conditioning it is beneficial to limit the controlled air space to the space used by people and save energy. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-09T02:17:54.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29107" PostTypeId="1" CreationDate="2019-07-09T13:02:31.073" Score="0" ViewCount="21" Body="&lt;p&gt;For an electronics device that I make I use a standard plastic (I think polystyrene) enclosure. This enclosure needs to be precision-drilled, ant I lack the dedicated equipment. What I think would work is to pour some kind of mould on the outside of the enclosure and bore it as needed. I would then reuse the mould for drilling other enclosures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And to the question - what material and thickness I should aim for for the said mould ?&lt;/p&gt;&#xA;" OwnerUserId="3412" LastActivityDate="2019-07-09T17:08:34.900" Title="mould for boring enclosures" Tags="|drilling|molding|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29109" PostTypeId="1" AcceptedAnswerId="33033" CreationDate="2019-07-09T15:24:30.927" Score="0" ViewCount="845" Body="&lt;p&gt;I want to customise a piece of fabric that is 100% polyester (exclusive of decorations, wash at 30º) with vinyl. My idea is to use a vinyl cutter (such as Silver Bullet) to cut a pattern on the sheet, remove the background, transfer to a transfer sheet, and use a heat press to adhere the design to the fabric.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen good results of this method on cotton and I am not sure if a heat press would work on polyester.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I transfer a digital design onto 100% polyester fabric?&lt;/p&gt;&#xA;" OwnerUserId="19384" LastActivityDate="2022-08-19T14:31:54.023" Title="Heat press of vinyl on 100% polyester fabric" Tags="|materials|heat-treatment|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29110" PostTypeId="1" CreationDate="2019-07-09T15:43:37.747" Score="-1" ViewCount="545" Body="&lt;p&gt;I have looked at many diagrams of the Chernobyl reactor and am wondering where the coolant comes from that enters the control rod channels when the control rods are raised. All the diagrams show coolant being pumped through the pipes that contain the fuel cells and then into the steam separators and back to the pumps (see enclosed diagram). All of the articles written about this reactor state that when the control rods are lowered, they displace coolant. None of the diagrams show additional coolant being pumped into the reactor vessel or being in any of the empty channels.&lt;a href=&quot;https://i.stack.imgur.com/tpP5p.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tpP5p.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21087" LastEditorUserId="12827" LastEditDate="2019-07-13T17:40:35.647" LastActivityDate="2019-07-13T17:51:54.013" Title="Is the core of an RBMK reactor submersed in coolant?" Tags="|nuclear-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29111" PostTypeId="2" ParentId="29107" CreationDate="2019-07-09T17:08:34.900" Score="1" Body="&lt;p&gt;the way this problem is usually solved is to precisely machine a chunk of metal to fill the  inside of the box snugly, and then clamp or bolt that chunk to the drill table. when you start drilling the box wall, the metal core inside it prevents the plastic from flexing. you continue drilling until the bit passes all the way through the plastic wall and engages the metal chunk. Then you withdraw the bit and pull the box off the core. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you cannot afford to have the core machined from metal, a good substitute is epoxy casting resin. you coat the inside of the box with &lt;em&gt;mold release&lt;/em&gt;, pour the resin straight into the box and let it harden, then strip the box off and there's your core. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By immersing a bolt into the soft epoxy so its head is in the epoxy and its threaded end sticks up out of it, you then have a convenient means of fastening the core to your drill table after the epoxy hardens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is common to blend metal or mineral powder with the epoxy to made it sturdier. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-07-09T17:08:34.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29112" PostTypeId="2" ParentId="29100" CreationDate="2019-07-09T17:11:12.153" Score="0" Body="&lt;p&gt;the handboook is probably the &lt;em&gt;McMaster-Carr Company&lt;/em&gt; parts catalog, which contains all sorts of mechanical gizmos and doodads available off-the-shelf for jobs like yours.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-07-09T17:11:12.153" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29113" PostTypeId="2" ParentId="29100" CreationDate="2019-07-09T17:55:53.840" Score="0" Body="&lt;p&gt;What you are asking is impractical and impossible to handle even in a scale that a normal person wrist can handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically anything that can deploy a load of 5lbs at a distance of 12 feet, if it has a handle that you describe in the order of 6 inches, will cause a torque of &lt;span class=&quot;math-container&quot;&gt;$$5*12ft/0.5ft=120lbs.ft\ $$&lt;/span&gt;  Which is beyond a regular man wrist power.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-09T17:55:53.840" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29114" PostTypeId="1" AcceptedAnswerId="29115" CreationDate="2019-07-09T18:16:13.357" Score="0" ViewCount="39" Body="&lt;p&gt;I have a quick question about the force needed to accelerate/decelerate an irregular object. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say I have a 3D object with the following general shape,&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YAZxr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YAZxr.png&quot; alt=&quot;Gripper Assembly on a Fixed Beam&quot;&gt;&lt;/a&gt;&#xA;This system is in general a gripper (gray) holding a load (yellow) and fixed onto a translating horizontal beam (blue).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking, to calculate the stopping force of the load and gripper, I just would need to find the center of mass of the load and gripper and apply Newton's Second Law with the expected acceleration/deceleration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does using a center of mass calculation with Newton's Second Law result in a large error from the actual system behavior?&lt;/p&gt;&#xA;" OwnerUserId="20847" LastActivityDate="2019-07-09T18:53:17.297" Title="Stopping Force of Irregularly Shaped Object" Tags="|mechanical-engineering|structural-engineering|structural-analysis|applied-mechanics|mechanical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29115" PostTypeId="2" ParentId="29114" CreationDate="2019-07-09T18:53:17.297" Score="2" Body="&lt;p&gt;As you said the basic deceleration is &lt;span class=&quot;math-container&quot;&gt;$$  \alpha =\frac{v_{initial}-v_{final}}{t}=\frac{v}{t}$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of secondary order magnitude are things like the tendency of the load to tilt forward an swing up under negative acceleration and that leads to checking for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The stability and stiffness of the beam and its mass compared to the mass of load and bracket.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The play of the bracket so as it does not permit the stopping degenerate into vibration and shimmying of the bearings.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The hammer action of the load at the end of its course.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-09T18:53:17.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29116" PostTypeId="1" CreationDate="2019-07-09T19:05:09.250" Score="3" ViewCount="73" Body="&lt;p&gt;I have a small self-built food processing unit (500sqft) but it is very labor-intensive. I want to make a larger unit (45,000sqft) with 100x more output using factory machinery (automation, conveyor belts, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current unit is composed of a lot of electrical components and plumbing systems; it's not top notch and beyond my expertise to scale into something larger. It is a food manufacturing and processing plant with many chemical inputs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not exactly sure who to contact regarding this build or where to begin my search. &lt;/p&gt;&#xA;" OwnerUserId="21092" LastEditorUserId="21092" LastEditDate="2019-07-10T09:32:29.410" LastActivityDate="2019-07-10T09:32:29.410" Title="What kind of professional(s) build/design factories?" Tags="|mechanical-engineering|electrical-engineering|manufacturing-engineering|industrial-engineering|" AnswerCount="0" CommentCount="3" ClosedDate="2019-07-09T19:18:53.047" ContentLicense="CC BY-SA 4.0" />
  <row Id="29117" PostTypeId="1" CreationDate="2019-07-09T23:26:59.110" Score="0" ViewCount="379" Body="&lt;p&gt;I am looking for a starting back of the envelope calculation to determine if an aluminum box with just free convection can dissipate 3.7W of internal heat generation or if I need to think about adding a heat sink into design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way I see it the heat generating electronics would be radiating heat via convection to internal air of box and then heat would travel into box and via free convection into atmosphere. I'm fine neglecting radiation and am not sure how to approach this problem after reading for the last couple hours on thermal stuff.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;T ambient = 50C&#xA;T max for electronics= 40C&#xA;T Internal starting =20C &#xA;Q generated by electronics = 3.7W&#xA;Box dims = 75mmx75mmx100mm , Area=.0412m^2, Vol=.0005625 m^3&#xA;Box thickness= 7mm&#xA;Box material is aluminum &#xA;R_aluminum=200&#xA;R_air =.0262&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help on how to approach this &lt;/p&gt;&#xA;" OwnerUserId="21096" LastActivityDate="2019-07-21T00:08:30.773" Title="Simple Thermal Analysis for sealed Aluminum box with 3.7 Watts of heat generation internally" Tags="|mechanical-engineering|mechanical|thermal-conduction|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29120" PostTypeId="2" ParentId="6682" CreationDate="2019-07-10T03:41:26.453" Score="1" Body="&lt;p&gt;&lt;strong&gt;ABS&lt;/strong&gt; has an acrid smell when heating is applied. The flame is yellow with blue edges and will not drip. ABS will continue to burn when the flame source is removed and the odor continues to be acrid, the flame color is the same yellow with blue edges and the material will drip. The rate of burning is slow and black smoke with soot is admitted into the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Polypropylene&lt;/strong&gt; smells acrid when a flame is applied and the flame is yellow in color. It will continue to burn after the flame source is removed giving off a sweet smell. The flame at this point is blue with a yellow tip. PP produces drips as it continues to burn slowly. When bent, polypropylene sheet will turn white on the crease line. Interestingly, PP floats in water (density &amp;lt; 1.0 g/cm3).&lt;/p&gt;&#xA;" OwnerUserId="21097" LastEditorDisplayName="user16" LastEditDate="2019-07-12T11:18:55.360" LastActivityDate="2019-07-12T11:18:55.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29122" PostTypeId="1" CreationDate="2019-07-10T09:03:31.630" Score="0" ViewCount="49" Body="&lt;p&gt;I would like to find the common volume between cylinders for different radii and at different angles (3-D). Can I do this using ANSYS design modeller? Or what can I use (preferably a free download for a student please) and how long would it take to compute one solution? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I essentially would like to create a lookup table for the common volume between different cylinders that look like this for example.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TD3mP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TD3mP.png&quot; alt=&quot;cylinders&quot;&gt;&lt;/a&gt;&#xA;Thank you for your help!&lt;/p&gt;&#xA;" OwnerUserId="21098" LastEditorUserId="21098" LastEditDate="2019-07-10T19:20:03.920" LastActivityDate="2019-07-10T19:20:03.920" Title="Software to find the common volume of intersecting cylinders?" Tags="|computer-aided-design|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29125" PostTypeId="1" CreationDate="2019-07-10T12:50:29.087" Score="1" ViewCount="22" Body="&lt;p&gt;The cryogenic fuel I'm using is methane at 100 K and 300,000 Pa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on my calculations, after compression, the pressure is 55 MPa and using the ideal gas equation the temperature is approx. 800 K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to keep the methane at a temperature of 300 K - 600 K preferably as close to 300 K as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am after any designs or solutions to keep the temperature low, like regenerative cooling, or is that just ridiculous?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The application is for a rocket, temperatures in the vicinity will be high.&lt;/p&gt;&#xA;" OwnerUserId="13506" LastEditorUserId="90" LastEditDate="2019-07-10T14:02:02.087" LastActivityDate="2019-07-10T14:02:02.087" Title="How to minimize temperature increase caused by compression of a cryogenic liquid in a multistage high pressure turbopump?" Tags="|fluid-mechanics|thermodynamics|aerospace-engineering|rocketry|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29126" PostTypeId="1" CreationDate="2019-07-10T17:59:51.503" Score="1" ViewCount="1064" Body="&lt;p&gt;All of the images of maglev trains that I've seen seem to have a continuous drive mechanism along the entire length of each car:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1mfqs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1mfqs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume this must be articulated some how, or the train wouldn't be able to go around curved track. How does this function? With two bogies on a traditional train car, the bogies only need to be able to rotate in place, with the train car being a linear line from one bogie to the next over the curved track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jfQlE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jfQlE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with however many different articulated sections per car here, how are they attached to the train so that it can turn?&lt;/p&gt;&#xA;" OwnerUserId="19347" LastEditorUserId="19347" LastEditDate="2019-07-10T18:13:08.320" LastActivityDate="2019-07-10T19:42:11.340" Title="How do maglev trains turn given that they seem to have more than two bogies per car?" Tags="|electromagnetism|rail|transportation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29127" PostTypeId="1" AcceptedAnswerId="29132" CreationDate="2019-07-10T18:07:39.903" Score="0" ViewCount="1580" Body="&lt;p&gt;Does there exist any type of Nut, that can be screwed on a threaded rod that can slipped in without having to go to the ends of the threaded rod?  I can sort of imagine a Nut that was a &lt;code&gt;U&lt;/code&gt; shape instead of &lt;code&gt;O&lt;/code&gt; shape, or perhaps two &lt;code&gt;C&lt;/code&gt; shapes that could interlock some how.  Does anything like this exist?&lt;/p&gt;&#xA;" OwnerUserId="21101" LastEditorUserId="21101" LastEditDate="2019-07-10T18:44:33.207" LastActivityDate="2023-09-19T00:03:59.083" Title="Fastening a Nut on threaded rod from the middle" Tags="|fasteners|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29128" PostTypeId="2" ParentId="29127" CreationDate="2019-07-10T18:23:30.083" Score="1" Body="&lt;p&gt;Yes, a threaded tube that has been cut in half length-wise and clamped or joined to the threaded section with nuts and bolts or welded as suitable.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-07-10T18:23:30.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29129" PostTypeId="2" ParentId="29127" CreationDate="2019-07-10T18:35:57.773" Score="2" Body="&lt;p&gt;In the world of machining, a lathe uses a feature called a half-nut. It's two half-nuts mounted on a slide in which they are separated to disengage the carriage from the lead screw, or brought closer together to engage the carriage to the lead screw. This stops or starts movement of the cutting tool on the work piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vcWgM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vcWgM.jpg&quot; alt=&quot;lathe half nut&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image above comes from &lt;a href=&quot;http://www.mini-lathe.com/Mini_lathe/Features/features_2.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Mini-lathe.com website&lt;/a&gt; and shows a crude version, which is all that is needed to accomplish the above-described task. The release/activation mechanism moves the upper half downward onto the lead screw (aka threaded rod, specialty version) and the lower half upward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are as many versions of half-nuts as there are versions of lathes using them.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-07-10T18:35:57.773" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29130" PostTypeId="2" ParentId="29126" CreationDate="2019-07-10T18:37:34.253" Score="1" Body="&lt;p&gt;There is a minimum radius used for a curve, these are wide and gentle turns and the &quot;slot&quot; in the train is wider than the top of the track to add to the clearance available.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-07-10T18:37:34.253" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29131" PostTypeId="2" ParentId="29126" CreationDate="2019-07-10T19:42:11.340" Score="1" Body="&lt;p&gt;There are guidance magnets and levitation magnets. The guidance magnets are designed to maintain the car alignment, never letting any physical contact. Ther is transverse inclination of the rails too, which helps reducing the curve of the turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The off center displacement for a turn radius of 2km at 3 axels of 12m spaced :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sqrt(2000^2 + 12^2)-2000= 0.036 = 3.6cm$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing this by two, splitting the offset between the three axels we need 1.8cm gap between the guidance magnets and the car vertical magnets. just to get an idea check the detail diagram.&lt;a href=&quot;https://i.stack.imgur.com/bkWV7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bkWV7.jpg&quot; alt=&quot;detail&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-10T19:42:11.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29132" PostTypeId="2" ParentId="29127" CreationDate="2019-07-10T21:41:25.937" Score="6" Body="&lt;p&gt;Granger Supply makes two different types:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sHOJf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sHOJf.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PEUBK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PEUBK.jpg&quot; alt=&quot;Caddy Threaded Rod Nut&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The round one is called A Steel Shaft Collar,two piece, Threaded . The second one is called a &amp;quot;CADDY THREADED ROD NUT&amp;quot;. They come in a number of sizes.&lt;/p&gt;&#xA;" OwnerUserId="21087" LastEditorUserId="21087" LastEditDate="2023-09-19T00:03:59.083" LastActivityDate="2023-09-19T00:03:59.083" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29133" PostTypeId="2" ParentId="29101" CreationDate="2019-07-10T21:59:20.077" Score="0" Body="&lt;p&gt;PA is force . F*L=PAL=PV is work. if you consider per Kg, then W= &lt;strong&gt;Pv&lt;/strong&gt;&#xA;When you write the energy balance equation at point 1 and 2, you write &lt;strong&gt;P1*v&lt;/strong&gt; left and &lt;strong&gt;P2*v&lt;/strong&gt; right along with other terms (internal, kinetic, potential). See, you get &lt;strong&gt;your desired term&lt;/strong&gt; by reorganizing: &#xA;&lt;strong&gt;0=(P2-P1)*v + (ke2-ke1) + (pe2-pe1) + (u2-u1).&lt;/strong&gt;&#xA;Considering no heat flow.&#xA;heat=work+change of internal/potential/kinetic energy. Well, that's the first law.&#xA;Think from Thermodynamic System perspective, the surrounding (upstream) does work on the system and the system does work on the surrounding (downstream). You can get the net work from difference&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It is interesting that unlike other work quantities, flow work is expressed&#xA;  in terms of properties. In fact, it is the product of two properties of the fluid.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now if you imagine this as combination property , it becomes easy to write energy balance equation. or otherwise you separately add the net work. &lt;strong&gt;(P2-P1)*v&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="21063" LastEditorUserId="21063" LastEditDate="2019-07-10T22:09:50.680" LastActivityDate="2019-07-10T22:09:50.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29137" PostTypeId="2" ParentId="29062" CreationDate="2019-07-11T08:52:25.863" Score="0" Body="&lt;p&gt;Because the compression in a turbojet is an aerodynamic effect caused by the rotating compressor blades.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lift and drag forces on an object moving through a fluid are proportional to the square of its relative velocity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F \propto v^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that velocity is linear with the rpm of the shaft because:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v = \omega \cdot r$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;hence, since work done is also linear with the applied force:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$W \propto F \propto \omega^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Thrust is not linear with the compression ratio, however, and also depends on an array of other factors. Thrust is the net effect of the Brayton cycle in the engine, and that cycle develops along an &lt;a href=&quot;https://en.wikipedia.org/wiki/Adiabatic_process#Ideal_gas_(reversible_process)&quot; rel=&quot;nofollow noreferrer&quot;&gt;adiabatic&lt;/a&gt; line during the compression step.&lt;/p&gt;&#xA;" OwnerUserId="10400" LastActivityDate="2019-07-11T08:52:25.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29138" PostTypeId="1" CreationDate="2019-07-11T09:59:30.013" Score="0" ViewCount="23" Body="&lt;p&gt;When selecting an actuator for a controls application - e.g. choosing a servo motor for a self-balancing robot - I understand that it's important for the actuators to have fast enough dynamics that they don't significantly impact the system's linearity, and thus ensure the control algorithms function as designed. However, how much faster is fast enough? Are there any &lt;strong&gt;rules of thumb&lt;/strong&gt; - c.f. Nyquist's Theorem in signal processing - for how I should choose my motor's rise time given that my controller's sample period is known?&lt;/p&gt;&#xA;" OwnerUserId="21067" LastActivityDate="2019-07-11T09:59:30.013" Title="How does a digital controller's sampling period dictate the system's actuation requirements?" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29139" PostTypeId="1" CreationDate="2019-07-11T11:04:14.617" Score="0" ViewCount="70" Body="&lt;p&gt;I'm in the processes of building something which requires I'll be able to control the independent opening/closure of two (small, ~10mm in diameter) different adjacent holes using one finger, and I'm trying to think of ways such a thing can be put together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The movement required from the finger shouldn't be too convoluted, so my original vision involves having the two holes open when there's no pressure applied to the button, and then when the button is pressed half-way through it closes only one hole while when it's pressed all the way down it closes both holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control in such a situation might be tough to perform with accuracy, so my other idea involves having the button's resting position rather at the middle of the available movement, allowing it to be moved up or down while the middle position leaves only one hole open, the upper position opens both and the lower position closes both (in this case the &quot;button&quot; won't be a button but rather a &quot;cup&quot; of sorts you can place your fingertip into).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed-open combination is unidirectional, so the possible closure patterns required are either both hole¹ and hole² open, hole¹ open and hole² closed, or both closed, but no requirement to have hole¹ closed while hole² is open.&#xA;They should be able to close the holes against the air pressure of exhalation (or less actually as there would be ventilation elsewhere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the hole-closing structure itself would involve padded surfaces dropping onto each hole to cover it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any tips how could such a mechanism be built?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT (additional info):&lt;/strong&gt;&#xA;I think some further clarification is required. The holes would be situated on a cylindrical surface. The controller should be placed at 90 degrees to the two holes (so it doesn't continue in the same line of direction), and the closing surfaces should open plenty so there's no obstruction in air flow, and possibly open to the side opposite the controller. Here's a rough sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ugW7e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ugW7e.jpg&quot; alt=&quot;Sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously this is not an actual mechanical design, just guidance for the wanted form.&lt;/p&gt;&#xA;" OwnerUserId="21107" LastEditorUserId="21107" LastEditDate="2019-07-11T18:15:38.983" LastActivityDate="2019-07-12T10:41:52.893" Title="Designing a 1-button system allowing for a differentiated control of 2 different valves?" Tags="|mechanical-engineering|design|gears|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29140" PostTypeId="1" CreationDate="2019-07-11T15:10:01.403" Score="0" ViewCount="148" Body="&lt;p&gt;I want to know how does Milkor MGL's trigger and cylinder work. In particular, I am interested in how exactly the trigger action switches the cylinder among the cartridge chambers. As far as i know, the trigger itself works in a basic double action principle, but, I simply, cannot figure by myself how it does manage to rotate the drum forward, it doesn't work like regular pistol revolver but in a rather different manner. The following video demonstrates the basic drum cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=KMzeIyqHW4Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=KMzeIyqHW4Y&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will greatly appreciate if you can give me some sort of diagram or schematic to make things clear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="21110" LastActivityDate="2020-02-15T10:09:38.470" Title="How does Milkor MGL Granade Launcher's trigger and cylinder mechanism work" Tags="|firearms|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29141" PostTypeId="1" AcceptedAnswerId="29144" CreationDate="2019-07-11T15:23:33.400" Score="1" ViewCount="122" Body="&lt;p&gt;I am reading the rocket propulsion elements book by sutton and I came across the following statement: &quot;Because the propellant tank has to fly. its mass is at a premium and the tank material is therefore highly stressed.&quot;&#xA;I am just learning, and have no strong technical background, so I will appreciate the explanation!&#xA;Thanks! &lt;/p&gt;&#xA;" OwnerUserId="14978" LastActivityDate="2019-07-11T15:54:30.607" Title="What does it mean that the mass of a propellant tank is at premium?" Tags="|structural-engineering|aerospace-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29142" PostTypeId="2" ParentId="26026" CreationDate="2019-07-11T15:28:39.483" Score="0" Body="&lt;p&gt;Answering my own question in hopes others will stumble upon it.&#xA;I did not find an acceptable way to directly export points off a spline.&lt;br&gt;&#xA;Attached is my two part solution.&lt;br&gt;&#xA;I was able to export the parameters that define the spline from Solidworks and then process that data with a small python script.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the VBA script to run in Solidworks that will export the data for a spline.&lt;br&gt;&#xA;In my particular application it is looking for two 3d sketches (centerLineFull and rackPosFull), each of these sketches has exactly 1 spline.&lt;br&gt;&#xA;The parameters of the spline are then output to a file that is formated to look like a python dictionary.&lt;br&gt;&#xA;You will need to modify all of this for your needs.. but you get the idea..&lt;br&gt;&#xA;[A lot of this was lifted from various Solidworks API help files. There are a lot of goof search terms in here that will lead to the correct places in the API help files ]&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    Attribute VB_Name = &quot;Macro21&quot;&#xA;Option Explicit&#xA;    Dim swApp As SldWorks.SldWorks&#xA;    Dim Part As SldWorks.ModelDoc2&#xA;    Dim swSelectMgr As SldWorks.SelectionMgr&#xA;    Dim swFeature As Feature&#xA;    Dim swSketch As Sketch&#xA;&#xA;    Dim swSketchSeg As SldWorks.SketchSegment&#xA;    Dim swCurveIn As SldWorks.Curve&#xA;    Dim varSplineParams As Variant&#xA;    Dim success As Boolean&#xA;    Dim swActiveSketch As Sketch&#xA;&#xA;    Dim SketchSegments As Variant&#xA;    Dim SketchSegment As Variant&#xA;    Dim swSketchSegment As SketchSegment&#xA;&#xA;    Dim FilePath As String&#xA;    Dim FileName As String&#xA;&#xA;&#xA;&#xA;Dim boolstatus As Boolean&#xA;Dim longstatus As Long, longwarnings As Long&#xA;Dim myModelView As Object&#xA;&#xA;&#xA;&#xA;&#xA;' Extract two integer values out of a single&#xA;' double value, assuming that a C int has 4 bytes&#xA;Type DoubleRec&#xA;    dValue As Double&#xA;End Type&#xA;Type Int2Rec&#xA;    iLower As Long&#xA;    iUpper As Long&#xA;End Type&#xA;&#xA;' Extract two integer values out of a single&#xA;' double value, by assigning a DoubleRec to the&#xA;' double value then copying the value over an&#xA;' Int2Rec and extracting the integer values&#xA;Function ExtractFields(dValue As Double, iLower As Integer, iUpper As Integer)&#xA;    Dim dr As DoubleRec&#xA;    Dim i2r As Int2Rec&#xA;    ' Set the double value&#xA;    dr.dValue = dValue&#xA;    ' Copy the values&#xA;    LSet i2r = dr&#xA;    ' Extract the values&#xA;    iLower = i2r.iLower&#xA;    iUpper = i2r.iUpper&#xA;End Function&#xA;Sub SplineDump(varSplineParams As Variant)&#xA;    Dim dTmpValue As Double&#xA;    Dim iNumKnots As Integer&#xA;    Dim iNumCtrlPts As Integer&#xA;    Dim iDimension As Integer&#xA;    Dim iOrder As Integer&#xA;    Dim iPeriodicity As Integer&#xA;    Dim iSplineArraySize As Integer&#xA;    Dim iSplineIndex As Integer&#xA;    Dim iVarIndex As Integer&#xA;    Dim i As Integer&#xA;    dTmpValue = varSplineParams(0)&#xA;    ExtractFields dTmpValue, iDimension, iOrder&#xA;    dTmpValue = varSplineParams(1)&#xA;    ExtractFields dTmpValue, iNumCtrlPts, iPeriodicity&#xA;    'Debug.Print &quot;Dimension &quot;; iDimension, &quot; Order&quot;; iOrder, &quot; NumCtrlsPts&quot;; iNumCtrlPts, &quot;Periodicity &quot;; iPeriodicity&#xA;    Print #1, &quot;'Dimension':&quot;; iDimension; &quot;,&quot;&#xA;    Print #1, &quot;'Order':&quot;; iOrder; &quot;,&quot;&#xA;    Print #1, &quot;'NumberOfControlPoints':&quot;; iNumCtrlPts; &quot;,&quot;&#xA;    Print #1, &quot;'Periodicity':&quot;, iPeriodicity; &quot;,&quot;&#xA;    Print #1, &quot;'KnotPoints':[&quot;&#xA;    iNumKnots = iOrder + iNumCtrlPts&#xA;    iVarIndex = 2&#xA;&#xA;    For i = 0 To (iNumKnots - 1)&#xA;    Print #1, varSplineParams(iVarIndex); &quot;,&quot;&#xA;    iVarIndex = iVarIndex + 1&#xA;    Next i&#xA;    Print #1, &quot;],&quot;&#xA;    Print #1, &quot;#End of Knot Points&quot;&#xA;    Print #1, &quot;&quot;&#xA;    Print #1, &quot;#Control Points...&quot;&#xA;    Print #1, &quot;'ControlPoints':[&quot;&#xA;    For i = 0 To (iNumCtrlPts - 1)&#xA;    Dim j As Long&#xA;    Dim X, Y, Z As Double&#xA;    X = varSplineParams(iVarIndex) * 1000#&#xA;    Y = varSplineParams(iVarIndex + 1) * 1000&#xA;    Z = varSplineParams(iVarIndex + 2) * 1000&#xA;    'Debug.Print X, Y, Z&#xA;    Print #1, &quot;[&quot;;&#xA;    Write #1, X, Y, Z;&#xA;    Print #1, &quot;],&quot;&#xA;    iVarIndex = iVarIndex + 3&#xA;    Next i&#xA;    Print #1, &quot;],&quot;&#xA;    Print #1, &quot;#End of Control Points&quot;&#xA;End Sub&#xA;Sub SelectStartFeature()&#xA;    Dim SketchArcs As Variant&#xA;    Dim SketchArc As Variant&#xA;    Dim SketchPoints As Variant&#xA;    Dim SketchPoint As Variant&#xA;    Dim swSketchPoint As SketchPoint&#xA;    Dim swArcType As SketchArc&#xA;    Dim Point(2) As Double&#xA;    Dim PointVariant As Variant&#xA;    Dim swXForm As MathTransform&#xA;    Dim swMathUtil As MathUtility&#xA;    Dim swMathPt As MathPoint&#xA;    Dim status As Boolean&#xA;    Dim work As String&#xA;    Dim i As Integer&#xA;&#xA;&#xA;    Print #1, &quot;#Start Features&quot;&#xA;    status = Part.Extension.SelectByID2(&quot;start&quot;, &quot;SKETCH&quot;, 0, 0, 0, False, 0, Nothing, 0)&#xA;    Debug.Print &quot;Selected Shetch 'start' is &quot;, status&#xA;    Set swFeature = swSelectMgr.GetSelectedObject6(1, 0)&#xA;    Set swSketch = swFeature.GetSpecificFeature2&#xA;    Set swXForm = swSketch.ModelToSketchTransform&#xA;    Set swXForm = swXForm.Inverse&#xA;    Set swMathUtil = swApp.GetMathUtility&#xA;&#xA;    Debug.Print &quot;Sketch is in 3d:&quot;; swSketch.Is3D&#xA;&#xA;    Debug.Print &quot;Number of arcs &quot;, swSketch.GetArcCount&#xA;    Print #1, &quot;'NumberArcsFound':&quot;, swSketch.GetArcCount; &quot;,&quot;&#xA;    SketchArcs = swSketch.GetArcs2&#xA;    Debug.Print &quot;Arc Center on Sketch&quot;, SketchArcs(12) * 1000, SketchArcs(13) * 1000, SketchArcs(14) * 1000&#xA;    'Create a Temporary point to use to transform from Sketch coor to model&#xA;    Point(0) = SketchArcs(12)&#xA;    Point(1) = SketchArcs(13)&#xA;    Point(2) = SketchArcs(14)&#xA;    PointVariant = Point&#xA;    Set swMathPt = swMathUtil.CreatePoint(PointVariant)&#xA;    Set swMathPt = swMathPt.MultiplyTransform(swXForm)&#xA;    Debug.Print &quot;Arc Center in Model&quot;, swMathPt.ArrayData(0) * 1000, swMathPt.ArrayData(1) * 1000, swMathPt.ArrayData(2) * 1000&#xA;    Print #1, &quot;'ArcCenter':[&quot;;&#xA;    Write #1, swMathPt.ArrayData(0) * 1000, swMathPt.ArrayData(1) * 1000, swMathPt.ArrayData(2) * 1000;&#xA;    Print #1, &quot;],&quot;&#xA;    Print #1, &quot;'NumberPointsFound':&quot;, swSketch.GetSketchPointsCount2; &quot;,&quot;&#xA;    Print #1, &quot;&quot;&#xA;    Debug.Print &quot;Number of points&quot;, swSketch.GetSketchPointsCount2&#xA;    i = 0&#xA;    SketchPoints = swSketch.GetSketchPoints2&#xA;    For Each SketchPoint In SketchPoints&#xA;    work = &quot;'Point&quot; + CStr(i) + &quot;':&quot;&#xA;    i = i + 1&#xA;    Set swSketchPoint = SketchPoint&#xA;    'Debug.Print swSketchPoint.Type, swSketchPoint.X, swSketchPoint.Y, swSketchPoint.Z&#xA;    If swSketch.Is3D Then&#xA;        Debug.Print swSketchPoint.Type, swSketchPoint.X, swSketchPoint.Y, swSketchPoint.Z&#xA;    Else&#xA;        Point(0) = swSketchPoint.X&#xA;        Point(1) = swSketchPoint.Y&#xA;        Point(2) = swSketchPoint.Z&#xA;        PointVariant = Point&#xA;&#xA;        Set swXForm = swSketch.ModelToSketchTransform&#xA;        Set swXForm = swXForm.Inverse&#xA;        Set swMathUtil = swApp.GetMathUtility&#xA;        Set swMathPt = swMathUtil.CreatePoint((PointVariant))&#xA;        Set swMathPt = swMathPt.MultiplyTransform(swXForm)&#xA;        Debug.Print &quot;Point Model&quot;, swMathPt.ArrayData(0) * 1000, swMathPt.ArrayData(1) * 1000, swMathPt.ArrayData(2) * 1000&#xA;        Debug.Print &quot;Point shetch&quot;, swSketchPoint.X * 1000, swSketchPoint.Y * 1000, swSketchPoint.Z * 1000&#xA;        Print #1, work;&#xA;        Print #1, &quot;[&quot;;&#xA;        Write #1, swMathPt.ArrayData(0) * 1000, swMathPt.ArrayData(1) * 1000, swMathPt.ArrayData(2) * 1000;&#xA;        Print #1, &quot;],&quot;&#xA;&#xA;    End If&#xA;&#xA;    Next&#xA;    Print #1, &quot;&quot;&#xA;&#xA;&#xA;    'Part.ClearSelection2 True&#xA;&#xA;End Sub&#xA;&#xA;&#xA;Sub main()&#xA;&#xA;    Set swApp = CreateObject(&quot;SldWorks.Application&quot;)&#xA;    Set Part = swApp.ActiveDoc&#xA;    Set swSelectMgr = Part.SelectionManager&#xA;&#xA;&#xA;&#xA;&#xA;    FilePath = CurDir$&#xA;    FilePath = Left(Part.GetPathName, InStrRev(Part.GetPathName, &quot;.&quot;) - 1)&#xA;    Debug.Print FilePath&#xA;    FileName = FilePath + &quot;_Table.txt&quot;&#xA;    Debug.Print FileName&#xA;    Open FileName For Output As #1&#xA;    Print #1, &quot;#Spline data&quot;&#xA;    Print #1, &quot;{&quot;&#xA;    Print #1, &quot;'FileName':' &quot;; Part.GetPathName; &quot;',&quot;&#xA;    Print #1, &quot;'Date':'&quot;; Now(); &quot;',&quot;&#xA;    Print #1, &quot;&quot;&#xA;&#xA;&#xA;    SelectStartFeature&#xA;&#xA;    Print #1, &quot;#Begin Center Line Spline&quot;&#xA;    Print #1, &quot;'CenterLineSpline': {&quot;&#xA;    Part.ClearSelection2 (True)&#xA;    success = Part.Extension.SelectByID2(&quot;centerLineFull&quot;, &quot;SKETCH&quot;, 0, 0, 0, False, 0, Nothing, 0)&#xA;    Part.SketchManager.Insert3DSketch (False)&#xA;    Set swActiveSketch = Part.GetActiveSketch2&#xA;    SketchSegments = swActiveSketch.GetSketchSegments&#xA;    For Each SketchSegment In SketchSegments&#xA;    Set swSketchSegment = SketchSegment&#xA;    Debug.Print swSketchSegment.GetName&#xA;    Debug.Print swSketchSegment.GetType&#xA;    If swSketchSegment.GetType = 3 Then&#xA;        Debug.Print &quot;This is type 3&quot;&#xA;        Exit For&#xA;    End If&#xA;    Next&#xA;    Set swCurveIn = swSketchSegment.GetCurve&#xA;    Print #1, &quot;'Spline Name':' &quot;; swSketchSegment.GetName; &quot;',&quot;&#xA;    Print #1, &quot;'Length':&quot;; swSketchSegment.GetLength * 1000#; &quot;,&quot;&#xA;&#xA;    ' False - do not want a cubic spline&#xA;    varSplineParams = swCurveIn.GetBCurveParams(False)&#xA;    SplineDump (varSplineParams)&#xA;    Part.SketchManager.Insert3DSketch (True)&#xA;    Print #1, &quot;},&quot;&#xA;    Print #1, &quot;&quot;&#xA;    Print #1, &quot;#End Center Line Spline&quot;&#xA;    Print #1, &quot;&quot;&#xA;&#xA;&#xA;    Print #1, &quot;#Begin Rack Spline Data&quot;&#xA;    Print #1, &quot;'RackSpline': {&quot;&#xA;    Set swActiveSketch = Nothing&#xA;    Set swSketchSegment = Nothing&#xA;    Set SketchSegments = Nothing&#xA;    Set swCurveIn = Nothing&#xA;    Part.ClearSelection2 (False)&#xA;    success = Part.Extension.SelectByID2(&quot;rackPosFull&quot;, &quot;SKETCH&quot;, 0, 0, 0, False, 0, Nothing, 0)&#xA;&#xA;    Part.SketchManager.Insert3DSketch (False)&#xA;    Set swActiveSketch = Part.GetActiveSketch2&#xA;    Set swSketchSegment = Nothing&#xA;    SketchSegments = swActiveSketch.GetSketchSegments&#xA;    For Each SketchSegment In SketchSegments&#xA;    Set swSketchSegment = SketchSegment&#xA;    Debug.Print &quot;in rack&quot;, swSketchSegment.GetName&#xA;    Debug.Print swSketchSegment.GetType&#xA;    If swSketchSegment.GetType = 3 Then&#xA;        Debug.Print &quot;This is type 3&quot;&#xA;        Exit For&#xA;    End If&#xA;    Next&#xA;    Print #1, &quot;'Spline Name':'&quot;; swSketchSegment.GetName; &quot;',&quot;&#xA;    Print #1, &quot;'Length':&quot;; swSketchSegment.GetLength * 1000#; &quot;,&quot;&#xA;    Set swCurveIn = swSketchSegment.GetCurve&#xA;    Set varSplineParams = Nothing&#xA;    varSplineParams = swCurveIn.GetBCurveParams(False)&#xA;    SplineDump (varSplineParams)&#xA;&#xA;    Part.SketchManager.Insert3DSketch (True)&#xA;    Print #1, &quot;}&quot;&#xA;    Print #1, &quot;#End Rack Spline Data&quot;&#xA;    Print #1, &quot;} #End of everything&quot;&#xA;&#xA;    Close #1&#xA;&#xA;&#xA;&#xA;End Sub&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Below is a python script that will read the output of above and give you as many points as you want along that spline.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    import scipy&#xA;from geomdl import BSpline&#xA;import math&#xA;#import easygui&#xA;from math import *&#xA;import numpy as np&#xA;import math&#xA;from functools import partial&#xA;np.seterr(divide='ignore', invalid='ignore')&#xA;&#xA;&#xA;filename = 'my_splines_Table.txt'&#xA;filehandle = open(filename,&quot;r&quot;)&#xA;infile = filehandle.read()&#xA;filehandle.close()&#xA;spd=eval(infile)&#xA;&#xA;CenterCurve = BSpline.Curve()&#xA;CenterCurve.rawdata = spd['CenterLineSpline']&#xA;CenterCurve.degree = spd['CenterLineSpline']['Dimension']&#xA;CenterCurve.ctrlpts = spd['CenterLineSpline']['ControlPoints']&#xA;CenterCurve.knotvector = spd['CenterLineSpline']['KnotPoints']&#xA;NumPoints = 250&#xA;for i in range(NumPoints+1): #Rember to generate the last point so you really get Numpoints + 1&#xA;    position = i  / 250&#xA;    point = CenterCurve.evaluate_single(position)&#xA;    print(&quot; Point &quot;,point)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;All of this is specific to my application. You will need to modify accordingly.&lt;br&gt;&#xA;There in no error checking or sanity checking.&lt;br&gt;&#xA;One thing to note is that when the part is rebuilt, the direction of the spline data may reverse. That is the start and end swap. All the data is correct but the python code my process it from end to start. For my application I have another reference point that is near the beginning of the spline. I check the start and end of the spline data to see which end is closer to the reference point. (That is what some of the extra bits are in the VBA script.)&lt;/p&gt;&#xA;" OwnerUserId="19402" LastActivityDate="2019-07-11T15:28:39.483" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29143" PostTypeId="1" CreationDate="2019-07-11T15:35:32.273" Score="0" ViewCount="170" Body="&lt;p&gt;In my country we generally boil milk before we drink. Often the milk heats up and spills from the pot. I want to detect milk boil before it spills off.I am thinking of a thermal camera solution that reads the temperature of liquid, but again it should be small enough to fit beside the gas stove. I have knowledge of python and arduino. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please give me some ideas which works without physically touching the milk.&lt;/p&gt;&#xA;" OwnerUserId="21111" LastActivityDate="2019-07-11T15:35:32.273" Title="How to detect milk spillage or milk boil?" Tags="|electrical-engineering|design|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29144" PostTypeId="2" ParentId="29141" CreationDate="2019-07-11T15:54:18.903" Score="6" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation&quot; rel=&quot;noreferrer&quot;&gt;rocket equation&lt;/a&gt; says that every bit of mass you carry on your rocket is very expensive, especially if that mass is not fuel. Designers of rockets are willing to spend lots of time and effort making sure that their rockets are as light as possible, hence &quot;mass is at a premium.&quot; This inevitably means that the tanks are as thin as is feasible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Stress_(mechanics)&quot; rel=&quot;noreferrer&quot;&gt;Stresses&lt;/a&gt; are the internal forces that are trying to tear solid objects apart. If two structures (like rocket fuel tanks) are carrying identical forces (like the pressure and weight of rocket fuel), the structure with thicker walls and bigger elements will have lower stresses because it has more area to spread the load out over. The thinner lighter structure is therefore more stressed, and closer to breaking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rocket structural designers have to design a structure that is just strong enough to carry the loads the rocket will see, but if they build it any stronger it will be heavier. There is a careful balancing act between safety and efficiency.&lt;/p&gt;&#xA;" OwnerUserId="15951" LastActivityDate="2019-07-11T15:54:18.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29145" PostTypeId="2" ParentId="29141" CreationDate="2019-07-11T15:54:30.607" Score="0" Body="&lt;p&gt;It means, the fuel tank's mass costs fuel to move it and of course that extra fuel has mass too and that mass needs more fuel. Premium here means extra fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a an effort to keep the mass of the fuel tank at the absolute safe minimum, hence allowing the tank material exposed to high stresses.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-11T15:54:30.607" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29146" PostTypeId="2" ParentId="29139" CreationDate="2019-07-11T16:27:12.777" Score="0" Body="&lt;p&gt;Lets say we have the hole two 3cm on the left side of the hole one and both on the surface of the exhale chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can have a 10cm long lever, hinged on the left side of hole 2 and a few centimetre to the left of it, making it possible for the lever to swing up and down around the hinge. On this lever you secure two sealing caps at appropriate location wrt to the two holes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These two caps once the lever is lowered will first block hole 2 and then both holes and lifting it back up will first open hole 1 then both holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This lever is attached on the other end to your finger cup which is secured in place by a simple spring and a guide groove with three positions.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Position one , the lever is in the middle,  holes 1 is open.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Position 2, the lever is down both holes are closed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Position3, lever up, both holes are open.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added a diagram after OP's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The default position of the lever is when hole 1 is open. you could change the positions as it may work better for your task. The rubber sealers are flexible and compressible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KIkNr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KIkNr.jpg&quot; alt=&quot;diagram&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-07-11T17:35:03.307" LastActivityDate="2019-07-11T17:35:03.307" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29149" PostTypeId="2" ParentId="29139" CreationDate="2019-07-11T20:04:02.700" Score="1" Body="&lt;p&gt;Similar mechanisms exist on some musical instruments (saxophone or flute). You have two small buttons attached to levers, button 1 closes hole one, button 2 closes hole 2. Both are adjactant, so the finger can easily slip from one to the other. To achieve your requirement, button 1 is equipped with a small cam that is also pushed by button 2 (but not vice versa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Several ways of doing this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TZ1SZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TZ1SZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either you add a small cam to the lower side of one button so the other button pushes that cam downward. Or you add a cam to the lever connecting the cup (the part covering the hole) to the button, so it pushes on the other lever. You can also add that cam to to the cup, so it acts on the other cup directly. These varaitions all achieve the same function: press on right button to close only one hole, press left button when closing both holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure you'll find all three variations and then some on a concert flute or saxophone, but to be frank I'm to lazy to snap and post pictures right now.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2019-07-12T10:41:52.893" LastActivityDate="2019-07-12T10:41:52.893" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="29150" PostTypeId="1" AcceptedAnswerId="29152" CreationDate="2019-07-11T20:46:13.530" Score="0" ViewCount="216" Body="&lt;p&gt;I have a 2D vibration sensor that can only measure in the X and Y directions, and it is mounted on top of a motor that is fixed into place, as shown in a rough sketch below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PzRNH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PzRNH.png&quot; alt=&quot;rough depiction of vibration sensor on motor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would I be able to work out the vibrations in the Z direction based on a cross-correlation of just the vibration in the X and Y directions?&lt;/p&gt;&#xA;" OwnerUserId="15712" LastActivityDate="2019-07-13T17:44:12.607" Title="Would a cross correlation between vibrations in the X and Y directions help determine vibrations in the Z direction?" Tags="|mechanical-engineering|vibration|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29151" PostTypeId="1" CreationDate="2019-07-11T21:19:39.943" Score="0" ViewCount="481" Body="&lt;p&gt;I'm trying to simulate a purifier in Aspen Plus in a cryogenic system. This purifier is just meant to change the pressure of the outlet stream based on the inlet flow and pressure (we are using the purest xenon money can buy and making it several orders of magnitude purer, so I'm just modeling it as pure xenon hence the purifiers only being pressure changers). I have tried using calculator blocks around different components (namely compressors and pipes) but none are allowing me to simulate a pressure drop the way I'd like. I figure the correct way to simulate this pressure drop would be with one of the custom user models, namely User2 (I've been eyeing it because I already have an excel sheet with the relevant formulae), however I'm sure any user model would work. I'm not exactly sure of how these user models work, however. Could anyone walk me through how to create a custom user model in Aspen, or point me to a tutorial? The section of the User Models documentation relevant to User and User2 hasn't been especially helpful to me.&lt;/p&gt;&#xA;" OwnerUserId="20819" LastActivityDate="2023-12-17T07:03:25.440" Title="Custom User Models in Aspen Plus" Tags="|chemical-engineering|simulation|process-engineering|computer-aided-design|pi-diagram|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29152" PostTypeId="2" ParentId="29150" CreationDate="2019-07-11T21:53:18.430" Score="2" Body="&lt;p&gt;Think it through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put a penny on a table.  Slide it away from you and back -- that's x.  Now slide it left and right -- that's y.  Now pick it up, without moving it forward, back, right or left -- that's z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're sensing x and y, will those change if you move the penny straight up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your sensor senses x and y, will it see z in any way shape or form?&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-07-11T21:53:18.430" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29153" PostTypeId="1" CreationDate="2019-07-11T22:45:03.870" Score="3" ViewCount="73" Body="&lt;p&gt;I was conducting some robotics tests on a old wind turbine today, and noticed strong magnetic fields around the steel tower. This magnetic field does not exist on other modern wind turbines at the same site. I am completely baffled by this, so I’m hoping I can find some answers here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These turbines are located in San Gorgonio Path near Palm Springs, CA. They are Nordtank 65 wind turbines, decommissioned about a decade ago. The towers are steel, bare, without paint. Overall they have very little rust as the climate is very dry and hot. The turbines still contain their generators, which contain magnets. But I don’t think these magnets are very strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using a decent mechanical compass, I can clearly see about 40 degrees of deflection as I walk towards the base of the tower from about 10 feet away. I tried this on all 6 decommissioned Nordtanks, all of them exhibited the same behavior. I tried on modern 0.5 MW turbines nearby while they are generating, but the max compass deflection I got was about 5 degrees at a distance about 5 inches away from the tower itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I first observed the magnetic field, my theory was that the turbine tower was still heating up from the morning sun. Due to Seebeck effect, there could be current flowing in the material as the base is cooled by the foundation buried into the ground. However the deflection on the compass did not stop even though the temperature stabilized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could have caused this? I would love to find out!&lt;/p&gt;&#xA;" OwnerUserId="21117" LastActivityDate="2019-07-29T16:20:56.470" Title="How do old wind turbine towers pickup magnetism overtime?" Tags="|steel|metallurgy|electromagnetism|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29154" PostTypeId="2" ParentId="29101" CreationDate="2019-07-12T00:17:12.090" Score="0" Body="&lt;p&gt;In your final sentence you say the work should be (P2-P1)*A. The units of your statement are force, which does not carry the same units as work. If you multiplied your statement by a distance, you would have the correct units. It just so happens to be that the product of pressure and specific volume at a control surface carries the same units as specific work. Since pressure is a force per unit area and specific volume is volume per unit mass, the unit volume and unit area cancel out to leave a length term. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow work is thought of as the work required to move the fluid through the control volume due to its own mass. In a real-life situation, the difference in pressure are influenced by many other forces then just the force exerted by the mass of the fluid. I hope this helps you understand the physical significance of flow work. Shihabus' answer gives the mathematical reason why flow work is a function of only thermodynamic properties. &lt;/p&gt;&#xA;" OwnerUserId="21118" LastActivityDate="2019-07-12T00:17:12.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29155" PostTypeId="1" CreationDate="2019-07-12T02:37:37.343" Score="2" ViewCount="44" Body="&lt;p&gt;I'm looking to measure the angular velocity of an arm as it moves. It seems to be a difficult thing to research, since the term 'angular velocity/frequency' is usually reserved for other areas of circuit analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I imagine something like a tachometer or a generator is most practical, but I can't find anything dealing with an incomplete rotation (i.e. about 90-120 degrees). Any help or push in the right direction would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="21119" LastActivityDate="2019-07-12T10:28:38.090" Title="What's the most practical way to measure angular velocity for small rotations?" Tags="|measurements|sensors|circuits|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29157" PostTypeId="1" CreationDate="2019-07-12T07:00:06.187" Score="2" ViewCount="621" Body="&lt;p&gt;It's my first time working with bearings, but this is a necessary part of my robotic arm project.  I have some $10 bearings that are 24mm OD.  I want to fit these tightly in a hole, ideally getting a press fit or interference fit (steel bearing, aluminum block).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ordered a reamer that's 24mm and plan to ream the hole on a lathe, but I am worried - what if the bearings are themselves OD undersized (I don't have any fancy measuring calipers by the way).  I think ideally I want a hole of maybe 23.99mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thinking ahead, if I run into this problem, could I cool the workpiece in ice before placing it on the lathe?  This way the workpiece would be reamed in a shrunken state, and then expand as it warms up to room temp?  The last thing I want to do is break a tool, but this seems like it could work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia says aluminum has a volumetric thermal expansion coefficient (av) of 69,  and for Silicon Carbide (the reamer is made of this) 8.31. &lt;/p&gt;&#xA;" OwnerUserId="21120" LastEditorUserId="12827" LastEditDate="2019-07-14T09:33:01.743" LastActivityDate="2019-07-14T09:33:01.743" Title="When reaming a hole to fit a bearing, can I get an underfit by icing the workpiece?" Tags="|machining|bearings|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29158" PostTypeId="2" ParentId="29157" CreationDate="2019-07-12T07:10:29.183" Score="2" Body="&lt;p&gt;If you cool the workpiece and the reamer is at &quot;normal&quot; temperature then your final hole will be larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you heat the workpiece and it cools down after reaming then the final hole will be smaller...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to know the size of the bearing before purchasing the correct reamer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I fitted new valve guides into a cylinder head once - not having a press I heated the cylinder head to about 200 deg C and put the guides in the freezer. They slid in a treat...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-07-12T07:10:29.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29159" PostTypeId="1" CreationDate="2019-07-12T07:43:39.527" Score="0" ViewCount="542" Body="&lt;p&gt;I'm currently trying to figure out the airflow of a gas turbine in m3/min.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought that i would be able to measure the windspeed over a known cross section on the intakte of the turbine and then air in would be the same as air out.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was then told that this is not true for gas turbines due to compression of some sort.&lt;br&gt;&#xA;If air mass in is not equal to air mass out, how do i then calculate/measure the airflow.&lt;br&gt;&#xA;For further info i have the following specs on the turbine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thrust in N&lt;br&gt;&#xA;EGT in °C&lt;br&gt;&#xA;Massflow in kg/s&lt;br&gt;&#xA;Exhaust gas velocity in km/h&lt;br&gt;&#xA;Power output in kW&lt;br&gt;&#xA;Fuel consumption in ml/min&lt;/p&gt;&#xA;" OwnerUserId="20237" LastActivityDate="2024-03-18T00:02:25.260" Title="How to calculate airflow in m3/min on a gas turbine outlet?" Tags="|airflow|turbines|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29160" PostTypeId="2" ParentId="29155" CreationDate="2019-07-12T10:28:38.090" Score="1" Body="&lt;p&gt;I'd use a video camera and attach a striped pattern to form a moire pattern which is easy to see in the video. Here's a link: [&lt;a href=&quot;http://www.unappel.ch/people/emin-gabrielyan/public/070804-multi-ring-moire-indicator/]&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.unappel.ch/people/emin-gabrielyan/public/070804-multi-ring-moire-indicator/]&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://web.archive.org/save/http://www.unappel.ch/people/emin-gabrielyan/public/070804-multi-ring-moire-indicator/&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&#xA;[&lt;img src=&quot;https://web.archive.org/web/20190712102656im_/http://www.unappel.ch/people/emin-gabrielyan/public/070804-multi-ring-moire-indicator/index_files/image035.gif&quot; alt=&quot;enter image description here&quot;&gt;]&lt;a href=&quot;https://web.archive.org/web/20190712102656im_/http://www.unappel.ch/people/emin-gabrielyan/public/070804-multi-ring-moire-indicator/index_files/image035.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;2&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16544" LastActivityDate="2019-07-12T10:28:38.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29161" PostTypeId="2" ParentId="29059" CreationDate="2019-07-12T11:40:37.440" Score="1" Body="&lt;p&gt;I’ve been working on and off on making a working navball  since 2013 and have gotten the mechanics working. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here’s the secret to getting it to work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For rotating electrical connections you use something called a slip ring. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To resolve the angle the ball has turned to you use an angular resolver. There are two types. Incremental and absolute. Incremental will show 0 degrees every time you start it up and will need to be told what angle it’s actually at. The absolute version is tons better because when you start it up it can instantly return the correct angle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started out working with steppers but ultimately it ended up too heavy. It’s simpler to use an absolute angle sensor with a small geared motor and use a PID  routine to align the ball to the correct angle. It turns them into a 360 degree servo motor... and the only way is to use slip rings like NASA did. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’m on the big problem of making the actual hemispheres. I think I might be able to do it but it’s harder than I thought. &lt;/p&gt;&#xA;" OwnerUserId="21123" LastActivityDate="2019-07-12T11:40:37.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29162" PostTypeId="2" ParentId="29157" CreationDate="2019-07-12T12:17:34.887" Score="1" Body="&lt;p&gt;You stated that the bearings are 24mm OD. Why would you be worried that they are undersized? I think your best bet, since you're not sure would be to get a scrap piece of stock and ream the 24mm hole and see if the bearing fits. If the bearing is 24mm it won't fit but you can emery the hole, easy to do on a lathe. Once your experimenting is done, start with the real piece. Good Luck.&lt;/p&gt;&#xA;" OwnerUserId="21087" LastActivityDate="2019-07-12T12:17:34.887" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29163" PostTypeId="1" CreationDate="2019-07-12T12:42:26.067" Score="1" ViewCount="205" Body="&lt;p&gt;I have a series of copper coils with ¼” cooling plates between each coil. The cooling plates will be manufactured from aluminum and cooled using copper tubing. There are seven in series, and a plate in between each but theoretically this should be able to be expanded to any number of these plates in series. A picture is below of the setup. I went through a similar exercise with a different configuration which was very similar, but I am not sure if I got the final process correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is current flowing through all of the copper coils and the resistance and current are known. What I originally came up with was that the final temperature of the copper would be &lt;code&gt;Tc=P*R_th +Tw&lt;/code&gt; where &lt;code&gt;R_th&lt;/code&gt; is the total thermal resistance and &lt;code&gt;Tw&lt;/code&gt; would be the temperature of the water. the thermal resistance would just be the addition of the thermal resistance due to the water (convection) as well as the resistance to travel the width of the insulated plate and the copper along the axis, but I am not sure how to go about this when all of them are in series and are simultaneously heating and cooling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting the convection coefficient for the coil cooling is something that I am not worried about, I am just trying to get the equation that would describe this. I am looking for the final temperature of the coil (red) after everything has settled to it's final temperature. I don't know if this can be achieved in steady state or if a transient process must be taken.&#xA;  &lt;a href=&quot;https://i.stack.imgur.com/jXwFK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jXwFK.png&quot; alt=&quot;The coil schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the cross section of the cooling plate in between each wire coil. The plates are in the white spaces shown. The black represents the circular cooling plate. The red is the geometry of the cooling tube which sits in each plate and carries flowing water.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wGYsV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wGYsV.png&quot; alt=&quot;Cooling Plate schematic&quot;&gt;&lt;/a&gt;&#xA;The schematic of heat generation and thermal resistances is down below, but would be in series with almost 7 others of the same variety. Note that the pipe carrying the water does not cover the whole area so it may not be beneficial to look at the problem as all the heat traveling through the piping (hence the two plate resistances) or if there is some way to reconcile that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xh5Fz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xh5Fz.jpg&quot; alt=&quot;Schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20953" LastEditorUserId="20953" LastEditDate="2019-07-15T15:37:49.810" LastActivityDate="2024-03-21T21:07:40.853" Title="Wire heating and heat transfer in a cylindrical pattern" Tags="|mechanical-engineering|thermodynamics|heat-transfer|thermal-conduction|convection|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29164" PostTypeId="2" ParentId="29159" CreationDate="2019-07-12T12:51:31.357" Score="0" Body="&lt;p&gt;The intake and exhaust are linked by the mass conservation equation, so the mass flow at the exhaust will be the same as at the intake plus any added fuel. Fuel mass is generally negligible as turbines tend to run very lean, but if you have it then you can use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have the mass flow at the exhaust and want the volumetric flow all you need is to link the mass and the volume, i.e. you need the density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this turbine is dealing with a fluid that behaves according to the ideal gas law, you could simply obtain the density from the temperature, which you also have. &lt;strong&gt;However&lt;/strong&gt; this disregards any compressibility effects of the fluid, effectively assuming the Mach number at the exhaust is close to zero: &lt;span class=&quot;math-container&quot;&gt;$M_8 = 0$&lt;/span&gt;. This can also be solved, but &lt;a href=&quot;https://web.stanford.edu/~cantwell/AA210A_Course_Material/AA210A_Course_Notes/AA210_Fundamentals_of_Compressible_Flow_Ch_10_BJ_Cantwell.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;requires more data than you seem to have&lt;/a&gt;; see equations 10.16 and 10.21 of the linked document, I'll type them in and elaborate once I have a minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, since you know the thrust and the mass flow, you could use &lt;a href=&quot;https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Effective_exhaust_velocity&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tsiolkovsky's equation&lt;/a&gt;, or more specifically the fact that: &lt;span class=&quot;math-container&quot;&gt;$$F_{thrust} = v_e \cdot \dot{m}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$v_e$&lt;/span&gt; is the effective exhaust velocity. Note that this is not the actual, measurable, exhaust velocity except for the case where the engine exhausts into a vacuum.&lt;/p&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="10400" LastEditDate="2019-07-12T13:03:36.193" LastActivityDate="2019-07-12T13:03:36.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29165" PostTypeId="2" ParentId="29157" CreationDate="2019-07-12T13:20:58.953" Score="1" Body="&lt;p&gt;Why do you want to do this the hard way by trying to cool or heat the component while it is on a lathe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To change the diameter of a 24mm in aluminum by 0.01mm, you need to change the temperature by about 20 degrees C, &lt;em&gt;and maintain the complete piece of metal at that temperature for the whole time it takes to ream the hole&lt;/em&gt; - and allowing for the heat generated by the reamer itself, of course. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to make a 23.99mm diameter hole, just use the correct size reamer to do the job. Problem solved!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-07-12T13:20:58.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29168" PostTypeId="1" CreationDate="2019-07-12T19:16:38.177" Score="1" ViewCount="126" Body="&lt;p&gt;I have assembled a benchtop torque measurement stand for evaluating electric motor and controller efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motor is 49kV , 5kW brushless outrunner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vZZBN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vZZBN.jpg&quot; alt=&quot;bldc torque stand&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Input power reading is taken from motor controller (VESC6, FOC). Current is limited to 140A, Input voltage is 48V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$RPM=480$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_{input}=5800W$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q_{calculated}=\frac{P_{input}*9.549}{RPM} = 117 Nm$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q_{assumed}=0.8*Q_{calculated}=94Nm$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&#xA;The torque stand consists of two load-cells holding the motor stator. Each cell  was calibrated and tested for 5g error and is rated for 10kg load. After the motor is installed, the weight reading is offset to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotor of the motor (shaft) is supported by the bearing and so the axial weights are compensated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The averaged mass readings from the load-cells at peak input power are:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$m_{topcell}=4328g$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$m_{botcell}=-4244g$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the torque arm: &lt;span class=&quot;math-container&quot;&gt;$$L_{arm}=0.065m$$&lt;/span&gt;&#xA;and Force measured: &lt;span class=&quot;math-container&quot;&gt;$$F_{cell}=\frac{(m_{topcell}-m_{botcell})}{1000}G=\frac{8572}{1000}*9.81=84N$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the output torque:&lt;span class=&quot;math-container&quot;&gt;$$ Q_{real}=F_{cell}L_{arm}=84*0.065=5.47Nm$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, &lt;span class=&quot;math-container&quot;&gt;$5.47 N$&lt;/span&gt; isn't even close. So either mass or torque arm is wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$L_{arm\_assumed}=\frac{Q_{assumed}}{F_{cell}}=\frac{94Nm}{84N}=1.11m$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;s&gt;So, the question is - what is my torque arm?&lt;/s&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;br&gt;&#xA;I have revised the approach and made &lt;span class=&quot;math-container&quot;&gt;$1.35m$&lt;/span&gt; lever. &#xA;The mass measured with digital scale was &lt;span class=&quot;math-container&quot;&gt;$1520g$&lt;/span&gt;, which is equal to &lt;span class=&quot;math-container&quot;&gt;$20.13Nm$&lt;/span&gt;.&#xA;If I plug this into another torque equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q=8.3\frac{I_A}{K_V}$$&lt;/span&gt;&#xA;I get &#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q=8.3\frac{140}{49}=23.7Nm$$&lt;/span&gt;&#xA;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, perhaps the true question I should be asking is &lt;br/&gt;&#xA;&lt;strong&gt;Why cannot I apply classic torque (&lt;span class=&quot;math-container&quot;&gt;$Q_{calculated}$&lt;/span&gt;) formula to an electric motor?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="10268" LastEditorUserId="10268" LastEditDate="2019-07-12T22:37:19.037" LastActivityDate="2024-03-18T10:06:46.017" Title="off-scale torque measured - what is my torque arm?" Tags="|motors|measurements|torque|power|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29169" PostTypeId="2" ParentId="29168" CreationDate="2019-07-12T19:42:27.400" Score="0" Body="&lt;p&gt;Whre are you measuring your torque arm from and to?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotational centre to the point of applied load resistance would be my suggestion, but that does not seem to be what you have at 6.5cm...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest I used was a long torque arm with masses to hang on it to return it to its static position, worked fine...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-07-12T19:42:27.400" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29170" PostTypeId="1" CreationDate="2019-07-12T19:51:30.067" Score="0" ViewCount="97" Body="&lt;p&gt;I am trying to create an air tight cable exit system for a fuel tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, an instrument is to placed in the tank, with its cable exiting through the top of the tank. This exit point needs to be air tight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, most of the cable glands i see on the market are specified as being water tight. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can i assume this to mean it is air tight as well ??&lt;/p&gt;&#xA;" OwnerUserId="19131" LastActivityDate="2019-07-12T20:25:54.343" Title="Can an enclosure be water tight and NOT air tight?" Tags="|piping|seals|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29171" PostTypeId="2" ParentId="29170" CreationDate="2019-07-12T20:25:54.343" Score="2" Body="&lt;p&gt;In general water tight does not mean air tight, while air tight does mean water tight assuming the material is not water soluble. Air contains some smaller molecules than water.  Either can diffuse through many different materials if given enough time though.  As a side note, if the fuel tank is supplying a motor then fresh air should be allowed in as the fuel level decreases or the motor will have a hard time pulling fuel out as a vacuum builds.&lt;/p&gt;&#xA;" OwnerUserId="20957" LastActivityDate="2019-07-12T20:25:54.343" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29172" PostTypeId="2" ParentId="14300" CreationDate="2019-07-12T20:33:40.783" Score="1" Body="&lt;p&gt;In addition to what @ericksonla describes, there is also an equal constraint in Fusion 360 sketching that does exactly what you wish. Maybe this got added since 2017.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/n3Lo6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/n3Lo6.png&quot; alt=&quot;snapshot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12827" LastEditorUserId="12827" LastEditDate="2019-07-12T20:40:33.180" LastActivityDate="2019-07-12T20:40:33.180" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29173" PostTypeId="2" ParentId="26634" CreationDate="2019-07-12T20:45:15.667" Score="0" Body="&lt;p&gt;Why are you using a contact set and not joints?   The common advice is to try to avoid contact sets since they are computationally expensive. Only use them in certain situations like proofing or fine tuning a contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is still relevant, can you share the design so it's easier to help?&lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2019-07-12T20:45:15.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29174" PostTypeId="2" ParentId="22796" CreationDate="2019-07-12T20:52:26.117" Score="0" Body="&lt;p&gt;Thread joints are for animation; sounds like you were using them for design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could you:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Have threads made to both parts; you can use section analysis to see that they match&lt;/li&gt;&#xA;&lt;li&gt;Place the parts together (the end result)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want the animation, start the thread joint from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think your problems were due to trying to land at a good end position, but if you take it the other way round, would it work?&lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2019-07-12T20:52:26.117" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29175" PostTypeId="2" ParentId="22796" CreationDate="2019-07-12T22:49:31.680" Score="0" Body="&lt;p&gt;For serious answers finite element analysis and physical testing are used. The oil industry uses hundreds of different threads for casing and tubing ; these connections are very highly loaded  , under very high pressures and need to be leak tight so the costs of FEM and testing  are warranted. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-07-12T22:49:31.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29176" PostTypeId="1" CreationDate="2019-07-13T01:03:32.913" Score="0" ViewCount="36" Body="&lt;p&gt;I am presently facing a problem in water plant . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a tank there are four pump placed at equal distance named A B C D .in this arrangement A and D  positioned  at near corner of the tank . I started A B C  pump at a same time also D kept for stand by purpose . &#xA;  In this arrangement the pump A gives less delivery than the other B C pumps. Where as all pump having same specification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pump locating at corner of tank disturb the delivery &lt;/p&gt;&#xA;" OwnerUserId="21128" LastActivityDate="2019-07-13T01:03:32.913" Title="water pump problems" Tags="|mechanical-engineering|fluid-mechanics|hydraulics|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29177" PostTypeId="1" AcceptedAnswerId="29181" CreationDate="2019-07-13T07:40:33.673" Score="0" ViewCount="181" Body="&lt;p&gt;I am trying to find the transfer function of a plant that contains numerous equipment including a heater, several tanks, pumps, membranes etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to design 2 PID controllers: the first one to control the temperature of the stream entering one of the membranes, and the second PID controller needs to maintain the solution concentration in one of the tanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen that there is a system identification tool in Matlab that I can use to find the plant transfer function. The problem is, I am not clear on exactly what the plant transfer function should describe. Should it represent the whole plant (e.g. all the tanks, pumps etc.) or only the part of the plant that I am interested in to be able to design the controllers (e.g. resulting in two different plant transfer function equations for each controller).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="21129" LastActivityDate="2019-07-14T00:59:37.887" Title="Identify plant transfer function to design PID controller" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29180" PostTypeId="2" ParentId="2125" CreationDate="2019-07-13T16:20:03.450" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I am new to mechanical engineering, although I have a scientific&#xA;  background (postgrad in Mathematics), and I (mostly) code for a&#xA;  living.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My observation is that there are plenty of analogues between how software is designed, tested for, coded (heh, in that order), and distributed - and how mechanical models are described in CAD. Maybe this is because most of what I do is parametric modeling, e.g. describing the computer a recipe on how to produce a model from parameters, that can then be tested.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You certainly sound to be on the right tracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with software tools, you need to pick and know your tools well, to work with them and not against them.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Meta note: this question is going against the StackOverflow &lt;a href=&quot;https://engineering.stackexchange.com/tour&quot;&gt;tour&lt;/a&gt;. It advices e.g. against opinionated answers and those &quot;.. with too many possible answers or that would require an extremely long answer&quot;. What this might mean is someone closes the entry eventually.&lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2019-07-13T16:20:03.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29181" PostTypeId="2" ParentId="29177" CreationDate="2019-07-13T16:30:37.357" Score="0" Body="&lt;p&gt;When you're working on systems design, the plant is the thing you're controlling, and it's up to you to figure out what that thing is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that your plant may not be my plant -- in fact, if you're buying a sub-system from me, that sub-system may be your plant, while to me it is my closed-loop system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the purposes of your control loops, for each controller the plant transfer function you're interested in is the one that describes the relationship between whatever plant input that your controller affects, and whatever plant output that is affected by your controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for example, one of your &quot;plants&quot; has a heater command as an input, and a temperature as an output.  The other one (I presume) has a flow rate as an input, and a solution concentrate of the desired reactant as an output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that if these two processes (temperature and concentration) are linked, so that each one affects the other, then you may not be able to design each PID controller independently.  In such a case you'd first need to worry about whether the coupling is strong enough to be an issue, and then what to do about it (which would be an answer to a different question).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastEditorUserId="19776" LastEditDate="2019-07-14T00:59:37.887" LastActivityDate="2019-07-14T00:59:37.887" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29182" PostTypeId="2" ParentId="29005" CreationDate="2019-07-13T17:32:26.457" Score="0" Body="&lt;p&gt;Wikipedia mentions shopping carts, explicitly:&lt;sub&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Caster_angle&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Caster_angle&lt;/a&gt;&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the extreme case, such as the caster wheel on a shopping cart, the system is undamped but stable, as the wheel oscillates around the 'correct' path.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The shopping cart wheels are not normally/ever casted, so the beginning of @ray-butterworth's effort is wasted in this case. He also mentions driving wheel a lot, which is unrelated to the question at hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where I think he hits the nail is these two paragraphs (with regard that the cart's swiveling wheels have trail):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;&lt;p&gt;On a vehicle with front-wheel steering, the centrifugal force also&#xA;  acts on the center of gravity and pushes all the wheels to the left.&#xA;  The rear wheels are fixed and so continue following the vehicle. The&#xA;  front wheels can turn, so as they push to the left against the ground,&#xA;  the ground pushes them to the right. With caster, or trail, this force&#xA;  provides negative feedback that makes the wheels return to their&#xA;  normal straight-forward position unless a steering wheel is forcing&#xA;  them to stay in a turned position.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;On a vehicle with rear-wheel steering, it is the front wheels that are&#xA;  relatively unaffected by the sideways force, and the rear wheels that&#xA;  react to it. The force of the road on the wheels makes them want to&#xA;  turn to the left. But the vehicle is turning because the rear wheels&#xA;  were already turned to the left, so in this case the centrifugal force&#xA;  provides positive feedback. The more the vehicle turns, the more the&#xA;  wheels tend to make it turn. Rather than forcing the vehicle to turn,&#xA;  the steering wheel must be used to prevent the vehicle from turning.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I wish he had made those two paragraphs the answer. But I also appreciate the deeper tour I got about how steering works with automobiles, fork lifts, trains, boats and airplanes.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Meta note: In the eyes of the &lt;a href=&quot;https://engineering.stackexchange.com/tour&quot;&gt;Engineering SE tour&lt;/a&gt;, the shorter and precise answers - maybe with links to blogs or articles that provide the full story - would be preferred.&lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2019-07-13T17:32:26.457" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29183" PostTypeId="2" ParentId="29150" CreationDate="2019-07-13T17:44:12.607" Score="0" Body="&lt;p&gt;What vibrations are you interested in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the vibrations somehow interlinked - what else beyond the question you placed to us do you know?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can constrain the question more, the situation may change. If you cannot, then @timwescott's answer is obviously the correct one.&lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2019-07-13T17:44:12.607" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29184" PostTypeId="2" ParentId="29110" CreationDate="2019-07-13T17:51:54.013" Score="0" Body="&lt;p&gt;This is just a random guess, but have you considered there could be an expansion chamber? I had one in a house I owned, connected with its heating pipes (also a closed liquid system).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9mtyM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9mtyM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Expansion_tank&quot; rel=&quot;nofollow noreferrer&quot;&gt;Expansion tank&lt;/a&gt; on Wikipedia.&lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2019-07-13T17:51:54.013" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29185" PostTypeId="2" ParentId="29157" CreationDate="2019-07-13T21:42:50.600" Score="1" Body="&lt;p&gt;In addition to the other answers about just using the correct sized reamer, If you are using a lathe, you can use a single point boring bar/tool instead of a reamer to bore the hole to the size. If on a milling machine, you can also just use a boring head/tool. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have to make multiple holes and time is of the essence, then, go with a reamer. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2019-07-13T21:42:50.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29186" PostTypeId="1" CreationDate="2019-07-14T13:18:12.943" Score="1" ViewCount="49" Body="&lt;p&gt;I want to design a motor system where a single stator coil is responsible for two rotors of neodymium magnets on its outside and inside, will this work, and what are the downsides I should be expecting? Any advice on perfecting the design will be very helpful. &lt;/p&gt;&#xA;" OwnerUserId="19370" LastActivityDate="2020-09-22T19:21:33.743" Title="Can I use a single stator coil to actuate/rotate two rotors inside and outside of the stator" Tags="|electrical-engineering|stepper-motor|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29189" PostTypeId="1" CreationDate="2019-07-14T21:10:44.737" Score="0" ViewCount="116" Body="&lt;p&gt;&lt;strong&gt;Background&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am working on a project to where I am trying to heat up air within a 2 inch tube, with high air flow.  I am using 2 inch tri clamp stainless steel tubes where I basically have a fan on one end and am trying to get an inline heater so that the result is a high flow rate and high temperature (around 400F - 500F).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I have tried so far&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have had some decent success by buying a cheapo heat gun from Harbor Freight and taking the nickel chromium array out of it, suspending it inside a 2 inch tube. I was able to get to about 400 F but after some time (an hour or so) the element seems to burn out and die. I seem to be so close to being able to achieve this goal but was hoping there is a more reliable method to heat the air without having to replace the element that often.  I would like to stay to electric heat sources if possible and it would need to be 120V, below 15A (standard house breakers in a kitchen).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am not asking for a particular product, but rather &lt;strong&gt;what electric heater types out there might work for my application&lt;/strong&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closest thing I could find was something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iXCCg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iXCCg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B077TPD4L7&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amazon link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one fits inside a 2 inch tri clamp tube but seems it needs to be in a liquid, not just with air flow around it? Would this more than likely just burn up too? &lt;/p&gt;&#xA;" OwnerUserId="14534" LastActivityDate="2019-12-18T20:03:20.537" Title="Heating hot air within a 2 inch tube" Tags="|heating-systems|heat-exchanger|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29190" PostTypeId="1" AcceptedAnswerId="29282" CreationDate="2019-07-15T08:16:20.727" Score="2" ViewCount="84" Body="&lt;p&gt;I have not taken any course in civil engineering. What I was taught was that a broad base would prevent something from toppling. So it makes me wondering instead of nailing the foundation of a building vertically deep down into the ground, if it would make sense to make multiple buildings share a horizontal foundation.&lt;/p&gt;&#xA;" OwnerUserId="5159" LastEditorUserId="5159" LastEditDate="2019-07-15T08:18:25.227" LastActivityDate="2019-07-21T03:19:16.950" Title="Can a horizontal foundation safeguard against earthquakes?" Tags="|structural-engineering|civil-engineering|structural-analysis|building-design|building-physics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29191" PostTypeId="1" CreationDate="2019-07-15T08:32:38.360" Score="0" ViewCount="468" Body="&lt;p&gt;Given:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;a reinforced concrete two-way slab at about 160 mm (the edges are cantilevered between the concrete walls)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the electricians buried the flexible electrical hose into the slab (red lines in the image below) and then covered the tubes with mortar (the channel are about 30 mm depth)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the mortar is a light one (probably supporting about 4 MPa or even less) and it covers the hoses superficially (1 to 3 mm over the top of the tubes)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;no rebars were cut&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the structure is not based on beams (the slab is sustained by precast concrete walls)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CE9GY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CE9GY.png&quot; alt=&quot;Red circuits&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ORg7x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ORg7x.png&quot; alt=&quot;Inside the concrete slab&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How does this affect the structure resistance of the reinforced concrete slab?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on your experience / simulations, could you provide an empirical percent regarding the &lt;strong&gt;new load capacity&lt;/strong&gt; based on the initial load capacity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that the load on the slab looks similar to this (where the vertical should look similar to the horizontal, including the red spots):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uUSsG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uUSsG.png&quot; alt=&quot;bending moments&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where red &amp;amp; yellow are for a top bending and blue is for a bottom bending.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: My purpose is to have the following stack over the slab for the permanent load (240 kg / m2):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;thin leveling screed =&gt; 40 kg / m2&lt;/li&gt;&#xA;&lt;li&gt;heating floor system based on water and insulated panels =&gt; 20 kg / m2&lt;/li&gt;&#xA;&lt;li&gt;sand-cement screed ~5 cm over the pipes (=&gt; ~6.5 cm) =&gt; 145 kg / m2&lt;/li&gt;&#xA;&lt;li&gt;adhesive + ceramic tiles =&gt; 35 kg / m2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;at which a normal useful load is added.&lt;/p&gt;&#xA;" OwnerUserId="21141" LastEditorUserId="21141" LastEditDate="2019-07-21T16:20:15.287" LastActivityDate="2019-07-21T17:01:35.723" Title="Impact of having flexible electrical hose buried into the reinforced concrete slab" Tags="|structural-engineering|concrete|reinforced-concrete|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29192" PostTypeId="1" CreationDate="2019-07-15T08:38:07.750" Score="0" ViewCount="167" Body="&lt;p&gt;I need to make a support (preferably metal) for a projector to be fixed on the wall. The projector size is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ot3ym.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Width: &lt;strong&gt;0.7m&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Length: &lt;strong&gt;1.1m&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weight: &lt;strong&gt;80kg&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone knows how i can solve this please let me know!&#xA;Thank you so much!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: This will be going to the back of the Musical theatre, the wall of the theatre is made of concrete block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lZpcg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lZpcg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks&lt;/p&gt;&#xA;" OwnerUserId="21142" LastEditorUserId="21142" LastEditDate="2019-07-22T07:09:42.200" LastActivityDate="2019-07-22T07:09:42.200" Title="Looking to design a cantilever support to hold up an 80kg projector" Tags="|structural-engineering|civil-engineering|structural-analysis|stresses|structures|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29193" PostTypeId="2" ParentId="29192" CreationDate="2019-07-15T09:38:23.557" Score="1" Body="&lt;p&gt;You could think of two L straps of either steel or aluminum similar to the attached photo. They should be 3mm thick, 20mm wide and 60cm long on each side. They have to be reinforced by a diagonal member of the same cross section welded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can be attached to a metal tray big enough to support the projector preferably with srapps to tie in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The two metal angles have two holes near the top and bottom to allow for proper anchors to the wall. The anchors need to resist 160kg pull out force. Like expansive bolts for concrete or lag bolts going through the studs.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/PsVus.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PsVus.jpg&quot; alt=&quot;bracket&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;After OP's edited their question and the fact that this projector is being used commercially I recommend these modifications.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use steel angle 4cm x 4cm by 5mm thick for legs of the bracket and the frame of the tray. Weld all corners.&lt;/li&gt;&#xA;&lt;li&gt;Use 3mm thick steel plate for the tray.   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-07-17T19:38:57.570" LastActivityDate="2019-07-17T19:38:57.570" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29194" PostTypeId="2" ParentId="7736" CreationDate="2019-07-15T13:34:59.007" Score="4" Body="&lt;p&gt;The diagram most closely related to the original question seems to be a &lt;strong&gt;kinematic diagram&lt;/strong&gt;. Kinematic diagrams are standardized in &lt;strong&gt;ISO 3952&lt;/strong&gt;, although it seems that most often people are just copying what they have seen without reading the standard (which is quite old, but then this is common anyway).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard is in 4 parts and covers symbols, and alternate pictograms for such things as: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Simple kinematic pairs such as used in bar linkages (part 1),&lt;/li&gt;&#xA;&lt;li&gt;gears and cams (part 2),&lt;/li&gt;&#xA;&lt;li&gt;geneva drives, ratchets, clutches and couplings (part 3),&lt;/li&gt;&#xA;&lt;li&gt;belt/chain drives, lead screws, flywheels, bearings (part 4).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we re-draw the image in the original question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lTJLH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lTJLH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;: Kinematic diagram, uses the same thin thick line style as normal engineering drawings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ISO 3952 standard does not have a motor symbol. That is because they are concerned with motions not dynamics. Either way it is a human readable format so you can always add your own clarifications. &lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2019-07-15T13:34:59.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29195" PostTypeId="1" CreationDate="2019-07-15T15:19:21.023" Score="0" ViewCount="232" Body="&lt;p&gt;I need to just get the volume from ANSYS Design Modeller but I would like a higher number of significant figures rather than just 5 (See picture).&#xA;How do I do this?&#xA;Thank you for your help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C3Ydx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C3Ydx.png&quot; alt=&quot;ANSYS screenshot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21098" LastActivityDate="2019-12-12T18:02:50.377" Title="ANSYS Design Modeller more significant figures" Tags="|design|ansys|computer-aided-design|ansys-workbench|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29196" PostTypeId="2" ParentId="29195" CreationDate="2019-07-15T15:49:16.470" Score="1" Body="&lt;p&gt;Since I have not comment rights, I'm gonna write this non-answer here (if someone with power reads this, please make this a comment). I believe all ANSYS licences should have access to &lt;strong&gt;SpaceClaim&lt;/strong&gt; (the other/newer CAD editor in the package), measuring the volume of a solid is straightforward from the &lt;strong&gt;&lt;em&gt;Measure&lt;/em&gt;&lt;/strong&gt; panel, which has the option to adjust the number of significant figures. I do not know how to do it from Design Modeller.&lt;/p&gt;&#xA;" OwnerUserId="18457" LastActivityDate="2019-07-15T15:49:16.470" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29197" PostTypeId="2" ParentId="29191" CreationDate="2019-07-15T16:21:39.133" Score="4" Body="&lt;p&gt;You haven't specified any material parameters, so for calculation purposes, I am going to assume a reinforcement yield strength of 500MPa and a cylinder compression strength for both concrete and mortar of 25MPa. As I am basically just guessing these values, you will have to adjust the conclusion accordingly, if I haven't been so lucky as to guess correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In bending failure, when the reinforcement yields and all concrete in tension cracks, the compression zone in concrete will have a height of approximately &lt;span class=&quot;math-container&quot;&gt;$\frac{\frac{\frac{\pi}{4}\left(4mm\right)^2}{100mm}\cdot500MPa}{0.8\cdot25MPa}=3mm$&lt;/span&gt;. (The factor of 0.8 is a correction factor for the shape of the stress-strain curve of concrete.) For a more correct value, this should be corrected for the safety factors, which will increase the compression zone a bit, but it will not matter much for the conclusion. Assuming we have a cover of at least, say, 10mm mortar on top of the hoses, there will be plenty of mortar above the hoses to transfer the compression, and the hoses themselves will not matter, as that part of the concrete would be fully cracked anyway and would not transfer any forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In shear failure, such a lightly reinforced slab will be even less sensitive to a few &quot;missing&quot; areas of concrete, as the hoses only weaken the slab locally. The actual calculation is rather more complex, but with a 4mm mesh the slab is unlikely to fail in shear before it fails in bending, so I am going to leave it as an &quot;exercise for the reader&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So most likely, it will not have any significant effect on the structural capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One potential issue is if the mortar, they used, is not equal strength to the regular concrete. This should not need to be a problems as mortars for these sort of usage scenarios are often made using ready-made mixes with a pretty high compression strength, but obviously I have no way of knowing if they chose the proper product here. And the mortar would obviously have to be appropriate for the environment in question.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastEditorUserId="15064" LastEditDate="2019-07-15T16:28:09.263" LastActivityDate="2019-07-15T16:28:09.263" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29198" PostTypeId="1" CreationDate="2019-07-15T16:48:11.217" Score="0" ViewCount="55" Body="&lt;p&gt;I have a system consisting of a hollow, rectangular cantilever beam (pictured below). I'd like to write the equilibrium stress equations &lt;em&gt;à la&lt;/em&gt; the Cauchy Stress notation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I account for the hollow section in the equation? How is it represented in a free-body diagram of the system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ve7lP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ve7lP.png&quot; alt=&quot;simple schematic of the cross sectional area of the beam.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21147" LastActivityDate="2019-07-15T16:48:11.217" Title="How to write the equilibrium stress equations for a hollow, rectangular beam?" Tags="|structural-engineering|materials|beam|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29202" PostTypeId="1" CreationDate="2019-07-15T19:42:50.340" Score="1" ViewCount="127" Body="&lt;p&gt;Does a screed applied over the slab increase the slab tension capacity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somehow it's counter intuitive because adding weight on the slab should tension it, but do the slab and the screed behave as a whole in this scenario?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What about a floating screed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the dependency between the slab load capacity (in N/m^2) and the total height of the slab / slab+screed (if they act as a whole)?&lt;/p&gt;&#xA;" OwnerUserId="21141" LastEditorUserId="21141" LastEditDate="2019-07-16T06:00:05.753" LastActivityDate="2019-07-20T23:01:19.240" Title="Does screed increase the slab tension capacity?" Tags="|structural-engineering|concrete|reinforced-concrete|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29203" PostTypeId="2" ParentId="29202" CreationDate="2019-07-15T20:48:53.553" Score="-2" Body="&lt;p&gt;Any steel mesh or netting if placed on the bottom of a slab following appropriate codes will increase the slab's tension capacity and reduce cracking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The are many reasons they don't do this though.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;concrete placement becomes difficult because  the larger aggregate won't pass through.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;concrete finishers can not walk and work on a screed or mess, they will damage it and displace the clearances.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-15T20:48:53.553" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29204" PostTypeId="2" ParentId="29190" CreationDate="2019-07-15T21:15:45.597" Score="2" Body="&lt;p&gt;The foundation pylons (caissons) are commonly used for penetrating deep through incompetent strata of soil to reach and embed into hard layers, or in retaining walls, not specifically for resisting earthquakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;seismic design of foundations are performed according to a geotechnical report of the site prepared by a soils engineer after he thoroughly investigates the site and nearby known fault lines, in collaboration with the geologist, structural engineer, city's grading department, state engineering department.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the grave consequences  of catastrophic failure of the large structures, many building departments require dynamic modeling and stress analysis of the foundation and structure exposed to the siesmic spectra of the 50 year anticipated earth quake. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-15T21:15:45.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29205" PostTypeId="1" AcceptedAnswerId="29207" CreationDate="2019-07-15T23:08:06.980" Score="1" ViewCount="87" Body="&lt;p&gt;A box is being lifted by a crane. Sometimes the box will be completely full, and sometimes it will only be partially full (indicated by the gray shaded regions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8TmRe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8TmRe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, where is the axis of bending in both cases? Does it follow the center of mass? Or does it always follow the geometric center?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Of course I hope the axis of bending will always be in the middle so it could take full advantage of the roof and floor to resist bending.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if it matters, the crane will actually lift the box at 4 points, not 2. And I moved them inwards from the edges just because I thought that would resist bending more, due to a shorter beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I know where the axis is, I can calculate the moment of inertia and use the deflection formula &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta = \dfrac{5qL^4}{384 E I}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As advised, I did this for a U-shape, but also for the square, for comparison. I had to choose concrete numbers so I picked a 3x3x12 box with wall thickness of 1 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hDqTL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hDqTL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the square, I found the formula &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_second_moments_of_area&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ I = a^4 - (a - 2t)^4 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ I = 3^4 - (3 - 0.02)^4 = 2.13849584$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the U-shape, I found the much more complex formula &lt;a href=&quot;http://www.efunda.com/math/areas/squarechannel.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and took it about the y-axis since that diagram was vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ I =  \frac{2ta^3 + (a - 2t)t^3}{3} - (2ta + (a - 2t)t)\frac{2ta^2 + (a - 2t)t^2}{2a^2 - 2(a - 2t)(a - t)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ I = \frac{2ta^3 + at^3 - 2t^4}{3} - \frac{(2ta + ta - 2t^2)(2ta^2 + at^2 - 2t^3)}{2a^3 - 2(a^2 - 3ta + 2t^2)} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ I = \frac{0.54000298}{3} - \frac{(0.0898)(0.180298)}{54 - 2(8.9102)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ I = apprx. 0.179553$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now I can finally use the deflection formula with both I's. I'm using steel so E = 200 GPa. The distributed load, q, is for a box 1/3rd full of crushed iron ore which has a density of 2,500 kg/m&lt;span class=&quot;math-container&quot;&gt;$^3$&lt;/span&gt;. q = 72,523 N/m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\delta = \dfrac{5qL^4}{384 E I}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\delta = \dfrac{5 * 72523 * 12^4}{384 * 200,000,000,000 * I}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\delta = \dfrac{7,519,184,640}{76,800,000,000,000 * I}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\delta = 0.00004578$&lt;/span&gt; m for the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\delta = 0.0005452766$&lt;/span&gt; m for the U-shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;...I have to conclude I did something wrong. It is hard to imagine a box of that size, with just 1-cm thick walls, would deflect by less than a millimeter when holding up iron ore, even at 1/3rd full.&lt;/p&gt;&#xA;" OwnerUserId="3510" LastEditorUserId="3510" LastEditDate="2019-07-21T04:05:07.443" LastActivityDate="2019-07-21T04:05:07.443" Title="Where is the axis of bending for a box lifted by a crane?" Tags="|mechanical-engineering|statics|deflection|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29206" PostTypeId="1" AcceptedAnswerId="29209" CreationDate="2019-07-15T23:21:19.837" Score="1" ViewCount="1841" Body="&lt;p&gt;I’m a graduate structural engineer with 4 years experience, working in the UK. A family friend is planning on doing a house extension which will require some structural calculations for steel beams. My question is:&#xA;What will building control in England stipulate in terms of the structural engineer’s qualifications, liability insurance etc? &#xA;Do you need to be a chartered engineer with insurance to undertake structural calculations in England? &#xA;I think my experience is sufficient for what is a relatively simple calculation pack but I’m not sure what the legal stance is on this in England. &lt;/p&gt;&#xA;" OwnerUserId="21151" LastEditorUserId="33" LastEditDate="2019-07-15T23:27:27.220" LastActivityDate="2020-07-15T15:15:16.493" Title="Do you need to be a chartered engineer to submit structural calculations to building control in the UK?" Tags="|structural-engineering|civil-engineering|licensure|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29207" PostTypeId="2" ParentId="29205" CreationDate="2019-07-16T01:46:07.663" Score="5" Body="&lt;p&gt;If by axis of bending you mean box functioning as a beam and also the top and walls and bottom of the box are integrated and work like a hollow rectangular beam, then the neutral axis is at the mid height of the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the content of the box is liquid, it is evenly distributed along the length of the box and should be considered distributed load. The level of liquid doesn't affect section properties of the box, such as its moment of inertia or location of its neutral axis.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-16T01:46:07.663" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29208" PostTypeId="1" AcceptedAnswerId="29217" CreationDate="2019-07-16T02:53:27.460" Score="0" ViewCount="227" Body="&lt;p&gt;Everything I've read on the Rankine cycle indicates that the feedwater pump is necessary for the cycle to work. However the steam cycle used in traditional household boiler / radiator systems seems to challenge this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cycle for the household boiler seems to be a closed loop cycle with just a heat source (boiler) and cold source (radiators). But those two pieces are sufficient to have steam flow through the radiators, condense, and then return to the boiler.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is my understanding of the boiler / radiator cycle correct?&lt;/li&gt;&#xA;&lt;li&gt;If so, and if the system was fitted with a turbine, could it produce any amount of power?&lt;/li&gt;&#xA;&lt;li&gt;If it produced any amount of power, how inefficient would it be compared to a Rankine cycle?&lt;/li&gt;&#xA;&lt;li&gt;Is there a name for the type of cycle in a boiler / radiator system?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="21153" LastActivityDate="2019-07-16T19:56:52.537" Title="Steam cycle for a boiler / radiator system vs Rankine cycle" Tags="|thermodynamics|heat-transfer|heating-systems|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29209" PostTypeId="2" ParentId="29206" CreationDate="2019-07-16T03:00:06.153" Score="2" Body="&lt;p&gt;IANAL, but the first step is to find whether you need planning permission for the work. Talk to your local council's planning department. They are usually helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do need to submit a planning application, you will have to create drawings in a standard architectural format for inspection by anyone who may potentially be affected by the work - e.g. the owners of neighbouring properties. The planning department will notify them of the application to give them opportunity to comment on it before it is approved. These drawings don't specify the technical details of the construction, but show the changes to the appearance of the existing buildings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you need planning permission or not, the work needs to comply with the UK building regulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://www.gov.uk/government/collections/approved-documents&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for (detailed) official guidance on how to do that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also want to contact the house insurers at the planning stage, rather than presenting them with a fait accompli at the end. They may wish to inspect the work in stages as it is carried out, or require copies of any planning department inspections. Failure to coordinate all the required inspections can cause significant delays to the work plan. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, the UK system doesn't care about the formal qualifications of the people involved, so long as the work meets the required standards. AFAIK there is no legal necessity for any form of insurance either on the work itself or the finished building - though it would be foolish not to have insurance against third party claims if there is any risk of a disaster damaging other people's property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will probably want some long-term guarantee of the work done should you want to sell the property - e.g. the National House-Building Council's standard guarantee is 10 years. That should be provided by the builder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the builder will be guaranteeing the work, they will need to be satisfied that the technical specifications for the work are sensible, rather than just &quot;doing whatever you tell them&quot; regardless of the risks involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also require final inspections if you have made major changes to the water, gas, or electricity installations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line of all this is that &quot;doing the calculations&quot; is often trivial compared with &quot;understanding how the system works&quot; - which is the expertise that professionals working in the field have.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-07-16T03:19:06.930" LastActivityDate="2019-07-16T03:19:06.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29211" PostTypeId="1" AcceptedAnswerId="29230" CreationDate="2019-07-16T04:19:39.043" Score="1" ViewCount="136" Body="&lt;p&gt;I was looking at laminar flow filters to aid in my mycology work I'm doing. Most filters use an aluminum extrusion with fins. I figured the options out there are pretty expensive for what you get and figured I could just 3d print my own straightener. Question is what shape do I use? I have seen honeycomb used and would be an easy option, but it really got me thinking about what shape would be optimal for this. We all know that when streamlining a positive body that a teardrop is close to optimal, but putting teardrops in a grid would yield a weird negative shape in the pattern. Any thoughts or ideas on this would be greatly appreciated, I plan on open sourcing the final design. Thanks&lt;/p&gt;&#xA;" OwnerUserId="21157" LastActivityDate="2019-07-17T12:59:47.420" Title="Laminar flow hood/air straightener" Tags="|fluid-mechanics|airflow|aerodynamics|airfoils|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29213" PostTypeId="2" ParentId="28817" CreationDate="2019-07-16T08:07:31.237" Score="-1" Body="&lt;p&gt;Back spilling can be found from &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; the total time to fill the mould&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're using top gate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T=\dfrac{AH}{a\sqrt{2gh}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$AH=$&lt;/span&gt; volume of mould cavity&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$a=$&lt;/span&gt; area of gate&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$h=$&lt;/span&gt; total height, which is the height of sprue and molten metal column above the sprue &lt;span class=&quot;math-container&quot;&gt;$(h_1)$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In top gate aspiration effect is to be considered. So to avoid aspiration effect, let &lt;span class=&quot;math-container&quot;&gt;$a_1=$&lt;/span&gt; the area of molten metal basin above sprue. Then&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dfrac{a}{a_1}= \dfrac{h_1}{H}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the area ratio in sprue to avoid aspiration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From eq, area change should be linear decreasing, but it should be taken curve to minimize friction and boundary layer separation.&lt;/p&gt;&#xA;" OwnerUserId="21159" LastEditorUserId="1832" LastEditDate="2019-07-19T01:46:12.013" LastActivityDate="2019-07-19T01:46:12.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29215" PostTypeId="2" ParentId="29163" CreationDate="2019-07-16T13:12:28.150" Score="0" Body="&lt;p&gt;Let’s assume that the tube is small in radius. Also assume the power input &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; to the surrounding coils is constant with angular position. Finally, assume that the plates and coils are essentially isolated to heat flow; all power input to the coils leaves only through the water cooling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these assumptions, the temperature profile in the water coil can be taken to be a function only of angular position according to the equation below with &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; in radians.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{T - T_i}{T_o - T_i} = \frac{\theta}{2\pi} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is independent of how much heat is added along the coil. The output temperature of the water &lt;span class=&quot;math-container&quot;&gt;$T_o$&lt;/span&gt; is found from an overall energy balance on a coil segment with water mass flow and specific heat capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{m}\tilde{C}_p\left(T_o - T_i\right) = P $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the coils are perfect thermal conductors, the angular temperature profile in the water tube will be duplicated in the coils.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you add convective heat flow out from the external wall of the coil, the final water temperature will be lower and the coils will have a radial temperature profile superimposed on the angular profile.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-07-16T13:12:28.150" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29216" PostTypeId="2" ParentId="29208" CreationDate="2019-07-16T13:14:36.377" Score="0" Body="&lt;p&gt;From my understanding, feedwater lines normally refer to a secondary fluid that is being heated by &quot;extra&quot; thermal energy in a cycle (see modified Rankine cycles). The boiler heats the water (normally keeping it in a liquid state in most modern residential applications) and either stores it in a tank or directly circulates it. This is the key difference between indirect and direct fired boiler systems (&lt;a href=&quot;https://www.energy.gov/energysaver/heat-and-cool/water-heating/tankless-coil-and-indirect-water-heaters&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.energy.gov/energysaver/heat-and-cool/water-heating/tankless-coil-and-indirect-water-heaters&lt;/a&gt;). &#xA;Turbines come in several different forms. A Pelton wheel is a type of turbine that operates with liquids only based on the movement of the liquid. Turbines in power gen applications operate with a superheated gas, which is probably not found even in steam heating systems. This fluid must be of high pressure and temperature to keep the steam quality above 1 for most of the expansion process. Thus, a turbine on a conventional hydronic heating system would not produce a lot of power, if any. It would probably damage a conventional turbine via liquid water ingestion. &#xA;There are some micro-combined heat and power systems that use the Rankine cycle to produce heat and electricity from the same heat source (boiler), but they are not common. Even then, the circulating fluid (the domestic hot water) is not the working fluid in the turbine. &lt;/p&gt;&#xA;" OwnerUserId="21118" LastActivityDate="2019-07-16T13:14:36.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29217" PostTypeId="2" ParentId="29208" CreationDate="2019-07-16T15:45:32.023" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is my understanding of the boiler / radiator cycle correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Boiler radiator cycles that are driven by free convection certainly exist, but most heating systems use one or more heating pumps to circulate the water (not steam). But the fundamental difference is that a rankine cycle extracts mechanical work from the steam, a heating system doesnt.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If so, and if the system was fitted with a turbine, could it produce any amount of power?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. The boiler in a steam engine is the part of the system at the highest pressure. Feed water will not flow into it without a pump.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a name for the type of cycle in a boiler / radiator system?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Thermodynamic cycles refer to cycles that extract work. A boiler/radiator system is no thermodynamic cycle in this sense.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-07-16T15:45:32.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29220" PostTypeId="2" ParentId="29208" CreationDate="2019-07-16T19:56:52.537" Score="0" Body="&lt;p&gt;Your heating system is not a Rankine cycle.  In fact it's not considered a thermodynamic cycle at all because all it does is transfer heat.  If you add a steam turbine you could possibly produce a tiny amount of mechanical power, but it would be very limited because the temperature and pressure are so low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you do that you'll quickly discover the problem - steam will be leaving the system in order to power the turbine.  That means you'll run out of water, so you have to condense the steam and pump the water back in, or vent it to the atmosphere and pump new water in.  In either case, you'll need the feedwater pump.&lt;/p&gt;&#xA;" OwnerUserId="6284" LastActivityDate="2019-07-16T19:56:52.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29222" PostTypeId="1" CreationDate="2019-07-17T01:22:25.753" Score="0" ViewCount="22" Body="&lt;p&gt;We have a drying process where we take a thin bread product and cool it in a room at 11 C and 90% relative humidity. The product comes in at 80 C and comes out at 22 C. There is air blowing across the product. Because of the design of the room (nearly a sealed box) it is hard to estimate the water gain of the air before and after passing the product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to size a dehumidifier to keep the cooling room at 60% relative humidity - however I need to estimate how much water I need to removing from the air. We would adjust the temperature to ensure that the product is 22 C at the end of the cooling process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can estimate how much evaporation is occurring at 90% relative humidity by doing a simple mass balance - however this will be higher at 60% relative humidity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering what the best approach is to take data from the 90% relative humidity condition and make a back of the envelope estimate of an evaporation rate at 60% relative humidity? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the moment I'm thinking of assuming the rate limiting step is mass diffusion between the air at the surface of the wrap (100% relative humidity) and the ambient humidity. If I do that then maybe the rate is merely proportional to the water kg of the air at the ambient humidity vs 100%?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="21174" LastActivityDate="2019-07-17T01:22:25.753" Title="Estimating rate of drying by comparing identical process with different ambient humidity" Tags="|chemical-engineering|cooling|evaporation|drying|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29226" PostTypeId="2" ParentId="29202" CreationDate="2019-07-17T11:06:23.133" Score="2" Body="&lt;p&gt;No, a screed does not increase a slab's capacity in any practical way. It is considered non-structural.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2019-07-17T11:06:23.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29227" PostTypeId="1" CreationDate="2019-07-17T12:07:32.623" Score="2" ViewCount="55" Body="&lt;p&gt;This question is related to the situation in an apartment in a reinforced concrete building that is approximately 30 years old.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ceiling is almost perfectly straight and level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the floor from which the existing screed was allegedly removed is lop-sided or crooked at many places, having differences between spots in the same room that go to 4 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that there were multiple layers of screed and only one was removed? Is there any way to determine if what I'm looking at is the reinforced concrete slab itself or another screed?&lt;/p&gt;&#xA;" OwnerUserId="21141" LastEditorUserId="21141" LastEditDate="2019-07-17T15:11:42.483" LastActivityDate="2019-07-29T00:03:39.773" Title="Crooked floor, but ceiling with no incline - could it possibly be a screed?" Tags="|structural-engineering|concrete|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29228" PostTypeId="1" CreationDate="2019-07-17T12:47:06.130" Score="2" ViewCount="45" Body="&lt;p&gt;What qualifies a energy method in the context of control theory?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Citing from a paper to give the context in which this expression came to my attention:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the challenging problems of the double inverted pendulum is its swing-up from the downward position to the upright position. The swing-up of the double inverted pendulum can be performed by using an energy method and by separating the swing-up into three steps: first stabilizing the first pendulum in the upright position, then swinging-up the second pendulum while stabilizing the first, and finally stabilizing both pendulums in upright position [11]. However, this control method requires considerable time before both pendulums reach the upright position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that in MPC their exist something like an energy cost function meaning that the terms in the cost function are the values with sign not the absolute values. Is the term related to that?&lt;/p&gt;&#xA;" OwnerUserId="20870" LastActivityDate="2019-07-17T12:47:06.130" Title="Meaning of energy method in control theory" Tags="|control-engineering|control-theory|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29229" PostTypeId="1" CreationDate="2019-07-17T12:48:18.807" Score="1" ViewCount="212" Body="&lt;p&gt;Trying to determine some information regarding the reinforced concrete slabs (reinforced with mesh) in an apartment, by looking through the holes in the middle of the rooms for lights, I've come across the following situation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;mesh located at 40 mm from the bottom of the ceiling in the middle of the room (visible in a plastic tube used for the light cables) and mesh located at 17 mm from the bottom of the ceiling nearer the edges.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What I'm seeing might be a different level of reinforcement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reinforcement is in 2 directions, but this is the information I can get from what I can see in existing holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Details&lt;/strong&gt;:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the slab height is 140 mm&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="21141" LastEditorUserId="21141" LastEditDate="2019-07-17T13:33:45.647" LastActivityDate="2023-12-23T23:08:41.053" Title="Reinforced mesh at different distance in a concrete slab?" Tags="|structural-engineering|concrete|reinforced-concrete|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29230" PostTypeId="2" ParentId="29211" CreationDate="2019-07-17T12:59:47.420" Score="2" Body="&lt;p&gt;Just doing a quick search, this NASA paper (&lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19810020599.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19810020599.pdf&lt;/a&gt;) indicates that no particular geometry has significant advantages for producing flow uniformity. I would stick to a well-tested geometry that is the simplest to produce with a 3D printer. Also, keep in mind the purpose of using honeycomb is to remove turbulence (vortices) from the flow via reducing the hydraulic diameter. Thus, so long as you don’t inflict a large pressure drop through the honeycomb (if you care about pressure drop), any shape with an appreciable reduction in flow hydraulic diameter will reduce the turbulent intensity. This paper seems to be useful as well: &lt;a href=&quot;https://conservancy.umn.edu/bitstream/handle/11299/108725/pr338.pdf?sequence=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://conservancy.umn.edu/bitstream/handle/11299/108725/pr338.pdf?sequence=1&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21118" LastActivityDate="2019-07-17T12:59:47.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29231" PostTypeId="2" ParentId="29229" CreationDate="2019-07-17T13:36:27.977" Score="0" Body="&lt;p&gt;This just sounds like the mesh was sagging between supports during the pour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Until you check across the width and length of the slab for the dimensions at several points there is not much else you can conclude.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-07-17T13:36:27.977" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29234" PostTypeId="1" CreationDate="2019-07-17T17:11:55.733" Score="1" ViewCount="28" Body="&lt;p&gt;I read something about defining hand gesture, which include &lt;strong&gt;mid-air hand gesture&lt;/strong&gt; (e.g a peace sign with the two fingers moving one after another), &lt;strong&gt;touch-based gesture&lt;/strong&gt; (e.g. on touch screen), and &lt;strong&gt;motion gesture&lt;/strong&gt; - manipulating the device (e.g. tilting a mobile phone sideway, pointing a mobile phone towards a person).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am not sure about how to categorise these gestures according to the following axis dimensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some definition I saw were;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Single-axis&lt;/strong&gt; Motion occurs around a single axis&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;No matter the gesture is done in the mid-air context or on a surface,&#xA;  if the movement is done only in one of the three coordinates, X, Y, or&#xA;  Z, it is single-axis.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Double-axis&lt;/strong&gt; &quot;If the gesture is performed on a surface, for example on a tabletop application, then the gesture has Double-Axis&#xA;  dimensionality.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Tri-axis&lt;/strong&gt; Motion involves either translational or rotational motion, not both. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&quot;If only either of translation or rotation occurs in a gesture which&#xA;  is performed in 3D space, it has tri-axis dimensionality.&quot;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Six-axis&lt;/strong&gt; Motion occurs around both rotational and translational axes&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;what if I flip down a phone along one axis, it can be considered a rotation in 3D space? if yes then it would be tri-axis even it is movement done in one single axis?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;for sliding a phone along a plane and in a straight line, is it a tri-axis motion as it is translation n 3D, or is it a single-axis movement as it moves along one axis only?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Any examples would be very helpful. Thank you very much!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;references&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*definition in quote is from here: &lt;a href=&quot;https://library.ndsu.edu/ir/bitstream/handle/10365/23110/Vafaei_Taxonomy%20of%20Gestures%20in%20Human%20Computer%20Interaction.pdf?sequence=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;TAXONOMY OF GESTURES IN HUMAN COMPUTER INTERACTION&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;definition not in quote signs is from here:&#xA;&lt;a href=&quot;http://doi.acm.org/10.1145/1978942.1978971&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://doi.acm.org/10.1145/1978942.1978971&lt;/a&gt;*&lt;/p&gt;&#xA;" OwnerUserId="21181" LastEditorUserId="8055" LastEditDate="2019-07-18T00:27:29.463" LastActivityDate="2019-07-18T00:27:29.463" Title="motion gesture and axis definition" Tags="|mechanical-engineering|sensors|computer-engineering|computer-hardware|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29235" PostTypeId="1" AcceptedAnswerId="29236" CreationDate="2019-07-17T17:13:33.640" Score="0" ViewCount="191" Body="&lt;p&gt;When I'm looking at images / videos about the load supported by reinforced concrete slabs, I always see a compression at the top of the slab in every point that varies from the center to the edges:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f4eoZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f4eoZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But isn't there a tension on the slab near the edges caused by the fact that the slab is supported by the walls?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, why I always see the first and not the second schema below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kqkmK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kqkmK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21141" LastActivityDate="2019-07-20T21:54:47.060" Title="Why not a reverse bending in the slab?" Tags="|structural-engineering|structural-analysis|reinforced-concrete|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29236" PostTypeId="2" ParentId="29235" CreationDate="2019-07-17T17:37:05.833" Score="2" Body="&lt;p&gt;the usual assumption is that the slab is &lt;em&gt;simply supported&lt;/em&gt; which means bending moments cannot be transferred from the slab to the walls. In the second sketch, the slab has &lt;em&gt;clamped&lt;/em&gt; or &lt;em&gt;cantilevered&lt;/em&gt; edges which &lt;em&gt;do&lt;/em&gt; support bending moments.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-07-17T17:37:05.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29237" PostTypeId="2" ParentId="29235" CreationDate="2019-07-17T17:42:34.033" Score="2" Body="&lt;p&gt;Either there is some problem with global settings of your software or you must have defined the slab simply supported on the edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you define the whole structure including adjacent bays and define the slab as two way or on way slab (as they are), you will see the moment on the supporting beams causing tension on top. And if you have cantilever slabs the have nothing but tension on top of the supporting beam and compression on the bottom of the slab's concrete&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only there are moment over the supporting edge beams but codes ask for the loading pattern on adjacent bays in a way that causes most severe negative and positive moments. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-17T17:42:34.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29238" PostTypeId="1" AcceptedAnswerId="29255" CreationDate="2019-07-17T21:22:16.607" Score="1" ViewCount="352" Body="&lt;p&gt;I have a quick question about the parallel axis theorem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say I have a box-like object rotated about the following pivot point&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lYgRd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lYgRd.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was wondering, by applying the parallel axis theorem, would the moment of inertia about the pivot point be the following,  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M = I_{cm} + M_d^2+M_h^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$I_{cm}$&lt;/span&gt; is the moment of inertia about the box's centroid and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is its mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, does the parallel axis theorem not hold true for two axis shifts?&lt;/p&gt;&#xA;" OwnerUserId="20847" LastEditorUserId="1832" LastEditDate="2019-07-19T01:51:53.763" LastActivityDate="2019-07-19T01:59:30.150" Title="Parallel Axis Theorem for x and y Shift" Tags="|mechanical-engineering|structural-analysis|applied-mechanics|statics|torque|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29239" PostTypeId="1" AcceptedAnswerId="29240" CreationDate="2019-07-16T22:08:15.610" Score="2" ViewCount="188" Body="&lt;p&gt;I am wondering if boiling water that is trapped inside of a semi-enclosed pipe would be a low-cost method of creating an underwater thruster that would function as a silent marine propulsion system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please reference the following conceptual drawing:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dgFFf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dgFFf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This drawing is showing a cross-sectional side view of a conceptual underwater thruster. There would be a copper pipe that would be partially positioned within a metal cylinder and within this metal cylinder there would be an AC induction heating element that would partially surround the closed end of the copper pipe. The other end of the copper pipe would be located outside of the metal cylinder and it would be open-ended. This heating element should create a rising water temperature gradient within the closed end of the copper pipe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The AC induction heating element would be turned on and would remain on so that it would continually boil the water. This will cause the water pressure (psi) inside the pipe to be higher that the ambient water pressure outside of the pipe/thruster and this imbalance should cause the thruster to move forward as indicated on the drawing. This thruster could be mounted externally to a ship or submarine's hull or it could be installed internally in the stern area of either vessel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main premise of this idea is that the boiling water will not be able to exit the copper pipe so there should be continual high water pressure inside the pipe and thus continual thrust. Also, there should always be water inside of the pipe because there should be a continual circulating flow of warm water exiting the copper pipe while colder outside water enters it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could boiling water inside a semi-enclosed pipe be a low-cost method for a silent marine propulsion system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the answers and comments I have received to original design above, I have redesigned this 'Boiling Water Thruster', shown in the new drawing below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FIUNJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FIUNJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this new underwater thruster design should provide silent propulsion for the ship/submarine, although the thrust would be low compared to propeller driven craft.&lt;/p&gt;&#xA;" OwnerUserId="33377" OwnerDisplayName="user26803" LastEditorDisplayName="user18610" LastEditDate="2019-07-28T12:43:21.727" LastActivityDate="2019-07-28T12:43:21.727" Title="Could boiling water inside a semi-enclosed pipe be a low-cost method for a silent marine propulsion system? (redesigned)" Tags="|fluid-mechanics|thermodynamics|design|propulsion|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="29240" PostTypeId="2" ParentId="29239" CreationDate="2019-07-16T22:52:30.367" Score="3" Body="&lt;p&gt;It can kinda work. Ish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your design is based, at its core, on the idea that warm water rises. Warm water only rises in cold water, so I am sceptical of the efficacy of simple thermal gradients. Nevertheless, bubble pumps have been in commercial use for many years, so if you get it just right I suppose it's possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you have some serious mass flow problems in your design. If I understand correctly, in your design you intend the flow path to be something like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1P3aT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1P3aT.png&quot; alt=&quot;naive pipe pump&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, this system is pretty hard to analyze simply. As the essential components are the pumping action and the flows, I think it will be easier to see on a simpler system that is effectively the same (the symbol is a generic pump):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/dM5Ho.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dM5Ho.png&quot; alt=&quot;equivalent mass flow system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(let right be positive)&#xA;I'm going to play a little loosy-goosy with mass and mass flow rate so as to not complicate the analysis with a whole lot of derivatives. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume you know that &lt;span class=&quot;math-container&quot;&gt;$F = ma$&lt;/span&gt;? In this system, there are two forces on the submarine: the force from sucking in the inlet water, which pushes to the left, and the force from pushing out the effluent water, which pushes to the right. Ok, so it is more useful to use the momentum forms of Newton's laws here:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F = \frac{dp}{dt}, \Delta p_{net} = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means he change of the momentum of the craft is equal to the change of momentum of the water. At the inlet pipe, the water is moving into the sub at some velocity &lt;span class=&quot;math-container&quot;&gt;$+v_{in}$&lt;/span&gt;, and some mass &lt;span class=&quot;math-container&quot;&gt;$m_{inlet}$&lt;/span&gt; is coming in. As &lt;span class=&quot;math-container&quot;&gt;$p = mv$&lt;/span&gt; and the momentum imparted to the sub is exactly opposite to that imparted on the water, so &lt;span class=&quot;math-container&quot;&gt;$$\Delta p_{in, sub} = -\Delta p_{in, water} = -V_{in} \cdot m_{in}$$&lt;/span&gt; It is similar at the outlet pipe, so &lt;span class=&quot;math-container&quot;&gt;$$\Delta p_{out, sub} = -\Delta p_{out, water} = +V_{out} \cdot m_{out}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, unless you are somehow generating water inside the craft, &lt;span class=&quot;math-container&quot;&gt;$m_{in}$&lt;/span&gt; &lt;em&gt;must&lt;/em&gt; equal &lt;span class=&quot;math-container&quot;&gt;$m_{out}$&lt;/span&gt;. Therefore, the total momentum imparted on the craft is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta p_{tot} = \Delta p_{in, sub} + \Delta p_{out, sub} = V_{out} \cdot m_{out} - V_{in} \cdot m_{in} = (V_{out} - V_{in})\cdot m_{water}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And there's the key problem with the design! If &lt;span class=&quot;math-container&quot;&gt;$V_{out} = V_{in}$&lt;/span&gt;, then $\Delta p = 0} and the craft cannot accelerate, no matter how fast you pump the water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok, now for some fluid mechanics. How can we make &lt;span class=&quot;math-container&quot;&gt;$V_{out} \ne V_{in}$&lt;/span&gt;?&#xA;Well, the mass &lt;span class=&quot;math-container&quot;&gt;$ \dot{m}$&lt;/span&gt; flowing through an orifice is proportional to area, velocity, and density:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{m} = A \cdot V \cdot \rho $$&lt;/span&gt;&#xA;because we know &lt;span class=&quot;math-container&quot;&gt;$\dot{m}_{in} = \dot{m}_{out}$&lt;/span&gt;, then &#xA;&lt;span class=&quot;math-container&quot;&gt;$$ A_{in} \cdot V_{in} \cdot \rho_{in} = A_{out} \cdot V_{out} \cdot \rho_{out}$$&lt;/span&gt;&#xA;Now while the would change from inlet to outlet if they are different temperatures, water only has a density change of 3% over its entire liquid range, so I am going to disregard that, so:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ A_{in} \cdot V_{in} = A_{out} \cdot V_{out} $$&lt;/span&gt;&#xA;Thus the only way that we can cause the craft to accelerate is to have a smaller outlet orifice than inlet (I'll let you work that one out). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we move back to your design, I can't see how you could possibly enforce that in a simple tube. The inlet and outlet streams are almost completely uncontrolled (they touch! you're going to have Rayleigh waves out the wazoo!) and while you could maybe do &lt;em&gt;something&lt;/em&gt; (put a mesh to restrict flow on the bottom?) I can't see an efficient one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, boiling is loud. Have you ever boiled a pot of water? It's certainly not silent. You would also need to get power from somewhere. If it's from a nuclear reactor (as you mention in a comment) those are also &lt;em&gt;really&lt;/em&gt; loud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're dead set on this bubble pump thing, I feel it would be much more feasible to simply put the inlet at the front:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c4qrz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c4qrz.png&quot; alt=&quot;bubble pump with inlet at front&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="Ash Pera" LastActivityDate="2019-07-17T03:35:52.290" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29241" PostTypeId="2" ParentId="29117" CreationDate="2019-07-18T00:55:21.480" Score="0" Body="&lt;p&gt;How is this box going to be supported? Don’t forget the base/supports of the box will be a path of heat transfer, even if it’s standing on legs or suspended. Secondly, which area are you referring to in the problem statement? The cube has several faces/planes. We would need more information on the specifics of the geometry and prior research/approaches tried to provide an adequate and concise answer. For now, I would begin by researching free convection and the thermal properties of the grade of aluminum you’ve chosen. &lt;/p&gt;&#xA;" OwnerUserId="21118" LastActivityDate="2019-07-18T00:55:21.480" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29242" PostTypeId="2" ParentId="29238" CreationDate="2019-07-18T04:30:36.157" Score="1" Body="&lt;p&gt;Parallel axis would apply if you first calculate the &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; of the box about an axis perpendicular to a line connecting the pivot point to the center of mass of the box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This I is not the same as the one fore a thin plate, the box has volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It gets a bit complicated, having to triple integrate in polar coordinate the &lt;span class=&quot;math-container&quot;&gt;$\text{d}v \cdot r^2$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2019-07-19T01:52:33.617" LastActivityDate="2019-07-19T01:52:33.617" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29245" PostTypeId="1" CreationDate="2019-07-18T10:23:13.727" Score="1" ViewCount="202" Body="&lt;p&gt;How does the load capacity for a reinforced concrete slab depend on the thickness of the slab?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose that we have the following constants:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;concrete quality (including the maximum compression resistance)&lt;/li&gt;&#xA;&lt;li&gt;same rebars (E.g.: mash of 4/100 mm)&lt;/li&gt;&#xA;&lt;li&gt;the rebars are at 35 mm from the bottom of the slab&lt;/li&gt;&#xA;&lt;li&gt;the load is constant in every point of the slab&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Firstly let's suppose that the slab has a thickness of:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;100 mm and supports a constant load (permanent: screed, ceramic tiles + temporary: objects, people) of 250 kg/m^2 (2.5 N/m^2).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;150 mm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;200 mm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How to determine the load capacity for points 2 and 3 based on point 1?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: The example is just to emphasize the idea. I'd like to have a more general solution.&lt;/p&gt;&#xA;" OwnerUserId="21141" LastActivityDate="2021-04-09T01:01:16.430" Title="Dependency between load capacity for reinforced concrete slab and thickness" Tags="|structural-engineering|structural-analysis|concrete|reinforced-concrete|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29247" PostTypeId="1" CreationDate="2019-07-18T14:48:57.690" Score="1" ViewCount="95" Body="&lt;p&gt;My guess is that they use brushless Motors that in turn powers the hydraulic pump.&#xA;But there must be some kind of valve that could help in the process. Still the burning question is just by adjusting the flow and pressure will it actually control precision required hydraulic equipment? Since there are lots of valves, which combination of valves can work? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there are many easy ways to control an electric actuator with feedback, using Arduino or simple &quot;pwm&quot; etc.&lt;/p&gt;&#xA;" OwnerUserId="8677" LastEditorUserId="8677" LastEditDate="2019-07-20T08:56:43.397" LastActivityDate="2019-07-25T11:13:47.250" Title="How are the modern hydraulic cylinders very precisely controlled?" Tags="|pumps|hydraulics|valves|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29248" PostTypeId="1" AcceptedAnswerId="29253" CreationDate="2019-07-18T17:20:48.043" Score="0" ViewCount="72" Body="&lt;p&gt;I am trying to test a piece of furniture based on a particular code, the code states that i can either drop a bag (with a specific shape) of 57 kg 1.2 in above the seat or I can use other methods as long as they can be shown to be equivalent. My company is interested on using vertical pneumatic cylinders instead since the test needs to be repeated 100 000 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: If i manage to get a piston that produces the equivalent of 57kg in force and i adjust its speed to match the speed at contact for a free falling bag (0.77 m/s), would this produce equivalent effects on my test sample?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if i am overlooking or neglecting something. Any advice will be appreciated&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks  &lt;/p&gt;&#xA;" OwnerUserId="21194" LastActivityDate="2019-07-19T16:43:14.763" Title="Emulate impact forces using pneumatic piston" Tags="|pneumatic|product-testing|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29249" PostTypeId="2" ParentId="29245" CreationDate="2019-07-18T17:23:38.010" Score="1" Body="&lt;p&gt;A very rough estimate to start designing is this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The changing of the depth of a concrete slab changes the resisting momen arm: &lt;span class=&quot;math-container&quot;&gt;$d-\frac{a}{2}$&lt;/span&gt; of the slab. With d, being effective depth and a, the depth of concrete stress block in compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In simply supported non prestressed slabs this distance changes&#xA;almost linearly, even though a, is not linear but the changes are within the estimate error tolerance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you multiply the slab depth by 20 % you approximately have 20% more moment resisting given all other factors are the same.&#xA;This is a rule of thumb tool to start estimating the first parameters of the slab design.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-18T17:23:38.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29251" PostTypeId="1" CreationDate="2019-07-18T19:03:56.317" Score="3" ViewCount="24" Body="&lt;p&gt;I have a plunger - that I'm using for glass forming - made out of a martensitic steel. Glass is at 1000°C and since martensite is a metastable phase, should I expect the properties of my plunger to change during its life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's the case how can I calculate how fast it is transforming?&lt;/p&gt;&#xA;" OwnerUserId="18457" LastActivityDate="2019-07-18T19:03:56.317" Title="Will a cyclic heat load transform my martensitic steel? How fast?" Tags="|materials|thermodynamics|steel|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29252" PostTypeId="1" CreationDate="2019-07-18T19:55:14.347" Score="0" ViewCount="26" Body="&lt;p&gt;I ran a 60 m seismic line. Energization with hammer. 12 (10 hz vertical 5 m spaced) geophones  and 5 shots (15 m spaced). Record length 0.5 s. Thanks to a borehole near a estrata of gravelly silt man-made (0.40 m) and very soft silty clay (to 5 m) and then clay improves its consistency deeper.&#xA;My seismic traces seemed to be ok, but at office the principal wave train gives 200 m/s and first arrivals refractions are that weak that can not be easily identified. I think the principal train wave shadows them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: ¿What is the meaning of a 200 m/s train wave in geophysics? &lt;/p&gt;&#xA;" OwnerUserId="21196" LastActivityDate="2019-07-18T19:55:14.347" Title="Wave train 200 m/s meaning" Tags="|geotechnical-engineering|seismic|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29253" PostTypeId="2" ParentId="29248" CreationDate="2019-07-18T21:25:22.523" Score="0" Body="&lt;p&gt;It's not a question of force, its momentum m*v in a non-elastic collision that remains constant. So you have to measure the mass of your pneumatic moving rod and make sure it complies with Newton's law in a way that its momentum is equal to the momentum of the bag at the time it touches the sofa. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After OP's comment, I added detail subscripts to identify moving parts, to clarify I mean the momentum of the pneumatic punch rod should be equal to the momentum of the bag at its final velocity just before it collides with the sofa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$m_{rod}*v_{rod}= m_{bag}*v_{bag}$$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will deliver the same impact to the sofa. As you said in your comment kinetic energy will not preserve.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-07-19T16:43:14.763" LastActivityDate="2019-07-19T16:43:14.763" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29254" PostTypeId="1" CreationDate="2019-07-18T22:15:53.537" Score="0" ViewCount="65" Body="&lt;p&gt;Last year we poured a RC suspended slab according to designs of a structural engineer. (Note we are just DIY home builders but have taken professional advice throughout.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slab was cast with three little 300mm diameter port-holes which we had planned to use for light, but for various reasons we wish to cover these up now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a steel plate over the top would be fine, but for waterproofing purposes I'd like to set a concrete square over the top, using a slurry to bond new to old concrete, and waterproofing the whole lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My concrete square would perform two functions: to provide an upstand to waterproof, and to cover the hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From a structural point of view, a piece of plywood would (hypothetically) be adequate, because over the top we will add 100mm insulation then hard finish like paving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm aware I may need to consider the mass of the concrete itself, so will need to add rebar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other possible solutions, like trying to seal a steel plate, then screeding over the top - but as we will be applying quite an expensive liquid membrane, then setting hard finish which I hope to last for 50 years on top, I'd like this to be a bullet-proof solution in terms of waterproofing. The fact they stick out of the ground is not a problem, as we will be adding insulation over the top of the remaining area before the paving stones. Although my idea would mean a reduction in area of insulation which is not idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a pic of the formwork (not complete) which shows square sides and the circle inside the hole is propped from underneath (NOTE: I have not yet adjusted prop height, when I do, the circles will be flush with the top of the concrete)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BazWK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BazWK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have also cut and bent rebar into 400mm lengths, with a right angle bend at each end and a further 100mm that could be doweled into the existing concrete to provide both reinforcement and prevent movement. I was going to use 4x lengths laid parallel per hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made no calculations, as I think this may be a trivial &quot;onsite solution&quot;. But I'm aware there will be an (albeit small) amount of suspended concrete here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone tell me if this is a sensible plan so far?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option might be to place a steel plate over the holes then add concrete as per my plan above.&lt;/p&gt;&#xA;" OwnerUserId="21197" LastEditorUserId="21197" LastEditDate="2019-07-18T22:39:55.363" LastActivityDate="2019-07-19T03:24:48.990" Title="How to reinforce concrete over a circular span" Tags="|concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29255" PostTypeId="2" ParentId="29238" CreationDate="2019-07-19T01:59:30.150" Score="1" Body="&lt;p&gt;Moment of inertia is based on both a point of reference &lt;strong&gt;&lt;em&gt;and&lt;/em&gt;&lt;/strong&gt; an axis of rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer to your question depends on the axis you're asking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the moment of inertia around the horizontal axis, you only care about the vertical shift (and visa-versa for the vertical axis, of course):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_{xx} = I_{cm} + M\cdot h^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any other axis, things get much more complicated, as suggested by &lt;a href=&quot;https://engineering.stackexchange.com/a/29242/1832&quot;&gt;@kamran's answer&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-07-19T01:59:30.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29257" PostTypeId="2" ParentId="29254" CreationDate="2019-07-19T03:24:48.990" Score="1" Body="&lt;p&gt;It's over kill, but yes it is fine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would just put to # 3 rebars dowelled inside the circles 6in OC. And fill the holes flush by  a couple of bags of ready-mix concrete and wet the circumstances with acrilic glue. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-19T03:24:48.990" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29259" PostTypeId="1" CreationDate="2019-07-19T13:53:20.927" Score="0" ViewCount="100" Body="&lt;p&gt;&quot;Tesla Model 3 owner died in a tragic accident with a semi truck. The Model 3 went under the truck’s trailer resulting “in the roof being sheared off as it passed underneath,” which is known as a “side underride” accident.&quot;&#xA;Is it possible to stop the car?&lt;/p&gt;&#xA;" OwnerUserId="21203" LastActivityDate="2019-07-21T02:20:57.177" Title="What should be the material and thickness of front pillars to stop car (~1,847kg and ~262 km/h) without killing the driver?" Tags="|mechanical-engineering|structural-engineering|materials|aerospace-engineering|car|" AnswerCount="2" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="29260" PostTypeId="2" ParentId="29259" CreationDate="2019-07-19T16:06:52.633" Score="0" Body="&lt;p&gt;At 262 km/h (163 mph) there are many other aspects to that particular crash type than just front pillars.  Even with pillars designed to stop the car many other factors could kill the driver.  Maybe look at NASCAR stock cars for a better idea of high speed safety.  Roll cages, Lexan instead of glass, helmets, seats/harness that severely restrict movement, etc.  They are designed from the ground up for high speed impact.&lt;/p&gt;&#xA;" OwnerUserId="20957" LastActivityDate="2019-07-19T16:06:52.633" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29262" PostTypeId="1" CreationDate="2019-07-19T20:01:20.460" Score="-1" ViewCount="99" Body="&lt;p&gt;Assuming that an electric bicycle and an electric scooter use the same motor and battery, which has more autonomy? that is, which makes more kilometers / miles per battery charge cycle? Obviously each vehicle logically has a different transmission and size of wheels.&lt;/p&gt;&#xA;" OwnerUserId="21207" LastActivityDate="2024-02-23T11:48:37.167" Title="Electric bicycle and an electric scooter" Tags="|mechanical-engineering|electrical|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29263" PostTypeId="2" ParentId="29262" CreationDate="2019-07-19T20:36:26.417" Score="0" Body="&lt;p&gt;It depends on the efficiency of the design, the wind drag, the quality of rejuvenative breaking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The factors most influential in performance are not the form factor alone. Also both the scooter and electrical bike do not necessarily need a transmission as the torque can be adjusted electronically. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-07-19T20:51:40.027" LastActivityDate="2019-07-19T20:51:40.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29264" PostTypeId="1" AcceptedAnswerId="29319" CreationDate="2019-07-19T23:49:48.930" Score="0" ViewCount="303" Body="&lt;p&gt;I am having a house built and the builder found organic soils during the excavation.  They removed the soils and then filled the cavity with 1x3 stone (see picture).  This is the corner where the sump crock is supposed to be and I am concerned that the digging of the hole for the sump will unsettle the stone.  The builder doesn't think so.  My question is should I be concerned about this or will disturbing the stone undermine the foundation.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UhA1e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UhA1e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="21208" LastActivityDate="2019-07-23T06:54:08.213" Title="3 x 1 stone under footing and sump crock" Tags="|civil-engineering|soil|foundations|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29265" PostTypeId="2" ParentId="29259" CreationDate="2019-07-20T00:01:42.187" Score="3" Body="&lt;p&gt;In high speed collisions essentially the human body fails under high g negative acceleration even if the structure of the car or cabin or test vehicle is strong enough to survive the collision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;strange things can happen that may not immediately be visible such as brain injury due to boiling of  Cerebrospinal fluid because of the vacuum created by rapid  deceleration of the skull during the crash when the race car helmet contacts the dashboard, (contracoup brain injury). This boiling will create craters in the brain. That is why there is helmet restraints to the back and sides of the seat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In design of structure of the cars for crash survivability the general strategy is to protect the occupants by placing sacrifice high energy absorbing structure around the cabin. these sacrifice collapsible structures are commonly the wheel and suspension structure, ribbed and perforated framing bars and studs, engine and other substantial components. they will bend and deform while absorbing the crash energy and providing precious time (in milliseconds) and room for the occupants to break the impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good restraints and padded potential contact surfaces, relocating objects which could be dangerous in a crash from the cabin, are other parts of the strategy.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An absolutely rigid structure is not part of the optimal strategy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is some tables on human crash tolerance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lJryD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lJryD.jpg&quot; alt=&quot;Human G acceleration tolerance&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ql0XI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ql0XI.jpg&quot; alt=&quot;Eiband chart of crash survivability&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-07-21T02:20:57.177" LastActivityDate="2019-07-21T02:20:57.177" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29266" PostTypeId="1" CreationDate="2019-07-20T02:25:38.277" Score="0" ViewCount="163" Body="&lt;p&gt;I have a structure that is used to support two shafts which are carrying relatively high loads. It is composed of two columns and a &quot;shaft supporting block&quot; in the centre (almost acting as a beam). The block is just a rectangular block of steel with two through holes in the centre. I want to analyse the stresses in the columns and block in order to verify if they are suitable and do not yield or fail. I need a relatively accurate solution and so I want to avoid approximate methods. The loads act at the bottom of each hole, in the centre of the thickness and therefore torsion is negligible or can be considered as 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought was to idealise the shaft supporting block as a beam and apply the moment distribution method. However, due to the two holes for the shafts to pass through, the block has a varying moment of inertia and therefore cannot (at least in my understanding,) be solved using moment distribution. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also thought of trying the column analogy method however, did not work due to the same above reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not have access to use finite element analysis (computer) and want to use standard structural analysis methods (e.g. moment distribution and not stiffness method). I have attached a rough sketch image of the real structure and idealised system for reference. &lt;a href=&quot;https://i.stack.imgur.com/aUPrX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aUPrX.jpg&quot; alt=&quot;Idealised Structure&quot;&gt;&lt;/a&gt;  &lt;a href=&quot;https://i.stack.imgur.com/NDEyI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NDEyI.png&quot; alt=&quot;Actual Structure&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20822" LastActivityDate="2023-12-26T21:02:40.290" Title="Analysis of Machine Frames With Cut Outs" Tags="|mechanical-engineering|structural-analysis|statics|machine-design|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29267" PostTypeId="1" CreationDate="2019-07-20T02:43:24.833" Score="0" ViewCount="156" Body="&lt;p&gt;An induction cooktop has a high frequency power source that is applied to a coil. I was wondering if you could take the high frequency electronics from a cooktop or even buy / design a high frequency power source and instead of a coil take a 50 or 200 foot long wire and attach to rebar to heat it to a temperature of only about 30 degrees Celsius in an environment that is -30 degrees Celsius?&lt;/p&gt;&#xA;" OwnerUserId="21210" LastActivityDate="2019-07-20T15:13:53.493" Title="Induction heating rebar in cold weather" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29268" PostTypeId="2" ParentId="29267" CreationDate="2019-07-20T04:17:12.277" Score="1" Body="&lt;p&gt;Induction heating works by making a big magnetic field.  The field induces currents to flow in the thing heated up, and the current flowing through the resistance of thing makes heat.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not just pass current through the rebar directly to heat it up?  This would be more efficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can't actually get the the rebar, because it is inside some concrete, then the induction method may be useful.  But, it's hard to make a magnetic field in a big volume, like hundreds of feet of rebar. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as the amount of power that you can deliver with induction heating, it is only limited by the amount of power it takes to create the magnetic field. Depending on the losses to the -30 degree ambient, maybe you could heat up the rebar that much.  You would have to model the heat flow and the magnetic fields for your particular situation.   &lt;/p&gt;&#xA;" OwnerUserId="19743" LastActivityDate="2019-07-20T04:17:12.277" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29269" PostTypeId="2" ParentId="29266" CreationDate="2019-07-20T05:05:36.433" Score="0" Body="&lt;p&gt;This is an indeterminate structure, meaning the moments and shears can not be determined by just the loads and reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you need the detailed measurements, thicknesses, the radius of the holes and define the posts as to their stiffness and dimensions and define supports at the base, whether they are roller, pin or fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However assuming pin supports and assuming your sketch is to scale, by inspection, it can be assumed two cantilever beams at the two sides imparting compression to the top half of the middle part and tension to the lower half.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you can solve the frame by Hardy Cross moment distribution method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is the Wikipedia link,&lt;a href=&quot;https://en.wikipedia.org/wiki/Moment_distribution_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiki moment&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-20T05:05:36.433" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29271" PostTypeId="1" CreationDate="2019-07-20T12:39:06.873" Score="1" ViewCount="31" Body="&lt;p&gt;I am using a Google Nexus 6 Mobile in India with Airtel as Network Provider. Recently Airtel has upgraded their Network to 2300 Mhz 4G. Whereas Google Nexus 6 Operates on Band 3 (1800Mhz) LTE. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I switch Mobile to LTE Mode - I am able to use 4G Network (Internet ) but there are no Incoming Calls. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I switch Mobile to 3G Network - I am able to get Incoming Calls but no 4G Network. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What can be the exact cause for this and possible solutions? &lt;/p&gt;&#xA;" OwnerUserId="21212" LastEditorUserId="7083" LastEditDate="2019-08-07T20:32:49.300" LastActivityDate="2019-08-07T20:32:49.300" Title="Mobile Frequency and Carrier Frequency not Matching" Tags="|electrical-engineering|telecommunication|wireless-communication|frequency-response|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29272" PostTypeId="2" ParentId="25650" CreationDate="2019-07-20T14:28:39.623" Score="0" Body="&lt;p&gt;Induction motors are amazing machines - they utilize several physical principles in a very elegant matter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your statement &quot;The field has to rotate at a slightly higher rate than the rotor&quot; is correct, however, there aren't any sensors. The physics is the only thing that makes it happens.&#xA;Lets deal with the classical case of a three phase motor, without using any variable-frequency drive. The induced magnetic field angular velocity around the stator axis is the same as the power supply frequency. &#xA;Seek for 3 phase electrical machines or rotating magnetic field in google for more information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ymFUJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ymFUJ.png&quot; alt=&quot;rotating magnetic field illustration from WIKIPEDIA&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without naming each phenomena specifically (look for Lenz's law and Lorentz force if you are carious about it), lets try to describe what is going on under the motor's hood:&#xA;The rotation of the magnetic field exposes the rotor to a time varying magnetic flux, which in turn induces electrical current along the rotor winding. Next, since the winding now conducts electrical current and also subjected to a magnetic field - a force is exerted on the winding, creating a torque around the rotation axis and causing the rotor to accelerate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is where the beauty of the motor design really come into play. As the rotor accelerates, its relative angular velocity with respect to the rotating magnetic field is gradually reduced. Is it clear? just to remind you that the magnetic field rotation speed is constant and dependent only on the supplying voltage frequency. &#xA;As the relative velocity is getting smaller, the induced current also decreases and so is the accelerating torque.&#xA;For each external load (torque) acting on the rotor, a different constant angular speed would be reached. At that speed the relative velocity between the magnetic field and the rotor will result in an internal torque equals to the external one.&#xA;This is why an induction motor will never rotate at the exact magnetic field speed (and therefor called asynchronous motor). If it was able to rotate at that speed, the relative velocity would be zero so no torque would be applied on the rotor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1  &lt;a href=&quot;https://i.stack.imgur.com/ymFUJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;: &lt;a href=&quot;https://i.stack.imgur.com/bANJf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/bANJf.png&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2019-07-20T14:28:39.623" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29273" PostTypeId="2" ParentId="29267" CreationDate="2019-07-20T15:13:53.493" Score="0" Body="&lt;p&gt;Search &quot; induction heat pipeline&quot; , there are references to heating for field welding. Also , pipelines have been heated full length by induction ( a conductor is buried next to the pipe) to warm contents to keep it fluid ; but a reference did not show in one search.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-07-20T15:13:53.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29274" PostTypeId="1" AcceptedAnswerId="29307" CreationDate="2019-07-20T15:58:41.947" Score="0" ViewCount="58" Body="&lt;p&gt;Suppose I want to implant a bio-chip that consumes a lot of power into my body, for example, a full-fledged CPU, and I don't want to lug around a heavy battery like the ones used in pacemakers, is it possible to be in Rome and do as the Romans do, i.e. draw and utilize the sugar inside my blood as fuel, like a huge living and breathing body cell or an organ like the liver? Let's assume I can put this device where there's a lot of blood flow, say the jugular vein.&lt;/p&gt;&#xA;" OwnerUserId="21213" LastActivityDate="2019-07-22T12:06:57.270" Title="Is it possible to power bio-chips with blood sugar?" Tags="|biomedical-engineering|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29276" PostTypeId="2" ParentId="29235" CreationDate="2019-07-20T21:54:47.060" Score="0" Body="&lt;p&gt;From your image, we can clearly see supports on two edges of the slab. It seems reasonable to assume this is your entire model (that is, that you haven't created a view where the rest of the structure is &quot;hidden&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being the case, the slab will behave entirely according to how you've defined the supports. By the slab's behavior, it seems you've defined the supports as pinned, not restricting rotations. This being the case, the slab is free to rotate around the supports and does so, such that no negative bending moment is generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a real structure, however, the slab's behavior is affected by the stiffness of its supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slabs are usually supported in one of two ways: either by walls or by beams (which are themselves supported by columns)&lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When slabs such as yours (which is only supported on two opposing edges) are supported by walls, the entire wall-slab-wall structure behaves like a column-beam-column frame. Just as the beam tries to deflect by rotating its supports (the tops of the columns) and the columns resist, generating negative bending moment on the beam, the walls resist the slab's rotation, generating the same negative moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, the columns/walls are flexible, so there is rotation at the beam/slab's edges. Just not as much as there would be if the connection were pinned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a slab is supported by a beam, however, is does basically behave as if it were a simple pinned support. That's because the slab rotating is felt by the beam as a rotation around it's own axis. That is, the beam feels torque (as well as bending moment from the vertical force transferred from the slab, of course). And beams are very flexible to torque, especially the types of beams commonly used in non-infrastructure construction. So the slab basically feels it can transfer its load to its support (the beam) and that it can rotate freely around that support: the very definition of a pinned support.&lt;sup&gt;2&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That being said, there's a catch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've so far only discussed a single slab being supported by walls and/or beams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if a wall/beam is supporting not just one slab, but two? That is, you have a slab to either side of the support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, in that case, the other slab's stiffness must also be taken into consideration. Making matters worse, two side-by-side slabs will usually be designed under the same loading conditions. So if one slab under load is trying to rotate a wall/beam one way, the other slab will be trying to rotate it the other way, most likely canceling that rotation out entirely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if two slabs share a supporting wall or beam, they are usually treated as being fully fixed on that edge.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;Slabs can also be directly supported by columns, but that's much less common and orders of magnitude harder to calculate.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; &lt;sub&gt;There is a slight effect from the beam's vertical deflection at midspan, but it is negligible and usually ignored.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-07-20T21:54:47.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29277" PostTypeId="2" ParentId="29191" CreationDate="2019-07-20T22:38:46.940" Score="1" Body="&lt;p&gt;While I agree with basically everything in &lt;a href=&quot;https://engineering.stackexchange.com/a/29197/1832&quot;&gt;@ingenørd's answer&lt;/a&gt;, I thought it best to add this answer, which is more optimistic overall, but also has some potentially really bad news.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Indeed, the areas which were cut out for the cables will have diminished capacity. I personally don't even think the strength of the mortar used to cover the hole is that relevant, since the lack of rebar to &quot;bind&quot; that mortar to the rest of the structure means the mortar almost certainly wouldn't act monolithically with the rest of the slab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the reduced lever arm would likely reduce the beam's strength by around 70%&lt;sup&gt;1&lt;/sup&gt; (read that footnote: this is the definition of a lazy, back-of-the-envelope calculation). Pretty significant, but likely covered by the safety factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since one should never go &quot;oh, well, that's what the safety factor's for!&quot;, we can also remember the effect of load redistribution. These areas have a lower capacity, but they're also far less rigid (around 60% as rigid)&lt;sup&gt;2&lt;/sup&gt;, so they'll need to resist a far lower load than the rest of the slab as well (though this equally means the rest of the slab will need to resist a larger-than-expected load, but that'd be distributed throughout the nearby slab).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now, the areas that concern me the most by far are the cut-outs along the walls (the areas of the bottom and right edges with the cutouts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this slab is supported by walls, there may be a non-trivial amount of negative bending moment near the supports. Since these cut-outs are transversal to the slab's direction of action (the slab near the bottom wall is mostly working in an up-down direction, this means the entire dimensioning cross-section of the slab has this reduced height.&lt;sup&gt;3&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making matters worse, 3&amp;nbsp;cm is pretty deep into a residential/office-building slab. If the cutout actually just cut right through the negative steel reinforcement, that could very well be catastrophic, since that'd mean the rebar was removed exactly where it's needed the most. And even if they didn't cut through the rebar, they almost certainly reduced its concrete cover, which could lead to long-term harm due to faster oxidation unless effective impermeabilization was performed.&lt;sup&gt;4&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; &lt;sub&gt;Result obtained in the most back-of-the-envelope way possible: adopting ingenørd's result of 3&amp;nbsp;mm of compression area and rounding it up to 4&amp;nbsp;mm, assuming we can discard the entirety of the cut-out (which seems likely), adopting a concrete cover of 20&amp;nbsp;mm, and &lt;span class=&quot;math-container&quot;&gt;$\phi10$&lt;/span&gt; rebar, we get an original lever arm of &lt;span class=&quot;math-container&quot;&gt;$160-20-5-1.5 = 133.5\text{ mm}$&lt;/span&gt;, versus a reduced lever arm of &lt;span class=&quot;math-container&quot;&gt;$160-20-5-30-2 = 103\text{ mm}$&lt;/span&gt;, or 77% of the original. And then rounded down because I felt like it.&lt;/sub&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; &lt;sub&gt;Again, just using the fact that moment of inertia is proportional to the cube of the height, so these areas would be &lt;span class=&quot;math-container&quot;&gt;$\dfrac{130^3}{160^3} = 54\%$&lt;/span&gt; as stiff. Though the stiffness is likely affected by the fact we're dealing with a two-way slab. Though that fact also improves load redistribution, I believe.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;3&lt;/sup&gt; &lt;sub&gt;Slabs are usually calculated on a &quot;per-meter&quot; basis. In the rest of the slab, this means that a &quot;one meter slab&quot; has a small 9&amp;nbsp;cm-wide cutout at the top, which doesn't sound too bad. In that bottom-right area, however, the entire &quot;one meter slab&quot; is shorter.&lt;/sub&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;4&lt;/sup&gt; &lt;sub&gt;Though, honestly, if they didn't cut through the rebar, I'd take that as a very reassuring piece of news. 3&amp;nbsp;cm would be an absurdly high concrete cover for non-infrastructure projects. So not only would it mean they probably won't cause the building to collapse, but that the designing engineers took all this into consideration (whether by agreement with the electricians or just previous experience) and placed their negative rebar accordingly. In which case it's also reasonable to assume they calculated the entire slab appropriately, taking the reduced cross-section into consideration as well.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-07-21T17:01:35.723" LastActivityDate="2019-07-21T17:01:35.723" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29278" PostTypeId="2" ParentId="29202" CreationDate="2019-07-20T23:01:19.240" Score="0" Body="&lt;p&gt;As mentioned by &lt;a href=&quot;https://engineering.stackexchange.com/a/29226/1832&quot;&gt;@AndyT's answer&lt;/a&gt;, screed doesn't increase load capacity. If anything, it lowers effective capacity since it is itself a load, so there's less capacity to support everything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is simple: the screed isn't bound to the structural concrete in any meaningful way. As a very rough example, think of an old-school phone book or a thick paper-back dictionary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you hold the book along its spine, the pages all sag. Indeed, they almost sag as much as if you were holding each page alone. However, if you hold the book along the top or bottom edges, the pages sag much less, especially near the spine. That's because the spine is literally binding the pages together, making them work in unison to resist the bending load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concrete slabs and screed basically behave like the pages when you hold the book along its spine: there's nothing there to help them work together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, it &lt;strong&gt;&lt;em&gt;is&lt;/em&gt;&lt;/strong&gt; possible to make the screen work with the concrete. To do so, you'd need to prepare the slab accordingly: take a pickaxe and/or to the dry concrete, removing the top layer of thin aggregates until you've exposed &quot;raw&quot; large aggregates. This will improve the friction between the concrete and the screed, but it's not enough. You'd also need to drill holes into the concrete and add properly-fixed rebar &quot;hooks&quot; to help transfer loads between the two materials. With this done with great care, you could consider the screed working with the rest of the structure (though not for the slab's and screed's self-weight, since the slab has already deformed to resist that load).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is, why would you want to do such a thing? It is hard work, and there's no reason to do it: the screed is usually made of far inferior-quality material, and if you need the added strength, why not just pour an extra-thick layer of concrete in the first place?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only mention the process because it actually is used on occasion. Just not with screed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's how concrete reinforcements are done on aging structures. You calculate the strength needed in a given old beam and then add the necessary steel reinforcement, a layer of high-quality concrete, and additional rebar drilled into the old beam to ensure the old and new concrete work in unison.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-07-20T23:01:19.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29279" PostTypeId="2" ParentId="29117" CreationDate="2019-07-21T00:08:30.773" Score="1" Body="&lt;p&gt;Consider the integrated circuit (IC) generating power &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; inside a box of surface area &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;. The energy balance equation is below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ P = \sigma A \left(\epsilon_{IC}T_{IC}^4 - \epsilon_{B}T_{B}^4 \right) + h_i A \left(T_{IC} - T_B\right) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term is the net radiation flow with the Stefan-Boltzmann constant, emissivities and temperatures. The second is the convection flow inside the box with the convection coefficient. Assuming that you know all materials related values, this is one equation with two unknown temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy balance equation from the box to the air is written as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sigma A \left(\epsilon_{IC}T_{IC}^4 - \epsilon_{B}T_{B}^4\right) + h_i A \left(T_{IC} - T_B\right) =  \sigma A \left(\epsilon_{B}T_{B}^4 - \epsilon_{a}T_{a}^4 \right) + h_a A \left(T_{B} - T_a\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy going in to the box from the inside equals the energy leaving the box from the outside. Assuming you know all the materials related values and the air temperature &lt;span class=&quot;math-container&quot;&gt;$T_a$&lt;/span&gt;, this is a second equation with the same two unknowns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In principle, the problem is solvable. In practice, the easiest first approach is to neglect all radiation terms. This step gives these two equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ P = h_i A \left(T_{IC} - T_B\right) =  h_a A \left(T_{B} - T_a\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows you that the temperature at the IC relative to the box will be balanced to first order by the ratio of internal to external convection coefficients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{\left(T_{IC} - T_B\right)}{\left(T_{B} - T_a\right)} =  \frac{h_i}{h_a}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the inside of the box is stagnate while the outside can be moving air, &lt;span class=&quot;math-container&quot;&gt;$h_i &amp;lt; h_a$&lt;/span&gt; in general. Therefore, &lt;span class=&quot;math-container&quot;&gt;$\left(T_{IC} - T_B\right) &amp;gt; \left(T_{B} - T_a\right)$&lt;/span&gt; in general. The IC will be hotter relative to the box than the box is relative to the air around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This first order approximation can be used as seed guesses to the full-blown equation with radiation. I might recommend a graphical solution in the full blown case.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-07-21T00:08:30.773" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29280" PostTypeId="1" CreationDate="2019-07-21T02:37:10.987" Score="1" ViewCount="203" Body="&lt;p&gt;I found this meme in a facebook group on engineering concepts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UBTYx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UBTYx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I selected A (steady and uniform). In selecting A I assumed the width of the river to remain constant along its path and no parameters change with time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why steady? Velocity doesn't change with time. It's steady.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why uniform? We get the same volume across any cross section because width of stream doesn't change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think people who selected any other option are confusing laminar with turbulent flow. This has nothing to do with it if you made the correct assumptions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my selection (A) and justification for it correct?&lt;/p&gt;&#xA;" OwnerUserId="622" LastActivityDate="2019-12-21T14:01:45.253" Title="Does turbulent river flow classify as uniform?" Tags="|mechanical-engineering|fluid-mechanics|mechanical|turbulence|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29281" PostTypeId="2" ParentId="29205" CreationDate="2019-07-21T02:51:18.920" Score="1" Body="&lt;p&gt;As implicitly mentioned in &lt;a href=&quot;https://engineering.stackexchange.com/a/29207/1832&quot;&gt;@kamran's answer&lt;/a&gt;, the neutral axis (which is what I understand by 'axis of bending') is constant regardless of the applied load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's because the neutral axis is a property of the structure resisting the load, and in this case the load does not help to &quot;carry itself&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the neutral axis will merely be equal to the position of the box's centroid. However, it is important to note the box should almost certainly be considered a U-shape. That's because the box's lid is unlikely to be sufficiently &quot;bound&quot; to the rest of the box to truly act monolithically with the rest. So you'd want to consider a lid-less box as your cross-section (though having to carry the weight of the lid, of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, U-shapes are very tricky to design under bending, due to the possibility of localized buckling of the &quot;wings&quot;. This is especially problematic Given that your U-shape needs to be &quot;belly-down&quot; (that is, looking like a U, not an &lt;span class=&quot;math-container&quot;&gt;$\Pi$&lt;/span&gt;), which means that the top of the beam (in this case, the wings) will be under compression. And given how low the center of gravity would be in this orientation, it'll be very high compression in the members the least equipped to withstand it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd honestly recommend talking to an engineer to calculate this for you (especially given that there are myriad other aspects we haven't taken into consideration, such as wind and dynamic forces), or perhaps renting boxes made for this situation instead of trying to make your own. Another &quot;solution&quot; would be to use more than two support points on each side of the box. Using many supports would drastically reduce the applied load.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-07-21T02:51:18.920" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29282" PostTypeId="2" ParentId="29190" CreationDate="2019-07-21T03:19:16.950" Score="3" Body="&lt;p&gt;Things topple when their center of mass moves past their supports.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a square box: if you push it over slightly and let go, it'll fall back onto its original position. But if you push it past 45 degrees (when the center of the box would be exactly over the supporting edge), it'll topple onto one of its other sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, obviously, the wider the base, the harder it is for something to topple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, that's not how earthquakes cause buildings to collapse. It's not that the earth moves the foundation so far that the building's center of gravity moves past it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Earthquakes damage buildings by moving the base very quickly. The law of inertia (an object at rest wants to stay at rest) means that though the base moves, the rest of the building doesn't want to. And once the rest starts moving one way, the earthquake is then pulling the foundation the other way. In the worst case, this leads to a positive feedback loop, with each earthquake movement feeding the rest of the building's movement (the so-called &quot;resonant frequency&quot;). This causes significant stresses in the structure, which can damage or even destroy the building if it's improperly designed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(For an example of how a low-energy earthquake at the resonant frequency is worth than a high-energy earthquake, see &lt;a href=&quot;https://youtu.be/6JwEYamjXpA?t=768&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-07-21T03:19:16.950" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29285" PostTypeId="2" ParentId="29247" CreationDate="2019-07-21T12:39:38.247" Score="1" Body="&lt;p&gt;&lt;strong&gt;Typical Hydraulic Servo Setup&lt;/strong&gt; (as used in structural testing): &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power source is usually a pump (e.g. a swash-plate pump) arranged to provide (up to a rated flow rate) at ~constant pressure. Hydraulic accumulators may be included to smooth out transient pressure drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The valves are typically directional control valves, so they can drive a piston/cylinder both ways. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Directional_control_valve&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Directional_control_valve&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For structural testing, the valves are usually piloted spool valves, electrically (coil) driven, precision ground (more often, lapped.) These valves are fairly linear (in terms of coil-current-input to flow-rate output) and impressive bandwidths (hundreds of Hz., esp. in flight controls like aircraft, missiles or booster rockets) but that linearity comes at the cost of flow leakages, on the order of a couple percent of the peak flow (per valve!) The valve drive is often dithered (strongly!) to minimize stiction (e.g. from particles in the fluid getting into the valve.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feedback transducers used vary, but for high-power/high-damage-if-anything-goes-wrong systems, analog sensing, especially built-right-into-the-device, ratiometric sensing (e.g. LVDTs) is useful, because such analog stroke feedback permits simple, robust (e.g. PID) control loops for position control. Load control is also common for systems with a reliable force sensor; cylinder loadcells are common.&lt;/p&gt;&#xA;" OwnerUserId="13014" LastEditorUserId="13014" LastEditDate="2019-07-25T11:13:47.250" LastActivityDate="2019-07-25T11:13:47.250" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29288" PostTypeId="1" AcceptedAnswerId="29293" CreationDate="2019-07-21T15:21:13.787" Score="2" ViewCount="139" Body="&lt;p&gt;To follow up with my &lt;a href=&quot;https://engineering.stackexchange.com/questions/29190/can-a-horizontal-foundation-safeguard-against-earthquakes&quot;&gt;previous question on the seismic design of buildings&lt;/a&gt;, I would like to ask if it makes sense to support buildings with strong thick 'crisscrossed' metallic frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the impression that an 'X-structured design' provides good structural integrity, as commonly seen in construction sites. Would it be wise to construct buildings out of a metallic 'skeleton' that is formed throughout by intersecting the diagonals of squares?&lt;/p&gt;&#xA;" OwnerUserId="5159" LastActivityDate="2019-07-21T17:33:23.587" Title="Can a crisscrossed metallic skeleton resist earthquakes for buildings?" Tags="|structural-engineering|civil-engineering|structural-analysis|building-design|building-physics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29289" PostTypeId="1" CreationDate="2019-07-21T15:33:52.177" Score="0" ViewCount="29" Body="&lt;p&gt;Given that the deeper you go into the ocean the higher the pressure, could this pressure difference between atmospheric (above water) and a certain depth in the ocean be used to run a stirling engine?&lt;/p&gt;&#xA;&lt;p&gt;The piston would have to have an open facing to both or either the water at a certain depth (high pressure side) or the atmospheric end (low pressure side.)&lt;/p&gt;&#xA;&lt;p&gt;In typical stirling designs the high pressure end uses heat in a fixed area which increases the pressure and moves the piston. What if the heat were the same and constant? Would that still work? The ocean depth would be a fixed pressure and that is why I ask.&lt;/p&gt;&#xA;&lt;p&gt;Instead of heat difference we would be using pressure difference between two areas to drive the stirling engine.&lt;/p&gt;&#xA;&lt;p&gt;Just a thot&lt;/p&gt;&#xA;&lt;h2&gt;edit&lt;/h2&gt;&#xA;&lt;p&gt;I appreciate the responses so I came up with a possible workaround for the high pressure side (in the ocean.)&lt;/p&gt;&#xA;&lt;p&gt;Let’s say for the high pressure side there is a open ended cylinder that is facing upward. In this cylinder, when the piston is fully extended upwards it goes halfway the distance of the cylinder length. On the downstroke it goes the opposite direction which would be towards the bottom of the ocean floor.&lt;/p&gt;&#xA;&lt;p&gt;So as some commenters have pointed out, when the high pressure piston has the upstroke, it is pushing against the depth of the ocean.&lt;/p&gt;&#xA;&lt;p&gt;A possible workaround may be to introduce air right before the upstroke on top the high pressure  piston so the piston is pushing a bubble of air out the end of the cylinder instead of the pressure of the water. The bubble would rise and be like a buffer until the piston reaches its furthest extension from which point the water fills the cylinder and the pressure difference from the atmosphere and the depth of the ocean drives the piston downwards again.&lt;/p&gt;&#xA;&lt;p&gt;I imagine a lot of cavitation would happen in such a setup so you would need very strong materials.&lt;/p&gt;&#xA;" OwnerUserId="13921" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-07-21T18:54:12.117" Title="Oceanic stirling engine: pressure difference between atmospheric and ocean depths would make engine work?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2019-07-21T17:18:27.337" ContentLicense="CC BY-SA 4.0" />
  <row Id="29290" PostTypeId="2" ParentId="29289" CreationDate="2019-07-21T16:23:09.320" Score="1" Body="&lt;p&gt;No. It will take work to push the apparatus down into the water far enough to perform compression work on the contents of the cylinder, meaning that you will not get that compression work &quot;for free&quot; and hence this scheme can't be used to drive the engine. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-07-21T16:23:09.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29291" PostTypeId="2" ParentId="29288" CreationDate="2019-07-21T16:25:42.007" Score="2" Body="&lt;p&gt;This is indeed commonly done to perform earthquake structural retrofits on buildings which do not meet current code and which are too valuable to tear down and completely replace. It is also common to insertut huge shock absorbers into the X-bracing structure to dissipate the earthquake energy. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-07-21T16:25:42.007" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29293" PostTypeId="2" ParentId="29288" CreationDate="2019-07-21T17:33:23.587" Score="1" Body="&lt;p&gt;earth quake is basically random vibration of the earth in several different ways, S wave, P wave etc. which is a complex and different subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X bracing is a way of making the structure stiff, so in short buildings combined with other measures is mostly an effective method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in high risers and tall building the critical thing is to mitigate the huge shears that X bracing imparts on the foundation and also make sure the building structure vibration caused the earth quake does not resonate with the vibration of site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in tall buildings the strategy is to design the building for earth quake and wind fluttering in a way that the natural frequency of the building is not close to the frequency of earthquake or wind turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why in tall buildings they sometimes use moment frame as opposed to X bracing. These are classified as passive methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some critical structures they insulate the building from the foundation by using rollers and shock absorbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in some new projects the have active smart earthquake resisting computer controlled hydraulic jacks and braces, which make the building sway, opposing and breaking the impulse of the earthquake.    &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-21T17:33:23.587" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29295" PostTypeId="1" AcceptedAnswerId="29305" CreationDate="2019-07-21T18:21:32.613" Score="1" ViewCount="137" Body="&lt;p&gt;I am currently designing a setup for an experiment which involves loading a weight to its extension arm, and by design this structure is not bolted to any surface so its free to move. I need to simulate whether this structure will tip over because of the torque caused by this weight. I looked at some FEA softwares such as patran  but I can only get static stress analysis, what kind of technique/software should I use to simulate this kind of motion in computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;a href=&quot;https://i.stack.imgur.com/tsZiC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tsZiC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21219" LastActivityDate="2019-07-23T15:12:19.350" Title="Structure analysis" Tags="|mechanical-engineering|structural-analysis|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29296" PostTypeId="2" ParentId="29189" CreationDate="2019-07-21T18:26:20.423" Score="1" Body="&lt;p&gt;Induction heat your pipe! Pipe is cheap to replace, and unlikely to burn out in a short time, as it has a rather large cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A white hot pipe insulated on the outside should provide plenty of hot air.&lt;/p&gt;&#xA;" OwnerUserId="12173" LastActivityDate="2019-07-21T18:26:20.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29297" PostTypeId="1" AcceptedAnswerId="29298" CreationDate="2019-07-21T19:26:01.390" Score="0" ViewCount="209" Body="&lt;p&gt;I am designing a PID controller for a system that requires several step changes in the reference variable (concentration of solute in a tank) over a set period of time. I can find the plant transfer function experimentally, by applying a step change in the input (flow rate) and measuring the resulting changes in concentration over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is not clear to me is: does the plant transfer function change each time I apply a different step change in the input variable (e.g. from 0.1 L/hr to 0.08 L/hr then after a while from 0.08 L/hr to 0.05 L/hr), or is the plant transfer function fixed for a given system (regardless of changes in the reference/input variable)? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="21129" LastActivityDate="2019-07-21T23:59:24.400" Title="Does the plant transfer function of a system change with variations in the reference variable?" Tags="|control-engineering|chemical-engineering|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29298" PostTypeId="2" ParentId="29297" CreationDate="2019-07-21T23:59:24.400" Score="0" Body="&lt;p&gt;Not enough detail, but in your case very possibly not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A transfer function is a description of a &lt;em&gt;linear system&lt;/em&gt;, using the Laplace transform (or the z transform if you're sticking to sampled time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The basic problem with the above is that most real-world systems are nonlinear.  We get around that by holding one eye closed and &lt;em&gt;pretending&lt;/em&gt; that our system is linear (if we're measuring its response), or by finding its linearization around an operating point (if we're designing from a mathematical model).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not up on the chemical engineering details, but I suspect that there are all sorts of nonlinearities in your system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the apparent transfer function that you extract from your measurements at one flow rate is considerably different than the one you extract at a significantly different flow rate, then yes, your linearized model of your nonlinear system's behavior at that different flow rate is, indeed, probably different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you've got the math chops for it, write out the differential equations that describe the reaction -- if they're nonlinear, then you can (A) expect that the transfer function is only an approximation, and (B) you can linearize them yourself around various operating points and see if the gains are different.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-07-21T23:59:24.400" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29299" PostTypeId="1" AcceptedAnswerId="29318" CreationDate="2019-07-22T00:50:52.853" Score="0" ViewCount="228" Body="&lt;p&gt;Why Bode plots are on jw axis when determining gain and phase margin?&#xA;Usually it is explained as 1+G(s)H(s) should not be zero and margins represent how far away from this we are. But for showing this (&quot;distance&quot; to G(s)H(s) = -1, i.e. 0dB, -180 degree phase combination) we use bode plot which are GH values on jw axis. I would rather expect it to be on the whole s plane.&lt;/p&gt;&#xA;" OwnerUserId="21221" LastEditorUserId="21221" LastEditDate="2019-07-22T23:27:00.670" LastActivityDate="2019-07-22T23:36:48.283" Title="Gain and phase margin" Tags="|control-theory|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29300" PostTypeId="2" ParentId="26155" CreationDate="2019-07-22T01:47:20.737" Score="0" Body="&lt;p&gt;&lt;strong&gt;Response to Question 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In low power applications, mAh helps determine the battery life. When determining the battery life for an application the mAh advertise in battery specification sheet is important, but more importantly how the mAh are derived. Different regions in the world use different criteria to determine the mAh. For example USA markets will use around 20mA load to derive the mAh for a battery where as the same manufacture will use a 30mA load for outside of USA. As a result the same battery will be listed as 1500 mAh battery for the US market and 1400 mAh for markets outside the US. Therefore if mAh is important for the application a dialog with the battery vendor will be very beneficial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a discharge curve vs service life chart for a Duracell Ultra 223 CRP2 battery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CIRvl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CIRvl.png&quot; alt=&quot;Discharge Curves for Duracell Ultra 223 CRP2 battery&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the above chart it note that the service life varies based on the discharge current. Therefore it is important to review the specification sheets to understand the battery capabilities as it relates to the specific application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Response to Question 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically there is no resistor inside the battery, but components use in battery contribute to a total effect resistance, also known as internal resistance. The two primary contributing components are electronic resistance and ionic resistance. Electronic resistance is made of material such as metal covers and material used to make contact. Electrochemical factors such as electrolyte conductivity, ion mobility and electrode surface area make up the ionic resistance component. Ionic resistance occurs slowly and effect on effective resistance varies base on various factors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; CR-P2 batteries are made by combining two CR123 batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://d2ei442zrkqy2u.cloudfront.net/wp-content/uploads/2016/04/Li223_US_OS.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Duracell Ultra 223 Lithium/Manganese oxide CRP2 battery&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.duracell.com/en-us/techlibrary/product-technical-data-sheets?region=262&amp;amp;type=299&quot; rel=&quot;nofollow noreferrer&quot;&gt;Duracell Product Technical Data Sheets&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sonybatteries.ee/lithium-photo-battery-cr-p2-6-v-sony.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sony Lithium Photo Battery CR-P2 6 V&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.powerstream.com/cr123a-tests.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Discharge tests of CR123A lithium manganese dioxide batteries&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://data.energizer.com/pdfs/batteryir.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battery Internal Resistance&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Comparison_of_commercial_battery_types&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comparison of commercial battery types&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://d2ei442zrkqy2u.cloudfront.net/wp-content/uploads/2016/03/16164724/2011-October-Ten-Things-you-should-Know.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;10 Things You Should Know About Designing with Batteries&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/5566/internal-resistance-of-batteries&quot;&gt;Internal resistance of batteries?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://data.energizer.com/pdfs/alkaline_appman.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alkaline Manganese Dioxide Handbook and Application Manual&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/348972/internal-resistance-of-a-battery&quot;&gt;Internal resistance of a battery&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-07-23T10:44:33.260" LastActivityDate="2019-07-23T10:44:33.260" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29301" PostTypeId="1" CreationDate="2019-07-22T02:01:11.113" Score="0" ViewCount="293" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/p0pF0.png&quot; alt=&quot;given the following views, project a third view in the direction of the arrow in first angle projections,&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fQJLW.png&quot; alt=&quot;given the following views, project two other views as shown&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First as a beginner in Technical drawing, I'm required to project two other views from the given views. In the first image the original object was not given and I'm required to produce the third view. I felt there should be the original object to be able to get dimension. Unlike the second drawing where the original object was shown so it becomes a little easier. &#xA;All I need is a correct way of approaching this kind of drawing in order to be guided accordingly.&#xA;Thanks in anticipation&lt;img src=&quot;https://i.stack.imgur.com/AGovl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="21222" LastEditorUserId="21222" LastEditDate="2019-07-22T11:51:37.827" LastActivityDate="2019-07-22T11:51:37.827" Title="Projection of a given view to produce another views" Tags="|technical-drawing|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29303" PostTypeId="2" ParentId="29295" CreationDate="2019-07-22T08:17:22.633" Score="1" Body="&lt;p&gt;Let's call the mass of weight W and the mass of base B, &#xA;the arm length extending outside of the base, L and the base width D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume 3/4 of mass of horizontal arm called A1 is outside of the base and the rest called A2 is over the base. And call the mass of the mast Ma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now all we have to do is check the balance of this machine about the edge of the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume the right edge of base D lies on the origin and the left side of the base on negative side of x axis.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;getting moment about origin &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$IF \ M_{origin} = [B*D/2+ Ma*D/2+(A2*D/4) -\ [W*L +A1*3/8L] &amp;gt;0 $$&lt;/span&gt; the moment is counterclockwise and the object is stable, if not it will tip over.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-07-22T16:35:30.413" LastActivityDate="2019-07-22T16:35:30.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29304" PostTypeId="1" CreationDate="2019-07-22T09:12:46.450" Score="0" ViewCount="363" Body="&lt;p&gt;How do I calculate the maximum force on a support beam if I know the weight of the roof? &#xA;I assume distribution will be even but I cannot find correct formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: 3 beams 60cm apart supporting a roof of 70kg/m2&lt;/p&gt;&#xA;" OwnerUserId="21224" LastActivityDate="2019-07-22T15:43:12.933" Title="Weight distribution roof construction" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29305" PostTypeId="2" ParentId="29295" CreationDate="2019-07-22T09:48:14.997" Score="0" Body="&lt;p&gt;As @kamran shows, you don't really need software for this - it's a fairly simple calculation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you wish to use a computer, I would select all the parts you have and look where the center of gravity for that combined object falls, laterally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is within the base's reach, your setup stands. If it is outside, it falls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Fusion 360, you find the center of mass under Inspect &gt; Center of Mass. Do realize that you must set the material values of the bodies right - that should be obvious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &lt;a href=&quot;http://help.autodesk.com/view/fusion360/ENU/?guid=GUID-EB8C9FD9-2CA6-4937-887D-0014E538BF2C&quot; rel=&quot;nofollow noreferrer&quot;&gt;Center of mass&lt;/a&gt; (Fusion 360 documents)&lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2019-07-22T09:48:14.997" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29306" PostTypeId="1" AcceptedAnswerId="29309" CreationDate="2019-07-22T10:16:43.477" Score="1" ViewCount="92" Body="&lt;p&gt;I have a very noisy air conditioner that have a 10kHz peak at 75 dB, the rest (fan/compressor/pump) are under the regulation limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to understand what is the measurement spectrum. I think 10kHz is out of the measurement spectrum, but I am not sure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the procedure to measure household appliances noise level?&lt;/p&gt;&#xA;" OwnerUserId="2070" LastActivityDate="2019-07-22T13:47:08.193" Title="Sound level emissions (IEC 60704-1) spectrum?" Tags="|acoustics|regulations|noise|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29307" PostTypeId="2" ParentId="29274" CreationDate="2019-07-22T12:06:57.270" Score="2" Body="&lt;p&gt;One possible way to do this &lt;em&gt;could&lt;/em&gt; be a &lt;a href=&quot;https://en.wikipedia.org/wiki/Microbial_fuel_cell#Power_generation&quot; rel=&quot;nofollow noreferrer&quot;&gt;microbial fuel cell&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has, apparently, been tested  ex-vivo: &lt;a href=&quot;https://www.researchgate.net/publication/236071090_A_pacemaker_powered_by_an_implantable_biofuel_cell_operating_under_conditions_mimicking_the_human_blood_circulatory_system_-_battery_not_included&quot; rel=&quot;nofollow noreferrer&quot;&gt;A pacemaker powered by an implantable biofuel cell operating under conditions mimicking the human blood circulatory system - battery not included&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another approach, which also works at least ex-vivo, is called Abiotic Biofuel Cell: &lt;a href=&quot;https://onlinelibrary.wiley.com/doi/abs/10.1002/elan.201400440&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pacemaker Activated by an Abiotic Biofuel Cell Operated in Human Serum Solution&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keywords here are biofuel cell and microbial fuel cells, knowing these you can google the rest yourself.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-07-22T12:06:57.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29308" PostTypeId="1" CreationDate="2019-07-22T13:05:46.327" Score="0" ViewCount="41" Body="&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a near 100% Pt wire (25 AWG / D ≈ 0.45mm) that I would like to attatch to a cold rolled steel (E235 / EN 1.0308) surface. Assume this surface is flat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Goal:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Keep wire attached to the surface for ~3 minutes while temperature&#xA;sinks from 800C to 500C. The wire is not carrying any load except its&#xA;own (max 20 cm).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;While:    &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Minimizing added heat capacity / mass&lt;/li&gt;&#xA;&lt;li&gt;Avoiding gases up to 1000C (so no&#xA;tape)&lt;/li&gt;&#xA;&lt;li&gt;Thermal conductivity&lt;/li&gt;&#xA;&lt;li&gt;Minimizing steel tempering (solidus&#xA;~1420C)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thoughts&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have considered soldering/brazing with a silver alloy (mechanically weak, but strong enough?). I soldered an iron wire onto the same surface with 60/40 Sn-Pb soldering wire in 20 seconds and achieved a solder strong enough to lift the entire test object (over one kilo) and bounce it around. Looking briefly at the metallic and atomic properties of Pb/Sn and Ag, I'm hoping silver will bond even stronger with the steel surface. I assume I should also be trying to dissolve/reduce the oxidation layer on my steel before and during the soldering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also been thinking an Al-Mn alloy can be easily made with a perfect melting temperature, achieve a BCC crystal structure (like my steel), and hopefully create a stronger bond to the steel. Is this a realistic approach? I have no experience with obtaining this, its price and the overall doability. I'm also unsure of whether corrosion makes an aluminum alloy unsuitable for this brazing technique.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have previously considered welding the platinum straight onto the steel, but I assumed this would have a major impact on the mechanical strength of my steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have a good way to solve this that is preferably swift, simple, repeatable and does not require expensive machinery (2000 USD+). I already have soldering, brazing and TIG/MIG/stick welding capabilities.&lt;/p&gt;&#xA;" OwnerUserId="21226" LastActivityDate="2019-08-22T01:11:51.273" Title="Carefully attaching Pt-wire to mild/carbon steel surface" Tags="|mechanical-engineering|materials|thermodynamics|chemical-engineering|steel|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29309" PostTypeId="2" ParentId="29306" CreationDate="2019-07-22T13:40:55.373" Score="1" Body="&lt;p&gt;&lt;strong&gt;What is the measurement Spectrum?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From BS EN 60704-1:2010+A11:2012 Page 10, Section 4.1 Paragraph 3&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The preferred noise emission quantity is the A-weighted sound power&#xA;  level, LWA, in decibels&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8SEM8m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8SEM8m.png&quot; alt=&quot;Acoustic Weighting Curves&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It is &lt;em&gt;not&lt;/em&gt; correct to assume that a peak at 10kHz is outside the measurement range&lt;/strong&gt; - you can reduce your 75dB measurement by -2.5dB according to the curve shown above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the procedure?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page 11 Paragraph 1&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to this standard, two principal methods exist, the direct&#xA;  method and the comparison method, as described in 4.2 and 4.3 below.&#xA;  These two methods can be used alternatively.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And then subsequently in Page 11 Section 4.2&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The direct method can be used only for measurements in qualified test environments&#xA;  according to ISO 3744 for free field conditions over reflecting plane(s), and according to ISO&#xA;  3743-2 for special reverberation test rooms. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and Page 11 Section 4.3&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The comparison method for measurement is explicitly described in ISO&#xA;  3743-1 and in ISO 3743-2.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2019-07-22T13:47:08.193" LastActivityDate="2019-07-22T13:47:08.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29310" PostTypeId="2" ParentId="29308" CreationDate="2019-07-22T15:28:43.270" Score="1" Body="&lt;p&gt;I think you are asking how to attach a Pt wire to carbon steel for a short time at 800 C /1472 F. I expect a plain carbon steel weld, like MIG, would do it. It may not melt the Pt during the few seconds of welding ; the Pt would be mechanically attached by the weld deposit . Keeping in mind that Pt thermocouples are routinely used to measure molten steel temperatures above 3200 F. Because your steel is &quot;cold rolled&quot; , not heat-treated , strength properties of the weld HAZ will not change significantly; trading cold rolled ( strength ?) for a fast normalize. PS - &quot;solidus&quot; has nothing to do with tempering steel.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-08-22T01:11:51.273" LastActivityDate="2019-08-22T01:11:51.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29311" PostTypeId="2" ParentId="29304" CreationDate="2019-07-22T15:43:12.933" Score="1" Body="&lt;p&gt;For a roof it'd be normal to assume even distribution of load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Don't forget to include applied load (e.g. snow) as well as the dead weight. And don't forget to include any safety factors required in your jurisdiction.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2019-07-22T15:43:12.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29312" PostTypeId="1" CreationDate="2019-07-22T16:15:00.223" Score="0" ViewCount="69" Body="&lt;p&gt;Say I have a light source at a distance of 1Km from a  light detector (photo detector). Right in front of the sensor, I keep a rotating reticle with some sort of pattern. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To better appreciate the situation take a look at the image below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Kv1o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Kv1o.png&quot; alt=&quot;https://www.ausairpower.net/XIMG/reticle-abc.png&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I want to simulate the detector output signal for various patterns on the reticle. For example &quot;reticle B&quot; from the above image would produce a square wave, and the &quot;reticle A&quot; would produce a  pulse train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any tool for doing such kind of simulation? Is there any python/matlab library?&lt;/p&gt;&#xA;" OwnerUserId="21228" LastEditorUserId="8055" LastEditDate="2019-07-22T18:28:59.463" LastActivityDate="2019-07-22T18:28:59.463" Title="Tool to simulate sensor data" Tags="|simulation|matlab|simulink|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29313" PostTypeId="1" CreationDate="2019-07-22T18:26:18.287" Score="2" ViewCount="33" Body="&lt;p&gt;I´ve been disassembling a lot of electronic devices in order to recycle them, separating plastics, electronic boards and bolts/other metal parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any proof of health hazards for doing it without any kind of protection? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I´m worried about possible substances used when building that could affect me by touching or inhaling them. &lt;/p&gt;&#xA;" OwnerUserId="21229" LastActivityDate="2019-07-22T18:26:18.287" Title="Health hazards of disassembling electronic devices?" Tags="|electrical-engineering|consumer-electronics|recycling|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29314" PostTypeId="1" CreationDate="2019-07-22T18:43:16.323" Score="3" ViewCount="42" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Preconsolidation pressure is the maximum effective vertical overburden stress that a particular soil sample has sustained in the past.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How far in the past does the soil remember its loading/the pressure it sustained? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, suppose that at a certain time the Preconsolidation pressure of a soil is &lt;span class=&quot;math-container&quot;&gt;$p_0$&lt;/span&gt; due to certain load it had bore in the past. Then does the Preconsolidation pressure &lt;span class=&quot;math-container&quot;&gt;$p_0$&lt;/span&gt; changes as the time passes given that the soil is left undistrubed? &lt;/p&gt;&#xA;" OwnerUserId="21230" LastEditorUserId="90" LastEditDate="2022-08-03T00:50:12.830" LastActivityDate="2022-08-03T00:50:12.830" Title="What is the effect of ageing on preconsolidation pressure?" Tags="|civil-engineering|geotechnical-engineering|soil|soil-mechanics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29318" PostTypeId="2" ParentId="29299" CreationDate="2019-07-22T23:36:48.283" Score="2" Body="&lt;p&gt;The idea behind Bode plot analysis is that you want to know how close the poles of the closed-loop transfer function are to the stability boundary.  If a pole is &lt;em&gt;on&lt;/em&gt; the stability boundary, then the closed-loop gain would calculate as infinite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of Laplace-domain analysis, the stability boundary is the &lt;span class=&quot;math-container&quot;&gt;$s = j \omega$&lt;/span&gt; line -- any pole that has a positive real part is outside of the stability region, and the system is unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you do an open-loop Bode plot, you are concerned with how close the open-loop gain gets to -1, because if it were &lt;em&gt;exactly&lt;/em&gt; -1 then there would be one or more poles sitting right on the stability boundary.  The degree to which the open-loop gain &quot;misses&quot; -1 is a measure of the degree to which the resulting closed-loop poles are far away from the stability boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that a drawback of the Bode plot is that you only know how far the closed-loop poles are from the stability boundary -- not whether they are actually stable.  For that you need to apply some intuition, or you need to make a Nyquist plot, and know how many unstable zeros the system has.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, too, that for a sampled-time system expressed in the z domain you can do a Bode plot by tracing &lt;span class=&quot;math-container&quot;&gt;$z = e^{j \omega \theta}$&lt;/span&gt; for &lt;span class=&quot;math-container&quot;&gt;$0 \le \theta \le \pi$&lt;/span&gt;, because that's the stability boundary in the z domain.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-07-22T23:36:48.283" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29319" PostTypeId="2" ParentId="29264" CreationDate="2019-07-23T06:54:08.213" Score="2" Body="&lt;p&gt;If the excavation has a slope of roughly 3 vertical to 1 horizontal, that's about 70 degrees with horizontal. An unloaded slope in stones, gravel or sand will not be stable at an angle of more than approximately 40 degrees with horizontal, so it is not possible to make the excavation without disturbing the soil. That angle is called the angle of internal friction and is considered a material parameters that describes the quality of the soil. The value of 40 degrees is an estimate of mine that I am guessing might be applicable to your stone fill, and it assumes a well-compressed fill of decent quality stones. And furthermore, it does not include any safety factors, so the slope of the excavation should be further reduced to about 35 degrees with horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of moving the sump crock away from the foundation, you could consider installing a small retaining wall.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2019-07-23T06:54:08.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29320" PostTypeId="1" CreationDate="2019-07-23T07:33:29.893" Score="0" ViewCount="30" Body="&lt;p&gt;I have a reaction where component X is required to react with formic acid producing a solid precipitate. The main problem is that component X has very low solubility in formic acid and the reaction takes almost 10 hours to reach completion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to improve the reaction time but all I can come up with as of yet:&#xA;1. run reactor at high temperature - 120 degC with reflux&#xA;2. use multiple smaller CSTRs to reduce mass/heat transfer limitations&#xA;3. ensure particle distribution of component X is neither too large nor too small causing it to float on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;are there any other simple techniques to improve this reaction? perhaps a different mixer/impeller design? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know of any catalysts that apply to this reaction and an investigation is not possible at this time. &lt;/p&gt;&#xA;" OwnerUserId="10923" LastActivityDate="2019-07-23T17:23:53.837" Title="Best reactor for solubility-limited reaction?" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29321" PostTypeId="1" CreationDate="2019-07-23T08:45:34.207" Score="0" ViewCount="93" Body="&lt;p&gt;Putting together a project, I've found this clutch but I've never seen this type before. It has a sort of flexible plate connecting the two parts, attached to both in two places corresponding to where the four teeth are. It seems to be working but I have no idea what torque it's ok for and how much I should compress it, if at all. Sorry for the shitty photo quality. &lt;a href=&quot;https://i.stack.imgur.com/p9MAA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p9MAA.jpg&quot; alt=&quot;photo of the coupling&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21231" LastActivityDate="2019-07-23T09:06:28.407" Title="I've never seen this type of flexible coupling, can someone tell me what it's called?" Tags="|mechanical-engineering|design|motors|mechanisms|coupling|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29322" PostTypeId="2" ParentId="29321" CreationDate="2019-07-23T08:56:42.360" Score="0" Body="&lt;p&gt;These drive clutches can be used to absorb drive train shocks and/or slight mis-alignments of the two shafts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are available in several sizes and capacities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flexible coupling or flexible drive coupling or flexible shaft couplings, see:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.skf.com/group/products/coupling-systems/flexible-shaft-couplings/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;enter link description here&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-07-23T08:56:42.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29323" PostTypeId="2" ParentId="29321" CreationDate="2019-07-23T08:59:08.133" Score="2" Body="&lt;p&gt;It could either be an &lt;em&gt;oldham coupling&lt;/em&gt;.  The photo's a bit blurry. Here's a web page on oldham couplings. &lt;a href=&quot;https://www.huco.com/products/couplings/oldham-couplings&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.huco.com/products/couplings/oldham-couplings&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2wrjF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2wrjF.jpg&quot; alt=&quot;Oldham Coupling&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.couplingtips.com/oldham-couplings/what-do-you-know-about-oldham-couplings/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.couplingtips.com/oldham-couplings/what-do-you-know-about-oldham-couplings/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or it could be a &lt;em&gt;Flexible flange coupling&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XbK8u.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XbK8u.jpg&quot; alt=&quot;Flexible flange coupling&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16544" LastEditorUserId="16544" LastEditDate="2019-07-23T09:06:28.407" LastActivityDate="2019-07-23T09:06:28.407" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29325" PostTypeId="1" AcceptedAnswerId="29794" CreationDate="2019-07-23T11:57:11.807" Score="1" ViewCount="137" Body="&lt;p&gt;I recently saw a tweet about &lt;a href=&quot;http://www.esa.int/spaceinimages/Images/2009/02/Hypervelocity_impact_sample&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hypervelocity Impact Sample&lt;/a&gt; and someone ask how is it possible to shoot a projectile at that speed and what happened to the used projectile when hitting the shield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found that it was probably shooted by a &lt;a href=&quot;https://www.nasa.gov/centers/wstf/site_tour/remote_hypervelocity_test_laboratory/two_stage_light_gas_guns.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Two Stage Light Gas Guns&lt;/a&gt; but I didn't found anything about the projectile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only found this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;They got their name because hypervelocity impact produces local&#xA;  pressures in the target and projectile that greatly exceed the&#xA;  material strength of these materials, the material behaves as if it&#xA;  has no strength, i.e., like a fluid, or hydrodynamic behavior&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20090010053.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Handbook for Designing MMOD Protection &lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y2Uzg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y2Uzg.png&quot; alt=&quot;HYPERVELOCITY IMPACT SAMPLE&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;What happened to the projectile ?&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Does the projectile melt with the shield ?&#xA;&lt;del&gt;At this speed, does the projectile melt before hiting the shield ?&lt;/del&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: deletion of the last question, test done in vacuum.&lt;/p&gt;&#xA;" OwnerUserId="21234" LastEditorUserId="21234" LastEditDate="2019-07-23T12:49:45.190" LastActivityDate="2019-08-24T06:10:27.313" Title="What happened to the projectile in Hypervelocity Impact tests?" Tags="|materials|aerospace-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29326" PostTypeId="1" AcceptedAnswerId="29331" CreationDate="2019-07-23T12:22:58.157" Score="2" ViewCount="26" Body="&lt;p&gt;I am wondering the following. What are the typical cycletimes (or timesteps), that are used in real industry-standard pid-controllers (for example for assembly-robots on a car-production-line). I didnt find anything specific in the literature, just the theory. Therefore I am asking here. Often, these information is not published by the vendors of these pid-servo-controllers, so maybe someone has some experience here. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="21236" LastActivityDate="2019-07-23T16:41:51.367" Title="Typical cycletime of an industry pid controller for servomotors" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29327" PostTypeId="1" CreationDate="2019-07-23T13:24:29.387" Score="0" ViewCount="1517" Body="&lt;p&gt;I am very confused by the three terms:lift and elevator and hoist&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they all technically same thing or different??&lt;/p&gt;&#xA;" OwnerUserId="22801" LastActivityDate="2019-07-31T19:31:45.030" Title="What is difference between lift and elevator and hoist?" Tags="|mechanical-engineering|structural-engineering|civil-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29328" PostTypeId="2" ParentId="29327" CreationDate="2019-07-23T13:33:19.227" Score="3" Body="&lt;p&gt;In terms of moving objects between levels, then not much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in terms of moving people, then lift and elevator are the English and American terms while a hoist is for goods only.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-07-23T13:33:19.227" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29329" PostTypeId="2" ParentId="29295" CreationDate="2019-07-23T15:12:19.350" Score="1" Body="&lt;p&gt;@kamran described how to analytically find the location of the C.G in your case. However, according to your comments I understand you want to play a little and gain experience with engineering programs. Am I Right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following @alephzero and @ Wasabi♦ comments, here you can find a way for implementing their approach on ANSYS WORKBENCH:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simplified model was created - containing two structures having different geometries. Each of these geometries is a one piece item made of a single material. This was done just for simplicity, of course you can apply the same approach on any super complicated assembly.&#xA;To make the examination even easier each item was simply supported to the ground as shown in the following screenshot:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/X9jou.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X9jou.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what you would like to do is to examine the support reactions. As long as the center of gravity is located between the two pin joints (no tip over), the two reactions will push the structure upwards. This is the case of the left scenario as shown on the following screenshot.&#xA;In case the center of gravity location falls out of the lower base limits, one of the reactions will be directed downwards - which means a tip over in your case. This is illustrated on the right side of the following screenshot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1Ss9g.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Ss9g.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As already discussed, Using a FEA tool just for analyzing the center of gravity location is quite a wasteful procedure. Generally, examining the reactions is used as a sanity checks for the finite element model.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2019-07-23T15:12:19.350" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29330" PostTypeId="2" ParentId="29325" CreationDate="2019-07-23T15:25:09.227" Score="0" Body="&lt;p&gt;In some tests, the Livermore crew fired their projectiles into piles of sandbags, as documented here: &#xA;&lt;a href=&quot;https://crgis.ndc.nasa.gov/crgis/images/8/85/TRP00308.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://crgis.ndc.nasa.gov/crgis/images/8/85/TRP00308.pdf&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I seem to recall reading that in later tests, they were firing into a bore shaft in a hill downrange, and that the bore was filled with carpet scrap (that variously compressed, shredded and/or melted), but I can't find any photos of that.&lt;/p&gt;&#xA;" OwnerUserId="13014" LastActivityDate="2019-07-23T15:25:09.227" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29331" PostTypeId="2" ParentId="29326" CreationDate="2019-07-23T16:41:51.367" Score="2" Body="&lt;p&gt;Without being at all facetious, it's the cycle time the engineer in charge of the control loop felt was best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It depends on the size of the motor, the requirements of the loop, the sophistication of the computing hardware, and how the motor is controlled.  I've written motor control loops with a cycle time of 100Hz, driving a current command to a transresistance amplifier.  I've written motor control loops with a cycle time of 10kHz, that repeated at each cycle of the PWM signal to the H-bridges of a class D amplifier, and that controlled the motor current cycle by cycle.  I've done quite a bit in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, you look at the fastest mode of the system that you're going to control (i.e., motor current, or motor speed, etc.), and you sample faster than that.  The rule of thumb is 10 times, but depending on the precision you need, and the difficulty of attaining high sampling rates, you may sample lower than that or much, much higher (5 is a practical lower limit, there is no practical upper limit, although the required numerical precision of your controller goes up linearly, or possibly as the square of the sampling rate).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-07-23T16:41:51.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29332" PostTypeId="2" ParentId="29320" CreationDate="2019-07-23T17:23:53.837" Score="1" Body="&lt;p&gt;I'm going to assume that you're utilizing the liquid phase based on your terminology. It sounds as though you're utilizing 100% liquid formic acid, and attempting to dissolve &quot;X&quot;, a solid or powder into it. I additionally am assuming that you're trying to achieve a continuous production flow as opposed to a batch reaction - If these are correct, then I have a few recommendations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Is it possible to utilize a common solvent of both &quot;X&quot; and Formic acid to continue reacting in the liquid phase? Hypothetical: perhaps the solubility of &quot;X&quot; in water is considerably greater than in formic acid, and a 50/50 mixture of formic acid and water may be able to hold considerably more of &quot;X&quot;. You mentioned that your desired product is an insoluble precipitate (in formic acid). You may be able to find a common solvent that still renders the product solid as insoluble. Even if this isn't the case, you may be able to get around this by increasing the reaction extent until reaching the solubility limit, at which point the precipitation will occur. There may be other hurdles to get around at this point (side reactions with solvent/selectivity, undesirable solubility limits of product etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) A second option would possibly be a fluidized bed reactor. This would be more complicated than a simple re-circulation fluidized bed, because your product precipitate would become intermingled with your feed reactant, &quot;X&quot;. A way to get around this would be to have an extended fluidized bed reactor as shown below (excuse my crude drawing)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/xwMa8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xwMa8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Vessel #1&lt;/strong&gt; would be utilized as a &quot;saturating vessel&quot; - the formic acid would be loaded up to the solubility limit with reagent &quot;X&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;At point #2&lt;/strong&gt;, an optional additional CSTR could be added to achieve some significant retention time if needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;At point #3&lt;/strong&gt;, a solids-liquid separator of some sort could be utilized to quickly remove the desired product and return an &quot;X&quot; depleted stream directly back to the saturation vessel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A few notes: If you're only attempting to do a batch reaction, this is much more complicated than necessary. This idea is intended for continuous production. Additionally, based on your 10 hour reaction time, it's highly likely that an additional CSTR would be needed. There will also be considerable engineering hurdles such as how to maintain an &quot;X&quot; bed, what is the optimum exit concentration of &quot;X&quot; in the depleted stream (it will likely be very high to increase rate of reaction, but that depends on the reaction mechanism and order). Additionally, depending on mass transfer rates, &quot;X&quot; particle mechanical recalcitrance and reaction rates, it may be entirely overkill to fluidize the bed in vessel #1, and a static solids contacting bed may be enough. Lastly, you may run into issues of product precipitation inside of vessel #1 - the critical factor here will be particle size &amp;amp; density differences of reagent &quot;X&quot; and the product. If the Stokes velocity/settling velocity of the product is greater than the reagent &quot;X&quot;, &lt;strong&gt;this design will not work as shown.&lt;/strong&gt; It would require considerable changes to entrap &quot;X&quot; in a sort of &quot;fixed bed plug flow solubilizer&quot; (not that this can't be done). Additionally, this was my &quot;gut reaction&quot; to your scenario. With more time and thought, a more elegant solution may present itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary benefit that I see from this design and added complexity is that by adjusting retention time, you can control the concentration of &quot;X&quot; in the depleted stream. In general, retaining high concentrations of all reagents in the reaction medium will increase the rate of reaction. &lt;/p&gt;&#xA;" OwnerUserId="8369" LastActivityDate="2019-07-23T17:23:53.837" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29333" PostTypeId="1" CreationDate="2019-07-23T18:19:12.373" Score="1" ViewCount="23" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/VquCY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Welding on some unknown Steel, I made the weld, and this stuff formed. It looks like a fast shutter speed shot of smoke or something for me but it's a solid material that has formed. It almost seems like millions of little Fibers at all flowed along with the hot air and gases coming off the weld. Has anyone seen this stuff before and does anyone have an idea what it is?&lt;/p&gt;&#xA;" OwnerUserId="21240" LastActivityDate="2019-07-23T18:19:12.373" Title="Flux Cored MIG on unknown steels, what is this stuff?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29334" PostTypeId="1" CreationDate="2019-07-23T19:48:30.623" Score="0" ViewCount="47" Body="&lt;p&gt;&lt;strong&gt;BROUGHT OVER FROM PHYSICS STACK EXCHANGE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the subject states I have hit a wall in a fluid mechanics associated problem in designing an experiment for my grad project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long story short, I'm an engineering undergrad who got a broad education in formal engineering topics (thermo, heat and mass, classical mechanics, systems, circuits, etc. etc.), but literally only did 1 fluid mechanics course. I am now a Masters student doing more biologic focused research. I am working with stem cells and I need to perform a flow experiment in which I run continuous fluid over my cells in a flow chamber at a specified flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is that I have been using a syringe pump as it was what the facility provided and it's less than optimal. Only 30 mL at a time and at my specified flow rate I end up having to change it a lot and therefore flow is discontinuous and not really up to the rigor of my experiment. I get ok results, but I think it can be better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to do, and my problem is, that I'd like to run a circulating system with a continuous pump. I just don't know how to figure out what pump to use. I have some support and money from the lab, and I'd prefer to be able to tweak the flow rate if possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To set up the problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using an Ibidi mu VI 0.4 chamber slide. Full literature available here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ibidi.com/channel-slides/57--slide-vi-04.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ibidi.com/channel-slides/57--slide-vi-04.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quick and dirty dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outer dimensions (w x l) 75.5 x 25.5 mm² Adapters Female Luer Number of channels 6 Channel volume 30 µl Channel height 0.4 mm Channel length 17 mm Channel width 3.8 mm Volume per reservoir 60 µl&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using their literature provided I have already calculated that I need to provide a flow rate of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4.73 mL/min&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which corresponds to a shear stress of: &lt;strong&gt;6 dyne/cm^2&lt;/strong&gt; (0.6 Pa)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my tubing dimensions are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L = 20 cm inner d = 1.6 mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a reservoir of cell media of maximum 50 mL which needs to sit beside the microscope in a heater to provide the maximum &quot;comfort&quot; and conditions that my cells like (37 degrees C)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the life of me I can't figure out how to determine an appropriate set up/pump to make this into a continuous flow, needed to pump/circulate for &lt;strong&gt;2 hours&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody at least put me on the right path to designing this?? I don't have the time to relearn all my fluids, but I don't mind brushing up on topical required reading to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can also attempt to make a diagram if you need a better visual representation. I hope I explained my problem thoroughly though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A frustrated grad student&lt;/p&gt;&#xA;" OwnerUserId="21242" LastActivityDate="2019-07-23T23:25:14.777" Title="Grad student needing help with a fluid mechanics issue in project" Tags="|mechanical-engineering|fluid-mechanics|pumps|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29337" PostTypeId="2" ParentId="29325" CreationDate="2019-07-23T23:16:43.957" Score="0" Body="&lt;p&gt;The projectile typically melts or vaporizes during hyper velocity impact due to the energy of impact being dissipated as heat during plastic deformation.  The physics is quite involved.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="3488" LastEditDate="2019-07-25T01:40:57.323" LastActivityDate="2019-07-25T01:40:57.323" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29340" PostTypeId="1" CreationDate="2019-07-24T00:26:00.097" Score="2" ViewCount="321" Body="&lt;p&gt;What is the maximum allowable crack size dimension in a post-tension concrete slab before it is considered a &quot;structural&quot; crack, in Arizona, USA? Is there a certain standard for this. I have tried looking for it in my state's ROC rules but they have nothing for what is considered to be a &quot;structural&quot; crack in a post-tension concrete slab. It's hard to give warranties without having specific measurements and standards to abide by.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: This is for a on-ground slab with an ABC foundation&lt;/p&gt;&#xA;" OwnerUserId="10348" LastEditorUserId="10348" LastEditDate="2019-07-24T01:24:06.293" LastActivityDate="2019-07-24T01:24:06.293" Title="Maximum allowable crack in Post Tension concrete" Tags="|concrete|reinforced-concrete|standards|prestressed-concrete|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29341" PostTypeId="1" CreationDate="2019-07-24T09:08:19.180" Score="0" ViewCount="155" Body="&lt;p&gt;Determine the change in dimensions of a rectangular bar of length 2&amp;nbsp;m, width 200&amp;nbsp;mm, and thickness 100&amp;nbsp;mm, when it is subjected to an axial pull of 20&amp;nbsp;kN in the direction of its length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take &lt;span class=&quot;math-container&quot;&gt;$E = 2\times10^5\text{ N/mm}^2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\mu = 0.3$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have gotten up to &lt;span class=&quot;math-container&quot;&gt;$\delta L = 0.01$&lt;/span&gt;. Then using &lt;span class=&quot;math-container&quot;&gt;$\mu = \dfrac{δb/b}{δL/L}$&lt;/span&gt;, I got &lt;span class=&quot;math-container&quot;&gt;$\delta b = \dfrac{\mu bL}{δL}$&lt;/span&gt;, which is 0.3×200×2000/0.01 = 12million!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know where I have mistaken but it is frustrating me.&lt;/p&gt;&#xA;" OwnerUserId="21248" LastEditorUserId="1832" LastEditDate="2019-07-24T23:18:14.330" LastActivityDate="2019-07-24T23:18:14.330" Title="Dimensions of a bar when it is subjected to axial pull" Tags="|mechanical-engineering|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29342" PostTypeId="2" ParentId="29280" CreationDate="2019-07-24T12:24:24.467" Score="1" Body="&lt;p&gt;I'm not sure if there is any further context to this meme, but I would have chosen B) Unsteady Uniform Flow. Since almost all real flows are unsteady and non-uniform, it is really an application of assumptions to characterize the flow as something simpler to analyze. I would assume a flood is highly transient, therefore, unsteady. There is probably not a constant flow source feeding the flood, rather a series of time-dependent, differently sized sources (streams, drainage basins, rain, etc). At the same time, the flow is turbulent. This means the velocity profile is relatively &quot;flat&quot; (&lt;a href=&quot;http://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/fprops/introvisc/node8.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/fprops/introvisc/node8.html&lt;/a&gt;), thus, justifying the uniform assumption. On a more macroscopic level, a flood is likely entirely turbulent through its domain, further justifying the assumption of relative uniformity. While the profile remains more or less flat, it will change in magnitude depending on the sources. On a microscopic level, the assumption of uniformity does not hold because of the turbulence (i.e. vorticies of varying sizes everywhere in the flow). &lt;/p&gt;&#xA;" OwnerUserId="21118" LastActivityDate="2019-07-24T12:24:24.467" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29344" PostTypeId="1" CreationDate="2019-07-24T14:11:30.397" Score="2" ViewCount="171" Body="&lt;p&gt;Concerning the big megawatt wind turbines, if there is such a big trouble with gearbox bearings failing much earlier than expected and costly maintanence or replacement of huge gearboxes, what is keeping people from placing the whole power unit(gearbox, generator, transformer, etc.) down on the ground, connecting the gearbox and turbine rotor with a chain-sprocket drive? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a belt or chain ,a certain margin of flexibility would compansate for misalignment of axis and absorb shock loads. Apart form delivery time, replacement of damaged parts would take much less time and effort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuously variable transmission systems are already being adapted to wind turbines so such a drive should not be an unrealistic option. Turbines with hydraulic drives are already being built with down tower power units. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this would have been done if it had more advantages but what is the big disadvantage here that overcomes such a saving of time and cost?&lt;/p&gt;&#xA;" OwnerUserId="21251" LastEditorUserId="320" LastEditDate="2019-08-24T01:51:17.397" LastActivityDate="2019-09-23T02:02:40.413" Title="Wind turbines with down tower power units. Why aren't they more common?" Tags="|gears|renewable-energy|wind-power|power-transmission|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29345" PostTypeId="2" ParentId="29344" CreationDate="2019-07-24T14:13:59.740" Score="0" Body="&lt;p&gt;The extra parts required (remote drive units, drive mechanism etc) is more costly than the cost of the servicing issues encountered when the units are up top.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-07-24T14:13:59.740" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29347" PostTypeId="2" ParentId="29344" CreationDate="2019-07-24T14:36:01.000" Score="0" Body="&lt;p&gt;Gear box failures were not unexpected by engineers , only environmentalists.  Over 15 years ago , at an American Society for Metals dinner meeting ( Houston chapter), I heard about the limited life of the gears and how difficult it was to get and install new ones. The speed increaser gears are very large so there were/are very few sources in the world capable of heat-treating them. The great majority of prime movers are high speed / modest torque : gas turbines,steam turbines, and internal combustion engines. Wind turbines are the exception ; they are low speed high torque. This combination puts great demands on the gears of a speed increaser - in particular the high torque. The speed increaser gears are at the limits of the steel in terms of bending load and contact fatigue load , so limited life was /is expected. Like driving a car 100 mph, the car will do it but not for nearly as far as driving at 65 mph. The rotational speed of the wind turbine must be increased to produce the proper electrical frequency with the limited diameter of the generator The suggested alternative requiring tons of chain and associated bearings, etc, would be a bigger problem. Also , how would the wind turbine rotate into the wind if it had a chain run to a stationary generator on the ground?&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-07-24T19:28:53.480" LastActivityDate="2019-07-24T19:28:53.480" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29348" PostTypeId="2" ParentId="29341" CreationDate="2019-07-24T14:43:17.157" Score="1" Body="&lt;p&gt;Let's do it over.&#xA;First we calculate the tension stress. &lt;span class=&quot;math-container&quot;&gt;$$\sigma=P/A=\frac{20kN}{100*200}=1N/mm^2  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then as per definition of Young modulus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \epsilon=\frac{\sigma}{E}=\frac{1N}{2×10^5}=0.000005 $$&lt;/span&gt;&#xA;and &#xA;&lt;span class=&quot;math-container&quot;&gt;$$ lateral\ shrinkage=\mu*0.000005= 0.00000166$$&lt;/span&gt;&#xA;And then&#xA;&lt;span class=&quot;math-container&quot;&gt;$$200*0.00000166= 0.000332mm  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the amount the 200mm side narrows is small, 33/100000mm.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-24T14:43:17.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29349" PostTypeId="1" CreationDate="2019-07-24T16:07:57.950" Score="0" ViewCount="11" Body="&lt;p&gt;Newbie question, moderately educated on RF but only the signal, not at all about the hardware. Looking to start experimenting on a concept that would require a dynamic shift over time from one frequency to another between the transmitter and receiver. When a part is listed as, for example, a 443Mhz transmitter, is that typically the only frequency it can output at? If so, what key word would I search for on a part site to find a dynamic frequency transmitter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried searching &quot;dynamic frequency transmitter&quot; on distributor sites such as Mouser Electronics with no apparent luck in the results.&lt;/p&gt;&#xA;" OwnerUserId="3746" LastEditorUserId="3746" LastEditDate="2019-07-24T16:13:06.500" LastActivityDate="2019-07-24T16:13:06.500" Title="Are dynamic frequency transmitters common? What's the terminology to use when searching for one?" Tags="|terminology|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29350" PostTypeId="1" CreationDate="2019-07-24T16:41:05.913" Score="0" ViewCount="305" Body="&lt;p&gt;I have answered the following questions, however I’m unsure if they’re correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) For this question do we consider gravity force, I have used it but I’m not sure if this invalidates my overal derivation. Since the flow is one-dimensional and in the x-direction, must we only consider the momentum and forces in that direction? So must I remove pg?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/LN75y.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/twDh3.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="21149" LastActivityDate="2024-03-31T22:00:55.080" Title="Fluid Mechanics - Fluid flow" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29351" PostTypeId="2" ParentId="29344" CreationDate="2019-07-24T17:52:35.653" Score="5" Body="&lt;p&gt;The math doesn't make any engineering sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 5MW turbine typically rotates at about 10 RPM, near enough 1 radian per second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming you had reasonably big sprockets on the chain drive, say 2 meters diameter (!!) the linear speed of the chain would be 1 meter /second, and the tension in the chain would be about 5MN or about &lt;em&gt;500 tons force&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something tells me that is not going to be more reliable than a gearbox - and of course if you want to reduce the chain tension by increasing the speed, then you still &lt;em&gt;have&lt;/em&gt; a gearbox at the top of the tower which has to transmit the full 5MW of power to the chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact the chain makes the loads on the bearings worse, not better. As well as the thrust force from the turbine, the top bearing now has to support a force of 500 tons &lt;em&gt;downwards&lt;/em&gt; to resist the tension in the chain. And of course the tower also needs to be strengthened, to support a turbine that effectively &quot;weighs&quot; 500 tons when it is working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe replacing the existing bearings if they fail is a better idea :)&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-07-24T18:01:43.930" LastActivityDate="2019-07-24T18:01:43.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29352" PostTypeId="2" ParentId="28664" CreationDate="2019-07-24T22:59:08.173" Score="1" Body="&lt;p&gt;Found out there called virtual sensors which are designed in the ecu program code and use a neural network to learn and then estimate oil pressure using input from other sensors.&lt;/p&gt;&#xA;" OwnerUserId="20731" LastActivityDate="2019-07-24T22:59:08.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29354" PostTypeId="1" CreationDate="2019-07-25T06:46:46.380" Score="0" ViewCount="151" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/FOG2q.jpg&quot; alt=&quot;enter image description here&quot;&gt;&#xA;Why cant we provide the bars cross bent in the beam column  joint...as the cracks are all diagonal...so inorder to avoid cracks why cant we provide the bars crossed in the joint instead of providing mess of vertical and horizontal steel?&lt;/p&gt;&#xA;" OwnerUserId="21257" LastActivityDate="2021-09-14T02:03:20.250" Title="Beam column joint reinforcement" Tags="|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29355" PostTypeId="1" CreationDate="2019-07-25T07:04:55.440" Score="0" ViewCount="456" Body="&lt;ol&gt;&#xA;&lt;li&gt;How to measure torque in a bicycle?&lt;/li&gt;&#xA;&lt;li&gt;how much newtons will produce by the pedal of the cycle ?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21258" LastActivityDate="2019-07-26T01:36:43.900" Title="Measuring Torque in a bicycle" Tags="|mechanical-engineering|automotive-engineering|applied-mechanics|torque|bicycles|" AnswerCount="1" CommentCount="4" ClosedDate="2019-07-27T20:30:53.567" ContentLicense="CC BY-SA 4.0" />
  <row Id="29356" PostTypeId="1" CreationDate="2019-07-25T07:49:39.273" Score="1" ViewCount="293" Body="&lt;p&gt;Fusion 360 has the &quot;browser&quot; (tree structure) and timeline to represent the CAD model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I realize I should have created a parent component for an already existing component, how to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Fjux1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fjux1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the picture, I wish to create a component &lt;code&gt;LockBogie&lt;/code&gt; that would wrap the already created component &lt;code&gt;Lock&lt;/code&gt; underneath it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;rewinded the timeline to prior of creation of &lt;code&gt;Lock&lt;/code&gt;&lt;/li&gt;&#xA;&lt;li&gt;created &lt;code&gt;LockBogie&lt;/code&gt; there so it would be available as a parent for &lt;code&gt;Lock&lt;/code&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then comes the dead end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first intuition was to edit the creation step of &lt;code&gt;Lock&lt;/code&gt; so I could have reparented it. There's no &quot;edit&quot; or &quot;reparent&quot; option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, I went to the browser. Maybe I can drag &lt;code&gt;Lock&lt;/code&gt; to be within &lt;code&gt;LockBogie&lt;/code&gt; (&lt;code&gt;LockBogie&lt;/code&gt; was the active component when I tried this). Nope. Dragging is not allowed, it seems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding a level of component hierarchy afterwards feels like a relatively common thing that should be no-fuss. If you have done it, please tell how you did it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Disclaimer: This question is particular to one tool. I will likely post it in the Fusion 360 forums as well, but wanted to probe how beneficial the Engineering SO could be in coming to solutions. If the general guidance states the site is for non-tool-related questions only, I will certainly remove this. :)&lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2019-07-25T07:49:39.273" Title="How to add a level to component hierarchy in Fusion 360?" Tags="|cad|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29358" PostTypeId="1" CreationDate="2019-07-25T09:08:53.933" Score="0" ViewCount="40" Body="&lt;p&gt;Can I buy individual (larger in size) e-paper capsules from somewhere? (for an electronics project for example) and is it even possible for such large e-paper capsules to be engineered?&lt;/p&gt;&#xA;" OwnerUserId="21259" LastActivityDate="2019-07-25T09:08:53.933" Title="Can e-paper capsules be made larger in size?" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29359" PostTypeId="1" CreationDate="2019-07-25T09:45:07.540" Score="0" ViewCount="215" Body="&lt;p&gt;Is there a good excel program somewhere or basic breakdown of how to implement a control loop in excel if trying to simulate the performance of a circuit?  I have data from an operating plant and am trying to simulate the control loop but I'm not getting the correct outputs for the loop inputs.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:   I think what i'm missing is basic. I have cv, mv, and sp and I'm trying to simulate the controller. I have the PID settings for the controller as well (I think it was Siemens). Gain is 0.8, Integral time is 15 seconds and i have data on a 2 second sample time (I realize this is low resolution). The issue i'm having is i'm looking at output between t1 and t2 and trying to pull the appropriate data to estimate the PID output but i am consistently getting a different delta and i think I'm just setting it up wrong mathematically.&lt;/p&gt;&#xA;" OwnerUserId="21260" LastEditorUserId="21260" LastEditDate="2019-07-25T10:29:58.770" LastActivityDate="2019-07-25T10:29:58.770" Title="How can I setup a controller simulator in excel?" Tags="|control-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29360" PostTypeId="1" CreationDate="2019-07-25T13:26:19.637" Score="2" ViewCount="731" Body="&lt;p&gt;started by designing the gear meshing, and now I need to calculate some parameters to determinate if this meshing works for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system considers an electric motor, which technical parameters are: &#xA; - Power supply: 220V/50 Hz&#xA; - Motor power: 180W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to what I been studying, I start by calculating the T input (no-load) of the electric motor.&#xA;For a 50 Hz three-phase supply and 2 poles: &#xA;T = 50(Hz) * 60 (s/min) * 2 (Neg-Pos pulse) / n° poles = 3000 RPM&#xA;Considering a slip speed of 6 - 7%, I end up with 2750 RPM (no-load)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tinput = 0.241384 (lbf.ft/min) * 5252 / 2750 = 0.46 (ft.lbf) = 0.57 Nm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, my question is; Which is the relation between gear ratio and RPM to find the T output? What do I need to calculate the Torque output?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that when the gear ratio increase, usually the rpm decreases and the torque increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to move a gate of 600 Kg with a speed between 18 mts/min - 20 mts/min.&lt;a href=&quot;https://i.stack.imgur.com/SBOvV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SBOvV.png&quot; alt=&quot;Calc&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14265" LastActivityDate="2019-07-26T12:20:10.377" Title="Torque and RPM calc" Tags="|mechanical-engineering|gears|mechanical|power-transmission|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29361" PostTypeId="2" ParentId="29360" CreationDate="2019-07-25T14:04:44.657" Score="2" Body="&lt;p&gt;The transmission ratio is related to the number of teeth on the worm gear, and the number of &lt;strong&gt;starts&lt;/strong&gt; on the worm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, &lt;span class=&quot;math-container&quot;&gt;$$i=\frac{z_{worm\ gear}}{z_{worm}}=\frac{25}{1}=25$$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that you have a reduction of &lt;span class=&quot;math-container&quot;&gt;$25:1$&lt;/span&gt;, so, assuming &lt;span class=&quot;math-container&quot;&gt;$100\%$&lt;/span&gt; efficiency, you would expect your output Torque to be 25 times higher than your input torque, and your output speed to be a twentyfifth of your input speed. &lt;span class=&quot;math-container&quot;&gt;$$T_{output}=25T_{input}\ ,\ \omega_{output}=\frac{\omega_{input}}{25}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This relationship will hold pretty much true for the speed ratio, although you will find that resistance from the gears and the load will reduce your input speed to be lower than you have calculated above. You can get an estimate for the expected motor speed by looking at its datasheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For torque, however, you need to account for lost power to e.g. friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to: &lt;a href=&quot;https://www.machinedesign.com/mechanical-drives/gear-efficiency-key-lower-drive-cost&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.machinedesign.com/mechanical-drives/gear-efficiency-key-lower-drive-cost&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Efficiency of a worm-gear speed reducer depends (in part) on its speed-reduction ratio. High-ratio units have a smaller gear-tooth lead (helix) angle, which causes more surface contact between them. This higher contact causes higher friction and lower efficiency. Typical worm-gear efficiencies range from 49% for a 300:1, double-reduction ratio, up to 90% for a 5:1, single-reduction ratio.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So, with your ratio of 25:1, it seems safe to guess at an efficiency around 80% as a starting point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at conservation of Power, including Losses,&lt;span class=&quot;math-container&quot;&gt;$$P_{in}=T_{in}\omega_{in}=T_{out}\omega_{out}+\left(0.2*P_{in}\right)$$&lt;/span&gt;&#xA;Therefore &lt;span class=&quot;math-container&quot;&gt;$$T_{out}=\frac{0.8*P_{in}}{\omega_{out}}=\frac{25\left(0.8*P_{in}\right)}{\omega_{in}}=\frac{20*P_{in}}{\omega_{in}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Re: your goal of moving a 600kg gate - you need to consider acceleration as well as just 'cruising speed' when doing your power calculations.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-07-25T14:04:44.657" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29362" PostTypeId="1" CreationDate="2019-07-25T14:19:11.953" Score="2" ViewCount="1103" Body="&lt;p&gt;I guess if we do this then, the size of the turbines could become smaller. &#xA;The condensor's size also might be reduced.&#xA;But the catch over here is the energy density of hot water is less then that of steam. This could be overcome by increasing the amount of water sent in. &lt;/p&gt;&#xA;" OwnerUserId="21264" LastActivityDate="2019-12-23T06:01:20.290" Title="Why don't we use hot water instead of steam to drive the turbines?" Tags="|turbines|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29363" PostTypeId="1" AcceptedAnswerId="29364" CreationDate="2019-07-25T15:37:01.610" Score="0" ViewCount="26" Body="&lt;p&gt;I want to buy a &quot;Swiss File #4&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I visited some local stores but noone knew what that was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this nice set of files. Can someone read what it says in the top?&#xA;Is some of these file equivalent to &quot;Swiss File #4&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If yes, i will buy this instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x5lN9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x5lN9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21265" LastActivityDate="2019-08-24T16:01:45.183" Title="Swiss File #4 Alternatives" Tags="|tools|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29364" PostTypeId="2" ParentId="29363" CreationDate="2019-07-25T15:45:40.707" Score="1" Body="&lt;p&gt;In the US, the wording you are looking for is &quot;Needle File Cut 4&quot;, I believe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Amazon search: &lt;a href=&quot;https://www.amazon.com/s?k=needle+file+cut+4&amp;amp;ref=nb_sb_noss_2&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/s?k=needle+file+cut+4&amp;amp;ref=nb_sb_noss_2&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Diamond file is not the same, having industrial diamond particles embedded into the surface, rather than teeth cut into the steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/File_(tool)#Diamond_files&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/File_(tool)#Diamond_files&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible that the files in your photo would work well for your application, but they are &lt;strong&gt;not&lt;/strong&gt; equivalent to a Swiss #4.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-07-25T15:45:40.707" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29365" PostTypeId="1" CreationDate="2019-07-25T15:56:44.350" Score="0" ViewCount="984" Body="&lt;p&gt;I am currently designing a torque sensor and am looking into the ability to use strain gauges as a instrument to measure strain and calculate torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current setup is two motors connected via a shaft. One of these motors will provide a torque and spin, while the other opposes and acts as a resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Iodqb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Iodqb.png&quot; alt=&quot;strain gauges on a rotating shaft&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strain gauges will be placed at opposing 45° angles to each other, one will measure the positive strain while the other will measure the negative (as shown in the image above).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZlQOB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZlQOB.png&quot; alt=&quot;Wheatstone Bridge example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gauges will be wired using a Wheatstone bridge, this will provide a voltage output and seems to be the recommended method of getting an output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this Vout translated to a strain and/or is a torque value calculated? &#xA;The end goal is to calculate a torque from the strain gauges.&lt;/p&gt;&#xA;" OwnerUserId="21267" LastActivityDate="2019-07-25T15:56:44.350" Title="Is it possible to calculate torque using strain gauges on a rotating shaft?" Tags="|mechanical-engineering|electrical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29366" PostTypeId="1" CreationDate="2019-07-25T18:55:45.103" Score="0" ViewCount="577" Body="&lt;p&gt;I am currently studying engineering thermodynamics from Cengel and Boles and there it is given that total energy of non flowing fluid per unit mass basis is given as &lt;span class=&quot;math-container&quot;&gt;$e=u+\frac{v^2}{2}+gz$&lt;/span&gt;.But if the fluid is non flowing how can it have macroscopic kinetic energy&lt;span class=&quot;math-container&quot;&gt;$\frac{v^2}{2}$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="19791" LastEditorUserId="19791" LastEditDate="2019-07-26T04:12:03.750" LastActivityDate="2024-03-18T00:50:05.717" Title="How can a non flowing fluid have kinetic energy?" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29367" PostTypeId="2" ParentId="29360" CreationDate="2019-07-25T19:31:05.203" Score="1" Body="&lt;p&gt;The gear ratio is the slope of worm gear per teeth of the round gear, so if one rotation of the worm gear turns two gears down on the round gear you have a ratio of&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{1}{(25/2)}= \frac{1}{12.5}.$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have to set this ratio according to the torque you need, same worm gear engaged to a round gear with more teeth is more torque, less speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then multiply this by the efficiency of your gear system, which has to do with the quality and finish of the gears, lubrication, and the friction in bearings and housing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as for the ratio of the torque,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume the gate takes 3 seconds to reach the speed of20meters/mins =1/3m/s so your acceleration is (1/3) /3=1/9ms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F=m* \alpha = 600*(1/9)= 66kgm= 645N force.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to figure out what mechanism you want to use to impart this force by using the torque of your electrical motor. Are you going to use a chain drive or crank arm. then you calculate the mechanical advantage of those.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-25T19:31:05.203" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29368" PostTypeId="2" ParentId="29362" CreationDate="2019-07-25T20:00:23.587" Score="1" Body="&lt;p&gt;Most people will probably say to &quot;Look at a steam table&quot;. While they're not wrong, I think what you're looking for is a less textbook answer, and to understand the abstract concept better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets look at what happens when you run cold water through a turbine, such as in a hydraulic dam power plant. The water in this case, an incompressible fluid, has a static head pressure created by the head-height of the water. Hydraulic power plant turbines, placed well below the top of the water height, utilize this head pressure to generate electricity. Now, what mechanism is used to generate that electricity? Is it heat? Does the water entering a hydraulic turbine come out of the other end colder? No, it certainly doesn't. So what mechanism is used? Is it the velocity of the water? Is it the pressure of the water? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, lets transfer over to a steam turbine. Don't worry too much about the type for right now - just think about a turbine blade using steam as the working fluid. What is driving the actual motion of the turbine blades? Is it heat? Well, we do know that the fluid changes temperature as it travels through the turbine.... but is that what is pushing those blades around, physically imparting the working fluid energy and creating a torque force on the blades? How does heat &quot;push&quot; the turbine? Hmm, doesn't make much sense does it... What other forms of potential energy are there, though? Kinetic? Sure! Oh, and pressure! Pressure is there too! Well shoot, which one is it? is it pressure or is it kinetic energy (velocity)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer is: it depends on the turbine design. Via Wikipedia: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Impulse turbines&lt;/strong&gt; change the direction of flow of a high velocity fluid or gas jet. The resulting impulse spins the turbine and leaves the fluid flow with diminished kinetic energy. There is no pressure change of the fluid or gas in the turbine blades (the moving blades), as in the case of a steam or gas turbine, all the pressure drop takes place in the stationary blades (the nozzles). Before reaching the turbine, the fluid's pressure head is changed to velocity head by accelerating the fluid with a nozzle.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Reaction turbines&lt;/strong&gt; develop torque by reacting to the gas or fluid's pressure or mass. The pressure of the gas or fluid changes as it passes through the turbine rotor blades.[3] A pressure casement is needed to contain the working fluid as it acts on the turbine stage(s) or the turbine must be fully immersed in the fluid flow (such as with wind turbines). The casing contains and directs the working fluid&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Turbine&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link to wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So - We know the working mechanism now behind a turbine. It's either Kinetic (velocity component), pressure (reactionary, mass or simply pressure component), or some combination of the two energies. So again we ask, &lt;strong&gt;Why don't we use hot water instead of steam to drive the turbines?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, if we're going to run hot water through a turbine, it has to have either a velocity or pressure potential energy. Where are we going to get that energy? Probably a pump right? But if we're using a pump to produce velocity or pressure head (kinetic or pressure component).... won't we just be using the same amount of energy to create that potential energy? We wouldn't be gaining anything! it would take just as much energy to pressurize/speed up the water as we would get in the form of useful work, except we would lose much to friction! That doesn't seem like it would work well... Well what about steam? Why would steam work better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take that same hot water, use that pump to pressure it up, but then &lt;strong&gt;ALSO&lt;/strong&gt; heat up the water to the point it turns into steam, then we just added more energy in the form of heat, but that heat was able to turn the water into steam, which is a volume increase of more than 1000x! Now remember that we have this fluid in an enclosed space, so volume is fixed. What happens instead? we produce either a &lt;strong&gt;pressure component&lt;/strong&gt; or we produce a &lt;strong&gt;velocity component&lt;/strong&gt; (or some of both). Hey, that's what we need to turn the turbine!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if we only utilized a pump and just pumped hot water around in circles, we wouldn't gain anything because we lack an additional &lt;strong&gt;pressure component&lt;/strong&gt; or &lt;strong&gt;velocity component&lt;/strong&gt;. The reason we utilize steam is that we can take that &lt;strong&gt;heat component&lt;/strong&gt; that isn't able to turn the turbine in its current form and we change it into a &lt;strong&gt;pressure component&lt;/strong&gt; or &lt;strong&gt;velocity component&lt;/strong&gt;, or a bit of both. Now, if you're feeling ambitious, you could pull up a P-V diagram of a Rankine cycle and see if you can draw what the alternative path would be if you never boiled the water into steam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;*disclaimer: there may be some details when it comes to the actual calculations that don't quite translate to what I've described here. If I made a mistake just point it out and I'll fix. Thanks :) &lt;/p&gt;&#xA;" OwnerUserId="8369" LastActivityDate="2019-07-25T20:00:23.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29370" PostTypeId="1" CreationDate="2019-07-25T22:46:00.393" Score="1" ViewCount="231" Body="&lt;p&gt;I am currently studying wind power and the power electronics used for it. In wind power a generator is driven by wind, thus the resulting power is of widely varying frequency and amplitude. The power grid, in turn, has strict requirements for the input power in terms of frequency, phaseshift and sinusoidal form. For this reason, power converters are today used routinely in wind power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The predominant way to get the power into the grid is to use an AC-DC converter followed by a DC-DC converter and a DC-AC converter. This seems rather complicated instead of using a single direct AC-AC converter. Why is the indirect conversion via the DC &quot;in-between&quot; route preferable?&lt;/p&gt;&#xA;" OwnerUserId="21269" LastEditorUserId="6767" LastEditDate="2019-07-28T19:33:26.473" LastActivityDate="2019-07-28T19:33:26.473" Title="Why/when is AC-DC-AC conversion superior to direct AC-AC conversion?" Tags="|electrical-engineering|power-electronics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29372" PostTypeId="2" ParentId="29354" CreationDate="2019-07-26T01:13:39.093" Score="1" Body="&lt;p&gt;We have created a very weak joint ready to break this way. like a perforated paper encouraging tear line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top and bottom of the column are connected via that X reinforcement which is susceptible to crack and rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rebar continuity into the joint ensures rigidity of the joint. in some joints we may even add additional straight vertical and horizontal bars parallel to continuous pars for extra strength of the joint.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-26T01:13:39.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29373" PostTypeId="2" ParentId="29355" CreationDate="2019-07-26T01:36:43.900" Score="2" Body="&lt;p&gt;Most bicycle cranks are 170mm and they vary from 165mm to 180mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Skilled bicyclist can apply up to 1.5 to 2 times their weight on the pedals in burst of power to pass a race competitor by leveraging the handles and using the entire frame of the bike as a lever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say a rider weighs 160lbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T = 160*1.5*0.56=136lbs.ft$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, a fit bicyclist can produce continuous 160-180 watts energy. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-26T01:36:43.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29374" PostTypeId="2" ParentId="29370" CreationDate="2019-07-26T03:53:37.300" Score="1" Body="&lt;p&gt;The wind turbines that I have worked on, and others I have read about, rotate the generator at the correct speed to produce the power at the correct voltage and frequency so it goes directly into the grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is why they control the pitch of the blades to keep with the tolerances stated by the power supply companies for voltage and frequency...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-07-26T03:53:37.300" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29375" PostTypeId="1" CreationDate="2019-07-26T05:06:46.083" Score="0" ViewCount="346" Body="&lt;p&gt;I'm trying to build structures to exclude some percentage of rainwater on experimental plots. I have three 20ft beam supports across the plot, spaced 3m apart and held in place by 4x4 posts that were stepped down in height by a total of 1ft across the three of them-creating a slope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However I realized that the polycarbonate roofing can't span the 3m gap between supports without sagging. In the specs they suggest support every 3ft, so I am looking for a way to achieve this support with my existing materials- roofing panels (6&quot; wide by 12ft) and pvc irrigation piping (depicted on the right below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the material specs: &lt;a href=&quot;https://images.homedepot-static.com/catalog/pdfImages/2c/2c6542a0-67b8-4463-b752-07be3f114eab.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Suntuf specs&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thought I had was creating rafters with the pvc, or creating additional support beams of pvc between the wood posts to shorten the span to 1.5m. I'd like to find a solution that uses minimal additional materials and has the potential to last for several years at least. I'm really trying to avoid any pooling or spilling of water, as it would kill the point of the experiment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any clever engineering solutions on how I can help the polycarbonate sheet span this 3m gap?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EPNn4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EPNn4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21271" LastEditorUserId="21271" LastEditDate="2019-07-26T14:41:39.867" LastActivityDate="2019-09-08T22:00:50.410" Title="How to get corrugated plastic roofing strips to span 3m for rainwater exclusion experiment without building additional ground supports?" Tags="|structural-engineering|civil-engineering|structures|plastic|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29376" PostTypeId="2" ParentId="29362" CreationDate="2019-07-26T05:17:12.690" Score="0" Body="&lt;p&gt;The way I like to think about it is what source of energy drives the process.  The function of a steam turbine is effectively to translate chemical energy into mechanical energy.  Heat is generated by the burning of some fossil fuel that then is used to heat up water. That water/steam is under a confined condition so that when the temperature increases, the volume is not allowed to increase at a comparable rate so pressure increases.  That pressure is then released in such a way that it converts some of the pressure into kinetic energy.  That kinetic energy is then imparted onto the turbine and converted into kinetic energy in the turbine.  The turbine then converts the kinetic energy into usable electrical energy via a generator.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The start point of the chain is the chemical energy.  The trick is converting the chemical energy into mechanical then into electrical.  By heating the water to steam the steam at temp will have pressure and the pressure is stored mechanical energy, which is available to use.  Heated water only gets stored thermal energy because pressure doesn't go up with temp for liquid water unless you're in one of the parts of the steam table where you're super heated and at extreme pressure so the water is technically not gaseous yet.  &lt;/p&gt;&#xA;" OwnerUserId="21260" LastActivityDate="2019-07-26T05:17:12.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29377" PostTypeId="2" ParentId="29350" CreationDate="2019-07-26T05:44:29.157" Score="0" Body="&lt;p&gt;It's been a while since fluid dynamics but gravity matters for the force balance.  In the force balance, you're going to have gravity, frictional forces from the surface/shear stress of the water, and a pressure applied to the fluid from some direction to complete the balance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't see anything in the question indicating the direction of flow but the easiest is probably going to be to assume that the flow is going downward or upward.  Gravity is acting in a downward direction so either a pressure or a frictional force has to balance with the gravitational force acting in the vertical direction or there's going to be acceleration in that direction.  Since the flow is steady, it's not accelerating.  If you make the flow upward then you need a pressure acting in the upward direction which then counters the force of the friction and gravity together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the flow is vertical you'll still have pressures along the side so as to ensure flow is moving in a direction but you can just say there's a pressure acting there and the two sides balance to zero making it meaningless in the force balance.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the flow is horizontal across the plate then you'll have a frictional force in that direction that you will have to account for by some pressure in addition to the vertical forces that remain in place.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd have to refer to a textbook to break down the shear stresses and velocity profiles as well as the equation you've used since I don't remember all of the terms offhand but everything should fit within the framework of that force balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Realistically whether the flow is horizontal or vertical it isn't harder it's just which terms you're claiming are at a balance with the others. &lt;/p&gt;&#xA;" OwnerUserId="21260" LastActivityDate="2019-07-26T05:44:29.157" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29378" PostTypeId="2" ParentId="29375" CreationDate="2019-07-26T05:56:08.993" Score="0" Body="&lt;p&gt;Cut strips from a sheet so that three strips glued together form a triangle - these may then be strong enough to reduce the sagging. Make the height of the triangle match the peak height of the sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, proper bracing is the only solution and that comes down to the design. Usually these panels have sufficient overlap, bracing and slope so they don’t have pooling issues when they sag slightly.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-07-26T05:56:08.993" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29379" PostTypeId="2" ParentId="19011" CreationDate="2019-07-26T06:30:03.267" Score="0" Body="&lt;p&gt;Several considerations:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cost &lt;/li&gt;&#xA;&lt;li&gt;Space: adding insulation makes the oven bigger on the outside or takes away space inside (if it shall fit into the same size cupboards an a non-insulated oven)&lt;/li&gt;&#xA;&lt;li&gt;You can't seal an oven hermetically, vapors need to escape anyway&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I think no. 2 is the biggest consideration. A typical domestic oven (with some insulation) will fit into an 60 cm slot in your kitchen and fits a 47 cm baking tray. How many cupcakes will fit into the oven on you picture, if you remove 7 cm on both sides?&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-07-26T06:30:03.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29380" PostTypeId="2" ParentId="29366" CreationDate="2019-07-26T08:48:41.677" Score="0" Body="&lt;p&gt;A non-flowing fluid can have kinetic energy if it is moving as a rigid body. For example, a full water bottle (no room for flow inside) is thrown through the air without rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is not likely the intent of the author for this situation.&lt;/p&gt;&#xA;" OwnerUserId="16805" LastActivityDate="2019-07-26T08:48:41.677" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29382" PostTypeId="1" CreationDate="2019-07-26T11:59:19.820" Score="0" ViewCount="56" Body="&lt;p&gt;Let's say that I have &lt;span class=&quot;math-container&quot;&gt;$14$&lt;/span&gt; regular matrices of order &lt;span class=&quot;math-container&quot;&gt;$4$&lt;/span&gt;.&#xA;I'd like to compare the data of each matrix &#xA;with all 13 others to see if there's a match&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an algorithm that can accomplish this task?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm more interested in the algorithm and not any particular&#xA;programming language.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT]&#xA;In response to the first comment, a match would be if all rows and columns of the matrix are the same, small 2x2 example.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; m1       m5&#xA;|00| === |00|&#xA;|00|     |00|&#xA;&#xA; m3       m1&#xA;|00| =\= |00|&#xA;|01|     |00|&#xA;&#xA; m2       m1&#xA;|10| =\= |00|&#xA;|01|     |00|&#xA;&#xA; m2       m4&#xA;|10| === |10|&#xA;|01|     |01|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It would continue like this for 16x16 but in the end I could get a list of matrix that maches&#xA;m1 matches m5&#xA;m2 matches m4&#xA;m1 doesn't match m3, etc...&lt;/p&gt;&#xA;" OwnerUserId="20730" LastEditorUserId="20730" LastEditDate="2019-07-28T07:15:48.533" LastActivityDate="2019-07-28T07:15:48.533" Title="algorithm to compare items in a set" Tags="|software|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29383" PostTypeId="2" ParentId="29360" CreationDate="2019-07-26T12:20:10.377" Score="1" Body="&lt;p&gt;You got good answers here, however, I think another little push is still needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest to examine the whole mechanism, end to end. Do you intend to drive the gate by a rack and pinion mechanism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lCQ2s.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lCQ2s.gif&quot; alt=&quot;rack and pinion illustration from wikipedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First we need to estimate the force, speed and power needed to drive the gate.&#xA;Let us conservatively assume a 0.1 friction coefficient between the gate and the ground (I guess the gate supported on wheels so practically the coefficient is much smaller) - that brings us to about 600 [N] needed for pushing it (600 [kg] times 9.8 [m/sec^2] times 0.1). &#xA;You also mentioned that the required linear velocity of gate is 0.33 [m/sec].&#xA;Putting it all together, the needed power for driving your gate is P = f*v = 600 [N] times 0.33 [m/sec] = ~200 [W]. &#xA;We still have not taken into consideration power losses along our system, so you should re-thinking about the choice of your motor. However, let's continue with the system analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know the size of your pinion or you have to adjust it yourself? &#xA;The rack and pinion interface convert the linear motion to a rotational motion. Thus, if the pinion radius is 24 [mm] (I purposely chose it to adjust the numbers) and the linear velocity is 0.33 [m/s] - the desired angular velocity of the pinion is 0.33 [rad/sec]/0.024 [m] = 13.75 [rad/sec] or 131 [RPM]. Similarly, the Torque the pinion has to deliver for the rack to apply 600 [N] is 600 [N] * 0.024 [m] = 14.4 [Nm].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @Jonathan R Swift explained, the worm gear ratio in your case is 1:25 so ideally the motor should deliver 0.576 [Nm] at 3275 [RPM].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we have to estimate the efficiency of both gear system. Efficiency of 0.95 is quite reasonable for spur gears while 0.5 efficiency would be a good guess for the worm (sometimes worm gears have even lower efficiencies). Furthermore, while the speed is not effected by the gear efficiencies, the motor shall supply higher torque in order to meet your requirement. Therefor, The needed motor torque is 0.575 [Nm] / (0.95*0.5) = 1.21 [Nm].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This calculation has to be repeated for the gate acceleration, as already discussed here by @kamran&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2019-07-26T12:20:10.377" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29384" PostTypeId="1" AcceptedAnswerId="29477" CreationDate="2019-07-26T12:46:02.810" Score="0" ViewCount="100" Body="&lt;p&gt;When I was in school, photosynthesis was taught as this amazing biological process that took light and water and turned carbon dioxide into oxygen and carbon. It was implied that this was limited to the domain of plants. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have we industrialised photosynthesis so that we can shine light on a test tube and have it brake cardbon dioxide into oxygen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;Is there an industrial process that makes use of photosynthesis?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="14103" LastActivityDate="2019-08-03T11:37:53.163" Title="Is there an industrial process that makes use of photosynthesis?" Tags="|chemical-engineering|chemistry|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29385" PostTypeId="2" ParentId="29382" CreationDate="2019-07-26T13:20:51.887" Score="1" Body="&lt;h2&gt;Brute force:&lt;/h2&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;for (i1=0; i1&amp;lt;array.len-1; i1+=1)&#xA;    for (i2=i1+1; i2&amp;lt;array.len; i2+=1)&#xA;        if (same(array[i1], array[i2]))&#xA;            print(i1, &quot;matches&quot;, i2)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;same(a,b)&lt;/code&gt; compares the two matrices.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Optimization:&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is an order N² algorithm, so for very large values of &quot;14&quot; and &quot;4&quot; you'd want to optimize the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some possibilities (I'm sure there are others, and perhaps better):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Create a vector of the checksums of each matrix, and call &lt;code&gt;same()&lt;/code&gt; only when the two elements have the same value.  This is still N² in the worst case, but in practice would potentially have far fewer comparisons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Do the same thing, but first sort the checksum vector to group elements with the same value together.  Then make comparisons only within each group.  (You'd also have to track the original position of each matrix in order to display the answer.)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't know what the elements of a matrix are, but if they are well ordered (e.g. the set of three items, &quot;&lt;em&gt;rock&lt;/em&gt;&quot;, &quot;&lt;em&gt;paper&lt;/em&gt;&quot;, &quot;&lt;em&gt;scissors&lt;/em&gt;&quot; isn't), you could sort the array (order N log(N)) and then a linear pass could find the consecutive equal arrays.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19930" LastEditorUserId="19930" LastEditDate="2019-07-27T00:15:30.360" LastActivityDate="2019-07-27T00:15:30.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29386" PostTypeId="1" AcceptedAnswerId="29391" CreationDate="2019-07-26T17:00:09.800" Score="1" ViewCount="80" Body="&lt;p&gt;Please refer to the following youtube video (pause it as soon as it starts)...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;strong&gt;USS William D Porter, the Unluckiest Ship in the Navy&lt;/strong&gt;&quot; by &lt;strong&gt;The History Guy&lt;/strong&gt;&#xA;&lt;a href=&quot;https://youtu.be/f9Gb4PakFTU?t=162&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/f9Gb4PakFTU?t=162&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a running clock super-imposed over the film which gives a HH:MM:SS:FF (where FF is the number of frames elapsed this second, 0-29).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How did 1940's technology achieve this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose I must concede that this counter may have been added years or even decades later, but the clocks on display here (there are several clips within this video that feature different styles of clocks) seem to &quot;blend in&quot; with the old footage which leads me to believe that they were added (perhaps in post-processing) during the war period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the astonishing ingenuity of wartime engineers I think this would be a relatively trivial task for them, but I would like to know how it was done.&lt;/p&gt;&#xA;" OwnerDisplayName="user6335" LastActivityDate="2019-07-26T21:49:26.230" Title="WW2 technology - How was this on-screen frame counter added to old film-media footage?" Tags="|engineering-history|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29390" PostTypeId="2" ParentId="29386" CreationDate="2019-07-26T20:22:07.373" Score="0" Body="&lt;p&gt;0-29 or 0-23?  If it was film, it would have been 24 frames per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how 1940's technology &lt;em&gt;did&lt;/em&gt; achieve this, but it &lt;em&gt;could&lt;/em&gt; be done either in post-processing, with a counter that runs at any speed, or the same stop-motion mechanism that advanced the film could advance a 24-position wheel with the numbers 0-23 on it, and flashed a light.  That wheel could have been geared to the seconds, minutes, &amp;amp; hours wheels.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-07-26T20:22:07.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29391" PostTypeId="2" ParentId="29386" CreationDate="2019-07-26T21:04:45.780" Score="3" Body="&lt;p&gt;If you right click on the YouTube video and set the playback speed to the minimum, if is obvious that the video has been converted from 24 FPS cine film to 30 FPS video for TV broadcasting. Every fourth frame in the original is repeated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sequence numbers correspond to the video frames (0-29 in each second) not to the original cine frames, and therefore must have been added when the cine film was converted. There is no indication when or why that was done, but first commercially successful video recorder with helical scanning heads (the Ampex 1000) was developed in 1953-54. It is unlikely that any video conversions made before that date would be in a format that was still playable, and they would probably have been of lower quality than the YouTube video. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, the time stamps could have been added at any time in the last 60 years or so, but were not part of the original film footage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At a superficial look, the video seems to simply repeat every 4th frame, rather than using the so-called &lt;a href=&quot;https://en.wikipedia.org/wiki/Three-two_pull_down&quot; rel=&quot;nofollow noreferrer&quot;&gt;3:2 pulldown&lt;/a&gt; interpolation that was used for telecine broadcasting. That fact might limit the time when the conversion might have been done, but of course &quot;obsolete&quot; equipment sometimes continues to be used long after there are &quot;better&quot; alternatives available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the conversion was done for historical archiving, the lack of interpolation may be a deliberate choice to preserve the original content of each film frame as accurately as possible, in case someone wants to examine it frame by frame for research purposes.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-07-26T21:49:26.230" LastActivityDate="2019-07-26T21:49:26.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29392" PostTypeId="1" CreationDate="2019-07-26T21:26:30.907" Score="0" ViewCount="689" Body="&lt;p&gt;I am trying to build a laminar flow box using instructions at these YouTube links for &lt;a href=&quot;https://www.youtube.com/watch?v=cDaJW77Wb5Q&quot; rel=&quot;nofollow noreferrer&quot;&gt;design&lt;/a&gt; and &lt;a href=&quot;https://www.youtube.com/watch?v=Sj4qk6I-hy8&quot; rel=&quot;nofollow noreferrer&quot;&gt;build&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They recommended getting a Dayton squirrel cage air blower model &lt;a href=&quot;https://electricmotorwarehouse.com/dayton-model-1tdu2-blower-1202-cfm-1390-rpm-115-230v-60-50hz-4c831/&quot; rel=&quot;nofollow noreferrer&quot;&gt;1TDU2&lt;/a&gt; rated at 1202 CFM. The performance sheet for this air blower ( CFM vs. in wg) can be seen at this &lt;a href=&quot;http://www.electricmotorwarehouse.com/content/Dayton/blower_specs/ITDU2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt; I've already purchased this... (can be returned)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am now considering buying one of these HEPA filters from Ebay:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A.&lt;/strong&gt; CAMFIL FARR. 855023669, PX-21.75-44.00-9-40-FU-00-00-0 MEGALAM PANEL HEPA FILTER (&lt;a href=&quot;https://www.ebay.com/itm/CAMFIL-FARR-855023669-PX-21-75-44-00-9-40-FU-00-00-0-MEGALAM-PANEL-HEPA-FILTER/361127918011?hash=item5414e6b9bb%3Ag%3Ax3YAAOSwOuBbvjIb%3Asc%3AFedExHomeDelivery%2123455%21US%21-1&amp;amp;LH_ItemCondition=3&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;) = 0.40 in W.G. / 100FPM / 554CFM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;B.&lt;/strong&gt; CAMFIL FARR 85502A548 MEGALAM PANEL P7-23.12-47.12-5-08-00-7D-00-0&#xA;(&lt;a href=&quot;https://www.ebay.com/itm/CAMFIL-FARR-85502A548-MEGALAM-PANEL-P7-23-12-47-12-5-08-00-7D-00-0/281516905913?hash=item418bb715b9%3Ag%3AxIoAAOSwuMFUfMWQ%3Asc%3AFedExHomeDelivery%2123455%21US%21-1&amp;amp;LH_ItemCondition=3&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;) = 0.58 in W.G. / 100FPM / 704CFM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;C.&lt;/strong&gt; CAMFIL FARR 855024774 MEGALAM PANEL P7-18.00-48.00-2-03-SR-00-00-0 PACK OF 2&#xA;(&lt;a href=&quot;https://www.ebay.com/itm/CAMFIL-FARR-855024774-MEGALAM-PANEL-P7-18-00-48-00-2-03-SR-00-00-0-PACK-OF-2/281516930978?hash=item418bb777a2%3Ag%3AzKcAAOSwkNZUfMmT%3Asc%3AFedExHomeDelivery%2123455%21US%21-1&amp;amp;LH_ItemCondition=3&amp;amp;autorefresh=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;link&lt;/a&gt;) = 0.71 in W.G. / 105FPM / 550CFM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My requirement for maintaining microbial sterility are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(i)&lt;/strong&gt; minimum 90 FPM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;(ii)&lt;/strong&gt; minimum 99.99% air purity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The help I seek is in the form of your answers to my questions below, please:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there is a large mismatch between air blower CFM rating and any of these filters recommended CFM rating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; Are any of the filters listed above matched with my air blower in terms of CFM rating? I guess not, but I want to be sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2.&lt;/strong&gt; Is matching of the air blower CFM and filter CFM done at zero pressure, or operating pressure drop water gauge (which varies across these 3 filters) ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt; FYI, I will be using a pre-filter to get rid of large particulate matter, that contributes additionally 0.3 in WG pressure drop&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4.&lt;/strong&gt; If at total WG pressure drop, the air blower CFM is X, and the rated air filter CFM is Y, then for maintaining filter integrity, I know X &amp;lt; Y and X = Y is OK (where X &gt; 90FPM). But how high can X get when X &gt; Y ? 20% , 30% , 100% - something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5.&lt;/strong&gt; What are the consequences of X &gt;&gt; Y ? Will it blow out the HEPA filter? And more air means exceeding capacity to capture particles, so more air, but partially dirty air, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6.&lt;/strong&gt; Can anyone help me with who to read and interpret a Camfil Farr HEPA filter's label, especially the description e.g. PX-21.75-44.00-9-40-FU-00-00-0, P7-23.12-47.12-5-08-00-7D-00-0, P7-18.00-48.00-2-03-SR-00-00-0 etc.?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;7.&lt;/strong&gt; I was told that for my application, a &quot;high flow&quot; (~1000 CFM) thicker HEPA filter (11 5/8 in) is better than a &quot;standard&quot; thinner one (5 7/8 in) with only ~ 500 CFM. Is this a valid piece of advice, given my goals?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;8.&lt;/strong&gt; From Ebay or elsewhere, do forum users have friendly advice for a (an equally or more) cost effective alternative to these Camfil Farr HEPA filters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks, in advance. Have a great weekend!&lt;/p&gt;&#xA;" OwnerUserId="21276" LastEditorUserId="21276" LastEditDate="2019-07-26T21:57:57.393" LastActivityDate="2019-07-26T21:57:57.393" Title="Matching Air blower CFM vs HEPA filter CFM, and ΔP in WG, for laminar flow design" Tags="|design|pressure|airflow|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29393" PostTypeId="1" AcceptedAnswerId="29398" CreationDate="2019-07-27T01:09:27.750" Score="1" ViewCount="53" Body="&lt;p&gt;My group made an RCA video/audio to RF modulator for a school project. The video is a NTSC composite video signal. Here is a clip of our tv receiving the output of the project on NTSC channel 2:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://drive.google.com/open?id=1-1xRHI8razjYvvptTsPsPBHt8T0z-yi5&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://drive.google.com/open?id=1-1xRHI8razjYvvptTsPsPBHt8T0z-yi5&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the video, we use an usb-to-hdmi adapter from Raycue, a hdmi-to-composite video/audio adaptor from Ozvavzk, and a tv from Thomson Consumer Electronics (model 19GT241).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made measurements of our audio and carrier signals using a Siglent SDS 1202X-E Oscilloscope and a $12 active probe (which is listed on eBay as &quot;RF Active Probe 0.1-1500 MHz&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I measured the audio carrier signal to be 55.2 MHz and I measured the video carrier signal to be 59.2 MHz. I did a FFT of the two signals - the audio's bandwidth looks narrow like a dirac-delta function and the video's bandwidth is about 500 KHz wide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the &quot;Pan-American television frequencies&quot; article on Wikipedia, if we are to transmit on channel 2, then our audio and video carrier signals should be tuned to 59.75 MHz and 55.25 MHz, respectively. I can't find information that supports the carrier frequencies of our project are alternative ones for channel 2. Why is our tv playing audio and video if the carrier frequencies in our RF project are not tuned according to the article?&lt;/p&gt;&#xA;" OwnerUserId="21275" LastActivityDate="2019-07-27T12:49:55.613" Title="Why is the tv playing if the carrier frequencies in our RF project appear unsupported?" Tags="|electrical-engineering|rf-electronics|radio|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29395" PostTypeId="1" CreationDate="2019-07-27T08:01:47.667" Score="0" ViewCount="62" Body="&lt;p&gt;I want to use a compressor to create a jet of air. At discharge from compressor how much CFM (Cubic Feet per Minute) are enough to create a force that can turn a heavy object (say 40 kg) at 3 rev per sec placed in front of air jet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just looking for a relationship between CFM and Jet Force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks. &lt;/p&gt;&#xA;" OwnerUserId="18407" LastEditorUserId="18407" LastEditDate="2019-07-27T08:08:29.510" LastActivityDate="2019-07-27T08:08:29.510" Title="Force of Jet from Air Compressor" Tags="|compressed-air|compressors|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29396" PostTypeId="1" AcceptedAnswerId="29594" CreationDate="2019-07-27T09:25:30.723" Score="1" ViewCount="62" Body="&lt;p&gt;When applying the moment distribution method also known as the Hardy Cross method, how do I take into account the self-weight of columns? In the case of the beams, I can just apply the weight load laterally. However, for the column the load is axial, therefore causing 0 fixed end moment even when there is a difference in the support bending moment reaction. &lt;/p&gt;&#xA;" OwnerUserId="20822" LastActivityDate="2019-08-11T00:43:56.513" Title="Self Weight in Moment Distribution (Hardy Cross)" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29397" PostTypeId="1" CreationDate="2019-07-27T11:45:54.883" Score="0" ViewCount="126" Body="&lt;p&gt;Online I found many guides and videos on how to calculate outputs of a planetary gearset but all of them have only one input. But how can I calculate output speed of carrier if ring and sun gear are moving at different speeds?&lt;/p&gt;&#xA;" OwnerUserId="21049" LastActivityDate="2019-07-27T11:45:54.883" Title="Is there an easy way to calculate carrier output of a planetary gear set which has two inputs?" Tags="|mechanical-engineering|gears|mechanisms|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29398" PostTypeId="2" ParentId="29393" CreationDate="2019-07-27T12:49:55.613" Score="1" Body="&lt;p&gt;The old analog signal TVs had a wide frequency range allocated to each channel.&#xA;Channel 2 for instance occupied everything between 54 MHz and 60 MHz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of effects of temperature and other conditions on the hardware (e.g. resistors and capacitors), the transmitter and the receiver could drift slightly from the exact frequency.&#xA;To compensate, the TV had a &quot;fine tuning&quot; control allowing one to adjust it to match the actual frequency.  The set you were using either had AFT (Automatic Frequency Control), or you manually used the fine tuning yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wikipedia's &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/Pan-American_television_frequencies&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pan-American television frequencies&lt;/a&gt;&quot; has this table, showing that your 55.2 and 59.2 were well within the channel 2 frequency range:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8MPMH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8MPMH.png&quot; alt=&quot;VHF low-band&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19930" LastActivityDate="2019-07-27T12:49:55.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29399" PostTypeId="1" CreationDate="2019-07-27T12:53:39.563" Score="0" ViewCount="29" Body="&lt;p&gt;I'm experimenting with rubbing a balloon on a silk (to get the electrons) and then touching the balloon to two small aluminium foil balls to see if they'll repel but they don't (if I do it with aluminium foil sheets, they stick to the balloon for some reason, and again, they don't repel). What's happening here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My final goal is to charge titanium dioxide powder by rubbing a charged balloon in the powder. (I don't know if that will work though)&#xA;Is that cheapest way to go about doing this?&lt;/p&gt;&#xA;" OwnerUserId="21281" LastActivityDate="2019-07-27T13:05:08.893" Title="How would I go about charging titanium dioxide for e-paper technology project?" Tags="|electrical-engineering|chemistry|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29401" PostTypeId="1" CreationDate="2019-07-27T14:26:43.643" Score="0" ViewCount="87" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5Vb8B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Vb8B.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that &lt;span class=&quot;math-container&quot;&gt;$P_{total}=P_{gauge}+P_{atmosphere}$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$P_{gauge}=P_{atmosphere}-P_{total}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_{atmosphere}=101.325 KPa$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$P_{gauge}=0.2 psig=1.38 KPa$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_{gauge}=(0.8 \times sin{30})(900)(9.807) - (900)(9.807)(h)=1.38 KPa$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies that &lt;span class=&quot;math-container&quot;&gt;$h=0.4m$&lt;/span&gt;, however the answer is &lt;span class=&quot;math-container&quot;&gt;$0.24m$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="17668" LastActivityDate="2019-07-27T17:43:44.190" Title="What Is The Liquid Depth Inside The Tank?" Tags="|fluid-mechanics|pressure|fluid|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29402" PostTypeId="2" ParentId="29401" CreationDate="2019-07-27T17:43:44.190" Score="1" Body="&lt;p&gt;Following your calculations, we equate 1.34 KPa to the height of the liquid then subtract it from 40cm=80cm*sin30. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ 1.38/101.325*1000/900=0.01513 \\ 40-15.13= 24.8cm $$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-27T17:43:44.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29403" PostTypeId="2" ParentId="15761" CreationDate="2019-07-27T18:16:41.350" Score="2" Body="&lt;p&gt;Water accumulating in the bottom of the bore is not a concern, as long as concrete is delivered at the bottom of the bore via a hose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concrete with the slump of less than 4inches with some plasticizers to minimize cement washing and floating to the surface of concrete will be used and because of its density; it will push the water up and not mix with it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the water has impurities or damaging ingredients like sulphates or chloride, then special admixtures must be used, which has to be designed by an specialist in coordination with a testing lab.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-27T18:16:41.350" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29404" PostTypeId="1" CreationDate="2019-07-27T20:53:00.787" Score="0" ViewCount="328" Body="&lt;p&gt;I am dealing with a statics problem in the following link. This question is from the courtesy of Engineering mechanics, Statics by R.C. Hibbeler, 13th edition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cBxwR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cBxwR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before applying vertical forces F and -F, 60 Ib of force causes 1.5 ft of stretch in springs. With the exercise of the two forces, springs stretch some more and this can be found from the geometry of the problem. This stretch is 0.27 ft at points C and D. Finally, again due to geometry F equals the force in one of the springs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I sum the initial stretch and the final stretch and multiply this with the stiffness of one spring but do not arrive at the true answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong here?&lt;/p&gt;&#xA;" OwnerUserId="15381" LastEditorUserId="1832" LastEditDate="2019-07-28T03:59:42.167" LastActivityDate="2019-07-28T03:59:42.167" Title="Statics poblem with a pre-stretched spring" Tags="|mechanical-engineering|statics|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29405" PostTypeId="1" AcceptedAnswerId="29406" CreationDate="2019-07-27T21:44:30.367" Score="0" ViewCount="77" Body="&lt;p&gt;In automatic transmissions some parts need to be moving with the engine in one moment and in the other they need to be stopped. Why don't transmisisons then use detent clutches to accomplish this?&lt;/p&gt;&#xA;" OwnerUserId="21049" LastActivityDate="2019-07-28T01:56:28.400" Title="Why aren't detent clutches used in car transmissions?" Tags="|mechanical-engineering|transmission|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29406" PostTypeId="2" ParentId="29405" CreationDate="2019-07-28T01:56:28.400" Score="2" Body="&lt;p&gt;Detent clutches use mechanical interference to force the halves of the clutch into hard engagement. The wear surfaces in this case are metal-on-metal which works well when the engagement duty cycle is low, as for example in the overrunning-clutch mechanism used in motorcycle starters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for purposes of shifting gear ratios in a car transmission, you do not want hard engagement because it jolts the car whenever a shift is made. Furthermore, the duty cycle of engagement and disengagement is high in a car transmission, which means the engagement surfaces will wear themselves out of tolerance before the desired service lifetime of the transmission as a whole has been met.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-07-28T01:56:28.400" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29407" PostTypeId="1" CreationDate="2019-07-28T09:00:47.117" Score="4" ViewCount="231" Body="&lt;p&gt;I have recently discovered flexure springs (see image below), and I am interested in using one in an amateur project I am building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I would prefer if I am able to adjust the &quot;stiffness&quot; of the flexure spring, to allow the spring to bear different loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Despite extensive searching online, I have not found a flexure spring design that provides flexible or adaptable stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to have a flexure spring, the stiffness of which, can be adjusted?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fkPki.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fkPki.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="932" LastActivityDate="2021-11-02T21:02:05.203" Title="Is an Adjustable Stiffness Flexure Spring possible?" Tags="|mechanical-engineering|design|springs|flexures|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29408" PostTypeId="1" CreationDate="2019-07-28T09:03:19.570" Score="0" ViewCount="23" Body="&lt;p&gt;My first thought was to charge the titanium dioxide particles by first charging a balloon by rubbing it on my hair and then touching the balloon to the titanium dioxide powder, but that would be ineffective and the charge will escape when the powder is submerged in the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better way would be to charge the titanium dioxide particles chemically after submerging them in the black-dyed water. How would I go about doing this safely, cheap and in home? (Maybe putting a non-dangerous alkaline in the final mix would charge the titanium dioxide?)&lt;/p&gt;&#xA;" OwnerUserId="21282" LastActivityDate="2019-07-28T09:03:19.570" Title="How do I charge titanium dioxide particles submurged in black-dyed water in home?" Tags="|chemical-engineering|chemistry|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29410" PostTypeId="1" CreationDate="2019-07-28T17:03:55.380" Score="0" ViewCount="112" Body="&lt;p&gt;Not being of an engineering background, I was first inclined to think that a door closer would reduce horizontal load on all hinges and not contribute any support in the vertical dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, bearing rotational mechanics in mind, is this necessarily true?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think, on any normal door:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The upper hinge would be under horizontal tension.&lt;/li&gt;&#xA;&lt;li&gt;The lower hinge would be under horizontal compression.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If a concealed door closer (similar to the model in the lower image) is installed into the frame below the center of mass of the door, would it exacerbate load on the hinges rather than protect them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/B9hM9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/B9hM9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concealed door closer, mounted in the jamb, providing constant spring-loaded compressive force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Q2tM5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Q2tM5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it contribute anything to the vertical load on the hinges?&lt;/p&gt;&#xA;" OwnerUserId="20926" LastEditorUserId="20926" LastEditDate="2019-07-28T17:29:24.803" LastActivityDate="2019-08-27T19:01:55.197" Title="Do concealed door closers exacerbate wear on a door's hinges?" Tags="|mechanical-engineering|structural-engineering|applied-mechanics|building-design|building-physics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29411" PostTypeId="1" AcceptedAnswerId="29412" CreationDate="2019-07-28T17:10:07.897" Score="1" ViewCount="135" Body="&lt;p&gt;Something simple is tripping me up in a design, and the problem can be reduced to a simple 'seesaw' setup: a rigid beam with pin joint and a weight hanging from each end. I have attached an image with dimensions and values of interest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4sGcH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4sGcH.png&quot; alt=&quot;Beam and Fulcrum&quot;&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pin joint is not centered in the beam so that one end is further from the joint than the other. The beam has been rotated to an angle, and the length of each side's projection onto the horizontal is known. My question is this: In analyzing this problem I end up showing that the torque induced by each weight is equal but the reaction induced at &lt;span class=&quot;math-container&quot;&gt;$F_2$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$F_1$&lt;/span&gt; does not equal &lt;span class=&quot;math-container&quot;&gt;$F_2$&lt;/span&gt;, and vice versa. Have I made a mistake in my analysis, and if so where?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Known: &lt;span class=&quot;math-container&quot;&gt;$A=30\ deg;\ R_1=10\ inch;\ R_2=5\ inch$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given: &lt;span class=&quot;math-container&quot;&gt;$F_1=10\ lbs;\ F_2=20\ lbs$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the vertical reaction force &lt;span class=&quot;math-container&quot;&gt;$F_{1,2}$&lt;/span&gt; induced at &lt;span class=&quot;math-container&quot;&gt;$F_2$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$F_1$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;span class=&quot;math-container&quot;&gt;$F_{1,tangent}=F_1\cos(A)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;span class=&quot;math-container&quot;&gt;$L_1=R_1/\cos(A)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) &lt;span class=&quot;math-container&quot;&gt;$\tau_1=F_{1,tangent}L_1=F_1R_1=100\ in.lbs$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) &lt;span class=&quot;math-container&quot;&gt;$L_2=R_2/\cos(A)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5) &lt;span class=&quot;math-container&quot;&gt;$F_{1,2,tangent}=\tau_1/L_2=100\cos(A)/R_2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6) &lt;span class=&quot;math-container&quot;&gt;$F_{1,2}=F_{1,2,tangent}\cos(A)=15\ lbs$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this result of &lt;span class=&quot;math-container&quot;&gt;$F_{1,2}=15$&lt;/span&gt; and going backwards yields &lt;span class=&quot;math-container&quot;&gt;$\tau_2=75\ in.lbs\neq\tau_1$&lt;/span&gt;. So it seems like I made an error, but I can't find it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To look at it another way, wouldn't this result suggest &lt;span class=&quot;math-container&quot;&gt;$F_2$&lt;/span&gt; cannot be greater than &lt;span class=&quot;math-container&quot;&gt;$15\ lbs$&lt;/span&gt; without breaking equilibrium. Yet when &lt;span class=&quot;math-container&quot;&gt;$F_2=20$&lt;/span&gt; the torque induced is &lt;span class=&quot;math-container&quot;&gt;$\tau_2=100=\tau_1$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so I keep going in circles. I could really use some help! Thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="21287" LastActivityDate="2019-07-28T23:23:22.417" Title="Simple Fulcrum Analysis" Tags="|mechanical-engineering|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29412" PostTypeId="2" ParentId="29411" CreationDate="2019-07-28T18:31:41.467" Score="0" Body="&lt;p&gt;We do not need to bother with the angle A, or you would have to consider the projection back and forth of the force F1 into its component orthogonal to the beam too.&#xA;The reaction of the F1 at the other end of the beam is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ 10lbs*10in/5in= 20lbs \ up $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This reaction is of course being canceled by F2, 20lbs pointing down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb vectors such as force can be applied at any point along their trace. So when you want to calculated their torque about a point you multiply the force by its distance from the fulcrum.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-28T18:31:41.467" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29413" PostTypeId="2" ParentId="29410" CreationDate="2019-07-28T18:55:02.247" Score="1" Body="&lt;p&gt;The tension and compression on the hinges are door mass, times half of door width ,B divided by the distance of upper and lower hinges, D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ m*(B/2)/D  $$&lt;/span&gt;&#xA;And each hinge shares m/3 as shear with the middle hinge having no horizontal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say concealed door closure is installed at x distance below the center hinge and applies a force F, so you have F*x = sum of the reaction of the three hinges. which is allocated most to lowers hinge and least to upper one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you add or subtract this reaction to the original hinge load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;remember if you have too much out-of-plane force at the swing of the door you want to account for that too. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-28T18:55:02.247" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29414" PostTypeId="1" CreationDate="2019-07-28T19:33:24.080" Score="2" ViewCount="505" Body="&lt;p&gt;I have accurate statistics about the total length of high-voltage (110-400 kV), medium-voltage (1-70 kV) and low-voltage (0-1 kV) transmission lines in my country, and they are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;High-voltage: 22 500 km&lt;/li&gt;&#xA;&lt;li&gt;Medium-voltage: 140 000 km&lt;/li&gt;&#xA;&lt;li&gt;Low-voltage: 240 000 km&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would like to obtain an accurate estimate of how much aluminum is used for the transmission lines, but so far have failed to find any. I assume all lines have approximately 3 conductors due to the 3-phase system (in some cases there may be a fourth one, but 3 is a good approximation). Thus, to calculate the amount of aluminum, I would need an accurate estimate of conductor radius for high, medium and low voltage networks. Obviously, I cannot climb to reach the transmission line and measure its size by a vernier caliper!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One source I found is &lt;a href=&quot;http://large.stanford.edu/courses/2010/ph240/harting1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://large.stanford.edu/courses/2010/ph240/harting1/&lt;/a&gt; where a figure says that optimal radius (one that minimizes total costs) for a 2.25 GW 3-phase 765 kV line is 4 centimeters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, for a total length of 402 500 km of transmission lines, there would be 402500000*pi*0.04^2 = 2023186 square meters of aluminum, or by weight 2023186 m3 * 2700 kg/m3 = 5462602200 kg of aluminum, per phase. Three phases would require 16.4 million metric tons of aluminum. Per capita, that would be approximately 3 metric tons, or 3000 kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel that estimate may be a bit on the high side, because I found a statistic that about 25 kg of aluminum is used per year per capita. Thus, it would require 120 years of production to manufacture 3 metric tons of aluminum per capita. I don't believe that all lines, including the low-voltage (and presumably low-current) lines would use 4 cm radius conductors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how much aluminum does a kilometer of high voltage (HV), medium voltage (MV) or low voltage (LV) transmission line use?&lt;/p&gt;&#xA;" OwnerUserId="12836" LastActivityDate="2019-07-30T16:13:57.793" Title="How much aluminum does a HV, MV or LV transmission line use?" Tags="|electrical-engineering|power-transmission|cost-estimation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29415" PostTypeId="2" ParentId="29411" CreationDate="2019-07-28T23:23:22.417" Score="0" Body="&lt;p&gt;As @Kamran stated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F1xL1=F2xL2&#xA;Therefore:&#xA;F2=[F1XL1] ÷L2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Radial calculations are a mechanical assumption when gravity is acting upon all equally and therefore discounted too. This means discounting angle A also from the equilibrium equation. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-07-28T23:23:22.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29416" PostTypeId="2" ParentId="29375" CreationDate="2019-07-28T23:41:37.507" Score="0" Body="&lt;p&gt;Looking at the sag I'd say space the supports one sixth of the length from the end of each sheet and bond the sheets together that the moment forces combined with each other cancel sag sufficiently. This way you only need to space the supporting beams 2m apart with hogging and suing moment forces cancelled by bonded sheets acting as the counterbalance to their own self weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: slope of entire arrangement should be such that mass of water does not remain on surface long enough to compromise anti-sag effects of bonding sheets together. Ultimately gravity will exceed structural strength due to material selection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slope should be about 1:24.&lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2019-07-29T03:13:54.173" LastActivityDate="2019-07-29T03:13:54.173" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29418" PostTypeId="2" ParentId="29153" CreationDate="2019-07-29T00:11:59.320" Score="1" Body="&lt;p&gt;It is a well established fact that ferrous metals become magnetised over time due to the earth's magnetism. The towers have been aligned with the field of torque sufficiently long enough to absorb electromagnetic energy that polarised the structures. This can be removed by exerting a substantial shock to the entire structure or heating it up beyond 768°C to realign the ionic crystalline bonds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cyclic heating and cooling process over decades has caused intensive stresses at molecular level that induce electronic field polarity exhibited by electromagnetism too. Rub 2 pieces of mild steel together long enough and they become magnetic. Try it with a couple dining knives. This is essentially what has been happening to the metal crystals all this time: friction. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2019-07-29T16:20:56.470" LastActivityDate="2019-07-29T16:20:56.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29419" PostTypeId="2" ParentId="28742" CreationDate="2019-07-29T01:01:29.997" Score="1" Body="&lt;p&gt;After a materials study of lithium ion batteries it was discovered that maximum lifespan was obtained by not discharging beyond 82% recharging up to 12% of maximum power capacity. Exceeding these levels proved that deposition of lithium metal crystals became detrimental to conductance capacity over time and even encouraged leaching of mineral from the solution without electronic current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After 300 cycles of 90%&gt; discharge to 100% charge the lifespans were critically affected but controlled charge inside 18% of max and min capacity extended life by as much as 72%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this they concluded that full discharge and complete recharge was detrimental to life expectancy of lithium ion batteries in regular use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Research white papers from Samsung research I've read agreed this research too. So to answer your question, don't let potential energy drop below 12% and, try not to maintain charge above 82% capacity. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2019-12-26T11:39:38.437" LastActivityDate="2019-12-26T11:39:38.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29420" PostTypeId="2" ParentId="28325" CreationDate="2019-07-29T03:22:42.710" Score="-2" Body="&lt;p&gt;According to Theoria Apotheosis on YouTube who is supposed to be a doctor of physics states that magnetic force is literally like an helix at the centrifugal point. The lines of force form a torus as seen by his invention that shows the exact field lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently accepted understanding is flawed which I have witnessed to be inaccurate in comparison. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-07-29T03:22:42.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29421" PostTypeId="2" ParentId="8533" CreationDate="2019-07-29T04:02:11.917" Score="0" Body="&lt;p&gt;After some thought it occurred to me that sedimentary data would exist showing previous flood events. To examine this would require soil samples to be taken where sediment has built up inside a river bend. The fine materials will remain on upper levels of each flood event and historic rainfall data may shed light on volume in catchment. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-07-29T04:02:11.917" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29422" PostTypeId="2" ParentId="14418" CreationDate="2019-07-29T04:13:45.910" Score="1" Body="&lt;p&gt;&quot;Why did he choose thicker arch sections at middle above mid-river and narrow sections at abutment supports?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greatest force is at the centre of the bridge, not the abutments. Since material is the same strength the thicker sections take higher stresses and therefore greater vertical forces which are transmitted down to the abutments and foundations. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-07-29T04:13:45.910" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29423" PostTypeId="1" CreationDate="2019-07-29T08:39:01.587" Score="1" ViewCount="258" Body="&lt;p&gt;I am doing a project where there is a plate with a hole on it where one ring will be mounted. The ring and plate need to be attached somehow. However, the ring must be able to rotate around the roll axis. I already have some ideas, but there are too complicated. (with bearing, etc) My questions is: what are some similar mechanisms that I can inspire? (The picture shown below is the basic demonstration of the project. The ring which is needed to be able to rotate is shown with the red color)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dKAWN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dKAWN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21290" LastEditorDisplayName="user21290" LastEditDate="2019-07-29T11:45:04.263" LastActivityDate="2019-07-31T10:58:34.927" Title="What kind of mechanisms are there to give an object free rotation?" Tags="|mechanisms|bearings|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29424" PostTypeId="2" ParentId="29423" CreationDate="2019-07-29T12:24:02.833" Score="0" Body="&lt;p&gt;Make the external surface of the red ring have a bearing track and the matching track on the interior of the black plate so you have a roller bearing created by the two items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or have the track on the red ring with 4 or even 8 holes drilled into the plate which have screws with a roller at the tip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sWi2T.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sWi2T.png&quot; alt=&quot;an idea:&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-07-29T12:47:20.020" LastActivityDate="2019-07-29T12:47:20.020" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29425" PostTypeId="2" ParentId="25769" CreationDate="2019-07-29T12:27:00.410" Score="0" Body="&lt;p&gt;According to university of Wolverhampton demolition masters degree literature, explosives are used where the shock is not likely to have detrimental effects on surrounding structures. Consideration of dust is important due to silicosis. The cost is high but relatively speaking not much different than other methods. Material from this method is usually a large mound which requires removal by the other two methods and also requires professionally qualified personnel to employ but they are very scarce and extremely well paid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cut and remove requires machine tools of higher rental costs but less expertise. Material may be removed to a specific site for disposal or reuse. There may be some hazardous material residue that should be dealt with outside the demolition area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crush and shift method is similar but dust and debris is less of a consideration than what lies underneath the foundations. With explosives there is need to consider the ground impact on subterranean services such as gas, water electricity and transportation tunnels. The other methods avoid this issue. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-07-29T12:27:00.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29426" PostTypeId="2" ParentId="28684" CreationDate="2019-07-29T12:37:53.147" Score="2" Body="&lt;p&gt;Take a look at a silicone gun mechanism. Simple and practical and does exactly what you are trying to invent. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-07-29T12:37:53.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29427" PostTypeId="1" CreationDate="2019-07-29T14:34:14.780" Score="2" ViewCount="219" Body="&lt;p&gt;The base current in the BJT is due to the recombination of holes and electrons. To neutralize the p region an electron leaves the p region. If we take it the other way round, the larger the base current the greater is the recombination, so the collector current should reduce, but &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;  &lt;span class=&quot;math-container&quot;&gt;$I_c/I_b$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$I_c=\beta I_b$&lt;/span&gt;, which shows that &lt;span class=&quot;math-container&quot;&gt;$I_b$&lt;/span&gt; is directly proportional to &lt;span class=&quot;math-container&quot;&gt;$I_c$&lt;/span&gt;.&#xA;I am searching for the physical essence of this thing. As we know that in MOSFET voltage controls the drain current by changing the channel width, similarly the BJT is said to be a current-controlled device, so I am eager to know how exactly the base current controls the collector current.&lt;/p&gt;&#xA;" OwnerUserId="21264" LastEditorUserId="7083" LastEditDate="2019-08-07T20:36:37.580" LastActivityDate="2019-08-08T10:43:12.820" Title="Why does collector current increase with base current in BJT?" Tags="|electrical-engineering|power-electronics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29428" PostTypeId="1" AcceptedAnswerId="29434" CreationDate="2019-07-29T15:11:06.377" Score="1" ViewCount="274" Body="&lt;p&gt;I general, if you model a system (pendulum for example) and you do not include damping into the system (maybe caused by friction): Is it generally possible to tune a PI or PID controller for an second-order-dynamical system that is undamped? Because without friction/damping no energy can leave the system, and therefore I am wondering, if PID-control can generally work for these systems or if this will lead always to oscilations or unstable control?&lt;/p&gt;&#xA;" OwnerUserId="21236" LastActivityDate="2019-07-30T18:58:06.993" Title="PID-control of undamped system" Tags="|control-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29430" PostTypeId="2" ParentId="3616" CreationDate="2019-07-29T16:04:30.920" Score="0" Body="&lt;p&gt;The noryl impellers of the pump to add to the load after the bearings and pump body gets hot. They expand, rub to the walls get over heated then melt. So not only the motor gets burnt out, but the pump too gets damaged.&lt;/p&gt;&#xA;" OwnerUserId="21292" LastEditorUserId="9" LastEditDate="2019-07-29T17:37:17.710" LastActivityDate="2019-07-29T17:37:17.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29431" PostTypeId="2" ParentId="6089" CreationDate="2019-07-29T16:30:38.547" Score="0" Body="&lt;p&gt;I use polycarbonate film which is used for binding manuals. Polyvinyl chloride sheets are equally good but when using whiteboard markets you must encaustic they are stressed within 24 hours to avoid nasty stains&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-07-29T16:30:38.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29432" PostTypeId="2" ParentId="25006" CreationDate="2019-07-30T13:15:05.830" Score="0" Body="&lt;p&gt;I was looking at the same thing. I think this answers it better: &lt;a href=&quot;http://accidont.ru/ENG/rodes.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://accidont.ru/ENG/rodes.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the diagram from the accepted answer:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Ai7Yp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ai7Yp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I think that they did not pull the control roads all the way up as in position II as that would make the system slower in getting more power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;given the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the up and down parts of the fuel rods are made of graphite (marked with purple in the diagram)&lt;/li&gt;&#xA;&lt;li&gt;there is a space between of 1.25m from the bottom of the graphite part of the control rod and the bottom of the graphite part of the fuel rod&lt;/li&gt;&#xA;&lt;li&gt;the water in that space acts as an inhibitor slowing down the reaction in the bottom part of the core and absorbing neutrons.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;when:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the control rod is lowered to the point where the previous gap that was filled with water is now displaced by graphite &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;then:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;you have a zone of only graphite at the bottom of the core, a hotspot of graphite&lt;/li&gt;&#xA;&lt;li&gt;graphite from the control rod and from the core itself will accelerate the fission&lt;/li&gt;&#xA;&lt;li&gt;as you continue lowering the control rod, the bottom of the core you will continue to accelerate the fission at the bottom of the core as there will always be graphite at the bottom until the control rods get in position IV&lt;/li&gt;&#xA;&lt;li&gt;it takes 18 seconds for the control rods to get in the IV position&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So i think that is why there is a power surge at the bottom of the core, but as i have no taring on this subject, and this only a theory i could be wrong.&lt;/p&gt;&#xA;" OwnerUserId="21299" LastActivityDate="2019-07-30T13:15:05.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29433" PostTypeId="2" ParentId="29414" CreationDate="2019-07-30T16:13:57.793" Score="1" Body="&lt;p&gt;Ok, since nobody answered, I investigated this more on my own. Note: the answer is just an estimate and might have incorrect assumptions. In particular, I'm assuming 400 kV means 400 kV line-to-line voltage and not 400 kV line-to-neutral voltage, and I'm also assuming the voltages and amperages are RMS and not peak. These assumptions, if incorrect, change the estimate only in a very minor way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;http://large.stanford.edu/courses/2010/ph240/harting1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://large.stanford.edu/courses/2010/ph240/harting1/&lt;/a&gt; there is a corona discharge threshold, which is given by the equation:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;U / 1.73 - 21.1 kV/cm * 0.95 * x * 1 * ln(d/x) = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where &lt;code&gt;U&lt;/code&gt; is the line-to-line voltage in kV, &lt;code&gt;1/1.73&lt;/code&gt; converts line-to-line voltage to line-to-neutral voltage, &lt;code&gt;x&lt;/code&gt; is the conductor radius in centimeters, and &lt;code&gt;d&lt;/code&gt; is the conductor spacing in centimeters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example: if &lt;code&gt;U = 400 kV&lt;/code&gt;, &lt;code&gt;d = 550 cm&lt;/code&gt;, this equation gives &lt;code&gt;x = 2.0655 cm&lt;/code&gt;, so approximately 2 cm conductor radius is needed for 400 kV lines with 5.5 meters of spacing or else corona discharge occurs and steals a lot of power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For high voltages, the conductor size is determined not by amperage, but rather by voltage, because of this corona discharge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://www.theseus.fi/bitstream/handle/10024/29401/Yli-Hannuksela_Juho.pdf?sequence=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;a thesis&lt;/a&gt;, about 1.5 meter spacing is used for 110 kV and 5.5 meter of spacing is used for 400 kV. We can assume 220 kV uses about 3.5 meters of spacing therefore. Plugging in these values, we get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;400 kV: 2 cm radius needed&lt;/li&gt;&#xA;&lt;li&gt;220 kV: 1 cm radius needed&lt;/li&gt;&#xA;&lt;li&gt;110 kV: 0.5 cm radius needed&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The thesis also gives in Appendix 1 typical aluminum cross sections:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Duck: 305 mm&lt;sup&gt;2&lt;/sup&gt;: 0.99 cm radius&lt;/li&gt;&#xA;&lt;li&gt;Finch: 565mm&lt;sup&gt;2&lt;/sup&gt;: 1.34 cm radius&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;We can probably be safe in saying that for 110 kV, it's not the corona discharge that limits the transmission line radius, but rather the resistive losses. The thesis gave no example of 0.5 cm radius conductor. So, we probably can bump up the 110 kV radius to 1.0 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have the information that the national grid operator has these transmission lines:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;400 kV: 4600 km&lt;/li&gt;&#xA;&lt;li&gt;220 kV: 2200 km&lt;/li&gt;&#xA;&lt;li&gt;110 kV: 7600 km&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The rest is 22500 km - 4600 km - 2200 km - 7600 km = 8100 km, which are the high voltage lines not operated by the national grid operator. We can probably be safe in saying that the lines not operated by the national grid operator are either 220 kV or 110 kV, so we get:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;400 kV: 4600 km @ 2 cm radius =&gt; 46.8 kilotons of aluminum used&lt;/li&gt;&#xA;&lt;li&gt;220 kV: 2200 km @ 1 cm radius =&gt; 5.6 kilotons of aluminum used&lt;/li&gt;&#xA;&lt;li&gt;110 kV: 7600 km @ 1 cm radius =&gt; 19.3 kilotons of aluminum used&lt;/li&gt;&#xA;&lt;li&gt;110 or 220 kV: 8100 km @ 1 cm radius =&gt; 20.6 kilotons of aluminum used&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, we can safely say the HV lines use about 92 kilotons of aluminum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How about MV and LV, then? Well, the amperage isn't probably reduced much from 110 kV, so&#xA;I assume they have conductor radius between 0.5 cm (242 kilotons of aluminum) and 1.0 cm (967 kilotons of aluminum). The thesis (page 25) says 1 mm2 cross section shouldn't carry more than 1 ampere, so a 30 kV medium voltage line would carry per phase 30 kV * 79 A = 2370 kW, or 30 kV * 314 A = 9420 kW, depending on the conductor radius (0.5 cm or 1.0 cm). This would give a power value between 7 MW - 28 MW, reasonable to power quite many homes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in summary, the transmission lines in this country are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;HV: 92 kilotons of aluminum&lt;/li&gt;&#xA;&lt;li&gt;MV and LV: between 242 and 967 kilotons of aluminum&lt;/li&gt;&#xA;&lt;li&gt;Total: between 334 and 1059 kilotons of aluminum (corresponding to between 61 and 193 kg per capita, which is between 2.4 and 7.7 years worth of aluminum production)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now we seem to be in the correct order of magnitude in the estimate.&lt;/p&gt;&#xA;" OwnerUserId="12836" LastActivityDate="2019-07-30T16:13:57.793" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29434" PostTypeId="2" ParentId="29428" CreationDate="2019-07-30T18:58:06.993" Score="4" Body="&lt;p&gt;Yes, as long as the plant is controllable, observable, and doesn't vary too much from the model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms of pole placement, a PID controller with a settable bandwidth on the derivative adds two poles, and has four settable parameters.  The system has four poles (two from the plant, two from the PID controller), so in theory, not only can you stabilize the system, you can put those poles wherever you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, you'd want to use a more robust design technique.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-07-30T18:58:06.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29435" PostTypeId="1" CreationDate="2019-07-31T10:08:23.917" Score="0" ViewCount="25" Body="&lt;p&gt;I am currently struggling with some lens equations. I hope this is the right place to ask such questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm trying to accomplish is to add a separate single lens to a camera setup and calculate the new focus plane (See the added image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dYRPc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dYRPc.jpg&quot; alt=&quot;Camera setup with added lens.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To calculate the new focus plane I am assuming the 16 mm Kowa lens (on the camera) to behave like a thin 16 mm lens. Using simple lens equations, I then calculate how adding the separate lens would shift the focal plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the results I get from these calculations do not correspond to what I see in my experimental setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? Does substituting the Kowa lens for a thin 16 mm lens in my calculations make that much of a difference? If so, how could I calculate this without knowing exactly what is inside of the Kowa lens?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Writing out my calculations here would make this a wall of text, so I left them out. If they are required though, I would be happy to add them.&lt;/p&gt;&#xA;" OwnerUserId="21306" LastActivityDate="2019-07-31T10:08:23.917" Title="How to calculate new focal plane position when adding a lens to a lens system?" Tags="|design|mathematics|optics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29436" PostTypeId="1" CreationDate="2019-07-31T10:30:53.587" Score="0" ViewCount="193" Body="&lt;p&gt;I am designing a new mobile robot platform and i need a generic solution for the robot in order to establish a proper mathematical modeling. My question is about calculation of the &lt;strong&gt;traction force&lt;/strong&gt; of the robot with respect to &lt;strong&gt;rotational speed&lt;/strong&gt; of the direct contact motor shafts which is mounted as the figure below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Szi6G.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Szi6G.jpg&quot; alt=&quot;Contact points of the vehicle, shafts and spherical wheel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a vehicle with &lt;strong&gt;3 contact points&lt;/strong&gt; with the ground. Two contact points are edges of steel cylindrical shafts with diameter &lt;strong&gt;d_c&lt;/strong&gt;. Other contact is provided by spherical wheel with diameter &lt;strong&gt;d_s&lt;/strong&gt; in order to provide static balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/okmUM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/okmUM.jpg&quot; alt=&quot;Related parameters are given on the diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cylindrical shafts are rotating with a high speed, &lt;strong&gt;n&lt;/strong&gt;. The angle between the shafts and the ground is &lt;strong&gt;a&lt;/strong&gt;. Coefficient of static friction is &lt;strong&gt;f&lt;/strong&gt;. Shaft is slipping on the ground because of high speed and low friction. The weight of the vehicle is &lt;strong&gt;w&lt;/strong&gt;. The distances between the contact points and the center line is &lt;strong&gt;b&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to calculate traction force &lt;strong&gt;T&lt;/strong&gt; with respect to rotational speed &lt;strong&gt;n&lt;/strong&gt; for a shaft mounted at a certain angle &lt;strong&gt;a&lt;/strong&gt; on a surface?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Given Parameters:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These parameters are just for example, i need a generic mathematical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d_c = 1 mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d_s = 10 mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b = 25 mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;n = 10000 rpm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a = 60 degrees&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f = 0.25&lt;/p&gt;&#xA;&#xA;&lt;p&gt;w = 0.1 kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Required:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;T(n, a) = ?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21307" LastActivityDate="2023-12-15T10:05:20.653" Title="How to calculate traction force w.r.t. rotational speed of a shaft mounted at a certain angle on a surface?" Tags="|mechanical-engineering|applied-mechanics|dynamics|torque|friction|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29437" PostTypeId="2" ParentId="29056" CreationDate="2019-07-31T10:46:15.870" Score="1" Body="&lt;p&gt;You can use &lt;strong&gt;Matlab Simulink.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think Simulink is the most satisfactory and easy way to build control systems with traditional block diagrams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mathworks.com/help/simulink/gs/simulink-block-diagrams.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mathworks.com/help/simulink/gs/simulink-block-diagrams.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some introductory block diagram models in Simulink:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ctms.engin.umich.edu/CTMS/index.php?example=Introduction&amp;amp;section=SimulinkControl&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ctms.engin.umich.edu/CTMS/index.php?example=Introduction&amp;amp;section=SimulinkControl&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example PID control block diagram with Matlab Simulink:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/382hh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/382hh.png&quot; alt=&quot;PID control block diagram with Matlab Simulink&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21307" LastActivityDate="2019-07-31T10:46:15.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29438" PostTypeId="2" ParentId="29423" CreationDate="2019-07-31T10:58:34.927" Score="2" Body="&lt;p&gt;Commercial display stands may give you an inspiration. You can simplify the idea for your project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example: &lt;a href=&quot;https://tr.aliexpress.com/item/32915896443.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://tr.aliexpress.com/item/32915896443.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b6J9i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b6J9i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21307" LastActivityDate="2019-07-31T10:58:34.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29439" PostTypeId="1" AcceptedAnswerId="29441" CreationDate="2019-07-31T12:20:32.483" Score="1" ViewCount="173" Body="&lt;p&gt;I am trying to determine the temperature of an underground concrete vault buried 10m underground. The vault has the following dimensions: 2640mm x 5995mm x 2640mm with a wall thickness of 250mm. The transformer has the following dimensions: 2083mm by 1702mm by 1257mm. It has heat losses of 2080W to 11000W. How can I determine the temperature inside the vault? How would adding mechanical ventilation change the temperature inside this vault? &lt;/p&gt;&#xA;" OwnerUserId="21308" LastActivityDate="2019-08-01T15:49:21.740" Title="How do I determine the temperature inside a transformer vault?" Tags="|mechanical-engineering|electrical-engineering|thermodynamics|heat-transfer|temperature|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29440" PostTypeId="2" ParentId="26654" CreationDate="2019-07-31T14:43:15.953" Score="1" Body="&lt;p&gt;This is a typical VWM (Virtual Work Method) problem in Mechanics. But it seems there are some missing.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Let's start with the energy equation:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;d/dt(PE + KE) = Net Power&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;There is no motion since it is in the static equilibrium. So, KE = 0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There is no external forces. So, Net power is also 0.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;PE = 0.5 * k_1 * s_1 - 0.5 * k_2 * s_2 + mgh&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;s_1: infinitesimal motion in the spring line 1&#xA;s_2: infinitesimal motion in the spring line 2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;!Missing part 1: Either mass of the beam or plane should be clarified!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's suppose there is no gravity or the system is established in a horizontal plane.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;d/dt(PE) = 0&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;d/dt(PE) = k_1 * s_1 + k_2 * s_2 = 0&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;So we need to relation with s_1, s_2 with r_1, r_2&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;At the end we convert your question into a geometry problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it seems that there is no correlation between these parameters. Something an extra angle parameter is also missing.&lt;/p&gt;&#xA;" OwnerDisplayName="user21307" LastEditorDisplayName="user21307" LastEditDate="2019-07-31T19:06:26.400" LastActivityDate="2019-07-31T19:06:26.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29441" PostTypeId="2" ParentId="29439" CreationDate="2019-07-31T16:05:25.060" Score="3" Body="&lt;h1&gt;Assumptions&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;Assume steady state and neglect all radiation terms. Assume the surroundings to the concrete wall is an infinite heat sink at a fixed (constant) temperature). Finally, neglect the heat transfer that occurs at the corners of the concrete box (set the internal and external areas of the box to be equal as far as heat transfer is concerned).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Formulations&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The energy balance from the transformer to the inside air of the concrete wall is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ P = h_i A_T (T_T - T_{ai}) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is the power supplied, &lt;span class=&quot;math-container&quot;&gt;$T_T, T_{ai}$&lt;/span&gt; are the transformer and internal air temperatures, &lt;span class=&quot;math-container&quot;&gt;$h_i$&lt;/span&gt; is the convection coefficient inside the box, and &lt;span class=&quot;math-container&quot;&gt;$A_T$&lt;/span&gt; is the area of the transformer. This equation says that power generated by the transformer is convected away to the inside air of the container. The convection coefficient &lt;span class=&quot;math-container&quot;&gt;$h_i$&lt;/span&gt; depends on how fast the air is moving (stagnate/free convection to forced convection). This gives you one equation with two unknowns (the temperatures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy balance from the air in the box to the concrete wall is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ P =  h_i A_T (T_T - T_{ai}) = h_i A_w (T_{ai} - T_{wi}) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, &lt;span class=&quot;math-container&quot;&gt;$T_{wi}$&lt;/span&gt; is the temperature of the internal wall and &lt;span class=&quot;math-container&quot;&gt;$A_w$&lt;/span&gt; is the area of the wall. This equation says that power generated by the transformer is convected away to the inside air of the container that is then convected to the box wall. It is one equation with two unknowns (the temperatures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy balance on the concrete wall without vents is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ k \frac{(T_{wi} - T_g)}{w} = h_i (T_{ai} - T_{wi}) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is the thermal conductivity of concrete, &lt;span class=&quot;math-container&quot;&gt;$T_g$&lt;/span&gt; is the (constant) ground temperature, and &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; is the wall thickness. This equation says that heat flow to the inside walls from the air is conducted away through the walls. It gives one more equation with the same two unknowns (the temperatures).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The energy balance on the concrete wall with vents is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ k (1 - f_v) \frac{(T_{wi} - T_g)}{w} + h_v f_v (T_{wi} - T_g) = h_i (T_T - T_{wi}) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this, &lt;span class=&quot;math-container&quot;&gt;$f_v$&lt;/span&gt; is the fractional area of the wall that contains vents and &lt;span class=&quot;math-container&quot;&gt;$h_v$&lt;/span&gt; is a pseudo coefficient for convection-type heat transfer through the vents. It also gives one more equation with the same two unknowns (the temperatures).&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Solution&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;In all cases without or with vents, you can combine the system to obtain two equations and two unknowns. The system can be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A starting point is to solve the flux equation without vents to find &lt;span class=&quot;math-container&quot;&gt;$T_{wi}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ P = k A_w \frac{(T_{wi} - T_g)}{w} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This says that the energy from the transformer goes through the walls by conduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use the value of &lt;span class=&quot;math-container&quot;&gt;$T_{wi}$&lt;/span&gt; and the equation &lt;span class=&quot;math-container&quot;&gt;$P = h_i A_w(T_{ai} - T_{wi})$&lt;/span&gt; to find &lt;span class=&quot;math-container&quot;&gt;$T_{ai}$&lt;/span&gt;. Continue with the heat transfer from transformer to air to find &lt;span class=&quot;math-container&quot;&gt;$T_T$&lt;/span&gt;. Finally, use these first guesses in the equation with vents to re-calculate the temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comparable problem is given &lt;a href=&quot;https://engineering.stackexchange.com/questions/29117/simple-thermal-analysis-for-sealed-aluminum-box-with-3-7-watts-of-heat-generatio/29279#29279&quot;&gt;at this link&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-08-01T15:49:21.740" LastActivityDate="2019-08-01T15:49:21.740" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29442" PostTypeId="2" ParentId="29436" CreationDate="2019-07-31T17:35:47.303" Score="0" Body="&lt;p&gt;You will have dynamic friction here, which is independent of the rpm of the rotating shaft, because it is slipping continuously. so you need to have the coefficient of dynamic friction, let's call it k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming each cylinder take 1/3 of the mass of your robot, it generates a traction of &lt;span class=&quot;math-container&quot;&gt;$ \ T=k*m/3  $&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-07-31T17:35:47.303" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29443" PostTypeId="2" ParentId="29327" CreationDate="2019-07-31T19:31:45.030" Score="0" Body="&lt;p&gt;&lt;strong&gt;Elevator&lt;/strong&gt; has a closed cab, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4Lw8X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4Lw8X.png&quot; alt=&quot;Elevator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Lift&lt;/strong&gt; has an open cab,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZtR9m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZtR9m.png&quot; alt=&quot;Lift&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Hoist&lt;/strong&gt; has no cab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ldzh7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ldzh7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21307" LastActivityDate="2019-07-31T19:31:45.030" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29444" PostTypeId="2" ParentId="19024" CreationDate="2019-07-31T20:09:09.647" Score="1" Body="&lt;p&gt;SPH is one of two different particle methods.  SPH has limitations as discussed above.  You may also want to look at Moving Particle Semi-Implicit method (MPS).  This method is better for incompressible fluids and is used in engines and gearboxes.  There is a case study and some videos on the website:  &lt;a href=&quot;https://enginsoftusa.com/Particleworks-CFD-Software.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://enginsoftusa.com/Particleworks-CFD-Software.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21312" LastActivityDate="2019-07-31T20:09:09.647" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29445" PostTypeId="1" AcceptedAnswerId="29447" CreationDate="2019-07-31T20:24:04.600" Score="0" ViewCount="307" Body="&lt;p&gt;I am having difficulty with two questions from a book I am studying about statics. I have seen the answers but it did not help me understand how the answers are derived. I want some advice on how to think about these types of problems and whether there are any tricks which help to solve them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1 (1st image);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small probe is gently forced against a circular surface with a vertical force F (the red vector). Determine the n- and t-components of this force as functions of the horizontal position S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gdjqU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gdjqU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2 (2nd image);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the x- and y-components of the tension T which is applied to point A of the bar OA. Neglect the effects of  the small pulley at B. Assume that R and alpha are known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p4k0Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p4k0Y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21313" LastActivityDate="2019-08-02T06:32:24.590" Title="Need help in deriving components of forces when they depend on lengths and angles. (2 questions)" Tags="|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29446" PostTypeId="1" AcceptedAnswerId="29458" CreationDate="2019-07-31T21:49:46.793" Score="1" ViewCount="238" Body="&lt;p&gt;I've been studying this topic for a couple of days now. I started by asking some questions regarding output Torque and gear ratios relations.&#xA;Now I came with the whole picture and some calculations that I been working on. My goal is to find out if I'm in the right direction, and also point-out relations between equations that I'm not sure of.&#xA;To make this more clear and clean, I divide the system into three sections; A, T and B.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZHtiM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZHtiM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In section A its the electric motor and worm/spur torque transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On another topic with a lot of help, I realise that the power provides by the motor is crucial to generate the output torque and angular velocity that your system requires. No matter the radius, ratios or N of teeth of the spur; if the power isn't enough, the gate won't move at the desired velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the problem in question: &#xA;My goal is to find out if this system will be able to move a gate of 600 kg with a velocity of 0.33 m/s, and how much time it will take to reach that velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The technical data of the motor is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P = 1/2 hp ; RPM = 1450 ; 220V/50Hz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SECTION A:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P = 0.5 hp = 373 (kg * m2 /s3) &#xA;Pin = Tin * win&lt;/p&gt;&#xA;&#xA;&lt;p&gt;win = 1450 rpm = 151.84 (rad/seg)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tin = 373 (kg * m2 /s3) / 151.84 (rad/seg) = 2.456 N*m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, of course, is in an ideal situation, without considering losses. (how can I add an estimation of electric motor losses?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worm/spur system has a gear ratio of 23:1. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the efficiency of the system of 80%, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tout = 23 * (0.8 * Pin) / win = 45.2 N*m &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, Tout also can be calculated by Tout = 0.8 * Pin / wout&lt;/p&gt;&#xA;&#xA;&lt;p&gt;angular velocity applied on the spur is:  wout = (0.8 * Pin) / Tout&lt;/p&gt;&#xA;&#xA;&lt;p&gt;wout = 0.8 * 373 (kg * m2 /s3) / 45.2 N*m = 6.6 rad/sec --&gt; 63 rpm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To consider real circumstances, its the moment of inertia has something to do within this case? I mean, is this Tout enough to accelerate the spur to 63 rpm??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SECTION T:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this section, I'm wondering what happens with force transmitted in the distance &quot;d&quot; to the pinion. I'm not sure if I need to consider losses in this section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SECTION B:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I have a lot of question about what's happening.&#xA;The Pinion has 17 teeth and is module 4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From gear design, I understand that module, and axial pitch determines the Lead and lead angle of the gear. This has a direct relation with the surface contact (involute) in the system and frictional forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the torque Tout the same as calculated for the worm/spur system? What about the radius of the pinion? If I increase the radius, the Torque will increase as well?&#xA;The radius of the pinion is = 0.025 (m)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tout = F * r &lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = 45.2 N*m / 0.025 m = 1808 N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that the force applied to the rack is 1808 N? Is this force enough to accelerate until 0.33 m/s?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = m * a = 600 kg * 9.8 m/s2 = 5880 N ; here I need to consider a friction factor according to the wheels and ground. (0.1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = 5880 N * 0.1 = 588 N; this means that I need to generate 588 N force to move the gate, but what about the acceleration? if the gate has 4 m, how much time will take to reach the 0.33 m/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need a last push to understand this system entirely.&lt;/p&gt;&#xA;" OwnerUserId="14265" LastActivityDate="2019-08-01T22:49:39.047" Title="Reduction System for Gate-opener motor" Tags="|mechanical-engineering|gears|torque|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29447" PostTypeId="2" ParentId="29445" CreationDate="2019-08-01T00:10:34.573" Score="0" Body="&lt;p&gt;In fact, these are just geometry problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The answer to the first question:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;F_n &amp;amp;= F\dfrac{(r^2 - s^2)^(1/2)}{r} \\&#xA;F_t &amp;amp;= F\dfrac{s}{r}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Figure shows the details of the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ATgJd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ATgJd.png&quot; alt=&quot;Solution of the first question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you think that way, you can solve the second question.&lt;/p&gt;&#xA;" OwnerDisplayName="user21307" LastEditorDisplayName="user21307" LastEditDate="2019-08-02T06:32:24.590" LastActivityDate="2019-08-02T06:32:24.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29448" PostTypeId="1" AcceptedAnswerId="29466" CreationDate="2019-08-01T01:54:50.350" Score="1" ViewCount="987" Body="&lt;p&gt;It appears to be a waste of electricity to keep a bank of refrigerators wide open with no doors or covers whatsoever on them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They seem to be just blowing cold air out into the supermarket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are not turned off at night when the store closes.  Even the &quot;green&quot; Whole Foods-type places do this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But is it really that inefficient to operate them that way? &#xA;  What is the relative efficiency between a refrigerator with a door and one without a door in a supermarket?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://img-s-msn-com.akamaized.net/tenant/amp/entityid/CCiQxr.img?h=416&amp;amp;w=624&amp;amp;m=6&amp;amp;q=60&amp;amp;u=t&amp;amp;o=f&amp;amp;l=f&amp;amp;x=750&amp;amp;y=399&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://img-s-msn-com.akamaized.net/tenant/amp/entityid/CCiQxr.img?h=416&amp;amp;w=624&amp;amp;m=6&amp;amp;q=60&amp;amp;u=t&amp;amp;o=f&amp;amp;l=f&amp;amp;x=750&amp;amp;y=399&quot; alt=&quot;Fridge without door&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a2zso.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a2zso.jpg&quot; alt=&quot;Fridge with door&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21318" LastEditorUserId="21318" LastEditDate="2019-08-02T18:25:54.130" LastActivityDate="2019-08-03T12:59:30.707" Title="Relative inefficiency of doorless fridges at supermarkets" Tags="|electrical-engineering|refrigeration|" AnswerCount="3" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29449" PostTypeId="1" CreationDate="2019-08-01T10:27:25.690" Score="0" ViewCount="318" Body="&lt;p&gt;I want to express the inertia tensor of a rotating rod (total length &lt;span class=&quot;math-container&quot;&gt;$L_1$&lt;/span&gt;) to use it in Lagrange mechanics for expressing the kinetic energy associated to the rotation with angular velocity &lt;span class=&quot;math-container&quot;&gt;$\dot{\theta}_1$&lt;/span&gt;. I know the moment of inertia regarding the center of mass &lt;span class=&quot;math-container&quot;&gt;$m_1$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$1/12\cdot m_1 L_1^2$&lt;/span&gt; and with the central axes theorem I obtain&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;J_{1yy} = J_{1zz} = 1/12\cdot m_1 L_1^2 + m_1 l_{c,1}^2&#xA;$$&lt;/span&gt;&#xA;So I can express the the kinetic energy as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;1/2 \cdot \dot{\theta}_1^2 (1/12\cdot m_1 L_1^2 + m_1 l_{c,1}^2)&#xA;$$&lt;/span&gt;&#xA;Is that right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotation is constrained to the horizontal plane around the origin (&lt;span class=&quot;math-container&quot;&gt;$\vec{x}_i = \vec{0}$&lt;/span&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gOqzz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gOqzz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20870" LastActivityDate="2020-08-25T19:02:20.343" Title="Mass inertia tensor for rod non-symmetric regarding coordinate system origin" Tags="|mechanical-engineering|applied-mechanics|inertia-matrix|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29450" PostTypeId="2" ParentId="29427" CreationDate="2019-08-01T10:51:31.917" Score="1" Body="&lt;p&gt;At a high level if you take an NPN BJT transistor for example, P region is the base of the transistor. The transistors P region is much lightly doped than the N region. Since the N region is heavily doped the there is more current flowing from N to P region. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vuOf2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vuOf2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the attach image shows the magnitude of the flow of current or electron by the size of the arrows. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally electrons are injected from the p-type base to the emitter which are called minority carriers. As indicated in the figure below in steady state the excess minority carrier concentration will have a straight line profile. The concentration of electrons will be lowest or zero at the collector side and highest at the emitter side. This is denoted by &lt;span class=&quot;math-container&quot;&gt;$n_p(0)$&lt;/span&gt; in the below figure.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D6LVE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D6LVE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The positive collector voltage &lt;span class=&quot;math-container&quot;&gt;$V_{CB}$&lt;/span&gt; is the cause for zero concentration at the collector side of the base. As a result the electrons are swept away across from the Collector Base depletion region. &lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-08-08T10:43:12.820" LastActivityDate="2019-08-08T10:43:12.820" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29451" PostTypeId="2" ParentId="29446" CreationDate="2019-08-01T11:37:48.683" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;This, of course, is in an ideal situation, without considering losses. (how can I add an estimation of electric motor losses?)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Firstly you need to specify the motor output torque according to the motor output speed. In order to do that you need to calculate the performance of AC Motor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Performance of the AC motor is variable. It depends on &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;full load speed (rpm),&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;full load torque (N*m),&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;phase type,&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;current output speed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In order to determine the performance of an AC motor you usually need a torque-speed curve diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;synchronous speed: 1450 rpm&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;full load torque: 2.456 N*m&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;220V/50 Hz is a typical single phase AC motor. But you also clarify the type of this single phase AC motor according to your design specifications. (These types are: Shaded pole, permanent-split capacitor, split-phase and capacitor-start). Each is unique in its physical construction and provides different starting and running characteristics. This is an example diagram which indicates torque-speed curves.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1ctQU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ctQU.png&quot; alt=&quot;Torque-Speed curves of different types of single phase AC motors&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to get full load torque from the motor, roughly you run the motor 85% of its maximum speed. This is approximately 1200 rpm as motor output speed.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;To consider real circumstances, its the moment of inertia has something to do within this case? I mean, is this Tout enough to accelerate the spur to 63 rpm??&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The important thing in order to run a mechanism at the first start not the inertia. It is static friction. For example: If you have zero-close static friction and high inertia for a machine, you need to wait longer in order to achieve required speed. In opposite, if you have high static friction and low inertia, you need to wait longer in order to first start but after that it starts to accelerate quickly. So, if you think your motor is weak for the first motion, using a capacitor may be a good idea. But in my opinion, it doesn't seem you face such a problem.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;In this section, I'm wondering what happens with force transmitted in the distance &quot;d&quot; to the pinion. I'm not sure if I need to consider losses in this section.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In this section T, you need to determine the specifications of bearings which holds the shaft. Only loss comes from bearings. But this loss quite negligible with respect to the gear meshes. Nevertheless, if you want to add this loss accurately, you can look at this website: &lt;a href=&quot;https://www.amroll.com/friction-frequency-factors.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amroll.com/friction-frequency-factors.html&lt;/a&gt;. The safest way in here is to consult the manufacturers for your design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other than these, your calculations seem correct.&lt;/p&gt;&#xA;" OwnerDisplayName="user21307" LastEditorDisplayName="user21307" LastEditDate="2019-08-01T22:49:39.047" LastActivityDate="2019-08-01T22:49:39.047" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29452" PostTypeId="1" CreationDate="2019-08-01T12:14:38.677" Score="1" ViewCount="84" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PrNpZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PrNpZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hello, so I'm looking for some help with part (a) of the question below. I've never seen an example like it before and I'm not sure where to start. How do I work would which is the missing member for the truss below? Any help would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="20365" LastActivityDate="2019-08-04T04:08:05.643" Title="Plane Truss- Missing Member" Tags="|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29453" PostTypeId="1" CreationDate="2019-08-01T12:42:52.683" Score="1" ViewCount="233" Body="&lt;p&gt;I search and read about Ferranti effect online, but none seems to be clear enough for me to understand how it actually happens. I must say, I do not have much prior knowledge about inductive/capacitive nature of the transmission lines - thus, it makes it harder to understand if the voltage drop is about these, or if it happens through the transmission line because of the no LOAD at the recieving end. I will appreciate if someone can explain &quot;ferranti effect&quot; giving enough, detailed explanations of the other factors as well (eg. current of the transmission line, etc.)&lt;/p&gt;&#xA;" OwnerUserId="21321" LastActivityDate="2021-07-24T04:04:40.487" Title="How does ferranti effect occur?" Tags="|electrical-engineering|power-transmission|transmission|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29454" PostTypeId="2" ParentId="29436" CreationDate="2019-08-01T14:26:09.497" Score="0" Body="&lt;p&gt;No, not even angle. If you have seen drag racing, they have high traction expensive tires. But when before the race they burn the tires to creat sticky hot surface, while creating clouds of smoke by revving the engine the car doesn't move forward. Because they have forced the tires to slip and skid. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-01T14:26:09.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29455" PostTypeId="2" ParentId="29449" CreationDate="2019-08-01T16:09:06.983" Score="1" Body="&lt;p&gt;Yes, you are right. Using parallel axis theorem, you add the moment of inertia by &lt;span class=&quot;math-container&quot;&gt;$m*r^2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes they use&lt;span class=&quot;math-container&quot;&gt;$ \ \omega^2 \ for \ \dot \theta^2  $&lt;/span&gt; as angular velocity. But it is in a context where the need to convert quantities to linear velocity$.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-01T16:09:06.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29456" PostTypeId="2" ParentId="29453" CreationDate="2019-08-01T16:36:55.003" Score="1" Body="&lt;p&gt;Here is my take on the ferranti effect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine a sea wall facing the ocean. Waves of height 1 meter are coming in toward shore and one of them strikes the wall. How high does the water splash vertically upward when the wave hits? If the wave splashes only one meter up (same height as the wave originally) then the wave is behaving as if it hadn't hit the wall at all, but we know it did, so the splash height has to be greater than the wave height. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wall represents an impedance discontinuity to the waves hitting it, and in the process of being reflected off the wall, the peak height of the wave exceeds its original amplitude. Reflections that occur off of impedance discontinuities in electrical systems will exhibit similar behavior, especially if the discontinuity is not lossy.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-08-01T16:36:55.003" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29457" PostTypeId="1" AcceptedAnswerId="39509" CreationDate="2019-08-01T16:46:17.797" Score="0" ViewCount="61" Body="&lt;p&gt;I'm dispensing around 1uL of volume with 1% accuracy. Although I'm confident in my pump, I'd like to be able to routinely confirm that I'm getting the exact volume I'm requesting every time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I reliably measure volume at this small scale? I've tried a precision scale, but the evaporation rate for the fluid is quick and I'm losing non-negligible volumes while waiting for the scale to settle. &lt;/p&gt;&#xA;" OwnerUserId="12913" LastActivityDate="2021-01-05T13:26:11.950" Title="How can I confirm nL Volume Measurement for Pump and Sensor Calibration" Tags="|chemical-engineering|sensors|pumps|flow-control|microfluidics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29458" PostTypeId="2" ParentId="29446" CreationDate="2019-08-01T18:01:26.967" Score="2" Body="&lt;p&gt;O.K.&#xA;After your &lt;a href=&quot;https://engineering.stackexchange.com/questions/29360/torque-and-rpm-calc&quot;&gt;last question&lt;/a&gt; we still have a couple of things to figure out. let's dig into it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly to my last answer, let's start our analysis at the load section (the gate). Its free body diagram may look as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q8Khy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q8Khy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pinion, from the rack &amp;amp; pinion mechanism generates the pushing force. As you correctly noticed, there is a friction resisting force. Last time we decided to conservatively define the friction coefficient as 0.1. It means that the total friction force would be 600 [kg] times 9.8 [m/sec^2] * 0.1 = 588 [N]. The net force, i.e. the difference between the pinion pushing force to the friction force is what drives the gate and causes it to accelerate. This difference divided by the mass results in the gate acceleration.&#xA;For instance, let's say you would like your gate to reach its final velocity (0.33 m/sec) after 4 seconds. Assuming constant acceleration, it would end up in 0.33 [m/sec] over 4 [sec] = 0.0825 [m/sec^2].&#xA;The net force for acceleration the gate should be 0.0825 [m/sec^2] times 600 [kg] = 49.5 [N].&#xA;This means that the pinion force is 49.5 [N] + 588 [N] = 637.5 [N].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Got it? Don't confuse between M&lt;em&gt;g which is the weight of the gate to M&lt;/em&gt;a which is the force needed to drive at a acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the pinion radius - we already discussed it last time. The rack &amp;amp; pinion mechanism does not multiply the input torque by a factor (like the worm gear), it just converts the torque into a force. This conversion factor is the pinion radius. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the inertia - It should be consider only when dealing with acceleration. Your constant speed calculation shall not take it into account.&#xA;I guess the pinion inertia would be negligible relatively to the reflected gate inertia. Try to compare between the two after you get all the basic ideas.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2019-08-01T18:01:26.967" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29459" PostTypeId="2" ParentId="29453" CreationDate="2019-08-01T18:22:33.920" Score="1" Body="&lt;p&gt;If you increased the frequency of the transmitted wave, eventually you would reach a frequency where the transmission line was resonating, with a node at one end and an antinode at the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you plot the transfer function of a resonant system in the frequency domain, the response gradually increases as the frequency changes from 0 to the resonance. So even at frequencies a long way from resonance (e.g. 50 or 60 Hz for a line a few km long) there is &lt;em&gt;some&lt;/em&gt; increase in amplitude compared with a non-resonant system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resonant frequency depends on the length of the line, and for a driving frequency a long way from resonance the shape of the transfer function explains the fact that the amplitude increase is proportional to the line length squared, and to the frequency squared.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-08-01T18:22:33.920" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29460" PostTypeId="1" CreationDate="2019-08-01T18:48:12.670" Score="0" ViewCount="67" Body="&lt;p&gt;I am trying to model an object in which there are two rings who are attached to each other and can rotate independently along the Y axis (see attached image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object needs to be machined and not too costly to produce, and solid (the rings won't detach if pulled onto).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have little knowledge in mechanical engineering but I'm working with an engineer and I want to give them the right pointers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QMpA6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QMpA6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20796" LastActivityDate="2019-08-01T18:48:12.670" Title="How can I model two attached rotating rings?" Tags="|mechanical-engineering|design|machining|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29461" PostTypeId="1" CreationDate="2019-08-01T19:26:44.730" Score="1" ViewCount="61" Body="&lt;p&gt;&lt;strong&gt;Background Information&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently looking to buy a house in the Southampton area in the UK and have noted there are quite a few houses which use a concrete panelling construction technique see example image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NxZoF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NxZoF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some are advertised as Reema construction which can be difficult / impossible to obtain a mortgage on.  I have read up on Reema construction and this has currently put me off any type of concrete panel type buildings currently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the major pros and cons of concrete panelling construction, other than the obvious it's cheap and quick to build?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a register in the UK to obtain information about the building type and construction which is accessible to the general public?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks inadvance,&#xA;Suze&lt;/p&gt;&#xA;" OwnerUserId="17526" LastActivityDate="2019-08-01T19:26:44.730" Title="House construction using concrete panels" Tags="|concrete|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29462" PostTypeId="1" CreationDate="2019-08-01T19:30:48.507" Score="0" ViewCount="881" Body="&lt;p&gt;Force due to drag at low velocities, is equal to some constant times negative velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_{d}=-c_{1}V$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The viscous damping coefficient equals decay constant divided by 2 times mass&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\gamma = \frac{c_{2}}{2m}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is &lt;span class=&quot;math-container&quot;&gt;$c_{1}$&lt;/span&gt; the same as &lt;span class=&quot;math-container&quot;&gt;$c_{2}$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is the drag force related to the viscous damping coefficient, what equation is there to relate them?&#xA;I think the relationship is linear but I'm not certain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For context, this is for a mass-spring system inside a beaker of water being damped by the friction of the water.&lt;/p&gt;&#xA;" OwnerUserId="21310" LastEditorUserId="1832" LastEditDate="2019-08-02T01:05:25.210" LastActivityDate="2024-02-07T18:05:25.730" Title="How to calculate the viscous damping coefficient given drag force" Tags="|mechanical-engineering|fluid-mechanics|drag|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29463" PostTypeId="2" ParentId="29462" CreationDate="2019-08-01T20:38:54.803" Score="0" Body="&lt;p&gt;What you have is not actually fully submersed drag equation, or else it would be correct to handle it as damping.&lt;/p&gt;&#xA;&lt;p&gt;The drag force on a submersed streamlined object is:&lt;span class=&quot;math-container&quot;&gt;$$ F_D=\frac{1}{2} \rho\cdot u^2C_DA  $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_{D} = $&lt;/span&gt;  the drag force&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \rho=$&lt;/span&gt; mass density of the fluid&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ u= $&lt;/span&gt; the flow velocity&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$C_D= $&lt;/span&gt; the drag coefficient (function of the Reynold's number)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A=$&lt;/span&gt;  is the crosssectional  area&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="19906" LastEditDate="2020-08-26T07:58:50.270" LastActivityDate="2020-08-26T07:58:50.270" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29464" PostTypeId="2" ParentId="29448" CreationDate="2019-08-01T21:10:55.190" Score="0" Body="&lt;p&gt;Most of those units as per the top image do have some type of insulating cover for the time the shops etc are closed. This is part of the closing process, and opening the next day...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lower image shows doors, which are, in many cases now, double glazed for efficiency as well.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-01T21:10:55.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29465" PostTypeId="2" ParentId="29448" CreationDate="2019-08-01T21:11:36.350" Score="0" Body="&lt;p&gt;You are right, refrigerators with door have approximately &lt;strong&gt;40% higher energy efficiency&lt;/strong&gt; on average. Doorless refrigerators are preferred when it is required extra convenience to consumers. For this type of refrigerators, in night, automatic or manual curtain (cover etc.) applications are already available for energy-saving.&lt;/p&gt;&#xA;" OwnerDisplayName="user21307" LastEditorDisplayName="user21307" LastEditDate="2019-08-01T21:18:27.040" LastActivityDate="2019-08-01T21:18:27.040" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29466" PostTypeId="2" ParentId="29448" CreationDate="2019-08-02T01:07:06.797" Score="3" Body="&lt;p&gt;The doorless models (this was the photo in the question when I answered it) aren't as inefficient as they appear to be. &lt;a href=&quot;https://i.stack.imgur.com/2wYGT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2wYGT.jpg&quot; alt=&quot;Original photo.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whenever someone opens a regular fridge door, its cold air pours out onto the floor and warm air replaces it inside the fridge.  If the doors are opened frequently, or held open for significant time, there is a lot of waste.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, the doorless models are designed to have a laminar flow of cold air from top to bottom.  Most of the cold air emitted at the top is sucked back in at the bottom, with far less mixing with the surrounding warm air than one might think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for products where temperature control isn't critical, and where many people are likely to take the product or to spend a long time looking and deciding, the doorless models are a good choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(They're also good from a marketing perspective, as they provide an excellent view of the product and don't have doors that fog up, but that's not the question.)&lt;/p&gt;&#xA;" OwnerUserId="19930" LastEditorUserId="19930" LastEditDate="2019-08-03T12:59:30.707" LastActivityDate="2019-08-03T12:59:30.707" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29468" PostTypeId="1" CreationDate="2019-08-02T08:26:33.947" Score="0" ViewCount="3197" Body="&lt;p&gt;I have a required clamping force for fastening two pieces of metal together, and am trying to work out if an m4 bolt will be sufficient. The pieces will be fastened without a nut, just a hole through one piece and a threaded hole in the aluminium block it needs clamping to.&lt;br&gt;&#xA;The clamping force that the fastener must be able to withstand is 3400N, so I have found that an m4 bolt should be suitable at grade 8.8 or higher (with an allowance for 75% of proof strength) &lt;a href=&quot;https://www.engineeringtoolbox.com/metric-bolts-minimum-ultimate-tensile-proof-loads-d_2026.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineeringtoolbox.com/metric-bolts-minimum-ultimate-tensile-proof-loads-d_2026.html&lt;/a&gt;.&lt;br&gt;&#xA;However, how can I verify that the thread in the aluminium plate will be able to withstand this force, or does the size of the hole not effect this (eg. if it can withstand it for an m6 bolt it will be fine with an m4)?&lt;br&gt;&#xA;Diagram to show hole placement and depth in aluminium.&lt;a href=&quot;https://i.stack.imgur.com/1iyPZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1iyPZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10180" LastEditorUserId="10180" LastEditDate="2019-08-02T09:57:30.600" LastActivityDate="2023-12-09T21:03:50.553" Title="Proof load for screwing into aluminium plate" Tags="|strength|bolting|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29469" PostTypeId="1" CreationDate="2019-08-02T09:07:39.723" Score="0" ViewCount="749" Body="&lt;p&gt;I'm studying the possibility to read an engine RPM by reading the vibrations of it. Basically an engine is &quot;oscillating&quot; on a car chassis on all 3 axes (X, Y, Z) but mostly on the Y one. When ignition is produced on a cylinder, then there should be like a &quot;spike&quot;. Considering for a 4 stroke engine that the crankshaft has 2 revolutions per ignition, if I can &quot;read&quot; ignition then I can do the math and get RPM.&#xA;Because my expertise in electronics is quite limited I ask here for an advice of how to approach this. Which accelerometer to use? What is range and sensitivity / resolution that I should look for?&#xA;Any advice / hint is really appreciated &#xA;Thank you in advance &lt;/p&gt;&#xA;" OwnerUserId="21328" LastEditorUserId="21328" LastEditDate="2019-08-02T09:13:50.367" LastActivityDate="2024-01-18T03:07:40.943" Title="Get engine rpm using accelerometer" Tags="|electrical-engineering|motors|automotive-engineering|" AnswerCount="2" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="29470" PostTypeId="2" ParentId="29468" CreationDate="2019-08-02T12:23:30.213" Score="0" Body="&lt;p&gt;You did not give any information regarding the choosing strategy of the bolt. Thus, we will have to assume you chose it correctly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are specifically asking for is the &quot;pull-out strength&quot; of the thread.&#xA;In case your bolt is quite strong (let's say, made of 180 ksi alloy steel) relatively to the plate material, it is indeed important to verify that the thread would not shear, or pulled out.&#xA;This pull-out force calculated as follows: the shear strength of the material X pi X thread diameter x thread depth x thread efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a simple one: pi * d is the perimeter of the thread, multiply it be thread depth will give you the total shear area. This area times the shear strength of the material (aluminum in your case) gives you the total shear force needed to pull this thread up.&#xA;Of course, all of this area works together against the bolt only theoretically. Practically, only a part of it actually works. This is why a thread efficient is added to the formula. Try 1/3 if you want to be on the very safe side. However, you may take a glance over &lt;a href=&quot;http://crosstools.com.au/helicoil-type-inserts-technical-study-of-strength-and-pullout.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; - by which a factor of 2/3 was found to be quite realistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as the clamping force is lower than the pull out force, your aluminum thread will survive.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2019-08-02T12:23:30.213" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29471" PostTypeId="2" ParentId="11962" CreationDate="2019-08-02T16:00:48.413" Score="0" Body="&lt;p&gt;The article below refutes theNamesCross' short answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.thetruthaboutguns.com/the-truth-about-barrel-length-muzzle-velocity-and-accuracy/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.thetruthaboutguns.com/the-truth-about-barrel-length-muzzle-velocity-and-accuracy/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at the charts and you will see that bullet quality probably has more of an effect on accuracy than barrel length.  But, it can also be seen that the shorted the barrel gets the greater the accuracy/repeatability of the shots.  This is due to shorter length barrels has less &quot;barrel whip&quot; than longer ones.&lt;/p&gt;&#xA;" OwnerUserId="21331" LastActivityDate="2019-08-02T16:00:48.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29472" PostTypeId="1" CreationDate="2019-08-02T23:01:33.163" Score="1" ViewCount="15" Body="&lt;p&gt;If we have the following forward converter:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Mujwp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Mujwp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when &lt;span class=&quot;math-container&quot;&gt;$Q_1$&lt;/span&gt; is on we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/koWLv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/koWLv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is it that the Diodes &lt;span class=&quot;math-container&quot;&gt;$D_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D_3$&lt;/span&gt; are off?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly when &lt;span class=&quot;math-container&quot;&gt;$Q_1$&lt;/span&gt; is off why is Diode &lt;span class=&quot;math-container&quot;&gt;$D_2$&lt;/span&gt; off?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0lElo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0lElo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21245" LastEditorUserId="7083" LastEditDate="2019-08-07T20:33:16.493" LastActivityDate="2019-08-07T20:33:16.493" Title="Forward converter currrent flow" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29473" PostTypeId="1" CreationDate="2019-08-03T03:23:25.547" Score="-2" ViewCount="44" Body="&lt;p&gt;So this is kind of like a Stirling engine. Would this work?&lt;/p&gt;&#xA;&lt;p&gt;A box that has a crankshaft, and two connecting rods that is under a vacuum (with no air inside of it.) The connecting rods are facing opposite each other and attached to the crankshaft on one end and then to a diaphragm on the other end which is part of the side of the box. The connecting rods are oriented such that when one connecting rod is fully extended it pushes the diaphragm outwards whilst the other connecting rod pulls its diaphragm it’s connected to inwards towards the inside of the box. The flexibility of the diaphragms and the outside atmospheric pressure and vacuum on the inside causes the reciprocating between the two. I am sure a better material could be used but to visualize the concept imagine the diaphragms are both made of rubber.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;I am not trying to make a perpetual motion machine&lt;/strong&gt; and&#xA;I know atmospheric pressure difference wouldn’t provide a ton of power in this setup.&lt;/p&gt;&#xA;" OwnerUserId="13921" LastEditorUserId="3505" LastEditDate="2021-12-28T21:58:21.480" LastActivityDate="2021-12-28T22:11:03.520" Title="Atmospheric motor concept plausible?" Tags="|motors|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29474" PostTypeId="2" ParentId="29473" CreationDate="2019-08-03T08:43:29.760" Score="1" Body="&lt;p&gt;It would stop with the crankshaft in the middle position: connecting rods angled sideways most, meaning distance between diaphragms shortest - least volume of vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's nothing providing work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's reduce the complexity to show how it does nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, remove the crankshaft, just connect the diaphragms with a single connecting shaft. Should work even better, not bending the connecting shafts sideways, right? No way to transfer work outside, but by itself, should move? If we want to limit the range of motion, we can add a spring at one end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next replace diaphragms with piston heads. An equivalent function mechanical component. You have vacuum between the pistons, air outside, the piston heads connected by a shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, since they are connected to each other rigidly, but may leak air through gaps in the sides, let's replace the shaft with a pipe the diameter of the pistons, protecting our precious vacuum inside, the pipe sliding against the walls of full metal chassis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we want to expose more surface or the piston to air, so we reduce the size of the chassis, leaving more of the piston exposed... until we've removed the entire chassis, the pressure in the middle of the piston equalized between top and bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you're left with a plain vacuum tank hanging on a spring. Why would it move again?&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2019-08-03T08:43:29.760" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29475" PostTypeId="1" CreationDate="2019-08-03T09:33:31.123" Score="0" ViewCount="158" Body="&lt;p&gt;I am sure you all remember those prehistoric computer mouses that had a ball under it. I'm looking for any points and tips on how to create a ball and socket wheel that has an almost frictionless connection with its casing and allows for no resistance in any direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To clarify, this wheel should be able to move essentially with a wide range of motion in an infinite amount of directions. I am hoping to use some rubber ball that would give an ideal friction against some smooth surfaces and the base case of this is that it will not move unless a force acts on it (unless it is on an angled plane of course).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know specifically if there are any technologies out there I can use for reference and if this concept is plausible, especially in simple mechanics, meaning no electronics, no moving parts or other fancy additions. Thank you for your help guys.&lt;/p&gt;&#xA;" OwnerUserId="21334" LastActivityDate="2019-08-03T10:18:49.627" Title="Ball and Socket Wheel Concept" Tags="|mechanical-engineering|structural-analysis|applied-mechanics|dynamics|prototyping|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29476" PostTypeId="2" ParentId="29475" CreationDate="2019-08-03T09:43:53.990" Score="1" Body="&lt;p&gt;You looking for an element in a ball table. Search for &lt;em&gt;ball table conveyor&lt;/em&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Jq65V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jq65V.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A castor is almost the same, but cheaper.&lt;/p&gt;&#xA;" OwnerUserId="16544" LastEditorUserId="16544" LastEditDate="2019-08-03T10:18:49.627" LastActivityDate="2019-08-03T10:18:49.627" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29477" PostTypeId="2" ParentId="29384" CreationDate="2019-08-03T11:37:53.163" Score="0" Body="&lt;p&gt;Thanks to @am304 who has not yet posted a full answer - but answered in a comment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the American Chemical Society Journal &lt;a href=&quot;https://www.acs.org/content/acs/en/pressroom/presspacs/2015/acs-presspac-april-29-2015/artificial-photosynthesis-could-help-make-fuels-plastics-and-medicine.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;we read&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The groups developed a stand-alone, nanowire array that captures light and with the help of bacteria, converts carbon dioxide into acetate. The bacteria directly interact with light-absorbing materials, which the researchers say is the first example of “microbial photoelectrosynthesis.” Another kind of bacteria then transforms the acetate into chemical precursors that can be used to make a wide range of everyday products from antibiotics to paints.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="14103" LastActivityDate="2019-08-03T11:37:53.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29478" PostTypeId="1" AcceptedAnswerId="29481" CreationDate="2019-08-03T15:09:14.393" Score="0" ViewCount="235" Body="&lt;p&gt;A friend and I were in a discussion and can't seem to find the answer to the question above. I said it was possible and he said it isn't and now we are both confused so could anyone help here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also would it be possible to spin carrier and sun gear while holding ring gear stationary?&lt;/p&gt;&#xA;" OwnerUserId="21049" LastEditorUserId="21049" LastEditDate="2019-08-03T15:22:01.817" LastActivityDate="2019-08-03T20:03:23.003" Title="In a planetary gearset is it possible to move both carrier and ring gear at the input speed while holding sun gear stationary?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29479" PostTypeId="1" CreationDate="2019-08-03T19:42:34.083" Score="0" ViewCount="47" Body="&lt;p&gt;It seems that measuring the modulus of elasticity of a material always involves the strength of materials or linear elasticity approach, where they make some experimental measurements for variables such as applied force, strain, deflection/deformation, etc... and then use some linear elasticity solver to back out the strain value.&lt;/p&gt;&#xA;&lt;p&gt;Are there any experimental devices that actually measure the modulus directly?&lt;/p&gt;&#xA;&lt;h1&gt;Edit 1:&lt;/h1&gt;&#xA;&lt;p&gt;I'm mostly interested in orthotropic materials. It's not very intuitive to me how Young's modulus is measured in orthotropic materials.&lt;/p&gt;&#xA;" OwnerUserId="19770" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-08-05T03:27:17.623" Title="Are there ways to measure modulus of elasticity with an experimental apparatus?" Tags="|elastic-modulus|solid-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29480" PostTypeId="1" CreationDate="2019-08-03T19:57:18.663" Score="1" ViewCount="79" Body="&lt;p&gt;I have a laser where I can adjust the focal length, however it has a large spot size. I am trying to shrink that down significantly with a microscope objective, how can I go about doing that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I put the laser out of focus, capture the light through the objective, then ensure the objective is at the right height and not out of focus? &lt;/p&gt;&#xA;" OwnerUserId="15332" LastActivityDate="2019-08-03T19:57:18.663" Title="Shrink a laser spot size" Tags="|optics|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29481" PostTypeId="2" ParentId="29478" CreationDate="2019-08-03T20:03:23.003" Score="2" Body="&lt;p&gt;No, if you hold the sun gear the carrier will turn at slower speed than the ring gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could have a setting where the ring gear is turning at the same speed as the input speed but then the carrier gear will turn slower or else it will break, because of the fact that is being spinned by the ring gear around the sun gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as to the second part of you question, again if you hold the ring gear the sun gear and carrier will spin at different speeds.&#xA;hear is a youtube clip, &lt;a href=&quot;https://www.bing.com/videos/search?q=planetry%20gear%20animation&amp;amp;docid=608052271535164301&amp;amp;mid=7CADA6E2BE3B0EB5CBDB7CADA6E2BE3B0EB5CBDB&amp;amp;view=detail&amp;amp;FORM=VIREHT&quot; rel=&quot;nofollow noreferrer&quot;&gt;planetary gear&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-03T20:03:23.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29485" PostTypeId="2" ParentId="29452" CreationDate="2019-08-04T04:08:05.643" Score="4" Body="&lt;p&gt;The first thing to note is the fact that, without some sort of bracing within BCEF, the structure is unstable. As-is, that square is completely flexible, and can flatten completely: CE can move down-and-to-the-right, transforming the square into a parallelogram and finally into a flat pancake, with C moving down to F and E moving 2&amp;nbsp;m below F (so BC becomes the same as BF, basically).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we will indeed need some bracing there. Part (a) of the question just asks whether that bracing should be placed at BE or CF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that the only &quot;problem&quot; with the structure is that it lacks any bracing in BCEF, theoretically there's no difference between either choice, since they both serve to brace that part of the structure. So you're just deciding what's the best choice between two valid options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two ways to think about this: what's the best layout for this specific element of the structure or what's the best layout for the entire structure. Usually, the proper way to think about it is for the entire structure. After all, if putting this brace in a way that's bad for it (that is, the piece needs to be heavier to withstand the load), but better for the entire structure, the cost-benefit is worth it: a single heavier brace to make the entire structure lighter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, however, the entire-structure view is a bit ambiguous (see below), so I'd make my decision thinking about what's best for the brace (well, for this simple exercise; in real life, I'd look at the entire-structure case more closely to see what was best).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So let's look at the problem both ways:&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What's best for the brace?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;It feels beyond the scope of this question, so I won't explain why (add a comment if you're interested, in which case I'll consider an edit). But a fun fact about such bracing is that the magnitude of the load is indifferent to the direction of the brace. The only thing that changes is the direction. That is, you choose whether the brace should be under tension or compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And that's the answer to the problem if you look at it this way. After all, the problem states this is a steel structure. And we know that steel behaves much better under tension than compression (due to the risk of buckling under compression). So we want to put that brace under tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now it's just a matter of determining which orientation will put the brace under tension. I find it quite intuitive that the answer is BE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, this is a structure under a force pointing mainly down and a bit to the left. As that force makes its way &quot;up the truss&quot;, it will get to BCEF as a series of internal forces, but the resultant of all those forces will still be basically down and to the left. Putting a brace in that direction means the brace will be naturally resisting that force in that same direction, which for such a brace would be tension. If the brace were instead CF, the force would need to climb up to C, where such a mainly-downward force becomes compression for the brace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if all you're looking at is what is best for the brace, BE is the clear winner.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;What's best for the structure?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This one's harder to determine by inspection, but it can still try to apply a bit of intuition. In this case, I find it easier to think about why CF is worse than BE, so let's think about it that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I basically explained the problem with a CF brace above, actually. As I said, the force will need to climb all the way up to C before the brace can help out. That already tells us that if the brace is at CF, then CE will suffer a greater load. After all, it'll need to carry the load which a BE brace would have already absorbed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that there's no brace on E means the entire horizontal force on the right-hand-side will need to be absorbed by EF. If the brace were in BE, it'd be able to absorb some of that horizontal load, making life a bit easier for EF.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After that, it gets a bit tricky to intuit the impact of the brace's orientation. But there's nothing stopping you from running a quick model to see the impact:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jztRE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jztRE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As suspected, a BE brace has tension, and leads to lower forces in CE and EF. However, it leads to much greater compression in BC and causes BF to suffer loads (a CF brace makes BF effectively useless).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what's better? A brace under compression which leads to lower compression (or none at all!) in other members but significatly higher tension in others? Or a brace under tension which leads to higher compression in some members but lower tension in others? &lt;em&gt;A priori&lt;/em&gt; it's a bit ambiguous, especially if you consider the fact that the brace is more susceptible to buckling than the horizontal members (since it's longer) and would be under a greater load. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Final decision&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In this case, I'd probably choose a BE brace: it's better for the brace and at least indifferent for the structure.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Diagram developed with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free educational 2D frame analysis tool.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-08-04T04:08:05.643" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29486" PostTypeId="1" CreationDate="2019-08-04T10:18:29.200" Score="2" ViewCount="123" Body="&lt;p&gt;I am trying to compute the optimal (a non-optimal solution would also work) size of carts that can move along guiding rails. Each cart is connected to the rails at two points. Each cart is also connected to the cart in front of it and the one behind it. It is a closed system. The carts can be connected directly as in picture (a) or there can be some arbitrary but constant size connector between them. In the first step I am just looking for a solution that works mechanically. It would also be great to know what is the best arrangement that allows the least force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In picture (a) I can find many solutions. However, my guiding rails have to have a shape as in (c). I can change the rails a little bit but they will have to have inside angles (red) and outside angles (yellow) as depicted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried experimentally to find a working solution through simulation, but is there an analytic way to compute which size of carts could move along these rails? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have a kinematics background and don't know what it is that I am looking for. Any pointer towards keywords/ papers where I can start my research would be very much appreciated. So far I am stuck at &quot;Polygon Effect&quot;. &lt;a href=&quot;https://i.stack.imgur.com/ax3nK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ax3nK.png&quot; alt=&quot;Rails&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21339" LastActivityDate="2023-12-13T08:01:30.157" Title="How to compute the size of a polygon moving along non-circular guide rails" Tags="|mechanical-engineering|kinematics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29487" PostTypeId="1" AcceptedAnswerId="29488" CreationDate="2019-08-04T11:05:00.500" Score="-1" ViewCount="256" Body="&lt;p&gt;This is from a design entrance that I am preparing for. I know it's a little different than a theoretical projection drawing that engineering books usually cover. It is more practical and visualization based question. I have tried really hard to find some resources to understand the approach to solve this kind of questions. I didn't find any. So, if someone decides to put it on hold. I would like to request them to attach some resource in the comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SndKZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SndKZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What would be the number of surfaces of the common part?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;My initial assumption was I need to first visualize the shape of the common part then count the number of surfaces. But, as people have told that is not necessary. Still, I would also like to know&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to visualize and draw the common part?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have tried to solve this myself but the best I can come up with this. I would also like to mention that  I am not an engineering student. But for my exam asks some questions from engineering drawing. I have a book for it written by someone from my country but that does not cover any union kind of question. I would be very much obliged if you show me the proper method to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KVg8v.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KVg8v.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21325" LastEditorUserId="21325" LastEditDate="2019-08-06T19:25:29.710" LastActivityDate="2019-08-06T20:15:44.913" Title="visualization of intersection of two objects" Tags="|mechanical-engineering|drafting|drawings|machine-design|" AnswerCount="1" CommentCount="4" ClosedDate="2019-08-04T21:34:53.810" ContentLicense="CC BY-SA 4.0" />
  <row Id="29488" PostTypeId="2" ParentId="29487" CreationDate="2019-08-04T12:02:13.863" Score="3" Body="&lt;p&gt;I don't want to spoil the fun of solving this for yourself so I offer the following suggestions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Redraw the assembly with all edges, including the hidden edges, shown as dotted lines.&lt;/li&gt;&#xA;&lt;li&gt;Draw solid lines on all the intersections.&lt;/li&gt;&#xA;&lt;li&gt;Join up any vertices which form edges not on the outer surfaces of the assembly.&lt;/li&gt;&#xA;&lt;li&gt;Count the faces.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Step 3 is the only tricky bit. In your example the uppermost edge of the horizontal element will form one such edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Post a sketch in an update to your question and we'll see if you've understood it.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0SNS7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0SNS7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Intersection of two prisms. The green lines are edges that already exist on the isometric view. The orange lines will be new edges created by the intersection operation.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2019-08-06T20:15:44.913" LastActivityDate="2019-08-06T20:15:44.913" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="29490" PostTypeId="1" AcceptedAnswerId="29493" CreationDate="2019-08-04T16:37:27.427" Score="1" ViewCount="1249" Body="&lt;p&gt;So the reason why torque converter is needed is because it disconnects engine from transmission. It has also extra benefit of multiplying torque but that is not 100% necessary. From what I have seen all automatics have some sort of constant input which is always connected to the shaft. Would it be possible to design an automatic transmission which wouldn't have constant input and therefore wouldn't need torque converter?&lt;/p&gt;&#xA;" OwnerUserId="21049" LastActivityDate="2019-08-05T02:03:37.543" Title="Is it possible to design automatic transmission without torque converter?" Tags="|mechanical-engineering|transmission|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29491" PostTypeId="1" AcceptedAnswerId="29496" CreationDate="2019-08-04T17:06:01.293" Score="1" ViewCount="4959" Body="&lt;p&gt;I'm learning a bit of stick welding since I keep needing it for certain projects. Total beginner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The latest thing I decided to make is a table similar to this one: &lt;a href=&quot;https://i.stack.imgur.com/5FZS8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5FZS8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The material is &lt;strong&gt;2mm&lt;/strong&gt; thick &lt;strong&gt;40x40mm&lt;/strong&gt; square tubing - don't know the exact metal type, should be &quot;regular&quot; steel that was a bit corroded from staying in the steel yard.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I tried using &lt;strong&gt;2mm&lt;/strong&gt; and &lt;strong&gt;1.6mm rutile electrodes&lt;/strong&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I tried the &lt;strong&gt;1.6mm at 55~ Amps AC&lt;/strong&gt; and the &lt;strong&gt;2mm at 60-65 Amps AC&lt;/strong&gt;. With higher or lower Amps, my electrodes started sticking.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;The angle I'm holding the electrode at is similar to this one: &lt;a href=&quot;https://i.ytimg.com/vi/qTaQf0iPH-g/maxresdefault.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.ytimg.com/vi/qTaQf0iPH-g/maxresdefault.jpg&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is happening that on straight flat sections I can spark and keep a straight line and don't burn a hole in metal. The weld is strong and consistent. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When trying to weld the two pieces together, on the outside and inside angles (see red circle) I keep burning through and bigger and bigger gaps keep opening because of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the inner corners I have not had this issue and I have successfully welded thicker metals with 2mm and 2.5mm electrodes without issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;: Edited for clarity that I'm not actually welding without melting metal :)&lt;/p&gt;&#xA;" OwnerUserId="21341" LastEditorUserId="21341" LastEditDate="2019-08-05T14:33:03.937" LastActivityDate="2019-08-05T14:33:03.937" Title="Arc-welding 2mm thick steel at an angle without burning-through" Tags="|metals|welding|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29493" PostTypeId="2" ParentId="29490" CreationDate="2019-08-04T23:18:28.373" Score="2" Body="&lt;p&gt;That is not why torque converters are used in automatic transmissions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are used because in addition to declutching the engine from the transmission at idle, they furnish &lt;em&gt;torque multiplication&lt;/em&gt;, in essence giving a 3-speed transmission a fourth &quot;gear&quot; that slips into and out of engagement without jerks and furnishes a smooth range of equivalent &quot;gear ratios&quot; without requiring inputs from the driver or the drive management computer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to design automatic transmissions without torque converters but they require servo-controlled clutches for starting and stopping, and you must also add at least one additional fixed gear ratio to the transmission to make up for the loss of same by elimination of the torque converter. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-08-04T23:18:28.373" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29494" PostTypeId="2" ParentId="29490" CreationDate="2019-08-05T02:03:37.543" Score="1" Body="&lt;p&gt;The original automatics did not have torque converters ; they mostly had fluid couplings . Chrysler in particular , had more than a half dozen different ones in the various cars. They were more like &quot;gear changers&quot; with an automatic clutch ; some still included a clutch pedal. Performance was very poor by todays standards. And there were overdrives that would make one shift . The GM Hydromatic was the first good automatic ( 1939 Olds and 1940 Cadillac as I remember) . It had a fluid coupling that had limited torque converter function - I forget the percentage. It could shift hard enough to burn rubber in the first 2 shifts ( with the V-8 engines).  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-08-05T02:03:37.543" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29496" PostTypeId="2" ParentId="29491" CreationDate="2019-08-05T02:40:21.527" Score="2" Body="&lt;p&gt;Welding always melts steel, you mean you are &quot;burning through&quot;, too much penetration. Wire feed welding -MIG is much easier to use than stick . MIG , also identified as flux core with or without gas and solid wire with gas - is the best choice for your job. Stick is difficult to learn by trial and error and more difficult with thin base metal as you have. Stick is not so bad with base metal of 1/2 in. and turn the amperage up high enough to prevent &quot;sticking&quot;.     Apparently you have the best electrode , 6013, that is the high titania/rutile type. Reverse polarity will give the lowest penetration ; more of the heat goes to melting the electrode and less heat into the work. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-08-05T02:40:21.527" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29497" PostTypeId="2" ParentId="29479" CreationDate="2019-08-05T03:27:17.623" Score="1" Body="&lt;p&gt;You will find a large number of standard tests for the elastic properties of anisotropic engineering materials (ususally composites) in the &lt;a href=&quot;https://www.astm.org/Standards/composite-standards.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASTM standards&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For rubber and other materials that can support large elastic stretches there are a few standard tests that can be extracted from &lt;a href=&quot;https://www.astm.org/Standards/rubber-standards.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rubber standards(Mechanical Test Methods)&lt;/a&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other standards are available for &lt;a href=&quot;https://www.astm.org/BOOKSTORE/BOS/TOCS_2017/04.08.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;soils and rocks&lt;/a&gt; and several specialized materials (e.g., foams, medical equipment, bones, etc.).&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2019-08-05T03:27:17.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29498" PostTypeId="1" AcceptedAnswerId="29499" CreationDate="2019-08-05T04:25:32.550" Score="2" ViewCount="64" Body="&lt;p&gt;My intuition on how a twist drill bit operates is that the cutting is done by the end of the drill bit and that the fluted section carries away material as it's cut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I look at twist drill bits, though, it appears that the edge of one of the flutes is sharpened. What purpose does this serve? Does this edge actually cut material?&lt;/p&gt;&#xA;" OwnerUserId="21349" LastActivityDate="2019-08-05T05:26:14.937" Title="Why is the fluted edge of a twist drill bit sharp?" Tags="|mechanical-engineering|machining|tools|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29499" PostTypeId="2" ParentId="29498" CreationDate="2019-08-05T05:18:38.670" Score="2" Body="&lt;p&gt;The edge is sharp to make sure the side of the hole is left smooth minimizing the cutting marks and to make sure that the resulting diameter is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if a precise diameter is required then a reamer is usually used to finish the hole.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-08-05T05:26:14.937" LastActivityDate="2019-08-05T05:26:14.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29500" PostTypeId="1" AcceptedAnswerId="29501" CreationDate="2019-08-05T05:57:12.993" Score="3" ViewCount="89" Body="&lt;p&gt;I noticed that the screw-on of my water tap often loosens itself up after I rotate the tap left and right enough times. Every time I rotate the tap left, it loosens the screw a bit, then if I rotate the tap right, the screw does not tighten itself to the same amount. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This, repeated enough times, loosens the screw enough so that the tap would leak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it might be solved if the screw would have a wavy thread. So, for example, instead of gently sloping downwards by 1 mm every revolution, it would slope up 1 mm for the first half, and down 2 mm for the second half. This way, the screw would lock itself in a local &quot;peak&quot; in the thread, and not loosen itself no matter if I rotate the tap left or right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there such a kind of screw? If so, what are they used for? If not, is it because they have some kind of mechanical weakness?&lt;/p&gt;&#xA;" OwnerUserId="21350" LastActivityDate="2019-08-05T06:28:10.317" Title="Why is the thread of the screw linear? Can it be wavy?" Tags="|mechanical-engineering|civil-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29501" PostTypeId="2" ParentId="29500" CreationDate="2019-08-05T06:28:10.317" Score="5" Body="&lt;p&gt;This idea doesn't work, because screwing the parts together will tend to strip the threads before the parts are fully squeezed together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead, engineers use different general methods to get the threads to lock and not unscrew themselves in the manner you experienced, as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) using screws or nuts with plastic inserts in them which interfere with the threads in the mating part. These are called &lt;em&gt;nylock&lt;/em&gt; fasteners or &lt;em&gt;lock nuts&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) using threads whose &lt;em&gt;diameter&lt;/em&gt; increases progressively along the length of the threaded portion, so as to generate progressively more interference as the parts are screwed together. these are called &lt;em&gt;tapered&lt;/em&gt; or &lt;em&gt;pipe threads&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) inserting a &lt;em&gt;lock washer&lt;/em&gt; between the parts with teeth that bite into the nut or screw head that discourage the parts from unthreading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) assembling the part with a glue applied to the threads that sets up and prevents the parts from being unscrewed. This glue is called &lt;em&gt;thread locker&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-08-05T06:28:10.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29503" PostTypeId="1" AcceptedAnswerId="29504" CreationDate="2019-08-05T07:45:57.040" Score="3" ViewCount="10864" Body="&lt;p&gt;I am designing a hydrophone, from the piezo buzzer. To hold the circular piezzo plate, I have designed a mechanical fixture. I need to use a washer/o-ring to make it leak-proof. I am confused about which one to use.&lt;/p&gt;&#xA;&lt;p&gt;Can you please explain the difference between washer and o-ring. What is the difference between them?&lt;/p&gt;&#xA;" OwnerUserId="21354" LastEditorUserId="90" LastEditDate="2023-08-23T09:01:10.807" LastActivityDate="2023-08-23T09:01:10.807" Title="What is the difference between washer and O-ring, what are their advantages of one over the other?" Tags="|mechanical-engineering|design|sensors|vibration|seals|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29504" PostTypeId="2" ParentId="29503" CreationDate="2019-08-05T08:26:12.140" Score="4" Body="&lt;p&gt;An o-ring usually (but not always) has a round section and is made of a compressible material such as rubber, neoprene, silicon etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A washer tends to have a square or rectangular section and is usually made of a harder material brass, copper, aluminium, steel etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not about advantages or disadvantages, but which is used will depend on the characteristics of the purpose, so if it is a delicate device at low pressure then a softer material can be used especially if the units are designed with matching channels for the sealing component to &quot;sit&quot; in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many times a copper washer can be used - some car sump plugs have copper washers for sealing, others just use machined steel faces...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to muddy the waters as it were, there are o-rings called Wills rings which are a hollow metal torus filled with an inert gas, these are capable of sealing at high temperature and pressure. One use was as part of an engine head gasket sealing the head to the block around the combustion chamber... &lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-08-05T19:35:01.000" LastActivityDate="2019-08-05T19:35:01.000" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29505" PostTypeId="1" CreationDate="2019-08-05T09:18:37.573" Score="0" ViewCount="302" Body="&lt;p&gt;I have an interesting problem here.&#xA;I have a box of certain dimensions. Inside of the box is some insulation and another box, from the same material as the outer box. I know everything possible about the materials possible.&#xA;Is there an equation to calculate that if there are for example -4 degrees Celsius outside, it will take XY minutes to cool the inside of the inner box by 1 degree?&#xA;I really don’t know where to start. Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="21355" LastActivityDate="2019-09-05T02:31:46.517" Title="Time to change temperature inside the box" Tags="|thermodynamics|thermal-conduction|thermal-insulation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29506" PostTypeId="1" CreationDate="2019-08-05T09:47:06.013" Score="1" ViewCount="181" Body="&lt;p&gt;I am writing a program to calculate chain lengths required for a particular sprocket pair when I noticed this&#xA;Consider a 9 teeth sprocket paired with an identical 9 teeth sprocket (image reference) the pitch circle is green and the initial position of the chain (used to calculate the chain length and number of links) is shown in BLUE. The total length is 285.75mm which comes out to be 18 link chain for 15.875mm chain pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I rotated the pitch polygon by 10 degrees (0.25 of the angle enclosed by 1 link which is 360/9 = 40deg) this is shown in orange. Then I drew the taut chain again (orange lines). Now if I measure the chain length, it is lesser than the calculated length before (by about 0.3mm)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this happening? Am I doing something wrong? This also means that if I had adjusted my centre distance to make the chain links a whole number in the second configuration, then as it rotates to the first configuration, the chain length increases..but we assume that the chain is inextensible &#xA;Added another drawing for reference (pardon the messy nature). The green curves are the sprocket and chain in initial position, the magenta curves are the rotated configuration. As we can see, the pink line between the 2 sprockets cannot accommodate the 5 chain pitches and is interfering with the RH sprocket teeth. Practically, the chain slackness will take care of this change in length. My question is about the theoretical aspect. Is it impossible mathematically to wrap such a chain around sprockets without encountering any changes in link length during rotation??&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fGere.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fGere.png&quot; alt=&quot;Reference Image&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nfuZL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nfuZL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21358" LastEditorUserId="21358" LastEditDate="2019-08-05T12:06:05.067" LastActivityDate="2019-08-05T12:06:05.067" Title="Chain length increasing during rotation of chain sprocket pair?" Tags="|mechanical-engineering|mechanisms|chain|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29507" PostTypeId="2" ParentId="29506" CreationDate="2019-08-05T10:00:57.590" Score="0" Body="&lt;p&gt;A chain is designed to sit on the root between the teeth at all times while it is in contact with the sprocket, so it will not get lower during rotation as you assume in your drawing.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-05T10:00:57.590" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29508" PostTypeId="2" ParentId="29506" CreationDate="2019-08-05T11:52:58.757" Score="1" Body="&lt;p&gt;You are not doing anything wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the chain is transmitting power, one side will be in tension and the other side will be slack, and the slack side can easily take up the changes in length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chains also &quot;stretch&quot; over time as the connections between the links wear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want the chain to be flopping about on the slack side, you use a chain tensioner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course if you have sprockets with more teeth, the size of the &quot;problem&quot; reduces.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-08-05T11:52:58.757" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29510" PostTypeId="1" AcceptedAnswerId="29519" CreationDate="2019-08-05T14:12:39.347" Score="2" ViewCount="166" Body="&lt;p&gt;I read about how an LCD screen works &lt;a href=&quot;https://www.explainthatstuff.com/lcdtv.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, and it made me wonder what actually happens when the screen shuts off. Based off of this, I can think of several possibilities:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Electricity is supplied to all the pixels/crystals, thus making them all appear dark. While this fits with the on/off explanations given in that article, it seems like a significant waste of energy to use extra energy when &lt;em&gt;not&lt;/em&gt; in use.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electricity is &lt;em&gt;not&lt;/em&gt; supplied to all the pixels, which would light them all up, but the backlight shuts off, so the screen appears dark.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electricity is maintained to the pixels that were receiving electricity when the screen was on, but the backlight shuts off so the screen appears dark.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I would be curious to know if different phases of the off screen are different (such as a off-screen-like screen saver versus a sleep mode or being shut off, etc.).&lt;/p&gt;&#xA;" OwnerUserId="21359" LastEditorUserId="21359" LastEditDate="2019-08-05T14:39:19.130" LastActivityDate="2019-08-05T18:51:01.513" Title="What happens when an LCD screen shuts off?" Tags="|computer-hardware|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29511" PostTypeId="1" AcceptedAnswerId="29528" CreationDate="2019-08-05T14:59:31.017" Score="0" ViewCount="201" Body="&lt;p&gt;What FEA software is capable of simulating static studies on orthotropic materials such as wood? I must be able to define mechanical properties in X,Y and Z. I would also like to have access to failure criteria such as tsai-hill or tsai-wu and probably even other custom criteria. My company uses solid edge simulation at the moment but i have not found a way to account for orthotropic materials, I also tried solidworks simulation on my side and I was able to perform most of what i need but i want to consider any additional options before i make a recommendation to my company to shift to a new software.&#xA;Any hints or advice would be greatly appreciated. &#xA;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="21194" LastActivityDate="2019-08-06T12:37:47.953" Title="FEA software for Orthotropic materials" Tags="|finite-element-method|solidworks|wood|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29512" PostTypeId="1" CreationDate="2019-08-05T15:09:07.577" Score="1" ViewCount="41" Body="&lt;p&gt;Looking for the answer to what kind of engine provides the thrust on the French hoverboard on which its inventor, Franky Zapata, recently crossed the English Channel, I find the same phrase repeated everywhere: it's a &quot;turbine engine&quot;. Exactly what is it? A turbojet? A turbofan? A turboprop? It is fuelled by paraffin (kerosene). Is all the thrust produced by propellers in a ducted fan configuration or does exhaust gas contribute? &lt;/p&gt;&#xA;" OwnerUserId="21360" LastEditorUserId="21360" LastEditDate="2019-08-05T15:21:05.983" LastActivityDate="2019-08-08T00:44:02.860" Title="What propulsion method: Franky Zapata hoverboard" Tags="|aerospace-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29513" PostTypeId="2" ParentId="29503" CreationDate="2019-08-05T16:00:54.293" Score="2" Body="&lt;p&gt;You can also use a gasket, which is typically soft sheet material, either rubber or paper.  They are often used when the interface you are sealing is a more complicated shape than a circle.  O-rings can do that too if you cut the groove properly.  The point of all of them when you are sealing is to make up for unevenness in the metal parts that are going together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My experience is that o-rings will take up much more unevenness than the other approaches, but they are more expensive to incorporate because they need a groove to sit in.  There are documents to show what sort of groove is best with a given cross section and material of o-ring.&lt;/p&gt;&#xA;" OwnerUserId="21362" LastActivityDate="2019-08-05T16:00:54.293" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29514" PostTypeId="1" AcceptedAnswerId="29518" CreationDate="2019-08-05T16:02:57.770" Score="1" ViewCount="1605" Body="&lt;p&gt;I'm revising for the UK LGV theory test. One of the questions goes something like this:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Which axle configuration is most effective at preventing petrol tankers from rolling over?&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;A. Tandem axles with double wheels&lt;/li&gt;&#xA;  &lt;li&gt;B. Tandem axles with air suspension&lt;/li&gt;&#xA;  &lt;li&gt;C. Tri-axles with single wheels&lt;/li&gt;&#xA;  &lt;li&gt;D. Tri-axles with double wheels&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The correct answer is apparently C, with the following explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The type of suspension fitted to a vehicle will influence its resistance to ‘roll-over’. Modern tri-axle semi-trailers fitted with single wheels on each side extend the tracking width available, making this the most stable configuration.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Why are single wheels better than double wheels?&lt;/li&gt;&#xA;&lt;li&gt;Is this specific to a tri-axle setup, or does it also apply to twin-axles or single-axles?&lt;/li&gt;&#xA;&lt;li&gt;Is this specific to trailers, or does this apply to rigid vehicles as well?&lt;/li&gt;&#xA;&lt;li&gt;If single wheels are better, why do so many vehicles have double-wheels?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="21361" LastActivityDate="2019-08-05T20:07:41.337" Title="Why are tri-axle trailers with single wheels more stable than double wheels?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29515" PostTypeId="1" CreationDate="2019-08-05T16:53:24.067" Score="2" ViewCount="41" Body="&lt;p&gt;As we know in a 4G network, the LTE give us a correlation about the SNR of user and the data rate of the user. This correlation is usually a table called Channel Quality Indicator (CQI) which is found in C&lt;a href=&quot;https://pdfs.semanticscholar.org/d738/479b73a487b1308ba18ac9a4d2e2a8df596a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;hannel Quality Indicator Feedback in Long Term Evolution(LTE) System&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:  Is there something similar like -LTE CQI- in 5G networks in NOMA(Non Orthogonal Multiple Access)?&lt;/p&gt;&#xA;" OwnerDisplayName="user21366" LastEditorUserId="7083" LastEditDate="2019-08-07T20:36:57.830" LastActivityDate="2019-08-07T20:36:57.830" Title="Non Orthogonal Multiple Acess and Channel Quality Indicator" Tags="|telecommunication|wireless-communication|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29518" PostTypeId="2" ParentId="29514" CreationDate="2019-08-05T18:41:01.800" Score="3" Body="&lt;p&gt;If the load is evenly distributed, then double wheels would be something like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[Mg/4] [Mg/4]              [Mg/4] [Mg/4]&#xA; \/     \/                   \/     \/&#xA;[tire] [tire]--------------[tire] [tire]&#xA;  |      |                   |      |&#xA;  |      |------(L_inner)----|      |&#xA;  |-------------(L_outer)-----------|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where single wheels would be something like:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[Mg/2]                            [Mg/2]&#xA; \/                                 \/&#xA;[tire]----------------------------[tire]&#xA;  |                                 |&#xA;  |-------------(L_outer)-----------|&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There's a reaction force at the road, so the &lt;em&gt;torque&lt;/em&gt; applied to the body is higher, because each side has a torque of &lt;span class=&quot;math-container&quot;&gt;$0.5L_{outer}\left(\frac{mg}{2}\right)$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$(0.5L_{inner} +0.5L_{outer})\left(\frac{mg}{4}\right)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tend to agree with &lt;a href=&quot;https://engineering.stackexchange.com/questions/29514/why-are-tri-axle-trailers-with-single-wheels-more-stable-than-double-wheels#comment53785_29514&quot;&gt;@ratchet freak&lt;/a&gt; about this feeling like a trick question/gotcha, though. I think there are aspects that are being ignored, like what happens to the suspension if the tanker actually starts to roll. If all of the load is put on one side, and there are only single tires, is the loading then sufficient to blow out the tires? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the crux of your other questions: by the logic above, there's more &quot;available&quot; restoring torque to correct a roll-over with single wheels, so then they'd be better in all applications - twin-axle, rigid vehicles, etc. However, you typically see heavy-duty vehicles with double wheels in order to distribute the load enough that it doesn't exceed the tire ratings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, I don't think the tire arrangement has much to do with tanker roll-over. I think a well-rested, qualified operator that understands the maximum safe speed is more important, along with a well-serviced brake system to allow the driver to slow the vehicle down in time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're to the point that you're depending on the specific tire configuration to save you, then you've probably already screwed up. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2019-08-05T18:41:01.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29519" PostTypeId="2" ParentId="29510" CreationDate="2019-08-05T18:51:01.513" Score="1" Body="&lt;p&gt;There's two different &quot;shut off&quot; methods: showing a black screen and actually powering off. To show a black screen, the backlight stays on but none of the pixels are energized. Light that gets polarized from the first filter doesn't change orientation (because the liquid crystals aren't energized) and so that light gets blocked by the second filter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turning off also de-energizes the pixels, but it also turns the backlight off. This is obviously more energy efficient, but also makes for darker black colors because the polarizing filters only stop &lt;em&gt;most&lt;/em&gt; of the light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most displays will have a single light source for the backlight, but some displays, like edge-lit displays, can control &lt;em&gt;regions&lt;/em&gt; of light and other displays use individual LEDs with each pixel. Both methods can vary the intensity of the backlight and this is sometimes called &quot;high dynamic range&quot; displays. You can read more about that &lt;a href=&quot;https://www.cnet.com/news/led-local-dimming-explained/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2019-08-05T18:51:01.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29520" PostTypeId="2" ParentId="29486" CreationDate="2019-08-05T18:54:03.087" Score="0" Body="&lt;p&gt;There's no uniform radius of curvature with your shape (c) so you can't have a closed series of rigid linkages that's capable of moving around the track. You can use a flexible link or cart as a tensioner, as &lt;a href=&quot;https://engineering.stackexchange.com/questions/29486/how-to-compute-the-size-of-a-polygon-moving-along-non-circular-guide-rails#comment53762_29486&quot;&gt;Jonathan R Swift mentioned&lt;/a&gt;, but otherwise it's not possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;:EDIT: - Consider this example of three carts/points on the track. In your (a) and (b) case, the radius of curvature is constant, so the arc length &lt;span class=&quot;math-container&quot;&gt;$s = r\theta$&lt;/span&gt; is constant, so the length of the chord that spans the endpoints of the arc is also constant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In case (c), you can see that there's no way for the bottom-right point to move down the right &quot;arm&quot; of the track because the bottom linkage, if it's rigid, isn't going to allow that. Similarly, the right linkage between the bottom-right point and the top point isn't going to expand and let the bottom-right point move down, either. The mechanism is jammed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It might &lt;em&gt;look&lt;/em&gt; like it's going to work if you make the links smaller but, as long as they are of some finite length &lt;span class=&quot;math-container&quot;&gt;$\ell&amp;gt;0$&lt;/span&gt;, they'll always be cutting a corner off and again, since the radius of curvature isn't constant, you'll find the length of track spanned when it's straight, for example, will be shorter than the length of track spanned when it's curved and so it'll still jam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you &lt;em&gt;need&lt;/em&gt; to have the chain be rigid then the &lt;a href=&quot;https://en.wikipedia.org/wiki/Tensioner&quot; rel=&quot;nofollow noreferrer&quot;&gt;tensioner&lt;/a&gt; needs to be incorporated into the track/rail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qd2er.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qd2er.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1633" LastEditorUserId="1633" LastEditDate="2019-08-05T22:52:34.410" LastActivityDate="2019-08-05T22:52:34.410" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29521" PostTypeId="2" ParentId="29514" CreationDate="2019-08-05T19:10:33.627" Score="1" Body="&lt;p&gt;Track width is one of the most important factors in stability and resistance against rollover in cars and trucks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why we see in race cars they even extend the wheels outside of the car body, or all street legal fast sport cars have flared out fenders to accommodated a wide track width stance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a double wheel axle the width of the wheelbase is from the center of the left double wheel to the center of the right double wheel which is approximately 20% shorter than a single wheel width, even though it has the benefit of distributing the wight over larger area of the pavement, hence applying less pressure on the road.As has been mentioned in @Chuck's answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other reason is when there is two wheel at each end when the truck is urged to roll as by a turn or a bump, the outside wheel's tire starts to compress into a more flattened shape but the the tire next to it winch is under much less torque because the fact that it is not far enough will bulge out a bit and exacerbate the situation on the outermost tire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you watch some videos of monster truck races which some have double-wheels you'll see this effect. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-08-05T20:07:41.337" LastActivityDate="2019-08-05T20:07:41.337" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29522" PostTypeId="2" ParentId="29505" CreationDate="2019-08-05T20:12:53.187" Score="1" Body="&lt;p&gt;The insulation will have a certain calculable thermal conductivity, in Watts per Kelvin (or the unit of measure of your choice).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inner box will have a certain calculable thermal capacity, in Joules per Kelvin.  That can be turned into Kelvins per Joule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Calculate the heat flux (in watts) into the inner box, note that a Watt is a Joule / second, and do some math.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the real world, the insulating material will have its own heat capacity, which will complicate things, and a rectilinear box will complicate things more -- a spherically symmetrical box would let you turn this into a 1-D problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in the real world, if there are any fluids involved (i.e., air), then there's a possibility of convection, at which point things go well over my head, and you'll find me in the lab with a bazillion thermocouples and an oven, figuring out answers experimentally.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-08-05T20:12:53.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29523" PostTypeId="1" AcceptedAnswerId="30013" CreationDate="2019-08-05T21:10:25.673" Score="2" ViewCount="8663" Body="&lt;p&gt;What are some materials that allow air to pass but not, or to a lesser extent,  water vapor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two materials I am aware of:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Teflon&lt;/strong&gt; is a material that allows water vapor to pass, but not air. Water itself can't pass.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What are some other sourceable substances which have this feature? What are some defining terms and measurements used to identify this property in a material, and how can the rate of water vapor / air transfer be manipulated in such a substance?&lt;/p&gt;&#xA;" OwnerUserId="9573" LastActivityDate="2019-09-09T01:02:23.723" Title="What is a material that allows air to pass but not water vapor?" Tags="|materials|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29524" PostTypeId="2" ParentId="29511" CreationDate="2019-08-05T21:15:33.443" Score="0" Body="&lt;p&gt;I would suggest Abaqus Standard or Abaqus Explicit (depending on the type of problem you are trying to solve).  Ansys Workbench should also have the features you need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cost will be around $20000 per year.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2019-08-05T21:15:33.443" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29526" PostTypeId="1" CreationDate="2019-08-06T04:25:39.177" Score="0" ViewCount="1190" Body="&lt;p&gt;I have a stepper motor on a 3D printer that heats up very bad because of the heated extruder (it is actually controlling the extruder). When this system starts, it works fine at the beginning, while the temperature is rising. But after the extruder has achieved its working temperature and stepper has gotten hot too, it start skipping steps. I am wondering whether it is possible for stepper to lose torque because of a high temperature? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my thinking:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the higher the temperature -&gt; the higher the internal coil resistance -&gt; the lower the current flowing through those coils -&gt; the lower the torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it make sense? What are &quot;okayish&quot; temperatures for steper motors like NEMA17?&lt;/p&gt;&#xA;" OwnerUserId="21373" LastActivityDate="2019-08-06T15:37:53.833" Title="Could stepper motor lose torque because of overheating?" Tags="|torque|stepper-motor|3d-printing|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29527" PostTypeId="1" CreationDate="2019-08-06T09:00:27.827" Score="0" ViewCount="373" Body="&lt;p&gt;Liquid propelled Rocket engines work on various cycles such as pressure-fed, pump-fed(gas generator,staged combustion, expander, etc) and these have various design trade-offs between complexity, reliability, operational simplicity, development cost(?) etc. What are the various trade-offs between these cycles?&lt;/p&gt;&#xA;" OwnerUserId="17722" LastEditorUserId="17722" LastEditDate="2019-08-12T13:27:27.470" LastActivityDate="2019-08-12T13:27:27.470" Title="What are the design trade-offs in the different rocket engine cycles for liquid propulsion systems" Tags="|rocketry|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29528" PostTypeId="2" ParentId="29511" CreationDate="2019-08-06T12:37:47.953" Score="0" Body="&lt;p&gt;Any of the major FEA systems (Nastran, Abaqus, Ansys, Adina, Comsol, etc) can do this. If you have a fairly limited requirements (e.g. only linear statics), look for the vendor with the cheapest entry-level or student-level version of the code which includes orthotropic material models.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-08-06T12:37:47.953" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29529" PostTypeId="2" ParentId="29526" CreationDate="2019-08-06T15:37:53.833" Score="1" Body="&lt;p&gt;There are a number of possibilities:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the higher the temperature -&gt; the higher the internal coil resistance -&gt; the lower the current flowing through those coils -&gt; the lower the torque.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possibilities that I can think of are &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the magnets temporarily losing magnetism (you say it works when it gets cool, so they aren't losing &lt;em&gt;all&lt;/em&gt; their magnetism permanently), &lt;/li&gt;&#xA;&lt;li&gt;the motor binding up because of differential expansion, either inside the motor or in your mechanism.&lt;/li&gt;&#xA;&lt;li&gt;your mechanism itself is binding up, and the motor is fine&lt;/li&gt;&#xA;&lt;li&gt;the motor has some internal protection (I doubt this, but you never know).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does it make sense? What are &quot;okayish&quot; temperatures for steper motors like NEMA17?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;NEMA17 is a size.  What a motor can actually stand depends on the motor itself.  Particularly now that there is a strong hobby market, I wouldn't trust a particular size of motor (especially if you got it surplus) to behave like other same-sized motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;would&lt;/em&gt; expect a decent industrial motor to work over an industrial temperature range (i.e., 0C or even -20C to +55C) -- but that's kind of a &quot;no true Scottsman&quot; fallacy -- if it doesn't work over the temperature range I want, I'll just say &quot;it's not decent then, is it?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, the motor works the way it works.  &lt;em&gt;If&lt;/em&gt; the manufacturer lives up to their promises, then you can look at the motor data and go with what it says.  If there's no data, that's a big tell that you can't trust the motor.  If you got the motor through surplus channels, or if the manufacturer isn't reputable, then all bets are off.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-08-06T15:37:53.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29530" PostTypeId="1" AcceptedAnswerId="29533" CreationDate="2019-08-06T16:21:23.787" Score="0" ViewCount="37" Body="&lt;p&gt;I need to move a machine, approx weight of 900Kg&#xA;I will put 3 tubes under the machine and roll the machine on the tubes&#xA;Holidays season, all industrial suppliers are closed around me and only have access to 2 kinds of tubes:&#xA;20mm diameter and wall thickness 1.5mm&#xA;or&#xA;25mm diameter and wall thickness 1.5mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is the thickness enough to support the weight?&#xA;Material is construction steel (we call it S235 here but I don't know the international name)&#xA;i would be interested in the formula to make the calcultion also&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="14673" LastActivityDate="2019-08-06T19:20:23.383" Title="Steel tube wall thickness for moving a machine" Tags="|steel|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29532" PostTypeId="1" CreationDate="2019-08-06T18:20:27.187" Score="1" ViewCount="102" Body="&lt;p&gt;How are automotive engine mounts tuned to behave differently for different frequencies, like they are soft for high frequency low amplitude vibrations and hard for low frequency high amplitude excitations..?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware of active or hydraulic mounts, but want to know how a regular mount is tuned to behave in such a way..?&lt;/p&gt;&#xA;" OwnerUserId="19588" LastActivityDate="2021-05-30T15:00:35.050" Title="Regarding automotive engine mounts" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29533" PostTypeId="2" ParentId="29530" CreationDate="2019-08-06T19:20:23.383" Score="0" Body="&lt;p&gt;It depends on the dimensions and geometry of your load and the evenness of the track. At first look it looks quite inadequate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as a basic rule the smaller the diameter the stronger is the tube. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're really out of alternative options, you may want to fill the smaller size tube with rich cement grout and use a 3/4 inch thick plywood between your load and rollers, such that you distribute the load to a wider contact surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure that when there is 2 rollers left under the load they share the load equally. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-06T19:20:23.383" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29534" PostTypeId="1" CreationDate="2019-08-06T20:19:08.350" Score="0" ViewCount="36" Body="&lt;p&gt;In the early days of steam engine powered ships, wouldn't a system that intermittently injects high pressure steam above a wide column filled with water, thereby displacing the water and expelling it through a nozzle connected at a 90° angle with the bottom of the column offer better efficiency than ungeared steam turbines? A valve would then open letting the steam escape and the water surge back in to the previous level and the cycle would repeat. &lt;/p&gt;&#xA;" OwnerUserId="18744" LastActivityDate="2019-08-07T00:04:26.070" Title="Turbineless steam propulsion for ships" Tags="|mechanical-engineering|steam|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29535" PostTypeId="1" AcceptedAnswerId="29536" CreationDate="2019-08-06T20:25:54.707" Score="0" ViewCount="40" Body="&lt;p&gt;If I understand it correctly in a CO&lt;sub&gt;2&lt;/sub&gt; laser engraver/cutter the laser itself is fixed in position and mirrors are used to redirect the laser to the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some mechanism used to keep the focal point on the material or is the focal point set to the midpoint of the machine's range? Wouldn't the beam be wider at the edges of the machines travel?&lt;/p&gt;&#xA;" OwnerUserId="21380" LastEditorUserId="3505" LastEditDate="2019-08-06T20:31:32.627" LastActivityDate="2019-08-06T21:06:02.570" Title="Laser engravers and focal points" Tags="|lasers|cutting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29536" PostTypeId="2" ParentId="29535" CreationDate="2019-08-06T20:47:32.987" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;... the laser itself is fixed in position and mirrors are used to redirect the laser to the material.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes. This system is called &quot;flying optics&quot;. The delicate laser is fixed in position and only the mirrors and lens move about. This simplifies the mechanical setup and reduces the inertia and required motor sizes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wouldn't the beam be wider at the edges of the machine's travel?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hUUjI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hUUjI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Divergence specification for Syntrad's &lt;a href=&quot;https://www.synrad.com/sites/default/files/2019-05/ti60%20Data%20Sheet%20US%204.10.19%20LoRes.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;ti60&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beam divergence can be quite small. The Synrad unit above has &amp;lt; 7 mrad which would give a divergence of up to 7 mm/m of beam length. The optical path (mirrors, etc.) at maximum range would have to have large enough mirrors and the final collimation lens diameter would have to cope with the enlarged beam. The other course of action would be to correct for the divergence at source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/axh1n.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/axh1n.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 2. See &lt;a href=&quot;https://www.scienceabc.com/innovation/what-is-a-laser-collimator.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Science ABC&lt;/a&gt; for example.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="13297" LastEditDate="2019-08-06T21:06:02.570" LastActivityDate="2019-08-06T21:06:02.570" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29537" PostTypeId="2" ParentId="29523" CreationDate="2019-08-06T21:08:15.577" Score="0" Body="&lt;p&gt;Pinhole-free silicone rubber membranes , ~.001 inches thick, can pass oxygen but will prevent water vapor from passing through. General Electric developed these types of membranes back in the 1960's but I don't know if anyone makes these or similar membranes anymore.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2019-08-06T21:08:15.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29539" PostTypeId="1" AcceptedAnswerId="29638" CreationDate="2019-08-06T22:58:54.010" Score="0" ViewCount="34" Body="&lt;p&gt;Hello dear engineering community.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a specific question about multiphase flows. I was wondering if anybody could help me with it, please.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume, I have a one dimensional water-air flow. Air is the dispersed phase. Let's say, it's a vertical flow. Water phase flow is incompressible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;QUESTION: if volume fraction (either of liquid or gas phase) is time dependent, then does it mean that that the flow is necessarily transient (either liquid or gas, or the entire mixture flow)? Or it can be steady?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="21383" LastEditorUserId="21383" LastEditDate="2019-08-13T23:25:54.650" LastActivityDate="2019-08-16T00:00:54.917" Title="How to infer from the volume fraction function whether a flow is transient or steady?" Tags="|fluid-mechanics|multiphase-flow|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29541" PostTypeId="2" ParentId="29534" CreationDate="2019-08-06T23:41:46.633" Score="1" Body="&lt;p&gt;If you did that, you've lost your steam. You have to keep it and run it to a condenser then back to the boiler. In the Titanic, the residual steam from the triple expansion piston engines was only 9 psi, down from about 215 psi. So I'm not sure what you had in mind by &lt;em&gt;high pressure steam&lt;/em&gt;. These gearless turbines were blowdown turbines. The steam volume at this point was just too great to keep adding cylinders to the engines. The turbines were more compact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What we are talking about, I hope - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=USawH8XxbdU&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=USawH8XxbdU&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But lets add a piston to separate the steam and the seawater. Now you can get your steam back, and you have what is called a &lt;a href=&quot;https://en.wikipedia.org/wiki/Free-piston_engine&quot; rel=&quot;nofollow noreferrer&quot;&gt;free piston engine&lt;/a&gt;. What you need to consider now is the size. The Titanic's turbine was 12 feet in diameter and expanded steam from 9 to 1 psig. How big would a two stroke free cylinder have to be to move the same amount of steam? How big would the exhaust valves have to be? They quit adding cylinders to the Titanic's piston engines when they could no longer manage the volume of steam and the heat losses from the piping would sap too much energy.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2019-08-07T00:04:26.070" LastActivityDate="2019-08-07T00:04:26.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29542" PostTypeId="1" AcceptedAnswerId="29585" CreationDate="2019-08-07T00:53:51.917" Score="1" ViewCount="137" Body="&lt;p&gt;I am trying to verify windfarmer generation estimates manually. To do so I am trying to manually calculate the estimated energy production (kWh) of a wind turbine. The information I have available is the &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Hourly weather data (wind speed, direction &amp;amp; air density at Hub Height, &#xA;and temperature) &lt;/li&gt;&#xA;&lt;li&gt;Power Curve of wind Turbine&lt;/li&gt;&#xA;&lt;li&gt;Thrust Coefficient of wind turbine&lt;/li&gt;&#xA;&lt;li&gt;Temperature de-ration curve of wind turbine&lt;/li&gt;&#xA;&lt;li&gt;Turbine specifications (blade radius, swept area, cut in &amp;amp; cut out speeds, Cp, Rated Power, etc)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;How would I use this data to estimate the potential annual generation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried multiplying hourly wind speeds * the specific value in the power curve, then de-rating the result based on the given temperature, However it seems to significantly overestimate the answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately I cannot share the data, but all I am looking for is a formula or calculation methodology. Any help be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="21385" LastActivityDate="2019-08-09T21:12:42.250" Title="How to estimate wind generation using power curve?" Tags="|electrical-engineering|energy|renewable-energy|wind-power|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29543" PostTypeId="2" ParentId="24438" CreationDate="2019-08-07T01:12:53.333" Score="2" Body="&lt;p&gt;In 1909, the same thing was done &lt;em&gt;twice&lt;/em&gt; on the Canadian Pacific Railway rail line through Kicking Horse Pass in the Rocky Mountains: &lt;a href=&quot;https://i.stack.imgur.com/EwtL3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EwtL3.jpg&quot; alt=&quot;Spiral tunnels at Kicking Horse Pass&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two circular tunnels (dashed white lines) were dug into the mountains, lengthening the route in order to decrease its steepness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is exactly the same situation and solution as in the original question.&lt;/p&gt;&#xA;" OwnerUserId="19930" LastEditorUserId="19930" LastEditDate="2019-08-08T14:23:30.137" LastActivityDate="2019-08-08T14:23:30.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29545" PostTypeId="2" ParentId="29532" CreationDate="2019-08-07T06:30:44.290" Score="0" Body="&lt;p&gt;There are many ways you can modify or design something to reduce resonance, and therefore reduce the risk of it being damaged by vibrations. In order to do so you need to know what frequency range the object will be experiencing, what it's made of, its dimensions, mass, etc., and can then do calculations (which are usually now done by computer software) to determine the resonant frequency of the object. You can then tune the object by changing some of the physical qualities mentioned above in order to move the resonant frequency out of the range of vibrations it might experience.&lt;/p&gt;&#xA;" OwnerUserId="21386" LastActivityDate="2019-08-07T06:30:44.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29549" PostTypeId="1" CreationDate="2019-08-07T11:56:32.697" Score="1" ViewCount="44" Body="&lt;p&gt;I read an article in a textbook which says stiffer rubber increases vibration damping and softer rubber increases vibration isolation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain this how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the difference between vibration damping and vibration isolation..?&lt;/p&gt;&#xA;" OwnerUserId="19588" LastActivityDate="2019-08-07T12:22:29.133" Title="Regarding rubber vibration properties" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29550" PostTypeId="2" ParentId="29549" CreationDate="2019-08-07T12:22:29.133" Score="3" Body="&lt;p&gt;Vibration isolators work because there is relative movement between different parts of the isolator. Softer rubber tends to create more relative movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vibration damping in rubber depends on the amount of strain energy transmitted through the damper. For the same vibration amplitude, higher stiffness tends to result in increased strain energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about the extreme situation where there is no connection at all between the two components (but something magical prevents the structure falling apart). The vibration isolation is perfect, because nothing is transmitted from one component to the other, but the amount of vibration damping is zero because nothing is absorbing energy from the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Warning: &lt;em&gt;general statements like the comment in your textbook can be misleading&lt;/em&gt; unless you think carefully about what is changing and what is kept constant. For example, changing the stiffness of part of the structure can change its vibration response (by change the resonant frequencies) independent of any effect on the system damping.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-08-07T12:22:29.133" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29553" PostTypeId="2" ParentId="13287" CreationDate="2019-08-07T15:34:11.790" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Plungers have solid plunger instead of a piston inside the cylinder cavity.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Plunger pumps produce pressures up to 200MPa, while piston pumps produce pressure at a maximum of 150Mpa.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;hope it helps to you useful&lt;/p&gt;&#xA;" OwnerUserId="21392" LastActivityDate="2019-08-07T15:34:11.790" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29554" PostTypeId="2" ParentId="13287" CreationDate="2019-08-07T17:54:33.673" Score="1" Body="&lt;p&gt;To the best of my knowledge, the main apparent difference between plunger and piston is that piston is short and plunger is long. You can see it on the picture in the @ericnutsch answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydraulic engineers employ plungers in different situations. Personally, I am only aware of two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First is pressure increase, as @GisMofx asked and as @PeerapongKladkrayang responded. Simplifying the computational schematic without loss of generality but emphasizing the considering effect, one can say that pressure produced by a pump equals to the force exerted on a piston/plunger rod divided by the piston/plunger area: &lt;span class=&quot;math-container&quot;&gt;$$p = \frac{F}{A}$$&lt;/span&gt; Let's consider two cylinders of the same crossectional area. Let's place a piston to the first cylinder and a plunger to the second one (as it is shown on the picture in the answer of @ericnutsch). Let's exert the same force, say &lt;span class=&quot;math-container&quot;&gt;$F = 100N$&lt;/span&gt;, on the rods of both of them. Let's say that the area of the cylinder and, therefore, the area of the piston is &lt;span class=&quot;math-container&quot;&gt;$A_{pist} = 10 m^2$&lt;/span&gt;; and the area of the plunger is two times smaller: &lt;span class=&quot;math-container&quot;&gt;$A_{plun} = 5 m^2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's compute what pressure we can obtain with piston and plunger in the same cylinder exerting the same force on them: &lt;span class=&quot;math-container&quot;&gt;$$p_{pist} = \frac{100N}{10m^2} = 10N/m^2$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$$p_{plun} = \frac{100N}{5m^2} = 20N/m^2$$&lt;/span&gt; See: in the same cylinder reducing the working crossectional area twice, I gain a double amplification in resulting pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, plunger allows me to increase working pressure without increasing the size of my cylinder. It may be helpful in some cases when I am restricted in space dedicated for cylinder installation. E.g., I am working on creation of a compact excavator. Everything needs to be as small as possible, providing as much resulting pressure as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tradeoff here is that I lose a lot of power on pumping liquid between the plunger and the inner cylinder wall. Hence, efficiency of a plunger pump should be smaller. It means that I can pump out less fluid than in a piston pump, and the outer power of plunger pump is smaller (it can be shown through some formulas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another application of plungers is in positive displacement pumps. Such as axial piston pumps, for instance. When a plunger moves in the cylinder, a liquid film between the plunger and the inner wall of the cylinder prevents the plunger from getting stuck. It is as if the liquid film supports the plunger just because the plunger is long. In turn, piston is short. And even if it had a supporting liquid film it would've easily gotten stuck, merely because it is short.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice, that distinguishing between plunger pump and piston pump is a product of engineering practice. It is not a scientific term. Therefore, there is no a strict general definition of a plunger and a piston. Many years ago, when engineers were designing early hydraulic devices, they figured out that it is a smart idea to reduce a working crossectional area in a hydraulic cylinder. That's it. If you tried and ask a fluid mechanics professor about piston/plunger definition - you would most likely get an answer that one is short and the other is long (as I did in my time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, notice that there is much more to add about the second application. But speaking about relative lengths of pistons and plungers - that's the only thing I can say.&lt;/p&gt;&#xA;" OwnerUserId="21383" LastEditorUserId="21383" LastEditDate="2019-08-07T18:07:37.910" LastActivityDate="2019-08-07T18:07:37.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29559" PostTypeId="1" CreationDate="2019-08-08T00:57:50.660" Score="-1" ViewCount="81" Body="&lt;p&gt;Is it possible to identify an electronics connector with the pinout diagram?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, this diagram shows an 18+2 pin configuration along with a key slot of some type at the top.  The diagram happens to have come from a recent model Nissan Pathfinder audio visual service manual, and is the connector responsible for connecting the amplifier to the speakers in the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xfSn9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xfSn9.png&quot; alt=&quot;pinout diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2020-09-01T04:06:10.643" Title="Identify electronics connector from pinout" Tags="|connections|electronics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29560" PostTypeId="2" ParentId="29559" CreationDate="2019-08-08T01:28:54.643" Score="-1" Body="&lt;p&gt;The pin out will be dependant on the equipment specified on the vehicle - some models have different levels of equipment due to marketing price points and the differences between models.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manufacturer may be identifiable by markings  on the connector itself. Worth looking at the connector in detail, sometimes the text / logo etc is small.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-08T01:28:54.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29561" PostTypeId="1" CreationDate="2019-08-08T02:21:28.170" Score="0" ViewCount="54" Body="&lt;p&gt;I am a very new student of engineering. I am in high school. The problem is my grandpa has an obsession with checking to see if the garage door is closed. So my dad is trying to get me to develop a solution. So far I have the idea of using photoelectric tripwire sensor to sense if the door is closed. But, I have no idea how to get the signal out of the sensor and to the lightbulb, or how to get the lightbulb to receive the signal.&lt;/p&gt;&#xA;" OwnerUserId="21394" LastEditorUserId="110" LastEditDate="2019-08-08T10:27:42.587" LastActivityDate="2019-08-08T12:26:01.657" Title="not sure how to use a signal from a sensor to make a lightbulb turn on or off" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29562" PostTypeId="2" ParentId="29561" CreationDate="2019-08-08T03:19:13.397" Score="4" Body="&lt;p&gt;Welcome to SE Engineering. If my dad gave me this task, I would do the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use any common variety reed switch sensor, you have probably seen them at Home Depot or your local hardware store. They come as a pair, you get a magnet assembly which mounts to the door and a molded reed switch which mounts to the door frame such that when the door opens the switch closes completing a separate circuit , in your case the separate circuit would be the coil of a relay which would turn a light on when the door is open. I will leave it for you to work out the electronic details, use LED's as your light source so you can use low voltages. You can find a myriad of such circuits online. You will have to play with the relative positions of the sensor and magnet so that the light only comes on when the door is not completely shut. Have fun !&lt;/p&gt;&#xA;" OwnerUserId="628" LastEditorUserId="628" LastEditDate="2019-08-08T03:25:53.483" LastActivityDate="2019-08-08T03:25:53.483" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29563" PostTypeId="1" CreationDate="2019-08-08T04:16:27.077" Score="1" ViewCount="245" Body="&lt;p&gt;I would like to show at high level that, in principle, the spectral efficiency (SE) of frequency division duplex (FDD) equals the SE of time division duplex (TDD) and both of them are half the SE of full duplex (FD).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Recall that in FD, the downlink (DL) and uplink (UL) transmissions utilize the whole spectrum for the whole time. On the other hand, in FDD these transmissions utilize half of the spectrum for the whole time and in TDD they utilize the whole spectrum half of the time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let us assume that an FD system transmits &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; bits in total (i.e., DL + UL) in &lt;span class=&quot;math-container&quot;&gt;$T_{\text{FD}} = T$&lt;/span&gt; seconds. Hence, the aggregated data rate is &lt;span class=&quot;math-container&quot;&gt;$R_{\text{FD}} = b/T$&lt;/span&gt; bits/s. The bandwidth utilized for the DL and UL transmissions is &lt;span class=&quot;math-container&quot;&gt;$B_{\text{FD}} = B$&lt;/span&gt; Hz. Therefore, the SE is &lt;span class=&quot;math-container&quot;&gt;$\text{SE}_{\text{FD}} = R_{\text{FD}} / B_{\text{FD}} = b/BT$&lt;/span&gt; bits/s/Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An equivalent FDD system would also need &lt;span class=&quot;math-container&quot;&gt;$T_{\text{FDD}} = T$&lt;/span&gt; seconds to send &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; bits and, therefore, would have an aggregated data rate of &lt;span class=&quot;math-container&quot;&gt;$R_{\text{FDD}} = b/T$&lt;/span&gt; bits/s. To this end, though, it would utilize a spectrum of &lt;span class=&quot;math-container&quot;&gt;$B_{\text{FDD}} = 2B$&lt;/span&gt; Hz. Thus, the SE of this FDD system would be &lt;span class=&quot;math-container&quot;&gt;$\text{SE}_{FDD} = R_{\text{FDD}} / B_{\text{FDD}} = b/2BT = \text{SE}_{FD}/2$&lt;/span&gt; bits/s/Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, an equivalent TDD system would need &lt;span class=&quot;math-container&quot;&gt;$T_{\text{TDD}} = 2T$&lt;/span&gt; seconds to send &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; bits and, therefore, would have an aggregated data rate of &lt;span class=&quot;math-container&quot;&gt;$R_{\text{FDD}} = b/2T$&lt;/span&gt; bits/s. For the DL and UL transmissions it would utilize a spectrum of &lt;span class=&quot;math-container&quot;&gt;$B_{\text{TDD}} = B$&lt;/span&gt; Hz. Hence, the SE of this TDD system would be &lt;span class=&quot;math-container&quot;&gt;$\text{SE}_{TDD} = R_{\text{TDD}} / B_{\text{TDD}} = b/2BT = \text{SE}_{FD}/2$&lt;/span&gt; bits/s/Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We see that, in principle, FDD and TDD have the same SE, which is half the SE of FD.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Is my reasoning correct or some of my assumptions is wrong? Does indeed TDD achieves an aggregated data rate that is half that of FDD (but, of course, the same SE as FDD since it uses also half of the bandwidth)?&lt;/p&gt;&#xA;" OwnerUserId="21395" LastActivityDate="2019-08-08T04:16:27.077" Title="Spectral Efficiency of FDD, TDD, and FD" Tags="|telecommunication|wireless-communication|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29564" PostTypeId="1" AcceptedAnswerId="29565" CreationDate="2019-08-08T05:10:38.863" Score="0" ViewCount="77" Body="&lt;p&gt;I purchased a pair of noise canceling headphones today and I was just wondering how exactly they are able to cancel out background noise? &lt;/p&gt;&#xA;" OwnerUserId="21396" LastActivityDate="2019-08-08T06:27:07.450" Title="How exactly do noise canceling headphones work?" Tags="|mechanical-engineering|electrical-engineering|manufacturing-engineering|process-engineering|computer-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29565" PostTypeId="2" ParentId="29564" CreationDate="2019-08-08T06:27:07.450" Score="2" Body="&lt;p&gt;Briefly they take the waveform of the ambient noise then produce the inverse waveform which is then generated. The two waveforms effectively cancel out leaving a much quieter ambient sound level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The important part is to be able to separate the wanted &quot;noise&quot; (music or voices) from the unwanted noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A search will give you several results that will provide much more detail.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-08T06:27:07.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29566" PostTypeId="2" ParentId="87" CreationDate="2019-08-08T11:30:51.927" Score="3" Body="&lt;p&gt;I know it's an old thread, but I'm putting my solution on here because I've just come across this issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Understanding the causes could help you deal better with hydrogen embrittlement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Overstressing:&lt;/strong&gt; &#xA;Primarily, extra stress put on material results in the form of blisters, cracks, hydride formation on the structures. In this case, the atomic hydrogen present in the atmosphere reacts with the material. This could lead to hydrogen embrittlement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electroplating Steel:&lt;/strong&gt;&#xA;Hydrogen embrittlement also occurs while plating and pickling due to the exposure of the parts to hydrogen. The only solution to prevent this baking the part immediately after plating. A typical temperature is to bake at 375ºF for 4 hours within 1 hour after plating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tensile strength &gt; 145 KSI:&lt;/strong&gt;&#xA;High-strength steels, titanium alloys and aluminium alloys are most vulnerable to hydrogen embrittlement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following resources may be useful &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.imetllc.com/training-article/hydrogen-embrittlement-steel/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydrogen Embrittlement of Steel&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://itafasteners.com/hydrogen-embrittlement-types-causes-and-solutions.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hydrogen Embrittlement - Causes and Solutions&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="21403" LastEditorUserId="8055" LastEditDate="2019-08-08T16:51:45.777" LastActivityDate="2019-08-08T16:51:45.777" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29567" PostTypeId="2" ParentId="29561" CreationDate="2019-08-08T12:26:01.657" Score="2" Body="&lt;p&gt;If you are very new to all of this, I think my recommendation would be to start with an electronics learner kit.  For example something like this: &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B0035XSZDI&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/Elenco-Electronic-Playground-Learning-Center/dp/B0035XSZDI&lt;/a&gt; (just an examples, there are many out there). Most of these will come will guided tutorials on wiring up basic circuits and explain how they work. Look for something that includes switches or sensors and lights/LEDs (i.e. all of the stuff you'll probably need for the garage door projects). Once you have some idea of how circuits work, then try the garage door projects.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-08-08T12:26:01.657" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29568" PostTypeId="1" CreationDate="2019-08-08T13:29:44.250" Score="0" ViewCount="404" Body="&lt;p&gt;When developing an ECR/ECN process, the question of when to roll a rev vs. when to create a new part number must be answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In researching this question, I've come across multiple &lt;a href=&quot;https://www.mcl.bz/blog/revision-or-new-part-number-f3-rule-helps-manufacturers-decide&quot; rel=&quot;nofollow noreferrer&quot;&gt;sources&lt;/a&gt; that state the principle of &quot;interchangeability&quot; as being the arbiter of this decision.  If the change is such that old and new parts can be co-mingled in the same parts bin, then only a drawing revision is required. If they must be kept separated because there is a change in Fit, Form, or Function (the F3 rule) that affects the next assembly, then the old and new parts must be kept separate and a new part number is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This seems reasonable on the face of it.  However, some things can change fairly often, such as firmware versions.  Strictly applying the rule would imply that any firmware version change requires a new part number, but this seems that it can rapidly become excessive as bugfixes and minor features are rolled out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be clear, I'm not asking about the version of the firmware itself - that is easily handled.  I'm referring to the PCBA into which the firmware is loaded.  We order our PCBAs from our vendor with firmware already loaded onto it.  For example, let's say that part number 12345 Rev A is the PCBA with firmware version 1.0.  If we then move to FW version 1.1 (for example, let's say we add another set of translation strings to support a new language in our menu system), I would argue that the mixing the old boards with the new would cause confusion in the case that the old boards are sent to someone who needs the new language.  So, the boards with FW v1.0 and FW v1.1 should not be mingled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the F3 rule then dictates that we should create a new part number with the new FW version, designate a new bin location to stock the parts, etc.  However, the rate at which seemingly minor changes in the FW (i.e. translations, bugfixes, etc) occur can lead to an explosion of part numbers, and consequently, warehouse bin locations, thus increasing cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone share an alternative for such a situation?&lt;/p&gt;&#xA;" OwnerUserId="21404" LastEditorUserId="110" LastEditDate="2019-08-08T21:29:41.233" LastActivityDate="2019-08-08T21:29:41.233" Title="New Part Number for every firmware update?" Tags="|drawings|engineering-managment|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29569" PostTypeId="1" AcceptedAnswerId="29570" CreationDate="2019-08-08T14:21:26.017" Score="0" ViewCount="106" Body="&lt;p&gt;I am a relatively recent BSME graduate working on reproducing the results of a technical paper to strengthen my ANSYS CFX skills. The paper is &quot;Effect of Section Thickness and Trailing Edge Radius on the Performance of NACA-65 Series Compressor Blades in Cascades at Low Speeds&quot; by Herrig, Emery, and Erwin. My goal is to reproduce pressure coefficient vs axial chord plots for one of the test cases using ANSYS CFX. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem Statement/Givens from Paper: The test was performed in a low-speed porous wall wind tunnel with an air average inlet speed of 95 ft/s. The particular airfoil geometry I am examining is an NACA 65-(12)12 (5 inch chord) at an angle of attack of 23.5 degrees with a solidity of 1.5. I made some assumptions not stated in the paper for my CFX model. I applied a free slip wall in place of the porous walls to capture the effect of boundary layer shedding. I also assumed the outlet static pressure was 1 atm and the inlet temperature is 15C. All other solid surfaces are applied as adiabatic walls with a no-slip condition. I am using the default k-epsilon turbulence model, isothermal heat transfer, and neglecting buoyancy effects. The pressure coefficient is to be determined along a mid-span plane on the surface of an airfoil far from the end walls (the third airfoil of the 5 I put in my cascade model). &lt;a href=&quot;https://i.stack.imgur.com/Cr7AW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cr7AW.png&quot; alt=&quot;Geometry produced in ANSYS&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Problem: The provided results from the paper have a minimum pressure coefficient of 0. My results, however, have a minimum ~1.05. I noticed all the pressure coefficient figures in the paper have a minimum at 0 and no negative values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Attempt at a Solution: I created a polyline at the intersection of the center blade surface and the mid-span plane. I wrote expressions in CFX Post to determine the area-averaged density, inlet velocity (although this is redundant), and static pressure. Then, I created a new variable to calculate the pressure coefficient. The calculation being performed is (P_static_local-P_static_inlet)/(1/2*rho_static_inlet*velocity_inlet^2). I've attached the results from the paper and my produced results. My trend seems to be mostly correct (save the waviness from producing the airfoil profile myself), however, the problem of the y-axis scale still remains. &lt;a href=&quot;https://i.stack.imgur.com/CXaCS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CXaCS.png&quot; alt=&quot;Results from Paper&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/MryAz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MryAz.png&quot; alt=&quot;Results from my Simulation&quot;&gt;&lt;/a&gt;. To complicate things further, the definition of the pressure coefficient they provide in the paper uses a local total pressure instead of a local static pressure, which changes the shape of the plot completely. &lt;a href=&quot;https://i.stack.imgur.com/TFJ4Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TFJ4Z.png&quot; alt=&quot;Variables from Paper&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/Bm5rk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bm5rk.png&quot; alt=&quot;My results using their definition of pressure coefficient&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know of a standard used to adjust the minimum pressure coefficient to 0? This is the only sort of logical explanation I can conjure at the moment. Of course, if anyone sees any glaring errors in my methodology please comment and let me know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your time&lt;/p&gt;&#xA;" OwnerUserId="21118" LastActivityDate="2019-08-08T14:41:08.613" Title="Pressure Coefficient with a Minima of 0 on Compressor Airfoils in Cascade" Tags="|fluid-mechanics|aerospace-engineering|aerodynamics|cfd|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29570" PostTypeId="2" ParentId="29569" CreationDate="2019-08-08T14:41:08.613" Score="1" Body="&lt;p&gt;In the report &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is the total pressure and &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the static pressure. See the page in the report before the one you attached in the OP.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With that definition, &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; will be zero at the stagnation point near the leading edge of the airfoil, and &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; will always be positive, as in their plots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;calculation being performed&quot; that you give for the pressure coefficient in the OP using two static pressures is not the same as the paper's definition.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-08-08T14:41:08.613" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29571" PostTypeId="2" ParentId="29568" CreationDate="2019-08-08T15:04:51.583" Score="2" Body="&lt;p&gt;If something is form-fit-function &lt;em&gt;backward&lt;/em&gt; compatible with an earlier rev., then usually you can just bump the rev -- depending on your customer base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most places that make changes as rapidly as you imply make the firmware field updatable, and track the product as &quot;Revision X hardware with revision Y firmware&quot; -- in no small part for the reasons you detail.  Even if you don't make the firmware field updatable, it's often wise to make it an easy update on final assembly, so that what you stock is the unprogrammed board, and loading firmware is done in some unit test step.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-08-08T15:04:51.583" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29574" PostTypeId="1" CreationDate="2019-08-08T21:10:46.973" Score="0" ViewCount="186" Body="&lt;p&gt;I'm working on a database which involves geometric design parameters, forces analysis and efficiency and losses for wormgear reduction design systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the tangential and axial forces applied to the gear reduction system, I need to relate the Sliding velocity with the friction factor for different materials combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this graphic chart in a technical paper. This graphic is exactly what I'm looking for, but I need a curve for steel worm and a polymer (Copolymer Acetal precisely) worm gear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9CcSD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9CcSD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be grateful if someone can help to find that material curve.&lt;/p&gt;&#xA;" OwnerUserId="14265" LastEditorUserId="14265" LastEditDate="2019-08-08T22:00:34.143" LastActivityDate="2020-05-05T03:05:45.527" Title="Friction factor for wormgear design materials" Tags="|mechanical-engineering|design|gears|friction|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29575" PostTypeId="1" CreationDate="2019-08-08T21:30:36.920" Score="0" ViewCount="343" Body="&lt;p&gt;What would be the typical efficiency of a large industrial single-stage squirrel-cage air blower?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example how many kilograms of ambient air (STP) would be moved per second by such blower which needs 20kW of electric power to turn its impeller (assuming 90% electric motor efficiency, make that 18kW of mechanical power at the impeller).&lt;/p&gt;&#xA;" OwnerUserId="21410" LastActivityDate="2019-08-08T22:25:52.517" Title="Squirrel-cage air blower efficiency" Tags="|airflow|industrial-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29576" PostTypeId="2" ParentId="29527" CreationDate="2019-08-08T22:01:38.497" Score="0" Body="&lt;p&gt;The question is somewhat vague, and so it is unclear whether you're asking about types/ categories of engines, or the actual cycle types. In general...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solid fuel rocket engines&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pros: &#xA;- Very simple and very reliable (no moving parts)&#xA;- High thrust&#xA;- Fuel is relatively safe for ground handling (often is just acrylic)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cons:&#xA;- Once they're lit... they're. No cutoff mechanism aside from burning through all the propellant&#xA;- Because of that (^), can be relatively inaccurate in terms of trajectory&#xA;- Typically non-reusable, but depends on housing material and characteristics&#xA;- Low ISP&#xA;- Large and heavy&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Liquid/ Bi-propellant/ Tri-propellant Engines&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pros: &#xA;- Moderate ISP&#xA;- Moderate/ High thrust&#xA;- Can be toggled on/ off&#xA;- Because of (^) can be used repeatedly&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cons:&#xA;- Much more complicated&#xA;- Much less reliable (many, many moving parts)&#xA;- Fuel (and overall system) is typically dangerous for personnel. Hydrazine a common liquid propellant is extremely toxic&#xA;- Requires 2 forms of media -- an oxidizer and a propellant&#xA;- Large and heavy&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electric Propulsion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pros: &#xA;- Extremely high ISP&#xA;- Designed for extremely long lifespans (and thus, highly reliable)&#xA;- Much smaller form-factor/ weight compared to liquid and solid engines&#xA;- Typically few to no moving parts&#xA;- Over long periods of time, can eventually reach velocities much much greater than that of the other forms of propulsion&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cons:&#xA;- Low thrust &#xA;- Because of (^) cannot be used for launch -- only in-space maneuvering&#xA;- Depending on the type, can be costly and complicated -- but this isn't always the case (such as with Phase Four's RF thruster)&lt;/p&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2019-08-08T22:01:38.497" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29577" PostTypeId="2" ParentId="29100" CreationDate="2019-08-08T22:12:14.627" Score="1" Body="&lt;p&gt;For mechanical devices, a good source of information is &quot;Mechanisms and Mechanical Devices Sourcebook&quot; by Sclater and Chironis (McGraw-Hill, 2007).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good book is &quot;507 Mechanical Movements - Mechanisms and Devices&quot; by Brown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A google search on &quot;mechanisms&quot; and &quot;machine design&quot; will throw up a lot of possibilities.  Beyond that, a patent search is a good idea.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2019-08-08T22:12:14.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29578" PostTypeId="2" ParentId="29575" CreationDate="2019-08-08T22:25:52.517" Score="1" Body="&lt;p&gt;These are called centrifugal fans and are typically from 60 to 90% efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are tested and rated by Air Movement and Control Association, AMCA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are usually tested under standard air conditions, which is clean, dry air at sea level 21 Celsius with barometric pressure of 29.92inches Mercury ( SCFM). The manufacturers usually have performance charts for different air pressure, temperature, density, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is Wikipedia link. &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Centrifugal_fan&quot; rel=&quot;nofollow noreferrer&quot;&gt;Centrifugal fan Wiki&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-08T22:25:52.517" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29579" PostTypeId="2" ParentId="29574" CreationDate="2019-08-09T01:59:21.990" Score="1" Body="&lt;p&gt;See &quot;&lt;a href=&quot;https://www.researchgate.net/profile/Christopher_Hooke/publication/223516306_Friction_and_wear_behaviour_of_acetal_and_nylon_gears/links/5b54b4cc45851507a7bdb613/Friction-and-wear-behaviour-of-acetal-and-nylon-gears.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Friction and wear behaviour of acetal and nylon gears&lt;/a&gt;&quot;, 2009, Wear 267 p. 639 for some torque-speed curves (acetal-acetal).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another source is &quot;Friction and wear of acetal: A matter of scale&quot;, 2005, Wear 259, p. 697.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A google scholar search will lead you to many other sources of useful data.  Whether you will get the data in the form of friction coefficient vs. sliding velocity is unclear from the papers I've looked at.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2019-08-09T01:59:21.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29581" PostTypeId="1" CreationDate="2019-08-09T14:51:22.513" Score="0" ViewCount="35" Body="&lt;p&gt;My knowledge in gears in limited to their types (bevel, spur, etc).  I am trying to combine the power/torque of two &lt;a href=&quot;https://hobbyking.com/en_us/25mm-shaft-universal-tri-motor-mount.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;DC motors&lt;/a&gt; using a differential gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a good handle on the idea and design from &lt;a href=&quot;https://engineering.stackexchange.com/a/2778/21402&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://www.quora.com/How-can-you-transfer-the-power-of-two-motors-onto-a-single-shaft&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I am having now is finding the correct differential gear to use (in term of shaft size, etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a PoC I don't mind using Lego or plastic.  Eventually these two motors will connect via another differential to two other motors (which have their own differential) for a total of 4 to hopefully turn the drive shaft of this &lt;a href=&quot;https://i.ebayimg.com/00/s/NTc2WDEwMjQ=/z/1zMAAOSwRB5c6AvE/$_72.JPG&quot; rel=&quot;nofollow noreferrer&quot;&gt;kettcar&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My overarching question is:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How/where do I choose the correct differential to use?&lt;/li&gt;&#xA;&lt;li&gt;How do I scale up/down the motor's output shaft to fit into whichever differential gear I end up using?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As to &quot;why&quot; do this:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Learning&lt;/li&gt;&#xA;&lt;li&gt;Learning&lt;/li&gt;&#xA;&lt;li&gt;Fun&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That said, if this approach is bat-crap crazy, then please feel free to say so and suggest an alternative.&lt;/p&gt;&#xA;" OwnerUserId="21402" LastActivityDate="2019-08-09T14:51:22.513" Title="How To Obtain The Correct Differential Gears With Motor Shaft Adapters" Tags="|motors|gears|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29582" PostTypeId="2" ParentId="29527" CreationDate="2019-08-09T15:12:27.327" Score="1" Body="&lt;p&gt;Disclaimer: I am not professional rocket scientist, information is primarily gathered through internet. This answer is provided with NO warranty.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This answer mainly focuses on liquid fuel engines.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Pressure-fed&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Very simple design that does not have pumps. Can be easily started. Chamber pressure (and thus the specific impulse) is very limited compared to other cycles. Mainly used in upper stages as low chamber pressure makes operation in atmosphere difficult.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Expander cycle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Simple design that uses cryogenic propellant to drive the pump (and cool the engine). Chamber pressure is similarly limited, also commonly used in upper stages.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Gas-generator cycle&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;More mechanically complex than the previously mentioned cycles, as it uses a pre-burner to drive the pump. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are several advantages to the gas-generator cycle over its counterpart, the staged combustion cycle. The gas generator turbine does not need to deal with the counter pressure of injecting the exhaust into the combustion chamber. This simplifies plumbing and turbine design, and results in a less expensive and lighter engine. - Wikipedia&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Chamber pressure is not as limited as in the previously mentioned cycles. It can be (and often is) used in first stage, as the low complexity makes developing larger engine easier.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Oxidizer-rich staged combustion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;A more advanced design. staged combustion allows greater chamber pressure and higher specific impulse, at the cost of heavier and more complex engine. This cycle is usually used by LOX/Kerosene engines, as the counterpart - Fuel-rich staged combustion cannot be used in LOX/Kerosene engines, due to extreme coking problem. Oxidizer-rich environment  proves to be extremely difficult to handle. Russia developed alloy materials to withstand the high-temperature and oxidizing environment.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Fuel-rich staged combustion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Similar to oxidizer-rich staged combustion, this cycle is mainly used by LOX/LH2 engines (and any other engine that does not have coking problem and wants high specific impulse). I guess the fuel-rich environment is easier to handle than the oxidizer-rich one.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Full-flow staged combustion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Extremely complex design that has the greatest specific impulse and mechanical complexity. The only built examples are RD-270 and Raptor. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Conclusion&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Complexities of engine cycles are usually positively correlated to development costs. Reliability depends on the exact design and technology of the engine (e.g. redundancy)&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Reference&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Encyclopedia Astronautica &lt;a href=&quot;http://www.astronautix.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.astronautix.com/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The English Wikipedia&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="21416" LastActivityDate="2019-08-09T15:12:27.327" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29583" PostTypeId="2" ParentId="29151" CreationDate="2019-08-09T20:23:09.730" Score="0" Body="&lt;p&gt;Which property package are you using? What type of mixture are you purifying? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because if it is non-ionic I would much better recommend you to use HYSYS and you can use a custom model in BVA and excel workbook for AspenHYSYS. &#xA;Calculator blocks and spreadsheets are often not the best options because they seek to solve the balance on its entirety and it favors the property package (which seeks to minimize error regardless of the results you seek). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would happily help if you give me conditions for the Xenon you are trying to use and the pressure drop you are attempting. &lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2019-08-09T20:23:09.730" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29584" PostTypeId="2" ParentId="29375" CreationDate="2019-08-09T21:11:48.280" Score="1" Body="&lt;p&gt;If you want to avoid re-auguring holes, you could go out and get some 10' (~3m) long 2&quot;X6&quot; (or 2X8).  You can run these in the direction of your corrugated plastic roofing.  Probably want at least 3 per panel, edges and middle, though you may get away with just edges.  Lay your Corrugated sheets on top of this.  The 2X6s should take care of the bending.   This would only be for rain loading.  no one walking on top or snow accumulation.  The deeper the section the more capacity you will get.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you only support the edges of the panel, they may sage in the other direction (perpendicular to the corrugations)&lt;/p&gt;&#xA;" OwnerUserId="6993" LastActivityDate="2019-08-09T21:11:48.280" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29585" PostTypeId="2" ParentId="29542" CreationDate="2019-08-09T21:12:42.250" Score="1" Body="&lt;p&gt;For an approximate estimate, use&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ P = \frac{1}{2}\rho C_p A u^3 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the air density, &lt;span class=&quot;math-container&quot;&gt;$C_p$&lt;/span&gt; is the power coefficient (read off the power curve that you have, where it will depend on the wind speed), &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is the swept area of the rotor, and &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; is the wind speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have hourly data, calculate this for each hour. That will give you a power in Watts (or kW, or MW), so you can then assume that that is the average output over the hour centred on the time of the weather measurements, and hence that the energy generated in that hour is the same number of Wh (or kWh, or MWh).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try to manually include temperature de-rating (I don't know exactly how this works, but if you have a curve you could apply it), but if your aim is to check that the Windfarmer result is sensible, this ought to get you pretty close.&lt;/p&gt;&#xA;" OwnerUserId="320" LastActivityDate="2019-08-09T21:12:42.250" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29586" PostTypeId="2" ParentId="26980" CreationDate="2019-08-09T21:20:47.953" Score="1" Body="&lt;p&gt;Part of the answer to this will be about considering how one optimizes the locations of wind farms and bridges. One of them likes to have a lot of wind. The other (well, the trucks on it) prefers to avoid this. So co-locating the two would necessarily involve compromises. That doesn't mean it's impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More pragmatically, a large wind turbine develops very large forces that try to tip it over: the tower has to take large lateral (sideways) loads. Bridge towers are typically designed for vertical, not horizontal load. My instinct here is you wouldn't be so much &quot;attaching wind turbines to the bridge&quot; as &quot;building a bridge along the line of some offshore wind turbines&quot;. Which in capex terms is probably going to cost nearly as much as building the two things separately, although the O&amp;amp;M cost of the wind farm should be substantially reduced.&lt;/p&gt;&#xA;" OwnerUserId="320" LastActivityDate="2019-08-09T21:20:47.953" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29587" PostTypeId="2" ParentId="28934" CreationDate="2019-08-09T23:08:36.633" Score="1" Body="&lt;p&gt;As you mentioned in your update you're going to want to start with a PCB edge mounted tactile switch. These can be quite small. So now what you've got to do is add a button.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A common solution would be to use a small plastic button with a&#xA;retaining flange. You've said you're worried about this rattling or&#xA;jamming, but 1. the travel is very short, so it won't jam. And 2. The&#xA;button will be so small that the rattling will be inaudible. The buttons in your prototype sound like they were designed with too much linear travel.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another option is to make the button part of the case, and retain it&#xA;using a living hinge. For a cheap product this is probably the way to&#xA;go, because it eliminates a part.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A 3rd, weirder option is to make the case flexible enough that the&#xA;button can be pressed &lt;em&gt;through it&lt;/em&gt;. Unfortunately this is not well suited to an edge button.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A flexible sticker or silicone coating can be added over the preceding assemblies to add some water/dust resistance.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-08-09T23:08:36.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29588" PostTypeId="1" CreationDate="2019-08-10T04:24:53.907" Score="2" ViewCount="21879" Body="&lt;p&gt;I don't know too much about USB-PD. I understood that USB-PD requires negotiation between the charger and the receiver. So probably my Windows laptop (Dell XPS 9570) should be able to know the current wattage input it's receiving. Windows probably also know since it's popping up saying slow charging. So I'm curious to know what wattage is my Dell is getting and is it getting the 100w it supposed to get?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My setup is a Dell XPS 9570 connected to Razer Core X Chroma eGPU. I've tried two cables, the Thunderbolt that came with the Razer Core X and the Apple Thunderbolt 3 0.8m, both of them still give that slow charging pop up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="21420" LastActivityDate="2023-12-21T16:07:25.920" Title="How to detect what wattage that my Windows laptop is receiving via USB-PD?" Tags="|power-electronics|computer-engineering|computer-hardware|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29590" PostTypeId="2" ParentId="29469" CreationDate="2019-08-10T16:13:25.883" Score="0" Body="&lt;p&gt;A piezo electric accelerometer would work here. Most general purpose accelerometers would be fine. Resolution is probably not that important if all you care about is the frequency. You'll also need a charge amplifier. A used accelerometer and charge amp could be had on eBay for pretty cheap. But then you'll also need either an oscilloscope/spectrum analyzer to read and display the signal (may be expensive even for a used one) or buy a data acquisition board and write software to do the signal processing (could be very challenging if you are a beginner).  Probably best route is to try to take a class (maybe at local community college) or buy an educational electronics starter kit. Either one will get you some experience and get you closer to being able to figure out this project &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-08-10T16:13:25.883" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29591" PostTypeId="1" CreationDate="2019-08-10T11:09:38.440" Score="1" ViewCount="245" Body="&lt;p&gt;I did not find any answers to this simple problem across the web, maybe seeking help in this forum helps:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, let's assume we have multiple windmills, each of them producing torque, but each of them produces different amounts of torque, because they are not equally exposed to wind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, how could one mechanically build some DIY gears and hacks to accumulate the energies/torques of the windmill-outputs together, without having back-effects to the windmills?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want ONE output gear, which has the summed-up energy of all input-gears (windmills), and when input-gear-1 would turn the output gear faster than the input-gear-2, input-gear-2 should also be able to add its torque to the output-gear (instead of acting rather as a brake for the output gear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't imagine that such a simple idea/problem has no solution yet in 2019, but I have to add that I am a complete noob in mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Important: I forgot to mention that I want to build this WITHOUT electricity or the use of any motors.&#xA;Think in terms of plastic-garden windmills. Little, wooden or plastic gears. Is it possible to achieve this with plain mechanics? (No electricity or magnetism involved?)&lt;/p&gt;&#xA;" OwnerDisplayName="ElectRocnic" LastActivityDate="2019-08-11T02:32:12.597" Title="How to accumulate torque of multiple independent gears?" Tags="|transmission|gears|torque|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29592" PostTypeId="2" ParentId="29591" CreationDate="2019-08-10T11:39:59.477" Score="0" Body="&lt;p&gt;We have a solution that is far more efficient in terms of material and space : each wind turbine produces electricity that can then be used in just about any amount necessary.&lt;/p&gt;&#xA;" OwnerUserId="10902" OwnerDisplayName="Solar Mike" LastActivityDate="2019-08-10T11:39:59.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29593" PostTypeId="2" ParentId="29591" CreationDate="2019-08-10T12:50:44.897" Score="2" Body="&lt;p&gt;There is a solution for this and it's called planetary gearset.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All Toyota hybrids (at least the traditional not-PHEV ones) have a planetary gearset that sums together the power produced by motor-generators and the internal combustion engine. With two motor-generators, it also allows infinitely finely adjustable CVT that is electrically adjusted. So, it replaces the traditional automatic transmission and doesn't need a torque converter anymore like traditional CVTs do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actually, technology very similar to this is used in differentials to distribute power from one engine to multiple wheels, but the technology to do the opposite, to combine power from multiple shafts would be basically the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In wind turbines, the problem is that the rotation speed of the wind turbine varies, whereas electrical grid is constant at 50 Hz or 60 Hz. So, an entirely different problem than what you claim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This problem is solved in &lt;a href=&quot;https://en.wikipedia.org/wiki/Doubly-fed_electric_machine&quot; rel=&quot;nofollow noreferrer&quot;&gt;doubly-fed electric machines&lt;/a&gt; where the magnetic field is produced by electricity and is made to rotate at a rate that is the difference between the turbine rotation speed and the electric grid &quot;rotation frequency&quot; (50 Hz or 60 Hz). Then the generator produces constant 50 Hz or 60 Hz even though the turbine may rotate at any speed. Also, you need an inverter only for the magnetic field, not for the entire electrical power produced by the machine. Much cheaper that way with wind turbines when compared to e.g. solar cells that produce DC and require an inverter for basically all of the power they produce.&lt;/p&gt;&#xA;" OwnerUserId="12836" OwnerDisplayName="juhist" LastActivityDate="2019-08-10T12:50:44.897" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29594" PostTypeId="2" ParentId="29396" CreationDate="2019-08-11T00:43:56.513" Score="2" Body="&lt;p&gt;I'm not sure what system of columns and beams you're describing, but the answer appears to be in the question: the self-weight of the columns has no effect on the support bending moment reactions.  You will use it in reactions at the column footings, but otherwise ignore it.&lt;/p&gt;&#xA;" OwnerUserId="21428" LastActivityDate="2019-08-11T00:43:56.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29595" PostTypeId="2" ParentId="29591" CreationDate="2019-08-11T02:32:12.597" Score="0" Body="&lt;p&gt;It's hard to tell what kind of answer is in scope because it isn't clear why you don't want one windmill to be a brake on another.  It's not like you're losing the power - the &quot;braking&quot; windmill is turning faster.  Any load you put on a windmill is going to cause such a braking effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have read that windmills work best under light torque - there's less wear on the gearing for one thing.  So maybe that's a reason.  Also, if you end up driving one windmill faster than the wind, that wouldn't be good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planetary gearset mentioned in another answer are one solution.  A planetary gearset sums the rotation of two rotating inputs (weighted according to gear ratios) into one rotating output.  The power load of the output is divided between the inputs according to the gear ratios and the characteristics of each input driver determines how that power allocation is divided between rotation speed and torque of that driver. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A torque converter such as is used on an automobile automatic transmission is another possibility.  Just as it allows an engine to run faster than the input shaft of the transmission, it would allow one windmill to run faster than another connected one, while giving the latter windmill's output a boost.  A  simple fluid coupling might also do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't want to convert to electric energy, but how about liquid or gas?  Have each windmill pump water into a pipe and run that water through a turbine.&lt;/p&gt;&#xA;" OwnerUserId="21428" LastActivityDate="2019-08-11T02:32:12.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29596" PostTypeId="1" AcceptedAnswerId="29603" CreationDate="2019-08-11T05:47:59.973" Score="1" ViewCount="106" Body="&lt;p&gt;I'm thinking of big halls, especially those that use chandeliers for light. The idea's simple: build a parabolic ceiling that uses mirrors, place the chandelier at the focus, and illuminate the entire hall with light. Since the light would be coming from the ceiling, it wouldn't be possible to &quot;stand in someone's light&quot; either, which would be great for e.g. exams. One would also be able to use just one light source for the entire hall. Why hasn't this been done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I'm missing some obvious reason but I can't place it. One guess could be that big parabolic mirrors are expensive, but we have such mirrors elsewhere (e.g. for telescopes, science museums). It could be that such a hall would look really bright if one looks upwards (at the ceiling), but that could be mitigated by, e.g., placing a one-way mirror between the ceiling and ground level that allowed light down but not up. What am I missing?&lt;/p&gt;&#xA;" OwnerUserId="21430" LastEditorUserId="19930" LastEditDate="2019-08-11T13:53:32.713" LastActivityDate="2019-08-11T18:48:50.807" Title="Why don't we have parabolic ceilings with mirrors &amp; a light at the focus?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29597" PostTypeId="1" AcceptedAnswerId="29599" CreationDate="2019-08-11T10:23:17.000" Score="1" ViewCount="50" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bWPP8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bWPP8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So i need to calculate the angular velocity of the dark disk, &lt;span class=&quot;math-container&quot;&gt;$w^*_3$&lt;/span&gt;.&#xA;In the master solution they state it is : &lt;span class=&quot;math-container&quot;&gt;$w^*_3 = w_3 + \overline{w_3}$&lt;/span&gt;&#xA;with :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$w_3 = \frac{|\dot{r}|}{4a}$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$|\dot{r}| = v_a + \overline{w_3}\cdot a$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$|\dot{r}| = v_b - \overline{w_3}\cdot 2a$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \Rightarrow|\dot{r}| = 2a\cdot (w_S+w_R)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\overline{w_3} = \frac{1}{a}(|r|-v_1)=\frac{1}{2a}(v_b-|r|)=\frac{1}{3a}(v_b-v_a)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \Rightarrow \overline{w_3} = 2w_r-w_s$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \Rightarrow w_3^* = \frac{5}{2}w_r-\frac{1}{2}w_s$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the question I have is that if I understood correctly,  &lt;span class=&quot;math-container&quot;&gt;$w^*_3$&lt;/span&gt; is the angular velocity of the disk itself (in its coordinate system) if you look at it from the disks center. So shouldn't be &lt;span class=&quot;math-container&quot;&gt;$\overline{w_3}$&lt;/span&gt; already be my &lt;span class=&quot;math-container&quot;&gt;$w^*_3$&lt;/span&gt;? Because it is the angular velocity with which my disk is actually spinning?(in it's coordinate system)&lt;/p&gt;&#xA;" OwnerUserId="21432" LastActivityDate="2019-08-11T16:38:11.870" Title="Question regarding calculation of angular velocity connected to coordinate transformation" Tags="|mechanical-engineering|structural-engineering|mathematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29599" PostTypeId="2" ParentId="29597" CreationDate="2019-08-11T16:26:24.443" Score="0" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega ^*_3$&lt;/span&gt; is the rotation rate of the disk in the frame pictured (equal to &lt;span class=&quot;math-container&quot;&gt;$2 * pi *$&lt;/span&gt; frequency  point &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; will appear in the 9 o'clock position). It is the sum of the orbit rate of &lt;span class=&quot;math-container&quot;&gt;$\omega _3$&lt;/span&gt;'s disc center about the system center in the frame pictured (&lt;span class=&quot;math-container&quot;&gt;$\omega_3$&lt;/span&gt;), plus the rotation rate about it's own axis in a reference frame where this axis is fixed (&lt;span class=&quot;math-container&quot;&gt;$ \bar\omega_3$&lt;/span&gt;). The first term considers the contact points with &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; fixed and the system rotating with disk &lt;span class=&quot;math-container&quot;&gt;$\omega_3$&lt;/span&gt;, so one turn of the center is one rotation of the disk, and point P makes one revolution. The second term considers  &lt;span class=&quot;math-container&quot;&gt;$\omega_3$&lt;/span&gt;'s axis fixed and the contact points free to run. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take &lt;span class=&quot;math-container&quot;&gt;$\omega_S$&lt;/span&gt; as given in each case, we get two different (linear) relations for &lt;span class=&quot;math-container&quot;&gt;$\omega_R$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\omega_3$&lt;/span&gt;. So &lt;span class=&quot;math-container&quot;&gt;$\omega_R = C_1*\omega_S$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\bar\omega_R= C_2*\omega_S $&lt;/span&gt; This lets us set up a general solution &lt;span class=&quot;math-container&quot;&gt;$\omega^*_3 = f(\omega_S,\omega_R)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_2$&lt;/span&gt; are solved using &lt;span class=&quot;math-container&quot;&gt;$\omega_S$&lt;/span&gt; and then they are used to compute &lt;span class=&quot;math-container&quot;&gt;$\omega^*_3$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2019-08-11T16:38:11.870" LastActivityDate="2019-08-11T16:38:11.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29600" PostTypeId="2" ParentId="28421" CreationDate="2019-08-11T17:38:03.170" Score="4" Body="&lt;p&gt;&lt;strong&gt;Step 1:&lt;/strong&gt; Draw the root locus of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lCrwe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lCrwe.png&quot; alt=&quot;Root locus of the system&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;Here you can see the two poles of your plant &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt; (marked with an x), at &lt;span class=&quot;math-container&quot;&gt;$p_1=-9$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$p_2=-1$&lt;/span&gt;, the pole of your controller &lt;span class=&quot;math-container&quot;&gt;$C(s)$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$p_c = 0$&lt;/span&gt; and the zero (marked with an o) at &lt;span class=&quot;math-container&quot;&gt;$z_c = -c$&lt;/span&gt; (for now just at a random location).&#xA;The purple squares indicate the poles of the closed loop system for a certain gain &lt;span class=&quot;math-container&quot;&gt;$K_p$&lt;/span&gt; and can only move over the blue lines (the root locus).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2:&lt;/strong&gt; Add the design requirements to your root locus.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Less than 10% overshoot&lt;/strong&gt; can be indicated by adding diagonal lines through the origin with a slope of &lt;span class=&quot;math-container&quot;&gt;$\frac{\Re}{\Im} = \pm\left|\frac{\ln(10\%)}{\pi}\right|\approx \pm 0.73$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\Re$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Im$&lt;/span&gt; represents the real and imaginary axis respectively. All the closed loop poles of your system have to be between these two lines in the left open half plane.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Rise time less than 1.2 seconds&lt;/strong&gt; does not have an analytic expression so cannot be translated directly to the root locus. My suggestion is to rewrite this requirement as a more conservative requirement on the settling time, such that the system is at least at 10% of it's final value within 1.2 seconds. The settling time requirement can be represented as a vertical line at &lt;span class=&quot;math-container&quot;&gt;$\Re = \frac{-\ln(0,1)}{t_s(10\%)} \approx -1.9$&lt;/span&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7xdKN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7xdKN.png&quot; alt=&quot;Root locus with design requirements&quot;&gt;&lt;/a&gt;&#xA;Here you can see the design requirements, the diagonal lines indicate the overshoot at 10% and the vertical line indicates the settling time of 1.2 seconds. To fulfill all the design requirements, all the poles of the closed loop system (purple squares) have to be in the white area. Hence for the current configuration were do not satisfy the design requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3:&lt;/strong&gt; is to tune the system. Since the purple block (pole) at the right cannot pass the zero and thus not fulfill the design requirements, we start with replacing the zero (e.g. at -1.5, thus &lt;span class=&quot;math-container&quot;&gt;$c = 1.5$&lt;/span&gt;). This will change the root locus significantly.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/p6H7Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p6H7Y.png&quot; alt=&quot;Root locus with replaced integrator zero.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we can start changing the gain &lt;span class=&quot;math-container&quot;&gt;$K_p$&lt;/span&gt; until all blocks are in the white area (e.g. &lt;span class=&quot;math-container&quot;&gt;$K_p = 2.4$&lt;/span&gt;)&#xA;&lt;a href=&quot;https://i.stack.imgur.com/PEYjo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PEYjo.png&quot; alt=&quot;Root locus with updated gain&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, the controller &lt;span class=&quot;math-container&quot;&gt;$C(s) = \frac{2.4(s+1.5)}{s}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 4:&lt;/strong&gt; Check if the designed controller fulfills the design requirements by taking the step response.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/47DsO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/47DsO.png&quot; alt=&quot;Step response of closed-loop system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be seen that the overshoot &lt;span class=&quot;math-container&quot;&gt;$D = 8\%$&lt;/span&gt; and the rise time &lt;span class=&quot;math-container&quot;&gt;$T_r = 0.593 -0.106 = 0.487$&lt;/span&gt; seconds thus all requirements are fulfilled with the current controller.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2019-08-11T17:38:03.170" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29601" PostTypeId="1" AcceptedAnswerId="31812" CreationDate="2019-08-11T18:08:52.150" Score="0" ViewCount="172" Body="&lt;p&gt;a long time back at school, I had a difficult excercise which was named X, Y and Z bodies.&#xA;They were drawings where both the front view and the top view were fully given. We had to find the view from the left plus the body in a sketch drawing.&#xA;Unfortunately, I don't remember the solutions anymore and I don't find out them anymore.&#xA;Could anyone help?&#xA;As an example, I added the Y body with below link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4GIoN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4GIoN.png&quot; alt=&quot;Y body&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ah, yes: the forms are squares and the middle of the Y is the middle of the squares - in both views.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help and hints in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best&#xA;Tom&lt;/p&gt;&#xA;" OwnerUserId="21433" LastEditorUserId="21433" LastEditDate="2019-08-11T18:34:05.217" LastActivityDate="2019-11-05T16:11:26.877" Title="Strange CAD excercises" Tags="|technical-drawing|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29603" PostTypeId="2" ParentId="29596" CreationDate="2019-08-11T18:48:50.807" Score="4" Body="&lt;p&gt;In the case of a mirrored paraboloid with a lamp at the focus, all the light rays would shine directly downward. Any object would have a sharply defined shadow directly below it; people would look sinister with dark eye sockets and nostrils. This is known as &lt;em&gt;hard&lt;/em&gt; light.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Contrast that with a light-colored ceiling (of any shape), which reflects light in many directions. This creates &lt;em&gt;soft&lt;/em&gt; light, and is more pleasant to be in.&lt;/p&gt;&#xA;" OwnerUserId="10706" LastActivityDate="2019-08-11T18:48:50.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29606" PostTypeId="1" CreationDate="2019-08-12T03:40:44.423" Score="1" ViewCount="386" Body="&lt;p&gt;First post here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm an undergraduate ME interested in fluid mechanics. I came across scramjets and I was interested about the engine. I want to calculate the mass flow rate of the engine and I made an attempt because it doesn't seem right and I can't find any resource explaining how to calculate the mass flow rate of a scramjet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I arrived at this using standard definitions: &lt;span class=&quot;math-container&quot;&gt;$ \dot m = (\frac {P}{RT}AM\sqrt{\gamma RT})  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is the absolute pressure of free stream (air, in this case)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the gas constant for air&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is cross sectional area  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; is the Mach number &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is absolute temperature &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; is ratio of specific heats &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be honest, I have no strong intuition about this since I'm only a junior undergrad who only has taken Thermo and Fluids (no heat transfer yet, compressible flow not in curriculum) but if you guys can steer me in the right direction I would be very happy. I sense that it should be more geometry dependent, maybe angle of attack should be considered as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you everyone!&lt;/p&gt;&#xA;" OwnerUserId="21437" LastActivityDate="2023-12-20T06:12:31.543" Title="Calculating mass flow rate of a scramjet engine?" Tags="|mechanical-engineering|fluid-mechanics|aerospace-engineering|aerodynamics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29607" PostTypeId="1" CreationDate="2019-08-12T07:02:43.423" Score="1" ViewCount="1273" Body="&lt;p&gt;I was going through a review of rotary compressor and i understood that the stagnation pressure and stagnation temperature both change in the rotor stage (as energy is being added).I also read that in the stator stage the stagnation temperature remains a constant but there is loss of stagnation pressure. If there is a loss in stagnation pressure shouldn't the stagnation temperature increase (I am guessing here that the loss is converted into heat)? (I am guessing this has to do something with expansion but I am not sure)&lt;/p&gt;&#xA;&lt;p&gt;additional info: &lt;a href=&quot;https://physics.stackexchange.com/questions/292589/why-does-a-diffuser-increase-pressure&quot;&gt;https://physics.stackexchange.com/questions/292589/why-does-a-diffuser-increase-pressure&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.physicsforums.com/threads/why-does-the-total-temperature-not-change-in-adiabatic-flow.812811/post-5113056&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.physicsforums.com/threads/why-does-the-total-temperature-not-change-in-adiabatic-flow.812811/post-5113056&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16275" LastEditorUserId="16275" LastEditDate="2022-09-17T15:58:53.717" LastActivityDate="2022-09-17T15:58:53.717" Title="Why stagnation temperature doesn't change in stator" Tags="|compressors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29609" PostTypeId="1" CreationDate="2019-08-12T10:09:25.860" Score="1" ViewCount="112" Body="&lt;p&gt;I have a configuration with two bolts and a point of force application (see figure).&#xA;I understand that, if perfectly rigid, the two parts depicted in the figure will behave identically in how they will transfer moment on the bolts from the point of force application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this specific case, there would be a torsion, that is moment on the upper bolt going forward and moment on the lower bolt going backward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: what options are there to keep the configuration (bolts, point of application of the force), but straighten the moment  on the bolts?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would having a shape like the left one made of a flexible material change anything?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oeOeU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oeOeU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;: to clarify, in the picture below i added, in red the resulting forces on the bolts using either of the two depicted rigid shapes, and the desired resulting forces on the bolts&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4D84w.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4D84w.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21441" LastEditorUserId="21441" LastEditDate="2019-08-12T20:02:46.033" LastActivityDate="2019-08-13T05:32:50.973" Title="Design a part to straighten moment and reduce torsion" Tags="|structural-engineering|structural-analysis|structures|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29612" PostTypeId="2" ParentId="29609" CreationDate="2019-08-12T14:26:07.333" Score="1" Body="&lt;p&gt;If you can have the bases extend up a little so that the center of the two holes is at level with the point of application of the force you won't have torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets call the lateral force on the screws &lt;span class=&quot;math-container&quot;&gt;$ F_A \text{ and } F_B \text{ when the load on the handle is } F_H$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call the up screw A and down screw B and the line section between them &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;. And let's call the projection of the point of force on this line x.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F_A= F_H\cdot\dfrac{L-x}{L} \text{ and } F_B= F_H * \dfrac{x}{L} \\&#xA;\text{and if } x = L-x \text{ then } x = \dfrac{L}{2} \\&#xA;\text{therefore } F_A = F_B \text{ and there is no torque.}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of the lateral loads on the bolts/screws, you always have a pull-out force on the screws.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After OP's comment saying they can not change the handle geometry but they could possibly add some bolts to the right of the point of application of force:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we add two more bolt with the order of sizes you have in mind at the right-hand side such that together with the original bolts they make a square of 10cmx10cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now let's calculate what is the maximum&lt;span class=&quot;math-container&quot;&gt;$ \ F_H $&lt;/span&gt; which is not going to create a negative reaction ( reaction pointing to the left).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By what you said the torque of the force on the handle is F times(5cm height of the handle offset+ 5cm from side of the base square to it CG=10cm,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque = &lt;span class=&quot;math-container&quot;&gt;$  F_H*10cm$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Horizontal bolt force without the effect of torque &lt;span class=&quot;math-container&quot;&gt;$V= F_H/4 $&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resistance of the bolts to the torque is the sum of the squares of the bolts distances to the center of the bracket: &lt;span class=&quot;math-container&quot;&gt;$ I_{baseplate}= \Sigma (F_{bolt}* r^2 )=4*F_B*50cm^2 $&lt;/span&gt; and each bolt contribution to resisting torque is &lt;span class=&quot;math-container&quot;&gt;$$ F_B= F_H *10*7.07/ 4*F_B*50cm^2 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The critical bolts are the upper right-hand bolt and the lower left-hand bolts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In those two bolts as long as &lt;span class=&quot;math-container&quot;&gt;$F_B \text{ due to torque }&amp;lt; \frac{F_H}{4}  $&lt;/span&gt; we are ok.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-08-13T05:32:50.973" LastActivityDate="2019-08-13T05:32:50.973" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29613" PostTypeId="2" ParentId="29609" CreationDate="2019-08-12T14:47:43.480" Score="3" Body="&lt;h2&gt;First a review of how to determine the direction of the supports&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;In these cases, it's always useful to imagine what would happen if there were no supports. In this case, what would happen if there were no bolts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noticed, one thing that'd happen is that the piece would rotate counter-clockwise. We don't want that to happen, so we can already tell that the supports will need to resist that rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the piece would obviously also move to the left. So the supports will need to resist that motion as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, we want to satisfy three conditions for any 2D object:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= 0 \\&#xA;\sum F_y &amp;amp;= 0 \\&#xA;\sum M_0 &amp;amp;= 0&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is, the sum of external forces (including reactions) in the horizontal direction must be zero, as must the sum of vertical forces and the sum of moments around any arbitrary point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's start with the first equation. Subscripts 1 and 2 describe the two screws, and I start by assuming both reactions point to the right, and are therefore positive:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum F_x &amp;amp;= R_{1,x} + R_{2,x} - F = 0 \\&#xA;\therefore F &amp;amp;= R_{1,x} + R_{2,x}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is pretty obvious: the two reactions must cancel out the applied horizontal force. But how the screws divide that load is still unknown (maybe they split it 50-50, maybe the top screw absorbs it all by itself, maybe the bottom one has a reaction greater than the applied force and the top one then has to cancel out that excess reaction... we don't know yet).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the second equation (now assuming both forces point up and are positive):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum F_y &amp;amp;= R_{1,y} + R_{2,y} = 0 \\&#xA;\therefore R_{1,y} &amp;amp;= -R_{2,y}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The applied force has no vertical component, so we only need to balance out the vertical components of the reactions. The difference in sign tells us they must be equal and opposite: maybe they're both zero, maybe they're a billion tons each (in different directions). Still to be determined.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for the last equation, calculating the moment around the bottom screw (remembering that moment is equal to force times perpendicular distance from the force to the selected point). I'll adopt &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; as the distance between the screws and &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; as the vertical distance from the force to the bottom screw. Forces which would cause a counter-clockwise rotation are positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum M_2 &amp;amp;= -R_{1,x} \cdot d + R_{2,x} \cdot 0 + R_{1,y} \cdot 0 + R_{2,y} \cdot 0 + F \cdot L = 0 \\&#xA;-R_{1,x} \cdot d + F \cdot L &amp;amp;= 0 \\&#xA;\therefore R_{1,x} = \dfrac{F \cdot L}{d}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that &lt;span class=&quot;math-container&quot;&gt;$R_{2,x}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_{2,y}$&lt;/span&gt; have a distance of zero and therefore are discarded for this calculation. This makes sense, since a force on a point doesn't cause rotation around that point. Likewise, &lt;span class=&quot;math-container&quot;&gt;$R_{1,y}$&lt;/span&gt; also has zero distance since we care about the &lt;strong&gt;&lt;em&gt;perpendicular&lt;/em&gt;&lt;/strong&gt; distance, and that force's &quot;line of action&quot; (it's projection) crosses through the other screw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we now know the direction of the force on the top screw: the opposite of what you drew, it will point to the right, countering the applied force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;span class=&quot;math-container&quot;&gt;$\sum F_x$&lt;/span&gt;, we can then see that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;R_{2,x} &amp;amp;= F - \dfrac{F \cdot L}{d} \\&#xA;&amp;amp;= F\left(1 - \dfrac{L}{d}\right)&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As drawn, &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is obviously greater than &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;, which would make this force &quot;negative&quot;, meaning our initial hypothesis that it was pointing to the right was incorrect: it actually points to the left, the same direction as &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;. This makes sense: the previous calculation told us the top screw's reaction was greater than the applied force, so the bottom screw now needs to compensate for that excess reaction.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;And now to answer your actual question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Notice that the entire calculation above didn't take the structure's shape into consideration in any way. Whether the beam was a diagonal or a U-shape or an M.C. Escher painting was irrelevant. No matter the shape of your beam, or whether you add more beams, the reaction will always be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can even add more screws between the existing ones and the conclusion won't change: more screws will reduce the load on each screw, but you'll still have some screws reacting to the right and others to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But notice the caveat in that previous paragraph: &quot;You can even add more screws &lt;strong&gt;&lt;em&gt;between the existing ones&lt;/em&gt;&lt;/strong&gt;&quot;. That's a hint to our eventual solution. And that hint points us back to the equation for &lt;span class=&quot;math-container&quot;&gt;$R_{1,x}$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$R_{1,x} = \dfrac{F \cdot L}{d}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned, the screws have reactions in different directions because &lt;span class=&quot;math-container&quot;&gt;$R_{1,x}$&lt;/span&gt; is actually greater than &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;. So, how can we fix that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, we just need to increase the spacing &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; between the screws, such that it is greater than &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;. If that happens, then &lt;span class=&quot;math-container&quot;&gt;$R_{1,x} &amp;lt; F$&lt;/span&gt;, therefore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$R_{2,x} = F\left(1 - \dfrac{L}{d}\right) &amp;gt; 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$R_{2,x}$&lt;/span&gt; is greater than zero, that means our initial hypothesis that it also points to the right was correct.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-08-12T14:47:43.480" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29614" PostTypeId="1" CreationDate="2019-08-12T15:29:45.117" Score="2" ViewCount="51" Body="&lt;p&gt;I'm rather new to the field of gear systems, and to the field of RC models to which this case applies. I'm trying to design a system where a single variable speed engine would drive two different propulsion systems (in this case, rows of oars on a model galley). I want to be able to steer the galley by means of differential rowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could use a double differential like in a tank, but it seems unnecessarily complicated, as I don't need to vary the turning radius continuously, in fact I only need three fixed configurations, RC-ed by a selector switch: both sides rowing, left only (turning right), right only (turning left). It could be go up to five if adding neutral turns left and right (the two sides rowing in opposite directions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A major additional difficulty is the need for synchronization: although it would only minimally affect performance, visually it's quite important to have both sides rowing together when going forward - this is also a point where a double differential would fail, in my understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone think of a relatively simple gear system to solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="21444" LastActivityDate="2020-01-09T21:02:30.307" Title="Gear switching for simple differential steering" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29615" PostTypeId="1" AcceptedAnswerId="29623" CreationDate="2019-08-12T16:24:47.973" Score="0" ViewCount="131" Body="&lt;p&gt;I have this old squirrel cage induction motor (nameplate below) that I just replaced the bearings on. When it runs, there is a rattling noise that gradually gets louder and then softer, repeating every ~5 seconds. &lt;a href=&quot;https://photos.app.goo.gl/oRZagiXwXN5VyAJo7&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here's a video of it in action.&lt;/a&gt; I don't think the motor made this noise before replacing the bearings, but I'm not sure. (I just bought it and the only time I ran it was briefly at the seller's house.) Does anybody know what might be the source of this rattle? I've put pictures of the motor and the rotor below, as well. Thanks!&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UqxBv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UqxBv.jpg&quot; alt=&quot;motor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HC6gH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HC6gH.jpg&quot; alt=&quot;nameplate&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/G83ig.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G83ig.jpg&quot; alt=&quot;rotor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21446" LastActivityDate="2019-08-13T00:39:52.697" Title="What is causing a pulsating, low-frequency vibration in an induction motor?" Tags="|mechanical-engineering|motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29616" PostTypeId="1" AcceptedAnswerId="29648" CreationDate="2019-08-12T16:27:03.933" Score="0" ViewCount="142" Body="&lt;p&gt;I am using Nema17 12 v motors with A4988 drivers for a project that drives an XY stage. I am programming the pulses to the STEP pin of the drivers in C and I can control the frequency and the shape of each pulse. I was wondering if anyone can tell me (or point me to a reference about) what is the relationship between pulse SHAPE and performance of the motor (as distinct from pulse FREQUENCY). By 'performance' I mean any effect on torque or the reliability of converting the step signal into a rotor step. For example, what would be the difference (for the same frequency) of applying these pulse shapes to the STEP pin of the driver:&#xA;&lt;code&gt;&#xA;a) -_____-_____-_____-_____ etc. (most of the cycle the signal is 'off'/'logical zero')&#xA;b) ---___---___---___---___ etc. (equal time for 'on' and 'off')&#xA;c) -----_-----_-----_-----_ etc. (for most of each cycle the signal is 'on'/'1')&#xA;&lt;/code&gt;&#xA;I am currently using form (b) for my pulses (as most examples I have seen do this) but I was wondering what advantage/disadvantage there may be for changing this. Many thanks in advance.  &lt;/p&gt;&#xA;" OwnerUserId="21445" LastActivityDate="2019-08-15T03:48:08.257" Title="Effect of step pulse shape on stepper motor function" Tags="|stepper-motor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29617" PostTypeId="1" CreationDate="2019-08-12T17:23:13.513" Score="2" ViewCount="6003" Body="&lt;p&gt;Just wondering if someone can please can give me some idea, as to how I can solve this question that is not related to current study. I haven't been able to work out where to start with it just yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Air flows parallel to the surface of a smooth flat plate 10m long. The boundary layer has zero thickness at the leading edge. The Reynolds number at the trailing edge of the plate is 10^7.&#xA;Calculate the total drag force due to skin friction on one side of the plate per unit width?&#xA;Assume that for a laminar boundary layer, up to Rex = 5_10^5, the skin friction coefficient is Cf = 1.328 (Rex)^-1/2 and for turbulent boundary layer Cf = 0.074 (Rex)^-1/5. Take the density of air as 1.2 kg/ m^3 and its dynamic viscosity as 1.8_10^-5 kg/ ms.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8013" LastEditorUserId="8013" LastEditDate="2019-08-12T17:31:22.847" LastActivityDate="2021-07-26T08:02:58.510" Title="How to Find Drag Force from Skin Friction" Tags="|aerodynamics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29618" PostTypeId="1" CreationDate="2019-08-12T17:25:07.430" Score="0" ViewCount="36" Body="&lt;p&gt;We need to build a small (1-2W) linear alternator for a project in engineering school. Is it possible to use finite element software to finetune our design instead of actually building real life models? If yes, which software should we use? We are using AutoCAD fusion 360 to design our 3D-model.&lt;/p&gt;&#xA;" OwnerUserId="21448" LastActivityDate="2019-08-12T17:25:07.430" Title="Is it possible to simulate the electric power output of a linear alternator using finite element analysis software?If yes, which?" Tags="|mechanical-engineering|electrical-engineering|generator|linear-motors|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29619" PostTypeId="2" ParentId="29614" CreationDate="2019-08-12T18:04:29.297" Score="1" Body="&lt;p&gt;I'm not sure if &quot;simple&quot; fits into this possible solution, but run this through the gray matter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your motor has an output gear which is connected to the left drive gear and the right drive gear. This results in those two gears turning in the same direction and if sized the same, at the same rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those gears are connected in a convenient manner to a set of planetary gear sets, one for each side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, planetary gears will provide reduction in a relatively small space, but they also allow for reversal using a braking mechanism. This will require some research on your part, but it's an easily observable action.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you attach the drive to the sun gear and lock the ring gear from rotating, the planet gears turn in one direction with a specific gear ratio, the calculation of which is left as an exercise to the reader. Those planet gears are often attached to a planet gear carrier, allowing you to direct the output in the manner desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, if you attach the drive to the sun gear and lock the planet carrier from rotating, you will get a different ratio (not necessarily) but you will get a reversed rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your description, you'd take the outputs from the rim to create a reciprocating motion, but that's a general assumption, because reciprocating motion isn't restricted by direction of rotation, but it's a good way to avoid top-dead-center complications.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-08-12T18:04:29.297" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29620" PostTypeId="2" ParentId="29617" CreationDate="2019-08-12T20:32:03.160" Score="3" Body="&lt;p&gt;The problem you cite in the question is a pretty standard homework problem in a graduate fluid mechanics course (it might be also an undergraduate course but less likely). Therefore, I would be reluctant to provide the solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is important to remember when dealing with this problem is that it is crucial to attend the lecture at which the subject was covered. Otherwise, it will be difficult to understand how to solve the problem since the books are not very clear on the subject.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding your question - where to start at - I would recommend Robert H. Nunn &quot;Intermediate fluid mechanics&quot; text book. In the 1989 edition (probably the first edition of the book) I would refer you to the paragraph 13.5 &quot;Flat plate boundary layer flows&quot; on p.255.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Brief introduction to the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was found that when a fluid flows over a flat plate, almost all flow velocity reduction happens in a narrow layer adjacent to the plate. This layer is called boundary layer. That is why in order to find drag force which fluid exerts on the plate, one should consider only narrow layer of the flow adjacent to the plate, i.e. boundary layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People have considered it. And revealed that drag force can very conveniently be calculated using &quot;skin friction coefficient&quot; &lt;span class=&quot;math-container&quot;&gt;$c_f$&lt;/span&gt; in general and &quot;friction drag coefficient&quot; &lt;span class=&quot;math-container&quot;&gt;$C_f$&lt;/span&gt; (or &lt;span class=&quot;math-container&quot;&gt;$C_D$&lt;/span&gt; sometimes) in particular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two formulas for friction drag coefficient: one if the ENTIRE boundary layer (over the entire length of the plate) is laminar; the other one is if the ENTIRE boundary layer is turbulent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was revealed that in most cases of flows over a plate, boundary layer can start as laminar and then at some length of the plate become turbulent. And then it is turbulent to the end of the plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to deal with it, if we have only formulas for entire laminar and entire turbulent boundary layers? There is one way (not very difficult) of dealing with it, about which one may read in the Nunn's book, for instance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear from the statement of your problem, that at some point the boundary layer of your flow converts from laminar to turbulent. It is clear because it is stated that at the trailing edge (i.e. at the end of the plate), &lt;span class=&quot;math-container&quot;&gt;$Re = 10^7$&lt;/span&gt; which is higher then critical &lt;span class=&quot;math-container&quot;&gt;$Re$&lt;/span&gt; at which conversion occurs: &lt;span class=&quot;math-container&quot;&gt;$Re_{crit} = 5·10^5$&lt;/span&gt;. Hence, you are dealing with a combined boundary layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Having found friction drag coefficient for a combined boundary layer (partially laminar - partially turbulent), one can find drag force as: &lt;span class=&quot;math-container&quot;&gt;$$F_D = C_f·(0.5·\rho · U_{free stream}^2·A_{plate})$$&lt;/span&gt; Here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; - density of the fluid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$U_{free stream}$&lt;/span&gt; - flow velocity long before it hits the plate&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A_{plate}$&lt;/span&gt; - plate's area&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NOTE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) &lt;span class=&quot;math-container&quot;&gt;$C_f$&lt;/span&gt; - is, actually, called either drag coefficient or friction drag coefficient (sometimes is denoted as &lt;span class=&quot;math-container&quot;&gt;$C_D$&lt;/span&gt;) - not skin friction coefficient as in the statement of your problem; friction drag coefficient is the coefficient written for the entire length of the plate, whereas skin friction coefficient (&lt;span class=&quot;math-container&quot;&gt;$c_f$&lt;/span&gt;) is written for a point on the plate; i.e. &lt;span class=&quot;math-container&quot;&gt;$C_f$&lt;/span&gt; is an integral of &lt;span class=&quot;math-container&quot;&gt;$c_f$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) You are given critical &lt;span class=&quot;math-container&quot;&gt;$Re$&lt;/span&gt; in order to determine at which point the boundary layer converts from laminar to turbulent. You will need it when you'll be calculating combined friction drag coefficient. In order to do it you will need to use the formula: &lt;span class=&quot;math-container&quot;&gt;$$x_{crit} = \frac{Re_{crit} · \mu}{\rho · U_{free stream}}$$&lt;/span&gt; Here: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; - dynamic coefficient of viscosity of the fluid (the thing which is called &quot;dynamic viscosity&quot; in your question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Re_{crit} = 5·10^5$&lt;/span&gt; - &lt;span class=&quot;math-container&quot;&gt;$Re$&lt;/span&gt; at which boundary layer becomes turbulent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x_{crit}$&lt;/span&gt; - length at which boundary layer becomes turbulent&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) You can find &lt;span class=&quot;math-container&quot;&gt;$U_{free stream}$&lt;/span&gt;, since you know &lt;span class=&quot;math-container&quot;&gt;$Re$&lt;/span&gt; at the end of the plate and the length of the plate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish you good luck in your endeavor to write an expression for &lt;span class=&quot;math-container&quot;&gt;$C_D$&lt;/span&gt; for a combined boundary layer. While you'll be doing it, I will be glad to answer you theoretical questions if any occur.&lt;/p&gt;&#xA;" OwnerUserId="21383" LastEditorUserId="21383" LastEditDate="2019-08-12T21:01:26.097" LastActivityDate="2019-08-12T21:01:26.097" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29621" PostTypeId="2" ParentId="28156" CreationDate="2019-08-12T21:20:03.423" Score="0" Body="&lt;p&gt;When talking about R values and holes, what you really should be measuring is the S value instead that bases on area. There are generally two techniques you can measure interior surface roughness of holes. One of them is to fill the hole with something and remove it, then measure that piece directly. Alternatively, you can also scan it optically to create a highly accurate 3D impression of the hole and measure Sz Sa Sp that way. R and S values are not equivalents, so care should be taken to review relevant literature on the subject when you are specifying a criteria to meet certain performance requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck!~&lt;/p&gt;&#xA;" OwnerUserId="13526" LastActivityDate="2019-08-12T21:20:03.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29623" PostTypeId="2" ParentId="29615" CreationDate="2019-08-13T00:39:52.697" Score="1" Body="&lt;p&gt;I listened to the recording and I suspect there's a loose bearing in the motor. Here's how to test it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the motor OFF, grab the end of the shaft and pull up and down on it as hard as you can. There should be NO wiggle at all in the shaft when you do this. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-08-13T00:39:52.697" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29624" PostTypeId="2" ParentId="29606" CreationDate="2019-08-13T00:55:19.090" Score="0" Body="&lt;p&gt;here is how to get started on this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for an ESTIMATE: we imagine the scramjet operating at steady-state cruise conditions with velocity V and an optimally-designed air inlet of area A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Optimal design of the air inlet means that a streamline in the flow right at the edge of the air inlet will terminate at the edge- this is the &quot;no-spillage&quot; condition. This means that all the air coming straight at the inlet will enter it and the primary shock wave at the inlet will reside somewhere inside the lip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these simplifications, the mass flow rate of air entering the inlet (and therefore traveling through the engine) will be (air density x air velocity x inlet area). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a perhaps gross simplification of reality but it will put you in the ballpark. The experts here can furnish you with better approximations. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-08-13T00:55:19.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29626" PostTypeId="1" AcceptedAnswerId="36199" CreationDate="2019-08-13T06:26:01.760" Score="1" ViewCount="563" Body="&lt;p&gt;I have a Proportional solenoid valve that uses a 0-30VDC for control (opening the orifice) I only have capability to provide 0-5V or 0-10V. I need to control the valve using my PC. Is there any DC Power supply device that I can control using PC to provide 0-30V (that I can change continuously as required during operation)? I have tried a voltage multiplier circuit. I am getting the required voltage but the Transistors (BC547) are burning out quickly due to over-current/heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a electrical/electronics engineer hence don't know if there is a better transistor or how to choose one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, searching for the PC controllable DC Power supply. if available.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/eIUsU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eIUsU.png&quot; alt=&quot;This is the circuit I have tested. But the BC547 transistor is burning out after 5 min.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19228" LastEditorUserId="19228" LastEditDate="2019-08-15T16:13:53.183" LastActivityDate="2020-06-11T07:24:59.023" Title="PC Controllable DC Power supply (0-30V)" Tags="|valves|electrical|circuit-design|electronics|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29627" PostTypeId="1" CreationDate="2019-08-13T09:35:02.047" Score="0" ViewCount="158" Body="&lt;p&gt;I'm designing a 3d print model of a ring that you can wear, but the outer surface of the ring can be dialed like a watch bezel.&#xA;Check this video to see how the bezel moves in grades so it stops on every minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=JQDZipgtan0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=JQDZipgtan0&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started a draft model as you see in this picture:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uDKaM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uDKaM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Those are the inner parts of the ring, but it doesn't seem to work as I expect because the inner gear is a bit loose and also hard to rotate because of friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm looking for, is the name of this mechanism (if it is known) or a link to a reference that explains such thing or even similar ready schematic that I can copy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PS: I found similar ring but I don't know how it works &#xA;&lt;a href=&quot;https://i.stack.imgur.com/IYSl1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IYSl1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21453" LastActivityDate="2019-09-12T15:49:16.917" Title="Create Bezel-like ring mechanism" Tags="|mechanical-engineering|gears|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29628" PostTypeId="2" ParentId="29627" CreationDate="2019-08-13T10:41:50.070" Score="1" Body="&lt;p&gt;Reviewing the question and the video leads me to believe that I erred in my answer below.  The photo of the cycloid gear threw me off in that direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the original question is answered by a detent mechanism, not a cycloid gear mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When one wishes to provide stops for rotation, a detent system is implemented. This can be as simple as a spring-loaded lever which engages a set of teeth on the gear, or can be more complex and more efficient by using a spring-loaded ball bearing engaging matching holes in the outer mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many common devices use a detent system. Consider when a knob is rotated and clicks are felt. Within the control would be at least one ball and spring and a number of holes to accept the ball, around the rim.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Portion below appears unrelated to original question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that the mechanism you seek may be known as a cycloid gear or cycloidal drive. The returns from my search using the first term add a line or two referencing the use of the device in clocks. I've noted that this device/mechanism has more parts than shown in your photo and as such, may require adaptation for your purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found in the past that this device is also used for pumping fluids (air/water/etc) as the volume between lobes varies appropriately during rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z3o5G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z3o5G.png&quot; alt=&quot;cycloid gear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2019-09-12T15:49:16.917" LastActivityDate="2019-09-12T15:49:16.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29629" PostTypeId="2" ParentId="29588" CreationDate="2019-08-13T10:49:50.313" Score="1" Body="&lt;p&gt;Based on the information provided your Dell XPS 9570 might not be designed fully utilize the USB-PD standard. The wattage of the power adapter with the circular jack will provide a good clue as to the maximum charging power. A brief review of &lt;a href=&quot;https://www.dell.com/en-us/shop/dell-laptops/xps-15-laptop/spd/xps-15-9570-laptop&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dell XPS 9570&lt;/a&gt; technical specification I was not able to determine maximum power capability of your laptop. Also there is also evidence based on some of references below that your laptop model is USB-C complaint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With regard to USB-PD standard, maximum current supported is 3 or 5 amperes. The current supplied is determine by the eMarker chip that is embedded in the USB-PD cable, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6jHDl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6jHDl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the Dell XPS 9570 is capable of supporting USB-PD maximum current capabilities then the USB-PD cables might not be able to support the maximum current capabilities. This can be identified by looking the cable specifications or using a e-Marker reader like the one below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NvKi4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NvKi4.jpg&quot; alt=&quot;E-marker reader&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system including the cable in this scenario support 20V and 5A. If interested I have provided a few references to help further investigate.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.elinfor.com/market/how-to-identify-the-usb-c-cables-with-or-without-e-maker-m-27&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to identify the USB-C cables with or without E-maker?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.renesas.com/us/en/support/technical-resources/engineer-school/usb-power-delivery-03-emarker-c-auth.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;USB PD Safety Implementation&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.dell.com/community/XPS/Dell-XPS-15-9570-USB-C-charging-specs/td-p/7342657&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dell XPS 15 9570 USB-C charging specs?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-08-14T09:52:16.697" LastActivityDate="2019-08-14T09:52:16.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29631" PostTypeId="1" AcceptedAnswerId="29655" CreationDate="2019-08-13T14:31:54.100" Score="0" ViewCount="1720" Body="&lt;p&gt;In HYSYS it is possible to create multiple lists of components using their proper fluid package to better calculate the properties of very different molecules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here I defined two lists:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;List 1&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Ft2uf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ft2uf.png&quot; alt=&quot;List 1 HYSYS&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;List 2&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/GCH8j.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GCH8j.png&quot; alt=&quot;List 2 HYSYS&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also defined the fluid packages I need and when I go to simulation and create a stream, I can only specify the composition for the first list of components as shown after. &lt;strong&gt;How should I do if I want to use the second list of components?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mWfTd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mWfTd.png&quot; alt=&quot;Stream HYSYS&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19799" LastActivityDate="2019-08-16T00:25:57.703" Title="How to use a specific list of components in a stream in ASPEN-HYSYS?" Tags="|chemical-engineering|software|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29632" PostTypeId="1" CreationDate="2019-08-13T15:39:43.070" Score="4" ViewCount="4047" Body="&lt;p&gt;The switches are so high it's not possible to see how they function on the ground. I really tried to look up information about this on the internet, but there's very little info available, and the info that is there is incomplete. I saw &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Switch-trolley.svg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this diagram&lt;/a&gt; but still don't understand how the switch works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lSESt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lSESt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seem simple to have a trolley pole follow the wire, it is simply U shaped and a spring pulls it up, so that force is applied against the wire. However I cannot possibly imagine how does the switches function, mechanically speaking. Also, how do they prevent a short-circuit at the switches' frog where both poles touches each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, back when tramways (UK: tram, US: streetcar or trolley) also had a trolley pole, did they have switches using the same mechanics as in a trolleybus switch?&lt;/p&gt;&#xA;" OwnerUserId="17069" LastEditorUserId="17069" LastEditDate="2019-08-14T08:32:00.277" LastActivityDate="2020-07-24T00:47:46.817" Title="How do trolleybus switches work?" Tags="|mechanical-engineering|electrical-engineering|transportation|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29634" PostTypeId="2" ParentId="26370" CreationDate="2019-08-13T17:29:51.850" Score="1" Body="&lt;p&gt;Companies such as &lt;a href=&quot;https://www.wmberg.com/berg/media/products/pdfs/M05F048.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;W.M. Berg&lt;/a&gt; make shaft spacers specifically for this purpose:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BYTdN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BYTdN.png&quot; alt=&quot;Snippet from W.M. Berg catalog page&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a 3mm ID bearing, Berg provides two thicknesses: 0.1mm and 0.15mm.&lt;/p&gt;&#xA;" OwnerUserId="3440" LastActivityDate="2019-08-13T17:29:51.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29635" PostTypeId="1" CreationDate="2019-08-13T18:32:21.783" Score="0" ViewCount="38" Body="&lt;p&gt;What type of bearing should I use to allow a ring to rotate around a fixed axle. Axle is horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rings inner diameter can be flush with the stationary rod or have a gap as needed. What type of bearing (or other device) would I use to allow the ring to spin freely?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I can get a basic bearing and maybe solder the ring onto the bearing but I want something more professional and elegant. Does there exist a two piece bearing set that you insert from either side of the rings inner diameter and connects together, and then can be slid onto the axle?&lt;/p&gt;&#xA;" OwnerUserId="21466" LastEditorUserId="282" LastEditDate="2019-08-19T20:56:56.200" LastActivityDate="2019-08-19T20:56:56.200" Title="What type of bearing should I use to allow a ring to rotate around a fixed axle" Tags="|bearings|friction|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29636" PostTypeId="1" CreationDate="2019-08-13T19:11:35.073" Score="3" ViewCount="273" Body="&lt;p&gt;Chapter 4, Part 3 of the MEPDG (Mechanistic Empirical Pavement Design Guide) states&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The 'effective' dynamic k-value represents the compressibility of all pavement layers beneath the PCC slab and base course&quot;. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, the Manual of Practice for the MEPDG states &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The effective k-value, therefore, essentially represents the compressibility of underlying layers (i.e. unbound base, subbase, and subgrade layers) upon which the upper bound layers and existing HMA or PCC layers are constructed.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I find these two definitions contradictory in regards of the inclusion or not of the base layer in the k-value estimation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)  Does the k-value represents all layers beneath the PCC slab or only all layers beneath the PCC slab and base?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Is the static backcalculated k-value from the FWD data the value I should enter directly as input in the Rigid Pavement Rehabilitation analysis in Pavement ME Design?&lt;/p&gt;&#xA;" OwnerUserId="21467" LastEditorUserId="18841" LastEditDate="2019-08-23T07:56:55.237" LastActivityDate="2019-08-23T07:56:55.237" Title="k-value Rigid Pavements" Tags="|pavement|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29638" PostTypeId="2" ParentId="29539" CreationDate="2019-08-13T23:25:36.010" Score="0" Body="&lt;p&gt;I think I've a partial answer to my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;1) Proof that liquid phase flow is steady if liquid volume fraction is time independent.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuity equation for incompressible liquid phase: &lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial \alpha_L}{\partial t}+\frac{\partial (\alpha_L·v_L)}{\partial x}=0$$&lt;/span&gt; Let's say that volume fraction is not time dependent and only spatially dependent; and assume that velocity is still spatially and time dependent. Then we'll have:&lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial (\alpha_L·v_L)}{\partial x}=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d \alpha_L}{dx}·v_L+ \frac{\partial v_L}{\partial x}· \alpha_L=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d \alpha_L}{dx}· \frac{1}{\alpha_L}=-\frac{\partial v_L}{\partial x}·\frac{1}{v_L}$$&lt;/span&gt;&#xA;Now we have that the entire left hand side is only spatially dependent, whereas the entire right hand side is assumed to be both spatially and time dependent. But it can't be. If we know for sure that the left hand side is not time dependent and is equal to the right hand side, then the right hand side must also not be time dependent: &lt;span class=&quot;math-container&quot;&gt;$$\frac{d \alpha_L}{dx}· \frac{1}{\alpha_L}=-\frac{d v_L}{d x}·\frac{1}{v_L}$$&lt;/span&gt;&#xA;&lt;strong&gt;Therefore, it follows that if liquid volume fraction is not time dependent then the liquid flow is steady.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same discussion should work for the gaseous phase (but I am not sure about that).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;2) Proof that liquid phase flow is transient if liquid volume fraction is time independent.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuity equation for incompressible liquid phase: &lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial \alpha_L}{\partial t}+\frac{\partial (\alpha_L·v_L)}{\partial x}=0$$&lt;/span&gt;&#xA; Let's say that volume fraction is both time and space dependent. Then assume that velocity can be only spatially dependent. Prove that it can't be. Let's do further math manipulations with liquid continuity:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial \alpha_L}{\partial t}+\frac{\partial \alpha_L}{\partial x}·v_L+\frac{d v_L}{d x}·\alpha_L=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{\alpha_L}·(\frac{\partial \alpha_L}{\partial t}+\frac{\partial \alpha_L}{\partial x}·v_L)=-\frac{d v_L}{d x}$$&lt;/span&gt;&#xA;Now, even if both of &lt;span class=&quot;math-container&quot;&gt;$\alpha_L$&lt;/span&gt; partial derivatives are constants and with the assumption that &lt;span class=&quot;math-container&quot;&gt;$v_L=f(x)$&lt;/span&gt;, the left hand side is still time dependent because &lt;span class=&quot;math-container&quot;&gt;$\alpha_L=f(x,t)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's write an integral formulation: &lt;span class=&quot;math-container&quot;&gt;$$v_L=-\int \frac{1}{\alpha_L}·(\frac{\partial \alpha_L}{\partial t}+\frac{\partial \alpha_L}{\partial x}·v_L)dx$$&lt;/span&gt;&#xA;From the integral formulation it is explicitly seen that since the integrand is time dependent, the result will also be time dependent. It means that if &lt;span class=&quot;math-container&quot;&gt;$\alpha_L=f(x,t)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$v_L$&lt;/span&gt; cannot be a function of space only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Therefore, it follows that if liquid volume fraction is time dependent then the liquid flow is transient.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same discussion should work for the gaseous phase (but I am not sure about that).&lt;/p&gt;&#xA;" OwnerUserId="21383" LastEditorUserId="21383" LastEditDate="2019-08-16T00:00:54.917" LastActivityDate="2019-08-16T00:00:54.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29642" PostTypeId="2" ParentId="29607" CreationDate="2019-08-14T12:01:06.213" Score="1" Body="&lt;p&gt;If the total temperature increased, energy would need to have been added to the flow via work or heat transfer. A change in stagnation temperature is directly related to work being taken or added to the flow (see energy equation/Euler work equation). Since the stator does not put work into the flow or extract work out, the total temperature will stay the same. Things become more nuanced when phenomena such as friction or secondary flows are involved. However, unless you have a crazy amount of friction, the total temperature of the flow will remain essentially the same in the stator. I am assuming you believe the temperature will increase from decreasing pressure because of the Ideal Gas Law. However, remember in compressible flow that the density is a variable and changes with pressure. Hopefully my answer is helpful for you.&lt;/p&gt;&#xA;" OwnerUserId="21118" LastActivityDate="2019-08-14T12:01:06.213" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29644" PostTypeId="2" ParentId="29632" CreationDate="2019-08-14T17:38:07.077" Score="1" Body="&lt;p&gt;We don't have trolleybuses in Ireland where I live so I've never had a good look at the systems used. I'm sure that many ingenious solutions have been invented to manage the switching but here's one:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The switches of the trolleybus network are controlled by the driver of the trolleybus. A switching contact rail is added to the trolleybus network before each switch. If the driver of the trolleybus wants to switch the trolleybus network switch, he operates a switch, which briefly switches on the trolleybus's heating system (a large power consumer). Current must be drawn when driving on the switching contact rail. If the driver of the trolleybus does not want to switch the trolleybus network switch, he operates another switch that switches off the high voltage system of the trolleybus. Thus no current is drawn when driving on the switching contact rail, and the trolleybus network switch does not switch. &lt;em&gt;Source: &lt;a href=&quot;http://www.obus-ew.de/e23.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;obus-ew.de&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This seems to be similar to many tram systems in that the driver makes the route selection rather than an automatic switcher. The article goes on to explain some of the problems encountered (some almost comical) with the system in Eberswalde, Germany, and you may find some of the other pages of interest.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2019-08-14T17:38:07.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29645" PostTypeId="2" ParentId="29632" CreationDate="2019-08-14T19:12:18.883" Score="2" Body="&lt;p&gt;In your diagram the switch boxes marked as (1) are critically important. As the bus begins a turn it pulls one trolley ahead of the other. If turning to the left, the right trolley is pulled slightly ahead of the left. If turning to the right, the left trolley is pulled slightly ahead of the right one. The sequence that the trolleys contact switch (1) tells the it how to switch the tracks in switches (2) and (4).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above came from a conversation I had with a San Francisco Muni tech many years ago when I worked as a tech at AT&amp;amp;T. He was down the street from a pole I was working on.&lt;/p&gt;&#xA;" OwnerUserId="12535" LastActivityDate="2019-08-14T19:12:18.883" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29646" PostTypeId="2" ParentId="26376" CreationDate="2019-08-14T22:21:07.600" Score="0" Body="&lt;p&gt;Google &quot;Controlling motors in the presence of friction and backlash&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a big subject, but the two essentials are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Use an inner velocity loop on the motor, if you can (i.e., don't put a PID on the position of the whole thing -- put a velocity PI on just the motor -- especially if the motor is geared).&lt;/li&gt;&#xA;&lt;li&gt;If there's friction, pulse the power to the motor slowly, with a drive greater than that magic 20% where you get no response.  Each time you pulse the motor it'll move a bit, and thus crawl in the direction you want.  It's not ideal, but it can be much better than sitting still forever, and then jumping (and overshooting) as happens with such an assembly under pure PID control.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-08-14T22:21:07.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29647" PostTypeId="2" ParentId="26376" CreationDate="2019-08-15T03:32:12.017" Score="1" Body="&lt;p&gt;Cascaded control may be a good solution, but there's something simpler that you could try first. Simply add a bit of logic after the PID loop to &lt;em&gt;skip&lt;/em&gt; from -20 to 20%.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;if(pid_output &amp;gt; 0%)      pid_output = (pid_output+20%)/1.2;&#xA;else if(pid_output &amp;lt; 0%) pid_output = (pid_output-20%)/1.2;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What your essentially doing is remapping the -100% to 100% output range of the PID controller onto the usable input range of the linear actuator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd throw a bit of deadzone in there as well to keep the system from oscillating. Like maybe if the pid output &amp;lt; +-5% just set it to zero.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-08-15T03:32:12.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29648" PostTypeId="2" ParentId="29616" CreationDate="2019-08-15T03:48:08.257" Score="0" Body="&lt;p&gt;With this driver, I think your changes will have no effect whatsoever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rising edge on the STEP line simply tells the driver to advance the motor firing sequence. It will advance and then hold and wait for the next step. How long you hold the signal high will not change anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a a chart from the datasheet. Notice that there are no vertical lines at the STEP falling edges. You can also see that when the edge occurs, nothing changes in the output waveform.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QDBK6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QDBK6.png&quot; alt=&quot;output current vs step signal chart&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There &lt;em&gt;are&lt;/em&gt; settings you can change that will change the pulse shape that the motor sees. For example micro-stepping and decay settings, but these are set elsewhere. &lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-08-15T03:48:08.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29649" PostTypeId="1" CreationDate="2019-08-15T08:58:05.053" Score="1" ViewCount="468" Body="&lt;p&gt;What is the meaning of power/energy density of an actuator?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Context:&#xA;As a novel kind of actuator, the dielectric elastomer ( DE )&#xA;demonstrates high-strain response [ 1 ] , high energy densities&#xA;( seventy times higher than conventional electromagnetic&#xA;actuators ) [ 2 ] , and high energy conversion ef ﬁ ciencies ( 60% –&#xA;90% [ 3 ]) . &lt;/p&gt;&#xA;" OwnerUserId="20870" LastActivityDate="2020-01-18T05:02:15.263" Title="Power/energy density of an actuator" Tags="|mechanical-engineering|electrical-engineering|actuator|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29651" PostTypeId="1" AcceptedAnswerId="29652" CreationDate="2019-08-15T17:16:52.173" Score="1" ViewCount="1411" Body="&lt;p&gt;The horizontal Force &lt;span class=&quot;math-container&quot;&gt;$A_x$&lt;/span&gt; is easy but how did they get the vertical forces &lt;span class=&quot;math-container&quot;&gt;$A_y$&lt;/span&gt; &amp;amp; &lt;span class=&quot;math-container&quot;&gt;$B_y$&lt;/span&gt;? I tried using similar triangles for the Force and length of the sides, somehow I can't manage to solve it. I wonder where I am making a mistake. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eVoUm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eVoUm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21486" LastEditorUserId="282" LastEditDate="2019-08-19T20:57:09.737" LastActivityDate="2019-08-19T20:57:09.737" Title="How do I find the vertical reaction forces?" Tags="|mechanical-engineering|statics|mechanical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29652" PostTypeId="2" ParentId="29651" CreationDate="2019-08-15T17:57:40.847" Score="2" Body="&lt;p&gt;There are multiple ways to solve this problem but I use the moment equilibrium around the points A and B. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$A_y 3l-q_0 4l^2=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the &lt;span class=&quot;math-container&quot;&gt;$A_y$&lt;/span&gt; reaction force from the equation above and use it again in the moment equilibrium around A this time: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$-4q_0l^2-B_y 3l=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also skip the second part, and use your intuition, since the structure is in equilibrium, the vertical forces should cancel each other out, otherwise the structure accelerates.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-08-15T17:57:40.847" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29653" PostTypeId="2" ParentId="24213" CreationDate="2019-08-15T18:58:56.807" Score="1" Body="&lt;p&gt;I would recommend using &lt;em&gt;Handbook of Hydraulic Resistance&lt;/em&gt; by I.E. Idelchik. This book is the literal bible for loss coefficient formulations. The book gives loss coefficient correlations based on test data. The image is an example of one chart provided for a tee configuration &lt;a href=&quot;https://i.stack.imgur.com/hcw1a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hcw1a.png&quot; alt=&quot;One of the many charts on tee and wye data&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21118" LastActivityDate="2019-08-15T18:58:56.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29654" PostTypeId="1" CreationDate="2019-08-15T20:39:24.220" Score="0" ViewCount="96" Body="&lt;p&gt;Both diesel and gasoline engines were developed in the late 19th to early 20th century. Diesel engines were for the most part of the 20th century used mostly for large applications in ships, trains, and for large vehicles such as trucks and busses, but gasoline engines remained the preferred option for personal cars during the 20th century. Only during the last decades of the 20th century did diesel engines begin to make a little bit of headroom in the personal car market. In the 80s more than &lt;a href=&quot;https://en.wikipedia.org/wiki/History_of_the_diesel_car#Early_20th_century&quot; rel=&quot;nofollow noreferrer&quot;&gt;10 percent of new cars sold in Europe were diesels&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What developments in diesel engines started to make them more attractive for personal cars? What made diesel powered personal cars go from 0% in the 50s to more than 10% in the 1980s? I'm specifically interested in if there were any technical advances that made diesels more suitable for personal cars, but I'm not sure if there were any.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(NB: after the signing of the Kyoto treaty in 1997 many European countries started to promote diesel a lot, and the share of diesel personal cars increased a lot. I am not interested in this part, but in the period before governments started to promote diesel cars substantially.)&lt;/p&gt;&#xA;" OwnerUserId="12424" LastActivityDate="2022-10-02T06:48:50.717" Title="What developments made diesel engines suitable for personal cars in the 1970s/1980s?" Tags="|engines|engineering-history|diesel|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29655" PostTypeId="2" ParentId="29631" CreationDate="2019-08-16T00:25:57.703" Score="1" Body="&lt;p&gt;You would need to create a whole new property package to use List 1 or 2 even if the EoS is the same as the initial one. There are a few times when this is impossible such as in Sulphur Recovery (Sulsim) or some Acid Gas properties. Also you should know you can share component lists with two property packages because there are times your reactor requires NRTL but your distillation tower requires CPA or it works fine with PR because you have to have the Efficiencies already measured out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to change component lists within the stream and keep the data from stream 1... You better create 2 streams one with Basis 1 and the other with Basis-2 each one with its own component list. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: Oh also to change the basis you do it in the Fluid Package cell in Worksheet/Conditions (Perhaps this was the answer to your question)&lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2019-08-16T00:25:57.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29656" PostTypeId="1" CreationDate="2019-08-16T00:30:05.933" Score="1" ViewCount="183" Body="&lt;p&gt;We are attempting to design a &lt;strong&gt;solar car&lt;/strong&gt;. The solar panel 4 1/2&quot; x 13&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like to break these problems into their simplest form, in order to eliminate the least amount of possibilities. So, the goal of the competition, as I understand it, is to transport the (provided) solar panel, and the (provided) electric motor down a 40 foot long track (the car is guided with fishing line as a guide wire).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The texture of the track is similar to the surface of a skateboard, although slightly rougher than that.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Of course there will be many other considerations besides the physical design of the car (gear ratios, lenses and/or mirrors for the panel, guide wire receiver), but right now I am focused on the car itself, and more specifically, the wheel orientation. I am wondering: &lt;strong&gt;What is the optimal wheel setup for this car on this track?&lt;/strong&gt; Some ideas I have found/had are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Less rotational inertia in wheels will mean more energy being invested into the car itself. Additionally, the car, overall, should remain light weight. So, less wheels, and less massive wheels will be better.&lt;/li&gt;&#xA;&lt;li&gt;This terrain is relatively rough. As such, wider and larger diameter wheels should be used (the idea that a LEGO car rolls down the driveway extremely bumpily, whereas a bicycle does not). So perhaps large (but light weight) &lt;strong&gt;barrel shaped wheels&lt;/strong&gt; would be most effective.&lt;/li&gt;&#xA;&lt;li&gt;The height of the fishing line guide wire can be chosen by us. So assuming we have a low friction guide wire receiver (like a lubricated, plastic drinking straw), it seems like &lt;strong&gt;we should set the guide wire in such a way to support a significant amount of the weight of the car.&lt;/strong&gt; (Due to the fact that the car is solar powered, it is extremely unlikely that it would &quot;peel out&quot;.) Should this supported weight be changing throughout the course of the track? Perhaps as the car moves along the track, the wire should support more and more weight, as the car will have less torque at those higher speeds, although, on the other hand, the sloped up rope could slow the car down.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;What would the most efficient wheel orientation/setup for such a car be?&lt;/p&gt;&#xA;" OwnerDisplayName="user21489" LastEditorDisplayName="user21489" LastEditDate="2019-08-16T00:51:46.200" LastActivityDate="2020-01-14T06:02:05.330" Title="What wheel setup on a small, solar powered car, would have the least rolling friction?" Tags="|mechanical-engineering|wheels|car|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29657" PostTypeId="1" CreationDate="2019-08-16T00:35:42.347" Score="1" ViewCount="34" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f792P.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f792P.png&quot; alt=&quot;oc&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain how the above boost control setup would work. The valve on the left uses a vac source applied to the top of the diaphram which would effectively help hold the spring closed at high boost levels providing high boost than a single valve setup would. I assume this would be used to give a large boost control range (similar to a CO2 boost controller). Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure what the D port on the actuator does though, I assume that D and E are both connected and would both work against the spring opening it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What control circuit would be required to drive the two valves. Would you need a split range controller with two different duty curves, or could you run them from a single controller with a single curve, eg run the two valves from a single PWM output?&lt;/p&gt;&#xA;" OwnerUserId="19646" LastActivityDate="2019-08-16T00:35:42.347" Title="Turbocharger boost control using two valves with a vac and boost reference for wide boost control range" Tags="|control-engineering|pressure|turbomachinery|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="29659" PostTypeId="1" CreationDate="2019-08-16T04:06:26.967" Score="0" ViewCount="67" Body="&lt;p&gt;So I am trying to write a high school research paper (3700 words) on the topic above. I am not sure if there is such a relationship, but I think intuitively we can agree that different materials will react differently to a meteor impact (especially in small lab size experiments since they are in what is called in the strength regime as opposed to gravity). I think a relationship exists because of the Newton-Laplace equation : &lt;a href=&quot;https://i.stack.imgur.com/tAtpJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tAtpJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So perhaps someone can shed light on how impact craters inform us about the target material's mechanical properties ?&lt;/p&gt;&#xA;" OwnerUserId="21490" LastActivityDate="2021-01-09T05:06:05.677" Title="Is there a relationship between depth/diameter/etc... of meteoric craters and the strength (eg:compressive strength) of the target material (sand...)?" Tags="|materials|civil-engineering|strength|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29661" PostTypeId="1" CreationDate="2019-08-16T06:07:37.560" Score="0" ViewCount="101" Body="&lt;p&gt;I come across a question-&lt;strong&gt;'How big an address space should I allocate for each ARM A53 core in a mobile processor?'&lt;/strong&gt; What is the answer for this?&#xA;Does it refer to memory management in SoC?&lt;/p&gt;&#xA;" OwnerUserId="21491" LastActivityDate="2023-12-23T22:06:30.763" Title="Memory management in for multiple cores of a processor" Tags="|computer-engineering|computer-hardware|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29663" PostTypeId="1" AcceptedAnswerId="36583" CreationDate="2019-08-16T07:53:30.473" Score="1" ViewCount="147" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6fAuR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6fAuR.png&quot; alt=&quot;illustration of wall&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a high and wide &lt;strong&gt;concrete wall&lt;/strong&gt; of say &lt;strong&gt;30 cm thickness&lt;/strong&gt;, covered with a sheet of &lt;strong&gt;vapor barrier&lt;/strong&gt; on one side, and a sheet of say &lt;strong&gt;30 cm thick&lt;/strong&gt; and effective &lt;strong&gt;insulation&lt;/strong&gt; on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;vapor barrier&lt;/strong&gt; completely blocks water vapor. The &lt;strong&gt;insulation&lt;/strong&gt; lets water vapor through and leads away any &lt;strong&gt;condensed&lt;/strong&gt; water downwards where it is &lt;strong&gt;drained away&lt;/strong&gt;. The concrete lets vapor vapor through, albeit slowly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air on the vapor barrier side (the inside of a hypothetical building) is &lt;strong&gt;25°C with 50% Relative Humidity&lt;/strong&gt;, and the air on the insulation side (the outside of a hypothetical building) is &lt;strong&gt;10°C with 100% RH&lt;/strong&gt;. See illustration above which shows the wall from above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to some vendors of basement insulation, a setup like this will cause the concrete wall to become relatively dry due to vapor pressure towards the insulation in which the water vapor will condense and allow the water to drip down and then be drained away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't make myself understand how this process works from a physical point of view. I thought an infinite source of water vapor spreads out until all air is saturated with water, so I would expect the wall to end up having 100% RH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the wall end up relatively dry, at say 70% RH? If yes, how exactly?&lt;/p&gt;&#xA;" OwnerUserId="16747" LastEditorUserId="16747" LastEditDate="2019-09-18T04:51:49.917" LastActivityDate="2020-07-07T13:37:06.450" Title="Can temperature gradient keep concrete wall below 100% RH in 100% RH atmosphere?" Tags="|building-physics|vapor-pressure|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29665" PostTypeId="1" AcceptedAnswerId="29682" CreationDate="2019-08-16T09:24:26.633" Score="2" ViewCount="317" Body="&lt;p&gt;I need to write a non-technical product requirements document (PRD) for a new machine that I want designed.&lt;/p&gt;&#xA;&lt;p&gt;I need to ensure that I have included all the salient information concisely and to the point.&lt;/p&gt;&#xA;&lt;p&gt;I have searched online for template PRDs but all of the documents I have come across, assume that the product is software so do not fit the needs of a PRD for a mechanical device.&lt;/p&gt;&#xA;&lt;p&gt;Can someone outline here, the structure and sections required for a PRD for a mechanical device (e.g. an engine), failing that, a link to such a document would suffice.&lt;/p&gt;&#xA;" OwnerUserId="932" LastEditorUserId="29410" LastEditDate="2021-02-28T06:48:55.150" LastActivityDate="2021-02-28T06:48:55.150" Title="What are the essentials of a good PRD (product requirements document) for a new mechanical device" Tags="|mechanical-engineering|design|product-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29666" PostTypeId="1" AcceptedAnswerId="29677" CreationDate="2019-08-16T09:50:23.217" Score="1" ViewCount="197" Body="&lt;p&gt;I know there are some aspects we can try to apply this surface area to volume ratio to say physical chemistry in order to see reactivity of chemicals. However I am very confused in my structural engineering elective course, it states that surface area to volume of high rise tower like structures is very important. I do see that the SA:V ratio of certain shapes in general give a very interesting and neatly simplified expression once computed, but I do not see what the application of SA:V ratio is when building structures of towers (Eiffel tower CN Tower etc...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help me explain the significance of SA:V ratio when building high rise structures (I would appreciate technical details as engineering is not my major of study i want to understand alittle better)? &lt;/p&gt;&#xA;" OwnerUserId="21492" LastActivityDate="2019-08-16T19:08:05.100" Title="Surface Area to Volume Ratio" Tags="|structural-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29667" PostTypeId="1" AcceptedAnswerId="29669" CreationDate="2019-08-16T10:28:43.187" Score="0" ViewCount="488" Body="&lt;p&gt;I'm wondering whether there are any notable losses in a typical automobile clutch when fully engaged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only losses that come to mind:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;material twisting with torque changes&lt;/li&gt;&#xA;&lt;li&gt;slipping due to imperfect friction&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Are these even significant and are there any more? What is a typical energy efficiency of a fully-engaged clutch?&lt;/p&gt;&#xA;" OwnerUserId="16802" LastActivityDate="2019-08-16T11:27:24.643" Title="Efficiency of a fully-engaged friction clutch" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29668" PostTypeId="1" CreationDate="2019-08-16T11:16:23.260" Score="1" ViewCount="979" Body="&lt;p&gt;I'm trying to find a suitable shock absorber for a small ATV with a specific value (to be determined later). But everywhere I look, no one seems to be giving the damping ratio of the shock absorbers. &#xA;I'm not expert in this field, so I'm wondering if I'm missing something, and that in reality I don't need the damping ratio to select a proper shock absorber, and that the damping ratio in fact can be controlled by some other means?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All I have is the intro to vibration degree course where the damping ratio was always given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help in advance.&lt;/p&gt;&#xA;" OwnerUserId="17951" LastActivityDate="2020-01-14T16:02:00.250" Title="Car's shock absorbers' damping ratio" Tags="|mechanical-engineering|automotive-engineering|vibration|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29669" PostTypeId="2" ParentId="29667" CreationDate="2019-08-16T11:27:24.643" Score="1" Body="&lt;p&gt;The effects the OP mentioned do exist and would probably be measurable in controlled conditions in a test rig. However they are negligible, because they only dissipate a tiny amount of energy &lt;em&gt;as the torque through the clutch changes&lt;/em&gt; with driving conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nearly all of the strain energy in the components that is stored as they twist up under torque load is recovered when the load is removed at they return to their undeformed shape. The amount of energy lost in one cycle of loading and unloading (independent of how long the cycle takes) is of the order of 0.001% to 0.01%, for metal components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the clutch is &lt;em&gt;continuously&lt;/em&gt; slipping as it transmits power, it will quickly fail because that is not how it is designed to operate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The analogous effects in bolted joints between components can dissipate a significant amount of energy and affect the mechanical damping of the system, but only if the load in the joint is continually changing at high frequency - e.g. something like a cylinder head bolt where the loads are changing in sync with the engine RPM, hundreds of times per second. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any energy losses in the clutch will be tiny compared with losses in the gearbox, for example. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-08-16T11:27:24.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29671" PostTypeId="2" ParentId="25123" CreationDate="2019-08-16T12:21:14.193" Score="1" Body="&lt;p&gt;An aluminium coating have a higher than 80% reflectivity in the wavelengths you want. Source: &lt;a href=&quot;https://www.layertec.de/en/capabilities/coatings/metallic&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.layertec.de/en/capabilities/coatings/metallic&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put the coating behind glass like in a common mirror, make sure to not use glass that absorbs UV lighting. Maybe you can use an aluminium coating without any glass in front.&lt;/p&gt;&#xA;" OwnerUserId="16747" LastActivityDate="2019-08-16T12:21:14.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29672" PostTypeId="2" ParentId="29661" CreationDate="2019-08-16T13:29:46.130" Score="0" Body="&lt;p&gt;Answers to this kind of question would mostly come from knowing the memory behavior of the software you're going to run on it. My advice would be to start with monitoring, then use eg. twice what seems to be a normal level. Without monitoring, you will never get past guessing. Which is quite okay for the initial values, btw. &lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2019-08-16T13:29:46.130" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29675" PostTypeId="2" ParentId="29656" CreationDate="2019-08-16T18:50:56.503" Score="1" Body="&lt;p&gt;For a power-limited application, the largest diameter wheel with the smallest contact patch with the pavement results in the most efficiency, because it minimizes bearing and rolling friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wheel inertia is only a factor at startup, when the wheel is accelerating up to speed. It does not affect the behavior of the car at constant cruise conditions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As pointed out by Solar Mike, your best choice is to go with a racing bicycle wheel, as it is already optimized for best efficiency, and they are available right off-the-shelf at a bike store. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorDisplayName="user21489" LastEditDate="2019-08-17T03:47:37.977" LastActivityDate="2019-08-17T03:47:37.977" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29676" PostTypeId="2" ParentId="29654" CreationDate="2019-08-16T19:01:14.707" Score="1" Body="&lt;p&gt;diesel power for passenger cars became more attractive as gasoline prices rose, but was hampered by the following issues: 1) they are noisier, 2) they vibrate more, especially at idle and low speeds, 3) they are harder to start in cold weather, 4) they produce less power-per-pound of engine than a gasoline engine, and 5) they pollute more- especially particulates and NOx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding cylinders (going to 5 or 6 instead of four) smoothed out the early diesels, better engine mounts kept the vibrations out of the frame, advanced glow plug technologies made them easier to start, adding turbocharging made them more powerful, and tailpipe emissions were reduced with things like computer control of injection cycles, stratified charge or precombustion chambers- but in the end, there wasn't any way to get them to meet the same emission standards as a gas engine without cheating (like VW), and the price and tax differences between diesel and gas dwindled away with time.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-08-16T19:01:14.707" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29677" PostTypeId="2" ParentId="29666" CreationDate="2019-08-16T19:08:05.100" Score="3" Body="&lt;p&gt;In high rise design surface area to volume ratio is a significant factor in the amount of energy used to keep the building air conditioned and useable for its design function. The larger the surface the more heat transfers either from the building to outside in the winter or vice versa in the summer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the first glance it would seem a cube would have the least surface to volume ratio and be an ideal shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However a cube has a big portion of its usable floor are near its core, far away from the exterior windows which provide natural lighting and ventilation. So the core of the building needs continuous lighting and ventilation. which adds to the energy consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The optimal building form factor and SA/V is one which would address all these concerns and find a balance between an energy efficient shell and energy needed for lighting and ventilation, etc.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Location and angle of orientation of the building with respect to sun rays, wind direction and snow drift.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The climate. Moderate Climate gives flexibility to have more surface to take advantage of natural light.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The extreme conditions the building envelope will likely be exposed to.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-16T19:08:05.100" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29678" PostTypeId="2" ParentId="29663" CreationDate="2019-08-16T20:19:56.913" Score="0" Body="&lt;p&gt;I have had to repair the concrete walls because of severe leaks and sometimes the replacement of a pipe, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were walls totally wet and even dark with water oozing out, and there were walls in much dryer condition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has to do with the quality and the aggregate grading of the concrete. Some concrete are dense and almost seal the infiltration of water vapors, some porous and so defective that they even bleed cement and expose the sand under sever humidity.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-16T20:19:56.913" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29679" PostTypeId="2" ParentId="29659" CreationDate="2019-08-16T20:54:21.063" Score="1" Body="&lt;p&gt;Look at the paper by Turtle, E. P., et al. &quot;&lt;a href=&quot;https://www.researchgate.net/profile/Gordon_Osinski/publication/253750582_Impact_Structures_What_Does_Crater_Diameter_Mean/links/0deec538cb4fbbf71c000000/Impact-Structures-What-Does-Crater-Diameter-Mean.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Impact structures: What does crater diameter mean&lt;/a&gt;.&quot; Large meteorite impacts III 384 (2005): 1-24. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading the paper, look at the list of references and try to locate the papers that seem to be most relevant from your local library (or e-mail the corresponding author).  That will give you an understanding of the complexity of the problem and the best one can do with a small amount of information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In particular, if you are considering impacts into granular materials, you can look at the introductory parts of Omidvar, Mehdi, Magued Iskander, and Stephan Bless. &quot;&lt;a href=&quot;https://www.researchgate.net/profile/Magued_Iskander/publication/259515946_Response_Of_Granular_Media_To_Rapid_Penetration/links/5a24a31faca2727dd87e545a/Response-Of-Granular-Media-To-Rapid-Penetration.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Response of granular media to rapid penetration&lt;/a&gt;.&quot; International Journal of Impact Engineering 66 (2014): 60-82.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another readable source is Collins, Gareth S., H. Jay Melosh, and Gordon R. Osinski. &quot;The impact-cratering process.&quot; Elements 8.1 (2012): 25-30.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastEditorUserId="1832" LastEditDate="2019-08-18T02:16:50.313" LastActivityDate="2019-08-18T02:16:50.313" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29680" PostTypeId="2" ParentId="26451" CreationDate="2019-08-16T21:46:04.543" Score="1" Body="&lt;p&gt;That was common over 100 years ago. A single steam engine, or water wheel , or electric motor was connected by flat leather belts to a series of drive shafts on the ceiling of the factory.  Each machine was connected by belt to a pulley on the ceiling ; pulley diameters were selected so that each machine ran at an appropriate speed. Speeds could be changed by slipping a belt to a different size pulley with out stopping . Belts were 6 to 12 in. wide .Some belts had guards some had only a railing . There must be books available but maybe not on the internet. I saw one in operation about 1960 ; my impression was that it was a dangerous pain in the butt system. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-08-16T21:51:10.880" LastActivityDate="2019-08-16T21:51:10.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29681" PostTypeId="1" CreationDate="2019-08-17T04:47:50.767" Score="1" ViewCount="45" Body="&lt;p&gt;I would to know what is (or how to calculate) the size (dia) if the smallest pipe to which I can weld a 1 inch pipe from the side of the bigger pipe perpendicular to its axis (like a tee).&lt;/p&gt;&#xA;" OwnerUserId="19228" LastActivityDate="2019-08-17T04:47:50.767" Title="What is the smallest size of pipe to which I can Weld a 1 inch pipe Perpendicular to the Axis?" Tags="|mechanical-engineering|manufacturing-engineering|welding|piping|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29682" PostTypeId="2" ParentId="29665" CreationDate="2019-08-17T09:55:06.947" Score="1" Body="&lt;p&gt;Before we start, just a word of caution that an engine is not only a mechanical device, they come with their own controllers (FADECs), and most of the time they have to integrate with some software (even when they operate through a throttle servo).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alright, I'll suggest two methodologies, both of them are applicable to many scenarios.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The Flexible Way&lt;/li&gt;&#xA;&lt;li&gt;The Well-Specified Way&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I've personally gone through both methodologies in aerospace and defense projects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Flexible way&lt;/strong&gt; can be used when you don't know the specifics of what you are doing,  e.g. you want to outsource the expertise of designing and integration of the subsystem. You can do this when you have good business partnerships with the subcontractor (who will develop and supply the Product). Several of the specifications of the design will eventually change during the project. Specifications may change because of the bigger system or because of the Product itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Flexible way, the Work Description can be more important than the Technical Specifications (=PRD, as you call it). So it's two documents you have to consider: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Work Description: HOW will they develop the PRODUCT (including how&#xA;will they interact with you)&lt;/li&gt;&#xA;&lt;li&gt;Technical Specifications: WHAT will the PRODUCT do, or What&#xA;properties will it have.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the flexible way, writing a brief document will suffice. Key Performance Parameters, and Key Characteristics/Properties could be sufficient. So the structure of the document can be as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;General Specifications&lt;/li&gt;&#xA;&lt;li&gt;Performance Requirements&lt;/li&gt;&#xA;&lt;li&gt;SubPart-A Requirements (Assume that you know the main part breakdown of the Product)&lt;/li&gt;&#xA;&lt;li&gt;SubPart-B Requirements&lt;/li&gt;&#xA;&lt;li&gt;Interface/Integration Requirements&lt;/li&gt;&#xA;&lt;li&gt;Other Requirements&lt;/li&gt;&#xA;&lt;li&gt;Appendix (e.g. CAD data for interface, technical drawings, draft interface control document, etc.)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Well-Specified way&lt;/strong&gt; is used when the Product to be developed is so important that you cannot risk changing the specifications on the road. It can be a mission critical subsystem or a performance critical item. For sake of completeness I'll refer you to the Defense Acquisition template. &lt;a href=&quot;http://acqnotes.com/acqnote/careerfields/system-requirements-document&quot; rel=&quot;nofollow noreferrer&quot;&gt;ref&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structure of a System Technical Specification document is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;ol&gt;&#xA;&lt;li&gt;Scope/Introduction    8&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;1.1. System or Subsystem Identification  8&lt;/li&gt;&#xA;&lt;li&gt;1.2. System or Subsystem Overview    8&lt;/li&gt;&#xA;&lt;li&gt;1.3. Document Overview   8&lt;/li&gt;&#xA;&lt;li&gt;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Applicable Documents    8&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;2.1. General 8&lt;/li&gt;&#xA;&lt;li&gt;2.2. Government Documentation    9&lt;/li&gt;&#xA;&lt;li&gt;2.2.1.   Government Specifications, Standards, and Handbooks 9&lt;/li&gt;&#xA;&lt;li&gt;2.2.2.   Other Government Documents, Drawings, and Publications  9&lt;/li&gt;&#xA;&lt;li&gt;2.3. Non-Government Publications 9&lt;/li&gt;&#xA;&lt;li&gt;2.4. Order of Precedence 10&lt;/li&gt;&#xA;&lt;li&gt;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;System or Subsystem Requirements    10&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;3.1. Required States and Modes   11&lt;/li&gt;&#xA;&lt;li&gt;3.2. System of Subsystem Functional Requirements 11&lt;/li&gt;&#xA;&lt;li&gt;3.2.1.   System or Subsystem Function    11&lt;/li&gt;&#xA;&lt;li&gt;3.3. System External Interface Requirements  11&lt;/li&gt;&#xA;&lt;li&gt;3.3.1.   Interface Identification and and Diagrams   11&lt;/li&gt;&#xA;&lt;li&gt;3.3.2.   Project Unique Identifier of Interface  12&lt;/li&gt;&#xA;&lt;li&gt;3.4. System Internal Interface Requirements  12&lt;/li&gt;&#xA;&lt;li&gt;3.5. System Internal Data Requirements   12&lt;/li&gt;&#xA;&lt;li&gt;3.6. Adaption Requirements   12&lt;/li&gt;&#xA;&lt;li&gt;3.7. Safety Requirements 13&lt;/li&gt;&#xA;&lt;li&gt;3.8. Security and Privacy Requirements   13&lt;/li&gt;&#xA;&lt;li&gt;3.9. System Environmenet Requirements    13&lt;/li&gt;&#xA;&lt;li&gt;3.10.    Computer Resource Requirements  13&lt;/li&gt;&#xA;&lt;li&gt;3.10.1.  Computer Hardware Requirements  14&lt;/li&gt;&#xA;&lt;li&gt;3.10.2.  Computer Hardware Resource Utilization Requirements 14&lt;/li&gt;&#xA;&lt;li&gt;3.10.3.  Computer Software Requirements  14&lt;/li&gt;&#xA;&lt;li&gt;3.10.4.  Computer Communications Requirements    14&lt;/li&gt;&#xA;&lt;li&gt;3.11.    System Quality Factors  14&lt;/li&gt;&#xA;&lt;li&gt;3.12.    Design and Construction Contraints  15&lt;/li&gt;&#xA;&lt;li&gt;3.13.    Personnel-Related Requirements  15&lt;/li&gt;&#xA;&lt;li&gt;3.14.    Training-Related Requirements   15&lt;/li&gt;&#xA;&lt;li&gt;3.15.    Logistics-Related Requirements  15&lt;/li&gt;&#xA;&lt;li&gt;3.16.    Other Requirements  16&lt;/li&gt;&#xA;&lt;li&gt;3.17.    Packaging Requirements  16&lt;/li&gt;&#xA;&lt;li&gt;3.18.    Statutory, Regulatory, and Certification Requirements   16&lt;/li&gt;&#xA;&lt;li&gt;3.18.1.  Statutory Requirements  16&lt;/li&gt;&#xA;&lt;li&gt;3.18.2.  Regulatory Requirements 16&lt;/li&gt;&#xA;&lt;li&gt;3.18.3.  Certification Requirements  16&lt;/li&gt;&#xA;&lt;li&gt;3.19.    Precedence and Criticality of Requirements  16&lt;/li&gt;&#xA;&lt;li&gt;3.20.    Demilitarization and Disposal   17&lt;/li&gt;&#xA;&lt;li&gt;&lt;ol start=&quot;4&quot;&gt;&#xA;&lt;li&gt;Verification Provisions 17&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;4.1. Verification Methods    17&lt;/li&gt;&#xA;&lt;li&gt;4.1.1.   Demonstration   17&lt;/li&gt;&#xA;&lt;li&gt;4.1.2.   Test    17&lt;/li&gt;&#xA;&lt;li&gt;4.1.3.   Analysis    17&lt;/li&gt;&#xA;&lt;li&gt;4.2. Inspection  17&lt;/li&gt;&#xA;&lt;li&gt;4.3. Special Verification Methods    17&lt;/li&gt;&#xA;&lt;li&gt;&lt;ol start=&quot;5&quot;&gt;&#xA;&lt;li&gt;Requirements Traceability   18&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;5.1. Traceability to Capability Document or System Specification 18&lt;/li&gt;&#xA;&lt;li&gt;5.2. Traceability to Subsystems Requirements 18&lt;/li&gt;&#xA;&lt;li&gt;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Appendix    18&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;li&gt;6.1. Appendix A: Acronyms and Definitions    18&lt;/li&gt;&#xA;&lt;li&gt;6.2. Appendix B: Key Performance Parameters/Key system Attributes    18&lt;/li&gt;&#xA;&lt;li&gt;6.3. Appendix C: Requirements Traceability Matrices  18&lt;/li&gt;&#xA;&lt;li&gt;6.4. Appendix D: Verification Matrices   19&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="1097" LastActivityDate="2019-08-17T09:55:06.947" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29683" PostTypeId="1" CreationDate="2019-08-17T10:45:28.277" Score="0" ViewCount="132" Body="&lt;p&gt;When discussing the stagnation flow properties we always discuss stagnation temperature, stagnation pressure, and stagnation enthalpy. Is there also a stagnation property called the stagnation volume i.e. the volume the fluid occupies when the fluid in the flow is brought to stagnation isentropically?   &lt;/p&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2019-08-18T06:00:39.910" Title="Is there something called the stagnation volume?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29684" PostTypeId="2" ParentId="29683" CreationDate="2019-08-17T11:47:33.973" Score="1" Body="&lt;p&gt;Stagnation volume is nothing but when the volume of a given gas or fluid has achieved a static condition. In case of stagnation the given mass achieves a state of constant zero or negligible velocity for practical cases. The entire kinetic energy energy is converted into potential energy. The pressure state can be any but its velocity must be zero.&#xA;For a more detailed analysis please refer to the attached link : &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://nptel.ac.in/courses/101103004/module4/lec3/1.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://nptel.ac.in/courses/101103004/module4/lec3/1.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21483" LastEditorUserId="10902" LastEditDate="2019-08-18T06:00:39.910" LastActivityDate="2019-08-18T06:00:39.910" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29685" PostTypeId="1" CreationDate="2019-08-17T12:21:51.130" Score="0" ViewCount="815" Body="&lt;p&gt;Our question is what size and alloy aluminum tube can be used to replace steel tube in a vertical load application. Our current spec is 1.5” OD STD steel tubing 7 feet long. We will use 6 poles attached to steel footings to support vertical stacking of containers that each weigh 1500 lbs. the bottom container will see 4500 lbs. from vertical stacking.  the containers are 45” wide and 87” long. There is one pole in each corner and 2 in the center on each side. Six tubes of steel 1.5” x 7’ steel is to heavy. Can we use aluminum tube round hex or square and what alloy ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="21500" LastActivityDate="2020-01-16T10:02:20.327" Title="Aluminum tube vs steel" Tags="|structural-analysis|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29687" PostTypeId="1" AcceptedAnswerId="29688" CreationDate="2019-08-17T13:45:32.907" Score="0" ViewCount="95" Body="&lt;p&gt;Nuclear power always require some way of converting the energy from nuclear to electrical or mechanical to be useful. For example, in a nuclear power generation facility, this is done through heat/steam which drives mechanical generators. What is the mechanism used to convert nuclear energy to rocket power?&lt;/p&gt;&#xA;" OwnerUserId="20755" LastActivityDate="2019-08-17T14:50:16.943" Title="How does a nuclear powered rocket engine work?" Tags="|rocketry|nuclear-engineering|nuclear-technology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29688" PostTypeId="2" ParentId="29687" CreationDate="2019-08-17T14:50:16.943" Score="2" Body="&lt;p&gt;In &lt;a href=&quot;https://en.wikipedia.org/wiki/Nuclear_thermal_rocket&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nuclear Thermal Rockets&lt;/a&gt; (NTRs) &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the heat from a nuclear reaction replaces the chemical energy of the propellants in a chemical rocket. In an NTR, a working fluid, usually liquid hydrogen, is heated to a high temperature in a nuclear reactor and then expands through a rocket nozzle to create thrust. The external nuclear heat source theoretically allows a higher effective exhaust velocity and is expected to double or triple payload capacity compared to chemical propellants that store energy internally.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="90" LastActivityDate="2019-08-17T14:50:16.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29690" PostTypeId="2" ParentId="29668" CreationDate="2019-08-17T15:20:07.907" Score="1" Body="&lt;p&gt;There are several companies that make adjustable dampers, of course, how much you want to adjust and also how much you want to spend are parameters that only you can decide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Companies like Bilstein, Spax, Fox and TFX can provide solutions based on the mass of the vehicle and the travel in a non-adjustable solution all the way to twin shocks with separate reservoirs with landing and extension rates controllable differently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fox can be found &lt;a href=&quot;https://www.ridefox.com/subhome.php?m=atv&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And TFX &lt;a href=&quot;https://tfxsuspension.com/shocks-quad-atv-tfx/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are others, but that should give you some directions to check out.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-17T15:20:07.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29694" PostTypeId="2" ParentId="24222" CreationDate="2019-08-17T23:36:48.640" Score="1" Body="&lt;p&gt;I’m thinking, based on what I see and what you describe, something like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;“Adjustable Axle Carrier (Arm)” (or Bracket)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It’s not too long of a name yet provides some detail while using some basic terms which represent its purpose. Limit the jargon unless it’s necessary or the context of the design or assembly warrants it. &lt;/p&gt;&#xA;" OwnerUserId="2559" LastActivityDate="2019-08-17T23:36:48.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29696" PostTypeId="1" CreationDate="2019-08-18T03:02:53.047" Score="0" ViewCount="71" Body="&lt;p&gt;Assuming induced supercavitation (like &lt;a href=&quot;https://en.wikipedia.org/wiki/Supercavitating_torpedo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Supercavitating torpedo&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Juliet_Marine_Systems_Ghost&quot; rel=&quot;nofollow noreferrer&quot;&gt;Juliet Marine Systems Ghost&lt;/a&gt;) reduces water surface area does that reduce friction enough to submerge a surface drive propeller without a reduction in thrust if the reduction in surface area equals what would be on the surface?&lt;/p&gt;&#xA;" OwnerUserId="21506" LastEditorUserId="282" LastEditDate="2019-08-19T20:57:20.040" LastActivityDate="2022-05-06T03:03:25.320" Title="Can a &quot;surface drive&quot; propeller be submerged if induced supercavitation reduces water surface area?" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29697" PostTypeId="1" CreationDate="2019-08-18T03:50:06.410" Score="0" ViewCount="231" Body="&lt;p&gt;I'm familiar with several books that deal with the analysis and design of simple mechanisms (such as Norton's Design of Machinery), including four-bar linkages, gear trains and cams. There are also books that deal with the design of machine elements, such as shafts, springs, gears, bearings and so on, taking mechanical properties of materials into considertion (like Shingley's mechanical engineering design).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I haven't been able to find a book about combining those individual machine elements into complex machines to perform a specific task. It seems very obvious to me that designing a milling machine, a wood chipper, a bottling machine or other such systems one would need more knowledge than simply the design of its individual elements. Many even require knowledge of thermodynamics, fluid mechanics or electronics to have been designed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any book you know of that deals with the design of machines that are more complex than simple four-bar mechanisms or gear trains, like the ones mentioned? Preferrably with the same mathematical rigor found in machine element books. I have also seen books dealing with engineering design in general (like Dieter's Engineering Design) but they are far too vague, only including a general oultine of design practices such as identifying customer needs, prototyping, documentation, etc. that apply to any engineering product, not only machines. &lt;/p&gt;&#xA;" OwnerUserId="18155" LastEditorUserId="18155" LastEditDate="2019-08-19T01:17:22.867" LastActivityDate="2019-08-19T01:17:22.867" Title="Any books about design of complex machines?" Tags="|machine-design|" AnswerCount="1" CommentCount="5" ClosedDate="2019-08-18T23:40:36.287" ContentLicense="CC BY-SA 4.0" />
  <row Id="29698" PostTypeId="2" ParentId="29697" CreationDate="2019-08-18T05:09:34.237" Score="1" Body="&lt;p&gt;The topics necessary warrant their own books and each topic is sufficiently large that the books already exist and are large already, so combining them into one book would mean you could not carry it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logical solution is to have books for each topic,: mechanics, thermofluids, tribolgy, control etc and you combine what you need according to the problem you face.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the problem faced is large or very challenging then teams of engineers get put together to combine the skills necessary.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-18T05:09:34.237" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29699" PostTypeId="1" CreationDate="2019-08-18T06:49:48.450" Score="1" ViewCount="47" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Electronic_stability_control&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia article for ESC&lt;/a&gt; describes the process as using hydraulic anti-lock brakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is it possible for electromagnetic braking to be used instead, with an electic motor/generator in each wheel?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect this configuration to reduce the components needed for ESC (especially for electic cars), as well as enabling ESC for smaller vehicles (where I've heard it's hard to fit), while lowering response time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So can ESC be implemented using electromagnetic braking from motors? If not, why? If it can, than why isn't it commonly used? Has it ever been attempted?&lt;/p&gt;&#xA;" OwnerUserId="21510" LastActivityDate="2021-01-09T11:05:20.533" Title="Can a car's Electronic Stability Control (ESC) operate using electromagnetic braking?" Tags="|automotive-engineering|regenerative-braking|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29700" PostTypeId="2" ParentId="29699" CreationDate="2019-08-18T07:04:49.673" Score="1" Body="&lt;p&gt;For &quot;classic&quot; cars ice and hydraulic brakes then you would need to add extra electro-magnetic brakes &amp;amp; control system adding to weight, complexity and of course cost. So, possible but not likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For hybrid cars that have motors on each wheel then the motor controllers and motors could be upgraded to suit, however there will probably need to be a separate braking system for reasons of safety ie legislation so using that for esc is more likely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For full-electric, then the response as per hybrid seems relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do you think esc is hard to fit on small cars? Most esc is a derivative of abs and is usually implemented in the ecu or with another small &quot;black box&quot; - not hard to fit... But most likely not fitted to small cars as they are built to a price-point so only the minimum is fitted. The up-market model may have it fitted.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-18T07:04:49.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29701" PostTypeId="1" CreationDate="2019-08-18T10:04:54.920" Score="1" ViewCount="26" Body="&lt;p&gt;If possible, could someone please compile a list of the engineering concepts associated with EV3 programming and Mindstorms robots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you - your answers are much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="21511" LastActivityDate="2019-08-18T10:04:54.920" Title="What engineering concepts are used in EV3 programming/ a Mindstorms robot?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="5" ClosedDate="2019-08-18T13:34:30.020" ContentLicense="CC BY-SA 4.0" />
  <row Id="29702" PostTypeId="1" CreationDate="2019-08-18T15:12:24.553" Score="0" ViewCount="142" Body="&lt;p&gt;Is there any National Pipe Thread Straight (non-tapered, requires O-ring or compression olive) that is compatible, even loosely, with a machine thread threaded rod? I want to put a reducer on the end of a pipe threaded with NPTS and reduce to like 1/4 inch NPTS nominal and have that reducer ride on a 5/16-18 threaded rod.  So it's like a shoring jack or lolly column, but really cheap, and the threaded rod can fit through a small clearance and guide the end of the pipe to the bearing point.&lt;/p&gt;&#xA;" OwnerUserId="5967" LastActivityDate="2021-06-09T15:02:35.207" Title="Straight Pipe Thread" Tags="|pipelines|threads|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29704" PostTypeId="2" ParentId="29696" CreationDate="2019-08-18T16:14:31.217" Score="1" Body="&lt;p&gt;The idea behind surface props is to get the appendage drag down to nothing. All the shafting and shaft support structure is out of the water. This means the propeller is a ventilating type. The hypertorpedo and Ghost are also ventilated designs. The difference lies in the vapor pressure and vapor composition, and the differences in hydrodynamics of the these mixed-phase flows. Ventilated flows don't have the same problems with cavitiation bubbles eroding the surface of the propeller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as propeller design goes, the devil is in the details. You have to account for static pressure (depth pressure). You have to account for the effective speed of sound in the different bubbly regions (the speed of sound in mixed phase flows can get stupid slow, so you end up with supersonic blade speeds and normal ogival blades don't work.) And you have to recognize that normal flow models just don't work. In order for bubbly flow models to work, you need to operate on a scale where for some representative length &lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt;, the flow properties change very little over this distance, but the distance is still very big compared to the bubble or cavity sizes. This condition simply isn't met in ventilated propellers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the basic idea is similar, you want cleavers, not ogives, but that is about all you can say. The rest is cut and try. Note that Ghost has the props out in front of the SWATH hulls. Ghost isn't anywhere near fast enough to want cavitation adapted props. And if memory serves, Hypertorpedos use props for cruise, but are basically just rockets at speed. Their prop regime is likely also noncavitating, but I'm not sure about that.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-08-18T16:14:31.217" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29705" PostTypeId="2" ParentId="29702" CreationDate="2019-08-18T19:26:33.290" Score="1" Body="&lt;p&gt;NPS as defined in ASTM A 530 is all tapered and has no &quot;O&quot; rings or torque shoulders.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-08-18T19:26:33.290" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29706" PostTypeId="1" CreationDate="2019-08-18T19:40:12.933" Score="1" ViewCount="38" Body="&lt;p&gt;I'm designing a custom part that needs an active shock absorber and space is restricted. I most likely won't be able to use a stock absorber and was thinking, can two squarish tubes be made tight enough?&lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2019-08-18T23:39:48.957" Title="Can hydraulics be used with a squarish piston?" Tags="|hydraulics|pistons|shock|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29707" PostTypeId="2" ParentId="29706" CreationDate="2019-08-18T23:39:48.957" Score="3" Body="&lt;p&gt;Assuming reasonable pressures, it should be workable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, keep in mind that the pressures will be higher at the corners and you'll run a higher risk of leakage / bursting.  Cylindrical tubes distribute the pressure evenly across the bearing surface and don't have that issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll also potentially have difficulty with fabricating a gasket to seal the space between the square tubes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the minimal difference in surface area between a cylindrical and square tube, and the ready prevalence of cylindrical tubes, it does beg the question as to why you feel a square tube is the more appropriate design choice.  &lt;/p&gt;&#xA;" OwnerDisplayName="user16" LastActivityDate="2019-08-18T23:39:48.957" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29709" PostTypeId="2" ParentId="29685" CreationDate="2019-08-19T09:16:15.097" Score="1" Body="&lt;p&gt;Aluminum columns usually fail under buckling when their slenderness ratio is greater than 99. using the formula K= L/r, where L is the free factored length and r is the radius of gyration. &lt;span class=&quot;math-container&quot;&gt;$r=(I/A)^{0.5}$&lt;/span&gt; In your case K may be smaller or bigger than one, needs more detail on the type of connections. We assume K=1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$r= 1/4\sqrt(D^2+d^2)= 1/4\sqrt(1.562+1/64)=0.505$&lt;/span&gt; Assuming 1/8 in tube thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case K=84in/0.505in= 165 , 165&amp;lt;99.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_{critical}=\pi^2\frac{EI}{(Kl)^2}$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I let you take over from here, but just eyeballing it, it looks ok.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-19T09:16:15.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29710" PostTypeId="1" AcceptedAnswerId="29717" CreationDate="2019-08-19T09:28:15.440" Score="0" ViewCount="875" Body="&lt;p&gt;Currently I'm doing a personal project, but I've some doubts about the calculation process. Well, I've a platform (2.30m length x 0.32m width). This platform is supported by 4 wheels with a diameter equals 120mm. This platform loads a motorbike. The maximum mass (platform + motorbike) that it will move is about 400kg. The surfaces contact are rubber with concrete (coefficient of friction 0.45). Rolling coefficient I think is 0.02). I'm uncertain wether to put 2 driven wheels or 4. The platform must travel a distance of 7 m. For the process I propose a speed of 0.1 m / s.&#xA;Well, How can I calculate the required torque to move this platform with a motorbike?&lt;/p&gt;&#xA;" OwnerUserId="21519" LastEditorUserId="4005" LastEditDate="2019-08-20T14:11:46.967" LastActivityDate="2019-08-20T14:11:46.967" Title="How to calculate required torque" Tags="|mechanical-engineering|mechanical|industrial-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29711" PostTypeId="1" CreationDate="2019-08-19T11:06:41.423" Score="0" ViewCount="60" Body="&lt;p&gt;I'm looking for a solution for keeping a vertical stack of small boxes under tension. I will use the device in the attached picture to hold the boxes which will have a lid on the top, and the boxes will be pushed towards the front by a linear actuator. So far I've thought of an extension spring which will be attached to the back end of the device and the bottom platform, but not sure if it's an optimal solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vLzpq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vLzpq.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;George&lt;/p&gt;&#xA;" OwnerUserId="21521" LastEditorUserId="21521" LastEditDate="2019-08-19T12:20:44.593" LastActivityDate="2019-08-19T12:20:44.593" Title="Mechanical solution for keeping a vertical stack under pressure" Tags="|mechanisms|mechanical|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="29713" PostTypeId="1" CreationDate="2019-08-19T13:06:24.490" Score="0" ViewCount="13" Body="&lt;p&gt;We need to connect two distinct heating networks on a wastewater treatment plant into one. So far I've only seen heating networks with one central heat source or one central manifold that all heat sources feed into, but this time we have distributed sources.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the state the system is now in:&#xA;network 1:&#xA;Natural gas boiler, several smallish consumers (two adjactant buildings)&#xA;network 2:&#xA;oil boiler, sewage gas CHP, several smallish and one big consumer (sveral buildings, spread aout about 100m apart)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Target state:&#xA;both networks (about 100m apart) are connected,&#xA;Manifold 1: &#xA;one additional gas CHP, one additional largish consumer&#xA;Manifold 2:&#xA;no change&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manifold 1 should supply power to manifold 2 when needed, never the other way round.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My idea is to have connect both hot and cold sides and have pumps (hot side manifold 1 -&gt; hot side manifold 2, cold side manifold 2 -&gt; cold side manifold 1). The &quot;hot&quot; pump will be controlled for delta T between both manifolds, the cold pump should ensure hydraulic balance but I dont know a good control scheme for this.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-08-19T13:06:24.490" Title="How to connect to heating networks both with sources and consumers into one network?" Tags="|heating-systems|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="29714" PostTypeId="1" CreationDate="2019-08-19T14:39:21.727" Score="0" ViewCount="119" Body="&lt;p&gt;I was told that PMMA is preferred over glass for windows in cold climates because of it's R-value being higher than glass.I wish there were a stick-on film made of glass! That could be used to cover PMMA windows, to eliminate the &quot;scratch&quot; factor if nothing else.  &lt;/p&gt;&#xA;" OwnerUserId="21523" LastActivityDate="2019-08-19T15:27:27.977" Title="pmma vs glass windows?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2019-08-19T17:50:45.923" ContentLicense="CC BY-SA 4.0" />
  <row Id="29715" PostTypeId="2" ParentId="29714" CreationDate="2019-08-19T15:07:31.310" Score="1" Body="&lt;p&gt;Hard to say about the preferences of the whole world. But i live in a what is generally considered to be a cold climate. Atleast here most window panes made out of glass. They simply are not single glass pane. But 3-4 separate pieces of glass separated by air for the outermost layer and possibly inert gas for the second layer. This increases the performance of windows significantly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The innermost pane is also often laminated with plastic. So in fact the pmma may be sandwitched between glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway, i would dispute PMMA is prefered. It is atleast not true in my location. But obviously can vary based on where you are, just that the claim sounds generally nonfactual. There can obviously be other locations like south pole where situation is wastly different.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PMMA can be coated with diamond too by the way, but I havent tested what happens to the coating when the temperatue changes radically.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2019-08-19T15:27:27.977" LastActivityDate="2019-08-19T15:27:27.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29716" PostTypeId="2" ParentId="7766" CreationDate="2019-08-19T19:46:28.383" Score="0" Body="&lt;p&gt;Whilst I am not an expert or remotely educated in this field of engineering, I am aware that the increase in speed is down to increase in electromotive power to each carriage. The diesel engines provide electricity to electric motors which drive the individual carriages. The &quot;go faster&quot; lever probably sets a lower resistance to the motors of each carriage and the dynamic power outout of the diesel motors adjusts automatically to reduced resistance to current flow. This in turn draws higher current and their greater energy consumption from the motors. This is why trains do not seem to &quot;change gear&quot; with acceleration. The engines kick up to the power output demand immediately and continue to deliver the load required until the electric motors bring equilibrium between power demand and kinetic motion. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-08-19T19:46:28.383" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29717" PostTypeId="2" ParentId="29710" CreationDate="2019-08-19T19:51:08.807" Score="0" Body="&lt;p&gt;Let's get a few things out of the way first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque or force have nothing to do with the speed, they interact with angular momentum or acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if the travel path is level you will be fine with just two driven wheels&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To figure the force needed let's say we apply enough force so that at the end of the 7m the speed changes from zero to 0.1m/s' as you mention and then the brakes engage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ x=\frac{v^2}{2a} \ ,\ a=\frac{v^2}{2x}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;span class=&quot;math-container&quot;&gt;$a=\frac{0.1^2}{2*7} =0.01/14=0.0007$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  F=ma =400kg*9.8*0.0007=2.8N\quad 2.8*1.02 =2.86\ \text{after the friction } $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this force should be divided by two for individual wheel and converted to torque applied at radius of 0.06m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ 2.86/(2*0.06)=24Nm \ \text{torqe/wheel}  $&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-19T19:51:08.807" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29718" PostTypeId="2" ParentId="18874" CreationDate="2019-08-19T20:09:55.010" Score="1" Body="&lt;p&gt;A laser would be the best method followed by repetitive razor slicing in ever deeper cuts to avoid frictional deformation. Another method would be cold cuts by freezing the silicone and using a regular mechanical saw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The key to doing it properly is removing deformation from the property of silicone when pressure is applied. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-08-19T20:09:55.010" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29719" PostTypeId="2" ParentId="18556" CreationDate="2019-08-19T20:14:16.427" Score="1" Body="&lt;p&gt;The problem here is moments of force. Rather than stiffening the fixings, add dampers like rubber washers with steel washers either side of the rubber dampers. This should absorb much of the instantaneous moment forces while providing vibration protection from adjacent mechanisms and fixings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fracture is due to strain hardening of the mild steel threaded section. Really what you want is a weaker replacement section attached to the steering rack. The break occurs where the concentration of forces exceeds the elastic modulus of the bolt resulting in what happened. Ideally you want a failure to be gradual so you have warning of imminent fatiguing failure. This would be designing the bent steel section as a thinner and more flexible bit so that it absorbs much of the shock forces at motion. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2020-01-24T01:00:23.877" LastActivityDate="2020-01-24T01:00:23.877" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29720" PostTypeId="1" CreationDate="2019-08-19T20:15:01.457" Score="0" ViewCount="108" Body="&lt;p&gt;I am currently working on designing an oscillating water column as part of a project. In summary:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;OWC are devices with a semi-submerged chamber or hollow open to the&#xA;  sea below, keeping a trapped air pocket above a water column. Waves&#xA;  force the column to act like a piston, moving up and down, forcing the&#xA;  air out of the chamber and back into it. This continuous movement&#xA;  force a bidirectional stream of high-velocity air, which is channelled&#xA;  through a Power-Take-Off (PTO). The PTO system converts the airflow&#xA;  into energy.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Oscillating_water_column&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Oscillating_water_column&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, the PTO system is a turbine, the rotation of which will power a rotary vane pump. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/le2yH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/le2yH.png&quot; alt=&quot;Number 2 is the pump rotor&quot;&gt;&lt;/a&gt;&#xA;(Number 2 is the pump rotor)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pump rotor and the turbine would be on the same axis and connected by a rod. In short, the rotation of the turbine would directly power the pump; electricity would not be produced in the process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been searching to see what turbines would be the most efficient, but I haven't found any information. What would be the most efficient turbine for this situation? Is it a Wells Turbine? Something like a bidirectional impulse turbine &lt;a href=&quot;https://i.stack.imgur.com/EnrYV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EnrYV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or something else?&lt;/p&gt;&#xA;" OwnerUserId="21524" LastActivityDate="2019-08-19T20:15:01.457" Title="Most efficient turbine for harnessing energy from bidirectional air flow?" Tags="|mechanical-engineering|fluid-mechanics|airflow|turbines|energy-efficiency|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29721" PostTypeId="1" CreationDate="2019-08-19T20:21:18.317" Score="0" ViewCount="1969" Body="&lt;p&gt;When we calculate the dimension of a foundation we always use the allowable SERVICE bearing capacity (lets call it &lt;strong&gt;qs&lt;/strong&gt;) and all the necessary checks are based on this value qs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why does the geotechnical report calculates the ultimate bearing capacity (lets call it &lt;strong&gt;qu&lt;/strong&gt;)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my own research I stepped upon an article that says that when the geotechnical engineer calculates the ultimate bearing capacity it calculates the ultimate load by multiplying the service loads with a safety factor (lets call it &lt;strong&gt;Fs&lt;/strong&gt;) now for a structural engineer (such as i am) we use different load combinations to calculate the ultimate load (especially when we have seismic and wind) in a simple case for example the most famous load combination in the ACI Code is 1.2D + 1.6L if in a certain case the Live Load is &gt;&gt;&gt;&gt; than the Dead Load the use of a simple Fs will not work to calculate the ultimate load combination.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that Explanation for this topic is feasible ? and what do you have other suggestion to the difference of qs and qu ? and where do we use each one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank You in Advance.&lt;/p&gt;&#xA;" OwnerUserId="10554" LastEditorUserId="90" LastEditDate="2022-08-03T00:48:44.563" LastActivityDate="2023-02-20T20:46:22.970" Title="Difference between Allowable Bearing capacity Service and Ultimate?" Tags="|civil-engineering|geotechnical-engineering|foundations|soil-mechanics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29722" PostTypeId="1" CreationDate="2019-08-19T20:38:26.170" Score="1" ViewCount="8164" Body="&lt;p&gt;The primary two functions of a Grade/Tie Beam in spread footing and/or pile caps foundation system is to support the Slab-On-Grade (S.O.G) and/or to take the moment from eccentric footing (or moment from lateral loads example wind and seismic).&#xA;While the pile cap level and depth may differ from one site and another the level of the slab-on-grade is fixed by architectural constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the most suitable level for a grade/tie beam is it at the slab-on-grade level or the footing level and what will happen if the beam is at the S.O.G level and the footing is subjected to moment?&lt;/li&gt;&#xA;&lt;li&gt;How to design such a beam, meaning what are the loads and how to calculate the lateral loads (especially shear and moment) from loads such as seismic and wind?&lt;/li&gt;&#xA;&lt;li&gt;In what Section of the ACI and Eurocode is the Tie/Grade Beam is mentioned and is it always required (meaning for all soil types and seismic zones ...)? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4GIK6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4GIK6.png&quot; alt=&quot;Tie/Grade Beam at different levels sections&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2019-08-19T21:31:14.607" Title="At what level is the Grade Beam most effective at pile cap level or at Slab-on-grade level?" Tags="|structural-engineering|civil-engineering|geotechnical-engineering|building-design|foundations|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29723" PostTypeId="1" CreationDate="2019-08-19T21:05:06.033" Score="0" ViewCount="26" Body="&lt;p&gt;Some cordless lawn mowers have a feature that causes the motor to cut off when the mower is rolled over a driveway or over a section of lawn where the grass is very sparse.  How does the mower sense this condition?&lt;/p&gt;&#xA;" OwnerUserId="21526" LastEditorUserId="110" LastEditDate="2019-12-24T13:57:39.503" LastActivityDate="2019-12-24T13:57:39.503" Title="How do cordless lawn mowers determine lawn density?" Tags="|sensors|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29724" PostTypeId="1" CreationDate="2019-08-19T21:06:17.147" Score="0" ViewCount="178" Body="&lt;p&gt;How to design a drop panel at a column in case the column above is rotated or not aligned to the below column (see image below), the code that i am using is the ACI Code if any reference for the design of such a case is available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another cases we usually use the drop panel to reduce the span between two columns and reduce the effects of shear on slab (especially punching shear), in case of seismic/wind resisting building is there any loads or changes to the design of these special elements and how do they affect the behavior of the building ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in Advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oewQa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oewQa.png&quot; alt=&quot;Drop panel plan view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2019-08-20T04:11:57.710" Title="Drop Panel Design Above Columns?" Tags="|structural-engineering|civil-engineering|design|structures|building-design|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29725" PostTypeId="1" CreationDate="2019-08-19T21:11:45.057" Score="0" ViewCount="398" Body="&lt;p&gt;Why in the design of vertical elements (columns and walls) for a structure against lateral loads (such as seismic and wind) we use a slab (or shell) with a stiffness modifiers (such as 0.25 according to ACI Code) to reduce the moment of inertia (example when we design vertical elements in ETABs and Robot), but in the simple design of the slab we do not use these modifiers (and a side question is: why don;t we include the lateral load of seismic and wind in the design of the slab)??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in Advance.&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2019-08-20T02:04:20.997" Title="Slab Stiffness Modifiers applied in Vertical Elements design but not in Slab design?" Tags="|structural-engineering|civil-engineering|design|building-design|seismic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29726" PostTypeId="1" CreationDate="2019-08-19T21:14:44.540" Score="0" ViewCount="3686" Body="&lt;p&gt;What is the fundamental difference between the ASCE Code and IBC Code and what is the objective of each one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in Advance.&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2019-08-19T22:37:03.993" Title="What is the Difference between ASCE and IBC Code?" Tags="|structural-engineering|civil-engineering|building-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29727" PostTypeId="1" CreationDate="2019-08-19T21:19:13.023" Score="0" ViewCount="640" Body="&lt;p&gt;what is the difference between the seismic design between ASCE and UBC code?&#xA;from load combinations to loads applied to the history between the sudden shift or change in the code. and which one of the codes does it result in over-reinforcement  (or overdesign) i mean what code for example will give column dimensions &gt;&gt; than the other one ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in Advance.&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2019-08-19T21:19:13.023" Title="Seismic Design Changes between ASCE and UBC Code?" Tags="|structural-engineering|civil-engineering|building-design|seismic|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29728" PostTypeId="1" CreationDate="2019-08-19T21:25:37.520" Score="1" ViewCount="3007" Body="&lt;p&gt;According to the ACI Code stiffness modifiers (modifiers used to reduce the moment of inertia) are used for each structural elements for example 0.7 for columns and 0.35 for beams... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who do seismic and/or wind design according to Eurocode where are these modifiers used and if they are not clearly available like in the ACI Code how does the code apply these modifiers ? does eurocode use a coefficient to reduce the compressive strength of the concrete instead of using these coefficient?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in Advance.&lt;/p&gt;&#xA;" OwnerUserId="10554" LastActivityDate="2023-05-27T09:55:53.470" Title="Difference between ACI Code and Eurocode in Stiffness modifiers?" Tags="|structural-engineering|civil-engineering|building-design|eurocodes|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29729" PostTypeId="2" ParentId="29722" CreationDate="2019-08-19T21:31:14.607" Score="2" Body="&lt;p&gt;The grade beam typically must be supported by competent soil. In some cases they are designed as per recommendation of soils report for specific tasks such as integrating soldier piles or supporting suspended structures such as utility rooms not getting support from their own foundation, or hillside structures overhanging over the slope. Then they are designed as beams with the caveat that they may be under the soils and need rebar cover as per concrete under the grade. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is the grade beams can and do transfer column or pylon moments and thus impart at times significant bearing force to the soil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The slab on grade is usually over the grade with humidity and moisture membrane of 4 inches of gravel and is not a critical component of the structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So grade beam is typically at a lower level unless it's Height is enough to join the slab.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-19T21:31:14.607" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29730" PostTypeId="2" ParentId="29726" CreationDate="2019-08-19T22:37:03.993" Score="1" Body="&lt;p&gt;Assuming you're referring to ASCE 7 &lt;em&gt;Minimum Design Loads and Associated Criteria for Buildings and Other Structures&lt;/em&gt;, it's important to note that this document is not a code. It is a standard, published by ASCE. This standard is referenced by the International Building Code, which has legal force when adopted by a jurisdiction.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2019-08-19T22:37:03.993" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29731" PostTypeId="2" ParentId="29262" CreationDate="2019-08-20T01:38:24.527" Score="1" Body="&lt;p&gt;Take a random electric bike and another electric motorbike, the base idea of designing them would be fairly similar: just a frame holding a battery and 2 wheels together. There is not that much difference between the two of them. Though I would say that the &quot;bias&quot; with electric motorbike is better (it is automatically assumed that it has a better range)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transmission... I would say that have fairly similar layout, though the electric bike (or some of them) has the option for &quot;pedalling&quot; (use just like a normal bicycle), and again, the pedals can also be removed, practically making the electric bike an electric motorbike.&lt;/p&gt;&#xA;" OwnerUserId="21520" LastActivityDate="2019-08-20T01:38:24.527" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29732" PostTypeId="2" ParentId="29725" CreationDate="2019-08-20T02:04:20.997" Score="1" Body="&lt;p&gt;If you read the commentary of the ACI code you you will understand that stiffness modifiers take into account the cracked section properties expected from cyclical loading. Slabs / diaphragms are a critical part of the lateral load resisting system. To ignore the slab design for lateral loads is a major mistake.&lt;/p&gt;&#xA;" OwnerUserId="19912" LastActivityDate="2019-08-20T02:04:20.997" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29734" PostTypeId="2" ParentId="29724" CreationDate="2019-08-20T04:11:57.710" Score="2" Body="&lt;p&gt;These conditions are called vertical structural discontinuity and or irregularity. Codes discourage this and usually have punishing load factors of up to 3 times tributary seismic, torsional or vertical design loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ductility of the structures suffers greatly by such discontinuity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the state of California where I used to work there were special forms that the engineer of the record had to fill pointing out the type and location of the discontinuity and confirm that special mitigative remedy has been employed to counter the loss of ductility and strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, any part of the structure that could cause out of range natural frequencies or stress concentration, in an earthquake vibration tends to draw huge shear concentration and act as a reciprocating saw cutting through the structure. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-20T04:11:57.710" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29735" PostTypeId="1" CreationDate="2019-08-20T06:27:23.027" Score="1" ViewCount="303" Body="&lt;p&gt;I did a Civil Engineering course some years ago and from my textbook I had this question. I found that a very interesting question, but unfortunately I haven't been able to find a solution to this. &#xA;I encourage people to put comments up, but ideally I would be appreciative if the comments are related as to how you do the calculation process to get to the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The half ordinates of the contour of the load water plane of a ship of 2,800 metric tons displacement, measured at intervals of 6 m, are 0.30, 3.33, 5.07, 6.18, 6.75, 6.90, 6.75, 6.18, 5.61, 4.47, 0.60 m respectively.&#xA;If the centre of gravity is 1.87 m above the centre of buoyancy, calculate the transverse metacentric height.&#xA;Sea water has a density of 1,025 kg/m^3.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, just wondering how you would best Explain the relation between metacentre and curve of buoyancy as vessel heals over.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qnEtk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qnEtk.png&quot; alt=&quot;My Calculations (Ship Transverse Metacentric Height)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8013" LastEditorUserId="8013" LastEditDate="2019-09-01T06:07:32.647" LastActivityDate="2019-09-05T00:59:27.953" Title="How to find Ship Transverse Metacentic Height" Tags="|stability|ships|" AnswerCount="0" CommentCount="12" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29738" PostTypeId="2" ParentId="29710" CreationDate="2019-08-20T08:33:49.787" Score="0" Body="&lt;p&gt;I put an answer but it don't reads correctly.. I put there:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My calculations are:&#xA;Total Weight (plaform + motorbike) = 4905 N&#xA;Radius of wheel = 0,06 m&#xA;Desired top speed = 0,1 m/s&#xA;Desired acceleration time = 3 s&#xA;Maximum incline angle = 5 degree&#xA;mu (rubber-concret) = 0,45&#xA;Crr = 0,02&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To determine the Total Tractive Effort (TTE):&#xA;TTE = RR + GR + FA where:&#xA;RR - Force necessary to overcome rolling resistance&#xA;GR - Force required to climb a grade&#xA;FA - Force required to accelerate to final velocity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RR = 4905 x 0,02 = 98,1 N&#xA;GR = 4905 x sin 5º = 427,5 N&#xA;FA = 4905 x 0,1 / (9,81 x 3) = 16,67 N&#xA;TTE = 542,27 N&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tw (wheel torque) = TTE x Rw (radius wheel) x RF (resistance factor)&#xA;Tw = 542,27 x 0,06 x 1,1 = 35,79 Nm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there correct? Thanks&lt;/p&gt;&#xA;" OwnerUserId="21519" LastActivityDate="2019-08-20T08:33:49.787" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29739" PostTypeId="1" CreationDate="2019-08-20T10:06:46.503" Score="1" ViewCount="37" Body="&lt;p&gt;I'm looking to measure and log the surface temperature of lithium ion cells (cylindrical and pouch). I tried using LM35 sensors with an Arduino, but, the drift was too much somehow. Any suggestions on how this system can be improved, as in - use different sensors or change the microcontroller?&lt;/p&gt;&#xA;" OwnerUserId="21530" LastActivityDate="2021-01-14T08:02:10.147" Title="Temperature logging of li-ion cells" Tags="|applied-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29740" PostTypeId="1" AcceptedAnswerId="29743" CreationDate="2019-08-20T10:56:54.983" Score="0" ViewCount="77" Body="&lt;p&gt;The Blast Furnace flue dust contains high rates of Zn (Zinc). I'd like to know the reason behind this. The chemical reaction of Zn inside blast furnace dust. Why it's rates becoming high?&lt;/p&gt;&#xA;" OwnerUserId="21502" LastEditorUserId="21502" LastEditDate="2019-08-21T05:11:12.477" LastActivityDate="2019-08-21T05:11:12.477" Title="Why do Blast Furnace flue dust generates high rates of Zn (Zinc)?" Tags="|chemical-engineering|metallurgy|chemistry|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29741" PostTypeId="1" CreationDate="2019-08-20T12:43:07.443" Score="0" ViewCount="32" Body="&lt;p&gt;If Sinter filter generates lots of Sodium Sulphate and then is there a way to make purification of this and which companies are already do this process?&lt;/p&gt;&#xA;" OwnerUserId="21502" LastActivityDate="2019-08-20T12:43:07.443" Title="Is it possible to do purification of Sodium Sulphate and which companies do this?" Tags="|chemical-engineering|chemistry|" AnswerCount="0" CommentCount="3" ClosedDate="2019-08-21T11:34:27.517" ContentLicense="CC BY-SA 4.0" />
  <row Id="29742" PostTypeId="1" CreationDate="2019-08-20T14:38:07.083" Score="1" ViewCount="156" Body="&lt;p&gt;I get that we are analysing the Piston in a non-inertial frame of reference, but the point of my question is that according to D'Alembert's Principle, whenever an inertial force comes in the line of motion of the object the net Force on the body becomes zero and the body is in a state of Dynamic Equilibrium. But even then, we write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_{net} = F_p -F_f - F_i$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$F_p =$&lt;/span&gt; Force due to combustion/pressure&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$F_f =$&lt;/span&gt; Frictional Force&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$F_i =$&lt;/span&gt; Inertial Force&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Due to the occurrence of inertial forces, shouldn't &lt;span class=&quot;math-container&quot;&gt;$F_{net} = 0$&lt;/span&gt;?&lt;/p&gt;&#xA;" OwnerUserId="21532" LastEditorUserId="1832" LastEditDate="2019-08-20T23:23:39.173" LastActivityDate="2019-08-20T23:23:39.173" Title="Inertial Forces while analysing forces on a Piston in a Slider - Crank Mechanism?" Tags="|mechanical-engineering|automotive-engineering|dynamics|mechanisms|building-physics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29743" PostTypeId="2" ParentId="29740" CreationDate="2019-08-20T16:02:34.260" Score="1" Body="&lt;p&gt;Not likely unless they use a very unusual ore. Blast furnaces generally do not use scrap. Scrap likely contains zinc ( die castings) . Electric arc remelt furnaces melt a high percentage of recycle containing zinc. There may still be open hearths or other furnaces used to melt recycle. So, furnaces other than blast furnaces are likely the source. Depending on the system , dust from different types of furnaces may be collected in one facility which may not be differentiated , leading to the idea that zinc is coming from a blast furnace.  Generally the mills remove as much zinc as possible from the recycle as I understand it causes deterioration of the refractories. After it is a few hours old ,zinc oxide is innocuous and is used in skin creams. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-08-20T16:02:34.260" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29744" PostTypeId="2" ParentId="29742" CreationDate="2019-08-20T16:50:04.387" Score="1" Body="&lt;p&gt;Yes, the inclusion of D'Alembert forces makes a dynamics problem into a static one. The equations of motion change from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F_{\rm gas} - F_{\rm friction} - F_{\rm conrod} = m_{\rm piston} a_{\rm piston} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sum F = F_{\rm gas} - F_{\rm friction} - F_{\rm conrod} - m_{\rm piston} a_{\rm piston} = 0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your example, you have forgotten the contribution of the connecting rod via the wrist-pin and thus you won't get net forces equals zero.&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2019-08-20T16:50:04.387" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29746" PostTypeId="1" CreationDate="2019-08-20T20:34:23.493" Score="0" ViewCount="34" Body="&lt;p&gt;This a somewhat speculative question; I would like to know whether it is feasible to additively manufacture large (&gt;100m) metal objects - from for example aluminum alloys - in a space environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Methods could include 3d-printing or hot/cold spray techniques, or perhaps a fine spray of molten metal onto a thin sheet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple shapes like spheres, cylinders, plates, beams are of interest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is some existing work, e.g. &lt;a href=&quot;https://www.nasa.gov/content/spiderfab&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/content/spiderfab&lt;/a&gt;, but nothing I can find focuses on manufacture of large simple objects using additive techniques. &lt;/p&gt;&#xA;" OwnerUserId="21535" LastEditorUserId="21535" LastEditDate="2019-08-20T22:45:31.017" LastActivityDate="2019-08-23T13:05:14.413" Title="Feasibility of additively manufacturing large metal objects in microgravity/high vacuum (space)" Tags="|aerospace-engineering|manufacturing-engineering|3d-printing|casting|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29748" PostTypeId="2" ParentId="29626" CreationDate="2019-08-20T21:23:16.303" Score="0" Body="&lt;p&gt;One good solutions is to use a &lt;a href=&quot;https://www.sparkfun.com/products/9670&quot; rel=&quot;nofollow noreferrer&quot;&gt;L298N Dual H-Bridge breakout board&lt;/a&gt;. There are plenty of of  these boards in the market. The Break out board looks something like the following&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YbhKi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YbhKi.jpg&quot; alt=&quot;Motor Driver 2A Dual L298 H-Bridge&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the image below it should be able to handle your voltage requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AXcpT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AXcpT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also the BC547 transistor has max current of 100mA. The Solenoid most likely is sourcing more than 100mA of currents. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;a href=&quot;https://i.stack.imgur.com/0wSIZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0wSIZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.teachmemicro.com/use-l298n-motor-driver/&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to Use L298N Motor Driver&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sparkfun.com/products/9670&quot; rel=&quot;nofollow noreferrer&quot;&gt;Motor Driver 2A Dual L298 H-Bridge&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sparkfun.com/datasheets/Components/BC546.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;BC546/547/548/549/550&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-08-20T23:10:15.023" LastActivityDate="2019-08-20T23:10:15.023" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29750" PostTypeId="2" ParentId="29649" CreationDate="2019-08-21T01:21:12.173" Score="2" Body="&lt;p&gt;Power density is defined and understood as the average power over the weight and/or volume of the actuator (depending on the unit). Here, it is understood as the effective power output divided by weight (or volume)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if your actuator is made from &lt;span class=&quot;math-container&quot;&gt;$2kg$&lt;/span&gt; of DE and delivers &lt;span class=&quot;math-container&quot;&gt;$5W$&lt;/span&gt; worth of power, then the power density is &lt;span class=&quot;math-container&quot;&gt;$5/2=2.5W/kg$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks @Phil Sweet for this comment. I'd add an example for power density per unit volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, the DE actuator is &lt;span class=&quot;math-container&quot;&gt;$5W$&lt;/span&gt; in output, and it has a volume of &lt;span class=&quot;math-container&quot;&gt;$0.02m^3$&lt;/span&gt;, the its power density will be &lt;span class=&quot;math-container&quot;&gt;$5/0.02=250W/m^3$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="21520" LastEditorUserId="21520" LastEditDate="2019-08-21T04:13:52.457" LastActivityDate="2019-08-21T04:13:52.457" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29753" PostTypeId="1" AcceptedAnswerId="29755" CreationDate="2019-08-21T13:04:19.003" Score="1" ViewCount="84" Body="&lt;p&gt;What is the difference between both of these terms? Especially in configuration and application?&lt;/p&gt;&#xA;" OwnerUserId="22801" LastEditorUserId="29410" LastEditDate="2021-02-28T06:49:08.307" LastActivityDate="2021-02-28T06:49:08.307" Title="Controller vs Micro controller?" Tags="|electrical-engineering|control-engineering|microelectronics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29754" PostTypeId="1" CreationDate="2019-08-21T17:29:52.617" Score="1" ViewCount="76" Body="&lt;p&gt;Estimates of area needed to be covered by solar panels to power all of the US using only solar are about equal to a 100 mile by 100 mile tract of land. But solar panels have a lifespan of only a few decades and need fossil fuels for their manufacturing. Was wondering why there are no solar manufacturing plants running on solar power and if such a facility will ever be feasible from a financial standpoint. How much time(in months/years) will it take for a solar plant powered by 10GW of solar panels(assume battery backup) to churn out 1 GW of solar panels?&lt;/p&gt;&#xA;" OwnerUserId="2340" LastEditorUserId="2340" LastEditDate="2019-08-28T03:55:28.977" LastActivityDate="2019-08-28T03:55:28.977" Title="Is it economical to use solar energy to manufacture solar panels?" Tags="|solar-energy|photovoltaics|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29755" PostTypeId="2" ParentId="29753" CreationDate="2019-08-21T17:31:10.760" Score="2" Body="&lt;p&gt;In pure control theory, everything is a mathematical entity.  So a controller is a mathematical entity that realizes a certain transfer function (or a certain time-dependent behavior, if there are multiple inputs or outputs involved).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A microcontroller is a term for a chip that contains a microprocessor core, memory, and peripherals, that can run stand-alone but for needing to be mounted to a board.  In other words, it's what would have been built onto a 200mm x 200mm square board in 1978, and would have cost \$200 in parts (in 1978 dollars), -- only microcontrollers hawe ten to one hundred times more of just about everything, and you can get them starting at about \$0.50 in quantity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To either confuse things or clarify: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can take a microcontroller and you can &lt;em&gt;implement&lt;/em&gt; a controller on it, using the microcontroller's peripherals, and make it part of a &lt;em&gt;physical&lt;/em&gt;, real-world control system.  This will rapidly lead to you being whacked upside the head with the difference between theory and practice, when you first turn your system on expecting good stable behavior from your plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you cannot take a controller from control theory and implement a microcontroller inside of it.  You could use some of the same language and implement a &lt;em&gt;model&lt;/em&gt; of a microcontroller (with much cursing and head-scratching from both your local control theorists and your local computer scientists).  But you couldn't realize a physical microcontroller.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-08-21T17:31:10.760" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29758" PostTypeId="1" AcceptedAnswerId="29759" CreationDate="2019-08-21T20:02:51.323" Score="1" ViewCount="59" Body="&lt;p&gt;We only have one adjusted FWD modulus value for HMA layers, but I can only see input options in the form of a grid of frequency and temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UICQb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UICQb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-08-21T20:02:51.323" Title="How can I use backcalculated modulus results from FWD testing to represent my existing condition using Pavement ME Design?" Tags="|pavement|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29759" PostTypeId="2" ParentId="29758" CreationDate="2019-08-21T20:02:51.323" Score="1" Body="&lt;p&gt;The FWD modulus results are an input for the existing asphalt layers. One value can be input and a grid is not necessary, as shown in your screenshot above. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should evaluate the backcalculated values for the asphalt layers and make sure the data you have is independent of temperature. The dynamic modulus of any asphalt layer is dependent on temperature and if the deflections are measured over a wide enough range of temperature, the backcalculated values can be grouped by time of day and temperature, as long as temperature is included in the FWD data file. These are engineering decisions and need to be made by the pavement engineer.&lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-08-21T20:02:51.323" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29760" PostTypeId="1" AcceptedAnswerId="29761" CreationDate="2019-08-21T20:50:41.003" Score="1" ViewCount="73" Body="&lt;p&gt;So far, I have only been able to model dry layers as I've only been able to represent the FDR layer with one modulus value.&lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-08-21T20:50:41.003" Title="How can we model a full depth reclamation (FDR) layer with emulsion (wet FDR layer) in a pavement design?" Tags="|pavement|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29761" PostTypeId="2" ParentId="29760" CreationDate="2019-08-21T20:50:41.003" Score="1" Body="&lt;p&gt;Please refer to the Manual of Practice for the Mechanistic-Empirical Pavement Design Guide (MEPDG).  The following is an extract from that manual:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Full-Depth Reclamation (In-Place Pulverization of Conventional&#xA;  Flexible Pavements) - Cold in-place recycling of the HMA and existing&#xA;  aggregate base layers, and hot in-place recycling of HMA. Cold&#xA;  in-place recycling as a rehabilitation strategy is considered&#xA;  reconstruction under the MEPDG design/analysis process and would be&#xA;  defined as a new flexible pavement. Hot in place recycling as a&#xA;  rehabilitation strategy is considered mill and fill with an HMA&#xA;  overlay of the existing flexible pavement. The thickness of the hot&#xA;  in-place recycled material is considered part of the HMA overlay, as&#xA;  well as the thickness of the milled material. Full-depth reclamation,&#xA;  however, was not included in the global calibration of AASHTOWare&#xA;  Pavement ME Design.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Mechanistic-Empirical Pavement Design Guide - A Manual of Practice 2015&lt;/strong&gt; &lt;strong&gt;(AASHTO)&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How you treat or simulate the FDR layer is an engineering decision. It depends on how the FDR layer is stabilized and designed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are adding an emulsion to the reclaimed layer, it can be simulated as a conventional asphalt layer and the pavement can then be simulated as new construction. If an emulsion is used and the layer is simulated as an asphalt layer for new construction, you need to determine whether the default or local calibration coefficients for fatigue cracking apply to this FDR layer.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This decision will depend on how that FDR layer with emulsion added is designed and constructed.  Some State agencies model the FDR layer as a high strength granular layer with a representative resilient modulus when a low amount of asphalt binder (4 percent or less) is used as the stabilizing material (Virginia for example). &lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-08-21T20:50:41.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29763" PostTypeId="2" ParentId="22520" CreationDate="2019-08-22T11:33:25.160" Score="0" Body="&lt;p&gt;It sure is possible but be prepared to face a lot of engineering challenges especially if you are planning to install all the equipment, mirrors etc by yourself. A lot of metalworking tasks used in the 16th &amp;amp; 17th centuries are no longer in use by firms but you can find hobbyist videos on YouTube doing that kind of thing. Common household mirrors have less than the required reflectivity index and all concentrated solar projects make use of heliostats that are designed to withstand harsh environments.&lt;/p&gt;&#xA;" OwnerUserId="2340" LastActivityDate="2019-08-22T11:33:25.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29764" PostTypeId="1" CreationDate="2019-08-22T05:39:51.623" Score="1" ViewCount="897" Body="&lt;p&gt;My friend and I are working on building a valveless pulsejet out of sheet metal. We initially built a small one (a few feet in length) using the below dimensions:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/72OBs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/72OBs.png&quot; alt=&quot;Dimensions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue was that we were unable to get it running without compressed air. When we had a nice stream of compressed air running in the intake, it would fire right up, but it would die when the air was removed. We hypothesized that our tolerances on the engine weren't very accurate, and so decided to make a bigger one where the tolerances wouldn't matter as much. We scaled all the dimensions by two. With the new engine, we were able to get it running with compressed air just fine like the first engine. However, we had the same issue. When the compressed air was removed the engine would die. We now had a different theory. We wondered if the fuel wasn't mixing well when it was injected, and what the compressed air was doing was mixing the fuel really well. To remedy this, we took the fuel line and crimped it at one end and drilled small holes in it along its length to try to create more of an even distribution of fuel. This also didn't work, and the engine died when the compressed air was removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I am wondering if anyone here has suggestions for what we can do to get this engine running without needing to hold a compressor to it the whole time! Below is a link to a bunch of pictures of the engine:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SoOcI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/SoOcI.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas are appreciated!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: We are using propane as fuel&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: With compressed air, we are able to get it to pulse correctly. It is not just a continuous whoosh. We watched some videos of pulsejets, and we know the correct sound.&lt;/p&gt;&#xA;" OwnerUserId="21559" OwnerDisplayName="andycate" LastEditorUserId="21559" LastEditDate="2019-08-24T04:00:15.367" LastActivityDate="2019-08-24T04:00:15.367" Title="Trouble with Valveless PulseJet" Tags="|combustion|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29766" PostTypeId="2" ParentId="29636" CreationDate="2019-08-22T20:06:49.390" Score="1" Body="&lt;p&gt;In relation to MEPDG/AASHTOWare Pavement ME Design, the resilient modulus values for all base (granular, asphalt, concrete or cement treated) and subgrade layers are used to calculate a single base modulus which represent all base layers and a subgrade dynamic k-value representing all the subgrade layers. These values are computed through backcalculation as mentioned in the NCHRP 1-37A main report and double letter appendices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short, the dynamic k-value is specific to the subgrade layers while the effective base modulus represents all base layers. This is why the computed base modulus is different than the resilient modulus specified for each individual base/subbase layer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your second question, the value entered for a rigid rehabilitation design using FWD data should be the dynamic k-value which is approximately two times the static k-value as stated in the software 1-37A report.&lt;/p&gt;&#xA;" OwnerUserId="20067" LastActivityDate="2019-08-22T20:06:49.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29767" PostTypeId="1" AcceptedAnswerId="29768" CreationDate="2019-08-22T20:40:47.137" Score="0" ViewCount="85" Body="&lt;p&gt;How are the integrated climatic model as discussed in &lt;a href=&quot;http://hdl.handle.net/2142/95718&quot; rel=&quot;nofollow noreferrer&quot;&gt;Enhanced Integrated Climatic Model Version 2.0&lt;/a&gt;,  and the thornthwaite moisture index (TMI) used to calculate erosion, suction, and frost heaving in Pavement Mechanistic-Empirical Design?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically interested in how precipitation affects pavement performance within Pavement Mechanistic Empirical Design (PMED) software.&lt;/p&gt;&#xA;" OwnerUserId="18841" LastEditorUserId="7083" LastEditDate="2019-09-07T17:31:01.783" LastActivityDate="2019-09-10T16:37:19.540" Title="What is the influence of precipitation on pavement performance in both jointed plane concrete pavements (JPCP) and asphalt concrete pavements?" Tags="|pavement|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29768" PostTypeId="2" ParentId="29767" CreationDate="2019-08-22T20:40:47.137" Score="2" Body="&lt;p&gt;Erosion, frost heaving, and suction are all ultimately caused by precipitation on the pavement. Precipitation is used in the following models:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The Enhanced Integrated Climatic Model (EICM) uses precipitation to calculate the monthly Thornwaite Moisture Index (TMI).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The TMI is used in calculating the equilibrium suction for the base and subgrade layers. The suction values are used to estimate water content within each sublayer which impacts the resilient modulus values.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The flexible roughness model uses the mean annual rainfall to calculate the frost and swell potential for the site factor parameter in the model.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The continuously reinforced concrete pavement (CRCP) model uses annual rainfall for the erosion calculation&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The Pavement ME Design software does not currently model precipitation infiltration into the pavement system from the surface. It assumes that the pavement system has adequate drainage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Resources:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Mechanistic Empirical Pavement Design Guide – A Manual of Practice&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;NCHRP 1-37A Documentation – Main document and double letter appendices&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;NCHRP Report 602 – Calibration and Validation of the Enhanced Integrated Climatic Model for Pavement Design &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18841" LastEditorUserId="18841" LastEditDate="2019-08-23T13:54:13.737" LastActivityDate="2019-08-23T13:54:13.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29769" PostTypeId="1" CreationDate="2019-08-23T02:05:27.103" Score="0" ViewCount="172" Body="&lt;p&gt;I've known of a few universities that include a class about VLSI design as a part of their control/automation engineering programs. I undestand VLSI design to be about creating integrated circuits with millions of logic gates. Is this actually ever done by control engineers in real life, and if so, in what sort of situations? Or is there any practical reason for taking a VLSI design course? I assumed microcontrollers, PLCs or PCs were used for practically all automation projects, and I can't really picture a situation in which a controls engineer would have to design a VLSI circuit him/herself, but I find the topic interesting and would gladly try to learn about it if it is actually useful.&lt;/p&gt;&#xA;" OwnerUserId="18155" LastEditorUserId="18155" LastEditDate="2019-08-23T02:14:21.933" LastActivityDate="2019-08-24T06:52:12.797" Title="Is learning VLSI design useful for a control/automation engineer?" Tags="|control-engineering|circuit-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29770" PostTypeId="2" ParentId="29769" CreationDate="2019-08-23T04:29:25.980" Score="1" Body="&lt;p&gt;Control/Automation engineer here. I think it's highly unlikely that you will ever need to design a semi-conductor chip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my experience the vast majority of Automation/Control jobs are based on PLCs, VFDs, PID Controllers, sensors, and relays. Even custom circuit boards are rare unless you're building something very exotic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even with high tech equipment like a robot arm, the circuits are &quot;hidden&quot; and instead a well defined electrical interface is broken out. You pay a premium, but this helps ensure long term reliability without re-inventing the wheel.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-08-23T04:29:25.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29771" PostTypeId="2" ParentId="29739" CreationDate="2019-08-23T04:40:08.177" Score="1" Body="&lt;p&gt;Analog can by tricky, because your signal can be distorted and the system will still work.. sort of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend switching to a digital sensor, that way you can avoid all the analog uncertainty. The DS18B20 is popular and quite accurate.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-08-23T04:40:08.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29772" PostTypeId="1" CreationDate="2019-08-23T05:49:34.923" Score="1" ViewCount="71" Body="&lt;p&gt;All the books I've used in my control systems class give a definition multiple-input, multiple-output systems and nonlinear systems, but all the techniques covered in the folloing chapters apply only to single variable LTI systems. The only occasions in which nonlinear systems are analyzed they are linearized from the very start in order to apply said techniques. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;What are some examples of real life situations in which one cannot use LTI systems, but rather a nonlinear or multivariable model? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What sort of techniques are used to analyze these systems, and what sort of math is required to understand them?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="18155" LastEditorUserId="6790" LastEditDate="2020-05-04T13:37:12.033" LastActivityDate="2020-05-04T13:37:12.033" Title="What are some practical applications of nonlinear control and multivariable control in the industry?" Tags="|control-engineering|nonlinear-control|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29773" PostTypeId="1" CreationDate="2019-08-23T08:10:20.537" Score="0" ViewCount="83" Body="&lt;p&gt;Suppose I have a matrix W. This matrix has an observability Gramian Q. Is there a way to see which changes in the matrix W will maximise the trace of Q? Is there a natural way to express maximising Q as a function of the entries of W? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!!&lt;/p&gt;&#xA;" OwnerUserId="21561" LastActivityDate="2019-08-23T08:10:20.537" Title="Maximising the trace of the observability Gramian" Tags="|control-theory|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29774" PostTypeId="2" ParentId="29764" CreationDate="2019-08-23T08:24:19.773" Score="-1" Body="&lt;p&gt;AIU pulse-jets, they require forward motion to keep supplying them with air (hence the need of the V1 to have starter rockets). In your case, the compressed air is supplying a flow of air which satisfies the engine's need for air. But when you stop the compressed air, then there is no air for the engine and it stops. What you need is something to supply the forward motion, which will keep the engine alive, which will supply the thrust to keep the forward motion alive . Try strapping it to a small airplane with a 1-tonne bomb in the front and see what happens .&lt;/p&gt;&#xA;" OwnerUserId="21562" LastActivityDate="2019-08-23T08:24:19.773" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29775" PostTypeId="1" CreationDate="2019-08-23T09:04:33.480" Score="1" ViewCount="270" Body="&lt;p&gt;i want to know about  the relationship between fatigue strength and tensile strength for thermoplastics especially PPSU (POLYPHENYLSULPHONE)@23 DEG.&lt;/p&gt;&#xA;" OwnerUserId="21563" LastActivityDate="2023-02-03T22:07:05.847" Title="is there any relationship between endurance limit and tensile strength for thermoplastics?" Tags="|materials|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29776" PostTypeId="2" ParentId="29775" CreationDate="2019-08-23T09:55:56.063" Score="0" Body="&lt;p&gt;The tensile strength of a material is how it resists tension and the difference between yield and failure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fatigue resistance is its behavior under cyclic loading which is more complicated, the amount of load and frequency makes a difference. The study of cyclic loading is “relatively” recent due to some airplanes failing due to cyclic loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are text books covering this in detail.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-23T09:55:56.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29777" PostTypeId="1" AcceptedAnswerId="29783" CreationDate="2019-08-23T10:10:12.880" Score="0" ViewCount="378" Body="&lt;p&gt;I would like a tig welder. It has to be able to weld 8mm thick steel flat bar so we can fix the front gate if it breaks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also want to be able to do stainless and aluminium  for projects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems my budget can only get me at most a 200A machine.. My budget is £500&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so would a 200A machine be enough?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also why do people say tig without argon isn't good?&lt;/p&gt;&#xA;" OwnerUserId="2054" LastActivityDate="2019-08-23T15:58:02.320" Title="200A tig welder enough to weld 8mm thick steel flat bar/stainless steel sheet/tubing" Tags="|welding|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29778" PostTypeId="2" ParentId="29777" CreationDate="2019-08-23T12:45:51.003" Score="0" Body="&lt;p&gt;Run a pass from one side doing 4mm and then do the other side, or build up by doing the root and then two further passes to fill the valley.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mostly depends on your welding skills - had an arc welder (DC) that would run 6mm rods and a 1/2” weld no problem... but I had a lot of practise with that, making sh1t rakes for tractors from broken truck halfshafts...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-23T12:45:51.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29779" PostTypeId="2" ParentId="29746" CreationDate="2019-08-23T13:05:14.413" Score="1" Body="&lt;p&gt;It appears that NASA has been conducting experiments with 3D printing in space since about &lt;a href=&quot;https://www.nasa.gov/sites/default/files/files/3D_Printing-v3.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;2014&lt;/a&gt;. Around this time is when a 3D printer was sent to the &lt;a href=&quot;https://www.nasa.gov/content/international-space-station-s-3-d-printer&quot; rel=&quot;nofollow noreferrer&quot;&gt;space station&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while this doesn't necessarily answer your question, it should help you to understand that this is current research. This means that there might not be a lot of other information available. Research has to be completed on smaller scale items before it can be considered for the larger-scale items that you are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are interested in hypothetical &quot;what-if?&quot; questions, those are likely better suited for &lt;a href=&quot;https://worldbuilding.stackexchange.com/&quot;&gt;Worldbuilding.SE&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="33" LastActivityDate="2019-08-23T13:05:14.413" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29780" PostTypeId="1" AcceptedAnswerId="29781" CreationDate="2019-08-23T13:48:54.123" Score="0" ViewCount="82" Body="&lt;p&gt;Can we build a building that could last thousands of years, if it was abandoned, in a humid continental climate? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this even possible with present knowledge?  How would such a structure be built? What foundation can actually withstand seasonal temperature changes, and precipitation, without requiring repairs after a couple decades at most?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't need to be very large, assume the size of a simple home, so 1500 square feet.&lt;/p&gt;&#xA;" OwnerUserId="21565" LastEditorUserId="21565" LastEditDate="2019-08-24T18:23:53.697" LastActivityDate="2019-08-24T18:23:53.697" Title="Can we build a building that could last thousands of years, if it was abandoned, in a humid continental climate?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2019-08-24T01:52:17.743" ContentLicense="CC BY-SA 4.0" />
  <row Id="29781" PostTypeId="2" ParentId="29780" CreationDate="2019-08-23T14:17:59.133" Score="5" Body="&lt;p&gt;In Mexico's hot and humid Yucatan peninsula are many examples of exactly what you are asking about, some dating back to 400 BC.&#xA;Many are abandoned in the middle of the rain forest, while others have been tidied up for tourists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most famous is &quot;&lt;a href=&quot;https://en.wikipedia.org/wiki/El_Castillo,_Chichen_Itza&quot; rel=&quot;nofollow noreferrer&quot;&gt;El Castillo, Chichen Itza&lt;/a&gt;&quot;, which has been declared &lt;a href=&quot;https://en.wikipedia.org/wiki/World_Heritage_Site&quot; rel=&quot;nofollow noreferrer&quot;&gt;a UNESCO World Heritage Site&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/thumb/5/51/Chichen_Itza_3.jpg/1920px-Chichen_Itza_3.jpg&quot; alt=&quot;Chichen Itza&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This one is only a thousand years old, but it could obviously continue surviving for many more centuries or millennia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We &lt;em&gt;can&lt;/em&gt; build structures that will last for thousands of years, but we generally choose not to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: &lt;strong&gt;How is it able to not fall apart like our structures? We construct a concrete foundation, its cracking almost immediately.&lt;/strong&gt; – J. M. Becker&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Coliseum in Rome was built over 2000 years ago, and most of it is still standing.  Much of it is made from concrete that has endured all this time.  The Romans expected it to last &quot;forever&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the equivalent of the anthropic principle.  The ancient societies &lt;em&gt;did&lt;/em&gt; build many flimsy structures, but the only examples that remain of their work for us to see are the few that &lt;em&gt;were&lt;/em&gt; well built.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's another factor to it.  Today's society is very different from what it used to be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Mayans built their structures as permanent monuments.  They were intended to last.  (Egyptian pyramids are nearly 5000 years old, but the original question asked for a &quot;humid climate&quot;, so I didn't use them as an example.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today's structures are built to make a profit for their builders, and to provide a quick and &lt;em&gt;modern&lt;/em&gt; building for their occupants.  In a few years they will be &lt;em&gt;old fashioned&lt;/em&gt; and replaced by another &lt;em&gt;trendy&lt;/em&gt; building.  There's no point in spending a fortune building something that will endure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Our current society and economy is based on this.  I use tools that were made 50 to 100 years ago, and they are still in good condition.  Today, unless I'm willing to spend an incredible price for high quality manufacturing, anything I buy now will be cheaply made in China and will need to be replaced in a year or two.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In today's dollars, years ago I could buy a tool for \$20–30, depending upon the quality I wanted.  Today I can buy a tool of similar quality for \$150, or a cheap version for $5.  There's nothing in between.  Most people choose the \$5 version, which means that the good quality versions must be made in small quantities, and so end up costing much much more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Household refrigerators that were made 50 years ago are still running fine (energy inefficient though), while most made 10 years ago are now scrap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When it comes to houses and office buildings, the situation isn't much different.&lt;/p&gt;&#xA;" OwnerUserId="19930" LastEditorUserId="19930" LastEditDate="2019-08-23T16:23:45.047" LastActivityDate="2019-08-23T16:23:45.047" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29783" PostTypeId="2" ParentId="29777" CreationDate="2019-08-23T15:58:02.320" Score="0" Body="&lt;p&gt;200 A is plenty of power for that relatively thin material. TIG stands for &quot; tungsten inert gas&quot; , without argon there is no inert gas so it is not TIG ( I don't know what it is). You need TIG for aluminum , although it requires practice. MIG is usually the first choice for steel and stainless . However , for MIG the &quot;inert gas&quot; can be replaced with CO2 or flux core wire.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-08-23T15:58:02.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29785" PostTypeId="1" CreationDate="2019-08-23T17:59:43.253" Score="0" ViewCount="35" Body="&lt;p&gt;I started with a book (pdf see here: &lt;a href=&quot;https://www.pdhonline.com/courses/m104/m104_new.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.pdhonline.com/courses/m104/m104_new.htm&lt;/a&gt;) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this book there is the following formulla found&#xA;&lt;span class=&quot;math-container&quot;&gt;$$f_0 = 170 / \sqrt{M_1 * t  / (1 + M_1 / M_2)}$$&lt;/span&gt;.&#xA;The page i refereing to is 4-2 (added below) but can also be found on this link(called page 28) &#xA;&lt;a href=&quot;http://buildingcriteria2.tpub.com/ufc_3_450_01/ufc_3_450_010032.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://buildingcriteria2.tpub.com/ufc_3_450_01/ufc_3_450_010032.htm&lt;/a&gt;&lt;br&gt;&#xA;For text related to this figure see page 27 of the same link.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SA7Am.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SA7Am.jpg&quot; alt=&quot;Figure 4-2 from the link&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume the situation it decibes is showed in situation B of figure 4-3&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0yW4P.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0yW4P.jpg&quot; alt=&quot;Figure 4-3 find on page 29&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not a difficult formula but still i have some difficulties with it... The for here most importand one is as follows. When try to convert the formula to metric i run into a unit problem...&#xA;So the unit version is as follows. &lt;span class=&quot;math-container&quot;&gt;$$f_0[1 / s] =  170[?] /  \sqrt{[lb / ft^2] * ([ft] / 12) / (1 + [lb / ft^2] / [lb/ft^2] )}$$&lt;/span&gt; &#xA;With &#xA;&lt;span class=&quot;math-container&quot;&gt;$(1 + [lb / ft^2] / [lb/ft^2]) = [-]$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$[lb / ft^2] * ([ft] / 12) = [lb / (ft * 12)]$&lt;/span&gt; resulting &lt;span class=&quot;math-container&quot;&gt;$1/[sec]  = 170[?] / \sqrt{[lb / ft]}$&lt;/span&gt; (i ignore the factor &lt;span class=&quot;math-container&quot;&gt;$1/12$&lt;/span&gt; for now).&lt;br&gt;&#xA;This would mean the unit (of the 170) of the numerator should be &lt;span class=&quot;math-container&quot;&gt;$\sqrt{[lb/ft]} /s$&lt;/span&gt; while i would expect that it should be related to the speed of sound(and thus in &lt;span class=&quot;math-container&quot;&gt;$ft/sec$&lt;/span&gt;. The denumerator is would have &lt;span class=&quot;math-container&quot;&gt;$ft$&lt;/span&gt; as unit).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is as this:&#xA;Where is this formula come from how is this formula build up. AND maybe a pointer to a paper where this formula comes from.&lt;/p&gt;&#xA;" OwnerUserId="21568" LastEditorUserId="21568" LastEditDate="2019-08-26T17:55:08.973" LastActivityDate="2019-08-26T17:55:08.973" Title="Formula natural frequency double wall" Tags="|vibration|sound-isolation|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29786" PostTypeId="1" AcceptedAnswerId="29797" CreationDate="2019-08-23T18:12:48.510" Score="0" ViewCount="81" Body="&lt;p&gt;When you're putting up a new building, you have to check the soil underneath for stability. How deep do you typically have to check? What's the extent of your &quot;due diligence&quot; when ensuring that a building won't collapse because of unstable ground? I imagine there might be different rules in different jurisdictions, but is there a typical, ballpark depth? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Relatedly: Has a building ever collapsed because of unstable conditions below the checked area?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm a total newbie, and I apologize for the many flaws you will be able to spot in this question.&lt;/p&gt;&#xA;" OwnerUserId="21570" LastEditorUserId="21570" LastEditDate="2019-08-23T18:38:30.990" LastActivityDate="2022-12-23T03:53:47.347" Title="How deep do you have to check for problems underneath a building site?" Tags="|structural-engineering|structural-analysis|building-design|building-physics|construction-management|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29787" PostTypeId="1" CreationDate="2019-08-23T19:05:15.577" Score="3" ViewCount="201" Body="&lt;p&gt;There is a metallical structure. I can estimate how much steel are in it. I would like to know how tall/strong a magnet should be to maintain a certain object to the metallical structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object is not that heavy, but it is submitted to multiple forces, like air flood, so it need to be sufficiently strong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing the force to handle, how could I know the size and optimal disposition of the magnet?&lt;/p&gt;&#xA;" OwnerUserId="21434" LastActivityDate="2023-03-07T21:03:30.503" Title="How to design a magnet to handle a known force?" Tags="|design|magnets|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29788" PostTypeId="1" CreationDate="2019-08-23T19:23:41.137" Score="0" ViewCount="84" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/20JvV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/20JvV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this called? I would say it is a shackle or I suppose, a turnbuckle, but it appears to have a right threaded rod on both sides and right threaded free nuts inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would just say it's a shackle or maybe a swivel, but there are some really important design properties that aren't specified by that.  I want to say something that is usually about 5 to 1 length width aspect ratio, has a hole in both ends, has a wide berth for inserting and tightening free nuts and seems to have some kind of in the field bending and maneuverability capability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/goS5U.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/goS5U.jpg&quot; alt=&quot;shackle_right&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iqRCa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iqRCa.jpg&quot; alt=&quot;shackle_left&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5967" LastEditorUserId="5967" LastEditDate="2019-08-23T19:43:41.303" LastActivityDate="2020-09-16T22:03:10.337" Title="identify shackle or swivel tensioner mechanism" Tags="|mechanisms|threads|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29790" PostTypeId="2" ParentId="29788" CreationDate="2019-08-23T19:32:32.073" Score="1" Body="&lt;p&gt;As only the right hand rod is threaded then tightening it (rotating clockwise looking from the left hand side) will increase the tension on the cable. All that appears to be needed is a bar that fits inside the turnbuckle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The left hand rod is freely rotating in the metal body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&#xA;The subsequent images seem to show examples that, imho, were put together (incorrectly...) from parts as available. If only one side is threaded then it works fine, if both sides are threaded then one has to be a left-hand thread otherwise it won’t shorten or lengthen the cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both are righthand threads then you have to find a way of only turning one of the nuts while keeping the other stationary.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-08-23T19:57:32.840" LastActivityDate="2019-08-23T19:57:32.840" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29792" PostTypeId="2" ParentId="29787" CreationDate="2019-08-23T21:11:58.453" Score="1" Body="&lt;p&gt;This will be very difficult to calculate analytically. Your best bet will be to simply experiment, or use the rule of thumb specs that the magnet manufacturer provides. Usually this comes in the form of holding force on a standard iron plate of some kind.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-08-23T21:11:58.453" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29794" PostTypeId="2" ParentId="29325" CreationDate="2019-08-24T06:10:27.313" Score="2" Body="&lt;p&gt;When the collision velocity of a projectile is fast and the target hard and unyielding the impact produces forces that are greater than the compressive yield strength of the projectile and target. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the projectile will collapse onto itself in plastic deformation while penetrating into a deepening crater in the target while converting its huge kinetic energy into heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This heat will liquify the projectile and crater into molten metal which can produce a cone of ejected red hot metal reflecting back and leaving traces of burn lines radiating out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The impact force,F , must be such that: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{F}{A_{projectile}} = m\frac{\delta v}{\delta t}/A= m\frac{v_{projectile}}{ \text{very small t}}/A&amp;gt; F_{\text{projectile yield strength}}  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the first rough explanation. Weird things happen in a fast collision. Extensive tests have been done by artillery and shell manufacturers and designers of armory shields. Albeit the data is proprietary trade secret.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-24T06:10:27.313" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29796" PostTypeId="2" ParentId="29769" CreationDate="2019-08-24T06:52:12.797" Score="0" Body="&lt;p&gt;Agree with Drew. I would also add that a knowledge of the popular robotic-style microcontroller boards and the software that runs on them (raspberry pi, arduino, beaglebone, etc.) would be far more useful to you.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-08-24T06:52:12.797" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29797" PostTypeId="2" ParentId="29786" CreationDate="2019-08-24T08:14:36.973" Score="2" Body="&lt;p&gt;Local governing agencies have the general geology information of the sites. And building departments usually recommend what kind if soils investigation and how many bores and how deep. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the soil condition is generally known to be good they even have pre approved sheets about the bearing capacity of the soil and type and depth of foundation required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the site needs investigation by a geologist and soils engineer, they are familiar with the local type of soils and will plan a strategy for location and depth of bors and type of tests to be done on the excavated samples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also in critical cases they have a deputy soils engineer on the job and they take continuous samples and if they deem they need to bore new holes and or modify the report to encorporate unlickly new emerging data they do it, while they are constantly in touch with the owner and building department.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-24T08:14:36.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29798" PostTypeId="1" CreationDate="2019-08-24T17:10:12.713" Score="-1" ViewCount="353" Body="&lt;p&gt;&lt;a href=&quot;https://blog.designerappliances.com/top-load-vs-front-load-washers/#topvsfront-clean&quot; rel=&quot;nofollow noreferrer&quot;&gt;Which cleans better?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jRozu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jRozu.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it just marketing or do they really clean/wash better due to some physical property or the way they work, I am asking how do they if they do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B I do not intend to indicate a single type. I am interested in averages.&lt;/p&gt;&#xA;" OwnerUserId="21576" LastEditorUserId="7083" LastEditDate="2019-09-07T16:59:12.577" LastActivityDate="2019-10-08T04:04:40.743" Title="From a mechanical point of view(Physics/Engineering) why are Front-Load washers said to clean/wash better than Top-Load washers on average?" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|applied-mechanics|mechanisms|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29799" PostTypeId="1" CreationDate="2019-08-24T20:56:51.910" Score="0" ViewCount="33" Body="&lt;p&gt;I need to remove 5&quot; x 48&quot; of the middle of a 6&quot;x12&quot;x16' header so I can clear the AC units on the top of my motorhome entering the carport. The header supports a 6/12 pitch roof with trusses.  What method of reinforcing should be accomplished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="21579" LastActivityDate="2019-08-24T21:07:02.843" Title="reinforce section of 6x12 header" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29800" PostTypeId="2" ParentId="29799" CreationDate="2019-08-24T21:07:02.843" Score="3" Body="&lt;p&gt;you need to hire an engineer and pull a permit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rough first idea is either to install posts under the header, or double it by attaching another identical on top of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both require attention to fasteners and connectors.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-24T21:07:02.843" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29801" PostTypeId="1" AcceptedAnswerId="29829" CreationDate="2019-08-24T21:16:28.750" Score="1" ViewCount="327" Body="&lt;p&gt;I have a 12&quot; flow meter that uses a float on a stem and whenever gas is supplied, the initial flow spikes the float to the top and gets wedged. The flow meter is plumbed downstream of a control valve and ball valve that is being used to regulate the flow at 40 cfm. The supply side is helium that can range between 60 - 80 psig and the outlet of the meter is plumbed to a chamber at half atmosphere, ~ 400 Torr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm aware that a ball valve should not be used to fine tune flow, that's why I need some clarification to properly correct this issue. I attached an image to help better explain my current setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vaB0e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vaB0e.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8936" LastActivityDate="2019-08-28T04:23:58.427" Title="Proper Flow Meter Installation" Tags="|flow-control|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29802" PostTypeId="2" ParentId="29801" CreationDate="2019-08-24T22:11:21.283" Score="0" Body="&lt;p&gt;How is the helium line upstream of the control valve? Any constrictions? Can you mount a manometer close to the control valve? How is the 80 psig supplied? Long line or tube? I have the suspicion that initial pressure in the helium line is high, then you open the ball valve and an iniial gush of helium wedges the floater of the flowmeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to install a manometer upstream of the first valve to check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for your real question, I would turn the control valve all the way shut, then open the ball valve all the way, then slowly open the control valve until my desired flow is established. The control valve should give you a far finer and more linear control over flow than the ball valve.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-08-24T22:11:21.283" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29803" PostTypeId="5" CreationDate="2019-08-25T03:40:19.207" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-08-25T03:40:19.207" LastActivityDate="2019-08-25T03:40:19.207" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29804" PostTypeId="4" CreationDate="2019-08-25T03:40:19.207" Score="0" Body="Vibration of mechanical systems is a field studied by mechanical engineers and civil engineers " OwnerUserId="17070" LastEditorUserId="17070" LastEditDate="2019-08-25T09:39:30.587" LastActivityDate="2019-08-25T09:39:30.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29805" PostTypeId="1" AcceptedAnswerId="29806" CreationDate="2019-08-25T05:54:59.197" Score="0" ViewCount="55" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wod2e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotor of fan is shown as a series of circles arranged in a circular manner. From youtube videos, I came to know that this is sort of a top view and the circles actually represent conductors through which current 'can' flow into the screen or come out of the screen (towards the viewers). &#xA;I wonder as to what happens to the current that arises in these conductors (due to the induction of EMF).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where do the electrons flow to ? There has got to be some path, which makes that path a conductor, and hence susceptible to be affected by lorentz force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why isn't that shown ?&#xA;(Diagrams would be highly helpful)&lt;/p&gt;&#xA;" OwnerUserId="21544" LastEditorUserId="21544" LastEditDate="2019-08-25T21:54:38.417" LastActivityDate="2019-08-25T21:54:38.417" Title="Why doesn't the flow of current through the conductors in the rotor of a fan affect it?" Tags="|motors|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29806" PostTypeId="2" ParentId="29805" CreationDate="2019-08-25T06:25:58.710" Score="0" Body="&lt;p&gt;Each of the circles or conductors are actually paired with another, probably diametrically opposite but this is not always the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the current is coming towards you on one circle then it will be going away from you on its pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is the path you are looking for as the linking connections are not shown as that diagram is a section drawn through the rotor.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-25T06:25:58.710" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29807" PostTypeId="1" AcceptedAnswerId="29808" CreationDate="2019-08-25T07:25:26.817" Score="0" ViewCount="179" Body="&lt;p&gt;Tractor carries more than his own weight on its trolley. The linkage between the tractor and the trolley is far away from the tractor. So, the whole weight of the trolley is on the back of the tractor. But why does not a tractor do wheelie with more weight on his behind?&lt;/p&gt;&#xA;&lt;p&gt;To make it understandable, here are some photos I designed:&lt;/p&gt;&#xA;&lt;p&gt;Tractor with empty trolley:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eZlS5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eZlS5.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Tractor with weight on its trolley must be like this but it doesn't happen.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/w0bkc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w0bkc.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;What is the reason for that?&lt;/p&gt;&#xA;" OwnerUserId="17768" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-08-25T07:37:23.257" Title="Why does a tractor not do wheelie with a lot of weight on the trolley?" Tags="|automotive-engineering|car|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29808" PostTypeId="2" ParentId="29807" CreationDate="2019-08-25T07:37:23.257" Score="4" Body="&lt;p&gt;Because the rear wheels are larger so the hitch connection point is below the centre line of the rear axle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if there is enough power then the tractor can do a wheelie anyway, have a look a tractor pulling competitions...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-25T07:37:23.257" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29810" PostTypeId="1" CreationDate="2019-08-25T09:07:10.700" Score="1" ViewCount="159" Body="&lt;p&gt;A capacitor is kept in the circuit of a fan to overcome the trouble of having torques in both directions. I would like to know if the capacitor could be replaced with an inductor without resulting in any change in operation of the fan ?&lt;/p&gt;&#xA;" OwnerUserId="21544" LastActivityDate="2019-11-01T23:25:20.010" Title="Wouldn't placing an inductor instead of a capacitor have the same effect in fans?" Tags="|motors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29811" PostTypeId="1" CreationDate="2019-08-25T10:14:08.953" Score="1" ViewCount="594" Body="&lt;p&gt;I have a cylinder with shell mesh. on its axis I have a beam with the same length of the cylinder. I want to create an RBE2 between each end edge nodes of the cylinder and the corresponding end node of the beam. how is this possible using APDL ?&lt;/p&gt;&#xA;" OwnerUserId="17070" LastActivityDate="2019-08-28T09:16:13.453" Title="How to create RBE2 between the end edges of a cylinder and end nodes of a beam element in ANSYS APDL?" Tags="|ansys|ansys-workbench|ansys-apdl|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29813" PostTypeId="2" ParentId="29810" CreationDate="2019-08-25T11:25:49.707" Score="3" Body="&lt;p&gt;I'll give you a quicky answer until an EE shows up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No, that won't work. The motor itself is mostly an inductor. To get a phase change you need the capacitor. At least it will be much more effective, smaller and cheaper. &lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-08-25T11:25:49.707" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29814" PostTypeId="2" ParentId="29798" CreationDate="2019-08-25T11:43:23.697" Score="0" Body="&lt;p&gt;Well , my guess is that in the vertical washing machine, the clothes are always inside the water and the water rotates with them during the wash. but in a horizontal one , the water always slushes back on the clothes and vice versa, I think this was your point&lt;/p&gt;&#xA;" OwnerUserId="17070" LastActivityDate="2019-08-25T11:43:23.697" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29815" PostTypeId="1" CreationDate="2019-08-25T12:48:59.240" Score="0" ViewCount="119" Body="&lt;p&gt;I am working with body wash products and there is a problem happening with the mixing process.&#xA;I use a 5000-liter tank to mix all the liquid materials including water, surfactant and preservatives. When I use a vacuum pump to add liquid materials from the bottom of tank, the content of preservative is A (g/100g). However, when I add them throug hthe top of tank, the content of preservative decreases. &#xA;I am really confused because everything (amount of materials, equipment, machines) used in 2 ways is the same. And I have no idea why the content decreases.&#xA;Please share your opinion. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="21585" LastActivityDate="2024-01-02T20:04:50.590" Title="What is the effect of different ways to adding liquid materials into a mixing tank?" Tags="|fluid-mechanics|chemical-engineering|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29817" PostTypeId="1" CreationDate="2019-08-25T13:56:11.507" Score="1" ViewCount="851" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fnpKi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fnpKi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I just do it by writing that the sum on y axis is 0 I get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$100+x=200$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x=100$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I try torque around the top:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x*800=200*500$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x=125$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="21424" LastActivityDate="2019-08-26T14:05:21.410" Title="When calculating a force, why do I get different result when I try to calculate via torque vs via sum of forces at an axis?" Tags="|mechanical-engineering|mechanical|solid-mechanics|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29818" PostTypeId="2" ParentId="29817" CreationDate="2019-08-25T14:23:52.140" Score="2" Body="&lt;p&gt;Essentially, the problem is poorly conditioned. As drawn, it cannot be in static equilibrium, which is why the equations of equilibrium are returning inconsistent solutions. We can confirm this by summing the moments about the point where force &quot;X&quot; is applied. Let's take counter-clockwise moments as positive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Sigma M = (-200*300) + (100*800) = 20,000 \neq 0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possibilities are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;One of the forces you have drawn is incorrect. For example, if the 100N load were also considered an unknown, we could solve for both reactions and ensure static equilibrium.&lt;/li&gt;&#xA;&lt;li&gt;The forces may be drawn correctly but there are additional reactions (namely, moment reactions) that have not yet been considered. If a moment reaction also existed at the point where force &quot;X&quot; is applied, we could use two equations of equilibrium and solve for both &quot;X&quot; and the moment, again ensuring static equilibrium.&lt;/li&gt;&#xA;&lt;li&gt;The element is not in static equilibrium, and the equations of statics don't apply. Additionally, while it is somewhat unclear from the sketch what the actual support/restraint conditions are, it's worth noting that at least 3 restraints are needed for a structure to be stable. Sometimes one of these support reactions will be zero for a particular load configuration, but in principle, all three are still necessary for stability. If Force &quot;X&quot; is the only support reaction, then the structure is unstable.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2019-08-26T14:05:21.410" LastActivityDate="2019-08-26T14:05:21.410" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29819" PostTypeId="2" ParentId="29817" CreationDate="2019-08-25T17:23:50.070" Score="0" Body="&lt;p&gt;Let's accept the 100N as the top force or reaction and go from there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in order to maintain equilibrium in this beam there are set of prescribed positions and forces, or pairs of (Y, F)that will work. Any other position with wrong conjugate force will break the balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curve containing all the conjugate pairs of (Y,F) has this equation which is just a statement of &lt;span class=&quot;math-container&quot;&gt;$ \Sigma M_{(Y_0) } =0$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Y*F-100N*800=0 \\ F=\frac{80000}{Y} $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is of the family of Y=1/X curves which appears in many physical concepts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if we plug 300 for y in this equation we get &lt;span class=&quot;math-container&quot;&gt;$F=80000/300=266.6N$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is obviously not what you have in your diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just by inspection we should be able to tell a force of 200N would work only at Y=400 and would not work at any other position.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-08-25T17:40:24.673" LastActivityDate="2019-08-25T17:40:24.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29820" PostTypeId="1" CreationDate="2019-08-25T18:19:36.573" Score="1" ViewCount="37" Body="&lt;p&gt;8) Consider an inextensible tank of 10 m3 of volume, isolated by an adiabatic &#xA;wrap. The tank contains nitrogen at 2 bar and 300 K. This tank is connected to a piping by which more N2 (at 300 K) is added to the tank until the internal pressure reaches 8 bar. At this moment, the admission valve between the tank and the piping is closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the final system temperature?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Resposta: T = 381,8 K)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I seriously can't seem to get this answer right. I can calculate the number of mols in the initial state, but how do I calculate the final number of N2 mols (and then the temperature)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I'm not that used to concept of adding a gas (at 300 K) to more gas at 300 K and the temperature increasing. Can you explain it to me?&lt;/p&gt;&#xA;" OwnerUserId="21589" LastActivityDate="2019-08-25T18:19:36.573" Title="Final temperature of an isolated vessel - Thermodynamics" Tags="|thermodynamics|heat-transfer|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29821" PostTypeId="2" ParentId="26537" CreationDate="2019-08-25T19:45:33.450" Score="1" Body="&lt;p&gt;The second equation is just the first equation with &lt;span class=&quot;math-container&quot;&gt;$K_i=0$&lt;/span&gt;. So the second equation would be a PD controller, while the first is PID controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One can derive the first equation by starting with common approximations for the integral &lt;span class=&quot;math-container&quot;&gt;$i[k]$&lt;/span&gt; and derivative &lt;span class=&quot;math-container&quot;&gt;$d[k]$&lt;/span&gt; terms&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;i[k] &amp;amp;= i[k-1] + T_s\,\frac{e[k] + e[k-1]}{2}, \tag{1} \\&#xA;d[k] &amp;amp;= \frac{e[k] - e[k-1]}{T_s}. \tag{2}&#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control signal could then also be constructed using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;u[k] = K_p\,e[k] + K_i\,i[k] + K_d\,d[k]. \tag{3}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first equation can be obtained using the difference of the control signal and substituting &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(2)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;u[k] - u[k-1] = &#xA;\left(K_p\,e[k] + K_i\left(i[k-1] + T_s\,\frac{e[k] + e[k-1]}{2}\right) + K_d\,\frac{e[k] - e[k-1]}{T_s}\right) - &#xA;\left(K_p\,e[k-1] + K_i\,i[k-1] + K_d\,\frac{e[k-1] - e[k-2]}{T_s}\right), \tag{4}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;namely expanding and simplifying this yields the same as your first equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;u[k] - u[k-1] = &#xA;\left(K_p + \frac{K_i\,T_s}{2} + \frac{K_d}{T_s}\right) e[k] + &#xA;\left(-K_p + \frac{K_i\,T_s}{2} - \frac{2\,K_d}{T_s}\right) e[k-1] + &#xA;\frac{K_d}{T_s} e[k-2]. \tag{5}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantage of using &lt;span class=&quot;math-container&quot;&gt;$(5)$&lt;/span&gt; over the combination of &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$(2)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$(3)$&lt;/span&gt; is that &lt;span class=&quot;math-container&quot;&gt;$(5)$&lt;/span&gt; uses less computations and memory. However for debugging the other implementation might be more convenient, since &lt;span class=&quot;math-container&quot;&gt;$i[k]$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$d[k]$&lt;/span&gt; would be directly available. This would make it for example easier to check for integral windup or whether the derivative is staying within certain desired bounds.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However as TimWescott mentioned, the derivative term from &lt;span class=&quot;math-container&quot;&gt;$(2)$&lt;/span&gt; can also be approximated with&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;d[k] = \frac{1 + \alpha}{T_s} (e[k] - e[k-1]) - \alpha\,d[k-1], \tag{6}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$-1\leq\alpha\leq1$&lt;/span&gt;. Namely for values close to zero the magnitude of the transfer function associated with &lt;span class=&quot;math-container&quot;&gt;$(6)$&lt;/span&gt; stays closer to that of a true derivative but the phase does drop significantly at high frequencies, while for values close to one the phase stays closer to 90° but the magnitude can increase a lot at high frequencies. It can be noted that &lt;span class=&quot;math-container&quot;&gt;$\alpha=1$&lt;/span&gt; is equivalent to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bilinear_transform&quot; rel=&quot;nofollow noreferrer&quot;&gt;bilinear transform&lt;/a&gt;. For negative values the transfer function looks like a high pass filter (pure derivative filtered by a low pass filter), which has the advantage of not amplifying high frequency noise. A similar thing could also be done for the integral, however the effects of an integral should only affect low frequencies, which is hardly affected by changing the parameter &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;span class=&quot;math-container&quot;&gt;$(6)$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$(2)$&lt;/span&gt; does not allow for &lt;span class=&quot;math-container&quot;&gt;$(4)$&lt;/span&gt; to be simplified down to &lt;span class=&quot;math-container&quot;&gt;$(5)$&lt;/span&gt;, due to the added pole from &lt;span class=&quot;math-container&quot;&gt;$(6)$&lt;/span&gt;, however one can write it into the following combined difference equation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;u[k] + (\alpha - 1)\,u[k-1] - \alpha\,u[k-2] = &#xA;\left(K_p + \frac{K_i\,T_s}{2} + \frac{K_d\,(1 + \alpha)}{T_s}\right) e[k] +&#xA;\left(K_p\,(\alpha - 1) + \frac{K_i\,T_s\,(\alpha + 1)}{2} - \frac{2\,K_d\,(1 + \alpha)}{T_s}\right) e[k-1] +&#xA;\left(-K_p\,\alpha + \frac{K_i\,T_s\,\alpha}{2} + \frac{K_d\,(1 + \alpha)}{T_s}\right) e[k-2]. \tag{7}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2019-08-25T19:45:33.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29822" PostTypeId="2" ParentId="29817" CreationDate="2019-08-25T20:43:35.723" Score="0" Body="&lt;p&gt;The answer is you have more variables than the equations, look at the general form of the problem: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/tfew5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tfew5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have in total five unknowns here and two equations of equilibrium: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y-z-x=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$-z(d+e)+yd=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The degree of freedom is equal to the difference between the number of variables and the number of equations here the degree of freedom is equal to three, however in your figure you assumed four known values, so at least one of the values in your problem is definitely incorrect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the number of equations are more than the number of variables you don't have any freedom and the system of equations are not solvable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above answers are correct, but I want to add something more. &#xA;If you treat the structure as a point mass then the force equilibrium should give you the right answer, but the structure here has a finite dimensions. &#xA;According to the second law of Newton (true for rigid bodies):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sum_iF_i=Ma_G$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sum of all external forces on a rigid body is equal to the mass of the body times the acceleration of its centroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The static equilibrium implies that the right hand side should be zero. However because of the finite dimensions of the body, not all the forces going through the centre of the mass so you have net non-zero torque around the instantaneous centre of rotation, so the body can rotates but doesn't translate.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-08-25T20:43:35.723" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29824" PostTypeId="1" AcceptedAnswerId="29834" CreationDate="2019-08-26T07:08:24.350" Score="0" ViewCount="346" Body="&lt;p&gt;According to &lt;a href=&quot;https://www.youtube.com/watch?v=P13Mau2VUWw&quot; rel=&quot;nofollow noreferrer&quot;&gt;What is Prestressed Concrete? - Practical Engineering&lt;/a&gt;, one way to reduce deflection in the steel rebar is to pre-stress the rebar by applying compressive stress in the member before service which is done by tensioning the steel rebar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Why does applying compressive stress to member decrease deflection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;especially with relation to the formula:&#xA;&lt;span class=&quot;math-container&quot;&gt;$y=\frac{FL^3}{48EI}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; is deflection&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; is force applied to member at &lt;span class=&quot;math-container&quot;&gt;$L/2$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is length of beam&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is elastic modulus&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; is second moment of area&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18276" LastActivityDate="2019-08-27T17:28:43.517" Title="Why does rebar in reinforced pre-stressed conrete reduce in deflection if compressive stress is applied to member?" Tags="|concrete|deflection|prestressed-concrete|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29825" PostTypeId="1" AcceptedAnswerId="29836" CreationDate="2019-08-26T07:22:18.513" Score="-1" ViewCount="680" Body="&lt;p&gt;When driving a car with manual transmission on start you slowly release the clutch so the car doesn't jerk. So clutch slippage here is not that bad if I understand this correctly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also couldn't dog clutch or synchromesh units be used as a mean of disconnecting the engine from the transmission? Then perhaps clutch could be used further down in a DCT to change between gears?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="21049" LastEditorUserId="21049" LastEditDate="2019-08-26T07:28:00.403" LastActivityDate="2019-08-26T17:33:54.100" Title="Is it ok for clutch to slip in DCT in certain situations?" Tags="|mechanical-engineering|mechanisms|mechanical|transmission|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29826" PostTypeId="2" ParentId="29825" CreationDate="2019-08-26T07:40:12.883" Score="0" Body="&lt;p&gt;Should take you no more than 3 seconds to engage the clutch fully, &lt;strong&gt;then&lt;/strong&gt; get your foot &lt;strong&gt;off&lt;/strong&gt; the clutch pedal...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some manual transmissions have hydraulic clutches inside the gearbox to enable changing between some gears (1&amp;amp;2, 3&amp;amp;4, 5&amp;amp;6 as those sets, while 2 to 3 etc need the use of the clutch pedal).&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-26T07:40:12.883" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29827" PostTypeId="2" ParentId="29815" CreationDate="2019-08-26T09:39:26.450" Score="0" Body="&lt;p&gt;It sounds like your &lt;strong&gt;additives are in suspension&lt;/strong&gt; not solution. Like a fruit juice drink. You take orange  powder stir it and it makes a nice drink. Leave it several hours without drinking it and you will return to a glass of water with orange stuff at the bottom. If the solution is not continually stirred the additives will sink to the bottom. &lt;/p&gt;&#xA;" OwnerUserId="21597" LastActivityDate="2019-08-26T09:39:26.450" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29828" PostTypeId="1" CreationDate="2019-08-26T10:02:17.080" Score="0" ViewCount="65" Body="&lt;p&gt;I have high frequency welding machine. During the welding time, I can see a wave in my lap and computer and also that time stuck my computer.Can you please help to solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="21598" LastActivityDate="2020-09-21T07:07:10.667" Title="Effect of high frequency welding near by the electronic things" Tags="|power-electronics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29829" PostTypeId="2" ParentId="29801" CreationDate="2019-08-26T10:05:57.147" Score="1" Body="&lt;p&gt;The flow meter needs to be installed such that it is always under pressure (in normal operation). That is it should be after the regulator but before the on/off valve. That way you will get no surge of gas entering an empty pipe and pushing up the float. It is quite a common problem.&lt;/p&gt;&#xA;" OwnerUserId="21597" LastEditorUserId="10902" LastEditDate="2019-08-28T04:23:58.427" LastActivityDate="2019-08-28T04:23:58.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29830" PostTypeId="2" ParentId="29817" CreationDate="2019-08-26T11:08:50.073" Score="0" Body="&lt;p&gt;The question seems to be incomplete. You haven't mentioned whether the system is in a static equilibrium or not. If I suppose it were the question then the conditions are incomplete as their net torque is not zero at each and every point. In such a case there will always be an angular velocity and an angular acceleration for the given which will be too much calculation involving when you take the angular acceleration into account.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a never a case where you will be getting two different answers for a same thing in classical mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please modify the question. Maybe then the community can help you out.&lt;/p&gt;&#xA;" OwnerUserId="21483" LastActivityDate="2019-08-26T11:08:50.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29831" PostTypeId="2" ParentId="29828" CreationDate="2019-08-26T12:22:41.677" Score="1" Body="&lt;p&gt;Check these two things immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A line &lt;strong&gt;filter on the computer supply&lt;/strong&gt;. The can be in the form of a plug that fits in the socket then you put your computer plug into the socket outlet on the filter. Ensure that your computer is on a different circuit (The circuit breaker or fuse for the computer is different to the one for the welder).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;strong&gt;earth of the welder&lt;/strong&gt;. The earthing point should be as close to the point of welding as possible so as to not allow any possibility of current flowing through any adjacent cables. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If these basics don't work then you will have to think about shielding and or separate supplies.&lt;/p&gt;&#xA;" OwnerUserId="21597" LastEditorUserId="10902" LastEditDate="2019-08-28T04:23:08.743" LastActivityDate="2019-08-28T04:23:08.743" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29832" PostTypeId="1" AcceptedAnswerId="29837" CreationDate="2019-08-26T13:23:17.763" Score="2" ViewCount="608" Body="&lt;p&gt;&quot;1 TR is equivalent to the rate of heat transfer needed to produce 1 U.S. ton (2000 lbs) of ice at 32F from water at 32F in one day, i.e., 24 hours.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While defining 1 Ton of Refrigeration (TR), why doesn't some specific value of pressure come into definition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't freezing water (to ice) related to pressure?&#xA;--&gt;If yes then why do we go away with it while defining 1TR ?&lt;/p&gt;&#xA;" OwnerUserId="21602" LastActivityDate="2019-08-28T06:59:45.140" Title="1 Ton of Refrigeration" Tags="|thermodynamics|refrigeration|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29833" PostTypeId="1" CreationDate="2019-08-26T15:07:37.030" Score="3" ViewCount="897" Body="&lt;p&gt;Would it make sense to integrate e(t) from t - c to t instead of 0 to t for the integral part of the PID controller?&#xA;What are the most common deviation from classic PID controller definition? &lt;/p&gt;&#xA;" OwnerUserId="7145" LastEditorUserId="7145" LastEditDate="2019-08-26T15:34:57.050" LastActivityDate="2021-06-12T07:49:24.943" Title="Limiting the integral to a time window in PID controller" Tags="|pid-control|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29834" PostTypeId="2" ParentId="29824" CreationDate="2019-08-26T16:11:56.393" Score="5" Body="&lt;p&gt;Prestress creates an upward deflection on the beam due to the combination of elastic strain, creep and shrinkage in the unloaded beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This negative deflection will bend back to a straight stance or at least minimize much of the deflection of the beam after the loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations of the deflection of a prestressed concrete beam depend on the curvature or geometry of the prestressed bars and crack depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this each case has to be analyzed depending on its configuration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are design aids that have many common rebar curves and bends. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-26T16:11:56.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29835" PostTypeId="2" ParentId="29832" CreationDate="2019-08-26T17:26:10.353" Score="0" Body="&lt;p&gt;Because that form of refrigeration work was defined for the convenience of air conditioning engineers, for whom the ambient pressure is always taken to be one atmosphere. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-08-26T17:26:10.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29836" PostTypeId="2" ParentId="29825" CreationDate="2019-08-26T17:33:54.100" Score="1" Body="&lt;p&gt;Clutches in car transmissions are designed on purpose to allow slip transiently, to make gear changes and standing starts smooth. This is not a bad thing. But they are not designed to slip all the time, which would wear them out quickly and generate lots of heat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is possible to use the synchronizers in a &lt;em&gt;motorcycle&lt;/em&gt; transmission to &quot;bang&quot; through the gears without using the clutch, but this is because the motorcycle weighs a lot less than a car and the synchronizers are stout enough relative to the weight of the bike to take this sort of punishment- for a while, anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is decidedly &lt;em&gt;not&lt;/em&gt; the case for cars and trucks, where this sort of abuse will ruin the synchronizers and eventually break the teeth off the gears.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-08-26T17:33:54.100" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29837" PostTypeId="2" ParentId="29832" CreationDate="2019-08-26T18:27:28.107" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;While defining 1 Ton of Refrigeration (TR), why doesn't some specific value of pressure come into definition? Isn't freezing water (to ice) related to pressure? If yes then why do we go away with it while defining 1TR?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Very little, if at all for the range of pressures likely to be encountered in air-conditioning applications where the term is used. Water and ice are virtually incompressible so the pressure makes very little difference. (I am unable to find a good graph of this at the moment.) Steam, on the other hand, is compressible and the pressure affects the latent heat of evaporation. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JxgM4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JxgM4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Before mechanical refrigeration was invented, keeping something cool when it was warm outside was the job of ice.  Ice would be collected during winter, transported to ice houses and stored until it could be sold for use in ice boxes. Image source: &lt;a href=&quot;https://www.fwdengineers.com/2017/05/23/a-ton-of-refrigeration/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Forward Engineers&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since ice was sold by weight and people had a concept of how much cooling a given weight would give it made sense to continue to use the &quot;ton of refrigeration&quot; measurement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 ton (refrigeration) = 3516.8 J/s (if spread out over 24 hours). That's 3.5 kW of cooling or 84 kWh. So at, say, 10 c/kWh that would cost 840 c per one ton ice cube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those not living in the colonies, 1 US ton = 907 kg. 1 tonne (1000 kg) would have a latent energy of fusion (if that's a valid term) of 92.6 kWh.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2019-08-26T22:22:26.680" LastActivityDate="2019-08-26T22:22:26.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29838" PostTypeId="2" ParentId="29833" CreationDate="2019-08-26T18:37:52.573" Score="3" Body="&lt;p&gt;Any question that asks &quot;would it make sense&quot; in control systems has the answer &quot;that depends on your plant&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of a limited time horizon integrator, you're turning your integrator from something that has infinite gain at DC into (essentially) a low-pass filter.  You lose the ability to completely cancel out a constant disturbance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I &lt;em&gt;think&lt;/em&gt; that there are some process-control methodologies that use a finite-horizon integrator to enhance stability, but I don't know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A similar method is to implement a so-called &quot;leaky integrator&quot;, i.e. a single-pole low-pass filter with DC gain equal to the reciprocal of its bandwidth: &lt;span class=&quot;math-container&quot;&gt;$H(s)=\frac{1}{s + \omega_0}$&lt;/span&gt;.  The leaky integrator &quot;looks&quot; like an integrator at frequencies above &lt;span class=&quot;math-container&quot;&gt;$\omega_0$&lt;/span&gt;, but has a bounded response.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-08-26T18:37:52.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29841" PostTypeId="1" CreationDate="2019-08-27T06:58:04.280" Score="1" ViewCount="249" Body="&lt;p&gt;I am confused about the difference between control systems and artificial intelligence, since both techniques can be used for automation.&lt;/p&gt;&#xA;&lt;p&gt;Please kindly try to give answer in a nutshell, an in depth or detailed answer may cause confusion because I am beginner.&lt;/p&gt;&#xA;&lt;p&gt;I have already put this question on another SE site at following link but the answer was detailed/in-depth.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://robotics.stackexchange.com/questions/19319/control-vs-artificial-intelligence&quot;&gt;https://robotics.stackexchange.com/questions/19319/control-vs-artificial-intelligence&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22801" LastEditorUserId="90" LastEditDate="2021-12-09T16:41:33.340" LastActivityDate="2022-05-09T08:57:09.450" Title="What is the difference between control systems and artificial intelligence in the robotic context?" Tags="|control-engineering|control-theory|robotics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29842" PostTypeId="1" CreationDate="2019-08-27T08:58:24.013" Score="1" ViewCount="52" Body="&lt;p&gt;A reinforced soil retaining wall is a system where a thin retaining wall is anchored into the ground it holds using plastic meshes. The wall surface is typically assembled of small concrete blocks (about 0.3 meters thick and similar length and height) and once every several rows of blocks are assembled ground is being put behind the blocks, leveled and compressed and plastic mesh is being put onto ground and onto blocks exposed horizontal surface, then the same is repeated until the entire wall height is built. The wall is either vertical or very steep.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens when the concrete wall wears out? If we disassemble it then the mass of the ground should erode quickly and collapse. How is the wall replaced once it wears out?&lt;/p&gt;&#xA;" OwnerUserId="374" LastEditorUserId="33" LastEditDate="2019-08-27T18:02:02.267" LastActivityDate="2019-08-27T18:02:02.267" Title="How are reinforced soil retaining walls replaced when they wear out?" Tags="|retaining-wall|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29843" PostTypeId="1" CreationDate="2019-08-27T10:34:33.387" Score="0" ViewCount="111" Body="&lt;p&gt;A topic that came up in a recent discussion was pushrod engines (specifically V2 configuration in a motorcycle) with four cams. After some googling, I was able to find an example in the Harley-Davidson Evolution engine (used in the Sportster models).&lt;/p&gt;&#xA;&lt;p&gt;Four distinct camshafts add a lot of complexity, because they need to be driven by separate gears. From what I've been able to find, the main reason to do that is in the different tappet construction, which requires the tappet to be directly &amp;quot;over&amp;quot; the camshaft:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Now it appears Harley may have decided to adopt roller-tipped sliding square or cylindrical tappets of the kind being used in large numbers by makers of radial aircraft engines—notably the Wright-Lawrance J-series engines, one of which would in 1927 power Charles Lindbergh’s solo flight across the Atlantic. Because these had to be located directly above the cam lobe, the simplest way to have the distinctive parallel pushrods that the A, B, and C-model Harley singles employ was to have a separate camshaft under each.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cycleworld.com/how-the-harley-davidson-sportster-motorcycle-came-to-have-four-cams/&quot; rel=&quot;nofollow noreferrer&quot;&gt;How the Harley-Davidson Sportster Came to Have Four Cams&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I don't really understand this explanation. Why does putting the tappet at an angle like dual- or single-cam engine do prevent using the aircraft engine technology? As far as I can imagine, those are also directly &amp;quot;over&amp;quot; the camshaft (meaning radially), they just reside at an angle that enables driving two rocker arms in the head.&lt;/p&gt;&#xA;&lt;p&gt;In fact, I was able to find numerous examples of HD engines with roller-tipped tappets that used only one camshaft. So if that's technically possible, what specifically was gained by the parallel tappet layout?&lt;/p&gt;&#xA;" OwnerUserId="8194" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2023-09-19T09:14:31.853" Title="What are the advantages of parallel pushrod configuration?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29844" PostTypeId="2" ParentId="29841" CreationDate="2019-08-27T11:12:43.313" Score="0" Body="&lt;p&gt;Robotics is not my specialist field but I will try my best to explain this.&lt;/p&gt;&#xA;&lt;p&gt;How you control any machine is via the control system. Obviously in Robotics this can be relatively simple or very complex depending on what the robot is doing. Control systems are today usually &amp;quot;controlled&amp;quot; by a program loaded into some sort of electronic interface device like a P.L.C. (Programmable Logic Controller). The P.L.C. will go through a series of steps looking at (information in) and activate outputs as the program requires.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Example if input a + b is present activate output 1&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The program is fixed unless you alter the program the device will always perform the same functions according to defined inputs being correct and will only perform duties within the constraints of the program.&lt;/p&gt;&#xA;&lt;p&gt;AI is similar in that it controls something but it has intelligence, it can learn.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Example if input a + b is present activate output 1&lt;/strong&gt; but last time it had no result because the machine had been switched to maintenance mode. So the A.I. self adjusts the program such that &lt;em&gt;if input a + b is present and maintenance mode is not selected activate output 1&lt;/em&gt;*&lt;/p&gt;&#xA;&lt;p&gt;The only problem with A.I is that the intelligence is only as good as the part of the program made to represent that intelligence.&lt;/p&gt;&#xA;" OwnerUserId="21597" LastEditorUserId="90" LastEditDate="2021-12-09T16:56:04.003" LastActivityDate="2021-12-09T16:56:04.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29846" PostTypeId="1" AcceptedAnswerId="29866" CreationDate="2019-08-27T13:26:18.880" Score="2" ViewCount="657" Body="&lt;p&gt;ESD-safe tweezers have a black coating on the handle. If I am grounded to a work bench, and normal tweezers are already conductive, then what is the point of the conductive black coating? Screw drivers sometimes have handles which may be insulators, but tweezers are all metal. Are they not as conductive as I am expecting? Maybe an oxide layer? &lt;/p&gt;&#xA;" OwnerUserId="21612" LastActivityDate="2019-08-28T12:22:18.113" Title="What's the point of ESD-safe tweezers?" Tags="|electrical-engineering|electrical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29847" PostTypeId="1" AcceptedAnswerId="29890" CreationDate="2019-08-27T14:38:59.983" Score="1" ViewCount="91" Body="&lt;p&gt;Firstly i am new to hydraulics, i want to build a hydraulic circuit that powers 4 cylinders at each leg of a Quad-Wheeled-Robot. The idea is to move the Chassis of the Robot up and down in all possible ways to avoid obstacles. &lt;strong&gt;I did design a system but i am not sure if it has everything i need is there anything that i need to add to it?.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Below is a picture of the simple design using only one actuator cylinder.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It has the following specs:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Dynamic Load:             4000 N bis 6000 N&lt;/li&gt;&#xA;&lt;li&gt;Motor:                    12V BLDC 1000W (Feedback)&lt;/li&gt;&#xA;&lt;li&gt;Pump:                     min 80 to max 150 Bar @ 3 to 5 L/min&lt;/li&gt;&#xA;&lt;li&gt;Tank:                     5 l&lt;/li&gt;&#xA;&lt;li&gt;Control Valve:            4x Proportional 4/3  Servo-Control Valves &lt;strong&gt;(closed Center)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;Zylinder:                 Stroke 200mm (Piston/Bore  32mm;&#xA;Rod:16mm)&lt;/li&gt;&#xA;&lt;li&gt;Check Valve and Unloading valve (if i need one)?&#xA;Also i have added a Cooling and Filter.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Updated Version all return lines going through Cooler&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ReSri.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ReSri.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21486" LastEditorUserId="21486" LastEditDate="2019-09-02T17:28:04.720" LastActivityDate="2019-09-02T17:28:04.720" Title="Help with simple hydraulic circuit to use for a four legged Robot" Tags="|hydraulics|valves|actuator|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29849" PostTypeId="2" ParentId="29824" CreationDate="2019-08-27T17:28:43.517" Score="3" Body="&lt;p&gt;Concrete is strong in compression but weak in tension. In tension, it will crack at a much lower stress than the allowable compression stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A reinforced concrete beam without any prestressing will generally have compression in concrete and tension in the rebars. If there are any significant tension stresses in the concrete, the concrete will crack and the rebars will take over, transferring the tension instead of the concrete. Since the part of the concrete which is cracked can no longer transfer any tension stresses, it will not contribute to the second moment of area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the concrete beam had a large compression force from prestressing, there would be no tension stresses and therefore also no cracking. That means the entire concrete cross section contributes to the second moment of area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in terms of the deflection formula, you quoted in the question, the answer is that the second moment of area is different, because we avoid the cracking.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2019-08-27T17:28:43.517" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29850" PostTypeId="2" ParentId="29842" CreationDate="2019-08-27T17:29:49.803" Score="1" Body="&lt;p&gt;If the concrete is only moderately damaged but the plastic meshes are intact, the most obvious approach is to cast a new concrete wall against the old concrete blocks. That would require drilling a large number of steel anchors into the concrete blocks first to be able to transfer forces between the plastic meshes and the new concrete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the plastic meshes are also damaged, it is a possibility to also drill ground anchors into the soil behind the retaining wall and grout them in order to fix them to the ground well behind the face of the wall. Using that sort of approach, you would essentially be constructing an entirely new retaining wall next to the old one.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2019-08-27T17:29:49.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29851" PostTypeId="1" CreationDate="2019-08-27T17:43:52.570" Score="2" ViewCount="41" Body="&lt;p&gt;I'm having trouble with a 74141 IC while building a nixie clock. At the moment I'm testing it with 10 leds and a 4K7 resistor, expecting them to light up one at a time as per schematic: &lt;a href=&quot;https://datasheet.octopart.com/NTE74141-NTE-Electronics-datasheet-32009451.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;NTE74141 Integrated Circuit TTL − BCD−to−Decimal Decoder/Driver 16−Lead DIP Type Package&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trouble is, both output 0 and 2 lights up no matter what input I set (by connecting ABCD input pins to 5V or ground). Is the IC corrupt or did I miss some detail?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Schematic:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/e6EqA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e6EqA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user21614" LastEditorUserId="8055" LastEditDate="2019-08-27T21:28:07.567" LastActivityDate="2019-08-27T21:28:07.567" Title="74141 BCD-to-decimal converter lighting wrong/multiple LEDs" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="2" ClosedDate="2019-08-27T21:28:34.283" ContentLicense="CC BY-SA 4.0" />
  <row Id="29853" PostTypeId="2" ParentId="14063" CreationDate="2019-08-27T22:27:28.930" Score="2" Body="&lt;p&gt;To create a pinned jointed truss structure you only need to define the points, which corresponds to the joints, and connect them using using lines that will be meshed with trusses elements. By definition, the truss elements have pinned joints, so you do not have to worry about modelling each of the defined joints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;! 2D Truss Analysis&#xA;!&#xA;/title, Truss&#xA;!&#xA;/PREP7 ! preprocessor phase&#xA;!&#xA;!&#xA;! define keypoints&#xA;!&#xA;K,1,   0, 0     ! keypoint, #, x, y&#xA;K,2,  50, 100&#xA;K,3, 100, 0&#xA;K,4, 150, 100&#xA;K,5, 200, 0&#xA;K,6, 250, 100&#xA;K,7, 300, 0&#xA;!&#xA;! define lines connecting kps&#xA;!&#xA;L,1,2&#xA;L,1,3&#xA;L,2,3&#xA;L,2,4&#xA;L,3,4&#xA;L,3,5&#xA;L,4,5&#xA;L,4,6&#xA;L,5,6&#xA;L,5,7&#xA;L,6,7&#xA;!&#xA;! element definition&#xA;!&#xA;ET,1,LINK1 ! truss element, type 1&#xA;R,1,100    ! real constant 1 &#xA;           ! Cross sectiona area: 100 mm^2&#xA;!&#xA;MP,EX,1,200e3 ! material property 1&#xA;              ! Young's modulus: 200 GPa&#xA;!&#xA;LESIZE,ALL, , ,1,1,1 ! specify divisions on unmeshed lines&#xA;LMESH,all ! mesh all lines&#xA;!&#xA;FINISH ! finish pre-processor&#xA;!&#xA;/SOLU ! enter solution phase&#xA;!&#xA;! Apply constraints at kps&#xA;!&#xA;DK,1,ALL,0 ! define a DOF constraint at a keypoint&#xA;DK,7,UY,0&#xA;!&#xA;! Apply loads&#xA;!&#xA;FK,1,FY,-100e3 ! define a force load to a keypoint&#xA;FK,3,FY,-100e3&#xA;FK,5,FY,-100e3&#xA;FK,7,FY,-100e3&#xA;!&#xA;SOLVE       ! solve the resulting system of equations&#xA;FINISH      ! finish solution-phase&#xA;!&#xA;! Enter postprocessor-phase&#xA;!&#xA;/POST1&#xA;PRRSOL,F             ! List Reaction Forces&#xA;PLDISP,2             ! Plot Deformed shape&#xA;PLNSOL,U,SUM,0,1     ! Contour Plot of deflection&#xA;ETABLE,SAXL,LS, 1    ! Axial Stress&#xA;PRETAB,SAXL          ! List Element Table&#xA;PLETAB,SAXL,NOAV ! Plot Axial Stress&lt;/code&gt;&lt;/p&gt;&#xA;" OwnerUserId="21619" LastEditorUserId="21619" LastEditDate="2019-08-29T14:55:46.347" LastActivityDate="2019-08-29T14:55:46.347" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29856" PostTypeId="1" CreationDate="2019-08-28T02:25:11.950" Score="0" ViewCount="91" Body="&lt;p&gt;The front shaft support bearing on my front deck lawnmower keeps burning out its seal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical scenario:  PTO coupling catches long strands of grass which wraps around the PTO coupling, and also around the splined shaft between the coupling and the bearing.  Friction heats up the plastic seal, it falls out, then the back up metal retainer falls out.  Soon after the bearing is history.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bearings are cheap, but it takes about 2 hours of awkward work to replace them. (I'm considering using a split pillow block to make the replacement easier.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only recently figured out the connection between long grass and bearing failure.  On this particular occasion, the failure was about 2 hours after the long grass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1:  What do I ask for to get a bearing with a seal that withstands more abuse?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2:  Are there other ways to protect the bearing seal?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit:  As per request:  Grasshopper 1995 721D&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Rear bearing never winds grass.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Mower is NOT used for a lawn, but to maintain aisles in a tree farm.  Some of this is done with a John Deere MX-6 rotary disk mower (6 foot wide; run by a 55 HP Deutz Tractor.  The grasshopper is in effect being used as a 'trim' mower, and is used for certain classes of weed control once a year, as well as mowing trails in the woods. (Snowberry, currants, raspberries, hazelnuts, baneberry, elderberry  It also mows lots of gopher hills.  Part of it's task too is to mow the oil lease access road twice a year -- another mess of clover, alfalfa, thistles, and grasses of types that embarrass  creeping red fescue as being a weakling come lately.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I am not trying for a manicured lawn.  I'm trying for a 'trimmed pasture'&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="8768" LastEditorUserId="8768" LastEditDate="2019-08-28T22:06:17.857" LastActivityDate="2019-09-02T15:48:28.727" Title="bearing keeps burning seals" Tags="|bearings|seals|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29857" PostTypeId="1" CreationDate="2019-08-28T03:14:52.330" Score="0" ViewCount="88" Body="&lt;p&gt;Is there a good way to systematically increase the accuracy and precision of a measuring tool using only mathematical means?  For example, average 10 measurements from the same tool can create a better measurement. Or using two independent tools and average them.  I don't know if measuring theory or statistical quality control or other subject can help.&lt;/p&gt;&#xA;" OwnerUserId="21620" LastActivityDate="2019-09-04T01:50:23.423" Title="accurate measurements with less accurate tools" Tags="|statistics|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29858" PostTypeId="2" ParentId="29832" CreationDate="2019-08-28T06:59:45.140" Score="0" Body="&lt;p&gt;The values for Standard Pressure are 1 atm or 101325 Nm-2 or 760 mm Hg or 101.325 kPa.&#xA;This pressure generally use in defining standards . Also for ideal condition the value of pressure taken into account is 1 atm . Hence it is to obvious to mention the value of standard there . &lt;/p&gt;&#xA;" OwnerUserId="21623" LastActivityDate="2019-08-28T06:59:45.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29859" PostTypeId="2" ParentId="29856" CreationDate="2019-08-28T07:48:46.703" Score="2" Body="&lt;p&gt;Clamp a split lock collar around the PTO shaft right next to the face of the bearing so grass cannot rub against the bearing seal. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-08-28T07:48:46.703" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29860" PostTypeId="2" ParentId="29856" CreationDate="2019-08-28T09:14:32.707" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Meticulously clean the mower after each use. This means hosing down the underside until no grass is left.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Let the thing cut the grass. If you are birdsnesting grass you aren't letting it cut. Slow down, and yes, sharpen the blades - every four hours if in deep grass.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Assuming this is a shaft running to a gearbox on the deck, make sure you don't have any loose hardware on the deck system, the lift system (check for worn bushings), loose/worn/missing outboard carrier bearing on the universal stub shaft. Grease the universal joints. These are normally a royal PITA to get at. Grease all chassis points every 50 hours and all blade spindles/deck casters/rollers every 10 hours.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;(I run mowers and fix them at the local golf course)&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-08-28T09:14:32.707" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29861" PostTypeId="2" ParentId="29811" CreationDate="2019-08-28T09:16:13.453" Score="1" Body="&lt;p&gt;Assuming &lt;code&gt;top_beam_n&lt;/code&gt; is a component with the upper end node of the beam, &lt;code&gt;top_cyl_n&lt;/code&gt; is a component with the upper end nodes of the cylinder, you can do:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;cmsel,s,top_beam_n,node&#xA;*get,num_top_beam,node,all,num,min&#xA;cmsel,s,top_cyl_n,node&#xA;cerig,num_top_beam,all,all&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kqZiJm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kqZiJm.png&quot; alt=&quot;CERIG&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12198" LastActivityDate="2019-08-28T09:16:13.453" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29862" PostTypeId="2" ParentId="29857" CreationDate="2019-08-28T09:52:19.287" Score="0" Body="&lt;p&gt;I think the Kalman filter can answer your question (it may be an overkill of your case but the folowing tutorial show you many interesting cases)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.kalmanfilter.net/default.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.kalmanfilter.net/default.aspx&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Kalman Filter is one of the most important and common estimation algorithms. The Kalman Filter produces estimates of hidden variables based on inaccurate and uncertain measurements. As well, the Kalman Filter provides a prediction of the future system state, based on the past estimations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="21626" LastActivityDate="2019-08-28T09:52:19.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29863" PostTypeId="1" CreationDate="2019-08-28T10:01:36.870" Score="0" ViewCount="167" Body="&lt;p&gt;I have a beam structure in the form of a Helix. How do I add the nodes positioned on this helix using ANSYS APDL ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to add the locations and IDs in separte sets. &lt;/p&gt;&#xA;" OwnerUserId="17070" LastActivityDate="2019-08-28T10:01:36.870" Title="How do I add the nodes' IDs and locations that are situated on a helix in space to a set in ADPL?" Tags="|ansys|ansys-apdl|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29864" PostTypeId="1" CreationDate="2019-08-28T10:14:08.213" Score="1" ViewCount="126" Body="&lt;p&gt;I have a rocket body with given Euler Angles, Quaternions and Angular-Velocities for given time steps. And I'd like to calculate the regular precession and nutation for every time step, so that I can plot the cange of Nutation-Angle over time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roll, pitch, yaw and Quaternions are given in inertial-frame, while Angular velocity is given in the Body-Frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason I read that the nutaion-angle is simply the given pitch. But I am not sure if that is really the angle I am looking for because the values dont seem to fit my prediction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to also plot the precession. So that it looks something like this &lt;a href=&quot;https://i.stack.imgur.com/2KTBB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2KTBB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for all suggestions.&lt;/p&gt;&#xA;" OwnerUserId="21624" LastActivityDate="2024-01-06T07:00:30.060" Title="Torque induced regular precession/ nutation calculation and visualization out of attitude Data" Tags="|aerospace-engineering|aerodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29865" PostTypeId="1" CreationDate="2019-08-28T12:11:14.770" Score="0" ViewCount="41" Body="&lt;p&gt;we know that MOR ( or model order reduction ) is used to reduce the size of heavy FEM models. but if there are so many faces, edges of the part that are in contact with another part in the model or within itself, is it usefull or wise to use MOR ?&lt;/p&gt;&#xA;" OwnerUserId="17070" LastActivityDate="2019-08-28T20:50:21.920" Title="Is Model order reduction good for complex models with a lot of contact points in them?" Tags="|mechanical-engineering|finite-element-method|simulation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29866" PostTypeId="2" ParentId="29846" CreationDate="2019-08-28T12:22:18.113" Score="4" Body="&lt;p&gt;The tweezers that you are describing have what's known as a Static Dissipative coating. This coating &lt;em&gt;is&lt;/em&gt; conductive, but it has a very high resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What this means is that they are able to dissipate any differential charge slowly/safely, without there being a very low impedance path to ground that might produce an arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Put simply, the tweezers will not prevent static discharge, but they will limit the peak current at which this occurs, which can be enough to save some (but not all) components from destruction.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-08-28T12:22:18.113" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29867" PostTypeId="2" ParentId="29865" CreationDate="2019-08-28T13:00:20.410" Score="1" Body="&lt;p&gt;We used to go for sections ie finding geometric similarity to reduce the model or to have the same number of cells more concentrated into a section for more detail. Also we would concentrate cells where we thought the interesting bits were likely to happen. Another method was to run a very coarse model to locate areas of interest...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relying totally on a &quot;button&quot; to reduce the model appropriately needs to be done with care or blind confidence...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-08-28T13:00:20.410" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29868" PostTypeId="1" CreationDate="2019-08-28T13:07:40.073" Score="0" ViewCount="45" Body="&lt;p&gt;A series DC motor has a high starting torque thanks to its field wiring consisting of thicker wires and fewer turns resulting in high starting torque. And a shunt DC motor do not have a high starting torque because of a field winding consisting of thinner wires and more turns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why couldn't wiring be made a choice depending on the use so that it does not matter which motor is being used ?&lt;/p&gt;&#xA;" OwnerUserId="21544" LastActivityDate="2019-09-01T03:14:32.097" Title="Does it matter the type of motor used ? Couldn't the wiring be changed to achieve high starting torque?" Tags="|motors|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29870" PostTypeId="1" CreationDate="2019-08-28T14:33:33.073" Score="0" ViewCount="761" Body="&lt;p&gt;When load increases on shunt DC motor, the speed decreases resulting in a downfall in back-EMF. This in turn, causes the net voltage, and hence the current to increase causing an increase in torque and thereby increasing the speed. Thus, it achieves a constant speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that there is no difference in the relations when it comes to series DC motors, except that the torque is related to the square of the current, why wouldn't a series DC motor be able to self regulate its speed ?&lt;/p&gt;&#xA;" OwnerUserId="21544" LastActivityDate="2019-09-01T02:33:52.950" Title="Why cannot series DC motors self regulate its speed?" Tags="|motors|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29873" PostTypeId="2" ParentId="29868" CreationDate="2019-08-28T19:27:38.630" Score="1" Body="&lt;p&gt;What you propose is commonly done: motor drive systems which allow the motor to be switched from shunt to series, while the motor is in use. You use the series wiring to start the machine and then above a certain speed you switch it to shunt. This is done in electric locomotives.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-08-28T19:27:38.630" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29874" PostTypeId="2" ParentId="29865" CreationDate="2019-08-28T20:50:21.920" Score="1" Body="&lt;p&gt;If parts of the model are &lt;em&gt;fixed&lt;/em&gt; in contact with each other, that does not have any particular influence on MOR.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand if the contact is not fixed (e.g. sliding surfaces in dynamics simulations) or there are nonlinear effects at the interface between parts of the model you need to be careful that the reduced model has the same physical behaviour as the unreduced model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even in that situation, you can often make large reductions in the parts of the model that &lt;em&gt;are&lt;/em&gt; linear. For example we routinely reduce FE models with about 300,000 degrees of freedom down to about 3,000 degrees of freedom and then do nonlinear analysis on the reduced model, and the validity of the reduced model results has been verified by experiments on the real world structures.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-08-28T20:50:21.920" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29875" PostTypeId="2" ParentId="29857" CreationDate="2019-08-28T21:08:30.860" Score="4" Body="&lt;p&gt;If you know that the errors in your measurements are truly random and have a normal distribution then you can improve the &lt;strong&gt;resolution&lt;/strong&gt; of a measurement by taking several measurements and averaging them. This technique is called &lt;strong&gt;oversampling&lt;/strong&gt; and is common in digital data acquisition systems where the hardware resolution may be low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You mentioned &lt;strong&gt;accuracy&lt;/strong&gt;, which is the difference between a measured value and the true value. This technique will not improve accuracy. If a measuring device is inaccurate it will still be inaccurate if you average a bunch of measurements. If you want accuracy you must calibrate your measuring tools against another tool that is known to have better accuracy...and so on back to NIST (in the U.S.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The word &lt;strong&gt;precision&lt;/strong&gt; usually implies &lt;strong&gt;repeatability&lt;/strong&gt;. I don't think you can get rid of something like long-term drift or temperature affects just by averaging multiple samples.&lt;/p&gt;&#xA;" OwnerUserId="20302" LastActivityDate="2019-08-28T21:08:30.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29876" PostTypeId="1" CreationDate="2019-08-28T22:37:07.910" Score="0" ViewCount="45" Body="&lt;p&gt;Assume you have a solid walled sphere with an opening of diameter x. The sphere has volume  0.02 litres and has air at 1.4 bars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What general equation would be appropriate to correctly size the diameter of the sphere's opening such that pressurized air in sphere would be released in exactly 40 seconds? End pressure should be 1 bar&lt;/p&gt;&#xA;" OwnerUserId="21634" LastEditorUserId="21634" LastEditDate="2019-08-28T23:43:01.027" LastActivityDate="2019-08-28T23:43:01.027" Title="Equation to calculate flow rate" Tags="|fluid-mechanics|valves|air|" AnswerCount="0" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="29877" PostTypeId="2" ParentId="29870" CreationDate="2019-08-28T22:40:11.367" Score="1" Body="&lt;p&gt;In a series-field DC motor the back EMF is proportional to the current times the motor speed.  In a frictionless motor this means that there is always some current below which the back-EMF is less than the applied armature voltage, so there is always some torque driving the motor to go faster.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So an ideas series-field motor simply never stops accelerating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An ideal series-field motor, driven by a constant voltage, has a torque that decreases with speed.  And in any &lt;em&gt;real&lt;/em&gt; motor, there is friction torque which is relatively constant with speed, and windage torque that increases with speed.  At some speed these parasitic torques  balance the ideal motor torque; the motor cannot go faster than that speed by itself.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because those parasitic torques are not well controlled, and because even a real series-field motor generally goes much faster than we'd like with constant voltage applied, the series-field motor is not considered to be self-regulating.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastEditorUserId="19776" LastEditDate="2019-08-29T14:54:31.230" LastActivityDate="2019-08-29T14:54:31.230" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29878" PostTypeId="2" ParentId="29868" CreationDate="2019-08-28T22:48:52.820" Score="0" Body="&lt;p&gt;Thank you for making me think about this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a real electromagnet, the ultimate strength of the magnetic field developed is a strong function of the amount of space filled by the wiring, the geometry of the armature, and the power dissipated in &lt;span class=&quot;math-container&quot;&gt;$I^2R$&lt;/span&gt; losses in the wiring.  If you take a motor whose field coil is packed full and rewind it with wire that has four times the area, then as long as you pack it full you'll use 1/4 the length of wire.  The resistance will go down by a factor of 16 (&lt;span class=&quot;math-container&quot;&gt;$4^2$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the same magnetic field, you'll need to drive four times the current -- which means that your &quot;new&quot; &lt;span class=&quot;math-container&quot;&gt;$I^2$&lt;/span&gt; is 16 times as much as your &quot;old&quot; &lt;span class=&quot;math-container&quot;&gt;$I^2$&lt;/span&gt;.  Since your new resistance is 1/16 as much, the overall &lt;span class=&quot;math-container&quot;&gt;$I^2R$&lt;/span&gt; losses will be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means -- on a basis of equal heat generated in the field winding, as long as you have a source of power for the field coils, it doesn't matter if it's series or shunt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the reason that shunt windings are good for starter applications is because you get more heat in the windings than the motor can stand for long periods, but the amount of heat in the field winding is self-regulating -- it's only super-high in that (hopefully short) period of time when the motor isn't moving; once things get going, the field winding current just naturally goes down, without needing the intervention of humans or even automatic circuitry.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-08-28T22:48:52.820" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29879" PostTypeId="2" ParentId="29864" CreationDate="2019-08-29T01:58:51.037" Score="0" Body="&lt;p&gt;Nutation is described differently for different systems. Orbital Mechanics generally uses the full-blown and fairly general treatment where nutation is defined as the difference between the average precessional vector's instantaneous location and the actual body axis of the object (think of a frame attached to the average precessional vector and consider the motion of the object's body axis relative to this). This may or may not be convenient for your dataset because there is no assumption that the instantaneous body rotation vector is identical to the body axis. Indeed for free-nutation, it never is. But your title asks about externally forced nutation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your comment about pitch = nutation sounds like it comes from gyroscopes. That is a different frame of reference because gyro's are operating in a quasi-constant gravity field (the field, if it changes at all, changes so slowly that it doesn't matter as far as rate calculations are concerned). In orbital mechanics, the gravity vector changes rapidly compared to other rates of interest, for example, the processional period of the earth is about 26,000 years, so the gravity vector is changing 26,000 times faster than the rate of interest. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you want to do is figure out how to express the nutation of the body about the average precessional vector. At this point, we need to know the specific Euler angle system you are using. Is it a true &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler_angles&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler&lt;/a&gt; system, or is it a Tait-Bryan angle set? According to this document (p 473) - &lt;a href=&quot;http://adsabs.harvard.edu/full/1992AREPS..20..469M&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://adsabs.harvard.edu/full/1992AREPS..20..469M&lt;/a&gt;, the conventional way to represent nutation is with two angles - &lt;span class=&quot;math-container&quot;&gt;$\Delta\epsilon$&lt;/span&gt;, the obliquity perturbation; and &lt;span class=&quot;math-container&quot;&gt;$\Delta\psi$&lt;/span&gt;, the longitude perturbation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's worth mentioning that precession rate, nutation, and axis wobble are all intertwined here. Including wobble complicates matters. This may not be a problem if only considering forced nutation. But keep in mind that the rotation vector and the body axis are different things.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2019-08-29T23:12:05.003" LastActivityDate="2019-08-29T23:12:05.003" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29880" PostTypeId="1" CreationDate="2019-08-29T04:11:11.357" Score="2" ViewCount="197" Body="&lt;p&gt;I have to drive several 12V servo motors. So I wanna to get 12V PWM Controller (12V Logic Controller). I have MCU as Atmel SAM3X8E arduino due board. It has 5V PWM outs. Is there any Controller IC available for this task?&lt;/p&gt;&#xA;" OwnerUserId="21636" LastEditorUserId="8055" LastEditDate="2019-08-29T17:25:30.333" LastActivityDate="2020-09-22T18:05:15.503" Title="How to drive 12V servo motors using 5V PWM or 3.3V PWM using ARM MCU?" Tags="|electrical-engineering|microelectronics|pwm|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29881" PostTypeId="1" CreationDate="2019-08-29T09:02:53.507" Score="0" ViewCount="53" Body="&lt;p&gt;I have created a 3D polyline and assigned a Beam profile to it. But I don't get the results from my Explicit analysis where this beam is in contact with a shell surface. I am afraid that the Beam has not enough DOFs as I would need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since the Explicit solver in Ansys is AutoDyn, I can not check what element types I have. where can I check them ?&lt;/p&gt;&#xA;" OwnerUserId="17070" LastActivityDate="2019-08-29T09:02:53.507" Title="Are the Beam elements created from a 3D polyline in Ansys SpaceClaim also 3D in their DOF?" Tags="|ansys|ansys-workbench|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29883" PostTypeId="2" ParentId="29880" CreationDate="2019-08-29T11:09:23.430" Score="1" Body="&lt;p&gt;Take a look at &lt;a href=&quot;https://store.arduino.cc/usa/arduino-motor-shield-rev3&quot; rel=&quot;nofollow noreferrer&quot;&gt;ARDUINO MOTOR SHIELD REV3&lt;/a&gt;. Here is what it looks like. If you have several motors then might have to stack several motor shields. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MEqwG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MEqwG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-08-29T11:09:23.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29884" PostTypeId="1" CreationDate="2019-08-29T13:32:09.357" Score="0" ViewCount="72" Body="&lt;p&gt;I already asked this question &lt;a href=&quot;https://math.stackexchange.com/questions/3337128/adaptive-feedforward-cancellation-afc-and-least-mean-squares-lms-for-periodi&quot;&gt;here&lt;/a&gt;, but since I did not get an answer yet I want to try it here as well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to implement an adaptive feedforward cancellation (AFC) for cancellation of the impact of periodic input disturbances on the output of a multiple-input single-output system. Filter weights are adapted using an LMS algorithm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Unfortunately I have problems understanding the details of the algorithm and I could not find any documents that give me concrete answers to my questions. In order to ask my questions here, I want to explain briefly what I know and what I did until now:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my setup there are two control inputs and one output. Furthermore, there are two sinusoidal disturbances with known frequency, but unknown amplitude and phase shift. One of those disturbances acts on each input. The basic structure of the control looks as shown here:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SchF2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Block diagram of the control including the LMS&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; is the secondary path (plant), which is an LTI system and which has the complex transfer function matrix&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\underline{\boldsymbol{S}}(j \omega) = &#xA;\begin{bmatrix}&#xA;\underline{\boldsymbol{S}}_1(j \omega) &amp;amp;&#xA;\underline{\boldsymbol{S}}_2(j \omega) \\&#xA;\end{bmatrix}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rewritten as real-value matrix, this is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\boldsymbol{S} = &#xA;\begin{bmatrix}&#xA;\text{Re}\{ \underline{\boldsymbol{S}}_1(j \omega) \} &amp;amp; &#xA;\text{Im}\{ \underline{\boldsymbol{S}}_1(j \omega) \} &amp;amp;&#xA;\text{Re}\{ \underline{\boldsymbol{S}}_2(j \omega) \} &amp;amp; &#xA;\text{Im}\{ \underline{\boldsymbol{S}}_2(j \omega) \} \\&#xA;-\text{Im}\{ \underline{\boldsymbol{S}}_1(j \omega) \} &amp;amp; &#xA;\text{Re}\{ \underline{\boldsymbol{S}}_1(j \omega) \} &amp;amp;&#xA;-\text{Im}\{ \underline{\boldsymbol{S}}_2(j \omega) \} &amp;amp; &#xA;\text{Re}\{ \underline{\boldsymbol{S}}_2(j \omega) \} \\&#xA;\end{bmatrix}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{d}$&lt;/span&gt; are the periodic input disturbances with &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\boldsymbol{d}(t) = &#xA;\begin{bmatrix}&#xA;w_{d,1,c} \, \cos(\omega t) + w_{d,1,s} \, \sin(\omega t)\\&#xA;w_{d,2,c} \, \cos(\omega t) + w_{d,2,s} \, \sin(\omega t)\\&#xA;\end{bmatrix}&#xA;=&#xA;\boldsymbol{X} \, \boldsymbol{w}_d &#xA; $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\boldsymbol{x} = &#xA;\begin{bmatrix}&#xA;\cos(\omega t) &amp;amp; \sin(\omega t)\\&#xA;\end{bmatrix}^\top&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\boldsymbol{X} = &#xA;\boldsymbol{I}_{\text{2x2}} \otimes \boldsymbol{x}^\top&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and with the vector &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{w}_d$&lt;/span&gt; containing the unknown coefficients of the harmonic disturbances:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\boldsymbol{w}_d = &#xA;\begin{bmatrix}&#xA;w_{d,1,c} &amp;amp; w_{d,1,s} &amp;amp; w_{d,2,c} &amp;amp; w_{d,2,s}&#xA;\end{bmatrix}^\top&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As said, the target is to cancel the impact of &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{d}$&lt;/span&gt; on the output &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; of the plant, which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\alpha_d = \boldsymbol{S} \, d&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;respectively&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\alpha_d = \boldsymbol{x}^\top \, \boldsymbol{S} \, \boldsymbol{w}_d&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the disturbances contain only a single frequency component, a narrow-band filter with input signals &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{x}$&lt;/span&gt;, weights &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{w}$&lt;/span&gt; and output &lt;span class=&quot;math-container&quot;&gt;$(\boldsymbol{X} \, \boldsymbol{w} )$&lt;/span&gt; is used fo cancelling &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{d}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With compensation, the output of the plant becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\alpha = \boldsymbol{x}^\top \, \boldsymbol{S} \, ( \boldsymbol{w}_d + \boldsymbol{w} ) &#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The optimal filter weigths should be found when the cost function &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;C = E\{\alpha^2(t)\} \approx \alpha^2(t)&#xA;$$&lt;/span&gt;&#xA;is minimized.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To achieve this, the filter weights are updated in direction of the negative gradient (steepest descent):&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;- \nabla C = - \frac{\partial \, C}{\partial \, \boldsymbol{w}}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\nabla C$&lt;/span&gt; is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{aligned}&#xA;\nabla C &amp;amp;= 2 \, \alpha^\top \, ( \nabla \alpha ) \\&#xA;&amp;amp;= 2 \, \alpha^\top \, \boldsymbol{x}^\top \, \boldsymbol{S} \\&#xA;\end{aligned}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{w}$&lt;/span&gt; and its derivative &lt;span class=&quot;math-container&quot;&gt;$\dot{\boldsymbol{w}}$&lt;/span&gt; are row vectors, the weights are updated using&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{\boldsymbol{w}}^\top = - \mu \, \nabla C \\&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A factor &lt;span class=&quot;math-container&quot;&gt;$\mu = \text{const.}$&lt;/span&gt; was introduced to adapt the step-size. By rearranging, this becomes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{aligned}&#xA;\dot{\boldsymbol{w}} &amp;amp;= - \mu \, (\nabla C)^\top \\&#xA;&amp;amp;= - 2 \, \mu \, (  \, \alpha^\top \, \boldsymbol{x}^\top \, \boldsymbol{S} )^\top \\&#xA;&amp;amp;= - 2 \, \mu \,  \boldsymbol{S}^\top \, \boldsymbol{x} \, \alpha&#xA;\end{aligned}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First question:&lt;/strong&gt;&#xA;There is the so called Filtered x-LMS algorithm (FXLMS). Using this algorithm, one filters the input signal &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{x}$&lt;/span&gt; using an model of &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{S}$&lt;/span&gt;. In my algorithm, &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{x}$&lt;/span&gt; is filtered by &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{S}^\top$&lt;/span&gt;. Otherwise, they are same I think. The FXLMS is discussed very often in literature. However, the algorithm that I'm refering to is found rarely. Why? At the moment I can't see any advantages or disadvantages if I compare them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Further on, I want to discuss the convergence rate of the algorithm.&lt;/strong&gt; The previously explained algorithm shows a frequence dependent convergence rate, as can be seen in following plot of my exemplary implementation:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ZvYNT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Convergence behavior&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there are documents claiming that convergence rate is frequency independent, if the derivative of the weights is calculated as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{\boldsymbol{w}} = - 2 \, \mu \,  \boldsymbol{S}^{-1} \, \boldsymbol{x} \, \alpha&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{\boldsymbol{w}} = - 2 \, \mu \,  \boldsymbol{S}^+ \, \boldsymbol{x} \, \alpha&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, I want to provide a plot which shows that this is true, at least for my test: &lt;a href=&quot;https://i.stack.imgur.com/e7DKQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;Convergence behavior&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I could only find few documents describing this algorithm and none of them answered following qeustion: (or at least I could not find the answer)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second question:&lt;/strong&gt; As far as is understood until now, the calculation rule using the inverse or Moore-Penrose pseudoinverse results from a normalization of &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{S}$&lt;/span&gt;. Is this correct? Why and in which cases is it useful to use &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{S}^{-1}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{S}^+$&lt;/span&gt; here? How can I derive the rule to calculate &lt;span class=&quot;math-container&quot;&gt;$\dot{\boldsymbol{w}}$&lt;/span&gt; using &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{S}^{-1}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{S}^+$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update:&lt;/strong&gt;&#xA;I think I made some progress regarding the second question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one substitutes &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; in&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{\boldsymbol{w}} = - 2 \, \mu \,  \boldsymbol{S}^\top \, \boldsymbol{x} \, \alpha&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this becomes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{aligned}&#xA;\dot{\boldsymbol{w}} &amp;amp;= - \underbrace{ 2 \, \mu \,  \boldsymbol{S}^\top \, \boldsymbol{x} \, \boldsymbol{x}^\top \, \boldsymbol{S} }_{\boldsymbol{K}} \, (\boldsymbol{w}_d + \boldsymbol{w}) \\&#xA;&amp;amp;= - \boldsymbol{K} \, \boldsymbol{w}_d - \boldsymbol{K} \, \boldsymbol{w}&#xA;\end{aligned}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This shows that &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{K}$&lt;/span&gt; can be interpreted as gain of the adaption. The adaption will have a frequency-independent convergence rate if the euclidean norm of the product &lt;span class=&quot;math-container&quot;&gt;$(\boldsymbol{S}^\top \, \boldsymbol{x} \, \boldsymbol{x}^\top \, \boldsymbol{S})$&lt;/span&gt; is equal to 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is achieved by dividing by the actual euclidean norm, which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;|| \boldsymbol{S}^\top \, \boldsymbol{x} \, \boldsymbol{x}^\top \, \boldsymbol{S} ||_2 = \frac{1}{ &#xA;\text{Re}\{ \underline{\boldsymbol{S}}_1 \}^2 + &#xA;\text{Im}\{ \underline{\boldsymbol{S}}_1 \}^2 + &#xA;\text{Re}\{ \underline{\boldsymbol{S}}_2 \}^2 + &#xA;\text{Im}\{ \underline{\boldsymbol{S}}_2 \}^2 }&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore it applies (in this case?):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{1}{ &#xA;\text{Re}\{ \underline{\boldsymbol{S}}_1 \}^2 + &#xA;\text{Im}\{ \underline{\boldsymbol{S}}_1 \}^2 + &#xA;\text{Re}\{ \underline{\boldsymbol{S}}_2 \}^2 + &#xA;\text{Im}\{ \underline{\boldsymbol{S}}_2 \}^2 } \boldsymbol{S}^\top = \boldsymbol{S}^+&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therewith it follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{aligned}&#xA;\dot{\boldsymbol{w}} &amp;amp;= - 2 \, \mu \,  &#xA;\frac{1}{ &#xA;\text{Re}\{ \underline{\boldsymbol{S}}_1 \}^2 + &#xA;\text{Im}\{ \underline{\boldsymbol{S}}_1 \}^2 + &#xA;\text{Re}\{ \underline{\boldsymbol{S}}_2 \}^2 + &#xA;\text{Im}\{ \underline{\boldsymbol{S}}_2 \}^2 } \, &#xA;\boldsymbol{S}^\top \, \boldsymbol{x} \, \boldsymbol{x}^\top \, \boldsymbol{S} \, (\boldsymbol{w}_d + \boldsymbol{w}) \\&#xA;&amp;amp;= - 2 \, \mu \, \boldsymbol{S}^+ \, \boldsymbol{x} \, \boldsymbol{x}^\top \, \boldsymbol{S} \, (\boldsymbol{w}_d + \boldsymbol{w})&#xA;\end{aligned}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't trust myself... so I would be grateful if anyone could tell me, whether all the stuff I wrote is (mathematically) correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, I did not fully understand how I can show, that frequency-independent convergence rate can be achieved by dividing by the euclidean norm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, I'm still missing an answer to my first question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt;&#xA;I asked the same question on &lt;a href=&quot;https://math.stackexchange.com/questions/3337128/adaptive-feedforward-cancellation-afc-and-least-mean-squares-lms-for-periodi&quot;&gt;Math Stackexchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="21639" LastEditorUserId="21639" LastEditDate="2019-08-30T13:03:38.677" LastActivityDate="2019-08-30T13:03:38.677" Title="Adaptive feedforward cancellation (AFC) and least mean squares (LMS) for periodic disturbance cancellation" Tags="|control-engineering|control-theory|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29885" PostTypeId="1" AcceptedAnswerId="29886" CreationDate="2019-08-29T14:20:26.567" Score="6" ViewCount="6982" Body="&lt;p&gt;One of my suspension bolts came undone while I was mountain biking recently, sadly I didn't notice.  I landed a big drop with only 2-3 threads left engaged, and unsurprisingly it's ruined them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rest of the threads (another 20 or so) are fine.  It's a blind hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the right size tap to cut the thread, but I don't know if that's the right thing to do.  Don't I risk being &quot;out of sync&quot; with the 20 good threads and causing problems with them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel there should be a tap that you insert deep into the bolt hole, and it then somehow expands into the known-good threads, before being able to be wound out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that does exist, please let me know what it's called and where I can get one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it doesn't, please give me advice on how best to fix this hole with the existing tap that I have is &lt;a href=&quot;https://www.parktool.com/product/frame-tap-tap-10&quot; rel=&quot;nofollow noreferrer&quot;&gt;Frame Tap&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vixDS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vixDS.jpg&quot; alt=&quot;Frame Tap&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21640" LastEditorUserId="1586" LastEditDate="2019-10-31T19:28:53.347" LastActivityDate="2020-08-28T19:59:13.923" Title="Fixing a blind bolt hole when the first 2-3 threads are ruined?" Tags="|threads|bolting|" AnswerCount="7" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29886" PostTypeId="2" ParentId="29885" CreationDate="2019-08-29T14:51:12.517" Score="10" Body="&lt;p&gt;I'm not a machinist, but I've tapped a few holes in my time. In my experience a tapered tap will handle this problem as the taper will reach down to the good threads and self align. Use some oil with the tap. I looked at your link and that tap doesn't have much taper so perhaps a more tapered tap might help if you have problems. There are any number of Youtube videos showing how to fix a tapped hole.&lt;/p&gt;&#xA;" OwnerUserId="9506" LastActivityDate="2019-08-29T14:51:12.517" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29887" PostTypeId="2" ParentId="29885" CreationDate="2019-08-29T15:06:37.647" Score="9" Body="&lt;p&gt;The product that you're describing (an 'expanding tap') doesn't exist, unfortunately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had a similar problem a few years back, and actually built my own out of a Sleeve anchor, like the one pictured below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you tighten the nut, it pulls a tapered cone inwards, pushing the sleeve apart.&#xA;I loosened the nut until the sides were parallel, shaved down the diameter using a lathe, except for a small portion near the tip, which I threaded the using a die matching the thread I was trying to fix. I then hardened the very tip of the sleeve, being careful not to harden too much as this would cause it to crack when expanding. If you are cutting aluminium this step may not be necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It wasn't pretty, and it wasn't in good enough condition to use a second time when it came out of the hole, but it formed a rough thread that was sufficient to allow me to align a 'regular' tap from the outside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fm57v.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fm57v.png&quot; alt=&quot;Sleeve Anchor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-08-29T15:06:37.647" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29889" PostTypeId="1" CreationDate="2019-08-29T16:43:47.567" Score="0" ViewCount="171" Body="&lt;p&gt;I am looking make a quick and dirty test ring using an off the shelf electromagnet. My force needs are small, less than a pound but I need to do so in a non contact manner. I know these magnets are strong but I was looking for a back of envelope formula to estimate holding force at various distances. I am thinking 1cm away from the surface of magnet. The hope is to get a 100+ lbs holding force magnets and still have a 1 or 2 lbs of hold a bit of distance away from the surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="11634" LastActivityDate="2019-08-30T19:33:02.673" Title="ElectroMagnetic Pulling Force fdropoff with distance" Tags="|design|modeling|electromagnetism|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29890" PostTypeId="2" ParentId="29847" CreationDate="2019-08-29T19:04:53.077" Score="1" Body="&lt;p&gt;If the first VMA 1.025 is acting as your pressure relief valve, you might be OK, depending on duty cycles/pressures/flowrates/etc...You do still build some heat flowing over that relief valve, but like above, depends on your system. I had a similar issue in a design recently so I feel the need to mention it. If you go with a dump, put the valve right after the pump, dump into a T/Y fitting before the radiator, just add a check valve before the T/Y on the other incoming side of the T/Y. With a center position, P and T blocked, you'll be building pressure/heat upstream of that valve with nowhere to go,not so much an issue in constant operation, but may be evident when the machine idles. If intermittent a great deal between operation and not, I would maybe cycle off the pump. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like the updated diagram, watch your filter though....you can filter your return oil coming out of your radiator into the tank, but your pump should not be puling oil through a filter, your going to want a strainer for that. I know, pretty similar names, but your strainer has much larger holes, never want to starve a pump for oil.&lt;/p&gt;&#xA;" OwnerUserId="11327" LastActivityDate="2019-08-29T19:04:53.077" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29892" PostTypeId="2" ParentId="24956" CreationDate="2019-08-30T01:10:41.463" Score="1" Body="&lt;p&gt;Generally speaking, there are 2 kinds of losses along the pipe, the &quot;friction&quot; with the pipe wall, and the &quot;friction&quot; at corner and junction. Note that the term &quot;friction&quot; might not be technically correct, but I use it because it is easy to understand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure loss along the pipe is &lt;span class=&quot;math-container&quot;&gt;$p(loss1)=1/2*ro*v^2*f*l/D$&lt;/span&gt; and the pressure loss at the corner is &lt;span class=&quot;math-container&quot;&gt;$p(loss2)=1/2*ro*v^2*E$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with ro as the density of the fluid and E is the &quot;curly&quot; letters denoted the loss coefficient at corner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your question only asks for loss at the junction &quot;Tee&quot;, so the &quot;E&quot; value is between 0.2 and 2.0. &lt;a href=&quot;http://fluid.itcmp.pwr.wroc.pl/~znmp/dydaktyka/fundam_FM/Lecture11_12.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Check this pdf file for more details&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Line flow, flanged E=0.2&lt;/li&gt;&#xA;&lt;li&gt;Line flow, threaded E=0.9&lt;/li&gt;&#xA;&lt;li&gt;Branch flow, flanged E=1.0&lt;/li&gt;&#xA;&lt;li&gt;Branch flow, threaded E=2.0&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Based on the drawing you provide, it is the line flow, and perhaps it is &quot;flanged&quot; (smooth pipe, for the lack of better word). Therefore, the value applied is 0.2. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diameters of the branches are half of the main line's diameter, therefore, the cross section of the main line A1 is 4 times as big as the branch ones. Applying the conservation of mass (assuming the density is unchanged), you have &lt;span class=&quot;math-container&quot;&gt;$4*A2*v0=A2*v1+A2*v1$&lt;/span&gt;. Solve this, and you have the final fluid speed in each branch pipe increased by 2 times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the fluid has changing density, well, then write down the Navier-Stokes equations and solve for the relationship between speed in the big pipe and speed in the small pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With known value for final speed, you can plug it in the equation at the beginning p(loss2) and solve for the needed value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, just a side note. If the question also asks for loss along the pipe, then you should use Moody Chart. The link I provide also explains how to use it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best of luck, mate.&lt;/p&gt;&#xA;" OwnerUserId="21520" LastActivityDate="2019-08-30T01:10:41.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29893" PostTypeId="2" ParentId="29885" CreationDate="2019-08-30T02:52:15.623" Score="5" Body="&lt;p&gt;You could drill out the hole to remove the munged threads down to the remaining good threads.  This should make it easier for the tap to get into the thread grove correctly.  Another thought would be to grind out the damaged threads if you could get a small cylindrical grinder. (Thinking Dremel tool here)&lt;/p&gt;&#xA;" OwnerUserId="21644" LastActivityDate="2019-08-30T02:52:15.623" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29894" PostTypeId="2" ParentId="29856" CreationDate="2019-08-30T12:41:27.557" Score="0" Body="&lt;p&gt;This is an incomplete answer, but may help other people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Learned lots yesterday.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Split pillow blocks aren't made in small sizes.  Surprised me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bearing supply house I went to (Motion canada) didn't have bearings with metal seals in the size I needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They also didn't have a narrow enough collar to be effective on the central bearing shaft.  It protrudes only about 1/4&quot; beyond the pillow block.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nisku Nut &amp;amp; Bolt did have washers that were about 1/16&quot; larger than the center shaft of the bearing, and a few thou too big to nest nicely in the ledge on the pillow block.  Some work with a sander fixed the latter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I placed the washer over the protruding bearing shaft, made sure it had clearance all the way around (10 to 20 thou?) and stuck it in place with bathtub silicon seal in the crack between the washer and the pillow block.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Potential failure modes:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The space between the face of the washer and bearing is on the order of 1 mm (40 thou) The washer is stationary, the bearing rotating.  The outer edge is more or less sealed, the inner edge of the washer open.  This may act as a dust collection centrifuge.  I'm debating whether to over grease this enough enough fill the space with grease.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the bathtub silicon lets go, I will have a washer ratling lose on the shaft between the bearing and the PTO.  I don't think this is a big worry.  There is nothing in that space to bind.&lt;/p&gt;&#xA;" OwnerUserId="8768" LastActivityDate="2019-08-30T12:41:27.557" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29895" PostTypeId="1" AcceptedAnswerId="29901" CreationDate="2019-08-30T12:49:41.963" Score="0" ViewCount="98" Body="&lt;p&gt;from the &lt;a href=&quot;http://T.%20Belytschko,%20L.%20Schwer:%20Large%20Displacement,%20Transient%20Analysis%20Of%20Space%20Frames;%20International%20Journal%20For%20Numerical%20Methods%20In%20Engineering;%20Vol.%2011,%2065-84%20(1977)&quot; rel=&quot;nofollow noreferrer&quot;&gt;paper in this link&lt;/a&gt; , which is the Belytscko beam. I could not figure out if it is a 3D beam or a 2D one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you don't have access to that paper, please check the &lt;a href=&quot;http://www.lstc.com/pdf/ls-dyna_theory_manual_2006.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;page 73 on this link&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XsWZb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XsWZb.png&quot; alt=&quot;Beam schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what confused me was the schematics on the original paper and the second link. in the beginning the general movements start from 3D but I don't see all the DOFs in the above image&lt;/p&gt;&#xA;" OwnerUserId="17070" LastEditorUserId="17070" LastEditDate="2019-08-30T13:18:19.040" LastActivityDate="2019-08-30T21:12:27.437" Title="Is Belytschko beam defined for 3D or 2D?" Tags="|structural-engineering|structural-analysis|finite-element-method|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29896" PostTypeId="2" ParentId="29885" CreationDate="2019-08-30T18:27:31.427" Score="3" Body="&lt;p&gt;The most common way to fix mangled threads, if they're really beyond fixing and if there's enough meat to open the hole a bit larger, is to use something like a &lt;a href=&quot;https://en.wikipedia.org/wiki/Threaded_insert#Helical_insert&quot; rel=&quot;nofollow noreferrer&quot;&gt;heli-coil&lt;/a&gt; (aka : threaded insert).  For a given thread, a helicoil will have a special oversized drill and tap which you use to clear out the mangled threads to a larger diameter and tap.  After threading in the helicoil you end up with the desired original thread diameter and pitch.&lt;/p&gt;&#xA;" OwnerUserId="12902" LastEditorUserId="12902" LastEditDate="2019-08-30T18:32:38.180" LastActivityDate="2019-08-30T18:32:38.180" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29897" PostTypeId="5" CreationDate="2019-08-30T19:23:31.473" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-08-30T19:23:31.473" LastActivityDate="2019-08-30T19:23:31.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29898" PostTypeId="4" CreationDate="2019-08-30T19:23:31.473" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2019-08-30T19:23:31.473" LastActivityDate="2019-08-30T19:23:31.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29900" PostTypeId="2" ParentId="29889" CreationDate="2019-08-30T19:33:02.673" Score="1" Body="&lt;p&gt;You will have to solve this by experimentation. There is not a simple equation for the reduction of magnetic force as an object is moved away from a magnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for this is that a magnetic field isn't a simple sphere. First imagine all the field lines that leave north side of the magnet and loop around to the south side. These lines actually form a circuit, a &lt;em&gt;magnetic&lt;/em&gt; circuit. Like an electrical circuit, the field lines will follow the most conductive pathways, for magnetism that would be pieces of iron and such. Unlike with electricity, even empty space will conduct a magnetic field to some extent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pieces of iron or other permeable material can shunt a magnetic field around some areas, or concentrate it in others. This makes for a field that is difficult to model with a simple algebra equation.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-08-30T19:33:02.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29901" PostTypeId="2" ParentId="29895" CreationDate="2019-08-30T21:12:27.437" Score="1" Body="&lt;p&gt;It is a 3D beam (6 degrees of freedom at each node).  The figure is 2D just to make things simpler.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2019-08-30T21:12:27.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29902" PostTypeId="1" CreationDate="2019-08-30T21:33:53.620" Score="0" ViewCount="352" Body="&lt;p&gt;I have 2 questions related to NASA CEA's Isp values in the output file.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When looking through the output of NASA's CEA tool, I am trying to understand the difference between its &quot;Ivac&quot; output and its &quot;Isp&quot; output. What is the difference between these two outputs?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;While the concept of a vacuum Isp makes sense to me, I don't understand why it is output in the context of  the inputs I put in; I gave the CEA code a finite value Pc/Pe number, meaning Pe was not 0. So what is Ivac mean in this context?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="9226" LastActivityDate="2019-08-30T21:33:53.620" Title="Isp Output of NASA CEA" Tags="|rocketry|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29903" PostTypeId="2" ParentId="29885" CreationDate="2019-08-30T22:51:26.820" Score="0" Body="&lt;p&gt;Use a 2nd Tap to clean the threads. Take out the tap and apply a thin layer of PTFE tape to the tap. Put the tap back in the hole pack the top of the hole with plastic metal. Let it set and wind out the tap. Flatten off the outer surface of the hole with a file run the tap down the hole again. Paint outer surface, Job Done. Sorry I cannot remember the trade name of the 3M's stuff. But I have posted a link to similar stuff.&#xA;&lt;a href=&quot;https://www.maxkote.co.uk/application-areas/metal-repair/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.maxkote.co.uk/application-areas/metal-repair/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note; Hand Tap come in sets of 3; taper, second and plug. Taper is the one with the pointed end. Plug is the one with flat end and almost no leading taper. Second looks a little like a plug but with more leading taper at the start.&lt;/p&gt;&#xA;" OwnerUserId="21597" LastEditorUserId="21597" LastEditDate="2019-08-31T22:58:12.287" LastActivityDate="2019-08-31T22:58:12.287" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29904" PostTypeId="1" CreationDate="2019-08-31T12:18:22.880" Score="0" ViewCount="137" Body="&lt;p&gt;So, I am working with FBG sensors and I have read that those sensors need to be re-calibrated every 10 years to get accurate measurements. But no where on the internet (AFAIK) is it explained why every 10 years it has to be re-calibrated and why.&lt;/p&gt;&#xA;" OwnerUserId="21567" LastActivityDate="2022-06-17T02:02:05.160" Title="Why are FBG sensors recalibrated every 10 years?" Tags="|electrical-engineering|sensors|calibration|" AnswerCount="2" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29906" PostTypeId="2" ParentId="29904" CreationDate="2019-08-31T18:42:28.617" Score="0" Body="&lt;p&gt;Fiber Bragg Grating sensors can reflect a particular wavelength of a light beam or a selection of wavelength in a fiber optic by having the sensors fiber refractive index grated according to specifications to reflect the particular wavelength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are used for many applications in fiber optics such as add drop multiplexing and precision measurement of heat and stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the are precision tools each manufacturer has its own criteria of calibration and set up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to a good educational YouTube video.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can click on the more link on the bottom of the video and read the physics involved and applications. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/apr-XUtjmBY&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fiber Bragg Grating link&lt;/a&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-08-31T18:42:28.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29907" PostTypeId="1" CreationDate="2019-08-31T20:28:35.997" Score="2" ViewCount="462" Body="&lt;p&gt;&lt;strong&gt;Background info:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;For decades CT scanners produced medical images from x-rays travelling through the tissues of the patient at different angles through analytical methods based on the &lt;a href=&quot;https://en.wikipedia.org/wiki/Radon_transform&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radon transform&lt;/a&gt; and &lt;a href=&quot;https://en.wikipedia.org/wiki/Projection-slice_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fourier slice theorem&lt;/a&gt;, and generally falling under the category of &lt;a href=&quot;https://www.aapm.org/meetings/99AM/pdf/2806-57576.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;filtered backprojection (FBP)&lt;/a&gt;, beautifully &lt;a href=&quot;http://www.ucdenver.edu/academics/colleges/medicalschool/departments/Radiology/About%20Us/Faculty/education-research-portal/Pages/filtbackproj.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;illustrated by Adam Kesner&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CBXap.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CBXap.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fpuxd.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fpuxd.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;and &lt;a href=&quot;https://en.wikipedia.org/wiki/Projection-slice_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LPZ4H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LPZ4H.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;This is nicely explained &lt;a href=&quot;https://www.youtube.com/watch?v=pZ7JlXagT0w&amp;amp;list=PLmecqweSELQPgUUG5eIvghR_tzUnK6lD8&amp;amp;index=5&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;These had limitations, such as the presence of sparse matrices with logarithmic transformations. The need for logarithmic transformation of the acquired data stems from the fact that the projected data &lt;span class=&quot;math-container&quot;&gt;$I_d$&lt;/span&gt; at a given angle is proportional to the exponential of the integral of the attenuation along the x-ray beam&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_d= I_0 \exp\left[- \int_0^d \mu(s; \bar E) \mathrm ds\right],$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; is the linear attenuation coefficient, which depends on location &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt; and effective energy &lt;span class=&quot;math-container&quot;&gt;$\bar E,$&lt;/span&gt; and hence, the log transformation is part of the generation of the 2D FT of the image. In iterative reconstruction the log transformation of the data comes about when calculating the maximum posterior estimate of the line integrals along different angles based on detected x-ray intensity, which is handled as log-likelihood problems. Sparse matrix issues may surface on the final image as artifact, which was basically addressed by increasing the dose delivered to the patient.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Iterative_reconstruction&quot; rel=&quot;nofollow noreferrer&quot;&gt;Iterative reconstruction&lt;/a&gt; came about as an attempt at applying computer-intensive optimization processes to handling problems such as the noise in the image, and with the ultimate goals of either improving the diagnostic image, or reducing dose. These &lt;a href=&quot;https://link.springer.com/article/10.1007/s00330-018-5810-7&quot; rel=&quot;nofollow noreferrer&quot;&gt;IR techniques&lt;/a&gt; seem to be referred as either &lt;a href=&quot;https://en.wikipedia.org/wiki/Iterative_reconstruction#Statistical_Reconstruction&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;hybrid&lt;/strong&gt; IR models (iDose (Philips) or ASiR (GE))&lt;/a&gt; or &lt;strong&gt;model-based&lt;/strong&gt; reconstruction (MBIR) (or &amp;quot;pure&amp;quot; IR). In both cases there is an objective or cost function and coefficients to be estimated, typically with a penalized regression model. This regression or optimization process is logically solved computationally (similar to gradient descent). The distinction between IR models is difficult to understand, but it seems as though initial IR products (ie &lt;a href=&quot;https://pubs.rsna.org/doi/full/10.1148/radiol.2015132766&quot; rel=&quot;nofollow noreferrer&quot;&gt;Philips iDose&lt;/a&gt; would improve images based on a penalized regression model for weightings of noisier measurements based on photon distribution statistics. MBIR, on the other hand, would take into consideration the x-ray anode and receptor geometry and optics to &lt;a href=&quot;https://www.ajronline.org/doi/10.2214/AJR.12.9424&quot; rel=&quot;nofollow noreferrer&quot;&gt;produce a &amp;quot;forward projection on an &lt;em&gt;assumed starting image&lt;/em&gt;.&amp;quot;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;In &lt;a href=&quot;https://aapm.onlinelibrary.wiley.com/doi/pdf/10.1002/mp.13123&quot; rel=&quot;nofollow noreferrer&quot;&gt;this great 2018 review paper on regularization techniques by Hao Zhang&lt;/a&gt;, the distinction between hybrid and model based seems to have been rendered obsolete in favor of SIR (Statistical Image Reconstruction) (please note that LDCT stands for low-dose CT scan):&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JtWD5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JtWD5.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;in contradistinction to FBP (&amp;quot;analytical&amp;quot;) or &lt;a href=&quot;https://ir.canterbury.ac.nz/bitstream/handle/10092/9851/12647037_Tang_etal_IVCNZ2012.pdf?sequence=1&amp;amp;isAllowed=y&quot; rel=&quot;nofollow noreferrer&quot;&gt;ART (&amp;quot;algebraic&amp;quot;)&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;p&gt;In general model-based IR seems to take statistical IR to a different level, incorporating neural networks, as well as adding more refined assumptions such as the optics of the receptors, etc. Largely, though, these are proprietary algorithms, which are not available to the public or the medical community.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Question:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;The question is in the end very basic, and makes reference to the terminology and general process used in the diagram found in &lt;a href=&quot;https://pubs.rsna.org/doi/full/10.1148/radiol.2015132766&quot; rel=&quot;nofollow noreferrer&quot;&gt;this review paper&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mXzV2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mXzV2.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;and specifically&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;whether the &lt;span class=&quot;math-container&quot;&gt;$\color{red}{\text{&quot;First image estimate&quot;}}$&lt;/span&gt; would (implied in the text) seem to be generated as an reconstruction process similar to FBP.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;If this is remotely correct, my understanding of the picture above could be reworded as follows:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The projection data gathered by the continuously spiraling x-ray beam passing through the patient forms the projected data or sinogram (which is visualized as a gray, vertically directed spiral or sine wave) containing the raw data. So far this is the &amp;quot;forward projection&amp;quot; (yellow arrow in the image below). The second step consists of&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;&amp;amp;\color{red}{\text{a backprojection}}\\&#xA;&amp;amp;\color{red}{\text{to get a rough picture}}\\&#xA;&amp;amp;\color{red}{\text{with the same methods}}\\&#xA;&amp;amp;\color{red}{\text{used in filtered backprojection (FBP) (?)}}&#xA;\end{align}$$&lt;/span&gt;&#xA;(corresponding to the red arrow in the picture below), which is later optimized progressively by pretending to extract from this rough picture a sinogram (&amp;quot;simulated fwd projection&amp;quot; in the image below), and comparing it to the initial sinogram:&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pZhX8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pZhX8.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;From everything above it would seem like hybrid IR algorithms would start off from a FBP of the raw data in the sinogram, while in the MBIR algorithms the first step would be simulated forward projection from a set of assumptions.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;So the most complete formulation of the question would be whether the &amp;quot;first image estimate&amp;quot; in hybrid IR algorithms is formed via FBP; and how the &amp;quot;simulated forward projection&amp;quot; is obtained in MBIR algorithms.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Online available references:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://link.springer.com/article/10.1007/s40134-012-0003-7&quot; rel=&quot;nofollow noreferrer&quot;&gt;Recent Advances in CT Image Reconstruction, J. Hsieh&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://ir.canterbury.ac.nz/bitstream/handle/10092/9851/12647037_Tang_etal_IVCNZ2012.pdf?sequence=1&amp;amp;isAllowed=y&quot; rel=&quot;nofollow noreferrer&quot;&gt;Using Algebraic Reconstruction in Computed Tomography, N. Tang&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://aapm.onlinelibrary.wiley.com/doi/pdf/10.1002/mp.13123&quot; rel=&quot;nofollow noreferrer&quot;&gt;Regularization strategies in statistical image reconstruction of&#xA;low-dose x-ray CT: A review, H. Zhang&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;http://incenter.medical.philips.com/doclib/enc/fetch/2000/4504/577242/577249/586938/587315/iDose4_-_Whitepaper_-_Technical_-Low_Res.pdf%3Fnodeid%3D8432599%26vernum%3D-2&quot; rel=&quot;nofollow noreferrer&quot;&gt;iDose Iterative Reconstruction Technique, Philips info&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.cockcroft.ac.uk/wp-content/uploads/2016/04/CT-Iterative_Reconstruction_Techniques.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;State of the Art: Iterative CT Reconstruction Techniques, L. Geyer&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://engineering.purdue.edu/%7Emalcolm/pct/CTI_Ch07.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Algebraic Reconstruction Algorithms&lt;/a&gt;  &lt;a href=&quot;https://www.dkfz.de/en/roentgenbildgebung/ct/ct_conference_contributions/BasicsOfCTImageReconstruction_Part2.pdf?m=1515503796&quot; rel=&quot;nofollow noreferrer&quot;&gt;Basics of Image Reconstruction&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21660" LastEditorUserId="21660" LastEditDate="2020-10-06T01:01:06.183" LastActivityDate="2020-10-06T01:01:06.183" Title="How is the first image estimate generated in CT scan iterative reconstruction (IR)?" Tags="|biomedical-engineering|medical-devices|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29908" PostTypeId="2" ParentId="15877" CreationDate="2019-08-31T20:31:34.153" Score="1" Body="&lt;p&gt;I'm not familiar with SAP nor MIDAS, however - It can be shown here how to meet your requirement in ANSYS.&#xA;For the same geometry, mesh, coordinate systems and material definitions - you can create different scenarios containing different supports and loads (called there 'solutions'). It is possible to solve each of them separately and even to apply solution combinations - for instance, calculating the total deflection when scenario A has 60% influence on the results and scenario B affects the remaining 40%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/slkE9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/slkE9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Image above illustrates a simple example of a steel beam being loaded totally differently in two cases. The results shows  a specific combination of the two scenarios.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2019-08-31T20:31:34.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29909" PostTypeId="2" ParentId="29870" CreationDate="2019-09-01T02:17:02.533" Score="1" Body="&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Why cannot series DC motors self regulate its speed?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Considering that there is no difference in the relations when it comes to series DC motors, except that the torque is related to the square of the current, why wouldn't a series DC motor be able to self regulate its speed ?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;The answer to why a series DC motor is not able to self regulate compared to a Shunt motor is a D.C. Shunt Motor has a high resistance field (excitation) winding and a D.C. Series Motor does not.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;To expand on this.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;1). Firstly your question is poorly defined&lt;/strong&gt; you should clearly state what type of equipment you are referencing. There are numerous types of D.C. Motors and variations on how the excitation can be achieved.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;2). Second your conclusion is somewhat off the mark&lt;/strong&gt;.You have failed to consider &lt;strong&gt;a) type of excitation&lt;/strong&gt; and &lt;strong&gt;B) the load variation&lt;/strong&gt;. Thus by  taking a simplistic view of how the series excited motor works with no relation to load &lt;strong&gt;you have created a completely false scenario&lt;/strong&gt; and &lt;strong&gt;compared that to a fictional shunt motor&lt;/strong&gt; that in your world has been built with the same physical characteristics of a series motor.&lt;/p&gt;&#xA;&lt;p&gt;With a series excited d.c. motor the load determines the speed at a given supply voltage and with no load the speed will continue to increase until the motor destroys itself. With a &lt;strong&gt;shunt excited motor the excitation circuit inherently has resistance added&lt;/strong&gt; into the circuit and except in exceptional circumstances will also have an additional resistance in circuit (usually variable for speed control). The inherent resistance comes from the &lt;strong&gt;smaller conductor size and increased number of turn used for the excitation coils&lt;/strong&gt;, which you cannot have in  a series excited motor. &lt;strong&gt;The resistance in the shunt excited motor winding's thus reduces the speed so that the motor  cannot reach a speed where it destroys itself&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;When load increases on shunt DC motor, the speed decreases resulting in a downfall in back-EMF. This in turn, causes the net voltage, and hence the current to increase causing an increase in torque and thereby increasing the speed. Thus, it achieves a constant speed.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;3). Thirdly you have told only half the story&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;But when we start loading the d.c. shunt motor, this decreases the speed of armature. This results in fall in back EMF.  This reduced back EMF causes a larger current to flow through the armature winding and larger current produces increased driving torque.&lt;/p&gt;&#xA;&lt;p&gt;Thus, the motor produces larger driving torque as it slows down. So motor will start running on the speed at which armature current is just sufficient to produce the required torque by the load.&lt;/p&gt;&#xA;&lt;p&gt;On the other hand, &lt;strong&gt;when the load on the  motor is reduced, the speed of armature increases&lt;/strong&gt; due to excess of driving torque. This increases the back EMF which results in decreased armature current and thus required torque is reached.  therefore speed of the Armature cannot indefinitely accelerate if all load is removed due to the back EMF and inherent resistance in the winding's of the shunt coil.&lt;/p&gt;&#xA;&lt;p&gt;In this way back &lt;strong&gt;EMF in a DC shunt motor regulates the flow of armature current according to the load requirement automatically&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Where does the back emf originate???&lt;/strong&gt; Why shunt field resistance is high in the case of dc shunt motor?&lt;/p&gt;&#xA;&lt;p&gt;Back emf is proportional to the rate of change of flux in the field winding. &lt;a href=&quot;https://www.electronics-tutorials.ws/inductor/inductor.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tutorials&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;rate of change of Flux in the field winding is proportional to the mmf in the winding.&lt;/p&gt;&#xA;&lt;p&gt;mmf = current passing through the winding * no of turns in the winding&lt;/p&gt;&#xA;&lt;p&gt;To attain the required mmf we can use more current and less turns ( or ) &lt;strong&gt;less current and more turns&lt;/strong&gt; .&lt;/p&gt;&#xA;&lt;p&gt;As the power loss in the field winding is considered it is....Power loss = Square of voltage / resistance&lt;/p&gt;&#xA;&lt;p&gt;As the voltage is constant for a high resistance it is possible to have less losses. So we prefer a &lt;strong&gt;high shunt field resistance&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;As the field resistance is high, current is less, so number of turns in the shunt field will be more to attain the required mmf.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://electrical-engineering-portal.com/4-types-of-dc-motors-and-their-characteristics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Types of D.C. Motors&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;In a series DC motor the field is connected in series with the armature.The field is wound with a few turns of large wire because it must carry the full armature current.&lt;/p&gt;&#xA;&lt;p&gt;In a shunt motor the field is connected in parallel (shunt) with the armature winding's and can be wired with thinner conductors as it does not take the the full armature current. It can also be easily reverse connected.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A characteristic of series motors is the motor develops a large amount of starting torque. However, speed varies widely between no load and full load. Series motors cannot be used where a constant speed is required under varying loads.&lt;/p&gt;&#xA;&lt;p&gt;Additionally, the speed of a series motor with no load increases to the point where the motor can become damaged. Some load must always be connected to a series-connected motor.&#xA;Series-connected motors generally are not suitable for use on most variable speed drive applications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The shunt-connected motor offers good speed regulation. The field winding can be separately excited or connected to the same source as the armature.&lt;/p&gt;&#xA;&lt;p&gt;An advantage to a separately excited shunt field is the ability of a variable speed drive to provide independent control of the armature and field.&#xA;The shunt-connected motor offers simplified control for reversing. This is especially beneficial in regenerative drives.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;Note&lt;/strong&gt; When I refer to separately excited it means the excitation field supply is independent from the motor supply.&lt;/p&gt;&#xA;" OwnerUserId="21597" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-09-01T02:33:52.950" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29910" PostTypeId="2" ParentId="25272" CreationDate="2019-09-01T02:20:03.960" Score="1" Body="&lt;p&gt;If you are asking for 2 different materials, the lightest and the strongest one, well...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lightest material is hydrogen gas (2g/mol, or 2g/24l at normal condition). The lightest pure metal is lithium (number 3 in the periodic table, it is even lighter than oxygen gas, by the way). According to &lt;a href=&quot;https://www.thoughtco.com/what-is-the-lightest-metal-608450&quot; rel=&quot;nofollow noreferrer&quot;&gt;this source&lt;/a&gt;, the lattice of nickel phosphorous tubes (Microlattice)  is the lightest one. No information is given on its precise number though&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://www.transmet.com/the-10-strongest-materials-known-to-man/&quot; rel=&quot;nofollow noreferrer&quot;&gt;source #2&lt;/a&gt;, graphene (practically carbon) is the strongest material known to mankind (of course, producing it is very expensive).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, if you ask for the lightest material among the strongest materials, based on a set requirement, then there is insufficient data to answer that.&lt;/p&gt;&#xA;" OwnerUserId="21520" LastActivityDate="2019-09-01T02:20:03.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29912" PostTypeId="2" ParentId="29868" CreationDate="2019-09-01T03:14:32.097" Score="1" Body="&lt;p&gt;&lt;strong&gt;A series DC motor has a high starting torque thanks to its field wiring consisting of thicker wires and fewer turns resulting in high starting torque. And a shunt DC motor do not have a high starting torque because of a field winding consisting of thinner wires and more turns.&#xA;But why couldn't wiring be made a choice depending on the use so that it does not matter which motor is being used ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The answer to this is extremely easy. The name of this type of motor is a compound motor.&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have existed for over a hundred years. Compound motors usually use both series and shut wound excitation fields. The series winding can be removed from circuit by an external contactor arrangement's when the motor has started, if required. As with Locomotives as posted by &quot;niels nielsen&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However true Compound motors use both winding's simultaneously and how the effects are inter-coupled has some very different effects on the motor and it's usage. &lt;a href=&quot;https://www.electrical4u.com/compound-wound-dc-motor-or-dc-compound-motor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Compound Motors&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://electrical-engineering-portal.com/4-types-of-dc-motors-and-their-characteristics&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://electrical-engineering-portal.com/4-types-of-dc-motors-and-their-characteristics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21597" LastActivityDate="2019-09-01T03:14:32.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29915" PostTypeId="1" CreationDate="2019-09-01T19:33:57.740" Score="0" ViewCount="292" Body="&lt;p&gt;I have been tasked with solving a series of linear equations in the form of A&lt;strong&gt;x&lt;/strong&gt; = &lt;strong&gt;b&lt;/strong&gt;, where &lt;strong&gt;b&lt;/strong&gt; is the zero vector. Specifically, I have a matrix E which is 4x7 and a column vector &lt;strong&gt;r&lt;/strong&gt; which is 7x1, as pictured:&lt;a href=&quot;https://i.stack.imgur.com/JDq1y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JDq1y.jpg&quot; alt=&quot;The E matrix&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W3dSN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W3dSN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must solve for E&lt;strong&gt;r&lt;/strong&gt; = &lt;strong&gt;0&lt;/strong&gt;.I'm having difficulty finding a function in GNU Octave that will solve this equation for me. I went to the documentation page at &lt;a href=&quot;https://octave.org/doc/v4.0.0/index.html#SEC_Contents&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://octave.org/doc/v4.0.0/index.html#SEC_Contents&lt;/a&gt; and failed to find any functions that would accept column vectors with variable elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found nothing of use in 1.2.4-Solving Systems of Linear Equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;linsolve() looked promising, but it doesn't work with variables as far as I can tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found nothing of use under 16-matrix manipulation, 18-linear algebra, nor 25-optimization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice or suggestions would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="21666" LastActivityDate="2019-09-01T22:15:46.443" Title="Solving a linear mass balance equation in the form of Ax = 0 with variables in vector x" Tags="|chemical-engineering|matlab|process-engineering|software|mass-transfer|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29916" PostTypeId="2" ParentId="29915" CreationDate="2019-09-01T20:15:38.230" Score="1" Body="&lt;p&gt;You don't need a computer program to solve this. You can almost do it in your head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last equation is satisfied by any values of &lt;span class=&quot;math-container&quot;&gt;$x, y, z$&lt;/span&gt; so you can forget about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you multiply out the first row of the matrix product, you get &lt;span class=&quot;math-container&quot;&gt;$$1 + 2.4 + x + 0 -5.6 + 0 + 0 = 0$$&lt;/span&gt;&#xA;which is pretty easy to solve for &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second row is just as easy to solve for &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;, and the third row gives you the value of &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have several sets of equations like this to solve with different coefficients in the &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; matrix, you can rearrange the equations in the form &lt;span class=&quot;math-container&quot;&gt;$$\begin{bmatrix} a_{11} &amp;amp; a_{12} &amp;amp; a_{13} \\&#xA;a_{21} &amp;amp; a_{22} &amp;amp; a_{23} \\ a_{31} &amp;amp; a_{32} &amp;amp; a_{33}&#xA;\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = &#xA;\begin{bmatrix} c_{11} &amp;amp; c_{12} &amp;amp; c_{13} &amp;amp; c_{14} \\&#xA;c_{21} &amp;amp; c_{22} &amp;amp; c_{23} &amp;amp; c_{24} \\ c_{31} &amp;amp; c_{32} &amp;amp; c_{33} &amp;amp; c_{34} &#xA;\end{bmatrix} \begin{bmatrix}d_1\\d_2\\d_3\\d_4\end{bmatrix}$$&lt;/span&gt; &#xA;and you should then be able to find some routines to multiply the &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt; matrices and then solve the equations.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A different way to solve this by computer is to consider all 7 elements of the vector as variables, and then add more equations which give them their known values. That would give a set of 7 equations like &lt;span class=&quot;math-container&quot;&gt;$$\begin{bmatrix}&#xA;1   &amp;amp; 1   &amp;amp; 1 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \\&#xA;2.3 &amp;amp; 3   &amp;amp; 0 &amp;amp; 1 &amp;amp; 2 &amp;amp; 0 &amp;amp; 3 \\&#xA;0.8 &amp;amp; 0.5 &amp;amp; 2 &amp;amp; 1 &amp;amp; 1 &amp;amp; 2 &amp;amp; 0 \\&#xA;0.3 &amp;amp; 0   &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 \\&#xA;1   &amp;amp; 0   &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xA;0   &amp;amp; 1   &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0 \\&#xA;0   &amp;amp; 0   &amp;amp; 0 &amp;amp; 0 &amp;amp; 1 &amp;amp; 0 &amp;amp; 0 \end{bmatrix}&#xA;\begin{bmatrix} a \\ b \\ x \\ y \\ c \\ -z \\ d\end{bmatrix} =&#xA;\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 2.4 \\ -5.6 \end{bmatrix}$$&lt;/span&gt; &#xA;where the three additional equations fix the valued of &lt;span class=&quot;math-container&quot;&gt;$a, b, c$&lt;/span&gt; (and the fourth equation, which was irrelevant for finding &lt;span class=&quot;math-container&quot;&gt;$x, y, z$&lt;/span&gt;, will give the value of &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;). &lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-09-01T22:15:46.443" LastActivityDate="2019-09-01T22:15:46.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29917" PostTypeId="1" CreationDate="2019-09-01T21:40:53.323" Score="0" ViewCount="270" Body="&lt;p&gt;I'm looking for examples of mechanisms that can constrain a hub rotating around a shaft from 0 to 360 degrees.  Rotation between 0 and 360 should be free.  Rotation &amp;lt; 0 and &amp;gt; 360 should be stopped.&lt;/p&gt;&#xA;&lt;p&gt;A few things I think could work but I am not not sure what this is called and searching on Google has not helped.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a simple pin would allow for 0..(360 - pin_width)&lt;/li&gt;&#xA;&lt;li&gt;a spiral would allow hard stops 0..360 but the pin would need to&#xA;float radially.&lt;/li&gt;&#xA;&lt;li&gt;a gear set of 2:1 would allow for pins but would add some complexity.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Thanks for any ideas folks might have and/or any names I might be able to use in a Google search.&lt;/p&gt;&#xA;" OwnerUserId="21668" LastEditorUserId="90" LastEditDate="2022-12-16T02:14:06.970" LastActivityDate="2022-12-16T02:14:06.970" Title="0 to 360 degress constrained rotation mechanism" Tags="|mechanisms|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29918" PostTypeId="1" CreationDate="2019-09-01T22:16:22.637" Score="1" ViewCount="112" Body="&lt;p&gt;When using IMU data (accelerations and angular velocities), many papers use a 4th order low-pass butterworth filter with a cutoff frequency of 5 Hz. However, none of them really justify why those filters have been chosen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone please explain how the order and cutoff frequency are chosen for a filter of this nature? My data has a sampling frequency of 125 Hz&lt;/p&gt;&#xA;" OwnerUserId="21669" LastActivityDate="2019-09-01T22:16:22.637" Title="How to choose order and cut-off frequency for low-pass Butterworth filter?" Tags="|sensors|acceleration|electronic-filters|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29919" PostTypeId="1" AcceptedAnswerId="29920" CreationDate="2019-09-02T02:47:03.250" Score="0" ViewCount="876" Body="&lt;p&gt;I intend to make an oscillating motion from a simple computer DC fan (say 5V). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: could a signal generator with a sinusoidal signal of (say voltage 5V  and frequency of 5 Hz) be used to power such a DC fan to produce the oscillating motion ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second question:&#xA;it is related to &lt;a href=&quot;https://engineering.stackexchange.com/questions/18427/oscillations-in-motor-control-systems&quot;&gt;Oscillations in motor control systems&lt;/a&gt;.&#xA;Will there be a damage for a low frequency and low voltage input to the fan ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For additional info, the load is lightweight, and the mechanical rotation is about ±45 degrees.&lt;/p&gt;&#xA;" OwnerUserId="19831" LastEditorUserId="19831" LastEditDate="2019-09-11T07:22:14.680" LastActivityDate="2019-09-11T07:22:14.680" Title="oscillation motion from DC motor (computer fan)" Tags="|electrical-engineering|motors|prototyping|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29920" PostTypeId="2" ParentId="29919" CreationDate="2019-09-02T04:14:54.623" Score="0" Body="&lt;p&gt;computer fans are driven with direct current but contain a circuit which chops the DC into pulses that sequentially excite sets of electromagnet coils inside the motor to make it turn. these motors are called &lt;em&gt;brushless dc&lt;/em&gt; motors and cannot be operated the way you want to. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can instead make a &quot;conventinal&quot; or &lt;em&gt;brushed DC&lt;/em&gt; motor vibrate the way you want simply by feeding it AC instead of DC. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better way to make vibrations out of electric signals is to feed them into the terminals of a &lt;em&gt;loudspeaker&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-02T04:14:54.623" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29921" PostTypeId="2" ParentId="29917" CreationDate="2019-09-02T05:00:40.170" Score="1" Body="&lt;p&gt;If you have an axially sliding collar with a helical groove, then the groove can give exactly 0 to 360...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The helix only needs to be sufficient that the ends of the groove don’t interfere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could also fix the collar axially and permit the pin to slide the necessary amount...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-09-02T05:43:43.973" LastActivityDate="2019-09-02T05:43:43.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29923" PostTypeId="1" CreationDate="2019-09-02T11:44:42.760" Score="0" ViewCount="30" Body="&lt;p&gt;My question is what is the most efficient way to convert 48V D.C. to 220V A.C. Not the easiest or most cost efficient but the most efficient electrically?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also if anyone has any cheats on the best way to resupply LED House light fittings that were originally 220v supply to 48V dc I would be grateful.&lt;/p&gt;&#xA;" OwnerUserId="21597" LastActivityDate="2019-09-02T11:44:42.760" Title="Solar power, Battery power to mains voltage" Tags="|electrical-engineering|solar-energy|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="29924" PostTypeId="1" CreationDate="2019-09-02T12:46:49.897" Score="0" ViewCount="33" Body="&lt;p&gt;Modern screwdriver bits, marketed for impact tool use, have &quot;torsion&quot; control, apparently implemented with a necking of the bit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this feature reduce the delivered peak torque?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Practically, does it reduce bit failure?&lt;/p&gt;&#xA;" OwnerUserId="17062" LastActivityDate="2019-09-02T12:46:49.897" Title="What are design features of impact type screwdriver bits?" Tags="|tools|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29925" PostTypeId="1" CreationDate="2019-09-02T15:17:33.557" Score="0" ViewCount="437" Body="&lt;p&gt;I need to compare two STL (Dental model Scanned) files and find the deviation between their points, SO this two STL files have different boundary, so i already used the &lt;a href=&quot;https://www.3dsystems.com/software/geomagic-control-x&quot; rel=&quot;nofollow noreferrer&quot;&gt;GEOMAGIC Control software&lt;/a&gt; for comparing STL files, one sample output file would be like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fX1ca.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fX1ca.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found these two like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.researchgate.net/post/Are_there_any_open_source_alternatives_for_geomagic&quot; rel=&quot;nofollow noreferrer&quot;&gt;Are there any open source alternatives for geomagic?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.researchgate.net/post/how_to_compare_quantify_populate_two_stl_models_having_shape_difference_to_see_result_as_color_stl_file_and_bar_telling_plus_and_minus_changes&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to compare/quantify (populate) two .stl models having shape difference to see result as color stl file and bar telling plus and minus changes?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;useful for finding free alternative software for Geomagic, So i installed the &lt;a href=&quot;http://www.danielgm.net/cc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cloudcompare&lt;/a&gt; in my Linux OS, and aligned two stl file  and used the &lt;strong&gt;&lt;em&gt;compute cloud/mesh distance&lt;/em&gt;&lt;/strong&gt; icon and get this output picture of stl which shown geomagic output above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZE1EP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZE1EP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in the &lt;strong&gt;CloudCompare&lt;/strong&gt; output i have 1 aria which i don't want to be.&#xA;How could i solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="2678" LastActivityDate="2024-01-10T00:02:40.393" Title="How to get similart Geomagic two stl file deviation in free Cloudcompare software?" Tags="|computer-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29926" PostTypeId="2" ParentId="29925" CreationDate="2019-09-02T15:23:20.190" Score="0" Body="&lt;p&gt;I found the reason, i must use other &lt;strong&gt;&lt;em&gt;stl&lt;/em&gt;&lt;/strong&gt; as references and the out put have this kind:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nhOZX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nhOZX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2678" LastActivityDate="2019-09-02T15:23:20.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29927" PostTypeId="2" ParentId="29856" CreationDate="2019-09-02T15:48:28.727" Score="0" Body="&lt;p&gt;I think the conclusion is probably wrong. You are assuming the seal falls out then the bearing gets wrecked. In my experience it is the other way round the wrecked bearing generates heat and the plastic seal falls out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The long grass should not be generating enough heat to damage the seal but it is probably putting load onto the bearing and wrecking it.&lt;/p&gt;&#xA;" OwnerUserId="21597" LastActivityDate="2019-09-02T15:48:28.727" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29928" PostTypeId="2" ParentId="29857" CreationDate="2019-09-02T16:07:02.963" Score="0" Body="&lt;p&gt;&lt;strong&gt;Is there a good way to systematically increase the accuracy and precision of a measuring tool using only mathematical means?&lt;/strong&gt; For example, average 10 measurements from the same tool can create a better measurement. Or using two independent tools and average them. I don't know if measuring theory or statistical quality control or other subject can help.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The question is a little confusing first you make a simple question &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;systematically increase the accuracy and precision of a measuring tool&#xA;  using only mathematical means &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;to which the &lt;strong&gt;answer is obviously No&lt;/strong&gt; then you proceed to talk about using more than one tool and doing averaging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So really the question should be &lt;strong&gt;Can measuring theory or statistical quality control  increase the accuracy and precision of  measurements?&lt;/strong&gt; To which again the &lt;strong&gt;answer is obviously No&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However what mathematical means can do is &lt;strong&gt;predict the percentage error and so the reliability of the measured values&lt;/strong&gt; and this can give you a theoretical (possible) answer using averages etc. However useful this may be &lt;strong&gt;it does not increase the accuracy and precision of the actual measurements.&lt;/strong&gt; and unfortunately never will. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to increase the accuracy and precision of the actual measurements it to use &lt;strong&gt;high precision instruments that are regularly re-calibrated against a known standard.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="21597" LastActivityDate="2019-09-02T16:07:02.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29930" PostTypeId="2" ParentId="23651" CreationDate="2019-09-02T18:43:09.223" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;So typically there is proportionality and a continuously varying (analogue) output.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct. This could be an analog voltage or a digital value.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In the case of a simple process like a domestic gas boiler, the boiler is either fully on or fully off. Or, say a cooling fan that can only be switched on or off to keep something cold.&#xA;  In these cases, can a PID controller be used at all? There is still an error term. Or would an alternative form of control be necessary?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It certainly could and is in many industrial control systems. Heating, for example, is very often controlled in this way using variable duty-cycle where full power is applied for a varying percentage of a fixed cycle time. This makes the power control system a simple on-off type which can be implemented using relays or SSRs (solid-state relays).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cCPiD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cCPiD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. SSRs (solid-state relays) allow rapid switching while allowing varying duty cycle for heating loads. Source: &lt;a href=&quot;http://lednique.com/opto-isolators-2/opto-triacs-zero-cross-work/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Opto-triacs, solid-state relays (SSR), zero-cross and how they work&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The duty cycle is determined by the the thermal response of the system. A room heating system may have a response time in tens of minutes so a long duty cycle (several minutes) may be appropriate. On the other hand, a heat sealing station for welding plastic films together may have a response time of seconds and a duty cycle of a second or two may be appropriate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If programming this control in a PLC the standard approach would be to use two timers.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Timer 1 is the period timer. Let's say it's set to 5 s.&lt;/li&gt;&#xA;&lt;li&gt;Timer 2 is the duty cycle timer. If the PID output is 25% then Timer 2's timeout value is set to 5 &amp;times; 0.25 = 1.25 s.&lt;/li&gt;&#xA;&lt;li&gt;The output turns on at the reset of the period timer (Timer 1) and turns off when&#xA;the duty cycle timer (Timer 2) reaches its timeout value.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;    T1  &#xA;+--|/|--------[T1 5000 ms]--&#xA;| &#xA;|   T1&#xA;+--|/|--------[T2 1250 ms]--&#xA;|&#xA;|   T2            HEAT &#xA;+--|/|------------( )-------&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 2. Pseudo PLC code.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;T1 runs and when it reaches 5000 ms it is energised, cuts its own feed and resets. The timer starts again.&lt;/li&gt;&#xA;&lt;li&gt;T2 resets every time T1 does and its is energised after the duty-cycle delay.&lt;/li&gt;&#xA;&lt;li&gt;The heat turns on until the duty-cycle timer is done.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Relays can be used for switching but when the switching period gets down below a minute or so mechanical wear becomes a problem. &#xA;SSRs solve the wear problem as they are solid-state and have no moving parts. Zero-cross types also eliminate both audible and electro-mechanical noise from the switching. (All the on periods in Figure 1 start on a zero-cross.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further reading:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/211403/ssr-not-switching-dimmed-load/211534#211534&quot;&gt;https://electronics.stackexchange.com/questions/211403/ssr-not-switching-dimmed-load/211534#211534&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2019-09-02T18:43:09.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29931" PostTypeId="1" AcceptedAnswerId="29933" CreationDate="2019-09-03T02:14:03.697" Score="0" ViewCount="153" Body="&lt;p&gt;I know that the simple way to measure the voltage stored in a lead-acid battery is to simply measure the positive and negative using a voltmeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my case, I think that my battery has a builtin charge controller and it is sealed like the picture below. How can I effectively measure it without opening?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0x7Hc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0x7Hc.png&quot; alt=&quot;asure &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gj8xr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gj8xr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is no way to measure it, how do I open this thing? I tried to open it using flat screwdriver and it leaves a dented mark.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ix9wX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ix9wX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3703" LastActivityDate="2019-09-04T11:36:02.163" Title="How can I measure the amount of voltage stored in a sealed lead-acid battery?" Tags="|battery|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29932" PostTypeId="1" CreationDate="2019-09-03T02:39:44.990" Score="0" ViewCount="17" Body="&lt;p&gt;I have a method developed to predict results of a signal that outputs an root mean squared error (RMSE) for each subject. Combining all 40 subjects, I get a mean RMSE of 5.5 with a standard deviation of 2.8. This is comparable to previous literature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I did a validation study using 5 subjects and obtained a mean RMSE of 12.4 with a standard deviation of 4.4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me these seem drastically different and I would not be able to conclude that the validation is good, but I want a more quantitative way to describe that. How would I go about this?&lt;/p&gt;&#xA;" OwnerUserId="21669" LastActivityDate="2019-09-03T02:39:44.990" Title="How to determine whether a validation study produces acceptable results?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29933" PostTypeId="2" ParentId="29931" CreationDate="2019-09-03T03:43:51.713" Score="1" Body="&lt;p&gt;The DC 12v output may or may not be direct from the battery, if it is current limited then probably not - you will need to check the spec sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top is probably screwed down as the pv controller is under there. The screws to get access are hidden under the orange graphic with all the labels, removing it without damage depends on how strong the adhesive is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way is to rub your thumb over the surface to find the screw holes then just uncover those... but if they fitted plastic hole covers you may not be lucky.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-09-03T03:43:51.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29936" PostTypeId="1" CreationDate="2019-09-03T08:24:43.350" Score="1" ViewCount="355" Body="&lt;p&gt;I'm trying to tune my satellite receiver, particularly to receive the signals of the Astra satellites, which are ASTRA 19.2E and ASTRA 2 at 28º East. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/buMTw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/buMTw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking at the picture the signal from ASTRA 28 should be focused on the UK (red circle which encloses Great britain) and ASTRA 19° on central europe (blue square). &#xA;What I observe in Italy, is the opposite: I receive a signal of 72% power strength from ASTRA 19 and of 92% from ASTRA 28°. How is that possible?&lt;/p&gt;&#xA;" OwnerUserId="21676" LastActivityDate="2019-09-03T08:24:43.350" Title="Is it better ASTRA 19.2°E or 28°, in central continental europe?" Tags="|telecommunication|wireless-communication|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29937" PostTypeId="1" CreationDate="2019-09-03T12:28:24.030" Score="0" ViewCount="24" Body="&lt;p&gt;How would I find out the optimum height of an ash stockpile in order to minimize diesel haulage costs, for a CAT haul truck?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My stockpile height is required to be between 25m to 50m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have worked out the fuel consumption and time taken from the start point to the haulage area, based on required speeds obtained from the rimpull-speed-gradeability curves for that particular CAT truck. But I'm not sure how to relate this to the calculating the optimum stockpile height? &lt;/p&gt;&#xA;" OwnerUserId="15545" LastActivityDate="2019-09-03T12:28:24.030" Title="Optimization of ash stockpile height to minimize haul truck fuel costs" Tags="|materials|civil-engineering|design|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29938" PostTypeId="2" ParentId="29931" CreationDate="2019-09-03T16:06:54.143" Score="1" Body="&lt;p&gt;The question you want to ask is &quot;how do I measure the amount of &lt;strong&gt;charge&lt;/strong&gt; stored in a battery&quot;, and the answer, for most sealed batteries is &quot;with great difficulty&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most battery &quot;fuel gauge&quot; applications track the battery charge state by watching the battery voltage, the current, and by integrating the current during charge and discharge.  And they still often get it wrong.  &quot;Battery charge monitoring&quot; is the subject you want to search for.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-03T16:06:54.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29939" PostTypeId="2" ParentId="29931" CreationDate="2019-09-03T17:13:55.967" Score="1" Body="&lt;p&gt;With a &lt;em&gt;sealed&lt;/em&gt; battery or one with a gelled electrolyte, you are stuck using the techniques listed by Tim above. In &lt;em&gt;nonsealed&lt;/em&gt; lead-acid batteries with liquid electrolyte, the &lt;em&gt;charge state&lt;/em&gt; of a cell can be determined by measuring the &lt;em&gt;specific gravity&lt;/em&gt; of the acid electrolyte, which varies according to how much lead is dissolved in it. Lots of dissolved lead means the battery has been in use and the electrolyte density goes up, very little lead means the battery is carrying a full charge and the density of the electrolyte is lower. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are clever, you can insert into the electrolyte a little plastic ball with a density slightly less than that of the electrolyte in a fully-discharged cell. The ball will sink when the battery is fully charged and float when it is almost fully discharged. If you put a clear plastic window in the top of the cell, you can deduce the charge state of the battery by looking for the ball through it. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-03T17:13:55.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29940" PostTypeId="1" CreationDate="2019-09-03T17:50:16.593" Score="0" ViewCount="68" Body="&lt;p&gt;Is there a dataset out there right now that shows the typical acceleration and angular velocity values for sensors attached to the thigh, shank, and feet for standard walking?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or literature/textbook that describes what these typical values should be?&lt;/p&gt;&#xA;" OwnerUserId="21669" LastEditorUserId="21669" LastEditDate="2019-09-03T18:13:28.030" LastActivityDate="2024-01-11T07:08:11.060" Title="Typical IMU accelerations and angular velocities for lower limbs" Tags="|mechanical-engineering|sensors|biomechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29941" PostTypeId="2" ParentId="29885" CreationDate="2019-09-03T18:09:47.683" Score="0" Body="&lt;p&gt;There are a few options. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general you need soomething like 1-1.5 times the bolt diameter's worth of thread to get full strength, with a steel bolt in a aluminium thread this may be a bit more but losing 2 or three with 20 left should be fine. In this case retapping the thread to clean up the mangled threads should fix the problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a decent depth of thread a taper tap on its own may be adequate but a bottomoimg tap may be worthwhile just for the sake of clearing out any swarf. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they are really mangled then it might also be worth drilling out the damaged threads completely. This will probaaly require using a drill stop to ensure that you don;t go too far into the intact thread and then retap to ensure you have a clean start to the remaining thread and also clear out any debris from teh original damage and the drilling process. Lubricate with a light oil or cutting fluid and then thoroughly flush out with something like degreasing spray.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is probbaly also worth using a threadlocking compound to prevent the problem happaning again,m this will also give a bit of extra support to the thread. &lt;/p&gt;&#xA;" OwnerUserId="4357" LastActivityDate="2019-09-03T18:09:47.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29944" PostTypeId="1" CreationDate="2019-09-03T18:51:10.500" Score="2" ViewCount="75" Body="&lt;p&gt;In Pavement ME, Version 2.5.5, why is there a tab for &quot;PCC Longitudinal Cracking&quot; calibration factors since longitudinal cracking prediction is not part of the current Pavement ME?&lt;/p&gt;&#xA;" OwnerUserId="21682" LastActivityDate="2019-09-03T21:53:27.917" Title="Pavement ME, Version 2.5.5 and PCC Longitudinal Cracking?" Tags="|pavement|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29946" PostTypeId="1" AcceptedAnswerId="29949" CreationDate="2019-09-03T20:11:37.337" Score="0" ViewCount="38" Body="&lt;p&gt;Soil compaction can occur through heavy vehicle traffic. I attached a picture (see below). There are three samples approx. to 5m distance to each side of a paved road. Vehicle traffic is sprodically (few times in a year, e.g. at harvest).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you think at the sampling points in the soil (0-30-cm depth) compaction could have occurred through vehicle traffic? The blue area is untilled, i.e. more aggregate stability. The grey part is tilled, i.e. soil aggregate structure is loose.* &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GfTVQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GfTVQ.png&quot; alt=&quot;road runs between two fields&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*The thing is, I noticed very high accumulation of soil organic matter under the blue area which also had lots of vegetation one yr prior to sampling -- unlike the grey area which has been cropland for long time. It is said that compaction can suppress mineralization of soil organic matter because of lower aeration in soil which inhibits decomposer communities.&lt;/p&gt;&#xA;" OwnerUserId="21684" LastActivityDate="2019-09-03T20:42:49.887" Title="Does this soil get compacted by occasional heavy vehicle traffic?" Tags="|pressure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29947" PostTypeId="1" CreationDate="2019-09-03T20:14:19.143" Score="1" ViewCount="42" Body="&lt;p&gt;I have a problem where there is an 80 Watt thermal load that needs to be cooled to maintain 80 Celsius. The base of the heat sink is 5&quot; x 7&quot; and is oriented vertically with cooling via natural convection. I am to provide the temperature differential of the heat sink vs. the thermal load based on the fin density.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Selected material: Aluminum 1050&lt;/li&gt;&#xA;&lt;li&gt;Fin shape: Rectangle &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To solve I found the film temperature, which helped me find the Beta (coefficient of volume expansion). I could then find the Rayleigh number, which in turn allowed me to find the optimal spacing between the fins. I selected a various amount of fin thicknesses and solved for the number of fins required based on its thickness. I then found the heat transfer coefficient. At this point I have everything required to solve the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Q = h(2 \cdot n_{fins} \cdot Length \cdot Height) \times (T_s-T_a)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am at a position where I don't know what to do. I don't understand what I am supposed to be comparing. I can find the temperature differential based on the information I found, but I don't know how the thermal load changes based on fin density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help in steering me in the right direction would be greatly appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="21685" LastEditorUserId="1592" LastEditDate="2019-09-04T02:52:55.860" LastActivityDate="2019-09-04T02:52:55.860" Title="Heat sink Design" Tags="|mechanical-engineering|heat-transfer|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29949" PostTypeId="2" ParentId="29946" CreationDate="2019-09-03T20:42:49.887" Score="0" Body="&lt;p&gt;30cm depth over 5m distance is approximately a cone with an angle of 3.5 degrees. This is outside the 45 degrees angle commonly assumed for influence of compaction or surcharge load. It means it could not benefit from traffic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anything this zone could have been negatively impacted. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Terzaghi this will by the bulging area of compression influence plane. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-03T20:42:49.887" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29950" PostTypeId="2" ParentId="29944" CreationDate="2019-09-03T21:53:27.917" Score="1" Body="&lt;p&gt;The PCC Longitudinal Cracking coefficients in Pavement ME Design correspond to the short jointed plain concrete pavement overlay of existing AC design type (SJPCP over AC). This design type and its calibration coefficients were introduced in version 2.3 in July 2016. These slabs are typically 5x5, 6x6, 7x7 thinner than conventional PCC designs. Their typical mode of failure is in the longitudinal direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that the longitudinal cracking calibration factors were included with the &quot;New Rigid&quot; calibration factors because the SJPCP design type does not fit well with the &quot;Bonded Rigid&quot; or &quot;Unbonded Rigid&quot; design types and their calibration factors. The bonded and unbonded factors are associated with PCC over PCC rehabilitation design types.&lt;/p&gt;&#xA;" OwnerUserId="20067" LastActivityDate="2019-09-03T21:53:27.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29951" PostTypeId="2" ParentId="11962" CreationDate="2019-09-04T01:30:05.360" Score="1" Body="&lt;p&gt;Something I would add to theNamesCross's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Barrel length provides extra velocity, which as name's said improves accuracy basically directly. There will be less drop and less wind drift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More powder could also be used, but there are limits, due to the maximum chamber pressure, maximum burn rate, and the speed of sound inside the combustion gas. Also heat and barrel wear would be increased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why do guns rarely have barrel lengths longer than say.. 24in? The reason is that there is are dramatically diminishing returns as the barrel length extends beyond 8 inches or so. If you check out the link Chris posted, you'll see that for a .308, the difference in velocity between a 16in and 20in barrel is only about 6%&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why don't most guns have stubby barrels?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Legality. In the US rifles must have a barrel longer than 16in, or the become classified as an evil &quot;short barrel rifle&quot;. These require a lot of paperwork to legally obtain.&lt;/li&gt;&#xA;&lt;li&gt;On a rifle, a barrel shorter than 12in may not have any advantage, since some space will required for a shooter to grip the rifle anyway.&lt;/li&gt;&#xA;&lt;li&gt;A longer barrel can add weight, and act as a stabilizer. Thereby increasing accuracy, even if the ballistics are not improved.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Based on the above, is fairly clear why most guns are the length they are. Rifles will usually be 12-16in, going up to 20 in extreme cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pistols have some other considerations. 1. They have to be light, so the chamber pressure is limited due to material strength. And 2. The barrel can't practically be longer than 6in or so. To get a good velocity out of a short, low pressure barrel, pistols use larger diameter bullets. These have more area for gas to push on so they accelerate over a shorter distance. This is ballistically disadvantageous, but it's not really an issue because pistols are designed for close range anyway.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-09-04T01:30:05.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29952" PostTypeId="2" ParentId="29857" CreationDate="2019-09-04T01:50:23.423" Score="0" Body="&lt;p&gt;One method which may work depending on what you're measuring, is to measure mulitples of a thing instead of just 1. For example measuring the thickness of a piece of paper by measuring 100 pieces and dividing the measurement by 100. If done correctly, this effectively makes your tool 100 times more accurate. Also works well when weighing small objects.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-09-04T01:50:23.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29953" PostTypeId="2" ParentId="29940" CreationDate="2019-09-04T01:58:02.853" Score="0" Body="&lt;p&gt;You could ballpark this in about 5 minutes by downloading an accelerometer logging app on your phone and duct taping it to various parts of your body.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-09-04T01:58:02.853" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29954" PostTypeId="1" CreationDate="2019-09-04T02:41:17.060" Score="0" ViewCount="72" Body="&lt;p&gt;I am struggling with a curve fitting calculation below. the part I am stuck with is arguably the simplest part, which is determining the coefficients &lt;span class=&quot;math-container&quot;&gt;$\\K_3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\\K_4$&lt;/span&gt;. It seems like it should be simple/straightforward but it's not clear cut at this stage. some clarity would be appreciated on what I am missing in terms of defining the coefficients of the equation assuming that &lt;span class=&quot;math-container&quot;&gt;$\\p = 5$&lt;/span&gt; just to keep it simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rbLLo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rbLLo.png&quot; alt=&quot;Formulation for calculating saturation curve part 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to calculate &lt;span class=&quot;math-container&quot;&gt;$\\I_s$&lt;/span&gt; using equation &lt;strong&gt;(II.10)&lt;/strong&gt; assuming that &lt;span class=&quot;math-container&quot;&gt;$\\I_h$&lt;/span&gt; = 0 and &lt;span class=&quot;math-container&quot;&gt;$\\I_s = I_0$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;if &lt;span class=&quot;math-container&quot;&gt;$\lambda\ = \lambda\ _m $&lt;/span&gt;  &lt;span class=&quot;math-container&quot;&gt;$\sin(\omega t)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\lambda\ ^5\ = \lambda\ _m^{5} $&lt;/span&gt;  &lt;span class=&quot;math-container&quot;&gt;$\sin^{5}(\omega t)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using the trigonometric power formula: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kCAli.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kCAli.png&quot; alt=&quot;Trigonometric power formula&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\lambda\ ^5\ = \lambda\ _m^{5} $&lt;/span&gt;  &lt;span class=&quot;math-container&quot;&gt;$\sin^{5}(\omega t)$&lt;/span&gt; = &#xA;&lt;span class=&quot;math-container&quot;&gt;$\dfrac{1}{16}(10\lambda_m^{5}sin^5(\omega t)-5\lambda_m^{5}sin(3wt)+\lambda_m^{5}sin(5wt))$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/auqDx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/auqDx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;note: I used wolfram alpha to complete this formula and couldn't figure out how to use subscript, for this equation&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$\lambda\ = \lambda\ _m $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could use some help with determining the coefficients &lt;span class=&quot;math-container&quot;&gt;$\\K_3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\\K_4$&lt;/span&gt;&#xA;what am I missing ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/auqDx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/auqDx.png&quot; alt=&quot; Formulation for calculating saturation curve part 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21687" LastEditorUserId="21687" LastEditDate="2019-09-07T00:27:43.083" LastActivityDate="2019-09-07T00:27:43.083" Title="Help with Engineering Mathematics, Struggling to determine coefficients using methods detailed in this research paper" Tags="|electrical-engineering|modeling|mathematics|systems-engineering|systems-design|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29955" PostTypeId="1" CreationDate="2019-09-04T04:34:19.953" Score="1" ViewCount="181" Body="&lt;p&gt;I'm looking for an elegant mechanism for restricting the movement of the wheel cradle in the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wheels are intended to be able to swivel around +-5° but not more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The initial idea is to have some kind of pin and a slot in the front plane. Is there an established way of making the axis itself somehow reduce the swivel?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The axis needs to be round in shape (no hump); there'll be tightening clips around it&lt;/li&gt;&#xA;&lt;li&gt;The forces to the rotational direction are minimal, i.e. just gravity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WIMbG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WIMbG.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2023-06-16T03:07:10.940" Title="Restricting rotational movement" Tags="|structural-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29956" PostTypeId="1" CreationDate="2019-09-04T05:05:31.600" Score="0" ViewCount="156" Body="&lt;p&gt;Major questions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are 0.028 locating tolerances divided by two for left/right shift?&lt;/li&gt;&#xA;&lt;li&gt;Do overall drawing tolerances apply to a thru hole with a locating tolerance of 0.028 (max material condition)?&lt;/li&gt;&#xA;&lt;li&gt;If a part is dimensioned off a theoretical center, does the overall dimension apply to the stackup?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QHHvu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QHHvu.jpg&quot; alt=&quot;Image of drawings &amp;amp; questions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21688" LastActivityDate="2019-09-04T05:05:31.600" Title="Is my tolerance stackup correct?" Tags="|tolerance|drawings|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29958" PostTypeId="1" CreationDate="2019-09-04T08:48:47.693" Score="0" ViewCount="215" Body="&lt;p&gt;In short, I'm looking for ways to reasonably precisely machine a &quot;+&quot;-axle hole, Lego compatible sockets, Lego compatible knobs, Lego compatible round peg/axle holes, and other cutouts for Lego-compatible parts on a small/hobby scale. Preferably by removing material from existing slightly over-size knobs / slightly under-size holes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been really enjoying making lego-compatible parts, which I mostly make with a 3D printer then modify by hand to achieve the desired fit. This is cumbersome and it's very hard to get good results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My FDM 3D printer isn't proving precise enough to produce a good fit for parts straight off the print bed. While I'm working on improving that, I'm looking for ways I can more easily &quot;finish&quot; or refine prints to produce a better fit for real Lego.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;&quot;+&quot;-axle hole&lt;/strong&gt;: Most importantly I'm looking for a way to cut a Lego-compatible &quot;+&quot;-shaped axle hole, or trim/expand an existing hole to improve fit. I just can't figure out a sensible way to cut this in a sufficiently uniform, reliable manner. Filing it out is just not an option. Tips for making helpful jigs or custom tools would be more than welcome.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Lego knob&lt;/strong&gt;: Knobs to go into Lego sockets are close to 4.9mm in diameter. I can not for the life of me figure out a good way to cut a nice accurate and smooth sided/topped 4.9mm knob from a slightly over-size knob printed in plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ideal results the knob should actually be slightly non-round (by a something like 100μm) so it deforms slightly into place, but I can't see that being practical to do at home. A nice round smooth fit would be a good start.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;socket to accept a Lego knob&lt;/strong&gt;: I haven't found any sensible way to do this. The knob diameter is 4.9mm. Most places that accept knobs aren't a round socket, they're usually 4 pegs, but round sockets a bit over 4.9mm are also used. A 5mm metric drill bit is too large and produces a loose, sloppy fit.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;smooth bore round axle hole&lt;/strong&gt;: A 3/16 inch imperial drill bit (4.76 mm) seems to be about perfect for a Lego round axle/peg shaft. So that much is working well, I can just print slightly undersized with thick walls, then drill it out and get a great result.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that there's a &lt;a href=&quot;https://www.cailliau.org/Alphabetical/L/Lego/Dimensions/More%20Dimensions/BBEditPreviewTemp.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;disturbingly thorough workup on Lego dimensions available here&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="21690" LastEditorUserId="21690" LastEditDate="2019-09-04T09:31:17.650" LastActivityDate="2019-09-04T09:31:17.650" Title="Cutting lego &quot;+&quot; cross-axle holes, lego knobs?" Tags="|machining|plastic|3d-printing|cutting|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="29960" PostTypeId="2" ParentId="29931" CreationDate="2019-09-04T09:46:03.180" Score="0" Body="&lt;p&gt;It is almost &lt;strong&gt;impossible to accurately know the value of charge stored in a battery&lt;/strong&gt;. The amount of charge a battery can hold varies with battery life, battery use, charging method and environmental issues.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Niel's&lt;/strong&gt; explains nicely how to know if the battery is fully charged (with liquid electrolyte filled batteries) and &lt;strong&gt;Tim's&lt;/strong&gt; tracking of the battery charge state by watching the battery voltage, and the current, likewise.&lt;/p&gt;&#xA;&lt;p&gt;However, both are &lt;strong&gt;only indicators of how much charge is in the battery. Not of &amp;quot;How much charge the battery can hold&amp;quot;&lt;/strong&gt; and, more to the point how long it will take to discharge.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Battery Lifetime &lt;a href=&quot;https://www.pveducation.org/pvcdrom/lead-acid-batteries/characteristics-of-lead-acid-batteries&quot; rel=&quot;nofollow noreferrer&quot;&gt;Characteristics of Lead Acid Batteries&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Over time, battery capacity degrades due to sulfation of the battery and shedding of active material. The degradation of battery capacity depends most strongly on the interrelationship between the following parameters:&lt;/p&gt;&#xA;&lt;p&gt;the charging/discharging regime which the battery has experienced&#xA;the DOD of the battery over its life&#xA;its exposure to prolonged periods of low discharge&#xA;the average temperature of the battery over its lifetime&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Measuring voltage is pointless as this will decrease as soon as load is applied. Even a nearly useless lead acid battery can, under certain circumstances, indicate a high voltage.&lt;/p&gt;&#xA;" OwnerUserId="21597" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-09-04T11:36:02.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29961" PostTypeId="1" AcceptedAnswerId="29964" CreationDate="2019-09-04T09:55:21.567" Score="0" ViewCount="43" Body="&lt;p&gt;A room 10&quot;x10&quot;x17&quot; lxbxh respectively to be heated 397F if a heater is placed inside the room (2 kw) how many heater are to be required to heat the room from room temperature to 397 F in 1 hour.&lt;/p&gt;&#xA;" OwnerUserId="21691" LastActivityDate="2019-09-04T15:48:13.373" Title="How many heaters needed" Tags="|thermodynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29962" PostTypeId="1" CreationDate="2019-09-04T12:14:58.990" Score="0" ViewCount="506" Body="&lt;p&gt;I'm designing a machine that I need to spin a horizontal disk at low RPM using a motor and gearbox. I'm not good at physics so I need help to find a motor/gearbox to fulfill my needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a horizontal disk, a shaft is connected at the middle of the disk, connected to the gearbox on top of disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are wheels under the disk to lower the friction and handle weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of the disk, there are meats.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the disk there is a blade. When the disk spins, cutted meats comes out of disk from bellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Disk and shaft's weight is 5kg, meat is 25kg (max), motor and gearbox are secured to the machine, Disk's friction to the machine is negligible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Required gearbox speed is 30 RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I decide which motor/gearbox is good for this project? I know the speed (30 RMP), how to calculate the needed Torque?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="21692" LastEditorUserId="21692" LastEditDate="2019-09-04T12:36:12.747" LastActivityDate="2024-01-12T06:05:45.750" Title="Calculating needed motor to spin a disk at low RPM" Tags="|mechanical-engineering|motors|gears|torque|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="29963" PostTypeId="2" ParentId="29962" CreationDate="2019-09-04T15:21:30.100" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6S4cA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6S4cA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Base on your description, is tis the system you are talking about?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First thing: Having just a single blade is just a waste, I recommend having a few more blades on it (more &quot;cutting&quot; per revolutions of the disc, or more &quot;cutting&quot; in the same time frame)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you need to find the first thing: cutting force, for example, based on experience, experiments or simply a value out of nowhere (it is acceptable for freshman projects, as long as it &quot;feels right&quot;). For example, you know that you can cut a slash of beef with a force of 100N. Due to your design, the blade is 0.1m from the center of the rotation (or from the centre of the shaft). Therefore, the torque is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T=F*r=100*0.1=10 Nm$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting 30 RPM into rad per second (basic unit), we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega = \pi  rad/s$&lt;/span&gt;, or approximatly 3.14159 rad/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power is equal to the product of torque and angular velocity, or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P=\omega*T = 3.14159 * 10 = 31.4159W$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the power output of the motor (with the gearbox included). The mechanical efficiency of the gearbox should be high, say 95%, which means the actual output of the motor (before the gearbox) is 33.069W, or just 33W because I hate long decimals. Continue to extrapolate the input of the motor, again we assume a 95% efficiency, leads to a 34.8W motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have it, a motor of 33W is &quot;just enough&quot; to cut the meat as you want. Now, you have to look for the technical data online, seeing which is the best motor for you (power is the closest match, range, size, weight...). You check for its speed limit, say it delivers 35W at 1600RPM. Now, you want to have an actual output of 30RPM (after gearbox). Therefore, the ratio must be&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$i=1600/30=53.33$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for a summary, with a crapload of assumption, you need a ~35W motor and a gearbox with 53.33 ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, a 53.33 decreasing ratio means that gonna be huge! Still, I make a lot of assumption here, which means &quot;making an &lt;strong&gt;ass&lt;/strong&gt; out of &lt;strong&gt;u&lt;/strong&gt; and &lt;strong&gt;me&lt;/strong&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main idea is just pure Math (secondary school math), so it should not be too hard... Hope it helps&lt;/p&gt;&#xA;" OwnerUserId="21520" LastActivityDate="2019-09-04T15:21:30.100" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29964" PostTypeId="2" ParentId="29961" CreationDate="2019-09-04T15:48:13.373" Score="1" Body="&lt;p&gt;You miss a piece of key information: the &quot;beginning&quot; temperature of the room. Oh, and for the love of God, what is wrong with SI units that you Americans never use them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Converting units&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;10&quot;=0.254m&#xA;17&quot;=0.4318m&#xA;397F=202.778C=475.778K&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Ok, from converting units makes me wonder the sanity of yours and your teacher. But considering that this is an Engineering SE, and we Engineers are never known to be sane in the first place...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption 1: The heater is just a single point - A room barely reaches 0.5m (just around your thigh level) will not be enough to put any regular heater in. Therefore, just assume this for a calculation exercise like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption 2: Humidity is ignore&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption 3: Ideal gas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption 4: Begin temperature as normal, or 25C, 298K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature change is&#xA;&lt;span class=&quot;math-container&quot;&gt;$\Delta T=475.778-298=177.778K$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat capacity of air: according to &lt;a href=&quot;https://www.ohio.edu/mechanical/thermo/property_tables/air/air_cp_cv.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt;, the heat capacity changes according to temperature. The same goes for density from &lt;a href=&quot;https://en.wikipedia.org/wiki/Density_of_air&quot; rel=&quot;nofollow noreferrer&quot;&gt;a reference source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, writing them in full would be long and tidious (and I doubt such is the case for a simple question like this), so just assume the 2 values would not be changed, therefore, &lt;span class=&quot;math-container&quot;&gt;$c=1.003$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\rho =1.225$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the air is&#xA;&lt;span class=&quot;math-container&quot;&gt;$m=\rho * V = 1.225*0.254*0.254*0.4318=0.034126kg$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total heat energy required (assumption number 6: No heat loss) is&#xA;&lt;span class=&quot;math-container&quot;&gt;$Q=m*c*\Delta T=0.034126*1.003*177.778=6.085kJ$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For one hour heating, the net heat output power is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P=Q/t=6.085/3600=1.69*10^{-3}kW=1.63W$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, this is significantly smaller than the heat output of 2kW of the heater. Something is wrong, either with my series of assumption, or with the way this question is given.&lt;/p&gt;&#xA;" OwnerUserId="21520" LastActivityDate="2019-09-04T15:48:13.373" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29966" PostTypeId="1" CreationDate="2019-09-04T16:17:14.143" Score="1" ViewCount="180" Body="&lt;p&gt;I'm trying to do dynamic analysis of an electric motor using Ansys Mechanical APDL.  The stator has 4 pole pairs and only 1 of them has windings. The rotor has 3 pole pairs with no windings. I want to calculate torque for different rotor positions [limited range of rotations] using a dowhile loop so i won't have to rotate the rotor myself [if i do so, it'll be a static analysis].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summery in prep7 section:&#xA;I have defined the materials I need. Then I have defined the shapes of my windings, rotor and stator and airgaps, and finally assigned materials and meshed everything for further steps.&#xA;my problem is within Dowhile Loop I believe because The command formats are correct as far as I know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, My University professor has debugged my original code and sent me this one and said that it work well on Ansys 9 installed on Windows XP. I use Ansys 15 and windows 10. Just in case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my source code. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;/prep7&#xA;&#xA;emunit,mks&#xA;csys,1&#xA;&#xA;et,1,plane53,0,,0&#xA;et,2,plane53,3,,0         !Coils&#xA;et,3,circu124,0           !External Resistor&#xA;et,4,circu124,4,0         !Independent Voltage Source, Piecewise Linear Load&#xA;et,5,circu124,5           !Stranded Coil (To FEA Domain)&#xA;&#xA;/pnum,area,1&#xA;&#xA;N,1                   ! CREATE NODES FOR CIRCUIT ELEMENTS&#xA;*REPEAT,7,1&#xA;&#xA;&#xA;!Defining Materials &#xA;   mp,murx,1,1             !air and rotor shaft&#xA;   TB,BH,2,,15             !stator and rotor iron&#xA;      tbpt,,146,0.5&#xA;      tbpt,,157,0.6&#xA;      tbpt,,168,0.7&#xA;      tbpt,,180,0.8&#xA;      tbpt,,194,0.9&#xA;      tbpt,,212,1&#xA;      tbpt,,234,1.1&#xA;      tbpt,,269,1.2&#xA;      tbpt,,329,1.3&#xA;      tbpt,,453,1.4&#xA;      tbpt,,797,1.5&#xA;      tbpt,,1895,1.6&#xA;      tbpt,,4476,1.7&#xA;      tbpt,,8595,1.8&#xA;      tbpt,,14510,1.9&#xA;   MP,murx,3,1             !Coil&#xA;   MP,RSVX,3,1e-18         !Resistivity of Coil&#xA;&#xA;R,3,0.7                   !Resistance Value&#xA;R,5,1                     !Symmetry Factor for Coil&#xA;&#xA;Vdc=100&#xA;Omega=1500                !RPM&#xA;turnonang=10&#xA;turnoffang=20&#xA;aa=0.5                    !The Factor Exhibiting the Fraction of 1 Degree for Rotation&#xA;&#xA;dt=1/(6*Omega)            !Time for 1 Degree Rotation&#xA;steptime=1000&#xA;R,4,Vdc,0&#xA;dt=aa*dt&#xA;timeindex=1e-9           !The Zero Moment&#xA;&#xA;*DIM,current_vector1,array,steptime+1,,,&#xA;*DIM,torque_vector1,array,steptime+1,,,&#xA;R,1,0.01*0.02,0.5*180,120e-3,-1,,&#xA;R,2,0.01*0.02,0.5*180,120e-3,1,,&#xA;&#xA;*afun,deg&#xA;&#xA;Rso=0.095&#xA;Rsy=0.08&#xA;Rsi=0.0482&#xA;Rro=0.0478&#xA;Rry=0.035&#xA;Rri=0.02&#xA;pi=acos(-1)&#xA;bethas=20&#xA;bethar=22.5&#xA;airgap=Rsi-Rro&#xA;&#xA;&#xA;!Defining stator and rotor keypoints&#xA;   k,111,Rsy,acos(Rsi*cos((pi/2)+(bethas/2))/Rsy),0               !stator pole keypoints &#xA;   k,112,Rsy,acos(Rsi*cos((pi/2)-(bethas/2))/Rsy),0&#xA;   k,113,Rsi,90+10,0&#xA;   k,114,Rsi,90-10,0&#xA;   k,115,Rro,90+(bethar/2),0                                        !rotor pole keypoints &#xA;   k,116,Rro,90-(bethar/2),0&#xA;   k,117,Rry,acos(Rro*cos((pi/2)+(bethar/2))/Rry),0&#xA;   k,118,Rry,acos(Rro*cos((pi/2)-(bethar/2))/Rry),0&#xA;   k,100,0,0&#xA;&#xA;&#xA;!Defining stator and rotor pole lines&#xA;   csys,0&#xA;   larc,111,112,100,rsy&#xA;   L,111,113                     !stator pole lines&#xA;   larc,113,114,100,Rsi&#xA;   L,112,114&#xA;   larc,115,116,100,Rro         !rotor pole lines&#xA;   L,115,117&#xA;   larc,117,118,100,Rry&#xA;   L,116,118&#xA;&#xA;&#xA;!defining Stator and rotor pole areas&#xA;   a,111,112,114,113           !stator pole area&#xA;   csys,1&#xA;   agen,8,1,,,,45,,&#xA;   pcirc,rso,rsy&#xA;   aadd,1,2,3,4,5,6,7,8,9&#xA;   a,115,116,118,117           !rotor pole area&#xA;   agen,6,1,,,,60,,&#xA;   pcirc,Rry,Rri&#xA;   aadd,1,2,3,4,5,6,7&#xA;&#xA;&#xA;!Defining winding areas&#xA;   rectng,-0.02,-0.01,0.049,0.069              !winding rectng&#xA;   rectng,-0.02,-0.01,-0.049,-0.069&#xA;   rectng,0.02,0.01,-0.049,-0.069&#xA;   rectng,0.02,0.01,0.049,0.069&#xA;&#xA;&#xA;!creating shaft and airgap areas &#xA;   pcirc,0,Rro&#xA;   pcirc,0,(Rro+Rsi)/2&#xA;   aovlap,8,5,6&#xA;   pcirc,(Rro+Rsi)/2,Rso&#xA;   pcirc,Rsi,Rso&#xA;asel,s,area,,1,6&#xA;asel,a,area,,10&#xA;aovlap,all&#xA;&#xA;NUMCMP,AREA             !Compress out unused area numbers&#xA;ALLSEL,ALL&#xA;&#xA;                        !coil areas 1,2,3,4&#xA;                        !shaft area 10&#xA;                        !stator area 23&#xA;                        !rotor area 14&#xA;                        !Outter airgap areas 6,15,16,17,18,19,20,21 ~ 22&#xA;                        !Inner airgap area 5,7,8,9,11,12 ~ 13&#xA;&#xA;!Assigning materials to stator, rotor, coil, shaft and air&#xA;   ASEL,S,area,,5,13      !assigning shaft and air materials&#xA;   ASEL,A,area,,15,22       &#xA;   AATT,1,1,1,&#xA;   ASEL,S,area,,23        !assigning stator and rotor material&#xA;   ASEL,A,area,,14&#xA;   AATT,2,1,1,&#xA;   ASEL,S,area,,1,2       !assigning coil materials&#xA;   AATT,3,1,2,&#xA;   ASEL,S,area,,3,4       !assigning coil materials&#xA;   AATT,3,2,2,&#xA;&#xA;&#xA;/PNUM,MAT,1&#xA;ALLSEL,ALL&#xA;APLOT&#xA;&#xA;!Meshing different Areas&#xA;   MSHAPE,1,2D   !Meshing of Inner Segments Included Shaft, Rotor, and Inner Part of Airgap&#xA;   SMRTSIZE,2&#xA;   ASEL,S,area,,7,14&#xA;   ASEL,A,area,,5&#xA;   AMESH,ALL&#xA;&#xA;   MSHAPE,1,2D     !Meshing of Outer Segments Included Stator and Outer Part of Airgap&#xA;   SMRTSIZE,3&#xA;   ASEL,S,area,,15,23,1&#xA;   ASEL,A,area,,1,4&#xA;   ASEL,A,area,,6&#xA;   AMESH,ALL&#xA;&#xA;!Defining Nodes Inside Coil Areas&#xA;   CSYS,0&#xA;!   N,1000001,-0.015,0.059,0&#xA;!   N,1000002,-0.015,-0.059,0&#xA;!   N,1000003,0.015,0.059,0&#xA;!   N,1000004,0.015,-0.059,0&#xA;&#xA;n1=node(-0.015,0.059,0)&#xA;n2=node(0.015,0.059,0)&#xA;n3=node(-0.015,-0.059,0)&#xA;n4=node(0.015,-0.059,0)&#xA;&#xA;ASEL,S,AREA,,14&#xA;ESLA,S&#xA;CM,ROTOR,ELEM&#xA;FMAGBC,'rotor'&#xA;&#xA;&#xA;!Rortating Inner Segments&#xA;   CSYS,1&#xA;   ASEL,S,area,,7,14&#xA;   ASEL,A,area,,5&#xA;   AGEN,,ALL,,,0,30-turnonang,0,,,1&#xA;&#xA;!Generating Constraint Equation at Rotor and Stator Interface&#xA;   ASEL,S,area,,13&#xA;   NSLA,S&#xA;   NSLA,S,1&#xA;   NSEL,R,LOC,X,Rro+0.25*airgap,Rro+0.75*airgap&#xA;   ASEL,S,area,,22&#xA;   ESLA,A,S&#xA;   CEINFT,0.25,,,,,,,0.05&#xA;&#xA;!Designing the external circuit&#xA;   Type,4&#xA;   Real,4&#xA;   E,2,1,3&#xA;&#xA;   Type,3&#xA;   Real,3&#xA;   E,2,4&#xA;&#xA;   Type,5&#xA;   Real,5&#xA;   E,4,5,n1&#xA;&#xA;   Type,5&#xA;   Real,5&#xA;   E,5,6,n2&#xA;&#xA;   Type,5&#xA;   Real,5&#xA;   E,6,7,n3&#xA;&#xA;   Type,5&#xA;   Real,5&#xA;   E,7,1,n4&#xA;&#xA;   ASEL,S,area,,1,&#xA;   ESLA,S&#xA;   NSLE,S&#xA;   CP,1,CURR,ALL&#xA;   CP,2,EMF,ALL&#xA;&#xA;   ASEL,S,area,,2,&#xA;   ESLA,S&#xA;   NSLE,S&#xA;   CP,3,CURR,ALL&#xA;   CP,4,EMF,ALL&#xA;&#xA;   ASEL,S,area,,3,&#xA;   ESLA,S&#xA;   NSLE,S&#xA;   CP,5,CURR,ALL&#xA;   CP,6,EMF,ALL&#xA;&#xA;   ASEL,S,area,,4,&#xA;   ESLA,S&#xA;   NSLE,S&#xA;   CP,7,CURR,ALL&#xA;   CP,8,EMF,ALL&#xA;&#xA;FINISH&#xA;&#xA;&#xA;/SOLU&#xA;   ANTYPE,4,new&#xA;   Time,1e-9&#xA;   CSYS,1&#xA;   LSEL,S,LOC,X,Rso&#xA;   DL,ALL,,AZ,0&#xA;   D,1,VOLT,0&#xA;   CNVTOL,CSG,,0.0000001,2,,&#xA;   ALLSEL,ALL&#xA;   SOLVE&#xA;&#xA;&#xA;cont=1&#xA;/post26&#xA;   NSOL,2,n1,CURR&#xA;   PRVAR,2&#xA;   *GET,ICUR1,VARI,2,rtime,timeindex&#xA;   current_vector1(cont)=ICUR1&#xA;   ttt=turnonang&#xA;   torque_vector1(1)=1e-10&#xA;   ICUR1=1&#xA;   steps=1&#xA;&#xA;   *Dowhile,ICUR1&#xA;      /Prep7&#xA;         *if,ttt,gt,turnoffang,then&#xA;            R,4,-Vdc&#xA;         *endif&#xA;&#xA;         ALLSEL,ALL&#xA;         CEDELE,1,1111111,,ANY       !Deletes Contraint Equations&#xA;&#xA;         ASEL,S,area,,7,14&#xA;         ASEL,A,area,,5&#xA;         AGEN,,ALL,,,0,-aa,0,,,1&#xA;&#xA;         ASEL,S,area,,13      !Generating Constraint Equation at Rotor and Stator Interface&#xA;         NSLA,S&#xA;         NSLA,S,1&#xA;         NSEL,R,LOC,X,Rro+0.25*airgap,Rro+0.75*airgap&#xA;         ASEL,S,area,,22&#xA;         ESLA,A,S&#xA;         CEINFT,0.25,,,,,,,0.05&#xA;&#xA;         ttt=ttt+aa&#xA;         cont=cont+1&#xA;         timeindex=timeindex+dt&#xA;         steps=steps+1&#xA;&#xA;      /SOLU&#xA;         ANTYPE,4,REST,steps-1,,1&#xA;         TRNOPT,full&#xA;         TIME,timeindex&#xA;         TIMINT,ON&#xA;         AUTOTS,off&#xA;         DELTIM,,,,1&#xA;         KBC,1&#xA;         TSRES,ERASE&#xA;         OUTRES,ALL,LAST,&#xA;         NEQIT,100&#xA;         CNVTOL,CSG,,0.0000001,2,,&#xA;         ALLSEL,ALL&#xA;         SOLVE&#xA;&#xA;      /POST1&#xA;         TORQSUM,'rotor'&#xA;         torque_vector1(cont)=-0.0873*torqvw&#xA;&#xA;      /POST26&#xA;         NSOL,2,n1,CURR&#xA;         PRVAR,2&#xA;         *GET,iCUR1,VARI,2,rtime,timeindex&#xA;         current_vector1(cont)=iCUR1&#xA;         ALLSEL,ALL&#xA;   *ENDDO&#xA;&#xA;   torque_vector1(cont,1,1)=1e-9&#xA;   *vlen,cont&#xA;   *DIM,rotor_position,table,cont,1,1&#xA;   *VFILL,rotor_position,ramp,turnonang,aa&#xA;   *DIM,current_vector,table,cont,1,1&#xA;   *DIM,flux_vector,table,cont,1,1&#xA;   *DIM,torque_vector,table,cont,1,1&#xA;&#xA;   *do,i,1,cont&#xA;      current_vector(i,1,1)=current_vector1(i,1,1)&#xA;      torque_vector(i,1,1)=torque_vector1(i,1,1)&#xA;   *enddo&#xA;&#xA;   *VPLOT,rotor_position,current_vector&#xA;&lt;span class=&quot;math-container&quot;&gt;```&lt;/span&gt;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="21694" LastActivityDate="2019-09-04T16:17:14.143" Title="Please help me with 2-D Dynamic analysis in ANSYS Mechanical APDL" Tags="|electrical-engineering|ansys-apdl|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29967" PostTypeId="1" CreationDate="2019-09-04T17:05:29.023" Score="0" ViewCount="78" Body="&lt;p&gt;What is the type of control system of door closer in attached photo? &lt;img src=&quot;https://i.stack.imgur.com/t9iNp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="22801" LastActivityDate="2020-02-02T14:23:48.213" Title="Identification of control system type? Open loop or close loop," Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29968" PostTypeId="2" ParentId="29967" CreationDate="2019-09-04T17:24:38.253" Score="0" Body="&lt;p&gt;As far as I understand, the device is not measuring anything. I think the door closer is an open-loop system that either always applies pressure on the door or applies more pressure on the door as you open the door and the natural steady-state position is door-closed position. neither of these systems doesn't use any measuring device to get feedback of the output.&lt;/p&gt;&#xA;" OwnerUserId="21694" LastActivityDate="2019-09-04T17:24:38.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29969" PostTypeId="1" CreationDate="2019-09-04T17:42:26.290" Score="2" ViewCount="3032" Body="&lt;p&gt;The problem of energy storage, especially with regard to renewables, remains unsolved. Or at least there do not seem to be any popular solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why not &lt;a href=&quot;https://en.wikipedia.org/wiki/Gravity_battery&quot; rel=&quot;nofollow noreferrer&quot;&gt;gravity batteries&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I love the idea, and they look like a beautiful and effective approach to the problem. The low-tech, simple principles grant it so many different region-taylored forms. Many of these designs could easily be low-cost. I could go on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there some shortcomings that I am missing here?&lt;br&gt;&#xA;I asked this &lt;a href=&quot;https://physics.stackexchange.com/questions/500115/why-arent-gravity-batteries-everywhere&quot;&gt;same question&lt;/a&gt; to the physics stack exchange, but I don't feel satisfied as to why they aren't more commonplace in renewable energy storage applications.&lt;/p&gt;&#xA;" OwnerUserId="21696" LastActivityDate="2019-10-09T14:54:35.017" Title="Why aren't gravity batteries everywhere?" Tags="|battery|renewable-energy|energy-storage|electrical-grid|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29970" PostTypeId="2" ParentId="29955" CreationDate="2019-09-04T17:42:37.037" Score="1" Body="&lt;p&gt;Ended up trying this: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8rs1s.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8rs1s.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pros:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;does not need extra pins / does not change looks&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Cons:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;may be difficult to guarantee the groove is always straight down (if the pin is screwed in)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Other suggestions are welcome and considered. :)&lt;/p&gt;&#xA;" OwnerUserId="12827" LastActivityDate="2019-09-04T17:42:37.037" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29971" PostTypeId="1" CreationDate="2019-09-04T17:46:46.427" Score="1" ViewCount="106" Body="&lt;p&gt;This steel is often used as a floor plate.  I could not find the Brinell hardness rating.&lt;/p&gt;&#xA;" OwnerUserId="21697" LastActivityDate="2023-03-22T17:03:45.433" Title="What is the Brinell hardness of ATSM A786 steel?" Tags="|steel|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29972" PostTypeId="2" ParentId="29971" CreationDate="2019-09-04T19:33:39.440" Score="1" Body="&lt;p&gt;I don't think there will be just one.  From &lt;a href=&quot;https://www.astm.org/Standards/A786.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;ASTM's web page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This specification covers carbon, low-alloy, high-strength low-alloy, and alloy steel hot-rolled floor plates for flooring, stairways, transportation equipment, and general structural purposes. Steel materials shall undergo heat and product analysis. Steel specimens shall also undergo tensile tests and shall conform to the required values of yield point, yield strength, and tensile strength.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It sounds like the specification is for a family of products, not for one specific  steel.  So &lt;em&gt;your&lt;/em&gt; ASTM A786 plate may be different from &lt;em&gt;mine&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-04T19:33:39.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29973" PostTypeId="2" ParentId="29969" CreationDate="2019-09-04T19:49:55.917" Score="6" Body="&lt;p&gt;So, &lt;a href=&quot;https://www.amainhobbies.com/gens-ace-3s-60c-lipo-battery-pack-w-ec5-connector-11.1v-5500mah-ga-b1134/p1001154&quot; rel=&quot;noreferrer&quot;&gt;this battery&lt;/a&gt; costs US$70, weighs 406 grams, and holds around 200,000 Joules of energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get the same energy from that weight as a &quot;gravity battery&quot; you'd have to pick it up over 49 &lt;strong&gt;kilo&lt;/strong&gt;meters into the air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There &lt;em&gt;are&lt;/em&gt; applications where picking stuff up and powering things as it falls is useful -- the gravity-powered lighting for 3rd-world countries is a great example, and there are some hydropower storage schemes that pump water uphill to a lake and then generate power as it drops.  But if you just take the trouble to do the math you'll see that &quot;big&quot; and &quot;unwieldy&quot; are definitely words that go with &quot;gravity energy storage&quot;.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-04T19:49:55.917" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29974" PostTypeId="1" AcceptedAnswerId="29991" CreationDate="2019-09-04T20:43:07.697" Score="1" ViewCount="249" Body="&lt;p&gt;This question, in short, is: Can a gas be compressed with high pressure liquid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a scenario, imagine a tank full of atmospheric gas at STP. Could something like a pressure washer pump (in this case water, but generally any) liquid into the tank thereby compressing the gas inside the tank? Pressure washers often reach pressures that are quite high (easily 2000+ psi depending.) This seems like a cheap (if inefficient) way to get high pressure gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to find any material on this idea, and I find it highly unlikely that I am the first person to think this. What this indicates to me is that this is NOT a good idea but the reason is not obvious to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the reason happens to do with the choice of gas/liquid, would this be solved with different materials? For instance, if the water is a problem for some reason could something like ln2 be used to compress nitrogen gas in some sort of specialized pump that could handle those type of temperatures?&lt;/p&gt;&#xA;" OwnerUserId="21698" LastActivityDate="2019-09-05T13:40:22.643" Title="Compressing gas with liquid" Tags="|compressed-air|compressors|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29975" PostTypeId="1" AcceptedAnswerId="30007" CreationDate="2019-09-04T20:46:43.340" Score="3" ViewCount="192" Body="&lt;p&gt;I have a pellet smoker, and I am customizing it for tighter temperature control. Currently, I am controlling how much material is burning to control temperature using a PID loop. I want to achieve tighter control by also slowing down or speeding up the fan in response to temperature changes. I'm looking to find an algorithm that would allow these two to work together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pellet smoker works by using an augur to feed wood pellets into a burn chamber. Temperature is controlled by turning the augur on or off to control the amount of material that is burning in the burn chamber, and varying the speed of a fan to control how much air is flowing to the burn chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jqFtA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jqFtA.png&quot; alt=&quot;Traeger Pellet Smoker example&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I am controlling temperature with a PID loop controlling only the augur motor. The fan is running at a fixed speed of 90% of its maximum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This leads to an acceptable temperature range, but I would like to achieve tighter temperature control by also using a PID loop on the fan speed rather than running it at a fixed speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can control the fan with a completely independent PID loop. But I was hoping someone can point me towards and algorithm that would get them working together since they are pretty tightly coupled, and the fan influences temperature faster, but also doesn't work at all if the augur isn't feeding enough material.&lt;/p&gt;&#xA;" OwnerUserId="21699" LastActivityDate="2019-09-06T15:38:07.767" Title="Controlling temperature using PIDs on two separate controllers" Tags="|control-engineering|pid-control|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29976" PostTypeId="1" AcceptedAnswerId="29983" CreationDate="2019-09-04T20:49:12.270" Score="0" ViewCount="67" Body="&lt;p&gt;A client has asked whether a container can be placed on an existing unreinforced concrete foundation of depth 100 mm. The grade of the concrete is unknown. How can the capacity (in kPa) of this plane concrete foundation be evaluated?&lt;/p&gt;&#xA;" OwnerUserId="10270" LastActivityDate="2019-09-05T12:18:10.540" Title="Capacity of plain concrete foundation on grade [Civil UK]" Tags="|civil-engineering|foundations|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29977" PostTypeId="2" ParentId="29971" CreationDate="2019-09-04T21:42:13.910" Score="1" Body="&lt;p&gt;I don't have current ASTM specs ( they are renewed every 5 years); However , it is very possible there is no hardness specified . There are tensile and yield strength and ductility requirements ; minimums  ( usually in a spec like this, no maximums are specified). One can get a pretty good estimate of the ( minimum ) hardness from conversion charts. For example : a minimum tensile of 60 ksi corresponds to a Brinell of about 105 ; a minimum tensile of 100 ksi corresponds to a BHN of 200. The yield strengths do not correlate as well to hardness. Conversion tables are available on the net  ( which will give better numbers than I took off a small chart).&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-09-04T21:50:12.130" LastActivityDate="2019-09-04T21:50:12.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29978" PostTypeId="1" AcceptedAnswerId="29981" CreationDate="2019-09-05T00:44:32.847" Score="1" ViewCount="223" Body="&lt;p&gt;I have lubricated the ball-bearings of two hand spinners (the anti-stress toys) and contrarily to my expectation the time they keep on spinning has been divided by four. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One fact I have noticed, the rolling bearings now makes almost no more noise at all: they were really noisy before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to explain myself why/how the energy in the two hand-spinners would dissipate so much faster ? It is contrary to my intuition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea ?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When I say divided by 4, it is an estimation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have an anti-smoking anti-stress technique that involves a short meditation, and I was using two hand-spinners, one in each hand for a duration of roughy 2 minutes, a bit more, (repeated numerous time for one year or so). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But after lubricating them this afternoon with WD-40, the duration falls to 30 seconds at most.&lt;/p&gt;&#xA;" OwnerUserId="2137" LastEditorUserId="2137" LastEditDate="2019-09-05T11:14:20.987" LastActivityDate="2019-09-05T11:14:20.987" Title="How/why lubricating a rolling bearing would make them spin less time (dissipate energy faster)?" Tags="|bearings|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29980" PostTypeId="2" ParentId="29505" CreationDate="2019-09-05T02:31:46.517" Score="1" Body="&lt;p&gt;There is also another way to do this. But of course, like all engineering exercises, there is a lot of things I have to assume (read: making an &lt;strong&gt;ass&lt;/strong&gt; out of &lt;strong&gt;u&lt;/strong&gt; and &lt;strong&gt;me&lt;/strong&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So first, I assume that the situation is 1D (easier to work with), then I ignore the heat capacity of the walls (boxes and insulation layer), I also assume the shape of the box is not important, no heat is lost from the inside (as in, if heat goes in, it is in). Along the way, if I make any more assumption, I will write it out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically speaking, the flow of heat is &quot;out&quot; (from hot to cold), or in this case, the &quot;cold stuff&quot; is going in the box (and the &quot;hot stuff&quot; is going out).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat flux (ie heat loss rate) is &lt;span class=&quot;math-container&quot;&gt;$\dot Q = k*A*\Delta T$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$\dot Q$&lt;/span&gt; as heat loss rate, k is the heat transfer coefficient, and A is the area. Here, we are working with heat flux (heat loss per unit of area), so &lt;span class=&quot;math-container&quot;&gt;$\dot q=k*\Delta T$&lt;/span&gt; shall be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have&#xA;&lt;span class=&quot;math-container&quot;&gt;$\frac{1}{k}=\sum \frac{1}{\alpha} + \sum \frac{\delta}{\lambda}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With&#xA;&lt;span class=&quot;math-container&quot;&gt;$\alpha $&lt;/span&gt; as heat transfer coefficient (hot to cold place, and vice versa, W/m^2/K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; is the thickness of the material layer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; is thermal conductivity (W/m/K)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case You plug in the data for three layers of materials (2 times of the boxes, and once for the insulation layer), you would have \frac{1}{k} and then, k. You know the original temperature of (air - ?) inside the inner box, so you can calculate the &lt;span class=&quot;math-container&quot;&gt;$\Delta T$&lt;/span&gt;. From here you can calculate the heat flux &lt;span class=&quot;math-container&quot;&gt;$\dot q$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But you also ask for the time to drop by 1 degree... I'm pretty sure that this one is a long formula with logarithm, exponent and stuff like that. Think about it, the colder you get, the less difference between you and the target, making flushing the heat out a bit harder. However, being engineers, we can &quot;approximate&quot; this my taking the average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making the next assumption that the system is static and the only energy considered is thermal energy, we calculate the internal energy of the gas&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$U=\frac{3}{2}nRT$&lt;/span&gt; with n as the number of mol of air, T is the beginning temperature in Kelvin, and R is 8.31 (8.3144 if you want more decimal). Do that again for the final temperaure (T=-4C), you would have the change in internal energy &lt;span class=&quot;math-container&quot;&gt;$\Delta U$&lt;/span&gt;. Because of my assumption of averaging above (or constant heat loss rate), we can calculate the time as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$t=\Delta U/ \dot q$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Divide this by the temperature difference &lt;span class=&quot;math-container&quot;&gt;$\Delta T$&lt;/span&gt; would give you the (super) time-average needed to decrease the temperature of the system by 1C or 1K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any criticism is welcome&lt;/p&gt;&#xA;" OwnerUserId="21520" LastActivityDate="2019-09-05T02:31:46.517" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29981" PostTypeId="2" ParentId="29978" CreationDate="2019-09-05T04:21:52.900" Score="6" Body="&lt;p&gt;The lubricant you put into the spinner bearings has both viscosity and surface tension, and it coats the rotating parts inside the bearing. Therefore, as the spinner bearing revolves, it churns the lube and makes and breaks air/lube interfaces- which dissipates energy and makes the spinner slow down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This energy dissipation mode is absent in the case of a dry bearing, in which case the main energy dissipation mode will be radiating away the noise generated when the bearing parts rattle around as the bearing spins. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This illustrates the fact that in a roller- or ball-bearing, the purpose of the lubricant inside them is not to reduce friction (the rolling elements do that) but to inhibit wear caused when the balls or rollers skid against the bearing races instead of rolling smoothly.  The lubricant adds a little bit of frictional drag but the tradeoff is worth it.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-05T04:21:52.900" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29982" PostTypeId="2" ParentId="29974" CreationDate="2019-09-05T04:28:36.727" Score="3" Body="&lt;p&gt;Yes indeed, air can be compressed by pressurized liquids directly as you suggest. However, there are problems that accompany this sort of scheme. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, if the gas is significantly soluble in the liquid, then the liquid will pick up gas while being pressurized which will then come forcibly foaming out of solution when the pressure is released.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, you have to be careful to prevent any liquid from finding its way into the piping that carries the compressed gas and vice versa. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For these reasons, it is common to place a thin rubber membrane between the gas and the liquid inside the tank where the compression takes place.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13363" LastEditDate="2019-09-05T06:22:08.960" LastActivityDate="2019-09-05T06:22:08.960" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29983" PostTypeId="2" ParentId="29976" CreationDate="2019-09-05T05:36:21.957" Score="2" Body="&lt;p&gt;In order to determine the capacity, you will need to know a couple of things.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Dimensions of the pad and container&lt;/li&gt;&#xA;&lt;li&gt;Bearing capacity of the soil under the slab.  You would normally receive this from a geotechnical report.&lt;/li&gt;&#xA;&lt;li&gt;The strength of the concrete.  This can be determined either through core testing or something called a &quot;schmidt hammer&quot;. The schmidt hammer can give you a ball park idea of the strength of the concrete.  However you need to calibrate it against something of a known strength first.&lt;/li&gt;&#xA;&lt;li&gt;You will want to check the concrete pad for crushing, shear.  You local building code should provide guidance on capacity checks for both bearing and shear.&lt;/li&gt;&#xA;&lt;li&gt;You will also want to look at unbalanced loads from the container, and potentially the container not being centered on the pad.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that many parts of the world will consider plain concrete (unreinforced) to be non-structural in nature.  That does not mean that it can't carry a load, just that its relatively weak compared to the structural loading reinforced concrete tends to take.  &lt;/p&gt;&#xA;" OwnerUserId="6993" LastEditorUserId="6993" LastEditDate="2019-09-05T12:18:10.540" LastActivityDate="2019-09-05T12:18:10.540" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="29984" PostTypeId="1" AcceptedAnswerId="30010" CreationDate="2019-09-05T05:56:27.740" Score="0" ViewCount="72" Body="&lt;p&gt;During truss fabrication, the truss member are usually connected by welding does behave like fixed joints not pin joint so how will that behave like truss without pin connection?&lt;/p&gt;&#xA;" OwnerUserId="21700" LastEditorUserId="110" LastEditDate="2019-09-05T10:01:22.153" LastActivityDate="2019-09-06T18:22:38.647" Title="Practical consideration of truss" Tags="|mechanical|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="29985" PostTypeId="1" AcceptedAnswerId="29994" CreationDate="2019-09-05T09:52:38.810" Score="0" ViewCount="62" Body="&lt;p&gt;For a DC motor, speed of the armature is inversly proportional to current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at the same time, we know that torque of the armature is directly proportional to the armature current. Since speed can be increased by applying torque, this means that speed is directly proportional to armature current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible ?&#xA;Where have I gone wrong ?&lt;/p&gt;&#xA;" OwnerUserId="21544" LastActivityDate="2019-09-05T15:30:53.040" Title="Contrradicting answers for relation of speed of a motor with current" Tags="|motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29987" PostTypeId="1" CreationDate="2019-09-05T10:51:50.043" Score="2" ViewCount="72" Body="&lt;p&gt;Throughout the trailer, hitch, and tow vehicle industry a recommendation of 10% tongue weight for a tow-behind trailer and 15% for a 5th wheel trailer seem universal. They have even been changed to 10%-15% and 15%-25% recently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did they come from? How were they determined? Was it to keep the trailer on the ball in all conditions? If so, a weight-distribution hitch dramatically increases the forces on the ball but there is no recommendation for lower tongue weights with a weight distributing hitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all the GVWR, GAWR, and GCWR numbers are thrown about, the only one with a standard seems to be GCWR with the relatively recent (last 10 years) SAE J2807 standard. And that standard has been widely criticized as not accounting for basic factors such as steering ratio and wheelbase when specs talk about steering wheel angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, where did the 10% recommendation come from?&lt;/p&gt;&#xA;" OwnerUserId="21701" LastEditorDisplayName="user16" LastEditDate="2019-09-05T18:33:02.337" LastActivityDate="2019-09-05T18:33:02.337" Title="What is the Source of the 10% Trailer Tongue Weight Recommendaton?" Tags="|mechanical-engineering|automotive-engineering|applied-mechanics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="29990" PostTypeId="2" ParentId="29975" CreationDate="2019-09-05T13:02:15.733" Score="2" Body="&lt;p&gt;Ideally, I would imagine there would be sufficient material in the burner and the fan speed would be the only thing to modulate to regulate temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that's the case, then I would have the target temperature as a reference input to the fan loop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, what you're left to look at is actual fan speed. As the material in the burner is consumed, the fan will have to work harder to stoke the fire up to an acceptable temperature. Conversely, if there's too much material, then the fan doesn't need to do a whole lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, considering that, then I would set the auger controller to achieve a target fan speed. I would start testing with a target fan speed reference of 50 percent, but you'll need to test because there's going to be some variation in how the fire is (or isn't) sustained at particular fan speeds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the fan speed modulates to regulate the temperature, and the auger moves to modulate the fan speed. As the fan speed creeps above the setpoint (the fire is dying), the auger feeds more material. &lt;/p&gt;&#xA;" OwnerUserId="1633" LastActivityDate="2019-09-05T13:02:15.733" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29991" PostTypeId="2" ParentId="29974" CreationDate="2019-09-05T13:40:22.643" Score="0" Body="&lt;p&gt;This is common in any farm water system.  A pressure tank is on a side branch of the line from the well to the farm water system.  At the cut-in pressure the well pump starts, and pushes water into the tank until the system reaches the cut out pressure.  Typically the two pressures are about 20 psi apart.  There is some adjustment available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water would absorb air under pressure, so some mechanism is in place to regulate the amount of air in the tank.  This was usually some combination of a float valve that released excess air, and a bleeder valve in the top section of well pipe that led some air back in the pipe so when the pump started, a slug of air was included.  I'm sure there are other ways.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failure of either of these would result in a waterlogged tank.  The well pump would cut it, almost immediately pressureize the system, cut off, and repeat as often as several times a second.  Hard on pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More commonly now there is either a bladder or a diaphragm in the tank to separate the water and the air.  These require no additional components.  Usually the tank is pressurized to about half the max pressure of the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have two in my farm system, one at the house and one near the far end of the irrigation line 1500 feet away.  A third of a mile of moving water has considerable momentum.  My solenoid operated valves were taking a beating when turning the water off.  I was popping connections from the pressure surge.  Adding a 25 gallon pressure tank meant that the surge would just overpressurize the tank somewhat.  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Water compressing air is used in hydraulic rams.  The momentum of a pipe full of water is halted with a 'clacker' valve.  The moving water loses it's momentum rushing into a air containing side chamber.  That air pressure then is used to move a small amount of the water to a higher elevation.  A well tuned system (Air volume, large in stream pipe, and small diameter discharge pipe have approximately equal volumes) can pump water to surprising elevations.  More info here: &lt;a href=&quot;https://en.wikipedia.org/wiki/Hydraulic_ram&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Hydraulic_ram&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;water pumped air can also be used as a way to generate energy.  Consider a vessel with an open bottom and two one-way valves on top.  When a wave passes it compresses the air in the vessel.  A valve opens and that compressed air goes to a storage tank.    The wave falls, that valve shuts, and another one opens admitting air into the top of the tank.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compressed air can then be used to operate a turbine.  The efficiencies of this are low.  6 foot waves would only generate about 3 psi.  There are better ways to extract energy from waves.&lt;/p&gt;&#xA;" OwnerUserId="8768" LastActivityDate="2019-09-05T13:40:22.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29992" PostTypeId="2" ParentId="29969" CreationDate="2019-09-05T14:05:57.317" Score="5" Body="&lt;p&gt;Consider 1 kWh = 1000 W * 3600 seconds = 3,600,000 J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 kg lifted 1 m = ~10J.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's use 1 metric tonne.  1000Kg.  So we get 10,000 J per meter of elevation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3,600,000 / 10,000 = 360 meters.  Around a thousand feet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ok. Getting a 15 cm diameter well dug and cased is on the order of \$20/foot.  So \$20,000 buys you a 1000 foot deep hole.  Use whatever length of 10 cm bar stock gets you a ton.  (In actual fact, using drill collar would be a prefab way to do this) Now all you need is some long cable, and a motor/generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But really:  Add a few more chunks of drill collar increase your weight up to, say 50T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have a battery that will store 50 kWh -- about 2 days use by a modern home -- and it cost you only somewhere between 20 and 40 grand.  Ball park a thousand bucks per kWh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;50 kWh is worth about \$4 at residential prices, about \$1.50 at wholesale prices.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Now we sort of do this with pumped hydro storage.  Water is a lot easier to work with than long slugs of steel pipe suspended by cables.  It also scales better.  Raise the rim of the reservoirs a few feet, and you can store a lot of extra tonnes of water.  (A 1 km2 lake 1 meter deep is a million tonnes of water.  The cycle efficiency is only about 70% or so, depending on how much you spend on big plumbing. (You put 10 kWh in and get 7 back next Tuesday)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pumped hydro is very workable.  If you have a mountain nearby.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Big LiON batteries are now getting down between 200 and 300 bucks per kWh. (&lt;a href=&quot;https://www.nrel.gov/docs/fy19osti/71714.pdf&quot; rel=&quot;noreferrer&quot;&gt;https://www.nrel.gov/docs/fy19osti/71714.pdf&lt;/a&gt; 2018 figures) There are other technologies in the wings that should cut this by another factor of 4 for stationary batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even those mongo batteries such as Tesla made for Australia pay off only when used several cycles per day.  They have addition payoff because they can prevent firing up gas generators to fill the gap as often -- takes about half an hour to bring a gas turbine online.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;TL:DR;  Gravity batteries are too big and bulky, and are too complicated to scale up easily.&lt;/p&gt;&#xA;" OwnerUserId="8768" LastEditorUserId="13311" LastEditDate="2019-09-05T14:38:11.223" LastActivityDate="2019-09-05T14:38:11.223" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="29993" PostTypeId="2" ParentId="29987" CreationDate="2019-09-05T15:14:42.493" Score="1" Body="&lt;p&gt;If you get the CG of the trailer too far back  the combination will be unstable, with the trailer fishtailing.  This fishtailing can be to the point of causing a crash.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you put too much tongue weight on the pull vehicle it'll be too heavily loaded -- so a really light trailer pulled by a heavy truck can have a far-forward CG.  Think of those hitch-mounted bicycle carriers as a super-extreme example of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This video: &lt;a href=&quot;https://www.youtube.com/watch?v=4jk9H5AB4lM&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=4jk9H5AB4lM&lt;/a&gt; (and yes, I know, link rot, etc. -- search on &quot;trailer stability video&quot; and you'll get something).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-05T15:14:42.493" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="29994" PostTypeId="2" ParentId="29985" CreationDate="2019-09-05T15:30:53.040" Score="5" Body="&lt;p&gt;For a DC motor, speed of the armature is inversely proportional to current.&lt;/p&gt;&#xA;&lt;p&gt;But at the same time, we know that torque of the armature is directly proportional to the armature current.&lt;/p&gt;&#xA;&lt;p&gt;Since speed can be increased by applying torque, this means that speed is directly proportional to armature current.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;How is this possible ? Where have I gone wrong ?&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Where you have gone wrong 1&lt;/strong&gt;&#xA;Firstly your question is poorly defined you should clearly state what type of equipment you are referencing. There are numerous types of D.C. Motors and variations on how the excitation can be achieved. Therefore I will ignore differential compound D.C. Motors.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Where you have gone wrong 2&lt;/strong&gt; is your statement Since &lt;strong&gt;speed can be increased by applying torque&lt;/strong&gt;. &lt;a href=&quot;http://lancet.mit.edu/motors/motors3.html&quot; rel=&quot;noreferrer&quot;&gt;Torque speed curve&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;In fact &lt;strong&gt;this is the exact opposite&lt;/strong&gt; of what happens. &lt;strong&gt;As speed increases the torque reduces.&lt;/strong&gt; As the the load is reduced and torque approaches zero the armature speed reaches maximum. because &lt;strong&gt;AS THE SPEED INCREASES THE BACK EMF INCREASES AND THEREFORE THE CURRENT REDUCES.&lt;/strong&gt; so the statement &lt;strong&gt;speed of the armature is inversely proportional to current&lt;/strong&gt; is correct.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.electronics-tutorials.ws/inductor/inductor.html&quot; rel=&quot;noreferrer&quot;&gt;Tutorials&lt;/a&gt; emf = inductance x rate of current change” and a circuit has an inductance of one Henry will have an emf of one volt induced in the circuit when the current flowing through the circuit changes at a rate of one ampere per second.&lt;/p&gt;&#xA;&lt;p&gt;One important point to note about the above equation. &lt;strong&gt;It only relates the emf produced across the inductor to changes in current&lt;/strong&gt; because if the flow of inductor current is constant and not changing such as in a steady state DC current, then the induced emf voltage will be zero because the instantaneous rate of current change is zero, di/dt = 0.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore it is obvious that as the speed of the armature increases so the rate of change of flux will increase and so the back emf also increase.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;However&lt;/strong&gt; if you were to say that the &lt;strong&gt;speed of a Electric locomotive can be increased by applying torque&lt;/strong&gt; that is correct, at least in the initial stages of movement.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;Where you have gone wrong 3&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Your statement torque of the armature is directly proportional to the armature current is also incorrect &lt;a href=&quot;https://www.yourelectricalguide.com/2017/09/dc-motor-characteristics-and-applications.html&quot; rel=&quot;noreferrer&quot;&gt;electrical guide&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;In fact &lt;strong&gt;T α Iaφ&lt;/strong&gt; Torque α Armature current x Field flux and thus &lt;strong&gt;the equation T= I? will depend on what type of motor you are referencing.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;On light loads, the &lt;strong&gt;torque produced by the series motor is proportional to the square of armature current&lt;/strong&gt; and hence curve drawn between torque and armature current up to magnetic saturation is a parabola. But after magnetic saturation flux φ is independent of excitation current and so torque is proportional to Ia and hence characteristics become a straight line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Whilst&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The flux of a &lt;strong&gt;shunt motor&lt;/strong&gt; is practically constant. Therefore, &lt;strong&gt;T α Ia&lt;/strong&gt; Torque produced is proportional to the armature current&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="21597" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-09-05T15:30:53.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29995" PostTypeId="1" CreationDate="2019-09-05T18:31:23.847" Score="0" ViewCount="92" Body="&lt;p&gt;What is the difference between bearing circle diameter and flange diameter in context of prevailing torque flange nut? I see that maximum value for flange diameter and minimum value for bearing circle diameter are defined while designing. Does this mean actual flange diameter on the nut can be any value between (including) these two values? &lt;/p&gt;&#xA;" OwnerUserId="21707" LastActivityDate="2019-09-06T09:30:10.597" Title="Flange nut design" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="29998" PostTypeId="1" CreationDate="2019-09-05T21:29:16.940" Score="0" ViewCount="64" Body="&lt;p&gt;Right now I am using nine components intended for home air ionizers in parallel, I'd like to build one big one.  I've heard that if one keeps the discharge voltage at -4kV, this produces negative ionization with a minimum of ozone, but I'd like to know more than that, e.g., where that voltage arises from, whether and how it changes with temperature and/or current applied, et cetera.  And I'd like to keep it as simple as possible, though I probably can still pull out my long-ago calculus if I must.  Suggestions?&lt;/p&gt;&#xA;" OwnerUserId="21710" LastActivityDate="2019-09-20T19:28:44.570" Title="How to minimize ozone production while ionizing air?" Tags="|circuit-design|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30001" PostTypeId="2" ParentId="29969" CreationDate="2019-09-06T01:10:41.127" Score="3" Body="&lt;p&gt;Are there some shortcomings that I am missing here?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Yes there are a couple of things you have missed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly I will deviate slightly from your question but I believe this is relevant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The need for &lt;strong&gt;localised storage&lt;/strong&gt; has only relatively recently &lt;strong&gt;become desirable&lt;/strong&gt;. With the increase of &lt;strong&gt;large scale Solar&lt;/strong&gt; and in particular &lt;strong&gt;wind farm projects&lt;/strong&gt; the need for localised energy storage has become a priority. Despite what the Sales people wish to tell you or what propaganda environmental groups or Governments wish to spout &lt;strong&gt;Renewable energy&lt;/strong&gt; &lt;strong&gt;IS NOT A SUCCESS&lt;/strong&gt;, because a &lt;strong&gt;large number of fossil fuel plants have to remain running 24/7&lt;/strong&gt; to take up the load on the GRID when the wind drops or the sunshine disappears. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hydro storage solutions cannot be compared as they are an older technology originally developed to solve the problem of water consumption in areas where the terrain was ideal but the flow of water was insufficient for hydro power generation. They also seldom meet the need for localised storage where the power is wind or solar generated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to repeat localised power storage is a &lt;strong&gt;relatively new requirement. Hence the lack of solutions&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The main points you have missed are the &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Time&lt;/strong&gt; The Big one of course is time. &lt;strong&gt;Gravity batteries are time limited&lt;/strong&gt;. The length of time a gravity storage unit can resupply the grid on loss of renewable energy supply is quoted as 8-16 hours by Energy Vault. This is a relatively short period of time compared with how long the wind could crease to blow and 16 hours of relatively low sunshine in winter months is not exceptional. Therefore &lt;strong&gt;Fossil powered generation would still be needed 24/7&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Capital costs&lt;/strong&gt; associated with building both the renewable energy sources and the localised storage. Whilst running costs for both are negligible (Charging of localised storage being affected by energy that would have &quot;gone to waste&quot;). However the interest payments on loans secured to build the facilities will add to the running costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Environmental costs&lt;/strong&gt; The effect of wind farms on the scenic environment has already become the focus of fierce opposition in many locations and Gravity storage solutions will be even more detrimental to the landscape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Location&lt;/strong&gt; Again linked with environmental but also with the renewable energy sources. Energy Vault depicts a gravity storage tower in the mist of a group of wind turbines. The siting of which would play havoc with the wind required to generate the power to be stored.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reliability&lt;/strong&gt; of Power Supplies. The increase in use of renewable energy will eventually bring the whole Power grid system into a state of Chaos. Traditional generation techniques cannot match the running costs of renewable energy sources. However renewable energy sources are not reliable. So in Countries where the electricity supply is not state owned how will corporate bodies manage these discrepancies? How will fossil fuel generation remain viable whilst most days of the year consumers are buying low cost, locally generated,  renewable energy? &lt;/p&gt;&#xA;" OwnerUserId="21597" LastEditorUserId="21597" LastEditDate="2019-09-06T05:04:28.300" LastActivityDate="2019-09-06T05:04:28.300" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="30003" PostTypeId="2" ParentId="29995" CreationDate="2019-09-06T09:30:10.597" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;The Bearing Circle Diameter is the diameter that &lt;em&gt;bears&lt;/em&gt; on the surface underneath (&lt;span class=&quot;math-container&quot;&gt;$d_w$&lt;/span&gt; on the drawing below). It is the diameter you would want to know when calculating contact area, friction, pressure, etc.&lt;/li&gt;&#xA;&lt;li&gt;The Flange diameter is the outer diameter of the flange (&lt;span class=&quot;math-container&quot;&gt;$d_c$&lt;/span&gt; on the image below) - it the the diameter that you would want to know if you were making sure the nut would fit in a counterbored hole&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mnobn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mnobn.png&quot; alt=&quot;Flange Nut - from ISO 4161:2012(E)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image is from ISO 4161:2012(E)&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-09-06T09:30:10.597" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30004" PostTypeId="1" CreationDate="2019-09-06T09:41:21.900" Score="-1" ViewCount="40" Body="&lt;p&gt;I was interested in this as a fun project. To be able to get the volume of air inhaled by a person has some cool applications I can think of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of a modified anemometer of sorts fitted in a tube that the person sucks / blows air through. Plotting the velocity of air against time and the area under the graph and multiplying that with the circumference of the tube would give volume of air inhaled?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/amp/s/makersportal.com/blog/2019/02/06/arduino-pitot-tube-wind-speed-theory-and-experiment%3fformat=amp&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/amp/s/makersportal.com/blog/2019/02/06/arduino-pitot-tube-wind-speed-theory-and-experiment%3fformat=amp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sort of thing attached to a tube through which a person sucks / blows. Then using the same logic to get the volume of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would either of these work? Are there better solutions?&lt;/p&gt;&#xA;" OwnerUserId="21712" LastActivityDate="2019-10-06T19:01:35.190" Title="Way to digitally measure volume of air inhaled by lungs?" Tags="|measurements|airflow|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30007" PostTypeId="2" ParentId="29975" CreationDate="2019-09-06T15:38:07.767" Score="2" Body="&lt;p&gt;&lt;strong&gt;Controlling temperature using PIDs on two separate controllers&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I am controlling how much material is burning to control temperature using a PID loop. I want to achieve tighter control by also slowing down or speeding up the fan in response to temperature changes. I'm looking to find an algorithm that would allow these two to work together.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So what is the real question are you looking for an algorithm to use as a second separate controller?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is one very basic problem with this approach. One controller will influence the other controller so you will have an unstable control cycle. For example as the fan reduced speed to reduce the temperature the argue speed would increase to raise the temperature back to it's original point, the fan would then reduce speed further to compensate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a common problem in air conditioning where both humidity and temperature control are needed one fights the other for dominance. Although the example you have given is more complicated some of the basics remain the same. You cannot use two independent PID controls. A special set of circumstances exist in which one control (I will call it a channel) needs to have priority control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is required is a two channel PID controller with a priority control option.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be honest I have never come a cross an algorithm for this that works successfully except in a dedicated controller like Eurotherm or someone similar would manufacture. 20 years ago they cost over 2000 usd. (you are paying for the development costs). Today I have no idea of the costs. Algorithms for PLC's from Siemens and the likes are next to useless and my old company asked a university to estimate a time and cost to develop such an algorithm  They estimated 2 years at which point the cost became irrelevant and the idea was scraped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Your problem is infinitely more complexed&lt;/strong&gt; than trying to control Temperature and Humidity. They are just two independent states that are physically related. So you can control each and just give priority to one. What you are trying to do is &lt;strong&gt;create 2 channels which are both based on temperature&lt;/strong&gt; so even though one channel could be given priority it will still produce conflict with the second channel and the result would be a totally unstable control cycle. I think even  Einstein would have considered this a considerable challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way forward that I can see is to decide on an alternative control strategy that provides two independent control channels, or scrap the idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However there is some light at the end of the tunnel If you had addressed this question to a cooking web site they may have given you better advice.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;&lt;strong&gt;A smoker with little airflow&lt;/strong&gt;, like some electrics, &lt;strong&gt;creates a larger stagnant bubble of air protecting the meat from smoke&lt;/strong&gt;. &lt;strong&gt;Rapid, random air movement breaks up the bubble. For example, a strong convection fan.&lt;/strong&gt;&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://genuineideas.com/ArticlesIndex/srasmokeparticles.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Without the smoke, you might as well be grilling.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21597" LastActivityDate="2019-09-06T15:38:07.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30008" PostTypeId="2" ParentId="25967" CreationDate="2019-09-06T17:25:11.303" Score="1" Body="&lt;p&gt;For standard layered-shell sections, you can obtain submatrices [ A ], [ B ], [ D ], and [ E ] via the SLIST command's FULL option or via the *GET command (Entity = SHEL).&lt;/p&gt;&#xA;" OwnerUserId="21716" LastActivityDate="2019-09-06T17:25:11.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30009" PostTypeId="2" ParentId="30004" CreationDate="2019-09-06T17:42:28.487" Score="0" Body="&lt;p&gt;The thing you are trying to build is called a spirometer, and there are already digital versions available.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Googling on the term should find you a wealth of information.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-06T17:42:28.487" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30010" PostTypeId="2" ParentId="29984" CreationDate="2019-09-06T18:22:38.647" Score="0" Body="&lt;p&gt;The geometry of the trusses are usually long narrow members carrying axial loads,connected to each other in triangles. And the connections are designed to let the axis of truss members pass trough a common center on the connection plate or connection bolted bracket. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this geometry there is no moment at the connections, even if they are rigidly connected. Therefore the connection can be assumed to be a pin connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be very small tertiary moments due to heat expansion or deflection but those are well within the tolerance range of the truss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That doesn't mean if a truss will be designed for extreme temperatures or large deflections the effect can be ignored.   &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-06T18:22:38.647" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30012" PostTypeId="1" CreationDate="2019-09-07T01:05:46.823" Score="1" ViewCount="345" Body="&lt;p&gt;I'm trying to figure out how to create a seal for a very low pressure gravity feed system. Essentially its a low pressure solenoid valve. Is there any equations to estimate the amount of force you would need on a o-ring to seal it from any leaks of pressure &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; its placed in a spring solenoid combination so I need to have this as low pressure as possible. Any ideas on what force I would need from the spring to seal a liquid the pressure is due to a height of about 40mm with a liquid like water. So if my math is right that's&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\rho g h = 997 kg/m^3 * 0.04m * 9.81m/s^2 = 391.22\ pascals&#xA;$$&lt;/span&gt;&#xA;at the o-ring so pretty small amount of pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also what material would you suggest for the O-ring I have considered, I have used 50 Durometer with no success and am thinking about moving to a silicone based O-Ring. I can't increase the force of the spring because a solenoid will need to overcome it and i'm limited on space.&lt;/p&gt;&#xA;" OwnerUserId="21725" LastEditorUserId="7083" LastEditDate="2019-09-07T17:30:40.297" LastActivityDate="2021-10-27T01:01:10.120" Title="O-Ring Design for low pressure applications" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30013" PostTypeId="2" ParentId="29523" CreationDate="2019-09-07T04:53:26.363" Score="3" Body="&lt;p&gt;I just check on Google and I find the following results&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.nitto.com/eu/es/about_us/brand/promotion/innovation/temish.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;TEMISH (TM) of Nitto &lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Silicone (yes, you read that right) (fixed for spelling mistake) - Apparently, the American company GE tested it out in 1950s&#xA;&lt;a href=&quot;https://i.stack.imgur.com/chybt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/chybt.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.dupont.com/brands/tyvek.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tyvek (R) of DuPont&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.gore-tex.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gore-Tex (R)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The stuff above can happen due to something called &quot;Molecular Sieve&quot;, find quick reading &lt;a href=&quot;https://en.wikipedia.org/wiki/Molecular_sieve&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Basically, the material concerned is porous (&quot;with holes&quot;), but these holes are very, very, very small, at the atomic level. The holes can block bigger molecules/atoms (like water - H2O), but allow air (say oxygen - O2) to pass through. Any material with sufficient small holes can &quot;block&quot; the water, and &quot;unblock&quot; air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, said is easier than done, because the water molecules have the size of nearly 3 angstroms (2.75 if you want more precise value), so... yeah. Building material like that is &lt;em&gt;possible&lt;/em&gt;, but &lt;em&gt;very hard&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="21520" LastEditorUserId="21520" LastEditDate="2019-09-09T01:02:23.723" LastActivityDate="2019-09-09T01:02:23.723" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30014" PostTypeId="1" CreationDate="2019-09-07T05:30:19.787" Score="2" ViewCount="1906" Body="&lt;p&gt;Recently I read an article about ABS brakes. In that, they repeated several times the term that traditional hydraulic brake replaced with ABS brakes. Why it so? The reason for choosing ABS is mentioned in that article. Is anybody knows about it? &lt;/p&gt;&#xA;" OwnerUserId="21726" LastActivityDate="2019-09-07T12:03:00.820" Title="What is the difference between hydraulic and ABS brakes?" Tags="|mechanical-engineering|hydraulics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30016" PostTypeId="2" ParentId="30014" CreationDate="2019-09-07T07:41:08.250" Score="2" Body="&lt;p&gt;The static friction coefficient of tires on dry road is approximately 0.7 up to 0.9 and on a wet road around 0.4. Remember as long as the tire is rolling even while the driver is pushing the brakes and the car is stopping with no slip this static coefficient applies and brings the car to a quick stop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you kick hard on your brakes and they lock, the tire skids and the steering becomes uncontrollable and the traction drops to very low kinetic levels in the range of 0.05 depending on the tire and road condition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ABS system controls the hydraulic brakes, stopping them just short of locking up and skidding, while permitting maximum brake force and maintaining control and steering. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-07T07:41:08.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30017" PostTypeId="1" AcceptedAnswerId="30024" CreationDate="2019-09-07T12:01:49.800" Score="0" ViewCount="97" Body="&lt;p&gt;From the finite element theory, what is remained a bit unclear for me, is that how do we know the location of gausian points where the integration is done for each specific type of element ? it is not indicated in many FEM books, but if someone could explain it here , it would be fantastic&lt;/p&gt;&#xA;" OwnerUserId="17070" LastActivityDate="2019-10-04T12:27:04.167" Title="How are the location of Gausian points determined in Finite element theory?" Tags="|mechanical-engineering|applied-mechanics|finite-element-method|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30018" PostTypeId="2" ParentId="30014" CreationDate="2019-09-07T12:03:00.820" Score="0" Body="&lt;p&gt;From my limited understanding, it can be understood as follow:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&quot;Traditional&quot; hydraulic brake: It is a simple mechanical system, the harder you slam or step on the brake pedal, the harder the brake clamp will stuck on the brake plate. It is a basic system, but it comes with a risk: In certain scenario, your wheel may be locked (ie stop rotating), but the moment of inertia of your car is too great, therefore, it keeps going forward. Needless to say, this is a &quot;problem&quot;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To solve the aforementioned problem, ABS (Anti-lock braking system) is built. For the ultra-simplified version (which is probably wrong), think the brake plate at the wheel is not constantly applying the maximum brake force, but using in a certain rhythm and/or progress.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The &quot;end-effect&quot; can be seen as follow&#xA;&lt;a href=&quot;https://i.stack.imgur.com/C0w0i.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C0w0i.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XXEHC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XXEHC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tXmJY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tXmJY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21520" LastActivityDate="2019-09-07T12:03:00.820" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30019" PostTypeId="2" ParentId="30012" CreationDate="2019-09-07T13:29:38.233" Score="0" Body="&lt;p&gt;Take a look at how irrigation valves work.  In a nutshell, the solenoid blocks a small orifice with a cone shaped plug, or the shaft blocks an orifice.  When the solenoid operates, the &quot;leak&quot; is used as pressure to operate the valve.  Start here:  &lt;a href=&quot;https://en.wikipedia.org/wiki/Solenoid_valve&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Solenoid_valve&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible source of inspiration would be the valve system on natural gas appliances.  Here, it's critical to not have leaks.  Pressures are on the order of what you mention here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you are using small forces, you may be able to get a better seal with an o-ring mounted in a groove on a shaft, and then use a long tapered bore. This allows the o-ring to wedge between the cylinder and the bore wall.  The price you pay is needing a longer solenoid stroke to provide sufficient clearance for the fluid to pass.&lt;/p&gt;&#xA;" OwnerUserId="8768" LastActivityDate="2019-09-07T13:29:38.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30020" PostTypeId="2" ParentId="29998" CreationDate="2019-09-07T13:53:46.553" Score="1" Body="&lt;p&gt;AFAIK generate your 4kV using off the shelf components.  The trick is getting the electrons off the wire, and into the air.  Sharp points are your friend here.  An electron at the tip is repelled by all electrons on the surface of the point behind it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, the more common practice is to use a very small diameter wire.  The tight radius of the wire gives the same effect. See the guts of a photocopier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Actual current is tiny.  Parasitic losses of  your circuit, and fans to move the ionized air will be much larger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thoughts on removing ozone:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;This is a ppm problem.  Would introducing a small amount of hydrocarbon along with a UV light source cause the ozone to recombine? Do the reaction in a mirror chamber so that the light has many chances to excite an ozone molecule.&lt;/li&gt;&#xA;&lt;li&gt;Ozone is used for water purification.  Implication is that it is soluble in water.  Don't know how the solubility compares to that of oxygen.  Use something the the stack scrubbers on power plants, where the air rises through a rain of water droplets just barely large enough to fall.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="8768" LastEditorUserId="8768" LastEditDate="2019-09-20T19:28:44.570" LastActivityDate="2019-09-20T19:28:44.570" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="30022" PostTypeId="1" CreationDate="2019-09-07T18:11:32.960" Score="0" ViewCount="715" Body="&lt;p&gt;We got a bunch of yellow thermocouple wire which the &lt;a href=&quot;https://www.omega.com/en-us/colorcodes&quot; rel=&quot;nofollow noreferrer&quot;&gt;Omega web site&lt;/a&gt; says is type K extension grade wire. They also sell thermocouple grade wire which looks to be brown.  What's the difference?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was speculating that the thermocouple grade is super pure and gives voltages that are in agreement with what's expected and extension grade will need to be calibrated but probably won't be too far off.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anybody have an experience with this?  How far off will it be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any ideas.&lt;/p&gt;&#xA;" OwnerUserId="4953" LastActivityDate="2019-10-08T07:01:50.537" Title="What's the difference between thermocouple grade and extension grade wire?" Tags="|temperature|thermocouple|wire|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30024" PostTypeId="2" ParentId="30017" CreationDate="2019-09-07T19:10:32.347" Score="0" Body="&lt;p&gt;As far as I know the location of integration points is governed by numerical method used for integration. I know only one method of numerical integration, Gauss Quadrature. For details of Gauss Quadrature, please see Wikipedia article or Introduction to Finite Element Method by J.N. Reddy.&lt;/p&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2019-09-07T19:10:32.347" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30025" PostTypeId="2" ParentId="30022" CreationDate="2019-09-07T20:19:11.997" Score="0" Body="&lt;p&gt;From the table info the &quot;extension&quot; grades seem to be different alloys ( likely lower cost). For K type , the extension alloy is limited to 392 F ; that is lower than most applications for K . I welded and installed many K couples ( long ago) and they all were used above 500 + F.  K thermocouples are Chromel ( A K A , Nichrome ) and Alumel ( 94- Ni , 2- Al, 3- Mn ). The table says the extension is &quot; nickel aluminum &quot; alloy, maybe that means it is Alumel. But I don't know how that affects the emf. So , I would stay with regular K unless your service temperature is below 392 F.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-09-07T20:19:11.997" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30026" PostTypeId="1" CreationDate="2019-09-07T20:24:11.850" Score="4" ViewCount="239" Body="&lt;p&gt;So I've been drilling standard 1/4&quot; aluminium in a small (250W) consumer drill press.  I've used a little oil to help, but I'm not sure if I'm doing it right.  The holes seem okay though. They're 10mm diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Judging by the swarf coming off the bit, am I doing it right or is there room for improvement? The edges of the swarf seem 'ripped' with little tags visible. And one side is dull, the other polished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ldobu.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ldobu.jpg&quot; alt=&quot;piccy&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2947" LastActivityDate="2019-09-08T21:25:19.817" Title="Is my aluminium drill swarf okay?" Tags="|machining|drilling|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30029" PostTypeId="2" ParentId="29798" CreationDate="2019-09-08T03:42:57.150" Score="1" Body="&lt;p&gt;All washing machines basically clean by causing water to flow around and through the clothing, mechanically removing embedded dirt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider the process from a symmetry point of view, there are obvious advantages to the front-load design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With top-loaders:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In large loads, the clothes at the bottom are continuously soaked, while near the top, especially when overloaded, the clothes might not be exposed to as much water.&lt;/li&gt;&#xA;&lt;li&gt;The clothes near the agitator will continually have a lot of water flowing past them, while clothes near the edges will experience far less current.&lt;/li&gt;&#xA;&lt;li&gt;Rinsing can require refilling the tub, possibly several times, and the rinse water isn't perfectly clean.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;With front-loaders:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;All clothes are repeatedly tumbled, so all end up soaking at the bottom for a while and then draining near the top for a while.&lt;/li&gt;&#xA;&lt;li&gt;All clothes experience the same amount of water flow.&lt;/li&gt;&#xA;&lt;li&gt;Rinse water runs through from top to bottom, so far less water is needed, and the rinse water is always clean.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, even if the two methods cleaned the same on the average (which they don't), the top-loader would have some clothes that aren't cleaned as well as others are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then combine that with the front-loader's more obvious advantages:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Only enough water to soak the clothes is needed, as opposed to enough water to completely cover them.&lt;/li&gt;&#xA;&lt;li&gt;Rather than having to refill the tub, possibly several times, clean water runs through from top to bottom, so far less rinse water is needed in a top-loader, and it rinses better.&lt;/li&gt;&#xA;&lt;li&gt;Only a little power is needed to fully tumble the clothes with a 180° rotation and then pause while the power of gravity causes the water to flow from top to bottom, as opposed to a lot of power needed to continuously run the agitator.&lt;/li&gt;&#xA;&lt;li&gt;Much higher spin speeds are possible, so the clothes will contain much less water when put into the dryer.&lt;/li&gt;&#xA;&lt;li&gt;Loads can be just as efficient and effective whether they are small or large.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The front-loader's disadvantages include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;One has to bend over to load and unload.&lt;/li&gt;&#xA;&lt;li&gt;More hand cleaning of the machine is needed.&lt;/li&gt;&#xA;&lt;li&gt;A run requires more time.&lt;/li&gt;&#xA;&lt;li&gt;The machine costs more to purchase.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;but they have nothing to do with how well the clothes get cleaned.&lt;/p&gt;&#xA;" OwnerUserId="19930" LastActivityDate="2019-09-08T03:42:57.150" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="30030" PostTypeId="1" CreationDate="2019-09-08T03:51:18.953" Score="0" ViewCount="573" Body="&lt;p&gt;software person here, venturing for the first time in the design of a mechanical system for a hobby project. Please note I have some experience in disassembling, fixing, reassembling mechanical systems, but design my own is a whole other thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, I would need a motor (that I know how to control) to rotate a vertical pole, or mast. The motor has a wormgear reduction box similar to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ACfLI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ACfLI.jpg&quot; alt=&quot;wormgear DC motor&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'd like to avoid is to put off-axis strain on the motor in case the pole is pushed sideways; also, the pole should be able to rotate as freely as possible to the torque from the motor does not have to be excessive. The weight pushing down from the pole should also be taken care of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first intuitive approach would be to use two thrust bearing devices similar to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oimOH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oimOH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so that the mast will be constrained on its axis and also the weight supported. I'd mount a gear around the mast with the motor with a smaller gear on its output shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice, should I consider anything else? Please, let me know if a more comprehensive drawing is needed to communicate what I want to achieve, I'll try to put one together.&lt;/p&gt;&#xA;" OwnerUserId="21735" LastActivityDate="2019-09-08T03:51:18.953" Title="Rotating vertical pole setup" Tags="|mechanical-engineering|motors|bearings|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="30031" PostTypeId="2" ParentId="30026" CreationDate="2019-09-08T05:35:20.393" Score="1" Body="&lt;p&gt;Agree with Blacksmith37. Note also that aluminum machining processes commonly use an oil-and-water emulsion called &quot;mouse milk&quot; as a lubricant, with a pump-driven feed nozzle aimed right at the hole, and a sump system that catches the used fluid and recycles it through a storage vessel and back to the nozzle again. If you are drilling thousands of holes and feeding the bits fast, you might consider this to prolong the life of your bits. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-08T05:35:20.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30032" PostTypeId="2" ParentId="30022" CreationDate="2019-09-08T06:59:18.373" Score="1" Body="&lt;p&gt;&lt;strong&gt;Introduction&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Thermocouple Wire that is used in a thermocouple from the point of sensing to the point of cold junction compensation (cjc end) where the signal is measured.&lt;/p&gt;&#xA;&lt;p&gt;A thermocouple is a sensor for measuring temperature, that consists of two dissimilar metals that are joined together at the sensing end. Different thermocouple types (e. G. J, K, T, E, etc) use different mixtures of metals in the wire. At the cjc end, the millivolt value provided by the thermocouple represents the difference in temperature of the sensing end as compared to the cjc end (also called the reference end).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Cable Colours,&lt;/strong&gt; you need to consult a cable colour coding guide like this one. &lt;a href=&quot;https://www.rjcables.com/thermocouple-extension-and-compensating-cables.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;R.J. Cables&lt;/a&gt; or this one &lt;a href=&quot;https://assets.omega.com/landing-pages/colorcodes/tc_colorcodes.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;T.C. colour codes&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Yellow does infact  refer to a K type thermocouple extension cable as opposed to Brown which is for a K type thermocouple compensating cable.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Cable types&lt;/strong&gt;, THERMOCOUPLE EXTENSION &amp;amp; COMPENSATING CABLES &lt;a href=&quot;https://www.rjcables.com/thermocouple-extension-and-compensating-cables.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;R.J. Cables&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Thermocouple extension &amp;amp; compensating cables, transmits, the cold junction of temperature sensor installed at the measuring point to the measuring instruments installed at remote location in control room.&lt;/p&gt;&#xA;&lt;p&gt;Extension cables are made from the same metals used in thermocouples, but for compensating cable, materials of similar thermoelectric properties as of thermocouple in use, are &lt;strong&gt;employed to reduce the cost&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="21597" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-09-08T06:59:18.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30033" PostTypeId="2" ParentId="30026" CreationDate="2019-09-08T12:37:47.493" Score="2" Body="&lt;p&gt;Looks just fine. Aluminum can do that. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long, stringy swarf(chips) is usually less preferable if it can be controlled.  What you want to avoid are those chips from getting compacted or clogged in the flutes of the drill. That will make it more difficult for you to drill the holes and eventually impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can try peck drilling your holes. The action of drilling down a short distance and then briefly pulling the drill upward slightly(distance upward depends on what works best in that situation). The upward retraction tends to break the chip away from the part making a smaller chip. Also, a bright drill(shiny/polished flutes(no black coating)) will help with the flow of the chips up the flutes and out of the way for more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here’s an animation:&#xA;&lt;a href=&quot;http://www2.me.rochester.edu/courses/ME204/nx_help/en_US/graphics/graphicLibrary/nx/manufacturing/milling/drilling/breakchip_demo1_anim.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;peck drill&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2559" LastEditorUserId="2559" LastEditDate="2019-09-08T21:25:19.817" LastActivityDate="2019-09-08T21:25:19.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30036" PostTypeId="1" CreationDate="2019-09-08T14:48:16.380" Score="1" ViewCount="12" Body="&lt;p&gt;Glass is brittle and easily damaged. So, a new kind of glass material is made, in which it's layers are made of hexagonal tablets (made of borosilicate glass) arranged in 1P4A or 5A, as happens in natural nacre. This synthetic glass proves to be more resistant to impact. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;More about is given in this link.&#xA;&lt;a href=&quot;https://science.sciencemag.org/content/364/6447/1260/tab-figures-data&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://science.sciencemag.org/content/364/6447/1260/tab-figures-data&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my question is, where does the terminology &quot;1P4A&quot; or &quot;5A&quot; come from? I cannot find it on the net.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also don't know if this is the right place to post this question. Please guide me.&lt;/p&gt;&#xA;" OwnerUserId="21742" LastActivityDate="2019-09-08T14:48:16.380" Title="Naming of 1P4A or 5A layer configurations" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30037" PostTypeId="2" ParentId="29907" CreationDate="2019-09-08T15:06:28.587" Score="1" Body="&lt;p&gt;It is difficult to tease out proprietary trade secrets from manufacturing companies; and it is further confusing to deal with misleading terms aimed at applications and clinical handling of these products. The main hurdle in understanding what's going on under the hood in these algorithms is the label &amp;quot;statistical algorithms&amp;quot; for hybrid systems. It turns out that although statistical methods are applied to hybrid models, they are mainly focused on eliminating noise, and addressing the problem of photon starvation (streak artifacts), which are particularly problematic in logarithmic data. These methods include better modelling of photon distribution, and penalized regression estimation of noise.&lt;/p&gt;&#xA;&lt;p&gt;The initial step in these hybrid methods can be a FBP that moves the sinogram (or raw data) into the image space, followed by progressive optimization steps always within image space. However, an empty initial image is also possible, as explain in &lt;a href=&quot;https://www.sciencedirect.com/science/journal/11201797/28/2&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;em&gt;Iterative reconstruction methods in X-ray CT&lt;/em&gt; by Marcel Beister, Daniel Kolditz, Willi A. Kalender Physica Medica Volume 28, Issue 2, April 2012, Pages 94-108&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The iteration process can be initiated with an empty&#xA;image estimate or using prior information, for example,&#xA;a standard FBP reconstruction or a volume of a similar&#xA;object. In general, the better the prior images match the&#xA;final images, the faster the process converges towards&#xA;a stable solution. The iterative process is finished when&#xA;either a fixed number of iterations is reached, or the&#xA;update for the current image estimate is considered small&#xA;enough or when a predefined quality criterion in the image&#xA;estimate is fulfilled.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jBy1j.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jBy1j.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;However, it is in model-based (latest generation) IR, aka MBIR, where statistics methods are utilized in full force, and both the signal and the noise are regressed through iterative approximations, incorporating a priori parametric information that takes into consideration such things as receptor geometry and optics.&lt;/p&gt;&#xA;&lt;p&gt;As in the case of hybrid models, the initial step is still a FBP to generate an initial &amp;quot;guess;&amp;quot; however, the critical distinction is the generation of synthetic sinograms (forward projection) from the reconstructed backprojection. In MBIR, hence, there is a back-and-forth optimization process that oscillates between image and data space.&lt;/p&gt;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://aapm.onlinelibrary.wiley.com/doi/pdf/10.1002/mp.13123&quot; rel=&quot;nofollow noreferrer&quot;&gt;Regularization strategies in statistical image reconstruction of low-dose x-ray&#xA;CT: A review, H. Zhang&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://pdfs.semanticscholar.org/31b4/3fcdd1a0fd9e9665925f23b16d918eed3e23.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Modelling the physics in the iterative reconstruction for transmission computed tomography,J. Nuyts&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://pdfs.semanticscholar.org/99e1/126f7f554b8ba02e3f5e0cacb04fe4b8f61a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Local basis-function approach to computed tomography, K. Hanson&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21660" LastEditorUserId="21660" LastEditDate="2020-09-30T20:17:54.400" LastActivityDate="2020-09-30T20:17:54.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30039" PostTypeId="1" CreationDate="2019-09-08T20:08:53.090" Score="0" ViewCount="224" Body="&lt;p&gt;I'm simulating different arrays for cylinders in a simple cross flow heat exchanger (cylindrical pipes) as shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8ztCJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8ztCJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inlet temperate(300K) is constant and the cylinders are at constant temperature(400K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm simulating different arrays for cylinders.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've calculated average Nusselt on each pipe then summed all of them and finally divided it by the number of cylinders (9). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives me an average Nusselt number for heat exchanger. Let's call it &lt;span class=&quot;math-container&quot;&gt;$Nu_{av}$&lt;/span&gt;.&#xA;&lt;span class=&quot;math-container&quot;&gt;$Nu_{av}=\frac{\sum Nu_{cylinder}}{9}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also calculated the heat transfer coefficient of heat exchanger using the following equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$\dot{m}c(T_o-T_i)=UA(T_s-T_i)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\dot{m}$&lt;/span&gt;: Mass flow of fluid passing the heat exchanger&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;: Specific heat of fluid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T_o$&lt;/span&gt;: Average outlet temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T_i$&lt;/span&gt;: Average inlet temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;: Heat exchanger coefficient &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;: Total surface of all cylinders&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T_s$&lt;/span&gt;: Cylinders surface temperature&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've simulated a lot of arrays. In some cases, the average Nusselt number(&lt;span class=&quot;math-container&quot;&gt;$Nu_{av}$&lt;/span&gt;) is bigger and in some others, the average outlet temperature and so, the heat exchanger coefficient(&lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt;) is bigger.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: Which one should be my criteria to compare cases? Average Nusselt number or heat exchanger coefficient?&lt;/p&gt;&#xA;" OwnerUserId="15941" LastEditorUserId="15941" LastEditDate="2019-09-08T20:30:22.757" LastActivityDate="2020-06-05T17:02:25.143" Title="Average Nusselt number or heat transfer coefficient to compare better heat transfer?" Tags="|heat-transfer|heat-exchanger|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30040" PostTypeId="1" CreationDate="2019-09-08T20:35:44.140" Score="0" ViewCount="354" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/brQii.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/brQii.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In both cases in the picture above torque is applied to the shaft, however in the first case the force &lt;span class=&quot;math-container&quot;&gt;$F_{c}$&lt;/span&gt; also produces bending across the shaft. So my question is, how do I know if a torque should be shown with a force like in the first example, or with a torque vector like in the second example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My best guess is that it depends on what produces the torque. Some ways do produce bending (like conic friction plates for example), while some don't.  &lt;/p&gt;&#xA;" OwnerUserId="21424" LastActivityDate="2019-10-10T00:00:07.187" Title="When does torque applied on a shaft produce bending and when does it not?" Tags="|mechanical-engineering|torque|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30041" PostTypeId="2" ParentId="30040" CreationDate="2019-09-08T20:48:36.603" Score="0" Body="&lt;p&gt;If you apply a force or several forces with the resultant vector offset from axis of a beam or shaft as in you first case it causes moment and torque and transversal shear and axial stress if it is not perpendicular to the axis of shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the forces resultant are such that they purely resolve to torque such as 4 equal forces acting around a circle perpendicular to the shaft equally at each 90 degree quadrant it will be just torque.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-08T20:48:36.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30042" PostTypeId="1" CreationDate="2019-09-08T22:26:20.977" Score="2" ViewCount="209" Body="&lt;p&gt;I could not find a Controls Stackexchange, I hope this is the right place, otherwise I would be glad to post it in the appropriate section. My question concerns the root locus method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a plant with some transfer function that I want to control with gain, I would arrive to such an equation,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$1 + \frac{K}{s(s+0.5)(s^2 + a s + 4)} = 0,$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is a given constant (think &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; for example), and &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; is a gain I will be able to choose. I want to draw the root locus by hand to have a rough idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I place my poles, the root locus on the real axis, count the branches (four go to infinity, no zeros), find the centroid then asymptotes, find the breakouts on the real axis, find the intersection of the root locus with the imaginary axis and find the the angle of departure from the two complex poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once I have done this, for &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; close to &lt;span class=&quot;math-container&quot;&gt;$\frac 12$&lt;/span&gt;, I find myself with a departure angle which is close to &lt;span class=&quot;math-container&quot;&gt;$-90^\circ$&lt;/span&gt; (for the upper pole), that is &quot;downwards&quot;. This seems ambiguous regarding which asymptote the branch leaving from the upper complex pole will eventually reach. I attach two plots to the post with values of &lt;span class=&quot;math-container&quot;&gt;$a = 0.5 \pm 0.1$&lt;/span&gt;: &lt;a href=&quot;https://www.wolframcloud.com/obj/om3/0.49.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$a = 0.49$&lt;/span&gt;&lt;/a&gt; and &lt;a href=&quot;https://www.wolframcloud.com/obj/om3/0.51.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;span class=&quot;math-container&quot;&gt;$a = 0.51$&lt;/span&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously now that I look at the plots, I can tell which branch goes where, and using some criterion like &quot;if the branch leaving the complex pole goes to the right, it will go on the right asymptote, and vice versa&quot;, you would indeed get the right answer. My question is,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a general way to tell which branch goes to which asymptote?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="21743" LastActivityDate="2019-09-09T14:12:14.003" Title="Root locus: ambiguity on which branch goes where?" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30044" PostTypeId="2" ParentId="30042" CreationDate="2019-09-09T02:45:41.477" Score="4" Body="&lt;p&gt;I think you're over-analyzing.  The root locus is an abstraction.  While it usually makes it easier to think of the poles as &quot;moving&quot;, you don't want to get too married to that thought.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It works better for me to just think of the poles as merging into one double pole (or triple, or n-tuple, for various contrived examples), and then breaking apart again into the asymptotic poles.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-09T02:45:41.477" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30046" PostTypeId="1" AcceptedAnswerId="30053" CreationDate="2019-09-09T12:35:28.827" Score="0" ViewCount="95" Body="&lt;p&gt;This site:(&lt;a href=&quot;https://www.motioncontroltips.com/faq-what-are-dc-shunt-motors-and-where-are-they-used/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.motioncontroltips.com/faq-what-are-dc-shunt-motors-and-where-are-they-used/&lt;/a&gt; ) explains the speed regulation of DC shunt motor as follows: If the load on the motor is increased, the armature rotation slows and back EMF is reduced, since back EMF is proportional to speed. With less back EMF voltage and a constant supply voltage the net voltage increases. The increase in net voltage results in an increase in armature current. Since torque is proportional to armature current, torque also increases. Finally, this increased torque allows the motor to increase its speed and compensate for the slowdown due to loading. Hence, a DC shunt motor is able to self-regulate its speed, and can be referred to as a constant speed motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that torque is inversly related to speed for a DC shunt motor, wouldn't this entire explanation be wrong as the last line mentions increase in speed due to increase in torque ?&lt;/p&gt;&#xA;" OwnerUserId="21544" LastActivityDate="2019-09-09T14:23:51.697" Title="Isn't this explanation of speed regulation in a DC shunt motor wrong?" Tags="|motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30048" PostTypeId="1" AcceptedAnswerId="30049" CreationDate="2019-09-09T12:46:40.397" Score="0" ViewCount="27" Body="&lt;p&gt;I'm planning a thought project for measuring the C02 in a room (at office namely), and communicating its results to a computer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I'm blocked at the early stage: which sort of sensor and which is the mechanism to measure the CO2 concentration in air ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I saw devices for sale, not communicating with computers, only LCD screens, and there were no indications on what is the system/mechanism used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any idea ?&lt;/p&gt;&#xA;" OwnerUserId="2137" LastEditorUserId="2137" LastEditDate="2019-09-09T12:52:59.840" LastActivityDate="2019-09-09T13:07:16.257" Title="What kind of sensors to be used to measure the carbon dioxyde concentration?" Tags="|gas|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30049" PostTypeId="2" ParentId="30048" CreationDate="2019-09-09T13:07:16.257" Score="1" Body="&lt;p&gt;I have used NDIR sensors, such as these:&#xA;&lt;a href=&quot;http://www.alphasense.com/index.php/products/ndir-safety/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.alphasense.com/index.php/products/ndir-safety/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's based on a light source which is flashed at a regular frequency and reflected on a series of mirrors before ending on a receiver. The sensor provides 2 signals: one &quot;reference&quot; signal, which remains constant regardless of the % of CO2, and one &quot;active&quot; signal which changes with the % of CO2 in the air. You have to calibrate the sensor with different % of CO2, and you can compensate for temperature and pressure variations.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-09-09T13:07:16.257" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30052" PostTypeId="2" ParentId="30042" CreationDate="2019-09-09T14:12:14.003" Score="0" Body="&lt;p&gt;To echo the answer by TimWescott, consider the situation when &lt;span class=&quot;math-container&quot;&gt;$a=0.5$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a00gY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a00gY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complex pole on the top goes to the left (green) and that on the bottom goes to the right (orange). This is just one of many possibilities. Both could have gone to right, or both to the left, or one left and the other right. All are correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the above case, there is no one correct way for a branch to go to an asymptote. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For cases &lt;span class=&quot;math-container&quot;&gt;$a=0.49$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$0.51$&lt;/span&gt;, the poles of the former are more to the right and so they tend to go off to the right. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tXyl7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tXyl7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/m6rSB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m6rSB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xuL46.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xuL46.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2019-09-09T14:12:14.003" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30053" PostTypeId="2" ParentId="30046" CreationDate="2019-09-09T14:12:51.207" Score="1" Body="&lt;p&gt;Considering that torque is inversely related to speed for a DC shunt motor, wouldn't this entire explanation be wrong as the last line mentions increase in speed due to increase in torque ?&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;torque is inversely related to speed&lt;/p&gt;&#xA;&lt;p&gt;This means&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The &lt;strong&gt;maximum Torque&lt;/strong&gt; is applied at &lt;strong&gt;Zero speed&lt;/strong&gt;, because torque is inversely related to speed.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;However&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;as speed increases, Torque reduces&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;until&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The required design speed and it's associated value of Torque are reached.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If additional load is applied &lt;strong&gt;speed reduces&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;as speed reduces, Torque increases&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Which then takes us back to where we started. The increased Torque will accelerate the speed, similar to what it did on starting, only in this case we are not increasing from zero or stall speed. But some point slower than the designed speed of the D.C. Shunt motor.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;strong&gt;as speed increases, Torque reduces&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;until&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The required speed and it's &lt;strong&gt;New associated value of Torque are reached&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Therefore if you relate this to the explanation. And view them as &lt;strong&gt;steps in a process not simultaneous events&lt;/strong&gt;,&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;The explanation is correct.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;this increased torque&lt;/p&gt;&#xA;&lt;p&gt;allows the motor to increase its speed and compensate for the slowdown due to loading&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="21597" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-09-09T14:23:51.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30054" PostTypeId="1" AcceptedAnswerId="30056" CreationDate="2019-09-09T14:56:34.357" Score="-1" ViewCount="593" Body="&lt;p&gt;I’m trying to find out if it’s possibly to form a cylinder or tube with a diameter of 1 inch using some kind of curling die for a press brake. I’ve seen dies that make small 360 degree bends like that but they always have additional length of sheet metal extending from the tubular section. Does anyone know if it’s possible to form just a small tube without additional material extending out? This tubing would be used for handles to lift something, similar to a handle bar on a bicycle or something. &lt;/p&gt;&#xA;" OwnerUserId="21749" LastActivityDate="2019-11-08T01:04:17.143" Title="Press brake tooling for 360 tube?" Tags="|mechanical-engineering|materials|design|steel|industrial-engineering|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30055" PostTypeId="1" CreationDate="2019-09-09T15:47:33.040" Score="0" ViewCount="689" Body="&lt;p&gt;I literally just have a thin steel wall that stands vertically with a load on top. The thin edges of the wall are oriented vertically with the load applied on top of the top thin edge, with the large planar faces of the wall facing horizontally to the sides. I thought there would be an easy calculation based on height, length, thickness (width) and a material parameter for the steel I’m using, but internet searches have yielded me no useful information as to the specific scenario I’m looking for. Is it not that simple to calculate wall strength based on a top loaded compressive force oriented vertically downward onto the height of the wall? Someone please help!&lt;/p&gt;&#xA;" OwnerUserId="21749" LastActivityDate="2019-10-09T20:01:35.220" Title="How do I calculate the compressive strength of a wall?" Tags="|structural-engineering|materials|structural-analysis|steel|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30056" PostTypeId="2" ParentId="30054" CreationDate="2019-09-09T16:02:10.843" Score="0" Body="&lt;p&gt;I have only seen it done on a continuous basis  . Flat stock is pulled through a few dies then final rollers press the two edges together to make a weld ( usually an ERW , electric resistance weld ). Most steel pipe up to about 24&quot; diameter is made this way. Small tubing , like 1 &quot;, is made that way. But some small tubing is made by starting with 4 &quot; pipe and cold drawing down to the desired size. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-09-09T16:02:10.843" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30058" PostTypeId="2" ParentId="30039" CreationDate="2019-09-09T16:37:59.070" Score="1" Body="&lt;p&gt;While this is not a direct answer, I would not use just the average Nusselt number or the overall heat exchanger coefficient. I would take this a step further and determine the heat exchanger efficiency, which is the ratio of heat transferred to the maximum amount of heat transferred. This way, you have a standardized basis to compare heat exchanger models. As SolarMike indicated, you should look at the pressure drop each heat exchanger inflicts on the flow. You could have an extremely efficient unit with a significant pressure drop. This would be detrimental from a system perspective. For example, a pump/compressor would need to input more work to compensate for the pressure loss or a downstream expander would suffer a loss of potential work. I highly recommend the book &lt;em&gt;Design of Fluid Thermal Systems&lt;/em&gt; by William S. Janna. &lt;/p&gt;&#xA;" OwnerUserId="21118" LastActivityDate="2019-09-09T16:37:59.070" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30059" PostTypeId="2" ParentId="30055" CreationDate="2019-09-09T19:23:23.933" Score="1" Body="&lt;p&gt;The analysis of your wall depends on many factors, is it supported at all sides or just the top and bottom or may be it is acting as a cantilever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also whether the supports are clamped or pinned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a very conservative start you can use &lt;span class=&quot;math-container&quot;&gt;$$ P_{Critical}= \frac{\pi^2EI}{(KL)^2}  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;K is the effective length factor and ranges from 2 for cantilever to 0.5 for fixed/fixed ends. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Roark’s Formulas for Stress and Strain by WARREN C. YOUNG&#xA;RICHARD G. BUDYNAS 7th edition has formulas on plate buckling on chapter 15, pp 715 to 730.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-09T19:23:23.933" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30060" PostTypeId="1" CreationDate="2019-09-09T19:47:30.823" Score="4" ViewCount="134" Body="&lt;p&gt;I have been asked to consider using RCC instead of traditional PCC on an upcoming project. For a rough order of magnitude estimate, can I assume that the pavement section design will be similar? I know that the mix design and construction methods will vary, but not sure if there will be any difference in the thickness of concrete or aggregate base.&lt;/p&gt;&#xA;" OwnerUserId="21752" LastActivityDate="2019-09-11T13:30:12.003" Title="Design Roller Compacted Concrete Pavement Section" Tags="|civil-engineering|concrete|pavement|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30061" PostTypeId="2" ParentId="8588" CreationDate="2019-09-09T20:10:21.217" Score="2" Body="&lt;p&gt;As you asked both, &amp;quot;How to make a conductive paint over PDMS?&amp;quot;, and, &amp;quot;How else can I make a conductive connection over the PDMS surface so that it can stick to it nicely?&amp;quot;, I will answer the latter question as your 'conductive paint' has been problematic for this application and there are already well established methods for conductive coatings for this particular material.&lt;/p&gt;&#xA;&lt;p&gt;One option is sputtering. Sputtering is a technique that is widely available and is a fairly inexpensive process for coating materials such as PDMS. Sputtered films also have an advantage over other coatings (such as your gum paint), in that these films typically have strong adhesion to the surface being coated and it is a non-destructive technique (e.g. it works well enough to be used in electron microscopy).&lt;/p&gt;&#xA;&lt;p&gt;Au, along with a host of other metals, are available for coatings, and they can be laid down in various thicknesses at rates from 10's to 1000's of nanometers per minute. In addition, this has been done many times before with this material and can give you the conductive surface you are looking for.&lt;/p&gt;&#xA;&lt;p&gt;Here is a wasp we coated awhile back for the microscopy labs:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LOUKd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LOUKd.png&quot; alt=&quot;The Au has a resistivity of a few ohms across the wasp, and could be reduced further with a thicker coating and a more uniform surface&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Au has a resistivity of a less than 2 ohms across the wasp, and could be sharply reduced with a thicker coating and a more uniform sample surface.&lt;/p&gt;&#xA;&lt;p&gt;Considering you have been painting the PDMS surface, I am inclined to assume minimal damage to the micro-structures is not an issue as is sometimes caused by this process.&lt;/p&gt;&#xA;&lt;p&gt;A second, and potentially easier option, is to simply use an Au, Ag or other conductive foil, although it is difficult to say if this is a good direction for your project as we have few details.&lt;/p&gt;&#xA;&lt;p&gt;Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="553" LastEditorUserId="553" LastEditDate="2021-08-26T19:11:30.580" LastActivityDate="2021-08-26T19:11:30.580" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30063" PostTypeId="1" CreationDate="2019-09-09T21:08:28.040" Score="1" ViewCount="392" Body="&lt;p&gt;One of the benefits of electric cars is that many parts can be eliminated, simplifying the overall system. Could EVs eliminate brakes as well by slowing the car down using reverse torque from the motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is a horrible idea for a number of reasons, but I'm wondering if there are any fundamental reasons why it couldn't be done. E.g. perhaps axles would fail from the torque required for extreme braking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I'm specifically asking about the case where the car provides a retarding torque using the motor that exceeds the amount that can be effectively used for regen braking. Or is regenerative braking torque limited by something that can't be solved with more power?&lt;/p&gt;&#xA;" OwnerUserId="21754" LastEditorUserId="21754" LastEditDate="2019-09-10T03:28:52.960" LastActivityDate="2019-09-10T03:28:52.960" Title="Could an EV reverse to brake?" Tags="|automotive-engineering|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30064" PostTypeId="2" ParentId="30063" CreationDate="2019-09-09T22:44:10.497" Score="1" Body="&lt;p&gt;As pointed out above in the comments, this is exactly how &quot;regenerative braking&quot; works: the motor is switched electrically so it works as a generator, and then it soaks up the kinetic energy of the moving car and pumps the resulting electric current into the battery, which gets charged up. while not 100% efficient, this process is good enough to significantly increase the useful range of an electric vehicle that does a lot of stopping and starting, and thus is an essential feature in all electric propulsion schemes currently out there.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-09T22:44:10.497" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30066" PostTypeId="2" ParentId="30063" CreationDate="2019-09-09T23:18:27.383" Score="-2" Body="&lt;p&gt;Lets consider Tesla model S with approximate weight of 2240kg, say the weight is distributed evenly between the front and rear axels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming new tires with friction coefficient of &lt;span class=&quot;math-container&quot;&gt;$ \mu=0.8 \ $&lt;/span&gt;the max friction force F, imparted on each wheel while braking to a full stop will be &lt;span class=&quot;math-container&quot;&gt;$$ F=\mu mg =(2240/4kg)*9.8m* 0.8/s^2 = 4390N  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which if divided by the radius of the wheel will be the torque needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T_{braking}=\frac{4390}{18*0.0254/2} =19512N.m$$&lt;/span&gt;&#xA;The max torque provided by the engine is &lt;span class=&quot;math-container&quot;&gt;$ \ 330N.m \ $&lt;/span&gt; per Tesla specs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we see the engine has to be nearly a 100 times more powerful to provide that much rejuvenating resistance force, even if they could come up with something near that power, disseminating the huge current created in just a few seconds will be a task of fire protection and other logistics. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-09-09T23:25:37.937" LastActivityDate="2019-09-09T23:25:37.937" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30067" PostTypeId="1" CreationDate="2019-09-09T23:37:11.580" Score="1" ViewCount="401" Body="&lt;p&gt;Both &lt;code&gt;bogie&lt;/code&gt; and &lt;code&gt;chassis&lt;/code&gt; seem to define the part of a wheeled vehicle that describes how the wheels are linked together. Are these terms interchangeable? if not how are they different?&lt;/p&gt;&#xA;" OwnerUserId="21755" LastActivityDate="2019-09-10T00:15:00.590" Title="What is the difference between a bogie and a chassis" Tags="|mechanical-engineering|robotics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30068" PostTypeId="2" ParentId="30040" CreationDate="2019-09-09T23:53:39.097" Score="1" Body="&lt;p&gt;Axial torque will never produce bending, just twisting. To get bending you need moments about the directions perpendicular to the shaft/beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your first figure the force &lt;span class=&quot;math-container&quot;&gt;$Fc$&lt;/span&gt; results in a moment along the vertical direction &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; because the force is offset in the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; direction, as well as a moment along the &lt;span class=&quot;math-container&quot;&gt;$z$&lt;/span&gt; direction because the force is offset in the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/YL6Ow.png&quot; alt=&quot;diagram&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="1051" LastActivityDate="2019-09-09T23:53:39.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30069" PostTypeId="2" ParentId="30067" CreationDate="2019-09-10T00:15:00.590" Score="2" Body="&lt;p&gt;A &lt;a href=&quot;https://en.wikipedia.org/wiki/Chassis&quot; rel=&quot;nofollow noreferrer&quot;&gt;chassis&lt;/a&gt; is the term commonly used to describe the overall structure or main structure of almost anything. The case for a computer can be described as the chassis, as it holds/supports/contains the components which comprise the computer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In vehicle terms, the chassis is the structure or frame to which wheels, suspension, engine, body, etc. are attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A bogie {wheel} (sometimes bogey) is considered a secondary load bearing wheel, often used to provide distribution of forces in a tracked vehicle such as a bulldozer. In railway terms, it is the non-powered wheel assembly that supports the rail cars and also can be used to join two cars while providing a pivot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider that the wheel alone is the bogey, the chassis that contains four of the wheels is also called the bogey, primarily in railway applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In tracked applications such as those used in robotics (ref: your tag), the bogey is a reference to each wheel that is non-driven or is a reference to the entire assembly of wheels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fDj2q.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fDj2q.jpg&quot; alt=&quot;tank bogey image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-09-10T00:15:00.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30070" PostTypeId="2" ParentId="30063" CreationDate="2019-09-10T01:09:03.077" Score="0" Body="&lt;p&gt;There are two ways that the &quot;reverse&quot; torque can be used. Technically speaking, of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first way is the &lt;strong&gt;literal&lt;/strong&gt; understanding of the phrase reverse. Instead of delivering the power to move the car forward, the motor now makes the car moving backward. This idea can be used for marine shipping with extremely heavy container ships, but I haven't heard any idea on it being used on cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second idea has been mentioned at the comments and the first answers: regenerative breaking. The motor in this case, instead of &quot;discharge mode&quot; (or normal working), works as a generator now. The efficiency was about &lt;a href=&quot;https://electrek.co/2018/04/24/regenerative-braking-how-it-works/&quot; rel=&quot;nofollow noreferrer&quot;&gt;60~70%&lt;/a&gt;. Still, it was enough to improve the deceleration and increase the range of the EV. &lt;a href=&quot;http://large.stanford.edu/courses/2016/ph240/brown1/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Some sources&lt;/a&gt; suggests that energy consumption can be decreased by 20% in certain scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope that help.&lt;/p&gt;&#xA;" OwnerUserId="21520" LastActivityDate="2019-09-10T01:09:03.077" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30071" PostTypeId="1" CreationDate="2019-09-10T01:29:18.833" Score="0" ViewCount="29" Body="&lt;p&gt;I have built a small, simple linear generator. It works, but I'm sure it could be more efficient because I didn't know how many windings of magnet wire to use.  I know there is a formula I could use to determine how many windings but have been unable to find it. Can someone provide me the formula or where I can find it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now for more details of the project in case that somehow helps. I have a cylinder about 2.5 cm long and about .4 cm inner diameter. I have a cylindrical neodymium magnet that just fits inside and is about .2 in height. Of course, when I hold my fingers at the ends and shake back and forth I get a small current. I would like to be able to light an LED (hopefully a super bright). I realize it will strobe on and off - which is exactly what I want. I haven't yet used it with an LED, but did connect it up to a multitester to see that I was generating a current. Perhaps I will need to upsize the magnet/change the cylinder size (diameter or length). I'm hoping I will not need to change things too much as I have space limitations. But I could change things a bit if needed. If anyone has thoughts, I would be happy to hear them, although the winding issue is more what I'm hoping to have answered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="21756" LastActivityDate="2019-09-10T01:29:18.833" Title="Windings for Linear Generator" Tags="|generator|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30072" PostTypeId="1" CreationDate="2019-09-10T07:51:40.430" Score="0" ViewCount="929" Body="&lt;p&gt;The calculations for a design I am doing are making me go round in circles.  I have simulated the design but I cannot seem to resolve this mathematically and it is really bugging me!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an image of the design - two levers with different moments connected to each other by a hinge at their ends.  Each lever's pivots can slide from left to right.  The system equilibrium is maintained by an actuator between the two levers exerting a static force.  This force is what I am trying to calculate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I need to work out the moment on each lever (easy enough) and resolve for a torque around the hinge point where the two levers are connected. From here I can calculate the force on the actuator using a free body diagram.  Calculating the moment around the hinge point however is proving tricky as it is not as simple as just adding the two moments together (so my simulations tell me).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note. Currently I am only interested in the static force to keep system in equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate the help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lo0dK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lo0dK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10086" LastActivityDate="2019-10-16T09:01:15.173" Title="Resolving moments and forces" Tags="|moments|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="30073" PostTypeId="1" CreationDate="2019-09-10T10:05:15.597" Score="0" ViewCount="108" Body="&lt;p&gt;I'm doing a project that involves knowing the average steady-state heat dissipation rate of a single cylinder air-cooled engine (through the engine wall and fins) at a given RPM. The problem is that I cannot seem to find any research regarding the same. Most of the data available out there revolves around the measurement of the peak heat flux across different components(piston, wall, etc.) at various crank angles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, even a rough value would be good enough.   &lt;/p&gt;&#xA;" OwnerUserId="10075" LastActivityDate="2019-09-10T16:41:35.553" Title="Relationship between Heat Flux and Engine RPM" Tags="|heat-transfer|engines|research|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30074" PostTypeId="2" ParentId="30072" CreationDate="2019-09-10T11:01:57.607" Score="1" Body="&lt;p&gt;With no dimensions at all, things get real messy, real fast here - putting some numbers in should make things a whole lot cleaner. I'm in a rush and haven't proofread. That said, here goes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aC2gn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aC2gn.png&quot; alt=&quot;Moments diagram labelled&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1 - find the reaction forces&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Taking moments about &lt;span class=&quot;math-container&quot;&gt;$R_1$&lt;/span&gt; gives &lt;span class=&quot;math-container&quot;&gt;$R_2=\frac{A(r-p)+B(r+q)}{r+s}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Taking moments about &lt;span class=&quot;math-container&quot;&gt;$R_2$&lt;/span&gt; gives &lt;span class=&quot;math-container&quot;&gt;$R_1=\frac{B(s-q)+A(s+p)}{r+s}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Step 2 - take moments about the hinge&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Looking left: &lt;span class=&quot;math-container&quot;&gt;$rR_1=pA+tF$&lt;/span&gt; therefore &lt;span class=&quot;math-container&quot;&gt;$F=\frac{r\frac{B(s-q)+A(s+p)}{r+s}-pA}{t}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Looking right: &lt;span class=&quot;math-container&quot;&gt;$sR_2=qB+tF$&lt;/span&gt; therefore &lt;span class=&quot;math-container&quot;&gt;$F=\frac{s\frac{A(r-p)+B(r+q)}{r+s}-qB}{t}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you put in values there, how does it compare to your simulation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Fixed mathematical errors after @Kamran pointed them out!!&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2019-09-16T08:19:58.827" LastActivityDate="2019-09-16T08:19:58.827" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30075" PostTypeId="2" ParentId="29919" CreationDate="2019-09-10T15:01:22.120" Score="1" Body="&lt;p&gt;Yes and no, maybe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computer fans are specialized, cheap brushless motors, with their own built-in controllers.  The ones that I took apart 30 years ago had 4-pole coil assemblies and an unknown number of poles in the magnets, and a 3-transistor circuit that actually made the rotation happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should be able to convert one to a galvonometer by disassembling it and driving the coils directly, in two pairs, in opposition.  Whether this will give you enough rotation is an unknown to me -- it depends on whether the magnet has two poles or six (either of which should work, for a fan).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as you don't drive the thing with too much current you shouldn't damage anything -- the two things to worry about are overheating the coils and demagnetizing the magnets.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-10T15:01:22.120" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30077" PostTypeId="2" ParentId="30073" CreationDate="2019-09-10T16:41:35.553" Score="1" Body="&lt;p&gt;Here is a way to get started with an estimate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, start with an e-mail to Briggs &amp;amp; Stratton and ask them. They are the biggest air-cooled engine manufacturer in the US. If anybody knows the answer, it's them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise, google on &quot;energy content in automobile engine exhaust&quot;. The difference in energy content in the exhaust and energy content of the fuel is the energy content lost by conduction out of the engine itself. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-10T16:41:35.553" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30078" PostTypeId="1" CreationDate="2019-09-10T17:03:24.557" Score="0" ViewCount="66" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NLsiV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NLsiV.png&quot; alt=&quot;**Part 1 of the example **&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QffZu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QffZu.png&quot; alt=&quot;Part 2 of the example. Can someone explain how can we find |Z| term and also from where does the cosine term arrive at the final equation. &quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Can someone explain how can we find |Z| term and also from where does the cosine term arrive at the final equation.&lt;/p&gt;&#xA;" OwnerUserId="21763" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-10-10T21:01:50.283" Title="Cannot understand an equation related to impedance. (Power system analysis)" Tags="|electrical-engineering|circuits|power-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="30079" PostTypeId="1" CreationDate="2019-09-10T17:32:26.560" Score="1" ViewCount="255" Body="&lt;p&gt;I know that if I have a thin rope of Young's modulus Y, then getting n of these ropes and holding them in parallel gives a Young's modulus of nY.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I was to braid/weave them, does this affect the Young's modulus? &lt;/p&gt;&#xA;" OwnerUserId="21764" LastActivityDate="2024-03-16T04:05:45.633" Title="Is the young's modulus of a braided rope stronger than the sum of its parts?" Tags="|elastic-modulus|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="30080" PostTypeId="1" CreationDate="2019-09-10T18:08:13.113" Score="0" ViewCount="714" Body="&lt;p&gt;I am wondering is it possible to have a M16 thread on a 15mm rod? How is this manufactured? Any links to more info or answer would be very helpful. &lt;/p&gt;&#xA;" OwnerUserId="21766" LastEditorUserId="17070" LastEditDate="2019-09-28T11:31:59.497" LastActivityDate="2019-09-28T11:31:59.497" Title="Could I use thread size M16 for a 15 mm diameter rod?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30081" PostTypeId="2" ParentId="30080" CreationDate="2019-09-10T19:09:39.120" Score="1" Body="&lt;p&gt;According to &lt;a href=&quot;https://www.cjwinter.com/blog/pitch-diameter-and-blank-diameter-defined&quot; rel=&quot;nofollow noreferrer&quot;&gt;CJ Winter&lt;/a&gt;, the act of rolling threads on a rod requires a blank with a diameter 0.051 mm smaller than the major thread diameter. That's not much of a difference from a 16 mm diameter rod (15.949 mm) and certainly not close to the 15 mm you've queried.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-09-10T19:09:39.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30082" PostTypeId="1" CreationDate="2019-09-10T19:12:21.857" Score="0" ViewCount="164" Body="&lt;p&gt;Say you have a metal rod with a square cross section (say 10cm by 10cm) of a certain length. You'd like to make it less prone to bending. Is it better to:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;keep it the way it is&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;remove material so that it has a T shaped cross section&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I wonder if removing material would actually make it more rigid from a bending perspective simply due to the T shaped cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="21767" LastActivityDate="2019-09-10T20:00:36.340" Title="bending rigidity of a metal rod" Tags="|mechanical-engineering|materials|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30083" PostTypeId="2" ParentId="30079" CreationDate="2019-09-10T19:29:28.100" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I know that if I have a thin rope of Young's modulus Y, then getting n of these ropes and holding them in parallel gives a Young's modulus of nY.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Young's modulus is a bulk property of a material, measured as the ratio of the deformation of a test piece of material to an applied stress.  Ideally it measures the same no matter the size of the test piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you are confusing &lt;a href=&quot;https://en.wikipedia.org/wiki/Young%27s_modulus#Not_to_be_confused_with&quot; rel=&quot;nofollow noreferrer&quot;&gt;Young's modulus&lt;/a&gt; with the spring constant of the rope (per Hooke's Law) -- and the spring constant of an &lt;em&gt;object&lt;/em&gt; will change based on how it is constructed, even while the Young's modules of the &lt;em&gt;material with which it is constructed&lt;/em&gt; will remain the same.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-10T19:29:28.100" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="30084" PostTypeId="2" ParentId="30082" CreationDate="2019-09-10T19:41:49.737" Score="1" Body="&lt;p&gt;What you want to do is look at the I value (I want to call it moment of inertia but this terms varies slightly depending where you are in the world) for the section.   If you are limiting the T section to the same maximum values as the square section, the T section is going to have a smaller I value.  Since your moment (bending) capacity is highly dependent on the I value, T section will correspondingly have a lower capacity compared to the square.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the advantages of using a T section is you may be able to reduce your material used.  If the T section has sufficient strength to meet your needs, then you would save the cost of the difference in the cross sectional areas.  Cost in terms of material in dollars as well as the cost of material in weight as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the nice things about the square shape is that it has the same bending capacity about X-X and Y-Y in both directions.  The T shape will have the same bending capacity about Y-Y in both directions, but its capacity will change about X-X depending which way its being bent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the things that will limit your T section compared to your Square section is something called Lateral Torsional buckling.  Basically as the unsupported length of the beam increases, the section will want to start to twist.  As some point this twisting action (Torsion) will cause the member to buckle sideways and the beam to fail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So of the two sections the square is stronger than the T.&lt;/p&gt;&#xA;" OwnerUserId="6993" LastEditorUserId="6993" LastEditDate="2019-09-10T20:00:36.340" LastActivityDate="2019-09-10T20:00:36.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30086" PostTypeId="1" CreationDate="2019-09-10T20:20:14.993" Score="0" ViewCount="577" Body="&lt;p&gt;as a child all of us have built a tower out of Lego. we built it as high as we could but then eventually it topples over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is there a formula to calculate how wide and heavy the base has to be for the tower not to topple&lt;/p&gt;&#xA;" OwnerUserId="2054" LastActivityDate="2019-09-11T04:41:37.637" Title="is there a formula to determine if a structure will topple over" Tags="|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30087" PostTypeId="2" ParentId="30078" CreationDate="2019-09-10T20:24:37.863" Score="1" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\underline Z$&lt;/span&gt; is, by definition, &lt;span class=&quot;math-container&quot;&gt;$\frac{\underline V}{\underline I}$&lt;/span&gt;.  So &lt;span class=&quot;math-container&quot;&gt;$\underline Z$&lt;/span&gt; is not shown, but it comes from the calculation for &lt;span class=&quot;math-container&quot;&gt;$\underline V$&lt;/span&gt; as a function of &lt;span class=&quot;math-container&quot;&gt;$\underline I$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The magnitude and phase angle is computed from &lt;span class=&quot;math-container&quot;&gt;$\underline Z$&lt;/span&gt;, and are basic complex number calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cosine term in the final equation comes from the definition of &lt;span class=&quot;math-container&quot;&gt;$i(t)$&lt;/span&gt; at the top of the example, which is in terms of &lt;span class=&quot;math-container&quot;&gt;$\cos \omega$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-10T20:24:37.863" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30089" PostTypeId="2" ParentId="30060" CreationDate="2019-09-10T21:33:33.823" Score="0" Body="&lt;p&gt;I am not an expert on roller compacted concrete but I did find a few links to reports on the topic. They are listed below.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.fhwa.dot.gov/pavement/pub_details.cfm?id=993&quot; rel=&quot;nofollow noreferrer&quot;&gt;FHWA Tech Brief on RCC&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://lib.dr.iastate.edu/cgi/viewcontent.cgi?referer=&amp;amp;httpsredir=1&amp;amp;article=1100&amp;amp;context=intrans_reports&quot; rel=&quot;nofollow noreferrer&quot;&gt;National Concrete Pavement Technology Center Guide for RCC Pavements&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.cement.org/cement-concrete-applications/paving/roller-compacted-concrete-(rcc)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Portland Cement Association RCC Page&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://rccpavementcouncil.org/category/resources/&quot; rel=&quot;nofollow noreferrer&quot;&gt;RCC Pavement Council Resources Page&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I hope you are able to find some useful information in the links above.&lt;/p&gt;&#xA;" OwnerUserId="20067" LastActivityDate="2019-09-10T21:33:33.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="30091" PostTypeId="2" ParentId="30086" CreationDate="2019-09-11T00:32:34.757" Score="3" Body="&lt;p&gt;Yes there is. Lets say you have a tall building with multiple stories and several unsymmetrical mass protrusions at different floors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To verify that your building is not going to topple:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;You find the center of mass of the buliding.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You drop a vertical line from this center of mass to the footprint of the building, which is usually the exterior extents of its foundation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If the vertical line fall within the footprint the building will not topple. If it falls outside of the footprint it is not stable and will topple on the side where it falls outside of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is called static overturning stability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many building codes require the vertical line to fall within an area cantered around the center of the geometry of that surface with sides parallel to footprint and 1/3 each side length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In siesmic areas this rule changes to  more conservative set of rules considering the dynamic behavior of the building and its tendency to potential rotational vibration or possibly resonance of buildings vibration with the ground motions spectra. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-11T00:32:34.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31091" PostTypeId="2" ParentId="30086" CreationDate="2019-09-11T04:41:37.637" Score="1" Body="&lt;p&gt;If your building is very tall and &quot;slender&quot;, then the rigidity also comes into play in a set of phenomenon called &lt;a href=&quot;https://en.wikipedia.org/wiki/Buckling#Columns&quot; rel=&quot;nofollow noreferrer&quot;&gt;slender column buckling&lt;/a&gt;. For a Lego tower, the tl;dr; is that at some point, the tower will start bending over and then topple due to the kind of situation that kamran described.&lt;/p&gt;&#xA;" OwnerUserId="22771" LastActivityDate="2019-09-11T04:41:37.637" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31093" PostTypeId="1" AcceptedAnswerId="31148" CreationDate="2019-09-11T07:06:11.373" Score="2" ViewCount="226" Body="&lt;p&gt;Recently I noticed a concrete &lt;a href=&quot;https://en.wikipedia.org/wiki/Speed_bump&quot; rel=&quot;nofollow noreferrer&quot;&gt;speed bumper&lt;/a&gt; created near my parking. On enquiring the purpose, the committee who took the decision told me that the purpose of the bumper is to safeguard an existing pipeline below the bumper from heavy vehicles. Some significant points:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Existing road in the complex is made up of concrete and is plain. The road has not suffered any damage from any heavy vehicle in past.&lt;/li&gt;&#xA;&lt;li&gt;The pipe is made up of polypropylene or some similar material. The diameter of pipe is about 6 inch. It is there to direct the water collected from roofs to rainwater harvesting reservoir.&lt;/li&gt;&#xA;&lt;li&gt;Speed bumper is solid and made up of concrete.&lt;/li&gt;&#xA;&lt;li&gt;Heavy vehicle here is a water tanker truck with capacity in multiples of kilolitres (Usually 5KL) filled to the top with water.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;&lt;BR&gt; Is it necessary to raise such a speed bump to prevent any harm to pipeline below concrete plain road? Is it helpful, has no effect or worse in avoiding impact on the pipeline below?&lt;/p&gt;&#xA;" OwnerUserId="22772" LastEditorUserId="22772" LastEditDate="2019-09-20T05:12:21.070" LastActivityDate="2019-09-20T05:12:21.070" Title="Can a speedbump safeguard a pipeline located below?" Tags="|civil-engineering|structural-analysis|structures|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31094" PostTypeId="2" ParentId="31093" CreationDate="2019-09-11T07:21:57.270" Score="0" Body="&lt;p&gt;If the pipe is buried sufficiently deep (more than 18&quot; for example) then no speed bump will be needed. Otherwise all the roads would be continuous speed bumps in many countries...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the pipe has to be close to the surface due to the fall it needs then yes a speed bump or hump may well be necessary to get the traffic over it without damage.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-09-11T07:21:57.270" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31095" PostTypeId="2" ParentId="30080" CreationDate="2019-09-11T07:31:33.757" Score="3" Body="&lt;p&gt;The minor diameter for an M16 Coarse Male thread is 13.55mm, therefore, yes - a rod of outer diameter 15mm could be threaded such that it would accept an M16 nut. The strength of this interface would be significantly reduced over a standard M16 thread/nut combination, however. Whether this matters depends on your application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you wish to add a threaded section to a smooth 15mm rod e.g. at one end, it may make more sense to turn down the diameter to 14mm at that end and thread it at M14. The overall diameter may go down, but the strength of the threaded connection will be higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for &quot;how this is manufactured&quot; - We need more information to provide a helpful answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For small scale, you could simply run a die down an 15mm smooth rod - if you use a lathe (powered down) to keep everything concentric you will get an acceptable thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you could turn down the outer diameter of an M16 rod to 15mm&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-09-11T07:31:33.757" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31096" PostTypeId="1" CreationDate="2019-09-11T10:26:03.293" Score="0" ViewCount="33" Body="&lt;p&gt;This is the equation for bandpass signal from transmitter. What are the values for different parameters present in this equation. For example, what should be the value of P, Sl(t), Fc etc. Also what is the meaning of j in this equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PAFhX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PAFhX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www1.aucegypt.edu/faculty/kseddik/Homepage/Publications_files/Elmossallamy_ICC_19_AmbNulNC.pdf#page=2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt; for context.&lt;/p&gt;&#xA;" OwnerUserId="22775" LastEditorUserId="1592" LastEditDate="2019-09-12T10:36:39.273" LastActivityDate="2019-09-12T10:36:39.273" Title="What does this bandpass signal equation indicates?" Tags="|wireless-communication|digital-communication|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31097" PostTypeId="1" AcceptedAnswerId="31120" CreationDate="2019-09-11T12:47:10.833" Score="1" ViewCount="117" Body="&lt;p&gt;How is the allowable number of axle-load to failure equation actually used in the software? Is there any documentation regarding the calculation of the tensile strains?&lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-09-12T19:51:57.800" Title="What is the &quot;allowable number of axle-load to failure equation&quot; used in Pavement ME Design?" Tags="|pavement|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31098" PostTypeId="1" AcceptedAnswerId="31104" CreationDate="2019-09-11T13:08:25.853" Score="4" ViewCount="1562" Body="&lt;p&gt;There's basically two ways train wheels can operate, the flanges could be either &quot;inner&quot; or &quot;outer&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8MhrA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8MhrA.png&quot; alt=&quot;Inner- and outer-style wheels&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Switches can be made in equivalent ways for inner and outer flanges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oPBa0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oPBa0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could expect that, just like some countries, as well as different railway companies within countries, picked up different rail gauges, they could have also varied between inner- and outer-flanges railways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We could imagine that, just like today some countries have left-hand and right-hand traffic, there would be inner and outer style rail wheels. Except this is not the case, the &quot;inner&quot; style is almost universal. Almost, because the &quot;outer&quot; style indeed did exist in railways' debuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ALZf7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ALZf7.jpg&quot; alt=&quot;Richard Trevithick railway (England)&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a technical reason that makes the &quot;inner&quot; style preferable ?&lt;/p&gt;&#xA;" OwnerUserId="17069" LastEditorUserId="17069" LastEditDate="2019-09-11T17:46:55.783" LastActivityDate="2019-09-11T21:01:27.337" Title="Is there any reason why &quot;inner-flange&quot; style trains are almost universal?" Tags="|mechanical-engineering|rail|transportation|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31099" PostTypeId="2" ParentId="30060" CreationDate="2019-09-11T13:24:19.800" Score="2" Body="&lt;p&gt;Pavement ME does not have an RCC pavement design module so using it to design RCC is not within the bounds of the design procedure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are talking general pavement performance principles, there is some similarity between RCC and undoweled jointed concrete pavement. You could use Pavement ME to look at the potential slab cracking for jointed concrete and expect some similarity with RCC.  Looking at roughness would be more of a stretch.&lt;/p&gt;&#xA;" OwnerUserId="22776" LastActivityDate="2019-09-11T13:24:19.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31100" PostTypeId="2" ParentId="30060" CreationDate="2019-09-11T13:30:12.003" Score="3" Body="&lt;p&gt;RCCPave is the pavement design software tool for RCC, this is the guide that supports the software.  &lt;a href=&quot;http://members.cement.org/EBiz55/ProductCatalog/Product.aspx?ID=1626&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://members.cement.org/EBiz55/ProductCatalog/Product.aspx?ID=1626&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can’t find a download for the software though, must need to contact ACPA.&lt;/p&gt;&#xA;" OwnerUserId="22777" LastActivityDate="2019-09-11T13:30:12.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31101" PostTypeId="2" ParentId="31098" CreationDate="2019-09-11T14:42:15.573" Score="4" Body="&lt;p&gt;On a fast curve the outer flange ( on the inside of the curve) will lift off the rail and the train will leave the tracks. While an inner flange is pushed down onto the rail giving more stability. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-09-11T14:42:15.573" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31102" PostTypeId="2" ParentId="30072" CreationDate="2019-09-11T18:58:57.700" Score="1" Body="&lt;p&gt;I take the liberty of borrowing to the very nice diagram Jonathan has drawn to save time, please let me know if you are not OK with this Jonathan and I delete it. thank you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sN5a0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sN5a0.png&quot; alt=&quot;diagram of the beam.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has made a couple of small errors which we are going to fix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are going to Ignore the hinge and the jack and handle the beam as a straight beam with a length &lt;span class=&quot;math-container&quot;&gt;$l = r+s.$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Taking moments about &lt;span class=&quot;math-container&quot;&gt;$ R_1$&lt;/span&gt; gives &lt;span class=&quot;math-container&quot;&gt;$ R_2= \frac{A(r-p)+B(r+q)}{r+s}$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Taking moments about &lt;span class=&quot;math-container&quot;&gt;$R_2$&lt;/span&gt; gives &lt;span class=&quot;math-container&quot;&gt;$R_1= \frac{B(s-q)+A(s+p)}{r+s}$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Now we get the free body diagram of R1 to the hinge, including the force F acting at arm t of the jack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Sigma M=0, \  t*F+ A*p-R_1*r=0 \\ tF= R_1*r-A*p \\ F= \frac{R_1*r-A*p}{t}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a double check we could draw the right hand FBD. They should be the same. I leave calculating the shear at the hinge to you.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-11T18:58:57.700" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31104" PostTypeId="2" ParentId="31098" CreationDate="2019-09-11T21:01:27.337" Score="7" Body="&lt;p&gt;If you leave off the flanges and just use cones that are solidly connected to an axle, then cones that taper outward (get &lt;em&gt;smaller&lt;/em&gt; as they go out) tend to be self-centering but undamped (or underdamped, I'm not sure which).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is because if the axle is offset, the wheel that's to the outside has a larger diameter, and tends to drive the axle to turn toward the center.  The &quot;outer flange&quot; design you should would do the opposite.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the &quot;inner flange&quot; picture that you present is mostly not relying on the flanges to keep the axle centered -- it's the taper of the wheels that does that.  The flange is there to damp oscillations, and for when the system is overwhelmed by circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There's a nice video of this on YouTube, but I'm too lazy right now to find it -- try searching on &quot;train tire design&quot;, or maybe those words with the word &quot;stability&quot; tossed in).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-11T21:01:27.337" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31105" PostTypeId="1" CreationDate="2019-09-11T23:46:17.877" Score="1" ViewCount="1095" Body="&lt;p&gt;Good day, &#xA;I would like to enquirer the followings: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the difference between studying materials science and materials engineering?&lt;/li&gt;&#xA;&lt;li&gt;Are there any jobs exclusive to those who have studied materials science/ materials engineering&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="22784" LastEditorUserId="110" LastEditDate="2019-09-12T10:38:12.397" LastActivityDate="2019-09-12T15:04:22.247" Title="Materials science and materials engineering" Tags="|materials|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31106" PostTypeId="1" CreationDate="2019-09-12T02:52:32.787" Score="-1" ViewCount="307" Body="&lt;p&gt;I am designing a liquid dispensing system for CNC machines that will top-up a mixed solution of coolant &amp;amp; water at the required concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Present concentration in the machine sump will almost always be richer than required, so I need to 'dose' the makeup liquid at a concentration of 2-3% to balance the concentration in the sump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water will be supplied under gravity through the plant's service line while the coolant shall be pumped from a barrel. I don't know if infeed of coolant can be automated as well since I've been trying to work on this for a while now and have not been able to come up with any viable solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done received some data from clients wherein I have concluded the following :-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Daily top-up is done once in the morning (start of the first shift).&lt;/li&gt;&#xA;&lt;li&gt;Top-up volumes range from 20-50 Litres.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So a daily top-up of 20-50 Litres @ concentration of 2-3% is what is required. Water flowing under gravity has to be mixed in-line with the coolant and the mixed solution supplied to the tank through a hose. I cannot have an intermediate mixing tank as there is severe crux for space, so the mixing has to be done in the pipes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Queries :-&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I have thought about using an Injection Quill to inject the coolant concentrate into the water pipe, but i am not sure if it will help.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Also, a dosing pump will dose in pulses, while the water flow is continuous. Can I get the right mixing efficiency in such a case?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Will Hall-Effect Turbine flow sensors be accurate enough for such an application to sense the flow-rate?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Are there any compact motorized valves I can use to throttle the flow of water?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I am planning to use a Capacitive Level Transmitter inserted into the tank to know the volume of liquid present and how much will be required.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The present concentration in the tank shall be fed into the HMI by the user. The PLC will then actuate the process with these parameters viz. makeup volume and concentration.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If someone can guide me as to how I can incorporate all this into a workable solution, I'd be forever grateful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;KV&lt;/p&gt;&#xA;" OwnerUserId="16020" LastActivityDate="2020-02-10T02:02:50.380" Title="Automating the Coolant Top-Up for CNC Machines" Tags="|machining|cnc|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31107" PostTypeId="2" ParentId="8977" CreationDate="2019-09-12T12:24:02.450" Score="2" Body="&lt;p&gt;The stress strain diagram of many materials is actually a curve on which there is no definite proportional limit. In such cases stress-strain proportionality is assumed to exist upto a stress at which the strain increases at a rate 50% or more than that shown by intial tangent to stress-strain diagram. If a line is drawn parallel to the intial straight line portion of stress-strain curve from 0.2% strain, it cuts the stress-strain curve at the point where strain rate increases at a rate 50% more than that shown by intial straight line. This is the concept behind chosing 0.2% proof stress as the yield stress. &lt;/p&gt;&#xA;" OwnerUserId="22789" LastActivityDate="2019-09-12T12:24:02.450" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31108" PostTypeId="1" CreationDate="2019-09-12T12:39:35.340" Score="0" ViewCount="3685" Body="&lt;p&gt;Household TV remote control use Infrared rays. Infrared rays have frequency from 300 GHz to 430 THz. What is the range of frequency used in TV remote?&lt;/p&gt;&#xA;" OwnerUserId="6614" LastActivityDate="2022-11-10T12:28:52.343" Title="What is the range of frequency used in TV remote?" Tags="|consumer-electronics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31109" PostTypeId="2" ParentId="31108" CreationDate="2019-09-12T13:08:01.110" Score="2" Body="&lt;p&gt;The most common type of IR diode used in things like remotes emits at a wavelength of 940nm. This corresponds to a frequency of 320THz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that this is the frequency of the light emitted by the diode.  The diode is pulsed on and off at about 40kHz (different manufacturers use different frequencies), and the receiver knows to only look for IR modulated at around that frequency. This is what makes it possible for the receiver to find the signal from the remote despite all the background emissions in a normal room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 40kHz is then turned on and off at about 1-2kHz. This on and off pattern tells the receiver which button has been pressed.&lt;/p&gt;&#xA;" OwnerUserId="12022" LastEditorUserId="86" LastEditDate="2019-09-13T13:26:03.890" LastActivityDate="2019-09-13T13:26:03.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31110" PostTypeId="2" ParentId="31105" CreationDate="2019-09-12T13:25:05.653" Score="1" Body="&lt;p&gt;I frame my answers based on the materials tetrahedron modified as published in the &lt;a href=&quot;https://www.nap.edu/catalog/758/materials-science-and-engineering-for-the-1990s-maintaining-competitiveness-in&quot; rel=&quot;nofollow noreferrer&quot;&gt;1990 Research Council Report in the US&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Materials science &lt;em&gt;defines&lt;/em&gt; relationships between nature (chemistry and structure) and properties (mechanical, environmental, thermal, electrical, optical, magnetic) or between preparation (synthesis and processing) and nature. Materials engineering &lt;em&gt;applies&lt;/em&gt; relationships between properties and performance (succeed/fail or min/max criteria) or between preparation and properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are two slides from my lectures for a course on the nature of materials (graduate level) to illustrate these statements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XLrlq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XLrlq.png&quot; alt=&quot;lecture slides on mat sci and mat eng&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The questions addressed in the disciplines are based on &lt;em&gt;why&lt;/em&gt; something is the way it is (science) or &lt;em&gt;how&lt;/em&gt; something can be used for an application (engineering). In a cynical view of the second slide above, one might say that materials scientists care nothing about where their material will be used (performance), they only care what is in it (nature), how it is made (preparation), and how it responds to a stimulus of some kind (properties). On the other hand, materials engineers care nothing about what the material really is at a fundamental level (nature), they only care whether making it (preparation) will give it the right responses (properties) so they can use it successfully in their application (performance). Unfortunately, some might say this cynical view has some level of truth, for example as shown by the cavalier attitude of some materials scientists about whether their materials have any real world applications at all or by the cavalier attitude of materials engineers to seek the fundamental reasons why their materials keep failing in use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The predominant view for jobs in industry in the US is derived from the perspective of materials engineering. Materials science is based more in academia or research labs. Studies classed as materials science also appear in chemistry and physics departments. In my view in the US over the last decade or so, more jobs exist classed solely as materials engineering than those that exist classed solely as materials science.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tangent to this, the predominate approach presented in the plethora of undergraduate textbooks on materials science/engineering starts from the perspective of materials engineers / metallurgists. Metals are the poster child of this group, followed by discussions of ceramics, a section in a chapter on semiconductors, and a chapter on polymers. Witness also the imbalance to metals engineering by specific examples of the snub to fundamental materials science in such textbooks by the lack of fundamental discussions about: the role of secondary bonds to define melting points of polymers, the role of HOMO/LUMO covalent bonding in molecules to explain band structures in semiconductors, or insights about dopant concentrations given in ppX scales translated to number density levels so as to calculate electrical conductivity. Finally, look also to this by the examples that present full chapters focused entirely on the various classes of engineering metals, various ways to process steel, various types of corrosion of steel in application systems, or TtT phase diagrams for steel.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-09-12T14:59:14.830" LastActivityDate="2019-09-12T14:59:14.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31111" PostTypeId="1" AcceptedAnswerId="31118" CreationDate="2019-09-12T13:36:28.513" Score="0" ViewCount="465" Body="&lt;p&gt;I come from Maths background, but recently I encountered problem where a channel is described as &quot;flat fading channel&quot;. I would like to make some conclusions based on that fact.  I suspect that for radio engineers this is self explanatory, but not for me. Do we prefer a channel to be flat fading vs selective fading? Can we even make the distinction here? If no what are the pros and cons of each?&lt;/p&gt;&#xA;" OwnerUserId="17980" LastActivityDate="2019-09-12T18:09:44.963" Title="Flat Fading vs Selective fading, which one is better?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31112" PostTypeId="1" CreationDate="2019-09-12T13:43:27.767" Score="0" ViewCount="119" Body="&lt;p&gt;I have watched videos on YouTube producing Aluminum powder by mechanical methods. Like &lt;a href=&quot;https://youtu.be/phMgqT6EsYc&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;. It uses Aluminum foil and coffee grinder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vMvAf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vMvAf.png&quot; alt=&quot;Aluminum foil on YouTube video&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SuHGn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SuHGn.png&quot; alt=&quot;Coffee grinder on YouTube video&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Qslmr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Qslmr.png&quot; alt=&quot;Aluminum powder produced by coffee grinder&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;https://youtu.be/s8_mo3at40M&quot; rel=&quot;nofollow noreferrer&quot;&gt;similar videos&lt;/a&gt; which use the same basic method but with some sort of ball mill:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WzkDC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WzkDC.png&quot; alt=&quot;Glass marbles as ball mill&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any reference to produce Aluminum powder by any metallurgic (non-mechanical) method like &lt;a href=&quot;https://en.wikipedia.org/wiki/Everett_Joel_Hall&quot; rel=&quot;nofollow noreferrer&quot;&gt;Everett Joel Hall&lt;/a&gt; method?&lt;/p&gt;&#xA;" OwnerUserId="15178" LastActivityDate="2019-09-12T13:43:27.767" Title="How to produce aluminium powder by E. J. Hall method" Tags="|chemical-engineering|metallurgy|metals|aluminum|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31113" PostTypeId="2" ParentId="31105" CreationDate="2019-09-12T15:04:22.247" Score="0" Body="&lt;p&gt;From a different point of view : Engineers get jobs in industry because they can solve practical problems. Scientists get jobs teaching and doing esoteric research. My first boss had a degree of &quot;metallurgy&quot; ( from MIT) , so a scientist. I had a degree of &quot;metallurgical engineer&quot;. He did not know how stuff worked such as the general differences between DC and AC electric. As an engineer ,I had 2 classes in electrical engineering, as a scientist he had none. He knew about about various metals being soluble in each other ; He did not know you could apply this principle and improve the ductility of some metals with a solution anneal ( dissolving brittle intermetallic compounds). He did do some clever stuff like distilling zinc from scrap ; he got the idea then hired an engineering firm to design and build the furnace. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-09-12T15:04:22.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31114" PostTypeId="2" ParentId="8977" CreationDate="2019-09-12T15:35:47.777" Score="0" Body="&lt;p&gt;Short answer : Because 0.2 % permanent strain ( offset) is what ASTM defined as the yield strength of steel.  This is a good &quot;rule of thumb&quot; definition for most steels , generally corresponding to a 0.5 % total strain. It is NOT the yield point ( unless by coincidence). Yield point is unique characteristic of some steels. Over 100 years ago it was used because it was easy to measure with no stress/strain graph; Also known as &quot;drop of the beam&quot; . I bet money I am the only one here who has seen &quot;drop of the beam&quot; tests conducted for material qualification.   The 0.5 % total strain ( 0.2 % permanent strain) is a good number for most metals and is widely defined as the yield strength. It is a problem for high strength steels so API defines the yield strength for high strength casing as up 0.8 % total strain depending on the grade of steel.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-09-12T15:35:47.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31118" PostTypeId="2" ParentId="31111" CreationDate="2019-09-12T18:09:44.963" Score="2" Body="&lt;p&gt;The names stem from the fact that communications engineers tend to think in the frequency domain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A flat fading channel is one where the entire signal is scaled by the fading, i.e. the signal is multiplied by a gain &lt;span class=&quot;math-container&quot;&gt;$a(t)$&lt;/span&gt; which (usually) varies more slowly than the bandwidth of the channel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A selectively fading channel is one where some time-varying, frequency-selective process is at work.  As an example of a baseband channel, this would be as if you're talking on the phone, and the frequencies from 1000Hz to 1200Hz were being randomly notched out and added back in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flat fading is easier to deal with, because the whole signal is unmolested other than being changed in amplitude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequency selective fading is much harder to deal with (although using OFDM would make it easier), because the impulse response of the channel is constantly changing, so not only is the signal distorted, but the equalizer that you'd need to reassemble it is constantly changing.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-12T18:09:44.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31119" PostTypeId="1" AcceptedAnswerId="31122" CreationDate="2019-09-12T19:43:28.410" Score="2" ViewCount="470" Body="&lt;p&gt;What are the likely causes of a stepped crack up and away from the edge of a lintel? See the diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EbDhn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EbDhn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10270" LastActivityDate="2019-09-12T22:42:43.847" Title="Cause of stepped crack up and away from lintel" Tags="|civil-engineering|masonry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31120" PostTypeId="2" ParentId="31097" CreationDate="2019-09-12T19:51:57.800" Score="2" Body="&lt;p&gt;The allowable number of axle-load repetitions to failure, &lt;span class=&quot;math-container&quot;&gt;$N_f$&lt;/span&gt; is used to calculate the incremental fatigue damage using Miner's law for each analysis interval (which is equal to one month, or two 15-day intervals if there is a freezing cycle in the month). ME pavement design procedure considers the following inputs in the damage calculation process:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Truck traffic: Different FHWA truck classes, different axle loads, tire configurations (single, tandem, tridem and quad axles), hourly and monthly adjustment factors, etc. &lt;/li&gt;&#xA;&lt;li&gt;Temperature variation within the pavement structure as a function of time. This is used to calculate asphalt mixture stiffness &lt;span class=&quot;math-container&quot;&gt;$E_{HMA}$&lt;/span&gt; which is used to calculate &lt;span class=&quot;math-container&quot;&gt;$N_f$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Pavement ME calculates the horizontal tensile strain at the bottom of AC layer (for bottom-up fatigue cracking cracking) and at the AC surface (for top-down fatigue or longitudinal cracking) using JULEA (linear elastic analysis procedure). Within each damage computation interval, the strains are calculated for each &lt;span class=&quot;math-container&quot;&gt;$i^{th}$&lt;/span&gt; group of axle loads (axle passes, &lt;span class=&quot;math-container&quot;&gt;$n_i$&lt;/span&gt;) at critical locations that are defined separately for different axle types. These strains are then used to calculate allowable load repetitions &lt;span class=&quot;math-container&quot;&gt;$N_{f,i}$&lt;/span&gt; for the &lt;span class=&quot;math-container&quot;&gt;$i^{th}$&lt;/span&gt; load group, and damage &lt;span class=&quot;math-container&quot;&gt;$D_i = n_i / N_{f,i}$&lt;/span&gt;. Cumulative damage due to all traffic is the sum of incremental damage caused by each load group in all analysis intervals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;National Cooperative Highway Research Program, “Guide for Mechanistic-Empirical Design of New and Rehabilitated Pavement Structures, Part 3, Chapter 4 - Design of New &amp;amp; Reconstructed Flexible Pavements.” NCHRP Project 1-37A Final Report, Washington, D.C. (2004). [Contains definition of critical strain computation locations]&lt;/li&gt;&#xA;&lt;li&gt;American Association of State Highway and Transportation Officials, “Mechanistic-Empirical Pavement Design Guide - A Manual of Practice”, 2nd edition, Washington D.C. (2015)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="22783" LastActivityDate="2019-09-12T19:51:57.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31121" PostTypeId="1" CreationDate="2019-09-12T22:12:34.713" Score="0" ViewCount="261" Body="&lt;p&gt;I need to compare two STL (Dental model Scanned) files and find the deviation between their points, SO this two STL files have different boundary, so i already used the GEOMAGIC Control software for comparing STL files.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found these two like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.researchgate.net/post/Are_there_any_open_source_alternatives_for_geomagic&quot; rel=&quot;nofollow noreferrer&quot;&gt;Are there any open source alternatives for geomagic?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.researchgate.net/post/how_to_compare_quantify_populate_two_stl_models_having_shape_difference_to_see_result_as_color_stl_file_and_bar_telling_plus_and_minus_changes&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to compare/quantify (populate) two .stl models having shape difference to see result as color stl file and bar telling plus and minus changes?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;useful for finding free alternative software for Geomagic, So i installed the &lt;a href=&quot;http://www.danielgm.net/cc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cloudcompare&lt;/a&gt; in my Linux OS, and aligned two stl file  and used the &lt;strong&gt;&lt;em&gt;compute cloud/mesh distance&lt;/em&gt;&lt;/strong&gt; icon and get this output picture of stl which shown geomagic output above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZE1EP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZE1EP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in Cloudcompare the Auto fit Aligning method don't work correctly and based of the user advises said which almost &lt;a href=&quot;https://www.cloudcompare.org/doc/wiki/index.php?title=Alignment_and_Registration#Fine_registration_with_ICP&quot; rel=&quot;nofollow noreferrer&quot;&gt;don't working&lt;/a&gt; correctly;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Sadly it's still very buggy and not well documented so you're likely&#xA;  to lose more time by using this method (even if it's 'automatic') than&#xA;  directly using one of the above. Maybe in the future if someone dares&#xA;  to dig in the code and fix it...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So i must Align the two model by selecting at least for point in reference model and the target model, so i want dose by using this mechanism (manually give some reference point to align process) give the output acceptable in compare with Geomagic control best Auto fit mechanism for aligning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be very useful to have more knowledge about this kind of aligning (manual selecting some reference point) with Auto fit aligning method because this works are for one academical research. for example i show the parameters for the manual aligning parameter in Cloudcompare Software  and using finally using Cloud registration:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aBGwg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aBGwg.jpg&quot; alt=&quot;gic &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as you can see the model finally has been setting aligned. in the cloud compare site about Cloud Registration method said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Automatic registration Fine registration with ICP For now, the only&#xA;  automatic method to very finely register two entities is the&#xA;  well-known Iterative Closest Point (ICP) algorithm. See the Fine&#xA;  registration (ICP) section.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So i have already used the &lt;a href=&quot;https://www.3dsystems.com/software/geomagic-control-x&quot; rel=&quot;nofollow noreferrer&quot;&gt;Geomagic Control&lt;/a&gt; for finding variance result vs couldcompare here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GGpcj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GGpcj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So dose this photo confirm the acceptable sameness of this two methods used for aligning used for variance Calculations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based of Dear Daniel suggestion i updated the result by calibrating the color of the result and the output shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oqseO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oqseO.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So does it means here we have some color changed in the deviation determined calculated by using CloudCompare Software with using Cloud Registration option with suggested Final Overlap 80% instead of 100%, but i think i must go through the those data outputs comparing with each others,am i right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As some additional request,If  possible, could you possibly explain the the Excel column meaning of the output data of this deviation result made by Cloudcompare, Shown below?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TdBht.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TdBht.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a lot.&lt;/p&gt;&#xA;" OwnerUserId="2678" LastEditorUserId="2678" LastEditDate="2019-09-15T08:23:39.520" LastActivityDate="2024-01-19T05:05:50.667" Title="Cad align with point vs auto fit at Cloudcompare via GEOMAGIC Control" Tags="|cad|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31122" PostTypeId="2" ParentId="31119" CreationDate="2019-09-12T22:42:43.847" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;What are the likely causes of a &lt;strong&gt;stepped&lt;/strong&gt; crack &lt;strong&gt;up and away&lt;/strong&gt; from the edge of a lintel?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Shearing causes stepped cracks and bending causes vertical cracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2hb6Y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2hb6Y.jpg&quot; alt=&quot;Shear vs. Bending Cracking&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SNei3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SNei3.jpg&quot; alt=&quot;Shear Cracking Remediation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &quot;&lt;a href=&quot;https://gharpedia.com/diagonal-cracks-brick-walls/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Diagonal Cracks in Brick Walls and Its Measures&lt;/a&gt;&quot; or &quot;&lt;a href=&quot;https://www.helifix.co.uk/products/remedial-products/helibar-remedial/&quot; rel=&quot;nofollow noreferrer&quot;&gt;HeliBar Remedial&lt;/a&gt;&quot; where the causes and methods for remediation are explained. Two different causes and means to prevent worsening of the problem.&lt;/p&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2019-09-12T22:42:43.847" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31123" PostTypeId="2" ParentId="31106" CreationDate="2019-09-12T23:43:22.917" Score="0" Body="&lt;p&gt;While it's possible to have a &lt;a href=&quot;http://www.flexxcool.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;tank of coolant concentrate&lt;/a&gt; and mix inline if you have so much work to do that automation is practical and manual checking is impractical then you should consider a &lt;a href=&quot;https://www.prab.com/fluid-filtration/guardian-coolant-recycling-systems/&quot; rel=&quot;nofollow noreferrer&quot;&gt;coolant recycling system&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you have noticed, using inline mixing is still going to require checking the machines; either manually or automatically. What you end up with is coolant with slowly deteriorating properties until you flush out periodically and occasionally have entirely fresh and properly mixed coolant. As a point of interest, lots of places (most places?) do exactly that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a recycling system you can remove mechanically dispersed fines and tramp oils which wear and gum up the equipment. Using a refractometer the concentration of purified and properly mixed coolant can be maintained all the time, instead of a varying and deteriorating coolant you can always have &lt;em&gt;good&lt;/em&gt; coolant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A quote from a website offering equipment to &lt;strong&gt;recycle&lt;/strong&gt; the coolant:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;The AC3 works exclusively in conjunction with the PRAB Guardian Central Coolant Recycling System. The Guardian uses a comprehensive coalescing filtration process, effectively removing mechanically dispersed tramp oils to 0.1% or less. The Guardian Coolant Recycling System has demonstrated a reduction in haul-away costs by 90%, and has lowered new coolant purchase up to 75%. &lt;strong&gt;Payback on equipment purchase is typically 6-9 months&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;A different website, &quot;&lt;a href=&quot;https://filtsol.com/main/index.php/products/coolant-recycling-system&quot; rel=&quot;nofollow noreferrer&quot;&gt;FSI - CRS - Coolant Recycling System&lt;/a&gt;&quot; offers a well explained breakdown of costs and savings: the cost of new coolant, hauling away used coolant, and disposal can be significant, let alone the wear and tear on the machines and slight variances that might result in the finished parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That second site has a photo of what a &lt;em&gt;simple&lt;/em&gt; system might look like and also offers a block diagram. They claim an 8 month payback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6oTYC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6oTYC.jpg&quot; alt=&quot;Coolant Recycling&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2019-09-12T23:43:22.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31124" PostTypeId="1" AcceptedAnswerId="31136" CreationDate="2019-09-13T00:40:59.980" Score="1" ViewCount="88" Body="&lt;p&gt;On a recent trip to Cape Elizabeth, Maine, I came across this fog signal:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/utCFA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/utCFA.jpg&quot; alt=&quot;photo of Cape Elizabeth fog signal&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears to have an electromagnetic diaphragm in the middle, surrounded by concrete and some very curiously shaped steel pieces.  Why is the steel surrounding the horn shaped the way it is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found some material on fog signal design, but nothing looks remotely close to what I see here.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Appendix:&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://uslhs.org/light_lists/lighthouse_list.php?id=43&quot; rel=&quot;nofollow noreferrer&quot;&gt;US Lighthouse Society - Cape Elizabeth (last updated 2004)&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.us-lighthouses.com/images/archive/c/cape-elizabeth-lighthouse-and-fog-signal.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.us-lighthouses.com/images/archive/c/cape-elizabeth-lighthouse-and-fog-signal.jpg&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cape_Elizabeth_Lights&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Cape_Elizabeth_Lights&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22797" LastActivityDate="2019-09-13T22:57:15.800" Title="Why is the Cape Elizabeth Fog Signal shaped like this?" Tags="|acoustics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31125" PostTypeId="2" ParentId="30054" CreationDate="2019-09-13T01:12:06.653" Score="0" Body="&lt;p&gt;To make handles (where liquid won't be flowing through the tube) simply crush tubing in a press. It's called tube flattening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F8oUr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F8oUr.jpg&quot; alt=&quot;Flattening Press&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Yi7lA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yi7lA.jpg&quot; alt=&quot;Flattening Press Results&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mZo2e.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mZo2e.png&quot; alt=&quot;Flattening Press&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LPhuI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LPhuI.jpg&quot; alt=&quot;Flattening Press&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NY66x.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NY66x.jpg&quot; alt=&quot;Crush Tubing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that the tubing is simply crushed to bend it into a handle shape using a press.&lt;/p&gt;&#xA;" OwnerUserId="14556" LastActivityDate="2019-09-13T01:12:06.653" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31127" PostTypeId="1" CreationDate="2019-09-13T03:08:31.247" Score="0" ViewCount="146" Body="&lt;p&gt;I have a question about general second-order system transfer functions. I notice two takes on this transfer function. One has a denominator of: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\tau^2 s^2 + 2\zeta\tau s + 1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; (tau) as they defined it - is 'natural period of oscillations' (not time constant), which is seen (for example) in the formula at this link &lt;a href=&quot;https://www.et.byu.edu/~tom/classes/436/ClassNotes/Class20(Second-Order).pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then there is another transfer function (seen in other control engineering textbooks) with a denominator of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$s^2 + 2\zeta w_n s + w_n^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if I divide the first expression by &lt;span class=&quot;math-container&quot;&gt;$\tau^2$&lt;/span&gt;, I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$s^2 + 2\frac{\zeta}{\tau}s + \frac{1}{\tau^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I can now compare &lt;span class=&quot;math-container&quot;&gt;$s^2 + 2\zeta w_n s + w_n^2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$s^2 + 2\frac{\zeta}{\tau}s + \frac{1}{\tau^2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Equating the constants of those two expressions above, I get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$w_n^2 = \frac{1}{\tau^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which boils down to &lt;span class=&quot;math-container&quot;&gt;$w_n = \dfrac{1}{\tau}$&lt;/span&gt; or&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\tau = \frac {1}{w_n}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the natural period of oscillations &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; is supposed to have units of time (eg. seconds). And angular frequency &lt;span class=&quot;math-container&quot;&gt;$w_n$&lt;/span&gt; is supposed to have units of radian per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual mathematical relation between angular frequency and period is supposed to be &lt;span class=&quot;math-container&quot;&gt;$w_n = 2\pi \frac{1}{\tau}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$1/\tau$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$f_n$&lt;/span&gt; (the cyclic frequency in Hz).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The comparison of the two second order transfer functions leads to &lt;span class=&quot;math-container&quot;&gt;$\tau = \frac{1}{w_n}$&lt;/span&gt;. But formally - the relation between &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$w_n$&lt;/span&gt; should actually be &lt;span class=&quot;math-container&quot;&gt;$\tau = 2\frac{\pi}{w_n}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is &lt;span class=&quot;math-container&quot;&gt;$\tau = \frac {1}{w_n}$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$\tau = \frac{2\pi}{w_n}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is - why is the above mathematical conflict occurring? A factor of &lt;span class=&quot;math-container&quot;&gt;$2\pi$&lt;/span&gt; out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm definitely missing something here. But would definitely like to understand the missing link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be that the &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; seen in &lt;span class=&quot;math-container&quot;&gt;$\tau^2 s^2 + 2\zeta\tau s + 1$&lt;/span&gt; has been incorrectly defined as the natural period of oscillations?&lt;/p&gt;&#xA;" OwnerUserId="19788" LastEditorUserId="19788" LastEditDate="2019-09-15T01:10:56.897" LastActivityDate="2019-09-15T01:10:56.897" Title="basic second order system transfer function - natural period and natural frequency" Tags="|control-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31128" PostTypeId="2" ParentId="31121" CreationDate="2019-09-13T08:58:28.700" Score="0" Body="&lt;p&gt;Can you change the color scale parameters so that it matches the one from Geomagic better?&#xA;- at least make it 'symmetrical' (see the second tab of the 'SF display params' section of the mesh properties)&#xA;- then change the saturation to +/- 1.76&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(and you can even create a custom color scale so as to have an even better control: see &lt;a href=&quot;https://www.cloudcompare.org/doc/wiki/index.php?title=Scalar_fields%5CColor_Scales_Manager)&quot; rel=&quot;nofollow noreferrer&quot;&gt;this CloudCompare link&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S.: when you use ICP, make sure to set the 'overlap' parameters to something around 80% as you don't have a perfect match between the two models here.&lt;/p&gt;&#xA;" OwnerUserId="22800" LastEditorUserId="6767" LastEditDate="2020-05-08T20:48:03.060" LastActivityDate="2020-05-08T20:48:03.060" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31134" PostTypeId="1" CreationDate="2019-09-13T16:34:03.793" Score="2" ViewCount="61" Body="&lt;p&gt;Suppose that we were to install bicycle pedals into a passenger car, say a &lt;a href=&quot;https://en.wikipedia.org/wiki/Honda_Civic_(eighth_generation)&quot; rel=&quot;nofollow noreferrer&quot;&gt;civic coupe&lt;/a&gt;, in such a way that pedaling is &lt;a href=&quot;https://en.wikipedia.org/wiki/Bicycle_performance#Mechanical_efficiency&quot; rel=&quot;nofollow noreferrer&quot;&gt;as mechanically efficient as riding an ordinary bike&lt;/a&gt;.  Further suppose that we have a transmission with whatever the appropriate gearing would be for actually moving the beast under pedal-power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the maximum speed of such a vehicle for an ordinary operator &lt;a href=&quot;https://en.wikipedia.org/wiki/Bicycle_performance#Energy_output&quot; rel=&quot;nofollow noreferrer&quot;&gt;who puts out about 200W&lt;/a&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following variables reduce the ambiguity of the problem... Please comment if more info is necessary:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Human power output 200W&lt;/li&gt;&#xA;&lt;li&gt;Sea level&lt;/li&gt;&#xA;&lt;li&gt;2019 Honda Civic Coupe&lt;/li&gt;&#xA;&lt;li&gt;Power lost to friction on gears and whatnot: 0%&lt;/li&gt;&#xA;&lt;li&gt;Power lost to road friction and air resistance: 100% at peak velocity&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE:&lt;/strong&gt; To be clear, I am not asking &lt;a href=&quot;https://engineering.stackexchange.com/help/dont-ask&quot;&gt;an open-ended &quot;what if&quot; type question&lt;/a&gt;.  Specifically, I am not open to arbitrary answers about possible things resulting from a pedal-powered car.  I am specifically interested in its speed limitations, which is a well-grounded physics problem.  This is a &quot;practical, answerable question&quot; in the sense that the force input via human pedal-power and the force applied to the car via friction and air resistance are measurable and reasonably well-known quantities.  Therefore, the maximum speed for a given car propelled by an average human operator should have a well-defined and unambiguous value.  In other words, there is one and only one numerical answer to this question, and it is therefore not open-ended.&lt;/p&gt;&#xA;" OwnerUserId="18937" LastEditorUserId="9" LastEditDate="2019-09-14T06:05:36.750" LastActivityDate="2019-09-14T06:05:36.750" Title="How fast could a human pedal a car?" Tags="|automotive-engineering|bicycles|" AnswerCount="0" CommentCount="1" ClosedDate="2019-09-13T17:20:38.880" ContentLicense="CC BY-SA 4.0" />
  <row Id="31135" PostTypeId="1" AcceptedAnswerId="31562" CreationDate="2019-09-13T19:02:51.580" Score="1" ViewCount="54" Body="&lt;p&gt;In v2.3 of the software, we noticed that in the rehabilitation flexible calibration tab, there are no standard deviation inputs for thermal cracking. They do appear in the new flexible rehabilitation tab. Was standard deviation for thermal cracking removed for flexible rehab designs? If so, when and why?&lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-10-17T18:32:19.687" Title="Pavement Mechanistic-Empirical Design application (v2.3) - Rehabilitation Flexible Calibration Factors" Tags="|pavement|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31136" PostTypeId="2" ParentId="31124" CreationDate="2019-09-13T22:49:57.370" Score="2" Body="&lt;p&gt;The steel structure will be designed as a diffraction grating to create a directional sound beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make a directional sound source, the area covered by the source needs to have linear dimensions several times bigger than the wavelength of the sound, and at low frequencies that means tens of meters. It is impractical to make the actual sound producing device as large as that, but the steel structure &quot;guides&quot; the sound to make the complete structure equivalent to a single loudspeaker cone the same size as the whole frontal area of the device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same principle is used on a smaller scale when creating loudspeaker &quot;stacks&quot; at rock concerts, etc. A tall narrow &quot;stack&quot; of speakers generates a sound field which covers a wide area horizontally (i.e. a wide enough angle to cover the whole of the audience) but a focussed beam in the vertical direction, so you are not wasting energy on sound radiating up into the air, or down into the ground close to the speaker location, but radiating all the sound &quot;horizontally&quot; into the audience area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally this is also why high-frequency loudspeakers are &lt;em&gt;small&lt;/em&gt; (only a few mm in diameter,) so that they do &lt;em&gt;not&lt;/em&gt; produce a directional beam of sound that can only be heard if you are positioned directly in front of the speaker.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-09-13T22:57:15.800" LastActivityDate="2019-09-13T22:57:15.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31137" PostTypeId="1" AcceptedAnswerId="31139" CreationDate="2019-09-14T06:00:53.090" Score="1" ViewCount="69" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;When I start the car, why do I hear the engine working when I'm not pressing the gas pedal?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How exactly does the torque and power of the gears work?&#xA;&lt;br&gt;&#xA;Torque = Radius * Force&#xA;&lt;br&gt;&#xA;Power = Torque * Speed&#xA;&lt;br&gt;&#xA;Do the different gears have different torque? They should, because they have difference radiuses, right? I'm a bit confused on how torque and power relates to the torque and what happens when we switch gears.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How automatic transmission cars detect what torque to put into the wheels and how do they put it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How electric cars detect what torque to put into the wheels and how do they put it?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21282" LastActivityDate="2019-09-14T08:11:45.280" Title="A few questions about cars" Tags="|mechanical-engineering|gears|torque|car|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31138" PostTypeId="2" ParentId="31137" CreationDate="2019-09-14T06:31:42.250" Score="0" Body="&lt;p&gt;Thankfully, I once took an Automotive engineering course. It was two years ago, but I still remember something... Of course, my answer is open for improvement, so if there is any constructive criticism, please go ahead&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The gas pedal is the controller for the throttle (think how much air is going into the engine), and this, in turn, controls the fuel injection. More pressing on the pedal means the throttle opens more, which means more air going in, and then, more fuel is input to use that excess amount of air.&#xA;That means, even without pedal pressing, the engine is still running (albeit with not as much as air as when you are pressing it). This is done by design, because the convential engines lack &quot;instant torque&quot; ability like electric vehicles (as in, you press the pedal, the engine will start and the car can move immediately)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The gear system in a car is used to &quot;decrease&quot; the rotational speed of the engine (say 2000 rounds per minute to about 70 rounds per minute), and with the acknowledgement that the power is (mostly) conserved, the torque value would increase between the output shaft of the engine and the wheel&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I don't remember precise details on the automatic transmission, but &lt;a href=&quot;https://www.youtube.com/watch?v=u_y1S8C0Hmc&quot; rel=&quot;nofollow noreferrer&quot;&gt;this guy has made a video on Youtube&lt;/a&gt;. However, you ask about how to detect &quot;torque at the wheel&quot;. I'm afraid that I don't know the answer. Heck, I'm not even sure if we can actually measure it directly (as in what is the net torque applied at the wheel, from both the engine and the environment at a given time). I only remember we calculate it, deriving it from the formula to gain the requirements for the driveline.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21520" LastActivityDate="2019-09-14T06:31:42.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31139" PostTypeId="2" ParentId="31137" CreationDate="2019-09-14T07:17:43.930" Score="0" Body="&lt;p&gt;The noise you hear is the engine running at the idle speed. To provide minumu energy for lubrication, charging the battery, air conditiong, water pump etc. In newer cars it has been eliminated for duration of intermittent stops to save fuel and pollute less. The computer will start the engine the moment you release the brakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gear box, manual and automatic,  function is to extend the range of engine torque. Because gasoline engines normal RPM is not spread enough to cover all the torque demand of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gears torque has to do with their arrangement in the gear box and as you said to the ratio of their RPM to that of the engine crankshaft. A gear turning 1/2 times faster provides 2 times more torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The automatic transmissions used to be controlled mechanically by the engine RPM, carburetor vacuum and rear axel RPM which has to do with the speed of the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But modern transmissions past 90s are controlled by a computer with many sensors, such as the air and engine temperatures, road condition, humidity, altitude, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electric cars, Tesla namely, don't have a transmission. They have all the torque range needed because they have wide range of RPM up to 20kRPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have a computer which converts the direct current of the battery to alternative 3 phase current whit variable frequency, read variable RPM, via an inverter. Completely controlled by the master computer and in many cases even override the driver to correct his or her errors. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-09-14T08:11:45.280" LastActivityDate="2019-09-14T08:11:45.280" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31140" PostTypeId="1" AcceptedAnswerId="31144" CreationDate="2019-09-14T09:43:48.783" Score="0" ViewCount="717" Body="&lt;p&gt;The homogeneous bar shown in Fig. P-106 is supported by a smooth pin at C and a cable that runs from A to B around the smooth peg at D. Find the stress in the cable if its diameter is 0.6 inch and the bar weighs 6000 lb.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YiQXT.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YiQXT.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When using the equilibrium of moment equation at point C, The Tension in the string is found to be 2957.13 lb.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6000(5) = 5(T) + 10(T)*(3/(34)^(1/2))&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I tried to use the equation of equilibrium for Vertical Forces. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;  6000 = T + T(3/(34)^(1/2))&#xA;&#xA;  T = 3961.714 lb&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GJrUC.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GJrUC.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any mistake in applying the equilibrium vertical condition? I'm having trouble finding any faults. The diagram for forces is given below&lt;/p&gt;&#xA;" OwnerUserId="20416" LastEditorUserId="20416" LastEditDate="2019-09-14T14:53:39.687" LastActivityDate="2019-09-14T17:04:23.613" Title="Calculating Forces using equations of equilibrium" Tags="|stresses|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31141" PostTypeId="1" AcceptedAnswerId="31142" CreationDate="2019-09-14T13:07:52.433" Score="9" ViewCount="14084" Body="&lt;p&gt;I was studying a book on control systems where I read that a block diagram is a specialized high-level type of flow chart.&lt;/p&gt;&#xA;&lt;p&gt;What is the difference between a normal flow chart and a block diagram?&lt;/p&gt;&#xA;" OwnerUserId="22801" LastEditorUserId="29410" LastEditDate="2021-02-28T06:48:13.450" LastActivityDate="2021-02-28T06:48:13.450" Title="Block diagram vs flow chart?" Tags="|control-engineering|control-theory|process-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31142" PostTypeId="2" ParentId="31141" CreationDate="2019-09-14T13:50:17.280" Score="11" Body="&lt;p&gt;Use Google's &lt;code&gt;define:&amp;lt;item to look up&amp;gt;&lt;/code&gt; feature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;block diagram&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;noun&lt;/em&gt;: block diagram; plural noun: block diagrams&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a diagram showing in schematic form the general arrangement of the parts or components of a complex system or process, such as an industrial apparatus or an electronic circuit.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;flow chart&lt;/strong&gt;&lt;br&gt;&#xA;&lt;em&gt;noun&lt;/em&gt;: flow chart; plural noun: flow charts; noun: flowchart; plural noun: flowcharts; noun: flow diagram; plural noun: flow diagrams&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a diagram of the sequence of movements or actions of people or things involved in a complex system or activity.&lt;/li&gt;&#xA;&lt;li&gt;a graphical representation of a computer program in relation to its sequence of functions (as distinct from the data it processes).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rUGOX.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rUGOX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. A sample block diagram of a computer. Source: [Code Cracker}(&lt;a href=&quot;https://codescracker.com/computer-fundamental/block-diagram-of-computer.htm&quot; rel=&quot;noreferrer&quot;&gt;https://codescracker.com/computer-fundamental/block-diagram-of-computer.htm&lt;/a&gt;).&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that while this gives a representation of the structure of a computer it gives no idea of program or firmware &lt;em&gt;flow&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7o5HR.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7o5HR.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 2. A &lt;a href=&quot;https://www.maximintegrated.com/en/app-notes/index.mvp/id/3569&quot; rel=&quot;noreferrer&quot;&gt;simplified flash boot-loader flowchart&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The boot-loader flowchart explains a &lt;em&gt;sequence&lt;/em&gt; but gives no idea of the structure of the device.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2019-09-14T14:12:05.630" LastActivityDate="2019-09-14T14:12:05.630" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31144" PostTypeId="2" ParentId="31140" CreationDate="2019-09-14T17:04:23.613" Score="0" Body="&lt;p&gt;You're on the right track but as noted by commenter Sam, the free body diagram for beam AC should include the support reactions from the pin at point C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/05nKbl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/05nKbl.jpg&quot; alt=&quot;Free Body Diagram of Beam AC&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, the general procedure using the equations of equilibrium will be:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sum moments about point C to solve for cable tension T&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Sigma M_C = -T(5m) - T \frac{3}{\sqrt{34}}(10m) + W(5m) = 0$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Sum forces in vertical direction to solve for the vertical reaction at point C, (Cy)&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Sigma F_y = T + T\frac{3}{\sqrt{34}} - W + C_y = 0$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Sum forces in the horizontal direction to solve for the horizontal reaction at point C, (Cx)&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Sigma F_x = T\frac{5}{\sqrt{34}} - Cx = 0$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Note that I just assumed directions for Cx and Cy in the sketch. If the equilibrium equations produce a negative value for either Cx or Cy it simply means the force acts in the opposite direction to the one I assumed.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2019-09-14T17:04:23.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31145" PostTypeId="1" CreationDate="2019-09-14T17:18:02.623" Score="1" ViewCount="29" Body="&lt;p&gt;In mandatory appendix 2 of ASME section VIII div. 1 the following calculation is given for calculating minimum bolt load required for operating conditions of a flange joint i.e. , under internal pressure:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$W_{m1}=\pi/4\times G^2\times P+2b\times\pi G\times mP$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where P is fluid pressure, G is gasket diameter, b is gasket contact width, m is gasket factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first term is clearly understandable, it is hydrostatic pressure force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second term is gasket force required to maintain a leak proof joint. But it should have been &lt;span class=&quot;math-container&quot;&gt;$b\times\pi G\times mP$&lt;/span&gt;, because area of a thin annulus of diameter G and thickness b is just &lt;span class=&quot;math-container&quot;&gt;$\pi G \times b$&lt;/span&gt; and not &lt;span class=&quot;math-container&quot;&gt;$\pi G \times 2b$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why there is the factor 2 in the second term?&lt;/p&gt;&#xA;" OwnerUserId="22815" LastActivityDate="2019-09-14T17:18:02.623" Title="Why a factor of 2 appears in gasket force calculation in ASME section VIII div. 1 flange design?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31148" PostTypeId="2" ParentId="31093" CreationDate="2019-09-15T05:35:58.973" Score="3" Body="&lt;p&gt;A speed bump not only does not help to protect a submerged pipe, but it may also even damage it more because of the impact of heavy truck axle hitting and falling down from the bump after it just has crossed it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An axel impact on hitting a flat landing area after dropping the face of a speed bump can be several times the weight of the axel. While an axel just passing on, will push the road just barely more than its weight depending on its speed and tire air pressure and some other small factors. We all experience the discomforting jerk on the suspension when we dismount a bump. Some people as a reflex brake partially after the tire rolls down the bump which makes things worse because of the car balking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way that something looking like a speed bump can help is when it spans the pipe, not installed before it and has been actually engineered to work as a bridge such as a bilaterally tapered concrete slab designed with foundations saddling the pipe and adequate rebars. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-15T05:35:58.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31150" PostTypeId="1" AcceptedAnswerId="31151" CreationDate="2019-09-15T06:39:47.037" Score="1" ViewCount="5258" Body="&lt;p&gt;I am trying to figure out the lifting mechanism of a manual Crank Adjustable Height Standing Desk. Please see the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MQBol.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MQBol.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is what it is inside the metal frame (check the red marked area)&#xA;that is causing the entire system go up and down when I move the crank? &lt;/p&gt;&#xA;" OwnerUserId="22824" LastActivityDate="2021-09-18T21:59:54.710" Title="Mechanism of a manual Crank Adjustable Height Standing Desk" Tags="|mechanical-engineering|hydraulics|mechanisms|actuator|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31151" PostTypeId="2" ParentId="31150" CreationDate="2019-09-15T07:14:17.640" Score="4" Body="&lt;p&gt;the crank handle is connected to the little black box, inside of which there are gears that rotate the metal shaft you can see running through the box and across the back of the table support. the right hand end of that shaft enters the support leg on the right which has another gearbox in it that rotates a shaft leading down the length of the support leg. That shaft has threads cut into it which engage a nut fixed to the lower portion of the support leg. rotating the crank handle then rotates the threaded shaft inside the leg which causes that shaft to push up or pull down on the upper half half of the support leg via the bearings inside the gearbox that is inside the top of the leg. This raises or lowers the table. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-15T07:14:17.640" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31152" PostTypeId="1" CreationDate="2019-09-15T08:02:10.467" Score="1" ViewCount="301" Body="&lt;p&gt;I did a Civil Engineering course some years ago and from my textbook I had a question on a pump with hoses. Unfortunately I haven't been able to find a solution to this yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pump feeds two hoses, each of which is 45 m long and is fitted with a nozzle. Each nozzle has a coefficient of velocity of 0.97 and discharges a 37.5 mm diameter jet of water at 24 m/s when the nozzle is at the same level as the pump. If the power lost in overcoming friction in the hoses is not to exceed 20 per cent of the hydraulic power available at the inlet end of the hoses,&#xA;calculate (a) the diameter of the hoses, taking f = 0.007, and (b) the power required to drive the pump if its efficiency is 70 percent and it draws its water supply from a level 3 m below the nozzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the calculations I have so far&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) Head lost in friction Hf = 4 * f * L * v^2/ D * 2g, where L = Length of hose, D = Hose diameter, v = hose velocity and f = friction coefficient&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If H = Head at inlet to pump and hoses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Head at nozzle Hp = H − 4 * f * L * v^2/ D * 2g&#xA;or Hp = Head at inlet to hoses - frictional head loss in each hose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If jet velocity = V and Cv = 0.97 Jet Velocity V=Cv √(2g ∗ Hp), where V = 24m/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hp = V^2/(2g ∗ Cv^2) = (24^2) / 2g * (0.97)^2, so Hp = 31.208 m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For continuity of flow, flow from hoses = flow from nozzle, so 1/4 Pi * D^2 * v = 1/4 * Pi * d^2 * V, where d = jet diameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v = V * (d2/D2) = 24 *(0.0375)^2 / D^2 = v = 0.03375 / D2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have then put this in the frictional head loss formula 4 * f * L * V2/D * 2g = 4 * 0.007 * 45 * (0.03375 D^-2)^2 / (D * 19.62) = 1.43521875 ∗ 10−3 ∗ D−4/ D ∗ 19.62 = 1.43521875 ∗ 10−3/ (D5 ∗ 19.62).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note 2g = 19.62&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If power lost in overcoming friction in hoses is not to exceed 20 per cent of hydraulic power available at inlet end of hoses, then H = 5 * 4 * f * L * v^2/ D2g&#xA;so Hp = 5 ∗ (4 * f * L * v2/ D * 2g) - (4 * f * L * v^2/ D2g) = 4 * (4 * f * L * v^2/ D * 2g) = 4 * 1.43521875 * 10^-3/ (D^5 * 19.62) &#xA;Hp = 5.740875*10^-3/ (D^5*2g)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So 2g * Hp = 5.740875 * 10^-3/ D^5 = 19.62 * 31.2018 = 612.1798m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D^5 = 5.740875 * 10^-3 / 612.179, so D = (5.740875 * 10^-3/ 612.1798)^0.2 = 0.09872m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;so Hose Diameter = 98.72 mm&lt;/strong&gt; &#xA;&lt;strong&gt;IN MY TEXT BOOK ANSWER WAS GIVEN AS 98.5mm, WOULD THIS BE DUE TO ROUNDING?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(b) Power from pump P = w * Q * Hp, &#xA;where w = unit weight of water, Q = flow through pipes and Hp = Head from pump&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to use Bernoulli equation with pump head term in it as well as head loss&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z1 + P1/pg + V1^2/2g + Hp = Z2 + P2/pg + V2^2/2g + Hf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this formula we assume that P1 ands P2 are atmospheric pressures at entry to the pump and exit of the nozzle, so P1 = P2 = 0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Z1 = 0 and Z2 = 3m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V1 = velocity in hoses and V2 = jet velocity from nozzle = 24m/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have to calculate velocity in hoses and then frictional head loss&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From before v = V * (d2/D2) so v = 24 * (0.0375^2)/ (0.09872^2) = 3.463m/s so V1 = 3.463m/s &#xA;Now we have to put V1 into frictional head loss formula Hf = 4  *f * L * v2/ D * 2g&#xA;so Hf = 4 * 0.007 * 45 * (3.463)^2 / (0.09872 * 19.62) = 7.80m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we have to put values in Bernoulli equation to find head from pump&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so using Z1 + P1/pg + V1^2/2g + Hp = Z2 + P2/pg + V2^2/2g + Hf&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0 + 0 + (3.463^2)/2g + Hp = 3 + 0 + (24^2)/2g + 7.8 Hp = 3 + (24^2)/2g + 7.8 - (3.463^2)/2g = 3 + 29.3578 + 7.8 - 0.611 Hp = 39.5468m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow from one hose Q = 1/4 Pi * D^2 * v = 1/4 * Pi * 0.09872^2 * 3.463 = 0.026507m^3/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power from one hose P = w * Q * Hp = 9810 * 0.026507 * 39.5468 = 10284W = 10.284kW as there are 2 hoses, total Power from hoses P = 10.284 * 2 = 20.567kW&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;as the pump has an efficiency of 70 per cent then Power from pump = 20.567/ 0.7 = 29.382kW&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IN MY TEXT BOOK THE ANSWER WAS GIVEN AS 31.1kW, IS THERE SOMETHING ELSE I SHOULD TAKE INTO CONSIDERATION?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8013" LastEditorUserId="8013" LastEditDate="2019-09-23T13:56:43.260" LastActivityDate="2019-09-23T13:56:43.260" Title="Calculate Power for pump" Tags="|fluid-mechanics|pumps|power|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31154" PostTypeId="1" CreationDate="2019-09-15T17:49:49.173" Score="1" ViewCount="1152" Body="&lt;p&gt;Ball bearing reduces friction to a large extent than sliding friction. Motors use a metal ring called bush which slides against the shaft when the shaft is rotating. If a force is exerted on the shaft in direction perpendicular to the direction of the shaft, the motor stops because of the so much friction present between the shaft and the bush. Here's a photo to be clear :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/bYiLF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... Whereas, if a ball bearing is used instead, the force won't be that much effective in that case because of the balls present in the ball bearing as they have to roll and not slide. For those thinking what a bush is, here is a photo:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zcWyP.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;br&gt;&#xA;&lt;sub&gt;&lt;em&gt;(Photo taken from Google, edited)&lt;/em&gt;&lt;/sub&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is the bush not replaced with a ball bearing?&lt;/p&gt;&#xA;" OwnerUserId="17768" LastEditorUserId="17768" LastEditDate="2019-09-15T19:46:16.847" LastActivityDate="2019-09-16T01:27:36.310" Title="Why is the Bush of a motor not replaced with a ball bearing?" Tags="|motors|friction|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31155" PostTypeId="2" ParentId="31154" CreationDate="2019-09-15T18:16:11.083" Score="1" Body="&lt;p&gt;The manufacturer chooses between a bush and a roller bearing based on the conditions that the motor will used in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the side loads are small or negligible and the speed within what they decide then they will use a bush.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are significant side loads and speed conditions then they will choose a roller bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other constraints are the available space for a bearing which is limited by the total size of the motor and the cost of parts and assembly...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some internal combustion engines that have roller bearings on the crankshaft but most engines rely on shell bearings with white metal, similar to a bush, but designed for forced lubrication.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-09-15T18:16:11.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31156" PostTypeId="2" ParentId="31154" CreationDate="2019-09-15T18:52:10.213" Score="1" Body="&lt;p&gt;Ball bearings are much more expensive than oil-impregnated bushings so they are only included in the motor design when the load or service conditions explained by Solar Mike require them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note also that although ball bearings are not needed in internal combustion engine crankshaft supports, some engine manufacturers (like Suzuki) will nonetheless use them so the engine will continue to run even if the oil pump fails during a race. So this is an example of a service condition that is independent of a load condition, in which including ball bearings is prudent.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-15T18:52:10.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31157" PostTypeId="1" CreationDate="2019-09-15T20:32:52.817" Score="1" ViewCount="580" Body="&lt;p&gt;Is there a rule of a thumb equation for the mass flow rate of a stationary axial compressor at sea level, using intake area and RPM? Or do other factors like blade geometry make too much of a difference for such an approximation to be possible?&lt;/p&gt;&#xA;" OwnerUserId="18744" LastEditorUserId="18744" LastEditDate="2019-09-15T20:39:39.237" LastActivityDate="2019-09-16T08:50:08.797" Title="Is there a way to calculate mass flow rate of an axial compressor?" Tags="|compressors|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31158" PostTypeId="1" AcceptedAnswerId="31163" CreationDate="2019-09-15T20:53:08.457" Score="0" ViewCount="151" Body="&lt;p&gt;I was trying to look for an L/D ratio published everywhere on the web and I can't seem to find it (for a project). Since I couldn't find it, I wanted to try calculating myself. Can't seem to find &lt;span class=&quot;math-container&quot;&gt;$C_l$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$C_d$&lt;/span&gt; though without some sort of paywall on research papers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The equations are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;L = &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2} C_{l} \cdot \rho v^2 A$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;D = &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2} C_{d} \cdot \rho v^2 A$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So L/D would just be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{L}{D} = \frac{C_l}{C_d}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or would the effective area be different for L and D, where L is the area of the top/bottom view of the plane, and D is the area of the front/back view?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only quantities I have are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v = Mach 0.92 = 315.62 &lt;span class=&quot;math-container&quot;&gt;$\frac{m}{s}$&lt;/span&gt;..............................top speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; = 0.003996 &lt;span class=&quot;math-container&quot;&gt;$\frac{kg}{m^3}$&lt;/span&gt;....................density of air at 40,000 ft (assuming it's flying at its maximum altitude)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I couldn't really find any area of the plane; only wingspan and vertical length, but that would give an inaccurate area of the plane. How would I determine this? If &lt;span class=&quot;math-container&quot;&gt;$\frac{L}{D}$&lt;/span&gt; is just &lt;span class=&quot;math-container&quot;&gt;$\frac{C_l}{C_d}$&lt;/span&gt; does anyone happen to know &lt;span class=&quot;math-container&quot;&gt;$C_l$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_d$&lt;/span&gt;? Thank you in advance.&lt;/p&gt;&#xA;" OwnerUserId="22828" LastEditorUserId="22828" LastEditDate="2019-09-15T21:08:38.190" LastActivityDate="2019-09-16T04:57:14.633" Title="How do you determine the L/D ratio of F-117 Nighthawk?" Tags="|aerospace-engineering|aircraft-design|drag|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31159" PostTypeId="1" AcceptedAnswerId="31160" CreationDate="2019-09-15T08:37:29.470" Score="0" ViewCount="53" Body="&lt;p&gt;I made a gimbal at home with PVC. Used some cement filled PVC pieces as hanging weight to make the phone steady. But when I shoot with it, I feel some tilting movements in the video due to the momentum of these weights. How can I reduce this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it really possible to make a gimbal at home which can produce some decently stabilised videos? &lt;a href=&quot;https://i.stack.imgur.com/IBNjH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;image of gimbal&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22830" OwnerDisplayName="sanoj" LastEditorUserId="22830" LastEditDate="2019-09-16T07:55:30.773" LastActivityDate="2019-09-16T07:55:30.773" Title="How to reduce tilting in self-made gimbal" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31160" PostTypeId="2" ParentId="31159" CreationDate="2019-09-15T19:32:19.753" Score="0" Body="&lt;p&gt;I'm going to guess at what's happening from your description.  In general, the &quot;moment&quot; on both sides of the pivot (weight x distance) should be close to equal.  You can have slightly more weight at the bottom so it wants to stay upright.  Inertia is what keeps things from tilting when you start or stop moving.  If you have weight just at the bottom (or a lot more weight, like a &quot;zero&quot; weight phone at the top and a concrete filled pipe at the bottom), you have a pendulum, and it will react to every motion.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For mass, you could fill both the top and bottom pipes with equal amounts of concrete.  Weigh them when the concrete has dried if you can measure that weight to the ounce, or use a balance and some additional small weights to see which one is heavier and by how much.  Your phone will add a small amount of weight to the top, as will the bracket you're using to hold it.  If the weight of one pipe will be a few ounces heavier than the weight of the other plus the phone and bracket, you're in business.  Otherwise, attach a small weight to the bottom pipe so it will be a few ounces heavier than the top including the phone and bracket.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, be precise matching the lengths of the arms to the top and bottom pipes.  The phone doesn't weight much, but if you mount it above the top pipe, the weight will be farther from the pivot, which will magnify it.  It would be better to mount it under the top pipe to minimize that, or even better, centered between the top and bottom pipes.  If you have a gimbal in multiple planes, that would complicate putting the phone on the inside.  You could compensate for the phone's distance by adding a little more weight at the bottom.&lt;/p&gt;&#xA;" OwnerUserId="12772" OwnerDisplayName="fixer1234" LastActivityDate="2019-09-15T19:54:36.347" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31161" PostTypeId="2" ParentId="31154" CreationDate="2019-09-16T01:19:46.097" Score="0" Body="&lt;p&gt;In addition to what @ solar mike and @ niels nielsen  said there are some properties of bushings that are better than rolling element bearings ( ball, roller,taper roller, pin, etc). A bushing can imbed some debris and keep working. They are normally chosen as thrust bearings over rolling element. And although generally cheaper than rolling element , they are chosen for application where cost is not a factor such as some large radial and axial compressors where solid silver bushings are used. You can get more info at GGB ( Glacier Garlock Bearing) and &#xA;&quot; tilting pad&quot; type bearings.   &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-09-16T01:27:36.310" LastActivityDate="2019-09-16T01:27:36.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31162" PostTypeId="2" ParentId="31159" CreationDate="2019-09-16T04:35:57.947" Score="0" Body="&lt;p&gt;I made one for my son's camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issue is usually with the bearings. I used 2 pairs of bearings taken from those &quot;spinners&quot; and put them on a threaded rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Used double nuts to &quot;just&quot; load the bearings to provide some damping and he had good results when he was taking videos while skiing...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-09-16T04:35:57.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31163" PostTypeId="2" ParentId="31158" CreationDate="2019-09-16T04:42:49.657" Score="0" Body="&lt;p&gt;Glide ratio, or glide angle meaning when you trim the plane to glide level with the engines off, is numerically equal to L/D ratio.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a start guess I would say with a wing load of 330kg/m2 and very small wing  aspect ratio it shouldn't be more than 12- 14.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This site has more info on F-117 Nighthawk. &lt;a href=&quot;https://jackryan.fandom.com/wiki/F-117_Nighthawk&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jackryan fandom&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-09-16T04:57:14.633" LastActivityDate="2019-09-16T04:57:14.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31164" PostTypeId="2" ParentId="31157" CreationDate="2019-09-16T05:34:40.920" Score="0" Body="&lt;p&gt;The design of the compressor makes all the difference. Shape of the blades, number of stages, design of the stators, and the material heat resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Increasing the pressure of the air increases it's temperature so the material properties is a constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a compromise between compressors designed for performance, and others designed for reliability. If one goes for maximum mass flow rate, the compressor could stall. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-16T05:34:40.920" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31165" PostTypeId="2" ParentId="31157" CreationDate="2019-09-16T08:50:08.797" Score="1" Body="&lt;p&gt;In theory, there is. Every fan can be described by characteristic numbers. One is the delivery number &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt;. It describes the ratio of acutall air flow to theoretical air flow, as given by area and speed and is defined as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \phi = \frac{4\dot{V}}{u\pi d^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$\dot{V}$&lt;/span&gt; as &lt;em&gt;volume&lt;/em&gt; flow rate in m³/s, u fan speed at circumference in m/s and d fan diameter in m. Knowing fan geometry and rpm, you can calculate u, knowing density, you arrive at &lt;span class=&quot;math-container&quot;&gt;$\dot{V}$&lt;/span&gt; so if you know &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; you could calcualte air-flow fom rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that you will be ahrd pressed to find delivery numbers for fans (They are usually not given in spec sheets IIRC).&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-09-16T08:50:08.797" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31167" PostTypeId="2" ParentId="13774" CreationDate="2019-09-16T11:55:56.517" Score="1" Body="&lt;p&gt;These phases can be quite ductile due to their layered structure at the atomic scale, naturally as others have suggested here, it depends on compositions and stoichiometry.&#xA;Under compressive conditions, high levels of performance can be achieved. This can be further enhanced by &lt;a href=&quot;https://www.researchgate.net/publication/304620073_Compressive_performance_and_crack_propagation_in_Al_alloy_Ti2AlC_composites&quot; rel=&quot;nofollow noreferrer&quot;&gt;fabrication of metal-max phase composities&lt;/a&gt;. Generally, crack propagation is the failure mechanism in Max phase materials, which exhibit plastic deformation prior to fracture, unlike most ceramics systems. &lt;/p&gt;&#xA;" OwnerUserId="22832" LastActivityDate="2019-09-16T11:55:56.517" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31169" PostTypeId="1" AcceptedAnswerId="31184" CreationDate="2019-09-16T15:15:22.330" Score="0" ViewCount="373" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8eMKJm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8eMKJm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A pinned system comprising bars of the same material and cross-sectional area is subjected to a vertical load P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am required to form a relationship between F&lt;sub&gt;AB&lt;/sub&gt; and P.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The first part of the solution is as such,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/detT3m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/detT3m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$cos\theta=\frac{\delta_{AB}}{\delta_{AC}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But how is the angle &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; the same after deflection? Shouldn't it be smaller?&lt;/strong&gt;&#xA;And I can't seem to form a relationship without assuming the theta as the same.&lt;/p&gt;&#xA;" OwnerUserId="16469" LastEditorUserId="13311" LastEditDate="2019-09-16T17:57:56.957" LastActivityDate="2019-09-17T15:24:51.617" Title="Pin joint system displacement question" Tags="|mechanical-engineering|structural-analysis|structures|beam|deflection|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31171" PostTypeId="1" CreationDate="2019-09-16T19:18:56.037" Score="3" ViewCount="231" Body="&lt;p&gt;I know the meaning of (instantaneous) “center of rotation” and “center of zero acceleration” for a rigid body. I know that they do not coincide (in general) and that their positions change with the motion of the rigid body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I know that for each point of the body we can define an instantaneous “center of curvature” related to the trajectory. My question is, how are these centers of curvature related to the “center of rotation” and “center of zero acceleration” of the whole rigid body? Is there any relation among them, in the most general case?&lt;/p&gt;&#xA;" OwnerUserId="22837" LastActivityDate="2024-01-25T00:08:56.100" Title="Center of rotation (rigid body)" Tags="|mechanical-engineering|kinematics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31172" PostTypeId="1" AcceptedAnswerId="31179" CreationDate="2019-09-16T19:33:41.157" Score="1" ViewCount="100" Body="&lt;p&gt;What kind of automatic transmission would be the best to connect an electric engine with the rear axle of a RWD car?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NFGZ1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NFGZ1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifications:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;200HP, 3000rpm electric motor&lt;/li&gt;&#xA;&lt;li&gt;top speed 220km/h&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The automatic transmission should have higher output RPM and be able to bring the car to top speed of 200km/h.&lt;/p&gt;&#xA;" OwnerUserId="18760" LastActivityDate="2019-09-17T03:21:32.950" Title="Electric engine and transmission" Tags="|gears|electric-vehicles|transmission|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31173" PostTypeId="1" CreationDate="2019-09-16T20:44:07.047" Score="0" ViewCount="91" Body="&lt;p&gt;I'm an avid fan of Cities: Skylines, and it has occurred to me that one of the types of interchanges that seems to fare very well with the simulated traffic is this three-layer diverging diamond (style) interchange. I realize this isn't a true DDI, because they require traffic lights in a staggered traffic flow on the same second level. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a name for this specific type of interchange, and do they work as well in real life as they do in simulated life?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kyDMl.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kyDMl.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://steamcommunity.com/sharedfiles/filedetails/?id=636634176&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://steamcommunity.com/sharedfiles/filedetails/?id=636634176&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22838" LastActivityDate="2019-09-16T20:44:07.047" Title="Need help identifying this three-layer diverging diamond-style interchange" Tags="|civil-engineering|bridges|traffic-intersections|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31174" PostTypeId="2" ParentId="31169" CreationDate="2019-09-16T21:02:08.717" Score="0" Body="&lt;p&gt;let's call the tension at the bars at B and D, T1 and at C, T. We see that &lt;span class=&quot;math-container&quot;&gt;$T_1=T\frac{ \sqrt3}{2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P = 2*T_1 +T= T +2 T\frac{\sqrt{3}}{2}=2.73205\cdot T$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The share of C bar from P load will be &lt;span class=&quot;math-container&quot;&gt;$$T=P\frac{1}{2.73205}  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \delta_c= \frac{Pl_c}{2.73205EA}= \frac{Pl\sqrt{3}/2}{2.73205EA} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The extended length of the C bar will be &lt;span class=&quot;math-container&quot;&gt;$l_c= Pl(\sqrt{3}/2) + \frac{Pl\sqrt{3}/2}{2.73205EA} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here you can find your extended lengths and angles of the frame after the loading. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The loadings and strains on B and D bar will fall in place obligingly.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-09-17T15:24:51.617" LastActivityDate="2019-09-17T15:24:51.617" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31175" PostTypeId="1" CreationDate="2019-09-16T22:00:44.403" Score="0" ViewCount="45" Body="&lt;p&gt;I'm working on an application where a panel rests on a flat plane, and is hinged along one edge with a bit less than 90 degrees of travel. I need to run a wire from the flat plane to the edge of the panel opposite the hinges. What I'd like is to have a small spool of the wire located near where the target edge meets the plane. The spool should let out wire as the panel slowly rises, and as it slowly falls it should pull the wire back in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would a product that accomplishes this be called? &lt;a href=&quot;https://i.stack.imgur.com/rsgNy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rsgNy.png&quot; alt=&quot;here&amp;#39;s an image to describe the situation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22839" LastActivityDate="2020-02-14T06:03:20.450" Title="I'm looking for a way to keep a wire coiled when it's not being pulled on, and retract when it isn't" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31178" PostTypeId="1" CreationDate="2019-09-17T02:09:12.917" Score="0" ViewCount="112" Body="&lt;p&gt;These question(s) could be potentially trivial, but its difficult to find visuals on CFD of VTOL aircraft. While watching a video of the f-35 performing its vertical take-off/hovering, I noticed the horizontal stabilizers, as well as the flaps being deployed down. Does this happen as a means to &quot;scoop&quot; as much air as possible, to help further push/lif the aircraft up?&lt;br&gt;&#xA;I'm also curious what the overall affect of airfoils on VTOL aircraft such as the harrier/f-35. Do airfoils affect hovering performance or only horizontal transition/flight? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!! &lt;/p&gt;&#xA;" OwnerUserId="22841" LastActivityDate="2019-09-17T03:43:28.570" Title="VTOL Aircraft Take-Off Mechanics Question" Tags="|cfd|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31179" PostTypeId="2" ParentId="31172" CreationDate="2019-09-17T03:21:32.950" Score="1" Body="&lt;p&gt;The essential question here is why or why not use a transmission in an electric car. Here are the underlying issues. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is commonly asserted that a DC electric motor is a constant-torque device, developing the same torque at standstill (0 RPM) that it does at its rated RPM (in this case, 3000). Advocates of the &quot;no-transmission&quot; position cite this as a reason why electric motors in cars do not need transmissions, but instead can be used in direct-drive mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the flaw in this reasoning is that power- the capacity to perform work at a certain rate- is the product of torque &lt;em&gt;and RPM&lt;/em&gt;, and it is this &lt;em&gt;product&lt;/em&gt; which accelerates the car. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that in the case of a motor which is starting from zero RPM to accelerate a car, its &lt;em&gt;power output&lt;/em&gt; at near-zero RPM is &lt;em&gt;near zero&lt;/em&gt;, and the motor does not develop its rated power output &lt;em&gt;until it is running at its rated RPM&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that to maximize the power output of the motor and thereby maximize the car's acceleration, we need to get the motor all the way up to 3000RPM as quickly as possible, which means &lt;em&gt;gearing the motor down&lt;/em&gt; when starting off from zero speed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, when we reach 3000RPM, we &lt;em&gt;upshift&lt;/em&gt; the transmission to a &quot;taller&quot; gear ratio, and shift again when the motor comes up to 3000RPM, and repeat until the drag force on the car body is equal to the force applied to the pavement by the rear wheels. this process is designed to keep the motor running at or near its maximum power point for as long as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, the fastest acceleration will be had if the transmission is capable of holding the motor right at 3000RPM throughout the car's acceleration from zero wheel speed to whatever its top speed is. This means that to &lt;em&gt;maximize acceleration&lt;/em&gt;, a transmission is necessary, and the best one will be a continuously-variable one that locks the engine speed at 3000RPM at all times.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-17T03:21:32.950" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31180" PostTypeId="2" ParentId="31178" CreationDate="2019-09-17T03:43:28.570" Score="1" Body="&lt;p&gt;Yes as you guess the flaps and stabilizer being down are to take advantage of the ground effect and scoop more air deflected by the ground at take off and also as you guessed again they help smooth transition into horizontal flight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the engines gradually rotate back to horizontal configuration, directing the thrust vector back the flaps and horizontal stabilizers angled down helps with additional lift and redundant authority of the controls. When enough lift from the wings and empennage is produced the flaps and stabilizers are retrieved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a softer and safer transition. Especially under the combat conditions where the pilot may lack perfect visual ques. However the VT is not an official part of the F-35's maneuvers. VL is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like you am an avid aviation hobbyist and currently fly a Cessna 172, but decades ago I did aerobatics and hang gliding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Her is a rendering buy Computer Fluid Design software. &lt;a href=&quot;https://i.stack.imgur.com/b8Vee.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b8Vee.png&quot; alt=&quot;CFD rendering&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-17T03:43:28.570" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31181" PostTypeId="2" ParentId="31175" CreationDate="2019-09-17T04:04:09.040" Score="0" Body="&lt;p&gt;you can use a small spring loaded roller for window shade. They cost in the range of $20 and many hardware stores custom cut them down to something like 6 inches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can remove the shade and wind your wire around it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is a photo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YQFgI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YQFgI.jpg&quot; alt=&quot;spring loaded roller blind.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-17T04:04:09.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31182" PostTypeId="2" ParentId="31175" CreationDate="2019-09-17T05:41:25.647" Score="1" Body="&lt;p&gt;Just bring the wire down to the hinge and have it go inline with the hinge so it rotates with the movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is done on car tailgates and works fine. Some wires fail but only after several years...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-09-17T05:41:25.647" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31183" PostTypeId="1" CreationDate="2019-09-17T08:22:47.310" Score="1" ViewCount="162" Body="&lt;p&gt;I am using AutoCAD 2016. In model space, I draw a colored 3D object. Then I made an isometric view of that object in the paper space (layout page). In the 'Appearance' section of 'Drawing View Editor' I selected the 'Shaded with visible lines' option as below.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O7oxP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O7oxP.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;In spite of that, in the layout page, the shade of the object, and its color is not displaying.&lt;/p&gt;&#xA;&lt;p&gt;I also make it confirm that in the 'Page Setup' window, the 'Display plot styles' checkbox is unchecked.&lt;/p&gt;&#xA;&lt;p&gt;I did many things to have shade colors in layout, but nothing became successful.&lt;/p&gt;&#xA;&lt;p&gt;Can anybody suggest me a solution?&lt;/p&gt;&#xA;" OwnerUserId="22843" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-09-17T08:22:47.310" Title="Why in paper space of AutoCAD 3D are colors of object shade not displaying" Tags="|autocad|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31184" PostTypeId="2" ParentId="31169" CreationDate="2019-09-17T09:55:11.253" Score="1" Body="&lt;p&gt;This is not a statically determinate system. To find how the load splits between the members, start by finding their stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the materials and cross-section areas are the same, the stiffness is inversely proportional to the length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Length of AC = &lt;span class=&quot;math-container&quot;&gt;$L\cos\theta$&lt;/span&gt;. &#xA;If the stiffness of AB is &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;, the stiffness of AC is &lt;span class=&quot;math-container&quot;&gt;$K/\cos\theta$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider a small downward displacement &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; at point A. The length of AB changes from &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$$\sqrt{L^2\sin^2\theta+ (L\cos\theta + x)^2} \approx \sqrt{L^2 + 2Lx\cos\theta}\\ \approx L + x\cos\theta$$&lt;/span&gt; to first order in &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, using the Binomial theorem .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the tension in AB is &lt;span class=&quot;math-container&quot;&gt;$Kx\cos\theta$&lt;/span&gt; and the downward component of the tension is &lt;span class=&quot;math-container&quot;&gt;$Kx \cos^2\theta$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tension in AC is &lt;span class=&quot;math-container&quot;&gt;$Kx/\cos\theta$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we have &lt;span class=&quot;math-container&quot;&gt;$P = 2Kx\cos^2\theta + Kx/\cos\theta$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_{\text{AC}} = Kx/\cos\theta$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P = F_{\text{AC}}(2cos^3\theta + 1)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the diagram &lt;span class=&quot;math-container&quot;&gt;$\cos\theta = \sqrt{3}/2$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$P = F_{\text{AC}}(3\sqrt{3}/4 + 1)$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. &lt;span class=&quot;math-container&quot;&gt;$F_{AC} = 0.435\,P$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-09-17T09:55:11.253" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31185" PostTypeId="1" AcceptedAnswerId="31188" CreationDate="2019-09-17T10:28:28.023" Score="0" ViewCount="137" Body="&lt;p&gt;The web of a H-section column has been completely corroded at the air-moisture interface with the concrete encasement leaving a hole in the web. The proposal is to break out the existing concrete and weld plates onto the web that will bridge over the hole in the web and will be welded on all four sides to good steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plates will be on both sides of the web. If the web is 8 mm what should the thickness of the plates be? At a minimum should they be 4 mm to reinstate the loss of section?&lt;/p&gt;&#xA;" OwnerUserId="10270" LastActivityDate="2019-09-17T14:51:57.870" Title="Plate thickness for repair of completely corroded web" Tags="|structural-engineering|civil-engineering|steel|welding|corrosion|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31186" PostTypeId="1" AcceptedAnswerId="31190" CreationDate="2019-09-17T11:46:39.047" Score="2" ViewCount="518" Body="&lt;p&gt;Should I consider them as a cylinder to find their inertia? Any tips will be helpful.&lt;/p&gt;&#xA;" OwnerUserId="22847" LastActivityDate="2019-09-17T23:20:43.990" Title="How to calculate the moment of inertia of a worm gear?" Tags="|gears|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31187" PostTypeId="2" ParentId="31171" CreationDate="2019-09-17T13:23:01.133" Score="0" Body="&lt;p&gt;Thought experiment - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given a free body &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; that is being acted on by forces, find its instantaneous center of rotation &lt;span class=&quot;math-container&quot;&gt;$P1$&lt;/span&gt; and its instantaneous center of zero acceleration &lt;span class=&quot;math-container&quot;&gt;$P2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this let us solve for the center of curvature for any point &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; on the body? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do so, we need the acceleration vector at &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; and the velocity vector at &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;. That lets us define the osculating circle to the path of &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; at that instant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given the coordinates of &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;, we know the direction of the velocity vector is perpendicular to the segment &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$P1$&lt;/span&gt;, but what is it's magnitude? We need the magnitude of the velocity at some other point (not &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;) to calculate it. The only other point is &lt;span class=&quot;math-container&quot;&gt;$P2$&lt;/span&gt;. But we know only it's velocity direction also, not it's magnitude. Similarly, we know the direction of the acceleration vector is perpendicular to the segment &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$P2$&lt;/span&gt;, but not it's magnitude, and we don't know the acceleration magnitude of &lt;span class=&quot;math-container&quot;&gt;$P1$&lt;/span&gt; either, so we can't derive it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So knowing the ICR and center of zero acceleration does not provide sufficient constraints to determine the ICC. We know the vector towards the ICC will be co-linear with segment &lt;span class=&quot;math-container&quot;&gt;$\overline{PP1}$&lt;/span&gt;. But we don't know the distance.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2019-09-17T13:39:16.913" LastActivityDate="2019-09-17T13:39:16.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31188" PostTypeId="2" ParentId="31185" CreationDate="2019-09-17T14:51:57.870" Score="1" Body="&lt;p&gt;You need to shore the structure first. Welding heat will lower the strength of the steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then you should use needle gun and grinder to bring the corroded material to bright metal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The patch plates should be thicker than 1/2 of the web thickness. H.S bolts are preferred method of fastening, but if welding is used it should be intermittent and done in alternate runs to avoid heat strains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The job must be primed and sealed before newly removed concrete is replaced with high strength repair concrete with proper acrylic admixture for strength and curing. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-17T14:51:57.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31189" PostTypeId="2" ParentId="29663" CreationDate="2019-09-17T18:55:53.353" Score="0" Body="&lt;p&gt;I am an electrical engineer but took some time to study pyschrometric charts which allow you to work out much of this stuff graphically quite quickly. I'm not an expert but here's my understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;This section of the answer is based on the incorrect understanding that it's 25&amp;deg;C outside and 10&amp;deg;C inside.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qyIzD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qyIzD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. The relevant points on a psychrometric chart. Chart source: Wikipedia Commons. Double-click for full resolution.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The outside RH is irrelevant as the membrane is impermeable. For that reason I've put the red dot on the 25&amp;deg;C, 0% RH point to avoid any confusion.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There's going to be a temperature gradient across the insulation and the wall. The water vapour at the barrier will be heated to the outside air temperature (or close to).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;The psychrometic chart tells us that your air contains 0.008 g H&lt;sub&gt;2&lt;/sub&gt; per g of dry air. This is about 8 g/kg or 10 g/m&lt;sup&gt;3&lt;/sup&gt; of air. I like to think of this as two teaspoons of water per m&lt;sup&gt;3&lt;/sup&gt; of air.&lt;/li&gt;&#xA;&lt;li&gt;As the air gets heated (follow the black arrow) the water content doesn't change but the RH decreases in your case from 100% to just below 40%.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;According to some vendors of basement insulation, a setup like this will cause the concrete wall to become relatively dry due to vapor pressure towards the insulation in which the water vapor will condense and allow the water to drip down and then be drained away.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I can't see why any condensation would be expected at the membrane. The dew point at that humidity ratio (water to air ratio) is 10&amp;deg;C so vapour should not condense on the membrane.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I thought an infinite source of water vapor spreads out until all air is saturated with water, so I would expect the wall to end up having 100% RH.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the first part of your sentence were true then the earth's atmosphere would be saturated everywhere. The humidity is regulated by the minimum temperature the air reaches. If the air cools to 0&amp;deg;C it will have about 4 g/kg humidity ratio. If it then warms up to 20&amp;deg;C the RH will fall to about 30%. (Again, note that the actual mass of water hasn't changed.)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will the wall end up relatively dry, at say 70% RH? If yes, how exactly?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I suspect that the RH will vary with the temperature gradient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe someone who actually knows what they're talking about can comment on this answer!&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One comment though: In my diagram, the outside is the 10° side. You seem to have assumed the outside is the 25° side.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Right-ho. Here we go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C7G7C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C7G7C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;Figure 2. The reversed situation.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Here we see that as the air gets closer to the vapour barrier its temperature rises towards 25&amp;deg;C. As it does so the RH drops from 100% to below 40% (black arrow). Condensation should not form because the RH is below the dew point. With my (limited) understanding I can't see why any condensation would occur in the concrete wall or insulation.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;What about condensation on the inside of the vapour barrier?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this to occur the RH has to rise to 100%. Since the humidity ratio won't be changing (the right axis reading on the chart will be constant) we would have to cool the vapour barrier to &amp;lt; 15&amp;deg;C (as shown by the orange arrow) before the air becomes saturated and condensation forms.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2019-09-18T17:31:30.043" LastActivityDate="2019-09-18T17:31:30.043" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31190" PostTypeId="2" ParentId="31186" CreationDate="2019-09-17T19:36:08.210" Score="2" Body="&lt;p&gt;I assume you mean the &lt;em&gt;worm&lt;/em&gt;, not the &lt;em&gt;worm gear&lt;/em&gt;, but I'll append a note about doing the worm gear at the end.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Actually do the integral; if nothing else, it'll justify all those math courses.&lt;/li&gt;&#xA;&lt;li&gt;Do the 3-D route (ick!)&lt;/li&gt;&#xA;&lt;li&gt;Model (on paper) the worm as a stack of disks of alternating diameters; half of them with diameter equal to the worm's minor diameter, and half with diameter equal to the worm's major diameter.  Then look up the moment of inertia of each of your modeled disks in a table, then finally multiply by the number of each size of disk.  This will slightly overestimate the moment of inertia, which probably isn't a bad thing.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you really mean worm &lt;em&gt;gear&lt;/em&gt;, then model two pieces; one a disk with diameter equal to the minor diameter (i.e. from trough to trough on the gear) with density equal to your worm gear material; the other as a shell with ID equal to the gear minor diameter and OD equal to the gear major diameter, and density equal to &lt;em&gt;half&lt;/em&gt; the gear's material density.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, it'll overestimate the moment of inertia, but again, that's probably not a bad thing.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-17T19:36:08.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31191" PostTypeId="1" CreationDate="2019-09-17T21:57:50.587" Score="0" ViewCount="295" Body="&lt;p&gt;In a typical 3D Equilibrium problem. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume you have two cables points supporting a mass. The two cables are K distance above origin and form a quadratic triangle with the mass which is a the same distance K from origin but in + y axis. A strut from origin interest the two cables at mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is. To find tension in two cables, do I need to find moment about origin or do I just need to equate the I,j and k components of the two tension forces to zero and solve then using Cramer's rule?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When do I need moment to find tension? When don't I need it?&lt;/p&gt;&#xA;" OwnerUserId="21634" LastActivityDate="2019-09-17T21:57:50.587" Title="Finding Tension in cable in 3D Equilibrium" Tags="|mechanical-engineering|statics|moments|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31192" PostTypeId="2" ParentId="31186" CreationDate="2019-09-17T23:20:43.990" Score="0" Body="&lt;p&gt;The thread's mass is half of the mass of a cylinder with a thickness of the difference between inner and outer. It's CG is 1/3 out from inner circle. 1/3*1/2= 1/6. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you calculate the I of a cylinder with a radius R = r inner + 1/6 thread depth, you're fine.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-17T23:20:43.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31194" PostTypeId="1" CreationDate="2019-09-18T03:22:33.243" Score="0" ViewCount="352" Body="&lt;p&gt;Consider the following consecutive, second order, irreversible reactions are carried out in a batch reactor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A + S Yield(k1) X&lt;/p&gt;&#xA;&#xA;&lt;p&gt;X + S yield (k2) Y&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One mole of A and two moles of S are initially charged to the batch reactor. Find the mole fraction of X remaining in solution after 75% of A is consumed. Assume k2=2k1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know some differential equations are needed but I'm not sure how to set this up.&lt;/p&gt;&#xA;" OwnerUserId="22853" LastEditorUserId="17070" LastEditDate="2019-09-27T20:55:47.493" LastActivityDate="2019-09-27T20:55:47.493" Title="How does reaction rate depends on concentrations in irreversible reactions?" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31195" PostTypeId="1" CreationDate="2019-09-18T04:54:06.203" Score="1" ViewCount="61" Body="&lt;p&gt;What is &lt;span class=&quot;math-container&quot;&gt;$V_{IH}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V_{IL}$&lt;/span&gt; in datasheets? I found High-level input voltage and Low-level input voltage on the datasheet?&lt;/p&gt;&#xA;" OwnerUserId="21636" LastEditorUserId="8055" LastEditDate="2019-09-19T06:35:10.763" LastActivityDate="2021-01-24T16:11:52.643" Title="Electronic component datasheet handling" Tags="|power-electronics|rf-electronics|consumer-electronics|microelectronics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31197" PostTypeId="1" CreationDate="2019-09-18T07:42:08.803" Score="1" ViewCount="102" Body="&lt;p&gt;I found these symbols in a electronic datasheet.. Are these all logic gates? or another?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IKoVT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IKoVT.jpg&quot; alt=&quot;What does this symbols mean?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21636" LastActivityDate="2019-09-18T11:06:39.497" Title="Electronic symboles in datasheets" Tags="|electrical-engineering|power-electronics|telecommunication|electrical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31198" PostTypeId="1" CreationDate="2019-09-18T08:27:40.533" Score="0" ViewCount="73" Body="&lt;p&gt;I have a MIMO system with a 2x2 transfer matrix, for example consider:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\begin{pmatrix}&#xA;\frac{2}{s+1} &amp;amp; \frac{1}{(s+1)(s+2)}\\ &#xA; \frac{1}{(s+1)(s+2)}&amp;amp;\frac{2}{s+2} &#xA;\end{pmatrix}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know how to characterize the influence that each channel has on another channel using Matlab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I would like characterize the effect of the coupling between the channels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that to see how much a channel influences another i could use RGA, but is there also a way to see this by looking at the step responses and the magnitudes of the frequency responses?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank's in advance.&lt;/p&gt;&#xA;" OwnerUserId="22780" LastEditorUserId="22780" LastEditDate="2019-09-18T13:06:12.743" LastActivityDate="2019-09-18T13:06:12.743" Title="effect of coupling in MIMO systems" Tags="|control-engineering|control-theory|matlab|transfer-function|systems-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31199" PostTypeId="2" ParentId="31197" CreationDate="2019-09-18T11:06:39.497" Score="1" Body="&lt;p&gt;Yes! The logic gates from a RS485/422. Most likely the signal are level shifted. Below is an except form &lt;a href=&quot;https://www.st.com/resource/en/datasheet/st485b.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Low power RS-485/RS-422 transceiver ST485B, ST485C&lt;/a&gt; datasheet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mhr05.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mhr05.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an another similar part. &lt;a href=&quot;https://www.renesas.com/us/en/www/doc/datasheet/isl32741e.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PROFIBUS Compliant 6kV VDE-Reinforced Isolated RS-485 Transceiver ISL32741E&lt;/a&gt;. The example circuit is converting signal 3.3V-5V and 5V-3.3V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CotFx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CotFx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-09-18T11:06:39.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31200" PostTypeId="2" ParentId="31195" CreationDate="2019-09-18T11:12:07.273" Score="1" Body="&lt;p&gt;You are correct. In the attached document shows minimum 2.0V and above is considered Logic High. Likewise maximum voltage of 0.8V and below is considered logic low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bOjOX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bOjOX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-09-18T23:12:53.557" LastActivityDate="2019-09-18T23:12:53.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31201" PostTypeId="1" AcceptedAnswerId="31204" CreationDate="2019-09-18T12:46:38.240" Score="0" ViewCount="453" Body="&lt;p&gt;A client wishes to use an existing steel beam to lift a 1 tonne load using a sling. The 3 m long beam is measured as UB 203x133 visually observed at both ends to be secure with end-plates with all four bolts present.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;calculation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For conservatism assume the lowest credible strength steel i.e. S275 steel and the lightest section i.e. UB 203 x 133 x 25. From the &lt;a href=&quot;https://www.steelforlifebluebook.co.uk/ub/ec3-ukna/buckling-resistance-s275/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blue Book&lt;/a&gt; for &lt;span class=&quot;math-container&quot;&gt;$L_{ef}$&lt;/span&gt; = 3 m, the design moment capacity &lt;span class=&quot;math-container&quot;&gt;$M_{Rd}$&lt;/span&gt; = 54.6 kNm (&lt;span class=&quot;math-container&quot;&gt;$c_1$&lt;/span&gt; = 1.00).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no dead load on the beam. Use partial safety factor &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; = 1.6 for imposed load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Design load &lt;span class=&quot;math-container&quot;&gt;$F_{Ed}$&lt;/span&gt; = 10 kN x 1.6 = 16&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By inspection beam is OK in shear. All 4 bolts present in each end plate so connection is OK. Check beam in bending.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Treat the beam as simply supported and assuming sling midspan:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$M_{Ed} = \frac{PL}{4} = \frac{16 \times 3}{4} = 12$&lt;/span&gt; kNm &amp;lt; 54.6 kNm &lt;span class=&quot;math-container&quot;&gt;$\therefore$&lt;/span&gt; lift OK&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please provide feedback on this assessment with a criticism or suggestions for improvement.&lt;/p&gt;&#xA;" OwnerUserId="10270" LastActivityDate="2019-09-18T19:32:42.340" Title="Assessment of steel beam for lifting to Eurocode 3" Tags="|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31202" PostTypeId="1" CreationDate="2019-09-18T13:26:47.303" Score="0" ViewCount="82" Body="&lt;p&gt;I am using Simufact from MSC Software to perform numerical simulations.&#xA;To be able to create my own sub-routines I need to have Fortran compiler installed.&#xA;Simufact manual mentions only Intel Fortran + Visual Studio. Both of them are really expensive to purchase.&#xA;I wil lbe guessing, the same method can be used for Marc and Fortran coupling&#xA;Is there a way to use open-source compilers for that? Can you guide me in this matter?&#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="20967" LastActivityDate="2019-09-18T13:26:47.303" Title="Simufact / MARC and Fortran" Tags="|finite-element-method|simulation|numerical-methods|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31203" PostTypeId="1" AcceptedAnswerId="31205" CreationDate="2019-09-18T19:03:34.297" Score="1" ViewCount="92" Body="&lt;p&gt;I want to set up steel truck fenders production line, but have no idea how to do that. Something like this picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit1:&#xA;I have two plans to form final curved shape , using a hydraulic press or using a customized rolling machine .&#xA;Final products must have no riples .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/DD7aB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="22858" LastEditorUserId="22858" LastEditDate="2019-09-20T07:06:41.507" LastActivityDate="2019-09-20T07:06:41.507" Title="How to form steel truck fenders" Tags="|mechanical-engineering|structural-engineering|metal-folding|" AnswerCount="1" CommentCount="3" ClosedDate="2019-09-20T01:58:01.833" ContentLicense="CC BY-SA 4.0" />
  <row Id="31204" PostTypeId="2" ParentId="31201" CreationDate="2019-09-18T19:16:03.153" Score="1" Body="&lt;p&gt;If you use the sling with a winch you have to consult the winch's manufacturer instructions usually printed on a tag or side of the winch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, they recommend a dynamic load factor of 3 to 4 which will make your factored load moment 40kN but still less than allowable 54.6kN.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ton is 9.81kN, but rounding up is ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the rest of your work seems to be ok.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-09-18T19:32:42.340" LastActivityDate="2019-09-18T19:32:42.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31205" PostTypeId="2" ParentId="31203" CreationDate="2019-09-18T19:50:17.530" Score="3" Body="&lt;p&gt;Your raw material will come in the form of a large spool of sheet metal containing a strip of steel hundreds of meters long. You will need an &lt;em&gt;unspooler&lt;/em&gt; to feed steel off the spool and a &lt;em&gt;straightener&lt;/em&gt; to take the curvature out of the steel strip. Then you need either a &lt;em&gt;punch press&lt;/em&gt; or a &lt;em&gt;shear&lt;/em&gt; to cut the sheet metal blanks to size and trim their corners. To put ripples or corrugations into the cut blanks, you will need a &lt;em&gt;roller die&lt;/em&gt; or a &lt;em&gt;rolling mill&lt;/em&gt;. To put a folded edge onto the sides of the blank you will need either a &lt;em&gt;bending brake&lt;/em&gt; or a set of &lt;em&gt;progressive dies&lt;/em&gt;. To bend the fenders into their final curved shape you will need a &lt;em&gt;sheet metal press&lt;/em&gt; and to smooth out any resulting ripples you will need an &lt;em&gt;ironing press&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To attach the mounting brackets you will need an &lt;em&gt;electric spot welder&lt;/em&gt;. Finally, you will need either an &lt;em&gt;electrostatic spray booth&lt;/em&gt; to paint them or a &lt;em&gt;powder coating rig&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-18T19:50:17.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31206" PostTypeId="1" CreationDate="2019-09-18T19:52:09.797" Score="0" ViewCount="201" Body="&lt;p&gt;Does anybody know of what magnitude should the Modulus Of Rigidity of a wool string should be? I know that neither wool or string are &quot;materials&quot; but if anybody can give me a rough estimate of a value I would be very thankful.&lt;/p&gt;&#xA;" OwnerUserId="22860" LastActivityDate="2019-09-18T19:52:09.797" Title="Modulus of Rigidity (Shear Modulus) of string" Tags="|materials|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31207" PostTypeId="1" AcceptedAnswerId="31288" CreationDate="2019-09-19T07:34:28.933" Score="1" ViewCount="1669" Body="&lt;p&gt;I have an &lt;code&gt;.inp&lt;/code&gt; file of Ansys that is basically a mesh, that could be imported into Ansys APDL and converting to &lt;code&gt;.cdb&lt;/code&gt; file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I also have another &lt;code&gt;.IGES&lt;/code&gt; file that I should import in the same model and assemble on the first one. How do I do that ?&lt;/p&gt;&#xA;" OwnerUserId="17070" LastEditorUserId="12198" LastEditDate="2019-09-25T06:33:08.323" LastActivityDate="2019-09-26T19:59:04.547" Title="How do I import an .inp and an IGES together into Ansys workbench?" Tags="|finite-element-method|ansys|ansys-workbench|ansys-apdl|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31208" PostTypeId="1" AcceptedAnswerId="31210" CreationDate="2019-09-19T07:58:30.570" Score="1" ViewCount="6036" Body="&lt;p&gt;I found this bouncy ball in my garage. It is too deflated to use and too inflated to store, so I would like to inflate it or deflate it. It doesn't have a standard valve for a bike pump: the only break in the plastic shell is this solid white plug that I was unable to remove.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rZ8HO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rZ8HO.jpg&quot; alt=&quot;Bouncy ball&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eLk5Y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eLk5Y.jpg&quot; alt=&quot;White plug on the bouncy ball&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is this plug, and can I inflate or deflate the ball?&lt;/p&gt;&#xA;" OwnerUserId="19384" LastEditorUserId="288" LastEditDate="2019-09-19T11:26:29.893" LastActivityDate="2020-06-23T20:10:33.943" Title="How to inflate or deflate a bouncy ball that lacks a standard valve?" Tags="|pumps|valves|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31209" PostTypeId="2" ParentId="15307" CreationDate="2019-09-19T09:09:28.127" Score="0" Body="&lt;p&gt;Thermal energy is, in part, obtained or collected in the breeder blanket concept. Here &lt;a href=&quot;https://www.researchgate.net/publication/263201067_A_Thermal_Discrete_Element_Analysis_of_EU_Solid_Breeder_Blanket_subjected_to_Neutron_Irradiation&quot; rel=&quot;nofollow noreferrer&quot;&gt;neutrons with high energy are used to produce heat and breed valuable tritium&lt;/a&gt; by interaction with lithium ceramic breeder pebbles. From interaction with a high energy neutron, Lithium fissions into helium and tritium. The lithium ceramics (usually &lt;a href=&quot;https://www.researchgate.net/publication/266857952_Solution_based_synthesis_of_mixed-phase_materials_in_the_Li2TiO3-Li4SiO4_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Li4SiO4 or Li2TiO3 or mixtures thereof&lt;/a&gt;) are present in the form of a pebble bed, cooled by helium. Heat, tritium and helium are produced within the bed, thus providing further fuel and useful heat for the generation of electricity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically to sum it up, the neutrons emitted in the fusion reaction serve to deliver energy to the external system that is isolated from the plasma (in which the fusion takes place).&lt;/p&gt;&#xA;" OwnerUserId="22832" LastEditorUserId="9" LastEditDate="2019-09-19T10:54:09.270" LastActivityDate="2019-09-19T10:54:09.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31210" PostTypeId="2" ParentId="31208" CreationDate="2019-09-19T10:20:32.130" Score="3" Body="&lt;p&gt;Yes, normally you can pull those plugs out - takes a strong grip or « folding » the ball to be able to push on the inner end of the plug...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a photo with such a plug taken out:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/w9wXX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w9wXX.jpg&quot; alt=&quot;Photo of plug out of the ball&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="19384" LastEditDate="2019-09-20T21:50:38.617" LastActivityDate="2019-09-20T21:50:38.617" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31211" PostTypeId="1" CreationDate="2019-09-19T10:42:29.860" Score="1" ViewCount="246" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Yx508.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Yx508.png&quot; alt=&quot;Here is the my circuit design for those who asked.&quot;&gt;&lt;/a&gt;I am simulating a pspice bandpass RC filter by cascading 2 opamps one HP and one LP. I am getting this bode plot at the end. Is the unity gain frequency the 2 frequencies at 0 dB?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cpuLn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cpuLn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22866" LastEditorUserId="22866" LastEditDate="2019-09-22T13:52:29.040" LastActivityDate="2019-09-22T13:52:29.040" Title="RC Bandpass Filter on pspice :unity gain frequency" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31213" PostTypeId="1" CreationDate="2019-09-19T13:56:25.610" Score="0" ViewCount="225" Body="&lt;p&gt;I have a tank that is filled with a viscous liquid (~3000 CPs). At the bottom of the tank there is a pipe that leads to a gear pump. There is 22&amp;quot; of pipe straight down from the tank followed by a slight bend (~ 20 degrees) and then 7&amp;quot; to the pump. The liquid level in the tank is variable (H).&lt;/p&gt;&#xA;&lt;p&gt;It is very important that the pump is not starved on the inlet side. The flow-rate out of the pump is 3000 grams per minute. The liquid has a specific gravity of 0.93.&lt;/p&gt;&#xA;&lt;p&gt;I want to know at what viscosity will I starve the pump. Conceptually, I can imagine that as the viscosity of the liquid increases there will be more and more resistance to it flowing from the tank through the pipe to the pump. To compensate for this the pump will have to work harder to pull the liquid through the pipe to achieve the same mass flow rate. At some point the pump will not be able to pull hard enough to overcome the resistance and it will starve. I found the following equations which seem to be the right way to go:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Q = \frac{\Delta P \pi  r^4}{8 \mu L}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The problem I have with the equation above is I do not know how to calculate the delta in the pressure. If I just use gravity (&lt;span class=&quot;math-container&quot;&gt;$\Delta P = \rho g \Delta H$&lt;/span&gt;) the flow rates I get are too small. I assume this is because it is not taking into consideration the negative pressure produced by the pump.&lt;/p&gt;&#xA;&lt;p&gt;Is there a way for me to calculate the hydraulic vacuum force of the pump? Can I put it in terms of pressure? Can I assume there is a pure vacuum inside the pump and say the pressure difference is -1 atm + the contribution from gravity? Is this something I should be able to look up in the manual for the pump?&lt;/p&gt;&#xA;&lt;p&gt;Any help or suggestions are appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="22869" LastEditorUserId="1293" LastEditDate="2020-10-15T03:19:11.123" LastActivityDate="2024-02-27T07:05:50.830" Title="Liquid delivery to a pump based on viscosity" Tags="|fluid-mechanics|pressure|pumps|liquid|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31214" PostTypeId="2" ParentId="31213" CreationDate="2019-09-19T14:06:59.417" Score="0" Body="&lt;p&gt;What you are investigating is - from the perspective of pumping system design - similar to cavitation. for your purpose &lt;span class=&quot;math-container&quot;&gt;$\Delta P$&lt;/span&gt; along the feed pipe is static pressure at pipe inlet minus &lt;em&gt;the minimum allowable&lt;/em&gt; pressure at the pump inlet. The first known, from atmospheric pressure, level in tank and specific gravity. Look up NPSH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is the vapor pressure of your medium at the specific temperature, if you fall below this pressure vapor bubbles will form at your pump inlet and the pump will run dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other thing that could happen is that the pump is not able to deliver to sufficient head on the pressure side, if the available head on the suction side is too low, then the pump may stall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So minimum allowable head on the suction side is the bigger of the two possible heads mentioned above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit to adress question for clarification:&lt;/em&gt; Forget, for a moment, the idea of suction. Only positive pressure exists. Fluids flow from high to low pressure. What does it mean for a pump to run dry? It means the flowrate &lt;em&gt;through&lt;/em&gt; the pump is higher then the flowrate &lt;em&gt;to&lt;/em&gt; the pump. In case of a gear pump, flow rate &lt;em&gt;through&lt;/em&gt; the pump is mostly a function of RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you could assume the absolute pressure at the inlet of the feed pipe and put that into the form above, This would be the flowrate if in place of your pump there's be a vacuum. This means, if you pump at or above this flowrate, there's actually be a vacuum - pump runs dry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, all fluids will form vapor at low enough pressure, from the perspective of your pump vapor is as bad as vacuum. That's why you want the pressure at the pump inlet &lt;em&gt;above&lt;/em&gt; the vapor pressure at that temp.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2019-09-20T19:38:11.223" LastActivityDate="2019-09-20T19:38:11.223" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31216" PostTypeId="1" CreationDate="2019-09-19T20:02:06.133" Score="0" ViewCount="33" Body="&lt;p&gt;Is it possible to create a gear train whose average speed is fixed, but whose speed constantly fluctuates? I'm imagining something that would ideally always complete a single revolution in fixed time, but move irregularly along the way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My interest in this problem is from the writings of Terry Pratchett, whose character Lord Vetinari has a clock which ticks irregularly, but keeps correct time. Numerous versions of this clock have been made using electronic control, but I'm interested in the mechanical possibilities of it.&lt;/p&gt;&#xA;" OwnerUserId="22874" LastActivityDate="2019-09-19T20:46:43.823" Title="Gear train with fixed average speed but irregular instantaneous speed" Tags="|gears|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31219" PostTypeId="2" ParentId="31216" CreationDate="2019-09-19T20:46:43.823" Score="1" Body="&lt;p&gt;So you could design an escape wheel with varying  spaces between the teeth but a fixed total duration.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-09-19T20:46:43.823" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31220" PostTypeId="1" AcceptedAnswerId="31223" CreationDate="2019-09-20T02:05:05.583" Score="0" ViewCount="287" Body="&lt;p&gt;I'm new to the FEA world and trying to learn the ropes through resources available online. I've imported a model into LISA FEA that I would like to analyze, however, when I press solve to conduct my analysis I'm getting an error message that states the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Error: Element 1 is collapsed. More than one local node share the same local node. &#xA;  Warning: Elements 395, 400, ... overlap each other.&#xA;  Failed&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Being so new to this, I do not have any idea as to how to interpret this. M question is: how can I tweak my model to get rid of these error messages. Namely, how do I &quot;uncollapse&quot; an element? How do I stop elements from overlapping?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although this is a LISA specific question, and I would appreciate advice as to how to overcome this directly through LISA, I am also open to general advice applicable to other types of FEA. If anyone could give me a few pointers as to how this issue is dealt with in FEA software, I would be greatly appreciative.&lt;/p&gt;&#xA;" OwnerUserId="19486" LastActivityDate="2019-09-20T10:57:00.947" Title="Issue with collapsed element in LISA Finite Element Analysis" Tags="|finite-element-method|software|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31221" PostTypeId="2" ParentId="18874" CreationDate="2019-09-20T03:34:41.637" Score="0" Body="&lt;p&gt;Here is a trick you can try. Cut the silicone edge so that the edge dimension (for lack of a better term) is a little smaller than you final finished dimension. Now make a temporary casting mold for the cut edge such that the inside edge of the mold adjacent to the cut edge forms a wall , the gap between the cut edge and the smooth mold wall is sized correctly. Now just cast in some silicone molding compound, the same silicone as your prototype. Because silicone bonds to silicone, you will have a perfectly smooth new edge.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2019-09-20T03:34:41.637" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31222" PostTypeId="1" CreationDate="2019-09-20T04:54:34.060" Score="0" ViewCount="76" Body="&lt;p&gt;I am a novice in control systems and just learned about PID controller. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So after watching tons of youtube videos I realized that the input of my plant is actually defined by the error. For the case of PID, when a gain multiplies to the error plus the integral of past errors plus the rate of change of error is the new input of my plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if my error is zero and plant works ideally? meaning there is no history of error being anything other than zero. So three terms have been and will be zero.  Does that mean the input of my plant is zero? That means that the desired output of each plant always should have an input equal to zero so that it would be equal to the setpoint? &lt;/p&gt;&#xA;" OwnerUserId="22876" LastActivityDate="2019-10-20T23:01:40.253" Title="Basic question about PID controller" Tags="|control-engineering|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31223" PostTypeId="2" ParentId="31220" CreationDate="2019-09-20T10:57:00.947" Score="1" Body="&lt;p&gt;I don't use LISA, but the most likely cause of both problems is that the geometry of the element mesh is invalid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;collapsed element&quot; message is saying that two (or more) nodes in element 1 have the same node numbers, or they are at the same position in space. For example if you have a 4-node &quot;rectangular&quot; element, you can't squish it down into a triangle shape by putting two &quot;corners&quot; at the same place, or you can't have a weird shape where two &quot;opposite&quot; sides of the element cross over each other, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;overlapping element&quot; message is saying that two (or more) elements cover the same area or volume. The elements have to fit together like building blocks, or tiles. They can't overlap in an arbitrary way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To see exactly what the problem is, display the model and highlight the element numbers in the error messages, and then figure out how to make a mesh that doesn't have those problems.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-09-20T10:57:00.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31224" PostTypeId="1" AcceptedAnswerId="31225" CreationDate="2019-09-20T11:52:24.597" Score="0" ViewCount="48" Body="&lt;p&gt;This is probably very well explained but I'm lacking the terminology knowledge and that's keeping me from finding the information I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a frame to which I need to attach a weight carrying arm. The arm will need to rotate vertically (1 degree of freedom) but will not be doing complete turns thus a bearing is overkill. The arm has a rectangular profile which is hollow inside, so I was planning to weld 2 small pieces of sheet metal to the frame that will form a bracket inside of which the arm will fit then drill a hole and put a bolt through but I'm afraid that due to the vertical rocking motion the bolt will quickly come loose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What would be the proper solution in this case?&lt;/p&gt;&#xA;" OwnerUserId="20333" LastEditorUserId="20333" LastEditDate="2019-09-20T11:58:58.570" LastActivityDate="2019-09-21T06:43:05.410" Title="Fastening rotating surfaces" Tags="|automotive-engineering|metallurgy|joining|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31225" PostTypeId="2" ParentId="31224" CreationDate="2019-09-20T12:25:06.423" Score="2" Body="&lt;p&gt;Long story short : In any moving part you need to use a bearing, however there are many, many types of bearings. The main two types are : &lt;a href=&quot;https://en.wikipedia.org/wiki/Plain_bearing&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Plain_bearing&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Ball_bearing&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Ball_bearing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also important is the direction of the bearing :&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Thrust_bearing&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Thrust_bearing&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you google on 'buy bearing' you can find them in all shapes and sizes on any technical supplier website. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyway if you insist on using the good old 'this is a bolt and can be used as a bearing' technique just use two nuts. &#xA;One nut as obvious the other as a counter nut tightened to the first one. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck! &lt;/p&gt;&#xA;" OwnerUserId="19798" LastEditorUserId="19798" LastEditDate="2019-09-21T06:43:05.410" LastActivityDate="2019-09-21T06:43:05.410" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31226" PostTypeId="1" CreationDate="2019-09-20T13:06:59.350" Score="0" ViewCount="60" Body="&lt;p&gt;I am not an engineer, just looking for a bit of advice. I hope that it is ok that I ask this here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a DC Motor from a treadmill. It is a DC 2.25hp, 180v, 4700 RPM, 7.5 amp motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to use it to power my small EMCO Lathe. I have an AC (in) to DC (out) variable speed controller. But the speed controller is outputting roughly 210 volts according to my multimeter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could this damage the motor if I connected it to the DC out of the variable speed controller seeing the motor says 180 volts? If so how do I go about finding a solution to this issue? Any information or suggestions would be greatly appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Damian&lt;/p&gt;&#xA;" OwnerUserId="22878" LastActivityDate="2019-10-21T00:02:10.453" Title="DC Motor Question" Tags="|electrical|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31227" PostTypeId="1" AcceptedAnswerId="31231" CreationDate="2019-09-20T16:45:27.553" Score="0" ViewCount="32" Body="&lt;p&gt;The exact reason why adding less than optimal level water but just enough water to drench and dive the whole sasuage in during boiling in a pot isn't the optimal quicking cooking method? My intiuion is that adding a swimming pool of water or a few drip of water won't cook the sasuage.&lt;/p&gt;&#xA;" OwnerUserId="252" LastActivityDate="2019-09-20T19:18:44.883" Title="Why adding less than optimal level water but just enough water to drench the sausage in general isn't the quickest way to cook?" Tags="|heat-transfer|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31228" PostTypeId="1" AcceptedAnswerId="31229" CreationDate="2019-09-20T16:46:42.840" Score="0" ViewCount="228" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vrXQk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vrXQk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Pressure_measurement#Bourdon_gauge&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia:&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Bourdon tubes measure gauge pressure, relative to ambient atmospheric pressure, as opposed to absolute pressure;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But looking at pictures such as the one above, it seems there is only one connection into the tube, the one where the gas whose pressure is being measured enters. It would seem that the deformation of the tube is only dependent on the pressure of the gas. So why does the Bourdon tube measure gauge pressure relative to atmosphere instead of absolute pressure?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2019-09-20T18:20:16.157" Title="Why does Bourdon tube measure relative pressure?" Tags="|pressure|measurements|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31229" PostTypeId="2" ParentId="31228" CreationDate="2019-09-20T18:20:16.157" Score="1" Body="&lt;p&gt;The atmosphere is acting on the outside surface of the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any internal pressure is acting against the atmospheric pressure, which means if you need absolute pressure readings you need to accurately determine the atmospheric pressure at that height and compensate to sea level as necessary.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-09-20T18:20:16.157" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31231" PostTypeId="2" ParentId="31227" CreationDate="2019-09-20T19:18:44.883" Score="1" Body="&lt;p&gt;As long as the sausage is completely immersed, you have optimized heat transfer between the water and the sausage, which will yield the shortest cooking time. Furthermore, by putting in no more water than necessary to accomplish this, you have minimized the heat-up time for the pot and water. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-20T19:18:44.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31232" PostTypeId="2" ParentId="31226" CreationDate="2019-09-20T19:22:32.503" Score="1" Body="&lt;p&gt;It probably will not, because the percentage difference between 180 and 210 is small compared to 210. But the way to tell for sure is to monitor the temperature of the motor case. Every motor has a maximum allowed service temperature, and as long as you stay below that, the motor will be OK. If you can find the manufacturer's name on the case, write them and ask for the max. T on that motor. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-20T19:22:32.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31233" PostTypeId="1" CreationDate="2019-09-20T20:05:44.290" Score="0" ViewCount="622" Body="&lt;p&gt;My fluid handling system requires a valve that will open at low pressures (2-6 PSI) and close at high pressures (8 PSI). Is there any commercially available valve that is suitable for this application?&lt;/p&gt;&#xA;" OwnerUserId="22881" LastActivityDate="2019-09-20T20:24:26.367" Title="Valve that opens at low pressure closes high pressure" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31234" PostTypeId="2" ParentId="31233" CreationDate="2019-09-20T20:24:26.367" Score="1" Body="&lt;p&gt;Search back-pressure regulators. Basic idea is they have a dome with a membrane diaphragm pressurized to the control pressure. If the incoming pressure is higher than the dome it pushes the membrane up and shuts the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a link to one. &lt;a href=&quot;https://www.equilibar.com/back-pressure-regulators/?gclid=CjwKCAjw8ZHsBRA6EiwA7hw_segT_ueYpTuKz-_ruca3aIWbnWMW4LKpG7-5ygorPrnxwlxGeznn4hoCnpsQAvD_BwE&quot; rel=&quot;nofollow noreferrer&quot;&gt;back pressure regulator&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-20T20:24:26.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31236" PostTypeId="2" ParentId="31222" CreationDate="2019-09-20T22:43:17.283" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But what if my error is zero and plant works ideally? meaning there is no history of error being anything other than zero. So three terms have been and will be zero. Does that mean the input of my plant is zero?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;That means that the desired output of each plant always should have an input equal to zero so that it would be equal to the setpoint?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, at least not if I understand what you're saying.  If the plant is &quot;perfect&quot; and linear, then for a setpoint that's not equal to zero, the error will start out high.  The controller will try to push the plant to the setpoint.  Eventually it'll get there, the error will asymptotically settle to zero, and the integrator term will asymptotically settle to whatever it needs to be to hold the plant output at the setpoint.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-20T22:43:17.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31237" PostTypeId="2" ParentId="31226" CreationDate="2019-09-20T23:23:56.467" Score="1" Body="&lt;p&gt;it's reasonable to expect that the motor controller uses pulse wave modulation, which sends electricity in brief, carefully timed pulses corresponding to the desired speed or a close approximation thereof. The full voltage provides the best torque and power (redundant?) while not running the motor at full speed. At short duration pulses, there would be no complications.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-09-20T23:23:56.467" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31238" PostTypeId="1" CreationDate="2019-09-21T04:29:58.940" Score="1" ViewCount="180" Body="&lt;p&gt;The image shown below is a sketch of the structure/frame that I am attempting to analyse using the Hardy Cross (Moment Distribution) Method. I need to find the bending moments, axial forces and shear forces on each of the individual members. After that, I am planning to apply the design equations according to codes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main reason I am stuck is due to the sway generated by the unsymmetric nature of the structure. I am unsure about where to apply the arbitrary sway load and also how the joint rotations are related (using displacement triangles). I need these details in order to analyse the structure using Moment Distribution with Sway. Software is not an option as I do not have access to it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pvA4p.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pvA4p.png&quot; alt=&quot;Geometry and Loading of Frame&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please demonstrate how to solve the structure (mainly find the bending moments).   &lt;/p&gt;&#xA;" OwnerUserId="20822" LastActivityDate="2019-09-21T04:29:58.940" Title="Analysis Of Planar Frame Using Hardy Cross Method (Moment Distribution) With Sway" Tags="|structural-analysis|statics|strength|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31239" PostTypeId="1" CreationDate="2019-09-21T07:48:31.537" Score="1" ViewCount="23" Body="&lt;p&gt;I'd like to know if I could get micelles to pick up microplastics in water. I read that microplastics are sometimes charged, so can we make the lyophobic end to pick up microplastics, so that they can then be collected on the water surface and be easily disposed? This is the idea I'm planning to pitch this for a science fair, is this actually feasible? I read that detergent micelles work well even in hard water, so can we use such detergent micelles to pick up microplastics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="22883" LastActivityDate="2019-09-21T07:48:31.537" Title="Can we engineer detergent micelles to pick up Microplastics?" Tags="|chemical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31240" PostTypeId="1" CreationDate="2019-09-21T09:03:00.533" Score="0" ViewCount="76" Body="&lt;p&gt;Here's a challenge - &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to add different sensors data (height,temperature,distance by laser) to pictures i take with a canon DSLR camera.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would it be possible to connect different sensors through a canon DSLR &quot;hot shoe&quot; or other camera input?&#xA;and then add the sensor information to the picture at the precise moment the picture was taken?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for example - have a temperature sensor connected to the camera, and add the temperature data to the image stored in the camera.&#xA;same way other data that comes with taken pictures such as geo location,shutter info, etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i need it to be VERY precise timing wise - exactly on the time the picture was taken.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;would love to hear your ideas,&#xA;Also looking for someone who can do this as paid project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Macgyver&lt;/p&gt;&#xA;" OwnerUserId="22884" LastEditorUserId="110" LastEditDate="2019-09-21T09:33:28.743" LastActivityDate="2019-10-22T19:02:40.367" Title="How to add external sensor data to pictures, canon DSLR, hot shoe, electronics" Tags="|electrical-engineering|computer-engineering|computer-hardware|consumer-electronics|electronics|" AnswerCount="0" CommentCount="14" ContentLicense="CC BY-SA 4.0" />
  <row Id="31241" PostTypeId="1" CreationDate="2019-09-21T16:08:36.610" Score="0" ViewCount="514" Body="&lt;p&gt;I have a bit of a newbie question. I've inherited a mechanical system, wherein the motor has a D-type output shaft with a diameter of 8mm. &#xA;(&lt;a href=&quot;https://www.ebay.co.uk/itm/DC-High-Torque-Worm-Reducer-Geared-Motor-Electric-Motor-with-Reduction-Gearbox/312731356078?hash=item48d03dcbae:m:mLcgZdBy9RmkbntdUcP9aDA&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ebay.co.uk/itm/DC-High-Torque-Worm-Reducer-Geared-Motor-Electric-Motor-with-Reduction-Gearbox/312731356078?hash=item48d03dcbae:m:mLcgZdBy9RmkbntdUcP9aDA&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motor is supposed to drive a component which has a sprocket (12.7 mm pitch, or 08b-1 standard, if you prefer.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been trying to find a suitable sprocket, however it seems the smallest bore I could find was 10mm, in short - I need to fit a large sprocket on a small shaft. I've been looking at taper bushes, collars, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any way I can increase the diameter of the motor shaft, so that I can install the sprocket via a grub screw? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is, perhaps, worth mentioning that I don't have access to a lot of tools, thus my machining capabilities are quite limited. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance. &lt;/p&gt;&#xA;" OwnerUserId="14581" LastActivityDate="2023-08-08T20:18:34.133" Title="Fitting a sprocket on a small motor shaft" Tags="|mechanical-engineering|machining|machine-design|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31242" PostTypeId="1" CreationDate="2019-09-21T18:08:55.633" Score="0" ViewCount="115" Body="&lt;p&gt;I am facing a difficult question and can't find answers anywhere else. What on the left is a hydraulic ram, it's pushing the box in the centre to move upward, while the four-bar linkage attached on box C is designed to regulate the centre box's motion and make sure it moves on a desired path. Assume all parts are rigid bodies except the hydraulic ram. O1, O2, B and G's location are known. At this instant, the ram provides an upward force which counterbalances the weight of the box, and it also gives a force pointing rightward, which is 1500N. The question is asking for the forces on each bar of the four-bar linkage, ignore the forces due to acceleration. Looks simple right? Just static load analysis. But my question is, where does the 1500N apply to? Does it apply to point G and B at the same time at 1500N? Or it only apply to G at 1500N? Any help is appreciated. I am not asking for the solution of the question, just where the 1500N applies to. Thanks. Diagram is here &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/sb5VH.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="22890" LastEditorUserId="1832" LastEditDate="2019-09-21T23:53:00.810" LastActivityDate="2019-09-21T23:53:00.810" Title="Static load analysis for four-bar linkage?" Tags="|hydraulics|linkage|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31244" PostTypeId="1" CreationDate="2019-09-21T21:05:15.483" Score="3" ViewCount="528" Body="&lt;p&gt;I am trying to find some type of mechanism where I can have a smooth but not too stiff rotation between two metal pieces. I was thinking of putting a shoulder screw through the pieces (the pieces will be side by side) and a nut at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A real world example I can think of is how when you're unfolding a ladder, the part that unfolds has a joint where rotation occurs. It's smooth and stiff enough that you can let go midway and the ladder wouldn't collapse inward, but not too stiff that you have to apply so much force to unfold the ladder. What do they use to do that?&lt;/p&gt;&#xA;" OwnerUserId="20752" LastActivityDate="2019-09-25T21:23:30.200" Title="What mechanism is available to rotate a metal piece on another metal piece?" Tags="|mechanical-engineering|mechanisms|metals|machine-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31245" PostTypeId="2" ParentId="31244" CreationDate="2019-09-21T21:32:54.993" Score="4" Body="&lt;p&gt;You have the right idea using a shoulder screw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In between the two pieces, use flat washers and a &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Belleville_washer&quot; rel=&quot;nofollow noreferrer&quot;&gt;springy Belleville washer&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A Belleville washer, also known as a coned-disc spring, conical spring washer, disc spring, Belleville spring or cupped spring washer, is a conical shell which can be loaded along its axis either statically or dynamically. A Belleville washer is a type of spring shaped like a washer. It is the frusto-conical shape that gives the washer its characteristic spring.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You can even stack them if needed.&#xA;&lt;hr&gt; &lt;a href=&quot;https://www.mcmaster.com/belleville-spring-washers&quot; rel=&quot;nofollow noreferrer&quot;&gt;They are available from McMaster-Carr&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="20985" LastEditorUserId="20985" LastEditDate="2019-09-25T21:23:30.200" LastActivityDate="2019-09-25T21:23:30.200" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31246" PostTypeId="1" CreationDate="2019-09-21T23:32:05.547" Score="0" ViewCount="68" Body="&lt;p&gt;So, for a cyclic corrosion test system, I need a nozzle or arrangement of nozzles that provides a flow rate of 45ml-140ml of salt solution per hour to an area 60cm square. All commercial nozzles I've been able to find with non-flat spray patterns have much higher flow volumes. Does anyone know if a specialty supplier for this type of equipment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I need to ensure that the area is saturated as evenly as possible - what sort of FEA or simulation program would be best for analyzing the distribution of mass/fluid in a spray arrangement?&lt;/p&gt;&#xA;" OwnerUserId="22892" LastActivityDate="2022-10-08T19:17:44.060" Title="Low-Flow Nozzles for Corrosion Spray Testing" Tags="|mechanical-engineering|fluid-mechanics|applied-mechanics|hydraulics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31248" PostTypeId="2" ParentId="31246" CreationDate="2019-09-22T00:22:33.833" Score="1" Body="&lt;p&gt;Why not just use a standard salt spray cabinet ? I am sure it can use different salts other than sea salt. Those we had were roughly 4 ft. cubes which may be big for your work , but may have future uses. And a standard unit results are easy to compare to literature and/or publish.  And as a standard unit it would have some resale value. I don't remember the particular manufacturer but it was some years ago and they have probably changed names. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-09-22T00:22:33.833" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31250" PostTypeId="2" ParentId="26891" CreationDate="2019-09-22T04:57:12.000" Score="0" Body="&lt;p&gt;There is no an absolute answer to your question... it always depends.&#xA;When you are considering to use plane or geodetic surveying, is because you are expecting that you will get differences that will affect to your work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1: You are working on a Skyscraper (Burj Khalifa). Which grid will you setup? It is less than 1 km^2... so plane? geodetic? No. You need to go further... and you need to consider the geoid...Why? because your Nadir (plumb) on the base is not the same on the last levels. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 2: You are working in Center America... in 250 km2 you can cross some countries and then you have 2 different sea levels... How will you handle it? It can not be fixed neither with plane or geodetic survey, as they can not absorv the level gradient on this area...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it always depends...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simple question: On &lt;a href=&quot;https://en.wikipedia.org/wiki/Guadarrama_Tunnel&quot; rel=&quot;nofollow noreferrer&quot;&gt;Guadarrama Tunnel&lt;/a&gt;, in order to create the geodetic grid on between both entries (29 km), how will you work???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For official grids (as they are usually very broad) geodetic surveying is used. When you need to increase the density in those networks, then plane geodesy. But those grids have an accuracy and precision that are ok for general mapping (just as reference), not for almost any work, just if you have a poor tolerance.&lt;/p&gt;&#xA;" OwnerUserId="22894" LastActivityDate="2019-09-22T04:57:12.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31251" PostTypeId="2" ParentId="26094" CreationDate="2019-09-22T05:06:21.000" Score="0" Body="&lt;p&gt;If it is a professional surveyor, when he/she create the DWG, on Units, he should work &quot;without units&quot;, so later you can choose your own units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, you can choose your own units on a DWG file, because the relevant issue is that proportion are ok, so once that you change the units, everything will fit if the survey was properly down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are not happy with it, you can just check one dimension (better as big as you can) and scale the DWG.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depends on the software you are using.&lt;/p&gt;&#xA;" OwnerUserId="22894" LastActivityDate="2019-09-22T05:06:21.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31252" PostTypeId="2" ParentId="10660" CreationDate="2019-09-22T05:21:28.223" Score="-1" Body="&lt;p&gt;By experience, it is not a good idea to take any standard value or manufacturer value. Just check your chain on 10 degrees and on 25 degrees for a very long distance, on the limit of your chain, with no arrow, so on a flat ground. And just make your numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you need to make a high precition work, you can not just take the values of one specification or one brand. You should check by yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So try to find one element with very small contraction or expansion, like a masonry wall and use this one as your base. Then, check the length of it with one low temperature and with a high temperature. If the wheater conditions are not the same, then record it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you have some repetitions with different conditions then you can make your numbers to check the expansion of your chain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The geodetic spanish base at Madridejos it was check with a 4 meters rod. It is a 14644 meters base... so it took 3 years to measure it in order to check the expansion of the invar rod. The accuracy was 2.5 mm for a +14 km baseline.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One manufacturer is giving this values as reference and as maximum deviation, but it is not the real one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow, check &lt;a href=&quot;https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20150018888.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;This&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22894" LastEditorUserId="22894" LastEditDate="2019-09-26T14:43:16.733" LastActivityDate="2019-09-26T14:43:16.733" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31253" PostTypeId="2" ParentId="3061" CreationDate="2019-09-22T05:35:01.860" Score="1" Body="&lt;p&gt;Depends on the accuracy you are looking for:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Solar observation. Traditional astronomical way... just place or use something very high (tower, column...) where you can know, exactly, which one is the base. F.e., on a cathedral tower, you can consider the top of it as you can know where you have the projection on the ground. Then follow the shade during the day... when you get the shortest length, it is pointing you North or South. Depends on the Hemisphere.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Star observation. Just follow one star with a theodolite or astrolab in different moments, until you get which ones are the maximum readings of it (as it is showing the limits of the star rotation against the Earth). Traditionally, the pole star was used as the declination (+89°19′8″) was very high, so the movement is very short. With a few readings, you will get the limits of the movement (by angles) and the center of it is the North. On the South Hemisphere, Crux is used.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;GyroTheodolites.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22894" LastEditorUserId="22894" LastEditDate="2019-09-22T07:29:06.467" LastActivityDate="2019-09-22T07:29:06.467" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31254" PostTypeId="2" ParentId="17045" CreationDate="2019-09-22T06:02:10.260" Score="1" Body="&lt;p&gt;It is not an easy answer... depends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, depends on the length of the tunnel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a short tunnel, let´s say 5 km, it is just a matter to create a grid on the surface, in top of the tunnel to give coordinates to both entries. Later on, the guide of the tunnel is made by total stations (theodolites anciently) by several methods: English traverse methodology (place benchmarks on pairs, make a traverse until the front of the tunnel in let´s say, par benchmarks and return back on the ones still not used... the problem of it is the turning on the front, but for 5 km is ok), English traverse plus axis (it is the same idea plus an open traverse on the axis of the tunnel, that one allows a better compensation and better results), traverse grids (instead to make observations on just traverse, the observations are made on small 4 point benchmarks grids... that ones is the best and most accurate methodology)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a long tunnel, the English traverse is not enough, and you should use the 2nd and 3rd methodology. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow... The study of it should be done before begin, because you can find some surprises... For example, on the Eurotunnel, this question was raised during the construction, and within a few kilometers, the deviation was huge (&gt;50 cm) in both entries (France and England). This deviation was corrected during the remaining tunnel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are another factors that are not considered, that it is not just direction... In long tunnels (&gt;25 km) the &quot;gravity&quot; (plumb) it is not the same, so, if you follow the level from both entries, the tunnel will not match on level, and that one should be corrected against the system you are using (normally ellipsoid).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gyrotheodolites are ok to know the North, but they can not guarantee coordinates, just alignment. So if your coordinates are not matching with a gyrotheodolite... what will you do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Fred... the &quot;plumb&quot; system fro tunnels it is just when you are accessing by shafts and when there is no way to place benchmarks or optical plumbs. It is very inaccurate and risky if you down know how to &quot;plumb&quot;. Vertical shafts are ok when you have small cover and you have a tunnel without water on top (undersea or under-river tunnels).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.fig.net/resources/proceedings/fig_proceedings/fig2010/papers/ts06h/ts06h_velasco_prieto_et_al_4542.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Read this&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22894" LastActivityDate="2019-09-22T06:02:10.260" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31260" PostTypeId="1" AcceptedAnswerId="31261" CreationDate="2019-09-23T06:21:38.583" Score="0" ViewCount="411" Body="&lt;p&gt;I have some homework question that i require some help understanding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) For the first question highlighted in red (referring to diagram viii), I know for sure that reactions at the supports is dependent on the length a because it determines the magnitude of moment it creates about the support and that it is not dependent on H because it is in the direction of the force(weight P) itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what I am uncertain about is the cable CE. Is this considered a two-force member in which tension at both ends will cancel each other out, hence varying the length L will not cause the reaction forces at A and B to change?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) For internal loadings, I am guessing that it will depend on the length L, because for internal loading, we are viewing the forces at the particular point and hence varying the angle of elevation will change the Rx and Ry component of the forces that the point and that it does not cancel out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also, is it true that for a two-force member that is in equilibirum, the net translational forces and rotational moment will be 0 throughout the entire segment?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jM0av.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jM0av.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22908" LastActivityDate="2019-09-23T12:16:18.680" Title="Concept on two-force member" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31261" PostTypeId="2" ParentId="31260" CreationDate="2019-09-23T08:05:33.470" Score="4" Body="&lt;p&gt;I don't want to take away the opportunity to work out the exact details for yourself, but I'll talk about the thought process I would use to approach these questions, and hopefully that will help clarify things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly, while I understand that the problem statement says to complete (a) before (b), I think at the point that you're getting stuck, it makes sense to sketch a few free body diagrams. It's part of the learning process, and as you do more and more of them you'll start developing a sort of intuition. My tendency is to run a couple quick 'thought experiments' either in my head or on paper, investigating two extremes. That's some of what I'll be doing below. Bear in mind that there are often multiple valid approaches to arrive at a solution - this is just one possibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Idea 1: Dependency of external reactions on structure geometry&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.1 Do external reactions depend on length 'a'?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps the quickest mental check of this one is to think about...what if 'a' were suddenly set to '2a'...would it change the support reactions? Given that the applied load is unchanged, we certainly can't say the support reactions would also be multiplied by 2. This suggests the support reactions aren't dependent on the numerical value of 'a' but on something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a thought experiment that will lead us toward what's really going on. Structure viii is a simply supported beam, so let's first imagine a more straightforward situation. What if we had a simply supported beam with single vertical point load at midspan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZUQHlm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZUQHlm.jpg&quot; alt=&quot;simply supported beam with midspan point load&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do we even need to know the quantitative beam length to determine the support reactions? No, we don't. We simply need to know the load is applied at midspan. That is, we need to know the &lt;em&gt;ratio&lt;/em&gt; of the beam lengths to the left and right of the load. This is enough to tell us that half the applied load will go to each support.  The same principle holds true for instances where the vertical load is offset from midspan. The length ratio (not the lengths themselves) is what determines the simply supported beam reactions. You'll see this when you complete part (b) of the assignment and go through the exercise of solving for the support reactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.2 Do external reactions depend on the vertical position 'H' of the applied load?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You're on the right track with this one. Mathematically I guess it relates to how we can slide a force along its line of action. My first thought is usually the physical intuition that whether I'm holding, say, a can of paint on a 6&quot; string or a 12&quot; string, I've got to support the same vertical load from the can of paint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EnnV3m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EnnV3m.jpg&quot; alt=&quot;cans of paint&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.3 Do external reactions depend on the height 'L' of the vertical member?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first glance, this one seems like a harder question, but the important thing to remember is that with support reactions we're investigating &lt;em&gt;external&lt;/em&gt; equilibrium. For this simply supported beam, we don't need to know what's happening inside the structure to find the external reactions. It's like how for a statically determinate truss, we can find the external reactions without investigating individual truss member forces or worrying about the specifics of truss geometry. So, for structure viii, we can solve for our support reactions using our equations of equilibrium without knowing 'L' at all. As with idea 1.1, this will probably become more apparent when you work through part (b) of the assignment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Idea 2: Dependency of internal loading on height 'L' of vertical member&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we're getting into internal equilibrium, so you're correct that the cable geometry matters. If we assume that the cable is continuous and that at the point where the cable connects to the beam, the cable is passing around some sort of frictionless pin - then we can say that the tension, T, at every point in the cable must be the same (and equal to the applied load). Changing height L of the vertical member change the angle of the cable. As you noted, when we hold T constant and vary the cable angle, the x- and y- components of T vary. This means the internal forces in beam segment CD and vertical member DE are also changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Idea 3: Two-force members and equilibrium&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any member in static equilibrium (two-force or otherwise) the equations of statics will be satisfied at &lt;em&gt;every&lt;/em&gt; point in the structure (including sections cut through a member). That is, &lt;span class=&quot;math-container&quot;&gt;$\Sigma F_x = 0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Sigma F_y = 0$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\Sigma M = 0$&lt;/span&gt;. Satisfying these equations is one way we can determine internal forces (shear, axial, moment) at points along a member.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2019-09-23T12:16:18.680" LastActivityDate="2019-09-23T12:16:18.680" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31264" PostTypeId="1" CreationDate="2019-09-22T18:32:24.987" Score="2" ViewCount="371" Body="&lt;p&gt;The words &quot;validation&quot; and &quot;verification&quot; frequently appear when reading the results of a paper. A Google search of these words left me even more confused:    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;verification&lt;/em&gt;: The process of establishing the truth, accuracy, or validity of something.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;validation&lt;/em&gt;: The action of checking or proving the validity or accuracy of something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Example&lt;/strong&gt;: I constructed a mathematical model of real-world system. I compared the model results with actual values and found that the model had an acceptable error.&lt;br&gt;&#xA;Did I verify or validate my model?&lt;/p&gt;&#xA;" OwnerUserId="22910" OwnerDisplayName="Neil" LastEditorUserId="22910" LastEditDate="2019-09-23T14:19:57.533" LastActivityDate="2019-09-24T08:47:46.593" Title="Difference between verification and validation" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31265" PostTypeId="2" ParentId="31264" CreationDate="2019-09-22T21:04:10.023" Score="0" Body="&lt;p&gt;You likely did both, but with different processes. You create a mathematical or computer model and you check that it matches the ideas that underlie it. In a computer model you check that you haven't made mistakes in the programming. In a mathematical model you check that you have used the right equations (or whatever) and evaluated them correctly. This is verification. I verify that the model is what I say it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other part requires matching the output of the model to some &quot;real world&quot; values. This is validation. But it is actually only &lt;em&gt;partial&lt;/em&gt; validation since the world isn't a deterministic machine and a different set of values from the world or a different &quot;run&quot; of the model might give wildly different values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Often validation of an experiment means that some one, usually someone else, has used a different experiment and come to the same conclusion. &quot;Yes, several studies show that children are shorter than adults.&quot; This is validation by replication, but not necessarily using exactly the same experimental design. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verification of a design is a process to determine that the design &lt;em&gt;can be expected&lt;/em&gt; to give valid results. A study that just &quot;guessed&quot; at the heights of people would be hard to verify. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Presumably you did both. You verified the design. You validated the results. &lt;/p&gt;&#xA;" OwnerUserId="22912" OwnerDisplayName="Buffy" LastActivityDate="2019-09-22T21:04:10.023" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31266" PostTypeId="1" CreationDate="2019-09-23T16:23:11.303" Score="0" ViewCount="45" Body="&lt;p&gt;Why is it that the only 2 books on CNC macro programming are by Peter Smid and one S K Sinha. Am I looking at this the wrong way? What am I missing? Is Macro programming one single thing that can be applied to CNC?&lt;/p&gt;&#xA;" OwnerUserId="22903" LastActivityDate="2019-09-23T16:23:11.303" Title="Limited resources for learning Fanuc Macro B programming, I need some direction as to how people learn this language" Tags="|cnc|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31267" PostTypeId="2" ParentId="31264" CreationDate="2019-09-23T18:03:47.840" Score="2" Body="&lt;p&gt;Here's an illustration to explain the difference. Suppose you were creating an accounting system for Enron, before it collapsed. The specification said you need to ensure that the output assumed that 2+2=5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Verification&lt;/strong&gt; is asking the system &quot;what is 2 + 2&quot; and getting the answer &quot;5&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Validation&lt;/strong&gt; is discovering that the Securities and Exchange Commission thought the correct answer should have been &quot;4&quot;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-09-23T18:03:47.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31268" PostTypeId="1" AcceptedAnswerId="36159" CreationDate="2019-09-23T19:16:06.037" Score="1" ViewCount="160" Body="&lt;p&gt;Suppose I have the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\dot{x} = Ax+Bu$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$y=Cx+Du$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the following Hamiltonian matrix:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$H=\begin{pmatrix}&#xA;A &amp;amp; \frac{1}{2}B^TB\\ &#xA; -CC^T&amp;amp;-A &#xA;\end{pmatrix}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the value of &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; which is the bound of the &lt;span class=&quot;math-container&quot;&gt;$H_{\infty}$&lt;/span&gt; norm, so it is the value such that &lt;span class=&quot;math-container&quot;&gt;$\left |T(j\omega)  \right |_{\infty }&amp;lt;\gamma$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that for the bounded real lemma, if the eigenvalues of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; have negative real part, and &lt;span class=&quot;math-container&quot;&gt;$I\gamma^2-DD^T&amp;gt;0$&lt;/span&gt;, then the Hamiltonian have no eigenvalues on the imaginary axis. I also know that if the eigenvalues of &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; have negative real part, then &lt;span class=&quot;math-container&quot;&gt;$\left |T(j\omega)  \right |_{\infty }&amp;lt;\gamma$&lt;/span&gt; holds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my question is : how do I find the value of &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been told the result is &lt;span class=&quot;math-container&quot;&gt;$\gamma=0.5$&lt;/span&gt; but I really can't get to this result. I have tried using the Shur's complement to see if this matrix is negative definite (so before doing that i switched sign to the Hamiltonian). In this way i thiught that if the A matrix is negative definite, it has all eigenvalues with negative real part, so the resulting value of &lt;span class=&quot;math-container&quot;&gt;$\gamma$&lt;/span&gt; form the computation would have been the searched value. But I don't find the desired result. Maybe I am missing a point and doing something wrong, or maybe I am complitely on the wrong path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody please help me? Thank's in advance.&lt;/p&gt;&#xA;" OwnerUserId="22780" LastActivityDate="2020-06-08T21:01:10.250" Title="value of $\gamma$ in the H-infinity norm" Tags="|electrical-engineering|control-engineering|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31271" PostTypeId="2" ParentId="31264" CreationDate="2019-09-24T08:47:46.593" Score="3" Body="&lt;p&gt;&lt;strong&gt;&lt;em&gt;Validation&lt;/em&gt;: Are we building the right product?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Verification&lt;/em&gt;: Are we building the product right?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Validation&lt;/strong&gt; is the process of making sure that you have objective evidence that user needs and intended uses are met. ... &lt;strong&gt;Verification&lt;/strong&gt; is typically making sure that you have objective evidence that specified requirements are met. It is usually done by tests, inspections, and in some cases analysis as well.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;i.e. when you are designing something, you will write a series of requirements relating to user needs. Validation is the check that those requirements actually meet the users needs, and then Verification is the check that you've actually met the requirements that you set out for yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, a requirement might be &quot;This device must be light enough for all users to pick up&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might then write a reqirement, such as &quot;The device must weigh &amp;lt; 30kg&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could Validate this, to show that your specified requirement actually meets the user need by doing research to find out how much weight your users could &lt;em&gt;acutally&lt;/em&gt; lift, or by referencing some existing research if the grip etc. is comparable. There will of course be other requirements (e.g. size, grip features), that also need to be included in order to ensure that the user need is met.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you've manufactured your device, you then Verify that it meets your requirements, by performing a physical test - weigh the device, and check that against your specification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some cases, certain features e.g. &quot;the device must be &amp;lt;1m long&quot; can be verified through a technical report (e.g. drawing review) without the need for a physical test. In this case e.g. if the device is supplied with a Certificate of Conformity confirming that it matches the drawings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the example that you wrote, you are performing a real world test, and then comparing this information to a value that was set by your previous research would be &lt;strong&gt;Verification&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Checking whether or not your mathematical model actually shows you something useful with regard to a user need would be &lt;strong&gt;Validation&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source: ASME &lt;a href=&quot;https://www.asme.org/topics-resources/content/validation-verification-for-medical-devices&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.asme.org/topics-resources/content/validation-verification-for-medical-devices&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-09-24T08:47:46.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31272" PostTypeId="1" AcceptedAnswerId="31373" CreationDate="2019-09-24T14:19:44.123" Score="0" ViewCount="296" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bQlsz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bQlsz.png&quot; alt=&quot;Modified Sine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to create an excel sheet that will use the modified sine equation above to output a 360° displacement chart when a user types in the acceleration, velocity, total distance traveled and start and stop angles. From what I've been told previously this is the equation I have to use, but my background is more on the excel side of things so I'm having trouble understanding exactly how to break this equation down into something I can use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone please explain how to solve this equation using the variables that will be available that I described above? Does this equation require any more or any less variables than what will be available?&lt;/p&gt;&#xA;" OwnerUserId="22921" LastActivityDate="2019-10-03T14:02:27.367" Title="Modified Sine Equation" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31275" PostTypeId="1" CreationDate="2019-09-25T04:54:01.457" Score="0" ViewCount="144" Body="&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=WQEbXD4Kqwc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is video of a purported &quot;perpetual motion machine.&quot;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9sMVB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9sMVB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you explain its behavior?&lt;/p&gt;&#xA;" OwnerUserId="3820" LastEditorUserId="3820" LastEditDate="2019-09-25T06:03:15.537" LastActivityDate="2019-09-25T10:34:43.483" Title="How to debunk this &quot;Perpetual Motion Machine?&quot;" Tags="|mechanisms|linear-motion|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31276" PostTypeId="2" ParentId="31275" CreationDate="2019-09-25T06:04:10.013" Score="3" Body="&lt;p&gt;Looking at the demo video, this machine looks &lt;strong&gt;fake&lt;/strong&gt; to me. The person who assembled and exhibited it in the video used some trick to move the machine. I believe there is a platform that can tilt and contains a background attached to it, a table on top of it, a demonstrator standing on the platform and a camera (And lighting) kept on this platform. This possible setup is shown in below drawing. (Apologies for poor drawing skills)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rhYY2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rhYY2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gives an illusion as if the black wheel is somehow transferring the energy to one of the large wheels and then storing it back from inertia. But looking at the demo carefully, it is evident that a lot of energy is wasted (while stopping and some from friction). Interestingly wheels are very smooth to minimize friction, but off course, it can not be zero.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;How to debunk this “Perpetual Motion Machine?”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It is not possible (in current state ;) of our universe) to reverse the entropy or in other words create free energy device.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Such tricky videos attract many views and generate good revenue for the channel.&lt;/p&gt;&#xA;" OwnerUserId="22772" LastEditorUserId="22772" LastEditDate="2019-09-25T06:26:51.300" LastActivityDate="2019-09-25T06:26:51.300" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31277" PostTypeId="1" CreationDate="2019-09-25T09:18:09.597" Score="0" ViewCount="39" Body="&lt;p&gt;I am creating a system that I drop from 20m to the ground and I want it to be solar powered. Weight is an issue so I am trying to minimise it as much as I can. How can I ensure that the solar panels remain intact and don't break upon being dropped if they are attached to this system? I was thinking foam insulating to both hold it in place and absorb the force, but are there any other suggestions?&lt;/p&gt;&#xA;" OwnerUserId="22924" LastActivityDate="2019-09-25T09:18:09.597" Title="How do I physically protect solar panels from snapping when dropped?" Tags="|mechanical-engineering|aerospace-engineering|photovoltaics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31278" PostTypeId="2" ParentId="31275" CreationDate="2019-09-25T10:34:43.483" Score="1" Body="&lt;p&gt;In my opinion, the device motion itself is not faked , but there may be camera trickery involved to hide an accomplice, acting as an additional energy source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at 0:49 seconds - he gives the blue wheel a spin - he is putting energy into the device. It is certainly believable that the inertia stored here would be sufficient to propel the device to the left.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every time the device is stopped at the LHS of the screen, the blue wheel momentarily goes out of frame - perhaps for an assistant to add to the spin - the shadows visible certainly suggest an extra pair of hands to me...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted by @alephzero, the device rolls to the right naturally at 0:22, suggesting that the table is slightly tilted in that direction. The manufacture needs only to hold the blue wheel slightly above the RH wheel without letting them touch, to allow it to roll to the right with gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarise:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Moving Left - energy supplied by human spinning the blue wheel&lt;/li&gt;&#xA;&lt;li&gt;Moving Right - rolling downhill&lt;/li&gt;&#xA;&lt;li&gt;Additional energy input provided when the blue wheel is out of frame.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-09-25T10:34:43.483" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31279" PostTypeId="1" CreationDate="2019-09-25T12:02:10.147" Score="1" ViewCount="1189" Body="&lt;p&gt;I did a Civil Engineering course some years ago and from my textbook I had a question on a pipe with a bend. &#xA;Now I think I have the part of the question on force correct, but I am not sure how the torque will be calculated in this situation. &#xA;Can someone please help and give me an idea as to how to work this out?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the question as an image with my calculations below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Cjz3T.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cjz3T.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8013" LastEditorUserId="8013" LastEditDate="2019-09-25T13:54:52.887" LastActivityDate="2024-02-05T21:06:30.773" Title="Find Torque and Force on Pipe with Bend" Tags="|fluid-mechanics|pipelines|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31280" PostTypeId="1" AcceptedAnswerId="31308" CreationDate="2019-09-25T12:09:51.713" Score="0" ViewCount="108" Body="&lt;p&gt;We know that &lt;a href=&quot;https://en.wikipedia.org/wiki/Blaise_Pascal&quot; rel=&quot;nofollow noreferrer&quot;&gt;Blaise Pascal&lt;/a&gt; built a &lt;a href=&quot;https://en.wikipedia.org/wiki/Pascal%27s_calculator#Carry_mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;mechanical calculator&lt;/a&gt; which had a carry mechanism called a sautoir. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PIr68.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PIr68.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see a video of it working &lt;a href=&quot;https://youtu.be/OpTAODDU1pg&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, and &lt;a href=&quot;https://youtu.be/-li5VWnqhmI&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We &lt;a href=&quot;https://books.google.com.au/books?id=piNNAAAAMAAJ&amp;amp;printsec=frontcover&amp;amp;source=gbs_ge_summary_r&amp;amp;cad=0#v=onepage&amp;amp;q&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;know that&lt;/a&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Charles_Babbage&quot; rel=&quot;nofollow noreferrer&quot;&gt;Charles Babbage&lt;/a&gt; wrote a book of &lt;a href=&quot;https://en.wikipedia.org/wiki/Mathematical_table#Tables_of_logarithms&quot; rel=&quot;nofollow noreferrer&quot;&gt;log tables&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In doing this, Babbage was inspired to create Log Tables mechanically, leading to his design of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Difference_engine#targetText=A%20difference%20engine%2C%20first%20created,small%20set%20of%20polynomial%20coefficients.&quot; rel=&quot;nofollow noreferrer&quot;&gt;Difference Engine&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: &lt;strong&gt;Did Charles Babbage's Difference Engine use the sautoir carry mechanism from Pascal's calculator?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="14103" LastActivityDate="2019-09-28T07:50:10.633" Title="Did Charles Babbage's Difference Engine use the sautoir carry mechanism from Pascal's calculator?" Tags="|gears|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31281" PostTypeId="2" ParentId="31279" CreationDate="2019-09-25T12:28:29.257" Score="0" Body="&lt;p&gt;It looks like you know how to calculate the forces due to a fluid flow? At the outlet, you must simply consider the momentum equation (Newton's Second Law), to find the axial force at said outlet - the torque then is simply this force multiplied by 1.7 (the perpendicular distance from the pivot, 'A').&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IDu6um.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IDu6um.jpg&quot; alt=&quot;Torque radius diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-09-25T12:28:29.257" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="31283" PostTypeId="1" AcceptedAnswerId="39575" CreationDate="2019-09-26T13:21:22.177" Score="0" ViewCount="386" Body="&lt;p&gt;I'm learning about how dog clutches work particularly in sequential transmissions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having a hard time understanding the principle which ensures dogs will always mesh correctly. I can easily imagine a situation in which dogs will collide and switching gears will not work such as when the speed of the input shaft matches the speed of the intermediary shaft and dogs are aligned together or when the engine is off and the vehicle is not moving then dogs can again collide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm designing my own clutch-less transmission using dog gears, because I thought they are most appropriate, so I'm trying to understand the principle, not so much the example cases I mentioned above.&lt;/p&gt;&#xA;" OwnerUserId="20333" LastEditorUserId="17070" LastEditDate="2019-09-27T20:55:42.783" LastActivityDate="2021-01-07T14:41:31.993" Title="How don't dog clutches collide?" Tags="|mechanical-engineering|automotive-engineering|gears|transmission|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31284" PostTypeId="2" ParentId="31283" CreationDate="2019-09-26T15:06:03.143" Score="1" Body="&lt;p&gt;Usually because dog clutches are engaged when the shafts are stationary - lathes are a good example as they change speeds when the work is stationary and the tool dis-engaged.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-09-26T15:06:03.143" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31285" PostTypeId="1" CreationDate="2019-09-26T17:01:03.863" Score="1" ViewCount="1208" Body="&lt;p&gt;Here is a Galantz flatbed microwave&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YqLwZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YqLwZ.jpg&quot; alt=&quot;fifth&quot;&gt;&lt;/a&gt;&#xA;The waveguide is smaller than a wavelength. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)Usually apertures of less than a wavelength are considered to block EM wave but not here, why ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)The width of the waveguide is 6 cm and the height about 2 cm, does that polarize the EM waves alonf the 6cm axis  ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the inside of the oven :&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oPW9A.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oPW9A.jpg&quot; alt=&quot;first image&quot;&gt;&lt;/a&gt;&#xA;3)What's the purpose of that metal bulge in front of the magnetron output ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's the bottom of the oven, the ceramic plate is lifted to reveal the wave stirrer&#xA;&lt;a href=&quot;https://i.stack.imgur.com/u65ey.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u65ey.jpg&quot; alt=&quot;second image&quot;&gt;&lt;/a&gt;&#xA;The wave stirrer is removed to show the opening for the waves entry and the shaft of the stirring plate&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6exe2.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6exe2.jpg&quot; alt=&quot;thid miage&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) can you explain the geometry of the disk and the reason designers chose to make the waves enter the oven from below the stirring disk?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4ygdr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4ygdr.jpg&quot; alt=&quot;fourth imge&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5926" LastActivityDate="2019-09-26T17:01:03.863" Title="Please explain this flatbed microwave design" Tags="|electromagnetism|microwaves|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31288" PostTypeId="2" ParentId="31207" CreationDate="2019-09-26T19:59:04.547" Score="1" Body="&lt;p&gt;You could import the &lt;code&gt;.inp&lt;/code&gt; file containing your mesh into &lt;em&gt;Ansys Mechanical APDL&lt;/em&gt;, then import your &lt;code&gt;.iges&lt;/code&gt; as well (through &lt;code&gt;File/Import/IGES&lt;/code&gt; or sometimes &lt;code&gt;Archive Model/Read&lt;/code&gt; in &lt;code&gt;/prep7&lt;/code&gt; works smoother), give your attributes to the different areas, mesh them, and export the whole model in &lt;code&gt;.cdb&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12198" LastActivityDate="2019-09-26T19:59:04.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31289" PostTypeId="1" AcceptedAnswerId="31290" CreationDate="2019-09-26T21:03:19.063" Score="0" ViewCount="60" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4mfLr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4mfLr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We're building this thing in AutoDesk Inventor and SolidWorks, yet I don't know what it is. Can someone please tell me? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="20289" LastEditorUserId="17070" LastEditDate="2019-09-27T20:55:52.443" LastActivityDate="2019-09-27T20:55:52.443" Title="How could I identify the object in the question body?" Tags="|mechanisms|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31290" PostTypeId="2" ParentId="31289" CreationDate="2019-09-26T21:36:16.613" Score="1" Body="&lt;p&gt;I suspect you're building this &quot;thing&quot; as an exercise to expose you to the various aspects of the software. As a beginner, you want a challenge in creating the part, but you don't want to be overwhelmed by a complex design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part includes a number of radii or fillets, a counter-bored hole as well as an ordinary hole and an offset on the tangs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design as pictured could be used for many different applications. I can envision it to provide a bearing mount (the counter-bored hole) for an axle. The offset conventional hole would permit the axle to be inset from the mounting surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is one of many possible applications for a very broad and vague design. Beyond the instructional aspect, the application is limited only by one's imagination and understanding of the technical world.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-09-26T21:36:16.613" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31293" PostTypeId="1" AcceptedAnswerId="31294" CreationDate="2019-09-27T02:12:37.167" Score="0" ViewCount="138" Body="&lt;p&gt;This is a general question about root locus plots. I notice that there is a source that says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;The root locus begins at the finite and infinite poles of G(s)H(s)&quot;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the part that says 'finite and INFINITE poles'. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should the part that says 'infinite' be omitted? The root locus only begins at the FINITE open loop pole locations, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the statement is indeed correct - about cases where a root locus may start at 'infinite' pole locations, then are there any examples of such cases?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks all!&lt;/p&gt;&#xA;" OwnerUserId="19788" LastEditorUserId="19788" LastEditDate="2019-09-27T02:20:10.560" LastActivityDate="2019-09-27T03:37:57.947" Title="root locus question - root locus 'begins' at finite open loop pole locations" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31294" PostTypeId="2" ParentId="31293" CreationDate="2019-09-27T03:37:57.947" Score="1" Body="&lt;p&gt;You can do a root locus plot of a system with more zeros than poles, even though such a thing is physically impossible. If you do, you find poles coming in towards your zeros, just as a system with an excess of poles has them running off to infinity.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-27T03:37:57.947" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31295" PostTypeId="1" AcceptedAnswerId="31393" CreationDate="2019-09-27T06:24:35.003" Score="-5" ViewCount="166" Body="&lt;p&gt;In industrial manufacturing lines, there is a a limit on the number of discarded parts per million.I need to have a sense about this, e.g. How different is 4 ppm from 1 ppm and how much does it indicate the improve in production quality?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me considering the price of a part as 1€, 1 ppm and 4 ppm seem so trivial&lt;/p&gt;&#xA;" OwnerUserId="17070" LastEditorUserId="17070" LastEditDate="2019-10-03T06:32:25.200" LastActivityDate="2019-10-05T17:43:15.773" Title="Talking in terms of 6 sigma, why 4 ppm would be very costlier than 1 ppm (part per milion)?" Tags="|manufacturing-engineering|industrial-engineering|statistics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31296" PostTypeId="2" ParentId="31295" CreationDate="2019-09-27T07:54:34.267" Score="1" Body="&lt;p&gt;1ppm is 4 times better than 4ppm. If 4ppm of parts are indicated it means that 0.0004% of these parts have to be discarded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is always better to have a lower number of discarded parts which means that 0.0001% of defects is better than having 4 times more.  Lower ppm (in this case) means less costs and higher productivity.&lt;/p&gt;&#xA;" OwnerUserId="5007" LastActivityDate="2019-09-27T07:54:34.267" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="31297" PostTypeId="1" CreationDate="2019-09-27T11:45:51.600" Score="1" ViewCount="197" Body="&lt;p&gt;I’m designing an RF device that should be able to use multiples sets of antennae, one a low-gain internal PCB antenna and the other an external high gain antenna attached via an SMA connector. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I’d like to be able to automatically detect when the external antenna is connected and disconnected (so that I can electronically switch between antennas) be able to switch to using it automatically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come across ways of doing this by adding a known resistance to the antenna and then simply detecting this resistance through the use of a potential divider. This however would require the antenna to be precisely calibrated and be a short circuit antenna. Ideally I would like to be able to detect any kind of antenna attached to the SMA connector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How might I go about doing this? Are there any other electrical properties of the antenna which could be exploited to detect its presence? (For example, would it be possible to somehow detect the capacitance of the antenna when it is connected or would this change in capacitance be insignificant?)&lt;a href=&quot;https://i.stack.imgur.com/EpqKk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EpqKk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22943" LastActivityDate="2019-09-27T11:45:51.600" Title="Methods for External Antenna Detection - RF systems" Tags="|rf-electronics|circuit-design|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31299" PostTypeId="2" ParentId="16475" CreationDate="2019-09-27T14:29:35.917" Score="0" Body="&lt;p&gt;I would say it's all a matter of power consumption and handling capacity. I will distinguish between two cases in which Uplink and Downlink (UL and DL from now on) are interchanged.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Satellite communication:&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If we take the typical Ka-band for satellite communication, we see that the &lt;strong&gt;UL&lt;/strong&gt; is around 30 GHz, while the &lt;strong&gt;DL&lt;/strong&gt; lays around 20 GHz. Satellites are constrained by their weight, being the cost around several thousands of euros per kilogram. For this reason, their equipment is also limited: they don't have big batteries and get their power essentially from solar panels. Thus, they cannot use big power for transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you very well stated, the attenuation is higher at higher frequencies. Since we have a power constrain, it seems logic to transmit in the lower frequency. &#xA;In the case of the UL, the so called &quot;ground segment&quot; consist essentially of several-dozen-meter-diameter antennas with whole buildings for the data processing and electrical supply, so the extra attenuation due to higher frequency is not as meaningful.&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Mobile communications&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In this case, the situation is just the opposite. The UL goes from the user phone to a fixed base station. Take, for example, any of the UMTS FDD frequency bands. You will check that the uplink frequencies are always smaller than downlinks'. In this case, it's the user's phone, which has a size constraint and relies on small batteries, while base stations usually have much better electric supplies with not such strigent requirements. Thus, it makes sense that the Uplink in mobile communications employs the lower frequencies, while the base station employs higher ones.&lt;/p&gt;&#xA;" OwnerUserId="22946" LastActivityDate="2019-09-27T14:29:35.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31300" PostTypeId="2" ParentId="31194" CreationDate="2019-09-27T14:32:47.423" Score="1" Body="&lt;h1&gt;Foundations&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The rate of change of amount (concentration) of any given component is a differential equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ r_j = \frac{dC_j}{dt} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The empirical rate law expression for a reaction defines how reaction rate depends on concentrations in the reaction expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ r_j = \pm k_{rxn} \Pi\ C_j^{n_j} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reversible reactions have forward and backward terms. Irreversible reactions have only forward terms. The reaction orders &lt;span class=&quot;math-container&quot;&gt;$n_j$&lt;/span&gt; are the stoichiometric coefficients of the component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a system of &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; components, we will have &lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt; independent differential equations.&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Specific Example&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;For component X in your system, you can obtain this expression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ r_X = \frac{dC_X}{dt} = k_1 C_A C_S - k_2 C_X C_S $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Summary&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;This gives first insight as to how to set up the problem. You will not need to consider the production rate of component Y. You will have three independent, first order, coupled differential equations. How this system is to be solved is entirely a different question of its own right. I recommend a numerical approach with graphical output. Two packages that take this approach are below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wavemetrics.com/project/KinSim&quot; rel=&quot;nofollow noreferrer&quot;&gt;KinSim&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.wavemetrics.com/project/KineticReactionDemo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kinetic Reaction Demo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Options using MatLab or Mathematica are likely posted on their sites as well.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-09-27T14:32:47.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31301" PostTypeId="1" CreationDate="2019-09-27T17:38:25.353" Score="0" ViewCount="29" Body="&lt;p&gt;Here is how the top of my grape press twists as I pull the car jack downwards (marked blue). I was wondering if adding another bracket (marked purple) on each side would solve the twist issue. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CAJ7O.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CAJ7O.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have attempted putting such plates but made of just 1&quot; plywood. Here they are in the image below. It made little to no difference.I think those steel plates, four of them, nicely bolted like such, could solve this problem. But where do I get the solid steel, who is going to cut and drill it for me... It might be a bit of a challenge, or maybe i can find something pre-made at the hardware store.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yzozS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yzozS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="3412" LastEditorUserId="3412" LastEditDate="2019-09-27T20:59:31.740" LastActivityDate="2019-09-27T20:59:31.740" Title="doubling the existent two brackets - will it help with my grape press?" Tags="|pressure|torque|wood|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31302" PostTypeId="2" ParentId="31289" CreationDate="2019-09-27T18:15:25.900" Score="1" Body="&lt;p&gt;It is a Z bracket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=z%20bracket&amp;amp;rlz=1C1GCEA_enCA851CA851&amp;amp;sxsrf=ACYBGNQyVIUw3sjDhx4SgP4k2X11wypl4A:1569607940286&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=0ahUKEwj8qN6dzfHkAhXJjFkKHaKuAFkQ_AUIEigB&amp;amp;biw=1536&amp;amp;bih=722&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google Search&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2019-09-27T18:15:25.900" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31303" PostTypeId="2" ParentId="31301" CreationDate="2019-09-27T19:47:40.523" Score="1" Body="&lt;p&gt;I would put plates at the corners, bolted where I've indicated (or choose your own spots -- just make it adequate, and think about how the wood is going to &quot;want&quot; to split when pressure is applied).  You should be able to use 1/8&quot; thick plain steel, and bolt them on with the same-sized bolts as are on your current brackets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would leave the current ones off -- you can already see that they've bent up at the corners.  I suspect that the top beam of your press was originally flush with the tops of the posts; it looks like it's gone up at least 1/4&quot; because of bendy brackets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N2ORb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N2ORb.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-27T19:47:40.523" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31304" PostTypeId="1" CreationDate="2019-09-28T05:28:27.553" Score="1" ViewCount="454" Body="&lt;p&gt;Most of the books that I referred to, used compensators to improve transient performances of a stable system using root locus and compensator design. I got a system which is actually unstable for all values of K (one pole is almost on imaginary axis). How can I change the stability of this by adding some poles or zeroes? For e.g., I can move the breakaway point on origin to left by adding some complex zeroes (trial and error). This way there will be some finite range of gain for which system is stable. Can someone point me to a good resource to learn this?&#xA;The TF whose root locus is plotted is: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{-95.17 s^3 - 8130 s^2 - 187.6 s - 7695}{0.1 s^5 + 1.292 s^4 + 3.138 s^3 + 2.179 s^2 + 0.016 s + 0.01}$$&lt;/span&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iIjMA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iIjMA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(There is one more finite zero far left which I couldn't fir in the screenshot and if it helps, this is a model related to some aircraft). Thanks for the help!&lt;/p&gt;&#xA;" OwnerUserId="22951" LastActivityDate="2020-10-22T19:04:00.447" Title="Adjusting root locus to get a stable system from unstable system" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31305" PostTypeId="1" CreationDate="2019-09-28T06:32:51.893" Score="0" ViewCount="47" Body="&lt;p&gt;Would it be possible to create a piece of metal, made out of an alloy or combination of alloys, that undergoes a shape-memory-esque change in structure in reaction to an electric current? Could it also react differently depending on the type or orientation/direction of the current? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ex:&#xA;A wire that, when electricity is passed through it one way, conracts/coils but extends when the currents flows in the opposite direction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;+__________- contracts&#xA;-__________+ expands&lt;/p&gt;&#xA;" OwnerUserId="22952" LastActivityDate="2019-09-28T06:32:51.893" Title="Electric Current Based Shape-Memory Alloy" Tags="|materials|metallurgy|electrical|alloys|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31306" PostTypeId="1" CreationDate="2019-09-28T06:53:49.947" Score="1" ViewCount="112" Body="&lt;p&gt;I did a Civil Engineering course some years ago and from my textbook I had a question on a bending pipe. &#xA;The purpose of this question is to see if the way that I have calculated this is correct, as my answer is slightly out from the textbook answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feedback will be much appreciated on my calculations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the question below &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xxvJW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xxvJW.jpg&quot; alt=&quot;Chapt 9 - Question 27&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/bfPC1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bfPC1.jpg&quot; alt=&quot;Chapt 9 - Question 27 Figure&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8013" LastEditorUserId="8013" LastEditDate="2019-09-28T07:19:57.913" LastActivityDate="2019-09-28T07:19:57.913" Title="Finding Horizontal and Vertical Force on Bending Pipe" Tags="|fluid-mechanics|pipelines|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31307" PostTypeId="1" CreationDate="2019-09-28T07:34:12.423" Score="1" ViewCount="435" Body="&lt;p&gt;I did a Civil Engineering course some years ago and from my textbook I had a question on boundary layer thickness. &#xA;So far I have been able to calculate the transition distance, but I am not sure how to calculate the boundary layer thickness from there.&#xA;The purpose of this question is to work out how to calculate the boundary layer thickness from what I have said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v8HeI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v8HeI.jpg&quot; alt=&quot;Fluid Mechanics - Chapt 14 Question 12&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8013" LastActivityDate="2024-02-04T18:08:10.950" Title="Boundary Layer Thickness for Flat Plate" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31308" PostTypeId="2" ParentId="31280" CreationDate="2019-09-28T07:50:10.633" Score="0" Body="&lt;p&gt;Short answer is no. Longer answer is that the Babbage Difference engine has taken the &lt;em&gt;concept of a sautoir&lt;/em&gt; and implemented something much more complicated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we have two wheels representing values in the calculation, and the gear in the middle that adds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cAatZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cAatZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then to carry numbers up - there was an external device:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eSPVv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eSPVv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that a single number is represented vertically on a column, so the values and the carrying device were replicated vertically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cMkrp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cMkrp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also this &lt;a href=&quot;http://www.meccano.us/counter/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Counter&lt;/a&gt; built by a person who built a &lt;a href=&quot;http://www.meccano.us/difference_engines/rde_2/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Difference  Engine&lt;/a&gt; built out of &lt;a href=&quot;https://en.wikipedia.org/wiki/Meccano&quot; rel=&quot;nofollow noreferrer&quot;&gt;Meccano&lt;/a&gt; - which implemented its own carry mechanism:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/P77LV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P77LV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This arm will engage between two of the double brackets on the detent assembly of the adjacent dial and cause it to advance by one unit.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bibliography&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=vdra5Ms__9s&quot; rel=&quot;nofollow noreferrer&quot;&gt;Babbage's Difference Engine No. 2, Part 2: The Calculation Section&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.meccano.us/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tim Robinson's Meccano Computing Machinery web site&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14103" LastActivityDate="2019-09-28T07:50:10.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31309" PostTypeId="2" ParentId="31307" CreationDate="2019-09-28T08:01:35.983" Score="0" Body="&lt;p&gt;Try using the following formulas for boundary layer thickness over a flat plate. These formulas are empirically derived.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For Laminar Boundary Layer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IDTDV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IDTDV.png&quot; alt=&quot;For Laminar Boundary Layer&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H0WBM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H0WBM.png&quot; alt=&quot;Local Reynolds Number&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;For Turbulent Boundary Layer&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1EjBg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1EjBg.png&quot; alt=&quot;Turbulent Boundary Layer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sources: Fundamentals of Aerodynamics by John D. Anderson&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: Updated the format of my answer as recommended by a moderator.&lt;/p&gt;&#xA;" OwnerUserId="22947" LastEditorUserId="22947" LastEditDate="2019-09-28T12:07:37.870" LastActivityDate="2019-09-28T12:07:37.870" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31311" PostTypeId="1" CreationDate="2019-09-28T15:28:05.003" Score="0" ViewCount="268" Body="&lt;p&gt;I've checked on quite a few places on the net and some say it occurs at the beginning of suction stroke and others say at the end of suction stroke. I believe it to be at the end of suction stroke because after exhaust some of heat absorbed by cylinder from combustion products remains and when suction is completed, the fresh charge will absorb some of that remaining heat thereby making the cylinder cooler. I'd like an explanation with graphs(ACTUAL T-S diagram, if possible would be much appreciated) and also citation of some credible sources if possible. I've uploaded a T-S diagram for SI engine but this is the ideal T-S diagram which like ideal P-V diagram(of otto cycle) misses the portion where scavenging and actual suction occurs after the expansion of remaining gases that remain at the end of exhaust stroke(this difference is clear from other two images of ideal and actual P-V diagrams). Also I'd like to know if there is any difference between its occurrence(the minimum temperature timing) in S.I and Diesel engines.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XxQyg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XxQyg.png&quot; alt=&quot;T-S diagram of an SI engine.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BNL73.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BNL73.jpg&quot; alt=&quot;Ideal P-V diagram of SI Engine&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/g6GEj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g6GEj.jpg&quot; alt=&quot;Actual diagram of SI Engine.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly this question is a multiple choice question of an exam with options:-&#xA;a) End of exhaust stroke    b) end of suction stroke    c) Beginning of suction stroke     d) beginning of exhaust stroke&lt;/p&gt;&#xA;" OwnerUserId="17427" LastEditorUserId="17427" LastEditDate="2019-09-28T16:03:23.050" LastActivityDate="2019-09-28T16:03:23.050" Title="When does the minimum temperature in a four-stroke cycle occur inside the engine cylinder?" Tags="|mechanical-engineering|thermodynamics|engines|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31312" PostTypeId="2" ParentId="31304" CreationDate="2019-09-28T16:16:41.523" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;How can I change the stability of this by adding some poles or zeroes? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In general, zeros tend to &quot;pull&quot; poles toward them.  So your double zero (with, presumably, high-speed poles off to the left, so that you have a physically workable system) is the correct thing to do.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can someone point me to a good resource to learn this?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Find books on &quot;classical&quot; control theory that were published before digital computers were readily available to engineers.  Root locust plotting was invented in the 1940's or 1950's (the seminal book, &quot;Control System Dynamics&quot; was published by the inventor of the root locus, Walter Evans, in 1954).  As a tool for actually doing formal and detailed synthesis of control loops, it's probably obsolete -- but as a tool for &lt;em&gt;understanding&lt;/em&gt; SISO control loop behavior, I think it's great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Opinion follows):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a device with enough computing power to get on the web, I believe that you would be better served to use that device for more modern methods of control system analysis and synthesis.  Bode plots and Evans (root locus) plots are great for designing SISO loops, and I use them all the time.  But I use root locus plots -- often literally in pen on the backs of envelopes -- to help me find a starting point, and then I use Bode plots on a computer to assess performance and robustness.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for analyzing stability and robustness of multiple-input systems, or for synthesizing controllers for such, you need more advanced methods.  If you're going to impress employers with your control systems chops, ditto.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-28T16:16:41.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31313" PostTypeId="1" CreationDate="2019-09-28T20:10:43.447" Score="0" ViewCount="86" Body="&lt;p&gt;Obviously it is a good career for indians. But it seems like ME is becoming EE and the adoption of various technologies has deskilled the job to the point where 90% of the available positions are outsourced or taken by h1bs. Is engineering still a good career with viable employment?&lt;/p&gt;&#xA;" OwnerUserId="22954" LastActivityDate="2019-09-28T21:37:01.950" Title="Is engineering still a good career for Americans?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="7" ClosedDate="2019-09-29T05:17:57.967" ContentLicense="CC BY-SA 4.0" />
  <row Id="31314" PostTypeId="2" ParentId="31313" CreationDate="2019-09-28T21:37:01.950" Score="1" Body="&lt;p&gt;I think the problem is most Americans are too lazy or more interested in having fun ( A K A liberal arts courses). Other than medicine; engineering is the most reliable area to get a good paying job . Now if you want to be a multi-millionaire , you need to invent some new kind of computer chip or develop a new social website. But if you want a good job with good money and , and at least I felt I was doing something worthwhile, engineering is great. I saw Americans become unwilling to put in the work starting in about 1970. Then it was the UK &quot;brain drain&quot; ; of course I did not realize it at the time. Our R &amp;amp; D department hired over a dozen PHDs from UK in a couple years. I am certain the American managers would have been happy to hire qualified Americans if they could find them. I often went to lunch with a Brit I worked with , so at lunch it was me and a half dozen guys who talked funny - ( what car has a bonnet ?). And today; My stepson just got a PHD in the the medical field ( not a medical doctor) and newsletter from his prestigious hospital containing promotion and advancement news had about 20 photos . Of the 20 about 3 looked like &quot;Americans&quot; ; again ,I expect the administrators would hire any qualified Americans they could find them . And, a little overkill, A grandson that my son and I encouraged to go into engineering , got way more &quot;friends&quot; on Facebook than semester hours of credit . After four years of fun and about 4 changes of majors , he needs about 2 years to graduate and I lost track of his current major , but I am sure he is having fun.   &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-09-28T21:37:01.950" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31315" PostTypeId="2" ParentId="8692" CreationDate="2019-09-29T03:47:13.130" Score="5" Body="&lt;p&gt;I’ll be installing a custom counterweight for my new garage door – it requires some clever engineering to make it work since a residential garage door doesn’t have a fixed weight: its track bends right above the door opening, and the part that is in the horizontal raceway provides no vertical load to the counterweight. So, as the door opens, it gets lighter and lighter.&lt;/p&gt;&#xA;&lt;p&gt;The solution I chose is to segment the counterweight: as it descends, the segments get caught on retainers so that the effective weight of the counterweight decreases the closer to the floor it is. Each segment weighs just 3 lbs, making installation easy, and there’s plenty of them so the operation will be smooth. A silicone washer will absorb the shock of the segments coming into contact with the perches and with the counterweight stack.&lt;/p&gt;&#xA;&lt;p&gt;Another solution I considered was a linkage (4-bar or two of them in series) to scale the load – not a problem with designing it; but manufacturing it would be too much effort for me.&lt;/p&gt;&#xA;&lt;p&gt;Another workable solution would be a cam that could even be manufactured out of wood, but I didn’t care enough to set things up to actually do that.&lt;/p&gt;&#xA;&lt;p&gt;Since the counterweight will be made of a stack of custom cast cement blocks, with attachment points for skate bearings as raceway rollers, it will have a small footprint and will be confined to an enclosure.&lt;/p&gt;&#xA;&lt;p&gt;I’m designing the system with a factor of safety of 5 after redundancy is lost, and the weight reduction system makes it inherently safe on the counterweight end of things: by the time the counterweight hits the bottom of the enclosure (a.k.a. the floor), it will weigh about a pound (from the point of view of the door) and that will be only the retainer plate for the cement blocks, as all the blocks will be resting on their padded perches throughout the travel range.&lt;/p&gt;&#xA;&lt;p&gt;I have figured a 3D printed last-resort brake for the garage door, so if both redundant cables snap on either side, the door will be immobilized by the sacrificial brake parts.&lt;/p&gt;&#xA;&lt;p&gt;The cables are going to be a redundant pair for each side of the door, with separate attachments to both the door and the counterweight stack, so that no single point of failure will affect both cables at the same time. The cables will be routed in a PVC pipe for confinement in case one snaps. The idlers for the cables are 11in diameter cast iron parts with suitable grooves that I found at a junkyard – I installed four ball bearings at each end of the cast shaft sleeve, and the shaft will be supported on both ends. This provides a decent bend radius for the oversized cable – twice as large as the minimum from the cable spec sheet. It also will look quite steampunk I think.&lt;/p&gt;&#xA;&lt;p&gt;I expect the counterweight system to be maintenance free and to easily outlast the house. I hate the garage door springs with a passion – I have dealt enough with them to just lose all respect for whoever came up with the idea, for the execution is so poor.&lt;/p&gt;&#xA;&lt;p&gt;I’ll also install my own opener – a junkyard worm gearbox I found in perfect condition, and a standard electric motor – both  oversized for the job. A simple and reliable cable tension “sensor” will provide overload and safety protection – a really simple system that will disconnect power to the motor and short the windings for electrodynamic braking should the cable get too taut or gets pulled too hard.. No electronics: just a spring loaded rocker with a roller, and two adjustable cams on the rocker to trigger a set of micro switches (again, redundant). The motor has soft start. The opto sensor “eye” will be a standard industrial Omron part that should last forever in the conditions I will be using it in.&lt;/p&gt;&#xA;&lt;p&gt;So, all in all a counterweight system is perfectly feasible if you want to make one yourself; but perhaps most people wouldn’t pay the cost if it was a commercial product. It would cost way more than a spring does. Unsightliness-wise, it will be neatly tucked in the space between each door and the side wall. Nobody will even notice it’s there if they won’t be told first :)&lt;/p&gt;&#xA;&lt;p&gt;I’ll probably update this answer once I get some pictures whenever I find time to install this thing.&lt;/p&gt;&#xA;&lt;p&gt;Update: installation is happening as I write this.&lt;/p&gt;&#xA;" OwnerUserId="22956" LastEditorUserId="22956" LastEditDate="2020-10-16T13:41:15.497" LastActivityDate="2020-10-16T13:41:15.497" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31316" PostTypeId="1" CreationDate="2019-09-29T05:06:03.683" Score="0" ViewCount="41" Body="&lt;p&gt;I am a little confused about how a 2-bit adder works. For 1-bit adder, my understanding of it is fine and I am able to draw out the truth table for it. However, if lets say I have a 2-bit adder, I can't seem to work out how the display shows the sum of both binary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example,&lt;br&gt;&#xA;1st full adder - A1 = 1, B1 = 0, Cin = 0 =&gt; sum = 1 and carry out = 0. The carry out then goes to another full adder hence, for the 2nd full adder A2 = 1, B2 = 0, Cin = 0 =&gt; sum = 1 and carry out = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this mean that my binary representation is 11? Based on looking at the both the sums, however I know that 01 + 01 = 10. I am pretty sure I am understanding something wrongly here and hopefully someone can help me out.&lt;/p&gt;&#xA;" OwnerUserId="22908" LastEditorUserId="110" LastEditDate="2019-09-29T11:01:18.317" LastActivityDate="2019-09-29T11:01:18.317" Title="Confusion about 2-bit adders" Tags="|electrical-engineering|circuits|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31317" PostTypeId="1" CreationDate="2019-09-29T06:38:28.043" Score="-1" ViewCount="127" Body="&lt;p&gt;Example an aluminium tube reinforced with steel tube inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;100*50*2mm thick aluminium with 80*40*3mm MS tube&lt;/p&gt;&#xA;" OwnerUserId="22959" LastEditorUserId="110" LastEditDate="2019-09-29T10:56:58.187" LastActivityDate="2021-03-23T06:00:15.083" Title="How to calculate the deflection of beam that is reinforced through formula?" Tags="|deflection|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31318" PostTypeId="2" ParentId="31279" CreationDate="2019-09-29T11:21:10.183" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u0JsX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u0JsX.jpg&quot; alt=&quot;Question 28&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am just working on the torque for part (b) and will have the rest of this answer up , as soon as I can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force to calculate torque is the force in the 45 degree direction from B to C.&lt;/p&gt;&#xA;" OwnerUserId="8013" LastActivityDate="2019-09-29T11:21:10.183" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31319" PostTypeId="1" CreationDate="2019-09-29T14:46:25.003" Score="1" ViewCount="402" Body="&lt;p&gt;For a scholarship exam I'm taking, it takes simple enough ideas and then messes with them in crazy ways. In this question four of the equilibrium equations I'd use to solve this are linearly dependent. Is there a 6th equation I can use to get a definition answer? Is it something to do with the fact the forces are exerted by springs? &#xA;&lt;a href=&quot;https://i.stack.imgur.com/P0PSA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P0PSA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/66keN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/66keN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22963" LastActivityDate="2019-10-02T23:40:18.533" Title="Is there a 6th equilibrium equation for a static beam?" Tags="|statics|moments|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31320" PostTypeId="2" ParentId="31319" CreationDate="2019-09-29T15:23:12.860" Score="0" Body="&lt;p&gt;It is a statically indeterminate structure so no application of static equilibrium equations will be sufficient to get the answer. But using the fact that the beam is infinitely stiff gives an ever quicker way: The beam will follow a perfectly straight line both before and after applying the load so the fundamental assumptions of Navier's formula for calculating the bending stress in a beam (plane cross section before bending remains plane after bending) are satisfied for the three springs as a beam cross section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I consider the springs as three points with an area of 1 each. Therefore this a cross section with section forces &lt;span class=&quot;math-container&quot;&gt;$N=W$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$M=\frac{WL}{6}$&lt;/span&gt; and the cross section properties of area &lt;span class=&quot;math-container&quot;&gt;$A=3$&lt;/span&gt; and section modulus &lt;span class=&quot;math-container&quot;&gt;$S=2\frac{L}{2}=L$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the spring forces in A, B and C resolve as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_A = \frac{N}{A} + \frac{M}{S} = \frac{W}{3} + \frac{WL}{6L} = \frac{W}{2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_B = \frac{N}{A} = \frac{W}{3}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_C = \frac{N}{A} - \frac{M}{S} = \frac{W}{3} - \frac{WL}{6L} = \frac{W}{6}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2019-09-29T15:23:12.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31321" PostTypeId="1" AcceptedAnswerId="31326" CreationDate="2019-09-29T15:37:11.437" Score="1" ViewCount="171" Body="&lt;p&gt;Around where I live there's these soapbox/gravity kart races (google &quot;carros de rolamentos&quot;) in which the wheels are built simply from ball bearings. As I was building my own, I began wondering how I could easily mount my large bearings onto my kart, and it occurred to me that an easy way to do it would be to get some smaller bearings inside the ones previously mentioned and mount the resulting pair onto the frame, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mvLi4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mvLi4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More importantly, I was under the impression that this would improve the performance of the kart in terms of speed. I am not completely sure though, hence why I am asking this question. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;On one hand,  I find that my older kart reaches some sort of terminal velocity when the inner and outer races of the bearings can only move so fast in relation to each other. So, the inner bearing would sort of take some load out of the larger one by rotating in the same direction, allowing the outer-outer race to rotate faster in relation to the inner-inner race (or, in the worst case scenario, I would get no gain but also no loss, unless the smaller bearing spins backwards). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On the other hand, I asked this to a student of mechanical engineering and he said he doubted I would get any gain, because there would be more overall friction, which does make some sense, but does not convince me much.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Also:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;We can assume that they will be well coupled together (I can file them if needed and will probably weld them afterwards).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I can't run any practical tests at the moment because I haven't bought the smaller bearings yet (I don't know if it's worth it)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;From my research, this kind of configuration has only been mentioned in some videos about fidget spinners, and sure enough, it seems to rotate fairly well.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="22962" LastActivityDate="2019-09-29T20:24:55.997" Title="Cascaded ball bearings" Tags="|mechanical-engineering|applied-mechanics|bearings|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31322" PostTypeId="2" ParentId="31319" CreationDate="2019-09-29T16:41:27.223" Score="1" Body="&lt;p&gt;Since the beam is rigid, the displacement and therefore the force in the middle spring must be the average of the two ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from left to right, we can write the forces as &lt;span class=&quot;math-container&quot;&gt;$A+B$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A-B$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vertical equilibrium gives &lt;span class=&quot;math-container&quot;&gt;$A = W/3$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking moments about the middle of the beam gives &lt;span class=&quot;math-container&quot;&gt;$2BL/2 = WL/6$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$B = W/6$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the forces from left to right are &lt;span class=&quot;math-container&quot;&gt;$W/2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$W/3$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$W/6$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-09-29T16:41:27.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31323" PostTypeId="2" ParentId="31317" CreationDate="2019-09-29T17:22:04.150" Score="0" Body="&lt;p&gt;You assume a new section by multiplying the area of the steel by the ratio of modules of elasticity of steel over aluminum while keeping the counter of messes at the same point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can just multiply the X dimension of the steel by n= Es/Ea, if the beam is loaded on Y direction. The tube will turn to an oval cylinder. Or you can just add a square section with area a = Atube * n - Atube at the center of the tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you calculate the second area moment, I , of this new section treating it as all aluminum and apply deflection equations as you'd to a simple beam.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-09-29T17:22:04.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31324" PostTypeId="2" ParentId="31321" CreationDate="2019-09-29T17:58:37.433" Score="0" Body="&lt;p&gt;You will increase the friction due to two sets of balls and lubricants with the races.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, this will cause more play in the system...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it worked and gave an advantage then it would be used by the F1 teams...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-09-29T17:58:37.433" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31325" PostTypeId="1" CreationDate="2019-09-29T19:17:37.877" Score="0" ViewCount="869" Body="&lt;p&gt;The question is straightforward. FEM is a third year course in mechanical engineering but not generally taught in electrical. Why is this?&lt;/p&gt;&#xA;" OwnerUserId="22965" LastActivityDate="2019-09-30T20:43:26.643" Title="Why is finite elements method not part of electrical engineering curriculum?" Tags="|finite-element-method|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31326" PostTypeId="2" ParentId="31321" CreationDate="2019-09-29T20:24:55.997" Score="1" Body="&lt;p&gt;Model airplane motors run at 50000 RPM with ball bearings; I don't think that your limit is due to speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, the assembly you picture will have the friction of, and be as rugged as, just using the inner bearing.  The outer will just be going along for the ride (and adding weight and expense to your assembly).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-09-29T20:24:55.997" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31327" PostTypeId="1" CreationDate="2019-09-30T03:18:00.460" Score="0" ViewCount="56" Body="&lt;p&gt;Could 3D objects be printed from a distance? Say, by pointing a laser at a stone wall?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not concerned with whether the requisite technology already exists or would be difficult to build, but rather, whether something like this could be built in principle, or if there is either a law of physics or a massive intractability that would prevent it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is possible, I'm interested in the theoretical limits. How far away can the material be (e.g. can we print something on the moon with a laser stationed on earth)? Can any material be used? Etc.&lt;/p&gt;&#xA;" OwnerUserId="22969" LastEditorUserId="17070" LastEditDate="2019-10-01T15:36:19.683" LastActivityDate="2019-10-01T15:36:19.683" Title="Could 3D printing be done at a distance e.g. by pointing laser?" Tags="|mechanical-engineering|manufacturing-engineering|3d-printing|" AnswerCount="0" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="31328" PostTypeId="2" ParentId="31325" CreationDate="2019-09-30T03:41:04.787" Score="1" Body="&lt;p&gt;Mechanical engineers have to deal all the time with things like stress distributions in objects of finite size and noninfinite strength or velocity, temperature, and pressure distributions in large amounts of flowing gases and liquids. Nothing quite like this exists in the field of electrical engineering. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-09-30T03:41:04.787" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31330" PostTypeId="1" AcceptedAnswerId="31370" CreationDate="2019-09-30T09:52:55.927" Score="0" ViewCount="68" Body="&lt;p&gt;I am writing a finite element code for my final year project of BS Mechanical Engineering. The geometry is an integration of several parts composed of different materials. I don't have exact values of material properties. However, I do have ranges of them. For example, for Steel1020, Young's modulus varies from 205-215 GPa. Should I use lower limit of Young's modulus or higher limit of Young's modulus, or average (geometric / arithematic)?&lt;/p&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2019-10-03T00:16:07.680" Title="Determination of Young's Modulus for use in Analysis" Tags="|mechanical-engineering|structural-engineering|structural-analysis|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31331" PostTypeId="2" ParentId="31330" CreationDate="2019-09-30T10:16:07.907" Score="1" Body="&lt;p&gt;Make it user definable and provide the info about the ranges as you stated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider whether you should limit the extremes of value...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-09-30T10:16:07.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31334" PostTypeId="1" AcceptedAnswerId="31336" CreationDate="2019-09-30T15:31:44.037" Score="1" ViewCount="2905" Body="&lt;p&gt;I've watched a few videos explaining how actuators work and it seems to me that a regular, non-blocking, actuator would freely move if not powered. But theory and practice are rarely the same thing.&#xA;I've never owned an actuator, and nobody on the entire internet have tried to push on an actuator that is not powered, hence this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give a bit more clarity, I'll describe my application and what I want to use the actuator for. There's this door that I often forget to close. I want the door to close after some time, if open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't need the actuator to ever open the door, just to close it, if open for more than, say 5 minutes. Edit: The closing mechanism has to be on the &quot;pull to close&quot; side of the door.&#xA;I want to be able to manually open and close the door, as that will still be the primary mode of operating the door.&#xA;The logic part will be done using an arduino and some sort of sensor that figures whether the door is open or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An actuator seems to fix my problem, but only if the shaft can freely move when not powered. So, can a cheap actuator &lt;a href=&quot;https://www.aliexpress.com/item/32984863011.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this&lt;/a&gt; be moved freely when not powered? If not, what would my options be?&lt;/p&gt;&#xA;" OwnerUserId="22976" LastEditorUserId="22976" LastEditDate="2019-10-02T11:03:11.073" LastActivityDate="2019-10-02T14:26:25.977" Title="Can linear actuators be moved manually?" Tags="|actuator|linear-motors|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31336" PostTypeId="2" ParentId="31334" CreationDate="2019-09-30T17:24:23.423" Score="2" Body="&lt;p&gt;Your primary question, if a linear actuator can be moved with the power off, will be no, unless you find a mechanism specifically created for that purpose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These actuators are effectively a pinion gear design with a very high reduction ratio. You'll note in the link you provided that heavy lift mechanisms such as a motorcycle lift and a medical bed are presented for this type of use. They work well because you can position the bed/lift top where you wish and stop without additional locking mechanisms, although you'll find most have a safety latch of some sort.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your application, closing a door, this type of linear actuator will be poorly suited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can envision a more mechanically complex arrangement which would use a linear actuator. Consider the manual release for a garage door opener. The chain and carrier for the opener can be considered the linear actuator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I disengage the manual release, then activate the button or remote, the door will remain closed. I can open and close manually the garage door without interference from the motorized mechanism, as long as I do not engage the latch system that is part of this garage door device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the linear actuator in fully-open mode has a pusher block attached, and remains in open mode, your door will operate normally. A sensor tells your system that the door is closed after the appropriate period.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, the device actuates the actuator and pushes the door closed, until the sensor informs the device that it is closed, at which point, the actuator returns to the original location, ready to push again on command.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be other approaches using clutches, cables, etc., but this is the one that popped into my alleged mind as an option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: similar method to my suggestion, based on new information regarding the placement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that the actuator is on the inside of the door, as specified. When idle (door closed), the actuator is retracted, to a point where the end of the ram is not in contact with the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the end of the ram is a magnet of sufficient strength, matching placement on the door of a metal panel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the door is ajar for the specified time, the ram/actuator will travel to the limit, &quot;collecting&quot; the door in the process. If that aspect is undesirable, a limit switch could be placed on the actuator, triggering motor reversal. Simple microswitches such as those used in 3D printers are inexpensive and easily found online.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the actuator contacts the door, either at full travel or with the switch activated, the magnet will pull the door closed as the ram retracts. When the door is closed, the additional space between the ram and the door will disengage the door, returning it to normal service.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If magnets are not desirable, a more complex design using the limit switch concept to trigger a latch and socket mechanism could be implemented. Additionally, a low-to-medium force detent mechanism might work, but that would require full extension to engage. Think of a horizontal plate on the door with a large diameter hole in the surface. The end of the ram would have a sphere that has to deflect or be deflected in order to engage the hole. Once engaged, reversal will pull the door closed until the retraction force pops the sphere free from the hole. Less reliable, less fussy to build, but still an option.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2019-10-02T14:26:25.977" LastActivityDate="2019-10-02T14:26:25.977" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31340" PostTypeId="2" ParentId="31325" CreationDate="2019-09-30T20:43:26.643" Score="0" Body="&lt;p&gt;The finite element method is used to solve problems where you need to find the electrical and magnetic fields.  The governing differential equations are the Maxwell's equations (either static or dynamic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any good engineering curriculum with have at least one course that deals with solving these equations (or their simplified forms) numerically.  The most popular numerical methods are finite differences, the method of moments, and finite elements.  However, undergraduate courses usually limit themselves to the first two and leave finite elements for graduate level courses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typical applications include computing the electromagnetic scattering (e.g., radar cross-sections) of objects, design of photonic crystals and metamaterals, reconstructing the electric field of the heart and brain from sparse samples, low level circuit and chip design, MEMS and radio-frequency engineering, wireless communication design, antenna design, remote sensing and subsurface scattering, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most popular multiphysics simulation software suites have paid large amounts to buy electromagnetic FEA software in recent years and they are used widely in industry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, recent developments in exterior calculus for finite elements have made it easier to deal with issues that were causes by the large null space of the div and curl operators of the field in quasistatic problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think FEA will become standard in undergraduate electrical engineering in a few years.&lt;/p&gt;&#xA;" OwnerUserId="3488" LastActivityDate="2019-09-30T20:43:26.643" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31341" PostTypeId="1" CreationDate="2019-09-30T21:04:12.107" Score="0" ViewCount="28" Body="&lt;p&gt;I'm trying to find a small approx 2.5&quot; touchscreen display that would be good for utilization in a small battery powered handheld instrument that has a frequently updating value that it reads.  e-ink/e-paper displays are low power but have refresh rate issues.  Other types of screens I'm finding use too much power to run off of a 9 volt for more than a day.   &lt;/p&gt;&#xA;" OwnerUserId="5034" LastActivityDate="2019-09-30T21:04:12.107" Title="small touchscreen for handheld instrument" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31342" PostTypeId="1" CreationDate="2019-10-01T01:41:08.847" Score="1" ViewCount="61" Body="&lt;p&gt;For instance, a compass can create a perfect circle, and two pins can be used to create an ellipse. Devices such as Spirographs can also be used to draw all sorts of continuous parametric curves. Is it then possible to construct other (simple) devices that may continuously draw parabolas? (Or hyperbolas, for that matter) If not, why not?&lt;/p&gt;&#xA;" OwnerUserId="22978" LastActivityDate="2019-10-01T01:41:08.847" Title="Does there exist a device which can construct a perfect parabola?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31343" PostTypeId="1" CreationDate="2019-10-01T06:42:11.723" Score="1" ViewCount="166" Body="&lt;p&gt;Let's say that there has been a handover from LTE to 3G of a particular UE. Once the handover is complete, the MME is no longer required to maintain the sessions/bearers that has been established for that particular UE. I have found the following call flow in a reference, which is not so reliable.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;1.  eNB  ⇒  MME     S1AP    S1      UEContextReleaseRequest&#xA;2.  MME  ⇒  SGW     GTPv2   S11     Release Access Bearers Request&#xA;3.  SGW  ⇒  MME     GTPv2   S11     Release Access Bearers Response&#xA;4.  MME  ⇒  eNB     S1AP    S1      UEContextReleaseCommand&#xA;5.  eNB  ⇒  MME     S1AP    S1      UEContextReleaseComplete&#xA;6.  SGSN ⇒  MME     GTP     Gn      SGSN context request&#xA;7.  MME  ⇒  SGSN    GTP     Gn      SGSN context response&#xA;8.  SGSN ⇒  MME     GTP     Gn      SGSN context acknowledgement&#xA;9.  MME  ⇒  MSC     SGSAP   SGs     SGsAP-EPS-DETACH-INDICATION&#xA;10. MSC  ⇒  MME     SGSAP   SGs     SACK SGsAP-EPS-DETACH-ACK&#xA;11. MME  ⇒  SGW     GTPv2   S11     Delete Session Request&#xA;12. SGW  ⇒  MME     GTPv2   S11     Delete Session Response&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Given the scenario, some of these calls make sense to me. For an instance, once the handover to 3G is done, there is no point of having the access bearers S1 between eNB and S-GW and S5/S8 bearer between S-GW and P-GW and so forth. So, the &lt;code&gt;Release Access Bearer Request&lt;/code&gt; in 2, and &lt;code&gt;Delete Session Request&lt;/code&gt; at 11 are sensible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I particularly don't understand is that, why are calls 6,7 and 8 there? What purpose do they serve? Or is this &lt;strong&gt;entire flow&lt;/strong&gt; blatantly wrong? If it is blatantly wrong, where can I find something sensible to read about what really happens? I have combed through the 3GPP specs for weeks now, but still couldn't find answers to the questions above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT-1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, after looking at some PCAPs, the purpose of the calls 6,7 and 8 is also clear. Upon falling back to 3G, the SGSN is going to require the UE context at MME, in a format that SGSN can understand. To an SGSN, MME looks like another SGSN. So it requests, the context from there over &lt;strong&gt;S3 interface&lt;/strong&gt;. However, the call flow above says the message exchange between MME and SGSN happens over Gn, that which contradicts with the the 3GPP spec, that which according to, the Gn interface is used to establish the connection between SGSN and GGSN. Now, the new question is going to be on the accuracy of the flow. Is this call flow accurate?&lt;/p&gt;&#xA;" OwnerUserId="22981" LastEditorUserId="22981" LastEditDate="2019-10-07T00:29:07.617" LastActivityDate="2019-10-07T00:29:07.617" Title="LTE to 3G IRAT Handrover: What happens afterward" Tags="|telecommunication|lte|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31345" PostTypeId="1" AcceptedAnswerId="31364" CreationDate="2019-10-01T09:22:44.897" Score="0" ViewCount="492" Body="&lt;p&gt;When analysing and then designing a rigid jointed, statically indeterminate structure, I can solve for all the forces and moments using the Hardy Cross (Moment Distribution) Method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I am unsure about how to solve the deflections on the beams and columns. I need these values in order to design based on serviceability limit states and verify deflection checks.    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you please demonstrate (or point out) a method to quickly and effectively find the displacements in a multi-storey, statically indeterminate structure. &lt;/p&gt;&#xA;" OwnerUserId="20822" LastActivityDate="2019-10-02T17:21:28.840" Title="Deflections Of Statically Indeterminate Structures" Tags="|structural-engineering|civil-engineering|structural-analysis|structures|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31346" PostTypeId="1" CreationDate="2019-10-01T09:26:53.500" Score="0" ViewCount="23" Body="&lt;p&gt;I am interested in pumping about 90 ml of water at 10 to 12 bars pressure in about 25 seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can this be done using electroosmotic pump? What is the energy consumption if it can be done..? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards &lt;/p&gt;&#xA;" OwnerUserId="19525" LastEditorUserId="17070" LastEditDate="2019-10-01T14:07:11.577" LastActivityDate="2019-10-01T14:07:11.577" Title="Could electro-osmotic pumping be used to pump 90 ml water at 10 bars in about 25 seconds?" Tags="|pumps|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31348" PostTypeId="1" CreationDate="2019-10-01T16:13:59.190" Score="1" ViewCount="100" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LbalV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LbalV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am making a robot for my college project. For now, the robot simply has to move forward, reverse, right and left. While the forward and reverse thing is very smooth but while taking a turn it wobbles continuously. I tried to add weight and slow down the speed but the vibration is still there. I need to know what I am doing wrong. Attaching the youtube links of the wobbling robot.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/3OqJjJY1anA&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wobbling robot during turning&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/GlL9NN1W6ns&quot; rel=&quot;nofollow noreferrer&quot;&gt;Smooth movement in the forward direction&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="19152" LastEditorUserId="8055" LastEditDate="2019-10-02T03:06:33.963" LastActivityDate="2019-10-02T21:44:05.340" Title="My robot is not able to maintain balance while turning" Tags="|mechanical-engineering|wheels|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31349" PostTypeId="1" CreationDate="2019-10-01T16:21:51.567" Score="2" ViewCount="138" Body="&lt;p&gt;I'm recreating an old mechanisme in 3d modeling software. The idea is that this triangular shape rotates around an off-center axis. This will create 4 phases. first (like in the picture) the construction L is down and stay stable in that position (which means the parts of the triangle thing that touch the construct L are partially circle. In the next phase the construction moves up, then it is stable again and in the last phase it moves down.&#xA;Questions are: what is the mathematics involved in the shape assuming it never looses contact with the construct L. Initially I made the sides out of 6 circular elements (arcs), but I'm not sure if that is even the correct shape. Once I have the shape, what is the position of L as function of the rotation of the object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/itQt5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/itQt5.png&quot; alt=&quot;wobble&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22990" LastActivityDate="2019-10-02T14:13:31.427" Title="recreating old mechanisme" Tags="|mechanisms|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31350" PostTypeId="2" ParentId="31349" CreationDate="2019-10-01T17:10:44.683" Score="3" Body="&lt;p&gt;I think you are looking for a &quot;curve of constant width&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6rqaa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6rqaa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Construction diagram for generating a curve of constant width from a triangle. Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Curve_of_constant_width&quot; rel=&quot;nofollow noreferrer&quot;&gt;Curve of constant width&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Curves of constant width can be generated by joining circular arcs centered on the vertices of a regular or irregular convex polygon with an odd number of sides (triangle, pentagon, heptagon, etc.).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2019-10-01T17:10:44.683" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31351" PostTypeId="2" ParentId="31348" CreationDate="2019-10-01T19:31:32.450" Score="1" Body="&lt;p&gt;Your connection of the wheels to the robot enclosure (suspension) is flimsy and has too much free play. It starts to jerk and gives enough kick energy to the flimsy box to keep it dancing. And the skidding of the wheels resonates with the vibration of the robot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To improve the stability without having to use electronics self balancing chips:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Use an integrated frame separate from the box, with wider distance between the wheels, wider track, more like a square footprint. It is essential that the wheels axels are substantial and have long extension, may be all the way to the opposite end.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Put diagonal bracing inside the box and use self locking screws to make it more a rigid frame, as it is it is too rickety, and side panels act like the pendulum to excite the vibration.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Try different size and material on wheels and may be a bit of toe out angle in the in the wheels because it seems the wheels' skidding is resonating with the box vibration natural frequency.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use a small flywheel on the motors to keep the torque output steady.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If practical change the geometry of robot to make it less narrow.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-01T19:31:32.450" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31353" PostTypeId="2" ParentId="31349" CreationDate="2019-10-01T21:10:58.853" Score="0" Body="&lt;p&gt;To create your triangular pick shape, you need two equilateral triangles, centered on each other.  Your spline should make contact with each of the corners of the triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ratio of lengths between the two triangles will give you the curvature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your ultimate problem is you don't have dimensions so you can do your math.  This SolidWorks drawing will give you the dimensions to build your slot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dimAq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dimAq.png&quot; alt=&quot;Pick of Destiny&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two rotations of Part C will give you the dimensions of your guideway.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2019-10-01T21:34:39.520" LastActivityDate="2019-10-01T21:34:39.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31355" PostTypeId="1" CreationDate="2019-10-02T00:10:09.017" Score="0" ViewCount="26" Body="&lt;p&gt;My son built a computer a few months ago with Gigabyte m/b and Ryzen 5 2600x. He bought the parts at a local Computer Super-Store and built it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I liked it so much I bought the parts at the same local Computer Super-Store and built one for myself. Had some trouble with bad h/w but Super-Store allowed returns and now my computer runs great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we were attempting to determine why my computer might be having trouble we noticed only one difference: his Ryzen 5 2600x was marked Rev. F and mine was marked as Rev. A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I returned the m/b and CPU I thought maybe I'd get a Rev. F also but again it just so happened that it was a Rev. A and everything does work fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'm very curious as to what the differences might be in those two Revs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you know what the differences are? Do you know where I might find the differences listed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, I Googled (actually duckduckgo) but didn't find anything even mentioning the Revs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even on the official site on the specific CPU details I don't see anything: &lt;a href=&quot;https://www.amd.com/en/support/cpu/amd-ryzen-processors/amd-ryzen-5-desktop-processors/amd-ryzen-5-2600x&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amd.com/en/support/cpu/amd-ryzen-processors/amd-ryzen-5-desktop-processors/amd-ryzen-5-2600x&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15264" LastActivityDate="2019-10-02T00:10:09.017" Title="Ryzen 5 2600x Rev. A versus Rev. F: What is difference in revs?" Tags="|computer-hardware|computer|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31356" PostTypeId="1" CreationDate="2019-10-02T01:19:29.017" Score="0" ViewCount="1317" Body="&lt;p&gt;I am trying to design and build a low cost tensile test machine for plastic specimens. Most machines I've seen measure strain data using an extensometer, either the clamp-on type or the noncontact type (using video). However, both options are expensive to implement, so I was wondering if it would be a good idea to measure crosshead movement vs. time to use it as elongation data, and the correlate it with cross-section area and force vs. time data to obtain a stress-strain curve. Would this be feasible, if extremely precise results are not required?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since so many machines use extensometers, I assume there must be a good reason not to use crosshead movement as elongation data. If so, why are crosshead movement and specimen elongation different? And how significant is said difference? Would much would test results be affected if crosshead movement were used as elongation?&lt;/p&gt;&#xA;" OwnerUserId="18155" LastActivityDate="2019-10-02T16:32:42.097" Title="Can crosshead movement be used to measure strain in tensile testing instead of an extensometer?" Tags="|materials|mechanical-failure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31358" PostTypeId="1" CreationDate="2019-10-02T10:09:43.240" Score="1" ViewCount="14" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Forward_osmosis&quot; rel=&quot;nofollow noreferrer&quot;&gt;Forward osmosis (FO)&lt;/a&gt; is an osmotic process that, like reverse osmosis (RO), uses a semi-permeable membrane to effect separation of water from dissolved solutes. The driving force for this separation is an osmotic pressure gradient, such that a &quot;draw&quot; solution of high concentration (relative to that of the feed solution), is used to induce a net flow of water through the membrane into the draw solution, thus effectively separating the feed water from its solutes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In a second step, the diluted draw solution is pumped through RO membranes, leading to desalinated water and a thickened draw solution to be recycled. The stated benefit is that the FO membrane faces far less scaling than in an RO process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VEFjK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VEFjK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;&lt;a href=&quot;https://www.osmotic-engineering.com/wp-content/uploads/2019/08/PER11-198.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Image source&lt;/a&gt;&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the electrolyte used as draw solution can be optimized so that no scaling on the RO membranes occurs. Since the manufactuers of these plants don't publish the composition of the draw solutions, a lot of research went into that direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have conceptual question on why FO is advantageuos at all: when water diffues through a membrane, the concentration of the solutes is higher near the membrane and this ultimately leads to scaling if hardness builders etc. are present. It is stated that this occurs far less when the diffusion is driven by osmotic pressure instead of pressure from a pump. &lt;strong&gt;Why is that so?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-10-02T10:09:43.240" Title="Why do forward osmosis membranes face less scaling?" Tags="|water-resources|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31359" PostTypeId="2" ParentId="31349" CreationDate="2019-10-02T10:46:33.933" Score="0" Body="&lt;p&gt;the answer to the question about the position of L versus the angle a is divided into six ranges for the angle :&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;0-60:       r1   &#xA;60-120:     (r1-r2)*cos(a-60)+r2   &#xA;120-180:    (r1-r2)*cos(a)+r1        &#xA;180-240:    r2        &#xA;240-300:    (r1-r2)*cos(a-60) +r1  &#xA;300-360:    (r1-r2)*cos(a)+r2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Where r1 is the radius of the bigger circle and r2 is the radius of the smaller circle. In my case in figure one of the answer from @transistor A, B, C, a,b,c form nice equal sided triangle so only two radii are involved. &lt;/p&gt;&#xA;" OwnerUserId="22995" LastEditorUserId="22995" LastEditDate="2019-10-02T14:13:31.427" LastActivityDate="2019-10-02T14:13:31.427" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="31360" PostTypeId="1" CreationDate="2019-10-02T10:59:58.450" Score="1" ViewCount="87" Body="&lt;p&gt;I would like to use a laser beam to weld small metal parts that are located within an air tight enclosure but want the laser head to be located outside the enclosure which would require the laser beam to pass through a glass window. Can laser beams used to weld metals pass through glass?&lt;/p&gt;&#xA;" OwnerUserId="23004" LastActivityDate="2019-10-02T23:01:33.317" Title="Laser beam welding" Tags="|welding|glass|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31362" PostTypeId="1" CreationDate="2019-10-02T16:08:32.737" Score="1" ViewCount="47" Body="&lt;p&gt;Traditionally, and continuing till the present day, wind turbines are conceived/presented as power plants which harvest the (macroscopic) kinetic energy carried by an externally available airflow, with corresponding &lt;a href=&quot;https://en.wikipedia.org/wiki/Betz%27s_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;upper bounds on how much can be harvested&lt;/a&gt;. Implicitly, this amounts on or relies on an assumption that it would be generally impossible to harvest the internal energy in the flowing gas by allowing it to exert &lt;a href=&quot;https://en.wikipedia.org/wiki/Work_(thermodynamics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;expansion work&lt;/a&gt; onto our system (in contrast to this being possible in steam turbines). Naive calculations, with airflow-parameters (speed, temperature, pressure) typical for our terrestrial atmosphere, suggest that one should be able able to harvest much more energy in this way than merely a fraction of the airflow's kinetic energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Q: What, if anything, prevents us from building wind-turbines that significantly exploit &quot;expansion work&quot;?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="6692" LastEditorUserId="6692" LastEditDate="2019-10-02T17:37:13.807" LastActivityDate="2019-10-02T17:37:13.807" Title="Conceiving a wind turbine as a thermodynamic machine" Tags="|fluid-mechanics|thermodynamics|wind-power|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="31364" PostTypeId="2" ParentId="31345" CreationDate="2019-10-02T16:31:46.417" Score="4" Body="&lt;p&gt;To rigorously solve for deflections of indeterminate structures, the most feasible method is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Direct_stiffness_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Direct Stiffness Method&lt;/a&gt; (aka matrix methods). This is generally the approach finite element method software packages (FTool, SAP2000, RISA, etc.) are implementing when dealing with line elements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For structures with a relatively small number of degrees of freedom it is possible to implement the direct stiffness method and arrive at a solution (somewhat) by hand. In my matrix methods class at university we generally solved the matrices using a calculator or by implementing a routine coded in MATLAB. So really, we were doing exactly what the FEM software does but more slowly and/or transparently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Prior to the advent of computers (including programmable calculators) engineers would have used approximate methods of analysis to ballpark deflections. One very simple example of this might be that if we know the load on a beam we can determine an upper-bound deflection by considering it to be simply supported. When doing a hand check of computer calculations, engineers today will still take this approach. That is, we make simplifying assumptions and get a back-of-the-envelope feel for whether the computer solution is appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computer-based analyses are fairly essential for the design of today's structures by today's design codes. The goal of hand calculations is rarely to exactly duplicate the computer solution. Rather, the goal is to use hand methods and engineering judgement to validate the computer solution.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastEditorUserId="2434" LastEditDate="2019-10-02T17:21:28.840" LastActivityDate="2019-10-02T17:21:28.840" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31365" PostTypeId="2" ParentId="31356" CreationDate="2019-10-02T16:32:42.097" Score="1" Body="&lt;p&gt;It is usually not accurate or precise enough ; primarily because of movement between the jaws or what ever device hold the specimens and the specimen. On metal samples , even an extensometer can make a little irregularity in the beginning of the strain curve as it &quot;seats-in&quot;. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-10-02T16:32:42.097" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31366" PostTypeId="2" ParentId="31360" CreationDate="2019-10-02T17:50:11.337" Score="2" Body="&lt;p&gt;Most weld optics have a protective glass that shields the (more expensive to replace) focusing lens from welding spatter. So there are definitely glasses that exist and can be used for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When designing your enclosure and/or the location of the metal parts during welding operation, try to let the laser beam propagate through the window as soon as possible after the focusing lens. The greater the distance, the higher the power density on the glass. Your design choices here of course depend on the focal length of the weld optic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also need active cooling for the window, depending on the power of your laser.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actual material depends on the wavelength of your laser and the power range, for simple (low power) applications, it is often BK7, for higher power quartz glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Important:&lt;/strong&gt; Such a laser protective glass is a special optical component that is particularly picked for power and wavelength. Do not use a random piece of window glass or similar without checking the absorption, transmittance and reflectance of the material at that wavelength. You don't want to direct a 2 kW disk laser on your protective glass and have half of it reflected back into the resonator or even worse, into the room you are standing in.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastEditorUserId="14629" LastEditDate="2019-10-02T23:01:33.317" LastActivityDate="2019-10-02T23:01:33.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31367" PostTypeId="2" ParentId="31348" CreationDate="2019-10-02T21:44:05.340" Score="0" Body="&lt;p&gt;You have a tall skinny skid-steer vehicle, and you're trying to run it on a surface with a high friction coefficient.  So the wheels are constantly grabbing and letting go, grabbing and letting go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try making either the front or rear pair of wheels castoring.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-10-02T21:44:05.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31368" PostTypeId="1" CreationDate="2019-10-02T23:33:16.887" Score="0" ViewCount="593" Body="&lt;p&gt;Our rocket team is trying to finalize the design of our injector for a small liquid rocket engine.  The current analysis challenge is verifying the correct pressure drop across the annular orifice leading to the combustion chamber which the Kerosene flows through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current equation we are working with is this one provided in Mathwork's documentation.&lt;br&gt;&#xA;&lt;a href=&quot;https://www.mathworks.com/help/physmod/hydro/ref/annularorifice.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mathworks.com/help/physmod/hydro/ref/annularorifice.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because we know the inner diameter of the annulus (the pintle body OD), and the properties of the liquid flowing through the reservoir leading to the injector orifice, we were able to calculate an orifice OD of 1.11inches. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason we are suspicious of this equation is because it makes reference to laminar flow, where we have a pretty high mass flow, and hence high flow velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any simple (undergraduate level) resources that can further inform us how to either directly calculate diameter, or could let us converge on a discharge coefficient and diameter through numerical methods?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!  Here is a quick picture of our injector if that helps.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KetC4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/KetC4.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14040" LastActivityDate="2019-10-02T23:33:16.887" Title="Flow through an Annular Orifice for Pintle Injector" Tags="|fluid-mechanics|rocketry|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31369" PostTypeId="2" ParentId="31319" CreationDate="2019-10-02T23:40:18.533" Score="0" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/a/31322/1832&quot;&gt;@alephzero has correctly described the solution&lt;/a&gt;: notice that since the beam is stiff, we can determine the proportion of deflections for each support. And since the force exerted by a spring is linearly proportional to its deflection, we also know the proportion of the reaction forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's also worth noting an important mistake you made in your work, and is the reason why you got that &quot;four of the equilibrium equations [...] are linearly dependent&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is simple: you can only have one sum-of-moments equation per structure. Using more such equations (as you have done) actually gives you no new information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, going back to the basics: for a 2D frame analysis, you only have three global equilibrium equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sum F_x = 0 \\&#xA;\sum F_y = 0 \\&#xA;\sum M = 0 \\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's it. If you have a more complex structure with more than three unknowns (or often two, when there's no horizontal force component to consider, such as in this case), you'll need to either be clever or hard-working.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, we got away with being clever. But if the supports weren't springs and/or the beam wasn't stiff, then we'd have to be hard-working and actually deal with the complicated math of hyperstatic structures. Don't worry, you'll learn about those soon enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then you'll discover how to solve them with software and forget everything.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-10-02T23:40:18.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31370" PostTypeId="2" ParentId="31330" CreationDate="2019-10-03T00:16:07.680" Score="2" Body="&lt;p&gt;As &lt;a href=&quot;https://engineering.stackexchange.com/a/31331/1832&quot;&gt;recommended by @Solar Mike&lt;/a&gt;, the most &quot;professional&quot; solution is to let the end-user decide. Give them some input mechanism (a textbox, slider or dropdown, for example) so they can define which value to adopt. This is what you'll see in basically any modeling software you'll find in the wild.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it's also very important to note the observations made in &lt;a href=&quot;https://engineering.stackexchange.com/questions/31330/determination-of-youngs-modulus-for-use-in-analysis/31331#comment56301_31330&quot;&gt;@alephzero's comment&lt;/a&gt;: it almost doesn't matter. The difference between 205 and 215&amp;nbsp;MPa is 4.5%. Given the errors involved in basically every step of an engineering design calculation*, that's peanuts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, that being said, if you really want to take a strict approach to this, you need to look at the worst case. This will often be to simply adopt the lowest value. But not always.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And in this case, it might very well not be. If you are sandwiching different materials, one will be stronger than the other(s). In this case, you'll probably want to adopt the highest stiffness for the material with the lowest yield strength, and visa-versa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This way, if these elements need to work together, the strong material will transfer the most force to the weaker material, increasing the likelihood that the weaker material yields first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then again, maybe the strong material would yield even earlier if it were stiffer (if it were right on the edge of yielding and the weaker material's increased stiffness is what's holding off the &quot;straw that'd break the camel's back&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So this is a non-trivial question which depends on the specific case being analyzed. So, if you want to be strict about this, you'd either need to look at the specific case to determine the worst case or (if your program doesn't take ages to run) just run the model with the different combinations of moduli for each material and select the worst case.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* Every step of the process is an approximation: materials aren't perfectly elastic, they'll likely have different properties along a beam, supports and connections aren't perfectly fixed/pinned/springy, beam dimensions will be different, the welder will come to work drunk that day, etc.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-10-03T00:16:07.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31371" PostTypeId="1" CreationDate="2019-10-03T11:10:55.647" Score="0" ViewCount="95" Body="&lt;p&gt;What is the definition of rate &lt;span class=&quot;math-container&quot;&gt;$(−ra)=−\frac{dCa}{dt}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$(−ra)=−1/V\frac{dNa}{dt}$&lt;/span&gt; ? I think the general one is the second one and first one is for constant volume reaction system. Is the above written rate equation only valid to batch reactor? If not can we use this in a PFR or MFR? In PFR we know the rate comes out to be &lt;span class=&quot;math-container&quot;&gt;$(−ra)=−\frac{dFa}{dV}$&lt;/span&gt; ,can we equate &lt;span class=&quot;math-container&quot;&gt;$−1/V\frac{dNa}{dt}=−\frac{dFa}{dV}$&lt;/span&gt;? If not then why? Is the rate equation not valid everywhere? Can we equate with CSTR as well ? eg &lt;span class=&quot;math-container&quot;&gt;$-\frac{dCa}{dt}=\frac{Fa_o-Fa}{V}$&lt;/span&gt; (considering it a constant volume reaction system CSTR)&lt;/p&gt;&#xA;" OwnerUserId="23022" LastActivityDate="2020-11-30T19:05:31.590" Title="chemical reaction engineeing" Tags="|chemical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31372" PostTypeId="2" ParentId="31371" CreationDate="2019-10-03T13:22:16.877" Score="-1" Body="&lt;p&gt;You are correct in saying that reaction rate &lt;span class=&quot;math-container&quot;&gt;$(−ra)=−\frac{dCa}{dt}$&lt;/span&gt; is for constant volume systems and that the more general form is or &lt;span class=&quot;math-container&quot;&gt;$(−ra)=−1/V\frac{dNa}{dt}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general reaction rate equation is the basis for all the types of reactors you mentioned, see the link below for the derivation explanation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.the-seventh-dimension.com/images/textlev/LEVENSPIEL%20Chemical%20reaction%20engineering-ch4-ch5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.the-seventh-dimension.com/images/textlev/LEVENSPIEL%20Chemical%20reaction%20engineering-ch4-ch5.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The form of the rate equation as shown should be valid everywhere within the vessel, so long as the limits of integration used to describe the situation are correct. If for example, a vessel has a significant dead volume due to the flows not being properly directed through the vessel (i.e. the &quot;well mixed&quot; assumption is not being satisfied), then the ideal results generated from this equation may deviate significantly from actual data.&lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2019-10-03T13:22:16.877" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31373" PostTypeId="2" ParentId="31272" CreationDate="2019-10-03T14:02:27.367" Score="0" Body="&lt;p&gt;You can define the &lt;code&gt;h&lt;/code&gt; and &lt;code&gt;beta&lt;/code&gt; constants, either in a cell or as a &lt;a href=&quot;https://www.ablebits.com/office-addins-blog/2017/07/11/excel-name-named-range-define-use/&quot; rel=&quot;nofollow noreferrer&quot;&gt;defined name&lt;/a&gt;. &#xA;After that, you can create a column with the &lt;code&gt;theta&lt;/code&gt; parameter and another column with the calculated &lt;code&gt;y&lt;/code&gt; parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for the &lt;code&gt;y&lt;/code&gt; should use the &lt;a href=&quot;https://support.office.com/en-us/article/if-function-69aed7c9-4e8a-4755-a9bc-aa8bbff73be2&quot; rel=&quot;nofollow noreferrer&quot;&gt;IF&lt;/a&gt; and &lt;a href=&quot;https://support.office.com/en-us/article/and-function-5f19b2e8-e1df-4408-897a-ce285a19e9d9&quot; rel=&quot;nofollow noreferrer&quot;&gt;AND&lt;/a&gt; functions. Suppose the &lt;code&gt;theta&lt;/code&gt; parameters are on column A. The formula should be something like&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;=IF(AND(0&amp;lt;=A1,A1&amp;lt;beta/8),h*(PI()/(4+PI()))*A1/beta - 1/4/(4+PI())*SIN(4*PI()*A1/beta),IF(AND(beta/8&amp;lt;=A1,A1&amp;lt;7*beta/8),h*(2*/(4+PI()))+PI()/(4+PI()))*A1/beta - 9/4/(4+PI())*SIN(4*PI()*A1/3/beta),h*(4/(4+PI()))+PI()/(4+PI()))*A1/beta - 1/4/(4+PI())*SIN(4*PI()*A1/beta) ) )&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To break down the formula, the IFs statements define which equation to use. &#xA;The criteria is accessed with the AND function, since you have a range on beta. The equation itself is straightforward.&lt;/p&gt;&#xA;" OwnerUserId="15487" LastActivityDate="2019-10-03T14:02:27.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31375" PostTypeId="1" CreationDate="2019-10-03T21:14:35.087" Score="0" ViewCount="35" Body="&lt;p&gt;In the 3rd step of cryogenic air separation, filtered air is purified to remove water, CO2, and hydrocarbons from the N2-O2-Ar stream. Im interested in knowing what happens with that CO2 and water. Can both resources be recovered in a useful form? Can the CO2 be purified/ are Praxair or others currently doing this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Context: Im doing research on a theoretical industrial process which uses pure CO2, water, and N2. I want to know if all these inputs can be obtained from a standard air separation facility (of course N2 can). Im particularly interested in CO2 recovery.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please note I'm not interested in buying gases. I'm working on a publication about a theoretical system to produce protein from air and solar power.&#xA;In your answer it would be helpful to provide links to papers or websites which give descriptions of the recovery process (energy costs etc).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="21741" LastActivityDate="2019-10-03T21:14:35.087" Title="Do companies recover CO2 and water from the purification step of cryogenic air separation, and how?" Tags="|compressed-air|air|fluid-filtration|cryogenics|recycling|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31376" PostTypeId="1" CreationDate="2019-10-03T23:39:47.043" Score="2" ViewCount="404" Body="&lt;p&gt;Last night while waiting for the train at South Yarra station in Melbourne, I saw these reflective triangles stuck on the sides of the train track rails. They were evenly spaced a few metres apart for as far as I could see down both ways of the tracks. They could only be seen on the far pair of tracks, and the tracks on the other side of the platform also had these little triangles on the far pair of tracks. I'm assuming that there are also triangles on the near pair, but stuck on the other side of the rails where I couldn't see them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/1/15/South_Yarra_platforms.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;a picture of the platforms and tracks&lt;/a&gt; on &lt;a href=&quot;https://en.wikipedia.org/wiki/South_Yarra_railway_station&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Wikipedia article for South Yarra Station&lt;/a&gt; from 2014 which shows they weren't present then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't work out what these reflective triangles might be used for. What is their purpose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hVwnO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hVwnOm.jpg&quot; alt=&quot;Reflective triangles on train track rails&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14330" LastActivityDate="2019-10-04T01:15:08.267" Title="What are these reflective triangles on the sides of train track rails for?" Tags="|rail|transportation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31377" PostTypeId="2" ParentId="31376" CreationDate="2019-10-04T01:15:08.267" Score="4" Body="&lt;p&gt;They are markers used for accurate position monitoring, without the need to close the tracks to trains and allow people to work close to them. All the markers can be scanned from a few fixed points (e.g. on the platforms) to create an accurate 3D picture of any movement of the track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the sleepers or ballast may have been replaced recently, or there might have been other civil engineering work nearby which could have disturbed the tracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similar markers are sometimes used as a reference points for train drivers (for example when shunting) with illumination from lights on the train, but that is less likely within a station where the platforms are presumably lit anyway when the station is in use at night.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;a href=&quot;https://www.berntsen.com/Surveying/Smart-Targets-Datums-Reflectors/RSAK80-Adapter-w-Smart-Angle-Targets&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.berntsen.com/Surveying/Smart-Targets-Datums-Reflectors/RSAK80-Adapter-w-Smart-Angle-Targets&lt;/a&gt;, for example.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-10-04T01:15:08.267" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31378" PostTypeId="1" CreationDate="2019-10-04T06:56:25.250" Score="0" ViewCount="562" Body="&lt;p&gt;This is a recommended follow on to one of my previous questions. It was stated that I could find the deflections of a beam by idealising it as a simply supported beam and then calculating accordingly. The loads laterally on the beam would cause a deflection perpendicular to the axis of the beam whereas the side-sway loads would cause deflections perpendicular to the column and thus the beam as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main question is: how do I find the magnitude of the deflection on a particular beam within a rigid jointed multi-storey, multi-bay frame like the one shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.buildinghow.com/portals/0/Book-B/5_3/image012.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://www.buildinghow.com/portals/0/Book-B/5_3/image012.jpg&quot; alt=&quot;Multi-stiorey frame&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I find the lateral deflection of each column and beam using traditional structural analysis methods? I can find the moments and all forces using moment distribution (Hardy Cross) method but I need the deflections in order to use serviceability limit states in design. I've also thought about the slope-deflection equation, however, I can only find the slopes at each end of the beams. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, how conservative are these results?&lt;/p&gt;&#xA;" OwnerUserId="20822" LastEditorUserId="20822" LastEditDate="2019-10-10T07:14:03.893" LastActivityDate="2019-10-10T07:14:03.893" Title="Deflection Of Beam In Structural Rigid Jointed Frame" Tags="|structural-engineering|structural-analysis|statics|structures|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31379" PostTypeId="1" CreationDate="2019-10-04T08:42:19.273" Score="0" ViewCount="170" Body="&lt;p&gt;The Hardy Cross method for analysis of fluid systems assumes that in inlet flows and outlet flows for the system are known. However, when one constructs such a system in real life, how do we ensure that the flow entering and exiting the system will be exactly equal to the parameters set initially? e.g. if the system is connected to a water tank for input, and has a loop with 3 outlets, and we assume that the input water flow is 300 l/s and the outlet flows are 100 l/s each. We decide all lengths and diameters as part of stating the initial condition to apply the Hardy Cross equation, and the method tells us what the flow rates will be based on the analysis. How do I know that the system will obey these parameters, and that the flow rates will match the initial assumptions I made. &lt;/p&gt;&#xA;" OwnerUserId="24853" LastActivityDate="2020-10-28T11:38:53.820" Title="How is the Outlet flow rate in a water system controlled?" Tags="|mechanical-engineering|fluid-mechanics|civil-engineering|water-resources|environmental-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31380" PostTypeId="2" ParentId="31379" CreationDate="2019-10-04T08:54:37.343" Score="0" Body="&lt;p&gt;You design the system for a demand that it should meet, 3x 100l/s in your example. If the demand is higher, the sytem will likely fail to deliver the actual demand. If the demand is lower, no problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the network side, you know the pressure you want to deliver the water at. You could supply valves at the tap that deliver only (say) 100l/s at 4 bar, but in most cases  actual water flow has to be managed on the consumer side and there are countless ways to do that: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;manually turning down the tap (if the consumer is a shower or something like that) &lt;/li&gt;&#xA;&lt;li&gt;If the consumer needs a constand water flow, they can use a combination of pressure regulator, simple flow meter and manual valve. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-10-04T08:54:37.343" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31381" PostTypeId="1" CreationDate="2019-10-04T10:02:12.617" Score="1" ViewCount="134" Body="&lt;p&gt;I am trying to determine the modulus of elasticity (Young's Modulus) for a rigid paper sheet. The issue I am having is that my results for the Young's Modulus change depending on what orientation (length extending out or width extending out) I use to measure it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material of the paper is not trivial, and as a result I am not able to find a datasheet on the material itself. Instead, I did some simple tests to calculate the Young's Modulus, especially since it is a thick sheet that behaves more like a sheet of metal than floppy paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I took the sheet, fixed one end, and measured the deflection at the other end (and other relevant parameters). I did this with both the length and the width extending freely over the fixed end. I accounted for curvature in the sheet when measuring the deflection in order to make the measurement as representative of an ideal beam as possible (there were some assumptions I made about the curvature, but I will not go into details about how I eliminated the effect of curvature in my deflection measurement unless I have to, as I believe the problem lies elsewhere). I then used the Euler-Bernoulli beam equations for distributed load, &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;, (in this case, the distributed load was due to gravity) to arrive at the Young's Modulus, &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$$E = \frac{qL^4}{8I\delta_{gravity}}$$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$q=$&lt;/span&gt; distributed load, &lt;span class=&quot;math-container&quot;&gt;$L=$&lt;/span&gt; the length of the paper extending over the fixed end, &lt;span class=&quot;math-container&quot;&gt;$I=$&lt;/span&gt; second moment of area, which changed depending on orientation, and &lt;span class=&quot;math-container&quot;&gt;$\delta_{gravity}=$&lt;/span&gt; deflection after removing effects of curvature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The results I achieved were consistent ( within 0.3 GPa of each other), but changed drastically, by almost an order of magnitude or more, when changing whether it was the width of the paper that was modeled as the &quot;length&quot; of my beam, or the length of the paper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering what caused this big change in the Young's Modulus, since I don´t know if it is typical for Young's Modulus to vary by such a large amount between orientations (10-90 times difference). Paper is an anisotropic material (thank you, alephzero), so it could be related to this fact. I believe I am missing some factor in the equation that is depenedent on orientation, such as a correction factor for structural stiffness due to sheet curvature, or something with the Poisson ratio, but of course, this belief can be misleading me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things to note:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material is foamboard, only the outside of the material is made of paper in the strictest sense. The inside is made of foam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I changed the distributed load, &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;, depending on the orientation, since the weight/length changes depending on whether the width or the length. I did the same thing for the second moment of area, &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second moment of area, &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt;, I modified the equation to more precisely model the shape of the curvature, but it did not change the value of &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; significantly.&lt;/p&gt;&#xA;" OwnerUserId="23029" LastEditorUserId="23029" LastEditDate="2019-10-04T13:51:07.147" LastActivityDate="2020-10-28T19:06:20.693" Title="Modification of Beam Equation for Thin Sheet" Tags="|mechanical-engineering|structural-analysis|beam|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31382" PostTypeId="2" ParentId="30017" CreationDate="2019-10-04T12:27:04.167" Score="2" Body="&lt;p&gt;If you are using standard Gauss quadrature, from the formulation, you determine the coefficients (weights of Gauss location) and their locations. You cannot arbitrarily decide their locations. Here are some examples: &lt;a href=&quot;http://edwilson.org/BOOK-Wilson/G-inter.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://edwilson.org/BOOK-Wilson/G-inter.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23031" LastActivityDate="2019-10-04T12:27:04.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31383" PostTypeId="2" ParentId="26641" CreationDate="2019-10-04T12:40:19.167" Score="-1" Body="&lt;p&gt;I think your convergence tolerance is not enough. Since you obtain first few solutions with low convergence requirements you are moving away from the real results for the next solution steps. From the code, it seems you are using unrealistic values. In example, total displacement is 5. The tolerance is 0.01, they are crude numbers. So it is not easy to determine which values must be used. Without proper use of realistic values, it is not easy for the model to converge. For a nonlinear problem, suppose you find disp1=0.01 with corresponding force1=0.01, then for disp2=0.02, assume an acceptable solution for force2=0.019, but if without properly adjusting tolerance, you may find force2=0.018, then the force3 result will definitely change based on the force2 value you obtained.&lt;/p&gt;&#xA;" OwnerUserId="23031" LastEditorUserId="23031" LastEditDate="2019-10-07T12:11:43.230" LastActivityDate="2019-10-07T12:11:43.230" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31385" PostTypeId="1" CreationDate="2019-10-04T15:45:45.557" Score="0" ViewCount="209" Body="&lt;p&gt;CNC milling causes residual stresses in metals.&#xA;So, after CNC milling of a metal part, say aluminium, do all manufacturers (I'm not referring to any specific manufacturer, just generally) heat treat such parts parts after milling?&lt;/p&gt;&#xA;" OwnerUserId="23033" LastActivityDate="2021-10-21T22:27:26.097" Title="Is CNC milling on metals always followed by heat treatment?" Tags="|manufacturing-engineering|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31386" PostTypeId="2" ParentId="31381" CreationDate="2019-10-04T17:17:01.160" Score="1" Body="&lt;p&gt;I would calculated the I being allocated to two different materials, foam and paper. one way of getting a sense of the difference between the two is to let the paper beam on its side flatwise, and measure the E of the paper by neglecting the contribution of foam, if it is not paper on both sides glue 2 back to back to make it so. Because in that configuration the paper is the farthest from neutral axis and most stressed. Then you can use this E to compare to the vertical beam position as the papers contribution. Because of the embedded errors in measurements of the weight and deflections extra care is needed. such as for cantilever just use a continuous length of the paper 3 lengths and leave 2 length in a frame or bracket, secured over a table or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use 10 sheets of paper bound together by paper tape at intervals to make the beam 10 times thicker, not to add to stiffness of the beam. Then divide you I by 10. this will average out small inconsistencies in fabrication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If practical add a concentrated load several times bigger than the weight of the paper at the end and calculate deflection due to both distributed and concentrated loads this takes care of possible uneven thickness of the paper. &lt;span class=&quot;math-container&quot;&gt;$ \ E= \frac{Pl^3}{3\delta_yI  }$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Try to look for shear warping or lateral buckling especially near the support, even if very small and hard to see, the out of plane secondary forces can produce twisting and torque which has to be counted for.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-10-04T18:33:06.063" LastActivityDate="2019-10-04T18:33:06.063" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31388" PostTypeId="1" CreationDate="2019-10-04T22:57:33.910" Score="0" ViewCount="33" Body="&lt;p&gt;Im expanding my 28' wide single story composite shingle  home.  I want to come out perpendicular off 1 plane of the roof to expand master bedroom, and then remove&#xA;the wall.  The foundation is crawl space with peer and beam foundation. Im installing an 18' 14x3.5 psl beam above the wall being removed that will have the 2x6 rafters and 2x6 ceiling joists nailed to the face.  Im in sf bay area with no snow load. The beam will rest on two opposite 90 degree corners of the foundation peer.  My question is do i need to reinforce the peer foundation at those corners to accomodate the load created at the peer and beam foundation corners as a result of removing the wall under beam.  I believe there may be 6' concrete columns under the concrete beam at the corners.&lt;/p&gt;&#xA;" OwnerUserId="23037" LastActivityDate="2020-10-29T03:02:10.253" Title="18' header beam installation for single story room addition" Tags="|foundations|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31389" PostTypeId="2" ParentId="31388" CreationDate="2019-10-05T01:44:42.253" Score="1" Body="&lt;p&gt;You need to go to your city building department and pull a permit. They usually have a code engineer who tells you what you need to do in detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your soil is ok and you are not in a hillside, it should not be too complicated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They expect not only foundation and columns modifications, but also a plan of addition of some shear walls or some kind of lateral load resisting premanufactured wall such as &quot;Simpson Shear Panel&quot; with its required new foundation to remedy the loss of lateral support due to removal of your wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may ask at the plan-check desk for their Type-5 remodeling handouts and approved test data sheet on the types of these premanufactured shear walls and their wind and seismic ratings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or else they ask you hire an engineer, and they may even ask you to upgrade the existing home up to code if the addition is bigger than their limit which triggers code upgrade.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-05T01:44:42.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31392" PostTypeId="1" CreationDate="2019-10-05T14:41:56.873" Score="0" ViewCount="64" Body="&lt;p&gt;I'm a civil engineering student, and it's my first year at the University of Applied Sciences, and I'm doing an exercise but can't figure out one of its parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question: what are the advantages and disadvantages of reinforcement outside the dikes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I've found only general information about dyke improvement, and not the specific information about the disadvantages and advantages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So any information related to this question by any of you will be helpful for me to be able to understand it and answer the question correctly.&lt;/p&gt;&#xA;" OwnerUserId="23041" LastEditorUserId="23041" LastEditDate="2019-10-05T19:05:48.803" LastActivityDate="2019-10-05T19:05:48.803" Title="Dyke improvement" Tags="|civil-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31393" PostTypeId="2" ParentId="31295" CreationDate="2019-10-05T17:29:49.130" Score="3" Body="&lt;p&gt;If we talk about quality and 6-sigma (which roughly means 1 ppm defects), the 1 ppm refers to test escapes, not test rejects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if a part only costs \$1, if testing fails to catch a bad part and we accidentally send it to a customer, where it fails in the field, the eventual cost (in failure analysis, corrective actions, documentation, time spent communicating with the customer, etc) can be much, much more than \$1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a benefit to demanding 6-sigma (1 ppm defects) or better quality from your vendors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you buy a \$1 part, and assemble it into a \$100 (or \$1000) product, where it causes the whole product to fail, then again the cost of that defect is much higher than the cost of the faulty part itself.&lt;/p&gt;&#xA;" OwnerUserId="234" LastEditorUserId="234" LastEditDate="2019-10-05T17:43:15.773" LastActivityDate="2019-10-05T17:43:15.773" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31396" PostTypeId="1" AcceptedAnswerId="31407" CreationDate="2019-10-05T19:53:58.540" Score="2" ViewCount="374" Body="&lt;p&gt;The concrete slab (?) supporting the swimming pool (≈ 10 x 10m x 1m) of my building has clearly a 'belly':&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/Uw3iK.jpg&quot; alt=&quot;concrete slab belly&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure whether this was built this way or evolved over time, however, my best guess is that after 25 years the heavy load of the swimming pool deformed the structure and even the pipes that goes right under it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;What is the technical term for this?  &lt;/li&gt;&#xA;&lt;li&gt;Is this a problem or not? Why?  &lt;/li&gt;&#xA;&lt;li&gt;Any action should be taken in order to prevent an accident?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3734" LastEditorUserId="3734" LastEditDate="2019-10-12T00:19:19.713" LastActivityDate="2019-10-12T00:19:19.713" Title="Concrete Slab Deformation" Tags="|civil-engineering|concrete|building-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31397" PostTypeId="2" ParentId="31396" CreationDate="2019-10-06T06:59:39.153" Score="0" Body="&lt;p&gt;That will fail eventually, concrete supports compression easily but not tension (something Barnes Wallis used to break the dams).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Get this inspected by a professional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One wonders if the situation has been exacerbated by increasing the level of water in the pool - is the level what the designer originally designed for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not only that was it correctly built to specification? Many situations exist where what the designer designed, the builder changed in terms of dimensions, quality (of mix for example) the reinforcing etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If that is your car, I would move it away...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one possibility is to drain it, cover it over and turn it into a dance hall or presentation / conference room.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-10-06T07:08:52.847" LastActivityDate="2019-10-06T07:08:52.847" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="31398" PostTypeId="1" CreationDate="2019-10-06T09:23:42.260" Score="0" ViewCount="381" Body="&lt;p&gt;I don't know if this is the correct forum as I can't find any suitable tags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We operate some of our equipment in fields inside tents.   Our big problem is thrips/thunder flies.  They somehow make it into the inside of the LCD monitors.  This happens to laptops too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is quite amusing and an the same time annoying that these creatures crawl around inside the screens.  If you press the monitor behind the thrip, it makes them move somewhere else.  If you press too close, it squashes the thrip and its corpse will be there forever.  Sometimes they just die in there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a bunch of monitors with dead thrips inside which are for field use only.  The main problem is when you look at data and wonder is that 112 or 1.2 because the thrip is over the 1 or .  You have to move the text up or down to make sure - wastes a lot of time.  Also when you are looking at charts, the thrip corpses obscure the display.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have tried sealing the edges with duck/duct tape but still the little buggers manage to get in.  The only way in I can think of is through the air vent which must not be covered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm just wondering what I can cover the vents with that will still allow airflow but stop the thrips from entering.&lt;/p&gt;&#xA;" OwnerUserId="2060" LastActivityDate="2019-10-06T09:23:42.260" Title="Sealing thrips out of monitors" Tags="|computer-hardware|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31399" PostTypeId="1" CreationDate="2019-10-06T13:38:09.413" Score="1" ViewCount="67" Body="&lt;p&gt;In the 50's and 60's, when anti aircraft missile technology was blooming, many designs used ramjet engines brought up to operating speed by rockets. This seems like a good idea, since ramjets have significantly higher specific impulse than rockets, increasing their range. Over decades however, it seems that the concept has been discarded in favor of pure rockets, with exception of anti shipping missiles. What could be the considerations that made engineers favor pure rockets? &lt;/p&gt;&#xA;" OwnerUserId="18744" LastEditorUserId="18744" LastEditDate="2019-10-06T14:20:22.170" LastActivityDate="2019-11-05T23:05:19.967" Title="Why did ramjet AA missiles fall out of favor over pure rockets?" Tags="|mechanical-engineering|aerospace-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31400" PostTypeId="2" ParentId="6101" CreationDate="2019-10-06T20:51:50.130" Score="1" Body="&lt;p&gt;If it is not in the solution annealed condition , aging ( I guess that is what you mean by tempering) will do about nothing. If is in &quot;as rolled&quot; condition it may have cooled quickly enough to be partly solution annealed; then it may harden slightly during aging. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-10-06T20:51:50.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31401" PostTypeId="1" AcceptedAnswerId="31408" CreationDate="2019-10-06T21:24:55.493" Score="0" ViewCount="62" Body="&lt;p&gt;If creep behavior can begin at 0.6T(melting) but, hot working for a pure metal starts at 0.4Tm, what is the explanation for this? &lt;/p&gt;&#xA;" OwnerUserId="23050" LastActivityDate="2019-10-07T16:22:07.923" Title="Creep behavior in metals" Tags="|materials|metals|creep|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31403" PostTypeId="1" CreationDate="2019-10-07T08:29:28.317" Score="0" ViewCount="99" Body="&lt;p&gt;We have an old stone built cottage.  We have a garage we are converting to a bathroom.  The garage roof is a barrel brick ceiling.  We need to break through from the house to the garage but need to know how to approach putting a lintel in where the roof is barrelled as to get the door height we would need to remove some bricks from the barrel   any advice or links to articles/you tube videos would be appreciated. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="23055" LastActivityDate="2024-02-13T19:07:15.857" Title="How to install a doorway through a stone wall where a barrel ceiling is on one side" Tags="|structural-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31404" PostTypeId="2" ParentId="31403" CreationDate="2019-10-07T09:14:17.093" Score="0" Body="&lt;p&gt;I would pin each stone/brick above where the lintel will be in a line and support that on both sides.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then remove and fit as necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seen it done and seen it in books as well... But the books were older...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can search on youtube but unless a professional has the time you may not find anything exactly identical.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-10-07T09:14:17.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31405" PostTypeId="1" CreationDate="2019-10-07T09:53:00.177" Score="2" ViewCount="79" Body="&lt;p&gt;I have a single axis analog capacitive acceleration sensor (from PCB Piezotronics) which is very reliable but expensive. Now I need more axis, unfortunately can not afford buying anymore. So I have ordered a cheaper and &quot;versatile&quot; alternative, digital SPI sensor (Murata) which is easier to interface without additional analog to digital converter, also it has 3 axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the datasheet it looks quite promising, but in reality it is not. I have noticed that the result is floating, even when the sensor is not moving.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/k8Wm7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k8Wm7.png&quot; alt=&quot;Digital sensor in silence&quot;&gt;&lt;/a&gt;&#xA;I made a FFT and histogram for both new and older sensor, putting them parallel and measuring, Murata looks strange.&#xA;Murata SPI: &lt;a href=&quot;https://i.stack.imgur.com/qZce2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qZce2.png&quot; alt=&quot;Digital sensor response&quot;&gt;&lt;/a&gt;&#xA;PCB analog:&lt;a href=&quot;https://i.stack.imgur.com/q4r0y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q4r0y.png&quot; alt=&quot;Analog sensor response&quot;&gt;&lt;/a&gt;&#xA;&lt;s&gt;Murata also measures a bit higher about 0.58Hz, when PCB is 0.55Hz.&lt;/s&gt; Edit: It is because the sample rate is not exactly the same. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think, is my approach is correct? How would you compare two sensors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Datasheet of &lt;a href=&quot;https://www.murata.com/-/media/webrenewal/products/sensor/pdf/datasheet/datasheet_scc2230-e02.ashx?la=en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Murata SCC2230-E02 sensor - Combined gyroscope and 3-axis accelerometer with digital SPI interface&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Datasheet of PCB sensor: &lt;a href=&quot;http://www.modalshop.com/filelibrary/3701G3FA3G.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.modalshop.com/filelibrary/3701G3FA3G.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23056" LastEditorUserId="23056" LastEditDate="2019-10-14T07:50:59.970" LastActivityDate="2019-10-14T07:50:59.970" Title="How to compare two acceleration sensors side-by-side?" Tags="|structural-analysis|sensors|vibration|acceleration|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31406" PostTypeId="1" CreationDate="2019-10-07T12:05:06.210" Score="0" ViewCount="124" Body="&lt;p&gt;I would like to build a shelf (1m width and 0.5m depth) using only four rectangular steel tubes on for legs.&#xA;Each legs are at the end of the rectangular tube. &#xA;I would like to put 500kg on the tubes, which are equally distributed.&#xA;How can I choose the width, height and wall thickness to ensure that the steel tubes are within a certain deflection?&lt;/p&gt;&#xA;" OwnerUserId="23058" LastEditorUserId="23058" LastEditDate="2019-10-08T18:17:04.363" LastActivityDate="2019-10-09T01:11:57.610" Title="How to size a rectangular steel tube to hold certain weight?" Tags="|steel|deflection|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31407" PostTypeId="2" ParentId="31396" CreationDate="2019-10-07T13:35:40.280" Score="4" Body="&lt;ul&gt;&#xA;&lt;li&gt;What is the technical term for this?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Bending downwards is called &lt;strong&gt;sagging&lt;/strong&gt;. (Technically &quot;sagging&quot; isn't moving downwards; it's bending in a shape that looks like the mouth on a smiley face).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing you'll see with concrete structures is &lt;strong&gt;creep&lt;/strong&gt;. This is where under constant load, the deflection increases over time.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is this a problem or not? Why?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Sagging deflection is inevitable. If you put a load on something it &lt;strong&gt;will&lt;/strong&gt; move. Whether this movement is big enough to be visible or not it another question, but it &lt;strong&gt;will&lt;/strong&gt; move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When designing something we often check for two different failures: ultimate strength and deflection. &lt;em&gt;(Note: There are plenty of other things we check for, this is a simplification.)&lt;/em&gt; Ultimate strength check = is it going to fall down. Deflection check = is it going to deflect so much that it scares people.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When assessing (i.e. checking) an existing structure, we generally only do ultimate strength checks; we don't worry so much about whether people are worried or not...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, your structure has clearly failed the deflection limits, as it's scared you! But that's no indication as to whether or not it has exceeded its ultimate strength limits, so its no indication of whether there's actually a problem or not.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Any actions should be done in order to prevent an accident?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Get a structural engineer to assess it. You've said in comments that the structure has been inspected due to this deformation - as long as that inspection has been carried out by a suitably experienced/qualified engineer, then you have no need to worry. I have no idea of how you satisfy yourself of their qualifications in your country.&lt;/p&gt;&#xA;" OwnerUserId="255" LastActivityDate="2019-10-07T13:35:40.280" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31408" PostTypeId="2" ParentId="31401" CreationDate="2019-10-07T16:22:07.923" Score="1" Body="&lt;p&gt;Hot work involves macroscopic deformation of the metal; no time is allowed for creep to take place so the concept of &quot;creep&quot; as such doesn't exist during hot work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The objective of hot work is to deform the workpiece to its desired shape at a temperature at which it happens to be easily malleable and will not strain-harden in the process. This means that the temperature you choose depends on exactly how soft you want it to get and how hot you need to get it so that solid-state diffusion through the lattice is more or less fully activated. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-10-07T16:22:07.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31410" PostTypeId="2" ParentId="31406" CreationDate="2019-10-07T20:15:15.007" Score="1" Body="&lt;p&gt;There are design aids which tabulate in detail all the section properties of hollow structural sections, HSS, and have online calculators and you find many if you google HSS sections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we disregard the roundness on the corners, which is not a big factor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We assume a maximum deflection of &lt;span class=&quot;math-container&quot;&gt;$\dfrac{\ell}{180} = \dfrac{1000}{180} = 5.5\text{ mm}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;M &amp;amp;=\frac{\omega l^2}{8} \\&#xA;&amp;amp;= 500/1000 * 1000^2/8 \\&#xA;&amp;amp;=62500kg-mm2&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adopting ASTM A36 steel with the following properties:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;E &amp;amp;= 200\text{ GPa} = 20400\text{ kg-mm}^2\\&#xA;F_a &amp;amp;= \dfrac{1}{2} F_y = \dfrac{1}{2} 210\text{ MPa} = \dfrac{1}{2} 21.4 = 10.7\text{ kg-mm}^2&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets pick up an HSS section with dimensions &lt;span class=&quot;math-container&quot;&gt;$$A_{base}, 2A_{height}, A/10_{thickness}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\sigma &amp;amp;= \frac{mc}{I} \\&#xA;I &amp;amp;= \frac{A\cdot(2A)^3 - 0.8A\cdot (1.6A)^3)}{12} = \frac{0.72A^4}{12} \\&#xA;\sigma &amp;amp;= 10.7 = \frac{62500A \cdot 12}{0.72A^4} = \frac{ 62500 \cdot 12}{0.72A^3} \\&#xA;A^3&amp;amp;=9786.7 \\&#xA;A &amp;amp;= 21\text{ mm, pick } 25\text{ mm}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to plug this &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and check its deflection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;y &amp;amp;= \dfrac{5\omega\ell^4}{384EI} \\&#xA;&amp;amp;=\frac{5 \cdot 0.5 \cdot 1000^4}{384 \cdot 20400 \cdot 23437} \\&#xA;&amp;amp;= 13.6\text{ mm} &amp;gt; 5.5\text{ mm}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That deflection is not okay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we pick one size larger and plug the numbers again until the deflection is less than 5.5&amp;nbsp;mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: We Used kg-force as opposed to N and millimeter versus meter and also we did not use a factor of safety, even though taking &lt;span class=&quot;math-container&quot;&gt;$F_a=1/2F_y$&lt;/span&gt; is some measure of that. Also we picked thickness as 1/10 base, which is not necessarily true, just as a roadmap. Check my numbers please. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2019-10-09T01:11:57.610" LastActivityDate="2019-10-09T01:11:57.610" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31413" PostTypeId="1" CreationDate="2019-10-08T07:36:58.850" Score="1" ViewCount="187" Body="&lt;p&gt;Diffusers in turbomachinery are used to transfer some of the kinetic energy of the fluid to pressure energy and to direct the fluid to the ducting/second stage/... I understand this and the mechanism behind it (geometry of diffuser changes the mass flux which creates a pressure gradient which decelerates the fluid) but I just cannot find any information how diffusers create more suction at the inlet of the pump/compressor. In laymans terms, how does the diffuser make the pump better for the end user? Is it more efficient?&lt;/p&gt;&#xA;" OwnerUserId="23069" LastActivityDate="2024-02-16T02:04:45.550" Title="Pressure recovery in turbomachinery" Tags="|pressure|turbomachinery|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31414" PostTypeId="1" AcceptedAnswerId="31427" CreationDate="2019-10-08T10:36:31.737" Score="0" ViewCount="255" Body="&lt;p&gt;I would like to try out a component that operates like a combination of the two pictures provided - spring loaded pull pin, with a pin finish like a tubular latch (chamfered/filleted finish).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do these exist?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If somebody can tell me their correct name or better still, provide me with a link to a manufacturer's site that would be great.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6HXXg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6HXXg.jpg&quot; alt=&quot;Spring Loaded Pull Pin&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yuprc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yuprc.jpg&quot; alt=&quot;Chamfered/Filleted Pin Finish&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23071" LastActivityDate="2019-10-10T15:05:44.023" Title="Spring Loaded Pin with Chamfered Pin Finish?" Tags="|springs|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31415" PostTypeId="1" AcceptedAnswerId="31425" CreationDate="2019-10-08T15:07:55.150" Score="5" ViewCount="8516" Body="&lt;p&gt;I'm designing a circular shaft that's going to carry a spread radial load of about 200kg and also transmit a 160Nm torque. I'm not very familiar with materials but what I'm guessing I'll be able to get at a reasonable price locally is some lower grade steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Replacing the spread load with a single force in the middle of the shaft I guess will look something like this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pUCcc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pUCcc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both points A and B are symmetrical so I'm taking in consideration only half of the shaft and calculating for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T = F\ell = 200 \cdot 0.5 = 100\text{ kNm}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is the torque at point &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; created by the force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tensile stress on the shaft will then equal&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$t = \dfrac{T}{W}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; is tensile stress and &lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt; is section modulus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a solid circular face the section modulus equals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$W = \dfrac{\pi d^3}{32}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have to assume either &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; in order to solve. I looked it up in &lt;a href=&quot;http://www.matweb.com/search/datasheet.aspx?bassnum=MS0001&amp;amp;ckck=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; table and it looks like the tensile yield strength of steel is 350&amp;nbsp;kPa so I'm taking a conservative estimate of 300&amp;nbsp;kPa for &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;d^3 &amp;amp;= \dfrac{32(T / t)}{\pi} \\&#xA;&amp;amp;= \dfrac{32 \cdot (100e3 / 300e6}{\pi} \\&#xA;&amp;amp;= 0.003397028 \\&#xA;\therefore d &amp;amp;= 0.150\text{ m}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously there is something wrong either in my logic or calculations because I'm getting unrealistic results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I went ahead and calculated the torsion stress as well with &lt;a href=&quot;https://engineering.stackexchange.com/a/17202/20333&quot;&gt;this formula&lt;/a&gt;. I found that &lt;a href=&quot;https://www.quora.com/What-is-the-shear-strength-of-mild-steel&quot; rel=&quot;nofollow noreferrer&quot;&gt;shear stress for steel is about 200-300&amp;nbsp;kPa&lt;/a&gt; so I'm again taking a conservative 200.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;d &amp;amp;= 1.72 \cdot \left(\dfrac{M}{\tau_{max}}\right)^{1/3} \\&#xA;&amp;amp;= 1.72 \cdot \left(\dfrac{160}{200e6}\right)^{1/3} \\&#xA;&amp;amp;= 0.0159 ~ 16mm&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which seems about right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of getting a 20&amp;nbsp;mm diameter profile but I would rather play it safe and ask for some help instead of just assume it's going to hold.&lt;/p&gt;&#xA;" OwnerUserId="20333" LastEditorUserId="1832" LastEditDate="2019-10-09T01:21:22.693" LastActivityDate="2020-08-05T12:06:32.220" Title="Calculating shaft diameter" Tags="|mechanical-engineering|materials|steel|strength|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31417" PostTypeId="1" CreationDate="2019-10-08T19:00:35.207" Score="0" ViewCount="145" Body="&lt;p&gt;I have a situation that I am hoping to solve with the use of electric motors, and a variety of controls. &#xA;I have a hydraulic winch which turns at speeds from 0-60 RPM.  This winch is coiling and uncoiling a rope.  Under normal operation, it would have anywhere between 200-300 pounds of pull on the rope it is coiling or uncoiling.&#xA;There is a situation, in our shop, where we would like to simulate the working conditions of our hydraulic winch.  I would like to attach another electric winch that would pull 200-300 pounds on the rope as the hydraulic winch is unspooling and also keep a pull of 200-300 pounds on the rope as the hydraulic winch in coiling.&#xA;What would the simplest way of achieving this be?&#xA;Thank you very much for your support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hello Everyone,&#xA;I would like to thank you all for your great input.&#xA;As much as I would love to hang a weight off the end of this winch, which would work out perfectly, there are situation where that setup would not work.&#xA;I need to be able to simulate this 200-300 pound pull.&#xA;The best way I can describe the situation is this.&#xA;Lets say you are on a boat traveling though the water pulling in a rope with a buoy attached in the middle of your rope.  As you are sailing though the water there is a constant pull on your rope but if you stop there will no longer be that constant pull on your rope.  If you are sailing and you winch in your rope, the buoy will reach the winch and get wrapped up nice and tight the way it is supposed to. If the boat stops, there will be no pull on the rope that is in the water allowing buoy to fall forward on the winch as it passes over the top of the winch.  This will cause big problems.&#xA;So with all that said, I am trying to build something that can be used in our shop and on the ship to retrieve the system if they have a propulsion failure.&#xA;So it needs to be relatively small, shouldn't be a problem with only 200-300 pounds pull and i can attached as much feedback \ encoders on either unit I want.&#xA;I have been looking at some oil shear technologies but would much rather do this electrically.&#xA;Thanks and I look forward to hearing more about your suggestions.&lt;/p&gt;&#xA;" OwnerUserId="23085" OwnerDisplayName="Tim" LastEditorUserId="23085" LastEditDate="2019-10-09T12:10:33.013" LastActivityDate="2019-10-09T15:02:44.383" Title="Constant force opposing hydraulic winch" Tags="|electrical-engineering|hydraulics|" AnswerCount="1" CommentCount="4" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31418" PostTypeId="1" AcceptedAnswerId="31420" CreationDate="2019-10-08T19:25:44.787" Score="2" ViewCount="706" Body="&lt;p&gt;Let'say the steel tank would be submerged deep in the ocean with 136 atm ambient pressure then the other tank with the same material would just be used to contain 136 atm pressure. How would the thickness of the wall differ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, if the tank is designed to be sunk deep in the ocean, would it be cheaper to design a tank that could contain 136 atm pressure then pressurize it at that pressure so that it would have zero forces acting on the tank's walls if the ambient pressure is also 136 atm?&lt;/p&gt;&#xA;" OwnerUserId="19011" LastActivityDate="2019-10-09T05:22:01.217" Title="What are the design difference of a steel tank that would withstand 136atm inside and 1atm ambient compared to 1atm inside and 136atm ambient?" Tags="|pressure|steel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31420" PostTypeId="2" ParentId="31418" CreationDate="2019-10-08T19:40:39.170" Score="4" Body="&lt;p&gt;It's the difference between holding up a weight by by hanging it from a cable vs. putting it on top of a column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tank that starts out round (ideally, if it starts out spherical) and is pressurized will tend to have its shape held by the pressure of the material inside.  Basically, the internal pressure will try to &lt;em&gt;make&lt;/em&gt; the tank round, so if that's the starting point, the internal pressure will, to some extent, help support the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a balloon -- the walls essentially don't have any bending strength, yet the balloon can hold pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A tank that's maintaining a vacuum has an unstable shape: it wants to suck down to nothing.  A perfectly spherical tank can withstand that, but it's not stable in buckling.  Any dent will tend to &quot;want&quot; to get bigger; this tendency will have to be opposed by walls that resist local bending even as they're getting compressed by the outside forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes, if you were going to sink a tank in the ocean, and if it otherwise made sense to do so, starting with it pressurized before sinking it would be a good idea.  Another sensible idea, if it fits what you're doing otherwise, would be to leave it open to the seawater and just let it fill as it sinks, then purge it once it's been installed.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-10-08T19:40:39.170" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31422" PostTypeId="1" AcceptedAnswerId="31442" CreationDate="2019-10-09T01:11:28.620" Score="1" ViewCount="480" Body="&lt;p&gt;I was given a challenge question from my teacher that I need some assistance in solving. The problem is to merely identify the two-force members. I understand the ideas of how to determine a two-force member (colinear forces, equal magnitude, etc.), but in this problem, they just aren't clicking for me. Thanks in advance for the help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WA7q8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WA7q8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18704" LastEditorUserId="1832" LastEditDate="2019-10-09T01:27:47.463" LastActivityDate="2019-10-10T01:28:15.170" Title="Statics Two Force Member Problem" Tags="|statics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31424" PostTypeId="2" ParentId="31418" CreationDate="2019-10-09T05:15:14.367" Score="2" Body="&lt;p&gt;when the tank is pressurized with 185atm it is very simple to calculate the thickness needed to hold the pressure.&#xA;Say the radius of the spherical tank is R. the max pressure is at a great circle passing through the center and is equal to &lt;span class=&quot;math-container&quot;&gt;$ \pi*R^2*186  $&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This pressure has to be supported by the tension stress on the thickness of the peremeter of the tank, t, at that circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$t = \frac{\pi R^2*186}{2 \pi R* \text{allowable tension stress of steel}}$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is a reasonably small number. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if the tank will be exposed to 185atm external pressure it is susceptible to catastrophic buckling and explosive crushing. so it has to be designed for moment resistance in case of buckling, which will turn out to be mostly following empirical methods which includes thick wall vessel design and possible use of rib stiffeners or use of thick hardened steel or special materials.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-10-09T05:22:01.217" LastActivityDate="2019-10-09T05:22:01.217" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31425" PostTypeId="2" ParentId="31415" CreationDate="2019-10-09T07:05:28.157" Score="2" Body="&lt;p&gt;According to a couple of textbooks on machine design and assuming that your FBD is correct, the maximum bending moment occurs at the centre of the shaft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a point load, it is equal to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$WL/4 = M$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where W is the point load and L is the length of your beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$(200*10^3)*1/4 = 50*10^3 Nm = 50*10^6 Nmm = M$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the maximum shear stress theory the combined torsional and bending stress is given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T_e = \sqrt{M^2 + T^2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But we also know that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ T_e = (\pi\div 16)*\tau*d^3$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where tau is the allowable shear stress for each material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the numbers into the first equation and then equate both of the &lt;span class=&quot;math-container&quot;&gt;$T_e$&lt;/span&gt; values. &#xA;You can then solve for d. This is one of the answers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also taking into consideration the bending moment using maximum normal stress theory,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ M_e = 0.5(M + \sqrt{M^2 +T^2}) $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When designing the shaft for the bending moment, we can apply the formula shown:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$M_e = (\pi \div 32)*\sigma_b*d^3 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\sigma_b$&lt;/span&gt; = Ultimate/Allowable Stress&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the numbers into the first equation &lt;span class=&quot;math-container&quot;&gt;$M_e$&lt;/span&gt; equation and then equate both of the &lt;span class=&quot;math-container&quot;&gt;$M_e$&lt;/span&gt; values. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can then solve for d. This is one of the answers. Find the largest diameter out of the first and second equation and use it. You should also apply some safety factor in the allowable stresses and maybe the diameter. If the load which you stated is uniformly distributed the maximum bending moment will also change which is contrary to your point load diagram. I am also unsure about the action of fatigue.&lt;/p&gt;&#xA;" OwnerUserId="20822" LastEditorUserId="20822" LastEditDate="2019-10-09T09:26:13.357" LastActivityDate="2019-10-09T09:26:13.357" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31426" PostTypeId="2" ParentId="31417" CreationDate="2019-10-09T07:38:41.940" Score="1" Body="&lt;p&gt;As others have said, an electric motor load solution could be quite expensive, but depending on your budget you could look into brushless DC motor torque control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the low rpm requirement you could get away with using a relatively small motor (something on the larger side of hobby size brushless motors) and gear it down with either a gear box or a belt system. You would want to look for a controller that has torque control which basically means it can sense the current through each phase and convert that current to a torque and use that in a feedback loop to maintain a set torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possibility cheaper control option would be to use a similar motor and reduction setup but to use a regular ESC to control the power to the motor and implement a  large torsion spring between your pulley drum and the inner drive shaft. This would allow you to use a relatively simple encoder feedback to measure the relative angle between the outer drum and drive which would translate to a certain torque on the spring. Maintaining that angle with the control loop would effectively keep a constant force on the rope.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you start down the path of one of these methods I would consult with others to verify the specs of your motor and controller choice. It can be very thermally demanding on a motor to maintain that force at very low or no speed and you would need a considerably beefier motor and controller to hold that static load indefinitely.&lt;/p&gt;&#xA;" OwnerUserId="23080" LastEditorUserId="23080" LastEditDate="2019-10-09T15:02:44.383" LastActivityDate="2019-10-09T15:02:44.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31427" PostTypeId="2" ParentId="31414" CreationDate="2019-10-09T07:51:22.903" Score="0" Body="&lt;p&gt;please note that (since there is a huge amount of manufacturers) it is not our job to look for your part. But as starting point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some sites where you can look for an index plunger. There is a ton of options but on first glance I couldn't make out the smae type you were looking for. It's likely that you have to resort to GisMofx's hack and grind/mill the bevel on the standard part. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.norelem.com/us/en/Products/Product-overview/Flexible-standard-component-system/03000-Spring-plungers-Indexing-plungers-Stops-Centring-positioning-components-Ball-lock-pins-T-slot-nuts/Indexing-plungers.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Norelem&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.ganternorm.com/en/products/3.1-Indexing-locking-blocking-with-pins-and-ball-shaped-elements/Indexing-plungers&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ganter&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.halder.com/gb/Products/Standard-Parts/Machine-and-Fixture-Elements/Index-Plungers-Index-Bolts&quot; rel=&quot;nofollow noreferrer&quot;&gt;Halder&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.carrlane.com/en-us/product/spring-loaded-devices/index-plungers&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carr-Lane&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'll extend this list to more sources whan people comment. &lt;/p&gt;&#xA;" OwnerUserId="23079" LastActivityDate="2019-10-09T07:51:22.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31431" PostTypeId="2" ParentId="29969" CreationDate="2019-10-09T14:54:35.017" Score="0" Body="&lt;p&gt;You are right, these seem like very simple storage systems; it does seem strange that we aren't using more of them. Until you look at the numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, we do already use gravity storage, pumped-storage hydroelectricity, but let us ignore that for now because it is only possible in the rare circumstances where geography complies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets instead take the example of a house-sized battery system, which can be installed virtually anywhere. By far the most well known current solution is the Tesla Powerwall 2, which can store 13.5kWh and costs on the order of $10k to purchase and install.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine a gravity system you could fit in your back yard, say, a 1m shaft (you could equally say tower, but you'll soon see why that is impractical) which you could throw a 3m shed over for safety and to house the winch system. What's the biggest weight you could practically get down that hole? Well, let's imagine its a 4m long, 1m wide cylinder of nice and cheap reinforced concrete, weighing in at 10T.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we want to store 13.5kWh using a 10T weight. Do the maths and assume something like 80% efficiency and you end up needing a shaft 620m long (and now you see why we ruled out a backyard tower).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add to the price of you nice cheap weight; digging a huge hole, a winch capable of reeling 620m of cable, control electronics, generator, reinforcement and waterproofing for your hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, clearly this system might work better if you had a really big winch, a really big weight and a really big, well-drained hole that's already there. Disused mineshafts come to mind. But even at scale, the cost is going to look rather silly against chemical batteries.&lt;/p&gt;&#xA;" OwnerUserId="23088" LastActivityDate="2019-10-09T14:54:35.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31432" PostTypeId="1" CreationDate="2019-10-09T14:57:28.173" Score="0" ViewCount="46" Body="&lt;p&gt;what is the most optimal cost effective diameter of a hyperloop housing vacuum tube housing hyperloops that can ferry ALL cargo currently travelling by air and sea to and from the US and Europe? maximum 30 tonnes&lt;/p&gt;&#xA;" OwnerUserId="12808" LastActivityDate="2019-10-09T16:11:40.280" Title="what is the most optimal cost effective diameter of a hyperloop tube ferrying ALL cargo between the US and EU" Tags="|vacuum|" AnswerCount="1" CommentCount="3" ClosedDate="2019-10-11T23:03:50.633" ContentLicense="CC BY-SA 4.0" />
  <row Id="31433" PostTypeId="1" AcceptedAnswerId="31437" CreationDate="2019-10-09T15:44:54.707" Score="0" ViewCount="590" Body="&lt;p&gt;Hello I am a simple rock climber, and would like to figure out how much force is applied to The lobe of a cam/the rock when the cam is loaded. So the cam lobe is circle the ramp or cam goes from 7/16”-11/16” I’m not sure that matters or if just the point of contact matters. If I ignore friction and those kind of nussance variables what is my formula if I 180lbs simply hang on the cam. (Is it a simple lever calculation)!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/2tKWf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="23089" LastEditorUserId="19776" LastEditDate="2019-10-09T16:25:29.353" LastActivityDate="2019-10-09T19:26:13.360" Title="Force applied to a cam follower" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31434" PostTypeId="2" ParentId="31432" CreationDate="2019-10-09T16:11:40.280" Score="1" Body="&lt;p&gt;Let's assume we are talking about transporting shipping containers, a ready unit of trade with well-established industry support via rail and road.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://www.joc.com/maritime-news/trade-lanes/trans-atlantic/trans-atlantic-carriers-cut-services-us-eu-trade-growth-slows_20180727.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Joc.com&lt;/a&gt;[1], EU sold the USA 1.3 million TEU (Twenty-foot Equivalent Units, read: containers) and the USA sold 0.68M TEU back. So we'll use the higher number and add some headroom for growth, let's say to 2M containers per year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, trade is not consistent, and unlike ships, you can't just order a bigger one when it suits, so our tunnel will need to be able to accommodate peaks of utilisation, so we'll double up to 4M TUE per year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now another issue, this one of our own making: we have in one fell swoop cut the surface transit time from EU to USA from several weeks to just a few hours. There are of course already airfreight routes which can do something similar, but the added potential for trade in fresh goods, or lean manufacturing this new form of transport would open up is hard to fathom. Let's double again to 8M TUE per year. This is our minimum rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The original &lt;a href=&quot;https://www.spacex.com/sites/spacex/files/hyperloop_alpha-20130812.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hyperloop Alpha&lt;/a&gt;[2] concept proposed peak rates of 2 pods per minute. We'll assume one pod = one container, in line with your 30T maximum rule. At peak rates then, a single lane could supply (365 days x 24 hours x 60 minutes x 2 TUE) = 1,051,200 TUE per year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So unfortunately 1 lane just isn't going to cut it; we need 8 lanes per direction for 16 total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your question asks how big do we make the vacuum tube to fit all of these in, which I believe misses a vital point. The potential for damage should any single pod experience a problem is huge when dealing with 30T travelling at 1000km/hr. With 16 separate pod lanes your chances for an incident rise by about x4. Far better to have them separate from each other. It also allows them to be built one at a time and have a much lower volume evacuated if each individual lane is stand-alone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we need 16 tubes, each big enough to accommodate a pod capable of carrying a shipping container. A container is about 2.5m tall and wide, so a rounded pod capable of carrying one internally would need a minimum internal diameter of 3.53m, but we'll add some additional area for the pod walls, to make this around 4m. The original Hyperloop called for a 60% capsule/tube diameter ratio to allow for airflow compression, so the internal diameter of each tube would need to be around 6.7m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the answer is 16 separate tubes, each 6.7m in diameter.&lt;/p&gt;&#xA;" OwnerUserId="23088" LastActivityDate="2019-10-09T16:11:40.280" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31435" PostTypeId="2" ParentId="31413" CreationDate="2019-10-09T16:27:25.350" Score="0" Body="&lt;p&gt;This website puts it pretty well: &lt;a href=&quot;http://bellgossett.com/hydronic-plumbing-accessories/pump-accessories/suction-diffuser/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://bellgossett.com/hydronic-plumbing-accessories/pump-accessories/suction-diffuser/&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without a suction diffuser, the pump inlet flow is subject to recirculation zones that produce pressure losses and a non-uniform velocity inlet. These recirculation zones can be the result of pipe bends and fittings. A suction diffuser &quot;straightens&quot; the flow and removes these flow structures. An uneven inlet pressure distribution and velocity profile inflicts uneven stresses onto the pump impeller. In absence of these stresses,the pump can operate more efficiently. Pressure losses are also detrimental to pumps since the inlet suction head is reduced. This can cause cavitation of the pump. Does that all make sense? &lt;/p&gt;&#xA;" OwnerUserId="21118" LastActivityDate="2019-10-09T16:27:25.350" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31436" PostTypeId="2" ParentId="31433" CreationDate="2019-10-09T17:01:28.377" Score="0" Body="&lt;p&gt;This is a hugely complicated question that's not going to be answered both accurately and briefly, because there's too many variables presented by the real world.  So I'm going to make some simplifying assumptions, to give you an idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm going to &lt;em&gt;present&lt;/em&gt; is a simple lever calculation, but it'll almost never apply to the real world.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say that the movable and fixed part of the cam get an equal amount of force on them when you shove it into a crack, and that the crack faces are perfectly parallel.  Then the free-body diagram of the cam in the crack would look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SftK0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SftK0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can immediately simplify that down to something you'd find in a first Statics class:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xS5lT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xS5lT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the beam (the diagonal line) is free to pivot, in order for it to not rotate the torques on it must be balanced.  This means that the condition &lt;span class=&quot;math-container&quot;&gt;$$\frac{F}{2} a = F_I b$$&lt;/span&gt; must be satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's pretty much it, for this ridiculously simplified case.  But crack angles will vary, cam designs will vary, you may have differing grips on one side of the crack than the other, etc., etc., etc..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that for a given design of cam (i.e., how much does it spread as it opens, which determines the angle of the contact point with respect to the pivot) and for a given inward or outward taper of crack, you can get an idea of how much outward push you'd get.  But even there, a crack that's got a rough enough surface with large enough irregularities would tend to catch the cam at a point where the effective taper is more favorable, leading to less outward force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good pessimistic estimate would probably be to assume that only one side or the other actually grips (so assume &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$F/2$&lt;/span&gt; in the calculations above), and that the taper of the crack is what it seems to be on a large scale (so assume no hidden advantages from catching on a high spot).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this really right you'd build a bunch of cams that are instrumented to measure the force on the rope and on the pivot, and then you'd go climbing, recording the forces on a datalogger and taking notes.  Be sure to do it on faces that are famous for having cracks that cams slip out of, and on ones that are famous for having cracks that are secure even if someone with five thumbs on each hand is putting in the cams.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-10-09T17:01:28.377" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31437" PostTypeId="2" ParentId="31433" CreationDate="2019-10-09T19:26:13.360" Score="0" Body="&lt;p&gt;The curve of a spring loaded cam is certain curve called exponential spiral with the unique property that the angle it touches the wall of the crack and a horizon remains the same (13-14 degrees) no matter haw much the cam expands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This small angle means the force applied by the cam to the crack is always &lt;span class=&quot;math-container&quot;&gt;$F_{horiz}= W_{weight}*cotan(13)= W*4.3$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which after many tests has proven to be the safest and most practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is the detail calculations. &lt;a href=&quot;http://www.valleygiant.com/cam_math.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;cam math.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-09T19:26:13.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31438" PostTypeId="1" AcceptedAnswerId="31870" CreationDate="2019-10-09T22:54:25.110" Score="1" ViewCount="730" Body="&lt;p&gt;Lately I've been watching some videos on machining and hydraulics as it always interested me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was toying with the idea of building a budget CNC mill with threaded rod for the axes, but the issue of backlash was bugging me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I thought about using hydraulic pistons instead. If used in conjunction with a  positive displacement pump and using a large hydraulic transmission ratio the issue of precision could be sorted out, and the incompressibility of the fluid would help to keep backlash to a minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched but found little info on hydraulic actuators for precision positioning, so I believe there's something I must be missing&lt;/p&gt;&#xA;" OwnerUserId="23092" LastActivityDate="2021-08-15T20:39:02.733" Title="Hydraulic linear actuators as CNC axis instead of screws" Tags="|mechanical-engineering|fluid-mechanics|hydraulics|machining|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31439" PostTypeId="1" AcceptedAnswerId="31500" CreationDate="2019-10-09T23:06:42.867" Score="0" ViewCount="680" Body="&lt;p&gt;I observed the rebuilding of a 12KW, 120V gas-powered generator.  I was amazed at how small the generator portion was (roughly the size of an electric motor you might find in a major home appliance).  The windings were pretty thin magnet wire, like you would find in an appliance motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting the 100A generator output to the breaker box requires heavy-duty wire.  That same current originates in the generator windings.  How is the thin wire in the windings able to carry 100A (which far exceeds what wire that size is rated for)? &lt;/p&gt;&#xA;" OwnerUserId="12772" LastEditorUserId="12772" LastEditDate="2019-10-10T00:22:39.307" LastActivityDate="2019-10-13T22:03:45.430" Title="How can thin wire in generator windings carry the high output current?" Tags="|generator|current|wire|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31440" PostTypeId="2" ParentId="31439" CreationDate="2019-10-09T23:54:41.970" Score="1" Body="&lt;p&gt;Note: Small Gauge mean the wire can carry a significant amount of current. Example per the attach AWG 4-5 can carry 100A. There are lot other factors need to be considered before selecting a wire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is guide line help you appreciate wire gauge vs current. This is purely a guideline. I suggest searching American Wire Gauge Charts vs Electrical Current load for additional information. Below are some links for you to get started.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IZLql.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IZLql.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.powerstream.com/Wire_Size.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wire Gauge and Current Limits Including Skin Depth and Strength&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/American_wire_gauge&quot; rel=&quot;nofollow noreferrer&quot;&gt;American wire gauge&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-10-09T23:54:41.970" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31442" PostTypeId="2" ParentId="31422" CreationDate="2019-10-10T01:28:15.170" Score="0" Body="&lt;h2&gt;None of these elements are two-force members&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Before anything, it's worth remembering what two-force members are: bars which only have two equal-and-opposite colinear forces applied, one at each extremity. Given these conditions, we also know the forces must be purely axial, with no shear component. Given they have no shear component, they'll also suffer no bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, it's worth mentioning that there are actually two ways of looking at that definition. One is strictly based on the structure itself, while the other considers the applied loading. The simple case which exemplifies the difference is a cantilever beam with only an axial force applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that beam a two-force member? If you only look at the structure, then &lt;strong&gt;&lt;em&gt;no&lt;/em&gt;&lt;/strong&gt;. It's obvious that under basically any other loading condition, this beam would present shear and bending moment. But if you only think of the currently applied forces, then &lt;strong&gt;&lt;em&gt;yes&lt;/em&gt;&lt;/strong&gt;: it only suffers equal-and-opposite colinear axial forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm of the opinion that the best way to think about it is by only looking at the structure: if you can come up with a classic loading pattern which generates bending moment, it's not a two-force member&lt;sup&gt;1&lt;/sup&gt;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So let's look at the structure ignoring its loads:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom beam AF is basically a simply-supported beam with a cantilever. If you put a vertical load anywhere along that span, the whole thing'll have shear and bending moments, so not a two-force member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bend which connects AF to the rest of the structure is obviously not a two-force member: axial loads aren't collinear. And even if you consider that as two separate members, their shared node is fixed. Therefore, any deflections will cause relative rotations and therefore bending moments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the &quot;box&quot; at the top, well, that's also a bunch of fixed beams. Any loading in that region will lead to bending moments all over the place.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;For the record, for those who prefer to consider the given loading, the same conclusion applies:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oO5LI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oO5LI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the bending moment diagram (ignore the actual values, I got lazy with the loading inputs). As we can see, all relevant members have bending moment and the box is unloaded (which in my book doesn't count as being a two-force member).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; For the pedants thinking of truss members with uniform loads applied, just remember the &quot;classic loading pattern&quot; for truss members is for uniform loads to be replaced by nodal forces. And nobody likes a pedant.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-10-10T01:28:15.170" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31443" PostTypeId="1" CreationDate="2019-10-10T04:54:06.633" Score="1" ViewCount="70" Body="&lt;p&gt;I wonder why do we get some times singularities (position to be infinity or velocity to be infinity) in the solutions while modelling a real system and in spite of taking all real parameter values of the system. We might have large value(of position or velocity)but why are we getting infinity as a solution. For example, at resonance in spring-mass damper and forced oscillations, at some conditions, I will end up having infinite positon. Why is this happening? In the case of fluid dynamics while using Point vortex models I will have infinite velocity at the centre of the vortex. Why is this happening? even though we follow the proper mathematical approach and always connecting math to the physical system. Suddenly we encounter this Why?. I accept that I don't know much of mathematics or physics .I'm interested in knowing physics and how math is helping us to know physics. This question might be absurd to others for me this is kind of amazing. This puzzled me for a long time.   &lt;/p&gt;&#xA;" OwnerUserId="23096" LastActivityDate="2019-10-12T23:48:53.253" Title="Why do we get singularities in a solutions while modelling a physical systems?" Tags="|modeling|mathematics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31444" PostTypeId="2" ParentId="31443" CreationDate="2019-10-10T06:24:08.150" Score="1" Body="&lt;p&gt;Because just about everything we calculate is a poor approximation of what really happens - we ignore some variables, make assumptions for others and then measure imprecisely as well.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-10-10T06:24:08.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31445" PostTypeId="2" ParentId="31438" CreationDate="2019-10-10T06:33:38.470" Score="1" Body="&lt;p&gt;Hydraulics can be precise, down to ~ 100µm. However:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;hydraulic pistons [...] used in conjunction with a positive displacement pump and using a large hydraulic transmission ratio the issue of precision could be sorted out, and the incompressibility of the fluid would help to keep backlash to a minimum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is not how this high precision is achieved in practice. AFAIK (I'm not in the field) high precision hydraulics are pre-packaged units of piston, valves and position sensor and control the piston for position. &lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-10-10T06:33:38.470" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31446" PostTypeId="1" CreationDate="2019-10-10T06:47:25.477" Score="0" ViewCount="35" Body="&lt;p&gt;I want construct a heating element by screen printing a flat coil pattern of conductive paste onto an insulated substrate. This conductive paste consists of a powder of conductive metal contained in a viscous epoxy paste. The pieces of metal are held together by the paste to form a conductive pathway. There are several options for the type of conductive metal filler - so far I have found silver, silver and copper mixed, carbon, graphite and nickel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What material characteristics should I be looking at in trying to assess the best filler material? I want to make a heating element that will heat up as quickly as possible with the least amount of electric current being put through the heating element. The max temperature of the heating element will be 100 C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am thinking that the two most important qualities for generating heat with the least electric current are Resistivity and Thermal Conductivity. Is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the specs on some pastes I have found. It seems like the Nickel paste would be best - is that right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nickel Volume Resistivity &amp;lt;0.004 ohm. cm Thermal Conductivity 1.44 W/M-K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Graphite Volume Resistivity 0.004 ohm. cm Thermal Conductivity 0.24 W/M-K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silver Volume Resistivity 3.00 E -3 ohm. cm Thermal Conductivity 1.45 W/M-K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carbon Volume Resistivity &amp;lt; 40 ohm. cm Thermal Conductivity 0.25 W/M-K&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silver/Copper Volume Resistivity 3.70 E-2 ohm. cm Thermal Conductivity 1.15 W/M-K&lt;/p&gt;&#xA;" OwnerUserId="23053" LastActivityDate="2019-10-10T06:47:25.477" Title="What material characteristics are most important for heating element material selection?" Tags="|materials|thermal-conduction|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31447" PostTypeId="1" CreationDate="2019-10-10T10:08:34.497" Score="1" ViewCount="188" Body="&lt;p&gt;I have a situation where ideally, I would like to resistance spot weld three rather than the usual two plates together (individual plate thickness approximately 0.7mm). Is this at all possible? If not, why not? Are there any precautions one should take?&lt;/p&gt;&#xA;" OwnerUserId="946" LastActivityDate="2019-10-10T10:26:13.123" Title="Is it possible to spot weld three rather than two plates?" Tags="|manufacturing-engineering|welding|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31448" PostTypeId="2" ParentId="31447" CreationDate="2019-10-10T10:26:13.123" Score="2" Body="&lt;p&gt;Yes, as per comment, but you have to check current density to make sure sufficient temperature is achieved otherwise the bond between some layers becomes weak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making, welding then testing several examples with variations of parameters is usually the way to proceed.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-10-10T10:26:13.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31449" PostTypeId="1" AcceptedAnswerId="31452" CreationDate="2019-10-10T13:21:59.343" Score="0" ViewCount="8353" Body="&lt;p&gt;For purposes of usage in home/auto and light industrial circumstances, what are the differences? What are the pros and cons, and suited uses, of each? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="23101" LastActivityDate="2019-10-10T16:42:34.703" Title="Silicone paste vs. silicone grease vs. silicone spray lube" Tags="|materials|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31450" PostTypeId="1" AcceptedAnswerId="31453" CreationDate="2019-10-10T13:56:35.083" Score="0" ViewCount="34" Body="&lt;p&gt;Isochoric means constant volume (In SI engine: piston should not move when the spark is ignited). In reality piston always keeps displacing up and down means volume changes. So, how does this make it a Isochoric Heat Addition in power stroke.&lt;/p&gt;&#xA;" OwnerUserId="23102" LastEditorUserId="23102" LastEditDate="2019-10-15T13:45:26.773" LastActivityDate="2019-10-15T13:45:26.773" Title="What practically is constant volume heat addition in SI piston engine?" Tags="|mechanical-engineering|pistons|" AnswerCount="1" CommentCount="4" ClosedDate="2019-10-11T23:03:13.973" ContentLicense="CC BY-SA 4.0" />
  <row Id="31451" PostTypeId="2" ParentId="31414" CreationDate="2019-10-10T15:05:44.023" Score="0" Body="&lt;p&gt;Thank you GisMofx and M.Bauer for the advice, those website will come in handy for future research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Norelem actually have one, which can be found &lt;a href=&quot;https://www.norelem.com/us/en/Products/Product-overview/Flexible-standard-component-system/03000-Spring-plungers-Indexing-plungers-Stops-Centring-positioning-components-Ball-lock-pins-T-slot-nuts/Indexing-plungers/03089-10-Indexing-plungers-with-rotation-lock-and-lead-in-chamfer.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also found a spring-loaded slide bolt latch from McMaster-Carr which will also suit what I need (&lt;a href=&quot;https://www.mcmaster.com/2206a21&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="23071" LastActivityDate="2019-10-10T15:05:44.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31452" PostTypeId="2" ParentId="31449" CreationDate="2019-10-10T16:42:34.703" Score="1" Body="&lt;p&gt;Silicone grease and paste are frequently meant the same thing. They are used for lubrication and because they wrap the area of application they protect the parts from rust and elements. When they are used in applications such as plumbing orings they have to be rated food grade, or for electrical uses they have to be rated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Silicone spray is mostly used as thin lubricant and locked tread breaker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes difference in consistency is just to solve access issues. There are sprays of silicone with different properties for different applications which are hard to reach.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-10T16:42:34.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31453" PostTypeId="2" ParentId="31450" CreationDate="2019-10-10T17:24:13.293" Score="1" Body="&lt;p&gt;the &lt;em&gt;rate&lt;/em&gt; at which heat is added is so fast that the piston has no opportunity to move during the process. Furthermore, when the piston is at the top of its stroke (which is where the heat is added) the crank-and-connecting rod mechanism yields a finite residence time of the piston in that position for a span of crank rotation, which means that even though the crank never stops &lt;em&gt;rotating&lt;/em&gt;, the piston spends a finite amount of time in an &lt;em&gt;almost motionless&lt;/em&gt; state at TDC. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-10-10T17:24:13.293" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31454" PostTypeId="1" AcceptedAnswerId="31458" CreationDate="2019-10-10T19:13:28.220" Score="0" ViewCount="41" Body="&lt;p&gt;I'm trying to reduce the rpm of a combustion motor. It operates at 2000 rpm with 3HP, and I need it to work at around 12 rpm (no matter the power). I've made the math with sprocket and chains, but the biggest sprocket is of 472 mm of diameter, which is too big. &#xA;Is there any cheap and easier way to achieve this?&#xA;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="23026" LastActivityDate="2019-10-11T15:12:19.153" Title="Trying to reduce RPM of a combustion motor" Tags="|motors|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31455" PostTypeId="1" AcceptedAnswerId="31456" CreationDate="2019-10-10T19:48:31.497" Score="0" ViewCount="44" Body="&lt;p&gt;If you're double-bagging some groceries, and one of the bags has tears in it while the other is intact, which bag should go on the inside?&lt;/p&gt;&#xA;" OwnerUserId="23107" LastActivityDate="2019-10-10T21:20:47.593" Title="Double-bagging problem: which bag on the inside?" Tags="|structural-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31456" PostTypeId="2" ParentId="31455" CreationDate="2019-10-10T21:20:47.593" Score="0" Body="&lt;p&gt;Put the torn bag inside, just to let it absorb the possible moisture of veggies and other freshly washed fruits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is interesting to note that some tears at critical locations where they could concentrate stress to outer bag can lead to tearing of the outer bag, while left alone by itself it could safely contain it's load. Same idea as the shallow scorelines the glazing guys put on a plane of glass to cut it.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-10T21:20:47.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31457" PostTypeId="1" CreationDate="2019-10-10T21:25:53.500" Score="1" ViewCount="315" Body="&lt;p&gt;I want to calculate brake torque from indicated torque for motorcycle.&lt;/p&gt;&#xA;" OwnerUserId="23109" LastActivityDate="2024-02-17T10:05:40.730" Title="Max torque given in engine specification is indicated torque or brake torque?" Tags="|mechanical-engineering|automotive-engineering|torque|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31458" PostTypeId="2" ParentId="31454" CreationDate="2019-10-10T22:34:05.297" Score="1" Body="&lt;p&gt;2000/12 is about 167, so you need to gear it down a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To a sensible 1st-order approximation, you don't want to gear down by more than 10:1 in any one stage (I'm not a mechanical engineer; if one of them speaks up feel free to believe them).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I were throwing this together from surplus parts, or even building a one-off machine or a short production run, I think I'd run belt drive to a gearbox with my desired total reduction.  So maybe 4:1 belt reduction to a 40:1 box, or 8:1 belt reduction to a 20:1 box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One stage of reduction on a belt drive to one more stage of reduction with a gear drive wouldn't be horribly impractical.  Because of the large torque multiplication, you'd need an enormous belt drive on the final stage which is why I'm suggesting gears or chains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you're building 10,000 a year, then you can afford to get an integrated motor/gearbox -- I suspect that's another story.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or I'd go with 200:1 reduction because it's a nice round number, and run the motor at 1200 RPM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep in mind that if the motor can do 3HP then the potential torque output at 12 RPM is around 1300 foot-pounds*, which is ginormous.  Size your shafts accordingly or put in torque-limiting clutches, and don't get in its way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;* For those who noted my earlier math error -- I did the torque calculation from the RPM and power, without regard to gear ratio (and I used a calculator).  So it should be correct.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastEditorUserId="19776" LastEditDate="2019-10-11T15:12:19.153" LastActivityDate="2019-10-11T15:12:19.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31460" PostTypeId="2" ParentId="31457" CreationDate="2019-10-10T23:29:23.500" Score="0" Body="&lt;p&gt;Brake torque usually means torque at the engine crankshaft, while there is no loss of power or toque due to transmission, powertrain or in your case the chain and gearbox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You would have to subtract those losses if the indicated torque is the wheel torque.   &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-10-10T23:36:34.820" LastActivityDate="2019-10-10T23:36:34.820" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31461" PostTypeId="1" AcceptedAnswerId="31462" CreationDate="2019-10-10T23:40:53.170" Score="1" ViewCount="429" Body="&lt;p&gt;I have a statics problem that I am running into some difficulties with. What is being asked here is to determine the support reaction at the points A and C. This would normally be no problem, but when I set up my equations (sum of the forces in the x-direction, sum of the forces in the y-direction, and total moment) I realize that there are too many unknowns for the problem to be solved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to assume something in order to eliminate one or two of the unknowns? (I am guessing it will be in the x-direction, but that's just a hypothesis.) Or should the problem be solved in a completely different way?&#xA;Thanks for the help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PWZk2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PWZk2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18704" LastActivityDate="2019-10-11T08:49:53.110" Title="Determining Support Reactions" Tags="|mechanical-engineering|civil-engineering|statics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31462" PostTypeId="2" ParentId="31461" CreationDate="2019-10-11T01:37:13.990" Score="2" Body="&lt;p&gt;The question has not defined any section properties so we assume all members are rigid. Then by inspection, we see this structure reduces to a 2 member frame AB and BC connected at a hinge at B and supported at pins on ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AB is at 45 degrees angle and BC is arctan(5/15) = arctan(1/3).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the OP correcting my arithmetic error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Sigma M_a= -10kN*10m + C_x*15m - C_y*25m =0 \quad 15C_x -25C_y=-100kNm \quad C_x=3 C_y\ \text{as per geometry of the triangle BCCh} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ 20C_y=-100kN\ , \\ C_y= -5kN \\ C_x=15kN $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Sigma Fx=0\ A_x= 15kN \ ,\\ \Sigma Fy=10=A_y-10-5=0, \ A_y= 15kN $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;check my numbers, please.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-10-11T04:02:50.110" LastActivityDate="2019-10-11T04:02:50.110" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31463" PostTypeId="1" CreationDate="2019-10-11T05:58:35.207" Score="0" ViewCount="72" Body="&lt;p&gt;I found this image of 2 piers of a Grider bridge with varying height. By any change could this be part of the design? The bridge is supposed to be a flyover for traffic decongestion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/f6JUF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/f6JUF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23113" LastEditorUserId="23113" LastEditDate="2019-10-11T07:00:01.370" LastActivityDate="2019-10-11T23:27:03.930" Title="Why do these piers have a height miss-match?" Tags="|bridges|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31464" PostTypeId="2" ParentId="31461" CreationDate="2019-10-11T08:49:53.110" Score="1" Body="&lt;p&gt;In this kind of questions, I believe that most of the attention should be focused on the general approach and the philosophy of attacking. Afterwards, each individual may proceed by writing its own equations and solving them in its prefer way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In our particular case, I guess you examined the whole structure, realized that there are 4 reactions (Ax, Ay, Cx and Cy) and thought to yourself - OK, here we have a statically indeterminate system (3 equations and 4 unknowns), so we must apply a unique technique in order to solve it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the trick here is hidden in the middle joint. The structure is divided into two parts (AB and BC) which are connected through a pin joint. This joint brings another two reactions to the table (Bx and By). You may first be tempted to think that &quot;hi, these are internal forces, they cannot add any useful information. I could alternatively create an artificial cut in any other place along the structure and look for the internal reactions there&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this would be a wrong way of thinking. If we cut our structure in any other zone, we would have to deal with 3 reactions in the cross section (two forces and a bending moment). The pin joint actually eliminates the bending moment which in turn allows us the solve the problem.&#xA;You may look at each part separately now. Part AB has 4 reactions and 3 equilibrium equations and the same is true for part BC. Since two of these reactions are the same (Bx and By) we end up with 6 equilibrium equations and 6 unknown (Ax, Ay, Bx, By, Cx and Cy). At this point we realize that we can solve the problem without applying sophisticated methods (using stiffness and calculating deflections and etc.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;@kamran showed you a good way of solving it. Just pay attention to the nice approach he used when writing the equations of part BC - Since there are no external forces exerted on it, the total reaction of joint C must be directed towards joint B. Otherwise, there would be a total moment applied on Joint B. The same is true, of course, to point C. By finding a relationship between Cy and Cx he was able to come up with the 4th equation you just look for.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2019-10-11T08:49:53.110" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31467" PostTypeId="1" AcceptedAnswerId="31469" CreationDate="2019-10-11T14:04:21.733" Score="0" ViewCount="24" Body="&lt;p&gt;In a paper in the context of controller design for DEAP actuators I came upon the following sentence:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Furthermore, the nonlinear behavior of the&#xA;  control plant requires either an adaptive or a robust design in&#xA;  order to ensure a precise and accurate control.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What is the relationship between a robust/adaptive controller and the nonlinearity of the control plant?&#xA;Isn't it somehow always necessary to design a robust controller no matter whether the control plant is linear or not?&#xA;Or are nonlinear plants in general more sensitive to parameter variations, e.g. nonlinear systems sensitive to initial conditions (chaotic behaviour)?&lt;/p&gt;&#xA;" OwnerUserId="20870" LastActivityDate="2019-10-11T15:18:12.753" Title="Nonlinear behaviour of control plant and adaptive or robust design" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31468" PostTypeId="1" CreationDate="2019-10-11T14:22:38.440" Score="1" ViewCount="45" Body="&lt;p&gt;Before I explain my problem I'd like to mention I'm still green and about 2 weeks into these topics. I learned most of it from youtube videos, wikipedia and other science sites, so I'll be explaining the smallest details in order to confirm I didn't mess anything up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have several questions regarding the P-N junction; the forming of a depletion region - the nature of ions created during the diffusion of carriers to be specific.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here's what I've figured out so far (please correct any inaccuracies):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;We dope the n region of the semiconductor with an element that has more electrons in it's valence shell than the semiconductor atoms. The added element &quot;wants to behave&quot; like the rest of the atoms in the crystal due to the rules of covalent bonds and gives away it's excess electrons. In doing so it becomes a positively charged ion, because it has less electrons than protons. Now there are a lot of free electrons in the conduction band of the n region which increases conductivity.&lt;/li&gt;&#xA;&lt;li&gt;We dope the p region of the semiconductor with an element that has less electrons in it's valence shell. The rules are the same except this time it wants to accept missing electrons to fit in the structure, so it becomes a negatively charged ion. Now there are a lot of electrons missing in the valence band of the p region which increases it's conductivity.&lt;/li&gt;&#xA;&lt;li&gt;a depletion region forms because the excess electrons in the n part are attracted to the holes in the p part. In doing so, the ions on both sides of the p-n border become neutral/complete again and they refuse to accept electrons at this point, so they create a barrier.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Am I right in thinking the depletion region forms because the dope ions have regained the balance between the amount of protons and electrons, so they repel electrons which are the only real charge carriers (By real I mean holes are just empty spaces for electrons, not some magical positive charge carriers that can move in a way that electrons do.) ?&lt;/li&gt;&#xA;&lt;li&gt;How come the depletion region even exists, shouldn't the depletion region atoms still share their electrons/holes with the rest of the structure?&lt;/li&gt;&#xA;&lt;li&gt;This one might be offtopic, but do atoms with 8 electrons in their valence shell accept or give away electrons? Is the octet rule a key concept to understand all of this fully, or maybe it has nothing to do with it?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Also, if you could recommend some good literature that can explain these topics thoroughly I'd be grateful.&lt;/p&gt;&#xA;" OwnerUserId="23118" LastEditorUserId="8055" LastEditDate="2019-10-11T22:35:36.530" LastActivityDate="2020-05-04T05:21:55.783" Title="P-N junction depletion region" Tags="|electrical-engineering|diffusion|semiconductor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31469" PostTypeId="2" ParentId="31467" CreationDate="2019-10-11T15:18:12.753" Score="1" Body="&lt;p&gt;The authors are assuming that you're going to make a linear or mostly linear controller after finding a linearized model of the plant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not that nonlinear plants are &lt;em&gt;sensitive&lt;/em&gt; to model variations, it's that for a certain class of nonlinear plant you can treat the plant as a linear time-varying system that has time-varying disturbances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in that sort of a design paradigm, you use some design process that works with time-varying plants: either you make an adaptive controller that's constantly discovering the up-to-date plant model and controlling it in some optimal way, or you make a robust controller that will control a linear time-varying plant that varies in the way your linearization does over the range of operating points of the real, nonlinear plant.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-10-11T15:18:12.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31470" PostTypeId="1" CreationDate="2019-10-11T16:30:29.823" Score="1" ViewCount="42" Body="&lt;p&gt;For improving the gear strength, which one is better based on manufacture and economy? Increasing module or increasing N? With the same surface strength.&lt;/p&gt;&#xA;" OwnerUserId="23120" LastActivityDate="2019-10-11T16:30:29.823" Title="gear design.improving gear strength" Tags="|mechanical-engineering|design|gears|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31471" PostTypeId="1" AcceptedAnswerId="31475" CreationDate="2019-10-11T16:47:29.320" Score="0" ViewCount="80" Body="&lt;p&gt;Since the voltage required to bridge the gap between the central electrode and the ground electrode of a spark plug can reach up to 40 000V and even more, my question is, how does the battery not die since the spark plug is grounded to it. Is all that energy lost in the ionization of the gas and thus the voltage reduced to ~12V?&lt;/p&gt;&#xA;" OwnerUserId="23121" LastEditorUserId="8055" LastEditDate="2019-10-11T22:30:54.720" LastActivityDate="2019-10-11T22:30:54.720" Title="What is the voltage at the ground electrode of a spark plug?" Tags="|electrical-engineering|battery|car|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31473" PostTypeId="2" ParentId="31443" CreationDate="2019-10-11T18:53:05.193" Score="1" Body="&lt;p&gt;Models fall into three classes: first-principled (theoretical), semi-empirical, and empirical. A theoretical model develops a mathematical expression for we can expect for the behavior of a system based on first-principled logic and analysis. The empirical model develops a mathematical expression for what we can observe about the behavior of a system by doing experiments on it. The semi-empirical tweaks theoretical equations to fit observations or tweaks empirical equations to include first principles as appropriate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By example, the ideal gas law was first developed as an empirical model using Boyles and Charles laws. It was later developed as a theoretical model using kinetic theory of gases. The van der Waals equation of state is a semi-empirical tweak of the ideal gas law to allow particles (that only existed using the kinetic theory) to have a volume and (attractive) interactions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First-principled models may not have immediate access to the full range of observable parameters. Hence, they may give analytical results that appear nonsensical or are not observed physically when the analytical model is used to make predications outside of the range where its parameters are valid. Empirical models never have an infinite supply of measurement values over all parameter space. Hence, they may give empirical results that appear nonsensical or are counter to actual observations when they are extrapolated outside the range of the original data used to create the model. Semi-empirical models can fail for either analytical or empirical reasons, as you might guess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Infinity is a mathematical concept. It is not a numerical value that can ever be experienced by anything real (OK, perhaps you are willing to wait &lt;em&gt;forever&lt;/em&gt; with the hope then truly to see infinity ... but that is a digression to another forum).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, no theoretical model that predicts results with or at &lt;span class=&quot;math-container&quot;&gt;$\infty$&lt;/span&gt; will ever be possible to validate in the life time of any real system. All empirical models will become less accurate as any of their parameters go to &lt;span class=&quot;math-container&quot;&gt;$\infty$&lt;/span&gt;. Semi-empirical ... again fails on either case as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We rationalize predictions made by theory at infinite values or results from theory that give infinite values as an indication that the model has reached its analytical limit (theoretical). We call inputs at or results at &lt;span class=&quot;math-container&quot;&gt;$\infty$&lt;/span&gt; as singularities. They are SINGULAR values (not multiple values), yet they are not measurable values. We state that the behavior of the model for the real system &quot;breaks down&quot; at singularities. At that point, we must accept that we have reached the limit of what we will ever be able to reliably predict using the given analytical model. Alternatively, we also may recognize that the real system will fail at the singularity, and if so, will fail in a way that we cannot predict exactly or precisely. A mechanical oscillator driven exactly at its resonance will shatter apart, and the theoretical model predicts that with a singularity. How does it shatter? Which directions do the pieces fly? Those are but two of the unpredictable questions for the real system at the singularity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When empirical models demonstrate singularities, we simply (figuratively) toss up our hands and admit &quot;The model was not to be used at that point anyway because it had no empirical support from prior observations.&quot;&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-10-12T12:33:26.037" LastActivityDate="2019-10-12T12:33:26.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31474" PostTypeId="1" CreationDate="2019-10-11T20:14:17.113" Score="1" ViewCount="48" Body="&lt;p&gt;I have seen three representation of stress concentration factor values &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;,&lt;span class=&quot;math-container&quot;&gt;$K_{IC}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K_t$&lt;/span&gt; now all that I know is &lt;span class=&quot;math-container&quot;&gt;$_I$&lt;/span&gt; in suffix represents opening mode of crack.The suffix &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; stands for tip.However I do not understand how the three are factually different.I would require some fundamental advice on the three. &lt;/p&gt;&#xA;" OwnerUserId="19123" LastActivityDate="2019-10-11T20:14:17.113" Title="What is the difference between $K$ ,$K_{IC}$ and $K_t$?" Tags="|mechanical-engineering|structural-engineering|materials|stresses|finite-element-method|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31475" PostTypeId="2" ParentId="31471" CreationDate="2019-10-11T21:17:56.067" Score="2" Body="&lt;ul&gt;&#xA;&lt;li&gt;All current flows in a closed loop circuit.&lt;/li&gt;&#xA;&lt;li&gt;The ignition coil primary current flows from the battery through the coil and returns to the battery through the chassis circuit.&lt;/li&gt;&#xA;&lt;li&gt;The ignition coil secondary current flows from the ignition HT terminal through the spark plug, across the gap and returns to the ignition coil through the chassis circuit.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2019-10-11T21:17:56.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31476" PostTypeId="2" ParentId="31463" CreationDate="2019-10-11T23:27:03.930" Score="3" Body="&lt;p&gt;It could be the tops are at just the same level, only with deeper beams to cover larger spans at the right hand side, except the last beam upfront is not installed yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The missing beam and the angle of perspective creates the illusion that the top is stepping down to the left.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-11T23:27:03.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31479" PostTypeId="1" CreationDate="2019-10-12T16:25:36.730" Score="0" ViewCount="100" Body="&lt;p&gt;Is it possible to build a building with 3 levels, with the 2nd and 3rd levels having areas with clear spans (no columns) of 130 ft length and 85 feet width? There can be columns outside of this clear span area and on the first (ground) level.&lt;/p&gt;&#xA;" OwnerUserId="23128" LastActivityDate="2020-07-20T16:04:00.310" Title="Maximum spans in multi-level building" Tags="|structural-engineering|civil-engineering|structural-analysis|building-design|" AnswerCount="3" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31480" PostTypeId="1" CreationDate="2019-10-12T16:27:39.753" Score="0" ViewCount="43" Body="&lt;p&gt;I have to do simple visualization or animation of movements for simple hand prosthesis. There are no validation or calculations required. I just want to show concepts working principles (movements mainly) with animation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the fastest way to do it?&lt;/p&gt;&#xA;" OwnerUserId="23127" LastEditorUserId="110" LastEditDate="2019-10-12T20:44:31.110" LastActivityDate="2020-10-19T23:20:48.143" Title="Simplest software for making 2D and 3D visualizations(animation)," Tags="|cad|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31481" PostTypeId="1" CreationDate="2019-10-12T18:17:46.557" Score="2" ViewCount="75" Body="&lt;p&gt;Are the effects of gravity and acceleration identical to the point that no sensor can tell the difference?&lt;/p&gt;&#xA;" OwnerUserId="23129" LastActivityDate="2019-10-13T01:00:37.727" Title="Can a sensor tell the difference between gravity vs. acceleration?" Tags="|sensors|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31482" PostTypeId="2" ParentId="31479" CreationDate="2019-10-12T18:42:57.083" Score="1" Body="&lt;p&gt;Yes, it is possible by using trusses, 2D or 3D. Or special configuration on adjacent span beams such that those beams can cantilever into the clear space by say 25' from each side thus reducing the effective span to 80' by 40' while using high strength concrete and post strees cable or some other expensive complex mixture of these. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One has to consider the very negative impact of wide spans with the top-heavy structure on seismic resistance of the building and extra height of the building to accommodate the depth of 12'-15' deep trusses.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-12T18:42:57.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31483" PostTypeId="2" ParentId="31480" CreationDate="2019-10-12T18:56:33.917" Score="1" Body="&lt;p&gt;Some years ago Auto Cad had a very powerful but intuitive module calles 3D studio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It could read as input Acad mesh files and also had its own simplified interface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a couple of hours, I could make richly rendered walkthrough clips of projects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These days there are numerous apps, many are freeware.&#xA;Here is a link to some. &lt;a href=&quot;https://www.google.com/search?q=freeware%20animation%20programs&amp;amp;rlz=1C1DIMA_enUS763US765&amp;amp;oq=freeware%20animtion&amp;amp;aqs=chrome.4.69i57j0l6.18736j1j7&amp;amp;sourceid=chrome&amp;amp;ie=UTF-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;freeware modeling&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-12T18:56:33.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31484" PostTypeId="2" ParentId="31479" CreationDate="2019-10-12T19:00:08.200" Score="0" Body="&lt;p&gt;Auditoriums and arenas have much larger clear spans ; It just takes large steel beams or trusses . The beam sizes will be strongly affected by the loads you plan for the upper floors.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-10-12T19:00:08.200" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31485" PostTypeId="1" AcceptedAnswerId="31487" CreationDate="2019-10-12T20:37:57.487" Score="2" ViewCount="72" Body="&lt;p&gt;I need a shaft that can be stretched along its length while also being resistant to torque enough to be a functional shaft (see pictures). I have a rotating sphere within another rotating sphere, hence the use of universal joints. But due to the universal joints I can't use a shaft that is long enough to slide up and down. You can see in my images where there's a shaft missing. I don't know of the existence of any rubber that is sturdy enough to withstand the torque applied while also being stretchy enough to span that gap. Perhaps there's other ways to achieve what I am trying to achieve without the use of universal joints (I am open to suggestions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hMhLk.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hMhLk.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vj1cb.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vj1cb.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps I could 3D print a sort of origami shaft that unfolds?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N6jan.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N6jan.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is getting to be a bit ridiculous lol what I'm trying to make shouldn't be rocket science&lt;/p&gt;&#xA;" OwnerUserId="23131" LastEditorUserId="23131" LastEditDate="2019-10-12T21:11:08.377" LastActivityDate="2019-10-12T21:27:54.157" Title="Is there such a material that is resistant to torque while easily stretching when under torsion?" Tags="|materials|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31487" PostTypeId="2" ParentId="31485" CreationDate="2019-10-12T21:27:54.157" Score="2" Body="&lt;p&gt;There are sliding shafts used for agricultural machinery commonly called a &quot;lemon shaft&quot; and they are available in several sizes.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-10-12T21:27:54.157" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31488" PostTypeId="2" ParentId="31443" CreationDate="2019-10-12T23:48:53.253" Score="0" Body="&lt;p&gt;When you model something you make some assumptions about the world. In empirical models you do not account for all variables, in first principles you lump things together etc. You therefore need to know what simplifications have been made inorder to understand your model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example quite many models fail to account for right propagation of forces over time. Now offcourse if you accout for even a really tiny delay or granularity of the universe that infinity wouldnt be there. But then it would be expensive and that singularity is a acceptable solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sometimes the mathematical underpinning is also not known well. For example in multibody dynamics we are unable to discern between several solutions since the method we use does not. Does this mean there is a better metod? Possibly but the method we use is quite good and works in most technical cases. Same thing regarding chaotic behaviour.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2019-10-12T23:48:53.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31489" PostTypeId="2" ParentId="31481" CreationDate="2019-10-13T01:00:37.727" Score="4" Body="&lt;p&gt;Imagine an accelerometer that is in a box, and held aloft by a rope which is attached to a pulley, balanced by another box of equal mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could either:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Cut the rope. The box will fall downwards, being accelerated by gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Push the box downwards with an acceleration of &lt;span class=&quot;math-container&quot;&gt;$9.81m/s^2$&lt;/span&gt;, from an ‘applied force’&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The accelerometer &lt;em&gt;can not&lt;/em&gt; distinguish between these two scenarios.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-10-13T01:00:37.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31493" PostTypeId="1" AcceptedAnswerId="32072" CreationDate="2019-10-13T10:12:42.903" Score="0" ViewCount="169" Body="&lt;p&gt;When analysing a structure using traditional methods (e.g. Hardy Cross, Slope Deflection, Force Method, etc.), it is a first-order elastic analysis. My question is, how would one analyse a statically indeterminate frame by hand but taking into account the second-order effects. This is more from an academic perspective and that is why I am asking for hand calculations using traditional methods or even direct stiffness method.   &lt;/p&gt;&#xA;" OwnerUserId="20822" LastActivityDate="2019-11-24T11:44:43.680" Title="Second Order Elastic Analysis" Tags="|structural-engineering|structural-analysis|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31495" PostTypeId="1" CreationDate="2019-10-13T12:27:40.417" Score="0" ViewCount="31" Body="&lt;p&gt;I am want to produce Willys' hood latches like links below&#xA;&lt;br/&gt;&lt;a href=&quot;https://www.kaiserwillys.com/media/catalog/product/cache/1/image/560x560/9df78eab33525d08d6e5fb8d27136e95/7/6/7601v_enlarged.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hood latch link 1&lt;/a&gt;&#xA;&lt;br/&gt;&lt;a href=&quot;https://images-na.ssl-images-amazon.com/images/I/31gj6IUCyRL._SX425_.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hood latch link 2&lt;/a&gt;&#xA;&lt;br/&gt;&lt;a href=&quot;https://picclick.com/Jeep-Willys-MB-GPW-windshield-to-hood-latch-271115518868.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hood latch link 3&lt;/a&gt;&#xA;&lt;br/&gt;What kind of metal shaping process is this? &#xA;Even I could not figure the mold out.&lt;/p&gt;&#xA;" OwnerUserId="23139" LastActivityDate="2019-10-13T12:27:40.417" Title="How to shape sheet metal like Willys hood latch?" Tags="|metal-folding|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31496" PostTypeId="1" CreationDate="2019-10-13T14:39:53.720" Score="0" ViewCount="270" Body="&lt;p&gt;By Googling I can find basically two types of steam tables for water. They are either &lt;em&gt;saturated steam tables&lt;/em&gt;, which I understand are for steam in saturated state. Here pressure are temperature are fixed together as saturated steam states form a curve in water phase diagrams. Then there are some tables for superheated steam, where temperature exceeds critical temperature for water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if I'm interested in values for water vapour/steam that is neither superheated nor saturated, that exists under the saturation curve in water phase diagrams?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are we specifically interested in saturated steam?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2020-03-09T15:28:26.713" Title="Are there steam tables for subsaturated steam?" Tags="|thermodynamics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31497" PostTypeId="2" ParentId="31496" CreationDate="2019-10-13T14:44:37.000" Score="0" Body="&lt;p&gt;Because for any mix of water vapor we re-calculate using (1-x), x being the dryness fraction.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-10-13T14:44:37.000" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31498" PostTypeId="1" CreationDate="2019-10-13T16:10:44.410" Score="0" ViewCount="190" Body="&lt;p&gt;I am working on structural analysis of an RC plane in a symmetrical maneuver. There are two types of loads on the structure a) aerodynamic loads and b) inertial loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have determined the aerodynamic loads by performing CFD analysis on Fluent. Mapping of aerodynamic loads from CFD mesh to structural mesh in MSC Patran is easy. The details of the load mapping are given in link [1].&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum inertial loads in any maneuver can be determined using analytical calculations or directly from different handbooks/standards including FAR 25 etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read somewhere that I can perform static analysis of unconstrained system using inertia relief. However, from what I have learnt from link [2], the inertial loads cannot be used in inertia relief analysis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question#1: Is it possible to perform the static structural analysis of an unconstrained object subjected to aerodynamic loads (which are nothing more than a pressure distribution) coupled with inertial loads? If yes, how?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question#2: If static structural analysis cannot be performed, do I have to perform a dynamics analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;LINKS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- &lt;a href=&quot;https://www.eng-tips.com/viewthread.cfm?qid=446687&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.eng-tips.com/viewthread.cfm?qid=446687&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2- &lt;a href=&quot;https://pdfs.semanticscholar.org/6e7c/9a0ba36d86ac73b3e2e213fa64e3b6c5b988.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://pdfs.semanticscholar.org/6e7c/9a0ba36d86ac73b3e2e213fa64e3b6c5b988.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2019-11-24T14:19:17.260" Title="Structural Analysis of an RC plane in flight" Tags="|structural-engineering|structural-analysis|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31499" PostTypeId="1" CreationDate="2019-10-13T20:07:27.487" Score="2" ViewCount="61" Body="&lt;p&gt;Manufacturers recommend keeping lithium batteries between 40 and 80 percent charged. I tried, but sometimes I let my phone get higher than that. Then I try to bring it down  soon as possible, either by playing graphic intensive apps or by charging some other device off my phone via USB-c. Is there a point in bringing it down once it already has overcharged. Which method is better/worse? &lt;/p&gt;&#xA;" OwnerUserId="23143" LastEditorUserId="110" LastEditDate="2019-10-13T22:04:50.947" LastActivityDate="2019-10-14T02:18:55.290" Title="Is it a good idea to intentionally discharge lithium batteries to 80%" Tags="|battery|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31500" PostTypeId="2" ParentId="31439" CreationDate="2019-10-13T22:03:45.430" Score="1" Body="&lt;p&gt;My question was based on an appearance that the thin winding wire seemed to violate the laws of physics, so there might be some other factors at play.  Neil_UK answers this in &lt;a href=&quot;https://electronics.stackexchange.com/a/325358/115681&quot;&gt;generator coils: max power given wire gauge&lt;/a&gt; on the Electrical Engineering site.  The bottom line is no.  It's mainly a matter of using multiple coils in parallel to manage the current handling requirements per coil.  Other factors, like the duty cycle and dissipating heat also come into play.&lt;/p&gt;&#xA;" OwnerUserId="12772" LastActivityDate="2019-10-13T22:03:45.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31501" PostTypeId="1" AcceptedAnswerId="31502" CreationDate="2019-10-13T23:43:46.280" Score="5" ViewCount="5450" Body="&lt;p&gt;How can I calculate the power and torque required for the motor on a wheeled robot or vehicle if a particular acceleration or movement up an incline is required?&lt;/p&gt;&#xA;" OwnerUserId="19778" LastEditorUserId="1832" LastEditDate="2019-10-14T02:09:13.377" LastActivityDate="2023-11-14T16:30:46.060" Title="How can I calculate the power and torque required for the motor on a wheeled robot/vehicle?" Tags="|motors|torque|wheels|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31502" PostTypeId="2" ParentId="31501" CreationDate="2019-10-13T23:44:03.410" Score="17" Body="&lt;p&gt;I laid everything out so you should only need to read it from top to bottom and look backwards for variables, never forward. I also tried to lay it out so hopefully you know where everything is coming from (as long as you have a basic understanding of power, torque, force, and friction...maybe even if you don't).&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \mu_{roll}$&lt;/span&gt; = coefficient of rolling friction for wheels (between 0 and 1)&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The most important unknown that needs to be determined or&#xA;estimated.&lt;/p&gt;&#xA;&lt;p&gt;For reference, a coefficient for rolling friction of 0.3 is already&#xA;very high and is for something like for soft wheels (which deform) on&#xA;a dirt road (which isn't flat or hard) where it is light enough it&#xA;won't sink in. Most of the time it should be more like 0.1 to 0.2&#xA;with it being lowest on smooth hard surfaces with smooth hard wheels.&lt;/p&gt;&#xA;&lt;p&gt;A coefficient of friction of 0.3 means rolling on wheels takes 30%&#xA;the force of just lifting it up. Ideally you want it to be zero. At&#xA;&lt;span class=&quot;math-container&quot;&gt;$\mu_{roll}&amp;gt;1$&lt;/span&gt; it is easier to just pick the thing and move it up rather than rolling it. Knowing this definition should help you an intuitive feel for things so a&#xA;value can be estimated. You will have to estimate or measure this&#xA;most important value or conservatively guess a worst case.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If measuring rolling friction (by pushing or pulling the vehicle to&#xA;determine the fraction of the weight that must be applied to slowly&#xA;budge the scooter on a horizontal surface) be aware of drive train&#xA;losses if it remains connected to the wheels during the test  which&#xA;can obfuscate the measurement for rolling friction.&lt;/p&gt;&#xA;&lt;p&gt;Drive train friction should technically be separate and not be lumped&#xA;with the rolling friction coefficient. This decreases accuracy in&#xA;inclined scenarios though where rolling friction decreases but drive&#xA;friction remains constant. There are also issues where the torque&#xA;through a gearbox is not reversible.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ v=$&lt;/span&gt; speed (&lt;span class=&quot;math-container&quot;&gt;$m \over s$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \theta=$&lt;/span&gt; angle of incline&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ m_{vehicle} = $&lt;/span&gt; mass of vehicle (&lt;span class=&quot;math-container&quot;&gt;$kg$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ g = $&lt;/span&gt; acceleration of gravity &lt;span class=&quot;math-container&quot;&gt;$ =9.81 \frac{m}{s^2}$&lt;/span&gt;...on Earth. I want to know if you're using this for something that isn't.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ W_{vehicle} = $&lt;/span&gt; weight of vehicle (&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ =m_{vehicle}\times g$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ W_{\perp vehicle} =$&lt;/span&gt; normal force (&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;)&lt;span class=&quot;math-container&quot;&gt;$=W_{vehicle}cos(\theta)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; = radius of driven wheel (&lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;)&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ N =$&lt;/span&gt; drivetrain reduction ratio (i.e. gear ratio) where &lt;span class=&quot;math-container&quot;&gt;$N&amp;gt;1$&lt;/span&gt; for reducing motor speed and increasing motor torque&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ n = $&lt;/span&gt; minimum number of motors engaged with ground.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;(i.e. Number of motors that are driving at least one wheel with&#xA;traction in your worst case conditions. For example, Mars Rovers have&#xA;one motor per wheel so for every wheel lifted off the ground, output for a motor is lost. But difficult to actually lift a rover wheel with them being on the end of pivoting leg suspension, but much easier to do with a conventional 4 wheel x 4 motor design with fixed wheels.&lt;/p&gt;&#xA;&lt;p&gt;Other robots may have multiple wheels ganged to a single per motor&#xA;via belt/chain/treads, so as long as one of those wheels is in&#xA;contact, the motor still propels the robot. For example, 6-wheeled vehicle where each side has a single motor chained/belted to all three wheels on that side will maintain motor output as long as at least one wheel on each side has traction.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\eta_{drivetrain} =$&lt;/span&gt; Drivetrain efficiency (between 0 and 1 for 0% to 100%)&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Good gearboxes can be 90%-95% efficient. Belt drives are pretty high&#xA;too. 80% is a good conservative estimate in general unless you are&#xA;using worm gears which have terrible efficiency. Sometimes gearbox&#xA;losses are fixed in which case you just add that wattage at the very&#xA;end of your calculations.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\eta_{motor} =$&lt;/span&gt; Motor efficiency (between 0 and 1 for 0% to 100%)&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;A good motor can be 80%-90% efficient or higher. 70% is a good&#xA;conservative estimate. 50% for a crappy motor.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you include motor efficiency, your power results will be the motor&#xA;electrical input power.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you do not include motor efficiency (i.e. &lt;span class=&quot;math-container&quot;&gt;$ \eta_{motor} = 1$&lt;/span&gt;),&#xA;then your final power numbers will be the motor output power.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F_{roll} =$&lt;/span&gt; force of rolling friction (&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;) (i.e. force required to maintain speed on level surface)&lt;span class=&quot;math-container&quot;&gt;$ = W_{\perp vehicle} \times \mu_{roll}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F_{incline} =$&lt;/span&gt; force required to overcome gravity on an incline (i.e. force required to maintain speed against gravity up an incline) (&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$=W_{vehicle}sin(\theta)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ a = $&lt;/span&gt; desired acceleration (&lt;span class=&quot;math-container&quot;&gt;$m \over s^2$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ = \frac{v}{t}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; is the number of seconds you are willing to wait to reach your desired speed&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F_{accelerate} = $&lt;/span&gt;force required to accelerate (i.e. after rolling friction and gravity have been nullified by other forces) (&lt;span class=&quot;math-container&quot;&gt;$N$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ = m_{vehicle} \times a $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega =$&lt;/span&gt; wheel angular velocity (&lt;span class=&quot;math-container&quot;&gt;$rad \over s$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ = \frac{v}{r}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$RPS =$&lt;/span&gt; wheel rotations per second &lt;span class=&quot;math-container&quot;&gt;$ = \frac{\omega}{2\pi}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$RPM =$&lt;/span&gt; wheel rotations per minute &lt;span class=&quot;math-container&quot;&gt;$ = RPS \times 60$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \tau_{roll} $&lt;/span&gt; = &lt;strong&gt;combined&lt;/strong&gt; wheel torque required to overcome rolling friction (i.e. torque required to maintain speed on a level surface) (&lt;span class=&quot;math-container&quot;&gt;$N \cdot m$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ = F_{roll} \times r_{wheel}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \tau_{incline} $&lt;/span&gt; = &lt;strong&gt;combined&lt;/strong&gt; wheel torque required to overcome gravity on an incline (i.e. torque required to maintain speed against gravity up an incline) (&lt;span class=&quot;math-container&quot;&gt;$N \cdot m$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ = F_{incline} \times r_{wheel}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \tau_{accelerate} = $&lt;/span&gt;&lt;strong&gt;combined&lt;/strong&gt; wheel torque required to accelerate (i.e. after rolling friction and gravity have been nullfied by other torques) (&lt;span class=&quot;math-container&quot;&gt;$N \cdot m$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ = F_{accelerate} \times r_{wheel}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \tau_{constant} = $&lt;/span&gt; &lt;strong&gt;combined&lt;/strong&gt; wheel torque required to maintain constant speed (&lt;span class=&quot;math-container&quot;&gt;$N \cdot m$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ = \tau_{roll} + \tau_{incline}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \tau_{wheel}= $&lt;/span&gt; total torque required (&lt;span class=&quot;math-container&quot;&gt;$N \cdot m$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ = \tau_{constant} +\tau_{accelerate}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \tau_{motors} = $&lt;/span&gt; &lt;strong&gt;combined&lt;/strong&gt; motor torque (&lt;span class=&quot;math-container&quot;&gt;$N \cdot m$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ = \frac{1}{\eta_{drivetrain}} \left(\frac{\tau_{wheel}}{N}\right)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \tau_{motor} = $&lt;/span&gt; torque &lt;strong&gt;per motor&lt;/strong&gt; (&lt;span class=&quot;math-container&quot;&gt;$\frac{N \cdot m}{motor}$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ = \frac{\tau_{motors}}{n}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; This is not strictly correct but it is about the best we can do to compensate for drivetrain friction. Technically, it should be: &lt;span class=&quot;math-container&quot;&gt;$ \tau_{motor} = \left(\frac{\tau_{wheel}}{N}\right) + \tau_{drivetrain}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\tau_{drivetrain}$&lt;/span&gt; is the drivetrain friction torque, but &lt;span class=&quot;math-container&quot;&gt;$\tau_{drivetrain}$&lt;/span&gt; is dependent on both RPM and load torque so it is really difficult to determine.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;At this point there are alternative ways to calculate the power, either using torques and angular velocities, or forces and linear velocities.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ P_{continuous} = $&lt;/span&gt; Total continuous power for all motors to maintain speed (&lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt;)&#xA;&lt;span class=&quot;math-container&quot;&gt;$ = \left(F_{roll} + F_{incline}\right)\times \left(v \times \frac{1}{\eta_{motor}} \times \frac{1}{\eta_{drivetrain}}\right)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ = \left(\tau_{roll} + \tau_{incline}\right) \times \left(\omega \times \frac{1}{\eta_{motor}} \times \frac{1}{\eta_{drivetrain}}\right)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;NOTE:&lt;/strong&gt; If we had a value for the drivetrain friction torque, &lt;span class=&quot;math-container&quot;&gt;$\tau_{drivetrain}$&lt;/span&gt; we could directly calculate the power loss due to drivetrain friction via &lt;span class=&quot;math-container&quot;&gt;$P_{drivetrain} = \tau_{drivetrain} \times \omega$&lt;/span&gt; and instead use &lt;span class=&quot;math-container&quot;&gt;$ P_{continuous} = \left(F_{roll} + F_{incline}\right)\times \left(v \times \frac{1}{\eta_{motor}}\right) + \left(P_{drivetrain} \times n\right)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$= \left(\tau_{roll} + \tau_{incline}\right) \times \left(\omega \times \frac{1}{\eta_{motor}}\right) + \left(P_{drivetrain} \times n\right)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We multiply by the number of motors because &lt;span class=&quot;math-container&quot;&gt;$P_{drivetrain}$&lt;/span&gt; is the power loss of a single drivetrain assembly, but each motors has its own drivetrain assembly while &lt;span class=&quot;math-container&quot;&gt;$P_{continuous}$&lt;/span&gt; is the total continuous power for all motors.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ P_{peak} = $&lt;/span&gt; Total peak power for all motors to accelerate (&lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$= P_{continuous} + \left(F_{accelerate} \times v \times \frac{1}{\eta_{motor}} \times \frac{1}{\eta_{drivetrain}}\right)$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ P_{continuous/motor} =$&lt;/span&gt; Continuous power per motor (&lt;span class=&quot;math-container&quot;&gt;$W \over motor$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ =  \frac{P_{continuous}}{n}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ P_{peak/motor} =$&lt;/span&gt; Peak power per motor (&lt;span class=&quot;math-container&quot;&gt;$W \over motor$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$ =  \frac{P_{Peak}}{n}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;No slippage is assumed. Static friction to initially get everything moving from standstill, wheel moment of inertia, and speed-dependent losses such as aerodynamic resistance or speed-dependent drive-train losses have been neglected.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;strong&gt;DC MOTOR SPECIFICATIONS:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;For DC motors, often torque-speed curves are not provided. Usually only the no-load RPM and stall torque are supplied. You do not use the stall-torque to size anything because this is the torque the motor produces at zero zero RPM while it is drawing enormous amounts of current and overheating. Nor do you want to use the no-load RPM which is what the motor spins at when no torque is applied. It also goes without saying that you cannot use the no-load RPM together with the stall torque in your calculations since they do not happen at the same time.&lt;/p&gt;&#xA;&lt;p&gt;However, not all is lost because for DC motors, the torque-speed curves are linear. As torque decreases from stall-torque to zero, the RPM increases from zero to the no-load RPM. Therefore, for the operating voltage for which the no-load RPM and stall-torque are supplied you can produce the torque-speed curve.&lt;/p&gt;&#xA;&lt;p&gt;What will not be on this torque-speed curve that is where it is best to run the motor at. But we can use rules of thumb which are accurate enough for many applications:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the point of maximum efficiency for permanent magnet DC motors tends&#xA;to be about 1/7th the stall torque and 6/7ths the no-load RPM&lt;/li&gt;&#xA;&lt;li&gt;the point of maximum power is 1/2 the stall torque and 1/2 the&#xA;no-load RPM&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Therefore, it is best to size your motor, wheel size, and gear/belt/chain reduction ratios and  so the most frequent, prolonged, and continuous mode of operation occurs at 1/7th stall torque and 6/7ths no-load RPM. Maintaining velocity, for example.&lt;/p&gt;&#xA;&lt;p&gt;Similarly, you also probably do not want to require more than half-stall torque from your motor for modes of operation that require accelerations over a prolonged period of time. Conversely, for brief tasks where additional torque is required it may be acceptable to exceed half stall-torque. It's up to you to judge how far beyond half stall-torque is acceptable for the tasks being performed. For example, for climbing over a bump or brief obstacle you might choose that 3/4 stall torque is acceptable but too much for accelerating up to speed during normal travel. Instead you choose that to be half stall-torque instead.&lt;/p&gt;&#xA;" OwnerUserId="19778" LastEditorUserId="19778" LastEditDate="2023-11-14T16:30:46.060" LastActivityDate="2023-11-14T16:30:46.060" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31503" PostTypeId="2" ParentId="31499" CreationDate="2019-10-14T01:01:01.750" Score="1" Body="&lt;p&gt;It is customary for reputed mobile phone manufactures to design electronics battery charging circuits to ensure that batteries are changed between 40% and 80%. This will address your concerns. By discharging the phone your reducing the battery capacity to less and 80%. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To extend battery life, avoid over heating your phone. High temperatures can also cause battery to degrade faster.  &lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-10-14T02:18:55.290" LastActivityDate="2019-10-14T02:18:55.290" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31506" PostTypeId="1" CreationDate="2019-10-13T18:08:16.303" Score="4" ViewCount="840" Body="&lt;p&gt;I need to calculate the required motor for a mobility scooter with the following user requirements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass of the scooter with passenger = 120kg&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Max Speed of scooter                   = 9Km/h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time from standstill to max speed      = 10s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;diameter of wheel is 0.203m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scooter can hande a slope of   5 degrees&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my working but it doesn's seem right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Force&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = ma +mg sin (5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = (120 x 0.25) + 2 x 9.8 sin (5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;F = 30 +1.7 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;F = 31.7 N&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Circumference of wheel&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 x PI x radius&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;= 0.63774 m&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;RPM&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RPM = Distance per hour travelled / circumference x60&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RPM = 9000 / 38.28 = &lt;strong&gt;&lt;em&gt;235.11 RPM&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Convert RPM to Rads&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;235 x 2x PI/ 60  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;= 24.61 rads&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Torque&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque = radius x force&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T = 0.105 x 31.7&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;T = 3.32 Nm&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Power  (mechanical)&lt;/strong&gt;&lt;br&gt;&#xA;= T x 24.61&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P(mech) = 79.2W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;= 0.079 kW&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An 80 W motor seems a bit low for a 120kG mobilty scooter to reach a top speed of 2.5 m/s with accleration of 0.25 m/s^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would somebody be able to verify if this motor size is correct or if not where the mistake is in the calculations. &lt;/p&gt;&#xA;" OwnerUserId="19967" OwnerDisplayName="daniel grisedale" LastEditorUserId="19967" LastEditDate="2019-10-31T21:34:31.473" LastActivityDate="2019-10-31T21:34:31.473" Title="Working out Motor torque" Tags="|motors|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="31507" PostTypeId="2" ParentId="31506" CreationDate="2019-10-13T19:37:35.923" Score="6" Body="&lt;p&gt;The biggest unknown you need to determine or guess is the coefficient of rolling friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For reference, a coefficient for rolling friction of 0.3 is already very high and is for something like for soft wheels (which deform) on a dirt road (which isn't flat or hard) where it is light enough it won't sink in. Most of the time it should be more like 0.1 to 0.2 with it being lowest on smooth hard surfaces with smooth hard wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A value of 0.3 means rolling on wheels on a horizontal surface takes 30% the force of just lifting it up. Ideally you want it to be zero. At &gt;1 it is easier to just pick the thing up rather than roll. Knowing this definition should help you get a intuitive feel for what the coefficient of friction should be. You will have to estimate or measure this most important value or conservatively guess a worst case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If measuring rolling friction (by pushing or pulling the scooter to determine the fraction of the weight that must be applied to slowly budge the scooter on a horizontal surface) be aware of drive train losses (i.e. gearbox friction) if it is present and linked to the wheels during the test  which can obfuscate the measurement. This should technically be included in drive efficiency term and not the rolling friction coefficient but may be easier to measure it along with rolling friction and just lump it all together as rolling friction. This decreases accuracy in inclined scenarios though where roling friction decreases but drive friction remains constant. If doing a lumped measurement and the gearbox was present but not the motor you can include motor efficiency separately using the drive efficiency term while leaving gearbox friction lumped in with rolling friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I laid everything out so you should only need to read it from top to bottom and look backwards for variables, never forward. I also tried to lay it out so hopefully you know where everything is coming from (as long as you have a basic understanding of power, torque, force, and friction...maybe even if you don't).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ m_{vehicle} = $&lt;/span&gt; mass of vehicle (kg)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ g = $&lt;/span&gt; acceleration of gravity &lt;span class=&quot;math-container&quot;&gt;$ =9.81m/s^2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ W_{vehicle} = $&lt;/span&gt; weight of vehicle (N) &lt;span class=&quot;math-container&quot;&gt;$ =m_{vehicle}\times g$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$v=$&lt;/span&gt; speed (m/s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mu_{roll}$&lt;/span&gt; = coefficient of rolling friction for wheels&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\theta=$&lt;/span&gt; angle of incline&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \eta = $&lt;/span&gt; drive efficiency (between 0 and 1 for 0% to 100%). Use 1 if you need output power (or in calculations of required output like torque). Use actual efficiency if you need input power&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F_{roll} =$&lt;/span&gt; force of rolling friction  &lt;span class=&quot;math-container&quot;&gt;$=W_{\perp vehicle} \times \mu_{roll}=W_{vehicle}cos(\theta)\times \mu_{roll}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ a = $&lt;/span&gt; desired acceleration &lt;span class=&quot;math-container&quot;&gt;$(m/s^2)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ F_{acceleration} = $&lt;/span&gt; ADDITIONAL force required to accelerate  &lt;span class=&quot;math-container&quot;&gt;$ =m_{vehicle} \times [a + gsin(\theta)] $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ r_{wheel} $&lt;/span&gt; = radius of driven wheel (m)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \tau_{roll} $&lt;/span&gt; = torque required to overcome rolling friction (i.e. to maintain constant speed)&lt;span class=&quot;math-container&quot;&gt;$ = F_{roll} \times r_{wheel}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \tau_{acceleration} = $&lt;/span&gt; ADDITIONAL torque required to accelerate &lt;span class=&quot;math-container&quot;&gt;$ =F_{acceleration} \times r_{wheel}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \tau_{total}= $&lt;/span&gt; total torque required to accelerate &lt;span class=&quot;math-container&quot;&gt;$=\tau_{roll}+\tau_{acceleration}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ P_{continuous} = $&lt;/span&gt; Continuous power to maintain speed &lt;span class=&quot;math-container&quot;&gt;$= F_{roll}   \times v \times \frac{1}{\eta}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ P_{peak} = $&lt;/span&gt; Peak power to accelerate &lt;span class=&quot;math-container&quot;&gt;$= [F_{roll} + F_{acceleration}] \times v \times \frac{1}{\eta}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Speed-dependent losses such as aerodynamic resistance or speed-dependent drive-train losses have been neglected.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="19778" OwnerDisplayName="DKNguyen" LastEditorUserId="19778" LastEditDate="2019-10-14T18:11:43.853" LastActivityDate="2019-10-14T18:11:43.853" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="31508" PostTypeId="1" CreationDate="2019-10-14T14:30:32.253" Score="1" ViewCount="80" Body="&lt;p&gt;A fluidic muscle is a pneumatic or hydraulic actuator, consisting of a flexible tube. When the tube is filled with a working fluid (pressurized air or hydraulic oil), it balloons out and shortens. In some applications, fluidic muscles are superior to pistons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In theory, you could imagine the same principle in reverse as a pump: Water flows into the tube under gravity, ballooning it out. The tube is then pulled apart, reducing its volume and ejecting the water. Check valves ensure proper water flow. Tbh, I don't see any advantage over other diaphragm pumps, but I could be wrong of course.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="10902" LastEditDate="2019-10-14T21:10:41.420" LastActivityDate="2023-12-31T17:51:06.680" Title="Fluidic muscle in reverse, as a pump. Does this exist?" Tags="|pumps|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31509" PostTypeId="1" AcceptedAnswerId="31510" CreationDate="2019-10-14T14:38:14.007" Score="3" ViewCount="333" Body="&lt;p&gt;I'm totally ignorant in this area, and probably didn't know how to properly google my question - sorry..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this specification here: (&lt;a href=&quot;https://si.shimano.com/pdfs/ev/EV-RH-IM10-2045B.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://si.shimano.com/pdfs/ev/EV-RH-IM10-2045B.pdf&lt;/a&gt;). My guess is that M10 is the about the metric thread specification (&lt;a href=&quot;https://en.wikipedia.org/wiki/ISO_metric_screw_thread&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/ISO_metric_screw_thread&lt;/a&gt;). But what is the meaning of &lt;code&gt;B.C. 3/8&quot;&lt;/code&gt; then?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question about this spec: (&lt;a href=&quot;https://si.shimano.com/pdfs/ev/EV-FH-2200-2202B.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://si.shimano.com/pdfs/ev/EV-FH-2200-2202B.pdf&lt;/a&gt;) - what would be the meaning of &lt;code&gt;6-11/32&quot;&lt;/code&gt; - it can't be a range, or can it?&lt;/p&gt;&#xA;" OwnerUserId="23147" LastActivityDate="2019-10-14T14:54:42.370" Title="What are the meanings of `B.C. 3/8&quot;` and `6-11/32&quot;`?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31510" PostTypeId="2" ParentId="31509" CreationDate="2019-10-14T14:54:42.370" Score="4" Body="&lt;p&gt;It looks like the hub axle has two different threads on it: M10 probably in the middle and 3/8&quot; either side of the middle section. Not sure what the &quot;B.C.&quot; refers to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F3XZ6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F3XZ6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is confirmed by the fact that the hub nuts (1), which go on the extremities are 3/8&quot;, whereas the lock nuts (18), which go on the middle section, are M10. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think &lt;code&gt;6-11/32&quot;&lt;/code&gt;means 6 inches and (or plus if you prefer) 11/32 of an inch. Indeed, if you do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;(6 + 11/32) * 25.4 = 161.77mm&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the spec says 161mm. You can apply the same logic for 16mm and 6-17/32:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8lN9S.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8lN9S.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-10-14T14:54:42.370" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31511" PostTypeId="1" CreationDate="2019-10-14T16:06:30.310" Score="1" ViewCount="128" Body="&lt;p&gt;I am trying to calculate what loading on a fixed roller will create a predetermined maximum deflection. I ran some calculations, but the values I'm getting are significantly lower than I would expect. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attached are the calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UOTQz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UOTQz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fZWZ5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fZWZ5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23148" LastEditorUserId="23148" LastEditDate="2019-10-15T13:28:08.213" LastActivityDate="2019-10-15T13:28:08.213" Title="How to calculate load on roller to create given deflection?" Tags="|mechanical-engineering|deflection|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31512" PostTypeId="1" CreationDate="2019-10-14T16:09:11.070" Score="1" ViewCount="56" Body="&lt;p&gt;I have a scenario where a valve q_in(t) is pumping water into a tank of height h(t) and area A, which is then pumped out by a second valve q_out(t) that has a resistance to water flow R.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I got this transfer function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \frac{H(s)}{Q_i(s)} = \frac{R}{A R s + 1} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where R= 1.5m(m^3/min) and A= 2m^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be grateful if someone could tell me how to get the value of Rs so I can plot the transfer function in Labview!&lt;/p&gt;&#xA;" OwnerUserId="23149" LastEditorUserId="23149" LastEditDate="2019-10-14T16:51:19.210" LastActivityDate="2019-10-14T16:51:19.210" Title="How to derive a value from a transfer function" Tags="|fluid-mechanics|control-engineering|transfer-function|labview|" AnswerCount="0" CommentCount="15" ContentLicense="CC BY-SA 4.0" />
  <row Id="31518" PostTypeId="1" CreationDate="2019-10-14T19:15:40.633" Score="0" ViewCount="114" Body="&lt;p&gt;How do I get the closed-loop transfer function of the system, fitted with a PID controller, if the input valve has a transfer function given by: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \frac{Q(s)}{P(s)} = \frac{Kv}{(Tv*S+I)} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where Kv is a constant of proportionality between the steady-state discharge and the pressure on the input valve. Tv is the time constant of the input valve. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure how I'm meant to use the input valve transfer function to work out the closed-loop system transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone help explain the steps?&lt;/p&gt;&#xA;" OwnerUserId="23149" LastActivityDate="2019-10-14T19:15:40.633" Title="How do I get the closed-loop transfer function of the system, fitted with a PID controller, if the input valve has a transfer function?" Tags="|control-engineering|pid-control|transfer-function|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31519" PostTypeId="1" CreationDate="2019-10-14T19:58:02.483" Score="0" ViewCount="19" Body="&lt;p&gt;I came across this problem a while ago and I'd like to solve it, but I'm not even sure where to start. How would I start this problem and what formulas do I need? I'm not asking for a solution, but that would be awesome. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Problem Statement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Gisj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Gisj.png&quot; alt=&quot;Problem Statement&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diagram&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LXtX3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LXtX3.jpg&quot; alt=&quot;Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23152" LastEditorUserId="8055" LastEditDate="2019-10-15T10:59:01.370" LastActivityDate="2019-10-15T10:59:01.370" Title="Problem with relative reference frames" Tags="|aerospace-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31520" PostTypeId="2" ParentId="31508" CreationDate="2019-10-14T20:02:47.617" Score="2" Body="&lt;p&gt;Every time you exhale, your ribs do a similar thing. They pull your rib cage in and down, making it a bit longer and narrower, pressuring the of air out of your lungs.&#xA;And then they relax and let the fresh air in. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="10902" LastEditDate="2019-10-14T21:11:09.607" LastActivityDate="2019-10-14T21:11:09.607" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31523" PostTypeId="1" CreationDate="2019-10-14T23:37:56.887" Score="0" ViewCount="106" Body="&lt;p&gt;So i have seen these super conductor rods around (which appear to be copper around titanium/niobium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am wondering (for a maker usecase) if its possible to create something similar by for example casting copper around rods of another material, i.e. iron, nickel, stainless steel, or even just titanium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the science behind this, is it possible to just cast 2 different metals together so they become fused?&lt;/p&gt;&#xA;" OwnerUserId="23154" LastActivityDate="2019-10-15T02:36:05.673" Title="Is it possible to cast copper around another metal" Tags="|materials|metallurgy|casting|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31524" PostTypeId="1" CreationDate="2019-10-15T01:38:33.730" Score="1" ViewCount="114" Body="&lt;p&gt;My RAM truck has a feature that intrigues me and I want to know how it works.  In auto-dimmer mode, the truck will automatically put on low beams headlights when there is on-coming traffic at night, as well as when it is behind another vehicle moving in the same direction at night.  However, it is smart enough to not go to low beams when my truck's headlights are reflecting back light off of signs.  My guess is that my headlights are putting out a unique digital signature, so when the sensor registers that digital signature from a reflection, it knows not to dim.  When the light is coming from another vehicle, it knows that that light is not its own, because that light does not have the digital signature.  Does anyone know?&lt;/p&gt;&#xA;" OwnerUserId="23156" LastActivityDate="2021-07-06T05:01:20.127" Title="Headlight auto-dimmer in newer cars" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31525" PostTypeId="2" ParentId="31523" CreationDate="2019-10-15T02:36:05.673" Score="1" Body="&lt;p&gt;Looking up superconducting cables on the net , I doubt very much that any casting is involved ( to produce the final cables). The figures show multiple concentric tubes , some tubes to carry the cooling fluid and other tubes are the conductors and copper and insulation... If it was necessary to combine the copper and conductor into a single tube ( in the bundle of tubes) , I expect the conductor wire would be put in the annulus between two copper tubes and cold drawn to make a relatively solid single tube... Titanium has unique properties and doesn't react well to being heated above 800 F; It absorbs N and O from the air which embrittle the Ti ; significant heating requires an atmosphere of argon or helium ( as used in welding Ti). &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-10-15T02:36:05.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31526" PostTypeId="1" AcceptedAnswerId="31528" CreationDate="2019-10-15T02:47:38.490" Score="0" ViewCount="317" Body="&lt;p&gt;This is a prework for a lab, I haven't seen things in the picture before and have no idea on how to transfer this to an equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pExay.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pExay.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23157" LastActivityDate="2021-07-06T10:06:20.560" Title="How to read the block diagram for the 2nd order integrator feedback system?" Tags="|signal|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31527" PostTypeId="2" ParentId="31524" CreationDate="2019-10-15T02:59:07.630" Score="1" Body="&lt;p&gt;It's a question of relative intensity. An oncoming headlight is beaming lots of photons into the light sensor in your truck, whereas the light reflected off of road signs (which might seem bright to your eyes) is a lot less intense. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bear in mind that the thermal time constant for a headlight filament is of order ~100's of milliseconds, which rules out digitally encoding the headlight output in a manner which would allow the sensor in your truck to tell the difference between a reflected beam and one originating from another vehicle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such techniques are however routinely used in the radars used in air-to-air missile targeting, where the search beam contains a time-stamped digital signature which gets compared to the return signal to verify it is &quot;real&quot; and not a spoofed version transmitted by the target plane which is intended to mislead the attacking plane's missiles. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-10-15T02:59:07.630" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31528" PostTypeId="2" ParentId="31526" CreationDate="2019-10-15T07:39:04.250" Score="1" Body="&lt;p&gt;The wavy symbols are integrators, which means that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dy}{dt} = x_1$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dx_1}{dt} = x_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the &quot;+&quot; sign gives you (the triangles are &quot;gains&quot;, which means multiply):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$x_2 = x(t) - a_0 y(t) - a_1 x_1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you combine all this together, you get your differential equation(s).&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-10-15T07:39:04.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31531" PostTypeId="1" AcceptedAnswerId="31532" CreationDate="2019-10-15T10:12:17.393" Score="1" ViewCount="790" Body="&lt;p&gt;What is the difference between perpendicular and transverse loads? I saw this on a forum for 3d printing, but I couldn't find any online explanations on what the difference between both measures of loading was.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NDysz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NDysz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19897" LastActivityDate="2019-10-16T21:08:26.783" Title="Difference between perpendicular and transverse loads?" Tags="|mechanical-engineering|3d-printing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31532" PostTypeId="2" ParentId="31531" CreationDate="2019-10-15T12:58:09.667" Score="1" Body="&lt;p&gt;3D Printed objects are made of layers, and are intrinsically weaker in one direction, as you can 'pull the layers apart'. In this case 'transverse' means 'in the vertical direction as built', and 'perpendicular' means 'in any of the horizontal directions as built'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph in question is with reference to infill patterns - these are generated in the slicer and can either be 2D (eg hexagonal tubes, the cross section is the same from one layer to the next) or 3D (eg cubes on their corners). This non-uniformity of the 2D patterns is why the patterns fail in compression (what the graph shows) differently according to the orientation. Imagine crushing a toilet roll tube from the top vs from the side.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2019-10-16T21:08:26.783" LastActivityDate="2019-10-16T21:08:26.783" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31533" PostTypeId="1" CreationDate="2019-10-15T14:10:40.107" Score="0" ViewCount="600" Body="&lt;p&gt;I am using a servo motor and controlling it via arduino. Servo motor is part of a control system and I have to use System Identification techniques in order to obtain the servo's transfer function. First step of the whole process is to collect input-output data of the servo's motion and use these data to perform system identification. I read and write data to the servo motor using the Hardware Package for Arduino of Matlab. My question is how to figure out the optimal test signals to enter as input to the servo due to the fact that servos are normally driven using PWM signals ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;: I have connected the servo with an Arduino and I currently control it via Matlab (I also can do it with Arduino IDE but I am generally using Matlab for my project). I have created a sine wave with maximum value 0.924 and minimum value 0.111 because this is the range at which my servo responds and changes its position. I am feeding these values at my servo and it performs an oscillation. Now what I want to acquire is the angle of the rotation of the shaft in degrees for each one of the input values or maybe the sine of the angle of rotation of the shaft. &lt;/p&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2019-10-18T11:41:09.817" LastActivityDate="2019-10-18T11:41:09.817" Title="How to obtain transfer function of servo?" Tags="|electrical-engineering|motors|matlab|transfer-function|servo|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31534" PostTypeId="1" CreationDate="2019-10-15T14:32:12.283" Score="1" ViewCount="100" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oNjGP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oNjGP.png&quot; alt=&quot;Problem set up&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to work out the angle subtended by the stretching of a conical spring and I have a solution but I'm not 100% sure about it so would appreciate some feedback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An Archimedes spiral spring is shown above with radii at various points. What I would like to come up with is an equation for the angle that each part of the spring gets stretched by as a function of the radius and the distance from rest &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. I know that for a normal spiral spring of constant radius &lt;span class=&quot;math-container&quot;&gt;$R_{ss}$&lt;/span&gt; if it is stretched by a small amount &lt;span class=&quot;math-container&quot;&gt;$dx$&lt;/span&gt; then it will subtend an angle &lt;span class=&quot;math-container&quot;&gt;$d\theta$&lt;/span&gt; therefore&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;dx=R_{ss}d\theta \\&#xA;\int_0^{x}dx=R_{ss}\int_0^\theta d\theta \\&#xA;\therefore \theta=\frac{x}{R_{ss}}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a spiral conical spring with radius &lt;span class=&quot;math-container&quot;&gt;$R_{cs}$&lt;/span&gt; that starts at &lt;span class=&quot;math-container&quot;&gt;$R_{cs0}$&lt;/span&gt; and grows to &lt;span class=&quot;math-container&quot;&gt;$R_{cs1}$&lt;/span&gt; what I did was integrate the radius:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;dx=dR_{cs}d\theta \\&#xA;\int_0^{x}dx=\int_{R_{cs0}}^{R_{cs1}} \int_0^\theta d\theta dR_{cs} \\&#xA;\therefore \theta=\frac{x}{R_{cs1}-R_{cs0}}&#xA;$$&lt;/span&gt;&#xA;It seems too simple though hence why I would like to check here what the thought is on the solution.&lt;/p&gt;&#xA;" OwnerUserId="18933" LastActivityDate="2023-02-27T06:08:36.070" Title="Arc subtended by stretching of a conical spiral spring" Tags="|springs|geometry|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31535" PostTypeId="1" CreationDate="2019-10-15T18:10:52.413" Score="0" ViewCount="158" Body="&lt;p&gt;The Euler-Bernoulli beam theory is based on the hypothesis that the beam deforms such that a straight line normal to the beam axis before deformation remains &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;inextensible&lt;/li&gt;&#xA;&lt;li&gt;straight&lt;/li&gt;&#xA;&lt;li&gt;normal to the beam axis&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;after deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find out the consequences of these assumption number 1 and 2 for displacement field. In the following discussion, &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; is the axial direction and &lt;span class=&quot;math-container&quot;&gt;$x_2$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$x_3$&lt;/span&gt; are in-plane directions. Similarly, &lt;span class=&quot;math-container&quot;&gt;$u_1$&lt;/span&gt; is the displacement along axial direction, and &lt;span class=&quot;math-container&quot;&gt;$u_2$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$u_3$&lt;/span&gt; are in-plane displacements. The coordinate system is placed at the centroid of the cross-section of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumption 1:&lt;/strong&gt; A straight line normal to the beam axis before deformation remains inextensible after deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a straight line &lt;span class=&quot;math-container&quot;&gt;$\overline{AB}$&lt;/span&gt; in the cross-section of the beam as shown in Fig. 1.&lt;a href=&quot;https://i.stack.imgur.com/eUkLD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eUkLD.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;. The length of the beam before deformation is given by&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$l_{AB}=\sqrt{(x_{2A} - x_{2B})^2 + (x_{3A} - x_{3B})^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, &lt;span class=&quot;math-container&quot;&gt;$x_{ij}$&lt;/span&gt; represent the ith-coordinate of jth point. For example, &lt;span class=&quot;math-container&quot;&gt;$x_{3B}$&lt;/span&gt; is displacement along &lt;span class=&quot;math-container&quot;&gt;$x_3$&lt;/span&gt; direction of point B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the length of line &lt;span class=&quot;math-container&quot;&gt;$\overline{AB}$&lt;/span&gt; after deformation is given by &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$l'_{AB}=\sqrt{(x_{2A} + u_{2A} - x_{2B} - u_{2B})^2 + (x_{3A} + u_{3A} - x_{3B} - u_{3B})^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the length of line &lt;span class=&quot;math-container&quot;&gt;$\overline{AB}$&lt;/span&gt; cannot change, i.e., &lt;span class=&quot;math-container&quot;&gt;$l_{AB} = l'_{AB}$&lt;/span&gt;, therefore, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sqrt{(x_{2A} - x_{2B})^2 + (x_{3A} - x_{3B})^2} = \sqrt{(x_{2A} + u_{2A} - x_{2B} - u_{2B})^2 + (x_{3A} + u_{3A} - x_{3B} - u_{3B})^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is possible if and only if &lt;span class=&quot;math-container&quot;&gt;$u_{2A} = u_{2B}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u_{3A} = u_{3B}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since, points A and B are arbitrary points on the cross-section, therefore, we can conclude that the &lt;span class=&quot;math-container&quot;&gt;$u_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u_3$&lt;/span&gt; displacement will be same for all points on the cross-section of the beam. In other words, &lt;span class=&quot;math-container&quot;&gt;$u_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u_3$&lt;/span&gt; are functions of &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$u_2(x_1, x_2, x_3) = u_2(x_1)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$u_3(x_1, x_2, x_3) = u_3(x_1)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assumption 2:&lt;/strong&gt; A straight line normal to the beam axis before deformation remains straight after deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is possible if and only if the line translate as rigid-body along axis &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; and rotate as a rigid-body about axes &lt;span class=&quot;math-container&quot;&gt;$x_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$x_3$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rigid-body translation: If the line moves as a rigid-body along &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt;, its displacement along &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; is given by &lt;span class=&quot;math-container&quot;&gt;$\overline{u_1}(x_1)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rigid-body rotation about &lt;span class=&quot;math-container&quot;&gt;$x_2$&lt;/span&gt;: If the line rotates as a rigid-body about &lt;span class=&quot;math-container&quot;&gt;$x_2$&lt;/span&gt;, the displacement of any point on the line along &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; is given by &lt;span class=&quot;math-container&quot;&gt;$x_3  \phi_2(x_1)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\phi_2(x_1)$&lt;/span&gt; is the angle of rotation about &lt;span class=&quot;math-container&quot;&gt;$x_2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rigid-body rotation about &lt;span class=&quot;math-container&quot;&gt;$x_3$&lt;/span&gt;: If the line rotates as a rigid-body about &lt;span class=&quot;math-container&quot;&gt;$x_3$&lt;/span&gt;, the displacement of any point on the line along &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; is given by &lt;span class=&quot;math-container&quot;&gt;$-x_2 \phi_3(x_1)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\phi_3(x_1)$&lt;/span&gt; is the angle of rotation about &lt;span class=&quot;math-container&quot;&gt;$x_3$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the total displacement along axis &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; is given by&#xA;&lt;span class=&quot;math-container&quot;&gt;$$u(x_1, x_2, x_3) = \overline{u_1}(x_1) + x_3 \phi_2(x_1) - x_2 \phi(x_1)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Owing to the reasoning mentioned above for both assumptions, are the conclusions made sensible and correct?&lt;/p&gt;&#xA;" OwnerUserId="17712" LastEditorUserId="17712" LastEditDate="2019-10-16T07:09:05.750" LastActivityDate="2021-03-09T13:05:00.593" Title="Consequences of Inextensibility of an in-plane line in Euler-Bernoulli Beam Theory" Tags="|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31536" PostTypeId="2" ParentId="31534" CreationDate="2019-10-15T20:41:48.330" Score="1" Body="&lt;p&gt;No, Its not right. If the cone becomes more, steep at the limit when &lt;span class=&quot;math-container&quot;&gt;$$   R_{cs1} \Longrightarrow R_{cs0} \quad \theta \Longrightarrow \infty $$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-15T20:41:48.330" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31537" PostTypeId="1" AcceptedAnswerId="31538" CreationDate="2019-10-15T20:49:59.717" Score="0" ViewCount="55" Body="&lt;p&gt;My question essentially is how much water to power green house LED lights. Specifics bellow :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Say I have a warehouse with a flat roof with a bunch of water tanks or the whole thing being a water tank. Using say 30 feet High and 20'Lx20'W. How much power could I generate utilizing the roof to store water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One lighting fixture could run 400 watts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was pondering these last few days on this, I've googled a couple of equations but not sure I used the right ones. Thanks for your input in advance.&lt;/p&gt;&#xA;" OwnerUserId="23167" LastActivityDate="2019-10-16T00:19:08.187" Title="How much rain water run off do I need to power an LED greenhouse" Tags="|electrical-engineering|electrical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31538" PostTypeId="2" ParentId="31537" CreationDate="2019-10-16T00:19:08.187" Score="2" Body="&lt;p&gt;Not very much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One liter of water has a mass of about 1kg.  The acceleration due to gravity is 9.8m/s^2.  One Joule is one Newton exerted over one meter.  So one liter of water falling one meter loses 9.8 Joules of potential energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your roof is 9 meters high, and has an area of 36 meters (6m * 6m).  If you cover that with a meter of water that's 36 cubic meters, or &lt;span class=&quot;math-container&quot;&gt;$36 \cdot 10^3$&lt;/span&gt; liters.  The average drop is 9.5 meters, so you get &lt;span class=&quot;math-container&quot;&gt;$w = (9.8 \mathrm{m/s^2})(9.5 \mathrm{m})(36 \cdot 10^3 \mathrm{l})(1 \mathrm{kg/l}) = 3.35 \mathrm{M J}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which sounds pretty impressive, until you do just a bit more math and discover that's 0.93 kilowatt-hours, or enough power for your 400W panel for about two hours, twenty minutes -- and that's assuming 100% energy conversion, where less than 50% is probably more realistic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's some &lt;span class=&quot;math-container&quot;&gt;$x^2$&lt;/span&gt; terms in there, but you're talking about needing a water tank that's taller than your greenhouse, and enough rain to keep it filled.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-10-16T00:19:08.187" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31540" PostTypeId="2" ParentId="31535" CreationDate="2019-10-16T11:50:04.303" Score="1" Body="&lt;p&gt;You are misunderstanding what the assumptions are.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only lines which are assumed to remain straight and normal to the axis (but not inextensible) are those perpendicular to the plane of neutral bending of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the beam bends in two planes, since the theory assumes small displacements and linear behaviour you can combine the deflections in each principal plane by superposition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deformed shape of the section changes because of Poisson's ratio for the material. Since the beam has tensile axial strain on one side of the neutral axis and compressive axial strain on the other side, it should be fairly obvious that Poisson's ratio will deform the cross section into some sort of trapezium shape. In fact the top and bottom faces end up curved, as in Figure 7.9 on page 12 &lt;a href=&quot;http://ocw.snu.ac.kr/sites/default/files/NOTE/Chapter%207%20-%20Solid%20mechanics_7.1-5.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;real&lt;/em&gt; assumption of E-B beam theory is ignoring the strain energy caused by shear strain deformations (the statements about straight lines used to derive the equations are just consequences of that  assumption) and therefore the deformation of the cross section is irrelevant for the theory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E-B beam theory doesn't answer the question &quot;what shape is the deformed cross section of the beam&quot; because it that is irrelevant to the bending along the length of the beam &lt;em&gt;if the assumptions made by the theory are valid,&lt;/em&gt; and for short and deep beams they are not valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a beam theory that  &lt;em&gt;does&lt;/em&gt; include shear effects, the cross section can deform, it does not remain perpendicular to the neutral plane of the beam, and in fact it does not even remain plane.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-10-16T11:50:04.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31541" PostTypeId="1" CreationDate="2019-10-16T12:05:53.187" Score="0" ViewCount="495" Body="&lt;p&gt;Most of finite element calculations are performed with an orthogonal mesh (Cartesian coordinate system). From this calculation, we get two values of required steel area per linear meter: &lt;span class=&quot;math-container&quot;&gt;$Ax, Ay$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$cm^2/m$&lt;/span&gt;) for each direction. However, my slab is circular, so the steel reinforcement have to be put in orthoradial direction. How do I convert the reinforcement from orthogonal to orthoradial?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="23173" LastActivityDate="2021-03-10T02:56:08.633" Title="Orthogonal to Orthoradial reinforcement" Tags="|structural-engineering|structural-analysis|steel|reinforcement|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31543" PostTypeId="1" AcceptedAnswerId="31544" CreationDate="2019-10-16T17:57:16.913" Score="2" ViewCount="234" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/PtbYw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how this is made and I don't know the name of the method to search for it to learn more.&lt;/p&gt;&#xA;" OwnerUserId="23177" LastActivityDate="2020-12-22T22:05:11.743" Title="What is the manufacturing method used to make aluminium C channel" Tags="|manufacturing-engineering|aluminum|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31544" PostTypeId="2" ParentId="31543" CreationDate="2019-10-16T18:50:34.063" Score="4" Body="&lt;p&gt;Aluminum sections are extruded from billets heated to 800-925F under high pressure and then pushed through a die.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the entire process temperature is carefully controlled, because depending on the aluminum alloy and performance expected heat stages at the beginning and exit out of die are crucial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is the diagram of the extrusion machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FCLMc.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FCLMc.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-16T18:50:34.063" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31546" PostTypeId="1" CreationDate="2019-10-16T21:27:53.417" Score="0" ViewCount="53" Body="&lt;p&gt;I guess that generating movement is harder then keeping a wheel rolling, but in terms of power consumption is it possible to compare the power consumption of a motor for driving wheels to that of a circuit that charges a capacitor and powers a solenoid? Running on high voltage solenoids can be very fast and powerful, and also pneumatic cylinders can be very fast.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say we talk about a 5g ball that is shot at a speed of 10m/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the preferred motor be different for different weights and speeds? E.g. will an actuator be the choice for heavy and slow situation? Why?  &lt;/p&gt;&#xA;" OwnerUserId="15977" LastActivityDate="2019-10-17T03:50:07.860" Title="Why do pitching machines use wheels and not solenoids, actuators, servos or peumatic cylenders?" Tags="|electrical-engineering|motors|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31547" PostTypeId="2" ParentId="31541" CreationDate="2019-10-16T23:38:45.003" Score="2" Body="&lt;p&gt;The fact that design of concrete slabs or beams uses straight reinforcing bars running orthogonally doesn't have much to do with the geometry of slab, beam or a deck, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has to do with the fact that straight bars can take large amount of tensile stress and with their counterpart compressive concrete take large moments, without any tendency to peel their way out of the embedding concrete. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you have a round deck or slab, it's better and more efficient to design them with orthogonal strait bars. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some projects architectural design specifies circular horizontal concrete beams for esthetic reasons. In those beams torque and twisting stresses and effects of lateral deflection make the design and construction complicated. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-16T23:38:45.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31548" PostTypeId="1" AcceptedAnswerId="31550" CreationDate="2019-10-17T02:42:57.120" Score="0" ViewCount="152" Body="&lt;p&gt;In my class I'm in, our lecturer presented frequency response methods such as the nyquist plot and bode diagrams. Then when designing a lead compensator later, he said one of the specifications should be a steady state error to a step response below a certain value. Why do we consider step responses? I have been reading ogata's book and the first sentence in the frequency response chapter says that frequency response is specifically for the case of sinusoidal inputs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It doesn't make sense to me to analyse the frequency response over a large range of &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; either since a non repeating signal has an infinite frequency anyway? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If someone could explain this it would be helpful, I can't post my lecture slides here, but I hope I was clear enough.&lt;/p&gt;&#xA;" OwnerUserId="23162" LastActivityDate="2019-10-17T10:39:27.420" Title="Why do we analyse frequency response for step inputs?" Tags="|control-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31549" PostTypeId="2" ParentId="31546" CreationDate="2019-10-17T03:50:07.860" Score="2" Body="&lt;p&gt;Two reasons.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The wheels take advantage of release energy of squeezing and compressing the ball. So they don't have to accelerate the ball to its final velocity; it wants to jump out by itself.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The turning wheel acts as a flywheel, preserving the angular momentum, reducing power needed to a smaller average.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-17T03:50:07.860" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31550" PostTypeId="2" ParentId="31548" CreationDate="2019-10-17T04:18:01.287" Score="3" Body="&lt;p&gt;We consider step responses because that's often a close approximation to the real-world behavior that we're designing for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We consider frequency response because that makes the design easier -- doing design in the frequency domain is a nicely structured process that translates reasonably well to the time-domain world that we live in.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-10-17T04:18:01.287" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31552" PostTypeId="2" ParentId="31548" CreationDate="2019-10-17T10:39:27.420" Score="0" Body="&lt;p&gt;In linear analysis, any real input can be decomposed into a set of step inputs, and the response is just the superposition of the step responses.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-10-17T10:39:27.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31554" PostTypeId="1" CreationDate="2019-10-17T13:31:36.480" Score="0" ViewCount="47" Body="&lt;p&gt;I want to put a vertical antenna on top of my roof. The antenna is basically 4 meters of PVC pipe and 5 meters of 19mm aluminium tubing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The roof at my house is corrugated sheet metal, and I want to avoid drilling (and even touching the current screws as much as possible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a render of my idea:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Pg2F.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Pg2F.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mast itself will be, of course, held by ropes. Will double sided tape (3M VHB) be able to prevent the base from sliding around during installation? Will the tape hold up to thermal expansion of the corrugated sheet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main idea of this support is being able to pivot the antenna in case of very strong winds, or servicing it (it has a coil in the middle).&lt;/p&gt;&#xA;" OwnerUserId="23187" LastActivityDate="2019-10-17T15:15:45.687" Title="Mounting an antenna base on a roof using only double sided tape?" Tags="|metals|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31557" PostTypeId="2" ParentId="31554" CreationDate="2019-10-17T15:01:58.913" Score="1" Body="&lt;p&gt;This is really hard to tell. But from experience, it probably wont hold. First, double sided tape is nowhere strong enough to hold a shaft that long securely - at least in the long run. The glue used on tapes wears off depending on exposure to heat, humidity, constant movement and UV - on a roof probably all of that is present.&#xA;So In your place, I'd think about other concepts. If you don't want to touch the roof as little as possible, consider putting the antenna onto a remote foundation next to your house for example.&#xA;But it's really hard to tell without knowing more and the people here probaby won't be able to help this way. Sorry&lt;/p&gt;&#xA;" OwnerUserId="23189" LastActivityDate="2019-10-17T15:01:58.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31558" PostTypeId="2" ParentId="31554" CreationDate="2019-10-17T15:15:45.687" Score="0" Body="&lt;p&gt;3M's outdoor tape is rather strong. I have attached a number of sports badges to a motor vehicle, which was then subjected to about five years of periodic car washes and Florida hot summers and &quot;cold&quot; winters. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The surface was thoroughly cleaned before application, as should be the case for your project. One can expect the tape to hold in the manner you suggest, although the thermal variations may weaken it a bit, eventually causing it to fail. Consider as an alternative method to provide additional reinforcement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clean and apply tape to the surfaces involved, but also provide for a retention system surrounding the base. For example, using rectangular blocks (steel is best to match thermal characteristics) placed around the perimeter of the base, spaced a bit free of the base, would allow for thermal expansion and contraction, while providing security that the base won't go out-of-column from the down-guys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EjMpf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EjMpf.png&quot; alt=&quot;mast base image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could undercut the blocks in such a way as to create additional retention, but that may be overkill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's important to consider to periodically inspect the mount, perhaps six months apart to ensure that it remains in place. &lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-10-17T15:15:45.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31559" PostTypeId="1" AcceptedAnswerId="31561" CreationDate="2019-10-17T15:37:26.020" Score="1" ViewCount="137" Body="&lt;p&gt;Why don't engineers build highways with corners such those on NASCAR speedways?&lt;/p&gt;&#xA;" OwnerUserId="18760" LastActivityDate="2019-10-17T17:31:08.383" Title="NASCAR corners on public roads" Tags="|highway-engineering|roadway|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31560" PostTypeId="2" ParentId="31559" CreationDate="2019-10-17T17:01:19.710" Score="2" Body="&lt;p&gt;The banking on speedways are designed for speeds in excess of one hundred miles per hour (160 kmh) and cannot maintain a vehicle position at speeds below a certain level. For ordinary traffic speeds, the banking does exist in some areas. The curves I've experienced at a motor speedway are so steeply banked as to be impossible to stand or walk. According to a quick search, angles in the general range of 30° are used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Daytona Speedway web site:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0qGbI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0qGbI.jpg&quot; alt=&quot;speedway banking&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I spent my high school years in northern Vermont USA and recall cycling to school through a curve on a 60 mph (96 kmh) roadway. The outside edge of the roadway was at a minimum, ten feet (3 meters) higher than the inside edge, and might have been much higher. More local to me is a set of lanes, two in each direction separated by a median. The outside edge of each segment is also raised by a yard (meter) or more compared to the inside edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These bank angles are proportional to the expected speeds, while not so severe that slower traffic would be sliding down the face. One can expect traffic anywhere to be able to stop without risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Today's multi-lane highways will have similar banks, with a more pronounced angle, as one expects speeds to be consistently high. If you consider that you are experiencing a slight angular force when driving through these banked curves, you don't feel the banking that would otherwise be apparent if you stopped and stepped out of the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should the nineteen sixties prophecy of one hundred plus mph traffic come true, it would be necessary to have more banked curves, but also to have restrictions on vehicles permitted to operate on such roads.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-10-17T17:01:19.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31561" PostTypeId="2" ParentId="31559" CreationDate="2019-10-17T17:31:08.383" Score="4" Body="&lt;p&gt;Public roads must be safe to handle slow, heavy traffic and be safe to pull over a stalled car under a snow storm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They should be flat enough to make it easy for emergency services such as ambulances, firemen to set up a triage and evacuate or help injured people out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are standards for road curves and turns and maximum slopes both uphill and sideways, and signage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Provision for safety at night and black ice conditions and safe visibility around the turns at night should be put in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They should be sloped gently to allow for surface runoff water to drain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Highway design is a specialty in civil engineering. A safe highway will save hundreds of lives over time.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-17T17:31:08.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31562" PostTypeId="2" ParentId="31135" CreationDate="2019-10-17T18:32:19.687" Score="1" Body="&lt;p&gt;The thermal cracking model in Pavement ME Design prior to v2.3 only considered the predicted transverse cracks for the overlay layer and not the transverse cracks that reflected through the overlay layer that originated in the existing layer. The standard deviation equations which correspond to the transverse cracking model only, were removed from the rehabilitation design calibration file when the new reflection cracking model was implemented (v2.3). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, the software calculates the total transverse cracking which a function of the reflected transverse cracking from the existing pavement and the thermal/transverse cracking from the new overlay layer. Therefore, the standard deviation equation in the software is now based on the total thermal/transverse cracking value instead. &lt;/p&gt;&#xA;" OwnerUserId="20067" LastActivityDate="2019-10-17T18:32:19.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31563" PostTypeId="1" AcceptedAnswerId="31568" CreationDate="2019-10-17T21:11:13.317" Score="0" ViewCount="1411" Body="&lt;p&gt;What kind of professional/architect/engineer builds agricultural systems, specifically automated hydroponic systems? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Things such as growing, cultivation, lighting, plumbing, harvesting, processing and packaging systems, all in one facility, much like a factory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if I wanted to setup such a facility, who would I have to consult?&lt;/p&gt;&#xA;" OwnerUserId="21092" LastActivityDate="2022-05-16T11:50:17.427" Title="What type of engineer/professional designs hydroponic farms?" Tags="|mechanical-engineering|structural-engineering|control-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31564" PostTypeId="1" CreationDate="2019-10-17T21:17:55.633" Score="3" ViewCount="4939" Body="&lt;p&gt;I'm designing a plastic holder that will secure a metal part with snaps. Which plastic will be able to take multiple removals and re-insertions without getting weaker (bent out of shape) over time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT 1]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following is my progression as I attempt to get the answer. Some of it is me thinking out loud.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've referenced these two documents and only found the following information:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;BASF Snap Fit Design Manual&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://web.mit.edu/2.75/resources/random/Snap-Fit%20Design%20Manual.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://web.mit.edu/2.75/resources/random/Snap-Fit%20Design%20Manual.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;(Page I1)&lt;/em&gt; Although snap-fits can be designed with many materials,&#xA;  the ideal material is thermoplastic because of its high&#xA;  flexibility and its ability to be easily and inexpensively&#xA;  molded into complex geometries. Other advantages&#xA;  include its relatively high elongation, low coefficient of&#xA;  friction, and sufficient strength and rigidity to meet the&#xA;  requirements of most applications.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I'm aware that thermosplastics are better. But which ones?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;(Page III-1)&lt;/em&gt; Rigidity can be increased either by using a&#xA;  higher modulus material (E) or by increasing the cross&#xA;  sectional moment of inertia (I) of the beam.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok so I need to find a list of plastic materials with higher modulus to make it stiffer. But does that correlate with repetitive loading?&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;(Page III-2)&lt;/em&gt; However, as the beam deflection increases, the beam&#xA;  stress also increases. This will result in a failure if the beam&#xA;  stress is above the yield strength of the material... The calculated stress or strain value should&#xA;  be less than the yield strength or the yield strain of the&#xA;  material in order to prevent failure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So the ideal plastic needs a high yield strength.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;(Page VI-2)&lt;/em&gt; Fatigue, or repetitive loading, is the third major cause&#xA;  of failure. Fatigue concerns primarily apply if hundreds or&#xA;  thousands of cycles are anticipated. While the design&#xA;  stress level might be well within the strength of the&#xA;  material, the repeated application of this stress can&#xA;  result in fatigue failure at some point in the future. &lt;strong&gt;Some polymers perform better than others in this regard,&#xA;  making them ideal candidates for snap-fits or living hinges&#xA;  that must flex repeatedly.&lt;/strong&gt; The first way to avoid a fatigue&#xA;  failure is to choose a material known to perform well in&#xA;  fatigue. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What are these ideal candidates? &#xA;It continues with the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;This can be done by comparing the so-called S-N&#xA;  curves of the materials, which show the expected number&#xA;  of cycles to failure at various stress levels and at different&#xA;  temperatures of exposure. The second way, still using the&#xA;  S-N curves, is to choose a design stress level, at the&#xA;  correct temperature, that results in the required number of&#xA;  load applications prior to failure. This method will usually&#xA;  be conservative since S-N curves are typically generated at much higher&#xA;  frequencies than would be anticipated for repeated&#xA;  application of a snap-fit assembly.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ok so I need to find these curves and compare.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Page IV-4 also contains this table showing allowable strain value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/01EEL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/01EEL.jpg&quot; alt=&quot;Plastic allowable strain value table&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This table is unclear to me as it appears to mix 70% and 100% yield strain values. If I'm reading it correctly, if the PEI value is 70%, it would be the best. If the PEI value is 100%, Acetal would be the best. Either way I think the list of compared materials could be longer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Bayer Material Science Snap Fit Joints for Plastics&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://fab.cba.mit.edu/classes/S62.12/people/vernelle.noel/Plastic_Snap_fit_design.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://fab.cba.mit.edu/classes/S62.12/people/vernelle.noel/Plastic_Snap_fit_design.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;(Page 4)&lt;/em&gt; In view of their high level of flexibility,&#xA;  plastics are usually very suitable materials&#xA;  for this joining technique.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, plastics.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;em&gt;(Page 11)&lt;/em&gt; The permissible deflection y (permissible&#xA;  undercut) depends not only on the shape but&#xA;  also on the permissible strain E for the material&#xA;  used.&#xA;  In general, during a single, brief snap-fitting&#xA;  operation, partially crystalline materials may&#xA;  be stressed almost to the yield point, amorphous&#xA;  ones up to about 70% of the yield strain.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So assuming a partially crystalline and amorphous plastic have the same yield point, the partially crystalline one would be better because it can flex further.&lt;/p&gt;&#xA;" OwnerUserId="15383" LastEditorUserId="15383" LastEditDate="2019-10-17T23:33:51.240" LastActivityDate="2019-10-18T09:36:21.827" Title="What plastic material is best for a snap-fit design?" Tags="|plastic|deflection|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31565" PostTypeId="2" ParentId="23547" CreationDate="2019-10-17T23:14:33.000" Score="2" Body="&lt;p&gt;We use the left bracket.  It looks the best in our opinion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1/16&quot; = 1[16 in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3/8&quot; = 3[8 in.&lt;/p&gt;&#xA;" OwnerUserId="23190" LastActivityDate="2019-10-17T23:14:33.000" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31566" PostTypeId="2" ParentId="31563" CreationDate="2019-10-18T04:39:51.817" Score="0" Body="&lt;p&gt;Any engineer who deals with fluids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the flow rate is determined (a biologist / plant specialist would do this...) then all else can be designed around other parameters - space, cost, size, storage etc etc&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-10-18T04:39:51.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31567" PostTypeId="1" AcceptedAnswerId="31575" CreationDate="2019-10-18T06:28:36.360" Score="0" ViewCount="579" Body="&lt;p&gt;A friend of mine asked me this question, but I don't know if it is possible or not. If it is possible, please show me how to construct it. If it is not possible, please prove it. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="23191" LastActivityDate="2019-10-18T13:41:54.850" Title="Is is possible to form a 3D object from the front, side, top views with a triangle, a square and a circle respectively?" Tags="|technical-drawing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31568" PostTypeId="2" ParentId="31563" CreationDate="2019-10-18T07:18:27.100" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Someone with a background in agriculture, biology or biotechnology can design the basic process and determine the principle mass and energy flows - how much area, how much artificial lighting if any, how much water, air changes if aneclosed building, nutrient or fertilizer dosing etc.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The basic design stage should allready consider the logistics of your plant, but also things like workplace health and safety&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Civil and structural engineers should design the main vessels and the general plant layout&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There are a number of works you specified that would be their own mechanical engineering speciality: HVAC but mostly processing and packaging equipment. Others, like plumbing/pumping can probalby be done by most MEs with some experience in plant design&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Electrical and control systems are on or two additional engineers you need.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In theory, the process designer designs the process, then the civil engineers size and place the vessels in their drawings, then the ME figure out the plumbing and at last the MEs show up to design the electrical systems for the machinery. In actuality, you will iterate through this process or some steps of it a few times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is about a dozen specialities you need. However, sometimes you can buy expertise along with equipment, this migh be the case with processing or packaging equipment or dosing stations for nutrients. You still need someone on your team with a basic understating to talk to them, however!&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For example, if I wanted to setup such a facility, who would I have to consult?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You either find a consulting engineering firm with a lot of experience in designing and planning (and building!) such a plant. Or you find anexperienced hydroponics expert (the first bullet point) and a consulting firm with experience in a relevant field like wastewater.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternativly, you find a company specialized in building and selling hydroponics farms. This will give you less choice over the final plant - the company will want to work with their preferred components and concepts, and crucially they will want to reuse as much egnineering work from previous projects as they can.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2019-10-18T12:45:46.653" LastActivityDate="2019-10-18T12:45:46.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31570" PostTypeId="1" CreationDate="2019-10-18T09:16:53.390" Score="0" ViewCount="951" Body="&lt;p&gt;I want to read some data from an .odb file, but when I open it using the C++ API I always get this error message:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;file.odb is from a more recent release of Abaqus.&lt;/p&gt;&#xA;&lt;p&gt;The current Abaqus installation must be upgraded before this output database can be opened.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;I also tried a newer version of Abaqus, but I got the same error. Is there a way to get the version from an odb file so that I don't have to try all Abaqus versions?&lt;/p&gt;&#xA;&lt;p&gt;My Abaqus version is 6.14-1.&lt;/p&gt;&#xA;" OwnerUserId="23192" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-10-18T09:16:53.390" Title="How to get version from an abaqus output database file?" Tags="|finite-element-method|abaqus|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31571" PostTypeId="2" ParentId="31564" CreationDate="2019-10-18T09:36:21.827" Score="3" Body="&lt;p&gt;I have designed parts which are snap-fit using Polypropylene. They work well, but the best advice I can give make one (or a few) in one or more material and do some life testing to figure out which material performs the best. That's how we determined which material and geometry of the snap-fit design worked the best.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the expected lifetime of the machine and typical usage, estimate how many removal/re-insertion operations the part is expected to do in its lifetime and do some accelerated testing. You can typically get a a few years' worth of usage in a few days, using an automated/motorised mechanism to perform the insertion and removal operations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Environmental factors such as humidity and temperature may also affect the longevity of the part. It may also be accepetable that the part has a finite lifetime, shorter than that of the machine on which it's fitted, and that it's replaced on regular basis, e.g. every year for a service/PM visit. This is more a commercial decision.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2019-10-18T09:36:21.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31572" PostTypeId="1" CreationDate="2019-10-18T10:41:03.640" Score="1" ViewCount="181" Body="&lt;p&gt;I want to place a swing hooks in place before the concrete is poured so that , i dont have to drill a hole later. My contractor is not ready to make these bolts with TMT bars itself.. what kind of hooks or bolts should i buy for this purpose and how should it be installed. I expect the swing to handle around 1000lbs of weight.&lt;/p&gt;&#xA;" OwnerUserId="23194" LastActivityDate="2021-11-07T02:04:55.477" Title="installing swing hook before concrete is poured" Tags="|concrete|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31573" PostTypeId="1" AcceptedAnswerId="31574" CreationDate="2019-10-18T13:13:01.887" Score="0" ViewCount="67" Body="&lt;p&gt;I'd like to know the tensile strain beneath the asphalt concrete (AC) layer in my design structure. Usually, the strain is used to calculate the number of axles (Nf) that the pavement structure can take before failure. Do you know how/where I can find that?&lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-10-18T13:13:01.887" Title="Does Pavement ME Design output the tensile strain beneath an AC layer?" Tags="|pavement|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31574" PostTypeId="2" ParentId="31573" CreationDate="2019-10-18T13:13:01.887" Score="0" Body="&lt;p&gt;The tensile strain beneath the AC layer is not an output in Pavement ME Design due to the vast number of calculations performed within the Asphalt Pavement Design System (APADS) analysis module. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The size of the output file is too large to output (3 – 10gb or more). The reason for the large file size is due to the analysis being performed for each month, sub-month, sub-seasons, vertical locations, all axle configurations, axle load spectra’s for each axle configuration, and all traffic wander locations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The “.fat” intermediate output file summarizes the monthly fatigue damage for the vertical depth and wander locations. The maximum fatigue damage value between all wander locations for the last vertical depth location is used for the bottom-up fatigue cracking calculations.&lt;/p&gt;&#xA;" OwnerUserId="18841" LastActivityDate="2019-10-18T13:13:01.887" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31575" PostTypeId="2" ParentId="31567" CreationDate="2019-10-18T13:41:54.850" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/oNZYu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oNZYu.png&quot; alt=&quot;object&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Created in Fusion 360, this is a 3D object with a cross section of a circle when viewed from above, a triangle when viewed from one side and a square when viewed from another side 90 degrees from the first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ortho views below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/59Puj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/59Puj.png&quot; alt=&quot;circle cross section&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TXeoH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TXeoH.png&quot; alt=&quot;square cross section&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U9BT8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U9BT8.png&quot; alt=&quot;triangle cross section&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is thus: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;sketch and extrude a square, creating a cube. select a face/plane,&lt;/li&gt;&#xA;&lt;li&gt;sketch and extrude a triangle, boolean intersect the two solids&lt;/li&gt;&#xA;&lt;li&gt;select a face/plane, sketch and extrude a circle, boolean intersect.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;This is the sequence in creating an ambigram, a three dimensional object in which the appearance changes when viewed from different (orthogonal) angles. &lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-10-18T13:41:54.850" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31576" PostTypeId="1" CreationDate="2019-10-18T13:44:36.657" Score="0" ViewCount="52" Body="&lt;p&gt;On surveys and site plans, I've often found treelines. These look like the below image, and serve to delineate in a very approximate way the areas full of trees and those that aren't:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LEUrr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LEUrr.png&quot; alt=&quot;treeline dividing an area of trees from a grass/meadow area&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standard for where the line should be drawn, in terms of either tree trunks or canopies? For example, in the image below, should line A approximately run down the canopies of the trees closest to the outside? Should line B approximately run down the trunks of those trees?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BsivM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BsivM.png&quot; alt=&quot;treeline with curved side highlighted in red with letter A, and other side highlighted in blue with letter B&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that these lines are all very approximate and one should never read too much into them. However, before approximating something, no matter how roughly, I'd like to know what the target to be aiming at is. (And likewise for interpreting someone else's approximation.)&lt;/p&gt;&#xA;" OwnerUserId="21409" LastActivityDate="2019-11-17T18:02:45.343" Title="Is there a standard for where treelines are drawn?" Tags="|drafting|surveying|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31577" PostTypeId="2" ParentId="31576" CreationDate="2019-10-18T17:46:09.530" Score="1" Body="&lt;p&gt;It depends on what the survey scope is. In some cities where certain types of trees must be protected, such as Oaktree in Ventura county, the trunks are marked and assigned an ID and canopies delineated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are a required part of any development application documents, such as Lot subdivision, zone change, etc.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-18T17:46:09.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31579" PostTypeId="2" ParentId="31572" CreationDate="2019-10-18T19:42:35.673" Score="1" Body="&lt;p&gt;I don't want to plug in commercial products here, but you can search for 5/8&quot; hold-down anchor bolts. They can take 3500-4000lbs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They require the ear of the bolt to cross the top slab reinforcement, but they need a 13inch min embedment in concrete, not more than 45 degrees slant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to check with your engineer if the slab is strong enough to take the dynamic load of 1000lbs, which usually has a load factor of 2.8.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-18T19:42:35.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31580" PostTypeId="1" CreationDate="2019-10-18T21:31:04.370" Score="-1" ViewCount="94" Body="&lt;p&gt;I'm an engineer who needs to do FEA for the first time.  I need to model chemical diffusion in a liquid in a container with specific geometry.  For now fluid is assumed to be not flowing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What features do I need to look for in FE software to get this done?  Diffusion doesn't fall under fluid dynamics does it?  It's not thermal (though I think it may be mathematically similar) or structural either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specific software suggestions are also welcome.  I suspect my needs are basic so ease of use would be nice too.&lt;/p&gt;&#xA;" OwnerUserId="23197" LastActivityDate="2022-08-18T05:06:35.603" Title="Finite Element software for chemical diffusion in liquid" Tags="|finite-element-method|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31581" PostTypeId="1" AcceptedAnswerId="31582" CreationDate="2019-10-18T22:35:52.463" Score="1" ViewCount="39" Body="&lt;p&gt;Does the quality of a microphone affect the frequency and intensity(dB) of the sound it can pick up?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do all microphones pick up sound as it is in real life? &lt;/p&gt;&#xA;" OwnerUserId="23198" LastActivityDate="2019-10-19T01:35:17.847" Title="Does the quality of a microphone affect the frequency and intensity(dB) of the sound it can pick up?" Tags="|structural-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31582" PostTypeId="2" ParentId="31581" CreationDate="2019-10-18T23:08:05.347" Score="1" Body="&lt;p&gt;Yes, fidelity of sound reproduction is one of the characteristics that varies from microphone to microphone, and it's considered better to have better fidelity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, by extension, no, not all microphones are alike in this regard.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-10-18T23:08:05.347" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31583" PostTypeId="2" ParentId="31581" CreationDate="2019-10-19T01:35:17.847" Score="1" Body="&lt;p&gt;TimWescott is right: not all microphones are alike in their output level and frequency response. Microphones with good frequency response are expensive in rough proportion to their bandwidth. In the recording industry, it is common to choose a specific microphone brand and type for recording a specific instrument or voice because its response matches the output of the source. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-10-19T01:35:17.847" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31584" PostTypeId="2" ParentId="31580" CreationDate="2019-10-19T05:00:36.490" Score="0" Body="&lt;p&gt;Check out Phoenics, sold by Cham.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Seen models of smoke through air in a city and blood through heart valves so it is very powerful, but not cheap...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are others StarCD etc&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2019-10-19T16:18:25.693" LastActivityDate="2019-10-19T16:18:25.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31585" PostTypeId="1" AcceptedAnswerId="31703" CreationDate="2019-10-19T05:08:26.607" Score="1" ViewCount="106" Body="&lt;p&gt;LMTD of a counter flow heat exchanger is 20°C and cold fluid enters at 20°c &amp;amp; hot fluid at 100°C, mass flow rate of cold fluid is twice that of mass flow rate of hot fluid. &lt;span class=&quot;math-container&quot;&gt;$C_p$&lt;/span&gt; of hot fluid is twice that of cold fluid's. How to find the exit temp. of Cold fluid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did it in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its 40°C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;LMTD &amp;amp;= \dfrac{\Delta T_1 - \Delta T_2}{\ln\left(\dfrac{\Delta T_1}{\Delta T_2}\right)} \\&#xA;\Delta T_2 &amp;amp;= \Delta T_1 \\&#xA;\therefore LMTD &amp;amp;= \dfrac{0}{0}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let &lt;span class=&quot;math-container&quot;&gt;$\Delta T_2 = x\cdot \Delta T_1$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;LMTD &amp;amp;= \dfrac{\Delta T_1 - x\cdot \Delta T_1}{\ln\left(\dfrac{\Delta T_1}{x\cdot \Delta T_1}\right)} \\&#xA;&amp;amp;= \dfrac{(1-x)\Delta T_1}{\ln\left(\dfrac{1}{x}\right)} \\&#xA;&amp;amp;= \dfrac{(1-x)\Delta T_1}{-\ln x}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Applying L'Hopital Rule:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;LMTD &amp;amp;= \lim_{x\rightarrow1} x \cdot \Delta T_1 \\&#xA;&amp;amp;= \Delta T_1 = \Delta T_2&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20 = Cold outlet Temp - (Cold inlet Temp = 20)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cold outlet Temp = 40&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, The Output Temp. of Hot Fluid is 80°C&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this approach correct?&lt;/p&gt;&#xA;" OwnerUserId="14845" LastEditorUserId="1832" LastEditDate="2019-10-26T19:52:41.700" LastActivityDate="2019-10-29T13:58:33.727" Title="Heat Exchangers" Tags="|heat-transfer|heat-exchanger|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31587" PostTypeId="1" CreationDate="2019-10-19T06:54:41.527" Score="1" ViewCount="65" Body="&lt;p&gt;I'm pursuing Mechanical Engineering and i need to learn AUTOCAD, is there any site where i can download free video tutorials&lt;/p&gt;&#xA;" OwnerUserId="23200" LastActivityDate="2019-10-20T04:52:42.777" Title="Where can i download AutoCad Tutorials?" Tags="|mechanical-engineering|autocad|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31588" PostTypeId="1" CreationDate="2019-10-19T07:59:25.300" Score="1" ViewCount="202" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zt82y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zt82y.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like to know about occurrence of earthquake events when we say that earthquake of 0.5 g will occur in 50 years in excedence of.&lt;/p&gt;&#xA;" OwnerUserId="23201" LastEditorUserId="8055" LastEditDate="2019-10-30T17:21:03.643" LastActivityDate="2022-08-16T08:04:40.460" Title="what means Probablity of event excedence of so many years" Tags="|structural-engineering|civil-engineering|geotechnical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31589" PostTypeId="1" CreationDate="2019-10-19T08:45:13.223" Score="0" ViewCount="87" Body="&lt;p&gt;Cold worked Fe-Ni rod is very brittle and not suitable for drawing. What would be the best heat treatment parameters for increasing the ductility of a Fe-Ni Alloy? Does increase in the Co content make it brittle? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="23202" LastActivityDate="2020-11-12T17:05:50.623" Title="Ductility of Fe-Ni Alloy" Tags="|materials|heat-treatment|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31590" PostTypeId="2" ParentId="31589" CreationDate="2019-10-19T15:18:52.390" Score="1" Body="&lt;p&gt;Because of the total mutual solubility , Ni -Fe alloys are more ductile than about any other ferrous combination. Like any metal , if heavily cold worked ,ductility is reduced. If carbon is added, ductility can be drastically reduced . Generally Ni-Fe alloys are used for unique magnetic properties. Not much else can be said without more information.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-10-19T15:18:52.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31591" PostTypeId="1" CreationDate="2019-10-19T16:11:52.500" Score="2" ViewCount="943" Body="&lt;p&gt;I would like to determine the mechanical output torque and power of a small motor without using expensive strain guages and similar and have come up with the idea of determining the maximum mass that the motor shaft can lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With reference to the image, if I suspend a mass over a pulley and find the maximum mass that the output shaft can lift (and wind the string around its shaft in the process), can I calculate the torque as: T= mgr where m is the mass in Kg, g=9.8 m/s2 and r is the radius of the shaft in metres?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this is correct then for the setup shown, and assuming the motor could lift no more than 25g, the value of T= 0.025x9.8x0.002= 0.00049 Nm. Then if the shaft is spinning at 1000 rpm, Power= 0.49W&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;a href=&quot;https://i.stack.imgur.com/QlwBi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QlwBi.jpg&quot; alt=&quot;arrangement for lifting a mass&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23203" LastActivityDate="2022-06-06T02:09:44.337" Title="Measuring shaft torque of a small motor" Tags="|torque|power|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31592" PostTypeId="2" ParentId="31591" CreationDate="2019-10-19T16:16:29.463" Score="0" Body="&lt;p&gt;I would suggest mounting the motor on bearings through its axis and then use a torque arm with masses to bring it back to the balance point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how classic dynamometers are designed. Then they use a load cell instead of masses to get an electronic reading...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-10-19T16:16:29.463" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31593" PostTypeId="2" ParentId="31587" CreationDate="2019-10-19T16:29:07.963" Score="2" Body="&lt;p&gt;If you are willing to pay monthly (first month free) then there are AutoCAD tutorials on LinkedIn Learning&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't have to use AutoCAD then Fusion360 is a very popular and equally capable programme that engineers and designers use and its free to use for startups and hobbyists and presumably students. As its cloud based the online support and comprehensive tutorials are all available online.&lt;/p&gt;&#xA;" OwnerUserId="23203" LastActivityDate="2019-10-19T16:29:07.963" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="31594" PostTypeId="2" ParentId="31591" CreationDate="2019-10-19T18:47:32.943" Score="1" Body="&lt;p&gt;This is roughly okay. except you want to make sure there is either no acceleration or if there is it is counted for by just adding the term ma with no g to the torque of the pully,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$T =mgr +mar$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the speed slower and measuring it easier one can choose a 1mm shaft and say 50 gr mass next to a vertical tape measure.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-10-19T19:51:38.670" LastActivityDate="2019-10-19T19:51:38.670" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31595" PostTypeId="2" ParentId="31588" CreationDate="2019-10-19T19:19:44.103" Score="0" Body="&lt;p&gt;I think this is the annual flood chart. The USGS earthquake hazard and probability maps are colored and have contour lines.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Exceedance probability in Y years. (This part is mathematical)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The expected number, n of exceedances in Y years is n = Y times r, the&#xA;  annual rate of exceedance. Assumption: The rate of earthquake&#xA;  occurrence in time is governed by the Poisson Law. Application: Under&#xA;  the Poisson Law, if you expect over some period of time n occurrences&#xA;  of “something”, the probability of 0 occurrences is e–n. If the&#xA;  “something” is exceedance of some ground motion, the probability of&#xA;  getting an exceedance is 1 – P(0). So, one can work backwards to find&#xA;  the annual rate of exceedance corresponding to “the probability of&#xA;  exceedance is 5% in 50 years.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Quote from USGS, learn basics. &lt;a href=&quot;https://earthquake.usgs.gov/hazards/learn/basics.php#what&quot; rel=&quot;nofollow noreferrer&quot;&gt;USGS link.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9rDSs.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9rDSs.gif&quot; alt=&quot;US earthquake map&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-19T19:19:44.103" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31597" PostTypeId="2" ParentId="31587" CreationDate="2019-10-20T04:52:42.777" Score="0" Body="&lt;p&gt;You can find many free courses if you search for it on YouTube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is one example: &lt;a href=&quot;https://thesourcecad.com/autocad-tutorials/&quot; rel=&quot;nofollow noreferrer&quot;&gt; AutoCad tutorials&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-20T04:52:42.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31598" PostTypeId="1" CreationDate="2019-10-20T10:41:27.633" Score="0" ViewCount="21" Body="&lt;p&gt;Use case: industrial stop valve placed in an another pipe welded perpendicularly making a T-shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make the question more specific let's consider 50mm inner diameter pipe for the linear actuator inside with electric motor and some gear drive (most probably needed).&#xA;The achieved torque should be continuously sustainable through the life-time of 20,000 hours. Rotation speed and power are of secondary importance while length should be reasonable, let's say: 1000mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I considered planetary, cycloidal, harmonic drives, magnetic gears and various direct drive motors with an option to stack them on the same shaft. The top value I found is achievable with electric motor and planetary gearbox: ~40Nm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can it be significantly higher, say 100Nm and what is the limit?&lt;/p&gt;&#xA;" OwnerUserId="23193" LastEditorUserId="23193" LastEditDate="2019-10-20T17:54:03.923" LastActivityDate="2019-10-20T17:54:03.923" Title="What is the top limit for torque per diameter ratio for electrical rotation mechanism working in a tube?" Tags="|motors|gears|torque|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31599" PostTypeId="1" CreationDate="2019-10-20T13:48:43.693" Score="1" ViewCount="60" Body="&lt;p&gt;I need help regarding the lwip stack with tiva c, I am building UDP to serial converter, the function is working correctly, but I have some concerns:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;When I receive the UDP packet buffer, I directly echoed it back if UART  is busy sending the previous UDP, is this correct because I am not sure what is the type of the allocated buffer?&lt;/li&gt;&#xA;&lt;li&gt;When UART is not sending, I put the data on array and send it to the UART function, is there any other efficient way to perform this ? i.e using ring buffer or any other techniques?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;the code for udp callback function for receiving packets:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;void Udp_Receive(void *arg, struct udp_pcb *pcb, struct pbuf *p,&#xA;               struct ip_addr *addr, u16_t port)&#xA;{&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;uint8_t *pui8Data;&#xA;uint32_t ui32Idx;&#xA;&#xA;//&#xA;// If packet more than 128 bytes drop it&#xA;//&#xA;if((p-&amp;gt;len) &amp;gt; 128)&#xA;  {&#xA;    pbuf_free(p);&#xA;    return;&#xA;  }&#xA;&#xA;//&#xA;// check UART and if it is sending just echo the UDP&#xA;//&#xA;if(uart_status == 1 )&#xA;{&#xA;       //&#xA;       // Send the echoed message.&#xA;       //&#xA;       udp_sendto(pcb, p, addr, 1234);&#xA;       pbuf_free(p);&#xA;}&#xA;else&#xA;{&#xA;    //&#xA;    // Received Packet information and send it to serial&#xA;    //&#xA;    udp_length = (p-&amp;gt;len);&#xA;    pui8Data = p-&amp;gt;payload;&#xA;    udp_to_serial_counter++;&#xA;    //&#xA;    // clear the payload array to receive new one.&#xA;    //&#xA;    for(ui32Idx = 0; ui32Idx &amp;lt; udp_length ; ui32Idx++)&#xA;       {&#xA;        udp_payload[ui32Idx] = 0;&#xA;        }&#xA;    //&#xA;    // fill the payload from buf into the array&#xA;    //&#xA;    for(ui32Idx = 0; ui32Idx &amp;lt; udp_length ; ui32Idx++)&#xA;       {&#xA;        udp_payload[ui32Idx] = pui8Data[ui32Idx];&#xA;       }&#xA;&#xA;    pbuf_free(p);&#xA;     //&#xA;     // Allocate a new pbuf for sending the response.&#xA;     //&#xA;     p = pbuf_alloc(PBUF_TRANSPORT,udp_length, PBUF_RAM);&#xA;     if(p == NULL)&#xA;     {&#xA;         return;&#xA;     }&#xA;     //&#xA;     // Copy the the received  packet payload into  the udp_payload&#xA;     //&#xA;     p-&amp;gt;payload = udp_payload ;&#xA;     p-&amp;gt;len = udp_length ;&#xA;     //&#xA;     // Send the echoed message.&#xA;     //&#xA;     udp_sendto(pcb, p, addr, 1234);&#xA;     pbuf_free(p);&#xA;     set_uart_send_status();&#xA;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The uart send function is :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UARTSend( uint8_t *pui8Buffer, uint32_t ui32Count)&#xA;{&#xA;    // Loop while there are more characters to send.&#xA;       while(ui32Count--)&#xA;       {&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;     while(!(ROM_UARTSpaceAvail(UART6_BASE)));&#xA;&#xA;     ROM_UARTCharPutNonBlocking(UART6_BASE, *pui8Buffer++);  &#xA;&#xA;   }&#xA;          while((ROM_UARTBusy(UART6_BASE)));&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;}&lt;/p&gt;&#xA;" OwnerUserId="23208" LastEditorUserId="110" LastEditDate="2019-10-20T16:39:42.637" LastActivityDate="2019-10-20T16:39:42.637" Title="Ethernet to serial converter using TM4C129EXL" Tags="|embedded-systems|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31600" PostTypeId="1" CreationDate="2019-10-20T20:00:16.443" Score="2" ViewCount="236" Body="&lt;p&gt;What is the rational for these two different 6x6 truck chassis arrangements? Is it merely a matter of intended distribution of weight across the chassis while in use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As in, two axles up front to support a heavy engine, two axles in the rear for heavy cargo, or for towing a trailer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kFI7e.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kFI7e.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KOtlC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KOtlC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19347" LastEditorUserId="19347" LastEditDate="2019-10-20T20:17:45.447" LastActivityDate="2019-10-20T20:17:45.447" Title="What is the rationale for a 6x6 truck chassis with 2 axles front vs. 6x6 chassis with 2 axles rear?" Tags="|automotive-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31601" PostTypeId="1" CreationDate="2019-10-20T21:09:54.443" Score="0" ViewCount="598" Body="&lt;p&gt;I'm attempting to choose a pressure regulator(for use with an inert gas) for a personal project I am working on. I know that the inlet pressure will be 2000 psi and outlet pressure of 300 psi. I also need the mass flow rate to be approximately 1 g/s. I've done some calculations with the perfect gas law and got a volumetric flow rate of 415.943 SCCM based on the inlet pressure. The Cv's I've calculated thus far does not seem correct as they are very small numbers. I'm wondering how to properly choose a pressure regulator that can fit these parameters and how to calculate the Cv properly. Any help is great!&lt;/p&gt;&#xA;" OwnerUserId="23211" LastActivityDate="2020-03-04T16:03:15.333" Title="How to calculate flow coefficient to choose a pressure regulator" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31602" PostTypeId="1" AcceptedAnswerId="32027" CreationDate="2019-10-20T21:38:33.920" Score="0" ViewCount="202" Body="&lt;p&gt;I am a graduate student trying to sharpen the direct stiffness method skills. However, I am stuck at the beginning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is given in the image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me to write down all the nodal and member forces with equations ?&lt;a href=&quot;https://i.stack.imgur.com/T1Tro.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T1Tro.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15772" LastEditorUserId="15772" LastEditDate="2019-10-21T08:44:57.670" LastActivityDate="2019-11-21T08:51:03.830" Title="2D- Truss Member Analysis" Tags="|civil-engineering|structural-analysis|finite-element-method|homework|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31603" PostTypeId="2" ParentId="31602" CreationDate="2019-10-21T04:13:39.980" Score="0" Body="&lt;p&gt;Let's call the length of square truss members L then the diagonal angle is 45 degrees (let me know if the diagram is not to scale).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Sigma M_1= -LF_{4,2}+ LF_{4,3}=0 \quad F_{4,2}=F_{4,3} \ pointing\ left \\ F_{3}= -F \ pointing \ right \quad member_{ 3,4}= \ in\ tension\quad R_3= F \ pionting\ left$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Sigma F_{v1}=0 \quad F_{v1}= F\ pointing\ down \ R_{v1}=F\ pointing\ up $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  \Sigma F_{h1}=0 \quad F_{h1}= F\ pointing \ left \quad R_{h1}=F\ pointing\ right \\ member \ 1,4 \ in \ compression \\ F_{1,4}= \sqrt{2}*F $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Members 1,2 and 2,4 are zero stressed.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-10-21T15:35:54.103" LastActivityDate="2019-10-21T15:35:54.103" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31604" PostTypeId="1" CreationDate="2019-10-21T04:30:04.347" Score="0" ViewCount="85" Body="&lt;p&gt;This problem contains multiple parts, and I will write my initial attempts at solving the problem under it's corresponding section&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we're given that the vapor pressures (described in Atm) of both the liquid and solid phases are as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ln(p_sat) = A - B/T + Dln(T)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are also given a table of coefficients for the two cases of the Solid-Gas and the Liquid-Gas transition:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Situation, A, B(K), D;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solid-Gas, 19.25, 15773, -0.755;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Liquid-Gas, 21.79, 15246, -1.255]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(There are no units given, just those numbers)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we are expected to determine:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a.) The normal boiling point&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;My initial thought was to work with the given equation and solve it for T, assuming p_sat to be 1atm (the normal boiling point) but ran into obvious algebraic difficulty in isolating T, and without introducing the Lambert W function (which I feel is beyond the scope of an introductory thermodynamics course) I'm not entirely sure what the best method to proceed is.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;b.) The triple point temperature&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In this case, my initial thought is to find the temperature at which the equation as written for both of the situations. That is, if there is no particular affinity for either the Solid-Gas or the Liquid-Gas transition it makes sense to me that the corresponding temperature would be the triple point.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;c.) The difference between the heat capacities of the liquid and the solid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d.) The heat of fusion at the triple point&lt;/p&gt;&#xA;&#xA;&lt;p&gt;f.) The slope of the liquid-solid boundary line at the triple point given densities of liquid and solid phases to be 5.66 and 7.14g/cm^3 respectively and if the atomic weight of the metal is 65.4.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any assistance would be very much appreciated. Even just pointers towards other equations I might be able to bring in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best,&#xA;-Richard&lt;/p&gt;&#xA;" OwnerUserId="23212" LastActivityDate="2021-07-13T06:05:55.677" Title="Given a known relation between the saturation pressure and temperature of an unspecified metal, how can we determine the following properties?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31605" PostTypeId="1" CreationDate="2019-10-21T08:18:10.340" Score="1" ViewCount="90" Body="&lt;p&gt;What is the effect of annealing on the defects present in the sample? Does it remove defects in the sample? Is there any chance that due to annealing the point defects present in the sample combine to form a large size void?&lt;/p&gt;&#xA;" OwnerUserId="23213" LastActivityDate="2020-11-15T08:01:35.183" Title="Effect of annealing on the defects" Tags="|materials|metallurgy|metals|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31606" PostTypeId="1" CreationDate="2019-10-21T10:58:46.800" Score="0" ViewCount="37" Body="&lt;p&gt;I was looking for data on ice resistance for these kinds of 1000 litres containers. I believe they are made of polyethylene with an outside aluminium structure. Given the physics of ice expansion, what could be the safe level of water inside if subjected to icing conditions ?&#xA;Does adding a half inflated bike tube inside could allow relieving pressure from ice ? &lt;/p&gt;&#xA;" OwnerUserId="17592" LastActivityDate="2019-10-21T18:06:26.440" Title="1m3 water container and ice damage prevention" Tags="|structural-engineering|mechanical-failure|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31608" PostTypeId="1" CreationDate="2019-10-21T13:24:15.937" Score="0" ViewCount="35" Body="&lt;p&gt;I need to cut a slot 4mm wide by 25mm long in aluminum tube with a 2.2mm wall thickness and a 35mm O.D.  The tube is 250mm in length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seemed a drill press with a fluting bit might suffice for this &quot;one-off&quot; task, in combination with a sliding table. But I don't have a lot of experience with bench tools and was advised that a drill press is not really designed to handle lateral stresses and that I should get a milling machine instead. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether or not a drill press could handle this task, what features make a milling machine better able to do it? Do milling machines have a different kind of spindle bearing? Higher bit-rotation speeds? More rigidity overall and tighter tolerances?&lt;/p&gt;&#xA;" OwnerUserId="21526" LastActivityDate="2019-10-21T15:23:41.903" Title="what design features allow a milling machine to withstand lateral pressures against the bit" Tags="|mechanical-engineering|metals|tools|cutting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31609" PostTypeId="2" ParentId="31608" CreationDate="2019-10-21T15:23:41.903" Score="1" Body="&lt;p&gt;It's accurate to suggest that a drill press is not designed to handle lateral forces on the bearings and spindle. Contrary to that, you'll see plenty of posts to various locations made by people who have added a drum sander to their drill press. This is certainly a lateral load on the spindle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a one-off task, it's likely that you would not cause excessive runout on the bearings and spindle, especially if you keep your load at the right level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For aluminum, you'd want a medium to high speed on the spindle and to use a lubricant such as kerosene (paraffin?) although some use WD-40. This lubricant prevents the aluminum from &quot;gumming up&quot; the flutes, effectively damaging the work piece.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should you decide to attempt this, create a strong jig to hold your workpiece. There is little-to-no chance of success if you attempt it by hand. You'd want a set of rails aside the tubing with a clamping mechanism to prevent rotation of the tube. Those rails should be mounted in a suitable sliding device to provide the necessary travel. Consider also to drill a start and end hole to facilitate the beginning of the operation, although the end hole isn't as important.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During the cutting operation, you should see slivers of aluminum, a sort-of cresent shape. If you don't get a uniform collection of slivers and instead get small particles, increase your feed rate and/or reduce the spindle speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also consider to cast about for a mini-mill group (groups.io) or a local mini-mill owner to assist you. If I learned of a maker in my area with this straight-forward task, I would certainly jump in to make the cut.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-10-21T15:23:41.903" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31610" PostTypeId="2" ParentId="31606" CreationDate="2019-10-21T16:38:10.480" Score="0" Body="&lt;p&gt;If you leave roughly 10% on top empty it should be okay. Ice increases in volume by approximately 8%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Leaving a half deflated bike tube will hardly help. Because ice formation starts from surface of the tank and by the time it gets to the tube it is too late.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the plastic containers can flex a bit to accommodate icing if you leave the top empty. But they eventually have a certain number of cycles an will crack after that.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-21T16:38:10.480" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31611" PostTypeId="2" ParentId="31606" CreationDate="2019-10-21T18:06:26.440" Score="1" Body="&lt;p&gt;Ice has a volume about 10% bigger than water, so an inner tube is too small to do anything - unless it floats, see below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Containers burst when the ice forms a plug and the trapped water then freezes and expands. One way to avoid that is float a beach ball in the tank. When the top of the water freezes over, the pressure beneath it will crack the ice round the ball and let some of the water get on top of the ice, rather than being trapped underneath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may find that a plastic tank is flexible enough so that this happens around the edges anyway - but it does no harm to provide another way out for the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You might also consider how much latent heat would be released if the entire tank froze solid. It would need to be in extreme conditions for a long time for that to happen. Presumably, the purpose of the tank is to be repeatedly filled and emptied, not just stay full of water and slowly freeze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of these tanks are also sold with external insulating jackets - they are sometimes used for hot water storage, but the heat insulation works just as well in either direction.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-10-21T18:06:26.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31612" PostTypeId="1" CreationDate="2019-10-21T18:51:07.800" Score="1" ViewCount="314" Body="&lt;p&gt;I have a circulation tank that runs a piping loop out and back in with a motor pump on it to keep fluid constantly agitated in it.  I want to add an inlet to the tank, so I can fill it without pouring liquid into it.  The way the line is set up, I have an option to fit the inlet between the tank and the motor pump.  And, I will need a priming pump to get fluid to the normal motor pump that's inline with the circulation tank.  Will a priming pump push past the circulation pump if I leave the priming pump on (if both are on, if just the one is on)?  Will the circulation pump pull through the priming pump if the priming pump is off? I think the way this will work is I use the priming pump until the line is filled with new fluid then turn it off and use the circulation pump to fill the rest of the tank.  I am worried that if one of two pumps in series is off, flow will be stopped altogether.  Are there any other possible problems I should consider when running pumps in series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gy724.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gy724.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23217" LastEditorUserId="23217" LastEditDate="2019-10-22T20:38:23.577" LastActivityDate="2024-03-31T19:06:15.757" Title="Question About Pumps in Series" Tags="|mechanical-engineering|fluid-mechanics|pumps|fluid|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31613" PostTypeId="1" CreationDate="2019-10-22T00:30:17.147" Score="0" ViewCount="129" Body="&lt;p&gt;Say we have a fixed cylinder with a radius of &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;. Looking at any cross-section, the shear stress at a radius (&lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;) is given by &lt;span class=&quot;math-container&quot;&gt;$\tau = G \gamma $&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the particles at the outer radius (&lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;) twist (or elongate) more than those at any inner radius (&lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;) so it makes sense that they would have more stress. However, I was wondering why it is that the outer radius particles twist more than the inner in the first place if the same force is applied at all radiuses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make it more clear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture a large circle being twisted due to a shear force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;. Why is it that a circle with a smaller radius will twist less due to the same shear force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt;? What is the reason behind this?&lt;/p&gt;&#xA;" OwnerUserId="23219" LastEditorUserId="23219" LastEditDate="2019-10-22T13:55:52.227" LastActivityDate="2021-05-24T20:07:25.803" Title="Why are parts of a cylinder closer to centre elongated less during torsion?" Tags="|structural-engineering|civil-engineering|statics|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31614" PostTypeId="2" ParentId="31613" CreationDate="2019-10-22T01:03:04.353" Score="0" Body="&lt;p&gt;First, picture two disks, placed parallel, and close together.  Draw a line through the center of each disk.  Line up the lines.  Are all the points on the lines the same distance apart?  Now rotate one disk by a few degrees.  Are all points on the lines the same distance apart?  Have the lines drawn further apart at the center of the disks, or at the edges?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, consider your statement &quot;why it is that the outer radius particles twist more than the inner in the first place if the same force is applied at all radiuses&quot;.  Why do you believe this to be true?  &lt;strong&gt;Can&lt;/strong&gt; it be true in an isotropic material?&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-10-22T01:03:04.353" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31615" PostTypeId="2" ParentId="31613" CreationDate="2019-10-22T02:02:01.933" Score="0" Body="&lt;p&gt;In fact the twist angle is assumed to be constant from the center of the cylinder out to its circumference.&#xA;If we call the twist caused strain at any distance r from the center &lt;span class=&quot;math-container&quot;&gt;$ds, \ then\ \theta= \frac{ds}{r} \quad and \ ds=\theta*r$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shear is &lt;span class=&quot;math-container&quot;&gt;$t=G*\epsilon= G*\theta*r \quad \text{or y in your equation}= G*yr$&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-22T02:02:01.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31616" PostTypeId="2" ParentId="31604" CreationDate="2019-10-22T03:25:32.903" Score="1" Body="&lt;p&gt;In order:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) The normal boiling point is when the pressure of the liquid is 1 atm. Graph the equation &lt;span class=&quot;math-container&quot;&gt;$Y = \ln(p) - (RHS)$&lt;/span&gt; (right hand side) to find the temperature where &lt;span class=&quot;math-container&quot;&gt;$Y = 0$&lt;/span&gt;. Alternatively, graph &lt;span class=&quot;math-container&quot;&gt;$p_{lv}$&lt;/span&gt; versus &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; and find when &lt;span class=&quot;math-container&quot;&gt;$p_{lv} = 1$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) The triple point occurs when the temperature and pressure of the solid, liquid, and vapor are all equal. Create a graph with &lt;span class=&quot;math-container&quot;&gt;$p_{sv}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$p_{lv}$&lt;/span&gt; as a function of &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt;. They will cross at the triple point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c) I would first challenge whether the question is asking for the difference in specific enthalpies not specific heat capacities. Consider X as either solid or liquid. Vapor pressures of X-gas transitions are defined by &lt;a href=&quot;https://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Clausius-Clapeyron equation&lt;/a&gt;. An empirical form is also found using &lt;a href=&quot;https://en.wikipedia.org/wiki/Antoine_equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Antoine equation&lt;/a&gt;. The transition enthalpy can be determined from a Clausius-Clapeyron form of &lt;span class=&quot;math-container&quot;&gt;$\ln(p)$&lt;/span&gt; as shown &lt;a href=&quot;https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Phase_Transitions/Clausius-Clapeyron_Equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Alternatively, heat capacity differences appear in the first reference in this paragraph near the end for a discussion on the second derivative of the phase transition lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d) The transition enthalpies from above will depend on temperature. Find the value at the triple point temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e) The &lt;a href=&quot;https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Physical_Chemistry_(Fleming)/8%3A_Phase_Equilibrium/8.3%3A_The_Clapeyron_Equation&quot; rel=&quot;nofollow noreferrer&quot;&gt;Clapeyron equation&lt;/a&gt; applies for a solid-liquid transition. This question asks for the value for &lt;span class=&quot;math-container&quot;&gt;$dp/dT = \Delta_{fus}\bar{H}/(T_{fus}\Delta_{fus}\bar{V})$&lt;/span&gt;, where molar values are used for the fusion enthalpy and volume changes. Molar volume is the inverse of moles per volume, which itself is the (mass) density divided by molar mass. The important hint here is to always write a &lt;span class=&quot;math-container&quot;&gt;$\Delta$&lt;/span&gt; as FINAL - INITIAL. Density decreases going from liquid -&gt; solid (fusion), so molar volume increases. Fusion is endothermic, so the enthalpy is positive. Review the differences between the slopes of the &lt;span class=&quot;math-container&quot;&gt;$dp/dT$&lt;/span&gt; lines for fusion of water (where solid ice floats) and almost any other liquid (where its solid sinks).&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-10-22T03:25:32.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31617" PostTypeId="2" ParentId="31605" CreationDate="2019-10-22T06:27:21.630" Score="2" Body="&lt;p&gt;Annealing activates solid-state diffusion which allows interstitials to find and occupy point vacancies. It also tends to release dislocation pileups at grain boundaries by enabling the growth of new, stress-free grains, for which the activation energy is at least partly furnished by the strain energy contained in the pileups. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The atom movements within the lattice are not sufficient to eliminate macroscopic voids of the sort caused by manufacturing errors (slag inclusions, folding voids, etc.). However, in the case of &lt;em&gt;sintering&lt;/em&gt; metal powders, it is possible to produce metal parts which start out microscopically porous in the &quot;green state&quot; (as cold-pressed powder) and wind up ~99% fully dense. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-10-22T06:27:21.630" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31618" PostTypeId="1" AcceptedAnswerId="31619" CreationDate="2019-10-22T12:33:02.960" Score="0" ViewCount="295" Body="&lt;p&gt;I have a bar which I want to move as if it was pivoted on a point beyond its end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;--------    *&#xA;&#xA;&#xA;         /&#xA;        /&#xA;       /&#xA;      /    &#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Is there a linkage, or other mechanism, which can let a bar move as if it were pivoted at a point beyond its end (the asterisk), yet not extend as far as the (virtual) pivot point?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mechanism can extend to the left, above and below the bar, but cannot extend towards the (virtual) pivot point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I have already thought of using two arcs to constrain the ends of the bar, but that seems prone to jamming.)&lt;/p&gt;&#xA;" OwnerUserId="23221" LastActivityDate="2019-10-22T16:28:03.317" Title="Is there a linkage or other mechanism which can simulate a pivot, without extending that far?" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31619" PostTypeId="2" ParentId="31618" CreationDate="2019-10-22T16:28:03.317" Score="1" Body="&lt;p&gt;Many of the car hoods or trunks do exactly what you are about to do. They open using a trapezoid link and a spring loaded shock absorber. They do this to give more space than a simple hinge, so the mechanic can work under the hood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some garage doors use the same mechanism and you can buy the parts from hardware stores.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-22T16:28:03.317" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31620" PostTypeId="1" AcceptedAnswerId="31621" CreationDate="2019-10-22T16:52:48.690" Score="1" ViewCount="52" Body="&lt;p&gt;I've been doing some reading into spinning wheels and old-fashioned sewing machines and so far as I can tell, they use a system similar to a piston in an engine. The main difference being, the oscillation is coming from a treadle. It got me wondering about something though. With the way the linkage works, rotation to oscillation should work fine at any speed. That being said, couldn't oscillation to rotation be messed up if the oscillations are happening too slowly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the connecting rod doesn't pass the middle point (where the rod is angled at 0°), another oscillation might cause the connection rod to push backwards, reversing the rotation. Also, if the rod is angled at 0° (almost exactly), the linkage might jam because the oscillation isn't being directed one way or another? If that is the case, the crank/wheel would need to have enough momentum to pass that spot when the next oscillation happens?&lt;/p&gt;&#xA;" OwnerUserId="23226" LastActivityDate="2019-10-22T17:54:08.407" Title="Do crank-sliders require some minimum amount of momentum to work in linear-to-rotary convertions?" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31621" PostTypeId="2" ParentId="31620" CreationDate="2019-10-22T17:54:08.407" Score="2" Body="&lt;p&gt;I've actually sat at a treadle sewing machine and made seams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, if they're going too slow they stop.  They have flywheels that lower that speed, but it's still there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that with a treadle machine you rock the treadle, which can both push and pull on the connecting rod.  With some pedal-operated machines (such as the spinning wheels I've seen) you only push on the pedal, and the connecting rod only pulls.  That works on a spinning wheel because you generally want the wheel going at a constant and fairly rapid rate.  With a sewing machine, you need finer control of the speed, and you want to be able to start at (almost) all positions and to go slower for delicate work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a treadle sewing machine, when you're ready to sew you use the hand-wheel on the machine to get it started (and over the top), then use the treadle to continue.  Starting it by hand means that you can feel which way you need to rock the treadle to continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least for the electric sewing machines that were in use through the 1970's, you did the same thing: the motors are generally poor at going slowly, and if you're doing delicate work you want direct control over the thing anyway -- so you use the wheel to get things started right, then the motor once you're sure you're ready to go.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-10-22T17:54:08.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31622" PostTypeId="1" CreationDate="2019-10-22T23:52:49.670" Score="0" ViewCount="40" Body="&lt;p&gt;I want to increase the evaporation rate in a tank by blowing warm air over the surface of the water in the tank; this will (hopefully) agitate the water, increase the airflow in the tank, and maybe heat the tank slightly. How could I do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main goal here is not to heat the tank, but to agitate the water in the tank. The fan/device/whatever works has to be somewhat small, in addition, and it should blow without human assistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for any and all help! &lt;/p&gt;&#xA;" OwnerUserId="23228" LastEditorUserId="23228" LastEditDate="2019-10-23T00:02:19.203" LastActivityDate="2019-10-23T00:28:15.547" Title="How could I automatically agitate water by blowing warm air over its surface?" Tags="|airflow|evaporation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31623" PostTypeId="2" ParentId="31622" CreationDate="2019-10-23T00:28:15.547" Score="1" Body="&lt;p&gt;If your primary objective is to increase evaporation, you are more likely to decrease the temperature of the water rather than increase it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can accept the reduction of temperature, consider that many otherwise-stagnant ponds use fountains to provide for evaporation and agitation to prevent algae growth. If your tank is large enough, a small aquarium pump and nozzle could provide additional exposure to the air. Depending on the current surface area, a number of such nozzles would further increase the flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a mini-fountain is unacceptable, a &quot;dribble&quot; pipe around the perimeter of the tank or criss-crossing the surface above the tank will also provide for increased exposure and evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details added via edit may be useful.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-10-23T00:28:15.547" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31626" PostTypeId="2" ParentId="31612" CreationDate="2019-10-23T06:02:15.320" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Will a priming pump push past the circulation pump if I leave the priming pump on (if both are on, if just the one is on)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If the flow rates of both pumps are closely matched, the circulation pump will pump the medium delivered by the priming pump. If the circulation pump is turned off, the medium will be pumped into the vessel through the outlet - the pump will pose a far higher resistance than a short piece of pipe, unless there's a check valve somewhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the flow rate of the circulation pump is higher, you will pump the fresh medium to the top of the tank + some from the tank, if the flow rate of the circualtion pump is lower part of the fresh medium will enter the tank through the bottom outlet (unless there's a check valve)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Will the circulation pump pull through the priming pump if the priming pump is off? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Possible, but not likely, as the flow resistance through the priming pump will be far higher than through the outlet pipe.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I think the way this will work is I use the priming pump until the line is filled with new fluid then turn it off and use the circulation pump to fill the rest of the tank. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I find this dubious. The line will always be filled, there will be no vacuum. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I am worried that if one of two pumps in series is off, flow will be stopped altogether. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Unless there's something we miss that should be no problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other possible problems I should consider when running pumps in series?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, I would be concerned about total pressure. In this case with the two open vessels (or are these pressure vessels?) this should be no issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You should check the documentation of your pumps or talk to the vendor or service person about behavior when tunred off. Centrifugal pumps  allow some medium to pass when not pumping, how much or how much resistance they pose depends entirely on the pump. PD pumps &lt;em&gt;can&lt;/em&gt; block the flow as well as a valve, or be driven by the flow, this depends entirely on the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to add:&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Both pmps centrifugal, there's a switch valve that turns off the line from the outlet to the entrance where Pump 2 enters the line.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;We can't know for sure without seeing the performance curves of the pumps and a system curve. I would size Pump 2 so that head &lt;em&gt;at the inlet of pump 1&lt;/em&gt; is about the same total head when pump 1 is recirculating, for the flowrate that pump 1 recirculates. Then you should find a stable point of operation for both pumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If pump 2 is undersized, pump 1 will cavitate, if pump 2 is oversized you will oeprate off it's BEP, waste energy and have a shorter pump lifespan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, you could simplify the setup by putting the line from pump 2 somehwere on the pressure side of pump 1, and have a check valve or switch valve between. &lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2019-10-24T08:21:52.117" LastActivityDate="2019-10-24T08:21:52.117" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31627" PostTypeId="1" AcceptedAnswerId="31631" CreationDate="2019-10-23T07:05:52.390" Score="0" ViewCount="118" Body="&lt;p&gt;I am watching this video(&lt;a href=&quot;https://www.youtube.com/watch?v=gVLhrLTF878&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=gVLhrLTF878&lt;/a&gt;) and beginning at around the 2:20 mark, it explains the significance of an expansion valve with the following ideas:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Restricts refrigerant flow to lower refrigerant pressure.&lt;/li&gt;&#xA;&lt;li&gt;Decreasing surrounding pressure &lt;span class=&quot;math-container&quot;&gt;$\lt$&lt;/span&gt; liquid(refrigerant) pressure implies boiling liquid.&lt;/li&gt;&#xA;&lt;li&gt;Decreasing surrounding pressure around a liquid allows the liquid to evaporate. The evaporation takes some of the kinetic energy from the liquid which consequently lowers the temperature of the liquid.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have listed the ideas as to how I have understood it, but number 2 and 3 seem contradictory to me. Maybe temperature and pressure do not have a linear relationship? Please, anyone, clarify any misunderstandings I have.&lt;/p&gt;&#xA;" OwnerUserId="23231" LastActivityDate="2019-10-23T23:56:43.130" Title="What is the idea behind an expansion valve?" Tags="|pressure|compressed-air|refrigeration|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31630" PostTypeId="1" CreationDate="2019-10-23T15:06:41.387" Score="0" ViewCount="35" Body="&lt;p&gt;Hello I was searching net but can find information. I am doing a hobbyist project. I would like use shaft with metric triangular thread as a leadscrew. Can I do that? I know ACME will be loads better but they are not easy to get my hands on where I live. &#xA;Also if it helps I am trying to create very miniature milling machine, controled by hand and maybe with a lathe built-in. I would of course not use it for anything too precise. Just for milling out in aluminium and wood for artistic purposes. Or to make precision tabletop drill ( here it shouldn't interfere right ? ) &lt;/p&gt;&#xA;" OwnerUserId="23210" LastActivityDate="2019-10-23T15:06:41.387" Title="M thread as leadscrew" Tags="|mechanical-engineering|threads|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31631" PostTypeId="2" ParentId="31627" CreationDate="2019-10-23T18:09:13.507" Score="2" Body="&lt;p&gt;Have you ever held your hand in front of a spray of atomized liquid from a can of liquid sprayer such as an air freshener?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's cold! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the molecules of liquid have consumed heat energy of the same liquid to pick up acceleration. They convert pressure and heat to the kinetic energy of the streaming out spray molecules. The same thing happens in an expansion valve.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-23T18:09:13.507" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31633" PostTypeId="1" CreationDate="2019-10-23T23:00:35.550" Score="0" ViewCount="20" Body="&lt;p&gt;I am trying to install a door motion sensor in a environment that will get very hot for our work. So the door could heat up to temperature of 150-200°F. We built the frames and door to handle the expansion but what can we do about the motion sensor? In guess with temp that hot it will expand causing the alarm to constantly go off.&lt;/p&gt;&#xA;" OwnerUserId="23236" LastEditorUserId="110" LastEditDate="2019-10-24T00:56:37.173" LastActivityDate="2019-10-24T00:56:37.173" Title="Will expanding metal disrupt my motion sensor?" Tags="|sensors|metals|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31634" PostTypeId="1" AcceptedAnswerId="31636" CreationDate="2019-10-23T23:30:54.810" Score="0" ViewCount="79" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mnUcu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mnUcu.png&quot; alt=&quot;diagram&quot;&gt;&lt;/a&gt;&#xA;To calculate the moment of inertia at o, I tried to apply (1/3mq^2) but I am not sure if it is correct. Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="17811" LastActivityDate="2019-11-23T02:01:00.113" Title="Moment of Inertia" Tags="|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31635" PostTypeId="2" ParentId="31627" CreationDate="2019-10-23T23:56:43.130" Score="0" Body="&lt;p&gt;Both boiling and evaporation involve changing a liquid to a gaseous state. Both are driven by the liquid/vapor system not being in equilibrium as far as the actual vapor pressure vs. The vapor pressure associated with the saturated state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boiling means the conversion process is happening throughout the bulk of the liquid and bubbles are forming. Evaporation is the same process, but it is happening on the surface of the fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The refrigerant compressor pumps vapor. Let's assume it operates on a constant volume basis. The pumping rate is proportional to vapor density at the pump inlet. Cooling the vapor at constant pressure increases density and flow rate. Increasing the pressure at constant temperature increases density and flow rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the expansion valve, fluid is being pulled through because the compressor is removing vapor from the exit side of the evaporator coil. The fluid at the expansion valve inlet is liquid. By design, you don't want vapor there. The exit of the expansion valve is mixed phase. There is a big pressure drop across the valve. As soon as the fluid enters the low pressure side, some of it flashes to vapor. The refrigerant will be in saturated equilibrium at the low-side pressure. Most of it is still liquid, but it is bubbly. The remaining liquid gets vaporized in the evaporator and cools the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A TXV valve like in the video has two inputs. There is a knob that adjusts the pressure drop at a given sense temperature (only to be adjusted by a qualified tech), and a temp bulb that senses the evaporator vapor temperature and changes the size of the passage in the valve. The result is a complicated balancing act.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the valve closes a bit (because the bulb senses too low a temp in the vapor line), the pressure difference increases. After a bit, the low side of the valve (bubbly liquid line) will be cooler and lower pressure. And the flow rate will be less as well. The density of the vapor on the low side will be less. This is the key result. By design, closing the valve a little results in a lower vapor density at the compressor inlet, reducing flow rate, and ensuring that only vapor is getting sent to the compressor. The lower flow rate means the vapor line temp rises, providing the system's negative feedback. You can't show this by just looking at the valve - you have to look at how the entire system responds to the valve, and it is designed so that it responds this way when working properly and properly charged. It may not work this way if there are problems in the system. Lots of these valves get replaced when they aren't bad, but the tech didn't find the real problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The TXV valve effects a change to the refrigerant flow rate. It senses the vapor line temperature, and changes the flow rate to keep the vapor line temperature above the saturation temperature. Most are pressure compensating so that in effect, they do a decent job of maintaining a fixed superheat in the low side vapor line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compare with AXV, automatic expansion valve - &lt;a href=&quot;https://neilorme.com/AEV.shtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://neilorme.com/AEV.shtml&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-10-23T23:56:43.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31636" PostTypeId="2" ParentId="31634" CreationDate="2019-10-24T00:53:18.037" Score="2" Body="&lt;p&gt;It looks like the question assumes the rod to be of negligible mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the rotational moment of inertia is just &lt;span class=&quot;math-container&quot;&gt;$I=mq^2 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may have confused the moment of inertia of a rod of mass m with length q.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-10-24T01:06:09.913" LastActivityDate="2019-10-24T01:06:09.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31637" PostTypeId="1" CreationDate="2019-10-24T04:24:08.483" Score="2" ViewCount="801" Body="&lt;p&gt;When studying about the axial compressor, I came across the term 'diffusion' being used to explain the conversion of kinetic energy to pressure, also terms like 'diffusion factor'. What is being diffused in the compressor?&lt;/p&gt;&#xA;" OwnerUserId="16275" LastActivityDate="2019-10-24T05:09:04.137" Title="What is diffusion in the context of compressor?" Tags="|compressors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31638" PostTypeId="2" ParentId="31637" CreationDate="2019-10-24T05:09:04.137" Score="4" Body="&lt;p&gt;An axial compressor accelerates a flow of air to high speed. All the molecules in that parcel of air have that speed superimposed on their random vibrational motion. The idea of a &lt;em&gt;diffuser&lt;/em&gt; is to get the fast-moving air to pile up against itself, slowing it down (i.e., decreasing its kinetic energy) and causing its pressure to increase (by increasing the speed of its random molecular motion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, the kinetic energy in that &lt;em&gt;common&lt;/em&gt; motion performs compression work on the air as it travels through the &lt;em&gt;diffuser&lt;/em&gt; increasing the &lt;em&gt;random&lt;/em&gt; velocity of its molecules. In an ideal diffuser, all the kinetic energy given to the fast-moving air by the axial fan is transformed into compression work, increasing the pressure of the air.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-10-24T05:09:04.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31639" PostTypeId="1" CreationDate="2019-10-24T09:07:09.437" Score="0" ViewCount="23" Body="&lt;p&gt;I'm trying to find out the term used for the practice of interchanging/relocating/swapping 2 machines to compensate the wear and tear caused by the operation frequency differences between them. This is a practice used in machine plants where a bunch of similar machines are being utilized and where some machines get used more than the others. For example coin store laundries do swap their most used machines with the ones getting less attention because they are located at the back of these stores.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm told that there's a term for this practice.&lt;/p&gt;&#xA;" OwnerUserId="4839" LastActivityDate="2019-10-24T09:07:09.437" Title="What's the term for interchanging a frequently used machine with a seldomly used one?" Tags="|terminology|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31641" PostTypeId="1" CreationDate="2019-10-24T11:23:57.277" Score="-1" ViewCount="29" Body="&lt;p&gt;Maybe something cool actually exists. Something clever. I was actually thinking of an hydraulic ram but im not sure it's a good idea. Another thing i saw was a one-way check valve. Would this work?&lt;/p&gt;&#xA;" OwnerUserId="23233" LastActivityDate="2019-10-25T04:05:05.407" Title="I need a valve siphon or device to empty a boat that doesnt need a battery or power" Tags="|waste-water-treatment|siphon|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31642" PostTypeId="2" ParentId="31588" CreationDate="2019-10-24T13:01:55.317" Score="1" Body="&lt;p&gt;Probability of Event Exceedance of a certain period of years, lets say within N years; is the probability of occurrence of the X-year-flood event, in which the peak flow the gained from statistic for a certain country or area, will be exceeded.&lt;/p&gt;&#xA;&lt;p&gt;Which in this case is flooding. And the Annual maximum daily flow was recorded in that particular year of occurrence, it served as a basic for calculating the probability of the event occurrence in the future period of N year.&lt;/p&gt;&#xA;&lt;p&gt;In the topic of civil engineering particularly, as we discuss, we may have to refer to the past data to justify the design limit states for a certain project or building.&lt;/p&gt;&#xA;&lt;p&gt;Say for a housing estate construction, we calculated the surface runoff into the lower region displaced by the construction of the housing estate (concrete, roads, etc). Then in designing the diverting drainage, we have to justify what is the size required to contain the flow rate of the runoff.&lt;/p&gt;&#xA;&lt;p&gt;Depending on what type of the projects, we may need to include the probability for exceedance for a certain year. Say 100 years for residence building, or 1000 years for a dam.&lt;/p&gt;&#xA;&lt;p&gt;Anyhow, the statistic itself does not justify the design, but an engineer who do the design. For an instance, he or she may have to include that the possibility of climate change would drastically increase the maximum flow, then design it differently rather than just following the data blindly.&lt;/p&gt;&#xA;" OwnerUserId="23239" LastEditorUserId="23239" LastEditDate="2020-11-24T03:39:20.610" LastActivityDate="2020-11-24T03:39:20.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31644" PostTypeId="2" ParentId="31479" CreationDate="2019-10-24T13:08:54.530" Score="0" Body="&lt;p&gt;The answer to this is that, engineers have to come out with different idea / creative solution to increase the span of the slab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe there are many other solutions in other places that not mentioned in the following:&#xA;1.) Cold-form steel trusses&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Post/Pre-tensioning concrete structure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Arc shape structures support with tensioning rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All which to depends on the architectural requirement and technology availability.&lt;/p&gt;&#xA;" OwnerUserId="23239" LastActivityDate="2019-10-24T13:08:54.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31645" PostTypeId="2" ParentId="31493" CreationDate="2019-10-24T13:15:29.047" Score="0" Body="&lt;p&gt;By hand? I believe you have to use matrixes to include all the formulae by superposition, then to solve the matrixes using our friendly Mathematics by hand instead of computer. &lt;/p&gt;&#xA;" OwnerUserId="23239" LastActivityDate="2019-10-24T13:15:29.047" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31646" PostTypeId="2" ParentId="29786" CreationDate="2019-10-24T13:23:46.997" Score="1" Body="&lt;p&gt;Practically, the geologist will do boring at different location at the site, we call it Bore Hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on their preliminary assessment, they will decide the depth of boring. Or by experience or trial &amp;amp; error. Whatever it is, the main thing we are looking for is hard layer. For it to be hard rock or hard dense sand layer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically we can get a soil profile at each Bore Hole containing the information as follow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.) Water Table&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Type of soil&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Strength of sample taken&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) SPT: number of blow recorded during boring&lt;/p&gt;&#xA;" OwnerUserId="23239" LastActivityDate="2019-10-24T13:23:46.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31647" PostTypeId="1" AcceptedAnswerId="31655" CreationDate="2019-10-24T13:29:34.157" Score="0" ViewCount="72" Body="&lt;p&gt;I've been looking at ways to re-concentrate a formic acid-water binary mixture with the VLE diagram below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vaJE8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vaJE8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in saying that to concentrate a solution of say , 40% formic acid by weight I could (excluding all recycle streams for simplicity) implement a distillation column at 1 bar to produce a distillate at the azeotrope (78%) and then pass this distillate into a second column operating at 0.1 bar to distill further to 95%?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aditionally, would operating the 1bar column for a distillate of say only 70% (past the 0.1 bar azeotrope) then performing the remaining concentration in the low pressure column result in any significant energy savings due to less steam required at the lower pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, if I was to START with a solution of 85% formic acid, already beyond both azeotropes, I could just perform simple distillation in a single column at either pressure point to obtain the 95% product?&lt;/p&gt;&#xA;" OwnerUserId="10923" LastActivityDate="2019-10-28T17:01:35.990" Title="Azeotropic Distillation - Am I interpreting this VLE diagram correctly?" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31648" PostTypeId="2" ParentId="29721" CreationDate="2019-10-24T13:38:15.967" Score="1" Body="&lt;p&gt;Let me raise an example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Ultimate Capacity, my slab thickness required is 80mm for a particular span X meter. Which my slab would not fail to carry the load into the beam and column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By Serviceability Capacity, my slab thickness required is 150mm for a particular span x meter. Which the deflection may not exceed the desired displacement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, one hand I have a sagging slab which looks like it is going to fall anytime, even hit my head when I am standing right at the middle underneath. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand I have a rather smooth and flat slab which look nice and gives me confident that it wont fall on my head &amp;amp; still have some space above me when I am standing underneath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In short,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Limit State Design, designers have to access both ultimate capacity and serviceability of a structure to serve its purpose!&lt;/p&gt;&#xA;" OwnerUserId="23239" LastActivityDate="2019-10-24T13:38:15.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31649" PostTypeId="1" CreationDate="2019-10-24T14:02:53.947" Score="1" ViewCount="75" Body="&lt;p&gt;Electrical engineer here. I'm looking to build an incredibly simple mechanism that in my mind involves kind of a pulley. Below is a diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WuTs0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WuTs0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it's essentially like a mechanical servo. I would like to be able to pull something, and cause another body (attached to an arm) to rotate. However, I don't know what kind of mechanism or part I need. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to build this. And I'm talking real simple here - like household items (popsicle sticks!) or hardware store (though I do have access to a 3d printer). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas would be thoroughly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="23243" LastActivityDate="2019-10-24T15:12:04.607" Title="Help me build a mechanical gizmo involving a pulley (I think)" Tags="|torque|experimental-physics|wheels|pulleys|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31650" PostTypeId="2" ParentId="31649" CreationDate="2019-10-24T15:04:25.357" Score="2" Body="&lt;p&gt;The simplest way to achieve this in my mind, is with two gears - one should have a drum attached with the rope wound around, and the other has the arm attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One benefit of this method, is that you can have the gears sized such that pulling the string a long way causes the output arm to rotate more slowly than the input gear - this will give you more control than a simple lever. You can add more gears to the chain, or attach a drum that is much larger than the initial driving gear to give you even more advantage - let me know if this is something that you would like to explore and I'll add more details - I'm in a hurry at the moment, sorry!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g1dnm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g1dnm.png&quot; alt=&quot;gear diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can make gears out of popsicle sticks, but give you have access to a 3D printer, you'd be pretty crazy not to use it in my opinion!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lfpnt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lfpnt.png&quot; alt=&quot;Popsicle stick gears&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2019-10-24T15:12:04.607" LastActivityDate="2019-10-24T15:12:04.607" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31651" PostTypeId="1" CreationDate="2019-10-24T16:23:34.457" Score="1" ViewCount="189" Body="&lt;p&gt;Suppose I have a material with an elastic modulus of E&lt;sub&gt;x&lt;/sub&gt; in the x direction and an Elastic modulus of E&lt;sub&gt;y&lt;/sub&gt; in the y direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I find the effective elastic modulus in an arbitrary direction (say for an angle theta from the x axis)?&lt;/p&gt;&#xA;" OwnerUserId="23247" LastActivityDate="2019-10-24T21:00:38.013" Title="Elastic Modulus in Arbitrary Direction for Orthotropic Material" Tags="|mechanical-engineering|elastic-modulus|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31652" PostTypeId="1" CreationDate="2019-10-24T17:05:09.727" Score="0" ViewCount="27" Body="&lt;p&gt;I'm trying to design a freely spinning wheel that will &quot;lock&quot; into a few different configurations. I'm planning on placing a rare earth magnet near a certain point on the wheel, and inserting bits of ferrous metal into the wheel so it will stop at those locations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two questions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Any thoughts on the general viability of this design, and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How can I find ferrous metal bits? I tried searching on hardware sites for pins and small dowels, but they're usually of a fairly nonmagnetic steel alloy. Are there any specific search terms I could use, or types of items that are highly magnetic that I could repurpose? &lt;/p&gt;&#xA;" OwnerUserId="23248" LastActivityDate="2019-10-24T17:31:15.673" Title="Designing a &quot;ratcheting&quot; spinning wheel with ferrous metal bits" Tags="|materials|magnets|product-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31653" PostTypeId="2" ParentId="31652" CreationDate="2019-10-24T17:31:15.673" Score="3" Body="&lt;p&gt;Many ordinary bolts and nuts are of magnetic steel, rather than stainless. If you have no complications with the threaded portions, you can cut away the heads to have your pins/dowels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you require smooth shaft type pins, there are longer bolts with only the end portion threaded, leaving the longer shank smooth. Off with the heads and you're set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, use smaller bolts and size your magnetic part based on the head of the bolt. Even conventional wood screws are magnetic, if you don't buy brass or stainless steel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visit your local hardware store with a magnet and you'll attract plenty of bolts/screws/nuts and some of the latter might not be of the hardware variety!&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-10-24T17:31:15.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31654" PostTypeId="2" ParentId="31651" CreationDate="2019-10-24T20:50:09.027" Score="1" Body="&lt;p&gt;The elastic modulus can be shown to be:&#xA;&lt;span class=&quot;math-container&quot;&gt;$\frac{E_x*E_y}{\sqrt{E_x^2sin^2(\theta)+E_y^2cos^2(\theta)}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the derivation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\epsilon_x=\frac{\sigma cos(\theta)}{E_x}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\epsilon_y=\frac{\sigma sin(\theta)}{E_x}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\epsilon=\sqrt{\epsilon_x^2+\epsilon_y^2}=\sigma\sqrt{\frac{cos(\theta)}{E_x^2}+\frac{sin(\theta)}{E_y^2}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$E=\frac{\sigma}{\epsilon}=\frac{E_x*E_y}{\sqrt{E_x^2sin^2(\theta)+E_y^2cos^2(\theta)}}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="23247" LastEditorUserId="23247" LastEditDate="2019-10-24T21:00:38.013" LastActivityDate="2019-10-24T21:00:38.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31655" PostTypeId="2" ParentId="31647" CreationDate="2019-10-24T21:53:46.263" Score="2" Body="&lt;p&gt;OP Quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I correct in saying that to concentrate a solution of say , 40% formic acid by weight I could (excluding all recycle streams for simplicity) implement a distillation column at 1 bar to produce a distillate at the azeotrope (78%) and then pass this distillate into a second column operating at 0.1 bar to distill further to 95%?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer: Yes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OP Quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Additionally, would operating the 1bar column for a distillate of say only 70% (past the 0.1 bar azeotrope) then performing the remaining concentration in the low pressure column result in any significant energy savings due to less steam required at the lower pressure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer: Significant is relative. But if the feed solution for the 1 bar column is 70 wt% as compared to 40 wt% in the first paragraph, you will save energy in operating at the 70 wt% specification because you will have less stages to get to the azeotrope and likely less vapor to generate to meet formic acid production rate targets. The lower pressure column will work and use the same amount of energy whether the feed to the first column is 40 wt% or 70 wt% because the second column always has azeotrope condition feed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OP Quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lastly, if I was to START with a solution of 85% formic acid, already beyond both azeotropes, I could just perform simple distillation in a single column at either pressure point to obtain the 95% product?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Answer: Yes&lt;/p&gt;&#xA;" OwnerUserId="10344" LastEditorUserId="10344" LastEditDate="2019-10-28T17:01:35.990" LastActivityDate="2019-10-28T17:01:35.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31656" PostTypeId="1" AcceptedAnswerId="31665" CreationDate="2019-10-24T22:06:22.650" Score="0" ViewCount="2310" Body="&lt;p&gt;Why is it that a total condenser and partial reboiler are used in a distillation column and it is not the other way around?&lt;/p&gt;&#xA;" OwnerUserId="23251" LastActivityDate="2019-10-25T12:56:36.307" Title="On partial reboiler and total condenser in distillation column" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31657" PostTypeId="1" CreationDate="2019-10-25T02:46:51.707" Score="0" ViewCount="22" Body="&lt;p&gt;Is anyone familiar with how to create a &quot;shaking&quot; fluidized bed. With vibrational amplitude in the 2&quot; range and Hz 15-20?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm constructing a vibrational fluidized bed dryer for the purpose of drying shredded coconut. This machine, as far as I can tell, is generally a continuous process - however my team and I are trying to construct a batch type system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two 1/2HP vibrating motors running of a VFD, mounted the the rigid frame at the center of the machine, which is attached to a series of 6 springs - 110 lb/in, 7n, 6&quot; coils evenly spaced throughout the structure. We have had success translating the motion evenly in the 40-50Hz range to achieve a fluidized bed. But it's our understanding that the fluidization will be superior in a shaking system. However when we run tests the coconut consistently goes to a certain side, depending on the vibrational force settings and the exact Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to translate this motion to the purely vertical plane and keep this amplitude consistent while still using springs? Or is a different dampening mechanism needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note, there are also about 20 15 lb/in springs added evenley throughout the structure to stablilize vibrations, are these necessary?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="23255" LastActivityDate="2019-10-25T02:46:51.707" Title="Translation of vibrational motion in the 15-20 Hz range" Tags="|fluid-mechanics|vibration|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31658" PostTypeId="2" ParentId="31641" CreationDate="2019-10-25T04:05:05.407" Score="1" Body="&lt;p&gt;You can buy a mechanical pump that you work by hand for this purpose. It has a cylinder with a piston inside, a suction hose, and a discharge hose. You dip the suction hose into the bilge water, aim the discharge hose overboard, and push the handle on the piston back and forth like a bicycle pump. This is called a &lt;em&gt;manual bilge pump&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-10-25T04:05:05.407" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31659" PostTypeId="1" AcceptedAnswerId="31662" CreationDate="2019-10-25T04:54:17.300" Score="1" ViewCount="264" Body="&lt;p&gt;Most electrical transformers that I am aware of have either copper or aluminum cable wound around an iron core.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the iron core be replaced with a copper core? Would there be any performance differences?&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="3505" LastEditDate="2019-10-25T21:28:53.393" LastActivityDate="2023-10-30T11:59:42.157" Title="Can the core of electrical transformers be made solely from copper?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31661" PostTypeId="1" CreationDate="2019-10-25T08:01:36.750" Score="0" ViewCount="267" Body="&lt;p&gt;I'm reading my book on strength of materials and it says that the stress in a point depends on the orientation of the section that we are observing, and since there are infinite number of section that can go through a point, there can be infinite different stresses in that point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is what is the real stress in that point, that we use in out calculations? Is it the largest one? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because obviously the &quot;real&quot; stress (if there is one) can't depend on the section we decide to look at, so I'm confused.&lt;/p&gt;&#xA;" OwnerUserId="21424" LastEditorUserId="21424" LastEditDate="2019-10-25T08:42:02.870" LastActivityDate="2019-11-25T00:01:55.323" Title="What is the actual stress in a certain point in a body?" Tags="|mechanical-engineering|strength|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31662" PostTypeId="2" ParentId="31659" CreationDate="2019-10-25T10:02:05.000" Score="6" Body="&lt;p&gt;It would not be a good idea at all - a transformer is effectively a magnetic circuit, so you need a low reluctance path for the magnetic flux to flow through. Copper has a high magnetic reluctance, so would not be a good choice.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Magnetic_reluctance&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Magnetic_reluctance&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Permeability_(electromagnetism)#Values_for_some_common_materials&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Permeability_(electromagnetism)#Values_for_some_common_materials&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="43645" LastEditDate="2023-10-30T11:59:42.157" LastActivityDate="2023-10-30T11:59:42.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31664" PostTypeId="2" ParentId="31661" CreationDate="2019-10-25T12:02:27.157" Score="1" Body="&lt;p&gt;Let's start getting rid of the confusion by talking about something simpler than &quot;stress&quot;, i.e. &lt;em&gt;force&lt;/em&gt;, and keep it simple by talking about two dimensions, not three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &quot;real&quot; force on an object acts in some definite direction - for example the weight of an object acts vertically downwards. But to do any calculations with forces, we usually want to &lt;em&gt;describe&lt;/em&gt; them using two components at right angles to each other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you have an object is on an inclined plane and you want to consider how it moves, or whether it slips if there is friction, you often want to describe its weight using components &lt;em&gt;parallel and perpendicular to the plane&lt;/em&gt;, not horizontal and vertical. You probably already learned how to &quot;resolve&quot; forces in this way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;real force&quot; (the weight) doesn't depend on how we choose to describe it, but the &lt;em&gt;description&lt;/em&gt; (the two components) does depend on the direction we choose for the components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same idea applies to other quantities, such as velocity, acceleration, etc. There is a name for the general concept - we call the &quot;real thing&quot; a &lt;em&gt;vector&lt;/em&gt; and its &quot;description&quot; is &lt;em&gt;the components of the vector&lt;/em&gt; in some &lt;em&gt;coordinate system&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same idea applies to stress. The &quot;real stress&quot; at a point in a body is more complicated than a vector. The name for the general concept is a &quot;second-order tensor&quot; (&quot;vector&quot; is just another name for a &quot;first-order tensor&quot;) but in a strength of materials course you probably won't learn much about the general concept of a tensor, only some specific results about stresses and strains. In 3-dimensional space, a vector has 3 components, and a 2nd order tensor has 9 (9 = 3&lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt;3). However, stress and strain are both &quot;symmetric&quot; tensors, which means there are only 6 independent components, not 9. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The engineering way to describe those 6 components is to consider a small rectangular block of material around a point, and think about the tension and shear forces on each face of the block. If we divide those forces by the area of the faces, we get a description of the three &lt;em&gt;direct stress&lt;/em&gt; and three &lt;em&gt;shear stress&lt;/em&gt; components at that point. (In two dimensions, there are two direct stresses and one shear stress, not three of each - other components are irrelevant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just like the with vectors, the size of the &lt;em&gt;numbers&lt;/em&gt; representing the same &quot;real stress&quot; depends on the orientation of the little block. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;rules&quot; about how the numbers change when you change the orientation of the block are a bit more complicated than finding the components of a vector, and I guess that is what you are currently learning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a basic problem in teaching all this. If you already know linear algebra at undergraduate level, the general idea of how to work with tensors should be straightforward, but you want to start doing practical engineering calculation with &quot;stresses and strains&quot; long before that point, so students get introduced to complicated-looking formulas and ideas like Mohr's circle as some sort of &quot;magic,&quot; without the general understanding of what is going on and &quot;why all this stuff works the way it does&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a second complication, in that engineers sometimes use the same words that mathematicians use, but with different meanings. For example it is common to write the six components of stress or strain in a row or column, which &lt;em&gt;looks like&lt;/em&gt; a &quot;vector with 6 components&quot;, even though it doesn't &lt;em&gt;behave like&lt;/em&gt; a vector mathematically. And you can then write an equation linking stress and strain where the material properties data &lt;em&gt;looks like&lt;/em&gt; a 6&lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt;6 symmetric matrix which (for deep reasons, not just good luck) happens to have 21 independent entries, corresponding to the 21 independent components of the &quot;real&quot; fourth-order tensor. (In two dimensions the 6's become 3's). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That notation is very convenient in practice, but it completely obscures what is &quot;really going on&quot; in the underlying physics.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-10-25T12:17:24.890" LastActivityDate="2019-10-25T12:17:24.890" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31665" PostTypeId="2" ParentId="31656" CreationDate="2019-10-25T12:56:36.307" Score="1" Body="&lt;p&gt;Partial reboiler is related to that fact that most times, there is material left in the bottom of the column. This can be because you need to maintain a level of liquid because the bottom take off is a purified stream you want (as product or waste treatment input) or there is not enough area to maintain the proper amount of boil up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total condenser is related to the fact that most times you want to condense the purified upper stream as that is the product (and since liquids have lower specific volumes than gases, it makes storage and transport more efficient) or you want a liquid stream into a second column or other downstream unit operation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is not a law of the universe that the partial reboiler and total condenser are the only way distillation can be done, but it is very common in current industries and so it is often discussed.&lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2019-10-25T12:56:36.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31666" PostTypeId="1" AcceptedAnswerId="31672" CreationDate="2019-10-25T16:12:07.693" Score="1" ViewCount="2334" Body="&lt;p&gt;I am confused whether a lead screw and a worm screw are the same exact thing. In my engineering workshop lab, they come up as separate things. For example, a worm screw is there in the indexing head of the milling machine (as part of the worm drive), while the lead screw is part of the feed mechanism in the milling machine and the lathe machine. They both look like the exact same thing. Can someone tell me what's the difference?&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Leadscrew&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Leadscrew&lt;/a&gt;&#xA;&lt;a href=&quot;https://en.wikipedia.org/wiki/Worm_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Worm_drive&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23266" LastActivityDate="2019-10-25T20:08:35.170" Title="Difference between lead screw and worm screw?" Tags="|mechanical-engineering|terminology|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31667" PostTypeId="1" CreationDate="2019-10-25T16:20:47.797" Score="0" ViewCount="251" Body="&lt;p&gt;Hi I use this circuit to detect phase loose, so if one of the phase lost , the phase loss signal will be high, the problem in this circuit are.&#xA;1-if the input voltage less at 50% it give phase loss , i did try to increase the resistors (R99,R100 and R102) this make work but it cause another problem&#xA;is when i have phase loss and if input voltage more than 380Vac , the circuit does not detect phase loss becuase all optocoupler are on.&#xA;So how can i make this circuit work right like if i loss one phase , the phase loss signal be high and is not turn off at high input voltage and also how can avoid getting phase loss when the voltage at less than 50%&lt;a href=&quot;https://i.stack.imgur.com/qyNhs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qyNhs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8764" LastActivityDate="2019-10-25T16:20:47.797" Title="phase loss detection" Tags="|power-electronics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31669" PostTypeId="1" CreationDate="2019-10-25T18:00:34.637" Score="0" ViewCount="194" Body="&lt;p&gt;I want to focus a 300W LED into a small area, about 5mm in diameter, much like focusing the sun's rays with a fresnel lens.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have to be like 2 meters away from the LED to focus the LED light with a fresnel lens. Why is that? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to focus the all of the LED's light somehow onto a small area?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it's possible, would it be akin to having a 300W laser? Mostly likely not, but why? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance for your answers. &lt;/p&gt;&#xA;" OwnerUserId="23267" LastActivityDate="2019-12-01T05:05:06.423" Title="Can you focus a 300W LED to a small area?" Tags="|optics|lasers|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31670" PostTypeId="2" ParentId="31661" CreationDate="2019-10-25T18:00:39.443" Score="0" Body="&lt;p&gt;Your book is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's imagine you have a vertical rod with a mass m hanging from it. Let's call the vertical axis Z.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we are interested in stresses on a point Q on the rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are going to assume the rod is an isotropic material like steel, which is the case for the majority of cases, meaning it has only one uniform young modulus and a unique Poisson ratio along all the 3 axis. If no we have to handle it differently by continuum mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the stresses on point Q depend on the orientation of the plane we are interested in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are interested in the stress on Q on a plane perpendicular to Z it is obviously, sigma= mg/A, area of the rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we are interested in stress in any plane parallel to the Z the stress is zero.  And if we are interested in stresses on Q along other angles we use Mohr circle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why we can use a cable with no lateral resistance to do the lifting more efficiently. Because the lateral stress is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In situations where we have at a given section interaction between several stresses, we find the principal stresses using the Mohr circle or the equations of Mohr.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-10-25T23:33:00.303" LastActivityDate="2019-10-25T23:33:00.303" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31671" PostTypeId="2" ParentId="31666" CreationDate="2019-10-25T18:15:16.560" Score="4" Body="&lt;p&gt;The lead screw and the worm (in a worm drive) seem similar in shape, but are different in what they interface with. A lead screw turns rotary motion into linear motion. A worm drive turns rotary motion into rotary motions (with a gear ratio). &lt;/p&gt;&#xA;" OwnerUserId="9506" LastActivityDate="2019-10-25T18:15:16.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31672" PostTypeId="2" ParentId="31666" CreationDate="2019-10-25T20:08:35.170" Score="3" Body="&lt;p&gt;One way to think of it is this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Worm, it's a short &quot;threaded cylinder&quot; with relatively few teeth. This rotates and in doing so, moves a worm wheel with much more teeth than it. Typically this results in a rotating motion as @Eirc Shain suggested, but it could also be used to drive a long Rack (see 'Rack and Pinion') in a linear way. If the rack has more teeth, and it's engaged &quot;from the side&quot; and not wrapped around the threaded cylinder, then that cylinder is a worm. While it may look like a thread form, it's actually more correct to think of them as elongated teeth, with the number of thread starts being effectively the number of teeth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, a lead screw is much longer and with more thread turns than the item that engages with it (usually a nut or a half nut), and that engagement wraps around the cylinder. This always outputs a linear motion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this make things any clearer?&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-10-25T20:08:35.170" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31673" PostTypeId="2" ParentId="31669" CreationDate="2019-10-25T20:17:25.403" Score="1" Body="&lt;p&gt;If you just want a highly focussed light source, consider a halogen car headlight bulb. The bulb itself contains a parabolic reflector to give a focussed beam. The Fresnel lens in the front of the actual headlight acts as a diffuser, rather than to focus the beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used a benchtop-sized test rig where a couple of those bulbs (12V 55W each) can heat up a small object (1mm diameter) to about 1000C in a few minutes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The individual semiconductor structures in an LED tend to be point sources which produce an omnidirectional radiation pattern (except where the rest of the structure physically gets in the way and blocks the radiation), which probably isn't what you want.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-10-25T20:17:25.403" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31674" PostTypeId="1" AcceptedAnswerId="31686" CreationDate="2019-10-25T22:55:48.497" Score="1" ViewCount="64" Body="&lt;p&gt;For an application of displacement measurement under 1 kHz, I have 3 options to choose from. The application is automotive. Unfortunately I don't have deep electronics knowledge and the catalogues contain so many abbreviated expressions in the names of their connectors, etc.&lt;/p&gt;&#xA;&lt;p&gt;What table of comparison in parameters would you suggest to classify them as best and to also justify them with my needs in this frequency range?&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The temperature range is up to 1000°C.&lt;/li&gt;&#xA;&lt;li&gt;The movement range is ±50mm&lt;/li&gt;&#xA;&lt;li&gt;Frequency : automotive range but let's say 1 kHz..&#xA;bolting can be the solution due to the temperatures.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I also have seen a table like this, but I would like to widen it to more parameters :&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AQZMg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AQZMg.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17070" LastEditorUserId="8055" LastEditDate="2020-10-12T22:22:33.273" LastActivityDate="2020-10-12T22:22:33.273" Title="How to compare multiple data acquisition systems?" Tags="|mechanical-engineering|automotive-engineering|vibration|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31675" PostTypeId="1" CreationDate="2019-10-26T02:51:17.580" Score="0" ViewCount="27" Body="&lt;p&gt;Given a lens-shaped frame with a single crossbar running parallel to the floor, how could we keep that object upright and stable on one of its endpoints using only wire?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape I'm describing is similar to theta: θ&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Except it's pointed at the top and bottom. See also &lt;a href=&quot;https://en.wikipedia.org/wiki/Lens_(geometry)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Lens (geometry)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can assume the frame is made from a flexible material, and that the entire frame is one contiguous unit. You can also assume we can freely affix any number of wires to any point on the object and any point on the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a follow-up, is there an efficient method of positioning the wires so we can keep the amount of wire needed as low as possible?&lt;/p&gt;&#xA;" OwnerUserId="23270" LastActivityDate="2019-10-26T07:46:46.453" Title="How to keep a symmetric-lens-shaped object upright using only wire?" Tags="|mechanical-engineering|structural-engineering|civil-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31676" PostTypeId="2" ParentId="31675" CreationDate="2019-10-26T04:28:49.517" Score="1" Body="&lt;p&gt;Put a ring around the top 1/3 of the object then have 4 guy wires coming down to two bars fixed to the crossbar, one on each side.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-10-26T04:28:49.517" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31677" PostTypeId="2" ParentId="31675" CreationDate="2019-10-26T07:46:46.453" Score="1" Body="&lt;p&gt;Look at pairs of glasses as they are wire holding a lens. &lt;/p&gt;&#xA;" OwnerUserId="16700" LastActivityDate="2019-10-26T07:46:46.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31681" PostTypeId="1" CreationDate="2019-10-27T01:44:45.280" Score="0" ViewCount="155" Body="&lt;p&gt;Normally, when you discretize continuous system with zoh method, your continuous system output with the corresponding discrete inputs passed through zero-order hold should be the same at sampling instants. But here in my code it is not the same, but I can't understand why? Here we have a second order system, and I apply discrete PID control to the discrete system and simulate it in for loop. Then I take the discrete input values and for simulating continuous time system output I accumulate (many) shifted and weighted (by discrete input values) step responses of the continuous system to get the output. Here is the code:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;import numpy as np&#xA;import matplotlib.pyplot as plt&#xA;import control.matlab as cn&#xA;&#xA;&#xA;sys = cn.tf([1, 1], [1, 4, 5])&#xA;Ts = 0.01&#xA;sysd = cn.sample_system(sys, Ts)&#xA;&#xA;print(sysd)&#xA;&#xA;# prints:&#xA;#&#xA;# 0.009851 z - 0.009753&#xA;# ---------------------&#xA;# z^2 - 1.96 z + 0.9608&#xA;&#xA;ref = np.ones(1100)&#xA;Kp = 10.&#xA;Kd = 1.&#xA;Ki = 1.&#xA;&#xA;y = []&#xA;y.append(0.)&#xA;u = []&#xA;u.append(Kp * (ref[0] - y[0] ) + Ki*(ref[0] - y[0] ) + Kd*(ref[0] - y[0]))&#xA;y.append(0.009851*u[0])&#xA;Acc_Intg = ref[0] - y[0]   # Accumulator for Sum of errrors for Integral term Ki.&#xA;&#xA;&#xA;for i in np.arange(1, 110):&#xA;    Acc_Intg = Acc_Intg + ref[i] - y[i]&#xA;    u.append(Kp * (ref[i] - y[i]) + Kd * ((ref[i] - y[i]) - (ref[i - 1] - y[i - 1])) + Ki * Acc_Intg)&#xA;    y.append(y[i]*1.96 - y[i-1]*0.9608 + u[i]*0.009851 - u[i-1]*0.009753)&#xA;&#xA;&#xA;T = np.arange(0, 0.8, 0.01)&#xA;u = np.array(u)&#xA;yc, TT = cn.step(sys, T)&#xA;yc = yc*u[0]&#xA;for i in np.arange(1, T.size -5):&#xA;    yaux, TT = cn.step(sys, T[0:(T.size - i)])&#xA;    yc[i:] = yc[i:] +  (u[i] - u[i-1]) * yaux&#xA;&#xA;&#xA;y = np.array(y)&#xA;plt.plot(np.arange(0, y.size * 0.01 - 0.0000001, 0.01), y)&#xA;plt.plot(T, yc)&#xA;plt.show()&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="21221" LastActivityDate="2019-10-27T01:44:45.280" Title="Continuous vs Discrete system simulation in Python" Tags="|control-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31683" PostTypeId="1" CreationDate="2019-10-27T14:08:36.967" Score="2" ViewCount="774" Body="&lt;p&gt;In a prior question it was established that any string or microinverter that is grid tied must put out sinusodial power. But there is also the task of being in sync with the grid. It seems from apperances that task for a single string inverter is far simpler than for an arrary of 28 microinverters on a 10kwh solar system to stay in step with each other let alone the grid. I can only imagine the analog nightmare of trying to cue 28 phonographs to play the same lp in unison and with variations in drive components and lp pressings. So what is the microinverter problem and solution to this symphonic nightmare of phase variance?&lt;/p&gt;&#xA;" OwnerUserId="23278" LastActivityDate="2021-07-19T19:07:55.717" Title="How do multiple enphase microinverters stay synced to the grid?" Tags="|power-electronics|power-engineering|solar-energy|photovoltaics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31686" PostTypeId="2" ParentId="31674" CreationDate="2019-10-27T19:22:44.737" Score="1" Body="&lt;p&gt;&lt;strong&gt;Edited&lt;/strong&gt; based on new information provided&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most important thing to start with is: what resolution DAQ do you need? 16 bit is fairly standard, cheaper ones will be 12 or 14 bit. more expensive ones will be 20 or 24 bit.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the specs you provided, Firstmark series 173-176, the conversion is 165 mm = supply voltage.  Max supply voltage is 35VDC, but you could use anything. You didn't provide the min frequency you are interested, so let's just assume you need all the way down to 0 Hz, i.e. needs to be DC coupled. Most basic DAQ systems would be able to read -10V to 10V.  You could get something that would read up 35V, you'll need something more expensive and complicated, so just assume you are keeping it simple and use 10V supply voltage.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A 16 bit ADC would set +32768 = 10V, and -32767 = -10V (note 2^16=65536, and we are using half that range for positive voltages and half for negative, hence 65536/2=32768... note that as I'm assuming DC coupled above, half the voltage range is unused... i.e. the DAQ system is setup to use -10 to +10, but you'll only be using 0 to 10V.  If you can assume AC coupled you can use the whole range, but in that case the minimum frequency you can read would be a few hertz, won't get DC).  So the smallest voltage you could read is 0.0003V.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the assumption that 10V=165mm, this means the smallest displacement you could read is 0.005 mm.  That would appear to meet your specifications.  So a 16 bit DAQ should meet your needs, you don't need to spring for anything more expensive than that.  On the other hand, a lot of cheaper systems might be 12 bit. Those will be 16 times worse resolution, i.e. minimum value you could read would be 0.08 mm. Probably would not work for you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you were also asking about &quot;40 kS/s&quot;, that's 40,000 Samples/second. That is not the maximum frequency that you can record however.  General rule of thumb would be divide sample rate by 2.5 to get the maximum frequency you can record, which is called the &quot;bandwidth&quot;.  In this case 40k/2.5 = 16 kHz.   You are referencing 7.2 kHz in the comments, which is less that I would have expected for a 40k sample rate.  But since you said that you only care about signals at 300 Hz or maybe 1kHz anyway, this will be plenty for you.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;while I'm thinking about it, another thing you should be looking at is whether the DAQ has built in anti-aliasing filters.  Virtually all of the good ones do, but some cheaper ones do not.  Unless you are 100% sure you know what you are doing, make sure you get one with an anti-aliasing filter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll also say that most of the big DAQ manufacturers have pretty helpful sales engineers.  Just call the company up and tell them what you are trying to do.  They can probably point you in the right direction.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="9275" LastEditDate="2019-10-28T02:06:59.247" LastActivityDate="2019-10-28T02:06:59.247" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31687" PostTypeId="1" CreationDate="2019-10-28T11:46:59.120" Score="0" ViewCount="188" Body="&lt;p&gt;I've often heard stress acts at a point and not on an entire cross-sectional area to which the load is applied normally.&lt;br&gt;&#xA;Im a little confused here, Let us consider an area of cross-section &lt;code&gt;A&lt;/code&gt; and let a load &lt;code&gt;P&lt;/code&gt; be applied normal to &lt;code&gt;A&lt;/code&gt; resulting in stress &lt;code&gt;s&lt;/code&gt;.&lt;br&gt;&#xA;Does this mean-  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;code&gt;s&lt;/code&gt; acts at a point on &lt;code&gt;A&lt;/code&gt; and not on the entire &lt;code&gt;A&lt;/code&gt;? If yes, why?&lt;/li&gt;&#xA;&lt;li&gt;Is the value of &lt;code&gt;s&lt;/code&gt; the same at all points on &lt;code&gt;A&lt;/code&gt;?&lt;/li&gt;&#xA;&lt;li&gt;Isn't &lt;code&gt;A&lt;/code&gt; significant to &lt;code&gt;s&lt;/code&gt;?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16605" LastActivityDate="2020-11-21T22:02:25.200" Title="Why does stress act a point instead of on an entire area normal to the stress?" Tags="|mechanical-engineering|materials|structures|stresses|mechanical-failure|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31688" PostTypeId="2" ParentId="31687" CreationDate="2019-10-28T14:16:30.193" Score="1" Body="&lt;p&gt;In general, the stress can change from point to point on a surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a flat surface with a uniform pressure load applied to it, the &lt;em&gt;normal&lt;/em&gt; component of the stress must equal the pressure everywhere to maintain equilibrium, but there can also be &lt;em&gt;shear&lt;/em&gt; stress components at the surface which are not constant over the whole surface. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, consider a cantilever beam with a uniform distributed load applied along its length. The shear stress at the surface (which is the same thing as the &quot;axial stress&quot; in the beam which you calculate using &lt;span class=&quot;math-container&quot;&gt;$My/I$&lt;/span&gt;) varies from zero at the free end to a maximum value at the fixed end. And when dealing with beams, you don't usually bother about the &lt;em&gt;direct&lt;/em&gt; stress component caused by the uniform load at all, because it is small compared with the axial stress.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-10-28T14:16:30.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31689" PostTypeId="1" AcceptedAnswerId="31737" CreationDate="2019-10-28T15:53:00.417" Score="3" ViewCount="7770" Body="&lt;p&gt;I am attempting to solve ANSYS mechanical (via workbench) using GPU acceleration, but I keep getting the following error: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;* WARNING *&lt;/strong&gt;                         CP =      79.594   TIME= 15:40:08&#xA;   The GPU accelerator capability is not valid when using the memory&lt;br&gt;&#xA;   saving option (MSAVE command) for the PCG solver.  The GPU accelerator&lt;br&gt;&#xA;   capability is disabled for this solution.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;How can I turn off the MSAVE option? It seems to be default for a model this size and I cannot find the way to turn it off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using V19.2.&lt;/p&gt;&#xA;" OwnerUserId="14498" LastActivityDate="2019-10-31T17:49:54.877" Title="Using GPU Acceleration in ANSYS Mechanical" Tags="|ansys|ansys-workbench|ansys-apdl|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31690" PostTypeId="2" ParentId="31683" CreationDate="2019-10-28T17:33:18.110" Score="1" Body="&lt;p&gt;A baseline grid-tie inverter samples the line voltage and produces current that is in phase with it and proportional to the maximum power it can squeeze out of the source (typically photovoltaic).  So it's actually the grid itself that the inverters are synchronizing to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can actually burden the grid with power it doesn't need, or make the job of regulating voltage at remote spots more difficult.  This can have a direct impact on the stability of the power grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do a search on &quot;smart grid tie inverters&quot; or &quot;advanced grid tie inverters&quot;.  There's a movement afoot in areas that are seeing a lot of alternative power sources to encourage or require power providers to have inverters that can be managed or otherwise be provisioned to be an assist to grid stability, rather than a burden.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-10-28T17:33:18.110" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31691" PostTypeId="2" ParentId="31687" CreationDate="2019-10-28T18:32:24.390" Score="1" Body="&lt;p&gt;Imagine you have a steel column as a solid square of 6-inch by 6-inch section carrying a load P sitting on a square base plate, 12-inch by 12-inch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normal stress on the base plate is not &lt;span class=&quot;math-container&quot;&gt;$ \ \sigma= \frac {P}{12^2} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stress due to the P on the baseplate will be a complex mixture of normal stress, shear stress and bending (moment) stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if we consider just the normal stress on the baseplate it follows a pattern of concentric roundish square contour lines which are degrading in intensity as they open farther.&#xA;I attach a crude diagram here, approximately showing the spread of stress contours. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yyyA3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yyyA3.png&quot; alt=&quot;stress under the column&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is why we consider the stress locally just for a point. In this case, we want to make sure the maximum normal stress is below the allowed stress, not the average stress.&#xA;The distribution of normal stress into the depth of the plate is a bit complicates but basically the stress opens into the depth following an exponential spiral profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In many of the structural members like beams, stress increases linearly vertically as the point is farther from neutral axis of the beam. There are however structural members much more sophisticated than a simple beam where the stress varies greatly on a different point of a section, eg, cranck shaft of a car engine.&#xA;Some members have almost uniform stress load across their cross-section, eg, a truss tension member at a point not very near its joint.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-28T18:32:24.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31694" PostTypeId="1" AcceptedAnswerId="49402" CreationDate="2019-10-29T00:35:06.963" Score="1" ViewCount="36" Body="&lt;p&gt;I'm a software engineer that works on custom embedded devices. I program these devices through a typical RJ-45 ethernet cable. In most cases, I can connect my computer to these devices through a typical network switch and update multiple devices at the same time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, some devices are not compatible with others, and will cause issues if connected. For example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u28Js.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u28Js.png&quot; alt=&quot;Isolated Network Design&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I cannot have traffic from Device A reach Device B, or vice versa. Typically, I only need to be connected to one at once, so my current solution is just moving an ethernet cable that is connected to my computer to Device A or Device B, but I'd like to improve that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking to either build or buy a togglable network switch, where the system can be in one of two states: Computer to Device A (and vice versa) or Computer to Device B (and vice versa). Basically, I minimally need to toggle ports on a network switch on and off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to find something like this online, so I was wondering if I can rig a switch up by splicing some network cables and toggling them in some way, but I'm worried about interference and potential byte corruption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any initial suggestions?&lt;/p&gt;&#xA;" OwnerUserId="23299" LastActivityDate="2022-01-24T05:00:57.107" Title="Network Switch with Togglable Ports" Tags="|embedded-systems|computer-hardware|networking|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31696" PostTypeId="1" CreationDate="2019-10-29T06:36:20.140" Score="0" ViewCount="2706" Body="&lt;p&gt;In the fourbar mechanism as in the following picture, I have to find the toggle position, the sliding distance and the time ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my understanding/plan to do this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The toggle positions (&quot;extreme ends&quot;) of the slider (block B) take place at 2 time, when link 3 and &quot;offset&quot; are aligned, and when link 3 and 2 are aligned.&lt;/li&gt;&#xA;&lt;li&gt;The sliding distance can be calculated via trigonometry from the above conclusion&lt;/li&gt;&#xA;&lt;li&gt;Assume the working cycle is the part where block B moves from left to right, I have to find the working angle of this movement at crank (&lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;), the time ratio would be &lt;span class=&quot;math-container&quot;&gt;$\frac{360-\beta}{\beta}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Am I right in my understanding so far? If not, where do I have it wrong. And then, how do I obtain the value for &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/El0I9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/El0I9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21520" LastActivityDate="2019-10-29T06:36:20.140" Title="Toggle position of a fourbar slider-crank" Tags="|mechanical-engineering|mechanisms|linkage|machine-design|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31697" PostTypeId="1" CreationDate="2019-10-29T07:32:05.123" Score="1" ViewCount="68" Body="&lt;p&gt;I've been wanting to design my own water carbonation device for a while.&#xA;For reference, I've been looking at devices like the Sodastream, which injects pure CO&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; into water to dissolve it with up to seven injections per bottle. The more injections, the higher the CO&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; contents. I have not actually built anything yet; for now I am just trying to understand how it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming that the device keeps the pressure constant, but the CO&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; doesn't dissolve all at once and needs multiple injections to get to the right carbonation level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assumption: the device opens the valve for a set amount of time with a known pressure, which releases a certain volume of CO&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; into the liquid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would assume that injecting the CO&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; creates a lot of small bubbles which increases the effective contacting surface area, leading to faster mass transfer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Through calculation and use of Henry's law + an approximation for the solubility constant I've found a temperature and pressure range for my desired carbonation levels: temperature 15 °C and pressure 2 bar CO&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; for a desired 4 g/L CO&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; (perhaps with variable pressure and/or temperature to control the carbonation levels).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a few questions related to this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;How can I apply my desired pressure and carbonation level to a direct injection method? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;How long would I need to inject, and how many times? Is there a relationship with injection pressure(preferably a formula if available)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the desired carbonation level isn't achieved after a single injection, how far would the injection pressure need to be raised?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is there any theoretical description of this process available? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Is direct injection of CO&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; an efficient method of carbonisation, or are there better/cheaper/more efficient ways to dissolve CO&lt;span class=&quot;math-container&quot;&gt;$_2$&lt;/span&gt; in water as fast as possible?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Edit: I have a background in analytical chemistry. Any engineering knowledge is self-taught. I never received (much) formal education on subjects like fluids/gases/physics, so I hope the information I've provided is applicable to some degree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(If you require any further information please let me know.)&lt;br&gt;&#xA;(originally posted on SE Physics, they recommended me to ask here)&lt;/p&gt;&#xA;" OwnerUserId="23305" LastEditorUserId="23305" LastEditDate="2019-10-29T07:39:33.720" LastActivityDate="2019-12-04T13:36:34.923" Title="Injecting CO$_2$ into water: how does injection work?" Tags="|pressure|gas|compressed-gases|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31698" PostTypeId="1" AcceptedAnswerId="32143" CreationDate="2019-10-29T08:02:44.783" Score="0" ViewCount="203" Body="&lt;p&gt;I have been drifting RWD (rear wheel drive) cars for some years now. I tend to think about every engineering solution that other drivers use and how does it work before applying it to my car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What has been baffling me from some time now is: &lt;strong&gt;Why using LSD (limited slip differential) results in bigger speeds when drifting?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Food for thought:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Welded differentials (100% lock) are very common in this motorsport since you are dealing with kinetic friction on rear tires anyway - which is constant regardless of their rotation speed (unless you match tire speed to ground and grip using static friction).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Open differentials (0% lock) on the other hand practically make drifting (powersliding) unusable as almost all of the engine torque is routed to unloaded wheel resulting in slowing down.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Slow motion drifting video: &lt;a href=&quot;https://www.youtube.com/watch?v=OG0cyjqDJCw&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=OG0cyjqDJCw&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="23304" LastActivityDate="2019-12-01T03:59:14.607" Title="Faster car drifting (motorsport) when using limited slip vs welded differential" Tags="|friction|wheels|power-transmission|traction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31700" PostTypeId="1" CreationDate="2019-10-29T11:10:05.380" Score="1" ViewCount="124" Body="&lt;p&gt;I'm trying a test a hydrofoil in a low Reynolds flume, but struggling to find a way to mount the hydrofoil in the flume and measure the lift and drag force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some idea of using a load cell but not sure how to set it up and connect it with to hydrofoil, as the hydrofoil span across the width of the flume. &lt;/p&gt;&#xA;" OwnerUserId="23308" LastActivityDate="2024-01-07T19:01:40.167" Title="Lift and drag force measurement in a flume" Tags="|fluid-mechanics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31702" PostTypeId="1" CreationDate="2019-10-29T12:47:06.250" Score="0" ViewCount="76" Body="&lt;p&gt;I just started using the Flowmaster program (ver 08.11.01.03 if relevant) to model a series of stormwater pipes which are almost certainly in a pressure flow situation based on a first round analysis using open channel flow which failed due to insufficient pipe capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I completed my model yesterday by modeling each pipe between structures and noted that the hydraulic grade line easily exceeds the grate elevation for a lot of my structures and is thus overflowing. In these situations, I sort of brute forced the program to model the next pipe section by assuming the downstream hydraulic head was equal to the height of the water column at the downstream structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn't ideal as it doesn't account for losses due to pipe bends, structures, and other elements that can cause additional losses. I have previously created an Excel spreadsheet to model these sorts of losses, but it was a very complex spreadsheet created for another company that I no longer have access to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that I previously made something like this with just Excel, it seems like Flowmaster should be capable of doing the same. Is it possible to use the program to model everything in the run in a single go so that losses are properly accounted for?&lt;/p&gt;&#xA;" OwnerUserId="9504" LastActivityDate="2019-10-29T12:47:06.250" Title="Can Flowmaster be used to model an entire pipe run?" Tags="|fluid-mechanics|civil-engineering|pressure|modeling|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31703" PostTypeId="2" ParentId="31585" CreationDate="2019-10-29T13:58:33.727" Score="0" Body="&lt;p&gt;Your approach is one method to solve &lt;a href=&quot;https://www.chegg.com/homework-help/questions-and-answers/balanced-heat-exchanger-one-capacitances-fluids-equal-e-effectiveness-affected-show-effect-q1665483&quot; rel=&quot;nofollow noreferrer&quot;&gt;a balanced heat exchanger&lt;/a&gt;. I might instead have simply divided the equation by &lt;span class=&quot;math-container&quot;&gt;$\Delta T_2$&lt;/span&gt; to obtain&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta T_{LMTD} = \Delta T_2 \left[\frac{R - 1}{\ln(R)} \right]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding the limit as &lt;span class=&quot;math-container&quot;&gt;$R \rightarrow 1$&lt;/span&gt; gives the same answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative approach with the above is to graph the numerator &lt;span class=&quot;math-container&quot;&gt;$R - 1$&lt;/span&gt; and denominator &lt;span class=&quot;math-container&quot;&gt;$\ln(R)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8dP6l.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8dP6l.png&quot; alt=&quot;Graph of numerator (red) and denominator (blue) of function&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graph shows in the limit as &lt;span class=&quot;math-container&quot;&gt;$R \rightarrow 1$&lt;/span&gt; that the ratio of numerator to denominator approaches unity.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2019-10-29T13:58:33.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31706" PostTypeId="2" ParentId="31698" CreationDate="2019-10-29T15:38:58.547" Score="0" Body="&lt;p&gt;Because with a lsd, when one wheel slips the other wheel actually turns faster, due to the way the differential works.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-10-29T15:38:58.547" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31707" PostTypeId="1" AcceptedAnswerId="33854" CreationDate="2019-10-29T17:01:31.347" Score="0" ViewCount="81" Body="&lt;p&gt;At my current role we have a hell of a time trying to get NPT threads to seal. We avoid them where we can but sometimes its just faster and cheaper to spec and buy parts with NPT threads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are using pipe dope and I fully trust in my techs skills. Is there any other methods or products out there that can help ensure a good seal ?&lt;/p&gt;&#xA;" OwnerUserId="14961" LastActivityDate="2020-03-27T19:47:27.147" Title="Leaky NPT fixes" Tags="|piping|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31708" PostTypeId="1" AcceptedAnswerId="31750" CreationDate="2019-10-29T17:57:15.923" Score="0" ViewCount="1869" Body="&lt;p&gt;A standard practice in many academic labs, which I'm told is &quot;bad practice&quot;, is to use RG58 cables (i.e. standard BNC cables) for power distribution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt; Why is this considered bad practice?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt; Is there a different sort of cable which is better for power distribution and compatible with BNC connectors?&lt;/p&gt;&#xA;" OwnerUserId="5103" LastEditorUserId="1586" LastEditDate="2019-10-31T18:10:13.670" LastActivityDate="2022-11-05T01:04:07.297" Title="Why is it bad to run power through BNC (RG58) cables?" Tags="|electrical-engineering|power-electronics|power-transmission|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31709" PostTypeId="1" CreationDate="2019-10-29T18:04:54.580" Score="0" ViewCount="67" Body="&lt;p&gt;I want to solve &lt;span class=&quot;math-container&quot;&gt;$V_o$&lt;/span&gt; in terms of &lt;span class=&quot;math-container&quot;&gt;$V_i$&lt;/span&gt; from  the following circuit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BSRtn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BSRtn.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R_1=95.7  \Omega, R_2=9.1  k\Omega, R_3=1.0  M\Omega$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the solution is form &lt;span class=&quot;math-container&quot;&gt;$V_o=kV_i$&lt;/span&gt; and I want the factor k.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R_{123}=R_1+R_{23}$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;span class=&quot;math-container&quot;&gt;$R_{23}= 9017.94$&lt;/span&gt; so the &lt;span class=&quot;math-container&quot;&gt;$R_{123}=9113.64\Omega$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do I get the factor k?&lt;/p&gt;&#xA;" OwnerUserId="23320" LastEditorDisplayName="user14407" LastEditDate="2019-10-29T22:23:57.133" LastActivityDate="2019-10-30T10:51:12.843" Title="Circuit analysis, voltage in and voltage out" Tags="|electrical-engineering|electrical|circuits|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31710" PostTypeId="1" CreationDate="2019-10-29T18:15:27.563" Score="1" ViewCount="241" Body="&lt;p&gt;I was wondering how to design this circuit in depth, since I have no idea what to do.&lt;a href=&quot;https://i.stack.imgur.com/5m0bF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5m0bF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23321" LastActivityDate="2023-04-15T04:09:36.157" Title="Designing a loaded voltage divider?" Tags="|electrical-engineering|design|power-electronics|circuits|circuit-design|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31711" PostTypeId="2" ParentId="31707" CreationDate="2019-10-29T18:31:29.667" Score="1" Body="&lt;p&gt;NPT are not intended to hold very high pressure , but I do not remember a number. The oil/gas industry uses proprietary threads for wells with pressure above roughly 500 psi ( depending on the company). NPT requires dope to plug the spiral leak-paths at the thread roots ; API pipe dope contains metal powders ( copper, zinc, and formerly lead) to block the leak paths ( API threads are similar to NPT , but not identical).  This assumes your NPT thread profile, pitch, diameter, taper, etc , conform to ASTM A 530. Common pipe dopes of teflon or traditional plumbers putty should be good to over 100 psi. Maybe a metal powder filled dope will help.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-10-29T18:31:29.667" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31712" PostTypeId="2" ParentId="28308" CreationDate="2019-10-29T19:32:27.200" Score="1" Body="&lt;p&gt;The first ring is barely useful, overly simplified, and certainly not representive of a real magnetic field. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second ring is possible but there is another option. Instead of the center being magnetic,  the fields enter one side and are mostly  conducted through the material to the other side and exit the magnet as seen here &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/6/6f/Gramme_Ring_Armature_-_Minimal_Core_Penetration.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;alternate example&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="23325" LastEditorUserId="23325" LastEditDate="2019-10-29T19:38:07.057" LastActivityDate="2019-10-29T19:38:07.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31713" PostTypeId="2" ParentId="31700" CreationDate="2019-10-29T20:51:50.373" Score="0" Body="&lt;p&gt;If practical install a shaft through the hydrofoil passing through its center of pressure into the walls of the flume continuing into two support donut load sensors. This way you can measure the lift and drag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you can use four pancake sensors two on each side. One set for lift, the other for drag.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find the center of pressure of the hydrofoil by trial and error or if it comes with a chart look up the chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I add a link to airfoils here which can be helpful. &lt;a href=&quot;https://aviation.stackexchange.com/a/19399/9265&quot;&gt;center of pressure airfoil&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-10-29T20:51:50.373" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31714" PostTypeId="1" CreationDate="2019-10-29T21:25:04.660" Score="1" ViewCount="65" Body="&lt;p&gt;I have a cad model of a plastic box that is made of two parts. Two parts are connected with resin isolation. I want to test if my part is at least in theory ip86 water-resistant. So is there any software for testing cad model waterproofing and isolation?&lt;/p&gt;&#xA;" OwnerUserId="20706" LastActivityDate="2019-10-29T21:25:04.660" Title="Is there any software for testing cad model waterproofing and isolation?" Tags="|cad|waterproofing|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31716" PostTypeId="2" ParentId="31710" CreationDate="2019-10-30T02:01:39.997" Score="0" Body="&lt;p&gt;Assume your design works, so you know that the voltages at A, B, and so on are the ones given in the problem.  Since you know the currents through the loads, you know some of the currents flowing out of the nodes A, B, C .... , With knowledge of the voltages, and some of the currents, you can figure out the other currents. Then you can solve for the resistors in the voltage divider that give you those currents.  (Note that the voltage at D in your diagram is 0,  you probably want D to be the voltage at the bottom of the battery. )  &lt;/p&gt;&#xA;" OwnerUserId="19743" LastActivityDate="2019-10-30T02:01:39.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31717" PostTypeId="1" CreationDate="2019-10-30T07:04:06.153" Score="0" ViewCount="360" Body="&lt;p&gt;In an LLE operation pilot plant, i saw that sieve trays are used instead of bubble cap and valve trays... In LLE operation, they require good mixing, so i was shocked why bubble cap and valve trays are not in used. I heard that sieve trays is commonly used in LLE plate tower too. Why is the benefit of using sieve trays in LLE operation other than the fact that it is easy to clean ? &lt;/p&gt;&#xA;" OwnerUserId="11561" LastActivityDate="2019-11-01T18:36:55.153" Title="Sieve trays used in Liquid liquid extraction tower instead of bubble cap and valve trays?" Tags="|chemical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31718" PostTypeId="1" AcceptedAnswerId="31755" CreationDate="2019-10-30T08:17:26.797" Score="2" ViewCount="678" Body="&lt;p&gt;For any testing that a part is suppose to pass a test, let's say the leakage test of a seal flange. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I individualize the cause? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know there is root cause analysis and for different cases I probably need Design of Experiment (DOE) analysis for considering all cases of test to find the root cause?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not seen any papers that talks about doe analysis and root cause analysis at the same time. &lt;/p&gt;&#xA;" OwnerUserId="17070" LastEditorUserId="1586" LastEditDate="2019-10-31T19:27:48.697" LastActivityDate="2019-11-01T20:52:10.817" Title="DOE analysis or root cause analysis?" Tags="|mechanical-engineering|mechanical-failure|product-testing|failure-analysis|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31719" PostTypeId="2" ParentId="31709" CreationDate="2019-10-30T10:51:12.843" Score="2" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R_{23} = {\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}}}$&lt;/span&gt; Resistors in parallel&lt;BR&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$R_{23} \approx 9.018KΩ$&lt;/span&gt; Plugging in the values&lt;BR&gt;&lt;BR&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$V_{out} = V_{in} \cdot {\frac {R_{23}}{R_{1} + R_{23}}}$&lt;/span&gt; &lt;a href=&quot;https://en.wikipedia.org/wiki/Voltage_divider&quot; rel=&quot;nofollow noreferrer&quot;&gt;Voltage division&lt;/a&gt;&lt;BR&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$V_{out} \approx   0.9895 \cdot V_{in}$&lt;/span&gt; Plugging in the values&lt;BR&gt;&lt;BR&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$k \approx 0.9895$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="22772" LastActivityDate="2019-10-30T10:51:12.843" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31721" PostTypeId="1" AcceptedAnswerId="31722" CreationDate="2019-10-30T15:59:40.550" Score="0" ViewCount="214" Body="&lt;p&gt;I want to calculate the voltage at node &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; respective to the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WbzrW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WbzrW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R_1=1.8\text{k}\Omega, R_2=3.8\text{k}\Omega, R_3=1.8\text{k}\Omega, R_4=5.8\text{k}\Omega, R_5=2.4\text{k}\Omega$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V1=4.3V, V2=2.3V$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that voltage &lt;span class=&quot;math-container&quot;&gt;$V=-V2+V1=2V$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$R_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_2$&lt;/span&gt; in series so &lt;span class=&quot;math-container&quot;&gt;$R_{12}=R_1+R_2=5.6\text{k}\Omega$&lt;/span&gt; then the current is &lt;span class=&quot;math-container&quot;&gt;$I=\frac{V}{R_{12}}=0.356\,\text{mA}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do I calculate the voltage at node &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; respective to the ground? Is it over &lt;span class=&quot;math-container&quot;&gt;$R_{45}$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$R_{3}$&lt;/span&gt;? &lt;/p&gt;&#xA;" OwnerUserId="23337" LastActivityDate="2019-11-01T15:23:33.800" Title="Voltage at node A respective to the ground" Tags="|electrical-engineering|electrical|circuits|circuit-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31722" PostTypeId="2" ParentId="31721" CreationDate="2019-10-30T16:39:51.967" Score="0" Body="&lt;p&gt;If you consider the RHS loop of your circuit and using &lt;span class=&quot;math-container&quot;&gt;$i_3$&lt;/span&gt; as the current going through &lt;span class=&quot;math-container&quot;&gt;$R_3$&lt;/span&gt; (in the same direction as &lt;span class=&quot;math-container&quot;&gt;$i$&lt;/span&gt;), &lt;span class=&quot;math-container&quot;&gt;$i_4$&lt;/span&gt; the current going through &lt;span class=&quot;math-container&quot;&gt;$R_4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$i_5$&lt;/span&gt; the current going through &lt;span class=&quot;math-container&quot;&gt;$R_5$&lt;/span&gt;, you have (assuming your polarity of &lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt; is correct):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ &#xA;-V_2 + V_1 + R_3 i_3 - V_A = 0\\&#xA;V_A = -R_4 i_4\\&#xA;V_A = -R_5 i_5\\&#xA;i_3 = i_4 + i_5&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which allows you to express &lt;span class=&quot;math-container&quot;&gt;$i_3$&lt;/span&gt; as a function of &lt;span class=&quot;math-container&quot;&gt;$V_A$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ i_3 = -V_A \left( \frac{1}{R_4} + \frac{1}{R_5} \right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you can use this in your loop equation to get &lt;span class=&quot;math-container&quot;&gt;$V_A$&lt;/span&gt; as a function of &lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ -V_2 + V_1 - R_3 V_A \left( \frac{1}{R_4} + \frac{1}{R_5} \right) - V_A= 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which gives you &lt;span class=&quot;math-container&quot;&gt;$V_A$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_A = \frac{V_1 - V_2}{1 + R_3 \left( \frac{1}{R_4} + \frac{1}{R_5} \right)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and if you plug the numerical values in you get &lt;code&gt;V_A = 0.9707V&lt;/code&gt;.&lt;/p&gt;&#xA;" OwnerUserId="334" LastEditorUserId="334" LastEditDate="2019-10-31T10:14:31.737" LastActivityDate="2019-10-31T10:14:31.737" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31723" PostTypeId="1" AcceptedAnswerId="33787" CreationDate="2019-10-30T17:36:24.593" Score="2" ViewCount="307" Body="&lt;p&gt;Using the sylvac UL 4 &#xA;Each time I invoke the (send data) function the sylvac LED panel reports '&lt;strong&gt;no.data&lt;/strong&gt;'.  To mention the obvious but nothing is received at the connected USB device.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is. . &lt;strong&gt;How do force the sylvac device to (have data) to send?&lt;/strong&gt;  Can anyone explain what this 'no.data' actually means from the calipers perspective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HU3U2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HU3U2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23340" LastActivityDate="2020-10-27T14:22:17.343" Title="Sylvac UL 4 - USB - (Send data) reports 'no.data'?" Tags="|measurements|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31724" PostTypeId="2" ParentId="31723" CreationDate="2019-10-30T20:53:15.367" Score="1" Body="&lt;p&gt;This might be a good place to start&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aBoXO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aBoXO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at &lt;a href=&quot;https://www.youtube.com/watch?v=RbufAzxTbeg&quot; rel=&quot;nofollow noreferrer&quot;&gt;IOT-Ready Sylvac Bluetooth® Profiles&lt;/a&gt;, and &lt;a href=&quot;https://www.youtube.com/watch?v=JR_BwDggBBc&quot; rel=&quot;nofollow noreferrer&quot;&gt;SYLVAC BLUETOOTH IOT 4.0 / KEYBOARD INPUT FUNCTION. WIRELESS DATA TRANSFER&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8055" LastEditorUserId="8055" LastEditDate="2019-10-31T15:45:40.987" LastActivityDate="2019-10-31T15:45:40.987" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31725" PostTypeId="2" ParentId="6196" CreationDate="2019-10-31T00:11:18.790" Score="0" Body="&lt;p&gt;There is not enough space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But they do manage to find enough space on the bigger ceramic components though.&lt;/p&gt;&#xA;" OwnerUserId="23343" LastActivityDate="2019-10-31T00:11:18.790" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31727" PostTypeId="2" ParentId="23288" CreationDate="2019-10-31T01:39:51.217" Score="1" Body="&lt;p&gt;An induction heater creates an oscillating magnetic field near the coil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;mechanism&quot; to damage an electronic component is that eddy currents are induced in the metal parts of the component.  These eddy currents heat up the part until something breaks.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the part breaks, yes, it will be due to magnetic induction.&lt;/p&gt;&#xA;" OwnerUserId="23343" LastActivityDate="2019-10-31T01:39:51.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31729" PostTypeId="2" ParentId="31718" CreationDate="2019-10-31T02:01:51.817" Score="2" Body="&lt;p&gt;Let assume a leak has been reported in the following pump system. As a result there is 1 inch of water on the floor. To identify the failure a root cause analysis (RCA) needs to be performed. To illustrate RCA few unrealistic conditions such as operating temperature range of -40C to 85C have been specified. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Kf1db.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Kf1db.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on the above image the flange consist of 3 different material with different thermal expansion co-efficient's. The RCA determines the rubber gasket as the cause of the failure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corrective action is to find a suitable replacement gasket. There are 20 different alternative gasket options. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Executing a Design of Experiment (DOE) can help determine the best alternative gasket. A list design parameters including the above mentioned unrealistic temperature specification would be input to develop an appropriate DOE. Good design parameters are once that can be measurable. The results DOE will provide a qualitative output to aid the decision making process.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an good example of list of parameter to consider for the DOE. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tKZYy.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tKZYy.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-10-31T10:56:56.420" LastActivityDate="2019-10-31T10:56:56.420" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="31730" PostTypeId="1" AcceptedAnswerId="34964" CreationDate="2019-10-31T09:09:23.303" Score="1" ViewCount="82" Body="&lt;p&gt;While my research about the untreated surface of moulded cast I read that the properties such as tensile strength, 0,2% proof stress and so on are different (much smaller) from the properties you can find in the tables. That’s because the samples have a treated surface to evaluate the material properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now my question is&lt;/strong&gt; how can I determine the thickness of the surface where my properties are much smaller?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The background is that I have a bore near the untreated surface loaded with hydraulic pressure and for safety I need a specific thickness of material around the bore. And because the material properties of the untreated surface are much smaller, I will not take this area into account to my calculations. If it is needed: the regarded cast is GJS 400.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="17780" LastActivityDate="2020-04-03T15:27:57.920" Title="(moulded) cast properties of the untreated surface" Tags="|mechanical-engineering|materials|metallurgy|casting|molding|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31731" PostTypeId="1" CreationDate="2019-10-31T12:55:12.763" Score="0" ViewCount="135" Body="&lt;p&gt;If we could be able to construct a gigantic hollow sphere, multiple elevators/chambers can glide in the interior of the sphere, accelerating and decelerating without colliding, providing artificial gravity of 1G. The sphere of course has to be gigantic so that the angular acceleration would be significantly less than the acceleration of the chamber, in order to minimize the coriolis effect. Could a solution like that to the artificial gravity problem be feasible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT]&#xA;Ok let’s simplify it a bit. Instead of a sphere, imagine a giant ring structure in space which stays still (doesn’t rotate). In the internal surface of the ring there is an elevator that can move along the circular path of the ring’s interior. The elevator starts to accelerate forward with 1G (following the circular path). Because of the huge proportions of the ring, the centrifugal force will increase but slowly. The key concept is to satisfy the following equation: forward acceleration + centripetal acceleration = 1G. So as we reach higher speeds, the forward acceleration of the elevator has to be reduced. When the  centripetal acceleration reaches 1G, the forward acceleration has to be 0. At this point the coriolis effect will be most severe. From that point the elevator decelerates but the equation: backwards acceleration + centripetal acceleration = 1G, still applies. Another important thing is that there is a distinct chamber that floats inside the elevator, so that it can rotate and stay in the same position for the person that is inside.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more concrete question is how big has to be that ring structure in order to provide 1g artificial gravity for 1 hour (30 minutes accelerating and 30 minutes decelerating the elevator)?&lt;/p&gt;&#xA;" OwnerUserId="23348" LastEditorUserId="23348" LastEditDate="2019-11-02T20:14:23.173" LastActivityDate="2023-12-12T14:01:45.190" Title="Artificial gravity with accelerating elevators enclosed in a gigantic sphere" Tags="|aerospace-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31733" PostTypeId="2" ParentId="31718" CreationDate="2019-10-31T14:21:42.507" Score="0" Body="&lt;p&gt;Is the question to use a system developed by bureaucrats ( DOE) or RCA developed by Deming, Juran , etc , over 70 odd years ? As a lifelong professional failure analyst I always looked for root cause; there is nearly always more than one cause for a significant failure. DOE came up with odd conclusions when doing failure analysis in areas of my experience of creep rupture and high temperature hydrogen attack ; that was years ago , maybe they are smarter now.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-10-31T16:44:36.333" LastActivityDate="2019-10-31T16:44:36.333" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31734" PostTypeId="1" AcceptedAnswerId="31768" CreationDate="2019-10-31T15:44:21.897" Score="1" ViewCount="145" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hymba.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hymba.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I am studying derivation of bernoulli equation from FM White's fluid mechanics and in that  pressure distribution is as shown in the image 1. I am unable to understand force calculation as shown in image .&lt;/p&gt;&#xA;" OwnerUserId="19791" LastEditorUserId="19791" LastEditDate="2019-10-31T16:04:22.180" LastActivityDate="2019-12-02T18:02:45.250" Title="Can anyone explain calculation of net force due to pressure in a streamtube of fixed control volume and infinitesmal size" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31735" PostTypeId="1" CreationDate="2019-10-31T16:12:39.040" Score="1" ViewCount="169" Body="&lt;p&gt;I'm currently studying passivity theorems of nonlinear systems, and I find it difficult to understand the difference between the different kinds of strict passiveness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any intuitive examples of the difference between state, output and input strictly passive systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Further, the passivity theorems use zero-state observability to prove global asymptotic stability for feedback-connection systems.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Definition 6.5, Khalil nonlinear systems)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Σ is zero-state observable if&#xA;no solution of x˙ = f(x,0) can stay identically in&#xA;S = {x ∈ R&#xA;n&#xA;|h(x,0) = 0} other than the trivial solution x(t) = 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This thread:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/8423/how-can-i-check-whether-a-nonlinear-system-is-zero-state-observable&quot;&gt;How can I check whether a nonlinear system is zero-state observable?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;States that the system state x and y should be 0 at the same time when u = 0. Why is this important?&lt;/p&gt;&#xA;" OwnerUserId="23351" LastActivityDate="2024-02-11T21:07:35.930" Title="Zero-state observability and strictly passive systems" Tags="|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31736" PostTypeId="1" CreationDate="2019-10-31T17:32:52.700" Score="1" ViewCount="20" Body="&lt;p&gt;Does the air cushion on which the original hyperloop idea envisioned by Elon Musk  require energy constantly to keep afloat, as the hovercraft does or would it be designed as the modern air suspension where energy requirement is once off, air is let into the  suspension cylinder, vehicle is floating in air then valves and seals take over.  &lt;/p&gt;&#xA;" OwnerUserId="12808" LastActivityDate="2019-10-31T17:32:52.700" Title="Does the air cushion on which the original hyperloop will sit require energy constantly or not" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31737" PostTypeId="2" ParentId="31689" CreationDate="2019-10-31T17:49:54.877" Score="3" Body="&lt;p&gt;I found the answer in the end with a lucky guess: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jl8OA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jl8OA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply enter &quot;MSAVE, off&quot; into the command line arguments in the advanced solver settings. &lt;/p&gt;&#xA;" OwnerUserId="14498" LastActivityDate="2019-10-31T17:49:54.877" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31738" PostTypeId="1" AcceptedAnswerId="31751" CreationDate="2019-10-31T18:15:59.117" Score="1" ViewCount="1443" Body="&lt;p&gt;Suppose i have 4 empty syringes. And by empty i mean they were having the air sucked out of them, and they &lt;strong&gt;&quot;contain vacuum&quot;&lt;/strong&gt;. Assume the head of the syringe is blocked so vacuum can be formed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to understand how to choose between this 4 different syringes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My goal is to have the &lt;strong&gt;piston reach the highest speed possible&lt;/strong&gt;. I don't know how to calculate what is better.. should i increase the diameter? - should i keep the diameter the same but make the syringe longer.. what is better?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When i release the piston - the atmospheric pressure will make the piston move pretty fast - and blow up the head of the syringe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a real world example. Just something for me to learn. I don't know what the piston friction with the walls is - but is the same for all 4 syringes so does not matter when comparing them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also i don't know the mass of the piston either.. i mean we can use a gut number like 0.25 kg. And the bigger piston is 1kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now how to calculate all this?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DMfJu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DMfJu.png&quot; alt=&quot;how to calculate the speed and force of the piston in a vacuum syringe?&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also i'm interested in forming an intuition of what to expect given changes to the design of this syringes rather then exact numbers. What happens if i increase the diameter 3 times.. is it better or worse .. that kind of thing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does it matter how long the syringe is? Can it happen that if i make it too long then the piston is actually slowed down? Is this piston constantly accelerating while is in the syringe and there is still vacuum in front of it .. or not actually? Can air to air turbulence slow things down? Where is the highest acceleration at? Maybe in the middle of the syringe ?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say this because if we go down to 1 mm tube, then capillarity is so high that you need to push on it like crazy to make things move. I'm thinking about what force is needed to squeeze that fluid in a medical equipment (needle + tube). So when it comes to tubes like this, maybe having a higher diameter is better?! Maybe this applies to air also?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can put the numbers in a calculator and get the result, but i'm interested more on how to think about this, not just the formula. How do you break it down in your mind - and not be wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But i do want to lean how calculate the exact numbers as well. Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="12155" LastActivityDate="2019-11-04T18:43:33.730" Title="How to calculate the speed and force of the piston in a vacuum syringe?" Tags="|pistons|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31740" PostTypeId="1" CreationDate="2019-10-31T18:27:52.330" Score="1" ViewCount="80" Body="&lt;p&gt;I am looking at specs of an air conditioner unit. It uses DC inverter technology and maximum current listed as 21.5A&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So are DC inverter motors considered resistive or inductive? What &lt;a href=&quot;https://en.wikipedia.org/wiki/Utilization_categories&quot; rel=&quot;nofollow noreferrer&quot;&gt;utilization category&lt;/a&gt; this device is in? AC-21, 22 or 23?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because I am looking for an isolating/safety switch and it is rated for AC-21 as 40A but 25A for AC-22 and 20A for AC-23&#xA;&lt;a href=&quot;https://www.katko.com/fi/tuote/kum-116u/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.katko.com/fi/tuote/kum-116u/&lt;/a&gt;&#xA;So it is difficult to decide if this switch is underrated for the air conditioning unit or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is what current rating should the isolating/safety switch should be for use with 21.5A DC inverter motor system? Obviously the answer also depends on the utilization category that the spec is based on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="23352" LastActivityDate="2019-10-31T18:27:52.330" Title="DC Inverter motor AC utilization category" Tags="|motors|power|safety|refrigeration|current|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31741" PostTypeId="2" ParentId="22230" CreationDate="2019-10-31T19:54:16.823" Score="1" Body="&lt;p&gt;In your force/momentum balance equations you always can include the pressure of the gas. The only time you can ignore it, is if the difference in &lt;span class=&quot;math-container&quot;&gt;$PA$&lt;/span&gt; is small compared to other forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example in a diffuser at the entrance you have a lower pressure and lower area than at the exit. This results in a significantly lower &lt;span class=&quot;math-container&quot;&gt;$PA$&lt;/span&gt; at the entrance, more than making up for the reward momentum loss, resulting in net forward thrust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could calculate the same thrust value by integrating the pressure across the wall of the diffuser. This might result in a more intuitive sense of why the diffuser gets pushed forward, while a converging nozzle will be pushed rearward (and the diverging portion of a supersonic nozzle will be pushed forward again)&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2019-10-31T19:54:16.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31743" PostTypeId="2" ParentId="25784" CreationDate="2019-11-01T12:03:32.337" Score="1" Body="&lt;p&gt;The pringling is caused by the positive (compression) hoop stress causing buckling around the perimeter. The positive hoop stress is created by the radial tension trying to pull the edges towards the middle as the center is displaced axially. Since there's not enough space for all the glass to move inwards, some of it goes up and some down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can see the same effect take place if you lay a piece of paper on top of a drinking glass and then push down in the middle. The center will move down and then the paper will buckle / fold such that some parts go up and some parts go down allowing the perimeter to move inward. As the material get thicker, the resistance to buckling goes up and folds/ bends get less extreme. This is why you'll only get a saddle rather than folding/ shattering.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastActivityDate="2019-11-01T12:03:32.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31744" PostTypeId="1" CreationDate="2019-11-01T12:24:13.610" Score="0" ViewCount="23" Body="&lt;p&gt;Would an engine that electrically generates ozone from the oxygen in air then uses it as fuel be efficient or possible? (This is not a Newtonian question – we’re talking chemical reaction not conservation of energy.)&#xA;Or what if ozone was used as an additive to the air intake for possible better fuel mileage on gas vehicles? (Those electric arc lighters produce ozone and are marketed as being wind proof so I wonder if an arc were safely maintained within the air intake if that would provide any boost?)&lt;/p&gt;&#xA;&lt;p&gt;Ozone will combust in certain concentrations and even when in a solution of water if temperature is raised to the combustion temperature (which is quite low.)&lt;/p&gt;&#xA;&lt;p&gt;As for the use of ozone as an energy source I believe the safest way to do so would be to make it on the fly and not store any amount that is not immediately used because it is a strong oxidizer and unstable.&#xA;Years ago I remember using a small van der graaf generator and recall that particular smell of ozone that is emitted while it runs. It doesn’t take long — maybe just a few seconds before I could smell the ozone in the air after turning on the machine. Also, with those electric arc lighters you can smell ozone being generated when you use.&lt;/p&gt;&#xA;&lt;p&gt;So ultimately I’m asking: would it be possible to use a mix of ozone and air as the source of combustion inside an internal combustion engine? Or what about an ozone and water solution plus air for use as combustion source for combustion engines? What would be the byproduct of this?&lt;/p&gt;&#xA;&lt;p&gt;Ozone as energy source for electric automobiles, jets, air planes etc?&lt;/p&gt;&#xA;&lt;h1&gt;Disclaimer: If you’re reading this please do not try at home. Ozone is dangerous at certain concentrations.&lt;/h1&gt;&#xA;" OwnerUserId="13921" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-11-01T12:24:13.610" Title="Ozone as an energy source?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31746" PostTypeId="2" ParentId="31730" CreationDate="2019-11-01T14:50:03.460" Score="0" Body="&lt;p&gt;I doubt there is significant cast surface affect in ductile cast iron. Other than roughness , the only surface affect should be decarburization . Cast irons have high carbon levels of about 3 % so losing a little carbon does not make much difference. Cast steels with much lower carbon content , such as 0.2 % ,can be much more affected by mold surface carbon loss. For an inexpensive material like cast iron , I expect the component is made thicker if there is any concern about strength.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-11-01T14:50:03.460" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31747" PostTypeId="1" AcceptedAnswerId="31753" CreationDate="2019-11-01T15:11:50.163" Score="0" ViewCount="196" Body="&lt;p&gt;I need a heated surface for a project and was thinking of using one of &lt;a href=&quot;https://www.amazon.co.uk/dp/B0758CQN6M/ref=cm_sw_r_cp_apa_i_5OeVDb9FW7WFR&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt;  PCB heater beds. But I'd need to drill a few holes in it - can I do this without stopping it working? Thanks&lt;/p&gt;&#xA;" OwnerUserId="11394" LastEditorUserId="6767" LastEditDate="2020-05-05T20:45:02.180" LastActivityDate="2020-05-05T20:45:02.180" Title="Drilling PCB 3d printer bed" Tags="|heating-systems|3d-printing|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31748" PostTypeId="2" ParentId="31721" CreationDate="2019-11-01T15:23:33.800" Score="0" Body="&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R_4 \parallel R_5 $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_3$&lt;/span&gt; form a voltage divider:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_A = \frac {R_4 \parallel R_5}{R_4 \parallel R_5 + R_3} \times (V_1 - V_2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or alternatively using series/parallel rules and Ohm's Law:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_A = \frac {V_1 - V_2}{R_4 \parallel R_5 + R_3}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$V_A = I_A \ (R_4 \parallel R_5)$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2019-11-01T15:23:33.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31749" PostTypeId="2" ParentId="31710" CreationDate="2019-11-01T15:47:56.527" Score="1" Body="&lt;p&gt;Your circuit works.  Proper ground connection.&lt;/p&gt;&#xA;&lt;p&gt;Your battery would be +12V -(-8V) = 20V.  Voltage at Point A = 12V, Point B = 9V, Point C = 4V and Point D = 0V.  The circuit is a good place to start.&lt;/p&gt;&#xA;&lt;p&gt;To get 12V at 6mA, you need &lt;span class=&quot;math-container&quot;&gt;$R_{L_3} = \frac {V}{I} = \frac {12V}{6mA} = 2kΩ$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Same process for 9V at 18mA and 4V at 4mA.  Three of the 7 resistors are known.  To get those voltages and currents, those are the resistors.&lt;/p&gt;&#xA;&lt;p&gt;The rest of the resistors must be selected to support those resistors. &lt;span class=&quot;math-container&quot;&gt;$R_1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$R_4$&lt;/span&gt; are in series (hense the name, loaded voltage divider) and each load resistor is in parallel with an individual series resistor.&lt;/p&gt;&#xA;&lt;p&gt;So load currents are 6mA, 18mA and 4mA.  We also need a little extra for &lt;span class=&quot;math-container&quot;&gt;$R_3$&lt;/span&gt;.  Make that 4mA, to make the math easier.&lt;/p&gt;&#xA;&lt;p&gt;So &lt;span class=&quot;math-container&quot;&gt;$I_T = 6mA + 18mA + 4mA + 4mA = 32mA$&lt;/span&gt;.  This is also current in &lt;span class=&quot;math-container&quot;&gt;$I_{R_4}$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;The voltage across &lt;span class=&quot;math-container&quot;&gt;$R_4$&lt;/span&gt; is: &lt;span class=&quot;math-container&quot;&gt;$V_{R_4} = 0 - (-8V) = 8V$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;So &lt;span class=&quot;math-container&quot;&gt;$R_4 = \frac {V_{R_4}}{I_T} = \frac {8V}{32mA} = 250Ω$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R_3$&lt;/span&gt;: At Point C, you want 4V and &lt;span class=&quot;math-container&quot;&gt;$R_3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_{L_1}$&lt;/span&gt; are in parallel. So:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_{R_3} = V_{R_{L_1}} = 4V - 0V = 4V$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This may seem obvious, but you need to do this for the following steps.  &lt;span class=&quot;math-container&quot;&gt;$V_C$&lt;/span&gt; - &lt;span class=&quot;math-container&quot;&gt;$V_D$&lt;/span&gt; = Voltage Drop across &lt;span class=&quot;math-container&quot;&gt;$R_3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Loads 1 to 3 are connected to Point D.  So their currents do not pass through &lt;span class=&quot;math-container&quot;&gt;$R_3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_{R_3} = I_T - I_{R_{L_3}} - I_{R_{L_2}} - I_{R_{L_1}} = 32mA - 6mA - 18mA - 4mA = 4mA$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The extra current I added.  Know current and voltage for &lt;span class=&quot;math-container&quot;&gt;$R_3$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R_2$&lt;/span&gt;: You want 9V at Point B and 4V at Point C.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_{R_2} = V_B - V_C$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$R_2$&lt;/span&gt; is connected to &lt;span class=&quot;math-container&quot;&gt;$R_3$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R_{L_1}$&lt;/span&gt;.  This means the current for those resistors must flow through &lt;span class=&quot;math-container&quot;&gt;$R_2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_{R_2} = I_{R_3} + I_{R_{L_1}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Know current and voltage for &lt;span class=&quot;math-container&quot;&gt;$R_2$&lt;/span&gt;.&lt;/p&gt;&#xA;&lt;p&gt;Only one left is &lt;span class=&quot;math-container&quot;&gt;$R_1$&lt;/span&gt;. Same process.  I leave this to you, but this will get you going in the right direction.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2020-11-25T23:04:17.020" LastActivityDate="2020-11-25T23:04:17.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31750" PostTypeId="2" ParentId="31708" CreationDate="2019-11-01T16:26:35.347" Score="4" Body="&lt;p&gt;Because it is not a power cable (Q1) using power connectors (Q2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BNC cable has a solid conductor and a ground braid to electrically shield and mechanically protect the conductor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Electrical reason:&lt;/strong&gt; Some ground braids are made of steel, which is not as good a conductor as copper, so voltage losses to the wire would be higher than copper wire.  Basically, wasted power.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current creates heat (&lt;span class=&quot;math-container&quot;&gt;$I^2 R$&lt;/span&gt; losses).  The ampacity rating of a power cable is determined by the maximum temperature the cable's insulation can handle.  The BNC wire was never meant to carry power, so the current flowing may cause the insulation to break down and cause a fire.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Safety reason:&lt;/strong&gt; The main reason BNC cable should not be used to send power is because of the connectors.  The outside of the metal BNC connector is uninsulated, so you have a shock risk, if you disconnect the load terminal while power is being applied.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Common sense reason:&lt;/strong&gt; You may know what you have done so are not at risk, but an unsuspecting person is &lt;strong&gt;NOT&lt;/strong&gt; expecting power on a signal cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you found an insulated BNC connector and added a warning sign, if it was disconnected a user may connect a normal connector negating your bad practice.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From: &lt;a href=&quot;https://yaoota.com/en-ng/product/generic-bnc-male-connector-uhf-walkie-talkie-antenna-28-inch-price-from-jumia-nigeria&quot; rel=&quot;nofollow noreferrer&quot;&gt;Generic BNC Male Connector UHF Walkie Talkie Antenna 2.8 Inch For Kenwood TK-388&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sw7g2m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sw7g2m.png&quot; alt=&quot;Insulated BNC Connector&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No one expects power over a BNC cable.  So power distribution over BNC cables is bad practice.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2019-11-01T19:46:35.377" LastActivityDate="2019-11-01T19:46:35.377" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31751" PostTypeId="2" ParentId="31738" CreationDate="2019-11-01T16:29:50.900" Score="1" Body="&lt;p&gt;If we ignore the friction and assume total vacuum ( both unrealistic) the force on the piston  with the larger diameter increases. But it is reasonable to assume the mass of the piston increases by the same ratio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F=ma$&lt;/span&gt; and say &lt;span class=&quot;math-container&quot;&gt;$KF=Kma$&lt;/span&gt; with K being the ratio of area of the cylinder, meaning acceleration doesn't change. of course if you have the mass of the piston differently just plug it in the above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And obviously the longer syringe will allow the work done on the piston increase, hence its final speed increase. &lt;span class=&quot;math-container&quot;&gt;$W=F*x  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final speed will be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ v_{final}= \sqrt{2xa} \  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the kinetic energy at the time of impact&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ E= 1/2 m_{piston}*v_{final} ^2$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-01T16:48:19.650" LastActivityDate="2019-11-01T16:48:19.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31752" PostTypeId="2" ParentId="31717" CreationDate="2019-11-01T18:36:55.153" Score="1" Body="&lt;p&gt;Sieve trays usually create less pressure drop, which is good for energy costs. Sieve trays are cheaper to make, so that's good for upfront capital requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bubble cap trays and valve trays may not provide enough liquid to liquid contact based on their design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being easy to clean may be a big enough reason to prefer the sieve tray, especially if one of the liquid phases has some solids or solids accumulate over time in the column. &lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2019-11-01T18:36:55.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31753" PostTypeId="2" ParentId="31747" CreationDate="2019-11-01T19:09:18.977" Score="1" Body="&lt;p&gt;If it is constructed like &lt;a href=&quot;https://reprap.org/wiki/PCB_Heatbed&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt; then at best you'll have dead stripes where you've drilled through conductors; at worst you'll short some adjacent stripes together and you'll (probably) get hot spots.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PCB layout software is cheap; you may want to see if you can get the part made as specified in the Wiki I link to (or use a service that uses gold flash rather than tin plate -- gold flash is much thinner, so it won't disturb the resistivity of the copper as much).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you can, design a board with holes.  Run an annular ring of copper around each hole, with stripes going into and out of the rings.  Current will conduct around the holes, then go back into stripes where it'll do it's job of heating up the board.  Without a lot of painful calculation the heating won't be perfectly uniform, but it should be pretty good -- or you can do a lot of painful calculation, and get it spot on.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-11-01T19:09:18.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31754" PostTypeId="1" CreationDate="2019-11-01T19:19:18.833" Score="0" ViewCount="36" Body="&lt;p&gt;Check out this 200-300 bar cylinders:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bwn39.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bwn39.jpg&quot; alt=&quot;200 to 300 bar high pressure cylinders &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it practically possible to build a bigger cylinder of say: &lt;code&gt;10m x 1m&lt;/code&gt; - and make it hold a pressure of &lt;code&gt;300 bar&lt;/code&gt; in a safe way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand that bigger cylinders might not exist due to the fact that are not needed - not practical/commercial in industry to have something bigger then a human can move around - but for what im thinking, a bigger cylinder might be needed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we build it with steel and using standard practices? Or we already reached some limits and this is why we settled for this cylinders with dimensions like in the picture? Sorry if this is a stupid question. Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="12155" LastActivityDate="2019-11-01T19:19:18.833" Title="Are we capable of building bigger high pressure cylinders?" Tags="|compressed-air|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31755" PostTypeId="2" ParentId="31718" CreationDate="2019-11-01T19:34:43.980" Score="1" Body="&lt;p&gt;I think you might be mixing up the intentions of the two analysis tools you you mentioned - Root Cause Analysis and Design of Experiments.   Consider thinking of it this way:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Root Cause Analysis is something you do in order to help identify, well, the &quot;Root Cause&quot; of some issue or problem.  There are multiple ways to go about doing a root cause analysis in the same way there are multiple way to clean your house. Over time, industry has developed a number of analysis &quot;tools&quot; that can be used to &lt;em&gt;help&lt;/em&gt; in figuring out a root cause.  Short of explaining each tool, for now I suggest just reading up a bit on some of the following techniques:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;5-Whys: Basically keep asking &quot;why&quot; it happened until you get to an answer that you can control&lt;/li&gt;&#xA;&lt;li&gt;Fishbone Diagram: visual tool to help map out the different sources that can contribute to the failure - machine related, material related, people related, environment related, etc.&lt;/li&gt;&#xA;&lt;li&gt;Pareto Charts, Scatter Charts, or other graphs: help to identify trends or interactions that might signal when or how a problem started.  &lt;/li&gt;&#xA;&lt;li&gt;Failure Modes and Effects Analysis (FMEA): A very methodical tool for evaluating and how various root causes/issues drive different failure modes and their effects which also considering their severity and a number of other factors.  This is what's considered a &quot;bottom-up&quot; approach to root cause analysis. &lt;/li&gt;&#xA;&lt;li&gt;Fault Tree Analysis (FTA): The &quot;top-down&quot; brother of an FMEA where a fault is traced down through all the possible chains of events that could lead to it. &lt;/li&gt;&#xA;&lt;li&gt;Anticipatory Failure Determination (AFD): A creative approach where you try to figure out HOW to actually MAKE the issue happen.  In your case, if you &lt;em&gt;wanted&lt;/em&gt; to MAKE seal to leak, how would you do it?  &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are all tools that are regularly used in risk based industries (think auto manufacturers, medical devices manufacturers, aerospace, etc.) but certainly in lots of other places as well. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the second topic you mentioned - Design of Experiments - while also an option to help with root cause determination, it more suited to helping test a theory, or develop a design or specification.  It's commonly used in determining how variation in different inputs can impact the output.  Consider something as simple as boiling an egg.  Your inputs could be a number of things: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Amount of Water&lt;/li&gt;&#xA;&lt;li&gt;Temperature of Water&lt;/li&gt;&#xA;&lt;li&gt;Time spent boiling&lt;/li&gt;&#xA;&lt;li&gt;Amount of salt added to water&lt;/li&gt;&#xA;&lt;li&gt;Age of the egg&lt;/li&gt;&#xA;&lt;li&gt;Size of the pot&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;A DOE is performed to analyze the compound interactions between this.  In a very high level summary the team would perform the following steps:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First decide what they wanted to consider a &quot;correctly&quot; boiled or &quot;good&quot; version of a hard boiled egg WAS. What are our specifications for what we WANT. &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then there are some ways to eliminate inputs or &quot;factors&quot; which don't really influence the outcome.  The more factors we have to evaluate the more complicated the DOE.  In this case, we can probably reasonably eliminate:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The size of the pot (unlikely to impact the finished egg - assuming of course its big enough to fit the egg and hold water)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;Age of the egg maybe?  Do we feel that this strongly influences the final product?&lt;/li&gt;&#xA;&lt;li&gt;Amount of water - as long as it's boiling (and we don't break the egg) then this probably doesn't matter. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This leaves us with 3 &quot;factors&quot; that we now want to evaluate (and which also seem reasonable):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Temperature of the Water&lt;/li&gt;&#xA;&lt;li&gt;Amount of salt added&lt;/li&gt;&#xA;&lt;li&gt;Time spent boiling. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Once a DOE is set up, we would define a number of different experiments where we boiled an egg while varying those parameters in known amounts. Eventually we'd end up with some data that we could statistically analyze to help us understand what input factors we want to control.  We'd end up finding that the temperature of the water is somewhat influenced by the amount of salt added. We'd also learn that the amount of time we needed to boil was correlated with the temperature.  Eventually, through our analysis, we'd likely come to the conclusion that while temperature is important, the amount we raise it by adding an acceptable amount of salt, doesn't impact the way our egg end up nearly as much as does the time we actually spend boiling it!  Not long enough and we end up with raw egg.  Too long and we turn it to dried rubber.  Our DOE has helped us identify that, when we now go to boil eggs, controlling the time we boil is the best way to guarantee us an egg made exactly the way we want it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, while this may seem a super trivial example, hopefully you can envision how the concept of a DOE as a &quot;tool&quot; could be extremely useful. And just as there were multiple different tools for performing RCA, the same exits for design analysis!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps to better understand how these tools relate. &lt;/p&gt;&#xA;" OwnerUserId="4897" LastEditorUserId="4897" LastEditDate="2019-11-01T20:52:10.817" LastActivityDate="2019-11-01T20:52:10.817" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31756" PostTypeId="1" CreationDate="2019-11-01T20:00:27.723" Score="-1" ViewCount="2698" Body="&lt;p&gt;[House Plan Image]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VEPiy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VEPiy.jpg&quot; alt=&quot;house plan image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My neighbor is not allowing us to wall plaster of the side outer wall.(Facing Neighbor side) How can I plaster the outer wall, which has a gap of 1 feet between our wall and neighbor enemy's wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Similar side wall with only bricks]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IaRG1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IaRG1.jpg&quot; alt=&quot;Similar side wall with only bricks&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please suggest the best technique to do side outer wall plastering.&#xA;Thanks in advance!!!!&lt;/p&gt;&#xA;" OwnerUserId="23364" LastEditorUserId="13297" LastEditDate="2019-11-01T20:21:27.267" LastActivityDate="2019-12-01T21:01:50.303" Title="Help! How to cement plaster outer wall. If gap between the two walls are 1 feet only?" Tags="|civil-engineering|masonry|reinforcement|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31757" PostTypeId="2" ParentId="31756" CreationDate="2019-11-01T20:17:16.680" Score="1" Body="&lt;p&gt;Depends on how wide your wall is. But there are pumps with small flexible hoses of 20 feet long. You can use a 20 foot pipe or handle and reach the spray nuzzle to the right space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to one manufacturer, &lt;a href=&quot;https://www.graco.com/us/en/contractor/product/24t834-toughtek-m680a.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;concrete pump.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-01T20:17:16.680" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31758" PostTypeId="1" CreationDate="2019-11-01T21:01:55.880" Score="0" ViewCount="121" Body="&lt;p&gt;I was doing the air flow in a 90 degree bend experiment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the first run, I connected the manometer to the tappings on the outer wall, then I turned off the bench and changed the connections to the inner wall. Then, I turned on the air flow again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each time I done this I got a different reading on the manometer for the airbox pressure and the atmospheric pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air flow settings weren't changed and no change in room temperature or nothing. It was like 10 minutes difference between each run.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this?&lt;/p&gt;&#xA;" OwnerUserId="18345" LastActivityDate="2023-12-12T03:05:30.720" Title="Air flow bench connected with a manometer. Atmospheric pressure of the manometer changes each time the bench is restarted. Why?" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31759" PostTypeId="2" ParentId="31758" CreationDate="2019-11-01T22:18:39.830" Score="0" Body="&lt;p&gt;Not an answer per se, so don't downvote - just too much for a comment without formatting...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Different readings for atmospheric pressure seems odd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do you have a barometer in the lab - usually mercury, 1 metre high with a vernier scale and a temperature correction chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are measuring the inside and outside values relative to atmospheric without an absolute reference of atmospheric then not sure what you are actually getting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps a diagram would help us all.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-11-01T22:18:39.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31760" PostTypeId="2" ParentId="29810" CreationDate="2019-11-01T23:25:20.010" Score="1" Body="&lt;p&gt;Short Answer:  The capacitor is used &lt;strong&gt;with&lt;/strong&gt; an inductor to create a rotating magnetic field.  There are &lt;strong&gt;two&lt;/strong&gt; coils (inductors) in a single-phase motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you apply single-phase ac to a coil, you get a magnetic field which is proportional to the current.  You apply a sine wave current and you get a sine wave magnetic field.  Starts at 0, goes to a maximum NS, back to 0, to a maximum SN, back to 0, repeat.  This will not create rotation.  If you apply that to a motor with a magnetized rotor, the rotor vibrates but does not rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On a three-phase motor, the application of three-phase power creates a 360° revolving magnetic field because the three coils (A, B &amp;amp; C) experience different phases of current.  As shown by linked image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://www.tpub.com/doeelecscience/electricalscience2143.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;AC Motor Theory&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JW1OH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JW1OH.png&quot; alt=&quot;Magnetic Field created by three phase current&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But single-phase current does not create a revolving magnetic field, so we have to cheat.  We apply the single-phase current to a winding &lt;strong&gt;AND&lt;/strong&gt; use a capacitor to phase shift the applied voltage to another angle and apply that to a second winding.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This second winding reaches it's peak magnetic field not at the same time, so it creates a torque in one direction to cause rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;http://electriciantraining.tpub.com/14177/css/Figure-4-3-Two-Phase-Rotating-Field-89.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electrical Training&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AfmNW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AfmNW.png&quot; alt=&quot;Single-Phase Magnetic Rotation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are different types which use capacitors to create a phase shift. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Capacitor Start, which has a capacitor and coil for starting. Centripetal switch disconnects coil as it gets up to full speed.&lt;/li&gt;&#xA;&lt;li&gt;Capacitor Run has a capacitor and coil in the circuit at all times.  Better torque.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But some designs, Shaded Pole use just inductors with different sized wires to create torque, but torque is significantly lower so it is only useful for light loads.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, two coils must be used to create rotation in a single-phase motor.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2019-11-01T23:25:20.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31761" PostTypeId="1" CreationDate="2019-11-02T00:05:48.073" Score="1" ViewCount="74" Body="&lt;p&gt;I designed a simple U shape 2mm sheet metal bent mount for a 4kg outdoors CCTV camera. The mount was accidentally produced from 5052-H32 Aluminum instead of 6061-T6. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it common to use 5052 aluminium for such structural use-case? Or it is better to reproduce the mount? A static linear analysis shows that there is a factor of safety of 1.6 with g factor of 12.&lt;/p&gt;&#xA;" OwnerUserId="23366" LastEditorUserId="90" LastEditDate="2019-11-02T02:40:21.327" LastActivityDate="2020-11-26T22:04:40.400" Title="Mounting outdoors camera using aluminum 5052" Tags="|structural-engineering|structural-analysis|structures|strength|aluminum|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31762" PostTypeId="2" ParentId="31731" CreationDate="2019-11-02T00:53:27.543" Score="0" Body="&lt;p&gt;TLDR: No this is not feasible.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's run some numbers.  Let's just say you want Earth normal gravity, 1g = 9.8 m/s^2.  Rour elevator starts at one side of your structure, and starts accelerating until it gets to the center.  At that point, it's going maximum speed. You need then to start decelerating. Then you'll get to the other edge of the sphere and reverse the process. So you'll have +1g for some time, and then -1g for some time, then +1g again, etc.. When you flip from +1g to -1g, what seems to be the floor becomes the ceiling and vice versa. This flipping is going to be the problem.  At best, it will be an inconvenience. At worst the person in the elevator will get nauseous and vomit.  The feasibility question comes down to how long can you go before you have to flip?  Let's just say that you wanted to make it 5 minutes between flips. The distance traveled under constant acceleration is given by &lt;span class=&quot;math-container&quot;&gt;$X = 0.5at^2$&lt;/span&gt;.  For a=9.81 m/s^2 and t=5*60, the radius of your sphere is going to be 441 kilometers or 275 miles. For reference the moon is about 1000 miles.  So your sphere needs to be one quarter of the size of the moon.  That's going to be impossible to build for a very long time.  Maybe technology will get there eventually, but not soon.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you could mitigate the size a bit by saying that you only want 1/2 of earth normal gravity, and you're willing to go 1 minute between flips. Flipping once a minute will be super inconvenient.  And the size is going to be 8.8 km or 5.5 miles long. Still way out of reach of current technology (for reference the ISS is just over 100 m long).  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-11-02T00:53:27.543" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31763" PostTypeId="2" ParentId="26638" CreationDate="2019-11-02T07:03:22.030" Score="-1" Body="&lt;p&gt;You can request contact displacements ('CDISP') in Field output requests.&#xA;Field output requests -&gt; F-Output-1 -&gt; Under the 'Contact' menu, select CDISP&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Vizualization module, select Primary and CDISP.&#xA;This will plot the contact displacements. &lt;/p&gt;&#xA;" OwnerUserId="23368" LastActivityDate="2019-11-02T07:03:22.030" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31765" PostTypeId="2" ParentId="30054" CreationDate="2019-11-02T16:25:00.323" Score="-1" Body="&lt;p&gt;You can do - the process called &quot;bumping&quot;. You can see how it works - with the backgauge of the press brake you push the metal and perform a lot of bends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the process to make light poles in industry. You will never get small diameters as 1 inch.&lt;/p&gt;&#xA;" OwnerUserId="23370" LastEditorUserId="1832" LastEditDate="2019-11-08T01:04:17.143" LastActivityDate="2019-11-08T01:04:17.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31767" PostTypeId="2" ParentId="31761" CreationDate="2019-11-02T17:16:15.580" Score="1" Body="&lt;p&gt;A quick check on the net shows the 5052 has a lower strength than the 6061 , mostly caused by heat-treatment. Tensile is 25% lower, yield is 30 % lower ; nominal properties.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-11-02T20:38:57.327" LastActivityDate="2019-11-02T20:38:57.327" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31768" PostTypeId="2" ParentId="31734" CreationDate="2019-11-02T17:23:05.687" Score="1" Body="&lt;p&gt;There seem to me to be three main things that appear initially confusing about the derivation. I believe they both have to do with the &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2} dpdA$&lt;/span&gt; term. I think the &lt;span class=&quot;math-container&quot;&gt;$-dp(A+dA)$&lt;/span&gt; term is quite self explanatory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the three main points are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Why is the term of the form &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2} dpdA$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) Why doesn't the &lt;span class=&quot;math-container&quot;&gt;$dA$&lt;/span&gt; term take into account the geometry of our streamtube (i.e. why doesn't the area use the formula for the side surface area of a frustum instead of using &lt;span class=&quot;math-container&quot;&gt;$dA$&lt;/span&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(3) Why doesn't the resultant force take into account the angle of deviation of the side of the streamtube when calculating the force, (i.e. why is the term &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2} dpdA$&lt;/span&gt; instead of &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2} dpdA \theta$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; is the angle of deviation from the axis of symmetry of the streamtube.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to part (1) is that when we remove the constant pressure &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; acting on all surfaces, we notice that we are left with a linearly increasing pressure, from the 'left' of streamtube, to the 'right'. If you are familiar with distributed loads from structural mechanics, the general formula for calculating the average force across a section with varying force, is &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}Fl$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; is the force distribution and &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; is the length of our section. This is analogous to how we treat the pressure, i.e. think of it as a 'pressure distribution' of sorts. The average pressure is then just &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}dp$&lt;/span&gt;. Note here that we merely assume a linear increase of the pressure across the volume. (I do not know the answer for why this is a valid assumption for full disclosure.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer part (2), This is simply because we are dealing with such small changes in area the actual geometry doesn't figure into the derivation. An extremely tiny change in surface area of an extremely tiny cube is exactly analogous to an extremely tiny change in area of a frustum for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to part (3) is simply that pressure is isotropic. Thus regardless of the angle, we know the force in the direction of &lt;span class=&quot;math-container&quot;&gt;$S$&lt;/span&gt; will still be &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{n}p\partial S$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{n}$&lt;/span&gt; is the unit vector in the specified direction, &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the pressure, and &lt;span class=&quot;math-container&quot;&gt;$\partial S$&lt;/span&gt; is the surface element. Have a look at this &lt;a href=&quot;http://www.atm.damtp.cam.ac.uk/people/mem/FLUIDS-IB/dyn.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.atm.damtp.cam.ac.uk/people/mem/FLUIDS-IB/dyn.pdf&lt;/a&gt; which will give a much better (and detailed) explanation than I have here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this covers the more odd circumstances (they were at least the most peculiar aspects of the formula to me ). Note I assumed the other term &lt;span class=&quot;math-container&quot;&gt;$-dp(A+dA)$&lt;/span&gt; is straightforward; it is simply an expression of force acting on the 'right' side surface of the element.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps somewhat.&lt;/p&gt;&#xA;" OwnerUserId="5469" LastActivityDate="2019-11-02T17:23:05.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31770" PostTypeId="1" CreationDate="2019-11-03T08:51:26.170" Score="0" ViewCount="263" Body="&lt;p&gt;I want to count human presence in a room through vibration sensing of concrete floor. For this I want an accelerometer with low frequency detection capability. I have used ADXL-345 which is not much sensitive for floor vibration sensing. Also piezo (see figure) sensors are not much sensitive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gdACB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gdACB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21033" LastEditorUserId="21033" LastEditDate="2019-11-04T11:01:08.263" LastActivityDate="2019-11-04T11:01:08.263" Title="which kind of accelerometer/vibration sensor are suitable for sensing vibration of building floor caused by human foot strike?" Tags="|vibration|" AnswerCount="3" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31771" PostTypeId="1" AcceptedAnswerId="31772" CreationDate="2019-11-03T09:02:58.810" Score="3" ViewCount="139" Body="&lt;p&gt;I am a recent electronics engineering graduate, currently legally working as an embedded software engineer in a company.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the future, I would like to become a freelancer, like signing up on a freelancing website and getting hired to do real projects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, as a side/weekend gig, I help engineering students with their diploma projects. It's obviously illegal but it's a very common practice around here. But frankly, it's a great opportunity for me to earn good money and learn my trade with real hands-on experience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the future, when looking for real customers, they will likely ask for a portfolio of my work. What am I going to show them? Some diploma projects of random people that I supposedly made? I'm afraid that would look very unprofessional.&lt;/p&gt;&#xA;" OwnerUserId="23373" LastActivityDate="2019-11-03T12:02:58.380" Title="How do I transition to legit freelancing, coming from a background of unethical projects?" Tags="|electrical-engineering|ethics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31772" PostTypeId="2" ParentId="31771" CreationDate="2019-11-03T12:02:58.380" Score="2" Body="&lt;p&gt;If you in the USA there are couple of good ways to start building a profile. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Volunteer at your local &lt;a href=&quot;https://www.firstinspires.org/robotics/frc&quot; rel=&quot;nofollow noreferrer&quot;&gt;FIRST robotics&lt;/a&gt; team. This is good place to get started learning new skills and using current skills&lt;/li&gt;&#xA;&lt;li&gt;Be an active member of the communities such as &lt;a href=&quot;https://engineering.stackexchange.com/questions&quot;&gt;engineering stackexchange&lt;/a&gt; where you can start building an online profile of you skills. You can asked embedded engineering type of question as well as answer engineering questions. &lt;/li&gt;&#xA;&lt;li&gt;You can contribute to open source projects, through platforms such as github. &lt;/li&gt;&#xA;&lt;li&gt;Solving a problem such as &lt;a href=&quot;https://engineering.stackexchange.com/questions/31770/which-kind-of-accelerometer-vibration-sensor-are-suitable-for-sensing-vibration&quot;&gt;Which kind of accelerometer/vibration sensor are suitable for sensing vibration of building floor caused by human foot strike?&lt;/a&gt; in open source platform help build a technical profile.&lt;/li&gt;&#xA;&lt;li&gt;Sharing you solutions via platform such as youtube, or other social media platform can help build a technical profile. &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://perscholas.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Par Scholas&lt;/a&gt; is great place to volunteer to share your technical skills, learn new skills and a build a technical project portfolio &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://turing.io/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Turing School of Software and Design&lt;/a&gt; is another good place to start building a technical profile. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://insights.dice.com/2017/01/11/tech-pros-volunteer-fun-ways/&quot; rel=&quot;nofollow noreferrer&quot;&gt;How Tech Pros Can Volunteer in Fun Ways&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.yearup.org/about-us/vision-mission-values/?location=national-us/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Year up&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://chicktech.org&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chick Tech&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://techbridgegirls.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Tech Bridge Girls&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.bizible.com/blog/volunteer-opportunities-for-tech-professionals&quot; rel=&quot;nofollow noreferrer&quot;&gt;10 Volunteer Opportunities For Tech Professionals To Transform Their Communities&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-11-03T12:02:58.380" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31775" PostTypeId="1" CreationDate="2019-11-03T15:15:13.980" Score="1" ViewCount="1573" Body="&lt;p&gt;A lifting device consists of an electric motor, a gearbox and a drum with a cable for lifting a mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The speed after the initial acceleration is &lt;span class=&quot;math-container&quot;&gt;$v=0.5\,\text{m/s}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gear efficiency &lt;span class=&quot;math-container&quot;&gt;$\eta=0.7$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum motor torque is &lt;span class=&quot;math-container&quot;&gt;$T=15 \, \text{Nm}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor inertia &lt;span class=&quot;math-container&quot;&gt;$J_m=0.005 \, \text{kgm}^2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Drum diameter &lt;span class=&quot;math-container&quot;&gt;$d = 0.5 \, \text{m}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor rotating speed after initial acceleration &lt;span class=&quot;math-container&quot;&gt;$n=2600\, \text{rpm}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The inertia of the drum &lt;span class=&quot;math-container&quot;&gt;$J_L$&lt;/span&gt; is considered small and thus neglected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to Calculate how much time it takes for the motor to accelerate the mass &lt;span class=&quot;math-container&quot;&gt;$m=510 \,\text{kg}$&lt;/span&gt; from standstill upwards to the given speed. Assume constant acceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2NtHZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2NtHZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know applying Newton's second law that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F-mg=ma$$&lt;/span&gt; and then we get the acceleration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$a=\frac{F-mg}{m}$$&lt;/span&gt; and the speed v&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$v=at$$&lt;/span&gt; so &lt;span class=&quot;math-container&quot;&gt;$$t=\frac{mv}{F-mg}$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motor rotates the geared system &lt;span class=&quot;math-container&quot;&gt;$\omega_M$&lt;/span&gt; and the geared system rotates the drum &lt;span class=&quot;math-container&quot;&gt;$\omega_L$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; causes the torque &lt;span class=&quot;math-container&quot;&gt;$T_L=mgr$&lt;/span&gt; on the rope drum. So the power of the load is &lt;span class=&quot;math-container&quot;&gt;$P_L=T_L\omega_L$&lt;/span&gt;. And therefore &lt;span class=&quot;math-container&quot;&gt;$\eta P_m=P_L$&lt;/span&gt; and there I get the &lt;span class=&quot;math-container&quot;&gt;$\omega_L$&lt;/span&gt; when &lt;span class=&quot;math-container&quot;&gt;$$\omega_L=\frac{\eta\, T_m\, \omega_m}{T_L}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do I get force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; so I can calculate the time or is there better way to do this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or can I calculate the acceleration &lt;span class=&quot;math-container&quot;&gt;$a = \frac{T}{J_{motor}}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="23320" LastEditorUserId="23320" LastEditDate="2019-11-03T15:29:25.217" LastActivityDate="2019-11-06T21:38:02.003" Title="Motor, a gearbox and a drum with a cable lifting a mass" Tags="|mechanical-engineering|control-engineering|gears|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31776" PostTypeId="1" CreationDate="2019-11-03T15:18:02.700" Score="1" ViewCount="39" Body="&lt;p&gt;I have to approximate the red function with Fourier harmonics. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Prt1O.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Prt1O.jpg&quot; alt=&quot;***enter image description here***&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am calculating complex &lt;strong&gt;Ck&lt;/strong&gt; the following way. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/21Hgv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/21Hgv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure what to put for period - &lt;strong&gt;T&lt;/strong&gt; (6 or 2pi or what?). What integral &lt;strong&gt;limits&lt;/strong&gt; to use (2/6*2pi and 4/6*2pi ??)?&#xA;The way I showed on the picture gives me invalid results... so how to fix it?&lt;/p&gt;&#xA;" OwnerUserId="23378" LastEditorUserId="23378" LastEditDate="2019-11-03T15:32:14.670" LastActivityDate="2019-11-03T19:09:16.237" Title="Fourier serie signal approximation" Tags="|electrical-engineering|telecommunication|signal-processing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31777" PostTypeId="2" ParentId="31770" CreationDate="2019-11-03T17:44:54.167" Score="1" Body="&lt;p&gt;As suggested in the comments PIR detector, ultrasonic sensor or micro phones are alternate to detect the motion or humans in floor. Accelerometer or piezo transducer might be the better option to detect human presence through vibration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional I will also take look at stain gauge or pressure sensors. The once that I am thinking might need to be connected at a pivot point. I am not sure how this will work on a concert floor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/10ov7.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/10ov7.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ej30c.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ej30c.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on working on similar problems having a good mechanical coupling is essential to get accurate and strong signal from the sensors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It appears that you are using hot glue to attach the ADXL 345 sensor module to the floor. My experience working with hot glue, is that hot glue tends to absorb some to the vibration. It is important to get a good mechanical coupling between the sensing element and the concrete floor. I would try to use Cyanoacrylates (super glue) to attach the ADXL 345 sensor module to the concrete floor to get a good mechanical coupling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: If you use Cyanoacrylates to obtain a good mechanical coupling you will not be able to remove the ADXL 345 sensor module from the floor without destroying the ADXL 345 sensor module.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.vishaypg.com/docs/11297/250lw.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;General Purpose Strain Gages—Linear Pattern&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.sciencedirect.com/topics/materials-science/electromechanical-coupling&quot; rel=&quot;nofollow noreferrer&quot;&gt;Electromechanical Coupling&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/3348/calculating-pitch-yaw-and-roll-from-mag-acc-and-gyro-data/3350#3350&quot;&gt;Calculating pitch, yaw, and roll from mag, acc, and gyro data&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-11-03T22:35:36.183" LastActivityDate="2019-11-03T22:35:36.183" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31778" PostTypeId="2" ParentId="31776" CreationDate="2019-11-03T19:09:16.237" Score="1" Body="&lt;p&gt;If you already know the explicit form of the function &lt;span class=&quot;math-container&quot;&gt;$f(x)$&lt;/span&gt;, then it's possible to find the periode by solving this equation &lt;span class=&quot;math-container&quot;&gt;$f(x) = f(x+P)$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a graphical representation of the function (finite), maybe by collecting the numerical values or other means then you have to follow the next two steps: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step one: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Trivial extension&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the general finite signal &lt;span class=&quot;math-container&quot;&gt;$f \in [[a, b) \rightarrow \mathbb{C}]$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$p= b-a \in \mathbb{R}$&lt;/span&gt; then its trivial extension is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$f_{triv} =\begin{cases}f(x)&amp;amp;x\in [a, b)\\ 0&amp;amp; otherwise \end{cases}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step two: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Periodic extension&lt;/strong&gt;&#xA; &lt;span class=&quot;math-container&quot;&gt;$$f_{per}  = \sum_{m \in \mathbb{Z}} f_{triv}(x-mp)$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and now you can talk about the periodic function and its Fourier representation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The periode is indeed six.  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-11-03T19:09:16.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31780" PostTypeId="2" ParentId="31770" CreationDate="2019-11-03T23:13:23.767" Score="1" Body="&lt;p&gt;Before you start trying to design the electronics, find out the range of what you are actually trying to measure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an order of magnitude, consider a very rough analysis of a 2 meter square concrete slab, 25mm thick and supported only at the edges, loaded with a 200Kg mass applied in a small area in the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertical deflection of the slab was about &lt;strong&gt;&lt;em&gt;0.005mm&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course a real concrete floor is going to be better supported than that simple model, so the deflections will be less. Also the model ignored any reinforcing bars in the concrete, which will also reduce the flexibility of the floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good luck trying to measure any deformation or acceleration with an amplitude of the order of micrometers, in the uncontrolled environment of a building. IMO the only option that might work would be something like a laser doppler vibrometer mounted on the ceiling - so long as you don't care about blinding the people in the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other ways of detecting humans, of course. The fact that they continuously emit a few hundred watts of heat would be pretty good place to start from when designing your system.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-11-03T23:13:23.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31781" PostTypeId="2" ParentId="31770" CreationDate="2019-11-03T23:46:02.230" Score="1" Body="&lt;p&gt;The vibration of floors can be in the order of millimeters, depending on the dynamic properties of the slab.  Vibration of modern large span slabs especially with shallow depth thanks to prestressed hi tensile strength bars under pedestrian steps is noticable. Dynamic impact of a person's step can be many times larger than their static weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most basic way is to cast a laser beam in small tangential angle onto a tray of water from 5 cm and video the vibration of the Ray bouncing of the surface  of the turbulenced tray onto a surface distanced 500 cm so you get 100x magnifying power. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-04T00:13:41.177" LastActivityDate="2019-11-04T00:13:41.177" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31782" PostTypeId="1" CreationDate="2019-11-04T01:53:56.963" Score="0" ViewCount="34" Body="&lt;p&gt;Not sure if an example is needed, but &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B009S750LA&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;here is one&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QAFsS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QAFsS.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm curious as to how the joint connected to the monitor itself &quot;remembers&quot; the last position the monitor was moved to. In my naiveté, I imagine the rotation being enabled by grease, so how is it that the monitor does not always rotate back towards the direction of gravity's pull?&lt;/p&gt;&#xA;" OwnerUserId="6737" LastEditorUserId="110" LastEditDate="2019-11-04T02:01:26.900" LastActivityDate="2019-11-04T07:17:18.123" Title="How does a monitor arm hold the monitor in place after rotating to a new position?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31783" PostTypeId="2" ParentId="29601" CreationDate="2019-11-04T03:12:34.857" Score="1" Body="&lt;p&gt;This is not a trivial question.  More trick, than straight-forward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only way to set two sides to look alike is if one face is non-linear.  Intersection appears to be at approximately 40% to 45% from front and top.  If it was at 50%, it would be a triangle from the sides. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have distorted the curve to illustrate it.  Top to intersection is straight.  Intersection to bottom is curved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hNpJL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hNpJL.png&quot; alt=&quot;Orthogonal Views.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2019-11-04T03:12:34.857" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31784" PostTypeId="1" CreationDate="2019-11-04T03:32:29.250" Score="0" ViewCount="74" Body="&lt;p&gt;I'm building a scale model for a proposed construction project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to simulate foundations operating in soil in a scale model?  What do people typically use?&lt;/p&gt;&#xA;" OwnerUserId="9350" LastEditorUserId="90" LastEditDate="2022-08-03T00:17:56.970" LastActivityDate="2022-08-03T00:18:21.587" Title="Modelling Soil in a Scale Model" Tags="|modeling|geotechnical-engineering|surface-modelling|soil-mechanics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31785" PostTypeId="1" CreationDate="2019-11-04T05:04:00.453" Score="2" ViewCount="241" Body="&lt;p&gt;If one were to search on Amazon for &quot;Test Indicator Flexible Holders&quot; you will turn up many items using this mechanism (please ignore the magnetic base; I'm not focusing on that part, just the red components and the knob):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ufkv4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ufkv4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can watch this video to see how they work:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/wVtT6xP_yjo&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/wVtT6xP_yjo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is clear from that video that, indeed, that single knob tightens down both (ball?) joints, both the joint mounted to the top of the magnetic base, but also the joint closest to the holder that is intended to hold the oscilloscope probe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is two-fold (primary/important question is first):&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Exactly how does this entire mechanism act to clamp both arms rigidly, in space? From the video and the pictures, it is not readily apparently what clamping action is at play here. My only guess is that there is some clamping action acting upon both links, but how can the single big black knob act to clamp all (ball?) joints?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Does this entire mechanism have some &quot;standard&quot; name in the industry? I know about &lt;a href=&quot;https://en.wikipedia.org/wiki/Ball_joint&quot; rel=&quot;nofollow noreferrer&quot;&gt;ball joints&lt;/a&gt;, but are these really just ball joints? I speculate that these aren't true ball joints, because of the single knob being sufficient to clamp them all rigidly, leading me to speculate that the combination of joints, linkage arms, and joints have some standard name.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="14357" LastEditorUserId="14357" LastEditDate="2019-11-04T15:16:56.050" LastActivityDate="2019-11-04T16:42:36.350" Title="How do Test Indicator Flexible Holders hold rigidly when tightened?" Tags="|mechanical-engineering|mechanisms|terminology|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31786" PostTypeId="2" ParentId="31782" CreationDate="2019-11-04T07:17:18.123" Score="1" Body="&lt;p&gt;All the pivot joints are “stiff” due to being tightened sufficiently with friction so once in place they don’t move - except for flexing of the mechanism.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-11-04T07:17:18.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31787" PostTypeId="1" CreationDate="2019-11-04T07:30:38.617" Score="0" ViewCount="487" Body="&lt;p&gt;I have a Scenario where my lock nut is tightend using a prevailing torque of 14 Nm (as mentioned by the manufacturer) on to the rotor placed between bearings (But the tightening direction and the rotor rotating direction is same unfortunately). The lock nut is loosened after particular life cycle of usage as the rotor rotates with a acceleration of 1230 rad/sec and every time it starts, the sudden acceleration loosens the lock nut. Is there a simple approch to calculate a appropriate prevailing torque for this Situation so that the lock nut does not loosen up when the rotor starts accelerating towards it's max RPM of 9400 in 1 second?&lt;/p&gt;&#xA;" OwnerUserId="14552" LastActivityDate="2024-03-20T08:07:45.977" Title="Break loose torque calculation of lock nut against rotating/acceleating bolts?" Tags="|mechanical-engineering|applied-mechanics|experimental-physics|industrial-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31788" PostTypeId="1" CreationDate="2019-11-04T07:46:47.210" Score="2" ViewCount="77" Body="&lt;p&gt;A typical folded bellows allows for axial expansion/compression, lateral offset, and angular deflection. Is there a bellows fold pattern that can accommodate a relatively large torsional rotation about the longitudinal axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Motivation&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking to prototype a waterproof (IPX67 or IPX68) flexible housing for DSLR cameras with large zoom lenses. A picture is worth a thousand words:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wENVW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wENVW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;&lt;a href=&quot;https://shuttermuse.com/author/dan-carr/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Dan Carr&lt;/a&gt; testing the RainCoat Pro 2 on a grizzly bear photo trip in British Columbia. From &lt;a href=&quot;https://shuttermuse.com/lenscoat-raincoat-2-review/&quot; rel=&quot;nofollow noreferrer&quot;&gt;shuttermuse.com&lt;/a&gt; Photo: &lt;a href=&quot;https://www.facebook.com/RowanThorntonPhotography/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Rowan Thornton&lt;/a&gt;&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photographer in the kayak is using a rain coat, which merely protects from rain. But the thought of my kayak turtling with $10k worth of camera and lens exposed makes me pucker. So I'd like to begin prototyping a flexible watertight bellows to surround the lens and mate with a flexible &quot;bag&quot; covering the camera body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, a regular bellows (like a bellows plunger) will allow for a zoom lens to accommodate changes in length. The problem is accommodating control of the zoom ring on the lens, which is torsional rotation of the outside of the bellows. I anticipate the maximum rotation is about 135° (+/- 65° or so from mid-zoom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a fold pattern to allow a cylindrical surface to be torsionally deformed like I'm describing?&lt;/p&gt;&#xA;" OwnerUserId="7022" LastActivityDate="2020-04-02T16:00:40.180" Title="Is there a bellows geometry that allows for torsional motion?" Tags="|mechanical-engineering|flexures|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31789" PostTypeId="1" CreationDate="2019-11-04T10:21:24.140" Score="2" ViewCount="76" Body="&lt;p&gt;We have a strange issue while trying to configure small router for bridging between available access point and ethernet network. While this AP is perfectly visible from desktops and mobile devices, it is not visible form the bridging router. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder how can it be? Visible WiFI it is just radio signal. Are there any advertised properties which can be filtered on?  &lt;/p&gt;&#xA;" OwnerUserId="23388" LastActivityDate="2020-07-31T14:02:50.317" Title="How can a Wi-Fi access point be visible from one device but not from another?" Tags="|wireless-communication|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31790" PostTypeId="1" CreationDate="2019-11-04T10:50:35.973" Score="1" ViewCount="1449" Body="&lt;p&gt;I am going to design a bowling machine. In this design, I'm going to use two motors that are being placed vertically and their shaft are attached with two horizontal identical wheels with a space of ball in between these two wheels.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/k1miv.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k1miv.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The wheen diameter is &lt;code&gt;360mm&lt;/code&gt; and the weight is &lt;code&gt;3kg&lt;/code&gt; each. the ball is &lt;code&gt;75mm&lt;/code&gt; dia and the weight is &lt;code&gt;160 grams&lt;/code&gt;. I need to calculate torque and rpm required to throw the ball at &lt;code&gt;170km/h&lt;/code&gt; velocity. The cricket pitch length is &lt;code&gt;22 meter = 2200mm&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Required Torque:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The required acceleration is a need to throw the ball at &lt;code&gt;170km/h&lt;/code&gt; means &lt;code&gt;47.22m/s&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$v^2=u^2+2fs$&lt;/span&gt; where intial velocity is &lt;span class=&quot;math-container&quot;&gt;$u=0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;span class=&quot;math-container&quot;&gt;$f = v^2/2s$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;accleration &lt;span class=&quot;math-container&quot;&gt;$f = 50m/s^2 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force required &lt;span class=&quot;math-container&quot;&gt;$ F = ma$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$m =$&lt;/span&gt; total weight of two wheels + wight of ball &lt;span class=&quot;math-container&quot;&gt;$= 6+0.160$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$= 6.016$&lt;/span&gt; kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;span class=&quot;math-container&quot;&gt;$F = 308N$&lt;/span&gt;  i.e, &lt;span class=&quot;math-container&quot;&gt;$154N$&lt;/span&gt; in each wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume rubber to leather kinetic friction &lt;span class=&quot;math-container&quot;&gt;$u = 0.4$&lt;/span&gt; it's not actually &lt;/p&gt;&#xA;&#xA;&lt;p&gt;kinetic Frictional force &lt;span class=&quot;math-container&quot;&gt;$Fr = 1.4 * 154 = 215.6$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Required torque &lt;span class=&quot;math-container&quot;&gt;$Ta = F * r$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$= 215.6 * (0.36/2) = 36.8 Nm$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Required RPM:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the linear velocity of the ball is &lt;span class=&quot;math-container&quot;&gt;$v=47.22m/s$&lt;/span&gt; &#xA;The velocity of the ball is an average velocity of two-wheel &lt;span class=&quot;math-container&quot;&gt;$v = (v1+v2)/2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the maximum velocity of the ball with no spin, the velocity of two-wheel must be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, &lt;span class=&quot;math-container&quot;&gt;$v1 = v = v2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ao, the angular velocity of each wheel should be &lt;span class=&quot;math-container&quot;&gt;$v = d × w × 0.001885$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, &lt;span class=&quot;math-container&quot;&gt;$w = v/(d * 0.001885)$&lt;/span&gt;RPM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$=2510$&lt;/span&gt;RPM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know I am right or wrong?&lt;/p&gt;&#xA;" OwnerUserId="23369" LastEditorUserId="23369" LastEditDate="2019-11-04T14:43:39.400" LastActivityDate="2019-11-04T14:43:39.400" Title="how to calculate required torque and rpm to through a 160 gram cricket ball horizontally at 160km/h velocity?" Tags="|mechanical-engineering|motors|automotive-engineering|torque|friction|" AnswerCount="0" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="31791" PostTypeId="2" ParentId="31789" CreationDate="2019-11-04T11:45:43.470" Score="1" Body="&lt;p&gt;Based on the post here are few things to look for and some publicly available open source tools. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First you might want to check the operating frequency, 2.4 GHz, 5GHz or both. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you are setting up a wireless network it is important to know surrounding in your case for the bridge router. There are many obstacles that can attenuate the wireless signal. Some of these are furniture, walls, metal objects etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I use wireshark as my wifi analyzer. Tool is a can be overwhelming. This tool is mostly text based. Below is image of the tool&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/i1qcX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i1qcX.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another tool to take a look at is wifi analyzer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LvEhU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LvEhU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are wifi analyzers for both iOS and Android. You might want do some research and find one that fits your needs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at these and let us know if these were helpful. &lt;/p&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-11-04T12:51:09.257" LastActivityDate="2019-11-04T12:51:09.257" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31792" PostTypeId="2" ParentId="31788" CreationDate="2019-11-04T14:19:54.973" Score="1" Body="&lt;p&gt;I don't see the neccesity for a bellows. I suggest you make an oversized tube from a waterproof, flexible plastic. Oversized means the diameter is significantly larger than the tele, then you should have enough flex. for the torsion. Be mindful that only the length ahead of the mount it really available for torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Take your time with accepting this answer, I have never designed such a system.)&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2019-11-04T14:19:54.973" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31793" PostTypeId="2" ParentId="31785" CreationDate="2019-11-04T14:36:03.843" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/12730/how-does-the-central-lock-mechanism-in-articulated-arm-magic-arm-works&quot;&gt;How does the central lock mechanism in articulated arm (magic arm) works?&lt;/a&gt; provides the answer to #1 for how the mechanism works.  And, due to now being able to search online for the right terminology (see #2 below), I also found &lt;a href=&quot;https://youtu.be/rOYCkSRMCx8&quot; rel=&quot;nofollow noreferrer&quot;&gt;a YouTube video&lt;/a&gt; that does a teardown of a &quot;Articulating Arm&quot; works &lt;a href=&quot;https://youtu.be/rOYCkSRMCx8?t=100&quot; rel=&quot;nofollow noreferrer&quot;&gt;at 100 seconds into the video&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For #2, I could not find a standard name for the mechanism, so I list several below (and it turns out that these are apparently not all the same mechanism). &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&quot;&lt;a href=&quot;https://www.hoffmann-group.com/US/en/hus/Metrology/Height-gauges,-measuring-stands/3-D-jointed-arm/p/442114&quot; rel=&quot;nofollow noreferrer&quot;&gt;3-D jointed arm&lt;/a&gt;&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;&lt;a href=&quot;https://www.bhphotovideo.com/c/product/616251-REG/Pearstone_9041660_9041670_8_3_21_1cm_Articulating.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Single-Point Locking Mechanism&lt;/a&gt;&quot; or &quot;Articulating Arm&quot;&lt;/li&gt;&#xA;&lt;li&gt;&quot;Triple-Jointed Arm&quot;: &lt;a href=&quot;https://www.amazon.com/s?k=Starrett%20660%20Base%20Indicator%20Holder%20With%20Triple-Jointed%20Arm&amp;amp;tag=duckduckgo-20&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amazon search for Triple-Jointed Arm&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&quot;Articulating Rosette Arm&quot;: &lt;a href=&quot;https://www.amazon.com/s?k=articulating%20rosette%20arm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amazon search for articulating rosette arm&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&quot;Magic Arm&quot;: &lt;a href=&quot;https://www.amazon.com/s?k=Magic%20Arm&amp;amp;ref=nb_sb_noss_2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Amazon search for magic arm&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;h2&gt;Difference between Articulating Arm, Magic Arm and Variable Friction Arm&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Apparently, a &quot;Variable Friction Arm&quot; is not the same as a &quot;Magic Arm&quot;, according to &lt;a href=&quot;https://youtu.be/kkfEMJA4vbU?t=48&quot; rel=&quot;nofollow noreferrer&quot;&gt;a YouTube video at 48 seconds in&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;00:48 Now, strictly speaking, what you see here is a variable&#xA;  friction arm, not a magic arm. Magic Arms tighten and loosen by&#xA;  swinging around a big ratchet. The problem is they tend to break&#xA;  quicker than the more sturdy knobs, so I prefer these versions.&#xA;  Don't worry though everyone calls the magic arms.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And, from &lt;a href=&quot;https://hoothollow.com/Tip-April-May%202009%20-%20Bogen%20Arms.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Bogen Articulated Arm and Super Clamp article&lt;/a&gt; we have a good explanation of the differences: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;While I've just described all three Arms, my recommendation for newcomers is the least expensive of the three, the Articulated Arm. In this model, there are three different controlling levers for controlling the movement of the wrist, elbow, and shoulder joints. While that may seem cumbersome compared to the one locking lever of the Magic Arm, or the variable friction knob of the Variable Friction Magic Arm, and it is a bit slower to use, those disadvantages are offset by the control you have. With either of the other two, you must have the Arm's position set exactly where you want it when you lock it into place. If an adjustment needs to be done, the lock must be loosened and the fine-tuning done, at the risk of the other joints going out of place.&#xA;  With the Articulated Arm, if I need to raise the flash a bit all I need do is loosen one joint -- perhaps the elbow, perhaps the wrist -- and move the flash into place. After doing so, I might find I'm still 'off' by a bit, and I might loosen another joint, maybe the shoulder this time, and do my fine-tuning. While this takes longer than doing the adjustment by loosening an Arm with just one knob, I think you can see that this also allows very precise and methodical adjustment changes.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If anyone has any information on any standardized name, that would be helpful, but at least I have names I can search for in order to source them from multiple suppliers:&lt;/p&gt;&#xA;" OwnerUserId="14357" LastEditorUserId="14357" LastEditDate="2019-11-04T16:42:36.350" LastActivityDate="2019-11-04T16:42:36.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31794" PostTypeId="1" AcceptedAnswerId="31804" CreationDate="2019-11-04T14:43:03.233" Score="0" ViewCount="428" Body="&lt;p&gt;I need to devise a way to ensure only NG passes through a pipe, and no condensates or other liquids. It is ok to shut down the pipeline if liquids would be detected. The dimensions involved are small and space limited.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a valve which shuts if liquid starts passing through but stays open for gas? Aside from a involved solution with a trap with a sensor attached, is there any other way to solve this?&lt;/p&gt;&#xA;" OwnerUserId="23389" LastActivityDate="2019-11-04T22:34:21.717" Title="Valve/Filter allowing through gas only, no liquid" Tags="|fluid-mechanics|gas|multiphase-flow|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31795" PostTypeId="1" CreationDate="2019-11-04T15:19:38.130" Score="1" ViewCount="32" Body="&lt;p&gt;I would like to know, which antenna 802.11 model, it your reach can be controlled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a some researches, I'm discovered a tool called &quot;iwconfig&quot;, native on some Operation systems Unix Like. It is used by own OS to manager interfaces wireless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some options of tool:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;interface essid {NNN|any|on|off}&#xA;interface mode {managed|ad-hoc|master|...}&#xA;interface freq N.NNN[k|M|G]&#xA;interface channel N&#xA;interface bit {N[k|M|G]|auto|fixed}&#xA;interface rate {N[k|M|G]|auto|fixed}&#xA;interface enc {NNNN-NNNN|off}&#xA;interface key {NNNN-NNNN|off}&#xA;interface power {period N|timeout N|saving N|off}&#xA;interface nickname NNN&#xA;interface nwid {NN|on|off}&#xA;interface ap {N|off|auto}&#xA;interface txpower {NmW|NdBm|off|auto}&#xA;interface sens N&#xA;interface retry {limit N|lifetime N}&#xA;interface rts {N|auto|fixed|off}&#xA;interface frag {N|auto|fixed|off}&#xA;interface modulation {11g|11a|CCK|OFDMg|...}&#xA;interface commit (Para salvar alterações)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I'm using an device USB TP-Link N TL-WN823N, 2.4 Ghz&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this devices is basic. I can only set some settings like, Mode of operation (ad hoc, monitor, manager), channel, essid...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe that I can control your reach, I should to manager the txpower or sensibility (sens). When I try to set it, the system returns that operation is not supported by device. Possibly is because this device is not dedicate to used like access point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I would like to know if are some antenna 802.11 that the reach can be controlled by txpower, sensibility or something&#xA;physical to break the reach &lt;/p&gt;&#xA;" OwnerUserId="23390" LastActivityDate="2019-11-04T15:19:38.130" Title="How I can controller the reach of WIFI antenna?" Tags="|wireless-communication|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31796" PostTypeId="2" ParentId="31723" CreationDate="2019-11-04T17:32:07.293" Score="0" Body="&lt;p&gt;Not sure if this is the sole cause for this (no.Data) error but. . .&#xA;Sitting the Bluetooth profile to (Hld) - (Virtual Keyboard) seems to have corrected this issue.  Measurement data arrives (in the app that has focus / cursor) as if this caliper were a keyboard attached to the PC. Very smooth and nice with a carriage return - linefeed appended.   &lt;/p&gt;&#xA;" OwnerUserId="23340" LastActivityDate="2019-11-04T17:32:07.293" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31799" PostTypeId="2" ParentId="31738" CreationDate="2019-11-04T18:43:33.730" Score="1" Body="&lt;p&gt;I think you're dismissing friction far too easily.&lt;/p&gt;&#xA;&lt;h2&gt;Frictionless&lt;/h2&gt;&#xA;&lt;p&gt;Since you asked for it here it is:&lt;/p&gt;&#xA;&lt;p&gt;The forces on the piston are pressure (from the atmosphere on the back of the piston) and friction. So we have:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P\,A=F=m\,a$$&lt;/span&gt;&#xA;Where &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; is pressure, &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; is area, &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; is mass, and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; is acceleration.&lt;/p&gt;&#xA;&lt;p&gt;Since the area, pressure, and mass are constant, the accerations must be as well. Thus we can use the equations of constant acceleration:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d=\frac12 a t^2$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\Delta V=a\,t$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Combining these two can give you final velocity in terms of distance and acceleration:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V=\sqrt{2\,d\,a}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now we can do some more interesting things if we make some assumptions about how the piston mass scales with different dimension syringes. For example, we could assume that the density of the syringe was constant, and thus the mass was proportional to the length, and proportional to area:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$m=\rho\,d\,A$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;if we plug that in we get:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P\,A=\rho\,d\,A\,a$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The areas cancel. So the cross section of the picton wouldn't matter at all if our assumptions were correct.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$a=\frac{P}{\rho\,d}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V=\sqrt{2\,d\,\frac{P}{\rho\,d}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The length/ distance cancels out too!&#xA;&lt;span class=&quot;math-container&quot;&gt;$$V=\sqrt{2\frac{P}{\rho}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Now, from your images it actually looks like your piston is the same length regardless of the length of your syringe. This would result in the mass only being proportional to area:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$m=\rho_2\,A$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V=\sqrt{2\,d\,\frac{P}{\rho_2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;So in this case: the longer the faster.&lt;/p&gt;&#xA;&lt;p&gt;However, speed might not be the factor you really care about. If you care about how hard the piston is going to hit, you might care more about impulse or kinetic energy. Ballistics usually measure impact energy when concerned with how strong an impact will be.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$KE = \frac12 m\,v^2 = d\, P \, A$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The energy that you impart into the piston is always equal to the volume of your vacuum multiplied by atmospheric pressure. This is true regardless of the mass of your piston.&lt;/p&gt;&#xA;&lt;h3&gt;Friction&lt;/h3&gt;&#xA;&lt;p&gt;It's likely that friction plays a significant role in the final velocity.&lt;/p&gt;&#xA;" OwnerUserId="1122" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-11-04T18:43:33.730" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31800" PostTypeId="1" CreationDate="2019-11-04T18:56:57.857" Score="1" ViewCount="81" Body="&lt;p&gt;I want to repair an outdoor lamp (see images below) however I can't get the glass container open. The methods I found on the internet are based on glue removers or based on heat but I'm not sure this will damage the print inside, as this is what I try to repair. Note that I think it is glued, however if this is not the case please let me know. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I tried so far:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Wringing it with a screwdriver (the image on the side shows a clear 'edge' on top, however I can't get really deep&lt;/li&gt;&#xA;&lt;li&gt;Applying moderate heat&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Goals:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;No damage the the circuit board inside&lt;/li&gt;&#xA;&lt;li&gt;No damage to the glass container (if unavoidable I prefer breaking the bottom part which does not contain LED's)&lt;/li&gt;&#xA;&lt;li&gt;Usage of common house tools or tools that are fairly cheap to get (and safe for me to use on a kitchen table)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Images:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom view&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dAgiT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dAgiT.jpg&quot; alt=&quot;Top view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Top view&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VBFt7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VBFt7.jpg&quot; alt=&quot;Bottom view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Side view&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iMNUF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iMNUF.jpg&quot; alt=&quot;Side view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Related:&#xA;&lt;a href=&quot;https://engineering.stackexchange.com/questions/22463/usticking-two-glass-elements-fused-by-vacuum&quot;&gt;Loosening glass that is vacuum&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23394" LastActivityDate="2019-11-05T01:30:13.433" Title="Seperating two glass pieces with a circuit board inside" Tags="|materials|glass|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31802" PostTypeId="2" ParentId="29601" CreationDate="2019-11-04T19:17:42.333" Score="1" Body="&lt;p&gt;One possible solution sketched below (the OnShape model is &lt;a href=&quot;https://cad.onshape.com/documents/20ee9d17b1f144ac6f38cbcd/w/5b25e44888c1837c5117797f/e/cfd6f906c94a6d179ff18b16&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1iQAy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1iQAy.png&quot; alt=&quot;Orthographic projections&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j5Rza.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j5Rza.png&quot; alt=&quot;3D&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11760" LastEditorUserId="11760" LastEditDate="2019-11-05T16:11:26.877" LastActivityDate="2019-11-05T16:11:26.877" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31803" PostTypeId="1" CreationDate="2019-11-04T19:43:49.293" Score="1" ViewCount="200" Body="&lt;p&gt;Sawstop table saw braking system works by putting a low voltage on the blade, and when it detects that voltage shorted to ground, the blade is retracted and slammed into an aluminum block.  The claim is that it can do this in the time it takes 3 tooth to pass, and their video corroborates this.  Looks like a 60 tooth blade.&lt;/p&gt;&#xA;&lt;p&gt;The big downside to this:  You have to replace the brake cartridge at around $60 US. and often replace or at least service the blade.  No one likes it when they lose a 200 dollar saw blade.&lt;/p&gt;&#xA;&lt;p&gt;At 3600 rpm, that's 60 rev per second, or 3600 teeth per sec.  To stop in 3 teeth is 3/3600 or 1/1200 of a second or not quite 1 ms.  This is 1/20 of a single voltage cycle.&lt;/p&gt;&#xA;&lt;p&gt;The second downside is getting false trips from either wet wood or from nails in the wood.&lt;/p&gt;&#xA;&lt;p&gt;So what alternatives are technically possible?&lt;/p&gt;&#xA;&lt;p&gt;Ran into a guy on another forum who claimed that a VFD driven 3 phase motor could stop a table saw blade fast enough that the arbor nut would spin off.&lt;/p&gt;&#xA;&lt;p&gt;Can a VFD do this in 1 ms without destroying the motor?  (I'm thinking in terms of transient voltages) Presumably you have a big fat power resistor to soak up the energy.&lt;/p&gt;&#xA;&lt;p&gt;The other method I was considering as a sawstop alternative:&lt;/p&gt;&#xA;&lt;p&gt;A pair of spring loaded (think high performance engine valve spring) brake pads on a set of callipers round the bottom of the blade.&lt;/p&gt;&#xA;&lt;p&gt;A formula 1 engine is limited to a max 18,000 rpm. 3.3 ms/rev. Valve closings must happen in  a small fraction of that time.  So the time to close is ball park correct.  Can it grip hard enough to stop the blade in 15 degrees of the blade's rotation?&lt;/p&gt;&#xA;&lt;p&gt;The forces on the blade are considerable.  You have to stop the motor too.  Typically you're talking a 3-5 HP. These are not pick up in one hand motors.&lt;/p&gt;&#xA;&lt;p&gt;Options for this:  My snowblower (84&amp;quot; wide, 48 hp PTO) has a linkage with two 4&amp;quot; diameter 1/8&amp;quot; steel plates.  They are linked with a common 1/4&amp;quot; diameter grade V machine  bolt.  Hit a chunk of firewood with the snowblower, the bolt sheers.   Usually takes me longer to find a replacement bolt than to be up and running again.&lt;/p&gt;&#xA;&lt;p&gt;The second way would be like my bushhog mower, which has a clutch between the gearbox and the blade. Some industrial sewing machines do this too.&lt;/p&gt;&#xA;&lt;p&gt;What ways can you stop a moving device fast enough to save injury, but not damaging the machine?&lt;/p&gt;&#xA;" OwnerUserId="8768" LastEditorUserId="8768" LastEditDate="2022-03-31T16:58:13.877" LastActivityDate="2023-12-22T00:08:30.917" Title="Alternatives to sawstop technology" Tags="|mechanical-engineering|applied-mechanics|safety|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31804" PostTypeId="2" ParentId="31794" CreationDate="2019-11-04T22:34:21.717" Score="0" Body="&lt;p&gt;What are you searching for is called: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Thermostatic expansion valve or (TXV).&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have one of those in your air conditioner. Because the compressor of AC's are designed to handel only vapour. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-11-04T22:34:21.717" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31805" PostTypeId="1" CreationDate="2019-11-04T23:32:35.737" Score="3" ViewCount="1779" Body="&lt;p&gt;I'm in need of help finding a third-order or higher system in which I can derive a transfer function. We have a class project in which we need to find a real-life example of the system that equates to a 3rd order system or higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem that I have is that I do not know what a third-order system looks like in real life. I can find a lot of examples of 2nd order systems (acceleration, velocity, and displacement). But no 3rd order! Help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Project guidelines:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 1:&lt;/strong&gt; Find a system of interest to you, discuss why this system is of specific interest to you and why this is a good topic for the class project;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 2:&lt;/strong&gt; Model the system in three forms: differential equations, transfer function, and state-space representation. Note if the system is nonlinear, please linearize it first;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 3:&lt;/strong&gt; Find and plot system output under step input and another input of your choice. Discuss the physical implications;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Step 4:&lt;/strong&gt; Study the system stability, discuss the physical meaning of instability in your case;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 5: Assume a negative unity feedback system,  two possible controllers: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ C_1(s)=k_0 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ C_2s= k_p + k_ds+k_i \frac 1s $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;• Design the two controllers, so that the system is stable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;• With all your designed control parameter values, what are the feedback system’s output under step input and another input of your choice? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;• What are the steady-state errors under the four cases (2 controllers, 2 inputs)? &lt;/p&gt;&#xA;" OwnerUserId="23385" LastActivityDate="2022-01-17T04:27:27.030" Title="Controls Project: Help needed finding 3rd order system" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31806" PostTypeId="2" ParentId="31800" CreationDate="2019-11-05T01:30:13.433" Score="1" Body="&lt;p&gt;You can cut it carefully with a hacksaw with fine tooth blade while it's held in a vise. Or cut it with a small grinder held in the vise, using a tablesaw guide.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-05T01:30:13.433" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31807" PostTypeId="2" ParentId="31784" CreationDate="2019-11-05T02:35:02.270" Score="1" Body="&lt;p&gt;Modeling soil behavior in a scaled-down lab system requires that the force of gravity be scaled up to maintain dynamic similitude. This requires the scaled system to be spun in a centrifuge at a speed which is mathematically derived from the linear scale-down factor.&lt;/p&gt;&#xA;&lt;p&gt;As pointed out by ShadowMan, there is an extensive literature on scaled modeling of soil behavior in response to, for example, earthquake loading.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="90" LastEditDate="2022-08-03T00:18:21.587" LastActivityDate="2022-08-03T00:18:21.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31808" PostTypeId="1" CreationDate="2019-11-05T04:05:42.653" Score="4" ViewCount="3593" Body="&lt;p&gt;I'd like to create something similar to a hydraulic press. I know I need some sort of motor, but I was curious about a fundamental question about motors when used in a system where the force required to continue motion increases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;General question: if a motor is under stress and the intended motion is impeded to some degree, does the motor get damaged?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Literal example: what happens to the motor of a ceiling fan if you grab the fan to immediately stop it? Does this damage the motor? Are there motors that expect resistance and either increase the torque or stop immediately?&lt;/p&gt;&#xA;" OwnerUserId="6737" LastActivityDate="2020-01-02T22:51:05.327" Title="Does a motor suffer damage when resistance to the intended motion is met?" Tags="|mechanical-engineering|motors|hydraulics|servo|" AnswerCount="6" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31809" PostTypeId="2" ParentId="31808" CreationDate="2019-11-05T05:03:40.350" Score="1" Body="&lt;p&gt;The fan starts to act like a resistor and soon will heat excessively, till the breaker cuts the power off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all of the electrical motors have built-in fuses and breakers which will act to protect them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Combustion engines usually are tougher and if the torque demand increases beyond theire output they just stop cranking and you need to reignite them. However before that they could damage transmission and bearings and mechanical links along the power train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jet engines if forced to work outside their power  parameters will stall, which used to be catastrophic. Modern jets have built-in smart electronics to protect them.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-05T05:03:40.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31810" PostTypeId="2" ParentId="31808" CreationDate="2019-11-05T07:04:28.903" Score="9" Body="&lt;p&gt;The answer depends on the specific type of motor you intend to use, but in general: when a motor is loaded down, it draws more current, which means its windings get hotter at the same time that the cooling fan attached to the motor shaft is being slowed down. if this heat is allowed to build up, it will eventually cause the insulation on the windings to fail and short-circuit themselves, in which case the motor is ruined and may catch fire. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A better answer can be provided for you if you can get a little more specific about what type of motor you are thinking of. Brushed DC? Brushless DC? AC?&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-11-05T07:04:28.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31812" PostTypeId="2" ParentId="29601" CreationDate="2019-11-05T12:05:55.277" Score="6" Body="&lt;p&gt;Here is an attempt at going for the least amount of faces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2hk0H.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2hk0H.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jcma1.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jcma1.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user12750" LastEditorDisplayName="user12750" LastEditDate="2019-11-05T13:22:11.990" LastActivityDate="2019-11-05T13:22:11.990" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31813" PostTypeId="2" ParentId="29601" CreationDate="2019-11-05T13:03:38.197" Score="2" Body="&lt;p&gt;The bottom right line is meant for transferring lines between the top view and the Left view. Same with the Front and Left vertical line. Ideally, the picture would also include dotted lines to indicate any hidden geometry inside or under the view, but let's assume that there aren't any hidden geometries - since there aren't any hidden lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would help indexing the corners in each view, to transfer the geometry properly, and knowing where each line crosses one from another view. Indicate also the &quot;bottom&quot; corners on the front view as &quot;under&quot; the upper corners in the top view. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw straight lines out from the points from the top view, until you reach the bottom right line. Switch directions upwards, to project it to the left side view. Project the same point from the front view as well, and where the two lines intersect from the same points in the left view plane, is where you have the location of that point in the left view.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can work through this exercise precisely, with steps when I get home from work in a few hours :)&lt;/p&gt;&#xA;" OwnerUserId="23403" LastActivityDate="2019-11-05T13:03:38.197" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31814" PostTypeId="2" ParentId="31808" CreationDate="2019-11-05T15:48:08.703" Score="6" Body="&lt;p&gt;It depends on the specific type of motor, as @nielsnelson mentions, but it also depends on what you use to drive the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, a motor that's designed to run continuously isn't going to take well to being stalled for any length of time, unless its drive voltage (and frequency, if it's an induction or synchronous machine) is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are motors that are designed to operate under continuous stall conditions under full voltage; there are motor drivers (VFDs for AC machines, servo drivers for DC motors) that can be set up to protect motors under continuous stall conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you can certainly design for the possibility of the motor stalling, but you have to do so from the outset.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-11-05T15:48:08.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31815" PostTypeId="2" ParentId="31808" CreationDate="2019-11-05T17:17:09.287" Score="2" Body="&lt;p&gt;There can be another case: if the motor is expected to stall time to time or experience an overload condition as a part of its routine, motor shaft can be equipped with a ball detent torque limiter instead of a fixed gear. That way, the motor can keep spinning in idle mode without notable damage while its working piece is blocked-by.&lt;/p&gt;&#xA;" OwnerUserId="18483" LastEditorUserId="18483" LastEditDate="2019-11-06T17:58:30.753" LastActivityDate="2019-11-06T17:58:30.753" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31816" PostTypeId="1" CreationDate="2019-11-05T18:03:54.020" Score="0" ViewCount="59" Body="&lt;p&gt;Is there an adhesive that works well with ceramic and is heat-resistant? The application is a teapot that should sustain temperatures up to 100C-105C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B000O60ZKK&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Devcon 5-minute epoxy&lt;/a&gt; worked, but easily detached when 100C water was poured in. This is getting hard to find for non-industrial uses.&lt;/p&gt;&#xA;" OwnerUserId="23407" LastActivityDate="2019-11-05T18:03:54.020" Title="What adhesive can I use for a teapot?" Tags="|adhesive|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31818" PostTypeId="1" CreationDate="2019-11-05T19:57:29.707" Score="0" ViewCount="47" Body="&lt;p&gt;My friend uses carbon fiber tow in creating a component for automobile machines. However the carbon fiber is actually much tougher than it needs to be. As a risk manager in software I am concerned that his carbon fiber tow poses an unnecessary health risk for his workers working with this material, due to the high &lt;a href=&quot;https://www.carbonfiber.gr.jp/english/material/safety.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;airborne asbestos-like&lt;/a&gt; risk that may present itself in the manufacturing process.  Of course he educates his workers on this risk and they all use proper protective gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, carbon fiber tow is prone to breaking and producing airborne particles when handled by hand, and I'd like to know whether &lt;a href=&quot;https://www.adafruit.com/product/1088?gclid=Cj0KCQiA2ITuBRDkARIsAMK9Q7NzgCw6jahUBvJFcIZTDHljlQnShpsj8l0SmfPHBrzJh7g6-4S8oc0aAgVWEALw_wcB&quot; rel=&quot;nofollow noreferrer&quot;&gt;stainless steel tow&lt;/a&gt; is &lt;em&gt;less&lt;/em&gt; prone to breakages. But I am not sure what quantitative measurement to examine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For his application, stainless steel tow is a perfect mechanical substitute. Thank you for not questioning whether it is a perfect substitute for his automobile part, because it is. The question here is whether stainless steel tow is more or less prone to breakage than carbon fiber tow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two metrics that I began trying to learn more about that could be valid here. The first is &quot;Fracture toughness&quot; and the other is &quot;tensile toughness&quot;. I am really just trying to get data on which is better, or how I could scientifically prove the stainless steel tow is less prone to breakages. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can assume that the diameter of each fiber within the tow is about 5-15 micrometers. What's the best data we can gather on the subject?&lt;/p&gt;&#xA;" OwnerUserId="9573" LastEditorUserId="9573" LastEditDate="2019-11-05T22:12:58.070" LastActivityDate="2019-11-08T20:40:34.153" Title="Measuring fracturability / airborne fuzziness of carbon fiber tow vs stainless steel tow" Tags="|mechanical-engineering|materials|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31820" PostTypeId="2" ParentId="31818" CreationDate="2019-11-05T20:41:38.893" Score="2" Body="&lt;p&gt;On the net ; 3/8 cable has 11,000 pound breaking strength. The figure shows the individual wire to be about 0.025 in. ( 7 X 19 cable). When I put that into a conversion , the result is 635 micrometers : Apparently more specifics are needed. Also, asbestos has a relatively unique crystal with &quot;hooks&quot; the catch on the cilia in the lungs. The cilia are then unable to move asbestos out of the lungs like ordinary &quot;dirt&quot;. giving rise to cancer .  Why do you think carbon fiber has this unique property ? Stainless wire , although very high strength  still has enough ductility not to &quot;shatter&quot; and disperse small particles into the air when it is pulled apart.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-11-05T20:41:38.893" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31821" PostTypeId="1" CreationDate="2019-11-05T21:24:25.627" Score="0" ViewCount="36" Body="&lt;p&gt;I have data with the following normalized values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unit vector for cylinder axis:&#xA;-0.00270666, -0.01226211,  0.9999212 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normalized coordinates for cylinder center&#xA;0.01412114, 0.01074875, 0.00017004&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normalized cylinder radius&#xA;0.12903491&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any program I can use to display a cylinder with the following parameters? I know of programs like OpenSCAD that accept height and radius for a cylinder which can them be translated,rotated, etc. but I'm not sure how to use the unit vector of the axis to get the height (I assume for rotation I can use trignometric functions to derive the angle of the cylinder).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="23411" LastActivityDate="2019-11-05T21:24:25.627" Title="Program for creating a cylinder given its axis, center and radius?" Tags="|cad|geometry|computer-aided-design|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31822" PostTypeId="2" ParentId="31808" CreationDate="2019-11-05T21:41:31.287" Score="0" Body="&lt;p&gt;Consider that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Torque_converter&quot; rel=&quot;nofollow noreferrer&quot;&gt;torque converter&lt;/a&gt; in an automatic transmission is explicitly designed to support this scenario.  Whereas, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Limited-slip_differential&quot; rel=&quot;nofollow noreferrer&quot;&gt;limited-slip differential&lt;/a&gt; is designed to handle the opposite scenario, where &lt;em&gt;insufficient load&lt;/em&gt; is encountered (on one of the drive shafts).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every motor can be protected from overload conditions in a motor-appropriate way.  Apparently, hydraulic presses have their own set of techniques for &lt;a href=&quot;https://www.thefabricator.com/thefabricator/article/stamping/recovering-from-press-overload-protection-systems-help-to-sense-overload-and-minimize-machine-tool-damage&quot; rel=&quot;nofollow noreferrer&quot;&gt;overload protection&lt;/a&gt;, often involving shunting of the hydraulic fluid via valves to remove power.  Note that the motor is not the only or even most expensive component which can be damaged by overload conditions.  In the context of hydraulic presses, the die is obviously a valuable component that you generally want to protect, even if your motor could easily survive abuse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conclusion: design safety in from the start, rather than thinking about how badly you can abuse one component or another.&lt;/p&gt;&#xA;" OwnerUserId="23412" LastActivityDate="2019-11-05T21:41:31.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31823" PostTypeId="1" CreationDate="2019-11-05T21:48:55.290" Score="0" ViewCount="96" Body="&lt;p&gt;We recently purchased a new Lathe - Haas with Netshare, Next Gen Control. We are copying a file to the Haas Drive's (User Data) folder from a network PC. The problem we're seeing is random and intermittent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When things are working smoothly we can repeatedly overwrite existing program files on the Haas drive's (User Data) folder without problem. However, occasionally when we try to delete or copy-over the file a Visual Basic (70 Permission Denied) error is displayed on the PC we're copying (from).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, it seems that something on the Haas device if holding onto this file, the file in it's (User Data) folder that we're trying to delete or copy over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know what might be causing this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This (URL below) is the error message we're seeing if of any interest&#xA;&lt;a href=&quot;https://docs.microsoft.com/en-us/dotnet/visual-basic/language-reference/error-messages/permission-denied&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://docs.microsoft.com/en-us/dotnet/visual-basic/language-reference/error-messages/permission-denied&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23340" LastActivityDate="2019-11-05T21:48:55.290" Title="Permission Denied copying to (User Data) dir -Haas Next Gen Control" Tags="|cnc|lathe|networking|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31824" PostTypeId="1" CreationDate="2019-11-05T22:55:22.017" Score="0" ViewCount="36" Body="&lt;p&gt;The gear box will have little torque as I am only using a 12 by 18&quot; wing which rotates. The two shafts I am using are 1/4 &quot;, one form the wing and the other going the to pivot for the wing to circle, with lift.  I can't find 1/4&quot; gears and I am not sure how to attach them to the two shafts so that they cause rotation in different directions.  I know only two gears of equal size are needed, but I never made a gear box before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.  Thanks in advanced!  &lt;/p&gt;&#xA;" OwnerUserId="23297" LastActivityDate="2019-11-07T19:30:04.640" Title="small gearbox for experimental wind mill" Tags="|mechanical-engineering|gears|applied-mechanics|mechanisms|wind-power|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31825" PostTypeId="2" ParentId="31399" CreationDate="2019-11-05T23:05:19.967" Score="1" Body="&lt;p&gt;I have worked on the edge of this technology. the simple answer is the performance of the scram jet doesn't make uo for  the added weight of the system.  The most recent work NASA has performed was the HYPER-X program, try this link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.nasa.gov/centers/armstrong/news/FactSheets/FS-040-DFRC.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.nasa.gov/centers/armstrong/news/FactSheets/FS-040-DFRC.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23297" LastActivityDate="2019-11-05T23:05:19.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31827" PostTypeId="1" CreationDate="2019-11-06T01:27:32.030" Score="1" ViewCount="701" Body="&lt;p&gt;While gain margin(s) and phase margin(s) in Bode Plots of open loop transfer functions are defined at particular frequencies (when they exist) --- such as phase margin is defined at frequencies where the dB gain cuts through 0 dB ------ may I ask if these particular frequencies (ie. gain margin frequencies and phase margin frequencies) are points of interest only because we can conveniently define them at these frequencies? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking that, if we choose a particular frequency in a Bode Plot where the dB gain curve does not cut through 0 dB, and where the phase curve does not cut through -180 degrees, then there would simply be TWO degrees of freedom, such that a particular combination of additional gain and additional phase would push the system to an unstable condition (or onset of instability).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do gain margins and phase margins also (in some way) cover other regions of the Bode Plot - aside from the very specific frequencies where the gain margin(s) and phase margin(s) are defined?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks all!&lt;/p&gt;&#xA;" OwnerUserId="19788" LastEditorUserId="19788" LastEditDate="2019-11-06T04:22:26.720" LastActivityDate="2019-11-07T00:07:31.427" Title="gain margin and phase margin - significance" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31828" PostTypeId="2" ParentId="31787" CreationDate="2019-11-06T08:56:34.580" Score="0" Body="&lt;p&gt;First of all I would recommend you the documenzt VDI 2230 - &quot;Systematic calculation of high duty bolted joints&quot;. The document is bi-lingual (German and English) and among some (german) books the source of my post.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all some design 'tips' which ma be useful depending on your application and the tools at hand:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Use higher tightening torque if possible. The goal of the torque is to get a clamping force bewteen the parts after all. But this is heavily influenced by friction and geometry of the connection. &lt;/li&gt;&#xA;&lt;li&gt;You can use loc-tide (or similiar product) to secure the connection. There are several grades of strength an this will certainly prevent slack.&lt;/li&gt;&#xA;&lt;li&gt;Use nord-lock washers (or similiar product) since they increase the the loosening torque due to their geometry. &lt;/li&gt;&#xA;&lt;li&gt;Avoid normal washers if possible since they increase the number of contact surfaces and hence increase settling in of the connection. In the end this means less clamping force and thus less friction which is preventing your nut from slackening.&lt;/li&gt;&#xA;&lt;li&gt;Use a bolt with a smaller shank. They are more flexible and can store more energy in the connection. Better for fatigue and tends less to settling in.&lt;/li&gt;&#xA;&lt;li&gt;Use special friction-increasing paste below your mating nut surface. This is used e.g. by cyclists to lower their required tightening torque on carbon frames. YOu can use abrasive powder mixed with bearing grease as well. BUT there may be longterm effects due to the small particles beeing crushed. &lt;/li&gt;&#xA;&lt;li&gt;If feasible, drill a hole for a cotter-pin. This only prevents loosing the nut but the connection may slip.&lt;/li&gt;&#xA;&lt;li&gt;Use a make-up method other than torquing it up. Good systems stretch the bolt with hydraulics to achieve the neccessary clamping force. The nut is solely put on Ä'hand-tight'. No preload (=clamping) force is lost due to friction in the thread.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If you want to calculate, this requires quite a lot of information. The problem depends on the following parameters:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;angluar acceleration acting on the nut. I would take the peak value since this is conservative.&lt;/li&gt;&#xA;&lt;li&gt;rotary moment of interia of the nut. With the peak acceleration this gives you the torque to loosen your bolt.&lt;/li&gt;&#xA;&lt;li&gt;geometry of the parts involved. You want your bolt to be flexible and the flange (rotor) to be stiff to make a durable connection.&lt;/li&gt;&#xA;&lt;li&gt;Friction coefficients µ of the material pair. This is also influenced by surface coatings and lubrication. There is a HUGEuncertainity connected with this parameter.&lt;/li&gt;&#xA;&lt;li&gt;The material and surface finish of nut and rotor. Rough surfaces tend to settle in/relax and reduce the clamping force in the connection after a fewe cylces.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="23079" LastEditorUserId="23079" LastEditDate="2019-11-11T22:16:05.120" LastActivityDate="2019-11-11T22:16:05.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31829" PostTypeId="1" CreationDate="2019-11-06T09:29:36.947" Score="1" ViewCount="1010" Body="&lt;p&gt;I've been struggling to this for some time now and can't really find the answer on the web, but: sometimes when I want to set circle inside of an angle, it just throws it out on the wrong side. Like on the photo below: I wanted to insert circle between two lines and trim it (so it would look like red arc at the end), where I choose 2 lines (TTR command) where the green stars are, but the circle got dropped on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's going wrong here, what's the matter?&#xA;I had my mouse (while entering radius) on the right side of the vertical line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aLnFr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aLnFr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="5797" LastActivityDate="2022-11-15T02:14:21.510" Title="Tan Tan Radius in AutoCAD doesn't put circle on right side" Tags="|autocad|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31831" PostTypeId="1" CreationDate="2019-11-06T13:19:33.257" Score="0" ViewCount="38" Body="&lt;h2&gt;Assumptions&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Assume a small bore hole with a few mm diameter in a hard plastic plate (e.g. POM), and further assume that after drilling the hole the surface of the plate is grinded with a disk with fine sand paper (e.g. 20 µm mean grit diameter). This would remove a possible larger radius from the edge of the bore hole and sharpen it to a certain extent. How will the final shape of the edge of the bore be affected by the grit diameter?&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Questions&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is it under this conditions reasonable to assume an approximate fillet radius of the bore hole's edge in the same range as the particle radius of sand paper?&lt;/li&gt;&#xA;&lt;li&gt;If not, is the expected fillet radius more likely to be larger or smaller than the particle radius?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="23415" LastEditorUserId="23415" LastEditDate="2019-11-06T15:16:41.937" LastActivityDate="2019-11-06T15:16:41.937" Title="Can the fillet radius of an edge shaped by grinding be guessed from the grit size of sandpaper?" Tags="|grinding|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31832" PostTypeId="1" CreationDate="2019-11-06T15:36:19.263" Score="0" ViewCount="91" Body="&lt;p&gt;I have developed a design for a river cleaning machine which can cut and collect weeds. The problem which is facing is how to rotate the circular blade which is attached in front of the machine. I mean first is to directly attach on a dc motor's shaft but there are less to zero options available for waterproof dc motor. Next is to shift the motor to another place and add a chain connection between the motor and the blade. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not a mechanical engineer. Can anyone help to figure out what should be the best way to solve this as I have to use nonwaterproof motors.&lt;a href=&quot;https://i.stack.imgur.com/XjAqg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XjAqg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The black circular spiky thing is the blade which is attached in front of the body. &lt;/p&gt;&#xA;" OwnerUserId="19152" LastActivityDate="2019-11-07T19:24:15.537" Title="How to rotate the blade?" Tags="|mechanical-engineering|motors|machining|waterproofing|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31833" PostTypeId="1" CreationDate="2019-11-06T15:40:21.777" Score="0" ViewCount="56" Body="&lt;p&gt;I've been looking into EV AC chargers for businesses (basically, any and all AC chargers up to 2 x 22kW). One thing I've noticed is that quite a few chargers look alike, sometimes to the point of just having different stickers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This raised two questions: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which companies actually design and manufacture EV AC chargers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do other companies add something to existing designs? Or do they put the chargers in a different jacket and then resell them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry if this is not the right place to ask this question. I wasn't quite sure where else to go to.&lt;/p&gt;&#xA;" OwnerUserId="23417" LastActivityDate="2021-03-30T21:05:30.570" Title="EV AC charging manufacturers" Tags="|electric-vehicles|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31834" PostTypeId="2" ParentId="31775" CreationDate="2019-11-06T16:22:40.883" Score="1" Body="&lt;p&gt;Since this mechanism is all rigidly coupled, it has one degree-of-freedom. One differential equation combined with boundary conditions is enough to describe the position, velocity and acceleration of the system. &lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Steps for main problem:&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Compute missing gear ratio as it is a useful parameter: &lt;span class=&quot;math-container&quot;&gt;$i=\frac{\omega_M}{\omega_L} = \frac{\frac{2\pi}{60}}{\frac{v}{r}}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Transform interia: In order to do derive the equation of motion you have to transform all inertia values to ONE position in the described 'force chain'. I take the hanging load as my transformation target as it is of linear kind of motion and hence more intuitive than rotary kind. The transformation itself is nicely derived in &lt;a href=&quot;http://www.mechanics.iei.liu.se/edu_ug/tmme11/reduced&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and many mechanics textbooks.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Transformed motor inertia &lt;span class=&quot;math-container&quot;&gt;$m_{M-&amp;gt;L} = \frac{J_M \cdot i^2}{r^2}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$[kg]$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Transformed drum inertia &lt;span class=&quot;math-container&quot;&gt;$m_{D-&amp;gt;L} = \frac{J_D}{r^2}$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$[kg]$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Load inertia &lt;span class=&quot;math-container&quot;&gt;$m_L$&lt;/span&gt; stays the same&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Transform forces acting on the mechanism: This is mainly the motor torque &lt;span class=&quot;math-container&quot;&gt;$T_M$&lt;/span&gt; which I assume is a constant. It is reduced by &lt;span class=&quot;math-container&quot;&gt;$\eta$&lt;/span&gt; which is the constant power loss in the geartrain.  There is also a static force &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; neccessary to prevent the load from falling down.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Transformed motor torque &lt;span class=&quot;math-container&quot;&gt;$F_{M-&amp;gt;L} = T_M \cdot i\cdot \eta$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$[N]$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Weight of load &lt;span class=&quot;math-container&quot;&gt;$G_L = m_L \cdot g$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Build differential equation according Newtons 2nd law: &lt;span class=&quot;math-container&quot;&gt;$ m \cdot a = \sum{F}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$(m_{M-&amp;gt;L} + m_{D-&amp;gt;L} + m_{L})  \cdot a = F_{M-&amp;gt;L} - G_L$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$(\frac{J_M \cdot i^2}{r^2} + \frac{J_D}{r^2} + m_L)  \cdot a = T_M \cdot i\cdot \eta - m_L \cdot g $&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Since all parameters are known you can solve for &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; which is a constant value. Integrating a over a time period &lt;span class=&quot;math-container&quot;&gt;$\Delta t_a$&lt;/span&gt; gives you the velocity &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; of the load, which is given above. Finally:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Delta t_a = \frac{v}{a} = \frac{v \cdot (\frac{J_M \cdot i^2}{r^2} + \frac{J_D}{r^2} + m_L)}{T_M \cdot i\cdot \eta - m_L \cdot g}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: The nominator of the fraction is the impulse of the system, the denominator the force trying to change the impulse. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;h3&gt;About transformation of inertia:&lt;/h3&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Assuming a lossless gearbox, there is conservation of power:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P_1 = T_1 \omega_1 = P_2 = T_2 \omega_2 $&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Basic kinematic relationship gives a fixed ratio between input and output:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{\omega_2}{\omega_1} = \frac{\dot\omega_2}{\dot\omega_1} = i$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Newtons 2nd law gives:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ J_1 \cdot \dot\omega_1 = T_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$ J_2 \cdot \dot\omega_2 = T_2$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;and you can write by dividing the equations above and using the gear ratio:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{T_2}{T_1} = \frac{\dot\omega_2}{\dot\omega_1} \cdot \frac{J_2}{J_1}= i \cdot \frac{J_2}{J_1} $&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Use the conservation of power and replace the ration of torques with the equation above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{T_2}{T_1} = \frac{\omega_1}{\omega_2} =&amp;gt; i \cdot \frac{J_2}{J_1} = \frac{1}{i} =&amp;gt; J_1 = i^2 \cdot J_2$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;When you crank a machine at side 2 with a gearbox with a ratio &lt;span class=&quot;math-container&quot;&gt;$i=2$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;you need &lt;em&gt;half&lt;/em&gt; the torque compared to cranking on the output shaft &lt;span class=&quot;math-container&quot;&gt;$T_2= \frac{1}{2}\cdot T_1$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;you need to crank &lt;em&gt;twice&lt;/em&gt; as much for a certain rpm at the load &lt;span class=&quot;math-container&quot;&gt;$\omega_2 = 2 \cdot \omega_1 $&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the inertia of the load (e.g. a flywheel) &lt;em&gt;you&lt;/em&gt; feel is cut down by factor of &lt;em&gt;four&lt;/em&gt; &lt;span class=&quot;math-container&quot;&gt;$J_2 = \frac{1}{2^2}\cdot J_1$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One could also say the moment of inertia is &lt;strong&gt;&lt;em&gt;reduced&lt;/em&gt;&lt;/strong&gt;, but only for this common case. If you sit on the output shaft, the motor inertia gets amplified. So probably it's better to speak of a &lt;strong&gt;&lt;em&gt;transformed&lt;/em&gt;&lt;/strong&gt; mass moment of inertia. In the case of a lossy gearbox, the torque gets decreased by the efficiency, but the general relation &lt;span class=&quot;math-container&quot;&gt;$J_1 = i^2 \cdot J_2$&lt;/span&gt; still holds true. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: For drive machines there is a gear ratio leading to an optimum regarding rotational acceleration: &lt;span class=&quot;math-container&quot;&gt;$ i_{opt} = \sqrt{\frac{J_2}{J_1}}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="23079" LastEditorUserId="23079" LastEditDate="2019-11-06T21:38:02.003" LastActivityDate="2019-11-06T21:38:02.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31835" PostTypeId="2" ParentId="31833" CreationDate="2019-11-06T18:42:05.693" Score="1" Body="&lt;p&gt;They are not the same, even while they meet the exact same standards minimum.&#xA;Basically level 1/2 chargers are actually not charging, they just deliver AC power safely to the onboard charger of the car. That's why some call them smart cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Level 3 chargers deliver very high amp DC directly to the battery and talk with the EV to control and modulate power delivery. They can charge much faster and are well suited for commercial fleets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a ranking of level 2 chargers by The Car-Bible magazine.  &lt;a href=&quot;https://www.carbibles.com/best-home-ev-charger/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Car-bible charger ranking&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is a good, detailed, Quora article about the in and outs of EV charging. &lt;a href=&quot;https://qr.ae/TW4A4x&quot; rel=&quot;nofollow noreferrer&quot;&gt;quora EV &lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-06T18:42:05.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31836" PostTypeId="2" ParentId="28677" CreationDate="2019-11-06T20:20:35.197" Score="2" Body="&lt;p&gt;You can take  have to use temperature dependent material properties for the e-modulus and Poisson's ratio. This scan be achieved with ´MPTEMP' followed by ´MPDATA´. Linear interpolation is used by the software. Example with fictious values:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;MPTEMP,1,     20,    100,   150,   400&#xA;MPDATA,EX,1,, 204e9, 197e9, 193e9, 178e9&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;There is also the option to input a polynomial of 4th degree, so you can fit the data with Scilab, Python or any other scientific software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more details please refer to the Ansys documentation.&#xA;Jet fuel can't melt steel beams.&lt;/p&gt;&#xA;" OwnerUserId="23079" LastActivityDate="2019-11-06T20:20:35.197" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31837" PostTypeId="1" CreationDate="2019-11-06T23:28:41.893" Score="0" ViewCount="30" Body="&lt;p&gt;I assume every materials laboratory has ways to test properties of new materials. Has anyone ever tested graphene's properties at shielding ionizing radiations? I guess It could be a good candidate due to tight packing of atoms&lt;/p&gt;&#xA;" OwnerUserId="23424" LastEditorUserId="23424" LastEditDate="2019-11-08T18:32:09.383" LastActivityDate="2019-11-08T18:32:09.383" Title="How good are graphene layers at shielding against ionizing radiation?" Tags="|materials|radiation|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31839" PostTypeId="2" ParentId="31827" CreationDate="2019-11-07T00:07:31.427" Score="1" Body="&lt;p&gt;First of all I find it more intuitive to find the gain and phase margins using the Nyquist plot instead of the Bode plot. However, you are correct that the gain and phase margins do give a bit of a narrow minded view on stability margins. Namely, one could construct a transfer function that is arbitrarily close to crossing the minus one point while still having great gain and phase margins, so a small change to both the gain and phase could make the system unstable. One such example would be the openloop transfer function&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;L(s) = \frac{0.21\,(2\,s + 1)}{s\,(s^2 + 0.8\,s + 1)} \cdot \frac{s^2 + 0.273\,s + 1.43}{s^2 + 0.135\,s + 1.43} \cdot \frac{s^2 + 0.04\,s + 1.56}{s^2 + 0.029\,s + 1.56} \cdot \frac{s^2 + 0.01155\,s + 1.592}{s^2 + 0.0091\,s + 1.592},&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where one could get closer to the minus one point by adding more and more inverse notch filters. From the Nyquist plot below it can be seen that &lt;span class=&quot;math-container&quot;&gt;$L(s)$&lt;/span&gt; has an infinite gain margin and 106 degrees of phase margin:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/13sij.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/13sij.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when applying both a gain and phase shift of only 1.05 and 8.11 degrees respectively the closed loop system (using unitary negative feedback) does already become unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An additional robustness measure one could use is the modulus margin, which is defined as the shortest distance from the minus one point to the open loop transfer function in the Nyquist plot. A more in-depth explanation can be found in &lt;a href=&quot;https://reader.elsevier.com/reader/sd/pii/S1474667017433441?token=4FC3E8CF07B4607A17759380316CBA83FE52C1E5A13D9725977CBA4D73D33DEE08559F37AF6CABEE8E64030C4FEF584E&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt; in section 4. It can also be shown that the inverse of the modulus margin is also equal to the peak of the magnitude of the sensitivity transfer function and that a given non-zero modulus margin also gives a lower bounds for the gain and phase margin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At my university they did teach the modulus margin, however I believe many other universities/control theory books only cover the gain and phase margin. Maybe this is they think it is easier for people to imagine how just the gain or phase of a system is changed for example due some amplification or time delay respectively. Or maybe it is because only these two terms became industry standards and the modulus margin did not. It can also be noted that in practice one probably does not often encounter systems like the above &lt;span class=&quot;math-container&quot;&gt;$L(s)$&lt;/span&gt;, so for most systems it does hold that a good gain and phase margin also implies a good modulus margin. However, your reasoning is correct that the gain and phase margin can sometimes fail and I would encourage you to also use the modulus margin.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2019-11-07T00:07:31.427" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31840" PostTypeId="1" CreationDate="2019-11-07T09:07:59.693" Score="3" ViewCount="348" Body="&lt;p&gt;Boeing 737NG aircraft are currently being checked for cracking in the &quot;pickle fork&quot; assembly that attaches the fuselage to the wings. The affected joint has 4 fasteners that attach the fork to the wing. It's intended mostly to resist shear loads, as the fuselage twists relative to the wing. Boeing's inspection data is available at &lt;a href=&quot;https://www.regulations.gov/document?D=FAA-2019-0711-0002&quot; rel=&quot;nofollow noreferrer&quot;&gt;the FAA.&lt;/a&gt; That includes this image:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/y1hG3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y1hG3.jpg&quot; alt=&quot;Boeing inspection image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is much discussion of this problem at the pilot's forum &lt;a href=&quot;https://www.pprune.org/rumours-news/625886-737ngs-have-cracked-pickle-forks-after-finding-several-jets-6.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;pprune.org&lt;/a&gt; Someone there &lt;a href=&quot;https://www.pprune.org/rumours-news/625886-737ngs-have-cracked-pickle-forks-after-finding-several-jets-13.html#post10595329&quot; rel=&quot;nofollow noreferrer&quot;&gt;states&lt;/a&gt; that the fasteners are meant to be an interference fit in holes that have been reamed and cold worked, and that the cracks have occurred because holes were drilled 6 thou oversize.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Why would this arrangement - 2 large fasteners and 2 small fasteners - have been chosen?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Boeing refer to the small fastener area as the &quot;fail-safe strap&quot;. &lt;strong&gt;In what way might this be regarded as fail-safe?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Where is it appropriate to use interference fit fasteners in these specially prepared holes?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Why would drilling the holes oversize lead to the observed cracking?&lt;/strong&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16606" LastActivityDate="2019-11-07T09:07:59.693" Title="Boeing 737NG pickle fork cracking. What determines the fastener layout?" Tags="|aircraft-design|shear|fasteners|" AnswerCount="0" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="31841" PostTypeId="2" ParentId="17570" CreationDate="2019-11-07T11:32:00.383" Score="1" Body="&lt;p&gt;You can use only C language for writing UDF. As UDF is an add on customization of Fluent, you have to use C language.  Because all APIs are exposed in C language only.&#xA;For automating ANSYS Fluent you can use scheme language. &#xA;In future ANSYS may provide multi language support, but as of now (ANSYS 19.2) there is support of only C language for writing UDFs.&lt;/p&gt;&#xA;" OwnerUserId="8627" LastActivityDate="2019-11-07T11:32:00.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31842" PostTypeId="1" AcceptedAnswerId="31843" CreationDate="2019-11-07T14:27:20.260" Score="0" ViewCount="60" Body="&lt;p&gt;In relation to &lt;a href=&quot;https://3dprinting.stackexchange.com/questions/5850&quot;&gt;this question about 3D printing&lt;/a&gt;, I wonder how much a 200 mm square glass plate heated only in a central 15 cm square region would bow upwards or downwards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the thermal expansion of glass, the linear expansion for 15 cm and 100°C deltaT would be 0.05 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the formulas provided in &lt;a href=&quot;https://planetcalc.com/1421/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://planetcalc.com/1421/&lt;/a&gt; I input 200 mm as chord length and 200.05 as arc length. The result is 2 mm height which seems a lot to me. Using 150 mm and 150.05 mm I get 1.7 mm which is still a lot.&lt;/p&gt;&#xA;" OwnerUserId="4544" LastActivityDate="2020-12-01T17:04:30.417" Title="How much an unevenly heated glass plate bows?" Tags="|glass|thermal-expansion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31843" PostTypeId="2" ParentId="31842" CreationDate="2019-11-07T15:21:54.710" Score="1" Body="&lt;p&gt;You are right that your numbers are over-estimates. They would be &quot;correct&quot; for a plate with no resistance to bending, which was only supported along two edges and heated across its full width between the unsupported edges.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a hard problem to produce a simple &quot;formula&quot;, because in the real situation the bending will be reduced by the stiffness of the glass, and because the deformed shape is not part of a circle but a bulge in the center of a square. In real life there will be some transition between the temperature of the heated part and the surrounding &quot;frame&quot; as well, and not a sudden temperature change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best way to get a theoretical answer would be to model this with finite element analysis. Alternatively, learn from other people's practical experience, which is quicker and cheaper than learning from your own practical experience!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW the simpler case of a circular plate was apparently studied as part of a PhD back in 1960, with this report as a spin-off: &lt;a href=&quot;http://naca.central.cranfield.ac.uk/reports/arc/rm/3245.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://naca.central.cranfield.ac.uk/reports/arc/rm/3245.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-11-07T15:21:54.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31846" PostTypeId="2" ParentId="31832" CreationDate="2019-11-07T19:24:15.537" Score="2" Body="&lt;p&gt;Small, waterproof electric motors are commonly available as &lt;em&gt;electric outboard motors&lt;/em&gt; that run silently on 12VDC and are used by sport fishermen and also to power small sailboats into and out of harbors. You'd convert them to saw use by removing the propeller and replacing it with a circular saw blade which has been modified to fit onto the propeller shaft, or by making a metal adaptor which allows the blade as-is to be mounted onto the shaft. This is the better option, since saw blade material is hard to machine. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-11-07T19:24:15.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31847" PostTypeId="2" ParentId="31824" CreationDate="2019-11-07T19:30:04.640" Score="1" Body="&lt;p&gt;Here is a simple &quot;hack&quot;. get you a derelict cordless electric drill which contains a two-speed transmission. the transmission will gear up the blade shaft to spin the motor shaft at a higher speed, and you'll have two choices of gear ratios to best match the speeds of the two components. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Better still (!!!) the motor in the drill will serve as a convenient electric generator!  All you have to do is disconnect the variable-speed trigger control and route the motor wires out of the drill housing to your diode pack and battery arrangement. You'll need the diodes to prevent the generator from being driven like a motor by the battery. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of that, the drill chuck in the electric drill can be used as-is to grasp the propeller shaft without having to machine a custom coupling. You just stick the shaft into the chuck as if it were a drill bit and tighten it down. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have fun!&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-11-07T19:30:04.640" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31848" PostTypeId="2" ParentId="31805" CreationDate="2019-11-07T21:48:28.000" Score="2" Body="&lt;p&gt;The &lt;a href=&quot;https://apmonitor.com/pdc/index.php/Main/ArduinoTemperatureControl&quot; rel=&quot;nofollow noreferrer&quot;&gt;Temperature Control Lab&lt;/a&gt; is a third order system with heater 1 as the system input (controller output) and temperature 2 as the process variable. There are model development and control development modules in MATLAB and Python.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CHQC2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CHQC2.gif&quot; alt=&quot;TCLab&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Model Development &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Simulate Step Response &lt;/li&gt;&#xA;&lt;li&gt;Convective Heat Transfer &lt;/li&gt;&#xA;&lt;li&gt;Radiative Heat Transfer &lt;/li&gt;&#xA;&lt;li&gt;Linearize Energy Balance &lt;/li&gt;&#xA;&lt;li&gt;First Order Plus Dead Time (Graphical) &lt;/li&gt;&#xA;&lt;li&gt;First Order Plus Dead Time (Regression) &lt;/li&gt;&#xA;&lt;li&gt;Impulse Response (Analytic Solution) &lt;/li&gt;&#xA;&lt;li&gt;State Space Simulation &lt;/li&gt;&#xA;&lt;li&gt;Second Order Response &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Higher (Third or Fifth) Order Response&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Control Development &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Controller Design &lt;/li&gt;&#xA;&lt;li&gt;P-Only Control &lt;/li&gt;&#xA;&lt;li&gt;PI Control &lt;/li&gt;&#xA;&lt;li&gt;PID Control &lt;/li&gt;&#xA;&lt;li&gt;PI Control Tuning &lt;/li&gt;&#xA;&lt;li&gt;PID Control Tuning &lt;/li&gt;&#xA;&lt;li&gt;PID with Feedforward &lt;/li&gt;&#xA;&lt;li&gt;Heater Actuator &lt;/li&gt;&#xA;&lt;li&gt;Temperature Sensor &lt;/li&gt;&#xA;&lt;li&gt;Block Diagram &lt;/li&gt;&#xA;&lt;li&gt;On/Off Control&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is the &lt;a href=&quot;https://apmonitor.com/pdc/index.php/Main/CourseSchedule&quot; rel=&quot;nofollow noreferrer&quot;&gt;schedule for the course&lt;/a&gt; I developed with TCLab as the hands-on portion of the class. The topics cover almost all of your required topics.&lt;/p&gt;&#xA;" OwnerUserId="23429" LastActivityDate="2019-11-07T21:48:28.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31849" PostTypeId="1" CreationDate="2019-11-08T01:22:36.913" Score="0" ViewCount="230" Body="&lt;p&gt;I have a machine which has belts with three pulleys, one of them is connected to the motors,the one in the middle moves to tighten the belt, and the third is idle pulley. This belt assembly sits on a plate and a strain gauge is attached to the plate. The function of the belts are such that they will pull a fiber by gripping on it and the strain gauge is for measuring the tension in the fiber.  The engineer who designed this is no longer available and I don't know how to convert the readings from the strain gauge into tension. I have attached the assembly image, can anyone tell me how to calculate tension in the fiber that is being pulled by the belts from the strain gauge measurements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if additional details are needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nD9dr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nD9dr.png&quot; alt=&quot;belts in blue and strain gauge in orange&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19110" LastActivityDate="2019-11-09T00:15:31.767" Title="How to configure strain gauges for tension measurement?" Tags="|mechanical-engineering|mechanisms|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31850" PostTypeId="2" ParentId="31849" CreationDate="2019-11-08T05:10:01.990" Score="0" Body="&lt;p&gt;Strain gauges come in a variety of configurations and are sensitive tools. They usually have an excitation voltage and an output voltage and have a certain GF, gauge factor, and are wired as full-bridge, half-bridge, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They need to be calibrated according to their manuals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have usually a gain of ~1000 for a 10 Volt excitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;they communicate to a PC computer via a PCMCIA&#xA;DAQCard after passing through an SCXI board or the SCXI board sends the signall through the computer parallel port to be processed by their App.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to have training and experience of working with them to know the effective range of strain and heat environment, the frequency of data acquisition, etc.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-08T05:10:01.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31851" PostTypeId="1" CreationDate="2019-11-08T05:14:09.110" Score="2" ViewCount="122" Body="&lt;p&gt;Just a quick question for you. Well hopefully. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A contractor has installed a retaining wall with colorbond fence on top. In the contract he stipulated that he would use 65mm x 65mm x 1.6mm powder coated steel posts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He has, by mistake, installed, 76mm x 38mm x 1.6mm powder coated steel posts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole fence is now leaning over and has failed. Now given that these posts are cemented into the sandy soiled ground, which of the posts would be stronger? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The posts have been installed standing up. I.E. the 76mm has the fence attached to it and the 38mm is facing each of the gardens. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help gratefully received. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="23437" LastActivityDate="2019-11-21T08:46:55.567" Title="Which post would be stronger?" Tags="|steel|strength|retaining-wall|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31852" PostTypeId="2" ParentId="31851" CreationDate="2019-11-08T05:56:02.733" Score="0" Body="&lt;p&gt;The thinner section (38mm) was not as strong as the post originally specified (68mm sq.).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has allowed the wind loading (guess) to cause the fence to fail.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-11-08T05:56:02.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31853" PostTypeId="1" AcceptedAnswerId="31858" CreationDate="2019-11-08T07:46:59.427" Score="2" ViewCount="223" Body="&lt;p&gt;Suppose I have a heat generator with a connected heat storage tank. If the heat generator remains the same and the heat storage tank is dimensioned larger, the running time of the heat generator increases due to the higher storage losses.&#xA;Why is this so?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Normally, the surface with exponent 2 and the volume increases with exponent 3 - should the heat storage tank not have fewer losses? &lt;/p&gt;&#xA;" OwnerUserId="23439" LastActivityDate="2019-11-08T19:10:57.803" Title="Storage losses of a heat storage tank" Tags="|thermodynamics|heat-transfer|heating-systems|energy-storage|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31854" PostTypeId="2" ParentId="31851" CreationDate="2019-11-08T08:15:16.467" Score="5" Body="&lt;p&gt;The incorrect posts are weaker as it is the dimension perpendicular to the fence which is critical. It sounds like you have a breach of contract here so you should ask them to replace with the originally specified posts without further charges to you.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-11-08T08:15:16.467" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31856" PostTypeId="1" CreationDate="2019-11-08T12:09:32.230" Score="0" ViewCount="37" Body="&lt;p&gt;In a highway Project, I have to design a road connecting to roads together. The map I have on AutoCAD is a elevation gridlines map and I have x, y and z coordinates&#xA;but I don't know how to get an elevation for a point that is not exactly on a clear grid point, as the points where the grid lines intersect, the elevation is given.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nMKUS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nMKUS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23441" LastEditorUserId="90" LastEditDate="2019-11-08T18:37:57.113" LastActivityDate="2019-11-09T00:49:45.560" Title="Getting the elevation of a specific point of a elevation gridline" Tags="|civil-engineering|cad|autocad|highway-engineering|computer-aided-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31858" PostTypeId="2" ParentId="31853" CreationDate="2019-11-08T19:10:57.803" Score="3" Body="&lt;p&gt;Heat loss due to conductivity of the walls of your container assuming the ambient outside temperature,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{\Delta Q}{\Delta t*A}=-k \frac{\Delta T}{\Delta x} =C_{constant}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the above equation, the left part is heat loss per unit of area per second and the right side is the thermal conductivity, K multiplied by the temperature gradient which is constant. x is the coordinate perpendicular to the temperature gradient of the tank wall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So every unit of the surface of the tanks loses heat at the same rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meaning the larger tank loses more heat because it has more surface area.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-08T19:10:57.803" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31859" PostTypeId="1" CreationDate="2019-11-08T19:11:57.000" Score="0" ViewCount="85" Body="&lt;p&gt;Since the Equation for the Angular velocity of Precession of a Spinning Disk is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\omega _{p}=\frac{m*g*r}{I*\omega _{spin}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;m -&gt; Mass of the spinning Disk. [&lt;span class=&quot;math-container&quot;&gt;$kg$&lt;/span&gt;]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;g -&gt; Acceleration due to Gravity. [&lt;span class=&quot;math-container&quot;&gt;$m/(s^2)$&lt;/span&gt;]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;r -&gt; Distance of the Centroid of disk from the pivot. [&lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I -&gt; Mass moment of inertia of the Disk. [&lt;span class=&quot;math-container&quot;&gt;$kg*m^2$&lt;/span&gt;]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega _{spin}$&lt;/span&gt; -&gt; Angular velocity of the disk. [&lt;span class=&quot;math-container&quot;&gt;$rad/s$&lt;/span&gt;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\therefore \, \, \omega _{p} (unit) = \frac{kg*\frac{m}{s*s}*m}{kg*m*m*\frac{rad}{s}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simplifying,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\omega _{p} (unit) = \frac{1}{rad*s}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shouldn't it be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\omega _{p} (unit) = \frac{rad}{s}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Answer that I got is 'Since radian is the ratio of lengths, it has no dimension. Which means the expression &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{rad*s}$&lt;/span&gt; Simplifies to &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{s}$&lt;/span&gt; and again could be written as &lt;span class=&quot;math-container&quot;&gt;$\frac{rad}{s}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, instead of &lt;span class=&quot;math-container&quot;&gt;$\frac{rad}{s}$&lt;/span&gt;, I could write &lt;span class=&quot;math-container&quot;&gt;$\omega _{p} (unit) = \frac{rev.}{s}$&lt;/span&gt;, which are similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I want to solve a problem with this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A gyroscope has a 1-kg disk that spins at 100 rev/s. The center of mass of the disk is 10 cm from a pivot which is also the radius of the disk. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the precession angular velocity.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;m = 1 [&lt;span class=&quot;math-container&quot;&gt;$kg$&lt;/span&gt;]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;g = 9.8 [&lt;span class=&quot;math-container&quot;&gt;$m/(s^2)$&lt;/span&gt;]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;r = 0.1 [&lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I (Disk) = &lt;span class=&quot;math-container&quot;&gt;$\frac{mr^2}{2}$&lt;/span&gt; = &lt;span class=&quot;math-container&quot;&gt;$5*10^{-3}$&lt;/span&gt; [&lt;span class=&quot;math-container&quot;&gt;$kg*m^2$&lt;/span&gt;]&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega _{spin}$&lt;/span&gt; = 100 [&lt;span class=&quot;math-container&quot;&gt;$rev/s$&lt;/span&gt;]&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On Solving, &lt;span class=&quot;math-container&quot;&gt;$$\omega _{spin} = 1.96 \, \frac{rev}{s}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;since, 1 rev/s = 2&lt;span class=&quot;math-container&quot;&gt;$\pi$&lt;/span&gt; rad/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\omega _{spin} = 12.315 \, \frac{rad}{s}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, instead I want to give the input for &lt;span class=&quot;math-container&quot;&gt;$\omega _{spin}$&lt;/span&gt; as &lt;span class=&quot;math-container&quot;&gt;$2 \pi * 100$&lt;/span&gt; [&lt;span class=&quot;math-container&quot;&gt;$rad/s$&lt;/span&gt;]. (Solving the problem again with all the above given data.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which gives &lt;span class=&quot;math-container&quot;&gt;$$\omega _{spin} = 0.31194 \, \frac{rad}{s}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is kinda Contradictory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since rev/s and rad/s are similar units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used the same method above to derive the unit of &lt;span class=&quot;math-container&quot;&gt;$\omega _{p} (unit) \,\, as \,\, \frac{rev}{s}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, please explain the catch in here..&lt;/p&gt;&#xA;" OwnerUserId="14845" LastEditorUserId="14845" LastEditDate="2019-11-14T18:42:38.337" LastActivityDate="2019-11-14T18:42:38.337" Title="Unit of Angular velocity of Precession" Tags="|applied-mechanics|" AnswerCount="1" CommentCount="2" ClosedDate="2019-11-09T03:26:14.323" ContentLicense="CC BY-SA 4.0" />
  <row Id="31860" PostTypeId="1" CreationDate="2019-11-08T20:38:39.987" Score="1" ViewCount="88" Body="&lt;p&gt;The Autocar E3 refuse truck has an energy recovery system developed by Parker Hannifin LLC that is expected to reduce fuel consumption by 50%. Pressurized fluid flows from carbon fiber-reinforced accumulator tanks to two hydrostatic motors that propel the vehicle forward. (The truck recharges the accumulators when it brakes.) The fuel cost for a regular refuse truck is $900 per month. How much can a private waste-hauling company afford to spend now on the recovery system, if it wants to recover its investment in 3 years plus a return of 14% per year, compounded semiannually? Assume no inter period compounding&lt;/p&gt;&#xA;&#xA;&lt;p&gt;İt is not a homework question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to solve it using factors, as the new system will save us half the cost of fuel per month, we can afford 450 USD per month on the new system. İt will be our annuity. I am not sure if I should use (P/A, %7,6) or find the effective interest rate first for a year than use it like (P/A,%14,xxx,3) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer for both formulas are different however this logic can be used for P/F,F/P so I don't understand.&lt;/p&gt;&#xA;" OwnerUserId="23447" LastEditorUserId="8055" LastEditDate="2019-11-10T18:58:19.663" LastActivityDate="2019-12-04T23:46:05.490" Title="Engineering economics - Factors P/A question" Tags="|engineering-economics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31861" PostTypeId="2" ParentId="31849" CreationDate="2019-11-09T00:15:31.767" Score="1" Body="&lt;p&gt;Here's how this would work.  The tension causes a change in strain. There is a certain calibration factor which tells you how much tension gives how much strain.  I think this is what you are looking for (please correct if wrong).  For simple shapes like a long circular cylinder, or a long rectangle, you could look up these relation in a textbook.  But for something like you've shown it will be much more complicated.  There are basically two ways to get this&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Detailed finite element model (e.g. ANSYS, NASTRAN, Abaqus, etc).&lt;/li&gt;&#xA;&lt;li&gt;Empirical calibration (i.e. apply a known tension to the part, for example using a hydraulic cylinder, and then read off the strain value)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Probably whoever designed this system has already done one of these two things.  The easiest thing is just to find that guy and ask him.  But if that's not an option then you'll have to it over yourself.  Unless you already know how to run finite element models, I don't recommend it.  It could take years to get good enough to get an accurate strain value on something like what you've shown.  So empirical calibration is probably your best option.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-11-09T00:15:31.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31862" PostTypeId="2" ParentId="31856" CreationDate="2019-11-09T00:49:45.560" Score="1" Body="&lt;p&gt;If you use ACad civil engineering, C3D, it automatically creates contour lines and surfaces from the grid data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way would be to draw circles from the point, P to the 4 grid points surrounding it and measure their radius, say r1,r2...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;then a good estimate is &lt;span class=&quot;math-container&quot;&gt;$ Z_P= \frac{Z1r1+Z2r2+Z3r3+Z4r4}{r1+r2+r3+r4} $&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-09T00:49:45.560" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31863" PostTypeId="2" ParentId="31859" CreationDate="2019-11-09T01:53:55.230" Score="0" Body="&lt;p&gt;As stated by &lt;a href=&quot;https://engineering.stackexchange.com/questions/31859/unit-of-angular-velocity-of-precession#comment57180_31859&quot;&gt;@Sam Farjamirad&lt;/a&gt;, a &lt;a href=&quot;https://en.wikipedia.org/wiki/Radian#Dimensional_analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;radian is actually a dimensionless quantity&lt;/a&gt;. It's simply a notation so we know what we're talking about, but it's actually basically meaningless. Therefore:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\dfrac{1}{\text{rad} \cdot s} \equiv \dfrac{1}{s}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2019-11-09T02:04:58.073" LastActivityDate="2019-11-09T02:04:58.073" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31865" PostTypeId="2" ParentId="14355" CreationDate="2019-11-09T07:35:37.970" Score="0" Body="&lt;p&gt;They are also used in transformators for changing an ac voltage to another value by inducing a current without any metal parts touching each other, just by the coils’ magnetic field. I&lt;em&gt;E=E&lt;/em&gt;I&lt;/p&gt;&#xA;&#xA;&lt;p&gt;İn your question it would work but the coils would generate more magnetic field because they have turns.&lt;/p&gt;&#xA;" OwnerUserId="23447" LastEditorUserId="23447" LastEditDate="2019-11-09T07:44:37.533" LastActivityDate="2019-11-09T07:44:37.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31866" PostTypeId="1" CreationDate="2019-11-09T14:50:42.073" Score="1" ViewCount="111" Body="&lt;p&gt;The optical encoder has a code wheel with a single track of slots and two photodetectors. The photodetectors are positioned so that they produce signals with a 90 degree phase difference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate how many slots there has to be in the code wheel in each of the following cases in order to measure the position of the linearly moving object with a resolution of 0.02 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 1:&#xA;motor + gear + ball screw (&lt;span class=&quot;math-container&quot;&gt;$p=8 \text{mm}$&lt;/span&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Case 2: motor + gear + conveyor belt (&lt;span class=&quot;math-container&quot;&gt;$d = 38 \text{mm} $&lt;/span&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in both cases the gear ratio is 6.5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it simple as this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{8}{0.02\cdot 6.5}=62$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{19}{0.02\cdot6.5}= 146$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so needed slot are 62 and 146.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this wrong and if it is how do I calculate this?&lt;/p&gt;&#xA;" OwnerUserId="23314" LastActivityDate="2019-11-09T14:50:42.073" Title="The optical encoder resolution" Tags="|mechanical-engineering|control-engineering|motors|gears|dynamics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31867" PostTypeId="1" CreationDate="2019-11-09T16:10:36.647" Score="0" ViewCount="80" Body="&lt;p&gt;I have been reading technical reports and whatnot about Chernobyl for some time, and I'm feeling that they-or their wording-is not quite letting me understand it. For lack of a better way to present my many questions, I am requesting an evaluation of one of my current understandings of the Chernobyl explosions; mainly if it is consistent with published research and what parts are and are not.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;01:23:44-Power in a hotspot in the lower southeast quadrant of the reactor causes these fuel channels to explode.&lt;/li&gt;&#xA;&lt;li&gt;The channels are depressurized and the resulting boiling of water forms steam that is released into the empty space between the channels and the graphite. Aiding greatly is the positive void coefficient kicking in at the broken area of the channel. &lt;/li&gt;&#xA;&lt;li&gt;01:23:46/7-This causes a lifting force on the reactor lid which lifts it up somewhat, jamming the control rods, tearing apart the remaining fuel channels (which are attached to it) releasing their steam in the same way as in #2.&lt;/li&gt;&#xA;&lt;li&gt;01:23:46/7-The lid moves so high it floats above the reactor hall floor. The steam bursts through the gap thus formed. This is the first explosion&lt;/li&gt;&#xA;&lt;li&gt;With the steam gone, pressure in the reactor drops and the lid begins settling down. Meanwhile, the loss of water is allowing an outrageous power surge.&lt;/li&gt;&#xA;&lt;li&gt;01:23:49/50-power is 300 gigawatts. A hydrogen or nuclear explosion blows the lid through the roof, and kills Valery Khodemchuk. This is the second explosion.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="23451" LastEditorUserId="23451" LastEditDate="2019-11-11T09:51:07.960" LastActivityDate="2019-11-11T09:51:07.960" Title="Is my understanding of the Chernobyl explosions correct?" Tags="|nuclear-engineering|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="31868" PostTypeId="2" ParentId="31851" CreationDate="2019-11-09T17:28:47.703" Score="1" Body="&lt;p&gt;Sue, but I'd suggest getting a lawyer, or at least consulting with one.  Mostly because lawsuits can hinge on seemingly the stupidest little things.  I'm not even sure you &lt;em&gt;want&lt;/em&gt; to bring up the fact that the posts are inadequate -- it should be enough that they are not what's stipulated in the contract.  If you &lt;em&gt;do&lt;/em&gt; bring it up, I'm pretty sure (and a lawyer would &lt;strong&gt;know&lt;/strong&gt;) that you either need a book of building standards that would demonstrate inadequacy, or an expert witness (who in the US would need to have a professional engineer's license; you'd need the equivalent where you are).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-11-09T17:28:47.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31869" PostTypeId="1" CreationDate="2019-11-09T18:44:56.607" Score="0" ViewCount="28" Body="&lt;p&gt;I have a swim platform on the back of my boat that extends 24” from the transom. It is supported by 5- 1” diameter stainless steel arms that are attached To the transom 13” below the platform and at 2” from the rear of the platform.&#xA;I would like to change the platform out for one that is 48” but would like to reuse the stock transom mounting points for the support arms. The new longer arms would therefore go from 13” below the platform back approx. 42” wher they would attach to the new platform.&#xA;Is there a way to calculate the new support arm angle and to determine if it will be enough support at the new reduced angle? &#xA;The width of the platform is 128” side to side and the support arms are evenly distributed across the width from about 6” in per side.&#xA;I’m here because I’m not an engineer or mathematician so bear with me if I left some crucial data out.&#xA;And if I do need to find a local engineer to call this, would it be a structural?&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="23452" LastActivityDate="2019-11-09T18:44:56.607" Title="Swim platform supports" Tags="|structural-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31870" PostTypeId="2" ParentId="31438" CreationDate="2019-11-09T18:59:33.380" Score="0" Body="&lt;p&gt;It would certainly be possible to make a machine like you are proposing, but there are good reasons why hydraulics aren't usually the optimal choice. There are exceptions to this, for example &lt;a href=&quot;https://piranhafab.com/press-brakes/&quot; rel=&quot;nofollow noreferrer&quot;&gt;CNC press brakes&lt;/a&gt; for metals or &lt;a href=&quot;https://www.baileigh.com/numerically-controlled-mandrel-benders-mb-nch-2-series&quot; rel=&quot;nofollow noreferrer&quot;&gt;CNC mandrel benders&lt;/a&gt; often have hydraulic actuators. The distinction is that usually the down sides of hydraulic positioning are only worth it if you need a lot of force to complete the operation. In your case with a hobby-scale CNC mill all of your travel axes can operate on very little force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some of the down sides to designing for hydraulic actuation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Speed: even smallish hydraulic cylinders require a significant volume of oil, and there are practical limits on the flow speed inside of hoses, so moving enough oil to get a hydraulic cylinder moving as fast as a lead screw or rack and pinion can move would require a large pump, large hoses, and large ports. All of this is possible, but very expensive as compared to a simple stepper or servo motor.&lt;/li&gt;&#xA;&lt;li&gt;Leakage: By there nature, pretty much all hydraulic valves involve some amount of leakage. Part of designing hydraulic systems is characterizing the amount of leakage and taking measures to compensate for it where the consequences are unacceptable. Again, it is possible to overcome these challenges, but buying high-precision valves, detailing counterbalance circuits, and maintaining the seals are all unnecessary costs.&lt;/li&gt;&#xA;&lt;li&gt;Controls: Hydraulic valves have moving parts that take time and power to actuate. They don't have perfectly linear responses with respect to time, voltage, or pressure differentials. As a result, achieving precise position control requires more careful control tuning and feedback. By contrast, for a hobbyist level machine using stepper motors, you can pretty much treat this entire challenge as a solved problem. Even at a higher level, configuring a VFD and controlling its setpoints is much easier than controlling a hydraulic valve with feedback.&lt;/li&gt;&#xA;&lt;li&gt;Cost: Aside from the ancillary challenges of integrating hydraulic actuators into machine design, the actuators themselves are simply more expensive. Hydraulic components are not produced in the same volumes that electric motors are, so there's not nearly as much of an economy of scale. In general hydraulic parts require more precise machining, more raw materials, and more detailed QC than motors, all of which translates into cost.&lt;/li&gt;&#xA;&lt;li&gt;Routing: Whereas an electric motor can be driven by one cable with power and control, hydraulic actuators need two hoses at a minimum for precise control in two directions. The hoses are much larger and stiffer than corresponding cables, and failures are more of a mess. When an actuator is being moved by another axis, the logistics of routing the hoses is a fairly big pain.&lt;/li&gt;&#xA;&lt;li&gt;Serviceability: Most users are better equipped to service electrical equipment than hydraulic equipment. A simple multimeter can allow a wide range of diagnostics on electrical equipment, whereas one needs a fair number of gauges and fittings to really find a problem in a hydraulic circuit. Even more than that, hydraulics are harder to disassemble since pretty much everything is a pressure component and can create a big mess.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="122" LastActivityDate="2019-11-09T18:59:33.380" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31872" PostTypeId="1" AcceptedAnswerId="31874" CreationDate="2019-11-10T15:05:08.103" Score="1" ViewCount="82" Body="&lt;p&gt;I'm not an engineer, just a novice hobbyist, so I'm not sure my terminology &quot;compression clamp&quot; is correct, but I'm attaching a drawing showing what I'm trying to accomplish. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plate will be placed over two rods and then &quot;screwed (partly) shut&quot; so that the plate will &quot;squeeze&quot; the rods and not budge under normal operational stresses.  The machine screw is inserted into the hole the pink arrow points at and threads into the hole the yellow arrow points at.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question: It seems to me that the threads will align properly if the gap between those two sections is cut in advance of the threading operation. Is that right? Or should the pink-arrow hole be left smooth-walled and only the yellow-arrow hole be threaded?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EfiRO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EfiRO.png&quot; alt=&quot;clamping plate&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21526" LastActivityDate="2019-11-10T17:07:30.010" Title="threading a compression clamp" Tags="|threads|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31874" PostTypeId="2" ParentId="31872" CreationDate="2019-11-10T17:07:30.010" Score="5" Body="&lt;p&gt;The pink Arrow hole should be left smooth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If both holes have threads when you turn the screw into the pink hole as soon as it passes the gap and reaches the yellow hole, it will keep the gap space constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turning the screw more will just drive it into both holes while keeping the gap at the same space till the head of the screw jams onto the lower surface of the pink hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any further attempts at tightening the screw will either damage the treads or the screw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pink hole should be left smooth and a right size washer used to allow the screw to tighten the bracket safely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or alternately you can use a half thread screw similar to this photo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6BWnI.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6BWnI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-10T17:07:30.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31876" PostTypeId="1" CreationDate="2019-11-11T01:59:57.233" Score="0" ViewCount="33" Body="&lt;p&gt;I wanted to know how the movement of the object floating near the boat would be like. I've assumed the following scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The floating object is at the location &lt;span class=&quot;math-container&quot;&gt;$(x, y)$&lt;/span&gt;. The boat starts moving from &lt;span class=&quot;math-container&quot;&gt;$(0, 0)$&lt;/span&gt; in the y-direction with a velocity &lt;span class=&quot;math-container&quot;&gt;$v_s$&lt;/span&gt;. The water is still.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking so far is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So due to the movement of the boat, there will flow of water let its velocity be &lt;span class=&quot;math-container&quot;&gt;$v_w$&lt;/span&gt;. Due to which a drag will be applied on the object which will be &lt;span class=&quot;math-container&quot;&gt;$C_d\rho v_w^2A/2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this, I'm unable to proceed further. Also, I think the object will oscillate with damping but I do not have the idea. So can I get help in how the motion of the object will be ?&lt;/p&gt;&#xA;" OwnerUserId="23461" LastActivityDate="2019-11-11T01:59:57.233" Title="Motion of an object due to movement of ship" Tags="|mechanical-engineering|fluid-mechanics|" AnswerCount="0" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31877" PostTypeId="1" AcceptedAnswerId="31889" CreationDate="2019-11-11T02:25:53.557" Score="2" ViewCount="66" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DWgGa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DWgGa.jpg&quot; alt=&quot;Air conditioner device installed near the ceiling&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost all air conditioners (AC) are installed on top. For house small office use, will be like the picture. For office with central air conditioner, the blowing come from the ceiling through duct. But however, still there are few occasion where the air conditioner is installed near the floor, like once I saw in a historical cathedral church, due to esthetic's consideration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question, is any thermodynamic reason so almost all air conditioners are installed on top? If any, what is that?&lt;/p&gt;&#xA;" OwnerUserId="17943" LastActivityDate="2019-11-11T21:29:37.217" Title="What is the thermodynamic consideration so almost all air conditioners are installed on top?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31878" PostTypeId="1" CreationDate="2019-11-11T03:30:29.720" Score="1" ViewCount="69" Body="&lt;p&gt;How does the AGV carrying payloads of huge heights as compared to it's size turn without turning the payload? Even if the body is attached to a center bearing, the bearing is very smooth then how come these bots are able to rotate so precisely?&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qcjaw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qcjaw.jpg&quot; alt=&quot;Alibaba's agv deployed in its warehouse&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19152" LastEditorUserId="3505" LastEditDate="2021-04-04T09:13:09.383" LastActivityDate="2021-04-04T15:27:55.200" Title="How does this AGV rotate without rotating the payload?" Tags="|mechanical-engineering|robotics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31879" PostTypeId="2" ParentId="31878" CreationDate="2019-11-11T05:07:09.960" Score="1" Body="&lt;p&gt;A platten or load table driven by hydraulics or an electric motor or large stepper motor will rotate the load to counter the movement of the driving unit.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-11-11T05:07:09.960" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31881" PostTypeId="1" CreationDate="2019-11-11T08:26:01.377" Score="1" ViewCount="147" Body="&lt;p&gt;I have a model on Abaqus that I've managed to run some 85 jobs on manually. Between jobs, only the loads vary, and I select these loads by &lt;strong&gt;resuming&lt;/strong&gt; what I need and &lt;strong&gt;suppressing&lt;/strong&gt; what I don't with my mouse (essentially, choosing k out of n pre-defined loads every time). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm new to scripting on Abaqus with Python, but I have been digging a fair amount through the official scripting guide, with no help. The task I'm working on requires me to extract, &lt;strong&gt;FOR EACH JOB&lt;/strong&gt;, all &lt;em&gt;active&lt;/em&gt; loads with their component magnitudes, and elements that they act on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Query 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Logically, one should be able to find job-specific loads from the corresponding odb file. But I'm unable to find forces by probing into it&#xA;&lt;em&gt;(odb.rootAssembly-&gt;steps-&gt;...?)&lt;/em&gt;. &lt;/li&gt;&#xA;&lt;li&gt;Going into the mdb &lt;em&gt;(mdb.model-&gt;steps-&gt;loadStates)&lt;/em&gt;, however, simply lists&#xA;down all the n pre-defined loads, which isn't very unhelpful. I need the k choices I make for each simulation. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Where do I find this information?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Query 2:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, how do I get for any element or node set, the elements or nodes in that set? (I believe nodes and elements are numbered in Abaqus)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I'll appreciate any help whatsoever. Thanks. &lt;/p&gt;&#xA;" OwnerUserId="23463" LastEditorUserId="23463" LastEditDate="2019-11-11T09:49:58.567" LastActivityDate="2019-11-11T09:49:58.567" Title="Extracting load info on Abaqus with Python" Tags="|mechanical-engineering|finite-element-method|aerospace-engineering|abaqus|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31882" PostTypeId="1" AcceptedAnswerId="31884" CreationDate="2019-11-11T09:48:41.973" Score="3" ViewCount="94" Body="&lt;p&gt;I want to replicate the kind of interface you see on inboard boats, pumps, or food blenders. Basically, I want put a motor in a sealed vessel, and have it turn a prop outside the vessel, in water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this usually done just with a layer of grease over a bearing? Are there any alternatives?&lt;/p&gt;&#xA;" OwnerUserId="23464" LastActivityDate="2019-11-11T17:42:43.833" Title="How are water-tight transmissions made, like the kind you see on inboard boats, pumps or food blenders?" Tags="|mechanical-engineering|waterproofing|transmission|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31883" PostTypeId="1" CreationDate="2019-11-11T12:22:39.903" Score="0" ViewCount="77" Body="&lt;p&gt;First Question&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does single split AC (the type with evaporator indoor and compressor outdoor) vent or exhaust air outside and thereby create negative air pressure in the room?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reason being I am trying to ventilate my home gym room through the door gap with a wall mounted supply fan in the living room outside the gym room, so the gym room needs to have negative air pressure. Installing an exhaust fan in the gym room defeats the purpose since i would like to work out in an air conditioned gym.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second Question (assuming the answer to the above is yes)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In another room (outside the living room as well), a master bedroom, i am planning to install another wall mounted supply fan but due to space issue, it has to be on the same wall as the air conditioner, right beside it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this then mean that the air vented by the AC compressor unit would immediately be sucked back into the supply fan, introducing the hot and high concentration of CO2 back into the room, defeating the purpose?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this then mean that having a bigger CFM supply fan in the living room would be better than having 2 smaller units one in living room and another in bedroom?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the unit in living room will not be affected by any AC and the air will still be able to travel to the gym room &amp;amp; bedroom via the small door gap provided that the AC is on and is able to create negative air pressure in the rooms?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="23465" LastActivityDate="2024-03-20T06:04:50.593" Title="Single Split Air Conditioner &amp; Supply/Exhaust Fan" Tags="|air-quality|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31884" PostTypeId="2" ParentId="31882" CreationDate="2019-11-11T17:42:43.833" Score="1" Body="&lt;p&gt;The seal you are looking for is commonly called an oil rotary seal ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found many brands and sizes with a quick search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the range of prices for a 1-inch shaft is around $5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/saCes.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/saCes.jpg&quot; alt=&quot;oil rotary seal&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-11T17:42:43.833" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31885" PostTypeId="2" ParentId="31883" CreationDate="2019-11-11T18:11:49.747" Score="0" Body="&lt;p&gt;The answer to the first part of your question is no, they do not vent the air out of the room. They recycle the air back into the room, or the newer single zone wall mounted heat pumps just fan the interior air through the air handling grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a need for ventilation it has to be provided by a separate fan.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer to the second part of your question is yes. A fan near the outlet of a wall AC will vent the cold fresh air right back out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should be placed outside of the downwash stream of the AC.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-11T18:11:49.747" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31886" PostTypeId="2" ParentId="31803" CreationDate="2019-11-11T18:55:00.803" Score="0" Body="&lt;p&gt;First alternative I've run into was suggested by Wasabi in a comment to my question.&lt;br&gt;&#xA;The video is here:  &lt;a href=&quot;https://www.youtube.com/watch?v=OdW7vhYYSdM&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=OdW7vhYYSdM&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This example uses a circular saw -- hand tool.  In essence they put an accelerometer chip to control the power and braking inputs to the saw. While it doesn't stop as fast as sawstop, it triggers the halt at the onset of kickback, where sawstop triggers on blade contact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A somewhat different approach could be made on a table saw:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mount strain gauges on the arbor mount. When kickback starts there is going to be an abrupt torque on the arbor, increasing the pressure on the mounts on the leading edge of the blade, and decreasing pressure on the trailing edge. On top of that the blade speed is going to drop abruptly, and the current rise.&lt;/p&gt;&#xA;" OwnerUserId="8768" LastActivityDate="2019-11-11T18:55:00.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31887" PostTypeId="1" CreationDate="2019-11-11T19:44:47.730" Score="1" ViewCount="280" Body="&lt;p&gt;I have a steel electrical box that is 15 cm x 15 cm x 10 cm. The weight of the box is 1kg. Inside the box is a relay that loses 30 Watts of heat while in operation (it's attached to a heat sink that limits the temperature rise to 2 &lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;C/W, though I don't think that matters).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to figure out the required volume of air surrounding the box in order to keep the temperature rise outside the box less than 10 degrees C (assume air temperature to start is 25 &lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about calculating this? I thought about using specific heat equation but I can't figure it out.&lt;/p&gt;&#xA;" OwnerUserId="23470" LastEditorUserId="90" LastEditDate="2020-05-10T18:37:52.267" LastActivityDate="2021-07-04T21:00:15.023" Title="How to calculate the volume of air required to keep electrical box cool?" Tags="|thermal-conduction|thermal-insulation|thermal-radiation|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31888" PostTypeId="1" CreationDate="2019-11-11T20:14:14.093" Score="0" ViewCount="293" Body="&lt;p&gt;I plan to mount a gripper mechanism on a drone to grip a payload. How do I calculate the force generated by the gripper to hold a given weight. What simulation software can be used to simulate the bite of this jaw and get the force &lt;/p&gt;&#xA;" OwnerUserId="23467" LastActivityDate="2019-11-11T22:06:34.690" Title="How do I calculate the bite force of a mechanical gripper jaw? Which software can be used to simulate the same?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31889" PostTypeId="2" ParentId="31877" CreationDate="2019-11-11T21:17:28.037" Score="5" Body="&lt;p&gt;The cold air has a higher density so it always settles down, by installing the AC on top, the cold air can interact faster with hotter air, so the cooling process is much faster. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you install the AC on the floor, the cold air, doesn't naturally travel upward, and even if it does, thanks to the density difference and the air flow of the AC, it then, would be less desirable for human being think about cold feet and hot upper body! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually the heat is coming from the outside  not the floor, windows for example, they pass the radiation, and simplify the conduction so the heat rate transfer is higher where the windows are installed. In a ten feet high room, an AC should be installed at 8 feet high, slightly above the windows, so the cold air can interacte faster with the hotter air.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also a technical issue here, duo to condensation you have to accommodate an extra duct to redirect the water to somewhere else, if you level the AC with floor then you need a pump to let the water flow, however by installing the AC at the recommended height, the gravity does the trick.   &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorDisplayName="user14407" LastEditDate="2019-11-11T21:29:37.217" LastActivityDate="2019-11-11T21:29:37.217" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31890" PostTypeId="2" ParentId="7955" CreationDate="2019-11-11T21:19:11.757" Score="0" Body="&lt;p&gt;NCBCLL is IBC Code for Non Concurrent Bottom Chord Live Load, used for uninhabitable/inhabitable attic space live loads @ 10/20 PSF through Alpine ITW&lt;/p&gt;&#xA;" OwnerUserId="23471" LastActivityDate="2019-11-11T21:19:11.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31891" PostTypeId="2" ParentId="31888" CreationDate="2019-11-11T22:06:34.690" Score="3" Body="&lt;p&gt;Depending on the gripper mechanism an the properties of the objects you want to lift, you can do it by hand calculation (ideal case). It would be much easier to give you advice if you gave information about the structure and material of object to be lifted (like rigid solids, compliant foamy things, or bulk like sand).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Now I'm assuming a rigid load object and a gripping mechanism to hold it using friction only.&lt;/strong&gt; You can use sum of forces/moment to calculate the forces your gripper excerts on the load. These are usually normal forces &lt;span class=&quot;math-container&quot;&gt;$F_N$&lt;/span&gt; acting perpendicular to the object and causing a maximum friction of &lt;span class=&quot;math-container&quot;&gt;$F_{fric} = \mu \cdot F_N$&lt;/span&gt;. This should be well above the weight of the object &lt;span class=&quot;math-container&quot;&gt;$m_L\cdot g$&lt;/span&gt; since this is the limit your gripper can &lt;em&gt;'bite'&lt;/em&gt;. Friction coefficients can vary widely and what works on a sunny day does not neccessarily work on rainy days or when covered in morning dew. As a standard loosely connected to this topic I recommend EN13155-&quot;Load-Carrying Equipment&quot;. There you can find information and safety factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Since you asked for software.&lt;/strong&gt; For typical gripping mechanisms I'd reccomend hand calcualtion first. Elsewise you have no feeling for the numbers you get by engineering software.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Multibody Dynamics Software MBD (for gripper mechanism, solid loads, little compliance, usually 3d modelling software, fast multibody solvers, expensive except for students) (&lt;a href=&quot;https://en.wikipedia.org/wiki/Multibody_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ANSYS MBD&lt;/li&gt;&#xA;&lt;li&gt;RecurDyn&lt;/li&gt;&#xA;&lt;li&gt;Simpack&lt;/li&gt;&#xA;&lt;li&gt;Adams&lt;/li&gt;&#xA;&lt;li&gt;MBDyn (free software)&lt;/li&gt;&#xA;&lt;li&gt;Comsol&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;General purpose modelling tools (for gripper mechanism and solid load, little compliance, multiple domains e.g. electrical, hydraulics and mechanism, mathematical approach, harder to get to results if there are no good libaries for your job)&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;OpenModelica (free)&lt;/li&gt;&#xA;&lt;li&gt;Dymola&lt;/li&gt;&#xA;&lt;li&gt;ANSYS Simplorer&lt;/li&gt;&#xA;&lt;li&gt;Matlab/Simulink &lt;/li&gt;&#xA;&lt;li&gt;MapleSim&lt;/li&gt;&#xA;&lt;li&gt;...&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If your goods and/ or your gripping mechanism is compliant (speak: deformations are of major importance) you probably have to use Finite Element Analysis tools (maybe coupled with software above, depending on your goals) FEA software like ANSYS can handle large deformations, semi-rigid assemblies, nonlinear material, frictional contact and gives really good results. BUT it can be very slow since the solver for FEA is different from MBD. This can still be a viable approach if you make good assumptions/simplications  for your model and if your assembly is lightweight. (I'd personally go for ANSYS since I have kind of love/hate relation with it ;-) )&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;ANSYS&lt;/li&gt;&#xA;&lt;li&gt;Patran&lt;/li&gt;&#xA;&lt;li&gt;Comsol&lt;/li&gt;&#xA;&lt;li&gt;Code Aster (free)&lt;/li&gt;&#xA;&lt;li&gt;... see &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_finite_element_software_packages&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt; for more&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;For background reading and Matlab I recommend &quot;Grundlagen und Methodik der Mehrkörpersimulation&quot; but don't know if an English version exists. For bulk goods you can use Discrete Element Software (DEM, &lt;a href=&quot;https://en.wikipedia.org/wiki/Discrete_element_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia here&lt;/a&gt;) like ANSYS RockyDEM to simulate the behaviour of sand, coal and so on...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps a bit. Clarifications welcome.&lt;/p&gt;&#xA;" OwnerUserId="23079" LastActivityDate="2019-11-11T22:06:34.690" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31893" PostTypeId="2" ParentId="31887" CreationDate="2019-11-12T02:30:46.483" Score="2" Body="&lt;p&gt;This question can only be answered after many tests as to the convective properties of the box, it's geometry with respect to the geometry of the void, ignoring radiation. Because convection which is going to be the major heat transfer mechanism is basically a complex movement of air flowing up in a widening plume, and losing its temperature.&#xA;Here is a link to imperical formulas for a single source convection from Engineering Toolbox site. &lt;a href=&quot;https://www.engineeringtoolbox.com/amp/convective-heat-source-d_1007.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But basically comparing your box to a 60 watts recessed light bulb housing which is usually an 8 inch by 8 inch cylinderical can, one could assume a 4- 6 inchclearance should be ok.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-12T02:30:46.483" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31894" PostTypeId="1" CreationDate="2019-11-12T05:23:28.857" Score="0" ViewCount="31" Body="&lt;p&gt;This is my project where I have created a river cleaning machine. It works fine but no actual calculations have been done regarding the power requirement of motors(A software engineer ). The machine is half-submerged in water.&lt;/p&gt;&#xA;&lt;p&gt;I want to know how to calculate the total amount for the force required to move this machine. Is there any formula that tells how much force is required to exceed the resistance of water and attain a certain speed.&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Kindly ignore the white PVC pipes attached on the side of the machine. I added that because of some balancing issues which I will solve and thus the pipes will be removed.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;My target is that even if I make a bigger model then using these same calculations I should be able to select the motor accordingly.&lt;/p&gt;&#xA;&lt;p&gt;Waiting for your amazing answers. :)&lt;a href=&quot;https://i.stack.imgur.com/BfWGS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BfWGS.jpg&quot; alt=&quot;River cleaning project&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19152" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-11-12T05:23:28.857" Title="How to calculate the force required to move this water bot with certain speed" Tags="|mechanical-engineering|water-resources|thrust|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31896" PostTypeId="1" AcceptedAnswerId="31898" CreationDate="2019-11-12T09:04:54.453" Score="7" ViewCount="7085" Body="&lt;p&gt;70% of the Earth's surface is covered in water, why can't a hydroelectric power plant be set up on each and every river? Is it that its too expensive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is electricity still a problem in countries that have abundant number of rivers? Why can't each and every river be associated with a hydro power plant through dams? What makes a river unsuitable?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My basic question is that Earth has got abundant amount of water. Then why is (hydro) electricity still scarce in some places? Is it that its too expensive?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My research shows that it might be expensive to set up but it is quite cheap in the long run compared to other power plants besides itself being low-maintenance.&lt;/p&gt;&#xA;" OwnerUserId="23475" LastEditorUserId="-1" LastEditDate="2019-11-13T09:34:55.697" LastActivityDate="2019-11-14T21:00:48.820" Title="Why is hydro-electric power still scarce in some places?" Tags="|water-resources|renewable-energy|" AnswerCount="7" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31897" PostTypeId="1" AcceptedAnswerId="31904" CreationDate="2019-11-12T09:31:56.220" Score="0" ViewCount="228" Body="&lt;p&gt;A tangent ogive body has a specific radius for a given length, so how would you determine length if only given the radius? &lt;/p&gt;&#xA;" OwnerUserId="17722" LastActivityDate="2019-11-12T16:25:35.860" Title="For a tangent ogive, given a radius, can you determine it's length?" Tags="|aircraft-design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31898" PostTypeId="2" ParentId="31896" CreationDate="2019-11-12T09:51:19.780" Score="29" Body="&lt;p&gt;Because most of the water is at the lowest point is can be, the ocean.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second issue is that you need quite a bit of vertical drop for water power to be a viable power source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are DIY hydro power setups but they include diverting part of a stream through a pipe into a turbine down the hill, but those tend to only be enough to power a single household and they require a suitable geography.&lt;/p&gt;&#xA;" OwnerUserId="288" LastActivityDate="2019-11-12T09:51:19.780" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31899" PostTypeId="1" CreationDate="2019-11-12T10:03:14.030" Score="0" ViewCount="72" Body="&lt;p&gt;What is the maximum torque the system can withstand at the shaft axis, considering a 2 Nm motor M driving the system through a gear drive and chain drive with an overall reduction ratio between 16:1 and 18:1, and an output speed of 6 rpm?&lt;/p&gt;&#xA;" OwnerUserId="23476" LastEditorUserId="13416" LastEditDate="2023-03-30T07:50:20.520" LastActivityDate="2023-12-25T11:05:25.633" Title="Compute the maximum resisting torque that can be overcome by the system at the shaft axis" Tags="|gears|mechanical|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31900" PostTypeId="1" CreationDate="2019-11-12T11:11:21.977" Score="3" ViewCount="173" Body="&lt;p&gt;Assume a physical object with arbitrary shape whose dimensions are on the order of 1 meter. &lt;strong&gt;How would one measure the surface area of the object?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may be many reasons for determining the surface area:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Plating or painting the object&lt;/li&gt;&#xA;&lt;li&gt;Determining rate of sublimation&lt;/li&gt;&#xA;&lt;li&gt;Determining &lt;a href=&quot;https://physics.stackexchange.com/questions/89452/does-the-strength-of-a-smell-depend-on-the-volume-or-surface-area-of-an-agglomer&quot;&gt;how strongly it smells&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Determining heat loss&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If more detail is needed, then assume the object is a rock or stone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This question is inspired by &lt;a href=&quot;https://physics.stackexchange.com/questions/512834/how-do-i-experimentally-measure-the-surface-area-of-a-rock&quot;&gt;a similar controversial question on Physics.SE&lt;/a&gt;. In that thread it was clear that those with a mathematical background felt that the question was &lt;a href=&quot;https://en.wikipedia.org/wiki/Coastline_paradox&quot; rel=&quot;nofollow noreferrer&quot;&gt;ill-defined&lt;/a&gt;, yet some creative answers were proposed. As this is really an engineering question, I feel that a generalized, better-defined version of the question would be on-topic here. See also &lt;a href=&quot;https://physics.meta.stackexchange.com/questions/12510/dupe-specific-physics-se-question-on-math-se-and-engineering-se-for-different-pe?noredirect=1#comment46920_12510&quot;&gt;the meta thread&lt;/a&gt; discussing the merits of asking this similar question on Engineering.SE.&lt;/p&gt;&#xA;" OwnerUserId="449" LastActivityDate="2019-11-13T18:25:19.470" Title="Determine surface area of arbitrary physical object" Tags="|surface-modelling|" AnswerCount="3" CommentCount="7" ClosedDate="2019-11-21T00:11:07.963" ContentLicense="CC BY-SA 4.0" />
  <row Id="31901" PostTypeId="2" ParentId="31896" CreationDate="2019-11-12T14:17:06.830" Score="10" Body="&lt;p&gt;There are two issues, harnessing and market.  You need both to be viable.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A market, makes most of the 70% of the land mass covered by water unusable, because even if you could harness the power, it is impossible to get the power to consumers.  There are no practical storage alternatives, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ocean buoys do harness ocean currents to meet their individual power needs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wave power does exist, but this is on a small scale, mostly experimental, close to land.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tidal power has great potential by either harnessing tides as they flow in and out or using dams to capture hide tide and harnessing outflow.  But they have to overcome damage by mother-nature or deal with impact on marine animals / environment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many rivers are ruled out because there are used by ships/boats, which would require locks to allow passage or damming would create a flood plane, which would impact existing infrastructure (farm land, communities).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other rivers are ruled out because of spawning fish.  Fresh-water spawning fish, like salmon, can pass dams by fish ladders, but dam flood plane is impacted because fish need calm areas to spawn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Countries with remote rivers (like Canada - 9% of land mass is fresh water - James Bay, Churchill Falls) have dammed them up and harnessed them.  This creates large amounts of power, which is shipped via transmission lines to urban regions.  They take up a lot of real estate, which impacts local communities, either having to relocate or impacting hunting / fishing / recreation territory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller rivers/lakes with a significant water head are harnessed as appropriate.  &lt;a href=&quot;https://en.wikipedia.org/wiki/List_of_generating_stations_in_Newfoundland_and_Labrador&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia list&lt;/a&gt; of generating facilities for Newfoundland and Labrador, Canada.  Not all are on the island of Newfoundland (Churchill Falls), but the hydro-electric facilities (~38 facilities) are not enough to supply the power needs of ~400,000 people.  This is supplemented by Diesel Turbine facility to meet peak power demand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Micro-hydro exists.  Any flow of water can potentially be harnessed.  Water-wheels to grind wheat.  There may be legislative hurdles to overcome.  After all you are impacting a community resource.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Canada, despite it's large geographic area and considerable hydro-electric facilities, still supplements peak power requirements with atomic, wind, solar and diesel (for remote regions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pumped Storage is another alternative.  Usually used to balance the grid load, pumped storage pumps water to a higher height during periods of low power usage to be delivered during peak power periods.  A nuclear power plant is kept at optimum efficiency, so during non-peak times, extra power is available to pump water uphill.  This makes sense because the extra Pumped Storage capacity is available in times when peak demand occurs (breakfast, lunch, supper).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To answer the question:  Most significant hydro-electric resources are either harnessed or planned to be harnessed.  There is a significant investment (money, resources) to harness these renewable resources, but the return on investment is worth the investment.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;The Grand Canyon would make an ideal dam site (remote, formed by water, no population [people or herds]), except for two things: rainfall &amp;amp; desert.  It does get some rain, but it is mainly dry because it is in a desert zone.  If it had the water flow, it would dammed and harnessed.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2019-11-13T13:02:19.417" LastActivityDate="2019-11-13T13:02:19.417" CommentCount="20" ContentLicense="CC BY-SA 4.0" />
  <row Id="31902" PostTypeId="1" CreationDate="2019-11-12T14:30:45.963" Score="0" ViewCount="33" Body="&lt;p&gt;I have a dxf file converted to dwg (unit: millimeters). There are 6 different drawings in the same file, How do I split those to different sheet. I wanted to take their consumption, as they are going to cut from a rexin 1.4 x 1 meter(w x h).&#xA;So I want to fill each drawing to sheet and get count. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Guide me a method, I am currently trying with autocad as they support the dxf file&#xA;Currently am trying with PATCH.dxf file (8x13 sizes in there)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://gofile.io/?c=FL2JGj&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://gofile.io/?c=FL2JGj&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="23462" LastActivityDate="2019-11-12T14:30:45.963" Title="How do split drawings from this without losing their dimension Autocad" Tags="|autocad|technical-drawing|drawings|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31904" PostTypeId="2" ParentId="31897" CreationDate="2019-11-12T16:25:35.860" Score="1" Body="&lt;p&gt;Looking at the the image below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$L=\sqrt{r^2-x^2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x=r-\left(\frac{D}{2}\right)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$L=\sqrt{r^2-\left(r-\left(\frac{D}{2}\right)\right)^2}=\frac{1}{2}\sqrt{-D (D - 4 r)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z9R0k.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z9R0k.png&quot; alt=&quot;Ogive Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-11-12T16:25:35.860" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31905" PostTypeId="1" CreationDate="2019-11-12T17:48:18.813" Score="2" ViewCount="233" Body="&lt;p&gt;I am trying to work out costs and order parts for a small liquid fuelled rocket I have designed. My oxidiser is liquid oxygen but I can’t seem to find anywhere to purchase it on the internet. Does anyone know where to buy liquid oxygen?&lt;/p&gt;&#xA;" OwnerUserId="23482" LastActivityDate="2019-11-14T09:55:36.367" Title="Buying Liquid Oxygen in Small Quantities" Tags="|design|chemical-engineering|aerospace-engineering|" AnswerCount="1" CommentCount="7" ClosedDate="2019-11-21T00:08:49.540" ContentLicense="CC BY-SA 4.0" />
  <row Id="31907" PostTypeId="2" ParentId="15094" CreationDate="2019-11-12T19:25:21.273" Score="0" Body="&lt;p&gt;Gravity is a force and a force has a direction.  So the opposite to parallel for a gravitational body is &lt;strong&gt;gravitational direction&lt;/strong&gt; or &lt;strong&gt;plumb&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://books.google.ca/books?id=4fiBDwAAQBAJ&amp;amp;pg=PA60&amp;amp;lpg=PA60&amp;amp;dq=%22gravitational%20direction%22%20parallel&amp;amp;source=bl&amp;amp;ots=FchEINY0eE&amp;amp;sig=ACfU3U1s_8xVrRDvSWz5kdx1hVQfprdT6w&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=2ahUKEwjFyfCMruXlAhWPZd8KHQzcDIcQ6AEwBXoECC4QAQ#v=onepage&amp;amp;q=%22gravitational%20direction%22%20parallel&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;An Introduction to Fluid Mechanics&lt;/a&gt; By Chung Fang&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Equation(3.1.3) indicates that a change in pressure can only take place in the&#xA;  direction &lt;strong&gt;parallel&lt;/strong&gt; to that of the gravitational acceleration. An increase in pressure is obtained if the direction from one point to another is the same as that of g, and vice versa. If the direction is orthogonal to the direction of g, no pressure change takes place. Thus, the free surface of water on the earth’s surface is always perpendicular to the &lt;strong&gt;gravitational direction&lt;/strong&gt;, for the pressure on the water free surface assumes a constant value, i.e., the pressure of the air above. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2019-11-12T19:25:21.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31908" PostTypeId="2" ParentId="31900" CreationDate="2019-11-12T19:50:42.583" Score="3" Body="&lt;p&gt;May sound not very precise but i take a stab on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wrap your object with wet thin paper, engineering tracing paper for example(Bienfang paper), let it dry with a hair dryer, then cut it into flat squares and triangles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Measure and add the surface area of the pieces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternately you can project a narrow beam of light so that it cast shadows of many angles on a screen which you can measure the main axis off the shadows and reassemble the geometry including the surface and volume.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-12T19:58:37.003" LastActivityDate="2019-11-12T19:58:37.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31909" PostTypeId="2" ParentId="31900" CreationDate="2019-11-12T20:14:22.980" Score="4" Body="&lt;p&gt;Photogrametry usually works quite well unless your object has some very complicated internal cavities. The good thing about photogrametey is that you can easily collect the data with any decent smartphone. Just remember that cameras lose scale so be sure to survey atleast one dimensions with a measuring tool. Or instrument the scan with a known object. There are lots of photogrametry packages even online ones that can triangulate for you. Then just open it in a 3D app eliminate extra faces and count the surface area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want more fancy methods you can use structured light scanners lidars and so on. Hell you can even use 6-12  pictures and milk.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastEditorUserId="1324" LastEditDate="2019-11-13T18:25:19.470" LastActivityDate="2019-11-13T18:25:19.470" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31910" PostTypeId="2" ParentId="28800" CreationDate="2019-11-12T21:13:10.710" Score="0" Body="&lt;p&gt;The best solution for adhering wires to things is to use rigidly attached anchor points, and then zip ties. Anchors that are part of the case, or are attached with screws are best. As I said in a comment, I haven't had good luck with those adhesive rectangle anchors. They're initially very strong, but after a few months I've seen them come dislodged by even the weight of a single Ethernet cable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of good mounting hardware:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mcmaster.com/cable-tie-mounts&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cable tie mounts&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Pi8BI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pi8BI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mcmaster.com/plastic-routing-clamps&quot; rel=&quot;nofollow noreferrer&quot;&gt;P-Straps&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/qpEmS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qpEmS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-11-12T21:13:10.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31911" PostTypeId="1" CreationDate="2019-11-13T06:41:02.230" Score="1" ViewCount="63" Body="&lt;p&gt;Can I use a conical head at the bottom for a small diameter steam manifold ( Schedule 40- 3 inch dia. and 2 inch dia. ) that needs to operate at high temperature (~450 deg Celcius) and High design pressure (25 bars = 363 psi)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since this is a steam distribution and condensate removal manifold, I want the (bottom) head to allow all liquid condensate to be drained via a trap at the bottom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can I use a conical head in such a high pressure case? Or are they reserved for lower pressure applications?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I stick with Elliptical head?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, which would be cheaper and easier to manufacture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in Advance.&lt;/p&gt;&#xA;" OwnerUserId="19228" LastActivityDate="2019-11-20T10:56:11.430" Title="Choosing Pressure Vessel head" Tags="|mechanical-engineering|pressure|process-engineering|pressure-vessel|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31912" PostTypeId="2" ParentId="28800" CreationDate="2019-11-13T06:59:21.247" Score="0" Body="&lt;p&gt;If you don't want to use a mechanical fastener, I would suggest &lt;em&gt;encapsulating&lt;/em&gt; them, by wrapping a thick adhesive around them and sticking it to the base material. You might try Sugru moldable silicone putty, or just a very thick (and non-corrosive!) RTV silicone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure to prepare surfaces well! Even slight fingerprints will trash most adhesive bond strengths!&lt;/p&gt;&#xA;" OwnerUserId="23490" LastActivityDate="2019-11-13T06:59:21.247" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31913" PostTypeId="2" ParentId="31905" CreationDate="2019-11-13T07:10:22.730" Score="1" Body="&lt;p&gt;In general, you can't buy cryogens on the internet because they're both hazardous, and perishable (as they boil off). This goes double for LOX, which does things like &lt;em&gt;turn leather gaskets and asphalt road surfaces into a shock-sensitive explosive&lt;/em&gt;. This caused the Bell X-1 Rocket Plane to mysteriously explode at least once. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vendors would be those such as Airgas, Linde, and Praxair, which also sell various compressed industrial gases and liquid nitrogen. However, you can expect to need a lot more credentials and restrictions than you would for the everyday gaseous oxygen, acetylene, argon, CO2, or even liquid nitrogen. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is likely that they will require a DOT-certified dewar to put it in, or that they will want to deliver one of those big ~50 gallon dewars with all the valves and hose fittings on the top. It is likely that they will impose other restrictions or want to see the kind of paperwork, training, etc that a business regulated under OSHA would be required to have. They may be flatly unwilling to deliver it anywhere that is not a business address -- and it is probably both illegal and unwise to carry any quantity of liquid oxygen in a passenger automobile. &lt;/p&gt;&#xA;" OwnerUserId="23490" LastEditorUserId="23490" LastEditDate="2019-11-14T09:55:36.367" LastActivityDate="2019-11-14T09:55:36.367" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31914" PostTypeId="2" ParentId="31899" CreationDate="2019-11-13T11:48:36.737" Score="0" Body="&lt;p&gt;As Yaniv Ben David states this post needs some clarification. Add a sketch and some more information. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming &quot;shaft axis&quot; is the output shaft and the transmission goes from fast at motor shaft to slow at output shaft the maximum torque is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ T_{out,max} = i \cdot T_{mot,max} = 32 Nm \dots 36 Nm $&lt;/span&gt; for 'typical' electric motors at low speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering a rough estimate 10% losses in the drive train, this means you get &lt;strong&gt;approx. 29 Nm&lt;/strong&gt; of torque on the output shaft.&lt;/p&gt;&#xA;" OwnerUserId="23079" LastActivityDate="2019-11-13T11:48:36.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31915" PostTypeId="1" AcceptedAnswerId="31924" CreationDate="2019-11-13T11:54:51.100" Score="5" ViewCount="310" Body="&lt;p&gt;I have a simulation model of a cylindrical hot water storage tank. The heat loss is determined by the surface area and the heat transfer coefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A configuration of the storage tank with 900 liters was validated with a real scenario. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I would like to test how the scenario changes when the storage tank gets bigger. Thereby my surface becomes bigger, but I don't know if the heat transfer coefficient changes as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does a larger storage tank usually improve the insulation so that the heat transfer coefficient changes?&lt;/p&gt;&#xA;" OwnerUserId="23439" LastActivityDate="2019-11-20T00:51:59.443" Title="Heat transfer coefficient of a cylindrical hot water storage tank depending on its size" Tags="|thermodynamics|heat-transfer|heating-systems|energy-storage|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="31916" PostTypeId="2" ParentId="31896" CreationDate="2019-11-13T13:50:16.477" Score="8" Body="&lt;p&gt;Rivers in a flat landscape are often unsuitable because you need a height difference to make the turbine work. Let's say a 1 meter height difference is required. There are many areas where the river's gradient is so shallow that raising the river's level by 1 meter would inundate everything for miles around, so you'd need to build a dam to contain the river. There comes a point where the dam's more expensive to construct than the value of the electricity you can extract from this height difference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's also the cost to shipping: every lock slows down traffic, and is another expensive bit of engineering to build and maintain. &lt;/p&gt;&#xA;" OwnerUserId="11645" LastActivityDate="2019-11-13T13:50:16.477" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31917" PostTypeId="2" ParentId="31896" CreationDate="2019-11-13T15:38:19.550" Score="2" Body="&lt;p&gt;Water is a valuable resource, getting more valuable by the minute. Simply locations that can build hydro power are scarce. In addition to significant ecological impact damming also changes how rivers flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a trivial thing. People who live in the basin you use as the hydro dam storage need to relocate. Worse people downstream may lose access to water, or atleast severly change their acess. This can cause problems with water tables etc. In addition people working in fishing and agriculture may be impacted. All of this is especially problematic if the river crosses a border. There are few places where this has caused tensions to rise to the level that may even trigger a war!  (Not much of a hydro plant if your neighbour bombs it when your getting finished)&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2019-11-13T15:38:19.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31919" PostTypeId="1" CreationDate="2019-11-13T16:19:41.180" Score="0" ViewCount="58" Body="&lt;p&gt;I have build my own bridge from paper and i want to know the elastic module for the bridge, my bridge is flat all over the way and cylindrical from the sides. &lt;/p&gt;&#xA;" OwnerUserId="23500" LastActivityDate="2019-11-13T18:26:29.937" Title="The elastic module for a paper bridge" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31920" PostTypeId="2" ParentId="31900" CreationDate="2019-11-13T16:53:48.710" Score="2" Body="&lt;p&gt;If budget is not an issue, there are 3D scanners with ‎0.1 mm of resolution.&lt;/p&gt;&#xA;" OwnerUserId="18457" LastActivityDate="2019-11-13T16:53:48.710" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31921" PostTypeId="2" ParentId="31896" CreationDate="2019-11-13T17:00:38.203" Score="4" Body="&lt;p&gt;In years past I would have been a fan of hydroelectric power but I have read&lt;sup&gt;1&lt;/sup&gt; about some of the downsides over the years. These include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Major ecological problems including blocking fish migration. Putting in fish ladders or stairs doesn't really work well as the fish can't find them with all the din of the spillways and turbines.&lt;/li&gt;&#xA;&lt;li&gt;Most silt up drastically reducing their reservoir capacity. I thinks 25 years or so is typical.&lt;/li&gt;&#xA;&lt;li&gt;More people have been killed by hydroelectric dam failures than by nuclear accidents. See &lt;a href=&quot;https://en.wikipedia.org/wiki/Dam_failure#List_of_major_dam_failures&quot; rel=&quot;nofollow noreferrer&quot;&gt;Major dam failures&lt;/a&gt; for example.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As a result, it seems, in North America there are many &lt;a href=&quot;https://en.wikipedia.org/wiki/Dam_removal&quot; rel=&quot;nofollow noreferrer&quot;&gt;dam removal programs&lt;/a&gt; to bring the rivers back to their more natural state.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt;I think &lt;em&gt;Something New Under the Sun: An Environmental History of the Twentieth-Century World&lt;/em&gt; (The Global Century Series) Paperback – April 17, 2001&#xA;by J. R. McNeill was one.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2019-11-13T17:00:38.203" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31922" PostTypeId="2" ParentId="31919" CreationDate="2019-11-13T18:26:29.937" Score="2" Body="&lt;p&gt;Elastic Modulus is a property of material, not the shape of the structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case it depends on the type of the paper you used and also weather the bridge walls are parallel to the paper fiber grain or perpendicular, or angled, because paper is not an isotropic material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most likely the paper you have used is composed of randomly mixed fibers, which means it does not have a consistent modulus of elasticity regardless of the direction of stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can test the modulus of elasticity of you paper by hanging a weight at the end of a narrow strip of the paper say 1/4 inch wide by 3 feet long and measure how much it elongates; then edit your question with these results.  we can then help with an estimate of the E. If you detect a grain repeat the test for strip parallel and perpendicular to that. you need to be judicial and careful in setup and measuring, may be first let the weight hang for few minutes and use a vernier.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-13T18:26:29.937" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31923" PostTypeId="2" ParentId="31915" CreationDate="2019-11-13T20:08:36.453" Score="1" Body="&lt;p&gt;In solid-fluid interfaces the heat transfer coefficient (HTC) is dominated by the properties of the fluid and flow, geometry has no effect on the HTC other than promoting zones where flow changes velocity/pressure/turbulence.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, for a simple geometry like a cylinder, the HTC will not change with the change in dimensions (if you keep the magnitude of the scaling within a reasonable range).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, if you increase the surface area you will dissipate energy quicker.&lt;/p&gt;&#xA;" OwnerUserId="18457" LastEditorUserId="18457" LastEditDate="2019-11-20T00:51:59.443" LastActivityDate="2019-11-20T00:51:59.443" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31924" PostTypeId="2" ParentId="31915" CreationDate="2019-11-13T21:28:34.740" Score="3" Body="&lt;p&gt;In a tank with no fluid movement and no heat source with steady ambient temperature, the temperature of fluid are layered strata roughly in onion layers but favoring the top of the tank, meaning the heat is gradually decreasing from the top in ellipsoid layers roughly following the geometry of the cylinder in you question, imagine inverte flames getting colder as they get near the tank walls and gently sliding up till they reach the top and flare open turn back down in a perpetuous circulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you increase the volume of your tank, it means you are providing new insulation for the hot core.&#xA;So in the larger tank the surface temperature will be less than the smaller tank while keeping similar distribution topography.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the ratio of heat loss will be smaller than jast smaller ratio of surface to volume of fluid.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-13T21:58:48.953" LastActivityDate="2019-11-13T21:58:48.953" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31925" PostTypeId="2" ParentId="31896" CreationDate="2019-11-13T21:42:22.680" Score="0" Body="&lt;p&gt;Geography is one thing (change in height as mentioned above, finding suitable areas to store large volumes of water, canyons are usually suitable places for hydro instead of rivers that doesn't change over height so countries such as Denmark and the Netherlands are at a disadvantage while places like Switzerland &amp;amp; India have large rivers that are concentrated)so  having a river that can continually provide a water flow that is substantial to keep a consistent pressure and not dry up all of a sudden is also important but damming it is a huge economic cost and investment as well. So plenty of things to consider. As well dams are more used during peak times when the load on the grid is high and water is then pumped back up when the load is low. Since dams can't hold a load for a long period of time.   &lt;/p&gt;&#xA;" OwnerUserId="23507" LastActivityDate="2019-11-13T21:42:22.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31926" PostTypeId="1" CreationDate="2019-11-13T23:47:04.677" Score="0" ViewCount="670" Body="&lt;p&gt;Question: Is 56 Nm enough torque for a delta robot with the below dimensions and specs? If not, what would be the minimum torque needed? Thank you in advance for your help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am working on a rotary delta arm design and needed help sizing the correct motor and gearbox. Below are the mechanical dimensions of the arm and application info.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Delta Arm Specs:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Base Radius (Center to Motor): 200 mm&lt;/li&gt;&#xA;&lt;li&gt;Primary Arm Length: 279.4 mm&lt;/li&gt;&#xA;&lt;li&gt;Secondary Arm Length: 441.96 mm&lt;/li&gt;&#xA;&lt;li&gt;End Effector Radius: 100 mm&lt;/li&gt;&#xA;&lt;li&gt;The arms are carbon fiber: Weight of arms is approx: 445 grams&lt;/li&gt;&#xA;&lt;li&gt;The gripping mechanisms weight between 300 grams to 1 kg&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Application Info:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Type: Pick and Place&lt;/li&gt;&#xA;&lt;li&gt;Max Payload: 3 kg&lt;/li&gt;&#xA;&lt;li&gt;Min picks per minute: 60&lt;/li&gt;&#xA;&lt;li&gt;Max pick per minute: 100&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I have been looking at using Teknic Clearpath motors with a planetary gearbox from Anaheim Automation. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Teknic &lt;a href=&quot;https://www.teknic.com/model-info/CPM-SCHP-2341S-ELNB/?model_voltage=75VDC&quot; rel=&quot;nofollow noreferrer&quot;&gt;NEMA 23 Motor&lt;/a&gt; Peak Torque: 5.8 Nm | Const (RMS) Torque: 1.2 Nm | Max RPM: 1860 with a 10 to 1 single stage gearbox &lt;a href=&quot;https://www.anaheimautomation.com/products/gearbox/planetary-gearbox-item.php?sID=625&amp;amp;pt=i&amp;amp;tID=1089&amp;amp;cID=30&quot; rel=&quot;nofollow noreferrer&quot;&gt;Anaheim Automation Gearbox&lt;/a&gt; Arc-Min: 3.  At gear output expecting: Peak Torque: 56.26 Nm and Const Torque: 11.64 Nm. RPM Expected: 186&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Here is a gif of the delta arm and its movement (Wikipedia) &lt;a href=&quot;https://en.wikipedia.org/wiki/Delta_robot#/media/File:DeltaRamki.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;Delta Arm Sketch&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23487" LastEditorUserId="23487" LastEditDate="2019-11-16T03:08:58.330" LastActivityDate="2019-11-16T03:08:58.330" Title="Delta Robot: Torque and Motor Sizing" Tags="|mechanical-engineering|motors|robotics|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31927" PostTypeId="1" AcceptedAnswerId="31937" CreationDate="2019-11-14T09:33:59.257" Score="0" ViewCount="79" Body="&lt;p&gt;On the image below there is a regular prism which is angled by &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;. &lt;a href=&quot;https://i.stack.imgur.com/AyUOn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AyUOn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The (black) mass m1 applies a torque &lt;span class=&quot;math-container&quot;&gt;$M_k$&lt;/span&gt;. The mass m2, which is the rest of the Body applies the torque &lt;span class=&quot;math-container&quot;&gt;$M_s$&lt;/span&gt; In the opposite direction. The black dot on the bottom right corner is the pivot point.&#xA;So i wanted to calculate the normal force distribution on the bottom surface which is completely touching the ground. &#xA;Because the force F1 has to be smaller than F2 right?&#xA;My final Goal is to calculate the friction of the prism.&#xA;So my question was: Does it matter if i calculate the normal Force by &lt;span class=&quot;math-container&quot;&gt;$ F_N = m * g$&lt;/span&gt;. Or do i have to integrate &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; over the whole bottom surface and thus derive the mean of the normal force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another problem is: If the prism is accelerated by a another torque &lt;span class=&quot;math-container&quot;&gt;$M_a$&lt;/span&gt; is applied. So you definitely cannot calculate the normal force by &lt;span class=&quot;math-container&quot;&gt;$m*g$&lt;/span&gt; anymore.&lt;/p&gt;&#xA;" OwnerUserId="23512" LastEditorUserId="23512" LastEditDate="2019-11-14T09:43:58.157" LastActivityDate="2019-11-15T05:26:21.230" Title="Normal Force distribution of Prism" Tags="|applied-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31928" PostTypeId="1" CreationDate="2019-11-14T10:03:16.897" Score="0" ViewCount="38" Body="&lt;p&gt;The last hundred years have seen a great degree of standardization of units, and the expression of units of measurement in standardized, instrument-agnostic, objective terms, often in the SI units. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I wonder why this has not happened in the case of the indentation hardness of solid matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems that the measurement of indentation hardness is fragmented into one scale per instrument, with one or more instruments being customarily used for each material, and it seems like the scales are mostly semi-qualitative and probably non-linear -- and they make it difficult to compare dissimilar materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Off the top of my head, I can name:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rockwell B: Used for soft steel, mostly.&lt;/li&gt;&#xA;&lt;li&gt;Rockwell C: Used for hardened steel and similar materials&lt;/li&gt;&#xA;&lt;li&gt;Brinell: One of the more quantitative-seeming scales, but uses an instrument that has some downsides. Used mostly for nonferrous metals and soft steel.&lt;/li&gt;&#xA;&lt;li&gt;Shore 00 and A durometers: Used for elastomers and soft gelatinous materials&lt;/li&gt;&#xA;&lt;li&gt;Shore D durometer: Used for &quot;hard&quot; plastic&lt;/li&gt;&#xA;&lt;li&gt;Knoop: Uses a weird-shaped indenter, usable for metals but mostly used for ceramics and other brittle stuff.&lt;/li&gt;&#xA;&lt;li&gt;Vickers: uses a pyramid-shaped indenter but not the same one as Knoop, apparently generates quasi-SI hardness measurements but isn't that widely used.&lt;/li&gt;&#xA;&lt;li&gt;Mohs Scale: Incredibly nonlinear empirical scale based on what materials can indent each other&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is it just because of the &quot;There are now 15 competing standards&quot; issue? Or is there a definitional problem in the concept of &quot;indentation hardness&quot; that makes it hard to reduce to a single value?&lt;/p&gt;&#xA;" OwnerUserId="23490" LastActivityDate="2019-11-14T10:03:16.897" Title="Why is there no widely-used standardized unit for indentation hardness?" Tags="|mechanical-engineering|materials|measurements|instrumentation|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31930" PostTypeId="1" CreationDate="2019-11-14T16:17:17.113" Score="1" ViewCount="48" Body="&lt;p&gt;When HVAC companies estimate the required size of a chiller unit for a greenhouse or an indoor farm, they reduce the estimated load by plant &quot;transpiration cooling.&quot; I understand how transpiration converts sensible heat to latent heat, however I don't understand how that would reduce the required size of a chiller unit.  If an artificial light is adding 4000 BTUs of energy into a grow room, the HVAC system still needs to remove that heat, regardless of the state it is in.  Transpiration does not remove heat from a room, it just transforms it.  &lt;/p&gt;&#xA;" OwnerUserId="23521" LastActivityDate="2019-11-14T16:17:17.113" Title="How does plant evapotranspiration reduce the required size of a chiller used to control the climate of an indoor farm?" Tags="|hvac|evaporation|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31932" PostTypeId="1" AcceptedAnswerId="31934" CreationDate="2019-11-14T17:04:49.823" Score="2" ViewCount="53" Body="&lt;p&gt;In the pharmaceutical environment, the drop height from the hopper to the area where it collects must be minimized. Tubes can be used to minimise the height. I know that it affects powder segregation. But do not understand the mechanism of it. How does it affect powder segregation ? As long as the flow is optimal (eg. no bridging), then why will powder still segregate if drop height is not minimal ?&lt;/p&gt;&#xA;" OwnerUserId="11561" LastActivityDate="2019-11-14T17:46:53.360" Title="what impact does drop height have on powder segregation?" Tags="|design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31934" PostTypeId="2" ParentId="31932" CreationDate="2019-11-14T17:46:53.360" Score="2" Body="&lt;p&gt;Small powder particles react with the air flow differently, such as small water droplets which never fall and remain suspended in the air and create haze. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a long shoot will give a chance to the powder to separate.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-14T17:46:53.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31935" PostTypeId="1" CreationDate="2019-11-14T18:01:05.473" Score="2" ViewCount="668" Body="&lt;p&gt;I guess this question would belong on an architecture page if Stack had one. &#xA;There is a 5 story building I was in recently that had fairly low ceilings (maybe about 8 feet) and a huge gap between floors (around 6 feet). Any specific reason for this design? All the pipes and electric cables could fit into a much smaller area, so it seems like either a waste of resources on making the building so tall, or they could have added more floors.&lt;/p&gt;&#xA;" OwnerUserId="23524" LastActivityDate="2023-06-16T06:13:22.440" Title="Huge gap between ceiling and floor above" Tags="|architecture|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31936" PostTypeId="2" ParentId="31935" CreationDate="2019-11-14T18:29:36.167" Score="2" Body="&lt;p&gt;the gap is usually referred to as a &lt;em&gt;mezzanine&lt;/em&gt;. It is through the mezzanine that computer cabling is strung and distributed into the cubicles above and below it via holes drilled through the floors and ceilings. In a building intended to live through several generations of computer technology and through several revisions of its floor plan, you'll need your IT guys to spend significant amounts of time pulling wires through the mezzanine, so you proportion it to allow easy human access.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-11-14T18:29:36.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31937" PostTypeId="2" ParentId="31927" CreationDate="2019-11-14T18:50:50.100" Score="0" Body="&lt;p&gt;You just need to find the CG of the parallelogram if there is no external torque, just add the overhanging mass of black triangle. This is typically called off-center column or P-Delta loading or a column with both axial and moment loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if there is external torque other than the black triangle mass you add it treating a+b as a beam with moment &lt;span class=&quot;math-container&quot;&gt;$M_k \ $&lt;/span&gt; applied at the right support of the beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So scaling your sketch we assume the height of the part as a, and then you have a base a+b with its CG at &lt;span class=&quot;math-container&quot;&gt;$$\bar X =\frac{1/2ab*b/2 +a^2*(b/2+a/2)+1/2ab*(a/2+2*b/2)}{1/2ab+a^2+1/2ab}  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the distribution of the forces on the base will be the sum of the normal weight divided by the base surface which is uniform stress, say q, and the triangular stress, say p, do to overturning moment which is  &lt;span class=&quot;math-container&quot;&gt;$ \sigma_{right}=\frac{6M( \bar X- 1/2(a+b))}{ (a+b/2)^2} \quad \sigma_{left}=0   $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we add these 2 stresses, q and the triangular p together and we get a trapezoid stress surface, with the right hand bigger than left hand stress.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I assumed the depth and mass density as 1.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-15T05:26:21.230" LastActivityDate="2019-11-15T05:26:21.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31938" PostTypeId="2" ParentId="31896" CreationDate="2019-11-14T21:00:48.820" Score="0" Body="&lt;p&gt;Hydroelectric power is (indirectly) solar and geothermal power:&#xA;Heat causes evaporation, temperature gradients create flow of air and water vapor to higher elevations, clouds form, rain falls, and hydroelectric power simply harvests the gravitational potential energy originally imparted by heat to the raindrops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, you can harvest hydroelectric power (a) anywhere it rains, and (b) anywhere between where it rains and a lower elevation so long as there is a continuous and preferred canal to supply the drainage from the high-rainfall areas.&lt;/p&gt;&#xA;" OwnerUserId="23525" LastActivityDate="2019-11-14T21:00:48.820" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31939" PostTypeId="2" ParentId="31935" CreationDate="2019-11-14T21:10:04.963" Score="1" Body="&lt;p&gt;Aside from the expenses, there is no reason why you shouldn't be able to raise the ceiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can safely move the ceiling to as close to cables and any plumbing as one inch. If extra height is essential for you, you can tastefully box in hard to move utilities such as plumbing P Traps. Many of the electrical circuitry and data cabling are fairly easy to lift up. Modern plastic electrical boxes are cheap and very easy to install, if need be. Potential ducting can be rerouted or changed with flat version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the ceiling itself if it is drywall over wood or light steel joists you just move everything up. If it is T bars and acoustic tiles just adjust the hanging wires.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have old ugly recessed lights or fan in the ceiling this is the time to install new smart LED lights and new fans.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-14T21:10:04.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31940" PostTypeId="2" ParentId="31935" CreationDate="2019-11-14T23:41:31.007" Score="0" Body="&lt;p&gt;It can also be required by a long clear span ; the depth is needed for beams or trusses. I am familiar with a building that has a clear span of 50 ft. using about 42&quot; between ceiling and floor.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-11-14T23:41:31.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31942" PostTypeId="1" CreationDate="2019-11-15T03:56:00.840" Score="2" ViewCount="540" Body="&lt;p&gt;Most of the windows in our rooms are mostly rectangular. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't we chose any other shape like circle, triangle or trapezium? Does it affects airflow through room? Is the reason is that they look fancy or some specific reason?&lt;/p&gt;&#xA;" OwnerUserId="23522" LastEditorUserId="6790" LastEditDate="2020-05-02T19:34:54.327" LastActivityDate="2020-05-02T19:34:54.327" Title="Why rectangular windows are so common?" Tags="|design|building-design|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31943" PostTypeId="2" ParentId="31942" CreationDate="2019-11-15T06:45:43.960" Score="1" Body="&lt;p&gt;There are a few reasons on which rectangle comes at optima:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ease of construction and operation (Opening/Closing/Sliding etc) which translates to less expensive.&lt;/li&gt;&#xA;&lt;li&gt;Easy to fit.&lt;/li&gt;&#xA;&lt;li&gt;Less material is wasted when you make the rectangular design.&lt;/li&gt;&#xA;&lt;li&gt;Easy to transport.&lt;/li&gt;&#xA;&lt;li&gt;Strength.&lt;/li&gt;&#xA;&lt;li&gt;One of the purposes of a window is to provide sunlight. A tall window means sunlight for more time as compared to a square window of the same area.&lt;/li&gt;&#xA;&lt;li&gt;Trivial: It is easier to calculate perimeter, area etc of rectangular shape for estimating material, effort etc and thus the cost.&lt;/li&gt;&#xA;&lt;li&gt;A rectangle is visually more pleasing than a square and has more attention span. This is the reason why televisions, books, paintings, photographs etc are mostly or always a rectangle.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;More popularity of rectangular shape means mass production and thus even lesser cost on top of the above advantages.&lt;/p&gt;&#xA;" OwnerUserId="22772" LastActivityDate="2019-11-15T06:45:43.960" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31944" PostTypeId="2" ParentId="31942" CreationDate="2019-11-15T07:30:42.330" Score="1" Body="&lt;p&gt;processing of long straight pieces of wood is an old industry called millwork.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is very easy to cut and plane and finish long straight pieces of wood or metal and assemble them into rectangular windows and doors. It is a fortunate thing that evergreen pine and similar species that are used for machining into lumber grow tall and straight, so the millwork done using them tends to warp very little and stay stable over centuries.&#xA;Most of the saws and finishing machines lend themselves to cut, extrude, long and straight continuous profiles. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This as opposed to round or arched window which is much harder to build and is very likely to warp after losing its moisture content and aging, has been one of the reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage of using rectangular is because they easily fit into any structure with minimum adjustment. and lest we forget we use a plumb to build our walls perfectly vertical to fight the gravity&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage is the property of the rectangular is to completely tile or cover a surface without leaving any left out areas, be it a house or a highrise or commercial building. It is a very basic way of standardization of windows in construction industry not much unlike standards for the doors, steps, pipe sizes,etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even now that technology can produce wide selection of shaep, the rectangular is the most practical and elegant. Many of turn of century architectural monument established the beauty of using large continuous panels of glass in thin rectangular aluminum frames. frameless glass windows, work seamlessly together because they are rectangular.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also as has mentioned in comments they are easier to pack and ship.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-15T07:30:42.330" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31946" PostTypeId="2" ParentId="15094" CreationDate="2019-11-15T12:55:49.843" Score="1" Body="&lt;p&gt;&lt;em&gt;&lt;sub&gt;Sorry I am only answering the title of your question, but this answer does not allow to go on and give you a recommendation for your second question (in the question body).&lt;/sub&gt;&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no &quot;&lt;em&gt;opposite&lt;/em&gt;&quot; to parallel, in engineering, in physics, or even in mathematics or phylosophy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not what you wanted to hear but bear with me. The notion of &lt;em&gt;opposite&lt;/em&gt; is only ever defined on a 1D domain. On a 1D domain (a line). If a word indicate a direction along this line, then a word can exist to describe the &lt;em&gt;opposite&lt;/em&gt; direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now take a 2D domain (let's start with a circle for now, but your non-parallel lines will fit there too). On the previous line, I only had 2 possible directions, each &lt;em&gt;opposite&lt;/em&gt; to each other, where I could navigate. Now starting from the center, there is an infinity of possible direction I can take (I can divide them in 360 degrees, 6.28 radian or any other metrics but the important thing is in any case there are more than 2 directions).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;parallel&quot; only indicate &lt;strong&gt;one&lt;/strong&gt; posible orientation (and 2 directions along this line). There is no &lt;em&gt;opposite&lt;/em&gt; to that orientation. Which one would it be, the orientation with an angle of 1°, an angle of 3°, an angle of 90°? None of the other orientation qualify for &quot;opposite&quot; (and none of them has an &quot;opposite&quot; either). There is no opposite to &quot;parralel&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the english language had defined a word exactly describing &quot;anything NOT parallel&quot;, you'd have the strongest candidate for the &lt;em&gt;opposite&lt;/em&gt; but even then it wouldn't exactly be true. One word would define a single point of the domain (of all the possible orientations) and the other would include the rest of the set (all the infinity of other possible orientations). I don't know if such an unbalance between 2 words would qualify them to be &lt;em&gt;opposite&lt;/em&gt;. I am not aware of such a word anyway (bar &quot;non-parallel&quot; if you consider that a word) ...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;That was the pedantic view. Now as an engineer, I sometimes use verbal shortcuts too, and in this context (as a verbal shortcut, knowingly not accurate but understandable enough for my interlocutors if provided in the right context ) I could see myself replying to &quot;we need these things parallel for reason xx ...&quot; with &quot;No it's the opposite, we need them &lt;em&gt;perpendicular/othogonal&lt;/em&gt; for reason zzz ...&quot;.&#xA;It would obviously be a gross misuse of language, but with context it may be understood fine because &lt;em&gt;parallel&lt;/em&gt; and &lt;em&gt;orthogonal&lt;/em&gt; are special orientations among the domain, with special properties that other angled orientation do not have. (and some of these properties between parallel and orthogonal could be considered opposite, although even that would require a bit of handwaving)&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;And just for fun (or to confuse people):&#xA;Say I have a trajectory, on a 2D domain, going along the positive side of the X axis (so &lt;em&gt;parallel&lt;/em&gt; to X axis). What is the &lt;em&gt;opposite&lt;/em&gt; of this object? Well for me if the orientation AND direction was defined, then the opposite object would be with the same orientation, but going in the &lt;em&gt;opposite&lt;/em&gt; direction. So it would be a line going along the negative side of the X axis.&#xA;Wait what ? ... It means that this object would also be parallel to X axis, and parallel to the initial object. It means, the opposite of a parralel trajectory is ... a parralel trajectory ... rhaaa, my brain is melting now ...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Last one: As said in comment, the premise of the question might even be unfounded. IF these 2 pillars are each exactly parallel to their local gravitation field, then they &lt;strong&gt;are parallel&lt;/strong&gt; to each other in the spacetime. Only earth gravity is warping the spacetime in which they reside so they &lt;em&gt;appear&lt;/em&gt; at an angle (yeah I thought &quot;at an angle&quot; is better than &quot;non-parallel&quot;, still don't have a good word for you I'm affraid).&lt;/p&gt;&#xA;" OwnerUserId="23532" LastEditorUserId="23532" LastEditDate="2019-11-15T13:02:46.597" LastActivityDate="2019-11-15T13:02:46.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31949" PostTypeId="1" AcceptedAnswerId="31952" CreationDate="2019-11-15T18:47:47.797" Score="1" ViewCount="1136" Body="&lt;p&gt;When we twist a shaft/beam/rod, shearing stresses are produced. The angle by which the end of the shaft rotates can be found as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$${\displaystyle \theta ={\frac {TL}{GJ}}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is torque, &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the length, &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; the shear modulus, and &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; a torsion constant depending on the geometry of the cross section. For circular cross section, &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt; equals the second moment of area. For non-circular shafts there is an effect called &quot;warping&quot; present where the cross sections don't remain planar during twisting; if we had a beam with a rectangular cross section and held it radially away from us and twisted it, the cross sections would bend towards us/away from us. Do I understand warping correctly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But shafts with circular cross sections do not experience this. Planar cross sections remain planar, and deform only in 2D. &lt;strong&gt;Why is this?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2019-11-16T00:11:29.313" Title="Why torsional warping does not occur for shafts with circular symmetry?" Tags="|torque|shear|deformation|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31950" PostTypeId="1" CreationDate="2019-11-15T21:02:11.057" Score="1" ViewCount="251" Body="&lt;p&gt;I'm trying to follow an example of this, but I get a wildly different answer than expected and I can't see where I'm going wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The example I've got, is for a 2.8 m steel column (RHS 150 x 100 x 5mm) fixed both ends with a simple static vertical central load. Couldn't be easier for a beginner. My calculation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Relevant properties (section properties taken from [here][1]) normalising length to metres:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Steel section&lt;/strong&gt;: RHS 150 x 100 x 5mm. &lt;em&gt;(unspecified hot or cold)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;L&lt;/strong&gt;: Free length = &lt;strong&gt;2.8 m&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;E&lt;/strong&gt;: 205 x 10&lt;sup&gt;9&lt;/sup&gt; = &lt;strong&gt;2.05 x 10&lt;sup&gt;11&lt;/sup&gt;&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;I (table)&lt;/strong&gt;: Least 2nd moment of area = 3.923 x 10&lt;sup&gt;6&lt;/sup&gt; mm&lt;sup&gt;4&lt;/sup&gt; = &lt;strong&gt;3.923 x 10&lt;sup&gt;-6&lt;/sup&gt; m&lt;sup&gt;4&lt;/sup&gt;&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;A (table)&lt;/strong&gt;: Area = 2373 mm&lt;sup&gt;2&lt;/sup&gt; = &lt;strong&gt;2.373 x 10&lt;sup&gt;-3&lt;/sup&gt; m&lt;sup&gt;2&lt;/sup&gt;&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;r (table)&lt;/strong&gt;: Least radius of gyration (as double check): 40.7mm = &lt;strong&gt;0.0407 m&lt;/strong&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;K&lt;/strong&gt;: Effective length factor = &lt;strong&gt;1 (fixed-fixed)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;s&lt;/strong&gt;: Slenderness = L/r = 2.8/0.0407 = &lt;strong&gt;68.8 (intermediate)&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EUler buckling force, calculated using I,K,L:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;π&lt;sup&gt;2&lt;/sup&gt; x (E.I) / (KL)&lt;sup&gt;2&lt;/sup&gt;&lt;br/&gt;= (3.142&lt;sup&gt;2&lt;/sup&gt; x (2.05 x 10&lt;sup&gt;11&lt;/sup&gt;) x (3.923 x 10&lt;sup&gt;-6&lt;/sup&gt;) / (1 x 2.8)&lt;sup&gt;2&lt;/sup&gt;&lt;br/&gt;= &lt;strong&gt;1.012 MN&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Euler buckling force, calculated using A,s (as double check):&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;π&lt;sup&gt;2&lt;/sup&gt; x (E.A) / (s&lt;sup&gt;2&lt;/sup&gt;)&lt;br/&gt;= π&lt;sup&gt;2&lt;/sup&gt; x (2.05 x 10&lt;sup&gt;11&lt;/sup&gt;) x (2.373 x 10&lt;sup&gt;-3&lt;/sup&gt;) / (68.8)&lt;sup&gt;2&lt;/sup&gt;&lt;br/&gt;&lt;strong&gt;= 1.014 MN&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the two calculations agree within the limits of the data table figures. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, the correct answer is apparently &lt;strong&gt;&lt;code&gt;Pc (compressive strength) = 124 N/mm&lt;/code&gt;&lt;sup&gt;&lt;code&gt;2&lt;/code&gt;&lt;/sup&gt;&lt;/strong&gt; = &lt;strong&gt;&lt;code&gt;294 kN&lt;/code&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The answer does confirm that slenderness is 68.8, and that the least radius of gyration is 40.7 mm, so I know I'm on the right track. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there's no actual full calculation, and the error isn't an obvious factor that might suggest the problem, so I don't actually understand where Ive gone wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;(The actual question is part of a longer worked example, so if I get stuck anywhere else I might need to update this. But for now, that's the point I'm stuck at... )&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="6366" LastEditorUserId="6366" LastEditDate="2019-11-17T07:46:48.413" LastActivityDate="2019-11-17T07:46:48.413" Title="Simple steel column buckling calc - where am I going wrong? (Eurocode)" Tags="|steel|columns|buckling|eurocodes|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="31951" PostTypeId="2" ParentId="31949" CreationDate="2019-11-15T22:08:57.870" Score="2" Body="&lt;p&gt;The reason is symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a circular rod under torsion, if you rotate the rod through any angle the warped shape has to look the same, because there is no &quot;special&quot; point around the circumference where you can start measuring angles from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the rod warps, the warped shape can only vary with the radius across the section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly, the warping must be the same at every cross section along the length of the rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now consider what the two ends of the rod might look like. Suppose you twist the rod with a clockwise torque, and one end of the rod warps to bulge outwards. The other end must therefore be bulging inwards. Now suppose you turn the rod round, end-to-end. The torque on the rod is still in the same clockwise direction, but now the inward and outward bulges are reversed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no physical reason why the rod should bulge one way and not the other since it is perfectly symmetrical and uniform. So the only possible &quot;warped shape&quot; that satisfies all these conditions is no warping at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course you can reach the same conclusion using 3D continuum mechanics, but learning how to deduce things using symmetry instead of solving systems of partial differential equations is a valuable practical tool in many situations!&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-11-15T22:26:46.043" LastActivityDate="2019-11-15T22:26:46.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31952" PostTypeId="2" ParentId="31949" CreationDate="2019-11-16T00:11:29.313" Score="2" Body="&lt;p&gt;Let's imagine a solid cylinder attached to a fix support at one end and free at other. We twist it at the other end by a torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will rotate about it's axis without warping, ( small angles condition).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we cut the cylinder with laser longitudinally in 4 cuts into 8 triangle sections, radiating from the center of the circle, so if you look at the cross section it look like a circle with 4 diagonal lines at 45 degrees. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we twist this cylinder with the same torque we see the triangles wind around each other like a cable and the end of them is not a plane any more. it looks like a concrete drill bit. The circular triangles have warped and if you measure the strain half of each triangle is shortened and half elongated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason is the strain energy stored into this axial deformation is more than what would be if there was only shear strain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if you weld the seams back together you deny the triangles the opportunity to wind about. So the plane of the rotated cylinder remains plane and the strain stored is just shear.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-16T00:11:29.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31953" PostTypeId="1" CreationDate="2019-11-16T00:12:20.657" Score="-1" ViewCount="79" Body="&lt;p&gt;I found the control for a quadcopter using the angular velocities or thrust power as the input, but I need a model where I can set the initial and final position (for example, from (0,0,1) to (0,1,1)). The inputs of the quadcopter are the linear velocities. I'm using a PID controller&lt;/p&gt;&#xA;" OwnerUserId="8089" LastActivityDate="2019-11-16T12:45:15.573" Title="Transfer function for a quadcopter with linear velocity inputs" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31954" PostTypeId="2" ParentId="31950" CreationDate="2019-11-16T06:23:26.273" Score="1" Body="&lt;p&gt;You've calculated the Euler buckling force, not the compressive strength. The part of the code you want is EN 1993-1-1 section 6.3.1, which specifies how to account for imperfections and safety factor in order to calculate the allowable load. The size of the imperfections depend on whether the profile is cold formed or hot finished and the relevant safety factor is defined in whatever national annex is applicable to your column.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2019-11-16T06:23:26.273" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="31956" PostTypeId="2" ParentId="31953" CreationDate="2019-11-16T12:45:15.573" Score="1" Body="&lt;p&gt;I guess you are searching for something like &lt;a href=&quot;https://ieeexplore.ieee.org/abstract/document/7321330&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;. Those kind of papers can be found in  &lt;a href=&quot;https://scholar.google.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;google scholar&lt;/a&gt;, e.g. searching &lt;em&gt;quadcopter lagrangian mechanic&lt;/em&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From your knowledge of PID, I assume you are some sort of engineer. If you are a trained electrical engineer, I'd recommend you read some book on &lt;a href=&quot;https://en.wikipedia.org/wiki/Hamiltonian_mechanics&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hamiltonian&lt;/a&gt;/&lt;a href=&quot;https://en.wikipedia.org/wiki/Lagrangian_mechanics&quot; rel=&quot;nofollow noreferrer&quot;&gt;(Euler-)Langrangian&lt;/a&gt; Mechanic, with which you should be able to understand said papers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I like &lt;a href=&quot;https://en.wikipedia.org/wiki/Classical_Mechanics_(Kibble_and_Berkshire_book)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kibble/Berkshire&lt;/a&gt;'s book. From that point on, you have non-linear ODEs and know what to do with them.&#xA;Does this help?&lt;/p&gt;&#xA;" OwnerUserId="23539" LastActivityDate="2019-11-16T12:45:15.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31957" PostTypeId="1" AcceptedAnswerId="32568" CreationDate="2019-11-16T13:01:02.730" Score="1" ViewCount="76" Body="&lt;p&gt;Lord Evulz just stole the Earth's atmosphere. Our hero must race his trusty old Mini - which he had already prepared to be airtight, to ensure the tires do not explode, with pre-packaged oxygen to supply his fuel with, etcetera... - to the last escape rocket before his air supply runs out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume regular Earth highway conditions - (with no traffic because everybody asphyxiated to death) - with the only difference that there is no air resistance. What are the limiting factors on the car's top speed? And what velocity might be attained?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the specific model matters much, take the 1977 British Leyland Mini 1000 - because it is Mr Bean's car and I fully expect him to find a way to drive it into orbit.&lt;/p&gt;&#xA;" OwnerUserId="23540" LastActivityDate="2019-12-28T15:46:14.110" Title="How fast could a Mini Cooper go in a vacuum?" Tags="|automotive-engineering|vacuum|car|" AnswerCount="2" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="31958" PostTypeId="1" CreationDate="2019-11-16T13:35:51.660" Score="0" ViewCount="150" Body="&lt;p&gt;They didn’t give me the radius of the wheel where external force F is acting .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I calculate the reactions at A,B .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When they didn’t give me the radius .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this question “ Wrong “ ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note the force F in the right down wheel &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone confirm that &lt;span class=&quot;math-container&quot;&gt;$ A_y = 0 , B_y = F $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/8qIN0.png&quot; alt=&quot;Question 4&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="23541" LastEditorUserId="23541" LastEditDate="2019-11-17T13:30:43.247" LastActivityDate="2019-11-19T22:53:32.570" Title="Not solvable question civil engineering" Tags="|structural-engineering|civil-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31959" PostTypeId="1" AcceptedAnswerId="31961" CreationDate="2019-11-16T13:56:22.540" Score="0" ViewCount="175" Body="&lt;p&gt;The picture (which I found on the web without any accompanying explanatory text) shows a home-made z-axis micro-adjust lead-screw add-on to the vertical milling attachment of a vintage hobbyist lathe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The greenish motor-and-spindle-mount casting rides up and down on the black post, sandwiched between two plates, which ride up and down on two smaller posts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't quite make out whether there is a threaded bushing of some kind where the arrows are pointing. How would that aspect of the design typically be engineered? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that if those plates were simply threaded, without some kind of insert, you could have a somewhat loose and sloppy sandwich.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bhs6Z.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bhs6Z.jpg&quot; alt=&quot;Z-axis micro-adjust lead-screw&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21526" LastActivityDate="2019-11-17T01:13:48.117" Title="how does this z-axis micro-adjust lead-screw design work" Tags="|mechanical-engineering|mechanisms|threads|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31960" PostTypeId="2" ParentId="31958" CreationDate="2019-11-16T15:36:45.267" Score="0" Body="&lt;p&gt;We have the force &lt;span class=&quot;math-container&quot;&gt;$ (\sqrt  2/2)* F \ $&lt;/span&gt; acting horizontally to the right at point H and horizontally to the left at point E , both horizontal components of the cord tension, F making a moment of &lt;span class=&quot;math-container&quot;&gt;$ \ M=(\sqrt  2/2)* F*a .$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we equate sum of moments about A to zero and find reaction at B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \Sigma M_A =0 \quad= \frac{(\sqrt  2/2)* F*a + F*2a}{2a} +R_B = 0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ -R_B=F+ \sqrt  2/4* F $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We ignored the radius of pulley as implied by absence of it in the diagram&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After checking my answer I realized there was an arithmetic error. So I used the simple way of just summing the moments about A to simplify it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two forces acting on the system: &lt;span class=&quot;math-container&quot;&gt;$Fv \quad and \ F_{diagonal}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \Sigma M_A=0 \quad F* \frac{(3 \sqrt 2* a)}{2*2a}+ &#xA;F* \frac{2a}{2a}+R_b=0 $$&lt;/span&gt;&#xA;Note:&lt;span class=&quot;math-container&quot;&gt;$  (3 \sqrt2/2*a)= \text{distance from A to  F(diag).}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  -R_b= 3 *\sqrt 2/4*F+F $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ R_A=3 *\sqrt 2/4*F  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: The sign reversal between the reactions indicates uplift reaction on A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We ignored the radius of pulley as implied by absence of its dimension in the diagram. Please check my arithmetic.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-18T07:56:31.053" LastActivityDate="2019-11-18T07:56:31.053" CommentCount="18" ContentLicense="CC BY-SA 4.0" />
  <row Id="31961" PostTypeId="2" ParentId="31959" CreationDate="2019-11-17T01:13:48.117" Score="1" Body="&lt;p&gt;It's difficult to determine from the photo which parts are moving and which parts are dependent on another part for specific movement. If you get proportionately smaller travel per revolution of the main knob, the other parts of the assembly could be a &lt;a href=&quot;https://en.wikipedia.org/wiki/Differential_screw&quot; rel=&quot;nofollow noreferrer&quot;&gt;differential screw mechanism&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differential screw mechanisms are used in micrometers, in which the main knob rotates a substantial amount and the spindle moves a fraction of the distance. It is a construction in which an inner rod is threaded to one specification and the outer rod is threaded to a slightly larger or smaller specification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've constructed a 3D model in which the travel of the spindle is 0.25 millimeters for each rotation of the main knob, which is threaded to 1.0 mm pitch, while the inner thread is 1.25, a difference of 0.25.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason for such a mechanism is that it is easier to build a 1.0 mm and a 1.25 mm pitch thread compared to a 0.25 mm pitch, which is what would be required if one desired to bypass the differential threading option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZAjMl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZAjMl.png&quot; alt=&quot;differential threading image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.thingiverse.com/thing:2279953&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thingiverse&lt;/a&gt; provided me the model to learn more of this construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OAcsO.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OAcsO.jpg&quot; alt=&quot;differential threading model from Thingiverse&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outer threads of this mechanism would be engaging the brackets you've noted in the photo. Note also the key and matching keyway between the inner threaded rod and outer sleeve. It's critical that the portion does not rotate in order to provide the fractional movement desired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your photo, it's logical to expect that only one of the brackets contains a differential mechanism, while the other provides an alignment bushing of sorts.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-11-17T01:13:48.117" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31962" PostTypeId="1" AcceptedAnswerId="31966" CreationDate="2019-11-17T04:04:34.427" Score="1" ViewCount="634" Body="&lt;p&gt;A friend and I were debating whether a biplane/propeller plane has a faster turning rate than jet aircraft. I argued that the propeller planes often do have better turning rates than jets, but I could not explain why. Does anyone know why this is?&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2021-11-14T04:25:03.730" Title="Turn rate of a jet vs. biplane" Tags="|mechanical-engineering|aerospace-engineering|aerodynamics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31964" PostTypeId="1" CreationDate="2019-11-17T12:55:09.640" Score="0" ViewCount="47" Body="&lt;p&gt;I'm a bit involved into building an actual airplane. And while the guys go conventional ways, I'm trying to look around to see other solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of that is to try and use and RC scale-down model (easier and cheaper to build) to test and find flaws in the design and also possibly to give pilots some training before flying the actual plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question is - what will and what will not be similar in a model's behavior (say 90 cm and 180 cm wingspan versions) compared to full scale airplane (about 9m wingspan). I've already found answers related to Reynolds number on Physics SE, so it seems aerodynamics test (if we do any on the model) should be run in a pressurized tube. It's also obvious that power-to-weight ration in a model will be much higher, but possibly that can be solved by depowering the engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I build a 1:10 or 1:5 scale RC model, will it in some way be useful for actual aircraft design, building and pilot training?&lt;/p&gt;&#xA;" OwnerUserId="23550" LastActivityDate="2019-11-17T13:22:57.570" Title="Using RC scale-down model to pre-test aircraft design" Tags="|aerospace-engineering|aerodynamics|aircraft-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31966" PostTypeId="2" ParentId="31962" CreationDate="2019-11-17T13:11:40.460" Score="3" Body="&lt;p&gt;A low speed biplane will have a smaller turn &lt;em&gt;radius&lt;/em&gt; than a jet, for the same reason that a car can make a smaller radius turn at low speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For &quot;normal&quot; non-aerobatic flight, the standard &lt;em&gt;time&lt;/em&gt; to make a horizontal turns is the same for all aircraft: 3 degrees per second, or 2 minutes for a full 360 degree circle. If a jet is flying at 5 times the speed of a biplane, the radius of the circle will be 5 times as big.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low speed aircraft may use a standard turn rate of 6 degrees per second not 3, and some very large aircraft use 1.5 degrees per second. Of course if air space is being shared by different aircraft, it is easier for air traffic control to avoid potential collisions if all the planes turn at the same rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The maximum rate of turn is limited by the centripetal acceleration of the plane. Sustained turns above about 5g are limited by the ability of the pilot to remain conscious, but partial turns up to 9g are possible. The maximum instantaneous turn rate for a military fast jet is typically about 20 degrees per second, which would theoretically give a 360 degree turn in 18 seconds if other factors didn't prevent that from being done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As well as the inability of the pilot to operate at sustained high g loads, the drag on the aircraft increases at very high turn rates, and the engine power may not be able to maintain constant speed, even flying a military jet with reheat. If the speed reduces too much, there is a high probability that the intended turn will become a stall or a spin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course the fastest &quot;turn&quot; in one sense of the word is an &lt;em&gt;intentional&lt;/em&gt; spin. Aircraft designed for aerobatics may be able to spin faster than &lt;em&gt;one complete turn per second&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-11-17T13:20:14.580" LastActivityDate="2019-11-17T13:20:14.580" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31967" PostTypeId="2" ParentId="31964" CreationDate="2019-11-17T13:22:57.570" Score="2" Body="&lt;p&gt;You need to decide between geometric similarity, kinematic similarity and dynamic similarity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to decide which have to be met, while allowing others to vary so you can get meaningful results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One change used is to test shapes in water instead of air as there is about a factor of 1000 between the two... This can provide meaningful results while keeping the models small enough to control and keeping the costs down. The difficulties of testing full scale items is usually a challenge - anyone have a wind tunnel which will fit the Airbus A380?&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-11-17T13:22:57.570" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31968" PostTypeId="1" CreationDate="2019-11-17T14:19:20.133" Score="2" ViewCount="136" Body="&lt;p&gt;I have the task of modeling the current to torque mapping for a given motor. I have an experimental set up where I can retrieve current, torque pairs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my initial approach was to model the relationship with a regression curve, but I realized that the motor certainly shows some kind of hysteresis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I be able to compute a current to torque mapping that also includes hysteresis from the data I obtained&lt;/p&gt;&#xA;" OwnerUserId="18808" LastActivityDate="2024-01-02T01:08:31.027" Title="How can I model a hysteresis curve from data?" Tags="|motors|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31969" PostTypeId="1" CreationDate="2019-11-17T15:11:45.843" Score="2" ViewCount="745" Body="&lt;p&gt;The Ziegler-Nichols method of tuning PID control loops has been around since the 1940's, and has been widely applied in many areas of engineering. I understand that it is really only a heuristic method of finding the right control parameters, and many others have been created over the years. However, when it was originally created, was there any mathematical reason for how the method worked. I am talking specifically about the open-loop step response method.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wPbWo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wPbWo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I cannot find any reason for why they decided upon the method they did. Is there any mathematical reason behind this?&lt;/p&gt;&#xA;" OwnerUserId="23552" LastActivityDate="2021-06-18T09:30:56.943" Title="What is the mathematical basis for the Ziegler-Nichols method for tuning a PID loop?" Tags="|control-engineering|pid-control|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31970" PostTypeId="2" ParentId="31962" CreationDate="2019-11-17T17:22:07.143" Score="0" Body="&lt;p&gt;Long ago , the IL Air National Guard flew F51 and F84 ; at summer camp they shared the same dedicated airfield. In the distance you could see an F84 line up on a straight approach to land; Occasionally an F51 would make a 90 degree turn and land in front of the F84. When the F84 made its large turn it was so far away that you couldn't see it unless you were looking that way with binoculars. The F51 were dramatically more maneuverable than the F84 . I think it is a safe guess the a biplane is more maneuverable than an F51 , so no comparison to a modern jet. However , the speeds are rather different . &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-11-17T17:22:07.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31971" PostTypeId="1" AcceptedAnswerId="31992" CreationDate="2019-11-17T20:00:32.107" Score="1" ViewCount="88" Body="&lt;p&gt;I'm looking for a few basic devices such as converting translational to rotational motion, 20:1 gearbox, etc. Found nothing at Home Depot. Could you suggest where does one buys/orders such components?&lt;/p&gt;&#xA;" OwnerUserId="3548" LastActivityDate="2019-11-19T21:31:51.187" Title="Where do I get basic mechanical components?" Tags="|mechanisms|" AnswerCount="2" CommentCount="4" ClosedDate="2019-11-21T00:12:34.460" ContentLicense="CC BY-SA 4.0" />
  <row Id="31972" PostTypeId="1" AcceptedAnswerId="31973" CreationDate="2019-11-17T20:18:59.053" Score="12" ViewCount="15057" Body="&lt;p&gt;The lower viscosity of water reduces friction.  Besides, the water can transfer heat way better than oils.  So why not use water as a lubricant in wind or hydro turbines, especially on equipment which is operating in the ocean ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems to be a perfect material.  It's the most abundant liquid on Earth, and it causes no ecological damage. &lt;/p&gt;&#xA;" OwnerUserId="17181" LastEditorUserId="23599" LastEditDate="2019-11-19T15:51:30.077" LastActivityDate="2019-11-19T16:05:40.240" Title="Why is oil used as the lubricant in power generators, while water is the most available, cheapest and accessible lubricant?" Tags="|mechanical-engineering|water-resources|" AnswerCount="4" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="31973" PostTypeId="2" ParentId="31972" CreationDate="2019-11-17T20:40:26.547" Score="59" Body="&lt;p&gt;The whole point of lubrication is to keep the two surfaces almost apart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water has low viscosity in comparison to oil-based lubricants, which means that it forms a thinner film (electrohydrodynamic film thickness) between the two surfaces; the thinner the film, the greater the chance of direct physical contact between the parts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the higher viscosity actually can be useful here. Another thing about water is that its viscosity doesn’t increase by pressure, unlike oil-based lubricants, as usually, the two surfaces are operating under pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oils are better vibration dampers than water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As @SolarMike mentioned in the comments, water is extremely corrosive and can also dissolve many destructive elements. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks to the capillary effect, water molecules can penetrate into micro-cracks on/beneath the surface of parts.  Once there, due to extreme pressure, those molecules can decompose into hydrogen and oxygen, and build up pressure in the cracks, which leads to crack propagation and failure. This phenomenon is called Hydrogen embrittlement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water also provides a very good environment for fungi and other living microorganisms, which decay and changes the PH of the water, usually a shift toward an acidic environment, which in turn also corrodes the steel. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorUserId="23599" LastEditDate="2019-11-19T15:51:22.203" LastActivityDate="2019-11-19T15:51:22.203" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="31974" PostTypeId="2" ParentId="31972" CreationDate="2019-11-17T21:32:45.047" Score="12" Body="&lt;p&gt;Sea water is about the worst thing in the world to put in a metallic mechanical device. As noted , corrosion would be horrendous ; dissimilar metals, stray currents , etc. Wind turbine gearing is an exceptionally demanding application ; EP ( extreme pressure ) lubes are needed with graphite and/or moly sulfide, etc. I am sure a major problem addressed by wind machine engineers for off shore applications , is sealing the lube systems to keep salt and moisture out. Not the least of problems is that water has little/no lubricity qualities like film strength. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-11-17T21:32:45.047" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31975" PostTypeId="2" ParentId="31971" CreationDate="2019-11-17T22:51:48.980" Score="1" Body="&lt;p&gt;If you are a company you should definitely go look on Misumi but they don’t sell to private.&#xA;Otherwise you could google for what you need and in most cases you will find some website selling it. I bought bearings, gearboxes and motors without issues of any kind.&lt;/p&gt;&#xA;" OwnerUserId="23555" LastActivityDate="2019-11-17T22:51:48.980" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31976" PostTypeId="1" CreationDate="2019-11-18T04:47:05.297" Score="0" ViewCount="74" Body="&lt;p&gt;There's a simple way to lift a platform a short distance:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Identical discs are mounted on a frame at the four corners of a&#xA;platform such that front pair of discs' axes of rotation are&#xA;colinear, as are those of the back pair.  The front and back axes&#xA;are parallel.&lt;/li&gt;&#xA;&lt;li&gt;Each corner of the platform (at rest) is mounted to its disc at the same offset (and angle) from the disc's center.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;When you lift one end of the platform, all of the discs have to turn identically, so the platform rises parallel to its at-rest position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the formal name (e.g. crank-slider) of this mechanism?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been searching the Internet for parts and keep coming up with harrows and cultivators (farm equipment for plowing).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt; I'm building a rack to store eight snowmobiles in two rows - one above the other.  I have maybe an inch to spare width-wise, so the solution has to be fairly thin, and the ceiling is already too low.  Fortunately, the top row can sit below the tallest bits of the bottom row (the windshields and handlebar loops can project into the upper level without being run over).  But the back halves of the upper decks need to lift 4-8&quot; to allow sleds to move into the lower level.  Discs are not only thin, but their rotation when the deck is fully lowered could be at an angle where raising the deck could combine pushing with lifting, rather than just pure lifting.  When raised, their rotation would be just past 12o'clock so they are &quot;locked&quot; by gravity, and &quot;unlocked&quot; with a slight pull.  (The decks would be unloaded when raised and lowered, else they would require power and vertical clearance that I don't have.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd rather find a manufactured thin-disc-on-a-thin-bearing but I have no idea what it would be called.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="23557" LastEditorUserId="23557" LastEditDate="2019-11-20T01:44:19.693" LastActivityDate="2019-11-20T21:19:46.463" Title="Name of this type of platform lift" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31978" PostTypeId="1" CreationDate="2019-11-18T08:33:03.403" Score="0" ViewCount="22" Body="&lt;p&gt;Not sure if this is the right place to ask this question, but anyways. I'm not very fond of circuits and stuff, but I've got a test question I really need to answer. The question is in the title, possible answer options are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Fibrous surface&lt;/li&gt;&#xA;&lt;li&gt;Temperature&lt;/li&gt;&#xA;&lt;li&gt;Thermal conductivity&lt;/li&gt;&#xA;&lt;li&gt;Velocity&lt;/li&gt;&#xA;&lt;li&gt;Opacity&lt;/li&gt;&#xA;&lt;li&gt;Average density&lt;/li&gt;&#xA;&lt;li&gt;Color&lt;/li&gt;&#xA;&lt;li&gt;Electromagnetic radio waves radiation (as far as I understand, the question suggests that we're speaking about waves produced by the object that we are measuring distance to itself)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Multiple choices are possible. I didn't manage to find an answer on the internet, so I hope someone can help me here.&lt;/p&gt;&#xA;" OwnerUserId="23563" LastActivityDate="2019-11-18T08:33:03.403" Title="What object characteristics may cause significant distance measurement deviations in Sharp optical sensors?" Tags="|sensors|distance-measurement|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31979" PostTypeId="2" ParentId="8807" CreationDate="2019-11-18T11:07:24.990" Score="0" Body="&lt;p&gt;When folding paper multiple times the radius of each layer at the fold is different. By applying pressure to the folded structure outer layers are subjected to a higher tensile force, than inner layers. The thicker the fold, the larger the radius, the more tensile stress on the outer layer. It happened to be that it ruptured (I.e. the stress in the outer layers became larger than ultimate tensile stress for the paper) on the 7th fold in this case.&#xA;But overall apparent &quot;brittleness&quot; isn't related to folding at all. This is essentially a composite material formed from multiple layers of highly compressed paper. The same effect would have been if the layers were pre-cut instead of folded.&lt;/p&gt;&#xA;" OwnerUserId="23565" LastActivityDate="2019-11-18T11:07:24.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31980" PostTypeId="2" ParentId="31976" CreationDate="2019-11-18T11:07:48.637" Score="1" Body="&lt;p&gt;I suspect you may be looking for a &quot;synchronized screw jack lift mechanism.&quot; Many sites exist when one searches using the above term. Some are not synchronized, allowing for tilt and translation and probably other movements, depending on the complexity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A &lt;a href=&quot;http://www.screw-jack.com/acme-screw-jack-for-table-lift.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;commercially oriented web page&lt;/a&gt; provided this image:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XPZPi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XPZPi.jpg&quot; alt=&quot;screw jack image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laser cutters use this method of bed movement as do many 3D printers. There are variations in the mechanical actuators as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: With the previous information as a foundation, consider that 3D printers use lead screws (pronounced leed) with follower nuts that are usually disk shaped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Vm5p7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Vm5p7.jpg&quot; alt=&quot;lead screw with follower&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a common part in other engineering realms, specifically machine tools and you may be able to find a diameter appropriate to your application, capable of handling the loads involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Laser cutters have tables which use the same system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9jWSL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9jWSL.jpg&quot; alt=&quot;laser cutter table lift mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For lasers and 3D printers, GT class belts and gears are used to rotate the screws. GT belts are available in a wide variety of sizes from 3 mm to 10 mm and possibly wider. If necessary, you could use a drive belt from any power transmitting device (motorcycle?) as long as the gear matches and can be secured to the lift screws.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2019-11-20T21:19:46.463" LastActivityDate="2019-11-20T21:19:46.463" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31982" PostTypeId="1" AcceptedAnswerId="31983" CreationDate="2019-11-18T13:58:11.933" Score="2" ViewCount="42" Body="&lt;p&gt;Is there any heat energy lost to the mechanical force of pushing a piston in a combustion engine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, would a natural gas burning as a flame produce more heat than the same amount of natural gas that was burned when driving a combustion engine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The context of this question is around fuel efficiency of natural gas water boilers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current understanding is that a naked natural gas flame is burning inside the unit, and heats water that is contained within a series of pipes, which in turn is cooled in a heat exchanger, and that same heat exchanger heats the incoming cool water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm wondering, would if it make any sense at all to burn that natural gas inside a combustion engine and use the heat of the combustion engine used to heat the water to the same extent as a naked flame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally my goal is to solve the question, can we heat water the same extent by burning the same amount of natural gas &lt;strong&gt;and&lt;/strong&gt; get free mechanical energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="23571" LastActivityDate="2019-11-18T14:24:02.040" Title="Recovering mechanical energy from natural gas water boiler" Tags="|thermodynamics|combustion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31983" PostTypeId="2" ParentId="31982" CreationDate="2019-11-18T14:24:02.040" Score="2" Body="&lt;p&gt;Check out combined heat and power systems... generates electricity and makes use of the waste heat to heat water - but is only really efficient when there is sufficient hot water demand ie heating load.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-11-18T14:24:02.040" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31985" PostTypeId="1" CreationDate="2019-11-18T16:35:39.340" Score="-2" ViewCount="41" Body="&lt;p&gt;I live in a small flat with all doors and windows sealed. The temperature outside is moderately cold. I assume the temperature in my flat is hotter because my body and the computer are emitting heating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I open the window slightly and turn on the air condition to a very high temperature will air move entirely from my house to outdoors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I want &quot;fresh&quot; air to come from the air condition and move outside as fast possible. I will need outdoors to act as a vacuum cleaner sucking up all the air from my flat rapidly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the effect to be exaggerated. When I put my finger on the slightly open window I want to feel the pressure sucking it towards outside.&lt;/p&gt;&#xA;" OwnerUserId="23578" LastActivityDate="2019-11-20T06:50:15.827" Title="How can I establish as much airflow from my house to outdoors as possible?" Tags="|thermodynamics|airflow|temperature|air|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31987" PostTypeId="1" CreationDate="2019-11-18T19:28:21.300" Score="1" ViewCount="89" Body="&lt;p&gt;I've got everything set. Material is aluminum alloy 6061. I've set the component contacts to no penetration. I have a contact set where both blocks meet and keep getting the same error. The bottom part is fixed. The head coming out has a no penetration contact with the top block. The only thing that works is by making it globally binded but I'm pretty sure that they are meant to not be binded&lt;a href=&quot;https://i.stack.imgur.com/vlFI2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vlFI2.png&quot; alt=&quot;objects&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23581" LastActivityDate="2019-11-18T19:28:21.300" Title="Solidworks static simulation having issues" Tags="|mechanical-engineering|statics|cad|solidworks|autocad|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="31988" PostTypeId="1" AcceptedAnswerId="31989" CreationDate="2019-11-18T23:12:58.797" Score="0" ViewCount="70" Body="&lt;p&gt;We usually use induction motors to turn electrical energy to mechanical, however synchronous motors are the machines on the other side of the grid (generation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A week ago we had a student visit to the hydroelectric and ... centrals and afterward we paid a visit to a steel company. I noticed that they use the combination of the synchronous and induction motors to do the exact same job, so the output power of the motors were exactly the same so the task and the hours of operation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why just not using induction motors ?&lt;/p&gt;&#xA;" OwnerUserId="23589" LastActivityDate="2019-11-18T23:38:57.823" Title="Why large industrial units combine induction and synchronous motors?" Tags="|electrical-engineering|motors|industrial-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31989" PostTypeId="2" ParentId="31988" CreationDate="2019-11-18T23:38:57.823" Score="0" Body="&lt;p&gt;As the name suggests, induction motors are inductive, it means they require inductive power input, or reactive power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with reactive power consumption is, if you max the monthly limitation out, the providers are going to charge you a lot.  Why? then this is another question. Hint: Think about the phase relationship in AC circuits. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The synchronous motors can receive active power from the grid, and at the same time (under some circumstances) generate reactive power and inject it into the grid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, by this method engineers can compensate for almost all reactive power that those induction motors consume so, the units don’t have to pay any fine to the provider, and providers will be happy.&lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-11-18T23:38:57.823" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="31990" PostTypeId="1" CreationDate="2019-11-19T00:41:56.623" Score="0" ViewCount="82" Body="&lt;p&gt;Curious as to what extent a shaft/pin will will follow a bore when being pressed in. &#xA;Considering fixturing for a project to hold a shaft straight, but most of what I have seen, the shaft will self align to an extent(as long as the angle is not too much). &#xA;Is there anything out there that says you can expect a shaft to align with the bore as long as the pressing angle is not more than x?&lt;/p&gt;&#xA;" OwnerUserId="23591" LastActivityDate="2019-11-19T00:41:56.623" Title="Pressing shaft into bore or bearing?" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31991" PostTypeId="1" AcceptedAnswerId="32005" CreationDate="2019-11-19T00:58:03.597" Score="3" ViewCount="62" Body="&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Kangaroo&quot; rel=&quot;nofollow noreferrer&quot;&gt;kangaroo&lt;/a&gt; has incredible jumping abiltiies. Its &lt;a href=&quot;https://en.wikipedia.org/wiki/Gastrocnemius_muscle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Gastrocnemius muscle&lt;/a&gt; is a major part of generating its jumping force. Has any engineer based a design, probably a vehicle, upon the kangaroo's locomotion system?  &lt;/p&gt;&#xA;" OwnerUserId="23590" LastActivityDate="2019-11-20T04:24:32.163" Title="Has the kangaroo's jumping ever been copied in human engineering?" Tags="|transportation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31992" PostTypeId="2" ParentId="31971" CreationDate="2019-11-19T01:52:54.727" Score="2" Body="&lt;p&gt;Here are my go to sources:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.pic-design.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pic Design&lt;/a&gt; (Good for gears, screws and other components)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.sdp-si.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stock Drive&lt;/a&gt; (Similar to Pic Design)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.wmberg.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;WM Berg&lt;/a&gt; (Good collection of parts. Particularly good for chain and belt drives and alternatives)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.mcmaster.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;McMaster-Carr&lt;/a&gt; (Broad range of products)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;a href=&quot;https://www.centuryspring.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Century Spring&lt;/a&gt; (Awesome for springs)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9506" LastEditorUserId="9506" LastEditDate="2019-11-19T21:31:51.187" LastActivityDate="2019-11-19T21:31:51.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31993" PostTypeId="2" ParentId="2081" CreationDate="2019-11-19T03:35:07.687" Score="1" Body="&lt;p&gt;Diesel engines can run on a wide variety of fuels, as long as the fuel provides some lubrication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Fuel_oil#Uses&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Fuel_oil#Uses&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Residual fuel oil is less useful because it is &lt;strong&gt;so viscous&lt;/strong&gt; that &lt;strong&gt;it has to be heated with a special heating system before use&lt;/strong&gt; and it may contain relatively high amounts of pollutants, particularly sulfur, which forms sulfur dioxide upon combustion. However, its undesirable properties make it very cheap. In fact, &lt;strong&gt;it is the cheapest liquid fuel available&lt;/strong&gt;. Since it requires heating before use, &lt;strong&gt;residual fuel oil cannot be used in road vehicles, boats or small ships&lt;/strong&gt;, as the heating equipment takes up valuable space and makes the vehicle heavier. Heating the oil is also a delicate procedure, which is impractical on small, fast moving vehicles. &lt;strong&gt;However, power plants and large ships are able to use residual fuel oil.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="20710" LastActivityDate="2019-11-19T03:35:07.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31994" PostTypeId="2" ParentId="12" CreationDate="2019-11-19T06:14:58.510" Score="1" Body="&lt;p&gt;If your watershed is in the U.S., you might find TR-55 useful. &quot;Technical Release 55 (TR-55) presents simplified procedures to calculate storm runoff volume, peak rate of discharge, hydrographs, and storage volumes required for floodwater reservoirs. These procedures are applicable in small watersheds, especially urbanizing watersheds, in the United States.&quot;  It can even be run on a calculator.&lt;/p&gt;&#xA;" OwnerUserId="23593" LastActivityDate="2019-11-19T06:14:58.510" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="31996" PostTypeId="1" CreationDate="2019-11-19T11:28:45.937" Score="0" ViewCount="45" Body="&lt;p&gt;What is the best way to design a circuit based on the following transfer function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T(s)=\frac{9216}{(s^2 + 48s + 9216)}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="23600" LastEditorUserId="18238" LastEditDate="2020-05-04T08:02:48.870" LastActivityDate="2020-05-04T08:02:48.870" Title="Design op-amp circuit from transfer function" Tags="|electrical-engineering|control-engineering|pid-control|transfer-function|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="31997" PostTypeId="1" AcceptedAnswerId="32019" CreationDate="2019-11-19T13:45:59.377" Score="0" ViewCount="403" Body="&lt;p&gt;with masses declared as grams and static stiffness as N/mm, how do I dimension them for Hz ?&lt;/p&gt;&#xA;" OwnerUserId="17070" LastActivityDate="2019-11-20T19:19:11.980" Title="How do I dimension the natural frequency formula?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="31998" PostTypeId="2" ParentId="31972" CreationDate="2019-11-19T15:44:50.003" Score="0" Body="&lt;h2&gt;&lt;em&gt;Lubricity&lt;/em&gt; is a material quality&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Look into the difficulties they have had removing sulfur from diesel fuel.  Sulfur was added (or not removed) to increase the lubricity of the diesel fuel, to improve freedom of motion in injection pumps.  That being missing is a hardship for older diesel engines whose pumps depend on it.  The fuel is basically the same in other respects, but for this lubricity factor.   This is easily restored with as little as 2% biodiesel.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So here you have these two liquids, mineral diesel and biodiesel - that are both seemingly the same.  But one is &quot;slick&quot; and the other not.  That property is &lt;strong&gt;lubricity&lt;/strong&gt;, and that is the very thing you need in a lubricating oil. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water has none.  &lt;/p&gt;&#xA;" OwnerUserId="11673" LastActivityDate="2019-11-19T15:44:50.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="31999" PostTypeId="2" ParentId="2081" CreationDate="2019-11-19T15:59:57.140" Score="0" Body="&lt;p&gt;Diesel and heater oil are the same thing , the tiny difference is the additive package ( a fraction of 1 % of the total). Diesel may have a better additive to prevent wax from crystallizing and plugging lines in cold weather , for example. From time to time in various locations there may be different limits on sulfur ; depending on the refinery this may or may not result in different S levels in the products. Heavy marine fuel is a different from diesel because marine engines have heaters to reduce the viscosity of heavy oils like &quot;bunker C&quot;.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-11-19T15:59:57.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32000" PostTypeId="2" ParentId="31972" CreationDate="2019-11-19T16:05:40.240" Score="4" Body="&lt;p&gt;Great question!  Water absolutely &lt;em&gt;is&lt;/em&gt; used as a lubricant in some power generating systems.  For example many Francis style turbines use their feed water as the primary lubricant and coolant of the bottom main bearing.  (In the old Loeffel turbines, that bearing was typically made of lignum vitae wood, and could last up to 100 years in near constant use).  Good engineering takes advantage of the cheap availability of water whenever applicable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But water freezes and expands, blowing apart your engines, so it's not suitable for every application.  This is enough of an everyday issue that automobiles are designed with &quot;freeze plugs&quot; and block heater ports and typically use toxic and environmentally unfriendly antifreeze chemicals in their liquid coolant systems.  If your generator fails in cold weather, you don't want it to freeze solid before you can manage to repair it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A freeze plug, also called a Welch plug, is a core hole plug purposely made weak enough that it will fail before the block cracks due to frozen coolant.  Core holes are left by the block casting process.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And although absolutely pure water is an electrical insulator in the laboratory, in real life it's not, because even the slightest amount of contamination makes water an excellent electrical conductor.  Oils, on the other hand, typically have consistently low conductivity until they are so loaded with metal contaminants that they have lost lubricity.  If your generator has an oil leak, the oil won't provide a conductive path to repair technicians and other system parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally, as others have already noted, water is very corrosive, especially seawater.  Since rust expands to five times the size of the steel it came from, you can't use any of the cheap iron alloys in water, which drives up the cost of materials pretty quickly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Petroleum oils are a superb lubricant, able to be purified and recycled many times.  At least as early as 1915 intelligent people were writing that burning fuels like petroleum is a criminal waste of a fantastically valuable and limited resource; it's robbing future generations in order to turn our shared treasure into atmospheric poisons.  It's nice to see people like yourself thinking about these things!&lt;/p&gt;&#xA;" OwnerUserId="23606" LastActivityDate="2019-11-19T16:05:40.240" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32002" PostTypeId="1" AcceptedAnswerId="32012" CreationDate="2019-11-19T16:49:39.670" Score="3" ViewCount="72" Body="&lt;p&gt;I am trying to prove with Matlab that if I have an improper system and I place poles at higher and higher frequencies the performances of the system improves. In particular I am considering the following two degree of freedom scheme:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/I5YwC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/I5YwC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$C_{f}= \frac{(1+s)(1+0.05s)^{2}]}{(1+\tau s)^{3}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My code is the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = tf('s');&#xA;P = 1/[(1+s)*(1+0.05*s)^2];&#xA;C = (s+1)/s;&#xA;&#xA;&#xA;tau_1 = 0.1;&#xA;CF_1 = [(1+s)*(1+0.05*s)^2]/((1+tau_1*s)^3);&#xA;&#xA;tau_2 = 0.01;&#xA;CF_2 = [(1+s)*(1+0.05*s)^2]/((1+tau_2*s)^3);&#xA;&#xA;tau_3 = 0.001;&#xA;CF_3 = [(1+s)*(1+0.05*s)^2]/((1+tau_3*s)^3);&#xA;&#xA;T1 = (C+CF_1)*P/(1+P*C);&#xA;T2 = (C+CF_2)*P/(1+P*C);&#xA;T3 = (C+CF_3)*P/(1+P*C);&#xA;&#xA;&#xA;figure;&#xA;bodemag(T1,'r',T2,'b',T3,'g'),grid&#xA;legend('tau = 0.1','tau = 0.01','tau = 0.001')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;so what I expected is that the performances with respect to the reference tracking increse as tau gets smaller, but if I do the Bode plot, what I get is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kT8fZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kT8fZ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from which I don't really see much of an improvement. Moreover if I change some values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TOWnj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TOWnj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is somenthing that to me does not makes sense because I should have that with &lt;span class=&quot;math-container&quot;&gt;$\tau =1$&lt;/span&gt; I should have better performances than with &lt;span class=&quot;math-container&quot;&gt;$\tau =10$&lt;/span&gt;, this because with &lt;span class=&quot;math-container&quot;&gt;$\tau =1$&lt;/span&gt; the pole is at higher frequencies that with &lt;span class=&quot;math-container&quot;&gt;$\tau =10$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody please help me solving this problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT] If I plot the step responses I see the same problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NkXPt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NkXPt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT 2]For completeness, I post the image of the step response for the first choise of tau's:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pJvua.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pJvua.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for these values of &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; there is a clear improvement for overshoot and a faster response. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried other values smaller than 1 for &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; and all of these show what I expected in the step response. While for values bigger than 1, I obtain something similar at the other situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does somenone know why this happens? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT 3] With the last one, so values of &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; smaller than 1, I have also noticed an increase in phase margin, so this should mean that the system is performing better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While, if I consider the values &lt;span class=&quot;math-container&quot;&gt;$\tau=1$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\tau=10$&lt;/span&gt; i get that for the first one the phase margin is 125 deg and for the second is 170 deg. So this should be in according to the fact that &lt;span class=&quot;math-container&quot;&gt;$\tau=10$&lt;/span&gt; performs better.&lt;/p&gt;&#xA;" OwnerUserId="22780" LastEditorUserId="22780" LastEditDate="2019-11-21T07:53:59.163" LastActivityDate="2019-11-21T09:11:48.260" Title="Problem with performances of a control scheme" Tags="|electrical-engineering|control-engineering|control-theory|matlab|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32003" PostTypeId="2" ParentId="31958" CreationDate="2019-11-19T18:49:13.500" Score="1" Body="&lt;p&gt;Well I had to scratch my head on this one and dust off more than a few cobwebs.  Now my understanding is that if you put a pulley on the ceiling and pass a rope through it to a 100 lbs weight and you hold that weight you are pulling on the free end of the rope with 100 lbs.  If you are pulling straight down, it means you are are putting 100 lbs and the block is pulling down at 100 lbs and the pulley is exerting 200 lbs on the ceiling.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are pulling parallel to the ceiling then there is a 100 lbs down force and a 100 lbs horizontal force. or if you prefer a 141.4 lbs force at 45 degrees downward from the ceiling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my understanding is not valid, then the rest of this answer is not valid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I dropped this problem into S-Frame and placed the sum of the component horizontal and vertical loads as nodal loads.  Because I can't just put 'a' in as a length I opted to make 'a' 10.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now provided my understanding of pulleys is correct, then this model confirms there is no vertical load at A (point 1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hAfYN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hAfYN.png&quot; alt=&quot;S-Frame Model&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is with the understanding that the vertical and horizontal components of th e 45 degree line break into roughly 70.7107 each&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sum of forces at point 7 is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Fy=100 down from vertical + 70.7107 down from diagonal&#xA;Fx=70 right from diagonal&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Sum of forces at point 4 is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Fy=100 up from vertical&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Sum of forces at point 5 is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Fy= 100 down from vertical + 70.7107 up from diagonal &#xA;Fx= 70.7107 left from diagonal&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="6993" LastEditorUserId="6993" LastEditDate="2019-11-19T22:53:32.570" LastActivityDate="2019-11-19T22:53:32.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32004" PostTypeId="2" ParentId="31997" CreationDate="2019-11-20T01:40:17.370" Score="-1" Body="&lt;p&gt;One Newton is equal to 101.971grams and one gram is 0.00980665N.&#xA;In free undamped vibration we have to convert grams to Newtons,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \omega_n = \sqrt \frac{k}{0.00980665*m_{grams}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  T= \frac{2\pi}{\omega_n} $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$  f_n=\frac{1}{T}= \frac{\omega_n}{2\pi}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Going over my answer after @John answer, I realize that I was wrong. I leave my answer should it be an example of how you can be wrong.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-20T19:19:11.980" LastActivityDate="2019-11-20T19:19:11.980" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32005" PostTypeId="2" ParentId="31991" CreationDate="2019-11-20T04:24:32.163" Score="1" Body="&lt;p&gt;If you mean wether we have similar structure and articulation of bones and muscles, yes we do, as your link indicates. And some of us by training can jump rather high.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is more than a muscle or a joint and muscles. Each of our species after long period of time has evolved for certain tasks, obviously.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think a kangaroo jumps by a nerve impulse which once triggered organizes an ordered sequence of muscle contraction and wave of impulse throughout it's entire body, like a set of struts and springs calibrated to achieve a mechanical feat. In fact all animals do this in sorts for locomotion and doing tasks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's in their genes, the path of nerve impulse, the call to instantaneous tightening of back and neck, prohibiting them to lag  limply as a damper, but  work as a wiplash and  and amplifier for the progression of smooth arc of jump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is part of the subject of the field of neural networking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used to run at my college, and we had learned that you run with your whole body not your legs.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-20T04:24:32.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32006" PostTypeId="2" ParentId="31911" CreationDate="2019-11-20T04:51:43.790" Score="1" Body="&lt;p&gt;I would advise staying away from the conical, as this is for low pressure applications. This would lead to more thermal flashing in the vessel. You would also want to consider sch 80 with high pressure steam and condensate applications.&lt;/p&gt;&#xA;" OwnerUserId="21724" LastActivityDate="2019-11-20T04:51:43.790" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32007" PostTypeId="1" CreationDate="2019-11-20T05:44:03.143" Score="-1" ViewCount="355" Body="&lt;p&gt;How do I change the coordinate system from two dimensional cartesian to two dimensional curve linear and polar by using Matrix?&lt;/p&gt;&#xA;" OwnerUserId="23613" LastEditorDisplayName="user14407" LastEditDate="2019-11-20T22:48:37.103" LastActivityDate="2019-11-20T22:48:37.103" Title="How do I change the coordinate system from x-y to n-t or r-theta by using Matrix?" Tags="|mechanical-engineering|dynamics|mathematics|mechanical|" AnswerCount="1" CommentCount="2" ClosedDate="2019-11-21T00:13:32.853" ContentLicense="CC BY-SA 4.0" />
  <row Id="32008" PostTypeId="2" ParentId="31985" CreationDate="2019-11-20T06:32:00.223" Score="0" Body="&lt;p&gt;As @Transistor has commented air conditioning systems don't add fresh air, they recycle the air. Because the return air going back to the air conditioner requires much less energy to be brought back to the thermostat setting then outside cold temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also modern system are very efficient and do their job by streamlining the system so that distribution of the conditioned air doesn't require a big fan/ blower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you feel you need more fresh air than what by code must be provided to your space, you can install a fan in a exterior wall location far from the air conditioner return so that your fan doesn't impede your air conditioner's job. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a note, sometimes these situations are a sign of asthma or some similar medical issue. you may want to ask a doctor.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-20T06:50:15.827" LastActivityDate="2019-11-20T06:50:15.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32009" PostTypeId="2" ParentId="32007" CreationDate="2019-11-20T09:42:10.660" Score="2" Body="&lt;p&gt;The coordinate transformation of polar to cartesian is given by: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\begin{pmatrix}&#xA;r \\&#xA;\theta&#xA;\end{pmatrix}&#xA;=&#xA;\begin{pmatrix}&#xA;\cos\theta &amp;amp; \sin\theta \\&#xA;-\sin\theta &amp;amp; \cos\theta&#xA;\end{pmatrix}&#xA;\begin{pmatrix}&#xA;x \\&#xA;y&#xA;\end{pmatrix}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a point mass, moving along a path in space, we can define an arbitrary point, and call it origine. Notice that the polar and the cartesian coordinate systems, have the same origine. And the &lt;span class=&quot;math-container&quot;&gt;$(x, y)$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$(r, \theta)$&lt;/span&gt; represent the position of the the point mass relative to that origine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tangent/normal (n-t) or curve linear coordinate system, moves with the point mass along the same path, notice that the origine of the new coordinate system now moves with the point mass. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, you want to express the position of the point mass in the curve linear coordinate in function of the old (cartesian) coordinates, but the position of the point mass coincide with the origine, so there is no explicit transformation here. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-11-20T09:42:10.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32010" PostTypeId="2" ParentId="31911" CreationDate="2019-11-20T10:56:11.430" Score="1" Body="&lt;p&gt;High pressure applications typically involve elliptic head. This is the shape with the lowest stress from all shapes possible. See the graph at the bottom of 6th page of this paper &lt;a href=&quot;https://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;url=https://pdfs.semanticscholar.org/bbce/eebd3ef103fe37d7f0185d193005d194e83c.pdf&amp;amp;ved=2ahUKEwielae8zfjlAhWWYisKHXqxBKUQFjAKegQIBhAB&amp;amp;usg=AOvVaw0s2shzOjwYd_dh04xFtZOI&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.google.com/url?sa=t&amp;amp;source=web&amp;amp;rct=j&amp;amp;url=https://pdfs.semanticscholar.org/bbce/eebd3ef103fe37d7f0185d193005d194e83c.pdf&amp;amp;ved=2ahUKEwielae8zfjlAhWWYisKHXqxBKUQFjAKegQIBhAB&amp;amp;usg=AOvVaw0s2shzOjwYd_dh04xFtZOI&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Especially important to minimize stress on the welding seam between the head and the cylinder (even if you have flanged connection, still ellipse is the safest way to go)&#xA;Liquid will fully drain from elliptical shape if you have the drain line to the trap right in the middle of the dish.&#xA;450degC is quite high too. Choose your material wisely. Yield stress is going to be much lower than at room temperature.&lt;/p&gt;&#xA;" OwnerUserId="23565" LastActivityDate="2019-11-20T10:56:11.430" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32011" PostTypeId="1" CreationDate="2019-11-20T12:48:45.347" Score="-1" ViewCount="115" Body="&lt;p&gt;Is it possible to derive velocity vector form position vector without unknown such as: position vector of A = (0.8)i + (1.2)j?&lt;/p&gt;&#xA;" OwnerUserId="23621" LastActivityDate="2019-11-20T14:08:03.017" Title="Position vector to Velocity vector" Tags="|mechanical-engineering|applied-mechanics|dynamics|mechanical|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32012" PostTypeId="2" ParentId="32002" CreationDate="2019-11-20T12:50:42.327" Score="4" Body="&lt;p&gt;Your closedloop crossover frequency (when the magnitude of &lt;span class=&quot;math-container&quot;&gt;$P(s)\,C(s)$&lt;/span&gt; is equal to one) lies at roughly 1 rad/s. This means that the feedback controller already causes the system to track reference signals that have a frequency content sufficiently below that. So adding feedforward that only approximates the inverse of the plant (&lt;span class=&quot;math-container&quot;&gt;$C_f(s)\,P(s)\approx1$&lt;/span&gt;) below the crossover frequency should not aid in the tracking performance, so that is why for both &lt;span class=&quot;math-container&quot;&gt;$\tau=1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\tau=10$&lt;/span&gt; the magnitude of the overall transfer functions both start to drop off around 1 rad/s, similar to if you wouldn't use any feedforward (&lt;span class=&quot;math-container&quot;&gt;$C_f(s)=0$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for &lt;span class=&quot;math-container&quot;&gt;$\tau=1$&lt;/span&gt; the magnitude of &lt;span class=&quot;math-container&quot;&gt;$C_f(s)\,P(s)$&lt;/span&gt; near 1 rad/s is still close to 0 dB, but the phase at 1 rad/s has already dropped to -135°, so the actual feedforward is actually doing partially the opposite of what the ideal feedforward would do and thus steering away from better reference tracking. For &lt;span class=&quot;math-container&quot;&gt;$\tau=10$&lt;/span&gt; the magnitude of &lt;span class=&quot;math-container&quot;&gt;$C_f(s)\,P(s)$&lt;/span&gt; near 1 rad/s is already close to -60 dB. So, even though the phase is already way lower compared when using &lt;span class=&quot;math-container&quot;&gt;$\tau=1$&lt;/span&gt;, the magnitude is so low that the feedback controller can counteract the little disturbance the feedforward term is causing near that frequency. So that is why &lt;span class=&quot;math-container&quot;&gt;$\tau=10$&lt;/span&gt; performs better (less overshoot) than &lt;span class=&quot;math-container&quot;&gt;$\tau=1$&lt;/span&gt;. Though, it can be noted that for &lt;span class=&quot;math-container&quot;&gt;$\tau=10$&lt;/span&gt; there is a 0.3 dB peak near 0.06 rad/s and an overshoot of 0.03 after 20 seconds, while the system with only feedback and no feedforward has no overshoot.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastEditorUserId="1765" LastEditDate="2019-11-21T09:11:48.260" LastActivityDate="2019-11-21T09:11:48.260" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32013" PostTypeId="2" ParentId="32011" CreationDate="2019-11-20T13:28:54.237" Score="0" Body="&lt;p&gt;If you only have a general postion vector that describes the path of the object, then it's impossible to derive the velocity vector. Here I try to demonstrate it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the point masses are dimensionless, we can consider two different point masses at the exact same location. Point mass &lt;span class=&quot;math-container&quot;&gt;$m_1$&lt;/span&gt; moves with velocity &lt;span class=&quot;math-container&quot;&gt;$v_1$&lt;/span&gt; so the point mass &lt;span class=&quot;math-container&quot;&gt;$m_2$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$v_2$&lt;/span&gt;, those masses share the same location but move with different speeds. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-11-20T13:28:54.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32014" PostTypeId="1" CreationDate="2019-11-20T13:33:36.833" Score="0" ViewCount="41" Body="&lt;p&gt;So if I have the number 17 ( 8 bit code), the signed-magnitude representation is : 00010001&lt;br&gt;&#xA;For -17 : 10010001&lt;br&gt;&#xA;For 1's complement 17 = 00010001&lt;br&gt;&#xA;For 1's complement -17 = 11101110&lt;br&gt;&#xA;For 2's complement 17 = 00010001&lt;br&gt;&#xA;For 2's complement -17 = 11101111&lt;br&gt;&#xA;Please I need someone to tell me if it is correct.&lt;br&gt;&#xA;Thank you in advance. &lt;/p&gt;&#xA;" OwnerUserId="23618" LastActivityDate="2019-11-20T21:33:23.353" Title="I just learned about signed magnitude ,1's complement and 2's complement representation and I want to make sure that I understood them" Tags="|electrical-engineering|design|data|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32015" PostTypeId="2" ParentId="32011" CreationDate="2019-11-20T14:08:03.017" Score="1" Body="&lt;p&gt;Simple answer is that you can't because you have to take derivative of the position vector with respect to time which would be zero in the case of the vector you described (It simply doesn't contain the temporal information). &lt;/p&gt;&#xA;" OwnerDisplayName="user23622" LastActivityDate="2019-11-20T14:08:03.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32016" PostTypeId="1" CreationDate="2019-11-20T15:29:53.827" Score="0" ViewCount="466" Body="&lt;p&gt;This is my first question here. I am having trouble finding the 2nd moment of area of a ring with a thickness 't' and radius 'r'. I seem to be a factor of 2 out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I do is I consider an area like a piece of pie, and say that this is dA. I believe&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$dA=r*t*dθ$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where r is the radius, t is the thickness and dθ is the small angle over which dA is encompassed. Then I integrate with respect to area:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\int r^2 dA$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\int_0^{2\pi} r^2 * r*t*d\theta$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\int_0^{2\pi} r^3*t*d\theta$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$2\pi*r^3*t$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is a factor of 2 out.&#xA;Any help would be appreciated&#xA;Thanks&#xA;Matt&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/a0POb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/a0POb.jpg&quot; alt=&quot;My Working outs&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23625" LastActivityDate="2019-11-20T22:41:08.053" Title="2nd Moment of Area of a ring" Tags="|materials|applied-mechanics|structures|moments|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32017" PostTypeId="2" ParentId="32016" CreationDate="2019-11-20T17:02:56.560" Score="0" Body="&lt;p&gt;You can't use the thickness as a constant. That is a circular section and has to be integrated. look at the shaded T on your diagram, the geometry is like a trapezoid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an easy way to do this which is subtracting the I of the inner disk from that of the outer disk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But following your way, we have.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$J_z= \int_{}^{} \int_{}^{} r^2 da= \int_{0}^{2\pi}\int_{r1}^{r2} r^2(rdrd\theta) =  \int_{0}^{2\pi}\int_{r1}^{r2}r^3drd\theta  \\ \int_{0}^{2\pi} \left [ \frac{r_2^4 -r_1^4}{4}  \right]d\theta = \pi/2(r_2^4-r_1^4)$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-20T22:41:08.053" LastActivityDate="2019-11-20T22:41:08.053" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32018" PostTypeId="2" ParentId="32016" CreationDate="2019-11-20T17:18:16.160" Score="2" Body="&lt;p&gt;You have done everything right! &#xA;You calculated the polar moment of inertia &lt;span class=&quot;math-container&quot;&gt;$J$&lt;/span&gt;, which is the sum of second moment of areas &lt;span class=&quot;math-container&quot;&gt;$I_i$&lt;/span&gt;. But here you have a symmetric object so: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$J = I_x+I_y$$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$$I_x=I_y=\frac{J}{2} = \pi r^3t$$&lt;/span&gt;. &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastActivityDate="2019-11-20T17:18:16.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32019" PostTypeId="2" ParentId="31997" CreationDate="2019-11-20T18:34:16.837" Score="2" Body="&lt;p&gt;Units of N, mm (for the stiffness k), and g (for the mass m) are not consistent. You need to convert these values to use in the basic equation for the frequency: &lt;span class=&quot;math-container&quot;&gt;$$f=\frac{1}{2*\pi} \sqrt \frac{k}{m}$$&lt;/span&gt; Knowing that &lt;span class=&quot;math-container&quot;&gt;$1 \, N = 1 \, kg*m/s^2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$1 \, m = 1000 \, mm$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$1 \, kg = 1000 \, g$&lt;/span&gt;, you get the following:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$f=\frac{1}{2*\pi} \sqrt \frac{N/mm*((1 kg*m/s^2)/N)*(1000 mm/m)}{g*(1 kg/1000 g)}$$&lt;/span&gt;&#xA;All of the units cancel except for &lt;span class=&quot;math-container&quot;&gt;$\sqrt {1/s^2}$&lt;/span&gt;. Doing the math gives the following result:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$f = \frac{1000}{2*\pi} \sqrt \frac{k \; in \; N/mm}{m \; in \; g}$$&lt;/span&gt;&#xA;where f is in units of cycles/sec = Hz.&lt;/p&gt;&#xA;" OwnerUserId="12984" LastEditorUserId="12984" LastEditDate="2019-11-20T18:41:56.147" LastActivityDate="2019-11-20T18:41:56.147" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32020" PostTypeId="2" ParentId="32014" CreationDate="2019-11-20T21:33:23.353" Score="2" Body="&lt;p&gt;Yup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can test your 2's compliment math on a calculator that displays binary.  For &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;-bit numbers, you negate the number by subtracting it from (positive) &lt;span class=&quot;math-container&quot;&gt;$2^n$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So -17, in 2's compliment, is 100000000 - 00010001 = 11101111.  Or subtract 17 from 256, and then display the result in binary.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-11-20T21:33:23.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32021" PostTypeId="1" AcceptedAnswerId="32066" CreationDate="2019-11-20T22:19:51.513" Score="-1" ViewCount="413" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BAoAz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BAoAz.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 40 kg load on the rope accelerates from 0 to 50km / h in 6 seconds. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rope is tightened with a force of 500kg while the whole assembly is on it.&#xA;The middle pulley is driven by a dc motor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to calculate the optimum angle between the middle (drive) pulley and the end pulley (pulley 2 on the picture) for driving without slipping or without too much load on the motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An angle between two pulleys that is too small would cause slippage on the rope, while too big angle would cause a higher load on the motor, and a higher power consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pulleys are made of aluminum and the rope has a coefficient of friction of 0.1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current angle between the two pulleys is 15 degrees, as seen in the picture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor spec are: 24V, 3560 oz/in stall torque, 4.3 HP  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need more information, please ask.&lt;/p&gt;&#xA;" OwnerUserId="19614" LastEditorUserId="19614" LastEditDate="2019-11-22T20:06:59.560" LastActivityDate="2019-11-23T17:57:42.547" Title="Calculating optimal angle on a rolling pulley wheel, which holds weight, on a horizontal cable" Tags="|motors|friction|pulleys|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32022" PostTypeId="1" CreationDate="2019-11-20T23:31:44.877" Score="0" ViewCount="1526" Body="&lt;p&gt;The top view looks at the top of the objet from the top of the object, the front view looks at the front of the objet from the front of the object. All seems logical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So why does the left-side view looks at the right of the object from the right of the object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm talking about first angle projection. I'm not entirely sure but I think I doesn't change anyway with third angle projection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who say it doesn't. It's in the words: yes, for the left-side view, the observer looks towards the left. But the side that he sees if he looks towards the left is the right-side of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we were following the same logic, front view would be the back-side of the object because the observer would be looking towards the front. But it isn't: for the front view, the observer looks towards the back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The naming isn't consistent. Either we name the view after the direction we are looking towards, or after the side of the object that we see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The position of the plane of projection doesn't change anything, it just changes were we draw it on the paper when we unfold the view. &lt;a href=&quot;https://i.stack.imgur.com/nvFnF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nvFnF.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23630" LastEditorUserId="23630" LastEditDate="2019-11-21T13:23:51.833" LastActivityDate="2019-11-21T19:03:23.513" Title="Why is the left-side view called like that?" Tags="|technical-drawing|" AnswerCount="1" CommentCount="11" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32025" PostTypeId="1" AcceptedAnswerId="32044" CreationDate="2019-11-21T06:51:02.477" Score="2" ViewCount="73" Body="&lt;p&gt;The amount of light that an object reflects is called its &lt;a href=&quot;https://en.wikipedia.org/wiki/Albedo&quot; rel=&quot;nofollow noreferrer&quot;&gt;albedo&lt;/a&gt;; a scale of 0-1, 0 being a black body ad 1 reflecting all light. This question is probably going to be closed [It is &quot;subjective&quot; ?? No idea how] so I may as well be brief. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the 1990's some scientists were suggesting that if you changed the albedo of an asteroid then it could be hotter on one side. That would change the curve of orbit. I have heard nothing since about this idea. I imagine that various ways would exist such as spraying the asteroid with paint or putting a solar panel on it. Is this possible? [I would ask if anyone is trying to do it but the rules of stack overflow say I can only ask one question.]  &lt;/p&gt;&#xA;" OwnerUserId="23590" LastActivityDate="2019-11-22T11:42:33.373" Title="Can you change an asteroid's albedo to take it off a collision course?" Tags="|aerospace-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32026" PostTypeId="2" ParentId="31851" CreationDate="2019-11-21T08:46:55.567" Score="1" Body="&lt;p&gt;First, there are different ways a post can fail.&#xA;For instance, the post could buckle (though rather unlikely, as the compressive load on the post isn't that high) failure of the footing, which is mostl depending on the embedment depth. However, I assume the most likely case is failure due to bending.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, for case 1, the cross-section is square, with sides &lt;span class=&quot;math-container&quot;&gt;$a=b=65mm$&lt;/span&gt;, which in turn relates to a CS-area &lt;span class=&quot;math-container&quot;&gt;$A=416 mm^2$&lt;/span&gt; and area moment of inertia &lt;span class=&quot;math-container&quot;&gt;$I_x=I_y=272'003 mm^4$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for case 2:&#xA;&lt;span class=&quot;math-container&quot;&gt;$a=76mm$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$b=38mm$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$A=365mm^2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$I_x=271'188mm^4$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I_y=91'848mm^4$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When comparing the two cross-sections, assuming a linear-elastic material behaviour, in case 2 the cross-sectional area (roughly corresponding to compressive strength, without considering buckling) is reduced by about &lt;span class=&quot;math-container&quot;&gt;$12\%$&lt;/span&gt;.&#xA;On the other hand, the &quot;strong&quot; moment of inertia &lt;span class=&quot;math-container&quot;&gt;$I_x$&lt;/span&gt; is about the same, though the &quot;weak&quot; moment of inertia &lt;span class=&quot;math-container&quot;&gt;$I_y$&lt;/span&gt; is reduced to about &lt;span class=&quot;math-container&quot;&gt;$34\%$&lt;/span&gt; of the square post's value.&#xA;This translates to the post having about one third of the specified posts bending resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus i think it is likely, that if a post with the dimensions specified in the contract was used, it would not have failed.&lt;/p&gt;&#xA;" OwnerUserId="8632" LastActivityDate="2019-11-21T08:46:55.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32027" PostTypeId="2" ParentId="31602" CreationDate="2019-11-21T08:51:03.830" Score="1" Body="&lt;p&gt;Because the specific value are not given, I will give a general overview of the method. When applying the direct stiffness method for trusses, the following steps must be used:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find the local stiffness matrix for each individual element as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\begin{bmatrix}AE/L &amp;amp; 0 &amp;amp; -AE/L &amp;amp; 0 \\0 &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\-AE/L &amp;amp; 0 &amp;amp; AE/L &amp;amp; 0 \\0 &amp;amp; 0 &amp;amp;0 &amp;amp;0\end{bmatrix}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need a beta matrix in order to convert the local coordinate system into global coordinates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\beta = \begin{bmatrix} cos\theta &amp;amp; sin\theta &amp;amp; 0 &amp;amp; 0 \\-sin\theta &amp;amp; cos\theta &amp;amp; 0 &amp;amp; 0\\0 &amp;amp; 0 &amp;amp; cos\theta &amp;amp; sin\theta \\0 &amp;amp; 0 &amp;amp;-sin\theta &amp;amp; cos\theta\end{bmatrix}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, &lt;span class=&quot;math-container&quot;&gt;$[\beta]^T[k_l][\beta] = [k_g]$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Complete the same process for all matrices and then assemble global matrices to create the system matrix [K]. Apply boundary conditions for the problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then solve for the nodal displacements by using the equation, &lt;span class=&quot;math-container&quot;&gt;$[K]^{-1}[F] = [U]$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where F is the forces at the joints and U is the nodal displacements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Expand [U] into elemental displacements &lt;span class=&quot;math-container&quot;&gt;$[U_{1,2,3}]$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply each by their own original beta matrices in order to produce &lt;span class=&quot;math-container&quot;&gt;$[\delta_{1,2,3}]$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Multiply each original stiffness matrices by their delta values like so:&#xA;&lt;span class=&quot;math-container&quot;&gt;$[k_l][\delta_{1,2,3}] = [F_{e 1,2,3,}]$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will produce the elemental forces. &lt;/p&gt;&#xA;" OwnerUserId="20822" LastActivityDate="2019-11-21T08:51:03.830" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32028" PostTypeId="2" ParentId="32022" CreationDate="2019-11-21T13:28:32.757" Score="3" Body="&lt;p&gt;This is a question of &quot;my left or your left&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine standing and facing somebody, as if looking in a mirror.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now imagine the person in front of you saying &quot;look left&quot;. Which way should you turn your head? it is their left, or your left? The choice you make affects what &quot;left&quot; means.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unlike a human, an object does not &lt;em&gt;have&lt;/em&gt; it's own point of view, therefore &lt;em&gt;your left&lt;/em&gt; is always the correct answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Take a look at the image below. The views on the middle row here are (from left to right): Left, Front, Right, Back. (The remainders are Top and Bottom).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MPnvK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MPnvK.png&quot; alt=&quot;3D Views&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine that the box is a human being, standing, and facing towards you. The front of the human is their face, &quot;their&quot; left would be on the right hand side, as you look at their front.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3D views are named &lt;strong&gt;from the perspective of the viewer&lt;/strong&gt; The Left view, is &lt;em&gt;on the left of the object&lt;/em&gt; when you look at the front, from the outside of the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Likewise, the back view is on the back of the object, as you look at the front of the object, from outside the object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simple fact is, that physical objects are looked at from the outside in, not the inside out, and the views are named accordingly&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2019-11-21T19:03:23.513" LastActivityDate="2019-11-21T19:03:23.513" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32029" PostTypeId="1" CreationDate="2019-11-21T14:27:06.497" Score="-2" ViewCount="1133" Body="&lt;p&gt;The best possible way to prevent the&#xA;bracket from failure is to ensure that it is enough rigid thereby ensuring absolute zero deflection at all points of the bracket. What is the view on the real world feasibility of this?&lt;/p&gt;&#xA;" OwnerUserId="23637" LastActivityDate="2019-11-22T21:53:34.650" Title="Zero Deflection" Tags="|mechanical-engineering|finite-element-method|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32030" PostTypeId="2" ParentId="32029" CreationDate="2019-11-21T15:11:18.377" Score="3" Body="&lt;p&gt;There is no material with zero deflection in the world. Every thing deflects, to different degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best possible way to prevent a bracket from failure is to design it for the task it must do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the load is a vibrating machine, it must be allowed to vibrate by with a frequency that is not likely going to resonate with that of the machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if the load is off center, it must be strong enough to resist the moment plus all the shear and normal forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the load is cyclic, it must be designed for fatigue, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it is part of a structure of an airplane it must be strong enough for all the above and be reliable under extreme temperatures and be light within the reason. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As commented by some, I thought a few points may help clarify my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Deflection is actually the way a bracket or any structure undertakes to support the load. Of course it has to be controlled. But a beam or slab or any other member of a structure as long as it is not deflected is not carrying any load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An archer knows to pull his bow to the most deflection to shoot his arrow faster and farther. Many huge suspension bridges find strength in predesigned deflection. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strength in engineering is the capacity to support static or dynamic loads or impact of objects while maintaining the geometry within accepted tolerances. it doesn't matter for this discussion the kind of deflection, or strain that has caused it. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-22T01:57:24.430" LastActivityDate="2019-11-22T01:57:24.430" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32031" PostTypeId="1" AcceptedAnswerId="33170" CreationDate="2019-11-21T16:02:30.133" Score="1" ViewCount="38" Body="&lt;p&gt;I need to find the response of my FTIR spectrometer. I do have a calibrated black-body source (area-emitter) which I measured at various temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system operates in inverse cm, cm-1. My approach so far was the following:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1) Measure spectra at temperatures &lt;span class=&quot;math-container&quot;&gt;$T_i$&lt;/span&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;2) Scale the spectra i.e. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the x-axis &lt;span class=&quot;math-container&quot;&gt;$x_{in~um} = 1e4/x_{in~cm-1}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;the y-axis, &lt;span class=&quot;math-container&quot;&gt;$y_{in~um} = y_{in~cm-1}/x_{in~cm}^2$&lt;/span&gt;&#xA;to account for the dispersion relation between measured energy spacing and desired wavelength spectrum. At this point an example for scaled measurement (blue) and theoretical blackbody (orange) at the same nominal temperature looks like this (CO2 and H2O lines are at the correct positions):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aYwfl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aYwfl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;3) calculate the response function &lt;span class=&quot;math-container&quot;&gt;$R=I_{BB}(T_i)/y(T_i)$&lt;/span&gt; (I think, here in a first approach I assume that the background radiation does not contribute significantly. I would have expected similar response functions for the different temperatures; however what I get looks like:&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RvlSl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RvlSl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &quot;response function&quot; drifts with temperature. I am unsure now if I have a misconception in the defintion of the response function, or if I picked up an artefact. I hope someone with more experience can help me out?&lt;/p&gt;&#xA;" OwnerUserId="23559" LastActivityDate="2020-02-10T14:57:57.910" Title="Calculate FTIR-system response functions from measurements of a calibrated black body" Tags="|measurements|optics|radiation|thermal-radiation|spectrometry|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32032" PostTypeId="1" CreationDate="2019-11-21T18:42:02.223" Score="1" ViewCount="112" Body="&lt;p&gt;In LLE physical processes, there is this level (interface) between the 2 components  (for example) that is being separated as shown in the picture. I know that the height of this level from the reference point (0%) tells us the efficiency of separation. I don't understand the mechanism of what this inferface level means and how does it tells us the efficiency of the process. What does it tells us if it is more or if it is less than the opitmal level (given by engineers) ?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/mvLhv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mvLhv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11561" LastActivityDate="2019-11-21T18:42:02.223" Title="Interface level between 2 components in Liquid-Liquid Extraction, What does it mean?" Tags="|chemical-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32033" PostTypeId="2" ParentId="5583" CreationDate="2019-11-21T18:55:13.090" Score="0" Body="&lt;p&gt;I would recommend using third party software for meshing like gmsh, gambit or fluent where you can import your geometry, edit it, define the topology and get the mesh.&#xA;Gmesh is free while the rest are not.&#xA;You might be able to find many tutorials on meshing in these softwares on youtube&lt;/p&gt;&#xA;" OwnerUserId="18714" LastActivityDate="2019-11-21T18:55:13.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32034" PostTypeId="2" ParentId="31498" CreationDate="2019-11-21T19:04:35.810" Score="1" Body="&lt;p&gt;Q1- No you cannot simulate a static simulation without any constrains, you need to define a constraint with respect to which calculations will be conducted.&#xA;I would suggest calculating all the forces in frame of the body, and then put it to rest and perform static structural analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2- It is your choice if you want to do a dynamic analysis, that totally depends on your need and skills.&#xA;You can go for static analysis if you can define the forces, drag, etc with respect to the plane, i.e in the lagrangian domain and not eulerian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, if you conduct a dynamic analysis that will convey a more clear picture and be of more use to you with regards to information and reliability.&lt;/p&gt;&#xA;" OwnerUserId="18714" LastActivityDate="2019-11-21T19:04:35.810" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32035" PostTypeId="1" CreationDate="2019-11-21T19:20:38.910" Score="3" ViewCount="661" Body="&lt;p&gt;I'm trying to do some calculations for a personal project.  I'm trying to understand column strength / buckling.  My questions relate to Effective Length Constant, which is based on Boundary Conditions (how the ends of the column are restrained) The four choices there are:  Free, Fixed, Guided or Pinned:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sampling...&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YsolB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YsolB.jpg&quot; alt=&quot;Sampling of End Conditions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And different end conditions affect the Effective Length Constant, used in buckling calculations:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fEByD.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fEByD.jpg&quot; alt=&quot;Effective Length Constants&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(edit.. ref: &lt;a href=&quot;https://www.amesweb.info/Beam/Column-Buckling-Formulas.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;source for the above Length Constants&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But my dilemma is I'm intending welded metal tubing without corner gussets.  I think with a gusset (and / or with cross bracing straps), I'd use Fixed-Fixed calculations, but simple butt welds that's not so clear.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have testing data or other experience on how to address end conditions for buckling calculations with a simple welded butt joint at the ends of a column under load?  &amp;lt;---- Here is my question for this posting... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just for reference, the kind of thing I'm investigating is a &lt;a href=&quot;https://www.discountramps.com/car-repair-ramps/p/CR-7214/&quot; rel=&quot;nofollow noreferrer&quot;&gt;drive on vehicle ramp.&lt;/a&gt;  Obviously lighter is better for handling and storage... And yes I'm thinking astronomically large safety margins, but I'd like to know the real load capability numbers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: &lt;a href=&quot;https://engineering.stackexchange.com/questions/28247/buckling-in-columns&quot;&gt;related reference.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, how does one address end conditions for welded in place column materials?   &lt;/p&gt;&#xA;" OwnerUserId="1567" LastEditorUserId="1567" LastEditDate="2019-11-22T06:08:15.707" LastActivityDate="2019-11-22T06:08:15.707" Title="Column Buckling Boundary Conditions -- Welded Tubing Practical Application?" Tags="|strength|solid-mechanics|columns|buckling|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32038" PostTypeId="1" CreationDate="2019-11-21T23:51:32.860" Score="1" ViewCount="22" Body="&lt;p&gt;I have a closed-loop system that represents the longitudinal dynamics of an aircraft with states x velocity, z velocity, pitch rate, pitch, and x and z inertial positions. The outputs of the system (where the system response is observed) are the pitch angle and z position. The realization of this particular system is completely reachable, and so I have placed closed-loop eigenvalues at -1.5, -2, -2.5, -3, -4, and -5. I have solved for the gain matrix that gives this desired behavior. In each modal space, the modal response is as desired; no oscillations and highly damped. However, when I simulate the system response to a random unit perturbation in the state space, the system response actually has oscillations. I figured that since all of the eigenvalues are strictly negative and real, that I should see no oscillation. Is the response I'm getting what's supposed to happen or am I doing something wrong? &lt;/p&gt;&#xA;" OwnerUserId="23644" LastActivityDate="2019-11-21T23:51:32.860" Title="Unexpected Closed-Loop System Response of Longitudinal Aircraft Dynamics" Tags="|control-engineering|matlab|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32042" PostTypeId="1" CreationDate="2019-11-22T06:02:00.887" Score="3" ViewCount="60" Body="&lt;p&gt;I am trying to plot the frequency response control effort of a control scheme in Matlab, and in particular I want to do this for different values of tau, which is a constant used to add a pole at high frequency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plot that i get is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kkgH9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kkgH9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in particular, this is the control sensitivity function of a 2 dof control scheme:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EUOks.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EUOks.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the code for obtaining the plot above is:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = tf('s');&#xA;P = 1/[(1+s)*(1+0.05*s)^2];&#xA;C = (s+1)/s;&#xA;&#xA;%2 DOF&#xA;&#xA;tau_1 = 0.1;&#xA;CF_1 = [(1+s)*(1+0.05*s)^2]/((1+tau_1*s)^3);&#xA;&#xA;tau_2 = 0.01;&#xA;CF_2 = [(1+s)*(1+0.05*s)^2]/((1+tau_2*s)^3);&#xA;&#xA;tau_3 = 0.001;&#xA;CF_3 = [(1+s)*(1+0.05*s)^2]/((1+tau_3*s)^3);&#xA;&#xA;Q1 = (C+CF_1)/(1+P*C);&#xA;Q2 = (C+CF_2)/(1+P*C);&#xA;Q3 = (C+CF_3)/(1+P*C);&#xA;&#xA;figure;&#xA;bodemag(Q1,'r',Q2,'b',Q3,'g'),grid&#xA;legend('tau = 0.1','tau = 0.01','tau = 0.001')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;my doubt is given by the fact that the control sensitivity function starts at zero for all the three cases. Shouldn't it start at an high value and then decrease? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case it does the opposite, so it starts at zero and grows until a value and stays constant there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I read this plot? Or it is complitely wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I plot the step response I obtain:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IBWeq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IBWeq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from it I zoomed to show better all the coloured lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody help me clarify my doubts? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="22780" LastActivityDate="2019-11-22T06:02:00.887" Title="Problem reading control effort frequency response plot" Tags="|electrical-engineering|control-engineering|control-theory|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32043" PostTypeId="1" CreationDate="2019-11-22T08:47:06.353" Score="3" ViewCount="65" Body="&lt;p&gt;I am trying to write a MATLAb code for determining the natural frequencies of a first-order shear deformation plate using Ritz method. I am using Legendre polynomials as trial functions in Ritz method, and my code works well for static analysis of plate. I have solved few structural problems with my code and compared the results with MSC NASTRAN. The results are accurate with maximum error of 5%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I try to find eigenvalues, at least one eigenvalue comes out to be negative. Also, the eigenvalues do not match with eigenvalues obtained from MSC NASTRAN (Modal Analysis).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible for a plate to have negative eigenvalues?&lt;/p&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2019-11-22T11:56:55.730" Title="Negative Eigenvalues in Ritz Solution of First-Order Shear Deformation Theory" Tags="|structural-engineering|numerical-methods|modal-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32044" PostTypeId="2" ParentId="32025" CreationDate="2019-11-22T08:55:09.557" Score="3" Body="&lt;p&gt;Yes, in theory, it is possible to use albedo modification as a way to influence the orbital parameters of a body through &lt;a href=&quot;https://en.wikipedia.org/wiki/Radiation_pressure&quot; rel=&quot;nofollow noreferrer&quot;&gt;solar pressure&lt;/a&gt;. This is the same principle used in solar sails, a propulsion method that relies on radiation pressure from the Sun to propel space probes and even adjust their attitude. The theoretical basis has been known since the late 19th century, the principle was experimentally demonstrated in 1899 by Lebedev and successfully deployed on a spacecraft in 2010 by the &lt;a href=&quot;https://en.wikipedia.org/wiki/IKAROS&quot; rel=&quot;nofollow noreferrer&quot;&gt;IKAROS&lt;/a&gt; mission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be applied to asteroids by simply making them more reflective, which will increase the effects of solar pressure. Over enough time, the minute but constant acceleration will add up to significant changes in their trajectory. If the aim is to avoid a planetary impact, it might not even be necessary to coat a specific side of the asteroid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://oaktrust.library.tamu.edu/bitstream/handle/1969.1/ETD-TAMU-2011-08-10184/GE-THESIS.pdf?sequence=2&amp;amp;isAllowed=y&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is a Master's Thesis&lt;/a&gt; on the related subject of designing a payload that can acomplish said albedo modification. From the abstract:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The development of the Surface Albedo Treatment System (SATS) onboard a&#xA;  spacecraft mission to the near earth asteroid (NEA) Apophis in 2012 is an innovative concept of deflecting NEAs from possible impact with the Earth through altering the &lt;a href=&quot;https://en.wikipedia.org/wiki/Yarkovsky_effect&quot; rel=&quot;nofollow noreferrer&quot;&gt;Yarkovsky effect&lt;/a&gt;, a non-secular force in the solar system due to uneven surface thermal emission most profoundly affecting small rotating bodies subjected to sunlight. Though this force is small, its magnitude can be dramatic if extended over a period of time and if it uses the close approach of an asteroid near Earth to magnify the perturbation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The payload dispenses colored powder called albedo changing particles (ACPs)&#xA;  onto the surface changing its albedo and indirectly the surface temperature which&#xA;  changes the Yarkovsky effect. This study gives an in-depth description of both&#xA;  computational and experimental parts of the design of this system with primary focus on&#xA;  initial ground test setup. The initial experiments proposed to design the SATS is outlined in detail and justified by the mission criterion of interest as well as modeling the actual dispersal on the surface. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="10400" LastEditorUserId="10400" LastEditDate="2019-11-22T11:42:33.373" LastActivityDate="2019-11-22T11:42:33.373" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32045" PostTypeId="2" ParentId="32043" CreationDate="2019-11-22T11:56:55.730" Score="3" Body="&lt;p&gt;The only way you can get negative eigenvalues is by including stress stiffness effects (sometimes called &quot;geometric stiffness&quot;), if there are compressive stresses which would cause the plate to buckle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There can be &lt;em&gt;zero&lt;/em&gt; eigenvalues if the plate can move as a rigid body, of course, and they might be calculated as &lt;em&gt;small&lt;/em&gt; negative numbers, but those should be a few orders of magnitude smaller that the first positive eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FWIW &quot;accurate to within 5%&quot; probably means &quot;wrong&quot;. Try to find a load case where the exact solution is known. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plotting the deformed shape for your negative eigenvalues might help to find the error - or at least convince you they really are wrong, if the shapes look crazy.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-11-22T11:56:55.730" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32046" PostTypeId="1" CreationDate="2019-11-22T12:04:05.260" Score="-1" ViewCount="381" Body="&lt;p&gt;I have trouble understanding the Brazing principle. What I know is that there are a couple of filler materials such as copper, silver, aluminim, etc....&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I remember from the past is that the filler material is different than the material where the two parts I would like to braze are made of (i.e. two stainless steel plates are brazed with copper or titanium filler).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to braze two metal parts with the same material for both sides (i.e. is it possible and/or recommended to braze two stainless steel plates with a stainless steel filler?)?&lt;/p&gt;&#xA;" OwnerUserId="5007" LastEditorUserId="5007" LastEditDate="2019-11-26T13:26:43.543" LastActivityDate="2019-11-26T13:26:43.543" Title="What materials can be used for Brazing" Tags="|materials|welding|brazing|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32047" PostTypeId="1" AcceptedAnswerId="32060" CreationDate="2019-11-22T12:06:49.093" Score="1" ViewCount="49" Body="&lt;p&gt;Exactly what are &quot;ceramics&quot;? I found a lot of similar &quot;definitions&quot; of that term online, e.g. on Wikipedia and &lt;a href=&quot;https://www.sciencelearn.org.nz/resources/1769-what-are-ceramics&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These (and other) sources essentially state that ceramics are non-metallic, non-organic materials. And they state the most prominent examples.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But these only seem to be necessary properties. &lt;strong&gt;Are there sufficient criteria based on which one can definitely decide whether a certain material is ceramic? Is this list of criteria comprehensive?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there even a widely accepted clear definition at all?&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-11-23T05:57:10.150" Title="Exact definition of ceramics" Tags="|materials|ceramics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32049" PostTypeId="1" CreationDate="2019-11-22T15:50:33.473" Score="2" ViewCount="24" Body="&lt;p&gt;In piezoelectric materials, two parameters are defined to construct relations between electrical and mechanical energy conversions. These parameters are the charge constant d and the charge constant g. In literature, it is always stated that  high g value is desirable for sensing applications, while high d value is desirable for actuation. This is proved as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;d is defined in two ways : &#xA;d= strain developed/ applied electric field &#xA;Based on this definition, large d constants relate to large mechanical displacements which are usually sought in actuators.&#xA;also d is defined as : &#xA;d= induced charge density/ applied stress &#xA;Based on this definition, however, I can conclude that high d value is also desirable for sensing applications, as more charge is induced per applied unit stress. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Regarding g constant, it is also defined in two ways &#xA;g= induced electric field/ applied stress &#xA;Output voltage is obtained by multiplying the calculated electric field by the thickness of ceramic between electrodes. High g constants favor large voltage output, and are sought after for sensors.&lt;br&gt;&#xA;The second definition is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;g= strain developed/ applied charge density. &#xA;Based on this definition, however, I can conclude that high g value is also desirable for actuating applications, as more strain is developed per applied unit charge density. &#xA;So I got confused of what is stated in the literature. Am I wrong ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, Based on the definitions of d and g : &#xA;d= induced charge density/ applied stress&#xA;g= induced electric field/ applied stress &#xA;I can say that g = epsilon* d ; (epsilon is the electric permittivity)&#xA;electric field  is equal to the charge density divided by the permittivity.&#xA;So what is the importance of defining 2 parameters?I think that defining one of them is adequate.&lt;/p&gt;&#xA;" OwnerUserId="9742" LastActivityDate="2019-11-22T15:50:33.473" Title="Why is it desirable to have high g value for sensing applications, while high d value is desirable for actuation in piezoelectricity?" Tags="|mechanical-engineering|electrical-engineering|materials|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32051" PostTypeId="2" ParentId="32046" CreationDate="2019-11-22T16:58:51.257" Score="1" Body="&lt;p&gt;Brazing process uses a filler metal with a lower melting temperature so the base metals are not melted ( as in welding). I guess soldering means the filler has a much lower melting temperature.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-11-22T16:58:51.257" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32054" PostTypeId="1" CreationDate="2019-11-22T19:44:05.457" Score="1" ViewCount="45" Body="&lt;p&gt;I have a Peltier's device of 10*10*3mm and the material is Bismuth Telluride. The hot side is 100 &lt;span class=&quot;math-container&quot;&gt;$^oC$&lt;/span&gt; and I want to dissipate the heat, and at the same time, the device should be portable i.e. it should not be part of the system so that I can't take it with me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried using a vapor chamber and heat pipes. Though I want something lightweight, reliable and effective.&lt;/p&gt;&#xA;" OwnerUserId="21336" LastEditorUserId="90" LastEditDate="2019-11-22T23:56:26.353" LastActivityDate="2019-11-22T23:56:26.353" Title="How can we dissipate the heat from a block of 10*10*3 mm with 100 degree Celsius without losing the portability of the block?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|heat-exchanger|cryogenics|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="32055" PostTypeId="2" ParentId="32029" CreationDate="2019-11-22T21:53:34.650" Score="1" Body="&lt;p&gt;Building on to what was previously said, when designing a bracket or mechanical fastener based connection, my major concern would most likely be the failure of the plate or the bolt/screw. For this, you may need to consider: bolt tearout, hole yielding or general yielding/failure of the plate. Furthermore, you may also need to consider the strength of the fastener.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For further information and the actual regulations, check the AISC steel construction manual or other relevant steel connection codes. &lt;/p&gt;&#xA;" OwnerUserId="20822" LastActivityDate="2019-11-22T21:53:34.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32056" PostTypeId="1" CreationDate="2019-11-22T22:01:09.917" Score="1" ViewCount="66" Body="&lt;h2&gt;Could anybody provide a resource for finding or researching historic construction methods?&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;My current need is for a typical vaulted or dome ceiling construction methods circa 1992 in the US but I think it would be a helpful resource to have for the community in general. It is easy to find modern typical methods but I am interested in how the methods and processes have changed over time.&lt;br&gt;&#xA;Any help is appreciated!  &lt;/p&gt;&#xA;" OwnerUserId="23652" LastActivityDate="2021-04-17T12:53:47.153" Title="historic construction method resources" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32057" PostTypeId="2" ParentId="31968" CreationDate="2019-11-22T23:19:32.800" Score="0" Body="&lt;p&gt;A very general way to model hysteresis is by a preisach model eg &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Preisach_model_of_hysteresis&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.m.wikipedia.org/wiki/Preisach_model_of_hysteresis&lt;/a&gt; it can get very computationally expensive though. Can you post some plots of your data and tell what you want to do with it? That might help get a better answer &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-11-22T23:19:32.800" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32058" PostTypeId="1" CreationDate="2019-11-23T01:14:20.413" Score="1" ViewCount="268" Body="&lt;p&gt;I'm interested in how to work out whether torsional effects can be ignored or not, in calculating the adequacy of a given steel beam in the following idealised eccentric scenario. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, equivalently, the criteria which must apply, for it to be appropriate to treat the beam and its UDLs as a simple question about vertical plane bending, needing only to consider the &lt;em&gt;actual vs design bending moment, deflection, and bearing support strength&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Rk07o.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rk07o.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Description of problem&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;The problem is idealised and static, so the available parameters are the beam's inherent properties, the UDLs Load 1 and Load 2, and the length (clear span) of the beam.  For simplicity, the beam can be deemed horizontal and both positionally and rotationally fixed at both ends. (It might be the top beam in a moment resisting frame, or both ends concreted into adjoining walls.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Load 2 is one side of a massive  and stiff horizontal rectangular object, such as a thick steel sheet or precast concrete floor, that's simply resting (unattached) on the bottom flange. (The other side of the sheet is simply resting unattached on a [relatively distant] parallel wall or second beam to the left). There are adequate stiffeners in the flange itself&lt;/p&gt;&#xA;&#xA;&lt;h1&gt;Question&lt;/h1&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;If Load 2 is small enough&lt;/strong&gt;, it will be contextually trivial and we can totally ignore the torsional element, and just treat the beam as subject to vertical bending only from Load 1.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;If Load 2 is large enough&lt;/strong&gt;, there will be a torsional dimension to the problem, requiring more complex calculations or numerical methods.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What method, criteria, or formula would be used to determine whether or not the torsional aspects of Load 2 can be ignored, for the purposes of calculating whether the beam is adequately strong for the setup and loads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it the case that enough stiffeners and the problem can be ignored?&lt;/p&gt;&#xA;" OwnerUserId="6366" LastActivityDate="2019-11-24T06:46:30.860" Title="What criteria must apply for torsional effects on this beam to be ignored in calcs?" Tags="|structural-engineering|structural-analysis|torque|beam|eurocodes|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32059" PostTypeId="1" CreationDate="2019-11-23T03:49:54.900" Score="1" ViewCount="137" Body="&lt;p&gt;I know momentum of the airflow inside jet engine generates thrust, but what about the pressure difference in front of and at the back of the engine? The air sucking intake at the front has lower than atmospheric pressure and the exhaust has higher than atmospheric pressure, does this pressure differential contribute to the thrust generated by the engine in adition to the momentum thrust?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The high pressure zone push the engine from the back side and low pressure zone pulls it from the front.&#xA;For example, if the exhaust has 1 square meter surface area, and the exhaust pressure is 2 atmosphere, then there is pressure equal to 10000 kilograms pushing it from the back since 1 atmosphere is ten thousands kilograms per square meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which of these two is true?&lt;/p&gt;&#xA;" OwnerUserId="10080" LastActivityDate="2019-11-25T00:25:26.390" Title="Does air pressure difference of intake exhaust contribute to jet thrust?" Tags="|airflow|thrust|jet|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32060" PostTypeId="2" ParentId="32047" CreationDate="2019-11-23T05:57:10.150" Score="2" Body="&lt;p&gt;The classic (materials science) definition of a ceramic is a compound that is held together by a network of &lt;em&gt;covalent bonds&lt;/em&gt;. The fact that these bonds are directional means it is &lt;em&gt;non-metallic&lt;/em&gt;- deformation by dislocation-mediated slip does not occur (so no ductility mechanisms are active at ambient temperature) and the substance is a poor conductor of both heat and electricity (since the electrons responsible for holding the material together are immobile).  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-11-23T05:57:10.150" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32061" PostTypeId="2" ParentId="32058" CreationDate="2019-11-23T07:58:44.390" Score="1" Body="&lt;p&gt;In your sketch, there seems to be no off-center loading so no torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if we had a slab sitting on say the lip of the lower flange then we had torque, &lt;span class=&quot;math-container&quot;&gt;$$t=w*x (\text{w is slab weight}/ ft \ , \  x= offset),$$&lt;/span&gt;&lt;br&gt;&#xA;This torque  would rotate the fix- fix (fix for rotation )beam by this equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \theta_{max-at- center}= \frac{t*L^2}{8*G*j} $$&lt;/span&gt;&#xA;G = Shear modulus&lt;/p&gt;&#xA;&#xA;&lt;p&gt;J = section polar moment of inertia or St Venant's section modulus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note at the similarity to M of a simply supported beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the diagram of the torque stress would be similar to that of the shear stress of a simply supported beam with uniformly distributed load: a straight line from,  &lt;span class=&quot;math-container&quot;&gt;$-tL/2 \ $&lt;/span&gt; max at left support to zero at the mid-span to &lt;span class=&quot;math-container&quot;&gt;$+tL/2  \ $&lt;/span&gt;max at the right support, depending on sign convention.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we would have to add the torque stress to the shear stress with sign and direction, a bit complicated. &lt;span class=&quot;math-container&quot;&gt;$\tau+v &amp;lt; allowable \ shear-stress  .\ $&lt;/span&gt; And depending on load combination multiply by appropriate factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MiujT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MiujT.jpg&quot; alt=&quot;diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;AISC webinar on designing beams for torsion has condensed useful material and real-life examples and links. &lt;a href=&quot;https://www.aisc.org/globalassets/continuing-education/quiz-handouts/designing-members-for-torsion.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Webinar&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-24T06:46:30.860" LastActivityDate="2019-11-24T06:46:30.860" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32063" PostTypeId="1" CreationDate="2019-11-23T14:35:30.593" Score="1" ViewCount="21" Body="&lt;p&gt;I'm not very good at physics, but I've got a problem that I really need to solve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the second picture you can see a pipe with flowing water. The point of this mechanism is to lit a LED whenever the water pressure becomes too high. We're using a comparator to track the pressure. The piston on the second picture is connected to the linear sliding potentiometer R1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S5fJBm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S5fJBm.png&quot; alt=&quot;Comparator&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/idMC2m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/idMC2m.png&quot; alt=&quot;Pipe&quot;&gt;&lt;/a&gt;&#xA;Known data:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$R2 = 5kΩ$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$R3 = 30kΩ$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Length of the potentiometer &lt;span class=&quot;math-container&quot;&gt;$d=50mm$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;When the potentiometer slider is at its lowest position, the voltage at its ends is &lt;span class=&quot;math-container&quot;&gt;$U=0V$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Voltage provided to the circuit is &lt;span class=&quot;math-container&quot;&gt;$U=5V$&lt;/span&gt; (excuse me if I'm not using the proper terminology, I'm not a native english speaker)&lt;/li&gt;&#xA;&lt;li&gt;Water density may be assumed equal &lt;span class=&quot;math-container&quot;&gt;$1000\frac{kg}{m^3}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Gravitational acceleration may be assumed equal &lt;span class=&quot;math-container&quot;&gt;$10\frac{m}{s^2}$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The problem is to calculate the minimum water pressure that would cause the LED to lit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I understand, the first thing to do would be to calculate the current at the negative end of the comparator. According to the Ohm's law, the current in that branch would be &lt;span class=&quot;math-container&quot;&gt;$I=\frac{U}{R2 + R3}=\frac{5}{35000}=\frac{1}{7000}A$&lt;/span&gt;. Now the voltage at the negative comparator end is &lt;span class=&quot;math-container&quot;&gt;$U=R2\times I=5000\times\frac{1}{7000}=\frac{5}{7}V$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now comes the part where I have no idea, what to do. My assumption is that if the voltage at the potentiometer's lowest point is 0V and at its highest point it is (probably) 5V, then it must become &lt;span class=&quot;math-container&quot;&gt;$\frac{5}{7}V$&lt;/span&gt; at &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{5}\times\frac{5}{7}=\frac{1}{7}$&lt;/span&gt; its length, so &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{7}\times50mm=\frac{50}{7}mm$&lt;/span&gt;. Even if these ideas are correct, I've got no idea, what to do next. I'd appreciate any help&lt;/p&gt;&#xA;" OwnerUserId="23563" LastActivityDate="2019-11-23T14:35:30.593" Title="Calculating the water pressure knowing the circuit that tracks it" Tags="|pressure|circuits|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32064" PostTypeId="2" ParentId="32059" CreationDate="2019-11-23T14:51:54.230" Score="2" Body="&lt;p&gt;The easiest way to view the problem is to draw a control volume around the engine that is big enough so that the air pressure is constant on the boundary surface of the control volume and the upstream air velocity is just the free-stream velocity everywhere on the upstream boundary surface of the control volume. The downstream velocity is faster, and the difference in the momentum across the boundary can be equated to net thrust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you start asking about the forces and pressure distributions over the engine components, you need a more detailed near field flow model. The simplest that helps with your question is the actuator disk model. A result is that half of the flow acceleration happens in front of the disk, and half happens downstream of the disk as the jet continues to contract. So if you have cowls and ducts around a jet engine, they are in a region of accelerating flow. Upstream, the pressure is lower than freestream pressure, and carefully designed intake ducts contribute to thrust. Or perhaps a better way to say it is that tranferring all the engine thrust via high speed shaft bearings is doing things the hard way. Better is to design the rotating components of jet so that they are largely thrust neutral, and let the fixed stators and cowls and nozzles produce most of the reaction thrust.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is an inlet duct effect, &lt;em&gt;but it is not in addition to the thrust computed for the momentum change, it is just a normal part of it&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2019-11-25T00:25:26.390" LastActivityDate="2019-11-25T00:25:26.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32065" PostTypeId="1" CreationDate="2019-11-23T16:36:45.210" Score="2" ViewCount="1952" Body="&lt;p&gt;This confusion arise because my teacher says that volume does not change under tensile/compressive stress within the elastic limit of the material ( consider metal here). But as far as I can see via applying Poisson's ratio it does ( otherwise lateral strain would be proportional to square root of longitudinal strain). So does the volume chaneg under tensile stress ? What might be the molecular basis of such? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I defined a quantity (cause it was simple to calculate) &lt;span class=&quot;math-container&quot;&gt;$ \frac {\Delta A}{A}$&lt;/span&gt; and then found that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;via **Poisson's ratio&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac {\Delta A}{A} = \alpha \epsilon _{lon}( \alpha \epsilon _{lon} +2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;whereas the one derived via assuming the volume to be constant (i.e., &lt;span class=&quot;math-container&quot;&gt;$AL = A_0 L_0$&lt;/span&gt;) was&#xA; &lt;span class=&quot;math-container&quot;&gt;$$ \frac {\Delta A}{A} = \frac {\epsilon _{lon}}{\epsilon _{lon} +1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here &lt;span class=&quot;math-container&quot;&gt;$ \alpha = \frac {\epsilon_{lat}}{\epsilon_{lon}}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user23622" LastEditorDisplayName="user23622" LastEditDate="2019-11-23T17:02:20.413" LastActivityDate="2019-11-23T18:37:32.753" Title="Does the volume change under the tensile or compressive stress?" Tags="|materials|civil-engineering|stresses|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32066" PostTypeId="2" ParentId="32021" CreationDate="2019-11-23T17:25:28.100" Score="1" Body="&lt;p&gt;Sorry if my explanation is simple (anything wrong please point out).&lt;br&gt;&#xA;With my understanding of your situation the right way should be something on this line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your objective velocity is &lt;em&gt;50Km/h (13,88m/s)&lt;/em&gt;&lt;br&gt;&#xA;To obtain this velocity in 6s you'd need calculate aceleration&lt;br&gt;&#xA; &lt;span class=&quot;math-container&quot;&gt;$$a=v/t$$&lt;/span&gt;&#xA; &lt;span class=&quot;math-container&quot;&gt;$$a = \frac{13,88[m/s]}{6[s]}$$&lt;/span&gt;&#xA; &lt;span class=&quot;math-container&quot;&gt;$$a = 2,315[m/s²]$$&lt;/span&gt;&lt;br&gt;&#xA;The force needed to obtain our aceleration is &#xA; &lt;span class=&quot;math-container&quot;&gt;$$F=m*a$$&lt;/span&gt;&#xA; &lt;span class=&quot;math-container&quot;&gt;$$F = 40[Kg] * 2,315[m/s²]$$&lt;/span&gt;&#xA; &lt;span class=&quot;math-container&quot;&gt;$$F = 92,6N$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To acelerate without slipping you need estatic friction force higher than your aceleration force so &lt;span class=&quot;math-container&quot;&gt;$F\mu_{est} &amp;gt; F$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F\mu_{est} = F_{motor} * \mu$&lt;/span&gt; ( &lt;span class=&quot;math-container&quot;&gt;$F_{motor}$&lt;/span&gt; for traction force on wheel)&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$92,6N = F_{motor} * 0,1$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$F_{motor} = 926N$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, for the angle you want we use trigonometry.&lt;br&gt;&#xA;Considering the cable pulling force is 500Kg total, I'll consider 2.500N traction in each side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$sin\theta = 930/2500$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\theta = 21,84º$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="23658" LastEditorUserId="23658" LastEditDate="2019-11-23T17:57:42.547" LastActivityDate="2019-11-23T17:57:42.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32067" PostTypeId="2" ParentId="32065" CreationDate="2019-11-23T18:20:53.843" Score="4" Body="&lt;p&gt;Yes, the volume changes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The relative change of volume &lt;span class=&quot;math-container&quot;&gt;$ΔV/V $&lt;/span&gt;of a cube due to the stretch of the material :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uBF73.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uBF73.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using &lt;span class=&quot;math-container&quot;&gt;$V = L^3$&lt;/span&gt; and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V + \Delta V = (L + \Delta L)\left(L + \Delta L'\right)^2$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;:&lt;span class=&quot;math-container&quot;&gt;$\frac{\Delta V}{V} = \left(1 + \frac{\Delta L}{L} \right)\left(1 + \frac{\Delta L'}{L} \right)^2 - 1$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the above derived relationship between &lt;span class=&quot;math-container&quot;&gt;$\Delta L$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\Delta L'$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;:&lt;span class=&quot;math-container&quot;&gt;$\frac {\Delta V} {V} = \left(1+\frac{\Delta L}{L} \right)^{1-2\nu} - 1$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and for very small values of &lt;span class=&quot;math-container&quot;&gt;$\Delta L \ and \ \Delta L'$&lt;/span&gt;, the first-order approximation yields:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac {\Delta V} {V} \approx (1-2\nu)\frac{\Delta L}{L}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For isotropic materials, we can use Lamé parameters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{1}{2} - \frac{E}{6K}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where K is bulk modulus and E is elastic modulus or Young's modulus.&#xA;&lt;a href=&quot;http://==%20Volumetric%20change%20==%20%20The%20relative%20change%20of%20volume%20&amp;#39;&amp;#39;%CE%94V&amp;#39;&amp;#39;/&amp;#39;&amp;#39;V&amp;#39;&amp;#39;%20of%20a%20cube%20due%20to%20the%20stretch%20of%20the%20material%20can%20now%20be%20calculated.%20Using%20%3Cmath%3EV%20=%20L%5E3%3C/math%3E%20and%20%3Cmath%3EV%20+%20%5CDelta%20V%20=%20(L%20+%20%5CDelta%20L)%5Cleft(L%20+%20%5CDelta%20L&amp;#39;%5Cright)%5E2%3C/math%3E:%20%20:%3Cmath%3E%5Cfrac%7B%5CDelta%20V%7D%7BV%7D%20=%20%5Cleft(1%20+%20%5Cfrac%7B%5CDelta%20L%7D%7BL%7D%20%5Cright)%5Cleft(1%20+%20%5Cfrac%7B%5CDelta%20L&amp;#39;%7D%7BL%7D%20%5Cright)%5E2%20-%201%3C/math%3E%20%20Using%20the%20above%20derived%20relationship%20between%20%3Cmath%3E%5CDelta%20L%3C/math%3E%20and%20%3Cmath%3E%5CDelta%20L&amp;#39;%3C/math%3E:%20%20:%3Cmath%3E%5Cfrac%20%7B%5CDelta%20V%7D%20%7BV%7D%20=%20%5Cleft(1+%5Cfrac%7B%5CDelta%20L%7D%7BL%7D%20%5Cright)%5E%7B1-2%5Cnu%7D%20-%201%3C/math%3E%20%20and%20for%20very%20small%20values%20of%20%3Cmath%3E%5CDelta%20L%3C/math%3E%20and%20%3Cmath%3E%5CDelta%20L&amp;#39;%3C/math%3E,%20the%20first-order%20approximation%20yields:%20%20:%3Cmath%3E%5Cfrac%20%7B%5CDelta%20V%7D%20%7BV%7D%20%5Capprox%20(1-2%5Cnu)%5Cfrac%7B%5CDelta%20L%7D%7BL%7D%3C/math%3E%20%20For%20isotropic%20materials%20we%20can%20use%20[[Lam%C3%A9%20parameters%7CLam%C3%A9%E2%80%99s%20relation]]%3Cref%3Ehttps://arxiv.org/ftp/arxiv/papers/1204/1204.3859.pdf%20-%20Limits%20to%20Poisson%E2%80%99s%20ratio%20in%20isotropic%20materials%20%E2%80%93%20general%20result%20for%20arbitrary%20deformation.%3C/ref%3E%20%20:%3Cmath%3E%5Cnu%20%5Capprox%20%5Cfrac%7B1%7D%7B2%7D%20-%20%5Cfrac%7BE%7D%7B6K%7D%3C/math%3E%20%20where%20%3Cmath%3EK%3C/math%3E%20is%20[[bulk%20modulus]]%20and%20%3Cmath%3EE%3C/math%3E%20is%20elastic%20modulus%20(or%20[[Young&amp;#39;s%20modulus]]).%20%20Note%20that%20isotropic%20materials%20must%20have%20a%20Poisson&amp;#39;s%20ratio%20of%20%3Cmath%3E-1%20%3C%20%5Cnu%20%3C%200.5%3C/math%3E.%20%20Typical%20isotropic%20engineering%20materials%20have%20a%20Poisson&amp;#39;s%20ratio%20of%20%3Cmath%3E0.2%20%3C%20%5Cnu%20%3C%200.5%3C/math%3E.%3Cref%20name=%22polymerphysics.net%22&quot; rel=&quot;nofollow noreferrer&quot;&gt;wikipedia link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-23T18:37:32.753" LastActivityDate="2019-11-23T18:37:32.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32068" PostTypeId="2" ParentId="32056" CreationDate="2019-11-23T18:51:52.367" Score="1" Body="&lt;p&gt;The restoration and maintenance of historic buildings is the subject of a post-graduate course for architects, engineers and construction managers.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;http://nps.gov/tps/standards/rehabilitation/rehab/masonry01.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;This link&lt;/a&gt; covers some aspects of small, historic building maintenance.&lt;/p&gt;&#xA;&lt;p&gt;The left panel has a chapter reference.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="28962" LastEditDate="2021-04-17T12:53:47.153" LastActivityDate="2021-04-17T12:53:47.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32069" PostTypeId="1" CreationDate="2019-11-23T19:59:01.893" Score="-1" ViewCount="71" Body="&lt;p&gt;I am required to find the temperature of the outside wall. I have rearranged an equation but it doesn't feel right&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ti and To are the inside and outside temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;U is the overall heat transfer coefficient. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A is the surface area of the wall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the inside 20m2 wall at 25 degrees is subjected to a heat rate of 10 kj/s and heat is transferred at the rate of 15w/cm2, determine the temp of the outer wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dQ}{dt} = UA(T_i-T_o)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{dQ}{dt} = 10Kj/s$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ U = 15W^o/cm^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ A = 20m^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T_i = 25^oC$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I will convert the units and use SI units.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$U=0.015 Kj/s$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$A = 20*10^4cm^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T_i = 298 K$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sub in numbers &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$10*10^3 = (0.015) * (20*10^4)*(298-T_0)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{10*10^3}{3000}=298-T_o$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$3.3=298-T_0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$294=-T_o$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly this is incorrect. Where am I going wrong here? I assume the units I am using are incorrect.&lt;/p&gt;&#xA;" OwnerUserId="20701" LastActivityDate="2019-11-23T20:43:34.090" Title="Heat Transfer Rate. Find Outside Temperature" Tags="|thermodynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32070" PostTypeId="2" ParentId="32069" CreationDate="2019-11-23T20:43:34.090" Score="0" Body="&lt;p&gt;without checking the whole calculation, there seems to be a sign error in the last line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$3.3 = 298-T_o \\  3.3-298=-T_o \\T_O =294.7  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's just a small sign error.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-23T20:43:34.090" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32072" PostTypeId="2" ParentId="31493" CreationDate="2019-11-24T11:44:43.680" Score="0" Body="&lt;p&gt;After completing some further research, I have come across a few approximate methods which were described in the book: Stability Design of Steel Frames. These include the two-cycle iterative method and modified slope deflection equations which can both analyse statically indeterminate structures while taking into account second-order effects. Thanks for all the answers. &lt;/p&gt;&#xA;" OwnerUserId="20822" LastActivityDate="2019-11-24T11:44:43.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32074" PostTypeId="2" ParentId="31498" CreationDate="2019-11-24T14:19:17.260" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Question#1: Is it possible to perform the static structural analysis of an unconstrained object &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes this is possible, it is called &quot;inertia relief&quot;.  It creates a state of static equilibrium by automatically applying forces/moments at the center of mass (or a specified node), such that the net force/moment on the structure is exactly zero.  Note that as this is still a static analysis, the assumption is that the acceleration of the object is zero. If you have an unconstrained object that is accelerating, then static analysis is not possible.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-11-24T14:19:17.260" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32075" PostTypeId="1" AcceptedAnswerId="32080" CreationDate="2019-11-24T15:25:13.577" Score="1" ViewCount="178" Body="&lt;p&gt;Some laps have patterns of holes drilled into them - two examples are below. In both cases the user claims that the holes (among other techniques) help ensure that the laps stays close to round when they're adjusted. I've seen it in enough places that I'm willing to believe there's some truth to it. Beyond vaguely saying that it redistributed the stress, I just don't see how though. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;How do these hole patterns help ensure roundness when the lap changes sizes?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples of the claims (time to start on the relevant sentence):&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=H2SeLcbcYkE&amp;amp;feature=youtu.be&amp;amp;t=188&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=H2SeLcbcYkE&amp;amp;feature=youtu.be&amp;amp;t=188&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=AEsaSN4LLU4&amp;amp;feature=youtu.be&amp;amp;t=178&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=AEsaSN4LLU4&amp;amp;feature=youtu.be&amp;amp;t=178&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Examples of the laps:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pKzI7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pKzI7.png&quot; alt=&quot;OD Lap&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gjLiV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gjLiV.png&quot; alt=&quot;ID Lap&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23661" LastActivityDate="2019-11-24T20:06:14.033" Title="Do the hole patterns drilled in laps really help ensure uniform expansion and contraction?" Tags="|machining|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32076" PostTypeId="1" AcceptedAnswerId="32082" CreationDate="2019-11-24T18:01:41.163" Score="1" ViewCount="346" Body="&lt;p&gt;I am trying to figure out what the exact term is for the clutch system used in a roll-up window shade. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought they were called centrifical clutches, but that typically gives me the opposite results like what is used in an automatic vehicle. Where speed causes it to engage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I want to look up some existing designs because I'm trying to create a product that uses one.&lt;/p&gt;&#xA;" OwnerUserId="23663" LastActivityDate="2019-11-25T06:11:37.850" Title="What is name of clutch system in a roller shade?" Tags="|mechanical-engineering|power-transmission|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32077" PostTypeId="2" ParentId="32076" CreationDate="2019-11-24T19:22:49.763" Score="1" Body="&lt;p&gt;A roller or inclined ramp clutch may be what you are talking about.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-11-24T19:22:49.763" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32078" PostTypeId="2" ParentId="32075" CreationDate="2019-11-24T19:51:20.427" Score="2" Body="&lt;p&gt;The function of the laps is diferent looking both vídeos.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first lap is for external diameter.&lt;br&gt;&#xA;The external piece (part with holes) distribute evenly the stress like you said. And the internal piece (tubing with a cut) redistributes the stress more evenly (in the vídeo he said the helical cut is better than a straight cut because the straight can get oval).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The way the geometry works is similar to bicicle wheels. The hub and rim are conected by the spokes and remain centered.&#xA;This lap works exactly like this. Using the screw in the external part will change the diameter of the rim, the spokes remain with the same distance, so the iner part (hub) need to acomodate the new geometry. The spokes apply exactly the same force, so the inner part adjusts equaly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second lap is for internal diameter &lt;em&gt;(this one I think the geometry is not perfect)&lt;/em&gt;.&lt;br&gt;&#xA;When you apply stress evenly the outer ring will open &lt;em&gt;(looks like not a perfect round part, but definetly closer)&lt;/em&gt;.&lt;br&gt;&#xA;In &lt;a href=&quot;https://youtu.be/AEsaSN4LLU4?t=940&quot; rel=&quot;nofollow noreferrer&quot;&gt;15:40 of the vídeo&lt;/a&gt; he says that lapping like he's doing the position of the piece need to change time from time to compensate if geometry is not perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to remember that laping is used to very precise and minimal changes in diameter, so the diference in tool geometry will be closely to zero.&lt;/p&gt;&#xA;" OwnerUserId="23658" LastActivityDate="2019-11-24T19:51:20.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32079" PostTypeId="2" ParentId="32075" CreationDate="2019-11-24T19:52:14.153" Score="0" Body="&lt;p&gt;For an object as &quot;chunky&quot; as a solid lap, tightening the adjustment screw will deform it out of flat because of Poisson's ratio (squeeze it one way, it swells up another way). By drilling those holes, the lap deforms more like a &lt;em&gt;hoop&lt;/em&gt; and less like a solid mass because they prevent compressive stresses from being transmitted towards the central axis. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-11-24T19:52:14.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32080" PostTypeId="2" ParentId="32075" CreationDate="2019-11-24T20:06:14.033" Score="0" Body="&lt;p&gt;You can think of the top picture as two semicircular curved beams joined at one end and with equal and opposite forces applied by the clamp at the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you apply a force at the end of a uniform cantilever beam, it doesn't bent into a circular arc but part of a cubic curve with more curvature at the root than at the tip. This will give an more constant curvature along the length of the &quot;beam&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To counteract that, the holes are smaller at the &quot;root&quot; of the beam than at the tip, which is equivalent to a change in the moment of inertia along the length of the beam - greater at the root (small holes) and smaller at the tip (large holes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second picture, a solid ring with no slots would simply be far too stiff to deform it by expanding the center. The slots and holes mean that the only continuous material which can support hoop stress is a thin ring at the outside of the disk.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-11-24T20:06:14.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32081" PostTypeId="1" CreationDate="2019-11-25T05:10:50.197" Score="1" ViewCount="371" Body="&lt;p&gt;I'm trying to figure this problem out, i know the correct answer, and i know the method, but for some reason i'm missing something and i'm just not getting there, any help would be greatly appreciated.&lt;/p&gt;&#xA;&lt;p&gt;Consider a Sluice Gate (lock gate) as pictured,&lt;a href=&quot;https://i.stack.imgur.com/joUZr.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/joUZr.gif&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;A brief Description is.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;A large reservour if fluid of depth &lt;span class=&quot;math-container&quot;&gt;$h_1$&lt;/span&gt; is held stationary in &lt;span class=&quot;math-container&quot;&gt;$x &amp;lt; 0$&lt;/span&gt; to the left of a vectical gate at the origin of width W. The gate is raised and fluid streams steadily through the opening between the bottom of the gate and the base of the reservoir.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Far upstream (the left of the gate) the fluid moves with constant velocity &lt;span class=&quot;math-container&quot;&gt;$u_1$&lt;/span&gt; in the x horizontal direction and uniformly through the depth &lt;span class=&quot;math-container&quot;&gt;$h_1$&lt;/span&gt;. far downstream,  (to the right of the gate) the fluid assumes a depth &lt;span class=&quot;math-container&quot;&gt;$h_2$&lt;/span&gt; and moves with constant velocity &lt;span class=&quot;math-container&quot;&gt;$u_2$&lt;/span&gt; in the horizontal x direction.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The job is to&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Derive a relation expressing conservation of mass.&lt;/li&gt;&#xA;&lt;li&gt;Use Bernoulli's equation along a suitable streamline in order to show two &amp;gt;possible choices of &lt;span class=&quot;math-container&quot;&gt;$\frac{h_2}{h_1}$&lt;/span&gt; are &lt;span class=&quot;math-container&quot;&gt;$\frac{h_2}{h_1} = 1$&lt;/span&gt; and &amp;gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{h_2}{h_1} =\frac{Fr^2+Fr\sqrt{Fr^2+8}}{4}$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$Fr$&lt;/span&gt; is the Froude number &lt;span class=&quot;math-container&quot;&gt;$Fr = \frac{u_1}{\sqrt{gh}}$&lt;/span&gt;&#xA;Then finally&lt;/li&gt;&#xA;&lt;li&gt;Find the net force on the raised gate&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;So,&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Deriving a relationship expressing conservation of mass is just understanding that as the flow empties from one side, it must rush into the other at the same rate,  and so we have&#xA;&lt;span class=&quot;math-container&quot;&gt;$$Q = u_1 A_1 = u_2 A_2 \implies Q = u_1 h_1W = u_2 h_2 W$$&lt;/span&gt;&#xA;as our relation.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;this has been bugging me quite a bit, i'll come back to this.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;with This question, i believe the correct method is to pick a control volume at about the sluice, then combine the above conservation of mass equation with bernoulli's equation for a steady flow. so&#xA;here i go...&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;We express Bernoulli's equation as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{2} \rho u^{2}_1+\rho g h_1 - \frac{1}{2}\rho u^{2}_{2}+\rho g h_2= p_{constant_1} - p_{constant_2} = P_{c}$$&lt;/span&gt;&#xA;where we have h is the elevation, &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; the density and u is the flow velocity. we have that the pressure also cancels out. and the above equation happens because bernoulli's is a constant along any streamline in the flow.&lt;/p&gt;&#xA;&lt;p&gt;now using this line of logic, i find myself coming up with &lt;span class=&quot;math-container&quot;&gt;$$F = \rho g \frac{\left(1-\frac{h_2}{h_1}\right)^3}{2\left(1+\frac{h_2}{h_1}\right)}$$&lt;/span&gt; which means im missing a step as the correct answer is actually&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_{net} = h_1^2 W\rho g \frac{\left(1-\frac{h_2}{h_1}\right)^3}{2\left(1+\frac{h_2}{h_1}\right)}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;This has been really bothering me for the last few, so any help would be greatly appreciated. Thank you for taking the time to read this.&lt;/p&gt;&#xA;" OwnerUserId="23133" LastEditorUserId="30651" LastEditDate="2021-08-27T21:59:11.080" LastActivityDate="2024-01-15T04:06:05.693" Title="Deriving the ideal equation for sluice gate problems - incompressible inviscid fluid" Tags="|fluid-mechanics|hydraulics|mathematics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32082" PostTypeId="2" ParentId="32076" CreationDate="2019-11-25T06:11:37.850" Score="1" Body="&lt;p&gt;It is a spring-loaded ratchet and pawl system. the pawl mechanism is unlocked by the centripetal force of the initial rolling of the shade by a user.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram. &lt;a href=&quot;https://i.stack.imgur.com/qwcNd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qwcNd.jpg&quot; alt=&quot;diagram.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-25T06:11:37.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32083" PostTypeId="1" CreationDate="2019-11-25T08:05:32.737" Score="0" ViewCount="116" Body="&lt;p&gt;For an existing Warehouse, with sprinkler system already installed:&#xA;We are adding 3 mezzanine floors, with around 27000 Sq. ft for each floor. While designing sprinkler system, is it necessary to add new Fire tanks, Fire pumps for sprinkler system once again or can we connect the sprinkler riser to the existing system. We are also adding around 12 Fire house box(3 per floor) which will be connected to the ring main. Is it okay?&lt;/p&gt;&#xA;" OwnerUserId="23668" LastActivityDate="2023-08-07T03:09:46.180" Title="Warehouse Expansion: Is a new fire pump and tank required" Tags="|fire-sprinkler|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32084" PostTypeId="1" CreationDate="2019-11-25T17:56:44.423" Score="1" ViewCount="95" Body="&lt;p&gt;So, I did a quick search around and didn't seem to find anything regarding this, so I thought perhaps I should just ask a question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Metals are malleable and will generally flow/deform if you roll it repeatedly. Localized concentrated loads will thus change the shape of a piece of metal bit-by-bit if exposed to repeated loading of the same/similar orientations (kind of like molding a piece of playdoh, with each repeated load only displacing a very minor amount of material). Presumably, flow should be the same in all directions of the contact point non-isotropically for a homogeneous material. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if there is eccentricity within the load, will that still hold true? If it does not, where is that critical limit for for X:B within the figure below that will produce some degree of eccentricity with the amount of metallic flow? Or am I thinking about this in a completely wrong way? See below example for what X and B refers to, and an example of what I am trying to figure out. Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Edit: I know flow will happen. What I am trying to understand is if load is eccentric, how will that affect flow behavior? Will it cause one side to flow more than the other IN THE ABSENCE of surface shear type loads? The below example will have all vertical loads (up/down) with assumed 0 lateral loads (left/right), and if you are just rolling it back and forth, will it flow more on the side closer to the edge just because it is closer to the edge? I don't think there are any formal theories but I can see it happening that way because flow ultimately is a localized plastic deformation that is constrained by surrounding material structure. If the surrounding material structure weakens, it somewhat makes sense that it will tend to have more plastic deformation in that direction. But is there some limit as to how far X has to be from the edge for it to be negligible? It's kind of like us saying eccentric loading on a concrete shallow foundation footing is okay if it is within the middle third. Hope that makes more sense what I am asking.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5bqHu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5bqHu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13526" LastEditorUserId="13526" LastEditDate="2019-11-27T15:54:17.257" LastActivityDate="2024-04-05T04:03:35.363" Title="Metallic flow behavior under eccentric loading conditions" Tags="|materials|metals|material-science|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32085" PostTypeId="2" ParentId="32084" CreationDate="2019-11-25T19:06:05.970" Score="0" Body="&lt;p&gt;Yes, it is still going to reshape and damage the wheel and the support. Maybe an example of unsymmetrical loading is a train wheel in relation to the tracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There has been extensive research and tests on that, to help the design the wheels and rails to lower the routine maintenance costs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1 Contact points&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wj5lc.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wj5lc.gif&quot; alt=&quot;contact points&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;2 Wear model of the flange.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dofz3.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dofz3.gif&quot; alt=&quot;flange wear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a brief but useful research on the  &lt;a href=&quot;https://link.springer.com/article/10.1007/s40864-017-0072-2&quot; rel=&quot;nofollow noreferrer&quot;&gt;railroad wheel.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-25T19:35:46.943" LastActivityDate="2019-11-25T19:35:46.943" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32087" PostTypeId="2" ParentId="32083" CreationDate="2019-11-25T21:00:03.780" Score="-1" Body="&lt;p&gt;Are you adding the addition without any permits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because not only you need to get approval from the fire-department for adding new tanks and fire pipes and Siamese connection, you need to meet all the new fire exits, signage, fire partitions, etc. Guess what your insurance company will say in case of a claim. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to get the structure checked for additional loading, electrical panels, and circuits, lighting, and airconditioning. handicap access, basically you have a substantial job and have to get it permitted.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-25T21:00:03.780" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32088" PostTypeId="2" ParentId="15075" CreationDate="2019-11-26T02:43:33.390" Score="0" Body="&lt;p&gt;If the system did not have an inert gas to purge all of the oxygen out, the system would be very hard to reach the desired micron level with in the vacuum. &lt;/p&gt;&#xA;" OwnerUserId="21724" LastActivityDate="2019-11-26T02:43:33.390" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32090" PostTypeId="1" CreationDate="2019-11-26T19:29:47.407" Score="-1" ViewCount="64" Body="&lt;p&gt;i have been researching deep dives. I understand that deep-divers have the same pressure &lt;em&gt;inside of their bodies&lt;/em&gt; as there is outside. Yet i still cannot understand how they do it. Even if they have the same pressure in their lungs and i assume their blood there is a large area that is not covered. I mean if i push down with the force of 100 kg on my finger, should it matter if i have 100 kg of pressure inside the blood in my finger? It would probably help a bit but i still don't understand why it doesn't get squeezed. Do deep divers feel a squeezing sensation? Does it become like swimming in syrup at deep depths? Honestly i don't understand how they do it with 100 kg pushing down on them... Even if the force of 100 kg is pushing inside them how come they can even move their arms??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also have another question. I get that as you add more and more water on top of you there will be more and more pressure, well, on top of you... But say for example under the arms of a diver. Is it the &lt;em&gt;exact&lt;/em&gt;(i guess i mean to ask if this has been tested rigourously) same pressure there? After all the shoulder is taking the pressure from above. Maybe a better example would be a closed diving-bell. The pressure below it is &lt;em&gt;exactly&lt;/em&gt; the same as on it's sides(?) And why would this be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally... As i understand it the pressure inside a closed diving bell is the same as the water pressure. But gas is easier to &quot;move around in&quot; -- or objects can more easily do work - to say it in scientific(:D) than when in liquids. Is this true or am i deluded?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are my questions. I realize perhaps i should make 3 separate posts but they are all aimed at understanding the same thing and are related to eachother so i hope it will be allowed to stand.&lt;/p&gt;&#xA;" OwnerUserId="23233" LastActivityDate="2019-11-27T21:18:30.703" Title="Trying to understand water pressure / air pressure" Tags="|pressure|pressure-vessel|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32091" PostTypeId="2" ParentId="32090" CreationDate="2019-11-26T21:17:00.977" Score="0" Body="&lt;p&gt;Our body is 60% water and this water permeates throughout the different tissues, keeping the outside and inside pressure at equilibrium.&#xA;At the maximum depth of deep diving, 40 meters, the hydrostatic water pressure is 4atm, which is far below the pressure our tissues can handle. just walking exposes your feet and toes tissues to more than that pressure. and as you said people push and press and lift weights with their hand all the time. Even small daily tasks such as flexing your arm to pull open a car door expose some tissue around your joints like wrist and elbow to high pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not getting into issues with nitrogen accumulation and rate of ascent because you said you already looked them up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The feeling of water getting harder to swim in or move around, say its thickness, in water, is called kinematic viscosity, which can be obtained by dividing the dynamic viscosity of a fluid by its density. But in case of water, because it is essentially an incompressible material, its kinematic viscosity doesn't change with depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your question regarding the pressure of water above and below the shoulder of the swimmer, the water pressure is the same in all directions, only varies by the level of the water. water is a liquid and liquids assume the shape of their container and maintain the same pressure at any given level regardless of the direction they apply that pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And yes normally the viscosity of gases in normal temperatures we experience daily is much less than the water so they can be pushed around easier or they can flow through a propeller much faster.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-26T21:48:50.420" LastActivityDate="2019-11-26T21:48:50.420" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32092" PostTypeId="1" CreationDate="2019-11-26T23:53:26.163" Score="1" ViewCount="200" Body="&lt;p&gt;I wonder if the presence of a gravity field, makes any difference in the properties of the finished parts? &#xA;That’s only if we can somehow, keep every other factor almost the same. &lt;/p&gt;&#xA;" OwnerUserId="23589" LastEditorDisplayName="user14407" LastEditDate="2019-11-27T00:03:33.410" LastActivityDate="2019-11-27T15:14:27.550" Title="Does gravity have any effect on metal casting?" Tags="|mechanical-engineering|aerospace-engineering|casting|production-technology|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32096" PostTypeId="2" ParentId="32092" CreationDate="2019-11-27T02:10:36.033" Score="1" Body="&lt;p&gt;It certainly does: any slag inclusions, for example, will tend to rise to the &quot;top&quot; surface of the casting, which hence furnishes an anisotropic distribution of inclusions in the finished part that depends on gravity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sprue or gate that feeds molten metal into the mold cavity is on the &quot;top&quot; side of the mold, then breaking the gate remnant off the finished part is one way of eliminating flaws in it. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-11-27T02:10:36.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32097" PostTypeId="1" AcceptedAnswerId="32139" CreationDate="2019-11-27T13:12:09.577" Score="0" ViewCount="548" Body="&lt;p&gt;I was reading some documents that said&#xA;25 wt. %, 500 ml&lt;em&gt;L&lt;sup&gt;-1&lt;/sup&gt;&#xA;as well as&#xA;20 g&lt;/em&gt;L&lt;sup&gt;-1&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think 25 wt. %, 500 mlL&lt;sup&gt;-1&lt;/sup&gt; is referencing 25% weight of 500 ml of water so that would be 125 ml of water&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20 g*L&lt;sup&gt;-1&lt;/sup&gt; would be 20 g per liter but since the reference is 500 ml then I should reduce this formula to 10 g to fit the 500 ml.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{NiSO}_4\cdot6\text{H}_2\text{O}\quad\left(20\ \text{g}\cdot\text{L}^{-1}\right)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\text{NH}_3\quad\left(25\ wt.\%,\ 500\ \text{ml}\cdot\text{L}^{-1}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="23681" LastEditorUserId="13311" LastEditDate="2019-11-28T10:52:51.957" LastActivityDate="2019-11-30T21:26:08.220" Title="Weight percentage and grams per liter" Tags="|measurements|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="32098" PostTypeId="2" ParentId="31969" CreationDate="2019-11-27T13:38:55.603" Score="1" Body="&lt;p&gt;This is supposed to be a comment rather than an answer, because I only reference &lt;em&gt;I understand that it is really only a heuristic method&lt;/em&gt;:&lt;/p&gt;&#xA;&lt;p&gt;You might also want to consider &lt;a href=&quot;https://en.wikipedia.org/wiki/Ackermann%27s_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ackermann's formula&lt;/a&gt; for determining the control parameters, which would be a &lt;em&gt;mathematical way&lt;/em&gt; with which you might find parallels to Ziegler-Nichols method.&#xA;If and when you do find parallels, please consider posting your findings here as I am very interested in the results. Thanks.&#xA;I hope it helps.&lt;/p&gt;&#xA;&lt;p&gt;EDIT (01/04/20): Note that Ackermann's formula is only valid for invertible controllability matrix (full rank), i.e. when your system is controllable in Kalman's sense.&lt;/p&gt;&#xA;" OwnerUserId="23539" LastEditorUserId="23539" LastEditDate="2021-06-18T09:30:56.943" LastActivityDate="2021-06-18T09:30:56.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32099" PostTypeId="1" AcceptedAnswerId="32100" CreationDate="2019-11-27T13:56:53.970" Score="1" ViewCount="2393" Body="&lt;p&gt;I have a discrete control system with known dead time and time constant, where a process variable is sampled at regular discrete time intervals and fed back to the controller. I want to find out an optimal, or at least good, sampling interval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I came accross a few websites saying that &quot;Sample time should be 10 times per process time constant or faster&quot;, e.g. &lt;a href=&quot;https://controlguru.com/sample-time-is-a-fundamental-design-and-tuning-specification/&quot; rel=&quot;nofollow noreferrer&quot;&gt;sample-time-is-a-fundamental-design-and-tuning-specification/&lt;/a&gt;&#xA;But I cannot find any theory supporting this statement, so I guess it may be a rule of thumb. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this statement correct? and can someone explain it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;====Edit====&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This Edit is after Jeffrey's answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to provide more details about my control system. It is a discrete system with a reference input. The goal is to keep the output close to the reference input (like a room temperature control system). The difference between the output and the reference input is the error fed to the controller, so the controller has control output accordingly. The system has a dead time, as well as a time constant, both at the scale of &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$10\mu s$&lt;/span&gt;. The fastest sampling interval of the output is at the sacle of &lt;span class=&quot;math-container&quot;&gt;$0.1\mu s$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is the fastest sampling rate good enough to achieve a stable control? &lt;/p&gt;&#xA;" OwnerUserId="23682" LastEditorUserId="23682" LastEditDate="2019-11-28T02:29:14.503" LastActivityDate="2023-11-16T08:44:31.330" Title="Rule of thumb for choosing a best sampling rate" Tags="|control-engineering|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32100" PostTypeId="2" ParentId="32099" CreationDate="2019-11-27T14:47:20.477" Score="3" Body="&lt;p&gt;Take a perfect analog signal input. Use a perfect controller with a zero time constant, an infinite look-ahead perspective to predict the upcoming signal, and no time constant to respond. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Nyquist-Shannon&lt;/a&gt; sampling theorem sets the lower bounds to resolve features in the signal. We must sample the output at a rate no less than 2x the function period. State this as below with &lt;span class=&quot;math-container&quot;&gt;$f^\star_c$&lt;/span&gt; as the sampling rate of the perfect controller and &lt;span class=&quot;math-container&quot;&gt;$f^\star_{fa}$&lt;/span&gt; as the rate of the function variation in the analog signal (each in units of Hz for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$f^\star_{c,min} = 2 f_{fa}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By example, an analog function with a cycle at 60 Hz requires a sampling rate of 120 Hz to capture correctly. Failing this metric, the perfect control system has a false representation of the true period of the signal to use to build its immediate response. Otherwise, the FFT of the incoming signal gives false representation for the otherwise perfect controller to use to predict where to go next.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a domain where the input signal is digital, we cannot measure data points that do not exist. The lowest sampling rate is the rate of the digital period of the signal input to the controller. We cannot sample at anything above 1x the signal rate. Given a digital signal rate of &lt;span class=&quot;math-container&quot;&gt;$f_{dp}$&lt;/span&gt;, regardless of the function period, the smallest sampling rate we can use on a perfect controller is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$f^\star_{c,max} = f_{dp}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When we have a digital signal that is running at 60 Hz, we can only sample at 60 Hz or below, never above. This is true regardless of the value of &lt;span class=&quot;math-container&quot;&gt;$f_f$&lt;/span&gt;, the frequency of the function in the signal itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The referenced example for the question shows a case for a pseudo-analog signal. In this case, the digital signal rate &lt;span class=&quot;math-container&quot;&gt;$f_{dp}$&lt;/span&gt; is shown to be larger than the function rate &lt;span class=&quot;math-container&quot;&gt;$f_f$&lt;/span&gt;. By example, we have an example of a case with a 1 Hz square wave signal output as a digital signal with samples at 60 Hz. Each square wave period over 1 s has 60 digital points in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, the minimum sample should default to the Nyquist-Shannon expression for the rate of the digital signal &lt;span class=&quot;math-container&quot;&gt;$f_{fd}$&lt;/span&gt; and should also respect the digital signal rate &lt;span class=&quot;math-container&quot;&gt;$f_{dp}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$f^\star_{c,min} = 2 f_{fd}\ \ \  f^\star_{c, max} = f_{dp}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, we should sample at least at 2 Hz and we cannot sample above 60 Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How does the factor of 10x arise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To carry this question further, we must assume that the frequency of the digital output is greater than the frequency of the underlying signal. We must either have a perfect analog signal fed to the controller or we must have &lt;span class=&quot;math-container&quot;&gt;$f_{dp} \gt f_{fp}$&lt;/span&gt;, perhaps at least by a factor of 10 or more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the reference that is quoted shows analysis only for &quot;perfect&quot; data with signal to noise &lt;span class=&quot;math-container&quot;&gt;$S/N = \infty$&lt;/span&gt;. In this case, we would expect only to need one data point per function cycle. We would resolve to the Nyquist–Shannon lower limit and the digital data rate upper limit. This has no factor of 10x anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where else can we look for an answer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, let's relax the assumption of perfect data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In data with noise, a general rule of thumb from uncertainty analysis is that a lower limit of 10 measurements is a reasonable approximation to obtain a mean from results that is comparable to what is obtained when sampling for large population statistics. A review of the &lt;a href=&quot;https://www.itl.nist.gov/div898/handbook/eda/section3/eda3672.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;student t-test values&lt;/a&gt; for example shows the 90% confidence value to obtain the sample mean from noisy data is 1.372. This is a (1.372 - 1.282)/1.282 = 7% difference from the infinite population value (the true mean).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I posit that the 10x value for sampling theory is a comparable rule of thumb for good sampling statistics using  controller with noisy digital input data. To obtain a working precision uncertainty of at least 7% from the controller that is sampling noisy data, sample at 10x over the function period. We might reason also that, as &lt;span class=&quot;math-container&quot;&gt;$S/N$&lt;/span&gt; in the data increases, the sampling rate should be increased beyond 10x of the period rate in the input signal &lt;span class=&quot;math-container&quot;&gt;$f_{fp}$&lt;/span&gt; (but can never exceed &lt;span class=&quot;math-container&quot;&gt;$f_{dp}$&lt;/span&gt; itself).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can next relax the assumption of a perfect controller. At that point, we should resolve the internal sampling of the system itself by similar uncertainty analysis rules. An uncertainty budget expression might be needed, and the outcome might be a comparable rule: The precision of the imperfect controller is affected by sampling rate in such a way that we state that 10x gives an acceptable controller precision even in the case of perfect input data (analog input with &lt;span class=&quot;math-container&quot;&gt;$S/N = \infty$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, here are references to well-structured discussions of sampling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://community.sw.siemens.com/s/article/digital-signal-processing-sampling-rates-bandwidth-spectral-lines-and-more&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://community.sw.siemens.com/s/article/digital-signal-processing-sampling-rates-bandwidth-spectral-lines-and-more&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.audiostream.com/content/sampling-what-nyquist-didnt-say-and-what-do-about-it-tim-wescott-wescott-design-services&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.audiostream.com/content/sampling-what-nyquist-didnt-say-and-what-do-about-it-tim-wescott-wescott-design-services&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-11-27T20:28:44.967" LastActivityDate="2019-11-27T20:28:44.967" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32101" PostTypeId="2" ParentId="32092" CreationDate="2019-11-27T15:14:27.550" Score="1" Body="&lt;p&gt;Redundant ; How would you pour with no gravity; the metal would just sit in the ladle after you turn it upside-down. Floating of slag/dross , and hydrogen gas ( always present to some degree) would not happen.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-11-27T15:14:27.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32102" PostTypeId="1" AcceptedAnswerId="32123" CreationDate="2019-11-27T17:35:14.763" Score="2" ViewCount="108" Body="&lt;p&gt;We're planning on shipping some items from the US to Australia as you cannot get them here, that being:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.chewy.com/sentry-stop-that-noise-pheromone-dog/dp/56486&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.chewy.com/sentry-stop-that-noise-pheromone-dog/dp/56486&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been in contact with a freight forwarder who can ship them to Australia, but obviously they will need to be shipped as a &lt;strong&gt;hazmat&lt;/strong&gt; item and go inside a particular sized box and be shipped by FedEx or DHL.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, as I understand it, the reason these are classified as hazmat is because  of the chance of bursting as well as being flammable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per the link above:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Contents under pressure. Do not puncture or incinerate container. Do not set on stove or radiator, expose to heat or store at temperatures above 120°F (49°C), as container may burst.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have no idea, but I assume they wouldn't be exposed to high temperatures such as that during shipping, but my question is...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming we got them ok and they didn't burst, do pressurised cans such as these still pose a risk to bursting in &quot;normal&quot; conditions AFTER they have been exposed to high heat? Such as if they &lt;em&gt;had&lt;/em&gt; been exposed to high heat during shipping, but we got them ok, would they still pose a risk to us with the possibility of bursting?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manufacturer provides a Material Safety Data Sheet on their website, which you have to email them or register to get it sent to you, but if it's of any interest you can download it &lt;a href=&quot;https://www.mediafire.com/file/y76wedx4pi2diqt/Sentry_Stop_That%21_Behavior_Spray_for_Dogs_and_Cats.pdf/file&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; from MediaFire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively you can find instructions to download it via &lt;a href=&quot;https://www.sentrypetcare.com/products/sentry-stop-that-noise-and-pheromone-spray/&quot; rel=&quot;nofollow noreferrer&quot;&gt;their site&lt;/a&gt; - see under &quot;PRODUCT SAFETY INFORMATION&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you don't want to download anything then you can see screenshots of the document I uploaded to an Imgur gallery here: &lt;a href=&quot;https://i.stack.imgur.com/44umr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/44umr.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note though, the above data sheet was provided via &lt;a href=&quot;https://www.sentrypetcare.com/products/sentry-stop-that-noise-and-pheromone-spray/&quot; rel=&quot;nofollow noreferrer&quot;&gt;their site&lt;/a&gt; that shows their old packaging, the link from Chewy above shows their new packaging, as far as I know the packaging is the only thing that has changed.&lt;/p&gt;&#xA;" OwnerUserId="23685" LastActivityDate="2019-12-01T20:26:42.417" Title="Do pressurised cans still pose a risk to bursting in normal conditions, but after being exposed to heat?" Tags="|pressure|chemical-engineering|gas|safety|compressed-gases|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32103" PostTypeId="1" CreationDate="2019-11-27T17:43:27.433" Score="1" ViewCount="1399" Body="&lt;p&gt;I am trying provide Customs with a tariff code for a helical spring of steel wire. The Harmonized Tariff Schedule asks for the &quot;cross-sectional dimension&quot; of the spring. I have been told that this is not the same as the diameter. Does anyone know how i can find the &quot;cross-sectional dimension&quot; of a spring with an inside diameter of 4.7 mm and an outside diameter of 5.3 mm? Not sure you would need it, but the height is 12.4 mm. Any help would be greatly appreciated. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="23686" LastActivityDate="2019-11-28T10:23:13.060" Title="Cross-Sectional Dimension of a Spring" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32104" PostTypeId="2" ParentId="32103" CreationDate="2019-11-27T19:28:22.153" Score="1" Body="&lt;p&gt;The cross section of the spring wire may be what they need. If so, and we call it &lt;span class=&quot;math-container&quot;&gt;$D_{wire}$&lt;/span&gt;, this can be calculated as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$D_{wire}=\frac{(D_{out}-D_{in})}{2}=\frac{(5.3-4.7)}{2}=0.3\text{ mm}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="13311" LastEditDate="2019-11-28T10:23:13.060" LastActivityDate="2019-11-28T10:23:13.060" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32105" PostTypeId="2" ParentId="32090" CreationDate="2019-11-27T21:18:30.703" Score="0" Body="&lt;p&gt;Everything comes down to Newton's law.  If you push on it, it moves.  So if it's not rigid, and you squeeze on it, then it'll shrink until the pressure inside is equal to the pressure outside.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I understand that deep-divers have the same pressure inside of their bodies as there is outside. Yet i still cannot understand how they do it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's not like it's done consciously.  Our tissues are flexible and so is our skin.  So any force on our skin must be balanced lest we collapse or blow up.  So if you squeeze on our outsides, then our skin flexes just enough that our insides get squeezed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if divers feel the pressure or not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But say for example under the arms of a diver. Is it the exact(i guess i mean to ask if this has been tested rigourously) same pressure there?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No.  The water at lower depths is holding up the water on top of it, so the pressure is necessarily higher.  There is, in fact, a pressure gradient of about 1/2 PSI per foot of depth.  So if you're six feet tall and floating upright, your feet experience 3PSI more than your head.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is why things float&lt;/strong&gt; -- put a buoyant thing into water, and the bottom surfaces see more force pushing up than the top surfaces get pushing down, and the net result is a force pushing up.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;And finally... As i understand it the pressure inside a closed diving bell is the same as the water pressure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That depends on the &quot;diving bell&quot;.  A true diving bell is open at the bottom, and pressurized so that the air doesn't all get crammed into the top.  But submarines and deep-sea submersibles are pressure vessels that keep their insides at 15PSI absolute through the strength of their hulls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But diving bell or submarine, yes, it's easier to work when you can just breath instead of using breathing apparatus.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-11-27T21:18:30.703" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32107" PostTypeId="2" ParentId="26527" CreationDate="2019-11-27T22:36:41.910" Score="0" Body="&lt;p&gt;I think the full answer is a little of both. When choosing to use CFD for any analysis there is always a trade-off between accuracy vs. cost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's take the rocket nozzle for example. The real physics of the problem include effects such as unsteadiness, turbulence which is inherently 3D, combustion, reacting chemicals for the combustion products etc. These can indeed be simulated with today's computers but as you add physics and complexity the computational cost rises. This sets the limit on the cost side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, let's say we really just want/need to understand if the nozzle is over/under expanded. Well then a simple 2D axisymmetric analysis (which is far cheaper computationally) will suffice. If we do care about things like how the reacting combustion products affect nozzle efficiency then those physics need to be included and consequently the computation cost rises as our level of accuracy increases. This sets the limit on the accuracy side. That is, what level of answer is &quot;good enough&quot;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rounding back to your question, it is a bit of both where the level of analysis is the result of the balance of cost and accuracy desired.&lt;/p&gt;&#xA;" OwnerUserId="8225" LastActivityDate="2019-11-27T22:36:41.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32108" PostTypeId="1" CreationDate="2019-11-28T06:26:23.993" Score="-1" ViewCount="29" Body="&lt;p&gt;i want to transmit 9v dc power overhead in 500 meter range. please tell me most suitable method for it. &#xA;i didn't tried it but i am making this system for operating 9 Volt router of wifi device.&lt;/p&gt;&#xA;" OwnerUserId="23688" LastEditorUserId="3505" LastEditDate="2019-11-28T07:11:44.617" LastActivityDate="2019-11-28T07:11:44.617" Title="Low voltage power transmission" Tags="|electrical|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32109" PostTypeId="2" ParentId="32108" CreationDate="2019-11-28T07:11:12.183" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;Measure the current required.&lt;/li&gt;&#xA;&lt;li&gt;Determine the minimum voltage (with safety margin) required to power the device correctly.&lt;/li&gt;&#xA;&lt;li&gt;Measure the output of your power supply.&lt;/li&gt;&#xA;&lt;li&gt;From 2 and 3 calculate the maximum voltage drop allowed.&lt;/li&gt;&#xA;&lt;li&gt;From 1 and 4 calculate the maximum cable resistance allowed.&lt;/li&gt;&#xA;&lt;li&gt;From 5 work out the wire size required. You can do this from cable tables or using an online calculator. Remember to check that you've included the return wire.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You may find that you require heavy, expensive cable. It may help you understand why power is transmitted at high voltage so that current can be reduced.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Please don't use block capitals in your question titles. It will get you some extra attention, but usually not the sort you want.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2019-11-28T07:11:12.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32110" PostTypeId="1" CreationDate="2019-11-28T17:11:11.763" Score="4" ViewCount="994" Body="&lt;p&gt;I have no background in thermodynamics or fluid dynamics, so please bear with me. Refer to the following figure:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/b4pOV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b4pOV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A working fluid with a flow rate of &lt;span class=&quot;math-container&quot;&gt;$1~kg/s$&lt;/span&gt; enters a heater at a pressure of &lt;span class=&quot;math-container&quot;&gt;$1~MPa$&lt;/span&gt; and temperature &lt;span class=&quot;math-container&quot;&gt;$150~^\circ C$&lt;/span&gt;, which raises its temperatures to &lt;span class=&quot;math-container&quot;&gt;$300~^\circ C$&lt;/span&gt;. It then enters a turbine, does work, loses pressure and temperature, and exits at &lt;span class=&quot;math-container&quot;&gt;$0.5~MPa$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$200~^\circ C$&lt;/span&gt; at the same flow rate. The bypass valve's purpose is to control the work output of the turbine by bypassing some of the working fluid from the inlet stream directly to the outlet stream. Suppose that we're free to make all the &quot;simplifying assumptions&quot; for this problem. Suppose also that initially the opening fraction of the bypass valve is &lt;span class=&quot;math-container&quot;&gt;$0\%$&lt;/span&gt; i.e., the valve is fully closed. My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How would the pressure and flow rate change as the bypass valve opening fraction is varied (&lt;span class=&quot;math-container&quot;&gt;$0\%$&lt;/span&gt; means fully closed, &lt;span class=&quot;math-container&quot;&gt;$100\%$&lt;/span&gt; means fully opened)? In particular, I'm interested to know them at points A, B, X, Y and E. &lt;/li&gt;&#xA;&lt;li&gt;Would a high pressure stream at Y intercepting the low pressure stream at E raise the pressure and flow rate of the latter? How could that be quantified?&lt;/li&gt;&#xA;&lt;li&gt;Could the valve's opening fraction be linearly related to flow rate and pressure? Meaning, could it be assumed that flow rate and pressure at point B would be halved if the valve was half-open (&lt;span class=&quot;math-container&quot;&gt;$50\%$&lt;/span&gt; opening fraction)? &lt;/li&gt;&#xA;&lt;li&gt;What would be the simplest way to model this valve? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Some questions might seem redundant, but I listed them to get a thorough response, for clarity and understanding. Also, all the numbers are arbitrary and any unreported quantities could be assumed. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="23691" LastEditorUserId="23691" LastEditDate="2019-11-29T13:29:45.637" LastActivityDate="2023-02-27T02:06:45.817" Title="Effect of bypass valve on flow rate and pressure" Tags="|fluid-mechanics|thermodynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32111" PostTypeId="1" CreationDate="2019-11-28T18:26:02.853" Score="2" ViewCount="71" Body="&lt;p&gt;I just started my BEng, so I actually don't feel ready to assess real cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Upper floor of my quite standard British terraced house.&#xA;We would like to hide an existing drainage pipe which is currently running along the wall.&#xA;To do that, I cannot find other ways but running it through the pictured floor joist (I already removed the plywood underfloor, as you can see).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/P6aIK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/P6aIK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Now, I am very wary of introducing possible structural weakenings when boring the joist: its sizes are 7cm (parallel to the floor) by 20cm (perpendicularly to the floor).&#xA;Since the outer diameter of the drain pipe is 43mm (even less if use solven weld), I should be fine according to the information I found from multiple sources on the internet (maximum 1/3 of the height of the joist, hole centered and away at least 2&quot; from the highest and lowest faces of the joist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'd really like to have further certainty for my specific case: maybe somebody who knows about these things could have a look at the picture and the data I provided and confirm this is the case?&#xA;Alternatively, should I hire a (real:) structural engineer for a proper consultation on this single issue? Or would that be overkill?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll be glad to hear anything you might have to suggest/explain/add.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="23694" LastActivityDate="2019-11-28T19:51:10.710" Title="Boring through joist" Tags="|structural-engineering|structures|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32112" PostTypeId="1" CreationDate="2019-11-28T18:26:54.907" Score="2" ViewCount="106" Body="&lt;p&gt;I only have these 3 data sets:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;rear right and left wheel speeds (they change with time)&lt;/li&gt;&#xA;&lt;li&gt;yaw rate (also changes with time)&lt;/li&gt;&#xA;&lt;li&gt;constant dynamic rolling radius&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How am I supposed to find the directional speeds of the vehicle? I know that yaw rate is equal to (the difference between the rear right and left wheel speeds)/(the distance between the rear wheels) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and&lt;/p&gt;&#xA;&#xA;&lt;p&gt;therefore the distance between the rear wheels should be equal to (the difference between the rear right and left wheel speeds)/(yaw rate) right? But when I work with the data sets and do this calculation, I see that the distance is not constant when it should be!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also, how to find the directional speeds with these given parameters only???&lt;/p&gt;&#xA;" OwnerUserId="23695" LastActivityDate="2019-12-04T17:40:41.473" Title="I have the yaw rate, rear wheel speeds, and the radius of tires, how to find x and y speeds?" Tags="|mechanical-engineering|control-engineering|kinematics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32114" PostTypeId="1" AcceptedAnswerId="32116" CreationDate="2019-11-28T19:33:24.537" Score="0" ViewCount="662" Body="&lt;p&gt;I am looking for borosilicate glass tubes which can be used to contain a vacuum, so the top is sealed and then there is a vacuum pump at the bottom where the edges are also sealed, effectively making a vacuum in a glass tube.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only thing to mention is that the walls need to be somewhat thicker than normal to protect against implosion due to the fact that there will be a vacuum inside, so I need an idea of how thick the glass needs to be bearing in mind that the inner diameter is 20 cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure outside the cylinder is atmospheric and the pressure inside is very low and can be taken to be 0.  Is there a formula or computation which would give me a good approximation for how thick the tube needs to be?  I also appreciate that there could be some shear stress due to the tube needing to be so thick.&lt;/p&gt;&#xA;" OwnerUserId="23697" LastActivityDate="2019-11-29T04:56:04.493" Title="Calculating the Thickness of a Glass Tube which contains a Vacuum" Tags="|pressure|stresses|vacuum|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32115" PostTypeId="2" ParentId="32111" CreationDate="2019-11-28T19:44:11.483" Score="3" Body="&lt;p&gt;You need to try to stay away from the joist supports. Because the shear is max there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would keep at least 15% of span away from the ends of the joist.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the notches in your photo to allow joist blocking need to be checked. Have they been over-cut, chipped corners, etc? is the wall double joisted? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you add two metal straps, say 12&quot; long with 10- 12D common nails near the top and bottom of the face of the joist over and under the pipe it adds strength. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-28T19:51:10.710" LastActivityDate="2019-11-28T19:51:10.710" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32116" PostTypeId="2" ParentId="32114" CreationDate="2019-11-28T23:24:27.823" Score="1" Body="&lt;p&gt;Let's call the diameter of your glass pipe, D, and assume the top and bottom of the tube are domes in a gentle transition from cylinder so we can ignor the local stress concentration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pressure, P, being radial around the pipe causes only compression and no moment or shear in the glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we cut the glass longitudinally we have two half cylinders such that the thickness, t, of the glass has to counter total pressure per unit length. Let's call the required stress of the glass, S.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t× S= P×D&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t= P×D/S= 1atm× D/S&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compression strength of normal glass is in the range of 1000psi, but it can vary depending on many factors so you check it, then you plug it in the above formula and use the required safety factor in your industry to get the thickness, t.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-11-29T04:56:04.493" LastActivityDate="2019-11-29T04:56:04.493" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32118" PostTypeId="1" CreationDate="2019-11-29T08:40:13.440" Score="0" ViewCount="1431" Body="&lt;p&gt;This is a follow up question of &lt;a href=&quot;https://engineering.stackexchange.com/questions/32099/rule-of-thumb-for-choosing-a-best-sampling-rate&quot;&gt;Rule of thumb for choosing a best sampling rate&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a discrete control system with a reference input. The goal is to keep the output close to the reference input (like a room temperature control system). The difference between the output and the reference input is the error fed back to the controller, forming a closed loop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system has a dead time, as well as a time constant, both at the scale of 1 to 10μs. I have found the open-loop transfer function &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt;, and derived that the bandwidth is at the scale of &lt;span class=&quot;math-container&quot;&gt;$10^4$&lt;/span&gt; rad/s. The fastest sampling interval of the output is at the sacle of 0.1μs.&#xA;The noisy of the input can be neglected, and the control accuracy is not very strict.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find out an optimal, or at least good, sampling interval. Till now I find two rules to choose the sampling interval:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Sample time should be at least 10 times smaller than the time constant, as I decribed in the previous question &lt;a href=&quot;https://engineering.stackexchange.com/questions/32099/rule-of-thumb-for-choosing-a-best-sampling-rate&quot;&gt;Rule of thumb for choosing a best sampling rate&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;The sampling frequency should be between 20 to 40 of the bandwidth, which can be found in many papers and books.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;As these two rules lead to different sampling interval (0.1μs to 1μs by rule 1 and ~10μs by rule 2), which rule should I take? &lt;/p&gt;&#xA;" OwnerUserId="23682" LastActivityDate="2019-11-29T08:40:13.440" Title="How to choose a best sampling rate, according to time constant or bandwidth?" Tags="|control-engineering|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="32119" PostTypeId="1" AcceptedAnswerId="32121" CreationDate="2019-11-29T09:52:00.197" Score="2" ViewCount="48" Body="&lt;p&gt;In my thesis, I deal with heating circuits, where several heat generators are present, e.g. a combined heat and power unit and a supporting boiler.&#xA;Usually, the boiler is placed in front of the heat storage tank, but I have also repeatedly seen heating circuits where the boiler is placed behind the heat storage tank. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the structure of the heating circuit of an &lt;a href=&quot;https://ieeexplore.ieee.org/document/6872090&quot; rel=&quot;nofollow noreferrer&quot;&gt;IEEE paper&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kviXO.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kviXO.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the advantages and disadvantages?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Do I understand correctly that if the boiler is in front of the heat storage tank, the water of the heat storage tank is heated and if it is behind it, it has its own reservoir of water which it heats?&lt;/p&gt;&#xA;" OwnerUserId="23439" LastActivityDate="2019-11-29T13:23:26.407" Title="Placement of supporting boiler in heating circuit" Tags="|heat-transfer|heating-systems|energy-storage|boilers|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32120" PostTypeId="2" ParentId="26628" CreationDate="2019-11-29T11:34:48.007" Score="1" Body="&lt;p&gt;The positive terminal is the leg connected to the led diode case, which is in contact with the metallic led holder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the board this leg is soldered on the opposite side of the components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More details here: &lt;a href=&quot;https://www.datasheetarchive.com/pdf/download.php?id=3986888527796ddb5524529252bea59f4ce640&amp;amp;type=P&amp;amp;term=Kingbor%2520LED&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.datasheetarchive.com/pdf/download.php?id=3986888527796ddb5524529252bea59f4ce640&amp;amp;type=P&amp;amp;term=Kingbor%2520LED&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23701" LastActivityDate="2019-11-29T11:34:48.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32121" PostTypeId="2" ParentId="32119" CreationDate="2019-11-29T13:23:26.407" Score="2" Body="&lt;p&gt;In the diagram shown, then if the water temperature of the storage tank falls below the set value, the aux boiler comes in to augment the water temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other possibilities include the use of a diverter valve so when the tank water temperature is too low, the valve changes to demand water heated by the boiler, if that boiler has a rapid response then direct, if not then it may well have a storage tank of its own.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-11-29T13:23:26.407" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32122" PostTypeId="2" ParentId="32110" CreationDate="2019-11-30T00:58:29.613" Score="1" Body="&lt;p&gt;I'll not get too deep, because my hidraulics basis is truly shallow.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Assuming all your pipes have the same section (possibiliting the same&#xA;flow rate of 1kg/s in all points).&lt;/li&gt;&#xA;&lt;li&gt;Disconsidering the termal effects, because the heated liquid will&#xA;expand and increase the pressure in the local (not too sure about&#xA;this part).&lt;/li&gt;&#xA;&lt;li&gt;Disconsidering possible negative effects in the case of reverse flow&#xA;of the fluid.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Your by-pass valve will control the flow (from 0% to 100%).&#xA;Like on eletrical sistem the fluid will take the easiest path everytime.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;How would the pressure and flow rate change as the bypass valve opening fraction is varied (0% means fully closed, 100% means fully opened)? In particular, I'm interested to know them at points A, B, X, Y and E. &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Your flow will not change if your section not change (here's exactly what the valve does).&lt;br&gt;&#xA;The pressure is fairly constant (not considering the termal effects). So, if your entry point is limited at 1MPa your exit point is limited the same (disconsidering the turbine that &quot;steals&quot; pressure in the sistem)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's simulate some cases:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1º&lt;/strong&gt; - Your valve is totally closed (0%). The readings will be:&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;A&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 1MPa &lt;del&gt;150º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;B&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 1MPa &lt;del&gt;150º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;C&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 1MPa &lt;del&gt;300º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;D&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 0,5MPa &lt;del&gt;200º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;E&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 0,5MPa &lt;del&gt;200º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;X&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 1MPa &lt;del&gt;150º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;Y&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 0,5MPa &lt;del&gt;200º&lt;/del&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2º&lt;/strong&gt; - Your valve is half open (50%). The readings will be:&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;A&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 1MPa &lt;del&gt;150º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;B&lt;/em&gt;&lt;/strong&gt; - 0,5Kg/s 1MPa &lt;del&gt;150º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;C&lt;/em&gt;&lt;/strong&gt; - ~0,5Kg/s 1MPa &lt;del&gt;300º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;D&lt;/em&gt;&lt;/strong&gt; - ~0,5Kg/s ~0,6MPa &lt;del&gt;200º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;E&lt;/em&gt;&lt;/strong&gt; - 1Kg/s ~0,6MPa &lt;del&gt;200º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;X&lt;/em&gt;&lt;/strong&gt; - 0,5Kg/s 1MPa &lt;del&gt;150º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;Y&lt;/em&gt;&lt;/strong&gt; - 0,5Kg/s ~0,6MPa &lt;del&gt;200º&lt;/del&gt;&lt;br&gt;&#xA;&lt;strong&gt;Obs.:&lt;/strong&gt; The aproximate values here need some real calculation for the true value, i've just put some numbers that I think resemble what should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3º&lt;/strong&gt; - Your valve is totally open (100%). The readings will be:&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;A&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 1MPa &lt;del&gt;150º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;B&lt;/em&gt;&lt;/strong&gt; - 0Kg/s (or close to this) 1MPa &lt;del&gt;150º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;C&lt;/em&gt;&lt;/strong&gt; - 0Kg/s (or close to this) 1MPa &lt;del&gt;300º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;D&lt;/em&gt;&lt;/strong&gt; - 0Kg/s (or close to this) 1MPa &lt;del&gt;200º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;E&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 1MPa &lt;del&gt;200º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;X&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 1MPa &lt;del&gt;150º&lt;/del&gt;&lt;br&gt;&#xA;&lt;em&gt;point &lt;strong&gt;Y&lt;/em&gt;&lt;/strong&gt; - 1Kg/s 1MPa &lt;del&gt;200º&lt;/del&gt;  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Would a high pressure stream at Y intercepting the low pressure stream at E raise the pressure and flow rate of the latter? How could that be quantified?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, there will be a pressure change at E, but this is affected more because of the flow from Y, and not because of the pressure.&#xA;Btw. The flow at E will be exactly the same unless your sistem lose/gain external flow from another point or from a pump.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;Could the valve's opening fraction be linearly related to flow rate and pressure? Meaning, could it be assumed that flow rate and pressure at point B would be halved if the valve was half-open (50% opening fraction)?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The flow rate will be linearly related at point B (not considering the termal effects). But the pressure will remain the same.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;4&quot;&gt;&#xA;  &lt;li&gt;What would be the simplest way to model this valve? &lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not so sure what you want with this question. By my understanding, you're asquing what is the simplest connection method to get a linear response in your turbine.&lt;br&gt;&#xA;And my sincere answer in this case will be, &lt;em&gt;&quot;I don't know&quot;&lt;/em&gt;. In my understanding the point you've chosen is the best point (no personal experience, so take with a grain of salt).&lt;/p&gt;&#xA;" OwnerUserId="23658" LastActivityDate="2019-11-30T00:58:29.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32123" PostTypeId="2" ParentId="32102" CreationDate="2019-11-30T01:52:45.590" Score="3" Body="&lt;p&gt;Assume that high temperature causes no chemical reactions in the contents or between the contents and the can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High temperature will increase the pressure in the can. Assume the can is fully unaffected in its mechanical integrity up to the point that it fails. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all of the above, when you get the can back at room temperature, it will have no change in its potential to fail or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One weak link will be the potential for increased fatigue at the seams on the can as the contents go to high pressure and back to lower pressure. Another weak link will be in the potential for over-stress on the plastic cap seal on the top of the container.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the contents react at high temperature and that temperature is exceeded, all bets are off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In summary, this system affords no absolute guarantee that it will have no increased probability to fail when temperature cycles to a high value and returns to room temperature. The biggest uncertainty is what exactly is meant by high temperature (i.e. 100 &amp;deg;C or 500 &amp;deg;C) and for how long (i.e. immediately or for a few minutes or for a few days). The practical issue is whether and to what extent your shipment will reach the bounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on what is stated on the Website, the failure seems to be immediate at a point at or just above 50 &amp;deg;C. I would have to say that whatever company is ready to accept these cans knows that they have to keep them below that temperature during shipping. But, what you might want to know is whether they were at a high temperature just below that for an extended time,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I might suggest to mark the cans before shipment with a paint or dye or wax that changes color irreversibly when a certain temperature is exceeded. Or include a temperature probe with the shipment that tracks temperature with time over the entire shipment. Then, you can decide at the receiving end of the shipment whether the cans where shipped to your satisfaction.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2019-12-01T20:26:42.417" LastActivityDate="2019-12-01T20:26:42.417" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32124" PostTypeId="1" CreationDate="2019-11-30T03:13:56.200" Score="0" ViewCount="14772" Body="&lt;p&gt;I have a second order system with open-loop transfer function&#xA;&lt;span class=&quot;math-container&quot;&gt;$G(s)={a\over(s+b)(s+c)}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$a,b,c$&lt;/span&gt; are already known.&#xA;This system has unity feedback.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am interested in the time consant of this system, but I need to know should I calculate based on the open-loop or closed-loop transfer function?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second question is how to calculate the time consant of a second order system? On this webpage (&lt;a href=&quot;https://apmonitor.com/pdc/index.php/Main/SecondOrderSystems&quot; rel=&quot;nofollow noreferrer&quot;&gt;Second Order Systems&lt;/a&gt;), it says a second order system may be the combination of two first order systems. If the time constant for the two first order system is &lt;span class=&quot;math-container&quot;&gt;$\tau_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\tau_2$&lt;/span&gt;, the time constant for the second order system is &lt;span class=&quot;math-container&quot;&gt;$\tau^2=\tau_1 \tau_2$&lt;/span&gt;. Can I compute according to this equation?&lt;/p&gt;&#xA;" OwnerUserId="23682" LastActivityDate="2020-04-29T11:52:55.130" Title="Time constant of second order system" Tags="|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32125" PostTypeId="1" AcceptedAnswerId="32135" CreationDate="2019-11-30T03:46:09.470" Score="0" ViewCount="56" Body="&lt;p&gt;Looking for a material that:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is transparent/clear - the clearer the better&lt;/li&gt;&#xA;&lt;li&gt;Can be made to about 80cm long x 8cm wide by somewhere between 2-5cm thick (likely 2-3cm, exact details will depend on the material's properties)&lt;/li&gt;&#xA;&lt;li&gt;Can be shaped on the cross section, either during manufacturing, or after&lt;/li&gt;&#xA;&lt;li&gt;Is at least a bit flexible - I'd estimate no more than 20 degree bend along the length, without too much force required - i.e. something that could be bent by hand, like a plastic ruler.&lt;/li&gt;&#xA;&lt;li&gt;Has a refractive index for visible light of 1.5 or above (higher is better)&lt;/li&gt;&#xA;&lt;li&gt;Is cheap, or at least not too expensive&lt;/li&gt;&#xA;&lt;li&gt;Is resistant to chipping or snapping&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Can anyone help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Partial matches to the above might be welcomed, but it would be useful to know why they are only partial matches. The key points are (1), (2), (3), (4) - other than those, the higher refractive index the better, the cheaper the better, the more resistant to chipping and snapping the better, but those are continuums rather than fixed points! The method of shaping also matters, ie something that can theoretically be shaped but in practice requires a really expensive or rare process, or hand shaping, is not practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I have access to injection molded Polyester resin, but I understand that might not be flexible.&lt;/p&gt;&#xA;" OwnerUserId="23706" LastEditorUserId="23706" LastEditDate="2019-11-30T16:30:18.793" LastActivityDate="2019-11-30T16:54:06.467" Title="Seeking transparent, somewhat flexible material" Tags="|materials|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32127" PostTypeId="1" CreationDate="2019-11-30T12:27:18.947" Score="0" ViewCount="201" Body="&lt;p&gt;Mechanics books always highlight that the main advantage of bearings is to reduce friction, especially when we talk about them in rotational motion applications. The example of ball bearings is always mentioned. The balls prevent and regulate the contact between the inner and the outer rings in which they reduce the friction &lt;strong&gt;between the two&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would that be the main philosophy behind bearings? If that is true, then would not be better off just to remove the whole setup and have the axis of rotation free from bearings so that there are not any inner or outers rings and therefore the friction problem would be removed from the beginning? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I always see that the friction treatment is secondary to a main purpose of bearings ( that I have not realized yet ). Should not be there a more fundamental reason of bearings? &lt;/p&gt;&#xA;" OwnerUserId="2467" LastActivityDate="2019-11-30T15:15:42.630" Title="What is the exact concept of bearings?" Tags="|mechanical-engineering|bearings|friction|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32128" PostTypeId="2" ParentId="32124" CreationDate="2019-11-30T14:12:17.837" Score="0" Body="&lt;p&gt;The equation you give, &lt;span class=&quot;math-container&quot;&gt;$\tau^2 = \tau_1\tau_2$&lt;/span&gt; is on the web page you linked to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However you need to understand that &lt;span class=&quot;math-container&quot;&gt;$\tau_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\tau_2$&lt;/span&gt; are &lt;strong&gt;&lt;em&gt;first order&lt;/em&gt;&lt;/strong&gt; time constants but &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; is a &lt;strong&gt;&lt;em&gt;second order&lt;/em&gt;&lt;/strong&gt; time constant. The first and second order constants do not have exactly the same physical interpretation, and they are not even in the same physical units!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-11-30T14:12:17.837" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32129" PostTypeId="1" CreationDate="2019-11-30T14:15:03.613" Score="0" ViewCount="4564" Body="&lt;p&gt;I know this is a basic question but google only gives formulas for more &quot;rounded&quot; shapes. How to calculate the minimal thickness for a flat wall knowing tensile strength of the material and the maximum pressure difference between two sides? &lt;/p&gt;&#xA;" OwnerUserId="18744" LastActivityDate="2022-10-06T17:16:35.063" Title="How to determine minimal wall thickness for a rectangular pressure vessel?" Tags="|pressure-vessel|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32130" PostTypeId="1" CreationDate="2019-11-30T14:51:01.833" Score="2" ViewCount="181" Body="&lt;p&gt;While watching a video of somebody flying an airplane over North Korea, I noticed in a river these two &quot;water fountains&quot;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bP3HJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bP3HJ.jpg&quot; alt=&quot;screenshot&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I vaguely recall seeing this previously, probably in the US, as well. My question is: What are they for? Just ornamental? Or do they have some kind of practical purpose?&lt;/p&gt;&#xA;" OwnerUserId="23711" LastActivityDate="2023-07-12T23:43:46.623" Title="What are these &quot;water fountains&quot; in the middle of a river in North Korea? (Probably also found elsewhere.)" Tags="|structural-engineering|civil-engineering|water-resources|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32131" PostTypeId="2" ParentId="32127" CreationDate="2019-11-30T15:13:43.790" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;so that there are not any inner or outers rings and therefore the friction problem would be removed from the beginning&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The axis needs to be supported somehow (i.e. in direct contact with the support) or it would be &quot;floating in space&quot;. As the later is impossible, you are forced to have the first. Usually you provide a lubricant between axis and support surfaces, but that lubricant won't be there in all possible operating scenarios, particularly during starting and stopping stages, effectively &lt;em&gt;increasing friction&lt;/em&gt;. There's also the problem of having a lubricant pressure big enough to overcome the forces at play, so the film of lubricant is as thick as needed (under dynamic loads). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So bearings are there to reduce friction wear to the axis or its support. The fact that they have their own friction is unavoidable, but something you can mitigate through careful design. &lt;/p&gt;&#xA;" OwnerUserId="17733" LastActivityDate="2019-11-30T15:13:43.790" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32132" PostTypeId="2" ParentId="32127" CreationDate="2019-11-30T15:15:42.630" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If that is true, then would not be better off just to remove the whole setup and have the axis of rotation free from bearings ... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Ball bearings change a system from sliding friction (as in a journal bearing) to rolling friction which is usually considerably less.&lt;/li&gt;&#xA;&lt;li&gt;Ball bearings prevent wear in one place as would occur with a horizontal rotating axis at the the point of support.&lt;/li&gt;&#xA;&lt;li&gt;Ball bearings can provide holding force in the axial direction - a drill-press for example - while providing lower rotational resistance than a plain bearing.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The term &quot;bearing&quot; is derived from the verb &quot;to bear&quot;; a bearing being a machine element that allows one part to support another. &lt;em&gt;Source: Wikipedia, &lt;a href=&quot;https://en.wikipedia.org/wiki/Bearing_(mechanical)&quot; rel=&quot;noreferrer&quot;&gt;Bearing (mechanical)&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2019-11-30T15:15:42.630" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32133" PostTypeId="1" CreationDate="2019-11-30T16:25:48.960" Score="1" ViewCount="11" Body="&lt;p&gt;How can I encode a quantized vector in matlab ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a part of a problem where I am asked to implement a communication block using matlab. &lt;/p&gt;&#xA;" OwnerUserId="23713" LastActivityDate="2019-11-30T16:25:48.960" Title="how to implement an encoder in matlab" Tags="|telecommunication|matlab|communication|digital-communication|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32134" PostTypeId="1" CreationDate="2019-11-30T16:52:43.167" Score="1" ViewCount="21" Body="&lt;p&gt;Can someone help with this question ? &#xA;it's a thermodynamic question based upon Sonttage book 7th edition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*)the question states :- &lt;/p&gt;&#xA;&#xA;&lt;p&gt;R-152a with a volume of 15 L and at 36 Celsius degrees with an equality of 80 % &#xA;can some one help to find the mass of the substance . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and there is another thing i need help with, (*when it comes to thermodynamic questions which includes springs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(*)What is the balance point of a spring in a spring - piston -cylinder system &lt;/p&gt;&#xA;&#xA;&lt;p&gt;appreciate your help guys  &lt;/p&gt;&#xA;" OwnerUserId="23714" LastActivityDate="2019-11-30T16:52:43.167" Title="Thermodynamics equality question" Tags="|thermodynamics|" AnswerCount="0" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32135" PostTypeId="2" ParentId="32125" CreationDate="2019-11-30T16:54:06.467" Score="1" Body="&lt;p&gt;There are optically clear silicone products often used for encapsulation. These are usually two part mixtures that are then cast. They are quite flexible. I doubt they are machinable so you would have to cast to shape using a form or mold of some kind. &lt;a href=&quot;https://www.quantumsilicones.com/2012/08/qsil-229lv/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; is an example, but there are many others. Just do a &lt;a href=&quot;https://www.google.com/search?q=optically%20clear%20silicone&amp;amp;oq=optically%20clear%20sili&amp;amp;aqs=chrome.0.0j69i57j0l4.5251j1j4&amp;amp;sourceid=chrome&amp;amp;ie=UTF-8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google search&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also found optically clear urethanes. &lt;a href=&quot;https://www.smooth-on.com/product-line/clear-flex/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is an example&lt;/a&gt;. Urethanes can be found with a range of stiffnesses.&lt;/p&gt;&#xA;" OwnerUserId="9506" LastActivityDate="2019-11-30T16:54:06.467" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32136" PostTypeId="2" ParentId="32129" CreationDate="2019-11-30T18:42:37.380" Score="2" Body="&lt;p&gt;The reason that you can't find anything on Google is that it is a bad idea to make a rectangular pressure vessel. You get stress concentrations at the corners which makes the vessel much weaker. This is bad because when pressure vessels fail, they release a huge amount of energy very suddenly. It only takes a little googling to find examples of fatalities due to pressure vessel failures. Pressure vessel design should not be attempted by inexperienced people. Hire a professional or just buy one off the shelf. &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-11-30T18:42:37.380" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32137" PostTypeId="2" ParentId="32129" CreationDate="2019-11-30T18:55:40.827" Score="0" Body="&lt;p&gt;There are two ways to deal with this question and both have to do with the deflection of plates (sides and top and bottom of the container) under the pressure load. The first alternative is to demand small deflection,1/2t, the second is allowing for large deflections say up to 20t to save on the material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the supports are the corners of the container and the can be approximated as a simple support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These have discussion and formulas among other references in Roark’s Formulas&#xA;for Stress and Strain&#xA;WARREN C. YOUNG&#xA;RICHARD G. BUDYNAS&#xA;Seventh Edition&#xA;McGraw-Hill pp 451 through 473. 11.10 Bending of Uniform-Thickness Plates with&#xA;Straight Boundaries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I quote the intro into the rectangular plates, but you got to search through the text to find your case. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In Ref. 35 experimentally determined deflections are given and&#xA;  compared with those predicted by theory. In Ref. 74 a numerical&#xA;  solution for uniformly loaded rectangular plates with simply&#xA;  supported edges is discussed, and the results for a square plate are&#xA;  compared with previous approximate solutions. Graphs are presented&#xA;  to show how stresses and deflections vary across a square plate.&#xA;  Chia in Ref. 91 includes a chapter on moderately large deflections of&#xA;  isotropic rectangular plates. Not only are the derivations presented&#xA;  but the results of most cases are presented in the form of graphs&#xA;  usable for engineering calculations. Cases of initially deflected plates&#xA;  are included, and the comprehensive list of references is useful.&#xA;  Aalami and Williams in Ref. 92 present 42 tables of large-deflection&#xA;  reduction coefficients over a range of length ratios a=b and for a&#xA;  variety—three bending and four membrane—of symmetric and&#xA;  nonsymmetric boundary conditions. Loadings include overall uniform&#xA;  and linearly varying pressures as well as pressures over limited areas&#xA;  centered on the plates.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-11-30T18:55:40.827" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32139" PostTypeId="2" ParentId="32097" CreationDate="2019-11-30T21:26:08.220" Score="0" Body="&lt;p&gt;No. The nickel sulfate hydrate needs to be dissolved in water so that you have 20g per litre. &#xA;The ammonia-water solution needs to be present as 500ml per litre. The 25% is just relating to the ammonia concentration to use (33% being the maximum at RTP).&lt;br&gt;&#xA;The ratio to use should be in your article as it looks like you're making an ammoniacal nickel precipitate.&#xA;However your question was using this example to obtain a generic understanding.  If you see a % sign, if it's telling you how to make a concentration,  it will usually have w/w, v/v or w/v after it(weight-weight, volume-volume and weight-volume respectively). &#xA;For this sort of question, I would post in Chemistry. &lt;/p&gt;&#xA;" OwnerUserId="11590" LastActivityDate="2019-11-30T21:26:08.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32140" PostTypeId="1" CreationDate="2019-11-30T23:08:56.007" Score="1" ViewCount="954" Body="&lt;p&gt;Let's assume a DMM is specified to have accuracy &quot;0.1% FS +/-0.02&quot; when used on the 50 V range. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I get that 0.1% FS means the true value will lie somewhere between measured value - 0.05 and measured value + 0.05. But what does +/- 0.02 mean here?&lt;/p&gt;&#xA;" OwnerUserId="23719" LastActivityDate="2019-12-02T10:33:46.067" Title="what does 0.1% FS +/- 0.02 mean?" Tags="|measurements|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32143" PostTypeId="2" ParentId="31698" CreationDate="2019-12-01T03:59:14.607" Score="0" Body="&lt;p&gt;To facilitate my explanation I'll follow the same scenario and the following definitions:&lt;br&gt;&#xA;I'll consider all explanation considering the car doing a left curve.&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$W_{rL} =$&lt;/span&gt; For Left Rear Wheel&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$W_{rR} =$&lt;/span&gt; For Right Rear Wheel&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$WD = $&lt;/span&gt; for Welded differential&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$LD = $&lt;/span&gt; for Limited-slip differential (btw Torsen is a type of &lt;span class=&quot;math-container&quot;&gt;$LD$&lt;/span&gt;)&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$OD = $&lt;/span&gt; for Open differential&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$r_v = $&lt;/span&gt; for rotation speed on the wheel&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$L_F = $&lt;/span&gt; for Lateral force (byproduct of the mass of the vehicle)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you start cornering what makes the car slip (drifting) is the resultant force from the sum of the force from rotation of the wheel and the lateral force resultant from the mass of the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the normal cornering (without drift) the &lt;span class=&quot;math-container&quot;&gt;$W_{rL}$&lt;/span&gt; will rotate slower than the &lt;span class=&quot;math-container&quot;&gt;$W_{rR}$&lt;/span&gt;. And the lateral force will be concentrated on the right side of the car.&lt;br&gt;&#xA;The &lt;span class=&quot;math-container&quot;&gt;$W_{rR}$&lt;/span&gt; will be subjected to more lateral force than &lt;span class=&quot;math-container&quot;&gt;$W_{rL}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &lt;span class=&quot;math-container&quot;&gt;$OD$&lt;/span&gt; the car will first accelerate &lt;span class=&quot;math-container&quot;&gt;$W_{rR}$&lt;/span&gt; until the wheel start slipping. On this moment almost all the power of the car goes to &lt;span class=&quot;math-container&quot;&gt;$W_{rR}$&lt;/span&gt; and your car will desacelerate until the slipping stops (and if you continue accelerating) there will be a loop without never slipping on &lt;span class=&quot;math-container&quot;&gt;$W_{rL}$&lt;/span&gt;, and so &lt;em&gt;no driftting here&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &lt;span class=&quot;math-container&quot;&gt;$WD$&lt;/span&gt; the car will start the drifting with the same speed on both wheels. In this case the minimum speed to get a true drift will need to surpass the minimum threshold from &lt;span class=&quot;math-container&quot;&gt;$W_{rL}$&lt;/span&gt;. But the &lt;span class=&quot;math-container&quot;&gt;$W_{rR}$&lt;/span&gt; will rotate with the same speed from &lt;span class=&quot;math-container&quot;&gt;$W_{rL}$&lt;/span&gt; (&lt;span class=&quot;math-container&quot;&gt;$r_v$&lt;/span&gt; is distributed 50% on &lt;span class=&quot;math-container&quot;&gt;$W_{rL}$&lt;/span&gt; and 50% on &lt;span class=&quot;math-container&quot;&gt;$W_{rR}$&lt;/span&gt;). But our lateral force is not the same with let's say &lt;span class=&quot;math-container&quot;&gt;$L_F$&lt;/span&gt;=20% on &lt;span class=&quot;math-container&quot;&gt;$W_{rL}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$L_F$&lt;/span&gt;=80% on &lt;span class=&quot;math-container&quot;&gt;$W_{rR}$&lt;/span&gt; (the correct &lt;span class=&quot;math-container&quot;&gt;$L_F$&lt;/span&gt; will vary acording with entry speed on curve and curve angle).&lt;br&gt;&#xA;The result in this case is that with &lt;span class=&quot;math-container&quot;&gt;$WD$&lt;/span&gt; the tire grip on &lt;span class=&quot;math-container&quot;&gt;$W_{rL}$&lt;/span&gt; is better than the grip on &lt;span class=&quot;math-container&quot;&gt;$W_{rR}$&lt;/span&gt; because of the inbalance of the sum of the forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With &lt;span class=&quot;math-container&quot;&gt;$LD$&lt;/span&gt; the car will start drifting first on &lt;span class=&quot;math-container&quot;&gt;$W_{rR}$&lt;/span&gt; like with &lt;span class=&quot;math-container&quot;&gt;$OD$&lt;/span&gt; &lt;strong&gt;but&lt;/strong&gt; our &lt;span class=&quot;math-container&quot;&gt;$LD$&lt;/span&gt; will garantee that some power remains in &lt;span class=&quot;math-container&quot;&gt;$W_{rL}$&lt;/span&gt; and the speed keeps increasing until &lt;span class=&quot;math-container&quot;&gt;$W_{rL}$&lt;/span&gt; starts slipping too.&lt;br&gt;&#xA;The moment the drifting start, the &lt;span class=&quot;math-container&quot;&gt;$LD$&lt;/span&gt; will distribute the &lt;span class=&quot;math-container&quot;&gt;$r_v$&lt;/span&gt; on both wheels to get a balance in the resultant force. So the values now will be something close to:&lt;br&gt;&#xA; - on &lt;span class=&quot;math-container&quot;&gt;$W_{rL}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$L_F$&lt;/span&gt;=20% + &lt;span class=&quot;math-container&quot;&gt;$r_v$&lt;/span&gt;=80%&lt;br&gt;&#xA; - on &lt;span class=&quot;math-container&quot;&gt;$W_{rR}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$L_F$&lt;/span&gt;=80% + &lt;span class=&quot;math-container&quot;&gt;$r_v$&lt;/span&gt;=20%&lt;br&gt;&#xA;With this combination the grip on the wheels will be the same and no lost power like with &lt;span class=&quot;math-container&quot;&gt;$W_{rR}$&lt;/span&gt; like with the &lt;span class=&quot;math-container&quot;&gt;$WD$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry by the lack of images to help the understanding.&lt;br&gt;&#xA;I'm new with stackexchange and I need to fiddle more with the commands to learn how to create some helping image and put here.&lt;/p&gt;&#xA;" OwnerUserId="23658" LastActivityDate="2019-12-01T03:59:14.607" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32144" PostTypeId="1" CreationDate="2019-12-01T04:26:31.243" Score="3" ViewCount="170" Body="&lt;p&gt;Buckminster Fuller has popularized &lt;strong&gt;Geodesic Domes&lt;/strong&gt; (invented by  &lt;em&gt;Dr. Walther Bauersfeld&lt;/em&gt; who based it on the use of convex polyhedrons), and thus, making buildings with a very large roofs that don't need pillars. &#xA;In popular science magazines it's said that if you just make the Geodesic Dome large enough, it would float by it self because the air pressure underneath the dome would be bigger than the weight pressing it down, and so, it would float.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: &#xA;&lt;em&gt;Has there ever been an serious attempt to prove that this was &lt;strong&gt;theoretically&lt;/strong&gt; or even &lt;strong&gt;practically&lt;/strong&gt; possible?&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the sense that has there ever been a serious paper published about this, at a university or institute? &lt;/p&gt;&#xA;" OwnerUserId="23722" LastEditorUserId="23726" LastEditDate="2019-12-01T21:20:19.500" LastActivityDate="2024-02-01T06:02:10.283" Title="Is a Geodesic Dome (fuller dome) that floats possible? What material or engineering effort would it take?" Tags="|mechanical-engineering|structural-engineering|civil-engineering|applied-mechanics|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32145" PostTypeId="2" ParentId="31669" CreationDate="2019-12-01T05:05:06.423" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have to be like 2 meters away from the LED to focus the LED light with a fresnel lens. Why is that?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are building an imaging system that images the LED onto another plane:  &lt;a href=&quot;http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lensdet.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lensdet.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance you need to be away from the LED depends on the focal length of your lens and the distance from the lens to where you are forming the image.  If you want to have the source closer, either get a shorter focal length lens, or move the plane you're focusing on further away.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a way to focus the all of the LED's light somehow onto a small area?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It depends on your definition of small.  The above link gives the equation for magnification of your image.  To have a lot of energy focused tightly, you'll need a magnification &amp;lt;&amp;lt; 1, which means putting the lens far from the source and close to the image.  Note that the smaller the source of the light, the smaller the area a given lense will be able to focus it down to.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If it's possible, would it be akin to having a 300W laser? Mostly likely not, but why?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It would not because lasers emit energy more densely than LEDs, meaning for a given lens, you will have a more intense spot than for an LED.  For focusing energy, there are two classes of laser that are relevant, single mode and multimode.  A single mode laser acts as if all of its energy is emitted from a single infinitely small point, and therefore when you image it down you can form an image a wavelength wide. A multimode laser appears to emit energy from a series of adjacent points, thus when focused down it behaves more like an LED, and cannot form a spot as small as a single mode laser.  &lt;/p&gt;&#xA;" OwnerUserId="23724" LastActivityDate="2019-12-01T05:05:06.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32148" PostTypeId="1" CreationDate="2019-12-01T12:34:06.470" Score="0" ViewCount="34" Body="&lt;p&gt;I want to control the flow of chemical proportionate to the forward speed of the vehicle( if the speed is more discharge is more vice-versa). Need to integrate vehicle speed with encoder and control the solenoid valve.&#xA;Is it possible to change duty cycle of PWM with respect to vehicle forward speed?&lt;/p&gt;&#xA;" OwnerUserId="23727" LastActivityDate="2019-12-01T12:34:06.470" Title="PWM -electronic solenoid valve" Tags="|electrical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32150" PostTypeId="1" AcceptedAnswerId="32289" CreationDate="2019-12-01T16:13:46.457" Score="0" ViewCount="53" Body="&lt;p&gt;Please explain the meaning of this statement with close reference to Precipitation hardening,&quot;The relative rates of nucleation and growth are controlled by temperature. The size and dispersion of the secondary phase is controlled by time, with coarsening occurring as time proceeds.&quot;What I do not understand is the difference between the usage of &quot;growth&quot; and &quot;size&quot; as different parameters. They say they are dependent on temperature and time respectively.Arent they basically the two sides of the same coin.Please explain me with close reference the the precipitation hardening in Al-4.5wt%Cu alloy.&lt;/p&gt;&#xA;" OwnerUserId="19123" LastActivityDate="2019-12-09T15:18:14.417" Title="Precipitation Hardening" Tags="|mechanical-engineering|materials|metallurgy|material-science|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32151" PostTypeId="1" CreationDate="2019-12-01T16:42:57.317" Score="1" ViewCount="680" Body="&lt;p&gt;I know, from Lyapunov criteria, that a system is stable (not asymptotically) if the system has eigenvalues with negative real part or it has eigenvalues with real part equals to zero, but in this case the algebraic multiplicity must be equal to the geometric multiplicity.&#xA;I don't understand why there's this condition when the real part is equal to zero. &lt;/p&gt;&#xA;" OwnerUserId="23729" LastActivityDate="2019-12-01T17:26:02.243" Title="Stability with eigenvalues' real part equal to zero" Tags="|control-theory|stability|eigenvalue-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32152" PostTypeId="2" ParentId="32151" CreationDate="2019-12-01T17:26:02.243" Score="2" Body="&lt;p&gt;If the algebraic multiplicity of an eigenvalue is greater than the geometric multiplicity, then the system matrix is not diagonalizable and there are vectors which are not linear combinations of the eigenvectors of the matrix.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you interpret the Lyapunov matrix as a &quot;generalized potential energy function&quot; (but not necessarily the real physical potential energy of the system) it should be clear why having a possible motion of the system which is not included in subspace spanned by the eigenvectors is going to be a problem, when trying to prove or disprove stability.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-12-01T17:26:02.243" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32153" PostTypeId="1" CreationDate="2019-12-01T18:28:52.657" Score="3" ViewCount="133" Body="&lt;p&gt;Typically the word &quot;journal&quot; refers to a collection of written records, experiences, or transactions. However, there is a common type of bearing known as a sleeve bearing, radial bearing, or journal bearing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The terms &quot;sleeve&quot; or &quot;radial&quot; are self-explanatory, but I'm interested in knowing more about the use of the term &quot;journal&quot; for this type of bearing. What is the historical (or mechanical) reason for this naming convention?&lt;/p&gt;&#xA;" OwnerUserId="11602" LastActivityDate="2019-12-01T18:28:52.657" Title="Etymology of the term &quot;journal bearing&quot;" Tags="|mechanical-engineering|motors|bearings|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32154" PostTypeId="1" CreationDate="2019-12-01T19:53:51.120" Score="1" ViewCount="26" Body="&lt;p&gt;I am dealing with mixed sensitivity controller design in SISO systems and I am comparing this approch with the classic approch. In particular, I have placed a weight for the sensitivity function in order to reach specific desired performances. After doing this, I have plotted the frequency responce of the control effort, and I have found this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LQQku.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LQQku.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so, what I have found is that the control effort when using the weight is higher the when we don't use any weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven' t found anything about why the control effort should be higher, but my opinion is that this happens because with a weight the controller has to do more work to force the system to have specific specifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this is just my opinion, because I haven't found any reference about this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone explain why the control effort is higher? Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT] To clarify what I am doing is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/glkdA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/glkdA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so I am placing a limitation on the bandwidth, on the resonance peak and on the low frequency, and this limitation is given by the red line, which is my weight, which I tuned with the desired specifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the plant is not non-minimum phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT 2] I can also plot the step response of the control effort:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/74hdc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/74hdc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;and also from here is evident that the control effort is higher.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But also I can see that both start from zero, and in particular, the blu line, so the one which identifies the mixed sensitiviy approch remains at a constant value in the step response.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why this happens? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="22780" LastEditorUserId="22780" LastEditDate="2019-12-02T18:12:49.460" LastActivityDate="2019-12-02T18:12:49.460" Title="Control effort in mixed sensitivity design" Tags="|electrical-engineering|control-engineering|control-theory|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32155" PostTypeId="2" ParentId="32130" CreationDate="2019-12-01T21:43:41.957" Score="3" Body="&lt;p&gt;They are merely ornamental.&lt;/p&gt;&#xA;&lt;p&gt;Near the top of the image you can see the base of &lt;a href=&quot;https://en.wikipedia.org/wiki/Juche_Tower&quot; rel=&quot;nofollow noreferrer&quot;&gt;Juche Tower&lt;/a&gt; (&lt;a href=&quot;https://goo.gl/maps/ogQ3B7AJQ8pQziAR8&quot; rel=&quot;nofollow noreferrer&quot;&gt;Google Maps&lt;/a&gt;), a monument in Pyongyang on the bank of the Taedong river.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Juche_Tower_%26_fountains.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fqUi0.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;(the two large fountains are curiously not visible in this picture, though)&lt;/p&gt;&#xA;&lt;p&gt;A &lt;a href=&quot;https://www.north-korea-travel.com/juche-tower.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;DPRK tourism website&lt;/a&gt; states the following (emphasis mine):&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The Juche Tower was built on the occasion of the 70th birthday of Kim Il Sung, by the River Taedong in 1982 (Juche 71). The tower has a 20 metre long torch which can be seen glowing throughout the night. [...]&lt;/p&gt;&#xA;&lt;p&gt;In front of it is a group of three people standing 30 metres high. Each person in the trio carry a symbol of the emblem of the Worker's Party of Korea; a hammer, sickle and brush. &lt;strong&gt;Two large fountains that rise up to 150 metres can be seen on the Taedong River.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Note the fountains are described in a way which can only be construed as ornamental.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-12-01T21:43:41.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32156" PostTypeId="2" ParentId="31580" CreationDate="2019-12-01T22:45:11.140" Score="0" Body="&lt;p&gt;Check &lt;a href=&quot;https://openfoam.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;openFOAM&lt;/a&gt; and in particular its solver interMixingFoam&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can visit &lt;a href=&quot;https://www.openfoam.com/documentation/guides/latest/api/interMixingFoam_8C.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.openfoam.com/documentation/guides/latest/api/interMixingFoam_8C.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And maybe contact the asker of this question in SE: &lt;a href=&quot;https://scicomp.stackexchange.com/questions/21040/how-to-model-the-mixing-of-two-fluids-in-a-container&quot;&gt;https://scicomp.stackexchange.com/questions/21040/how-to-model-the-mixing-of-two-fluids-in-a-container&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="17733" LastActivityDate="2019-12-01T22:45:11.140" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32159" PostTypeId="2" ParentId="32130" CreationDate="2019-12-02T07:22:51.910" Score="1" Body="&lt;p&gt;These are ornamental. There are, however, surface areators that are sometimes used to areate large water bodies, as a form of water treatment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how one can look like from far above, in the lower right arm of this lake. AFAIK this is the polishing pond to the wastewater treatment plant on the other side of the small road (&lt;a href=&quot;https://goo.gl/maps/uDp3Yj8P6Low771y9&quot; rel=&quot;nofollow noreferrer&quot;&gt;location is here&lt;/a&gt;):&#xA;&lt;a href=&quot;https://i.stack.imgur.com/smJ37.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/smJ37.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But high water jets don't serve this purpose and waste energy, so surface areators look very different from these jets. This is one from close up in action:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FrC8D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FrC8D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could be that the machine in the image above is the same type as in the map snippet (&lt;a href=&quot;https://www.enviropro.co.uk/entry/39029/Spaans-Babcock/O2-Max-surface-aerators/&quot; rel=&quot;nofollow noreferrer&quot;&gt;image source&lt;/a&gt;).&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2019-12-02T07:30:13.873" LastActivityDate="2019-12-02T07:30:13.873" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32160" PostTypeId="2" ParentId="2333" CreationDate="2019-12-02T08:25:17.283" Score="1" Body="&lt;p&gt;The forged composite material used by Lamborghini and Calloway (a.k.a. &quot;forged carbon&quot;) has a higher modulus and tensile strength than other composite materials like glass fibers and even traditional carbon fiber composites made with twill or satin weaves. The &quot;randomness&quot; of the fiber orientation in the &quot;forged carbon&quot; composite produces a stronger part in terms of flexion resistance, Young's Modulus and tensile strength. The method used by Lamborghini for things like driving compartment shells and certain structural and suspension components is comprised of pre-fabricated sheet material made of a vinyl ester resin and short length (i.e. chopped) carbon fibers which are pressed (where the &quot;forged&quot; term comes into play) into molds under pressure and heat (similar to how pre-preg is used). The small-random fibers also allow for more complex shapes to be formed as compared to traditional carbon fiber twill fabrics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.lambolab.org/wp-content/uploads/03research/pub/05chop/2011-ASC-montreal-forged-suspens-ICE.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.lambolab.org/wp-content/uploads/03research/pub/05chop/2011-ASC-montreal-forged-suspens-ICE.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23734" LastActivityDate="2019-12-02T08:25:17.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32162" PostTypeId="2" ParentId="32140" CreationDate="2019-12-02T10:33:46.067" Score="2" Body="&lt;p&gt;The ±0.02 is an additional uncertainty voltage that needs to be added, regardless of the full scale value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you would end up with &lt;strong&gt;&quot;measured value&quot; ± 0.07 V&lt;/strong&gt; on the 50 V range.&lt;/p&gt;&#xA;" OwnerUserId="20228" LastActivityDate="2019-12-02T10:33:46.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32164" PostTypeId="1" CreationDate="2019-12-02T11:38:11.807" Score="1" ViewCount="98" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HXXjJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HXXjJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to design a compensator to improve the yaw rate (&lt;span class=&quot;math-container&quot;&gt;$\dot{r}$&lt;/span&gt;) control performance of an aircraft. The lateral dynamics for this aircraft is given in steady space form as following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pIpKl.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pIpKl.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where A (4x4) and B(4x1) are both known. I am also given a set of closed loop and observer poles (6x1). I tried to construct a set of equations to incorporate the washout filter and rudder actuator (&lt;span class=&quot;math-container&quot;&gt;$\delta_{r}$&lt;/span&gt;) by creating the steady-space model for the aircraft, obtaining the transfer function, and finding the closed loop transfer function of the system including the rudder actuator and washout filter. In order to perform pole placement in MATLAB using the place() function I needed to have a state space model, so I used the tf2ss() to obtain the new state space model. However, the A matrix in this case was 11x11, but my system has only 6 states. Also, the rank of the observability matrix was 6. I am really confused on what I should do next to compute the compensator's transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The programming language I am using is MATLAB, so if you have any suggestions please include the MATLAB code. &lt;/p&gt;&#xA;" OwnerUserId="23737" LastEditorUserId="23737" LastEditDate="2019-12-02T11:45:32.663" LastActivityDate="2019-12-02T11:45:32.663" Title="Design compensator to improve yaw rate control performance for aircraft" Tags="|mechanical-engineering|control-engineering|control-theory|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32165" PostTypeId="1" CreationDate="2019-12-02T11:41:30.093" Score="-1" ViewCount="1669" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/k1Hr1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess Cp = dH/dT= dq/dT=dS&#xA;Thus SdT = integral of CpdT.&#xA;From that ∆H is given and we can calculate ∆G.&#xA;However, entropy for this reaction should be -7.4J/molK and I am getting like 26J/molK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also S is not ∆S which is a big flag in this working.&lt;/p&gt;&#xA;" OwnerUserId="23738" LastEditorUserId="16981" LastEditDate="2019-12-02T14:05:58.773" LastActivityDate="2022-04-25T13:34:48.517" Title="Calculating Entropy from Gibbs Free Energy?" Tags="|thermodynamics|chemical-engineering|material-science|homework|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32167" PostTypeId="2" ParentId="32165" CreationDate="2019-12-02T13:56:45.600" Score="1" Body="&lt;p&gt;Entropy change is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta S = \int \frac{\delta q_{rev}}{T} $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Under an assumption that the phase transformation is reversible, this becomes&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta_{pt} S = \frac{\Delta_{pt} H}{T_{pt}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Enthalpy and entropy depend on temperature as&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta H(T_2) -  \Delta H(T_1) = \int C_p dT $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta S(T_2) -  \Delta S(T_1) = \int \frac{C_p}{T} dT $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Gibbs energy change is&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta G = \Delta H - T \Delta S $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;The Gibbs energy change of the phase transformation at the transformation conditions is calculated directly from the enthalpy change and transition temperature.&lt;/p&gt;&#xA;&lt;p&gt;To calculate the value at a different temperature, you must first calculate the enthalpy and entropy changes at the different temperature.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2022-04-25T13:34:48.517" LastActivityDate="2022-04-25T13:34:48.517" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32168" PostTypeId="1" CreationDate="2019-12-02T16:10:33.047" Score="1" ViewCount="46" Body="&lt;p&gt;&lt;em&gt;Apologies if this is not the correct SE for this sort of question. But, this is the only place I've seen an appreciable number of CAD questions. If this isn't appropriate for engineering.se, a pointer to a more appropriate site would be greatly appreciated.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this spline model of a car fender in 3DSMAX:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cxLmR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cxLmR.png&quot; alt=&quot;3DSMAX spline model of a fender&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to figure out how to turn this into something I can use in a CAM package like Fusion 360 or Inventor. I'd considered just building the model in Fusion, but there are a number of compound curves that are difficult to describe parametrically. (Also, my MAX skills are a lot better than my Fusion skills.)&lt;/p&gt;&#xA;" OwnerUserId="13612" LastActivityDate="2019-12-02T16:10:33.047" Title="Getting From 3D a model to CNC router" Tags="|cad|cnc|surface-modelling|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="32171" PostTypeId="1" CreationDate="2019-12-02T19:21:50.020" Score="1" ViewCount="34" Body="&lt;p&gt;I'm thinking of what mechanical mechanism can be used to alternate between two different lifting heads, an adhesive and a non-adhesive side, attached to an arm, so that either head comes forward depending on the position of the arm. E.g. moving it down brings one forward, and moving it back up brings the other one forward. I have explained in detail in attached diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ipdPL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ipdPL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9580" LastActivityDate="2019-12-02T19:21:50.020" Title="Mechanism to alternate between two lifting ends" Tags="|mechanisms|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32172" PostTypeId="1" AcceptedAnswerId="32174" CreationDate="2019-12-02T21:59:52.177" Score="0" ViewCount="260" Body="&lt;p&gt;I am studying the mixed sensitivity design for control systems, and I have seen that the control effort has limitations given by the bandwidth of the process and by the bandwidth of the sensitivity function. In particular I have seen that if we plot the control sensitivity function, we have that if I am above the bandwith of the process I have an increase of the control effort, while if I am below it decreases. And also I have a lower limitation given by the bandwidth of the sensitivity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't understand why. Also, does the controller bandwidth impose some limitations? Can please someone explain this to me? Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT] I recognize that my question was to shallow, so I will try to do more reasonings and add more argumentations as I search for answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will use an approch based on the mixed sensitivity for my discussion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have the following plant and controller:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = tf('s');&#xA;G = 10/((s+10)*(s+1));&#xA;K1 = 0.5/s;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Now I define the following mixed sensitivity controller, where I used a weight for the sensitivity in order to achieve my desired specifications, which are a 60 dB attenuation at high frequency, a cross over frequency of 0.66353 and a maximum resonance peak of 2 dB. I also inroduce a weight for the control effort, which I use to limit the amount of effort the control has to do. In particular, I use a constant weight, so by doing so I am hopefully trying to limit the control effort to increase too much:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W_bs2 = 0.66353;   %same cross over frequency as the previuos point&#xA;M = 2;           %peak of the sensitivity&#xA;A = 0.001;      %attenuation&#xA;Ws2 = (s/M + W_bs2)/(s+W_bs2*A);%sensitivity weight&#xA;%bodemag(1/Ws),grid;&#xA;Wu = tf(1);&#xA;[K2,CL,GAM2] = mixsyn(G,Ws2,Wu,[]);  %define the controller with the mixed &#xA;sensitivity&#xA;display(GAM2);&#xA;K2 = minreal(K2);  %define a minimal order controller&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;some of the comments may not make sense because they refer to some previous code.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I define the control effort:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Q1 = K2/(1+G*K2); %control effort&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and I plot the control effort and the weight of the control effort, which should impose a limitation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jgu7m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jgu7m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in this case I used a control effort weight equal to 1, but probably an higher value would have been better.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the bandwith of the controller is 0.9879, so if I go above it the control effor shoulb be increasing instead of decreasing. Let's try:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I define a second sensitivity weight which gives an higher cross over frequency to the control  effort when defining the controller:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;W_bs3 = 2;   %same cross over frequency as the previuos point&#xA;M = 2;           %peak of the sensitivity&#xA;A = 0.001;      %attenuation&#xA;Ws3 = (s/M + W_bs3)/(s+W_bs3*A);%sensitivity weight&#xA;%bodemag(1/Ws),grid;&#xA;Wu = tf(1);&#xA;[K3,CL,GAM3] = mixsyn(G,Ws3,Wu,[]);  %define the controller with the mixed &#xA;sensitivity&#xA;display(GAM2);&#xA;K3 = minreal(K3);  %define a minimal order controller&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and if I plot the control effort of it and confron it with the previous result:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hT36v.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hT36v.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the second mixed sensitivity controller gives me the yellow line, while the first gives me the red line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I am not sure I am going to the right direction, but hopefully I am getting closer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone help me? Thanks again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT 2] So,after a lot of searching and thinking I think the point is that I haven't really understood the concept of bandwisth and how it influences the system. I am in particular looking at sensitivity bandwidth and complementary sensitivity bandwidth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point, my idea is that the sensitivity bandwidth an the complementary sensitivity bandwidth should be limited by the fact that &lt;span class=&quot;math-container&quot;&gt;$S+T=1$&lt;/span&gt;. But still not really clear what happens. So it should be that if S is large, T cannot be too large. But is the bandwidth related to this? &lt;/p&gt;&#xA;" OwnerUserId="22780" LastEditorUserId="22780" LastEditDate="2019-12-04T15:23:32.233" LastActivityDate="2019-12-04T15:23:32.233" Title="Limitations on the control effort" Tags="|electrical-engineering|control-engineering|control-theory|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32173" PostTypeId="2" ParentId="10660" CreationDate="2019-12-02T22:14:24.220" Score="1" Body="&lt;p&gt;I have a K&amp;amp;E LOVAR 100 ft base tape. And the ‘brochure’ specs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;LOVAR per K&amp;amp;E is quoted at 0.44ppm/C. NIST discusses the difference in CTE amongst commercial “alloys of steel” as being only a few percent +/- and negligible. Steel is approximately 12-13ppm/C. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, 0.44ppm +/- “a few percent” of 0.44ppm I believe is the answer to your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In several textbooks, field experience with Invar tapes temperature correction was unnecessary. Error in tension was a greater source of uncertainty.  I have made careful measurements at near 0C and sunny conditions with tape temperature of 45+, over 180 m, 600 ft, using a digital load cell for tension. I can neither confirm or deny less than 1ppm. However, I applied 0.44ppm and the residual error was less than the temperature correction. &lt;/p&gt;&#xA;" OwnerUserId="7220" LastActivityDate="2019-12-02T22:14:24.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32174" PostTypeId="2" ParentId="32172" CreationDate="2019-12-03T00:30:29.400" Score="1" Body="&lt;p&gt;The reason the closed-loop bandwidth is related to control effort is pretty clear: once you're outside the bandwidth of the plant, the higher the frequency of the command, the higher the controller command amplitude (i.e., controller effort) needs to be to get the same result.  So it should just flow from intuition that if you're pushing the system loop closure above the plant's bandwidth that the controller effort will go up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not getting your comment about the sensitivity bandwidth being linked to control effort -- there's more to it than that, because sensitivity is at least in part a function of the plant itself.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Also, does the controller bandwidth impose some limitations?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not in theory-land.  There's an implicit assumption that you're going to define a controller of arbitrary bandwidth, then give it to some poor slob (possibly me) to implement; and that your job is then done.  While real-world designs sometimes &lt;strong&gt;do&lt;/strong&gt; feature drivers or sensors of limited bandwidth, it's more common that there are real-world limitations on how hard you can push the physical devices (or how much you can ask of power supplies); that limits the available control effort.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-12-03T00:30:29.400" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32175" PostTypeId="1" CreationDate="2019-12-03T06:38:24.030" Score="2" ViewCount="176" Body="&lt;p&gt;I am currently a project intern, working on Light Fidelity (Li-Fi). Most part of the project deals with development of software for Renesas Micro controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a simple doubt regarding &lt;strong&gt;timer and interrupt programming&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a part of my project, I have to generate an interrupt as soon as the timer overflows. I am trying hard and finding it difficult to achieve it . Also. I need to know about &lt;strong&gt;how to calculate the timer overflow value, so that an interrupt is generated.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be great help,if these questions are solved and explained in detail.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="23744" LastEditorUserId="1586" LastEditDate="2019-12-24T13:57:01.807" LastActivityDate="2019-12-24T13:57:01.807" Title="Embedded C Programming" Tags="|electrical-engineering|embedded-systems|electronics|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32176" PostTypeId="1" CreationDate="2019-12-03T09:31:03.577" Score="1" ViewCount="1771" Body="&lt;p&gt;Recently, I have bought &lt;a href=&quot;https://www.amazon.co.uk/Magnetic-Charging-Multiple-Micro-Light-Type-Data-Fast/dp/B07HRSFRLM&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; product. It is basically a USB cable, with a magnetic end, and a tiny attachment that goes into the mobile device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plugged it into my laptop cooling stand (which has USB hub), and then plugged the tiny &quot;adapter&quot; into my phone. The magnetic end of the cable was free, and it stuck onto the metal leg of my work table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It caused short circuit and burnt the fan. I only realized this after I plugged it into my laptop, and the same thing caused the laptop to restart immediately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a photo of the gadget.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8i00u.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8i00u.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason behind this? The leg of my table is not connected to my computer, obviously. So, how come this USB cable was able to burn my fan, and restart my laptop?&lt;/p&gt;&#xA;" OwnerUserId="23745" LastEditorUserId="23745" LastEditDate="2019-12-03T21:58:50.053" LastActivityDate="2021-02-05T01:40:31.440" Title="Why does this magnetic charger cause short circuit when it sticks to the metal leg of my table?" Tags="|electrical-engineering|electromagnetism|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32177" PostTypeId="2" ParentId="32175" CreationDate="2019-12-03T10:00:52.473" Score="0" Body="&lt;p&gt;The following link might help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.renesas.com/us/en/support/technical-resources/engineer-school/mcu-programming-peripherals-04-interrupts.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;mcu-programming-peripherals-04-interrupts.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has information about how to use interrupts (very similar to Arduino).&#xA;However, for the timer I only see references to functions like &lt;code&gt;milli&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you mention, the best way is to set up a timer which automatically sets the value of the time, after the interrupt is generated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this,  you need to set the relevant registers, which can be found in the datasheet of your MCU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, it works something like this:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Chose a timer peripheral (if there are more, high likely)&lt;/li&gt;&#xA;&lt;li&gt;Set the timer value (this can involve multiple registers, like a value, multiplier, something to repeat interrupts or a one-shot, or a comparator).&lt;/li&gt;&#xA;&lt;li&gt;Start the interrupt&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17482" LastActivityDate="2019-12-03T10:00:52.473" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32178" PostTypeId="1" AcceptedAnswerId="32179" CreationDate="2019-12-03T12:35:53.290" Score="0" ViewCount="62" Body="&lt;p&gt;I'm doing my first machine design for a project and I have a Honda GX160 combustion motor which has the following specs for power and speed:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/WUNKx.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WUNKx.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to use 0.24 HP at 2000 rpm. Is this possible? Or I can just get 2000 rpm at 2.1 HP as the NET POWER curve shows?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm using this motor because it's the only one I have available for this project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(It would also work for me at 1300 rpm if it has at least 0.24 HP, but I don't think its recommended)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="23026" LastActivityDate="2019-12-03T14:04:25.997" Title="Reading Motor Datasheet" Tags="|motors|machine-design|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32179" PostTypeId="2" ParentId="32178" CreationDate="2019-12-03T14:04:25.997" Score="1" Body="&lt;p&gt;The chart shows the maximum power with the throttle wide open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can run at lower powers simply by having the throttle partly closed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use a low power output like 0.24HP, you will probably need some sort of automatic control system to keep the RPM down to 2000. A simple throttle lever with manual adjustment might be too sensitive to adjust successfully.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A small engine like this may not run smoothly below 2000RPM. Any sudden change in the load might cause it to stall, unless you add a bigger flywheel to your design.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-12-03T14:04:25.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32182" PostTypeId="1" CreationDate="2019-12-03T15:51:00.113" Score="0" ViewCount="73" Body="&lt;p&gt;What would be the best way to calculate velocity of gas from a three phase separator? I know that operating temperature is 75F, pressure 1000 psia, pipe diameter 20 inch, and pipe length to the point where velocity needs to be known is 45 inch. I'm having some trouble determining pressure drop over such a small distance and was wondering if there is a better way I could work this out. &lt;/p&gt;&#xA;" OwnerUserId="23751" LastActivityDate="2019-12-03T15:51:00.113" Title="Calculate gas velocity from separator" Tags="|chemical-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32183" PostTypeId="1" AcceptedAnswerId="32185" CreationDate="2019-12-03T16:39:24.117" Score="0" ViewCount="90" Body="&lt;p&gt;Consider a four-wheel steering vehicle in which:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The front wheels follow Ackerman condition&lt;/li&gt;&#xA;&lt;li&gt;The rear wheel are rotated with a fixed angle (&lt;em&gt;dir&lt;/em&gt; == &lt;em&gt;dor&lt;/em&gt; in the picture below, and such angles do not change if the driver moves the steering wheel)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Actually, this is similar to a classic front wheel steering vehicle, but the rear wheels are not &quot;straight&quot;, they are angled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of trajectory would this vehicle follow? Is there any closed-form equation to describe it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VA8cv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VA8cv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23752" LastActivityDate="2019-12-03T18:46:16.170" Title="Car trajectory with angled rear wheels (non-Ackerman)" Tags="|automotive-engineering|wheels|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32185" PostTypeId="2" ParentId="32183" CreationDate="2019-12-03T18:46:16.170" Score="0" Body="&lt;p&gt;It means there is still some side slipping and dragging of the wheels if the rear tires and front tires are not all aligned to track circles around one common center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case, the rear wheels axis are parallel and never meet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a vehicle needs for steering at both the rear axle as well as the front, then the geometry of the steering linkage both at the front and the rear axle should direct inner wheels to turn more so that all the wheels axis meet at the center of rotation of the vehicle. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-03T18:46:16.170" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32187" PostTypeId="1" CreationDate="2019-12-04T04:51:40.123" Score="2" ViewCount="38" Body="&lt;p&gt;Several years ago, I was visiting the Advanced Photon Source at Argonne National Laboratory, and I remember that in one of the working areas there were some large cylinders suspended in the air above the electronics racks by wire.  Our tour guide had mentioned that the researcher had been an acoustic engineer and put them up himself to dampen the constant loud drone of the facility.  I was skeptical before I walked under it and, despite the area being quite open and unshielded acoustically to the machinery, the cylinders had done a near perfect job of dampening the sound.  Now that I've been working in facilities with similar noise issues, I'm wondering if this is a common solution or if there is a name for these simple noise dampening devices? &lt;/p&gt;&#xA;" OwnerUserId="23759" LastActivityDate="2019-12-05T04:14:07.207" Title="Is there a technical name for a sound dampening suspended cylinder in loud indoor environments?" Tags="|acoustics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32188" PostTypeId="2" ParentId="32187" CreationDate="2019-12-04T07:27:06.777" Score="2" Body="&lt;p&gt;I don't know if there is a specific name for that. But it is a known fact that sound-absorbing fabrics and foams especially if placed in acoustical focal points of noisy spaces large enough to promote reverberation, will help the attenuating of noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These decorative objects have to meet fire and safety codes and be easy to wash and maintain. And they scale and align in a prescribed way. They usually are composed of 3 components: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 - the sound-absorbing foam &lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 - the wireframe structure&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 - the cover fabric which is sound transparent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several manufacturers that help design these devices and sell them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is the catalog of one manufacturer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1cHf4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1cHf4.jpg&quot; alt=&quot;suspended absorbers&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-04T07:39:07.300" LastActivityDate="2019-12-04T07:39:07.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32189" PostTypeId="1" CreationDate="2019-12-04T10:00:00.323" Score="0" ViewCount="34" Body="&lt;p&gt;I would like to know why axis can be controlled independently in a quadcopter. I mean, I was told that you can control &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; independently but I couldn't find a demonstration&lt;/p&gt;&#xA;&lt;p&gt;Edit: the testbench that is used when you want to see how the response of a quadcopter would be is a one arm quadcopter (i.e, just 2 propellers like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/FiTHU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FiTHU.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;That testbench can be used because you can control how the response of the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; axis is and then use the &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; axis. And, when you put both axis together, each axis response would be as they were in the testbench&lt;/p&gt;&#xA;&lt;p&gt;How can we demonstrate that?&lt;/p&gt;&#xA;&lt;p&gt;Or, the question would be: why you can use a one-arm testbench for a quadcopter?&lt;/p&gt;&#xA;" OwnerUserId="8089" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-12-05T14:47:42.960" Title="Control each axis separately - quadcopter" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32192" PostTypeId="1" AcceptedAnswerId="32200" CreationDate="2019-12-04T14:45:27.317" Score="0" ViewCount="331" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mEugv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mEugv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you consider a motor attached to a spur gearbox with a reduction ratio of say &lt;span class=&quot;math-container&quot;&gt;$r=50:1$&lt;/span&gt; which is then back-driven so that the motor acts as a generator, what would be the effective moment of inertia of the system? If the motor has a rotor inertia of say &lt;span class=&quot;math-container&quot;&gt;$1*10^{-5} kgm^2$&lt;/span&gt; and the gearbox has a mass inertia of &lt;span class=&quot;math-container&quot;&gt;$1*10^{-6} kgm^2$&lt;/span&gt; would I be correct in calculating the overall inertia through &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$I_{total}=r^2I_{rotor}+I_{gearbox}=0.025 kgm^2$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="23762" LastEditorUserId="90" LastEditDate="2019-12-04T22:29:57.327" LastActivityDate="2019-12-06T11:26:52.297" Title="Impact of Gearbox on Motor Inertia" Tags="|mechanical-engineering|electrical-engineering|motors|applied-mechanics|mathematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32193" PostTypeId="1" AcceptedAnswerId="32205" CreationDate="2019-12-04T15:17:37.480" Score="1" ViewCount="139" Body="&lt;p&gt;I have been asked to provide some of our technical docs in both metric and US units.  For the most part, that is easy enough:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;approx 1cm → approx 1/2&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;500μm → 0.02&quot; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;But then I start getting stuck:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;50μm → 0.002&quot; or 2 thou or 2 mils? Is there a common symbol for thou/mills?&lt;/li&gt;&#xA;&lt;li&gt;20nm → ???&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I imagine that people who care about nanometers have mostly gone metric now anyway, but it would still be good to provide US measurements for everything. What units are commonly used for lengths smaller than 0.001&quot; in the US, and what symbols are used for them?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it varies across different fields, the relevant ones would be optics, precision machining and mechanical engineering.&lt;/p&gt;&#xA;" OwnerUserId="12022" LastActivityDate="2019-12-05T01:58:22.457" Title="US customary units below 1 mil/thou?" Tags="|distance-measurement|unit|international|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32195" PostTypeId="1" AcceptedAnswerId="32197" CreationDate="2019-12-04T15:37:39.203" Score="1" ViewCount="580" Body="&lt;p&gt;Belleville disc springs are usually spec'd with a free height, loaded height, and loaded force.  From this information you can calculate the average spring rate of a belleville spring.  However, what I would like to know if if the force is actually linear (or nearly linear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume it is not, but don't know what curve the force might follow.  Is there an equation that can predict the force-curve (plot of force versus height) of a belleville spring given ID, OD, free height, and thickness?&lt;/p&gt;&#xA;" OwnerUserId="23767" LastActivityDate="2019-12-04T16:01:00.493" Title="Is the Spring Rate for a Belleville Spring Linear?" Tags="|mechanical-engineering|springs|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32196" PostTypeId="1" CreationDate="2019-12-04T15:56:06.053" Score="1" ViewCount="143" Body="&lt;p&gt;Imagine I have a combustion chamber with a nozzle. There is a high pressure oxygen bottle and a high pressure gaseous fuel bottle connected to the chamber. Would the fuel efficiency be lower if I performed combustion in a series of many very short burns than if I emptied the tanks in one go? &#xA;Since at the end of any burn there's a period where pressure decreases, I imagine that the total impulse of the engine will be lower for the intermittent combustion since it will go through many of these periods instead of just one, leading to increased fuel consumption.&lt;/p&gt;&#xA;" OwnerUserId="18744" LastEditorUserId="6790" LastEditDate="2020-05-04T01:22:26.260" LastActivityDate="2023-05-19T10:37:31.333" Title="Inefficiency of intermittent burn in rockets" Tags="|mechanical-engineering|rocketry|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32197" PostTypeId="2" ParentId="32195" CreationDate="2019-12-04T16:01:00.493" Score="1" Body="&lt;p&gt;The curve is not linear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$s = $&lt;/span&gt; actual deflection (mm)&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$h_o = $&lt;/span&gt; total possible deflection (mm)&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$F = $&lt;/span&gt; actual load (N)&lt;br&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$F_c = $&lt;/span&gt; designed limit load (N)  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YI9yT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YI9yT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://www.spirol.com/library/sub_catalogs/dsc-Deflection_and_Load_Characteristics_us.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here&lt;/a&gt; you can see the original archive with the ploted graph from one industry (I usualy use this material for reference about these type of spring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know the exact formula. But the reasoning should be (this is my guess):  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;From 0% Force to ~70% force should be affected mostly by the&#xA;compression between the inner diameter and outer diameter.&lt;/li&gt;&#xA;&lt;li&gt;From ~70% Force to 100% should be affected mostly by bending.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="23658" LastActivityDate="2019-12-04T16:01:00.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32200" PostTypeId="2" ParentId="32192" CreationDate="2019-12-04T16:59:29.277" Score="1" Body="&lt;p&gt;That should be correct. The rotational velocities and accelerations in the motor, caused by turning the output shaft, increase by r. Along with the acceleration, the reaction torque in the motor increases by the same amount. &#xA;The reduction ratio will also apply on this torque and and hence increase it again by r at the output shaft. &lt;/p&gt;&#xA;" OwnerUserId="23772" LastEditorUserId="90" LastEditDate="2019-12-04T22:32:17.427" LastActivityDate="2019-12-04T22:32:17.427" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32201" PostTypeId="2" ParentId="32112" CreationDate="2019-12-04T17:40:41.473" Score="1" Body="&lt;p&gt;You're making the assumption, that all wheels are rolling without slip. Is it a front wheel driven vehicle? Slip could cause the deviations you see.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what directional speed are you looking for? Forward velocity? At the mid plane? Use the average of the two longitudinal speeds of the wheels (wheel rate/radius). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lateral? Should be zero of you have no slip. If you have slip, you need to know how much, else you can't do much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry, I have no credits to make this a comment.&lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2019-12-04T17:40:41.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32202" PostTypeId="2" ParentId="19911" CreationDate="2019-12-04T20:11:28.893" Score="2" Body="&lt;p&gt;From 1,000 yards away the sound will not seem extremely loud, but in the right conditions, should be audible. I would go with 3/8&quot; thickness for durability as well as producing a nice &quot;ping&quot;, as well as metal chain links to hold it up. If you use rubber, you may dampen the sound, and if you mount right to the wood, that's when divots may form on the gong. Just bolt it to maybe 6 inches of chain, and then the chain to the wood. You will want a bit of swing, to help with ricochets, and to help extend the life of the gong. I would also personally use a rectangle, say 24w X 36h. Hope this helps!&lt;/p&gt;&#xA;" OwnerUserId="23773" LastActivityDate="2019-12-04T20:11:28.893" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32204" PostTypeId="1" CreationDate="2019-12-04T23:13:08.257" Score="0" ViewCount="986" Body="&lt;p&gt;Why does combustion happen at constant pressure in gas turbines but at constant volume in pulsejets? I need to ensure my design adds heat at constant volume. Is it because combustion starts from essentialy the escape point in pulsejets with the flame front acting like a wall? &lt;/p&gt;&#xA;" OwnerUserId="18744" LastActivityDate="2019-12-05T04:00:23.157" Title="Why does combustion happen at constant pressure in jet engines?" Tags="|mechanical-engineering|thermodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32205" PostTypeId="2" ParentId="32193" CreationDate="2019-12-05T01:58:22.457" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;50μm → 0.002&quot; or 2 thou or 2 mils?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've seen both 0.002&quot; and 2 mils.  On a drawing it would always be 0.002&quot;.  In a specification document it could be either. I've never seen 2 thou written in a formal specification (but I have heard people say it in the shop). But that might just be my experience. There could be variation from industry to industry.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is there a common symbol for thou/mills?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I'm not aware of any.  We just always wrote &quot;mils&quot; (not &quot;mills&quot;)&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What units are commonly used for lengths smaller than 0.001&quot; in the US, and what symbols are used for them?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Colloquially, I know that people in precision machine shops refer to 0.1 mils (0.0001&quot;) as &quot;tenths&quot; (as in one ten thousandth of an inch).  I've never seen that written in a formal specification document though, just as talking between machinists.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For US customary units, there's nothing smaller than an inch, you just start adding more zeros, applying prefixes, or using scientific notation. For example, it would be common to specify surface roughness in microinches.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;20nm → ???&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I would write it as &quot;7.87e-7 inches&quot;&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-12-05T01:58:22.457" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32206" PostTypeId="2" ParentId="32204" CreationDate="2019-12-05T04:00:23.157" Score="2" Body="&lt;p&gt;A Brayton-cycle gas turbine compressor presents a large mass flow of air to the combustors at a pressure that varies with the speed of the blades and their pitch. For the moment we assume that both of these are constant. Heat is added to the compressed air by burning fuel with it in the combustors and the resulting hot gas is free to expand out towards and then through the power turbine. This means that as the combustion process is occurring, the volume of a parcel of gas getting heated in a combustor is expanding and since the direction of that expansion is toward the exhaust nozzle, flow continuity requires that each parcel of gas traversing the combustors must accelerate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this way, the gas turbine's heat addition process occurs at variable volume but more or less constant pressure. (Since the gas flowing through the engine is being accelerated, a force is being exerted on it. The resulting reaction force is the thrust output of the engine.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a pulsejet, the heat addition process is not a constant: it occurs at separated intervals, and the explosion of a discrete charge of fuel &amp;amp; air happens so fast that the heat addition is modeled as an almost instantaneous event, with no time for the gases to begin expanding while the combustion occurs. With no time for expansion, the heat addition therefore occurs at more or less constant volume. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-12-05T04:00:23.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32208" PostTypeId="2" ParentId="29070" CreationDate="2019-12-05T04:09:39.920" Score="2" Body="&lt;p&gt;Beeswax is a naturally-occurring hotmelt glue, and the earliest (artificial) hotmelt glue compositions contained beeswax as one constituent. Now there are hundreds of different modern hotmelts available in which synthetic and natural waxes and resins are blended together to yield a broad range of melt temperatures and glue line strengths. I recommend you have a look at some hotmelt manufacturers' brochures &amp;amp; technical data; perhaps you can find a suitable beeswax replacement to test in your application. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, note that an important quality of adhesives for musical instrument repair is the ability to undo them with heat and moisture to facilitate future repairs. The worst possible thing you can do to a clarinet, guitar, saxophone or violin is to repair it with an epoxy, cyanoacrylate or urethane glue because those will make it impossible to disassemble. So, be cautious!&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-12-05T04:09:39.920" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32209" PostTypeId="2" ParentId="32187" CreationDate="2019-12-05T04:14:07.207" Score="0" Body="&lt;p&gt;In music recording studios, those tubular absorbers are common. I hear them referred to as &quot;sound tubes&quot; but I do not know if there is another name for them. I recommend you look at some sales literature for manufacturers of &lt;em&gt;acoustic treatments for sound recording facilities&lt;/em&gt; to find out.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-12-05T04:14:07.207" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32210" PostTypeId="1" AcceptedAnswerId="32230" CreationDate="2019-12-05T04:56:01.123" Score="0" ViewCount="56" Body="&lt;p&gt;I'm looking for a data that what is the typical application range for a common uni directional strain gauge like used in weight scales? &#xA;I've read about around 200 ppm in a research gate article, can I measure 40 ppm with a cheap setup (arduino nano plc, industrial grade op amp, soldered connections)?&lt;/p&gt;&#xA;" OwnerUserId="15534" LastActivityDate="2019-12-06T01:14:14.847" Title="what is the range of the measurable relative elasticity with a commonly used strain gauge" Tags="|measurements|deformation|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32211" PostTypeId="1" CreationDate="2019-12-05T06:59:21.987" Score="1" ViewCount="131" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5iTm4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5iTm4.png&quot; alt=&quot;Air conditioner&amp;#39;s coil&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/v6Sls.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v6Sls.jpg&quot; alt=&quot;Blower of an air conditioner&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/YaFPN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YaFPN.jpg&quot; alt=&quot;Blower of air conditioner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I observed the fan on external units of air conditioners does not force air towards the hot coil containing hot liquid to make it cooler. Instead it draws air from from the direction of the coil it and blows it away. In my opinion, it would be more effective if the air from the fan was blown toward the coil rather than the fan is drawing air from the coil and blowing it away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then my question, why?&lt;/p&gt;&#xA;" OwnerUserId="17943" LastEditorUserId="90" LastEditDate="2023-10-30T19:37:38.580" LastActivityDate="2023-10-30T23:07:32.577" Title="Why do the fan or blower of the Air Conditioner not face the coil flowing hot liquid?" Tags="|thermodynamics|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32212" PostTypeId="1" CreationDate="2019-12-05T07:14:07.017" Score="1" ViewCount="874" Body="&lt;p&gt;I'm a student. Now, I am learning to use LM5030 IC to design a Push-Pull converter. So, I read LM5030's datasheet and can not understand the function of COMP pin (Pin 3). The red line is Feedback circuit, input its connect to VOUT, output its connect to COMP pin.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ts5gF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ts5gF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Following the info in the datasheet, it's said: &quot; The comparator polarity is such that 0 V on the COMP pin will cause a zero duty cycle&quot;. The duty cycle is to switch on/of 2 MOSFET. And in above figure, the output of the feedback circuit will make COMP = 0 when VOUT &gt; 0, it means the duty cycle for 2 MOSFET will be zero.&lt;br&gt;&#xA;That makes me can not understand. Did I miss something? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S: &lt;a href=&quot;https://www.ti.com/lit/ds/symlink/lm5030.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;LM5030 datasheet&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="23780" LastEditorUserId="23780" LastEditDate="2019-12-11T01:23:28.360" LastActivityDate="2019-12-11T01:23:28.360" Title="What 's the function of COMP pin(Pin 3) in LM5030 - Push Pull IC?" Tags="|electrical-engineering|feedback-loop|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32213" PostTypeId="2" ParentId="32196" CreationDate="2019-12-05T07:16:24.890" Score="0" Body="&lt;p&gt;The built-up to momentum in the mass of gas flowing out of the nuzzle and building the design exit Mach and stability of the combustion is better left alone from the disruption of on-off cycles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Avoiding flame blow out in the combustion chamber and continuity in vector thrust navigation lend better to long bursts.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-05T07:16:24.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32217" PostTypeId="1" CreationDate="2019-12-05T12:43:32.540" Score="1" ViewCount="272" Body="&lt;p&gt;&lt;a href=&quot;https://www.maxongroup.com/maxon/view/product/251601&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.maxongroup.com/maxon/view/product/251601&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.maxongroup.com/maxon/view/product/gear/spur/gs45/301171&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.maxongroup.com/maxon/view/product/gear/spur/gs45/301171&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that this motor-gearbox combination is backdriven so that the motor acts as a generator. My goal is to try and make a relatively accurate simulation of the losses/damping that occur due to the fact that it's driven in reverse. From what I understand, the total inertia of the motor felt by the output of the gearbox (now used as an input) can be calculated by the equation below where &lt;span class=&quot;math-container&quot;&gt;$I_{gearbox}$&lt;/span&gt; refers to the mass inertia of the gearbox.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$I_{total}=r^2I_{rotor}+I_{gearbox}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that motor cogging torque needs to be considered as does the friction in the gearbox, but I am unsure as to how this is done. My impression is that the most resistance comes from the friction within the spur gearbox as it is occuring to the point that it takes grown man strength to get it turning at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there specially designed gearboxes which are designed to work somewhat equivalently in reverse as well? I know that &quot;planetary gearheads up to two stages or spur gearheads&quot; should work but the efficiency is still low. &lt;a href=&quot;https://drive.tech/en/stream-content/dc-motors-as-generators&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://drive.tech/en/stream-content/dc-motors-as-generators&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or would I be better off making a simple gear train with two/three spur gears? The driving torque would be quite low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any assistance&lt;/p&gt;&#xA;" OwnerUserId="23762" LastEditorUserId="23762" LastEditDate="2019-12-06T12:44:31.887" LastActivityDate="2019-12-06T19:21:36.500" Title="Driving Gearbox in Reverse" Tags="|mechanical-engineering|electrical-engineering|gears|applied-mechanics|torque|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32218" PostTypeId="2" ParentId="32212" CreationDate="2019-12-05T12:55:14.880" Score="1" Body="&lt;p&gt;As indicated in LM5030 datasheet COMP is output to an error amplifier. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circuit for the LM3411 is as follows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j5Pel.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j5Pel.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The COMP output from the LM5030 is an input to the &lt;a href=&quot;http://www.ti.com/product/LM3411&quot; rel=&quot;nofollow noreferrer&quot;&gt;LM3411&lt;/a&gt; which is an precision secondary regulator. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The LM3411 is a low-power fixed-voltage (3.3 V or 5 V) precision shunt regulator designed specifically for driving an optoisolator to provide feedback isolation in a switching regulator.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-12-05T12:55:14.880" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32219" PostTypeId="2" ParentId="32189" CreationDate="2019-12-05T14:47:42.960" Score="1" Body="&lt;p&gt;With that testbench you are only testing the tracking performance of the rotation around a given axis. The dynamics of each of those rotations is dominated by the inertia around that axis, also known as &lt;a href=&quot;https://en.wikipedia.org/wiki/Moment_of_inertia&quot; rel=&quot;nofollow noreferrer&quot;&gt;moment of inertia&lt;/a&gt;. The moment of inertia contribution of point masses around a given axis can defined as &lt;span class=&quot;math-container&quot;&gt;$I=\sum_i m_i\,r_i^2$&lt;/span&gt;, with &lt;span class=&quot;math-container&quot;&gt;$m_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$r_i$&lt;/span&gt; the mass and (shortest) distance to the rotation axis of each point mass respectively. The removed arms of the quadcopter would contribute to this inertia. However, its contribution will be very small, since the mass of those arms would be very close the axis of rotation and thus &lt;span class=&quot;math-container&quot;&gt;$r_i$&lt;/span&gt; would be very small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can also be noted that in-flight the propellers, on the arms that are removed on your testbench, would be spinning. This spinning would add &lt;a href=&quot;https://en.wikipedia.org/wiki/Angular_momentum&quot; rel=&quot;nofollow noreferrer&quot;&gt;angular momentum&lt;/a&gt; which would also cause &lt;a href=&quot;https://en.wikipedia.org/wiki/Precession&quot; rel=&quot;nofollow noreferrer&quot;&gt;precession&lt;/a&gt;. The propellers used on a quadcopter often are relatively small and light weight. Therefore, their moment of inertia and also their angular momentum is small. This causes the effect of precession to be small as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, removing an arm does change the dynamics of the rotation around a given axis. However, these changes are small compared to the remaining dynamics.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2019-12-05T14:47:42.960" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32220" PostTypeId="2" ParentId="32211" CreationDate="2019-12-05T14:57:58.893" Score="1" Body="&lt;p&gt;You are correct. In terms of fan efficiency, it is more efficient to force cool air onto or through a hot zone, than to draw hot air from a hot zone and expel it to the atmosphere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the external unit of split system air conditioners (heat pumps) you have to consider other things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The current arrangement may be easier and/or cheaper to manufacture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, under the current arrangement, the external units of split system air conditioners is compact and thus requires occupies less space where they will be located. Also, the fan is located at the front of the unit, away from the wall. When it expels the hot air from around the coil it does so freely, without any obstacles impeding the air flow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the fan to blow cooler air onto the coil two situations are possible. Firstly the fan could be at the front of the unit drawing unimpeded air from the atmosphere. It then blows that air onto the coil forcing the hot air around the coil to move. That forced hot air will have to negotiate any internal obstacles within the unit and could hit the wall of the building, which may restrict the movement of the air away from the air conditioning unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other possibility is to locate the fan at the rear of the unit drawing in air from the narrow gap between the wall of the building and the unit, which is restriction to airflow, and forcing it onto the coil. Again, that forced hot air will have to negotiate any internal obstacles within the unit - a further restriction to airflow - before exhausting to the atmosphere, at the front of the unit.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2019-12-05T14:57:58.893" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32222" PostTypeId="2" ParentId="32211" CreationDate="2019-12-05T17:22:00.407" Score="0" Body="&lt;p&gt;Putting the fan behind the coil makes it harder to stick a finger into the fan arc. Positioning the fan as close as possible to the coil and shrouding it on the sides minimizes the efficiency difference mentioned in the comments above. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-12-05T17:22:00.407" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32223" PostTypeId="2" ParentId="32211" CreationDate="2019-12-05T17:24:21.313" Score="0" Body="&lt;p&gt;Putting the fan behind the coil makes it harder to stick a finger into the fan arc. It also directs the air blast directly over the fan motor, improving its cooling. Positioning the fan as close as possible to the coil and shrouding it on the sides minimizes the efficiency difference mentioned in the comments above.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-12-05T17:24:21.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32225" PostTypeId="1" CreationDate="2019-12-05T20:42:49.283" Score="0" ViewCount="436" Body="&lt;p&gt;I am trying to size a servo motor and my basic physics and math is so rusted that I can't figure out how to go about it?&#xA;The scenario is as follows:&#xA;I need to lift an object in the vertical direction against gravity. The object has a mass of 30kg. It is lifted to a height of 1m. It takes 5 sec to lift the object. I want to select a servo motor which is capable of doing this task. I have two options for the mechanism, rack and pinion (R&amp;amp;P) or lead screw (LS).&#xA;I want the lifting to be done in three sections&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Acceleration of 250 mm/s^2 for 1 sec. distance covered is 125mm&lt;/li&gt;&#xA;&lt;li&gt;Constant velocity of 250 mm/s for 3 sec. distance covered is 750mm.&lt;/li&gt;&#xA;&lt;li&gt;Deceleration of 250 mm/s^2 for 1 sec. distance covered is 125mm.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;First thing is the power of the motor. From basic physics I know that&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Weight (force) = mass x g = 30 x 10 = 300N &#xA;Work = Force x Distance = 30 x 1 = 300Nm &#xA;Power = Work/time = 300/5 = 60W&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is the base Power if I add in some friction and a safety factor I would say that 100W-120W should be enough. But I have serious doubts about this calculation. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;Is the power calculation same when I have a Rack and Pinion or a Lead&#xA;  Screw?&lt;/li&gt;&#xA;  &lt;li&gt;The downwards force due to gravity in case of Lead screw is acting at&#xA;  an angle on the screw threads so should I be resolving that weight&#xA;  component?&lt;/li&gt;&#xA;  &lt;li&gt;What happens to the equations in case of downward motion?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was doing some online research and many factors came up in sizing of servos. For example the Inertia ratio, velocity profile, max torque, RMS torque [1].&#xA;I found a forum post which suggested calculating motor RPM based on linear distance and pitch of the lead screw [2]. &#xA;Then calculating the motor torque using this relationship &lt;code&gt;Torque = 30 x Power / pi x RPM [3]&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But this formula is not accounting for the required velocity and acceleration&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;4&quot;&gt;&#xA;  &lt;li&gt;Also how can I find the inertia ratio for a vertical moving load? Do I&#xA;  need to calculate the inertia of my payload when the payload is moving&#xA;  in linear direction only?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I want to understand the basic principles to the sizing so I can size the motor for a belt or a rack and pinion or any other mechanism. Any help or guidance is appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: [1] I was trying to follow this two part video tutorial &lt;a href=&quot;https://www.youtube.com/watch?v=4MaGqSQfYOk&amp;amp;t=210s&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=4MaGqSQfYOk&amp;amp;t=210s&lt;/a&gt;&#xA;[2] &lt;a href=&quot;https://www.physicsforums.com/threads/calculate-torque-power-needed-to-lift-a-4-kg-load.661367/#post-4211630&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.physicsforums.com/threads/calculate-torque-power-needed-to-lift-a-4-kg-load.661367/#post-4211630&lt;/a&gt;&#xA;[3] &lt;a href=&quot;http://wentec.com/unipower/calculators/power_torque.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://wentec.com/unipower/calculators/power_torque.asp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and some other sources&#xA;&lt;a href=&quot;https://www.orientalmotor.com/technology/motor-sizing-calculations.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.orientalmotor.com/technology/motor-sizing-calculations.html&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.motioncontroltips.com/tips-for-sizing-a-servo-motor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.motioncontroltips.com/tips-for-sizing-a-servo-motor/&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.motioncontroltips.com/faq-how-do-i-calculate-the-inertia-of-a-servo-driven-system/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.motioncontroltips.com/faq-how-do-i-calculate-the-inertia-of-a-servo-driven-system/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23789" LastEditorUserId="23789" LastEditDate="2019-12-05T21:33:25.823" LastActivityDate="2019-12-06T15:57:06.637" Title="Sizing an AC Servo Motor for lifting pallets" Tags="|motors|torque|servo|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32227" PostTypeId="2" ParentId="32225" CreationDate="2019-12-05T21:36:45.827" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is the power calculation same when I have a Rack and Pinion or a Lead Screw?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes and no.  Either power calculation needs to take the efficiency of the gearing into account, but you can pretty much count on the efficiency of the rack &amp;amp; pinion being much better than the lead screw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, lead screws (and worm gears) are often used &lt;em&gt;because&lt;/em&gt; they have poor efficiency.  Such screw-based gears often have an efficiency less than 50%, and if a gear train has an efficiency less than 50%, then when you take torque off of the input shaft the output will lock; this provides a built-in safety mechanism (or, in the case of driving a large inertial load, a built-in mechanism for destroying gearboxes when the output shaft stops and the load just keeps on going).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The downwards force due to gravity in case of Lead screw is acting at an angle on the screw threads so should I be resolving that weight component?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Not if you have the efficiency as a percent.  The torque you need to apply to the mechanism is the torque you'd apply to a frictionless mechanism divided by the efficiency.  So for your 30kg mass moving at 0.25m/s, the upward force is roughly 300N (taking &lt;span class=&quot;math-container&quot;&gt;$g = 10 \mathrm{m / s^2}$&lt;/span&gt; because I'm lazy), the power exerted on the mass is then 75W, so the power exerted on a 40% efficient lead screw would be 187.5W.  You'd then need to use the distance/turn spec of the screw to turn that into a torque.  The calculation for the rack &amp;amp; pinion is the same except that the efficiency will be different.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What happens to the equations in case of downward motion?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then the gearbox losses help to brake the motion, instead of hindering.  This means that the behavior is asymmetrical.  I don't have the exact numbers, but for the case of efficiency greater than 50%, you don't need much torque, and for the case of efficiency less than 50%, you actually need to &quot;push&quot; the geartrain to work against friction.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-12-05T21:36:45.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32228" PostTypeId="1" CreationDate="2019-12-05T22:12:30.060" Score="-1" ViewCount="59" Body="&lt;p&gt;can someone explain why the highlighted area (at Mb=0) is &lt;strong&gt;-&lt;/strong&gt; 5x2 and not positive? Since the rest of the loads(and the rest of that very same load) pointing down were documented with a +, &#xA;I dont understand why this part is calculated with a minus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2AV4r.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;: &lt;a href=&quot;https://i.stack.imgur.com/2AV4r.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2AV4r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you! &lt;/p&gt;&#xA;" OwnerUserId="23790" LastActivityDate="2019-12-06T01:37:12.413" Title="Beam with internal hinge - Reactions help" Tags="|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32229" PostTypeId="2" ParentId="32228" CreationDate="2019-12-05T22:59:10.163" Score="1" Body="&lt;p&gt;Because by convention they have set clockwise moment as positive. We are taking moments so we consider the sign of the direction of the rotation not the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inspecting line you have highlighted: the sum of moments about B, we see they have assigned negative to the reaction C (-7.c) implying counterclockwise as negative.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So with this convention, the moment 5*2*1 should be negative because it is counterclockwise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would not make any difference in the final reactions, it's just that usually we assume the counterclockwise moments as positive.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-06T01:37:12.413" LastActivityDate="2019-12-06T01:37:12.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32230" PostTypeId="2" ParentId="32210" CreationDate="2019-12-06T01:14:14.847" Score="0" Body="&lt;p&gt;I think it will come down to the quality of your amplifier. Let's put some numbers together.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With 10 volt excitation (typical) and a gage factor of 2.2 (typical), and full bridge wiring as per line 9 in this reference (&lt;a href=&quot;https://www.hbm.com/en/7163/wheatstone-bridge-circuit/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.hbm.com/en/7163/wheatstone-bridge-circuit/&lt;/a&gt;, there are other ways to do a full bridge but this is common), 40e-6 in/in would give an output of 0.5 mV. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other to make a good measurement, you'll want a minimum resolution on your analog to digital converter to at least 10 times less than this (i.e. if your minimum resolution is 4e-6, then you can distinguish 40e-6 from 36e-6 or 44e-6, but no finer. That's probably the minimum acceptable and might want something a lot better depending on your requirements).  So let's stay you need to be able to read 0.05 mV (i.e. 50 microvolts) with your ADC. Averaging multiple values can help increase resolution, but only to a point. you need a decent enough resolution to start with.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Arduino nano has a 10 bit ADC with a 5V range (&lt;a href=&quot;https://store.arduino.cc/usa/arduino-nano&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://store.arduino.cc/usa/arduino-nano&lt;/a&gt;).  So 1 least significant bit is 5 / 1024 ~= 5 mV.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus you can read 5 mV and you want to be able to read 0.05 mV.  So your amplifier needs to be able to provide an amplification of 100:1 (without introducing any non-linearity, distortion, noise, etc).  You'll also need to keep the noise on the input side of the op amp to on the order of 0.05 mV (or else the op-amp will just amplify the noise along with the signal).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A professional quality strain gage amplifier which would meet those specs (e.g. Omega DMD-466) is going to run \$400-\$500 or more.  I doubt you'll be able to meet the same performance with a &quot;cheap&quot; setup, although I won't say for sure that you can't do it.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-12-06T01:14:14.847" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32232" PostTypeId="1" CreationDate="2019-12-06T04:24:20.807" Score="-1" ViewCount="416" Body="&lt;p&gt;Today, I have been reading a lot about sewer systems on Wikipedia, in the hope that I would find information or at least photos showing what I think of as sewers, namely somewhere under the streets in a city where you can climb down through manholes with ladders and then walk around on little pavements down there, with rivers of sewer water flowing calmly around and the walls being made of rocks or something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's what I think of when I hear &quot;sewer&quot;. But I found none of that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read about ancient sewers (with very few photos or illustrations of any kind) as well as saw a photo of some modern sewer pipe which definitely doesn't let you walk around in it. Nowhere could I see the &quot;real&quot; sewers, which I associate with Victorian-era England.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I wonder: why did they have to make them so large-scale? Why do humans need to go down there? And if they had to back in the day, then why don't they now, or in ancient times?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A ton of details are missing in my understanding of sewers. I've also always wondered how they could afford to do this everywhere in a large city. It seems like an insane amount of work just to transfer away some turds. Yes, I realize it's a crucial necessity to prevent diseases and provide for modern/convenient living, but still, it seems so wasteful (no pun intended) to build them so &quot;roomy&quot;. Maybe that was only done for certain sections or something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I especially wonder how in the world they were able to do this when the city was already made. Seems like you'd have to tear down all the houses if you want to make massive sewer systems, and then rebuild all the houses. Did they really do that?&lt;/p&gt;&#xA;" OwnerUserId="23792" LastActivityDate="2020-07-07T11:01:28.933" Title="Why are/were sewers made large enough for somebody to walk down in there?" Tags="|structural-engineering|fluid-mechanics|building-physics|engineering-history|fluid-filtration|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32233" PostTypeId="2" ParentId="32232" CreationDate="2019-12-06T05:57:31.977" Score="2" Body="&lt;p&gt;Some things to consider regarding Victorian era sewers.&lt;/p&gt;&#xA;&lt;p&gt;How much effluent do humans produce each day and how many people are there and thus how large does a sewerage system need to be to accommodate that quantity of effluent?&lt;/p&gt;&#xA;&lt;p&gt;The other thing to consider is the technology available at the time and the ease and cost of construction. Allied to this is the need for maintenance and repair of the system and how maintenance will be undertaken.&lt;/p&gt;&#xA;&lt;p&gt;At the time, brick lined tunnels were an effective and safe means of constructing a sewer, when human labor was abundant and cheap and some tasks were not mechanized.&lt;/p&gt;&#xA;&lt;p&gt;Also, will the sewer system need to accommodate other things that may intentionally of unintentionally enter the system that could cause blockages or damage the system. &lt;a href=&quot;https://en.wikipedia.org/wiki/Fatberg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Fatbergs&lt;/a&gt; have become a recent issue for sewage systems.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="-1" LastEditDate="2020-07-07T11:01:28.933" LastActivityDate="2020-07-07T11:01:28.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32234" PostTypeId="2" ParentId="32232" CreationDate="2019-12-06T07:06:41.110" Score="2" Body="&lt;p&gt;The few important reasons is that the digging was done manually by laborers, who like the miners had to work by getting in and needed cavernous spaces to breath and also needed to move the dirt out on makeshift tracks and wagons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modern engineering which strives for efficiency and least damage to echo system, designs the minimal  working size pipes which can be placed by just pushing through or drilling or extruding the pipes leading to sanitary treatment centers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for maintenance  they use digital scopes and routers and sump pumps mostly built onto a sanitation truck as opposed to sending men into the tunnel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Centralized monitoring stations capable of detecting and rerouting a clogged pipe, etc, with new efficient piping have made it possible to have much more effective system with much less cost.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-06T07:06:41.110" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32235" PostTypeId="1" AcceptedAnswerId="32254" CreationDate="2019-12-06T08:57:49.137" Score="1" ViewCount="338" Body="&lt;p&gt;I need to work on a project where multiple objects need to exchange trough wireless protocol and they are near to High frequency TIG welding machine.&#xA;All objects works by pair (one emitter and one receiver) and they are up to 20 pairs in the same area (something like 30 or 50 square meter)&#xA;Each pair do not have to interfer with other pair of objects and they have to not be interfered by the TIG welding machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I better use WIFI or Bluetooth protocol?&#xA;Range is not an issue as each object of a pair is not far from the other object of the same pair.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For what I found, High frequency TIG welding is given to 1Mghz &lt;/p&gt;&#xA;" OwnerUserId="14673" LastActivityDate="2022-02-28T06:39:51.793" Title="High frequency TIG welding and Bluetooth or WIFI frequency" Tags="|welding|wifi|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32238" PostTypeId="1" CreationDate="2019-12-06T12:57:16.220" Score="8" ViewCount="7461" Body="&lt;p&gt;I have a steel coil spring that is used by compression. It is too strong, and I would like to reduce its strength by some fraction, ideally keeping its length. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reducing it to roughly half would be nice. Uncontrolled heating and cooling in air would make it very unelastic, I think, and shorter after the first compression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can be manually handled. (The dimensions are: cylindrical with diameter 3.5 mm, length 60 mm, and round wire wire diameter 0.5 mm.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be interested in the theory, even if it's something that I can not do, like precisely controlled heat treatment. I have only two of them, it can be very slow.&lt;/p&gt;&#xA;" OwnerUserId="711" LastActivityDate="2019-12-14T17:37:10.277" Title="Can I weaken a coil spring consisting of spring steel?" Tags="|steel|springs|elastic-modulus|heat-treatment|" AnswerCount="8" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32240" PostTypeId="2" ParentId="32238" CreationDate="2019-12-06T13:38:30.097" Score="2" Body="&lt;p&gt;I would consider perhaps the easiest way would be to change the leverage...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Failing that, I would get a new spring made with the details necessary and spare one 10% above and below your &quot;new&quot; value.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-12-06T13:38:30.097" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32241" PostTypeId="2" ParentId="32238" CreationDate="2019-12-06T13:45:16.147" Score="7" Body="&lt;p&gt;You could try a bath of concentrated acid. As long as you could maintain circulation so the concentration was fairly constant, a spring presents a very uniform cross-section and should be dissolved at a consistent rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend keeping one of the springs as a reference so that you can measure the result after a short period and adjust your time correspondingly to obtain the desired result.&lt;/p&gt;&#xA;" OwnerUserId="23088" LastActivityDate="2019-12-06T13:45:16.147" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32242" PostTypeId="1" CreationDate="2019-12-06T14:39:47.770" Score="0" ViewCount="488" Body="&lt;p&gt;I'm a bit confused with the wording of the question. What does it mean by expressing the bending moment and the deflected shape of BC as a function of z? Thank you&lt;img src=&quot;https://i.stack.imgur.com/yUm3Y.png&quot; alt=&quot;![enter image description here&quot;&gt;]&lt;a href=&quot;https://i.stack.imgur.com/yUm3Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z1rhw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z1rhw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23799" LastActivityDate="2020-01-05T13:39:21.923" Title="Deflected Beam Shape as a Function of z" Tags="|civil-engineering|statics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32243" PostTypeId="2" ParentId="32175" CreationDate="2019-12-06T15:22:27.797" Score="1" Body="&lt;p&gt;This response is based on RX130 Capacitive touch development board as shown below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5ucps.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5ucps.jpg&quot; alt=&quot;Renesas RX130 Capacitive touch development board&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;IDE: e2Stuido v7.5 and above&lt;/li&gt;&#xA;&lt;li&gt;Toolchain: Renesas RXC toolchain&lt;/li&gt;&#xA;&lt;li&gt;Board: RX130 Capacitive touch development board&lt;/li&gt;&#xA;&lt;li&gt;MCU: RX130 R5F5130ADFN&lt;/li&gt;&#xA;&lt;li&gt;Debugger: E1 &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The steps below are to help setup the RX130&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Set up clocks HOCO Clock and VCC to 5.0&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vumdR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vumdR.png&quot; alt=&quot;Set up VCC&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;make sure other clocks a deselected&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZA9ns.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZA9ns.png&quot; alt=&quot;Select the clocks&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Select the Compare Match Timer and add software module&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Up2wJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Up2wJ.png&quot; alt=&quot;Select Software Component&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3ZeAq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3ZeAq.png&quot; alt=&quot;Add Module&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KSf7F.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KSf7F.png&quot; alt=&quot;Compare Match Timer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;Configure the Compare Match Timer (CMT)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZQFGS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZQFGS.png&quot; alt=&quot;Name the CMT&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3BOfT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3BOfT.png&quot; alt=&quot;Configure CMT&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above are all the step you need to configure the Compare Match Timer. There are steps that have been left out to be concise. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To test the CMT setup a GPIO port connected to a LED and toggle the LED when CMT has rolled over and interrupt is generated&lt;/p&gt;&#xA;" OwnerUserId="110" LastActivityDate="2019-12-06T15:22:27.797" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32244" PostTypeId="2" ParentId="32225" CreationDate="2019-12-06T15:57:06.637" Score="0" Body="&lt;p&gt;After giving this some thought I think now I understand more clearly what to do.&#xA;There is three torques associated to the three stages of motion.&#xA;1. The accelerated motion a = 250mm/s^2&#xA;2. The constant speed motion a = 0&#xA;3. The deceleration with a = -250mm/s^2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first two torques are of interest for me as they will give me maximum and nominal torque. (The deceleration section will depend on the efficiency of my gear train. But in no case will that torque be more than the acceleration torque.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My load is 300N, to cause an acceleration of 0.25 m/s^2 in the upward direction I need a net force of 0.25x30 = 7.5N. So the total upward force needed is 300 + 7.5 N = 307.5N.&#xA;Using the gear train efficiency I can then calculate, as pointed out by TimeWescott, the actual force needed to generate this acceleration and then the turn ration to find the torque. This will be by maximum torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For constant speed motion the required upward force will be same as the load. The rest of the calculation for nominal torque will be the same.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the inertia ratio I will find the load inertia from this formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque = (Jload +Jmotor) x alpha&#xA;Max torque is calculated and linear acceleration (0.25) can be converted to angular acceleration (alpha) for a lead screw. Jmotor can be found in the datasheet. I can then find the Jload and use it to find the inertia ratio of load and motor.&lt;/p&gt;&#xA;" OwnerUserId="23789" LastActivityDate="2019-12-06T15:57:06.637" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32245" PostTypeId="2" ParentId="32242" CreationDate="2019-12-06T16:03:25.717" Score="0" Body="&lt;p&gt;You need to find the right expression that explains the bending moment along the z dimension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something like for &lt;span class=&quot;math-container&quot;&gt;$z=2meters$&lt;/span&gt; (your limit is &lt;span class=&quot;math-container&quot;&gt;$-2m&amp;lt;=z&amp;lt;=4m$&lt;/span&gt;) the bending will be max. And for &lt;span class=&quot;math-container&quot;&gt;$z=4meters$&lt;/span&gt; the bending will be 0mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll not put the equation here because this looks like a homework question so you can find with your own material.&lt;/p&gt;&#xA;" OwnerUserId="23658" LastEditorUserId="23658" LastEditDate="2019-12-06T16:19:12.870" LastActivityDate="2019-12-06T16:19:12.870" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32246" PostTypeId="2" ParentId="32242" CreationDate="2019-12-06T16:33:21.693" Score="2" Body="&lt;p&gt;In this question, the axes x and y are in the plane of the cross-section of the beam and z is along the length of the beam pointing horizontally to the right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reaction forces are,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \Sigma M_C=0 \quad  20kN*4*2+ 30kN*6-B_V*4=0 \\ B_v = 340kN/4 =85kN \\ C_v= 80+30-85=25kN $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment on the cantilever part is &lt;span class=&quot;math-container&quot;&gt;$$ M= 30kN*( z+2) \ assuming\ z_B =0  $$&lt;/span&gt;&#xA;And I let you do the deflection,by either double integration or area moment.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-06T19:45:34.043" LastActivityDate="2019-12-06T19:45:34.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32247" PostTypeId="2" ParentId="32081" CreationDate="2019-12-06T17:08:01.753" Score="0" Body="&lt;p&gt;I think the missing point is that the force you want is not in the unit N/m^3, as the term&#xA;(1−h2/h1)^3/2*(1+h2/h1) = K  is unit less so in order to get the correct answer you should find the force as point load by (pgh* K * [Area = h W ])  as pgh is the pressure at the bottom &#xA;pgh* K * h W = pgW h^2 &lt;/p&gt;&#xA;" OwnerUserId="18593" LastActivityDate="2019-12-06T17:08:01.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32248" PostTypeId="2" ParentId="32238" CreationDate="2019-12-06T17:42:07.027" Score="4" Body="&lt;p&gt;By &quot;strength&quot; do you mean stiffness/modulus? The modulus does not change with tensile or yield strengths. Make it with wire of 1/2 the diameter for 1/2 the modulus. Acid will hydrogen stress crack hardened steel ( This condition has many other names like &quot;embrittlement&quot;.). Make it out of aluminum and get 1/3 the modulus or titanium and get 2/3 the modulus ( approximately).  Doing your own heat-treatment could work ; simple enough to try : Heat to orange in a gas flame ( 1600 F +/-) hold time - a couple seconds , IMMEDIATELY quench into cold salt water. Temper at 300 to 400 F ,somewhere around an hour is traditional. Good for carbon and most low alloy steels. High alloy or tool steel -another story. The heat-treatment risk is the exceptionally small thickness which could partially decarburize in a few seconds at 1600 F. Commercially it would be heat-treated in a protective atmosphere or vacuum , possibly with induction heating. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2019-12-06T17:51:58.483" LastActivityDate="2019-12-06T17:51:58.483" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32249" PostTypeId="2" ParentId="7901" CreationDate="2019-12-06T18:44:48.027" Score="1" Body="&lt;p&gt;As an aside, if you want to bend and play with it. Here's a tip. Get a pot of soap, a gas blow torch and sit by a bath. Smear the part with soap. Heat with the blow torch until the soap turns black, then drop it in the cold bath. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on your air temperature you've now got a few hours to bend it to your desired shape. once down leave it in the airing cupboard or somewhere else nice and toasty for a day or so. The alloy will then regain it's stiffness. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you've done is melt the nickel and copper crystalline structures within the aluminium. Now it will fell, bend and act like pure simple aluminium. Once you give it time the copper and nickel will reform the internal structure and you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Welcome to home heat treatment. it's exactly the same way I saw piper making skin panels in Vero Beach. Heat the cut to shape skin panels, drop them in a tank then bend them onto the back and clico rivet. Then they put the parts out in the Florida sunshine. Job done.  &lt;/p&gt;&#xA;" OwnerUserId="23802" LastActivityDate="2019-12-06T18:44:48.027" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32250" PostTypeId="2" ParentId="32238" CreationDate="2019-12-06T19:08:29.787" Score="3" Body="&lt;p&gt;Probably just another hypotetical idea: &#xA;If you squeeze the wire from circular into a square shape, it will have about 54% of the original rate. If you squeeze it slightly more into a rectangle, it will at some point become 50%. &#xA;And 1/2 the diameter of the wire does not gie you 1/2 the rate. It would give you 1/16 of the rate. 85% of the diameter gives you 1/2 the rate.&lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2019-12-06T19:08:29.787" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32251" PostTypeId="2" ParentId="32235" CreationDate="2019-12-06T19:12:40.940" Score="1" Body="&lt;p&gt;I may be missing something important, but both Bluetooth and Wifi operate in the &gt;GHz bands. &#xA;With 1MHz you shouldn't remotely get any impact. &#xA;That only true of course, if the welding machine does not produce any higher frequencies as a side effct. &lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2019-12-06T19:12:40.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32252" PostTypeId="2" ParentId="32217" CreationDate="2019-12-06T19:21:36.500" Score="1" Body="&lt;p&gt;If you want a hight reduction ratio (such as 1:50 as mentioned in your other question), don't use spur gears. These are self-locking.&#xA;Lock after cycloidal drives: &lt;a href=&quot;https://en.wikipedia.org/wiki/Cycloidal_drive&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Cycloidal_drive&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your rpm's are low, the unbalanced property might not hurt so much.&lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2019-12-06T19:21:36.500" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32254" PostTypeId="2" ParentId="32235" CreationDate="2019-12-06T20:31:06.013" Score="3" Body="&lt;p&gt;Don't worry about the frequency of the applied power.  Worry about the fact that a spark gap is a fairly efficient way (and can be a &lt;em&gt;really&lt;/em&gt; efficient way) of generating radio frequency radiation.  It's the way Heinrich Hertz made &lt;a href=&quot;https://en.wikipedia.org/wiki/Heinrich_Hertz#Electromagnetic_waves&quot; rel=&quot;nofollow noreferrer&quot;&gt;the first man-made radio waves&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would test.  I would start by getting a spectrum analyzer in the same room as your arc welder and making estimates of the signal degradation from that, then I would actually do some end-to-end tests.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some areas of study -- and RF propagation is one of them -- where &lt;strong&gt;all&lt;/strong&gt; computational answers end up being inexact because the interaction with the environment is too convoluted.  Unless you're an absolute RF genius, you just want to take measurements and experiment.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-12-06T20:31:06.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32255" PostTypeId="2" ParentId="32238" CreationDate="2019-12-06T21:46:24.483" Score="3" Body="&lt;p&gt;Another hypothetical way to reduce the amount of material in the spring would be to fix it in a jig stretched to double its normal length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, heat it sufficiently to anneal it and remove the tension in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, re-temper it and cut it in half, back to its original length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But just buying a new spring would be a much simpler method.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-12-06T21:46:24.483" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32256" PostTypeId="1" AcceptedAnswerId="32279" CreationDate="2019-12-07T02:23:22.123" Score="-2" ViewCount="46" Body="&lt;p&gt;no code involved and no technical equipment is going to involved in experiment and implementation for business/tech startup that is, in brief, a way for me to exercise during my business studies!  Building branded CROSSFIT exercise equipment with readily available materials.  With reasons, I chose each material following, bicycle tires (cheap rubber known for durability), milk cartons (hoping to combine with chemicals to create ultimate durable and form factor retention), filler&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Creating a recipe will take some effort I realize, also i did previous research and there are two routes to go, 3dimensional structures make plastic being recycled more durable mechanically, but the better opportunity is to melt rubber into ingots, treat the milk cartons with low-budget chemicals, and then combine the three aforementioned parts into weight lifting plates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Help if you can, Im looking for the low budget route that will maximize output with low effort.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What chemicals make plastic more durable with form factor rentention the most possible, and what lubricants and filler will make the combination as durable as possible. &lt;/p&gt;&#xA;" OwnerUserId="23804" LastActivityDate="2019-12-09T01:09:50.470" Title="Will specific fillers or chemicals make simple plastic milk cartons more durable, enough to be used as weightlifting plates" Tags="|recycling|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32259" PostTypeId="1" CreationDate="2019-12-07T15:50:06.760" Score="1" ViewCount="27" Body="&lt;p&gt;I am new to the &lt;strong&gt;AutoDesk AutoCAD&lt;/strong&gt; drawing software. I would be really appreciated it if someone could help me out. I will attach the image of the content.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found 3D drawing from the internet, which is &lt;strong&gt;STP&lt;/strong&gt; file extension. I imported the 3D content into AutoCAD. It is a Cessna plane drawing actually. But, the problem is the plane is not colored. How can I &lt;strong&gt;color&lt;/strong&gt; it since it has multiple layers on the drawing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, after the coloring process, I wish to &lt;strong&gt;get png&lt;/strong&gt; files from some specific angles to be able to use it in my 2D drawing. How can I manage to do that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/BZIoU.png&quot; alt=&quot;Image of the 3D Cessna Plane&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="23809" LastEditorUserId="1832" LastEditDate="2019-12-07T22:24:40.200" LastActivityDate="2019-12-07T22:24:40.200" Title="I have 3D model ready in Autocad. I would like to color it and extract a 2D png file from it to use" Tags="|autocad|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32260" PostTypeId="1" CreationDate="2019-12-07T16:11:34.310" Score="2" ViewCount="37" Body="&lt;p&gt;Let us assume we have a motor with a bearing which is exhibited to wear. Over time, we can observe higher vibration, increased noise and the motor speed reduces due to the bearing fault. In the end, the bearing fails and the motor is not able to provide its specified function (drive something) anymore (called failure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My Issue: I am not sure what would be the right name of the failure mode in this case. I found a lot of sources that call this the &quot;&lt;em&gt;bearing failure mode&lt;/em&gt;&quot;. However, according to ISO Condition Monitoring 13372:2012 a failure mode is the &quot;observable manifestation of a fault&quot;. So, I think in this case observable are the loss of speed, increased noise and the higher vibrations and I would call my failure mode: &quot;&lt;em&gt;higher vibratons, increased noise and speed loss&lt;/em&gt;&quot;. In this case, &quot;bearing failure&quot; would be the name of the failure cause. Furthermore, there are other causes that result in the same observalbe properties and so one failure mode can have several root causes, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is right and why? I do not want to over complicate this, but I need some clearity here. It is also interesting regarding FMECA. This is a &quot;Failure mode effects and criticality analysis&quot; in which the concept of a potential failure mode plays a big role. Is there a good and reliable source (literature) with examples how to differentiate what a failure mode is? What is the best practice?&lt;/p&gt;&#xA;" OwnerUserId="14138" LastEditorUserId="14138" LastEditDate="2019-12-08T21:58:40.440" LastActivityDate="2019-12-08T21:58:40.440" Title="Is a motor failure caused by a defect bearing considered to be a failure mode or is it higher vibration and speed loss in FMEA/FMECA?" Tags="|mechanical-failure|failure-analysis|maintenance|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32261" PostTypeId="1" CreationDate="2019-12-07T17:00:31.597" Score="-1" ViewCount="76" Body="&lt;p&gt;I wonder whats the usage of orifice/orifice plate in Water Jet Cutting Machines. To increase pressure(due to a smaller diameter with respect to the entrance cross-sectional area) or to convert to higher velocity water or both?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;thanks!&lt;/p&gt;&#xA;" OwnerUserId="23621" LastActivityDate="2020-09-28T23:06:02.007" Title="Uses of orifice in Water Jet Cutting?" Tags="|mechanical-engineering|fluid-mechanics|machining|energy|cutting|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32262" PostTypeId="1" CreationDate="2019-12-07T18:16:00.540" Score="3" ViewCount="44" Body="&lt;p&gt;In our digital computers class, the professor mentioned that the outter tracks of a magnetic disk is more reliable because the outter bit are &quot;bigger&quot;. Isn't the packing density of a magnetic disk uniform? If we compare this with a optical CD, the inner tracks are more reliable. Why is this the case?&lt;/p&gt;&#xA;" OwnerUserId="23812" LastActivityDate="2019-12-07T18:16:00.540" Title="Why are the outter tracks of a magnetic disks more stable?" Tags="|computer-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32263" PostTypeId="2" ParentId="32261" CreationDate="2019-12-07T20:06:07.430" Score="1" Body="&lt;p&gt;The use of the orifice plate is to optimize the jet diameter and adapt it to the possible abrasive material or the job detailing finesse requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The orifice which is normally of Tungsten carbide and in the industry sometimes called jewel does not control the pressure. pressure and velocity are controlled by powerful pumps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Uses a high velocity stream of Ultra High Pressure Water 30,000–90,000 psi (210–620 MPa) which is produced by a high pressure pump with possible abrasive particles suspended in the stream.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Quote from Wikipedia.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3V5pL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3V5pL.png&quot; alt=&quot;diagram of water jet cutter, Wikipedia&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-07T20:06:07.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32264" PostTypeId="1" AcceptedAnswerId="32267" CreationDate="2019-12-07T20:31:29.437" Score="-2" ViewCount="141" Body="&lt;p&gt;In Rocket Lab's presentation video, &lt;a href=&quot;https://www.youtube.com/watch?v=eZE043cKcD4&amp;amp;t=21s&quot; rel=&quot;nofollow noreferrer&quot;&gt;at around 00:21&lt;/a&gt;, you can see some engineer work with CAD software. Here's a screenshot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YQoj9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YQoj9.png&quot; alt=&quot;CAD software used by Rocket Lab&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone recognize which software they're using? Asking for a friend.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2019-12-07T21:47:33.473" Title="What CAD software is this?" Tags="|cad|software|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32266" PostTypeId="2" ParentId="32238" CreationDate="2019-12-07T21:19:15.353" Score="2" Body="&lt;p&gt;While I think the concentrated acid solution is brilliant, I'd like to suggest a refinement. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Rather than use just any old acid, you should specifically use a &lt;em&gt;chemical etchant&lt;/em&gt;.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically, a chemical etchant's nucleophile strength does not come from Hydrogen, so it should have the benefit of much lower -if not entirely eliminated- hydrogen enbrittlement. (At least theoretically, anyway.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good etchant for this might be Ferric Chloride, [FeCl3].  Copper(II) Chloride [CuCl2] is another possibility.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Getting an even and precise etch will also be up to idiosyncrasy's of the exact setup you try and run. Chemical concentration, bath agitation, cleanliness of the work, temperature, ...&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Another, related, option is to use electrochemistry instead of direct chemistry. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Specifically, use the reverse of electrowinning, electroetching.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of the pros of this method is that, with controlled current + time, you could fairly easily calculate and control the exact mass of the material removed by the relationship between the unit of electric charge (coulombs), and the unit of molar mass (moles.) This extrapolates well to controlling the exact change in spring's mechanical property. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With direct chemistry, concentration + temperature + time = mass of material removed. Even if you precisely dial in the last two variables, the first one is going to be a ever changing target, as the work will dilute/poison the bath.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Probably the biggest con of this (aside from requiring a semi-sophisticated electronic power supply) is that surface irregularities will become much more of a problem than with chemical methods. In particular, any insulator will be practically inert to etching with the electrochemical method. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Conversely, an insulator might still be attacked by a pure chemical method. e.g. Any oil on the surface might become esterfied to something soluble, be freed from the surface, thus allowing the etchant to get at the actual work piece.&lt;/p&gt;&#xA;" OwnerUserId="23815" LastEditorUserId="23815" LastEditDate="2019-12-08T07:49:29.257" LastActivityDate="2019-12-08T07:49:29.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32267" PostTypeId="2" ParentId="32264" CreationDate="2019-12-07T21:47:33.473" Score="2" Body="&lt;p&gt;I think the hint comes from the tab that says SOLIDWORKS in the left column, near the top of the block of actions. It's not visible in this screen shot, but is clearly visible in the video linked in the post.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-12-07T21:47:33.473" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32268" PostTypeId="2" ParentId="32238" CreationDate="2019-12-08T00:47:31.897" Score="0" Body="&lt;p&gt;A possible way you could fix this, if you have access to something like a blowtorch, or better yet, a blacksmithing forge: you can heat the steel up until it is glowing orange turning yellow, and then find a place to set the spring down while it cools without anything touching it. If you allow it to cool slowly enough, the spring will become annealed, softer and more malleable and flexible.&lt;/p&gt;&#xA;" OwnerUserId="23817" LastActivityDate="2019-12-08T00:47:31.897" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32270" PostTypeId="1" AcceptedAnswerId="32298" CreationDate="2019-12-08T11:13:41.343" Score="0" ViewCount="43" Body="&lt;p&gt;Original Question was &lt;a href=&quot;https://earthscience.stackexchange.com/questions/18625/how-much-does-the-weight-of-urban-structures-buildings-affect-the-compaction&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in Van, Turkey. Van City is situated on an alluvial plain beside Lake Van. General geological structure of the area can be seen on pages 43-44 in &lt;a href=&quot;http://www.dsi.gov.tr/docs/sempozyumlar/van-g%C3%B6l%C3%BC-hidrolojisi-ve-kirlili%C4%9Fi-konferans%C4%B1-van.pdf?sfvrsn=2&quot; rel=&quot;nofollow noreferrer&quot;&gt;this conference book&lt;/a&gt;. One of the images is below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0gFEm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0gFEm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the article in the link, sediment layers has a slope of 15-30 degrees toward the lake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the construction of every building, the pressure over the sediments increases. Obviously, this increase is greater in downtown area which has a lot of apartment blocks (mainly 5 to 7 story high reinforced concrete buildings). Downtown area is roughly in the middle of the plain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to know that how much does the weight of urban structures (buildings) affect the compaction (permeability/porosity/density?) of alluvial sediments below a city and how does this affect the flow of underground water?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much the stress caused by the weight of the building decreases with depth? (of course if at least a broad value can be given to help to explain the size of the change). I wonder to understand that either the effect of the weight of urban structures &lt;strong&gt;is generally negligible, or not.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="18213" LastEditorUserId="90" LastEditDate="2022-08-03T00:47:31.847" LastActivityDate="2022-08-03T00:47:31.847" Title="How much does the weight of urban structures (buildings) affect the compaction (permeability, porosity, density) of alluvial sediments below a city?" Tags="|geotechnical-engineering|soil|soil-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32271" PostTypeId="1" CreationDate="2019-12-08T11:55:52.830" Score="1" ViewCount="87" Body="&lt;p&gt;My original theoretical consideration was using the electric motor's rotor as drive shaft in a car. It led me to this question: &quot;What will happen if we place a couple of separate, small stators around a long rotor instead of 1 large stator around a short rotor?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a biology teacher. A motor design like this seems useful to me when I consider the structure of skeletal muscles. A skeletal muscle is constituted of a lot of long cells/fibers. Our body controls the force produced by the muscle by controlling the number of contracting fibers at a time. When we want to pick up a pen on the table, a few fibers are recruited; but when we want to pick up the table all fibers are recruited to generate enough force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess that via changing the number of the active stators, a motor like this can be more efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any electric motor design like this? Or, is it possible and if it is, does it make sense from the points of efficiency and weight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AasXa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AasXa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18213" LastEditorUserId="18213" LastEditDate="2019-12-08T18:27:43.040" LastActivityDate="2024-03-16T11:02:55.350" Title="What will happen if we place a couple of separate, small stators around a long rotor instead of 1 large stator around a short rotor?" Tags="|electrical-engineering|motors|electric-vehicles|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32273" PostTypeId="1" CreationDate="2019-12-08T14:09:56.760" Score="0" ViewCount="115" Body="&lt;p&gt;How would I write a Cost Engineering support plan which reflects the Project Control Cycle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A wine company is wondering if they should invest in another factory. Only this time it will be in the USA. This will be their first factory in the USA and they are wondering what the cost will be and what kind of steps you could take to support the business case through the different project phases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the steps that need to be taken to successfully support a business case decision?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is I haven't done anything like that before and I don't know what the structure of the report should be and to which things I should focus on. Any help would be greatly appreciated!&lt;/p&gt;&#xA;" OwnerUserId="23823" LastEditorUserId="9770" LastEditDate="2020-01-13T14:32:20.133" LastActivityDate="2021-02-06T16:07:25.733" Title="How to write a Cost engineering support plan?" Tags="|civil-engineering|cost-estimation|cost-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32274" PostTypeId="1" CreationDate="2019-12-08T19:03:40.637" Score="1" ViewCount="178" Body="&lt;p&gt;I'm willing to take my own shot at the construction of a blast supercharged valveless pulsejet, a holy grail among DIY jet enthusiasts.&#xA;However, since my idea is not very compatible with the geometry of most popular pulsejet engines, im going to have to build a custom valveless pulsejet first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To that end, I need to know these basic design priniciples:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The ratio of the main exhaust pipe length to intake pipe length&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The ratio of main exhaust pipe diameter to intake pipe diameter&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The appropriate amount of flare in the main exhaust pipe&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any properties of combustion chamber that have to be kept in a defined relationship to the two pipes&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I know there's a remarkable amount of variation in existing designs, therefore im only asking for a rule of a thumb for choosing those parameters.&#xA;I would have asked on the pulsejet forum but im still awaiting account activation&lt;/p&gt;&#xA;" OwnerUserId="18744" LastActivityDate="2019-12-08T19:03:40.637" Title="Basic dimension relationships in a valveless pulsejet engine" Tags="|mechanical-engineering|propulsion|jet|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32275" PostTypeId="2" ParentId="32271" CreationDate="2019-12-08T19:05:49.653" Score="0" Body="&lt;p&gt;A genius by the name of  Nicolai Tesla conjured the idea of a shaft of an engine capable of delivering a variable range of power/torque with variable speed all controlled by demand, from scratch in his mind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;He even thought of a system of delivery of electrical power suitable for this purpose, three phase AC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These engines are called a variable frequency drive, similar to what you see in a battery drill. The more you push the trigger back the faster the drill spins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another good example is the engine of a Tesla care. By changing the frequency and current fed to the engine they control the RPM and torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of adding stators to the shaft, they make that same stotor flexible&#xA;that causes the rotor to turn faster and or deliver more torque.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-08T19:05:49.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32276" PostTypeId="1" CreationDate="2019-12-08T22:10:25.690" Score="2" ViewCount="195" Body="&lt;p&gt;Logged in a little time ago i have a new question. So I am trying to mesh an arifoil geometry using ANSYS meshing. Airfoil number is NACA 84-M profile. Running  2-D simulation my main objective is to find a good enough mesh for a specific task, which has to do mainly with active flow control, using a tiny jet nozzle located at the top back of my airfoil section. &lt;a href=&quot;https://i.stack.imgur.com/peJR5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/peJR5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;The outline of my airfoil sectin has been devided into 4 main sections. &#xA;Lower part &#xA;Upper front part&#xA;Jet nozzle part (only made of 2 points)&#xA;Upper back part (from jet nozzle through trailing edge of the airfoil)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding my outter part of my flow domain i have used a C-type structured mesh seen below.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/VsCbj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VsCbj.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I left the nozzle spot and my main concern is the fact that i want to use a different face meshing for that specific spot. Trying to alter the number of cells due to the fact that the number should be enough to simulate flow phenomena. &#xA;    Shall I use different meshes for the bottom and both of the upper two sections of my geometry? (remeber that the hole airfoil has been bi-sected into 4 different categories, described previously).&#xA;    Total chord length has been dimensionalized. Points arranged in the range of [0 to 1]. So the total chord length is 1m, or 1mm, it doesnt play a role. &#xA;If you have many articles in consideration, please reply, anything should be much appreciated.&#xA;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="23826" LastActivityDate="2019-12-08T22:10:25.690" Title="Airfoil meshing using ANSYS" Tags="|fluid-mechanics|aerospace-engineering|ansys|cfd|meshing|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32278" PostTypeId="1" CreationDate="2019-12-09T00:22:10.550" Score="-1" ViewCount="83" Body="&lt;p&gt;I have a transfer function below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$H(s)=(s^2+(1/2)s+(1/2))/(s^2+s+1)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using this, I have to design an RLC circuit with that transfer function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first hand, I did some inverse Laplace Transform, but it didn't seem to be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm stuck in here not knowing how to implement that circuit only with a Transfer Function...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any small hints or clues would be appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="23828" LastActivityDate="2019-12-09T23:35:31.657" Title="Designing an RLC circuit with specific transfer function" Tags="|transfer-function|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32279" PostTypeId="2" ParentId="32256" CreationDate="2019-12-09T01:09:50.470" Score="1" Body="&lt;p&gt;So if I understand what you want to do correctly, your plan is to melt down bicycle tires, melt down milk cartons, and then combine those two with some third thing, and create a new product.  Your question is what should that third thing be.  If that is incorrect please edit the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your biggest problem here is that you can't melt down bicycle tires.  The process of creating them is irreversible.  Just google &quot;can I melt down bicycle tires&quot; and you'll get a bunch of pages explaining why you can't do it.  To quote one of the top hits: &quot;it's like un-baking bread&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now milk cartons, you could melt down.  Those are usually just HDPE.  You can modify HDPE by adding other components to tweak certain properties.  However, I doubt you'll be able to do it cheaply, and you certainly won't be able to do it without a lot of &quot;technical equipment&quot; and specialized knowledge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sorry to rain on your parade but I don't think what you have in mind is feasible.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-12-09T01:09:50.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32281" PostTypeId="1" AcceptedAnswerId="32282" CreationDate="2019-12-09T04:19:36.737" Score="2" ViewCount="664" Body="&lt;p&gt;I was thinking of building a free standing structure out of plumbing pipe that would go flush against the wall to hang jackets and other items.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attached is a sketch of the general thought of what the structure would look like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lr8KI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lr8KI.png&quot; alt=&quot;Sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picturing it in my head, it seems like it should be all good. But then I just got curious about that actual math to get a rough idea of what force it would take to tip it over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is assuming that the pipe is strong enough to hold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I started looking into the calculations for levers, but it seems to be like there are a couple levers chained together. At that point it became beyond me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Note: I plan on taking other safety precautions like attaching to the wall. But I am more curious theoretically now.&lt;/em&gt;&lt;/p&gt;&#xA;" OwnerUserId="23830" LastActivityDate="2019-12-09T07:00:10.660" Title="What calculation to find how much force before tipping over?" Tags="|mechanical-engineering|structural-engineering|structures|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32282" PostTypeId="2" ParentId="32281" CreationDate="2019-12-09T06:32:13.113" Score="1" Body="&lt;p&gt;This structure as it is is not going to tip over no matter how much force you apply on the top where you have the arrow at 12 inches.&#xA;Unless the force is so much that the pipe is going to break or crush under the compression stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the arrow falls within the footprint of your hanger. Let's do a quick calculation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you hang a weight of P lbs from the arrow and let us call the point on the floor at the left corner A and the one on the right corner at the end of 18inch B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the moments about A and knowing that for equilibrium the sum of moments should be zero, we have:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  \Sigma M_a=0 ,\quad P*12+ B_v*18=0 \\ B_v=- \frac{P*12}{18}=- \frac{2}{3}P$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We note that a reaction vertically pointing up ( by the negative sign) by the amount of 2/3 of any weight we hang on top is sufficient to keep the hanger balanced. And &lt;span class=&quot;math-container&quot;&gt;$\ A_v = -\frac{1}{3}P $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real life though if you hang a really heavy object from the arrow, your vertical pipe can bend forward under the bending moment, or the corner on the bottom may give a bit so the hanging point moves horizontally forward and then the hanger's balance can become shaky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a general rule whenever you have the weight's vertical projection fall inside the footprint of the support the structure is balanced. It could be a huge weight of a high-rise building, still, it is vertically balanced and will not tip over. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of the buildings, there are some safety factors that have to be applied depending on the use and type of the building.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-09T07:00:10.660" LastActivityDate="2019-12-09T07:00:10.660" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32283" PostTypeId="1" CreationDate="2019-12-09T09:06:30.080" Score="-1" ViewCount="58" Body="&lt;p&gt;I have seen bigg cargo, and other ships, and I like their structure very much., a question come sin my that why the ship structure is like this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/zEVii.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain what was reason for this. I am not an engineer is this field but still I am interested to know why is it so. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:) I mean, why the shape of ship is in this given in the picture. &lt;/p&gt;&#xA;" OwnerUserId="23832" LastEditorUserId="23832" LastEditDate="2019-12-09T18:20:34.460" LastActivityDate="2019-12-09T18:20:34.460" Title="How the design of ship help to survive in the water" Tags="|mechanical-engineering|fluid-mechanics|marine-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32284" PostTypeId="1" AcceptedAnswerId="32286" CreationDate="2019-12-09T11:55:25.197" Score="3" ViewCount="323" Body="&lt;p&gt;I am trying to calculate the internal forces and their nature in this truss, however, I cannot seem to do so as both the method of joint and the method of section do not work. I think this is because there are not enough conditions of equilibrium that can be used. Is there any way to actually calculate the forces in the members even if this is the case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, at joint A, there are 3 unknown members, and all the members acting through it are only concurrent at that specific point, making it impossible for me to use the equilibrium condition of the sum of moments being 0.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Y5fcu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y5fcu.png&quot; alt=&quot;The diagram of the truss&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be extremely appreciated if someone could inform me about how to calculate the forces in the members of this truss. Thank you.&lt;/p&gt;&#xA;" OwnerUserId="23834" LastEditorUserId="23834" LastEditDate="2019-12-09T12:02:19.513" LastActivityDate="2021-10-26T20:28:57.587" Title="Truss analysis of internal forces in members" Tags="|structural-engineering|structural-analysis|statics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32286" PostTypeId="2" ParentId="32284" CreationDate="2019-12-09T13:48:27.850" Score="2" Body="&lt;p&gt;The structure is not statically determinate, unless you remove bar v. The internal forces will depend on the relative stiffness of the members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The practical way to do the analysis is make a finite element model.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-12-09T13:48:27.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32287" PostTypeId="2" ParentId="32150" CreationDate="2019-12-09T14:19:00.757" Score="1" Body="&lt;p&gt;You got time &amp;amp; temperature switched.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;temperature, pressure&lt;/p&gt;&#xA;" OwnerUserId="1915" LastActivityDate="2019-12-09T14:19:00.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32288" PostTypeId="2" ParentId="32283" CreationDate="2019-12-09T14:51:16.803" Score="1" Body="&lt;p&gt;This is the end part of the ship during the construction and is not finished yet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This part when finished will be called Stern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The stern houses the rudder and the propellers and follows the keel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is designed to have a streamlined narrow shape under the water to help reduce the drag of the ship and vibration due to the rotation of the propellers and the reaction of the rudder. It is designed to deliver a smooth stream of water to the propellers to reduce turbulence. Hence the horizontal structural steel stiffeners are wider her for strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top deck which is dry and above the water level flares out wide to accommodate more surface for containers or volume for the oil if this will be a tanker. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a diagram from the site www.marineinsight.com.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MwY9l.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MwY9l.png&quot; alt=&quot;diagram from, https://www.marineinsight.com/ &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-09T14:57:21.067" LastActivityDate="2019-12-09T14:57:21.067" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32289" PostTypeId="2" ParentId="32150" CreationDate="2019-12-09T15:18:14.417" Score="1" Body="&lt;p&gt;More precipitation sites initiate the higher the temperature. So you want a higher temp to have many sites.  The precipitates grow faster ( become coarse) at higher temperature. But the objective is many precipitates ( high temp.) with small size ( low temp). So the temperature selection is a compromise. I think some newer precipitation stainless steels have a two temperature age : start high , then a second lower temperature. Digressing ; one of my favorite metallurgical terms &quot;anti-phase domain&quot; is the undefined zone ( several atoms wide) between the matrix and the precipitate.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-12-09T15:18:14.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32290" PostTypeId="1" CreationDate="2019-12-09T16:53:30.360" Score="1" ViewCount="237" Body="&lt;p&gt;Although both aeroplane and rocket burn fule and eject smoke but&#xA;Why aeroplane moves and runs horizontally before take off ?why does not it takes off at its original/initial place just like rocket? &lt;/p&gt;&#xA;" OwnerUserId="22801" LastActivityDate="2019-12-17T10:05:44.660" Title="Rocket vs aeroplane? Difference in take off?" Tags="|aerospace-engineering|aerodynamics|rocketry|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32291" PostTypeId="1" AcceptedAnswerId="32297" CreationDate="2019-12-09T17:28:49.253" Score="1" ViewCount="171" Body="&lt;p&gt;Does a house heating system affect indoor air humidity? &#xA;Or to be more precise, does it affect your perception of how dry the air in the house feels (for instance dry is your mouth waking up)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I searched for this but could only find explanations from furnace company websites)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every (lay)person I ask this seems to think house furnaces make the air drier (at least the perceived dryness). One friend claims that the air gets even dryer than the outside air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intuition is that heating can only increase the evaporation inside the house (from the little surface liquid water sources (shower, taps) there may be), therefore making humidity go up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect people think this because furnaces are only on in the winter when the outside air is dry leading to dry air inside as air exchanges. Thus it could be due to this spurious correlation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible explanation is that the higher air temperature increases the air's water carrying capacity, which in the absence of an abundant water source, extracts water from your body faster. Therefore increasing perceived dryness. Even if these two countervailing effects are true, can we use first principles to bund which effect dominates?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT (after seeing the answers and discussions below): &#xA;Perhaps perceived humidity is too loose to define. What if we define the water loss of a human over a period as our measure? But my ideal answer would have two numbers to compare: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Baseline_loss = how much water does a person looses sleeping in a cold sealed room (say at 1C temperature) with 100% relative humidity and no air movement. &lt;/li&gt;&#xA;&lt;li&gt;Total_loss = how much water does a person loses if the same sealed&#xA;room was heated up to 22C (a nice in-home temperature).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Then Heating_induced_loss = Total_loss- Baseline_loss&lt;/p&gt;&#xA;" OwnerUserId="23840" LastEditorUserId="23840" LastEditDate="2019-12-13T16:22:04.707" LastActivityDate="2019-12-13T16:22:04.707" Title="Does a house heating system affect perceived and true indoor air humidity?" Tags="|heat-transfer|heating-systems|dehumidification|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32293" PostTypeId="1" AcceptedAnswerId="32294" CreationDate="2019-12-09T18:53:32.307" Score="1" ViewCount="295" Body="&lt;p&gt;I am trying to mate a gear with a shaft via a key.key coincident with gear and cocentericity of gear with shaft cannot both be made. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bDn97.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bDn97.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23679" LastEditorUserId="8055" LastEditDate="2019-12-10T11:37:39.793" LastActivityDate="2019-12-10T11:37:39.793" Title="solidworks mating gear with shaft problem" Tags="|mechanical-engineering|gears|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32294" PostTypeId="2" ParentId="32293" CreationDate="2019-12-09T19:28:36.093" Score="2" Body="&lt;p&gt;Your issue is that you're assuming the key and key way will have exactly the same dimensions. In reality the key will have to be slightly smaller than the keyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What yo want to do is start with the &lt;strong&gt;concentricity&lt;/strong&gt;, then use the &lt;strong&gt;width&lt;/strong&gt;, or &lt;strong&gt;parallel&lt;/strong&gt; mate to line up the key and keyway. This way you're not over-constraining the assembly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solidworks sometimes lets you get away with that, but even when it does it's not a great idea and it can break unexpectedly later.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-12-09T19:28:36.093" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32295" PostTypeId="2" ParentId="32290" CreationDate="2019-12-09T19:47:25.913" Score="4" Body="&lt;p&gt;A plane relies on aerodynamic lift to stay airborn. This is much more efficient than using raw vertical thrust like a rocket does. The downside is that planes always need to be moving perpendicular to the direction of gravity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pretty much all planes except for fighter jets do not have enough thrust to take off vertically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For traveling from point A to B on land that's fine, since the path is perpendicular to gravity anyway. If you wanted to go straight up though, a plane would be forced to fly in a large spiral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the short answer is: Planes take off horizontally because they have to. Rockets can take off in any direction, but they're not efficient.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2019-12-09T19:47:25.913" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32296" PostTypeId="2" ParentId="32278" CreationDate="2019-12-09T20:04:31.600" Score="0" Body="&lt;p&gt;The first thing you should do is try to find out what a RLC circuit even looks like with the variables you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{q(t)}{C}+L q''(t)+R q'(t)=u(t)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the Laplace &lt;span class=&quot;math-container&quot;&gt;$$\mathcal{L}(Q)\to \frac{\text{C} \left(\text{L} q'(0)+\text{L} q(0) s+q(0) R+\mathcal{L}(U)\right)}{\text{C} \text{L} s^2+\text{C} R s+1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solving this gives you a general function of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$q(t)\to c_1 e^{\frac{1}{2} t \left(-\frac{\sqrt{C R^2-4 L}}{\sqrt{C} L}-\frac{R}{L}\right)}+c_2 e^{\frac{1}{2} t \left(\frac{\sqrt{C R^2-4 L}}{\sqrt{C} L}-\frac{R}{L}\right)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing to the solution of the Transfer function and its solution:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$H(s)=\frac{s^2+\frac{s}{2}+\frac{1}{2}}{s^2+s+1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta (t)+\frac{1}{12} e^{-\frac{1}{2} \left(1+i \sqrt{3}\right) t} \left(i \left(\sqrt{3}+3 i\right) e^{i \sqrt{3} t}-i \sqrt{3}-3\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be able to get yourself closer to your circuit design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have access to mathematica and System modeler (or matlab and simulink) you could also just design exactly the circuit to your specific functions. Keep in mind your transferfunction which is in the form &lt;span class=&quot;math-container&quot;&gt;$H(s)=\frac{U(s)}{Y(s)}$&lt;/span&gt; has an input..which is why the results don't look the same, and where that &lt;span class=&quot;math-container&quot;&gt;$\delta (t)$&lt;/span&gt; shows up and other bits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this &lt;a href=&quot;https://arxiv.org/pdf/1711.10245.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; you can read more about the differential equations, and how they relate that to your design specifics.&lt;/p&gt;&#xA;" OwnerUserId="14318" LastEditorUserId="14318" LastEditDate="2019-12-09T23:35:31.657" LastActivityDate="2019-12-09T23:35:31.657" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32297" PostTypeId="2" ParentId="32291" CreationDate="2019-12-09T20:18:27.303" Score="4" Body="&lt;p&gt;Your last statement is basically correct:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Another possible explanation is that the higher air temperature increases the air's water carrying capacity, which in the absence of an abundant water source, extracts water from your body faster. Therefore increasing perceived dryness.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Heating air does &quot;dry it out&quot; in a practical sense. For example, a food dehydrator does not have any water storage mechanism, all it does is warm the ambient air up and blow it over the food. Your furnace can do exactly the same thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's talk about why that is. Humidity is commonly measured as RH or &quot;relative humidity&quot;, which is expressed as a percentage. 100% RH means the air is holding as much water as it can. Anytime the RH is less than 100, the air will steal water from everything it touches due to evaporation. The lower the RH the faster the evaporation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next thing that's important to know is that the amount of water air can hold increases &lt;em&gt;dramatically&lt;/em&gt; with temperature. So if we take air and heat it, the amount of water in it stays the same, but the capacity of the air increases. This warm air now wants a lot more water than it has, so it will steal water from everything it touches. The technical way to say this is that RH is water/water_capacity, by increasing the water capacity we've decreased the RH; all without doing anything to the water itself!&lt;/p&gt;&#xA;" OwnerUserId="20490" LastEditorUserId="20490" LastEditDate="2019-12-10T22:38:13.720" LastActivityDate="2019-12-10T22:38:13.720" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32298" PostTypeId="2" ParentId="32270" CreationDate="2019-12-09T20:26:54.943" Score="0" Body="&lt;p&gt;In many cases, the weight of the building on the soil is roughly equal to the weight of the layer of the soil excavated for the foundation, subterranean parking, utility rooms, storage, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume the buildings are made of 20cm thick concrete slabs, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$0.2*1m/2400kg.m^3= 480kg/m2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and a live load of &lt;span class=&quot;math-container&quot;&gt;$100kg/m^2 \quad total=580kg/m^2 \quad 580*6floors= 3480kg/m2$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We estimate the soil weight is around &lt;span class=&quot;math-container&quot;&gt;$1600kg/m^2$&lt;/span&gt; which is not &#xA;unreasonable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume the depth of foundation and parking 3m, excavation. &lt;span class=&quot;math-container&quot;&gt;$  3*1600= 4800kg/m^2$&lt;/span&gt; which is even a bit larger than the weight of the building added to the soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also many sites need grading to create a flat pad for building and its surrounding landscaping and access roads. And because usually grading is done favoring cut rather fill, that takes some more weight from the top of the soil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, these kinds of jobs require a soil and geology report and one of the reasons for the report is checking into your concern. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-09T20:54:01.763" LastActivityDate="2019-12-09T20:54:01.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32300" PostTypeId="2" ParentId="32290" CreationDate="2019-12-09T23:49:36.233" Score="0" Body="&lt;p&gt;Rockets are carrying their own oxygen. Airplanes rely on the oxygen in the atmosphere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also as has been mentioned in @drew's answer airplane engines have been optimized for flying in the atmosphere with continuous power on but very small relative thrust, a range of between 20% to 30% of the weight of the airplane. While rockets must provide 100+% thrust for the lift off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are new technologies that are breaking this mold like solar wind rockets or vertical take off planes with rotating wings or military jets which have hovering capabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the two major differences in majority of the cases is rockets do not need air either for developing lift or for oxygen, while planes need air to generate lift and use it's oxygen to burn their fuel.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-09T23:49:36.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32301" PostTypeId="1" AcceptedAnswerId="32303" CreationDate="2019-12-10T03:15:56.590" Score="8" ViewCount="1772" Body="&lt;p&gt;My task is to create a 3D Sketch of a trombone like the one below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1UcTV.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1UcTV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pp4Je.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pp4Je.png&quot; alt=&quot;the end of the trombone&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My professor recommended defining the points 1-4 and putting a spline through them. I did that, but my resulting design looks weird. It curves inward between points 1 and 2 before curving out towards the rest of the points. I'm not sure what I'm doing wrong, and I don't know if I have enough given constraints to curve it outward manually myself by editing the spline&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VhGYC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VhGYC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/jKVqL.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jKVqL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23845" LastActivityDate="2019-12-11T13:08:06.890" Title="SOLIDWORKS - 3D Sketch of Trombone" Tags="|mechanical-engineering|design|cad|solidworks|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32302" PostTypeId="1" AcceptedAnswerId="32313" CreationDate="2019-12-10T07:02:00.563" Score="0" ViewCount="752" Body="&lt;p&gt;I am having trouble understanding how indoor temperature plays into what a healthy indoor relative humidity level should be.  Most sources I have come across searching the topic online provide a range based on the outdoor temperature.  For example, according to &lt;a href=&quot;https://www.thermastor.com/relative-humidity-and-your-home/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Therma-Stor&lt;/a&gt;, &quot;The ideal relative humidity for health and comfort is about 40–50%. In the winter months, it may have to be lower than 40% RH to avoid condensation on the windows.&quot;  I understand this is important for preventing condensation that can lead to mold and other problems, but I have not seen condensation forming on any windows and I think the air may be too dry in my case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I live in a home that was built in 1938.  It is not well-insulated and there have been several additions in its history.  It is currently winter in my location, so the furnace is running.  I do not feel any drafts coming from outside, so I believe most of the heat escapes due to the poor insulation -- the closer I am to exterior walls, the colder it feels.  The thermostat in my home is set to 64 degrees Fahrenheit (17.8  &lt;span class=&quot;math-container&quot;&gt;$^oC$&lt;/span&gt;).  Increasing the temperature beyond that has a minimal effect on the downstairs temperature, while the upstairs temperature increases drastically.  I have looked at the air ducts, but I cannot find any place where I am able to adjust the airflow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The outdoor temperature ranged from 19-49 degrees Fahrenheit (-7 to 9.4 &lt;span class=&quot;math-container&quot;&gt;$^oC$&lt;/span&gt;) today, with a current outdoor temperature of 21 degrees Fahrenheit (-6 &lt;span class=&quot;math-container&quot;&gt;$^oC$&lt;/span&gt;) and humidity of 79%.  I just measured the temperature and humidity in my bedroom, which is downstairs, and it is 57 degrees Fahrenheit (13.9 &lt;span class=&quot;math-container&quot;&gt;$^oC$&lt;/span&gt;) with a humidity of 37%.  In the office, which is upstairs, the temperature is 64 degrees Fahrenheit (17.8 &lt;span class=&quot;math-container&quot;&gt;$^OC$&lt;/span&gt;) with a humidity of 35%.  I placed the thermometer/hygrometer in the center of each room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering the temperature of my bedroom is quite low, I am assuming that the 37% humidity level is pretty dry -- the thermometer/hygrometer also provides a reading of &quot;Low&quot; next to the humidity level.  It seems that most sources I have found assume a more reasonable indoor temperature in the range of 68-72 degrees Fahrenheit (20 to 22 &lt;span class=&quot;math-container&quot;&gt;$^oC$&lt;/span&gt;).  I do have a portable humidifier I could run in my room, and I am wondering what would be an appropriate humidity level under these outdoor and indoor conditions? Is there some sort of an equation I could use, maybe something similar to how the thermometer/hygrometer determines when to display a reading of &quot;Low,&quot; &quot;OK,&quot; or &quot;High?&quot;&lt;/p&gt;&#xA;" OwnerUserId="23846" LastEditorUserId="90" LastEditDate="2019-12-11T02:41:59.743" LastActivityDate="2019-12-11T02:41:59.743" Title="How does indoor temperature factor in when determining ideal relative humidity in a home?" Tags="|hvac|temperature|air|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32303" PostTypeId="2" ParentId="32301" CreationDate="2019-12-10T09:35:04.850" Score="12" Body="&lt;p&gt;See the .gif below!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0CTx8.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0CTx8.gif&quot; alt=&quot;make drag handle horizontal&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: You may need to enable visibility of these handles if they're not on by default - it depends on your default profile/settings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IBXZc.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IBXZc.png&quot; alt=&quot;Turn on handles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-12-10T09:35:04.850" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32304" PostTypeId="1" CreationDate="2019-12-10T12:30:17.953" Score="1" ViewCount="67" Body="&lt;p&gt;I have a question about design automation or generative design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a company that creates pumps. Their work-flow is following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)  create design of a pump using CAD&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2)  create mesh for computer simulation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3)  perform simulation (usually fluid dynamics)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4)  using results of a simulation, decide if the design is good enough&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5)  if not, go back to step 1&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are doing all these steps by hand. I was wondering if there is any possibility to automate any of these steps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example algorithm or software that takes the design and tries to change important parameters (for example length or width of something) based on simulation results automatically (= performs some kind of optimization of the given parameter). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this something that generative design can accomplish? I am completely new to this field (however I have a background in CFD simulations). I will I really appreciate any help, introduction or link, so I can learn the basics.    &lt;/p&gt;&#xA;" OwnerUserId="23847" LastActivityDate="2020-01-23T19:29:21.533" Title="Design automation using fluid dynamics simulations" Tags="|fluid-mechanics|design|cad|computer-aided-design|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32305" PostTypeId="1" CreationDate="2019-12-10T14:04:30.630" Score="1" ViewCount="340" Body="&lt;p&gt;i designed a gearbox with 3 shafts.i wanted to do sth so shafts only have rotating movement and not any other type of movement.i cant find any type of mate that can do that,any suggestions?&lt;a href=&quot;https://i.stack.imgur.com/lBhkQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lBhkQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x4giH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x4giH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23679" LastEditorUserId="23679" LastEditDate="2019-12-10T15:14:46.703" LastActivityDate="2019-12-10T15:40:57.060" Title="prevent shaft from moving in solidworks" Tags="|mechanical-engineering|solidworks|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32307" PostTypeId="2" ParentId="32305" CreationDate="2019-12-10T14:50:52.997" Score="1" Body="&lt;p&gt;See the .gif below showing how you might set these mates up. &lt;strong&gt;You need to add another part or a sketch into your assembly with axes in the correct places, and then mate your shafts to be concentric to those axes.&lt;/strong&gt; Here I have set the parts coincident to the front plane to also limit their axial movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Av9WG.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Av9WG.gif&quot; alt=&quot;Shaft constraint with master model&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2019-12-10T15:40:57.060" LastActivityDate="2019-12-10T15:40:57.060" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32309" PostTypeId="1" CreationDate="2019-12-10T15:48:01.277" Score="-1" ViewCount="93" Body="&lt;p&gt;I am searching for help with simulating a physical audio device and get its input from the OS (mainly Windows and Linux). Couldn't find any information about this so far. The idea is to block the default audio output device, simulate one, get the actual audio output and transmit it over the local network. Would really appreciate ideas on how I can realize that. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="23849" LastActivityDate="2019-12-13T11:41:18.917" Title="C++ simulate physical audio output" Tags="|software|audio-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32313" PostTypeId="2" ParentId="32302" CreationDate="2019-12-10T22:34:38.967" Score="3" Body="&lt;p&gt;In the world of heating/ventilating/air conditioning engineering (a.k.a. &quot;HVAC&quot;) there is a number called the &lt;em&gt;comfort index&lt;/em&gt; (which you can search on) which takes both humidity and temperature into account to determine where most humans would feel comfortable. This will tell you the range of optimum humidity percentages to shoot for at a given temperature setting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the use of the humidifier: generally speaking, at lower temperatures, the more humidity, the better your comfort. I recommend adding humidity until you just start getting condensation on your windows, then back off by 10%. Note that mold problems commonly arise in humid houses during the winter with the main culprit being chests of drawers and tall wardrobes backed up against exterior walls. The narrow gap between the furniture and the (poorly-insulated) wall and the poor circulation there will cause the temperature of the air in it to drop below the condensation point and moisturize the wallboard, leading to the rampant growth of mildew on the wall in a pattern which exactly duplicates the outline of the chest or wardrobe. It's a good idea to regularly check for mold behind the furniture, and also to pull those things a few inches away from the walls to improve circulation and prevent condensation. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-12-10T22:34:38.967" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32314" PostTypeId="2" ParentId="32291" CreationDate="2019-12-10T22:43:44.637" Score="2" Body="&lt;p&gt;To summarize the comments given here: During the winter, the cold air outside can hold only a small amount of water in solution, which means it takes very little moisture in cold air to get it to 100% relative humidity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then, when your furnace draws in some of that cold outside &quot;makeup&quot; air, it heats it up and its relative humidity then drops significantly- and you begin feeling &quot;dry&quot; because the warm air in your house begins drawing moisture out of your skin, clothing, furniture, plants, goldfish bowl, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of this, it is common for houses to feel uncomfortably dry in the winter unless you put a kettle of water on the stove to simmer whenever the heat is running.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-12-10T22:43:44.637" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32315" PostTypeId="2" ParentId="32304" CreationDate="2019-12-11T01:32:33.980" Score="2" Body="&lt;p&gt;TO do this efficiently you need an analysis program than can automatically calculated the sensitivities of your design variables to the parameter changes, i.e. mathematically a matrix where the &lt;span class=&quot;math-container&quot;&gt;$i,j$&lt;/span&gt; term is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\dfrac{\partial\, \text{parameter}_i}{\partial\, \text{variable}_j}$$&lt;/span&gt; for all the variables and parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That might mean you need a different CFD program from the one you are currently using.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have done this sort of thing for structural analysis but I'm not a CFD guy. I expect a Multiphysics application like comsol would be able to handle it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One problem you will have to solve is how to come up with a numerical measure of &quot;the best design&quot;, especially if you are using the typical engineering design methodology trading off several parameters against each other until the result &quot;seems about right&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For non-trivial sized problems, calculating all the sensitivities from a single analysis is more or less essential, and it adds very little to the cost of each analysis run, so long as the code has been written to do it. Otherwise the number of different analysis runs will be prohibitively large, if you have more than one or two design variables you want to change. The &quot;sensitivity matrix&quot; allows the optimizing part of the code to change all the design parameters at once for each successive run, which is probably &lt;em&gt;more&lt;/em&gt; efficient than what an experienced designer can do by hand.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-12-11T01:38:18.947" LastActivityDate="2019-12-11T01:38:18.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32316" PostTypeId="1" CreationDate="2019-12-11T05:31:53.310" Score="1" ViewCount="25" Body="&lt;p&gt;I'm aware that polymers like most plastics can go through stress cycles indefinitely as long as the covalent bonds are not ruptured. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, I hear that even metals like steel and titanium do have limited life-spans no matter how small the stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm quite interested in the application of diamond-like carbon coatings to materials. Some of these coatings can even be flexible. For example, I'm beginning to see DLC being applied to headphone diaphragms, which have to be a little flexible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So while the plastic diaphragm that the DLC is bonded to may have an indefinite service life, I'm not sure if the same is true for the DLC coating itself. Would anyone be able to give me an insight into whether or not a DLC coating would have an indefinite life like polymers, or if it has a certain amount of cycles it can go through before it starts to deteriorate? &lt;/p&gt;&#xA;" OwnerUserId="16018" LastActivityDate="2019-12-11T05:31:53.310" Title="Would Amorphous Diamond-like Carbon coatings have a fatigue cycle limit?" Tags="|materials|fatigue|material-science|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32317" PostTypeId="1" CreationDate="2019-12-11T09:08:36.070" Score="-2" ViewCount="72" Body="&lt;p&gt;What elements affect the outcome of a raw egg drop in a random container?&lt;/p&gt;&#xA;&lt;p&gt;I’ve considered air resistance, mass, size of container, size and type of padding. Are there other elements?&lt;/p&gt;&#xA;" OwnerUserId="23859" LastEditorUserId="13311" LastEditDate="2021-01-04T07:59:40.257" LastActivityDate="2021-01-05T05:28:11.710" Title="Predict egg drop outcome in random container" Tags="|materials|design|structural-analysis|control-theory|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32318" PostTypeId="2" ParentId="32317" CreationDate="2019-12-11T10:56:34.210" Score="3" Body="&lt;p&gt;Whether the egg breaks is a function of how much pressure is put on the egg. The same as if you had an egg on the table and started to push down on it with your hand - eventually the force will be high enough to cause the shell to break. If you used a pin instead of your hand then the force required would be lower as its spread over a smaller area.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, assuming you can find through experimentation an expression for how much force it takes to break an egg for a given surface area (likely not constant due to the curvature of the shell), then all you need to work out is the peak force of impact, and the area of application.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Area is influenced by the type of padding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Force is tricker - we need to look at the deceleration of the egg from its final speed before impact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final speed will be affected by the mass, and any air resistance (is there a parachute?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak deceleration will depend on the design of any crumple zones. Does the container bounce or just come to a standstill on the ground? A longer total deceleration phase will mean a lower peak force and a safer egg.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-12-11T10:56:34.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32319" PostTypeId="1" CreationDate="2019-12-11T12:43:03.367" Score="1" ViewCount="688" Body="&lt;p&gt;Suppose a driver gear has larger radius than a driven gear. For example, we have a pinion with a main disk, where the main disk is larger than the pinion and the pinion is a follower. This is the case in many fishing reels that are &quot;baitcaster&quot; models, such as the following picture diagram,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IyPHZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IyPHZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am wondering if a larger gear ratio will result in more torque supplied to the driven gear? Specifically, suppose we have two reel models such that the first has a &lt;span class=&quot;math-container&quot;&gt;$7$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; main drive gear to pinion radius, and the second has a &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; ratio. Will the &lt;span class=&quot;math-container&quot;&gt;$5$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; have less torque being supplied to the pinion gear? &lt;/p&gt;&#xA;" OwnerUserId="23863" LastActivityDate="2022-12-17T23:49:43.523" Title="If a driver gear has larger radius than a driven gear, will a larger gear ratio result in more torque supplied to the driven gear?" Tags="|gears|applied-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32320" PostTypeId="1" CreationDate="2019-12-11T12:44:09.880" Score="2" ViewCount="83" Body="&lt;p&gt;I notice that when &lt;a href=&quot;https://www.customcrateengines.com/engines/ford.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;custom car engines&lt;/a&gt; are sold they nearly always have a big name type name like &quot;Ford&quot; or &quot;Chevy&quot;. Is this because the block and crankshaft comes from those companies and the custom engine company makes everything else, or for some other reason? Why can't custom engine companies make the whole engine? &lt;/p&gt;&#xA;" OwnerUserId="1672" LastActivityDate="2019-12-16T16:21:58.033" Title="Why do custom engines typically specify a type like &quot;Ford&quot; or &quot;Chevy&quot;?" Tags="|automotive-engineering|engines|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32321" PostTypeId="1" CreationDate="2019-12-11T12:54:21.720" Score="1" ViewCount="62" Body="&lt;p&gt;My coworkers and I are finally testing our agriculture sensor on the field. As you can imagine, the moisture and dust are important variables to keep aside from the electronics inside our product. Therefore we now want to take into consideration the sealing more seriously. &#xA;The case of our product is made of Polypropylene, and a gasket generates the seal. We want to purchase an adjustable torque screwdriver for the sealing, but I don't have any idea of the torque range we need. Of course, we want to create an airtight seal and avoid damaging the case plastic thread.&lt;/p&gt;&#xA;" OwnerUserId="14265" LastActivityDate="2019-12-12T06:04:00.713" Title="How to select the correct Bolt/screw Torque for sealing?" Tags="|mechanical-engineering|torque|plastic|seals|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32322" PostTypeId="2" ParentId="32301" CreationDate="2019-12-11T13:08:06.890" Score="1" Body="&lt;p&gt;In addition to my original answer, I thought it worth adding an example of how you might model the full trombone. I noticed that in your screenshot you had &quot;Plane 3&quot; visible, indicating to me that you've been adding reference geometry outside of the three base planes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This model can be made using just two sketches and four features, with all of the dimensions within those sketches being exactly copied from the reference drawing in their positioning. This removes a lot of potential for error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1) Sweep path: I have taken the flared end to be the datum, and built a sketch following the path of the tube. The drawing is dimensioned to the centreline so this is easy to match. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YsGGfm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YsGGfm.png&quot; alt=&quot;sweep path&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 2) Sweep body: Select the path, and Sweep a body. Select &quot;Circular Profile&quot; and enter in the outer diameter of the tube [I'm aware the drawing in in inches and my model is in millimetres - I didn't think it worth changing my settings just for this demo!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vcAx2m.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vcAx2m.png&quot; alt=&quot;Sweep Body&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 3) Bell Profile: We covered this in the original answer. The only thing to add, is that I've set the left-most anchor point to be coincident with the tube surface. This means that if you were to edit the tube diameter that we set in Step 2, then this sketch would automatically update. &lt;em&gt;This&lt;/em&gt; is where defining the curve of the bell as a parabola instead of with coordinate points would come into its own, as the whole curve would be able to adjust to accommodate the change. Note again how all dimensions are linked to the same datum that we picked before.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6V4Nam.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6V4Nam.png&quot; alt=&quot;Bell Profile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 4) Revolve Bell: Pretty self explanatory, revolve. It's best practice to put a centreline into your sketch, and SolidWorks should recognise this as the Axis of Revolution automatically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tWCgrm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tWCgrm.png&quot; alt=&quot;Revolve Bell&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 5) Shell Body - shell the body to the specified wall thickness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8jQCWm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8jQCWm.png&quot; alt=&quot;Shell Body&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 6) Clean Rim: Because the drawing specifies that the rim has a horizontal external diameter, rather than being perpendicular to the spline, this meas that the shell command leaves a small 'step' on the inside of the rim. This is not an error - try reducing the wall thickness down to a smaller value to see where this comes from. Using the 'Delete and Patch' tool will remove these two faces, and extend the inner surface of the bell to meet the external diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YNKLcm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YNKLcm.png&quot; alt=&quot;Clean Rim&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 7) Admire your work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0jyJW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0jyJW.png&quot; alt=&quot;Full Demo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-12-11T13:08:06.890" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32324" PostTypeId="2" ParentId="32320" CreationDate="2019-12-11T14:25:12.293" Score="4" Body="&lt;p&gt;There is no reason in principle why a custom engine builder couldn't cast a block themselves. Everything done by the level of automation in this process &lt;em&gt;could&lt;/em&gt; by done by hand, given enough highly skilled craftsmen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=5oUDzkkdkpQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=5oUDzkkdkpQ&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether there are customers who would want to pay for a hand-made block which costs as much as the rest of the car is another question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other issue is that if you want to replace the engine in a production car, rather obviously you want something which has the same mechanical interfaces to everything else in the car. So unless you want to cast it from a different material for some reason, you might as well start from a standard block that already fits.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-12-11T14:25:12.293" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32325" PostTypeId="1" CreationDate="2019-12-11T16:09:07.497" Score="0" ViewCount="47" Body="&lt;p&gt;I think I once used a dictionary of engineering terms that was either a European effort or one of an (global) international standardization body - but I can't remember...&lt;/p&gt;&#xA;" OwnerUserId="20699" LastEditorUserId="20699" LastEditDate="2019-12-11T17:50:31.723" LastActivityDate="2020-01-10T18:03:30.503" Title="Dictionary of standardized engineering terms?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32326" PostTypeId="2" ParentId="32325" CreationDate="2019-12-11T16:11:14.850" Score="1" Body="&lt;p&gt;I saw &quot;European EN, DIN, ISO, IEC and VDA Standards&quot; somewhere and then found it using &lt;a href=&quot;https://duckduckgo.com/?q=iec+dictionary&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://duckduckgo.com/?q=iec+dictionary&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.electropedia.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.electropedia.org/&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="20699" LastActivityDate="2019-12-11T16:11:14.850" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32327" PostTypeId="1" AcceptedAnswerId="32328" CreationDate="2019-12-11T16:29:00.010" Score="-2" ViewCount="1174" Body="&lt;p&gt;could anybody help me by providing an internal involute spline data sheet similar to the attached picture for an 1-3/8&quot; splined shaft with 21 teeth, 16DP and 30 degree pressure angle.&#xA;Thanks for your help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/twAhf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/twAhf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23869" LastActivityDate="2019-12-11T19:10:10.303" Title="Internal involute spline 1.375 - 21T" Tags="|mechanical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32328" PostTypeId="2" ParentId="32327" CreationDate="2019-12-11T19:10:10.303" Score="1" Body="&lt;p&gt;From Machinery's Handbook for a flat root side fit spline. There are a few other different styles, this is what I use. This is a class 5 tolerance (typical industry level) per ANSI B92.1-1970&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tJBjZ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tJBjZ.png&quot; alt=&quot;21T16P30A&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2019-12-11T19:10:10.303" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32329" PostTypeId="1" CreationDate="2019-12-11T20:16:16.117" Score="1" ViewCount="25" Body="&lt;p&gt;Is there an arrangement of components the takes a source of gas with oscillating pressure and outputs a steady near constant pressure stream? &lt;/p&gt;&#xA;" OwnerUserId="18744" LastActivityDate="2019-12-12T07:48:57.767" Title="Fluid dynamics' equivalent to current smoothing" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32330" PostTypeId="2" ParentId="32320" CreationDate="2019-12-11T21:46:33.680" Score="2" Body="&lt;p&gt;First, because designing a car engine takes an incredible amount of engineering.  It's not something that someone just sits down and does in an afternoon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second, because a &quot;custom built&quot; XYZ engine (302 Ford, 427 Chevy, whatever) is going to be much closer to a bolt-in replacement to a same-type engine.  A typical 65-75 small block Chevy engine won't even physically fit into the engine compartment of a typical 65-75 Ford, and visa-versa -- the companies arranged their cars differently, and the oil sumps are at opposite ends of the engines.  So even if you didn't get murdered by Ford fans for doing it, it would be a job and a half to put a small-block Chevy into a Boss 302 Mustang -- and ditto for a 302 Ford into a Camaro.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third, there's this thing called &quot;brand loyalty&quot;.  Even though I'm not possessed of much of it, if I have some nice Chevy, I'm probably going to want to put a Chevy engine in it.  Ditto Ford with Ford or Dodge with Dodge.  You might find me putting a small-block Chevy into a Henry J, though.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-12-11T21:46:33.680" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32331" PostTypeId="2" ParentId="32329" CreationDate="2019-12-12T03:22:53.943" Score="1" Body="&lt;p&gt;This is a common problem in the field of &lt;em&gt;noise control&lt;/em&gt; (within the larger field of &lt;em&gt;engineering acoustics&lt;/em&gt;). Do you have access to any of the standard textbooks on engineering acoustics? You'll most likely find your answer there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also please note that there is a device called a &lt;em&gt;noise suppressor&lt;/em&gt; or &lt;em&gt;muffler&lt;/em&gt; which is commonly used to eliminate or minimize the generation and propagation of noise from a pressurized air outlet. If you search on these terms you will uncover the websites of the firms that design and sell them, and find out how to solve your problem.&#xA;.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-12-12T03:22:53.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32332" PostTypeId="2" ParentId="32321" CreationDate="2019-12-12T06:04:00.713" Score="3" Body="&lt;p&gt;Use a dial gauge and an adjustable torque wrench / screwdriver, then measure the compression of the seal compared to the torque applied - keep increasing torque setting until compression is sufficient for purpose without exceeding the thread capacity.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-12-12T06:04:00.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32333" PostTypeId="2" ParentId="32319" CreationDate="2019-12-12T07:39:52.353" Score="2" Body="&lt;p&gt;Actually it will be the opposite. In any case the smaller the driving gear the higher the torque applied to the driven gear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think about the trade off between speed and torque. If you change the gear ratio so your output rpm is higher, then you can assume that you have lost some mechanical advantage as a trade off, ie. your torque is reduced.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course any time you make a driving gear larger, that means it will rotate the driven gear more for every rotation of the drive gear, and vice versa when you make it smaller.&lt;/p&gt;&#xA;" OwnerUserId="23080" LastActivityDate="2019-12-12T07:39:52.353" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32334" PostTypeId="2" ParentId="32329" CreationDate="2019-12-12T07:48:57.767" Score="0" Body="&lt;p&gt;You can add a few bypass pipes with a constraint such as a fin or a stator or a positive blocking valve, spring loaded, that will intake the tip of pressure surge amplitude and deliver it muted farther down the stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A comparable liquid regulator would be one with a slanted ramp skimming the flow at the average volume and merging it back down stream.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-12T07:48:57.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32335" PostTypeId="1" CreationDate="2019-12-12T09:08:40.217" Score="2" ViewCount="161" Body="&lt;p&gt;I'm working with solidworks and trying to make a plane at a certain distance and angle from another plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just for some context, I want to do something like this:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/3E9C9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3E9C9.png&quot; alt=&quot;m&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My plan was to create a plane at the reference and then make a sketch on the new plane and convert entities so that the feature on the bottom right extends to the back. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know I can do it like this, and extrude out the sketch, then make another sketch on the side face:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/oH4ml.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oH4ml.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But Im wondering if theres a more direct/efficient way?&lt;/p&gt;&#xA;" OwnerUserId="23845" LastActivityDate="2019-12-13T15:13:28.667" Title="Trying to get a plane at a certain distance and angle" Tags="|mechanical-engineering|cad|solidworks|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32336" PostTypeId="1" CreationDate="2019-12-12T09:38:54.843" Score="1" ViewCount="123" Body="&lt;p&gt;&lt;strong&gt;Exercise&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The building is the following:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4LOw0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4LOw0.png&quot; alt=&quot;Building&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The exercise is to basicly calculate everything (forces, reactions, ...) and draw several free body diagrams, given &lt;span class=&quot;math-container&quot;&gt;$a,h,\vec{G},(\vec{g})$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt; is a frame, &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; is a bar, &lt;span class=&quot;math-container&quot;&gt;$6$&lt;/span&gt; is a heavy disc with weight &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; and the rest parts are rods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now my problem is drawing the free body diagram for the component 6. I got the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fCkvj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fCkvj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would result in:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\sum F_x=0 : cos(\alpha)4=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\sum F_y=0 : -y_3-y_5-sin(\alpha)4-G=0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\sum M_{(3)}=0 :  aG+3a\cdot (y_5+sin(\alpha)4)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However the first equation would result in the force &lt;span class=&quot;math-container&quot;&gt;$4=0$&lt;/span&gt;, which doesn't really make any sense for me. If I think about it I would say, that rod 4 should take some force and make the building stronger... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you tell me, wehere my problem is? :)&#xA;Thanks&lt;/p&gt;&#xA;" OwnerUserId="23876" LastActivityDate="2020-01-13T02:01:40.353" Title="Simple Free Body Diagram" Tags="|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32337" PostTypeId="2" ParentId="32335" CreationDate="2019-12-12T10:16:07.683" Score="1" Body="&lt;p&gt;You can model the part at 90 degrees, and then use the Move Face tool to rotate the feature about the top connecting edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;N.B. You can select features (as well as single faces) when picking which faces to move.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This whole model can be made with four features, and no additional planes or reference geometry beyond the base planes.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;an extrude from above for the 'L' bracket&lt;/li&gt;&#xA;&lt;li&gt;A cut from the front to add slots and to round off the end.&lt;/li&gt;&#xA;&lt;li&gt;An extrude from the front for the 'notched' plate, shown blue in the image below.&lt;/li&gt;&#xA;&lt;li&gt;Move face, to rotate it into position, shown yellow in the image below.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Again, the key point to take away from this is &lt;em&gt;editability&lt;/em&gt;. You should be able to change any one dimension without the rest of the model needing to be manually updated. Selecting the Extrude feature (made in step 3) as the input for the Move face (step 4) means that even if you edit the extrude to have extra holes or to remove a slot, the move face will still work. Manually selecting the faces may cause this feature to fail as new faces appear or old faces are removed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YemSA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YemSA.png&quot; alt=&quot;Rotate Face Demo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-12-12T10:16:07.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32338" PostTypeId="1" CreationDate="2019-12-12T20:15:12.847" Score="-1" ViewCount="143" Body="&lt;p&gt;I am finding it difficult to get consensus on the definition of design wind speed for aeolic turbines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it was the velocity that corresponds to the maximum power output, but I have already heard that it could be the velocity that corresponds to lambda (tip speed ratio) equals 7.&#xA;What is this really about?&lt;/p&gt;&#xA;" OwnerUserId="7947" LastEditorUserId="16486" LastEditDate="2019-12-19T18:22:38.520" LastActivityDate="2019-12-19T20:31:50.717" Title="What is and how do you calculate the wind velocity design for wind turbines?" Tags="|turbines|renewable-energy|wind-power|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32340" PostTypeId="1" AcceptedAnswerId="32343" CreationDate="2019-12-13T01:51:27.927" Score="2" ViewCount="506" Body="&lt;p&gt;I'm trying to make an equation driven conical spiral, you can download the file &lt;a href=&quot;https://www.dropbox.com/s/14u0kdhxeui9px8/Coil%20spring%20no%20disk.SLDPRT?dl=0&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. I know that I can use the spiral tool to achieve this but there are reasons why I want to do it like this. I have five global variables that control the characteristics of the spring: the initial radius ri, the final radius rf, height H, number of coils nc, and diameter of wire d (it's a square wire so the diameter is the width/height of the square). The equations for generating the coil are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x=(&quot;ri&quot;-t*((&quot;rf&quot;-&quot;ri&quot;)/&quot;H&quot;))*cos(t*((2*pi*&quot;nc&quot;)/&quot;H&quot;))&#xA;&#xA;y=(&quot;ri&quot;-t*((&quot;rf&quot;-&quot;ri&quot;)/&quot;H&quot;))*sin(t*((2*pi*&quot;nc&quot;)/&quot;H&quot;))&#xA;&#xA;z=t&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;with limits from t=0 to t=H. The problem is that when I change the variables, the spring does not rebuild, even when I use Ctrl+Q several times. I have no idea why this is. I have another small coil above the spring which I'm going to use to make an end coil and it seems to update ok but the main coil doesn't. Any help would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="18933" LastActivityDate="2019-12-13T12:11:59.530" Title="Equation driven conical spring not rebuilding properly in Solidworks 2019" Tags="|solidworks|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32342" PostTypeId="2" ParentId="32309" CreationDate="2019-12-13T11:41:18.917" Score="0" Body="&lt;p&gt;On a linux system, all devices are items in the filesystem. Ther're organized under /dev&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can do read and write operations on them. Hence, reading from an audio device is just reading from the &quot;file&quot;. Try &quot;ll /dev/snd&quot; to enumerate available devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure, but you may get away without programming, if you pip the output of you device on one computer into a device on the other computer, with just system tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On windows: I don't know. Probably way more complicated, but others may know better. &lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2019-12-13T11:41:18.917" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32343" PostTypeId="2" ParentId="32340" CreationDate="2019-12-13T12:11:59.530" Score="1" Body="&lt;p&gt;I can't answer your original question re: why using CTRL+Q did not successfully rebuild for you when using an equation driven curve - because when I downloaded your file, I could not recreate the error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I can do is demonstrate how you could apply your global variables to the built-in helix/spiral tool, such that the you can edit your variables in exactly the same way, and it automatically rebuilds whenever you edit a global variable value. (SW sends a CTRL+B when you press enter in the value/equation box).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.imgur.com/MgukWXh.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/MgukWXh.gif&quot; alt=&quot;Editing Global Variables on Coil Spring&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not just suggesting this 'just because I can' - it's a more robust way to model this object. Your supplied model using the equation curve takes almost &lt;strong&gt;four times&lt;/strong&gt; as long to rebuild. (I have deleted features in your model after the initial sweep to keep it a level playing field)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QoLnD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QoLnD.png&quot; alt=&quot;Rebuild times for helix/spiral vs equation curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here is my file to download if you would like: &lt;a href=&quot;http://www.filedropper.com/conicalspringglobalvariablesdemo&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.filedropper.com/conicalspringglobalvariablesdemo&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-12-13T12:11:59.530" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32344" PostTypeId="1" CreationDate="2019-12-13T13:25:45.247" Score="1" ViewCount="254" Body="&lt;p&gt;For sizing purposes I'm trying to calculate head loss for a level spreader outlet for water, where the water is being supplied from below by pressure. It's essentially a reverse slot drain—instead of water flowing down the slot and leaving, it's entering from below and leaving up through the top of the slot. There's no outlet pipe; everything leaves through the slot for the purposes of this calculation. Here's a sketch showing two views of it (I left off the inlet pipe):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/C825J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C825J.png&quot; alt=&quot;two views of slot weir showing path of water and relevant dimensions&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the water leaves the slot, vertically, I know it can be modeled as weir flow, so I can calculate the height it would have to reach for a given flow rate for a sharp rectangular weir of its length. From that I know I need at least enough head/water height to reach that height above the slot: &lt;span class=&quot;math-container&quot;&gt;$D_{pipe} + H_{slot} + H_{weir}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My question is,&lt;/strong&gt; how do I model the friction losses from the slot itself? I imagine the width of the slot to be significant, at least for small enough widths. I found a rectangular submerged orifice equation from &lt;a href=&quot;http://irrigation.wsu.edu/Content/Calculators/Water-Measurements/Rectangular-Submerged-Orifices.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;this site&lt;/a&gt; (using cfs &amp;amp; square feet):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Q = 0.61A \cdot \sqrt{64.4 \cdot H} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and there are definite similarities: the top of the slot in my problem is also a rectangular opening that, once enough water is flowing, will have water on all sides. However, it's turned 90 degrees, so I feel that the above equation might not be appropriate. I imagine the equation above takes into consideration the need for additional head to drive enough horizontal flow, which might be different from the vertical flow in the slot I'm considering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's an appropriate way to calculate the head loss from the slot opening itself?&lt;/p&gt;&#xA;" OwnerUserId="21409" LastActivityDate="2019-12-13T13:25:45.247" Title="Head loss from rectangular slot orifice flowing up?" Tags="|hydraulics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32345" PostTypeId="1" AcceptedAnswerId="32347" CreationDate="2019-12-13T14:35:38.070" Score="1" ViewCount="480" Body="&lt;p&gt;I want to attach a 12 toothed spur gear with a bore diameter of 6mm onto the shaft of this &lt;a href=&quot;https://www.maxongroup.com/maxon/view/product/488607&quot; rel=&quot;nofollow noreferrer&quot;&gt;motor&lt;/a&gt; (6mm diameter). How would I go about doing this? From what I see, the shaft is smooth so is my only option to find a gear with a set screw? Or is it sufficient to attach a gear with no set screw and simply rely on the friction of the bore? The torques will be low. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any websites that one would recommend for finding gears like this? I'm finding it impossible to find gears with a module of 1.0, bore diam. of 6mm, and a thickness of less than 20mm to fit onto the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated. &lt;/p&gt;&#xA;" OwnerUserId="23762" LastActivityDate="2019-12-13T15:10:57.860" Title="Attaching Gear onto Motor" Tags="|mechanical-engineering|motors|gears|power|power-transmission|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32346" PostTypeId="2" ParentId="32345" CreationDate="2019-12-13T14:49:27.950" Score="2" Body="&lt;p&gt;What are the respective tolerances on the “6mm”?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either it will be a press fit which will be sufficient to handle the torque and shock loading or it will not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, then you have to fit set screws and even machine a flat on the shaft.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-12-13T14:49:27.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32347" PostTypeId="2" ParentId="32345" CreationDate="2019-12-13T15:09:36.153" Score="3" Body="&lt;p&gt;Here you go: &lt;a href=&quot;https://uk.rs-online.com/web/p/spur-gears/8787907/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://uk.rs-online.com/web/p/spur-gears/8787907/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c7lHc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c7lHc.png&quot; alt=&quot;1 Mod 12 Tooth Gear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble is that a 1MOD 12 Tooth gear has a pitch diameter of 12mm, i.e.the bottom of the teeth is less than 12mm, so once you drill a 6mm hole through the middle there's really not much material left at all to make a hub from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the specific requirement or constraint that means you need this specific gear on that specific motor? You may find there is a more robust and elegant solution if you can be more flexible in your approach, but without the information, I can't help any further.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-12-13T15:09:36.153" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32348" PostTypeId="2" ParentId="32345" CreationDate="2019-12-13T15:10:57.860" Score="2" Body="&lt;p&gt;Boston Gear or McMaster-Carr might have suitable off the shelf gears.&#xA;You could also press the gear onto a hub that has a set screw. I would avoid pressing anything directly onto a motor shaft, you could bend the rotor and/or damage the internal bearings.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2019-12-13T15:10:57.860" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32354" PostTypeId="1" AcceptedAnswerId="32356" CreationDate="2019-12-13T17:10:12.890" Score="-1" ViewCount="38" Body="&lt;p&gt;Hello I am building a door and I have taken a large rectangle piece of wood for the base.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the top-right point of the rectangle I am planning to connect it with the top right point of a  large rectangle piece of wood  vertically which would be the door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will create a rotation axis correct or am I wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I usually trust myself before building something like this but this time I am undecided , i dont know so I came to this site for help whatever.&lt;/p&gt;&#xA;" OwnerUserId="23890" LastEditorUserId="23890" LastEditDate="2019-12-13T17:20:43.003" LastActivityDate="2019-12-13T20:24:31.697" Title="Door functional or not?" Tags="|applied-mechanics|" AnswerCount="1" CommentCount="7" ClosedDate="2019-12-13T23:18:18.283" ContentLicense="CC BY-SA 4.0" />
  <row Id="32355" PostTypeId="1" AcceptedAnswerId="32357" CreationDate="2019-12-13T19:55:28.937" Score="-1" ViewCount="64" Body="&lt;p&gt;I'm doing a project that involves spinning up a (roughly) 2.28 kg object to 5000 rpm (via a DC motor) and then letting it continue rotating under its own influence (without any more input from the motor).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See my image:&lt;img src=&quot;https://i.stack.imgur.com/JwgAB.png&quot; alt=&quot;![enter image description here&quot;&gt;]&lt;a href=&quot;https://i.stack.imgur.com/JwgAB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For my project to work, friction and other forms of resistance must be reduced to a minimum. My initial thought was simply to to use the DC motor itself, but in my experience, most such motors generate excess internal friction. And clearly, a motor that goes on spinning without electrical input will become a generator, thus creating more resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does anyone have any suggestions? Preferably something that I could simply buy.&lt;/p&gt;&#xA;" OwnerUserId="23883" LastActivityDate="2019-12-13T20:59:47.280" Title="Designing A Freely Spinning Object" Tags="|design|motors|prototyping|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32356" PostTypeId="2" ParentId="32354" CreationDate="2019-12-13T20:24:31.697" Score="0" Body="&lt;p&gt;If I understand you correctly, your idea would work. I would just try to focus on what you're using to connect the two components with. Because of the structure of your design, I'd recommend a Hinge Bolt as there will be a lot of stress on whatever is used. I'd also suggest making sure the lower part is firmly connected to whatever is beneath it, perhaps screwed into place, as you don't want the whole thing to flip over. I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="23883" LastActivityDate="2019-12-13T20:24:31.697" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32357" PostTypeId="2" ParentId="32355" CreationDate="2019-12-13T20:59:47.280" Score="1" Body="&lt;p&gt;Option 1: A plain old ordinary motor with a clutch of some sort that is completely non-contact when it is disengaged.  If losses when you're spinning the thing up don't matter too much, then a rubber drive wheel on a motor that engages the outside circumference of your object (or a wheel on the axle on top) would work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Option 2: DC motor have &quot;excess&quot; losses because they have permanent magnets in them, that continue to cause losses in the armature iron as they spin.  So use a motor that doesn't have permanent magnets, such as a wound-field DC motor (hard to get, unless you get a universal motor), or an induction motor driven from a VFD to get you above synchronous speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be a trade-off here: most inexpensive motors of this sort are still not aerodynamically perfect, and may not be happy at 5000 RPM.  Moreover, finding suitable motors that are both off the shelf and as small as your application demands will be a challenge.  You'll have to shop around for a suitable motor (and possibly VFD), and you may not be able to avoid aerodynamic losses unless you're going to operate this in a vacuum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that if I were designing this for production and could specify the motor, I'd want a small induction machine or a variable-reluctance motor -- probably the induction machine.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-12-13T20:59:47.280" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32358" PostTypeId="2" ParentId="32273" CreationDate="2019-12-13T21:40:13.610" Score="1" Body="&lt;p&gt;Cost engineering is &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;the engineering practice devoted to the management of project cost,&#xA;  involving such activities as estimating, cost control, cost&#xA;  forecasting, investment appraisal and risk analysis.&quot; &quot;Cost Engineers&#xA;  budget, plan and monitor investment projects. They seek the optimum&#xA;  balance between cost, quality and time requirements.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;With Regard the project control cycle, &lt;a href=&quot;https://www.pmi.org/pmbok-guide-standards/foundational/pmbok&quot; rel=&quot;nofollow noreferrer&quot;&gt;Project Management Body Of Knowledge&lt;/a&gt; is a good place to get started. Below is short overview get to started. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kwBrV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kwBrV.jpg&quot; alt=&quot;Five Phases of Project Management&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some other items to consider are  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Location&lt;/strong&gt;: Following a short list of regions to consider. Finger Lakes, NY, Willamette Valley, OR, Sonoma County, CA, Napa Valley, CA, Charlottesville/Central Virginia, Santa Barbara County, CA, Walla Walla, WA.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Resources&lt;/strong&gt;: This includes labor, raw material, land, regulations (Federal, state and local)&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Regulations&lt;/strong&gt;: It is important to take a look are Federal, State and Local regulation as they could have impact on cost.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Tariffs&lt;/strong&gt;: With on going trade negotiations between USA and other countries it might good idea to review if the project might be impacted by Tariffs&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Communication&lt;/strong&gt;: Language barriers also can have impact on the project. You might have language translator if the medium of communication is something other than English.&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Measurement Units&lt;/strong&gt;: Unfortunately the is still following &lt;a href=&quot;https://en.wikipedia.org/wiki/United_States_customary_units&quot; rel=&quot;nofollow noreferrer&quot;&gt;United States customary units&lt;/a&gt;, while the rest of the world has moved on. This could have an effect on the project. &lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Risk&lt;/strong&gt;: Some type of risks to consider are  Default risk, Inflation risk, Maturity risk and Liquidity risk&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Default risk&lt;/strong&gt; refers to the likelihood that you will fail to get your money back or receive the return you are due. Assessing this risk&#xA;  means deciding if the company in which the investment is made is&#xA;  likely to default. Startup companies typically have high default risk&#xA;  for example.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Inflation risk&lt;/strong&gt; addresses the chances that the predictable overall economic rise in prices for goods and services will cause the&#xA;  investment to lose some of its value. Investment in long-term ventures&#xA;  presents greater inflation risk than short-term investing.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Maturity risk&lt;/strong&gt; refers to the gamble taken when investors tie up their cash in investments that require a set period. When money is&#xA;  tied up in long-term investments, it cannot be used for other, perhaps&#xA;  more profitable, purposes.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;Liquidity risk&lt;/strong&gt; refers to the possibility that an investment may need to be sold to free up cash. An investment's liquidity is how&#xA;  readily it can be converted back into cash prior to maturation, or if&#xA;  that's even possible. The easier it is to sell off assets when needed,&#xA;  the lower the liquidity risk.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.thrillist.com/travel/nation/best-us-wine-regions-to-visit-finger-lakes-napa-valley-sonoma-valley&quot; rel=&quot;nofollow noreferrer&quot;&gt;America's Best Wine Regions to Visit, Ranked by Sommeliers&lt;/a&gt; &lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.smartsheet.com/blog/demystifying-5-phases-project-management&quot; rel=&quot;nofollow noreferrer&quot;&gt;Demystifying the 5 Phases of Project Management&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Cost_engineering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cost engineering&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-12-13T21:46:00.950" LastActivityDate="2019-12-13T21:46:00.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32359" PostTypeId="1" CreationDate="2019-12-13T22:49:14.577" Score="2" ViewCount="432" Body="&lt;p&gt;I'm wondering if anyone has any experience cutting aluminum foil on a very cheap CNC or laser cutter. I have a lot of 3d printers but haven't yet purchased a CNC router. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am talking about aluminum foil about as thick as Reynold's wrap (brand named aluminum foil). So, that's about 0.2mm thick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have another use case where I may want to cut an aluminum sheet that is about 1.0mm tall. Do you think it would be able to cut that, in multiple passes even?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To what extend can a cheap CNC router cut aluminum?&lt;/p&gt;&#xA;" OwnerUserId="9573" LastEditorUserId="9573" LastEditDate="2019-12-13T23:21:47.043" LastActivityDate="2019-12-18T22:27:04.907" Title="Can a 3018 or 1610 CNC router cut aluminum (foil)?" Tags="|cnc|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32360" PostTypeId="2" ParentId="32336" CreationDate="2019-12-13T23:51:14.980" Score="3" Body="&lt;p&gt;Your result is correct: member 4 is what's known as a &lt;a href=&quot;https://en.wikipedia.org/wiki/Zero_force_member&quot; rel=&quot;nofollow noreferrer&quot;&gt;zero-force member&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This result is inescapable given it's definition as a rod which can only take in axial loads. Since it's on a diagonal, any axial loads on it will generate a horizontal component. However, at the bottom node you've studied, there are no other members which can resist that horizontal component. So you have no choice: rod 4 can't have any axial force at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it did, that would mean the node is unstable, with a net horizontal force. And, by Newton's laws of motion, where there's a force, there's an acceleration. So if rod 4 had any internal force, it'd mean this isn't a structure but a mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One escape would be if we can assume the disk 6 can take a structural load. In such a case, rod 4 would be under tension, generating a leftward force which would be resisted by disk 6's compression. But we can't make that assumption since we don't know anything about disk 6. It could well be made of lead: very heavy, but very weak.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the by, the Wikipedia page above is worth a read, since it describes how zero-force members aren't actually useless in the real world:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Reasons for Zero-force members in a truss system&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;These members contribute to the stability of the structure, by providing buckling prevention for long slender members under compressive forces&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;These members can carry loads in the event that variations are introduced in the normal external loading configuration&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And pro-tip, don't use numbers to describe the forces from rods (for instance, writing &lt;span class=&quot;math-container&quot;&gt;$4 = 0$&lt;/span&gt; to describe your conclusion about the forces in rod 4). It'll just get you confused between what's a rod and what's a multiplication constant. Instead, use &lt;span class=&quot;math-container&quot;&gt;$F_4 = 0$&lt;/span&gt; or something of the sort. Never use numbers for variables; always use letters (Latin or Greek) with optional subscripts.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-12-13T23:51:14.980" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32361" PostTypeId="1" AcceptedAnswerId="32362" CreationDate="2019-12-14T03:25:28.810" Score="2" ViewCount="20" Body="&lt;p&gt;Looking at a part like this&#xA;LM2596&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is rated for 3Amps with a voltage range from 3.3-12 volts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does that mean that this voltage regulator will only output a maximum of 3amps? &lt;/p&gt;&#xA;" OwnerUserId="23892" LastActivityDate="2019-12-14T03:37:39.540" Title="Maximum amps from a switching voltage regulator" Tags="|power|power-electronics|power-engineering|regulations|current|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32362" PostTypeId="2" ParentId="32361" CreationDate="2019-12-14T03:37:39.540" Score="2" Body="&lt;p&gt;That is an oversimplification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the data sheet &lt;a href=&quot;http://www.ti.com/lit/ds/symlink/lm2596.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.ti.com/lit/ds/symlink/lm2596.pdf&lt;/a&gt; there are two relevant features, internal over-temperature protection and a maximum current limiter (Typical value 4.5A, minimum 3.6A).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bottom line is that trying to get more than 3A &lt;em&gt;continuous&lt;/em&gt; current from the regulator will probably not end well. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-12-14T03:37:39.540" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32363" PostTypeId="1" CreationDate="2019-12-14T03:46:06.890" Score="1" ViewCount="21" Body="&lt;p&gt;Consider an ideal adiabatic compressor, ignoring the remainder of the thermodynamic cycle. In this case, the heat flow &lt;strong&gt;q&lt;/strong&gt; into and out of the compressor itself is zero, by definition of adiabatic compression. The output shaft work then is:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$h_{out} - h_{in} = w_{shaft}$$&lt;/span&gt;&#xA;However, if we then consider the case of a compressor which experiences no losses &lt;strong&gt;excepting&lt;/strong&gt; thermal inertia and heating. It follows logically that there would be losses to the output work due to heat energy imparted to the compressor itself. My question is, suppose we have two identical compressors, one which is made of Copper, and one which is made of Steel. If we ignore phase changes and assume that both materials can infinitely absorb heat. Will the losses imparted on the exhaust of the compressors be different in any way? Just because Copper will increase in temperature more than Steel for the same input of energy, does this have any impact on the efficiency of a compressor?&lt;/p&gt;&#xA;" OwnerUserId="23222" LastActivityDate="2019-12-14T03:46:06.890" Title="Ignoring Mechanical Material Properties and Design Implications, in a Compressor, does the Thermal Density of the Blades Affect Efficiency?" Tags="|thermodynamics|compressors|engines|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32365" PostTypeId="1" CreationDate="2019-12-14T08:06:29.333" Score="1" ViewCount="74" Body="&lt;p&gt;This is related to my recent question&#xA;&lt;a href=&quot;https://engineering.stackexchange.com/questions/32238/can-i-weaken-a-coil-spring-consisting-of-spring-steel?noredirect=1&amp;amp;lq=1&quot;&gt;Can I weaken a coil spring consisting of spring steel?&lt;/a&gt;. The answer here could create an additional answer there, but I do not expect that. The difference to that question is that this is about changes in the actual spring properties, so much that there could be a fully theoretical answer with no relation to empiric coil behaviour. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What changes in the properties of a coil spring if I stretch it so much that it causes plastic deformation, and cut it to restore the length?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I take a steel coil spring with a length without load of 100 mm. I extend it so much that it plastically deforms, and has a new rest length of 150 mm. Now, I cut the spring to the length of 100 mm. If it had 20 windings before, it now has about 13 windings. &#xA;It does not need to be a steel spring, if a more homogeneous material simplifies it. An answer could even use an abstract material that may or may not &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is different?&lt;br&gt;&#xA;&lt;strong&gt;How have the spring properties changed?&lt;/strong&gt;&lt;br&gt;&#xA;The mass is reduced to 2/3, how does that change the resonance frequency? Is it influenced by other factors?&lt;/p&gt;&#xA;" OwnerUserId="711" LastEditorUserId="711" LastEditDate="2019-12-14T10:14:15.213" LastActivityDate="2019-12-14T17:30:59.517" Title="Overextending and cutting a coil spring" Tags="|steel|modeling|springs|elastic-modulus|coil-spring|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32366" PostTypeId="2" ParentId="32365" CreationDate="2019-12-14T17:30:59.517" Score="2" Body="&lt;p&gt;Stiffness:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If you cut the spring by 1/2, it will have double stiffness (spring rate). This means your 2/3 springs rate will be 3/2 of the original rate. This is independent of the length of the spring, it is onyl the number of coils, that cause the difference. &lt;/li&gt;&#xA;&lt;li&gt;If you deform a material into plastic deformation, depending on the material anything can happen. &lt;/li&gt;&#xA;&lt;li&gt;For significant elongations, also the diameter of the coil will be reduced significantly (you can easily try this with a ball pens spring). This will increase the stiffness by the factor of whicht the coild diameter is reduced. 1/2 coild diameter, 2/1 stiffness. &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The eigenfrequencies of the spring itself (note plural) change. For two reasons: increased stiffness and reduced wheight. Both will increase the eigenfrequencies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you assemble the spring with other bodies, the eigenfrequencies of the system will also depend on the other bodies masses and how the're connected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any way, the spring will become stiffer, this is opposite of what you wanted in your other question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have an idea for the other question, that I will put there.&lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2019-12-14T17:30:59.517" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32367" PostTypeId="2" ParentId="32238" CreationDate="2019-12-14T17:37:10.277" Score="1" Body="&lt;p&gt;Another way to decrese the stiffness without making the spring longer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Making it longer would obviously decrese the stiffness. And adding other springs in series would do also. &#xA;You could add springs in series and put them one in another. The inner springs should be shorted and connected end-to-end. You could have three springs in total. The second connected to then end of the first, going back inside. At some point (say half of the outer springs lengt) you append another spring, which is again inside the second. It should be long enough to reach out of the first one. Well, you may have to cut the first one a little bit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't expect this will be useful, but it really depends on factors as your packaging, available of springs, what you really want to achieve...&lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2019-12-14T17:37:10.277" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32368" PostTypeId="2" ParentId="21466" CreationDate="2019-12-14T17:39:25.337" Score="1" Body="&lt;p&gt;&lt;strong&gt;No&lt;/strong&gt;, for both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can and do routinely produce high quality standard coil springs on the industrial scale, and are used to produce custom springs in the same quality.&#xA;And we are not talking about outliers like defective individual springs or spring batches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea itself that a contemporary industrial spring could be regularly defective, or in some other way not just a working spring, is irritating to me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Item 2. is clear cut, but in 1., you ask whether it is even possible an insignificant change exists that could be significant in the product of the competitor. I do not think there is a weak version of spring degradation, but there is something related that is interesting: A part made of soft, annealed steel that is subject to deformation similar to a spring, would become harder from the deformation. Work hardening results in much harder steel, but hard and brittle, not hard and elastic. It's almost like a bad spring would fix itself. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So from the engineering perspective: No, they do not degrade, confidently without even checking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there is an interesting aspect to it that can suggest it is not just fraud against a competitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a psychological effect unrelated to the material of the springs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;It is the placebo effect.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting part is that the placebo effect is something much more powerful beyond what is common knowledge. It has nothing to do specifically with replacing prescription drugs by inert substances. That is only one of the ways how to make something happen only because the patient thinks some external cause exists, like a medication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People have a, roughly speaking, so personal and intense relationship with this particular use of a standard coil spring that an attribution that is not obviously wrong can be stable in a subculture tradition.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The placebo effect can easily work to change how strong a spring seems to be. And in a way, the spring becomes legitimately weaker for the person, he does perceive it.&lt;/p&gt;&#xA;" OwnerUserId="711" LastEditorUserId="711" LastEditDate="2019-12-14T22:14:38.870" LastActivityDate="2019-12-14T22:14:38.870" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32369" PostTypeId="2" ParentId="21466" CreationDate="2019-12-14T18:04:09.003" Score="0" Body="&lt;p&gt;Springs do degrade in length and load. How much and how fast depends on various factors. One very obvious factor is the design, which causes for the same displacement at the grip a different strain in the material. The higher the strain, the higher the wear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the spring wire a bending beam. You can generate the same second moment of inertia (which is one factor of generating spring stiffness) with a flat and wide cross section and a high and narrow one. The last one however will generate more strain in the material. &#xA;Or, compare a two thinner wires with one thicker one. Or if you have a thicker, but longer wire (more coils). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That said, I also do not think, that any quality manufacturer would deliver a design, that shows an effect of degradation. &#xA;But I woulnd't be surprised if you find them. For anything that people have manufactured, you can find examples of ultra low quality. Besides design, also material can be just too crappy. Just, using this to sell something may only show how much this manufacturer just escaped this ultra low quality business.&lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2019-12-14T18:04:09.003" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32370" PostTypeId="2" ParentId="19205" CreationDate="2019-12-14T19:09:52.813" Score="0" Body="&lt;p&gt;This is more an answer to your problem, than the actual question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When in the world of professional and commercial CAE (where Simulink is), the natural way to implement the controller for the helicopter would be software like Simulink, or Activate, or similar. This is a strong indication if the controller is complex, most often if you develop a commercial grade controller.&#xA;The natural way to simulate the helicopter is an MBD (multi-body-dynamics) software. It usually comes with prostprocessing software, that allows you to plot any values or render a 3d animation. This is a strong indication if you develop a helicopter, that you turned into a prototype and that should actually fly at some point. MotionView/MotionSolve, ADAMS, RecurDyn, Simpack, ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FEM can back the MBD model with elastic bodies. If the helicopter is not an RC type, then you probably want the rotor blades modeled like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If one of the two is a simplyfied model, you can do it in the other software. This is like trying to do the helicopter in Simulink. But, the same can be done the other way round. The helicopter model in MBD could contain some differential equations for the controller. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Industries use both at the same time. MBD and controls can be connected via: &#xA;a) Cosimulation, where at finite time steps, values are passed between the models. When designing controllers, this is not so wrong, because digital controllers also have to work at finite time steps. &#xA;b) FMU's can be used to intergrate one model in the other. The FMU implements only the equations, the software of the other part integrates/solves these. For example: You use MBD to simulate the helicopter and choose this as host. You export the FMU of the controller from the controls software and import it in the mbd software. The mbd software will solve the equations of the controller along with the equations of the mbd-model. This will prevent the solvers from interchanging values at time steps or use values from last time step.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using professional software might take away some of the thrill of developing it on your own, but if you're after that, you should probably choose the python approach, that Alexander proposed. I am not sure how much simulink would be adding to that, if you're in a educational situation.&lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2019-12-14T19:09:52.813" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32371" PostTypeId="1" CreationDate="2019-12-14T19:54:07.940" Score="1" ViewCount="93" Body="&lt;p&gt;I'm just getting-into melting-down cans to make aluminum-based slingshots, hammers and the like.  In my ignorance I didn't realize that copper/brass were also things that I could process which has left me hoping to learn:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;should I be looking to make an alloy or using 1 metal and, if 1, which would be optimal?  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;are there any &quot;additional treatments&quot;, such as quickly cooling the hot metal-product into a block of ice, or perhaps having the casting box on a vibrating-panel to further 'settle' things?  All I know to do for increased-strength is to make as-clean-as-possible a mix IE scoop as much slag/skim from the top of the molten metal as possible!&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;is there enough variance among aluminum (or non-alloy metal products) to make it worthwhile to source specific-types, IE maybe aluminum from a car-engine part will make a stronger/better ingot than one made from a heap of aluminum cans?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;A hammer and a sling-shot are my 1st goals with this, so while I'm not the best at knowing 'strength-types' I don't think &quot;tensile&quot; strength will be as important as yield-strength or compressive-strength but not sure, if I've gotta choose between optimizing for the slingshot OR the hammer, the hammer is far more important (and I'm not interested in working with lead to do this!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks a ton for any insight on this am in the middle of diving-in, learning I could do this at-home was truly a christmas-present LOL :)&lt;/p&gt;&#xA;" OwnerUserId="23895" LastEditorUserId="23895" LastEditDate="2019-12-14T20:28:58.897" LastActivityDate="2022-10-05T22:20:40.723" Title="Is there any way to improve the strength of simple at-home aluminum/copper/brass castings (sand-casting hammers)" Tags="|metallurgy|strength|aluminum|casting|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32372" PostTypeId="1" CreationDate="2019-12-14T21:56:35.803" Score="3" ViewCount="99" Body="&lt;p&gt;I own a small farm which apparently lands in the middle of 2 small mountains therefore is a big wind tunnel.  The wind gets pretty insane here.  Wondering if there is a structure I can put around my yard that would reduce the wind but be more cost effective than building a 10ft brick wall around my entire property.&lt;/p&gt;&#xA;" OwnerUserId="1033" LastActivityDate="2019-12-14T22:39:49.437" Title="Structure other than a wall to reduce wind?" Tags="|structural-engineering|retaining-wall|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32373" PostTypeId="2" ParentId="32371" CreationDate="2019-12-14T22:08:50.740" Score="1" Body="&lt;p&gt;Copper with aluminum is called aluminum bronze in the trade; no zinc. Up to about 10% aluminum and a couple % iron . That system will get to about 300 BHN / Rc 30; a respectable hammer - not ductile. Manganese bronze is copper with 30 % zinc and a few % Al , Mn, and iron , also a very strong system for copper alloys . Copper Development Association will have more info. On the otherhand , I can't imagine how you could make them in the &quot;backyard&quot;. I used to work at a smelter that produced these and other alloys and even with full foundry and lab facilities the melter would keep his fingers crossed on the aluminum and manganese bronzes because it was not a simple matter of getting chemistry in the right ranges but also getting ratios correct  such as Al/Fe, Al/Mn , and others.  I suggest staying with copper + 9 % aluminum , no zinc, and see what happens. If you get in the ballpark , aluminum bronze has a narrow freezing range and requires a large riser to get a sound casting. I hate to say it but both aluminum and manganese bronzes have a &quot;martensite&quot; reaction and can be heat-treated but not nearly as much affect as steel heat treatment. For now just stay with melting. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-12-14T22:08:50.740" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32374" PostTypeId="2" ParentId="32372" CreationDate="2019-12-14T22:39:49.437" Score="5" Body="&lt;p&gt;If you think about fast growing trees, you'll get a wind break that isn't particularly expensive or labor intensive. Farmers in the center of the continent of the USA historically used this method to reduce the amount of topsoil lost to such things. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.hobbyfarms.com/5-great-trees-to-use-for-windbreaks/&quot; rel=&quot;noreferrer&quot;&gt;HobbyFarms&lt;/a&gt; recommends Eastern Red or Northern White cedar for windbreaks, although the former is a shorter breed, while the latter can reach one hundred feet, somewhat outside your desired range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QWM7k.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QWM7k.jpg&quot; alt=&quot;windbreak image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If vegetation isn't in your future, mimicking a stand of trees may provide similar results. Pole-type structures with &quot;branches and leaves&quot; will break up the force of wind.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2019-12-14T22:39:49.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32376" PostTypeId="1" AcceptedAnswerId="32383" CreationDate="2019-12-15T11:30:34.493" Score="20" ViewCount="12230" Body="&lt;p&gt;How many general methods are there for transferring electricity from the railway to a train? I could see that some trains are connected by a pantograph and some have a third rail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other methods? What is the general engineering principle behind each? What are the basic differences (pros and cons) of each method? I was looking for these questions but could not find a good review. If anyone has a reference or could answer them kindly do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, is it possible to build a small model of such a system and scale it up? I am not talking about a commercial model train. I am aiming at self-designing the same engineering principals and demonstrating them on a scalable small model. If so, I would be happy to have a nice tutorial reference. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am focusing on &lt;a href=&quot;https://en.wikipedia.org/wiki/Electric_multiple_unit&quot; rel=&quot;noreferrer&quot;&gt;Electric multiple units (EMU)&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="15952" LastEditorUserId="5554" LastEditDate="2019-12-16T21:12:37.457" LastActivityDate="2019-12-25T19:05:01.683" Title="What are the engineering principles for a train to get electricity from the railway" Tags="|rail|" AnswerCount="7" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32377" PostTypeId="1" CreationDate="2019-12-15T13:11:17.797" Score="2" ViewCount="397" Body="&lt;p&gt;&lt;strong&gt;Context:&lt;/strong&gt;&#xA;I have a lead screw rod, nut and a pulley aligned vertically. The nut is attached on top of the pulley so their respective bores are aligned and the rod is arranged so that it is threaded through the nut but its diameter is smaller than the pulley bore so no contact is made between rod and pulley.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the pulley spines the nut turns, because the pulley and nut are in a fixed position the rod moves up and down via the spinning nut and through the pulley bore. Essentially a linear actuator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt;&#xA;My problem is how to support the rod from lateral load. Please could you comment on the validity of my proposed solutions and propose your own if you have something more effective&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Potential Solutions:&lt;/strong&gt; A: The most obvious to me would be to have a linear guide rail with a linear bearing in parallel and a secondary nut on the threaded rod that is not fixed like the first nut so that as the rod moves up and down the secondary nut provides support via the guide rail.  The issue with this is it predetermines the height of the mechanism as it will be has high as the linear guide rail, I would like the height to be as small as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;B: Put a linear bearing through the rod and fix it's position to the wall of the actuator so that as the rod moves up and down it moves through the linear bearing which provides support via the wall. The issue with this is I think linear bearings are designed for smooth rods and the gaps in the threaded rod could allow the ball bearings to flick out under load, is this a correct assumption ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;C: Have 3-4 roller bearings arranged around the outside of the rod, so that the rod moves between the bearings. The issue is I don't know if there is enough contact area to be effective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Diagrams:&lt;/strong&gt;&lt;br&gt;&#xA; &lt;a href=&quot;https://i.stack.imgur.com/b9A1c.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b9A1c.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I appreciate what I am essentially looking for is a spline shaft, but that is not something I have access to so I am trying to mimic the linear aspect using a regular lead screw&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All help appreciated.&lt;/p&gt;&#xA;" OwnerUserId="23897" LastActivityDate="2024-01-25T04:04:10.483" Title="How to solve this linear motion problem?" Tags="|mechanical-engineering|actuator|linear-motion|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32381" PostTypeId="2" ParentId="32376" CreationDate="2019-12-15T16:57:11.220" Score="9" Body="&lt;p&gt;In terms of safety for the public at large, an overhead cable &amp;amp; pantograph would be used outside in the open so as to prevent anyone, or any forms of wildlife or suburban pets or farm animals from inadvertently being electrocuted by coming into contact with a third rail at ground level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where a third rail system is used on the Earth's surface a dedicated rail corridor, either elevated or barricaded, as proposed for &lt;a href=&quot;https://en.wikipedia.org/wiki/Maglev&quot; rel=&quot;nofollow noreferrer&quot;&gt;maglev&lt;/a&gt; systems, needs to be used to prevent issues associated with potential electrocution of people &amp;amp; animals &amp;amp; subsequent issues if the dead or injured entities are on the rails when a train arrives - such as possible derailment of the train or further injury to the entity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People &amp;amp; animals are unlikely to wander around underground train tunnels thus the chances of being electrocuted by a third rail is less than on the Earth's surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, underground rail tunnels have limited cross-sectional dimensions - for geotechnical and establishment cost reasons. An overhead electric cable and pantograph would create problems &amp;amp; would need to be very compact. To alleviate such problems it is easier to have a third rail for underground rail systems.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="20116" LastEditDate="2019-12-17T17:37:54.050" LastActivityDate="2019-12-17T17:37:54.050" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="32383" PostTypeId="2" ParentId="32376" CreationDate="2019-12-15T20:06:00.543" Score="19" Body="&lt;p&gt;Pantograph and third rail are pretty much it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineering principles: both have a conducting surface on the train (moving) in contact with the stationary rail/wire, in both cases you need a material that's resilient and conductive. The contact strip is a &lt;a href=&quot;https://www.researchgate.net/publication/4355737_Condition_monitoring_of_pantograph_contact_strip&quot; rel=&quot;noreferrer&quot;&gt;wear material&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Differences: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;overhead catenary wires are flexible and will &lt;a href=&quot;https://www.mts.com/cs/groups/public/documents/library/mts_2010192.pdf&quot; rel=&quot;noreferrer&quot;&gt;move around&lt;/a&gt; when a train drives underneath them. At &lt;a href=&quot;https://en.wikipedia.org/wiki/Brecknell_Willis_high_speed_pantograph&quot; rel=&quot;noreferrer&quot;&gt;high&lt;/a&gt; speeds, the &lt;a href=&quot;https://zhangjin881001.wordpress.com/2015/03/15/pantograph-technology-in-hsr/&quot; rel=&quot;noreferrer&quot;&gt;pantograph&lt;/a&gt; can set up waves in the catenary wire, if these waves aren't damped sufficiently, the pantograph will start bouncing.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;third rail is limited to low voltages to reduce the risk of arcing between the rails. The third rail systems I know of (UK), use &lt;a href=&quot;https://en.wikipedia.org/wiki/Railway_electrification_in_Great_Britain&quot; rel=&quot;noreferrer&quot;&gt;750 V DC&lt;/a&gt; which limits the amount of power you can draw without excessive losses, so no heavy or fast trains. Overhead wiring can use 25 kV, which has much lower resistive losses. &lt;a href=&quot;https://en.wikipedia.org/wiki/File:Europe_rail_electrification_en.svg&quot; rel=&quot;noreferrer&quot;&gt;Several power systems&lt;/a&gt; are in use for overhead catenary systems, from 1500V DC to 25 kV AC 50 Hz. This patchwork of systems makes it difficult to run international trains in Europe, although with the introduction of high-power electronics it's become easier to build locomotives that can run on 2-4 different systems. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Third_rail&quot; rel=&quot;noreferrer&quot;&gt;This Wiki page&lt;/a&gt; does a decent job of discussing the advantages and disadvantages of third rail systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overhead catenary is more susceptible to damage (from e.g. falling trees, vehicles that ignore the height limit), but has less risk of electrocution on e.g. level crossings. Electrocution risk can be mitigated by interrupting the third rail on crossings, but that has problems of its own (interruptions in the power supply).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scaling up/down: &#xA;you can demonstrate the basic principles. In fact, third rail and overhead catenary are both available in standard scale model trains. The pros and cons are more difficult to demonstrate in a model: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;scale model catenary is much less flexible than full size: the wire thickness of the scale model is not in proportion, so the wire is much stiffer. &lt;/li&gt;&#xA;&lt;li&gt;scale models use 12V, which has no electrocution risk and arcing is much less of an issue than in the full-size system.&lt;/li&gt;&#xA;&lt;li&gt;Due to the much lower speeds and shorter running times, wear in a scale model is much less of an issue than in the full-size system.&lt;/li&gt;&#xA;&lt;li&gt;due to the much lower weight, power interruptions are more of an issue in the scale model than in the full-size system.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11645" LastEditorUserId="11645" LastEditDate="2019-12-17T12:47:45.753" LastActivityDate="2019-12-17T12:47:45.753" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="32384" PostTypeId="1" AcceptedAnswerId="49855" CreationDate="2019-12-16T02:51:18.247" Score="1" ViewCount="620" Body="&lt;p&gt;What are the nominal voltage ratings and operating tolerances for 60-Hz electric power systems above 100 volts up to a maximum system voltage of 1 kV in Brazil?&#xA;I am looking for a document like ANSI C84, that describes typical distribution service and equipment nameplate voltage ranges and tolerances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know about 60% of the population is served by a 127 Volt system and a minority is served by a 220V system and both systems use the same type N plug.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example is the 127 Volt system split phase 127/256 like North American 120/240?&#xA;Is the 220 Volt system split phase 220/440 ?&#xA;What are the standard three phase voltages, 220Y127, 380Y220 ?&#xA;Thank you&lt;/p&gt;&#xA;" OwnerUserId="23901" LastActivityDate="2022-02-19T17:06:20.070" Title="Nominal voltage ratings in Brazil?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32385" PostTypeId="1" CreationDate="2019-12-16T04:46:48.230" Score="0" ViewCount="130" Body="&lt;p&gt;I am an electrical engineering student that wishes I could have taken some civil engineering courses. Anyways I'm trying to design a pvc structure that I will use as a somewhat complicated room divider. I have the basic design drawn out, but I'm not sure how to design for stability and loading (eg, how many supporting beams to build). I'm familiar with Solidworks and should be able to sketch up a design to do stress testing if that's something useful? I did some basic mechanics from first year, and know how to setup equilibrium equations but I'm not sure if that would be of any help in the design process. I'm open to learning new mathy thing and would love to treat this as a learning experience to solve a practical problem. A nudge in the right direction would be appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="23903" LastActivityDate="2021-09-06T12:10:10.937" Title="Designing PVC structure" Tags="|civil-engineering|design|structural-analysis|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32386" PostTypeId="2" ParentId="32377" CreationDate="2019-12-16T08:35:41.203" Score="0" Body="&lt;p&gt;You mentioned availability, so it sounds like you want to get away with the components you have. While there may not be a solution to that, I want to share my idea, how to overcome the &quot;contact surface&quot; problems in B and C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yould you mill or cut a notch along the lead screw rod, such that at three angles 120° between each noth, there is a smooth surface along the rod. This would in no way prevent the nut from performing as required. The Roller bearings in C would roll in these notches. &#xA;You may want the shape of the notch to match the roller bearings, such that there is a line contact, rather than point contact. But, this really depends on the lateral load, that you have. &lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2019-12-16T08:35:41.203" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32387" PostTypeId="2" ParentId="32385" CreationDate="2019-12-16T08:47:41.333" Score="1" Body="&lt;p&gt;The analysis of the structure will greatly vary depending upon the loads, application, section size, etc. Expanding the description of the structure will greatly help in providing specific steps in designing. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first step would be to estimate the loads on your structure (you most likely may need a safety factor). Your second step would be to have a rough guess at how many members, supports and size sections are required to sustain the loads and then analyse it by hand or using Solidworks. Making use of the stresses found, verify if the members you have selected provides adequate strength. Else design accordingly from first principles and complete the analysis again. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be required to reiterate the process a couple of times to get the ideal solution. Also considering it is a room divider, major design procedures will most likely not be required.&lt;/p&gt;&#xA;" OwnerUserId="20822" LastActivityDate="2019-12-16T08:47:41.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32389" PostTypeId="1" CreationDate="2019-12-16T11:06:52.223" Score="0" ViewCount="939" Body="&lt;p&gt;There are &quot;analog&quot; TIG welders where the parameters are directly controlled by rotary knobs and buttons. For example &lt;a href=&quot;https://www.tbws.co.uk/shop/jasic-pro-tig-200p-acdc-analog-inverter-welder/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; one. Then there are &quot;digital&quot; ones where parameters are controlled by one rotary encoder and several buttons. Link &lt;a href=&quot;https://www.eastwood.com/tig-200-digital.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Besides the actual difference in the user interface, does a digitally controlled welder have any general benefits over an analog one when it comes to welding performance? (Or perhaps vice-versa?)&lt;/p&gt;&#xA;" OwnerUserId="23905" LastEditorUserId="23905" LastEditDate="2019-12-16T12:59:23.743" LastActivityDate="2023-12-26T07:06:45.920" Title="What are the benefits of a digitally controlled TIG welder?" Tags="|welding|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32390" PostTypeId="2" ParentId="32389" CreationDate="2019-12-16T12:08:07.190" Score="0" Body="&lt;p&gt;Most rotary knobs are resistive potentiometer that have moving mechanical parts that tend to wear out over time and also change the electrical characteristics. This will affect the performance of your analog TIG welder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is an image of the internal of an rotary knob&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r68HE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r68HE.jpg&quot; alt=&quot;Internal of rotary knob&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With rotary knobs it is hard dial in the settings. Due to wear and tear their is additional variations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively digital controls has minimal mechanical components are designed to last a long a period of time with minimal wear and tear. Below is an image of a digital membrane switch. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1dyII.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1dyII.jpg&quot; alt=&quot;Example of a digital membrane switch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These membrane switches will always command the an accurate settings. This is will improve the performance of the welder. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a cross section of a digital membrane switch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SLFfJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SLFfJ.png&quot; alt=&quot;Cross section of digital membrane&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the step sizes it is possible to accurately adjust the settings. It is much harder to achieve high accuracy with analog rotary knobs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also digital controls switches tend to be cheaper compared to matching rotary components. This is also an advantage. The electronics technology is advancing every day, this transitioning over to digital can provide better products are cheaper price. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally digital TIG welders had additional features to improve the performance. I suggest checking out the below references for details&lt;/p&gt;&#xA;&#xA;&lt;p&gt;References:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/oXOEYGmLXG0?t=71&quot; rel=&quot;nofollow noreferrer&quot;&gt;TIG Welder Buyer's Guide - TIG 200 Digital, TIG 200 AC/DC or TIG 200 DC Overview&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://youtu.be/OdGf1yiZ2lk&quot; rel=&quot;nofollow noreferrer&quot;&gt;TIG 200 Digital Welder Features Brief Overview: Pulse, Spot Weld Timer, 2T &amp;amp; 4T&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-12-17T00:13:20.120" LastActivityDate="2019-12-17T00:13:20.120" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32391" PostTypeId="2" ParentId="32376" CreationDate="2019-12-16T12:37:58.653" Score="1" Body="&lt;p&gt;S-Bahn Berlin and similar use a third rail with alternating mount points, around 60 cm above the rails, they can be mounted that passengers are not in danger - and allow for lower train profile - especially when building such a rail-system in existing city structures. As far as I know the underground trains in Berlin use a similar system&lt;/p&gt;&#xA;" OwnerUserId="23850" LastActivityDate="2019-12-16T12:37:58.653" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32392" PostTypeId="1" AcceptedAnswerId="32556" CreationDate="2019-12-16T14:36:15.800" Score="0" ViewCount="521" Body="&lt;p&gt;Both circulation and vorticity have to do with rotation of a fluid element.How are vorticity and circulation are related.&lt;/p&gt;&#xA;" OwnerUserId="22829" LastEditorUserId="22829" LastEditDate="2019-12-16T14:57:17.157" LastActivityDate="2019-12-27T10:52:46.153" Title="Is it possible to have circulation without vorticity?" Tags="|fluid-mechanics|fluid|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32393" PostTypeId="2" ParentId="32320" CreationDate="2019-12-16T15:51:10.543" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Is this because the block and crankshaft comes from those companies and the custom engine company makes everything else, or for some other reason? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Usually, yes. Custom engines come in various &lt;a href=&quot;https://www.torquecars.com/tuning/tuning-stages.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;stages of tuning&lt;/a&gt;. Stage 1 is pretty close to a stock engine, with just a few components replaced. Those replaced components have to fit the original engine. &#xA;The more power you want to generate, the more components you need to replace. They still all need to fit the original engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For extreme applications (top-fuel drag racing), you can get engines where every component including the block has been custom-made and all that's left of the original engine is the main dimensions (distance between piston centerlines, distance between crankshaft main journals etc). These main dimensions remain because that means many of the parts are compatible across their entire catalog: a billet crankshaft that fits stage 3-7 is cheaper to make than a crankshaft that's unique to your 'everything is custom' stage 7 offering.  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Why can't custom engine companies make the whole engine?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They can, it's just very expensive, and usually not necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big car companies produce engines by the thousand, and they have an assembly line set up to do this quickly and efficiently. Custom engine companies don't have the production volume to be able to afford this, so they have to spend more manual labor and machine time to build an engine. Things like the engine block are complicated to manufacture (requires big infrastructure for iron/aluminium casting, or a very expensive 5-axis CNC machine to mill the whole block). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So they only build the parts they really need to. A stage 1 tuning package doesn't require a new block, because the standard block has enough margin built in to handle the extra power. &lt;/p&gt;&#xA;" OwnerUserId="11645" LastEditorUserId="11645" LastEditDate="2019-12-16T16:21:58.033" LastActivityDate="2019-12-16T16:21:58.033" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32395" PostTypeId="2" ParentId="21018" CreationDate="2019-12-16T17:32:32.187" Score="0" Body="&lt;p&gt;How about using a Kinect or similar device ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a 3D camera that outputs a real-time depth measurement of the whole scene... pretty fast and low latency (designed for gaming)... if you have depth info for the whole scene you can use it to control your focus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an example of a kinect coupled with a projector:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=bA4uvkAStPc&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=bA4uvkAStPc&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12907" LastActivityDate="2019-12-16T17:32:32.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32396" PostTypeId="1" CreationDate="2019-12-16T17:37:59.800" Score="-1" ViewCount="38" Body="&lt;p&gt;If a building or big structure is situated on top of flat ground without any of its part inside the ground, should the building be safe? Let's say our building is a big (as big as we can imagine) rubic's cube made with mordern day building materials.&lt;/p&gt;&#xA;" OwnerUserId="23909" LastEditorUserId="61" LastEditDate="2019-12-17T07:51:31.007" LastActivityDate="2019-12-19T00:39:37.367" Title="Earthquake proof building" Tags="|structural-engineering|" AnswerCount="1" CommentCount="6" ClosedDate="2019-12-22T21:12:47.883" ContentLicense="CC BY-SA 4.0" />
  <row Id="32397" PostTypeId="2" ParentId="32376" CreationDate="2019-12-16T19:07:53.237" Score="2" Body="&lt;p&gt;In scale model trains, the 2 rails used for running the train also power the train.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are a variety of scales that use this method: O, HO, N, Z, G, and others. Some use a 3rd rail model or overhead lines either as an alternative method or to emulate the trains they are modelling. The brand Lionel typically uses a 3 rail track and is very similar to O scale trains in size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These trains are often small enough to be run using low DC voltage, where direct skin contact between the rails gives either no, little, or minimal shock value. More advanced, newer trains use Digital Command Control (DCC) systems, this is an AC power system which acts as both the AC power source and has a digital signal embedded in it. The decoder in the train decodes the addressed commands and converts the AC to DC power for locomotive power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some larger scale models that can actually carry humans use a small gas, diesel, or oil engine. I'm sure you can use an electric motor, such as something from an electric car, so that's just a minor engineering detail. Getting more voltage and current to the motor through the track might be an issue, though, which is why larger trains don't put electricity though the load bearing wheels. I don't know all the specific details, but putting electricity through a bearing has issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm answering this way because just as scaling up something like this has drawbacks, so might scaling down another system. Using a 3rd rail or an overhead carriage might not work well for a smaller system, so it'd be easy to consider doing what I've mentioned, only to find out later that scaling it back up doesn't work as intended either. I haven't done research on this, so I don't know if there's a cut-off point or overlap where one system starts to fail and another system suddenly becomes useful. There may also be a speed factor, where a high speed method could be completely different from a low or medium speed method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My intention here is to make you aware of a system not mentioned by others as well as letting you know that which type of system may depend entirely on what size or speed you want to make your project, so that you will have a good idea as to which system might work best for your intended use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd suggest that you not only ask what systems currently exist, but also ask what worked for people in the size and speed range you are looking at. Asking that 2nd part of the question might end up being another question on SE, seeing as you already have the first part in this question. Linking or otherwise referencing the two questions would be a relevant way to ask that 2nd question.&lt;/p&gt;&#xA;" OwnerUserId="16686" LastEditorUserId="-1" LastEditDate="2019-12-24T19:06:19.390" LastActivityDate="2019-12-24T19:06:19.390" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32398" PostTypeId="2" ParentId="32396" CreationDate="2019-12-16T19:30:30.140" Score="1" Body="&lt;p&gt;Even if a building is just sitting on the ground which is not acceptable for many reasons, it needs to be designed for horizontal, P wave, and vertical, S wave, vibrations of the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The friction between concrete and compact soil is between 0.60 and 0.40 depending on the type of soil. So a 4 story structure with an estimate of 3000kg weight per square meter will experience approximately 1500kg per square meter lateral force during an earthquake.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the foundation of the building is concrete the forces of the earthquake can transfer by shaking of the ground to the foundation. Then the foundation will move laterally with the ground, while the inertia of the floors above will hold them back, causing the structure to vibrate and undergo large stresses due to moment and shear which may eventually lead to distruction of the building. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also vertical earthquake wave, S wave, these can break the structure same way a ship caught in cross waves break apart. The building designer has to consider that.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="16486" LastEditDate="2019-12-19T00:39:37.367" LastActivityDate="2019-12-19T00:39:37.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32399" PostTypeId="1" CreationDate="2019-12-16T19:44:09.067" Score="1" ViewCount="62" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/iusMd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iusMd.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is it possible to find all the support reactions of this continuous beam using fourth order equation without using principle of superposition?&lt;/p&gt;&#xA;" OwnerUserId="23910" LastActivityDate="2019-12-16T19:44:09.067" Title="Finding reactions of a continuous beam" Tags="|structural-analysis|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32400" PostTypeId="1" AcceptedAnswerId="32412" CreationDate="2019-12-16T20:16:23.693" Score="2" ViewCount="433" Body="&lt;p&gt;I want to make a sprung latch mechanism which keeps a door in place until a certain force is applied to the door. In the diagram attached, the force  the door must release at is Fd, and the latch spring force necessary to hold it until Fd is reached is Fs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are these:&#xA;1. How do I work out what force spring I need? &#xA;2. Assuming the contact is frictionless, and that distance d stays the same, does it make a difference what the angle j is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many many thanks for any help!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zMsrJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zMsrJ.png&quot; alt=&quot;latch free body diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11394" LastActivityDate="2019-12-18T05:21:04.230" Title="Latch Mechanics" Tags="|mechanical-engineering|applied-mechanics|mechanisms|mechanical|solid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32403" PostTypeId="2" ParentId="32376" CreationDate="2019-12-17T09:40:50.293" Score="3" Body="&lt;p&gt;The newest generation of trains can be battery-powered, recharging only on or near stations. This is made possible by lithium-ion batteries like those used in electric cars, combined with regeneration when slowing down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The benefit is that only a small fraction of the track needs to be electrified, saving on infrastructure costs. Stations are located in population centers where electricity is already available.&lt;/p&gt;&#xA;" OwnerUserId="669" LastActivityDate="2019-12-17T09:40:50.293" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32404" PostTypeId="2" ParentId="32290" CreationDate="2019-12-17T10:05:44.660" Score="1" Body="&lt;p&gt;Vertical-takeoff aircraft do exist. A lot of research has gone into this (mainly in the 1960s), but it all led to one conclusion: there's no commercial case for vertical-takeoff jet aircraft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Only 4 VTOL (vertical takeoff and landing) aircraft have gone into production: the Harrier, the F-35 (B version only), the V-22 Osprey and the Yak-38. These are all military aircraft created for a specific niche: the need to operate from small bases either on land or on a ship. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is an example of a &lt;a href=&quot;https://www.secretprojects.co.uk/threads/hawker-siddeley-vtol-stol-concepts.3153/&quot; rel=&quot;nofollow noreferrer&quot;&gt;civilian VTOL design&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/faN0r.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/faN0r.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;as you can see, there's a massive increase in the number of engines compared to a normal airliner. All of these engines are used for only a few minutes per flight, they are dead weight and drag the rest of the time, making operation of a VTOL airliner much more expensive than a conventional aircraft. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;VTOL operation is also very loud, so the only advantage of a VTOL aircraft (doesn't need a large airfield so it could land closer to its destination city) falls foul of noise regulations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rockets are the other way round: they need massive thrust to get up to the speed required for orbit. This also turns out to be enough thrust to take off vertically. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rocket only spends a few minutes inside the atmosphere. So while it is possible to add wings to a rocket to take off horizontally, those wings would be dead weight for most of the flight. Rockets operate close to the edge of what's physically possible, and dead weight is the last thing you want. &lt;/p&gt;&#xA;" OwnerUserId="11645" LastActivityDate="2019-12-17T10:05:44.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32406" PostTypeId="1" AcceptedAnswerId="32410" CreationDate="2019-12-17T10:20:50.810" Score="3" ViewCount="85" Body="&lt;p&gt;I've been asked to modify an existing zimmer/walking frame for a tall person. Bashing the bits of metal is the easy bit, just use tube to make longer legs to replace the original short removable adjustable ones, using tube that's actually thicker wall than the original. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tricky bit is now they've asked me for a new weight rating, given that the frame is larger. The original frame was rated for use by people up to 180kg. It would be quite easy to say that as the new height is &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, we should derate the weight by &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;, or by &lt;span class=&quot;math-container&quot;&gt;$x^2$&lt;/span&gt;, or by &lt;span class=&quot;math-container&quot;&gt;$x^3$&lt;/span&gt;, but justifying a PIDOOMA for the exponent or the weight is more difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the expected failure modes of the frame? Here's a picture of it after fitting with the extended legs, the lowest 500mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/brQq4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/brQq4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm assuming two failure types are relevant, prompt failure resulting in the user ending up on the ground, and increased wear meaning a weakening after several years' use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For prompt failure, if I assume that all joints are hinges, then the weakest appear to be on the cross member holding the two side frames together, which is why I guess there are corner braces there. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment on those hinges, should the user slump sideways on the frame, will be increased slightly by the longer legs, but will still be dominated by the width of the frame, so not even by a factor of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. By my reckoning, it's the smallest height of the triangle formed by the width and height of the walker, and a diagonal from one handle to the floor contact on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the new legs are stronger than the original, and use the original spring pins into the same size holes, there should be no reason at all to expect any reduction in simple vertical strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fretting moment at the area of tube overlap where the adjustable legs couple to the frame, held in place by spring pins, will increase directly in proportion to the adjustable leg length. However, the overlap region has been increased in length, more than proportionately, to reduce the contact force caused by this moment. So it may be no worse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A colleague has suggested a simple reduction to a nice round 100kg, which would certainly be conservative. My concern is that I wouldn't be able to justify it, and that it may exclude other potential users for no good reason.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there industry standard tests for this sort of thing? How would the original manufacturer have rated it for 180kg users?&lt;/p&gt;&#xA;" OwnerUserId="8092" LastActivityDate="2019-12-17T15:02:35.343" Title="Failure modes of a zimmer frame" Tags="|structural-analysis|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32408" PostTypeId="1" CreationDate="2019-12-17T13:38:41.763" Score="1" ViewCount="291" Body="&lt;p&gt;In mechanics, we have the so called compatibility conditions, which quarantee that when a body deforms, the strains are &quot;compatible&quot; in such a way to no discontinuities or gaps for inside the body as it deforms. Intuitively, this means that infinitesimal pieces of material connected in the undeformed body stay connected, and won't end up inside one another.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Compatibility_(mechanics)&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;..In the continuum description of a solid body we imagine the body to be composed of a set of infinitesimal volumes or material points. Each volume is assumed to be connected to its neighbors without any gaps or overlaps. Certain mathematical conditions have to be satisfied to ensure that gaps/overlaps do not develop when a continuum body is deformed. A body that deforms without developing any gaps/overlaps is called a compatible body. Compatibility conditions are mathematical conditions that determine whether a particular deformation will leave a body in a compatible state.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Reading the Wikipedia page, these equations/conditions are developed:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{\partial^2\epsilon_{11}}{\partial x_2 ^2}-2\frac{\partial^2\epsilon_{12}}{\partial x_1 \partial x_2}+\frac{\partial^2\epsilon_{22}}{\partial x_1 ^2}=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are derived by stating that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;..there are six strain-displacement relations that are functions of only three unknown displacements. This suggests that the three displacements may be removed from the system of equations without loss of information.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Firstly, sure, the above statement does make sense, but what does it have to do with compatibility, as I described above? Why is it that when a few variables are removed from the system, we are now guaranteed to have a deformed state where no material overlaps or develops discontinuities?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of other material that I've read online just state the same intuitive idea of compatibility, and then just state that the deformation is compatible when the strain is &lt;em&gt;integrable&lt;/em&gt; and proceed to find conditions ensuring that. Alternatively,  it is proven that the solution to this equation can be found:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\epsilon=\frac{1}{2}[\nabla u +(\nabla u)^T]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, sure, we need to know that this equation can be solved to have a sensible strain field, but how do we know the solution strain field is &lt;em&gt;compatible&lt;/em&gt;, in the intuitive sense described at the beginning?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why simply being able find a solution displacement field, or being able to integrate the strain field imply that the strain is compatible?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2024-04-06T16:06:05.753" Title="Understanding compatibility of strains" Tags="|applied-mechanics|stresses|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32409" PostTypeId="1" CreationDate="2019-12-17T14:27:12.753" Score="1" ViewCount="20" Body="&lt;p&gt;I'm trying to find a sensor that can measure distance between sensor and some sort of detectable material (metal or small wireless chip) size of no more than a penny.&lt;br&gt;&#xA;It should be able to measure through fabric as thick as (~40cm) with max distance of 2 meters and accuracy 0.3%.&#xA;What are my option if any?&lt;/p&gt;&#xA;" OwnerUserId="23918" LastActivityDate="2019-12-17T14:27:12.753" Title="Accurate distance sensor via soft materials" Tags="|sensors|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32410" PostTypeId="2" ParentId="32406" CreationDate="2019-12-17T14:53:53.933" Score="3" Body="&lt;p&gt;Your last paragraph is the key one: Yes, there industry standard tests, and they were able to rate it for 180kg by passing those tests. You can do theoretical calcs until the cows come home to make sure you pass those tests first time, but &lt;strong&gt;you must undertake and pass these tests prior to adding a rating sticker&lt;/strong&gt; or you open yourself right up to liability.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked at BS EN ISO 1119-1:1999 &quot;Walking aids manipulated by both arms - Requirements and Test Methods&quot; Part 1: Walking frames&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have picked out some key points, but you should read the whole document (or local equivalent) yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One walking frame shall be tested, through Stability, Static Load, Fatigue, and Static Leg-strength in that order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Stability: Forward, Backward and Sideways tests have different tilt angls, then a vertical load of 250N is applied to the centre of the hand grips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Static Load: A load of 1,500N is applied, with an additional 15N per kilogram of maximum user weight above the standard maxium of 100kg. You would need to test to 2,700N (+/-2%) to demonstrate equivalence&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fatigue: A cyclic load of 800N plus 8N per kg over 100kg (1,440 for equivalence) applied downwards on the handles, 200,000 times, slower than 1Hz.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Static Leg-strength: Applied to each of the legs in turn, applied a minimum of 10mm further out than the lower end of the leg, and directed towards the centre of the frame. done by removing the rubber foot, and putting in a tightly fitting plug up to 100mm inside the leg. Height adjustment of frame must be at maximum. The standard &lt;em&gt;requires&lt;/em&gt; a loading force for this of 300N, applied gradually over 2s, and held for minimum of 5s. &lt;strong&gt;but&lt;/strong&gt; it also recommends testing at 500N, to provide a safety factor &quot;for reasonable foreseeable misuse&quot;.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2019-12-17T15:02:35.343" LastActivityDate="2019-12-17T15:02:35.343" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32412" PostTypeId="2" ParentId="32400" CreationDate="2019-12-17T21:27:47.863" Score="1" Body="&lt;p&gt;The force of resistance of the latch varies. at the time the door touches the latch it is zero and when it is pushed fully back it is k*d.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;why F*tan j is the horizontal component? after OP's comment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we draw the perpendicular, &lt;span class=&quot;math-container&quot;&gt;$F_d*cos j,$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_d* Sin j \ , $&lt;/span&gt; components of Fd on the surface of latch, the hypotenuse of the triangle is the tan of that angle and is the horizontal component factor of the Fd.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have ignored the round corner of the door on the diagram and assumed it's a sharp point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;End of edit.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ F_d*tanj=k*d   \\F_d=\frac{k*d}{tanj }$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steeper the angle J the less force is needed to push the door open.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-18T05:21:04.230" LastActivityDate="2019-12-18T05:21:04.230" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32413" PostTypeId="1" CreationDate="2019-12-18T00:06:08.363" Score="-1" ViewCount="84" Body="&lt;p&gt;What is the flow path of water immediately at the stern of a large, deep draft ship (like a container ship 250+m long for example)? Does it fill in the area displaced by the hull from all sides equally? or does more water flow in from the bottom or sides?&lt;/p&gt;&#xA;&lt;p&gt;This was prompted by a ridiculous video of a guy accidentally pulling the dead man key out of his jetski as he reaches out to try and touch a large commercial freighter resulting in him capsizing next to the ship as it passes.&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=L3c9IKvfmsQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=L3c9IKvfmsQ&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;At the core of the question is, based on the replacement fluid flow around the stern of the ship, is it ever likely that he would get sucked in behind the ship and risk being killed by the prop?&lt;/p&gt;&#xA;&lt;p&gt;If someone has a diagram or animation that would be most helpful&lt;/p&gt;&#xA;&lt;p&gt;Edit - For an example, lets assume the ship's hull looks about like this&#xA;&lt;a href=&quot;https://i.stack.imgur.com/hYG7y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hYG7y.png&quot; alt=&quot;Ship Hull Rear&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23932" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-12-18T19:51:59.230" Title="Describe the replacement flow of water at the stern of a large ship" Tags="|fluid-mechanics|ships|naval-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32414" PostTypeId="1" CreationDate="2019-12-18T06:55:47.403" Score="0" ViewCount="83" Body="&lt;p&gt;we are developing a product and would like to use silicone, as it has good texture and durability, however it is an insulator. We would like heat to be able to pass through the silicone rapidly. Does anyone know of any alternatives to silicone that can conduct heat and is widely available?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="23934" LastActivityDate="2019-12-19T16:15:55.550" Title="Alternative to silicone that is non insulating" Tags="|material-science|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32415" PostTypeId="1" AcceptedAnswerId="32418" CreationDate="2019-12-18T10:09:47.217" Score="6" ViewCount="2722" Body="&lt;p&gt;I want to get a rough estimate for the transmission efficiency of two different gear trains. One is a one-stage gear train with 11.67:1 ratio (140/12 teeth) and the other is a compound two-stage gear train with overall gear ratio of 18:1 (72/12 + 36/12 teeth). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read in a forum that one can expect a 10% loss per stage of gear train in this case but I wanted a second opinion. Any help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="23762" LastActivityDate="2019-12-19T13:00:43.933" Title="How Much Efficiency is Lost in a Gear Train" Tags="|mechanical-engineering|gears|energy-efficiency|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32416" PostTypeId="1" CreationDate="2019-12-18T11:35:33.660" Score="2" ViewCount="2023" Body="&lt;p&gt;As I have been working through examples of construction projects, I've noticed that in nearly every case of a structural column (sample size considers only homogeneous material columns, no composite structures) the materials chosen to construct the beam will yield before the beam design chosen will experience Euler buckling. I may not be considering all use cases but is it easier (or more likely, safer) to design for yield failure modes than for buckling failure modes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps manufacturing tolerances on mass produced steel beams are sloppy enough compared to material composition that it requires a larger Factor of Safety?&lt;/p&gt;&#xA;" OwnerUserId="23222" LastActivityDate="2019-12-21T21:44:11.643" Title="Buckling vs. Yielding Failure of Columns" Tags="|statics|beam|mechanical-failure|buckling|yield-point|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32417" PostTypeId="1" CreationDate="2019-12-18T11:47:14.817" Score="2" ViewCount="4147" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hsEcR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hsEcR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the King Post and &quot;Truss Web&quot; or &quot;braces&quot; are there to make the whole truss &quot;stronger&quot;, but can someone explain what are they actually supporting if there is no load on the bottom chord/tie beam? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do a truss analysis with loads on the &quot;upper&quot; nodes, the kingpost and webbing are 0 members. I'm guessing you can't actually do a truss analysis on this (altho it is called a truss) because the majority of forces are bending forces? the braces are there to keep the rafters from bending and the kingpost is there to keep the bottom chord/tie beam from sagging?&lt;/p&gt;&#xA;" OwnerUserId="23937" LastEditorUserId="23937" LastEditDate="2019-12-18T13:54:53.803" LastActivityDate="2022-10-07T15:47:22.300" Title="King post roof truss analysis" Tags="|structural-engineering|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32418" PostTypeId="2" ParentId="32415" CreationDate="2019-12-18T12:58:15.103" Score="6" Body="&lt;p&gt;Generally gear trains with lower ratios are more efficient due to reduced amount of sliding action between the gear teeth in mesh, so the compound gear train would probably be a better solution. That also depends on the quality of the gears, center distances, bearings, lubrication and thermal considerations, but short answer is two stage would probably be more efficient. If all of the components are of &quot;good&quot; quality, you could expect 95% or better efficiency.&#xA;AGMA standard 917 goes into a lot more detail if you want to do analytics.&#xA;Edit- Caveat: typically gears with fewer than 24 teeth have modified geometry to avoid undercutting the involute profile, so this could affect sliding action as well. If you change to a finer diametral pitch or module you could avoid that.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastEditorUserId="23871" LastEditDate="2019-12-18T13:08:52.180" LastActivityDate="2019-12-18T13:08:52.180" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32419" PostTypeId="2" ParentId="32416" CreationDate="2019-12-18T14:03:09.710" Score="3" Body="&lt;p&gt;Yes, it is better to design for yield than buckling because Euler buckling is a result of instability as opposed to overloading force.&#xA;Gamma f is determined from Eurocodes which are from extensive research and materials testing. This means yield is safer than buckling because buckling is not predictable nor easily counteracted in design. It also would mean the engineer had not applied the correct factor of safety for the structure or that the moments of force were incorrectly accounted for. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2019-12-21T21:44:11.643" LastActivityDate="2019-12-21T21:44:11.643" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32420" PostTypeId="2" ParentId="32416" CreationDate="2019-12-18T17:33:57.273" Score="4" Body="&lt;p&gt;The basic Euler buckling formula is unrealistic because it implies perfect geometry and perfect alignment of the loads. Therefore a large empirical safety factor is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also a potentially catastrophic failure mode, in the sense that the buckled column supports no load at all, unlike plastic failure in compression for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are situations where it makes sense to design light weight structures with buckling failure at the limiting loads, but civil engineering is not one of them - there are far too many unknowns.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-12-18T17:33:57.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32421" PostTypeId="2" ParentId="32414" CreationDate="2019-12-18T17:49:29.407" Score="1" Body="&lt;p&gt;The easy answer is to load the silicone with silver, copper, or aluminum oxide dust to a high volume percentage. This will furnish conduction paths for heat- however, it also renders the cured silicone less flexible and tear-resistant. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2019-12-18T17:49:29.407" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32422" PostTypeId="2" ParentId="32413" CreationDate="2019-12-18T19:12:20.633" Score="1" Body="&lt;p&gt;The water stream flow on the back of ship depends on two main factors, ignoring waves and possible wake shock. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The shape and curve of the wet surface on that section. The more streamlined, the less turbulent flow and ideally less rushing of the water.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The depth which causes larger hydrostatic pressure. That is the reason they avoid creating a bulkhead at the stern, while providing a linear flow path to feed the propellers.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-18T19:12:20.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32423" PostTypeId="2" ParentId="32417" CreationDate="2019-12-18T20:40:28.640" Score="2" Body="&lt;p&gt;These are usually prefabricated trusses and should be able to handle all potential variety loading on a roof. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;In typical light framing structures, rafters are placed at 16&quot; O.C. which will cause moments and will load the web members.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Wind load depending on the angle will stress the king post, the web members and even can cause uplift.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The bottom cord usually supports the ceiling wight through ceiling joists.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The roof may have unsymmetrical seasonal loads such as snow piling up on wind leeward, or rooftop installation of chimineas, HVAC, vents, and live load of the repair crew. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All these and cases not mentioned above will cause loading on all different members of a typical roof truss.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-18T20:40:28.640" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32424" PostTypeId="2" ParentId="32416" CreationDate="2019-12-18T21:00:44.457" Score="1" Body="&lt;p&gt;The code has prescribed how to design a column, wood, steel alloy, or many common materials, depending on their effective length and slenderness, the radius of gyration, and the eccentricity of axial load and type of connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Short columns fail under the crushing, mid-range columns should be considered for both crushing and buckling, long columns have to be checked for buckling. Each code has a set of load combinations and each of these combinations has its own way of handling the ultimate buckling load.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-18T21:00:44.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32425" PostTypeId="2" ParentId="22911" CreationDate="2019-12-18T21:04:57.377" Score="2" Body="&lt;p&gt;Old topic but... to be sure you have the correct info : The IGV positions are given by the seperate lines as indicated at the top (continious, dashed, dotted and dash-dot) At these IGV positoins, tests are done at several outlet diffuser positions (the numbers at the bottom of these lines in the curves) Example; There are several dashed lines (IGV pos 04) and these have different numbers at the bottom (outlet diffuser vane position). Basically, IGV pivots the curve around the 0 flow max head point and outlet diffusers move the whole curve horizontally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The numbers are simply the numbers of the actuator arm position (one for the IGV and one for the diffuser), that is on the outside of the machine. You can see them in the left os the picture in the link for the IGV &#xA;&lt;a href=&quot;https://www.howden.com/Howden/media/Howden/img/products/compressors/single-stage-turbo-compressor.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.howden.com/Howden/media/Howden/img/products/compressors/single-stage-turbo-compressor.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ASME PTC-10 is the test protocol used at this specific test for warranty on flow/pressure and power consumption.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.asme.org/codes-standards/find-codes-standards/ptc-10-performance-test-code-compressors-exhausters&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.asme.org/codes-standards/find-codes-standards/ptc-10-performance-test-code-compressors-exhausters&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The curves you see here is a test curve of the machine, a KA44SV where KA44 is the size, S is for outlet diffuser vanes, V for IGV's, from testing (which was done for every machine before it left the factory) in Helsingor, Denmark. The IGV's were used to compensate isentropic head for off-design conditions (determined by pressure measurement, inlet temp measurement and diffuser position), the diffuser vanes for flow control. This way, they got a pretty constant high efficiency over the whole flow range, all the way down to ~45% from the max flow at a constant pressure, at a constant rpm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can not really determine flow/pressure from these to find a duty point in practice. These curves are used to program IGV position control in the PLC so when installed, the IGV goes to the optimal position automatically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is also another curve that comes with testing, that shows you the efficiency of the machine over the complete working range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HV-Turbo used to be the world leader in centrifugal compressors for wastewater aeration. They made them from the KA2 (starting at ~ 45 kW) to the KA100 (max. 2.7 MW). They were taken over by Siemens (approx. 2005) and they sold of the whole package to Howden in 2018 I believe. Factory in Helsingor is gone sadly.&lt;/p&gt;&#xA;" OwnerUserId="23940" LastEditorUserId="23940" LastEditDate="2019-12-18T21:37:06.863" LastActivityDate="2019-12-18T21:37:06.863" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32426" PostTypeId="2" ParentId="32359" CreationDate="2019-12-18T22:21:00.977" Score="1" Body="&lt;p&gt;Aluminium foil will probably shred. Stick it on the back of a cardboard,maybe? I would rather use a blade or scissors.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For 1 mm aluminium you should try to use low feed rate (x and y axis speed) and depth of cut (z depth, which corresponds to number of passes) and iterate till you get desired finish and productivity of the machine and tools. Pair it with a coolant of either compressed air (eye protection required), water or oil (check if the machine has a way to drain these fluids properly). Use a cutting tool designed for cutting metals as wood cutting tools might damage prematurely and will put undesired loads the machine.&lt;/p&gt;&#xA;" OwnerUserId="23941" LastEditorUserId="23941" LastEditDate="2019-12-18T22:27:04.907" LastActivityDate="2019-12-18T22:27:04.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32427" PostTypeId="2" ParentId="32417" CreationDate="2019-12-18T22:30:55.100" Score="1" Body="&lt;p&gt;If the truss is prefabricated with metal plates at the joints as in your picture, it isn't a truss any more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pin jointed trusses are a very useful educational tool to teach people how to resolve forces and moments by hand, but that doesn't mean real-world 21st-century structures are built that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Medieval wood framed structures were &lt;em&gt;literally&lt;/em&gt; pin-jointed, as in &lt;a href=&quot;https://en.wikipedia.org/wiki/Timber_framing#/media/File:Chevilles_en_bois_dans_une_charpente_ancienne.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this Wikipedia image&lt;/a&gt;, but that was then.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-12-18T22:30:55.100" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32428" PostTypeId="2" ParentId="32415" CreationDate="2019-12-18T23:01:10.400" Score="3" Body="&lt;p&gt;In the book &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/0262731541&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bicycling Science&lt;/a&gt; by David Gordon Wilson, it says the following:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;When new, clean, and well-lubricated, and when sprockets with a&#xA;  minimum of 21 teeth are used, a chain transmission is highly efficient&#xA;  (at a level of maybe 98.5 percent or even higher)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;(Third Edition, chapter 9: &lt;em&gt;Mechanics and Mechanisms: Power Transmission&lt;/em&gt;, page 318)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It goes on to describe that efficiency inevitably suffers due to wear, dirt, and poor maintenance and adds &quot;it is not known by how much the efficiency of chain drives decreases in adverse conditions,&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Later, on page 342 it provides some other factors which influence efficiency:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;&lt;p&gt;Transmission efficiency decreases as the size of the rear sprocket&#xA;  is reduced.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;Efficiency diminishes as the amount of torque transferred (or chain tension) is decreased.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The maximum efficiency attained is at relatively high power (175 W) and low pedal rpm (60) and in the lowest gear (meaning the&#xA;  largest- diameter rear sprocket, with twenty-one teeth) and is just&#xA;  over 98 percent.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The additional losses because of chain offset (the two sprockets not being in line) are negligible.&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;li&gt;&lt;p&gt;The type of lubrication, or even whether there is lubricant present, has almost no effect on efficiency&lt;/p&gt;&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally some typical gearing ratios are matched against some other factors to estimate efficiency:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Table 9.3:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OAGtT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OAGtT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Based on personal riding experience I would find it hard to believe a 20% loss was not an overestimate, but I have no other sources with quantified data to back that up.&lt;/p&gt;&#xA;" OwnerUserId="6489" LastEditorUserId="6489" LastEditDate="2019-12-18T23:36:00.713" LastActivityDate="2019-12-18T23:36:00.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32429" PostTypeId="2" ParentId="3176" CreationDate="2019-12-19T00:21:25.943" Score="0" Body="&lt;p&gt;You would need a dual O ring system with sufficient spacing between the O rings. As for a sealant, use a silicon based lubricant. The interference pressure for would need to exceed the pneumatic pressure in the pipes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for picking the pipes in varying positions, an helical cam system would suffice with dual spring-loaded bearing balls that keep it either locked or open through a rotation (twisting) motion. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-12-19T00:21:25.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32430" PostTypeId="2" ParentId="32338" CreationDate="2019-12-19T00:56:06.057" Score="0" Body="&lt;p&gt;From what I understand of turbine dynamics, the pressure gradient is what determines mechanical motion. Creating a directional gradient for lateral motion perpendicular to pressure differential is what causes the turbine blades to move. Where the gradient is increased by larger surface area, wind velocity may be lower for action to appear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Surprisingly, the biplane of Great War era was able to stay aloft at considerably lower speeds than single fixed wing craft. This principle applies to turbines would result in bi-bladed or even Tri-bladed turbines variable of motion at lower wind velocities with the risk that higher winds would possibly become deleterious to structural integrity of the blades. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-12-19T00:56:06.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32431" PostTypeId="1" AcceptedAnswerId="32483" CreationDate="2019-12-19T03:12:30.150" Score="0" ViewCount="233" Body="&lt;p&gt;In this tutorial &lt;a href=&quot;https://www.youtube.com/watch?v=EVL7TzCde8I&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a current limiting application where two op amps are being used to limit the max output current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first differential opamp with 4 equally sized resistor values; connected to the output through a shunt resistor to measure the current draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The output from this opamp is then fed into the non inverting input of the second opamp while the inverting input of the opamp has a variable resistor to limit the output current.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone explain the ratio of this second opamp inverting input resistor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lets say I wanted to limit the current from 1-5 amps or 0-1 amps or even 2-20amps.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the ratio?&lt;/p&gt;&#xA;" OwnerUserId="23946" LastActivityDate="2019-12-22T12:59:01.973" Title="current limiting through opamp" Tags="|current|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32432" PostTypeId="1" CreationDate="2019-12-19T03:50:50.627" Score="3" ViewCount="49" Body="&lt;p&gt;I am trying to find estimates of the minimum kinetic energy required to impact the surface of asphalt roads as a first-order impact assessment of road functionality to rockfall (i.e. boulders hitting the road and creating a hole/crater). What I am really interested in is some sort of fragility function relating the kinetic energy or dynamic pressure with a measure of impact – e.g. probability of &quot;perforation&quot; or hole on the road surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't found much on that in the Natural Hazard literature and was wondering if anyone here could suggest any Engineering reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="23943" LastEditorUserId="23943" LastEditDate="2019-12-20T04:20:42.440" LastActivityDate="2019-12-20T04:20:42.440" Title="What is the kinetic energy required to impact the surface of asphalt roads?" Tags="|mechanical-engineering|civil-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32438" PostTypeId="1" CreationDate="2019-12-19T11:44:14.160" Score="2" ViewCount="100" Body="&lt;p&gt;I am emulating a throttle handle from a locomotive. See the diagram below for what I am trying to create.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its a throttle handle with 9 'positions' or notches (0-8 power). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The handle will probably pivot on the potentiometer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need some ideas on how to create mechanical notches, so it clicks into each throttle position or the user knows when they change to a different position. I cannot find much inspiration on the web, as other similar systems seem 'sealed'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xdB42.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xdB42.jpg&quot; alt=&quot;Proposed handle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23922" LastActivityDate="2019-12-25T19:13:02.463" Title="Notching a throttle handle" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32440" PostTypeId="2" ParentId="32415" CreationDate="2019-12-19T13:00:43.933" Score="0" Body="&lt;p&gt;Assuming you mean gears and not chain and sprocket systems, it really depends on the size and cost and duty of the gear reduction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest case is a colinear drop box that has a propshaft input and a propshaft output with no side loads at all. For modest ratios, the reduction should be around 3 percent of the &lt;em&gt;maximum rated torque&lt;/em&gt;. For example, if the gearbox is rated for 1000 ft lbs, because your 600 ft lb diesel requires a 30% derating, and you are operating at around 400 ft lbs most of the time, you still loose 3% of 1000, or 30 ft lbs. That's 7.5%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other load factors and physical arrangements can further reduce this. Overhung loads (side load associated with a spocket on the input or output shaft) may require additional bearings or a lower rating and tend to be a bit less efficient.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-12-19T13:00:43.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32441" PostTypeId="2" ParentId="32392" CreationDate="2019-12-19T13:24:17.503" Score="0" Body="&lt;p&gt;The &lt;strong&gt;Biot-Savart Law&lt;/strong&gt; can be interpreted as the relationship between the velocity induced by a vortex tube and the strength of the vortex tube, i.e. its circulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See: &lt;a href=&quot;https://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: you need to be a little careful in ascribing causes and effects due to the unfortunate adjective &quot;induced&quot; in induced velocity. In classical fluid mechanics there is no action-at-a-distance, so vorticity cannot cause changes to fluid velocities at remote locations.&lt;/p&gt;&#xA;" OwnerUserId="1908" LastActivityDate="2019-12-19T13:24:17.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32442" PostTypeId="2" ParentId="32438" CreationDate="2019-12-19T13:25:14.667" Score="7" Body="&lt;p&gt;A spring loaded ball in the handle and matching machined detents in the casing is a classic solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A raised bump, or dimple (from the other side) in a piece of spring steel can also work and can take up less space compared to a spring and ball.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-12-19T13:25:14.667" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32443" PostTypeId="1" CreationDate="2019-12-19T15:31:41.153" Score="1" ViewCount="123" Body="&lt;p&gt;I am studying control systems, and I have found that if I consider I control scheme where I filter the reference , if the zero of the pre-filter is too close to the imaginary axis, we have a degradation of the performances. For example, I have plotted this in Matlab:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/buznc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/buznc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which I moved towards the imaginary axis by increasing a constant value tau_num. The closest zero to the imaginary axis is relative to tau_num = 0.7, which infact has the worst performances, since it is present a resonance peack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thus, so far I have demonstarted this to be true just with Matlab, but I don't understand the reason for which this happens. I have found just that it is present a limit from the physical components, for which the zero cannot be too close to the dominant pole, but I haven't found anything else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody please help me? Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[EDIT] I am working with the following scheme:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/c0aHL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c0aHL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which I have a prefilter &lt;span class=&quot;math-container&quot;&gt;$C_r(s)$&lt;/span&gt;, and the transfer functions I am using are the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;P = 1/[(1+s)*(1+0.05*s)^2];&#xA;C = 10*(s+1)/s;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and for the pre-filter I am considering different values, which are shown in the plot above, which are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;tau_d1 = 0.9;&#xA;tau_num1 = 0.1;&#xA;&#xA;tau_d2 = 0.9;&#xA;tau_num2 = 0.4;&#xA;&#xA;tau_d3 = 0.9;&#xA;tau_num3 = 0.7;&#xA;&#xA;C_r_1 = (1+tau_num1*s)/(1+tau_d1*s);&#xA;C_r_2 = (1+tau_num2*s)/(1+tau_d2*s);&#xA;C_r_3 = (1+tau_num3*s)/(1+tau_d3*s);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;and the closed loop transfer functions are:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T1 =  C_r_1*C*P/(1+C*P);&#xA;T2 =  C_r_2*C*P/(1+C*P);&#xA;T3 =  C_r_3*C*P/(1+C*P);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;which I plot in this way:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;figure;&#xA;bodemag(T1,'b',T2,'r',T3,'g'),grid&#xA;legend('tau_num = 0.1','tau_num = 0.4','tau_num = 0.7');&#xA;title('amplitude of frequency response as zero gets slower')&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="22780" LastEditorUserId="22780" LastEditDate="2019-12-19T18:07:19.777" LastActivityDate="2019-12-19T18:07:19.777" Title="Why a zero too close to the imaginary axis gives bad berformances?" Tags="|electrical-engineering|control-engineering|control-theory|systems-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32444" PostTypeId="2" ParentId="32414" CreationDate="2019-12-19T16:15:55.550" Score="0" Body="&lt;p&gt;Toyochem has a product called LIOELM™ FTS series — Thermal Conductive Adhesive Sheets.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They claim it is flexible but has high thermal conductivity (maximum 10W/m･K).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.toyo-chem.com/en/products/electronics/lioelm_fts.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.toyo-chem.com/en/products/electronics/lioelm_fts.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23088" LastActivityDate="2019-12-19T16:15:55.550" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32445" PostTypeId="1" AcceptedAnswerId="32447" CreationDate="2019-12-19T18:37:57.440" Score="0" ViewCount="155" Body="&lt;p&gt;Assume there is a rectangular beam of length L resting on a rigid horizontal surface. The beam is of a real homogeneous material, say aluminum. Assume the surface is infinitely rigid. I apply a vertical force F on each end of the beam, clamping the beam to the surface.&#xA;1) How do the clamp forces on the surface distribute along the beam?&#xA;2) What would be the shape of the curve for the bottom of the beam that would result in equal distribution of forces?&#xA;This has real applications in gluing of large flat objects together, where standard clamps cannot reach the interior areas.&#xA;Thanks&#xA;Kevin&lt;/p&gt;&#xA;" OwnerUserId="23960" LastActivityDate="2019-12-20T18:22:06.840" Title="load distribution" Tags="|statics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32446" PostTypeId="2" ParentId="32338" CreationDate="2019-12-19T20:31:50.717" Score="0" Body="&lt;p&gt;Wind speed is the speed of wind coming/hit the propeller surface perpendicularly, or inline to the turbine's axis for horizontal axis wind turbine (HAWT), or 90 degree perpendicular to axis of a vertical axis wind turbine (VAWT).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For HAWT (calculation for VAWT is quite different), the kinetic power of the wind is: half of the air density (rho) times cross-section (A) times power of 3 of wind-velocity (V^3). Or:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kinetic power = 0.5*Rho&lt;em&gt;A&lt;/em&gt;V^3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where:&#xA;A is 2*pi*r, with r is the blade's length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, kinetic power of wind is proportional to the air density (rho), affected significantly by the blades length, and affected very significant by the wind speed as it has power of 3 (multiplied three times).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you are thinking about tip speed of the blade is another thing. It is related to the power efficiency of a propeller of an airplane or a fan. It will be related to reynold number, stall, laminar, etc. Mach 0.8 so far is claimed the most effective tips speed. Mean, if you have a big fan and the fan axis or shaft you connected to an engine to power it, if you force the fan speed more than mach 0.8 (0.8 of sound speed), you will lost much power (in horse power) without gaining more force generated by the fan (or propeller as commonly in airplane). So, the airplane speed or tip speed is maintained maximum mach 0.8, not more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clearly, the wind speed you are asking is the speed of the wind hitting or coming to the wind turbine blades, perpendicularly to the cross section, NOT the speed of the tip.&lt;/p&gt;&#xA;" OwnerUserId="17943" LastActivityDate="2019-12-19T20:31:50.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32447" PostTypeId="2" ParentId="32445" CreationDate="2019-12-19T20:36:57.917" Score="0" Body="&lt;p&gt;Let's say you want to apply a uniform load of q over the length of the beam pressuring equally along the length. clamps are holding the ends with the force of F= qL/2 at each end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine there was no floor supporting the beam and these were the reactions of a distributed loading, &lt;span class=&quot;math-container&quot;&gt;$q= 2F/l.$&lt;/span&gt; This q is the pressure you want to apply uniformly upward from the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That load would deflect the beam into a parabola with maximum deflection of &lt;span class=&quot;math-container&quot;&gt;$5qL^4/384EI, \\ \text{and the equation of the beam's centerline deflection is}  \ y= -\frac{qx}{24EI}(L^3-2Lx^2+x^3)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This curve is the deflection of a simply supported beam under the uniformly distributed load of q, which looks like a rope hanging at 2 ends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to camber your beam so that the centerline of it is deflected down by the above eq. and form it by the above deflection curve to have the floor distribute a uniform load, q, to it when it is clamped and forced down to perfect straight line touching the surface.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-20T18:22:06.840" LastActivityDate="2019-12-20T18:22:06.840" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32450" PostTypeId="1" CreationDate="2019-12-20T00:04:02.953" Score="1" ViewCount="178" Body="&lt;p&gt;I'm working on an application which requires varying the speed of a conveyor that will frequently need to stop and start.&#xA;More specifically, we need to first set the speed of the belt, and then have frequent cycles of 3-4 seconds on and then stop for 3-4 seconds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had initially thought a simple 3-phase conveyor with a VFD (for speed adjustment) and then a contactor (for start/stop) would be fine... But a few google searches seemed to indicate that this would probably damage the VFD in a short order, since the VFD needs a few seconds to start correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there a canonical way to set up this sort of system ? The conveyor load is low (few kgs) and the speed is also low (around 1m / 10 seconds).&lt;/p&gt;&#xA;" OwnerUserId="23962" OwnerDisplayName="Petrov" LastEditorUserId="6767" LastEditDate="2019-12-23T19:17:51.790" LastActivityDate="2019-12-23T19:17:51.790" Title="Variable speed conveyor / frequent stop and start" Tags="|conveyor|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32451" PostTypeId="2" ParentId="32450" CreationDate="2019-12-20T00:54:18.587" Score="0" Body="&lt;p&gt;I think the best choice is a programmable VFD with computer controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a custom choice with a variety of interfaces ;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keypad Control (Run/Stop)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 Wire Control (Fwd/Stop)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 Wire Control (Fwd/Rev/Stop)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Run/Stop Serial Comms (RS485)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequency Ref. from Keypad&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequency Ref. 0-10V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequency Ref. Potentiometer&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequency Ref. 4-20mA&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequency Ref. Serial RS485&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frequency Ref. Pulse Input&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PID Feedback Signal 4-20mA&lt;/p&gt;&#xA;" OwnerUserId="8847" LastActivityDate="2019-12-20T00:54:18.587" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32452" PostTypeId="2" ParentId="32450" CreationDate="2019-12-20T03:36:36.183" Score="1" Body="&lt;p&gt;A lot of VFDs are programmable to the extent that you describe. You can program several preset speeds and send a command to the VFD that will tell it to start the motor, accelerate at a preset rate to a preset speed. When you want the motor to stop, you send command is sent to the VFD, and it will decelerate the motor to a stop at another preset deceleration rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The difficult part will be to determine if the motor can accelerate and decelerate as quickly as you would like. However, if it can be done, a VFD can likely do it as quickly as starting and stopping with a contactor and with less stress on the motor and mechanical system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Start, stop and preset speed selection commands can be by contact open/close or by serial communication. If you need to have more than two or three preset speeds they can be sent by analog signal or serial communications. Some VFDs may store more than two or three.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is best to set the system up so that the fastest operation ever required is  obtained at or near the speed of the motor when operating at the 50 or 60 Hz power frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is selecting a VFD, consider what brands are popular in your area of the world. Also consider what brands are offered by local suppliers that are trustworthy and helpful. Download and read the manual before buying. You need a general-purpose VFD, not a VFD that is designed for fans and pumps.&lt;/p&gt;&#xA;" OwnerDisplayName="user6431" LastActivityDate="2019-12-20T03:36:36.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32453" PostTypeId="1" AcceptedAnswerId="32477" CreationDate="2019-12-20T07:43:13.647" Score="3" ViewCount="2724" Body="&lt;p&gt;I would like to control a 1-phase 3HP motor using a SSR ( the reason being, I need to switch it on/off using a signal (0-5VDC).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are normal SSRs (ex:Fotek 40A model) capable of handling the starting  current of a 3HP motor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the protection devices that I need to use in the circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess a Motor MCB is required. Is anything else required?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All opinions/suggestions are welcome.&#xA;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="19228" LastActivityDate="2021-06-03T16:20:05.923" Title="AC Motor On/OFF Control using SSR instead of contactor" Tags="|electrical-engineering|control-engineering|motors|pumps|electrical|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32454" PostTypeId="1" AcceptedAnswerId="32457" CreationDate="2019-12-20T08:11:51.523" Score="3" ViewCount="390" Body="&lt;p&gt;So for any structure, simply supported beam, three hinge frame, cantilever etc. where there is a uniformly distributed load. For example, for a simply supported beam with uniform load of w along the length of x. Why is the bending moment equation w(x)(x/2)? I understand that the magnitude of the force is given by w(x), but is the reason for the (x/2) is because the magnitude acts halfway through the load? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I know that shear force and bending moment are related, with the shear force being the derivative of the bending moment, so I can just integrate the shear force equation to get w(x)(x/2), but I want to confirm through fundamentals as to why the equation is as it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks.&lt;/p&gt;&#xA;" OwnerUserId="23676" LastActivityDate="2019-12-20T08:51:28.043" Title="Why is the bending moment equation for a uniformly distributed load (w) across a length x on a structure given by w(x)(x/2)?" Tags="|moments|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32455" PostTypeId="1" AcceptedAnswerId="32469" CreationDate="2019-12-20T08:22:02.220" Score="0" ViewCount="62" Body="&lt;p&gt;I need to hold small pieces of wood in place to be drilled by a mini computer-numerical control (CNC) mill. I have a &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B07459P7BF&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;book stand&lt;/a&gt; (no affiliation) that has two &quot;springed pegs&quot; fastened with screws:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Springed pegs allow very thick book&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think these would do very well and I could not find such springed pegs separately. Are they available under a different name?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Update&lt;/strong&gt;: My current solution prevents the pieces from shifting horizontally by slotting them in a pocket with their exact size with 5 mm of depth. I have been holding the pieces in place with double-sided tape and it works. A milling job takes about a minute and I have hundreds of pieces, so I am looking for a quicker solution. The machine uses a downcutter that presses the wood into the bed, so I think light tension would be enough.&lt;/p&gt;&#xA;" OwnerUserId="19384" LastEditorUserId="19384" LastEditDate="2019-12-20T09:05:45.540" LastActivityDate="2019-12-20T22:40:13.273" Title="Where to find springed pegs with screws for a CNC mill?" Tags="|manufacturing-engineering|cnc|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32456" PostTypeId="2" ParentId="32455" CreationDate="2019-12-20T08:35:02.057" Score="2" Body="&lt;p&gt;The standard solution is to use milling &lt;a href=&quot;https://littlemachineshop.com/products/product_category.php?category=11&quot; rel=&quot;nofollow noreferrer&quot;&gt;clamps&lt;/a&gt;. These consist of a screw that fits in the T-slots on the milling table, and a &lt;a href=&quot;https://www.mcmaster.com/milling-machine-clamps&quot; rel=&quot;nofollow noreferrer&quot;&gt;sturdy metal fitting&lt;/a&gt; that holds down the work. They're available in various forms to fit different situations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Spring clamps would have to be massive to provide enough force to keep the work from sliding around. But your solution helps with that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jzXfU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jzXfU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The item you're looking for is a &lt;a href=&quot;https://weihuispring888.en.made-in-china.com/product/ljdEqJhPEucO/China-Custom-Flat-Spring-Clip.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;flat spring clip&lt;/a&gt; with a &lt;a href=&quot;https://www.mcmaster.com/thumb-screw-knobs&quot; rel=&quot;nofollow noreferrer&quot;&gt;thumb screw&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="11645" LastEditorUserId="11645" LastEditDate="2019-12-20T09:24:54.393" LastActivityDate="2019-12-20T09:24:54.393" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32457" PostTypeId="2" ParentId="32454" CreationDate="2019-12-20T08:51:28.043" Score="3" Body="&lt;p&gt;consider an infinitesimally small section of a beam from coordinate &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$x_1-dx$&lt;/span&gt; loaded with a distributed load w. The moment of this load about point &lt;span class=&quot;math-container&quot;&gt;$x=0 \  is \\ M= w*x*dx.$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now if we have the load area extended from &lt;span class=&quot;math-container&quot;&gt;$x_0 \ to\  x,$&lt;/span&gt; we need to integrate the moment over the length of beam loaded with w: &lt;span class=&quot;math-container&quot;&gt;$$M= \frac{1}{2}wx^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or as you noted because the CG of load rectangle w*x is at its middle the moment is the area times the distance of CG to &lt;span class=&quot;math-container&quot;&gt;$x_o$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ M= wx*x/2=wx^2/2$&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-20T08:51:28.043" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32458" PostTypeId="1" CreationDate="2019-12-20T10:50:40.200" Score="3" ViewCount="1138" Body="&lt;p&gt;The question is already in the title, but I guess some context might be interesting:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm living in Munich (Germany). It is the area in Germany which is probably most expensive to live. Even the surrounding area is way more expensive. Building higher is only partially possible due to construction regulations (e.g. in the center nothing is allowed to be higher than the &lt;a href=&quot;https://en.wikipedia.org/wiki/Munich_Frauenkirche&quot; rel=&quot;nofollow noreferrer&quot;&gt;Frauenkirche&lt;/a&gt;; roughly 100m).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most desired are the areas close to &lt;a href=&quot;https://en.wikipedia.org/wiki/S-train&quot; rel=&quot;nofollow noreferrer&quot;&gt;S-train stations&lt;/a&gt;, especially close to the ones that are highly frequented (I'd guess about one train every 2 minutes or so).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost at the center, there are buildings next to the train tracks. For example &lt;a href=&quot;https://goo.gl/maps/qWvdxmXu29fqyknz8&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Munich Google office&lt;/a&gt;. However, I have never seen buildings over the tracks (except for bridges). And the tracks cover a lot of space which is super central. To give you an impression, look at the screenshot below:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;North of the tracks, east of the yellow-colored bridge, you can see the Google office. the building isn't huge, but it's not small either.&lt;/li&gt;&#xA;&lt;li&gt;The big building in the top of the screenshot is &lt;a href=&quot;https://de.wikipedia.org/wiki/Paketposthalle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Paketposthalle&lt;/a&gt;. It covers an area of 20.000 m².&lt;/li&gt;&#xA;&lt;li&gt;We're talking about 2200 EUR/m² to 4200 EUR/m² (&lt;a href=&quot;https://stege-immo.de/blog/grundstuecke-in-munchen/&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LCzwB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LCzwB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this reason, I wonder why I haven't seen any buildings directly over train tracks. Is insulation against the noise the problem? Is safety a problem, e.g. if a train derails? Is it the vibrations that might damage the house? Building safety exits? Is there not enough area below where one could build the support structures for the building?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are way more expensive areas around the world than Munich. I guess Hong Kong / Tokyo / Shanghai might be one extreme example. Do they have buildings over train tracks? I'm not talking about underground ... but thinking about it, I'm not quite sure if there would be a difference if the city grows over the train tracks.&lt;/p&gt;&#xA;" OwnerUserId="3061" LastActivityDate="2021-10-19T12:54:51.710" Title="Why is building houses over train tracks way more difficult / expensive?" Tags="|civil-engineering|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32459" PostTypeId="2" ParentId="32455" CreationDate="2019-12-20T11:45:18.930" Score="2" Body="&lt;p&gt;As you have the &quot;pocket&quot; already, one method would be to have holes in the pocket base and use a vacuum cleaner to &quot;suck&quot; the piece down.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-12-20T11:45:18.930" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32460" PostTypeId="1" AcceptedAnswerId="32461" CreationDate="2019-12-20T13:37:23.197" Score="2" ViewCount="43" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MXicV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MXicV.png&quot; alt=&quot;Cross-section of cylinder&quot;&gt;&lt;/a&gt;&#xA;The situation I'm imagining is the one shown above, which is the cross section of a cylinder made of alternating layers of material. The material axes are shown where 1 and 2 are in the same plane and 3 is parallel to the stacking direction. If a cylinder composed of this material is subjected to a torque around its central axis as shown, would it be subject to axial stresses and therefore result in torsional warping? I read &lt;a href=&quot;https://engineering.stackexchange.com/questions/31949/why-torsional-warping-does-not-occur-for-shafts-with-circular-symmetry&quot;&gt;here&lt;/a&gt; and in textbooks that circular members don't have torsional warping but they are usually assumed to be made of isotropic materials. Would this be the case for this type of cylinder too?&lt;/p&gt;&#xA;" OwnerUserId="18933" LastActivityDate="2019-12-20T16:20:19.167" Title="Would an orthotropic cylinder subjected to torsion have warping?" Tags="|materials|applied-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32461" PostTypeId="2" ParentId="32460" CreationDate="2019-12-20T16:20:19.167" Score="1" Body="&lt;p&gt;Yes, it will warp.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine the blue layer material stiffer, with higher Young modulus E and shear modulus G and the white material as less stiff. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The blue material will be the dominant material in reacting to the torsion and will warp in developing both St Venanat shear stresses and warping stress which are axial tensile and compressive stresses and additional shear stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZrExl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZrExl.png&quot; alt=&quot;warped torsion&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Torsion causing warping in the blue laminates.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kxlaM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kxlaM.png&quot; alt=&quot;axil stresses and shear due to torsion.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Additional axial stresses due to the warping.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The white layer will undergo similar deformation forced by the blue material deformation and will be stressed less, with a wavy step like undulations at the intersection line with blue material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In these diagrams, I have not shown the white material for clarity.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-20T16:20:19.167" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32462" PostTypeId="1" AcceptedAnswerId="32465" CreationDate="2019-12-20T17:31:56.117" Score="1" ViewCount="62" Body="&lt;p&gt;I am looking at a control system which has an unstable pole in the process. The transfer functions of process and controller are the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;G = 10/((s+10)*(s-1));&#xA;K1 = 4*(s+1)/s;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where G is the process and K1 is the controller. I have found with the Routh Criterion that the closed loop is stable for &lt;span class=&quot;math-container&quot;&gt;$K&amp;gt;1.125$&lt;/span&gt; and so I have chose the value 4 since it gives good performances. Now, when I plot the root locus of the closed loop:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;T1 = K1*G/(1+K1*G)&#xA;figure;&#xA;rlocus(T1)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I get the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0GS1h.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0GS1h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I don't undestand is from where the zero that cancels the unstable pole comes from, since the closed loop transfer function is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{(40(s + 1))}{(40 - 10s + 40s + 9s^2 + s^3)}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can somebody help me?&lt;/p&gt;&#xA;" OwnerUserId="22780" LastActivityDate="2019-12-20T19:28:10.113" Title="Doubt on zero pole cancellation" Tags="|control-engineering|control-theory|systems-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32463" PostTypeId="1" CreationDate="2019-12-20T17:58:25.297" Score="2" ViewCount="61" Body="&lt;p&gt;I have a project where I'd need to measure small voltage (0.05 mV) output of a wheatstone bridge with an arduino mini pro.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/32210/what-is-the-range-of-the-measurable-relative-elasticity-with-a-commonly-used-str&quot;&gt;Based on my previous question most probably it can't be done with its analogue input directly.&lt;/a&gt;&#xA;&lt;a href=&quot;https://engineering.stackexchange.com/questions/32210/what-is-the-range-of-the-measurable-relative-elasticity-with-a-commonly-used-str&quot;&gt;what is the range of the measurable relative elasticity with a commonly used strain gauge&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is that can a &lt;a href=&quot;https://www.mouser.com/datasheet/2/813/hx711_english-1022875.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;hx711 module&lt;/a&gt; measure so small voltages? &#xA;Based on the calculation in the question it has the resolution of 3 uV from 0 to 5V excitation voltage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can estimate that what resolution I can expect in practice?&lt;/p&gt;&#xA;" OwnerUserId="15534" LastEditorUserId="110" LastEditDate="2019-12-24T14:04:47.183" LastActivityDate="2019-12-24T14:04:47.183" Title="How can I calculate the smallest voltage a given adc can read?" Tags="|measurements|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32464" PostTypeId="2" ParentId="32458" CreationDate="2019-12-20T18:49:48.127" Score="4" Body="&lt;p&gt;Basically you'd first need to wrap a tunnel around a working track with the least amount of interruption, with all the access, egress, safety Mesures, utilities, drainage, etc. And then try to create a new city grid of roads, buildings, utilities over it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It will be impossibly expensive, and very disruptive to daily life of people in the neighborhood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a rule of thumb, demolishing something and building from scratch is always much easier and cost-effective than trying to work around an existing, working project.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-20T18:49:48.127" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32465" PostTypeId="2" ParentId="32462" CreationDate="2019-12-20T19:28:10.113" Score="3" Body="&lt;p&gt;rlocus() takes the &lt;em&gt;open loop&lt;/em&gt; transfer function as an argument, not the closed loop. i.e. G not T. Look at the documentation for rlocus here &lt;a href=&quot;https://www.mathworks.com/help/control/ref/rlocus.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mathworks.com/help/control/ref/rlocus.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-12-20T19:28:10.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32467" PostTypeId="1" AcceptedAnswerId="32470" CreationDate="2019-12-20T20:25:36.127" Score="2" ViewCount="42" Body="&lt;p&gt;I am currently studying a robotics paper, where a normalized frobenius norm of a matrix is calculated as part of determining the condition number of the corresponding linear system of equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Am5HJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Am5HJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/22Z6C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/22Z6C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\mathbf{W}$&lt;/span&gt; is defined as &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\mathbf{W} = \frac{1}{n} \mathbf{I}$$&lt;/span&gt; and inserting this definition into &lt;span class=&quot;math-container&quot;&gt;$\left|\left|\mathbf{J}\right|\right|$&lt;/span&gt; results in &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\left|\left|\mathbf{J}\right|\right| = \sqrt{\text{tr}\left(\mathbf{J}\mathbf{W}\mathbf{J}^{T}\right)} =\sqrt{\text{tr}\left(\mathbf{J}\frac{1}{n}\mathbf{I}\mathbf{J}^{T}\right)} = \sqrt{\frac{1}{n}\text{tr}\left(\mathbf{J}\mathbf{I}\mathbf{J}^{T}\right)} = \sqrt{\frac{1}{n}\text{tr}\left(\mathbf{J}\mathbf{J}^{T}\right)} .$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The normalization factor is therefore independet of &lt;span class=&quot;math-container&quot;&gt;$\mathbf{J}$&lt;/span&gt;. Therefore, what is the purpose of scaling &lt;span class=&quot;math-container&quot;&gt;$\left|\left|\mathbf{J}\right|\right|$&lt;/span&gt; down by a factor proportional to its dimension? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In terms using &lt;span class=&quot;math-container&quot;&gt;$\kappa(\mathbf{J})$&lt;/span&gt; to optimize a mechanism, this would mean that the respective value of each iteration would simply be scaled down by constant factor, and therefore not influence the results. Am I missing something here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="8674" LastEditorUserId="8674" LastEditDate="2019-12-21T01:19:59.337" LastActivityDate="2019-12-21T01:19:59.337" Title="Purpose of normalization of Jacobian in a parallel manipulator" Tags="|design|robotics|kinematics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32468" PostTypeId="1" CreationDate="2019-12-20T22:38:55.760" Score="1" ViewCount="559" Body="&lt;p&gt;In short, the question is: How the EPC affects an automatic transmission?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm planning to build an automatic transmission controller module, TCM for 4F27E Ford transmission. So far I haven't found an aftermarket transmission controller that would work for this transmission, hence this project. For now I would like to raise a discussion about EPC. I will then write a control software for the TCM based on what I learn about EPC and other solenoids.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an automatic transmission, EPC solenoid controls the pressure in the hydraulic circuits affecting the line pressure. There is a steady state pressure, that is the amount of pressure required to hold applied clutches and bands without slipping after the shift has completed and for steady use. The line pressure controls the shift, too low pressure will cause soft and long shifts; too high will cause harsh shift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Great theory, but...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: does the line pressure affects the torque converter? The 4F27E (and similar) has a dedicated solenoid to control the torque converter clutch (TCC) but normally is an on-off solenoid, is it possible achieve the increase of speed with fairly constant RPM, commonly observed in automatic transmission, by varying the EPC? (or this is purely done by the torque converter characteristics?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Continuing the question: as the theory says, high pressure causes harsh shift. Does the TCM need to action on the EPC to reduce the line pressure to the required level before gear shift?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More: The line pressure varies with the engine RPM (or more precisely, speed of input shaft connected to the hydraulic pump). Does that means the line pressure is function of RPM x EPC x temperature, or there are other variables?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers.&lt;/p&gt;&#xA;" OwnerUserId="23965" LastActivityDate="2019-12-20T22:38:55.760" Title="Electronic Pressure Control, EPC" Tags="|mechanical-engineering|automotive-engineering|mechanical|transmission|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32469" PostTypeId="2" ParentId="32455" CreationDate="2019-12-20T22:40:13.273" Score="2" Body="&lt;p&gt;Not what you're asking for, but maybe even more suitable?  These are also available in a low-profile version.  Google &quot;toggle clamp&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gPJef.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gPJef.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or these?  Google &quot;spring plunger&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9mgVo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9mgVo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-12-20T22:40:13.273" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32470" PostTypeId="2" ParentId="32467" CreationDate="2019-12-21T00:54:28.460" Score="2" Body="&lt;p&gt;It is because you are using the trace of the matrix, which is equal to the &lt;em&gt;sum&lt;/em&gt; of the singular values.  The ideal &lt;span class=&quot;math-container&quot;&gt;$\mathbf{JJ^T}$&lt;/span&gt; has singular values all equal to 1, but for an &lt;span class=&quot;math-container&quot;&gt;$n \times n$&lt;/span&gt; matrix this sum would be equal to &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;, not &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, and the text is defining the ideal matrix norm as 1.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2019-12-21T00:54:28.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32471" PostTypeId="1" AcceptedAnswerId="32475" CreationDate="2019-12-21T13:38:10.390" Score="1" ViewCount="45" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X9jqz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X9jqz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do they do? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aux contacts are normally open and the device is MR PQM 500 reactive power compensation system.&lt;/p&gt;&#xA;" OwnerUserId="23966" LastEditorUserId="23966" LastEditDate="2019-12-22T06:45:57.813" LastActivityDate="2019-12-22T06:45:57.813" Title="What's the function of those coils there?" Tags="|electrical-engineering|power-electronics|electrical|power-engineering|electronics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32472" PostTypeId="1" CreationDate="2019-12-21T13:59:56.100" Score="1" ViewCount="75" Body="&lt;p&gt;I have an &quot;DaVinci IQ&quot; portable vaporizer. Since the battery quickly runs out of power, I often keep it lying next to me, plugging from the computer into its USB-to-Mini-USB (or maybe it's Micro-USB -- unsure about that detail), often beyond the point where all the five LEDs are lit, indicating that it's fully charged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know where I heard this, but growing up with &quot;early&quot; rechargeable batteries, it was always a &quot;thing&quot; that you were supposed to stop charging them as quickly as possible after they are fully charged, or else &quot;bad things&quot; supposedly happen. I don't know what exactly those bad things were, but I'm pretty sure this was true at some point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: is it still true? Should I unplug it ASAP once it's fully charged, or does it no longer make any difference whatsoever?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And, if the answer is that this still has to be done, why is this? Why can't the device &quot;stop sending in electrons&quot; once it's fully charged, since it clearly can detect that?&lt;/p&gt;&#xA;" OwnerUserId="23967" LastActivityDate="2019-12-22T11:40:07.687" Title="Is it still a bad idea to keep modern electronics plugged into the charger cable for longer than necessary?" Tags="|power-electronics|battery|consumer-electronics|electronics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32475" PostTypeId="2" ParentId="32471" CreationDate="2019-12-21T17:32:13.613" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;The aux contacts are normally closed ...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is unlikely. I suggest that they are normally open and close before the main contacts do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You haven't provided a link for the MR PQM 500 system but I suspect that it is a capacitor bank with a controller which switches the contactors to switch in capacitance as required to bring the power factor closer to unity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is problem when switching in capacitors. If the mains voltage is close to peak at the instant the contacts close a very high current will flow with resultant arcing, etc. because the discharge capacitor appears like a short-circuit. To prevent this a special arrangement is employed:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The auxiliary contact block at the front is driven by the main contactor movement but the auxiliary contacts close slightly before the main ones.&lt;/li&gt;&#xA;&lt;li&gt;The three coils on top are inductors which have enough inductance to limit the current into the discharged capacitor to a safe value. This is enough to bring the voltage up so that the main contacts don't see the big surge they otherwise would.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2019-12-21T17:32:13.613" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32476" PostTypeId="1" AcceptedAnswerId="32478" CreationDate="2019-12-21T17:58:48.313" Score="1" ViewCount="58" Body="&lt;p&gt;How much steam is emitted from an electric arc furnace in a unit of time? I have not been able to find any data on this which is strange considering the amount of water that is used to cool an electric arc furnace.&lt;/p&gt;&#xA;" OwnerUserId="7538" LastActivityDate="2019-12-21T20:29:43.843" Title="Electric Arc Furnace Steam Production" Tags="|electrical-engineering|structural-engineering|steel|welding|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32477" PostTypeId="2" ParentId="32453" CreationDate="2019-12-21T19:18:24.620" Score="4" Body="&lt;p&gt;&lt;strong&gt;Yes, but...&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's not great data available on the Fokotek products, but in general switching HP sized motor loads with solid state relays is a fairly expensive proposition. Because the voltage and current values don't cross zero at the same time, motor loads (inductive loads) are harder to switch than resistive loads. 3HP is just on the cusp where SSRs are still practical if you don't mind spending a little money, but you have to be careful about product selection. Here are some things that can help you figure out what a given device is rated for when you read the datasheet:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Products that are intended to switch voltage to motors will be rated with a horsepower capacity in addition to an amperage. For examply, &lt;a href=&quot;http://www.crydom.com/en/products/catalog/cw-series-24-ac-panel-mount.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cryodon&lt;/a&gt; does make SSRs that are rated for a 3HP motor load at 120VAC (I'm assuming you're in the US using residential power, switching is slightly more efficient at higher voltages,) but note that the nominal current rating of the contactor is 125 Amps!&lt;/li&gt;&#xA;&lt;li&gt;On the more technical end, switching capacities will actually be rated according to a specific standard. The relay needs to handle a different load at the time of switching for a light bulb, a transformer, and a motor all of the same nominal amperage. For equipment marketed to the US, you'll often find the rated switching current defined for each &lt;a href=&quot;https://en.wikipedia.org/wiki/Utilization_categories#Table&quot; rel=&quot;nofollow noreferrer&quot;&gt;IEC utilization category&lt;/a&gt; that the device is intended for. Your motor probably falls under AC-3 so you would look for a specific rating in that category and compare it to the HP or Full Load Amperage (FLA) of your motor.&lt;/li&gt;&#xA;&lt;li&gt;Sometimes listings will be shown with respect to a specific standard as well, eg UL508 or IEC 60947. In this case the datasheet should indicate specifically the type of motor and the voltage that the HP rating applies to.&lt;/li&gt;&#xA;&lt;li&gt;In general, if it doesn't have an HP rating or a specific utilization category, and nothing on the data sheet specifically says it can be used on motor loads, you should assume that it cannot be. Depending on the size it may work for a while, but you are likely to damage the switching element over time and it will eventually fail. In the world of motors, many failure modes can be dangerous, including an inability to stop the motor and a resistive short causing a fire in the switching element.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As far as other protective devices, you are correct that you should use a dedicated circuit breaker to protect from a short in the motor. In addition to this you should* have a thermal overload device. This device is meant to protect the motor in case of a stall where there's no short circuit, but the motor is drawing it's full amperage for too long without being able to cool itself. The trip current is much lower than the circuit breaker, but the time over which this device averages data is much longer. Motors are available with thermal overloads built into them, and separate thermal overload devices are also available (usually as an accessory for a contactor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your application involves a safety concern when it operates, then even if you use an SSR for routine operations, your lockout/tagout and/or emergency stop functions should be accomplished using a more positive means of power disconnection (probably a disconnect switch for the LOTO and a mechanical contactor for E-stop.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the main thing driving you to the SSR is the control voltage available from your top-level controller (perhaps an arduino?) there are a few options that may be better choices for you:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A conventional contactor with a thermal overload, and a &lt;a href=&quot;https://www.sparkfun.com/products/15093&quot; rel=&quot;nofollow noreferrer&quot;&gt;smaller relay&lt;/a&gt; to step up your 5VDC signal. Your controller would control a small relay with a 5V or less coil voltage. The load side of the small relay would be connected to your Line Voltage at 120VAC (or whatever your motor uses) but instead of powering the motor, it would power the coil on a traditional motor contactor (&lt;a href=&quot;https://www.automationdirect.com/adc/shopping/catalog/motor_controls/iec_contactors_-z-_starters_-z-_overloads/fuji_contactors_-z-_overloads/32_to_50_amp/sc-e2s-110vac&quot; rel=&quot;nofollow noreferrer&quot;&gt;this is just one example&lt;/a&gt;). In this configuration the small relay (often called a pilot relay or a PLC relay) on has to be rated for the coil consumption current of the relay so 1.1 amps AC-14 or AC-140 in the example I linked to. This solution also allows you to use a conventional &lt;a href=&quot;https://www.automationdirect.com/adc/shopping/catalog/motor_controls/iec_contactors_-z-_starters_-z-_overloads/fuji_contactors_-z-_overloads/32_to_50_amp/tk-e2-2600&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermal overload&lt;/a&gt; accessory without having to mount it separately in your enclosure. Note that since most thermal overloads and contactors are designed for 3 phase motors, and include a mechanism to detect loss of a single phase, you'll need to feed the same current through all 3 paths on the contactor. In practice, this is easy, you just run one leg of your power in through two paths in series. &lt;/li&gt;&#xA;&lt;li&gt;Alternately, you could use a small variable frequency drive. A VFD is basically 3 SSRs and a top-level controller, and it would give you lots of options for control, and typically includes monitoring that counts as your thermal overload detection. THe biggest down side is that VFDs are pretty much only available for 3-phse motors, so if you already have a single phase motor you might not be able to use a VFD. VFDs are available that accept single phase inputs, but still output three phase. VFDs are designed to be controlled my outside sources, so communicating over a 5 volt interface or even something more sophisticated will be straightforward. Additionally, VFDs will reduce the amount of inrush power that your motor pulls off of the line voltage supply, which can start to be a problem in motors of this size.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;*(Whether this is strictly required or not depends on your jurisdiction and some other details, but it's definitely a good practice if you don't like fire.)&lt;/p&gt;&#xA;" OwnerUserId="122" LastActivityDate="2019-12-21T19:18:24.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32478" PostTypeId="2" ParentId="32476" CreationDate="2019-12-21T19:25:06.500" Score="2" Body="&lt;p&gt;I‘m not quite sure i understand the question, assuming everything is fine, there should be no steam at all. These kind of systems can violently explode if there is a runaway steam production...as quoted &lt;a href=&quot;https://spraycooled.tsg.bz/wp-content/uploads/EAF-Explosions-A-Deadly-but-Preventable-Problem.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&#xA;(Warning pdf)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a pressurized cooling system: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;...A two-square-inch hole in a tubular panel results in more than 16,000 gallons of water spilled into the furnace in just one hour...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Using spray cooling, there is a loss of considerably less...only 5 gallons per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since water expands around 1700~x when turning to steam, one could calculate approx how much steam that would produce, assuming all of the water doesn‘t flash to stream at once and explode, but generally there shouldn‘t be any steam at all....or are you asking soemthing else?&lt;/p&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2019-12-21T19:25:06.500" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32479" PostTypeId="1" AcceptedAnswerId="32488" CreationDate="2019-12-21T19:32:58.843" Score="1" ViewCount="128" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X2IJv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X2IJv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/wfl0C.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wfl0C.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understand correctly, one of the star's teeth is bent into one of 4 nut's slots to secure the nut. What happens if none of the teeth align with the slots? Do you unscrew the nut a bit? Or it is somehow mathematically made that there will always be one slot that aligns properly?&lt;/p&gt;&#xA;" OwnerUserId="21424" LastActivityDate="2019-12-22T14:00:08.487" Title="What happens if a star lock washer doesn't align properly with the nut?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32480" PostTypeId="2" ParentId="32476" CreationDate="2019-12-21T20:29:43.843" Score="1" Body="&lt;p&gt;No steam at all from the furnace ; even the charge is preheated so that it is dry. Water may be used to cool the power transformers.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-12-21T20:29:43.843" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32481" PostTypeId="1" CreationDate="2019-12-21T20:45:36.200" Score="1" ViewCount="517" Body="&lt;p&gt;(This question comes from a Chemical Engineering background, but basic Math and Programming knowledge is also required. I hope it still falls in the scope of engineering.stackexchange.com  If not, please move.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume a (slow hydrolysis) first-order reaction of a component D in a CSTR, which is described by the well-known ODE:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d[D]}{dt} = \frac{\dot{V}_{in}}{V_{liq}}([D]_{in} - [D]) - k_1 [D]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$[D]$&lt;/span&gt; is concentration of D, &lt;span class=&quot;math-container&quot;&gt;$\dot{V}_{in} = q_{in}$&lt;/span&gt; is input volumetric flow, &lt;span class=&quot;math-container&quot;&gt;$V_{liq}$&lt;/span&gt; is the reactor volume, and &lt;span class=&quot;math-container&quot;&gt;$k_1$&lt;/span&gt; is the first order rate constant. The reactant is dissolved in water and converted to two gaseous components whose solubility is practically 0. Thus, the formation rate is proportional to the reaction rate of D, &lt;span class=&quot;math-container&quot;&gt;$k_1[D]$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have put this in a small Matlab/Octave (main) script:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;% Define basic parameters&#xA;V_liq   = 2500;     % m^3&#xA;k_1     = 0.320;    % 1/d&#xA;D_in    = 180;      % kg/m^3&#xA;q_in    = 40;       % m^3/d&#xA;Y_M     = 420;      % m^3/t [D]&#xA;Y_N     = 389;      % m^3/t [D]&#xA;%% Part 1: Steady-state calculation&#xA;D_steady    = (D_in*q_in/V_liq)/(q_in/V_liq + k_1);       % kg/m^3&#xA;q_gas_M     = V_liq*Y_M*k_1*D_steady*1e-3;  % m^3/d&#xA;q_gas_N     = V_liq*Y_N*k_1*D_steady*1e-3;  % m^3/d&#xA;&#xA;[D_steady, q_gas_M/24, q_gas_N/24] % Second element should = 120 (m^3/h)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ans =&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;8.5714   120.0000   111.1429&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Now, I am trying to &lt;strong&gt;validate my dynamic simulation&lt;/strong&gt; by passing a constant feed vector. The feed occurs every hour, with the fresh liquid replacing the same amount of reaction liquid, thus maintaining &lt;span class=&quot;math-container&quot;&gt;$V_{liq}$&lt;/span&gt; constant. (Strictly speaking, this is not a CSTR anymore, rather a semi- or quasi-CSTR.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have put the model equations in separate function (using a for-loop for the feed):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function resultsArray = reactorModel1(D_feed, initialValue, parameters, D_in)&#xA;    % Simulate gas production per hour with algebraic reactor model&#xA;    % Feed is solved via a for-loop&#xA;&#xA;    hoursTotal  = length(D_feed);&#xA;    k_1         = parameters(1);&#xA;    Y_M         = parameters(2);&#xA;    Y_N         = parameters(3);&#xA;    V_liq       = parameters(4);&#xA;    resultsArray = zeros(hoursTotal, 3);&#xA;    t           = 1/24;&#xA;&#xA;    liquid_feed = D_feed/(D_in*1e-3);&#xA;&#xA;    initialValue4Model0 = (initialValue*(V_liq - liquid_feed(1))*1e-3 ...&#xA;        + D_feed(1))*1e3/V_liq; % kg/m^3&#xA;    resultsArray(1, 1) = initialValue4Model0*exp(-k_1*t);&#xA;    % Simple for-loop with feed as vector per hour&#xA;    for pHour = 2:hoursTotal&#xA;        initialValue4Model = (resultsArray(pHour-1, 1)*(V_liq - liquid_feed(pHour))*1e-3 ...&#xA;            + D_feed(pHour))*1e3/V_liq; % kg/m^3&#xA;        resultsArray(pHour, 1) = initialValue4Model*exp(-k_1*t);&#xA;    end&#xA;    resultsArray(:, 2) = V_liq*Y_M*k_1*resultsArray(:, 1)*1e-3; % m^3/d&#xA;    resultsArray(:, 3) = V_liq*Y_N*k_1*resultsArray(:, 1)*1e-3; % m^3/d&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Which is called in the second part of my main script, where I (wrongfully?) assume &lt;span class=&quot;math-container&quot;&gt;$[D](t=0) = [D]_{steady}$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%% Part 2: Dynamic calculation with constant feeding&#xA;parameters  = [k_1, Y_M, Y_N, V_liq];&#xA;t_max       = 168; % hours&#xA;D_feed      = q_in*D_in/24*1e-3*ones(t_max, 1);&#xA;initialValue = D_steady;&#xA;resultsArray = reactorModel1(D_feed, initialValue, parameters, D_in);&#xA;[resultsArray(end, 1), resultsArray(end, 2:3)/24]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ans =&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;8.5223   119.3119   110.5056&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you can see, all three components at &lt;span class=&quot;math-container&quot;&gt;$t_{max}$&lt;/span&gt; are &lt;em&gt;lower&lt;/em&gt; than in steady-state. If you take a look at the progression of D (&lt;code&gt;figure; plot(resultsArray(:, 1))&lt;/code&gt;), you see it slightly decreasing over time. However, all parameters (liquid volume, kinetic constant, flowrate of feed, concentration of D in water) are the same between steady-state and dynamic simulation!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: Is there a way to **analytically determine &lt;span class=&quot;math-container&quot;&gt;$[D](t=0)$&lt;/span&gt; ** so that &lt;span class=&quot;math-container&quot;&gt;$[D](t)$&lt;/span&gt; becomes constant? (As stated above in the comment, I need &lt;code&gt;q_gas_M = 120&lt;/code&gt;, since only gaseous flows are continuously measured.)&lt;/p&gt;&#xA;" OwnerUserId="23969" LastEditorUserId="16981" LastEditDate="2019-12-23T13:24:14.343" LastActivityDate="2024-01-25T03:07:05.820" Title="How to calculate initial concentration of first-order reaction in CSTR?" Tags="|chemical-engineering|modeling|simulation|matlab|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32482" PostTypeId="2" ParentId="8525" CreationDate="2019-12-21T21:16:30.577" Score="0" Body="&lt;p&gt;What I would suggest is add a p cut in the flange of diameter slightly larger than the thickness of the horizontal flange. With that you maintain dimensions and manage a tolerance fit without introducing a weakness that may lead to web failure. A square cut could be catastrophic. Once the beam is in place it is only a matter of diagonal positioning to align with the correct final destination without wiring about tolerances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a prevailing precaution against the web tearing, I would consider welding a diagonal strengthening rib above the angular T beam vertical cut or add a flange to the end of the beam where it is full size up to the middle of the web before the p cut position. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-12-21T21:16:30.577" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32483" PostTypeId="2" ParentId="32431" CreationDate="2019-12-22T02:07:13.880" Score="0" Body="&lt;p&gt;Below is the schematics that is discussed in your question. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YbEP0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YbEP0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the resistor for the opamp in the bottom right hand corner can be any type of potentiometer, example it could be a 1K multi turn potentiometer like the one below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MrtVH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MrtVH.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is better representation of the above circuit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zCSqe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zCSqe.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the circuit work for ampere rangers outside of 0-1A you might have resize the resistors on the TLC272 opamp. This would require some work and I will leave that for you do the analysis and present to the community for feedback. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively I would try the following circuit with the LM358 for additional ampere rangers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/v4Pyy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v4Pyy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest taking a looking at &lt;a href=&quot;https://www.youtube.com/watch?v=ywFpNdvSC7A&quot; rel=&quot;nofollow noreferrer&quot;&gt;Adjustable Switching Power Supply Using LM2576 [Buck Converter, CC-CV]&lt;/a&gt; for details on how this circuit works. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.bourns.com/pdfs/3296.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;3296 - 3/8 ˝ Square Trimpot® Trimming Potentiometer&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.smbaker.com/lm2576-constant-voltage-constant-current-switching-power-supply&quot; rel=&quot;nofollow noreferrer&quot;&gt;LM2576 constant voltage / constant current switching power supply&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.pcbway.com/blog/technology/How_to_build_an_adjustable_switching_power_supply_using_LM2576_[Buck_Converter__CC_CV].html&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to build an adjustable switching power supply using LM2576 [Buck Converter, CC-CV]&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/product/TLC272/technicaldocuments&quot; rel=&quot;nofollow noreferrer&quot;&gt;TLC272 Dual Single Supply Operational Amplifier&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lit/ds/symlink/tlc272.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;TLC272, TLC272A, TLC272B, TLC272Y, TLC277 LinCMOS PRECISION DUAL OPERATIONAL AMPLIFIERS&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lit/ug/sbou192/sbou192.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;DIYAMP-SC70-EVM&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lit/ds/symlink/lm2576.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;LM2576xx Series SIMPLE SWITCHER® 3-A Step-Down Voltage Regulator&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.pcbway.com/project/shareproject/How_to_build_an_adjustable_switching_power_supply_using_LM2576.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;How to build an adjustable switching power supply using LM2576?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2019-12-22T12:59:01.973" LastActivityDate="2019-12-22T12:59:01.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32484" PostTypeId="1" AcceptedAnswerId="32522" CreationDate="2019-12-22T06:07:53.660" Score="1" ViewCount="67" Body="&lt;p&gt;These parts were salvaged from a special mechanical calibration rig. I don't see anything that looks like that in McMaster-Carr, but perhaps I don't know where to look. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aluminum part is probably custom but the rod (300-series stainless steel) and the plastic (acetal, I think) threaded part with the brass handle look like standard parts. The aluminum part has an internal thread so the depth of the plastic part can be adjusted by screwing it in or out. The plastic part has a conical lead-in and a bore that fits the stainless steel rod with some play. I believe it acted as a pivot with springs to take out the play. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2RSfz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2RSfz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1276" LastActivityDate="2019-12-24T21:11:12.287" Title="What are these mechanical components called? (assuming they are standard components)" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="32486" PostTypeId="2" ParentId="32472" CreationDate="2019-12-22T11:40:07.687" Score="1" Body="&lt;p&gt;AHh the battery problems of devices and the plug in myths.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So to get past this on the get go, modern electronics using fancy chemistries monitor the voltages of the batteries being charged. Once they are charged to their ideal values, the electronics turn off the charging entirely, or drastically reduce it. (trickling is often not in use either, ofc this depends on battery chemistry) Without these electronics you'll get a thermal runaway and they will burn/blow up as see &lt;a href=&quot;https://www.youtube.com/watch?v=YuKF8XfCVKQ&amp;amp;feature=youtu.be&quot; rel=&quot;nofollow noreferrer&quot;&gt;in the link&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So as long as the internal monitoring of any electronics is working fine, then you really don't have anything to work about. If they for some reason don't, then you can have problems very quickly. I leave my laptop plugged in almost exclusively and it's battery is fine. (it's a macbook, so I couldn't take the battery out anyways). As I am sure lots (most) of people leave things plugged in for 'too long' and I haven't heard of any exploding batteries in the news lately. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless, however for safe practice, it's probably a 'best practice' to remove them when you can. Batteries are on borrowed time regardless, eventually they die and need to be replaced, it seems yours are the case. &lt;/p&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2019-12-22T11:40:07.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32488" PostTypeId="2" ParentId="32479" CreationDate="2019-12-22T14:00:08.487" Score="2" Body="&lt;p&gt;See &lt;a href=&quot;https://www.skf.com/us/products/bearings-units-housings/bearing-accessories/lock-nuts/requiring-keyway/installation-removal/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.skf.com/us/products/bearings-units-housings/bearing-accessories/lock-nuts/requiring-keyway/installation-removal/index.html&lt;/a&gt; for installation instructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you noted, the spacing can be set so that things are always pretty close. There are 19 tabs on the washer and four slots on the nut. The slot in the tread shown in figure 3 is wide enough to allow the washer to rotate slightly to relative to the bearing race, and this means the washer can always have one tab bent over.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A Starlock washer is a different item entirely - the teeth are on the inside and serve to limit axial movement of the washer in one direction, when placed over a circular shaft.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2019-12-22T14:00:08.487" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32489" PostTypeId="1" CreationDate="2019-12-22T14:16:27.303" Score="3" ViewCount="31" Body="&lt;p&gt;I’m curious about the sound made by some types of high powered electric motors and related control devices. Many (but not all) large electrical motor systems make high pitched sounds that correlate to acceleration. How to describe this sound? It is essentially a mid-high tone that rises in pitch as the drive system accelerates, and drops somewhat in deceleration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One example is subway trains and streetcars, specifically modern electronically controlled electric motor unit “EMU” trains. As operators add power to move the trains forward, the tone rises in proportion to motor speed (and therefore overall vehicle speed). &#xA;Older pre-war trolleys seem not to make these high tones, or if they do, they are drowned out by other noises. This seems to be associated with modern high power electronic switching devices, and not necessarily with the motors they drive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other electronic devices that make high tones are VFD motor controls, and going back a few years, camera strobe flash units that made a high rising tone as they recharged.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To generate a sound, some physical matter has to vibrate, and push adjacent matter until landing on an eardrum. I’d really appreciate knowing the precise source of these particular tones (i.e.: going beyond “well, high power electronics make sounds”). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 1: generally which specific types of component vibrate, and what force causes the vibration (e.g.: capacitors, thyristors, magnetism, etc)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question 2: are there associated structural considerations? Does the vibration cause any strain or mechanical fatigue to the device itself or surrounding wiring or mounting surfaces? Other interesting effects?&lt;/p&gt;&#xA;" OwnerUserId="23974" LastEditorUserId="3505" LastEditDate="2019-12-22T19:04:53.510" LastActivityDate="2019-12-22T19:18:35.733" Title="Motor control sounds" Tags="|electrical-engineering|acoustics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32490" PostTypeId="2" ParentId="26987" CreationDate="2019-12-22T14:27:05.333" Score="0" Body="&lt;p&gt;I think I've got what you are asking.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most probably response is. At the end of the radial holes (the big ones) there's another radial hole matching with the diameter of the axial holes (the litle ones).&lt;br&gt;&#xA;And this could be machined either with universal milling machine with a divider or with a cnc machining center (most probably).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Without understanding the whole sistem I can't be sure but the probability like this higher because the sistem should work like on car's piston sistem where the gas and oxigen enter and there's one way valves both in inlet and outlet.&lt;br&gt;&#xA;Looking at the image looks like all holes should be for only one function (inlet or outlet). And on the other side there should be the other function holes (probably in the exact same alignment.&lt;/p&gt;&#xA;" OwnerUserId="23658" LastActivityDate="2019-12-22T14:27:05.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32491" PostTypeId="2" ParentId="28430" CreationDate="2019-12-22T14:54:13.250" Score="1" Body="&lt;p&gt;First of all, I don't know the tecnology so this is my analisys based on what looks like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These cannister looking should be travel limiters like &lt;em&gt;@Ward&lt;/em&gt; pointed out.&#xA;The reason of the wheels and springs misaligned should be to protect the real springs inside of the tank. The sistem probably is like a lever arm and the fine parts are inside of the tank (like spring and shock absorber) to protect from environment damage (tanks need to pass on some very agressive places and need to stop to do maintenance in enemy territory should be a very bad idea).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your first model there's no limiters on some wheels probably because of the other wheel upside. But probably inside the tank the axle of these wheels probably are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;With an internal limiter.&lt;/li&gt;&#xA;&lt;li&gt;With the wheel axis connected with the closest wheel (when the closest wheel bumps in the limiter this wheel bumps too).&lt;/li&gt;&#xA;&lt;li&gt;With the wheel axis connected with the closest wheel to work in reverse (when the other wheel go up, this wheel go down) to avoid loose tracks. &lt;em&gt;(I think the major probability lies here)&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="23658" LastActivityDate="2019-12-22T14:54:13.250" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32493" PostTypeId="1" AcceptedAnswerId="32499" CreationDate="2019-12-22T18:13:12.947" Score="1" ViewCount="50" Body="&lt;p&gt;They say that in the near future high cosmic radiation is possible. In addition, the Earth’s magnetic field is likely to weaken.&#xA;I want to create something like a camera that would protect against electromagnetic radiation. What will be effective for this? Useful articles, real examples? I want to protect equipment, for example, my phones, watches, if at least somehow.&#xA;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="23977" LastActivityDate="2019-12-23T09:59:11.977" Title="Protect equipment against electromagnetic radiation" Tags="|electromagnetism|radiation|" AnswerCount="1" CommentCount="3" ClosedDate="2020-01-12T00:16:02.783" ContentLicense="CC BY-SA 4.0" />
  <row Id="32495" PostTypeId="2" ParentId="32489" CreationDate="2019-12-22T19:18:35.733" Score="3" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 1: generally which specific types of component vibrate, and what force causes the vibration (e.g.: capacitors, thyristors, magnetism, etc)?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Most of the sound is going to come from coils - inductors and motor windings. The current through them will cause attraction or repulsion between the turns of the coil and cause them to flex a little. The iron magnetic circuits will also experience some forces and generate some audible sounds.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 2: are there associated structural considerations? Does the vibration cause any strain or mechanical fatigue to the device itself or surrounding wiring or mounting surfaces? Other interesting effects?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I don't know for sure but I expect it would and motor windings are carefully packed into the windings and a varnish applied to lock everything in place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the change in sound: I suspect that you are describing a change in harmonics due to pulse-width modulation of the motor power. Check the next time you hear it if the fundamental frequency is staying the same but the tone is changing - rather like singing a note and changing the shape of your mouth and oral cavity or a guitar wah-wah pedal. You may also try a spectrum analyser / frequency meter on your phone to study the phenomena.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hk1qg.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hk1qg.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. DART EMU. Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Dublin_Area_Rapid_Transit&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Dublin Area Rapid Transit (Ireland) uses a 1500 V DC supply generated from the 50 Hz mains. To avoid interference the signalling track circuits are 83.3 Hz and the DC chopper on the EMUs was (originally) 427 Hz as far as I remember. This is all to avoid false energisation of the AC track-circuits.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2019-12-22T19:18:35.733" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32497" PostTypeId="1" CreationDate="2019-12-23T00:46:02.603" Score="2" ViewCount="764" Body="&lt;p&gt;I designed a battery charging, protection, and boost circuit for a project I am working on. However, I made a complete noob mistake and forgot to find the overcurrent value for the battery protection IC I am using, and it turns out the overcurrent value is too low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The manufacturer makes multiple protection ICs in the same package size, so I can likely just swap it out with a new IC with a higher overcurrent value. However, overcurrent is specified as overcurrent voltage, and I am not sure how to convert this into an actual current draw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need the protection IC to allow charging current of at least 1A (no more than 2A), and a discharge current of at least ~1.5A (and no more than 2A). The battery is a 2200mAh li-po cell so this should be perfectly within spec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the datasheet for the IC (I am using the AP9101CAK6-ANTRG1 variant):&#xA;&lt;a href=&quot;https://www.diodes.com/assets/Datasheets/AP9101C.pdf?src-supplier=Digi-Key&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.diodes.com/assets/Datasheets/AP9101C.pdf?src-supplier=Digi-Key&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is what I have tried:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I changed the value of R2 (from the example circuit, R5 in my circuits) to 1k from 2.7k . I noticed no difference in the allowed current draw.&lt;/li&gt;&#xA;&lt;li&gt;I shorted P- to Battery - to bypass the MOSFETS, this allowed for higher current draw, which confirmed that it is a problem with the protection IC and not the boost converter.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I also am having trouble with getting the battery to charge, I tried connecting the battery directly to the charge IC and charging worked normally, so again there is a problem with the protection circuit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If it helps at all, I have attached my circuit schematics below&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kUdxh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kUdxh.png&quot; alt=&quot;Charge Controller Schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/88v9q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/88v9q.png&quot; alt=&quot;Protection and Boost Converter Schematic&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help would be greatly appreciated as I have struggled with this for several hours, but I think I lack the background knowledge to figure this out.&lt;/p&gt;&#xA;" OwnerUserId="23978" LastActivityDate="2023-03-16T08:19:02.653" Title="How can I calculate overcurrent from over voltage for the AP9101C IC" Tags="|electrical-engineering|circuits|computer-engineering|circuit-design|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32498" PostTypeId="2" ParentId="26987" CreationDate="2019-12-23T07:48:24.950" Score="1" Body="&lt;p&gt;That part is likely two different parts, one of them pressed inside the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each part could be made in many different ways. How they were produced highly depends on what material they were made from. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;High part number productions will often start with a ‚near net shape’ mold, and will be cast to almost what the final part is in dimension this saves fabrication time and material costs. Even highly complicated engine parts or entire blocks will be cast if they can be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this the parts will go into machining, if this is again, a high production part, they will likely be highly likely fully automated cnc machines. 4 Axis would be enough for that part with only a single part change for final contour. Depending on the part and tolerances, they may be pressed together first then machined or any combination. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After this they may go into whatever post processing that needs to be done, like pressing, hardening, plating, cleaning etc. &lt;/p&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2019-12-23T07:48:24.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32499" PostTypeId="2" ParentId="32493" CreationDate="2019-12-23T09:45:14.783" Score="3" Body="&lt;p&gt;There are 2 effects to look out for:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Magnetic induction into wiring from e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Coronal_mass_ejection&quot; rel=&quot;nofollow noreferrer&quot;&gt;coronal mass ejections&lt;/a&gt;. The electrical grid is already &lt;a href=&quot;https://www.nbcnews.com/mach/space/how-we-ll-safeguard-earth-solar-storm-catastrophe-n760021&quot; rel=&quot;nofollow noreferrer&quot;&gt;largely protected&lt;/a&gt; against these events. To protect equipment in your home, unplug it.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Radiation damage, in the event that Earth's magnetic field collapses. In computers, this manifests first as &lt;a href=&quot;https://en.wikipedia.org/wiki/Single-event_upset&quot; rel=&quot;nofollow noreferrer&quot;&gt;Single Event Upsets&lt;/a&gt;: components in a chip that change their state due to being hit by a charged particle. This can corrupt stored data and disrupt operations. Most of these are temporary (the component still works), but occasionally a component can be destroyed if the charged particle has enough energy and the feature size of the chip is small enough.&lt;/p&gt;&#xA;&lt;p&gt;There are radiation-tolerant designs (built for the military and for spaceflight applications), but this is not something you can retrofit to an existing design. Your best bet is to disconnect the equipment and bury it - a few meters of water or soil &lt;a href=&quot;https://what-if.xkcd.com/29/&quot; rel=&quot;nofollow noreferrer&quot;&gt;should be enough&lt;/a&gt; to protect your equipment.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;There is &lt;a href=&quot;https://en.wikipedia.org/wiki/Earth%27s_magnetic_field#Future&quot; rel=&quot;nofollow noreferrer&quot;&gt;no indication&lt;/a&gt; Earth's magnetic field will collapse. Field strength is decreasing at the moment, but that variation is within the normal range.&lt;/p&gt;&#xA;" OwnerUserId="11645" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2019-12-23T09:59:11.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32500" PostTypeId="1" CreationDate="2019-12-23T11:29:06.783" Score="1" ViewCount="404" Body="&lt;p&gt;From what I read, tape decks auto-reverse by having a sensor keep track of an encoder. When the &lt;a href=&quot;http://home.iitk.ac.in/~amit/courses/371/adarsh/contents.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;take up pulley&lt;/a&gt; stops turning, the logic board reverses the direction of the motor (and does some other stuff to pickup different &lt;a href=&quot;https://en.wikipedia.org/wiki/Cassette_tape#Track_width&quot; rel=&quot;nofollow noreferrer&quot;&gt;tracks&lt;/a&gt; on the tape).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my art project the rotating reels are the only necesseary parts, and the exact speed of the tape doesn't matter. All I want to do, is rotate one reel and then move the input gear to the other reel when the tape has run out on one side. It just needs to rotate forever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assume it would be somewhat similar to &lt;a href=&quot;https://youtu.be/esVq6jfTigM&quot; rel=&quot;nofollow noreferrer&quot;&gt;this mechanism&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y8ayC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y8ayC.png&quot; alt=&quot;Mechanism for converting two-way to one-way rotation&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to use as few parts as possible, and no electronics (except for the small motor that drives the mechanism). Is this at all possible?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not yet seen the perfect solution, but I did find this cassette tape adapter:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wo2d8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wo2d8.jpg&quot; alt=&quot;Cassette tape adapter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From the Wiki page on this type of &lt;a href=&quot;https://en.wikipedia.org/wiki/Cassette_tape_adapter&quot; rel=&quot;nofollow noreferrer&quot;&gt;adapter&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;To simulate tape movement, a system of gears or a drive belt connects the tape player's drive motor (via the take-up spindle) to a wheel inside the adapter. This wheel rides against the detection wheel to simulate tape movement. Because the wheel never stops spinning, the deck never senses an end-of-tape and never tries to reverse the tape.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think my solution would look similar to the inside of that adapter, but then stuck to the back of a normal cassette. It will provide the drive motor, also rotate the spindles that the reels are on. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I'm looking for is a gear mechanism that will 'walk' to the reverse gear when the first gear stops turning (like in the image above). Does that make sense to anyone?&lt;/p&gt;&#xA;" OwnerUserId="23982" LastEditorUserId="23982" LastEditDate="2019-12-27T14:39:00.357" LastActivityDate="2024-03-04T13:11:20.283" Title="Is there a way to auto-reverse cassette tape with gears only?" Tags="|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32501" PostTypeId="2" ParentId="32500" CreationDate="2019-12-23T11:49:02.460" Score="1" Body="&lt;p&gt;You could have the orange and green gears permanently in mesh mounted on a plate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then if you rotate that plate then the orange gear become un-meshed from the blue gear and the green gear comes into mesh - this will reverse the direction of the orange gear. The rotation may be achieved by a small solenoid...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Turning the motor off then on may also be good.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2019-12-23T11:49:02.460" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32503" PostTypeId="1" CreationDate="2019-12-23T15:04:57.153" Score="1" ViewCount="585" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ver8w.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ver8w.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The figure shows input and output variables for both longitudinal and lateral stability transfer functions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I completely understand what each term stands for but I was wondering what would be the best way to physically interpret the relations. I am having trouble visualizing  the relationships.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your reference:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\eta$&lt;/span&gt; = elevator control&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt; = thrust pertubations&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\xi$&lt;/span&gt;= alieron control&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt; = rudder control&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$u,v,w$&lt;/span&gt; = translation velocity in &lt;span class=&quot;math-container&quot;&gt;$x,y,z$&lt;/span&gt; axis respectively.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$p,q,r$&lt;/span&gt; = angular rate in &lt;span class=&quot;math-container&quot;&gt;$x,y,z$&lt;/span&gt; axis respectively.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\phi,\theta, \psi$&lt;/span&gt; = roll, pitch and yaw respectively&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="23988" LastActivityDate="2019-12-23T23:34:52.547" Title="Aircraft Longitudinal and Lateral Transfer Function Terms" Tags="|control-engineering|control-theory|aerospace-engineering|aircraft-design|stability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32504" PostTypeId="1" AcceptedAnswerId="32507" CreationDate="2019-12-23T16:41:38.067" Score="3" ViewCount="145" Body="&lt;p&gt;Being a non-engineering type, I'm trying to calculate what power and torque I would need from a DC stepper motor to execute a particular task.  I'm certain that my question will open me up to potential observations regarding my lack of knowledge in this field. That said, if your critical observations use humor to make the point, I will appreciate them for that. However, I do expect to gain some good knowledge from this group.  Here goes nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am building a device that will be used to lift, and hold 200 pounds approximately 2 feet in elevation.  The mechanical elements of this device are 100% the same as a standard worm-shaft driven, scissor jack for a motor vehicle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Envision a letter &quot;A&quot; as the two arms on the top of the scissor, and the left arm is anchored to the substrate, and the right arm is attached to the threaded block that travels along the worm-shaft. As the worm-shaft is rotated, the threaded block will travel along its axis which will in-turn cause the &quot;A&quot; to get taller and skinnier. This achieves the lift needed for the device.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 1:&lt;/strong&gt;&lt;br&gt;&#xA;In order to achieve the required power and torque to lift and hold 200 lbs without having to manually turn the worm-shaft (as one would when jacking up a car), should I couple the motor to the worm shaft concentrically, or use one gear on the motor shaft, and one on the worm-shaft?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question 2:&lt;/strong&gt;&lt;br&gt;&#xA;Depending on the direct-couple method, or the gear ratio method, how do I determine the power and torque of the motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I really appreciate any help I can get with this challenge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7xyVs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7xyVs.jpg&quot; alt=&quot;Scissor Jack&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23989" LastEditorUserId="8055" LastEditDate="2019-12-24T13:56:16.607" LastActivityDate="2019-12-24T13:56:16.607" Title="Determining the required specs of Stepper Motor" Tags="|torque|stepper-motor|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32507" PostTypeId="2" ParentId="32504" CreationDate="2019-12-23T21:53:26.863" Score="0" Body="&lt;p&gt;The length of each arm of your jack should be comfortably bigger &lt;span class=&quot;math-container&quot;&gt;$12&quot;(inch)&amp;gt; 2ft/2$&lt;/span&gt; because if they are close to 12&quot; they lock at horizontal position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We chose 18&quot; and set the pegs at the lowest level so the arm is 20 degrees when not deployed meaning the jack can expand from 6&quot; to 36&quot; with a travel course of 30&quot; which is more than you need but is more practical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal force at the worm-screw is &lt;span class=&quot;math-container&quot;&gt;$$F=200/2*cot(20)= 100lbs*2.74= 274.7 lbs$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From here on F becomes smaller if we lift the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the diameter of your worm-screw is 1/2&quot; and its pitch is 1/32&quot;, meaning the slope of the thread is approximately &lt;span class=&quot;math-container&quot;&gt;$$ 1/2*\pi*32= 1/50$$&lt;/span&gt;  and the force on it is  &lt;span class=&quot;math-container&quot;&gt;$$f_{screw}=274.7/50= 5.49 lbs \\ \text{and the torque needed is}\\ T=5.49*1/4&quot;= 1.373 lbs.ft $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the case with no friction and 100% efficiency. I would use a load factor of 1.5 and check the manufacturer's recommendation.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-23T22:10:06.460" LastActivityDate="2019-12-23T22:10:06.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32508" PostTypeId="1" CreationDate="2019-12-23T21:56:28.300" Score="2" ViewCount="103" Body="&lt;p&gt;In &lt;a href=&quot;https://engineering.stackexchange.com/questions/32416/buckling-vs-yielding-failure-of-columns&quot;&gt;this question&lt;/a&gt; for example (in the first answer) it is said that the Euler critical load formula is unrealistic, because it assumes perfect geometry and perfect alignment of the load. I've found similar statements from other sources as well. &lt;strong&gt;What is meant by these?&lt;/strong&gt; I've seen &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%27s_critical_load&quot; rel=&quot;nofollow noreferrer&quot;&gt;some derivation&lt;/a&gt; of the Euler buckling formula, but there is no mention of the assumption of &quot;perfect geometry&quot; or that the load needs to be &quot;perfect aligned&quot;. The Euler buckling formula seems to follow quite directly from Euler beam theory. So could somebody clarify a bit what do these assumptions mean? Thank you!&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2019-12-23T22:46:23.420" Title="What is meant by &quot;perfect geometry&quot; in the context of Euler buckling?" Tags="|structural-engineering|buckling|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32509" PostTypeId="2" ParentId="24415" CreationDate="2019-12-23T21:58:35.410" Score="0" Body="&lt;p&gt;Tin electro plate would be easy to do ,so likely cheaper than nickel despite the high price of tin. .&lt;/p&gt;&#xA;&lt;p&gt;Or, avoid plating and use German Silver; brass with about 10 to 20 % Nickle that gives it silver color.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2020-09-19T21:53:37.660" LastActivityDate="2020-09-19T21:53:37.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32510" PostTypeId="2" ParentId="32508" CreationDate="2019-12-23T22:07:07.033" Score="4" Body="&lt;p&gt;It means exactly what it says. The geometric tolerances in the beam are exactly zero. The material is perfectly isotropic and homogeneous. The end load is distributed across the surface of the end of the beam in a way which is perfectly consistent the Euler beam theory and does not produce any local stress concentrations at the end. Etc, etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These &quot;perfect&quot; assumptions are no different to what is made in most simple structural analysis, and often they are not very important. In fact &lt;a href=&quot;https://en.wikipedia.org/wiki/Saint-Venant%27s_principle&quot; rel=&quot;nofollow noreferrer&quot;&gt;St Venant's principle&lt;/a&gt; was originally an empirical statement that &quot;these details usually don't matter&quot;, though it was later made mathematically rigorous.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason these imperfections &lt;em&gt;do&lt;/em&gt; matter in the Euler buckling formula is the basic assumption that &lt;em&gt;before the beam buckles, it acts as a perfectly rigid body.&lt;/em&gt; If it is slightly curved, or the load is slightly offset from the center, that assumption is not true and the basic assumption of the theory is not valid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you apply a gradually increasing load to a slightly curved beam, it will &quot;bow&quot; outwards in one preferred direction. Even if the material is stiff and the length of the beam &lt;em&gt;measured around the curve&lt;/em&gt; remains almost constant, the &lt;em&gt;straight-line distance&lt;/em&gt; between the ends will change as the curvature increases, and therefore the applied load is doing work &lt;em&gt;before&lt;/em&gt; buckling takes place, which is ignored by the Euler theory. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course it is possible to formulate a theory of buckling which includes this type of effect, but it is not very useful in practice because the &lt;em&gt;size&lt;/em&gt; of the effects is unknown in a real world situation. For example, even if a beam was manufactured close to &quot;perfectly straight&quot;, it may be bent and/or twisted into a different (unknown) shape when it is assembled into the rest of the structure.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2019-12-23T22:14:51.363" LastActivityDate="2019-12-23T22:14:51.363" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32511" PostTypeId="2" ParentId="32508" CreationDate="2019-12-23T22:46:23.420" Score="1" Body="&lt;p&gt;In Euler's buckling formula the bifurcation to the left or right buckling is symmetrical and will lead to a stable post-buckling curve in the column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But an eccentric loading or a defect in material, even though can be solved analytically, in many cases the buckling favors one side and the curve may not be stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the diagram of some variety of buckling compared to Euler's.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from the book Exact Solutions for Buckling of Structural Members&#xA;By C.M. Wang, C.Y. Wang.&#xA;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mvR4p.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mvR4p.png&quot; alt=&quot;buckling diagrams.&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-23T22:46:23.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32512" PostTypeId="2" ParentId="32503" CreationDate="2019-12-23T23:34:52.547" Score="1" Body="&lt;p&gt;The object of the game is find an expression for the forces, X,Y,Z; and the moments &lt;em&gt;L,M,N&lt;/em&gt; on the airframe expressed as a function of the state vector  {&lt;em&gt;x,y,z, phi,theta,psi, u,v,w, p,q,r&lt;/em&gt;} and the control vector {aileron,elevator,rudder,flap,thrust}. This is done by first picking a steady trim state and linearizing the dynamic system about this trim state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dotted line in the block suggests that a common separation of variables scheme is being used that simplifies the influence of the controls on the forces and moments. For this scheme to be valid, you need geometric symmetry, aerodynamic symmetry (&lt;em&gt;v,p,r,phi&lt;/em&gt; = 0), and negligible onboard angular momentum (engines). Then you can construct a pair of functions where the rate of change of {&lt;em&gt;u,w,q,theta&lt;/em&gt;} depends only on {&lt;em&gt;u,w,q,theta&lt;/em&gt;} and {thrust,flap,elevator}(called the longitudinal dynamics subset); the rate of change of {&lt;em&gt;v,p,r&lt;/em&gt;} depends only on {&lt;em&gt;v,p,r,phi&lt;/em&gt;} and {aileron,rudder}; {&lt;em&gt;phi&lt;/em&gt;} only on {&lt;em&gt;v,p,r,phi&lt;/em&gt;} (together, the lateral dynamics subset); and the position update (x,y,z,psi) depends on all eight prior terms but not the control vector (the navigation subset).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See &lt;em&gt;Flight Vehicle Aerodynamics&lt;/em&gt; by Mark Drela, chapter 9. I lifted this info from that book. It also lists expressions for all the elements in the matrices. Those expressions (there is a whole raft of them) is where the physical interpretation comes from in terms of plugging in the vehicle characteristics and seeing how things shake out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For instance, the lateral dynamics subset has a total of 24 elements. 7 are zero, one is purely geometrical, and one is a constant, leaving 15 that are dependent on the vehicle properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See also &lt;em&gt;Dynamics of flight - Aircraft Stability and Control&lt;/em&gt; by Etkin &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This PDF from Stanford by Roberto Bunge appears to have the same development, but I haven't been through it all the way. &lt;a href=&quot;http://adl.stanford.edu/sandbox/groups/aa241x/wiki/e054d/attachments/c53d7/Aircraft%20Flight%20Dynamics%202015_04_13.pdf?sessionID=62f441d3fcc6b4014c66ce9aa5d732f561008d30&quot; rel=&quot;nofollow noreferrer&quot;&gt;Stanford Aircraft Flight Dynamics PDF&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2019-12-23T23:34:52.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32513" PostTypeId="1" AcceptedAnswerId="32569" CreationDate="2019-12-24T01:06:21.373" Score="0" ViewCount="168" Body="&lt;p&gt;When analysing and designing a structure for the serviceability limit state the deflections of each beam must be found. When considering a multistory rigid jointed structure (as below) and applying first-order elastic analysis, what method must be applied in order to find the magnitude of the maximum lateral displacements on each beam (e.g. &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; on the deflected shape)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially my question is, how do I find the value of &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; on beam BE from the deflected structure.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in idealising the beam as the simply supported beam with the correct concentrated moments on either side (acting as fixed support).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/8vpIy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8vpIy.png&quot; alt=&quot;Example Of Structure To Find Deflection&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20822" LastEditorUserId="20822" LastEditDate="2019-12-24T05:55:31.863" LastActivityDate="2019-12-28T17:44:13.717" Title="Deflection In Multistory Structure" Tags="|structural-engineering|structural-analysis|deflection|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32514" PostTypeId="2" ParentId="32513" CreationDate="2019-12-24T03:28:19.127" Score="2" Body="&lt;p&gt;In US, The American Society of Civil Engineers (ASCE) Structural Engineering Institute (SEI) publishes Minimum Design Loads for Buildings and Other Structures (ASCE/SEI 7-10).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section 12.8.3 Vertical Distribution of Seismic Forces, details the vertical distribution of seismic forces. It roughly applies the total design base shear of the building in a triangular weighted pattern using the height of the story under consideration as a multiplication factor for the shear of the story divided by total base shear of the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can download the book for free from the State of California, just search for ASCE 7-10 download from CA gov.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-24T08:47:37.567" LastActivityDate="2019-12-24T08:47:37.567" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32516" PostTypeId="2" ParentId="32376" CreationDate="2019-12-24T12:19:01.613" Score="4" Body="&lt;p&gt;The concepts and various technologies have been known and tried and documented over the last 140 years ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To give an overview:&#xA;First, for DC and one-phase AC, you need two conductors (or &quot;connections&quot;) between the fixed installations and the locomotive. For three-phase AC, you need three conductors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The advantages and disadvantages cannot be summed up shortly - there are too many aspects, from purely electrical to maintenance to safety to reliability (weather!) to economic to aesthetic (see &quot;split rail&quot; below) etc.etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are, off my head, nearly one-and-a-half dozen variants:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Siemens' first railway used both tracks for DC supply (one in, one out). One of the disadvantages: If someone gets to stand on both rails, then ...&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Two slotted overhead tubes(!) - first electric streetcar near Vienna (Mödling to Hinterbrühl) in 1883.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A two-rail overhead track, on which one runs a small &quot;car&quot; that is pulled by the locomotive.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Centered single overhead wire with contact from below (most widely used today). Disadvantages are of course problems with wind; the thin wire, which is problematic especially with DC systems (see e.g. Italy's system); the propagation speed of the wire's movement (which must be significantly higher than the train's speed - gets problematic with high speed lines like France's TGV); and the reflection of the wire's movement at fixed points (the whole system can be modelled as a wave propagation system, with &quot;mirrors&quot; at fixed points - leading to complex, and in certain cases dangerous, wave phenomena).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Off-center single overhead wire; typical for mining railways, where it mst be possible to fill cars from above. Also used on a Swiss line, because the center location is used by a main line overhead line.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;3-phase system with two overhead wires, third phase via tracks. Northern Italy had a large &quot;trifase&quot; system for many decades, Matterhorn railway uses it up to now.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;3-phase with 3 wires above each other, current collected sideways - used at experimental German railway that first reached more than 200 km/hour in 1903.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;(Off-center) Third rail at about rail level, with contact on upper side (typical for southern England).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Centered third rail (between tracks) with contact from above.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Third rail at about rail level, with contact on side (e.g. S-Bahn Hamburg).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Third rail at about rail level, with contact from below (new standard for third rails, for safety reasons; e.g. subways in Vienna and Munich).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Third and fourth rail (London Tube, for electro-chemical reasons).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A third rail below one of the running rails, which is split in its middle for lowering a contact shoe. Was used in Vienna before WW I on the innermost streetcar lines, as the emperor (and others) did not want to see overhead lines purely for aesthetic reasons. There is a video on youtube showing streetcars around 1910 with this system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Overhead third rail - used e.g. in Simplon tunnel (for clearance reasons with truck transport); at Salzburg main station (for aesthetic reasons); on Vienna S-Bahn (I think for maintenance reasons).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;By a cable on a drum - works only for quite short distances; typical e.g. for cranes.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Continuous induction - e.g. Bombardier's &quot;Primove&quot; (coils are only energized when vehicle is above).&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Induction at some segments or stops, stored in batteries or (newer) capacitors. Used experimentally in some street cars.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Of course, you can &quot;abstract&quot; these various methods to &quot;something touched higher up; something touched somewhat lower; a cable; or induction&quot;, if you want. But that would be a layman's idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H.M.&lt;/p&gt;&#xA;" OwnerUserId="23997" LastEditorUserId="23997" LastEditDate="2019-12-24T13:24:07.407" LastActivityDate="2019-12-24T13:24:07.407" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32517" PostTypeId="2" ParentId="32458" CreationDate="2019-12-24T13:15:01.250" Score="3" Body="&lt;p&gt;Check out Wien Franz Josefsbahnhof. From the &lt;a href=&quot;https://de.wikipedia.org/wiki/Wien_Franz-Josefs-Bahnhof&quot; rel=&quot;nofollow noreferrer&quot;&gt;German Wikipedia&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Die letzten Gebäude auf dieser Überplattung wurden Anfang der 1990er Jahre fertiggestellt, vor allem das sogenannte Universitätszentrum Althanstraße (UZA) mit dem ehemaligen Standort der Wirtschaftsuniversität Wien (UZA 1) sowie den bestehenden Geowissenschafts-, Pharmazie- und Biologieinstituten der Universität Wien (UZA 2-5).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or, in English:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The latest buildings to be built on that superstructure were built at the beginning of the 1990s, among them the so-called University Centre Althanstraße (UZA) housing departments of Vienna's University of Economics (UZA 1) as well as geoscience, pharmaceutical and biology departments of the University of Vienna (UZA 2-5).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They are quite unhappy with these buildings, but only because they were built shoddily back then. I studied there when I was younger - no problem with the trains below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm sure there are many more places where buildings are erected over railways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: I tried to find more information. It seems that the idea of superstructures - &quot;Überplattung&quot; (as it is called in Austria) or &quot;Überbauung&quot; (general German term) - was en vogue in D/A/CH from the 1970s onwards to maybe the 1990s, but then fell out of favor - I do not know why, but most publications are from before 2000. There are a few remarks in this century about Stuttgart 21 (that that would have been a possibility to keep the older tracks), Regensburg and some places in Austria - but none of these is more than an idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Strange ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;... the railways don't seem to have big reservations; check out e.g. the parking deck in Krems a.d. Donau (Lower Austria) which is built squarely over the station tracks - which do have quite some shunting for the nearby Danube harbor.&lt;/p&gt;&#xA;" OwnerUserId="23997" LastEditorUserId="23997" LastEditDate="2019-12-29T10:36:11.533" LastActivityDate="2019-12-29T10:36:11.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32521" PostTypeId="2" ParentId="26578" CreationDate="2019-12-24T20:38:28.857" Score="1" Body="&lt;p&gt;This question is asking you to describe how buoyancy works using cylinders because their volumes are easy to calculate. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, some things we know:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;to be buoyant means an object has displaced a mass of fluid equal to it’s own mass. &lt;/li&gt;&#xA;&lt;li&gt;density is expressed as mass/volume, so if you know a fluid’s mass and density then you can find its volume by dividing mass by density or &lt;span class=&quot;math-container&quot;&gt;$\rho = m / V$&lt;/span&gt;, therefore &lt;span class=&quot;math-container&quot;&gt;$V = m / \rho$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;you know the mass of fluid being displaced - it’s exactly the mass being added to the inner cylinder &lt;/li&gt;&#xA;&lt;li&gt;the volume of a cylinder is its cross sectional area multiplied by its height, or &lt;span class=&quot;math-container&quot;&gt;$V = A \cdot H$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If you’re with me so far then putting all this together for your question looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;V_{water} &amp;amp;= \dfrac{m}{\rho} \\&#xA;A_{total} &amp;amp;= A_{outer} - A_{inner}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And given #4 above:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_{water} = A_{total} \cdot H_{\delta}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by area to solve for change in height:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$H_{\delta} = \dfrac{V_{water}}{A_{total}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And substituting known values:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$H_{\delta} = \dfrac{\frac{m}{\rho}}{A_{outer} - A_{inner}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Did that explanation help you understand where area fits in?&lt;/p&gt;&#xA;" OwnerUserId="24005" LastEditorUserId="1832" LastEditDate="2019-12-28T15:48:39.453" LastActivityDate="2019-12-28T15:48:39.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32522" PostTypeId="2" ParentId="32484" CreationDate="2019-12-24T21:11:12.287" Score="0" Body="&lt;p&gt;Okay, I found an engineer who was familiar with the way they were used and it's not particularly exciting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The three parts were screwed in to immobilize the motion platform while the cryostat etc. was being loaded and unloaded via a crane, then screwed out to free the motion. There was no deliberate damping other than that created by driving the electromagnetic shakers through a low-impedance source. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Frankly, I like the other ideas better. &lt;/p&gt;&#xA;" OwnerUserId="1276" LastActivityDate="2019-12-24T21:11:12.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32523" PostTypeId="1" CreationDate="2019-12-24T22:31:43.953" Score="1" ViewCount="1437" Body="&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/HcXIS.jpg&quot; alt=&quot;Image of aquarium stand&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I purchased this aquarium and stand secondhand, and a few people I’ve spoken to have been suspicious of the stand’s ability to hold the aquarium when filled with water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tank holds roughly 55 gallons. The legs are 1/2” diameter cold rolled steel. The lower frame is about 8” from the ground. The upper frame is 30” from the ground. Both frames are welded to the legs. The upper and lower frame are angled steel, approximately 1” x 1” x 1/8”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confident that this stand can hold the 600lb filled tank. Can this stand reliably hold 800-1000 lbs? &lt;/p&gt;&#xA;" OwnerUserId="24008" LastEditorUserId="24008" LastEditDate="2019-12-25T00:07:19.683" LastActivityDate="2019-12-25T02:34:01.357" Title="How much weight can this aquarium stand hold?" Tags="|steel|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32524" PostTypeId="2" ParentId="32523" CreationDate="2019-12-25T01:39:51.383" Score="2" Body="&lt;p&gt;It looks like it can hold 1000 lbs.&#xA;Except it seems it can be vulnerable against an inadvertent impact by a person walking or bumping to it with something in their hand. Wich can twist the legs loose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a stability measure they have welded the legs in a gentle slant to the 2 angle frames.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a double check you wand to check the welds. If they  are solid and continued along the 1 inch side of the angles and if when you hold the legs and try to twist them or tap them they don't feel jerky, it should be ok. Hi pitch tone of the legs when banged with a light hammer is a good sign.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately the best test is to load the table with 10 sacks of sand 96lbs and shake it.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-25T01:39:51.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32525" PostTypeId="1" AcceptedAnswerId="32544" CreationDate="2019-12-25T02:09:47.683" Score="1" ViewCount="3637" Body="&lt;p&gt;My friend has proposed that the buildings in the images below have been sinking into the ground over time and that they're not planned with basements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He thinks that the bottom of the building was once been at ground level and that some of them have windows and doors left from the past. I'm skeptical to this, I can't possibly think that a building can sink two floors down in the ground.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He's arguing that so many old buildings with basements have &quot;misplaced&quot; windows too close to ground level. I'm thinking that it's not possible to have windows or doors underground, as that would pose a risk for mold, water damage, and shattered windows, not to mention the fact that a building would sink 3 or more meters into the ground, how would the engineers not be able to design a foundation that is not able to support the weight of the building.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We really need an answer to this, we're not engineers, and can't find any information online. Please leave sources as well if possible, thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an image &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TDCql.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TDCql.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More images can be found at &lt;a href=&quot;https://i.stack.imgur.com/ex9l3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/ex9l3.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: When doing a reverse image search for one of the images, it leads me to some conspiracy theory called mud flood, it's the most stark image of them all where it looks like the building has sunken 8 meters into the ground or more it can't have possibly been built like this. &lt;a href=&quot;https://www.google.com/search?q=mud%20flood&amp;amp;hl=en-SE&amp;amp;sxsrf=ACYBGNRoTlujntkxR-pdhKhcoca-r2hFqg:1577290787017&amp;amp;source=lnms&amp;amp;sa=X&amp;amp;ved=0ahUKEwjT6MSImtHmAhXuAhAIHXkFAa8Q_AUIDSgA&amp;amp;biw=1858&amp;amp;bih=926&amp;amp;dpr=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reverse image search on google&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dzv8L.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dzv8L.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24010" LastEditorUserId="24010" LastEditDate="2019-12-25T16:25:45.847" LastActivityDate="2024-02-13T03:57:20.400" Title="Buildings sinking into the ground" Tags="|structures|building-design|" AnswerCount="4" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32526" PostTypeId="2" ParentId="32523" CreationDate="2019-12-25T02:34:01.357" Score="1" Body="&lt;p&gt;Like the first week quiz in &quot;strength of materials&quot; . 8 legs at 1/2 &quot; diameter = about 1.5 in. square inches. Cold rolled might be 50 ksi , but maybe it is hot rolled , so use a yield about 40 ksi ( much lower than the tensile ) . So I would not put more than 60,000 pounds ( 30 tons) on the stand. BUT , failure will be geometric instability or buckling , so it would be best to limit the load to about 30.000 pounds. On the other hand I have a 30&quot; long similar stand with 0.31&quot; ( 5/16 ) legs and it has held up 700 lb for decades ( deep 40 gal on top and a long 20 gal on the bottom).&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-12-25T02:34:01.357" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32529" PostTypeId="2" ParentId="32525" CreationDate="2019-12-25T07:12:04.367" Score="0" Body="&lt;p&gt;I found this youtube video about the same situation on Moscow Polytechnic excavation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?time_continue=125&amp;amp;v=Vp8JTRb_Vhw&amp;amp;feature=emb_logo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Moscow polytechnic Youtube clip.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Excavations of the building of the Polytechnic Museum in Moscow on Lubyanka Square on September 26, 2017. The bricks go deep underground. History is written by order of the authorities, history is changed by politicians, history is rewritten by scientists. The true story is many times distorted and redone.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Apparently it is done for political, or other indoctrination reasons. Seems like there are many sites like this layin around in Russia.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2019-12-25T07:12:04.367" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32530" PostTypeId="1" CreationDate="2019-12-25T08:31:05.700" Score="-1" ViewCount="221" Body="&lt;p&gt;I very often hear how &quot;stupid&quot; the Roman number system (I, II, III, IIII/IV, V, VI, etc...) was, and I'm often prepared to agree, except for low numbers, where it can look very beautiful for titles and stuff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I'm wondering if the reason I find it &quot;stupid&quot; is just that I grew up with the &quot;Arabic&quot; (0, 1, 2, 3, ...) system, and thus I'm just not used to the Roman one, or if the Roman one truly is inferior.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems strange to me that the clever and wise Romans, who build all those amazing buildings and temples and things, would have come up with such a silly and impractical number system. It doesn't add up to me, somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be that it did work for the kind of math they needed for their, by today's standards, primitive contraptions, but it would have &quot;fallen apart&quot; as soon as they had reached a certain point where more and more intricate machines are built and bigger and more numbers need to be crunched constantly?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if such had happened, I'm sure they would have either changed the number system or realized that it can be used perfectly fine with very large numbers, and in fact is much easier to work with thanks to this or that smart method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I never see Roman numbers in any kind of complex math context. It's 100% Arabic numbers, including for the internals of database software and stuff like that.&lt;/p&gt;&#xA;" OwnerUserId="24012" LastActivityDate="2019-12-25T19:03:18.280" Title="If the Roman number system was so bad, how did the Romans build all that relative high-tech?" Tags="|engineering-history|" AnswerCount="2" CommentCount="5" ClosedDate="2019-12-28T00:44:00.530" ContentLicense="CC BY-SA 4.0" />
  <row Id="32532" PostTypeId="2" ParentId="32530" CreationDate="2019-12-25T11:21:55.547" Score="4" Body="&lt;p&gt;What makes you think the ancient Romans did scientific or engineering analysis or used any form of mathematical analysis? Most of the mathematics currently used by engineers was developed much later - 17th to 18th Centuries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ancient Greeks understood &lt;span class=&quot;math-container&quot;&gt;$a^2 + b^2 = c^2$&lt;/span&gt;, in relation to right angled triangles (&lt;a href=&quot;https://en.wikipedia.org/wiki/Pythagorean_theorem&quot; rel=&quot;nofollow noreferrer&quot;&gt;Pythagoras' Theorem&lt;/a&gt;). They knew about ellipses by using two vertical sticks embedded in a sand pit and drawing and ellipse using a stick constrained by a rope around the two vertical sticks. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Cartesian_coordinate_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Cartesian system of co-ordinates&lt;/a&gt; only became widely known after Rene Descartes published the idea in 1637.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Euclidean_geometry&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euclidean geometry&lt;/a&gt; is about the relationship of shapes that are drawn. The mathematics behind it all came much later. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineering done by the ancients (Babylonians, Greeks, Romans, Egyptians &amp;amp; the ancient Chinese) was the result of trial and error experimentation - try something and it works, do it again; if not do ascertain why it failed and try something else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Development of engineering came about by pushing limits and observing the results and building upon accumulated knowledge.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2019-12-25T11:21:55.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32535" PostTypeId="2" ParentId="32530" CreationDate="2019-12-25T19:03:18.280" Score="1" Body="&lt;p&gt;The Romans (and all other older civilizations) never used numeric mathematics for their engineering. The materials used (mainly stone and wood) are very hard to treat mathematically; and are quite forgiving in reality, so there is no need for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What they did use, was geometry for aesthetic reasons - but no numbers are needed to do geometry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Engineering mathematics essentially started with Hooke's law and formulas built on it for statics; together with logarithms which made multiplication and exponentation much easier to handle. All that happened in the 17th and early 18th century.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H.M.&lt;/p&gt;&#xA;" OwnerUserId="23997" LastActivityDate="2019-12-25T19:03:18.280" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32536" PostTypeId="2" ParentId="32376" CreationDate="2019-12-25T19:05:01.683" Score="2" Body="&lt;p&gt;I didn't see anyone mentioning a keel system.  In this system which was used in the past by DC Transit (DC = District of Columbia, nothing to do with voltage), a keel extends from beneath the car through a slot midway between the tracks.  A shoe on the bottom of the keel contacts the power rail.  This was designed for public safety and the absence of an overhead wire contact.  In some respects it must be similar to no. 13 listed above by H.M. Muller.&lt;/p&gt;&#xA;" OwnerUserId="24021" LastActivityDate="2019-12-25T19:05:01.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32537" PostTypeId="2" ParentId="32438" CreationDate="2019-12-25T19:13:02.463" Score="1" Body="&lt;p&gt;Just buy a step switch and use it as your axle - it has the notches ready-made and costs a few dollars, at most.&lt;/p&gt;&#xA;" OwnerUserId="23997" LastActivityDate="2019-12-25T19:13:02.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32538" PostTypeId="2" ParentId="26631" CreationDate="2019-12-25T20:19:13.603" Score="0" Body="&lt;p&gt;A simple weighted ratchet fulfills your specification. Gravity is the &quot;symmetry breaker&quot; in that case. So what else do you implicitly assume that excludes a ratchet?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And: Your plunger thing does not work, I think: Unless the &quot;connection wires&quot; are elastic, you either cannot shove through the plunger; or it can move back - unless you assume that gravity keeps one of the two pallets in the slots - but then, you would not need the &quot;wires&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;H.M.&lt;/p&gt;&#xA;" OwnerUserId="23997" LastActivityDate="2019-12-25T20:19:13.603" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32540" PostTypeId="1" CreationDate="2019-12-26T04:03:34.740" Score="4" ViewCount="69" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hIM4O.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hIM4O.jpg&quot; alt=&quot;A fan&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is my fan in my office. Sometimes I used it if my air conditioner (AC) has problems, or I my self have problems with my health due to the freon. What I observed is that the heat (or probably temperature) will be reduced if I turn on the fan (I mean, there will be wind blow there) even if the room is remain segregated from outside (no air interchanged with outside).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then my question, what is the explanation for heat reduction due to wind blowing?&lt;/p&gt;&#xA;" OwnerUserId="17943" LastEditorUserId="90" LastEditDate="2019-12-26T10:51:40.323" LastActivityDate="2019-12-26T10:51:40.323" Title="What is the (thermodynamic) explanation for heat/temperature reduction due to wind?" Tags="|thermodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32541" PostTypeId="2" ParentId="20998" CreationDate="2019-12-26T05:19:33.967" Score="0" Body="&lt;p&gt;In the laboratory you might use the triaxial test or oedometer test. In the field you may be able to use the come Penetrometer test, Standard Penetration test or even dilatometer test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technically, from my 1st year uni understanding, soil doesn't have an elasticity modulus without water. Since all clay has water by adsorbtion, it is the water tensile stress being measured. This calls into question whether you need to examine the liquid limit of soil too. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2019-12-26T05:19:33.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32542" PostTypeId="2" ParentId="32540" CreationDate="2019-12-26T05:41:42.303" Score="4" Body="&lt;p&gt;Nope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The temperature will rise. &#xA;The reasons are: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a. You will be burning your food and breath which will tend to increase the temperature. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b. The electric energy not 100% converted to create magnetic flux in the coil. Instead, some of the electric energy is converted into heat by the coil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c. The fan blades are transferring their mechanical energy (pressure and kinetic energy) to the air to move around you. This motion of air will be taken to rest at some distance from the fan. While air is going to rest it loses the kinetic energy, converting it to heat by means of friction. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;By all means, the room temperature only increases!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But why do you feel cold? &#xA;The airflow on your skin carries away the heat from your skin continuously (sweat is evaporated and this phase change draws heat from your skin).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the reason you feel chilliness when air flows around you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you switch off the fan then, you won't feel this chilliness but, you will feel warmness suddenly. &lt;/p&gt;&#xA;" OwnerUserId="13721" LastEditorUserId="90" LastEditDate="2019-12-26T10:47:52.017" LastActivityDate="2019-12-26T10:47:52.017" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32544" PostTypeId="2" ParentId="32525" CreationDate="2019-12-26T12:14:29.030" Score="2" Body="&lt;p&gt;In London many buildings are built with multi-level basements below ground level. This looks like one such building where the surrounding earthworks have been removed prior to commencement of engineering works. I suspect that the vertical reinforcing steel spikes are where load-bearing walls are going to be placed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second image it looks like a French urban development where concrete piles have been driven deep into the ground to prevent lateral substance of adjacent ground. The columns are capped with concrete retaining walls to resist horizontal load pressures. The open construction site appears to be cleared for underground parking for a major building project. There doesn't appear to be any subsidence of any buildings there at all. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="16486" LastEditDate="2020-01-10T23:30:54.313" LastActivityDate="2020-01-10T23:30:54.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32545" PostTypeId="2" ParentId="28325" CreationDate="2019-12-26T13:48:10.727" Score="1" Body="&lt;p&gt;Magnetic field lines are always continuous loops. And divergence of magnetic field would imply the presence of a magnetic monopole, which despite extensive search for their presence, have not been found.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some confusion may arise because permanent magnets are often modelled mathematically using a distribution of magnetic monopoles; because it makes the math easier (scalar vs vector potential), and one can achieve the same fields outside the magnet with such an approach as with the more physically correct current loop distribution. But this is just a mathematical convenience, and at best confusing when trying to answer questions about the field inside the magnet. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;TLDR: In the absence of physical magnetic monopoles, field lines are always continuous loops.&lt;/p&gt;&#xA;" OwnerUserId="14899" LastActivityDate="2019-12-26T13:48:10.727" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32547" PostTypeId="1" CreationDate="2019-12-26T21:45:25.377" Score="13" ViewCount="5435" Body="&lt;p&gt;I am struggling to understand why it is recommended to push start a car in 2nd gear rather than 1st - in terms of the physics involved, I fail to see a reason why this would be the case. I have compiled where I am at currently with a for/against list for each option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasons why 1st gear would be better&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The engine has a greater mechanical advantage, so an engine that is not firing correctly, is sputtering and not producing as much torque as a normally running engine is less likely to be forced into a stall than in 2nd gear.&lt;/li&gt;&#xA;&lt;li&gt;For a push speed of X, the engine will be spinning at a higher RPM than in 2nd gear and I assume a higher RPM gives the engine a better chance of starting?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasons why 2nd gear would be better&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;It would give the people pushing the car a greater mechanical advantage than 1st gear, which may make it easier to keep pushing the car while attempting to start it. However, I'm not sure how important this is, as the aim is surely to get the engine running on its own, so continuing to rely on the force provided by the people pushing the car, even after the clutch has been engaged, already suggests the push start has not been successful?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If the push start is unsuccessful and the car comes to a halt, it should do so over a greater distance/time in 2nd gear than 1st, thereby reducing the risk of injury to the people pushing the car, which could happen if the car stops suddenly.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reasons why gear selection should be irrelevant&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Assuming the retarding effect of each engine stroke is a constant and the push speed (and therefore momentum) is the same in both cases, then surely the number of strokes before the car comes to a halt should be the same in both cases. As mentioned above, I assume the likelihood of a successful start is probably related to RPM. However, if it is not and is simply a fixed chance that the engine will start during each stroke, then the likelihood of the engine starting should be directly related to the number of strokes, thereby suggesting there is no difference between the two options? - While the car may stop sooner in 1st than 2nd, the number of rotations of the engine should be the same. &#xA;Admittedly, higher RPM is likely to increase engine braking and therefore starting in 1st gear may mean greater retardation per stroke, possibly favouring 2nd gear in terms of increasing the number of strokes before the car stops. Having said that, given the engine does a lower RPM for each MPH in 2nd than 1st, then the car has to slow down more in 1st than 2nd to reach the speed where the RPM will be too low for the engine to run without stalling, which could in fact go in favour of suggesting that 1st gear will in fact have more potentially successful strokes than 2nd. Overall, it seems as though these benefits would approximately counteract the negatives for each gear in this case.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In terms of the physics involved, I fail to see an obvious reason why being in 2nd gear will be beneficial. Have I missed anything? Or is this more of an old wives' tale than actually being something that will impact the likelihood of a successful push start?&lt;/p&gt;&#xA;" OwnerUserId="20042" LastEditorUserId="-1" LastEditDate="2019-12-28T00:41:53.127" LastActivityDate="2019-12-29T23:07:44.513" Title="Is there a reason (in terms of the physics involved) why it is better to push start a car in 2nd gear rather than 1st?" Tags="|automotive-engineering|engines|" AnswerCount="8" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32548" PostTypeId="2" ParentId="32547" CreationDate="2019-12-26T22:16:05.483" Score="3" Body="&lt;p&gt;I never heard of pushing a car is 1st; have you tried it ? In the good old days when cars were not so reliable , I started more than a few in 3 rd ( of 3 speed ). Occasionally started one in second by letting out the clutch after it was moving then quickly pushing the clutch back in and hoping it started instead of sliding the wheels. Maybe with a small engine ( 2 L) with low comprression it would be possible. In the low gears the engine turns over many RPM faster the the wheels which is very difficult , if even possible to do. In top gear the engine is only turning over at 3 or 4 times the RPM of the wheels.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-12-26T22:16:05.483" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32549" PostTypeId="2" ParentId="32547" CreationDate="2019-12-26T22:19:00.327" Score="15" Body="&lt;p&gt;It depends on the car.  If it's a big displacement high performance engine, then you may not be able to get the rear wheels to turn unless you're in the highest gear.  If it's got an itty bitty engine, then push-starting in 1st may work best.  While I don't have personal experience, because weight transfers to the front on deceleration a front wheel drive car probably has an advantage in this regard, too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Back when I drove unreliable cars with manual transmissions 2nd or 3rd (in a 1969 Toyota Corolla, with 1100cc of pure power under the hood) seemed to work best, particularly in gravel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Note that the method is to get the car moving when it's out of gear or with the clutch pressed in, then put it in gear and pop the clutch out suddenly.  Even on a little car, in second you can expect to squeak the tires before the engine catches.  Note also that while starting going down a hill may seem attractive, there's nothing worse than being at the &lt;em&gt;bottom&lt;/em&gt; of said hill without yet having started the car.)&lt;/p&gt;&#xA;" OwnerUserId="19776" LastEditorUserId="19776" LastEditDate="2019-12-27T15:33:52.437" LastActivityDate="2019-12-27T15:33:52.437" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="32550" PostTypeId="2" ParentId="32547" CreationDate="2019-12-26T22:30:04.593" Score="0" Body="&lt;p&gt;You have a better gear transmission ratio between meters and rounds per minute.&#xA;In the second gear you need more meters for the same amound of rounds the shaft turns.&#xA;You get therfore more power on you engine.&#xA;It helps with compression of gases and friction.&lt;/p&gt;&#xA;" OwnerUserId="24031" LastActivityDate="2019-12-26T22:30:04.593" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32552" PostTypeId="2" ParentId="32547" CreationDate="2019-12-27T01:53:38.283" Score="7" Body="&lt;p&gt;A higher gear ratio means that less force is needed to turn over the engine by pushing the car. Aside from the issue of the tires slipping, humans are more likely to be able to maintain the speed of the car long enough to allow the engine to start if it is in a higher gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are tow-starting the car, the same thing applies: in a higher gear, there is less &quot;jerk&quot; on the tow rope when you release the clutch, which is less likely to break the rope or cause the towing car to skid or stall. Tow-starting a car at 20 or 30 mph in 3rd gear is easier than trying to do it at 5 or 10 mph in 1st or 2nd. If the towing car stalls and the towed car rear-ends it, this is not a good way to keep your friends!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2019-12-27T01:53:38.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32554" PostTypeId="2" ParentId="3616" CreationDate="2019-12-27T09:48:55.257" Score="0" Body="&lt;p&gt;As the submersible pump is designed to be located inside water, the water will be natural cooler to the pump. If it is not located inside the water, clearly it will be overheat immediately just after started. Unless you modify the pump by adding cooling oil inside the tube/body, it will be just fine. But keep in mind that adding oil inside the body will add more drag. Drag of oil is probably thousand time of drag of air (check viscosity's ratio). But if you only intend to use the motor, then just open the casing and give external fan to cool the rotor, and give the suitable load to substitute the water load (i.e: fan with longer and heavier blades). I myself are very common to just take a motor from one purpose and use it to another purpose.&lt;/p&gt;&#xA;" OwnerUserId="17943" LastActivityDate="2019-12-27T09:48:55.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32556" PostTypeId="2" ParentId="32392" CreationDate="2019-12-27T10:52:46.153" Score="0" Body="&lt;p&gt;No. If no vorticity then, no circulation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mathematically, &#xA;The flux of vorticity is circulation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \Gamma = \int\int \omega~ ds$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So,  &lt;span class=&quot;math-container&quot;&gt;$\Gamma \rightarrow $&lt;/span&gt; 0   when &lt;span class=&quot;math-container&quot;&gt;$\omega \rightarrow $&lt;/span&gt; 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Physically, vorticity is just not a rotation of a fluid element. But it is the rotation of that element about its own axis (spin). (Please know about free/forced vortex for a kind of imagination).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, circulation is non-zero only when there is a flux of vorticity. &lt;/p&gt;&#xA;" OwnerUserId="13721" LastActivityDate="2019-12-27T10:52:46.153" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32558" PostTypeId="1" CreationDate="2019-12-27T16:32:10.377" Score="1" ViewCount="64" Body="&lt;p&gt;Say I have a container completely full of water with a volume of 3000 gallons and I'm filling this container with water at a rate of 8.2 gallons per minute, it's already full at 200 PSI, how do I calculate how much pressure I've gotten up to at 10 sec, 20 sec, etc?&lt;/p&gt;&#xA;" OwnerUserId="24047" LastEditorUserId="24047" LastEditDate="2019-12-27T18:22:34.383" LastActivityDate="2019-12-27T22:36:24.220" Title="How do I Calculate Pressure at Multiple Points in Time while Filling a Container with Water?" Tags="|fluid-mechanics|pressure|pumps|hydraulics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32559" PostTypeId="2" ParentId="32547" CreationDate="2019-12-27T18:57:24.420" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;For a push speed of X, the engine will be spinning at a higher RPM than in 2nd gear and I assume a higher RPM gives the engine a better chance of starting?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It could be possible that in second gear, since the push speed of X gives us lesser rpm of the engine providing less of a jerk when starting, causing lesser mechanical stress on the power train.&lt;/p&gt;&#xA;" OwnerUserId="24049" LastEditorUserId="1832" LastEditDate="2019-12-28T00:47:44.810" LastActivityDate="2019-12-28T00:47:44.810" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32560" PostTypeId="2" ParentId="32558" CreationDate="2019-12-27T21:03:14.727" Score="1" Body="&lt;p&gt;we need to assume some parameters to start building a model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let's say your container is a steel sphere with thickness, t =  0.25&quot; to make things a bit easier with a volume of 3000 gallons. So converting to cubic ft, we get &#xA;&lt;span class=&quot;math-container&quot;&gt;$$V= 3000*0.133=401ft^3 \\ R=\sqrt[3]{3/4*V/\pi} =4.574'$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we calculate the stress of this steel container for 200psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \sigma = \text{200*area of sphere/ t*circumfrence  }=\frac{200*(4.54*12)^2*\pi}{ t*2\pi 4.574*12 }\\=100*4.574*12*4=21955psi $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every 10 seconds you are increasing the volume of the container (water is incompressible) by a ratio of &lt;span class=&quot;math-container&quot;&gt;$(8.2/6)/3000 =0.00045  $&lt;/span&gt; this will increase the radius of the container by the &lt;span class=&quot;math-container&quot;&gt;$R_{expander}= R_{Initial}*(1+\sqrt[3]{0.00045}) \ $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And stress will increase on the container by a ratio of &lt;span class=&quot;math-container&quot;&gt;$\ R_{expanded}^2/R_{expanded}^3.$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can calculate the new stress using &lt;span class=&quot;math-container&quot;&gt;$ E*\epsilon=\sigma .  $&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then plug in the new stress in the formula and calculate the pressure. I suspect the container will explode soon, considering steel's huge Young modules.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically your container will act like a balloon we fill with water. I&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-27T22:36:24.220" LastActivityDate="2019-12-27T22:36:24.220" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32562" PostTypeId="2" ParentId="31957" CreationDate="2019-12-27T23:23:11.437" Score="2" Body="&lt;p&gt;In an atmosphere, the top speed is probably limited by air resistance more than anything, but in a vacuum it would be limited by other factors. Rolling resistance is a major effect that would definitely limit speed. As discussed in the comments, engine rpm would also be a limit if using stock gear ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to a quick google search, the top speed of the Mini is 81 mph stock- actually quite impressive for a British economy car from the 70s. Whether the limiting factor proves to be rolling resistance or engine redline, it won't increase much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would expect less than 100mph, probably like 95 before the decreased power at high rpms, rolling resistance, and internal friction(wheel bearings, CV joints, transmission) conspire to prevent you from reaching higher speeds. This, of course, assumes that your engine hasn't melted down yet from lack of cooling, but I trust Mr Bean to have installed a sufficient system(ice tank in the cabin?)&lt;/p&gt;&#xA;" OwnerUserId="24052" LastActivityDate="2019-12-27T23:23:11.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32563" PostTypeId="1" AcceptedAnswerId="32581" CreationDate="2019-12-27T23:47:05.573" Score="1" ViewCount="56" Body="&lt;p&gt;I have the coordinates of 3 points thrgouh which, a circle should pass . Having the coordinates of the points in 3D, how could I have the coordinates of the center of circumscribed circle ?&#xA;also : if one of the points has some deviations and causes a circumscribed circle couldn't pass through the 3 points, is there a way to determine the required coordinates of the third point in a way that the circle could be constructed to find the deviation in space ?&lt;/p&gt;&#xA;" OwnerUserId="17070" LastActivityDate="2019-12-30T08:17:33.133" Title="How could I get the coordinates of the circumscribed circle giving the 3 points coords?" Tags="|applied-mechanics|mechanisms|geometry|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32564" PostTypeId="2" ParentId="32547" CreationDate="2019-12-28T08:55:39.507" Score="0" Body="&lt;p&gt;First a disclaimer: Bump starting a modern car is typically risky, as I have mentioned in comment above. Do it at your risk.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe the optimal gear depends on circumstances. For the engine, just RPM matters. From RPM PoV, you can push the car slowly at 1st gear or more quickly at 2nd gear. If the RPM is the same, it does not matter for the engine. It can however matter for the source of pushing:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;If you push the car by a human, I would probably go for 1st gear,&#xA;as people cannot run much quickly. Also, it lowers the risk of&#xA;sudden change of car speed.&lt;/li&gt;&#xA;&lt;li&gt;If you push the car by another car, I would probably go for a&#xA;similar gear as the pushing car have.&lt;/li&gt;&#xA;&lt;li&gt;If you push the car downhill, it probably depends on the&#xA;declination.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From safety point-of-view, it is likely safer to bump start the car at lower speed, as you can be distracted by watching your RPM. You are less likely you will reach a curve/crossroad on the road. Also, if you are towed by a car that stalls, you are more likely to be able to stop soon enough.&lt;/p&gt;&#xA;" OwnerUserId="24053" LastEditorUserId="90" LastEditDate="2019-12-28T09:56:33.047" LastActivityDate="2019-12-28T09:56:33.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32566" PostTypeId="1" CreationDate="2019-12-28T11:43:36.890" Score="1" ViewCount="17" Body="&lt;p&gt;Not sure if there is any software that would model the movements of physical parts, or if even they can move with the given configurations ( software engineer background, no idea the lingo used here ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, at the start  &lt;a href=&quot;https://www.youtube.com/watch?v=EdCX05flrWM&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; it shows modelling of gears simulated, do CAD programs have this capability or this is something else? are there programs that are able to determine if parts were made (for example) from plastic, then they would break, or if they were made of steel then they would require too much torque that the stepper motor wouldn't be able to handle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, what are the current limitations of modelling and simulation software for mechanical systems?&lt;/p&gt;&#xA;" OwnerUserId="24058" LastActivityDate="2019-12-28T11:43:36.890" Title="Visualization/Verification software for interacting parts e.g. Gears etc" Tags="|modeling|simulation|terminology|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32567" PostTypeId="1" CreationDate="2019-12-28T12:32:58.977" Score="3" ViewCount="47" Body="&lt;p&gt;I purchased a step down transformer from a junk yard. I tested the transformer and it works but I am confused as to the secondary winding.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The primary side has these markings&#xA;G 0V 100V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The secondary has these markings&#xA;37V 18V 0V 6V 0V&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the secondary the markings&#xA;37V 18V 0V are all connected. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 6V and 0V on the right hand side are connected as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I connect the transformer and try to take readings from the secondary, I get mostly 6-8V between any two points on the secondary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Visual&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;      Primary&#xA;    G 0V 110V&#xA;------------------&#xA;------------------&#xA;     Secondary&#xA; 37V 18V 0V [space] 6V 0V&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Anyone have any idea what's going on with this transformer?&lt;/p&gt;&#xA;" OwnerUserId="23946" LastActivityDate="2019-12-28T12:32:58.977" Title="Understanding transformer secondary winding" Tags="|electrical-engineering|power|ac|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32568" PostTypeId="2" ParentId="31957" CreationDate="2019-12-28T15:46:14.110" Score="1" Body="&lt;p&gt;&lt;strong&gt;TLDR:&lt;/strong&gt; Top speed increases from 81 to 89 mph. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air drag force is 0.5 x air density x drag area * speed^2. &#xA; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Automobile_drag_coefficient&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Automobile_drag_coefficient&lt;/a&gt;).  Therefore drag power is 0.5 x air density x drag area * speed^3.  The cube term on speed is very important, i.e. doubling speed requires 8 times more power to overcome air resistance.  At highway speeds, air resistance becomes dominant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to &lt;a href=&quot;https://www.automobile-catalog.com/auta_details1.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.automobile-catalog.com/auta_details1.php&lt;/a&gt;, the 1977 Mini has an &lt;em&gt;Estimated&lt;/em&gt; drag area of 0.788 m^2 &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Doing the math, this means that at the top speed of 81 mph, overcoming air drag requires about 31 HP. According to same website, the 1977 Mini has 39 net HP.  Therefore, at top speed, 79% of the engine's power is being used to overcome air resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in theory, removing air resistance would give a huge amount of extra available power.  However, based on the same site again , max horsepower occurs at 4750 RPM, whereas the engine redline is 5200 rpm.  Therefore, only a modest increase of 9% is possible.  Top speed would be 89 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To achieve a higher max speed, you would need to redesign the gearbox to take advantage of the reduced air drag. I.e. give a higher max velocity for the same engine speed by using a different gear ratio.  &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2019-12-28T15:46:14.110" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32569" PostTypeId="2" ParentId="32513" CreationDate="2019-12-28T17:44:13.717" Score="1" Body="&lt;p&gt;From a straightforward, academic perspective (by which I mean disconsidering seismic and other code-enforced conditions), your approach is either exactly or roughly correct (depending on your definition, as described below).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, you must first calculate the entire structure and determine all the internal forces. As you've stated, that will include both the loads directly applied to BE but also any nodal loads transmitted to BE from the columns. This will give allow you to make a perfectly accurate representation of the internal forces on the beam and therefore of the deflections it faces &quot;in isolation&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, what your analysis is currently missing is that the beam actually &lt;strong&gt;&lt;em&gt;isn't&lt;/em&gt;&lt;/strong&gt; in isolation. It belongs to a larger structure which is also deforming, and this deformation also applies what can be effectively considered an imposed displacement to the nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this particular case, both nodes of BE will be moved to the right, while the left-most node will also move up and the right-most node, down. This means that BE's &quot;straight line representation&quot; in the deformed configuration is actually slanted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And so it becomes a matter of definitions and possibly codes. What is the deflection to be considered when checking the servicibility limit state?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm Brazilian and therefore follow code NBR 6118 for concrete structures and NBR 8800 for steel structures. Both of these&lt;sup&gt;1&lt;/sup&gt; are unclear as to which deflection to use, stating only that elements should be &quot;analysed as isolated, simply-supported beams under loading&quot;&lt;sup&gt;2&lt;/sup&gt;. I interpret this as allowing your method of disconsidering imposed deflections, but it's not air-tight (imposed deflection could be considered a load, but they do use the word &quot;isolated&quot;...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another interpretation is to consider the relative displacement of the nodes: both nodes drop by some amount, but that cancels out, so the only thing that might increase users' unease (the basis for SLS) is the additional deflection of one of the nodes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or you might want to consider the total deflection of both nodes: after all, if your columns are very &quot;compressible&quot;, your users might become uncomfortable if high variable loading causes the ceiling to drop by half a meter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That depends on your relevant code (and possibly how you interpret it). If you need to consider these imposed deflections, then you're basically forced to use the stiffness matrix method: you can certainly do it by hand, but it'll just give you a headache.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See below the model I used to look at this problem, and the three ways you can interpret the deflection of BE:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The top isolated beam is your method, ignoring the imposed displacements&lt;/li&gt;&#xA;&lt;li&gt;The middle beam adopts the relative displacement of the nodes, concentrated on the right node&lt;/li&gt;&#xA;&lt;li&gt;The bottom beam adopts the full displacement of both nodes and is equivalent to the actual result obtained from the frame (the slight discrepancy is a rounding error due to my inputs).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hAW00.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hAW00.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; and therefore the Eurocode as well, since they're the inspiration for the Brazilian codes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;2&lt;/sup&gt; That's my own literal translation from the Portuguese.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;Diagrams obtained with &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, a free 2D frame analysis program.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2019-12-28T17:44:13.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32571" PostTypeId="2" ParentId="32547" CreationDate="2019-12-28T18:55:57.677" Score="3" Body="&lt;p&gt;Having pushed many cars on many occasions, on snow as well as gravel and tarmacadam. I can categorically state;  on gravel 3rd best nothing lower than 2nd, on snow 4th you can try 3rd and you may be lucky, that's if you have chains or studs on,  on Tarmacadam 2nd gear, no lower. Downhill you could use 1st but I would still recommend 2nd. The reason is traction, work it out for yourselves. All I can say is I've done it many many times and glad I took notice of the old chaps, otherwise I’d still be pushing now. You can start it in reverse on a slight down hill, but always remember it is always down to traction, whether or not you have any.&lt;/p&gt;&#xA;" OwnerUserId="24062" LastEditorUserId="3672" LastEditDate="2019-12-29T03:14:25.667" LastActivityDate="2019-12-29T03:14:25.667" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32574" PostTypeId="2" ParentId="32547" CreationDate="2019-12-29T23:07:44.513" Score="0" Body="&lt;p&gt;First let's review:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In 1st, you get small wheel revs per engine rev&lt;/li&gt;&#xA;&lt;li&gt;In 4th, you get large wheel revs per engine rev&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Invert that:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In 1st, you get large engine revs/wheel rev =&gt; large engine braking force&lt;/li&gt;&#xA;&lt;li&gt;In 4th, you get small engine revs/wheel rev =&gt; small engine braking force&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;All else being equal, we want the most engine revs we can get, which also translates in to faster engine rpm which will aid starting.  However, we have a limit:  our tires can only handle so much engine braking force before they will skid instead of roll on the pavement.  For most cars (generalizing here!) first gear will make the tires skid, while 2nd gear won't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As mentioned by @Barrie, you can vary this general rule taking into account the size of then engine (thus engine braking force) and the slipperiness of the ground (pavement/dirt/sand/gravel/snow).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, nearly all cars (&amp;amp; motorcycles) are now fuel-injected, so if you have no juice for the fuel-pump you can't start the car no matter how well you try to bump-start it!&lt;/p&gt;&#xA;" OwnerUserId="24074" LastActivityDate="2019-12-29T23:07:44.513" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32575" PostTypeId="1" CreationDate="2019-12-29T23:31:23.520" Score="-1" ViewCount="38" Body="&lt;p&gt;I am not looking for a specific number of course. I know that it may perhaps vary with different vehicles. However, I'm trying to figure out how to theoretically calculate the minimum necessary frictional force. In other words, what variables determine how much kinetic frictional force is required to maintain traction? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not asking for how to find frictional force, but how to know whether a vehicle can or cannot operate on certain roads with varying coefficients of kinetic friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess a main &quot;sub-question&quot; would be is min. friction requirement independent or dependent of mass? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, do smaller vehicles which lesser mass have smaller min. friction requirements than bigger vehicles? Is it really determined by mass or rather the normal force, which can change in magnitude in an incline? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example, can a truck travel with sufficient traction with the same min. frictional force required for a bicycle?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there an equation relating mass or normal force to determine the min. frictional force required? &lt;/p&gt;&#xA;" OwnerUserId="20849" LastEditorUserId="20849" LastEditDate="2019-12-29T23:56:57.367" LastActivityDate="2019-12-30T03:55:36.407" Title="What is the minimum frictional force required for a vehicle to operate smoothly, retain traction, and prevent any skids?" Tags="|control-engineering|automotive-engineering|friction|traction|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32576" PostTypeId="1" CreationDate="2019-12-30T03:52:53.280" Score="-1" ViewCount="113" Body="&lt;p&gt;I am a student and is making power project using Push-Pull transformer. I used the Reference design: &lt;a href=&quot;http://www.ti.com/tool/TIDA-01352&quot; rel=&quot;nofollow noreferrer&quot;&gt;TI Designs: TIDA-01352&lt;/a&gt;. I stuck to find 2 replaceable Push-Pull transformers as in suggesting transformers(750340744 &amp;amp; 750343547 of Wurth Electronik). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are two datasheets of the transformer I want to replace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;750343547: &lt;a href=&quot;https://www.we-online.de/katalog/datasheet/750343547.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.we-online.de/katalog/datasheet/750343547.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;750340744: &lt;a href=&quot;https://www.we-online.com/catalog/datasheet/750340744.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.we-online.com/catalog/datasheet/750340744.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried to find on Mouser, Digikey, etc. Could you suggest me to find there transformers with above similar specifications?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P/S: I sent to Wruk Electronik to ask, but they want more information about name project, company name, my job, the price I want, etc. It made me confused. That is reseason why I want to replace them with similar specifications  &lt;/p&gt;&#xA;" OwnerUserId="23780" LastActivityDate="2021-05-09T04:18:03.760" Title="Finding replaceable push pull transformer with similar specifications" Tags="|electrical-engineering|power-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32577" PostTypeId="2" ParentId="32575" CreationDate="2019-12-30T03:55:36.407" Score="1" Body="&lt;p&gt;basically the acceptable coefficient of friction friction requirement is not dependant on mass because the heavier the vehicle is the harder in presses on the road, and so the more it grips.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what all types and sizes of road vehicles use rubber tyres.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;how much actual friction is required depend on what is being done, an accelerating vehicle needs to provide the acceleration through the force the tyres put into the road, while one traveling in a straght line on a flat sufrace needs to only compensate for drag.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Can a truck travel with sufficient traction with the same min. frictional force required for a bicycle?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;only if the car has the same mass as the bicycle, or possibly if the car performs much less extreme manouvers.  insufficient friction is what allows skids.&lt;/p&gt;&#xA;" OwnerUserId="24075" LastActivityDate="2019-12-30T03:55:36.407" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32578" PostTypeId="2" ParentId="32576" CreationDate="2019-12-30T04:05:47.800" Score="0" Body="&lt;p&gt;On digikey you'll find that sort of thing under &quot;pulse transformers&quot;,&#xA;on element14 under &quot;coupled inductors&quot;&#xA;on mouser under &quot;coupled inductors&quot; again.&lt;/p&gt;&#xA;" OwnerUserId="24075" LastActivityDate="2019-12-30T04:05:47.800" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32579" PostTypeId="1" AcceptedAnswerId="32592" CreationDate="2019-12-30T06:01:00.477" Score="-1" ViewCount="47" Body="&lt;p&gt;Around what percent of energy provided by gasoline is lost as heat due to rolling friction in vehicles? Is this number higher for larger vehicles likes trucks than smaller vehicles like bikes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately, I want to know if energy loss due to ground friction is a problem and whether an attempt to decrease rolling friction can help to resolve the inefficiency? &lt;/p&gt;&#xA;" OwnerUserId="20849" LastActivityDate="2020-01-01T06:53:11.360" Title="What is the average energy loss due to rolling friction in contemporary motor vehicles?" Tags="|automotive-engineering|heat-transfer|friction|traction|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32581" PostTypeId="2" ParentId="32563" CreationDate="2019-12-30T08:17:33.133" Score="2" Body="&lt;p&gt;Solve the following four equations; here, A, B and C are the 3D-vectors of your three points. M is the center of the circle (what you are looking for), r is its unknown radius. x means &quot;vector product&quot;.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[1] (A-M)² = r²&#xA;[2] (B-M)² = r²&#xA;[3] (C-M)² = r²&#xA;[4] ((C-A)x(C-B))(C-M) = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The first three equations say that the distance from the each point to the center must be r. These three conditions will, in 3D, have as solution a straight line of the centers of all spheres that go through points A,B,C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To find the center of a &lt;em&gt;circle&lt;/em&gt;, one must force the solution to the plane where A,B,C lie. There are many ways to accomplish this; the following one seems the easiest one: First, one creates the plane's normal vector: Two vectors in the plane are C-A and C-B, and their cross product is a normal vector. To force C-M (and hence M) also into the plane, the scalar product with normal vector must be 0 (so that they are at right angles, i.e., the normal vector of the plane is orthogonal to C-M).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A=(1,1,1), B=(2,3,1), C=(3,3,2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;M=(x,y,z)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are the four equations:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[1] (1-x)² + (1-y)² + (1-z)² = r²&#xA;[2] (2-x)² + (3-y)² + (1-z)² = r²&#xA;[3] (3-x)² + (3-y)² + (2-z)² = r²&#xA;&#xA;(2,2,1) x (1,0,1) = (2,-1,-2), so&#xA;[4] 2(3-x) - (3-y) - 2(2-z) = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;With all parentheses expanded:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;[1] 1-2x+x² +  1-2y+y² + 1-2z+z² = r²&#xA;[2] 4-4x+x² +  9-6y+y² + 1-2z+z² = r²&#xA;[3] 9-6x+x² +  9-6y+y² + 4-4z+z² = r²&#xA;[4] 6-2x    -  3+y     -4+2z     = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;To get rid of all the squares in [1]...[3], subtract subsequent equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[1]-[2] is a linear equation in x,y,z&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-3+2x -8+4y       = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;[1]-[3] is also a linear equation in x,y,z&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;-8+4x -8+4y -3+2z = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;[4] is anyway a linear equation in x,y,z&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;6-2x  - 3+y -4+2z = 0&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;This is now a simple system of three linear equations with three variables, which is easy to solve. The solution yields the coordinates of M:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;x = 13/6, y = 5/3, z = 11/6&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;r can be computed from any of [1]...[3]:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;r = square root of ((1-13/6)² + (1-5/3)² + (1-11/6)²) = square root of 5/2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And as joojaa pointed out above, you can always find a circle through 3 points (unless they are on a straight line, which will make the equations above unsolvable).&lt;/p&gt;&#xA;" OwnerUserId="23997" LastActivityDate="2019-12-30T08:17:33.133" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32582" PostTypeId="2" ParentId="29023" CreationDate="2019-12-30T17:03:14.817" Score="1" Body="&lt;p&gt;To accurately measure the time of a system is no causal task. You cannot taken samples and extrapolate since you will be ignoring far too many variables such as&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;material thickness&lt;/li&gt;&#xA;&lt;li&gt;linear acceleration/deceleration&lt;/li&gt;&#xA;&lt;li&gt;Jerk&lt;/li&gt;&#xA;&lt;li&gt;Material heat absorption&lt;/li&gt;&#xA;&lt;li&gt;Curve accel/decel&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The best way to do this would be to read generated gcode and calculate your time with a script that will read the NC or GCODE that was generated and as pointed out in comments, getting machine specific settings, this benefits the fact that the operator who knows the material best will knowingly or subconsciously choose the correct parameters and gcode settings (which will adjust for workpiece specific properties and you will read the machine time ahead of cutting. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The machine will follow this code as accurately as it can and your script will calculate the time based on that with minimal error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After creating a code that is accurate to a minute or seconds, can you then optimize visa experimenting and interpolation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Have a look at the &lt;a href=&quot;https://github.com/prusa3d/PrusaSlicer&quot; rel=&quot;nofollow noreferrer&quot;&gt;prusa slic3r github&lt;/a&gt; for a program that does the same for 3d printing. Which as of 30.12.19, can quite accurately predict printing time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your system is more ‚manual‘ and/or half automated, you can create a electronic (a voltsge divider for example) that gives you a digital high/low every time the laser is on/off, and you simply write a timer that runs. After taking hundreds of samples one could write script that predicts cutting time , but it won‘t be particularly accurate as soon as you start producing something against the ‚norm‘.&lt;/p&gt;&#xA;" OwnerUserId="14318" LastEditorUserId="14318" LastEditDate="2019-12-30T18:44:19.930" LastActivityDate="2019-12-30T18:44:19.930" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32583" PostTypeId="1" AcceptedAnswerId="32594" CreationDate="2019-12-30T21:47:00.253" Score="0" ViewCount="133" Body="&lt;p&gt;I've been reading the &lt;a href=&quot;https://www.comsol.com/blogs/what-is-geometric-nonlinearity/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comsol post on geometric nonlinearity&lt;/a&gt;. The article introduces situations where the small angle approximation and linear analysis of structures gives inaccurate results. For example, the case of a beam attached from both ends is discussed, specifically when the deformation is so large that the horizontal supporting force must be taken into account and the stiffness of the beam is no longer constant. Buckling also comes up. Here is a quote:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Buckling, or the loss of stability when the load reaches a certain critical value, is caused by geometrically nonlinear effects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why is buckling caused by geometrically nonlinear effects?&lt;/strong&gt; What is meant by this statement?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2020-01-01T19:45:54.017" Title="Why is buckling caused by nonlinear geometry?" Tags="|structural-engineering|buckling|" AnswerCount="1" CommentCount="2" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32584" PostTypeId="1" CreationDate="2019-12-30T22:05:18.760" Score="1" ViewCount="35" Body="&lt;p&gt;I am attempting to model the acoustic behaviors of a circular membrane (drum head). I know that software like ANSYS will give me the modes of vibration pretty easy from a simulation. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what if I also want the Q (quality factors) ie. decay properties of the various modes? Is there any way to determine that? Can ANSYS do this? If so, how? If not, which software can provide it and any basic starter tips on how to get it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please excuse me if this is a very basic question. I am not an engineer of course but I am doing my best to figure out things as I go for my goal. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="16468" LastActivityDate="2020-01-01T12:55:19.377" Title="Estimating the quality factor (decay properties) of an object's vibrational modes?" Tags="|mechanical-engineering|vibration|acoustics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32585" PostTypeId="2" ParentId="32584" CreationDate="2019-12-31T03:56:06.947" Score="1" Body="&lt;p&gt;There are at least 3 factors that influence the damping&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Internal material damping within the membrane. For a single ply drum head, this is probably small, but if you are using a multi-ply head (e.g. the batter head on a typical snare drum), it will be larger and could get pretty complicated to model&lt;/li&gt;&#xA;&lt;li&gt;Friction at the interface of the drum head to the drum. Friction is very hard to model in general, and its hard to predict ahead of time.&lt;/li&gt;&#xA;&lt;li&gt;Energy radiated to the surrounding air as sound.  From the point of view of the membrane, the energy carried away by the sound decreases the energy of the membrane and thus its damping energy.  This is an acoustics / fluid structure interaction problem. It should be possible to model this and calculate the values. I used to do similar calculations for beams and the theory should be similar for membranes.  However, it might involve fairly advanced math.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Which of these three is dominant probably depends on what modes you are talking about.  For some modes and some drums, you might be able to neglect one of these mechanisms. However, they can all get pretty complicated. If you are not an engineer, its probably easiest to just run an experiment and measure the decay rates experimentally.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastEditorUserId="9275" LastEditDate="2020-01-01T12:55:19.377" LastActivityDate="2020-01-01T12:55:19.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32586" PostTypeId="1" AcceptedAnswerId="38314" CreationDate="2019-12-31T03:59:09.550" Score="0" ViewCount="329" Body="&lt;p&gt;I have tried to compare two courses, and there seems to be some overlap, but not as much. Which method is better when controlling a nonlinear system? Also, what is the main difference between the two?&lt;/p&gt;&#xA;" OwnerUserId="24083" LastEditorUserId="6790" LastEditDate="2020-07-31T20:17:45.213" LastActivityDate="2020-10-22T15:07:07.580" Title="Which course to take? Optimal control? Nonlinear control?" Tags="|control-engineering|control-theory|optimal-control|nonlinear-control|" AnswerCount="2" CommentCount="2" ClosedDate="2020-10-22T17:43:02.080" ContentLicense="CC BY-SA 4.0" />
  <row Id="32587" PostTypeId="2" ParentId="32586" CreationDate="2019-12-31T10:51:55.137" Score="1" Body="&lt;p&gt;In control systems, the main focus is the design of a controller for machines or robots, here we mainly deal with linear systems application of linear control theory.&#xA;While non-linear systems is an advanced topic, where we deal with advance and mathematically more complex systems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If its ur first course in control systems or automatic control. Go for the linear control system, but if you have a basic background of controls then you can go for non-linear control.&lt;/p&gt;&#xA;" OwnerUserId="18714" LastActivityDate="2019-12-31T10:51:55.137" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32588" PostTypeId="1" CreationDate="2019-12-31T14:58:28.373" Score="-1" ViewCount="27" Body="&lt;p&gt;I want to calculate the forces applied on a bar holding the wings, and there is a load connected to the middle of the bar. My question is, how are the forces act on the bar, supposed the wings are rectangular (I'm not sure if it matters). should I treat it as the lifting force is spreaded along the bar and the load pulling down in the middle of the bar?&lt;/p&gt;&#xA;" OwnerUserId="24086" LastActivityDate="2019-12-31T21:29:07.340" Title="How do I calculate the forces on a glider's foundation?" Tags="|aerodynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32589" PostTypeId="1" CreationDate="2019-12-31T15:17:26.957" Score="0" ViewCount="80" Body="&lt;p&gt;How would I go about comparing the efficiency of different heater block shapes for the outlined system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, in the attached diagram there is a cylindrical heater cartridge (A) that is used to transfer heat to the cylindrical polymer (B) through the heater block(C). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U9NL4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U9NL4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(A): Heater Cartridge&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(B): Polymer liquid flow, carrying heat out&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(C): Heater block, aluminium&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(D): Temperature Sensor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This setup is for the heater block of a 3D printer. The aim being to provide constant heat to the polymer (B) that flows through the system. The aluminium block acts as a thermal storage to reduce temperature fluctuation in the polymer filament as the flow rate changes. The temperature sensor (D) reacts to changes and controls the heater cartridge in order to try and maintain constant temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wish to redesign this heater block to find the most efficient shape for this function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question is how would I calculate the efficiency of different shapes for this purpose? Losses to the surrounding air will have to be taken into account as the surface area will change with different shapes. Different shapes having different surface area to volume ratios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I must admit I don't know where to start with this problem, it has been quite some time since I've had any formal thermodynamics education. Any answers, pointers or links to resources are greatly appreciated. Thank you!&lt;/p&gt;&#xA;" OwnerUserId="23791" LastEditorUserId="23791" LastEditDate="2019-12-31T15:56:53.230" LastActivityDate="2020-01-10T04:25:31.910" Title="Method for comparing efficiency of heater block designs?" Tags="|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32590" PostTypeId="2" ParentId="25003" CreationDate="2019-12-31T16:10:45.817" Score="1" Body="&lt;p&gt;I don't see a factor for he energy to collapse the elastomeric tube. For many flow rates this could be the largest energy consumption.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2019-12-31T16:10:45.817" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32591" PostTypeId="2" ParentId="32588" CreationDate="2019-12-31T21:01:03.820" Score="1" Body="&lt;p&gt;The lift distribution across the span of the wing is due to the different pressures under the wing and over the wing. in most roughly rectangular wings this distribution of the lift pressure decreases as measured further from the center/the fuselage of the plane. If the wing is swept back and in twisted or has a delta shape things get more complicated. Here is a graph of wing load versus span position for low Re numbers at an angle of attack of 6 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BXaNj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BXaNj.png&quot; alt=&quot;loca lift&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very rough estimate of the distribution of the loads on your bar would be to apply the pressure in a trapezoid pattern depending on the shape of your glider.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And as you mention you can apply the load hanging from the center. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;NASA has a java app to calculate the wing loads here, Foilsim III .&lt;a href=&quot;https://www.grc.nasa.gov/WWW/K-12/freesoftware_page.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Foilsim lll&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2019-12-31T21:29:07.340" LastActivityDate="2019-12-31T21:29:07.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32592" PostTypeId="2" ParentId="32579" CreationDate="2020-01-01T06:53:11.360" Score="0" Body="&lt;p&gt;There are 2 standards. One of them is called towing resistance and means obviously the slow towing force on flat solid pavement required to keep a loaded wheel rolling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is a fraction of wheel load and usually for cars and trucks it is in the range of 0.0045 to 0.008 and for bicycles with 120psi tire pressure 0.002.  if we multiply this by 100 we get the percent factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some tire manufacturers produce performance tires with less rolling resistance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See the table in this Wikipedia article.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.m.wikipedia.org/wiki/Rolling_resistance&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia rolling resistance.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-01T06:53:11.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32593" PostTypeId="1" CreationDate="2020-01-01T16:08:04.307" Score="2" ViewCount="254" Body="&lt;p&gt;Say we have a plan of a building as follows and it is required to be designed using DDM (Direct Design Method). since the moment needs to distributed laterally into a column and middle strips and according to ACI 8.4.1.5 &quot;A column strip is a design strip with a width on each side of a column centerline equal to the lesser of 0.25ℓ2 and 0.25ℓ1.&quot; but in this case, both L1 and L2 have different spans and i dont know what to choose.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/itHON.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/itHON.png&quot; alt=&quot;&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="13708" LastActivityDate="2021-09-23T00:07:00.630" Title="What will be the the width of the column strip in the case of 2 different spans in the same direction?" Tags="|structural-engineering|civil-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32594" PostTypeId="2" ParentId="32583" CreationDate="2020-01-01T19:45:54.017" Score="2" Body="&lt;p&gt;The real world doesn't know anything about approximations you make in mathematical models of it. If always behaves in a fully 3-dimensional way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For Euler buckling of a column, you are really studying &lt;em&gt;perturbations&lt;/em&gt; about the steady stress condition, which is just an axial direct stress in the column. You want to find the end load at which a small perturbation becomes unstable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you try to make a mathematical model that only includes first-order approximations to the behaviour of the perturbed structure, the results say that the column is stable for any compressive load. Since that doesn't correspond to what happens in real life, the over-simplified model is just wrong, and useless.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an example of why first-order approximations don't always work, consider a bar of length &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; along the &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; axis, from &lt;span class=&quot;math-container&quot;&gt;$x = 0$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$x = L$&lt;/span&gt;, with no loads applied to it. The strain and stress in the bar is obviously zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, rotate the bar through 180 degrees, so the end at &lt;span class=&quot;math-container&quot;&gt;$x = L$&lt;/span&gt; is now at &lt;span class=&quot;math-container&quot;&gt;$x = -L$&lt;/span&gt;. Obviously the strain and stress in the bar is still zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you analyse this as a simple column, you have changed the original length of &lt;span class=&quot;math-container&quot;&gt;$+L$&lt;/span&gt; to a length of &lt;span class=&quot;math-container&quot;&gt;$-L$&lt;/span&gt;, creating an axial strain of &lt;span class=&quot;math-container&quot;&gt;$-2$&lt;/span&gt;, and therefore there is a huge compressive stress in the bar. Clearly something went wrong with the math!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make a model with the correct behaviour (or at least, &lt;em&gt;closer&lt;/em&gt; to the correct behaviour than &quot;completely wrong&quot;) you have to include second-order effects. For example, if the column is perturbed into a curved shape, &lt;em&gt;the point of application of the end load position moves corresponding to the deformed shape of the column.&lt;/em&gt;  If the end of the column moves &quot;sideways&quot; from the axis, that means the end load is now applying a bending moment to the base of the column as well as an axial load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, since the perturbed shape of the column is a curve, and the length of the center line of the column does not change &lt;em&gt;measured along the curve&lt;/em&gt;, that implies that the end of the column moves &lt;em&gt;axially&lt;/em&gt; as well as sideways. That means that the applied load does work (= the appled force &lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt; the axial component of the displacement).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem with understanding most &quot;simple&quot; derivations of Euler buckling is that you meet them &lt;em&gt;before&lt;/em&gt; you have learned the general concept of how to model arbitrary (large) displacements of a body. Therefore, the derivation often seems to be &quot;pulling rabbits out of a hat&quot; in order to get to the accepted answer, but doesn't provide a convincing explanation of why it was done that way.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An article like your blog link doesn't really help. It doesn't say anything that is &lt;em&gt;wrong&lt;/em&gt;, but it can't replace a university-level course on continuum mechanics that is a follow up to the &quot;small displacements and small strains&quot; approximations that are made in most first courses.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-01-01T19:45:54.017" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32595" PostTypeId="2" ParentId="32593" CreationDate="2020-01-01T21:16:50.520" Score="1" Body="&lt;p&gt;The adjacent span ratio disqualify this project for DDM.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit is 0.66  or 1.3 ratio. Here we have 7.5/5= 1.5.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-01T21:16:50.520" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32596" PostTypeId="1" CreationDate="2020-01-01T21:48:15.333" Score="-1" ViewCount="67" Body="&lt;p&gt;I have a PCB maker at school, but no one really knows how it works, because it's a pretty old thing. I asked for the instruction manual and it turns out, the manual is also pretty old. If i want to import my &lt;strong&gt;gerber&lt;/strong&gt; file the manual says I should import these files:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IK0rN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IK0rN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, I'm not really sure if this is the right place to ask, but I don't have a &lt;strong&gt;.TOP .BOT .BOA&lt;/strong&gt; or &lt;strong&gt;.DRD&lt;/strong&gt;. These are the files in my gerber zipfile:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KtXpJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KtXpJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, my question, which files am I supposed to import to the PCB. (By the way, the PCB maker is a LPKF Protomat s62)&lt;/p&gt;&#xA;" OwnerUserId="23704" LastActivityDate="2020-01-01T22:39:22.613" Title="Which gerber file to use for pcb maker" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32597" PostTypeId="2" ParentId="32596" CreationDate="2020-01-01T22:39:22.613" Score="1" Body="&lt;p&gt;I think the thing to know here is that the file extensions of Gerber files are not standardized. Some programs use the extension to indicate the layer, and others just include the layer in the filename. To this day, when you upload Gerbers to PCB fab services, they often mis-identify which layers are which.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With all the extension confusion, it's sometimes difficult to know if your files even &lt;em&gt;are&lt;/em&gt; gerbers. For example, the drill file is usally not a gerber, but rather a &quot;NC drill&quot; file type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The easiest way to handle this is not worry about the file extensions too much, and just make sure you have your own consistent system for keeping track of what's what.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Almost every PCB manufactuerer will require the following (using a 2 layer board as an example):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Top Silk Screen Gerber&lt;/li&gt;&#xA;&lt;li&gt;Top Solder Mask Gerber&lt;/li&gt;&#xA;&lt;li&gt;Top Copper Gerber&lt;/li&gt;&#xA;&lt;li&gt;Bottom Copper Gerber&lt;/li&gt;&#xA;&lt;li&gt;Bottom Solder Mask Gerber&lt;/li&gt;&#xA;&lt;li&gt;Bottom Silk Gerber&lt;/li&gt;&#xA;&lt;li&gt;Board Outline Gerber&lt;/li&gt;&#xA;&lt;li&gt;NC Drill File&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The extension doesn't matter, a gerber is a gerber (except the NC drill file).&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2020-01-01T22:39:22.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32598" PostTypeId="1" CreationDate="2020-01-01T23:48:58.630" Score="-2" ViewCount="1989" Body="&lt;p&gt;For a number of years now, I have been buying instant coffee products. Since the main brand has been Nescafe, this question is specifically about their products, even though I vaguely recall that the other brands also did this same thing. (I can't swear on it, though.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you buy a jar of &quot;Nescafe Gold&quot;, for example, and unscrew the top, you are met with a thin but firm protection layer, obviously intended to increase the shelf life for the product. However, there isn't any part of this which allows human fingers to pull this protection layer up easily, in one go. In fact, every single time I open a new bottle, I'm scared to get little pieces of that protection layer into the actual instant coffee, since the only way I have found to actually unseal it reasonably safely is to take a tea spoon and carefully knock a hole in the middle and then start pulling away the pieces, hoping that it doesn't make a mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It &lt;strong&gt;always&lt;/strong&gt; makes a mess.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There seriously doesn't seem to be any way to open it in a &quot;sane&quot; and &quot;obvious&quot; way. I'm absolutely baffled by how this can be, and I've been trying to imagine scenarios in my head where they sat in a board room somewhere and decided to design it this way (and keep it for a very long time) in spite of the problems and frustrations and intangible badwill it would produce. Not to mention that they probably use the product themselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried numerous alternate ways to open it, including cutting along the edges, but nothing is efficient and safe, as one would expect from such a popular and well-funded instant coffee product. It almost seems like they want you to open as few packages as possible, which is the opposite of their best business interest...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone have any facts about this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: It seems like the responses so far don't understand what I'm asking at all. I'm simply wondering why they don't have a little extra bit of the exact same material as they are currently using &quot;overflowing&quot;, for a human to be able to &quot;grab on&quot; to something and thus open the seal without making a mess. Why require each customer to fiddle around with knives and stuff? With so many things having been simplified to an extreme degree in recent decades, why is this particular product still using what seems to be the &quot;version 1.0&quot; of vacuum sealers? You would still see if somebody has been tampering with it. It would just be much easier to actually open.&lt;/p&gt;&#xA;" OwnerUserId="24090" LastEditorUserId="1832" LastEditDate="2020-01-04T00:52:43.877" LastActivityDate="2021-01-20T17:29:33.807" Title="Why do Nescafe's (and others'?) instant coffee jars have an inner seal with no &quot;grip&quot; for easy removal?" Tags="|vacuum|product-engineering|seals|" AnswerCount="2" CommentCount="2" ClosedDate="2021-01-20T17:41:48.940" ContentLicense="CC BY-SA 4.0" />
  <row Id="32599" PostTypeId="2" ParentId="32598" CreationDate="2020-01-02T01:16:33.120" Score="1" Body="&lt;p&gt;Maybe this is actually what they design it for: to show immediately if someone tampered with the seal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Big brands have to be prepared for acts of arson, sabotage, lunacy, sick people seeking pleasure out of hurting strangers, disgruntled employees, etc.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-02T01:16:33.120" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32600" PostTypeId="2" ParentId="16782" CreationDate="2020-01-02T01:42:50.687" Score="-1" Body="&lt;p&gt;Hurstwic has been doing some interesting work with basic iron/steel smelting. Not an answer to your question, but, Yes, you can smelt your own Iron/Steel at home with relative safety (your mileage may vary). You'd have to figure out what additional properties you'd need to add in the smelt, and the refinement work would take additional research, but it's probably more possible than others on here have thought, to make a variety of steels at home (which is how steel was made 1,000+ years ago anyway).&lt;/p&gt;&#xA;" OwnerUserId="24091" LastActivityDate="2020-01-02T01:42:50.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32601" PostTypeId="1" CreationDate="2020-01-02T05:25:36.290" Score="-2" ViewCount="208" Body="&lt;p&gt;I am using the PID block in Simulink and limiting the output to (+/-)5. The plant becomes unstable and untunable after putting these limits. The plant is stable and tuneable otherwise if no limits are applied. &#xA;The output from the PID tuner goes beyond 10 when no limits are applied and I want to limit the output to 5. &lt;a href=&quot;https://i.stack.imgur.com/893iU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/893iU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="24092" LastActivityDate="2020-01-02T18:04:04.927" Title="Limiting PID controller's output making the plant unstable and untunable" Tags="|electrical-engineering|control-engineering|pid-control|matlab|simulink|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32603" PostTypeId="1" AcceptedAnswerId="32604" CreationDate="2020-01-02T06:23:34.000" Score="1" ViewCount="3848" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xIzOe.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xIzOe.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt; could there ever be a structure that would require something like this? &lt;/p&gt;&#xA;" OwnerUserId="24093" LastActivityDate="2021-10-14T08:22:24.947" Title="Could a bolt this size actually be used on a project?" Tags="|mechanical-engineering|design|structural-analysis|steel|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32604" PostTypeId="2" ParentId="32603" CreationDate="2020-01-02T12:28:26.087" Score="4" Body="&lt;p&gt;From my location it appears the image is broken, however I can assume with my imagination and the comments it’s a pretty big bolt.&lt;/p&gt;&#xA;&lt;p&gt;Exceptionally large bolts are surprisingly not that uncommon. They often show up in oil industry to bolt flanges and for use in huge structures. But they are usually custom low production bolts designed for a specific use case and not mass manufactured. M180 is the largest I’ve ever seen in a bolt list to purchase from a supplier.&lt;/p&gt;&#xA;&lt;p&gt;24” nuts was “common” for a time, however this has mostly changed to higher quality smaller diameter alloys and bolts.&lt;/p&gt;&#xA;&lt;p&gt;At one point there were 98” diameter, and nearly 30’ long bolts made in the 60s&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Penrith Engineering Works in Clydesdale Scotland in 1967. The bolts measured 27' 4&amp;quot; long each and had a diameter of 4' 2&amp;quot;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As far as i remember it was an experiment to bolt oil tankers together that failed miserably.&lt;/p&gt;&#xA;&lt;p&gt;As mentioned in the oil industry it is common to use large diameter bolts, as much as 20” (w510) can be “easily” purchased from ITH for example.&lt;/p&gt;&#xA;&lt;p&gt;There are of course large problems with large bolts (fitting, yes?) handling, manufacturing, weight, transportation, actually bolting them, etc etc And often when the design requires such large bolts then one will generally choose a different (permanent) connection method.&lt;/p&gt;&#xA;" OwnerUserId="14318" LastEditorUserId="90" LastEditDate="2021-10-14T08:22:24.947" LastActivityDate="2021-10-14T08:22:24.947" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32606" PostTypeId="2" ParentId="32576" CreationDate="2020-01-02T13:36:33.983" Score="0" Body="&lt;p&gt;For the current transformer(750340744), I found a replaceable one: &lt;a href=&quot;https://www.murata-ps.com/pub/data/magnetics/kmp_5300c.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;53100C&lt;/a&gt; with turn ratio: 1:100; primary current rating 10A. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the push-pull transformer(750340744), I found a replaceable one: &lt;a href=&quot;https://kr.mouser.com/datasheet/2/336/P766-1144949.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;PH9385.015NL&lt;/a&gt;. But I still wonder about turns ratio. The datasheet said that: Turns Ratio&#xA;(1:4):(8:5) = 1:5. In my design, pin 2 ties to pin 3 and pin 7 ties to pin 6. And I want Turns Ratio (1-2):(8-7) = 1:5 ; (1-2):(6-5)=1:5; (1-2):(3-4) = 1:1. So, whether this transformer is what I need? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jUtYi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jUtYi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23780" LastActivityDate="2020-01-02T13:36:33.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32608" PostTypeId="1" AcceptedAnswerId="32614" CreationDate="2020-01-02T14:52:03.403" Score="2" ViewCount="559" Body="&lt;p&gt;what is the physical interpretation of poles and zeros of a mechanical system?&#xA;I'm asking things like the meaning if I have only 0 or only complex numbers. And about the number of the poles and zeros, it should be equal? What can I infer about the stability?&lt;/p&gt;&#xA;" OwnerUserId="13700" LastActivityDate="2020-01-02T17:13:53.407" Title="what is the physical interpretation of poles and zeros of a mechanical system?" Tags="|control-engineering|control-theory|applied-mechanics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32609" PostTypeId="1" AcceptedAnswerId="32637" CreationDate="2020-01-02T14:53:38.167" Score="1" ViewCount="67" Body="&lt;p&gt;This is an HVAC mechanical engineering question on a document from a manufacture where they convert units of electrical energy to therms of natural gas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Carrier has writeups for building energy simulation modeling, and &lt;a href=&quot;https://www.shareddocs.com/hvac/docs/1004/public/0a/hap_ehelp_031.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this link&lt;/a&gt; is a PDF for a writeup using the HAP software.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;The example on page 2 as a high level feasibility they convert units of NG to kWh:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In a certain locale the price for electricity is 0.0960 dollar per&#xA;  killowatt hour and the cost of natural gas is 0.937 dollar per therm.&#xA;  If we convert the gas price to use units of kWh, we get a gas price of&#xA;  0.0320 $/kWh. Therefore a kWh of electricity is three  times as expensive as a kWh of natural gas. If the average annual COP of the&#xA;  DHRC is more than three times that of the gas boiler, then heating&#xA;  cost will be reduced by using the DHRC. If the COP of the DHRC is less&#xA;  than three times that of the boiler, heating cost will increase. Note:&#xA;  This is a single example for illustration. Energy prices vary widely&#xA;  by location and over time. You should use your own local utility&#xA;  prices to assess heat recovery viability for your projects.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I know 1 therm = 29.3001 kWh or 1 kWh = 0.03413 therm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what’s the math behind coming up with “convert the gas price to use units of kWh, we get a gas price of 0.0320 $/kWh”?&lt;/p&gt;&#xA;" OwnerUserId="24098" LastActivityDate="2020-01-04T03:14:09.950" Title="energy unit conversions" Tags="|mechanical-engineering|hvac|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32610" PostTypeId="2" ParentId="10289" CreationDate="2020-01-02T15:30:19.020" Score="-1" Body="&lt;p&gt;You can use steel reinforcement bar couplers to connect. Do not you think that this will be much easier? After all, you do not need to use welding equipment, only portable hydraulic pressing equipment. Simplicity and high assembly speed are advantages.&lt;/p&gt;&#xA;" OwnerUserId="24100" LastEditorUserId="1832" LastEditDate="2020-01-04T00:55:24.420" LastActivityDate="2020-01-04T00:55:24.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32613" PostTypeId="1" CreationDate="2020-01-02T16:57:18.360" Score="-1" ViewCount="42" Body="&lt;p&gt;Let's say I have a beam supported exactly in the middle (like a seesaw) and a distributed force applied on it across the beam. What is the way to calculate the displacement, moment, shear...etc of the beam?&lt;/p&gt;&#xA;" OwnerUserId="24086" LastActivityDate="2020-01-02T19:45:18.100" Title="SINGLE supported beam" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32614" PostTypeId="2" ParentId="32608" CreationDate="2020-01-02T17:13:53.407" Score="4" Body="&lt;p&gt;The poles and zeros of a LTI plant or system in relation to their eigenvalues determine how quickly a system will stabilize or destabilize and if it oscillates. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When looking at these in the complex plane, the Re(a+ib) determine how quickly a system will exponentially stabilize or destabilize (the left vs right half plane) while the Im(a+ib) determine how the system oscillates. (High/low freq and amplitude)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poles and zeros do not have to be equal, however imaginary systems always appear in pairs namely a+ib and a-ib. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A system is absolutely stable if all poles are at the left half plane =&amp;lt;0. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other stability descriptions can be named depending on specific criteria such as conditional, marginal, or unstable or chaotic depending on where poles lie. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can find a lot of infos on the web about this, such as t&lt;a href=&quot;https://web.mit.edu/2.14/www/Handouts/PoleZero.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;his pdf from MIT.&lt;/a&gt; (warning pdf)&lt;/p&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2020-01-02T17:13:53.407" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32615" PostTypeId="1" AcceptedAnswerId="32638" CreationDate="2020-01-02T17:14:21.543" Score="1" ViewCount="114" Body="&lt;p&gt;I have a workflow of two functions. Both of them aggregate the items that are receiving and only the first is adjustable. I mean that I can decide how many items to aggregate on the first function &lt;code&gt;F1&lt;/code&gt; in order to not exceed the capacity (bandwidth) of the channel that connects to the second function &lt;code&gt;F2&lt;/code&gt;. In other words, I control the throughput that the second function is receiving items. If I aggregate more at &lt;code&gt;F1&lt;/code&gt; I decrease the throughput. If I aggregate less at &lt;code&gt;F1&lt;/code&gt; I increase the throughput. My goal is to tune the system for its maximum throughput while the input workload varies and I have a limited channel between &lt;code&gt;F1&lt;/code&gt; and `F2. This it the picture of my system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qSNe6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qSNe6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data stream of items (workload) that &lt;code&gt;F1&lt;/code&gt; receives varies along the time. I want to control the throughput output of &lt;code&gt;F1&lt;/code&gt; in order to not surpass the channel capacity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am planning to use PID control to decide how many items to aggregate on the first function before to emit the items to the second function. I am a newbie in control systems. As far as I understood I have to start with the Proportional controller (P) which is basically the current error. Then I have to decide if I will use the Integral controller (I) which is the historic cumulative value of the errors. And then I have to decide if I will use the Derivative controller (D) which is based on its current rate of change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For me, it is not clear yet if I really need the Integral and Derivative controllers. I mean, if I find the error between the actual throughput and the desired throughput I just set the first aggregate function to aggregate items in order to have the desired throughput. In my mind this is enough, isn't it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something could influence is the time that I take to aggregate items on the first function. Suppose that I have a slow processor. But, in my use case what is the insight that I have to have in order to use the Integral or the Derivative controllers?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&#xA;Felipe&lt;/p&gt;&#xA;" OwnerUserId="24103" LastEditorUserId="24103" LastEditDate="2020-01-03T15:13:14.023" LastActivityDate="2020-01-04T11:40:02.080" Title="Is it necessary to use PID control in my use case?" Tags="|control-theory|pid-control|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="32617" PostTypeId="2" ParentId="32601" CreationDate="2020-01-02T18:04:04.927" Score="0" Body="&lt;p&gt;Not all systems are tunable in all ranges. You can find via the Nyquist plot the stability ranges for the ranges of a gain the system can be stable. This can be plotted and &lt;a href=&quot;https://www.mathworks.com/help/control/ref/nyquist.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;analyzed in matlab&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;H = tf([2 5 1],[1 2 3]);&#xA;nyquist(H)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2020-01-02T18:04:04.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32618" PostTypeId="2" ParentId="32613" CreationDate="2020-01-02T19:45:18.100" Score="0" Body="&lt;p&gt;Because of the symmetry, you have zero slope over the support and if you draw the FBD at the support you see it is like a cantilever beam with half of the length of the beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call the uniformly distributed load &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; and length &lt;span class=&quot;math-container&quot;&gt;$2L$&lt;/span&gt; for convenience.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ M_{max}=\frac{wL^2}{2} \; and\; V{max}= 2wL  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum moment and shear happen over the support. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I let you draw the moment and shear diagrams.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-02T19:45:18.100" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32619" PostTypeId="1" CreationDate="2020-01-02T20:33:43.650" Score="3" ViewCount="767" Body="&lt;p&gt;This is probably a fairly basic question for seasoned Siemens NX users, but I can't find a solution anywhere.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a large assembly and I would like to set certain components to wire frame view, while still viewing other parts as solid bodies. I know it's possible, because apparently I've done it before, but I can't remember or find how I did it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JfuDF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JfuDF.png&quot; alt=&quot;Wire-frame limit switch in assemble&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can set the entire assembly to wire frame easily enough, but I would like to set only certain parts to be. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, when I go into the part (the limit switch in this case) the part default is NOT wire frame but a solid body. I'm assuming (hoping) there is a setting in the assembly that I can toggle off/on per part as needed. Solidworks had something very similar, but I'm not using NX and am still inexperienced with the available tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hopefully this is an acceptable forum to be asking this!&lt;/p&gt;&#xA;" OwnerUserId="10590" LastActivityDate="2020-01-02T20:33:43.650" Title="Wireframe parts in Siemens NX" Tags="|cad|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32620" PostTypeId="2" ParentId="31808" CreationDate="2020-01-02T22:17:03.017" Score="0" Body="&lt;p&gt;On a typical hydraulic press, there is negligible danger of motor overloading. However your instincts are correct. Consider the operating sequence of a press. An electric motor drives a pump. The pump drives oil into a cylinder. The cylinder extends until no more motion is possible. What happens after that? Oil must escape somewhere, or pressure will build continuously until some part of the system fails, whether that is a (1) hose bursting, (2) pump housing cracking, (3) cylinder seals rupturing, (4) electric motor overloading, etc... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;All modern hydraulic presses include a relief valve for this reason. Once hydraulic pressure reaches some maximum setting (typically between 100 - 350 bar), the relief valve opens to allow oil to escape back to the reservoir. Thus all components are protected from overloading, including the electric motor. The relief valve setting is typically adjustable - therefore it can be set appropriately for many sizes of electric motor within a wide range of kW ratings. Happy to discuss further if you have questions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of a typical press circuit. &lt;a href=&quot;https://www.coalhandlingplants.com/hydraulic-system/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.coalhandlingplants.com/hydraulic-system/&lt;/a&gt;. Ignore the brief explanations for the symbols in that article. They are not very clear. I only link that article because the first circuit shows the 8 basic components correctly laid out. Here are some better-explained symbols: &#xA;&lt;a href=&quot;https://www.hydraproducts.co.uk/Blog/how-to-read-hydraulic-circuits&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.hydraproducts.co.uk/Blog/how-to-read-hydraulic-circuits&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24104" LastEditorUserId="24104" LastEditDate="2020-01-02T22:51:05.327" LastActivityDate="2020-01-02T22:51:05.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32621" PostTypeId="1" AcceptedAnswerId="32633" CreationDate="2020-01-03T00:30:42.207" Score="1" ViewCount="25" Body="&lt;p&gt;I am interested in putting a small (3&quot;x3&quot;x2&quot;) assembly under X-ray, following a shock test, in order to evaluate failures of an internal ceramic piece. Unfortunately, this assembly utilizes quite a bit of potting material (epoxy) and as such, cannot easily be disassembled. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering reaching out to some local dentist practices to see if I can purchase X-ray time from them, however, I'm concerned that the material makeup of my assembly will not allow adequate penetration and will result in a poor image. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The general makeup of the assembly from the outside and moving inwards is...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0.125&quot; thick aluminum outer casing&lt;/li&gt;&#xA;&lt;li&gt;0.25&quot; to 0.5&quot; foam-like epoxy &lt;/li&gt;&#xA;&lt;li&gt;some small (but strong) magnets&lt;/li&gt;&#xA;&lt;li&gt;a ceramic &quot;skeleton&quot; structure (this is what I'm most interested in evaluating for failures)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Should note that the magnets are not necessarily in the line-of-sight of the ceramic, so 'penetration-ability' of those is a non-issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, regarding the aluminum, with a HVL of 0.06&quot; at 120 keV (&lt;a href=&quot;http://www.sprawls.org/ppmi2/RADPEN/&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;), I'm worried that I may not be able to sufficiently penetrate the outer shell and produce a good image.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;MY QUESTIONS ARE:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;strong&gt;Penetration aside: Is it reasonable to suspect that I will be able to gather useful information and identify fractures in the ceramic?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;strong&gt;Google tells me that the typical energy capability of a dentistry x-ray is &amp;lt;200 keV. For someone with experience in the radiation field, do you expect this would be sufficient to penetrate my test article and produce a good image?&lt;/strong&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17105" LastActivityDate="2020-01-03T20:11:51.717" Title="Determining If A Test Article Is Suitable For X RAY Following Shock Testing?" Tags="|structural-analysis|failure-analysis|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32622" PostTypeId="2" ParentId="270" CreationDate="2020-01-03T01:45:50.523" Score="5" Body="&lt;p&gt;In addition to the other answers, water is commonly used as a primary working fluid in industrial hydraulic systems &lt;em&gt;with&lt;/em&gt; additives to prevent the issues mentioned above. Typical applications include furnaces, foundries, plastic extrusion, etc. Anywhere that standard hydraulic oil represents a major fire or contamination hazard, yet operating costs do not justify purchasing synthetic fluids for $1100+/barrel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Factory hydraulics are ideal for water-based fluids because the environment is controlled. Atmospheric temperature is typically maintained within 15C-30C year-round. So boiling and freezing are non-issues. Oil properties can be continuously monitored for changes in pH, composition, or particulate levels. New water can be regularly added to the system to replace losses from evaporation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The percentage of water in a typical mixture may be between 30 - 80%. As the fraction of water increases, the viscosity of the mixture decreases. Low viscosity is the primary challenge of water-based hydraulics. Modern hydraulic pumps (axial-piston, radial, vane, etc) are designed with internal leakage passages for cooling. Somewhere between 2-6% of total flow is &quot;leaked&quot; or intentionally siphoned away from the flow outlet for cooling of the pump's sliding surfaces. The flow rate through these passages is obviously dependent on viscosity. The problem with running 100% water is that viscosity falls below 1.0 cSt at 20C. Compare that to 46 cSt viscosity of standard hydraulic oil. The machining tolerances of pump surfaces made for pure water would be extremely fine (&amp;lt;1 micron), or leakage could siphon away the majority of total flow &lt;em&gt;and&lt;/em&gt; surfaces could frequently experience contact due to the thin film of water separating them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Current manufacturing tolerances &lt;em&gt;in series production&lt;/em&gt; are simply not that good yet. There are specialized pump and valve manufacturers like Oilgear and Dynex-Rivett with more expensive product lines targeted at the extreme end of the water-based fluid market (80-95% water). There are also products like the Danfoss PAH pumps which can operate at reduced pressures with pure tap water. However a PAH costs approx 25x more than an equivalent-displacement pump for standard fluids.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Source - my former boss spent years trying to redesign standard axial-piston pumps for 100% water&lt;/p&gt;&#xA;" OwnerUserId="24104" LastActivityDate="2020-01-03T01:45:50.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32623" PostTypeId="2" ParentId="25125" CreationDate="2020-01-03T03:27:02.310" Score="1" Body="&lt;p&gt;Great question with many potential applications. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The short answer to your question is YES. A hydraulic motor can absolutely be coupled to a pump in a useful way. You would effectively have made a hydraulic transformer. You are exchanging low flow @ high pressure for high flow @ low pressure, with some transmission loss.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What kind of efficiency to expect? Let's say for example that you couple a bent-axis piston motor @ 5000 psi with a vane pump @ 400 psi. Expect approximately 93% efficiency from the bent axis unit and 80% from the vane unit. So a net efficiency of 0.93 * 0.80 = 0.74 for the transformer. Therefore if you input 10 gpm flow to the bent-axis side, expect approx 92 gpm flow through the vane pump. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Note that there is another efficiency loss from the high-pressure pump which was supplying your bent-axis unit. Let's estimate around 18% of energy lost between that pump and whatever control valves might be involved. A more efficient design would use an electric motor to replace the bent-axis unit. However in many applications, it is not practical to install an electric motor for this purpose. Further, the high-pressure supply pump may continue to spin regardless of how much flow is actually required. Installing the additional electric motor would not provide much additional efficiency benefit in that case. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Regardless of the efficiency discussion, the entire topic has many many applications for &quot;2-speed&quot; devices requiring high flow @ low pressure, up to a certain distance. Then briefly requiring low flow @ high pressure, to complete the actuation. One classic example is the compacting cylinder used in recycling machines. As this cylinder extends, pressure slowly builds due to the elastic nature of the recycled material (eg. cardboard, plastic). Pressure increases rapidly towards the end of cylinder extension, as air pockets in the material are reduced. Pressure will peak at the end of the stroke. Finally, the cylinder will retract under negligible load.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lastly here is an article I wrote going into much more detail on this topic for various different industrial and mobile hydraulic applications. &#xA;&lt;a href=&quot;https://www.fluidpowerworld.com/white-paper-design-concepts-for-power-limiting-open-circuit-architectures/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.fluidpowerworld.com/white-paper-design-concepts-for-power-limiting-open-circuit-architectures/&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="24104" LastEditorUserId="24104" LastEditDate="2020-01-03T15:53:48.680" LastActivityDate="2020-01-03T15:53:48.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32624" PostTypeId="1" CreationDate="2020-01-03T04:31:48.823" Score="1" ViewCount="73" Body="&lt;p&gt;From wikipedia, there's &lt;a href=&quot;https://en.wikipedia.org/wiki/Linkage_(mechanical)#/media/File:Func_Geen_Log(u).gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;this animated GIF of a linkage&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sS2DA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sS2DA.jpg&quot; alt=&quot;Here&amp;#39;s a frame from the animation:&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question: How does it calculate log? All I see there are trig functions and circles. &lt;code&gt;log&lt;/code&gt; doesn't show up in geometric maths, does it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friend thinks it's just a (lucky) approximation.&lt;/p&gt;&#xA;" OwnerUserId="24112" LastActivityDate="2024-03-31T17:39:54.510" Title="Why does this linkage calculate log?" Tags="|linkage|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32625" PostTypeId="2" ParentId="8118" CreationDate="2020-01-03T10:54:16.347" Score="0" Body="&lt;p&gt;Hi to the next person to read this,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The preceding appears to be all correct, however, it appears to be for a solid composite beam. If you are using a USB board as a core, you must rather do composite beam calculations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I landed on this post because I wanted to refresh my memory to calculate composite beams for a snow yatch I am building (like an iceboat but on special skis I have laminated). The cross beam is 12 feet, laminated with 6oz unidirectional ( I presume 140GPA, thanks wwarniner!) and woven 6oz 45 degree twill for torsional rigidity and capping (this is more complex with poisson ratios and that stuff (?). Anyway, I have 400 lbs centre load, simply supported and I will pre-stress for extra stiffness (laminate tensile side flat, then compression side with twice the curvature desired. Loads up the core in compression and adds loads of Stiffness! I am looking to know roughly how many layers Of UD I will need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The process is as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;You calculate the total moment of inertia for each rectangular area consdering the distance from the neutral axis that really pays off!!! That is why surface layers in a sandwich make it so light and stiff.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You then calculate the contribution of each area multiplied by it's stiffness (E) for the flex. Soft materials and layers near the neutral axis contribute insignificantly, as opposed to surface layers, particularly UD in my case. Note: A honeycomb or foam core as you can imagine contributes insignificantly, all the work is done by the extreme fibres. However, you must be concerned about core shear in this case)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can do the same for strength using UFM or UTS (ultimate tensile strength)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Reminders: Rectangle I = b*h^3/12, Z = I/c. (C is distance to extreme fibre, that is h/2). Right? (This is all from memory, so pls  chk 2 b sure)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is what I remembered from the Gibbs and Cox composite design bible I used many years ago. Once you use it, you ré member the principles for life ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, if anyone would like the most delightful engineering read ever, I strongly recommend a book I read about the development of all this in war times, &quot;The Science of Strong Materials&quot; by a British engineer by the name of J.E.Gordon. Composite calculations came from plywood calculation for aircraft) Bloody good read if I do say so meeself ;-)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&#xA;Paul Isabelle,&#xA;Industrial designer and tinkerer&lt;/p&gt;&#xA;" OwnerUserId="24114" LastEditorUserId="24114" LastEditDate="2020-01-03T11:16:39.733" LastActivityDate="2020-01-03T11:16:39.733" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32626" PostTypeId="1" AcceptedAnswerId="32627" CreationDate="2020-01-03T12:44:13.660" Score="1" ViewCount="80" Body="&lt;p&gt;I am using Nastran for buckling analysis, and I'm new to all of this so please bear with me. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a linear buckling analysis, I understand that you must first apply any compressive load in the static subcase. Then solve the eigenvalue buckling problem in the second subcase using the first static subcase. The buckling load is then the amount of compressive load applied multiplied by the resulting eigenvalue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I now want to apply a second fixed force of exactly 50N laterally and see what the buckling load would be with this additional 50N force. Now if I apply both the compressive load and the lateral load to the static subcase, and then solve the eigenvalue buckling using that subcase, the resulting buckling load varies a lot depending on the compressive load applied in the static subcase. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example 1:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Apply 1N compressive force, eigenvalue is 100. So buckling load is 1N*100 = 100N&lt;/li&gt;&#xA;&lt;li&gt;Apply 10N compressive force, eigenvalue is 10. So buckling load is 10N*10 = 100N&#xA;This makes sense, however:&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Example 2:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Apply 1N compressive force and 1N lateral force, eigenvalue is 90. So buckling load is 1N*90= 90N.&lt;/li&gt;&#xA;&lt;li&gt;Apply 10N compressive force and 1N lateral force, eigenvalue is 9.5. So buckling load is 10N*9.5=95N, which is higher than 90N. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This just proves that I am doing something wrong, and I am hoping someone can clarify what it is.&#xA;Any help would be appreciated, thank you!&lt;/p&gt;&#xA;" OwnerUserId="24115" LastActivityDate="2020-01-03T14:08:08.610" Title="How would I do a buckling analysis with an additional fixed known lateral force?" Tags="|mechanical-engineering|structural-engineering|finite-element-method|buckling|nastran|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32627" PostTypeId="2" ParentId="32626" CreationDate="2020-01-03T13:52:58.877" Score="2" Body="&lt;p&gt;In your example 2, Nastran is scaling &lt;em&gt;both&lt;/em&gt; loads as part of the buckling load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your first case, you have buckling with &lt;span class=&quot;math-container&quot;&gt;$90\times 1 = 90$&lt;/span&gt;N axially plus &lt;span class=&quot;math-container&quot;&gt;$90\times 1=90$&lt;/span&gt;N laterally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second, you have buckling with &lt;span class=&quot;math-container&quot;&gt;$9.5\times 10 = 95$&lt;/span&gt;N axially plus &lt;span class=&quot;math-container&quot;&gt;$9.5\times 1 = 9.5$&lt;/span&gt;N laterally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From example 1 you have a buckling load of 100N axially plus zero laterally.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could plot a graph with those three points, and estimate the axial load when the lateral load is 50N. Then set up the Nastran input for those values and repeat if necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternatively, since you have now bracketed the answer between 90N and 95N, you could run analyses with 91+50, 92+50, 93+50 and 94+50, and estimate when the buckling factor is exactly 1.0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a general situation, this sort of loading is likely to be incompatible with the assumptions of linear buckling. You could model it as a nonlinear analysis. First ramp up the side load from 0 to 50N, then increase the axial load and see what happens. You may find the structure doesn't &quot;buckle&quot; in the sense of a catastrophic instability, but just gets more flexible as you increase the axial load and lateral deflection increases.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2020-01-03T14:08:08.610" LastActivityDate="2020-01-03T14:08:08.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32630" PostTypeId="2" ParentId="24717" CreationDate="2020-01-03T19:02:04.353" Score="1" Body="&lt;p&gt;To summarize your problem description, you have ...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;3 HP, 1200 RPM electric motor&lt;/li&gt;&#xA;&lt;li&gt;40 MPa (400 bar) pump rated 4 LPM @ 40 RPM&lt;/li&gt;&#xA;&lt;li&gt;Gearbox ratio TBD between motor and pump&lt;/li&gt;&#xA;&lt;li&gt;Desired operation of 20MPa (200bar) at 4 LPM&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;First, we can make a few observations from the problem data. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;The full-load torque of your electric motor. Note that this number is a nominal rating. Typical general-purpose induction motors can run at 10-15% higher torque than the full-load rating without damage, as long as cooling air flow is not obstructed. &#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_{motor} = 3 \ hp * {1 \ kW \over 1.341 \ hp} = 2.24 \ kW$$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M_{motor} = {P_{motor}\over2*\pi*{n/60}} = 17.8 \ Nm$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;&lt;p&gt;Where did you find a pump with those specs? It's possible in theory but I've never seen anything close. A typical 4 LPM positive displacement pump would have a max speed over 1800 rpm. This avoids the need for a gearbox. Rexroth, Parker, Hydac, Eaton... many options available for pumps. Gear-type (200 bar) or radial piston (400 bar) pumps are typically the most cost-effective at your flow rate. Here is one catalog for example - &#xA;&lt;a href=&quot;https://dc-us.resource.bosch.com/media/us/products_13/product_groups_1/industrial_hydraulics_5/pdfs_4/re11263.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://dc-us.resource.bosch.com/media/us/products_13/product_groups_1/industrial_hydraulics_5/pdfs_4/re11263.pdf&lt;/a&gt; &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;In case you really do have a pump along those lines... here is the power transmitted through the hydraulic fluid:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_{flow} = p*Q*{100000 \ Pa \over 1 \ bar}*{1 \ m^3/s \over 60000 \ lpm} = 1.33 \ kW$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Choose a typical pump efficiency of 85% and you have the power required to spin the pump: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$P_{pump} = {P_{flow} \over 0.85} = 1.57 kW$$&lt;/span&gt;&#xA;This is safely within the nominal motor power rating of 2.24 kW, including whatever gearbox efficiency you want to assume. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Lastly we can verify that motor torque is also sufficient to drive the pump at 40 rpm. &#xA;&lt;span class=&quot;math-container&quot;&gt;$$M_{pump} = {P_{pump}\over2*\pi*{n/60}} = 374.8 \ Nm$$&lt;/span&gt;&#xA;So your gearbox would need to amplify torque by&#xA;&lt;span class=&quot;math-container&quot;&gt;$${M_{pump} \over M_{motor}} = {374.8 \ Nm \over 17.8 \ Nm} = 21.1 \ min \ gear \ ratio$$&lt;/span&gt;&#xA;Which is accomplished with plenty of overhead thanks to the speed ratio:&#xA;&lt;span class=&quot;math-container&quot;&gt;$${n_{motor} \over n_{pump}} ={1200 \ rpm \over 40 \ rpm} = 30.0 \ actual \ gear \ ratio$$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Units: &lt;span class=&quot;math-container&quot;&gt;$$P[kW], \ \ p[bar], \ \ n[rpm], \ \ Q[lpm], \ \ M[Nm]$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="24104" LastActivityDate="2020-01-03T19:02:04.353" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32631" PostTypeId="2" ParentId="18444" CreationDate="2020-01-03T19:52:03.733" Score="1" Body="&lt;p&gt;Symbol 23 is just a plugged cavity. The two incoming hydraulic lines are both blocked. No flow is exchanged between them. The designer's intent for the cavity is future use. Various cartridge valves can be inserted into the cavity for either pressure control, flow control, directional control, or other purposes. You can search &quot;DCP-2 cavity plug&quot; or see this link for a typical example: &lt;a href=&quot;https://www.doering.com/pdf/DCP-2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.doering.com/pdf/DCP-2.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24104" LastEditorUserId="24104" LastEditDate="2020-01-03T20:18:04.153" LastActivityDate="2020-01-03T20:18:04.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32633" PostTypeId="2" ParentId="32621" CreationDate="2020-01-03T20:11:51.717" Score="2" Body="&lt;p&gt;X-rays are a poor way to test for cracking in a ceramic. A crack in a ceramic part which has not shattered to pieces is a very small feature, of order ~0.001&quot; wide or less, which commonly-used x-ray equipment cannot resolve. Second, a tiny crack in a ceramic part does not present a significant difference in x-ray absorptivity, inelastic scattering, or diffraction  between the cracked and uncracked regions to the incoming x-rays which means that a tiny crack will not produce &lt;em&gt;x-ray contrast&lt;/em&gt; in the imaging apparatus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition, the materials which enclose your ceramic part will have the effect of attenuating and scattering the incoming x-rays which degrade their ability to resolve useful contrast in that portion of the &quot;sandwich&quot; you need to find cracks in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In light of this, you would be better off trying another sort of nondestructive testing technique than x-radiography.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-01-03T20:11:51.717" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32634" PostTypeId="1" AcceptedAnswerId="32636" CreationDate="2020-01-03T20:17:34.063" Score="0" ViewCount="304" Body="&lt;p&gt;Consider the following problem from Hibbeler's Engineering Mechanics, Statics. (13th edition)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rwIjV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rwIjV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the associated free body diagram and equilibrium equations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Lv0Wh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Lv0Wh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is solved by equating &lt;span class=&quot;math-container&quot;&gt;$N_B$&lt;/span&gt; to zero. Let's say I was asked to determine all support reactions and the force in cable CD (as in the picture, not requiring the ramp to just start being lifted) It seems that 3 equilibrium equations are not sufficient, as there are 4 unknowns &lt;span class=&quot;math-container&quot;&gt;$F\, ,N_B\,,A_x\,and\, A_y $&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What renders this statics problem unsolvable, is it statically indeterminate?&lt;/p&gt;&#xA;" OwnerUserId="15381" LastEditorUserId="8055" LastEditDate="2020-01-04T00:53:55.237" LastActivityDate="2020-01-04T15:55:08.840" Title="Statical equilibrium in 2D" Tags="|statics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32635" PostTypeId="1" CreationDate="2020-01-04T02:12:51.313" Score="2" ViewCount="25" Body="&lt;p&gt;I frequently use the Equivalent Full Load Hour (EFLH) estimation method to calculate the expected energy usage for commercial HVAC equipment.  While not ideal, it works well enough to given an order of magnitude for the energy savings that we can expect if and when equipment is replaced.  The EFLH can usually be approximated well enough form available tables for different building types in various locations, and the efficiency, usually in the form of SEER or EER, is typically right in the product data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While working on a large manufacturing plant, I came across a slew of 50-ton Trane Voyager rooftop package units. The high efficiency versions of these units have 5 stages, but do not provide any indication of traditional efficiencies that I am used to seeing throughout their literature.&lt;br&gt;&#xA;&lt;a href=&quot;https://www.trane.com/content/dam/Trane/Commercial/global/products-systems/equipment/unitary/rooftop-systems/voyager-commercial-27-1-2-to-50-tons/RT-PRC033L-EN_05112019.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.trane.com/content/dam/Trane/Commercial/global/products-systems/equipment/unitary/rooftop-systems/voyager-commercial-27-1-2-to-50-tons/RT-PRC033L-EN_05112019.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a straightforward way to go about calculating energy usage for equipment like this?&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2020-01-04T02:12:51.313" Title="Energy/Efficiency Calculations for Large Multi-stage Package HVAC" Tags="|hvac|cooling|energy-efficiency|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32636" PostTypeId="2" ParentId="32634" CreationDate="2020-01-04T02:29:54.287" Score="0" Body="&lt;p&gt;Your question is a bit confusing. If the ramp is resting on the pier at point B They usually relax the cable with zero force to allow for the ramp to move up and down a little to accommodate the gentle swell of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, theoretically, if the cable is in tension and there is a reaction at B as well this will be an indeterminate structure and one would need to consider section properties and stresses of the cable and the ramp beam as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the entire ramp would vibrate with a forced principal mode shape passing the two supports at C and A if the cable is loaded with tension. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-04T02:29:54.287" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32637" PostTypeId="2" ParentId="32609" CreationDate="2020-01-04T03:14:09.950" Score="3" Body="&lt;p&gt;I think the problem is just that you're expecting the solution to be more elegant; it looks like you already have it pretty much figured out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rOzlo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rOzlo.png&quot; alt=&quot;equation with terms dropped out&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2020-01-04T03:14:09.950" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32638" PostTypeId="2" ParentId="32615" CreationDate="2020-01-04T11:40:02.080" Score="3" Body="&lt;p&gt;Even if you can tune a function as per your needs, using proportional control(P) only you can never fully eliminate error from the system. Error is inherent in P control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using I control along with P, you can fully eliminate the error in the system, but now your system has a specific transient response. D control may be used if you want to control the transient response of the system, you want it faster or slower, how much overshoot you can manage to achieve a faster response, that depends on your system and need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of error in the system using only the P control depends on system parameters like the OpAmps you might be using. So if the error is tolerable, you need not add I and D. But if you need zero error at least I is needed, while D will let you vary the speed of the response&lt;/p&gt;&#xA;" OwnerUserId="18714" LastActivityDate="2020-01-04T11:40:02.080" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32639" PostTypeId="1" AcceptedAnswerId="32645" CreationDate="2020-01-04T12:02:26.363" Score="38" ViewCount="9379" Body="&lt;p&gt;I recently visited a railway museum with a lot of Soviet-era locomotives. The picture below shows the wheels on a locomotive from the 1930s or 1940s. Some of the wheels are thicker on one side (as it it were a partially filled cup). What is the purpose of that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JbStW.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JbStW.jpg&quot; alt=&quot;Picture of the wheels of a Soviet-era locomotive&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Now that I am looking at the picture again I wonder if they are balancing the weight of the coupling rods in some way.)&lt;/p&gt;&#xA;" OwnerUserId="24129" LastEditorUserId="6767" LastEditDate="2020-01-06T18:45:31.910" LastActivityDate="2020-01-07T23:50:49.097" Title="What is the purpose of these “partially filled in” locomotive wheels?" Tags="|rail|wheels|" AnswerCount="6" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32640" PostTypeId="2" ParentId="32639" CreationDate="2020-01-04T12:05:49.000" Score="31" Body="&lt;p&gt;Those are balance weights against the joints used for the rods linking the wheels together.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-01-04T12:05:49.000" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32641" PostTypeId="2" ParentId="32634" CreationDate="2020-01-04T15:55:08.840" Score="1" Body="&lt;p&gt;Imagine there is no rope - determinate.&#xA;Imagine there is no floor - determinate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you gradually increase tension in the rope the reaction force at B will gradually fall as the tension rises, until it reaches zero. If the tension rises further it will start to move. So, you see the reaction force is directly driven by the tension, provided the tension is with the range between zero and when the bridge starts to lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So yes, it's indeterminate in this interim state, but there is not any reason to find the static equilibrium forces in this state since the worst case forces are in the two determinate states that I described above.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2020-01-04T15:55:08.840" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32642" PostTypeId="2" ParentId="32639" CreationDate="2020-01-04T16:18:09.693" Score="-2" Body="&lt;p&gt;These act similar to the flywheel of an engine, keeping the steam engine running smoothly, especially at the start of rolling when there is not much rolling momentum in the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are positioned such that when the links are nearly expanded fully and there is a chance of wheel lock-up at the dead point, the gravity of the ballasts encourage the wheel to keep rolling and not lock up, or worse, back up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some comments/ answers, I think a few points need to be clarified.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If these were meant to be just ballast for the wheels and counter the weight of joints or links attached to them, they needed to be installed off the plane of the wheel by the same amount as those links. Because as they are they create a considerable imbalance on the wheel if we look at the wheel's cross-section. A rotating wheel with an attached off-plane mass will not be balanced by adding weight in its plane; it will tend to turn about an axis which passes through the slanted axis of symmetry, therefore wobble.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The fact that they are elongated circular segments, as opposed to just a cylindrical shape similar to the mass of the links, helps them have greater angular momentum, hence additional rotational energy (parallel axis angular momentum: the sum of spinning momentum and rotation of CG about an axis momentum)   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-01-05T05:39:10.827" LastActivityDate="2020-01-05T05:39:10.827" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32643" PostTypeId="5" CreationDate="2020-01-04T17:05:58.273" Score="0" Body="&lt;p&gt;Building Information Modeling (BIM) is a digital representation of physical and functional characteristics of a facility. It is used in modern building modeling software such as Autodesk Revit to provide extended information about equipment and other modeling components beyond simple dimensions.  BIM encompasses the dimensional fit of elements into drawings so that they can appear as they would in real life and provides additional meta-data such as:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Electrical requirements&lt;/li&gt;&#xA;&lt;li&gt;Network Requirements&lt;/li&gt;&#xA;&lt;li&gt;Manufacturer&lt;/li&gt;&#xA;&lt;li&gt;Maintenance information&lt;/li&gt;&#xA;&lt;li&gt;O&amp;amp;M manual links&lt;/li&gt;&#xA;&lt;li&gt;Physical connection requirements&lt;/li&gt;&#xA;&lt;li&gt;Flow rates&lt;/li&gt;&#xA;&lt;li&gt;Feed rates&lt;/li&gt;&#xA;&lt;li&gt;Lifespan&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="9770" LastEditorUserId="9770" LastEditDate="2020-01-04T21:14:31.217" LastActivityDate="2020-01-04T21:14:31.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32644" PostTypeId="4" CreationDate="2020-01-04T17:05:58.273" Score="0" Body="Building Information Modeling (BIM) is a digital representation of physical and functional characteristics of a facility." OwnerUserId="9770" LastEditorUserId="9770" LastEditDate="2020-01-04T19:53:08.347" LastActivityDate="2020-01-04T19:53:08.347" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32645" PostTypeId="2" ParentId="32639" CreationDate="2020-01-04T21:47:04.560" Score="55" Body="&lt;p&gt;Those are &lt;a href=&quot;https://en.m.wikipedia.org/wiki/Counterweight&quot; rel=&quot;noreferrer&quot;&gt;&lt;em&gt;counterweights&lt;/em&gt;&lt;/a&gt;. They work exactly the same as those lead counterbalance weights on the wheels of your automobile. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they left those out, then those connecting rods and bearings would create an out-of-balance condition, and the wheels would vibrate at higher speeds. That could very well damage the wheels. But as a couple of others have nicely pointed out, &lt;em&gt;the violent shaking could derail the train&lt;/em&gt;.&lt;/p&gt;&#xA;" OwnerUserId="20985" LastEditorUserId="20985" LastEditDate="2020-01-07T00:34:58.697" LastActivityDate="2020-01-07T00:34:58.697" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32646" PostTypeId="2" ParentId="4056" CreationDate="2020-01-05T00:17:23.743" Score="3" Body="&lt;p&gt;The wavelength of the light with the dissociation energy of epoxy phenyl oxygen is 300-310 nm. Over time, when exposed to this wavelength, it will become opaque and yellow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The solution is simple: add a layer which absorbs strongly the UVa (it has a high absorptivity in these wavelengths), like oxanilidies, and this will inhibit the opaque degradation.&lt;/p&gt;&#xA;" OwnerUserId="24132" LastEditorUserId="9" LastEditDate="2020-01-06T19:54:36.937" LastActivityDate="2020-01-06T19:54:36.937" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32647" PostTypeId="2" ParentId="29617" CreationDate="2020-01-05T02:54:49.760" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KNvuR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KNvuR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Solution&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;1)&lt;/em&gt; Upstream velocity is not given, but it goes into the drag force expression. Hence, free stream velocity is necessary to find: &lt;span class=&quot;math-container&quot;&gt;$$Re_L=\frac{\rho · U_{free stream} · L}{\mu}=10^7$$&lt;/span&gt; (Re at the end of the plate where x=L is given: &lt;span class=&quot;math-container&quot;&gt;$10^7$&lt;/span&gt;). From here free stream velocity can be found as: &lt;span class=&quot;math-container&quot;&gt;$$U_{free stream} = \frac{Re_L · \mu}{\rho · L}=\frac{10^7 · 1.8·10^{-5}\frac{kg}{m·s}}{1.2\frac{kg}{m^3} · 10m}=15\frac{m}{s}$$&lt;/span&gt;&#xA;&lt;em&gt;2)&lt;/em&gt; Just to get appreciation of the boundary layer structure (and to formally prove that we do in fact deal with a mixed laminar + turbulent flow), let's find &lt;span class=&quot;math-container&quot;&gt;$x_{crit}$&lt;/span&gt; - position along the plate where initially laminar boundary layer converts to a turbulent boundary layer. &lt;span class=&quot;math-container&quot;&gt;$Re_{crit}$&lt;/span&gt; is given, upstream velocity is found in the item 1, hence &lt;span class=&quot;math-container&quot;&gt;$x_{crit}$&lt;/span&gt; can be found from the expression for &lt;span class=&quot;math-container&quot;&gt;$Re_{crit}$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$$x_{crit} = \frac{Re_{crit} · \mu}{\rho · U_{free stream}}=\frac{5·10^5 · 1.8·10^{-5}\frac{kg}{m·s}}{1.2\frac{kg}{m^3} · 15\frac{m}{s}}=0.5m$$&lt;/span&gt;&#xA;One can see on the schematic that conversion occurs pretty close to the leading edge. Hence, the effect of initial laminar boundary layer can be neglected in general. The entire boundary layer can be assumed turbulent and expression for turbulent boundary layer (which is given in the statement of the problem) can be safely used.&#xA;Still, we shall proceed without this assumption.&lt;br&gt;&#xA;&lt;em&gt;3)&lt;/em&gt; Drag coefficient for the mixed laminar-turbulent boundary layer needs to be found now. &lt;strong&gt;If &lt;span class=&quot;math-container&quot;&gt;$Re_{crit}$&lt;/span&gt; is assumed to be equal &lt;span class=&quot;math-container&quot;&gt;$5·10^5$&lt;/span&gt; (which is given in the statement of the problem) and neglecting effects of the surface roughness, drag coefficient can be found as follows:&lt;/strong&gt; &lt;span class=&quot;math-container&quot;&gt;$$C_{f_{total}}=C_{f_{turb_{0-&amp;gt;L}}}-\frac{1750}{Re_L}$$&lt;/span&gt; Here: &lt;span class=&quot;math-container&quot;&gt;$C_{f_{total}}$&lt;/span&gt; - the drag coefficient of interest, &lt;span class=&quot;math-container&quot;&gt;$C_{f_{turb_{0-&amp;gt;L}}}$&lt;/span&gt; - drag coefficient as if the entire boundary layer were turbulent, &lt;span class=&quot;math-container&quot;&gt;$\frac{1750}{Re_L}$&lt;/span&gt; - amendment taking into account half a meter of the laminar boundary layer at the trailing edge. &lt;span class=&quot;math-container&quot;&gt;$C_{f_{turb_{0-&amp;gt;L}}}$&lt;/span&gt; can be found according to the expression given in the statement of the problem: &lt;span class=&quot;math-container&quot;&gt;$$C_{f_{turb_{0-&amp;gt;L}}}=\frac{0.074}{Re_L^{0.2}}$$&lt;/span&gt; Also, &lt;span class=&quot;math-container&quot;&gt;$Re_L$&lt;/span&gt; - &lt;em&gt;Re&lt;/em&gt; number at the end of the plate - is given in the statement of the problem. Overall, drag coefficient is equal to: &lt;span class=&quot;math-container&quot;&gt;$$C_{f_{total}}=\frac{0.074}{Re_L^{0.2}}-\frac{1750}{Re_L}=\frac{0.074}{(10^7)^{0.2}}-\frac{1750}{10^7}=0.00278$$&lt;/span&gt;&#xA;&lt;em&gt;4)&lt;/em&gt; Now, drag force acting on one side of the plate can be found as follows: &lt;span class=&quot;math-container&quot;&gt;$$F_D = C_{f_{total}}·(0.5·\rho · U_{free stream}^2·A_{plate})$$&lt;/span&gt; Here, &lt;span class=&quot;math-container&quot;&gt;$A_{plate} = L·b$&lt;/span&gt; is the plate's area, where &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; is the plate's length and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; is the plate's width. Therefore, drag force per unit width can be found as follows:  &lt;span class=&quot;math-container&quot;&gt;$$\frac{F_D}{b} = C_{f_{total}}·(0.5·\rho · U_{free stream}^2·L)=0.00278·0.5·1.2\frac{kg}{m^3}·(15\frac{m}{s})^2·10m=3.753\frac{N}{m}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Answer:&lt;/strong&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$3.75\frac{N}{m}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="21383" LastEditorUserId="21383" LastEditDate="2020-01-14T17:57:04.223" LastActivityDate="2020-01-14T17:57:04.223" CommentCount="16" ContentLicense="CC BY-SA 4.0" />
  <row Id="32648" PostTypeId="2" ParentId="32639" CreationDate="2020-01-05T03:23:43.270" Score="16" Body="&lt;p&gt;The oversized counterweights on the second set of wheels are to help balance the additional inertia of the pistons, piston rods and crossheads, which are directly connected to the wheels on this axle. (In a former life I was a museum conservator and rebuilt, and operated, steam traction engines and a locomotive.) ronford&lt;/p&gt;&#xA;" OwnerUserId="24133" LastActivityDate="2020-01-05T03:23:43.270" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32649" PostTypeId="1" AcceptedAnswerId="32653" CreationDate="2020-01-05T03:59:52.357" Score="1" ViewCount="817" Body="&lt;p&gt;I want to do a non-linear analysis (possibly SOL 400) on Nastran. I used PyNastran (python module) to set-up the BDF file for a Linear buckling analysis (sol 105). However I feel like it might be too difficult to set everything up correctly with non-linear analysis due to the little information I found. Tutorials on the subject all recommend a pre-processor like Patran, which I don't have. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know a free pre-processor that can actually produce the nastran input file (BDF)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know of salome-platform.org, but my understanding is that it cannot produce the BDF file. Or is there another solution to my problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help to solve my problem would be appreciated. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="24115" LastActivityDate="2020-01-05T15:18:45.130" Title="Free pre-processor for Nastran?" Tags="|structural-engineering|structural-analysis|applied-mechanics|finite-element-method|nastran|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32650" PostTypeId="1" CreationDate="2020-01-05T08:31:59.427" Score="0" ViewCount="170" Body="&lt;p&gt;Does 200lb/in of load on a S12x31.8 over 15 feet seem reasonable? I'm having a tough time trusting the numbers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two story house with an attic as below. Dotted lines show load bearing walls W1 and W2. There is currently a single wood 4x6 rough hewn girder running the entire 36 feet bowing 1-2.5 inches between lally columns spaced 8ft apart (see the red highlighted areas with greatest negative deflection (up to 3&quot; from level). I've started raising slowly (1/4&quot; every 4-6 days), and I'm thinking just replacing the wood girder with steel in the basement (unfinished). And using 24x24x18&quot; poured, rebar reinforced supports over packed calcareous sand which should support 9000-12000PSF. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I am guessing that center girder needs to be rated to 160-190 lb/in. For S12x31.8, my numbers are showing I can span up to 15 feet (180in) and still maintain an acceptable L/360 deflection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unit Load - q : 200 (lb/in)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total Load : 36000 (lb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Length of Beam - L : 180 (in)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment of Inertia - I : 218 (in4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modulus of Elasticity - E : 29000000 (psi)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance of extreme point off neutral axis - y : 6 (in)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Support Force - R1 : 18000 (lb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Support Force - R2 : 18000 (lb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum Stress - : 22294 (psi)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum Deflection - : 0.432 (in)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is attractive because I can use two supports and span the entire structure, perhaps using three separate sections. Does this seem right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm considering running three evenly spaced girders using three sections each, because I'm worried there needs to be more support for the tired floor joists. See the image below option 2. This would seem to make 120lb/in more than adequate and even single center load @ 10 tons looks like it would be acceptable (especially near W1, which would fall middle of the girder). I'm having a hard time believing these numbers. Does this seem right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total Load : 20000 (lb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Length of Beam - L : 144 (in)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moment of Inertia - I : 218 (in4)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modulus of Elasticity - E : 29000000 (psi)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance of extreme point off neutral axis - y : 6 (in)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Support Force - R1 : 10000 (lb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Support Force - R2 : 10000 (lb)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum Stress - : 19817 (psi)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum Deflection - : 0.197 (in)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NxWYN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NxWYN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yzGh6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yzGh6.png&quot; alt=&quot;Options for girder replacement&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24134" LastActivityDate="2020-01-05T21:53:09.773" Title="Maximum I-beam girder S12x31.8 span" Tags="|structural-engineering|civil-engineering|structures|beam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32651" PostTypeId="1" CreationDate="2020-01-05T13:38:23.750" Score="0" ViewCount="35" Body="&lt;p&gt;Whenever i see videos of two trucks pitting with each other to see who has a stronger, i always see comments that sounds like:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;If you load the other truck more it would beat the other one&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;That truck is an all wheel drive and the other is 4 wheel drive&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It got me thinking how to you calculate the optimum weight for a given torque of a four wheeled vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To make everything simple lets assume that out truck is a Tesla truck where each motor can generate torque equally on all the wheels and the wheels are evenly spaced. If the the motor on each wheel can lets say generate a 1000kg-cm of torque how would i find the optimal weight so that the truck can push to its maximum potential. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My follow up question would then be. What happens if the other truck has double the torque at half the weight which one will be able to push better? What would then be the formula for the pushing power? &lt;/p&gt;&#xA;" OwnerUserId="17493" LastEditorUserId="17493" LastEditDate="2020-01-05T17:35:24.843" LastActivityDate="2020-01-05T17:35:24.843" Title="Formula for finding the optimal weight for a given torque and vice versa" Tags="|mechanical-engineering|torque|friction|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32652" PostTypeId="2" ParentId="32242" CreationDate="2020-01-05T13:39:21.923" Score="0" Body="&lt;p&gt;1.) Find all the reaction forces first, then draw a SFD diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.) Express the shear force using &lt;strong&gt;z&lt;/strong&gt; as variable, you will get your f(z).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Then calculate the bending moment, then draw a BMD diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.) Integrate the shear force equation from 2.) into a bending moment equation f'(z) with a constant &lt;strong&gt;C&lt;/strong&gt; at the back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.) Identify the constant C in the bending moment equation with the maximum moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;5.) Integrate the bending moment equation from 4.) into a deflection equation f''(z) with a constant &lt;strong&gt;C&lt;/strong&gt; at the back.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;6.) Identify the constant C in the deflection equation, substitute back into f''(z).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you are done, a deflection equation f''(z) in terms of z.&lt;/p&gt;&#xA;" OwnerUserId="23239" LastActivityDate="2020-01-05T13:39:21.923" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32653" PostTypeId="2" ParentId="32649" CreationDate="2020-01-05T15:18:45.130" Score="2" Body="&lt;p&gt;As far as I know, there is no free preprocessor for Nastran.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the docs (&lt;a href=&quot;https://docs.salome-platform.org/5/smesh/user/importing_exporting_meshes_page.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://docs.salome-platform.org/5/smesh/user/importing_exporting_meshes_page.html&lt;/a&gt;) Salome will export to Nastran format, but it's just the mesh (nodes and elements). It won't do the case controls and all the nonlinear solution control parameters and all the other things in the deck&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want the full deck, I think your options are patran, hypermesh, NX (or write the deck yourself in a text editor... That's free but big learning curve) &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2020-01-05T15:18:45.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32654" PostTypeId="2" ParentId="32651" CreationDate="2020-01-05T15:21:58.537" Score="1" Body="&lt;p&gt;It is more question of the friction than torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The truck with the greater torque will be pushed back the moment its tires start to skid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque on a tire will be delivered to the pavement very effectively as long as static friction conditions are involved (no slipping). But in the case of slipping the kinetic friction is the active force and most of the work of the torque goes to heat and burning the rubber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In most of the tires on a paved dry road, the kinetic friction coefficient is around 0.3 to  o.4 and disastrously lower on wet roads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This implies a truck with a smaller torque if loaded say two times as much on each tire as the truck with higher torque will win.  &lt;span class=&quot;math-container&quot;&gt;$$ F_{friction}= weight*\mu $$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-05T15:21:58.537" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32655" PostTypeId="2" ParentId="32650" CreationDate="2020-01-05T18:21:48.923" Score="2" Body="&lt;p&gt;You need an engineer to calculate your project and submit it to the authorities for approval.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The beam calculations seem correct but the supports and connections need to be designed and may call for the structural upgrade of the whole house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concentrated loads in the middle of a beam have different code and a load factor of 2.8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before you do anything talk to the permit authority. They have a free code and planning advisory service.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-01-05T21:53:09.773" LastActivityDate="2020-01-05T21:53:09.773" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32656" PostTypeId="1" CreationDate="2020-01-05T18:47:49.187" Score="4" ViewCount="797" Body="&lt;h2&gt;The problem&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;The problem is to find the coefficients for the P, I and D terms of a PID controller used to regulate the object that minimize the the integral of the squared error (ISE):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Q = \int_0^\infty \epsilon^2(t) \ dt  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The object being controlled is described by the transfer function:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ G(s) = \frac{1}{s^3 + 6s^2 + 5s} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;My attempt&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;My idea was to find to find the expression for error in the s-domain, then perform an inverse Laplace transform to get the expression for the error in the time domain, then integrate the error from &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$\infty$&lt;/span&gt; and then find the values for the terms &lt;span class=&quot;math-container&quot;&gt;$K_1, K_2, K_3$&lt;/span&gt; that will minimize the function. I apologize for the mistakes if I've made any as my knowledge of control theory is very basic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I managed to do:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transfer function of a PID controller is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ K(s) = k_1 + \frac{k_2}{s} + k_3 s $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$H(s)$&lt;/span&gt; is the overall transfer function, &lt;span class=&quot;math-container&quot;&gt;$W(s)$&lt;/span&gt; is the input, let's say it's the unit step function, the expression for error in the s-domain is, if I'm correct:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E(s) = W(s) - H(s)W(s)$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$E(s) = W(s)[1 - H(s)]$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$E(S) = \frac{1}{s}[1 - H(s)]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The overall transfer function will be given by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ H(s) = \frac{K(s)G(s)}{1 + K(s)G(s)} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After substitution and simplification:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ H(s) = \frac{k_3 s^2 + k_1 s + k_2}{k_3 s^2 + k_1 s + k_2 + s^4 + 6 s^3 + 5 s^2} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, substituting for the expression for error:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$E(S) = \frac{1}{s}[1 - \frac{k_3 s^2 + k_1 s + k_2}{k_3 s^2 + k_1 s + k_2 + s^4 + 6 s^3 + 5 s^2}]$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used Matlab &lt;code&gt;ilaplace&lt;/code&gt; command to calculate the inverse Laplace transform of &lt;span class=&quot;math-container&quot;&gt;$E(s)$&lt;/span&gt; and got:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \epsilon(t) = \frac{\delta(t)}{t} - \frac{(exp[(- k_2 - t^2 k_3 - 5t^2 - 6t^3 - t^4)(t^2 + 6t + 5)]}{t} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know if I've arrived at the correct result, what seems weird to me is that the resulting expression for &lt;span class=&quot;math-container&quot;&gt;$\epsilon(t)$&lt;/span&gt; doesn't contain any &lt;span class=&quot;math-container&quot;&gt;$k_1$&lt;/span&gt; term. But assuming this is correct, I would finish by finding the values of parameters &lt;span class=&quot;math-container&quot;&gt;$k_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$k_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$k_3$&lt;/span&gt; that minimize this integral: &lt;span class=&quot;math-container&quot;&gt;$Q = \int_0^\infty \epsilon^2(t) \ dt  $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Question&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Now, I would be grateful if you could verify whether my approach and solution is correct and describe how you would find the values of the parameters that minimize this integral using Matlab i.e. what commands and how you would use them. I'm wondering if lack of closed-form solution to the integral would make finding the parameters impossible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or maybe, there is a simpler method of solving this problem, perhaps by avoiding computing the integral.&lt;/p&gt;&#xA;" OwnerUserId="24139" LastEditorUserId="24139" LastEditDate="2020-01-05T18:54:31.047" LastActivityDate="2021-10-07T14:03:40.520" Title="Finding coefficients for a PID controller that minimize the integral of the squared error (ISE)" Tags="|control-engineering|control-theory|pid-control|matlab|transfer-function|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32659" PostTypeId="1" CreationDate="2020-01-06T00:19:43.480" Score="1" ViewCount="75" Body="&lt;p&gt;the cold line from evaporator to compressor will gain tons of heat on a super hot day even with insulation. if they put the compressor right outside the evaporator, both of the long lines will be hot and there will be a bit of free cooling. why don't they do something so obvious? there must be a very good reason right?&lt;/p&gt;&#xA;" OwnerUserId="24144" LastActivityDate="2020-01-06T22:25:22.947" Title="why not put the compressor right next to the evaporator?" Tags="|compressors|heat-exchanger|evaporation|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32660" PostTypeId="2" ParentId="26117" CreationDate="2020-01-06T04:18:14.893" Score="1" Body="&lt;p&gt;As Phil said in the comments, the short answer is yes regarding a two-stage system. As an added bonus, during the rest of the year besides the really hot time, the system will probably also keep the house more consistent since it will be able to operate at a lower capacity rather than cycling off. &#xA;At 11 years old, however, unless you never change filters or live right on a large body of saltwater, I would hope your compressor is not already on the verge of dying off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a side note to some of the other details in the message, you mention the attic insulation as R-32. If that was blown in insulation, sometimes the thickness can be misleading if it seemed too much or if it is fiberglass batting, moisture will compromise it. In addition to the insulation on your walls, you should also look closely at your ductwork. If it is insulated running through a hot attic, your 3 ton AC might only be delivering 2-2.5 tons to the registers. &lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2020-01-06T04:18:14.893" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32661" PostTypeId="2" ParentId="32659" CreationDate="2020-01-06T06:05:11.023" Score="3" Body="&lt;p&gt;The evaporator is usually in the area to be cooled ie where the people work, read etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the compressor takes space and tends to be noisy it is located away from that area. That makes the system in the « active » area smaller and quieter.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-01-06T06:05:11.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32662" PostTypeId="2" ParentId="32639" CreationDate="2020-01-06T08:03:08.430" Score="7" Body="&lt;p&gt;These counterweights are found on all piston-driven steam locomotives (not just Soviet designs), and even a few types of diesel locomotive using connecting-rod drive.  As other answers note, their purpose is to balance the weight and momentum of the connecting and drive rods, which would otherwise cause heavy vibration when running at speed, to the extent of lifting the wheels off the track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a diesel example; the weights are relatively small here, but they can be seen on all three wheels to the lower left of the axle.  On this shunting locomotive, the traction motors are connected to the front and rear axles, while the centre axle is driven through the connecting rods which also prevent any single axle from slipping under load.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sjNvU.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sjNvU.jpg&quot; alt=&quot;BR Class 08 shunter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On this more elegant steam loco, designed for express passenger service, there are both strengthening webs on the crank side of the wheel, and balance counterweights on the opposite side.  There's actually a second pair of cylinders between the frames, driving the leading axle, so the counterweighting is different for all three pair of driving wheels.  Part of the counterweighting for the leading axle is built into the inner crank, rather than having it all on the wheels.  Here you can also see that the carrying wheels on the leading bogie are not counterweighted, because they don't have any cranks.&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aLVmi.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aLVmi.jpg&quot; alt=&quot;GWR Castle&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24148" LastActivityDate="2020-01-06T08:03:08.430" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32663" PostTypeId="1" CreationDate="2020-01-06T08:42:44.213" Score="0" ViewCount="1900" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yvjDL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yvjDL.png&quot; alt=&quot;Formula for cantilever beam with distributed equal force&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hey, I make a project in which I use a cantilever beam with distributed equal load across the whole beam and I'm using the formulas above to calculate the displacement, moment... &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My specifications for the beam are:&#xA;4.5m long aluminum 2024 beam, 4x3 cm walls (the force is applied on the 4cm side), 2mm thickness. and a young's modulus of 70 GPa. a force of 40 N/m is applied&lt;/p&gt;&#xA;&#xA;&lt;p&gt;from my calculations the moment of inertia is 37 mm^4, meaning the maximum displacement should be 0.6m at X=4.5 and the maximum moment is 400 N-m at X=0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I put the results in the formula for bending stress, the maximum bending stress is 162 MPa, while the yield stress for this type of aluminum is around 300 MPa.&#xA; &lt;a href=&quot;https://i.stack.imgur.com/36KXr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/36KXr.png&quot; alt=&quot;Bending stress formula&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;something here doesn't seems right for me. the beam is bent 0.6m but does not exceed the yield point, maybe I did a mistake somewhere?&lt;/p&gt;&#xA;" OwnerUserId="24086" LastActivityDate="2020-01-06T18:07:31.317" Title="Problem with cantilever beam and yield strength" Tags="|structural-engineering|beam|yield-point|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32664" PostTypeId="1" AcceptedAnswerId="32695" CreationDate="2020-01-06T12:21:20.040" Score="5" ViewCount="483" Body="&lt;p&gt;Suppose I have a cogeneration plant that runs on natural gas and produces heat and electricity. The heat effeicieny is &lt;span class=&quot;math-container&quot;&gt;$\eta_{th} = 0.4$&lt;/span&gt; and the electricity efficiency is &lt;span class=&quot;math-container&quot;&gt;$\eta_{el} = 0.4$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The burning of natural gas produces 215 g/kWh of CO2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the calculation of the CO2 emission of the generated electricity from the cogeneration plant, one could now calculate &lt;span class=&quot;math-container&quot;&gt;$\frac{215}{0.4} = 537.5$&lt;/span&gt; g/kWh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I would like to calculate the CO2 emissions for the electricity generation of the cogeneration plant, taking into account that CO2 emissions are saved because another heat generator has to generate less heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A natural gas boiler with an efficiency of &lt;span class=&quot;math-container&quot;&gt;$\eta_{th} = 0.9$&lt;/span&gt; is taken as a reference for heat generation. So the 1 kWh of heat is generated with &lt;span class=&quot;math-container&quot;&gt;$\frac{215}{0.9} = 238.88$&lt;/span&gt; g of CO2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the CO2 emissions for the electricity generation in this case?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;My calculation attempt:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First I would calculate how much CO2 emssions the boiler has for 0.4 kWh:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$238.88 \cdot 0.4 = 95.55$&lt;/span&gt; g of CO2&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I would subtract this value from the previously calculated co2 emission of the electricity:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$537.5 - 95.55 = 441.95$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The electricity generated by the cogeneration plant would then have a CO2 emission of 441.95 g/kWh.&lt;/p&gt;&#xA;" OwnerUserId="23627" LastEditorUserId="23627" LastEditDate="2020-01-07T13:11:55.177" LastActivityDate="2020-01-08T08:26:03.380" Title="Calculation of the specific CO2 emission of a cogeneration plant for the produced electricity" Tags="|electrical-engineering|energy|heating-systems|energy-efficiency|power-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32667" PostTypeId="1" AcceptedAnswerId="32670" CreationDate="2020-01-06T15:33:44.703" Score="2" ViewCount="1035" Body="&lt;p&gt;Recently I have seen a question as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimension of the hole is &lt;span class=&quot;math-container&quot;&gt;$50 +.02/-.00$&lt;/span&gt; mm and shaft is &lt;span class=&quot;math-container&quot;&gt;$50 +.02/-.00$&lt;/span&gt; mm.The minimum clearance is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is .00 but the correct answer as per the website is -.02&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where am I going wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me.&lt;/p&gt;&#xA;" OwnerUserId="24160" LastActivityDate="2020-01-06T18:09:02.393" Title="Difference between min/max clearance and min/max tolerance and relation between them" Tags="|tolerance|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32668" PostTypeId="2" ParentId="26117" CreationDate="2020-01-06T15:45:08.377" Score="1" Body="&lt;p&gt;The typical solution in Houston is to have two AC units for a two story house. A lower cost , efficient way to cool the upper rooms is with window units , if permitted in your location. They are relatively efficient and cheap . I have a $ 100 unit that easily cools a 13 X 13 room room addition.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-01-06T15:45:08.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32670" PostTypeId="2" ParentId="32667" CreationDate="2020-01-06T18:09:02.393" Score="7" Body="&lt;p&gt;The minimum clearance occurs when the largest possible shaft meets the smallest possible hole. That would be in this case a shaft of &lt;span class=&quot;math-container&quot;&gt;$50.02 mm$&lt;/span&gt; in a hole of &lt;span class=&quot;math-container&quot;&gt;$50.00 mm$&lt;/span&gt;. The difference between the two is the &lt;span class=&quot;math-container&quot;&gt;$-0.02 mm$&lt;/span&gt; from the solution you found.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2020-01-06T18:09:02.393" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32671" PostTypeId="2" ParentId="32458" CreationDate="2020-01-06T21:47:06.710" Score="1" Body="&lt;p&gt;Who owns the land the tracks are running on?  In North America, I would refer to the land as the Rail Operator's Right of Way.   I would hazard a guess that most rail companies want to just focus on the business of running trains and for the most part are not that interested in real estate that does not pertain to them.  Building a building over tracks will usually involve some sort of disruption to rail operations.  When material falls off buildings (eg falling concrete) its a potential hazard for track operation.  When it comes time to replace the building, demolition plans will most likely disrupt rail service again.  While admittedly deals can be struct with the Right of Way Owners that would generated another revenue stream for them, it may not always be in their interest as it takes focus away from running trains.&lt;/p&gt;&#xA;" OwnerUserId="6993" LastActivityDate="2020-01-06T21:47:06.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32672" PostTypeId="2" ParentId="32659" CreationDate="2020-01-06T22:25:22.947" Score="2" Body="&lt;p&gt;What you describe is quite common where there isn't any problem with power and noise. It requires a bit more sophistication to manage the superheat. And yes, you need to quiet the thing some. I replaced a '60s era 20 ton compressor in a restaurant's dining room AC closet while they were serving dinner. This is also a good idea near the coast to keep the salt off the thing and protect the expensive compressor from hurricanes. I've only seen it on bigger systems, though.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2020-01-06T22:25:22.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32674" PostTypeId="1" AcceptedAnswerId="32746" CreationDate="2020-01-07T01:59:22.180" Score="4" ViewCount="953" Body="&lt;p&gt;I am trying to calculate the maximum braking force/deceleration for a bicycle.&#xA;Note: the question assumes &lt;strong&gt;only the front brake is used&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Free body diagram:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qlNjm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qlNjm.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/pYLXE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pYLXE.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Typo:&lt;/strong&gt; in the image above, &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$F_{a,\ max(dry)} = g X_3 / h$&lt;/span&gt;&amp;quot; is wrong, since it is an acceleration and it should be written as &amp;quot;&lt;span class=&quot;math-container&quot;&gt;$a_{max(dry)} = g X_3 / h$&lt;/span&gt;&amp;quot;&lt;/p&gt;&#xA;&lt;p&gt;For dry tarmac where I know there is enough grip to get the rear wheel to lift I managed to calculate the max deceleration (result=6.8m/s^2). (this is the deceleration where all weight is transferred onto the front wheel (rear wheel will start lifting up at higher deceleration)).&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;From the moment equilibrium equation (using the front tire contact patch as the origin) I got the following equation for maximum deceleration before the rear tire starts lifting off (assuming there is enough grip):&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1cgVz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1cgVz.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;I used this to calculate max deceleration in dry conditions&lt;/p&gt;&#xA;&lt;p&gt;However I am now trying to calculate the max deceleration in wet weather (lower coefficient of friction) where the front wheel slipping should be considered as an option.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;I tried expressing Nf (normal force on front wheel contact patch) in terms of a (deceleration) and got the following result:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5y8ht.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5y8ht.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;Which when simplified results in:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fnwFD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fnwFD.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;br /&gt;&#xA;With this formula I can calculate Nf&lt;/p&gt;&#xA;&lt;p&gt;Assuming that there won't be enough grip (in the wet) to make the rear wheel lift up.. I can then use Nf to determine Fbr (braking force).&#xA;To determine maximum deceleration (assuming wheel slip occurs before rear lift up) I can use the equilibrium of forces in x-direction:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;ΣFx=0=Fa-Fbr =m&lt;em&gt;a-μ&lt;/em&gt;Nf --&amp;gt; a=μ*Nf/m&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Are my assumptions/calculations correct? If so: is there a better/faster way of doing this?&#xA;To check if the rear wheel lifts up or if the tire slips first I will compare the max deceleration for slipping tire scenario and max deceleration for lifting rear wheel scenario and see which deceleration has a lower deceleration value to see which scenario will happen first&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;When braking some weight is transferred from the rear to the front wheel, but if the deceleration is too high the front wheel will slip.&#xA;Gently increasing brake pressure will lessen the chance of front wheel slipping...&lt;/p&gt;&#xA;&lt;p&gt;How should I go about determining max deceleration without the front wheel slipping, given a certain friction coefficient μ ?&#xA;I can calculate the max braking force without slipping for a given Normal force on the front wheel but &lt;em&gt;&lt;strong&gt;as soon as you start braking this normal force changes due to weight redistribution&lt;/strong&gt;&lt;/em&gt; and I don't know how to account for this.&lt;/p&gt;&#xA;&lt;p&gt;Thank you very much for any input! It is greatly appreciated&lt;/p&gt;&#xA;" OwnerUserId="10328" LastEditorUserId="4544" LastEditDate="2021-06-30T15:54:45.943" LastActivityDate="2021-06-30T15:54:45.943" Title="Calculate max deceleration without front wheel slipping (bicycle)" Tags="|statics|dynamics|friction|bicycles|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32676" PostTypeId="1" CreationDate="2020-01-07T05:08:07.233" Score="-1" ViewCount="128" Body="&lt;p&gt;Recently I have come across a question as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the eye tip of a centrifugal impeller, blade velocity is 200 m/s while the uniform axial velocity at the inlet is 150 m/s. If the sonic velocity is 300 m/s, then the inlet Mach number of flow will be?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My answer is &lt;span class=&quot;math-container&quot;&gt;$\frac{150}{300} = .5$&lt;/span&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;but the correct answer is &lt;span class=&quot;math-container&quot;&gt;$0.83$&lt;/span&gt;, with an explanation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$M = \frac{\sqrt{(blade\ velocity)^2 + (axial\ velocity)^2}}{sonic velocity} = 0.83$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: when we are considering the inlet mach why are we considering the blade velocity also why not just inlet flow velocity?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me, Thanks.&lt;/p&gt;&#xA;" OwnerUserId="24160" LastActivityDate="2020-01-07T07:15:25.957" Title="Inlet Mach No for a centrifugal pump?" Tags="|pumps|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32677" PostTypeId="2" ParentId="32676" CreationDate="2020-01-07T07:15:25.957" Score="1" Body="&lt;p&gt;There is a velocity triangle relating the velocities at the blade - this changes as the shape of the blade changes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both components of the fluid have to be taken into account: the velocity of the fluid at the inlet and the axial velicity to get the outlet velocity.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-01-07T07:15:25.957" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32678" PostTypeId="1" CreationDate="2020-01-07T08:27:30.570" Score="1" ViewCount="40" Body="&lt;p&gt;So I am involved in an introductory course in mathematics for pilots. I have the following problem I would like to give them, as it can be a practical application of integration&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rUKNF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rUKNF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;However, does there exists -- however how rudimentary -- real life example where such a process naturally occur? I was thinking about Reciprocating engine / piston engine for example Wright R-3350 Duplex-Cyclone which was used during the second world war. However, I am not sure that the process that occurs in the piston is actually an isoterm process. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;em&gt;So my question is does there exists a real life example of an engine / piston / motor (preferably related to aviation) which can relate to the problem above?&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="24177" LastActivityDate="2020-01-07T09:45:34.720" Title="Real life example of motor/piston involving an isoterm process?" Tags="|motors|compressed-gases|pistons|aviation|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32679" PostTypeId="1" AcceptedAnswerId="32680" CreationDate="2020-01-07T08:33:41.830" Score="2" ViewCount="715" Body="&lt;p&gt;Is it possible to model a heat pump cycle without using tools like EES, but using Excel? I am considering to use Refprop application as an add-in in Excel, and model a heat pump cycle. Another alternative would be to create a P-h database for various fluids, and go from very complicated series of equations to find pressure, temperature, and density values.&lt;/p&gt;&#xA;" OwnerUserId="24176" LastActivityDate="2020-01-07T08:50:50.380" Title="Modelling a refrigeration cycle in Excel" Tags="|thermodynamics|modeling|refrigeration|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32680" PostTypeId="2" ParentId="32679" CreationDate="2020-01-07T08:50:50.380" Score="2" Body="&lt;p&gt;You can do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I set up a sheet using vlookup to interpolate through fluid characteristics and because the source data I used (from steam tables) had a 5 degree increment I had to build an interpolation function using several vlookups / I will let you do that as it is good experience.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-01-07T08:50:50.380" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32681" PostTypeId="2" ParentId="32678" CreationDate="2020-01-07T09:45:34.720" Score="1" Body="&lt;p&gt;We assume isothermal as having to include all the heat gains and losses to the gas during the cycle would be too onerous and add little to the accuracy of the result.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-01-07T09:45:34.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32683" PostTypeId="2" ParentId="28732" CreationDate="2020-01-07T16:58:29.863" Score="1" Body="&lt;p&gt;Following up on Sam's categories of pumps -&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;There are industry examples of your first category - rotating swash plate + stationary barrel pumps. Here is one animated example of a pump that I've used personally - &lt;a href=&quot;https://www.youtube.com/watch?v=sP3gc4b8Z5k&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=sP3gc4b8Z5k&lt;/a&gt;. The relevant part of the video starts at 0:48. Pardon the marketing material. The pump's barrel &lt;em&gt;must&lt;/em&gt; remain stationary because each piston chamber is connected directly to an outlet check valve. As they show at 1:30 in the video, the concept of a stationary barrel allows for separate flows from a specialized cover plate, which isolates 1 or more piston chambers from the others. A similar &quot;isolation&quot; concept is used by Bosch-Rexroth, Hydac, and HAWE for multi-chambered radial-piston pumps.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;This is the most common variety of axial piston pump. The barrel / housing (valve plate) clearance does indeed allow leakage for pump cooling as you describe. Note that this is only 1 of 4 typical leakage sources in a rotating-barrel pump... &#xA;&lt;a href=&quot;https://i.stack.imgur.com/ArIHk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ArIHk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(&lt;span class=&quot;math-container&quot;&gt;$Q_{lp}$&lt;/span&gt;) piston / bore clearances, (&lt;span class=&quot;math-container&quot;&gt;$Q_{ls}$&lt;/span&gt;) slipper / swashplate clearances, (&lt;span class=&quot;math-container&quot;&gt;$Q_{lv}$&lt;/span&gt;) barrel / valve plate clearance, and (not pictured) swash plate control system return oil. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Good description on the bent-axis category. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Overall your post categorizes axial-piston pumps correctly into the 3 general groups. However some of the descriptions are only true in specific situations. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;For pump category 1, you mention 120C as an upper operational limit. 120C is an extreme upper limiting point for petroleum-based hydraulic oil used in &lt;em&gt;any&lt;/em&gt; type of axial piston pump. However there is no single number for temperature limit. It completely depends on the working fluid used with any specific pump design. Each fluid has a unique relationship between temperature and relevant fluid properties - viscosity, density, compressibility. Also, each pump may have interfaces with unique clearances optimized for a particular fluid. Pump failure often results from viscosity falling below some functional limit (1.0-10 cSt) due to temperature rise. Also pump failure can be caused by temperature alone in the case of elastomer seal degradation. Lastly note that bent-axis type pumps (category 3) may survive longest at elevated temperatures because they are designed without piston slippers for axial load support. Less sliding components that can fail.. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Category 1 is not inherently simpler or more efficient than the other categories. It's true that the barrel / valveplate sliding interface is eliminated in Category 1. However multiple new valves are introduced for periodic sealing of each piston chamber. Additional bearings are needed to support the side-load on the rotating swashplate. Also &lt;em&gt;far&lt;/em&gt; more expense is involved to achieve variable-displacement operation. Lastly the weight of the rotating group has negligible influence on power consumption. Component weight only affects the accel/decel time needed to change speeds. Steady-state power consumption is a function of viscous fluid shear on surfaces, along with any power diverted for displacement control purposes. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="24104" LastActivityDate="2020-01-07T16:58:29.863" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32684" PostTypeId="2" ParentId="32639" CreationDate="2020-01-07T17:59:12.517" Score="2" Body="&lt;p&gt;Slightly different answer - aside from the hammer effect of having out of balance connecting rods, there is also a different problem set. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the left and right hand rods are at connected at 90 degrees wrt each other (deliberately, as this is how you avoid the dead center starting problems referred to above), what happens is that as they rotate, the whole engine starts to sway from side to side due to the unbalanced rotating load. If you happen to hit a resonant frequency with the rails, the engine (and train) can derail entirely. The faster you go, and the more driving wheels that the engine has, the worse the problem. 2-10-4 (&quot;Texas&quot;class) locomotives were notorious for having this problem, for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, having one side 'pushing' whilst the other is 'pulling' can cause the axle boxes to work loose and this will severely increase wear on the axle boxes, wheels and brakes etc and in extreme cases, it can cause the loco chassis to fail (bend or even break) due to repeated flexing. Since both Berkshire (2-8-4) and Texas locomotives were designed to pull heavy freight loads, this can also cause the loco to 'fishtail' at any speed, which causes (different) problems again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rick&lt;/p&gt;&#xA;" OwnerUserId="24185" LastActivityDate="2020-01-07T17:59:12.517" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32685" PostTypeId="1" AcceptedAnswerId="32688" CreationDate="2020-01-07T18:34:09.180" Score="2" ViewCount="180" Body="&lt;p&gt;Is there any meaning of using Kalman Filter for cases when you do not have good statistical model of the system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you have a drone and it has IMU sensor and GPS sensor. You do not have any statistical information, like variance, covariance, etc. If you want to estimate your current position, is there any advantage of using Kalman filter over just averaging the two sensors? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the above problem, if in practice the Kalman filter has some advantage, what are the reasons behind it that I did not consider in my above example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="21221" LastEditorUserId="6790" LastEditDate="2020-05-04T08:02:26.713" LastActivityDate="2020-05-04T08:02:26.713" Title="Kalman filter for sensor fusion — what is the advantage?" Tags="|sensors|robotics|kalman-filters|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32686" PostTypeId="1" AcceptedAnswerId="32757" CreationDate="2020-01-07T18:44:24.737" Score="3" ViewCount="4221" Body="&lt;p&gt;I have received a CO2 cylinder from an online seller fitted with a protective cap over the valve:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/COqRM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/COqRM.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand these should unscrew, but this one appears to be stuck very firmly.  Is there a standard or recommended way to remove these safely?  The biggest challenge seems to be securing the cylinder itself to stop it slipping while attempting to turn the cap with a strap wrench or similar.  Any tips?&lt;/p&gt;&#xA;" OwnerUserId="10132" LastActivityDate="2022-05-28T06:37:31.300" Title="How to remove protective valve cap from CO2 cylinder?" Tags="|mechanical-engineering|safety|pressure-vessel|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32687" PostTypeId="2" ParentId="32686" CreationDate="2020-01-07T18:47:39.453" Score="4" Body="&lt;p&gt;Use two strap wrenches or an equivalent - one on the tank and one on the cap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You won't crush the tank :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you need to make one, then a length of 4*2 and some old car seat belt bolted to one end works well enough - it did when I had to do something similar.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2020-01-07T19:41:44.010" LastActivityDate="2020-01-07T19:41:44.010" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32688" PostTypeId="2" ParentId="32685" CreationDate="2020-01-07T19:38:31.797" Score="3" Body="&lt;p&gt;&lt;strong&gt;You can't just &quot;average the two sensors&quot;&lt;/strong&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acceleration readings are essentially meaningless unless you know their direction, you can't know their direction unless you have the platform orientation, and you can't know the platform orientation from the IMU (unless it has a &lt;em&gt;reliable&lt;/em&gt; magnetometer).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the IMU-alone case, your system basically has the information that you'd have if you were stuck inside of Einstein's elevator.  The system knows -- with errors -- how rapidly its velocity (not position) is changing, and how rapidly -- with errors -- it is turning.  That's &lt;em&gt;all&lt;/em&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best that you can do is to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Maintain a 3D orientation estimate by integrating the gyro readings -- basically this gives you a rotation from your vehicle frame to North East and Down, or whatever your preferred &quot;global&quot; reference frame is.&lt;/li&gt;&#xA;&lt;li&gt;Maintain a 3D velocity estimate in your global reference frame by rotating the accelerometer readings into the global frame with the orientation estimate, and then integrating.&lt;/li&gt;&#xA;&lt;li&gt;To complicate things, you need to know your local acceleration due to gravity and you need to correct for it in the vertical direction -- so any errors in your estimate of &quot;down&quot; translate immediately into an offset in the effective acceleration reading in the North or East directions.&lt;/li&gt;&#xA;&lt;li&gt;Maintain a 3D position estimate in your global reference frame by integrating the velocity readings.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;You can say at this point &quot;Yes, but I'm going to keep everything in the vehicle frame&quot; -- fine, but life will be just as complicated and inaccurate.  You're not going to gain any information by doing so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have a 9-state estimate (rotation, velocity, position, each in 3D), with a rather nasty nonlinear relationship between rotation and the rest.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that any error in starting state, plus noise, drift or offset in the IMU readings, gets integrated multiple times.  It's pretty much twice for accelerometer readings, but rather complicated for gyro readings because the orientation estimate is used to rotate the accelerometer readings.  It is very difficult to do.  In general, with typical consumer-grade IMUs, the state estimate will be utterly useless within seconds, &lt;strong&gt;if&lt;/strong&gt; it starts out perfect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There actually &lt;em&gt;are&lt;/em&gt; (or at least were) purely inertial navigation systems in the 1960's and 1970's that used this technique -- but they needed hugely expensive IMU hardware, they needed very careful pre-launch calibration, they needed maps of the local graviometric variations for their route, and even the ones in strategic bombers and submarines were backed up by people (or machines, in the case of nukes) taking celestial readings (star or sun shots with a sextant).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For typical consumer-grade IMU hardware, with perfect starting calibration, you can free-run for maybe a second or so before the error gets into the meters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Need for a statistical model&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the specific case of a drone (or anything else) with an IMU and GPS, you can side-step the need for a statistical model of the vehicle, or at least you can make a model without prior knowledge of the vehicle's motion.  I've done this, and it works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, you treat the IMU output as the system input (just treat the system as having mass subject to accelerations and rotations), then you compare the system's position to the GPS readings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your model is &lt;span class=&quot;math-container&quot;&gt;$\dot x = f \left ( x, \hat u \right)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$y =  x_{position}$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$\hat u(t)$&lt;/span&gt; is the output of the IMU.  (Note that this is the state-space representation for a system that is nonlinear in the state evolution, but linear in the output).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only statistical variation you need to account for is the IMU and GPS error.  For the IMU you let &lt;span class=&quot;math-container&quot;&gt;$\hat u(t) = u(t) + w(t)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$w(t)$&lt;/span&gt; is your expected IMU noise, and &lt;span class=&quot;math-container&quot;&gt;$u(t)$&lt;/span&gt; is the actual body motion.  Similarly, for the GPS you let &lt;span class=&quot;math-container&quot;&gt;$\hat y_{position} = y_{position} + r(t)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$y_{position}$&lt;/span&gt; is your actual position, &lt;span class=&quot;math-container&quot;&gt;$\hat y_{position}$&lt;/span&gt; is your actual GPS reading, and &lt;span class=&quot;math-container&quot;&gt;$r(t)$&lt;/span&gt; is your expected GPS error.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So there is a &quot;statistical model&quot;, but it's one that you can gain from a knowledge of the properties of the IMU and of GPS.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Orientations from the fusion&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because you're keeping track of everything with that covariant matrix, the Kalman filter will sort out varying accelerations into the platform orientation, position, and velocity.  It doesn't actually have to know anything about the platform dynamics, because those are being directly read by the IMU.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can reason this out intuitively.  When you add GPS into the mix, the system is now in the same situation as you would be if you were stuck inside of Einstein's elevator &lt;em&gt;but with a position readout&lt;/em&gt;.  If you're in Einstein's elevator &lt;strong&gt;and you know that you are not moving relative to the earth&lt;/strong&gt; then you know that the &quot;down&quot; that your accelerometer is telling you is also &quot;true down&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you know which way is up.  Let's say that you're watching your position readout, your gyro, and your accelerometer.  You notice some tugs on the elevator -- your gyro says you haven't moved, the accelerometer vector has tilted toward the front wall and then the back, and the GPS is now reporting that you are some distance east of where you started.  &lt;strong&gt;Now&lt;/strong&gt; because you know that you moved straight east &lt;strong&gt;and&lt;/strong&gt; that the sideways acceleration was in the direction of the front of the elevator.  So you can face the front of the elevator, say &quot;I am pointing to the East&quot;, and you have successfully oriented yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is what the Kalman filter is doing with it's seemingly inadequate statistical model -- if you get &lt;span class=&quot;math-container&quot;&gt;$f(x, u)$&lt;/span&gt; right, because of the way the covariant matrix ends up evolving, just sitting still will rapidly let the Kalman know which way is down (because it's not moving, yet the accelerometer is reporting acceleration &quot;upward&quot;).  Then, if the Kalman sees a changing lateral acceleration coupled with motion in the GPS reporting, it'll deduce lateral direction.  Moreover, this will continue to happen -- the combination of GPS and acceleration will &lt;em&gt;always&lt;/em&gt; give information about orientation along the line of acceleration, so if the acceleration direction continually changes, the Kalman filter will know orientation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's probably refinements that could be had by actually modeling the platform dynamics and using the control inputs to the platform to increase the information flowing into the filter -- but I've done it the way I describe above, and it worked great (after considerable work).&lt;/p&gt;&#xA;" OwnerUserId="19776" LastEditorUserId="19776" LastEditDate="2020-01-12T17:19:38.107" LastActivityDate="2020-01-12T17:19:38.107" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="32691" PostTypeId="1" CreationDate="2020-01-08T01:22:32.830" Score="1" ViewCount="35" Body="&lt;p&gt;I’m attempting to make a model for medical professionals to practice ultrasound guided procedures. What substances or materials would be adequate to project a picture similar to human tissue? I have a few ideas but want to pick your brains. Thank you&lt;/p&gt;&#xA;" OwnerUserId="24196" LastActivityDate="2020-01-08T06:26:37.133" Title="What type of material can I use to make a model used for medical ultrasound practice?" Tags="|material-science|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32692" PostTypeId="2" ParentId="26844" CreationDate="2020-01-08T04:44:16.880" Score="1" Body="&lt;p&gt;It is the Kelvin-Planck Statement which gives your answer:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;&lt;em&gt;It is impossible to construct a device which operates on a cycle produces work continuously while interacting with a single Thermal Energy Reservoir&lt;/em&gt;&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;or in lay man terms: Heat being a low grade energy, we cannot completely convert it into work and some heat has to be rejected.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now have a look at the following pic. you should get a fair understanding of your question:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/O7F0m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O7F0m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24160" LastActivityDate="2020-01-08T04:44:16.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32693" PostTypeId="2" ParentId="32691" CreationDate="2020-01-08T06:26:37.133" Score="1" Body="&lt;p&gt;There is a substance called &lt;em&gt;ballistic gel&lt;/em&gt; that is used by bullet manufacturers to shoot bullets into, to test their behavior when passing through flesh. Since it is designed to mimic the bodies of large mammals, perhaps it would work for your purposes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;it is sold as a dry powder to which water is added, after which it sets up like very stiff jell-o. By adding lumps of stuff like barium sulfate to it, you can vary its density locally and thereby get ultrasound reflections off it. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-01-08T06:26:37.133" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32694" PostTypeId="1" CreationDate="2020-01-08T06:53:03.237" Score="1" ViewCount="129" Body="&lt;p&gt;It seems that use of the Robertson charts (Robertson, P.K. [1990] &quot; Soil Classification using the cone penetration test&quot; Canadian Geotechnical Journal 27(1), 141-8) to classify soil type based on CPT results is widely accepted practice in the Netherlands. I cannot, however find any prescriptions in the current Eurocode which explicitly allows this. I have checked all relevant standards:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;EN 1997.1&lt;/li&gt;&#xA;&lt;li&gt;EN 1997.2&lt;/li&gt;&#xA;&lt;li&gt;NEN-EN ISO 22476-1&lt;/li&gt;&#xA;&lt;li&gt;ISO 14688-1 and ISO 14688-2&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some of these standards are perhaps not the most recent and others do not contain the national annexes, so perhaps that is my problem. However, I cannot find a clear stipulation which allows or prescribes use of the above charts to actually determine soil &lt;em&gt;type&lt;/em&gt; based on CPT results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could anyone indicate why this is allowed? Am I missing something in the standard? If it is common practice how does it not contradict the standard?&lt;/p&gt;&#xA;" OwnerUserId="4808" LastEditorUserId="90" LastEditDate="2022-08-03T00:19:13.333" LastActivityDate="2023-12-26T02:01:50.197" Title="Does Eurocode allow use of Robertson chart to determine soil type based on CPT results?" Tags="|geotechnical-engineering|soil|eurocodes|soil-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32695" PostTypeId="2" ParentId="32664" CreationDate="2020-01-08T08:26:03.380" Score="2" Body="&lt;p&gt;Your basic logic is sound: You compare the case of the cogeneration plant + boiler vs. boiler alone. Your math looks correct to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where did you get the value 215 g/kWh from? All (german language!) sources I found give 202g/kWh for natural gas, which may well be because actual of differences in grid gas composition. When I do the math myself I arrive at 197 g/kWh if using the lower heating value, only 178 g/kWh for the HHV. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Heat_of_combustion&quot; rel=&quot;nofollow noreferrer&quot;&gt;difference between HHV and LHV&lt;/a&gt; could be crucial since many boilers today are condensing boilers (that utilize the heat of evaporation of the exhaust water) but few to none cogeneration plants are, so you should possibly use different carbon intsities for both.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, the effiencies for the cogeneration plant don't look like real life numbers but are not too far of (I'm just now looking at a datasheet for a small CHP with 38% and 43% respectivly, in my experience this is a rather high &lt;span class=&quot;math-container&quot;&gt;$\eta_{el}$&lt;/span&gt;)&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2020-01-08T08:26:03.380" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32696" PostTypeId="1" AcceptedAnswerId="32700" CreationDate="2020-01-08T09:46:37.417" Score="0" ViewCount="947" Body="&lt;p&gt;Two reservoirs are connected by two pipes with different diameters in series. The elevation of Reservoir A is 40m and Reservoir B is 35m to respective water surfaces. The first pipe is 500 m long with a diameter of 0.075m and the second pipe is 2000m long with a diameter of 0.1m. What is the flow between the two reservoirs? The absolute roughness is 0.01 for both pipes and teh pipe elevation is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XSxpG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XSxpG.png&quot; alt=&quot;Schematic of the Problem Outlined&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I have tried so far is by trial and error in the Moody Diagram find a Friction Factor to use with the Darcy-Weisbach Equation. However, I cannot get a number matching my first guess of the Reynolds Number. The pipes have different diameters so that means I have to do a trial for each pipe. Also when calculating I cannot use the Head loss of 5m but of 1m since that is the correspondent Head Loss for that length of pipe? How do I go about solving for the problem above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your help.&lt;/p&gt;&#xA;" OwnerUserId="24198" LastActivityDate="2020-01-08T21:33:47.800" Title="Determining steady flow between two reservoirs with two different pipes in between" Tags="|civil-engineering|hydraulics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32697" PostTypeId="1" CreationDate="2020-01-08T17:32:03.833" Score="1" ViewCount="606" Body="&lt;p&gt;I am trying to understand Kirchoff Plate  Theory and I am unable to understand the difference between bending and twisting moment for a plate. In a beam structure, if moment is along the axis of the beam, it creates twist, and if the moment is normal to the axis of the beam, it results in the bending of the beam. However, how do we differentiate between moments when dealing with plates.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2020-01-08T20:57:21.533" Title="Understanding Bending and Twisting Moments in a Plate" Tags="|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32699" PostTypeId="2" ParentId="32697" CreationDate="2020-01-08T20:57:21.533" Score="1" Body="&lt;p&gt;Bending moment creates axial compressive and tensile stresses above and below the mid-surface and shear normal to mid-surface parallel to surface loading, q. In the most basic case roughly like a plate supported at the end acting as a wide beam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Twisting moment causes in plate shear and normal stresses, I attach a very simple diagram for that. This is a case of twisting on a differential section of a cylinder plate under torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attache the diagram here. &lt;a href=&quot;https://i.stack.imgur.com/DmE7Y.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DmE7Y.jpg&quot; alt=&quot;twisting&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-08T20:57:21.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32700" PostTypeId="2" ParentId="32696" CreationDate="2020-01-08T21:33:47.800" Score="0" Body="&lt;p&gt;Here's how I'd attack the question:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;write down the Darcy Weisbach equation, with two terms on the right sides for the two different pipes. Head losses add up.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;exchange flow velocity for a term dependent on volume flow and pipe diamter in both terms on the right side.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The pipe is really long and really thin and the head is really small. Start by using the laminar flow friction factor (64/Re)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Because of this you could solve this algebraically but I'd use any numerical tool (excel target value search is fine). &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Check the actual resultant Re in both pipe segments, if the assumption (laminar flow) holds, if not take the friction factor from the moody chart and reiterate.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="61" LastActivityDate="2020-01-08T21:33:47.800" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32704" PostTypeId="1" CreationDate="2020-01-09T08:12:40.070" Score="2" ViewCount="38" Body="&lt;p&gt;In my control equation, there is a term &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{\partial \rho u}{\partial t}+\frac{\partial \rho u^2}{\partial x}+\cdots=\frac{\partial p}{\partial x}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to discretize it in staggered grid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For convenience, I include a picture of a staggered grid:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7iCmf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7iCmf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If u is replaced by a scalar variable like &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt;, we can simply integrate it in the scalar CV and write &lt;span class=&quot;math-container&quot;&gt;$p_P-p_P^o$&lt;/span&gt; in which superscript &lt;span class=&quot;math-container&quot;&gt;$o$&lt;/span&gt; means old time value. However in staggered grid, we store speed at cell face &lt;span class=&quot;math-container&quot;&gt;$w,e,n,s$&lt;/span&gt; of a scalar CV then how can we write it in discrete form? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I don't know how should we integrate this equation because there are two types of variables involved.&lt;/p&gt;&#xA;" OwnerUserId="24210" LastEditorUserId="24210" LastEditDate="2020-01-09T12:33:30.480" LastActivityDate="2020-01-09T12:48:19.420" Title="How to deal with time derivative of speed toward time in staggered grid in CFD?" Tags="|cfd|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32705" PostTypeId="2" ParentId="32704" CreationDate="2020-01-09T10:01:36.497" Score="1" Body="&lt;p&gt;I can't believe that I forgot in SIMPLE we first treat &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; as known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The we should integrate this equation over the u-control volume. Because the time term no relation to &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;, so it's simply multiplied by &lt;span class=&quot;math-container&quot;&gt;$\Delta V$&lt;/span&gt;. Thus we can write &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\rho (u_w-u_w^o)\Delta V$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="24210" LastEditorUserId="10902" LastEditDate="2020-01-09T12:48:19.420" LastActivityDate="2020-01-09T12:48:19.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32708" PostTypeId="1" CreationDate="2020-01-09T19:47:49.380" Score="-1" ViewCount="37" Body="&lt;p&gt;I have a query regarding the calculation of the so called “Modal Participation Factor time signal” from a combination of &lt;/p&gt;&#xA;&#xA;&lt;p&gt;a) transfer functions (amplitude and phase for each eigenmode) obtained from &quot;unit&quot; accel loading in an FEM model &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and &lt;/p&gt;&#xA;&#xA;&lt;p&gt;b) real accel-time test data. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to somehow combine the above 2 to obtain the so called &quot;modal participation factor time history&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can some-one point me in the right direction?&#xA;Best Regards&lt;/p&gt;&#xA;" OwnerUserId="14216" LastActivityDate="2020-01-09T22:24:20.370" Title="Construct &quot;Modal Coordinates Time History&quot; from transfer function and test accel-time data" Tags="|acceleration|frequency-response|eigenvalue-analysis|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32710" PostTypeId="2" ParentId="32708" CreationDate="2020-01-09T22:11:11.483" Score="2" Body="&lt;p&gt;I assume you will be doing this in a software, matlab or Abaqus for example, it’s not particularly easy thing to do with your given info, but. Perhaps this example method will help&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://abaqus-docs.mit.edu/2017/English/SIMACAEBMKRefMap/simabmk-c-rodlindynamic.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Abaqus&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;Infos at &lt;a href=&quot;https://wiki.csiamerica.com/display/kb/Time-history+analysis&quot; rel=&quot;nofollow noreferrer&quot;&gt;csiamerica&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.mathworks.com/matlabcentral/fileexchange/30866-modal-time-history-analysis-of-structures&quot; rel=&quot;nofollow noreferrer&quot;&gt;Matlab function&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I am unsure how directly your data types will fit, but these are some links to get you started&lt;/p&gt;&#xA;" OwnerUserId="14318" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2020-01-09T22:24:20.370" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32711" PostTypeId="1" CreationDate="2020-01-09T23:04:36.680" Score="1" ViewCount="80" Body="&lt;p&gt;I have 10 cylindrical &lt;strong&gt;neodymium permanent magnets&lt;/strong&gt; about 5mm in diameter, 5 mm long, embedded in a thermoplastic polymer.    I want to &lt;strong&gt;demagnetize them&lt;/strong&gt; without removing them, or heating them above 60 C - they must remain motionless, +/- 1 mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the magnets are NOT all aligned in the same direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can move a demagnetizer within 5 inches of the assembly, or place the entire (1m x .3m x .2m) assembly within a demagnetizer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes - I thought about melting the plastic and replacing them - but all the other stuff must stay invariant :( &lt;/p&gt;&#xA;" OwnerUserId="24217" LastActivityDate="2020-01-09T23:30:23.270" Title="Demagnetize neodymium permanent magnets at &lt; 60 degrees C" Tags="|temperature|magnets|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32712" PostTypeId="1" AcceptedAnswerId="32714" CreationDate="2020-01-09T23:27:11.607" Score="2" ViewCount="572" Body="&lt;p&gt;I see these frequently on tall power lines like the one in the image below, sometimes a single one at each end, sometimes pairs as are shown here. Sometimes on one side of the tower but not the other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I vaguely remember reading once that they are used to reduce swinging in high winds, and where I live typhoons come regularly.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What are these called?&lt;/li&gt;&#xA;&lt;li&gt;Are they used to reduce swinging in high winds?&lt;/li&gt;&#xA;&lt;li&gt;How do they do this? &lt;/li&gt;&#xA;&lt;li&gt;When is a pair used instead of a single; why not just use a bigger one?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://www.wired.com/story/iran-apt33-us-electric-grid/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wired&lt;/a&gt; original: &lt;a href=&quot;https://i.stack.imgur.com/pLy9u.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/pLy9u.jpg&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ejd3a.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ejd3a.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QJWgN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QJWgN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6264" LastActivityDate="2020-01-12T09:11:59.557" Title="How do those &quot;dipole-shaped&quot; weights attached to sections of electrical power lines work?" Tags="|mechanical-engineering|fluid-mechanics|vibration|aerodynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32713" PostTypeId="2" ParentId="32711" CreationDate="2020-01-09T23:30:23.270" Score="1" Body="&lt;p&gt;According to a very comprehensive presentation from &lt;a href=&quot;https://www.kjmagnetics.com/blog.asp?p=temperature-and-neodymium-magnets&quot; rel=&quot;nofollow noreferrer&quot;&gt;KJ Magnetics&lt;/a&gt;, it is likely that you will not be able to accomplish your objective.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The graphs and text indicate that you will have to know the correct composition information of your magnets and that you can only partially de-magnitize at 80°C and higher. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even for those compositions that will accept a reduction of magnetism at 80°C, when the heat is removed, some magnetism will remain. I was intrigued to read that for some magnets, temperatures as high as 1000°C is required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jtdQe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jtdQe.png&quot; alt=&quot;heat vs magnetism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-01-09T23:30:23.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32714" PostTypeId="2" ParentId="32712" CreationDate="2020-01-10T01:17:48.343" Score="5" Body="&lt;p&gt;According to the &lt;a href=&quot;https://en.wikipedia.org/wiki/Stockbridge_damper&quot; rel=&quot;noreferrer&quot;&gt;wiki&lt;/a&gt;, these are called Stockbridge dampers. The design is tuned for a specific length and one might also expect a specific mass/material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weights are mounted on flexible arms, typically a length of stiff cable, which allow the assembly to absorb energy that might otherwise be imparted to the longer power lines, or more accurately, that kinetic energy (movement) is removed from the lines, preventing oscillation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Directly from the wiki:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The Stockbridge damper targets oscillations due to aeolian vibration;&#xA;  it is less effective outside this amplitude and frequency range.&#xA;  Aeolian vibration occurs in the vertical plane and is caused by&#xA;  alternating shedding of vortices on the leeward side of the cable. A&#xA;  steady but moderate wind can induce a standing wave pattern on the&#xA;  line consisting of several wavelengths per span. Aeolian vibration&#xA;  causes damaging stress fatigue to the cable and represents the&#xA;  principal cause of failure of conductor strands. The ends of a&#xA;  power line span, where it is clamped to the transmission towers, are&#xA;  at most risk. The effect becomes more pronounced with increased&#xA;  cable tension, as its natural self-damping is reduced.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-01-10T01:17:48.343" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32715" PostTypeId="2" ParentId="32589" CreationDate="2020-01-10T04:25:31.910" Score="1" Body="&lt;p&gt;Here is a first order analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the input heat flow &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_{in}$&lt;/span&gt; is distributed between heating the polymer &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_{p}$&lt;/span&gt; and loss to the air &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_{air}$&lt;/span&gt;. We can write a basic energy balance as below.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{q}_{in} = \dot{q}_{p} + \dot{q}_{air}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air loss is controlled by the convection coefficient and the area of the block as &lt;span class=&quot;math-container&quot;&gt;$\dot{q}_{air} = h_{air}A\Delta T$&lt;/span&gt;. Use this to write efficiency as the ratio of heat to the polymer divided by heat input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\varepsilon = \dot{q}_{p}/\dot{q}_{in} = 1 - \left( h_{air} A \Delta T / \dot{q}_{in}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system will be more efficient to heat the polymer when the convection coefficient around the block is decreased. It will also be more efficient to heat the polymer when its area is smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The shape of the block appears in a calculation of the Biot number factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$Bi = h_{air} (V/A) / k $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has the ratio volume to surface area as well as the thermal conductivity of the block. The smaller the value of Bi, the closer the system is to a uniform temperature profile throughout. Values of &lt;span class=&quot;math-container&quot;&gt;$Bi \approx 0.1$&lt;/span&gt; or less is the typical cutoff for lumped analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This all suggests that smaller is better. However, this neglects the heat transfer from the block to the polymer. In this case, smaller is not necessarily better. To achieve the same desired temperature change of the polymer in a shorter channel, a slower mass flow rate of the polymer will be required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where you go next depends on the details.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this sets a starting point.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2020-01-10T04:25:31.910" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32717" PostTypeId="2" ParentId="3928" CreationDate="2020-01-10T09:44:02.297" Score="-1" Body="&lt;p&gt;I believe the operation of a dry cabinet is much simpler than that. It just raises the internal temperature to a couple of degrees above ambient which increases the capacity of the air inside to hold vapour. This prevents water from condensing onto the contents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No moisture is moved to the outside so not much going on really. &lt;/p&gt;&#xA;" OwnerUserId="24222" LastActivityDate="2020-01-10T09:44:02.297" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32718" PostTypeId="1" CreationDate="2020-01-10T10:15:37.527" Score="4" ViewCount="63" Body="&lt;p&gt;I wish to rotate a short cylinder (thick disc) by 150 degrees. I chosed to use a double torsion spring for this (as the picture below). The picture shows only one side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if the above design (cylinder and spring) would actually work. I have no knowledge in springs..&#xA;Do you have alternative suggestions for the design of cylinder and spring&lt;a href=&quot;https://i.stack.imgur.com/25VAK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/25VAK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;? &lt;/p&gt;&#xA;" OwnerUserId="24223" LastActivityDate="2020-01-10T15:40:23.420" Title="torsion spring applicaiton" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32720" PostTypeId="1" CreationDate="2020-01-10T12:15:06.040" Score="1" ViewCount="67" Body="&lt;p&gt;I have an existing water supply system for the cattle on my farm whereby 6 troughs located in different fields are all fed from a common header line.&#xA;The pressure in that line ranges from 3-5 bar (It’s well water supplied from a submersible pump).&#xA;Water is supplied to each trough based on the water levels using a mechanical ball-cock system.&#xA;Currently I need to visit each of those 6 troughs each day to manually add a liquid mineral solution (to provide extra nutrients to the cattle at this time of year).&#xA;That solution comes in the form of plastic drum.&#xA;Currently I distribute about 350ml of that solution in total daily which covers 60 cows&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to automate this process in some way so that I can feed the solution directly into the header line.&#xA;What would be the best way of doing this to control the supply of the nutrient solution?&#xA;I only want to add the solution when there is flow in the header line. &lt;/p&gt;&#xA;" OwnerUserId="24224" LastActivityDate="2020-01-10T13:45:09.813" Title="How to add a controlled amount of a nutrient solution to a water line" Tags="|fluid-mechanics|chemical-engineering|pumps|process-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32721" PostTypeId="2" ParentId="32720" CreationDate="2020-01-10T12:21:07.337" Score="1" Body="&lt;p&gt;Then a type of pump that can do precisely controlled delivery is called a peristaltic pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is basically a tube compressed by rollers and as the rollers roll over the tube a precise amount gets delivered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will have to check the max working pressure.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-01-10T12:21:07.337" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32722" PostTypeId="2" ParentId="32720" CreationDate="2020-01-10T13:45:09.813" Score="2" Body="&lt;p&gt;Our water purification system uses an injection pump, in this case a diaphragm style pump. It has a control for volume dispensed during operation. The operation is limited to when water is flowing by a water flow switch. When water is moving through the system, a small hat is lifted and closes a reed switch, activating a relay in the controller, which then operates the pump.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm confident that one can create a similar design using off-the-shelf components, as this particular unit, configured by a local plumbing supply house, was overpriced. Water flow switches are not uncommon now, but when I made the purchase, I was mostly ignorant of such things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Amazon, as an example:&#xA;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B07G2F7NKJ&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;water flow switch&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4WakJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4WakJ.jpg&quot; alt=&quot;water flow switch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-01-10T13:45:09.813" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32723" PostTypeId="2" ParentId="32718" CreationDate="2020-01-10T15:40:23.420" Score="2" Body="&lt;p&gt;Yes. To reduce wear, material fatigue, and strength needed to turn it, you should use a couple coils of the spring, leave enough play so it won't seize when tensioned, and obviously the disc needs to sit on an independent bearing/bushing/hinge to turn, spring only providing tension - it's no replacement for a bearing.&lt;/p&gt;&#xA;" OwnerUserId="203" LastActivityDate="2020-01-10T15:40:23.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32724" PostTypeId="1" CreationDate="2020-01-10T16:06:36.040" Score="0" ViewCount="246" Body="&lt;p&gt;I'm currently trying to figure out a cheap way to build a cocktail machine. &#xA;For now the machine should have about 20-30 possible input liquids. These liquids should remain in their original bottle and should electrically be pumped up when needed. (I know I could use gravity instead of pumps, but the bottles should be stored in a cabinet under the machine itself.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At first I was looking for a good type of pump to use for this project and I think peristaltic pumps would be ideal because then I could be sure that the liquids haven't come in contact with anything but the hose. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then I tried to find the cheapest pumps of this type available, but unfortunately the cheap ones are still about 5€ and they would only do 150ml/min. I think 500ml/min would be the absolute minimum required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I'm thinking that maybe there is a cheaper way and I just don't know the components. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my idea would be that maybe I could just buy one big pump (maybe &lt;a href=&quot;https://www.aliexpress.com/item/4000175544443.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;something like this&lt;/a&gt;) and then connect all 20-30 hoses to it using a bunch of adapters (maybe a bunch of &lt;a href=&quot;https://www.aliexpress.com/item/32707672027.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt;).&#xA;Then I would just need cheap way to electrically close all but one hose at a time. So I was wondering if there if some sort of electric valve that is very cheap (maybe 1-2€) and doesn't come into direct contact with the liquid. I was thinking maybe some sort of electromagnet that squishes the hose or a DC motor that uses some sort of screw mechanism to bend the hose until nothing can pass through anymore.&#xA;I should also mention that I'm going to use something like an ESP8266 or maybe an Arduino to control the whole machine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any such component?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit:&lt;/strong&gt;&#xA;As suggested in the comments it might be a better idea to simply pump air into the bottles to get the liquid out. This way I can run completely separated hoses avoiding the problem that liquids may flow back into the wrong bottle (which would especially suck for milk and cream etc.).&#xA;I found really &lt;a href=&quot;https://www.aliexpress.com/item/4000567064321.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;cheap air valves&lt;/a&gt; that I could use for this purpose. &#xA;I could power/control them using a &lt;a href=&quot;https://www.aliexpress.com/item/4000246379046.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;relay board like this&lt;/a&gt; which I'd attach to an Arduino. Then I'd then just connect a compressor or CO2 container at a reasonable pressure to the valve input.&#xA;Sounds like this could work to me. But what if these valves aren't perfectly clean? I should probably have some sort of air filter behind each of these, right? Any ideas how I could accomplish that in a cheap way?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have intention of using this machine commercially btw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit2:&lt;/strong&gt;&#xA;Another problem with the air idea is that I'd have to buy a compressor for that. And they are loud and expensive form what I can tell and besides that they are completely overkill. I doubt I'd even need a 1 bar of pressure. &#xA;Is there maybe some sort of cheap low power (but therefore very quiet) compressor that I could use?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit3:&lt;/strong&gt;&#xA;Could I build an air filter &lt;a href=&quot;https://www.instructables.com/id/Make-an-Air-Purifier-on-the-cheap/&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this one&lt;/a&gt; but instead of using a PVC tube, simply use the hoses that go into the bottles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit4:&lt;/strong&gt; &#xA;Bottles that are carbonated in the first place could be another challenge because they would just push out liquid automatically over time unless I build a mechanism that detaches the compressor and then opens all valves that that no bottle can build up pressure while the machine is idling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit5:&lt;/strong&gt;&#xA;Maybe a quiet fish tank air pump could be used instead of a compressor. &#xA;I found some that are rated:&#xA;throughput: 192 liters / hour&#xA;power: 2 watts&#xA;pressure (MPa): &gt;0.012&#xA;noise: &amp;lt; 40 db&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think 0.012 MPa would be 0.12 bars or 1.74 psi. But I have no understanding of how much that actually is and if it would be enough.&lt;/p&gt;&#xA;" OwnerUserId="24227" OwnerDisplayName="Forivin" LastEditorUserId="24227" LastEditDate="2020-01-11T23:51:13.207" LastActivityDate="2020-01-11T23:51:13.207" Title="Using one electrical pump for multiple hoses" Tags="|fluid-mechanics|pumps|valves|compressed-air|" AnswerCount="0" CommentCount="17" ContentLicense="CC BY-SA 4.0" />
  <row Id="32725" PostTypeId="2" ParentId="32694" CreationDate="2020-01-10T18:26:51.503" Score="0" Body="&lt;p&gt;The place to look is in EN1997-2 and as far as I can tell it does not specify how to determine soil type from CPT results. It also does not specify that such a method is a nationally determined parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means your are outside the codes and it is your own responsibility to ensure that the method you use is sufficiently reliable. And there is no well-defined criteria for how to do that, but a method that 1) is written by and expert in the field, 2) is generally considered to be best practise and 3) has been tested on a large number of cases, should make the cut. In short, you are allowed to use Robertson's charts because everyone else is using Robertson's charts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't checked the Dutch national annex to EN1997-2, as I don't already have a copy and I'm not going to buy one for the purposes of this question, but it is not possible to specify a method for something that is not a nationally determined parameter in the national annex. The only thing, they can do, is provide &quot;supplementary information&quot; that a given method is sufficiently accurate or similar. They cannot require you to use a specific method or outrule a method. Therefore, the charts are allowed, regardless of what the national annex says on the topic. (But they could potentially make it a lot easier to answer the question of whether the charts are okay to use.)&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2020-01-10T18:26:51.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32726" PostTypeId="2" ParentId="32694" CreationDate="2020-01-10T23:41:27.410" Score="0" Body="&lt;p&gt;Eurocode 7 and 8 document soil classification which may be found &lt;a href=&quot;https://soilpropertytesting.com/EURO_CODE_7.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; and &lt;a href=&quot;https://www.rcsolver.com/en/eurocode-8-ground-types&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; respectively. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2020-01-10T23:41:27.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32727" PostTypeId="1" AcceptedAnswerId="32744" CreationDate="2020-01-11T02:02:46.467" Score="1" ViewCount="806" Body="&lt;p&gt;I am trying to control a mechanical system with some transfer function T (output is position &amp;amp; input is voltage). I have implemented a state feedback controller in order for the output of the system to reach a certain position. The actuator of the system is an RC Servo Motor which is actuated by PWM signals. I want to convert the output of the controller into proper PWM in order to control the system. Are there any guidelines which I should follow or is there a specific way in order to manage to convert the signal ?&lt;/p&gt;&#xA;" OwnerUserId="18238" LastActivityDate="2020-01-14T19:37:32.230" Title="How to convert output of controller (control signal) into PWM signal?" Tags="|mechanical-engineering|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32729" PostTypeId="1" AcceptedAnswerId="32743" CreationDate="2020-01-11T05:50:59.090" Score="0" ViewCount="2335" Body="&lt;p&gt;As in the first image, &#xA;Moment at any point between C and B should be 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when I try to solve it by substituting equal and opposite forces at end C so that anti-clockwise moment of &lt;span class=&quot;math-container&quot;&gt;$F(l-a)$&lt;/span&gt; and force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; act at C, moment diagram comes out different.&#xA;Why is there discrepancy?&#xA;&lt;img src=&quot;https://i.stack.imgur.com/riis7.jpg&quot; alt=&quot;Actual Bending Moment and shear force diagram&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NfmYh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="21733" LastEditorUserId="21733" LastEditDate="2020-01-12T09:46:52.717" LastActivityDate="2020-01-12T15:35:30.073" Title="Problem while solving for the moment of a cantilever beam" Tags="|structural-engineering|beam|moments|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32730" PostTypeId="2" ParentId="19308" CreationDate="2020-01-11T10:14:15.797" Score="0" Body="&lt;p&gt;The moment at the hinge is sought:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M=-R_1\times X_1+W_2\times X_2-W_3\times X_3+W_4\times X4$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where the sign of the moment is &lt;span class=&quot;math-container&quot;&gt;$+CCW$&lt;/span&gt;:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;P_1&amp;amp;=\text{reaction left}\\&#xA;P_2&amp;amp;=\text{weight left}\\&#xA;P_3&amp;amp;=\text{weight right}\\&#xA;P_4&amp;amp;=\text{reaction left}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$X_1$&lt;/span&gt; through &lt;span class=&quot;math-container&quot;&gt;$X_4$&lt;/span&gt; are the absolute values of the distances from the hinge.&lt;/p&gt;&#xA;&lt;p&gt;Moment at the hinge is the sum of the moments about it.&lt;/p&gt;&#xA;" OwnerUserId="24230" LastEditorUserId="28917" LastEditDate="2020-10-04T15:04:12.943" LastActivityDate="2020-10-04T15:04:12.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32733" PostTypeId="1" AcceptedAnswerId="32734" CreationDate="2020-01-11T21:19:06.960" Score="4" ViewCount="1100" Body="&lt;p&gt;Does the stacking of peltiers device shows promising results in achieving the cryogenic temperature? I have tried doing the experiment, the overall temperature of the system increased around 80ºC and then the dissipation of that heat was the prime issue? Heatsinks usually fail to dissipate this high amount of heat. &#xA;Is there any alternative technology to achieve cryogenic temperature electrothermally without producing much amount of heat? &#xA;Can any modifications in stacking help?&lt;/p&gt;&#xA;" OwnerUserId="21336" LastActivityDate="2022-12-16T02:03:24.793" Title="Can a Peltier's device (i.e.Thermoelectric cooler) achieve cryogenic temperature efficiently?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|refrigeration|cryogenics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32734" PostTypeId="2" ParentId="32733" CreationDate="2020-01-11T23:53:53.603" Score="4" Body="&lt;p&gt;There is no way to get around the physics in this case. The fundamental issue here is that peltier effect refrigerators are individually extremely inefficient, and stacking them in essence &quot;stacks&quot; their inefficiencies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why are they inefficient? Because the material they are made from ideally should have zero electrical resistance and infinite thermal resistance &lt;em&gt;at the same time&lt;/em&gt;. No known substance can exhibit both of these characteristics, and so we are left with &quot;compromise&quot; material sets which do allow for the device to function, but most of the electrical energy that is put into the system gets wasted as heat.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-01-11T23:53:53.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32735" PostTypeId="2" ParentId="29025" CreationDate="2020-01-11T23:59:07.673" Score="2" Body="&lt;p&gt;There are a lot of variables to consider for a thermal analysis, but none of them require pressure as a boundary condition.  As alephzero indicated in their comment, this is function of heat source and sink.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the most basic situation, steady-state conduction, you could use the equivalent resistance equation: &lt;span class=&quot;math-container&quot;&gt;$$Q = (T_1-T_2)/R$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; is heat transfer rate, &lt;span class=&quot;math-container&quot;&gt;$T_1$&lt;/span&gt; is heat source, &lt;span class=&quot;math-container&quot;&gt;$T_2$&lt;/span&gt; is heat sink, and &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; is the thermal resistance which is related to material and geometry. As the equation shows, pressure is not a factor.&lt;/p&gt;&#xA;" OwnerUserId="24234" LastActivityDate="2020-01-11T23:59:07.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32736" PostTypeId="2" ParentId="26471" CreationDate="2020-01-12T00:18:51.390" Score="1" Body="&lt;p&gt;Have you considered using an orifice plate in conjunction with your valve?  You could use a larger diameter pipe and valve and insert a plate with an orifice drilled to the specific diameter to achieve your desired pressure drop.  This would allow you to use the valve fully open and fully closed and avoid adding a long run of piping.&lt;/p&gt;&#xA;" OwnerUserId="24234" LastActivityDate="2020-01-12T00:18:51.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32737" PostTypeId="2" ParentId="32712" CreationDate="2020-01-12T08:50:31.237" Score="2" Body="&lt;p&gt;In order to supplement &lt;a href=&quot;https://engineering.stackexchange.com/a/32714/6264&quot;&gt;@fred_dot_u's answer&lt;/a&gt; and add a few details on the nature of the problem that these devices address I'll add a bit of additional material.&lt;/p&gt;&#xA;&lt;p&gt;For a description of a numerical model of how Stockbridge dampers work and ANSYS simulation results, see for example &lt;a href=&quot;https://journals.sagepub.com/doi/pdf/10.1177/0954406212452064&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aeolian vibration of a single conductor with a Stockbridge damper&lt;/a&gt;, Oumar Barry, Donatus CD Oguamanam and Der Chyan Lin, Proc IMechE Part C: J Mechanical Engineering Science, 227(5) 935–945, DOI: 10.1177/0954406212452064&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Also, from  J-L Lilien's &lt;a href=&quot;http://www.tdee.ulg.ac.be/userfiles/file/Vibrations_eoliennes_intro.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Power Line Aeolian Vibrations&lt;/a&gt;.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Aeolian vibration is a low amplitude (conductor diameter) high frequency (5 to 150 Hz) phenomenon. Aeolian vibration is one of the most important problems in transmission lines because it represents the major cause of fatigue failure of conductor strands or of items associated with the support, use, and protection of the conductor. In this phenomenon, conductor strand fatigue failures occur at the suspension clamps or at the clamps of the other devices installed on the conductor such as spacers, spacer dampers, dampers and other devices.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KiBEL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KiBEL.png&quot; alt=&quot;Fig 2.1 typical broken strand due to aeolian vibration after removal of suspension clamp&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Fig 2.1 typical broken strand due to aeolian vibration after removal of suspension clamp.&lt;/p&gt;&#xA;&lt;p&gt;Forces induced by vortex shedding are the cause of this type of vibration (Blevins1990, Buckner 1968, Claren et al 1969 &amp;amp; 1974).&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kWrP7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kWrP7.png&quot; alt=&quot;Fig 2.2 Flow visualization of vortex shedding&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Fig 2.2 Flow visualization; identification according to (row, column): (1,1) R=1.1 (Taneda), (1,2) R=9.6 (Taneda), (2,1) R=26 (Taneda), (2,2) R=140 (Taneda), (3,1) R=2000 (Werlé &amp;amp; Gallon), (3,2) R=10000 (Corke &amp;amp; Nagib)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The fatigue mechanism&lt;/p&gt;&#xA;&lt;p&gt;The fatigue in power line conductors are caused by fretting between different layers&#xA;of strands at their contact area (fig 5.1). Some cracks appear and propagate. The&#xA;stick/slip elliptical contact area is depending on the radius of curvature taken by the&#xA;conductor. During alternating bending stress (due to oscillations) that radius of&#xA;curvature is changing continuously. It is particularly important near clamp ends at the&#xA;last point of contact.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rZXd9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rZXd9.png&quot; alt=&quot;Fig 5.1 the inter-strand contact area where first crack may appear (Epri-1979)&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Fig 5.1 the inter-strand contact area where first crack may appear (Epri-1979)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;http://preformed.com/images/pdfs/Energy/Transmission/Motion_Control/VORTX_Vibration_Damper/EN-ML-1007-4AeolianViBook.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Aeolian Vibration Basics&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/G1dFn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G1dFn.png&quot; alt=&quot;Figure 14 – VORTX™ Damper&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Figure 14 – VORTX™ Damper&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gpGwv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gpGwv.png&quot; alt=&quot;Figure 16 – Damper Response Test Set-Up&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Figure 16 – Damper Response Test Set-Up&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RmyrT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RmyrT.png&quot; alt=&quot;Figure 17 – Damper Response Curve&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Figure 17 – Damper Response Curvee&lt;/p&gt;&#xA;&lt;p&gt;Figure 17 shows the power (in watts) delivered to the damper by the shaker (vertical axis) over a range of frequencies. The results are shown for three of the same damper design, along with the average. This plot shows four distinct peaks in the power curve. These represent the resonant frequencies of the damper weights. Each weight has two resonant frequencies: one where the furthest end of the weight has the maximum movement, and the other where the end nearest the clamp has the maximum movement (higher frequency). The VORTX Damper shown in the curve has two different weights, each with different resonant frequencies, which accounts for the four peaks in the curve&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="6264" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2020-01-12T09:11:59.557" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32738" PostTypeId="1" CreationDate="2020-01-12T10:49:31.460" Score="-1" ViewCount="49" Body="&lt;p&gt;Hummvee has front lights located far from edges, closer to each other. That makes it harder to see edges of the vehicle in the dark.&lt;/p&gt;&#xA;" OwnerUserId="24239" LastActivityDate="2020-01-12T11:25:08.007" Title="Why HMMWV's lights aren't located at the edges?" Tags="|design|car|lighting|" AnswerCount="1" CommentCount="1" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32740" PostTypeId="2" ParentId="32738" CreationDate="2020-01-12T11:25:08.007" Score="2" Body="&lt;p&gt;The lights are mounted within specifications for positioning as required by laws in countries where they are used/sold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, designers position them relating to the use the vehicle is likely to see, so if it is used through woods etc them the lights get mounted farther inboard so they get a bit more protection.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-01-12T11:25:08.007" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32743" PostTypeId="2" ParentId="32729" CreationDate="2020-01-12T15:35:30.073" Score="2" Body="&lt;p&gt;This is cheating a bit, but let's start backwards by looking at the correct diagrams you've already shown us and comparing them to your solution. This will hopefully let you notice something which hints at the problem with your solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your solution is the bending moment equation &lt;span class=&quot;math-container&quot;&gt;$M = F(\ell - a) - Fx$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, what is the bending moment equation for the correct moment diagram? Think about it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Well, that's a trick question. What I should have asked is: what are the bending moment &lt;strong&gt;&lt;em&gt;equations&lt;/em&gt;&lt;/strong&gt; (plural!) for the correct moment diagram?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is an obvious discontinuity in the slope of the bending moment diagram: it is linear from A to B, and then constant at 0 from B to C. That's my point: you could never solve this with a single equation, you'll always need two: one from A to B, another from B to C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what's the source of that discontinuity? Well, for that we need to remember that bending moment is the integral of shear force. Since you haven't asked about the correct shear diagram, I'm going to assume you understand why it looks as shown in the correct answer (simply put, it's constant and equal to &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; between the support and the force, and zero elsewhere).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if the shear equation is discontinuous, that means the derivative (slope) of the bending moment equation will also be*. And since the shear equation gives us the slope of the bening moment equation at any point, we can trivially see that the slope of the line will be equal to &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; from A to B, and zero from B to C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given this, we can derive the two bending moment equations (here using &lt;span class=&quot;math-container&quot;&gt;$x=0$&lt;/span&gt; at A):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$M = \begin{cases}&#xA;Fx - Fa &amp;amp;\text{if }x\in[0, a]\\&#xA;0 &amp;amp; \text{otherwise}&#xA;\end{cases}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* Note that a discontinuous slope means the slope can change instantaneously, but the value of the bending moment equation on both sides of the discontinuity must be equal; the slope is discontinuous, not the value.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-01-12T15:35:30.073" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32744" PostTypeId="2" ParentId="32727" CreationDate="2020-01-12T16:42:00.687" Score="4" Body="&lt;p&gt;Your controller output needs to be converted to the PWM signal your RC motor can understand and what your controller can output, so you can use a simple linear function like &lt;span class=&quot;math-container&quot;&gt;$y(x)=m*x+b$&lt;/span&gt; to achieve this, your x-axis being your position, and your PWM in bits (8, 10, 12, etc) along the y-axis. &#xA;&lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; if you happen to be off hazy in your linear functions is your zero position of your RC motor, which as it crosses the y axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An example converting amps to pwm in Wiring might look like this: &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;// signal between -2.2 and 2.2 amps&#xA;highPWM = 256;&#xA;lowPWM = 0;&#xA;// convert Amps to PWM signal&#xA;float atp(float sgnl) {&#xA;  float thing;&#xA;  thing = (((highPWM) - (lowPWM)) / 4.4) * sgnl + nullPWM;&#xA;  return thing;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However this isn‘t the only functional way to program such a conversion and of course your signal may not be linear which will require a different function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on which way you write it, some languages also have built in functions such as Map() which will do the math for you. &lt;/p&gt;&#xA;" OwnerUserId="14318" LastEditorUserId="14318" LastEditDate="2020-01-12T18:09:13.927" LastActivityDate="2020-01-12T18:09:13.927" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32745" PostTypeId="2" ParentId="32733" CreationDate="2020-01-12T17:27:00.680" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Does the stacking of peltiers device shows promising results in achieving the cryogenic temperature?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you consider actually working in devices that are working in satellites to be &quot;promising&quot;, then yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I have tried doing the experiment, the overall temperature of the system increased around 80ºC and then the dissipation of that heat was the prime issue? Heatsinks usually fail to dissipate this high amount of heat.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Then you aren't thinking your design through properly.  You need a multi-stage cooler, and going from the cold end to the hot end, each stage needs to be larger than the last.  The ones I saw looked like pagoda roofs, with a small cooler at the cold end, and a cooler with over 100 times the area at the hot end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's been way to long for me to remember the specifics, but you need to start with what you want to keep cold, what temperature and heat flux you need to deal with.  Then design a stage that takes that heat out most efficiently, and call it your new &quot;thing I want to cool&quot;, and use that as input for your next stage.  Keep going until you're done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I was shown these devices we looked at all the complications, looked at the fact that we were doing something that was earthbound and therefor close to maintenance facilities, and decided &quot;eh, we'll stick with moving parts&quot;.&lt;/p&gt;&#xA;" OwnerUserId="19776" LastActivityDate="2020-01-12T17:27:00.680" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32746" PostTypeId="2" ParentId="32674" CreationDate="2020-01-12T17:36:32.890" Score="2" Body="&lt;p&gt;At a quick readthrough, your calculations seem to be correct and they also seem to answer the hard parts of your question. I also doubt the calculations can be made significantly simpler than your version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a recalculation of the &quot;wet condition&quot; as follows:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;With no lift of the rear wheel, there will moment equilibrium around the center of mass, i.e. &lt;span class=&quot;math-container&quot;&gt;$N_R x_2 + F_{BR}h = N_F x_3$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;With no vertical acceleration of the rear wheel, we also have vertical equilibrium, i.e. &lt;span class=&quot;math-container&quot;&gt;$F_G = N_R + N_F$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Assume we are breaking hard enough to reach the friction coefficient of wheel and ground, &lt;span class=&quot;math-container&quot;&gt;$F_{BR} = \mu N_F$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;Solving for the front wheel normal force gives &lt;span class=&quot;math-container&quot;&gt;$N_F = \frac{F_G x_2}{x_1-\mu h}$&lt;/span&gt; as you already got. This is normal force adjusted for weight redistribution caused by braking.&lt;/li&gt;&#xA;&lt;li&gt;So to calculate the corresponding deceleration, simply check the assumption that the rear wheel stays on the ground, &lt;span class=&quot;math-container&quot;&gt;$N_F &amp;lt; F_G$&lt;/span&gt; and calculate &lt;span class=&quot;math-container&quot;&gt;$a=\frac{\mu N_F}{m}$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2020-01-12T17:36:32.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32747" PostTypeId="1" CreationDate="2020-01-12T18:28:41.753" Score="-1" ViewCount="108" Body="&lt;p&gt;I'm new here and would like to ask where the cavitation would happen in my design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;a href=&quot;https://i.stack.imgur.com/63gn3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/63gn3.png&quot; alt=&quot;Rim Drive Technology&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24244" LastActivityDate="2020-01-14T09:23:06.943" Title="The Cavitation Problem" Tags="|hydraulics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32748" PostTypeId="2" ParentId="32747" CreationDate="2020-01-12T19:07:27.300" Score="0" Body="&lt;p&gt;Cavitation happens where there is pressure drop usually near the eye of an impeller of a pump or after sharp turns in the flow caused by the blades or the geometry of the pump on the interior, low pressure, side of the bend. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They can even happen in our blood-stream sometimes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The partial vacuum encourages the formation of bubbles (cavitation) that travel with the flow until they reach a high-pressure zone where the may instantaneously implode and damage the surface or part near to this zone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two types of cavitation: suction and discharge. You need to look for suction near the inlet and discharge cavitation inside the pump housing near the discharge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simulation software also can predict the cavitation area. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-01-12T19:14:43.963" LastActivityDate="2020-01-12T19:14:43.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32749" PostTypeId="2" ParentId="25146" CreationDate="2020-01-12T22:11:05.723" Score="1" Body="&lt;p&gt;You cannot determine the operating voltage of a coil in this way unfortunately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way I can think of to determine the voltage of an unmarked relay would be to connect it to and adjustable power supply (if DC) or variac (if AC). Then simple turn up the voltage until the relay seems to &quot;click&quot; reliably. I would then turn it up a bit more to provide an operating margin.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2020-01-12T22:11:05.723" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32750" PostTypeId="2" ParentId="32727" CreationDate="2020-01-12T22:33:07.040" Score="2" Body="&lt;p&gt;First, go take a look at some documentation on the servo PWM signal. The gist of it, is that a PWM signal is a pulse between 1000us and 2000us, repeated every 20ms. The signal is interpereted as 0 when the pulse it 1500us. More than that is positive, and less negative (or the reverse, can't remember).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How you convert your output signal to this format will depend on the device. On an arduino there are numerous examples on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are trying to control the servos position, and the servo takes a &lt;em&gt;position&lt;/em&gt; signal in the form of PWM. So all you have to do is convert your position value into servo position and send that to the servo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways to do this, so I will suggest a simple one. Use a table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply move the servo through it's range, and at several points record the servo position using your system as well as the pulse width you're sending. This creates a table converting your position value directly to a PWM value. Between these points you can interpolate.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastEditorUserId="20490" LastEditDate="2020-01-14T19:37:32.230" LastActivityDate="2020-01-14T19:37:32.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32751" PostTypeId="2" ParentId="28087" CreationDate="2020-01-13T04:24:20.010" Score="1" Body="&lt;p&gt;Type 21 flanges are &quot;integral flanges&quot; and in my experience these are completely manufacturer dependent.  This grouping would include long weld neck (LWN), heavy barrel (HB), equal barrel (EB), or any other custom offering.  I've experienced where two NPS 2&quot; long weld neck flanges on the same pressure vessel built around 1994 had different thicknesses because they were procured from two different suppliers.  I think this is less common now as manufacturers have consolidated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe this is why the EN1092-1 refers to component standards, which I read to mean manufacturer's standard.&lt;/p&gt;&#xA;" OwnerUserId="24234" LastActivityDate="2020-01-13T04:24:20.010" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32756" PostTypeId="1" CreationDate="2020-01-13T11:51:14.540" Score="0" ViewCount="105" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/36tFA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/36tFA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As shown in the image, I have done a vibrating wire type strain gauge and pressure cell (not shown in the image) instrumentation on full scale form. the objective is to measure the lateral pressure of the concrete on the form while concrete was poured and until later initial setting time. Though the interest was to measure only the concrete pressure. As an additional information I measured strain too. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a doubt in the measured pressure. I checked the calibration and raw data. It seems everything is OK. But the measured values are far less than the expected theoretical value. Fortunately, We have concrete strain measurement. But unfortunately I do not know how to use the strain data to back calculate and get the lateral pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a standard formula available to calculate lateral concrete pressure from the measurements of lateral concrete strain?&lt;/p&gt;&#xA;" OwnerUserId="15386" LastEditorUserId="15386" LastEditDate="2020-01-14T03:56:35.353" LastActivityDate="2023-12-18T13:07:50.973" Title="Is there a standard formula available to calculate lateral concrete pressure from the measurements of lateral concrete strain?" Tags="|concrete|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32757" PostTypeId="2" ParentId="32686" CreationDate="2020-01-13T14:45:57.637" Score="0" Body="&lt;p&gt;I tried very large strap wrenches, a heat gun, the lot.  In the end, nothing short of an &lt;strong&gt;angle grinder&lt;/strong&gt; would do the job.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be careful with the depth of your cut, to avoid hitting the valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bB1mc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bB1mc.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/BERw0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BERw0.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would not try this on any cylinders containing flammable gases or oxygen!&lt;/p&gt;&#xA;" OwnerUserId="10132" LastActivityDate="2020-01-13T14:45:57.637" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32758" PostTypeId="2" ParentId="32747" CreationDate="2020-01-13T15:52:10.373" Score="1" Body="&lt;p&gt;Your pump is an axial flow pump. I'm not an axpert, but from skimming literature it appears that the most relevant form of cavitation ins common ax flow pumps appears at the tips of the blades (where speed is highest, possibly aided by shear with the pump housing, tip vortex cavitation is the search term). Since the inlet of ax flow pumps is typically fairly wide, this is not the palce where you'd expect caviation. So in your design, I would investigate the areas directly upstream and downstream of the rotorspecifically between the static vanes and the guiding vanes and near the outter edges of the rotor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another common point where cavitation occurs is the inlet. However in your design there appears to be no real reduction in free diameter along the pump, so the axial component of flow velocity should be pretty constant. My &lt;em&gt;guess&lt;/em&gt; is that the inlet won't see much cavitation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But at the end of the day these are (not very well) educated guesses and you need to do CFD and physical experiments to verify your design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit to add:&lt;/strong&gt;&lt;br&gt;&#xA;The OP stated in a (removed) comment that CFD is out of their reach. The (in many ways beter!) alternative is a physical test. There's no other way to determine the characteristics of  of a novel impeller. The good news is that you can somewhat extrapolate from from a small prototype using affinity laws (Though I assume your pump will always be pretty small, because of the seals)&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2020-01-14T09:23:06.943" LastActivityDate="2020-01-14T09:23:06.943" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32759" PostTypeId="1" CreationDate="2020-01-13T16:06:45.497" Score="2" ViewCount="226" Body="&lt;p&gt;We have a project that requires us to use a qty of 4, Ø2&quot; yoke roller bearings mounted two per side on a trolley that roll on a steel track. The maximum total weight that we estimate on the track at any given time while in service will be between 8k to 12k lbs. This will be distrubuted too the 4 rollers which each have a capacity of 8,000 lbs each. So there would about 2k-3k per roller. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will be using standard yoke roller bearings that have an outer ring hardness at 58-62 Rockwell C or 58-62 Rc. Our Track is hardened to 52-58 Rc. Knowing that the track has a hardness less than that of the bearing, wouldn't it stand to reason that the track will wear out before the bearing? This is unacceptable for this project.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need a solution so we can make the bearings the wear item on this project. The track would take many man hours to replace , where a bearing would take very little time to replace and cost much less than a new track.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would need to increase the track hardness to slightly higher than the bearing or lower the hardness of the bearing to make the bearing the wear item instead of the track. The downside to this is that I would sacrifice bearing L10 Life by lowering the hardness of the outer ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We would love some help solving this issue if anyone has had any experience with this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;****UPDATE 1/14/2020*****&#xA;Since I posted this question, I have spoken with quite a few other mechanical and manufacturing engineers who specialize in bearing and track design. They all pretty much explained me the same thing that using a a track that is 40-58 Rc will work just fine with a yoke roller that has an outer ring hardness of 58-62 which is the industry standard for that type of bearing. The bearing internal rolling elements and seals will wear out long before the track or the OD of the outer. Being that the track is 52-58 Rc it should endure many yoke roller swap outs over the years with the loads and application this is being used with. I'd like to thank everyone for your time and effort put into helping us come to a conclusion of this issue.&lt;/p&gt;&#xA;" OwnerUserId="18027" LastEditorUserId="18027" LastEditDate="2020-01-14T13:34:38.670" LastActivityDate="2020-01-14T13:34:38.670" Title="How do I determine what hardness to change the bearing or the track?" Tags="|mechanical-engineering|stresses|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32760" PostTypeId="2" ParentId="32759" CreationDate="2020-01-13T18:00:30.687" Score="1" Body="&lt;p&gt;You could try pressing the yoke rollers into plastic sleeves, essentially giving the yokes plastic ODs and thus a softer surface. We tried experimenting with this a few years ago at a bearing manufacturer I worked at. Typical materials are Acetal resin and Nylon 6/6.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's also possible only the bearing races are heat treated, leaving the OD would be softer. I would verify that before going any further.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hard yoke ODs could also cold work the rails making them slightly harder where they contact, that might be enough on it's own to solve your problem.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-01-13T18:00:30.687" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32761" PostTypeId="1" AcceptedAnswerId="32762" CreationDate="2020-01-13T18:19:27.910" Score="2" ViewCount="81" Body="&lt;p&gt;I am looking for a linear pneumatic cylinder to raise and lower a fixed object. Everything will be submerged in oil (for electrical purposes). I am new to engineering and pneumatics in particular, and I don't know if I can completely submerge the cylinder without damaging it.&lt;/p&gt;&#xA;" OwnerUserId="24253" LastActivityDate="2020-01-13T18:38:17.083" Title="Can I submerge a pneumatic cylinder in oil?" Tags="|mechanical-engineering|pneumatic|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32762" PostTypeId="2" ParentId="32761" CreationDate="2020-01-13T18:38:17.083" Score="1" Body="&lt;p&gt;If it is a double-acting cylinder then possibly fine, BUT you will need to check the seals are not going to be damaged by the oil and make sure that the pressure inside the cylinder, on both sides of the piston, is greater than the oil pressure so oil is less likely to make its way past the seals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A single acting cylinder needs to be able to fill or expel air on the low pressure side of the piston which will be an issue when submerged.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-01-13T18:38:17.083" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32763" PostTypeId="1" CreationDate="2020-01-13T20:28:27.497" Score="1" ViewCount="134" Body="&lt;p&gt;I'm writing a fusion360 adding to model windmill blades. My first goal is to model the &lt;strong&gt;NREL Phase IV&lt;/strong&gt; in which the airfoils (S809) definitions come with &lt;strong&gt;&lt;em&gt;thickness&lt;/em&gt;&lt;/strong&gt; and also &lt;strong&gt;&lt;em&gt;span station&lt;/em&gt;&lt;/strong&gt; (wich I¿dont know what is):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZtY8n.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZtY8n.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right now, I'm able to calculate the max thickness. as shown in the next picture as the difference between the max point on the upper surface(y-direction), and the min point(y-direction) of the lower surface. I'm not really sure if this is correct, and then I don´t know how to modify the entire airfoil with the new thickness definition in the table above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BVFtm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BVFtm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="2211" LastActivityDate="2020-01-13T20:28:27.497" Title="How to modify Airfoil thickness" Tags="|airfoils|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32764" PostTypeId="2" ParentId="32759" CreationDate="2020-01-13T20:42:08.413" Score="2" Body="&lt;p&gt;Are you are using the outer race as a &quot;wheel&quot;. Sounds like an unusual design , maybe a steel tire should be put over the outer race. The steel tire could easily be any hardness you want. It would be poor judgement to mess with the rolling element bearing hardness. The heat-treatment has been developed over decades. One factor is the residual compressive stresses developed on the surface ( by New Departure years ago) to significantly extend fatigue life. Another is a very small amount of retained austenite ( as I remember it also improves life but I forgot the mechanism ). There are other proprietary factors like double tempering. Improvised heat-treatment will affect these factors , likely adversely.  Sounds like a big chance to take because someone thinks there may be excessive wear.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-01-13T20:42:08.413" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32765" PostTypeId="2" ParentId="32756" CreationDate="2020-01-13T20:58:01.363" Score="0" Body="&lt;p&gt;Poured concrete after the initial plastic stage changes volume for a large number of reasons. Some changes are small like autogenous volume change ( 0.001 range), some large which can cause cracks and strain in the concrete like plastic shrinkage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many studies on the concrete Volume change and ways to mitigate its undesired effects and take advantage of desired effects.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forms will deflect a bit and give rise to the expansion of the concrete as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hypothetically though ignoring all the above effects, one could try to establish the stiffness of the concrete forms by measuring the formwork under the load and finding the counterpart of its &quot;modulus of elasticity&quot;, &lt;span class=&quot;math-container&quot;&gt;$E_{formwork}$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then one can use the &lt;span class=&quot;math-container&quot;&gt;$$ \sigma=\epsilon E_{formwork} $$&lt;/span&gt;&#xA;This will be the stress of the concrete due to its confinement by the forms. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-01-13T21:06:07.183" LastActivityDate="2020-01-13T21:06:07.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32766" PostTypeId="1" CreationDate="2020-01-13T21:07:31.303" Score="1" ViewCount="71" Body="&lt;p&gt;I have been experimenting with a furnace at my house and am looking to attempt the more difficult process of creating bismuth-oxide based optical glass for the purposes of creating optical elements (think lenses and reflective glasses). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm having trouble finding what the raw material inputs to create the best mixture should be, as well as best practices for an amateur such as myself. Most of the literature on how to produce good results are either pay walled or too esoteric for me to understand. I was able to find a &lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3344235/#!po=90.0000&quot; rel=&quot;nofollow noreferrer&quot;&gt;helpful paper&lt;/a&gt;, with the following excerpt: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The raw materials of tellurium dioxide, TeO&lt;sub&gt;2&lt;/sub&gt; (Aldrich 99.5%), and&#xA;  bismuth (iii) oxide Bi&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt; (Aldrich 99.0%) were used to synthesize the &#xA;  glass sample. The glass sample was prepared by using a melt quenching&#xA;  method [32,33]. The bismuth (iii) oxide, x = 5, 8, 10, 12, 15 in mol%&#xA;  was added into the tellurite oxide and was weighed using a digital&#xA;  weighing machine with an accuracy of ±0.01 g and then mixed together&#xA;  by using a mortar and pestle. The mixture was poured into an alumina&#xA;  crucible and put into an electric furnace set at 100 °C for a period&#xA;  of 30 min. The alumina crucible was used because it can withstand high&#xA;  temperatures and does not react with the raw materials during the&#xA;  melting process as opposed to porcelain crucibles. The temperature was&#xA;  then increased by 10 °C/min until 900 °C and the then maintained for 2&#xA;  h. When the melting process was completed, the molten liquid was cast&#xA;  into a stainless steel cylindrical shape mould which had been&#xA;  preheated at 400 °C for 30 min. The sample was annealed at 400 °C in a&#xA;  second furnace for 1 h, then the furnace was turned off. The glass&#xA;  sample was cut at a thickness of about 2 mm for the required&#xA;  measurements [33].&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I have pulled a few assumptions from this paragraph which it would be helpful for me to verify before testing this out:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm assuming the passage &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;the glass sample was prepared by using a melt quenching method [32,33]. The bismuth (iii) oxide, x = 5, 8, 10, 12, 15 in mol% was added into the tellurite oxide&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;is simply stating that the end concoction consists of a combination between 85-95% TeO&lt;sub&gt;2&lt;/sub&gt; and 5-15% Bi&lt;sub&gt;2&lt;/sub&gt;O&lt;sub&gt;3&lt;/sub&gt;, correct?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;What is the purpose of annealing for in this context? My end case for the result pieces are simply for prototyping, but my guess is I will have to polish the pieces with cerium oxide without destroying them so they can not be too brittle. Is annealing required?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I'm lost for understanding what the &lt;code&gt;quenching method&lt;/code&gt; is, results online are sparse and the linked papers are pay walled. What is the process for this like? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Are there other things I should know? I have &lt;a href=&quot;http://devil-forge.com/gas-furnaces/37-fb1s.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;this furnace&lt;/a&gt; and have had success with simple metals but not sure what else I need to know for something a bit more chemically involved like optical glass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="24254" LastEditorUserId="90" LastEditDate="2020-01-14T03:48:24.480" LastActivityDate="2020-01-14T03:48:24.480" Title="How to furnace-melt bismuth oxide into optical glass?" Tags="|chemical-engineering|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32771" PostTypeId="1" AcceptedAnswerId="35348" CreationDate="2020-01-14T14:59:42.033" Score="3" ViewCount="116" Body="&lt;p&gt;Power output of nuclear reactors is controlled by control rods that sit in between the fuel rods: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are pulled out to increase fission rates - and lowered to decrease them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control rods are made from neutron absorbing materials - such as cadmium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When lowered completely, so many neutrons are absorbed, that no chain reaction is maintained. Pulling them out increases the neutrons that are available for fission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The lower tips of the fuel rods will almost always be exposed to neutrons, so these should burn down much faster (have their uranium  fissioned) than the upper parts that rarely have the control rods removed from them. Is there some engineering going on to have the fuel rods burn down at the same rate along their whole length?&lt;/p&gt;&#xA;" OwnerUserId="24262" LastActivityDate="2021-04-04T02:37:09.847" Title="How do nuclear power plants prevent fuel rods from depleting nonuniformaly?" Tags="|nuclear-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32772" PostTypeId="1" CreationDate="2020-01-14T15:20:10.180" Score="1" ViewCount="341" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9DmlI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9DmlI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;I need to develop a mechanism - Where in I have a shaft that needs to rotate at different speeds. Lets say at a higher RPM clockwise and a lower RPM anti-clockwise. &#xA;This shaft is attached to my machine, so I can't have 2 shafts. &#xA;If i have 2 motors to the same shaft, I'm afraid there'll be feedback on the other motor, while one is running.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, like the image attached. Both the motors are connected to the lead screw. The motor (and the lead screw attached to it) on the left rotates at a higher rpm and this motion is used to lift my system. Similarly from the other side but to drop it quicker (the reason for the speed difference). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas? Thanks :)&lt;/p&gt;&#xA;" OwnerUserId="24263" LastEditorUserId="24263" LastEditDate="2020-01-14T15:40:26.423" LastActivityDate="2020-01-14T15:40:26.423" Title="I need a mechanism/method to control a shaft that can rotate at different speeds. (faster clockwise and slower anti-clockwise)" Tags="|mechanical-engineering|gears|mechanical|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32773" PostTypeId="2" ParentId="25223" CreationDate="2020-01-14T17:51:19.710" Score="1" Body="&lt;p&gt;The quench all depends on the grade of steel that you are working with. Each steel is specially formulated to be used for a specific type of application. As others have stated you need to narrow down your material then investigate from that point. I don't know how anyone could answer this vague of a question with any accuracy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following list gives a brief description what each Grade of tool steel is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A Series: Air Hardening, Cold Work Tool Steel, Air Quench Only&lt;/li&gt;&#xA;&lt;li&gt;D Series: High Carbon, High Chromium, Cold Work Tool Steels, All D&#xA;grade except D3 are quenched in either air or salt bath&lt;/li&gt;&#xA;&lt;li&gt;L Series:Low Alloy Special Purpose Tool Steels, Use many different&#xA;methods to quench including molten salt baths for some&lt;/li&gt;&#xA;&lt;li&gt;P Series: Mold Steels, most use oil or brine to quench.&lt;/li&gt;&#xA;&lt;li&gt;H Series: Hot Work Tool Steels, Molton Salt Bath, Atmosphere or&#xA;Vacuum to quench.&lt;/li&gt;&#xA;&lt;li&gt;T Series: Tungsten High-Speed Tool Steels, these are quenched in air,&#xA;oil or molten salt.&lt;/li&gt;&#xA;&lt;li&gt;M Series: Molybdenum High-Speed Tool Steels, these are quenched in&#xA;air, oil or molten salt.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Most of these materials are tested by time and usage for decades and most heat treat companies have perfected the processes they use to ensure a good finished product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every type of steel will react different to one specific heat treatment. This why there are so many methods. Each method is used for a specific purpose. So to answer your question broadly, no brine cannot be used for all types of tool steel and still have a good product. Some will crack some will distort. each type of steel needs to be quenched so the steel can accurately cool down at a specific rate to get the desired end result for each type of steel.A brine bath will not do this for all tool steels.&lt;/p&gt;&#xA;" OwnerUserId="18027" LastActivityDate="2020-01-14T17:51:19.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32774" PostTypeId="1" AcceptedAnswerId="32788" CreationDate="2020-01-15T03:16:36.240" Score="2" ViewCount="115" Body="&lt;p&gt;I have a plant &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt; which gives position and velocity as output&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ G = [G_{ru}(s)\quad G_{vu}(s)]^T $$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$G_{ru}$&lt;/span&gt; is the integral of &lt;span class=&quot;math-container&quot;&gt;$G_{vu}$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; is the input to the system. Now if I want to control the system with a proportional-derivative law I can write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ u = C e_r $$&lt;/span&gt;, with &#xA;&lt;span class=&quot;math-container&quot;&gt;$$C = (k_p +s k_d)$$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$e_r$&lt;/span&gt; is the error with respect to the reference position (e.g., a step command),&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can build the sensitivity and complementary transfer functions as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ S = (1+G_{ru}C)^{-1} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T = G_{ru}C(1+G_{ru}C)^{-1} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I instead consider the system as having 2 outputs, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I define a new controller as &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$C_{2} = [k_p\quad k_d]$$&lt;/span&gt;, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the corresponding sensitivities as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$S_2 = (I_2 + GC_2)^{-1}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$T_2 = GC_2(I_2 + GC_2)^{-1}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;we can see that &lt;span class=&quot;math-container&quot;&gt;$S_2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T_2$&lt;/span&gt; are now &lt;span class=&quot;math-container&quot;&gt;$2 \times 2$&lt;/span&gt; transfer matrices. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is: why &lt;span class=&quot;math-container&quot;&gt;$T_2(1,1)$&lt;/span&gt; differs from &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; since they have the same controller and represent the same thing (in this case the how the position behaves given a step command)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding bode plots are depicted here below. Only the dc gain is the same, but the transient is quite different. Why does this happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qN2dR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qN2dR.png&quot; alt=&quot;frequency domain&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PRWNu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PRWNu.png&quot; alt=&quot;step response&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="17008" LastActivityDate="2020-01-15T17:16:00.670" Title="analysis of PD controller vs static-gain" Tags="|control-engineering|control-theory|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32775" PostTypeId="1" AcceptedAnswerId="32791" CreationDate="2020-01-15T03:50:53.757" Score="-1" ViewCount="545" Body="&lt;p&gt;I am trying to find the forces in hinches &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;, &#xA;&lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$D$&lt;/span&gt;. The exact solution is not important for me. I introduced the following internal forces to solve the problem: &lt;span class=&quot;math-container&quot;&gt;$A_y$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$B_x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$B_y$&lt;/span&gt; in B, &lt;span class=&quot;math-container&quot;&gt;$C_x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_y$&lt;/span&gt; in &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;, a force &lt;span class=&quot;math-container&quot;&gt;$F_{DE}$&lt;/span&gt; in the direction &lt;span class=&quot;math-container&quot;&gt;$\vec{DE}$&lt;/span&gt;. But I think I am missing something. I have currently 6 unknowns. But I need equilibrium for the subsets ABE (grey), ABF (yellow) and CDG (yellow). Are there any other internal forces / torques I should introduce?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9G6ey.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9G6ey.png&quot; alt=&quot;&quot;&gt;Pruning shears&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/jsXKy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jsXKy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First subsystem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$C_x + F_{DE} \cos(45) = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$C_y + F_{DE} \sin(45) + 20= 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_{DE} \sin(45) \times 25 + 20 \times 150= 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second subsystem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$B_x - C_x = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$B_y - C_y - A_y - 20= 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$-F \times 150 + A_y \times 60 - C_x \times 30 = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Third subsystem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$-B_x - F_{DE} \cos(45)  = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$A_y -B_y - F_{DE} \sin(45)= 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$-A_y \times 60+ F_{DE} \sin(45) \times 55  = 0$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have 9 equations but only 6 unknowns. Are there any linear interdependencies between the equations?&lt;/p&gt;&#xA;" OwnerUserId="24266" LastEditorUserId="24266" LastEditDate="2020-01-15T15:51:47.597" LastActivityDate="2020-01-15T19:21:57.547" Title="Forces on pruning shears" Tags="|statics|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32776" PostTypeId="1" AcceptedAnswerId="32781" CreationDate="2020-01-15T04:51:37.297" Score="2" ViewCount="7848" Body="&lt;p&gt;In solidworks , is it possible to trim a part against another part in assembly level&lt;/p&gt;&#xA;" OwnerUserId="24267" LastActivityDate="2020-11-24T15:13:40.407" Title="Trim parts against parts in assembly level solidworks" Tags="|solidworks|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32777" PostTypeId="1" CreationDate="2020-01-15T04:57:44.150" Score="1" ViewCount="317" Body="&lt;p&gt;Do grey cast iron expand from freezing point to room temperature (cooling process)?&lt;/p&gt;&#xA;" OwnerUserId="24268" LastActivityDate="2020-01-16T01:53:15.403" Title="grey cast iron from freezing point to room temperature (cooling process)" Tags="|mechanical-engineering|manufacturing-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32778" PostTypeId="1" AcceptedAnswerId="32846" CreationDate="2020-01-15T06:35:39.040" Score="-1" ViewCount="299" Body="&lt;p&gt;I had/have a 3D camera back in the day, and I gotten a few pictures. I kinda stopped using it since there wasn't anything I could do with the MPO files. However, the camera might gain a new life if I can turn pictures into a CNC carving by grayscale the MPO files. Does anyone know if this is possible and where it is?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW I wasn't sure if this is the right subgroup to ask this to. If it isn't, then please kindly direct me to the sub group I should ask such a question to.&lt;/p&gt;&#xA;" OwnerUserId="24269" LastActivityDate="2020-01-20T16:14:10.363" Title="Is it possible to grayscale a MPO file (3D camera) for a CNC carving" Tags="|3d-printing|cnc|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32781" PostTypeId="2" ParentId="32776" CreationDate="2020-01-15T09:54:36.787" Score="2" Body="&lt;p&gt;Yes, it is possible. The cut must be added to the feature tree of the part to be trimmed, but the location of the body remains defined within the assembly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wouldn't usually recommend this as best practice, however, since there are a lot of interlinking references. (Suggesting an alternate referencing structure is outside the scope of this question, however).&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;In the assembly, click Edit Component for the part to be cut&lt;/li&gt;&#xA;&lt;li&gt;Insert - Feature - Cavity&lt;/li&gt;&#xA;&lt;li&gt;Select the part that you wish to do the cutting with.&lt;/li&gt;&#xA;&lt;li&gt;Hide the cutting tool if necessary. You can't suppress or delete it, as the cavity feature is referencing it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Xrioa.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Xrioa.gif&quot; alt=&quot;Assembly Boolean Demo&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2020-01-15T09:54:36.787" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32782" PostTypeId="1" CreationDate="2020-01-15T11:00:02.887" Score="1" ViewCount="354" Body="&lt;p&gt;I would like to construct an aluminium profile frame which includes connections with stiff fixed angles at 60 and 120 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What options do exist for creating those connections, preferably without having to further process the profiles themselves?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many ways for connecting elements at 0 or 90 and sometimes 45 and 135 degrees, but even with exact cuts at different angles (which I would like to avoid) there do not seem to be many tools for creating other connections available.&lt;/p&gt;&#xA;" OwnerUserId="24272" LastActivityDate="2020-01-15T11:00:02.887" Title="How do I to connect aluminium profiles with 60 and 120 degree angles?" Tags="|aluminum|connections|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32784" PostTypeId="1" AcceptedAnswerId="32786" CreationDate="2020-01-15T15:40:39.370" Score="1" ViewCount="75" Body="&lt;p&gt;I am putting together a basic machine for “pad printing” extremely small and precise parts. The design is very much the same as most homemade drill presses and consists of linear rods, flanged bearings and compression springs, but in place of a drill bit is a 30mm conical silicone rubber pad/stamp. The object being stamped will be positioned underneath on a precision XYR stage. A basic hand lever will be used to press the the rails/block downwards, applying minimal pressure. Scale will be approximately 10-15” tall. The blocks will be made of aluminium at this point and will be mounted to a very large solid base. Screws will he used to fix each component together. My goal, once set up, is for the pad to land in the exact same spot every time and with absolutely no movement or variation. It needs to be very rigid and with no slop.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like to know what diameter linear rods I should use, and if it makes much difference. I was originally looking at 40mm but this seems large?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I use standard steel rods? I have also considered introducing a ball spline shaft to prevent rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure what size or type screws would be best suited for holding this thing together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any other factors I am overlooking that will help out? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have come up with a few different layouts and I may be overthinking it. Do any of them make a difference to the overall result and accuracy of the press? Is there any particular direction I should be taking?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you &lt;a href=&quot;https://i.stack.imgur.com/66HMX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/66HMX.jpg&quot; alt=&quot;concepts&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24274" LastEditorUserId="24274" LastEditDate="2020-01-15T15:55:44.583" LastActivityDate="2020-01-16T09:10:14.073" Title="Help with linear rod drill press like design/materials" Tags="|springs|linear-motion|coil-spring|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32785" PostTypeId="2" ParentId="32777" CreationDate="2020-01-15T16:05:12.163" Score="1" Body="&lt;p&gt;The thermal expansion coefficient for grey cast iron is ~ 10.8 in/(in*degR) see: &lt;a href=&quot;https://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;list of expansion coefficients&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would indicate that expansion is expected from ambient temperature to melting temperature so the reverse would be to expect shrinkage while cooling to ambient temperature.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slightly related note, I was taught that gray cast iron is more ductile than regular cast iron so it is less susceptible to shrinkage cracking while it is cooling, for whatever that may be worth.&lt;/p&gt;&#xA;" OwnerUserId="24234" LastActivityDate="2020-01-15T16:05:12.163" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32786" PostTypeId="2" ParentId="32784" CreationDate="2020-01-15T16:12:03.150" Score="2" Body="&lt;p&gt;it may be necessary to construct more than one design (known as iteration) in order to refine your final objective. &quot;Very rigid and with no slop&quot; is a rather general description. &quot;A tolerance of 0.10 mm&quot; would be more precise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding of pad printing is that one requires the impression pad to be re-inked from a source located near the destination. This suggests that your design requires a rotating head, which is another location of &quot;slop&quot; in the construction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you do not require maximum efficiency, you could get away with having the re-inker in the same location as the parts to be inked, with suitable alignment guides to ensure that the precision is maintained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you require the re-inker to be offset, your drawings don't show a precise enough option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally speaking, for the most precision, use rectangular/linear assemblies and avoid cylindrical mounts for anything moving. My mini-milling machine has a rectangular column, while my drill press has a cylindrical column, as the drill press does not offer the precision of the mill. Related to this build question, the milling machine does not rotate about the z-axis and would not be capable of re-inking in the offset manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One of your drawings shows a pair of vertical rods. This approximates a rectangular support structure, but allows for torsion of the rods in the rotation of the z-axis. The 40 mm diameter referenced is not going to be &quot;too big&quot; for this type of structure, based on the description of the precision required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unrelated to this build, have you considered photo-lithography to accomplish your objective?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many details omitted from your design description, but enough of a basic question that this post may give you assistance.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2020-01-16T09:10:14.073" LastActivityDate="2020-01-16T09:10:14.073" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32787" PostTypeId="2" ParentId="32481" CreationDate="2020-01-15T17:15:36.807" Score="0" Body="&lt;p&gt;(Partly answering my own question)&#xA;It appears that if I use a different solution method, I actually get the expected result for &lt;em&gt;constant&lt;/em&gt; input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The alternative solution consists in solving the whole ODE (including mass transport terms) symbolically, and then modifying the resulting algebraic equation to accept &lt;code&gt;t&lt;/code&gt; and &lt;code&gt;q_in&lt;/code&gt; as &lt;em&gt;vectors&lt;/em&gt;, rather than just scalars.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function resultsArray = reactorModel2(D_feed, initialValue, parameters, D_in)&#xA;    % Simulate gas production per hour with algebraic reactor model&#xA;    % Feed is solved as part of an algebraic equation (symbolic solution of CSTR ODE)&#xA;&#xA;    hoursTotal  = length(D_feed);&#xA;    t           = (1:hoursTotal).'/24;&#xA;    q_in        = D_feed/D_in*1e3*24; % m^3/h&#xA;&#xA;    k_1         = parameters(1);&#xA;    Y_M         = parameters(2);&#xA;    Y_N         = parameters(3);&#xA;    V_liq       = parameters(4);&#xA;    resultsArray = zeros(hoursTotal, 3);&#xA;    % Symbolic solution of CSTR ODE&#xA;    resultsArray(:, 1) = (D_in*q_in + exp(-(t.*(q_in + V_liq*k_1))/V_liq).*(q_in*(initialValue - D_in) ...&#xA;        + V_liq*k_1*initialValue))./(q_in + V_liq*k_1); % kg/m^3&#xA;&#xA;    resultsArray(:, 2) = V_liq*Y_M*k_1*resultsArray(:, 1)*1e-3; % m^3/d&#xA;    resultsArray(:, 3) = V_liq*Y_N*k_1*resultsArray(:, 1)*1e-3; % m^3/d&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The corresponding call from the console&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;resultsArray2 = reactorModel2(D_feed, initialValue, parameters, D_in);&#xA;[resultsArray2(end, 1), resultsArray2(end, 2:3)/24]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;gives&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;ans =&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;8.5714  120.0000  111.1429&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;When I compare both solutions graphically, you can see that they slightly differ. Not sure why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vlQdM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vlQdM.png&quot; alt=&quot;compareBothSolutionMEthodsWithConstantInput&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, as soon as I start using a &lt;em&gt;dynamic&lt;/em&gt; input vector (maintaining the overall sum of fed amount constant), Method 1 gives far more plausible results than Method 2. But this should be adressed in another question, I guess.&lt;/p&gt;&#xA;" OwnerUserId="23969" LastActivityDate="2020-01-15T17:15:36.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32788" PostTypeId="2" ParentId="32774" CreationDate="2020-01-15T17:16:00.670" Score="4" Body="&lt;p&gt;Both the (negated) output and reference  are needed to calculate the error. So in the scalar case the reference also gets multiplied by &lt;span class=&quot;math-container&quot;&gt;$s\,k_d$&lt;/span&gt;, while &lt;span class=&quot;math-container&quot;&gt;$T_2(1,1)$&lt;/span&gt; only considers the contribution of &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$y_1$&lt;/span&gt;. In order to get the correct results you should also add the contribution of &lt;span class=&quot;math-container&quot;&gt;$s\,r$&lt;/span&gt;, which could be expressed as &lt;span class=&quot;math-container&quot;&gt;$s\,T_2(1,2)$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$T_2(2,2)$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2020-01-15T17:16:00.670" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32789" PostTypeId="1" AcceptedAnswerId="32790" CreationDate="2020-01-15T18:43:31.807" Score="1" ViewCount="47" Body="&lt;p&gt;I have a stepper motor linear actuator assembly &lt;a href=&quot;https://www.amazon.co.uk/Akozon-Automatic-Actuator-Aluminum-Effective/dp/B07JYGRFFL/ref=asc_df_B07JYGRFFL/?tag=googshopuk-21&amp;amp;linkCode=df0&amp;amp;hvadid=375410926273&amp;amp;hvpos=1o2&amp;amp;hvnetw=g&amp;amp;hvrand=9849643731108924450&amp;amp;hvpone=&amp;amp;hvptwo=&amp;amp;hvqmt=&amp;amp;hvdev=c&amp;amp;hvdvcmdl=&amp;amp;hvlocint=&amp;amp;hvlocphy=1006948&amp;amp;hvtargid=pla-633082499162&amp;amp;psc=1&amp;amp;tag=&amp;amp;ref=&amp;amp;adgrpid=77488740915&amp;amp;hvpone=&amp;amp;hvptwo=&amp;amp;hvadid=375410926273&amp;amp;hvpos=1o2&amp;amp;hvnetw=g&amp;amp;hvrand=9849643731108924450&amp;amp;hvqmt=&amp;amp;hvdev=c&amp;amp;hvdvcmdl=&amp;amp;hvlocint=&amp;amp;hvlocphy=1006948&amp;amp;hvtargid=pla-633082499162&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this for example&lt;/a&gt;, but the stepper shaft seems able to move axially back and forth - hence when I change direction it has to do a few rotations, forcing the ballscrew against one end or the other, before the rail platform itself starts moving. Why could this be happening and what can I do to fix this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone has a diagram showing the internals of a NEMA stepper motor that would be really helpful also. I've seen many on the internet but none showing me enough to understand this problem. Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="11394" LastActivityDate="2020-01-15T19:05:34.207" Title="How to constrain stepper motor inside case?" Tags="|mechanical|stepper-motor|actuator|linear-motion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32790" PostTypeId="2" ParentId="32789" CreationDate="2020-01-15T19:05:34.207" Score="2" Body="&lt;p&gt;Constrain the driven shaft and use a coupling to the motor.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-01-15T19:05:34.207" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32791" PostTypeId="2" ParentId="32775" CreationDate="2020-01-15T19:21:57.547" Score="2" Body="&lt;p&gt;Writing down a system of equations for all the unknowns is generally not the best strategy.  Instead, look for strategic opportunities to apply one equation of equilibrium, to one component, where it will involve one unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A good starting point is often to sum moments about some location that is in line with one or more of your unknowns.  That way, those unknowns are not part of the equation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if you look at part CDE, and you take the moment about C, you only have to deal with force DE as an unknown.  With one equation, you solve for one unknown, force DE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, stay with part CDE.  Sum horizontal forces, and that will give you the horizontal component of the force at C.  Sum vertical forces and that will give you the vertical component of the force at C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you know all the forces that touch part CDE, so move on to different part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, use the strategy of summing moments at a location that is in line with one or more unknowns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at part ABF, and sum moments about B.  You know the forces at point C, so the only unknown will be the force at point A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep going, and you will get all the unknowns.&lt;/p&gt;&#xA;" OwnerUserId="24276" LastActivityDate="2020-01-15T19:21:57.547" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32792" PostTypeId="2" ParentId="32777" CreationDate="2020-01-15T22:12:23.493" Score="1" Body="&lt;p&gt;No and yes; and no. Pearlitic grey freezes as austenite plus graphite. It shrinks as it cools until the austenite transforms to pearlite , this causes expansion over a very short temperature range . Then the pearlite and graphite continue to shrink. This is a higher strength grey iron. The other grey iron ( most common): Ferritic iron probably freezes as ferrite and graphite and shrinks all the way to room temperature ( no phase transformation).( There is a chance there is some austenite to ferrite transformation at high temperature like 1600 F if small amounts of austenite are formed during freezing). Nodular (ductile) iron is about the same . Malleable iron is a different story.White cast iron is very hard and brittle ,has few uses , mostly it is made into malleable iron by heat-treatment&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2020-01-16T01:53:15.403" LastActivityDate="2020-01-16T01:53:15.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32794" PostTypeId="1" CreationDate="2020-01-16T12:48:24.330" Score="2" ViewCount="638" Body="&lt;p&gt;When we measure the undrained strength of clay via CPT or triaxial test for instance, is there a way to correlate this to drained strength? How? I ask because measuring drained strength takes very long time but undrained strength can be measured quickly&lt;/p&gt;&#xA;" OwnerUserId="24283" LastEditorUserId="90" LastEditDate="2022-08-03T00:46:53.977" LastActivityDate="2022-08-03T00:46:53.977" Title="Drained and undrained shear strength of clay correlated?" Tags="|geotechnical-engineering|soil-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32795" PostTypeId="1" AcceptedAnswerId="32796" CreationDate="2020-01-16T12:56:25.677" Score="1" ViewCount="151" Body="&lt;p&gt;I'm trying to cut some bendable flaps on aluminium sheets for hooking/clipping onto some fabric. The flaps I have in mind look like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FFklU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FFklU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are these common features? Does anyone know the geometry of the cutout? I wish SolidWorks had them as standard features.&lt;/p&gt;&#xA;" OwnerUserId="4302" LastActivityDate="2020-01-16T13:13:15.710" Title="Bendable flaps on sheet metal" Tags="|design|cad|metals|cutting|metal-folding|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32796" PostTypeId="2" ParentId="32795" CreationDate="2020-01-16T13:13:15.710" Score="1" Body="&lt;p&gt;They will need to be sized appropriate to the load that will be applied, which means that they are not “standard”.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need to make sure the part that bends is sufficiently strong and will have sufficient life when it is flexed often.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-01-16T13:13:15.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32797" PostTypeId="2" ParentId="29702" CreationDate="2020-01-16T13:16:54.237" Score="0" Body="&lt;p&gt;British Standard parallel pipe threads aren't tapered (as the name implies), BSPP for short. That might work for you.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-01-16T13:16:54.237" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32799" PostTypeId="2" ParentId="32794" CreationDate="2020-01-16T16:29:33.767" Score="2" Body="&lt;p&gt;They are certainly correlated but there is no simple relationship. A commonly used rule of thumb, where I live, is &lt;span class=&quot;math-container&quot;&gt;$c'=0.1c_u$&lt;/span&gt; for overconsolidated clays and zero otherwise, but reality is much more complex. See for example &lt;a href=&quot;https://www.geo.dk/media/1209/correlation-between-drained-shear-strength-and-plasticity-index-of-undisturbed-overconsolidated-clays_final-after-1-review_b_sorensen-okkels.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt;. To the best of my knowledge, no accurate and generic relationship has been found.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2020-01-16T16:29:33.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32801" PostTypeId="2" ParentId="18556" CreationDate="2020-01-17T00:10:02.083" Score="2" Body="&lt;p&gt;to prevent the nut from unscrewing, you must use either 1) a thread-locking compound (Loctite or equivalent) on the threads during assembly, 2) split lock washers, 3) an aircraft-grade castellated nut-and-safety-wire assembly, or 4) a nut with a nylon insert to prevent unscrewing (Nylock or equivalent).&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-01-17T00:10:02.083" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32802" PostTypeId="1" CreationDate="2020-01-17T02:59:34.537" Score="0" ViewCount="42" Body="&lt;p&gt;Most architectural companies have a template of blueprints they reuse for every project. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an owner-builder, I need to make these blueprints from scratch&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On page 1, I have to list all applicable 2020 Residential Building codes for California&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I find a summary list of applicable building codes I can put on the first page &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you,&lt;/p&gt;&#xA;" OwnerUserId="24289" LastActivityDate="2020-01-17T21:39:02.760" Title="Where can I find a list of building codes to put on a set of blueprints for a addition" Tags="|structural-engineering|civil-engineering|architecture|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32804" PostTypeId="1" CreationDate="2020-01-17T06:27:15.700" Score="0" ViewCount="55" Body="&lt;p&gt;Hello I am researching first millennium BC Nile freighter rudders. They sailed upstream and came  back down with the current. The Greek writer Herodotus tells us that going downstream they were towed by a raft  and towed a stone behind, no doubt to provide steeridge way. Most had twin quarter rudders but a recent excavation had two closely spaced holes in the keel.  Amongst other possibilities they may have been used for a pair of rudders with blades on one side only with one facing forward and the other facing aft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In use they may have acted as a reversable rudder and skeg. The rearward blade was lashed amidships when going downstream with the ship presenting her stern to the current and the forward rudder would be used. When sailing upstream the forward rudder would be lashed and the rearward one used. Overall length of the rudders is unknown as is the profile. Do you think the hydrodynamics could act to make it function as a single foil given that the blades were flat.  The  vessel was around 26M in length &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q&lt;/p&gt;&#xA;" OwnerUserId="24291" LastEditorUserId="24291" LastEditDate="2020-01-21T05:56:50.813" LastActivityDate="2020-01-21T05:56:50.813" Title="First millennium BC Egyptian rudder systems" Tags="|fluid-mechanics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32805" PostTypeId="1" CreationDate="2020-01-17T06:54:02.600" Score="1" ViewCount="172" Body="&lt;p&gt;In my project, I have to control the linear motion in one of the two modes. It will  be either in position control or load control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;for the displacement control, The servo motor takes the quadrature encoder input as a feedback signal to control the displacement and getting the very accurate displacement control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same way, for the load control mode, I want to control the motor, based on the load cell signal as a feedback input. I could not find a possibility of to do so without having an intermediate controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a servo drive, there is a torque mode control mode. But there is no feedback input in the drive. In this case, I think, the servo drive is controlling the motor torque may be through the current consumption of the motor. Rather, is it possible to control the motor based on load cell signal as a feedback input? If there is a solution to connect load cell input for the torque control, then intermediate controller for load control can be avoided.&lt;/p&gt;&#xA;" OwnerUserId="15386" LastEditorUserId="15386" LastEditDate="2020-01-20T07:12:06.857" LastActivityDate="2020-01-20T07:12:06.857" Title="Is it possible to have the load cell signal as feedback to the servo drive of the servo motor?" Tags="|motors|linear-motion|servo|linear-motors|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32806" PostTypeId="1" AcceptedAnswerId="32807" CreationDate="2020-01-17T09:59:11.843" Score="3" ViewCount="45" Body="&lt;p&gt;Let's assume I have a cogeneration plant that produces heat with a nominal output. The heat is given into a heat storage tank, which is filled from above. The maximum flow temperature of the cogeneration plant is 85 °C, if this occurs an emergency shutdown takes place. The maximum water temperature in the heat storage tank is 95 °C.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GcuNy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GcuNy.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it theoretically possible in this case to reach a water temperature of over 85 °C only with the cogeneration plant?&lt;/p&gt;&#xA;" OwnerUserId="23627" LastActivityDate="2020-01-17T10:18:54.170" Title="Question about flow temperature and storage temperature in a cogeneration plant" Tags="|electrical-engineering|energy|heating-systems|energy-efficiency|power-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32807" PostTypeId="2" ParentId="32806" CreationDate="2020-01-17T10:18:54.170" Score="4" Body="&lt;p&gt;when the cogeneration plant switches of upon reaching 85°C flow and without another heat source, there's no way to reach 95°C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you sure the switch-off happens at 85°C flow, and not at 75°C return &lt;em&gt;to&lt;/em&gt; the cogernetion plant(this is a typical value for Otto-cycle cogeneration plants for the emergency cooler to kick in)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have other heat sources and want to achieve a higher flow temp. you can use a setup with flow like this: cold site of storage -&gt; cogeneration plant -&gt; boiler -&gt; hot side of storage. &lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2020-01-17T10:18:54.170" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32808" PostTypeId="1" CreationDate="2020-01-17T10:21:30.927" Score="-1" ViewCount="82" Body="&lt;p&gt;I am new in learning C++. I have used this code to show an image, but there is an error that I do not understand! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the simple code:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;include iostream&lt;/p&gt;&#xA;&#xA;&lt;p&gt;include conio.h&lt;/p&gt;&#xA;&#xA;&lt;p&gt;include opencv2/core.hpp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;include opencv2/highgui.hpp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;include opencv2/opencv.hpp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;include opencv2/imgproc.hpp&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using namespace cv;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;using namespace std;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;int main()&lt;/p&gt;&#xA;&#xA;&lt;p&gt;{&#xA;    std::cout &amp;lt;&amp;lt; &quot;Hello World!\n&quot;;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Mat img = imread(&quot;lena.jpg&quot;);&#xA;&#xA;namedWindow(&quot;input image&quot;, WINDOW_NORMAL);&#xA;&#xA;imshow(&quot;image&quot;, img);&#xA;&#xA;waitKey(0);&#xA;&#xA;return 0;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;}&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7X9yK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7X9yK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;when I run, system.cpp will open and &quot;Exception Unhandled&quot; will be apear!!!&#xA;here is a picture of that, may you help please!&lt;/p&gt;&#xA;" OwnerUserId="24294" LastActivityDate="2020-01-17T12:28:18.033" Title="how can I solve this error in c++: Exception Unhandled?" Tags="|biomedical-engineering|" AnswerCount="1" CommentCount="2" ClosedDate="2020-01-18T00:50:51.430" ContentLicense="CC BY-SA 4.0" />
  <row Id="32809" PostTypeId="2" ParentId="32808" CreationDate="2020-01-17T12:28:18.033" Score="1" Body="&lt;p&gt;I have found the solution!&#xA;I did these three steps and everything is ok now!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1)  I restored my system through: start/create a restore point/ system protection/ system restore/next/next...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) I work with visual studio 2019 and I have two kind of libraries: v14 and v15, so I add both directories into the libraries direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) copy my code to new project!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;after these steps, the code works properly :D&lt;/p&gt;&#xA;" OwnerUserId="24294" LastActivityDate="2020-01-17T12:28:18.033" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32810" PostTypeId="1" CreationDate="2020-01-17T12:35:57.503" Score="1" ViewCount="32" Body="&lt;p&gt;Since ceramics are resilient to shear loads, could one make locking lugs in a gun out of them or will tensile loads also be present leading to failure of such lugs? Also, could those tensile loads be decreased by decreasing the aspect ratio of the lugs? &lt;/p&gt;&#xA;" OwnerUserId="18744" LastActivityDate="2020-01-17T20:01:20.017" Title="Are ceramic materials suitable for making locking lugs for guns?" Tags="|material-science|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32811" PostTypeId="2" ParentId="32802" CreationDate="2020-01-17T14:43:37.663" Score="2" Body="&lt;p&gt;The same Building Department that you eventually submit your plans to.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In California other than very small single story additions, you find it more practical to contract a professional.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any addition to second floor requires a complete structure upgrade.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any addition in a hillside site requires a complete geology/soils report.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would start from the planning department, they are trained to walk you through the process.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-17T14:43:37.663" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32813" PostTypeId="1" CreationDate="2020-01-17T16:38:59.960" Score="1" ViewCount="594" Body="&lt;p&gt;I'm having an issue when I draw some elements. They are been drawn away from the Z=0, is causing to be non-coplanar with other elements. So, I cannot make fillets or can find crossing points. I don't want to keep flatten them or setting Z=0 manually. Does anyone know how to start drawing in Z=0 by default? Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="24297" LastActivityDate="2021-10-30T17:01:05.063" Title="Autocad start drawing away from the Z=0 (non-coplanar)" Tags="|autocad|technical-drawing|drawings|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32814" PostTypeId="1" CreationDate="2020-01-17T19:34:55.487" Score="2" ViewCount="29" Body="&lt;p&gt;I'm looking for alternatives to measure the moisture content in a concrete mixer machine. The ones on the market are expensive, like Hidro-Mix for apparent reasons; mixers are a hostile environment for sensors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any information on alternative methods to measure and visualise the humidity inside a mixer or have an economical sensor brand for this purpose is welcome. &lt;/p&gt;&#xA;" OwnerUserId="14265" LastActivityDate="2020-01-17T19:34:55.487" Title="Moisture sensor for Mixers" Tags="|control-engineering|sensors|concrete|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32815" PostTypeId="2" ParentId="32810" CreationDate="2020-01-17T20:01:20.017" Score="1" Body="&lt;p&gt;ceramics exhibit very poor impact load resistance (low &lt;em&gt;toughness&lt;/em&gt;) and are extremely notch-sensitive, making them unsuitable for use in gun actions. What you want instead is something that can absorb and sustain impact loads without damage, which argues in favor of steel which has been surface-hardened instead. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-01-17T20:01:20.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32816" PostTypeId="1" CreationDate="2020-01-17T20:26:00.307" Score="1" ViewCount="13" Body="&lt;p&gt;(Sorry if there's a better place for this. It's my first question)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm doing a final work for a Bachelor in Electronics and my work consists in making a fuel meter for a car. Its connected via OBD2 port using CAN protocol.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My problem is i can't seem to find anywhere how to ask for the Mass Air Flow (MAF). The only thing i found was trouble codes eg. faulty MAF or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help is really apreciated, as im running out of time for my project and i've found little information so far.&lt;/p&gt;&#xA;" OwnerUserId="24298" LastActivityDate="2020-01-17T20:26:00.307" Title="(Mechanics/Electronics) Anybody know DTC code for asking Air flow measurement on OBD2?" Tags="|electronics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32817" PostTypeId="1" CreationDate="2020-01-17T20:48:58.747" Score="1" ViewCount="29" Body="&lt;p&gt;So I'm working on a personal project, and I need to figure out how much 12 gauge wire (which is ~2mm thick) I would need to cover a cone shaped rubber core, which has a surface area of roughly 3,417,810 mm^2. I have no idea where to start with figuring it out and would appreciate the help!&lt;/p&gt;&#xA;" OwnerUserId="24299" LastActivityDate="2020-01-17T20:48:58.747" Title="How can I find the length of 12 gauge wire needed for a cone of if I know it's surface area" Tags="|electrical-engineering|electrical|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32818" PostTypeId="2" ParentId="32802" CreationDate="2020-01-17T21:39:02.760" Score="0" Body="&lt;p&gt;Can pick sections from this text document and paste into plans&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://archive.org/stream/2019californiare00unse_0/2019californiare00unse_0_djvu.txt&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://archive.org/stream/2019californiare00unse_0/2019californiare00unse_0_djvu.txt&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24289" LastActivityDate="2020-01-17T21:39:02.760" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32819" PostTypeId="1" AcceptedAnswerId="32829" CreationDate="2020-01-18T03:20:08.587" Score="-1" ViewCount="48" Body="&lt;p&gt;My prof is the type who makes up his own questions (rather than pulling them from textbooks). This is his latest work... Anyone got any ideas on how to solve this? He assures me that you don't need velocity or chord or additional info (like Mach numbers) to figure it out and I keep ending up with 3 unknowns with only 1 or 2 equations! Here's the question:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;High-performance gliders have very large aspect ratios in order to&#xA;  soar efficiently (for reasons we will discuss later in the course). A&#xA;  1:30 scale glider model with an aspect ratio of AR=30 is tested in a&#xA;  water tunnel, and found to have a lift coefficient of CL=0.8 at a&#xA;  Reynolds number of Re=2.0*10^6. Based on this result, what is the&#xA;  maximum mass of a full-scale glider, operating in air at the same&#xA;  Reynolds number, that can maintain level flight? Give your answer in&#xA;  kg. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Acceleration due to gravity: g=9.81 m/s^2&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For water: ρ=1000 kg/m3 and μ=1.0*10^-5 kg/ms&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;For air: ρ=1.2 kg/m3 and μ=1.9*10^-5 kg/ms&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="23678" LastEditorUserId="23678" LastEditDate="2020-01-18T16:30:47.607" LastActivityDate="2020-01-19T01:34:07.670" Title="Aerodynamics - How can you solve for mass when velocity &amp; chord are not given?" Tags="|mechanical-engineering|aerodynamics|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32823" PostTypeId="1" AcceptedAnswerId="32824" CreationDate="2020-01-18T07:07:20.423" Score="1" ViewCount="338" Body="&lt;p&gt;I am currently trying to model the buckling behaviour of a cylinder using sol 400 nonlinear static analysis using Nastran. I am using the arc-length (modified Riks) method as apparently it can increment load and displacement in both positive and negative directions, as opposed to Newtons method which only increases the load or displacement increment. However, my problem is that the analysis always stops after a certain load due to convergence problems. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried changing many of the parameters such as MINALR/MAXALR ( min/max arc-length ratio between increments) and MXINC(max increments), however I still don't get any good results. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I feel like I am missing a parameter that enables the load to decrease after a maximum load has been reached. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am aware that load control and displacement control is possible, but my understanding is that arc-length method uses neither of these, and is sort of a combination of the two. My model has a force applied without any enforced displacement. Is this maybe the problem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any other ideas which might help me?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="24115" LastActivityDate="2020-01-18T11:44:50.830" Title="For arc-length method in nastran, how to analyse post-buckling behaviour with load decrease" Tags="|structural-engineering|structural-analysis|aerospace-engineering|buckling|nastran|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32824" PostTypeId="2" ParentId="32823" CreationDate="2020-01-18T11:44:50.830" Score="1" Body="&lt;p&gt;If your model is very symmetrical (e.g. a circular cylinder with an axial compressive load, modeled with a regular grid of elements not an automatically generated irregular grid) you can hit problems because when you reach the instability, there are several possible unstable paths and no way for the program to choose between them. (In the circular cylinder example, the cylinder could theoretically buckle in any lateral direction.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One solution is to make the model unsymmetrical, so it prefers to buckle in one particular direction. For example, apply a small side load in the direction you want it to buckle.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-01-18T11:44:50.830" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32826" PostTypeId="1" CreationDate="2020-01-18T22:48:29.920" Score="1" ViewCount="295" Body="&lt;p&gt;How do I calculate the deflection and stress of a beam that is rotating freely (for example in space) when a moment is being applied to it, for example by rocket motors at the ends of the beam?&lt;/p&gt;&#xA;" OwnerUserId="24305" LastActivityDate="2020-01-19T13:41:37.683" Title="Deflection and stress in a freely rotating beam" Tags="|beam|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32827" PostTypeId="2" ParentId="32826" CreationDate="2020-01-19T00:25:02.617" Score="0" Body="&lt;p&gt;If the moment applied starts from zero and grows up to the final level in a short period of time, such as a perturbation, it will cause the rod to vibrate roughly similar to diapason or if it is a rod its first mode of vibration with trace a pattern like X in the space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the moment applied holds steady after initial impact the rod will vibrate and rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Duhamel intgral can be applied in simple cases, but in real world situations like the effect of impact of a car crash into the road guard rail the use FEM methods is preferred.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-19T00:25:02.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32828" PostTypeId="1" AcceptedAnswerId="32833" CreationDate="2020-01-19T00:28:52.177" Score="1" ViewCount="90" Body="&lt;p&gt;I have a electronics assembly company, and we run two assembly lines. They consist of several machines, with the heaviest being around 4000lbs, on 6 metal casters. We currently have the line on a concrete slab, inside a large converted warehouse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The concrete slab is badly spalled, and apparently didn't have a vapor barrier when originally poured. If we leave anything heavy on the floor (cardboard box), it will be dripping with water the next day. We maintain a humidity level of 45%, which probably brings more water up also. We are located in Florida, so have a high water table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The landlord is willing to work with us to fix this issue, and a contractor recommended we do a vapor barrier on the slab, then a grid of 2x4's, then tongue and groove plywood. On the plywood we would install ESD safe vinyl tiles. They said this is common in gyms and dance studio's. They use rubber strips under the 2x4's to help dampen noise and vibration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's hard to find information on this method, is this suitable for this amount of weight, in an industrial environment? I really don't see of many other options if this isn't acceptable to do. I would assume we would do the 2x4's either 6&quot; or 12&quot; on center.&lt;/p&gt;&#xA;" OwnerUserId="3839" LastActivityDate="2020-01-19T13:52:04.377" Title="Raised plywood false floor for heavy equipment usages" Tags="|concrete|wood|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32829" PostTypeId="2" ParentId="32819" CreationDate="2020-01-19T01:34:07.670" Score="1" Body="&lt;p&gt;I discovered the solution method through more banging head against a wall:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You have to do a double-substitution of velocity and chord, together, and you only need to focus on the prototype.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Solve Reynolds equation for u&lt;em&gt;c (u&lt;/em&gt;c = Re*(mu)/(rho))&#xA;Use Aspect ratio to find area in terms of chord only (S = c^2 * (AR))&#xA;Plug area into lift force equation&#xA;Rearrange to get (u&lt;em&gt;c)^2 in the equation and solve using values for (rho), (u&lt;/em&gt;c), C_L, and (AR).&lt;/p&gt;&#xA;" OwnerUserId="23678" LastActivityDate="2020-01-19T01:34:07.670" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32830" PostTypeId="1" AcceptedAnswerId="32840" CreationDate="2020-01-19T02:20:16.227" Score="1" ViewCount="78" Body="&lt;p&gt;I want to design an artificial leg for disabled people. The idea is design of a Device for patients with problems on one side of the leg from the knee down. This device first detects intent to move through movement of the hip.  The knee flexes when it detect force in the ankle foot. Then it decides the type of motion needed for the foot by using force sensors placed on the sole of the healthy foot. These force sensors are placed on three different regions; heel, metatarsal and toe. Using the human gait pattern in relation to force and time. This device can segment the pattern of the person’s gait and analyze the required gait of the disabled ankle-foot so it may assist the patient to move appropriately. This device also assists in telerehabilitation as it records the patients gait data and sends it to a medical professional for efficient follow up and prevention of further injury or fall. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have little knowledge on where to start and I want to take courses. I know I need mechanical and electrical engineering lessons. However, I don't know which courses to take and where to take them. Any recommendations?&lt;/p&gt;&#xA;" OwnerUserId="24307" LastActivityDate="2020-01-20T12:18:27.260" Title="what to learn to design mechatronics device that move" Tags="|mechanical|robotics|electrical|" AnswerCount="3" CommentCount="1" ClosedDate="2020-01-20T21:51:29.700" ContentLicense="CC BY-SA 4.0" />
  <row Id="32831" PostTypeId="2" ParentId="32830" CreationDate="2020-01-19T03:51:48.530" Score="1" Body="&lt;p&gt;For the mechanical side, you should consider these:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Engineering Statics&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Engineering Dynamics&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Strength of materials&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Machine design&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Control systems&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mechatronics&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculus&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Linear algebra&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Differential equations&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;CAD&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These are pretty core mechanical engineering courses. Not sure what EE courses you should take, but hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-01-19T03:51:48.530" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32832" PostTypeId="2" ParentId="32826" CreationDate="2020-01-19T13:41:37.683" Score="2" Body="&lt;p&gt;To do this, first calculate the linear and angular acceleration of the beam produced by the applied loads, assuming it is a rigid body.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then calculate the &lt;em&gt;distributed&lt;/em&gt; inertia loads that are equivalent to the acceleration (i.e. mass &lt;span class=&quot;math-container&quot;&gt;$\times$&lt;/span&gt; acceleration for each point in the body).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, apply the &lt;em&gt;opposite&lt;/em&gt; of the inertia loads plus the applied loads. The total load on the system is now zero, so you can fix it at any point (e.g. its center of mass) and there will be no reaction where you fix it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a example (ignoring rotation, to keep it simple) consider a rocket modeled as a uniform bar of length &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;, cross section &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt;, and mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;, with an axial force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; applied at one end. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rigid body acceleration is &lt;span class=&quot;math-container&quot;&gt;$F/m$&lt;/span&gt; in the axial direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corresponding inertia loads are a uniform distributed force &lt;span class=&quot;math-container&quot;&gt;$F/L$&lt;/span&gt; applied along the length of the beam. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, you do a stress analysis with the force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; at the end, plus a distributed load &lt;span class=&quot;math-container&quot;&gt;$-F/L$&lt;/span&gt; along the length. The result is that the internal axial stress varies linearly, from &lt;span class=&quot;math-container&quot;&gt;$F/A$&lt;/span&gt; at the point where it is applied, down to zero at the other end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computer software may have an option to do all this automatically, called &quot;inertia relief&quot; or something similar.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-01-19T13:41:37.683" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32833" PostTypeId="2" ParentId="32828" CreationDate="2020-01-19T13:52:04.377" Score="3" Body="&lt;p&gt;The use in gyms and dance studios is not a good reference. Those floors are designed for dynamic impact loads from people dropping gym equipment and dancers jumping around. They need to be able to absorb the impact energy by deflecting under load, and therefore the plywood is introduced as a &quot;soft&quot; element which is able to bend with relative ease so the dancers don't hurt their knees and ankles when landing. The distance between the 2x4's is not just a maximum distance but also a minimum distance to prevent the floor from becoming too stiff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is generally not the best way to design floors for heavy machinery especially if the machinery is able to induce vibrations in the underlying structure. Then it will be dancing around. But assuming the machinery won't be moved around all that much, it is an option to create hardpoints to support it — supports that don't rely on the stiffness of the plywood. And by reducing the distance between the 2x4's and increasing the thickness of the plywood, it is possible to design it for fairly large point loads. So it is possible to do it right, but don't accept a design that has been copy-pasted from a gym or dance studio. Demand that is redesigned for the weight as well as any vibrations of your machinery and it should be okay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this type of use, I would normally recommend an epoxy-based coating applied directly on the concrete, but in your case it may be impossible to get the concrete dry enough for the coating to stick properly. So maybe a bitumen-based membrane on top of the concrete, a concrete screed on top of the membrane to protect it and then an epoxy-based coating on top of the screed could be an alternative approach capable of supporting very heavy machinery.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2020-01-19T13:52:04.377" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32834" PostTypeId="1" CreationDate="2020-01-19T19:20:46.577" Score="2" ViewCount="138" Body="&lt;p&gt;I came across the infamous control handle assembly in two different books, referencing the views differently. One of the had it like the following photo&lt;a href=&quot;https://i.stack.imgur.com/QT3lb.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QT3lb.jpg&quot; alt=&quot;One of the two labeling of the two views of the control handle body&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other book labelled the side view in this photo as the elevation (front view).&#xA;The question here is how to determine the correct choice of views? I remember from school that this will also determine the dimensioning on the views (in which direction numbers are put, and their relative positions' to the dimension lines), so if somebody could point me to the standard used in settling this issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One colleague told me the elevation should be the view with the greatest dimension, but I think the given example doesn't totally agree with this point. I mean there should be a decisive factor because in many machine parts it is not obvious which is which.&lt;/p&gt;&#xA;" OwnerUserId="24310" LastActivityDate="2020-01-19T19:20:46.577" Title="Differentiating between Two Orthographic Projections" Tags="|technical-drawing|machine-design|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32835" PostTypeId="1" CreationDate="2020-01-19T22:15:37.533" Score="1" ViewCount="25" Body="&lt;p&gt;**Background **&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm struggling finding how to calculate the pressure that occurs within my system and then once calculated I need to reduce that pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My system&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 1 inch copper pipe that is 9 inches long.  Inside that pipe I have a 3/4&quot; heating element designed for heating liquids with heat transfer oil surrounding it and threaded into the 1 inch pipe.  My desired temperature is about 450 degrees F. The system is a closed system and the fluid isn't moving so of course as I heat the oil up, pressure will increase.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will soldered copper connections handle this pressure?  Is there a way to calculate the maximum pressure at 450 degrees F?  I believe once I know that, a pressure relief valve should help with this problem? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for the help and feel free to ask clarification questions. &lt;/p&gt;&#xA;" OwnerUserId="14534" LastActivityDate="2020-01-19T22:15:37.533" Title="Oil pressure inside copper tube?" Tags="|thermodynamics|heat-transfer|pressure|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32836" PostTypeId="1" AcceptedAnswerId="32857" CreationDate="2020-01-19T22:54:50.993" Score="0" ViewCount="573" Body="&lt;p&gt;There are the &quot;consumer&quot; negative ion detectors which are about 50 USD, reviews appear to indicate that they are not measuring very much at all.  Then there are the professional ones which are $800 and up.  As I understand it the customary standard measurement is in millions of ions per cubic centimeter; the precision needed is not great, 25% or 20M/cm3 will do, range 10 to 300 would be very helpful.  Any ideas?&lt;/p&gt;&#xA;" OwnerUserId="21710" LastActivityDate="2020-10-17T18:05:55.557" Title="Simple negative ion detector circuit?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32837" PostTypeId="1" CreationDate="2020-01-19T23:08:08.330" Score="3" ViewCount="95" Body="&lt;p&gt;I've been trying for months to find a way to heat moving air inside a tube to high temperatures.  The closest to success I ever had was using an element from a heat gun. It would achieve the desired temperature but over time would burn out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This has led me to believe making my own coil is probably the best approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Parameters&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my desired temperature in the tube is about 450 degrees F with around 400 CFM going through it. At 1500 watts of power,  what diameter and length nickel chromium wire would I need for this? &lt;/p&gt;&#xA;" OwnerUserId="14534" LastEditorUserId="9770" LastEditDate="2020-01-20T21:52:46.817" LastActivityDate="2020-07-18T23:03:30.347" Title="Nickel Chromium Wire for high temperatures" Tags="|heating-systems|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32838" PostTypeId="2" ParentId="32837" CreationDate="2020-01-20T00:03:53.447" Score="0" Body="&lt;p&gt;Nichrome , very similar to Inconel 600 and 601 : Is/are the standards ,actually  about the only choice ; some manufacturers may put different names on the same stuff.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-01-20T00:03:53.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32839" PostTypeId="2" ParentId="32830" CreationDate="2020-01-20T07:11:33.293" Score="1" Body="&lt;p&gt;To add to @jko's answer:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;robotics mechanism design&lt;/p&gt;&#xA;&#xA;&lt;p&gt;robotics control system design, or generalized dynamic system controls theory&lt;/p&gt;&#xA;&#xA;&lt;p&gt;human biomechanics&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-01-20T07:11:33.293" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32840" PostTypeId="2" ParentId="32830" CreationDate="2020-01-20T07:33:26.870" Score="1" Body="&lt;p&gt;You will need to do some math and sizing calculations along the way so by all means check out the course suggestions in the other answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of the time, engineering isn't about inventing something new but finding a solution that also fits to your problem (like a specific type of sensor or bearing). Courses on robotics or prosthetics building should give you many hints about what parts etc. exist. Additionally, read up as much as you can on existing prosthetics how they go about things, try to spend some time at a relevant trade fair and have things shown and explained to you. Spend time with people using prosthetics and try to understand their gripes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm in a completely different field but to me, discussions on why &lt;em&gt;this&lt;/em&gt; solution was chosen over &lt;em&gt;that&lt;/em&gt; solution are often enlightening.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lastly, never underestimate manufacturability: Can the thing I design actually be built?&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="10902" LastEditDate="2020-01-20T12:18:27.260" LastActivityDate="2020-01-20T12:18:27.260" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32841" PostTypeId="1" CreationDate="2020-01-20T07:47:22.193" Score="7" ViewCount="3365" Body="&lt;p&gt;There is a heat storage tank that is directly loaded from the top and the heat is also taken from the top. &#xA;The colder water from the heating circuit return flow enters the heat storage tank at the bottom. This creates a layered water temperature in the heat storage tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are three temperature sensors inside the heat storage tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/b8Rfd.png&quot; width=&quot;514&quot; height=&quot;248&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to calculate the stored energy via these three temperature sensors?&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Edit - Calculation Attempt according to Solar Mike:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$V_{storage} = 1000\ l = 1\ m^3$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$t_{top} = 82\ ^\circ C$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$t_{middle} = 70\ ^\circ C$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$t_{bottom} = 55\ ^\circ C$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$t_{average} = \frac{82+70+55}{3} = 69\ ^\circ C$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Q = M \cdot c_{H_2O} \cdot \Delta T = V_{H_2O} \cdot p_{H_2O} \cdot c_{H_2O} \cdot \Delta T $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ \ \ = 1\ m^3 \cdot 1.163\ \frac{kWh}{m^3K} \cdot (69\ ^\circ C - 20\ ^\circ C) = 56.987\ kWh$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does this calculation give me a fairly correct value for the stored energy?&lt;/p&gt;&#xA;" OwnerUserId="23627" LastEditorUserId="23627" LastEditDate="2020-01-21T14:01:06.300" LastActivityDate="2022-10-11T13:02:20.297" Title="Calculation of the stored energy for a heat storage tank" Tags="|thermodynamics|energy|heating-systems|power-engineering|energy-storage|" AnswerCount="2" CommentCount="10" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32844" PostTypeId="1" CreationDate="2020-01-20T15:20:51.643" Score="1" ViewCount="42" Body="&lt;p&gt;What are the best foundation type for a construction in the city and earthquake prone area?&#xA;If bored pile foundation is the best, which type is the best, square, circle? what are the considerations of that type?&lt;/p&gt;&#xA;" OwnerUserId="14641" LastActivityDate="2020-01-20T16:19:04.513" Title="The best foundation type?" Tags="|structural-engineering|geotechnical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32846" PostTypeId="2" ParentId="32778" CreationDate="2020-01-20T16:14:10.363" Score="0" Body="&lt;p&gt;This was answered at another place. There is 2 programs that I know about now.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://triaxes.com/articles/from-stereo-pair-to-lenticular-print/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://triaxes.com/articles/from-stereo-pair-to-lenticular-print/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.depthmask.com/priceen.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.depthmask.com/priceen.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;none so far is that great when it comes to the details&lt;/p&gt;&#xA;" OwnerUserId="24269" LastActivityDate="2020-01-20T16:14:10.363" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32847" PostTypeId="2" ParentId="32844" CreationDate="2020-01-20T16:19:04.513" Score="2" Body="&lt;p&gt;It is the right foundation type that is the best. Not necessarily pylons or any one size fits all solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Design of foundation is a process tightly tied to the type of soil, geology of the site, building usage and many more factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When all these factors are worked together and the loads on the foundation are determined, there is a determination as to the type of foundation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This can be a combination of footings and grade beams, mat foundation, or if the competent soil strata is deep, pylons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In some structures a combination of all these is used because either the geology is not uniform or the structure massing and height varies.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-20T16:19:04.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32849" PostTypeId="2" ParentId="25235" CreationDate="2020-01-20T19:55:22.737" Score="0" Body="&lt;p&gt;Yes, it is possible to design an international smart energy grid that can transmit DC current for long distances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The earliest work on this that I'm aware of is from Gregor Giebel and Gregor Czisch, from about 2000. Czisch's thesis, written at Kassel University, was based on his cost-optimisation model looking at the trade-offs between storage, extra generation and long-distance transmission. You can find more about the work - some is in German, some in English - at &lt;a href=&quot;https://transnational-renewables.org/Gregor_Czisch/veroeffentlichungen.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;his website&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;HVDC transmission losses are low, of the order of 2-3% per 1000 km.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the work of Czisch and Giebel, both PV and wind had dropped in cost significantly, so it's now much cheaper than it was to over-build them both, and curtail when necessary, and that reduces the need for both short-term storage and interconnection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nevertheless, there is still a role for long-distance transmission: weather systems get up to about 2000km across, so there is value in having the network span areas larger than that. &lt;a href=&quot;https://spectrum.ieee.org/energywise/energy/renewables/noaa-model-finds-renewable-energy-could-be-deployed-in-the-us-cost-effectively-without-storage-to-cut-carbon-emissions&quot; rel=&quot;nofollow noreferrer&quot;&gt;Chris Clack and Sandy Macdonald&lt;/a&gt; did similar analysis for the USA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the need for long-term storage is still there. Storage hydro systems are of the right scale for this requirement - Europe has storage hydro which is of the order of 180 TWh, which is the right ballpark. However, it's geographically concentrated in a few places, whereas the need for storage is spread out at continental scale. So a high-capacity continental-scale grid is needed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The larger the grid, the more climatic systems it spans. And the more climatic systems it spans, the lower the storage requirements for all demand covered by that grid. A grid that spans Europe, the Middle East and North Africa could provide huge benefits to everyone in that area.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2020-01-20T19:55:22.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32850" PostTypeId="1" AcceptedAnswerId="32851" CreationDate="2020-01-20T22:45:26.727" Score="2" ViewCount="261" Body="&lt;p&gt;First time asking question here.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;I'm not a mechanical engineer!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a simple question that I can't explain the answer to it.&#xA;I have a double gear (2 concentric gears with different radius) and I don't understand why the torque is the same for both.&#xA;I found in the internet that the torque should be the same because they are cocentric, but don't understand why.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gear Example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7V1jF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7V1jF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24318" LastActivityDate="2020-01-21T01:12:39.967" Title="Torque between concentric gears" Tags="|gears|torque|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32851" PostTypeId="2" ParentId="32850" CreationDate="2020-01-21T01:12:39.967" Score="3" Body="&lt;p&gt;Consider, for example, the shaft passing through the two gears has a torque of 10Nm and the diameter of the larger gear is twice as the diameter of small gear, &lt;span class=&quot;math-container&quot;&gt;$ R_{large}= 2*r_{small} $&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tangential force on the circumference of the large gear is then &lt;span class=&quot;math-container&quot;&gt;$\frac{10Nm}{R}=\frac{10Nm}{2r}\quad$&lt;/span&gt;  but the tangential force on the small gear is twice as much&lt;span class=&quot;math-container&quot;&gt;$ =\frac{10Nm}{r} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the torque produced by the small gear is &lt;span class=&quot;math-container&quot;&gt;$  =\frac{10Nm}{r}*r=10Nm.$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the torque produced by the big gear is &lt;span class=&quot;math-container&quot;&gt;$=\frac{10Nm}{2r}{2r}=10Nm.  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So regrdless of the diameter of the gears the torque on them is equal. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-21T01:12:39.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32852" PostTypeId="2" ParentId="32841" CreationDate="2020-01-21T04:32:14.293" Score="1" Body="&lt;p&gt;It is difficult to calculate the heat capacity because we have two regimens contributing to the temperature gradient inside the tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Heat conductivity of the water establishes a temperature gradient descending from the core of the tank to the tank wall which would cause slow convection up, and advection by the agitation of the circulating pump which causes a fast and likely turbulent flow of hot water down and cold water up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The sensors' readings are not very helpful because of the spiral flow mixing the hot and cold water and disturbing a uniform predictable pattern of water circulation in the tank one would never know if the reading is of a random plume of water or a steady read.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A very rough estimate would be modeling the core temperature an average of the difference between the heat generator's output and the three sensors'.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the power of the circulation pump things will vary by a large amount.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend using simulation software.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-21T04:32:14.293" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32853" PostTypeId="2" ParentId="3928" CreationDate="2020-01-21T10:31:55.893" Score="0" Body="&lt;p&gt;I believe warmer air has more capacity to hold vapour so instead of condensing onto the items inside the cabinet, the moisture stays in the air. I have 2 cabinets and neither have any outlet so the water vapour can't go anywhere. Also they are relatively cheap to buy as have a simple design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They seem pretty effective though as mine run at the value I set (45% rh) with no noise and consume around 10 Watts.&lt;/p&gt;&#xA;" OwnerUserId="24222" LastActivityDate="2020-01-21T10:31:55.893" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32854" PostTypeId="1" CreationDate="2020-01-21T11:52:47.220" Score="-1" ViewCount="41" Body="&lt;p&gt;Elastic waves in structural elements are described by &#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;    \rho \frac{\partial^2 u}{\partial t^2} &amp;amp;= (2 \mu + \lambda) \frac{\partial^2 u}{\partial x^2} + \mu \left( \frac{\partial^2 u}{\partial y^2} +  \frac{\partial^2 u}{\partial z^2} \right) + (\lambda+\mu) \left( \frac{\partial^2 v}{\partial x \partial y} + \frac{\partial^2 w}{\partial x \partial z} \right) \\&#xA;    \rho \frac{\partial^2 v}{\partial t^2} &amp;amp;= (2 \mu + \lambda) \frac{\partial^2 v}{\partial y^2} + \mu \left( \frac{\partial^2 v}{\partial x^2} +  \frac{\partial^2 v}{\partial z^2} \right) + (\lambda+\mu) \left( \frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 w}{\partial y \partial z} \right) \\&#xA;    \rho \frac{\partial^2 w}{\partial t^2} &amp;amp;= (2 \mu + \lambda) \frac{\partial^2 w}{\partial z^2} + \mu \left( \frac{\partial^2 w}{\partial x^2} +  \frac{\partial^2 w}{\partial y^2} \right) + (\lambda+\mu) \left( \frac{\partial^2 u}{\partial x \partial z} + \frac{\partial^2 v}{\partial y \partial z} \right) &#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a thin plate, these equations result in lamb-waves. However, imagine that we make the plate much thinner such that we end up in a very thin sheet. Is it then possible to assume that there is no variation over the thickness (&lt;span class=&quot;math-container&quot;&gt;$\partial / \partial z =0$&lt;/span&gt;)? If there is no variation over the thickness (z-direction) of the thin sheet, then the equations become&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align}&#xA;    \rho \frac{\partial^2 u}{\partial t^2} &amp;amp;= (2 \mu + \lambda) \frac{\partial^2 u}{\partial x^2} + \mu \frac{\partial^2 u}{\partial y^2} + (\lambda+\mu) \frac{\partial^2 v}{\partial x \partial y}  \\&#xA;    \rho \frac{\partial^2 v}{\partial t^2} &amp;amp;= (2 \mu + \lambda) \frac{\partial^2 v}{\partial y^2} + \mu \frac{\partial^2 v}{\partial x^2} + (\lambda+\mu)\frac{\partial^2 u}{\partial x \partial y}  \\&#xA;    \rho \frac{\partial^2 w}{\partial t^2} &amp;amp;= \mu \left( \frac{\partial^2 w}{\partial x^2} +  \frac{\partial^2 w}{\partial y^2} \right) &#xA;\end{align}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Question&lt;/strong&gt;: Is this a real case? Is this form of the equations used for real systems/structures (if yes, which ones)? Does it have a specific name like plane stress, antishear strain, quasi 2D elastodyanmics ...? &lt;/p&gt;&#xA;" OwnerUserId="24322" LastActivityDate="2020-01-22T01:19:17.270" Title="Elastic waves in very thin plate" Tags="|structural-engineering|structural-analysis|structures|dynamics|acoustics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32855" PostTypeId="2" ParentId="22646" CreationDate="2020-01-21T12:18:21.363" Score="2" Body="&lt;p&gt;The constant rotational speed is the first excitation frequency, mostly referred to as 1P. The second excitation frequency is the rotor blade passing frequency: NbP in which Nb is the number of rotor blades: 2P for a turbine equipped with two rotor blades, 3P for a three bladed rotor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;source: &lt;a href=&quot;https://ocw.tudelft.nl/wp-content/uploads/Soft_soft___Berlijn_2002.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://ocw.tudelft.nl/wp-content/uploads/Soft_soft___Berlijn_2002.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24323" LastActivityDate="2020-01-21T12:18:21.363" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32856" PostTypeId="1" CreationDate="2020-01-21T15:28:40.963" Score="0" ViewCount="105" Body="&lt;p&gt;My Inverter has 3 MPPT inputs. For some design simplifications (Involving ATS), i would like the three arrays to have separate PV+ but join the PV- wires together at a bus-bar and take the Inverter negative DC input from there.&lt;a href=&quot;https://i.stack.imgur.com/McCnO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/McCnO.png&quot; alt=&quot;effective layout of the setup&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will this have an effect on the performance of MPPT operating independently? or should i keep the PV- wires separate?&lt;/p&gt;&#xA;" OwnerUserId="24324" LastActivityDate="2020-01-21T16:55:00.903" Title="Joining the negative PV wires of 3 solar arrays for 3 MPPT controllers in an inverter?" Tags="|solar-energy|photovoltaics|solar|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32857" PostTypeId="2" ParentId="32836" CreationDate="2020-01-21T15:29:29.080" Score="0" Body="&lt;p&gt;The standard measurement for a genuine, &lt;em&gt;high-density&lt;/em&gt; negative ion generator is 1,000,000 negative ions (anions) per cm3 &lt;em&gt;at a distance of 1 meter from the ionizer&lt;/em&gt;. That is the level that Dr. Michael Terman of Columbia University used in his well-known study of the treatment of depression using negative ions. I believe he has a patent on that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for an accurate ion counter, look at the &lt;a href=&quot;https://www.alphalabinc.com/product/aic/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Alpha Lab air ion counter&lt;/a&gt;. That's what we used in &lt;a href=&quot;https://web.archive.org/web/20180826040944/http://negativeiongenerators.com/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;our negative ion generator business&lt;/a&gt; (which we might sell because we are retiring).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the web, you'll find several variations of a two-transistor negative ion detector schematic. It can be quite useful; the LED just starts to glow at ~9000 ions/cm3. (In fact, our IDS-2 ion detector was based on that circuit. Google is your friend.) And the model IG-133* series we manufactured --that generated that ion density-- made its LED just glow &lt;em&gt;at 8' away&lt;/em&gt; from the ion emitter on those models. Therefore, it was quite useful in determining whether a particular ionizer was actually a high-density ionizer or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also make a simple but useful one with an NE-2 neon lamp in parallel with a .001 uF disc ceramic capacitor. Solder a 6&quot; piece of bare wire to one end, and hold the other end between your thumb and forefinger. Unless a static charge has built up on your body, it will slowly blink within 6&quot; from a high-density ionizer. We used to give these (IDN-1) away with every ionizer purchase.&#xA;&lt;hr&gt;&#xA;This is &lt;strong&gt;not&lt;/strong&gt; spam. &lt;strong&gt;We no longer sell any of these products.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="20985" LastEditorUserId="20985" LastEditDate="2020-01-21T15:45:32.967" LastActivityDate="2020-01-21T15:45:32.967" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32858" PostTypeId="2" ParentId="32856" CreationDate="2020-01-21T16:55:00.903" Score="1" Body="&lt;p&gt;Without seeing the inverter, if the maker provides 3 separate pv- then I would keep them separate. If they could be joined together, then the maker would have done so.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-01-21T16:55:00.903" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32859" PostTypeId="1" AcceptedAnswerId="32861" CreationDate="2020-01-21T18:53:00.563" Score="1" ViewCount="1542" Body="&lt;p&gt;I'm a computer scientist doing a dissertation in robotics and I'm designing a mechanical arm that mirrors a humans, as such I am wanting to pivot the entire arm from a &quot;shoulder&quot; using a stepper motor and some gearing, I was looking at holding torques for different stepper motors and realised that I'm not actually sure how to calculate the force that is going to be put on the arm and the motor when the arm is straight out perpendicular, is this the same as the holding torque? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think part of my confusion is I am not to sure what terms to actually search, so sorry if this has already been answered before.&lt;/p&gt;&#xA;" OwnerUserId="24326" LastActivityDate="2020-01-21T20:18:27.327" Title="Calculating the torque required to move and hold an arm parallel to the floor" Tags="|mechanical-engineering|gears|robotics|stepper-motor|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32860" PostTypeId="1" AcceptedAnswerId="32862" CreationDate="2020-01-21T18:54:24.773" Score="1" ViewCount="74" Body="&lt;p&gt;I need help designing simple circuits that resonate around specific frequencies. Basically I am looking for recommendations as to what size capacitor, inductor, and resistors I should get. Sorry if this is not the right forum - I am new to all of this!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will use them to model neuronal resonance in an undergrad course. Each student group would get a breadboard + components, simple audio / function generator (adjusting input frequency) and measure resulting AC voltage with a multimeter. They will estimate the resonant frequency range by noting which input frequency produces the highest voltage response on the multimeter, and then change out components of the circuit to see how their properties affect this output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was hoping to create circuits that resonate around 8 Hz, 16 Hz, and anywhere between 30-35 Hz. These don't have to be exact.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;" OwnerUserId="24325" LastActivityDate="2020-01-21T20:01:51.157" Title="Designing RLC circuits at specific frequencies for teaching use" Tags="|circuit-design|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32861" PostTypeId="2" ParentId="32859" CreationDate="2020-01-21T19:15:00.120" Score="1" Body="&lt;p&gt;To paraphrase &lt;a href=&quot;https://www.controleng.com/articles/stepper-motor-torque-basics/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.controleng.com/articles/stepper-motor-torque-basics/&lt;/a&gt;, holding torque is the amount of torque required to turn a stationary rotor. As long as this is higher than the torque provided by your &quot;arm&quot; when it is extended, it should remain in the position you set it. To determine the maximum torque needed, multiply the weight of your arm by the distance from the point of rotation to it's center of mass. This will give the N-m or lbs-ft required to raise the arm (your operating torque), this is &lt;strong&gt;different&lt;/strong&gt; than the holding torque.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-01-21T19:15:00.120" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32862" PostTypeId="2" ParentId="32860" CreationDate="2020-01-21T20:01:51.157" Score="2" Body="&lt;p&gt;Assuming that you have an &lt;span class=&quot;math-container&quot;&gt;$RLC$&lt;/span&gt; circuit, the resonant frequency will be&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ f_0 = \frac{1}{2\pi\sqrt{LC}} $$&lt;/span&gt;&#xA;so, theoretically, you can pick either &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; arbitrarily and then calculate the other for a given resonant frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, you want circuits that are resonant at low frequencies. For reasonable values of &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; you will need very large &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt;. As an example of something similar you could look at the crossover networks used with subwoofer speakers. In the real world, high-value inductors also tend to have a high parasitic resistance, because they are made with many turns of wire. This will tend to make the resonance appear less ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have an audio generator or function generator you may not be able to drive the low impedances that you will get at resonance. You will need &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; to be at least 50&amp;Omega;, perhaps several hundred ohms. For students to be able to observe a significant change in behavior at resonance, the total impedance of the &lt;span class=&quot;math-container&quot;&gt;$RLC$&lt;/span&gt; should be significantly greater than &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt; when close, but not at, resonance. This will also push you to large inductors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest that you try calculating some values and then do some shopping to see what you can purchase. When I have done these lab exercises I used circuits that were resonant in the tens of kilohertz range, and the components were fairly easy to find and easy for students to work with.&lt;/p&gt;&#xA;" OwnerUserId="20302" LastActivityDate="2020-01-21T20:01:51.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32863" PostTypeId="1" AcceptedAnswerId="32868" CreationDate="2020-01-21T20:11:27.130" Score="-1" ViewCount="130" Body="&lt;p&gt;it is known that the stability of a system with negative feedback may be analized through the Nyquist stability criterion, which is based on observing the number of turns around the point (-1;0) in the complex plane.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But, what if we want to apply the same criterion for a positive feedback system? Is it correct or should be modified in some ways?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can we say also for positive feedback that the system is unstable if and only if there are at least 1 pole with positive real part?&lt;/p&gt;&#xA;" OwnerUserId="24327" LastActivityDate="2020-01-22T04:21:48.167" Title="Nyquist stability criterion for positive feedback" Tags="|control-engineering|control-theory|systems-engineering|stability|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32864" PostTypeId="2" ParentId="32859" CreationDate="2020-01-21T20:18:27.327" Score="1" Body="&lt;p&gt;When the arm is straight out it has a torque &lt;span class=&quot;math-container&quot;&gt;$T= mg*L \quad $&lt;/span&gt;with L the distance between arm's CG and rotation center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your stepper motor should provide this torque to keep the arm in balance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if you need to rotate the arm up and down you need an additional torque such that if you rotate the arm with angular acceleration, &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$T_{rotation}=I\alpha$$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$  I= 1/3mL^2$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-21T20:18:27.327" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32865" PostTypeId="1" CreationDate="2020-01-21T22:12:41.213" Score="1" ViewCount="47" Body="&lt;p&gt;In &lt;a href=&quot;http://read.pudn.com/downloads166/ebook/758109/PCI_Express_CEM_1.1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;the PCIe specification&lt;/a&gt; (see page 75), the &quot;short arm&quot; of the bracket has two cutouts; one for the slot that the retaining screw fits in, and another square cutout that has additionally had the sides bent back at a diagonal angle. Actual brackets also have these features. However, at least some slot covers omit the diagonally-angled bends.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the purpose of this cutout, and especially, the bent bits?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For bonus points, what is the purpose of the round hole at the opposite side of the bracket that some brackets have, and some don't?&lt;/p&gt;&#xA;" OwnerUserId="21536" LastActivityDate="2020-01-21T22:12:41.213" Title="What purpose do the &quot;angled tabs&quot; on a PCIe bracket serve?" Tags="|computer-hardware|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32866" PostTypeId="2" ParentId="32854" CreationDate="2020-01-22T01:19:17.270" Score="1" Body="&lt;p&gt;I've not looked at your specific equations, but in general if you take a plate and make it very thin, that's called a &lt;strong&gt;&lt;em&gt;membrane&lt;/em&gt;&lt;/strong&gt;.  A membrane is to a plate what a string is to a beam.  It has no bending stiffness, and can only carry load through in-plane tension.  You can definitely have traveling waves in membranes.  E.g. a drum head is a membrane.  See &lt;a href=&quot;https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membrane&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membrane&lt;/a&gt; for just one example.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2020-01-22T01:19:17.270" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32868" PostTypeId="2" ParentId="32863" CreationDate="2020-01-22T04:21:48.167" Score="1" Body="&lt;p&gt;You could just multiply &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; with minus one, change the positive feedback to negative feedback and proceed with the standard analysis for the Nyquist stability criterion. Here &lt;span class=&quot;math-container&quot;&gt;$H$&lt;/span&gt; is the transfer function in the feedback path as shown in &lt;a href=&quot;https://www.electronicshub.org/wp-content/uploads/2015/07/Negative-feedback-signal1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;this image&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2020-01-22T04:21:48.167" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32869" PostTypeId="1" CreationDate="2020-01-22T13:39:30.673" Score="1" ViewCount="1084" Body="&lt;p&gt;Okay, bit of a weird one.  Imagine an object of known mass in free fall, tethered by a wire rope to an infinitely stiff sky hook. When that wire rope eventually tightens, what will be the stopping force on said object, i.e. what is the dynamic tension experienced by the wire?&lt;/p&gt;&#xA;&lt;p&gt;Using linear motion equations we can determine the final velocity of the object.  If the stopping time or distance were known then we could calculate the object's deceleration and therefore the force.&lt;/p&gt;&#xA;&lt;p&gt;However, how does one determine the deceleration? I thought about using Hooke's law or the wire's strain as a way to determine it's elongation - but one cannot work these out without the force!  Also I can't find any reliable info on a wire's stiffness. Is there a rule of thumb for something like this?&lt;/p&gt;&#xA;&lt;p&gt;Any help or pointers are gratefully received.&lt;/p&gt;&#xA;" OwnerUserId="10086" LastEditorUserId="19906" LastEditDate="2021-03-18T12:24:34.050" LastActivityDate="2021-03-18T13:49:13.177" Title="Stopping force on wire rope from a falling object attached to it" Tags="|structural-analysis|dynamics|linear-motion|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32870" PostTypeId="2" ParentId="32869" CreationDate="2020-01-22T15:10:08.647" Score="0" Body="&lt;p&gt;Let's say you have a wire of steel with the cross-sectional area &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; and length &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; (for instance, say 20 km). Then, the strain energy of the stretched wire is equal to the kinetic energy of the falling mass once it has reached a depth beyong &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; from the sky hook. Therefore, &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}mv^2+mgh=\frac{1}{2}A E_{steel}\epsilon_{steel}^2 \ $&lt;/span&gt; since the work done by strain is a triangular area. We can then calculate for &lt;span class=&quot;math-container&quot;&gt;$\epsilon = \epsilon_{steel}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\Delta x=\epsilon \cdot 20km.$&lt;/span&gt; And as has been mentioned in the comment by @Jonathon R swift, the mass will bounce back up.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="24822" LastEditDate="2021-03-18T13:18:32.797" LastActivityDate="2021-03-18T13:18:32.797" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32873" PostTypeId="1" AcceptedAnswerId="33470" CreationDate="2020-01-22T17:55:36.657" Score="4" ViewCount="39" Body="&lt;p&gt;On my air tool compressor or scuba gear you can see or hear a diaphragm regulator sloughing off a tiny amount of gas to the atmosphere when you adjust the regulator down.  Does a natural gas regulator do the same if there is variation in the supply line pressure?&lt;/p&gt;&#xA;" OwnerUserId="5967" LastActivityDate="2020-03-02T19:18:58.653" Title="gas regulator exhaust gas during diaphragm adjustment?" Tags="|compressed-gases|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32874" PostTypeId="1" CreationDate="2020-01-22T21:27:49.077" Score="1" ViewCount="178" Body="&lt;p&gt;I've always been very annoyed by those regular single beeps that smoke detectors start doing increasingly often as the batteries start to run out, to warn the user to replace the batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lately, I've been observing how a bunch of kids seem to sit in a room and just ignore those beeps completely, even after I point it out and ask them to really listen. They can't hear them. I seriously believe that they just don't physically hear them. Kind of like those dog whistles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this backed up by science? Do kids under a certain age literally not hear those &quot;low battery&quot; warning beeps? Is this maybe per design, even? If so, why? Isn't it important that as many humans as possible hear this so that they replace the old batteries with new ASAP?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's really frustrating to me when they don't even hear it, but I hear it &lt;em&gt;very&lt;/em&gt; well.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;I already asked this question a few days ago in &quot;Biology&quot;, but it was &quot;off-topic&quot;.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="24338" LastEditorUserId="1832" LastEditDate="2020-01-22T23:20:27.480" LastActivityDate="2020-01-23T04:49:37.597" Title="Do kids under a certain age not hear the warning beeps that smoke detectors make when they are running out of battery power?" Tags="|electrical-engineering|electrical|audio-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32875" PostTypeId="2" ParentId="32874" CreationDate="2020-01-23T04:49:37.597" Score="5" Body="&lt;p&gt;Unless they have hearing loss, young people tend to be able to hear a larger range of sound frequencies than &lt;a href=&quot;https://www.rehab.research.va.gov/jour/05/42/4suppl2/gordon-salant.html&quot; rel=&quot;noreferrer&quot;&gt;older people&lt;/a&gt;, particularly &lt;a href=&quot;https://www.scientificamerican.com/article/bring-science-home-high-frequency-hearing/&quot; rel=&quot;noreferrer&quot;&gt;higher frequencies&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Some security companies recently began manufacturing machines designed to emit an annoying sound that prevents teenagers from loitering outside stores and shops. Teens are effectively driven away, but many adults can’t hear this sound at all! Things got even more interesting when some kids realized they could turn this technology on their elders by making the sound into a ringtone. Students with the ringtone can receive text message notifications during class without many teachers or administrators suspecting a thing.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="90" LastActivityDate="2020-01-23T04:49:37.597" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32876" PostTypeId="1" CreationDate="2020-01-23T09:02:33.717" Score="1" ViewCount="95" Body="&lt;p&gt;Power steering on a car is a form of torque amplifier: the driver applies X Nm (or ftLb) of torque, and the power steering adds Y torque -  but maintaining the same angle of turn.  But what's the approximate ratio of Y:X for the power steering on a typical domestic car? Is it e.g 10:1 ?&lt;/p&gt;&#xA;" OwnerUserId="24340" LastActivityDate="2020-01-24T11:19:25.667" Title="By approximately what ratio does power steering (on a car) amplify the torque applied?" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32878" PostTypeId="1" CreationDate="2020-01-23T14:15:27.440" Score="2" ViewCount="149" Body="&lt;p&gt;I have a CAD model designed in FreeCAD and need to export it to a STEP file (neutral CAD format) to be used in numerical anaysis software. I would like to assign names to selected surfaces in order to later define analysis boundary conditions on them. Is there a way to&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Define surface names directly in FreeCAD so that they are kept by the STEP export and seen by the analysis pre-processor after import, or&lt;/li&gt;&#xA;&lt;li&gt;Manually edit the STEP file and inject the names at approproiate places&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Regarding the second option, I am able to identify the relevant surface and the definition looks like this&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#885 = CYLINDRICAL_SURFACE('',#886,53.);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;I tried adding a name between the apostrophes but this didn't seem to make any difference&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;#885 = CYLINDRICAL_SURFACE('mySurfaceName',#886,53.);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;A link to a detailed description of the STEP format would be very useful too.&lt;/p&gt;&#xA;" OwnerUserId="24342" LastActivityDate="2020-01-23T14:15:27.440" Title="Assign a name to surface in CAD STEP file" Tags="|finite-element-method|cad|cfd|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32879" PostTypeId="1" AcceptedAnswerId="32884" CreationDate="2020-01-23T14:22:01.443" Score="1" ViewCount="57" Body="&lt;p&gt;Imagine you are in the mountains and it is very cold. The walls are colder than the air, so air close to the wall will lose heat to the wall, and by losing heat it might lose its humidity (water condenses to the wall). Now I know this is bad for the paint on the wall. However, in some sense, it is heating up the wall, which will lessen the cooling effect of the air. Please tell me if my logic is correct.&#xA;Also, if the walls are now wet, could they allow for more condensation? Or it is impossible because all the surface is covered in water? &#xA;PS: House without heating nor insulation.&lt;/p&gt;&#xA;" OwnerUserId="24343" LastEditorUserId="24343" LastEditDate="2020-01-23T14:33:08.027" LastActivityDate="2020-01-24T17:24:36.283" Title="Wall Humidity Condensation" Tags="|thermodynamics|heat-transfer|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32880" PostTypeId="1" AcceptedAnswerId="32891" CreationDate="2020-01-23T14:19:16.213" Score="2" ViewCount="201" Body="&lt;p&gt;If I'm in the wrong section of stack exchange, please help me go to the right place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to find out if this combination of motor shaft and clamps will work out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found these clamps/bushings&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.maedler.de/product/1643/1621/spannsaetze-com-b-bohrung-6-bis-100mm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.maedler.de/product/1643/1621/spannsaetze-com-b-bohrung-6-bis-100mm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I'm curious if they will work out with the kind of shaft on this motor &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://mehatron.rs/step-motor-nema-34-86x155-60a-130nm-sa-duplom-osovinom&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mehatron.rs/step-motor-nema-34-86x155-60a-130nm-sa-duplom-osovinom&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem is that the shaft of that motor is machined and the clamp relies on the full surface of the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By the way, those clamps work for the motors with a shaft containing removable key ( for example MiGE 130ST-M15015 ).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aliexpress is full of these shaft couplings, but i do not believe that they are sufficient to hold these kinds of torque levels, for example :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.aliexpress.com/item/32967906911.html?spm=a2g0o.productlist.0.0.2f2c46a1ZWZH4p&amp;amp;algo_pvid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd&amp;amp;algo_expid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd-9&amp;amp;btsid=deae1df8-64c1-479e-9d65-e967ef04e93f&amp;amp;ws_ab_test=searchweb0_0,searchweb201602_8,searchweb201603_53&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.aliexpress.com/item/32967906911.html?spm=a2g0o.productlist.0.0.2f2c46a1ZWZH4p&amp;amp;algo_pvid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd&amp;amp;algo_expid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd-9&amp;amp;btsid=deae1df8-64c1-479e-9d65-e967ef04e93f&amp;amp;ws_ab_test=searchweb0_0,searchweb201602_8,searchweb201603_53&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;or these :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.aliexpress.com/item/32956726953.html?spm=a2g0o.productlist.0.0.2f2c46a1ZWZH4p&amp;amp;algo_pvid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd&amp;amp;algo_expid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd-8&amp;amp;btsid=deae1df8-64c1-479e-9d65-e967ef04e93f&amp;amp;ws_ab_test=searchweb0_0,searchweb201602_8,searchweb201603_53&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.aliexpress.com/item/32956726953.html?spm=a2g0o.productlist.0.0.2f2c46a1ZWZH4p&amp;amp;algo_pvid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd&amp;amp;algo_expid=0a5001e7-7d75-4397-9a01-0e5d64cde7cd-8&amp;amp;btsid=deae1df8-64c1-479e-9d65-e967ef04e93f&amp;amp;ws_ab_test=searchweb0_0,searchweb201602_8,searchweb201603_53&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The good thing about these sahft couplers is that they can be made to fir the shaft perfectly, with 2 screws for the 2 machined parts i was talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The forces are around 10 Nm and I want to avoid any kind of slipping.&lt;/p&gt;&#xA;" OwnerUserId="24349" OwnerDisplayName="Curious Guy" LastEditorUserId="24349" LastEditDate="2020-01-23T23:08:47.553" LastActivityDate="2020-01-24T18:34:33.170" Title="Motor shaft mounting with clamps and bushings" Tags="|motors|coupling|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32881" PostTypeId="2" ParentId="32879" CreationDate="2020-01-23T15:08:56.157" Score="2" Body="&lt;p&gt;One consideration that is being overlooked in your description is mass. The mass of the wall and the mass of the cold mountain are substantially greater than the mass of air that is in contact with the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, the air will have convection currents, which will move the airflow in such a way as to bring new moist air into contact with the wet wall. The wet wall also is subject to gravity, causing the water to flow downward, making available convection and conduction for the &quot;incoming&quot; air. There's really no limit to the condensation relative to &quot;heating&quot; the wall, if the outside temperature remains lower.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As long as there is a disparity of temperature between the outside and inside of the house, there will be airflow and depending on the relative humidity, condensation. If there is no influx of drier air to the house, the condensation may cease.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-01-23T15:08:56.157" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32882" PostTypeId="2" ParentId="32304" CreationDate="2020-01-23T19:29:21.533" Score="1" Body="&lt;p&gt;I have done something very similar to what you mention using two workflows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Ansys Workbench for a relatively complex simulation (structural optimization of a metal structure under high temperature conditions where the material starts to behave viscous)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Python scripts managing an external FE simulation code. The Python code basically generated a set of design variables using Differential Evolution and the response was computed with the FE code&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21516" LastActivityDate="2020-01-23T19:29:21.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32883" PostTypeId="1" CreationDate="2020-01-23T19:36:11.260" Score="2" ViewCount="44" Body="&lt;p&gt;Is there a standardized file format for parametrised solid modelling? I am familiar with STEP and IGES file formats for saving solid models, but AFAIK they don’t store information regarding parameters, so that the another person (not the one who created the file) can modify some parameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that every proprietary file formate (Solidworks, Inventor and so on) stores the parameters, but I am interested in a non proprietary file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="21516" LastActivityDate="2020-01-23T19:36:11.260" Title="Is there a standardized file format for parametrised solid modelling?" Tags="|computer-aided-design|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32884" PostTypeId="2" ParentId="32879" CreationDate="2020-01-24T06:27:47.520" Score="3" Body="&lt;p&gt;Certainly we could calculate the heat gain to the wall due to condensation (a glass of ice water will melt faster in humid climates due to the heat gain from condensation).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice, though, this will be insignificant compared to typical winter cooling loads.  Also, you will never heat the wall above the dew point of the air (temperature at which water condenses out of the air), which is likely in winter to be quite low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will never be able to actually heat the room this way because to heat the room you need the temperature of the wall to be above the temperature of the air in the room (heat always moves from high temperature to low temperature).  Once the wall is hotter than the air next to it, no water will ever condense from the air.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastEditorUserId="90" LastEditDate="2020-01-24T17:24:36.283" LastActivityDate="2020-01-24T17:24:36.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32885" PostTypeId="2" ParentId="32841" CreationDate="2020-01-24T07:06:57.527" Score="0" Body="&lt;p&gt;I feel like this may be an X-Y problem.  You are asking for the energy stored in a vessel, but what will you do with the information once you know?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Energy in water&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A glass of room temperature tap water has an energy content, but few would find this information useful.  Engineers tend to look at the change in energy either put in to water or that can be taken from it.  Like how much home heating we could produce from a quantity of hot water, or how many gallons of 65 degree water would it take to give a  45 degree C shower. From a pure physics standpoint, the total energy in liquid water at atmospheric pressure is the energy required to heat it from absolute zero to its melting point as ice, the energy to melt the ice, and the energy required to heat it to its current temperature.  But this isn't very useful because we are unlikely to encounter extremely low temperatures and so it would be difficult to extract all that energy.  Remember that heat always flows from hot to cold.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Useful energy in water&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your calculation shows that a temperature change of 49 degrees C will be an energy change of 57 KW-hr.  But what is that exactly?  You wont be able to run a 1000 watt blow drier for 57 hours using that energy.  You could use it to heat up some other volume of water, or possibly to heat up a room using a heat exchanger, but it becomes hard to extract all of that energy when the temperatures converge.  You use 20 degrees in the calculation, but don't describe what that represents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What we need&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what we need is to know what you are trying to do with this heated water.  Is the circle in the diagram taking out heat and then pumping cold water back into the system? Once we know what you want to do with all the hot water we can better answer how much capacity the water has to do it. Also, let us know how accurate you need the answers to be.  Moon shots and swimming pool heaters have different accuracy requirements  ;^)&lt;/p&gt;&#xA;" OwnerUserId="24351" LastEditorUserId="24351" LastEditDate="2020-01-25T05:10:08.283" LastActivityDate="2020-01-25T05:10:08.283" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32886" PostTypeId="1" CreationDate="2020-01-24T08:49:00.970" Score="0" ViewCount="86" Body="&lt;p&gt;Regarding the similarities, I know that the complete similarity is not always possible because all geometric, kinematic, dynamic similarity can not be met simultaneously. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because &lt;strong&gt;both&lt;/strong&gt; geometric and dynamic similarity can &lt;strong&gt;not&lt;/strong&gt; be achieved at the same time? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's the reasons? Please give some examples and explanations.&lt;/p&gt;&#xA;" OwnerUserId="23837" LastEditorUserId="23837" LastEditDate="2020-01-24T09:01:30.520" LastActivityDate="2020-01-27T00:48:47.273" Title="Complete Similarities" Tags="|fluid-mechanics|automotive-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32887" PostTypeId="2" ParentId="32876" CreationDate="2020-01-24T11:00:35.227" Score="1" Body="&lt;p&gt;I had a go at measuring it - on my Toyota Auris. I used a handheld digital device (for weighing luggage), which conveniently records the maximum weight sensed in a measurement.  I strapped it around the spoke of the steering wheel in a fixed location, just inside the rim, and made sure that I was always applying the force tangentially.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My measurements were all taken with the vehicle stationary (not while driving!) and with the car parked on gravel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the power steering on, turning the wheel seemed to take a force of anywhere between 5 and 20 N (so maybe 1-4 Nm torque). With the power steering off I got readings of between 15 and 25 N. Not great data, but it seems to suggest that the amplification from the power steering is &lt;em&gt;of the order of&lt;/em&gt; 10:1. Though, I am not very confident in this result, and I have to say that my subjective &lt;em&gt;impression&lt;/em&gt; was that it was greater than this&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If anyone can provide a more specific figure based on manufacturer's design specs-  for any domestic car  -  I would still appreciate it. &lt;/p&gt;&#xA;" OwnerUserId="24340" LastEditorUserId="24340" LastEditDate="2020-01-24T11:19:25.667" LastActivityDate="2020-01-24T11:19:25.667" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32890" PostTypeId="2" ParentId="32886" CreationDate="2020-01-24T18:10:11.003" Score="1" Body="&lt;p&gt;Complete similarity (Similitude) is not always possible because complete modeling is not always possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Geometric Similarity&lt;/strong&gt; is easy.  We just need the model to be a good scale representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Kinematic Similarity&lt;/strong&gt; is harder because we need the test conditions to mirror expected system conditions.  Can you mirror expected temperature, pressure, and speed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Dynamic Similarity&lt;/strong&gt; is also hard because the boundary conditions and boundary layer characteristics need to match between model and actual.  Dynamic similarity needs Kinematic Similarity (similar forces are required to make the dynamics similar), and Kinematic needs Geometric (or else we could not compare forces).  As a result, we cannot achieve dynamic without geometric.  Dynamic Similarity is extremely hard to achieve in some flow conditions, like high-speed aerodynamics where the fluid can no longer be considered incompressible.  But even in low-flow conditions it can be difficult to achieve, which is why the topic is complex, and why flow modeling in general is difficult.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastEditorUserId="24351" LastEditDate="2020-01-27T00:48:47.273" LastActivityDate="2020-01-27T00:48:47.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32891" PostTypeId="2" ParentId="32880" CreationDate="2020-01-24T18:34:33.170" Score="2" Body="&lt;p&gt;I am going to say that no, your chosen coupling would not work.  Based on the K-K cross-section in the drawing, there would be a large portion of the shaft that is not in contact with the coupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe what you need is a coupling where it is held against a flat in the shaft by a set screw, like the one here (though in your size):  &lt;a href=&quot;https://www.omc-stepperonline.com/shaft-coupling/5mm-6mm-flexible-coupling-18x25mm-cnc-stepper-motor-shaft-coupler-st-fc02.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.omc-stepperonline.com/shaft-coupling/5mm-6mm-flexible-coupling-18x25mm-cnc-stepper-motor-shaft-coupler-st-fc02.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will need to match your torque requirements to the coupling as well.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-01-24T18:34:33.170" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32892" PostTypeId="1" AcceptedAnswerId="32900" CreationDate="2020-01-24T18:49:45.457" Score="2" ViewCount="7815" Body="&lt;p&gt;I am stuck on this problem for a while:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a square-shaped camera, with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1000 pixels on vertical side&lt;/li&gt;&#xA;&lt;li&gt;1000 pixels on horizontal side&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I want to know the number of meters per pixel depending of the distance, I mean:&#xA;If I have a 1 meter long object which is 100 meters far from the camera, how many pixels of the resulting picture will be occupied by the object?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure if it is useful, but we can estimate the length and size of the camera:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;0.05 meters vertical&lt;/li&gt;&#xA;&lt;li&gt;0.05 meters vertical&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;How to calculate the number of meters per pixel as defined above?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From first answers and the given formula, I am wondering:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Does the formula depends of how far from the lens is the pixel device?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;SECOND EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have this data: The Optical Form Factor of the camera is: 1/2.5''&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it related to the lens focal length? I did not find a definition on wikipedia and only &lt;a href=&quot;https://blogs.cisco.com/datacenter/qsfp-dd-an-evolutionary-approach-to-400gbe-interconnect&quot; rel=&quot;nofollow noreferrer&quot;&gt;this blog article&lt;/a&gt; that I don't understand&lt;/p&gt;&#xA;" OwnerUserId="21434" LastEditorUserId="21434" LastEditDate="2020-01-25T12:23:01.430" LastActivityDate="2020-01-26T03:40:02.540" Title="How to calculate meters per pixel for a given camera?" Tags="|mathematics|optics|geometry|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32894" PostTypeId="2" ParentId="32892" CreationDate="2020-01-24T19:01:19.490" Score="1" Body="&lt;h1&gt;We don't know enough yet, but we're close&lt;/h1&gt;&#xA;&#xA;&lt;p&gt;We can't determine the answer without knowing the magnification provided by the camera lens.  What we need to know is how much your object fills up the sensor.  Once we know how much of the sensor is taken up by the object, we can use simple ratios and division to determine the object-dimension-per pixel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can give you the calculations, though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let n be the percent of the sensor that the meter-long object takes up.  We look at the display and in the simplest fashion measure what percent the object takes up (we will assume the display shows 100% of the sensor pixels).  We could also use lens attributes, but it would be harder.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the object takes up n% of the sensor, which means it takes up n x 1000 pixels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once we know how many pixels, we divide them into the length of your object and that will give us the object length per pixel.  This would be the maximum resolution possible of the object.  Actual resolution would be affected by the fidelity of the camera lens.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-01-24T19:01:19.490" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32895" PostTypeId="1" CreationDate="2020-01-24T19:25:21.450" Score="-1" ViewCount="1084" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jr2BE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jr2BE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to determine the point where the maximum stress is located. Also, I want to know how this kind of questions is answered.&lt;/p&gt;&#xA;" OwnerUserId="24358" LastActivityDate="2020-01-26T21:01:05.737" Title="Maximum stress point" Tags="|stresses|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32896" PostTypeId="2" ParentId="32895" CreationDate="2020-01-24T19:41:10.590" Score="0" Body="&lt;p&gt;The answer is A. It's mainly because of the elastic modulus of those materials. The plastic is very flexible and can't transfer the force to the end of the hole. So the force is transfered at the outer end. If the materials would be interchanged, the maximum stress would be located at the end of the hole.&lt;/p&gt;&#xA;" OwnerUserId="24360" LastActivityDate="2020-01-24T19:41:10.590" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="32897" PostTypeId="2" ParentId="32895" CreationDate="2020-01-25T00:08:12.813" Score="1" Body="&lt;p&gt;let's say the coefficient of bounding between the plastic rod and concrete is q and the diameter of the plastic rod is d, meaning every square inch of the plastic embedded in the concrete has q lbs adhesion, thus &lt;span class=&quot;math-container&quot;&gt;$q\pi d $&lt;/span&gt; for every inch length of embedment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore the bound between the plastic rod and concrete starts from zero at point A and linearly increases to the level where the bound becomes equal to maximum tension the plastic rod can take. Assuming the maximum tensile strength of the plastic rod &lt;span class=&quot;math-container&quot;&gt;$ F_p $&lt;/span&gt; this point is located at an embedment length of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$L_{embement}=\frac{F_p*Area\ of\ rod}{q\pi d}= \frac{F_p\pi d^2/4}{q\pi d}=\frac{F_pd}{4q}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This point is independent of the points B, C, D on the graph. It is pre-determined by the properties of the plastic rod. but it certainly is not at point A, because there the concrete stress is zero.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I attach a graph of concrete bounding stress here. The stress in rod depends on its embedment length.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The axial tensile stress in the rod at point A is the sum of cylindrical shear stress plus the cone shear stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&#xA;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I added the narrowing of the rod under the Poisson's effect to clarify why shear stress at point A is zero. Notice that due to axial elongation under tension the rod gets a bit narrower through a transition region from the Point A inward.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sROey.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sROey.jpg&quot; alt=&quot;diagram of bounding stress&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-01-26T21:01:05.737" LastActivityDate="2020-01-26T21:01:05.737" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32898" PostTypeId="1" AcceptedAnswerId="32903" CreationDate="2020-01-25T03:41:02.103" Score="2" ViewCount="73" Body="&lt;p&gt;So, I have been doing research into the use of Insulated &lt;em&gt;Composite&lt;/em&gt; Concrete Forms (ICCFs) for small-scale construction, and that has led me to looking at the Rastra ICCF system, as it has been in production for a long time, is relatively well documented, and is versatile enough to answer a variety of building design needs; in particular, Rastra parts can not only be used to build &lt;em&gt;walls&lt;/em&gt;, but &lt;em&gt;horizontal assemblies&lt;/em&gt; (floors/ceilings/roofs) as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the published documentation for Rastra ICCFs stops short of providing span charts for their use in horizontal assembly construction.  I contacted Rastra's technical support regarding this omission, and they sent me a PDF file that included the following drawing and determination-table (span chart) for Rastra floor systems using Trigon-type reinforcing girders and supplemental reinforcement laid in a bed of Rastra end elements, then finished with a concrete pour:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gYoM1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gYoM1.png&quot; alt=&quot;drawing of Rastra floor using end elements&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1NEeD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1NEeD.png&quot; alt=&quot;Rastra floor span table&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of this is clear enough (although I'm not sure how to get my paws on prefabricated reinforcing girders of the style they used, given that I'm across the pond from them); however, it does not answer one question for me: are the loads given across the top of the posted table inclusive or exclusive of the floor assembly's unit weight (inherent deadload)?&lt;/p&gt;&#xA;" OwnerUserId="964" LastActivityDate="2020-01-25T08:03:29.087" Title="Are the design load values in this floor determination table (span chart) inclusive or exclusive of the given floor weight (inherent dead load)?" Tags="|structural-engineering|building-design|reinforced-concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32900" PostTypeId="2" ParentId="32892" CreationDate="2020-01-25T07:07:10.913" Score="1" Body="&lt;p&gt;Independent of what you are imaging, the ratio of object dimension to it's image dimension is a function of the lens focal length and the distance to the object.   &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Knowing the sensor dimensions allows you to determine if the image will &quot;fit on the sensor&quot;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Knowing the sensor pixels per distance allows you to express the distance in pixels.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image dimension / lens-focal-length  = Object dimension / distance to object.&lt;/strong&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So eg a 1m long object at 100 metres from the lens, with a 200mm lens, will result in &lt;/p&gt;&#xA;&#xA;&lt;p&gt;ID / 0.2m = 1m / 100m  or&#xA;image dimension = 1 x 0.2 / 100 = 2 mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sensor is a full frame &quot;35mm&quot; sensor of typically 24 x 35 mm size, in landscape mode, and the object is eg a 1m vertical fencepost, then the 2mm image will occupy 2/24th = 8.333%n of the sensor height. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the sensor is a 4000 x 6000, 24 megapixel sensor then the image of the 1m fence[post will be 4000 pixels x 2/24 = 333 pixels tall on the sensor.&lt;br&gt;&#xA;So 1 pixel is 1/333 of a meter at 100m at 200mm focal length = 3 mm at the object,&lt;br&gt;&#xA;Which is awesomely impressive.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Each pixel on the sensor is 24mm / 4000  = 6 um.&lt;br&gt;&#xA;Which is also awesomely impressive. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bhpm0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bhpm0.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;In the image above, &quot;Height&quot; to base size is the same for both triangles.&lt;/strong&gt;    &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So: Image-height / focal-length  = Object height / distance to subject.&lt;/strong&gt;&lt;br&gt;&#xA;So Image-height  = focal-length x Object height / distance to subject.**  &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Note:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edited - I had a factor of 10 error initially.&lt;br&gt;&#xA;I KNEW my 200mm lenses were not THAT good, so went back and found the error. &lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2020-01-26T03:40:02.540" LastActivityDate="2020-01-26T03:40:02.540" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32901" PostTypeId="2" ParentId="19035" CreationDate="2020-01-25T07:19:37.060" Score="4" Body="&lt;p&gt;Year's ago I took out the inside of my Zippo and closed the top, then slid a tight piece of heat shrink over the lighter case. Heated it till the heat shrink was tight the took a Razer knife and cut along 3 side's leaving the hinged area in contact. Refilled the cartridge with fluid put back in the lighter lasted a lot longer on one fill up plus it was gripper look pretty cool and keep the lighter looking New the lid closed nicely never got wobbly all my friends caught on and did the same, I've use the same lighter for 30+ year's in till it got misplaced or someone wanted for them self. For real it really worked. (Remember to take the cartridge out before heating)&lt;/p&gt;&#xA;" OwnerUserId="24362" LastActivityDate="2020-01-25T07:19:37.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32902" PostTypeId="1" CreationDate="2020-01-25T08:00:31.470" Score="-1" ViewCount="37" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9gw69.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9gw69.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I just need a hint , I know it's out of context but that's how I found it in a Facebook group, I thought someone might have seen this before, I study automation and I know that Kp is the proportional gain in a PID controller&lt;/p&gt;&#xA;" OwnerUserId="9357" LastActivityDate="2020-01-25T13:06:37.417" Title="Any idea about what does this image represent in control systems?" Tags="|control-engineering|pid-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32903" PostTypeId="2" ParentId="32898" CreationDate="2020-01-25T08:03:29.087" Score="3" Body="&lt;p&gt;The load values in the table is described as the design load with no further qualifiers and that will generally mean the total load including all dead load and live load as well as safety factors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would usually be problematic to specify the capacity exclusive of the inherent dead load because multiple load safety factors need to be checked, that is both a set of factors with high safety on live load combined with lower safety on dead load as well as a high safety on dead load combined with a lower safety on live load, so the critical safety factor on the inherent dead load would depend on the size of the live load and non-inherent dead load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since you write that you're across the pond from the manufactorer, I'll add just for good measure that since material safety factors etc. vary from country to country, such a table is only valid for the jurisdiction it was intended to be used in.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2020-01-25T08:03:29.087" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32904" PostTypeId="1" CreationDate="2020-01-25T09:25:34.287" Score="-1" ViewCount="37" Body="&lt;p&gt;I am trying to learn soundproofing on &lt;a href=&quot;https://en.wikipedia.org/wiki/Soundproofing&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki&lt;/a&gt; (please let me know if there exists any other oninle tutorials).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are several basic approaches to reducing sound: increasing the distance between source and receiver, using noise barriers to reflect or absorb the energy of the sound waves, using damping structures such as sound baffles, or using active antinoise sound generators.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;consider this situation, there is a noisy machine which generate lots types of noise in room A where I am not allowed to do anything, I am living in room B, where I am allowed to do something for soundproofing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9ci51.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ci51.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the approaches including increasing the distance, using active antinoise sound generators seems not to be suitable for my situation, so, noise barriers, damping structures are the approaches available, right? what I am mising? what esle I could use?&lt;/p&gt;&#xA;" OwnerUserId="24363" LastActivityDate="2020-01-25T20:23:25.883" Title="What approaches for soundproofing for this situation?" Tags="|audio-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32905" PostTypeId="2" ParentId="32902" CreationDate="2020-01-25T13:06:37.417" Score="0" Body="&lt;p&gt;That looks like a trajectory on a phase plane&lt;/p&gt;&#xA;" OwnerUserId="21516" LastActivityDate="2020-01-25T13:06:37.417" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32906" PostTypeId="2" ParentId="32904" CreationDate="2020-01-25T13:12:22.623" Score="0" Body="&lt;p&gt;&quot;Noise&quot; or sound level is just waves of air pressure. So anything to reduce air pressure. Noise level is reduced by increasing the distance from the source following the inverse square law (same sound intensity force but over a larger area). Sound baffling equipment work as well by just absorbing the sound waves instead of reflecting them around an enclosure. Ever notice a how a room seems &quot;louder&quot; after you take all the furniture out? The bare floor and walls reflect a lot more sound waves than furniture. So adding some carpet and soft chairs or wall hangings would help (or just straight up acoustic tiles). Check out anechoic chambers for extreme ideas.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-01-25T13:12:22.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32907" PostTypeId="2" ParentId="26631" CreationDate="2020-01-25T13:28:36.843" Score="1" Body="&lt;p&gt;You need &lt;em&gt;something&lt;/em&gt; to introduce an asymmetry. First try with friction.&lt;/p&gt;&#xA;&lt;p&gt;Let's start based on a (disc detainer lock](&lt;a href=&quot;http://www.lockwiki.com/index.php/Disc_detainer&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.lockwiki.com/index.php/Disc_detainer&lt;/a&gt;) idea and modify the mechanism. We don't care about locking some outer device, we want to latch the key in permanently, so the sidebar goes away and so does the only spring, attached to it. We don't care about the outer gates, just about the keyway area.&lt;/p&gt;&#xA;&lt;p&gt;We leave three discs, and configure them like this:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;bottom-mostdisc (3) is fixed to the body, keyway unmodified.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;middle disc (2) has keyway modified, its sides cut to such angles the key with zero-cut has 30 degrees of play in it. Due to it being bound, this is all the range of rotary motion the key can make, but initially turned 30 degrees back, so the key inserted engages it immediately, turning it turns the disc.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;top disc (1) is bound to the middle disc, rotating with it. Its keyway is unmodified from normal disc detainer (fits zero-cut snugly, the key can rotate in it freely only as much as the matching cut in the key allows)&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The key's cut matching disc 3 is 30 degrees, meaning this is the range of motion the fixed disc allows the key to rotate. Additionally, it's narrower than others, less play on the sides of the disc (spacer area) meaning pulling/pushing on the key will apply tension to  that one, fixed disc first without applying force to other discs.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;key position 2 is zero-cut, which would normally mean the disc has no play and always turns with the key. But our modified keyway will only engage with the key in the 'locking' direction. Turnng the key back won't reach it.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;key position 1 is another 30 degree cut. That means the key in reality never engages disc 1. It can turn the range allowed by disc 1, and touch disc 3 at each extreme of its motion, but not move it.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Now after inserting the key and turning it, pulling it will engage disc 3 against the 'blank' zero-cut area past key's cut 3. If you turn the key back, so the blank tip can be pulled through the disc 3's keyway, you'll press tke key's position 2's zero-cutagainst disc 1... and since we have the tip already within keyway of disc 3 (narrower cut!) it won't budge sideways so you can't move disc 1.&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;Approach 2: use randomness.&lt;/p&gt;&#xA;&lt;p&gt;Again, let's start with something like a disc detainer core, but more modified than before. The key is just a cylinder with a big number of notches cut by a lathe around its sides. The notches have rectangular profile, no sloped walls. The discs of the core are smaller than the core by the same amount as depth of the notches. Their central holes (keyway) match the diameter of the 'key'. And there are spacers in between, matching the uncut areas of the key (plus tolerance).&lt;/p&gt;&#xA;&lt;p&gt;Upon inserting the 'key', just give it a shake or a whirl. The discs will slide around randomly, engaging the notches in the 'key'. Only aligning them all centrally would release the key, but even if one or two align that way naturally, there's still ten or so that won't. The chance random motion would align them all into a perfectly centered stack releasing the key is astronomically small.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2020-06-24T08:16:29.393" LastActivityDate="2020-06-24T08:16:29.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32908" PostTypeId="1" CreationDate="2020-01-25T14:19:39.450" Score="2" ViewCount="48" Body="&lt;p&gt;I can not find a clear explanation for when NOT to use Bessel's correction and use N instead of N-1. As I understand it, Bessel's correction would apply to things such as clinical trials, sampling from a production line, voter polling, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am performing an experiment analyzing mass flow rate into 3 entry points. I want to analyze the deviation of the distribution into those 3 entry points and compare it to an ideal perfectly even distribution of a (relatively) known mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I know the &quot;true population mean&quot; in this circumstance - its an even distribution so if I reduce everything to percentages, it would be X = 1/3 = 0.333. I believe I need to use N instead of N-1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The complicating factors are that while I think I know the starting mass for the experiment, my scale has an instrument error associated with it. Also my measurement of mass flow rate versus time has multiple measurement instruments associated with it as well!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I guess if I boil down the question.... I need to figure out if I need to use N or N-1 for standard deviation. Does the error inherent to the measurements necessitate N-1 or does knowing my &quot;ideal distribution&quot; allow me to use N?&lt;/p&gt;&#xA;" OwnerUserId="24365" LastEditorUserId="90" LastEditDate="2020-01-25T17:19:49.283" LastActivityDate="2020-01-25T17:19:49.283" Title="When NOT to use Bessel's Correction for Standard Deviation" Tags="|statistics|flow-control|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32910" PostTypeId="2" ParentId="10341" CreationDate="2020-01-25T15:24:56.307" Score="0" Body="&lt;p&gt;From &quot;Thermodynamics: an engineering approach&quot;, by Yunus A. Çengel, Michael A. Boles:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a piston–cylinder device containing liquid water at 20°C and 1 atm pressure being heated. Under these conditions, water exists in the liquid phase, and it is called a &lt;strong&gt;compressed liquid&lt;/strong&gt;, meaning that it is &lt;em&gt;not about to vaporize&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As more heat is transferred, the temperature keeps rising until it reaches 100°C. At this point water is still a liquid, but any heat addition will cause some of the liquid to vaporize. A liquid that is &lt;em&gt;about to vaporize&lt;/em&gt; is called a &lt;strong&gt;saturated liquid&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once boiling starts, the temperature stops rising until the liquid is completely vaporized. Any heat loss from this vapor will cause some of the vapor to condense. A vapor that is &lt;em&gt;about to condense&lt;/em&gt; is called a &lt;strong&gt;saturated vapor&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the temperature of the vapor is, let us say, 300°C; and if we transfer some heat from the vapor, the temperature may drop somewhat but no condensation will take place as long as the temperature remains above 100°C (for P = 1 atm). A vapor that is &lt;em&gt;not about to condense&lt;/em&gt; (i.e., not a saturated vapor) is called a &lt;strong&gt;superheated vapor&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="24367" LastActivityDate="2020-01-25T15:24:56.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32912" PostTypeId="1" AcceptedAnswerId="32913" CreationDate="2020-01-25T16:10:22.073" Score="1" ViewCount="68" Body="&lt;p&gt;Consider a mass of about &lt;span class=&quot;math-container&quot;&gt;$130 \pm 10 $&lt;/span&gt; kg, falls from height, let say &lt;span class=&quot;math-container&quot;&gt;$2.5 $&lt;/span&gt; m. &#xA;In order to dampen the vibration and sound, I use silencer or drop pads. I was wondering if introducing the macroscopic projections, helps to reduce the thickness of the shock absorber in comparison to the classic flat surface shock absorbers. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/c38BE.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/c38BE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Wj7hQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Wj7hQ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After reading Kamran's answer: &#xA;I think the impact energy results in transversal deformation of the projected surface, that is in contrast to the vertical compression of the flat surface.  So there would be less energy transmitted to the floor.  &lt;/p&gt;&#xA;" OwnerDisplayName="user14407" LastEditorUserId="24426" LastEditDate="2020-01-31T08:11:38.153" LastActivityDate="2020-01-31T08:11:38.153" Title="Does the surface roughness play any role in shock absorbing?" Tags="|mechanical-engineering|materials|dynamics|shock|impact|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32913" PostTypeId="2" ParentId="32912" CreationDate="2020-01-25T16:25:12.710" Score="0" Body="&lt;p&gt;No, look at it this way: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;you are removing material that would otherwise help break the fall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All it does is give the mass a chance to penetrate deeper into the drop pad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be ok if you add it to existing pad for a thicker pad if you are looking for less noise and less stress to the floor. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-25T16:25:12.710" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32914" PostTypeId="2" ParentId="32904" CreationDate="2020-01-25T18:43:18.283" Score="0" Body="&lt;p&gt;You can try to use heavy, lush carpeting, thick solid fabric for drapery, bulky furniture on rubber leg bases, and even heavy wallpapers. wear rubber sole shoes as a vibration barrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your space is structurally connected to the other room, unfortunately, much of the base spectrum of noise will be transferred through the foundation or beams and walls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may also consider using an active noise canceling headset at the noisiest times.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-01-25T20:23:25.883" LastActivityDate="2020-01-25T20:23:25.883" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32915" PostTypeId="1" CreationDate="2020-01-26T03:06:47.617" Score="1" ViewCount="681" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lNtsG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lNtsG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we look at a cross section at any point, stress will be the same &lt;span class=&quot;math-container&quot;&gt;$\sigma=F/A$&lt;/span&gt; and the hooke law says that &lt;span class=&quot;math-container&quot;&gt;$\sigma=E*\epsilon$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means that the deformation will be constant throughout the beam, which obviously can't be correct. So what am I missinterpreting here?&lt;/p&gt;&#xA;" OwnerUserId="21424" LastActivityDate="2020-01-26T18:39:23.687" Title="Is a beam is subject to axial load, does it mean that deformation is the same in each point?" Tags="|solid-mechanics|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32916" PostTypeId="2" ParentId="32915" CreationDate="2020-01-26T08:09:50.503" Score="2" Body="&lt;p&gt;Strain, &lt;span class=&quot;math-container&quot;&gt;$\epsilon = \Delta L/L$&lt;/span&gt;, is the ratio of deformation, not the deformation. Deformation along the beam even with the same &lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt; is proportional to the length of the section we consider, like deformation for one foot of the beam is half of the deformation of two feet. It means deformation is not constant. The stress &lt;span class=&quot;math-container&quot;&gt;$\ \sigma  $&lt;/span&gt; is constant no matter which cross-section we consider. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can think of the beam like a spring under compression. It compresses uniformly along its length but if we consider only 1/10 of its length the compression is 1/10 of total compression of the spring. And deformation for that section is 1/10th of the total deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had confused the strain with deformation. I corrected my answer to clear this mix-up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is the deformation that varies with section length, not the strain, apologies. And thanks to @JohnHoltz for commenting on that.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-01-26T18:39:23.687" LastActivityDate="2020-01-26T18:39:23.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32918" PostTypeId="1" AcceptedAnswerId="32919" CreationDate="2020-01-26T12:56:27.683" Score="-1" ViewCount="270" Body="&lt;p&gt;Consider this image of a loop shaping design:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ghIlY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ghIlY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here which is the bandwidth of the system? I have a doubt because until now I have considered as bandwidth to the system the one of the complementary sensitivity function, but I heard someone sayong that it is not a good measure for the bandwidth, so it is better to consider as bandwidth the one of the sensitivity function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody clarify this?&lt;/p&gt;&#xA;" OwnerUserId="22780" LastActivityDate="2020-01-26T15:11:28.957" Title="Doubt on bandwidth of a control system" Tags="|control-engineering|control-theory|systems-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32919" PostTypeId="2" ParentId="32918" CreationDate="2020-01-26T15:11:28.957" Score="2" Body="&lt;p&gt;There are many definitions of bandwidth. Typically it is determined from the open loop transfer function and not from the sensitivity or complementary sensitivity functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a SISO feedback loop, with plant &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; and controller &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;, a common definition of bandwidth is the 0 dB crossover frequency of the open loop transfer function &lt;span class=&quot;math-container&quot;&gt;$PC$&lt;/span&gt;. Assuming that the sensitivity can be written as &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{1 + PC}$&lt;/span&gt; this is also (roughly) the frequency where the sensitivity peaks. &lt;/p&gt;&#xA;" OwnerUserId="11869" LastActivityDate="2020-01-26T15:11:28.957" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32920" PostTypeId="2" ParentId="32915" CreationDate="2020-01-26T16:20:22.740" Score="0" Body="&lt;p&gt;It will be the same at each point while in the elastic stress range ( about 0.5 % strain for typical steels). When it reaches yield it can become non-uniform. For a research project on torsional strain; Strain was uniform ( indicated by ink lines on the polished specimen) until yield , then all plastic strain occurred in a small zone. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-01-26T16:20:22.740" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32921" PostTypeId="2" ParentId="32915" CreationDate="2020-01-26T16:24:30.627" Score="3" Body="&lt;p&gt;You seem to be mixing up strain and deformation. The strain, &lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt; or change in length per length, is clearly uniform throughout the beam, and therefore the deformation is not. You get the deformation by integrating the strain, so the deformation varies linearly with zero at one end and maximum at the other.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2020-01-26T16:24:30.627" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32923" PostTypeId="1" CreationDate="2020-01-26T19:34:45.800" Score="2" ViewCount="53" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KhElA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KhElA.png&quot; alt=&quot;part file giving cirpattern instructions&quot;&gt;&lt;/a&gt;&lt;a href=&quot;https://i.stack.imgur.com/kme99.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kme99.png&quot; alt=&quot;screen-shot of assembly file.. there isn&amp;#39;t anything to see, it&amp;#39;s totally ordinary..&quot;&gt;&lt;/a&gt;something really funky happened &lt;a href=&quot;https://i.stack.imgur.com/kme99.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kme99.png&quot; alt=&quot;successful cirpattern in part (?!? :-D)&quot;&gt;&lt;/a&gt;recently, trying to track down what actually happened.  I created a sweep on a 3d-sketched path, then mirrored it accross a plane so that the initial sweep and  the mirrored sweep share some of the same volume.  Then I tried to circular pattern the two of them around a central axis.  ordinarily when I try to do this I get the error &quot;only merging features can be patterned&quot;.  however, one time doing this, I was researching how to do it, then, on giving up, I clicked something and bada-boom the whole thing worked.  with a catch.  it seemed like solidworks had automatically created an assembly with the same name, imported the part into it, and carried out the operation that I wanted it to do, but in the assembly, not the part.  I swear I didn't do all these  things myself.  now, I'm trying to figure out how it  happened, because, obviously, super useful!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the record, I've found a work-around. creating a solid surface with which to connect all the sweeps, then sketching the end profile of the sweep on the solid surface fixes the problem, and no assembly is necessary.  so  you could say that this is a superior solution anyway because it avoids the extra file getting generated, but I'm still super curious about the other thing that happened.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second edit:  I tried to reproduce the error for a commentator.. and.. failed.  it worked!  exactly the way I wanted it to!  Right in the part!  So.. we still have  a mystery, but it isn't clear if it's worth trying to solve..&lt;/p&gt;&#xA;" OwnerUserId="13081" LastEditorUserId="13081" LastEditDate="2020-01-27T11:23:19.593" LastActivityDate="2020-01-27T11:23:19.593" Title="how to automatically create an assembly from a part" Tags="|solidworks|" AnswerCount="0" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="32925" PostTypeId="1" CreationDate="2020-01-26T21:28:40.690" Score="1" ViewCount="70" Body="&lt;p&gt;What is the strongest actuator (that can exert a controlled amount of force &amp;amp; be able to start/stop) that an infinitely advanced civilization (e.g. only limited by the laws of physics) could create, and how would it work?&#xA;The size and mass must be macroscopic (e.g. human-sized).&lt;/p&gt;&#xA;" OwnerDisplayName="user24374" LastEditorDisplayName="user24374" LastEditDate="2020-01-27T16:02:22.987" LastActivityDate="2020-01-27T16:02:22.987" Title="What is the maximum actuator strength allowed by physics?" Tags="|experimental-physics|actuator|" AnswerCount="2" CommentCount="11" ClosedDate="2020-02-01T19:26:54.590" ContentLicense="CC BY-SA 4.0" />
  <row Id="32926" PostTypeId="1" CreationDate="2020-01-26T23:07:50.293" Score="1" ViewCount="57" Body="&lt;p&gt;I have about (this is experimental) 0.75 to 1 liter of water. Ive found I can use 400W of power through peltiers to reduce water flowing through at 1L/minute from 32C to 20C in about 1 to 1.5 minutes through a 0.25 lpm flow rate tube that cools by absorbing 8C per minute (thus maintaining a roughly 20C temperature). There is also other power consumed by fans and voltage converters but for this I'm simply addressing the water temperature. I assumed at first this would be Wh (watt hours required) to achieve this but my pack DOD (depth of discharge) isn't holding up. I have calculated everything. Resistance, amps being pulled, voltage to meet temperature settings, temperature probed before and after cooling. Does anyone have a clue what I should expect from a 3s15p battery pack with 480Wh  and only feeding the peltiers 7.5v from a 11.1A source (forget the loss from regulators and fans...they are accounted for through a separate calculation and power source)&lt;/p&gt;&#xA;" OwnerUserId="24375" OwnerDisplayName="Paul Pease" LastActivityDate="2020-01-26T23:15:40.220" Title="I hope this is the correct forum for temperature related question" Tags="|thermodynamics|temperature|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32928" PostTypeId="2" ParentId="32925" CreationDate="2020-01-27T00:44:05.243" Score="0" Body="&lt;p&gt;There are no limits on such forces according to physics, per se.  The limites tend to be based on strengths of materils, manufacturing capability, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Archimedes said 2300 year ago, &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt; “Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. ”&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;We could theoretically build an actuator that placed its fulcrum on a star, and used the energy of galaxies to move it.  The maximum actuator strength would thus be defined by what it could push against, how large it could be made, and what forces could act upon it.  With a sufficiently advanced civilization with unlimited access to resources and technology, the only limits become how much of the universe's resources and energy could be put to use in such a system.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-01-27T00:44:05.243" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32929" PostTypeId="2" ParentId="32925" CreationDate="2020-01-27T02:20:36.220" Score="1" Body="&lt;p&gt;You can make an actuator arbitrarily (e.g. infinitely) strong providing you make it big enough, and this would be equivalent to using an arbitrarily large number of regular strength actuators, in parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want that force limited to a specific area, instead of spreading it as wide as needed, your one hard limit comes from general relativity. Your energy density (as output of the actuator), converted to its mass equivalent (according to &lt;span class=&quot;math-container&quot;&gt;$E=mc^2$&lt;/span&gt;) must not exert gravitational force resulting in collapse into a black hole.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will be more other limits of engineering nature before that - e.g. if you let your actuator collapse into neutronium, it's unlikely to retain shape and function any more. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Long before that you'll exceed tensile strength of all possible materials and have it simply break, either flow or snap, although this could be overcome with using generated fields (electric, magnetic) as &quot;materials&quot; of its structure, bypassing common material engineering woes - although at that point your actuator ceases to be a material object and becomes a construct built from energy. It's not as sci-fi as it sounds; the entirety of &quot;high energy side&quot; plumbing of the VASIMR ion engine is built like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, while commonly an actuator is understood as a reusable mechanical device like a piston or a servo, the definition can be stretched quite far, and you'll find single-use pyrotechnic actuators where extreme momentary forces are required, and even currently there's nothing stopping you from basing one upon a &lt;a href=&quot;https://www.businessinsider.com/fastest-object-robert-brownlee-2016-2?IR=T&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermonuclear explosion&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="203" LastEditorUserId="203" LastEditDate="2020-01-27T02:25:59.390" LastActivityDate="2020-01-27T02:25:59.390" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32930" PostTypeId="1" CreationDate="2020-01-27T05:48:56.710" Score="0" ViewCount="279" Body="&lt;p&gt;Why does blast furnace (in Iron &amp;amp; Steel works) top gas temperature stays low and what are the fixes for that to increase without using a burner?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The top gas temperature goes below 80 degrees (celcius) and the system requires a burner to increase the level of temp. Why we would need a burner, there should be a different way for not to decrease its temperatures.&lt;/p&gt;&#xA;" OwnerUserId="24364" LastEditorUserId="24364" LastEditDate="2020-01-27T07:07:37.613" LastActivityDate="2020-01-27T07:07:37.613" Title="How to increase blast furnace top gas temperature?" Tags="|mechanical-engineering|chemical-engineering|metallurgy|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32931" PostTypeId="1" CreationDate="2020-01-27T06:27:09.250" Score="1" ViewCount="40" Body="&lt;p&gt;I want to design a shaft for a motor test rig. The motor and generator are 48V and 1.5kW and the maximum torque produced by the motor is 18Nm. RPM varies from 2900-3100. What shall be my next steps? I also intend to use Lovejoy coupling.&lt;/p&gt;&#xA;" OwnerUserId="24379" LastActivityDate="2020-01-27T16:13:38.390" Title="Shaft Design for Motor Test Rig" Tags="|design|motors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32932" PostTypeId="1" AcceptedAnswerId="32933" CreationDate="2020-01-27T10:16:05.220" Score="1" ViewCount="65" Body="&lt;p&gt;I have a transfer function with a RHP zero in &lt;span class=&quot;math-container&quot;&gt;$s=+1$&lt;/span&gt;, and I am trying to show in Matlab thath this limitation is present. In particular, initially I had a MIMO system, which I decoupled and then I started doing this analysis on the two separate SISO system that came out of the decoupling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My code is the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = tf('s');&#xA;G = [2/(s+1) 3/(s+2);1/(s+1) 1/(s+1)];&#xA;&#xA;&#xA;d_12 = -(G(1,2)/G(1,1));&#xA;d_21 = -(G(2,1)/G(2,2));&#xA;D_simp = [1 d_12;d_21 1];  %decoupler&#xA;F1 = G*D_simp;    &#xA;&#xA;k1 = 5;&#xA;k2 = 2;&#xA;k3 = 50;&#xA;&#xA;loop1 = loopsens(F1(1,1),k1/s);&#xA;loop2 = loopsens(F1(1,1),k2/s);&#xA;loop3 = loopsens(F1(1,1),k3/s);&#xA;&#xA;figure;&#xA;bodemag(loop1.Ti,'r',loop2.Ti,'b',loop3.Ti,'g'),grid&#xA;figure;&#xA;bodemag(loop1.Si,'r',loop2.Si,'b',loop3.Si,'g'),grid&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;where I am trying to increase the bandwidth of the system by increasing the gain. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The expected result is that if I increase the bandwidth , I should have worse performances as the bandwidth increases, but what happen is exaclty the opposite:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ldMRl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ldMRl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;can somebody please help me solve this problem? &lt;/p&gt;&#xA;" OwnerUserId="22780" LastActivityDate="2020-01-27T11:47:36.983" Title="Problem with RHP zero in transfer function" Tags="|electrical-engineering|control-engineering|control-theory|systems-engineering|signal-processing|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32933" PostTypeId="2" ParentId="32932" CreationDate="2020-01-27T11:47:36.983" Score="3" Body="&lt;p&gt;This is because for all three of of your choices for the gain the closed loop system is unstable. Namely, for just an integrator as controller the gain should be below 1.5 in order for the closed loop system to be stable and all of your gains are above that.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2020-01-27T11:47:36.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32934" PostTypeId="1" CreationDate="2020-01-27T13:28:17.080" Score="1" ViewCount="334" Body="&lt;p&gt;A weir can be used to measure flow rate, from the known relationship between height of the water level above the weir crest, weir length and flow rate (assuming the water level downstream is below the weir crest). However, the flow coeffiecent depends on the conditions, among other things on wether the flow is aerated from below. In low flow conditions, the water flow may cling to the weir resulting in another flow coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ESRyd.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ESRyd.png&quot; alt=&quot;DWA-M 181 - flow conditions at different flowrates&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the DWA-M 181 straight weirs can achieve a decent accuracy for flow ranges &lt;span class=&quot;math-container&quot;&gt;$Q_{max}/Q_{min}$&lt;/span&gt; = 25 to 30, if the weir is built according to the ISO 1438. The crucial element is areation of the water jet, according to the DWA-M 181. What does ISO 1438 (or other similarly authoritative voices) say about straight weirs and how to build to achieve decent accuracy over a wide range?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the weir will be mounted into a rectangular concrete channel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Image source is the DWA-M 181 that mostly points to the ISO 1438)&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2020-01-27T15:46:52.997" LastActivityDate="2020-01-27T16:36:58.773" Title="How to construct a weir for Volume flow measurement" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32936" PostTypeId="2" ParentId="32934" CreationDate="2020-01-27T15:38:39.453" Score="1" Body="&lt;p&gt;Used to do this for students for a lab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once you have a nice flow of water handy, nice small stream is good. Then you make a suitable notch either v-notch or a rectangular notch and you can get the coefficients for 60degrees which is easy to make from some suitable wood.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then with some waterproof sheet you can make a &quot;seal&quot; attached to the wood notch which can be pushed to the sides and floor of the river bed so that all the flow goes through the notch. We used to weight this down with rocks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then all you have to do is measure accurately - make sure the wood notch is upright otherwise that will introduce an extra error. Working out the velocity is inherent with error - playing &quot;poo sticks&quot; is one way to get an approximation of the velocity - you will need this when designing the notch as it cannot overflow, you want the flow to be mid-range on the notch really.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to have a smaller and a larger weir to cope with low flow issues then that is possible - just design a suitable blanking plate for the notch not in use...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2020-01-27T16:36:58.773" LastActivityDate="2020-01-27T16:36:58.773" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32937" PostTypeId="2" ParentId="32934" CreationDate="2020-01-27T15:56:10.090" Score="0" Body="&lt;p&gt;Documentation about weirs states that low flow will be problematic due to the problem you describe. Notched weirs try to compensate for this, but there are still limits.  Could we design a complex system that managed flow across a large range?  Sure. But the beauty of weir measurement is the simplicity.  If you want accurate flow measurement from very low to very high, we need something other than a weir.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-01-27T15:56:10.090" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32938" PostTypeId="2" ParentId="32931" CreationDate="2020-01-27T16:13:38.390" Score="1" Body="&lt;p&gt;Most engineers would just take the motor and generator shafts and use one or the other if the two need to be located farther apart than a coupling would provide. 2 HP is nothing from an industrial standpoint, but that is a fairly high speed, and vibration becomes an issue.  The type of coupling will matter if we also have misalignment between the driver and driven units. If there are other design considerations (weight, vibration, etc.), we might get into the details of shaft design to build a hollow shaft, or a vibration-damping system.  A long heavy shaft also starts to add radial stress to the bearings beyond rotor weights, so we might want to support the shaft with an intermediate bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I realize this doesn't answer you question directly, but it's hard to design without detailed specifications.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-01-27T16:13:38.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32939" PostTypeId="1" CreationDate="2020-01-27T18:09:21.317" Score="1" ViewCount="2955" Body="&lt;p&gt;I am trying to calculate the flow rate of Nitrogen through a cylindrical pipe. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I only need a ballpark value, and the only information I have is the inlet pressure as 15 psig (assuming the same pressure throughout the pipe, so there is no pressure drop), the pipe diameter as 6&quot; and material which is stainless steel. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried applying Bernoulli and Poiseuille's equations but I since there is no pressure drop and I do not have a pipe length I am unsure how to use these equations.&lt;/p&gt;&#xA;" OwnerUserId="24382" LastActivityDate="2020-01-27T18:09:21.317" Title="How to calculate the flow rate of a gas through a pipe?" Tags="|mechanical-engineering|fluid-mechanics|pressure|hydraulics|" AnswerCount="0" CommentCount="5" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32940" PostTypeId="1" CreationDate="2020-01-27T18:30:11.007" Score="3" ViewCount="97" Body="&lt;p&gt;Since the title is a mouthful, here goes an explanation and a sketch I've drawn a so that I can describe it correctly:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to design a rotating arm. The rotational part will be a small pipe, installed vertically. It will be spatially restricted by two bearing, each in an end and they will be fixed to a larger structure. The arm itself is an I beam that's relatively long. I'm trying to calculate the minimum cross section of the pipe as well as the correct bearing forces. The rotating arm will be subjected to a vertical force in its end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/73IzS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/73IzS.png&quot; alt=&quot;FBD of my problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I already have: I believe that once the arm is deflected, each bearing will suffer horizontal and vertical forces. I also believe that the horizontal reactions are such that they neutralize the vertical force where the bodies meet, such as that: &lt;span class=&quot;math-container&quot;&gt;$2*R_F*H = F*arm_{length}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suppose the vertical reaction will be just &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; in the lower bearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried using a software to determine the support reactions and fooling around with it I've figured that sometimes the support can have a moment reaction in it and I have no clue what would this situation would be and where is this moment coming from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I correct in my assumptions?&lt;/p&gt;&#xA;" OwnerUserId="24383" LastEditorUserId="24383" LastEditDate="2020-01-28T12:39:51.800" LastActivityDate="2020-01-28T12:39:51.800" Title="What are the correct reactions of a rotating arm connected to a vertical pipe restricted by bearings at its ends?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32941" PostTypeId="2" ParentId="32940" CreationDate="2020-01-27T19:51:54.750" Score="1" Body="&lt;p&gt;You are correct that there will be both vertical and horizontal forces on the bearings.  These do not cancel each other out, though, and must be accounted for separately.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The vertical will be equal in total to &lt;strong&gt;F&lt;/strong&gt; (I will take your 6kN to be F).  This force will be borne by one or both bearings, as long as one or both have thrust bearing capability.  Don't forget the weight of the shaft and the I beam cantilever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There will also be horizontal forces caused by the moment.  The moment is equal to &lt;strong&gt;L * F&lt;/strong&gt; located at the H/2 point on the shaft.  This will result in a force of &lt;strong&gt;(L * F) / (H/2)&lt;/strong&gt; on each bearing, in opposite directions.  Again, we need to also account for the weight of the cantilever beam at L/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At this point we have the data needed to construct a load diagram showing forces downward on the bearing(s), a moment on the shaft that will try to impart an S shape into it, and the horizontal loads on the bearings from the moment.  Wiki has a fine explanation of the process.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-01-27T19:51:54.750" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32942" PostTypeId="2" ParentId="32940" CreationDate="2020-01-27T21:07:10.200" Score="0" Body="&lt;p&gt;If the rotation is gentle enough not to cause any considerable amount of centripetal force on the shaft or swinging motion both horizontally and vertically on the end of the beam or pendulum vibration of the beam or there is no dynamic loading, which are all very likely according to the configuration of your I beam and the swivel mechanism of the shaft to happen, then your approach is ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would check even in the most basic case for the natural frequency of the beam and the shaft and make sure they don't resonate. I can see the beam under certain conditions flailing wildly and become unstable.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-27T21:07:10.200" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32943" PostTypeId="1" AcceptedAnswerId="32944" CreationDate="2020-01-27T21:11:11.253" Score="2" ViewCount="81" Body="&lt;p&gt;For brushless DC motors we have hall sensors to keep track of rotor; however, for stepper motor we don't know the actual position of the rotor and therefore while changing pulses with some speed it may happen that the rotor cannot keep up with the rotating magnetic field and lags behind. Then some unexpected behaviors may occur. How do we make sure that this is not going to happen? In practice, does it happen just above some speed value, i.e. below that speed everything is ok and no such problem occurs? Or what are some other aspects to consider for avoiding such scenarios.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="21221" LastActivityDate="2020-01-28T03:55:32.173" Title="Stepper motor speed control problem" Tags="|control-engineering|motors|pid-control|stepper-motor|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32944" PostTypeId="2" ParentId="32943" CreationDate="2020-01-28T03:55:32.173" Score="4" Body="&lt;p&gt;Although the advertised advantage of a stepper is that it can be driven &quot;open loop&quot; (no position or velocity feedback required for accuracy, instead you just keep count of the steps) this is only true when the speed with which the step signals are being fed to it are below a certain rate threshold. Beyond that threshold, the stepper armature begins to lag behind the steps and position errors then occur. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, the &quot;stepper approximation&quot; only holds when the compliance and inertia of the load fall within certain bounds. Outside those bounds, the relationship between the number of steps sent to the stepper and its angular position is broken and position errors again result. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This limits the use of steppers to certain applications where the stepper speed and its load are well-specified to be within the stepper's &lt;em&gt;accuracy envelope&lt;/em&gt;, as specified by the manufacturer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Outside that envelope, the designer must use a DC motor and shaft encoder, plus a closed-loop control system that uses position reports from the encoder to &quot;juice&quot; the motor in such a manner as to minimize or eliminate position errors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This stepper-versus-DC motor-and-encoder tradeoff is well-known within the design community. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-01-28T03:55:32.173" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32947" PostTypeId="1" CreationDate="2020-01-28T08:24:26.027" Score="6" ViewCount="1769" Body="&lt;p&gt;I'm aware that answers may vary and it really does depend on the company/country where the PID was drawn but I have checked back for the last 30 years and found no drawing legend that describes what this symbol is and was hoping someone could point me in the correct direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N6DMM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N6DMM.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The symbol being &quot;GOS±&quot;&lt;/p&gt;&#xA;" OwnerUserId="10923" LastEditorUserId="10923" LastEditDate="2020-02-03T06:29:13.420" LastActivityDate="2020-05-23T22:06:38.073" Title="P&amp;ID Symbol identification, unknown meaning" Tags="|control-engineering|chemical-engineering|" AnswerCount="3" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="32948" PostTypeId="1" CreationDate="2020-01-28T09:05:41.023" Score="1" ViewCount="73" Body="&lt;p&gt;I assume this wood framing is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Wall_stud&quot; rel=&quot;nofollow noreferrer&quot;&gt;wall stud&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/O9pC1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O9pC1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My room looks like as follow with concrete walls&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/OH6fk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/OH6fk.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to soundproof external noise into my room (red circle), do I need to build wall stud?&lt;/p&gt;&#xA;" OwnerUserId="24363" LastEditorUserId="24363" LastEditDate="2020-01-28T09:28:30.380" LastActivityDate="2020-01-29T02:01:06.303" Title="In order to soundproof external noise, do I need to build wall stud on concrete walls?" Tags="|home-improvement|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32951" PostTypeId="1" AcceptedAnswerId="32960" CreationDate="2020-01-28T14:08:18.730" Score="1" ViewCount="219" Body="&lt;p&gt;A piston pump has a swash plate that pivots back and forth to control the flow rate of the hydraulic fluid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a version of this pump where the swash plate rotates like a cam instead of the pistons rotating?&lt;/p&gt;&#xA;" OwnerUserId="24389" LastActivityDate="2020-01-29T01:58:36.193" Title="Rotating Swashplate on Piston Pump" Tags="|pumps|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32953" PostTypeId="2" ParentId="32948" CreationDate="2020-01-28T16:40:34.300" Score="1" Body="&lt;p&gt;The way I have seen it is to use 2 X 6 on each end and floor and ceiling of the wall then build two separate walls  with 2 X 4.  Stagger the 2 X 4 in the walls so the only solid contact for sound conduction are the 2 X 6 ends. Fill the space ( now 5.5&quot; instead of 3.5&quot; ) with insulation. The result is little solid contact ( at the ends) for conduction and a thicker sound absorbing insulation. I was describing a partition wall , but essentially you could put the same thing against a concrete wall ; The important factor , you don't want 2 X 4 touching both wall surfaces as they would conduct sound.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2020-01-29T02:01:06.303" LastActivityDate="2020-01-29T02:01:06.303" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32954" PostTypeId="1" AcceptedAnswerId="32955" CreationDate="2020-01-28T17:50:16.930" Score="0" ViewCount="102" Body="&lt;p&gt;I have the following transfer function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$G(s)=\frac{-s^3 +s}{s^4+3s^3+2s^2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this is a transfer function of a plant with two poles in the origin. So I want to desing a controller with two poles in the origin in order to have zero steady state error. So the controller I want to desing is of the type:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$C(s)=\frac{k}{s^2}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;now, I want to find the k such that the closed loop is stable, but I am having some problems in doing so. I am using the Routh criterion, and I have that the closed loop is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{-k(s-1)}{s^4+2s^3-ks+k}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the table for the Routh criterion:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;|&lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;|&lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$2$&lt;/span&gt;|&lt;span class=&quot;math-container&quot;&gt;$-k$&lt;/span&gt;|&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{k}{2}$&lt;/span&gt;|&lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{-k^{2}-2k}{\frac{k}{2}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and where I have omitted terms it means that there is a zero and the bar &lt;span class=&quot;math-container&quot;&gt;$|$&lt;/span&gt; has been used to say that these numers are in different columns.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By doing this I obtain that the closed loop is stable if &lt;span class=&quot;math-container&quot;&gt;$k&amp;gt;0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$k&amp;lt;-4$&lt;/span&gt;, but if I use these values I have that the closed loop is unstable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And also I would like to ask if there is a simpler way to find the value of &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; instead of using the Routh Criterion, for example using Matlab.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong?&lt;/p&gt;&#xA;" OwnerUserId="22780" LastActivityDate="2020-01-29T08:46:47.757" Title="Problem finding value of the gain such that the closed loop is stable" Tags="|control-engineering|control-theory|systems-engineering|stability|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32955" PostTypeId="2" ParentId="32954" CreationDate="2020-01-28T18:38:42.877" Score="2" Body="&lt;p&gt;To solve this problem I would:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Sketch the block diagram of the control network.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Derive the required closed loop transfer function in terms of &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C(s)$&lt;/span&gt;.  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Substitute the expressions for &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C(s)$&lt;/span&gt; to obtain the closed loop transfer function in terms of &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Calculate the poles of the closed loop transfer function using your preferred method. Alternatively, you could use the Routh criterion to evaluate stability.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;It should not matter which closed loop transfer function you select (e.g. sensitivity, complementary sensitivity, etc.), they should all have the same closed loop poles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To do this in Matlab use &lt;code&gt;tf&lt;/code&gt; to define your plant and controller transfer functions; then use &lt;code&gt;feedback&lt;/code&gt; to calculate the closed loop transfer function (or better use the expression for the closed loop transfer function in terms of &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C(s)$&lt;/span&gt; derived above). You can then calculate the closed loop poles from this expression using the &lt;code&gt;pole&lt;/code&gt; command.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To get an expression in symbolic form you will likely have to use the symbolic toolbox. &lt;/p&gt;&#xA;" OwnerUserId="11869" LastEditorUserId="11869" LastEditDate="2020-01-29T08:46:47.757" LastActivityDate="2020-01-29T08:46:47.757" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="32956" PostTypeId="1" AcceptedAnswerId="33127" CreationDate="2020-01-28T23:08:51.370" Score="1" ViewCount="483" Body="&lt;p&gt;In a PE (polyethylene) foam factory there have been some incidents with static electricity that have created minor fires. Fire is a concern in such a factory, as PE is highly flammable, and even more as PE foam uses propane or butane gas in its manufacturing process (that's what creates the bubbles in the foam).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issues we've been having are mostly related to the thinnest of the foams we produce, which is 1 mm thick. It's formed by extrusion into a flat fabric of about 1m width and we form rolls of 350 linear meters. The machine's speed is about 78 meters / minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The foam is rolled over cardboard cores in a torque machine (the shaft is chrome-coated steel). The plant is located in a very dry, cold place (humidity is often around 40-60% and temp ranges between 10-28°C).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've gathered the following observations while analysing this problem:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;While the roll is below 45cm in diameter, the static electricity is negligible (I don't have an instrument to measure it, I just get my arm near the roll and see if the hair is drawn towards the foam).&lt;/li&gt;&#xA;&lt;li&gt;Once the diameter gets beyond the 45cm diameter the static is noticeable, to the point of firing little sparks (when it reaches the max diameter).&lt;/li&gt;&#xA;&lt;li&gt;The static builds up first near the sides of the roll, even though it can be felt in the center too. All of the incidents result from a spark igniting the roll &lt;em&gt;in any of the sides&lt;/em&gt;, never in the center.&lt;/li&gt;&#xA;&lt;li&gt;The foam goes through a series of rolls (puller, thkcness measurment) before being formed into a roll. Most of those rolls are metallic (steel) and in direct contact with both the fabric and the chasis of the machine (grounded). I can't notice static in the fabric as it leaves this series of puller rolls.&lt;/li&gt;&#xA;&lt;li&gt;There's static buildup no matter if the foam is in contact with the shaft or not (sometimes it moves sideways, leaving the cardboard core and &quot;touching&quot; the shaft).&lt;/li&gt;&#xA;&lt;li&gt;I suspect some slipping between each layer of foam while the roll is formed (due to the shaft rotating slightly faster than the foam is fed), but I can't confirm (the mechanism is synchronized by a PLC).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My hypothesis is that the static buildup is being caused by friction of the foam with the surrounding air, at a rate that is faster than its ability to release it through the shaft (its only contact with ground).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EvgFN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EvgFN.png&quot; alt=&quot;PE foam roll&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions&lt;/strong&gt;:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Shall I discard the idea that the static buildup is being caused by friction between layers of the foam in the roll (as they're the same material, theoretically they're not apart in the triboelectric series)?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other than spraying the area (or even the foam) with any watery solution to increase the conductivity of the air, what would you suggest to keep the sparks from jolting and causing fires?&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;2A. If a watery solution spray is the best solution there is, what makes a good mixture?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2B. Would a metallic side plate in contact with the shaft and the side of the roll be a good, safe solution? If so, what material would you use?  &lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;3&quot;&gt;&#xA;&lt;li&gt;How could I measure the static electricity built up into the roll to ascertain a &quot;safe limit&quot;?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17733" LastActivityDate="2020-02-07T13:39:27.110" Title="Static electricity in PE foam rolling" Tags="|electrical|fire|polyethylene|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32957" PostTypeId="5" CreationDate="2020-01-28T23:12:53.677" Score="0" Body="" OwnerUserId="-1" LastEditorUserId="-1" LastEditDate="2020-01-28T23:12:53.677" LastActivityDate="2020-01-28T23:12:53.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32958" PostTypeId="4" CreationDate="2020-01-28T23:12:53.677" Score="0" Body="Plastic (material): materials consisting of any of a wide range of synthetic or semi-synthetic organic compounds that can be molded into solid objects. This includes thermos-softening &amp; thermos-hardening plastics. " OwnerUserId="17733" LastEditorUserId="90" LastEditDate="2020-01-29T05:34:39.670" LastActivityDate="2020-01-29T05:34:39.670" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32959" PostTypeId="1" AcceptedAnswerId="32969" CreationDate="2020-01-29T01:29:20.173" Score="1" ViewCount="39" Body="&lt;p&gt;&lt;a href=&quot;https://www.thespruce.com/wonderboard-backer-board-1822682&quot; rel=&quot;nofollow noreferrer&quot;&gt;WonderBoard&lt;/a&gt; is a cement-based backer board mainly used for tiling in wet places. Specifically, its components are Portland cement, expanded polystyrene (EPS) beads, aggregates, and fiberglass reinforcement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks like WonderBoard is also a trademark/&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RkJnm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RkJnm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a way to find who holds the trademark?&lt;/p&gt;&#xA;" OwnerUserId="24363" LastActivityDate="2020-01-29T17:13:23.897" Title="Is there a way to find who holds the trademark?" Tags="|home-improvement|" AnswerCount="1" CommentCount="3" ClosedDate="2020-02-01T19:20:38.483" ContentLicense="CC BY-SA 4.0" />
  <row Id="32960" PostTypeId="2" ParentId="32951" CreationDate="2020-01-29T01:58:36.193" Score="1" Body="&lt;p&gt;Here are two categories of pumps with non-orbiting pistons: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Axial piston pumps with rotating swash plate.&lt;br&gt;&#xA;Operating principle at 0:48 - &lt;a href=&quot;https://www.youtube.com/watch?v=sP3gc4b8Z5k&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=sP3gc4b8Z5k&lt;/a&gt;&lt;br&gt;&#xA;Commercial examples:&lt;br&gt;&#xA;Dynex Rivett PFXXXX and PVXXXX series&lt;br&gt;&#xA;Oilgear PFBX and PFCX series  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Radial piston pumps.&lt;br&gt;&#xA;Operating principle at 0:15 - &lt;a href=&quot;https://www.youtube.com/watch?v=2T-6jDm_ebI&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=2T-6jDm_ebI&lt;/a&gt;&lt;br&gt;&#xA;Commercial examples:&lt;br&gt;&#xA;Bosch-Rexroth PR4-XX series&lt;br&gt;&#xA;HYDAC HRK series&lt;br&gt;&#xA;HAWE R series  &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Need more info about inlet/outlet pressure, flow, temperature, working fluid, noise limits, motor speed/power/torque, geometric constraints, etc... to recommend a specific pump. &lt;/p&gt;&#xA;" OwnerUserId="24104" LastActivityDate="2020-01-29T01:58:36.193" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="32961" PostTypeId="1" CreationDate="2020-01-29T02:18:39.243" Score="0" ViewCount="29" Body="&lt;p&gt;A cable drum is driven by 3 electric motors via a bull gear arrangement. Gear ratio, total motor inertia (including pinion gears) and total drum inertia are known. From the motor variable speed drives, the motor angular acceleration and torque are measured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can the following formulae be used to calculate how much torque the drum will produce? (And in turn calculate the cable tension)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\tau_{out} = {R\tau_{in}-\alpha(I_{motor}+{I_{drum}\over R^2}) }$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\tau_{out}:$&lt;/span&gt; Torque out from drum, &lt;span class=&quot;math-container&quot;&gt;$R:$&lt;/span&gt; Gear ratio, &lt;span class=&quot;math-container&quot;&gt;$\tau_{in}$&lt;/span&gt; Torque supplied by motors, &lt;span class=&quot;math-container&quot;&gt;$\alpha:$&lt;/span&gt; Motor angular acceleration, &lt;span class=&quot;math-container&quot;&gt;$I_{motor}:$&lt;/span&gt; Inertia of motors and pinion gears, &lt;span class=&quot;math-container&quot;&gt;$I_{drum}:$&lt;/span&gt; Inertia of drum and bull gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the formulae does not account for transmission efficiency. Where in the formulae would this fit in?&lt;/p&gt;&#xA;" OwnerUserId="24385" LastActivityDate="2020-01-29T02:18:39.243" Title="Torque output of transmission when accounting for moment of inertia" Tags="|transmission|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32962" PostTypeId="1" CreationDate="2020-01-29T03:47:23.617" Score="0" ViewCount="328" Body="&lt;p&gt;When mixed with water how does gypsum act as a binder in creating concrete?&lt;/p&gt;&#xA;" OwnerUserId="16653" LastActivityDate="2022-09-27T18:25:32.297" Title="How does Gypsum Concrete work?" Tags="|civil-engineering|concrete|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32964" PostTypeId="1" CreationDate="2020-01-29T14:41:24.987" Score="2" ViewCount="186" Body="&lt;p&gt;Oversampling means to sample at significantly more than the Nyquist Rate. I can well understand it in signal re-construction: more samples can improve the ADC's resolution, signal-to-error ratio, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when in a control system, I have some difficulities understanding it. Assume that the frequency of the process variable is &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt;, so the Nyquist Rate is &lt;span class=&quot;math-container&quot;&gt;$2f$&lt;/span&gt;, and we sample at &lt;span class=&quot;math-container&quot;&gt;$2f$&lt;/span&gt; (the controller will take action upon each sample). Maybe we don't get a stable process variable, so we decide to improve the sampling rate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, we sample at &lt;span class=&quot;math-container&quot;&gt;$20f$&lt;/span&gt;, a sampling frequency 10 times higher (also the controller takes action upon each sample), and this time we obtain a stable process variable. However, because the controller action is also 10 times faster, the frequency of the process variable is no longer &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt;, but much higher, say &lt;span class=&quot;math-container&quot;&gt;$f'$&lt;/span&gt;.  Now the Nyquist Rate becomes &lt;span class=&quot;math-container&quot;&gt;$2f'$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Comparing sampling frequency &lt;span class=&quot;math-container&quot;&gt;$20f$&lt;/span&gt; and new Nyquist rate &lt;span class=&quot;math-container&quot;&gt;$2f'$&lt;/span&gt;, they may be very close, so we cannot call it &quot;oversampling&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my understanding towards oversampling and Nyquist rate correct?&lt;/p&gt;&#xA;" OwnerUserId="23682" LastActivityDate="2020-02-20T23:58:51.347" Title="Understanding oversampling in control system" Tags="|control-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32965" PostTypeId="1" AcceptedAnswerId="32979" CreationDate="2020-01-29T15:46:23.433" Score="1" ViewCount="100" Body="&lt;p&gt;Consider we have the third order plant described by the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ y = G(s)u $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\ G(s) = \frac{b_{2}s^{2}+b_{1}s+b_{0}}{s^{3}+a_2s^2+a_1s+a_0} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that all parameters &lt;span class=&quot;math-container&quot;&gt;$\ a_0,a_1,a_2,b_0,b_1,b_2 $&lt;/span&gt; are unknown then we can develos a linear parametric model of the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ z = θ^{*Τ}φ $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ θ^*=[a_2\ a_1\ a_0\ b_2\ b_1\ b_0]^T $&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\ φ = [-\frac{s^2}{Λ(s)}y \ \ \ -\frac{s}{Λ(s)}y \ \ \ -\frac{1}{Λ(s)}y \ \ \ -\frac{s^2}{Λ(s)}u \ \ \ -\frac{s}{Λ(s)}u \ \ \ -\frac{1}{Λ(s)}u]^T $&lt;/span&gt; ,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ z = \frac{s^3}{Λ(s)}y $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, let's suppose that the parameters &lt;span class=&quot;math-container&quot;&gt;$\ a_2,a_1,a_0 $&lt;/span&gt; are known and their values are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ a_2 = 3 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ a_1 = 1 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ a_0 = 2 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can we develop a linear parametric model of the plant considering these parameters known and as a result the vector &lt;span class=&quot;math-container&quot;&gt;$\ θ $&lt;/span&gt; should be &lt;span class=&quot;math-container&quot;&gt;$\ θ^* = [b_2 \ \ \ b_1 \ \ \ b_0]^T $&lt;/span&gt; ? &lt;/p&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-03-15T05:34:26.247" LastActivityDate="2020-03-15T05:34:26.247" Title="Develop linear parametric model of a third order plant" Tags="|mechanical-engineering|control-engineering|control-theory|transfer-function|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32967" PostTypeId="1" AcceptedAnswerId="32972" CreationDate="2020-01-29T16:01:34.237" Score="1" ViewCount="165" Body="&lt;p&gt;I have a network of stainless steel pipes that are distributing water throughout a building. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know the flow rate of water entering the system is 600 IGPM, but am unsure of how this water will be distributed. I have attached a simplified schematic of what the pipe network looks like. &lt;a href=&quot;https://i.stack.imgur.com/jB4P0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jB4P0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it safe to assume the flow rate throughout this network will be the same, even though the diameters change, based on the continuity equation?&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A1*V1 = A2*V2&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="24382" LastActivityDate="2020-01-29T19:06:12.020" Title="Distribution of water through a pipe network" Tags="|mechanical-engineering|hydraulics|fluid|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32969" PostTypeId="2" ParentId="32959" CreationDate="2020-01-29T17:13:23.897" Score="2" Body="&lt;p&gt;The (r) symbol indicates this is a registered trademark, and as such has registered with the US Patent and Trademark Office. The USPTO maintains a public searchable &lt;a href=&quot;http://tmsearch.uspto.gov/bin/gate.exe?f=tess&amp;amp;state=4807:luw4a3.1.1&quot; rel=&quot;nofollow noreferrer&quot;&gt;database&lt;/a&gt; of all registered marks. A little trial and error reveals the pertinent search term here, is &quot;Wonder Board&quot; including the space. This &lt;a href=&quot;http://tmsearch.uspto.gov/bin/showfield?f=doc&amp;amp;state=4807:luw4a3.2.2&quot; rel=&quot;nofollow noreferrer&quot;&gt;mark&lt;/a&gt; is registered to Modulars, Inc. This same corporation appears to hold the mark &quot;wonder-board&quot; as well, dating from the 70s and 80s.&lt;/p&gt;&#xA;" OwnerUserId="2434" LastActivityDate="2020-01-29T17:13:23.897" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32970" PostTypeId="1" CreationDate="2020-01-29T17:35:52.350" Score="0" ViewCount="72" Body="&lt;p&gt;Is there any efficient way of using a large volume but low pressure (&amp;lt;2 bar) stored compressed air, either to drive an engine or some other application?&lt;/p&gt;&#xA;" OwnerUserId="24406" LastEditorUserId="3505" LastEditDate="2020-06-27T20:28:31.250" LastActivityDate="2020-06-28T06:33:07.347" Title="Compressed Air Engine" Tags="|automotive-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32971" PostTypeId="2" ParentId="32970" CreationDate="2020-01-29T18:17:39.147" Score="0" Body="&lt;p&gt;welcome on this SE. Your question might be a little vague for Engineering SE, but here is a possibility:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use this volume of air as a refrigerant: You put it at ambient temperature, wait for the air volume to come to the ambient temperature, and then you put it close to a volume to cool down: the difference of pressure will cool down the volume of air, and heat transfer will do the rest.&lt;/p&gt;&#xA;" OwnerUserId="21434" LastActivityDate="2020-01-29T18:17:39.147" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32972" PostTypeId="2" ParentId="32967" CreationDate="2020-01-29T19:06:12.020" Score="1" Body="&lt;p&gt;The flow rate through the network will most definitely not change only with the area of the pipes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As water flows through a system, its pressure (often referred to as head) is reduced by friction along the way and changes in elevation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pipe friction is a function of pipe diameter, pipe length, fluid characteristics, and pipe wall smoothness.  Every bend in a pipe also introduces additional head loss.  The head loss due to this directly affects the flow rate through the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the key thing that will affect flow through the network is the usage by each end user.  I doubt that these pipes just empty into rooms.  So the flows in each branch pipe will be determined by the outlet of the pipes.  In a non-industrial use we can probably assume that water requirements fluctuate greatly.  For industrial use we can determine flows by direct measurement (buckets, etc), or flow measurement devices in the pipes.  If we had static pressures throughout the system we might be able to back in to flows, given that we know the total.  I feel like this might have to be an iterative process.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-01-29T19:06:12.020" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32973" PostTypeId="1" CreationDate="2020-01-29T19:11:05.680" Score="0" ViewCount="943" Body="&lt;p&gt;I have watched &lt;a href=&quot;https://www.youtube.com/watch?v=6KC-a97joew&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video&lt;/a&gt; where the speaker explained how the vertical spin of a quadcopter is stabilized, and how it could be unstabilized to make the drone rotate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a screenshot of the video, with:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In green: motors&lt;/li&gt;&#xA;&lt;li&gt;In red and blue : rotation sense of the motors&lt;/li&gt;&#xA;&lt;li&gt;In brown: arrows for the force exerted by the motors on the quadcopter (3rd newton law)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YNNmu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YNNmu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not understand that much the principles, so I would like an explanation with this question: What is the formula of the force momentum created by one of quadcopter's motors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking is:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I imagine it is not the value of the torque used by the motor to rotate, otherwise it would be independant of the speed of the motor.&lt;/li&gt;&#xA;&lt;li&gt;It should not be either the lift force created by the motor (momentum would be zero in respect of the force vector's orientation).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What bothers me in the video is: Why are there only two sides of each motor where there is a force (brown arrows)? There should be four since the torque is applied all around the axis. And if there are four forces on four sides (i.e. infinite forces around the axis), then how is the difference of force (and thus torque) created?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then how is this torque (on the drone's vertical axis) created by each of the motors?&lt;/p&gt;&#xA;" OwnerUserId="21434" LastEditorUserId="21434" LastEditDate="2020-01-29T20:54:00.753" LastActivityDate="2024-02-10T22:07:30.953" Title="How to calculate the torque created by a rotating motor?" Tags="|motors|torque|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32974" PostTypeId="2" ParentId="32947" CreationDate="2020-01-29T22:38:31.517" Score="6" Body="&lt;p&gt;This is an instrument symbol, they are read in the following way, &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;a href=&quot;https://blog.projectmaterials.com/instrumentation/pid-symbols/&quot; rel=&quot;noreferrer&quot;&gt;ISA Symbology&lt;/a&gt;&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The symbology of the instrument it is placed in a circle of 7/16 “or 12 mm (see below), was in the top there is the functional identification (of the instrument or device) composed with max 5 letters, and in the bottom there is identification of the loop (measurement and/or control) consists of a progressive number of 3 or 4 numbers, followed by a literal suffix to identifies two or more identical instruments inserted in the same chain and therefore having the same functional identification&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;First letter, (what is being measured)&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;Following Letters, which can be passive, active, and a modifier&lt;/li&gt;&#xA;&lt;li&gt;The Bottom numbers are which loop they run on.&lt;/li&gt;&#xA;&lt;/ol&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So in your case:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;G - Is specifically User choice, which in this case may mean Gas.&lt;/li&gt;&#xA;&lt;li&gt;O - Orifice or a restriction &lt;/li&gt;&#xA;&lt;li&gt;S - Switch&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="14318" LastEditorUserId="14318" LastEditDate="2020-01-29T22:46:04.037" LastActivityDate="2020-01-29T22:46:04.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32975" PostTypeId="1" CreationDate="2020-01-29T23:42:58.220" Score="0" ViewCount="57" Body="&lt;p&gt;I have a need for a pump that can suck a vapor from a 0.1 atm environment, and flow into a 1 atm environment. I know that there are electrical pumps that can do this. But are there any pumps like this that can be operated manually? (powered by hand or body weight)&lt;/p&gt;&#xA;" OwnerUserId="24409" LastActivityDate="2020-01-30T00:36:32.327" Title="Does a pump like this exist?" Tags="|vacuum|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32976" PostTypeId="2" ParentId="32975" CreationDate="2020-01-30T00:36:32.327" Score="2" Body="&lt;p&gt;I've owned a hand-held pistol-type vacuum pump, useful for testing distributor advance (yeah, long time ago). A quick search shows the inexpensive versions are capable of 0.6-0.7 atm, not quite what you seek.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional searching (reading from the links) allowed me to discover a similar design, listed as capable of 0.1 atm. If you require to remove vapor from this environment, you'd need a pump to go slightly lower in order to have a proper differential.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not sure of your presentation. How did the environment reach 0.1 atm, or are you wishing to take a 1 atm chamber down to 0.1 atm? If the latter, and the volume is not too great, the one I've found may work for you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At US$259, it should be a better performer than the thirty or forty dollar versions.&#xA;&lt;a href=&quot;https://www.powerprobe.com/web-store-1/lrhv90-vacuumpressure-kit?rq=vacuum%20pump&quot; rel=&quot;nofollow noreferrer&quot;&gt;hand vacuum pump from PowerProbe&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jAdna.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jAdna.jpg&quot; alt=&quot;PowerProbe vacuum pump&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the terms &quot;manual vacuum pump&quot; provided these results. No manipulation of terms provided for a foot operated vacuum pump, although references were found for converting a bicycle floor pump to vacuum. It's unlikely that such a pump would reach 0.1 atm, however.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-01-30T00:36:32.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32978" PostTypeId="1" CreationDate="2020-01-30T08:21:21.433" Score="-1" ViewCount="18" Body="&lt;p&gt;Basically put a light coat of activated carbon on the collector, because surface area is so great, but I cant find info on its ability to collect solar energy, esp. compared to other materials.   Im guessing not because its probably a poor thermal conductor.&lt;/p&gt;&#xA;" OwnerUserId="24411" LastActivityDate="2020-01-30T16:12:11.313" Title="Would using activated carbon in a solar air collector help due to increased surface area?" Tags="|solar-energy|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="32979" PostTypeId="2" ParentId="32965" CreationDate="2020-01-30T12:55:31.150" Score="1" Body="&lt;p&gt;As &lt;code&gt;fibonatic&lt;/code&gt; commented the proper linear parameterization in terms of the unknown &lt;span class=&quot;math-container&quot;&gt;$\ b_2,b_1,b_0 $&lt;/span&gt; parameters is as follows: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ y = G(s)u  $&lt;/span&gt;  is in terms of a differential equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ \dddot{y} + a_2\ddot{y}+a_1\dot{y}+a_0y = b_2\ddot{u}+b_1\dot{u}+b_0u $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the fact that we only have measurements of the input &lt;span class=&quot;math-container&quot;&gt;$\ u $&lt;/span&gt; and output &lt;span class=&quot;math-container&quot;&gt;$\ y $&lt;/span&gt; of the system, we can't use their derivatives. As a result, we filter each term with a third order stable filter (poles of filter need to be negative) &lt;span class=&quot;math-container&quot;&gt;$\ Λ(s) = s^3 + λ_1s^2 + λ_2s + λ_3 $&lt;/span&gt; where the coefficients &lt;span class=&quot;math-container&quot;&gt;$\ λ_1,λ_2,λ_3 $&lt;/span&gt; are chosen though the poles of the filter. Filtering the above differential equation results in no use of differentiators and produces the following equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ \frac{s^3}{Λ(s)}y \ + a_2\frac{s^2}{Λ(s)}y \ + a_1\frac{s}{Λ(s)}y + a_0\frac{1}{Λ(s)}y \ = b_2\frac{s^2}{Λ(s)}u \ +b_1\frac{s}{Λ(s)}u \ + b_0\frac{1}{Λ(s)}u $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's define &lt;span class=&quot;math-container&quot;&gt;$\ z = \frac{s^3}{Λ(s)}y \ + a_2\frac{s^2}{Λ(s)}y \ + a_1\frac{s}{Λ(s)}y + a_0\frac{1}{Λ(s)}y \ \  $&lt;/span&gt; since &lt;span class=&quot;math-container&quot;&gt;$\ Λ(s),a_2,a_1,a_0,y $&lt;/span&gt; are known. Our equation now becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ z = [b_2 \ \ b_1 \ \ b_0] [\frac{s^2}{Λ(s)}u \ \ \frac{s}{Λ(s)}u \ \ \frac{1}{Λ(s)}u]^T  $&lt;/span&gt; which is in the form: &lt;span class=&quot;math-container&quot;&gt;$\ z = θ^{*Τ}φ $&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$\ θ^{*} = [b_2 \ \ b_1 \ \ b_0]^T $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ φ = [\frac{s^2}{Λ(s)}u \ \ \frac{s}{Λ(s)}u \ \ \frac{1}{Λ(s)}u]^T $&lt;/span&gt;. So, now we have the linear parametric model in terms of the unknown parameters &lt;span class=&quot;math-container&quot;&gt;$\ b_2, b_1,b_0 $&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Correspondingly we can come up with the linear parametric model in terms of the parameters &lt;span class=&quot;math-container&quot;&gt;$\ a_2,a_1,a_0 $&lt;/span&gt; being unknown and the parameters &lt;span class=&quot;math-container&quot;&gt;$\ b_2,b_1,b_0 $&lt;/span&gt; being known. Following the same procedure, the linear parametric model is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ z = θ^{*Τ}φ $&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$\ z = \frac{s^3}{Λ(s)}y \ - b_2\frac{s^2}{Λ(s)}u \ - b_1\frac{s}{Λ(s)}u - b_0\frac{1}{Λ(s)}u \ \  $&lt;/span&gt; since &lt;span class=&quot;math-container&quot;&gt;$\ Λ(s),b_2,b_1,b_0,y $&lt;/span&gt; are known, &lt;span class=&quot;math-container&quot;&gt;$\ θ^{*} = [a_2 \ \ a_1 \ \ a_0]^T $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ φ = [-\frac{s^2}{Λ(s)}y \ \ -\frac{s}{Λ(s)}y \ \ -\frac{1}{Λ(s)}y]^T $&lt;/span&gt;. So, now we have the linear parametric model in terms of the unknown parameters &lt;span class=&quot;math-container&quot;&gt;$\ a_2, a_1,a_0 $&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-01-31T10:34:44.143" LastActivityDate="2020-01-31T10:34:44.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32980" PostTypeId="1" AcceptedAnswerId="32981" CreationDate="2020-01-30T13:26:05.187" Score="2" ViewCount="593" Body="&lt;p&gt;I've heard this term used a lot in literature but I don't understand what the technique is they are referring to. Please help me out here. &lt;/p&gt;&#xA;" OwnerUserId="253" LastActivityDate="2020-01-30T23:55:31.220" Title="What does &quot;bedding-in&quot; mean in casting technology?" Tags="|mechanical-engineering|manufacturing-engineering|casting|production-technology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32981" PostTypeId="2" ParentId="32980" CreationDate="2020-01-30T15:07:51.203" Score="4" Body="&lt;p&gt;According to p.33 of &lt;a href=&quot;http://www.iitg.ac.in/engfac/ganu/public_html/Metal%20casting%20processes_1.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this presentation&lt;/a&gt; 'bedding in' is a process of packing the molding sand by ramming the sand around and under the pattern until the sand is tightly packed and even with the parting line. There is also a glossary with foundry and casting terms defined &lt;a href=&quot;http://www.atlasfdry.com/glossaryb.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. This is used when the parts to be cast are quite large, often as a step in pit molding or floor molding. A description of pit molding is found at &lt;em&gt;&lt;a href=&quot;https://www.coursehero.com/file/p1nqcj7/1219-CLASSIFICATION-OF-MOLDING-PROCESSES-Molding-processes-can-be-classified-in/&quot; rel=&quot;nofollow noreferrer&quot;&gt;coursehero.com&lt;/a&gt;&lt;/em&gt;: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;12.20.3 Pit Molding Usually large castings are made in pits instead of drag flasks because of their huge size. In pit molding, the sand under the pattern is rammed by bedding-in process. The walls and the bottom of the pit are usually reinforced with concrete and a layer of coke is laid on the bottom of the pit to enable easy escape of gas. The coke bed is connected to atmosphere through vent pipes which provide an outlet to the gases. One box is generally required to complete the mold, runner, sprue, pouring basin and gates are cut in it.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Floor bedding is discussed in &lt;em&gt;MANUFACTURING PROCESSES&lt;/em&gt; By J. P. KAUSHISH. An excerpt from Google Books (also found &lt;a href=&quot;https://books.google.com/books?id=FDIfTrE3BjUC&amp;amp;pg=PA205&amp;amp;lpg=PA205&amp;amp;dq=bedding%20in%20pit%20molding&amp;amp;source=bl&amp;amp;ots=dx6k6G1tcL&amp;amp;sig=ACfU3U1_uNOYHfTENFOufxmGjQORg9ZhJw&amp;amp;hl=en&amp;amp;ppis=_e&amp;amp;sa=X&amp;amp;ved=2ahUKEwim3JvNvqznAhVulHIEHXyACyUQ6AEwFHoECAsQAQ#v=onepage&amp;amp;q=bedding%20in%20pit%20molding&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nv1UA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nv1UA.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9682" LastEditorUserId="9682" LastEditDate="2020-01-30T23:55:31.220" LastActivityDate="2020-01-30T23:55:31.220" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32982" PostTypeId="1" AcceptedAnswerId="33328" CreationDate="2020-01-30T16:00:11.560" Score="0" ViewCount="1945" Body="&lt;p&gt;I have a doubt concerning the closed loop frequency responce of a control system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose I have the following situation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zu0iP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zu0iP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which is plotted the frequency response of a system with different badwidths. In this case I know that in generale the performances are good as the bandwidth of the closed loop increases and I know that the price to pay to increase the bandwidth is a high control effort, but is there some other drawbacks in increasing the bandwidth of a system? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking abouth the noises, which are at high frequencies, this if the bandwidth is at high frequencies, and the system does not roll off fast, the noises enter into the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I am not sure about what I said.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that as the badwidth increases, also the reference tracking gets better, but I also see that the roll off is not smooth, I so I was interpreting also this as a fact that there are worse performances with respect to the noise rejection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, I know that there are some limitation of increasing the bandwidth of a system. So, considering a stable and minimum phase system, I know that the limit in bandwidth ig given by the hardware. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if I increase the bandwidth is there some other limitation? By doing some experimentaition I have see that the disturbance rejection gets worse after a certain frequency, is it due to noise?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can somebody please help me?&lt;/p&gt;&#xA;" OwnerUserId="22780" LastEditorUserId="22780" LastEditDate="2020-01-30T16:30:49.660" LastActivityDate="2020-02-20T23:52:29.677" Title="What are the drawbacks of inceasing the bandwidth of the system?" Tags="|electrical-engineering|control-engineering|control-theory|systems-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32983" PostTypeId="2" ParentId="32978" CreationDate="2020-01-30T16:12:11.313" Score="0" Body="&lt;p&gt;No, the carbon would just oxidize away. You add sunlight and air (including oxygen) to your carbon so you have all the ingredients to produce CO2. &lt;/p&gt;&#xA;" OwnerUserId="669" LastActivityDate="2020-01-30T16:12:11.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32984" PostTypeId="2" ParentId="32973" CreationDate="2020-01-30T18:10:57.993" Score="0" Body="&lt;p&gt;IMO, the 4 forces can be summarised as two forces. Since there are 2 X forces and two Y forces (in the XYZ coordinates), the 2 X forces can be represented by one effective X force. Same to Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope I've answered your question. &lt;/p&gt;&#xA;" OwnerUserId="5408" LastActivityDate="2020-01-30T18:10:57.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32986" PostTypeId="1" CreationDate="2020-01-30T21:47:00.773" Score="0" ViewCount="781" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VvYQE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VvYQE.png&quot; alt=&quot;in a separate, new, clean file, it works&quot;&gt;&lt;/a&gt;this should work.. surely..  I even tested it in another file with another solid body present but not connected..  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;moved the operation to the very beginning of the design tree, suppressed everything else, same problem.  it must be something about the file..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kgvHr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kgvHr.png&quot; alt=&quot;the problem&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13081" LastEditorUserId="13081" LastEditDate="2020-01-30T23:55:22.363" LastActivityDate="2020-01-30T23:55:22.363" Title="&quot;Intermediate profile #2 could not be solved. Consider editing the relations and dimensions of the original sketch.&quot;" Tags="|solidworks|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="32987" PostTypeId="2" ParentId="32973" CreationDate="2020-01-31T01:59:16.037" Score="0" Body="&lt;p&gt;The four propellers basically act as rotating wings and follow the same aeronautic laws, except more complex. They create lift, have to fight the drag and can stall, a grossly simplified version of the complex dynamics of a propeller. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The diagram you have is just to help illustrate the narrative of the clip and is not complete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because of the variation of the speed from the center to the tip, the angle of attack is not uniform along the length of the blade, it is twisted and get more flat near the tip. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some sites for hobbyist quad-copter makers and there are some free courses on the design of the propellers. If you search for aeronautic courses you will find many, it can get mathematically involved. NASA has a reference &lt;a href=&quot;https://www.grc.nasa.gov/www/k-12/airplane/propeller.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here.&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the link to a site with simplified but insightful, detailed coverage of torque, thrust, power, with diagrams and charts and easy to follow worked out examples. &lt;a href=&quot;http://www.mh-aerotools.de/airfoils/index.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;prop design&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-01-31T01:59:16.037" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32988" PostTypeId="1" CreationDate="2020-01-31T05:20:14.513" Score="5" ViewCount="189" Body="&lt;p&gt;The satellite view of &lt;a href=&quot;https://www.google.com/maps/@-38.2321329,145.1238585,1255m/data=!3m1!1e3&quot; rel=&quot;noreferrer&quot;&gt;this overpass (on Google Maps)&lt;/a&gt; shows some banded earthworks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fPxrA.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fPxrAm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When driving by, you can see that &lt;a href=&quot;https://www.google.com/maps/@-38.2321607,145.1252278,3a,75y,106.7h,91.93t/data=!3m6!1e1!3m4!1sxbS8WffadTMKn-rrhkNJPQ!2e0!7i13312!8i6656&quot; rel=&quot;noreferrer&quot;&gt;each band is an elongated triangular shape (street view)&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wavT6.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wavT6m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is that these earthworks act as a wind break by making the wind turbulent, so as to minimise the force of cross winds on the overpass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is that what they're for? What are they called?&lt;/p&gt;&#xA;" OwnerUserId="14330" LastActivityDate="2020-02-07T18:08:04.123" Title="What are these banded earthworks near a freeway overpass for and what are they called?" Tags="|highway-engineering|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="32989" PostTypeId="2" ParentId="15991" CreationDate="2020-01-31T05:33:41.087" Score="1" Body="&lt;p&gt;In my experience you should always dig when creating detention ponds. This allows you to position it at the best location available to you incase water levels run high. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this purpose, you may want to build the banks up higher than you actually intend them to go, and then leave a gap where you plan for any excess water to escape. You'll need to create a path leading away from the pond toward a creek, stream or river. You'll also need the local government's permission to do so before you introduce a new tributary to the local water table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on how big the pond is I would also recommend a very good PVC liner or if ita a bigger detention pond some kind of &lt;a href=&quot;https://tankliners.com.au/&quot; rel=&quot;nofollow noreferrer&quot;&gt;tank liners&lt;/a&gt; will do the trick.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a really good guide here also: &lt;a href=&quot;https://www.hunker.com/13424245/how-to-create-a-retention-pond-for-water-runoff&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.hunker.com/13424245/how-to-create-a-retention-pond-for-water-runoff&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;good luck with your project&lt;/p&gt;&#xA;" OwnerUserId="24424" LastActivityDate="2020-01-31T05:33:41.087" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32990" PostTypeId="1" AcceptedAnswerId="32991" CreationDate="2020-01-31T05:59:43.840" Score="-1" ViewCount="56" Body="&lt;p&gt;I ask this because I want to get into robotics(mainly the building of it), so I wanted to go to college to learn mechanical engineering but now that I have found out so many different types of engineering(other than just mechanical and electrical, which are what I thought where the only engineering degrees) I don't know for sure which one would be best for me.&lt;/p&gt;&#xA;" OwnerUserId="24426" LastEditorUserId="24426" LastEditDate="2020-01-31T06:42:18.397" LastActivityDate="2020-01-31T09:22:36.033" Title="Would mechanical engineering work robotics and if not what field does?" Tags="|robotics|" AnswerCount="2" CommentCount="8" ClosedDate="2020-02-01T19:27:31.820" ContentLicense="CC BY-SA 4.0" />
  <row Id="32991" PostTypeId="2" ParentId="32990" CreationDate="2020-01-31T06:47:25.927" Score="0" Body="&lt;p&gt;Mechanical engineering is an excellent choice for a person interested in robotics. In the mechanical engineering field, you will learn about:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-modeling the dynamics of mechanical systems and control systems theory &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-strength of materials and materials engineering, including materials selection&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-kinematics of mechanisms &lt;/p&gt;&#xA;&#xA;&lt;p&gt;-machine design, including bearings, cams, levers, gearboxes, limit switches, screw drives, belt drives and the motors that power them&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-manufacturing methods, including machining, welding, casting, and bolted assemblies.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-01-31T06:47:25.927" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="32992" PostTypeId="2" ParentId="32947" CreationDate="2020-01-31T07:32:22.853" Score="2" Body="&lt;p&gt;Morbo explains the logic quite well, however I disagree regardin the meaning of the letters. According to the IEC 62424 (which would specify H/L instead of + -) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;G means distance or position is measured&lt;br&gt;&#xA;O is local display or display on SCADA&lt;br&gt;&#xA;S means switch&#xA;+- wiould refer to the two positions something switches on&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since all the GOS+- operate on valves I'd say that these are the limit switches, to display valve position on the SCADA and supply position data for control purposes.&lt;/p&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2020-05-23T22:06:38.073" LastActivityDate="2020-05-23T22:06:38.073" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32993" PostTypeId="1" CreationDate="2020-01-27T09:21:02.687" Score="0" ViewCount="331" Body="&lt;p&gt;&lt;a href=&quot;https://www.envirosafetyproducts.com/resources/dust-masks-whats-the-difference.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;A &quot;respirator with a rating of P100, has been tested to be proof against oil, and tested to filter 99.97% of all particles 0.3 microns in diameter or larger.'&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;As you know, &lt;a href=&quot;https://www.quora.com/It-is-said-that-most-of-the-atom-is-empty-space-How-can-space-be-empty-If-space-is-not-empty-what-is-this-space-in-the-atom&quot; rel=&quot;nofollow noreferrer&quot;&gt;an atom is 99.9999% empty space&lt;/a&gt;. So even if a respirator's edge is touching your face, why can't pathogens cross and sneak past the face seal? See the green arrows in my pictures below.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;To wit, how can we guarantee that the gap between your face and a respirator's edge is smaller than any virus or pathogen? &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.sfgate.com/california-wildfires/article/N95-P100-mask-respirator-camp-fire-explained-13399136.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;N95, P100: What do all these mask numbers mean and how do I know it's keeping me safe? - SFGate&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the mask isn’t sealed around your face, you’re just breathing through the edges, and it doesn’t filter much at all.  Men with beards cannot get the full benefit of a mask, and many men have looked at shaving it off for the sake of their lungs.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;3M writes in this &lt;a href=&quot;http://multimedia.3m.com/mws/mediawebserver?mwsId=SSSSSufSevTsZxtUOxmG4x_SevUqevTSevTSevTSeSSSSSS--&quot; rel=&quot;nofollow noreferrer&quot;&gt;handy but rather complex guide&lt;/a&gt;: “Do not use tight-fitting respirators or loose-fitting facepieces with beards or other facial hair or conditions that prevent direct contact between the face and the edge of the respirator.”&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The left is a &lt;a href=&quot;https://www.aliexpress.com/item/32603948219.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;3M P100 respirator&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RWUES.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RWUES.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerDisplayName="user24573" LastActivityDate="2020-02-02T14:26:36.920" Title="How exactly do respirators seal out pathogens?" Tags="|safety|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32995" PostTypeId="2" ParentId="32990" CreationDate="2020-01-31T09:22:36.033" Score="0" Body="&lt;p&gt;Mechatronics is the relatively new engineering discipline that combines mechanical and electrical engineering, with robotics right at the forefront.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Mechatronics&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Mechatronics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-01-31T09:22:36.033" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32996" PostTypeId="2" ParentId="13554" CreationDate="2020-01-31T09:49:34.973" Score="0" Body="&lt;p&gt;Just because India has used it for a century does not mean it meets European safety standards. As with any construction material, a factor of safety should applied. In this instance it is good practice, developed over centuries of mistakes and fatalities, to imbue buildings with failsafes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Were an infant lying on the floor to be hit on the head by such a tile from a height of 2m or more, the consequences could be life threatening and life changing injuries would likely result. Consider the legal implications of that worst-case scenario.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with all structural design approaches, we look at the structure integrity for the purpose of safe and durable design. Clearly these tiles are durable undisturbed but should the be disturbed that may not be the case.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would suggest application of fibre reinforcement by way of a mesh. If the tiles are fired and designed to be waterproof then steel mesh but if they are pressed and dried below 85°C, a plastic mesh might do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In thinking of sustainable design, use of recycled materials may be a better solution. A simple sheet of paper has great tensile properties. Gypsum board is often reinforced this way. Woven paper strips may be a good option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let me know what you think  &lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2020-01-31T09:49:34.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="32997" PostTypeId="2" ParentId="32973" CreationDate="2020-01-31T09:52:58.850" Score="0" Body="&lt;p&gt;The arrows on your diagram are intended as a visualisation aid but, in this instance, you are better off without them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque is (effectively) conserved.&lt;br&gt;&#xA;Applying a torque impulse to a body &quot;in free fall&quot; is the rotational equivalent of applying an straight line impulse to a body in free fall and Newton's laws apply - the body will rotate under the effects of the impulse until an equal and opposite torque is applies. This may be via another equal and opposite torque impulse, or via eg aerodynamic drag, or some combination. Be wary of the special case of lithobraking. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a vacuum (where your propellors don't do much good) the torque imparted to the motor rotor relative to the stator is matched by opposite sense torque in the body. Your motor will rapidly try to self destruct, so lets add some air.&lt;br&gt;&#xA;Air accelerated by the propellor will have torque conveyed to it by the rotor, but transfer is not 100% efficient. In a single rotor vehicle (eg classic helicopter) you need to counteract the residual torque and this is usually done with a tail rotor. The long moment arm of the tail rotor reduces the force required.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a quad rotor machine with two rotors rotating in each 'direction' and at identical power and speed there is zero net torque. Operate one rotor at higher or lower speed and the net torque becomes non zero. For control purposes it is probably better to alter two rotors with the same rotational direction.&lt;br&gt;&#xA;Once you have differing clockwise and anticlockwise torques the next difference is applied to the structure as a whole and, just as with the classic helicopter, it must rotate if there is no restoring force to oppose it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The amount of force has an upper limit but the actual value is indeterminate without additional information.  The torque produced by a motor is (to about a 2% approximation)&lt;br&gt;&#xA;RPM x torque = power&lt;br&gt;&#xA;(torque in kg.m, power in watts).&lt;br&gt;&#xA;or torque = power/RPM.&lt;br&gt;&#xA;Electrical power input and RPM are known (or knowable) to the flight controller.&lt;br&gt;&#xA;However, the electrical power is not fully efficiently converted to air driving power so the available torque at the motor mount will be less than this maximum amount.&lt;br&gt;&#xA;A delta reduction or increase in two diagonally opposed propellors will result in some net torque on the aircraft somewhat less than Power1/RPM1 - Power2/RPM2.&lt;br&gt;&#xA;(with sign of torque depending on delta power sign and rotation direction) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;A little playing&quot; probably allows the system designers to have a good idea of what will be achieved. As these aircraft invariably have 3D gyroscopes the achieved rotation rate can be monitored. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;__________________________&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;ADDED:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;... if one motor has higher torque than other, and thus higher &quot;counter torque&quot;, the counter torque will create on the center of the quadcopter a rotation centered on the most powerful motor, and thus instead on rotating on a vertical axis, the center of mass of the drone (at the center of the quadcopter) will drift rather than rotate. Am I wrong?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are wrong.  I said &quot;torque is conserved&quot;. You are saying torque is converted to linear motion. On what basis would you think that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, no - not centred on the most powerful (or different motor) - but about the centre of the force couple based on overall forces on the machine and location of the c of g and it's moment of inertia and ... . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that I said &quot; ... A delta reduction or increase in two diagonally opposed propellors &lt;strong&gt;will result in some net torque on the aircraft&lt;/strong&gt; somewhat less than Power1/RPM1 - Power2/RPM2. ...&quot; -&gt; Anyone designing the drone to rotate using this system will ALMOST CERTAINLY do it by providing balanced torque change across a diagonal with diagonally opposed two motors  - so the rotation will be about the Drone's centre (if CofG and aerodynamic and centre  of inertia are all at the same point (as they tend to be (approximately) for a reasonably symmetrical drone. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even better may be (name motors ABCD around periphery) to Torque UP motors A &amp;amp; C and torque down B &amp;amp; D to get rotating torque overall about the aircraft's centre with the same lift. ) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BUT, if you do change the torque on one motor you will get net applied torque and the craft WILL rotate. Exactly where it rotates about depends on the vector sum of the various forces and location of centres of aerodynamic pressure, inertia, ... .    BUT trying to work out exactly what it rotates about is an excessive complication - if it is travelling say North at say 20 kph, if it rotates as it flies you would not visually be able to tell WHERE it was rotating about as the rotation velocities would sum with the travelling velocity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You seem to want to insist that torque causes linear motion and not rotation. You have no good basis for this belief.&lt;/p&gt;&#xA;" OwnerUserId="399" LastEditorUserId="399" LastEditDate="2020-02-01T11:49:42.107" LastActivityDate="2020-02-01T11:49:42.107" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="32999" PostTypeId="1" CreationDate="2020-01-31T12:06:21.190" Score="1" ViewCount="154" Body="&lt;p&gt;I am a civil engineering undergraduate. I have an interest in Biotropic Design and sustainable energy and am working on a new conceptual design. I would like to know what analytics may be done to prove the efficiency of the system  in a &lt;em&gt;proof of concept&lt;/em&gt; for local government. I don't know how to calculate any of the results that I will likely be asked such as:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What volumetric size is the system likely to occupy?&lt;/li&gt;&#xA;&lt;li&gt;What is the estimated power output? &lt;/li&gt;&#xA;&lt;li&gt;How much is it going to cost to run?&lt;/li&gt;&#xA;&lt;li&gt;How much water will it need to run? &lt;/li&gt;&#xA;&lt;li&gt;What pressure needs to be maintained to keep the system free of condensation before the turbine? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Add any other questions you think you'd ask if you were expected to spend £5 million on building a new powerstation for your city and answer it with the necessary equations to back up a positive attitude towards getting the job done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system works on the principle of negative pressure as opposed to energy intensive positive pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two tanks in a closed system. Tank A is the wet tank and Tank B is the dry tank. A contains water at nominal room temperature but this may vary depending on weather conditions and may decrease to no less than 278K. Tank B temperature is unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Between A &amp;amp; B is a turbine that unrestrained operates at an estimated 99.9% efficiency {take this as fact}. I would like to calculate energy output at 50% efficiency under load producing 1GW minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tank A pressure is 3psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can choose the minimum size tank to maintain that will maintain negative pressure maximum of 10psi for 24 hour period. The maximum size is immaterial as it will tie up with another system for another purpose that is kept in negative pressure for permanent operation at a maximum of 8psi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A proof of concept research system has already been made and works. It uses a Tesla turbine. The Tesla design is being improved but there are patent design modifications I cannot reveal with that aspect. Tank A measures 348K and system depressurised to 25 inches of mercury (ultimately 28.5inches). Tank B is easily 278K, if not lower. The turbine output is estimated at 1kW. It is a closed system. Eventually it will be designed to extract thermal energy from naturally heated water. That water may be up to 380K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to get water to boil at 272K and keep a constant supply available. The incoming water will have a minimum temperature of 278K.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know anything about any equations relating to this question. &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How can I calculate the amount energy extracted with all the above&#xA;information?&lt;/li&gt;&#xA;&lt;li&gt;What equipment would I need to measure energy and power?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;If there is anything else you think should be added to the list, do edit and suggest rather than comment. I would like this question to become a permanent example for reference.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(no, this is not a university exam question.)&#xA;&lt;a href=&quot;https://youtu.be/JvRCB_HhseY&quot; rel=&quot;nofollow noreferrer&quot; title=&quot;Cold Steam Research&quot;&gt;Cold Steam Research&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="16486" LastActivityDate="2020-01-31T22:43:25.863" Title="Energy calculations for steam turbine" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|steam|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33000" PostTypeId="2" ParentId="26773" CreationDate="2020-01-31T12:09:58.037" Score="0" Body="&lt;p&gt;Yes, we are working on creating a new type of power supply using &lt;a href=&quot;https://youtu.be/JvRCB_HhseY&quot; rel=&quot;nofollow noreferrer&quot;&gt;cold steam and negative pressure&lt;/a&gt;. There are two tanks with a Tesla turbine between them. The links between the tanks are evacuated to 25 inches of mercury (0.85 bar) with water in the primary tank that boils rapidly and the steam passes through the turbine to produce mechanical energy such as would be applied to standard steam turbines. The difference is that this system has one moving part and can operate bearingless almost regardless of the turbine mass.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The peak performance turbine output unloaded is estimated at 1kW. So a thousand of these would generate 1 MW but there are plans underway to increase efficiency and tank volume adding disk size to nozzle ratios to discover the best output configuration. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect that a ratchet-type centrifugal clutch flywheel may help with fluctuations in output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have asked some questions regarding &lt;a href=&quot;https://engineering.stackexchange.com/q/32999/16486&quot;&gt;this turbine&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="16486" LastEditorUserId="3505" LastEditDate="2020-01-31T19:28:11.713" LastActivityDate="2020-01-31T19:28:11.713" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33001" PostTypeId="1" CreationDate="2020-01-31T14:14:46.370" Score="0" ViewCount="317" Body="&lt;p&gt;I got a brick-wall house.  The wall is 25cm thick.  I want to add windows.  There are 2 ways of adding a window:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Adding the window on the inside side of the wall, so that if the aluminium frame of the window is 5cm deep, it means that there will be a 20cm of the wall exposed as &quot;outside of house area&quot;.  &lt;/li&gt;&#xA;&lt;li&gt;Another option is to install the window on the outside side, so that the 20cm of the wall will be inside of the house, where I can make better use of it.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I like option (2), but the problem is that I heard rumours that (2) causes water leakage and that I better follow (1).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am really doubtful of this rumour, and I think the real issue might be that the installation is not done properly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Is the rumour correct?  And why?&lt;/li&gt;&#xA;&lt;li&gt;Any special installations procedures for (2) to avoid water leakage?&lt;/li&gt;&#xA;&lt;li&gt;If I adopt (2), should I leave some small wall-lip on the outside?  E.g. instead of making the window flush against the outer wall, I pull it inwards by, say, 2cm, so that the after installing the 5cm-thick aluminum frame, 18cm is inside the house, 2cm outside the house?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="24429" LastActivityDate="2021-06-25T07:03:55.343" Title="How to properly fit almuminum glass window without leakage?" Tags="|aluminum|seals|waterproofing|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33002" PostTypeId="2" ParentId="25399" CreationDate="2020-01-31T16:39:09.317" Score="0" Body="&lt;p&gt;There is a generic trick for converting displacement constraints into forces. This works in a wide range of applications including in this case. Though there are other methods too, and you dont really need this in abacus.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make something else constrain your object then put a simple spring connection between the actual constraint. Make this spring very very stiff. This then acts as if the thing was moved by displacement was a force, now all forces get solved. &lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2020-01-31T16:39:09.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33004" PostTypeId="1" CreationDate="2020-01-31T18:51:24.540" Score="0" ViewCount="26" Body="&lt;p&gt;I have the following coupled equations for the conservation of mass and momentum of a compressible fluid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;\begin{equation}&#xA;\rho_t + (\rho u)_z = 0,&#xA;\end{equation}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;(\rho u)_t + (\rho u^2)_z + \tfrac{1}{k}\rho_z = \left(\frac{dp}{dz}\right)_f + \rho g&#xA;$$&lt;/span&gt;&#xA;over the domain &lt;span class=&quot;math-container&quot;&gt;$[0, 2L]\times[0, \infty)$&lt;/span&gt; for density and velocity &lt;span class=&quot;math-container&quot;&gt;$\rho = \rho(z, t)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$u = u(z, t)$&lt;/span&gt; respectively. Note that &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;g = g(z) = \begin{cases}&#xA;9.81 &amp;amp; \text{if } 0 \le z \le L&#xA;\\&#xA;-9.81 &amp;amp; \text{if } L &amp;lt; z \le 2L&#xA;\end{cases}&#xA;$$&lt;/span&gt;&#xA;and that &lt;span class=&quot;math-container&quot;&gt;$\left(\frac{dp}{dz}\right)_f$&lt;/span&gt; is a frictional pressure term that is non-linear in &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The weak variational problem is: &lt;span class=&quot;math-container&quot;&gt;$\textit{find $u\in V_1, \rho\in V_2$ such that}$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{equation}&#xA;\int_0^{2L} v\rho_t - v_z\rho u dz = 0&#xA;\end{equation}&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\int_0^{2L}q(\rho u)_t - q_z\rho u^2 - \tfrac{1}{k}q_z\rho - q\left(\frac{dp}{dz}\right)_f - q\rho gdz = 0&#xA;$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\textit{for all test functions $v, q$ in spaces $V_1$, $V_2$ respectively.}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am then discretising these equations in time using an implicit backward Euler discretisation for implementation in the finite element solver software Firedrake (this is a standard approach when using Firedrake).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem I'm having is that the Firedrake solver does not converge with the inclusion of &lt;span class=&quot;math-container&quot;&gt;$\left(\frac{dp}{dz}\right)_f$&lt;/span&gt;, and gives the wrong solution when &lt;span class=&quot;math-container&quot;&gt;$\left(\frac{dp}{dz}\right)_f$&lt;/span&gt; is excluded.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in order to narrow down the causes of this divergence, my question is: are there specific spaces for &lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$V_2$&lt;/span&gt; that should be used for this kind of variational problem?&lt;/p&gt;&#xA;" OwnerUserId="20748" LastActivityDate="2020-01-31T18:51:24.540" Title="Defining appropriate test function spaces for the finite element solution of Euler's fluid equations" Tags="|fluid-mechanics|finite-element-method|numerical-methods|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33005" PostTypeId="2" ParentId="32988" CreationDate="2020-01-31T19:16:55.337" Score="1" Body="&lt;p&gt;Visually appealing, natural shaped barriers to decrease noise and light from a heavy use freeway.&lt;/p&gt;&#xA;&lt;p&gt;From: &lt;a href=&quot;https://www.peninsulalink.com.au/Images/ContentEditor/Documents/Freeway%20Design%20&amp;amp;%20Landscaping.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Peninsula Link - Information Sheet&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BE8ht.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BE8ht.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;The landscape design also includes major shaped earth features, sometimes known as &lt;strong&gt;land-forming&lt;/strong&gt;, at the Bungower Road and the Mornington-Tyabb Road interchanges.&lt;/p&gt;&#xA;&lt;p&gt;Along with providing interesting visual features along the roadway these land forms also act as &lt;strong&gt;natural barriers to light and sound&lt;/strong&gt;.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;The view from Mornington-Tyabb access ramp at is more representative of the actual barrier.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bBOz5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bBOz5.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;From &lt;a href=&quot;https://www.peninsulalink.com.au/Images/ContentEditor/Documents/Managing%20Traffic%20Noise.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Penisula Link - Managing road traffic noise&lt;/a&gt;&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Managing traffic noise on the freeway is a &lt;strong&gt;key aspect&lt;/strong&gt; of the Peninsula Link design.&lt;/p&gt;&#xA;&lt;p&gt;Where the acoustic model for Peninsula Link predicts that the noise limits in this policy are likely to be exceeded, noise mitigation measures such as noise walls and earth mounds will be applied to reduce traffic noise.&lt;/p&gt;&#xA;&lt;p&gt;The choice of noise mitigation measure depends on the space available, aesthetics, visual impact and the practicality of building noise walls within the road reserve.&lt;/p&gt;&#xA;&lt;p&gt;The height and location of the noise walls and earth mounds on Peninsula Link have been determined using the acoustic model.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;They had space available, so they used shaped earth mounds to block noise and light from the expressway from the surrounding countryside.  Better than unsightly noise walls.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/maps/@-38.2330515,145.1247161,3a,75y,31.18h,98.54t/data=!3m6!1e1!3m4!1sU5rjlbtNjg5T90E3NlNxPw!2e0!7i13312!8i6656!5m1!1e2&quot; rel=&quot;nofollow noreferrer&quot;&gt;View from Bungower overpass from 2013&lt;/a&gt; showing tent-like structure of earth-works.  They clearly tower over freeway blocking view and sound of freeway from access roads (and surrounding countryside).&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Sq7qf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Sq7qf.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;p&gt;The earth mounds appear to be all the same shape, same angles, same peak to peak distance, parallel to intersecting road, with increasing number of planted trees off of mound peak.  They are seeded with ground cover, no trees or bushes.  Both intersecting roads are overpasses, which would cause amplification of sound as vehicles pass underneath.&lt;/p&gt;&#xA;&lt;p&gt;This is too specific.  If you take what they say as guidance.  They ran acoustic models of noise levels that would be produced and this is the solution to cancel out the prominent frequencies that would be produced by traffic.&lt;/p&gt;&#xA;&lt;p&gt;A noise wall is just a noise wall.  They have used them all over the expressway in residential areas.  Mounds shaped as triangles is unique. They even have an artistic rendition of the Bungower mounds.  This is some highway engineer's pride and joy.  Something they can show their grand-kids.  Their signature.&lt;/p&gt;&#xA;&lt;p&gt;It would be easier to flatten out the land and plant trees.  That would block noise and visual impacts.  This is used of other areas of the road.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2020-02-07T18:08:04.123" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="33007" PostTypeId="1" CreationDate="2020-01-31T20:33:28.967" Score="1" ViewCount="38" Body="&lt;p&gt;There might be a proper term for what I'm looking for, but forgive me, I'm not a professional engineer.  I have a small cylinder with compressed air that connects using push-to-connect's (like &lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B01MQYJ9T6&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;these&lt;/a&gt;).  I am looking for a device that, like a solenoid, takes the pressurized air as an input and then can switch which port the air comes out of.  The difference is that I want to have more than two output ports.  Something in the range of 4-8 would be nice.  I will be controlling this with an Arduino.  Does such a device exist?  If so, what's it called?&lt;/p&gt;&#xA;" OwnerUserId="24431" LastActivityDate="2020-02-01T07:09:12.360" Title="Need to route air, like a solenoid does, but with many output directions" Tags="|pneumatic|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33008" PostTypeId="1" AcceptedAnswerId="33015" CreationDate="2020-01-31T21:16:12.027" Score="0" ViewCount="57" Body="&lt;p&gt;I am an engineering student. I want to get into microscopy research and bio-science. So I want to buy a fair enough compound microscope. I searched for them and there are so many variants from different manufacturers. So, I couldn't select one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My budget is $1000. I know that sounds way too overkill for entry level, but just say that this is total amount I have to spend for a good microscope. I am currently in India and associated price for variants of major manufacturer is not mentioned here so if anyone can mention price(as in India), would be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which manufacturer and which variant should you recommend specifically?&lt;/p&gt;&#xA;" OwnerUserId="17939" LastActivityDate="2020-08-01T00:06:35.593" Title="Compound Microscope recommendation" Tags="|optics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33009" PostTypeId="2" ParentId="32962" CreationDate="2020-01-31T21:40:41.187" Score="1" Body="&lt;p&gt;When I first started in consulting engineering on of my first jobs was inspecting the leveling of condominium ground floor slab on grade floors that has sunk about 300 mm (1') from where they were supposed to be.  It turned out the ground underneath had settled a lot more and there was a huge void  under the slabs in these units.  It was fixed using two products.&lt;/p&gt;&#xA;&lt;p&gt;One they called foamy concrete which today I think I hear it referred to as cellular concrete and the other they called self leveling concrete.  Both were very similar in nature.  All the sand and aggregate that you would normally see in a concrete mix were removed and essentially replaced with gypsum powder.  Incredibly fine particle size compared to sand.  It makes the concrete in both case essentially behave like water.&lt;/p&gt;&#xA;&lt;p&gt;To support the slab, they drilled two holes in the slab in opposite corners.  They shoved the hose nozzle in one and and started to pump foamy concrete into the void until it started to come out the opposite hole.  Let it sit and settle for a bit and repeat the process a couple of more times until it no longer settle back into the hole.  What makes the foamy concrete foamy is they actually blow air into the concrete mixture which causes the concrete liquid mix to turn into foam.  Its relatively weak compared to structural concrete, probably around 1 MPa.  But considering bearing capacities tend to be measured in kPa, it is plenty strong to support a residential slab on grade.&lt;/p&gt;&#xA;&lt;p&gt;In order to level the concrete floor back up to its original position, they used &amp;quot;self leveling&amp;quot; concrete.  This was basically the foamy concrete without the air in it.  The just poured that out of the hose creating a giant puddle on top of the slab.  The concrete really seemed to have the consistency of chocolate milk.  They just kept pouring it until the top of the puddle was where the slab should have been.&lt;/p&gt;&#xA;&lt;p&gt;The gypsum powder is allows the mixture to essentially behave like water, and reduced the cost of just using cement and water.&lt;/p&gt;&#xA;&lt;p&gt;So to sum up, using gypsum can allow you to change the viscosity of the concrete essentially turning it into water. It also allows you to use less cement compared to just mixing cement and water.&lt;/p&gt;&#xA;" OwnerUserId="6993" LastEditorUserId="90" LastEditDate="2021-11-28T05:13:29.650" LastActivityDate="2021-11-28T05:13:29.650" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33011" PostTypeId="1" AcceptedAnswerId="33171" CreationDate="2020-01-31T22:08:41.157" Score="2" ViewCount="274" Body="&lt;p&gt;I'm trying to help a student with a project. The task is to generate CNC G-Code from a Solidworks hole table export (CSV) using VBA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is the export format is going to be a difficult to parse. Samples include the following:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TAG,X LOC,Y LOC,SIZE&#xA;A1,15,50,&amp;lt;MOD-DIAM&amp;gt; 10 &amp;lt;HOLE-DEPTH&amp;gt; 10&#xA;B1,30,80,&amp;lt;MOD-DIAM&amp;gt; 8 THRU ALL&#xA;&#xA;M10 - 6H THRU ALL&#xA;C2,134.50,18,&amp;lt;MOD-DIAM&amp;gt; 8.50 THRU ALL&#xA;&#xA;&amp;lt;HOLE-SPOT&amp;gt;&amp;lt;MOD-DIAM&amp;gt; 23.79 &amp;lt;HOLE-DEPTH&amp;gt; 7.50&#xA;D2,75,24.50,&amp;lt;MOD-DIAM&amp;gt; 13.50 THRU ALL&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Each tool size seems to be assigned a letter and a sequence number. That's fair enough and leads to the hope that we could sort by first letter. The problem is that the metric tap sizes follow the same format. I'm sure there are other problems the more we dig.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are we missing a trick here on the export or has anyone any ideas? I can't find any Solidworks documentation on the matter.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Solution&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Based on Jim S's answer below it became clear that complex holes (tapped and counterbore, for example) are split by CR-CR-LF. This is easy enough to filter out in VBA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jNcT5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jNcT5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. A screen grab of the code as displayed by Notepad++ with View | Show Symbol | Show end of line turned on. These three lines should all be one.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Code snippet 1&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Private Sub CommandButton1_Click()&#xA;    'From http://codevba.com/office/read_text_file_into_string_variable.htm#.XkHkHM77R1I&#xA;    Dim strFilename As String: strFilename = Application.ActiveWorkbook.Path &amp;amp; &quot;\HoleTablecsv&quot;&#xA;    Dim strFileContent As String&#xA;    Dim iFile As Integer: iFile = FreeFile&#xA;    Open strFilename For Input As #iFile&#xA;       strFileContent = Input(LOF(iFile), iFile)&#xA;    Close #iFile&#xA;    a = Replace(strFileContent, vbCr &amp;amp; vbCrLf, &quot; &quot;)&#xA;    Debug.Print a&#xA;End Sub&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2020-02-11T00:05:08.913" LastActivityDate="2020-02-11T00:05:08.913" Title="Solidworks hole table export data format" Tags="|solidworks|cnc|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33012" PostTypeId="2" ParentId="33007" CreationDate="2020-01-31T22:14:19.103" Score="2" Body="&lt;p&gt;I'm sure someone has invented such a thing, but you could do the same with a distribution header/pipe and multiple outlet solenoids.  Your software would determine which solenoid were turned on or off. This would have the exact number of outlet solenoids you desire.   A device to do this as a single entity would likely just be a block with several solenoids, each controlled separately.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-01-31T22:14:19.103" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33014" PostTypeId="2" ParentId="32999" CreationDate="2020-01-31T22:43:25.863" Score="1" Body="&lt;p&gt;The energy that is available to use in any pressure/temperature combination steam is easily determined from a steam table. Steam can be passed through a turbine at theoretically any pressure/temperature as long as the exhaust of the turbine is at a lower pressure than the inlet.  This is the basic steam cycle from Thermo 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The issues will become how you keep the exhaust pressure that low, and also how you get rid of the exhaust steam.  Condensing the steam back to water so it can be pumped either out or back to the source tank will require cooling it, so you will need a cooling medium below the steam temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to your questions, number one would be you telling the local government how it is actually supposed to work.  If there is a patent then the patent describes the process, and you cannot maintain it as a secret.  There is almost no chance of anyone other than a sucker giving you money for a process when you won't tell them how it is supposed to work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to how big, how much energy, etc, that is all based on the output of a kg of water, and then scaled based on how much work your input energy will provide.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Costs will be a function of all the design inputs.  If I understand what you are trying to do, I suggest that getting water to boil at 272K will be several orders of magnitude more difficult than getting it to boil at 348K. Getting water to boil at that low a pressure will require almost a perfect vacuum, which is difficult to achieve at scale.  And then we need an even more perfect vacuum at an even lower temperature to exhaust the steam from the turbine. &lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-01-31T22:43:25.863" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33015" PostTypeId="2" ParentId="33008" CreationDate="2020-02-01T01:56:01.707" Score="0" Body="&lt;p&gt;You are going to need to try these for yourself and your application. For example: If you will be mostly using 50 X to 100 X , you don't need an oil immersion lens.  For inexpensive scopes I liked Unitron . But a quick look on the net shows some things have changed since I used them ; Also, my experience was inverted stage/metallurgy. I recommend a binocular. A good light is essential. Don't get more magnification than you need ; a fuzzy 1000X is not as useful as a sharp 100X . I have no clue about the India market or prices, But Nikon , Canons and Olympus were excellent . Many other excellent names like Zeiss , Reichert and Wilde. A used brand name is better than a new unknown maker , in my opinion . Today I expect a video display is very common ,however I am not familiar with them. You will need to look on the net or in a library and learn about things like &quot;exit pupil&quot;. Good luck.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2020-02-01T15:56:59.507" LastActivityDate="2020-02-01T15:56:59.507" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33016" PostTypeId="2" ParentId="33001" CreationDate="2020-02-01T03:52:04.587" Score="1" Body="&lt;p&gt;If your brick wall is constructed correctly it should have an air/moisture barrier inside the wall. Then you need to attach the flashings of your aluminum window to that barrier which is usually a continuous membrane by window weatherproofing tape available at home improvement centers and install a waterproof sill with proper details on the bottom of the window to shed the rain out. the location of that barrier in the wall determines where you install the window.&#xA;Here is a detail &lt;a href=&quot;https://i.stack.imgur.com/1ITaH.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1ITaH.jpg&quot; alt=&quot;detail&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your brick wall is not up to standards and doesn't have a moisture barrier, you have to cut a rough opening and then cut a groove inside the thickness of the wall 5cm all around the opening and insert the flashing or flange of the aluminum windows into this grove and seal it with caulking all around. Again you need to install galvanized flashing to shed the rain on the outside face of the wall both on top and on the bottom of the window over the stone/ concrete sill. You need to hire a mason with experience in these types of work, or you may end up having a permanent leaking problem. In this case, you can install the window close to the outside face of the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Attention to providing an uninterrupted all-around seal with metal flashing and caulking is imperative.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-02-01T03:52:04.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33017" PostTypeId="1" AcceptedAnswerId="33036" CreationDate="2020-02-01T05:16:49.160" Score="-1" ViewCount="47" Body="&lt;p&gt;The other day, some restauraunts in the union were closed &quot;due to steam outage.&quot; I had seen the venting of course, but had thought little about why the school has a steam system. As best I can find, steam is usually used for heat distribution from a central heat plant; more efficient, etc. But the average yearly low temperature here is 57 degrees. What purpose does the system serve?&lt;/p&gt;&#xA;" OwnerUserId="24435" LastActivityDate="2020-02-02T19:35:52.983" Title="Why does a Gulf Coast campus need an active steam network?" Tags="|civil-engineering|heating-systems|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33018" PostTypeId="1" CreationDate="2020-02-01T05:35:27.937" Score="1" ViewCount="39" Body="&lt;p&gt;I am trying to create a prototype for a glitter shooting mechanism. It is worth mentioning that I have no experience with mechanical engineering or fluid dynamics. I think the below image describes the situation very well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r0szM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r0szM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fan is considerably wider than the diameter of the pipe so I just visually designed a connecting pipe from the fan to the PVC pipe. The goal is to have the fan blow glitter up through the longer section of pipe and out the top. The glitter will initially be stored in the bottom of the bend, not blocking the air passage completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, the issue I see is that a lot of air is coming back out through the inlet shooting almost sideways out once it leaves the fan. I was considering adding a pipe section with the same dimensions as the fan to direct the flow into the inlet and prevent the air from squeezing out through the sides (i.e. forcing it against the air being sucked in). Would this work?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I maximize the ability of the fan to push glitter given access to more PVC pipe and a 3D printer?&lt;/p&gt;&#xA;" OwnerUserId="24436" LastEditorUserId="24436" LastEditDate="2020-02-01T14:28:52.087" LastActivityDate="2020-02-01T14:28:52.087" Title="Design of fan pipe for propelling glitter" Tags="|mechanical-engineering|fluid-mechanics|design|airflow|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33019" PostTypeId="2" ParentId="33007" CreationDate="2020-02-01T07:09:12.360" Score="2" Body="&lt;p&gt;One tube with a long axial slot which tightly fits inside another which has holes spaced such that rotating the internal tube connects to each hole in turn.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One issue is when you need to get to hole 12 from hole 1 all the hole in between may be activated for a short time, cutting the flow may be a solution.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-02-01T07:09:12.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33021" PostTypeId="1" CreationDate="2020-02-01T12:37:05.330" Score="2" ViewCount="129" Body="&lt;p&gt;I am using MSC Nastran for nonlinear static analysis (sol 400). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Step 1 contains a small perturbation load laterally on a thin-walled cylinder. &#xA;Step 2 contains a very large compressive load. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually when I do this, the Load Step increments from 0 to 1 for the first step. Then for the second step, the load increments from 1 to a number under 2 (basically until it stops because the structure fails.) So in this second step, if the indicated load step is 1.3, then the actual compressive load in that load step is 0.3*applied load. (0.3 because take away 1.0 as that is from the previous step.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is now I am in a situation where the load step starts decreasing the moment the second step starts (right after LOAD STEP = 1.0). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the arc-length method, so the load can decrease, but it shouldn't decrease this early. I want at least some compressive load until it starts decreasing. Because my understanding is that the compressive load in the second step is negative, and is therefore actually causing tension instead of compression. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So does anyone know what might be wrong and how to solve this problem?&#xA;I would appreciate any kind of help, thank you!&lt;/p&gt;&#xA;" OwnerUserId="24115" LastActivityDate="2020-02-01T12:37:05.330" Title="In a nonlinear static two-step subcase, why does the load step decrease after the first step?" Tags="|mechanical-engineering|structural-engineering|structural-analysis|aerospace-engineering|nastran|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33023" PostTypeId="1" AcceptedAnswerId="33024" CreationDate="2020-02-01T19:45:07.737" Score="0" ViewCount="61" Body="&lt;p&gt;Good afternoon, I have recently ordered a normally closed relay to fix on my car's daylight-running lights. When the main lights come on the DLR should switch-off. Unfortunately amazon has delivered instead of a normally closed 2 normally open relays (!!?). Is there a clever way to emulate the normally closed relays functionality with two normally open relays?&#xA;Thanks in advance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;regards&#xA;perry&lt;/p&gt;&#xA;" OwnerUserId="24440" LastActivityDate="2020-02-01T21:55:39.803" Title="Emulate normally closed relay with 2 normally open relays" Tags="|electrical-engineering|power-electronics|car|electronics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33024" PostTypeId="2" ParentId="33023" CreationDate="2020-02-01T21:07:13.503" Score="1" Body="&lt;p&gt;You have a chance. This might work as a temporary solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A1oUW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A1oUW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Switching DLRs off with an NO when headlights are on.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How it works:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;When the ignition is on and the headlights are off the headlamps will provide a low enough resistance to allow the RLY to pick up. The normally open (NC) contact will close and the DLRs will turn on.&lt;/li&gt;&#xA;&lt;li&gt;When the headlamps are on there will be 12 V on both sides of the relay so no current will flow and the relay will drop out switching off the DLRs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uDuC9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uDuC9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*Figure 2. Modification to detect both dip and high beam status.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The DLRs will turn on only if both DIP and HIGH are off.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Note that in both circuits I've powered the relay from a line switched by the ignition. This will avoid draining the battery when the ignition is turned off.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2020-02-01T21:55:39.803" LastActivityDate="2020-02-01T21:55:39.803" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33026" PostTypeId="1" AcceptedAnswerId="33035" CreationDate="2020-02-02T05:37:40.827" Score="0" ViewCount="173" Body="&lt;p&gt;Could a Stirling Engine replace the cooling system of an engine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Instead of having a clutch and fan attached to the main drive shaft to cool an engine, could a Stirling Engine and clutch share that drive shaft to cool and add torque? What problems may arise?&lt;/p&gt;&#xA;" OwnerUserId="20183" LastEditorUserId="20183" LastEditDate="2020-02-02T05:44:21.407" LastActivityDate="2023-11-07T07:20:02.387" Title="Stirling Engine Cooled Engine" Tags="|mechanical-engineering|fluid-mechanics|thermodynamics|design|heat-transfer|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33028" PostTypeId="1" CreationDate="2020-02-02T12:57:30.190" Score="1" ViewCount="25" Body="&lt;p&gt;One of the advantages of fluid bed drying especially in the pharma industry is the “Turbulence in the bed causes mild attrition to particles thus producing more spherical free flowing product.” &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is from the textbook and I don’t understand the meaning behind it. Can anyone explain?&lt;/p&gt;&#xA;" OwnerUserId="11561" LastActivityDate="2020-02-02T15:33:31.337" Title="Advantages of Fluid Bed Drying" Tags="|drying|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33029" PostTypeId="2" ParentId="29967" CreationDate="2020-02-02T14:23:48.213" Score="2" Body="&lt;p&gt;This a spring loaded dampening system that one can describe in a form of &lt;span class=&quot;math-container&quot;&gt;$\ddot{x} + 2 D \omega_{0} \dot{x} + \omega_{0}^2= 0$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's simply a piston and spring with a mechanical leverage system that pulls the door closed, since it will try to return to it's most stable position, This is stable open loop system with poles certainly on the left half plane. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is an example of a norton 1600, but a google image search will turn up all sorts of different implementations&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y2V3j.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y2V3j.jpg&quot; alt=&quot;norton 1600 door closer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no external force beyond one opening the door, no electronics, sensors or electricity and no controlforce. &lt;/p&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2020-02-02T14:23:48.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33030" PostTypeId="2" ParentId="32993" CreationDate="2020-02-02T14:26:36.920" Score="1" Body="&lt;p&gt;The mentioned P100 particulate filter approximately corresponds to a P3 filter according to European standards.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A half mask with a P3 filter has a nominal protection factor of 48.&lt;/li&gt;&#xA;&lt;li&gt;A full face mask with a P3 filter has a nominal protection factor of 1000. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This difference already illustrates that the protection is not only determined by the quality of the filter but also by the face seal leakage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that these nominal protection factors are tested in a laboratory. The values should not be used for the assessment of a real workplace.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;There are many reasons for not using nominal protection factors for assessing the likely protection at the workplace. The laboratory tests do not represent the activities covered in workplace situations; the laboratory tests only involve a small number of individuals, who cannot represent a significant portion of the population of respiratory protective device wearers in the workplace; the individuals selected for the test panel are likely to be well-trained and are familiar with the test procedures; the standards allow for disallowing people who do not pass the initial screening test needed before the total inward leakage test is carried out.&lt;br&gt;&#xA;  – European Standard EN 529 – &lt;em&gt;Respiratory protective devices – Recommendations for selection, use, care and maintenance – Guidance document&lt;/em&gt; (2005)&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;For a realistic assessment, it is better to use so-called assigned protection factors, which depend on applicable national regulatory requirements. Typically, these values correspond to the level of respiratory protection that can realistically be expected to be achieved in the workplace by 95 % of adequately trained and supervised wearers using a properly functioning and correctly fitted respiratory protective device.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A half mask with a P3 filter has an assigned protection factor of 30 (Germany) or 20 (UK).&lt;/li&gt;&#xA;&lt;li&gt;A full face mask with a P3 filter has an assigned protection factor of 400 (Germany) or 40 (UK).&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If better protection factors are needed, you should use a power assisted filtering devices, which has the advantage that the pressure inside the facepiece may remain above the ambient air pressure.&lt;/p&gt;&#xA;" OwnerDisplayName="user24447" LastActivityDate="2020-02-02T14:26:36.920" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33031" PostTypeId="1" CreationDate="2020-02-02T15:18:35.370" Score="2" ViewCount="182" Body="&lt;p&gt;I created fillets/blending for a cube corner in 3 cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the first case, the 3 radius are equal. The blending surface is a sphere patch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the second case, two radius are smaller than the third. The blending surface is a torus patch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for the third case, one radius is smaller than the other two. I cannot figure out the algorithm generating the blending patch. I did a lot of search for papers on surface/edge blending, but didn't find the algorithm for this case. Are there any document or paper on the algorithm for this case?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could someone give me some hint?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The STEP and picture files for the three cases are attached.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/p0g2R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/p0g2R.png&quot; alt=&quot;fillet case 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/85hTX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/85hTX.png&quot; alt=&quot;fillet case 2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xjvZk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xjvZk.png&quot; alt=&quot;fillet case 3&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24448" LastActivityDate="2020-02-02T15:18:35.370" Title="Cube vertex/corner blending/fillet algorithm" Tags="|cad|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33032" PostTypeId="2" ParentId="33028" CreationDate="2020-02-02T15:33:31.337" Score="2" Body="&lt;p&gt;It means that the particles rub and bounce against each other, which removes any “odd” bits that may otherwise cling to the proper surface of the particle.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-02-02T15:33:31.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33033" PostTypeId="2" ParentId="29109" CreationDate="2020-02-02T18:09:16.320" Score="0" Body="&lt;p&gt;I got another similar shirt, 100% polyester, to see if it could work. I found the maximum temperature for polyester at &lt;a href=&quot;https://www.thespruce.com/select-correct-ironing-temperature-for-fabrics-2146186&quot; rel=&quot;nofollow noreferrer&quot;&gt;this page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;Iron Setting Temperatures in Celsius and Fahrenheit&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;If your iron uses a different scale or you want to know more exacting temperatures for ironing different types of fabric, follow these guidelines:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;...&lt;/p&gt;&#xA;  &#xA;  &lt;ul&gt;&#xA;  &lt;li&gt;Polyester: 148 C/300 F&lt;/li&gt;&#xA;  &lt;/ul&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I found duration for heat pressing &lt;a href=&quot;http://wiki.makespace.org/Equipment/TShirtPress&quot; rel=&quot;nofollow noreferrer&quot;&gt;at this page&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Suggested settings: temp=160 time=16 (seconds). Varies with different vinyls.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So I used 20 seconds at 148ºC. I cut the design by hand or on a Silver Bullet Vinyl Cutter. The result works very well:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eJThN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eJThN.jpg&quot; alt=&quot;First shirt with hand-cut design&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/10jRc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/10jRc.jpg&quot; alt=&quot;Second shirt with computer-cut design&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19384" LastActivityDate="2020-02-02T18:09:16.320" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33035" PostTypeId="2" ParentId="33026" CreationDate="2020-02-02T19:24:08.400" Score="2" Body="&lt;p&gt;A Stirling engine is a way to remove heat from a heat source.  So  theoretically you could attach a Stirling engine to a radiator-like device and extract the heat using the engine.  We could then use the output of the engine to do work.  The most likely output would be an electric generator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trying to use the Stirling engine to add power directly to the crankshaft would likely be futile.  The power differentials would be large, and Stirling Engines don't have a good way to change speeds based on user (the driver) input.  The most likely result would be a gas engine spinning a Stirling engine and messing up the Stirling cycle completely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;While theoretically this could be done, the reality is that Stirling engines aren't very good at getting rid of heat, and are large clunky devices for what they do.  There is a reason they are not in widespread use.  If you had a lot of space and money to build Stirling engines around a stationary regular engine then your idea could work.  I fear, however, that this will never be a financially viable solution.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-02-02T19:24:08.400" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33036" PostTypeId="2" ParentId="33017" CreationDate="2020-02-02T19:35:52.983" Score="0" Body="&lt;p&gt;Steam networks are used for heating (there are still heaters in use on the gulf coast), creating hot water, and any other energy loads a university might need.  I suspect that the restaurants in question didn't have hot water.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-02-02T19:35:52.983" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33037" PostTypeId="2" ParentId="26191" CreationDate="2020-02-02T21:38:52.827" Score="0" Body="&lt;p&gt;I am working on this same exercise, and I don't have the answer yet, but I did find (on p28 of &lt;a href=&quot;http://ecee.colorado.edu/~ecen5616/WebMaterial/06%20Instruments%20and%20Optical%20Invarient.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this pdf&lt;/a&gt;) this useful illustration showing how the marginal and chief ray can be defined for an object at infinity. (Information which really should be included in the course!)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sgsDI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sgsDI.jpg&quot; alt=&quot;marginal and chief ray for object at infinity&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/KKsAi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KKsAi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24453" LastEditorUserId="-1" LastEditDate="2020-02-04T00:30:13.383" LastActivityDate="2020-02-04T00:30:13.383" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33038" PostTypeId="2" ParentId="24623" CreationDate="2020-02-02T22:02:56.687" Score="2" Body="&lt;p&gt;Nature will always try to minimize the number of dislocations and grain-boundaries, whereas in the case of vacancies, there is an optimum number that will lead to the lowest Gibbs energy.&lt;/p&gt;&#xA;" OwnerUserId="24454" LastActivityDate="2020-02-02T22:02:56.687" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33039" PostTypeId="1" AcceptedAnswerId="33046" CreationDate="2020-02-02T23:17:47.333" Score="3" ViewCount="1294" Body="&lt;p&gt;I am building a desoldering gun. While the pump I am using provides decent suction, I would like a quick powerful vacuum burst to prevent the solder from solidifying in the nozzle rather than in the chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I figure there must be the equivalent of a capacitor where a vacuum is held in reserve and released when needed. Does such a device exist? If so what is it called. Finally, are they available in small sizes, say the size of a tennis ball?&lt;/p&gt;&#xA;" OwnerUserId="2248" LastActivityDate="2020-02-04T15:43:22.430" Title="What is the equivalent of a capacitor in pneumatics?" Tags="|airflow|vacuum|pneumatic|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33040" PostTypeId="2" ParentId="33039" CreationDate="2020-02-02T23:52:11.803" Score="4" Body="&lt;p&gt;The equivalent of a capacitor in terms of pneumatics would be a chamber such as a metal sphere the size of a tennis ball. It would be necessary to have an external vacuum pump to evacuate the air prior to the quick release.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Auto tire houses use the reverse concept to seat a new tire on the rim. A cylindrical tank is loaded to substantial pressure by a compressor. The nozzle of the tank is shaped to fit the edge of the wheel rim and actuated by a ball valve, releasing a huge volume of air into the tire, assisting the seating process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your concept, you'd want to have a similar ability of opening the vacuum chamber, which may be more difficult than creating or finding a suitable chamber. For the application you describe, you can get away with a cylinder of some wall thickness but some experimentation may be required to determine the overall volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd expect that you'd have a permanently (in-use) attached vacuum pump at one end, the nozzle to apply to the solder to be removed at the other end and the release mechanism in the middle. After each use, the pump would have to remove the air, creating a bit of delay.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may discover that a similar product using a vacuum pump exists:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.tequipment.net/Hakko/FR-301/Desoldering-Equipment/Rework-Stations/&quot; rel=&quot;nofollow noreferrer&quot;&gt;desoldering tool&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6GKtx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6GKtx.png&quot; alt=&quot;desoldering tool&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may have a chamber, or a pump with sufficient volume to accomplish the task.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-02T23:52:11.803" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33044" PostTypeId="1" AcceptedAnswerId="33055" CreationDate="2020-02-03T02:12:02.213" Score="-1" ViewCount="44" Body="&lt;p&gt;an &lt;a href=&quot;https://engineering.stackexchange.com/q/33041/24455&quot;&gt;acoustic anechoic chamber&lt;/a&gt; has mainly 3 features&lt;/p&gt;&#xA;&#xA;&lt;p&gt;stopping sound getting in; &#xA;stopping sound getting out;&#xA;reduce echoey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a term to describe a room only used to stop sound getting in?&lt;/p&gt;&#xA;" OwnerUserId="24455" LastActivityDate="2020-02-03T21:52:56.087" Title="Is there a term to describe a room only used to stop sound getting in?" Tags="|terminology|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33046" PostTypeId="2" ParentId="33039" CreationDate="2020-02-03T08:50:03.070" Score="3" Body="&lt;p&gt;The device is called a &quot;receiver&quot; in pneumatics - this is basically the tank you see on any garage compressor, or much larger and more elaborate setups in industrial settings.  Its function can be modelled almost exactly like a capacitor.  It stores energy by accumulating air inside of it (the integral of the incoming flow, just as charge is the integral of current), and will maintain a pressure (analogous to voltage) in proportion to its volumetric capacity after the external pressure source is removed.  It even acts as a low-pass filter, essentially smoothing out the inconsistent input from the compressor, in fact this is the reason why garage compressors almost always have a tank attached.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a lot of cases in the industrial world, receivers are seen at the point of usage as well, for example a valve that must close quickly, a pneumatic conveying line, or an air cannon.  In those cases it's analogous to a capacitor used to store energy for a very high current but short discharge, such as a camera flash.&lt;/p&gt;&#xA;" OwnerUserId="24459" LastEditorUserId="24459" LastEditDate="2020-02-04T15:43:22.430" LastActivityDate="2020-02-04T15:43:22.430" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33047" PostTypeId="1" CreationDate="2020-02-03T09:02:13.990" Score="-1" ViewCount="667" Body="&lt;p&gt;Trying to come up with a MATLAB code for calculation of heat transfer due to incident radiation on a building's walls, would the solution involve using the conduction equation only? What would be the boundary conditions used here?&lt;/p&gt;&#xA;" OwnerUserId="24460" LastActivityDate="2020-02-03T18:55:07.520" Title="How do you calculate heat transfer due to solar radiation for a multi-layer wall given the incident solar radiation and wall specs?" Tags="|mechanical-engineering|civil-engineering|thermodynamics|heat-transfer|solar-energy|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33049" PostTypeId="1" CreationDate="2020-02-03T12:29:44.467" Score="1" ViewCount="80" Body="&lt;p&gt;According to basic electrical machinery theory, all transformers experience power losses due to winding resistances, eddy currents, leakage flux and hysteresis. So when a load is connected to, say, a transformer with a 13800/240 V rating, the output voltage of  a transformer is slightly less than 240 V, for example 225 V. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, when a real transformer has a rating of, for example, 13800/240 V etched on its plate, is this rating just the ideal rating based on the turns ratio N1/N2, or is the transformer built with a slightly different turns ratio to compensate for losses, so that at full load it ACTUALLY delivers 240V?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the same thing for the power rating etched on the plate. Is it the ideal rating or does it take into account losses at full load?&lt;/p&gt;&#xA;" OwnerUserId="18155" LastActivityDate="2020-02-03T17:07:16.640" Title="Does the voltage rating on transformer plates account for losses?" Tags="|electrical-engineering|power-transmission|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33050" PostTypeId="2" ParentId="33047" CreationDate="2020-02-03T13:13:50.683" Score="3" Body="&lt;p&gt;How complex this is depends on your needs. Once you know the solar flux, you need to calculate how much of that energy is absorbed by the wall (a function of the surface - how much is reflected? This itself may be a function of the angle of incidence, which will vary through the day...). This will heat up the outside of the wall, and it will start to conduct through to the building... But... There will be heat lost from the external surface of the wall, which reduces the total energy that makes it to the inside of the building. This will be lost as Convection and Radiation. Convection depends on the surface and the wind, Radiation depends on the temperature difference between the wall and the black-body temperature of the sky. Both of these values will vary on a minute-to-minute basis throughout the day... And all of this is before you even think about the variable convection on the inside of the building, and how the temperature difference with the air there will also vary etc...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, if you want to calculate a single point in time, then sure - it's not so bad. Just do an energy balance. But if you want to model how the building behaves over the course of a day? things get a bit more tricky.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2020-02-03T18:55:07.520" LastActivityDate="2020-02-03T18:55:07.520" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33051" PostTypeId="1" AcceptedAnswerId="33089" CreationDate="2020-02-03T13:54:23.843" Score="1" ViewCount="121" Body="&lt;p&gt;I need a source of air pressure for an air bearing (3D printed cylinder that expands air upwards like an air hockey table). I would like to be able to control the flow of air. I don't have a lab air supply or advanced equipment, this will basically be a homemade build.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An off the shelf electric pump might be a good supply source. What do I need to regulate the flow?&lt;/p&gt;&#xA;" OwnerUserId="24463" LastActivityDate="2020-02-05T15:58:54.067" Title="How can I build a regulated pressurised air supply?" Tags="|airflow|pumps|compressed-air|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33052" PostTypeId="2" ParentId="33051" CreationDate="2020-02-03T14:51:28.763" Score="1" Body="&lt;p&gt;A quick search for &quot;how does a gas regulator work&quot; resulted in a clear explanation from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Pressure_regulator&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki world&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pfqet.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pfqet.png&quot; alt=&quot;single stage regulator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Verbatim explanation:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Single stage regulator&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;High pressure gas from the supply enters into the regulator through&#xA;  the inlet valve. The gas then enters the body of the regulator, which&#xA;  is controlled by the needle valve. The pressure rises, which pushes&#xA;  the diaphragm, closing the inlet valve to which it is attached, and&#xA;  preventing any more gas from entering the regulator.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The outlet pressure and the inlet pressure hold the diaphragm/poppet&#xA;  assembly in the closed position against the force of the large spring.&#xA;  If the supply pressure falls, it is as if the large spring compression&#xA;  is increased allowing more gas and higher pressure to build in the&#xA;  outlet chamber until an equilibrium pressure is reached. Thus, if the&#xA;  supply pressure falls, the outlet pressure will increase, provided the&#xA;  outlet pressure remains below the falling supply pressure. This is the&#xA;  cause of end-of-tank dump where the supply is provided by a&#xA;  pressurized gas tank.[citation needed] With a single stage regulator,&#xA;  when the supply tank gets low, the lower inlet pressure causes the&#xA;  outlet pressure to climb. If the spring compression is not adjusted to&#xA;  compensate, the poppet can remain open and allow the tank to rapidly&#xA;  dump its remaining contents. In other words, the lower the supply&#xA;  pressure, the lower the pressure differential the regulator can&#xA;  achieve for a given spring setting.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The outlet side is fitted with a pressure gauge. As gas is drawn from&#xA;  the outlet side, the pressure inside the regulator body falls. The&#xA;  diaphragm is pushed back by the spring and the valve opens, letting&#xA;  more gas in from the supply until equilibrium is reached between the&#xA;  outlet pressure and the spring. The outlet pressure therefore depends&#xA;  on the spring force, which can be adjusted by means of an adjustment&#xA;  handle or screw.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wpHTl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wpHTl.png&quot; alt=&quot;two stage regulator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Double stage regulator&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Two stage regulators are two single stage regulators in one that&#xA;  operate to reduce the pressure progressively in two stages instead of&#xA;  one. The first stage, which is preset, reduces the pressure of the&#xA;  supply gas to an intermediate stage; gas at that pressure passes into&#xA;  the second stage. The gas now emerges at a pressure (working pressure)&#xA;  set by the pressure adjusting control knob attached to the diaphragm.&#xA;  Two stage regulators have two safety valves, so that if there is any&#xA;  excess pressure there will be no explosion. A major objection to the&#xA;  single stage regulator is the need for frequent adjustment. If the&#xA;  supply pressure falls, the outlet pressure may change, necessitating&#xA;  adjustment. In the two stage regulator, there is improved compensation&#xA;  for any drop in the supply pressure.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've disassembled the regulator for a helium balloon inflator tank and found it to be virtually identical to the image. For balloon inflation, precise regulation is not important and the regulator was single stage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was comprised of a number of o-rings, seating the metal diaphragm to the machined parts and of course, the needle valve adjustment mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The failure of the regulator requiring repair was erosion of the o-rings and easily repaired. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could construct a DIY version using a thin sheet of spring steel, but the critical aspect of the build would be to ensure the enclosure will hold the expected pressure. One would not expect a 3D printed pressure chamber to manage any appreciable pressure, although an explosion of PLA or ABS would be less severe than that of steel or other metals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Off the shelf items for the needle valve assembly might be a carburetor kit for a mower engine or similar powerplant. Experimentation is warranted for the DIY side of things.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-03T14:51:28.763" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33053" PostTypeId="2" ParentId="33049" CreationDate="2020-02-03T16:51:02.080" Score="1" Body="&lt;p&gt;Ideal transformer - transformation ratio&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\alpha = \frac {V_1} {V_2} = \frac {N_1} {N_2} = \frac {I_2} {I_1}$$&lt;/span&gt;&#xA;Apparent Power:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$S_1 = S_2$$&lt;/span&gt;&#xA;Real transformer:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\alpha = \frac {V_1} {V_2} = \frac {N_1} {N_2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$S_1 &amp;gt; S_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The kVA rating on transformers deals with the output of the secondary side or &lt;span class=&quot;math-container&quot;&gt;$S_2$&lt;/span&gt;.  There has to be a greater input power (and current) on the primary side (&lt;span class=&quot;math-container&quot;&gt;$S_1$&lt;/span&gt;) to deal with losses.  To get full-load 100kVA at 95% efficiency means 105.26kVA must be put in.  The nameplate would show 100kVA.  This is the full-load output of the transformer.  Actual kVA output will depend upon load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$S_1$&lt;/span&gt; must be greater than &lt;span class=&quot;math-container&quot;&gt;$S_2$&lt;/span&gt; (100kVA) to deal with losses inside the transformer.  Since &lt;span class=&quot;math-container&quot;&gt;$V_1$&lt;/span&gt; cannot increase, greater &lt;span class=&quot;math-container&quot;&gt;$S_1$&lt;/span&gt; means &lt;span class=&quot;math-container&quot;&gt;$I_1$&lt;/span&gt; must increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;13800/240 V means that when you apply 13800V at 60Hz to the primary, you get 240V at the secondary.  You would measure 240V at the terminals of the secondary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The secondary feeds a load.  Assume voltage at load is 225V.  This is external to the losses of the transformer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;15V was lost to the wires.  This voltage loss depends upon current (load) and wire size (cross-sectional area) and length of feeder.  Larger current, smaller wire and longer feeder means greater voltage loss.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2020-02-03T17:07:16.640" LastActivityDate="2020-02-03T17:07:16.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33054" PostTypeId="1" CreationDate="2020-02-03T20:10:02.343" Score="1" ViewCount="22" Body="&lt;p&gt;I have a piece of company developed equipment that uses a pot mounted on a rotating base plate with a 0.15m torque arm and load cell, a mixing paddle is used to stir the mixture while the controller calculates torque from the motors rpm and force applied to the load cell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is considering a servo viscometer and many other digital viscometers calculate viscocity based upon torque can this be achieved using the setup with what principles would you require?&lt;/p&gt;&#xA;" OwnerUserId="16818" LastActivityDate="2020-02-03T20:10:02.343" Title="how do you calculate viscocity from torque arm load cell measurements?" Tags="|motors|measurements|torque|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33055" PostTypeId="2" ParentId="33044" CreationDate="2020-02-03T21:52:56.087" Score="2" Body="&lt;p&gt;Acoustic test booth &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, acoustic tests require special spaces. They should be conducted in a place where the sounds can be isolated and any unwanted noises are kept outside. This is where acoustic test booths come in.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.enoisecontrol.com/acoustic-test-booth-chamber/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.enoisecontrol.com/acoustic-test-booth-chamber/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Specifically, a hemi-anechoic chamber has some sound-reflective inside surfaces, typically the floor, in order to simulate operating conditions.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2020-02-03T21:52:56.087" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33057" PostTypeId="2" ParentId="22909" CreationDate="2020-02-04T05:15:04.230" Score="1" Body="&lt;p&gt;This post is a little dated but thought someone might benefit from another angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;@Solar Mike is correct.  The reason you overflow frequently is because your exit valve is undersized for in inlet flows you're specifying.  The only remedy is to limit your inlet flows to the range that the exit valve can handle, OR, increase your valve size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You don't specify if you're simulating a real exit valve or a virtual one. If real, then there should be a specification sheet where you can calculate the maximum flow, at 4m differential press and 100%Cv (maximum open valve position). Then limit your inlet flows to this value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the valve is virtual, and you're only approximating a plant scenario, then size your valve using a simplified valve equation,   &lt;span class=&quot;math-container&quot;&gt;$F_{Valve}= k  \cdot C _{V}  \cdot \sqrt {h}$&lt;/span&gt;. For example, setting the max flow to 100% at 100%Cv, and at 4meters yields    &lt;span class=&quot;math-container&quot;&gt;$k=\frac {F_{Valve}}{C _{V}  \cdot \sqrt {h}}= \frac {100}{100 \cdot \sqrt {4}}= \frac {1}{2}=0.5$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will insure against overflowing the tank. The following table shows how valve flow varies with both head and valve opening.  It also shows that @kamran is correct in that flow is a function of 2 variables, so you can't really calculate the value directly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using water, a mass balance and volume balance are equivalent (in this context) due to incompressibility.  And specifying steady-state mandates FlowIn = FlowOut = ValveFlow.  You end up with FlowIn = ValveFlow.  The mass balance equation becomes  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$FlowIn = \frac {1}{2} \cdot C_v \cdot \sqrt {h}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem here is that there are 3 variables and only 1 equation.  Thus, to specify the system you have to fix 2 values.  Since you want to calculate the Cv, then you must fix FlowIn and TankLevel, h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final operational equation @Abrar is looking for is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$C_{v}=\frac {F_{In}}{k  \cdot \sqrt {h}}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously, with the inlet flow seemingly fixed, you can also fix the valve opening, then calculate the tank height.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YWQBm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YWQBm.jpg&quot; alt=&quot;Valve Flow = Function of h, Cv&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24434" LastActivityDate="2020-02-04T05:15:04.230" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33058" PostTypeId="1" CreationDate="2020-02-04T06:20:21.837" Score="1" ViewCount="166" Body="&lt;p&gt;Ok, so I am pretty confused at the moment with a question that I have been given to work out by my lecturer and was wondering if I could get some guidance with it. I will attach images of the question and diagram of the circuit while trying to say how I believe it should be worked out. I have not yet had the classes for this question yet but wanted to get a better understanding of it before I go into the class. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I guess my main question would be this, is there a calculation that I could use to get a measurement of &lt;span class=&quot;math-container&quot;&gt;$I_b$&lt;/span&gt; reliably without having to randomly guess numbers to get to 10 micro Amps and with regards to the results I have currently would you say I am on the right track? As I feel like I am complicating the question more than I should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question says &quot;Using Figure 8 adjust VBB so that the base current is set at 10 micro Amps, and complete table 4.&quot; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Straight away I see that &lt;span class=&quot;math-container&quot;&gt;$I_b$&lt;/span&gt; going to the base of the emitter has to be reading 10 micro Amps, I have a multi meter in place to make sure it is the reading I am getting but to do this I have to change the VBB voltage to make the multi meter read the right amount of Amps. So for me I have literally been spending 5 minutes trying to find out what 10 micro Amps is every time and id assume there is a much easier way to find this than spending 5 minutes doing 0.556677, 0.556679, 0.556683 etc until I find the right value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I had also took note that I am given VCE which tells me if I change VCC to say 0.6V I will get the same reading on the VCE meter of 600mV and I have to work out &lt;span class=&quot;math-container&quot;&gt;$I_c$&lt;/span&gt; which is the measurement on the multi meter from the transistor collector to VCC.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For some reason the attaching a imgur link wont work through the link section so I will leave the link below for the gallery of the 4 images.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 1&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SzgJz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SzgJz.jpg&quot; alt=&quot;Image 1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Image 2&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/A3Wf6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A3Wf6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/o39yI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://i.stack.imgur.com/o39yI.jpg&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24468" LastEditorUserId="8055" LastEditDate="2020-02-05T13:02:59.420" LastActivityDate="2024-03-28T17:07:20.943" Title="BJT Output Characteristics Question" Tags="|electrical-engineering|measurements|circuits|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33059" PostTypeId="1" CreationDate="2020-02-04T09:50:15.100" Score="4" ViewCount="236" Body="&lt;p&gt;I am curious as to how different is precast concrete design than conventional cast-in situ reinforced concrete design as far as structural engineering is concerned. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To my mind, they should be the same, because precast is essentially using the same materials as cast-in situ, with the same material characteristics. The only difference between them is just one is casted in batch in a factory, and another is casted on site. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Am I right? What does Euro Code or ACI say about it?&lt;/p&gt;&#xA;" OwnerUserId="3353" LastActivityDate="2020-03-13T19:08:47.527" Title="The design of precast concrete-- are they the same with cast-in situ design?" Tags="|structural-engineering|concrete|reinforced-concrete|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33060" PostTypeId="1" CreationDate="2020-02-04T11:14:34.957" Score="1" ViewCount="98" Body="&lt;p&gt;At the moment the whole analysis stops when 1 subcase fails to converge. How can I keep the analysis going such that all subcases are are run, even if they all might produce an error?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm pretty sure there is a command for this to write in the case control section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All help would be appreciated, thanks!&lt;/p&gt;&#xA;" OwnerUserId="24115" LastActivityDate="2020-02-04T11:14:34.957" Title="In Nastran, how to ignore fatal error due to failed convergence and move on to next subcase?" Tags="|structural-analysis|finite-element-method|aerospace-engineering|nastran|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33061" PostTypeId="1" AcceptedAnswerId="33093" CreationDate="2020-02-04T12:17:59.833" Score="2" ViewCount="322" Body="&lt;p&gt;I have a very long beam which is subjected to a torque (I have the moments and shear forces with respect to X along the beam). I can calculate the critical load when the beam would bend if it was loaded axially, however I don't know how to calculate the critical torque on the beam before it &quot;buckles&quot; by twisting. How would I go about calculating this critical load/torque?&lt;/p&gt;&#xA;" OwnerUserId="24305" LastActivityDate="2021-03-02T03:39:47.293" Title="Twist &quot;buckling&quot; critical load" Tags="|stresses|beam|buckling|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33062" PostTypeId="2" ParentId="33058" CreationDate="2020-02-04T12:26:45.253" Score="0" Body="&lt;p&gt;Below is the equation for &lt;span class=&quot;math-container&quot;&gt;$I_c$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$I_c=\beta I_b$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; is unique for the transistor. The value can be found in the transistor datasheet &lt;a href=&quot;https://www.st.com/resource/en/datasheet/cd00003228.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;BC107B&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In datasheet the parameter &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; is usually labeled as &lt;span class=&quot;math-container&quot;&gt;$h_{FE}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;span class=&quot;math-container&quot;&gt;$\beta$&lt;/span&gt; value for BC107B from the datasheet is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KGDAo.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KGDAo.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In short you are creating a graph similar to the image below for the BC107B transistor&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fsdS1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fsdS1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://electronics.stackexchange.com/questions/270393/industry-values-for-beta-%CE%B2-in-bjt-current-amplification&quot;&gt;Industry Values for Beta (β) in BJT current amplification&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/29427/why-does-collector-current-increase-with-base-current-in-bjt&quot;&gt;Why does collector current increase with base current in BJT?&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2020-02-04T23:28:19.217" LastActivityDate="2020-02-04T23:28:19.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33064" PostTypeId="1" AcceptedAnswerId="33066" CreationDate="2020-02-04T16:13:24.507" Score="1" ViewCount="195" Body="&lt;p&gt;&lt;a href=&quot;https://royalsocietypublishing.org/doi/10.1098/rspa.2013.0689&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://royalsocietypublishing.org/doi/10.1098/rspa.2013.0689&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A chain of some sort goes upwards, turns in a half circle (according to the drawn diagram), and goes back down.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/aSDuW.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aSDuW.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Focusing on the half circle, the equation given by the paper to account for its centripetal force is&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sdtLr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sdtLr.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;In our initial analysis, we assume the curved region at the apex of the fountain is small enough, and has tight enough curvature, that the centripetal acceleration is much larger than the gravitational acceleration (v2/r≫g). The centripetal acceleration is then provided by the tension in the chain. If the (local) radius of curvature is r and the tension in the curved region is Tc&quot; Also the lambda is mass per unit length&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is Tc divided by r in the left half of the equation? If it was force per unit length then I would expect Tc to be divided by (pi)r instead of just r.&lt;/p&gt;&#xA;" OwnerUserId="24473" LastActivityDate="2020-02-04T16:53:25.637" Title="calculation of tension in curved chain for chain fountain (Mould effect)" Tags="|mechanical-engineering|applied-mechanics|mechanical|pulleys|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33065" PostTypeId="1" CreationDate="2020-02-04T16:22:58.160" Score="1" ViewCount="91" Body="&lt;p&gt;I have a question which seems like an easy one, but I couldn't find anything on Internet to be 100% sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The free space path loss equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$FSPL=20\text{log}(d)+20\text{log}(f)+20\text{log}(\frac{4\pi}{c})$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if the receiving antenna is at the same height as the transmitting antenna, &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is just the horizontal distance between the two. But if we now move the receiving antenna up and down (without increasing &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;), the straight line between the two antennas is this:&lt;a href=&quot;https://i.stack.imgur.com/mb1S9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mb1S9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the path loss equation becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$FSPL=20\text{log}\bigg(\sqrt{(h_{T_x}-h_{T_r})^2+d^2}\bigg)+20\text{log}&#xA;(f)+20\text{log}(\frac{4\pi}{c})$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is this correct? Maths seems to agree, but I don't know if I am not missing somthing obvious here. Thanks.&lt;/p&gt;&#xA;" OwnerUserId="17980" LastActivityDate="2020-02-04T20:52:05.490" Title="How does free space path loss change with receiver's antenna's height" Tags="|power-transmission|transmission|signal|radio|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33066" PostTypeId="2" ParentId="33064" CreationDate="2020-02-04T16:53:25.637" Score="2" Body="&lt;p&gt;Consider a small element of chain with length &lt;span class=&quot;math-container&quot;&gt;$dl$&lt;/span&gt; at the top of the arc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Its mass is &lt;span class=&quot;math-container&quot;&gt;$\lambda\,dl$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is curved through an angle &lt;span class=&quot;math-container&quot;&gt;$dl/2r$&lt;/span&gt; on each side of the vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the downward force on it is &lt;span class=&quot;math-container&quot;&gt;$2T_c\sin(dl/2r) \approx T_c\,dl/r$&lt;/span&gt; (the first 2 is because there are two free ends of the element)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The acceleration downwards is &lt;span class=&quot;math-container&quot;&gt;$v^2/r$&lt;/span&gt; (ignoring gravity).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Newton's second law gives the equation in the paper.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-02-04T16:53:25.637" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33068" PostTypeId="1" CreationDate="2020-02-04T18:01:51.497" Score="-1" ViewCount="177" Body="&lt;p&gt;I am a software engineer with no knowledge of mechanical engineering.  I need to buy or build a rotating disc which can rotate in 30 degree steps only i.e each time I try to rotate it, it should take a 30 degree step and stay fixed in that place. Hope it makes sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching on Google I could find only info about electrical rotary encoders.  Unable to figure out how to search for mechanical devices I need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess is these sort of devices must already exists. Can please share any info you may have on this and provide suggestions on buying or building this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wd4BT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wd4BT.png&quot; alt=&quot;Drawing&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update: As shown in above pic, on the surface of the disc, there would be LEDs of different color, spaced about 30 degrees apart (or 45 degrees). Each time I want to point one specific light to the sensor. So from time to time, I want to manually rotate the disc to make the required led point towards the sensor (i.e some led pointed exactly to sensor and no intermediate position) and be it locked there.&lt;/p&gt;&#xA;" OwnerUserId="24474" LastEditorUserId="9" LastEditDate="2020-02-06T11:17:57.757" LastActivityDate="2020-02-06T11:17:57.757" Title="Design suggestions for Mechanical rotating disc with 30 degrees step" Tags="|mechanical-engineering|mechanisms|mechanical|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33070" PostTypeId="2" ParentId="30012" CreationDate="2020-02-04T20:04:41.807" Score="1" Body="&lt;p&gt;Buna-N is a typical o-ring material and will seal at low pressure, standard is 70A. Make sure you lubricate them before installation, or they can tear and won't seal anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Apple Rubber or Parker Hanifin both have extensive o-ring application handbooks that will size everything for you. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Size your spring to get the adequate compression recommended here: &lt;a href=&quot;https://www.applerubber.com/src/pdf/seal-design-guide.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.applerubber.com/src/pdf/seal-design-guide.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-02-04T20:04:41.807" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33071" PostTypeId="2" ParentId="33065" CreationDate="2020-02-04T20:52:05.490" Score="1" Body="&lt;p&gt;Your equation is correct for &lt;em&gt;your&lt;/em&gt; definition of &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;.  However, your definition of &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; seems incorrect.  &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; is supposed to the be distance between the transmitter and the receiver, that is the distance on a straight line from the transmitter to the receiver.  In your diagram you have (I think incorrectly) defined &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; as the horizontal distance on some plane, presumably the surface of the Earth, that the transmitter and receiver are not both on.  Remember, this is free space that we're talking about, so it doesn't make sense to talk about horizontal distance on a reference plane.&lt;/p&gt;&#xA;" OwnerUserId="20993" LastActivityDate="2020-02-04T20:52:05.490" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33072" PostTypeId="2" ParentId="33068" CreationDate="2020-02-04T21:31:43.997" Score="0" Body="&lt;p&gt;Your question may be closed due to being rather vague and open-ended. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your goal can be accomplished with stepper motors and appropriate controllers, with servo motors and controllers, and is also likely with a strictly mechanical approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider to edit your question to reflect a more detailed methodology. If you do not require automation, indexing mechanisms are common.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did a quick search for indexers but found the best results on Thingiverse, an online repository for 3D printed models. One of the models I found was somewhat more complex than expected, a scroll cam indexer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The YouTube has a &lt;a href=&quot;https://www.youtube.com/watch?v=2atiOFA-qnQ&quot; rel=&quot;nofollow noreferrer&quot;&gt;one minute video&lt;/a&gt; of one such device, which rotates 60° for each turn of the driving shaft. The one I found on &lt;a href=&quot;https://www.thingiverse.com/thing:277159&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thingiverse&lt;/a&gt; appears to have eight pins, for 45° movement per rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, the Thingiverse entry does not provide source files, which would allow one to modify the parameters to achieve your desired rotation. The concept is clear, however.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the non-3D world, there are a few images for scroll cam indexers:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/k5her.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/k5her.jpg&quot; alt=&quot;scroll cam indexer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One could extrapolate from the above image that a suitably sized worm gear assembly could provide what you require. The input worm gear would have to turn many times to make a 30° rotation, but the worm gear could be driven by a set of gears to increase the single crank turn to that required of your structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kxVkn.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kxVkn.jpg&quot; alt=&quot;24:1 worm gear reduction&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This images shows what is described as a 24 to 1 reduction assembly. One turn of the crank gives you 1/24th of a rotation, while you need 1/12th for your objective. This means the input shaft need have only a 2:1 increase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such a design is relatively mechanically simply, if it meets your unstated restrictions.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-04T21:31:43.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33074" PostTypeId="1" CreationDate="2020-02-05T03:22:14.373" Score="1" ViewCount="677" Body="&lt;p&gt;I'm trying something fancier..  hopefully it will be clear from the screen-shot.. linear and circular patterns work.  now, however, I'd like to spiral the cuts.  first try isn't working..  it gave me an error at one point, but I didn't comprehend it, and it hasn't shown up again.&lt;a href=&quot;https://i.stack.imgur.com/lwvtu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lwvtu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is a second screen-capture presenting the problem.  I try to select the outside of the cylinder as the surface that the pattern should be normal to, and it doesn't work..&lt;a href=&quot;https://i.stack.imgur.com/T4SwH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T4SwH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;partial solution:  I tried the following (somewhat lame, it gave me no sound?) tutorial and this time it worked:  &lt;a href=&quot;https://www.youtube.com/watch?v=X9xSMBdqcr4&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=X9xSMBdqcr4&lt;/a&gt; .  two things I've learned:  the first one.. is rather trivial but irks me for some reason.  a helix can't be created from an existing sketch,  you have to create a fresh sketch just for it.  this seems.. wasteful of time and effort, particularly considering the second thing I've learned:  in order for the curve driven pattern to work, the helical path -must- be on the surface to which the object to be patterned is to be  normal...  so the question would become, why is solidworks built this way.  is  there a reason for it, or is  it just an oversight? :-)&lt;/p&gt;&#xA;" OwnerUserId="13081" LastEditorUserId="13081" LastEditDate="2020-02-09T12:34:51.353" LastActivityDate="2020-02-09T12:34:51.353" Title="how to pattern a cut-loft around a helix" Tags="|solidworks|" AnswerCount="1" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="33076" PostTypeId="1" CreationDate="2020-02-05T06:06:03.007" Score="0" ViewCount="116" Body="&lt;p&gt;I'm trying to build a small pick and place system using a suction cup and a 12V vacuum pump. The problem I'm having is that I'm not getting a good enough seal between the cup and the object, and I read that it could be improved by increasing airflow. I've also read that placing pumps in parallel can increase airflow, but is there a limit to the number of pumps that can be placed in parallel? Would the pumps overheat as a result of using too many pumps in parallel? The pump I'm using has a datasheet vacuum pressure of -0.08Mpa and an airflow of 16L/min. The pump itself is composed of 2 pumps in parallel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any advice appreciated.&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="4302" LastEditDate="2020-02-05T07:50:22.533" LastActivityDate="2020-02-05T07:50:22.533" Title="Increasing suction air flow with multiple pumps" Tags="|pressure|vacuum|vacuum-pumps|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33077" PostTypeId="1" CreationDate="2020-02-05T07:07:48.277" Score="-1" ViewCount="30" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AoSu7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AoSu7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm talking about this one. On what principle does it work? The only thing i can think of is that you press the shaft with the screws and then it works based on friction but I have no idea if it is correct.&lt;/p&gt;&#xA;" OwnerUserId="21424" LastActivityDate="2020-02-05T08:17:50.413" Title="How does a flexible CNC coupler work?" Tags="|coupling|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33078" PostTypeId="2" ParentId="33077" CreationDate="2020-02-05T08:17:50.413" Score="2" Body="&lt;p&gt;Those four screws are usually tightened against a flat or flats machined on the two shafts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That prevents any slipping and possible damage to either shaft.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-02-05T08:17:50.413" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33079" PostTypeId="2" ParentId="26701" CreationDate="2020-02-05T09:14:29.270" Score="0" Body="&lt;p&gt;To answer my own question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula for hydraulic radius, velocity and flow remains the same. But what is important to note is the wetted perimeter, &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the flow is partial flow, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P=2h+b$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When the flow is full flow,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$P=2h+2b$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(because now the top part of the box is also &quot;wetted&quot;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This explains why the &quot;strange&quot; behavior how when &lt;span class=&quot;math-container&quot;&gt;$D:RISE$&lt;/span&gt; approaches &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\frac{Q}{Q_f}$&lt;/span&gt; can go way beyond &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;, and when &lt;span class=&quot;math-container&quot;&gt;$D:RISE=1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\frac{Q}{Q_f}$&lt;/span&gt; goes back to &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt;. &lt;/p&gt;&#xA;" OwnerUserId="3353" LastActivityDate="2020-02-05T09:14:29.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33080" PostTypeId="1" CreationDate="2020-02-05T09:27:48.840" Score="-1" ViewCount="30" Body="&lt;p&gt;I live on the east coast of Australia, where a Hyperloop faces the challenges that we have a mountain range pretty much right up against the coast, where most people live. There's very little flat space to build on, or so I have heard or read somewhere. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyhow Is also read recently that the International Energy Agency released a report saying offshore wind could/should be a big part of decarbonising the power grid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found myself imagining a string of offshore wind towers up and down the easy coast, then I imagined a Hyperloop that used those as support columns, avoiding the difficult coastal terrain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would that work?&lt;/p&gt;&#xA;" OwnerUserId="24483" LastActivityDate="2020-02-05T10:39:43.723" Title="Could a string of large offshore wind towers support a Hyperloop tube?" Tags="|structures|" AnswerCount="1" CommentCount="2" ClosedDate="2020-02-08T01:23:37.203" ContentLicense="CC BY-SA 4.0" />
  <row Id="33081" PostTypeId="2" ParentId="33080" CreationDate="2020-02-05T10:39:43.723" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Could a string of large offshore wind towers support a Hyperloop tube?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;It's a well defined [tm] engineering problem well within the physical capabilities of available materials and methods, so, yes, they could.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the practicality of integrating the two very different systems is vanishingly small.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Offshore wind turbine towers are spaced at very large distances apart compared to the structures that would be necessary to support a practical hyperloop track. Adjusting WT tower placement so their foundations and part of their structures could be jointly utilised is conceptually feasible, but the percentage of hyperloop supports suitable for joint use would be a small percentage of the total, and the added constraints of routing and location imposed on both systems would be likely to be significant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A hyperloop track could be used to provide support for a large number of smaller wind turbines along the while route, but the economics of wind turbines greatly favour the largest and tallest market proven* systems available.&lt;br&gt;&#xA;(*There are always larger and taller systems seeking to prove themselves).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It seems likely that a &quot;hyperloop&quot; could be economically be built in a land based tunnel than on an above surface marine platform.  Security, and mechanical stability are both significant requirements. While subterranean tunnel systems face challenges from earthquakes, this is a well known and understood (if not well predictable) phenomenon and paid response sensor systems are liable to render a tunnel system no more dangerous than a marine above surface platform. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;From 'memory' - figures could be researched and added if of interest:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;China operates a large network of high speed rail - mostly above ground, but still subject to derailment effects from earthquakes and other land movement. Speeds are around (from memory) 150+ kph for the Z trains and 200-250 kph for the D trains, with faster speeds in some sections. While well below hyperloop speeds the challenges are similar. I have heard of the very occasional report of Chinese rail incidents but they seem to be as rare as one would hope. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over middle and longer distances a D train achieves trip times about half that of a car over the same general route. (I've traveled in D trains a few times). &lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2020-02-05T10:39:43.723" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33082" PostTypeId="1" AcceptedAnswerId="33097" CreationDate="2020-02-05T12:34:27.193" Score="-2" ViewCount="166" Body="&lt;p&gt;Will epoxy resin dissolve a silicone spatula, cake icing spatula?  Will it work well enough to get the job done without contaminating the resin?&lt;/p&gt;&#xA;" OwnerUserId="5967" LastActivityDate="2020-02-05T22:57:15.253" Title="will silicone spatula be damaged by epoxy resin?" Tags="|composite-resin|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33084" PostTypeId="2" ParentId="33074" CreationDate="2020-02-05T13:16:39.257" Score="2" Body="&lt;p&gt;I've pretty much answered this in the comments, but didn't want to hyperlink an image:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Make your loft into an un-merged solid, that can be used as a cut tool later.&lt;/li&gt;&#xA;&lt;li&gt;Pattern bodies instead of features, this is &lt;em&gt;always&lt;/em&gt; more reliable&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another thing to note - you can use the helix feature directly as your path, there's no need for a 3D sketch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t0PGK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t0PGK.png&quot; alt=&quot;helix loft pattern&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2020-02-05T13:16:39.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33086" PostTypeId="1" CreationDate="2020-02-05T15:00:33.637" Score="4" ViewCount="62" Body="&lt;p&gt;In old scientific apparatus for electrical experiments, and more particularly in instruments such as galvanometers and torsion balances, we often need to hold one end of a thread in place. Examples include the following (particularly large photos are given as links):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/4/4a/Galvanometer-MHS_229-IMG_3875-gradient.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://upload.wikimedia.org/wikipedia/commons/4/4a/Galvanometer-MHS_229-IMG_3875-gradient.jpg&lt;/a&gt; (Very large)&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://catalogue.museogalileo.it/images/cat/oggetti_944/1560_3283_1933-011_944.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://catalogue.museogalileo.it/images/cat/oggetti_944/1560_3283_1933-011_944.jpg&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://catalogue.museogalileo.it/gallery/DemonstrationModelAstaticGalvanometer.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://catalogue.museogalileo.it/gallery/DemonstrationModelAstaticGalvanometer.html&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Galvanometer (courtesy of museo galileo, &lt;a href=&quot;https://catalogue.museogalileo.it/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6iwuB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6iwuB.jpg&quot; alt=&quot;Nobili differential galvanometer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one description, from museo galileo, &lt;a href=&quot;https://catalogue.museogalileo.it/object/NobilisLargeAstaticGalvanometer.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IZTcz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IZTcz.jpg&quot; alt=&quot;Nobili astatic galvanometer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A pair of astatic needles is suspended from the fine silk wire attached to a torsion mechanism at the top of the pillar. The needles can be raised or lowered by means of a complex screw-operated scissors-type mechanism.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;In Coulomb's torsion balance, we get the following drawing, and description:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dmbo7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dmbo7.jpg&quot; alt=&quot;Coulomb balance&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The top, n° 1, bears a knob b, the pointer io, and the clasp of&#xA;  suspension, q ... The clasp q (figure 2, n° 1) has approximately the shape of&#xA;  the extremity of a solid mechanical pencil clamp [porte-crayon], which can be&#xA;  tightened by means of the annulus q. The clasp of this porte-crayon holds the end of a filament of very thin silver. The other end of the filament of silver is fixed (fig. 3) at P, by the clasp of a cylinder Po of copper or of iron, whose diameter is but a [ligne], and whose end P is split and forms a clasp which is tightened by means of the collar Φ.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I've been reviewing these sorts of mechanisms for a while, and Coulomb's description (the clasp of a mechanical pencil) is the closest I've found to a proper description. But with mechanical pencils, as I know them, the clasp (e.g.a two-part split piece of plastic in a modern one) supports a fixed diameter of lead, typically much thicker than the width of a silk thread. As well, it's not clear from my (limited) research what mechanisms were used for &quot;porte-crayons&quot; in that time period. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking to recreate some of these instruments in some way, e.g. with modern methods of fabrication and using modern mechanisms available. But these thread-holders are giving me a fair bit of trouble. &lt;strong&gt;Is there any resource available where I can get mechanical drawings of such mechanisms, or is there a common &quot;household item&quot; which uses these sorts of clasps that I could look at? Does anyone have any ideas about how these would have worked?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="18257" LastActivityDate="2020-02-05T15:00:33.637" Title="How do these thread-holder mechanisms work?" Tags="|mechanisms|engineering-history|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33087" PostTypeId="1" CreationDate="2020-02-05T15:09:59.200" Score="0" ViewCount="114" Body="&lt;p&gt;I've spent a large amount of my life thinking back on past times, both long ago and &quot;recently&quot;, meaning ~1950s to me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A lot of things make sense, but some things just fundamentally baffle me. One of those things is the seemingly total disregard for safety when driving cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;People were not idiots a few decades ago. Well, at least not the ones who invented and manufactured large machines such as cars. They &lt;em&gt;must&lt;/em&gt; have very quickly, long before the 1950s, had tons of evidence of people dying in traffic accidents, or getting seriously hurt. How did it not occur to them to have some kind of belt or any other safety measure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's not like the safety belt wasn't invented, and used in various military contexts. And even if that weren't the case, it's a very obvious invention. We aren't talking about hi-tech sensors and air bags all around the passengers and refined safety measures like that -- we're talking about a basic belt around the humans so that they don't come flying through the window and become a pulp of blood and guts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And those seats, leaving the whole head part &quot;free&quot;, seems just silly as well. Who felt safe driving like that? Even at a low speed, you can feel it in your neck when there is an unexpected stop and there is nothing supporting your back of the head. These two things seem like they would've been implemented at the latest &lt;em&gt;months&lt;/em&gt; after the modern car was invented in what I assume must've been around 1900 -- &lt;em&gt;long&lt;/em&gt; before the 1950s when they (well, some) seemingly started to even consider human lives and health in the making of cars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can understand if they didn't bothered with seat belts in a wagon powered by one or two horses, but even then, they probably had some kind of handle or something to grab on to. When you introduce powerful engines, well, it just seems to go without saying that you would have to think about the consequences.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible that I'm somehow over-estimating the amount and seriousness of car crashes and traffic accidents in general? Did people really not care whatsoever about whether they were alive or dead or in a hospital for the rest of their lives, as long as they got to drive a car without any safety measures whatsoever? I don't understand this mentality, and I'm having a hard time understanding it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, many other things can usually be explained in a satisfactory way, but this especially just doesn't make sense to me.&lt;/p&gt;&#xA;" OwnerUserId="24487" LastActivityDate="2020-02-05T23:31:44.880" Title="How is it possible that it took until the 1950s or even longer before they started using car belts and neck-supported seats?" Tags="|safety|car|" AnswerCount="4" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33088" PostTypeId="1" CreationDate="2020-02-05T15:25:40.943" Score="2" ViewCount="23" Body="&lt;p&gt;In a design for a heated block I am working on I only have the space for a 6mm X 10mm hole for my heating element to be inserted into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Originally I had planned to use a 40W heater cartridge but the smallest I can find is 6mm x 20mm and the extra 10mm that would stick out the block would be inconvenient as I want to insulate the area around this from the heat.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a good alternative to a heater cartridge that would fit my specs?&lt;/p&gt;&#xA;" OwnerUserId="23791" LastEditorUserId="23791" LastEditDate="2020-02-05T16:18:04.417" LastActivityDate="2020-02-05T16:18:04.417" Title="Alternative to heater cartridge for 10mm deep hole?" Tags="|heating-systems|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33089" PostTypeId="2" ParentId="33051" CreationDate="2020-02-05T15:58:54.067" Score="1" Body="&lt;p&gt;As was mentioned in the comments, you can get off-shelf air compressors, which are quite good at regulating pressure. Now if you want to regulate the flow rate, that becomes a bit more tricky.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your first (cheap ~ £10) option is to go for something like this (see &lt;a href=&quot;https://www.smc.eu/en-eu/products/as-standard-w-built-in-one-touch-fitting~134135~nav&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.smc.eu/en-eu/products/as-standard-w-built-in-one-touch-fitting~134135~nav&lt;/a&gt; for details):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/brIl7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/brIl7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will allow to set your flow rate to a constant value &lt;strong&gt;for a given pressure&lt;/strong&gt;, but you'll have to measure the flow rate in some other way. Also, if the pressure changes, your flow rate will also change, even if the setting on the speed controller hasn't changed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your second and more expensive option, is to go for a mass flow controller, such as this one (see &lt;a href=&quot;https://www.bronkhorst.co.uk/en/products/gas_flow_meters_and_controllers/elflow_select/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.bronkhorst.co.uk/en/products/gas_flow_meters_and_controllers/elflow_select/&lt;/a&gt; for more details):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/StnjH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/StnjH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now this will allow you to measure and control the flow rate to whatever value you desire and maintain that flow rate regardless of inlet pressure variations. This is controlled by a software running on your PC and communicating with the instrument via a serial cable (you'll need a serial to USB converter). Now this will set you off ~£1,200 so not cheap, but it's pretty fool-proof.&lt;/p&gt;&#xA;" OwnerUserId="334" LastActivityDate="2020-02-05T15:58:54.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33090" PostTypeId="2" ParentId="33087" CreationDate="2020-02-05T16:27:16.417" Score="2" Body="&lt;p&gt;I started driving in the UK in the 1960s, so this is &quot;first hand&quot; experience of what you find hard to imagine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are ignoring the fact that typical driving speeds were lower, general car handling performance (i.e. acceleration/braking/cornering - drum brakes, cross-ply tires, and rear wheel drive were all standard) was much lower, traffic density was lower, and (at least in the UK) there were virtually no modern-design high-speed roads - the national motorway network only reached 1000 miles of road in 1972.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To summarize, any fool can get in a modern car and within 30 seconds have a head-on crash with an immovable object (or worse, an oncoming vehicle) at a speed of 70 mph or more, and since there are plenty of fools who &lt;em&gt;do&lt;/em&gt; drive modern cars like that, the cars need safety systems to mitigate the consequences. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you tried driving a 1950s-era car the same way, you would have slid sideways of the road (relatively safely, except for a damaged ego) long before you got into a life threatening situation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider the spec of a typical UK small family car from the end of the 1950s: the Ford Anglia 105. Four-speed manual gearbox (no synchromesh on 1st and reverse). Max engine power, 40 bhp. Max torque, 53 ft-lb. Max speed, 73mph. Acceleration: 0-60mph in 30 seconds. Drum brakes on all four wheels. Not a vehicle designed to be even &lt;em&gt;capable&lt;/em&gt; of having modern high speed road accidents!&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2020-02-05T16:41:49.473" LastActivityDate="2020-02-05T16:41:49.473" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33091" PostTypeId="2" ParentId="32947" CreationDate="2020-02-05T16:33:58.563" Score="1" Body="&lt;p&gt;According to my experience the symbol GOS stands for the valve stem position switch (limit switch), GOS+ = valve open (stem in high position), GOS- = valve closed (stem in lowposition). In old P&amp;amp;ID's letter G denoted position.&lt;/p&gt;&#xA;" OwnerUserId="21367" LastActivityDate="2020-02-05T16:33:58.563" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33092" PostTypeId="2" ParentId="33087" CreationDate="2020-02-05T18:11:02.280" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;There wasn't as many catastrophic accidents. Our body disintegrate s around 20g. There were not many crashes that intense.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;There were not enough public awareness and expectation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The technology to design a strong path of force from the belt to it's bracket and through the seat frame to car frame was not well developed.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The consumer safety standards were not universally accepted. And manufacturers did not share their internal research results of course.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Today many of even economy cars have safety features like automatic collision avoidance, multiple airbags, driver alertness alarms, because it cost next to nothing and society knows of them and demands them.  I bought for my old airplane a satellite position reporting radar for 1/10 of what it used even a couple of years ago. We are making smart things more efficiently. And we have learned to expect no compromise when it concerns our safety.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-02-05T18:11:02.280" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33093" PostTypeId="2" ParentId="33061" CreationDate="2020-02-05T20:01:31.447" Score="1" Body="&lt;p&gt;There is no buckling under torsion in a beam, there is lateral web buckling under vertical loads under certain circumstances.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A beam under pure torque will gradually twist due to St Venant and warping torsion until it yields with no sudden loss of strength as in buckling under axial load, however, under combined axial stress and torque it will buckle and needs to be analyzed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is the section torque moment under pure torsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T=G*I_t \phi^ \prime -EI\phi^ { \prime \prime \prime }$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T   is the torsional moment at the cross-section.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\phi^ \prime \ and\ \phi^ { \prime \prime \prime }  $&lt;/span&gt; are first and third derivateves WRT X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$I_w $&lt;/span&gt; is the warping constant&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ I_t $&lt;/span&gt; is ST Venant's torsional constant.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-02-05T20:01:31.447" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33094" PostTypeId="2" ParentId="33087" CreationDate="2020-02-05T20:36:31.173" Score="1" Body="&lt;p&gt;In the US, the NHTSA, National Highway Traffic Safety Administration, was formed in 1966.  Prior to this there really wasn't a regulatory agency in the US for such things - the world was a much more laissez faire place then.  Innovations in safety came from the manufacturers, who noticed and wanted to do something about deaths rates.  Volvo introduced the 3-point belt in 1958, but it wasn't until well  into the 70s that cars came with them standard as part of the default belt.  Mercedes pioneered crumple zones.  US cars in the 60s often had safety belts as options, not even standard equipment.  We had a 1973 Ford that had shoulder belts strapped up in the overhead over the windows and you had to take them down and connect them to use them.  They were terrible.  No give, so you couldn't even reach the radio.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Old cars were death traps.  Anyone who gives their kid a 60s Mustang to drive is giving them a bomb.  Gas tanks behind the rear seat, solid steering columns that extended past the front wheelbase, no side impact protection, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's really the same thing with head restraints and airbags.  It took government regulations to make them happen, because people tend to just not care.  There was a national outcry about the extra expense of airbags when they were introduced.  A car without them would be unthinkable today.  It wasn't until the 80s that wearing seat belts even became widespread.  The popular opinion of the time was, &quot;I want to be thrown free of the accident,&quot; a view since proven as wholly terrible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lately, the US Insurance Institute for Highway Safety has led the way for changes.  Smaller side windows (protection against SUV side impacts) and larger A pillars (offset frontal crash protection) are the direct result of their tests and manufacturers wanting better rated vehicles.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-02-05T20:36:31.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33095" PostTypeId="1" CreationDate="2020-02-05T21:57:27.517" Score="1" ViewCount="35" Body="&lt;p&gt;I didn't make it obvious enough below that I was interested mostly in infrared, as the current use case is measuring power radiated from warm objects and from the ground. As pointed out in comment by @gerrit (thank you):&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;You may want to clarify that you're interested in the Earth surface, so about 0.3 – 100 µm will cover everything you need.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;In my naivety I first asked a related question over at Earth Science &lt;a href=&quot;https://earthscience.stackexchange.com/questions/19127/how-much-energy-does-earth-surface-radiate-at-night-per-square-meter&quot;&gt;here&lt;/a&gt;. Ultimately, I would like to know how much electromagnetic energy power I could detect from an object or a surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a home project, is there a way to measure how much EM power is being radiated from a nearby object? It's fine, preferable even, to ignore energy that radiates away from the detector and only measure what is captured by a nearby detector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At work we have things like spectrum analyzers and power meters and such that measure power at different frequencies in wired communications, and that's similar to what I'm after, but I'm looking for a way to measure EM radiation, mostly infrared power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I could wave my hand and magically have exactly what I want pop into my hand, it would be a 1 meter square flat device with 1 side measuring EM power so I could hold it just above the ground and get a result of &quot;1 watt&quot; (1 watt per meter squared, since the device is 1 meter), or hold it next to a tree or a person and measure power coming off of them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is both for my own use (both curiosity and possible future personal projects), and also as a science project for my kids since my oldest one is experimenting with measuring power with a multimeter and solar panel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there is a type of device that does this, then that's great and the name of the device would be helpful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If this kind of measurement requires a process (eg: &quot;use device such-and-such to measure power at lots of points and then use method so-and-so to integrate across a square meter&quot;) then that is acceptable as long as it can be done by a hobbyist and doesn't require expensive tools worth thousands of dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;How do you measure electromagnetic power over a surface area to find the watts per square meter?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="16752" LastEditorUserId="16752" LastEditDate="2020-02-06T17:12:14.360" LastActivityDate="2020-03-07T19:01:20.177" Title="Measuring EM radiation power over surface area" Tags="|measurements|power|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33096" PostTypeId="2" ParentId="33087" CreationDate="2020-02-05T22:19:12.503" Score="0" Body="&lt;p&gt;Seat belts were considered racing equipment. You couldn't get insurance for a car that was hot-rodded to include features like roll bars and seat belts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Oh, Neck supports get in the way of hats (but they were invented in 1921 in the form of the removeable headrest). People wore hats back then— you know— bowler hats. Cars were designed to accommodate bowler hats until the mid '50s. The pork-pie hat came into existence because of MGs and other little ragtops. The car seat was supposed to look and feel like your living room sofa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And let's not forget the solid steel dashboards that would crack you skull open like an egg. Tucker introduced the first padded dash in the 1948 Tucker Torpedo. Advocates had already been calling for them for 20 years by then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an interesting read when you have some time - &#xA;&lt;a href=&quot;http://mvhap.org/mvhappdfs/styling.vs.safety.noteworthy.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://mvhap.org/mvhappdfs/styling.vs.safety.noteworthy.pdf&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2020-02-05T23:31:44.880" LastActivityDate="2020-02-05T23:31:44.880" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33097" PostTypeId="2" ParentId="33082" CreationDate="2020-02-05T22:57:15.253" Score="1" Body="&lt;p&gt;Silicone molds are often used for epoxy resin casting, so I don't see any problems.&lt;/p&gt;&#xA;" OwnerUserId="24491" LastActivityDate="2020-02-05T22:57:15.253" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33099" PostTypeId="1" CreationDate="2020-02-06T00:55:14.543" Score="1" ViewCount="17" Body="&lt;p&gt;we are using the &lt;a href=&quot;https://www.danielgm.net/cc/&quot; rel=&quot;nofollow noreferrer&quot;&gt;CloudCompare&lt;/a&gt; for comparing two mouth STL model for Final doctoral theses, like works done in &lt;a href=&quot;https://www.researchgate.net/publication/319698918_Mandibular_Surgical_Navigation_An_Innovative_Guiding_Method&quot; rel=&quot;nofollow noreferrer&quot;&gt;this paper&lt;/a&gt;(&lt;a href=&quot;https://www.researchgate.net/profile/Vincenzo_Abbate3/publication/319698918_Mandibular_Surgical_Navigation_An_Innovative_Guiding_Method/links/59fb57d20f7e9b9968b9bff5/Mandibular-Surgical-Navigation-An-Innovative-Guiding-Method.pdf?_sg%5B0%5D=mHsXK3vdmqqMtk8QEsoozmteG5F8kJewDn-ZcM0rcMzenPXOG1xt0q3mvLQvdk5ifFOwlahUnDnTeMHj90E68g.rfPtUCDRIU_NM2xQps_mlq_CmuSzua12c2_q-3HkvEeImgguvUIf-n6ps28cDFpk4r5E9VFnJHDtZetPcBHIAQ&amp;amp;_sg%5B1%5D=8sFiJcM7BsFO_hnehQ2W48wiyfwA2B8UFMZivaW1F7Z-YSaoLb3HS4GOCZsTeoIKXyt2QqsLg4WQa7Hz3lgjU3-7D-T8P__Vy--dR3CvaDBa.rfPtUCDRIU_NM2xQps_mlq_CmuSzua12c2_q-3HkvEeImgguvUIf-n6ps28cDFpk4r5E9VFnJHDtZetPcBHIAQ&amp;amp;_iepl=&quot; rel=&quot;nofollow noreferrer&quot;&gt;pdf&lt;/a&gt;) shown blow:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ETISn.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ETISn.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the instruction said &lt;a href=&quot;https://www.cloudcompare.org/forum/viewtopic.php?t=3452&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VdhZB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VdhZB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so i have tried to insert the STL file and use this align method :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.postimg.cc/QdQCbWTS/result-9.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.postimg.cc/QdQCbWTS/result-9.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so the output are only 256 numbers by this columns names:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pcyts.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pcyts.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so asked how to increase the result number of compared output result in CSV file and is this methods is right for accessing to the compared this two models?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;also what is the means of the csv colums means whis are like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5waGs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5waGs.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i have uploaded the two STL files here, for more information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://s7.picofile.com/file/8387291350/stl_files.zip.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://s7.picofile.com/file/8387291350/stl_files.zip.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your attention.&lt;/p&gt;&#xA;" OwnerUserId="2678" LastEditorUserId="2678" LastEditDate="2020-02-06T09:06:38.997" LastActivityDate="2020-02-06T09:06:38.997" Title="how to get more number variance compared as the csv output file of two model stl files in Cloudcompare" Tags="|modeling|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33100" PostTypeId="1" CreationDate="2020-02-06T02:29:31.957" Score="4" ViewCount="31" Body="&lt;p&gt;I'm trying to figure out a correlation between correct feed rating and the feed force. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can a specific feed force be declared to create the correct feed rate per material per tool speed?&lt;/p&gt;&#xA;" OwnerUserId="20275" LastEditorUserId="20275" LastEditDate="2020-02-09T03:07:43.710" LastActivityDate="2020-02-09T03:07:43.710" Title="Is constant force a useful parameter for correct drilling operation?" Tags="|drilling|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33101" PostTypeId="1" AcceptedAnswerId="33112" CreationDate="2020-02-06T04:20:56.150" Score="2" ViewCount="383" Body="&lt;p&gt;When I happened to see pictures of a highway in the desert, the question arose in my mind. The highway will probably be buried if a sandstorm rages. So is there any solution to this problem?&#xA;&lt;a href=&quot;https://www.sciencenews.org/blog/wild-things/algal-blooms-created-ancient-whale-graveyard&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sciencenews.org/blog/wild-things/algal-blooms-created-ancient-whale-graveyard&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://www.sciencenews.org/wp-content/uploads/2014/02/wt_LaFamilia_South_CerroBallena_free.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="24493" LastEditorUserId="13311" LastEditDate="2020-02-06T07:19:14.603" LastActivityDate="2020-02-06T17:40:36.973" Title="How is a highway prevented from being buried by sand in the desert?" Tags="|civil-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33102" PostTypeId="1" CreationDate="2020-02-06T05:23:42.487" Score="0" ViewCount="30" Body="&lt;p&gt;I am trying to run a CFD simulation for an ASTAR nozzle but cannot get the solution to converge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the steps I took to try and run the simulation. I made the geometry in Solidworks and imported the Parasolid file into the geometry (careful to to select 2D for the analysis type) and generated it. In the meshing I applied edge sizings (with bias toward the walls) and face meshing to get a decent looking mesh with skewness below 0.6 for all parts. For the setup conditions I did:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;General: changed type to density-based&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Models: turned on Energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Materials: in the fluid air I changed density to ideal-gas and viscosity to sutherland (with default settings)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Boundary conditions: set outlet to pressure-inlet at pressure of 3e5 Pa, pressure-outlet at 0 Pa, and walls are no-slip condition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I left all the other settings as is, initialized it and ran it for 500 iterations, but it just blew up and I'm not sure what could be going wrong. Any ideas or tips on what to check for this? I've attached an archived work file that has all the setup and information in case this proves useful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Files can be found here: &lt;a href=&quot;https://studentcommunity.ansys.com/thread/solution-diverges-for-converging-diverging-nozzle/?postbadges=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://studentcommunity.ansys.com/thread/solution-diverges-for-converging-diverging-nozzle/?postbadges=true&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24290" LastActivityDate="2020-02-06T05:23:42.487" Title="Solution diverges for Converging-Diverging Nozzle" Tags="|fluid-mechanics|ansys|cfd|fluent|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33103" PostTypeId="2" ParentId="33095" CreationDate="2020-02-06T08:28:33.370" Score="1" Body="&lt;p&gt;This would be extremely difficult without specifying the portion of the EM spectrum you wanted to measure.  Gamma rays will have different detectors than radio waves than infrared heat waves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;IR detectors use lenses and basically a camera sensor to detect the heat coming off of an object.  A directional gamma detector would need a method to differentiate direction, since a handheld detector wouldn't have enough shielding to prevent readings from behind.  For radio waves would probably want a reflecting dish to amplify the signal.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-02-06T08:28:33.370" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33105" PostTypeId="1" CreationDate="2020-02-06T09:42:13.187" Score="-1" ViewCount="151" Body="&lt;p&gt;Why inner radius of the housing washer and shaft washer is different in thrust bearing?&#xA;I checked on SKF's website dimensions of a thrust bearing. The inner radius of the housing washer is different than the shaft washer inner radius. Can any you tell why is that?&lt;/p&gt;&#xA;" OwnerUserId="15698" LastActivityDate="2020-02-06T13:06:58.533" Title="Why inner radius of housing washer and shaft washer is different in thrust bearing?" Tags="|thrust|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33106" PostTypeId="2" ParentId="33105" CreationDate="2020-02-06T13:06:58.533" Score="0" Body="&lt;p&gt;The race with a tighter inner diameter is meant to press onto a shaft and transmit the thrust load from the shaft into the bearing elements. The housing race has a looser bore and is meant to clear the shaft so it does not accept any thrust from the shaft, this race is typically a press into your housing.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-02-06T13:06:58.533" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33107" PostTypeId="1" CreationDate="2020-02-06T13:44:37.327" Score="1" ViewCount="310" Body="&lt;p&gt;General Electrics boasts that its Evolution series diesel-electric locomotives are capable of dynamic braking. Dynamic braking uses the electrical traction motors on the locomotive axles - motors are switched into generator mode and convert kinetic energy of the locomotive into electricity which is then directed either into large resistors (most common) or into charging batteries.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds good but a typical scenario is having two or three diesel-electric locomotives pulling a train of one hundred carriages. A typical locomotive has up to ten axles and weighs up to two hundred tons, so three locomotives have up to thirty axles and weigh up to six hundred tons. They together pull around one hundred carriages and each of those has four axles without any motors (so unable to implement dynamic braking) and weighs around one hundred tons each (when fully loaded). One hundred carriages together weigh about ten thousand tons total - about 16 times more than the locomotives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it looks like if the locomotives use dynamic braking and the carriages don't use any braking at the same time then the whole train should &quot;compress&quot; and either derail or collapse into the locomotives because of inertia. So the carriages should use air brakes each time when the locomotives use any brakes so that the whole train slows down evenly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means that dynamic braking can only be used to slow down the locomotives, not the entire train and the carriages have to use air brakes every time the locomotives use any brakes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can dynamic braking be used to slow down the entire train or does it only slow down the locomotives?&lt;/p&gt;&#xA;" OwnerUserId="374" LastActivityDate="2020-02-06T17:36:02.167" Title="How does dynamic braking work with railway locomotives?" Tags="|mechanical-engineering|electrical-engineering|rail|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33108" PostTypeId="1" CreationDate="2020-02-06T15:05:31.900" Score="-1" ViewCount="419" Body="&lt;p&gt;I have a flexure problem. One of its ends is fixed to a frame (say F1) which is stationary (no rotation &amp;amp; no translation) and the other end is also fixed to another frame (say F2) which can move in translatory motion (rotation fixed &amp;amp; translation free). The flexure here is assumed to be a slender column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On top of the flexure, F2 has a weight &lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt; and is moved with a force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; in the horizontal direction causing a displacement of &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; whereas at the bottom the flexure is held rigidly by F1 and doesn't allow any rotation or translation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ynMng.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ynMng.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to find the displacement &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; caused due to the forces &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$W$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please help me solve this. Also, I request you to write the moment equation I need to consider here... and anything else I need to know.&lt;/p&gt;&#xA;" OwnerUserId="24497" LastEditorUserId="1832" LastEditDate="2020-02-08T01:21:11.187" LastActivityDate="2020-02-08T01:21:11.187" Title="Flexure Deflection - Both ends fixed (one translation free)" Tags="|structures|aerospace-engineering|deflection|columns|flexures|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33109" PostTypeId="2" ParentId="28549" CreationDate="2020-02-06T15:40:49.377" Score="1" Body="&lt;p&gt;Since there are no answers in the last 9 months, how about a guess? The Widget is a coring drill and an electromagnet. The electromagnet holds the drill against the bomb casing during the drilling operation. The cooling fan cools the fins around the electromagnet. The copper colored shaft is the drill head which can drill only as deep as allowed by the collar (looks like about 1 inch max.). After drilling, the Widget would be removed (looks heavy, thus the two lifting handles) and the bomb could be back filled with acid to neutralize the internal explosives, arming mechanism, etc. I would think this method might be used in an urban area where it is expedient to neutralize the bomb rather than attempting to move it to a save detonation area.&lt;/p&gt;&#xA;" OwnerUserId="24499" LastActivityDate="2020-02-06T15:40:49.377" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33111" PostTypeId="2" ParentId="33107" CreationDate="2020-02-06T17:36:02.167" Score="1" Body="&lt;p&gt;Dynamic braking is used in combination with the air brakes in the rest of the train. As such it is capable of only capturing a portion of the kinetic energy that needs to be dissipated in order to stop the entire train. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For carefully-applied braking, the &quot;jack-knifing&quot; collapse of a &quot;compressed&quot; railway car string is avoidable. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-02-06T17:36:02.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33112" PostTypeId="2" ParentId="33101" CreationDate="2020-02-06T17:40:36.973" Score="5" Body="&lt;p&gt;The usual solution is to build a fence with lots of slats in it about 20 feet away from the edge of the road. The fence slows down the airflow and thereby causes the sand in it to fall out and accumulate right behind the fence, thereby keeping most of it off the road. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since most of the sand carried by the wind scoots along close the surface of the ground, the fence does not have to be more than 4 or 5 feet tall to catch most of the sand. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the fence is about half-buried in sand, its effectiveness diminishes and the sand around it has to be plowed away. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-02-06T17:40:36.973" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33113" PostTypeId="2" ParentId="33108" CreationDate="2020-02-06T18:05:21.053" Score="0" Body="&lt;p&gt;Depending on the slenderness of your beam you may need to iterate this several times to get close to the final deflection, Y.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, we find the deflection due to lateral load F and then we apply the Pdelta moment and find the secondary shear, v and add it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deflection, Y due to force F is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Y= \frac{Fl^3/8}{3EI}=\frac{Fl^3}{24EI}  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same as the deflection of a cantilever beam with half the length subjected to F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we add &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Sigma M_{base}=0\quad W * Y = V * L$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ V = W * Y / L$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;plugging this V we get a secondary deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Y= \frac{Vl^3}{24EI}  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we add the two deflections and do the same steps again until the difference between the results is within our tolerance limit.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-02-06T18:05:21.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33114" PostTypeId="1" AcceptedAnswerId="33115" CreationDate="2020-02-06T18:25:38.863" Score="1" ViewCount="316" Body="&lt;p&gt;I was wondering why motorcycle brakes have so much more stopping power than bicycle hydraulic brakes (Than can be found on mountain bikes for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rotor has a bigger size which gives better leverage for the brake pads and thus more stopping power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;But I don't understand the following:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the moment (force times lever length) exerted on the brake lever (by the rider's hand) is the same (is this a correct assumption) and the total amount of lever travel is the same... the amount of clamping force the hydraulic brake's caliper's pistons exert on the pads must be equal (I think)(since the same amount of energy is used at the lever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is the free travel (the distance the pads have to travel before they hit the disc) much smaller on a motorcycle disc brake resulting in a shorter (free throw) of the lever which means there is more percentage of the lever travel that can be used to 'build pressure' meaning a smaller piston can be used which would give more clamping force at the pads?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The brake pads are much larger on motorcycles but then again because of the increased surface area one would have to exert more force on the bigger brake pads to get the same pressure/area. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Are my conclusions incorrect?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is the pad material a lot softer on the motorcycle brake pads?&lt;/li&gt;&#xA;&lt;li&gt;What design changes would enable higher clamping force with the same lever throw/force exerted on lever (if any design changes could accomplish this)?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Another example (light motorcycle vs heavy motorcycle):&#xA;my motorcycle (Suzuki gn250, 140 ish kg wet weight) has a hydraulic front (disc) brake. A Honda Goldwing (very heavy motorcycle , 380 kg wet weight) also has a hydraulic front (disc) brake. &lt;strong&gt;What would be the main differences that enable the Goldwing's front brake to have so much more power?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another question:&#xA;The Goldwing has 2 discs/2 calipers in the front. There is more pad material indeed but the same thought (more surface area of pads means more force needs to be exerted in total on the brake pads in order to get the same force/area on the contact patch between pads and disc surface). &#xA;&lt;strong&gt;Is even softer material used on the brake pads for 2 rotor front brake systems?&lt;/strong&gt;&#xA;Or am I missing something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some disc brakes have 2 piston calipers, some have 4 piston calipers. It is suggested that the 4 piston caliper variants have more stopping power..&#xA;Is it true that the 4 piston calipers have more stopping power in general? If so, what is the reason behind this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="10328" LastEditorUserId="10328" LastEditDate="2020-02-06T19:19:47.290" LastActivityDate="2022-12-13T14:47:13.407" Title="Hydraulic Brakes - Why motorcycle brakes more powerful than bicycle brakes?" Tags="|hydraulics|mechanical|friction|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33115" PostTypeId="2" ParentId="33114" CreationDate="2020-02-06T19:15:51.420" Score="2" Body="&lt;p&gt;Disk brake efficiency is often tied to what is called swept area. This is the surface that is covered by the brake pad as the wheel rotates. I suspected that disk brakes are more effective with a larger swept area, which is true but not for the reason one might expect. It is allowed that a larger swept area provides for greater mass to be cooling when not under the pads, but that's not the primary factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bicycles are frequently designed to be as light as possible, for reasons related to limited human power. This means smaller diameter disks, smaller pads and smaller swept area. It matters not how much power the hand can apply to the lever if there is insufficient material to accept the force. The smaller diameter means a smaller (shorter) lever arm when speaking of mechanical advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motorcycles have substantially more power and can afford to have a few more pounds of rotating weight, including doubling up the number of disks. This increases the friction area providing for more powerful braking. Accordingly, larger diameter means a longer lever arm as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reverse is true, however. A lightweight bicycle and rider need not have toss-you-over-the-handlebars braking, even though they do, when operated poorly by the rider. Motorcycles can also perform this acrobatic feat, but with the mass of a motorcycle and rider, more braking is necessary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the same reason, four pads on a disk are going to apply more braking force than two. Calipers in disk brakes operate by floating against the surface of the disk, reducing almost to zero the distance traveled by the pad when braking. This means that the majority of travel of the hydraulic system is placed into friction. Old-style drum brakes still in use on the rear wheels of some vehicles have to be adjusted to keep the shoes clear of the braking surface, &quot;eating up&quot; some of the force by traveling to contact the drum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pad softness/hardness/compostion usually means something in terms of pad life and damaging wear to the disks. A soft pad may provide great, grippy braking and wear out in half the time of a harder pad, while the super hard pads will still grip well when properly bedded, but may chew away disk material in short order, requiring expensive disk replacement. Foreign material between the pad and disk will accomplish the wear without providing braking benefits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's advised to use the manufacturer's recommendation for pads on a specific disk. When replacing a disk with a more exotic version, pad recommendation is always required.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In response to the commented question about insufficient material: Your hand can apply enough force to brake a heavy motorcycle and if you're on a bicycle, your hand does not have less force as a result. Applying more force than the bicycle brakes can accept results in something else going wrong. The calipers could be forced apart, the hydraulic tubing used could rupture, or your muscles will just hurt more than they should. I had hydraulic brakes on a tandem cycle and the response and power was wonderful and they were rim brakes, not disks. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't have an answer to the question about doubling pad area, as the physics involved in calculating such a complex system is beyond my understanding. With hydraulics, the increased area of the pistons in the calipers would mean a corresponding increase in the travel of the hand brake master cylinder.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2020-02-06T22:46:22.817" LastActivityDate="2020-02-06T22:46:22.817" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33116" PostTypeId="1" AcceptedAnswerId="33126" CreationDate="2020-02-06T19:21:42.787" Score="1" ViewCount="267" Body="&lt;p&gt;I've got a flashlight with a 3.7 V 18650 cell that charges with a 3.5 mm barrel plug.  They provided adapter that says 4.2 V +/- 0.5.  Can I use a 5 V USB to barrel plug adapter to charge it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE&#xA;So the adapter clearly says +- .5 v.  Is this an obvious misprint. It must be .05 right?&lt;/p&gt;&#xA;" OwnerUserId="5967" LastEditorUserId="5967" LastEditDate="2020-02-06T21:16:14.653" LastActivityDate="2020-02-07T17:30:32.777" Title="Can I charge 18650 Li-ion cell from 5 V USB?" Tags="|pid-control|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33117" PostTypeId="2" ParentId="33116" CreationDate="2020-02-06T19:42:55.610" Score="1" Body="&lt;p&gt;Check the current rating on the existing power adapter. The amperage it draws should not be greater than the source you are considering to use with the 5v usb to barrel plug adapter. If, for example, your existing adapter lists 300 ma and your available source is 1000 ma, you will not overload that new source. If the existing source lists 1.2 a or 1200 ma and your new source is 1000 ma, it will overheat and likely fail, possibly in a dangerous manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is typical of these small units to have charging times measured in hours, implying low current draw.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-06T19:42:55.610" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33118" PostTypeId="1" CreationDate="2020-02-06T22:09:40.810" Score="1" ViewCount="33" Body="&lt;p&gt;I am trying to derive a heat budget model to estimate the temprature of very small a stationary vertical cylinder (1mm) under environmental conditions. Imagine a wooden stick. Simply put, I have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R = H&#xA;InSW * albedo + sigma * emissivity * Tcylinder^4 = rho_air * cp * (Tcylinder - Tair)/rH.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;R = sum of radiative components&lt;/li&gt;&#xA;&lt;li&gt;H = convective components (latent heat and conduction ignore)&lt;/li&gt;&#xA;&lt;li&gt;InSW = Incident solar radiation observed by a pyranometer&lt;/li&gt;&#xA;&lt;li&gt;albedo = albedo of the cylinder&lt;/li&gt;&#xA;&lt;li&gt;sigma = ste-boltz constant&lt;/li&gt;&#xA;&lt;li&gt;emissivity = emissivity of the cylinder to longwave IR&lt;/li&gt;&#xA;&lt;li&gt;Tcylinder = temperature cylinder&lt;/li&gt;&#xA;&lt;li&gt;rho_air = density of the air&lt;/li&gt;&#xA;&lt;li&gt;cp = heat capacity of the air&lt;/li&gt;&#xA;&lt;li&gt;rH = resistance to convection&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is the basis of a heat budget model, but how do I incorporate the fact that its a cylinder? I've managed to find information about how to deal with rH with the forced and natural convections, so what I'm more concerned about is the radiative components. Is it alright for me to leave them like this or do I need to start accounting for the geometry and its potential impact on the intercepted light and resistance to radiative transfer?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="19472" LastEditorUserId="19472" LastEditDate="2020-02-06T22:15:23.687" LastActivityDate="2020-02-07T02:22:37.780" Title="How to consider the dimension of a cylinder when computing its radiative budget?" Tags="|mechanical-engineering|heat-transfer|radiation|convection|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33119" PostTypeId="2" ParentId="33116" CreationDate="2020-02-06T23:06:06.763" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;It must be .05 right?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;On a 4.2 V PSU that would be 1.2% accuracy. For a cheap power supply that would be unlikely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Lithium-based cells require careful charging and discharging. I suspect that the power supply designers used the following criteria:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Cheap.&lt;/li&gt;&#xA;&lt;li&gt;Cheap.&lt;/li&gt;&#xA;&lt;li&gt;Must charge the battery - even if it takes a long time.&lt;/li&gt;&#xA;&lt;li&gt;Mustn't cause too many to catch fire due to overcharging.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Most devices such as mobile phones use a 5 V power supply (not a &quot;charger&quot;) to power the device and the charger or charge management system is built into the phone. It will limit the charge voltage and current to safe values and terminate charging when &quot;full&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would be very careful with this. Check out &lt;a href=&quot;https://batteryuniversity.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Battery University&lt;/a&gt; for a wealth of information on all types of batteries.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2020-02-06T23:06:06.763" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33121" PostTypeId="2" ParentId="33118" CreationDate="2020-02-07T02:22:37.780" Score="2" Body="&lt;p&gt;You certainly need to take account of the shape of the object somehow.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a simple example, consider a sphere of radius &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; in space where everything except solar radiation can be ignored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cross section area of the sphere which is absorbing solar radiation is &lt;span class=&quot;math-container&quot;&gt;$\pi r^2$&lt;/span&gt; but the surface area which is emitting it is &lt;span class=&quot;math-container&quot;&gt;$4 \pi r^2$&lt;/span&gt;. So the radiative heat balance equation looks something like &lt;span class=&quot;math-container&quot;&gt;$4 \pi r^2 T_{\text{sphere}}^4 = \pi r^2 T_{\text{sun}}^4$&lt;/span&gt; and in the equilibrium state (assuming the sphere has high conductivity so the surface temperature is uniform) will be &lt;span class=&quot;math-container&quot;&gt;$T_{\text{sphere}}= T_{\text{sun}}/\sqrt 2$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a unsymmetrical body like a cylinder, the amount of radiation absorbed will depend on the orientation of the cylinder relative to the sun, but the amount emitted will not.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-02-07T02:22:37.780" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33122" PostTypeId="1" CreationDate="2020-02-07T02:47:26.363" Score="2" ViewCount="69" Body="&lt;p&gt;Average hot air balloons have a diameter of about 18 m. The largest  out there is roughly 32 m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, are there any reasons other than fuel concerns that would prevent having a larger balloon, say around 50-60 m as a lifting body? I.e. problems with air mixing, pressure differentials, etc in such a large volume? &lt;/p&gt;&#xA;" OwnerUserId="24507" LastActivityDate="2020-02-08T00:43:50.677" Title="What engineering obstacles would need to be overcometo make a hot air balloon bigger than the largest made?" Tags="|aircraft-design|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33125" PostTypeId="2" ParentId="33122" CreationDate="2020-02-07T05:41:36.547" Score="2" Body="&lt;p&gt;As the volume of a hot air balloon is increased, so increases its surface area- through which the burner heat is lost. Since the volume increases faster than the area for a spherical balloon, more volume means more lift and lower heat losses for that amount of lift. However, more volume also means more fuel burned to fill the balloon to stay aloft, and beyond a certain balloon size the burners have to be running continuously to maintain lift. That determines the practical maximum size of a hot air balloon for a standard set of burners. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This also means that in general, a balloon is designed for a certain payload specification. This sets the balloon volume and hence the fuel burn rate, and from that comes the burner capacity in BTU's. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastEditorUserId="13363" LastEditDate="2020-02-07T17:44:36.513" LastActivityDate="2020-02-07T17:44:36.513" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33126" PostTypeId="2" ParentId="33116" CreationDate="2020-02-07T09:24:36.997" Score="0" Body="&lt;p&gt;I think maybe, definitely NOT do this.  Measurement shows the provided adapter/charge controller ranges 4.0 to 4.3 v and current at 200-220 mA.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging into an un charge controlled 5 v 1A power supply immediately drives the current to 1.1 A.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Certainly leading to overheating, possibly venting and leaking, venting leaking and lithium fire, rapid venting, explosion, and then fire.  Pick your poison.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;UPDATE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the adapter is rated at 500 mA; when connected to a 1 A USB load tester it supplies only 250ma.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, someone tell me, where is the charge controller, in the flashlight or in the wal wart?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;FINAL HOPEFUL UPDATE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a disturbing development after charging at 4 v for 180 mAH the voltage across it dropped to 3.0 and it started getting hot, AND the flashlight is only intermittently shining any light at all.&lt;/p&gt;&#xA;" OwnerUserId="5967" LastEditorUserId="5967" LastEditDate="2020-02-07T17:30:32.777" LastActivityDate="2020-02-07T17:30:32.777" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33127" PostTypeId="2" ParentId="32956" CreationDate="2020-02-07T13:39:27.110" Score="0" Body="&lt;p&gt;Since asking the question I went to study the subject a little bit more. I want to share my conclusions, hoping that they could be of help to someone else.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Static buildup can't be caused by friction (or contact) between layers of the same material. They don't make a triboelectric pair. Most likely, the static is being generated by the foam slipping in the surrounding air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This brings us to the second point: try to humidify the air. That could be accomplished by spraying the area, but more technically by installing humidifiers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In addition to installing the side plate you mentioned (a good conductor, such as copper is the obvious pick, as it doesn't require any mechanical properties), it's a good idea to ensure that the chassis of the rolling machine is properly grounded (as well as those structures in which the material slips).&lt;/p&gt;&#xA;" OwnerUserId="17733" LastActivityDate="2020-02-07T13:39:27.110" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33128" PostTypeId="1" CreationDate="2020-02-07T13:53:58.383" Score="1" ViewCount="29" Body="&lt;h2&gt;Context&lt;/h2&gt;&#xA;&lt;p&gt;A plastics factory with installed power of about 500 kVA. Most of the consumption is represented by motors and resistances.&lt;/p&gt;&#xA;&lt;h2&gt;Problem&lt;/h2&gt;&#xA;&lt;p&gt;There are several (8+ a day) power interruptions, lasting usually less than 50 seconds each. There's a diesel power plant and an automatic transfer (of which they have damaged two already). Usually the plant &amp;quot;enters&amp;quot; when the power is reestablished, which doesn't solve the problems caused by the interruption and instead causes the transfer to suffer. In very unstable days, they prefer to override the transfer and have it in &amp;quot;manual&amp;quot; to avoid damage. Interruptions beyond 18 seconds require a complete stop of some of the machines and the consequent generation of considerable waste.&lt;/p&gt;&#xA;&lt;h2&gt;Proposed solution&lt;/h2&gt;&#xA;&lt;p&gt;There are large UPS (500+ kVA), often used to serve data centers and the like. Their prices range around USD 200k for a 500 kVA one. The idea is to set up the transfer to take on for larger outages (say in excess of 120 or 150 seconds) and to have the UPS as the main source for shorter ones.&lt;/p&gt;&#xA;&lt;h2&gt;Doubts and Questions&lt;/h2&gt;&#xA;&lt;p&gt;I'm not sure if the inductive nature of the load and their start peaks can be effectively handled by the UPS. I don't know if it's &lt;em&gt;robust enough&lt;/em&gt; for the task at hand (in lack of a better word). What protections should I consider for it?&lt;/p&gt;&#xA;&lt;p&gt;In one UPS specs sheet I read that it can handle 150% load for one minute (that will be 750 kVA) and 125% (625 kVA) for 10 minutes. Does that mean that it can keep the plant running for those times? What should I consider to avoid the UPS not performing to expectation?&lt;/p&gt;&#xA;&lt;p&gt;Do you see a better solution?&lt;/p&gt;&#xA;" OwnerUserId="17733" LastEditorUserId="90" LastEditDate="2022-08-17T22:38:58.387" LastActivityDate="2022-08-17T22:38:58.387" Title="Solving short electricity interruptions with an UPS" Tags="|electrical-engineering|electrical|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33129" PostTypeId="1" AcceptedAnswerId="33178" CreationDate="2020-02-07T14:34:21.220" Score="1" ViewCount="732" Body="&lt;p&gt;For comparison: in Euler's formula, we calculate the critical force such that when the column is subjected to it, it buckles. As far as I understand, the yield strenght of the material doesn't really matter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when facing eccentric load, secant formula is used and a maximum strain is calculated, instead of a critical force. Does it makes sense then to use yield strength as a maximum admissible strength? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: also, when derivating the formula for the maximum strain, one already considers the compressible strain and the bending combined. Thus, when calculating the maximum strain, I don't need to further make any considerations about their interations, correct?&lt;/p&gt;&#xA;" OwnerUserId="24383" LastEditorUserId="24383" LastEditDate="2020-02-07T14:39:28.517" LastActivityDate="2020-02-11T03:31:45.220" Title="What's the usefulness of maximum stress calculated on Secant's formula for buckling?" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33130" PostTypeId="2" ParentId="33122" CreationDate="2020-02-07T15:13:42.267" Score="2" Body="&lt;p&gt;You need to figure out how to land a large balloon in conditions with no wind, without the passengers ending up underneath the collapsed balloon.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In windy conditions, you also need a larger landing area to avoid damaging the collapsed balloon, and at some size point you will need machinery to pack it for transportation because of its own weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The largest passenger carrying balloon (as of 2017) was around 40m tall, which gives some idea how much ground area is required to operate it, remembering that balloons are not steerable so you don't have the option of landing at a purpose-built site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, the only real purpose of having a bigger envelope is to carry more load, and eventually you hit design problems trying to make a completely flexible structure like a balloon strong enough.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-02-07T15:13:42.267" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33134" PostTypeId="1" CreationDate="2020-02-07T17:20:40.297" Score="0" ViewCount="42" Body="&lt;p&gt;So I have a set of data, units purchased per month over a 4 or 5 year period.  I have about 3000 different items, but here are 2 examples of what Ive done:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Item 1: Mean 24 STDEVA 18&#xA;It has purchases practically every single month of each year.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Item 2: Mean 12 STDEVA 15&#xA;It has at least 1 or 3 months with purchases of 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do understand that stdev doesnt treat 0's differently than it does outliers.  So for example, if most numbers are 1, then a 0 now and then wont increase your std dev.  But if most numbers are 100, then a 0 now and then will increase it, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, how can I determine which items are safer to purchase (inventory) such that I wont be so affected by large stdev?  Iow, 18 of 24 is still a pretty big number.  Its telling me that 68% of all values in that set are +/-18 units away from 24.  That could be as low as 6 or as high as 42.  On the other hand, 15 of 12 is just as bad.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How should I use the std dev to determine my safety level of purchases for a given year?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set 1: 14,80,14,52,12,71,23,12,32,9,14,28,2,9,13,74,25,62,18,58,19,21,14,11,6,25,37,1,46,48,83,29,8,26,50,15,26,66,19,65,93,4,16,98,62,24,0,0,6,5,33,0,1,70,0&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Set 2:&#xA;4,4,13,30,22,50,16,4,60,16,31,35,10,10,10,8,10,23,0,3,18,0,10,8,37,73,2,11,9,18,4,0,3,11,19,3,7,0,3,0,0,2,0,4,4,14,11,0,0,0,0,4,0,0,0&lt;/p&gt;&#xA;" OwnerUserId="14955" LastActivityDate="2020-02-07T20:42:42.947" Title="Interpreting Std Deviation" Tags="|mathematics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33135" PostTypeId="1" CreationDate="2020-02-07T18:19:29.473" Score="1" ViewCount="398" Body="&lt;p&gt;&quot;Candlesticks&quot; for this question are really just a stand-in for axisymmetric silver, brass, etc structures, possibly hollow and made of thin sheet material?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IBh4d.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IBh4d.jpg&quot; alt=&quot;Candlestick&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xdcu0.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xdcu0.jpg&quot; alt=&quot;Candlesticks&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RRSCB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RRSCB.jpg&quot; alt=&quot;Chamberstick&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my research, today we would use metal spinning for sheet material, possibly casting, or turning with a lathe from solid stock are also possible. Hammer forming has certainly been around a long time. But in the late 1700's and early 1800's, it seems like metal spinning was not in common use, and these candlesticks are so ornate that they couldn't have been done only on a lathe. What techniques were commonly used? Where can I learn more about this type of work? &lt;/p&gt;&#xA;" OwnerUserId="18257" LastActivityDate="2020-02-07T22:05:53.473" Title="How were these candlesticks made in the 1800s?" Tags="|metals|engineering-history|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33136" PostTypeId="2" ParentId="33135" CreationDate="2020-02-07T19:49:32.313" Score="0" Body="&lt;p&gt;Because I'm actively researching this topic, I'll put all of my thoughts in this temporary answer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The crafts of &lt;strong&gt;goldsmithing,&lt;/strong&gt; &lt;strong&gt;silversmithing&lt;/strong&gt;, and &lt;strong&gt;brownsmithing&lt;/strong&gt; (coppersmithing) would be closely related to the manufacture of these sorts of implements. Techniques involved (and still found today) include:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rolling (manufacture of sheets)&lt;/li&gt;&#xA;&lt;li&gt;Wire-drawing (manufacture of wires)&lt;/li&gt;&#xA;&lt;li&gt;Wrought work (forming)&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Raising&lt;/li&gt;&#xA;&lt;li&gt;Planishing&lt;/li&gt;&#xA;&lt;li&gt;Repousse&lt;/li&gt;&#xA;&lt;li&gt;Chasing&lt;/li&gt;&#xA;&lt;li&gt;Embossing&lt;/li&gt;&#xA;&lt;li&gt;Stamping&lt;/li&gt;&#xA;&lt;li&gt;Pressing&lt;/li&gt;&#xA;&lt;li&gt;Soldering&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;Casting&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;As seen &lt;a href=&quot;https://www.history.org/foundation/journal/Winter03-04/foundry.cfm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, &lt;strong&gt;candlesticks can conventionally be made by casting in a foundry&lt;/strong&gt;. They can be made by hammering techniques as well if the design is simple enough. They are often made in separate halves, which are then soldered together and polished.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very high quality implements can be made by raising and soldering alone. This is shown by videos such as &lt;a href=&quot;https://www.youtube.com/watch?v=hJOx7c_qwys&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;. From the same channel, silver pieces made in parts are also shown, &lt;a href=&quot;https://www.youtube.com/watch?v=-YrskkXJJuc&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this&lt;/a&gt;. Other techniques in silversmithing that are still technically within the 1800's timeframe can be found in Gee's &lt;em&gt;The Silversmith's Handbook&lt;/em&gt; (1886). Here, more specific tools such as the sparrow-hawk, and the snarling tool, are described. Other techniques mentioned include swaging, dooming, spinning, and stamping, which is cited as coming into general use around 1851. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fastening of elements could be accomplished by e.g. soldering, riveting, pinning, etc, and threaded fasteners were mass produced beginning in the early 1800's. These methods have not changed dramatically. This is relevant as e.g. a rivet appears to be used in the third &quot;chamberstick&quot; shown in the question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When classical methods were inconvenient, or when mass production is needed, hand-tools are replaced by machining tools. Lathes powered by steam, water wheels, or even horses, were available and used for turning as one would expect. Metal spinning would be a particularly appealing choice, which grew in popularity towards the end of the 19th century; very nice implements can be made with this method to obtain common axisymmetric shapes. A great example of silver candlesticks being made with spinning in a simple way &lt;a href=&quot;https://www.youtube.com/watch?v=4cjLFKXD8OA&quot; rel=&quot;nofollow noreferrer&quot;&gt;is shown here&lt;/a&gt;. &lt;/p&gt;&#xA;" OwnerUserId="18257" LastActivityDate="2020-02-07T19:49:32.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33138" PostTypeId="2" ParentId="33134" CreationDate="2020-02-07T20:42:42.947" Score="2" Body="&lt;p&gt;The two data illustrations both have very large s.d.'s relative to the mean. Therefore the upper predicted limit for future data is very large. This single method of analysis does not seem to be adequate for your purpose. Perhaps the data needs to be tied to months of the year, or to your frequency of advertising, or to the seasons, or to whatever is driving the large variation. &lt;/p&gt;&#xA;" OwnerUserId="24517" LastActivityDate="2020-02-07T20:42:42.947" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33139" PostTypeId="2" ParentId="33135" CreationDate="2020-02-07T22:05:53.473" Score="2" Body="&lt;p&gt;A couple casting techniques not mentioned were &quot;cores&quot; and &quot;lost wax&quot; , both are old. A sand core inside the casting would make it hollow ; for copper alloys and silver I think a 1/4&quot; wall thickness is likely. And for lost wax , the desired item is made in wax , a ceramic cement is poured around it , solidified , then put into furnace to melt and burn out the wax. Then pour molten metal into the hollow. I would add that threaded fastener manufacturer has changed dramatically in about 50 years with the advent of rolled threads and gas carburizing furnaces. As you say spinning certainly looks attractive : For the low power needed for thin copper and silver alloys , could a foot powered lathe ( where it rotates alternately clockwise and counter-clockwise) be possible for spinning ? &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-02-07T22:05:53.473" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33140" PostTypeId="2" ParentId="33122" CreationDate="2020-02-07T22:31:09.607" Score="2" Body="&lt;p&gt;What you need to do is look closely at the physics of the inflated envelope and how the shape is determined. Let's limit the variables and hold the temperature differential constant as we make the balloon bigger. As the envelope height dimension grows, the pressure differential between the inside and outside grows. So that if the envelope tension is the same, the Gaussian curvature will increase. That's not what you want, you want the curvature to decrease in order to make the balloon bigger. So now you have to &lt;s&gt;decrease&lt;/s&gt; increase the envelope tension in proportion to the square of the size change in order to maintain dimensional similitude. You want most of the envelope tension to come from the payload support tendons, and little from the hoop tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So you have to solve a fairly complicated but reasonably well conditioned set of equations that account for -&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Envelope weight (lbs/sqft) as a function of size and temperature difference. &lt;/li&gt;&#xA;&lt;li&gt;Envelope shape as a function of construction details, payload, temperature, and size.&lt;/li&gt;&#xA;&lt;li&gt;Envelope thermal performance wrt fuel system cost and weight.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;You can definitely set this up in Excel and use the add-in solver to optimize the envelope parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As sanity check, if we start with a balloon with 20 tendons at 25# tension each, and double the linear scale, we now need 40 tendons to maintain panel size, and need to tension them at 100# to keep similarity with the original. Thus lift (total tension) goes with volume, as we would expect for similar shapes.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2020-02-08T00:43:50.677" LastActivityDate="2020-02-08T00:43:50.677" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33141" PostTypeId="1" AcceptedAnswerId="33142" CreationDate="2020-02-07T23:34:00.790" Score="1" ViewCount="342" Body="&lt;p&gt;I am making a project using 10mm aluminum extrusions. I am trying to connect all three extrusions like the picture below shows and I am going to make my own bracket (in red) out of aluminum since the variety of choices is very scarce for this 10mm size, but I will include a picture of something similar so you can get an idea of what I'm making. I am wondering which arrangement is stronger and more rigid- the left one, or right one? Or is there an even better way to do it? &lt;a href=&quot;https://i.stack.imgur.com/vdQv3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vdQv3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/yKgvV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yKgvV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8974" LastEditorUserId="8974" LastEditDate="2020-02-07T23:42:24.377" LastActivityDate="2020-02-09T00:02:30.223" Title="What is the strongest way to fasten 8020 extrusions in this arrangement?" Tags="|structural-engineering|beam|building-design|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33142" PostTypeId="2" ParentId="33141" CreationDate="2020-02-08T00:15:48.260" Score="2" Body="&lt;p&gt;If you consider the intersecting areas for the left side extrusions as being more evenly distributed, you should be able to expect that the load distribution will also be more even in that configuration. What's equally important is that you require equal strength over the three pieces. In the right side image, if the strength required on the angled portion is less, you can expect similar distribution of forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gusset method you are considering is a common joining mechanism for extrusions primarily because of its ease and strength. You can increase the strength incrementally by adding straps on the inside of the angles attached with the channel nuts.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-08T00:15:48.260" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33145" PostTypeId="1" CreationDate="2020-02-08T15:11:41.213" Score="3" ViewCount="107" Body="&lt;p&gt;I am trying to implement a simple LQR controller in MATLAB for a purely deterministic system. The code is shown below:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;%% Continuous Time&#xA;clear all; close all; clc;&#xA;% Parameters&#xA;n = 2;&#xA;m = 1;&#xA;A = [1 1; 0 1];&#xA;B = [0.5; 1];&#xA;C = [1 0];&#xA;Q = eye(2);&#xA;QT = Q;&#xA;R = 10;&#xA;x0 = [1; 0];&#xA;T = 10;&#xA;N = 50;&#xA;&#xA;% Backwards recursion for P(t), t = T -&amp;gt; 0&#xA;P0 = QT;&#xA;[tout,Pvecout] = ode45(@(t,Pvec) mRiccati(t,Pvec,A,B,Q,R,n),[0 T], P0(:));&#xA;Pvecout = flip(Pvecout);&#xA;&#xA;% Flip (and interpolate) to get P(t), t = 0 -&amp;gt; T&#xA;t = linspace(0,T,N);&#xA;deltat = t(end) - t(end-1);&#xA;P = [];&#xA;for i = 1:N&#xA;    Pvecoutk = interp1(tout,Pvecout,t(i));&#xA;    P(:,:,i) = reshape(Pvecoutk,n,n);&#xA;end&#xA;&#xA;% Simulate dynamics with (time-varying) LQR controller&#xA;x = x0;&#xA;xk = x0;&#xA;u = [];&#xA;% Linearized dynamics (!)&#xA;Ak = eye(2) + deltat * A;&#xA;Bk = deltat * B;&#xA;for i = 1:N-1&#xA;    Kk = inv(R) * B' * P(:,:,i);&#xA;    uk = -Kk * xk;&#xA;    xk = Ak * xk + Bk * uk;&#xA;    x = [x, xk];&#xA;    u = [u, uk];&#xA;end&#xA;&#xA;% Plot results&#xA;figure; hold on; grid on;&#xA;plot(t,x(1,:)); plot(t,x(2,:));&#xA;figure; hold on; grid on;&#xA;plot(t(1:end-1),u);&#xA;&#xA;%% Useful Functions&#xA;function dPdtvec = mRiccati(t,Pvec,A,B,Q,R,n)&#xA;    P = reshape(Pvec,n,n);&#xA;    Pdot = (P * A + A' * P - P * B * inv(R) * B' * P + Q);&#xA;    dPdtvec = Pdot(:);&#xA;end&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Instead of solving the discrete time Riccati Equation (which I have already implemented), I am trying to do everything in continuous time to get the solution. Thus, as you can see at the bottom, I have a function that computes the matrix Riccati ODE, and this gets fed into MATLAB's od45 solver, with the usual initial condition &lt;span class=&quot;math-container&quot;&gt;$P(T) = Q_N$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$Q_N$&lt;/span&gt; is the terminal cost matrix. This gives me &lt;span class=&quot;math-container&quot;&gt;$P(t)$&lt;/span&gt;, which checks out well. The problem is that when I go and simulate the dynamics, they diverge for some reason. I am not sure why. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interestingly enough, if I use the the &lt;em&gt;constant&lt;/em&gt;, steady-state gain &lt;span class=&quot;math-container&quot;&gt;$K_{\infty} = R^{-1}B^\intercal P(0)$&lt;/span&gt;, then everything works out nicely, but if I use time-varying gains (which are more accurate!), the solution doesn't work out. I think it may be due to some kind of discretization errors when simulating the dynamics, but I really don't know. Feel free to copy and past the script; any thoughts would be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="15049" LastActivityDate="2020-02-08T15:11:41.213" Title="LQR Implementation in MATLAB" Tags="|control-engineering|control-theory|matlab|optimal-control|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33146" PostTypeId="1" CreationDate="2020-02-08T16:27:45.217" Score="3" ViewCount="102" Body="&lt;p&gt;I want to control the temperature of a steel mold about 300 mm square by 1000 mm high. The mold has ten electic cartridge heaters installed in it. There are ten corresponding thermo-couples, one near each heater. Each heater/thermocouple pair is a zone. The simple solution is to use ten PID loops, one for each zone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that the zones interact, and heat from one zone's heater heats the other zones too. The closer the zones are to each other the more they affect each other. The output of each PID loop will be a disturbance to each other PID zone. The specification is to control each zone to &lt;span class=&quot;math-container&quot;&gt;$\pm 1 {^\circ} C$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A control system using proportional controllers (P term only) in each zone was tried. The temperature of each zone oscillated, far exceeding the &lt;span class=&quot;math-container&quot;&gt;$\pm 1 {^\circ} C$&lt;/span&gt; requirement. The system would never settle. This isn't suprising with strict proportional control. I expect that a PID control would work better, but I don't know what the interactions will do.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will multiple interacting PID control zones stabilize? Is there a better approach to controlling the system?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: All zone set-points are the same temperature. They all need follow a linearly increasing temperature profile from &lt;span class=&quot;math-container&quot;&gt;$20^\circ C$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$180^\circ C$&lt;/span&gt; over twenty hours. The internal cavity of the mold has a very irregular shape.&lt;/p&gt;&#xA;" OwnerUserId="24525" LastEditorUserId="24822" LastEditDate="2020-03-02T12:29:08.477" LastActivityDate="2020-03-02T12:29:08.477" Title="Temperature control with interacting PID loops" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33147" PostTypeId="2" ParentId="8619" CreationDate="2020-02-08T18:04:44.500" Score="0" Body="&lt;p&gt;I was looking at this site to try and figure out how a syphon might break other than by an air leak.  However I might like to make a suggestion!  I use a siphons to do a trickle water drain from my aquarium.  In my refugium we have a toilet type bulb float that kicks on when the water gets too low.  On the other side of the refugium I have 2 wells (one inside the tank and one outside the tank) with a syphon between them...  the well on the inside of the tank has holes in it so water from the refugium can flow into it.  They are at the lowest level I want the water in the refugium to reach.  Outside the tank I have a well that has one drain hole with a tube inserted that is lower than the lowest water level in the refugium.  I hope you are following this ok.  What happens is water trickles out of the refugium into the inside well.  The syphon carries water to the outside well (which never runs dry).  The outside well drains water to the drain but only can do it until water reaches the level of the water of the hole in the outside well.  I have a shut off valve in the final drain to be able to stop the whole process!  My issue has been somehow the syphon gets broken from the turn off valve to the drain.  But I think I have solved that issue by reading this! I can send pictures if it will help!  There are no pumps involved at all!  Only gravity and strategic placements!&lt;/p&gt;&#xA;" OwnerUserId="24526" LastActivityDate="2020-02-08T18:04:44.500" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33149" PostTypeId="1" CreationDate="2020-02-08T22:01:06.603" Score="1" ViewCount="87" Body="&lt;p&gt;Comparing the official fuel efficiency figures for several cars, I was surprised to discover that the &lt;a href=&quot;https://en.wikipedia.org/wiki/Smart_Fortwo&quot; rel=&quot;nofollow noreferrer&quot;&gt;Smart Fortwo&lt;/a&gt; has a weight between 750 to 800 kg and needs 4.3 L/100 km, but the &lt;a href=&quot;https://en.wikipedia.org/wiki/Volkswagen_Polo&quot; rel=&quot;nofollow noreferrer&quot;&gt;VW Polo&lt;/a&gt; with a weight of 1650 kg reaches 3.8 L/100 km. That is, the Smart has half the weight, but still a worse fuel efficiency. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When driving up and downhill (but not on a flat surface) does one cancel the other? That is, driving uphill with more weight costs more fuel, but you need less fuel when going downhill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are the major fuel efficiency improvements rather on other places, like better tires, better injection, better streamlining and hence lower weight cannot outweigh poor performance in these fields?&lt;/p&gt;&#xA;" OwnerUserId="5534" LastActivityDate="2020-07-09T12:29:22.330" Title="How much weight has weight in fuel efficiency?" Tags="|energy-efficiency|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33150" PostTypeId="2" ParentId="33149" CreationDate="2020-02-08T22:15:34.837" Score="0" Body="&lt;p&gt;Driving downhill or breakings only generate energy in cars that have regenerative systems, battery, transducer, electrical motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weight may not be as big a factor in fuel consumption as the other factors such as a streamlined body, efficient drivetrain, and better tire design.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Weight to engine power ratio counts mostly in allowing faster acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Sigma f=0 \quad F_{engine}- F{friction} = ma \quad a= \frac{F_{engine}- F{friction}}{m}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means same engine will produce more acceleration in the lighter cae.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-02-09T01:59:53.373" LastActivityDate="2020-02-09T01:59:53.373" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33151" PostTypeId="2" ParentId="33141" CreationDate="2020-02-09T00:02:30.223" Score="2" Body="&lt;p&gt;I would say the configuration on the right would be the best if you cut the top (vertical) extrusion at a 45 degree angle at the bottom where it meets the &quot;45 degree strut&quot;. This way all the struts will be flush mounted together. Then you can add side straps and joining plates have the same geometry as the strut-intersection area. &lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2020-02-09T00:02:30.223" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33152" PostTypeId="1" CreationDate="2020-02-09T00:30:08.283" Score="-2" ViewCount="49" Body="&lt;p&gt;I'm doing an article on a company whose Title V Operating Permit is currently under review by state regulators. (See biographical note below.) See also my specific questions, numbered below. Most important are #4 &amp;amp; #7 &amp;amp; #8.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm trying to arrive at some measure of the company's baghouse filtration efficiency. The company melts heavy metals such as Cd, Cr, As &amp;amp; Pb in a batch heated in a 2500F furnace. The company and state regulators agree that 98% of the Cr that is emitted is Cr(VI).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There were several stack emission tests done for various HAPs; I'm focused on Cr. There are two issues at hand that may have underestimated the facility's Cr emissions for different products on an on-going basis. One is the fact that, violating the state's instructions, the company made a test product that used only ~10% of the Cr used in its most Cr-heavy product. That's one way the emissions test underestimates the company's potential actual production emissions of Cr going forward. It makes products colored with ten-times more Cr.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second problem is that the emissions test used chromium oxide (Cr2O3), a very refractory type of Cr that is subject to forming unmelted &quot;stones&quot; in the batch. Thus, some significant amount of the Cr neither entered into solution nor was potentially vaporized and off-gassed off the melt to enter baghouse filtration -- some small portion of that to be emitted to ambient air. The company also uses sodium dichromate (Na2Cr2O7) as a coloring agent. In fact, it had many hundreds of pounds of Na2Cr2O7. on-site as reflected in state Fire Marshal data. But instead of using the far less refractory and more soluble Na2Cr2O7 for the test melt, it used Cr2O3. The fact that Cr2O3 remains 'inert,' in effect, in the unmelted stones, also potentially underestimates the amount of Cr emitted when Na2Cr2O7 is melted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such was the test-melt that yielded stack emissions data. Through the application of U.S. EPA's AEROMOD modler to arrive at dispersion levels, such low-ball emission data may have yielded an artificially high chromium daily use allowance. And, in fact, the company has already received permission to use more than 350 lb/day of chrome. Which seems like a lot for the single furnace that uses most of the Cr. (Should the company protest that it won't use 350 lb/day, note that its formal request to regulators was for permission to use almost 600 lb/day of chrome. And, to repeat, it's agreed that any emissions are Cr(VI), a rampant carcinogen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such is the context in which I query the good experts of Stack Exchange. Here's my questions: Is it possible, given the data in the public documents I link to below, to arrive at some estimate of the company's baghouse filtration efficiency?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The documents linked just below include the source test report, the dispersion modeling report and especially the company's Usage Rate Request -- that one is DOCUMENT A. Does that data permit you to arrive at a viable estimate for its filtration system's efficiency rating?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 'semiotics' of everything I've encountered in my past month of research suggest that the filters are at most MERV 16. And MERV 16 filtration would not reassure neigbors justly concerned about Cr6, Cd, Pb &amp;amp; As, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that the documents below are all public documents. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Cr test used 8.05 lbs of pure chrome, Cr2O3.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to the chrome-use request (DOCUMENT A), the lb CR VI emitted/lb Cr used = 7.59 X 10-6 lb/per lb used. It would seem we could multiply that number by 8.05 for the total amount of Cr6 emitted in the test. CORRECT?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOCUMENT A: See the consultant's request for allowable chrome use here -- &lt;a href=&quot;https://www.oregon.gov/deq/FilterDocs/BullseyeChromiumUsageTechMemo.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;&lt;em&gt;see here, page 3&lt;/em&gt;&lt;/strong&gt;&lt;/a&gt; for the equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOCUMENT B: Here's DEQ's review of the source test -- &lt;a href=&quot;https://www.oregon.gov/deq/FilterDocs/BullseyeGlassTestReportReview2017.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;&lt;em&gt;see here, Cr data, page 2.&lt;/em&gt;&lt;/strong&gt;&lt;/a&gt; Page 3 contains data on baghouse pressure drop and the like.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOCUMENT C: Here's DEQ's discussion of the air modeling report &lt;a href=&quot;https://www.oregon.gov/deq/FilterDocs/BullseyeModelingReview.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;&lt;em&gt;-- see here.&lt;/em&gt;&lt;/strong&gt;[3]&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;DOCUMENT D: Here's the Title V Operating Permit application &lt;a href=&quot;https://www.oregon.gov/deq/FilterDocs/bullseyeTitleVapp.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;strong&gt;&lt;em&gt;-- see here&lt;/em&gt;&lt;/strong&gt;&lt;/a&gt;. It's worth noting that on page 17, the company states the Efficiency Rating for Baghouse West (which controls 18 separate 2500F furnaces) is &quot;99+%.&quot; To my understanding, that statement doesn't really work. That is, a MERV 16 filter is 95% effective, perhaps a bit more. And the next step up, a HEPA filter is 99.97%. So 99+% has no real standing with any group establishing efficiency standards.And, again, at a public meeting and in documents, the company has never claimed HEPA filters on its baghouse. Had it spent to the money for HEPA, it certainly would have been shouting it from the rooftops. Also see pgs 75 to 77 of DOCUMENT D linked in this paragraph for more hard data.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given the data in these documents, can you state (within some range) the facility's baghouse filtration efficiency rating?&lt;/li&gt;&#xA;&lt;li&gt;The permit application -- DOCUMENT D above, pg. 17 -- notes a 99+% Efficiency Rating. But, as has been pointed out to me, in the absence of sizing of the PM to be filtered, is that sort of a meaningless statement? (After all, the cheapest filter can have 99+% rating if it's asked to stop large rocks or some such.) So does that &quot;99+%&quot; rating have no real meaning?&lt;/li&gt;&#xA;&lt;li&gt;What is the aerodynamic diameter -- approximately -- of a chromium particle off-gassing off a 2500F glass-melt? A range? An estimate of the size? Way down in nanometers?&lt;/li&gt;&#xA;&lt;li&gt;Or is question #3 the wrong question since the chromium coming off the melt will be vaporized into a gas? Will the vapor phase gaseous Cr pass through the baghouse and be emitted into the air?&lt;/li&gt;&#xA;&lt;li&gt;Or is the system designed so that the gaseous Cr will condense to particulate form?&lt;/li&gt;&#xA;&lt;li&gt;If so, can we estimate the size of the PM? A few dozen nanometers perhaps?&lt;/li&gt;&#xA;&lt;li&gt;If many pounds of Cr are used per day, according to Fire Marshal data, a lot of it will be Na2Cr2O7, which is more volatile and less refractory than the Cr2O3 used in the test melt. Either way, do nearby receptors, of which there are many, including schools and day-care centers, have reason for concern? (Almost entirely all the Cr emitted is CrVI.)&lt;/li&gt;&#xA;&lt;li&gt;How much would it cost to boost this baghouse filtration system up to a HEPA level?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="24527" LastEditorUserId="1832" LastEditDate="2020-02-10T00:58:56.200" LastActivityDate="2020-02-10T00:58:56.200" Title="Public data re possible estimate of baghouse efficiency. Plus characteristics of chromium emissions. Thanks!" Tags="|mechanical-engineering|manufacturing-engineering|safety|environmental-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33153" PostTypeId="1" CreationDate="2020-02-09T04:42:00.930" Score="1" ViewCount="690" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/utAmU.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/utAmU.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking for a software in which I can make images of this kind.&lt;/p&gt;&#xA;" OwnerUserId="21424" LastActivityDate="2020-02-11T17:03:45.883" Title="In what (free) software can I draw bodies in a 3D coordinate system?" Tags="|mechanical-engineering|software|drawings|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33154" PostTypeId="2" ParentId="33152" CreationDate="2020-02-09T06:30:31.757" Score="0" Body="&lt;p&gt;The only way to determine baghouse efficiency would be a particulate sample at the inlet and outlet.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-02-09T06:30:31.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33157" PostTypeId="2" ParentId="33153" CreationDate="2020-02-09T11:58:10.333" Score="2" Body="&lt;p&gt;For Windows, Linux and Mac, &lt;a href=&quot;http://solvespace.com/index.pl&quot; rel=&quot;nofollow noreferrer&quot;&gt;Solvespace&lt;/a&gt; will allow one to create three dimensional objects in a parametric as well as free-form manner. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tVflT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tVflT.png&quot; alt=&quot;solvespace example image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Text and dimensions are part of the program's features. The web site provides tutorials and other useful reference material.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you did not have the requirement for dimensions, &lt;a href=&quot;http://www.openscad.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;OpenSCAD&lt;/a&gt; would have been a good choice. There may be a library for OpenSCAD which supports dimensioning, but I doubt the arrows would be part of that option.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-09T11:58:10.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33158" PostTypeId="2" ParentId="33146" CreationDate="2020-02-09T13:59:19.940" Score="2" Body="&lt;p&gt;What you need is call a multi input multi output (MIMO) state space controller. There's lot of tutorials on this online if you search. Short summary is that you'll build several matrices that describe the influence of each heater on every sensor. E.g. Response at sensor 3 due to a step change in power at heater 7. From this matrix your controller will be able to set the power to all heaters in order to reduce error at all sensors simultaneously &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2020-02-09T13:59:19.940" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33160" PostTypeId="1" CreationDate="2020-02-09T16:25:46.277" Score="2" ViewCount="55" Body="&lt;p&gt;A &lt;a href=&quot;https://www.youtube.com/watch?v=1UsdId5Oddk&quot; rel=&quot;nofollow noreferrer&quot;&gt;flame arrester&lt;/a&gt; is basically a heat exchanger in front of the tubulure. It conducts the heat of flame to some other place (to the liquid and/or the container especially if the container is built from a metal material), thereby extinguishes the flame before getting into the container. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I built a simple flame arrester with a tea strainer in front of the bottle and dumped some drops of denaturated alcohol into the fire. A controlled experiment demonstrated that the flame arrester works and if there is no flame arrester, an explosion is inevitable. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I don't think I should use this approach in a real-life situation because I can't know how much flame the arrester could arrest in a real case. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How can I test a proper flame arrester if it is produced correctly? &lt;/p&gt;&#xA;" OwnerUserId="20275" LastActivityDate="2020-02-09T21:54:36.610" Title="What is the test procedures for a flame arrester?" Tags="|safety|fire|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33161" PostTypeId="1" CreationDate="2020-02-09T18:51:36.827" Score="3" ViewCount="64" Body="&lt;p&gt;I have a square panel of 100 LEDs (a series of LED strips).  I would like to use some sort of optics setup to make this panel emit light that looks like it's from a single source at infinity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible with a diffuser / fresnel lens type setup?&lt;/p&gt;&#xA;" OwnerUserId="24512" LastActivityDate="2020-02-15T20:42:28.173" Title="How to make a panel of LEDs appear to be a single source infinitely far away" Tags="|optics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33164" PostTypeId="2" ParentId="33161" CreationDate="2020-02-09T22:55:25.610" Score="1" Body="&lt;p&gt;Any image or object placef at the focal point of a convex lens will by projected out to infinity on a cylindrical beam of light. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using an objective convex lens to focus the LED lights on the focal point of the projection lens will do that.  the smaller this objective image the closer the outgoing beam  of light to light coming from a single source far away.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-02-09T22:55:25.610" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33166" PostTypeId="1" CreationDate="2020-02-10T09:07:41.877" Score="2" ViewCount="299" Body="&lt;p&gt;I calculated the flow around an object using OpenFOAM's &lt;code&gt;simpleFoam&lt;/code&gt; and wanted to evaluate it's drag coefficent. Actually the drag coefficient is in a reasonable range for the object so I thought the simulation went well. However when I looked at the flow fields they appear to have some strange diagonal artefacts in them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mesh I used looks like this where the flow is coming from the left. It is refined near the object&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LXV3H.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LXV3H.png&quot; alt=&quot;Mesh&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The resulting pressure field looks for some reason like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DqGDJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DqGDJ.png&quot; alt=&quot;Resulting pressure field&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the velocity field like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NZUQ5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NZUQ5.png&quot; alt=&quot;Resulting velocity field&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flow is coming from the left with a uniform velocity of ~0.03. This corresponds roughly to a Reynolds number of 800 for the object. The calculation was done in laminar mode and the following interpolating schemes where used:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;ddtSchemes&#xA;{&#xA;    default         steadyState;&#xA;}&#xA;&#xA;gradSchemes&#xA;{&#xA;    default         Gauss linear;&#xA;    grad(U)         cellLimited Gauss linear 1;&#xA;}&#xA;&#xA;divSchemes&#xA;{&#xA;    default         none;&#xA;    div(phi,U)      bounded Gauss linearUpwindV grad(U);&#xA;    div(phi,k)      bounded Gauss upwind;&#xA;    div(phi,epsilon) bounded Gauss upwind;&#xA;    div(phi,omega)  bounded Gauss upwind;&#xA;    div((nuEff*dev2(T(grad(U))))) Gauss linear;&#xA;}&#xA;&#xA;laplacianSchemes&#xA;{&#xA;    default         Gauss linear corrected;&#xA;}&#xA;&#xA;interpolationSchemes&#xA;{&#xA;    default         linear;&#xA;}&#xA;&#xA;snGradSchemes&#xA;{&#xA;    // Korrigiert für Non-Orthogonalität&#xA;    default         corrected;&#xA;}&#xA;&#xA;wallDist&#xA;{&#xA;    method meshWave;&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;For the solvers the following set-up was used:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;solvers&#xA;{&#xA;    p&#xA;    {&#xA;        solver          GAMG;&#xA;        tolerance       1e-8;&#xA;        relTol          0.01;&#xA;        smoother        GaussSeidel;&#xA;    }&#xA;&#xA;    Phi&#xA;    {&#xA;        $p;&#xA;    }&#xA;&#xA;    U&#xA;    {&#xA;        solver          smoothSolver;&#xA;        smoother        GaussSeidel;&#xA;        tolerance       1e-8;&#xA;        relTol          0.1;&#xA;        nSweeps         1;&#xA;    }&#xA;&#xA;    k&#xA;    {&#xA;        solver          smoothSolver;&#xA;        smoother        GaussSeidel;&#xA;        tolerance       1e-8;&#xA;        relTol          0.1;&#xA;        nSweeps         1;&#xA;    }&#xA;&#xA;    epsilon&#xA;    {&#xA;          solver          smoothSolver;&#xA;          smoother        GaussSeidel;&#xA;          tolerance       1e-08;&#xA;          relTol          0.1;&#xA;            nSweeps         1;&#xA;    }&#xA;&#xA;    omega&#xA;    {&#xA;        solver          smoothSolver;&#xA;        smoother        GaussSeidel;&#xA;        tolerance       1e-8;&#xA;        relTol          0.1;&#xA;        nSweeps         1;&#xA;    }&#xA;}&#xA;&#xA;SIMPLE&#xA;{&#xA;    nNonOrthogonalCorrectors 0;&#xA;    consistent      yes;&#xA;}&#xA;&#xA;potentialFlow&#xA;{&#xA;    nNonOrthogonalCorrectors 10;&#xA;}&#xA;&#xA;relaxationFactors&#xA;{&#xA;    equations&#xA;    {&#xA;        U               0.9;&#xA;        k               0.7;&#xA;        epsilon         0.7;&#xA;        omega           0.7;&#xA;    }&#xA;    fields&#xA;    {&#xA;        p               0.8;&#xA;    }&#xA;}&#xA;&#xA;cache&#xA;{&#xA;    grad(U);&#xA;}&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;What could cause these strange diagonal patterns?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As suggested by the comments, I increased the distance between the object and the surrounding walls. However I still got the same issue with the diagonal fragments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fkhLf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fkhLf.png&quot; alt=&quot;Velocity field with distant walls&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5Afaa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5Afaa.png&quot; alt=&quot;Pressure field with distant walls&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10513" LastEditorUserId="10513" LastEditDate="2020-02-16T00:31:28.263" LastActivityDate="2020-02-16T00:31:28.263" Title="OpenFOAM simpleFoam calculation of flow around object leads to strange result" Tags="|cfd|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33167" PostTypeId="1" CreationDate="2020-02-10T11:08:29.527" Score="3" ViewCount="27" Body="&lt;p&gt;I am new here, and I'm new to Arduino. I am making an attempt to control a high voltage (~48V), high current (~6A), bipolar (4 Wires) stepper motor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand how to drive a stepper motor, and I have found the bipolar schematics on the Arduino library, but none of the solutions handle a motor this big.  Most drivers use the SN754410 or L293 from what I've seen.  I have also noticed stacking of the SN754410 chips for higher current ratings, but I suppose I would need 6 of them, and they still will not handle 48V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would I have to build my own dual h-bridge from NPN Transistors or &lt;a href=&quot;https://famaga.ch/catalog/behlke/order-109737&quot; rel=&quot;nofollow noreferrer&quot;&gt;MOSFETS&lt;/a&gt;?  Does anyone have an example circuit that has some protective features, such as the back current diodes.  Would this be enough protection to avoid damage to the Arduino?  I would also like to keep this relatively cost efficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!  I hope I made everything clear.&lt;/p&gt;&#xA;" OwnerUserId="24546" LastActivityDate="2020-02-10T11:08:29.527" Title="High Voltage, High Current, Bipolar (4 Wires) Stepper Motor Control" Tags="|stepper-motor|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33168" PostTypeId="1" CreationDate="2020-02-10T14:47:12.950" Score="4" ViewCount="2243" Body="&lt;p&gt;I wished to know the composition of Arizona 1gal plastic jugs that I have.&#xA;The bottom is stamped PP COMPATIBLE and has a #7 OTHER RIC code.&lt;/p&gt;&#xA;&lt;p&gt;I understand PP recyclable would be #5 triangle.  #7 means it doesn't fit in the plastic recyclable categories 1-6.&lt;/p&gt;&#xA;&lt;p&gt;They're great bottles and would be nice to reuse for water storage and so forth.  Therefore the inner layer composition, safety and leaching become my concern.&lt;/p&gt;&#xA;&lt;p&gt;It feels and looks like PP, so my guess is it's #7 because of the (polystyrene?) label that is glued right on.&lt;/p&gt;&#xA;&lt;p&gt;On 1/13/2020, I wrote to hello@drinkarizona.com, which was the inquiry address given at their Contact Us webpage.  Below is the letter&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Hi,&lt;br&gt;&#xA;Can you tell me the composition of your #7 bottle such as the gallon&#xA;ice tea comes in? Your FAQ page seems to imply it's a special grade of&#xA;PP, at least in part..&lt;br&gt;&#xA;&lt;a href=&quot;https://www.drinkarizona.com/faqs&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.drinkarizona.com/faqs&lt;/a&gt; &lt;br&gt;&#xA;I am most concerned what is the inner layer, what is its composition, and&#xA;whether there are added plasticizers such as BPA and BPS.&lt;br&gt;&#xA;Thank you,&lt;br&gt;&#xA;X&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;As of 2/10/2020 I have received no reply.&lt;/p&gt;&#xA;&lt;p&gt;I do not have a photo of the Arizona bottom handy, but here is a pic of another bottle,&#xA;this one a 6oz. juice bottle, which is also #7 OTHER and has the same &amp;quot;G&amp;quot; in rectangle logo that appears on the Arizona bottle.  Searching this emblem and major bottle manufacturers I was unable to identify the maker and inquire that way.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/W51oc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/W51oc.jpg&quot; alt=&quot;G-bottle&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Update: Here are pics of an Arizona 1g jug:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/BsPaK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BsPaK.jpg&quot; alt=&quot;Ariz 1g front&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UvKmP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UvKmP.jpg&quot; alt=&quot;Ariz 1g bottom&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;What are Arizona plastic gallon jugs made of?&lt;/p&gt;&#xA;" OwnerUserId="24550" LastEditorUserId="24550" LastEditDate="2021-01-15T10:27:26.703" LastActivityDate="2023-10-31T03:08:31.727" Title="What is the composition of Arizona Ice Tea 1gal plastic bottles stamped #7 OTHER and PP COMPATIBLE?" Tags="|plastic|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33169" PostTypeId="1" CreationDate="2020-02-10T14:49:05.970" Score="1" ViewCount="28" Body="&lt;p&gt;I am evaluating the specifications of an FTIR instrument (Bruker Vertex) and came across the following statements/items:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Signal-to-Noise 5 sec sample, 5 sec reference: &gt;10,000:1 (&amp;lt;4.34 x 10-5&#xA;  AU noise) peak-to-peak @ 2,000 cm-1, 4 cm-1 spectral resolution and&#xA;  standard optical components&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Signal-to-Noise, typical 5 sec: 16,000:1 (2.71 x 10-5 AU noise)&#xA;  peak-to-peak 1 min: 60,000:1 (7.2 x 10-6 AU noise) peak-to-peak Note:&#xA;  Achieved typical Signal-to-Noise peak-to peak values are statistical&#xA;  numbers which depend on the applied measurement parameters and&#xA;  laboratory working conditions.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I was wondering how this should be understood. I would assume, that in order to specify the SNR you have to state how much input the system gets. &#xA;(Sth alike: For  X mW/cm-1 &lt;strong&gt;coupled into&lt;/strong&gt; the system, at frequency Y,  at resolution Z the SNR is ...  )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, here it appears as if this was an &quot;absolute&quot; value (apparently depending on integration time). Is 10'000:1 this the ratio between saturation signal level and noise i.e. dynamic range? (This would only be of limited use...) And what could the &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;5 sec reference&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;mean?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;UPDATE&lt;/strong&gt;: I may have been misguided by what I am trying to measure (emission of light source, without reference to the reference lamp in the system). Could it be, that 10'0000:1 refers to the signal level of the reference lamp vs. the noise level? Can anyone confirm?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="23559" LastActivityDate="2020-02-10T14:49:05.970" Title="Meaning of SNR in Instrument Specifications" Tags="|optics|signal|spectrometry|noise|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33170" PostTypeId="2" ParentId="32031" CreationDate="2020-02-10T14:57:57.910" Score="1" Body="&lt;p&gt;For anyone coming later:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I ended up using a procedure described here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kevin C. Gross, Phenomenological model for IR emission from high explosive detonation fireballs&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In one sentence: I ended up measuring the blackbody spectrum at increasing temperatures. Then fitting a linear relation between measured and theoretical value for emission intensity (per wavelength) to obtain system gain and offset. So far, this leads to consistent results when measuring validation samples. Thanks @ Jeffrey J Weimer for the tip.&lt;/p&gt;&#xA;" OwnerUserId="23559" LastActivityDate="2020-02-10T14:57:57.910" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33171" PostTypeId="2" ParentId="33011" CreationDate="2020-02-10T16:08:54.387" Score="2" Body="&lt;p&gt;The problem is that the CSV export is including the new line characters that are part of the hole callout.  So this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A1,2.07,9.78,&amp;lt;MOD-DIAM&amp;gt; .257 THRU&#xA;5/16-18 UNC  THRU&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;should really look like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;A1,2.07,9.78,&amp;lt;MOD-DIAM&amp;gt; .257 THRU 5/16-18 UNC  THRU&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;You can perform a string replace of any pair or new line characters with some delimiter of your own (say, {new-line}) to make parsing easier, and you can get data like this:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;TAG,X LOC,Y LOC,SIZE&#xA;A1,2.07,9.78,&amp;lt;MOD-DIAM&amp;gt; .257 THRU{new-line}5/16-18 UNC  THRU&#xA;A2,2.53,6.28,&amp;lt;MOD-DIAM&amp;gt; .257 THRU{new-line}5/16-18 UNC  THRU&#xA;B1,4.72,5.03,&amp;lt;MOD-DIAM&amp;gt; .197 THRU{new-line}M6X1.0 - 6H THRU&#xA;B2,5.88,8.49,&amp;lt;MOD-DIAM&amp;gt; .197 THRU{new-line}M6X1.0 - 6H THRU&#xA;B3,7.74,6.57,&amp;lt;MOD-DIAM&amp;gt; .197 THRU{new-line}M6X1.0 - 6H THRU&#xA;C1,6.05,3.38,&amp;lt;MOD-DIAM&amp;gt; .266 THRU{new-line}&amp;lt;HOLE-SINK&amp;gt;&amp;lt;MOD-DIAM&amp;gt; .531 X 82°&#xA;C2,7.09,9.74,&amp;lt;MOD-DIAM&amp;gt; .266 THRU{new-line}&amp;lt;HOLE-SINK&amp;gt;&amp;lt;MOD-DIAM&amp;gt; .531 X 82°&#xA;C3,7.69,2.61,&amp;lt;MOD-DIAM&amp;gt; .266 THRU{new-line}&amp;lt;HOLE-SINK&amp;gt;&amp;lt;MOD-DIAM&amp;gt; .531 X 82°&#xA;C4,9.63,8.89,&amp;lt;MOD-DIAM&amp;gt; .266 THRU{new-line}&amp;lt;HOLE-SINK&amp;gt;&amp;lt;MOD-DIAM&amp;gt; .531 X 82°&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="3440" LastActivityDate="2020-02-10T16:08:54.387" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33172" PostTypeId="1" CreationDate="2020-02-10T19:02:43.833" Score="1" ViewCount="35" Body="&lt;p&gt;I'm attempting to design a system that consists of 4 secondary(ABCW), 1 primary(Film) tank, and a pump. The idea being to Transfer the contents of each secondary tank to the primary one, then back to the source tank before repeating the process for each of the remaining secondary tanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The simplest way to do this would be with a 2 way pump, but they seem to be very expensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I've been trying to design around a uni-directional pump, but the simplest layout I've come up with requires 3-4 solenoid valves to route the flow back in the reverse direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nEEq6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nEEq6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(Dark squares represent solenoids, large box is the pump)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Not knowing the first thing about hydraulics, I'm hoping this isn't the optimal design. Could anyone point me in the right direction? Or show me an affordable ≥5L/min bi-directional DC pump?&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Edit: the volume of the tanks is ~1L&lt;/li&gt;&#xA;&lt;li&gt;Edit 2: Affordable = Ideally under $100, not over 200&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="24542" LastEditorUserId="24542" LastEditDate="2020-02-10T22:15:13.667" LastActivityDate="2020-02-10T23:42:21.843" Title="Bi-directional liquid transfer" Tags="|hydraulics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33173" PostTypeId="1" CreationDate="2020-02-10T19:11:57.920" Score="-1" ViewCount="213" Body="&lt;p&gt;I have a problem with the expression of a measurement result.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a resistance decade box used as variable resistor in order to balance a Wheatstone Bridge. Let's call this resistance decade box &lt;span class=&quot;math-container&quot;&gt;$R_c$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value of &lt;span class=&quot;math-container&quot;&gt;$R_c$&lt;/span&gt; balancing Wheatstone bridge is &lt;span class=&quot;math-container&quot;&gt;$5000.0\ \Omega$&lt;/span&gt; (the resolution of the decade box is &lt;span class=&quot;math-container&quot;&gt;$0.1\ \Omega$&lt;/span&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The uncertainty of resistance decade box, depending on the value of &lt;span class=&quot;math-container&quot;&gt;$R_c$&lt;/span&gt;, is in my case, &lt;span class=&quot;math-container&quot;&gt;$0.001\ \Omega$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the uncertainty have to be expressed with only one significant digit, and since the resolution of the decade box is minor than &lt;span class=&quot;math-container&quot;&gt;$0.001\ \Omega$&lt;/span&gt;, how can I express the value of &lt;span class=&quot;math-container&quot;&gt;$R_c$&lt;/span&gt; with its uncertainty?&lt;/p&gt;&#xA;" OwnerUserId="24556" LastEditorUserId="1832" LastEditDate="2020-02-11T00:43:48.707" LastActivityDate="2020-02-11T19:09:54.223" Title="Expression of uncertainty measurement" Tags="|measurements|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33175" PostTypeId="2" ParentId="33172" CreationDate="2020-02-10T23:42:21.843" Score="2" Body="&lt;p&gt;That wouldn't be an uncommon industrial use.  You would need two inlet and two exit valves at the pump to direct flow in or out the way you want.  Plastic 1/4 inch solenoid valves are cheap, manual Ball valves are cheaper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for something that small, a simple gear pump would work, and is bi-directional.  Just connect it to a reversing motor.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-02-10T23:42:21.843" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33176" PostTypeId="1" AcceptedAnswerId="33243" CreationDate="2020-02-11T00:38:56.727" Score="-1" ViewCount="129" Body="&lt;p&gt;Consider the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ y(s)=b\frac{e^{−sτ}}{s+a}u(s) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where  a,b are unknown, constant parameters and &lt;span class=&quot;math-container&quot;&gt;$\ τ&amp;gt;0 $&lt;/span&gt; is an unknown delay constant. How can I obtain a plant parametric model for this system in the following form ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ z = \theta^{*T}\phi \ + η $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which satisfies the following condition:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\tau = 0 \Rightarrow  η = 0 $&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-05-10T17:46:19.213" LastActivityDate="2020-05-10T17:47:14.987" Title="How to obtain plant parametric model of given system?" Tags="|control-engineering|control-theory|modeling|mathematics|transfer-function|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33178" PostTypeId="2" ParentId="33129" CreationDate="2020-02-11T03:31:45.220" Score="1" Body="&lt;p&gt;We need to check the column for both the maximum stress with the secant formula and critical buckling load if the eccentricity is small and the column is slender and tall. As we know critical buckling load in medium and short columns gives wrong, over the yield strength results, so we always check for P/A&amp;lt; (allowable stress) too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The secant formula already calculates the maximum stress under the combination of the axial load and moment stress due to the bending of the column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7t15W.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7t15W.gif&quot; alt=&quot;secant formula&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-02-11T03:31:45.220" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33181" PostTypeId="2" ParentId="10809" CreationDate="2020-02-11T09:46:28.660" Score="1" Body="&lt;p&gt;Oils absolutely leak out. (I found this link searching for solution) They leak out in the cabin (sometimes even during delivery, with the seal on) &#xA;I always carry my &quot;first aid kit oil set&quot; with me, and every time some leaks out in the cabin luggage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The worse experience: I took a few bottles up on a hike in Peru (up to 5200 altitude) They were all empty when I tried using them. I believe that the pressure pushes out the oil through the plastic (the small hole) and the black top doesn't seal enough to stop the liquid from coming out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anybody advise me, how I could store my oils when I travel, please. I tried putting plastic under the black screw top. It may help a bit, but doesn't solve the problem.   Thanks&lt;/p&gt;&#xA;" OwnerUserId="24564" LastActivityDate="2020-02-11T09:46:28.660" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33182" PostTypeId="1" CreationDate="2020-02-11T09:49:04.090" Score="2" ViewCount="144" Body="&lt;p&gt;We have one imported strain gauge load cell, which shown open circuit in the input impedance. I removed the protective gel and corrected the wiring. Now the load cell is working. But I want to cover the strain gauge with the gel for the protection against moisture and dust. I want to know the product/technical name of the gel. please see the image. the blue arrow pointing the material that I am talking about.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NmnH6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NmnH6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="15386" LastEditorUserId="15386" LastEditDate="2020-02-12T06:19:57.817" LastActivityDate="2020-02-12T06:19:57.817" Title="What is the gel used for covering the strain gauge in the load cell?" Tags="|sensors|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33185" PostTypeId="2" ParentId="33173" CreationDate="2020-02-11T14:22:50.890" Score="3" Body="&lt;p&gt;Device uncertainty is composed of two factors: calibration or offset and scale or measurement. The former is accuracy and the latter precision.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first factor &lt;span class=&quot;math-container&quot;&gt;$\pm 0.1\ \Omega$&lt;/span&gt; is likely the scale/measurement uncertainty. It states the degree to which you can repeatably measure a value. Alternatively, it expresses the range within which measured values will report the same value on the device. By example, for your case, input values between 99.9 to 100.1 (rounded up here, it is really 100.0&lt;span class=&quot;math-container&quot;&gt;$\bar{9}$&lt;/span&gt;) will report as 100.0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an aside, the term &quot;resolution&quot; is perhaps a misnomer. When this is the step size of the tick marks, then your scale/measurement uncertainty is half this value, or &lt;span class=&quot;math-container&quot;&gt;$\pm 0.05\ \Omega$&lt;/span&gt;. In this case, input values between 99.95 and 100.04&lt;span class=&quot;math-container&quot;&gt;$\bar{9}$&lt;/span&gt; will report as 100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second factor &lt;span class=&quot;math-container&quot;&gt;$\pm 0.001\ \Omega$&lt;/span&gt; is likely quoted as a percentage of the setting. Assume that your device has absolutely no offset errors (i.e. an input value of zero reads exactly zero). This value says that your device propagates internal errors to read any value above zero with this level of accuracy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The final report of total uncertainty uses the rule of summation of variances. Variances are the squares of uncertainty values. So, the total is a quadrature sum of calibration and measurement:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ S_T^2 = \sigma^2_C + \sigma^2_M$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The total in this case is &lt;span class=&quot;math-container&quot;&gt;$\sqrt{0.1^2 + 0.001^2} = 0.1$&lt;/span&gt;. The total is dominated by the scale / measurement uncertainty. Use &lt;span class=&quot;math-container&quot;&gt;$\pm 0.05$&lt;/span&gt; when you mean to use the step size of the device readings for the first factor.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2020-02-11T19:09:54.223" LastActivityDate="2020-02-11T19:09:54.223" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33187" PostTypeId="1" CreationDate="2020-02-11T15:39:46.207" Score="6" ViewCount="199" Body="&lt;p&gt;I'm wondering if there are materials good at transmitting sound but is thermally insulating? I'm trying to figure out a good material in an engine exhaust application which will pose a barrier to the high-temp exhaust flow, but allow the exhaust noise to pass through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm particularly interested in low frequency sound only - 50 Hz to 350 Hz range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following cross-section image defines the application I have in mind a bit better. I'm trying to develop active noise control for engine exhaust noise. I'm hoping to use commercially available loudspeakers and so need to separate it from the main exhaust flow's heat &amp;amp; humidity. The material I'm looking for here is labelled &quot;Material X&quot;. Sound (green arrows) will be transmitted through it, but the exhaust flow (white arrows) cannot move through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6Oz4r.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6Oz4r.png&quot; alt=&quot;problem definition&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is somewhat opposite to &lt;a href=&quot;https://engineering.stackexchange.com/questions/7520/are-there-any-materials-known-to-be-sound-insulating-but-thermally-conductive&quot;&gt;this&lt;/a&gt; question.&lt;/p&gt;&#xA;" OwnerUserId="24569" LastEditorUserId="6790" LastEditDate="2020-05-04T18:25:46.310" LastActivityDate="2020-10-01T20:03:45.563" Title="Are there materials that are good at transmitting sound but thermally insulating?" Tags="|materials|acoustics|" AnswerCount="3" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33189" PostTypeId="1" AcceptedAnswerId="33202" CreationDate="2020-02-11T16:06:54.137" Score="1" ViewCount="554" Body="&lt;p&gt;I'm writing a story set in NYC and trying to describe certain industrial objects. I'm having trouble with specific technical terms for the items that I've marked w/ arrows. E.g. I don't want to use the generic term &quot;support&quot; or &quot;wall&quot; if there's a more specific word like viaduct, truss, pier, girder, silo, etc. Also specific materials (are they steel or iron, or a specific kind of design,style, type). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;After extensive research, I'm still not confident which technical terms apply to these specific things. Hoping a civil or other engineer might be able to shed some light. p.s. for the Smith-9th elevated train, what's this black material and when/why was it put on if anyone knows).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;R&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4TfPt.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4TfPt.jpg&quot; alt=&quot;278W in Brooklyn&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/soDUy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/soDUy.jpg&quot; alt=&quot;278W in Brooklyn&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/yDuil.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yDuil.jpg&quot; alt=&quot;Brooklyn Bridge&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Rz6qf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rz6qf.jpg&quot; alt=&quot;Smith-9th Street Station, Brooklyn&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uuANQ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uuANQ.jpg&quot; alt=&quot;Smith-9th Street Station, Brooklyn&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nakq4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nakq4.jpg&quot; alt=&quot;Smith-9th Street Station, Brooklyn (also, what&amp;#39;s this black material and when/why was it put on if anyone knows)&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/rwXSd.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rwXSd.jpg&quot; alt=&quot;Baysite Fuel Depot&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uWoXc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uWoXc.jpg&quot; alt=&quot;Oil depot and cement company&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24570" LastActivityDate="2020-02-12T17:23:03.897" Title="Vocabulary: NYC Elevated Highway &amp; Trains, Industrial Park" Tags="|structural-engineering|civil-engineering|structures|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33190" PostTypeId="2" ParentId="33153" CreationDate="2020-02-11T17:03:45.883" Score="1" Body="&lt;p&gt;You could try using &lt;a href=&quot;https://www.freecadweb.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;FreeCAD&lt;/a&gt;, it's a 3D modelling software. It's more geared towards making a 3D assembly than making drawings though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option is &lt;a href=&quot;https://inkscape.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Inkscape&lt;/a&gt; which is a vector graphics program.&lt;/p&gt;&#xA;" OwnerUserId="21516" LastActivityDate="2020-02-11T17:03:45.883" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33191" PostTypeId="2" ParentId="25545" CreationDate="2020-02-11T17:05:25.903" Score="0" Body="&lt;p&gt;Instead of having a hard 0/1 binary boundary condition try having a smoth and continuous one, try using the &lt;em&gt;tanh()&lt;/em&gt; function for instance&lt;/p&gt;&#xA;" OwnerUserId="21516" LastActivityDate="2020-02-11T17:05:25.903" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33192" PostTypeId="2" ParentId="33189" CreationDate="2020-02-11T17:41:02.173" Score="2" Body="&lt;p&gt;Supports on bridges or over-heads are called trusses in general.   Made up of diagonal, vertical, and horizontal members (those adjectives get nouned a lot, diagonals for example).  Some of those are just support columns. In California they tend to fill and/or surround them with concrete.  I-beams when they are shaped like an I or an H.  Some of those are just walls or over-hangs.  The red thing at the gas station is a pipe, probably fire water for sprinklers.  Might be AFFF, foam for oil fires.  The last picture is storage tanks, use bin or storage bin, hopper tank.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-02-11T17:41:02.173" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="33195" PostTypeId="2" ParentId="17769" CreationDate="2020-02-12T04:41:21.487" Score="2" Body="&lt;p&gt;How tall the wall needs to be depends on a few factors as others have mentioned in the comments. Since you've provided relatively vague design criteria, I cannot give a definite answer. However, &lt;a href=&quot;http://noisetools.net/noisecalculator2&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; tool is quite useful to determine the effectiveness of a wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The multi-spectrum mode might prove useful for you.&lt;/p&gt;&#xA;" OwnerUserId="24569" LastActivityDate="2020-02-12T04:41:21.487" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33196" PostTypeId="1" CreationDate="2020-02-12T04:42:05.450" Score="1" ViewCount="50" Body="&lt;p&gt;When a copper sheet is rolled into cylinder, it tends to unroll itself. Is there any name for this? &lt;/p&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2020-02-12T12:21:11.783" Title="&quot;Springiness&quot; of rolled copper sheet" Tags="|materials|solid-mechanics|material-science|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33197" PostTypeId="2" ParentId="22252" CreationDate="2020-02-12T04:51:04.193" Score="1" Body="&lt;p&gt;If I'm understanding your question correctly, you want to sum up levels given as 1/3 octave bands into 1/1 octave band resolution. This can be accomplished by decibel summation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$L_\Sigma = 10 * log(10^\frac{L1}{10} + 10^\frac{L2}{10} + 10^\frac{L3}{10})$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; are levels in dB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please see &lt;a href=&quot;http://www.sengpielaudio.com/calculator-octave.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; for an online summation tool.&lt;/p&gt;&#xA;" OwnerUserId="24569" LastActivityDate="2020-02-12T04:51:04.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33198" PostTypeId="1" CreationDate="2020-02-12T09:40:33.480" Score="1" ViewCount="2999" Body="&lt;p&gt;The Diagram below shows a lap joint where two plates are held together by a bolt. The applied force (F) is &lt;span class=&quot;math-container&quot;&gt;$42~\text{kN}$&lt;/span&gt; and the shear stress is &lt;span class=&quot;math-container&quot;&gt;$593~\frac{\text{N}}{\text{m}^2}$&lt;/span&gt; Assuming that the joint will be held together by a standard metric bolt, determine the minimum diameter permissible and identify the standard bolt diameter that will need to be used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cWZ9B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cWZ9B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is my answer is this correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sum f_x=0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V = \frac{p}{1} = \frac{42}{1} = 42kN$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\text{shear Stress} = 593N/m^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$593^2 = 351.649 Kn$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac {42000}{351.649} = 119$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ a = \frac{pi}{4} D^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$D = \sqrt\frac{4A}{pi}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$D = \sqrt\frac{(4) (119)}{pi}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ = 12.309$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the correct answer, and the correct way to complete this question?&lt;/p&gt;&#xA;" OwnerUserId="14101" LastEditorUserId="14101" LastEditDate="2020-02-12T12:33:57.793" LastActivityDate="2020-02-12T12:33:57.793" Title="Determine Bolt dia from Shear stress and force" Tags="|stresses|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33199" PostTypeId="2" ParentId="33198" CreationDate="2020-02-12T10:33:54.513" Score="3" Body="&lt;p&gt;Your units don't match, and you are calculating a required area of 17.7m², not a diameter of 17.7mm. I guess, the shear stress should have been an allowed stress in MPa, not Pa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why do you divide the force by 4? Are there four bolts even though your diagram only shows one?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When converting the required area to a bolt diameter, you also need to consider whether the shear cut is placed in the shank of the bolt or in the thread.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And finally you need to check the contact stress between bolt and plate (the bearing resistance), and adjust for the eccentricity of the load in a single lap joint with just the one bolt.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2020-02-12T10:33:54.513" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="33200" PostTypeId="2" ParentId="33196" CreationDate="2020-02-12T11:29:31.137" Score="1" Body="&lt;p&gt;Spiral torsion springs and constant-force springs behave in the same manner, although that's by design. A material's stiffness is what makes it want to resume its least stressed state. See Hooke's law.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastEditorUserId="10902" LastEditDate="2020-02-12T12:21:11.783" LastActivityDate="2020-02-12T12:21:11.783" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33202" PostTypeId="2" ParentId="33189" CreationDate="2020-02-12T13:21:31.937" Score="3" Body="&lt;p&gt;There are a whole lot of terms in your photos.&lt;/p&gt;&#xA;&lt;h3&gt;Photo 1:&lt;/h3&gt;&#xA;&lt;p&gt;That vertical structure is referred to as a &lt;strong&gt;PIER&lt;/strong&gt;.  Another style of pier which as far as I know is a series of piles and a pier cap is referred more specifically as a &lt;strong&gt;BENT&lt;/strong&gt;.  Piers perform the job of supporting the bridge between abutments by carrying the vertical (and horizontal loads) to ground.  The vertical element of the pier the arrow is pointed at I would classify as a &lt;strong&gt;PIER SHAFT&lt;/strong&gt; or &lt;strong&gt;PIER COLUMN&lt;/strong&gt;.  The material being steel plate.  The horizontal piece at the top attached to the leg, is referred to as the &lt;strong&gt;PIER CAP&lt;/strong&gt;.  It just is to support the girders and transfer their loads to the shafts.  The shaft carry the load down to its footings, normally below ground, which are normally spread footings or piles.&lt;/p&gt;&#xA;&lt;p&gt;Also of note in this photo are the &lt;strong&gt;GIRDERS&lt;/strong&gt;.  There are 4 of them  These are the main longitudinal structural shapes supporting the bridge.  Girders are normally either steel, concrete, or wood.  They can also come in a variety of shapes, I, box, and trapezoidal being some of the more common ones.  In this case you have steel plate I girder.  Their job is to carry the vertical load of the deck to its supports (abutment or pier normally).  This particular bridge is a bit of an odd ball as it has additional smaller longitudinal members running parallel with the girders.  I would call these smaller members &lt;strong&gt;STRINGERS&lt;/strong&gt;.  They are transferring the vertical load from the deck to the Diaphragms.&lt;/p&gt;&#xA;&lt;p&gt;Above the pier cap you will see a transverse steel plate member running between girders and it tappers as it extends out past the exterior girder. This member is referred to as &lt;strong&gt;DIAPHRAGM&lt;/strong&gt;. The diaphragm performs several tasks.  The main being to help redistribute the load between girders so the girders deflect more evenly, or behave as a group.  The other job they do is help prevent twisting of the girders.  Not that big of an issue on short straight bridges, but can be more of an issue on curved and longer spans.  At the abutments and piers, the diaphragms also tend to be a bit beefier.  This allows them to be used as jacking points.  This allows the bridge to be lifted slightly to allow bearings to be replace and minor repairs if any being performed around the bearings.  The diaphragms will also help transfer some of the horizontal loads the bridge deck may experience. Due to the stringers that are being supported by the diaphragms, one could argue that the diaphragms in this case are actually &lt;strong&gt;FLOORBEAMS&lt;/strong&gt;.  I would personally hesitate to call them that though as that is a term more often used with truss style bridge as opposed to a slab on girder bridge which is what I was assuming was shown in photo 1.&lt;/p&gt;&#xA;&lt;p&gt;Beneath the girders you will notice some small steel members in an X pattern.  This is commonly called &lt;strong&gt;CROSS BRACING&lt;/strong&gt;  These are normally L shaped sections, but can also be flat plate or rods to name a couple of other options.  Their designed normally so one diagonal will be in tension and the other diagonal is in compression.  The compression is assumed to overload the capacity of the diagonal and it will deflect/buckle out of shape and therefore not carry significant structural load. However the members will work in tension quite nicely.  Since tension resistance is pretty much based on pure cross sectional area, it does not really matter if the member is a rod, L-section, plate or a cable in theory (never seen one as a cable though).&lt;/p&gt;&#xA;&lt;p&gt;The portion outside of the outer most girder (exterior girder) is referred to as the &lt;strong&gt;CANTILEVER&lt;/strong&gt;, bridge deck cantilever, deck cantilever, etc.  It gets the cantilever name by the way the forces from it are transferred back to the exterior girder.&lt;/p&gt;&#xA;&lt;h3&gt;Photo 2:&lt;/h3&gt;&#xA;&lt;p&gt;The arrow pointing up is pointing at a portion of what appears to me to be a cantilevered bridge deck.  If you are referring more to the under side of the deck as opposed to the extension of the deck from the longitudinal support then the term I would use is &lt;strong&gt;SOFFIT&lt;/strong&gt;.  Soffit means the under side of the deck or ceiling.&lt;/p&gt;&#xA;&lt;p&gt;The arrow pointed to the left I would simply call a wall.  It appears very solid.  I am not aware of a special term for it and will add that it is not common for a bridge as it adds significant weight.  Bridge design tends to try and keep bridges light.&lt;/p&gt;&#xA;&lt;p&gt;The arrow pointed to the right I would refer to as a &lt;strong&gt;BARRIER&lt;/strong&gt;.  There are 3 common forms of barriers.  Parapet wall which looks very rectangular and is usually made of solid concrete.  It can be topped with a pedestrian handrail if appropriate.  Barrier wall which has beveled traffic face with a toe portion. Its similar to the parapet wall except its design is such that the car tire will impact first normally and it helps redirect the car slightly before the rest of significant impact occurs.  The third type is an Open railing system.  In this case you have an open railing system and more specifically a post and railing system&lt;/p&gt;&#xA;&lt;h3&gt;Photo 3:&lt;/h3&gt;&#xA;&lt;p&gt;The left arrow is pointed at the end of a &lt;strong&gt;CROSS BEAM&lt;/strong&gt;.  This is one element that makes up a truss style bridge.  Its job is to transfer horizontal load to the adjacent truss.  It also helps improve the torsional stiffness of the bridge so the truss does not twist and collapse over long spans.  Truss bridges where you have the tops of the truss connected and the traffic passes through the middle of the truss are referred to as Through Trusses.  There is a shorter style of truss bridge where there are no transverse members attached at the top.  It has a couple of names I am aware of:  Pony Truss and Saddle Truss.&lt;/p&gt;&#xA;&lt;p&gt;The right arrow is pointed at a member in a truss system call a &lt;strong&gt;VERTICAL&lt;/strong&gt;.  As pointed out by others, The vertical plan of a truss is made up of 4 main types of elements.  Verticals which as the name suggests go up and down.  Diagonals that go from top to bottom on some angle. Top chord which is the upper usually horizontal members.  Bottom chord which is the lower usually horizontal member.  Where the chords connect with with vertical or diagonals is usually referred to as a Node, though that comes up more for structural analysis purposes, and may just use connection in the field.  Normally the top chord is in compression and the bottom chord is in tension.&lt;/p&gt;&#xA;&lt;p&gt;Also of interest in this photo is that the transverse member instead of being an I beam or Channel, or Angle (L-section), is what I would call a built up member.  Without taking too close a look I would guess it is a pair of Channels (C-Sections) back to back and stitched together with baton plates.  &lt;a href=&quot;https://engineering.stackexchange.com/questions/15140/is-there-a-specific-name-for-diagonally-braced-truss-members-like-those-on-the-q&quot;&gt;This question&lt;/a&gt; has a more detailed description of this member make up.&lt;/p&gt;&#xA;&lt;p&gt;Also note that small roughly 45 degree short member at the end of the transverse beam that connects to the vertical.  This is commonly referred to as a knee brace.  Its job is to help stiffen the connection of the transverse member to the vertical.  This is what helps with the overall torsional stiffness of the bridge.&lt;/p&gt;&#xA;&lt;h3&gt;Photo 4:&lt;/h3&gt;&#xA;&lt;p&gt;This is a &lt;strong&gt;DIAGONAL&lt;/strong&gt; as discussed in an earlier photo.  If this is the last member of the truss, meaning it's lower end is resting on a pier or an abutment, then you may also refer to this as the end diagonal.  It will be the diagonal carry the largest axial load of all the bridge diagonals.  As a result the end diagonal will usually be a larger member than interior diagonals.&lt;/p&gt;&#xA;&lt;h3&gt;Photo 5:&lt;/h3&gt;&#xA;&lt;p&gt;This is an &lt;strong&gt;END DIAGONAL&lt;/strong&gt;.  See description of photo 4.&lt;/p&gt;&#xA;&lt;h3&gt;Photo 6:&lt;/h3&gt;&#xA;&lt;p&gt;The lower arrow is simply a truss or frame being used as a pier to support the bridge.  If the pier trusses were all interconnected, you could refer to this as trestle.  Though trestles are more commonly made out of wood and more often than not rail bridge.  There are definitely exceptions to this though.&lt;/p&gt;&#xA;&lt;p&gt;The upper arrow looks like its some sort of wall.  I would suspect non-structural and more like a debris fence.  Really hard to say though with out getting different angles and closer look.&lt;/p&gt;&#xA;&lt;p&gt;The black material looks like some sort of wrap used to either catch debris and keep it from falling on people and property below.  If if was only on the truss members I might guess it would be netting to prevent birds from nesting and crapping on the bridge.  Bird crap can be quite corrosive for steel.  Oh I just thought of this, if that structure is slated for demolition, the black covering could be a blast net / mat intended to contain the shrapnel / flying debris caused by the use of explosives in explosive demolition.&lt;/p&gt;&#xA;&lt;h3&gt;Photo 7:&lt;/h3&gt;&#xA;&lt;p&gt;The upper arrow seems to be pointed at some red pipe.  I would assume this is steel pipe and it is painted red to make it stand out so people know its part of the fire suppression system as opposed to the plumbing for the roof drain.  It is not a structural member.&lt;/p&gt;&#xA;&lt;p&gt;The lower arrow is pointed at a beam.  I would probably refer to this as a cross beam.  I may also call it a brace as it is connecting the two columns.&lt;/p&gt;&#xA;&lt;h3&gt;Photo 8:&lt;/h3&gt;&#xA;&lt;p&gt;The arrow on the left is pointed at what I would call storage tanks.  If its oil or gasoline I would refer to them as fuel tanks.  notice the wall around the tanks.  This is a safety precaution.  Those walls make a second tank or bath tub if you will.  odd are the volume inside that external wall is close to 1.25-1.5 times the volume of the inner tank.  The outer wall should contain the contents of the tank in the event of a failure/leak of the inner tank.&lt;/p&gt;&#xA;&lt;p&gt;The arrow on the right pointed at the round metal structure I would call those structures hoppers or silos.  This is really depended on what is being stored inside them.  Dry contents like sand or grain I would definitely use one of these two terms.&lt;/p&gt;&#xA;&lt;p&gt;The other arrow pointed at the rather square structure I am not sure what I would call it from this angle other than a building or structure.&lt;/p&gt;&#xA;&lt;h3&gt;Materials&lt;/h3&gt;&#xA;&lt;p&gt;For the most part, metal used in bridges in north america will all be structural steel.  The strength of the steel has changed significantly over the years, but it is all steel to the best of my knowledge.&lt;/p&gt;&#xA;" OwnerUserId="6993" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2020-02-12T17:23:03.897" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33203" PostTypeId="1" CreationDate="2020-02-12T16:39:18.760" Score="1" ViewCount="21" Body="&lt;p&gt;I've looked for answers without success, so turn to the experts. At air temperature, solid-phase sodium dichromate (Na2Cr2O7) -- or the equivalent, sodium dichromate dihydrate (Na2Cr2O72H2O) -- is introduced to a 2500F glass furnace. It can be 5% or more of a ~1200 lb batch comprised largely of silicon dioxide (SiO2). Total time in the furnace = 16 h.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much of the Na2Cr2O7 goes into solution in glass as a coloring agent. But some portion will off-gas off the glass-melt, transforming into a gas phase.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #1: Does the Na2Cr2O7 turn into a liquid first and then reach its boiling point and then evaporate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #2: Or does it turn directly from a solid to a gas -- sublimation? Note that, mixed in the ~1200 lb batch, it goes from air temperature directly into a 2500F furnace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Either way, gas-phase Na2Cr2O7 vapor particles rise up off the melt. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The furnace exhaust stream is estimated to be greater than 750F.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #3: At what temperature will the Na2Cr2O7 vapor particles desublimate -- turn back -- from a gas to solid Particulate Matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #3.A. Will they return to solid phase prior to reaching the baghouse filters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #4 What size might these vapor particles be? A range in nanometers -- 30 or 50, maybe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #5: Gas phase particles are not captured by normal baghouse filtration, correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #6. Filters pre-coated with lime are effective at controlling Hydrofluoric acid, for example. (Tons of it are used at this facility.) Is there any filter coating or treatment effective at capturing vapor-phase chromium?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question #7 Is there any way at all to control vapor-phase Cr off a 2500F glass melt?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've scoured the literature for many an hour with no luck. Now I'm turning to the experts, starting with you good folks. Many thanks. &lt;/p&gt;&#xA;" OwnerUserId="24527" LastActivityDate="2020-02-12T16:39:18.760" Title="Questions on Na2Cr2O7 turning to a vapor phase in 2500F furnace. And possible ways, if any, to filter it" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33204" PostTypeId="2" ParentId="31385" CreationDate="2020-02-12T18:17:49.540" Score="4" Body="&lt;p&gt;No. CNC machining is done in a huge variety of industries and applications, many of which do not benefit from heat treatment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;CNC machining - and metal cutting in general - is an incredibly broad technology. It is used frequently in demanding industries and applications, and many of these (aerospace, performance sports, power generation) do benefit from heat treatment post-processes. But CNC machining is also used in low value and low volume applications, and heat treatment is often overkill or even destructive there. For instance, CNC machining is used to create sacrificial parts (like soft jaws) which are used once and then discarded; it would be a waste of time and resources to heat treat these, and doing so might result in damage to the final part (which is meant to be held gently in a &lt;em&gt;soft&lt;/em&gt; material).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if you narrowed your question to a specific industry or application, the reality is that heat treating (or any other technology, for that matter) is only applied to a subset of the parts in that field.&lt;/p&gt;&#xA;" OwnerUserId="24577" LastActivityDate="2020-02-12T18:17:49.540" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33205" PostTypeId="2" ParentId="26948" CreationDate="2020-02-12T18:34:17.023" Score="0" Body="&lt;p&gt;EDIT: I misread the original post and suggested a solution that provides 1:2 gearing &lt;em&gt;in one direction only.&lt;/em&gt; In other words, a gearing system that works in forward but freewheels in reverse. Reading the question over, this is not what the OP requested.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;I would recommend splitting your two requirements out into separate subassemblies:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;A 1:2 speed ratio, easily achievable as you suggested via planetary gears&lt;/li&gt;&#xA;&lt;li&gt;One-way driving, which you can use a ratcheting mechanism for.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For ratcheting assemblies, I'd suggest something that looks like DT Swiss's &quot;Star drive&quot; mechanism for bicycle freehub assemblies. See image 4 &lt;a href=&quot;https://patents.google.com/patent/US20080006500A1/en?q=DT&amp;amp;q=Swiss&amp;amp;oq=DT%20Swiss%20&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;; the entire assembly contains just one spring and a set of identical toothed discs to obtain reliable one-way drive. A mechanism like this one can be designed in-line with your shafts, has a low part count, and is proven to work well with outer diameters of about 25 mm (and my guess is that it could go a lot smaller).&lt;/p&gt;&#xA;" OwnerUserId="24577" LastEditorUserId="24577" LastEditDate="2020-02-12T21:25:03.333" LastActivityDate="2020-02-12T21:25:03.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33206" PostTypeId="2" ParentId="26948" CreationDate="2020-02-12T21:19:50.660" Score="1" Body="&lt;p&gt;An &quot;orbitless&quot; drive can achieve this in a similar envelope as a planetary:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/hERmvD_Fe1o&quot; rel=&quot;nofollow noreferrer&quot;&gt;Orbitless demo&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;Nutating&quot; gears can also provide similar results.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A fixed planet carrier planetary will have opposing rotation between the sun and ring, &lt;strong&gt;if&lt;/strong&gt; there is an odd number of planets in line between the sun and ring (most typical planetary applications). If you have an even number of planets on each leg of the carrier and the carrier is fixed, your ring and sun will rotate in the same direction. Crude diagram below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NIKrF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NIKrF.png&quot; alt=&quot;multi planet planetary&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;Your sun gear will need to have half as many teeth as your ring gear to achieve your 2:1 ratio, that means the planets will need to combine to have half as many teeth as the sun (eg 20 tooth planets, 80 tooth sun, 160 tooth ring).&#xA;At a &lt;em&gt;minimum&lt;/em&gt; you should have 9 teeth on a gear, meaning your smallest possible ring gear would have a 0.6&quot; pitch diameter with 120 diametral pitch (very fine) teeth. In my experience the minimum number of teeth you should have is 24 (regardless of pitch) to avoid undercutting. This would make your smallest ring gear 192 tooth, making your pitch diameter 1.6&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is your space constraint? Also, your loads and speeds are what should determine your pitch and face widths as well.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastEditorUserId="23871" LastEditDate="2020-02-12T21:45:45.667" LastActivityDate="2020-02-12T21:45:45.667" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33207" PostTypeId="1" AcceptedAnswerId="33208" CreationDate="2020-02-12T21:21:22.293" Score="2" ViewCount="146" Body="&lt;p&gt;I am having trouble understanding how to approximate the solution to this problem using the Ritz method and the weak form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d^2u}{dx^2} - u=0; \ \ x \in [0,1]$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$u(x=0)=0; \ \ \frac{du}{dx} \bigg|_{x=1} = 20$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We multiply the strong form of the equation by a weight function &lt;span class=&quot;math-container&quot;&gt;$w$&lt;/span&gt; and integrate over the domain by parts to get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \int_0^1w\left[\frac{d^2u}{dx^2} - u \right] dx = \int_0^1 w \frac{d^2 u}{dx^2} dx - \int_0^1wu \ dx$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\int_0^1 \frac{du}{dx} \frac{dw}{dx} dx - \int_0^1 wu \ dx - 20 w(1) = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is the weak form. Now, I am confused because when I try to change this into a matrix equation, I practically try to solve&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\int_0^1 \frac{du}{dx} \frac{dw}{dx} dx = \int_0^1 wu \ dx + 20 w(1) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I try to use, say &lt;span class=&quot;math-container&quot;&gt;$$u\approx\hat{u}=a_1 \phi_1 + a_2 \phi_2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should be able to form a linear problem &lt;span class=&quot;math-container&quot;&gt;$$\mathbf{K} \vec{a} = \vec{b}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\mathbf{K}$&lt;/span&gt; is a &lt;span class=&quot;math-container&quot;&gt;$2 \times 2$&lt;/span&gt; matrix obtained from the left hand side of the equation where &lt;span class=&quot;math-container&quot;&gt;$w \to \phi_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$u \to \phi_j$&lt;/span&gt; to get the entries &lt;span class=&quot;math-container&quot;&gt;$K_{ij}$&lt;/span&gt;, but this substitution would not work to obtain the components of  &lt;span class=&quot;math-container&quot;&gt;$\vec{b}$&lt;/span&gt; since I this substitution gives me terms of &lt;span class=&quot;math-container&quot;&gt;$\phi_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\phi_j$&lt;/span&gt; and this makes no sense since &lt;span class=&quot;math-container&quot;&gt;$\vec{b}$&lt;/span&gt; is indexed only by one subindex. I know I am not understanding something, but this was my professor's explanation and I am confused.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much for your help!&lt;/p&gt;&#xA;" OwnerUserId="24580" LastEditorUserId="24580" LastEditDate="2020-02-12T22:10:05.847" LastActivityDate="2020-02-12T22:10:05.847" Title="How to use the Ritz method with the weak form to approximate solution of differential equation" Tags="|finite-element-method|numerical-methods|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33208" PostTypeId="2" ParentId="33207" CreationDate="2020-02-12T22:00:24.030" Score="2" Body="&lt;p&gt;Everything in your OP looks correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think you have just got into a muddle with the notation, and what is known and what is unknown.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\phi_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\phi_2$&lt;/span&gt; are known functions, i.e. you &lt;em&gt;choose&lt;/em&gt; two sensible functions for the problem you want to solve, to give a &quot;good&quot; approximation to the solution you expect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can differentiate them with respect to &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. Let &lt;span class=&quot;math-container&quot;&gt;$\dfrac{d\phi_i}{dx} = \phi'_i$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you let &lt;span class=&quot;math-container&quot;&gt;$w = \phi_1$&lt;/span&gt;, your integral equation becomes &lt;span class=&quot;math-container&quot;&gt;$$\int_0^1 (a_1 \phi'_1 + a_2 \phi'_2)\phi'_1\,dx - \int_0^1(a_1\phi_1 + a_2\phi_2)\phi_1\,dx = 20 \phi_1(1).$$&lt;/span&gt;&#xA;You can evaluate the integrals &lt;span class=&quot;math-container&quot;&gt;$\int\phi_1^2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\int \phi_2\phi_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\int {\phi'_1}^2$&lt;/span&gt;, and &lt;span class=&quot;math-container&quot;&gt;$\int \phi'_2\phi'_1$&lt;/span&gt; (either analytically or numerically) since the functions &lt;span class=&quot;math-container&quot;&gt;$\phi_i$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\phi'_i$&lt;/span&gt; are known.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You then get an equation of the form &lt;span class=&quot;math-container&quot;&gt;$$K_{11} a_1 + K_{12} a_2 = b_1$$&lt;/span&gt; where you know the &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$b$&lt;/span&gt; terms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Similarly letting &lt;span class=&quot;math-container&quot;&gt;$w = \phi_2$&lt;/span&gt; gives you the equation &lt;span class=&quot;math-container&quot;&gt;$$K_{21} a_1 + K_{22} a_2 = b_2.$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are the two rows of the matrix equation &lt;span class=&quot;math-container&quot;&gt;$\mathbf{K}\vec a = \vec b$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2020-02-12T22:05:31.777" LastActivityDate="2020-02-12T22:05:31.777" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33209" PostTypeId="2" ParentId="32110" CreationDate="2020-02-12T22:01:10.683" Score="1" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I would read into the concept of the discharge coefficient and its relationship to flow, pressure, density, and area &lt;a href=&quot;http://www.valvias.com/discharge-coefficient.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.valvias.com/discharge-coefficient.php&lt;/a&gt; This concept is complicated by compressible effects. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Mass flow coming into the system must equal the mass flow going out. According to your diagram, there are no leak paths between A and E. Therefore, the flow at E will not change based on Y. The pressure will change though because of the different fluids mixing. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;See the equation in the first link for a relationship between flow and pressure. The relationship is non-linear. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Many 1D flow solvers have methods to model valves, but equations such as the ones provided in the link can suffice. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="21118" LastActivityDate="2020-02-12T22:01:10.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33210" PostTypeId="2" ParentId="29140" CreationDate="2020-02-12T23:30:41.180" Score="1" Body="&lt;p&gt;The drum is spring-loaded, so you basically &quot;crank&quot; it by hand after loading rounds;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then it just rotates by the spring tension after every shot (gas pressure releases the drum lock on top)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a nice explanation: &lt;a href=&quot;https://youtu.be/BRb2iFkwDC4?t=623&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/BRb2iFkwDC4?t=623&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24582" LastEditorUserId="24582" LastEditDate="2020-02-12T23:41:15.890" LastActivityDate="2020-02-12T23:41:15.890" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33211" PostTypeId="1" AcceptedAnswerId="33217" CreationDate="2020-02-13T10:00:42.013" Score="0" ViewCount="5024" Body="&lt;p&gt;&lt;strong&gt;What i am looking for:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for a formular to calculate the water output in &lt;span class=&quot;math-container&quot;&gt;$l/s$&lt;/span&gt;.&#xA;My goal is to match a pump (pumping the water back up) with the following setup, to keep it going with as few corrections as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Where i am at the moment:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently i have Torricelli's law for an ideal system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But with this i am not sure if i have to take &lt;span class=&quot;math-container&quot;&gt;$h = h_2 - h_3$&lt;/span&gt; or &lt;span class=&quot;math-container&quot;&gt;$h = h_2 - h_3 + h_1$&lt;/span&gt;?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Restrictions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pipe is flexible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The friction of the pipe.&lt;/li&gt;&#xA;&lt;li&gt;The angles.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;My setup:&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Rfcey.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Rfcey.jpg&quot; alt=&quot;Sketch of the setup.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;An upper water tank with a water level &lt;span class=&quot;math-container&quot;&gt;$h_2$&lt;/span&gt; mm (which is consistant!).&lt;/li&gt;&#xA;&lt;li&gt;A lower water tank.&lt;/li&gt;&#xA;&lt;li&gt;A pipe with it's intake &lt;span class=&quot;math-container&quot;&gt;$h_3$&lt;/span&gt; mm above the upper Tank bottom. And it's outlet &lt;span class=&quot;math-container&quot;&gt;$h_1$&lt;/span&gt; mm below the upper tank bottom.&lt;/li&gt;&#xA;&lt;li&gt;The pipes diameter is &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; mm.&lt;/li&gt;&#xA;&lt;li&gt;The pipes length &lt;span class=&quot;math-container&quot;&gt;$l_1$&lt;/span&gt; mm.&lt;/li&gt;&#xA;&lt;li&gt;The water flow is started by sucking (like stealing fuel).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="24586" LastEditorUserId="24586" LastEditDate="2020-02-19T12:16:45.130" LastActivityDate="2020-02-19T12:16:45.130" Title="Calculating the flow rate of a siphon system" Tags="|fluid-mechanics|pipelines|fluid|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33212" PostTypeId="1" CreationDate="2020-02-13T10:21:29.433" Score="2" ViewCount="661" Body="&lt;p&gt;A double shear butt joint is shown below in the diagram 4 plates are held together by two bolts. The applied force is 42KN. Determine the shear stress within the joint&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dF5Fg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dF5Fg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;a &amp;amp;= 2 \frac{\pi}{4} d^2 \\&#xA;&amp;amp;= 113.0973355 \\&#xA;\frac{42}{2} &amp;amp;= 21 \\&#xA;\frac {21}{\pi/4\cdot12} &amp;amp;= 20.05352283\text{ kN/mm}^2&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="14101" LastEditorUserId="14101" LastEditDate="2020-02-16T14:27:16.410" LastActivityDate="2020-02-16T14:27:16.410" Title="Double Shear Stress butt joint" Tags="|structural-engineering|stresses|shear|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33216" PostTypeId="1" CreationDate="2020-02-13T14:36:06.050" Score="1" ViewCount="36" Body="&lt;p&gt;I am designing an outdoor sensor and I need to test it for continuous outdoors service. One of the test I wish to perform is a solar radiation test for the actinic effect (like procedure 2 in MIL-STD-810). I want to examine how the sensor will behave under continuous thermal radiation from the sun. &#xA;The problem is that MIL-STD-810 states that I have to make a 56 days long test, which is not practical. If I go to civil standards like IEC60068-2-5 then they do not suggest a test duration at all. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would you test an outdoors unit to validate it's integrity under long time solar radiation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&#xA;Eli&lt;/p&gt;&#xA;" OwnerUserId="23366" LastActivityDate="2020-02-13T14:36:06.050" Title="Solar Radiation Environmental Test Duration" Tags="|mechanical-engineering|environmental-engineering|product-testing|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33217" PostTypeId="2" ParentId="33211" CreationDate="2020-02-13T16:16:43.743" Score="0" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Currently i have Torricelli's law for an ideal system.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You could have started out with Bernoullis Equation, but it doesn't really matter here.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;But with this i am not sure if i have to take h=h2−h3 or h=h2−h3+h1?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Neither, if the free jet exiting the pipe is not continous, then I'd say you would use &lt;span class=&quot;math-container&quot;&gt;$\Delta h = h_1+h_2$&lt;/span&gt;. I don't see any reason to place the pipe so far above the surface level of the lower container that this would be a problem, so I will answer that the height is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/706J8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/706J8m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It can even dip into the lower container. The height difference is always the difference between the two water surfaces, it does not matter where exactly you place the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I studied the so called steam filament in fluid dynamics, so we image a streamline (green) from point 0 to 3 and follow it. (hence I recommended Bernoulli as a starting point)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tf3Rx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tf3Rxm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I'd like to imagine the parts that cancel each other out, e.g. the inverted U shaped top section, the fluid needs to go up a little and go down the same distance. Overall there is no height difference then.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I marked all parts that cancel out each other and I get:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nAxl7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nAxl7m.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is consistent with what I said earlier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;I know this is a very over simplyfied way to regard the problem, however I thought the visualization may help here. You can calculate everything with bernoulli and get the same result.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The velocity of the surface.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What surface? You said they are constant, i.e. the containers are rather large (one of the assumptions of Toricelli btw). If not you need to modify your calculations bearing in mind that:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$v = \frac{\partial h}{\partial t}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and need to use the version of &lt;a href=&quot;https://ocw.mit.edu/courses/mechanical-engineering/2-25-advanced-fluid-mechanics-fall-2013/inviscid-flow-and-bernoulli/MIT2_25F13_Unstea_Bernou.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;bernoulli for instationary flow&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The angles.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They don't matter, see above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;edit&lt;/strong&gt; in regards to Scott Dunnington's comment, I need to clarify that my approach of course neglects friction in the pipe (as you did with your assumptions).&#xA;However you can extend the Bernoulli equation with a term that accounts for these additional pressure drops.&#xA;It probably won't matter, but just for  the sake of complecity.&lt;/p&gt;&#xA;" OwnerUserId="1632" LastActivityDate="2020-02-13T16:16:43.743" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33218" PostTypeId="2" ParentId="33187" CreationDate="2020-02-13T21:35:09.403" Score="1" Body="&lt;p&gt;From the diagram you have provided us it appears to me that trying to &quot;brute force&quot; a noise cancellation scenario with a single loudspeaker at right angles to the flow of the exhaust gas is highly inefficient. If I may be so bold as to suggest a better way to approach the problem, you might want to consider using a sequence of passive impedance elements (expansion chambers) to muffle the sound first and apply active noise cancellation transducers to each chamber in the array, a divide and conquer approach so to speak. An excellent reference for designing passive muffler systems can be found in NASA Technical Note D-7309, &quot; An Improved Method For Design Of Expansion Chamber Mufflers With Application To An Operational Helicopter &quot;, Tony Parrott, author. By combining active and passive noise cancelling elements, you should be able to dramatically reduce the extreme thermal/sound transmitting materials requirements for your system . &lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2020-02-13T21:35:09.403" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33219" PostTypeId="1" CreationDate="2020-02-14T06:19:02.163" Score="-1" ViewCount="107" Body="&lt;p&gt;By laser projection I mean the laser beam that is projected onto the photosensitive drum in the laser printer. The printer receives digital image data from a computer processes it and projects it onto the photosensitive drum using laser beams.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I want to understand is:  &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;How does the conversion from digital image data to a laser beam projection actually happen? What is the process? What are the steps involved? What are the components (eg. electronic circuits) involved?&lt;/li&gt;&#xA;&lt;li&gt;Does the laser beam project the entire image at once onto the drum like a slide projector projects a entire slide onto the screen or does the laser beam like a pencil &quot;draw&quot; the image onto the drum like how a child would draw a picture on a paper using a pencil?&lt;/li&gt;&#xA;&lt;li&gt;If the laser beam draws like a pencil, how is the laser guiding mirror that directs the laser beam onto the drum able to move so fast that a page is printed every few seconds?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="16605" LastEditorUserId="16605" LastEditDate="2020-02-14T07:03:15.453" LastActivityDate="2020-02-14T17:36:25.110" Title="How is digital image-data converted to a laser projection in laser printers?" Tags="|electrical-engineering|lasers|consumer-electronics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33220" PostTypeId="2" ParentId="33187" CreationDate="2020-02-14T07:56:50.920" Score="0" Body="&lt;p&gt;AFAIK, sound transmission is a matter of the interface between materials, not just the materials themselves. Especially going from a light/low-density material to a heavier tends to reflect a lot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'd leave out &quot;Material X&quot; and attach the duct directly to exhaust pipe. The thin pipe wall should vibrate just fine with your speaker. The duct itself and the air inside will be the main heat conductors. For the duct walls, you want a dense material - I suggest stainless steel or any other highly alloyed metal (&quot;bas metals&quot; are good at conducting heat, highly alloyed metals not).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next step, you need to find the operating temperatures of your exhaust and the allowable temeprature of your speaker. You can calculate or  approximate the k value for the duct and the air inside, then you know how much heat the speaker will have to radiate away (don't forget the heat the speaker produces, may be neglible).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both thermal convection/conduction and accoustics are full of weird effects at interface and surprises. Expect to spend quite some time working with prototypes before you get it right.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2020-02-14T07:56:50.920" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33221" PostTypeId="1" CreationDate="2020-02-14T12:32:35.037" Score="2" ViewCount="89" Body="&lt;p&gt;In a week from now, I am going to teach a one-time class to first-year engineering students about 3D geometry, more specifically: distances between points and lines/planes, and the common perpendicular between two skew lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any cool engineering use cases (construction, chemistry, physics, computer science...) that I can use to illustrate some of these mathematical problems? I have looked a while on the internet, but cannot seem to find any good ones. And all textbooks I have consulted stay on the mathematical side of the spectrum, without applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="24601" LastActivityDate="2020-02-15T22:51:00.370" Title="Planes, lines and distances: looking for engineering problems" Tags="|mathematics|education|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33222" PostTypeId="2" ParentId="33221" CreationDate="2020-02-14T14:33:36.340" Score="2" Body="&lt;p&gt;Doing piping layout for a new chemical processes utilizes 3D modeling software a lot of the time now. I would think trying to show how pipes can and can't physically be laid out could use those concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical situation is that one large pipe starts from a source somewhere, and a number of smaller pipes branch from the large pipe and can run in any direction, having to cross over/under other pipes. Trying to run a new piping network in the midst of an existing piping network can get pretty challenging.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The situation I see that might be plausible to set up is to make a fictional pipe network and then analyze how (or if it's even possible) to run a new header and branch piping for an expansion project.&lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2020-02-14T14:33:36.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33223" PostTypeId="1" CreationDate="2020-02-14T15:51:28.410" Score="1" ViewCount="175" Body="&lt;p&gt;In the Alfred Hitchcock Presents episode &quot;Togeather&quot;, S03E15, at 17:17, you see a drunk man leaving a telephone without putting it on its handle, with the person he was talking to begging him to &quot;break the connection&quot; because he &quot;needs to call somebody else&quot;. The drunk man ignores the voice and seemingly forgets all about ever talking on the telephone, leaving the building. The person on the other end is stuck in a room with just a telephone and is trying to get out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this not a plot hole? Would not a telephone call, now as well as back then, automatically &quot;terminate&quot; immediately or shortly after one of the two &quot;puts the handle in its place&quot;? Why would he be worried about the other side not putting on their handle? Why would that matter?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This makes me wonder. If you could truly indefinitely keep a line open for anyone who calls you, by simply not hanging up on your end, the potential for abuse is endless. I refuse to believe that this was ever the case without some really good proof.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The costs you could force upon helpless callers alone would be enough, but imagine if they need to call the cops or something really important... It simply cannot be the case.&lt;/p&gt;&#xA;" OwnerUserId="24604" LastActivityDate="2020-03-07T04:56:19.757" Title="Would not a telephone call (from the late 1950s) auto-disconnect if one of the parties &quot;closes the handle&quot;?" Tags="|telecommunication|engineering-history|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33225" PostTypeId="2" ParentId="33223" CreationDate="2020-02-14T16:36:40.223" Score="5" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;If you could truly indefinitely keep a line open for anyone who calls you, by simply not hanging up on your end, the potential for abuse is endless. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's the way the POTS (plain old telephone system) worked. The caller controlled the line.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This &quot;feature&quot; was exploited by criminals who called the police stations from phone boxes and ripped the handsets out of the call boxes so that the police lines were held busy during a heist.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I refuse to believe that this was ever the case without some really good proof.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;That's your choice.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to prove it then make a call from a landline telephone and ask the other end to hang up for an hour. Keep the line open and they should hear you when they pick up the handset again.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The costs you could force upon helpless callers alone would be enough, but imagine if they need to call the cops or something really important... &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The charges would apply to the caller, not the called.&lt;/li&gt;&#xA;&lt;li&gt;The caller can hang up at any time and will get the dial tone on handset pickup.&lt;/li&gt;&#xA;&lt;li&gt;The called party will not be able to make or receive calls unless they have the ability to place a call on hold to make a second call.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;It simply cannot be the case.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Oh yes it can.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2020-02-14T16:36:40.223" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33226" PostTypeId="2" ParentId="33219" CreationDate="2020-02-14T17:36:25.110" Score="3" Body="&lt;p&gt;I'll answer the questions in reverse order.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;3&quot;&gt;&#xA;  &lt;li&gt;If the laser beam draws like a pencil, how is the laser guiding mirror that directs the laser beam onto the drum able to move so fast that a page is printed every few seconds?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/h7cqP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/h7cqP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. A simplified drawing of the laser and mirror arrangement. Source: &lt;a href=&quot;https://computer.howstuffworks.com/laser-printer7.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;How Stuff Works&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mirror in many cases just rotates continuously causing the beam to sweep from one side to the other in very rapid succession and always in the same direction. You can see something similar in many barcode scanners. This very simple arrangement requires no acceleration or deceleration of the mirror.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol start=&quot;2&quot;&gt;&#xA;  &lt;li&gt;Does the laser beam project the entire image at once onto the drum like a slide projector projects a entire slide onto the screen or does the laser beam like a pencil &quot;draw&quot; the image onto the drum like how a child would draw a picture on a paper using a pencil?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;One 600 dpi (0.06 mm) dot at a time building up to one line at a time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;ol&gt;&#xA;  &lt;li&gt;How does the conversion from digital image data to a laser beam projection actually happen? What is the process? What are the steps involved? What are the components (eg. electronic circuits) involved?&lt;/li&gt;&#xA;  &lt;/ol&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This will vary but the general idea will be:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If the printer has a byte per pixel then 2&lt;sup&gt;8&lt;/sup&gt; = 256 grey-scale levels are possible.&lt;/li&gt;&#xA;&lt;li&gt;Synchronise the memory reads with the mirror at the start of each line.&lt;/li&gt;&#xA;&lt;li&gt;Read a byte of memory and adjust the laser power to suit.&lt;/li&gt;&#xA;&lt;li&gt;Read the next byte and at the right time adjust the laser power to suit. &lt;/li&gt;&#xA;&lt;li&gt;Repeat until the page is complete.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The speeds involved are quite impressive. If we take a modest 600 dpi printer that can print six pages per hour (10 s per page) we get about 11 &amp;times; 8 &amp;times; 600&lt;sup&gt;2&lt;/sup&gt; = 31 million pixels for a letter / A4 page. For a 10 s print that works out at 31 ns per pixel or a data transfer rate of 3.2 MB/s.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2020-02-14T17:36:25.110" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33227" PostTypeId="2" ParentId="33223" CreationDate="2020-02-14T17:42:31.940" Score="2" Body="&lt;p&gt;This is still done by scammers who want to trick somebody into believing there is something wrong with the internet connection of their PC, and that some malware on their PC can disconnect their phone line permanently.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To &quot;prove&quot; the phone is disconnected, they ask the victim to hang up and call them back. They simply leave their own phone off the hook, so when the victim picks up his/her phone the call is still in progress and the victim can't get a dial tone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The scammers will then call back and start asking for the victim's personal information that &quot;they need to fix the problem&quot;.&#xA;It only works on landline phones, of course.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-02-14T17:42:31.940" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33228" PostTypeId="2" ParentId="33221" CreationDate="2020-02-14T17:45:14.673" Score="1" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9ufxa.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9ufxa.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Deceptively complex? Image source: &lt;a href=&quot;https://www.dailymail.co.uk/sciencetech/article-2049258/New-electricity-pylons-T-shaped.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Daily Mail&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Analysis of the classic cable pylon alone would probably take a full class.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For any of your students that fall in love with these structures then I recommend &lt;a href=&quot;https://pylons.org/&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Pylon Appreciation Society&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2020-02-14T18:01:18.740" LastActivityDate="2020-02-14T18:01:18.740" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33230" PostTypeId="1" CreationDate="2020-02-14T19:04:12.027" Score="1" ViewCount="85" Body="&lt;p&gt;I need to install a replacement air handler in a facility that has an area where a large volume of particulates are in the air.  Unfortunately, the previous designer tried to get away with the cheapest design possible and just put in pleated filters rather than the bag filters that were an option in the existing RTU.&lt;br&gt;&#xA;In the new unit, I would like to put in bag filters, but realized that even if they allow more throughput for dirty air, they will still probably need to be cleaned out more frequently.  Thinking about this reminded me of a baghouse on a silo that I specified awhile back and it had the functionality to effectively backwash the bag filters by running air in reverse to clean off the filter.&#xA;Is anyone aware of such a design being available for HVAC where, perhaps, a damper would flip temporarily and change the flow through the filter to run backwards and it would route that airstream as exhaust that would be dumped to the outside periodically?&lt;/p&gt;&#xA;" OwnerUserId="9770" LastActivityDate="2020-02-19T10:33:59.757" Title="HVAC Bag Filter Backwash" Tags="|airflow|hvac|air-quality|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33231" PostTypeId="2" ParentId="33230" CreationDate="2020-02-14T19:49:00.440" Score="1" Body="&lt;p&gt;Ignoring the concept that &quot;backwashing&quot; a filter as exhaust effectively negates the value and purpose of having filters, consider instead of a constantly clogging filter system, one that separates the particulates from the primary air stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a shop vac which has pleated filters. The filter is easily clogged by sawdust from a table saw, band saw, disk sander, etc. I purchased a high quality cyclone separator which causes the incoming laden air to travel in a circle. The particles are forced to the outside by inertia, pulled downward by gravity against a sloping inner chamber, which increases the velocity until the particulates reach the much narrower opening in the bottom, falling into the static chamber.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air flow, being lighter, is able to travel to the center of the conic chamber and upward as part of the vacuum exhaust forces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This diagram below courtesy of &lt;a href=&quot;http://canvas.pantone.com/gallery/4424943/External-Cyclonic-Filter-for-Vacuum-Cleaner&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://canvas.pantone.com/gallery/4424943/External-Cyclonic-Filter-for-Vacuum-Cleaner&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/N2bji.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/N2bji.png&quot; alt=&quot;cyclonic filter for vacuum cleaner&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The device I purchased removes easily the fine particles of which powdercoating powder is made. It's possible that the design effectiveness is related to the force available from the vacuum cleaner motor. If a larger volume and smaller velocity is involved, it may be necessary to have a larger diameter conic section to establish a greater separation force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another advantage of this type of separation device is that the vacuum effectiveness is not reduced by the quantity of particulates removed, while a filter weakens the vacuum as the cells become clogged. One must stop the vacuum periodically to dump the chamber, but the same applies for changing or cleaning filters.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-14T19:49:00.440" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33232" PostTypeId="2" ParentId="33221" CreationDate="2020-02-14T20:03:33.520" Score="1" Body="&lt;p&gt;Just of list of things that come to mind:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Piping for oil refineries &amp;amp; water distribution systems for domestic&#xA;water supply &amp;amp; hydro-electric power stations.&lt;/li&gt;&#xA;&lt;li&gt;Multi storey car parks - above or below ground.&lt;/li&gt;&#xA;&lt;li&gt;Underground mines using sub horizontal tunnels, vertical to sub&#xA;vertical shafts &amp;amp; spiral ramps.&lt;/li&gt;&#xA;&lt;li&gt;The design of roads in hilly or mountainous terrain. If the terrain&#xA;is not too steep roads can just go over the terrain. In steep terrain&#xA;&lt;a href=&quot;https://matadornetwork.com/read/terrifying-switchback-roads-drive/&quot; rel=&quot;nofollow noreferrer&quot;&gt;switch back roads&lt;/a&gt; are needed.&lt;/li&gt;&#xA;&lt;li&gt;Piping &amp;amp; ventilation systems in high rise builds&lt;/li&gt;&#xA;&lt;li&gt;Intercontinental ballistic missile interception, as was proposed in&#xA;the 1980s - &lt;a href=&quot;https://en.wikipedia.org/wiki/Strategic_Defense_Initiative&quot; rel=&quot;nofollow noreferrer&quot;&gt;Strategic Defense Initiative&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2020-02-14T20:13:37.167" LastActivityDate="2020-02-14T20:13:37.167" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33233" PostTypeId="2" ParentId="33230" CreationDate="2020-02-14T20:41:42.503" Score="1" Body="&lt;p&gt;There's nothing stopping you from doing a reverse air flow cleaning system.  I think the type of particulate and filter media type would determine how well it would work.  I would think the problem is where does the backflow material go?  A big plenum you have to sweep out regularly maybe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are replacing the air handler entirely, pulse jet or vibrating bags might get you there.  Again, type of particulate matters a lot here for how well it can be made to fall off of the bags.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As with almost everything, how well these things work are going to be proportional to the cost.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-02-14T20:41:42.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33234" PostTypeId="1" CreationDate="2020-02-15T01:41:10.300" Score="0" ViewCount="42" Body="&lt;p&gt;I have 0.13 mm stainless steel wire flattening pinch rollers&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NrEks.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NrEks.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;that are mechanically pressed by two bolts. I need to press them very hard to get good flattened wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is, the bearings got very very noisy and they broke and the pinch rollers shaft was broken, so how I can solve this problem?&lt;/p&gt;&#xA;" OwnerUserId="24608" LastEditorUserId="90" LastEditDate="2020-02-15T19:28:20.970" LastActivityDate="2020-02-15T19:28:20.970" Title="Failure of wire flattening pinch rollers" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33235" PostTypeId="2" ParentId="33234" CreationDate="2020-02-15T07:04:29.840" Score="2" Body="&lt;p&gt;You need to fit larger bearings or double up bearings or use double race bearings. Larger shafts should also be considered.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design seems classic of your device but the loading on the bearings should be considered.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2020-02-15T07:21:54.257" LastActivityDate="2020-02-15T07:21:54.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33236" PostTypeId="1" AcceptedAnswerId="33238" CreationDate="2020-02-15T08:23:37.153" Score="2" ViewCount="46" Body="&lt;p&gt;I am trying to find the exact tool which i assume is a modified t ruler to purchase but i cannot see any exactly like it. Any help would be appreciated:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Picture attached:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/6yE38.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6yE38.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is adjustable and has a fixed right angle piece on top so it is square with the material you are marking. All the t rulers i find only are essentially a standard ruler with an adjustable moving piece. I need the T ruler with the t at the top and then the adjustable moving part.&lt;/p&gt;&#xA;" OwnerUserId="24611" LastActivityDate="2020-02-15T11:13:51.250" Title="Trying to find this exact type of T Ruler?" Tags="|mechanical-engineering|tools|" AnswerCount="1" CommentCount="1" ClosedDate="2020-02-16T00:11:12.050" ContentLicense="CC BY-SA 4.0" />
  <row Id="33237" PostTypeId="2" ParentId="29140" CreationDate="2020-02-15T10:09:38.470" Score="0" Body="&lt;p&gt;As eyeballpaul said there is a small gas piston in the MGL that releases the drum lock when a round is fired.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a screenshot from World of Guns:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aKdwh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aKdwh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(World of Guns is a game that allows you to disassemble and view the operation of hundreds of guns.)&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2020-02-15T10:09:38.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33238" PostTypeId="2" ParentId="33236" CreationDate="2020-02-15T11:13:51.250" Score="2" Body="&lt;p&gt;I'd consider this to be an identification question related to engineering, rather than a product recommendation question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A search using &quot;scriber and depth gauge&quot; presents a large number of inappropriate devices and a smaller number of what appear to be the right device:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jI6Hk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jI6Hk.png&quot; alt=&quot;scriber depth gauge&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As one might note, this is not of the same scale, but appears to be an accurate match.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=dcO-WNlVnh8&quot; rel=&quot;nofollow noreferrer&quot;&gt;This video&lt;/a&gt; presents one method to construct a scriber and depth gauge or the same smaller scale. One could expect it is thus possible to build at a larger scale.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Searching for larger versions was fruitless, but only a short period of time was used for the search.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-15T11:13:51.250" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33239" PostTypeId="1" CreationDate="2020-02-15T17:41:23.280" Score="0" ViewCount="96" Body="&lt;p&gt;I want to edit some STL projects with Inventor. According Autodesk documentation, since Inventor is not able to edit mesh models, the most practical way without installing other software in my situation, seems opening the STL with a program able to export to DWG, then importing the DWG in Inventor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to open the STL model with 3DS Max but using the export to DWG function the resulting DWG seems a skeleton of the model rather than the DWG version of the original model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Example&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Original STL model&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tfewg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tfewg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Exported DWG model&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2Qj4g.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2Qj4g.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How could I avoid this problem?&lt;/p&gt;&#xA;" OwnerUserId="24614" LastActivityDate="2020-02-15T17:41:23.280" Title="3D Printing Models exported to DWG from STL using 3DS Max have missing surfaces" Tags="|3d-printing|autodesk-inventor|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33240" PostTypeId="2" ParentId="33230" CreationDate="2020-02-15T18:18:36.380" Score="0" Body="&lt;p&gt;Pardon my illustration (I didn't find an image online) but this was an innovation I heard of recently as part of a course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As dust accumulates on the bag, it gets heavier. This causes the spring keeping the bag taut to extend, and the bag loses its tautness. This causes the dust to fall off, and spring is able to pull the bag taut again.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think there was some mechanism to give a 'jerk' to the bag to make the dust fall off faster but it evades my memory.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sWjEF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sWjEF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12513" LastEditorUserId="12513" LastEditDate="2020-02-19T10:33:59.757" LastActivityDate="2020-02-19T10:33:59.757" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33241" PostTypeId="1" CreationDate="2020-02-15T19:19:24.267" Score="1" ViewCount="306" Body="&lt;p&gt;For a rod in compression, the max shear stress at a point is &lt;span class=&quot;math-container&quot;&gt;$\pm\sigma_x/2$&lt;/span&gt;, i.e. half the compressive stress applied at either end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This value is obtained using the equation for &lt;span class=&quot;math-container&quot;&gt;$\sigma_\theta$&lt;/span&gt;, the longitudinal stress at any point in a material if it is subjected to a set of stresses (equivalently represented by normal and shear stresses), then rotated through an angle &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt;, while these same stresses remain in place. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, does this max shear stress in a rod in compression refer to when the rod is rotated by 45 degrees such that it experiences max shear stress under the same compressive load? Or is this a value that the rod experiences at all times as a result of weaknesses in the material along different axes? Because at an angle of 0 degrees, there wouldn't be any shear stress in a perfectly homogeneous rod under compression, would there?&lt;/p&gt;&#xA;" OwnerUserId="24617" LastActivityDate="2020-07-15T00:03:00.240" Title="Physical Significance of Max Shear Stress in a Rod in Compression" Tags="|mechanical-engineering|stresses|shear|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33242" PostTypeId="2" ParentId="33241" CreationDate="2020-02-15T19:59:54.390" Score="1" Body="&lt;p&gt;What you are referring to is the shear stress in Mohr circle at a plane rotated counterclockwise 45 degrees from the horizontal &lt;span class=&quot;math-container&quot;&gt;$ \sigma_n $&lt;/span&gt; axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It means if we consider a differential small cube of the rod under just plane compression, normal stress, at a plane with an angle of 45 the average &lt;span class=&quot;math-container&quot;&gt;$ \sigma_n = \frac{ (o+\sigma_{n\ max})}{2 }= \tau_{xy\ max} $&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It does not refer to rotating the actual rod.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-02-15T23:53:53.343" LastActivityDate="2020-02-15T23:53:53.343" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33243" PostTypeId="2" ParentId="33176" CreationDate="2020-02-15T20:04:28.317" Score="2" Body="&lt;p&gt;Solution to the linear parameterization of the given system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ y = \frac{b}{s+a}e^{-sτ}u = \frac{b}{s+a}e^{-sτ}u \ + \ \frac{b}{s+a}u \ - \ \frac{b}{s+a}u = \frac{b}{s+a}u \ + \ \frac{b}{s+a}(e^{-s\tau} \ - \ 1)u  $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ sy = -ay+bu+b(e^{-s\tau}-1)u $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We filter both sides of the equation with a stable first order filter &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Λ(s) = s + l_1, \ l_1 &amp;gt;0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the final parametric model is now:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ \frac{s}{Λ(s)}y = -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the last equation can be written in the form: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ z = \theta^{*T}\phi + η $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\ z = \frac{s}{Λ(s)}y, \ \theta^* = [a \  b], \ \phi = [-\frac{1}{Λ(s)}y \ \  \frac{1}{Λ(s)}u]^T $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ η = \frac{b}{Λ(s)}(e^{-sτ}-1)u $&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that for &lt;span class=&quot;math-container&quot;&gt;$\ \tau = 0 \to η=0.  $&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The new input vector of the parametric system &lt;span class=&quot;math-container&quot;&gt;$\ z $&lt;/span&gt;, can be derived directly from the input of the initial system &lt;span class=&quot;math-container&quot;&gt;$\ y. $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: Slightly different approach&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{s}{Λ(s)}y = -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point, we can come up with a linear parametric model which contains the initial input vector &lt;span class=&quot;math-container&quot;&gt;$\ y $&lt;/span&gt; following the procedure below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Λ(s) = s + l_1 \Rightarrow s = Λ(s) - l_1, \ l_1&amp;gt;0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{Λ(s) - l_1}{Λ(s)}y = -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u \Rightarrow $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{Λ(s)}{Λ(s)}y - \frac{l_1}{Λ(s)}y =   -a\frac{1}{Λ(s)}y+b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u \Rightarrow $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ y = -(a-l_1)\frac{1}{Λ(s)}y +b\frac{1}{Λ(s)}u+ \frac{b}{Λ(s)}(e^{-s\tau}-1)u  $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the last equation can be written in the form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ y = \theta^{*T}\phi + η $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\ \theta^* = [(a-l_1) \ \  b], \ \phi = [-\frac{1}{Λ(s)}y \ \  \frac{1}{Λ(s)}u]^T $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ η = \frac{b}{Λ(s)}(e^{-sτ}-1)u $&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that for &lt;span class=&quot;math-container&quot;&gt;$\ \tau = 0 \Rightarrow η=0.  $&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-05-10T17:47:14.987" LastActivityDate="2020-05-10T17:47:14.987" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33244" PostTypeId="2" ParentId="33161" CreationDate="2020-02-15T20:42:28.173" Score="0" Body="&lt;p&gt;You need a proper Fresnel lens, and then diffusion. This video explains it very well (but it does build on a tv display panel, rather than a build from scratch):&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=8JrqH2oOTK4&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=8JrqH2oOTK4&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1011" LastActivityDate="2020-02-15T20:42:28.173" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33245" PostTypeId="2" ParentId="33221" CreationDate="2020-02-15T22:51:00.370" Score="1" Body="&lt;p&gt;There are lots of uses. Distance between two points comes up so often that i couldnt even count the uses. One particularily cool way to demonstrate it is raytracing, or sphere tracing. Raytracing is all about finding line plane intersections. And sphere tracing about closest point on a primitive. While raytracing seems like a gimic for graphics it is a valid simulation for many optics problems, as well as any number of other stochastic physics simulations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many engineering problems today use CAD. But the CAD application couldnt work without intersection finding. You can find lots of interesting examples in analytical geometry books for engineers. These are nolonger commonly thaught since our CAD allready handle those. But the examples in books are as valid and interesting today as they were yesterday. Look one up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Distance between 2 points can be a demonstrated well by iteratively finding a solution to a problem that you dont know how to solve analytically. But can use gradient descent optimisation to solve by minimizing the distance. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But there are more good examples.&lt;/p&gt;&#xA;" OwnerUserId="1324" LastActivityDate="2020-02-15T22:51:00.370" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33246" PostTypeId="1" CreationDate="2020-02-15T22:51:43.577" Score="2" ViewCount="112" Body="&lt;p&gt;I'm trying to complete a project for my solidworks course and it involves nanoelectronics. The idea is to create a mesh array of field effect transistors (FETs) that will monitor the firing of neurons in a microfluidic device. My design is inspired by science articles like this one:&#xA;&lt;a href=&quot;https://iopscience.iop.org/article/10.1088/1361-6528/ab3534&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://iopscience.iop.org/article/10.1088/1361-6528/ab3534&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've already had some problems with using Solidworks for nano-groove technology as 72k features is too much for my computer to handle, but there won't be a need for as many nanotubes and circuits for the mesh. I can make a 3D image of what I want, but my professor wants me to make an electrical simulation to verify my design works. My professor is also not too available to me at the moment, so I'm limited on asking him what to do. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's no need to worry about how I will shape and orient the nano-scale wires. I just need to know how I would build a new electronic part from scratch. If I just make copper and carbon tubes, is that going to be enough to properly simulate? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some images to help clarify what I want to make:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0v2tj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0v2tj.png&quot; alt=&quot;FET image&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XiDwc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XiDwc.jpg&quot; alt=&quot;Mesh Electronics&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UnOlU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UnOlU.jpg&quot; alt=&quot;Mesh Electronics&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/SefKP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SefKP.png&quot; alt=&quot;Mesh Circuits&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="19886" LastEditorUserId="19886" LastEditDate="2020-02-17T21:43:51.190" LastActivityDate="2020-02-17T21:43:51.190" Title="How do I create nano-scale mesh electronics in Solidworks?" Tags="|cad|solidworks|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33247" PostTypeId="1" CreationDate="2020-02-15T23:46:02.557" Score="3" ViewCount="105" Body="&lt;p&gt;I'm working through the derivation for the Kalman filter using orthogonal projections. It's pretty intuitive that if we want to generate the best linear estimate for an &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt;-vector &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; using some &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt;-vectors &lt;span class=&quot;math-container&quot;&gt;$y_0, \dots, y_t$&lt;/span&gt;, we would take the orthogonal projection of &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; onto the subspace &lt;span class=&quot;math-container&quot;&gt;$\mathbb{R}^m$&lt;/span&gt; spanned by the vectors &lt;span class=&quot;math-container&quot;&gt;$y_i$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the Kalman filter case, it seems that we assume &lt;span class=&quot;math-container&quot;&gt;$n&amp;gt;m&amp;gt;t$&lt;/span&gt;. That is, a lower dimensional observations &lt;span class=&quot;math-container&quot;&gt;$(y)$&lt;/span&gt; than states &lt;span class=&quot;math-container&quot;&gt;$(x)$&lt;/span&gt; we are trying to estimate. Similarly, we are trying to build a recursive estimator that uses additional observations &lt;span class=&quot;math-container&quot;&gt;$y_i$&lt;/span&gt; such that at timestep &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; we have &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; observation vectors and our subspace is of dimension &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: what happens when &lt;span class=&quot;math-container&quot;&gt;$t=m=n$&lt;/span&gt;? This would mean, I think, that the orthogonal projection is meaningless and it wouldn't help your filter to gain new estimates... is this correct?&lt;/p&gt;&#xA;" OwnerUserId="24619" LastActivityDate="2020-02-15T23:46:02.557" Title="Kalman Filter by Orthogonal Projection" Tags="|control-engineering|control-theory|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33248" PostTypeId="1" CreationDate="2020-02-16T04:54:26.020" Score="0" ViewCount="44" Body="&lt;p&gt;I'm building a milk pasteuriser. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The milk in the cooling tank has a starting temperature of &lt;em&gt;4&amp;deg;C&lt;/em&gt;. It should then be heated to &lt;em&gt;72&amp;deg;C&lt;/em&gt; for &lt;em&gt;30 sec&lt;/em&gt;, then cooled back down again to &lt;em&gt;4&amp;deg;C&lt;/em&gt; as quickly as possible (if I understand the process correctly).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've got 2 bulk tanks, one with a heating fluid of 90&amp;deg;C and another with cooling fluid of 4&amp;deg;C.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is this: &lt;strong&gt;What should the length of each of the copper induction coils be to heat/cool the fluid to the appropriate temperature for the required length of time.&lt;/strong&gt; &lt;em&gt;(I'm looking for a formula if such exists)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;**Side note: I'm not an engineer. I'm a home hobbyist with background in maths/programming.&lt;/p&gt;&#xA;" OwnerUserId="20987" LastEditorUserId="20987" LastEditDate="2020-02-16T13:03:37.560" LastActivityDate="2020-02-16T13:03:37.560" Title="What should length of copper coil be to heat fluid from Temp A to Temp B?" Tags="|heating-systems|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33250" PostTypeId="2" ParentId="24608" CreationDate="2020-02-16T17:29:08.057" Score="0" Body="&lt;p&gt;This is how the whole backward vector &lt;span class=&quot;math-container&quot;&gt;$\ z $&lt;/span&gt; is computed using matlab &lt;code&gt;ode45(...)&lt;/code&gt; solver:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;tspan - 0:0.01:10; &#xA;&#xA;eqtn = @(t,x,y) [x(2)-interp1(tspan(:),y(:),t) ; -3*x(2)-2*x(1)+3*interp1(tspan,y(:),t)];&#xA;x = [0;0];&#xA;[~,phi_1] = ode45(@(t,x)eqtn(t,x,y),tspan, x);&#xA;&#xA;&#xA;eqtn = @(t,x,y) [x(2) ; -3*x(2)-2*x(1)-interp1(tspan,y(:),t)];&#xA;x = [0;0];&#xA;[~,phi_2] = ode45(@(t,x)eqtn(t,x,y),tspan, x);&#xA;&#xA;&#xA;eqtn = @(t,x,u) [x(2) ; -3*x(2)-2*x(1)+interp1(tspan,u(:),t)];&#xA;x = [0;0];&#xA;[~,phi_3] = ode45(@(t,x)eqtn(t,x,u),tspan, x);&#xA;&#xA;z = [phi_1(:,1) phi_2(:,1) phi_3(:,1)];&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;It is assumed that the vectors &lt;span class=&quot;math-container&quot;&gt;$\ y \  \&amp;amp; \  u $&lt;/span&gt; are known and represent the output and the input of the original system correspondingly. &lt;/p&gt;&#xA;" OwnerUserId="18238" LastActivityDate="2020-02-16T17:29:08.057" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33252" PostTypeId="1" CreationDate="2020-02-16T18:48:06.410" Score="0" ViewCount="182" Body="&lt;p&gt;I'm confused by this. There is a moment of inertia for a shape, let's say circle:&#xA; &lt;a href=&quot;https://i.stack.imgur.com/J0ypa.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J0ypa.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you make it 3D, you have a new moment of inertia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/aB7kO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/aB7kO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any connection between these two? &lt;/p&gt;&#xA;" OwnerUserId="21424" LastActivityDate="2020-02-17T06:34:20.847" Title="Are there different moments of inertia for shapes and bodies?" Tags="|mechanical-engineering|moments|solid-mechanics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33253" PostTypeId="1" CreationDate="2020-02-16T19:26:01.073" Score="-1" ViewCount="192" Body="&lt;p&gt;I have some batteries labelled as Extra Heavy Duty with a picture of a trashcan and the letters Pb for lead, and an x over the trashcan which I believe means not to discard in the trash.  The batteries also say not for retail sale, so I suspect that they were the original batteries in some children's toys from which I pulled them out.  I found on the web that the likely chemistry is zinc chloride, and I don't see any lead in the chemical reaction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the source of the lead?  Is it just impurities in the zinc, as can occur with zinc-galvanized hardware?  I know that in the galvanization process lead is added to the kettles for metallurgical reasons.  Is there something similar going on here?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="24628" LastActivityDate="2020-02-17T17:56:37.650" Title="why is there a lead (Pb) warning on extra heavy duty batteries?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33254" PostTypeId="1" AcceptedAnswerId="33256" CreationDate="2020-02-16T21:11:44.870" Score="3" ViewCount="79" Body="&lt;p&gt;I want to build a shelf that can I can place near a wall and still be able to rotate it around 180 degrees. I therefore would want to mount the shelf on a rotating pole and to reduce the needed distance to the wall would have a sliding mechanism that lets me slide the shelf off-center to one side (see the following sketch).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Bes9S.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bes9S.jpg&quot; alt=&quot;Sketch of the sliding mechanics&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I built the following prototype out or cardboard to verify the idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Aqnex.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Aqnex.jpg&quot; alt=&quot;Cardboard Prototype&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So given I can get the center pole to be stable enough (for example by attaching it to the wall) is it feasible to have such a sliding mechanism that can hold a weight of around 40kg in the off-center position and still slide? Am I missing any other critical problems here? Any hints for similar existing parts or mechanisms would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="24626" LastActivityDate="2020-02-17T02:01:23.987" Title="Is it feasible to mount a 40kg shelf on a rotating and sliding pole?" Tags="|structural-engineering|stresses|wood|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33255" PostTypeId="2" ParentId="33187" CreationDate="2020-02-17T00:24:01.767" Score="1" Body="&lt;p&gt;In doing a quick scan of Google for your answer, I came across an article on a ring type device developed by scientists at Boston University that passes air but blocks sound in certain frequency ranges. Sounds like it could be a breakthrough for muffler design, go check it out.&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2020-02-17T00:24:01.767" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33256" PostTypeId="2" ParentId="33254" CreationDate="2020-02-17T00:47:20.057" Score="2" Body="&lt;p&gt;Based on your nicely constructed model, it appears that two axes of movement will be required. The vertical axis can be addressed by a pair of bearing assemblies securely mounted to the top and bottom of the center pole:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q6fv5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q6fv5.png&quot; alt=&quot;bearing assembly&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may be necessary to find a different type of assembly which allows you to secure the inner race to the pole, but the image serves the purpose of this discussion. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this pillow block bearing which has a setscrew on the inner race:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TSGkM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TSGkM.png&quot; alt=&quot;pillow block assembly&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These types of bearings are usually designed to tolerate axial loading but for light duty, slow speed operation will probably hold well enough. You may also find some other bearing types suited for thrust loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With a suitable bearing assembly on each end of the pole, one could create a trolley to attach to the bearing assembly. This trolley would ride in a track internal to the shelf structure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ejl2q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ejl2q.png&quot; alt=&quot;roll-off track wheel&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This image above is for a rolling roof mechanism. It would be necessary to have four of these at each bearing assembly (at the corners?) and matching rails above and below both bearings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;None of this is to scale and a 40 kg load may require large scale components. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've been building a CNC router which uses v-shaped wheels, riding in aluminum extrusions with matching v-grooves. This permits the carriage to travel from one end of the gantry to the other, but it's not carrying 40 kg, either.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3D printers use linear bearings which provide similar movement, but again, not 40 kg loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance to the wall will be based on the spacing of the travel wheels from the axle center as well as the size of the wheels. I can envision using in-line skate wheels with matching tracks, providing suitable load capacity. One could get away without using the wheel portion and having bearing contact directly on the travel rails, if there is sufficient clearance.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-17T00:47:20.057" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33257" PostTypeId="2" ParentId="33254" CreationDate="2020-02-17T02:01:23.987" Score="1" Body="&lt;p&gt;What you are looking for is already available as 70&quot; TV mount, pull down over fireplace, in many Internet shopping sites for prices ranging between $50-100.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I add a photo just as an example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1i10b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1i10b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-02-17T02:01:23.987" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33258" PostTypeId="2" ParentId="33252" CreationDate="2020-02-17T06:34:20.847" Score="2" Body="&lt;p&gt;The general idea in both cases is similar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first case, we are basically summing the differential areas multiplied by the square of their distance from the axis of rotation, neutral axis. This is tightly related to the concept of the linear relationship of stress and strain in the flexural properties of beams and Young's modulus. This is usually called the second area moment or second-moment of inertia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ I_x=\int y^2dA $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second case, as shown in your second diagram, We are summing the differential mass elements of a solid (doesn't have to be a solid, liquid, gas or plasma is fine) again multiplied by the square of their distances from the axis of rotation, to get the inertia for angular momentum and the concepts of rotational energy, work, and preservation of angular momentum and the entire spectrum of related physics.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$  I=\int r^2dm$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-02-17T06:34:20.847" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33261" PostTypeId="1" AcceptedAnswerId="33326" CreationDate="2020-02-17T11:34:04.770" Score="0" ViewCount="1397" Body="&lt;p&gt;We've been requested to find the resonant frequencies of a 3rd order plant in class and were presented to the following method to do so:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/FCFC5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FCFC5.png&quot; alt=&quot;Example included&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, when using the &lt;span class=&quot;math-container&quot;&gt;$\ \omega_r $&lt;/span&gt; formula, to find both of the resonant freq's do I need to place &lt;span class=&quot;math-container&quot;&gt;$\ \omega_{n1} $&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$\ ζ_{1} $&lt;/span&gt; to get &lt;span class=&quot;math-container&quot;&gt;$\ \omega_{r1} $&lt;/span&gt; (and &lt;span class=&quot;math-container&quot;&gt;$\ -\omega_{n2} $&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$\ ζ_{2} $&lt;/span&gt; to get &lt;span class=&quot;math-container&quot;&gt;$\ \omega_{r2} $&lt;/span&gt;),&#xA;or I am supposed to use the same &lt;span class=&quot;math-container&quot;&gt;$\ ζ_{2} $&lt;/span&gt; (of the Pole) for both, like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \omega_{r1} = f(\omega_{n1}, ζ_{2}) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \omega_{r2} = f(\omega_{n2}, ζ_2) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope the question is clear enough, thanks!&lt;/p&gt;&#xA;" OwnerUserId="24634" LastEditorUserId="18238" LastEditDate="2020-02-17T15:50:50.353" LastActivityDate="2020-02-20T22:32:43.850" Title="Control System; Finding resonant frequency of 3rd order" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33263" PostTypeId="1" CreationDate="2020-02-17T16:37:15.207" Score="2" ViewCount="70" Body="&lt;p&gt;I recently learned about traveling wave reactors (TWRs) on the &lt;a href=&quot;https://www.netflix.com/title/80184771&quot; rel=&quot;nofollow noreferrer&quot;&gt;Netflix Bill Gates special&lt;/a&gt;, as an example of a cutting-edge technology Gates is funding which has the potential solve several environmental problems.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The TWR is a nuclear fission reactor which sounds a &lt;em&gt;bit&lt;/em&gt; too good to be true (via &lt;a href=&quot;https://en.wikipedia.org/wiki/Traveling_wave_reactor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia&lt;/a&gt;):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Uses spent fuel from other fission reactors, meaning it doesn't require mining of new uranium&lt;/li&gt;&#xA;&lt;li&gt;&lt;em&gt;Consumes&lt;/em&gt; the spent fuel from these reactors, meaning it could be used to dispose of this nuclear waste&lt;/li&gt;&#xA;&lt;li&gt;Doesn't produce any of its own toxic byproducts&lt;/li&gt;&#xA;&lt;li&gt;Can run in containment for decades, with no need to add/remove fuel, reducing operational costs and risk to operators&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So, what gives? &lt;strong&gt;Why don't these exist yet?&lt;/strong&gt; The Wikipedia article (and the Netflix special) mention some political challenges that have delayed construction of a demonstration reactor, but I wasn't able to find any additional detail on technical challenges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like fusion reactors (which have both political and financial support, and never seem to get any closer to existing), I assume there must be some specific, unsolved technical challenges.&lt;/p&gt;&#xA;" OwnerUserId="17004" LastActivityDate="2021-04-02T18:45:26.933" Title="What are the key technical challenges to development of traveling wave reactors?" Tags="|energy|nuclear-engineering|nuclear-technology|nuclear-reprocessing|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33264" PostTypeId="2" ParentId="33253" CreationDate="2020-02-17T17:56:37.650" Score="1" Body="&lt;p&gt;If the battery has a lead warning on it, its chemistry is lead/lead sulfate and both the plates in the electrolyte and the jumpers between cells are made of lead. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-02-17T17:56:37.650" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33266" PostTypeId="1" CreationDate="2020-02-17T18:49:04.173" Score="1" ViewCount="20" Body="&lt;p&gt;I am searching for some software to simulate the behaviour of soft bodies like cheese in an quantitive way. For example I want to know the forces in the body during the simulation. &#xA;I thought of programming one by myself but this would probably take way too long. And I'm confused about all the programs I found on google.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So does anyone know some software to simulate things like cheese?&lt;/p&gt;&#xA;" OwnerUserId="23512" LastActivityDate="2020-02-17T18:49:04.173" Title="Software for Soft Body Dynamics" Tags="|simulation|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33267" PostTypeId="1" CreationDate="2020-02-17T20:08:39.797" Score="1" ViewCount="167" Body="&lt;p&gt;I'm trying to calculate the reliability of a system that has proven to be a bit more complicated than I thought. Here is the block diagram for the system:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/MisEM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MisEM.png&quot; alt=&quot;Block Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Every single component in this diagram is a Line Replaceable Unit with a mean downtime of 2 hours. If any of A-F fail, that's a failure for the &quot;subsystem&quot; indicated by the dotted line. The entire system will successfully work as long as only 1 of the dotted line, A-F, &quot;subsystems&quot; has failed. Because of the mean downtime of 2 hours, if two of the dotted line subsystems fail within 2 hours of each other, the system fails. This subsystem highlighted in green is not particularly difficult. What's throwing me off is the inclusion of components G1 and G2, routers that fail when either A (UPS) or B (DC power) fails. This makes up the subsystem highlighted in red. The entire system fails if both routers (components G1 and G2) fail, within 2 hours of each other of course.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, I thought that the reliability of the whole system can be calculated with Bayes Theorem which would give the reliability of the system as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reliability of green subsystem * probability that G1 and G2 do not fail&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PLUS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reliability of red system * probability that another of the dotted line subsystems does not fail&lt;/p&gt;&#xA;&#xA;&lt;p&gt;PLUS&lt;/p&gt;&#xA;&#xA;&lt;p&gt;reliability of red system * probability that the other red subsystem does not fail&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, I think I'm vastly misunderstanding Bayes Theorem and maybe the system as a whole. Can someone lead me in the right direction to how the inclusion of these G components affects the system reliability? Maybe I just need to re-read up on Bayes Theorem?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="24641" LastActivityDate="2020-02-17T20:08:39.797" Title="Calculating reliability of a parallel redundant system" Tags="|systems-engineering|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33268" PostTypeId="1" CreationDate="2020-02-18T00:29:37.763" Score="-1" ViewCount="82" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Strowger, an undertaker, was motivated to invent an automatic&#xA;  telephone exchange after having difficulties with the local telephone&#xA;  operators, one of whom was the wife of a competitor. He was said to be&#xA;  convinced that she, as one of the manual telephone exchange operators,&#xA;  was sending calls &quot;to the undertaker&quot; to her husband.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Source: &lt;a href=&quot;https://en.wikipedia.org/wiki/Strowger_switch&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Strowger_switch&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Isn't it more than a little bizarre that an &lt;em&gt;undertaker&lt;/em&gt; comes up with this highly complicated, highly technical, new machine invention? And the quote makes it out as if nobody else had even thought of this, as if just &quot;thinking about it&quot; magically makes it appear as a real thing. As if telephone companies and tons of scientists had not been thinking about it since the advent of the telephone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;He conceived his invention in 1888&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I always thought that they had manual (human) telephone exchange operators well into the 1960s and beyond in most of the world, even as late as in the 1980s in many places? And this is talking about 1888? Am I not reading this correctly? Am I confused somehow?&lt;/p&gt;&#xA;" OwnerUserId="24644" LastActivityDate="2020-02-18T01:56:29.380" Title="How is it possible that an *undertaker* invented the automatic telephone exchange? And as early as in 1888?" Tags="|telecommunication|engineering-history|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33269" PostTypeId="2" ParentId="33268" CreationDate="2020-02-18T01:42:09.140" Score="3" Body="&lt;p&gt;You may be misunderstanding what &quot;undertakers&quot; actually did in this time period. Even in the 1950s in the UK when I was growing up, in rural areas the local undertaker not only did the carpentry and joinery to make his own coffins, but also did house repairs and general engineering work on farm machinery etc, and quite likely did a bit of furniture making and similar crafts as well. The reason was the practical one - there simply were not enough people living (and dying) in the area to support a &quot;full time undertaker&quot; business at prices which the population could afford!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to his biography on Wikipedia, Strowger was the son of a miller, so he would certainly have been involved with machinery since childhood. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't think it is incredible for somebody with that type of background to invent a device which is basically a rotary switch. If a local farmer had ever asked Strowger to fix a broken reaper-binder (invented fifteen years before 1888) he would have already seen a device that used some of the same basic engineering ideas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For comparison, Tivadar Puskás, the inventor of the telephone exchange and the multiplex switchboard (in the same year as Strowger's invention) was a Hungarian immigrant who started work in the USA as a travel agent, and then a gold miner. The modern idea that &quot;you need a PhD to invent anything new&quot; didn't exist back then.&lt;/p&gt;&#xA;" OwnerUserId="683" LastEditorUserId="683" LastEditDate="2020-02-18T01:56:29.380" LastActivityDate="2020-02-18T01:56:29.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33271" PostTypeId="2" ParentId="12705" CreationDate="2020-02-18T03:27:54.663" Score="1" Body="&lt;p&gt;Wire mesh and rebar are roughly the same strength ; 55,000 psi yield ( both are available in various grades ).  A heavy wire , 8 gage ( American Wire Gage) is 0.123&quot; diameter . For an ordinary house , 0.5 &quot; diameter rebar is common ( mine also has some 0.625&quot; and 0.375&quot; ) . Spacing is roughly the same , 6 in. between wire or rebar. So in a mesh structure there is 0.123&quot; diameter steel and in a rebar structure there is 0.5&quot; diameter steel in the same space. Without doing all the arithmetic; there is no way mesh is close to the strength of rebar. Mesh is used because it requires much less labor to install.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-02-18T03:27:54.663" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33272" PostTypeId="1" CreationDate="2020-02-18T03:46:05.473" Score="0" ViewCount="72" Body="&lt;p&gt;Has something like a hardware replicator ever been built? Something akin to &lt;a href=&quot;https://en.wikipedia.org/wiki/Self-replicating_spacecraft&quot; rel=&quot;nofollow noreferrer&quot;&gt;von Neumann probes&lt;/a&gt;, only it doesn't have to be for space exploration?&lt;/p&gt;&#xA;&lt;p&gt;When I say &amp;quot;replicator&amp;quot;, in this context I mean a piece of hardware that can create more instances of itself.&lt;/p&gt;&#xA;&lt;p&gt;If it has not been built, how might one go about it? At first, I was thinking that the first unit would contain a computer with a program telling it what raw materials in the environment to look for and use. But then I realized that this means the very first instance would need to have fine enough motor control to build a new machine with a new computer in it. That's seems awfully complex for a first instance of a replicator (after all, self-replicating molecules are much simpler).&lt;/p&gt;&#xA;&lt;p&gt;So there seems to be an optimum somewhere in the middle between the following two: making the first replicator too simple may result in its lacking the ability to replicate, but if we make it too complex then it will be too hard for it replicate. But any approach that comes to mind to build this machine seems too hard for replication.&lt;/p&gt;&#xA;&lt;p&gt;PS: Engineering noob so not sure how to tag this!&lt;/p&gt;&#xA;" OwnerUserId="22969" LastEditorUserId="22969" LastEditDate="2020-07-02T19:08:22.947" LastActivityDate="2021-11-08T21:33:16.670" Title="Has a replicator ever been built?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33274" PostTypeId="1" AcceptedAnswerId="33285" CreationDate="2020-02-18T09:37:17.323" Score="2" ViewCount="3082" Body="&lt;p&gt;DB9 female port often have female slots for two screws to secure the male plug onto it. What is the type of this screw?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Word 'screw' is mentioned 7 times in Wikipedia DB-9 article but not the type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Same port is a COM port on computer. Old school mouse port too.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At least Farnell, parts distributor have these screws categorized under: d-sub-jack-screws, and many items say M3 that being the type of the thread and the size both, 3 millimeters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I have an unknown screw that fits into the mentioned female slot, and I have a legit M3 type screw, and the threads are way denser on the standard M3 screw. They have the same thickness tho.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two leftmost ones fit, and are of the type of which I want to know the designation, the one on the right is M3&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PA4R1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PA4R1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14435" LastEditorUserId="14435" LastEditDate="2020-02-18T09:51:02.170" LastActivityDate="2022-09-05T23:04:42.533" Title="Screw type associated with COM port/DB-9" Tags="|computer-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33275" PostTypeId="2" ParentId="33274" CreationDate="2020-02-18T09:46:07.937" Score="0" Body="&lt;p&gt;It says this in one of the screw definitions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thread Size - Imperial  : 4 - 40&lt;/p&gt;&#xA;" OwnerUserId="14435" LastEditorUserId="14435" LastEditDate="2020-02-18T09:58:56.687" LastActivityDate="2020-02-18T09:58:56.687" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33276" PostTypeId="1" CreationDate="2020-02-18T09:49:18.173" Score="2" ViewCount="90" Body="&lt;p&gt;we have a alkaline solution with Cl content of 3500 ppm, and pH=10 ,my question is if the carbon steel material is proper for this solution or not?&lt;/p&gt;&#xA;" OwnerUserId="24647" LastActivityDate="2020-02-18T16:46:31.327" Title="carbon steel corrosion in chloride solution" Tags="|corrosion|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33279" PostTypeId="1" CreationDate="2020-02-18T14:29:02.680" Score="2" ViewCount="680" Body="&lt;p&gt;I am trying to extend a metal wire mesh over a large distance, while tensioning it to limit the deflection of the mesh due to the wind as much as possible. After doing some research I came across Roark's Formulas for Stress and Strain, which has a number of formulas regarding the deflection of flat rectangular plates under a variety of support conditions. In particular, I have been trying to use this formula, for a plate under a uniform transverse distributed load, with tension applied to its short edges:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/gw5UP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gw5UP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this is assuming it is a solid plate, not a welded wire mesh. I have tried to use these equations, but with the distributed load from the wind only applied to the percentage of the area taken up by the wires of the mesh. The other problem I have run into while using theses formulas is that there are only values of the three coefficients for values of P/Pe up to 5, and I would like to investigate the effects tensioning the mesh would have with higher values of this ratio. There are other ways of preventing the mesh from deflecting besides tensioning it, but I would like to investigate this method, as well as learn about the behavior of the mesh in a broader sense.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any formulas/methods of calculating the deflection of a mesh due to wind out there? Or is there a way to use Roarks to solve this problem? Any advice or direction to additional resources is appreciated!&lt;/p&gt;&#xA;" OwnerUserId="24653" LastEditorUserId="24653" LastEditDate="2020-02-18T14:37:10.053" LastActivityDate="2024-03-29T06:05:35.443" Title="How to determine the deflection of a wire mesh under wind load?" Tags="|mechanical-engineering|deflection|meshing|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33280" PostTypeId="1" AcceptedAnswerId="33281" CreationDate="2020-02-18T15:20:14.383" Score="2" ViewCount="3248" Body="&lt;p&gt;I'm looking to determine the difference between the two yield strengths on a Matweb material property table for &lt;a href=&quot;http://www.matweb.com/search/datasheet.aspx?matguid=77b8eca6a7324775827b768f95637200&quot; rel=&quot;nofollow noreferrer&quot;&gt;DuPont Performance Polymers Zytel® 105F BK010 PA66&lt;/a&gt;. The top is for &quot;50% RH; ISO 527-1/-2&quot; per the comments and the bottom is &quot;DAM;ISO 527-1/-2&quot;. I need to know what 50% RH and DAM mean so I can make an informed decision. Any help would be appreciated.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x7AiG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x7AiG.png&quot; alt=&quot;Image from Matweb Table&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18926" LastActivityDate="2020-08-07T13:32:36.723" Title="What does 50% RH mean?" Tags="|materials|plastic|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33281" PostTypeId="2" ParentId="33280" CreationDate="2020-02-18T16:13:14.907" Score="6" Body="&lt;p&gt;DAM means &quot;dry as manufactured&quot; or &quot;dry as moulded&quot;, i.e. the material is in its moisture-free state immediately after manufacture. The DAM moisture content is typically less than 0.2%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;50% RH&quot; means the material has absorbed moisture from an environment with 50% relative humidity until it has reached an equilibrium state.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-02-18T16:13:14.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33283" PostTypeId="2" ParentId="33276" CreationDate="2020-02-18T16:46:31.327" Score="2" Body="&lt;p&gt;It depends on how long you want it to last to decide if it is &quot;proper&quot;. Generally the solution sounds like seawater ( although seawater pH is around 8 +).Corrosion will be strongly affected by factors not presented: oxygen, velocity, temperature, etc.As a wild guess a corrosion rate of one millimeter per year is possible.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-02-18T16:46:31.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33284" PostTypeId="2" ParentId="33279" CreationDate="2020-02-18T19:32:18.213" Score="0" Body="&lt;p&gt;There are several issues that you need to look into.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Codes have wind pressure charts, not at this very low level though.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The mesh, its wind stagnation behavior, and structural properties depend on the geometry of the mesh and its opening dimensions and the twisted knots or spot welds used to assemble it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say you somehow find all this data.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An approximate solution is to assume a deflection shape similar to the diagram below &lt;span class=&quot;math-container&quot;&gt;$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1&#xA; $$&lt;/span&gt;and solve it by using strip concrete slab method on FEM software or energy methods using the hyperbolic surface as a guess deflection shape.  Just don't expect it can be done in a few days.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/13GEI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/13GEI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-02-18T19:32:18.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33285" PostTypeId="2" ParentId="33274" CreationDate="2020-02-18T19:34:26.913" Score="5" Body="&lt;p&gt;4 - 40 UNC ASME/ANSI (US customary units rather than Imperial)&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.google.com/search?q=4-40+unc+thread+dimensions&quot; rel=&quot;nofollow noreferrer&quot;&gt;Thread Specifications&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The final two digits are the number of turns per inch, so the pitch is 0.025&amp;quot; (US: 25 mils, UK: 25 thou).&lt;/p&gt;&#xA;" OwnerUserId="24581" LastEditorUserId="24581" LastEditDate="2021-12-10T21:06:03.133" LastActivityDate="2021-12-10T21:06:03.133" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33286" PostTypeId="1" CreationDate="2020-02-18T20:44:08.810" Score="2" ViewCount="40" Body="&lt;p&gt;I’m trying to create the following extrude.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MmRpm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MmRpm.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have created a crankshaft based on the following sketch:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZdAPC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZdAPC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I then removed material as follows, which is placed on the rear surface of the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YtBPc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YtBPc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The material is removed successfully, however my final result has a thick extruding circle around the top of the crankshaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cnyi5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cnyi5.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong ? &lt;/p&gt;&#xA;" OwnerUserId="19105" LastActivityDate="2020-02-18T20:44:08.810" Title="Creating an extrude in creo parametric 5" Tags="|cad|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33295" PostTypeId="1" AcceptedAnswerId="33346" CreationDate="2020-02-19T10:09:33.057" Score="2" ViewCount="3151" Body="&lt;p&gt;Looking at a model of the HMS Ark Royal I found myself intrigued by the prominent 'poles' lining the flight deck. Some investigation on Google reveals that they can apparently be lowered to lay horizontally, presumably to keep them out of the way of any aircraft landing or taking off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/T8Wna.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/T8Wna.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Source: &lt;a href=&quot;https://www.hydro-international.com/content/article/discovering-h-m-s-ark-royal&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.hydro-international.com/content/article/discovering-h-m-s-ark-royal&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only hint of a purpose that I found was an old Google Group where someone mentions they are radio antenna. Unfortunately, I could not find a more authorative source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question remains, what are these?&lt;/p&gt;&#xA;" OwnerUserId="24665" LastEditorUserId="24665" LastEditDate="2020-02-19T10:14:48.140" LastActivityDate="2020-02-26T21:24:23.827" Title="What are the 'poles' off the side off this aircraft carrier's flight deck?" Tags="|mechanisms|marine-engineering|ships|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33296" PostTypeId="1" AcceptedAnswerId="33301" CreationDate="2020-02-19T10:11:08.080" Score="2" ViewCount="89" Body="&lt;p&gt;I’m working on a DIY project and need some assistance. I have a single motor that can lift a load of 500 pounds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to use 4 smaller motors to lift the same load as long as the 4 smaller motors equal the same torque as the single motor?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This isn’t a question about gearing or placing the 4 motors on a single shaft....each of the smaller motors would work independently to lift the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thought is this... You have a 3’ lever on a fulcrum situated at a given distance (let’s say the fulcrum is exactly in the center of the 3’ lever. 100 foot pounds is placed on the lever, but the load does not lift. Is it possible to place another lever against the load with the exact same conditions as the 1st lever. Then, both levers are given 100 foot pounds....will the load lift with the added mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is hypothetical of course.... I would love to understand this.&lt;/p&gt;&#xA;" OwnerUserId="24667" LastActivityDate="2020-02-19T15:55:05.917" Title="Lifting a Mass with one or multiple motors" Tags="|torque|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33298" PostTypeId="1" AcceptedAnswerId="33299" CreationDate="2020-02-19T12:22:05.653" Score="1" ViewCount="135" Body="&lt;p&gt;If I have a uniform rod &lt;code&gt;--&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then I push on each end toward the center &lt;code&gt;&amp;gt;--&amp;lt;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So it bends &lt;code&gt;&amp;gt;^&amp;lt;&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What shape does it make? Is it a parabola, a catenary, or something else?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assume constant stiffness along the rod. I'm using the rod to draw an aerodynamic shape, and I'd like to know what shape I'm drawing. I've read about parabolas and catenaries, but I don't know the answer to this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers!&lt;/p&gt;&#xA;" OwnerUserId="12760" LastActivityDate="2020-02-19T13:12:40.760" Title="What shape does a rod bend into when you push the ends together?" Tags="|aerodynamics|drafting|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33299" PostTypeId="2" ParentId="33298" CreationDate="2020-02-19T13:12:40.760" Score="1" Body="&lt;p&gt;For a slender column, pinned at both ends, the typical shape  would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y(x)=\delta_{max}\sin\left(\frac{\pi x}{L}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\delta_{max}$&lt;/span&gt; is the maximum deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;see &lt;a href=&quot;https://www.continuummechanics.org/columnbuckling.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.continuummechanics.org/columnbuckling.html&lt;/a&gt; for more details.&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2020-02-19T13:12:40.760" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="33300" PostTypeId="1" CreationDate="2020-02-19T13:57:05.150" Score="1" ViewCount="23" Body="&lt;p&gt;After an a object is formed through direct laser sintering, as a post-processing procedure the object is sent for hot-isostatic pressing, In hot isostatic pressing the sintered object is placed in a mold and then pressure and heat are applied.&lt;br&gt;&#xA;I was wondering what would go wrong if the bare object without the mold  was made to undergo pressing? Remember the sintered object already has a shape acquired during sintering so it may need a mold to get a specific shape during pressing.   My assumption- In hot isostatic pressing the object might get softened and hence deformed due to heat and thus the requirement of a mold to keep the object firm. Is this assumption true??&lt;br&gt;&#xA;And apart from deformation are there any other undesirable effects caused to the un-molded object during hot isostatic pressing?&lt;/p&gt;&#xA;" OwnerUserId="16605" LastEditorUserId="16605" LastEditDate="2020-02-19T18:48:31.207" LastActivityDate="2020-02-19T18:48:31.207" Title="Why is a mold required in isostatic pressing after performing Sintering" Tags="|mechanical-engineering|metallurgy|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33301" PostTypeId="2" ParentId="33296" CreationDate="2020-02-19T15:55:05.917" Score="2" Body="&lt;p&gt;Ignoring losses, the total force on the object is all that matters.  So 4 motors each supplying 25% of the torque would be exactly equivalent to 1 motor at 100% torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This applies everywhere.  Two people each exerting 50 pounds can lift a 100 pound object;  two levers double the force applied; two locomotives double the pulling power, etc.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-02-19T15:55:05.917" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33303" PostTypeId="1" AcceptedAnswerId="33304" CreationDate="2020-02-19T17:59:18.140" Score="1" ViewCount="117" Body="&lt;p&gt;When you compare single wall tent to double wall tent usually you will hear short summary &quot;double wall tent prevents condensation from building up&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So in most fair comparison it should work like this -- I have imaginary single wall tent (i.e. tent with only one layer, the rainfly) and using it will cause condensation to build up. OK.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now let's add to this tent inner tent (mesh layer) and this prevents condensation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hmm... I don't see it. I mean, condensation occurs when the warm air is cooled down to the point it no longer can hold steam. I guess the mesh barrier can slow down the process (i.e. you would have warm space inside, middle space between layers, and the cold air outside), the warm air from inside not so soon reaches the outer layer. But eventually it will reach it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Second guess would be that inner mesh gets a little wet, and by this it helps the air to hold the rest of the amount of steam. But this effect I would say is a condensation anyway, just not visible because of the nature of the mesh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So is there any other effect that contributes to the notion &quot;inner tent prevents condensation&quot;?&lt;/p&gt;&#xA;" OwnerUserId="20299" LastEditorUserId="20299" LastEditDate="2020-02-19T18:13:01.860" LastActivityDate="2020-02-19T18:13:28.083" Title="The physics behind the condensation in tents?" Tags="|steam|air|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33304" PostTypeId="2" ParentId="33303" CreationDate="2020-02-19T18:13:28.083" Score="2" Body="&lt;p&gt;They don't prevent condensation. They just reduce the chances of condensation getting onto your kit down to practically nothing, if you do it right.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The condensation stays on the inside of the outer layer. Your kit is on the inside of the inner layer. The condensation doesn't cross the gap between the inner and outer layer. It runs down the inside of the outer layer. And then it can't pond inside the inner layer.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2020-02-19T18:13:28.083" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33307" PostTypeId="1" CreationDate="2020-02-19T22:26:14.697" Score="3" ViewCount="287" Body="&lt;p&gt;Like a lot of people nowadays, I'm fitting a log fire into the back of a camper van but can't find the minimum height above the roof for the chimney. Ideally, it should be as low as possible for aerodynamic and roof clearence purposes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Building regs state that the pipe should be 2' above the highest point on the roof and I've seen people blindly follow this with their conversions (which is ridiculous) but nobody says why. Air flow? Safety for thatched roofs? The most recent paper I've found on the subject was published in 1946 and doesn't address this issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The venting for the stove is entirely different anyway with a 30cm cubed combustion chamber, a 130mm internal sized flue, a 500cm straight pipe to the roof and a 3-5Kw output. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there some formula for a minimum chimney height above the roof? Or a university compartment specializing in combustion?&lt;/p&gt;&#xA;" OwnerUserId="24671" LastActivityDate="2020-02-19T22:26:14.697" Title="What is the minimum height of a chimney needed in a camper?" Tags="|heating-systems|building-physics|fire|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33308" PostTypeId="2" ParentId="32837" CreationDate="2020-02-19T22:57:21.493" Score="1" Body="&lt;p&gt;Nichrome wire &lt;em&gt;is&lt;/em&gt; the correct material to use for this. That's almost certainly what the heat gun element is made from anyway.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You may be able to increase the longevity by using more nichrome wire for the same power, thus reducing the wire temperature. You will need to increase the wire diameter, or run strands in parallel to maintain the same wattage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to the popularity of vaping, assortments of nichrome wire in various gauges can be purchased on amazon and ebay.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2020-02-19T22:57:21.493" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33311" PostTypeId="1" CreationDate="2020-02-20T03:53:52.157" Score="1" ViewCount="1517" Body="&lt;p&gt;On large cargo ships (dry bulk carriers, or tankers) there are main and auxiliary engines.  Auxiliary engines can be internal or external combustion (generators vs boilers) and generate electricity to power the ship's equipment (cargo pumps, cargo cranes and all other machinery).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have read that tankers can have pumps that are powered by the main engine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a number of questions about the engine usage related to powering equipment:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Generally do ships have either boiler(s) or generator(s) as auxiliary engines, or could they have both for different purposes?&lt;/li&gt;&#xA;&lt;li&gt;Do newer ships more commonly use generators instead of boilers?&lt;/li&gt;&#xA;&lt;li&gt;Are larger equipment (cranes or cargo pumps) powered by the auxiliary engines or is it common to have these powered by the main engine? &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="24676" LastEditorUserId="24676" LastEditDate="2020-02-20T04:46:31.927" LastActivityDate="2020-02-21T00:06:36.830" Title="Auxiliary engines on cargo vessels, what types are used for what purpose?" Tags="|marine-engineering|ships|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33314" PostTypeId="2" ParentId="26094" CreationDate="2020-02-20T06:53:45.390" Score="0" Body="&lt;p&gt;It was probably drawn in “model space” and then converted to “paper space” to fit the drawing sheet size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ask the draftsman to print to a specific scale. (It May take a few sheets, but you can fit together.)&lt;/p&gt;&#xA;" OwnerUserId="23859" LastActivityDate="2020-02-20T06:53:45.390" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33315" PostTypeId="1" CreationDate="2020-02-20T08:57:58.260" Score="1" ViewCount="222" Body="&lt;p&gt;I saw a video where a person was measuring hardness of a material with Shore durometer type D and was comparing it to steel. The screen on durometer showed 96.5 HD and I was just wondering, how much would that be in Brinells? (or alternatively Rockwellss or Vickers)&lt;/p&gt;&#xA;" OwnerUserId="24679" LastEditorUserId="110" LastEditDate="2020-09-09T13:51:34.100" LastActivityDate="2020-09-09T13:51:34.100" Title="Shore durometer type D to brinells" Tags="|materials|material-science|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33318" PostTypeId="1" CreationDate="2020-02-20T13:20:03.990" Score="-1" ViewCount="83" Body="&lt;p&gt;I would like to know whats the relationship between these two, is it linear? or what? also i would like to know the mathematical relationship between these two. Thank you&lt;/p&gt;&#xA;" OwnerUserId="24684" LastActivityDate="2020-02-20T14:18:48.313" Title="What is the relationship between Rotational speed (RPM) and the speed of a drone (Air speed)" Tags="|dynamics|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33319" PostTypeId="1" CreationDate="2020-02-20T14:16:19.257" Score="0" ViewCount="38" Body="&lt;p&gt;I was reading about characteristic of a comparator.&#xA;It states&#xA;Comparator indicator device should be such that the readings are obtained in least possible time and for this magnification system should be such that the readings are dead beat.And there should be no backlash&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone kindly explain what is dead beat reading and an example about blacklash in comparator&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="24685" LastActivityDate="2020-02-20T14:16:19.257" Title="What is dead beat reading?" Tags="|mechanical-engineering|measurements|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33320" PostTypeId="2" ParentId="33318" CreationDate="2020-02-20T14:18:48.313" Score="1" Body="&lt;p&gt;If you are referring to rotational speed (rpm) of the Unmanned Air Vehicle's (UAV) motors versus the air speed of the UAV, there are other variables involved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This includes the specific geometry of the UAV, all-up weight including selected battery, aerodynamics and number of motors installed. The propeller diameter and pitch is critical to the operation of the UAV and will change the performance of the vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many factors to complicate a mathematical relationship. I suspect that designers use a method involving trial and error as much as anything mathematical.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-20T14:18:48.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33321" PostTypeId="1" CreationDate="2020-02-20T14:54:20.727" Score="1" ViewCount="63" Body="&lt;p&gt;Currently, I am trying to find rotameter calibrated for R134A gas flow. However, most companies only sell rotameter or variable area flowmeter for water/Air/Nitrogen flow. Is it possible to use any air rotameter for R134A and convert the scale reading to corresponding refrigerant flow through the use of rotameter conversion equation(shown below) ? Will the converted mass flow rate be accurate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Qs = Qi * sqrt((Ts&lt;em&gt;Pa)/(Ta&lt;/em&gt;Ps))&#xA;Where Qs = true SCFM&#xA;Qi = indicated flowrate ('SCFM')&#xA;Ts = calibration temp (R)&#xA;Ta = actual temp (R)&#xA;Pa = actual pressure (psia)&#xA;Ps = calibration pressure (psia)&#xA;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="24686" LastActivityDate="2020-02-20T14:54:20.727" Title="Can any Air rotameter be used for refrigerant gas by using equation conversion?" Tags="|refrigeration|compressed-gases|compressible-flow|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33323" PostTypeId="1" CreationDate="2020-02-20T17:42:11.937" Score="1" ViewCount="79" Body="&lt;p&gt;I have a 10mm metal disk that is 1mm thick.  It is easily bent when a force is applied (green arrow) to the unsupported edge.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SCErh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SCErh.png&quot; alt=&quot;Base Shape&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to add a strut, but at what angle?  The part is weight-sensitive, so to keep it simple the strut will be a 1mm wire (single cast part).  A strut with a shallow angle looks like it will buckle, but increasing the angle adds to the overhang.  How do you find the balancing point that gives the most strength?  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rPc4Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rPc4Z.png&quot; alt=&quot;Strut Angles&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I&lt;/p&gt;&#xA;" OwnerUserId="24687" LastActivityDate="2020-02-20T17:42:11.937" Title="Strut Angle Vs Overhang" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33325" PostTypeId="1" CreationDate="2020-02-20T22:01:08.290" Score="-1" ViewCount="32" Body="&lt;p&gt;So my father has this patent, and I'm not entirely sure how a certain aspect of it is supposed to be working. I understand how it can be useful, but I don't understand how such a seemingly simple concept has not yet been applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The part labelled 119 is an eccentric cam, and as it rotates it can advance or retard the timing of the valve opening. Does it achieve this by changing the leverage? He claims it's &quot;pre-loading&quot; the spring so that the rocker assembly doesn't need as much force to open the spring, and thus it opens sooner. But, don't springs have a constant force?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm not familiar with patent drawings, but I'm fairly sure there is an additional piece holding up part 101 that has what was described as a &quot;float spring&quot; , that will not allow 101 to pivot until it is bottomed out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've asked him these questions, but he's not an educated man and I'm no engineer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://patents.google.com/patent/US9732639B1/en&quot; rel=&quot;nofollow noreferrer&quot;&gt;Link to Patent&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pBysL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pBysL.png&quot; alt=&quot;Image of Patent US9732639B1&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tldr; how does part 119 advance/retard valve timing?&lt;/p&gt;&#xA;" OwnerUserId="24691" LastEditorUserId="24691" LastEditDate="2020-02-20T22:07:07.743" LastActivityDate="2020-02-20T22:50:01.673" Title="Understanding a Variable Lift Valve Train Patent" Tags="|engines|timing|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33326" PostTypeId="2" ParentId="33261" CreationDate="2020-02-20T22:32:43.850" Score="1" Body="&lt;p&gt;The resonance is caused by the poles, therefore &lt;span class=&quot;math-container&quot;&gt;$\omega_r = \omega_{n_2} \sqrt{1-2\zeta_2^2} \approx 2.9462$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The zeros do not cause a resonance, but an antiresonance.&#xA;Thus, the antiresonance frequency is located at&#xA;&lt;span class=&quot;math-container&quot;&gt;$\omega_{ar} = \omega_{n_1} \sqrt{1-2\zeta_1^2}\approx 0.9592$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can verify this by drawing the Bode Diagram&lt;a href=&quot;https://i.stack.imgur.com/VHeeA.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VHeeA.png&quot; alt=&quot;Bode Diagram&quot;&gt;&lt;/a&gt;&#xA;Note that there is a small deviation between the values obtained using the formula and the Bode Diagram created with MATLAB.&#xA;The small error is caused by the poles and zeros being close to each other, while the used formula only holds for a pure second order system.&#xA;The further away the poles are from each other, the smaller the effect and the better the formula will estimate the (anti)resonance frequency.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2020-02-20T22:32:43.850" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33327" PostTypeId="2" ParentId="33325" CreationDate="2020-02-20T22:50:01.673" Score="1" Body="&lt;p&gt;The rotation of shaft 119 with the cam changes the load on the valve springs, but it also changes the location of the fulcrum of the lever assembly parts 101, 103 and 104.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the bulk of the cam moves closer to the pivot 104, the travel at the right end of 101 will be increased compared to the travel when the bulk of the cam is on the opposite side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can exaggerate the movement by considering that the cam is all the way to the left. Pushing up on 103 or 104 only a small amount causes the other end to travel a great distance. With the cam all the way to the right, 103/104 movement would have to be substantial.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In levers and other mechanical devices, it's sometimes easier to get a grasp on the action by enlarging or exaggerating it for a moment.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-20T22:50:01.673" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33328" PostTypeId="2" ParentId="32982" CreationDate="2020-02-20T23:05:44.873" Score="1" Body="&lt;p&gt;The three main drawbacks for a high bandwidth are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Noise amplification&lt;/strong&gt;.&#xA;The higher the bandwidth, the more you are going to act on high frequent stuff in your error signal.&#xA;Noise typically dominates at the higher frequency and therefore the system will start reacting to it.&#xA;Therefore, the noise will not be rejected  / ignored but amplified.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Robustness&lt;/strong&gt;.&#xA;In general, a high closed loop bandwidth will reduce the robustness margins (e.g. phase margin, modulus margin and gain margin).&#xA;If you have tight margins, you can increase them with some advanced filtering (e.g. notch filter).&#xA;However, the accuracy of how 'good' you can tune these filters depends on the model accuracy.&#xA;Thus, if there is a mismatch between your model and the setup, or the setup changes over time, this can have a negative effect on your robustness margins and can even cause instability.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;strong&gt;Large control signals&lt;/strong&gt;.&#xA;In general, a high bandwidth requires a high gain.&#xA;When the error is relatively large, this results in large control signals.&#xA;Then the input signal may be saturated, resulting in nonlinear plant behavior, which may be undesired.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="8216" LastEditorUserId="8216" LastEditDate="2020-02-20T23:52:29.677" LastActivityDate="2020-02-20T23:52:29.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33329" PostTypeId="2" ParentId="32964" CreationDate="2020-02-20T23:58:51.347" Score="2" Body="&lt;p&gt;I think you're messing up two different things.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, you have your process, which takes place at a frequency &lt;span class=&quot;math-container&quot;&gt;$f_p$&lt;/span&gt;.&#xA;If you sample by &lt;span class=&quot;math-container&quot;&gt;$f_s = 2 f_p$&lt;/span&gt; you are able to observe the behavior of process up to the frequency &lt;span class=&quot;math-container&quot;&gt;$f_p$&lt;/span&gt;.&#xA;The Nyquist frequency is the minimum frequency at which a finite bandwidth signal needs to be sampled to retain all of the information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if the controller needs to to react on the behavior observed at &lt;span class=&quot;math-container&quot;&gt;$f_p$&lt;/span&gt;, the controller bandwidth should be at least &lt;span class=&quot;math-container&quot;&gt;$f_p$&lt;/span&gt;, while the sampling frequency needs to be much higher.&#xA;According to (Franklin, 2015, p.613), the sampling rate should be at least 20 times the bandwidth of the controller. However, best performance is obtained when the sample rate is greater than 25 times the controller bandwidth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Reference:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Franklin, G., Powell, J., &amp;amp; Emami-Naeini, A. (2015). Feedback control of dynamic systems (Seventh ed.). Boston: Pearson.&#xA;mentions that &lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2020-02-20T23:58:51.347" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33330" PostTypeId="2" ParentId="33311" CreationDate="2020-02-21T00:06:36.830" Score="1" Body="&lt;p&gt;There is no one way to do things.  Ship design is continually evolving.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, cargo ships are low and slow, as in powered by large 2-stroke diesel engines with big propellors and slow speeds.  Higher speed vessels use smaller 4-stroke engines with smaller propellors and higher speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditionally, propulsion has been diesel, but as ship sizes get larger, companies seek energy efficiency and specialiaty requirements (noise, passenger comfort, etc.), diesel-electric is being embraced (diesel to electrical propulsion).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have two requirements: propulsion and auxiliary loads.  Auxiliary powers everything else.  Diesel-drives have diesel propulsion (2 or 4 stroke) and 4-stroke auxiliaries.  Diesel-electric drives have 4-stroke gensets (diesel generator) generating AC power and large electric motors for propulsion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ships use 50Hz or 60Hz auxiliaries.  A rough estimate would be 5% of propulsion power would power auxiliaries.  A lot of energy goes up the stack.  Heat exchangers recovery this heat to heat boilers, which can be used as required.  In principle, this energy could be used to generate the electricity powering some or all of the power for the auxiliaries.  The problem being synchronizing this power with 50Hz or 60Hz auxiliary diesel generated power.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Yes to all possibilities.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;And this sort of answers this also.  As ships get larger and IMO forces energy efficiency on ship owners, diesel-electric ships become the predominant choice.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Again, this would be vessel specific, but most cargo pumps are electrical.  Electrically driven pumps are easier to control, but diesel-engine pumps have more capacity, with one cavaet, they burn more fuel.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;So FiFi monitors (long distance fire-fighting pumps putting out up to 2 tons/s - FiFi 3 = 9600 m3/hr @ 70m height) are typically main diesel-engine driven pumps.  They put out enough water to drive ships backwards over 3 knots.  In this case, it makes sense to use large capacity of the propulsion engines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cargo pumps are typically electrically driven, so the load can be provided by auxiliaries or even shore power (burn no fuel = cheaper electricity).  This is more of an efficiency or scale issue.  You can hammer a nail with a sled hammer, but you don't want to be holding the nail.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;8-16 electrically driven cargo pumps vs 1 or 2 diesels running at low efficiencies.  Diesels have the best fuel consumption around 75% to 85% of MCR (Manufacturers Continuous Rating). Power up auxiliaries and match load on auxiliaries to load required by pumps.  Propulsion engines would be below 25% load.  &lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2020-02-21T00:06:36.830" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33331" PostTypeId="1" CreationDate="2020-02-21T03:53:16.770" Score="6" ViewCount="63" Body="&lt;p&gt;I'm designing a device that uses impact to remove ice from sidewalks. I have a question about how to best design the impact mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can design it to deliver several light impacts or fewer harder impacts. The energy will be the same in both cases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which is the best way to go?&lt;/p&gt;&#xA;" OwnerUserId="24692" LastActivityDate="2020-02-22T15:26:29.747" Title="Is a single large impact better than several small impacts for breaking ice" Tags="|mechanical-engineering|building-physics|impact|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33332" PostTypeId="1" CreationDate="2020-02-21T04:26:12.333" Score="0" ViewCount="139" Body="&lt;p&gt;Cracked gases usually are quenched after cracking reactions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why for lighter gases like ethane, indirect quenching is used, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But for cracking of heavier naphtha or gas oil, most cooling is performed by direct quench. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that cracked heavy gas oil contain a larger amount of fuel oil and vice versa but why.&lt;/p&gt;&#xA;" OwnerUserId="11561" LastActivityDate="2020-02-21T15:36:37.307" Title="Difference between light and heavy cracking of hydrocarbons?" Tags="|chemical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33333" PostTypeId="1" CreationDate="2020-02-21T04:54:13.707" Score="-1" ViewCount="29" Body="&lt;p&gt;How to make comments on the book of Patankar, titled 'f' in 1980? Is it really worthy for beginners in the field of numerical heat transfer to read it?&#xA;Besides, after finishing reading this book, what else learning materials should I review if I want to conduct a good simulation in duct convective heat transfer? &lt;/p&gt;&#xA;" OwnerUserId="24694" LastActivityDate="2021-01-13T19:01:45.207" Title="Comments on Patankar's book Numerical heat transfer and fluid flow" Tags="|fluid-mechanics|heat-transfer|simulation|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33334" PostTypeId="2" ParentId="33333" CreationDate="2020-02-21T05:05:41.997" Score="0" Body="&lt;p&gt;Read the material for the simulation package you are using, also check to see if there are other relevant books: Simonson Heat Transfer may be worth a look - at least I found it very useful and it is still in my personal library.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can’t comment on Patankar...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-02-21T05:05:41.997" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33335" PostTypeId="1" CreationDate="2020-02-21T06:36:42.453" Score="-1" ViewCount="52" Body="&lt;p&gt;There are hydraulic cylinders with the size of a train wagon, or a smaller building, like this one: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/pjqe1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/pjqe1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But what are they used for? Could you please give a few examples in the world: perhaps metal processing factories, mining machines (guess)? Maybe provide a link to a few pictures on their applications, or cases where they are assembled into something larger?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="24697" LastEditorUserId="8055" LastEditDate="2020-02-21T19:45:27.017" LastActivityDate="2021-01-13T18:01:35.157" Title="What exactly are Large Hydraulic Cylinders used for?" Tags="|actuator|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33336" PostTypeId="2" ParentId="33335" CreationDate="2020-02-21T06:42:44.927" Score="1" Body="&lt;p&gt;Have been used on offshore oil rigs to get the platform level compared to the sea bed.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-02-21T06:42:44.927" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33337" PostTypeId="2" ParentId="33332" CreationDate="2020-02-21T07:03:02.403" Score="1" Body="&lt;p&gt;Not only fuel but consider the properties of each gas, cracked gases when cooled usually turn to an organic fluid, so to extract that excess water you just need to cool those things to room temperature. Direct quenching in lighter gases will mean you need more energy to get rid of the water. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can go ahead and try to design direct quenching of light gases and heavy gases in Hysys, then try to size them (yeah keep a constant flow for both) and you can see which is the cheaper option. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Direct quenching is more efficient because &quot;there is no better area of contact than the medium itself&quot; but if separating them is tough you tend to move away from it. &lt;/p&gt;&#xA;" OwnerUserId="17922" LastActivityDate="2020-02-21T07:03:02.403" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33338" PostTypeId="1" AcceptedAnswerId="33345" CreationDate="2020-02-21T07:47:18.163" Score="0" ViewCount="47" Body="&lt;p&gt;I wanted to modify my kneading machine. In fact it can work for at maximum 5 minutes, and then I have to let it cool for 20 minutes (it is a very cheap model), and it is a problem for me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I was thinking of adding a sort of cooling system, for instance a fan (like that used in motherboards). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lX6jN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lX6jN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking of removing the red &quot;hat&quot; of the machine and replacing it with a bigger hat, so that there will be space for the cooling system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have the following questons:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Do you agree with me with the position of &quot;cooling ports&quot; from with the air will enter and exit (IN, OUT)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) How can I realize the bigger hat? Which material? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) How can I connect the bigger hat to the machine?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) I was thinking of doing this modification because it is a cheap model and because I think that the only problem of using it for more than 5 minutes is the excessive heat. So if we solve it with a proper cooling system, I can use it for instance for 20 minutes. Do you agree with me about this last statement?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No problems about the electrical supply of the fan because I will build an external supply circuit for it.&lt;/p&gt;&#xA;" OwnerUserId="24327" LastActivityDate="2020-02-21T19:45:06.343" Title="Building a cooling system for a kneading machine?" Tags="|mechanical-engineering|materials|manufacturing-engineering|cooling|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33339" PostTypeId="1" CreationDate="2020-02-21T12:02:35.073" Score="0" ViewCount="454" Body="&lt;p&gt;I’m new to the nonlinear static simulation in SOLIDWORKS… and therefore I’m trying to understand the advanced options in SOLIDWORKS Simulation. And here the “singularity elimination factor” is mentioned.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found this link - &lt;a href=&quot;https://www.javelin-tech.com/blog/2014/04/singularity-elimination-factor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.javelin-tech.com/blog/2014/04/singularity-elimination-factor/&lt;/a&gt; - with a short explanation about the SEF. But still there are some general questions for me to clarify and I hope you can help me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Questions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The proposed modification of the value SEF enables the simulation to terminate. Values closer to 0 are more effective. But how is this adjustment of SEF affecting the results? Is there anything to be aware of?&lt;/li&gt;&#xA;&lt;li&gt;This article mentions four points (stiffness singularities, incremental strain being too large, incremental rotations being too large, contact iterations non-converging) which may cause the simulation fail. How can I relate which of the four points is responsible for the simulation fail?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for your help.&lt;/p&gt;&#xA;" OwnerUserId="17780" LastActivityDate="2020-02-21T12:02:35.073" Title="Using the „Singularity elemination factor“ in SOLIDWORKS Simulation" Tags="|solidworks|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33341" PostTypeId="2" ParentId="33332" CreationDate="2020-02-21T15:36:37.307" Score="1" Body="&lt;p&gt;Cat crackers in the US do not quench gas-oil products except incidentally by the steam stripping of the catalyst. Hot gas oil and hot catalyst are combined and quickly cracked, then the vapor products are separated from the catalyst in the reactor Some directly and some by steam stripping. All product and steam goes to vapor recovery. Catalyst goes to the regenerator where small remaining amounts of hydrocarbon and coke are burned to reheat the catalyst.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-02-21T15:36:37.307" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33342" PostTypeId="1" CreationDate="2020-02-21T16:27:30.943" Score="0" ViewCount="31" Body="&lt;p&gt;I am doing R&amp;amp;D work and am in need of terminals that can withstand 1,000 degrees Celsius, but cannot seem to find anything anywhere.  The highest temperature ratings I have found are around 600 degrees Celsius.  The terminals will be used in a furnace like enclosure.  Any guidance would be appreciated.  &lt;/p&gt;&#xA;" OwnerUserId="24706" LastActivityDate="2020-02-21T16:27:30.943" Title="Extreme temperature quick disconnect terminals" Tags="|electrical-engineering|material-science|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33344" PostTypeId="2" ParentId="33331" CreationDate="2020-02-21T19:13:45.287" Score="4" Body="&lt;p&gt;Multiple smaller, light impacts are better. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heavy concentrated impact can damage and crack the already fragile pavement which has likely lost most of its resilience due to the freezing temperatures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, a multiple teeth kind of a roller dragged behind or ahead of a snow shovel truck will break the ice layer into more uniform, smaller shards, turn them into a sort of sand or gravel, easier to mop or lift by a machine.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-02-21T19:13:45.287" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33345" PostTypeId="2" ParentId="33338" CreationDate="2020-02-21T19:45:06.343" Score="3" Body="&lt;p&gt;The motors used in inexpensive, occasional-use appliances often do not have cooling fans in them because they are not designed for continuous operation. This means that cutting holes in the top case of your mixer might not help get the heat out of the motor windings- unless you added a fan for this purpose. That would be a time-intensive proposition, requiring several hours. You then have to ask yourself the question of what is an hour of your time worth?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general, if your appliance hits its thermal cutout after 5 minutes of operation, it is the wrong tool for the purpose and should be replaced by one of higher capacity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;An alternative solution is to buy a second mixer of the same type, and let the one cool down while using the other.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-02-21T19:45:06.343" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33346" PostTypeId="2" ParentId="33295" CreationDate="2020-02-21T19:48:20.817" Score="2" Body="&lt;p&gt;They are HF radio masts that are hinged down for flying operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My initial thoughts they were radio direction finders or airframe recovery booms, but my research does not support it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In port they are up.  With planes in the air, they are down.  Four port, four starboard.  Two forward, six stern.  They are connected by cables.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://www.the-blueprints.com/blueprints/ships/carriers-uk/4038/view/hms_ark_royal_1939/&quot; rel=&quot;nofollow noreferrer&quot;&gt;the-blueprints.com&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/GuuxP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/GuuxP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://external-preview.redd.it/WuW3Z-PY_RxXHWT7gWhAtiixqGqIOehyCg0WzSCtfMQ.jpg?auto=webp&amp;amp;s=d9091df063954c286b5f544bd8e6c12b0de09b87&quot; rel=&quot;nofollow noreferrer&quot;&gt;Deployed as vessel was sinking (Higher quality image).&lt;/a&gt;  You can clearly see end curves down towards water.  Deployed they are below flight deck.  Clearly seen from YouTube video &lt;a href=&quot;https://youtu.be/Rn77MzyWs9M?t=39&quot; rel=&quot;nofollow noreferrer&quot;&gt;With The 'ark Royal' (1940)&lt;/a&gt; as a plane takes off.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VF00p.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VF00p.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are substancial as seen from YouTube video &lt;a href=&quot;https://youtu.be/oTsR8IMpQzU?t=307&quot; rel=&quot;nofollow noreferrer&quot;&gt;HMS Ark Royal - 14 de Noviembre de 1941&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vxdIX.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vxdIX.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The timing is around when Radio Direction Finding was introduced (which is why I answered the question) and RDF was used by the HMS Rodney supporting the Arc Royal.  The Arc Royal was built before the Illustrious.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From: &lt;a href=&quot;https://ethw.org/Radio_Location_Goes_to_Sea_-_Chapter_2_of_Radar_and_the_Fighter_Directors&quot; rel=&quot;nofollow noreferrer&quot;&gt;Radio Location Goes to Sea - Chapter 2 of Radar and the Fighter Directors&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The new-construction carrier HMS Illustrious was commissioned in April 1940, and would be the first RN carrier to have RDF. Later, while exercising with her air squadrons off Bermuda, Illustrious tried using RDF to guide fighter intercepts. HMS Formidable would be the second RN carrier to have RDF, commissioning in October 1940.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;These antennas are common to carriers.  The next Arc Royal and the Illustrious have them too.  All carriers of the period and later have them.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SX2CY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SX2CY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://books.google.co.uk/books?id=ROE7BAAAQBAJ&amp;amp;pg=PA64&amp;amp;lpg=PA64&amp;amp;dq=uss%20essex%20anatomy&amp;amp;source=bl&amp;amp;ots=RizN2wUWBm&amp;amp;sig=suZbWrpKwN4_kT4xHdvxzhkEssE&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0CF4Q6AEwDGoVChMIipvOydv0xgIVpCnbCh1O3Qyj#v=onepage&amp;amp;q=radio&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;Essex Class Aircraft Carriers of the Second World War&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Lexington (CV-16) ... SBD Dauntless dive-bombers are marshalled at either end of the flight deck, and the &lt;strong&gt;radio masts&lt;/strong&gt; are hinged down for flying operations.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://forums.appleinsider.com/discussion/11888/&quot; rel=&quot;nofollow noreferrer&quot;&gt;anyone serve on an aircraft carrier or in the know about them?&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I served as an electronics tech onboard the USS Nimitz for three years and those were my babies.  Those are two piece 35 foot HF whip antenna.  They are in the down position during flight operation (and generally when underway) to keep jet wings safe.  They are manually lifted when in close confines.  They have nothing to do with HFDF or fall arrest.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;And finally, you can trust the &lt;a href=&quot;https://i.pinimg.com/originals/33/8e/00/338e004f4bf9ad8e3d62b639907e5a43.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;model world&lt;/a&gt; because they really do do their research.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WfzWY.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WfzWY.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://sketchfab.com/3d-models/ark-royal-3ce8cf93dc8b4779bc373849a4d32388&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sketchfab 3d Model of Arc Royal.&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1592" LastEditorUserId="1592" LastEditDate="2020-02-26T21:24:23.827" LastActivityDate="2020-02-26T21:24:23.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33351" PostTypeId="1" CreationDate="2020-02-22T09:42:41.527" Score="1" ViewCount="55" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CI26T.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CI26T.jpg&quot; alt=&quot;picture of the part&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you design a mold to create that part? The only way I can think of is to make the parting line horizontal to the part instead of vertical. But is there any way to design the mold so that the parting line is vertical like in the picture?&lt;/p&gt;&#xA;" OwnerUserId="24714" LastEditorUserId="110" LastEditDate="2020-02-22T12:49:20.510" LastActivityDate="2020-02-22T21:39:47.743" Title="Designing an injection molding mold for a part with a certain parting line and ejector pin location" Tags="|plastic|molding|injection|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33353" PostTypeId="2" ParentId="33331" CreationDate="2020-02-22T15:26:29.747" Score="-2" Body="&lt;p&gt;Several small ones.&#xA;Think of the Tesla truck broken window.&lt;/p&gt;&#xA;" OwnerUserId="5408" LastActivityDate="2020-02-22T15:26:29.747" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33354" PostTypeId="1" CreationDate="2020-02-22T16:46:59.870" Score="0" ViewCount="436" Body="&lt;p&gt;I’m trying to assemble a pistol pin into a piston using creo parametric 5.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I select the surface of the pin and the inside hole of the pistol this works fine and say that it is partially contrained.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, when I select the planes front on the pin and the piston it says invalid constraint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/FagcB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/FagcB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first part of the image works fine but not the second part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any ideas why this may be ? &lt;/p&gt;&#xA;" OwnerUserId="19105" LastActivityDate="2020-02-22T16:46:59.870" Title="Invalid constraints in creo" Tags="|computer-aided-design|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33356" PostTypeId="2" ParentId="33351" CreationDate="2020-02-22T21:39:47.743" Score="1" Body="&lt;p&gt;Yes, this can be done as follows. The multiplicity of through-holes in the sidewalls of the part can be created using what's called a &lt;em&gt;slide&lt;/em&gt; populated with small square pins that are moved in by hydraulic pressure or a mechanical cam to shut off against the male core in the mold. The part is extracted from the mold by withdrawing the slides and stripping the finished part off the core. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The pins are machined with standard draft angles on their sides to accomodate the particular type of resin being molded, and the standard practices of the molding shop. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Good design usually includes pockets in the core into which these pins fit closely, to get good shutoff action without leaving a thin skin of plastic blocking the hole (which is called &lt;em&gt;flash&lt;/em&gt;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The actuation means inside the mold is generically referred to as a &lt;em&gt;mechanism&lt;/em&gt; or a &lt;em&gt;cam&lt;/em&gt; and a mold having slides actuated by hydraulics is called a &lt;em&gt;hydraulic&lt;/em&gt;.  A mold having a mechanism, cam or hydraulic system built into it is referred to as &lt;em&gt;complex&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-02-22T21:39:47.743" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33358" PostTypeId="1" CreationDate="2020-02-22T23:20:50.260" Score="1" ViewCount="32" Body="&lt;p&gt;I am currently designing a symmetric parallel mechanism that can be modeled as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LbeJh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LbeJh.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here, &lt;span class=&quot;math-container&quot;&gt;$\tau^j$&lt;/span&gt; denotes actuator moments and &lt;span class=&quot;math-container&quot;&gt;$k_i$&lt;/span&gt; are torsion springs constants. For a specific point in the workspace of the system, the actuator moments as well as the torsion angles &lt;span class=&quot;math-container&quot;&gt;$\varphi_i^j$&lt;/span&gt; are known. It is of interest to determine possible sets of torsion spring constants &lt;span class=&quot;math-container&quot;&gt;$\mathbf{k}$&lt;/span&gt; that would result in the torsion angles &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{\varphi}$&lt;/span&gt;, given the actuator moments &lt;span class=&quot;math-container&quot;&gt;$\boldsymbol{\tau}$&lt;/span&gt;. The links can be assumed rigid.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, I am looking for the relationship between the actuator moments, torsion spring constants and torsion angles when the mechanism is in static equilibrium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current approach is to look at the total energy of the system. Since the potential energy of a torsion spring is &lt;span class=&quot;math-container&quot;&gt;$$ U = \frac{1}{2}k\varphi^2 = \frac{1}{2}\tau\varphi, $$&lt;/span&gt; the total energy of the system in static equilibrium can be written as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\frac{1}{2}(\tau^1\varphi_1^1 + \tau^2\varphi_1^2 + \tau^3\varphi_1^3) = \frac{1}{2}(k_1(\varphi_1^1)^2 + k_1(\varphi_1^2)^2+k_1(\varphi_1^3)^2+k_2(\varphi_2^1)^2 + k_2(\varphi_2^2)^2+k_2(\varphi_2^3)^2+k_3(\varphi_3^1)^2 + k_3(\varphi_3^2)^2+k_3(\varphi_3^3)^2)$$&lt;/span&gt;&#xA;or&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \sum_{j=1}^3\tau^j\varphi_1^j = \sum_{i=1}^3\sum_{j=1}^3k_i(\varphi_i^j)^2. $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the right way to model this system, or is my approach omitting some important observation?&lt;/p&gt;&#xA;" OwnerUserId="8674" LastActivityDate="2020-02-22T23:20:50.260" Title="How do I correctly model this system of torsion springs?" Tags="|modeling|springs|energy|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33359" PostTypeId="1" AcceptedAnswerId="33378" CreationDate="2020-02-23T02:32:44.017" Score="-1" ViewCount="67" Body="&lt;p&gt;I'm trying to create a place for these PCBS to mount onto:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://rads.stackoverflow.com/amzn/click/com/B072Z7Y19F&quot; rel=&quot;nofollow noreferrer&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.com/ELEGOO-Prototype-Soldering-Compatible-Arduino/dp/B072Z7Y19F/ref=pd_cart_vw_crc_1_2/133-3052301-7665733?_encoding=UTF8&amp;amp;pd_rd_i=B072Z7Y19F&amp;amp;pd_rd_r=75e160a5-3830-464f-8ad1-cb79cd1c127b&amp;amp;pd_rd_w=rUo6r&amp;amp;pd_rd_wg=K9U0z&amp;amp;pf_rd_p=3ffc9231-b582-4361-9816-b39fdbe2f421&amp;amp;pf_rd_r=12SMA97QX1HJGFXAQERM&amp;amp;psc=1&amp;amp;refRID=12SMA97QX1HJGFXAQERM&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but I don't know where to position the mounting holes for the PCB so that it can be&#xA;screwed onto a standoff because the documentation for them only states how long each side of it is. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already bought the PCBs and have a circuit soldered to one of them so I can't buy better documented ones. I've currently tried printing screw holes at varying distances to test for the ideal distance but the plastic bends too much to be reliably compared to the ideal unbending mounting board, so I would need to waste dozens of grams of plastic per failed hole distance test to make thick prints to accurately test.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone here have a solution to this?&lt;/p&gt;&#xA;" OwnerUserId="24722" LastActivityDate="2020-02-24T09:47:33.540" Title="How do you create a mounting device for a PCB if the PCB's mounting hole positions aren't documented?" Tags="|cad|3d-printing|prototyping|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33360" PostTypeId="1" AcceptedAnswerId="33367" CreationDate="2020-02-23T02:38:00.050" Score="-1" ViewCount="32" Body="&lt;p&gt;I am curious how taking the most basic display such as this OLED one that works with the ardunio is driven: &#xA;&lt;a href=&quot;https://i.stack.imgur.com/PqHTp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PqHTp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;a href=&quot;https://i.stack.imgur.com/deitO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/deitO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;i.e. We are using just a clock and data line to drive a 128x64 pixel display. The pixels containing the OLEDs with TFTs are arranged in a cross-hatch design, but what is the link between them? There are probably shift-registers to cut down on the number of pins, correct? Any guidance is helpful, thanks. &lt;/p&gt;&#xA;" OwnerUserId="15332" LastEditorUserId="15332" LastEditDate="2020-02-23T02:43:00.180" LastActivityDate="2020-02-23T19:02:33.123" Title="How is the most basic display driven?" Tags="|electrical-engineering|computer-engineering|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33361" PostTypeId="1" CreationDate="2020-02-23T14:45:55.190" Score="3" ViewCount="1067" Body="&lt;p&gt;I want to model a DC motor in order to eventually do some simulations (temperature/cooling related) and I'm wondering what is the best way to model the motor windings?&#xA;I've seen a few video's on how others have done it but these only model 1 layer of windings onto the stator in the video's I've found so far (here is one of the videos: &lt;a href=&quot;https://www.youtube.com/watch?v=j_7m2-e9MFw&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=j_7m2-e9MFw&lt;/a&gt; )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is the reference image for the motor I am trying to model:&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/lhpzT.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lhpzT.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;I'll only need to model one of the windings (1 of 36 if i counted correctly) since I can use a circular pattern.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: I could perhaps model a simplified version of the coils for the simulation (for example using rings of copper instead of spirals) if it would be significantly easier to model and give results that are not too far off.&#xA;Or should I perhaps model sleeves of copper (of the approximate size/mass of the copper windings) instead of the individual windings&#xA;? Would this still give me a somewhat accurate result? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Background: I want to design a liquid cooling system for this motor and want to simulate the effectiveness of several different designs using Solidworks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The windings have multiple layers (and the ones in the YouTube videos I found only have 1 layer). How would I go about creating the 'multilayered spiral shape' I need in order to create the winding model using sweep command? Or is sweep not the best option for this model? if so what modeling method would you recommend?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first layer of the 'spiral' will go from left to right, the next layer from right to left, the following from left to right again and so on. The same way as the wire in this following video is wound around a spool : &lt;a href=&quot;https://www.youtube.com/watch?v=j_7m2-e9MFw&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=j_7m2-e9MFw&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My current plan of approach is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Draw a 3d sketch of the first (bottom layer, the one that is wound directly against the stator) layer of winding(creating the 3d sketch in the same way as in the video ( &lt;a href=&quot;https://www.youtube.com/watch?v=GBNAnqolQ-I&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=GBNAnqolQ-I&lt;/a&gt; )). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It should look something like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BnPNw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BnPNw.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;image source: &lt;a href=&quot;https://www.youtube.com/watch?v=GBNAnqolQ-I&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=GBNAnqolQ-I&lt;/a&gt; (at 12:46)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;2)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Add a fillet (round the corners) to all of the corners of this 3d sketch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should look similar to this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q93N6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q93N6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Copy the 3d sketch and scale up the copy such that it's new location matches the location of the second winding layer (I'll have to scale up the sketch in 2 of 3 axis only since the width of the winding doesn't change depending on which winding layer I'm modeling)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mirror the 3d sketch such that the new winding layer is wound in the opposite direction as the previous layer (from left to right instead of from right to left and visa versa)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;5)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Repeat this process until the desired amount of layers is achieved&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;6)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Connect the several copies (aka layers) of the initial 3d sketch together to create a single 3d sketch out of them&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;7)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Use sweep feature along 3d sketch to create a 3d model of the windings&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;There are a couple of issues with this method though... :&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I ever decide to change the number of winding layers or thickness of the winding wire or radius of the stator (the part that the wire wraps around) I will have to do a lot of 'cleanup work' to the winding model in order to not get intersection of the windings with each other (if i were to decrease stator radius without decreasing wire thickness) or get way too much spacing in between the wires (if i were to for example reduce wire thickness without changing the stator's 3d model manually (stator = the thing the wire wraps around). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most optimal solution would be one where the amount of 'turns' in each layer would be automatically adjusted based on the wire thickness, the width of the spiral itself and the predetermined spacing between individual wires.&#xA;And what would be even better is if I were to change the wire thickness.. all the layers would 'shift' upwards when increasing wire thickness and downward when decreasing wire thickness (since the same amount of layers with thicker wire results in more overall thickness).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Perhaps this is not possible though and I should accept that I'll have to remodel/retouch the winding whenever I change any of these parameters.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope that explanation was somewhat clear. If not please let me know and I'll try to clarify.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there something I can improve to my method of approach or an entirely different approach which would be better? What approach would you yourself use if you were to model this ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any input is greatly appreciated! Thank you!&lt;/p&gt;&#xA;" OwnerUserId="10328" LastEditorUserId="10328" LastEditDate="2020-02-23T20:33:03.000" LastActivityDate="2020-02-24T09:42:19.630" Title="How to model Motor Windings in Solidworks?" Tags="|modeling|cad|solidworks|technical-drawing|drawings|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33365" PostTypeId="1" AcceptedAnswerId="33374" CreationDate="2020-02-23T17:45:17.913" Score="1" ViewCount="23" Body="&lt;p&gt;I read that wine bottle produces sound when u blow sideways onto its rim because &#xA;some frequencies of the white noise produced in your mouth gets amplified when they match the natural frequency of the bottle. But is the tight neck of the bottle necessary for this to happen? If I blow into a closed end tube with uniform width, can't I produce the same effect? Why do I need the small opening?&lt;/p&gt;&#xA;" OwnerUserId="24473" LastActivityDate="2020-02-24T08:22:20.213" Title="How do cavity resonators differ from tubes of uniform width?" Tags="|fluid-mechanics|airflow|compressed-air|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33367" PostTypeId="2" ParentId="33360" CreationDate="2020-02-23T19:02:33.123" Score="2" Body="&lt;p&gt;Pretty much every display is multiplexed. Typically each column is turned on (pulled to positive supply, for example) in turn and the rows for the pixels to be turned on in that column are switched to the opposite polarity (pulled to negative supply in this example). Usually the sequence happens at high speed so that there is no noticeable flicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For very small displays the rows and columns can be driven by the microcontroller. For larger displays this would require too many pins on the controller so the data to be displayed is somehow shifted in serially and shift-registers and latches hold the data within the display. This can have an advantage for displays that change relatively infrequently because the microcontroller can send the data to the display and then forget about it while it gets on with more important stuff. Meanwhile the display can continuously refresh based on the last data it received.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;We are using just a clock and data line to drive a 128 &amp;times; 64 pixel display. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The simplest implementation of this display would require 128 + 64 pins to drive it. Obviously this is not possible on an Arduino.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The pixels containing the OLEDs with TFTs are arranged in a cross-hatch design, but what is the link between them? There are probably shift-registers to cut down on the number of pins, correct?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2020-02-23T19:02:33.123" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33369" PostTypeId="1" CreationDate="2020-02-23T20:13:08.263" Score="3" ViewCount="54" Body="&lt;p&gt;I am posed with a scenario where a hot water solar thermal system has been setup. The system comprises of a cold water tank, heating arrays and then a hot water tank. The system is closed loop and water always keeps recirculating. The system uses CPC Evacuated vacuum tubes in the link below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://andyschroder.com/CPCEvacuatedTube/TechnicalSpecifications/&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://andyschroder.com/CPCEvacuatedTube/TechnicalSpecifications/&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As a part of the system's primary loop, water is drawn from the cold water tank, passed through the array and it arrives into the hot water tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the secondary loop is used to consume the energy and cool down the water in this process. In the secondary loop, Water is drawn from the hot water tank, passed via a heat exchanger and it arrives back into cold water water tank.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the heat demand from the secondary loop is consistant and higher then the supply, everything works in harmony and the hot water tank temperature stays in control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problems comes in when the secondary loop no longer requires heat energy(due to a downstream low demand). Since the primary loops keeps generating energy, water in the hot water tank exceeds 95C and drain procedures are activated to prevent steam generation/excessive pressures.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is exagerated by the fact that the downstream system's demand pattern is very inconsistent and the demand maybe zero for several days frequently(so I know the thermal system has been over designed but downsizing permanently is not an option)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Question:&#xA;Considering the situation above, what is the safest way to stop power generation from the arrays.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that just stopping the recirculation would cause hot water residing inside the vacuum tubes to heat up beyond boiling point and pop PRVs on the array. Over some time this would also mean the copper tubes inside the vacuum tubes would scale and deteriorate aswell. Is this thought correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A solution that came to my mind was to stop the recirculation when the hot water tank has very high temperature and use compressed air to purge the water stagnant inside the array? Is this a good idea?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If air purging could be practical, I wanted to understand if there is a type of valve that would allow water to go though while preventing compressed air to go through? If such a mechanical valve exists, water could be purged through different parallel arrays without investing into automatic valves per array.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking forward to hearing your thoughts on how would you tackle this problem.&lt;/p&gt;&#xA;" OwnerUserId="24729" LastActivityDate="2020-02-23T21:17:51.117" Title="How to periodically shut down a recirculating hot water solar thermal system" Tags="|mechanical-engineering|heat-transfer|heating-systems|valves|heat-exchanger|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33370" PostTypeId="2" ParentId="33369" CreationDate="2020-02-23T21:17:51.117" Score="2" Body="&lt;p&gt;I would not mess around like this at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would make a secondary heat dump (old radiator in the garden, swimming pool...) and divert excess heat there.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-02-23T21:17:51.117" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33372" PostTypeId="2" ParentId="22183" CreationDate="2020-02-24T02:55:20.423" Score="0" Body="&lt;p&gt;To enlarge a bit upon Carl Witthoft's answer, which I upvoted, remember that the swivel mechanism in the skateboard's wheel trucks is designed to provide a certain (fixed) amount of steering action for a given amount of board tilt. At ordinary board speeds (a brisk run) the board tilt provides the rider with approximately the right amount of lean angle to make the turn &lt;em&gt;coordinated&lt;/em&gt; (as it would be for instance on a bicycle in a turn). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The rider has no way to decouple his lean angle from the board tilt angle to produce a coordinated turn at speeds different from the board's &quot;design speed&quot;- meaning that at slower speeds, the rider will tend to fall towards the inside of the turn and at greater speeds he will tend to fall towards the outside of the turn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At high speeds, fighting this tendency puts the board into a speed wobble in which the upper part of the rider's body does not track the movements of the board while the lower part does. Mayhem ensues. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-02-24T02:55:20.423" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33374" PostTypeId="2" ParentId="33365" CreationDate="2020-02-24T03:41:55.417" Score="0" Body="&lt;p&gt;This phenomenon is known as flow-induced resonance.  We learned about this in the submarine Navy, as water flowing over an open hole in the deck will produce a tone, and making tones is a bad thing for subs.  The only links I could find for this were to scientific papers about the effect.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You personally need the small opening because you can't produce enough flow across the bottle if the opening is large.  But a large enough flow would work on any size hole/tube/bottle combination, although the resulting resonance may be outside a human's perceivable frequency range (20 Hz to 20 kHz for young healthy individuals).&lt;/p&gt;&#xA;" OwnerUserId="24351" LastEditorUserId="24351" LastEditDate="2020-02-24T08:22:20.213" LastActivityDate="2020-02-24T08:22:20.213" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33375" PostTypeId="1" CreationDate="2020-02-24T08:33:23.087" Score="3" ViewCount="178" Body="&lt;h2&gt;The application&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I want to implement the control of a photovoltaic inverter. The goal would be to keep the consumption of a microgrid (the power absorbed from the grid) to a constant value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I can only measure the grid load and the PV inverter setpoint. I cannot measure the PV production.&lt;/li&gt;&#xA;&lt;li&gt;My measuring frequency is about 1s&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Some numbers: let's say that I have a load varying (quite fast) between &lt;code&gt;500&lt;/code&gt; and &lt;code&gt;1000W&lt;/code&gt; and a &lt;code&gt;500W&lt;/code&gt; solar panel, and I want to keep the power absorbed from the grid at &lt;code&gt;~500W&lt;/code&gt; (this is my &lt;em&gt;target&lt;/em&gt;). Thus, if the measured grid load is at &lt;code&gt;769W&lt;/code&gt;, the PV inverter should provide &lt;code&gt;769-500=269W&lt;/code&gt; (&lt;em&gt;inverter setpoint&lt;/em&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This would seem trivial, in the sense that I can control the inverter and tell it to provide exactly &lt;code&gt;269W&lt;/code&gt;. However, due to several delays (communication latency, inverter command reception time and the time it takes to adjust to the new inverter setpoint), it might take several seconds for the new command to actually take place: by the time the &lt;code&gt;269W&lt;/code&gt; setpoint is reached, the new requested value will probably be different (as the &lt;em&gt;measured load&lt;/em&gt; varies).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;The algorithm&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I tried implementing a PID-like algorithm to fix this, and it seems to work more or less fine. However, compared to normal PIDs, there is a key difference. In my algorithm, I measure the grid load (PV = Process Variable) and calculate the &lt;strong&gt;correction&lt;/strong&gt; that needs to be applied to the inverter setpoint (SP) based on how far the PV is from the target (T):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$E = PV - T$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$C_P = k_P \times E$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$C_I = k_I \times \int_{0}^{now} E(t) \, dt&#xA; $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$C_D = k_D \times \frac{\partial E}{\partial t}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$newSP = SP - (C_P + C_I + C_D) $&lt;/span&gt;,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$k_P$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$k_I$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$k_D$&lt;/span&gt; are tuning constants and &lt;span class=&quot;math-container&quot;&gt;$newSP$&lt;/span&gt; is my new setpoint (control signal) for the PV inverter. This is different from the usual PID process, since I calculate a &lt;strong&gt;correction&lt;/strong&gt; to apply to the current setpoint, and not the setpoint directly. I can do this since the PV and the SP refer to the same physical quantity, and it seems to make more sense this way. This also actually reduces or eliminates the need for an integral component, since by design we have no steady-state error that needs to be compensated for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I do not have any PID training, nor I have tried producing a first-order plus deadtime model of this system. I might be &lt;em&gt;completely&lt;/em&gt; wrong.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Testing&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;I've been testing this setup and it works well for this application, better than I could make a normal PID work (but to be fair I have zero experience on tuning a PID). As expected, I found that in this algorithm the integral component is mostly harmful, as it only leads to oscillations: the P and I components do not need to balance each other anymore like on a normal PID, since the P correction goes to 0 once we reach the desired target, and the setpoint becomes stable. Thus, I have found myself using mostly a PD control.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;h2&gt;Questions&lt;/h2&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Does this PID-like algorithm make any sense in this case? In particular, the choice to act on the correction value instead of the PV value itself?&lt;/li&gt;&#xA;&lt;li&gt;Is this a standard control method? I would like to find some base-level theory on this.&lt;/li&gt;&#xA;&lt;li&gt;Is this application more suited to some other control mechanism?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="24602" LastActivityDate="2020-03-04T11:49:00.600" Title="Does this PID-like algorithm make any sense for a solar inverter?" Tags="|control-engineering|control-theory|pid-control|photovoltaics|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33377" PostTypeId="2" ParentId="33361" CreationDate="2020-02-24T09:42:19.630" Score="-1" Body="&lt;p&gt;I'm can't offer much advice on the thermal analytics part - don't have the specific knowledge.&#xA;However, your modeling problem has a simple solution. You can define variables in the Solidworks model, and use them (or equations derived from them) in your drawings.&#xA;Take a look here:&#xA;&lt;a href=&quot;https://help.solidworks.com/2018/English/SolidWorks/sldworks/HIDD_EQUATION_MANAGER.htm?id=4e1fdbad62f645edaff8e62eb3c9684c#Pg0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://help.solidworks.com/2018/English/SolidWorks/sldworks/HIDD_EQUATION_MANAGER.htm?id=4e1fdbad62f645edaff8e62eb3c9684c#Pg0&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24734" LastActivityDate="2020-02-24T09:42:19.630" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33378" PostTypeId="2" ParentId="33359" CreationDate="2020-02-24T09:47:33.540" Score="0" Body="&lt;p&gt;Measure them using a caliper.&#xA;The best method will be to measure the inner and outer distance between two holes. The center-to-center distance is the average of the two measurements.&lt;/p&gt;&#xA;" OwnerUserId="24734" LastActivityDate="2020-02-24T09:47:33.540" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33379" PostTypeId="1" CreationDate="2020-02-24T10:38:08.737" Score="0" ViewCount="1252" Body="&lt;p&gt;Why does increasing derivative gain &lt;span class=&quot;math-container&quot;&gt;$K_d$&lt;/span&gt; in a  PID controller loop lead to a reduction in overshoot and an increase in rise time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is not a homework question, it is a question to improve my understanding. I know that the output of the derivative path of the controller is &lt;span class=&quot;math-container&quot;&gt;$K_d \times \frac{de}{dt}$&lt;/span&gt;, but I don't understand why increasing &lt;span class=&quot;math-container&quot;&gt;$K_d$&lt;/span&gt; would lead to the rise time increasing and less of an overshoot as when &lt;span class=&quot;math-container&quot;&gt;$\ K_d $&lt;/span&gt; is greater, the controller output would be greater, thus leading to an increased overshoot? &lt;/p&gt;&#xA;" OwnerUserId="24557" LastEditorUserId="18238" LastEditDate="2020-03-18T05:58:39.807" LastActivityDate="2020-03-18T05:58:39.807" Title="Increasing Kd in PID control loop" Tags="|control-engineering|chemical-engineering|control-theory|pid-control|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33380" PostTypeId="1" CreationDate="2020-02-24T14:31:47.677" Score="1" ViewCount="49" Body="&lt;p&gt;I was looking at how the water cools my body when it evaporates off it. Then I thought the specific heat capacity could be used to calculate how much energy (i.e heat) would be taken out of a room when water that absorbed its heat is transferred out of the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried and researched all I could, but as I am not an physicist, I can't really get calculations together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a setup with one mister, which is simply a device that atomizes water to small drops (I have no clue how small), and a large fan (it draws air) on the ceiling to &quot;suck&quot; up the vapour and transfer it to the open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Room size = 2m x 2m x 3m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mister sprays 1L of water per minute (16.67ml per second), as tiny drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am getting confused, as some water would probably vaporise and some would remain water. I think that the small drops have a large surface area, so absorb energy fast, and thus turn into gas (steam?). I can't understand how to calculate the heat energy transfer from the room to the water and steam, how to use ratios. All this is getting messed up the more I think of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone direct me what I must calculate, as I'm really dumbfounded as of now.&lt;/p&gt;&#xA;" OwnerUserId="24715" LastEditorUserId="1832" LastEditDate="2020-02-26T14:21:05.593" LastActivityDate="2020-02-26T14:21:10.323" Title="Calculation for lowering temperature in a room… room-fridge with water mist?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33382" PostTypeId="2" ParentId="33379" CreationDate="2020-02-24T14:43:40.777" Score="1" Body="&lt;p&gt;It is basically the same as a damper on a car.&#xA;It will counteract on any change in &quot;velocity&quot;.&#xA;Hence, it will decrease the overshot, but it will also try to resist any initial change of velocity, thus increase the rise time.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2020-02-24T14:43:40.777" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33384" PostTypeId="1" AcceptedAnswerId="33385" CreationDate="2020-02-24T16:43:13.277" Score="-1" ViewCount="29" Body="&lt;p&gt;Why do we keep speed of turbofan of turbofan jet engine, low.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what is the maximum thrust a turbofan jet engine., how much can thrust a turbofan jet engine produce with just compressed air.&lt;/p&gt;&#xA;" OwnerUserId="24740" LastActivityDate="2020-02-24T18:37:13.327" Title="Effect of speed of turbo fan in bypass jet engine" Tags="|mechanical-engineering|fluid-mechanics|applied-mechanics|aerospace-engineering|jet|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33385" PostTypeId="2" ParentId="33384" CreationDate="2020-02-24T18:37:13.327" Score="1" Body="&lt;p&gt;The fan speed is kept low because the blades in it must not exceed the speed of sound during operation, otherwise the efficiency of the fan will go down and noise generation will go up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am not sure your second question is meaningful; all jet engines generate thrust by discharging a large mass flow rate of compressed air from their tailpipes. They compress that air by burning fuel and capturing some of the heat energy thus produced in a turbine which then spins the compressor.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-02-24T18:37:13.327" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33386" PostTypeId="1" CreationDate="2020-02-24T20:10:49.103" Score="-1" ViewCount="266" Body="&lt;p&gt;Let me start off by saying that this is strictly NOT a homework question, so before you mark it as &quot;homework&quot;, think.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was looking at how the water cools my body when it evaporates off it. Then I thought the specific heat capacity could be used to calculate how much energy (i.e heat) would be taken out of a room when water that absorbed it's heat is transferred out of the room.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tried and researched all I could, but as I am not an physicist, I can't really get calculations together. Every time, I put down calculations using specific heat capacity, mass, temperature change, energy taken in/out, and then everything looks out of place.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a setup with one mister, which is simply a device that atomizes water to small drops (I have no clue how small), and a large fan (it draws air) on the ceiling to &quot;suck&quot; up the vapour and transfer it to the open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Room size = 2m x 2m x 3m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mister sprays 1L of water per minute (16.67ml per second), as tiny drops.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Temperature outside = 40 Celsius&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am getting confused, as some water would probably vaporise and some would remain water. I think that the small drops have a large surface area, so absorb energy fast, and thus turn into gas (steam?). I can't understand how to calculate the heat energy transfer from the room to the water and steam, how to use ratios. All this is getting messed up the more I think of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can someone direct me WHAT I must calculate, as I'm really dumbfounded as of now. Please, no closing the question for God's sake. I asked in multiple SE forums, but all were closed. Is this how useful SE gets?&lt;/p&gt;&#xA;" OwnerUserId="24715" LastActivityDate="2020-02-25T16:46:10.557" Title="How can water mist lower a room's temperature?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33387" PostTypeId="1" CreationDate="2020-02-24T22:43:27.107" Score="3" ViewCount="134" Body="&lt;p&gt;@Involutius contributed an &lt;a href=&quot;https://engineering.stackexchange.com/questions/13852/involute-gear-curve-when-root-diameter-falls-below-base-diameter&quot;&gt;article&lt;/a&gt; on the equations that describe the root profile of gears which are cut by a trapezoidal-profiled hobbing tool. From that article, I was able to do a slightly different derivation to get accurate curves for the involute as well as the trochoidal root curve, for both normal and profile-shifted gears. Thanks so much, Involutius!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I have another question regarding gear tooth profiles. The entire discussion of the root curve profile is based on a conventional (and, for that matter, a flat trapezoidal) hobbing gear cutter. I'm interested in somewhat more exotic gear applications, and I wonder if the newer power skiving machines can cut, say, negatively profile-shifted spur pinion without the severe impact to tooth strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Simply put, my question is, &quot;Does power skiving have a dramatically different result for root gear profiles than conventional hobbing?&quot;&lt;/p&gt;&#xA;" OwnerUserId="24645" LastActivityDate="2024-03-05T21:02:00.217" Title="Gear profiles (involute and root) generated by power skiving" Tags="|gears|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33388" PostTypeId="1" CreationDate="2020-02-25T00:09:08.383" Score="1" ViewCount="322" Body="&lt;p&gt;I have the data from a tensile test including axial displacement, force, and strain, as well as E1, E2, Nu12, and Nu21 for the carbon fiber prepreg. I am trying to use this axial data to calculate Gxy for a 4-ply laminate with the angular orientations [10/-10/-10/10]. How can this be done.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am confused since the Q matrices and Q bar matrices depend on Gxy so I don't know how to use those if we are solving for Gxy. Also, I do not have any data from the transverse directions, only in the direction of the applied load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The test coupon I used was 12&quot; long by 2&quot; wide.&lt;/p&gt;&#xA;" OwnerUserId="24743" LastActivityDate="2020-02-25T00:09:08.383" Title="How can I calculate the shear modulus (Gxy) for the material of a composite laminate [10/-10/-10/10] using test data?" Tags="|mechanical-engineering|structural-engineering|materials|composite|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33390" PostTypeId="1" CreationDate="2020-02-25T11:11:43.627" Score="-1" ViewCount="87" Body="&lt;p&gt;What books and online materials exist that explain the mechanical design of a bioreactor? Can someone recommend any?&lt;/p&gt;&#xA;" OwnerUserId="24749" LastActivityDate="2020-02-25T13:31:12.190" Title="Mechanical design of a bioreactor" Tags="|mechanical-engineering|systems-design|machine-design|biogas|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33391" PostTypeId="2" ParentId="33387" CreationDate="2020-02-25T13:08:33.087" Score="0" Body="&lt;p&gt;Tooth bending strength is more dependent on tooth geometry than actual manufacturing processes. You can get customs tools for either hobbing or skiving to achieve whatever root geometry you want although they will be more expensive than off the shelf tools.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-02-25T13:08:33.087" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33392" PostTypeId="2" ParentId="33390" CreationDate="2020-02-25T13:31:12.190" Score="1" Body="&lt;p&gt;Bioreactors: Process and Analysis, by Tapobrata Panda&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.co.uk/books?id=HVGSfMa062kC&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://books.google.co.uk/books?id=HVGSfMa062kC&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BJRIN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BJRIN.png&quot; alt=&quot;Bioreactors Book Cover&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;This book also covers the mechanical design considerations for the various kinds of bioreactors.&quot;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2020-02-25T13:31:12.190" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33393" PostTypeId="1" AcceptedAnswerId="33396" CreationDate="2020-02-25T15:17:40.660" Score="0" ViewCount="55" Body="&lt;p&gt;I used minutely sampled wind speed data for one year to calculate the yield power from specific wind turbines to be used in electric vehicles charging station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am supposed to calculate the number of electric vehicles that can be charged within each day given the total constant energy from wind turbine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many batteries can I charge using a 100 kW supercharger if I have energy of 45 MWh for one day? Battery capacity is 50 kWh, it takes 21 min to fully charged (or up to 80% for example) using 100 kW supercharger.&lt;/p&gt;&#xA;" OwnerUserId="24752" LastActivityDate="2020-12-13T06:44:12.657" Title="number of electric vehicles that can be charged with specific amount of energy" Tags="|wind-power|electric-vehicles|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33394" PostTypeId="1" AcceptedAnswerId="33401" CreationDate="2020-02-25T15:18:14.373" Score="-1" ViewCount="89" Body="&lt;p&gt;According to Wikipedia (I know) a parallel manipulator is a single end effector driven by several serial chains. Does this include a simple scissor lift mechanism? And is there a better definition of what a parallel mechanism actually is?&lt;/p&gt;&#xA;" OwnerUserId="24751" LastActivityDate="2020-03-02T08:01:25.990" Title="Is a scissor mechanism considered a parallel manipulator" Tags="|mechanical-engineering|mechanisms|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33395" PostTypeId="2" ParentId="33386" CreationDate="2020-02-25T16:46:10.557" Score="2" Body="&lt;p&gt;The answer to your question is explicitly furnished in a cooling effectiveness chart contained in the &lt;em&gt;Trane Handbook Of Air Conditioning&lt;/em&gt;, in the chapter dealing with &lt;em&gt;evaporative cooling&lt;/em&gt;. This chart has a complex family of curves in it and axis scales on it which will yield the temperature depression (cooling effect) as a function of the ratio of water to air flow rate, water source temperature, outside air temperature, and relative humidity. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;No calculations are required. You just select the input parameters, follow the lines, and their intersection point on the chart gives you the answer. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-02-25T16:46:10.557" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33396" PostTypeId="2" ParentId="33393" CreationDate="2020-02-25T17:15:22.847" Score="1" Body="&lt;p&gt;There isn't enough information in the question to give a reliable answer.  But here are some possible crude answers, just in case you are being tested on basic arithmetic and the relationship between power and energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 kWh is an amount of energy equivalent to 1 kW for 1 hour. It's also  the amount of energy equivalent to 10 kW for 0.1 hour, or 0.01 kW for 100 hours. It's just constant power multiplied by time; or, in a more complex but more realistic case, power integrated over time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1 MWh = 1000 x 1 kWh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are 24 x 3 = 72 lots of 20 minute intervals in a day. That means that a single 100kW charger could, at most, charge 72 vehicles for 20 minutes each in a day. This would be 24 hours x 100 kW = 2,400 kWh = 2.4 MWh, assuming that the batteries' states of charge would allow constant charging at the full 100 kW. That's way in excess of your 45 MWh available from the wind turbine each day on average.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way to look at it, is to assume that the number of chargers is not a limit - if demand increases, more will get built. In that case, if in a day you have 45 MWh available, and each charge is 33.33 kWh (100 kWh x 20 minutes), then you can charge 45,000 / 33 which is around 1,350 vehicles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a somewhat more realistic problem to the solution , you need to know when power is demanded, and when it is supplied. There will be trade-offs between not using power when it's generated, having chargers standing around unused most of the time, having people waiting for a charger, and building battery storage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But actually, in reality, you just wouldn't do it this way at all. You plug the wind turbines into the grid, and sell the power to the market. You plug the chargers into the grid, and buy power from the market. That way, all the needs of balancing supply and demand are carried out by the system operator, and they can do it far more economically than anyone else, because they're aggregating everyone's demand and everyone's supply.&lt;/p&gt;&#xA;" OwnerUserId="163" LastActivityDate="2020-02-25T17:15:22.847" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33397" PostTypeId="1" CreationDate="2020-02-25T17:23:19.283" Score="2" ViewCount="25" Body="&lt;p&gt;I'm using Aspen Custom Modeler to create a model to be used in Aspen Plus and Aspen Plus Dynamics, and I'm having trouble figuring out how to code the mass balance, energy balance, etc. I'm essentially coding up a pump, but a pump which takes both flow and inlet pressure as arguments to determine the outlet pressure. My model code is as follows.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Model XePump&#xA;&#xA;  In1 as input MainPort;&#xA;  Out1 as output MainPort;&#xA;&#xA;  // Internal&#xA;  Pdiff             as press_diff               (description:&quot;Pressure difference&quot;);&#xA;  Svar              as fraction                 (1.0,fixed,description:&quot;Operating speed parameter&quot;);  &#xA;&#xA;  Pdiff = 1.0 * In1.P * (0.16169605386 - In1.F) * Svar;&#xA;  Out1.P = In1.P + Pdiff;&#xA;  Out1.F = In1.F;&#xA;&#xA;End&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;The errors I'm getting are in Aspen Plus, the model doesn't have any outlet flow rate, despite explicitly setting it to be equal to the inlet flow rate. Any tips?&lt;/p&gt;&#xA;" OwnerUserId="20819" LastActivityDate="2020-02-25T17:23:19.283" Title="Custom Models in Process Engineering" Tags="|modeling|process-engineering|software|pi-diagram|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33398" PostTypeId="1" CreationDate="2020-02-25T17:28:33.457" Score="2" ViewCount="132" Body="&lt;p&gt;I've been racking my head around calculating the amount of heat loss still water will experience through a run of insulated piping. The water in the tank is maintained at roughly 550°F saturated, and a standpipe measures the water level some distance away. However, most of the example problems and literature I can find assume some movement of water through the piping, and here, we know the water will be still. I'm assuming this needs to be setup in a way that we can equate the Qin to be equal to Qout, with Qout being losses due to radiation and convection, and we should be able to find the temperature at the end of the pipe if we know all of the other parammeters, but I'm not sure how to setup my Qin. Any ideas? Thank you in advance.&lt;a href=&quot;https://i.stack.imgur.com/Ntno9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ntno9.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24755" LastActivityDate="2020-02-25T19:49:10.470" Title="Heat Loss through a run of insulated piping" Tags="|heat-transfer|pipelines|piping|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33400" PostTypeId="2" ParentId="33398" CreationDate="2020-02-25T19:34:50.737" Score="2" Body="&lt;p&gt;Have a look at heatsink calculations used in power electronics. The thermal loss calculation is quite analogous to an electrical circuit. This will all be much simpler in SI units rather than your quaint &amp;deg;F but I've left the numbers there for now.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Voltage &amp;lrarr; Temperature above ambient (&amp;deg;C or K)&lt;/li&gt;&#xA;&lt;li&gt;Current &amp;lrarr; Heat flow (W)&lt;/li&gt;&#xA;&lt;li&gt;Resistance &amp;lrarr; Thermal resistance (&amp;deg;C/W or KW&lt;sup&gt;-1&lt;/sup&gt;) &lt;/li&gt;&#xA;&lt;li&gt;Ground &amp;lrarr; Ambient temperature (&amp;deg;C or K)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bU2nu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bU2nu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. A thermal model built using &lt;a href=&quot;http://circuitlab.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;CircuitLab&lt;/a&gt;&lt;sup&gt;1&lt;/sup&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the values I've chosen the simulator shows that the junction of pipe X and Y should be at 358&amp;deg;F and a heat throughput of almost 6 W.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suggest that you make up a resistor model as complicated as you require, measure the temperatures at each junction and calculate the heatloss on some section that you know the ratings for and play with the resistor numbers until the model agrees with your readings. Ultimately you're interested in the current through Rx.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;sup&gt;1&lt;/sup&gt; CircuitLab is not free but if you head over to SE's &lt;a href=&quot;http://electronics.stackexchange.com&quot;&gt;Electrical Engineering&lt;/a&gt; you can play with it for free. There's a button on the question editor toolbar.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2020-02-25T19:49:10.470" LastActivityDate="2020-02-25T19:49:10.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33401" PostTypeId="2" ParentId="33394" CreationDate="2020-02-25T20:44:25.757" Score="1" Body="&lt;p&gt;A parallel mechanism is one where multiple &quot;chains&quot; connect the moved part to the ground.  The parallel portion references that each chain (really a set of actuators and linkages) can move separately at the same time, and each is required to move in separately to achieve the desired platform orientation.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your two legs are a parallel mechanism.  For a dog, its 4 legs are a parallel mechanism.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A serial mechanism is a single leg - it extends, retracts, and twists to provide the needed input to the body.  The position of each joint is dependent on the supporting joint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A scissor lift is really just a simple mechanical advantage device that lifts a platform up and down only.  We could define it as a limited version of a parallel mechanism where the 2 serial mechanisms are linked, but I don't think the definition fits well.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastEditorUserId="24351" LastEditDate="2020-03-02T08:01:25.990" LastActivityDate="2020-03-02T08:01:25.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33403" PostTypeId="1" CreationDate="2020-02-25T22:59:30.370" Score="2" ViewCount="58" Body="&lt;p&gt;Recoiless guns have breeches that allow gas to escape out of the back of the gun. Therefore, theres no sealing action to be performed by the casing. Why is ammunition with perforated cases used, when a propellant bag (like in other types of artillery) affixed to the shell would be cheaper an lighter? Is it necessary to build up the pressure for combustion of the propellant? &lt;/p&gt;&#xA;" OwnerUserId="18744" LastEditorUserId="18744" LastEditDate="2020-02-25T23:14:27.327" LastActivityDate="2020-02-25T23:14:27.327" Title="Why do recoilless rifle ammunition still have casings?" Tags="|firearms|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33404" PostTypeId="1" CreationDate="2020-02-26T03:24:58.197" Score="1" ViewCount="107" Body="&lt;p&gt;I am working on a geoengineering project that, at the moment, seeks to use many mass drivers to project lunar regolith (lunar soil) or other materials into space to create targeted shading effects on the earth for specific purposes. The material would intersect the moon allowing for easy recollection of material for quick reuse. There are other space infrastructure implications that would allow sharing of resources and a lot that can be done to increase efficiency. The problem is I can't find anywhere to help with some of the basic calculations. I know that mass drivers can be 80-90% energy efficient on the higher end and the project will be in an extremely low atmosphere environment. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would like some help with the basic math to calculate energy needs. Even links to find the information and explore it would be welcomed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The idea is to constantly project small units, no more than 10lb but probably smaller, approximately 3km/second from a mass driver (the small size as I understand it should allow a faster rate of projecting material which should make the whole process easier). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How much energy would it take to take per pound of material to travel per km/second assuming 80% energy efficiency? Browny points if you can give an estimation of the refractory period between projections or point me to how to calculate that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P.S. This is a big complex project and I always get random questions going in every direction. I'm used to it and happy to answer any questions or add them to my list of questions that need answering, but please provide useful feedback instead of blank criticism and recognize that I only have so many characters per comment to respond.&lt;/p&gt;&#xA;" OwnerUserId="24747" LastEditorUserId="24747" LastEditDate="2020-02-26T03:41:28.997" LastActivityDate="2020-03-01T20:06:16.667" Title="How much energy do mass drivers need to project 1lb of material 1km/second" Tags="|mechanical-engineering|aerospace-engineering|mathematics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33405" PostTypeId="1" CreationDate="2020-02-26T05:41:20.120" Score="-1" ViewCount="36" Body="&lt;p&gt;As i read on some website of nasa, that velocity of the gas flowing from nozzle of the jet is responsible for thrust,&#xA;But ky confusion is that, as the velocity increases presure should drop at nozzle, and asnthe velocity at the compressor is low, presure must be high there.&#xA;And at the time when the gas flow through nozzle, walls of nozzle must experience presure in direction of that of the gas flowing out of the jet, all the forces produced by presure diffrence and that on nozzle is in opposite firection of the movement of the jet,&#xA;&lt;strong&gt;Then how does a jet produce a thrust in opposite direction and how velocity of gas flowing out of the nozzle of the jet affect the thrust produced by it.&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="24740" LastActivityDate="2020-03-01T20:15:06.157" Title="How a turbo jet engine produces thrust" Tags="|mechanical-engineering|automotive-engineering|aerospace-engineering|power-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33407" PostTypeId="1" CreationDate="2020-02-26T09:22:25.340" Score="1" ViewCount="32" Body="&lt;p&gt;Growing up I would have these  RC Car, they had the cover which eiher a cover of a truck or any brand of cars, I could take this cover off and then change the batteries of the rc car or just check the components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These covers came with the car, had an elastic plastic form and were quite strong&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Moving to my question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What process are those covers made ?&#xA;molding or the other options.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have built my own rc car and looking for a place I can order my own customized cover for the car.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design of the cover, I made myself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried 3d printing but to be honest I am not really happy with the results, due to the fact that I have to use sandpaper to smoothen it and it is quite heavy.&lt;/p&gt;&#xA;" OwnerUserId="24762" LastActivityDate="2020-02-26T10:20:05.997" Title="Rc Car BodyShell created" Tags="|plastic|molding|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33409" PostTypeId="2" ParentId="33407" CreationDate="2020-02-26T10:20:05.997" Score="2" Body="&lt;p&gt;Many/most radio control vehicle bodies are formed by using a vacuum forming mechanism. A buck is created, the shape to match the body, placed on a perforated surface. A sheet of thin plastic is heated and placed over the buck, simultaneous with applying vacuum to the perforated surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This action draws the softened plastic around the buck. In a few moments, the plastic is cooled, the vacuum is released and the sheet is trimmed from the desired shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a makerspace in your area, you may discover someone with a vacuum forming machine. They are reasonably easy to construct and can use an ordinary shop or house vacuum. The  heating of the plastic sheet can be more difficult without a specially constructed device, but there are YouTube videos showing people who use household ovens and hot air guns to soften the plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gVkCY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gVkCY.png&quot; alt=&quot;vacuum forming machine&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above image is a hand-made device currently offered on eBay and serves as a representative sample of a vacuum forming machine. Not shown is the heating system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A more sophisticated DIY version can be found on &lt;a href=&quot;https://www.instructables.com/id/Vacuum-Forming-Rig/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Instructables&lt;/a&gt;. There are quite a few DIY options on that site, most of which include references for the heating portion.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-02-26T10:20:05.997" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33410" PostTypeId="1" CreationDate="2020-02-26T10:57:52.830" Score="-2" ViewCount="77" Body="&lt;p&gt;How much thrust can I produce with a turbofan jet engine without injection of fuel and with just compression of air, which will be heated by convection if initial temperature of air of 20 &lt;sup&gt;o&lt;/sup&gt;C and its temperature after compression is 1400 &lt;sup&gt;o&lt;/sup&gt;C, which is then diffused through nozzle of the jet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to produce approximately 6.8 kN of thrust.&lt;/p&gt;&#xA;" OwnerUserId="24740" LastEditorUserId="90" LastEditDate="2020-02-28T12:41:57.963" LastActivityDate="2020-02-28T12:41:57.963" Title="Thrust by turbo fan jet engine" Tags="|mechanical-engineering|chemical-engineering|aerospace-engineering|process-engineering|power-engineering|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="33411" PostTypeId="1" CreationDate="2020-02-26T12:54:37.500" Score="-1" ViewCount="57" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/74aMJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/74aMJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;Everything i've tried so far is probably wrong. I tried using 40 as the traverse angle and 30 as the azmuth angle to find the cosine of the euler angles, and then multiplied all three by the force's magnitude to seperate the force. How would you solve a problem like this step by step? How do you seperate the force into components and what do you do to them? &lt;/p&gt;&#xA;" OwnerUserId="24765" LastActivityDate="2021-01-13T20:04:45.447" Title="How do you replace forces and place them in another location?" Tags="|statics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33412" PostTypeId="1" CreationDate="2020-02-26T13:01:31.523" Score="1" ViewCount="26" Body="&lt;p&gt;Complete hobbyist here. I have two questions I have a problem figuring out.&#xA;I have hard time finding information about axis and types of connections they have. For example take this worm gear reducer: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hs9Lq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hs9Lq.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It has axis connection with small extrusion on one side, where motors usually have shaft with one flat side. I found this adapter :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sCzWI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sCzWI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;on the site it's called &lt;code&gt;Axle sleeve&lt;/code&gt; but I haven't found this part by this name elsewhere, so maybe there is another more precise term?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a beginner friendly resource I can study to get more familiar with this system? What both types of connections are called etc.&lt;/p&gt;&#xA;" OwnerUserId="24766" LastActivityDate="2020-02-26T13:01:31.523" Title="Axis connection types" Tags="|motors|coupling|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33414" PostTypeId="2" ParentId="2764" CreationDate="2020-02-26T15:58:06.030" Score="0" Body="&lt;p&gt;The sad faces are where the hand insert components go. If any sad faces are showing, the operator hasn't finished their job.&lt;/p&gt;&#xA;" OwnerUserId="24769" LastActivityDate="2020-02-26T15:58:06.030" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33417" PostTypeId="2" ParentId="33411" CreationDate="2020-02-26T17:51:58.563" Score="0" Body="&lt;p&gt;Using 40 as the traverse angle and 30 as the azmuth angle to find the cosine of the euler angles, and then multiplied all three by the force's magnitude to seperate the force. The find the cross product of the r distance vector to a (.55i + .4j -.2j), to find the moment vector at a. The force has the same magnitude in each component, but also opposite signs. &lt;/p&gt;&#xA;" OwnerUserId="24765" LastActivityDate="2020-02-26T17:51:58.563" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33420" PostTypeId="2" ParentId="14982" CreationDate="2020-02-27T01:06:10.100" Score="0" Body="&lt;p&gt;It sounds like you just need a small &lt;a href=&quot;https://tankliners.com.au/&quot; rel=&quot;nofollow noreferrer&quot;&gt;liner&lt;/a&gt; on the plate itself. Foam sealants are typically used when the space that needs sealing is greater than 1/4inch. For smaller spaces, waterproof caulking may be used instead. Just Do not allow the foam sealant to come into contact with skin; it adheres to the skin and can pull off the top layer of the epidermis when removal is attempted.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Other quick and easy temporary measures for stopping pipe leaks include wrapping waterproof tape over the bad spot or rubbing the hole with a stick of special compound. Applying epoxy paste or inserting a self-tapping plug into the hole are other alternatives. When using waterproof tape, be sure to dry the pipe thoroughly before you start wrapping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's an interesting site I found about the topic too:&#xA;&lt;a href=&quot;https://haynes.com/en-us/tips-tutorials/how-trace-interior-water-leaks-your-car&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://haynes.com/en-us/tips-tutorials/how-trace-interior-water-leaks-your-car&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24425" LastActivityDate="2020-02-27T01:06:10.100" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33424" PostTypeId="1" AcceptedAnswerId="33429" CreationDate="2020-02-27T13:39:36.913" Score="2" ViewCount="334" Body="&lt;p&gt;I'm stuck half way through construction of parametric conical helix antenna on a dieletric. I have created the intersection line but I'm struggling to make it a planar surface on top of the conical object as shown in the picture attached. I also have attached the Solidworks file.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jnmtp.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jnmtp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24778" LastEditorUserId="90" LastEditDate="2020-02-27T21:17:29.620" LastActivityDate="2020-03-02T10:14:18.483" Title="Solidworks - parametric conical helix antenna on a dielectric" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33425" PostTypeId="2" ParentId="33375" CreationDate="2020-02-27T22:36:22.623" Score="2" Body="&lt;p&gt;&lt;strong&gt;The algorithm&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The standard control system is depicted below and has the following signals&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bXfcK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bXfcK.png&quot; alt=&quot;Control system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;reference signal: &lt;span class=&quot;math-container&quot;&gt;$r(t)$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;error signal: &lt;span class=&quot;math-container&quot;&gt;$e(t) = r(t) - y_m(t)$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;input/control signal: &lt;span class=&quot;math-container&quot;&gt;$u(t) = K_p e(t) + K_i \int_0^t e(\tau)\mathrm{d}\tau + K_d \frac{\mathrm{d}e(t)}{\mathrm{d}t}$&lt;/span&gt; &lt;/li&gt;&#xA;&lt;li&gt;output signal: &lt;span class=&quot;math-container&quot;&gt;$y(t)$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;measured signal: &lt;span class=&quot;math-container&quot;&gt;$y_m(t)$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using your notation, we get&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$r(t) = T$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$y(t)$&lt;/span&gt; = PV&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$e(t) = r(t) - y(t) = T - PV = -(PV - T) = -E$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$u(t) = K_p e(t) + K_i \int_0^t e(\tau)\mathrm{d}\tau + K_d \frac{\mathrm{d}e(t)}{\mathrm{d}t} = -C_P - C_I - C_D = - (C_P + C_I + C_D)$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;So the first difference is the sign change you introduce by setting the error as &lt;span class=&quot;math-container&quot;&gt;$E = PV - T$&lt;/span&gt; instead of the default notation &lt;span class=&quot;math-container&quot;&gt;$E = T - PV$&lt;/span&gt;.&#xA;This is just a minor difference and does not affect the performance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second difference is in how you define the control input.&#xA;The default method would be &#xA;&lt;span class=&quot;math-container&quot;&gt;$$u(t) = K_p e(t) + K_i \int_0^t e(\tau)\mathrm{d}\tau + K_d \frac{\mathrm{d}e(t)}{\mathrm{d}t}.$$&lt;/span&gt;&#xA;However, you're expressing the control input in discrete time.&#xA;The most straight forward method to translate this continuous type controller to a discrete time controller at sample &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$u(k) = K_p e(k) + K_i \sum_{l = 0} ^k e(l) T_s + K_d \frac{e(k) - e(k-1)}{T_s},$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$T_s$&lt;/span&gt; is the sampling time.&#xA;However, you could imagine that this controller would not have the same performance as this just a simple discretization method (forward Euler).&#xA;There are many different discretization methods, but I'll not bother with it since I came across this site: &lt;a href=&quot;http://bestune.50megs.com/typeABC.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://bestune.50megs.com/typeABC.htm&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This site shows three discrete PID implementations (type A,B,C).&#xA;You're controller implementation&#xA;&lt;span class=&quot;math-container&quot;&gt;$$u(k) = u(k-1) + (C_P + C_I + C_D)$$&lt;/span&gt; looks very similar to the once used there.&#xA;However, they determine &lt;span class=&quot;math-container&quot;&gt;$C_P$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$C_I$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_D$&lt;/span&gt; in a different manner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think the base-level theory you're searching for is mostly on controller implementation and discrete time /digital (tuning and) control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Tuning&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The implemented controller is a PID controller, however this does not mean that you have to use all three terms.&#xA;Depending on the system dynamics and the performance demands you can choose between the following controllers: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;P&lt;/li&gt;&#xA;&lt;li&gt;PI&lt;/li&gt;&#xA;&lt;li&gt;PD&lt;/li&gt;&#xA;&lt;li&gt;PID.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If there is an integrator present in your system dynamics, it would be a bad choice to choose a PI or PID controller (since the I-term is already embedded in your system dynamics).&#xA;Based on the description of your testing results I assume that there is an integrator present in the system.&#xA;Therefore, you should go for a P or PD controller instead (thus set &lt;span class=&quot;math-container&quot;&gt;$K_I = 0$&lt;/span&gt;).&#xA;The choice between P or PD is fairly easy.&#xA;If you increase the gain (&lt;span class=&quot;math-container&quot;&gt;$K_P$&lt;/span&gt;), while having &lt;span class=&quot;math-container&quot;&gt;$K_D = 0$&lt;/span&gt; and you get overshoot, you should use the PD controller (since the D-action adds damping).&#xA;If there is no overshoot, just stick to the P controller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have problems finding proper values for &lt;span class=&quot;math-container&quot;&gt;$K_P$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$K_D$&lt;/span&gt;, you could use some tuning rules e.g. &lt;a href=&quot;https://en.wikipedia.org/wiki/Ziegler%E2%80%93Nichols_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;the Ziegler Nichols tuning method&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Choosing a better control strategy to cope with the long delay and low sampling frequency is hard if you do not have any knowledge about your system.&#xA;To overcome the delay you should have some prior knowledge about how your system is going to react on the input, since measuring the output includes the delay.&#xA;Hence, you should move to Model Predictive Control (MPC), where the most simplest form would be the &lt;a href=&quot;https://en.wikipedia.org/wiki/Smith_predictor&quot; rel=&quot;nofollow noreferrer&quot;&gt;Smith predictor&lt;/a&gt;.&#xA;An other possibility is to estimate the disturbance in advance (thus before measuring it) and already start reacting to it by means of feed-forward.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastEditorUserId="8216" LastEditDate="2020-03-04T11:49:00.600" LastActivityDate="2020-03-04T11:49:00.600" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33426" PostTypeId="1" CreationDate="2020-02-27T23:13:52.803" Score="3" ViewCount="518" Body="&lt;p&gt;As you all now, a typical stress-strain curve for concrete under compression can be seen in any reinforced concrete design textbook. The decreasing part of the curve (post-peak) can be obtained by testing the specimen with displacement controlled test machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why do we prefer the strain controlled compression test instead of the stress controlled test?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We use the stress-strain curve to calculate ultimate strength capacity of the section. The reinforced concrete member which is designed according to the stress-strain curve is subjected to load such as dead loads, earthquake loads etc during its lifetime. I think that the strain controlled tests are not appropriate to model dead loads because they are constant. The stress controlled stress strain curves might be more appropriate for the dead loads and earthquake. &lt;/p&gt;&#xA;" OwnerUserId="24785" LastEditorUserId="90" LastEditDate="2020-02-28T12:39:23.190" LastActivityDate="2020-02-28T12:39:23.190" Title="Why do we use the strain-controlled test for the compressive strength of concrete?" Tags="|materials|civil-engineering|stresses|concrete|reinforced-concrete|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33427" PostTypeId="2" ParentId="33410" CreationDate="2020-02-28T01:25:02.040" Score="0" Body="&lt;p&gt;There's at least 10 parameters you need to know before you could calculate thrust.  What is your bypass ratio?  What is the mass flow?  What is the ambient temperature and pressure?  What is the forward velocity?  what is the compression ratio?  etc, etc, etc.  You really need either a textbook on jet engine performance, or take a class.  NASA has a very crude simulator that might get you started &lt;a href=&quot;https://www.grc.nasa.gov/WWW/K-12/airplane/ngnsim.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.grc.nasa.gov/WWW/K-12/airplane/ngnsim.html&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2020-02-28T01:25:02.040" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33428" PostTypeId="1" CreationDate="2020-02-28T16:08:14.113" Score="0" ViewCount="91" Body="&lt;p&gt;I'm using an NPN transistor to pull the gate of a P-channel FET low. I'm driving the base with 3.3v and a 10k series resistor. I have a 100k resistor in series with the collector and the gate of the P-channel FET which is then pulled up to 5.4v with another 100k resistor. So my base current is about 20x my collector current. Any trouble with that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ufrOG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ufrOG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24790" LastActivityDate="2020-03-07T12:27:12.977" Title="can base current be far less than collector current?" Tags="|electrical-engineering|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33429" PostTypeId="2" ParentId="33424" CreationDate="2020-02-28T19:20:07.187" Score="4" Body="&lt;p&gt;See below for how I set up my Global Variables, and for all equations that I used. The only formulae that actually needed calculations are the taper angle for the helix (calculated using basic trigonometry from the Min/Max diameters of the cone, and the height) and the height of the helix (the height of the final object, plus 2x the pitch. This allows it to be cut to height and have a smooth runout - I hope this is correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NEQ7W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NEQ7W.png&quot; alt=&quot;Equation Editor View&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The steps that you need to take to model this item are as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Set up your global variables &lt;em&gt;before&lt;/em&gt; you start modelling.&lt;/li&gt;&#xA;&lt;li&gt;I renamed 'Top Plane' as 'Base Plane', since this will be the base of the cone. I then inserted a plane called &quot;Max Height&quot;, set to an offset from the Base Plane, defined by the global variable &quot;Height&quot;. Always define your planes or axes as early in the history tree as possible.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Set up your sketch for the cone. I started with an infinite length centreline - this allows us to use diameter dimensions, rather than needing to divide the global variables by 2. The cone is defined by two horizontal construction lines, one coincident with the origin (and base plane), and the other with the 'Max Height' plane. Note that the &quot;Height&quot; is not dimensioned here, it's driven by the location of the plane. I have added an extension past the Base and Max Height planes equal to the &quot;Pitch&quot; global variable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vSxjym.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vSxjym.png&quot; alt=&quot;cone sketch&quot;&gt;&lt;/a&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Revolve the Cone - It should automatically select the infinite centreline as the axis of rotation.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Generate your Helix. I started a sketch on the base face of the cone, and used convert entities to make a circle coincident with the bottom edge. You can't type equations directly into the helix tool (like you can with regular dimensions), so you need to just type a starting point, before moving to the Equation editor to add values as shown in my first image. I set it to be defined by Pitch and Height, with taper enabled. I put my Cone profile on the right plane so that a start angle of 0 deg will be coincident with the Cone Profile Sketch. You can see in the image below how to select the helix definition - if height and revolution is more appropriate, you can select it here, and adjust your global variables as required.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RzuMR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RzuMR.png&quot; alt=&quot;how to select helix defintion&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;6&quot;&gt;&#xA;&lt;li&gt;Set up your Antenna profile sketch. This should be coincident with the Cone profile, and the start of the Helix. I have set the width and thickness equal to their respective global variables. Note how the angle of this is defined by the Cone Profile sketch, which is in turn driven by the Max/Min diameters.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ayndR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ayndR.png&quot; alt=&quot;antenna profile&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;7&quot;&gt;&#xA;&lt;li&gt;Sweep the Antenna along the Helix. Uncheck 'Merge result' in the options to keep this as a separate body from the cone.&lt;/li&gt;&#xA;&lt;li&gt;Cut off the bottom of both parts - using Cut with Surface and the Base Plane&lt;/li&gt;&#xA;&lt;li&gt;Cut off the top of both parts - using Cut with Surface and the Max Height Plane&lt;/li&gt;&#xA;&lt;li&gt;Sit back and admire your handiwork&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gUQSB.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gUQSB.gif&quot; alt=&quot;parametric helix antenna&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2020-03-02T10:14:18.483" LastActivityDate="2020-03-02T10:14:18.483" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33430" PostTypeId="1" AcceptedAnswerId="33431" CreationDate="2020-02-28T20:06:14.290" Score="-1" ViewCount="78" Body="&lt;p&gt;When I open Paint, I change the size to 500x500. It appears more as a rectangle than a square. Has anyone else had this happen? Is there an explanation? Not sure if this is a thing that differs on different computers.&lt;/p&gt;&#xA;" OwnerUserId="24793" LastActivityDate="2020-02-28T20:27:52.570" Title="Are pixels not squares and more like rectangles?" Tags="|computer-engineering|computer|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33431" PostTypeId="2" ParentId="33430" CreationDate="2020-02-28T20:18:27.653" Score="3" Body="&lt;p&gt;Looks like a square to me - are you sure you selected the pixel checkmark, and also unchecked the &quot;Maintain aspect ratio&quot; box before typing in your second '500'?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QlHZlm.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QlHZlm.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/LkJmo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LkJmo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/K4Lzs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K4Lzs.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastEditorUserId="13311" LastEditDate="2020-02-28T20:27:52.570" LastActivityDate="2020-02-28T20:27:52.570" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33433" PostTypeId="1" AcceptedAnswerId="33439" CreationDate="2020-02-28T23:44:13.677" Score="1" ViewCount="54" Body="&lt;p&gt;I want to keep the pressure inside a balloon (effectively - a rubber vessel) constant, as the balloon is compressed. I'm wondering if a regulator like &lt;a href=&quot;https://in.rsdelivers.com/product/norgren/r72g-2gk-rmn/norgren-pneumatic-regulator-g-1-4/2390670&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt; (&lt;a href=&quot;https://en.wikipedia.org/wiki/Pressure_regulator?wprov=sfla1&quot; rel=&quot;nofollow noreferrer&quot;&gt;wiki page here&lt;/a&gt; showing mechanism - I believe) can do it? They're designed to regulate a higher relatively constant pressure (eg large volume at 30bar) down to &amp;lt;1bar in a smaller vessel to keep that constant. I think these are one way tho - if the pressure starts increasing in the outlet the inlet will close off, rather than dissipate that increase back to the source, whereas I want to effectively use it to do exactly that. Would this work? Or is there a better way to do this? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for all help given!&lt;/p&gt;&#xA;" OwnerUserId="11394" LastActivityDate="2020-02-29T07:15:39.353" Title="Regulating pressure down inside vessel as it increases" Tags="|pressure|pressure-vessel|pneumatic|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33434" PostTypeId="1" CreationDate="2020-02-28T23:57:45.703" Score="2" ViewCount="142" Body="&lt;p&gt;I am posting this here because of the multidisciplinary knowledge needed to tackle this issue. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Problem:&lt;/strong&gt; I have to make an electromechanical device that is required to have controllable (downward) piston strokes, that can up to deliver 500N of force upon impact with a body. The piston's stroke length can be flexible, but it needs to deliver at least 500N when it hits something and go back up to its starting position. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The real difficulty:&lt;/strong&gt; The first problem is meeting the required frequency of this operation. Each full cycle (starting position &gt; travel down &gt; strike &gt; come back up) needs to happen at least twice a second (100 - 120 cycles per minute). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Work done so far:&lt;/strong&gt; Basic math tells me that this should be doable with 200-300W of power accounting for losses. However, I have an electrical engineering background with very little mechanical/kinematics/dynamic loads knowledge. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a 1hp DC brushed motor, that's driven through PWM and I have control over the motor's operation. However, I would really appreciate some input on what kind of set up you would use to transfer the rotary motion to linear motion. The project has a $1000 budget and currently there are the following options: &#xA;1. Crank piston style (Rotor turns the crank connected to a linear shaft&#xA;2. Carry ball and screw system &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor Specs: Operated at 24V, 5600RPM no load, 0.6Nm Torque. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would gladly take any mathematical suggestion too! Theoretical analysis tends to get quite complex as the piston does not have to constantly move with 600N (No load to pull back) but only when it's on the downward stroke. &lt;/p&gt;&#xA;" OwnerUserId="24796" LastActivityDate="2020-02-29T23:41:14.207" Title="Design Input: Rotary motion to Linear motion, What system should I choose?" Tags="|mechanical-engineering|electrical-engineering|motors|linear-motion|pistons|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33435" PostTypeId="2" ParentId="33434" CreationDate="2020-02-29T06:31:05.977" Score="1" Body="&lt;p&gt;If budget was not an issue you could design a crank / piston assembly from scratch...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As it is an issue then consider re-purposing a small single cylinder engine - lawnmower or chainsaw perhaps. Consider lubrication as well...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-02-29T06:31:05.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33436" PostTypeId="1" CreationDate="2020-02-29T06:45:07.973" Score="-1" ViewCount="48" Body="&lt;p&gt;The system consists of two disks. The upper disc is connected to the pipe. The disk below is a plane.&#xA;The gas flows out radially from the disc gap. A formula for finding the increment of the radial volume and the distance r from the center of the disc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean that the greater the distance r from the center of the disc, the larger the volume of the gap with r as the radius.  From the center, what is the formula for the increase in the radial volume of the gap and r?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose the flow of gas flowing from the pipeline is &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt;, the radius of the pipeline is &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, and the height of the gap is &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$h &amp;lt;&amp;lt; R$&lt;/span&gt;.  It is assumed that the gas is an ideal gas with no viscosity and no temperature change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The calculation method I think of is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\lim_{\Delta r\to 0}{\cfrac{(r+\Delta r)^2πh-r^2πh}{\Delta r}}=2rπh$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;correct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U0SB7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U0SB7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;When h is constant, &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; is a function of &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt;.  When &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; has an increment &lt;span class=&quot;math-container&quot;&gt;$\Delta r$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; also has an increment &lt;span class=&quot;math-container&quot;&gt;$\Delta V$&lt;/span&gt; . I want to know the relationship between &lt;span class=&quot;math-container&quot;&gt;$\Delta V$&lt;/span&gt;  and &lt;span class=&quot;math-container&quot;&gt;$\Delta r$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="19873" LastEditorUserId="19873" LastEditDate="2020-03-04T02:13:22.087" LastActivityDate="2020-03-04T02:13:22.087" Title="A formula for finding the increment of the radial volume and the distance r from the center of the disc" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="11" ClosedDate="2020-03-07T13:04:22.957" ContentLicense="CC BY-SA 4.0" />
  <row Id="33437" PostTypeId="2" ParentId="33434" CreationDate="2020-02-29T06:58:29.377" Score="2" Body="&lt;p&gt;Obviously cranks and pistons have been used for hundreds of years for this.  Your other option would be fast-acting motors or hydraulic rams.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But a 500N (or is it 600?) force isn't what happens when something hits an object.  Hitting an object imparts kinetic energy, and a 1 HP motor just doesn't have a lot to give because you have to speed something up so it can be slowed down when it hits.  Newtons would be a force, like compressing a spring, and even then you need to say how much power you are going to put into whatever you're hitting. So 5/600 N over what distance?  Also, if you are really planning on slinging something around that hits like 5 sledgehammers even a little bit a tiny 1 HP motor isn't going to cut it.  Whatever you're banging around is going to beat the mechanism to hell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This doesn't seem like a good application for an electric motor, because all the power is in the downstroke, which will slow the motor down and drive up the amps.   Maybe a large flywheel would help.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-02-29T06:58:29.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33438" PostTypeId="2" ParentId="33433" CreationDate="2020-02-29T07:00:58.173" Score="2" Body="&lt;p&gt;The only way to keep pressure constant as volume is reduced is by let out the same amount of liquid/gas as the volume contracts.  That is exactly what a pressure reducing valve does.  As long as it has the capacity to vent at the rate you need you should be fine.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-02-29T07:00:58.173" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33439" PostTypeId="2" ParentId="33433" CreationDate="2020-02-29T07:15:39.353" Score="2" Body="&lt;p&gt;Have a second separate balloon that expands as the first is compressed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will also keep the pressure constant as the volume is increased if that is part of your requirement.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-02-29T07:15:39.353" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33441" PostTypeId="1" CreationDate="2020-02-29T13:09:22.810" Score="0" ViewCount="182" Body="&lt;p&gt;&lt;strong&gt;I am trying to create a 3d sketch similar to this (that is equation driven)&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5vbX6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5vbX6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;(the image is a screenshot from a tutorial but the sketch in the tutorial is not equation driven unfortunately). image source: &lt;a href=&quot;https://www.youtube.com/watch?v=GBNAnqolQ-I&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=GBNAnqolQ-I&lt;/a&gt; (at 12:46)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;SW file:&lt;/strong&gt; &lt;a href=&quot;https://drive.google.com/open?id=1VuQUzYxyjPer2lX7t0EAsDkeTctPBQJj&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://drive.google.com/open?id=1VuQUzYxyjPer2lX7t0EAsDkeTctPBQJj&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I made a sketch of a few points using a linear sketch pattern (amount of instances and spacing are both equation driven). The points that can be seen in the images below in my sketch will be used as the 'corners' of the 3d sketch .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my explanation is unclear/incomplete please let me know and I'll try and fix it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to have the same sketch on another parallel plane, I tried copying the sketch over to another plane but unfortunately when I change the equation variables the original linear sketch pattern sketch (on the original plane) does change (amount of instances/spacing) but the copied sketch (on the parallel plane) does not. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also tried using 'convert entities' but unfortunately I cannot (or I don't know how to) select an entire sketch instead of individual points/lines/shapes etc using 'convert entities'. If I'm not able to select the entire sketch using 'convert entities' I am quite sure that when the number of instances of my original sketch changes the sketch on the second plane created using 'convert entities' will not have these instances added.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I could manually redraw the sketch (using the same equations/variables for the amount of instances in the linear sketch pattern and for the spacing in the linear sketch pattern) but I think there should be a better/more efficient way of doing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I doing wrong? What could be a possible solution to this issue?&#xA;Should I not use a loft for creating this model? is this the right approach?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Some images:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First sketch drawn on original plane:&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/45FuV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/45FuV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA; (the sketch = the 36-ish points on the line on the left side of the left plane in the image)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sketch copied over to second plane:&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/6piUc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/6piUc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;This looks as if it has worked, however as soon as I change the variables in my equations the following happens (the original sketch's linear sketch pattern changes (the amount of instances/the spacing change) as they should according to the corresponding equations) but the copied sketch does not change at all.. which is an issue since I need all of the copies to change whenever I change the original.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/L74OI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L74OI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DpfR6.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DpfR6.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;background: I am trying to model DC motor windings using loft feature (as explained in the following post &lt;a href=&quot;https://engineering.stackexchange.com/questions/33361/how-to-model-motor-windings-in-solidworks&quot;&gt;How to model Motor Windings in Solidworks?&lt;/a&gt;). I am trying to create the points required for the 3d sketch that can be seen in 'step 1' of the post I linked above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I am trying to achieve:&lt;/strong&gt; (original sketch (in the green circle) to be copied over to the 3 lines (in the red circles) and all copies to change (number of instances/spacing) when i change the original sketch (by changing my equation variables).&#xA;&lt;a href=&quot;https://i.stack.imgur.com/0Xh5x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0Xh5x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT: I just thought of an issue with the approach explained above;&lt;/strong&gt; if I were to change the variables in my equations which would result in more points per sketch, the 3d sketch which i would draw by hand wouldn't get extra segments added automatically so i don't think it's a good approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternative approach (current plan of attack): &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;1) Draw 1 'spiral' (4 points) using 3d sketch (dimensions driven by equations/variables)&lt;/li&gt;&#xA;&lt;li&gt;2) use loft feature to create solid body out of this sketch&lt;/li&gt;&#xA;&lt;li&gt;3) linear pattern the loft feature (amount of instances in pattern and pattern spacing once again determined by variables/equations)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This way the model should update properly whenever the variables are changed in my equations and I won't have to do any manual fixing (I think). Any thoughts/suggestions/criticism regarding this approach? If so please let me know.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="10328" LastEditorUserId="10328" LastEditDate="2020-02-29T13:25:51.777" LastActivityDate="2020-02-29T13:25:51.777" Title="Solidworks - Copy equation driven linear pattern sketch to other plane" Tags="|solidworks|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33442" PostTypeId="1" CreationDate="2020-02-29T15:27:43.363" Score="3" ViewCount="214" Body="&lt;p&gt;I just discovered in my parents' basement a Sprite can from 1995* and also a Coca-Cola can probably from the same year. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both cans are unopened and have no visible damage or holes. The Coca-Cola can feels &quot;normal&quot;, but &lt;strong&gt;the Sprite can is empty!&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uUdki.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uUdki.jpg&quot; alt=&quot;Sprite can from 1995&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZhDxS.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;You can hear in my video&lt;/a&gt;&lt;/strong&gt; that there is no liquid sloshing around in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't find any way that the soda could have escaped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We also don't see any mess near where the cans were, but I don't know for certain that the cans stayed in the same place for 25 years, so maybe there could have been a mess of liquid leaked out somewhere else if the cans had been stored somewhere else earlier. But nothing feels sticky or looks like there has been a leak of any kind.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are possible explanations for how a carbonated drink could disappear from a sealed aluminum can?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;*I had kept it as a &quot;collector's item&quot; from when the Houston Rockets won their second championship. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: I originally posted to &lt;a href=&quot;https://physics.stackexchange.com/q/532731/46543&quot;&gt;https://physics.stackexchange.com/q/532731/46543&lt;/a&gt; , but the moderators closed the question and said it belonged in Engineering. Also, I followed advice of commenters there and refrigerated the can and then later submerged it in warm water, but as &lt;a href=&quot;https://i.stack.imgur.com/22OZP.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;you can see in this video, there were no bubbles&lt;/a&gt; revealing any holes! &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, many people suggested that the can &lt;em&gt;never&lt;/em&gt; had liquid in it, but that seems extremely unlikely since neither I nor anyone in my family remember ever encountering &lt;em&gt;any&lt;/em&gt; can &lt;em&gt;ever&lt;/em&gt; that was sealed but empty.&lt;/p&gt;&#xA;&#xA;&lt;h2&gt;UPDATE&lt;/h2&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/7kaHj.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Video of opening the 25-year-old Coca Cola can&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://i.stack.imgur.com/RK4IN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;Video of opening the 25-year-old Sprite can that felt empty&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="23609" LastEditorUserId="23609" LastEditDate="2020-03-03T20:09:23.517" LastActivityDate="2020-03-07T19:03:47.703" Title="How is a 25-year-old can of soda now empty without having been opened or poked?" Tags="|structural-engineering|materials|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33444" PostTypeId="1" CreationDate="2020-02-29T18:49:01.273" Score="1" ViewCount="45" Body="&lt;p&gt;In the YouTube video &lt;a href=&quot;https://www.youtube.com/watch?v=zjBbfcK3Y8Y&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ductility Test&lt;/a&gt;, I saw a black substance undergo a ductility test. It appeared to be as ductile as silly putty although silly putty is just a very viscous liquid. It seems that it can keep being pulled thinner and thinner. I couldn't find the name of the substance in the video. I just want to know the name of that substance. Once I learn the name of that substance, I can ask another question about the properties of that substance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll ask whether it's a solid unlike silly putty. I know it started off as a liquid but I'll ask whether it got turned into a solid or just a very viscous liquid. I'll ask &quot;If it's a solid, what is its yield strength?&quot; If it has a higher yield strength than the typical chromium steel, it could be very revolutionary because it cannot sustain crack propagation and therefore cannot be weakened much by the presence of a crack.&lt;/p&gt;&#xA;" OwnerUserId="6419" LastEditorUserId="6419" LastEditDate="2020-02-29T19:27:10.640" LastActivityDate="2020-02-29T20:06:03.770" Title="What is the name of the substance that underwent the ductility test in the YouTube video Ductility Test?" Tags="|materials|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33445" PostTypeId="2" ParentId="33444" CreationDate="2020-02-29T20:06:03.770" Score="1" Body="&lt;p&gt;If you compare your video with &lt;a href=&quot;https://www.youtube.com/watch?v=TE8zYxUJHt0&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=TE8zYxUJHt0&lt;/a&gt; which uses a very similar test procedure, the material is probably the same: bitumen.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-02-29T20:06:03.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33446" PostTypeId="1" CreationDate="2020-02-29T20:30:20.330" Score="-1" ViewCount="131" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/25q2i.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/25q2i.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;enter image description here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZPDUV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZPDUV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sHm52.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sHm52.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I am having issue solving this problem, I solved one part by the fact that in the given distance of 2L , 2 tubes are involved so I can find the heat conducted per unit length.&#xA;Whereas I don't understand how to use fourier's law to compute the temperature at the insulation point even though I know the maximum temperature is going to be at the insulation point. Please give me any suggestions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/DxeoN.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DxeoN.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ROTBZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ROTBZ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have seen your answer and edited, Its not possible for the heat to flow in &#xA;x direction under steady state conditions because all of the absorber plate is&#xA;receiving the same amount of radiation flux from the sun hence why would&#xA;heat flow in x.&#xA;Temperature should be maximum along the lower surface of the absorber plate &#xA;at the insulation points, I've made an analogy to fluid flow, its like you have made 2 holes in the bottom of a tank of water and water is flowing out.&#xA;I've attached 2 more pages to show my intuition.&lt;/p&gt;&#xA;" OwnerUserId="18319" LastEditorUserId="18319" LastEditDate="2020-03-02T02:06:29.880" LastActivityDate="2020-03-02T02:06:29.880" Title="Heat Transfer Problem , 1 Dimensional Conduction Extended Surfaces" Tags="|thermodynamics|heat-transfer|thermal-conduction|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33447" PostTypeId="2" ParentId="33434" CreationDate="2020-02-29T23:41:14.207" Score="1" Body="&lt;p&gt;Use a screw press mechanism and reducing gearbox. Then it becomes an issue of calculating the force multiplication of the screw and gearbox mechanism to get 500 newtons. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It won't be fast, but it will provide the force you require.&lt;/p&gt;&#xA;" OwnerUserId="24692" LastActivityDate="2020-02-29T23:41:14.207" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33448" PostTypeId="1" CreationDate="2020-03-01T03:12:39.660" Score="1" ViewCount="29" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MjiwK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MjiwK.jpg&quot; alt=&quot;Feeder line to 100 needle orifices&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider a water distribution setup consisting of a feeder line blocked at one end but having 100 needle outlets all the same size. The feeder line cross sectional area is much greater than a needle cross sectional area so the flow from any needle is very close to one hundredth the overall flow into the feeder line. I think the feeder line would act like a massive reservoir which is being topped up at the entry point. Lets say that the flow into the feeder line is 1L/sec. Then the flow out through any needle is 10mL/sec. Assume that the system is in steady state with all needles happily delivering their 10mL/sec.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario A&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the entry point of the feeder line we introduce 100mL of red dyed water. Assume that this does not affect the water characteristics other than it's now red. How is this red water distributed through the needles? Intuitively, I'm thinking that the first needles (the lower numbered ones) will get more of the red water than the later needles. But is this true? Does anyone have any mathematics that could describe this? I briefly considered that it was like an impulse function; but this would only be so if the steady state of the system changed. It doesn't. It's still water with the same characteristics, just red. It's like being able to tag electric current with a colour - it's still electric current...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Scenario B&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the entry point to the feeder line we install a water switch. The water can be switched to either blue water or red water. If we have the switch set to BLUE and we have all needles delivering their 10mL/sec of blue water and then switch to RED. How long does it take the system to be delivering 10mL/sec of red water from all of the nozzles? Assume that the switching action has no effect on the system itself other than changing the colour of the water. I'm hoping to find some sort of mathematics to describe this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Otherwise I might have to build the thing and see what happens!&lt;/p&gt;&#xA;" OwnerUserId="24807" LastActivityDate="2020-03-01T03:12:39.660" Title="Water distribution through N x Needle orifices" Tags="|fluid-mechanics|modeling|hydraulics|numerical-methods|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33449" PostTypeId="1" CreationDate="2020-03-01T10:52:02.370" Score="0" ViewCount="152" Body="&lt;p&gt;I need to identify the seal/gasket shown below so I can order a replacement. I can measure the OD, ID and Height, but I'm not sure what to call it to narrow down my search.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What type of seal is this? What are the parameters I should be searching for?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't need an exact match, but something close, as the use is very low-pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is used in a food-safe context with 100C water at near atmospheric pressure. There is no lubrication and is meant to be washed with food-safe detergents.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seal has a flat side on the &quot;bottom&quot; that sits against a screw cap. The &quot;top&quot; side of the seal has two small lips that create the seal when the cap is screwed on. The inner diameter is flat and meets the bottom profile at a right angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seal does &lt;strong&gt;not&lt;/strong&gt; sit in a groove or channel in the cap, but just stretches around the OD of the threads and sits on a flat surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The seal has an OD of 63mm, an ID of 59mm and a height of 2mm. --These are my best attempt at measuring this as I am unsure how to properly measure a seal like this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Add OD, ID, Height&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Top side&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/myKY6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/myKY6.jpg&quot; alt=&quot;Interface side - two lips&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;Lip Profile of Top Side&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/QsGCq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QsGCq.jpg&quot; alt=&quot;Profile of interface side&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;Bottom Side&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/UBurZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UBurZ.jpg&quot; alt=&quot;Bottom side&quot;&gt;&lt;/a&gt;&#xA;&lt;strong&gt;Inner Diameter face&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XMoca.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XMoca.jpg&quot; alt=&quot;Inner Diameter&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24811" LastEditorUserId="24811" LastEditDate="2020-03-02T08:32:42.613" LastActivityDate="2024-03-12T05:01:30.173" Title="Help identifying seal/gasket" Tags="|seals|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33450" PostTypeId="1" AcceptedAnswerId="33456" CreationDate="2020-03-01T11:48:24.177" Score="-1" ViewCount="32" Body="&lt;p&gt;The vehicle running at Hill station.. Same time vehicle moves back ward direction in manual transmission system. So avoid the problem.. I think ratchet pinion mechanism suitable! or not?&lt;/p&gt;&#xA;" OwnerUserId="24812" LastEditorUserId="24812" LastEditDate="2020-03-01T11:58:06.060" LastActivityDate="2020-03-01T21:48:48.413" Title="Problem for backwards moving vehicle in hills" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33451" PostTypeId="1" CreationDate="2020-03-01T11:54:15.297" Score="2" ViewCount="67" Body="&lt;p&gt;I have a few brackets that are 1.8mm thick with 4.5mm countersunk holes &lt;a href=&quot;https://www.aliexpress.com/item/32760951479.html?spm=a2g0o.detail.1000023.9.54c3301cdNyQzf&quot; rel=&quot;nofollow noreferrer&quot;&gt;like this&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DZhh1m.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DZhh1m.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The countersunk machine screws that I've found have head heights of at least 2.2mm (some 3.2mm). This is taken from &lt;a href=&quot;https://www.mcmaster.com/screws&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mcmaster.com/screws&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eVkp4.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eVkp4.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone know what sort of countersunk screws should be used with these brackets? I can't seem to find any common ones with a head height of ~1.5mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="4302" LastEditDate="2020-03-01T13:23:17.773" LastActivityDate="2020-03-01T13:23:17.773" Title="What screw to use with 1.8mm thick countersunk brackets?" Tags="|mechanical-engineering|design|" AnswerCount="0" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33452" PostTypeId="1" AcceptedAnswerId="33453" CreationDate="2020-03-01T17:46:06.523" Score="0" ViewCount="1605" Body="&lt;p&gt;Distinguish between dead center and live center in a center lathe tool when this terms are used in the context of work holding in a lathe &lt;/p&gt;&#xA;" OwnerUserId="24815" LastEditorUserId="24815" LastEditDate="2020-03-01T17:48:48.837" LastActivityDate="2024-01-30T03:53:24.360" Title="Distinguish between dead center and live center" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33453" PostTypeId="2" ParentId="33452" CreationDate="2020-03-01T17:49:30.743" Score="5" Body="&lt;p&gt;A center holds the work during rotation at the tailstock end of the lathe. A live center is mounted in bearings and rotates with the work, while a dead center does not rotate - the work rotates about it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Live center Rotates while dead centre center does not rotate.. &lt;/p&gt;&#xA;" OwnerUserId="24814" LastActivityDate="2020-03-01T17:49:30.743" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33454" PostTypeId="2" ParentId="33404" CreationDate="2020-03-01T20:06:16.667" Score="1" Body="&lt;p&gt;The bare bones answer is energy = (mass x velocity²)/2. This means your projectile will have a kinetic energy, once accelerated, of e=(0.453kg x 1E3² m/sec)/2 = 226824 joules.&#xA;But also, you must add the energy required to accelerate the mass. This derives from &#xA;f=ma, so you have to know the launching force and time permitted to arrive at your final velocity. Note also that you are hoping to project at a velocity of 3 km/sec. The velocity is a squared term in the energy equation. Therefore don't just multiply the above answer by 3. Allow for the force of gravity (more energy). Allow for efficiency (more energy).&lt;/p&gt;&#xA;" OwnerUserId="24817" LastActivityDate="2020-03-01T20:06:16.667" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33455" PostTypeId="2" ParentId="33405" CreationDate="2020-03-01T20:15:06.157" Score="0" Body="&lt;p&gt;It is not only the velocity of the gas that produces thrust, but the mass of the gas per second that is being jetted from the nozzle. Since energy = mass * velocity², this energy drives the jet forward (action produces and equal and OPPOSITE reaction). Blow up a balloon and let it go; same principle.&lt;/p&gt;&#xA;" OwnerUserId="24817" LastActivityDate="2020-03-01T20:15:06.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33456" PostTypeId="2" ParentId="33450" CreationDate="2020-03-01T21:48:48.413" Score="0" Body="&lt;p&gt;There is a system called a &quot;hill holder&quot; in cars which engages the parking brake on a hill while stopped. When the clutch is engaged to drive up the hill, the parking brake is slowly released by the mechanism. Manual transmission Subarus had this feature years ago. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-03-01T21:48:48.413" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33457" PostTypeId="2" ParentId="33446" CreationDate="2020-03-02T00:18:10.077" Score="0" Body="&lt;p&gt;Your mistake is simply in making the statement &lt;span class=&quot;math-container&quot;&gt;$dT/dx = 0$&lt;/span&gt;. The statement &lt;span class=&quot;math-container&quot;&gt;$dT/dx$&lt;/span&gt; is for an insulating wall, and neither the radiation surface nor the tube locations are insulating. The top surface has the maximum temperature &lt;strong&gt;AND&lt;/strong&gt; it has an incoming heat flux. The surface at the tubes has the minimum temperature and has an exiting heat flux. As a final hint, the steady state profile across the plate will be linear &lt;span class=&quot;math-container&quot;&gt;$dT/dx$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastActivityDate="2020-03-02T00:18:10.077" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33458" PostTypeId="1" AcceptedAnswerId="33459" CreationDate="2020-03-02T07:14:18.060" Score="-1" ViewCount="258" Body="&lt;p&gt;I have a technical drawing, but I can't understand the &quot;h2&quot;. The &quot;h2&quot; points to a hole, and I don't understand it. Can you please give me an answer? Thanks!&lt;/p&gt;&#xA;" OwnerUserId="24825" LastActivityDate="2020-03-02T07:53:04.500" Title="What does &quot;h2&quot; mean on technical drawing?" Tags="|technical-drawing|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33459" PostTypeId="2" ParentId="33458" CreationDate="2020-03-02T07:53:04.500" Score="4" Body="&lt;p&gt;H2 refers to the tolerance of the hole - whether it’s an upper or lower case letter shows if it’s a hole or shaft tolerance. In this case, as you mention it’s on a hole, I would expect it to be “H6”, which means for an 6.00mm hole, it would be acceptable if manufactured in the range 6.00-6.03mm in diameter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.ytimg.com/vi/c8TKftViusQ/maxresdefault.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2020-03-02T07:53:04.500" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33460" PostTypeId="1" CreationDate="2020-03-02T09:16:19.027" Score="-1" ViewCount="81" Body="&lt;p&gt;I'm stuck with how to calculate the external torque which must be applied to prevent the aircraft rotating about axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/y9I5f.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/y9I5f.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The imagine shows an aircraft which has a jet engine with a rotor that has an angular velocity of 6000 rpm and a moment of inertia of 10 kgm^2 (both about 0Z). If the aircraft moves into a dive by rotating about 0X at 3°/s, what external torque must be applied to the aircraft about 0Y in order to prevent it rotating about this axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the torque required T = I⋅(Ω×ω) where I is the Moment of inertia, ω is the rotor angular velocity in rad/s and Ω is the precession rate but I'm confused about how to calculate these.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!                 &lt;/p&gt;&#xA;" OwnerUserId="23799" LastActivityDate="2020-03-05T00:01:05.693" Title="How to Calculate Torque to Prevent Rotating" Tags="|mechanical-engineering|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33461" PostTypeId="2" ParentId="33449" CreationDate="2020-03-02T13:06:01.473" Score="0" Body="&lt;p&gt;Looks like a shaft seal or oil shield (which can be used for fluids besides oil, water/steam in your case).&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-03-02T13:06:01.473" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33462" PostTypeId="1" AcceptedAnswerId="33492" CreationDate="2020-03-02T13:19:24.423" Score="1" ViewCount="587" Body="&lt;p&gt;I have been struggling with the following problem for a few days now and would greatly appreciate some assistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm new to control systems and have very little experience with PD controllers and never heard of the pole placement method so please correct me where I'm wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The main problem I seem to be having is that there are three poles &lt;span class=&quot;math-container&quot;&gt;$\ s(s+2)(s+5) $&lt;/span&gt; all other examples i have done have had only two &lt;span class=&quot;math-container&quot;&gt;$\ (s+2)(s+5) $&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I don't see how I can you the standard formulas I have been using:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ s^2+2ζω_ns+ω_n^2 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ s^2+(K_d-1)s+(K_p+1) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like I said I'm very new to this so please forgive me if anything I've said is completely wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance for any assistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mOk4g.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mOk4g.png&quot; alt=&quot;Control systems Question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24828" LastEditorUserId="18238" LastEditDate="2020-03-23T15:08:29.027" LastActivityDate="2020-05-10T02:49:54.483" Title="Proportional-Derivative compensator design" Tags="|control-engineering|control-theory|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33465" PostTypeId="1" AcceptedAnswerId="33467" CreationDate="2020-03-02T16:38:58.443" Score="1" ViewCount="160" Body="&lt;p&gt;I have a question regarding inertia and friction between gears.&#xA;I have a gear which is connected to the rod at one end and an electric motor which is connected to the other end. I know the torque of the rod and I want to calculate the torque of the motor I need.&#xA;I have been told that I can't lay on transmission ration only and I also have to include the calculations of the inertia and friction between gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can please someone explain me how should I do it (every example will appreciated).&lt;/p&gt;&#xA;" OwnerUserId="24318" LastActivityDate="2020-03-02T17:27:32.000" Title="Inertia and friction between gears" Tags="|mechanical-engineering|gears|friction|transmission|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33466" PostTypeId="2" ParentId="33436" CreationDate="2020-03-02T16:46:41.077" Score="2" Body="&lt;p&gt;Since area between the plates is &lt;span class=&quot;math-container&quot;&gt;$A=\pi \cdot r^2$&lt;/span&gt; and volume is &lt;span class=&quot;math-container&quot;&gt;$V=A \cdot h$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$V=\pi \cdot r^2 \cdot h$&lt;/span&gt;. Am I missing something you are trying to do?&lt;/p&gt;&#xA;" OwnerUserId="24835" LastEditorUserId="1632" LastEditDate="2020-03-03T05:24:06.060" LastActivityDate="2020-03-03T05:24:06.060" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33467" PostTypeId="2" ParentId="33465" CreationDate="2020-03-02T17:27:32.000" Score="2" Body="&lt;p&gt;To turn your gear from rest, you'll need to overcome the inertia of your entire system (rod and gear).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Torque = inertia x angular acceleration. You'll need to figure out what angular (rotational) acceleration you need/can get from your motor. You can approximate inertia with I = (1/2) x mass x radius^2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can approximate gears being 95% efficient if you're using spur gears (straight teeth). This means 5% of the energy put in is being lost as friction.&#xA;So your required motor torque = (operating torque + acceleration torque)/ 0.95 efficiency [friction].&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-03-02T17:27:32.000" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33468" PostTypeId="1" CreationDate="2020-03-02T17:39:33.803" Score="1" ViewCount="129" Body="&lt;p&gt;Does anyone have application experience with harmonic gear drive efficiencies as opposed to planetary or parallel axis reduction systems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm looking into drives with about a 25:1 reduction, and it seems the efficiencies go down as reduction ratio decreases.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interested in using something with a smaller footprint than a multi-stage planetary, if possible.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-03-02T17:39:33.803" Title="Efficiency of harmonic gear drives" Tags="|gears|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33469" PostTypeId="1" CreationDate="2020-03-02T18:00:36.007" Score="-1" ViewCount="90" Body="&lt;p&gt;Suppose that I am given an equation of some positional vector in the Cartesian coordinate system that is based on some theta shown right below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/D2yoY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/D2yoY.png&quot; alt=&quot;Question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am asked to find the linear acceleration based on these positional vectors (in terms of the variables), so naturally I derived twice for each of these equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for example for x1''(t) I tried (though I am not clear if I did this properly):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x_1''\left(t\right)=\:f''\left(\theta \left(t\right)\right)\cdot \theta ''\left(t\right)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if my x1(t) equals:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x_1\left(t\right)\:=\:-r_1cos\left(\theta _1\left(t\right)-90^{\circ} \right)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would my x1''(t) be &lt;span class=&quot;math-container&quot;&gt;$r_1cos\left(\theta _1\left(t\right)-90^{\circ} \right)\cdot \theta _1'\left(t\right)+r_1sin\left(\theta _1\left(t\right)-90^{\circ} \right)\cdot \theta _1''\left(t\right)$&lt;/span&gt; ???&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for x2''(t) what would be a proper way of deriving the equation in terms of the variable now there are multiple variables?? Would it be something like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$x_2\left(t\right)=\:f''\left(\theta _1\:\left(t\right),\theta _2\left(t\right)\right)\cdot \:\theta _1\:''\left(t\right)\cdot \theta _2''\left(t\right)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question seems very basic (and it honestly is )and I am asking how to derive twice properly with variables but dynamics problem seems to have its own notations for everything. Please helpmeout&lt;/p&gt;&#xA;" OwnerUserId="24837" LastEditorUserId="24837" LastEditDate="2020-03-02T18:10:14.113" LastActivityDate="2020-03-05T01:03:36.153" Title="Dynamics - Deriving Linear acceleration based on angle theta" Tags="|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33470" PostTypeId="2" ParentId="32873" CreationDate="2020-03-02T19:18:58.653" Score="0" Body="&lt;p&gt;According to a valued associate from Washington DC Gas, residential regulators can vent small amounts of raw natural gas to the surrounding environment and cannot be enclosed.  Any enclosed regulator say, in a ship, must have active ventilation measures.  Interior appliances like water heaters are regulated from the external regulator and do not vent.&lt;/p&gt;&#xA;" OwnerUserId="5967" LastActivityDate="2020-03-02T19:18:58.653" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33473" PostTypeId="2" ParentId="33428" CreationDate="2020-03-03T00:11:09.587" Score="1" Body="&lt;p&gt;Kirchhoff's Voltage Law (KVL) must be satisfied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are fixiated on:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \beta = \frac {I_C}{I_B}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So &lt;span class=&quot;math-container&quot;&gt;$I_C$&lt;/span&gt; must be larger than &lt;span class=&quot;math-container&quot;&gt;$I_B$&lt;/span&gt;, but this implies the transistor is operating in the active region.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;KVL input side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_{BB} - V_{R_B} -V_{BE} = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_B = \frac {V_{BB} - V_{BE}}{R_B} = \frac {3.3V - 0.7V}{10k\Omega} = 260 \mu A$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;KVL output side:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V_{CC} - V_{R_{C_1}} - V_{R_{C_2}} - V_{CE} = 0$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_C = \frac {V_{CC} - V_{CE}}{R_{C_1} + R_{C_2}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume &lt;span class=&quot;math-container&quot;&gt;$V_{CE} = 0$&lt;/span&gt;, we can get &lt;span class=&quot;math-container&quot;&gt;$I_{C_{Max}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_C = \frac {V_{CC}}{R_{C_1} + R_{C_2}} = \frac {5.4V}{100k\Omega + 100k\Omega} = 27 \mu A$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So based upon biasing resistors, the maximum &lt;span class=&quot;math-container&quot;&gt;$I_C$&lt;/span&gt; is 1/10 of &lt;span class=&quot;math-container&quot;&gt;$I_B$&lt;/span&gt;.  Transistor is in the cut-off region.&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2020-03-03T00:11:09.587" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33474" PostTypeId="1" CreationDate="2020-03-03T05:12:00.960" Score="-1" ViewCount="133" Body="&lt;p&gt;Please give calculation for peak power consumption wrt whichever is efficient.&lt;/p&gt;&#xA;" OwnerUserId="17945" LastEditorUserId="17945" LastEditDate="2020-03-03T06:44:00.230" LastActivityDate="2020-03-03T19:33:43.410" Title="Running single Peltier Module at high power vs running multiple peltier Modules at low power? Which is more efficient?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ClosedDate="2020-03-07T13:06:41.443" ContentLicense="CC BY-SA 4.0" />
  <row Id="33476" PostTypeId="1" AcceptedAnswerId="33482" CreationDate="2020-03-03T07:18:20.350" Score="5" ViewCount="15631" Body="&lt;p&gt;I am thinking of building a indirect Peltier AC to cool a room of approx size 30m&lt;sup&gt;3&lt;/sup&gt;. Single/multiple Peltier modules will be used to cool a insulated box with water inside. The box will be kept outside the room. A pump will circulate the  cold water of the box in a pipe which will go inside the room. Inside the room, The pipe will be connected to something similar to a radiator, but rather than throwing heat into the air, it will throw cool air.&#xA;Also, I want to achieve maximum efficiency, by correctly balancing the current, voltage and temp difference.&#xA;Pumping heat out of the Peltiers will most probably not be a problem as they will be water cooled.&#xA;Is this practically possible?&lt;/p&gt;&#xA;" OwnerUserId="17945" LastEditorUserId="90" LastEditDate="2022-07-19T11:28:01.553" LastActivityDate="2022-07-19T11:33:54.943" Title="Can Peltier Modules be practically used to cool a room?" Tags="|mechanical-engineering|thermodynamics|hvac|refrigeration|thermocouple|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33479" PostTypeId="1" AcceptedAnswerId="33481" CreationDate="2020-03-03T09:12:18.067" Score="1" ViewCount="86" Body="&lt;p&gt;As you can see in the below shematic, the magnetic water conditioners can be used for the pipes of incoming process water on sinter plants to reduce surface tension of water and get fuel consumption benefits. But I researched and couldn't find any type of this conditioners/armatures .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wHutw.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wHutw.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look at first article about this: &lt;a href=&quot;https://www.steeltimesint.com/contentimages/features/KumarIronMaking2.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.steeltimesint.com/contentimages/features/KumarIronMaking2.pdf&lt;/a&gt;&#xA;This looked unrealistic to some engineers that magnetic water conditioner is not really exist thing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But.. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please also read: &lt;a href=&quot;https://www.sciencedirect.com/science/article/pii/S187538921400532X/pdf?md5=3846d794d248117cceb8e706f4b1489f&amp;amp;pid=1-s2.0-S187538921400532X-main.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sciencedirect.com/science/article/pii/S187538921400532X/pdf?md5=3846d794d248117cceb8e706f4b1489f&amp;amp;pid=1-s2.0-S187538921400532X-main.pdf&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above article looks more realistic, so is this magnetic charging chute can reproduce for water as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there any industrial-scale magnetic water conditioner exist?&lt;/p&gt;&#xA;" OwnerUserId="24364" LastEditorUserId="24364" LastEditDate="2020-03-04T11:44:20.270" LastActivityDate="2020-03-04T11:44:20.270" Title="What kind of magnetic water conditioner can industrial sinter plant use?" Tags="|mechanical-engineering|fluid-mechanics|chemical-engineering|mechanical|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33480" PostTypeId="1" AcceptedAnswerId="33485" CreationDate="2020-03-03T18:04:50.500" Score="-1" ViewCount="46" Body="&lt;p&gt;Does anyone know where I can find Heating tape that has these same leads or know what they are called? It is meant to be plugged into the second picture attatched.&lt;a href=&quot;https://i.stack.imgur.com/ahQQf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ahQQf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;a href=&quot;https://i.stack.imgur.com/G5ybU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/G5ybU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LXsZz.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LXsZz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24706" LastEditorUserId="24706" LastEditDate="2020-03-03T18:18:27.467" LastActivityDate="2020-03-03T20:50:38.113" Title="Heating tape that has leads like picture below" Tags="|electrical-engineering|heating-systems|heat-treatment|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33481" PostTypeId="2" ParentId="33479" CreationDate="2020-03-03T18:55:30.900" Score="1" Body="&lt;p&gt;The mystical claims for magnetic water conditioning have been given a special technical term to describe them, used only by real hip engineers. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is &lt;em&gt;jive- _ss baloney&lt;/em&gt;.  &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-03-03T18:55:30.900" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33482" PostTypeId="2" ParentId="33476" CreationDate="2020-03-03T19:22:28.163" Score="6" Body="&lt;p&gt;Peltier thermoelectric cooling modules can be used to cool anything you like. It will not be economical compared to a standard off the shelf air-conditioner (R-134A compressor cycle). If you source the modules, pumps, heat sinks and fans cheap and ignore your labor cost you may be able to come out at roughly the same manufactured cost a regular air conditioner. The cost of operation however will be considerably more expensive as the efficiency of themoelectric modules is much less.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So a thermoelectric cooler would not be &quot;practical&quot; for everyone, but if you had other constraints, it may make it a good solution.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Such a device could be made to run silent (as long as the fans and&lt;br&gt;&#xA;pumps were very quiet)&lt;/li&gt;&#xA;&lt;li&gt;Can run off of variable amounts of power (for&#xA;example variable power output from a solar array)&lt;/li&gt;&#xA;&lt;li&gt;If the system is properly designed they can last much much longer than a compressor (no moving parts)&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Here are some notes on the efficiency. Here is a clip from the &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermoelectric_cooling#Performance&quot; rel=&quot;noreferrer&quot;&gt;Thermoelectric Cooling Wiki&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In refrigeration applications, thermoelectric junctions have about 1/4&#xA;  the efficiency compared to conventional means (they offer around&#xA;  10–15% efficiency of the ideal Carnot cycle refrigerator, compared&#xA;  with 40–60% achieved by conventional compression-cycle systems&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://en.wikipedia.org/wiki/Carnot_cycle&quot; rel=&quot;noreferrer&quot;&gt;Carnot efficiency&lt;/a&gt; can be improved by lowering the temperature of sink (location you are discharging the heat). There are some options here since the design permits storing heat in water. Running the cooler at night when the air temp is colder permits the cooler (and a compressor unit) to run more efficiently. Similarly you could put water storage on the hot side of the modules as well, allowing that water to circulate at night and provide a cooler sink than the air would during the day.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And just because themoelectric modules are not comparatively efficient today, does not mean it will always be the case. Material science is always pushing the limits. Solid state devices will only get better and better as new materials and techniques are developed.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2020-03-03T19:22:28.163" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33483" PostTypeId="2" ParentId="33474" CreationDate="2020-03-03T19:33:43.410" Score="0" Body="&lt;p&gt;A thermoelectric cooling module has a thermal efficiency of some fraction of &lt;a href=&quot;https://en.wikipedia.org/wiki/Carnot_cycle&quot; rel=&quot;nofollow noreferrer&quot;&gt;Carnot efficiency&lt;/a&gt;. The &lt;a href=&quot;https://en.wikipedia.org/wiki/Thermoelectric_cooling#Performance&quot; rel=&quot;nofollow noreferrer&quot;&gt;thermoelectric cooling wiki&lt;/a&gt; puts them at 10 to 15% of Carnot efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running themoelectric modules in a series (stacked) configuration would be very disadvantageous as the waste heat also has to be pumped.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running thermelectric modules in a parallel (side by side) configuration would be advantageous because it would permit more heat transfer surface area on both sides of the module, which gives smaller temperature differences, which gives a higher Carnot efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Running multiple themoelectric modules in &lt;strong&gt;parallel&lt;/strong&gt; is more efficient than a single module for the same energy input.&lt;/p&gt;&#xA;" OwnerUserId="2819" LastActivityDate="2020-03-03T19:33:43.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33484" PostTypeId="2" ParentId="33442" CreationDate="2020-03-03T20:14:00.463" Score="0" Body="&lt;p&gt;Cans are stored under pressure. Left alone for a long time the acids, citric, phosphoric, etc, under pressure which can be augmented by heat, will eat away small fissures faster than they do under atmospheric pressure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This will allow the contents to gradually escape.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-03T20:14:00.463" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33485" PostTypeId="2" ParentId="33480" CreationDate="2020-03-03T20:50:38.113" Score="3" Body="&lt;p&gt;When searching for 1100 watt heating tape, a number of returns match almost exactly the image you've provided. A few of the manufacturers provide NEMA separable molded plugs and also offer bare leads. The bare leads would be useful in this case, as you can add your own 90° banana plugs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dn8aO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dn8aO.png&quot; alt=&quot;right angle enclosed banana plug&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To locate the correct plug, I used &quot;right angle enclosed banana plug&quot; as the search term. The image and source information came from &lt;a href=&quot;https://www.zoro.com/zoro-select-banana-plug-right-angle-1000vac-red-5txc0/i/G1470156/feature-product?gclid=EAIaIQobChMIrMzDg5b_5wIViMDICh1c_wWLEAQYAiABEgIGnPD_BwE&quot; rel=&quot;nofollow noreferrer&quot;&gt;Zoro&lt;/a&gt; online. I had better returns using shrouded in place of enclosed. Many sources. Be sure to check the current draw for the tape you'll use and the connector you'll select.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-03-03T20:50:38.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33486" PostTypeId="1" CreationDate="2020-03-04T06:30:56.083" Score="1" ViewCount="1121" Body="&lt;p&gt;The RCC column was cut in my absence due to passage of pipe for bathroom fittings. The thickness is about 1.5 in. Will it hamper the stability of structure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/q7Fod.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/q7Fod.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lnl8W.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lnl8W.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="11926" LastEditorUserId="1832" LastEditDate="2020-03-07T13:49:22.397" LastActivityDate="2020-03-08T02:46:57.733" Title="Is it safe to slightly cut the edge of RCC column for passage of pipe in bathroom?" Tags="|structural-engineering|structural-analysis|structures|building-design|reinforced-concrete|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33487" PostTypeId="2" ParentId="31601" CreationDate="2020-03-04T13:08:46.117" Score="1" Body="&lt;p&gt;Don't let the small &lt;span class=&quot;math-container&quot;&gt;$C_v$&lt;/span&gt; values throw you off.  Many pressure regulators for research applications work well in the well below 1E-5, 1E-6, and even much lower.  You may already have a good calculator, but this &lt;a href=&quot;https://www.engineeringtoolbox.com/flow-coefficients-d_277.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;EngineeringToolBox&lt;/a&gt; page has a gas &lt;span class=&quot;math-container&quot;&gt;$C_v$&lt;/span&gt; calculator that is easy to use.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I also question if you have resolved what will control the mass flow.  Can you clarify whether you primary  objective is flow control?  It could be you need a pressure reducing regulator to protect the mass flow device, but, as suggested above, a PRR will not control the flow for you.  &lt;/p&gt;&#xA;" OwnerUserId="14790" LastEditorUserId="9" LastEditDate="2020-03-04T16:03:15.333" LastActivityDate="2020-03-04T16:03:15.333" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33491" PostTypeId="1" CreationDate="2020-03-04T20:06:37.027" Score="1" ViewCount="316" Body="&lt;p&gt;I am designing an enclosure that holds a lens (clearance fit) and a photodiode on a PCB. The alignment of the photodiode to the lens must be pretty tight, thus I am using a boss on the enclosure (also a clearance fit) to position the PCB so that the photodiode is directly underneath the lens. (see first image).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SCcQB.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SCcQB.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do you address clearance fit gaps when doing tolerance loops? Is it good practice to shift everything to its worst case position when doing loops like so?:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/urkFL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/urkFL.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24864" LastEditorUserId="24864" LastEditDate="2020-03-04T20:12:23.817" LastActivityDate="2020-03-04T20:12:23.817" Title="Tolerance Analysis on clearance holes" Tags="|mechanical-engineering|mechanical|tolerance|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33492" PostTypeId="2" ParentId="33462" CreationDate="2020-03-04T23:21:48.337" Score="2" Body="&lt;p&gt;First of all let's consider the aspect of dominant poles. You have to control the plant:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ G(s) = \frac{1}{s(s+2)(s+5)} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poles of this plant are located at : &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$p_1 = 0$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$p_2 = -2$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$p_3 = -5$&lt;/span&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Dominant poles in general are the poles which are closer to the imaginary axis and define the response (behaviour) of the system. In this particular system the poles that are closer to the imaginary axis are: &lt;span class=&quot;math-container&quot;&gt;$\ -2,0 $&lt;/span&gt;, so we can say that they dominate the dynamic behaviour of the system. It is always good if possible to model a system by a second order transfer (while obtaining the steady state behaviour) due to the simplicity of the math involved in the deisgn process. In order to reduce the system‘s order and simultaneously maintain the steady state response we can bring this system up to the system: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ G(s) = \frac{\frac{1}{5}}{s(s+2)} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can use MATLAB to check the steady state responses of the two systems. These are shown at the graph below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/UiLOU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/UiLOU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that both systems have the same steady state responses. Now, we continue with the design process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;First Step:&lt;/strong&gt; Find the closed loop transfer function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ G(s) = \frac{1/5}{s(s+2)} \rightarrow Plant $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ G_c(s) = K_p+K_ds \rightarrow PD \  Controller $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the closed loop transfer with unity feedback is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ H(s) = \frac{0.2(K_p+K_ds)}{s^2+s(2+0.2K_d)+0.2K_p} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The characteristic polynomial (denominator of the closed-loop transfer function) is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ s^2+(2+0.2K_d)s+0.2K_p = 0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Second Step:&lt;/strong&gt; Pick the gains of the controller with respect to the system's design specifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ ζ = 0.707 $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ t_s = \frac{4}{ζω_n} \Rightarrow 2=\frac{4}{ζω_n} \Rightarrow ω_n = 2.8289 \ rad/sec $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Third Step:&lt;/strong&gt; Obtain the desired characteristic polynomial with respect to the desgin specifications:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ s^2+2ζω_ns+ω_n^2 = 0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ ζ: $&lt;/span&gt; Damping Ratio &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ ω_n: $&lt;/span&gt;  Natural Frequency&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which by plugging in the vaues for &lt;span class=&quot;math-container&quot;&gt;$\ ζ $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ ω_n $&lt;/span&gt; becomes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ s^2 + 4.0001s + 8.0027 = 0 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Fifth Step:&lt;/strong&gt; Match the coefficients of the system's characteristic polynomial with the corresponding ones of the desired characteristic polynomial. By doing so we come up with the values of &lt;span class=&quot;math-container&quot;&gt;$\ K_p $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ K_d $&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ K_p = 40.0135 $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ K_d = 10.0005 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Sixth Step:&lt;/strong&gt; Check the step response to verify the requirements. By plugging in the values for the gains &lt;span class=&quot;math-container&quot;&gt;$\ K_p $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ K_d $&lt;/span&gt; to the closed loop transfer function we obtain the following step response of the system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lsqnz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lsqnz.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is shown that the settling time is: &lt;span class=&quot;math-container&quot;&gt;$\ t_s=1.86sec $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you put these particular gains and simulate the initial third order system, you will find out that the settling time is larger than the settling time of the equivalent second order system. Below is the graph of simulation of the initial system (blue line) and the equivalent second order system (red line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XbgNt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XbgNt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens because the third pole of the initial system &lt;span class=&quot;math-container&quot;&gt;$\p_3 = -5 $&lt;/span&gt; is not that far away so we can ignore it. It is near the dominant poles which means that it also affects considerably the dynamic behaviour of the system. However, it is important to notice that at steady state the two systems reach the same amplitude which means that the equivalent system we obtained is correct. In general, we can approach a third order by a second order system when the third pole of the third order system is sufficiently further away than the dominant poles. For example consider the following system: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ H(s) = \frac{1}{s(s+2)(s+30)} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The poles of this system are: &lt;span class=&quot;math-container&quot;&gt;$\ 0, -2, -30 $&lt;/span&gt;. The third pole of this system is sufficiently bigger than the two dominant poles. This system is approached by the second order system:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ G(s) = \frac{\frac{1}{30}}{s(s+2)} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The step responses of these systems, which are similar, are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/BSq3f.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/BSq3f.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we follow the same design process, we obtain the following PD controller: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ G_c(s)=240.3213+60.0631s $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if we simulate the initial third order system and the equivalent second order system in series with this PD compensator (and unity feedback), we obtain the following graph:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Y8fCI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Y8fCI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is obvious that the step responses of the two systems are almost identical. The difference in settling time is &lt;span class=&quot;math-container&quot;&gt;$\ 0.03 $&lt;/span&gt; seconds, which is neglectible. In conclusion, the aspect of dominant poles is one hundred percent correct when there is a pole which is enough bigger than the poles which are close to the imaginary axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below, I place some MATLAB code which produces the above results.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = tf('s');&#xA;Kp = 240.3213; % Or any gain&#xA;Kd = 60.0631% Or any gain&#xA;initial = tf([0 0 0 1],[1 32 60 0]); % Or any system&#xA;equivalent = tf([0 0 1/30],[1 2 0]); % Or any equivalent&#xA;pd_compensator = Kp+Kd*s;&#xA;initial_comp = series(pd_compensator,initial);&#xA;equivalent_comp = series(pd_compensator,equivalent);&#xA;closed_initial = feedback(initial_comp, 1);&#xA;closed_equivalent = feedback(equivalent_comp, 1);&#xA;&#xA;figure&#xA;step(closed_initial)&#xA;hold on&#xA;step(closed_equivalent)&#xA;hold off&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-05-10T02:49:54.483" LastActivityDate="2020-05-10T02:49:54.483" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33493" PostTypeId="2" ParentId="33460" CreationDate="2020-03-04T23:43:41.900" Score="0" Body="&lt;p&gt;Angular momentum L is &lt;span class=&quot;math-container&quot;&gt;$ L= I\omega = 10kgm^2*2\pi*6000/60= 2\pi*6000kgm^2 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \frac{d\phi}{dt}=3/s=0.016radian/s= \frac{1}{L} \tau $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \tau= 0.016*2\pi*6000=2\pi*100kgm^2 $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This torque must act at XZ plane 90 degrees to dive angle and in your coordinate shown is negative. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a real situation, one needs to consider the angular momentum of the entire plane and the change in drag and lift. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because the numbers are round I did the calculations in my head, check my work. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW. I am a private pilot and fly a single-engine, piston, Cessna 172. I have never had to consider this. Albeit we apply yaw correction by rudders for controlling the propeller's differential angle of attack.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-03-05T00:01:05.693" LastActivityDate="2020-03-05T00:01:05.693" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33494" PostTypeId="2" ParentId="33469" CreationDate="2020-03-05T01:03:36.153" Score="0" Body="&lt;p&gt;Looks like you’re doing 2d joint kinematics! We can think of this as an exercise in the chain rule. We can apply the following for any function describing position w.r.t time, &lt;span class=&quot;math-container&quot;&gt;$\theta$&lt;/span&gt; just happens to be the problem at hand.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Linear Velocity&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Linear Velocity is the first derivative of position with respect to time, which can be computed as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{dx_1}{dt} = \frac{df}{dt} = \frac{\partial f}{\partial \theta} \frac{d\theta}{dt} + \frac{\partial f}{dt}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial f}{\partial \theta} \frac{d\theta}{dt}$&lt;/span&gt;  is the portion of linear velocity caused by rotation, and &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial f}{dt}$&lt;/span&gt; is the linear velocity of the mechanism independent of rotation. So if your mechanism were mounted on a car, the first term is the velocity of the point &lt;span class=&quot;math-container&quot;&gt;$x_1$&lt;/span&gt; if the car were stationary, and the second term is the velocity of the car. It sounds like &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial f}{dt} = 0$&lt;/span&gt; for your problem.&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Linear Acceleration&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;Differentiating again, we use everybody’s favorite rule, the product rule.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{d^2x_1}{dt^2} =  \frac{d^2f}{dt^2} =  (\frac{d}{dt}\frac{\partial f}{\partial \theta} \frac{d\theta}{dt} + \frac{\partial f}{\partial \theta} \frac{d^2\theta}{dt^2}) + \frac{\partial^2f}{\partial t}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the same analogy: &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial^2f}{\partial t}$&lt;/span&gt; is the acceleration of the car, and the parenthesized pair of terms are the contribution of angular acceleration and angular velocity to linear acceleration. Recall that a constant circular motion implies an inward acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; does not vary directly with time, every term is zero except &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial f}{\partial \theta} \frac{d^2\theta}{dt^2}$&lt;/span&gt;, the angular acceleration term, &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With many arguments (thetas in your case), we can write:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{d^2x_n}{dt^2} =  \frac{d^2f}{dt^2} =  \sum_{n=1}^{N} (\frac{\partial^2f}{\partial \theta_n \partial t} \frac{d\theta_n}{dt} + \frac{\partial f}{\partial \theta_n} \frac{d^2\theta_n}{dt^2}) + \frac{\partial^2f}{\partial t}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="3374" LastActivityDate="2020-03-05T01:03:36.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33495" PostTypeId="1" CreationDate="2020-03-05T01:32:15.230" Score="0" ViewCount="39" Body="&lt;p&gt;My background is in electrical engineering so I am something of a novice when it comes to pneumatic systems, but I have been tasked with designing a sampling system that draws an extremely low flow at low pressure with a diaphragm pump (0.25 lpm) which is then read by a flow meter (0..1 lpm) before passing through an oxygen sensor. The trouble is that the pressure pulsations from the pump cause a very inaccurate reading at the flow meter. We have since discovered that implementing a makeshift latex reservoir between the pump and flow meter almost eliminates the problem completely, however this is not a very industrial solution. After further research I found that a similar solution in hydraulic systems is known as a hydraulic accumulator, but I can't seem to find an equivalent product for pneumatic systems. Is there some obvious solution that I am missing or does no such product exist?&lt;/p&gt;&#xA;" OwnerUserId="24865" LastEditorUserId="13297" LastEditDate="2020-03-05T09:34:55.340" LastActivityDate="2020-03-05T09:34:55.340" Title="How to achieve pulsation damping in pneumatic rather than hydraulic system?" Tags="|mechanical-engineering|pneumatic|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33497" PostTypeId="2" ParentId="33495" CreationDate="2020-03-05T02:53:50.317" Score="0" Body="&lt;p&gt;you can add a very flexible blader with a light ballast weight pushing it down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You just have to change your zero datum to the new level.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-05T02:53:50.317" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33498" PostTypeId="1" CreationDate="2020-03-05T03:21:07.470" Score="1" ViewCount="104" Body="&lt;p&gt;It's easy to obtain a room-scale radiant heater (of the order of 700-800W per bar, eg &lt;a href=&quot;http://www.goldair.com.au/product-catalogue/heating/radiant-heaters/2400w-3-bar-radiant-heater-gir335&quot; rel=&quot;nofollow noreferrer&quot;&gt;this three-bar heater&lt;/a&gt;) that operates very simply and easily - you turn it on, it consumes electricity, it generates heat. Convenient, cheap, simple. But they don't seem to be made in lower wattages, and my inexpert brain is wondering: what would happen if one of these were to have some electronics added to run it only part of the time? For instance, let's say I selected 75% power on one bar; it would close the circuit for 750ms, then cut the power for 250ms, repeat indefinitely.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any critical downsides to a system like this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would this actually consume (roughly) 75% of the power and produce 75% of the heat, or would something in the behaviour of this kind of heating element change this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why don't heaters like this exist, with full option for any power level you choose?&lt;/p&gt;&#xA;" OwnerUserId="24868" LastActivityDate="2020-03-05T18:33:42.380" Title="Bar heater operating on a duty cycle - possible problems?" Tags="|electrical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33499" PostTypeId="1" CreationDate="2020-03-05T03:30:42.093" Score="2" ViewCount="42" Body="&lt;p&gt;I've run some stress simulations but I keep getting some unexpected results. I have 3 parts connected using 4 bolts as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YvYdK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YvYdK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A torque of 1N is applied to the hole on the right to simulate a hung weight. After running several stress simulations, the two highest points of stress concentration appear to locate at the bottom of the part with the hole, specifically at the end of the fillet on the outer surface:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yISB2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yISB2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H18Cb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H18Cb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, there are two hotspots along the end of the fillet -- equal distance away from the edges. Does anyone know what might cause the stress at these two locations?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here's an overlay of the mesh over the stress hotspot:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/U3k7p.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/U3k7p.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="4302" LastEditDate="2020-03-05T11:34:25.230" LastActivityDate="2020-03-05T11:34:25.230" Title="Unexpected stress concentration results" Tags="|mechanical-engineering|stresses|solidworks|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33500" PostTypeId="2" ParentId="33495" CreationDate="2020-03-05T04:24:50.977" Score="1" Body="&lt;p&gt;There are pneumatic accumulators, but they introduce the following problem:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are doing a chemical assay on the material being sampled by your apparatus, adding &lt;em&gt;compliance&lt;/em&gt; to eliminate pulsate flow usually means adding a fairly large reservoir into the flow path, so that the pulses get absorbed by the compliance of the gas inside that reservoir. This will also delay the measurements, in that the reservoir must become fully charged with the sampled gas flow before it flows to your measurement device. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would recommend a &lt;em&gt;peristaltic pump&lt;/em&gt; to pull your samples out of the process flow instead of a diaphragm pump. They introduce less pulsation into the sample flow.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-03-05T04:24:50.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33501" PostTypeId="1" CreationDate="2020-03-05T10:31:08.843" Score="2" ViewCount="205" Body="&lt;p&gt;We all  know that in an indirect (closed loop) hydronic system, an expansion tank is needed ,to deal with thermal expansion of the circulating water, that can be either open expansion or closed expansion tank however I've mostly seen indirect solar water heating systems with closed expansion tank.&#xA;Can we use open expansion tanks in this type of system?&lt;/p&gt;&#xA;" OwnerUserId="23043" LastActivityDate="2024-01-06T20:01:55.323" Title="Can we use an open expansion tank in solar water heating system?" Tags="|solar-energy|thermal-expansion|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="33504" PostTypeId="2" ParentId="3883" CreationDate="2020-03-05T12:31:37.217" Score="0" Body="&lt;p&gt;We might arrive at the answer &lt;span class=&quot;math-container&quot;&gt;$\text{12.3%}$&lt;/span&gt; for OMC, and &lt;span class=&quot;math-container&quot;&gt;$2001.273 \dfrac{kg}{m^3}$&lt;/span&gt; for the MDD. The reason maybe why the author's answer for MDD is &lt;span class=&quot;math-container&quot;&gt;$2020 \dfrac{kg}{m^3}$&lt;/span&gt; is due to the application of &lt;span class=&quot;math-container&quot;&gt;$\text{95% confidence}$&lt;/span&gt; on the answer; i.e.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\displaystyle &#xA;2001.273 \pm \text{95%} = [1982.261, 2020.285] &#xA;$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that's where the &lt;span class=&quot;math-container&quot;&gt;$2020$&lt;/span&gt; might come from.  ~&lt;/p&gt;&#xA;" OwnerUserId="24871" LastActivityDate="2020-03-05T12:31:37.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33506" PostTypeId="1" CreationDate="2020-03-05T15:14:40.677" Score="-2" ViewCount="106" Body="&lt;p&gt;Became president of a rocket club. Since I'm kind of new to this whole internship/job stuff (haven't had any in the past) so I'm wondering if this would contribute towards it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, what should I do to really land one as a president of a rocket club? I know ofc that I will have to do typical rocketry stuff, having some workshops, inviting experts in rocketry for talks, hosting rocket events in my uni. But besides these things that I have mentioned. What else would I have to improve my chances of getting an internship/job in an aeronautical company?&lt;/p&gt;&#xA;" OwnerUserId="24874" LastActivityDate="2020-03-05T16:48:57.517" Title="As a president of my university rocket club. Would it be easier for me to get an internship/job in an aeronautical company?" Tags="|aerospace-engineering|" AnswerCount="3" CommentCount="2" ClosedDate="2020-03-07T13:01:40.613" ContentLicense="CC BY-SA 4.0" />
  <row Id="33507" PostTypeId="2" ParentId="33506" CreationDate="2020-03-05T15:24:47.530" Score="0" Body="&lt;p&gt;It definitely helps, but your relevant coursework and grades are a large (if not larger) factor. Apply early and to as many opportunities as you can.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there any competitions with other school's &quot;rocket clubs&quot;? Winning awards at those would help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your school has a career center, utilize that as well to polish your resume.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-03-05T15:24:47.530" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33509" PostTypeId="2" ParentId="33506" CreationDate="2020-03-05T16:19:11.827" Score="0" Body="&lt;p&gt;leadership is important and your being president shows leadership.  It's the kicker over other candidates with similar classes and grades.  &lt;/p&gt;&#xA;" OwnerUserId="21312" LastActivityDate="2020-03-05T16:19:11.827" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33510" PostTypeId="2" ParentId="33506" CreationDate="2020-03-05T16:48:57.517" Score="1" Body="&lt;p&gt;In itself this can be either good or bad. If you can use it to demonstrate an active interest in rocketry, plus some leadership and organizational skills, that is obviously good.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But don't try something like this just &quot;because it will look good on your application&quot;. Don't forget that if you get an interview at an aerospace company, quite likely your interviewer will have spent some time working on &quot;real&quot; rockets and also have some management experience, so expect to end up talking about what you did &lt;em&gt;in detail.&lt;/em&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is usually obvious to an experienced interview whether you are just &quot;saying what you think they want to hear&quot; and/or talking technical BS, or whether you &lt;em&gt;really&lt;/em&gt; contributed something to the club.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-03-05T16:48:57.517" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33511" PostTypeId="1" CreationDate="2020-03-05T17:41:42.317" Score="2" ViewCount="291" Body="&lt;p&gt;What sensor, using what physical principle, can penetrate a closed rectangular object like the trailer of a truck and tell if the space is empty or loaded?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why is this question relevant? One of the biggest problems of maintaining &amp;amp; building roads is to understand how much load is transported over that road per time increment. A loaded truck can bring much more damage to a road than many cars can (a 18 wheeler can have the impact of over nine thousand cars). Current in-road, at-speed weighing technology can cost millions. The reason for this questions is trying to find a way to use existing low cost sensors (or a combination of multiple sensors) at the roadside to assist with estimating the weight of a truck. If we know the truck/ trailer combination is empty or full, we typically know the lowest and maximum load weight. And then can estimate the average load on this road per time increment. This is much better than current vehicle counts and is commercially very interesting.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks for staying on topic and trying to point me in the right direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Initial thoughts: It is easy if you can visually see the load (like on a logging truck). It is hard if it is a closed curtainsider, bulk load or container truck from a stationary roadside measurement unit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Computer vision could be used to compare the bounce of the cab, analyze the suspension; use thermal camera to detect higher temperature of tyres or axis; acoustics of empty vs loaded. What other sensors/ physical principles should be tried like measure on an uphill stretch: slower truck of same vehicle class will be loaded truck.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, many thanks for sharing your thoughts on what physical principle could be tried to tackle the problem to detect if a confined space is empty or filled, from the outside without weighing it. An approximation with an acceptable error rate is fine.&lt;/p&gt;&#xA;" OwnerUserId="24877" LastActivityDate="2020-06-25T20:06:13.393" Title="What sensors/methods can be used from the roadside to determine if a truck is loaded or empty - without weighing it?" Tags="|mechanical-engineering|electrical-engineering|sensors|computer-engineering|ultrasound|" AnswerCount="5" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33512" PostTypeId="2" ParentId="33498" CreationDate="2020-03-05T18:33:42.380" Score="1" Body="&lt;p&gt;Electric heaters are &lt;code&gt;LR circuit&lt;/code&gt; Inductance and resistance, controlling heaters with operation duty cycle will be far expensive then using simple use a high &lt;code&gt;LR coil&lt;/code&gt;. switching controlled power supply will introduce skips or surge voltage into the supply line. hence not advisable. more power switching mean more surges. Moreover various county has certain government regulation around it to lower inductive loads by adding parallel capacitors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;off-topic note for enthusiast : In terms of fluid mechanics it is called Hydraulic shock&lt;/p&gt;&#xA;" OwnerUserId="24878" LastActivityDate="2020-03-05T18:33:42.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33513" PostTypeId="2" ParentId="33511" CreationDate="2020-03-05T18:42:07.067" Score="2" Body="&lt;p&gt;Ultra sound sensors could be used, you know the general size of all available container sizes and the material it is made of create a lookup table for such empty containers returning sounds. sound returned from a loaded truck will be different. It will not give you &lt;code&gt;mg&lt;/code&gt; though.&lt;/p&gt;&#xA;" OwnerUserId="24878" LastActivityDate="2020-03-05T18:42:07.067" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33514" PostTypeId="2" ParentId="33511" CreationDate="2020-03-05T22:26:16.497" Score="0" Body="&lt;p&gt;I got pulled over and was issued a speeding ticket fifty-some years ago in the middle of nowhere, and there were no guns then to clock you.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I asked the cop how he estimated my speed, which by the way was pretty accurate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He said he positioned himself after a certain curve on the road and used a vertical road sign as a plumb, and by checking the degree of the cars lateral tilt due to the centripetal force he could tell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can put a camera on a similar position, the loaded trucks will sway more. One can establish a cross table of loads and tilts for different trucks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Very practical and almost no maintenance.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-05T22:26:16.497" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33516" PostTypeId="1" CreationDate="2020-03-06T00:21:36.437" Score="-1" ViewCount="478" Body="&lt;p&gt;I'm trying to use a battery to activate an op amp schmitt trigger but learned that there needs to be both a positive and negative voltage supplied to Vcc+ and Vcc-. I've seen a lot of schematics online that show using a secondary voltage source for this purpose but I can't do that due restrictions of my circuit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: Here are extra details to clarify the circuit:&#xA;input battery voltage: 14.8V-12.3V(14.8V at max charge, 12.2V max depleted voltage before the battery starts taking damage)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 2: add a better labeled diagram and added second full schematic the op amp schmitt trigger is supposed to be a part of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Trigger voltages: I want the op amp to send a signal while 12.2V &amp;lt; Vss and stop at 12.2 &gt;= Vss. (ignore my earlier comment saying 12.3V is the shutoff point, read my schematic wrong)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3rNdT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3rNdT.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I create both a negative and positive voltage for the vcc+ and vcc- connections of the op amp with only a singular voltage source?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YfWRC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YfWRC.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24722" LastEditorUserId="24722" LastEditDate="2020-03-06T20:14:09.893" LastActivityDate="2020-03-07T13:00:24.157" Title="How do you create a negative and positive voltage(-6V and 6V) with a singular voltage source?" Tags="|electrical-engineering|circuits|" AnswerCount="4" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33517" PostTypeId="2" ParentId="33511" CreationDate="2020-03-06T02:45:20.353" Score="1" Body="&lt;p&gt;Measure the deflection of the suspension springs in the trailer. The greater the load, the lower the trailer will sit. You prepare a calibration chart using known weights loaded on a given brand of trailer and repeat for the most popular brands. Then you can detect how heavily the trailer is loaded by how much closer to the ground it is than it would be if empty. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other way to do this is to measure the area of the tire's contact pad on the pavement surface, and measure the tire pressure. From these you can solve explicitly for the load that that tire is carrying. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-03-06T02:45:20.353" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33518" PostTypeId="2" ParentId="33516" CreationDate="2020-03-06T02:49:16.553" Score="0" Body="&lt;p&gt;This is easy, if you are using batteries. For example, if the circuit requires -6V-0-+6V, you get two 6 volt batteries, wire them in series, and bring out a wire from their junction point. The junction wire serves as the 0 volt point, the top of the battery stack is then at +6 volts and the bottom is at -6 volts, relative to the junction wire. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-03-06T02:49:16.553" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33519" PostTypeId="1" CreationDate="2020-03-06T03:25:24.947" Score="1" ViewCount="45" Body="&lt;p&gt;Is there a standards organization or methodology to determine the elastic material properties of common materials?  What constants are normally used to define these properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More specifically, the set of equations for linear elasticity can be written as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\nabla\cdot\sigma + {F} = \rho\ddot{{u}}\\&#xA;{\varepsilon} =\tfrac{1}{2} \left[{\nabla}{u}+({\nabla}{u})^\mathrm{T}\right]\\&#xA;{\sigma} = {C}:{\varepsilon}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; - Stress tensor&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\epsilon$&lt;/span&gt; - Strain tensor&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$u$&lt;/span&gt; - Displacement&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; - Stiffness tensor&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Using Voigt notation, we can write the stiffness tensor as a &lt;span class=&quot;math-container&quot;&gt;$6\times 6$&lt;/span&gt; symmetric matrix&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;C = \begin{bmatrix}&#xA;C_{11} &amp;amp; C_{12} &amp;amp; C_{13} &amp;amp; C_{14} &amp;amp; C_{15} &amp;amp; C_{16}\\&#xA; &amp;amp; C_{22} &amp;amp; C_{23} &amp;amp; C_{24} &amp;amp; C_{25} &amp;amp; C_{26}\\&#xA; &amp;amp; &amp;amp; C_{33} &amp;amp; C_{34} &amp;amp; C_{35} &amp;amp; C_{36}\\&#xA; &amp;amp; &amp;amp; &amp;amp; C_{44} &amp;amp; C_{45} &amp;amp; C_{46}\\&#xA; &amp;amp; &amp;amp; &amp;amp; &amp;amp; C_{55} &amp;amp; C_{56}\\&#xA; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; C_{66}&#xA;\end{bmatrix}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, this matrix &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; defines the elastic material properties of a material, but I've been unable to find any reference or database that lists materials in this format.  I've found a few random materials that list their properties in terms of something like the bulk and shear modulus, which then map to this tensor as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;C = \begin{bmatrix}&#xA;\lambda + 2\mu &amp;amp; \lambda &amp;amp; \lambda &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA; &amp;amp; \lambda + 2\mu &amp;amp; \lambda &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA; &amp;amp; &amp;amp; \lambda + 2\mu &amp;amp; 0 &amp;amp; 0 &amp;amp; 0\\&#xA; &amp;amp; &amp;amp; &amp;amp; \mu &amp;amp; 0 &amp;amp; 0\\&#xA; &amp;amp; &amp;amp; &amp;amp; &amp;amp; \mu &amp;amp; 0\\&#xA; &amp;amp; &amp;amp; &amp;amp; &amp;amp; &amp;amp; \mu&#xA;\end{bmatrix}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\lambda$&lt;/span&gt; - First Lame parameter&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\mu$&lt;/span&gt; - Shear modulus&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$\lambda + 2\mu/3$&lt;/span&gt; - Bulk modulus&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;However, this only works for isotropic materials.  Generally speaking, we need all 21 elements of &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;.  As such, I didn't know if there is a standards organization or methodology for determining these properties beyond the shear and bulk modulus.  Alternatively, if shear and bulk are the only parameters really needed, how are these determined and what organization standardizes these tests?&lt;/p&gt;&#xA;" OwnerUserId="6904" LastEditorUserId="6904" LastEditDate="2020-03-09T15:50:05.087" LastActivityDate="2020-03-09T16:23:54.370" Title="Is there a standards organization or methodology to determine the elastic material properties of common materials?" Tags="|mechanical-engineering|structural-engineering|materials|elastic-modulus|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33521" PostTypeId="2" ParentId="19068" CreationDate="2020-03-06T09:53:42.143" Score="1" Body="&lt;p&gt;I believe that the Borda-Carnot equation is only valid for incompressible flows, while you mention that the injector orifice is choked. While the flow becomes incompressible pretty quickly after the orifice, at this precise point, it is not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I think you should use 1D isentropic nozzle theory (wikipedia offers a good start on the topic).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the injector orifice is choked and there is no shock in the injector channel or downstream of it, then the pressure ratio should be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;P_{res}/P_{a} = \left(\frac{\gamma+1}{2}\right)^{\frac{\gamma}{\gamma-1}}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with &lt;span class=&quot;math-container&quot;&gt;$P_{res}$&lt;/span&gt; the tank pressure, &lt;span class=&quot;math-container&quot;&gt;$P_a$&lt;/span&gt; the ambient pressure, &lt;span class=&quot;math-container&quot;&gt;$\gamma = \frac{c_p}{c_v}$&lt;/span&gt; the ratio of heat capacities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's assuming you are injecting gas into gas. If the fuel in the tank is liquid, then the situation is different because of evaporation.&lt;/p&gt;&#xA;" OwnerUserId="10381" LastActivityDate="2020-03-06T09:53:42.143" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33522" PostTypeId="1" AcceptedAnswerId="33541" CreationDate="2020-03-06T10:13:05.997" Score="3" ViewCount="129" Body="&lt;p&gt;I'm translating a document in which there is a mention of &lt;strong&gt;обрешетки&lt;/strong&gt; used for preparing goods (medical equipment of large size) for shipment. This term derives from reshetka (решетка) - lattice, and means &quot;custom-made frame of wooden planks built around a load destined for shipping&quot;. It's kind of a rough crate built around the product intended for shipping, not pre-made.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder what the proper term in English might be. Multitran &lt;a href=&quot;https://www.multitran.com/m.exe?l1=2&amp;amp;l2=1&amp;amp;s=%D0%BE%D0%B1%D1%80%D0%B5%D1%88%D0%B5%D1%82%D0%BA%D0%B0&quot; rel=&quot;nofollow noreferrer&quot;&gt;provides a long list of options&lt;/a&gt;, but without describing them much, so it's hard to distinguish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's a whole &lt;a href=&quot;http://docs.cntd.ru/document/1200011144&quot; rel=&quot;nofollow noreferrer&quot;&gt;National Standard document in Russian&lt;/a&gt; that describes how these &quot;frames&quot; must be built, with many pictures. However, it provides the wrong translation there in English: &quot;roof boarding&quot; (because the same Russian term is also used for roof lattice). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/NOewG.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NOewG.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fHhF6.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fHhF6.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm submitting my translation in an hour, so I'll stick with descriptive &quot;wooden frames&quot; but I'd be glad to learn the formal term if it exists. &lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="349" LastEditDate="2020-03-07T14:43:16.190" LastActivityDate="2020-03-07T14:43:16.190" Title="What do we call these &quot;transport frames made of wood&quot; (обрешетки in Russian)" Tags="|terminology|transportation|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33523" PostTypeId="1" CreationDate="2020-03-06T12:34:12.413" Score="1" ViewCount="95" Body="&lt;p&gt;I'm currently working on a setup where at some operations my motor has to drive a shaft up to the motor's maximum operating speed (12,000rpm) and a torque of ~0.14Nm but at other conditions this shaft can drive the motor up to 40,000 rpm. However, since this is outside the motor's operating range, and the shaft doesn't transmit any power (i.e. the torque on this shaft is zero), I want to decouple the motor from the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been looking for electromagnetic clutches but so fare I only found clutches with a maximum operating speed of 20,000rpm. Do you know if such clutches, of other mechanisms, exist with maximum speeds of 40,000rpm and higher? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you advance!&lt;/p&gt;&#xA;" OwnerUserId="24885" LastActivityDate="2020-03-06T12:34:12.413" Title="Clutch for high-speed operation (&gt;40 000 rpm)" Tags="|mechanical-engineering|mechanisms|robotics|actuator|transmission|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33525" PostTypeId="2" ParentId="6756" CreationDate="2020-03-06T14:12:18.287" Score="2" Body="&lt;p&gt;The answer provided by Dave is correct but it covers only a very specific case of no-friction and constant speed. There are three forces acting on a vechicle which must be accelerated from 0 to v speed over a slope:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mxvZb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mxvZb.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cr is the rolling friction coefficient, around 0.01 for cars on asphalt, so probably something less for bikes. Don't know for skateboards, but they have forul large&amp;amp;flat wheels, so maybe it's more similar to car coefficient than bike coefficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Being it a skateboard over a slope, it has a very slow speed, so we can neglect air drag, but for a generic vehicle you must also take it into account:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F_D = \frac 1 2  \rho  C_D  A  v^2 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \rho = air density = 1.22 kg/m3 $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ C_D = \text{drag coefficient (around 0.8 for bike/human, 0.2-0.3 for cars)}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ A = \text{Section area or Frontal area, m2}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$v = \text{speed, m/s}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Inertia force resists to acceleration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found a calculator: it's designed for robots, so it disregards both wheel and air drags:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.robotshop.com/community/blog/show/drive-motor-sizing-tool&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.robotshop.com/community/blog/show/drive-motor-sizing-tool&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a much more complex&amp;amp;complete simulator, but I don't understand how to input custom data for motor:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://vesc-project.com/calculators&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://vesc-project.com/calculators&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24889" LastActivityDate="2020-03-06T14:12:18.287" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33526" PostTypeId="1" CreationDate="2020-03-06T14:41:19.413" Score="1" ViewCount="30" Body="&lt;p&gt;I am trying to solve the system of ODEs below. I am applying the finite differences to z, so I get a system of ODE that I can solve with a solver like ode45.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uOXzz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uOXzz.png&quot; alt=&quot;ODE system&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that under the conditions that interest me integration is too slow (time step is very small) and I get no result. I would like to know what is the underlying problem and if there is any way to solve it. I have tried: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;using different solvers (tried all of them)&lt;/li&gt;&#xA;&lt;li&gt;using dimensionless variables&lt;/li&gt;&#xA;&lt;li&gt;using the pdepe function of Matlab to solve the PDE system directly without applying finite differences myself. Same problem occurs.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Full code (using the parameters that do not work)&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;function transport_model()&#xA;global parameter L Di epsb Q Cfeed K tfinal npz npt F ui h&#xA;&#xA;parameter = [0.044511*432.75 432.75];    % isotherm parameters&#xA;L = 0.4;            % m, column length&#xA;epsb = 0.367;       % column bulk porosity&#xA;ui = 4e4;        % m/s, velocity&#xA;Cfeed = 200;     % feed concentration&#xA;K = 0.0782          % s-1, mass transfer coefficient&#xA;tfinal = 400;      % min, final time for calculation&#xA;npz = 50;           % number of discretization points in z&#xA;npt = 20;&#xA;&#xA;F = (1-epsb)/epsb;&#xA;tspan = 0:tfinal/(npt-1):tfinal;&#xA;y0 = zeros(2*npz,1);&#xA;h = L/(npz-1);&#xA;&#xA;sol = ode15s(@sedo, tspan, y0);&#xA;t = sol.x;&#xA;C1 = sol.y(1:npz,:);&#xA;q1 = sol.y((npz+1):end,:);&#xA;&#xA;plot(t, C1(end,:))&#xA;xlim([0 tfinal])&#xA;ylim([0 Cfeed])&#xA;&#xA;&#xA;&#xA;function DyDt = sedo(t,y)&#xA;global Cfeed K  npz F ui h&#xA;&#xA;N = npz;&#xA;y(1) = Cfeed;&#xA;DyDt = zeros(2*N,1);&#xA;&#xA;% Forward finite differences&#xA;DyDt(1) = -ui * 1/(2*h)*(-y(3)+4*y(2)-3*y(1)) - F * K * (y(1) - ilangmuir(y(N+1)));&#xA;DyDt(N+1) = K * (y(1) - ilangmuir(y(N+1)));&#xA;&#xA;% Central finite differences&#xA;for i=2:N-1&#xA;    DyDt(i) = -ui * 1/(2*h)*(y(i+1)-y(i-1)) - F * K * (y(i) - ilangmuir(y(N+i)));&#xA;    DyDt(N+i) = K * (y(i) - ilangmuir(y(N+i)));&#xA;end&#xA;&#xA;% Backward finite differences&#xA;DyDt(N) = -ui * 1/(2*h)*(3*y(N)-4*y(N-1)+y(N-2)) - F * K * (y(N) - ilangmuir(y(N+N)));&#xA;DyDt(2*N) = K * (y(N) - ilangmuir(y(N+N)));&#xA;&#xA;&#xA;&#xA;function c = ilangmuir(q)&#xA;% langmuir solved for c&#xA;global parameter&#xA;&#xA;a = parameter(1);&#xA;b = parameter(2);&#xA;&#xA;c = q /(a - b * q);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;However, if I try to model a different case (different conditions) I get the correct response. So I am assuming this is a numerical problem caused by the conditions I am trying to model and not an error in the model. For instance, if I use the conditions below I get the correct response. &lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;parameter = [0.044511*432.75 432.75];    % isotherm parameters&#xA;L = 0.4;            % m, column length&#xA;epsb = 0.367;       % column bulk porosity&#xA;ui = 0.0056;        % m/s, velocity&#xA;Cfeed = 0.0025;     % feed concentration&#xA;K = 0.0782          % s-1, mass transfer coefficient&#xA;tfinal = 4000;      % min, final time for calculation&#xA;npz = 50;           % number of discretization points in z&#xA;npt = 20;&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/47AML.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/47AML.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24891" LastActivityDate="2020-03-06T14:41:19.413" Title="Slow integration of system ODE under specific conditions in adsorption problem" Tags="|chemical-engineering|modeling|matlab|mass-transfer|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33527" PostTypeId="2" ParentId="31496" CreationDate="2020-03-06T14:52:07.470" Score="1" Body="&lt;p&gt;Not sure this is universal but in my field of interest (steam turbines), we are interested in saturated steam because that is what we get in the last stages of the low pressure turbines. In our case, moisture is detrimental to the performance of the turbine, so it is important to be able to predict the moisture content aka liquid fraction aka (1-steam quality).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturated steam tables &lt;em&gt;at thermodynamic equilibrium&lt;/em&gt; are one way to get this information.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, we could be (and are) interested in steam in a &lt;em&gt;subcooled&lt;/em&gt; state, i.e. steam that remains dry at temperatures lower than the saturation temperature. This state is in fact often encountered in steam turbines, because the flow is expanded so fast that steam &quot;doesn't have time&quot; to condense at equilibrium. It remains dry, until the temperature/pressure conditions reach a threshold called the Wilson line. Then, condensation happens and thermodynamic equilibrium is restored.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8PCmg.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8PCmg.png&quot; alt=&quot;wilson line in enthalpy/entropy diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The properties of subcooled steam are usually extrapolated from the properties of superheated steam.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Whether you are interested in saturated steam at equilibrium, or subcooled steam, the reference data for the industry is the one from the International Association for the Properties of Water and Steam (&lt;a href=&quot;http://www.iapws.org/relguide/IF97-Rev.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.iapws.org/relguide/IF97-Rev.html&lt;/a&gt;)&lt;/p&gt;&#xA;" OwnerUserId="10381" LastActivityDate="2020-03-06T14:52:07.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33528" PostTypeId="1" CreationDate="2020-03-06T15:08:20.993" Score="-2" ViewCount="37" Body="&lt;p&gt;So I'm the president of my unis rocket club. The thing is, petroleum jobs got money and although I would apply for aeronautical companies first, just as a backup. I wanna apply for the petroleum companies. Do you guys think that this can work out? If it can't work out, may I know why? And if maybe, what I should do then?&lt;/p&gt;&#xA;" OwnerUserId="24874" LastActivityDate="2020-03-06T19:45:53.393" Title="As a president of a rocket club for my university. If I apply for internship/job for a petroleum company. Would this work out?" Tags="|aerospace-engineering|" AnswerCount="1" CommentCount="1" ClosedDate="2020-03-07T04:34:04.037" ContentLicense="CC BY-SA 4.0" />
  <row Id="33529" PostTypeId="2" ParentId="10767" CreationDate="2020-03-06T15:41:22.810" Score="1" Body="&lt;p&gt;Among mechanical engineers, such as me, a very general rule is about 10 feet per second.  &lt;a href=&quot;https://www.engineeringtoolbox.com/flow-velocity-water-pipes-d_385.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineeringtoolbox.com/flow-velocity-water-pipes-d_385.html&lt;/a&gt; That's when water erosion of metal surfaces starts being significant, particularly when the flow turns, such as in elbows, Tees, and curves in general.  It's also when friction loss becomes noticeable and for some elaborate systems of large extent, pumping costs go up.  Whether the flow is turbulent isn't of consequence in itself, and it would be nearly impossible to avoid it in most practical systems.  Yes, it may be surprising, but water erosion can be a problem, and in general that's different from cavitation, which occurs when the pressure in a liquid becomes lower than the liquid's vapor pressure.  Most of that is avoided by keeping the inlet of a pump, for instance, above the Net Positive Suction Pressure (NPSH) specification for that pump. Cavitation can also occur in other applications, such as propellers with liquids. And again, the design parameter there is fluid pressure, not velocity. &lt;/p&gt;&#xA;" OwnerUserId="14623" LastEditorUserId="14623" LastEditDate="2020-03-08T15:40:16.650" LastActivityDate="2020-03-08T15:40:16.650" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33530" PostTypeId="1" AcceptedAnswerId="33532" CreationDate="2020-03-05T17:31:26.770" Score="15" ViewCount="509" Body="&lt;p&gt;I keep seeing these large rectangular structures on job sites they put in the ground. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/uJSXm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uJSXm.jpg&quot; alt=&quot;image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm guessing something for strengthening the surface structures?&lt;/p&gt;&#xA;" OwnerUserId="24895" OwnerDisplayName="v15" LastEditorUserId="1832" LastEditDate="2020-03-07T13:42:33.313" LastActivityDate="2020-03-07T13:42:33.313" Title="What is this large steel structure used in excavation?" Tags="|civil-engineering|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33531" PostTypeId="2" ParentId="33530" CreationDate="2020-03-05T17:44:03.427" Score="13" Body="&lt;p&gt;It is shoring for trenching work.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If a trench collapses on a worker the weight and force of the material will lead to crush injuries.  Even if you are able to dig someone out from a collapse, crush injuries can lead to crush syndrome ending with kidney failure and death.  &lt;/p&gt;&#xA;" OwnerUserId="7755" OwnerDisplayName="Fresh Codemonger" LastActivityDate="2020-03-05T17:44:03.427" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33532" PostTypeId="2" ParentId="33530" CreationDate="2020-03-05T17:47:12.817" Score="26" Body="&lt;p&gt;It is a trench shield. It gets placed in a trench after the trench is dug to prevent workers from being hurt or killed in the event of a trench collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/rL7Mc.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/rL7Mc.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This picture from GMC trench shield shows a partially collapsed trench with a shield installed that would protect the workers installing the blue brute pipe.&lt;/p&gt;&#xA;" OwnerUserId="21087" OwnerDisplayName="JACK" LastActivityDate="2020-03-05T17:47:12.817" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33533" PostTypeId="1" CreationDate="2020-03-06T15:42:30.783" Score="2" ViewCount="31" Body="&lt;p&gt;In an ABS system there are sensors on each of the wheels which determine the angular velocity in order to pulse the brake of an individual wheel which registers a diverging reading indicating brake lock-up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My understanding is that when one of the angular velocities is significantly lower than the others, indicating that the wheel is sliding or about to slide, the system kicks in on that wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is, what would the best way of determining quantitatively:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) which of the wheels is in danger (some multiple of std dev from the mean of all wheels?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) how much deviation would be permissible, because there is always going to be a variation when the car is turning and we would not want to activate the ABS for such a situation.&lt;/p&gt;&#xA;" OwnerUserId="24892" LastActivityDate="2020-03-06T15:42:30.783" Title="Determining the threshold condition of an ABS brake system" Tags="|automotive-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33534" PostTypeId="2" ParentId="33528" CreationDate="2020-03-06T19:45:53.393" Score="2" Body="&lt;p&gt;To repeat my previous answer, but in blunter terms since you ask the same question again:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Interviewers don't have any interest in what fancy titles you give yourself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The interesting thing is what you, and the club you claim to be president of, &lt;strong&gt;&lt;em&gt;actually did.&lt;/em&gt;&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Nobody expects a would-be petroleum engineer to have spent his/her spare time digging for oil. What counts is doing &lt;em&gt;something&lt;/em&gt; that is practical engineering, or managing a practical engineering project. &lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-03-06T19:45:53.393" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33535" PostTypeId="2" ParentId="33516" CreationDate="2020-03-06T20:11:23.590" Score="0" Body="&lt;p&gt;In case, the current requirement is low you can tap the zero volt from the mid point of a uniform restrictive potentiometer &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QWwRI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QWwRI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: energy loss will be significant &lt;/p&gt;&#xA;" OwnerUserId="24878" LastActivityDate="2020-03-06T20:11:23.590" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33537" PostTypeId="1" CreationDate="2020-03-06T21:40:10.377" Score="1" ViewCount="25" Body="&lt;p&gt;This is a general question, so sorry if it breaks the rules. However, I think it is very important as more often than not someone else has already solved a similar problem that can be applied to solve the problem at hand or at least give clues on how to proceed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, I was trying to figure out a way to distribute fluid flow evenly across several openings in a channel. Luckily a short google search revealed an article ”Flow distribution in manifolds” which was precisely what I was looking for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What do you think is the most efficient method of finding similar problems that have already been solved? &lt;/p&gt;&#xA;" OwnerUserId="20027" LastActivityDate="2020-03-06T21:40:10.377" Title="What is the best way to find solved problems similar to the one at hand?" Tags="|dynamics|fluid|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33538" PostTypeId="1" CreationDate="2020-03-06T21:55:31.140" Score="0" ViewCount="33" Body="&lt;p&gt;What is the best way to set up a dielectric strength test?  I am planning on using two threaded copper rods for adjustable movement that will receive the voltage.  In between the two rods will be the dielectric (wafer/thin film).  Any insight on how to create a platform for testing consisting of two copper rods and a place to put the dielectric would be appreciated.  &lt;/p&gt;&#xA;" OwnerUserId="24706" LastEditorUserId="24706" LastEditDate="2020-03-09T17:21:16.050" LastActivityDate="2020-03-09T17:21:16.050" Title="Best way to go about &quot;HiPot&quot; testing setup" Tags="|mechanical-engineering|electrical-engineering|systems-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33539" PostTypeId="2" ParentId="33511" CreationDate="2020-03-07T02:49:00.647" Score="0" Body="&lt;p&gt;Roughly tell between full and empty by measuring inertia. &#xA;If the road surface is slightly twisted, then the truck will sway left and right a bit. Even with different suspension systems I can tell the difference when a truck drives over a particular stretch of road. Not sure how to automate detection.&lt;/p&gt;&#xA;" OwnerUserId="24898" LastActivityDate="2020-03-07T02:49:00.647" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33540" PostTypeId="2" ParentId="33516" CreationDate="2020-03-07T03:35:48.660" Score="1" Body="&lt;p&gt;Below is a simple circuit that I have used to create dual power supplies with a single power source for prototyping purposes. Depending on the current draw size resistors and capacitors. There are elegant solutions widely available. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/0ukJX.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/0ukJX.png&quot; alt=&quot;Simple dual supply from a single source&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An elegant solution would be to use a MAX1044 / ICL7660 Switched-Capacitor Voltage Converter. Below is a high level diagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PsgGK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PsgGK.png&quot; alt=&quot;Max1044 / ICL7660 high level diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Alternate solution would be MAX660. Below is a high level application diagram. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8dHzz.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8dHzz.png&quot; alt=&quot;MAX660 application diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additional solution would be to use LTC1044/7660. Below is a high level circuit diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5wxmh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5wxmh.png&quot; alt=&quot;LTC1044 / 7660 Switch capacitor voltage Converter circuit diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;References:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://datasheets.maximintegrated.com/en/ds/ICL7660-MAX1044.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;MAX1044/ICL7660 Switched-Capacitor Voltage Converter&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;http://www.ti.com/lit/ds/symlink/max660.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;MAX660 Switched Capacitor Voltage Converter&lt;/a&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.analog.com/media/en/technical-documentation/data-sheets/LTC1044_7660.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;LTC1044/7660 Switched Capacitor Voltage Converter&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastEditorUserId="110" LastEditDate="2020-03-07T12:05:57.257" LastActivityDate="2020-03-07T12:05:57.257" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33541" PostTypeId="2" ParentId="33522" CreationDate="2020-03-07T04:38:06.387" Score="4" Body="&lt;p&gt;The English term for this is definitely &lt;strong&gt;&lt;em&gt;crate&lt;/em&gt;&lt;/strong&gt;.&lt;/p&gt;&#xA;" OwnerUserId="20985" LastActivityDate="2020-03-07T04:38:06.387" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33542" PostTypeId="2" ParentId="33223" CreationDate="2020-03-07T04:56:19.757" Score="0" Body="&lt;p&gt;This was the case in Australia in the late 70's at least.&#xA;The phones were pulse dialling then, in my case with an automatic exchange.&#xA;You could call someone, converse, and the person you called would hang up. &#xA;If you kept holding the handset (because you were concentrating hard on writing notes) you could talk to them again if they picked the phone up.&#xA;Of course the calls were all charged by time, with a tiny noise every 3 minutes when you would be charged.&lt;/p&gt;&#xA;" OwnerUserId="24898" LastActivityDate="2020-03-07T04:56:19.757" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33543" PostTypeId="1" AcceptedAnswerId="33544" CreationDate="2020-03-07T06:59:59.267" Score="-1" ViewCount="25" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1BLFk.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1BLFk.jpg&quot; alt=&quot;&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would i resolve the forces so they are acting on the bottom joints.&lt;/p&gt;&#xA;" OwnerUserId="24901" LastActivityDate="2020-03-07T08:05:47.717" Title="What to do when the forces on a truss is not on the joints" Tags="|mechanical-engineering|statics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33544" PostTypeId="2" ParentId="33543" CreationDate="2020-03-07T08:05:47.717" Score="2" Body="&lt;p&gt;We consider each member that has mid-span loads as a simply supported beam and resolve the loads on the beam to vertical reactions and those reactions will be loads at those joints. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then solve the truss with the midspan loads removed and reactions added as joint loads.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After we find all the member axial forces and supports' reaction we Solve that member as a simply supported beam with appropriate axial compression or tension and find the moments and shears on that member.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, the first bottom member on the right, say we call the support B and the joint to its right A:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \Sigma M_B=0 \ 312*5.05/6.7= 235kN =P_A \rightarrow \ and \ 312-235= 77=P _B  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we add these two loads to joints A and B and remove them from mid-span of this member. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we solve the truss and find the axial load for AB which should be tension and solve it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to pay extra attention to those members with compression, checking for buckling with both the secant method and Euler's critical buckling load.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-07T08:05:47.717" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33545" PostTypeId="2" ParentId="33059" CreationDate="2020-03-07T11:39:33.777" Score="2" Body="&lt;p&gt;Fundamentally the design principles are the same - the concrete doesn't much 'care' if it is cast in its final location or somewhere else then moved when it has cured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In Eurocode (specifically EN1992-1-1, ‘Eurocode 2: Design of concrete structures- Part 1-1: General rules and rules for buildings’) there is a section ‘Additional rules for precast concrete elements and structures’ which has supplementary rules. That is, the majority of the design is the same, but you get some additional rules to follow.&#xA;The additional rules cover transient situations, temporary and permanent bearings and the detailing of connections and joints between members.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The transient rules are mainly because in a precast yard the manufacturer wants to strike elements quickly – well before they reach their final strength, so you’ll be lifting and shifting elements when the concrete is only partly cured. The code doesn’t actually have specific rules to follow, it just says that transient conditions should be “considered specifically” (clause 10.2, referring to 10.1.1)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The connections and joints parts do have some specific rules to tell you how to handles such matters as making a slab by joining individual planks, i.e. how you quantify the behaviour of typical connections between the planks.  The codes has rules about treating half-joints, types of bearings, anchorage of reinforcement at support and so on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Overall, section 10 is nine pages out of 230 in the code, so relatively little and minor additional material.&lt;/p&gt;&#xA;" OwnerUserId="902" LastActivityDate="2020-03-07T11:39:33.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33546" PostTypeId="2" ParentId="33428" CreationDate="2020-03-07T12:27:12.977" Score="0" Body="&lt;p&gt;As stated by @StainlessSteelRat the transistor is operating in the cut off region. Below is graph from the datasheet showing the relationship between &lt;span class=&quot;math-container&quot;&gt;$V_{CE}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$I_B$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I_C$&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g7PBe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g7PBe.png&quot; alt=&quot;Vce Ib and Ic relationship for 2SC4617&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;References: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.onsemi.com/pub/Collateral/2SC4617-D.PDF&quot; rel=&quot;nofollow noreferrer&quot;&gt;2SC4617G NPN Silicon General Purpose Amplifier Transistor&lt;/a&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="110" LastActivityDate="2020-03-07T12:27:12.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33547" PostTypeId="2" ParentId="33516" CreationDate="2020-03-07T13:00:24.157" Score="2" Body="&lt;p&gt;Partial answer. More later - maybe!&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/YfWRC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/YfWRC.png&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. OP's circuit.&lt;/em&gt;&lt;/p&gt;&#xA;&lt;p&gt;There are a number of problems with your circuit.&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Your op-amp VCC+ is fed through a 5.6 MΩ resistor. The &lt;a href=&quot;https://www.diodes.com/assets/Datasheets/AS358-358A-358B.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;datasheet&lt;/a&gt; says that the quiescent current, I&lt;sub&gt;CC&lt;/sub&gt;, is 0.7 mA. From Ohm's Law we can see that you would need a supply of V = IR = 0.7m × 5.6M = 3920 V just to power up the op-amp. With a 12 V battery you're a tad short of adequate voltage.&lt;/li&gt;&#xA;&lt;li&gt;Zener diodes will drop voltage when reverse biased &lt;em&gt;and&lt;/em&gt; current is flowing through them. You have them connected directly to the inverting input of your op-amp which has a very high input impedance and has an input bias current of 20 nA and the datasheet doesn't specify whether this is in or out. This won't be enough to to cause diode breakdown.&lt;/li&gt;&#xA;&lt;li&gt;You seem to be trying to decrease the inverting input by 3.6 + 4.7 × 2 = 13 V. Presumably this is to compare with the voltage at the non-inverting input but the voltage at this input will switch between 4 V and 8 V, depending on the state of the op-amp output.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;You really can't just cobble together circuits like this without better understanding.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I'm trying to use a battery to activate an op-amp Schmitt trigger but learned that there needs to be both a positive and negative voltage supplied to Vcc+ and Vcc-.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;No. Single-ended supply, V+ and GND, will be fine for your application.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;I've seen a lot of schematics online that show using a secondary voltage source for this purpose but I can't do that due restrictions of my circuit.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;You can power the circuit from the battery being monitored.&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Here are extra details to clarify the circuit: input battery voltage: 14.8V-12.3V(14.8V at max charge, 12.2V max depleted voltage before the battery starts taking damage)&lt;/p&gt;&#xA;&lt;p&gt;Trigger voltages: I want the op amp to send a signal while 12.2V &amp;lt; Vss and stop at 12.2 &amp;gt;= Vss.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;We can write this more clearly as:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Output enabled when V&lt;sub&gt;SS&lt;/sub&gt; ≥ 12.6 V.&lt;/li&gt;&#xA;&lt;li&gt;Output disabled when V&lt;sub&gt;SS&lt;/sub&gt; ≤ 12.2 V.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;The difference between the two is the &lt;em&gt;hysteresis&lt;/em&gt; and you need to check that the values you choose are correct. If you set the hysteresis to finely then the act of switching on the load will cause the battery voltage to sag which will switch the load off which will allow the battery voltage to rise which will cause the load to switch on again which ...&lt;/p&gt;&#xA;&lt;p&gt;When you get a proper comparator circuit and decide on the level of hysteresis you want you can do the calculations. Texas Instruments have a &lt;a href=&quot;http://www.ti.com/lit/ug/tidu020a/tidu020a.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comparator with Hysteresis Reference Design&lt;/a&gt; application note but it may be a bit advanced for you. DayCounter's &lt;a href=&quot;https://www.daycounter.com/Calculators/Comparator-Hysteresis-Calculator.phtml&quot; rel=&quot;nofollow noreferrer&quot;&gt;Comparator Hysteresis Calculator&lt;/a&gt; may be more suitable.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2020-03-07T13:00:24.157" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33548" PostTypeId="1" CreationDate="2020-03-07T15:06:54.497" Score="3" ViewCount="146" Body="&lt;p&gt;I have this simple design to lift an access door, but I am looking for the motor specs to do this heavy duty job. This is a simple design, I have added some information of the time, length of the part and weight that I am looking to move.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;the time factor that I am looking for this is, 3sec to 6 sec speed for the lift.&lt;/li&gt;&#xA;&lt;li&gt;the door weight is 250lbs&lt;/li&gt;&#xA;&lt;li&gt;The door length is 7’6”&lt;/li&gt;&#xA;&lt;li&gt;Door access will move in 90 degree in a horizontal to vertical motion.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Everything is made of steel (frame and access door).&#xA;Looking for a motor and gear box that can handle that work in a daily basis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If any more info is needed to help me out to find a motor that handle this work, i am available to provide it.&lt;img src=&quot;https://i.stack.imgur.com/jshqE.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="24906" LastEditorUserId="24906" LastEditDate="2020-03-09T00:30:02.913" LastActivityDate="2020-03-18T21:03:15.430" Title="What motor torque specs that is suited for a horizontal to vertical 90 degree lift?" Tags="|mechanical-engineering|electrical-engineering|motors|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33549" PostTypeId="1" AcceptedAnswerId="33550" CreationDate="2020-03-07T17:11:21.620" Score="1" ViewCount="58" Body="&lt;p&gt;This kind of a sleeve from cardboard put upon a cardboard box is called &lt;strong&gt;&lt;a href=&quot;https://www.multitran.com/m.exe?l1=2&amp;amp;l2=1&amp;amp;s=%D0%BE%D0%B1%D0%B5%D1%87%D0%B0%D0%B9%D0%BA%D0%B0&quot; rel=&quot;nofollow noreferrer&quot;&gt;обечайка&lt;/a&gt;&lt;/strong&gt; (obechaika) in Russian. What do we call them in English? I was told on Proz.com that we call them &amp;quot;cardboard shells&amp;quot;, is that so?&lt;/p&gt;&#xA;&lt;p&gt;I was translating a text in which medical products were packaged into cardboard boxes with such additional shells to provide additional protection.&lt;/p&gt;&#xA;&lt;p&gt;Such shells provide additional strength but sometimes such shells are made from decorative paper with fancy design and are put around, for instance, a box of chocolates to make it look more refined. These are also &lt;strong&gt;obechaikas&lt;/strong&gt; in Russian.&lt;/p&gt;&#xA;&lt;p&gt;Here, a cardoard box is slid inside an &lt;strong&gt;obechaika&lt;/strong&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/tEOHf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tEOHf.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/m6Ak3.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/m6Ak3.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;Here's another use of the term &lt;strong&gt;obechaika&lt;/strong&gt;, a decoratice paper shell around a food product in a plastic container:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z9RSc.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z9RSc.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;In this usage &lt;strong&gt;obechaika&lt;/strong&gt; is more frequently called &lt;strong&gt;shuber&lt;/strong&gt;, for some reason. The word is yet absent from Russian dictionaries, might be a new coinage.&lt;/p&gt;&#xA;" OwnerUserId="349" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2020-03-07T17:20:28.853" Title="What do we call these corrugated board sleeves put upon cardboard boxes to provide additional strength?" Tags="|terminology|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33550" PostTypeId="2" ParentId="33549" CreationDate="2020-03-07T17:20:28.853" Score="4" Body="&lt;p&gt;I would use “cardboard sleeve”, and it would apply to both your examples.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-03-07T17:20:28.853" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33551" PostTypeId="1" CreationDate="2020-03-07T17:21:11.420" Score="0" ViewCount="745" Body="&lt;p&gt;What is the importance and relevance of equilibrium and compatibility equations in elasticity.  What problem do they solve?&lt;/p&gt;&#xA;" OwnerUserId="24749" LastActivityDate="2020-03-07T19:42:38.590" Title="Importance of equilibrium and compatibility equations in elasticity" Tags="|solid-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33552" PostTypeId="2" ParentId="33551" CreationDate="2020-03-07T18:23:51.153" Score="1" Body="&lt;p&gt;The equilibrium equations mean every small element of the material is in equilibrium , i.e. the forces balance at every point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The compatibility equations mean the deformed material is &quot;continuous&quot; everywhere, i.e. it doesn't have any internal holes, cracks, or overlapping regions.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-03-07T18:23:51.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33553" PostTypeId="2" ParentId="33442" CreationDate="2020-03-07T19:03:47.703" Score="1" Body="&lt;p&gt;It never had liquid in it. It was either a display can or a collectors can.&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-07T19:03:47.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33554" PostTypeId="2" ParentId="33551" CreationDate="2020-03-07T19:36:53.753" Score="0" Body="&lt;p&gt;Equilibrium is statics means the member under consideration is not accelerating, moving or turning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;e.g., On a simply supported beam with a length L with a load P at L/2 if the right-hand support provides any value less than P/2 resistance the beam will rotate down.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Compatibility means there is a unique geometrical displacement, rotation, deflection at any point in a structure. e.g., if we cut a beam at a point to study its free body diagram and this section under stresses has rotated an angle A, and a deflection delta, the other part of the beam adjacent to it has an identical angle of rotation and deflection. This property is used to drive additional equations to help solve the stresses and forces on a member.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-03-07T19:42:38.590" LastActivityDate="2020-03-07T19:42:38.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33556" PostTypeId="1" CreationDate="2020-03-07T23:51:35.070" Score="1" ViewCount="86" Body="&lt;p&gt;I want to calculate the friction and inertia between 2 bevel gears with straight teeth.&#xA;I know how to calculate those parameters between regular gears with straight teeth.&#xA;Is there any difference for bevel gears?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks,&lt;/p&gt;&#xA;" OwnerUserId="24318" LastActivityDate="2020-03-07T23:51:35.070" Title="Bevel gear friction and inertia calculation" Tags="|mechanical-engineering|gears|friction|power-transmission|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33557" PostTypeId="2" ParentId="33486" CreationDate="2020-03-08T02:46:57.733" Score="1" Body="&lt;p&gt;Would you believe if someone told you it's ok? I think not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Because there is not enough information. One needs to know the loads on this column, it's function, what rebars and concrete mix are used in it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Generally if the rebars are not damaged you can repair the cut with appropriate epoxy mix applied after properly cleaning and preparing the notch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many of the building supply places have patching tool and epoxy by reputable manufacturers. One needs to follow their instructions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would seek advice from a professional regardless. The entire workmanship and the floor holes look like a careless job.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-08T02:46:57.733" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33559" PostTypeId="1" CreationDate="2020-03-08T14:21:59.920" Score="1" ViewCount="196" Body="&lt;p&gt;I have used 250 watt DC motor, for DIY electric potters wheel. Motor rated with RPM 3300 and geared reduction to 330 RPM. The issue I am facing is wobbling of the motor head. This issue is with the motor head/shaft and not the disc. How to fix it?&lt;/p&gt;&#xA;" OwnerUserId="24909" LastActivityDate="2020-03-08T14:21:59.920" Title="Wobbling DC motor head" Tags="|mechanical-engineering|electrical-engineering|motors|gears|bearings|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33560" PostTypeId="1" AcceptedAnswerId="33564" CreationDate="2020-03-08T16:57:03.927" Score="-1" ViewCount="30" Body="&lt;p&gt;I am designing phone cases for 3D Printing. I am fairly new to the field, however I cannot find design specifications for any phone at all, i.e. exact measurements for the positioning of cameras, buttons, ports, etc. Can anybody recommend somewhere where I can find these documents? Also, if this question is in the wrong forum, can a link to the correct forum be linked please.&lt;/p&gt;&#xA;" OwnerUserId="24910" LastActivityDate="2020-03-08T22:56:43.520" Title="Where can I get design specifications for phones?" Tags="|design|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33561" PostTypeId="1" CreationDate="2020-03-08T18:31:25.707" Score="0" ViewCount="44" Body="&lt;p&gt;I am attempting to drive a device with a motor, but also require water input.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there such a device that serves as both a drive motor with a water pump through the motor's shaft?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In essence, it would have the functions of both a motor and a water pump in one.&lt;/p&gt;&#xA;" OwnerUserId="24912" LastActivityDate="2020-03-08T18:31:25.707" Title="Motor with Water Pump Through Shaft?" Tags="|motors|pumps|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33563" PostTypeId="1" AcceptedAnswerId="33567" CreationDate="2020-03-08T21:18:07.367" Score="2" ViewCount="46" Body="&lt;p&gt;I've found this video on YouTube:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/hWK8Gv0JHqk?t=220&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/hWK8Gv0JHqk?t=220&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Watch from 3:40 to 4:20.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like to learn more about this linkage that makes that link alternate between hitting and missing the sliding link.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've searched for the animated movements from &lt;a href=&quot;http://507movements.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;507 Mechanical Movements&lt;/a&gt; superficially and found nothing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="24914" LastActivityDate="2020-03-09T01:00:07.630" Title="What is this bistate linkage?" Tags="|linkage|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33564" PostTypeId="2" ParentId="33560" CreationDate="2020-03-08T22:56:43.520" Score="1" Body="&lt;p&gt;You'll highly unlikely find any exactly drawings, as mentioned in the comments. These are usually trade secrets. However, you can pull out a caliber and measure the phone you have in mind yourself and make a few prototypes, continually reiterating until you get the results you want.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classic &lt;a href=&quot;https://en.wikipedia.org/wiki/Reverse_engineering&quot; rel=&quot;nofollow noreferrer&quot;&gt;Reverse Engineering&lt;/a&gt; :)&lt;/p&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2020-03-08T22:56:43.520" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33565" PostTypeId="1" CreationDate="2020-03-09T00:28:44.890" Score="1" ViewCount="32" Body="&lt;p&gt;I have pulled some CAT6 through the ceiling from one room to a nearby cupboard, for line level audio signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See wiring: &lt;a href=&quot;https://i.stack.imgur.com/JLzaB.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JLzaB.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't used a balun or DI box, and there isn't TOO much hum and hiss, it's manageable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But something strange is happening with the audio from Alexa, it sounds really bad: a lot of hiss and distortion. When you turn the volume down, the signal to noise is so bad that the hiss is really high, also the distortion doesn't really disappear it just gets quieter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I unplug the Alexa and swap it out for audio from my phone, it sounds really clean. So you'd think it's a problem with the Alexa. But...&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I tried another Alexa and the same problem happens.&lt;/li&gt;&#xA;&lt;li&gt;I turned the Alexa up (and mixer input down) and vice versa, this didn't help&lt;/li&gt;&#xA;&lt;li&gt;So then I plugged the Alexa directly into the amp and it sounds really good&lt;/li&gt;&#xA;&lt;li&gt;I plugged the Alexa back and tried it in a battery pack, rather than into the USB power supply, and the problem is still there.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;What is the reason for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Should I be using baluns, or a DI box or something like that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've heard it's fine sending line level signals over CAT6 - indeed with the phone plugged in, it's all good.&lt;/p&gt;&#xA;" OwnerUserId="21197" LastActivityDate="2020-03-09T05:46:38.623" Title="What causes the hiss from one audio source but not another over long cable?" Tags="|audio-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33566" PostTypeId="2" ParentId="33565" CreationDate="2020-03-09T00:49:46.150" Score="0" Body="&lt;p&gt;verify if the cable is making some contact or is open, verify if the plug on both end has a little chip or debris that is distorting the signal.&#xA;The Alexa device has no output signal filter&#xA;Here is a link for a device that could help:&#xA;&lt;a href=&quot;https://images.app.goo.gl/wy59kAzKFgh9gUZg6&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://images.app.goo.gl/wy59kAzKFgh9gUZg6&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24906" LastActivityDate="2020-03-09T00:49:46.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33567" PostTypeId="2" ParentId="33563" CreationDate="2020-03-09T01:00:07.630" Score="1" Body="&lt;p&gt;I believe what you are seeing is a pendulum mounted on a sliding bracket. The weight with the thumbscrew indicates that it can be tuned to match the required movement to engage the V-shaped &quot;socket&quot; into which it engages on alternating strokes. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Slowed down to one-quarter speed, the video shows that on one stroke, the pendulum is swinging left, while the sliding mechanism is pushing right, with inertia causing the engaging lever to rise. In the next half-stroke, the sliding mechanism pushes left  with the lever in the socket, preventing inertial movement of the lever.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next half-stroke to the right disengages the lever from the socket, gravity causing the lever to fall simultaneously with the pendulum causing the lever to fall with sufficient momentum to prevent the next left sliding stroke from engaging the socket.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I searched for various terms attempting to locate a specific term with no success.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe one can reproduce this mechanism by experimenting with varying angles between the lever and the pendulum arm and by varying distance on the pendulum weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd love to know what purpose the alternating actuation serves. One thinks it could be a two stroke engine system of sorts.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-03-09T01:00:07.630" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33569" PostTypeId="2" ParentId="33565" CreationDate="2020-03-09T05:46:38.623" Score="2" Body="&lt;p&gt;I will first assume you do not have any of the wires crossed here. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is unlikely that the cause is in the CAT6 cable, but here's how to test this: send the alexa signal (red) thru the cable to the mixer. Then, instead of plugging them into the mixer, unplug the OUTPUT (blue) lines from the mixer and connect them to the alexa (red) wires. Then plug the blue cables into your amp. the signal path is alexa-&gt;red cable-&gt;blue cable-&gt;amp and play a file. I'll bet you get no distortion, no hiss, and a strong signal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Furthermore, I'll bet the line level input on your mixer is expecting a much stronger signal at its input terminals than alexa is set up to deliver. If possible, compare the signal amplitude from the roku unit with that from alexa on a 2-channel oscilloscope. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-03-09T05:46:38.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33570" PostTypeId="1" CreationDate="2020-03-09T06:13:10.230" Score="0" ViewCount="350" Body="&lt;p&gt;I would like to know if there is someway i can figure out if the car wheel is rotating in clockwise or anticlock wise direction.&#xA;Im working on a project which determines if car rolled back on a hill before the user accelerates forward.&#xA;I tried this using gps co=ordinates to determine if there is roll back. As this roll back could be in cms before user moves forward, im not getting reliable results. &#xA;The only way i can think of now is to depend on wheel rotation. if the wheel rotates backward on slope i would know that it did roll back a little. So is there a way for me to determine if wheel rotated in clock wise or anti-clock wise direction through CAN bus interface or something.&#xA;Any help here would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="24919" LastActivityDate="2020-03-11T01:14:41.810" Title="Determine if car is moving forward or backward" Tags="|mechanical-engineering|motors|gears|wheels|car|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33571" PostTypeId="1" CreationDate="2020-03-09T10:30:34.860" Score="1" ViewCount="98" Body="&lt;p&gt;In&#xA;Fema national US&amp;amp;R response system - Structrual collapse technicican training manual&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It says:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;&lt;em&gt;There are also vertical loads generated in a structure by earthquake shaking, but as mentioned previously, these forces rarely overload the vertical load resisting system. Earthquake induced vertical forces have caused damage to heavy concrete structures with high dead load compared to design live load. These vertical forces also increase the chance of collapse in concrete frame buildings due to either increased or decreased compression forces in the columns. (Increased compression that overloads columns or decreased compression that reduces column bending strength.)&lt;/em&gt;&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I didn't understand the second part of last sentence in paranthesis, where it says &quot;....or decreases compression that reduces column bending strength&quot;.... what does this mean and why?&lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2020-03-10T09:59:53.623" Title="column bending stress under vertical earthquake load" Tags="|civil-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33572" PostTypeId="2" ParentId="33571" CreationDate="2020-03-09T11:16:33.360" Score="2" Body="&lt;p&gt;For a symmetrical column with maximum bending stress &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt; and compressive stress &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt;, the combined stresses range from &lt;span class=&quot;math-container&quot;&gt;$C+B$&lt;/span&gt; on one side of the column to &lt;span class=&quot;math-container&quot;&gt;$C-B$&lt;/span&gt; on the other side.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If &lt;span class=&quot;math-container&quot;&gt;$C$&lt;/span&gt; is reduced, then &lt;span class=&quot;math-container&quot;&gt;$C-B$&lt;/span&gt; reaches the design limit (in tension) for a smaller value of &lt;span class=&quot;math-container&quot;&gt;$B$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-03-09T11:16:33.360" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33574" PostTypeId="1" CreationDate="2020-03-09T13:10:02.137" Score="1" ViewCount="167" Body="&lt;p&gt;I'm looking for a good reference book on the topic of fluid dynamics/hydraulics systems engineering.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'd like a reference for simlpe/estimate calculations on things like major losses (pipes), minor losses (bends, orifices, expansions, contractions, etc). The media that I most frequently work with are water, glycol, hydrogen and air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that CFD and exact value calculations can be quite cumbersome in fluid dynamics, and that is why I am looking for an engineering reference book for estimate calculations that can help simplify the calculations when I need to make system design estimates (&gt;100 different calculations).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please let me know if you have any suggestions!&lt;/p&gt;&#xA;" OwnerUserId="24922" LastActivityDate="2020-03-09T13:10:02.137" Title="Reference book for hydraulic systems design" Tags="|fluid-mechanics|thermodynamics|pressure|hydraulics|systems-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33575" PostTypeId="2" ParentId="33570" CreationDate="2020-03-09T13:45:54.443" Score="1" Body="&lt;p&gt;A quick search for various terms based on &quot;how to read wheel sensors for direction,&quot; and &quot;how to read wheel sensors&quot; and a few other variations resulted in a mish-mash of answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are &lt;a href=&quot;https://www.ecumaster.com/products/wheel-speed-to-can/&quot; rel=&quot;nofollow noreferrer&quot;&gt;devices available&lt;/a&gt; that connect to wheel sensors and output a canbus data stream. The manual for the linked device references that the device can be connected parallel with an existing ABS on the vehicle. The manual also contains a link to free download software to read the device on a computer. I was unable to determine if direction is part of that stream, but I suspect that it is, based on the results of &lt;a href=&quot;https://www.knowyourparts.com/technical-resources/hubs-bearings/wheel-speed-sensors/&quot; rel=&quot;nofollow noreferrer&quot;&gt;another search&lt;/a&gt;:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Magneto Resistive (Active Wheel Speed Sensors)&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;A magneto resistive sensor has a sensing element that contains an&#xA;  output module that is mounted on a substrate and magnetic material.&#xA;  These magnetic rings are often located on the seal of the bearing. The&#xA;  magneto resistive sensor’s main advantage is the ability to sense&#xA;  direction and operate with a wide air gap. The sensor is more stable&#xA;  in high-vibration conditions and operates more efficiently than the&#xA;  reluctance sensor.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;&lt;strong&gt;The wheel speed sensors can detect the direction of rotation&lt;/strong&gt; of the&#xA;  reluctor ring and zero speed. The sensor’s compact size allows it to&#xA;  be incorporated into a hub bearing utilizing an alternating magnetic&#xA;  pole reluctor ring that is built into the rubber seal. A magneto&#xA;  resistive wheel speed sensor power input can range from 12V-5V from&#xA;  the ABS controller.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;The sensor produces a square wave signal with a high signal of 1.93&#xA;  volts, and a low signal of 0.64 volts. The high signal must cross&#xA;  1.29V and the low signal must cross 0.97V to low. The main advantage is that the amplitude of the signal does not change.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Additional information suggests that one can use an oscilloscope to read the signal from the wheel sensor, the result of which would be a square wave. I'd expect that the spacing of the square wave may determine the direction of rotation but that's barely-founded conjecture. It's also possible that a pair of square waves are generated by the sensor and the difference between the two determine direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another &lt;a href=&quot;https://www.searchautoparts.com/motorage/training/active-wheel-speed-sensors&quot; rel=&quot;nofollow noreferrer&quot;&gt;link I pursued&lt;/a&gt; more or less confirms my conjecture, by stating that the newer devices contain two sensors offset from each other:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;The sensor assembly consists of two sensing elements mounted&#xA;  side-by-side with an amplifier chip built into the sensor assembly.&#xA;  The output from each sensing element rises and falls the same way as&#xA;  with a magnetic pickup, and the amplifier converts that to digital&#xA;  high-low voltage signals. Also, as before, the high-low switching&#xA;  frequency is proportional to wheel speed, and because the signal is&#xA;  always present, zero wheel speed can be detected. But because the&#xA;  sensing elements are next to each other, the two voltage signals are&#xA;  always slightly out of phase: one rises or falls just a few degrees&#xA;  after the other. That's how the sensor detects the direction of&#xA;  rotation. If the signal from element A lags behind the signal from&#xA;  element B, the wheel is turning clockwise. If B lags behind A, the&#xA;  wheel is turning counterclockwise. The exact same out-of-phase&#xA;  dual-sensor technique is used in radio knobs.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-03-09T13:45:54.443" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33576" PostTypeId="2" ParentId="31496" CreationDate="2020-03-09T15:28:26.713" Score="0" Body="&lt;p&gt;Water vapor below saturation is the vapor pressure of liquid water.  By definition, saturated steam is steam for which the vapor pressure equals boiling point pressure/temperature.  Vapor pressure below the boiling point is easily determined by the appropriate vapor pressure tables/graphs/calculators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saturated steam is important because it is the state before steam is heated above boiling.  So: boiler exit conditions for saturated steam boilers, superheating boilers before superheating, condenser conditions, and some turbine conditions.  Enthalpy of vaporization, h&lt;sub&gt;fg&lt;/sub&gt;, is the heat required to fully vaporize water at the boiling point.  It is what most heat is used for in making steam. &lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-09T15:28:26.713" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33577" PostTypeId="2" ParentId="33571" CreationDate="2020-03-09T16:05:47.273" Score="1" Body="&lt;p&gt;In a concrete column with bending moment and compression, the tensile stress is reduced and the compression stress is increased by the shift of the neutral axis location. However, this condition is OK because for ductility concrete members are designed to fail on the tensile stress because of the large stretch the steel re-bars take before the yield point. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But when during the earthquake the compression decreases the tensile stress goes back up and because of the lateral forces of the quake there is likely even more moment applied to the column and more tensile stress which could be in excess of design stress, refer to the diagram.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/RXb5O.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/RXb5O.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-09T16:05:47.273" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33580" PostTypeId="1" CreationDate="2020-03-10T07:21:17.550" Score="1" ViewCount="26" Body="&lt;p&gt;Is there an (international) standard which determines the condition when a car seat belt locks? Which standard is it? Does the standard determines a speed in m/s of the belt, such that it locks if this speed is exceeded or does it determine an acceleration in m/s^2? How are the values?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there also a standardized testing procedure to ensure that the values are satisfied?&lt;/p&gt;&#xA;" OwnerUserId="24929" LastEditorUserId="24929" LastEditDate="2020-03-10T07:44:42.090" LastActivityDate="2020-03-10T07:44:42.090" Title="Standards for safety car seatbelts locking condition" Tags="|safety|car|standards|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33582" PostTypeId="2" ParentId="33571" CreationDate="2020-03-10T09:59:53.623" Score="0" Body="&lt;p&gt;So the answer is for this instant: When there is already a lateral earthquake force which may crate tension on a column, MINUS gravity compression, PLUS the &lt;strong&gt;UPWARDS&lt;/strong&gt; acting earthquake vertical force. (Or sometimes there can be moments on top of columns even before earthquake due to load distributions on beams which would put one side of column in tension already and when an UPWARDS earthquake vertical force act on it, it will put further tension demand on column)&lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2020-03-10T09:59:53.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33583" PostTypeId="1" CreationDate="2020-03-10T11:54:41.203" Score="2" ViewCount="59" Body="&lt;p&gt;I am looking for a solution for mounting a shelf that holds a pressure sprayer to precast Concrete panel with a Foam core insulation sandwiched in the middle.   total weight to support is 300 pounds, The shelf has 6 holes to mount with located in 2 rows of 3. The panel is  8 inch total (3 inches then 2-inch foam panel then 3 inches). ( diameter of holes in Shelf is 1/2 inch so I think 3/8 inch anchors are appropriate.  If anyone could send a link to a product that would be great.   Engineering data / Submittal type info would be very helpful as well.   I need to validate that the ANchors are appropriate for precast concrete.  Thanks in advance! &lt;/p&gt;&#xA;" OwnerUserId="24932" LastActivityDate="2020-03-11T01:06:54.207" Title="Need Anchors for Pre-cast Concrete Panel with Insulation Inside" Tags="|structural-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33584" PostTypeId="1" AcceptedAnswerId="33585" CreationDate="2020-03-10T12:24:24.293" Score="1" ViewCount="91" Body="&lt;p&gt;Not sure if anyone would know this, but was discussing it with friends, and it seems like quite the puzzle. In short, how is nuclear fuel in reactors handled?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Logically, the nuclear fuel must be transported to the site, then placed into the reactor, then - once fully used - removed. But given the strength of the radioactivity of the elements, how on earth is this done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the Chernobyl disaster as an example, the guys who had to clear up the mess weren't able to be close to the source for more than ca. 60 seconds without it being deadly. Even with lead protection, electrical circuits broke, so robots seem not to be an option. Just seems like a really difficult task to be able to transport, install, remove and dispose of the uranium without there being major hurdles.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Anyone know who it is done?!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, on a related topic: if a nuclear power station is decommissioned, how is the core dismantled? How can you safely remove the irradiated graphite, for example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My friends and I spent ages debating how it could be done and we would love to know the answer!!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="24933" LastActivityDate="2020-06-16T02:36:10.847" Title="How is nuclear fuel dealt with?" Tags="|nuclear-engineering|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33585" PostTypeId="2" ParentId="33584" CreationDate="2020-03-10T13:05:28.917" Score="1" Body="&lt;p&gt;The spent fuel rods stay first for around five years under water, in a Spent fuel pool (SPF), until radiation and heat reduced enough that they can be transported. For transport, they get placed in a special capsule. In case of the CASTOR (&lt;strong&gt;CA&lt;/strong&gt;sk for &lt;strong&gt;S&lt;/strong&gt;torage and &lt;strong&gt;T&lt;/strong&gt;ransport &lt;strong&gt;O&lt;/strong&gt;f &lt;strong&gt;R&lt;/strong&gt;adioactive material), made by German GNS, these capsules consist of a 40 cm thick cast iron wall, which again is filled with neutron moderator rods. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trick to reduce the impact of the radiation during loading the rods into the capsules is pretty simple: do it all under water. Water happens to provide very good radiation shielding. The capsule gets moved with a crane into the water pool, the rods are placed in it, and then the closed capsule gets pulled out again. After drying, the capsules can be loaded on trains and get moved on.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2020-03-10T13:05:28.917" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33586" PostTypeId="1" AcceptedAnswerId="33608" CreationDate="2020-03-10T18:12:38.150" Score="-1" ViewCount="171" Body="&lt;p&gt;I'm trying to calculate the bending moment distribution on a fixed-fixed Euler-Bernoulli beam. It is known that interelement nodes show different values for the bending moment. What is customary to do with this discrepancy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Im doing the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;M_z =\begin{bmatrix}-6/L^2 + 12x/L^3 \\-4/L + 6x/L^2 \\ 6/L^2 - 12x/L^3 \\-2/L + 6x /L^2 \end{bmatrix}^T\begin{bmatrix} w_1 \\ \theta_1 \\w_2 \\ \theta_2\end{bmatrix}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This returns the moment distribution along the element, where the 1st vector is the second derivative of the shape functions and the 2nd one is the computed displacements.&lt;/p&gt;&#xA;" OwnerUserId="24936" LastActivityDate="2020-03-28T15:43:48.683" Title="Bending moment in Euler-Bernoulli finite Element" Tags="|mechanical-engineering|structural-analysis|stresses|beam|finite-element-method|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33587" PostTypeId="1" CreationDate="2020-03-10T20:18:03.300" Score="0" ViewCount="63" Body="&lt;p&gt;Do we consider possible effects of fire, when we do structural design for reinforced concrete buildings? &#xA;But for steel buildings we do?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also for reinforced concrete buildings, the thermal expansion difference will not be issue because thermal expansion coefficient of concrete and reinforced steel is very close to each other?&lt;/p&gt;&#xA;" OwnerUserId="24283" LastEditorUserId="24283" LastEditDate="2020-03-10T20:39:09.890" LastActivityDate="2020-03-11T19:15:04.410" Title="Any effect of fire in structural design of buildings?" Tags="|structural-engineering|civil-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33588" PostTypeId="2" ParentId="33587" CreationDate="2020-03-10T20:42:05.377" Score="1" Body="&lt;p&gt;Not directly. Fire protection is a separate part of the design process and code. Structural design assumes that the protection is provided as prescribed by the codes.&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-10T20:42:05.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33589" PostTypeId="1" CreationDate="2020-03-10T20:52:53.653" Score="1" ViewCount="82" Body="&lt;p&gt;I’m a student and have to create a machine that can launch a small bowling ball to knock over some pins.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking that I could use elastic potential energy to launch the ball but I’m not sure if I should use springs or elastic bands. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I don’t have a lot of space to work with because it can only be 0,5m x 0,5m x 0,5m. And for that reason I am leaning towards springs but I’m not entirely sure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance, if you need more information about the project just ask.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: For the project I have to launch a ball 12 meters away and knock over 5 bowling pins. I am not allowed to use pneumatics or explosives. The ball has to stay on the ground and the goal is to knock over the pins like in a bowling match so I need a way to aim my system to hit a different pin when I only hit a few of them, but I already have an idea for this. &#xA;I thought about only using a ramp but when I calculated the speed it would have it would not be enough to consistently knock over bowling pins. Then I thought about adding springs or elastic bands or something like that to give it some extra speed. &lt;/p&gt;&#xA;" OwnerUserId="24938" LastEditorUserId="24938" LastEditDate="2020-03-11T22:22:04.343" LastActivityDate="2020-03-11T22:22:04.343" Title="Launching a ball with springs or with elastic bands" Tags="|mechanical-engineering|springs|" AnswerCount="0" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="33590" PostTypeId="1" AcceptedAnswerId="33591" CreationDate="2020-03-10T22:21:57.303" Score="1" ViewCount="49" Body="&lt;p&gt;I’m building a pitching machine that has an 8” wheel (16” OD tire) that is directly mounted to a 1/2” motor shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most similar questions reference a gearbox or pulpy system, but my situation is directly mounted to a 12v motor shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Due to motor/fabrication design, the closest I can get my magnetic rpm sensor is a radius of 5” from the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So known variables for the calculation will be RPM at a diameter of 10” and the OD of the tire at 16” all on the same wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the speed (MPH) of the tire?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would imagine the OD of the tire runs slower than the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this makes sense.&lt;/p&gt;&#xA;" OwnerUserId="24939" LastActivityDate="2020-03-11T00:48:46.330" Title="Calculate MPH on the tire based on RPM of the hub?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33591" PostTypeId="2" ParentId="33590" CreationDate="2020-03-10T23:05:52.660" Score="1" Body="&lt;p&gt;Whatever RPM your sensor reads is the RPM of the 16&quot; tire as well. You say the tire is directly mounted on the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;E.g., if your sensor is reading 100RPM your tire is turning the same and the ball is ejected theoretically, depending on geometry of the shoot and ignoring spitting added acceleration,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ v= 16/12 *\pi*100= 133*\pi (ft/m) $&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-03-11T00:48:46.330" LastActivityDate="2020-03-11T00:48:46.330" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33592" PostTypeId="1" AcceptedAnswerId="33593" CreationDate="2020-03-11T00:47:32.020" Score="1" ViewCount="139" Body="&lt;p&gt;I have some issues regarding a lead screw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's used as the main driving system for a scissor lift which requires about 2Nm torque to operate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the exact part if it matters: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.amazon.co.uk/TOOGOO-Optical-Housings-Aluminum-Coupling/dp/B07PFQKQV6/ref=sr_1_2?keywords=200mm+opticalis&amp;amp;qid=1583887143&amp;amp;sr=8-2-spell&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.co.uk/TOOGOO-Optical-Housings-Aluminum-Coupling/dp/B07PFQKQV6/ref=sr_1_2?keywords=200mm+opticalis&amp;amp;qid=1583887143&amp;amp;sr=8-2-spell&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;due to space constraints in the design I have to use a universal joint and place the motor at a very slight angle to clear a few things:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.amazon.co.uk/sourcing-map-Rotatable-Universal-Steering/dp/B07H9ZKBSB/ref=sr_1_4?keywords=6mm%2Bto%2B8mm%2Buniversal&amp;amp;qid=1583887292&amp;amp;sr=8-4&amp;amp;th=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.amazon.co.uk/sourcing-map-Rotatable-Universal-Steering/dp/B07H9ZKBSB/ref=sr_1_4?keywords=6mm%2Bto%2B8mm%2Buniversal&amp;amp;qid=1583887292&amp;amp;sr=8-4&amp;amp;th=1&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of how much I tighten the grub screws, they will eventually come loose (only on the lead screw end, not the motor end) and slide over the the lead screw. I've tightened them so much that even one of them became stripped and I had to replace it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a bottle of medium strength loctite, does using that for the grub screws solve the issue or are there better ways of doing it? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another idea was using a file to make a flat spot on the lead screw but I'm not sure how well that'd work or if it makes things worse. Since I only have 1 lead screw, I thought it'd be best to ask before destroying it.&lt;/p&gt;&#xA;" OwnerUserId="24941" LastActivityDate="2020-03-11T14:47:08.607" Title="Driving a lead screw using a universal joint" Tags="|mechanical-engineering|materials|design|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33593" PostTypeId="2" ParentId="33592" CreationDate="2020-03-11T01:02:32.873" Score="1" Body="&lt;p&gt;Loctite works well and this application, while not the 'normal' application, I would expect to give good results. If it's necessary to be sure that your solution works over time, I would go with a physical connection. The trouble with the flat spot idea is to locate it properly and not mess up the treads such that the parts can not be screwed together. I would try assembling and then using a marker through the set hole mark the spot so that you can see it after dissembling. The threads can be repaired by careful filing between them with a small triangular shaped file. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or, perhaps put the Loctite into the bottom of the set screw and the set the screw. This would have the same affect as you proposed but instead of removing material to get the set screw to grab, you're adding material, so to speak&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-11T01:02:32.873" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33594" PostTypeId="2" ParentId="33583" CreationDate="2020-03-11T01:06:54.207" Score="1" Body="&lt;p&gt;This is a situation that should be reviewed by a structural engineer. More about preventing damage to the panel then about support of the item.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For anchor information, check out &lt;a href=&quot;https://www.strongtie.com/mechanicalanchoringproducts_anchoringsystems/category&quot; rel=&quot;nofollow noreferrer&quot;&gt;Simpson Mechanical Anchors&lt;/a&gt;. The Titen HD is a common choice&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-11T01:06:54.207" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33596" PostTypeId="1" CreationDate="2020-03-11T11:14:32.497" Score="0" ViewCount="1239" Body="&lt;p&gt;I have to answer a few questions on transfer functions using Matlab. The first question, which I solved without Matlab, gives a time response graph for an LR circuit, and asks me to find the first order transfer function. I ended up with:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ G(s) =\frac{2}{s+2} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The next question says &quot;determine the CLTF if the system has unity negative feedback and calculate the new values for &lt;span class=&quot;math-container&quot;&gt;$\  τ $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ k $&lt;/span&gt;. I'm stuck with this part - I know that the general CLTF for unity feedback is: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ G_c(s)= \frac{G}{1+GH} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and I know that &lt;span class=&quot;math-container&quot;&gt;$\ H = 1 $&lt;/span&gt; because of the unity feedback.&#xA;This is as far as I can get, so any help with this is appreciated! &#xA;Parts I'm struggling with are:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Is &lt;span class=&quot;math-container&quot;&gt;$\ G(s) $&lt;/span&gt; in the unity feedback system the same as the &lt;span class=&quot;math-container&quot;&gt;$\ G(s) $&lt;/span&gt; I worked out already? These two questions are part of the same question but I can't tell if they follow on from each other or if they're separate. This is all the info the questions give so I can't think what else &lt;span class=&quot;math-container&quot;&gt;$\ G(s) $&lt;/span&gt; should be in the feedback system.&lt;/li&gt;&#xA;&lt;li&gt;I obtained a CLTF for the system using &lt;span class=&quot;math-container&quot;&gt;$\ G(s) = \frac{2}{s+2} $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ H(s) = 1 $&lt;/span&gt;, and got &lt;span class=&quot;math-container&quot;&gt;$\ G_c(s) = \frac{2}{s+4} $&lt;/span&gt;, but as I said above I'm not sure I'm using the correct value for &lt;span class=&quot;math-container&quot;&gt;$\ G(s) $&lt;/span&gt;, because when I try to work backwards to find  &lt;span class=&quot;math-container&quot;&gt;$\ τ $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ k $&lt;/span&gt;, I get the same values as before.&lt;/li&gt;&#xA;&lt;li&gt;Most importantly - if these values are wrong, which I believe they are, how do I use Matlab or ServoCad to obtain new values for &lt;span class=&quot;math-container&quot;&gt;$\ τ $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ k $&lt;/span&gt;? Where τ is the time constant and k is the gain.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="24943" LastEditorUserId="24943" LastEditDate="2020-03-11T12:01:03.540" LastActivityDate="2020-03-12T22:45:20.057" Title="How to find closed loop transfer function and use it to identify τ and k?" Tags="|control-engineering|matlab|transfer-function|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33598" PostTypeId="2" ParentId="33596" CreationDate="2020-03-11T12:14:25.203" Score="3" Body="&lt;p&gt;The general form of a transfer function for a first order system is the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T(s) = \frac{K}{\tau s+1} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ K \rightarrow $&lt;/span&gt; DC Gain of the system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ \tau \rightarrow $&lt;/span&gt; Time constant of the system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The above form can also be written in another way as described below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T(s) = \frac{K}{\tau s+1} = \frac{b_0}{s+a_0} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By matching the parameters of these two forms we obtain the formulas for the system parameters &lt;span class=&quot;math-container&quot;&gt;$\ K $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ \tau $&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ b_0 = \frac{K}{\tau} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ a_0 = \frac{1}{\tau} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the case where the transfer function is: &lt;span class=&quot;math-container&quot;&gt;$\ G(s) = \frac{2}{s+2} $&lt;/span&gt;, it is obvious that the system parameters (by working out the math) are: &lt;span class=&quot;math-container&quot;&gt;$\ \tau = \frac{1}{2} $&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\ K = 1 $&lt;/span&gt;. Following, the same procedure for the closed loop transfer function: &lt;span class=&quot;math-container&quot;&gt;$\ G_c(s) = \frac{2}{s+4} $&lt;/span&gt;, we obtain the following values for the system parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \tau = \frac{1}{4} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ K = \frac{1}{2} $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is how someone can obtain the values for the system parameters by working out the math and using fundamental control theory. Below, I place some MATLAB code in order to do the same (Control System Toolbox is needed):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;s = tf('s');&#xA;open_loop = 2/(s+2);&#xA;open_loop_dc_gain = dcgain(open_loop);&#xA;closed_loop = feedback(open_loop,1);&#xA;closed_loop_dc_gain = dcgain(closed_loop);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In order to find out the values for the time constants, you can check out this question of the mathworks community:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mathworks.com/matlabcentral/answers/362850-finding-time-constant-and-steady-state-error-with-simulink&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mathworks.com/matlabcentral/answers/362850-finding-time-constant-and-steady-state-error-with-simulink&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-03-12T22:45:20.057" LastActivityDate="2020-03-12T22:45:20.057" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33599" PostTypeId="2" ParentId="33592" CreationDate="2020-03-11T12:56:08.363" Score="2" Body="&lt;p&gt;There are a range of options for such a connection.  I would put loctite for a plain grub screw at the bottom of the list.  Depending on the safety aspects of the joint breaking loose, it might be very important for it not to slip.  Your other options (in rough order of reliability):&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Grub screw to a flat-spotted shaft&lt;/li&gt;&#xA;&lt;li&gt;Grub screw into a drilled and tapped hole&lt;/li&gt;&#xA;&lt;li&gt;Through-hole with a tapered pin&lt;/li&gt;&#xA;&lt;li&gt;keyway and key&lt;/li&gt;&#xA;&lt;li&gt;Spline fit or other shape-matched fit&lt;/li&gt;&#xA;&lt;li&gt;welded-on connectors (like an automobile)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-11T12:56:08.363" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33601" PostTypeId="2" ParentId="33587" CreationDate="2020-03-11T15:09:17.293" Score="1" Body="&lt;p&gt;Fire protection is important in large building design. For steel I have seen hollow steel columns filled with antifreeze/water, also intumescent coatings are used . Also concrete covering steel ,this concrete may also provide strength. And the sprinkler systems which includes a gravity feed tank at high elevation. Of course local codes take priority but you can exceed them.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2020-03-11T19:15:04.410" LastActivityDate="2020-03-11T19:15:04.410" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33607" PostTypeId="1" CreationDate="2020-03-12T12:41:13.540" Score="1" ViewCount="606" Body="&lt;p&gt;I understand the equivalency between the MSD and RLC circuits.  Mass ~ resistance, spring ~ inductor, damper ~ capacitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the equations of the form&#xA;MSD:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ m\ddot{x} + c\dot{x}+ kx = F    $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;RLC:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ L\ddot{i}  + R\dot{i} + \frac{1}{C}i = V    $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was thinking a bit about what the chemical engineering equivalent would be for a &quot;plant circuit&quot;.  For a capacitor, I thought it could be equated to a tank with a inlet at the top and the outlet at the opposite end (also at the top).  It would fill without outputting anything until it is full.  However, if the inlet stops, the outflow stops and thus doesnt work as a capacitor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For resistance/mass, i thought the tank size might be the best representation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea for an inductor/damper.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After some more thinking, it became clear that a single tank cannot possible approximate such a system, and it has to be a dual tank system.  However, this complicates the ODE to such a point where a equivalency is not intuitive at all!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would be awesome if someone could shed some light on this for me.  I might be missing something somewhere.&lt;/p&gt;&#xA;" OwnerUserId="1553" LastActivityDate="2020-03-12T22:31:07.067" Title="Chemical engineering equivalancy to the spring-mass-damper/RLC circuit" Tags="|mechanical-engineering|electrical-engineering|fluid-mechanics|chemical-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33608" PostTypeId="2" ParentId="33586" CreationDate="2020-03-12T12:44:44.873" Score="0" Body="&lt;p&gt;I found the solution: just evaluate at x = &lt;span class=&quot;math-container&quot;&gt;$L_{element}/2$&lt;/span&gt; and the exact solution comes out. These points are called super convergent.&lt;/p&gt;&#xA;" OwnerUserId="24936" LastActivityDate="2020-03-12T12:44:44.873" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33609" PostTypeId="1" CreationDate="2020-03-12T13:00:06.010" Score="1" ViewCount="52" Body="&lt;p&gt;In the buildings designed with strong column weak beam concept, the columns will remain elastic, their plastic deformation will be limited. but this is not true at the base of the building. i mean i understand why at the base of building earthquake force is largest but why do we let plastic deformation there?&lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2020-12-07T18:04:00.440" Title="column yielding under earthquake" Tags="|structural-engineering|civil-engineering|structural-analysis|structures|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33610" PostTypeId="2" ParentId="33607" CreationDate="2020-03-12T15:59:27.933" Score="1" Body="&lt;p&gt;It's not necessarily chemical engineering, but fluid and thermal dynamics have similar but not 100% equivalent systems: &lt;a href=&quot;https://en.wikipedia.org/wiki/System_equivalence&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/System_equivalence&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-03-12T15:59:27.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33611" PostTypeId="1" CreationDate="2020-03-12T16:34:11.277" Score="1" ViewCount="232" Body="&lt;p&gt;I'm attempting to model a crank-rocker mechanism (see picture). A motor is connected to the crank side and is driving the mechanism with a torque &lt;span class=&quot;math-container&quot;&gt;$\tau_{motor}$&lt;/span&gt;. A load is attached to the rocker side which applies a torque &lt;span class=&quot;math-container&quot;&gt;$\tau_{load}$&lt;/span&gt;. The moment of inertia on the motor side is &lt;span class=&quot;math-container&quot;&gt;$J_\theta$&lt;/span&gt; and the moment of inertia on the load side is &lt;span class=&quot;math-container&quot;&gt;$J_\varphi$&lt;/span&gt;. I want to find the equivalent moment of inertia &lt;span class=&quot;math-container&quot;&gt;$J_{eq}$&lt;/span&gt; that the motor experiences from &lt;span class=&quot;math-container&quot;&gt;$J_\varphi$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$J_\theta$&lt;/span&gt; combined. &lt;a href=&quot;https://i.stack.imgur.com/53jMC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/53jMC.png&quot; alt=&quot;system overview&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have used Freudenstein's equation to find the relationships &lt;span class=&quot;math-container&quot;&gt;$\varphi(\theta)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\frac{\partial \varphi}{\partial \theta}(\theta)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a gearbox (with constant gear ratio) the solution would be &lt;span class=&quot;math-container&quot;&gt;$J_{eq} = (\frac{N1}{N2})^2 J_\varphi + J_{\theta}$&lt;/span&gt; (see &lt;a href=&quot;https://electronics.stackexchange.com/questions/325607/dc-motor-differential-equation?noredirect=1&amp;amp;lq=1&quot;&gt;https://electronics.stackexchange.com/questions/325607/dc-motor-differential-equation?noredirect=1&amp;amp;lq=1&lt;/a&gt;).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is &lt;span class=&quot;math-container&quot;&gt;$J_{eq}(\theta)$&lt;/span&gt; in my case? My intuition is that the relationship should become something like &lt;span class=&quot;math-container&quot;&gt;$J_{eq}(\theta) = ( \frac{\partial \varphi}{\partial \theta}(\theta))^2 J_\varphi + J_{\theta}$&lt;/span&gt;. However, I have not been able to prove this.&lt;/p&gt;&#xA;" OwnerUserId="24959" LastEditorUserId="24959" LastEditDate="2020-03-12T23:38:35.310" LastActivityDate="2020-03-12T23:38:35.310" Title="Equivalent moment of inertia in a crank-rocker mechanism" Tags="|mechanisms|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33612" PostTypeId="2" ParentId="33609" CreationDate="2020-03-12T17:01:11.753" Score="1" Body="&lt;p&gt;A building designed by the seismic code will resist the seismic spectra with no structural damage. There may be some minor damage to partitions or furniture banging around.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the code provides for extra ductility in the structure to be flexible enough under an exceptionally strong earthquake greater than design spectra to deform in a way that alerts the occupants of eminent danger and provides time for them to vacate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depending on the use of the building this design parameter varies and the amount of acceptable damages becomes smaller.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Such that in hospitals and critical buildings sometimes the columns are insulated from the foundation by hydraulic rollers or other means. Allowing the building to float on the foundation escaping the ground's shakes to a large extent. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-12T17:01:11.753" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33613" PostTypeId="2" ParentId="33607" CreationDate="2020-03-12T20:23:45.120" Score="1" Body="&lt;p&gt;Such a plant system could certainly be modeled as differential equations, but a chemical engineering equivalent would be concerning reaction rates and heat effects, not fluid flows, which I consider to be mechanical engineering (obviously there is overlap).  I am not aware of equivalent fluid system equivalents to all of these.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Mass/inductance: resists changes in direction.  I suppose that pulsing fluids would have this but I've never seen fluid system modeled this way.  Fluids do have inertia but no one cars about forces on them, at least in plants.&lt;/li&gt;&#xA;&lt;li&gt;Spring/resistor: differing force according to position. This would be height portion of the head loss equation&lt;/li&gt;&#xA;&lt;li&gt;Damper/capacitor: resistance to movement according to velocity.  This is head &#xA;loss due to friction.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Most fluid process systems run at steady state.  Velocities don't changes or if they do it tends to be limited in its effects.  Speed up a pump, open a valve, or fill a tank.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastEditorUserId="24351" LastEditDate="2020-03-12T22:31:07.067" LastActivityDate="2020-03-12T22:31:07.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33615" PostTypeId="1" CreationDate="2020-03-12T20:44:35.417" Score="7" ViewCount="7602" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Hjjjp.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Hjjjp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So suppose I want to make a single shelf protruding from a wall and to place a heavy object on it (Figure 1). A shelf like this would fail (assuming no internal mounting that penetrates into the wall). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, in an ideal scenario, I would want to place a vertical column/beam directly below the shelf to help support the weight on top (Figure 2).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet this isn't always ideal due to other objects below the shelf. So in this case I've seen angle braces being used below the shelf (Figure 3). Now generally, from what I've seen, these angle braces are roughly 45 degrees from vertical. Why is that the common angle? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I haven't been able to find any calculations online to determine the ideal or maximum angle here. What would be the maximum (from vertical) angle that a shelf like this could support? I understand that there are many variables at play including the distance that the shelf hangs from the wall, the thickness and strength of the shelf material, the weight of the object on top of it, etc., but is there a rough calculation that could be used?&lt;/p&gt;&#xA;" OwnerUserId="15383" LastActivityDate="2023-05-01T02:12:51.310" Title="What is the maximum angle of a shelf angled brace?" Tags="|mechanical-engineering|structural-engineering|structural-analysis|" AnswerCount="4" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33617" PostTypeId="2" ParentId="33615" CreationDate="2020-03-12T22:30:02.657" Score="6" Body="&lt;p&gt;There is no maximum angle.  Theoretically a shelf could be supported solely via cantilever (a support coming straight out from the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the angle in use is the angle from the wall to the support, the compression stress on the support will be the load divided by the cosine of the angle.  At straight up and down, the angle is zero, and the cosine is 1, so the force is equal to the load.  As the support moves up the wall, the compression force on the support will rise (practically until the force creates a bending moment in the support for small angles).  These calculations in this case are simple, but can get complex in many-membered supports.  This is the basis of the Engineering Class &lt;em&gt;Statics&lt;/em&gt;, which all Civil and Mechanical Engineering students take, usually one of the very first engineering classes taken.  At its core, it is resolving vector loads (the angled support) into and out of vertical and horizontal loads using sines and cosines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the limits involve the strength of the support and the desired space limitations.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-12T22:30:02.657" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33618" PostTypeId="2" ParentId="33615" CreationDate="2020-03-12T22:36:18.460" Score="12" Body="&lt;p&gt;The angle is arbitrary. One can decide which angle is more practical.&lt;/p&gt;&#xA;&lt;p&gt;However, it comes at a cost: the more the angle from the vertical the more pulling and pushing force on the anchors or escrows.&lt;/p&gt;&#xA;&lt;p&gt;Let's say your box is 100lbs such that its center of the gravity is one foot away from the wall.&lt;/p&gt;&#xA;&lt;p&gt;We check the pullout force on the top anchor/s which happens to be equal to the push-in force on the bottom anchor/s&lt;/p&gt;&#xA;&lt;p&gt;We check 2 cases for the angle is 30 and 60 degrees.&lt;/p&gt;&#xA;&lt;p&gt;For 30 degrees from vertical Fanchor is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \Sigma M=0 \ \rightarrow\ 100*12 -\cos30*\mathit{F_a}=0  \ \rightarrow\  \mathit{F_a}=\frac{100*12}{0.866*12=10.39}=115.4 \text{lbs}$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;For 60 degrees Fanchor is:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \Sigma M=0 \ \rightarrow\  100*12 -\cos60*\mathit{F_a}=0  \ \rightarrow\  \mathit{F_a}=\frac{100*12}{6}=200\text{lbs} $&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;We note that the higher angle needs a stronger anchor.&lt;/p&gt;&#xA;&lt;p&gt;We also have a vertical load on each anchor equal to 1/2 of the shear which is 50lbs, but this load is constant and independent of the angle of the diagonal brace.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="-1" LastEditDate="2023-05-01T02:12:51.310" LastActivityDate="2023-05-01T02:12:51.310" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33619" PostTypeId="2" ParentId="33615" CreationDate="2020-03-12T22:49:55.400" Score="2" Body="&lt;p&gt;Here's a way to think about it without the math&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The supporting angle is the same as a full depth bracket but with the 'unneeded' stuff removed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You need the brace because your shelf bracket is too 'skinny'. If you made the bracket deeper, how deep would you make it? That's your question in a nutshell.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW: I'm a structural engineer, if that makes a difference&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-12T22:49:55.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33620" PostTypeId="2" ParentId="31508" CreationDate="2020-03-13T00:35:17.420" Score="0" Body="&lt;p&gt;If you could come up with an application for it, surely it could be used as a pump.  But it wouldn't be terribly efficient, because you need to power a linear actuator to make it work, probably via a hydraulic pump, when you could have just used a pump and motor to start.  And if you did have a cheap/free source of reciprocating energy, you would also need to beat the efficiency and cost of the hundred of years old hand well pump (piston with flap), diaphram pump, or simple reciprocating piston pump (which typically uses check valves).  The efficiency of returned energy after deformation is probably not as good as just acting on the fluid directly.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-13T00:35:17.420" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33621" PostTypeId="2" ParentId="33584" CreationDate="2020-03-13T03:40:05.153" Score="0" Body="&lt;p&gt;Freshly-prepared fuel rods (low-enriched uranium) are not lethally radioactive and do not require extraordinary handling precautions. However, once they have been put into service, extremely radioactive fission products begin to accumulate in them which renders them deadly by the time they are &quot;burned up&quot; and due for removal from the reactor core. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Water is nowhere near as effective at shielding as lead for example, but it is cheap and does an excellent job of carrying away heat released during decay. So you just use lots of it.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-03-13T03:40:05.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33622" PostTypeId="1" CreationDate="2020-03-13T07:45:40.997" Score="4" ViewCount="133" Body="&lt;p&gt;While watching random videos shot in the USA (in the present time), I've noticed these little weird outdoors boxes:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yrheK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yrheK.jpg&quot; alt=&quot;XXXX&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the purpose of these? They look like they might be related to ventilation or something, but they are pretty far away from the nearby buildings. We certainly have no such &quot;mystery boxes&quot; where I live, and I don't even have a good guess what they might be for.&lt;/p&gt;&#xA;" OwnerUserId="24961" LastActivityDate="2020-03-13T17:31:19.553" Title="What are these outdoors &quot;mystery boxes&quot; for in the USA?" Tags="|mechanical-engineering|electrical-engineering|structural-engineering|heat-transfer|airflow|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33623" PostTypeId="1" CreationDate="2020-03-13T08:25:23.557" Score="3" ViewCount="480" Body="&lt;p&gt;Is this L-type corner window correctly structured? We used 18 rebars in total, 6 for each window. This building will soon have an RCC lintel. &lt;a href=&quot;https://i.stack.imgur.com/O8Bia.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/O8Bia.jpg&quot; alt=&quot;The corner window is split into 3 segments and each segment is reinforced with 6 rebars 3 on each side &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24962" LastActivityDate="2020-03-13T16:34:19.533" Title="Is this corner window correctly structured?" Tags="|civil-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33624" PostTypeId="1" CreationDate="2020-03-13T09:18:38.480" Score="3" ViewCount="456" Body="&lt;p&gt;There is &lt;a href=&quot;https://duckduckgo.com/?q=germany&amp;amp;t=h_&amp;amp;iax=videos&amp;amp;ia=videos&amp;amp;iai=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DCohu_9IhlYc&quot; rel=&quot;nofollow noreferrer&quot;&gt;an airport bus (Germany)&lt;/a&gt; which has it's rear wheels covered by the fender. Even front wheels have strange shape of fender around them (tilted toward front). What purpose of that?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/cptKI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cptKI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24239" LastEditorUserId="24239" LastEditDate="2020-03-14T06:45:39.633" LastActivityDate="2020-03-14T06:45:39.633" Title="Why rear wheels of the bus are covered?" Tags="|transportation|car|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33625" PostTypeId="2" ParentId="33622" CreationDate="2020-03-13T09:40:21.100" Score="4" Body="&lt;p&gt;Almost anything can be in those pedestals, but the more common inhabitant would be cable television related. The smaller ones are for subscriber taps, while the larger ones contain amplifiers or interface devices. Fiber optic cables are used for longer distances, while there are interface devices to convert from fiber to coaxial feeder lines to the subscriber taps and to the house.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The design of the one in the photo is likely cable television. Others you'll see can be telephone service devices, but are usually metal boxes and more angular.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-03-13T09:40:21.100" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33626" PostTypeId="2" ParentId="33623" CreationDate="2020-03-13T16:22:55.487" Score="1" Body="&lt;p&gt;It is structurally a weak point and requires some kind of mitigation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The corners of any brick wall structure are where forces from deferent angles meet and have to be resolved. In most of these types of buildings, we see this concern has been addressed by installing a concrete column on the corner or have a built-in concrete column hidden in the wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would add two steel channels, for example, MC 8x20, on either side the windows on the inside and bolted them to the walls at every 16&quot; by 1/4 inch bolts and let them be attached at the bottom to the foundation and on the top to a rafter, concrete roof or any structural member strong enough. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-03-13T16:34:19.533" LastActivityDate="2020-03-13T16:34:19.533" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33627" PostTypeId="2" ParentId="33624" CreationDate="2020-03-13T16:39:55.953" Score="2" Body="&lt;p&gt;Either or both: aerodynamics and style&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For aerodynamics, they would fully cover the front wheel as well but that would not allow for the wheel to turn.&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-13T16:39:55.953" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33628" PostTypeId="1" CreationDate="2020-03-13T16:59:04.430" Score="1" ViewCount="131" Body="&lt;p&gt;From my experience, I've been calculating RSS tolerances by squaring all relevant tolerances, summing them up and then square rooting the sums... ie: &lt;strong&gt;RSS=[Σ(tol^2)]^0.5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've also heard of other methods to calculate RSS including dividing by 3, ie: &lt;strong&gt;RSS=3*[Σ(tol/3^2)]^0.5&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the main difference between these two methods?&lt;/p&gt;&#xA;" OwnerUserId="24864" LastActivityDate="2020-03-13T16:59:04.430" Title="RSS Tolerance calculation and usage" Tags="|mechanical-engineering|mechanical|tolerance|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33629" PostTypeId="2" ParentId="33624" CreationDate="2020-03-13T17:19:35.300" Score="1" Body="&lt;p&gt;They may do it for the safety of passengers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Passengers in the tarmac usually have been exhausted and disoriented after a long flight and possible jet-lag. They have at times small children tagging along with both hands holding luggage and carry on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They need to be shielded from any moving part, tires, cavities, as well as the obvious spraying of water by the wheel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The front wheels are usually not in the path of foot traffic.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-13T17:19:35.300" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33630" PostTypeId="2" ParentId="33622" CreationDate="2020-03-13T17:31:19.553" Score="0" Body="&lt;p&gt;I am with the Charter cable here in Los Angeles and there is a box very similar to this on the sidewalk near my house that is their connection box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One time we had problem with our connection and their tech had to change a board on that box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;He said that somebody had broken into the box to steal parts but alarms must have shaken him. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-13T17:31:19.553" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33631" PostTypeId="1" CreationDate="2020-03-13T17:34:33.063" Score="1" ViewCount="62" Body="&lt;p&gt;I need help dimensioning a screw-beam system of a press. I built one prototype but the beam is bending too much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Below is a sketch of my press. In the middle it has a fixing screw which presses downwards on a beam (3D printed plastic) that presses down my Device Under Test (DUT). The problem is that the beam bends too much so it doesn’t apply pressure evenly downwards on the DUT.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(I used plastic instead of a harder material because I have access to a 3D printer but no access to a workshop)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pltvh.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pltvh.jpg&quot; alt=&quot;Sketch of beam in press&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I did some research and found in the book “Mechanics of Materials” of Gere and Timoshenko following formulae for displacement &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\delta = \frac{F\cdot L^3}{48 \cdot E \cdot I}&#xA;$$&lt;/span&gt;&#xA;F = Force downward&#xA;L = Length of beam&#xA;E = Young's Modulus&#xA;I = Moment of inertia&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And for the moment of inertia:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;I = \frac{b \cdot h^3}{12}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So that I have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\delta = \frac{F\cdot L^3}{4 \cdot E \cdot b \cdot h^3}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I believe I can specify a maximum displacement &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; of let’s say 0.1 mm and the compute the necessary beam height &lt;span class=&quot;math-container&quot;&gt;$h$&lt;/span&gt; or choose another material with a higher Young’s modulus E&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;My first question is: How do I compute the force F that the screw is foing downwards? I tried looking in Shigley’s “Mechanical engineering Design” Chapter about screws but couldn’t find anything. &lt;/li&gt;&#xA;&lt;li&gt;My second question is: Is my dimensioning approach (to specify a maximum bending) correct?&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="21516" LastActivityDate="2020-03-14T05:42:43.447" Title="Dimensioning of beam pressed by screw in assembly" Tags="|mechanical-engineering|applied-mechanics|beam|fasteners|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33632" PostTypeId="2" ParentId="33624" CreationDate="2020-03-13T18:08:52.440" Score="0" Body="&lt;p&gt;An airport bus is unlikely to &quot;need&quot; efficient high speed aerodynamics, but it does need to be highly visible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the still picture, note the lights which are evenly spaced along the length of the bus, and that one of them is in exactly the same position as the rear wheel. The front two lights seem to be mounted on the front fairing as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Accidents where a luggage cart driver tries to drive through the middle of a parked bus because he thought the lights at each end were two separate short vehicles are preventable!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-03-13T18:08:52.440" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33633" PostTypeId="2" ParentId="33631" CreationDate="2020-03-13T18:59:11.887" Score="1" Body="&lt;p&gt;Force F is related to torque and the pitch of the screw:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let's call the radius of your screw r and the tread pitch N and torque t.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  F= \frac{\tau/r}{(2\pi r/N )}   $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the answer to the 2nd part is the numbers are ok, but these are for very small angles of deformation, things around a few degrees, and L should bee the distance between the supports as Jko said.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-03-13T20:00:08.293" LastActivityDate="2020-03-13T20:00:08.293" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33634" PostTypeId="2" ParentId="33059" CreationDate="2020-03-13T19:08:47.527" Score="2" Body="&lt;p&gt;The connections of precast concrete requires more specialized design and also more specialized installation crew in the field. The joints in precast concrete construction are the most critical items. Poorly designed or installed joints can lead to progressive collapse. Also the temporary loads during construction can be critical for the design and installation sequence of precast concrete. &lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2020-03-13T19:08:47.527" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33637" PostTypeId="1" AcceptedAnswerId="33649" CreationDate="2020-03-13T20:46:02.053" Score="2" ViewCount="499" Body="&lt;p&gt;When I cut aluminum bar or t-slot extrusion on my bandsaw I am not happy with how &quot;square&quot; this edge is.  I don't have a milling machine at home, and I have been told grinding aluminum is a very bad idea.  &lt;/p&gt;&#xA;" OwnerUserId="21120" LastActivityDate="2020-12-22T21:18:56.227" Title="How can I square off a piece of aluminum bar after cutting without a milling machine?" Tags="|aluminum|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33638" PostTypeId="1" AcceptedAnswerId="33662" CreationDate="2020-03-13T20:55:34.880" Score="1" ViewCount="1282" Body="&lt;p&gt;I have a real mechanical system (ball &amp;amp; beam) which I want to control. The system is approximated by the second order transfer function: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T(s) = \frac{b_0}{s^2+a_1s+a_2} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and in state-space form:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{x_1} = x_2 $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{x_2} = -a_1x_2-a_2x_2+b_0u $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the parameters &lt;span class=&quot;math-container&quot;&gt;$\ b_0,a_1,a_2 $&lt;/span&gt; have been identified, so they are known. I have designed two controllers. The first one is a full state feedback controller described by the equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ u = -Kx+k_rr $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the term &lt;span class=&quot;math-container&quot;&gt;$\ k_rr $&lt;/span&gt; is because I want input reference &lt;span class=&quot;math-container&quot;&gt;$\ (r) $&lt;/span&gt; tracking. The other controller is a dynamic state feedback controller desctibed by the equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ u = -Kx-k_iz $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ \dot{z} = x_1-r $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Both controllers are doing well but the only problem I have is when the ball reaches the reference point there is a bit of oscillation of the system (the beam oscillates) in order to keep the ball at the desired point. I have found out that the oscillations occur due to the measurement of the ball's position and not because of the gains of the controllers because this happens even if I make these gains smaller. The sensor used to measure the ball's position on the beam is a linear potentiometer and it appears that the measurements include some unwanted noise which results to rapid changes of the control signal which leads to the oscillations. I have tried to reduce the noise by using a low pass filter with the below transfer function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F(s) = \frac{15}{s+15} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The block diagram of the control loop is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/SxGHV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/SxGHV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The oscillations have been reduced but not achieved a satisfactory level. Is there any way to modify these controllers in a way to affect these oscillations and reduce them ? Would a PID-controller perform better due to the existence of the derivative of the error ? Or should I design a better filter for the noise reduction which should also be fast (because the system is functioning in real time) enough ? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I place some plots for two different cases (PD-controller gains) where the actual position (output) is shown compared to the reference and the control signal. Note that the sudden changes to the real system's output are due to external disturbances I apply in order to check if the system goes back to the reference. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;First Case (&lt;span class=&quot;math-container&quot;&gt;$\ K_p = 3.9352 $&lt;/span&gt; &amp;amp; &lt;span class=&quot;math-container&quot;&gt;$\ K_d = 2.248 $&lt;/span&gt;):&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EBYJC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EBYJC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vM2iC.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vM2iC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Second Case (&lt;span class=&quot;math-container&quot;&gt;$\ K_p = 2.3536 $&lt;/span&gt; &amp;amp; &lt;span class=&quot;math-container&quot;&gt;$\ K_d = 1.6423 $&lt;/span&gt;):&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/In7jJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/In7jJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/x0I5B.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/x0I5B.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-03-22T21:15:09.183" LastActivityDate="2021-12-31T04:04:09.367" Title="How to reduce oscillations (rapid changes of control signal) when controlling a real system, which occur due to noise in measurement from sensors?" Tags="|control-engineering|control-theory|measurements|sensors|noise|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33639" PostTypeId="2" ParentId="33615" CreationDate="2020-03-14T00:32:58.217" Score="0" Body="&lt;p&gt;What is the maximum &lt;em&gt;weight&lt;/em&gt; for a shelf with angled braces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With two of these it's 1000 pounds &quot;when properly installed&quot;, which means filling all three holes with three inch exterior grade screws or better, directly into studs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/JGA50.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JGA50.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(&lt;a href=&quot;https://www.lowes.com/pd/Style-Selections-15-63-in-Shelf-Bracket/1090555&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The angle looks to be bout &lt;strong&gt;60 degrees&lt;/strong&gt;. I don't think I've ever seen one any less, at least, not one that I'd install, like the ones that are stamped with a curve to provide stability and sometimes don't even have a brace.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Construction of the bracket itself and its fasteners will dictate its application. Notice how the bracket has a vertical brace and ~1/8&quot; thickness throughout, instead of a flimsy stamped curve laying horizontal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maximum weight? Whatever the manufacturer says. Safety factor? As large as possible. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I do closets with these (when the entire old system fell down because it &lt;em&gt;wasn't&lt;/em&gt; brackets like these), it's one bracket every 16&quot; center into a stud with 3&quot; Deckmate screws. I don't get callbacks after installing these. And I don't install anything that you couldn't &lt;em&gt;do pull-ups on&lt;/em&gt;, because fully loaded and grabbed while falling and injuring you isn't going to be on &lt;em&gt;my&lt;/em&gt; conscious.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://diy.stackexchange.com/a/163779/23295&quot;&gt;In what orientation should this L-bracket be when hanging shelves?&lt;/a&gt; : Kept as high as possible so that you don't hit your head on it and to provide the most clearance below it (as pictured above).&lt;/p&gt;&#xA;" OwnerUserId="15179" LastActivityDate="2020-03-14T00:32:58.217" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33640" PostTypeId="2" ParentId="13463" CreationDate="2020-03-14T00:57:58.667" Score="0" Body="&lt;p&gt;Just ignore it. Wood is resilient and will redistribute load. Also, there are large safety factors involved. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm going to guess that you're a newer engineer. I used to think and do this too, it was hard for me to let go of thinking about these type of small details. If you doing this, it's a good, it means that you're a good engineer in my book, as you think of and understand the details, more than most. After a while you're start to see how and when to be concerned about these things and when to know that it's not an issue&lt;/p&gt;&#xA;&#xA;&lt;p&gt;BTW, the inside faces of the studs are not braced and you do need to consider that for F'c&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-14T00:57:58.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33641" PostTypeId="2" ParentId="8525" CreationDate="2020-03-14T01:07:51.523" Score="1" Body="&lt;p&gt;The short answer is yes, and it's expected that there is clearance for constructability. The Steel Erection manual allows for these types of tolerances. Did you try asking your steel fabricator what is typical? Or check your general notes, that is where you will find this type of information. Or ask your structural engineer. If you mess up, you don't want to have to tell the jury that you got your advice from a bunch of yahoos online :)&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-14T01:07:51.523" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33642" PostTypeId="1" CreationDate="2020-03-14T03:40:51.147" Score="3" ViewCount="1066" Body="&lt;p&gt;If I have a solid material, let's say two bricks, one of which is thicker and I do modal/vibration testing. What would be the relation between thickness and natural frequency of any of them, is there any equation that can explain it, or does it depend on the material itself?&lt;/p&gt;&#xA;" OwnerUserId="24972" LastEditorUserId="90" LastEditDate="2020-03-14T10:18:02.870" LastActivityDate="2020-09-10T02:18:06.620" Title="Relation between natural frequency and the thickness" Tags="|vibration|frequency-response|modal-analysis|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33643" PostTypeId="2" ParentId="33642" CreationDate="2020-03-14T04:27:56.417" Score="2" Body="&lt;p&gt;We know for undamped harmonic vibration&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\omega= angular velocity \space, L=length of brick \ , m=mass \ , \\ I= \text{second moment of area of the brick section } \ , \ h= \text{the height of the brick section} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \omega = \sqrt {k/m }$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And  &lt;span class=&quot;math-container&quot;&gt;$ k=I/L $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  I =bh^3/12$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So everything else being equal the thicker brick will have larger &quot; I &quot; hence larger k, therefore, a shorter period and larger natural frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ydy8T.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ydy8T.jpg&quot; alt=&quot;free free brick vibration&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-03-14T14:29:49.733" LastActivityDate="2020-03-14T14:29:49.733" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33645" PostTypeId="2" ParentId="33631" CreationDate="2020-03-14T05:42:43.447" Score="3" Body="&lt;p&gt;You are on track with your beam calculations, but I doubt you will be happy with the results, just because a 3-D printed part is unlikely to act the way a cast or milled solid plastic part would.  You would be well served to just buy aluminum bar stock and cut it with a hack saw.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As to the screw force produced, I doubt you will be happy with that either.  Much of the torque will go to thread friction instead of down into the beam. Without friction, a screw is modeled as an simple inclined plane, so force exerted would be given by dividing the force produced by the sine of the thread angle (not the angle of the threads, but of the pitch, in metric threads the arctan of pitch/diameter).  The force would be the torque divided by the radius, to the mid-point of the threads.  Greased threads would be appropriate, and you could use a scale to calculate your friction losses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I recommend using something like a beam that you can place weights on to give you a force driving downward onto your testing beam.  It can then act directly without worry of friction.  Loosely anchor the beam on one side of your press, have a connector going down to your testing device, and extend the beam over the other side. Known weights will then produce a known force, and you could also change the lever by placing weights on different notches along the beam. &lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-14T05:42:43.447" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33646" PostTypeId="1" CreationDate="2020-03-14T10:30:26.850" Score="5" ViewCount="75" Body="&lt;p&gt;On this satellite image, there are pipelines extending from a terminal at sea to an LNG (Liquid Natural Gas) storage facility.  Why do the pipelines, at regular intervals, deviated from a straight line?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a simple way to prevent / dampen shock waves from passing through the liquids from one end to the other and damaging equipment on the other end?  Could you use baffles in the pipes instead?  Would baffles significantly increase drag whilst these deviations increase resistance only slightly (due to longer pipe and forcing fluid to change direction multiple times)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sDWKx.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sDWKx.png&quot; alt=&quot;regular deviations of three 90 degree turns in LNG pipeline on Grain island&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.google.com/maps/@51.4402401,0.6996781,959m/data=!3m1!1e3&quot; rel=&quot;noreferrer&quot;&gt;https://www.google.com/maps/@51.4402401,0.6996781,959m/data=!3m1!1e3&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="20676" LastActivityDate="2020-03-14T10:30:26.850" Title="Why do some pipelines have regular 90 degree deviations?" Tags="|fluid-mechanics|chemical-engineering|" AnswerCount="0" CommentCount="1" ClosedDate="2020-03-14T12:53:42.073" ContentLicense="CC BY-SA 4.0" />
  <row Id="33649" PostTypeId="2" ParentId="33637" CreationDate="2020-03-15T00:07:56.823" Score="3" Body="&lt;p&gt;I used to build structures  using the 80/20 aluminum extrusions , I used a mitre saw with a good aluminum cutting blade lubricated with stick wax (or any good cutting fluid). Makes beautiful square cuts with minimal burring. Be careful applying the stick wax to the rotating blade, otherwise your mitre saw will become a finger removing device by default  :-(&lt;/p&gt;&#xA;" OwnerUserId="628" LastActivityDate="2020-03-15T00:07:56.823" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33651" PostTypeId="1" CreationDate="2020-03-15T08:31:56.487" Score="1" ViewCount="193" Body="&lt;p&gt;So we all know the basics behind specific speed Ns of hydro-turbines.&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$N_s = RPM \cdot  \frac{ \sqrt{(P)} }{ H^{1.25}}$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where P is the power output and H is the head. The theory is that once you know the Ns for that model of turbine, you are able to find out either the RPM, power output or head if you are given absolute values for 2 of the 3 mentioned variables.&lt;/p&gt;&#xA;&lt;p&gt;Let's say I have a turbine with a known Ns. If my head increases but my power output remains constant due to say, reduced flow, then the RPM will increase since Ns is fixed. So far, so good.&lt;/p&gt;&#xA;&lt;p&gt;However, if my head remains constant but my power output increases due to increased flow, then my RPM must decrease since Ns is fixed. But if we think about it that doesn't make sense...the RPM must be increased for a larger power output, no? Is there something wrong with my understanding of Ns?&lt;/p&gt;&#xA;" OwnerUserId="24982" LastEditorUserId="19906" LastEditDate="2020-12-11T05:52:55.103" LastActivityDate="2023-12-27T06:08:56.097" Title="Some questions about the specific speed (Ns) of hydro turbines" Tags="|turbines|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33652" PostTypeId="1" CreationDate="2020-03-15T08:57:54.103" Score="5" ViewCount="28185" Body="&lt;p&gt;I've noticed this a few times, the last time today:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VrT3h.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VrT3h.jpg&quot; alt=&quot;tank&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's a big cylinder-shaped &quot;tank&quot; (for water?) outside of the front part of the truck. What's the purpose of this thing?&lt;/p&gt;&#xA;" OwnerUserId="24983" LastActivityDate="2020-03-16T19:05:44.360" Title="What are those large cylinder &quot;tanks&quot; attached to the outside of trucks for?" Tags="|car|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33653" PostTypeId="2" ParentId="33652" CreationDate="2020-03-15T10:00:01.417" Score="7" Body="&lt;p&gt;They are external &lt;a href=&quot;https://fleet-airfilters.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;air filters&lt;/a&gt; for the engine.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="-1" LastEditDate="2020-03-16T11:39:58.477" LastActivityDate="2020-03-16T11:39:58.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33656" PostTypeId="1" CreationDate="2020-03-15T16:21:32.720" Score="1" ViewCount="36" Body="&lt;p&gt;I have a linear repeat such  that for up to  11 repeats, the  flex operation works perfectly.  for 12 and  more, the operation goes haywire.  is there an alternate way of doing it that avoids this problem? &#xA;the operation  is as follows:  &lt;a href=&quot;https://i.stack.imgur.com/v5ico.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/v5ico.png&quot; alt=&quot;the operation&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;this is the operation for 11 repeats (works perfectly) &lt;a href=&quot;https://i.stack.imgur.com/oTStp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/oTStp.png&quot; alt=&quot;up  to 11 repeats, works great&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;this is what I get for anything beyond 11 repeats:  &lt;a href=&quot;https://i.stack.imgur.com/Fdl5g.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Fdl5g.png&quot; alt=&quot;beyond 11, it fails&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;the basic patern is as follows: &lt;a href=&quot;https://i.stack.imgur.com/QvQqs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QvQqs.png&quot; alt=&quot;basic patern&quot;&gt;&lt;/a&gt;  it is first linearly repeated (up to 11 times, apparently), then circular patterned around an axis, then the flex (twist) operation is applied.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit:  it has been suggested to use weldmets.  I'm now getting the hang of this,  but it still has problems..&lt;a href=&quot;https://i.stack.imgur.com/VRcb2.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VRcb2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  ..sigh..  my order of operations:  &lt;a href=&quot;https://i.stack.imgur.com/8QFSN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8QFSN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13081" LastEditorUserId="13081" LastEditDate="2020-03-19T21:53:29.863" LastActivityDate="2020-03-19T21:53:29.863" Title="flexing operation fails when structure is too long" Tags="|solidworks|" AnswerCount="0" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="33657" PostTypeId="2" ParentId="33651" CreationDate="2020-03-15T17:13:36.623" Score="0" Body="&lt;p&gt;Judging by the question, apparently not!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Okay, this isn't a trivial thing to visualize, but when you say flow increases, head stays the same, power increases, and Ns stays the same, what has to happen to the geometry of the rotor? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ns being constant implies here that the two rotors being talked about are geometrically similar. The one with the larger flow is bigger. To operate at the same head, it will run at a lower rpm.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2020-03-15T17:13:36.623" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33661" PostTypeId="1" CreationDate="2020-03-15T20:32:45.910" Score="2" ViewCount="39" Body="&lt;p&gt;I'm doing this special project where I have to make concrete using a Martian simulant provided from a commercial source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I measured it as 500 grams initially, then submerged it in water for a day. It came back as 727 grams, meaning that it gained 200 grams of water, as it is now in its saturated surface dry state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$A = \dfrac{W_{SSD} - W_{OD}}{W_{OD}}\times100\%$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Following this equation to measure the absorption value, it would mean that the value would be around 50%, which is very high. Usually, normal aggregates on Earth have an absorption rate of around 10%!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there anything that I may have been doing wrong? I am very skeptical of my results&lt;/p&gt;&#xA;" OwnerUserId="16653" LastEditorUserId="1832" LastEditDate="2020-03-15T20:47:53.527" LastActivityDate="2020-03-16T00:12:57.130" Title="Can aggregate gain almost 200 grams of water when submerged in water for a day?" Tags="|materials|civil-engineering|concrete|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33662" PostTypeId="2" ParentId="33638" CreationDate="2020-03-15T23:15:37.153" Score="1" Body="&lt;p&gt;You can change the type and/or order of the lowpas filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your current filter has a -3 dB reduction at 15 rad/s (~2.4 Hz) and decreases with a -1 slope every decade.&#xA;This means that at 150 rad/s (~24 Hz) the amplification is still -10 dB (factor 0.1), which is not that much.&#xA;Therefore you could change the order of the filter, e.g.&#xA;&lt;span class=&quot;math-container&quot;&gt;$$F(s)=\frac{1}{(\tau\cdot s + 1)^n} $$&lt;/span&gt;&#xA;with &lt;span class=&quot;math-container&quot;&gt;$\tau = \frac{1}{15}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; the desired order.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, this filter will have some disadvantages when going for higher orders, therefore you could consider the Butterworth filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the noise is present at a certain frequency, you should use a Notch filter to suppress signals at the selected frequency band.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2020-03-15T23:15:37.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33663" PostTypeId="2" ParentId="33661" CreationDate="2020-03-16T00:12:57.130" Score="1" Body="&lt;p&gt;Clay can absorb almost 40% water at saturation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Amount of water aggregate can hold depends on many factors, e.g., its size (the smaller the more surface) and texture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many methods for measuring the saturation of a soil sample but they require established labs. I assume you have measured your sample correctly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, considering the huge difference in the development process on the Earth and Mars and totally different erosion systems and how authentic your simulant is, there are too many variables to be looked into.  &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-16T00:12:57.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33666" PostTypeId="1" AcceptedAnswerId="33690" CreationDate="2020-03-16T09:21:52.693" Score="1" ViewCount="79" Body="&lt;p&gt;I'm looking to put gradual pressure on a syringe.&lt;/p&gt;&#xA;" OwnerUserId="24993" LastEditorUserId="24993" LastEditDate="2020-05-07T11:11:46.600" LastActivityDate="2020-05-07T11:11:46.600" Title="Changing (alternating non linear) screw motion" Tags="|mechanical-engineering|fluid-mechanics|pressure|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="33671" PostTypeId="2" ParentId="33652" CreationDate="2020-03-16T14:52:16.947" Score="2" Body="&lt;p&gt;Donaldson type air filters.air intakes through those filters directly through the turbo charger,which is spun up via the exhaust side of the engine through a separate intake port and turbine vanes. The cold air is compressed as it&#xA;Is passed into a charge air cooler,radiator like in its construction,and condensed,so more oxygen is available to the engine...in simple terms.without the air filters the engine would be ground to dust by dust and fine particles.&lt;/p&gt;&#xA;" OwnerUserId="24999" LastActivityDate="2020-03-16T14:52:16.947" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33672" PostTypeId="2" ParentId="33666" CreationDate="2020-03-16T17:46:54.570" Score="0" Body="&lt;p&gt;Just a basic crude idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you can have the piston screw threads installed on a hexagonal shaft in 6 sections (not a continuous tooth but 6 individual teeth or spikes) around a  shaft that has 6 separate faces and is flexible so it can change its diameter by letting these faces move in/out. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the female thread in the syringe shaped like a cone, a funnel, getting wider as it gets deeper, but with the same pitch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the slope of the thread pitch will be less as it gets deeper and the piston will move less/ a full rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;my apologies for the poor quality sketch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MKQdJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MKQdJ.jpg&quot; alt=&quot;flexible pump&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-16T17:46:54.570" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33675" PostTypeId="2" ParentId="33652" CreationDate="2020-03-16T18:23:31.963" Score="0" Body="&lt;p&gt;Definitely air filter housings. Contrary to popular belief they are not chrome. They are stainless steel. There is very little if any chrome on nicer big trucks anymore. Chrome will crack due to rock chips an water, road grime and road salt will eventually cause the chrome plating to peel off the steel. Stainless bumpers are double the price but unless you are a jacked driver and mess it up it will last forever and stay shiney.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the days of 55 stay alive most trucks had the air filter housing on the passenger side and either one or two fuel filter housings on the drivers side.&lt;/p&gt;&#xA;" OwnerUserId="25004" LastActivityDate="2020-03-16T18:23:31.963" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33676" PostTypeId="1" CreationDate="2020-03-16T18:32:08.263" Score="2" ViewCount="49" Body="&lt;p&gt;A response surface that I have obtained from a DOE analysis with multiple variables, show small irregularities when I plot 2D variable-vs-variable curves . like somewhere it should be exponentially and only reduce, it changes the gradient in some points. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I assuem it's from the impreciseness of the fitting of the Response surface. How much effort shouldI put to obtain a more precise curve ? &lt;/p&gt;&#xA;" OwnerUserId="17070" LastActivityDate="2020-12-15T17:21:59.737" Title="How much should I put effort to refine a Response surface?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33678" PostTypeId="1" CreationDate="2020-03-16T19:44:23.520" Score="1" ViewCount="564" Body="&lt;p&gt;I want to build an antenna rotator for a 1M dish. How do I calculate the required torque for the motor of the elevation axis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's a sketch of the problem:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/Cgwhj.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Cgwhj.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the &quot;bottom&quot; of the dish is the pivot point, what would be the required torque to rotate the dish around this pivot point? Obviously the motor will have a series of reductions to reach the required torque and speed. The last stage a worm gear with a turn ratio enough to prevent rotation from the antenna weight when no power is being applied.&lt;/p&gt;&#xA;" OwnerUserId="23187" LastActivityDate="2024-03-26T11:03:45.407" Title="How to calculate the required torque for an azimuth/elevation antenna rotator?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33680" PostTypeId="1" CreationDate="2020-03-16T20:29:11.440" Score="1" ViewCount="33" Body="&lt;p&gt;I would like to estimate the time needed to open a linear single-acting pneumatic actuator.&#xA;The actuator's piston is loaded with a spring pushing downwards and the operating fluid (air or nitrogen) pushes upwards; other forces like weight and friction are present. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The operating fluid is provided by a large tank (ideally infinite) with stagnation pressure &lt;span class=&quot;math-container&quot;&gt;$P_\infty$&lt;/span&gt; and stagnation temperature &lt;span class=&quot;math-container&quot;&gt;$T_\infty$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the tank is connected directly to the actuator, balance of forces, with fluid force depending on actuator inlet mass flow rate, leads to an ODE which I can solve numerically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since tank and actuator are actually connected by a long pipe, &lt;span class=&quot;math-container&quot;&gt;$\frac{length}{diameter}\approx 1000$&lt;/span&gt;, I need to estimate the time to reach the equilibrium state in the pipe (&quot;pressurise the pipe&quot;). Initial conditions for pipe and actuator are &lt;span class=&quot;math-container&quot;&gt;$P|_{t=0}&amp;lt;P_\infty$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$T|_{t=0}=T_\infty$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Does anyone know a way to estimate the time required to reach the equilibrium state in the pipe, maybe based on some dimensional argument or some typical time related to the pipe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not, I must determine the evolution of pressure, temperature and mass flow rate inside the pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have been reading literature for compressible flow in pipes, usually coming from natural gas distribution applications, but the problem of transient compressible flow in a pipe, be it isothermal or non-isothermal, leads to a system of PDEs. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The model verification examples I found led me to suppose that the PDE was &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;requiring some sort of boundary condition also at the exit of the pipe. Since I do not know the pressure / temperature / mass flowrate entering the actuator in advance, I am missing something.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;but since the PDE is hyperbolic, it actually requires boundary conditions at the inlet only, besides initial conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any help or suggestion would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="25000" LastEditorUserId="25000" LastEditDate="2020-03-18T15:57:26.940" LastActivityDate="2020-03-18T15:57:26.940" Title="Linear pneumatic actuator opening time - pressurise connection pipe" Tags="|fluid-mechanics|pipelines|pneumatic|actuator|compressible-flow|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33681" PostTypeId="2" ParentId="33676" CreationDate="2020-03-16T22:31:49.620" Score="3" Body="&lt;p&gt;Well, this depends entirely on what you plan to do with your response surface.  We have no idea what you want to do with it, so we can't answer that.  Here's what you should do: ask yourself, &quot;what is the worst thing that could happen if this response surface is not accurate enough?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are building a nuclear reactor, and the answer is &quot;hundreds of people will die&quot; then you might want to spend another few years on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are writing the physics engine for a video game, and then answer is &quot;the physics in the game will be slightly unrealistic&quot; then you might not need to spend any more time on it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Somewhere in between might be a project where the answer is &quot;nobody gets hurt but my company could lose $X&quot;.  In that case spend time on it in proportion to the value of X versus the cost of your time in improving it.&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2020-03-16T22:31:49.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33683" PostTypeId="1" CreationDate="2020-03-17T01:18:27.057" Score="1" ViewCount="552" Body="&lt;p&gt;Is surgical and medical mask based on static electricity? By static electricity I mean like 3M N95 mask with at least two different layers of positive and negative static electrical layer. I googled and find that someone says medical mask filters only by physical fiber to trap particles physically, while some others say it also attract particles by fibers with static electricity to increase breath flow rate. However, it seems not true because medical mask seems like have only three layers, outer, inner, and middle, which is seemly impossible to form a static electricity mask. Unlike 3M N95 mask has 5 layers, outer, positive charged, separator, negative charged, inner, which is more trust worthy. BTW, surgical and medical masks are basically the same beside outer layer. Surgical is blood-proof, and medical is not. And, I ask because of corona virus. If without static electricity, then it is possible to reuse only if I do not damage it physically, to protect myself from potential public risk.&lt;/p&gt;&#xA;" OwnerUserId="25006" LastEditorUserId="12754" LastEditDate="2020-03-17T10:44:26.043" LastActivityDate="2020-03-17T10:44:26.043" Title="Is surgical and medical mask based on static electricity?" Tags="|materials|safety|material-science|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33684" PostTypeId="2" ParentId="33678" CreationDate="2020-03-17T02:00:30.990" Score="0" Body="&lt;p&gt;we just do it for gravity and assume 90% efficiency for the gear, or if they have a datasheet look that up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of the dish being perforated in certain angles it could act as a wing and create flutter, lift,  destructive vibration. So we need to use a light truss to stiffen the dish.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call the dish mass m and its depth H, and assume by just eyeballing the parabola's CG axis is at H*0.45. and say we want the torque &lt;span class=&quot;math-container&quot;&gt;$  \tau \ $&lt;/span&gt; to deflect the dish from horizontal to 90 degrees up vertical.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  \tau_{max }= m *0.45H/9*(1/0.9)=0.055mH$&lt;/span&gt; &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-17T02:00:30.990" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33685" PostTypeId="2" ParentId="33680" CreationDate="2020-03-17T03:49:10.770" Score="1" Body="&lt;p&gt;Flow rate of the gas into the cylinder will change according to the pressure drop.  So first you have to model flow rate through the pipe to the cylinder, which you can probably model as a plenum.  I would assume this is a differential equation version of head loss flow.  So as the mass of air enters the cylinder you have the pressure increase in the cylinder, which reduces the flow through the pipe.  If you want precision here then you should also model opening of the valve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now as pressure builds to greater than the load side of the cylinder you need to model cylinder movement against spring, load, friction, and inertia.  So another differential equation, except now it effects the cylinder pressure as well since expansion of the cylinder changes the pressure increase.  Depending on pressures and pipe sizes, the inertia of the load might be important, or the pipe might be the limiting factor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Sounds like a matrix solution to me.  Or you could build a couple and time them.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-17T03:49:10.770" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33687" PostTypeId="1" CreationDate="2020-03-17T09:49:57.627" Score="1" ViewCount="50" Body="&lt;p&gt;First off - this project is still in the early planning stages. The basic scope is that a plant want's to switch from using conc. sulphuric acid (~98%) in its process (nitration) to oleum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I questioned why they would do this and their reasoning is supply shortage of sulphuric acid causing downtimes. Apparently an alternative supplier is not possible and they want to use oleum to supplement their shortfalls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm conducting a high-level review of this idea. &#xA;From my brief assessment it would seem that in most cases sulphuric acid and oleum are interchangeable, and, in cases where equipment is designed for compatibility with sulphuric acid it should also be compatible with oleum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the points that have I have flagged so far...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;storage tanks&lt;/strong&gt; -- material compatibility seems fine, but generation of fumes (SO3) may not be compatible with the scrubbing system&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;pumps/handling --&lt;/strong&gt; oleum (depending on concentration) can have significantly different viscosity and freezing points which in the local climate might cause issues.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;nitration&lt;/strong&gt; -- nitration reaction kinetics would likely be affected when switching to oleum...this would require labscale investigation and cannot be predicted accurately. The presence of dissolved SO3 might affect the reaction?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;reconcentration&lt;/strong&gt; -- there is currently a system in place designed to reconcentrate spent sulphuric acid (vacuum distillation), what changes would be required to re-concentrate oleum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be appreciated if anyone with experience could contribute other concerns or comment on the above.&lt;/p&gt;&#xA;" OwnerUserId="10923" LastEditorUserId="10923" LastEditDate="2020-03-17T10:03:37.630" LastActivityDate="2020-03-17T20:57:21.020" Title="Considerations for switching from Conc. Sulphuric acid to Oleum" Tags="|materials|chemical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33688" PostTypeId="1" AcceptedAnswerId="33699" CreationDate="2020-03-17T14:38:06.813" Score="1" ViewCount="364" Body="&lt;p&gt;I've been trying, as a thought exercise, to run some calculations on a floor jack-like mechanism that is expected to lift 130kg (287 lbs). Dimensions and setup are as per the image below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XpAYi.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XpAYi.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 130kg load is to be positioned at the end of the boom, to simplify things, I assume the load would be at 90 degrees relative to the boom. My understanding, is that the force would be greatest then, anyway...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's my approach to try and size the motor:&#xA;1. Find the torque at point A, due to the load on the end of the boom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F = ma = 130\cdot10 = 1300\text{ N}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I specify &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; as 10, rather than 9.8. So, 130&amp;nbsp;kg load exerts a force of 1300&amp;nbsp;N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque, then at point A would be &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\tau = rF\sin\theta = 0.5 \cdot 1300 \sin(90) = 650\text{ Nm}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol start=&quot;2&quot;&gt;&#xA;&lt;li&gt;I don't know what force the linear actuator needs to produce - x. I proceed to calculate what applied force at point B would yield a torque greater than 650&amp;nbsp;Nm. Let's assume a 10% safety margin, so 715&amp;nbsp;Nm needed at point B.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The angle will go from 90 degrees down, let's say to 30 degrees at the max lift point of the boom. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With reduced angle, the torque is also diminished, so at the highest point, the linear actuator's force applied will need to be (at 30 degree angle relative to the boom)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$715 = rF\sin\theta = 0.1F\sin(30°) \therefore F = 23,833\text{ N}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sounds like a ridiculously high force rating for an electric linear actuator. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my approach correct? I'm quite certain I'm missing something important. Your help will be greatly appreciated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT&lt;/strong&gt;&#xA;Clearly, the above setup isn't ideal for an electrical linear actuator. The other lifting setup I can think of, is a mobile crane, like in the pic below. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Taking the same size and parameters as above, the force needed by the actuator is much more palatable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/eflbJ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eflbJ.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume force is applied at a distance of &lt;strong&gt;40cm&lt;/strong&gt; from the point of rotation and an initial &lt;strong&gt;30 degree&lt;/strong&gt; angle. The lifting mechanism needs to counter a torque of &lt;strong&gt;715Nm&lt;/strong&gt; (calculated above). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$715 = rF\cos\theta = 0.4F\cos(30°) \therefore F = 2,102.94\text{ N}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this correct? (ignoring the counter-weights on the actual crane) &lt;/p&gt;&#xA;" OwnerUserId="14581" LastEditorUserId="14581" LastEditDate="2020-03-18T10:39:32.137" LastActivityDate="2020-03-18T10:39:32.137" Title="Sizing a motor for a floor jack-like mechanism" Tags="|mechanical-engineering|torque|linear-motors|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33689" PostTypeId="1" CreationDate="2020-03-17T14:41:19.487" Score="0" ViewCount="20" Body="&lt;p&gt;Suppose a train is running at 30kmph and suddenly the power is switched off. How do we calculate the reducing velocity at intervals and what factors will affect the change in it?. Further how much time will it take to stop from that position and the distance it will cover before coming to halt?&lt;/p&gt;&#xA;" OwnerUserId="25010" LastActivityDate="2020-03-17T14:41:19.487" Title="Train decelerations in no power condition" Tags="|friction|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33690" PostTypeId="2" ParentId="33666" CreationDate="2020-03-17T18:07:43.077" Score="4" Body="&lt;p&gt;Cams offer a simple way to make a variable motion profile.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/swtoS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/swtoS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. A simple cam-operated contraption can be built out of wood.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Syringe.&lt;/li&gt;&#xA;&lt;li&gt;Syringe holder. This really needs a slot from one edge to allow easy changing of the syringe.&lt;/li&gt;&#xA;&lt;li&gt;Drive-rod guide lower.&lt;/li&gt;&#xA;&lt;li&gt;Drive-rod guide upper.&lt;/li&gt;&#xA;&lt;li&gt;Drive rod. This and the two guides ensure that the syringe only sees vertical motion on its plugner.&lt;/li&gt;&#xA;&lt;li&gt;The drive cam. The radius on this changes to give the motion profile you require for a constant velocity rotation of the cam.&lt;/li&gt;&#xA;&lt;li&gt;Cam axle.&lt;/li&gt;&#xA;&lt;li&gt;Cam handle.&lt;/li&gt;&#xA;&lt;li&gt;Assembly supports.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Instructions for use:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Rotate cam handle anti-clockwise to the horizontal position (9 o'clock).&lt;/li&gt;&#xA;&lt;li&gt;Lift the drive-rod and insert the syringe.&lt;/li&gt;&#xA;&lt;li&gt;Rotate cam handle clockwise at even speed to 3 o'clock position.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Design:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Get out your protractor, some paper and a pencil.&lt;/li&gt;&#xA;&lt;li&gt;Measure the stroke of your syringe. This is the difference between the cam's minimum and maximum radius.&lt;/li&gt;&#xA;&lt;li&gt;Draw a semicircle on the paper with a radius of, say 50 mm.&lt;/li&gt;&#xA;&lt;li&gt;Draw radii every 10&amp;deg; from the centre out past the semi-circle.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;On each 10&amp;deg; line mark a point as many mm away from the semi-circle as you want the syringe depressed at that angle. So, for an even dispense on a 90 mm syringe you would have a point at 50, 55, 60, ... 135, 140 mm giving you a stroke of 90 mm in 180&amp;deg;. Draw that first as your reference and then you should be able to work out how to get the profile you want. Your profile just has to:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;start on the same 0&amp;deg; start point as we've constructed above.&lt;/li&gt;&#xA;&lt;li&gt;be always increasing.&lt;/li&gt;&#xA;&lt;li&gt;finish on the 180&amp;deg; point we've constructed above.&lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Have fun.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The image was created with &lt;a href=&quot;http://onshape.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;OnShape&lt;/a&gt; which is an online, in-your-browser 3D CAD package. Use is free if you are prepared to make your designs public. It's all rather amazing!&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2020-03-17T18:15:11.140" LastActivityDate="2020-03-17T18:15:11.140" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33691" PostTypeId="1" CreationDate="2020-03-17T18:25:57.730" Score="0" ViewCount="33" Body="&lt;p&gt;I am working on an acid filtration system and I need to be able to pressurize the effluent from the diaphragm pump such that the acid can be forced through a filter. I am not quite sure if a diaphragm pump can pressurize the influent and also be able to adjust the pressure. Can they do this? Or is there another pump that I should look for? &lt;/p&gt;&#xA;" OwnerUserId="25012" LastActivityDate="2020-03-17T20:25:45.470" Title="Can air-powered diaphragm pumps pump out the effluent at a higher pressure than inputed?" Tags="|pumps|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33692" PostTypeId="1" CreationDate="2020-03-17T18:43:49.357" Score="1" ViewCount="31" Body="&lt;p&gt;I have a fastener with an accessibility problem.  At the time of installation, the fastener is at the bottom of a crowded box with obstructions varying stiffness (Metal Plates and harnessing).  It is possible, but very inconvenient to position and drive the fastener using a &lt;a href=&quot;https://www.grainger.com/product/6TFG2&quot; rel=&quot;nofollow noreferrer&quot;&gt;pick up tool&lt;/a&gt;, then drive the fastener with a long set of 1/4&quot; drive socket extensions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking that a better solution would be a permanently installed drive tool which extended the fastener to the top of the box.  The solution I am imagining would leave the fastener placed permanently, or rather at an earlier stage of assembly, then allow the fastener to be driven without having to fish for the fastener from the top of the box.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Design Considerations&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The grip length of the fastener is unchanged&lt;/li&gt;&#xA;&lt;li&gt;Driving the fastener is the primary consideration&lt;/li&gt;&#xA;&lt;li&gt;A secondary consideration is positioning, the fastener remains in a slot and needs to be positioned in the slot to capture the thread on the far end.   &lt;/li&gt;&#xA;&lt;li&gt;A bonus would be if the coupling allowed the screw to be driven without changing the depth of rod where it is being driven.  This would seem to require a coupling which could trasfer torque and also be telescoping.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;To assist in imagination, I'll say that this is an 10-32x2&quot; socket head fastener, driven from 18&quot; away.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible Solutions:&#xA; - Use an adhesive to bond a hex driver to the fastener then run a dedicated socket or bit extension to where it can be reached at the top of the box.&#xA; - Weld a steel rod to the fastener.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Looking for better ideas and/or products which would make this easy to implement.&lt;/p&gt;&#xA;" OwnerUserId="2716" LastActivityDate="2020-03-17T18:43:49.357" Title="Permanant extension for Fastener" Tags="|design|fasteners|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33693" PostTypeId="2" ParentId="33691" CreationDate="2020-03-17T20:25:45.470" Score="1" Body="&lt;p&gt;In general, a diaphragm pump is a constant volume pump that can raise the pressure of the fluid as it is discharged from the pump, as shown in it's pump curve. You can try to find a pump that outputs the needed pressure, or have equipment after the pump to adjust the fluid's pressure.&lt;/p&gt;&#xA;" OwnerUserId="10344" LastActivityDate="2020-03-17T20:25:45.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33695" PostTypeId="1" CreationDate="2020-03-17T21:39:20.513" Score="1" ViewCount="53" Body="&lt;p&gt;I have a MDOF FEM model in which I perform a modal transient analysis based on the modal eigenfrequencies extracted in a prior step.&#xA;I wish to excite only the first natural frequency of the system via a half sine impulse of 10G. I do not want to see contributions from other modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this possible? I'm not sure how to define the time abscissa value which correlates to the first natural mode. Any suggestions will be appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Best Regards&lt;/p&gt;&#xA;" OwnerUserId="14216" LastActivityDate="2020-03-17T21:39:20.513" Title="Half Sine Impulse in a modal Transient Analysis" Tags="|shock|impact|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33697" PostTypeId="1" AcceptedAnswerId="33716" CreationDate="2020-03-18T02:45:43.473" Score="-1" ViewCount="84" Body="&lt;p&gt;I'm looking for a very very thin wire that does not bend easily. Which kind of material should I look for? The thinnest the wire the better but copper &lt;a href=&quot;https://imall.com/product/50m-0.01mm-Insulating-Copper-Wire-BGA-Motherboard-Fingerprint-Maintenance-Fly-Line-Solder-for-iPhone/banggood.com/1369196/530-225138/en&quot; rel=&quot;nofollow noreferrer&quot;&gt;can be found&lt;/a&gt; in a very thin format (0.01mm) but it bends with very few force applied to it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the least ductile and thinnest wire material I can buy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="25017" LastEditorUserId="25017" LastEditDate="2020-03-18T05:57:07.600" LastActivityDate="2020-03-19T07:24:07.560" Title="What is the less ductile and thinnest wire material?" Tags="|materials|wire|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33698" PostTypeId="2" ParentId="33379" CreationDate="2020-03-18T04:58:41.337" Score="0" Body="&lt;p&gt;The derivative path of a PID controller &quot;predicts&quot; the future error regarding the variable which needs to be controlled and reach a certain reference point. More specifically, the outcome of the derivative path is analogous to the rate of change of the error:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ error = reference - measurement $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example, if the error is quickly decreasing then the output of the derivative would produce a negative value which will then be added to the summation of the outputs of the proportional and integral paths resulting in a reduction of the total control signal. The reduction of the control signal generally means a reduction on the rate of change of the controlled variable (for example the velocity, if position needs to be controlled). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The gain &lt;span class=&quot;math-container&quot;&gt;$\ K_d $&lt;/span&gt; corresponds to the weight the control engineer adjusts to this behavior regarding the derivative path. Increasing the derivative gain &lt;span class=&quot;math-container&quot;&gt;$\ K_d $&lt;/span&gt; means that it is important for the system to react more aggressively to the rate of change of the error. Increasing &lt;span class=&quot;math-container&quot;&gt;$\ K_d $&lt;/span&gt; means the output of the derivative path is larger which, as mentioned above, will reduce the total amount of the control signal, the value of rate of change of the controlled variable (velocity) will decrease and as a result the system will approach more slowly towards the desired reference point &lt;span class=&quot;math-container&quot;&gt;$\ \rightarrow $&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$\ t_r $&lt;/span&gt; increases (lower velocity) &amp;amp; &lt;span class=&quot;math-container&quot;&gt;$\ \%OS $&lt;/span&gt; decreases (controller slows down the system before it reaches the reference point). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the derivative path of the PID controller determines how the system approaches the reference point, whether this happens fast or slowly, and correspondingly reduces or increases the total control signal following the below pattern:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;error decreases &lt;span class=&quot;math-container&quot;&gt;$\ \rightarrow $&lt;/span&gt; output is negative&lt;/li&gt;&#xA;&lt;li&gt;error increases &lt;span class=&quot;math-container&quot;&gt;$\ \rightarrow $&lt;/span&gt; output is positive&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;and the amplitude of the output depends on how fast the error changes. &lt;/p&gt;&#xA;" OwnerUserId="18238" LastActivityDate="2020-03-18T04:58:41.337" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33699" PostTypeId="2" ParentId="33688" CreationDate="2020-03-18T05:23:49.767" Score="1" Body="&lt;p&gt;You determined the torque at A correctly.  You can get the force at B using the same equations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;650 Nm / .1 m = 6500 N.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, it's a lot, 5 times the load, but  not close to where you ended up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The force will go down with the angle, meaning you should be dividing the force by the cosine, not multiplying by the sine (.5), but then you actually used .3 instead of sin(30).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can look at the device as a simple machine.  If you want to directly apply force at B, you are on the wrong end of the lever, and you will be giving yourself mechanical &lt;em&gt;dis&lt;/em&gt;advantage.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A typical jack like this overcomes the load by using hydraulic pressure to convert long throws of the handle into small movements of the lifted load.  Other options for jacks are screw scissors, which use the screw as a simple plane to trade large amounts of rotary motion into linear motion, or ratchet jacks which just use levers to move the load into a higher ratcheted position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you want to use some type of motor, you would typically determine a method of mechanical advantage that let you lift the load within the desired time period.  A earth-moving crane wouldn't notice your load because it's a huge machine designed for moving tons at a time.  Similarly, a clockwork-like mechanism would take years to move such a load.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastEditorUserId="24351" LastEditDate="2020-03-18T05:31:22.953" LastActivityDate="2020-03-18T05:31:22.953" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33700" PostTypeId="1" AcceptedAnswerId="33701" CreationDate="2020-03-18T05:54:03.410" Score="-1" ViewCount="59" Body="&lt;p&gt;I'm designing a block for creating turbulence inside a traditional furnace. I would like to know about a material which is economical and can be put into operation for more than 2 years.&lt;/p&gt;&#xA;" OwnerUserId="25018" LastActivityDate="2020-03-18T06:09:36.753" Title="What material can withstand continuous working temperature of 1000 degree Celsius?" Tags="|materials|manufacturing-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33701" PostTypeId="2" ParentId="33700" CreationDate="2020-03-18T06:09:36.753" Score="1" Body="&lt;p&gt;Most industrial furnaces are lined with &lt;a href=&quot;https://theconstructor.org/building/fire-bricks-properties-types-uses/29377/&quot; rel=&quot;nofollow noreferrer&quot;&gt;fire bricks&lt;/a&gt; to ensure long operating times at high temperatures. Depending on the properties of the bricks, particularly the grade of the &lt;a href=&quot;https://en.wikipedia.org/wiki/Fire_clay&quot; rel=&quot;nofollow noreferrer&quot;&gt;fire clay&lt;/a&gt;, temperatures up to 1640 &lt;span class=&quot;math-container&quot;&gt;$\small\sf{^o}$&lt;/span&gt;C can be tolerated.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Fire clay contains two major constituents- silica and alumina, of which, the silica percentage varies from 60 to 70% and alumina varies from 25 to 35%&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="90" LastActivityDate="2020-03-18T06:09:36.753" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33702" PostTypeId="1" AcceptedAnswerId="33823" CreationDate="2020-03-18T07:07:29.163" Score="3" ViewCount="1555" Body="&lt;p&gt;I am testing out different materials with different specific heat capacities and their correlation with soundproofing (e.g. wood, cork, styrofoam, etc.) but I can't seem to find the specific heat capacity of cardboard anywhere (the average S.H.C. of regular corrugated cardboard).&lt;/p&gt;&#xA;" OwnerUserId="25019" LastActivityDate="2020-11-14T00:00:39.250" Title="What is the specific heat capacity of corrugated cardboard?" Tags="|thermodynamics|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33703" PostTypeId="1" CreationDate="2020-03-18T08:01:53.867" Score="1" ViewCount="65" Body="&lt;p&gt;Suppose that Joe and Sue are neighbors out on the country. The year is 1950. But it could also be 1975. Or even 1999. Even though they live next to each other, and can see each other through the windows, there is still a non-trivial distance between them, especially in winter time with lots of snow and coldness. They don't fancy constantly putting on clothes and walking over to knock on the other's door to ask something.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, they have a habit of calling each other on the telephone for convenience when they need to ask something and don't feel like going outside for whatever reason. They did consider using two tincans with a string between the houses, but they thought it looked too childish, and they couldn't figure out a good way to signal for the other person to pick up their tincan. So they use the normal telephone.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regardless of the year, they are using the &quot;POTS&quot; (Plain Old Telephone System).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When Joe calls Sue or Sue calls Joe, being neighbors, would the signal first travel down to some telephone company far away, and then back to the other person? Or would they get a direct connection going straight through the telephone wire right to the other house and into their telephone?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've tried to understand how telephones work all my life, but never quite managed, so I really don't know if this is a stupid question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It sounds reasonable to me that, given the extremely short distance between the two parties, the signal will travel instantly between them, not triggering any kind of &quot;log&quot; at the telephone company's equipment (thus being free of charge) and also won't allow anyone to tap into this call remotely, unless they are right there between their houses with a physical thing hooked onto the telephone poles' wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a correct assumption by me? Would it be free and private due to the extreme proximity? And if so, when would it stop being free and private? How many houses or what distance triggers a different &quot;route&quot; and the ability for remote wiretapping?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, please note that I'm not talking about whatever digital system is in place now, but what used to be the case with the old telephone system.&lt;/p&gt;&#xA;" OwnerUserId="25020" LastActivityDate="2020-03-18T18:23:08.400" Title="With the &quot;plain old telephone system&quot;, would two neighbors be able to call each other privately?" Tags="|telecommunication|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33704" PostTypeId="2" ParentId="33703" CreationDate="2020-03-18T10:18:55.583" Score="1" Body="&lt;p&gt;Two neighbors living in a city or major town would be able to call each other the &quot;&lt;a href=&quot;https://www.mentalfloss.com/article/62876/10-aspects-old-telephones-might-confuse-younger-readers&quot; rel=&quot;nofollow noreferrer&quot;&gt;Plain Old Telephone System&lt;/a&gt;&quot;. In rural areas however they would have been able, but there would have used the telephone differently to city people because rural regions tended to have &lt;a href=&quot;https://en.wikipedia.org/wiki/Party_line_(telephony)&quot; rel=&quot;nofollow noreferrer&quot;&gt;party lines&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Party lines provided no privacy in communication and when the line was being used by one person, no-one else connected to the line could use the line to make another call until the first person ended their call. However, any on the neighbors could listen to the conversation. Depending on location, &lt;a href=&quot;https://techchannel.att.com/play-video.cfm/2012/6/6/AT&amp;amp;T-Archives-Party-Lines&quot; rel=&quot;nofollow noreferrer&quot;&gt;party lines&lt;/a&gt; exited into the mid 1980s.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2020-03-18T10:46:22.557" LastActivityDate="2020-03-18T10:46:22.557" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33706" PostTypeId="1" AcceptedAnswerId="33718" CreationDate="2020-03-18T11:04:41.310" Score="2" ViewCount="88" Body="&lt;p&gt;I am using GA and MCMC for estimating the parameters of a transfer function. GA returns results that each parameter has one value when minimizing the error function, and MCMC returns the whole chain results for each parameter after the burnin, which gives a statistical distribution of the possible values for that parameter. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first question is: What is the benefit of using MCMC over GA? Which one is better than other in what aspects? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My second question is: Do I get the mean value of each parameter from the distribution of the chain in order to get the fit for the transfer function? If so, do I do that with mean taking the mean value of the distribution? what if the distribution is skewed?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you in advance &lt;/p&gt;&#xA;&#xA;&lt;p&gt;sorry I am new here, I didn't know how to tag this question.&lt;/p&gt;&#xA;" OwnerUserId="25024" LastActivityDate="2020-03-19T10:15:19.017" Title="Compare Genetic Algorithms and Markov Chain Monte Carlo in their statistical properties in parameter estimation task" Tags="|control-engineering|transfer-function|statistics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33707" PostTypeId="1" AcceptedAnswerId="33723" CreationDate="2020-03-18T13:17:58.200" Score="0" ViewCount="35" Body="&lt;p&gt;I have biomedical data for analysis and I am using Genetic Algorithm for parameter estimation for the model of the data. As I read about the system identification, I saw terms like dynamical system. I read further and I am still confused whether the electrical impedance readings of a muscle tissue is considered a dynamical system or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it a nonlinear system? how can I tell? is there a direct rule on whether a biological tissue is a nonlinear system or not?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance&lt;/p&gt;&#xA;" OwnerUserId="25024" LastActivityDate="2020-03-19T23:20:31.330" Title="Is biological tissue a Dynamical System and a Nonlinear System?" Tags="|biomedical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33710" PostTypeId="2" ParentId="12184" CreationDate="2020-03-18T15:56:31.240" Score="4" Body="&lt;p&gt;Since I wasn't satisfied with the answers here, I did some studying, and I thing I can explain, what enables the torque multiplication.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will follow the fluid starting from entering the pump, and explain how it will apply torque to each of the parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, the fluid will enter pump at the center, which will both move the fluid through through the pump, to the outer edge, as well as add rotation to the fluid in the direction of the pump. This will naturally apply a torque to the pump in the direction opposite to the rotation of the pump (this torque is the torque coming from the engine).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, the rotating fluid will move from the outer edge of the pump to the outer edge of the turbine and enter the turbine. In the turbine it's blades are shaped to reverse the rotation of the incoming fluid, while it is pushed to the center. This interaction will apply a torque to the turbine (and from there to the transmission) in the direction of the rotation of the incoming fluid (opposite to engine torque).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Last, comes the critical part to the torque multiplication, the stator. The stator is in the center of the torque converter, between the turbine and the pump. The stator blades will once again reverse the rotation of the fluid coming from the turbine, to match the rotation of the pump. This will apply a torque to the stator in the direction opposite of the one to the turbine (the stator is fixed in place, though, as it's name tells us, so it will not rotate). Exiting the stator, the fluid will repeat it's cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, what is it, that allows the mismatch between the input torque and output torque. It is the torque applied to the stator. At an equilibrium, all the torques in the system will add to zero, so the torque applied to the engine must be exactly the opposite of the sum of the torques applied to the stator and the output. Since the output torque and the stator torque are opposite, the output torque has to be larger than the input torque (by the torque applied to the stator to be exact)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps someone to understand the black magic of torque multiplication. For me, at least, it was a mystery for far longer than I'm willing to admit.&lt;/p&gt;&#xA;" OwnerUserId="25026" LastActivityDate="2020-03-18T15:56:31.240" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33711" PostTypeId="2" ParentId="33703" CreationDate="2020-03-18T18:23:08.400" Score="3" Body="&lt;p&gt;Joe and Sue's calls would go to their exchange where the lines were terminated on an automatic exchange. There is (generally) no electronics between their wall socket and the exchange circuits so there was no way of switching calls.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It was all invented by a funeral undertaker so it couldn't be that hard - could it?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Video links:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=owAjCeHRgMA&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Telephone Exchange (1982) British Telecom Schools Film&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=xZePwin92cI&amp;amp;t=319s&quot; rel=&quot;nofollow noreferrer&quot;&gt;AT&amp;amp;T Archives: The Step-By-Step Switch&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2020-03-18T18:23:08.400" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33713" PostTypeId="2" ParentId="33548" CreationDate="2020-03-18T21:03:15.430" Score="1" Body="&lt;p&gt;So if I understand the question correctly, we are raising a &quot;drawbridge type&quot; door via a motor/gearbox at the pivot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we assume that the door is a uniform shape/density then it has a mass of 113kg (250 lbs) acting at a distance of 1.14m (7'6&quot; / 2) from the pivot. This produces a torque of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$1.14\mathrm{\:m} \times 113\mathrm{\:kg} \times 9.81\mathrm{\:m/s^2} = 1264\mathrm{\:Nm} $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So your motor-gearbox combination will need to produce this amount of torque at zero-speed just to hold the door stationary at floor level. As the door closes the torque required to hold it in each position will reduce - to zero at closed (vertical). However in order to actually close the door some additional torque is required to accelerate it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although we know the maximum acceptable time period to close through &lt;span class=&quot;math-container&quot;&gt;$90^\circ$&lt;/span&gt; (6 seconds) the additional torque required will depend on the level of acceleration at each point in the move (the 'move profile'). For ease of calculation lets assume constant acceleration throughout the 6 seconds; so for the door to rotate through &lt;span class=&quot;math-container&quot;&gt;$\pi/2\mathrm{\:rad}$&lt;/span&gt; (90&lt;span class=&quot;math-container&quot;&gt;$^\circ$&lt;/span&gt;) in 6s then the maximum rotational velocity of the door needs to be &lt;span class=&quot;math-container&quot;&gt;$\pi/6\mathrm{\:rad\:s^{-1}}$&lt;/span&gt;. To achieve that in 6s of constant acceleration needs an acceleration of &lt;span class=&quot;math-container&quot;&gt;$\pi/36\mathrm{\:rad\:s^{-2}}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Achieving that requires a torque calculated based on the moment of inertia, which for our door is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\frac{1}{3} \times 113\mathrm{\:kg} \times 2.28\mathrm{\:m^2} = 196\mathrm{\:kg\:m^2}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Meaning that the additional torque required for acceleration is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$196\mathrm{\:kg\:m^2} \times \pi/36\mathrm{\:rad\:s^{-2}} = 17.1\mathrm{\:Nm}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This move profile will mean maximum speed is at the end of movement, i.e. the door is slamming pretty hard! So in actual fact we probably want to accelerate faster at the start and then decelerate to stationary at the end. This will require a higher initial acceleration (and therefore more torque). On the other hand the mass of the door being supported is decreasing through the closure, so the available torque for accelerating will be increasing. This means that a move profile with low initial acceleration, then increased acceleration through mid-closure and a short period of deceleration at the end, would probably be the best option.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In practice these type of systems are typically not closed-loop controlled (i.e. apply power to the motor and let it do it's thing), with aspects like damping at the end of the stroke being added mechanically. This makes it difficult to know the exact move profile, but from our calculations it seems likely that around &lt;span class=&quot;math-container&quot;&gt;$1300\mathrm{\:Nm}$&lt;/span&gt; of torque from our system is probably the minimum we could get away with. In addition we need to be able to achieve a minimum rotational speed of &lt;span class=&quot;math-container&quot;&gt;$\pi/6\mathrm{\:rad\:s^{-1}}$&lt;/span&gt;, or 12 seconds per complete rotation (5 rpm). This means the motor needs to have a minimum power of:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\pi/6\mathrm{\:rad\:s^{-1}} \times 1300\mathrm{\:Nm} = 681\mathrm{\:W}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we have ignored lots of stuff like mechanical friction you'd probably want to be looking at something more like &lt;span class=&quot;math-container&quot;&gt;$1\mathrm{\:kW}$&lt;/span&gt; with a suitable gearbox to achieve the torque/speed specs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As noted in the comments, springs/counter weights or actuating the door from a different point (e.g. cables connected to the non-pivot end) may also make the motors job considerably easier.&lt;/p&gt;&#xA;" OwnerUserId="157" LastActivityDate="2020-03-18T21:03:15.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33715" PostTypeId="1" CreationDate="2020-03-18T21:23:03.477" Score="2" ViewCount="45" Body="&lt;p&gt;I have parameterized the following common bulkhead tank, which has cylindrical walls (easy enough) and ellipsoidal bulkheads. The upper tank is full of a fluid of density 800 kg/m^3 and the lower is full of a fluid of density 1141 kg/m^3. Dimensions are in m, internal tank pressure (in green) is in 10s of kPa. I assume the shared bulkhead is perfectly rigid &amp;amp; no pressure is transmitted between tanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to calculate the principal stress components in the ellipsoidal bulkheads so that I can obtain full-body radial and longitudinal stress profiles. I can't find any information on how to do this, or (better yet) a generalized approach to determining surface stresses from pressure in non-spherical or non-cylindrical situations. I'm not an engineer by training &amp;amp; I have trouble knowing where to start with this. Help?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5K1Hx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5K1Hx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25032" LastActivityDate="2020-03-18T21:23:03.477" Title="How does one calculate the principial component stresses in an ellipsoidal tank due to head pressure?" Tags="|structural-engineering|fluid-mechanics|modeling|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33716" PostTypeId="2" ParentId="33697" CreationDate="2020-03-19T07:24:07.560" Score="2" Body="&lt;p&gt;For microscopic electrical probe pins, gold-plated &lt;em&gt;tungsten&lt;/em&gt; is used because its stiffness modulus is large- which provides best resistance against buckling under compressive stress.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-03-19T07:24:07.560" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33717" PostTypeId="1" CreationDate="2020-03-19T09:37:05.757" Score="3" ViewCount="14798" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lNfDJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lNfDJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm reading the &lt;a href=&quot;https://en.wikipedia.org/wiki/Flexural_modulus&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wikipedia page&lt;/a&gt; on flexural modulus. It is said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Ideally, flexural or bending modulus of elasticity is equivalent to the tensile modulus (Young's modulus) or compressive modulus of elasticity. In reality, these values may be different, especially for polymers. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;But how can they be different? Young's/elastic modulus is measured in the tensile test, by pulling a sample of the material and measuring the stress/strain response. In the bending test we bend the sample instead, but the fibers at the bottom (as seen in the picture) are in tension. Thus, if we inspect as small sample at the bottom of the beam, it is subject to tensile forces only and it should therefore behave in the same way as if the sample was being subjected to the tensile test. How does the sample know it is part of a beam being bent instead of being a part of a beam being pulled? &lt;strong&gt;How can the two moduli be different?&lt;/strong&gt; The article mentions polymers specifically, but do the moduli differ for metals as well?&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2021-07-27T19:03:50.337" Title="Why would the flexural modulus be different than the elastic modulus?" Tags="|material-science|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33718" PostTypeId="2" ParentId="33706" CreationDate="2020-03-19T10:15:19.017" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 1: What is the benefit of using MCMC over GA? Which one is better than other in what aspects?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As you state, if you are using the &lt;em&gt;Markov Chain Monte Carlo (MCMC)&lt;/em&gt; properly, it will give you the complete probability distribution of your estimated parameters.&#xA;This gives you knowledge about the quality of your estimation, but this is a computationally expensive method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;em&gt;Genetic Algorithms (GA)&lt;/em&gt; will converge to a minimum of your cost function, which can be local minimum.&#xA;Furthermore, there is no knowledge about the quality of the estimation, rather than the cost-function value.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Question 2: Do I get the mean value of each parameter from the distribution of the chain in order to get the fit for the transfer function? If so, do I do that with mean taking the mean value of the distribution? what if the distribution is skewed?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If taking the mean-value of the distribution is a good approximation of your parameter depends on the type of distribution. I would state the in general the bin with the highest probability is the best approximation of your parameter.&#xA;However, if you applied MCMC to estimate your parameters, you can use this technique again to estimate the system response.&lt;/p&gt;&#xA;" OwnerUserId="8216" LastActivityDate="2020-03-19T10:15:19.017" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33719" PostTypeId="1" CreationDate="2020-03-19T16:31:59.620" Score="2" ViewCount="49" Body="&lt;p&gt;I am running a SAP calculation on a proposed building which will be utilising a 8.5 kW ASHP for space and water heating. I am also running a comparative analysis against a mains gas boiler to show the reduction in dwelling emission rate (DER). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have noticed that when specing the heat pump in my model the TER shoots up from 14.49 kgCO&lt;sub&gt;2&lt;/sub&gt;/yr/m&lt;sup&gt;2&lt;/sup&gt; (in the gas powered model) to 21.5 kgCO&lt;sub&gt;2&lt;/sub&gt;/yr/m&lt;sup&gt;2&lt;/sup&gt;. This is having a pretty massive impact on the percentage improvement of DER over TER between models. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can any one explain why this happens? I cannot find an explanation elsewhere and I would hate to overstate the potential improvement the heat pump brings. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers&lt;/p&gt;&#xA;" OwnerUserId="25039" LastEditorUserId="90" LastEditDate="2020-03-19T18:27:15.693" LastActivityDate="2020-03-19T18:27:15.693" Title="Increase in target emission rate (TER) when using a air source heat pump (ASHP) in standard assessment procedure (SAP) calculation" Tags="|modeling|heating-systems|building-design|building-physics|renewable-energy|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33721" PostTypeId="1" AcceptedAnswerId="33722" CreationDate="2020-03-19T17:12:51.113" Score="1" ViewCount="114" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3mdkW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3mdkW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When a crack has been initiated in a material, further loading of the material causes stress concentrations at the crack tip. This high stress causes the material to break at the tip and the crack grows. As the material is loaded again, another stress concentration occurs and the crack grows again. &lt;strong&gt;But why do we need to unload the material in between?&lt;/strong&gt; Why doesn't the crack just propagate immediately from the initial notch? &lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2020-03-19T18:19:12.710" Title="Why does fatigue crack growth require unloading?" Tags="|materials|stresses|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33722" PostTypeId="2" ParentId="33721" CreationDate="2020-03-19T18:19:12.710" Score="2" Body="&lt;p&gt;When the material &quot;breaks&quot;, it's the inter-molecular bonds that are breaking. Once they are able to form a new bond to the next available molecule the crack propagation stops and stress is reduced. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the load is removed, the material is deformed slightly from it's previously unloaded state. Adding load again creates a similar stress compared to the first load cycle causing another cycle of fracture.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Look up brittle vs ductile fracture, those sources will explain why some materials fail instantly while others fail after high number of cycles.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-03-19T18:19:12.710" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33723" PostTypeId="2" ParentId="33707" CreationDate="2020-03-19T23:20:31.330" Score="1" Body="&lt;p&gt;Generally speaking, any mathematical model that is time dependent will be considered a 'dynamic' system, some engineering descriptions will say dynamic systems are described as systems having the ability to store and release energy. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From &lt;a href=&quot;https://en.wikipedia.org/wiki/Dynamical_system&quot; rel=&quot;nofollow noreferrer&quot;&gt;Wiki&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In mathematics, a dynamical system is a system in which a function describes the time dependence of a point in a geometrical space. Examples include the mathematical models that describe the swinging of a clock pendulum, the flow of water in a pipe, and the number of fish each springtime in a lake.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;As far as nonlinear things go. Most real world systems are nonlinear. However, it's difficult to say immediately whether something is, before one goes on to model it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But to give an example from the modelling of a tentacle:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/2gUVi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2gUVi.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Biological real world systems get complicated fast, and end up being inherently, nonlinear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However we often linearize such systems around a operating point (&lt;a href=&quot;https://en.wikipedia.org/wiki/Taylor_series&quot; rel=&quot;nofollow noreferrer&quot;&gt;Taylor Series&lt;/a&gt; for example) once we have a clear understanding of what it may be. Will will drastically reduce your problem with a given amount of (hopefully) small error. &lt;/p&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2020-03-19T23:20:31.330" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33725" PostTypeId="1" CreationDate="2020-03-20T02:22:48.907" Score="1" ViewCount="197" Body="&lt;p&gt;I have recently purchased an ODrive servo controller made for hobby BLDC motors.  Since I had a BLDC motor at home, I did not purchase a dual shaft motor when I got the ODrive.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is if I am to use this BDLC motor as a servo, which side should I affix the encoder to, and which side should I affix to the axis I am planning to rotate (the load)?  I have an 8mm shaft on one side and I have a plate with 4 threaded holes on the other side, and need to choose.  I can 3d print anything needed but do not have access to any milling machine or lathe.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you can see the two parts of the motor, along with the 8mm shaft on one end.  The &quot;STATOR&quot; is on top.  The &quot;ROTOR&quot; is on the bottom.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/i8OER.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i8OER.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here you see the back side of the motor (the &quot;ROTOR&quot;), and note that the shaft does not extend sufficiently to mount something to it, but that there are 4 threaded holes where something can be fixed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ka8VD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ka8VD.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are a couple encoders I have access to (just as FYI, not necessarily the focus of the question).  They are a SICK industrial style and a CUI amt102-v rotary encoder:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Ss9sH.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Ss9sH.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TQl0W.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TQl0W.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21120" LastActivityDate="2020-03-20T02:22:48.907" Title="Given a BLDC motor with one shaft, where should I attach my encoder?" Tags="|motors|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33726" PostTypeId="1" CreationDate="2020-03-20T06:50:44.193" Score="1" ViewCount="586" Body="&lt;p&gt;I understand what a heatsink does, but I cannot wrap my head around how the newly announced Xbox Series X's heat-sink chassis works. From &lt;a href=&quot;https://www.xbox.com/en-US/consoles/xbox-series-x?&amp;amp;ef_id=CjwKCAjwsMzzBRACEiwAx4lLGw4_-dzkecsu7yo-wVffHHtZCipIx5eNgPnYJLLlWrBq5cTQTn2BsRoC7YAQAvD_BwE:G:s&amp;amp;OCID=AID2000034_SEM_ZTFFI6uc&quot; rel=&quot;nofollow noreferrer&quot;&gt;Xbox&lt;/a&gt;:&lt;/p&gt;&#xA;&lt;blockquote&gt;&#xA;&lt;p&gt;Heat-sink chassis&lt;/p&gt;&#xA;&lt;p&gt;A heat sink is a non-electronic, passive mechanical component that has a vital role in the reliability and performance of the Xbox Series X and is integrated directly into the chassis of the console. This unique component of the parallel cooling architecture helps lend to the distinctive design of the console.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/M39pZ.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/M39pZ.jpg&quot; alt=&quot;enter image description here&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;The reason it perplexes me is because it seems to be lacking two main factors to be a heatsink. It does not have rows of fins or spikes to maximize surface area to the air, and it does not appear to maximize the surface area in contact with the hottest sections of the circuit boards (GPU/CPU). Is this just marketing fluff?&lt;/p&gt;&#xA;" OwnerUserId="15383" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2020-04-07T23:40:43.543" Title="How does the Xbox Series X &quot;Heat-sink chassis&quot; work?" Tags="|thermodynamics|heat-transfer|airflow|cooling|heatsink|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33731" PostTypeId="1" AcceptedAnswerId="33735" CreationDate="2020-03-20T17:25:01.053" Score="4" ViewCount="704" Body="&lt;p&gt;This is a follow-up on my previous question here: &#xA;&lt;a href=&quot;https://engineering.stackexchange.com/questions/33688/sizing-a-motor-for-a-floor-jack-like-mechanism&quot;&gt;Sizing a motor for a floor jack-like mechanism&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The floor jack setup is clearly inefficient when it comes to electric actuators. Something similar to a mobile crane would work better, as I'd benefit from the mechanical advantage of the lever formed between point of rotation and actuator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: can the actuator be anchored behind the point of rotation of the lifting arm, e.g. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/qilFf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/qilFf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;NB: 400 mm is the linear actuator's install length. The stroke is 300mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If my understanding is correct, the system should have significant mechanical advantage?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;EDIT&lt;/em&gt;&#xA;Would this setup be similar to what you see on skip trucks? E.g. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/Bpvi2.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Bpvi2.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="14581" LastEditorUserId="14581" LastEditDate="2020-03-21T09:21:50.800" LastActivityDate="2020-03-21T11:51:37.367" Title="Linear actuator behind lifting boom" Tags="|mechanical-engineering|torque|linear-motion|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33732" PostTypeId="1" CreationDate="2020-03-20T20:40:48.597" Score="2" ViewCount="495" Body="&lt;p&gt;Here's an animation of a gearbox and clutch from Wikipedia.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I learned to drive recently, and was surprised at how the clutch worked, engaging gradually as I moved the pedal by fractions of a millimetre.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How is this possible? It looks like the gears would be pretty quickly stripped away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PSyGX.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PSyGX.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25056" LastActivityDate="2020-03-22T15:58:16.130" Title="How can a clutch engage gradually without wearing away very quickly?" Tags="|mechanical-engineering|gears|transmission|" AnswerCount="4" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33733" PostTypeId="2" ParentId="33732" CreationDate="2020-03-20T21:21:36.247" Score="1" Body="&lt;p&gt;This just shows gears and synchromesh... no clutch operation at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The clutch pressure plate moves some 5 millimetres or so but the pedal quite a bit more. The plate will slip until sufficient pressure is applied by the pressure plate. This is where the rapid damage is caused by those who “ride” the clutch ie use the clutch pedal as a footrest or slip the clutch too much especially when reversing or manoevering...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you get experience you can easily feel the bite point and be able to move off smoothly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you get to the point where you can change gear and it is all so smooth the passengers don’t notice - long gone are the comments “did you fill up with kangaroo juice?”&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2020-03-20T22:12:24.807" LastActivityDate="2020-03-20T22:12:24.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33734" PostTypeId="2" ParentId="33732" CreationDate="2020-03-20T21:28:29.430" Score="5" Body="&lt;p&gt;As suggested by the comments, there are a couple of missing factors in this animation. The first is there is no clutch being shown. A typical transmission involves a means of disengaging the power to the transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other factor that is missing is that the parts are rotating (other than from a standing start). The parts being coupled are rotating at different speeds and only need to match up with each other. In a standing start environment, having the clutch engaged means the first set of couplings are not spinning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As referenced, an automotive transmission also has synchronizing devices and can be described as synchromesh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The devices internal to the transmission create rotational pressure to the gear that is about to be engaged, bringing it up to speed or slowing it to match. There is a clear and &lt;a href=&quot;https://www.youtube.com/watch?v=0Bqs-oHBBQk&quot; rel=&quot;noreferrer&quot;&gt;concise video online&lt;/a&gt; about this process.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4q66J.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4q66J.png&quot; alt=&quot;synchromesh image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.matfoundrygroup.com/News%20and%20Blog/How_Does_Synchromesh_Work&quot; rel=&quot;noreferrer&quot;&gt;Another site&lt;/a&gt; which includes also a video is even more illustrative of the process and resembles your animation. One aspect of this particular link of note is that more clearly represents the clutch-like action of the synchronizer ring.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lkUXT.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lkUXT.png&quot; alt=&quot;another synchronizer image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the case of older motor vehicles, prior to the development of synchronizing devices, an upshift was performed by shifting to neutral, removing one's foot from the accelerator pedal (slowing the engine speed) and the clutch which also slowed the input transmission shaft, allowing for a closer match to the next gear. The clutch would be again engaged with minimal clatter/damage/wear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For downshifts, one shifted into neutral, foot off the clutch, apply pedal pressure to the accelerator to speed up the input to the transmission, then clutch again and engage the lower gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many semi-trailer tractors did not have synchronizing for years after the implementation in automobiles. As a youth, I would listen for the downshifts of semi-trucks approaching stop signs and traffic lights. The gnashing of transmission teeth would tell you the driver was not yet skilled in matching input/output speeds.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For such a simple mechanism, much is accomplished for today's manual transmission operator.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-03-20T21:28:29.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33735" PostTypeId="2" ParentId="33731" CreationDate="2020-03-20T23:26:57.880" Score="2" Body="&lt;p&gt;Initial thoughts? I'm not a mechanical engineer but I think you'd be disappointed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/AMRja.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/AMRja.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. A quick pictorial analysis shows that there is almost no reason for the lift arm to move up.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, most of the effort in the linear actuator would be taken up by a force vector from C to A with very little of the effort going into rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VeM4q.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VeM4q.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 2. An online structural analysis program such as that offered by &lt;a href=&quot;http://structural-analysis.com&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://structural-analysis.com&lt;/a&gt; may be useful.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This quick little mock-up uses a pin (1) at the left end of your 400 mm lifting arm with 1 kN applied to the arm offset by 100 mm. The rotation is constrained by a roller joint (2). We can see that only 1/3 of the force is going to do any useful work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll have to play around with the scaling on this. It's not optimised for sub-meter parts so you may need to scale up for good visibility.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/baqsl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/baqsl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 3. At the start of the lift the ram is pulling the crank (green) with a starting angle of 45&amp;deg; or so - and the chains aren't even tight yet. This isn't too bad.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IUAXp.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IUAXp.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 4. The situation appears to be worse when lifting off the truck as the crank (green again) angle is very low. Note, however, that the lift has been flattened against the skip to minimise height and that again the chains are slack. There is no load initially and by the time the chains are taut the crank will be at about 45&amp;deg;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The skip truck design seems to be well thought out.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2020-03-21T11:51:37.367" LastActivityDate="2020-03-21T11:51:37.367" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33736" PostTypeId="2" ParentId="33731" CreationDate="2020-03-21T04:09:39.253" Score="1" Body="&lt;p&gt;Let's call the point where the hinge is welded to the boom D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Scaling CD from your diagram we pick 30mm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That means the mechanical leverage of this gadget is approximately &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F_{boom}= F_{actuator}\frac{30}{400}= 0.075F{actuator}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This gets even worse because we need to consider the reduction of leverage due to the location of D, a nearly ratio of 1/20 efficiency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And on top of that, the boom will lock up after a small rotation where the actuator will hit the boom  &lt;span class=&quot;math-container&quot;&gt;$\theta=\arctan(30/144)= 11.7degrees $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By increasing the length of CD and changing the location of the pivots we can improve the leverage. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-21T04:09:39.253" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33737" PostTypeId="1" AcceptedAnswerId="33739" CreationDate="2020-03-21T11:37:35.150" Score="92" ViewCount="47114" Body="&lt;p&gt;If you let a color TV display something from an unplugged display port, it displays snow, right? Randomly varying black or white pixels on the whole screen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My guess was that the television was decoding the noise from the unplugged wires as a video signal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming this, it makes sense that B&amp;amp;W TVs make B&amp;amp;W snow (they would decode the noise as being a B&amp;amp;W video signal). But then &lt;strong&gt;color TVs&lt;/strong&gt; should make &lt;strong&gt;colored snow&lt;/strong&gt;, right? Shouldn't they decode the noise as being a color video signal?&lt;/p&gt;&#xA;" OwnerUserId="25061" LastEditorUserId="6767" LastEditDate="2020-03-22T16:27:06.843" LastActivityDate="2020-03-27T14:12:25.310" Title="Why do color TVs make black and white snow?" Tags="|electrical-engineering|signal-processing|" AnswerCount="4" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="33738" PostTypeId="1" AcceptedAnswerId="33742" CreationDate="2020-03-21T14:13:01.080" Score="2" ViewCount="142" Body="&lt;p&gt;How does the sine part i.e. &lt;span class=&quot;math-container&quot;&gt;$\dfrac{\ell\sin(\theta)}{2}$&lt;/span&gt; appear in the calculation of shear flow (&lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt;)? I mean, if I'm not mistaken, the sine part is the distance from the x-axis to point D, but how is it derived? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dyg1G.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dyg1G.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25062" LastEditorUserId="1832" LastEditDate="2020-03-21T17:40:33.773" LastActivityDate="2020-03-21T19:09:23.680" Title="Shear flow for unequal leg channel" Tags="|mechanical-engineering|structural-engineering|beam|solid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33739" PostTypeId="2" ParentId="33737" CreationDate="2020-03-21T14:31:50.757" Score="133" Body="&lt;p&gt;If the receiver does not detect the sub-carrier for the &quot;colour burst&quot; signal which is transmitted during the horizontal blanking period the receiver switches on the &quot;colour-killer&quot; circuit so the set reverts to black and white mode. The colour-burst signal - 8 to 10 cycles of 3.85 MHz - is unlikely to be generated by random noise.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/laNEg.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/laNEg.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. The colorburst signal is transmitted on the &quot;back porch&quot; between the horizontal blanking pulse and the start of that line's luminance signal. The colorburst signal is used to synchronise the QAM (quadrature amplitude modulation) oscillator which can hold its frequency accurately enough for the remainder of the cycle and decode the I and Q information. Source: &lt;a href=&quot;https://www.maximintegrated.com/en/design/technical-documents/tutorials/7/734.html&quot; rel=&quot;noreferrer&quot;&gt;Maxim tutorials&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There's plenty of information on the web.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Links:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;a href=&quot;https://www.youtube.com/watch?v=InrDRGTPqnE&quot; rel=&quot;noreferrer&quot;&gt;Compatible Color: The Ultimate Three-For-One Special&lt;/a&gt; by Technology Connections is highly recommended.&lt;/li&gt;&#xA;&lt;li&gt;Signal Processing Stack Exchange: &lt;a href=&quot;https://dsp.stackexchange.com/questions/46401/why-is-tv-static-noise-always-black-and-white&quot;&gt;Why is TV static noise always black and white?&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;li&gt;Reenigne: &lt;a href=&quot;https://www.reenigne.org/blog/why-is-tv-static-in-black-and-white/&quot; rel=&quot;noreferrer&quot;&gt;Why is TV static in black and white?&lt;/a&gt;.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2020-03-23T19:47:29.023" LastActivityDate="2020-03-23T19:47:29.023" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33742" PostTypeId="2" ParentId="33738" CreationDate="2020-03-21T19:09:23.680" Score="2" Body="&lt;p&gt;The calculation you are showing concerns the shear force in the top flange. The shear stress is calculated at a horizontal distance of &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt; from B. To use Zhuravskii's shear stress formula, you need the first moment of area of that part of the top flange with length &lt;span class=&quot;math-container&quot;&gt;$l$&lt;/span&gt;. It has an area of &lt;span class=&quot;math-container&quot;&gt;$tl$&lt;/span&gt; and its centroid has a distance to the &lt;span class=&quot;math-container&quot;&gt;$X$&lt;/span&gt;-axis of the distance from the axis to B, &lt;span class=&quot;math-container&quot;&gt;$|Y_B|$&lt;/span&gt;, plus the distance from B to the center of the bit of flange measured parallel to the &lt;span class=&quot;math-container&quot;&gt;$Y$&lt;/span&gt;-axis, i. e. &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{2}l\sin\theta$&lt;/span&gt;. So that gives a term of &lt;span class=&quot;math-container&quot;&gt;$tl\left(|Y_B|+\frac{1}{2}l\sin\theta\right)$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2020-03-21T19:09:23.680" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33744" PostTypeId="2" ParentId="26094" CreationDate="2020-03-21T22:25:22.037" Score="1" Body="&lt;ul&gt;&#xA;&lt;li&gt;Measure the physical distance between two known points in the garden - the further apart, the better. Let's call this 'M'.&lt;/li&gt;&#xA;&lt;li&gt;Measure the same distance on the scaled drawing. Let's call this 'S'.&lt;/li&gt;&#xA;&lt;li&gt;Scale the drawing by &lt;span class=&quot;math-container&quot;&gt;$ \frac M S $&lt;/span&gt; and you should now have a 1:1 drawing.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2020-03-21T22:25:22.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33745" PostTypeId="2" ParentId="33732" CreationDate="2020-03-22T00:19:34.210" Score="2" Body="&lt;h2&gt;You missed an important part: The clutch&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Your intuition is exactly right that the gears would quickly shred themselves if that was the only thing bringing them into synchronization.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you hadn't noticed, your drawing accounts for only the shifter in the middle of the console. That drawing does not cover that pedal to the left of the brake - the one you were commenting on.  This animation does: It's the huge orange disc.  It is disconnecting the entire powertrain from the engine.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Dh4TF.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dh4TF.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;http://wheelsandroads.com/simple-animation-answers-everything-clutch-works/&quot; rel=&quot;nofollow noreferrer&quot;&gt;src&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the clutch in, and shifted into neutral, the green gear and blue (here) shaft are disconnected &lt;em&gt;at both ends&lt;/em&gt;, and continue spinning only because of their own inertia. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, if you're doing a normal up-shift into the next taller gear, the gears will match up when the blue shaft is spinning slower than it did in the last gear. The blue shaft only weighs a couple of pounds, and will soon spin down all by itself, slowing through and past the optimum speed for sync-up into the next gear.  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Now if you know what you're doing, you can catch it at the instant when the speeds match, and it will mesh right into the next gear.  &lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;However if you are not that fast, there is a mechanism called a &quot;synchromesh&quot; attached to the gear, which will drag down or speed up the blue shaft to get it to the right speed.  It only has to spin up a couple pounds of blue shaft, so it's very lightly built. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The synchromesh is a limited-slip device, and is covered thoroughly in fred_dot_u's answer. It's only designed to spin/slow the couple of pounds of blue shaft; it's not for use as a clutch. So use the clutch pedal! &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The blue shaft is meshed/engaged to the output shaft... but the blue shaft (or to be more precise, the green gear) is &lt;em&gt;not&lt;/em&gt; in sync with the engine.  That's taken care of easily by the orange clutch, when you let your foot off the clutch pedal!  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;That clutch is rated to start a fully loaded vehicle and trailer up a San Francisco hill. Speeding/slowing the engine to match is a trivial job for it.  &lt;/li&gt;&#xA;&lt;li&gt;You can help further by manually syncing engine speed, but it's not that important. &lt;/li&gt;&#xA;&lt;/ul&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Modern fuel-injected engines with long intake runners can manufacture useful torque at quite low RPMs.  As a result, in the hands of a competent driver, a clutch can take a lot less wear &amp;amp; tear than it used to, and could potentially last the life of the car.   While that's a fine thing to aspire to, it's not that big a deal if you clutch wet and sloppy. They're not &lt;em&gt;that&lt;/em&gt; hard to access to replace. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Clutches sit between the engine and transmission. You must pull the transmission off the engine to service or exchange the clutch.  That's not a simple job, because you have to get driveshafts out of the way (a bigger annoyance of FWD cars)... but it's also not a particularly brutal job, and it can be DIY'd.   Unlike a modern automatic transmission, a manual gearbox is liftable. You can pick up an econobox transmission with one hand.&lt;/p&gt;&#xA;" OwnerUserId="11673" LastEditorUserId="11673" LastEditDate="2020-03-22T00:42:04.037" LastActivityDate="2020-03-22T00:42:04.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33748" PostTypeId="1" CreationDate="2020-03-22T10:39:20.800" Score="0" ViewCount="76" Body="&lt;p&gt;Does anyone know the required strength of the mounting points of a car seat? I'm working on a new sliding mechanism and don't want to damage the integrity of the seat mounting. &quot;Child seats&quot; have rather poisoned the search results for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;No points for &quot;don't do it&quot; or &quot;buy one in&quot; - what I'm building simply isn't available - a 2-3' track extension into the rear of the vehicle. The extension is only while the vehicle is stationary so it's really down to the material which will probably be a little weaker than the thick, folded steel of the regular runners. I'm just looking for a baseline I can work from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mike.&lt;/p&gt;&#xA;" OwnerUserId="24671" LastActivityDate="2020-04-04T09:09:05.610" Title="Car seats mounting requirements" Tags="|automotive-engineering|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33749" PostTypeId="2" ParentId="6958" CreationDate="2020-03-22T10:42:25.633" Score="0" Body="&lt;p&gt;I had the very same doubt, but the things got clear once I looked an animation &lt;a href=&quot;https://upload.wikimedia.org/wikipedia/commons/f/f5/Radial_engine_timing-small.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://upload.wikimedia.org/wikipedia/commons/f/f5/Radial_engine_timing-small.gif&lt;/a&gt;.&#xA;Since the 2 plates where all the rods are attached isn't blocked on the crankshaft, but instead it's mounted on bearings, if you had all the rods on these 2 plates running on bearings, at any fire of any cylinder these 2 plates would try to rotate freely over the crankshaft and mess up the entire engine. So you need 1 fixed con-rod in order to avoid this and actually discharge the force on the crankshaft.&lt;/p&gt;&#xA;" OwnerUserId="25081" LastActivityDate="2020-03-22T10:42:25.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33752" PostTypeId="2" ParentId="33732" CreationDate="2020-03-22T15:58:16.130" Score="0" Body="&lt;p&gt;It depends on what you mean by &quot;gradually&quot; ? For an automobile if it slips (much) less than a second , wear is minimal. As a boy I had an oversize clutch , I let it slip ( less than a second, but too long ) and burned off the lining in a month. I slipped it because spinning the back wheels seemed too aggressive - wrong! Slippage , especially between gears should be very brief. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-03-22T15:58:16.130" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33753" PostTypeId="2" ParentId="33748" CreationDate="2020-03-22T16:03:53.377" Score="2" Body="&lt;p&gt;In the US there are codes and standards covering hundreds of pages of coded as to type and materials and standard tests. And codes for even the huge testing machines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically it's a huge liability can of worms.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are codes on chemistry, fire resisting, structural strength, alergy tests, bolts and weld, you name it, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can start the search from :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;49 CFR § 571.209 - Standard No. 209; Seat belt assemblies.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many hyperlinks on the code that you should check.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also need to consult with a specialised attorney. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the link. &lt;a href=&quot;https://www.law.cornell.edu/cfr/text/49/571.209&quot; rel=&quot;nofollow noreferrer&quot;&gt; 49CFR&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-22T16:03:53.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33754" PostTypeId="1" CreationDate="2020-03-22T17:04:32.273" Score="1" ViewCount="726" Body="&lt;p&gt;I'm doing a project for thermodynamics where I have to measure the power (work over time) of a Stirling engine for varying temperature differences. I don't necessarily need an exact answer (as in: ... Watt), but a &quot;relative&quot; measurement would also work (i.e. when the temperature difference between the hot end and the cold end is doubled, the work increases by a factor of ...).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've thought of filming the engine run in slow-mo, and then using the RPM of the Stirling engine to calculate power, but I could not quite work out how to do this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions?&#xA;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="25088" LastEditorUserId="90" LastEditDate="2020-03-22T18:27:47.093" LastActivityDate="2023-09-06T02:11:57.400" Title="How to measure the power output of a stirling engine" Tags="|mechanical-engineering|thermodynamics|measurements|experimental-physics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33755" PostTypeId="1" CreationDate="2020-03-22T18:15:22.850" Score="4" ViewCount="108" Body="&lt;p&gt;I am not very familiar with the mechanics of neither a car nor a medical ventilator. Considering global coronavirus pandemic and shortage of medical ventilators in many countries, I saw some news suggesting that automakers are starting to manufacture medical ventilators, or the leaders of countries are requesting them to do so (e.g. &lt;a href=&quot;https://abcnews.go.com/Business/automakers-offer-build-ventilators-us-faces-critical-shortage/story?id=69689489&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt; or &lt;a href=&quot;https://www.cnet.com/roadshow/news/general-motors-ventilators-ventec-coronavirus-spread/&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;). So I was wondering if there is some overlap or similarity in &lt;strong&gt;production engineering&lt;/strong&gt; of a car and a medical ventilator that makes automakers as the best candidate for making medical ventilators. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In other words, &lt;strong&gt;from an engineering viewpoint&lt;/strong&gt;, what makes automakers suitable for engineering and manufacturing medical ventilators at this moment?&lt;/p&gt;&#xA;" OwnerUserId="25089" LastActivityDate="2020-03-25T23:04:20.213" Title="Why automakers are/should be making medical ventilators?" Tags="|automotive-engineering|biomedical-engineering|product-engineering|medical-devices|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33756" PostTypeId="1" CreationDate="2020-03-22T19:07:35.830" Score="0" ViewCount="30" Body="&lt;p&gt;We have a body that goes underwater (let's say a whale) with magnetometer and accelerometer data. I also have GPS, date, depth, pressure and temperature data.&#xA;I want to know if there is any simple way (Matlab code most preferably) to combine data from magnetometer and accelerometer to find a way to estimate speed of the body (whale). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have thought of 2 ways to perform this but I don't know how to proceed:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Differentiate depth to determine speed of descend&lt;/li&gt;&#xA;&lt;li&gt;Use acceleration and magnetic data to determine pitch, roll, yaw (I prefer this method).&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Please feel free to provide a link if there is an already-existing matlab code for all of this. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Feel free to ask question so I can further explain. No gyroscope data&lt;/p&gt;&#xA;" OwnerUserId="25094" LastEditorUserId="110" LastEditDate="2020-03-23T00:00:23.580" LastActivityDate="2020-03-23T00:00:23.580" Title="How to design a method to estimate vx=speed in direction of swimming" Tags="|measurements|sensors|acceleration|positions-measurement|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33757" PostTypeId="2" ParentId="33755" CreationDate="2020-03-22T19:29:44.097" Score="3" Body="&lt;p&gt;I think you could hack any poppet valve engine block to create a rapid pressure swing adsorbtion machine, but better to start with a block designed as a compressor in the first place. Lubricating oil sequestering would be an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most people don't really understand the relationship between manufacturing, logistics, and assembly. They probably think they can just switch the assembly line over, but where are the parts coming from?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A friend of mine worked with an eye surgery group that did a lot of free work in third world countries. He was the lead guy who arrived a couple weeks ahead of the surgeons; and he had to build the surgery from scratch with zero budget - just using local talent and resources. He had saddlemakers build a bunch of bellows ventilators from wood and leather, just like the old smithy bellows. They were used with intubation under general anesthesia. It's not rocket science.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2020-03-22T19:29:44.097" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33758" PostTypeId="1" AcceptedAnswerId="33760" CreationDate="2020-03-22T20:14:15.250" Score="0" ViewCount="99" Body="&lt;p&gt;Consider we have a second order system (plant) described by the following transfer function:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ P(s) = \frac{b_0}{s^2+a_1*s+a_2} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which is controller by a PD-controller:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ C(s) = K_p+K_ds $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to study the impact that the external noise has to the system, I have come across the so called noise sensitivity function which is described by the below equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ N(s) = \frac{C(s)}{1+P(s)C(s)} $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, deriving &lt;span class=&quot;math-container&quot;&gt;$\ N(s) $&lt;/span&gt; produces an improper transfer function with degree of numerator being &lt;span class=&quot;math-container&quot;&gt;$\ 3 $&lt;/span&gt; and of denominator being &lt;span class=&quot;math-container&quot;&gt;$\ 2 $&lt;/span&gt;. This results, for example, to the imcapability of obtaining a step response in order to study the behaviour. Is the definition of the noise sensitivity function wrong or am I missing something ? &lt;/p&gt;&#xA;" OwnerUserId="18238" LastActivityDate="2020-03-23T01:10:29.213" Title="Noise sensitivity transfer function" Tags="|control-engineering|control-theory|transfer-function|noise|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33760" PostTypeId="2" ParentId="33758" CreationDate="2020-03-23T01:10:29.213" Score="2" Body="&lt;p&gt;This is because your controller is also improper. A common way to correct this is to add a low-pass filter to the derivative (effectively making it a high-pass filter). It can be noted that the PD controller could then also be seen as a lead-lag filter.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, if you are only interested in the step response you could also use that a step is the integral of an impulse. So you could add an integrator to the system, which raises the order of the denominator, and use impulse instead of step on the causal modified system.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2020-03-23T01:10:29.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33761" PostTypeId="2" ParentId="33737" CreationDate="2020-03-23T01:35:09.113" Score="16" Body="&lt;p&gt;The short and academically correct answer: a random noise signal is random. The lack of organization precludes that there is no separate color carrier present, and the set goes, as is already answered, into b/w mode. For split seconds there might be randomly something that looks like a color carrier might be present, then there might be a quick colorish flicker.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;yes, you do not see wrong: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;while the generated signal is strictly b/w there are reasons that the academical requirement for b/w noise is not really happening: technical reasons and physiological reasons.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A little ramble of why we still see sometimes color in real random noise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;physiological reasons: 'optical illusion'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;reasons that have to do with improper function of an analog TV&lt;/p&gt;&#xA;&#xA;&lt;p&gt;reasons that have to do with improper function of analog to digital conversion that can make random noise appear tinted on a digital TV (a 'should not happen' that happens actually all the time on cheap TVs, there is a reason for 'cheap' and 'costly' digital TVs...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's go, or just stop here if you like... everything has basically been said, this is just for your enjoyment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;pink artifacts: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;'snow', when we see it: &#xA;our eyes do add color tints , as they do add 'correlations'. Some see faces and stuff (TV people...maybe talk to someone about it... ;-D  )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;there are some optical illusions out there that move some b/w patterns and we (at least I do, don't have my bionic eyes yet..) do see color although there is none at all on the moving pattern&#xA;google for 'optical illusions'&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Technical reasons why 'should be b/w noise' may have some hues and colored pixels:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;color in random noise on an analog color TV (or computer monitor):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Reminding: noise from 'no signal' or from a blank analog magnetic medium on an analog TV cannot have a color signal which would require organization of the noise into frames of certain durations ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are instances where a color TV will render random noise (b/w) as color: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The color convergence is out of alignment and the three (r/b/g) guns focus slightly off center of the color triple on the screen (instead of 100% r/b/g it gives a little tint), either always, or per pixel and randomly.&#xA;Stationary tints mean you need to demagnetize, random tints per pixel mean there is noise on the beam control.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&quot;I see color noise on a digital TV...&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A digital TV has no noise, it works or not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A rendered analog signal (from a VCR empty tape or from an analog tuner / receiver) will be b/w at the origin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the analog signal is converted to a digital signal and here trouble can lurk in the detail:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the conversion from analog to digital, the above rules for analog signals do apply: no blanking period means no color signal went in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your analog inputs to a digital TV or monitor can come from&lt;/p&gt;&#xA;&#xA;&lt;p&gt;dual tuners: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;even if there are no analog TV stations you might have an analog signal from some arcade games, from a camera abusing of an analog TV channel, or an old VCR you light into your digital TV via analog channel 2 or 3 (you need to know what a rotary dial phone is to understand this....). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again: academically these should be strictly b/w after conversion from analog to digital.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some reasons how conversion imprecision can add color to 'white' noise:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bad converters used to convert from analog to digital can include 'dust' in the rapidly generated digital pixel value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is a software issue in the converter. &#xA;A 0 is not exactly zero , a one is a little less than all bits on, and some bits are picked sometimes in the middle. &#xA;When I say here and from here on '0' and '1' I am talking about 'all bits 0' and 'all bits 1' in the per pixel color depth value. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bad video a/d converter designs use FP units on fast and cheap general purpose chips: again: 0 is not exactly zero ever and 1 is at most .999999. they do some mantissa tricking to process, e.g., a 24 bit pixel signal through an FP unit which is a general purpose, cheaper (and faster, vectorized) chip than a real video converter chip.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This creates a hue that is different per pixel and per instance (the same digital pixel lights up in slightly different hues each time it is used).&lt;/p&gt;&#xA;" OwnerUserId="25099" LastActivityDate="2020-03-23T01:35:09.113" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33762" PostTypeId="1" CreationDate="2020-03-23T02:35:26.670" Score="1" ViewCount="85" Body="&lt;p&gt;My grandfather and I are running some experiments down the river, however we don't have expertise in engineering so I bring up this question to the community and hope someone could help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How many kWh a turbine with the following features would generate?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Jp7NYm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Jp7NYm.jpg&quot; alt=&quot;Just a simple prototype&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/XxH4Zm.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XxH4Zm.jpg&quot; alt=&quot;Just a simple prototype&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These images are just a simple prototype which does not apply some of the following features:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Shape: Reel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Length: 1.50m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diameter: 1.20m. (Including its palettes and axis which has the thickness of 20cm.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;10 Longitudinal/helical palettes with length of 1.50m., thickness of 2cm. and height of 50cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Total weight: 50kg. (Palettes and axis included for torque boost.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow rate (amount flowing through the device): 3m³/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;River conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Flow rate: 209.000 m³/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Speed: 2 meters per second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Depth: 40m.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be installed 2 meters beneath the river surface, placed horizontally.&lt;/p&gt;&#xA;" OwnerUserId="25070" LastEditorUserId="25070" LastEditDate="2020-03-27T23:29:47.453" LastActivityDate="2020-03-27T23:29:47.453" Title="How many kWh this turbine would generate?" Tags="|mechanical-engineering|structural-engineering|control-engineering|hydraulics|turbines|" AnswerCount="0" CommentCount="13" ContentLicense="CC BY-SA 4.0" />
  <row Id="33763" PostTypeId="1" CreationDate="2020-03-23T05:28:25.590" Score="-2" ViewCount="48" Body="&lt;p&gt;I was solving a mechanics sum&lt;/p&gt;&#xA;&#xA;&lt;p&gt;An athlete running a 100m run starts from rest and reaches the maximum speed in a distance of 15 m. He runs the remaining distance with that velocity and reaches the finish in 9.6. Calculate the initial acceleration and maximum velocity acquired by the athlete.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kindly help me to understand the concept of this sum&lt;/p&gt;&#xA;" OwnerUserId="24685" LastEditorUserId="1832" LastEditDate="2020-03-23T15:39:21.953" LastActivityDate="2020-03-23T15:39:21.953" Title="Maximum velocity of an athlete" Tags="|mechanical-engineering|mechanisms|kinematics|" AnswerCount="1" CommentCount="3" ClosedDate="2020-03-23T15:38:34.050" ContentLicense="CC BY-SA 4.0" />
  <row Id="33764" PostTypeId="2" ParentId="33763" CreationDate="2020-03-23T07:11:35.310" Score="1" Body="&lt;p&gt;The runner has two types of movement: accelerating from 0 to &lt;span class=&quot;math-container&quot;&gt;$ v_c \quad a= \text{acceleration}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call the constant speed Vc. So the average speed on the first 15m is half of this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$9.6=\frac{15}{Vc/2}+\frac{85}{Vc}=\frac{2*15+85}{Vc}=115/Vc  \\ 9.6Vc=115 \quad Vc=115/9.6=11.97m.s$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-23T07:11:35.310" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33765" PostTypeId="2" ParentId="33754" CreationDate="2020-03-23T11:26:14.907" Score="2" Body="&lt;p&gt;You need a load to calculate power. Meaning you need to move something or generate something.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Traditional steam and internal combustion engines used brakes to calculate torque, which was then converted to power.  Modern automobile dynomometers uses either a load cell to measure the torque, or measure the speed of the dyno cylinder and calculate the power it takes to accelerate the mass of the cylinder. My guess is that unless you build or buy a large Stirling engine, any load you out on it will make it stop.  The typical demonstration devices you can buy to run on top of your coffee mug will not work well enough for this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that powering a generator is your best bet.  You will need a large enough engine to generate a measurable amount electricity.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-23T11:26:14.907" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33766" PostTypeId="1" CreationDate="2020-03-23T12:17:32.240" Score="1" ViewCount="55" Body="&lt;p&gt;I'm trying to understand in depth the behavior of progressive scan TV monitors (LCD/LED - Digital Television). From what we already know, the tv's screen scans the image row by row (as rolling shutter cameras does), sequentially from top to bottom. However, I would like to know a basic time schedule that can help me answer the following questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What is the definition of the duration of the &lt;strong&gt;scanning of one row&lt;/strong&gt;, and how is it related to the &lt;strong&gt;response time&lt;/strong&gt; of the screen?&lt;/li&gt;&#xA;&lt;li&gt;Does the scanning of one frame happens before the previous frame scan ends? In other words, Does this scanning occurs in an &lt;strong&gt;overlapped&lt;/strong&gt; manner between consecutive frames? Or there is a wait time between consecutive frames, so the screens holds one frame during a &lt;strong&gt;short amount of time&lt;/strong&gt; before starting the transition to the next frame. In this case, what is the definition of that time?.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="25106" LastEditorUserId="25106" LastEditDate="2020-03-24T11:35:15.567" LastActivityDate="2020-03-24T22:00:44.183" Title="How progressive scan televisions works in depth?" Tags="|telecommunication|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33769" PostTypeId="2" ParentId="33748" CreationDate="2020-03-23T15:56:40.897" Score="1" Body="&lt;p&gt;Going from the seat belt papers, the highest figure mentioned is 46,000N which is kind of helpful in that it gives me a starting point on raw materials but not manufactured solutions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I may have to modify my design to bypass the problem completely. A hefty, steel latch on either side of the seat transferring the force from the base to the seat mounts (both known to be within spec) should mean that whatever changes I put in between do not become the weakest link in the assembly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;[Edit:]&#xA;I've discovered a work-around. Unwin rails (aka L-Track) is the stuff they put in the back of ambulances so is certified for collisions. They also lock by simply dropping a washer into the required position so 4 linear actuators will do the trick. In addition, a relatively simple linkage will allow them to fold down when not in use.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9iN3B.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9iN3B.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;Now I'm trying to figure out whether 3mm, T6 aluminium C channel can take a 300lb load without tearing at the bolt holes, if not I may have to go with steel.&lt;/p&gt;&#xA;" OwnerUserId="24671" LastEditorUserId="24671" LastEditDate="2020-04-04T09:09:05.610" LastActivityDate="2020-04-04T09:09:05.610" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33771" PostTypeId="1" CreationDate="2020-03-23T18:38:08.243" Score="-1" ViewCount="45" Body="&lt;p&gt;I am working on a one-wheel project, I would like to use an electric go kart motor chain driven to a 30&amp;nbsp;cm diameter wheel and get 90&amp;nbsp;kg at least 13&amp;nbsp;m/s with an acceleration of 3 m/s&lt;sup&gt;2&lt;/sup&gt;, I would like to be able to climb 20% grade hills. Can someone help me calculate how much motor wattage is required so I can decide on motor and battery?&lt;/p&gt;&#xA;" OwnerUserId="25111" LastEditorUserId="1832" LastEditDate="2020-03-24T20:31:27.483" LastActivityDate="2020-03-24T20:32:39.373" Title="how to determine how much speed a motor will produce under load?" Tags="|mechanical-engineering|electrical-engineering|motors|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33776" PostTypeId="1" AcceptedAnswerId="33777" CreationDate="2020-03-23T20:33:36.107" Score="1" ViewCount="176" Body="&lt;p&gt;I should have produced a custom designed bottle, but metal molds are some expensive and because we don't require too many products (we just need 100-200 piece), I need a way to reduce the cost. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I thought that, maybe I can use wooden mold for this. But I don't have experience about that. Is there anyone who has tried/used this method or anyone who has knowledge about that? Is it correct method?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Specs&lt;/strong&gt;: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Bottle thickens can be same as any water bottle (you can see examples &lt;a href=&quot;https://www.google.com/search?q=5%20Liter%20Natural%20Rectangular%20bottle&amp;amp;sxsrf=ALeKk00WlBGtRl2qDDokZT2_JJ0tzVzVpA:1584994438295&amp;amp;source=lnms&amp;amp;tbm=isch&amp;amp;sa=X&amp;amp;ved=2ahUKEwixpNyztLHoAhUTj3IEHS_8C6AQ_AUoAXoECGMQAw&amp;amp;biw=1532&amp;amp;bih=673&quot; rel=&quot;nofollow noreferrer&quot;&gt;in this search&lt;/a&gt;).&lt;/li&gt;&#xA;&lt;li&gt;Color can be clear or white (clear is preferred)&lt;/li&gt;&#xA;&lt;li&gt;Some inlet and outlet necks needed (so mold needs to be some complex)&lt;/li&gt;&#xA;&lt;li&gt;Bottle will be 5 to 7 liters with the shape of like rectangular prism&lt;/li&gt;&#xA;&lt;li&gt;Material will be derivative of polyethylene probably&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1488" LastActivityDate="2020-03-23T22:31:08.987" Title="Can wooden mold be used for blow molding?" Tags="|product-engineering|molding|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33777" PostTypeId="2" ParentId="33776" CreationDate="2020-03-23T21:15:45.033" Score="2" Body="&lt;p&gt;You will have some of the same considerations of creating a mold when using wood as when using metal. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Underhangs/undercuts are to be avoided.&#xA;Corners are to be radiused, no sharp edges.&#xA;Alignment of mold halves must be ensured.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your listed considerations, it doesn't appear that a wood mold, especially one made from hardwood would be a problem. With such a small item count, one can expect that you also do not need full automation. Manual processing of the bottle making will certainly save money.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Consider that you'll need someone to create the mold halves for you. If you have an existing positive model, it can be used to create the tool paths for a CNC router, although the size you've referenced would mean a router with a substantial thickness capacity. Hobby CNC routers are often limited to around 100 mm or so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finding someone to make the molds in wood will be your most challenging component.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Added below in response to comment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Manual processing means opening the mold by hand, inserting the envelope of plastic, closing and clamping the mold and activating the injection of air. Placing the mold into a heated environment is also included in this sequence, as is removing it and de-molding the part.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is to contrast an automated processing system where the mechanicals are constructed to be mostly hands-off. Push the button and everything happens at hundreds of parts per hour.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With respect to the type of wood, you'll likely have to experiment. Close-grained hardwoods will be strongest and most heat resistant as well. You'll have to test for heat based on the melting point or glass-transition temperature of your selected plastic. As with metal molding, pre-heat the mold prior to inserting the plastic, use appropriate safety gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Mold lifetime and other durability considerations will again depend on your selection of wood, as well as local availability of this material. As you've noted, you have CNC capability, which means a dozen molds aren't going to be much more expensive than one, in terms of material cost.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2020-03-23T22:31:08.987" LastActivityDate="2020-03-23T22:31:08.987" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33778" PostTypeId="1" AcceptedAnswerId="33796" CreationDate="2020-03-23T22:35:20.363" Score="6" ViewCount="1944" Body="&lt;p&gt;My generator (a Fermont MEP-803A, 4 cyl diesel, 10kW rated) can run in three distinct modes:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Single-phase 120V, in which one terminal is hot, and another is neutral&lt;/li&gt;&#xA;&lt;li&gt;Single-phase 120V/240V, in which two terminals are hot, and another is neutral. The difference between the terminals is 240V, and each one to neutral is 120V.&lt;/li&gt;&#xA;&lt;li&gt;Three-phase 208Y, in which three hot terminals are each 208V apart from each other, and each one is 120V from the neutral terminal.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I need to run a bunch of lights that all need 120V. So the obvious thing would be to just use the 120V mode. But I'm thinking it may be better (more efficient maybe? less vibration maybe?) to run it in three-phase mode, with the load balanced among the three hot terminals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My thinking, based on my coarse understanding of generators, is that in three phase operation, the torque should be more homogeneously distributed over the course of a single turn of the engine, so there should be less vibration and more equality in effort among the cylinders. In short, I'm guessing that three-phase generator operation would have the same advantages three-phase motors have over single-phase motors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is my thinking correct here? If not, why not? Conversely, are there any disadvantages to three-phase operation over single-phase operation?&lt;/p&gt;&#xA;" OwnerUserId="25141" LastEditorUserId="25141" LastEditDate="2020-03-23T23:11:42.820" LastActivityDate="2020-03-26T23:45:38.790" Title="Three-phase versus single-phase operation of a generator" Tags="|torque|vibration|energy-efficiency|generator|diesel|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33779" PostTypeId="2" ParentId="33778" CreationDate="2020-03-23T23:23:05.047" Score="6" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;But I'm thinking it may be better (more efficient maybe? less vibration maybe?) to run it in three-phase mode, with the load balanced among the three hot terminals.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4TsG7.gif&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4TsG7.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. The load is constant through the generator cycle. Source: T. Davies - website not found.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;My thinking, based on my coarse understanding of generators, is that in three phase operation, the torque should be more homogeneously distributed over the course of a single turn of the engine, so there should be less vibration and more equality among in effort among the cylinders. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You are correct. The beauty of three-phase is that the torque presented by a balanced three-phase load is constant through the full cycle.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In short, I'm guessing that three-phase generator operation would have the same advantages three-phase motors have over single-phase motors.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wpkJE.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wpkJE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 2. The three-phase voltage waveforms are 120&amp;deg; apart. (Graphic by author.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/l3VGO.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/l3VGO.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 3. The &lt;strong&gt;power&lt;/strong&gt; produced by each phase will be given by P = I&lt;sup&gt;2&lt;/sup&gt;R so the powers are all positive. (This assumes a resistive load.) Note that the sum is constant which means that the load on the generator is constant through the full cycle. (Graphic by author.)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Is my thinking correct here? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Conversely, are there any disadvantages to three-phase operation over single-phase operation?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Three-phase is efficient in its use of copper. You will need four conductors rather than two but they will be lighter. Technically, if you balance the load well you can reduce the neutral to zero or size it for your worst case imbalance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only downside is a slight increase in complexity and the need for fusing on three phases instead of one.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2020-03-23T23:23:05.047" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33780" PostTypeId="2" ParentId="33771" CreationDate="2020-03-23T23:26:24.777" Score="0" Body="&lt;p&gt;A cart,  once it gets to steady speed will have to fight only  tires kinetic friction which is a number in the order of 1%, mechanical friction, drag produced the air passing around it. .&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All of these factors depend on your design and the efficiency of the individual components.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Typically electric carts use one kilowatt power.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many vendors of 48 volt motors and batteries and cart assemblies have some data on the performance of their kits.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-23T23:26:24.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33781" PostTypeId="1" CreationDate="2020-03-24T03:33:43.527" Score="1" ViewCount="95" Body="&lt;p&gt;I tried to apply full state feedback on DC motor (with no load) and it works fine in the simulation on Simulink&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But the question is if I have a load applied on the motor (let’s say constant load or any thing else) the system in this case has two inputs voltage and torque load (Tl).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can control only one input (voltage) and I have no control on the other input (Tl), So in this case what can I do in order to be able to perform full state feedback ?&lt;/p&gt;&#xA;" OwnerUserId="25163" LastActivityDate="2022-06-05T13:00:05.110" Title="Applying full state feedback on DC motor with torque load" Tags="|control-theory|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33782" PostTypeId="1" AcceptedAnswerId="33784" CreationDate="2020-03-24T03:53:05.707" Score="1" ViewCount="2412" Body="&lt;p&gt;I have a well with a 1.5 HP pump. The pump is rated for 240V input. At the moment, the only nearby available power is 4-wire 208Y 3phase. I am considering connecting two of the three phases to the well pump, so it receives 208V where it is expecting 240V.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To what extent, and by what mechanism, will it hurt the pump if it is powered by 208V instead of 240V? Are there warning signs I can observe (like something electrical I can measure, or something I can observe about the flow of water) that would indicate whether this lower voltage is causing problems?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;More generally, when an AC motor says it's rated for a certain voltage, without specifying a range, how tight is that calibration? Like, what is the &quot;probably not much worse&quot; voltage range? My intuition says &quot;ohh eee ah around 15%&quot; but I'm realizing I have no actual basis for that sentiment.&lt;/p&gt;&#xA;" OwnerUserId="25141" LastEditorUserId="25141" LastEditDate="2020-03-24T14:56:17.567" LastActivityDate="2020-03-24T17:11:43.890" Title="Running a well pump rated for 230V on 208V power" Tags="|electrical-engineering|motors|torque|pumps|current|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33784" PostTypeId="2" ParentId="33782" CreationDate="2020-03-24T09:46:25.617" Score="2" Body="&lt;p&gt;The 1.5 HP motor is built to draw a specific amperage at full load at 240V.  Because power equals voltage times current, a lower than rated voltage will draw more amps through the motor.  So if your supply is 208, your current will increase by a ratio of 240/208, or about 15%.  Low voltage/high amps is a common failure mode for hand power tools being used with undersized extension cords.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the motor is under a light load, you will likely never notice.  But if it is running at full load, you might find it overheating, tripping overloads, or just generally failing early.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastEditorUserId="25141" LastEditDate="2020-03-24T17:11:43.890" LastActivityDate="2020-03-24T17:11:43.890" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33785" PostTypeId="2" ParentId="33778" CreationDate="2020-03-24T10:00:33.683" Score="1" Body="&lt;p&gt;@Transistor's answer is quite complete and correct, but I feel a bit more complex than what you require.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only difference in how you pull power off of your generator is in how the output windings are connected to a plug.  Each of the three windings will have the same rating whether you pull 1, 2, or all 3 phases.  So if your total load (the bunch of lights) is less than the single phase rating it really doesn't matter at all.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason it doesn't matter is that your diesel prime mover will never notice how the power is being made off the generator.  Three phase power is great and works very well, but in this case the diesel engine is the thing that makes power in strange pulses.  Evening out the load in nice 120 degree phases won't matter a bit to an internal combustion engine.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-24T10:00:33.683" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33786" PostTypeId="1" CreationDate="2020-03-24T11:42:34.493" Score="4" ViewCount="76" Body="&lt;p&gt;Why is preferred on the construction of big stadium roofs and of cooling towers, per example?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the advantage of this geometric figure that makes it more economical?&lt;/p&gt;&#xA;" OwnerUserId="5825" LastActivityDate="2020-03-24T14:09:32.617" Title="Why is a paraboloid hyperboloid shape preferred on some structures?" Tags="|structural-engineering|structures|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33787" PostTypeId="2" ParentId="31723" CreationDate="2020-03-24T12:04:41.150" Score="1" Body="&lt;p&gt;As briefly mentioned above, you need to either connect the caliper to one of the software &quot;masters&quot; offered by Sylvac, or use the caliper in &lt;code&gt;HId mode&lt;/code&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Select your model&lt;/strong&gt; from this page: &lt;a href=&quot;https://www.sylvac.ch/download-en/manuals/hand-tools&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sylvac Hand Tools&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Download the appropriate manual&lt;/strong&gt;; the other answers here do not refer to your appropriate model manual.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Enable the HId mode&lt;/strong&gt; through a sequence that might look like this below. Please note that some calipers don't have HId mode.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XrBF8.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XrBF8.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25181" LastActivityDate="2020-03-24T12:04:41.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33788" PostTypeId="2" ParentId="33786" CreationDate="2020-03-24T14:09:32.617" Score="4" Body="&lt;p&gt;Assuming that you mean a &quot;Hyperbolic Paraboloid&quot;, the answer is that it's easy to make from common construction materials due to the fact that the surface can be made up of straight lines.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other reasons to choose a shape like this over e.g. a flat or pitched roof (aesthetics, drainage, etc.), but I take your question to mean &quot;why choose this specific mathematical shape, rather than something else similar (that apparently ticks all the same boxes)?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Z7xwK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Z7xwK.png&quot; alt=&quot;Hyperbolic Paraboloid&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2020-03-24T14:09:32.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33790" PostTypeId="1" AcceptedAnswerId="33791" CreationDate="2020-03-24T15:39:31.853" Score="-1" ViewCount="635" Body="&lt;p&gt;I'm new to Control Systems. I've been given this block diagram and I would like to find the overall transfer function. How can I do it?&#xA;I was thinking about simplifying the diagram by placing in parallel C(s) and R(s), but I don't think it is the right approach.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/e1WI7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/e1WI7.png&quot; alt=&quot; &quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25192" LastActivityDate="2020-03-24T15:59:47.260" Title="Find the transfer function of a basic block diagram" Tags="|control-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33791" PostTypeId="2" ParentId="33790" CreationDate="2020-03-24T15:59:47.260" Score="0" Body="&lt;p&gt;This is actually quite simple when one understood the concept, I'll try to make the start and you can solve the rest from there on your own.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You want to get a function that describes &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt; in terms of &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt;, so you just work your way back from the end. Let's call the edge that goes into the &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt; block &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;.&#xA;Then we can write for the last part of the block diagram: &lt;span class=&quot;math-container&quot;&gt;$$ y = G(s) \cdot x$$&lt;/span&gt;&#xA;Now we can move on further through the diagram, by replacing &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; with what it actually is, a sum of the edges that come out of &lt;span class=&quot;math-container&quot;&gt;$C(s)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$R(s)$&lt;/span&gt;. From here it's mostly rinse and repeat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Eventually you will have a function that contains only your input and, in case of a feedback like here, also your output, which you can then solve for &lt;span class=&quot;math-container&quot;&gt;$y$&lt;/span&gt;. This yields the transfer function.&lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2020-03-24T15:59:47.260" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33792" PostTypeId="1" CreationDate="2020-03-24T16:44:26.197" Score="2" ViewCount="2018" Body="&lt;p&gt;Both as a kid and as an adult, I've encountered elevators lined up next to each other, or single ones, which have two buttons on each floor (outside of the elevator):&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;^&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;And:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;V&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;Pressing one of them makes it light up. Not understanding what the arrows/pointers symbolize, I always pressed both. They would stay lit until the elevator arrived. I have no idea if I made myself wait longer or caused issues for others by doing this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As far as I can tell, there is nothing whatsoever which explains what the buttons do. No instructions of any kind. This was apparently considered so obvious that even the dumbest person could figure it out. But clearly, they had not thought about me...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, the &lt;em&gt;most likely&lt;/em&gt; way in which they are meant to work is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pressing &quot;up symbol&quot; means &quot;I want to go up&quot;, and pressing &quot;down symbol&quot; means &quot;I want to go down&quot;. That is, you are just &lt;em&gt;implying&lt;/em&gt; that you want it to arrive at your floor ASAP, but the information you input is where you want to go. Correctly telling the elevator system this information supposedly helps it perform optimally for everyone, and potentially even shortens your own wait time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;However, if that's what it means, then why am I allowed to press &lt;em&gt;both&lt;/em&gt; buttons, which then both stay lit? Shouldn't it then disable the other button since you pressed the first? Since you can press both, that gives zero useful information to the elevator system. Unless that means that there are people standing in line who want to go up and others who want to go down... but that sounds too complex of a setup.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible interpretation is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pressing &quot;up symbol&quot; means &quot;I want the elevator to go up, stopping at my floor, because it's currently below my location&quot;, and pressing &quot;down symbol&quot; means &quot;I want the elevator to go down, stopping at my floor, because it's currently above my location&quot;. Since each person can see where the elevator currently is (on a separate series of lights, above the elevator), this is quite possible, albeit perhaps slightly less likely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So... which is it? Or does it work in some entirely different way? Am I the only person in the world who manages to get confused by those buttons?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Wouldn't the best design be to have just &lt;strong&gt;&lt;em&gt;one&lt;/em&gt;&lt;/strong&gt; button per elevator entrance which simply means &quot;bring the elevator here ASAP&quot;? At least that cannot be misunderstood.&lt;/p&gt;&#xA;" OwnerUserId="25194" LastActivityDate="2020-04-24T01:12:15.520" Title="When an elevator has two buttons next to it, ^ and v, what do they actually mean?" Tags="|mechanical-engineering|electrical-engineering|structural-engineering|design|building-design|" AnswerCount="4" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33793" PostTypeId="1" CreationDate="2020-03-24T16:50:19.557" Score="1" ViewCount="283" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HvzHc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HvzHc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I create two helixes, identical except for two different diameters of wind.  connecting their ends, I create a spiral slide as a surface.  I then create an hour-glass like revolved surface, then use the intersection of these two surfaces to create a helix that starts out with larger diameter, gets smaller, then gets large again.  I then create a sketch at the end of this and use the two sketches to define a spiraling sweep.  the problem is, the sweep doesn't twist properly as it follows the helix.  I would like it to twist such that if a surface of the sweep was tangent to the original hour-glass like revolved surface, it stays that way all the way along.  I think this must be possible, but as you can see in the image, that is not what it's doing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hTd6D.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hTd6D.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13081" LastEditorUserId="1832" LastEditDate="2020-03-24T20:27:20.357" LastActivityDate="2020-03-24T20:27:20.357" Title="how to twist a sweep around a spiral in solidworks" Tags="|solidworks|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33794" PostTypeId="2" ParentId="33792" CreationDate="2020-03-24T17:13:29.053" Score="1" Body="&lt;p&gt;The up arrow button is for when you want to go up in the elevator.  Then down arrow is when you want to go down.  If you press only the one you want, the elevator will stop on the way up or down only, instead of opening once for one way and then again when it passes the other way.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-24T17:13:29.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33795" PostTypeId="2" ParentId="33792" CreationDate="2020-03-24T17:41:39.117" Score="5" Body="&lt;p&gt;The elevator control system stores various bits of information in a logical fashion. The system will be optimised to maximise the utility of the lift. This generally means moving the maximum number of people to their desired destinations in the minimum time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The logic is smart. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If someone on the tenth floor presses the &amp;dtrif; button the elevator will run to the 10th floor.&lt;/li&gt;&#xA;&lt;li&gt;If someone presses the &amp;utrif; button on an intermediate floor before the elevator passes that floor it will stop to collect the passenger.&lt;/li&gt;&#xA;&lt;li&gt;If someone presses the &amp;dtrif; button while the elevator is going up the system will ignore the request until the elevator is going down.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pressing &quot;up symbol&quot; means &quot;I want to go up&quot;, ... That is, you are just implying that you want it to arrive at your floor ASAP, but the information you input is where you want to go.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes, ASAP is always assumed subject to the first-come, first-served sequencing.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Correctly telling the elevator system this information supposedly helps it perform optimally for everyone, and potentially even shortens your own wait time.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Correct. You must have caused a lot of delay through your life so far. Hopefully we can avoid further loss of life.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;However, if that's what it means, then why am I allowed to press both buttons, which then both stay lit? Shouldn't it then disable the other button since you pressed the first? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. You may want to go up and someone else wants to go down. The elevator will stop and the passenger needs to determine which direction it is running before stepping in.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Since you can press both, that gives zero useful information to the elevator system. Unless that means that there are people standing in line who want to go up and others who want to go down... but that sounds too complex of a setup.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Nope. That's the way it works. Yes, elevators are complex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another possible interpretation is:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Pressing &quot;up symbol&quot; means &quot;I want the elevator to go up, stopping at my floor, because it's currently below my location&quot;, and pressing &quot;down symbol&quot; means &quot;I want the elevator to go down, stopping at my floor, because it's currently above my location&quot;. Since each person can see where the elevator currently is (on a separate series of lights, above the elevator), this is quite possible, albeit perhaps slightly less likely.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No, the user doesn't have to do the optimisation. The control logic does that.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Am I the only person in the world who manages to get confused by those buttons?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Probably not.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Wouldn't the best design be to have just one button per elevator entrance which simply means &quot;bring the elevator here ASAP&quot;? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;No. Imagine everyone leaving the apartment to go to work in the morning. The optimum sequence is for everyone to press the down button when they arrive at the lift, let the lift run non-stop to the top and then pick everyone up on the way down.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At least that cannot be misunderstood.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Ummm ...&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2020-03-24T17:41:39.117" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33796" PostTypeId="2" ParentId="33778" CreationDate="2020-03-24T17:59:12.703" Score="2" Body="&lt;p&gt;While the other answers here are compelling, they're only half the truth.  If you consult your manual you'll find that the generator load windings are reconfigured when using the mode selection switch to one of three output modes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IheLc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IheLc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In either mode, the full output of the generator is available, either as 104A@120V(1ph, one circuit), 52A@120/240(1ph/split, two circuits), or 34A@120/208(3ph, three circuits).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you select position #2 (1ph @ 120V) then you can place your entire load between L3 and LO without worrying about balancing anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The only reason to select three-phase mode is if you actually have a three phase load to drive.  It has no other effect on the available power of the generator, but using it in that mode does require you to manually balance your single phase loads.  Choose the mode that fits your load.  If you only need 120V 1ph service and you need all 104A for single phase 120V loads then choose that mode.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, if you run single phase loads in 3-phase mode or in 120/240 split phase mode then you &lt;strong&gt;DO&lt;/strong&gt; have to worry about keeping the phases reasonably balanced yourself by evenly distributing loads to the L1/L2/L3 terminals (in 3-phase mode), or to the L1/L3 terminals (in 120/240V split phase mode).&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&lt;strong&gt;Operator's manual, 2-18&lt;/strong&gt;  &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;When using single phase connections, always attempt to balance loads between terminals(do not connect all loads between one terminal and LO). Failure to observe this caution can result in damage to generator set.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you have a 3ph load you need to power, obviously you need to select three-phase mode, and a 3-phase load will naturally be balanced if you're not adding any single phase loads on top of it.  If you're only driving single phase loads, save yourself the headache and just use mode position #2 (1ph, 120V).  That way all your loads simply go between LO and L3 and you don't have to worry about balancing anything.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As always, the lesson is - &lt;a href=&quot;https://greenmountaingenerators.com/manuals-and-support/mep-803a-manuals/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Read your manuals&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="12902" LastEditorUserId="12902" LastEditDate="2020-03-26T23:45:38.790" LastActivityDate="2020-03-26T23:45:38.790" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="33797" PostTypeId="2" ParentId="33766" CreationDate="2020-03-24T22:00:44.183" Score="1" Body="&lt;p&gt;The question is still a little confused after the edits.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm trying to understand in depth the behavior of progressive scan TV monitors (LCD/LED - Digital Television). &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The subject is really only of interest because our new digital systems have to be able to receive an analog signal the format of which was entirely designed around the available display technology of the time. Interlaced scan was used on cathode ray tube (CRT) television to sort out a couple of problems:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;The low frame rate gave a noticable flicker as the phosphors had faded by the time the next refresh came along. &lt;/li&gt;&#xA;&lt;li&gt;Slower phosphors could be used but these gave increased motion blur.&lt;/li&gt;&#xA;&lt;li&gt;Interlaced displays used a short-decay phosphor but drew the odd and even lines on alternate scans. The result was that all areas of the screen were partially refreshed at twice the frequency.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;This is largely irrelevant for digital LCD/LED TVs as the display logic holds the pixels on from frame to frame without fading. The digital receivers now have to receive and digitise the analog transmissions for display on a matrix monitor.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;What is the definition of the duration of the scanning of one row, and how is it related to the response time of the screen?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/WgGBP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/WgGBP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. The NTSC composite colour waveform. Source: &lt;a href=&quot;https://www.maximintegrated.com/en/design/technical-documents/tutorials/7/734.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;Maxim&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The duration is simply the time it took for the electron beam to scan one line of the TV. As a rough estimate divide one second by the frame rate and again by the number of lines. &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Does this scanning occurs in an overlapped manner between consecutive frames? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The system couldn't support an overlap. After all the horizontal lines were transmitted the relatively long vertical retrace pulse was given, the vertical deflection waveform was reset and the next frame began.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Or there is a wait time between consecutive frames, so the screens holds one frame during a short amount of time before starting the transition to the next frame. In this case, what is the definition of that time?.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The digital display can be designed to do it in a number of ways. If it was me (and I know little about the topic) I would digitally buffer a full frame (both odd and even interlaces) in the graphics card and then have the graphics card output the frame in non-interlaced mode. Remember that each pixel will remain on until updated, unlike the fade-out of the CRT display.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2020-03-24T22:00:44.183" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33798" PostTypeId="1" CreationDate="2020-03-25T00:55:35.000" Score="14" ViewCount="11861" Body="&lt;p&gt;Screenshot of a video where they are featured in the background: &lt;a href=&quot;https://i.stack.imgur.com/fRENp.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fRENp.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They appear to have no electronics built in, such as an electric charge station, or a parking ticket dispenser or something like that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They appear to just be large, solid metal poles that are roughly placed with one per parking space in front of the store.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they a safety measure? So that nobody can drive like a madman into the store? Is that really so common as to have them waste all that money on &quot;safety poles&quot; everywhere?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are they meant for bicycles to lean against them? For people to hang out in that area? Resting against them? Are they purely ornamental? Are they considered aesthetically pleasing?&lt;/p&gt;&#xA;" OwnerUserId="25197" LastActivityDate="2020-03-29T02:46:55.247" Title="What are those large poles for that are always in front of US stand-alone store buildings?" Tags="|structures|building-design|" AnswerCount="8" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="33799" PostTypeId="2" ParentId="33798" CreationDate="2020-03-25T01:13:23.130" Score="4" Body="&lt;p&gt;They are to prevent &quot;smash and rob&quot;. A situation where someone steals a vehicle and drives through a store front to rob the store. Usually an accomplice drives a second vehicle to escape. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-03-25T01:13:23.130" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33800" PostTypeId="2" ParentId="33798" CreationDate="2020-03-25T04:43:45.467" Score="38" Body="&lt;p&gt;They are called bollards.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Their function is to provide safety for pedestrians from traffic, distracted drivers who may encroach into the sidewalk, or runaway cars e.g a driver who inadvertently leaves his/her car in neutral without the emergency brake set.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many municipalities require them and have codes and specifications defining their dimensions and strength.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="-1" LastEditDate="2020-03-26T05:05:23.840" LastActivityDate="2020-03-26T05:05:23.840" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="33801" PostTypeId="1" CreationDate="2020-03-25T11:30:50.737" Score="0" ViewCount="133" Body="&lt;h1&gt;My problem:&lt;/h1&gt;&#xA;&lt;p&gt;I'm trying to model fatigue disbond growth in Abaqus. So far I've tried this using multiple approaches with mixed results.&lt;/p&gt;&#xA;&lt;h1&gt;What I've tried:&lt;/h1&gt;&#xA;&lt;p&gt;I first started with modelling a static general step. I've tried:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Interaction with fracture properties, I was not able to propegate a crack:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;both with and without a &amp;quot;debonding with vcct&amp;quot; crack under interactions-&amp;gt; specials-&amp;gt; crack&lt;/li&gt;&#xA;&lt;li&gt;both with and without material defined with traction seperation damage&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;Interaction with cohesive behaviour with damage, I was able to propegate a crack:&lt;/li&gt;&#xA;&lt;li&gt;X-fem with material defined with traction seperation damage, I was able to propegate a crack:&#xA;&lt;ul&gt;&#xA;&lt;li&gt;both using no contact property and a vcct fracture property&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I was initially hoping to use the debond using vcct because it seems like the thing to use for my case but I've set that aside for the time being.&lt;/p&gt;&#xA;&lt;p&gt;I then moved on to modelling fatigue using a direct cyclic step. It's not something I've used before so I might have done this incorrectly. I've tried to figure it out using tutorials as well as reading the documentation but I'm not sure it's correct. The settings I used are:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;cycle time period: 1 (standard setting)&lt;/li&gt;&#xA;&lt;li&gt;incrementation: both automatic and fixed&lt;/li&gt;&#xA;&lt;li&gt;increment size: ranging from 1 to 0.05, I've figured out that this is the step size one cycle is modelled in, but I'm not sure how the accuracy of this relates to the fatigue modelling&lt;/li&gt;&#xA;&lt;li&gt;maximum number of iterations: 5 (instead of 200 in order to save computational time)&lt;/li&gt;&#xA;&lt;li&gt;number of fourier terms: have not touched this&lt;/li&gt;&#xA;&lt;li&gt;max allowable temperature change per increment: 1 to 1e-5 I think this is used to choose the optimum step size, not sure what a reasonable number is&lt;/li&gt;&#xA;&lt;li&gt;fatigue: I include low level cycle fatigue analysis&lt;/li&gt;&#xA;&lt;li&gt;cycle increment size: anything from 10 to 1000&lt;/li&gt;&#xA;&lt;li&gt;maximum number of cycles: anywhere from 1000 to 100000&lt;/li&gt;&#xA;&lt;li&gt;damage extrapolation tolerance: 1 I have no idea what this does so I haven't touched it&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;I've created an amplitude with:&lt;/p&gt;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;frequency 6.28 (loading and unloading once per cycle)&lt;/li&gt;&#xA;&lt;li&gt;starting time 0&lt;/li&gt;&#xA;&lt;li&gt;initial amplitude 0.5&lt;/li&gt;&#xA;&lt;li&gt;A 0&lt;/li&gt;&#xA;&lt;li&gt;B 0.5 (combined with the intial amplitude this should mean the loading is from 0 to 1 I think)&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&lt;p&gt;Unfortunately I have not been able to get the cohesive interaction to disbond under fatigue loading. I think I might be supposed to change some of the keywords, but I've not been able to change them successfully.&lt;/p&gt;&#xA;&lt;p&gt;The x-fem approach did work with the fatigue loading. However, I originally did this using a velocity constraint. When I switched to loading or a displacement constraint the crack did not propegate, even with loads significantly higher than the case with the velocity constraint.&lt;/p&gt;&#xA;&lt;h1&gt;Where I'm currently at:&lt;/h1&gt;&#xA;&lt;p&gt;I've got a model that sometimes cracks when using a direct cyclic step using X-fem with a vcct interaction. I'm very confused and don't know what I could try next.&lt;/p&gt;&#xA;&lt;h1&gt;My question:&lt;/h1&gt;&#xA;&lt;p&gt;How can I fix my model such that I can model disbonding under fatigue loading? I understand helping like this might not be easy so any resources that you could point me towards would be very much appreciated.&lt;/p&gt;&#xA;" OwnerUserId="20195" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2020-03-25T11:30:50.737" Title="Fatigue modelling in Abaqus" Tags="|abaqus|fatigue|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33802" PostTypeId="1" AcceptedAnswerId="33812" CreationDate="2020-03-25T12:12:04.293" Score="3" ViewCount="1094" Body="&lt;p&gt;The stiffness of a uniformly loaded cantilever beam is given as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/sBiyI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sBiyI.png&quot; alt=&quot;enter &amp;lt;imag{e description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am &lt;strong&gt;trying to derive the stiffness&lt;/strong&gt;, but I am facing a problem. I would be glad, if someone could help me out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What I have done so far:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;From my understanding, the stiffness &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is defined as:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$k = \frac{F}{\delta}$$&lt;/span&gt;&#xA;where &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; is the force that moves the beam at a certain point by &lt;span class=&quot;math-container&quot;&gt;$delta$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hence, I calculate the tip deflection and the equivalent point load at the tip and use those two values to get the stiffness.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that the deflection curve is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y(x) = \frac{p}{24EI}(6L^2x^2-4Lx^3+x^4)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tip deflection becomes: &lt;span class=&quot;math-container&quot;&gt;$$\delta_{tip} = y(L) = \frac{p}{24EI}(6L^4-4L^4+L^4) = \frac{pL^4}{8EI}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the equivalent force at the tip, I do the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;I calculate the equivalent point load&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Then I calculate the equivalent tip load from the point load (moment = 0 around fixed end)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vfd4h.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vfd4h.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But then I end up with the following stiffness:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$k=\frac{\frac{1}{2}pL}{\frac{pL^4}{8EI}} = \frac{4EI}{L^3}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;which is wrong...&lt;/strong&gt; &lt;/p&gt;&#xA;" OwnerUserId="9039" LastActivityDate="2020-03-25T18:50:17.770" Title="Derivation of the stiffness of a uniformly loaded cantilever beam" Tags="|mechanical-engineering|statics|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33804" PostTypeId="1" AcceptedAnswerId="33806" CreationDate="2020-03-25T14:45:22.367" Score="9" ViewCount="1876" Body="&lt;p&gt;Before I start, sorry if this doesn't make sense English is not my first language. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my country electricity is nationalized so there is only one energy company and one grid. When you build new house you just connect your house cables to the gird.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is in country/region where there are multiple companies providing energy, does each company have its own grid/cables/wires reaching homes? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And what happens if there is a new company, does it have to build it is own grid?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are there multiple transmission lines for each company?&lt;/p&gt;&#xA;" OwnerUserId="20414" LastEditorUserId="20414" LastEditDate="2020-03-25T15:06:37.397" LastActivityDate="2020-03-26T13:01:05.137" Title="How do electric companies supply electricity to homes in countries where energy is privatized?" Tags="|electrical-engineering|electrical|" AnswerCount="5" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33805" PostTypeId="2" ParentId="33804" CreationDate="2020-03-25T15:07:33.883" Score="1" Body="&lt;p&gt;In the UK the grid is owned by one entity and power suppy companies sell to customers in the areas they service.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The power companies may or may not be generating power...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In fact, EDF is one company that sells in the UK and it is based in France.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This causes price choices so customers can shop around for the best deal for their needs.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-03-25T15:07:33.883" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33806" PostTypeId="2" ParentId="33804" CreationDate="2020-03-25T16:41:35.683" Score="8" Body="&lt;p&gt;Works similarly in the US (compared to UK responses). Natural monopolies are allowed to exist for utilities as it wouldn't make sense to have 3 or 4 different sewer lines running to everyone's home.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distribution company that owns the grid may or may not produce electricity itself. But there is (usually) only one power line that connects to a user.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastEditorUserId="23871" LastEditDate="2020-03-25T19:10:00.300" LastActivityDate="2020-03-25T19:10:00.300" CommentCount="11" ContentLicense="CC BY-SA 4.0" />
  <row Id="33807" PostTypeId="2" ParentId="33804" CreationDate="2020-03-25T17:07:23.607" Score="6" Body="&lt;p&gt;In the UK there are three different parts of the complete system: generation, transmission and distribution, and supply (sales) to customers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many companies that generate electricity, varying in size from national (and even international) companies down to small organizations that operate a few wind turbines at one location. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are two parts to the transmission and distribution system. The &quot;National grid&quot; is a single system for long-distance transmission between large generating plants, operating at high voltages. This feeds local &quot;distribution&quot; networks operating at lower voltages and eventually connecting to individual customers (i.e. houses and factories).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electricity supply companies buy power from the generating companies, pay the transmission and distribution companies to transport it, and sell it to individual customers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Usually, domestic customers are only connected to one distribution network and have one electricity meter. If a customer wants to change supplier (e.g. to get a better price) nothing is changed physically. The customer's meter readings are used by the new supplier instead of the old one. The supply company knows the total amount of electricity it is selling (using the meter readings) and buys that quantity from the generator companies. The buying price changes continuously depending on supply and demand, but the supplier company sets its own price to sell to customers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some large (national) companies may operate both as both generation companies and supply companies but the two functions are logically separate.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, the UK supply network does not actually end at the borders of the UK - for example there are links to France, Belgium, and the Netherlands which can operate in either direction depending on supply and demand. On average, around 5% of the electricity used in the UK is actually generated in Europe.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-03-25T17:07:23.607" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33809" PostTypeId="2" ParentId="33802" CreationDate="2020-03-25T18:15:11.240" Score="1" Body="&lt;p&gt;you may not substitute distributed load with a concentrated load for deflection. in your case the deflection of the original loading if we call &lt;span class=&quot;math-container&quot;&gt;$ \  \omega L=P \ $&lt;/span&gt; is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\delta_{max}=\frac{\omega L^4}{8EI}=\frac{P L^3}{8EI}  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;but if we substitute and place P at the center of the beam&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \delta_{max}=\frac{5P L^3}{48EI}=5/6 \frac{P L^3}{8EI}  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As we see the substitution has lead to 1/6 reduction in the reflection and that's is how it should be because of the fact the parts of the distributed load past the center of the beam are more effective in bending it than those on nearer to the support with less moment.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-25T18:15:11.240" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33810" PostTypeId="2" ParentId="33798" CreationDate="2020-03-25T18:25:34.190" Score="3" Body="&lt;p&gt;For safety. They are to protect whatever is behind them from cars. They're very strong and will stop or slow down a car that would otherwise cause a lot of damage or risk to life.&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-25T18:25:34.190" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33811" PostTypeId="2" ParentId="33792" CreationDate="2020-03-25T18:32:47.313" Score="0" Body="&lt;p&gt;They are just to tell the elevator which way you want to go. If you want to go up then push the up symbol ^, then, if the elevator is already on a trip going down past your floor, it won't stop. It will stop on its way back up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This saves a lot of time. You don't get an elevator going in the wrong direction, and those already on the trip don't have to stop at every floor where someone wants to get a trip, it stops only for those going in the same direction&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can push both because you might not be the only want that wants to use the elevator. If someone else wants to use it, they might want to go in the opposite direction as you and in that case the would push the other button so that both are lit up.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's polite to only push the button for the direction that you want to go. It's polite to also wait until the elevator is going in your direction to get on, even if it stops for the person going the other direction, wait for it to come back in your direction. We all like our personal space, so only being in the elevator when you need to is polite&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-25T18:32:47.313" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33812" PostTypeId="2" ParentId="33802" CreationDate="2020-03-25T18:50:17.770" Score="3" Body="&lt;p&gt;You came so close to the answer, you ran right past it!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you've said, the stiffness coefficient is defined as force over distance: how much force you need to apply to deflect an object (a beam or a spring, for example) in a given way a certain distance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you correctly derived, the deflection at the point of a beam under a distributed load is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta = \dfrac{pL^4}{8EI}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, that's actually basically the answer right there, actually. It's saying that if you apply a distributed load &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; over a beam, it will deflect in a given way a certain distance &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt;. That's really close to our definition of a stiffness coefficient above, except that here we're talking about a distributed load instead of a force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, how can we convert this distributed load to a force? You actually tried too hard. All we have to do is calculate the total load &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; created by the distributed load, which you correctly derived as&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$P = pL$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And we're done. There's no need to perform any sort of transformation to the force's location or anything. If we just plug that into the deflection equation, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;\delta &amp;amp;= \dfrac{PL^3}{8EI} \\&#xA;\therefore \dfrac{P}{\delta} &amp;amp;\equiv k = \dfrac{8EI}{L^3}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Huzzah.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing to remember here is that stiffness coefficients are unique to the exact loading and boundary configuration given, and to the specific point they are calculated for. If you convert a distributed load into a concentrated one and calculate the resulting deflection, you'll get a different stiffness coefficient (after all, a beam under distributed loading deflects differently from one under a concentrated load). If you use that same beam with the same loading, but calculate the deflection at midspan, you'll get another stiffness coefficient. None of these coefficients are wrong, they're just correct for different cases.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So calculating the stiffness coefficient is the easy step: calculate the deflection at the desired point, sum up the total force applied throughout the entire structure, and then divide the deflection by that.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-03-25T18:50:17.770" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="33813" PostTypeId="2" ParentId="33798" CreationDate="2020-03-25T20:06:29.370" Score="2" Body="&lt;p&gt;Other answers have covered a lot of the original use cases but nowadays terrorism is one of the big reasons why you see them being put up just about everywhere.  According to this article, the more study ones are designed to stop a loaded truck up to 80 mph: &lt;a href=&quot;https://www.sdmmag.com/articles/97143-anti-terrorism-bollard-manufacturer-helps-secure-2-iconic-us-locations&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.sdmmag.com/articles/97143-anti-terrorism-bollard-manufacturer-helps-secure-2-iconic-us-locations&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know for certain but I believe the Nice, France terror attack was a huge driver in their proliferation.&lt;/p&gt;&#xA;" OwnerUserId="25227" LastActivityDate="2020-03-25T20:06:29.370" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33814" PostTypeId="1" CreationDate="2020-03-25T20:29:33.763" Score="1" ViewCount="70" Body="&lt;p&gt;I have an 11th order transfer function in the continuous domain (simplified linearized model of a grid-forming converter) and I'm looking for a way to characterize it with a single number for evaluation of a parameter sweep. I've so far been using settling time because that's a typical requirement in grid codes, but there are discontinuities in a parameter sweep (when a peak moves past the threshold the function jumps). That for example makes the use of search algorithms difficult and also makes it harder to define an optimum over all operating points (the location of the discontinuities of course varies depending on the operating point). There also wouldn't really be a point in linearizing if I end up evaluating something that I can extract from a nonlinear model :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I'm looking for something that is similar to settling time, but without the discontinuities. I have several pole pairs in close proximity to each other and to zeros, so simply evaluating the damping ratio of the dominant pole pair won't do. I know that I can weigh pole pairs by frequency, but that would still neglect the zeros. I need to somehow incorporate the zeros as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope my problem is comprehensible. Any hint in the right direction is appreciated!&lt;/p&gt;&#xA;" OwnerUserId="25226" LastActivityDate="2023-12-25T22:06:00.713" Title="Evaluate &quot;damping&quot; of higher-order transfer function - something similar to settling time without the discontinuities" Tags="|control-engineering|transfer-function|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33815" PostTypeId="2" ParentId="33798" CreationDate="2020-03-25T20:42:49.630" Score="9" Body="&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Bollard&quot; rel=&quot;noreferrer&quot;&gt;Bollards&lt;/a&gt; and other defensive oriented structures significantly predate the 2016 Nice attacks. They have been in place at the construction of many buildings in the US and across the world for decades including government and multi-national corporation buildings. You probably just never noticed them. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is my opinion that you see more of the type above due to the rising prevalence of reckless and distracted driving. The kind of bollard shown above will stop cars and small trucks as long as they are not traveling at a significant rate of speed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jJSfo.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jJSfo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;NYC Federal Reserve&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QHjqC.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QHjqC.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;IBM Armonk&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/kevEz.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/kevEz.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Herbert Hoover Building Washington DC&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/X5SQo.jpg&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/X5SQo.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Arc de Triomphe&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many other methods are used as well including fences with raised stone or concrete bases and berms or trenches depending on the available terrain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.buildings.com/article-details/articleid/13693/title/protect-buildings-from-vehicle-attacks&quot; rel=&quot;noreferrer&quot;&gt;Protect Buildings from Vehicle Attacks [2012]&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25229" LastActivityDate="2020-03-25T20:42:49.630" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33816" PostTypeId="2" ParentId="33737" CreationDate="2020-03-25T21:10:23.523" Score="1" Body="&lt;p&gt;I disagree.  Every color TV I got (and I repaired a lot of them in the '70's and '80's) displayed colored snow when tuned to blank air.  In fact this was a much touted &quot;feature&quot; mentioned in consumer magazines when there was nothing interesting to watch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The trouble with newer sets is instead of &quot;working&quot; and displaying what they receive, they are &quot;too smart&quot; and &quot;refuse to work&quot;.  This makes them useless for tasks for which they are not specifically designed, such as displaying stereo video (with shutter glasses).  I must resort to heavy, cumbersome &quot;boob tubes&quot; for this.&lt;/p&gt;&#xA;" OwnerUserId="25231" LastActivityDate="2020-03-25T21:10:23.523" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33817" PostTypeId="2" ParentId="33804" CreationDate="2020-03-25T21:54:12.620" Score="2" Body="&lt;p&gt;I doubt there is any place where the provision of electricity is completely privatized. Take California, for example. There's Pacific Gas &amp;amp; Electric, which is a corporation that is owned by stockholders. (The terms &quot;public&quot; and &quot;private&quot; get confusing here; since the stock is publicly traded, it's a &quot;public&quot; company, but since it is owned by individuals rather than the government, it's a &quot;private&quot; company). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, there's also the Public Utility Commission, which is a government entity that has wide discretion over PG&amp;amp;E. It can veto price increases, levy fines, order diversification, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;PG&amp;amp;E owns the power lines, and they ultimately supply the electricity to consumers. However, the situation regarding the &lt;em&gt;production&lt;/em&gt; of electricity is much more complicated, and I have only a vague understanding. My understanding is that PG&amp;amp;E is both a power production and power distribution company, but there are also companies that are just power production companies, and PG&amp;amp;E is required to allow them to offer their services to consumers. So if someone sets up a wind farm and wants to sell their electricity, PG&amp;amp;E has to let them use their power lines. There was a thing with Enron which, again, I only vaguely understand it, but my understanding is that Enron was supplying energy to PG&amp;amp;E, and they were manipulating the price of energy to gouge consumers.&lt;/p&gt;&#xA;" OwnerUserId="12911" LastActivityDate="2020-03-25T21:54:12.620" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33818" PostTypeId="2" ParentId="33798" CreationDate="2020-03-25T22:29:06.723" Score="29" Body="&lt;p&gt;They are called Bollards. But these aren't &lt;em&gt;defensive&lt;/em&gt; as suggested. &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;Look closer.  There's no curb.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;This is a new style of store (from the architecture) and &lt;strong&gt;curbs have been eliminated for ADA reasons&lt;/strong&gt; (benefit of wheelchairs). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As such, there is literally nothing to prevent a car from drifting forward and hitting the building; &lt;em&gt;it's not even clear where you should stop the nose of your car&lt;/em&gt;. There's clearly intended to be a sidewalk there, but &lt;strong&gt;lots of people&lt;/strong&gt; will simply pull right up and block the sidewalk.  I see it all the time, and often have to walk around the offending car. Hence, the bollards. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The benefit (ADA-wise) is that wheelchair users and other limited-ability people &lt;em&gt;are not confined to using the &quot;handicap spaces&quot; - and can park anywhere and directly access the sidewalk.  This is consistent with US &quot;ADA 2.0&quot;&lt;/em&gt; doctrine that people in wheelchairs should be able to use the same route as everyone else.  &lt;/p&gt;&#xA;&#xA;&lt;h2&gt;These bollards are &lt;em&gt;not&lt;/em&gt; a fortification.&lt;/h2&gt;&#xA;&#xA;&lt;p&gt;Keep in mind, OP's example is an ordinary convenience store.  It lacks any particular national importance; it's not a gathering point for &lt;em&gt;tens&lt;/em&gt; of people let alone &lt;em&gt;thousands&lt;/em&gt;. &lt;a href=&quot;https://en.wikipedia.org/wiki/Ram-raiding&quot; rel=&quot;noreferrer&quot;&gt;Ram-raiders&lt;/a&gt; are not an issue here, since that is to penetrate locked doors in empty stores, and &lt;strong&gt;these stores are open 24x7&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here's how you can tell. There are clearly marked parking spaces, &lt;em&gt;and the bollards are set on the center-line of the parking space&lt;/em&gt;.  The design really depends on cars at least making some attempt to park inside the lines. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bare minimum width of a parking space is 8.5 feet (2600mm), because you need at least 24&quot; (600mm) width to open the car door.  Well the bollards are typically &amp;lt;12&quot; (300mm)wide, and these look like 6&quot; trade size pipe (168.3mm).  That means with the standard &quot;guard a curbless sidewalk&quot; bollards, &lt;strong&gt;you can easily drive a car between them&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, completely worthless as a fortification. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you look at bollards &lt;em&gt;as&lt;/em&gt; fortifications, there's less than 5 feet (1500mm) between them. Further, this 6&quot; pipe wouldn't do - even filled with concrete, it's just too flimsy. &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;* ADA 2.0 isn't really a thing, I just made that up as a slang way to talk about it. What is true is that there was a major shift in ADA doctrine, as they discovered that (gosh) wheelchair users rarely travel alone - and often travel with others in wheelchairs, and that tends to overwhelm &lt;em&gt;limited&lt;/em&gt; accommodations like lifts tacked onto stairs. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Initially, ADA was to provide access &lt;em&gt;at all&lt;/em&gt;. Abled people could climb the magnificent steps, and wheelchair users go around the back of the (older) building by the dumpsters and people shooting up, and push a buzzer and wait 5 minutes to have a guard escort them through the non-public areas, and it's 15 minutes before they actually catch up with their friends. &lt;/li&gt;&#xA;&lt;li&gt;The new doctrine is that wheelchair users are equals. You're not allowed to build the grand staircase, and you cannot create barriers that would send wheelchair users on a different path.  Curbless construction is consistent with the new doctrine. &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="11673" LastActivityDate="2020-03-25T22:29:06.723" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="33819" PostTypeId="2" ParentId="33755" CreationDate="2020-03-25T23:04:20.213" Score="1" Body="&lt;p&gt;Automotive manufacturing encompasses a wide range of types of devices, many of which are broadly similar in terms of their structure and function to medical devices. e.g. processing electronics, actuators, sensors, feedback displays. They are also reasonably similar in the size of devices they make - this sort-of makes sense from an abstract perspective, as in both industries they are providing functions directly to humans.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where they differ is in the scale and cost of production, with automotive being several orders of magnitude higher volume and cheaper. As the bottleneck in medical device production is likely to be volume and (perhaps to a lesser extent) cost, and the need for automotive parts is probably also diminished right now, it makes sense to at least investigate whether automotive production facilities could be short-term switched to produce medical devices.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the type and model of automotive parts changes reasonably regularly I'd guess that retooling might be possible in time, but that there may be difficulties in cleanliness and certification/validation for medical use. However if the choice is between a uncertified ventilator or no ventilator I know which I would choose.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(caveat - all the above based on engineering speculation, I am not directly involved in these industries myself)&lt;/p&gt;&#xA;" OwnerUserId="157" LastActivityDate="2020-03-25T23:04:20.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33820" PostTypeId="1" CreationDate="2020-03-25T23:10:20.390" Score="1" ViewCount="48" Body="&lt;p&gt;What is the process of applying dye to anodized surfaces (specifically aluminum)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is electrophoretic deposition (EPD) just the process used to apply the dye to anodised surfaces?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or is it a different type of coating?&lt;/p&gt;&#xA;" OwnerUserId="17383" LastActivityDate="2020-03-25T23:10:20.390" Title="How is EPD different / related to anodising?" Tags="|materials|chemical-engineering|surface-preparation|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33823" PostTypeId="2" ParentId="33702" CreationDate="2020-03-26T01:40:07.270" Score="1" Body="&lt;p&gt;I don't think the specific head capacity matters much when you are trying to do sound insulation. Characteristics related to sound interactions matter more, like shape, or sound propagation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is lots of info: I have no idea why I spent time answering this question.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substance                                       Specific Heat - cp - (J/kgoC)&#xA;Cellulose, cotton, wood pulp and regenerated    1300 - 1500&#xA;&lt;a href=&quot;https://sciencing.com/insulators-keep-hot-water-10027507.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat stored is then cp(cb)*M(cardboard)*T(K)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Paper   cp= 1336 J/(kg*K)&#xA;&lt;a href=&quot;https://www.engineeringtoolbox.com/specific-heat-capacity-d_391.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;location of cp&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Corrugated cardboard is a stiff, strong, and light-weight material made up of three layers of brown kraft paper. In 1884, Swedish chemist, Carl F. Dahl, developed a process for pulping wood chips into a strong paper that resists tearing, splitting, and bursting.&#xA;&lt;a href=&quot;http://www.madehow.com/Volume-1/Corrugated-Cardboard.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;kraft paper&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kraft. Kraft comes from the German word meaning strong. Kraft paper is made from at least 80% sulfate wood pulp. It is course and exceptionally strong, making it well suited for a packaging substrate.&#xA;&lt;a href=&quot;https://www.sciencedirect.com/topics/materials-science/kraft-paper&quot; rel=&quot;nofollow noreferrer&quot;&gt;Composition&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What the other 20% is is hard to find. The essential elements for making kraft pulp are wood fibre, water, chemicals and heat.&#xA;&lt;a href=&quot;https://www.catalystpaper.com/products/how/pulp&quot; rel=&quot;nofollow noreferrer&quot;&gt;Source&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://sciencing.com/insulators-keep-hot-water-10027507.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;General thermal characteristics&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6285" LastActivityDate="2020-03-26T01:40:07.270" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33824" PostTypeId="1" CreationDate="2020-03-26T05:05:33.547" Score="1" ViewCount="134" Body="&lt;p&gt;im importing my 2D sketch from AutoCAD and trying to extrude base in Solidworks. I realized that when I tried to extrude a shape, it's connected to another shape, and it's extruding together with it whi&lt;a href=&quot;https://i.stack.imgur.com/C4Wwe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/C4Wwe.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;ch I don't want. How to solve this issue. I have attached a picture of two shapes which extruding together. &lt;/p&gt;&#xA;" OwnerUserId="24778" LastActivityDate="2020-03-27T09:28:49.513" Title="extruded Boss/Base error" Tags="|solidworks|autocad|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33825" PostTypeId="1" CreationDate="2020-03-26T05:45:17.010" Score="-1" ViewCount="198" Body="&lt;p&gt;I am working on a package collection system and I was wondering how I could do the mechanics of a door that slides up and down. I have thought of using pulleys with string or linear actuators, but all of those are quite unreliable. I would like to hear your ideas.&lt;/p&gt;&#xA;" OwnerDisplayName="user25241" LastActivityDate="2020-03-26T12:24:01.860" Title="What would you call a sliding door that slides up and down and how would you make it's mechanincs?" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33826" PostTypeId="2" ParentId="5494" CreationDate="2020-03-26T05:54:10.853" Score="2" Body="&lt;p&gt;If this is your first time working with structural steel sections, or simply looking for accurate data on their mechanical properties, find the official &quot;Handbook of Steel Construction&quot; for your region. Here in Canada it is &quot;Canadian Institute of Steel Construction (CISC): Handbook of Steel Construction&quot; and in the USA it is the &quot;American Institute of Steel Construction (AISC): Steel Construction Manual&quot;. I am not sure about other countries but they pretty much all follow the same title format and are either called &quot;The Handbook of Steel Construction&quot; or &quot;Steel Construction Manual&quot;. It should be fairly easy to find the official version for your region if you search for that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As someone who stumbled across this thread trying to research these same questions, I know how hard it can be to find reliable answers. So I cannot stress this enough. GET YOURSELF A COPY!! This book will literally answer every question you have and I really wish I would have found it sooner.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cheers, eh.&lt;/p&gt;&#xA;" OwnerUserId="25242" LastEditorUserId="25242" LastEditDate="2020-03-26T06:04:19.477" LastActivityDate="2020-03-26T06:04:19.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33827" PostTypeId="2" ParentId="33825" CreationDate="2020-03-26T06:17:47.333" Score="0" Body="&lt;p&gt;Take look at garage doors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many tried and tested systems. Some are sectional, some solid. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example my garage door is 60 years old and is still in perfect condition. I had to change the remote opener a couple of times.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are industrial heavey service doors mostly made of steel to make them hard to break open.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have some heavy lifters mostly pulling the door up by chains.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the new commercial service doors have electronic transmitters with layers of codes to be able to set by the manager of the store and automatically change according to daily schedule to deffrenciate admittance to proper employee/s. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-26T06:17:47.333" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="33829" PostTypeId="1" CreationDate="2020-03-26T07:27:29.357" Score="-1" ViewCount="70" Body="&lt;p&gt;Assume that I have an object in 3D, and I want to have a controller that stabilizes its orientation (also referred to as &quot;attitude&quot;). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To be sure we are clear: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;attitude is measured in &quot;degrees&quot; or &quot;radians&quot; &lt;/li&gt;&#xA;&lt;li&gt;attitude rate is measured in &quot;degrees/s&quot; or &quot;radians/s&quot;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The controller is made of two cascaded control loops:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;attitude controller: the error controlled by this controller is the difference between the &lt;em&gt;reference attitude&lt;/em&gt; and the &lt;em&gt;real attitude&lt;/em&gt;&lt;/li&gt;&#xA;&lt;li&gt;attitude rate controller: the error controlled by this controller is the difference between the  reference &lt;em&gt;attitude rate&lt;/em&gt; and the &lt;em&gt;real attitude rate&lt;/em&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The attitude controller is made only of a P term, and therefore it is not hard to tune. However, the attitude rate controller has all three terms P, I, and D.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the well-known &lt;a href=&quot;https://en.wikipedia.org/wiki/Ziegler%E2%80%93Nichols_method&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ziegler-Nichols method&lt;/a&gt; to tune the attitude rate controller. However, I am not sure that theoretically, what I am doing is correct. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: is it possible to tune the P, I, and D terms of the attitude rate controller while observing the step response by giving a command in attitude? To apply this procedure, should I provide an attitude rate command instead? (The main problem is I cannot give an attitude rate command because my object would start to rotate and this is not practical). FYI and for completeness: my object is a drone, but I don't think it matters too much regarding my question.  &lt;/p&gt;&#xA;" OwnerUserId="18666" LastActivityDate="2021-01-12T11:02:40.260" Title="Tuning a cascaded attitude and attitude rate PID controller" Tags="|control-theory|pid-control|robotics|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="33830" PostTypeId="2" ParentId="19004" CreationDate="2020-03-26T12:09:52.893" Score="2" Body="&lt;p&gt;The Tasman bridge in Hobart, Tasmania has a pile 265ft below mean water level. &#xA;This on the western side, I understand, so is not connected to the highest span.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://www.parliament.tas.gov.au/history/tbstats.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://www.parliament.tas.gov.au/history/tbstats.htm&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25254" LastActivityDate="2020-03-26T12:09:52.893" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33831" PostTypeId="2" ParentId="33825" CreationDate="2020-03-26T12:24:01.860" Score="1" Body="&lt;p&gt;I would call such a door a vertically sliding door, as opposed to a laterally or horizontally sliding door.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regarding the mechanism, weights, pulleys and either cords or chains are used for &lt;a href=&quot;https://en.wikipedia.org/wiki/Sash_window&quot; rel=&quot;nofollow noreferrer&quot;&gt;sash windows&lt;/a&gt; would work for doors as well. &lt;a href=&quot;https://www.mightonproducts.com/products/sash-window-hardware/sash-balances&quot; rel=&quot;nofollow noreferrer&quot;&gt;Sash balances&lt;/a&gt; are a alternative to cords and counter weights.&lt;/p&gt;&#xA;" OwnerUserId="90" LastActivityDate="2020-03-26T12:24:01.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33832" PostTypeId="2" ParentId="33804" CreationDate="2020-03-26T13:01:05.137" Score="1" Body="&lt;p&gt;In the UK, the grid is divided into the transmission network (high voltage; long distance) and the distribution networks (lower voltage networks; connects directly to the consumer and small generators). The transmission network is owned by National Grid (in England and Wales) and the distribution networks are owned by a different companies that operate in different regions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your electricity supplier doesn't need to own any physical infrastructure. They buy ahead of time the amount of generation they expect their customers to need on the electricity market. Physical location of customers and generators doesn't matter; the grid is treated like a collective &quot;pool&quot; of power and you have to put in as much in as your customers will take out. Wind forecasting allows wind farms to sell generation ahead of time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;National Grid then look at the generation profile and make adjustments to ensure that circuits are not overloaded, generation matches demand and various other constraints are met. Engineers in the control room can see prices for each generator to increase or decrease generation. The price for fossil fuel generators to decrease generation is usually negative because they save fuel but still get paid for the electricity they would've generated. For wind farms it is positive because they loose their renewable energy incentives.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electricity flows into and out of the transmission network are all metered. A company called Elexon does the accounts with those readings. As a generator or electricity supplier, if your generation or demand didn't match what you contracted for, you pay money according to the difference. This money pays for National Grid to balance the system as above.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I don't know how Elexon work out how much your electricity supplier's customers consumed given that suppliers share the distribution networks. I assume it must be based on estimates for households and meter readings for large consumers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Essentially owning infrastructure (power lines, cables etc.) is a separate business from being an electricity supplier.&lt;/p&gt;&#xA;" OwnerUserId="25255" LastActivityDate="2020-03-26T13:01:05.137" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33833" PostTypeId="2" ParentId="33798" CreationDate="2020-03-26T13:08:43.947" Score="3" Body="&lt;p&gt;Bollards in Switzerland have been placed to prevent and hinder drivers from parking vehicles with 2 wheels on the pavement or to stop them from parking their cars in front of shops.&lt;/p&gt;&#xA;" OwnerUserId="25257" LastEditorUserId="90" LastEditDate="2020-03-26T19:22:58.013" LastActivityDate="2020-03-26T19:22:58.013" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33834" PostTypeId="1" CreationDate="2020-03-26T14:21:50.390" Score="0" ViewCount="104" Body="&lt;p&gt;I have a straight-line distance of 250 km between two towers and want to find the length of the transmission line that is between these two towers while accounting for the &lt;strong&gt;sag&lt;/strong&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The line should be larger than 250 km. Given that the estimated sag is 60 m, how do I find the real transmission line? Can you provide me with references?&lt;/p&gt;&#xA;" OwnerUserId="25262" OwnerDisplayName="neurotech" LastEditorUserId="90" LastEditDate="2020-03-30T04:10:19.787" LastActivityDate="2020-08-27T05:04:25.423" Title="Transmission line length with sag" Tags="|transmission|" AnswerCount="1" CommentCount="10" ContentLicense="CC BY-SA 4.0" />
  <row Id="33835" PostTypeId="1" AcceptedAnswerId="33838" CreationDate="2020-03-26T15:58:37.037" Score="1" ViewCount="591" Body="&lt;p&gt;Currently I'm designing a tracked vehicle for a school project. It has to carry a weight and be able to overcome some obstacles which are about 3/4th the height of the tank. What I'm stuck at, is calculating the required torque for the motors. &#xA;&lt;a href=&quot;https://i.stack.imgur.com/NsYWO.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NsYWO.png&quot; alt=&quot;Basic layout with variables. The upper right wheel is going to be powered.&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is the basic layout of the tank with some variables. The wheel in the upper right is going to be powered.&#xA;Given that there will be enough traction between the tracks and the obstacle, how can I accurately calculate the required torque for the motor of this tracked vehicle to climb the obstacles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in advance!&lt;/p&gt;&#xA;" OwnerUserId="25269" LastActivityDate="2020-03-26T18:41:20.207" Title="Required torque of climbing tracked vehicle" Tags="|mechanical-engineering|torque|electric-vehicles|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33836" PostTypeId="2" ParentId="33834" CreationDate="2020-03-26T16:22:23.410" Score="1" Body="&lt;p&gt;There are to problems with the question.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Over 250 kilometers earth has 5 kilometers curve. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;the weight and tension in the cable are prohibiting to normal materials. W*125000/60 = approximate tension.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Otherwise there are many references online to solve a catenary self weight cable length. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-03-26T17:36:38.990" LastActivityDate="2020-03-26T17:36:38.990" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33838" PostTypeId="2" ParentId="33835" CreationDate="2020-03-26T18:41:20.207" Score="0" Body="&lt;p&gt;The max torque at the time of the climbing is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's call the power wheel radius .r.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T=mg*sin(theta)*r&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the track passes the hump of non powered wheel the tank will flip back, because it's CG falls back outside of the track footprint. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-26T18:41:20.207" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33840" PostTypeId="1" CreationDate="2020-03-26T21:26:23.783" Score="0" ViewCount="139" Body="&lt;p&gt;For every action, there is an equal and opposite reaction. Sample case me doing a push up I force myself to move up by pushing against a rigid body (the floor).&lt;/p&gt;&#xA;&lt;p&gt;There goes I can not push myself sideways by thrusting my hands forward but as the case of a pushup. Or if I hit the water with an open fist it feels painful because the movement causes the liquid to feel like a rigid body (instantiating rapid opposing motion on a body).&lt;/p&gt;&#xA;&lt;p&gt;So with that, moving to rockets, their motion (in earth) is caused by a heavy surge of air simulating the hard body (with the open fist slap on water logic) and when enough causes the rocket to move in the other direction.&lt;/p&gt;&#xA;&lt;p&gt;My current logic in this argument is for the opposite reaction to occur, there is an opposing opposite matter that causes makes the opposite reaction to recache in the other direction?&lt;/p&gt;&#xA;&lt;p&gt;As I am trying to understand how something would move in a vacuum. In the case of a bird also it pushes air down for it to go up... My mechanical physics is not that great I am tackling this in a more logic-based approach.So, a case scenario definition of all the definitive forces and direction of reaction would be well appreciated and is this logic correct (for a motion to occur there must be something resisting movement).&lt;/p&gt;&#xA;" OwnerUserId="25285" LastEditorUserId="90" LastEditDate="2023-06-14T05:22:42.737" LastActivityDate="2023-06-15T00:06:42.230" Title="What causes opposite motion in Newton's third law, or is the motion matter independent?" Tags="|linear-motion|" AnswerCount="5" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33841" PostTypeId="2" ParentId="33840" CreationDate="2020-03-26T22:33:53.773" Score="1" Body="&lt;p&gt;Rather than think of it as 'resisting', think of it as force. In a vacuum, you are correct that there is nothing to push against, nothing resisting. The resisting, as you put it, comes from pushing out something that was taken into space with the rocket, the gas from the rocket motor. In force terms, it takes force to push that gas out. The force is in balance. The force that is pushed out the end of the rocket, the gas in the engine, is the same amount of force that is pushing the rocket forward. &lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-26T22:33:53.773" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="33842" PostTypeId="2" ParentId="33840" CreationDate="2020-03-27T00:18:26.640" Score="0" Body="&lt;p&gt;Imagine you are a young astronaut, my favorite is Niel Armstrong, doing some repairs outside on your spaceship antenna and once you finish your job, to your horror, you realize your personal rocket is jammed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cool as ice and confidently you reach for the safety cord on your waist.  Now you really shudder; the safety rope connection, which accidentally became disconnected when you were fighting your way with the big heavy wrench, is floating in the space some 20 feet away and is slowly going away.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;you are racking your brains, what to do??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then the spark ignites in your brain: Newton's third law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You grab the heavy wrench from your belt and toss it violently in the opposite direction of the safety rope into the empty space.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am sure you could see it in your mind. What's going to happen?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Edit&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I decided to plug in some crude numbers here and respond to some comments by the OP. we assume there is no air, total vacuum and obviously no friction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;let's say the astronaut has done enough spacewalk that he knows to lean horizontally with his feet directed to the safety cord and throughs the wrench straight out from as close to his heart as to receive the least torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's say the force of his hand on the wrench is 30kgs and the length his hand swings before releasing the wrench is 75cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As per Newton's third law his hand, hence his entire body, absorbs exactly equal force in the opposite direction.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming the astronaut with his spacesuit weighs 120kgms the force will accelerate his body back towards the safety cord.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ W_{work}=E_{kinetic-energy}= \quad F*x=30kg*0.75m  =22.5kg.m$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;speed of astronaut towards the cord is= V and can be calculated by his kinetic energy&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ E=22.5=1/2mV^2 \rightarrow V^2=22.5*2/m=45/120=0.375m/s \\ V=\sqrt{.375}=0.61m/s$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And the astronaut will reach the cord in 20/.61= 32 seconds, without any friction or rocket. Just by ejecting the wrench and using the reacting force to his advantage.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-03-28T02:11:34.543" LastActivityDate="2020-03-28T02:11:34.543" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="33844" PostTypeId="1" AcceptedAnswerId="34873" CreationDate="2020-03-27T04:51:43.590" Score="-1" ViewCount="124" Body="&lt;p&gt;I want to design a robust control system using the internal model design specifications. The block diagram is the one shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nbo1B.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nbo1B.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am trying to obtain the transfer function &lt;span class=&quot;math-container&quot;&gt;$\ \frac{Y(s)}{R(s)} $&lt;/span&gt; in order to acquire the characteristic polynomial of the closed loop system but for some reason I am stuck. This is what I have done so far:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ E_{a}(s) = R(s) - Y(s) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ U(s) = E_{a}(s)G_{c}(s) - KX(s) $$&lt;/span&gt; ( &lt;span class=&quot;math-container&quot;&gt;$\ U(s) $&lt;/span&gt;: control input to the process &lt;span class=&quot;math-container&quot;&gt;$\ G(s) $&lt;/span&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Y(s) = U(s)G(s) = E_{a}(s)G(s)G_{c}(s)-KX(s)G(s) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\ $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Y(s) = R(s)G(s)G_{c}(s) - Y(s)G(s)G_{c}(s) - KX(s)G(s) $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$\  $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ Y(s) = \frac{G(s)[R(s)G_{c}(s)-KX(s)]}{1+G(s)G_{c}(s)} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From this point, I really don't know how to continue. Obviously the term that confuses me is &lt;span class=&quot;math-container&quot;&gt;$\ X(s) $&lt;/span&gt;. So, now how should I proceed and obtain the overall transfer function ?&lt;/p&gt;&#xA;" OwnerUserId="18238" LastActivityDate="2020-03-29T16:56:29.397" Title="How to obtain transfer function of control diagram with Internal Model Control?" Tags="|control-engineering|control-theory|transfer-function|robust-control|" AnswerCount="1" CommentCount="12" ContentLicense="CC BY-SA 4.0" />
  <row Id="33847" PostTypeId="2" ParentId="33824" CreationDate="2020-03-27T09:28:49.513" Score="2" Body="&lt;p&gt;The reason these two shapes are extruding together, is, because in the .DWG file, they &lt;em&gt;are connected to each other&lt;/em&gt;. See what happens when you zoom right in?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/F3tfU.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F3tfU.gif&quot; alt=&quot;zoom error&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What you need to do in this instance, is either fix your original .DWG file, or simply use the .DWG as a reference for a new SolidWorks sketch. I have shown the process for this below. N.B. I was unable to use coradial for both circles, because, they don't exactly share a centrepoint! This is yet another error with the .DWG - I set the inner circle coincident to one of the points to fix this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.imgur.com/AJvSoWO.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/AJvSoWO.gif&quot; alt=&quot;manual fix process&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2020-03-27T09:28:49.513" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33849" PostTypeId="1" CreationDate="2020-03-27T12:17:18.963" Score="2" ViewCount="50" Body="&lt;p&gt;Let's consider a FE buckling calculation without plasticity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there theoretically any chance that the non-linear analysis will give a higher buckling load than the linear one?&lt;/p&gt;&#xA;" OwnerUserId="8592" LastActivityDate="2020-03-27T12:17:18.963" Title="Can the critical buckling load from non-linear FE calculation be higher than the one from linear?" Tags="|finite-element-method|buckling|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33850" PostTypeId="1" CreationDate="2020-03-27T13:58:56.423" Score="2" ViewCount="85" Body="&lt;p&gt;I am trying to control a real inverted pendulum by means of a DC motor, for this I must model the torque of the motor, the work I've done in this direction is the following (which I believe is right):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/scTMs.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/scTMs.jpg&quot; alt=&quot;Derivation of model&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Something I don't get is, both the motor back-emf constant (&lt;span class=&quot;math-container&quot;&gt;$k_e$&lt;/span&gt;) and the torque constant (&lt;span class=&quot;math-container&quot;&gt;$k_t$&lt;/span&gt;) are supposed to be the same, at least as far as I've read on several places. Is it safe to assume both are the same (&lt;span class=&quot;math-container&quot;&gt;$k_e = k_t = K$&lt;/span&gt;), in which case, I can obtain the constant by measuring the angular speed and the voltage applied to the motor, and dividing both, right?&lt;/p&gt;&#xA;" OwnerUserId="25306" LastEditorUserId="1832" LastEditDate="2020-03-27T15:52:02.470" LastActivityDate="2020-03-27T17:13:26.640" Title="Modeling the torque of a real motor" Tags="|control-engineering|motors|torque|modeling|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33851" PostTypeId="2" ParentId="33737" CreationDate="2020-03-27T14:12:25.310" Score="2" Body="&lt;p&gt;The question was well answered by &quot;Transistor&quot; on Mar 21 at 14:31. The only part missing was that the other colour tv systems of the past, known as PAL (Phase Alternate Line) and SECAM (Séquentiel Couleur à Mémoire) worked in a very similar fashion, both had a colorburst signal at 4.43 MHz, superimposed on a blanking portion of a horizontal synchro pulse. PAL with phase alternating between 135° and 225° from line to line. SECAM did not care about the phase since it used FM modulation for colour information.&#xA;The important part is that all the analog tv receivers had a circuit designed to detect the presence of color signal, with noise immunity at the top of the requirements list.&#xA;So when the signal is very week or noisy (or missing) the analog color tv set reverts back into a black and white mode.&lt;/p&gt;&#xA;" OwnerUserId="25304" LastActivityDate="2020-03-27T14:12:25.310" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33852" PostTypeId="2" ParentId="15691" CreationDate="2020-03-27T14:40:56.543" Score="2" Body="&lt;p&gt;In such a case following multiple solutions are suggested.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Tension piles are the best solution. Raft often creates uplift pressure if the water table is high or may vary season to season.&lt;/li&gt;&#xA;&lt;li&gt;Now the cheapest one: Place pipes at various and multiple positions in the raft vertically to release the pressure. These pipes shall be interconnected above the raft in an appropriate slope towards a harvesting tank. Water will rise in the pipe which can be dropped in the tank from where can be pumped out for gardening. No pressure shall be created under the raft. This pipe network can be embedded under the floor of the basement.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;One more important point: place a thick layer of stones and stone dust underneath the raft. It will reduce the capillary action in the soil and water will tend to flow horizontally instead of vertically. Allow a horizontal passage of water below the raft. It can be easily done by placing half perforated pipes network below the raft in the layer of stones a d guide the water outside the building area.&lt;/p&gt;&#xA;" OwnerUserId="25309" LastEditorUserId="1832" LastEditDate="2020-03-27T15:49:40.973" LastActivityDate="2020-03-27T15:49:40.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33853" PostTypeId="2" ParentId="33850" CreationDate="2020-03-27T17:13:26.640" Score="3" Body="&lt;p&gt;Your general model for an electric motor could be for example, as follows.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$J \phi ''(t)=\text{I} \tau -R \phi '(t)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with J as your [kg m^2] inertia, I in [A] amps, and R your friction...at the moment, I can't remember the unit is, and &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; your torque constant [Nm/A]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a simple linear ODE and is solved also fairly easily.&#xA;With initial conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\phi (0)=0,\phi '(0)=\phi _0$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\phi (t)\to \frac{J e^{-\frac{R t}{J}} \left(\text{I} \tau -R \phi _0\right)-\text{I} J \tau +\text{I} R t \tau +J R \phi _0}{R^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is of course your &lt;em&gt;angle&lt;/em&gt; dependant on time, but we want the change, or namely the velocity.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we take the derivative, &lt;span class=&quot;math-container&quot;&gt;$\frac{d\phi}{dt}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\omega(t)=\frac{i R \tau -R \left(\text{i$\tau $}-R \phi _0\right) e^{-\frac{R t}{J}}}{R^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we take some a data trace we have from playing the motor at different amperages, we can calculate or fit R and J if they're unknown, like for example:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IrMf5.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IrMf5.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;With the values being {R -&gt; 0.0000186966, J -&gt; 0.000138553} for that particular example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in your case, you can find your torque constant if you know the rest. However, generally you can find that in a motor data sheet...as it appears your don't have one, you can go about it this way. Are you using some kind of software to be able to read such things, matlab, mathematica, or scripted something yourself? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would highly suggest trying to get a data trace and fit the data to &lt;span class=&quot;math-container&quot;&gt;$\omega(t)$&lt;/span&gt; rather than messing around with the maths. &lt;/p&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2020-03-27T17:13:26.640" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33854" PostTypeId="2" ParentId="31707" CreationDate="2020-03-27T19:47:27.147" Score="2" Body="&lt;p&gt;With good quality threads, 3 wraps of PTFE tape can seal NPT connections above 1,000 psi. &#xA;SAE J1926 straight thread o-ring ports will seal a lot easier though. I've also used this company in the past: &lt;a href=&quot;https://www.highpressure.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.highpressure.com/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-03-27T19:47:27.147" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33856" PostTypeId="1" CreationDate="2020-03-28T02:23:20.607" Score="3" ViewCount="580" Body="&lt;p&gt;I have a mechanical engineering question that I can seem to figure out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically I have an object this is being pulled down by two forces. Each force is 10N in the downward direction. There is a nut welded to the bottom of a plate and a threaded rod runs through the plate and into the nut.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I can't figure out how much torque will be needed on the threaded rod to lift the the object. Let's ignore friction for the time being. A diagram is included to show what I've got.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It would be really helpful if I could get some insight into how to setup the problem. I did well in my college physics courses but I'm was an applied math major.. 8-) And they really didn't talk about how to convert rotational work into translational work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/hm5ql.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/hm5ql.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="23942" LastActivityDate="2020-03-28T08:22:50.623" Title="torque required to lift an object" Tags="|torque|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33857" PostTypeId="2" ParentId="33856" CreationDate="2020-03-28T03:01:38.967" Score="0" Body="&lt;p&gt;Let's call the diameter of the screw d and and its pitch N (number of threads/ meter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we ignore the friction the screw will act just like a ramp or slope of a ramp, and assuming the slope smaller than a few degrees we can say &lt;/p&gt;&#xA;&#xA;&lt;p&gt;sin (alpha) =alpha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The torque you need has to be bigger than&lt;/p&gt;&#xA;&#xA;&lt;p&gt;T= 20/ (N&lt;em&gt;pi&lt;/em&gt;d/)Newton. m&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-03-28T08:22:50.623" LastActivityDate="2020-03-28T08:22:50.623" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33858" PostTypeId="2" ParentId="33856" CreationDate="2020-03-28T04:33:40.603" Score="2" Body="&lt;p&gt;Since we're not having to deal with friction, conservation of energy is our best friend. There's no need to &quot;convert rotational work into translational work&quot;, merely observe that they must be equal in a stable object. So&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$W_F = W_T$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$W_F$&lt;/span&gt; is the work done by the applied forces and &lt;span class=&quot;math-container&quot;&gt;$W_T$&lt;/span&gt;, by the torque.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;A note before we continue: a screw's pitch is usually defined as the axial distance between two threads. However, it can actually be given in two different ways: metric screws usually give you the pitch as a distance (i.e. 1.5&amp;nbsp;mm), while inch-based screws tend to be given in terms of threads-per-inch.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://upload.wikimedia.org/wikipedia/commons/0/00/Lead_and_pitch.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the derivation below I'll be using pitch as a distance.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;So, if we think of a single rotation of the screw, the plate will move along the rod by a length equal to the screw's pitch &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; (the linear distance between two threads).* Therefore, we know that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$W_F = Fd$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the rotational work, we can imagine it is created by a force &lt;span class=&quot;math-container&quot;&gt;$f$&lt;/span&gt; applied tangent to the nut, rotating the screw once around the rod's spiral. We can save ourselves some work and consider the length of that spiral equal to the circumference of the rod. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;W_T &amp;amp;= 2\pi rf \\&#xA;&amp;amp;= 2\pi T&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The last steps are now trivial:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;W_F &amp;amp;= W_T \\&#xA;Fd &amp;amp;= 2\pi T \\&#xA;\therefore T &amp;amp;= \dfrac{Fd}{2\pi}&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you instead have the pitch defined in threads-per-distance, the modification is trivial:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T = \dfrac{F}{2\pi d}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where the torque's distance unit is the distance unit in the pitch's threads-per-distance (usually threads-per-inch, therefore inches).&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* Strictly speaking, the dimension &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; to be used is the &lt;em&gt;lead&lt;/em&gt;, not the pitch. The lead is the axial distance covered by one rotation around the screw, while the pitch is the distance between two threads. In most screws, these values will be the same (as in the left side of the image above) and pitch is the more common term, which is why I used it above. However, when using multi-start screws, the lead will be greater than the pitch (right side of the image), and &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt; should be adopted as the lead.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2020-03-28T04:42:53.467" LastActivityDate="2020-03-28T04:42:53.467" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33859" PostTypeId="1" CreationDate="2020-03-28T08:53:29.173" Score="0" ViewCount="102" Body="&lt;p&gt;During this covid-19, I was thinking about designing a low cost simple ventilator using a blower. There is a limit of the inspiratory and expiratory pressures, the inspiratory pressure being higher. Can anyone tell me how can I adjust the pressures? I also don't know how to control the fraction of the inhaled oxygen (FiO2). Will an oxygen concentrator work in this case? &lt;a href=&quot;https://blog.arduino.cc/2020/03/17/designing-a-low-cost-open-source-ventilator-with-arduino/&quot; rel=&quot;nofollow noreferrer&quot;&gt;If anyone wishes to look into the plan, it is in this link.&lt;/a&gt; &lt;/p&gt;&#xA;" OwnerUserId="25323" LastEditorUserId="90" LastEditDate="2020-03-28T09:21:26.227" LastActivityDate="2024-01-09T05:05:50.743" Title="How can the air pressures exiting from a blower be adjusted?" Tags="|chemical-engineering|biomedical-engineering|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="33860" PostTypeId="1" AcceptedAnswerId="34867" CreationDate="2020-03-28T10:46:17.077" Score="-1" ViewCount="65" Body="&lt;p&gt;I bought a 200x90 sq.cm. bed and assembled it. The bed frame is assembled out of hollow iron rods. After using for a few days I started having back pain. So I checked the bed and found that some of those rods got bent in the middle and that made the whole frame uneven. I want to strengthen it, however I don't have much stuff at hand, and we are in a lockdown due to coronavirus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have not thrown out the wooden planks from my older bed. So I thought putting them on the bars will distribute the load and may help it. My question is what is the pattern that will provide maximum strength? I arranged them in a way (see picture) and it already fills quite solid. I will appreciate any feedback. For information, I don't have an electronic drill machine or any such thing; also I don't know how to use them.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for reading, it is my first post here. :) Note: I also posted this question in home improvement stack exchange, just fyi.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/uouMA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/uouMA.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25325" LastActivityDate="2020-03-29T02:17:38.233" Title="Strengtheing a weak bed frame" Tags="|statics|stability|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="33861" PostTypeId="1" CreationDate="2020-03-28T13:08:54.047" Score="1" ViewCount="263" Body="&lt;p&gt;The downside of using single-acting return spring pneumatic cylinders is the decreasing output force through a full stroke due to the opposing spring force. I'm wondering how would you design a spring return pneumatic cylinder which produces an increasing output force? Would you simply need to increase the input pressure?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lXmX0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lXmX0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="25328" LastActivityDate="2020-03-29T06:29:05.700" Title="Increasing Force Output on a Spring Return Pneumatic Cylinder" Tags="|mechanical-engineering|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="33862" PostTypeId="2" ParentId="33798" CreationDate="2020-03-28T13:42:49.847" Score="1" Body="&lt;p&gt;At Sprint, the telecom company's headquarters in Overland Park, KS when they built their billion $$$ campus in 2001, the CEOs and Executive building had a separate entrance and these bollards. It was expressly there to prevent truck and car bombs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For those who only think of 9/11, &lt;a href=&quot;https://www.historynet.com/1983-beirut-bombing-the-blt-building-is-gone.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;remember the truck bombing by Iran in Beirut, Lebanon in the 1980s which killed 241 US marines and soldiers&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Never Forget.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Marine sentries initially paid little attention to the Mercedes truck.&#xA;  Heavy vehicles were a common sight at the airport, and in fact the BLT&#xA;  was expecting one that day with a water delivery. The truck circled&#xA;  the parking lot, then picked up speed as it traveled parallel to a&#xA;  line of concertina wire protecting the south end of the Marine&#xA;  compound. Suddenly, the vehicle veered left, plowed through the&#xA;  5-foot-high wire barrier and rumbled between two guard posts.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;By then it was obvious the driver of the truck—a bearded man with&#xA;  black hair—had hostile intentions, but there was no way to stop him.&#xA;  The Marines were operating under peacetime rules of engagement, and&#xA;  their weapons were not loaded. Lance Corporal Eddie DiFranco, manning&#xA;  the sentry post on the driver’s side of the truck, soon guessed the&#xA;  driver’s horrifying purpose. “He looked right at me…smiled, that’s&#xA;  it,” DiFranco later recalled. “Soon as I saw [the truck] over here, I&#xA;  knew what was going to happen.” By the time he managed to slap a&#xA;  magazine into his M16 and chamber a round, the truck had roared&#xA;  through an open vehicle gate, rumbled past a long steel pipe barrier,&#xA;  threaded between two other pipes and was closing on the BLT barracks.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Sergeant of the guard Stephen Russell was alone at his&#xA;  sandbag-and-plywood post at the front of the building but facing&#xA;  inside. Hearing a revving engine, he turned to see the Mercedes truck&#xA;  barreling straight toward him. He instinctively bolted through the&#xA;  lobby toward the building’s rear entrance, repeatedly yelling, “Hit&#xA;  the deck! Hit the deck!” It was futile gesture, given that nearly&#xA;  everyone was still asleep. As Russell dashed out the rear entrance, he&#xA;  looked over his shoulder and saw the truck slam through his post,&#xA;  smash through the entrance and come to a halt in the midst of the&#xA;  lobby. After an ominous pause of a second or two, the truck erupted in&#xA;  a massive explosion—so powerful that it lifted the building in the&#xA;  air, shearing off its steel-reinforced concrete support columns (each&#xA;  15 feet in circumference) and collapsing the structure. Crushed to&#xA;  death within the resulting mountain of rubble were 241 U.S. military&#xA;  personnel—&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="25329" LastEditorUserId="25329" LastEditDate="2020-03-29T02:46:55.247" LastActivityDate="2020-03-29T02:46:55.247" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34861" PostTypeId="2" ParentId="33861" CreationDate="2020-03-28T15:00:26.787" Score="1" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I'm wondering how would you design a spring return pneumatic cylinder which produces an increasing output force? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think you mean &quot;produces an &lt;em&gt;increased&lt;/em&gt; output force&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The counter-force exerted by the spring at maximum stroke (or any point during the stroke) is fixed and determined by the spring parameters such as spring set, and spring rate. (See Lee Spring's &lt;a href=&quot;https://www.leespring.co.uk/blog-spring-design-a-guide.asp&quot; rel=&quot;nofollow noreferrer&quot;&gt;A guide to spring design&lt;/a&gt; for some terminology.) So the actual force at the shaft will be F&lt;sub&gt;shaft&lt;/sub&gt;&amp;nbsp;=&amp;nbsp;F&lt;sub&gt;air&lt;/sub&gt;&amp;nbsp;-&amp;nbsp;F&lt;sub&gt;spring&lt;/sub&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It stands, therefore, that by increasing the air pressure that the shaft force can be increased.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Would you simply need to increase the input pressure?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Yes.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2020-03-28T15:00:26.787" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34862" PostTypeId="2" ParentId="33586" CreationDate="2020-03-28T15:43:48.683" Score="0" Body="&lt;p&gt;If you have a fixed-fixed beam then you always get zero internal force and displacement because of zero nodal displacement. If displacement vector is zero then bending moment is also zero, because of approximate nature of finite element. I know two ways to fix this discrepancy:&#xA;- split element into two and add node in middle&#xA;- consider the distributed load effect in internal force using double integration method etc.. (for example code look at method &lt;code&gt;GetLoadInternalForceAt()&lt;/code&gt; at this file: &lt;a href=&quot;https://github.com/BriefFiniteElementNet/BriefFiniteElement.Net/blob/master/BriefFiniteElementNet/ElementHelpers/EulerBernoulliBeamHelper.cs&quot; rel=&quot;nofollow noreferrer&quot;&gt;EulerBernoulliBeamHelper.cs&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="8108" LastActivityDate="2020-03-28T15:43:48.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34863" PostTypeId="2" ParentId="33861" CreationDate="2020-03-28T17:29:41.047" Score="0" Body="&lt;p&gt;there are a couple of ways to increase the output:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Because the force of the spring is proportional to its deformation, you could use a cylinder with additional room for the spring so that even when the piston is fully deployed the spring is only partially pushed back.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You can use a softer spring with a sealed bladder of air on the back assisting it.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-28T17:29:41.047" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34865" PostTypeId="1" AcceptedAnswerId="34866" CreationDate="2020-03-28T22:19:47.577" Score="1" ViewCount="39" Body="&lt;p&gt;If I already have a small DC axial 2m3/h fan blowing air at something, what will happen if I add a larger more powerful 6m3/h DC axial fan blowing from behind that one (i.e. its the larger fan first, then then smaller, then the object to blow air at)? Will the smaller one just hinder the airflow from the first, or are their outputs cumulative? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks!&lt;/p&gt;&#xA;" OwnerUserId="11394" LastActivityDate="2020-03-28T22:53:59.303" Title="Cumulative effect of two axial fans" Tags="|fluid-mechanics|airflow|compressed-air|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34866" PostTypeId="2" ParentId="34865" CreationDate="2020-03-28T22:53:59.303" Score="1" Body="&lt;p&gt;It depends on their aerodynamic properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Jet engines are multiple fans as well, but they can and do different functions. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some work together to increase the pressure and reduce the speed of stream.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some even drive energy from the stream of air.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In your case similar situations can happen.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The combination can increase the flow or the pressure. Or it may cause the entire stream to rotate in the cylinder and stall the system. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-28T22:53:59.303" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34867" PostTypeId="2" ParentId="33860" CreationDate="2020-03-29T02:17:38.233" Score="2" Body="&lt;p&gt;I would take the wooden slats from the old bed and cut them so they can stand on end and be trapped between the steel tubing and the floor, thereby supporting the load that the steel tubing by itself cannot. If you wish, you can cut a round notch in one end of the slat so as to capture the tubing; this will help prevent the slats from popping out of position. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-03-29T02:17:38.233" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34868" PostTypeId="1" AcceptedAnswerId="34870" CreationDate="2020-03-29T04:16:28.707" Score="3" ViewCount="182" Body="&lt;p&gt;I am doing a Finite Element analysis of a continuum. In my problem definition I use these units: &#xA;mm for length, kPa for applied stress, and MPa for material stifness (Young's modulus). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;With these units, I get my output in: kPa for stress and micron for displacement, which is what I prefer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I have to add Mass Density information to my problem. To benchmark my setup, I am trying to reproduce an old textbook example, in which the Mass Density is given in units of: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\\lb-sec^2 / in^4\,&#xA;$$&lt;/span&gt;&#xA;What would be the equivalent of this in my adopted SI unit system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My impulse is to use:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\\kN / mm^3\,&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any suggestions or advice will be appreciated!&lt;/p&gt;&#xA;" OwnerUserId="26332" LastActivityDate="2020-12-22T05:47:57.443" Title="Unit Conversion for Mass Density" Tags="|mechanical-engineering|finite-element-method|dynamics|experimental-physics|unit|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="34869" PostTypeId="2" ParentId="33861" CreationDate="2020-03-29T06:29:05.700" Score="1" Body="&lt;p&gt;You cannot use a return &lt;em&gt;spring&lt;/em&gt; system without reducing the output force as the spring gets energy put into it.  You could design an equivalent return system, however, using something other than a spring:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;a pulley that moves a weight vertically against gravity.  The force of the weight would be constant regardless of the stroke position.&lt;/li&gt;&#xA;&lt;li&gt;a reservoir fitted to the non-pressurized side that has liquid that would rise in level and be vented to atmosphere.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I'm sure you could come up with other similar ideas.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-29T06:29:05.700" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34870" PostTypeId="2" ParentId="34868" CreationDate="2020-03-29T09:17:22.477" Score="4" Body="&lt;h1&gt;First Part&lt;/h1&gt;&#xA;&lt;p&gt;Simplifying the&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;1 \ lbf = 1 \ slug .ft/sec^2&#xA;$$&lt;/span&gt;&#xA;then&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;1 \ lbf \ sec^2 / in^4 = 1 \ (slug .ft/sec^2 ). (sec^2/in^4) = 12 \ slug /in^3&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;12 \ slug / in^3 * (14.6 kg / 1\ slug)*(in/25.4 \ mm)^3 = 0.0107 \ kg/mm^3&#xA;$$&lt;/span&gt;&#xA;Note that that each &lt;code&gt;slug&lt;/code&gt; equals to 14.6 &lt;code&gt;kg&lt;/code&gt; so &lt;code&gt;(1 slug / 14.6 kg)&lt;/code&gt; equals to &lt;code&gt;1.0&lt;/code&gt; and any expression multiplied with 1.0 will still matematically equal to original expression.&lt;/p&gt;&#xA;&lt;p&gt;then&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;1.0 \ kg/mm^3  = 93.53 \ lbf \ sec^2 / in^4 &#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;we name above as eq1&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Second Part&lt;/h1&gt;&#xA;&lt;p&gt;Changing shape of the target formula:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;1 \ kN / mm^3 = 1 \ kN / mm^3 * (1000 \ N/1 \ kN) * (1 \ kgf / 10 \ N) = 100 kgf/mm^3&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;again note that&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;(1000 \ N/1 \ kN) = (1 \ kgf / 10 \ N) = 1.0&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Also i took &lt;span class=&quot;math-container&quot;&gt;$ kgf = 10 N $&lt;/span&gt; which is approximate, correct amount is &lt;span class=&quot;math-container&quot;&gt;$ kgf = 9.81 N $&lt;/span&gt;&#xA;finally&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;kN / mm^3 = 100 \ kgf/mm^3&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;1.0 \ kgf/mm^3 = 0.01 \ kN / mm^3&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;we name above as eq2&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Third part&lt;/h1&gt;&#xA;&lt;p&gt;&lt;code&gt;kg&lt;/code&gt; and &lt;code&gt;kgf&lt;/code&gt; units are not consistent units on the paper, but looks equal on non scientific conversations. For example someone can say this chair mass is 10kg or it's weight is 10kgf, both refers to same chair with same mass and weight. so if we conventionally assume: &lt;code&gt;1 kgf = 1 kg&lt;/code&gt;, eq1 and eq2 can be combined like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;kgf/mm^3 = 0.01 \ kN / mm^3 = 93.53 \ lbf \ sec^2 / in^4 &#xA;$$&lt;/span&gt;&#xA;(Note: base on comments, &lt;code&gt;1 kgf&lt;/code&gt; equals &lt;code&gt;9.8 N&lt;/code&gt;, but we assumed &lt;code&gt;1 kgf = 10 N&lt;/code&gt;&lt;/p&gt;&#xA;&lt;h1&gt;Final part&lt;/h1&gt;&#xA;&lt;p&gt;so&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;1.0 \ kN / mm^3 = 9353.0 \ lbf \ sec^2 / in^4 &#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;or&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;1.0 \ lbf \ sec^2 / in^4  = (0.00010691) \ kN / mm^3 &#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;Which are the answers&lt;/p&gt;&#xA;&lt;hr /&gt;&#xA;&lt;h1&gt;Notes&lt;/h1&gt;&#xA;&lt;p&gt;I am not sure about the assumption of &lt;code&gt;1 kgf = 1 kg&lt;/code&gt; in third part, but the rest should not have problem. Anyways this answer could be considered as a suggestion.&lt;/p&gt;&#xA;" OwnerUserId="8108" LastEditorUserId="8108" LastEditDate="2020-12-22T05:47:57.443" LastActivityDate="2020-12-22T05:47:57.443" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="34871" PostTypeId="1" AcceptedAnswerId="34880" CreationDate="2020-03-29T09:27:56.253" Score="0" ViewCount="282" Body="&lt;p&gt;Apologies if I have done something wrong, first time posting on a stack exchange website.&#xA;&lt;br&gt;&#xA;&lt;br&gt;&#xA;Four forces act perpendicular to the surface of a circular slab (assume mass of slab is 0 and middle of slab is at 0,0,0)(all forces in -y direction) &#xA;&lt;br&gt;&#xA;Force 1 is located at 5,0,0 and has a magnitude of 77.4kN&#xA;&lt;br&gt;&#xA;Force 2 is located at 0,0,5 and has a magnitude of 48.6kN&#xA;&lt;br&gt;&#xA;Force 3 is located at 0,0,-5 and has a magnitude of 24.4kN&#xA;&lt;br&gt;&#xA;Force 4 is located at -5,0,0 and has a magnitude of 97.6kN&#xA;&lt;br&gt;&#xA;What is the coordinates of the resulting force (x,y,z) to two decimal places?&#xA;&lt;br&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KXhYo.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KXhYo.jpg&quot; alt=&quot;Visual Representation of Question&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;My working currently and my initial answer&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IKmS5.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IKmS5.jpg&quot; alt=&quot;Working and answer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br&gt;&#xA;&lt;br&gt;&#xA;The given answer is (0.49,0,-0.41) and I don't know where I went wrong.&#xA;&lt;br&gt; &#xA;&lt;br&gt;&#xA;After talking to my lecturer, she said the question was supposed to be determining the z coordinate only, which is why it would say the answer was correct when 0.49 was put first, meaning that the answers I calculated were correct.&lt;/p&gt;&#xA;" OwnerUserId="26334" LastEditorUserId="26334" LastEditDate="2020-03-30T10:40:05.587" LastActivityDate="2020-03-30T10:40:05.587" Title="Resultant force coordinate for parallel 3d forces acting on a slab" Tags="|statics|moments|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34872" PostTypeId="2" ParentId="33859" CreationDate="2020-03-29T15:24:02.017" Score="0" Body="&lt;p&gt;I am going to answer the question on lowering the pressure, but I will not encourage you to use it on any patient due to the fact that professional medical equipment requires rigorous tests, using a homemade medical device would pose a large hazard on the patient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since your goal is to build a low cost ventilator, I believe a simple adjustable valve can lower the pressure effectively by draining part of the supplied gas away. &lt;/p&gt;&#xA;" OwnerUserId="25299" LastActivityDate="2020-03-29T15:24:02.017" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34873" PostTypeId="2" ParentId="33844" CreationDate="2020-03-29T16:56:29.397" Score="0" Body="&lt;p&gt;Well, after a while this is what I finally came up with. I have tested the results in simulation environment as well as on the real system and the produced control behaviour was what I expected. Firs of all, I rewrote the block diagram in the form shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;a href=&quot;https://i.stack.imgur.com/F55SR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F55SR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The whole procedure of the transfer function is shown below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The state variables of the plant process are &lt;span class=&quot;math-container&quot;&gt;$\ x_1, \ x_2 $&lt;/span&gt; and they are equal to: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ x_1 = y \rightarrow X_1(s) = Y(s) $$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ x_2 = \dot{y} \rightarrow X_2(s) = sY(s) - y(0) $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering zero initial conditions, the transfer function of the inner loop is (simple feedback system):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ H(s) = \frac{G_p(s)}{1+k_1G_p(s)+sk_2G_p(s)} $$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now it all comes down to the outer loop being again a simple feedback loop:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T(s) = \frac{G_c(s)H(s)}{1+G_c(s)H(s)} $$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="18238" LastActivityDate="2020-03-29T16:56:29.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34874" PostTypeId="2" ParentId="33840" CreationDate="2020-03-29T17:33:46.053" Score="1" Body="&lt;p&gt;This is one of Newton's Axioms. I.e., strictly speaking the rule does not need to be able to be derived from some other physical law, but an empirical fact supported by a huge number of actual and possible experiments.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This axiom (together with the second of Newton's Axioms), leads to the inertial principle - the total momentum of a closed system is constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To avoid the reactive force, you don't need &quot;vacuum&quot; but rather &quot;no other body that exerts force on you(r system)&quot;:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;A rocket works in open space (vacuum) as it does within earth atmosphere because the gas/smoke it ejects, forms the bodies it interacts with - the farther an amount of gas is ejected, the more the rocket is driven forward.&lt;/li&gt;&#xA;&lt;li&gt;For your push-ups, you need the the floor underneath (so you can push the earth a bit more downwards). If you are an astronaut floating in space, a push-up is only a push-apart of your space suit (and won't help your workout :-) ). &#xA;To maintain a constant total momentum, earth is pushed a bit down while you are pushed up (if you push yourself up by about 0.1 m, earth is pushed &quot;down&quot; by about 10&lt;sup&gt;-22&lt;/sup&gt; m, which is less than the width of an atom's nucleus.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;=&gt; If you want to move in vacuum, you must either have some &quot;contactless&quot; force to some peer object that is beyond this vacuum (e.g., flying in your own orbit around the sun, which attracts you by gravitation even through vacuum), or you have to eject something in the opposite direction (as your rocket does for you).&lt;/p&gt;&#xA;" OwnerUserId="26343" LastEditorUserId="90" LastEditDate="2020-03-30T04:08:20.737" LastActivityDate="2020-03-30T04:08:20.737" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34875" PostTypeId="1" CreationDate="2020-03-29T19:17:47.880" Score="1" ViewCount="81" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dLEMx.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dLEMx.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi guys, &#xA;I am solving an exercise for a course of Computational Continuum Mechanics. Briefly, my question is how can i compute the infinitesimal strain tensor at different nodes of the finite element mesh ?  I try to explain: I solved an exercise and I have a perfectly working code in MATLAB ( I attach it at the end, you can copy and paste it and run it if you want.) which compute the displacement vector field in the simple case of a rotation of pi/4 of a rigid body making use of the FEM under the assumption of a liner constitutive behavior of the material.&lt;a href=&quot;https://i.stack.imgur.com/JCRcY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/JCRcY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now I am asked to do the following :&#xA;&quot;Compute in the finite element program the infinitesimal strain tensor at several different nodes of the finite element mesh. Compare these values with those obtained analytically in the previous question.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yet I don't quite catch what I am supposed to do. I know I can compute the strain with something like this ( with &lt;span class=&quot;math-container&quot;&gt;$e_{0}$&lt;/span&gt; the orientation in the initial state) &lt;span class=&quot;math-container&quot;&gt;$ \begin{equation}\epsilon(\overrightarrow{e_{0}})=\overrightarrow{e_{0}} \cdot \boldsymbol{\epsilon} \cdot \overrightarrow{e_{0}}\end{equation}$&lt;/span&gt;. I also know that the infinitesimal strain is defined as &lt;span class=&quot;math-container&quot;&gt;$\begin{equation}\varepsilon=\frac{1}{2}\left[\boldsymbol{F}+\boldsymbol{F}^{T}\right]-\boldsymbol{I}\end{equation}$&lt;/span&gt; and I am indeed able to compute it afterwards.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;I know that the strain tensor should be a local property defined point by point but I cannot think of a way to formally evaluate it at a mesh node&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Someone can help me out ? &lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;% PARAMETERS&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;% geometric:&#xA;H = 10; % height, in mm&#xA;L = 10; % length, in mm&#xA;% prescribed rotation angle and rotation tensor&#xA;theta = pi/4;&#xA;% material properties, in MPa&#xA;G     = 1000;&#xA;kappa = 3000;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;% define the basis&#xA;ee = cartesianbasis2d('e1', 'e2');&#xA;e1 = ee(1);&#xA;e2 = ee(2);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;% determine material coefficient tensor&#xA;I   = identity(2, ee);&#xA;I4  = identity(4, ee);&#xA;I4S = 1/2 * (I4 + rtranspose(I4));&#xA;C   = kappa * I*I + 2*G * (I4S - 1/3*I*I);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;% make FE mesh&#xA;m1 = 3; % number of elements in horizontal direction&#xA;m2 = 3; % number of elements in vertical direction&#xA;[x, conn] = femgrid([ L  H ], [ m1  m2 ], ee, 'ElementType', 'Linear');&#xA;m = size(conn, 1); % number of elements in the mesh&#xA;n = size(x, 1);    % number of nodes in the mesh &lt;/p&gt;&#xA;&#xA;&lt;p&gt;% coordinates of integration points &#xA;% (in the coordinate system of the master element)&#xA;% and the weight factors&#xA;xi = [ -1/sqrt(3)*e1 - 1/sqrt(3)*e2&#xA;        1/sqrt(3)*e1 - 1/sqrt(3)*e2&#xA;        1/sqrt(3)*e1 + 1/sqrt(3)*e2&#xA;       -1/sqrt(3)*e1 + 1/sqrt(3)*e2 ];&#xA;w  = [ 1&#xA;       1&#xA;       1&#xA;       1 ];&lt;/p&gt;&#xA;&#xA;&lt;p&gt;% assemble system matrices in element loop&#xA;K = zeros(2, ee, n, n); % zero matrix of nxn 2nd order tensors in basis ee&lt;/p&gt;&#xA;&#xA;&lt;p&gt;for e = 1:m % loop over all elements&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;iie = conn(e, :); % nodes of the current element e&#xA;xe  = x(iie);     % coordinates of these nodes&#xA;&#xA;% compute element matrices in integration point loop&#xA;Ke = zeros(2, ee, 4, 4); % zero matrix of 4x4 2nd order tensors in basis ee&#xA;&#xA;for k_integration_point = 1:4 % loop over integration points &#xA;&#xA;    xi1 = dot(xi(k_integration_point), e1); % x1 coordinate of the current integr. point&#xA;    xi2 = dot(xi(k_integration_point), e2); % x2 coordinate of the current integr. point&#xA;&#xA;   % column of the shape functions&#xA;    Ne = [ 1/4*(1-xi1)*(1-xi2)&#xA;           1/4*(1+xi1)*(1-xi2)&#xA;           1/4*(1+xi1)*(1+xi2)&#xA;           1/4*(1-xi1)*(1+xi2) ];&#xA;&#xA;    % column of the gradient of the shape functions&#xA;    % with respect to the local coordinates of the mater element&#xA;    gradxiNe = [ -1/4*(1-xi2)*e1 - 1/4*(1-xi1)*e2&#xA;                  1/4*(1-xi2)*e1 - 1/4*(1+xi1)*e2&#xA;                  1/4*(1+xi2)*e1 + 1/4*(1+xi1)*e2&#xA;                 -1/4*(1+xi2)*e1 + 1/4*(1-xi1)*e2 ];&#xA;    % Jacobian&#xA;    J = gradxiNe' * xe; &#xA;    % column of the gradient of the shape functions&#xA;    % with respect to the global coordinates of the mesh&#xA;    gradNe = dot(inv(J), gradxiNe); &#xA;&#xA;    % element matrix and right hand side  ??????????????????&#xA;                                        Ke = Ke + w(k_integration_point)*dot(gradNe,C,gradNe')*det(J);     &#xA;&#xA;end % end of integration point loop&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;% assembly&#xA;    K(iie, iie) = K(iie, iie) + Ke;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;end % end of element loop&lt;/p&gt;&#xA;&#xA;&lt;p&gt;% apply boundary conditions by partitioning:&#xA;bottom = [ 1:(m1+1) ]; % nodes on the bottom boundary of the mesh&#xA;left   = [ (m1+2):(m1+1):(m1+1)&lt;em&gt;(m2) ]; % nodes on the left boundary of the mesh&#xA;right  = [ 2&lt;/em&gt;(m1+1):(m1+1):(m1+1)*(m2) ]; % nodes on the right boundary of the mesh&#xA;top    = [ ((m1+1)&lt;em&gt;m2+1):(m1+1)&lt;/em&gt;(m2+1) ]; % nodes on the top boundary of the mesh&#xA;% nodes where essential b.c. are applied&#xA;iiu = [bottom left right top];&#xA;% the remaining nodes&#xA;iif = setdiff(1:n, iiu);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Kuu = K(iiu, iiu);&#xA;Kfu = K(iif, iiu);&#xA;Kuf = K(iiu, iif);&#xA;Kff = K(iif, iif);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;% rotation tensor corresponding to given angle theta&#xA;Q = cos(theta)*e1*e1 - sin(theta)*e1*e2 + sin(theta)*e2*e1 + cos(theta)*e2*e2;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;% prescribed displacements of the boundary nodes&#xA;% according to the given rotation tensor Q&#xA;uu  = dot( (Q-I),x(iiu));&#xA;ff  = zeros(1, ee, length(iif), 1);&lt;/p&gt;&#xA;&#xA;&lt;p&gt;% solve the system for unknowns&#xA;                                        uf = Kff(ff-dot(Kfu,uu));&#xA;u = zeros(1, ee, n, 1);&#xA;u(iiu) = uu;&#xA;u(iif) = uf;&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;                                    fu = dot(Kuu,uu)+dot(Kuf,uf);&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;f = zeros(1, ee, n, 1);&#xA;f(iiu) = fu;&#xA;f(iif) = ff;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;figure(1)&#xA;clf&#xA;% plot the displacement solution&#xA;quiver(x, u, 0);&#xA;axis equal&#xA;hold on&#xA;%femplot(x, conn, 'Color', 'k')&#xA;femplot(x+u, conn, 'Color', 'r');&#xA;hold off&lt;/p&gt;&#xA;" OwnerUserId="26344" LastEditorUserId="26344" LastEditDate="2020-03-30T08:47:59.513" LastActivityDate="2020-03-30T08:47:59.513" Title="Linear FEM: Infinitesima Strain tensor at different element mesh" Tags="|finite-element-method|solid-mechanics|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34877" PostTypeId="2" ParentId="34871" CreationDate="2020-03-29T19:58:02.963" Score="0" Body="&lt;p&gt;I haven't looked at your specific numbers but on quick review it appears that your approach is correct by summing the moments (force x distance) in each of the two directions, and that both directions are independent from each other.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two things&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;To be an engineer, you have to learn to trust YOURSELF. The option that you are right is always an option. The book could be wrong. Find another way to do the math to double check your answer or confirm the books answer&lt;/li&gt;&#xA;&lt;li&gt;An engineer solves problems, yes? Or how about this: engineers create solutions. The former is passive, the later is assertive. You are doing this problem passively 'by the book.' However, you can be assertive and find your own approach and use it if it is valid. I'll use your question as an example. Note that the distances are all the same... figured out where I'm going yet? The distance is not a variable in this problem, it's a fixed number. Therefore, the problem is a simple algebra problem. All you need is the ratios of the loads and then apply that to the distance. If the two loads equal themselves, the ratio is 1:1 and the coordinate is directly between them, since the distance is the same for both from the origin, then the resultant is at 0. What happens when the ratio is not 1:1, on what side is the resultant and how far over? If nothing else, with this, you can come up with a very good approximate answer in seconds, in your head!&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-29T19:58:02.963" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34878" PostTypeId="1" CreationDate="2020-03-29T20:50:48.637" Score="2" ViewCount="586" Body="&lt;p&gt;I have a simple wall mount I want to analyze using hand calculations (pure beam bending if possible) that looks like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XjCiI.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XjCiI.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to be able to model it as if it was in a state of pure bending with a point load at the very end of the single bar section. The two bar section is fixed to the wall. However, I'm a bit confused with how to treat the &quot;pin&quot; (for lack of a better word) that connects the two sections together. Obviously, it's purpose is to transfer load to the two bars mounted to the wall, but I'm not sure if it would be a poor assumption to assume 50% of the load is transferred to each bar, and that the only force transferred through the pin is a straight downwards force and a moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know it's probably a pretty basic/simple problem, but I was hoping someone could provide some insight on how I could deal with the pin and the two bars using hand calculation. Thanks!&lt;/p&gt;&#xA;" OwnerUserId="26346" LastActivityDate="2022-01-19T17:01:15.113" Title="Structural Analysis of a TV Mount" Tags="|structural-engineering|structural-analysis|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34880" PostTypeId="2" ParentId="34871" CreationDate="2020-03-29T21:33:03.187" Score="0" Body="&lt;p&gt;We start by the forces along the x-axis about D and call the resultant distance from D, X.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\Sigma M_{about\ D}=0 \rightarrow  X*(97.6+24.4+77.4+48.6  ) \\= 73*5+77.4*10 \quad \rightarrow X=\frac{365+774}{248}=4.59$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  5-4.59=0.41$&lt;/span&gt; X is 0.41 meters to the left of the origin &lt;span class=&quot;math-container&quot;&gt;$ \ X=-0.41m \quad $&lt;/span&gt; as per the book.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The same steps apply to the Y-axis.&#xA;Please check my numbers. also, I assumed the radius 5m not 6m as shown, because you asked 5m.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-29T21:33:03.187" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34882" PostTypeId="2" ParentId="34878" CreationDate="2020-03-29T23:30:42.043" Score="1" Body="&lt;p&gt;Let's say your tv is 50lbs and the middle lever is 24&quot;long and 6&quot; high and the distance the hinge or shaft extends beyond on top and bottom before they reach the support members on the wall is also 6&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The moment is $  M= PL= 50*2' =100lbsft \$&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the shear is 50 which is most likely only supported by the lower arm of the support. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This moment will also cause pulling, tensile stress on top which wants to bend or cut the shaft, and compression on the bottom which again wants to push the shaft in.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;if we call the tensin T and compression C &lt;span class=&quot;math-container&quot;&gt;$  \quad T=C=100/(6/12)=200lbs $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thes forces need to be supported by the two brackets on the wall. a shear of 50 lbs and pulling on the top and pushing on the bottom of 100lbs each, assuming they are also 6&quot; wide.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-29T23:30:42.043" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34883" PostTypeId="1" CreationDate="2020-03-29T23:55:29.017" Score="1" ViewCount="635" Body="&lt;p&gt;the &lt;a href=&quot;https://en.wikipedia.org/wiki/Bristol_Motor_Speedway&quot; rel=&quot;nofollow noreferrer&quot;&gt;Bristol Motor Speedway&lt;/a&gt; is 0.533 mi long racetrack. The road is banked and circular. The straights are banked to 6–10° and turns to 26–30°. I have made a small diagram for ilustration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XXbq9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XXbq9.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Before the cars fully enter into the turn, they have to pass through a section of the track where the banking graually progresses from 10° to 30°. Does anyone knows how is determined the lenght of the road between two levels of banking?&lt;/p&gt;&#xA;" OwnerUserId="24735" LastActivityDate="2020-04-02T19:24:52.873" Title="Banking progression in NASCAR tracks" Tags="|cad|highway-engineering|roadway|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="34884" PostTypeId="1" CreationDate="2020-03-30T00:28:45.947" Score="1" ViewCount="21" Body="&lt;p&gt;I am studying Gas dynamics and came across the lift angle approximation as based on Skelett Theorie (in german). I have done google searches on finding this theory but no luck on any information. Can anyone perhaps in Aerodynamics help? Thanks.&lt;/p&gt;&#xA;" OwnerUserId="24243" LastActivityDate="2020-03-30T00:28:45.947" Title="Skelet Theorie, can anyone help?" Tags="|aerodynamics|compressed-gases|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34885" PostTypeId="1" AcceptedAnswerId="34893" CreationDate="2020-03-30T02:09:27.780" Score="1" ViewCount="1572" Body="&lt;p&gt;I'm interested in printing a working two degree of freedom planetary gear set. I searched around and found a paper from University of Maryland in 1988 (&lt;a href=&quot;https://i.stack.imgur.com/vim9v.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;source&lt;/a&gt;). I am having difficulty translating the information into a real-world model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The goal is to make a system that allows two independent inputs to affect the speed and torque of a single output. As the paper points out, a differential system doesn't quite make this possible.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Although mechanical systems with multiple inputs or multiple outputs have existed for many years, generally, they have been used as a series of one-DOF devices rather than as true multi-DOF mechanisms. For example, the automotive bevel-gear differential, a two-DOF mechanism with one input and two outputs, is made-up of a one-DOF gear train in series with its gear box. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I tried analyzing a standard differential system but I don't believe you can alter it to use two independent inputs. The six link systems described in the paper seem to be the right direction, and so I would like to understand the graph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the figure from the paper:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vim9v.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vim9v.png&quot; alt=&quot;Excerpt from paper&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And here is my attempt to interpret graph 6-1-1:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K9PkF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K9PkF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think this is clearly incorrect because elements 5 and 6 have no effect on the system. Could anyone help me interpret the graph?&lt;/p&gt;&#xA;" OwnerUserId="21287" LastActivityDate="2020-04-01T06:38:06.613" Title="Gear train using two inputs and one output. Understanding six link epicyclic gear topology - Interpreting displacement graphs" Tags="|mechanical-engineering|gears|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34886" PostTypeId="2" ParentId="34883" CreationDate="2020-03-30T02:36:25.287" Score="0" Body="&lt;p&gt;They have some prescribed transition curves as an intermediary between two bends with different radius. In this case, one of the bends is straight with an infinite radius.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bank angle is a linear function of R, the radius of the track, not proportional to the distance from the end of the straight section. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So right after the straight patch transitions into the curve the angle barely increases as the radius is very large. As the radius of the track approaches the tight turn, the change in the bank angle increases. By the time the intermediary curve turns int the circular arc of the track, the angle is equal to 30 degrees.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-03-30T02:36:25.287" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34887" PostTypeId="1" AcceptedAnswerId="34889" CreationDate="2020-03-30T04:22:19.970" Score="1" ViewCount="386" Body="&lt;p&gt;I might be wrong here. The solid modeling done in CAD like Catia, NX, Solidworks,etc can be taken as input in CNC and the desired shape and size of piece can be obtained.The CAD itself convert the model into G -code.So why do we need to learn CNC programming if CAD can do it? Stay safe.&lt;/p&gt;&#xA;" OwnerUserId="22829" LastActivityDate="2021-10-21T20:02:27.280" Title="What is a purpose of learning G-code?" Tags="|mechanical|cnc|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34889" PostTypeId="2" ParentId="34887" CreationDate="2020-03-30T07:38:22.660" Score="4" Body="&lt;p&gt;Because some programs for some operations can be written quicker than the time it takes to produce the equivalent CAD file. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also use of variables makes the program more useful ie a program to produce a cylinder can be written to take diameter as the controlling argument while producing many CAD files for various different diameters takes time.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-03-30T07:38:22.660" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34890" PostTypeId="1" CreationDate="2020-03-30T08:48:39.443" Score="1" ViewCount="61" Body="&lt;p&gt;I am currently working on a design for a beer pump (liquid dispenser) of (15 kg approx.), that will be clamped to a table with 2 clamps. This component has a handle attached to the top, where approximately 150 N will be applied each time the beer is to be poured. I need to determine how much clamping force is required for the component to stay rigid into the table without falling. I am new to this field, hoping someone can guide me towards the right answer.&lt;/p&gt;&#xA;" OwnerUserId="26352" LastEditorUserId="90" LastEditDate="2020-03-30T09:46:40.283" LastActivityDate="2020-03-30T09:46:40.283" Title="Clamping Force required" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="34891" PostTypeId="1" CreationDate="2020-03-30T00:55:19.807" Score="-1" ViewCount="89" Body="&lt;p&gt;I have a crossbow with a 220lb draw pull. I am designing a mechanism to draw the string back 12&quot; as quickly as possible using a 1/2&quot; ball lead screw connected to a 12 to 24v DC high torque motor and guided by two linear slides. I need to know how many KG of torque will be needed to draw the 220Lb string back in around two seconds? Or is it even possible?&lt;/p&gt;&#xA;" OwnerDisplayName="Lynn" LastActivityDate="2020-03-30T10:50:31.477" Title="Torque needed to draw back 220Lb's of weight?" Tags="|torque|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="34892" PostTypeId="2" ParentId="34891" CreationDate="2020-03-30T10:15:51.697" Score="2" Body="&lt;blockquote&gt;&#xA;  &lt;p&gt;I have a crossbow with &#xA;  - 220lb (100 kg) draw pull.&lt;br&gt;&#xA;  - 12&quot; (0.3m) draw distance &#xA;  - 1/2&quot; (12.5 mm) ball lead screw &#xA;  - 12 to 24v DC high torque motor &#xA;  - guided by two linear slides. &lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;I need to know how many KG of torque will be needed to draw the 220Lb string back in around two seconds? &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;You &lt;strong&gt;really&lt;/strong&gt; want to know motor power. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy = force x distance.&lt;br&gt;&#xA;Power = Energy/time. &#xA;Draw force = 220 lbf = 100kg = 980  newton.&lt;br&gt;&#xA;Draw stroke = 12&quot; = 300mm = 0.3M &#xA;g = 9.8 (or 10 :-) )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Energy here is ABOUT F x D&lt;br&gt;&#xA;= 100 kg x g x 0.3m = 100 x 9.8 x 0.3 ~= 300 Joules.&lt;br&gt;&#xA;To do that in 2 seconds &#xA;Power = energy/time = 300 J / 2s = 150 Watts.&lt;br&gt;&#xA;Say 200 - 250 Watts to be safe due to losses.&lt;br&gt;&#xA;Not trivial. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Motor spec will depend on lead screw pitch (gearing) but Wattage is as above.&lt;br&gt;&#xA;For an eg 1mm pitch thread motor must turn draw / pitch &#xA;= 300mm/1mm = 300 turns in 2 seconds = 9000 RPM.&lt;br&gt;&#xA;With a 12V motor that is a motor with 'kV' (= RPM/Volt) of&lt;br&gt;&#xA;RPM/V = 9000/12 = 750 RPM per volt = 750 kV.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;RC model motors with 250 W rating and 750 kV speed are commonly available.&lt;br&gt;&#xA;A brushed motor would probably be cheaper.&lt;br&gt;&#xA;For a 250 Watt brushed motor RPM at 12V may be quite a lot below 9000 RPM -  Adjust lead screw pitch to suit.&lt;br&gt;&#xA;eg a 3mm pitch lead screw will require&lt;br&gt;&#xA;300 / 3 = 100 turns &#xA;in 2 seconds = 50 turms/second = 3000 RPM (as you'd expect).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Cordless drill:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As Solar Mike noted -  a cordless drill may be a good solution.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Power ratings well above 150W - 250W are available.&lt;br&gt;&#xA;These are usually not the cheap ones :-(  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine loaded chuck speed in RPS.&lt;br&gt;&#xA;eg 140 RPM (low speed setting) = 2.333 RPS&#xA;Winding drum circumference = draw distance / time / RPS&#xA;= 300mm / 2 / 2.333  = 64 mm&lt;br&gt;&#xA;~= 20mm dia drum. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;____________________________________&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Battery:&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using 12V battery at 250 Watt, I = P/V ~= 250/12 (or less) ~= 20A&lt;br&gt;&#xA;This is within the capability of good quality power rated 18650 LiIon cells.&lt;br&gt;&#xA;4 x 18650 Lion = 12 - 16 V.&lt;br&gt;&#xA;More V = less current.&lt;/p&gt;&#xA;" OwnerUserId="399" OwnerDisplayName="Russell McMahon" LastEditorUserId="399" LastEditDate="2020-03-30T10:50:31.477" LastActivityDate="2020-03-30T10:50:31.477" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34893" PostTypeId="2" ParentId="34885" CreationDate="2020-03-30T13:50:38.083" Score="1" Body="&lt;p&gt;Your model looks mostly accurate, your gap between gears 5 and 6 isn't large enough though. What are you treating as your inputs and your outputs? If you start your analysis like that I think you'll see the effect 5 and 6 have on the system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Machinery's Handbook has some basic examples of epicyclic systems with 2 inputs. &lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-03-30T13:50:38.083" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="34895" PostTypeId="1" CreationDate="2020-03-30T15:33:16.093" Score="1" ViewCount="76" Body="&lt;p&gt;In terms of fully saturated vapour and liquid for water. Why in a real world heat pump cycle for example may the water not exactly reach these saturation lines as indicated in an ideal cycle? &lt;/p&gt;&#xA;" OwnerUserId="26356" LastEditorUserId="90" LastEditDate="2020-03-30T17:10:57.323" LastActivityDate="2020-03-30T18:53:27.020" Title="Fully saturated liquid and vapour" Tags="|thermodynamics|pumps|heating-systems|vapor-pressure|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34896" PostTypeId="2" ParentId="31613" CreationDate="2020-03-30T17:03:57.547" Score="1" Body="&lt;p&gt;Looking at your final question, it's important to make sure one thing is clear: a bigger cross-section (i.e. cylinder) will always twist &lt;em&gt;&lt;strong&gt;less&lt;/strong&gt;&lt;/em&gt; than a smaller one.&lt;/p&gt;&#xA;&lt;p&gt;Yes, the outermost fiber of the cross-section will be rotated more than internal fibers, but the outermost fiber of a large cross-section will rotate less than the outermost fiber of a small cross-section. And therefore obviously the inner fiber of the large cross-section at a distance from the center equal to the smaller cross-section's radius will rotate even less.&lt;/p&gt;&#xA;&lt;p&gt;That's because the rotation is proportional to the stress on each fiber of the cross-section. The stress is a reaction to the torsion, so Newton's Third Law tells us that the resultant &amp;quot;inner torsion&amp;quot; created by the stress must be equal to the external torsion. This inner torsion is equal to the product of the stress at a given point by its distance from the center, integrated over the entire cross-section. A larger cross-section can distribute that &amp;quot;inner torsion&amp;quot; over a larger area, so the resultant stress (and therefore deformation, by &lt;a href=&quot;https://en.wikipedia.org/wiki/Hooke%27s_law#Tensional_stress_of_a_uniform_bar&quot; rel=&quot;nofollow noreferrer&quot;&gt;Hooke's Law&lt;/a&gt;) is lower than for a smaller cross-section.&lt;/p&gt;&#xA;&lt;p&gt;As for why the outermost fibers deform more than the inner ones, there are a few ways to think about it:&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;We just assume it does. This is similar to the assumptions of &lt;a href=&quot;https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory&quot; rel=&quot;nofollow noreferrer&quot;&gt;Euler-Bernoulli Beam Theory&lt;/a&gt;, one of which is that sections that are parallel remain parallel under bending. Likewise, we assume that cross-sections under torsion behave as if suffering a solid-body rotation, with the relationships between all points within the cross-section being maintained. In a cylinder, this naturally implies that the more distant fibers must suffer more deformations: a point near the center barely has to move to rotate 1°, while one at the edge must move a lot more.&lt;/p&gt;&#xA;&lt;p&gt;Obviously, this isn't a blind assumption: experimental data shows this is very close to the truth.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If you think of rotations in terms of angles, it just kind of makes sense. As described above, if you are rotating the whole cross-section by a certain angle, of course the outer fibers will be more deformed. The only other option would be to say that the angle of rotation isn't equal across the cross-section: i.e. that instead of all fibers rotating 1° and therefore having different deformations, that they have equal deformations and therefore different rotations.&lt;/p&gt;&#xA;&lt;p&gt;Again, a good reason to say this isn't the case is because experimental data tells us it isn't. It would also lead to weird questions like &amp;quot;how to interpret the deformation at the center?&amp;quot;&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2020-03-30T17:03:57.547" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34897" PostTypeId="1" AcceptedAnswerId="34899" CreationDate="2020-03-30T18:37:49.287" Score="3" ViewCount="247" Body="&lt;p&gt;I'm programming a drawing plugin for autocad to draw cross-section in beams and ribs. I need to know if it is worth it to add (to my plugin) the feature to add more than two rebar rows, or if I'll be covering the most use cases with two.&lt;/p&gt;&#xA;" OwnerUserId="26358" LastEditorUserId="33" LastEditDate="2020-04-02T14:35:25.707" LastActivityDate="2020-04-02T14:35:25.707" Title="How common is it to use more than two Rebar rows in a beam (bottom or top rebar)?" Tags="|structural-engineering|civil-engineering|autocad|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34898" PostTypeId="2" ParentId="34895" CreationDate="2020-03-30T18:53:27.020" Score="1" Body="&lt;p&gt;In the real world, processes don't stop at exactly the right phase point.  Real world processes are based on the heat transfer that actually occurs, so it will be (for example) less than 100% vapor, or be slightly superheated.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-30T18:53:27.020" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34899" PostTypeId="2" ParentId="34897" CreationDate="2020-03-30T19:53:03.573" Score="1" Body="&lt;p&gt;Depends on the context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Standard residential and office buildings usually don't require much rebar, so two layers should be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Structures under greater loads (i.e. bridges), however, can often have more than two rows. This can obviously include residential/office buildings with more unusual architecture requiring wide spans between columns, etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, I'll leave it for you to determine whether designing for only two rows will satisfy the needs for your project.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-03-30T19:53:03.573" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34900" PostTypeId="1" AcceptedAnswerId="34906" CreationDate="2020-03-30T22:16:46.363" Score="-1" ViewCount="305" Body="&lt;p&gt;Now that Medtronic has open sourced (&lt;a href=&quot;http://newsroom.medtronic.com/news-releases/news-release-details/medtronic-shares-ventilation-design-specifications-accelerate&quot; rel=&quot;nofollow noreferrer&quot;&gt;WIP&lt;/a&gt; reportedly for its completeness) the design of a ventilator, the following question to experts in medical engineering with specialization on lung ventilators.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What could be blocking obstacles, if any, for a typical DIY shop with say mid-range equipment to replicate such a device following these specs?&lt;/p&gt;&#xA;" OwnerUserId="11687" LastActivityDate="2020-03-31T00:35:41.653" Title="Complexity of Medtronic respirator vs. DIY" Tags="|medical-devices|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34902" PostTypeId="1" AcceptedAnswerId="34903" CreationDate="2020-03-30T23:15:32.087" Score="10" ViewCount="2378" Body="&lt;p&gt;On &lt;a href=&quot;https://en.wikipedia.org/wiki/Buckling&quot; rel=&quot;noreferrer&quot;&gt;Wikipedia&lt;/a&gt;, buckling is defined as follows:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;In engineering, buckling is the &lt;em&gt;sudden&lt;/em&gt; change in shape of a structural component under load such as the bowing of a column under compression or the wrinkling of a plate under shear. If a structure is subjected to a gradually increasing load, when the load reaches a critical level, a member may suddenly change shape and the structure and component is said to have buckled. &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Why do we define buckling as a &lt;em&gt;sudden&lt;/em&gt; change in shape? This definition implies that a column is perfectly straight until we apply a load over a critical limit, after which the column suddenly curves sideways. But in real life, columns are not exactly straight and loads are not applied exactly to the center lines of columns so there is a bending moment on the column (and therefore everywhere in the column) for &lt;em&gt;any&lt;/em&gt; load, not just a load above a certain limit.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Classical Euler buckling assumes a perfectly ideal column, and for that buckling appears as a bifurcation solution after a critical load is reached (and not before), but a real life column is not ideal and any eccentricity on the column or the load means that a bending moment is technically present for any load, however small.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;So are real life instances of buckling not strictly speaking buckling according to this definition?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2020-04-02T04:37:58.500" Title="Definition of buckling as a sudden deformation" Tags="|structural-engineering|buckling|" AnswerCount="8" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34903" PostTypeId="2" ParentId="34902" CreationDate="2020-03-30T23:50:24.560" Score="7" Body="&lt;p&gt;Yes, buckling is a sudden change in shape.  A unevenly made column that bows as a result of increasing stress would just experience deformation.  Or a metal bar bent by a &quot;strongman.&quot;&#xA; Buckling is different from regularly occurring plastic deformation.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-30T23:50:24.560" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34904" PostTypeId="2" ParentId="34897" CreationDate="2020-03-31T00:02:13.893" Score="2" Body="&lt;p&gt;Very common. It's also very common to have two or more rows at each of the top and bottom.&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-31T00:02:13.893" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34905" PostTypeId="2" ParentId="34902" CreationDate="2020-03-31T00:08:55.183" Score="1" Body="&lt;p&gt;Because it's a failure. Once it starts, it weakens the structure and, assuming the load is consistent, the strength is reduced and it can now buckle easier in a feedback cycle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Longer spans of wood can 'buckle' and not be a strength issue. Just yesterday I answered the question of what sized piece of wood to use to span between two posts and that would not carry any load, it was purely aesthetic. The failure mode would be for the top of the member, say a 2x4, to roll over. However, this isn't the formal definition of buckling because the member wasn't a compression member. Or is it? The top of the 2x4 is in compression and it has buckled to the side.&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-03-31T00:08:55.183" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34906" PostTypeId="2" ParentId="34900" CreationDate="2020-03-31T00:35:41.653" Score="3" Body="&lt;p&gt;A typical DIY shop would fall over the electronics immediately.  This is more a piece of electronics than a mechanical device.  First of all, the shop would need the source of the chips which as far as I can tell aren't even referenced.  This is a set of assembly drawings, not complete product specifications. It tells you how to build a circuit board, but certainly not the processor at the heart of it. Second, the shop would need a software arm to write the software for the product, which is also not given.  What is given is a set of software &lt;em&gt;Requirements&lt;/em&gt;, which is very different from software design specifications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe a manufacturer of similar devices could use this to aid their design and manufacturing process, but a DIY shop has no chance.&lt;/p&gt;&#xA;" OwnerUserId="24351" LastActivityDate="2020-03-31T00:35:41.653" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="34907" PostTypeId="2" ParentId="34902" CreationDate="2020-03-31T00:38:44.020" Score="3" Body="&lt;p&gt;Real structures do suffer from buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes, real structures are never perfect. Columns aren't perfectly vertical, cross-sections aren't perfectly consistent throughout the span, the material isn't perfectly homogenous, and the loads aren't perfectly centered. That is all true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for loads lower than the true buckling load (which is much smaller than the result from the Euler equation), all of these imperfections merely lead to deformations. If these deformations are lateral, the deformations will increase the applied bending moment, further increasing deformations. But this feedback cycle has a limit at which the second-order effects stabilize and we have a solid column.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, a single grain of sand over the true buckling load will cause a sudden and &lt;strong&gt;&lt;em&gt;immediate&lt;/em&gt;&lt;/strong&gt; deformation of theoretically infinite amplitude. Obviously, real columns don't deform infinitely, they simply collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For an intuitive visual prop, squeeze a raw strand of spaghetti between two fingers. Depending on how you apply the force from your fingers, you can very well deform it laterally in a controlled manner. As you slowly increase the force you apply, the strand will bow further and further out. And then suddenly it'll snap. In this particular example, the collapse mechanism would probably be via buckling, but I think it's a useful visual distinction between &quot;stable deformations&quot; (including second-order effects) and buckling.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-03-31T00:38:44.020" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34908" PostTypeId="2" ParentId="34902" CreationDate="2020-03-31T00:40:15.683" Score="6" Body="&lt;p&gt;Defining buckling as &quot;a sudden change of shape&quot; seems to be mixing up cause and effect IMO.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What happens is that the load-deflection path of the structure &lt;em&gt;bifurcates&lt;/em&gt; when the load reaches a critical value, and the structure then follows whichever branch requires the least energy.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since the stiffness of the two possible paths are usually several orders of magnitude different, it &lt;em&gt;looks&lt;/em&gt; as if there is a &quot;sudden deformation&quot;. On the other hand if you apply a controlled &lt;em&gt;displacement&lt;/em&gt; to deform the structure instead of a load, there may be no sudden deformation at all, but the &lt;em&gt;force&lt;/em&gt; you are applying suddenly decreases. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can also have situations where the post-buckling behaviour is stable and the buckling will reverse when the load is removed - for example &quot;diagonal buckling&quot; in thin plates subjected to shear loads, where the plate buckles (and wrinkles) along the direction of the minimum (compressive) principal stress, but still carries load in the direction of the maximum (tensile) principal stress.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-03-31T00:40:15.683" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34909" PostTypeId="2" ParentId="34902" CreationDate="2020-03-31T01:26:45.657" Score="1" Body="&lt;p&gt;A column or, a bar, even an empty can of soda may have imperfections and or a nonsymmetrical loading and carry a load and deform without buckling. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buckling happens when the load exceeds the critical loading and the deformation is sudden and large and can continue even without increasing the load anymore.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A column may and usually does have defects, such as drilled holes for bolts connecting to other members, residual stresses from manufacturing, and can still support loads. But it will buckle yet after it reaches its critical load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A can of soda, if we load it by placing control weights on a metal plate on top of it, will deform imperceptively until the buckling load is reached, then it will be crushed in like an accordion suddenly and will keep wrinkling even if we don't increase the load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Buckling happens at a stress level called bifurcation point. where the total elastic and plastic energy in the buckled member is less than it would be if the member kept straining.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-03-31T02:09:16.457" LastActivityDate="2020-03-31T02:09:16.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34910" PostTypeId="2" ParentId="34902" CreationDate="2020-03-31T02:46:43.953" Score="0" Body="&lt;p&gt;TL:DR Buckling is a statics phenomena. Collapse is a dynamic phenomena.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;So are real life instances of buckling not strictly speaking buckling according to this definition?&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;They may originate from a buckling event, but then things get complicated at a rate that isn't consistent with how we define buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What's missing from your definition is that buckling is defined as a statics problem, and collapses are inherently dynamic in nature. In classic buckling, the jump to a new geometry happens very fast compared to any change in the geometry at the load points. Buckling in a column happens at a fixed length and some of the stress is instantly released. If forces reappear, the length continues to shorten and the collapse will continue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dynamic impact strength of a thin cylinder can be much higher than predicted by buckling theory if the duration of the impact is so short the deformation jump can't get itself organized (think billiard cues on a break and bow arrows hitting a bone). The critical buckling load is very sensitive to the rate of load application and release after buckling has occurred. &lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2020-03-31T02:46:43.953" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34911" PostTypeId="1" AcceptedAnswerId="35458" CreationDate="2020-03-31T07:40:50.193" Score="-1" ViewCount="215" Body="&lt;p&gt;I was looking at the following form of the Euler equations:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial \vec{v}}{\partial t}+(\vec{v} \cdot \nabla) \vec{v}+\frac{1}{\rho} \operatorname{grad}(p)=\vec{k} \quad \Longleftrightarrow \quad \frac{\partial v_{i}}{\partial t}+\sum_{j=1}^{3} \frac{\partial v_{i}}{\partial x_{j}} v_{j}+\frac{1}{\rho} \frac{\partial p}{\partial x_{i}}=k_{i}, \quad i=1,2,3$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't understand the notation of the gradient (&lt;span class=&quot;math-container&quot;&gt;$(\vec{v} \cdot \nabla) \vec{v}$&lt;/span&gt;)&#xA; Because I only know the gradient for scalar valued functions from &lt;span class=&quot;math-container&quot;&gt;$\mathbb{R^n} \rightarrow \mathbb{R}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It looks to me as if the gradient of a vector field is defined as the Jacobian. Is this correct?&lt;/p&gt;&#xA;" OwnerUserId="26364" LastEditorUserId="12754" LastEditDate="2020-04-01T22:52:18.677" LastActivityDate="2020-04-29T11:20:43.757" Title="Euler-Equations in gradient form" Tags="|fluid-mechanics|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34912" PostTypeId="2" ParentId="34902" CreationDate="2020-03-31T12:39:17.117" Score="-1" Body="&lt;p&gt;This is probably something where a video is worth a thousand words. Check out some videos of actual buckling tests.  Do you see how the deflection increases slowly at first as they pour in the sand and then suddenly the deflection increases a lot? That's buckling &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://youtu.be/l_HlbF4EoJs&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://youtu.be/l_HlbF4EoJs&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="9275" LastActivityDate="2020-03-31T12:39:17.117" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34913" PostTypeId="1" CreationDate="2020-03-31T15:21:54.973" Score="1" ViewCount="25" Body="&lt;p&gt;A group of us are sewing face masks to prevent the spread of sputum. I've learned that the lining of an N95 mask is made of polypropylene.  Polypropylene is also a common interfacing product for sewing.  I want to identify polypropylene in non woven breathable &quot;fabric&quot; products that I have on hand around the house, for example: some &quot;plastic&quot; grocery bags are breathable and, I suspect, made of polypropylene.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there a simple test?  The burn test?  &lt;/p&gt;&#xA;" OwnerUserId="26368" LastActivityDate="2020-03-31T15:21:54.973" Title="I want to test non woven fabrics to see if they are polypropylene, the lining material of N95 masks" Tags="|biomedical-engineering|product-engineering|waterproofing|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34915" PostTypeId="1" CreationDate="2020-03-31T20:56:39.343" Score="-1" ViewCount="341" Body="&lt;p&gt;When analysing a beam, my book starts off by saying that the displacement in a beam is linear:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;u=a1+a2x&lt;/p&gt;&#xA;&#xA;&lt;p&gt;u-displacement&lt;/p&gt;&#xA;&#xA;&lt;p&gt;x-distance from the left end of the beam to the point of interest.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it doesn't explain how do we know it is linear. In a beam you could argue that it's obvious, but in some more complex finite elements, like triangle, displacement is described by some polynomials. How do we know it's not some exponential function for example?&lt;/p&gt;&#xA;" OwnerUserId="21424" LastActivityDate="2020-03-31T21:29:21.583" Title="How do we know the displacement function in finite elements method?" Tags="|finite-element-method|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34916" PostTypeId="2" ParentId="34887" CreationDate="2020-03-31T21:00:35.267" Score="2" Body="&lt;p&gt;Well if engineers didn't learn G-code, why even learn what tool will be operated? Does it matter if its a lathe or mill or multi-axis something? It just produces a part that's an exact match of the CAD model, right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think however this assumes too much perfection and reality in the abstraction of a CAD model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the very least as a teaching device it is useful to learn G-code to gain understanding of the capabilities and limitations of the machine tool which is going to be operated. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Secondly, it is useful to learn about manufacturing because many engineers will go on to design &amp;amp; invent very similar machines themselves. So it serves as an example as well as a tool to make use of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I think more generally -- design optimization should include feedback from the manufacturing process so that a useful part can be produced without extra costs - costs including runtime, material waste, tool wear, etc. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although CAD systems may be capable of abstracting away many of the details of how the G code would actually operate the machine I don't think you would be able to gain as much appreciation for these details if you only ever treated it as a black box.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(And for that matter more generally there are all sorts of &quot;tools&quot; in engineering that you could just use without really understanding them - engineering formulas are after all mostly approximations of physics, you just can plug in the numbers. But engineers are trained to understand at least several levels deeper that what you might need to just minimally scrape by with. This is done in order to recognize the limitations of these tools and generally be able to make sound decisions.)&lt;/p&gt;&#xA;" OwnerUserId="6489" LastActivityDate="2020-03-31T21:00:35.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34917" PostTypeId="2" ParentId="34915" CreationDate="2020-03-31T21:29:21.583" Score="1" Body="&lt;p&gt;To create useful a finite element model, the elements have to satisfy several conditions, for example:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;If you refine the mesh, in the limit the stress and strain within one element will be approximately constant. Therefore the shape functions must be able to represent constant stress and strain. As a special case, they must be able to represent &lt;em&gt;zero&lt;/em&gt; strain - i.e. arbitrary rigid body motion, including rigid body rotation of the element in 2D or 3D models.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The shape functions must be &lt;em&gt;geometrically compatible&lt;/em&gt; across the element edges or faces, otherwise the model will contain cracks or holes when the structure deforms.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A finite element model should give the same results if the complete mesh, loads, and boundary conditions are translated and rotated as a rigid body. For example the beam element should &quot;work&quot; in any arbitrary orientation in space, not just along the X axis of the coordinate system.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Another very desirable property is that solutions automatically satisfy the so-called &quot;patch test&quot;. Suppose you model a simple region (e.g. a rectangular area or volume) and apply boundary conditions where the correct solution is constant stress or strain inside the volume. The FE model should represent this exactly for &lt;em&gt;any arbitrary distorted mesh&lt;/em&gt; that fills the complete area - not just for a &quot;regular&quot; mesh consisting of rectangles or rectangular bricks.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;These requirements mean that the shape functions &lt;em&gt;must&lt;/em&gt; contain constant and linear polynomial terms, and the easiest way to formulate an element is to make &lt;em&gt;all&lt;/em&gt; the terms in the shape functions polynomials (possibly including quadratic and cubic terms as well as linear ones.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some types of elements which do include other &quot;special&quot; types of stress and strain. For example, there are elements which can model the infinite stress field at the end of a crack in the material more efficiently than by using a mesh with very small elements around the end of the crack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incidentally, making FE models that check out these conditions numerically is a good way to test the code of a new element that you have written.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-03-31T21:29:21.583" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34918" PostTypeId="2" ParentId="34897" CreationDate="2020-03-31T21:40:06.327" Score="0" Body="&lt;p&gt;I have seen 4 layers at top and 6 layers in the bottom of some deep beams in a high-rise hospital building.&lt;/p&gt;&#xA;" OwnerUserId="26372" LastActivityDate="2020-03-31T21:40:06.327" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34919" PostTypeId="1" CreationDate="2020-03-31T22:07:11.540" Score="2" ViewCount="44" Body="&lt;p&gt;I'm interested in using linear quadratic regulators (LQR) to model prey avoidance behaviors. I want to adapt the LQR algorithm to incorporate dynamics and state costs that maximize the distance from a fixed point (instead of minimize the distance; 'repulsion' instead of 'attraction'), eg like the prey maximizing the distance from a predator or the predator predicting the trajectory of the prey.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This might only work in a subset of cases (eg with something like trace(Q) &gt; 0; or with some kind of state-dependent trajectory tracking). I'm interested in whether LQR can do this at all, or whether I'll have to add non-linearities (and the tractability of these non-linearities).&lt;/p&gt;&#xA;" OwnerUserId="26373" LastActivityDate="2020-03-31T22:07:11.540" Title="avoidance with LQR" Tags="|control-engineering|control-theory|optimal-control|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34920" PostTypeId="1" CreationDate="2020-03-31T22:37:31.493" Score="2" ViewCount="727" Body="&lt;p&gt;In many sources, such as &lt;a href=&quot;https://www.continuummechanics.org/columnbuckling.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;, it is said that buckling failure is independent of the material's strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in many building codes, yield stress of the material is a factor that changes the allowed buckling load. &lt;a href=&quot;https://en.wikipedia.org/wiki/Perry-Robertson_formula&quot; rel=&quot;nofollow noreferrer&quot;&gt;The Perry-Robertson formula&lt;/a&gt; for example, which is the basis for buckling curves in Eurocode has yield stress as a parameter. The formula is &lt;a href=&quot;http://www.colincaprani.com/files/notes/Column%20Notes.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;derived&lt;/a&gt; by assuming initial curvature on a beam and solving for the allowed load so that yield stress is not reached in the cross section of the beam. So clearly yield stress affects the magnitude of allowed load against buckling. &lt;strong&gt;So why is it correct to say that buckling has nothing to do with yielding?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2021-08-31T05:23:21.440" Title="Why is buckling said to be independent of material strength?" Tags="|structural-engineering|buckling|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34921" PostTypeId="1" CreationDate="2020-03-31T23:06:56.107" Score="3" ViewCount="96" Body="&lt;p&gt;With recent debates, one frequently hears that it's currently not possible for car manufacturers to meet the 55 MPG energy efficiency rating required by 2025.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, is there something physically impossible about meeting this standard? Is it achievable within the next 5 years? If not, what is preventing it?&lt;/p&gt;&#xA;" OwnerUserId="26374" LastEditorUserId="26374" LastEditDate="2020-04-01T03:12:53.590" LastActivityDate="2020-08-29T05:00:45.083" Title="Can we currently achieve 55 MPG on Non-Hybrid vehicles using internal combustion engines?" Tags="|combustion|fuel-economy|" AnswerCount="1" CommentCount="8" ContentLicense="CC BY-SA 4.0" />
  <row Id="34922" PostTypeId="1" CreationDate="2020-03-31T23:32:04.987" Score="2" ViewCount="61" Body="&lt;p&gt;Sorry for such a basic question, but it is a safety concern that brings me here:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We have a hook installed on the ceiling. From it a rope chair (&lt;a href=&quot;https://www.google.com/search?hl=en&amp;amp;client=firefox-a&amp;amp;rls=com.ubuntu::official&amp;amp;um=1&amp;amp;q=hammock%20chair&amp;amp;oq=&amp;amp;tbm=isch&quot; rel=&quot;nofollow noreferrer&quot;&gt;examples&lt;/a&gt;) hangs.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming good form and no wild swinging, will doing pull-ups on this hook place a greater stress on it than me sitting on the rope chair?&lt;/p&gt;&#xA;" OwnerUserId="10805" LastActivityDate="2020-04-01T03:15:44.243" Title="basic physics: does pulling on a hook place greater stress on it than passively hanging from it?" Tags="|mechanical-engineering|structural-engineering|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34924" PostTypeId="2" ParentId="34922" CreationDate="2020-04-01T02:18:41.690" Score="1" Body="&lt;p&gt;I would expect the max stress to be slightly higher when sitting in the chair because of the fact that some people tend to plop down into them. Doing pull ups will create a lot more cycles of near max load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that wood, if that is what it is connected to, has the unusual ability to better resist shorter term loads. An impact load, with a very short duration, doubles it! Even snow loads, that last much longer but are not there all the time, increase the ability to carry the load by 25%.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I doubt that you have much to worry about, assuming that the hook was appropriately sized and attached for the chair.&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-04-01T02:18:41.690" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34925" PostTypeId="2" ParentId="34921" CreationDate="2020-04-01T02:23:36.477" Score="1" Body="&lt;p&gt;It's more than possible. The 2020 Toyota Prius already claims it with 58 on the highway and 56 combined highway and city.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limit is higher. &lt;a href=&quot;https://dothemath.ucsd.edu/2011/07/100-mpg-on-gasoline/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Here is an article&lt;/a&gt; that covers a lot of details&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-04-01T02:23:36.477" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="34926" PostTypeId="2" ParentId="34920" CreationDate="2020-04-01T02:32:10.367" Score="2" Body="&lt;p&gt;I think that your first link covers it. Stress or strain virus the -strength- of the materiel is not part of the buckling equation. This equation provides the upper limit, it is going to buckle at this point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The strength can also be based on the yield strength if the material is stressed to that point before reaching the Euler buckling point. This is the limit based on material strength. If it is lower than the Euler, then it controls design&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then design strength will be lower because of factors of safety, etc&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-04-01T02:32:10.367" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34927" PostTypeId="2" ParentId="34922" CreationDate="2020-04-01T03:15:44.243" Score="1" Body="&lt;p&gt;When you sit down you are lowering yourself slowly and also your butt muscles (gluteus maximus, gluteus medius, gluteus minimus) which are strong and big cushion and break your impact so one could assume roughly you momentarily impart 1.2-1.5 times your weight.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But depending on how fast you swing down from your pull up if you are not experienced to control your descent you could easily Apply 2 times your weight at the bottom deadlock point when falling to a dead stop. and the fingers muscles are not too fat to add any padding.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-01T03:15:44.243" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34928" PostTypeId="1" CreationDate="2020-04-01T05:20:22.570" Score="2" ViewCount="67" Body="&lt;p&gt;With the hype around COVID-19 DIY technology, everyone seems to circle around the same concepts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm trying to think outside the box and I was wondering:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it possible to controllably create an aerogel that is extremely permeable and also captures virus particles?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is in essence a Yes/No question, but I'd love to get some mid-lengthy answers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it's a very interesting concept.&lt;/p&gt;&#xA;" OwnerUserId="1299" LastActivityDate="2020-04-01T05:20:22.570" Title="Aerogel to filter virus particles" Tags="|materials|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34929" PostTypeId="2" ParentId="34885" CreationDate="2020-04-01T06:38:06.613" Score="0" Body="&lt;p&gt;After help from jko I was able to put together a working model for a six link epicyclic gear system. It is essentially three standard planetary gear mechanisms tied together, each with a sun, four planets on a carrier, and a ring gear. I color coded the important interconnections:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) Blue: The left and right planet arrangements share a common carrier. This double-sided carrier is also the main axis of rotation. The two parts making up this carrier mesh and their rotation is coupled.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) Yellow: The middle and right planet arrangements share a common sun gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) Red: The left sun gear is also the middle carrier.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fNTNs.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fNTNs.png&quot; alt=&quot;6 Link Mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21287" LastActivityDate="2020-04-01T06:38:06.613" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34931" PostTypeId="1" AcceptedAnswerId="34933" CreationDate="2020-04-01T12:26:46.433" Score="1" ViewCount="95" Body="&lt;p&gt;Sometimes engines are described as being intended for a particular orientation, e.g., &quot;The 4.6 L; 278.6 cu in (4,565 cc) LD8 (VIN &quot;Y&quot;) is a transverse V8 for front-wheel drive cars.&quot; What exactly does that entail? Why would an engine function differently with transverse instead of longitudinal mounting?&lt;/p&gt;&#xA;" OwnerUserId="26378" LastActivityDate="2020-05-31T18:14:29.547" Title="How do automotive engines differ for longitudinal vs. transverse applications?" Tags="|automotive-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34933" PostTypeId="2" ParentId="34931" CreationDate="2020-04-01T12:55:47.883" Score="1" Body="&lt;p&gt;The design of the engine ie a 90 degree v takes more space than a 60 degree v then the total length of the engine plays a part as well as how the transmission is attached.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-01T12:55:47.883" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34935" PostTypeId="2" ParentId="34920" CreationDate="2020-04-02T00:24:17.697" Score="3" Body="&lt;p&gt;A column can be considered &quot;failed&quot; if it buckles. The equation for the critical buckling load for a slender column contains the material's elastic modulus but not its yield strength. In this sense, the load-to-buckle is independent of the yield strength. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-04-02T00:24:17.697" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34936" PostTypeId="2" ParentId="34911" CreationDate="2020-04-02T00:55:18.157" Score="1" Body="&lt;p&gt;This is operator notation, and the triangle operator in question is &lt;strong&gt;Del&lt;/strong&gt; (or Nabla). It isn't the &lt;strong&gt;Grad&lt;/strong&gt; operator, which is the &lt;strong&gt;Del&lt;/strong&gt; operator (by its lonesome) appearing in front of a &lt;em&gt;scalar&lt;/em&gt; function. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The dot product of a vector and &lt;strong&gt;Del&lt;/strong&gt; is a scalar like any other dot product. So you have a scalar multiplied by a vector.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By itself, I don't know of any name for &lt;em&gt;just that operator part&lt;/em&gt;, but it is the normal notation for these terms in the &lt;a href=&quot;https://physics.stackexchange.com/questions/473758/convective-derivative-vs-total-derivative&quot;&gt;Material Derivative&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://medium.com/@ravikalia/nabla-the-thread-linking-grad-jacobian-and-hessian-operators-8747bd0b6e17&quot; rel=&quot;nofollow noreferrer&quot;&gt;Jacobian&lt;/a&gt; (or its transpose, I couldn't figure out which) has the similar nine terms, but is a dyadic 3 X 3 form. This is a 3 x 1 vector with each member having three terms.   &lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2020-04-02T01:33:37.587" LastActivityDate="2020-04-02T01:33:37.587" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34937" PostTypeId="1" CreationDate="2020-04-02T03:24:20.113" Score="-1" ViewCount="58" Body="&lt;p&gt;The goal is to scale up a general aviation aircraft by 15%. Dynamic similitude is the primary goal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have reviewed research on scaling aircraft models but it seems most of it is focused on creating geometric scaled models and using various methods to interpret the data and make it applicable to the full scale aircraft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am seeking advice for how to create a scaled aircraft where the primary objective is dynamic similitude, with no regard given to geometric similitude. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The aim once again is to scale up an existing aircraft design while keeping the flight characteristics as similar as possible.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you for your time.&lt;/p&gt;&#xA;" OwnerUserId="25002" LastActivityDate="2021-01-14T19:04:10.377" Title="Achieving dynamic similitude by sacrificing geometric similitude?" Tags="|fluid-mechanics|aerospace-engineering|aerodynamics|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34938" PostTypeId="2" ParentId="34902" CreationDate="2020-04-02T04:37:58.500" Score="0" Body="&lt;p&gt;In the real world, and even in most moderately complex numerical simulations, you are right: columns are not perfectly straight and any load applied will be slightly off center and apply a moment in all situations. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;As an exercise take a ruler and push the ends together with your hands. At some point the center of the ruler will bow out in one direction or the other. If you maintain that same force the ruler will remain in the same shape (static equilibrium). Think about this from an energy perspective: while pushing on the ends of the ruler when it is straight there is zero internal energy being stored in the ruler because no displacement occurs. When the stick suddenly flexes your hands are now closer than they were before and you have applied force over that distance, this necessarily means you have done work on the system. That work is conserved as energy stored elastically in the ruler. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The ruler could be bent to the same state purely by applying moments on either end.  During the initial phase of loading, by pure compression, before buckling the amount of moment needed to to get to this same state is offset by the increasing compression force. Eventually enough compression occurs that the additional moment needed approaches zero. You can derive theoretical limits to this for perfect geometries to your hearts content. In the real world due to a variety of factors this limit may be significantly less if you actually tested it to its limit. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some of the potential deformed shapes that are possible can have huge deformations and wild shapes. If your structure was rubber you might be able to see that in the real world. For things like a steel I-beam buckling it will &quot;buckle&quot; according to these theories up until it yields and becomes plastic deformation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Yes in the real world pure buckling according to the theories exists albeit with some correction factors needed for meaningful engineering decisions .. and overbuilding everything is always the best option. &lt;/p&gt;&#xA;" OwnerUserId="26386" LastActivityDate="2020-04-02T04:37:58.500" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34939" PostTypeId="2" ParentId="34937" CreationDate="2020-04-02T05:35:12.253" Score="0" Body="&lt;p&gt;This is actually a bit more complicated than first appears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just one step is to maintain the Reynolds number&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$\text{Re} = \frac{v\cdot l\cdot\rho}{\mu}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;v = flow speed&lt;/p&gt;&#xA;&#xA;&lt;p&gt;l = characteristic length along the flow path, like wing chord&lt;/p&gt;&#xA;&#xA;&lt;p&gt;ρ = fluid density&lt;/p&gt;&#xA;&#xA;&lt;p&gt;μ = dynamic viscosity of the fluid&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and that means to get the same Reynolds number on a wing scaled down by four you need 4 times faster wind. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In real wind tunnels, they know this means supersonic speeds. So they run the test with a slower wind but with correction equations many times built into the hook's strain digitizer. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically because of nonlinear relation between air inertial properties and its dynamic viscosity a wing or a plane can not be modeled linearly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second step is the Mach ratio, relative speed of wing to speed of sound.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One way to make the scale behave similarly is to increase the air pressure hence density by reverse off scaling factor. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are some other factors to watch for.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This link may help you as a starting point.&#xA;&lt;a href=&quot;https://aviation.stackexchange.com/q/52113/9265&quot;&gt;Scaling of planes&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-04-02T06:01:13.113" LastActivityDate="2020-04-02T06:01:13.113" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34941" PostTypeId="1" CreationDate="2020-04-02T09:10:53.753" Score="3" ViewCount="38" Body="&lt;p&gt;I'm doing some work on a domestic water system, mostly 15 or 22mm copper pipe with compression fittings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The work includes replacing various stop-cocks and drain valves some of which are probably 30 or more years old. Some are jammed solid, some turn easily and some turn eventually after squirting with oil and waggling back and forwards for a while.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Most of them rely on brass to brass contact, maybe with a fiber washer, and some gland packing to maintain a seal around the shaft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to replace all these items, if my main criteria is working life expectancy, what should I replace these with? More or less the same thing? Or new ball valves with polished steel ball in a nylon housing? Bearing in mind that whatever I use is only likely to be turned on average, once every few years?&lt;/p&gt;&#xA;" OwnerUserId="20630" LastEditorUserId="6767" LastEditDate="2020-11-01T16:16:09.600" LastActivityDate="2020-11-01T16:16:09.600" Title="What domestic water valve is least likely to jam over time?" Tags="|hydraulics|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="34942" PostTypeId="1" AcceptedAnswerId="34944" CreationDate="2020-04-02T09:31:10.243" Score="1" ViewCount="89" Body="&lt;p&gt;I have a project in which I have to use a up to 2.5mm diameter needle with a 90 degree (L shape) angle similar than the one in the following picture:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/4QWTL.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4QWTL.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I say similar because there is only the tip of the needle is only 1mm since the 90 degree angle&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I need to find some kind of material I can get a wire of that goes inside the needle that will bend so it moves back and forward throughout the L shape of the needle and strong enough so it can push a springed metal present at the 2 or 3 mm from the tip of the needle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the dilemma that if the material is too stiff it won't easily go back and forward because of the L shape but if it is not enough it won't have enough strength to make the spring to compress on the tip end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought about a wooden snake alike solution but since the wire must be so thin I can't find anything to make it of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ttpPr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ttpPr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Maybe you have a better idea. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you so much in advance&lt;/p&gt;&#xA;" OwnerUserId="25017" LastActivityDate="2020-04-02T14:00:59.460" Title="Looking for thin flexible and strong material" Tags="|materials|stiffness|material-science|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34943" PostTypeId="1" CreationDate="2020-04-02T10:35:24.353" Score="1" ViewCount="75" Body="&lt;p&gt;I have a question regarding shockwaves. So as far as my knowledge, a flow moving at Ma= 1 will create a normal shockwave (&lt;span class=&quot;math-container&quot;&gt;$arcsin(1/Ma)$&lt;/span&gt;) , and a flow moving faster will cause an oblique shockwave. So can a  normal shockwave occur if the flow is diverged with angle Beta, if it was close to speed of sound?&lt;/p&gt;&#xA;" OwnerUserId="24243" LastActivityDate="2024-01-09T19:00:45.107" Title="Normal Shockwaves occurrence at cone tip with Angle Theta" Tags="|fluid-mechanics|aerospace-engineering|aerodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34944" PostTypeId="2" ParentId="34942" CreationDate="2020-04-02T14:00:59.460" Score="1" Body="&lt;p&gt;Here is an idea: try using a bundle of very thin wires and fuse the ends together at the tip. You could prototype this with small gauge, multi-strand, electrical wire. Solder the ends together so you have a solid tip for pressing. You can buy stainless steel wire down to 50 microns very inexpensively and fuse the ends together if corrosion is a concern.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will get the combined axial stiffness of the bundle because the fused tip along with the constraint of the needle tip will keep the wires bundled together while it pushes. The wires will be able to turn the corner more easily because they are able to move relative to one another in bending. It's important that you do not fuse both ends or you will loose some of this advantage. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could try something like this: &lt;a href=&quot;https://www.mcmaster.com/34235t27&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mcmaster.com/34235t27&lt;/a&gt; (nylon coated)&#xA;or this: &lt;a href=&quot;https://www.mcmaster.com/3461t63&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mcmaster.com/3461t63&lt;/a&gt; (uncoated)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are braided bundles of 133 wires with an OD of 0.94 mm. I would suggest the uncoated version for most flexibility.  &lt;/p&gt;&#xA;" OwnerUserId="26386" LastActivityDate="2020-04-02T14:00:59.460" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34946" PostTypeId="1" AcceptedAnswerId="34955" CreationDate="2020-04-02T15:49:11.770" Score="2" ViewCount="88" Body="&lt;p&gt;What is the cause of turbojets'/gas turbines' high power to weight ratio compared to piston engines? They certainly use a generally less efficient thermodynamic cycle. &lt;/p&gt;&#xA;" OwnerUserId="18744" LastActivityDate="2020-04-02T20:56:33.667" Title="What gives gas turbine/turbojet engines such high energy density?" Tags="|mechanical-engineering|turbomachinery|jet|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34947" PostTypeId="1" CreationDate="2020-04-02T16:08:51.410" Score="-1" ViewCount="33" Body="&lt;p&gt;A water pipe has an inner diameter of 20 mm. How many liters of soap runs through the tube in ten minutes when the speed is 1.5 meters per second&lt;/p&gt;&#xA;&#xA;&lt;p&gt;what i know so far:&#xA;Q = flow rate is in cubic metres per second (or minute, etc). It means the volume of flow that passes through something in unit time. &quot;Unit&quot; time, for example if I say 10m3/s, it means 10 cubic metres of something passed through in 1 second.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now the formula Q=AV (known as continuity equation) is used to find Q. But how do i find it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;is this the answer: 3,14 * 20*20 / 4 = 100 =&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.14 * 100 = 314 = 0,314 meters&lt;/p&gt;&#xA;&#xA;&lt;p&gt;0,314 meters * 1.5 m/s = 0,471&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How i calculated it&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The area of the pipe is 0.314 m2. The flow rate can be found in 1.5m/s using the formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q = Av&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q = (0.314 m2)(1.5m/s)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q = 0.471 m3/s&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flow rate can be converted to liters per second using: 1 m3/s = 1000 L/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q = 47.1 L/s then i just multiplie this by 60 * 10 ?&lt;/p&gt;&#xA;" OwnerUserId="26393" LastEditorUserId="26393" LastEditDate="2020-04-02T18:29:06.487" LastActivityDate="2020-04-02T18:29:06.487" Title="A question about fluid mechanics" Tags="|fluid-mechanics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="34952" PostTypeId="2" ParentId="34947" CreationDate="2020-04-02T18:28:22.500" Score="1" Body="&lt;p&gt;I'm posting this as an answer because it answers the question, but I'm still unclear as to where you're having trouble...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You are trying to obtain how many liters (volume) go through the pipe over ten minutes (time).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You know the continuity equation &lt;span class=&quot;math-container&quot;&gt;$Q = A \cdot v$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt; is a measure of volume over time. So, for example, if you know that &lt;span class=&quot;math-container&quot;&gt;$Q = 1\text{ liter/second}$&lt;/span&gt;, I hope it's clear how to calculate how many liters you get in ten minutes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So we now know the first thing you need to do is calculate &lt;span class=&quot;math-container&quot;&gt;$Q$&lt;/span&gt;. Thankfully, you know the speed &lt;span class=&quot;math-container&quot;&gt;$v$&lt;/span&gt; (a given in the question), and you have the diameter, so calculating the area &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; shouldn't be too hard either.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-04-02T18:28:22.500" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34953" PostTypeId="2" ParentId="34883" CreationDate="2020-04-02T19:24:52.873" Score="3" Body="&lt;p&gt;In highway design, this is called a superelevation and spiral transition: &lt;a href=&quot;https://en.wikipedia.org/wiki/Track_transition_curve&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Track_transition_curve&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The length of the spiral is based upon the design speed. For USA highway design, guidance is provided in the AASHTO green book &lt;em&gt;A Policy on Geometric Design of Highways and Streets&lt;/em&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Considering that text is not available for free, state highway departments may include AASHTO related information in their posted roadway design manuals.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is a link to the Arizona Dept of Transportation Manual: &lt;a href=&quot;https://azdot.gov/sites/default/files/2019/06/2014-roadway-design-guidelines.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://azdot.gov/sites/default/files/2019/06/2014-roadway-design-guidelines.pdf&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Section 202.3 - Superelevation Transisition on page 200-9 discusses the design of length of the spiral transition.&lt;/p&gt;&#xA;" OwnerUserId="26396" LastActivityDate="2020-04-02T19:24:52.873" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34954" PostTypeId="1" AcceptedAnswerId="35062" CreationDate="2020-04-02T20:49:50.670" Score="1" ViewCount="78" Body="&lt;p&gt;Problem 3.6 in Khalil's Nonlinear Control: &lt;em&gt;Use given Lypunov candidate function to prove that the origin is exponentially stable&lt;/em&gt;. The system is &#xA;&lt;span class=&quot;math-container&quot;&gt;$$\dot{x}=\begin{bmatrix}x_2\\-h(x_1)-2x_2\end{bmatrix},\hspace{0.5cm}h(x_1)=x_1\left(2+\frac{x_1^2}{1+x_1^2}\right)$$&lt;/span&gt;&#xA;and the given Lyapunov candidate function is&#xA;&lt;span class=&quot;math-container&quot;&gt;$$V(x)=\int_0^{x_1}h(s) ds+\frac{1}{2}\left(x_1+x_2\right)^2$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To prove exponential stability, we need to find four positive constants, &lt;span class=&quot;math-container&quot;&gt;$k_1,k_2,k_3,a$&lt;/span&gt;, such that&#xA;&lt;span class=&quot;math-container&quot;&gt;$$k_1\lVert x \rVert^a \le V(x) \le k_2\lVert x \rVert^a$$&lt;/span&gt;&#xA;and&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\dot{V}(x)\le -k_3\lVert x \rVert^a$$&lt;/span&gt;&#xA;We can write the Lyapunov function more explicitly as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$V(x)=\frac{3x_1^2}{2} + \frac{1}{2}\left(x_1+x_2\right)^2 - \frac{1}{2}\ln(1+x_1^2)$$&lt;/span&gt;&#xA;and its time derivative as&#xA;&lt;span class=&quot;math-container&quot;&gt;$$\dot{V}(x)=-\left(\frac{x_1}{\sqrt{2}} + \frac{x_2}{\sqrt{2}}\right)^2 - \frac{3x_1^2}{2}-\frac{x_2^2}{2}-\frac{x_1^4}{1+x_1^2}$$&lt;/span&gt;&#xA;It is relatively clear that we can use &lt;span class=&quot;math-container&quot;&gt;$a=2$&lt;/span&gt;, and that there exists some &lt;span class=&quot;math-container&quot;&gt;$k_1,k_2,k_3$&lt;/span&gt; that will hold.&#xA;But how do we find suitable values for the &lt;span class=&quot;math-container&quot;&gt;$k_i$&lt;/span&gt;'s ?&lt;/p&gt;&#xA;" OwnerUserId="1145" LastEditorUserId="6790" LastEditDate="2020-05-04T13:36:26.907" LastActivityDate="2020-05-04T13:36:26.907" Title="Bounds to prove exponential stablity for given Lyapunov function" Tags="|control-engineering|control-theory|mathematics|nonlinear-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34955" PostTypeId="2" ParentId="34946" CreationDate="2020-04-02T20:56:33.667" Score="3" Body="&lt;p&gt;Turbojets and turbofans support extremely large mass flow rates of air &amp;amp; fuel through them, relative to their weight. (The majority of the work performed by burning that fuel is taken up by running the compressor stage of the engine and what's left over generates thrust.) In comparison, piston engines cannot support similar fuel burn rates relative to their weight.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-04-02T20:56:33.667" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34956" PostTypeId="1" AcceptedAnswerId="34982" CreationDate="2020-04-02T21:43:21.617" Score="0" ViewCount="208" Body="&lt;p&gt;Wikipedia has a good &lt;a href=&quot;https://commons.wikimedia.org/wiki/File:Gearbox_4gears.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;animation&lt;/a&gt; of a manual transmission, but something irks me.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/QSMcf.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/QSMcf.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I understood correctly, in the animation, the input shaft is in green, and the output shaft is in cyan. The dog clutches (synchromesh in modern cars) are therefore between the free gears and the output shaft. This forces both the engine and the gear assembly to change speed in a gear shift. If the synchromesh were between the input shaft and the free gears, only the input shaft (therefore the engine) would need to change speed during a gear change. I don't see anything after a quick Google that suggests things are different in many other gearboxes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason for this that I'm apparently missing? Acting as a flywheel is the only that crosses my mind.&lt;/p&gt;&#xA;" OwnerUserId="6589" LastEditorUserId="3505" LastEditDate="2020-04-03T10:37:25.647" LastActivityDate="2020-04-03T23:55:23.977" Title="Why does the synchromesh in manual transmissions act on the output shaft?" Tags="|car|transmission|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34959" PostTypeId="2" ParentId="34956" CreationDate="2020-04-03T04:29:16.467" Score="1" Body="&lt;p&gt;That is a simple animated drawing, getting a gearbox out of a recent car and stripping it would show you that most gearboxes rarely have dog clutches.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They have synchromesh rings - usually a bronze based material that has a conical surface to match a cone machined on the steel gear - this works to match the gears to the same speed by friction as the gears are brought into mesh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Synchros have been used for many years and were, for a long time, only fitted to third and fourth gears. Commonly called “crash” boxes. As manufacturng processes, costs and the driving population changed then synchro was fitted to all gears except reverse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The synchro works and has no effect on the engine / flywheel because the clutch has disconnected the engine form the input shaft during the gear change.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit to answer based on change to question:&#xA;Then the output shaft is solidly connected to the axle / wheels (unless it is a Saab of a certain vintage) so the synchro makes sure the internals change speed to match the output shaft by friction as mentioned before.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2020-04-03T08:08:30.763" LastActivityDate="2020-04-03T08:08:30.763" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="34961" PostTypeId="1" CreationDate="2020-04-03T06:27:51.183" Score="-1" ViewCount="87" Body="&lt;p&gt;So I am studying Gas dynamics and I know the math regarding Normal and Oblique shockwaves, but I am not sure why would one occur and not the other. Is there a rule when normal shocks will occur or when oblique?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would appreciate some help.&lt;/p&gt;&#xA;" OwnerUserId="24243" LastActivityDate="2020-04-29T10:56:32.663" Title="What determines if a shockwave is normal or oblique" Tags="|fluid-mechanics|aerodynamics|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34962" PostTypeId="1" CreationDate="2020-04-03T06:40:42.513" Score="1" ViewCount="137" Body="&lt;p&gt;Can I use polymer clay to make a mold and then melt HDPE plastic in it in the oven? My goal is to be able to dig a fairly uncomplicated shape in a polymer clay block and then fill that mold with small HDPE plastic filaments. I'm hoping that the melted plastic would fill the mold with the help of gravity pushing it down or I can get it out to apply pressure a few times. Will the polymer clay melt? Do I have to bake it beforehand? If clay doesn't work, what other easily found and cheap substitute exists? Thank you.&lt;/p&gt;&#xA;" OwnerUserId="21282" LastActivityDate="2020-04-03T06:40:42.513" Title="Can I use polymer clay to make a mold and then melt HDPE plastic in it in the oven?" Tags="|plastic|molding|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34963" PostTypeId="1" CreationDate="2020-04-03T09:52:06.713" Score="1" ViewCount="15" Body="&lt;p&gt;There is a method using in automotive industry which is called &lt;a href=&quot;https://www.wikiwand.com/en/Hemming_and_seaming&quot; rel=&quot;nofollow noreferrer&quot;&gt;roller hemming&lt;/a&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a project about this method but before beginning the project I need to learn the standards(EN,ANSI or ISO) which are related to &lt;strong&gt;roller hemming&lt;/strong&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is that &lt;em&gt;how can I learn what are these standards&lt;/em&gt;. When I read some articles about this method unfortunately I couldn't find an answer.&lt;/p&gt;&#xA;" OwnerUserId="26402" LastActivityDate="2020-04-03T09:52:06.713" Title="How can I learn the standards(EN,ANSI,ISO) which is related to &quot;roller hemming method in automotive industry&quot;?" Tags="|mechanical-engineering|automotive-engineering|standards|rolling|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34964" PostTypeId="2" ParentId="31730" CreationDate="2020-04-03T10:44:09.940" Score="0" Body="&lt;p&gt;For the overview I drawed two cast samples. One has a cast surface with varying strength values over the depth – this curve is only qulitative. And on the right there is a sample without cast surface. The strength values are constant along the depth.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/z13TC.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/z13TC.png&quot; alt=&quot;cast with surface and without&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deeper the cast surface is the more loss of the strength values arises. With the strength values I mean tensile strength and yield stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cast surfaces have a specific depth which is not easy to determine generally because the depth is shape dependent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The convection within the melting affects also the thickness of the cast surface. A well marked convection minimises the thickness of the cast surface. Therefore it is recommended to do a simulation of your cast to get proper values for your calculation… Otherwise you have to guess the thickness. It can be 0.2 mm up to ?? mm depending of the shape of your cast and the possibility to allow convection within your melting while solidification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And now the calculation topic:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;If you are taking the whole cast or only the inner cast (without cast surfaces) into account – it‘s up to you…&lt;/li&gt;&#xA;&lt;li&gt;When you are only taking the inner cast you can use the full strength values for your calculations&lt;/li&gt;&#xA;&lt;li&gt;When you want to use the whole cast it is recommended to use a reduction factor for your strength values.&lt;/li&gt;&#xA;&lt;li&gt;In both cases you have to rethink proper the choice of your saftyfactor&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="17780" LastActivityDate="2020-04-03T10:44:09.940" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34965" PostTypeId="1" AcceptedAnswerId="34977" CreationDate="2020-04-03T11:39:15.897" Score="2" ViewCount="52" Body="&lt;p&gt;I'm planning to drive a &lt;a href=&quot;https://www.ebay.com/itm/New-Hiwin-MGN9H-Linear-Guides-MGN-Series-Linear-Bearings-30mm-to-1190mm-Long-/172789096756&quot; rel=&quot;nofollow noreferrer&quot;&gt;small linear guide&lt;/a&gt; (4cm long) with a servo in a setup like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Pswbu.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Pswbu.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if the ball bearings inside would sustain too much stress due to the vertical forces exerted by the links whilst driving it linearly. AFAIK ball bearings can support reasonable radial load, but I'm aware that these guides are usually driven by lead screws instead. Is it a good idea to drive the guide the way presented above?&lt;/p&gt;&#xA;" OwnerUserId="4302" LastEditorUserId="4302" LastEditDate="2020-04-03T12:07:13.600" LastActivityDate="2020-04-03T21:24:13.100" Title="Linear guide ball bearings" Tags="|mechanical-engineering|bearings|linear-motion|servo|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34966" PostTypeId="1" CreationDate="2020-04-03T11:45:19.250" Score="2" ViewCount="58" Body="&lt;p&gt;I have a basic design of two arms which need to be brought together. But I tried the design using a test setup and it did not work successfully. Here is a diagram of my design - &#xA;&lt;a href=&quot;https://i.stack.imgur.com/bEuVQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bEuVQ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What I observed is that arm 2 moves up before arm 1. I want both of them to move towards each other to cause a compression, and away from each other to cause expansion. Is there any way in which that could be achieved using minimal design requirements? &lt;/p&gt;&#xA;" OwnerUserId="20285" LastEditorUserId="77" LastEditDate="2020-04-03T13:28:01.820" LastActivityDate="2020-04-03T21:08:56.150" Title="How to make pivoting arms move symmetrically towards each other?" Tags="|applied-mechanics|pulleys|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34967" PostTypeId="2" ParentId="34966" CreationDate="2020-04-03T13:38:18.920" Score="2" Body="&lt;p&gt;I can think of two options:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) have the arms connected by gears at the pivot point - when one is moved so will the other, in fact one could be fixed...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) connect the arms with a threaded rod (needs left &amp;amp; right hand threads) and use a pulley to rotate the rod - then the arms will move according to the  pitch of the thread - actually 1 turn will move twice the pitch.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2020-04-03T13:52:46.857" LastActivityDate="2020-04-03T13:52:46.857" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34968" PostTypeId="2" ParentId="34966" CreationDate="2020-04-03T14:32:08.973" Score="0" Body="&lt;p&gt;You could try 2 bar linkages, it's basically what you're already trying but without a pulley and unequal displacement.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-04-03T14:32:08.973" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34969" PostTypeId="1" AcceptedAnswerId="34994" CreationDate="2020-04-03T15:04:59.580" Score="1" ViewCount="1422" Body="&lt;p&gt;I'm making a Z (coffee table) linkage with two interlocking C channels and a couple of steel linkage bars. While moving, they aren't load bearing but when extended, they will take a weight of around 300lbs. I need to figure out whether 3mm T66 aluminium can take that weight without tearing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The entire weight will be spread across two bolts (10mm with bushing) and the top half of that so that's Pi x 10mm diameter / 2 because the weight is all on the top half x 2 bolts - around 31.4mm x 3mm thick = 94.2mm squared. I've found &lt;a href=&quot;https://www.makeitfrom.com/material-properties/6060-T66-Aluminum/&quot; rel=&quot;nofollow noreferrer&quot;&gt;tables of properties&lt;/a&gt; for T66 which give figures in MPa (megapascals?) but that's where my half-remembered high school physics gives out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I calculate the maximum load across a given area from these properties?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: To include sketch.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/nahfP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nahfP.png&quot; alt=&quot;Unwin rails&quot;&gt;&lt;/a&gt;&#xA;The block to the left is the (fixed) seat base, along the top are Unwin rails along which the seat can slide. When not needed, the linkage around the 4 bolts can rotate it away into the floor. When they are raised (as shown), the seat+occupant, for which I am allowing 300lbs) can sit directly over the hinged section. Ignoring the rail overlap, that means that the full weight is supported by the top of the two round pivots (actually, since it's made from C channel and there are one of these on each side, that's 2x2x2x10mm bushings in 3mm aluminium).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this makes the question clearer.&lt;/p&gt;&#xA;" OwnerUserId="24671" LastEditorUserId="24671" LastEditDate="2020-04-03T22:56:38.103" LastActivityDate="2020-04-05T04:17:48.327" Title="Calculating tensile strength for aluminium" Tags="|aluminum|material-science|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="34970" PostTypeId="1" CreationDate="2020-04-03T15:24:38.087" Score="1" ViewCount="771" Body="&lt;p&gt;I have a pressure regulator to maintain a specified pressure in a vessel that also contains a liquid. The regulator can pressurize or relieve pressure of the vessel. I want to prevent the liquid from flowing back to the regulator while keeping the gas flowing both ways. I can imagine a device that would do that but can't find a readily available one on the market. Any ideas?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought simplifying my problem was gonna help but now I'm thinking that sharing the exact context of my situation might help you help me.&#xA;So, this situation is happening on a counter-pressure beer canning machine (cheers!) The pressure regulator maintains a set pressure of CO2 in the machine's can filling tank. The level of beer is this tank is regulated by an actuated globe valve and the beer transfers from the BBT to the machine's tank by a differential pressure between the two tanks. During operation, the regulator has to supply and relieve pressure from the machine's tank (two-way flow). My issue is that if for any reason the beer supply valve doesn't close completely, the beer level rises up to the CO2 supply port, eventually overflows right into the regulator and ruins it...so as a safety, I want to install a check valve that would only close in the presence of liquid (beer) while allowing gas (CO2) to flow both ways the rest of the time.&lt;/p&gt;&#xA;" OwnerUserId="26406" LastEditorUserId="26406" LastEditDate="2020-04-06T14:05:23.490" LastActivityDate="2020-04-06T14:05:23.490" Title="Is there a type of check valve that allows gas to flow in both direction but stops liquid from flowing back?" Tags="|gas|valves|liquid|" AnswerCount="3" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34971" PostTypeId="2" ParentId="31730" CreationDate="2020-04-03T15:27:57.920" Score="0" Body="&lt;p&gt;There is no significant loss of cast surface strength in ductile cast iron . In fact the ASM Handbook ( 9th ed.) shows higher strength near the surface , the amount of strengthening depends on grade and thickness of casting. Basically caused by the cooling rate; faster cooling = more strengthening.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-04-03T15:27:57.920" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34972" PostTypeId="2" ParentId="34970" CreationDate="2020-04-03T16:58:17.527" Score="2" Body="&lt;p&gt;If the flow rate is low for the gas then a ball that floats on the liquid will work - similar to how most wet vacs stop liquid getting into the motor...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-03T16:58:17.527" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34974" PostTypeId="2" ParentId="34970" CreationDate="2020-04-03T20:36:14.867" Score="1" Body="&lt;p&gt;Any form of gravity trap. For examplea piece of vertical pipe with a large enough diamter will work - liquid gathers at the bottom, diameter need to be large enough to let bubbles pass through the fluid, avoiding plug flow.  &lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2020-04-03T20:36:14.867" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34975" PostTypeId="2" ParentId="34966" CreationDate="2020-04-03T21:08:56.150" Score="2" Body="&lt;p&gt;There are many ideas which have worked for different gears&lt;/p&gt;&#xA;&#xA;&lt;p&gt;like this drafting compass with a threaded shaft and a roller:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/cTRWF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/cTRWF.png&quot; alt=&quot;drafting compas&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this pantograph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DvSO3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DvSO3.png&quot; alt=&quot;pantograph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-03T21:08:56.150" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34976" PostTypeId="2" ParentId="34970" CreationDate="2020-04-03T21:21:04.270" Score="1" Body="&lt;p&gt;A &quot;steam trap&quot; should do what you want. There are a few types with different characteristics .&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-04-03T21:21:04.270" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34977" PostTypeId="2" ParentId="34965" CreationDate="2020-04-03T21:24:13.100" Score="0" Body="&lt;p&gt;If you limit the retracted angle of the arm to &lt;span class=&quot;math-container&quot;&gt;$\ a&amp;lt;60^{\circ}  $&lt;/span&gt; it should be ok.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would limit the maximum vertical load at two times the horizontal push. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  F_y = \  \leq 2F_x$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-03T21:24:13.100" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34980" PostTypeId="1" CreationDate="2020-04-03T22:08:42.893" Score="1" ViewCount="281" Body="&lt;p&gt;Essentially I have a 100 lbs. pile driver with an eyelet for lifting it on top. (It's on rails) I want to set up and automated machine to lift and drop the driver at a variable pace. Preferably I would like to use a cheap electric motor. I can use pulleys, and/or a block and tackle system to reduce the amount of work the motor needs to do. I can route/3D print materials if need be. I have looked at mechanical systems and setups all over the internet, but have yet to find a simple effective solution for what I am trying to do.&lt;/p&gt;&#xA;" OwnerUserId="26413" LastActivityDate="2020-04-04T12:03:31.053" Title="What's a simple machine to lift and drop a 100 lbs weight repeatedly using an electric motor?" Tags="|mechanical-engineering|motors|pulleys|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="34982" PostTypeId="2" ParentId="34956" CreationDate="2020-04-03T23:55:23.977" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/EMEQi.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/EMEQi.png&quot; alt=&quot;Honda B Series&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Image from &lt;a href=&quot;https://www.j-k-tuning.com/Transmission/BTrans2.php&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.j-k-tuning.com/Transmission/BTrans2.php&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can put them anywhere you want to (and people do). But it is easier to put them on big gears than little ones. If your four speed was working the other way around with the green as an output shaft, all the gears would have to have the inverse ratios, and you would have all your syncros on little gears.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the image above, you can see both shafts carry shift forks in this model.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2020-04-03T23:55:23.977" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34983" PostTypeId="1" CreationDate="2020-04-04T00:23:16.937" Score="-1" ViewCount="56" Body="&lt;p&gt;Is there a relation between engine size and its power to weight ratio? For example if you shrink the engine to half the size will it have exactly half the power?&lt;/p&gt;&#xA;" OwnerUserId="9199" LastActivityDate="2020-04-05T00:53:00.437" Title="Relation between engine size and its power to weight ratio" Tags="|mechanical-engineering|energy-efficiency|engines|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34984" PostTypeId="2" ParentId="34983" CreationDate="2020-04-04T01:15:24.950" Score="2" Body="&lt;p&gt;No, more like 1/8. Power (and weight) is most closely related to volume, which changes with the cube of the linear dimensions.&lt;/p&gt;&#xA;" OwnerUserId="86" LastActivityDate="2020-04-04T01:15:24.950" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34985" PostTypeId="2" ParentId="34980" CreationDate="2020-04-04T01:29:39.970" Score="0" Body="&lt;p&gt;You can have your motor attached to a vertical chain rolling around 2 pulleys.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This chain with several  brackets constantly moves  a one hundred lbs metal ballast which is assembled on the shaft of your mechanical hammer and releases it on top. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It drops down repeatedly, banging you jack-hammer onto the job. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-04T01:29:39.970" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34986" PostTypeId="1" CreationDate="2020-04-04T01:59:19.140" Score="1" ViewCount="45" Body="&lt;p&gt;I have a pneumatic cylinder with a stripped fitting, which causes a slow leak. I took it to a machine shop and he tried to re-thread it but there wasn't enough good metal left to get a good thread. He told me there wasn't enough room for an oversize fitting and suggested I try liquid thread sealant. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unfortunately, this hasn't worked. There is still a slow leak.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is for a DIY RV loft type bed that uses 4 pneumatic cylinders to lift the bed up toward the ceiling, which will never need to support more than about 25 LBS or so - just the weight of the bed.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm thinking I should maybe try Blue Block or epoxy, but hoping to get some advice first.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Much thanks...Jon&lt;/p&gt;&#xA;" OwnerUserId="26415" LastActivityDate="2020-04-04T01:59:19.140" Title="How to repair stripped, leaky pneumatic fitting?" Tags="|aluminum|threads|pneumatic|" AnswerCount="0" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="34987" PostTypeId="1" CreationDate="2020-04-04T02:31:24.530" Score="1" ViewCount="15" Body="&lt;p&gt;I would like to create a boss using a spline with many points, then scale it using a design table.  The catch is that I don't want to manually set dimensions (in x,y, and z) for each of the points of the spline, for adding to the table.  I would just like -all- the dimensions of all the sketches to be automatically added, if I so desire.  does solidworks have  this functionality?  I can understand that this  might result in a night-mare, but then again, it might not be so hard to group the dimensions by x, y, and z for each point on the spline..&lt;/p&gt;&#xA;" OwnerUserId="13081" LastActivityDate="2020-04-04T02:31:24.530" Title="how to automatically add non-specified dimensions to a design table" Tags="|solidworks|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34988" PostTypeId="1" AcceptedAnswerId="35003" CreationDate="2020-04-04T06:05:26.650" Score="1" ViewCount="38" Body="&lt;p&gt;I am reading a textbook on fundamentals of vibrations and it has a very brief introduction to shock spectra. It mentions that &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;For comparison purposes, it is customary to use the response of an&#xA;  undamped single degree of freedom system to obtain the shock spectrum.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So is it that we always find the maximum response of an undamped single DOF system to get the shock spectrum? If yes, then how does it help us, because can't the response of the structure which I am analysing (say a building) be widly different from that of a single DOF system? For example, consider two shocks. The shock which seems &quot;stronger&quot; (higher maximum response in general) when using a single DOF system may actually be weaker when using my structure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, just saying an undamped single DOF system seems vague because response to the same shock may differ for different single DOF systems. Is there a specific single DOF system that is used here (like the mass-spring system)?&lt;/p&gt;&#xA;" OwnerUserId="26419" LastActivityDate="2020-04-05T03:11:28.847" Title="Is shock spectrum always obtained for an undamped single degree of freedom system?" Tags="|structural-engineering|structural-analysis|vibration|shock|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34989" PostTypeId="1" CreationDate="2020-04-04T06:22:17.400" Score="0" ViewCount="88" Body="&lt;p&gt;We have all seen PVT diagrams. From my understanding one should measure pressure, volume and temperature on or near the outside of the system. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Volume: maybe by displacing water initially and then cameras.&lt;/li&gt;&#xA;&lt;li&gt;Temperature: I have some thermometers lying around. But since the system can have different temperatures on different parts of its boundary I don't know how many places I need to measure it.&lt;/li&gt;&#xA;&lt;li&gt;Pressure. I don't know how to measure this. Maybe if I hang the system on a hook I can know the pressure at the point it was hung from, but all the other points have me stumped.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Have someone made PVT diagrams and know how it is done?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I found some methods to measure pressure in fluids:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;fiber optical submersible pressure sensor&lt;/li&gt;&#xA;&lt;li&gt;pitot tube + manometer&lt;/li&gt;&#xA;&lt;li&gt;prandtl tube + manometer&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="26420" LastEditorUserId="26420" LastEditDate="2020-04-12T18:47:26.367" LastActivityDate="2020-04-12T18:47:26.367" Title="How to measure and create PVT diagram" Tags="|thermodynamics|measurements|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="34990" PostTypeId="1" CreationDate="2020-04-04T06:57:16.923" Score="2" ViewCount="120" Body="&lt;p&gt;Process Engineering Newbie here.&#xA;What kind of Steam Traps are best suited for condensate removal in super-heated steam line or manifold?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The super-heated steam is to be supplied to a process reactor and will be at a temperature of around 390 to 420 degree Celsius.&#xA;Also the system is supposed to run at different pressures ranging from 1 to 19 barg. I understand that this is not an optimal system. This is a research project and we are developing a proof-of-concept Prototype and is not for deployment in the industry. Hence, the wide range of pressure and high temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope experienced engineers in the forum can guide me on how to select a suitable type of steam-trap.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks in Advance!&lt;/p&gt;&#xA;" OwnerUserId="19228" LastActivityDate="2020-04-04T06:57:16.923" Title="What's the best type of steam Trap for Superheated Steam?" Tags="|mechanical-engineering|process-engineering|pressure-vessel|steam|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34992" PostTypeId="2" ParentId="34969" CreationDate="2020-04-04T09:40:53.303" Score="0" Body="&lt;p&gt;Okay, trying to refresh my knowledge with a trawl through Wikipedia and online physics text books.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The yield strength is the point where the material can return to its normal shape after deforming under stress so I guess that's what I'm looking for. For this material, it's 170 MPa. While we're at it, let's go metric on the mass to avoid mixing units so 300lb = 136Kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As stated earlier, all of the weight is applied to the top half of 2x10mm x 2 sides per C channel x 2 units per seat. &#xA;Circumference = pi x d&#xA;Half the circumference = pi x d / 2 = pi x 10mm / 2 x 3mm thick = 47mm2 per hole (approx). &#xA;2 holes per side per mount = 8 bolts = 376mm2 total.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the question now is &quot;Which is greater, 170MPa or 136Kg spread across 376 square millimeters?&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to Wikipedia, 1 Pascal = 1Kg/m2, so the load is 136 / 0.376 = 362Pa.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So by my reckoning when the load is measured in units and the capacity is measured in hundreds of mega units, I should be fine.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right? I'm open to corrections on any of this. Google-reckoning can only get you so far.&lt;/p&gt;&#xA;" OwnerUserId="24671" LastEditorUserId="24671" LastEditDate="2020-04-04T09:49:52.377" LastActivityDate="2020-04-04T09:49:52.377" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34993" PostTypeId="2" ParentId="34980" CreationDate="2020-04-04T12:03:31.053" Score="1" Body="&lt;p&gt;100 lbm =~ 45 kgm.&lt;br&gt;&#xA;Wattage =~ kg x metres_vertical x 10 / time_seconds.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So per metre of head,&lt;br&gt;&#xA;power = 450 Watts / seconds_to_lift.&lt;br&gt;&#xA;A battery electric drill (or even a mains drill) with a &quot;drum&quot; driven by the chuck can be as low geared as you wish. You can also buy geared motors. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At say 120 RPM = 2 RPS (drill on low speed screw driving mode) say with a 25mm / 1 inch  dia drum&lt;br&gt;&#xA;~= 80 mm circumference = 160mm vertical/second&lt;br&gt;&#xA;= 6s for 1m lift so about 450/6 = 75 Watts. Say 100 Watts plus.   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Smaller dia drum = slower = less watts. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Release could be electromagnet.&lt;br&gt;&#xA;Could use endless 'rope' with 'carriages' or return carriage each time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A threaded rod driven by a drill or geared motor is also a potential low cost and 'not too hard&quot; solution.&lt;/p&gt;&#xA;" OwnerUserId="399" LastActivityDate="2020-04-04T12:03:31.053" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34994" PostTypeId="2" ParentId="34969" CreationDate="2020-04-04T13:26:25.283" Score="2" Body="&lt;p&gt;I confirm that in your case the yield stress is an adapted limit value.&#xA;However, there is a unit error in your computation. You forgot to convert the area in m&lt;sup&gt;2&lt;/sup&gt; which means that the results is not 362 Pa but 362x10&lt;sup&gt;6&lt;/sup&gt; Pa = 362 MPa. &#xA;Assuming that your reasoning is correct to compute the effort, your are above the limit of 170 MPa.&lt;/p&gt;&#xA;" OwnerUserId="26427" LastEditorUserId="90" LastEditDate="2020-04-05T04:17:48.327" LastActivityDate="2020-04-05T04:17:48.327" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="34995" PostTypeId="1" CreationDate="2020-04-04T14:19:38.430" Score="2" ViewCount="152" Body="&lt;p&gt;I recently found the following link &lt;a href=&quot;https://anglingai.com/224b-4-4-blood-line-swim-bait-mold/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://anglingai.com/224b-4-4-blood-line-swim-bait-mold/&lt;/a&gt;.&#xA;There you will see that they sale hundreds of different aluminum molds related to fishing. These molds are really inexpensive (from 50 up to 200 USD). How they are producing them? I though that molds cost thousands of dollars.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am currently in progress on designing a product and I will need a mold for the enclosure of the product for low production runs (no more than 200-300 items per month). The enclosure will need to be either from ABS or some other kind of strong material that is not brittle. Probably also, the material will need to somehow be manually injected in order to reduce the total cost.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where can I find someone to produce me low cost molds (like the above) for a product design that I will create with Fusion 360? Can this be done or I am really naive? I am mostly interested for the outside finish of the enclosure. Inside it there will be electronics that will be potted with special type of resin in order to protect them from the elements. As such, the inside will be as simple as possible.&lt;/p&gt;&#xA;" OwnerUserId="16386" LastActivityDate="2020-04-10T11:38:32.393" Title="Low cost molds. Can such molds be produced?" Tags="|machining|cnc|molding|" AnswerCount="5" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="34996" PostTypeId="2" ParentId="34995" CreationDate="2020-04-04T16:17:35.023" Score="1" Body="&lt;p&gt;The auto industry does/did used molds of zinc ( Zamack , kirksite) to make steel forming molds for limited numbers. I expect if you make a clay ( or equal) model of the part . Then  pour molten zinc around it ( melts a little under 800 F). You will need parting lines , gates , etc, in the mold to get the clay out and your plastic in.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-04-04T16:17:35.023" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34997" PostTypeId="2" ParentId="34995" CreationDate="2020-04-04T17:12:59.483" Score="2" Body="&lt;p&gt;You could look into &lt;a href=&quot;https://www.protolabs.com/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Protolabs&lt;/a&gt;. They provide a range of services from 3D printing, CNC machining, short run injection molding using aluminum tooling, production molding and more. They are really fast. They have an &lt;a href=&quot;https://www.protomold.com/ProtoQuote.aspx&quot; rel=&quot;nofollow noreferrer&quot;&gt;online quoting and ordering system&lt;/a&gt; which includes an analysis of the moldability of your part. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low cost aluminum tooling is cheaper and faster to make then traditional steel molds, but surface finish isn't as good and the number of parts that can be made before the mold wears out is much lower. Just having a mold isn't enough as you still need a molding machine to create parts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just to be clear, I have no association with Protolabs other than once being a customer. Undoubtedly there are other good shops. &lt;/p&gt;&#xA;" OwnerUserId="9506" LastEditorUserId="9506" LastEditDate="2020-04-10T11:38:32.393" LastActivityDate="2020-04-10T11:38:32.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34998" PostTypeId="2" ParentId="24601" CreationDate="2020-04-04T18:26:25.690" Score="1" Body="&lt;p&gt;A device used to heat domestic potable water to about 120 &lt;sup&gt;&lt;span class=&quot;math-container&quot;&gt;$\circ$&lt;/span&gt;&lt;/sup&gt;F (50 &lt;sup&gt;&lt;span class=&quot;math-container&quot;&gt;$\circ$&lt;/span&gt;&lt;/sup&gt;C) is usually called a hot water heater. To heat water above that temperature the device used is called a boiler.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hot water boilers are usually required to heat hot water systems and can operate from low pressures to 160 psig (1100 kPa) max.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Steam boilers provide steam to a steam system that is used for heating only.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Low pressure steam boiler or heating steam boiler can operate to a maximum pressure of 15 psig (103 kPa). High pressure steam boiler (as per ASME) operates at pressures over 15 psig (103 kPa).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;High pressure and high temperature hot water boilers, the ones exceeding the above, are found in Manufacturing plants, hospitals etc. These types of boilers will need heavy duty non pressurized expansion tanks. &lt;/p&gt;&#xA;" OwnerUserId="26429" LastEditorUserId="90" LastEditDate="2020-04-05T04:13:53.360" LastActivityDate="2020-04-05T04:13:53.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="34999" PostTypeId="2" ParentId="34995" CreationDate="2020-04-04T19:47:21.953" Score="2" Body="&lt;p&gt;Judging from the pictures, they are CNC machined out of solid aluminium.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That isn't an expensive process, so long as you have the CNC machine. A basic one that isn't junk for sale on E-bay would cost a few thousand dollars, which isn't a big deal for a commercial company if their selling price amounts to hundreds of dollars per hour of machining time.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-04-04T19:47:21.953" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="35000" PostTypeId="1" CreationDate="2020-04-05T00:10:46.703" Score="0" ViewCount="36" Body="&lt;p&gt;I know this depends on a variety of factors including overall traffic volumes, peak traffic volumes, types of vehicles, and type of road, but how exactly do traffic engineers determine how many lanes a section of road should contain and when the number of lanes should be increased? &lt;/p&gt;&#xA;" OwnerUserId="12615" LastActivityDate="2020-04-05T21:59:01.910" Title="What determines when a road should be widened?" Tags="|transportation|highway-engineering|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35001" PostTypeId="2" ParentId="34995" CreationDate="2020-04-05T00:45:13.067" Score="1" Body="&lt;p&gt;cheap molded parts are made by sharing space in a larger mold used to make parts for many customers at once. such a setup is called a MUD mold; each individual customer owns an &lt;em&gt;insert&lt;/em&gt; in the master mold and when the molding shop runs the mold they open up the gates to those mold cavity inserts whose customers desire parts at that time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The parts exit the mold &quot;treed&quot; to one another on a common &lt;em&gt;sprue&lt;/em&gt;. this is then inserted into a press with razor knives that detach each part from the sprue and catch it in its own box. the individual boxes are then sent to the customers. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-04-05T00:45:13.067" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35002" PostTypeId="2" ParentId="34983" CreationDate="2020-04-05T00:53:00.437" Score="2" Body="&lt;p&gt;as you shrink a heat engine, its volume (in which power is developed) falls faster than its surface area (through which losses occur) which means a small engine will inherently be lossier than a large engine of the same design. This scaling relationship is universal, which means that engine designers prefer one large engine instead of two smaller ones. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-04-05T00:53:00.437" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35003" PostTypeId="2" ParentId="34988" CreationDate="2020-04-05T03:11:28.847" Score="1" Body="&lt;p&gt;Not necessarily, many shock response spectra are created for systems under the impulse of a shock with damping.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However many codes allowing seismic design using the response spectra with an estimated 100-year or whatever design earthquake require the assumption of the three spectra &lt;span class=&quot;math-container&quot;&gt;$ S_d, \ S_v, \, and \ S_a $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;related as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  S_v= \omega S_d\\ S_a= \omega^2 S_d$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And this is valid only for a single degree of freedom system with no damping.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-05T03:11:28.847" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35004" PostTypeId="1" CreationDate="2020-04-05T07:16:01.737" Score="3" ViewCount="42" Body="&lt;p&gt;I exported a model of the mechanism created in SolidWorks to SimMechanics in order to explore the possibilities of control. To solve the problem, i need to know the angular position of the mobile platform. In reality, I can do this with the help of a sensor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to calculate these angles in SimMechanics?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/axMEu.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/axMEu.gif&quot; alt=&quot;Structure&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5BrfM.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5BrfM.png&quot; alt=&quot;Mechanism&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="24680" LastActivityDate="2020-04-05T07:16:01.737" Title="How to measure / calculate Tait-Bryan angles of a moving platform in SimMechanics" Tags="|mechanical-engineering|simulation|solidworks|kinematics|simulink|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35006" PostTypeId="1" AcceptedAnswerId="35009" CreationDate="2020-04-05T08:47:18.937" Score="0" ViewCount="70" Body="&lt;p&gt;See video:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=75Z7-gmd_qk&amp;amp;t=226&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=75Z7-gmd_qk&amp;amp;t=226&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This does not make sense to me. As far as I got it, bottom pipe is isolated from top 2 and just controls valve of the top 2 pipes. So, when air from bottle goes into left &quot;liar&quot; trough top 2 tubes of the right one it should just stop there. There is no way air can get to the middle one and also return to the right one to close/open it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope somebody can explain this to me, because I'm starting to loose my mind a bit over this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used Transistor's schematic to illustrate my frustration better:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/s2FC3.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/s2FC3.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See, in the video it's clearly seen that input air pipe goes only into A cylinder's input pipe. B and C does not receive air at all! Also it seems that B and C are closed initially, while A is open. And I can't really see any springs that you mentioned either. This is what really frustrates me. How B and C can operate if they are not even connected to air supply?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ctBXY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ctBXY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;OK, I'm feeling really embarrassed now. Are this tubes actually fused together? The problem is, Theo holds his finger over it almost all the duration of operation and it really seems that this 2 tubes are separate... In this case, Transistor's schematic is correct and my problem is solved.&lt;/p&gt;&#xA;" OwnerUserId="26436" LastEditorUserId="26436" LastEditDate="2020-04-05T17:46:41.293" LastActivityDate="2020-04-07T09:19:06.330" Title="How does 3 &quot;liars&quot; in Strandbeests work?" Tags="|fluid-mechanics|pneumatic|" AnswerCount="3" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="35007" PostTypeId="1" AcceptedAnswerId="35049" CreationDate="2020-04-05T09:00:42.563" Score="3" ViewCount="569" Body="&lt;p&gt;(&lt;a href=&quot;https://drive.google.com/drive/u/0/folders/1G6pZJOcN0G8Zk7YzqGfa1C8KK2aCzxId&quot; rel=&quot;nofollow noreferrer&quot;&gt;solidworks files here&lt;/a&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've made a small assembly thet mooves smoothly manually, but when I try to moove it the same way is motion study, it collides is a strange way... what can be the cause of that?&lt;a href=&quot;https://i.stack.imgur.com/A5Xl9.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/A5Xl9.png&quot; alt=&quot;manually&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xocd0.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xocd0.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26435" LastEditorUserId="26435" LastEditDate="2020-04-05T13:39:12.810" LastActivityDate="2020-04-07T10:22:55.473" Title="Solidworks motionstudy/motion study causes parts to &quot;jump&quot;" Tags="|mechanical-engineering|cad|solidworks|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="35008" PostTypeId="2" ParentId="35006" CreationDate="2020-04-05T09:27:06.100" Score="0" Body="&lt;p&gt;one of them (consider it randomly) will overcome the others in the initial relese of pressure. instead of thinking about it as input and output I fined it more comfertable to look at it as an electronic switch- the upper tube is &quot;+&quot;, the middle is &quot;-&quot; and the bottom is &quot;info/data&quot; (kind of like a transistor). anyway, here's a picture that I hope will help:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/vUpcf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vUpcf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26435" LastEditorUserId="26435" LastEditDate="2020-04-05T13:41:12.303" LastActivityDate="2020-04-05T13:41:12.303" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35009" PostTypeId="2" ParentId="35006" CreationDate="2020-04-05T09:51:13.560" Score="2" Body="&lt;p&gt;Draw the schematic!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/625lh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/625lh.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. (a) The circuit. (b) Step 1. (c) Step 2.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The circuit relies on slight differences in timing of the various components due to differences in friction and tolerance to get started. As shown in 1(a) the circuit is just three pilot operate valves. The springs push the spools into the position shown and all valves are &quot;open&quot; (they will pass air). Note that before air is connected each valve is feeding through to the pilot of the next valve.&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;In 1b the air has been turned on. All three pilots were being given air but B has changed over first. This cuts the pilot air for C ensuring that it stays in the open position and supplies air to the pilot of A.&lt;/li&gt;&#xA;&lt;li&gt;When A turns on it cuts the supply to B. B returns to the open position turning on C and cutting the supply to A.&lt;/li&gt;&#xA;&lt;li&gt;A returns to open and turns on B again.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The sequence should run B - A - C - B - A - C - B - A ...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZaSnF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZaSnF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 2. A more physical representation of the circuit.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Note that the air pressure on top tends to drive the valves open. This replaces the springs in my Figure 1.&lt;/li&gt;&#xA;&lt;li&gt;The &quot;pilot&quot; actuators on the bottom have a larger diameter so they will always beat the top side and be able to close the valves.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2020-04-05T18:07:59.270" LastActivityDate="2020-04-05T18:07:59.270" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35010" PostTypeId="1" CreationDate="2020-04-05T12:26:40.050" Score="1" ViewCount="42" Body="&lt;p&gt;I want to melt a bed of plastic powder so that the powder bed once melted and then cooled forms a solid sheet of plastic.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to melt the powder I'm thinking of passing a nichrome (Ni80/Cr20) wire over the bed without the wire touching the bed. With the direction of motion of the wire perpendicular to the length of the wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The heat from the wire should radiate from the wire and pass into the plastic powder in order to melt the powder. Is this possible? Is it possible to to melt plastic powders through heat radiation from a nichrome wire.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Conditions:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;As an alternative to melt the bed with radiation, one could heat the bed's platform. But I can't do that because the  bed's platform is made of plastic and might begin to melt if I heat it. So heating through radiation is the only viable option I've thought of. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Bed dimensions: 30 cm X 15 cm.   &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;I can use a wire of any diameter. Single wire or multiple wire in parallel.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Melting temperature of plastic powder  used is 270-300°C.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Any distance between the wire and the powder bed is possible. Close distance between the wire and powder is preferred.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;If not nichrome any other material can do the job.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="16605" LastEditorUserId="16605" LastEditDate="2020-04-05T12:38:03.160" LastActivityDate="2020-04-05T21:13:06.587" Title="Melt plastic powder bed with radiant heat?" Tags="|electrical-engineering|thermodynamics|heat-transfer|radiation|thermal-radiation|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35011" PostTypeId="1" CreationDate="2020-04-05T12:42:28.887" Score="1" ViewCount="50" Body="&lt;p&gt;&lt;a href=&quot;https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6797553/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6797553/&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.hatchbox3d.com/collections/pla&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.hatchbox3d.com/collections/pla&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are 3D printing masks that need to be used by medical staff treating COVID19 patients.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first link is an overview of the material we are using for masks in general.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The second is the specific product we are making masks out of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Quesions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1- To sterilize; can we safely do 120C for 30 minutes without the mask losing shape? What about 70C for 30 minutes? Will it affect integrity and water/air proof &quot;ness&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2 - Should we sterilize in humid air? dry air? again to maintain shape and impermeability&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3 - will this be stable with exposure to 1) Vapor H2O2 @ &gt;480 ppm for 30 minutes? 2) 1:9 bleach:water immersion? 3) household disinfectants (google Lysol)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We are making masks from this material and would be happy to recieve insights from people in the field. &lt;/p&gt;&#xA;" OwnerUserId="26438" LastEditorUserId="26438" LastEditDate="2020-04-05T13:11:00.953" LastActivityDate="2020-04-07T17:14:10.267" Title="Stability of 3D-printed mask after sterilisation" Tags="|mechanical-engineering|materials|strength|stability|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35012" PostTypeId="1" CreationDate="2020-04-05T16:42:28.380" Score="0" ViewCount="29" Body="&lt;p&gt;Are there any succesful solid fueled internal combustion engines? Particularily, ones that produce shaft power (not rockets and not ramjets)&lt;/p&gt;&#xA;" OwnerUserId="18744" LastActivityDate="2020-04-05T16:54:04.107" Title="Are there succesful solid fueled internal combustion engines?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35013" PostTypeId="2" ParentId="35012" CreationDate="2020-04-05T16:54:04.107" Score="4" Body="&lt;p&gt;over the years, people have tried running ICE's on &lt;em&gt;coal dust&lt;/em&gt;. It is possible, but it's difficult to meter the dust into the engine's intake to get the proper mixture, and it's difficult to transport the dust from a fuel tank into the engine compartment.&lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-04-05T16:54:04.107" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35014" PostTypeId="1" CreationDate="2020-04-05T18:50:08.963" Score="-1" ViewCount="479" Body="&lt;p&gt;I was editing assembly, then selected one part and presses &quot;Edit part in place&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to get out of this mode back to assembly editing?&lt;/p&gt;&#xA;" OwnerUserId="24724" LastActivityDate="2020-04-07T16:38:47.213" Title="How to exit from editing part in place?" Tags="|solidworks|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35015" PostTypeId="1" AcceptedAnswerId="35017" CreationDate="2020-04-05T19:08:48.483" Score="1" ViewCount="601" Body="&lt;p&gt;V is a rotation factor to account for the difference in ball rotations for outer ring rotation vs. inner ring rotation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;V = 1 for inner ring rotation&#xA;V = 1.2 for outer ring rotation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Shigley's Mechanical Engineering Design - Ninth Edition&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the reason behind this?&lt;/p&gt;&#xA;" OwnerUserId="26440" LastActivityDate="2020-04-05T19:38:26.740" Title="Why rotation factor of bearing for inner ring is 1 and for outer ring is 1.2?" Tags="|mechanical-engineering|design|bearings|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35017" PostTypeId="2" ParentId="35015" CreationDate="2020-04-05T19:38:26.740" Score="2" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zk36C.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zk36C.gif&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 1. Source: Wikipedia's &lt;a href=&quot;https://en.wikipedia.org/wiki/Ball_bearing&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ball-bearing&lt;/a&gt;.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance rolled on the inner ring is the same as the distance rolled on the outer ring. Due to the difference in circumference the ball will pass the same point on the inner ring about twice as often as it passes the same point on the outer ring (for these dimensions).&lt;/p&gt;&#xA;" OwnerUserId="3505" LastActivityDate="2020-04-05T19:38:26.740" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35018" PostTypeId="1" CreationDate="2020-04-05T20:31:51.483" Score="-1" ViewCount="302" Body="&lt;p&gt;title says it all.. scaling in solidworks with design tables usually involves providing the table with all the dimensions of the sketch, then producing alternate configurations within the table.  is there a way to instead specify the overall scaling of the part?  right now, I'm creating new parts for each different scaling (there aren't -that- many.. so far..)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit:  I think this is the answer:  &lt;a href=&quot;https://forum.solidworks.com/thread/79957&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://forum.solidworks.com/thread/79957&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13081" LastEditorUserId="13081" LastEditDate="2020-04-05T20:37:51.737" LastActivityDate="2020-04-20T03:13:43.233" Title="can I scale my part in solidworks using a design table, but without scaling through any sketches?" Tags="|solidworks|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35019" PostTypeId="2" ParentId="35010" CreationDate="2020-04-05T21:13:06.587" Score="2" Body="&lt;p&gt;Any diameter wire, multiple parallel heated elements describes a toaster, an electric oven, any number of radiant heaters already in existence. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your problem is not whether you can do it with such a system, but more of a question if you can avoid melting the plastic bed on which the powder sits.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It's unlikely that you can regulate the heat in such a precise manner as to have it cease to exist at the interface between the powder and the bed. If you are able to use a ceramic or metal bed, you should be able to make such a system work. If you are able to locate a plastic for the bed that exceeds the 270-300°C reference, you may be able to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You could possibly create a liquid cooling system for the plastic bed, but it will also cool the powder via conduction negating the melting process. If you can tolerate a partial melt, with some powder remaining un-melted and insulating the bed, it should work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The bed is the problem, not the heat.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-04-05T21:13:06.587" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35020" PostTypeId="2" ParentId="35000" CreationDate="2020-04-05T21:59:01.910" Score="2" Body="&lt;p&gt;The &lt;em&gt;Highway Capacity Manual&lt;/em&gt; contains specific analysis methods for determining acceptable levels of service. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/Level_of_service_(transportation)&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://en.wikipedia.org/wiki/Level_of_service_(transportation)&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Available funding and rights-of-way also determine the priority of improvements. &lt;/p&gt;&#xA;" OwnerUserId="26396" LastActivityDate="2020-04-05T21:59:01.910" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35021" PostTypeId="1" CreationDate="2020-04-05T23:59:59.233" Score="1" ViewCount="104" Body="&lt;p&gt;I'm considering a potential product but I'm not sure about the core technology I'd need.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fundamental idea is that there would be a cable (or something similar) that would be pulled.  But as it was pulled with X force, the device (mechanically and/or electronically) would provide equal (but slightly less) resistance.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if it's pulled with 10 pounds, it would respond with (for example) 9 pounds of resistance (so the net result would the cable moving at 1 pound pressure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it would also need to react in real-time (more or less instantly), as the pull force would be constantly changing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is there already some kind of device that does that  I'm not an engineer, more on an entrepreneur, so I'm a bit out of my depth here.  Thanks!&lt;/p&gt;&#xA;" OwnerUserId="26442" LastActivityDate="2024-01-16T23:01:35.163" Title="A device to provide real-time proportional resistance to variable force" Tags="|machine-design|machine-elements|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35022" PostTypeId="1" CreationDate="2020-04-06T00:23:21.143" Score="-1" ViewCount="260" Body="&lt;p&gt;Roads in the USA use asphalt, like every other country.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, the &lt;strong&gt;&lt;em&gt;sidewalks&lt;/em&gt;&lt;/strong&gt; don't use asphalt. Instead, they use enormous, gigantic square blocks of some kind of stone material. Where I live, the sidewalks consistently use asphalt, just like the road. (They are not on the same level, and separated by a kind of long, rectangular stone.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why did they go for this way in the USA? What's the benefit of using those flat square rocks instead of asphalt, specifically for the sidewalk? If they are so good, why not also use them for the road as well?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since they already use asphalt for the road, it seems like it would be the most convenient and logical to also use asphalt for the sidewalk, but they go out of their way to (I assume) laboriously place these plates everywhere for people to walk on.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One thing I remember somebody saying in the 1990s was that he wished that the sidewalks would be like in the USA because they were better for skateboards for some reason, but I doubt that the ones in charge cared that much about skaters...&lt;/p&gt;&#xA;" OwnerUserId="26443" LastActivityDate="2020-04-06T03:32:22.427" Title="Why do the sidewalks on USA roads use big, massive stone plates instead of asphalt?" Tags="|roadway|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35024" PostTypeId="2" ParentId="35022" CreationDate="2020-04-06T03:32:22.427" Score="4" Body="&lt;p&gt;Slipform paving isn't that big a deal, some basic ground preparation, then follow the line. The contraction joints are added to the wet concrete either by the machine or by workers with hand tools. As the concrete cures, it contracts, and the pavement cracks at the joints. Concrete is much harder and holds up to edging and other abuse that asphalt simply can't handle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Concrete lasts much longer than asphalt, and sidewalks are very expensive to replace once all the landscaping is in, so the more durable option is usually the way to go. Asphalt road surfaces can be repaired much more cost effectively. You do see some asphalt sidewalks where fussy landowners aren't an issue, such as at schools and business parks.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/fx57V.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fx57V.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2020-04-06T03:32:22.427" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35025" PostTypeId="1" AcceptedAnswerId="35044" CreationDate="2020-04-06T06:57:23.340" Score="-1" ViewCount="137" Body="&lt;p&gt;Problem statement: 2 kg of air at 500 kPa expands adiabatically in a closed system until its volume (V) is doubled and it's temperature become equal to that of surrounding which is at 100 kPa, 5°C. For the process determine&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;maximum work &lt;/li&gt;&#xA;&lt;li&gt;change in availability&lt;/li&gt;&#xA;&lt;li&gt;irreversibilty&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For air take &lt;span class=&quot;math-container&quot;&gt;$c_v= 0.718\text{ kJ/kg K}$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$u = c_v T$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$PV=mRT$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$R= 0.287\text{ kJ/kg K}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the solution, maximum work is solved using state equation including entropy which I have no doubt about. But my doubt is: &lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;why can't we use integration &lt;span class=&quot;math-container&quot;&gt;$p\partial V$&lt;/span&gt; for an adiabatic process to find maximum work here?&lt;/li&gt;&#xA;&lt;li&gt;as per the question, final temperature is same as that of surrounding and final volume is 2 times of initial volume also it is an adiabatic process. Final pressure using &lt;span class=&quot;math-container&quot;&gt;$PV^{1.4}=\text{constant}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$PV=mRT$&lt;/span&gt; are giving different result for the final pressure of the system.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="26446" LastEditorUserId="1832" LastEditDate="2020-04-06T13:30:05.050" LastActivityDate="2020-04-09T17:24:16.787" Title="Maximum work from the process" Tags="|thermodynamics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35026" PostTypeId="2" ParentId="35021" CreationDate="2020-04-06T07:15:47.013" Score="0" Body="&lt;p&gt;You can wind your cable around a flywheel connected to a shaft rotating a  fan in a high viscosity grease. Like an automatic transmission.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Geared to correct proportion. This mechanism is basically a friction machine but doesn't create much wear and tear. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-06T07:15:47.013" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="35028" PostTypeId="1" AcceptedAnswerId="35066" CreationDate="2020-04-06T10:34:13.687" Score="1" ViewCount="39" Body="&lt;p&gt;So to the point. In our project (Robotics project) we mostly use sheet metal and PLA printed parts to build. But we want to level up and be ready for an advanced run and even low-volume production of the product. Right now we're in the last days of an advanced prototype (Phase II) . My question is, what do you guys recommend to level up the design and start to be ready for low-volume production (15 - 20 pieces)? What are the tips and tricks when designing a more finished product? For example, The two things we're considering to change are CNC machining the plastic parts that have high loads on them, and for other 3D printed parts we are considering using different technology than FDM printing. Right now plastic injection molding is not an option, we are not in mass production. Carbon composites are not an option too. I will be glad to hear from you about more ways of bringing the product closer to an industrial one and start thinking about low-volume production. Any article/guide will be great!&lt;/p&gt;&#xA;" OwnerUserId="20231" LastEditorUserId="20231" LastEditDate="2020-04-06T13:21:15.570" LastActivityDate="2020-04-08T07:16:56.397" Title="From advanced prototype to low-volume production" Tags="|machining|robotics|industrial-engineering|product-engineering|production-technology|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35030" PostTypeId="1" AcceptedAnswerId="35041" CreationDate="2020-04-06T13:04:49.780" Score="5" ViewCount="156" Body="&lt;p&gt;The picture below shows a veritcal &lt;a href=&quot;https://en.wikipedia.org/wiki/Joint_(building)&quot; rel=&quot;noreferrer&quot;&gt;construction joint&lt;/a&gt; of an approximately 50 years old reinforced concrete industrial structure. The expansion joint was likely originally sealed with a &lt;a href=&quot;https://www.raygrahams.com/products/138890-flexcell-expansion-joint-2440mm-x-100mm-8-x-4.aspx&quot; rel=&quot;noreferrer&quot;&gt;flexcell&lt;/a&gt; which due to age and water ingress has deterioriated and come away from the joint exposing a peforated metal pipe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is does any one know why the pefortated pipe was installed there originally approximately 50 years ago? Does it have a drainage purpose? The joint should have been &lt;a href=&quot;https://en.wikipedia.org/wiki/Waterstop&quot; rel=&quot;noreferrer&quot;&gt;waterstopped&lt;/a&gt;. Is this a second line of defence? If so where do you expect it to drain to? Has anyone seen a similar detail to this before?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/J1mOC.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/J1mOC.png&quot; alt=&quot;perforated pipe within construction joint&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10270" LastEditorUserId="10270" LastEditDate="2020-04-06T19:29:55.500" LastActivityDate="2020-04-07T00:37:36.630" Title="Why is there this perforated pipe inside this vertical construction joint in a reinforced concrete wall?" Tags="|civil-engineering|concrete|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35031" PostTypeId="1" CreationDate="2020-04-06T15:19:22.187" Score="0" ViewCount="438" Body="&lt;p&gt;I have a beam with rectangular cross section, but rather an strange shape in the length wise direction as seen below:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Tnqvt.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Tnqvt.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I have taken out the symmetry part of it:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/K8Ns1.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/K8Ns1.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And it actually has some nice radius's due to metal forming:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dXdYe.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dXdYe.png&quot; alt=&quot;beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the beam is fixed at 0 and symmetric about 5. The Beam sees a Distributed load at s4. The distributed load is actually a gasket so when my beam deflect I actually need to iterate to make sure I reach a steady condition. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason why I would like analytic solution is for understanding of whats going on. The problem is a coupled problem, below the beam is a fluid flowing and also the beam has electric current flowing through it, so its all coupled and I would like to experiment with different materials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I made a function that represent my beam in Mathcad:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mf5CD.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mf5CD.png&quot; alt=&quot;piece curve&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried calculating the moment along the curve by saying that the axial force can be neglected and that the fixed(welded end) can rotate, it might not be 100% accurate but I assumed this will be close enough to the real world:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/mfiCh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/mfiCh.png&quot; alt=&quot;moment&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and the moment seems to make sense:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/8JaLR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/8JaLR.png&quot; alt=&quot;moment diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;this also fits with FEM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/IkYvN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IkYvN.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that I can not figure out how to calculate the deflection:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bJTtl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bJTtl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can you not simply take the derivative of the moment or something to get the deflection?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried to solve this but it gives me  completely wrong answer and not a nice symmetric deflection curve, I also need to guess the slope at the beginning of the beam and make sure it's the same as the end (6.03225deg). So I have the curve of the beam (inside/outside i.e outside -inside= thickness) as well as the moment and the Young's modulus and inertia of the cross section, so I should be able to calculate the deflection right?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7Wek7.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7Wek7.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26451" LastEditorUserId="26451" LastEditDate="2020-04-07T11:44:08.490" LastActivityDate="2020-04-08T08:14:28.087" Title="Calculate deflection of a beam with an arbitrary shape" Tags="|mechanical-engineering|finite-element-method|moments|deflection|" AnswerCount="1" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="35032" PostTypeId="2" ParentId="15884" CreationDate="2020-04-06T16:09:05.703" Score="1" Body="&lt;p&gt;Someone proposed an original method. The result is binary (it doesnt give you a percentage) but it might gives you a rough idea if UV pass through the glass or not: &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Testing can be one with a simple trick that will require access to a&#xA;  UV flashlight. You can find an affordable UV flashlight online or&#xA;  borrow one. Once you have it, go with it into a darkroom together with&#xA;  the sunglasses and paper money or a credit card. Turn on the&#xA;  flashlight and beam it on the credit card or money. Various strange&#xA;  symbols that you do not usually see in ordinary light will become&#xA;  visible. They include a line on the paper money and letters on credit&#xA;  cards. These are watermarks that the government and banks use to&#xA;  identify counterfeits.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;source &lt;a href=&quot;https://www.quora.com/How-can-I-tell-if-my-sunglasses-are-UV-protected/answer/Ronak-Furia-2&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ronak Furia, on quora&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="6641" LastActivityDate="2020-04-06T16:09:05.703" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35033" PostTypeId="1" CreationDate="2020-04-06T16:22:42.107" Score="1" ViewCount="577" Body="&lt;p&gt;For statically &lt;strong&gt;indeterminate&lt;/strong&gt; trusses, which method is better? Force or displacement method? And why?&#xA;I read somewhere that displacement method is not suitable for trusses but I must verify. &lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2020-04-06T21:07:31.633" Title="Which method is better to solve indeterminate trusses, force or displacement method?" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35035" PostTypeId="2" ParentId="22833" CreationDate="2020-04-06T17:43:12.980" Score="0" Body="&lt;p&gt;Here is the derivation for your problem - Note the two states and the trick used to get the second state&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZgvdR.jpg&quot; alt=&quot;&quot;&gt;&lt;/p&gt;&#xA;" OwnerUserId="26452" LastEditorUserId="1832" LastEditDate="2020-04-06T18:34:04.793" LastActivityDate="2020-04-06T18:34:04.793" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35037" PostTypeId="1" CreationDate="2020-04-06T18:48:43.090" Score="-1" ViewCount="99" Body="&lt;p&gt;I would like to design a part, matching other parts of an assembly and I don't want to remember sizes. Can I just restrict editing to some part and/ or create new part from within existing assembly?&lt;/p&gt;&#xA;" OwnerUserId="24724" LastActivityDate="2020-04-06T19:24:05.497" Title="Is it possible to edit part inside assembly simultaneously seeing an assembly or some of it?" Tags="|solidworks|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35038" PostTypeId="2" ParentId="35037" CreationDate="2020-04-06T19:24:05.497" Score="0" Body="&lt;p&gt;You can edit one component while in assembly mode. Just select the desired component, then click &quot;Edit Component&quot;. If you alter an mating surfaces you'll have to go through and renew the mates reference surfaces.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To create a new part within the assembly, you can select insert components-&gt;new part. That will insert a blank part you can then edit as desired.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-04-06T19:24:05.497" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35039" PostTypeId="2" ParentId="35033" CreationDate="2020-04-06T21:07:31.633" Score="2" Body="&lt;p&gt;Depends on your definition of &quot;better&quot; and on the specific truss being analyzed.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The displacement method is useful in that it is agnostic to such things, statically determinate and indeterminate structures are identical, as far as the method is concerned. Obviously, indeterminate structures will have more complicated solutions, but the basic working is all the same. This is why almost all structural analysis software uses the displacement method.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For structures which are only slightly externally indeterminate (by one or two degrees of freedom) and which aren't very complex (or which can be decomposed into simple sub-structures), the force method might be useful if you're doing things by hand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For internally indeterminate structures... I honestly don't remember how to use the force method, so I'd certainly use the displacement method.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;In case a refresher is needed on internal and external indeterminacy: external indeterminacy means you can't find the support reactions using only the equilibrium equations; internal indeterminacy means you can't solve for the members' internal forces using only equilibrium equations.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-04-06T21:07:31.633" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35040" PostTypeId="1" CreationDate="2020-04-07T00:08:46.443" Score="2" ViewCount="39" Body="&lt;p&gt;I'm reading some old microcontroller code for a PID algorithm. There's a strange function (to me) that seems to work like this:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$output = \frac{1}{n}input + \frac{n-1}{n}previous\_output$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The value &lt;span class=&quot;math-container&quot;&gt;$n$&lt;/span&gt; varies. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Over several iterations, it will eventually output a value very close to the input, but sudden jumps in the input create a lag that takes a few cycles to catch up. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This function is used in multiple places, for instance:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;it's used on the setpoint when calculating the error - the current process variable is subtracted from f(setpoint) (where 'f' is the function I described above) instead of the true setpoint. This seems to produce a weighted set point, where the weight is inversely proportional to the rate of change - is that correct?&lt;/li&gt;&#xA;&lt;li&gt;it's also used in another component of the PID when comparing the &lt;em&gt;true&lt;/em&gt; error to the previous value. The previous value is put through this function, instead of using the true previous value. In this case, it's harder to understand it's purpose.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;Is this a common function used in PID controllers? I haven't found a description that quite seems to explain this, but if it might be a case of just not knowing the right keyword to search for. &lt;/p&gt;&#xA;" OwnerUserId="26455" LastActivityDate="2020-04-07T00:08:46.443" Title="Is this function common in PID controllers?" Tags="|pid-control|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35041" PostTypeId="2" ParentId="35030" CreationDate="2020-04-07T00:24:47.167" Score="3" Body="&lt;p&gt;Yes, it is a hydrostatic pressure relief pipe (part of the drainage).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Behind all the retaining walls, swales, subdivision's low land lots, you need to have a trench or channel filled with aggregate and a perforated pipe leading to surface runoff drainage as a minimum.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Aggregate mus be stratified from 1.5&quot; near the pipe to sand 18&quot; out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I suspect in your photo corrosion has washed the aggregate out.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-04-07T00:37:36.630" LastActivityDate="2020-04-07T00:37:36.630" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35042" PostTypeId="1" AcceptedAnswerId="35060" CreationDate="2020-04-07T00:30:46.100" Score="1" ViewCount="144" Body="&lt;p&gt;I want to manually measure the dimension of a pattern printed on a flat surface, with an accuracy of about 0.1mm. For simplicity sake, let's say this is a black rectangle, printed on a white background, and i want to measure its width and height. The dimension can be between 20cm and 80cm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried with a vernier caliper, but placing the caliper accuratly enough on the rectangle borders is impossible, especially since the caliper jaws do not lie flat on the surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What kind of measuring instrument could i use? Is there something dedicated to this task, or a way to use a standard instrument to do that?&lt;/p&gt;&#xA;" OwnerUserId="26456" LastEditorUserId="26456" LastEditDate="2020-04-07T19:07:01.483" LastActivityDate="2020-04-07T21:03:01.393" Title="Measuring a pattern size on a flat surface" Tags="|mechanical-engineering|measurements|distance-measurement|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="35044" PostTypeId="2" ParentId="35025" CreationDate="2020-04-07T03:30:15.003" Score="0" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;If you simply integrate the internal energy change from pressure-volume changes you are assuming that going from state 1 -&gt; 2 is fully reversible (isentropic). The way the question has been created makes it irreversible, so there is an increase in entropy in your system. If the system changed reversibly your thoughts would be correct. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;For the same reason as above. the relationship where you are using the ratio of specific heats (Cp/Cv) in the exponent is only valid for an isentropic process. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That dang entropy gets us every time ... &lt;/p&gt;&#xA;" OwnerUserId="26386" LastActivityDate="2020-04-07T03:30:15.003" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35045" PostTypeId="2" ParentId="35042" CreationDate="2020-04-07T07:07:53.093" Score="2" Body="&lt;p&gt;I would give a shot at image processing, if there is a good contrast between the surface you want to measure and the background.&#xA;To do so you need to take a picture of the surface that includes an object of known size for calibration. &#xA;Then, open source software like &quot;imageJ&quot; are able to perform the measurement.&#xA;The accuracy of this method strongly depend on the possibility to have a photograph that is parallel to the surface to be measured. However, this can be partially circumvented by taking several photos, repeating the sizing, and considering the mean value.&lt;/p&gt;&#xA;" OwnerUserId="26427" LastActivityDate="2020-04-07T07:07:53.093" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35047" PostTypeId="1" AcceptedAnswerId="35061" CreationDate="2020-04-07T07:51:32.163" Score="1" ViewCount="338" Body="&lt;p&gt;When double-checking/confirming results produced by structural analysis software, many engineers recommend verifying and spot-checking key or critical members if the entire structure is not analysed. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, for rigid jointed structures with sway, it is not possible (at least in my understanding) to analyse individual members. This means that hand calculation methods such as moment distribution or slope deflection cannot be applied directly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;How would I go about analysing those members in a sway frame?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also, to what storey of an indeterminate structure would you analyse purely by hand calculations without computer software when considering code requirements?&lt;/p&gt;&#xA;" OwnerUserId="20822" LastActivityDate="2020-04-07T18:34:54.447" Title="Analysis Of Statically Indeterminate Structures By Hand" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35048" PostTypeId="2" ParentId="35006" CreationDate="2020-04-07T09:19:06.330" Score="0" Body="&lt;p&gt;o.k. i think you'r just confused because they have different tops, and the tubs are hard to follow, but they are just all connected to the bottle, as i draw in my first answer&#xA;&lt;a href=&quot;https://i.stack.imgur.com/w8BPY.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/w8BPY.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26435" LastActivityDate="2020-04-07T09:19:06.330" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35049" PostTypeId="2" ParentId="35007" CreationDate="2020-04-07T10:22:55.473" Score="1" Body="&lt;p&gt;General points to note: The &quot;Standard&quot; mates are more robust than the Mechanical mates, and should be used wherever possible. A cam should have one continuous face, you can use the fitspline tool to create this.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See below for how I fixed your model:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.imgur.com/bJk9p4V.gif&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.imgur.com/bJk9p4V.gif&quot; alt=&quot;fix Cam Screencap&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13311" LastActivityDate="2020-04-07T10:22:55.473" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35050" PostTypeId="1" AcceptedAnswerId="35920" CreationDate="2020-04-07T10:34:28.383" Score="0" ViewCount="34" Body="&lt;p&gt;I'm going to buy Thomas vacuum cleaner with Aqua filter and use it to sand walls a bit with grinder. And after finishing the decoration, use it as an ordinary cleaner. May construction plaster, putty sanding harm it? Or it is OK to use the household vacuum for small amount of grinding?&lt;/p&gt;&#xA;" OwnerUserId="26462" LastEditorUserId="26462" LastEditDate="2020-04-07T11:03:47.647" LastActivityDate="2020-05-23T18:07:34.530" Title="Can I use household vacuum cleaner with AquaFilter as construction/industrial one" Tags="|grinding|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35051" PostTypeId="2" ParentId="35050" CreationDate="2020-04-07T10:56:43.153" Score="1" Body="&lt;p&gt;I doubt it will last.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Professional units are designed for higher duty hours with more suction etc which means better bearings and greater power for example.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The home use equipment is not designed to be used at that type of duty hours - possibility of overheating etc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A simple example is a toaster - buy one for a home and it is designed to do 4 or 8 slices per morning per day. Put that same toaster in a canteen feeding 200 workers per morning for breakfast and it might last 2 days... home use it can last 5 years.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-07T10:56:43.153" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35052" PostTypeId="1" AcceptedAnswerId="35053" CreationDate="2020-04-07T11:51:16.720" Score="-1" ViewCount="37" Body="&lt;p&gt;Hello all I have a magnet in my car to stick my phone to the dashboard. But I changed phone and now my new phone needs some metal stuck to the case in order to attract to the magnet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I read online these metals are called ferromagnetic (iron, steel, and others).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if any of you fine folks can suggest me a place I can butcher some from something I might already have at home (or can be easily found). At the moment we have the lockdown and most shops are shut. I can't figure it out on my own haha.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just need like an inch or 2 square piece of metal that will  stick to my magnet in the car. Sadly, I don't have great tools, but might be able to borrow some rudimental hand tools.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for reading and any help!&lt;/p&gt;&#xA;" OwnerUserId="26463" LastActivityDate="2020-04-07T12:00:27.077" Title="Cheap / Easy source of small amount of ferromagnetic metal" Tags="|magnets|" AnswerCount="1" CommentCount="2" ClosedDate="2020-04-07T13:25:50.840" ContentLicense="CC BY-SA 4.0" />
  <row Id="35053" PostTypeId="2" ParentId="35052" CreationDate="2020-04-07T12:00:27.077" Score="1" Body="&lt;p&gt;A steel washer would work well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Find one about 1&quot; in diameter and that should be sufficient.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you have a file or sanding tool then round the edges off.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Make sure the surfaces you are gluing are &quot;keyed&quot; ie scratched to provide a better joint.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-07T12:00:27.077" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="35054" PostTypeId="2" ParentId="29059" CreationDate="2020-04-07T15:02:12.377" Score="0" Body="&lt;p&gt;Have you considered using rollers attached to motors to rotate the ball? Look at old school computer mice with ball wheels. Instead of using sensors to track position, use rollers to drive your sphere. Add a 3rd motor perpendicular to the other 2 to get your 3rd axis of rotation. Nonius encoder with hall effect sensors can provide very accurate absolute angular position. No belts, no gears, no differentials.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Fewer components = fewer possibilities of error.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-04-07T15:02:12.377" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="35055" PostTypeId="1" CreationDate="2020-04-07T15:26:06.183" Score="0" ViewCount="54" Body="&lt;p&gt;I understand the relationship of force method and displacement method and how we use them in matrices in FEM. &#xA;How about virtual work? Is it used in matrices and FEM too?&lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2020-04-07T15:26:06.183" Title="How does virtual work method relate to matrix method or FEM in structural analysis?" Tags="|structural-engineering|structural-analysis|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35056" PostTypeId="2" ParentId="35014" CreationDate="2020-04-07T16:38:47.213" Score="3" Body="&lt;p&gt;Deselect this option and it will return to the assembly. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XX9yS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XX9yS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26386" LastActivityDate="2020-04-07T16:38:47.213" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35057" PostTypeId="1" AcceptedAnswerId="35064" CreationDate="2020-04-07T16:43:55.950" Score="3" ViewCount="165" Body="&lt;p&gt;The reinforced concrete beam (approximate dimensions 36&quot; x 18&quot;) shown in the image below has a row of circular holes at 3/5ths of the height of the cross-section at approximately 300 mm centres. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is why does this concrete beam have these holes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My first thought is that these holes were due to the &lt;a href=&quot;http://www.siteright.net/tie-bar-and-tie-bar-connector-p-88.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;tie bars&lt;/a&gt; providing lateral support to the formwork vertical walls. However, they appear to be too frequent for that to be true.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/LasbW.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/LasbW.jpg&quot; alt=&quot;row of circular holes through concrete beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10270" LastActivityDate="2020-04-07T20:28:44.187" Title="Why is there a row of circular holes in this concrete beam?" Tags="|civil-engineering|beam|reinforced-concrete|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35058" PostTypeId="2" ParentId="34995" CreationDate="2020-04-07T16:54:55.813" Score="2" Body="&lt;p&gt;These types of molds are for casting and curing. There is basically zero injection pressure so they can be small and require minimal machining. You can keep them closed with paper binding clamps. You can see from the photo they took a piece of stock aluminum, faced it with a fly cutter and machined the cavity. Easy and Inexpensive.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When you inject thermoplastics, like ABS, the injection pressures can be 1000 atmospheres. This is required because of the high viscosity and fast filling times required before the plastic solidifies. The molds are very thick to withstand the internal pressure and clamping loads required. Aluminum tools generally are for short runs and hard chromed tool steel is required for high volumes or abrasive plastics. The mating surfaces of the mold have to be near perfect to prevent flashing so they are machined in many steps and hand finished and polished. A simple aluminum injection or compression mold can cost tens of thousands.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For your application you could get away with using aluminum for the mould but at small scale thermoplastic injection moulding is difficult and expensive. You could consider casting the enclosure with a mineral filled epoxy or similar. The best option if you can make it work in your design is vacuum forming. I have done lots of short run enclosures using vacuum formed hair-cell textured abs. It hides all the defects on the surface and won't show scratches. ABS is very soft, you can scratch it with your finger nail. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The best source for standard sizes of enclosure I have ever found is poly-case. If you have different connectors that need to come out from the case they have options with blank face plates that can be cut to accommodate your connectors. &#xA;&lt;a href=&quot;https://www.polycase.com/abs-enclosures&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.polycase.com/abs-enclosures&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26386" LastActivityDate="2020-04-07T16:54:55.813" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35059" PostTypeId="2" ParentId="35011" CreationDate="2020-04-07T17:14:10.267" Score="1" Body="&lt;p&gt;You are probably using an FMD type printer. No matter how good your printer, or the density of the print, you will have porosity. You need the temperature to denature the proteins of the virus. If someone is wearing a mask all day breathing hot air into it, especially with PLA, there will be absorption into the plastics. I think you cannot assume that a chemical method will infiltrate into all pores to destroy viral particles. Also you would need to be aware of the residual chemical left behind from the process in the plastic. PLA is very hygroscopic. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;See this article: &quot;For gravity displacement sterilizers the penetration time into porous items is prolonged because of incomplete air elimination. This point is illustrated with the decontamination of 10 lbs of microbiological waste, which requires at least 45 minutes at 121°C because the entrapped air remaining in a load of waste greatly retards steam permeation and heating efficiency.&quot;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.cdc.gov/infectioncontrol/guidelines/disinfection/sterilization/steam.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.cdc.gov/infectioncontrol/guidelines/disinfection/sterilization/steam.html&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26386" LastActivityDate="2020-04-07T17:14:10.267" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35060" PostTypeId="2" ParentId="35042" CreationDate="2020-04-07T18:28:05.147" Score="6" Body="&lt;p&gt;I had a plastic ruler at one time in my life and it looked something like Figure 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3Yotv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3Yotv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;Figure 1. Overview.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gQMHF.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gQMHF.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 2. Close up of precision grid.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9RJ9X.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9RJ9X.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Figure 3. In-use.&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How to use:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Align the right edge to the nearest whole cm less than the width of the item to be measured.&lt;/li&gt;&#xA;&lt;li&gt;In the precision section, find the closest intersection between a sloping line and a horizontal line.&lt;/li&gt;&#xA;&lt;li&gt;Add the three numbers. In this case it's 30 + 5 + 0.7 mm. = 35.7 mm.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I drew this in a few minutes on a CAD program. Vertical spacing is 2 mm. Print it off on laser-compatible transparent film on a decent laser printer and you should be in business. If you mirror the image before printing you will print on the &quot;back&quot; of the film and will have no parallax error when the film is pressed on the part. To make a longer scale you can tape several together.&lt;/p&gt;&#xA;" OwnerUserId="3505" LastEditorUserId="3505" LastEditDate="2020-04-07T21:03:01.393" LastActivityDate="2020-04-07T21:03:01.393" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="35061" PostTypeId="2" ParentId="35047" CreationDate="2020-04-07T18:34:54.447" Score="3" Body="&lt;p&gt;In my professional experience, I've found few opportunities to perform literal &quot;by hand&quot; verification.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless you're dealing with simple structures (or structures which can be simplified as such), it gets quite complicated and if you perform any mistakes you won't know if it's because of the model or your &quot;by hand&quot; calculation. And I personally have a hard time identifying (non-obvious) mistakes in my own calculations when I'm not even sure they're there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That being said, the hard part of &quot;by hand&quot; verification of non-trivial structures is the actual structural analysis. Thankfully, the world has many different computer programs that do this exceedingly well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So I check my work in one structural analysis program by creating another model in another program. This other model, however, is usually much simpler: hell, my tool of choice was &lt;a href=&quot;https://www.ftool.com.br/Ftool/&quot; rel=&quot;nofollow noreferrer&quot;&gt;Ftool&lt;/a&gt;, which only does 2D frame analysis. So if I was trying to double-check a 3D program, I'd have to recreate the model in Ftool using slices, manually calculating the applied loads (usually using &lt;a href=&quot;https://debug.pi.gr/default.aspx?ch=557&quot; rel=&quot;nofollow noreferrer&quot;&gt;simple distribution areas&lt;/a&gt;), considering the support conditions (it's much easier to be explicit and intentional about your supports and hinges in 2D than 3D), etc. It's important not to get too detailed in this model: don't try to model a complex support condition using master-slave relations or whatever; make the best simplification you can: fully hinged or fully fixed (or if absolutely necessary, do both and get an average of the results).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And then I let Ftool do its job and give me the internal forces, which I can then compare to the 3D model.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they're close enough, great. It's much harder to get the boundary conditions wrong in a 2D model (where you explicitly define what they are at a single point, while the 3D model may require you to apply conditions to multiple points to represent the same effect, giving you room for error), so if they're close but actually wrong, it's because you explicitly messed up (i.e. in both models you forgot that that one beam there is actually hinged to the column, not fixed... a mistake you likely would've made even if doing the verification fully by hand). Getting someone else to take a look at your 2D model should further reduce the odds of that happening. And the odds of you having made different mistakes in each model that cancel out to give similar wrong results are infinitesimal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If they're different, it's far easier to double-check that you didn't mess up your 2D model, and then compare the results to see what's similar and what's different, helping you identify your mistake (while doing it by hand risks a screw up in one element of the stiffness matrix leading you to get everything wrong).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This offers an efficient verification which is quite close to &quot;by hand&quot; -- you are still the one defining the structure's loads, supports and dimensions, in a format (2D) where it's much harder to accidentally mess up the model -- while still being efficient and honestly less error-prone than literally doing it by hand.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-04-07T18:34:54.447" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35062" PostTypeId="2" ParentId="34954" CreationDate="2020-04-07T19:24:04.550" Score="1" Body="&lt;p&gt;You don't necessarily have to find &lt;em&gt;exact&lt;/em&gt; constants &lt;span class=&quot;math-container&quot;&gt;$k_1,k_2,k_3$&lt;/span&gt;, only need to show that there exists some &lt;em&gt;positive&lt;/em&gt; constants. In your example above, I can say that &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{V} \leq -\bigg(\frac{x_1 + x_2}{\sqrt{2}}\bigg)^2 = -\frac{1}{2}\|x\|^2,  &#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;span class=&quot;math-container&quot;&gt;$k_3 = 1/2$&lt;/span&gt; would work. For the function &lt;span class=&quot;math-container&quot;&gt;$V(x)$&lt;/span&gt; itself, we know that it is positive definite. Note that the log term is &lt;em&gt;always&lt;/em&gt; negative so &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;V(x) \leq \frac{3}{2}x_1^2 + \frac{1}{2}\|x\|^2 = 2\|x||^2 - \frac{3}{2}x_2^2 \leq 2\|x\|^2,&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so &lt;span class=&quot;math-container&quot;&gt;$k_2 = 2$&lt;/span&gt; would work. For &lt;span class=&quot;math-container&quot;&gt;$k_1$&lt;/span&gt;, you can just pick an arbitrarily small positive constant that satisfies the lower bound.&lt;/p&gt;&#xA;" OwnerUserId="15049" LastActivityDate="2020-04-07T19:24:04.550" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35063" PostTypeId="1" CreationDate="2020-04-07T19:25:26.267" Score="1" ViewCount="249" Body="&lt;p&gt;Suppose I have the state space system &lt;span class=&quot;math-container&quot;&gt;$a\dot{x} = -x + \frac{1}{k}h(x) + u$&lt;/span&gt;, and output &lt;span class=&quot;math-container&quot;&gt;$y = h(x)$&lt;/span&gt;, where &lt;span class=&quot;math-container&quot;&gt;$a,k &amp;gt; 0$&lt;/span&gt;. The only information I'm given about &lt;span class=&quot;math-container&quot;&gt;$h(x)$&lt;/span&gt; is that &lt;span class=&quot;math-container&quot;&gt;$h\in[0,k]$&lt;/span&gt;. I want to show that the system is passive using the storage function &lt;span class=&quot;math-container&quot;&gt;$V(x) = a\int_{0}^{x}h(\sigma)\ d\sigma$&lt;/span&gt;. The definition of a passive system is &lt;span class=&quot;math-container&quot;&gt;$u^\intercal y \geq \dot{V}$&lt;/span&gt; for all &lt;span class=&quot;math-container&quot;&gt;$(x,u)$&lt;/span&gt;. Thus,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\dot{V} = ah(x) \dot{x} = h(x) (-x + h(x)/k + u) = \frac{1}{k}h(x)(h(x) - kx) + h(x) u.&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Rearranging this gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;h(x)u = yu = \dot{V} + \frac{1}{k}h(x)[kx-h(x)].&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In order to prove passivity, I need to show that &lt;span class=&quot;math-container&quot;&gt;$\frac{1}{k}h(x)[kx - h(x)] \geq 0$&lt;/span&gt;, for all &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt;. However, all I'm given is that &lt;span class=&quot;math-container&quot;&gt;$h\in[0,k]$&lt;/span&gt; so how can this be true in general? Am I missing something really simple here?&lt;/p&gt;&#xA;" OwnerUserId="15049" LastActivityDate="2024-01-26T23:05:50.713" Title="Passivity in State Space Systems" Tags="|control-engineering|control-theory|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35064" PostTypeId="2" ParentId="35057" CreationDate="2020-04-07T20:28:44.187" Score="3" Body="&lt;p&gt;I've not seen them before but I think that they are surely penetrations to run wiring. They on top for reasons including&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;close to the lid where wiring would be normally attached between the beams&lt;/li&gt;&#xA;&lt;li&gt;in a place of lower bending stress, near the compression zone, not in the tension zone were reinforcing would be located or concentrated&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I suppose that they could be for air flow, a venting of sorts.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In any case, I believe that they are not there as a design feature of the beam. That is, these holes serve no purpose for the beam itself.&lt;/p&gt;&#xA;" OwnerUserId="24759" LastActivityDate="2020-04-07T20:28:44.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35065" PostTypeId="1" CreationDate="2020-04-07T21:58:23.690" Score="1" ViewCount="212" Body="&lt;p&gt;I would like to perform several FEM simulations on a given mesh with ABAQUS, where only the orientation of an element set would change in between.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Currently, I'm just generating several inp files (see an example below) and I run them one after the other. But this has the inconvenience to reload every time the mesh, which takes time due to its large size.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would there be a way to load the mesh only once, perform the simulations, save the results in a file, modify the orientation (orientation label going from ORI_1 to ORI_X), reperform the simulation, save in another file and so on?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thank you in advance,&#xA;Regards,&#xA;R.B.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;*HEADING&#xA;TEST&#xA;*INCLUDE,INPUT=Mesh.inp&#xA;*INCLUDE,INPUT=PARAMETERS_MATRIX.inp&#xA;*Solid Section, elset=AllE, material=Matrix, ORIENTATION=Matrix&#xA;*INCLUDE,INPUT=PARAMETERS_CRYSTAL.inp&#xA;*INCLUDE,INPUT=ORIENTATIONS_LIST.inp&#xA;*Solid Section, elset=MID, material=CRYSTAL, ORIENTATION=ORI_123&#xA;&#xA;*Step, NLGEOM=NO, name=Step-1&#xA;*Static &#xA;1., 1, 1e-05, 1.&#xA;*INCLUDE,INPUT=MESH_BC.inp&#xA;*Output, field&#xA;*element output, POSITION=CENTROIDAL, elset=AllE&#xA;S,E,EVOL&#xA;*End Step&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="26467" LastActivityDate="2023-12-29T20:01:55.250" Title="How to repeat on ABAQUS FEM simulations without reloading the mesh?" Tags="|finite-element-method|simulation|abaqus|feedback-loop|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35066" PostTypeId="2" ParentId="35028" CreationDate="2020-04-08T06:57:04.077" Score="1" Body="&lt;p&gt;I believe your primary goal when transitioning from single prototypes to small batch production should be to improve manufacturability, and ensure that your design is scalable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Examples of parts/systems that may need to be revised:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Parts that were made without drawings, for example by match drilling, or &quot;freestyling&quot;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Parts that are not easily obtainable in quantity&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Parts/systems that have known issues, which were manageable for a single prototype.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Systems that require excessive manual labor for manufacture. For example hand soldered cable harnesses may need to be replaced with off-the-shelf options.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Parts that are functional, but not aesthetically pleasing may need to be replaced&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;I would recommend focusing less on methods, and more on results. Instead of asking &quot;how do I level up&quot;, I would ask: &quot;if I started making 20 of these right now, what problems would I run into&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;To address the issue of 3D printed components. First ask yourself what the problem with them is, is it Aesthetics? Strength? Time? Cost? There is nothing inherently wrong with using 3D printed parts on a product.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are other options, but they will all be more expensive than 3D printing (for 10-20 units) so you will probably want to have a good reason for choosing something else. Some alternatives:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;CNC, good looking and strong parts, but very expensive.&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Laser cutting / water jet. Almost as cheap as 3d printing, strong materials, looks good. But the parts will need to be re-designed to be 2D&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Resin molding. Not economical, in my experiance this is only used to simulate what an injection molded part will look like.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Other 3D printing. There are a lot of types, such as sintered plastic/metal, resin, etc that may be able to solve problems that your FDM PLA prints are having.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;" OwnerUserId="20490" LastEditorUserId="20490" LastEditDate="2020-04-08T07:16:56.397" LastActivityDate="2020-04-08T07:16:56.397" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35067" PostTypeId="2" ParentId="35031" CreationDate="2020-04-08T08:14:28.087" Score="1" Body="&lt;p&gt;I have a couple of comments to your calculation:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;The axial force is essential to calculating the bending moment and is important for more than the axial deformation, which is likely to be small in this case. If a beam makes a 90 degrees bend, the axial force and shear force &quot;switch places&quot;, i. e. the axial force immediately before the bend is equal to the shear force immediately after the bend. Therefore, as you ignore the axial force, you are calculating the shear force and bending moment for a straight beam without any bends.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;As Wasabi has noted in a comment, you are missing the bending moment at the ends. (The integration constant from integrating the shear force.) Your FEM-model shows a global maximum here, which is the expected result.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;You don't give complete equations for both deflection and angular rotation, both the one you give for the deflection isn't correct. The curvature is &lt;span class=&quot;math-container&quot;&gt;$\kappa(x) = \frac{M(x)}{EI}$&lt;/span&gt; and the angular rotation is &lt;span class=&quot;math-container&quot;&gt;$\alpha(x) = \int_0^x \kappa(x) dx + C$&lt;/span&gt; with C being an integration constant. You seem to be using this, so so far so good. To calculate the deflection (ignoring the axial deformations for simplicity), you would use &lt;span class=&quot;math-container&quot;&gt;$\frac{d\delta}{dx} = \alpha$&lt;/span&gt;. For a straight beam, you could use &lt;span class=&quot;math-container&quot;&gt;$\delta(x) = \int_0^x \alpha(x) dx + C_2$&lt;/span&gt; with &lt;span class=&quot;math-container&quot;&gt;$C_2$&lt;/span&gt; being another integration constant, which I will note is not what you are doing. (You need to integrate the curvature twice to get a deflection.) As the beam isn't straight, the calculation becomes more complicated and you will need to use a vectorized version: &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; is a deformation perpendicular to the beam and will change direction around the bends. In orders words, the underlying assumption of the integration, that &lt;span class=&quot;math-container&quot;&gt;$\delta(x+dx) = \delta(x) + \alpha dx$&lt;/span&gt;, is simply not true (in the non-vector version), because &lt;span class=&quot;math-container&quot;&gt;$\delta(x)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\alpha dx$&lt;/span&gt; are deformations in two different directions.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In conclusion, the current approach of your calculations isn't more accurate than approximating the beam with a straight beam that is fixed in both ends. You should be able to look up a complete solution for such a regular, straight beam in a decent textbook of your choice, so even though you almost certainly have considered this already, I'll just add that it might be worth the effort to compare this crude approximation to your FEM-model just to check just how crude it is.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2020-04-08T08:14:28.087" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35068" PostTypeId="1" AcceptedAnswerId="35073" CreationDate="2020-04-08T13:16:52.060" Score="2" ViewCount="354" Body="&lt;p&gt;I need to calculate whether a polycarbonate sheet can withstand a certain maximum force. The force is an impact force on any one point of the sheet at any given time. All I have are the Young's modulus and the dimensions of the sheet. Any ideas how I can go about calculating the maximum force the sheet can withstand? This polycarbonate plate is placed vertically and held on all edges by a frame.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="26474" LastEditorUserId="26474" LastEditDate="2020-04-08T14:21:06.707" LastActivityDate="2020-04-09T00:24:02.770" Title="Calculate whether a polycarbonate plate can withstand a certain force" Tags="|structural-engineering|structures|stresses|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35069" PostTypeId="1" AcceptedAnswerId="35074" CreationDate="2020-04-08T16:14:32.267" Score="2" ViewCount="36" Body="&lt;p&gt;I have been working on a question, and my lecturer has hinted me with this equation that doesn't include forces due to weight?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$C_{M_x}=C_{M_{ac}}+C_L(x/c−x_{ac}/c)$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What he told us was that due to the derivation of the equation we can ignore weight with this equation? However when taking the moment of course one must take weight into account correct? Is it included in the lift coefficient?&lt;/p&gt;&#xA;" OwnerUserId="26475" LastEditorUserId="26475" LastEditDate="2020-04-09T04:33:59.087" LastActivityDate="2020-04-09T04:33:59.087" Title="Why is it when taking the moment coefficient relative to c.o.p., the weight is ignored?" Tags="|aerospace-engineering|aerodynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35070" PostTypeId="2" ParentId="14329" CreationDate="2020-04-08T17:42:17.907" Score="1" Body="&lt;p&gt;I figured it out.  First right click your mesh in the project tree and then click &quot;Clear Generated Data&quot;.  Then, right click, say your &quot;Node Move&quot;, there should be a &quot;Delete&quot;, click &quot;Delete&quot;, then right click your &quot;Mesh Edit&quot; and there should also be a &quot;Delete&quot;, click that.&lt;/p&gt;&#xA;" OwnerUserId="26476" LastActivityDate="2020-04-08T17:42:17.907" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35072" PostTypeId="1" CreationDate="2020-04-08T21:35:34.110" Score="0" ViewCount="66" Body="&lt;p&gt;I'm struggling to see how Integral gain is used for motor positioning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Suppose we have an AC motor (with a 2048PPR encoder on the shaft) connected to a winch which drives an object forwards and backwards. At 0 speed and 0 torque, there is no movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, using Proportional gain, we can drive the motor to a position and as it approaches the setpoint, the output to the motor decreases. The motor hopefully lands close to the setpoint.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we use Integral gain, where the error accumulates, when we reach the setpoint the value of I is still high and the motor overshoots (even with very little ki). This wouldn't be a problem if the motor required a non-zero voltage to maintain its position (e.g. using the motor as a hoist).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Integral may get very small, causing the motor to stop, but wouldn't it then rise up again after some time (depending on Ki), overshoot, and just oscillate around the setpoint infinitely? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A basic solution I've come up with is to use Integral gain to &quot;push&quot; the motor to its setpoint (effectively correcting the steady-state error), and when the motor is &quot;close enough&quot; to the setpoint, set the output to 0.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this a commonly used method? Are there any other approaches?&lt;/p&gt;&#xA;" OwnerUserId="26478" LastActivityDate="2020-04-09T01:39:33.993" Title="Using Integral gain for motor positioning" Tags="|motors|control-theory|pid-control|" AnswerCount="1" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="35073" PostTypeId="2" ParentId="35068" CreationDate="2020-04-09T00:24:02.770" Score="6" Body="&lt;p&gt;For impact force calculations you need some more information such as the shear modules and the material properties like toughness and tensile yield and shear yield.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Also you need to define a physical object that will impact the sheet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the object is much more rigid than the sheet and pointed it will go through like a bullet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Imagine an impact force has caused your sheet to behave like trampoline table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You will see a cone shape deflection under the load opening up to a wide sloped flange depression.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the very sharp bottom of this cone most of the stress is tensile due to cylindrical elongation of this segment.&#xA;And some crushing at the center.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as we move up most of deformation is due to punching shear on concentrically widening rings with a bit of biaxial bending moment.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The military shield and armoire manufacturers  have done extensive research on the competition between the pentration of mortar and resistance of shields.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are limiting your limit to elastic range, your force has to have a nominal minimum diameter and you can assume an arbitrary dynamic factor of 3 and check Roark's handbook of engineering formulas for some static forces and edge support conditions. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-09T00:24:02.770" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35074" PostTypeId="2" ParentId="35069" CreationDate="2020-04-09T00:35:35.750" Score="3" Body="&lt;p&gt;These are the dimensionless coefficients. The equation computes the &lt;span class=&quot;math-container&quot;&gt;$C_{m_x}$&lt;/span&gt; at some point not at the aerodynamic center by adding the &lt;span class=&quot;math-container&quot;&gt;$C_{m_{ac}}$&lt;/span&gt; to a (nondimensional) force times a (nondimensional) distance. Note that the &lt;span class=&quot;math-container&quot;&gt;$C_{m_{ac}}$&lt;/span&gt; is independent of angle of attack. To the first order, it is a function of the camber line. But the &lt;span class=&quot;math-container&quot;&gt;$C_l$&lt;/span&gt; is a function of angle of attack. So the &lt;span class=&quot;math-container&quot;&gt;$C_{m_x}$&lt;/span&gt; varies with angle of attack.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you are looking for where the weight is &quot;hiding&quot;, changes in weight are reflected in changes to &lt;span class=&quot;math-container&quot;&gt;$C_l$&lt;/span&gt; in level flight.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2020-04-09T00:35:35.750" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35075" PostTypeId="2" ParentId="35072" CreationDate="2020-04-09T01:39:33.993" Score="1" Body="&lt;p&gt;when setting the proportional gain and the integration constant, it is common practice to include just enough resistance in the model so the control loop is critically damped. this prevents the system from oscillating or &quot;hunting&quot; about the setpoint and represents the optimum PID response regime. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is also common practice to define a position error window or tolerance band so that when the position lands within the window, the PID system is disabled and replaced by a lock-and-hold system. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-04-09T01:39:33.993" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35076" PostTypeId="1" AcceptedAnswerId="53262" CreationDate="2020-04-09T06:53:49.637" Score="2" ViewCount="802" Body="&lt;p&gt;I have continuous A-weighted sound pressure level (SPL) values at 1 minute time resolution (&lt;code&gt;LAeq,1min&lt;/code&gt;). I would like to calculate &lt;code&gt;LAeq&lt;/code&gt; SPL at &lt;code&gt;half-hourly&lt;/code&gt; and &lt;code&gt;hourly resolutions&lt;/code&gt; using these values. Can someone suggest on how this could be achieved?  &lt;/p&gt;&#xA;" OwnerUserId="16435" LastActivityDate="2023-12-16T10:00:19.790" Title="How do I calculate hourly LAeq sound pressure level using 1-mint LAeq values?" Tags="|acoustics|noise|" AnswerCount="2" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35080" PostTypeId="1" CreationDate="2020-04-09T12:49:53.633" Score="1" ViewCount="104" Body="&lt;p&gt;I have a flat piece of wood shaped like this: &lt;a href=&quot;https://i.stack.imgur.com/42MkE.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/42MkE.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;And it is turning around axle.&#xA;What i would like to know is, if it is rotated e.g. at 1000 RPM, how much power would that take?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know that drag force should be equal to this:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;F_D=\frac{1}{2}C_D\rho_{air}Av^2&#xA;$$&lt;/span&gt;&#xA;And power:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;P = Fd&#xA;⋅&#xA;v&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using P air = 1.2 and Cd = 1.1 (i don't know if this is correct) . &lt;/p&gt;&#xA;&#xA;&lt;p&gt;First i calculated speed based on outer radius(27cm), but because of V^2 that was a big mistake, now i have split horizontally (or vertically if you are watching the picture) in parts of 0.5cm, and then calculated power for each part individually and then add them together, but i am getting what i think is lower value than what it should be (in range of 150-200W), i am getting ~87W at 660RPM. I am guessing i have a wrong approach to this problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT 1:&#xA;Here is python code how i calculated power with my way and also with Yaniv way.&lt;/p&gt;&#xA;&#xA;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;RPM = 500&#xA;RPS = RPM / 60&#xA;def estimatedPower(RPS):  # RPS = revolution per second&#xA;    finalPower = 0&#xA;    forceSmallPiece = 0&#xA;    forceBiggerPiece = 0&#xA;    for i in range(0, 54):&#xA;        radius = i/2&#xA;        if radius &amp;gt; 11:&#xA;            A = 0.0008&#xA;            V = radius/100*2*3.1415 * RPS&#xA;            F = (1.1*1.2*A*V*V)/2&#xA;            forceBiggerPiece += F&#xA;        else:&#xA;            A = 0.000375&#xA;            V = radius/100*2*3.1415 * RPS&#xA;            F = (1.1*1.2*A*V*V)/2&#xA;            forceSmallPiece += F&#xA;        P = F * V&#xA;        finalPower += P*2    # *2 is because there are 2 blades&#xA;    print(&quot;power: &quot;,finalPower)&#xA;    return finalPower, forceSmallPiece, forceBiggerPiece&#xA;&#xA;finalPower, forceSmallPiece, forceBiggerPiece = estimatedPower(RPS)&#xA;&#xA;&#xA;AngularSpeed = RPM * 2 * 3.1415/60&#xA;&#xA;Fd = forceBiggerPiece&#xA;Fd1 = forceSmallPiece&#xA;&#xA;&#xA;torque1 = Fd * 0.19&#xA;torque2 = Fd1 * 0.055&#xA;&#xA;torque = (torque1 + torque2) * 2    # *2 because there are 2 fan blades&#xA;p = torque * AngularSpeed&#xA;&#xA;print(&quot;Power with torque: &quot;,p)&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;" OwnerUserId="26483" LastEditorUserId="26483" LastEditDate="2020-04-09T16:43:05.057" LastActivityDate="2020-04-09T18:24:12.970" Title="Required power to spin fan at certain RPM" Tags="|power|drag|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35081" PostTypeId="2" ParentId="35080" CreationDate="2020-04-09T14:27:50.510" Score="1" Body="&lt;p&gt;First, pay attention to the fact that the total force acting on the wood is zero.&#xA;Since your device is turning, it would be much easier to calculate the power as torque times the angular velocity: P = tau * w.&#xA;When Tau is the total torque around the rotation axis (Nm) and w is the item&#xA;angular speed (rad/sec).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the drag force is coming from a forced flow having a velocity much greater than your wood linear velocity at the ends - I suggest to calculate the torque as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Deal only with one half of the wood and then multiply the result by&#xA;two &lt;/li&gt;&#xA;&lt;li&gt;Divide the taken half into two areas - a 16x16 cm square and a&#xA;11x7.5 rectangle&lt;/li&gt;&#xA;&lt;li&gt;For each area, calculate the drag force according to&#xA;your formula &lt;/li&gt;&#xA;&lt;li&gt;Assume the force on each area acts exactly on its&#xA;center. Figure what are resulted torques are by multiplying the&#xA;forces by their distance from the rotation axis (19 cm and 5.5 cm&#xA;respectively)&lt;/li&gt;&#xA;&lt;li&gt;Sum up both torques and multiply by two as mentioned&#xA;in step #1 - since there are two symmetric halves of that wood&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;That's it. Remember that 1000 RPM equals roughly to 105 rad/sec - and are ready to go.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, since the drag force in your case is due to the rotation itself - we need another approach. lets attack it in two alternative ways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The numerical approach:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Deal only with one half of the wood and then multiply the result by&#xA;two &lt;/li&gt;&#xA;&lt;li&gt;Divide the taken half into two areas - a 16x16 cm square and a&#xA;11x7.5 rectangle&lt;/li&gt;&#xA;&lt;li&gt;divide the radius to small sections (dx), in each calculate the&#xA;torque as Fd times r. When r is the distance from the rotation axis&#xA;to specific section. Pay attention - the area of each section is dx&#xA;times its height (16 cm and 7.5 cm). In your example you took the&#xA;whole rectangle area..&lt;/li&gt;&#xA;&lt;li&gt;Sum up both torques and multiply by two as mentioned&#xA;in step #1 - since there are two symmetric halves of that wood&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The analytic approach:&lt;/strong&gt;&#xA; You have to integrate the force times the distance over the whole distance. Lets right down the force formula in its differential form:&#xA; &lt;span class=&quot;math-container&quot;&gt;$F=\int 0.5 \cdot\rho\cdot C_d \cdot(w\cdot r)^2\cdot l\cdot dr $&lt;/span&gt;&#xA;Note that &lt;span class=&quot;math-container&quot;&gt;$l\cdot dr $&lt;/span&gt; is the differential representation of A. &#xA;The same way, the torque formula can be written down as:&#xA; &lt;span class=&quot;math-container&quot;&gt;$Torque=\int 0.5 \cdot\rho\cdot C_d\cdot (w\cdot r)^2\cdot l\cdot r\cdot dr $&lt;/span&gt;&#xA;Yep, we just multiply the force by r - which is the distance from each infinitesimal section to the rotation axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we integrate the torque formula from 0 to a total radius R we will get:&#xA; &lt;span class=&quot;math-container&quot;&gt;$Torque= 0.125\cdot \rho\cdot C_d \cdot w^2 \cdot l\cdot R^4$&lt;/span&gt;&#xA;Check the units - we get &lt;span class=&quot;math-container&quot;&gt;$\frac{kg*m^2}{sec^2}$&lt;/span&gt; which equals to Nm - the units of torque. You can now use this formula to calculate both torques and go on. Just be careful to deal correctly with the boundaries of the far away area - since it is not integrated from 0.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastEditorUserId="16860" LastEditDate="2020-04-09T18:24:12.970" LastActivityDate="2020-04-09T18:24:12.970" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="35082" PostTypeId="1" CreationDate="2020-04-09T15:16:15.820" Score="1" ViewCount="57" Body="&lt;p&gt;I'm looking to log some information on a piece of equipment I have running at home. It displays the information I want - operating rpm and some other settings - on seven-segment displays. In an ideal world, I'd like to log the value of this setting continuously, or at intervals. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The obvious thing to do would be to open it up and interface with the onboard electronics somehow, but I don't have much experience with this and am worried about breaking something. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;A non-invasive solution a friend suggested was to use a camera to read the value off the screen. They figured if a camera were to be placed adjacent to the display and either recorded video or took pictures of the display periodically, these video/images could be automatically analysed for the values of the seven-segment displays and the results logged. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have some experience with image analysis and reckon analysing brightness of certain pixels corresponding to active sections of the display relative to its background could be used to find the value from the display, as long as the camera/display were in a fixed position.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This idea sounds pretty good to me but based on some google research I can't find much information on similar schemes. Has anyone tried something similar, and have any pointers? Would something like this be possible? Is there a better way of implementing it? &lt;/p&gt;&#xA;" OwnerUserId="26485" LastActivityDate="2020-04-09T15:16:15.820" Title="Reading a value off a screen with a camera" Tags="|process-engineering|data|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35083" PostTypeId="1" CreationDate="2020-04-09T15:29:22.817" Score="1" ViewCount="38" Body="&lt;p&gt;I have a problem with thermodynamics.&#xA;I have this question:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$q_{12} = ?$&lt;/span&gt; Steam heating in constant volume. &lt;span class=&quot;math-container&quot;&gt;$p_1 = 1.76$&lt;/span&gt; bar. Steam quality: &lt;span class=&quot;math-container&quot;&gt;$x_1 = 0.7$&lt;/span&gt;;  &lt;span class=&quot;math-container&quot;&gt;$T_2 = 200$&lt;/span&gt; Celsius.&#xA;Where ’1’ is the current state of the steam, and ’2’ is the goal state.&#xA;What I need for this task is: &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$h_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$s_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$p_1 = 1.76$&lt;/span&gt; bar, &lt;span class=&quot;math-container&quot;&gt;$v_1$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$u_1$&lt;/span&gt;&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$h”$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$s”$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$p”$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$v”$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$u”$&lt;/span&gt; &lt;span class=&quot;math-container&quot;&gt;$t”$&lt;/span&gt; (the line between the liquid-vapor and vapor region)&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;span class=&quot;math-container&quot;&gt;$h_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$s_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$p_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$v_2$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$t_2 = 200$&lt;/span&gt; Celsius&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;My question is that how could I calculate this from a steam table? I know that this is isochoric changing, so &lt;span class=&quot;math-container&quot;&gt;$v =$&lt;/span&gt; constant, but I couldn’t find any solution on the internet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks guys! I really appreciate it. It makes me mad for days :)&lt;/p&gt;&#xA;" OwnerUserId="26486" LastEditorUserId="8216" LastEditDate="2020-04-09T18:19:44.590" LastActivityDate="2020-04-09T18:19:44.590" Title="Thermodynamics, how can i do this with steam?" Tags="|thermodynamics|steam|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35084" PostTypeId="2" ParentId="35025" CreationDate="2020-04-09T16:55:53.287" Score="1" Body="&lt;p&gt;1) The maximum mechanical work of a process is not the same as the actual work done on/by the system. The former is calculated using the assumptions:&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;reversible&lt;/li&gt;&#xA;&lt;li&gt;ideal gas&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In the reversible case,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \Delta U = 0 \rightarrow w = -q\ \ \mathrm{IUPAC\ form,\ for\ Clausius\ form\ } w = q $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next, the starting equation for work is &lt;span class=&quot;math-container&quot;&gt;$w = -\int p_{ext}\ dV$&lt;/span&gt;. Only when we say that the process is reversible can we state that &lt;span class=&quot;math-container&quot;&gt;$w = -\int p\ dV$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) &lt;strong&gt;For a reversible adiabatic processes with ideal gases&lt;/strong&gt;, we can substitute into the mechanical work using &lt;span class=&quot;math-container&quot;&gt;$pV^\gamma = $&lt;/span&gt;constant. An adiabatic reversible process is also always isentropic. An isentropic process is not always a reversible adiabatic one. An isentropic process also does not always mean that we are using an ideal gas. In summary, we cannot say that we can use &lt;span class=&quot;math-container&quot;&gt;$pV^\gamma = $&lt;/span&gt;constant for the work of any generic isentropic process. This is a shortcut that ignores the three founding rules for the substitution (an adiabatic, reversible process on an ideal gas).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally, regardless of whether the real process is reversible or irreversible, we can always write&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta U = \int T\ dS - \int p\ dV $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All terms in the above are state functions, and the change &lt;span class=&quot;math-container&quot;&gt;$\Delta U$&lt;/span&gt; does not care about the path.&lt;/p&gt;&#xA;" OwnerUserId="16981" LastEditorUserId="16981" LastEditDate="2020-04-09T17:24:16.787" LastActivityDate="2020-04-09T17:24:16.787" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35085" PostTypeId="1" CreationDate="2020-04-09T17:06:17.197" Score="1" ViewCount="11" Body="&lt;p&gt;exactly what the title says.  and, what is the easiest way of doing this?  I can do it clunky and manually by first creating a design table with only one dummy variable then adding the colour colums..&lt;/p&gt;&#xA;&#xA;&lt;p&gt;edit: this &lt;a href=&quot;https://help.solidworks.com/2018/English/SolidWorks/sldworks/t_inserting_file_design_table.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://help.solidworks.com/2018/English/SolidWorks/sldworks/t_inserting_file_design_table.htm&lt;/a&gt; , with the correction that the A2 cell must be named &quot;family&quot;, as described here (not where I originally found it) &lt;a href=&quot;https://help.solidworks.com/2011/English/SolidWorks/sldworks/LegacyHelp/Sldworks/Configuration/Formatting_a_Design_Table.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://help.solidworks.com/2011/English/SolidWorks/sldworks/LegacyHelp/Sldworks/Configuration/Formatting_a_Design_Table.htm&lt;/a&gt; is what I was looking for..&lt;/p&gt;&#xA;" OwnerUserId="13081" LastEditorUserId="13081" LastEditDate="2020-04-09T18:02:27.417" LastActivityDate="2020-04-09T18:02:27.417" Title="can a design table be created -only- for colours in solidworks?" Tags="|solidworks|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35086" PostTypeId="1" CreationDate="2020-04-09T18:34:15.357" Score="-1" ViewCount="392" Body="&lt;p&gt;I can find it when the transfer function is written in standard form but I don't know how to find it if it is written differently. For example &lt;span class=&quot;math-container&quot;&gt;$G(s)=\dfrac{1}{s(s+1)^3}$&lt;/span&gt;, is this order three or four?&lt;/p&gt;&#xA;" OwnerUserId="26487" LastEditorUserId="18238" LastEditDate="2020-04-10T11:37:03.137" LastActivityDate="2020-04-10T11:37:03.137" Title="Order of a system" Tags="|control-theory|transfer-function|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35087" PostTypeId="1" AcceptedAnswerId="35088" CreationDate="2020-04-09T22:03:05.040" Score="1" ViewCount="71" Body="&lt;p&gt;I want to calculate the improvement in flow rate that can be achieved with changing the tubing size when the tubing is sub-millimeter size. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I originally approached this using a simplification of Poiseuille's law which says &lt;span class=&quot;math-container&quot;&gt;$\Delta Q=\Delta r^4$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, lets say I'm changing from 0.5 to 1.5 millimeter tubing.  That gives &lt;span class=&quot;math-container&quot;&gt;$\Delta r=0.050$&lt;/span&gt;, which when raised to the 4th power is a tiny number (6.25E-05). This seems wrong intuitively, based on the &lt;em&gt;relative&lt;/em&gt; increase in size (i.e. new tube is 300% of the original tube)   &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If I convert to micrometers first, the result jumps to 6.25E06, which seems really really large. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What am I missing here, because this doesn't seem like the right approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="2598" LastActivityDate="2020-04-10T01:22:27.007" Title="Relative improvement in flow from diameter change with sub-milliimeter tubing" Tags="|fluid-mechanics|flow-control|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35088" PostTypeId="2" ParentId="35087" CreationDate="2020-04-10T01:22:27.007" Score="1" Body="&lt;p&gt;You made a small mistake if we are going to use Poiseuille's law.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$   Q_{final}= Q_{initial}\Delta r^4 \rightarrow  \Delta Q =(\frac{r_{final}}{r_{itial}})^4$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Plugging your numbers,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ Q_{final}=Q_{initial} (0.75/0.25)^4 \\ =Q_{initial}3^4=81Q_{initial}$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-10T01:22:27.007" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35090" PostTypeId="1" AcceptedAnswerId="35092" CreationDate="2020-04-10T01:49:10.320" Score="2" ViewCount="383" Body="&lt;p&gt;I am looking at designing a rack and pinion system that will be laser cut out of some plastic. While the cad system that I am using has a spur gear generator, I was looking at the design of the rack gears. I thought that it needed an involute type profile, but many of the references that I see have a flat profile instead of an involute type profile. For example, this metal rack gear from McMastercarr has a seemingly flat profile.&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/jm7vq.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/jm7vq.png&quot; alt=&quot;mcmastercar&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Is this common? Is the reason that it has a flat profile due to the fact that the radius of the base circle for racks is theoretically infinite?&lt;/p&gt;&#xA;" OwnerUserId="26490" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2020-04-17T04:14:20.767" Title="Does the rack for a rack and pinion often have or need to have the involute profile?" Tags="|mechanical-engineering|design|gears|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35091" PostTypeId="2" ParentId="31887" CreationDate="2020-04-10T06:38:30.297" Score="2" Body="&lt;p&gt;If I get you right, there is a box having a small device that dissipates 30 Watts of heat continuously into the air. The air is blocked into a box, no way in or out. There are no fans/blowers trying to cool the device itself nor circulating the air inside the box. The size of the box is to be defined and the air temperature inside the box has to be figured out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the air volume were infinite - we could assume that its ambient temperature is fixed (25 deg), approximate a natural heat convection coefficient (something around 7-10 watts &lt;span class=&quot;math-container&quot;&gt;$\frac{Watt}{m^2K}$&lt;/span&gt;) and find out the temperature over the heatsink by using the well-known  &lt;span class=&quot;math-container&quot;&gt;$q=hA(\Delta T)$&lt;/span&gt; equation. When q is the dissipated heat, A is the total area of heat convection and &lt;span class=&quot;math-container&quot;&gt;$\Delta T$&lt;/span&gt; is the difference between the ambient temp and the temp over the heatsink surface (assuming it is uniformly distributed).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, in your case the temperature of the air inside the box will rise up since its volume is finite. In the long run, the steady state, the total amount of heat flowing out of the box must be exactly 30 watts and the air temperature will converged to a constant value.&#xA;Choosing the desired walls are is a bit tricky procedure: Assuming the walls have a constant uniform temperature of 25 degrees, and guessing the air temperature is something between 25 deg to the heatsink temp - we can find the total area needed for allowing the 30 watt heat to flow outside the box under those assumptions. It is a kind of back and forth calculation for getting an order of magnitude. For whatever area you finally choose, you can find out the air temperature close to the walls by using  &lt;span class=&quot;math-container&quot;&gt;$q=hA(\Delta T)$&lt;/span&gt;. This time A is the walls area. We can take an assumption that the air temperature inside the box is uniform and having that value.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now you have a decent approximation of the air temperature and you can go on finding the heatsink temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Of course, this is only an approximation. For more precise calculation you may use a CFD tool.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Using the specific heat and heat capacity are needed only if you want to analyze transient problem. I.e - checking the temperature behavior during time before reaching the steady state.&lt;/p&gt;&#xA;" OwnerUserId="16860" LastEditorUserId="16860" LastEditDate="2020-04-10T08:49:59.380" LastActivityDate="2020-04-10T08:49:59.380" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35092" PostTypeId="2" ParentId="35090" CreationDate="2020-04-10T07:22:44.657" Score="3" Body="&lt;p&gt;Standard (circular) involute gears have a stationary contact normal. It is usually a line, that is tangential to the base radii of the two gears. When you replace one gear with a rack, then for the meshing to be the same on the gear, the rack needs to be compliant to the given contact normal. From the stationary contact normal you can easily follow, that the contact surface must be flat. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The observation of an infinite base radius (on the rack) may be misleading, because the (theoretical) involute of the racks (infinite) base circle would be a point. &lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2020-04-10T07:22:44.657" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35094" PostTypeId="2" ParentId="35086" CreationDate="2020-04-10T10:49:57.857" Score="2" Body="&lt;p&gt;There are two (between others) characteristics that define the transfer function of the system and moreover the system itself. &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Order &lt;span class=&quot;math-container&quot;&gt;$\ \rightarrow $&lt;/span&gt; Defined by the maximum power of the laplace variable &lt;span class=&quot;math-container&quot;&gt;$\ s $&lt;/span&gt; in the denominator&lt;/li&gt;&#xA;&lt;li&gt;Type &lt;span class=&quot;math-container&quot;&gt;$\ \  \rightarrow $&lt;/span&gt; Defined by the number of poles at the origin (only) of the transfer function. Poles of a system are the roots of the denominator of the transfer function.&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;In this particular example, by working out easily the math of the denominator, you will find out that the order of the system is: &lt;span class=&quot;math-container&quot;&gt;$\ 4 $&lt;/span&gt;. For some extra tiny analysis, you can see that this system is of type &lt;span class=&quot;math-container&quot;&gt;$\ 1 $&lt;/span&gt;. You can also find relevant information at this page by using matlab to find out the order of your system. However, I encourage you at least in the beggining to find it out by yourself:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.mathworks.com/help/control/ref/order.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.mathworks.com/help/control/ref/order.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although I answered your question, I advise you to first do your own search and try to figure out the solution to your problem on your own before posting here since this was a pretty simple question and could be answered easily by doing a quick google search. &lt;/p&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-04-10T11:23:41.313" LastActivityDate="2020-04-10T11:23:41.313" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35099" PostTypeId="1" CreationDate="2020-04-10T17:35:48.663" Score="-1" ViewCount="389" Body="&lt;p&gt;From what I read, concrete has much lower compressive strength than ceramics. So why aren't we using ceramics to build? Is it because ceramic failure is difficult to predict? Because it's hard to shape? Expense? Or the low tensile strength has a role?&lt;/p&gt;&#xA;" OwnerUserId="26499" LastActivityDate="2020-04-12T17:46:04.517" Title="can ceramic be used as a building material?" Tags="|concrete|building-design|building-physics|ceramics|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35100" PostTypeId="2" ParentId="35099" CreationDate="2020-04-10T17:49:23.597" Score="4" Body="&lt;p&gt;structural ceramic does have some applications in things like engine parts, but its cost and manufacturing difficulty makes it a less ideal choice for something as large as a building foundation&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The thing about concrete is that you can mix it with water, and pour it into a space to make a foundation.  Ceramics have to be kiln fired at 1300 + C for hours. It's just not practical to make a house-sized kiln when concrete is good enough.&lt;/p&gt;&#xA;" OwnerUserId="6488" LastEditorUserId="6488" LastEditDate="2020-04-12T17:46:04.517" LastActivityDate="2020-04-12T17:46:04.517" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35101" PostTypeId="2" ParentId="35099" CreationDate="2020-04-10T18:50:27.940" Score="4" Body="&lt;p&gt;One of the foundational principles of structural engineering is obviously safety. And one of the foundational aspects of safety is the ability to show warning signs prior to any accidents. In the case of structural engineering, that means designing structures to crack or sway or groan or what-have-you prior to collapse. In fact, when designing beams, one of the checks that must be done is that the beam doesn't deflect too much: if the beam has too much of a &quot;belly&quot;, people walking under it will naturally and understandably feel uncomfortable. So we design beams not to have excessive deflections to avoid that discomfort in the building's users but also because we &lt;strong&gt;&lt;em&gt;want&lt;/em&gt;&lt;/strong&gt; users to be uncomfortable if they see a beam with excessive deflections or a column which is starting to show cracks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Excessive deflections and active cracks* are the building's way of screaming &quot;&lt;a href=&quot;https://youtu.be/uG_KHjd_PSc?t=81&quot; rel=&quot;nofollow noreferrer&quot;&gt;get out&lt;/a&gt;&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why the most common construction materials are steel, reinforced concrete and wood. All of these materials have significant plastic deformations prior to collapse, as seen by their stress-strain diagrams:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.chegg.com/homework-help/questions-and-answers/4-stress-strain-curves-tensile-measurement-three-different-categories-materials-ceramics-p-q25329184&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/93PHW.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;br&gt;&#xA;&lt;sup&gt;A: ceramics; B: metals; C: polymers (concrete also has a vaguely similar shape, though it is obviously much stronger). Click through to source.&lt;/sup&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here we can see the difference in the behavior of ceramics (A) vs. steel (B) and concrete (similar enough to C for our purposes).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you increase the load (stress) in steel or concrete, the slope of their diagram gets shallower. This means that the structure is deflecting more and more for every increment in load. Eventually, at the end of their respective diagrams, the material will collapse, but you'll have huge deflections before that happens, which should give people plenty of warnings.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a good visualization of this effect, see &lt;a href=&quot;https://www.youtube.com/watch?v=6ycbDCnoO8M&quot; rel=&quot;nofollow noreferrer&quot;&gt;this video of a concrete beam deflection test&lt;/a&gt;. The first crack appears when the beam deflects 8&amp;nbsp;mm (around 1:40 in the video, though I only spot it around 2:10). When the beam finally collapses (4:55), it's deflected 250&amp;nbsp;mm and has been showing massive cracks and even spitting chunks of concrete for ages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For ceramics, however, the slope is constant and almost vertical: you increase the load, the structure barely deflects, all is fine. You increase the load, the structure still barely deflects, all is fine. You increase the load... and now you've passed the material's limit stress and it suddenly collapses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if the beam in the video linked above were ceramic, it'd go from basically no deflection to collapse in an instant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's the difference between ceramics and the materials we tend to use for construction: &lt;strong&gt;&lt;em&gt;ceramics are &lt;a href=&quot;https://en.wikipedia.org/wiki/Brittleness&quot; rel=&quot;nofollow noreferrer&quot;&gt;brittle&lt;/a&gt;&lt;/em&gt;&lt;/strong&gt;: they have barely any elastic deformation and no plastic deformation. You apply loads to ceramic and it either basically doesn't move or it instantly collapses. Steel and concrete, on the other hand, are ductile: they have noticeable elastic deformations and significant plastic deformations prior to collapse.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, unless you're dealing with very specific situations, usually requiring ceramics' specific thermodynamic properties, as mentioned in &lt;a href=&quot;https://engineering.stackexchange.com/a/35100/1832&quot;&gt;@Mohammad Athar's answer&lt;/a&gt;, safety demands that ceramics be avoided, so that the structure may die not with a bang but with a prolonged scream of &quot;get out&quot;.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;&lt;sub&gt;* as opposed to inactive cracks (those that aren't growing) which can be caused by initial settling or other movements but which are no longer a threat to the building's structural integrity.&lt;/sub&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastEditorUserId="1832" LastEditDate="2020-04-10T19:04:59.233" LastActivityDate="2020-04-10T19:04:59.233" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="35105" PostTypeId="1" CreationDate="2020-04-11T00:51:07.013" Score="2" ViewCount="44" Body="&lt;p&gt;I'm reading a book on robotics, and it has a different section on each term. The way both concepts are worded and phrased mathematically make them seem like they both constrain the configuration space of a closed loop mechanism in more or less the same way. I'm not sure if I'm missing something, but they seem like the same thing.&lt;/p&gt;&#xA;" OwnerUserId="26508" LastActivityDate="2020-04-11T14:19:05.453" Title="Are Loop Closure Equations and Holonomic Constraints the same thing?" Tags="|mechanical-engineering|mathematics|robotics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35107" PostTypeId="1" AcceptedAnswerId="35121" CreationDate="2020-04-11T03:33:00.617" Score="1" ViewCount="365" Body="&lt;p&gt;I am designing a home CNC mill for machining nonferrous metals.  I plan to use linear guides to support either a Y-axis table or the 2 legs of a Y-axis moving gantry.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear guides are used to provide low friction linear motion along one axis while resisting forces in other directions (both rotational and translational).  They come in many sizes as shown below, and must be mounted to very flat surfaces and adjusted for parallelism with respect to a &quot;reference&quot; guide.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/VhW3R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/VhW3R.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a number of identical linear guides at my disposal that are 12mm in width which is on the smaller end of the spectrum (although they are lower in height profile). I am curious if using more than 2 linear guides (3 or 4 all in parallel) is beneficial (provided that I mount them all in parallel).  Are there any drawbacks I should think about when considering using more than 2 linear guides to guide something?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason I ask is (1) I don't want to waste a linear guide if this is a &quot;weakest link&quot; situation, and (2) using 2 linear guides is by far the most common I've seen in commercial designs for linear stages.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/bxauJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/bxauJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Someone might ask &quot;if you're worried about rigidity why dont you go buy some bigger stronger linear rails&quot;?  I guess that's also part of my same question - could using multiple smaller linear guides additively achieve the same effect as a bigger stronger linear guide (while at the same time maybe  being lower profile which I think should decrease the moment arm for the roll axis)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ZwCBV.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ZwCBV.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here are some diagrams from newports website if they help for reference: &lt;a href=&quot;https://www.newport.com/t/manual-positioning-basics&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.newport.com/t/manual-positioning-basics&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;straightness deviation:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vy62g.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vy62g.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;tilt rigidity:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/eH0OP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eH0OP.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;load capacity:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ChG3r.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ChG3r.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21120" LastEditorUserId="21120" LastEditDate="2020-04-11T06:23:22.610" LastActivityDate="2020-04-11T22:29:29.193" Title="Can I use more than 2 linear guides to support a linear stage?" Tags="|linear-motion|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35108" PostTypeId="1" CreationDate="2020-04-11T05:06:26.213" Score="-1" ViewCount="195" Body="&lt;p&gt;I have a problem regarding following two structures.&lt;br&gt;&#xA;&lt;strong&gt;&lt;em&gt;First one:&lt;/em&gt;&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/7y8E8.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7y8E8.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;&lt;em&gt;Second one:&lt;/em&gt;&lt;/strong&gt;&lt;br&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/ovvRF.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ovvRF.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What arrangemet is better ? Why ?&lt;br&gt;&#xA;Actually I have no idea which one is better.&lt;/p&gt;&#xA;" OwnerUserId="26509" LastActivityDate="2020-04-13T15:26:24.997" Title="What arrangement of web members is better?" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ClosedDate="2020-04-13T15:30:00.307" ContentLicense="CC BY-SA 4.0" />
  <row Id="35109" PostTypeId="2" ParentId="35099" CreationDate="2020-04-11T05:57:02.233" Score="1" Body="&lt;p&gt;It is very common to use ceramics as a building material in the form of bricks. As the process of making ceramics requires heating to a very high temperature, it is not practical to make it in anything larger than these small units. And not all bricks are ceramic; not surprisingly, ceramic bricks tend not to be a common building material in areas where the appropriate types of clay aren't plentiful.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem of ceramics being a brittle material as discussed in Wasabi's answer is also solved by using the material in small units with a much weaker material (mortar) inbetween. When the mortar is sufficiently weak compared to the bricks, the mortar will always fail before the bricks do and the structure as a whole becomes able to crack massively prior to a total failure, making it safer overall.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2020-04-11T05:57:02.233" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="35110" PostTypeId="2" ParentId="35108" CreationDate="2020-04-11T05:58:57.620" Score="0" Body="&lt;p&gt;The one with diagonals opening out up and out gfrom the center, number one.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By inspecting the members we immediately say that the diagonal members are 1.41 times longer than the horizontal and vertical members.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The longer members with the same cross-section perform better in tension than compression, due to slenderness factors and buckling concerns. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the first configuration, the diagonals are under tension and will just need to be checked for pure tension.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the second configuration, the diagonals are under compression, we would need to make sure the diagonals are strong against in-plane and out of plane buckling.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="1832" LastEditDate="2020-04-13T15:26:24.997" LastActivityDate="2020-04-13T15:26:24.997" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35111" PostTypeId="1" CreationDate="2020-04-11T07:16:28.950" Score="-1" ViewCount="65" Body="&lt;p&gt;I have a system of second-order ODEs &lt;br/&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\mathbf{M\ddot{x} + C\dot{x} + Kx = f}&#xA;$$&lt;/span&gt;&#xA;I want to know some good numerical methods to solve this system of the equation given the initial conditions at time &lt;span class=&quot;math-container&quot;&gt;$t = T$&lt;/span&gt; that is I am given &lt;span class=&quot;math-container&quot;&gt;$x(T)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\dot{x}(T)$&lt;/span&gt; and I want to solve for &lt;span class=&quot;math-container&quot;&gt;$x$&lt;/span&gt; between &lt;span class=&quot;math-container&quot;&gt;$t = [0, T)$&lt;/span&gt;&lt;br/&gt;&#xA;I have tried to solve this with the Newmark method which is stable with a positive time step but with a negative time step, Newmark fails miserably. &lt;/p&gt;&#xA;" OwnerUserId="26511" LastActivityDate="2020-04-11T23:38:48.960" Title="How to solve a system of second order ODE from time t = T to t = 0" Tags="|structural-engineering|structures|finite-element-method|dynamics|solid-mechanics|" AnswerCount="3" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35112" PostTypeId="1" CreationDate="2020-04-11T08:59:51.340" Score="-1" ViewCount="31" Body="&lt;p&gt;I have the following situation. My cat goes outside every day, the problem is I have no way of telling when he is back around the house unless I go outside myself. &#xA;I need some small wireless device I can attach to his collar that would alert me when he is in the range of my WiFi. GPS locators are just too big for a cat and movement detectors are not really working. I need something that's at max 1cm in size and simple. A wireless mini camera could do the trick, it would start streaming when in the vicinity of my WiFi, but the problem is the privacy laws in my country. If the cat entered someone's property with the camera, I could be sued. So I need an alternative, some simple small signal emitting device or something that can be picked up by my WiFi so that I know he's around, nothing more.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I should mention that I have no idea how these stuff actually work, so something I've said here could sound dumb I guess.&lt;/p&gt;&#xA;" OwnerUserId="26512" LastEditorUserId="26512" LastEditDate="2020-04-11T09:07:37.393" LastActivityDate="2020-04-11T12:14:32.503" Title="Small wireless proximity device to alert me when my cat gets into my WiFi range" Tags="|wireless-communication|signal|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35113" PostTypeId="1" CreationDate="2020-04-11T10:08:17.403" Score="3" ViewCount="332" Body="&lt;p&gt;In Derek Seward's &lt;a href=&quot;https://books.google.com/books?id=FR3nBgAAQBAJ&quot; rel=&quot;nofollow noreferrer&quot;&gt;Race Car Design&lt;/a&gt; and in some of the videos on race car braking systems, such as &lt;a href=&quot;https://www.youtube.com/watch?v=N2NyxdBNFe0&quot; rel=&quot;nofollow noreferrer&quot;&gt;this one&lt;/a&gt;, it is mentioned the use of two master cylinders, one for the front brakes and one for the back.&#xA;The bias between front and back being set using different lever length between the pedal and each master cylinder.&lt;/p&gt;&#xA;&lt;p&gt;Why is this configuration preferred over a single master cylinder of same surface area and a bias valve? Is it safety requirement to keep both front and back hydraulic systems entirely separated or is there any technical benefit?&lt;/p&gt;&#xA;" OwnerUserId="26515" LastEditorUserId="6790" LastEditDate="2020-08-03T18:46:59.997" LastActivityDate="2023-07-10T18:43:08.350" Title="Why do race cars — such as Formula 1 cars — use two brake master cylinders?" Tags="|mechanical-engineering|automotive-engineering|" AnswerCount="2" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="35114" PostTypeId="2" ParentId="35113" CreationDate="2020-04-11T10:33:03.753" Score="1" Body="&lt;p&gt;Two cylinders are often needed due to the fluid input needed by the multi-pot calipers used front and rear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The control of balance is also handy.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-11T10:33:03.753" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35115" PostTypeId="2" ParentId="35112" CreationDate="2020-04-11T12:14:32.503" Score="1" Body="&lt;p&gt;If your objective is only to determine device in range, and you've located a suitable wireless camera to meet your size requirements, such a device could be used without privacy concerns if you painted over the lens. You could also simply place the camera in a completely solid enclosure. Your receiver would change from a no-signal status to a black/blank screen status, providing the notification you require.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Additionally, you may consider to find a software package that polls your network and advises you when a device is added, activated or vanishes from the network. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your size restrictions may be too severe, though, as you also have to consider a battery to power the device and a battery smaller than a centimeter square is going to have extremely short capacity.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-04-11T12:14:32.503" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35116" PostTypeId="1" CreationDate="2020-04-11T12:53:52.423" Score="0" ViewCount="74" Body="&lt;p&gt;I'm confused about the Lagrangian of a piezoelectric material. I've seen multiple sources which seem to contradict each other.&#xA;The way I personally approach the problem is the following (&lt;em&gt;&lt;span class=&quot;math-container&quot;&gt;$\underline{note}$&lt;/span&gt;: I'm ignoring the work done by external forces and charges for now, since they are not relevant to the discussion&lt;/em&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$L = K - U $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; is the kinetic energy, and &lt;span class=&quot;math-container&quot;&gt;$U$&lt;/span&gt; is the potential energy which consists of mechanical strain energy &lt;span class=&quot;math-container&quot;&gt;$U_m$&lt;/span&gt; + electrical energy &lt;span class=&quot;math-container&quot;&gt;$U_e$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$K = \int\frac{1}{2}\rho \dot u_i \dot u_i dV$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$U_m = \int\frac{1}{2}S_{i}T_{i} dV$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$U_e = \int\frac{1}{2}E_{i}D_{i} dV$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\rho$&lt;/span&gt; is the density, &lt;span class=&quot;math-container&quot;&gt;$\dot u_i$&lt;/span&gt; is the velocity, &lt;span class=&quot;math-container&quot;&gt;$S_i$&lt;/span&gt; is the strain tensor (Voigt notation), &lt;span class=&quot;math-container&quot;&gt;$T_i$&lt;/span&gt; is the stress tensor (Voigt notation), &lt;span class=&quot;math-container&quot;&gt;$E_i$&lt;/span&gt; is the electric field, &lt;span class=&quot;math-container&quot;&gt;$D_i$&lt;/span&gt; is the electric displacement field. This results in the Lagrangian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  L = \frac{1}{2}\int(\rho \dot u_i \dot u_i - S_{i}T_{i} - E_{i}D_{i}) dV$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The problem is that most references write the Lagrangian as:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  L = \frac{1}{2}\int(\rho \dot u_i \dot u_i - S_{i}T_{i} + E_{i}D_{i}) dV$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;with the sign in front of the electrical energy reversed. Examples:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://scmero.ulb.ac.be/Publications/Thesis/Piefort01.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;Example 1&lt;/a&gt;: if you go to page 56, you'll see they derive the Lagrangian from the electrical enthalpy, and I don't get why it gives a different result (in the form of a + sign in front of the electrical energy).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://books.google.com.lb/books?id=lCePDwAAQBAJ&amp;amp;pg=PA278&amp;amp;dq=piezoelectric%20%22lagrangian%22&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=0ahUKEwjqwp3bzN3oAhUBxYUKHVooDGQQ6AEILzAB#v=onepage&amp;amp;q=piezoelectric%20%22lagrangian%22&amp;amp;f=true&quot; rel=&quot;nofollow noreferrer&quot;&gt;Example 2&lt;/a&gt;: same result, with the constitutive relations substituted in the Lagrangian.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is twofold:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) Where is the error in my approach in that it yields the wrong sign in front of the electrical energy?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) In the main book I'm studying, which is &lt;em&gt;Engineering Analysis of Smart Material Systems.&lt;/em&gt; Donald J. Leo, 2007, if you go to &lt;a href=&quot;https://books.google.com.lb/books?id=jGAvKfLykNsC&amp;amp;pg=PA218&amp;amp;lpg=PA218&amp;amp;dq=The%20tilde%20over%20a%20symbol%20denotes%20the%20fact%20that%20the%20strain%20and%20electric%20displacement%20are%20expressed%20in%20terms%20of%20the%20local%20piezoelectric%20coordinates%20and%20not%20the%20global%20coordinates.%20The%20local%20coordinate%20system%20for%20each%20piezoelectric%20element%20can%20be%20transformed%20to%20the%20global%20coordinates%20through%20the%20transformations&amp;amp;source=bl&amp;amp;ots=5tyv1NmGD0&amp;amp;sig=ACfU3U2coUWoCyy1dRJ25BpxoEszouMnTQ&amp;amp;hl=en&amp;amp;sa=X&amp;amp;ved=2ahUKEwjQhc_02t7oAhVs8-AKHf4rAXMQ6AEwAHoECAsQKA#v=onepage&amp;amp;q=The%20tilde%20over%20a%20symbol%20denotes%20the%20fact%20that%20the%20strain%20and%20electric%20displacement%20are%20expressed%20in%20terms%20of%20the%20local%20piezoelectric%20coordinates%20and%20not%20the%20global%20coordinates.%20The%20local%20coordinate%20system%20for%20each%20piezoelectric%20element%20can%20be%20transformed%20to%20the%20global%20coordinates%20through%20the%20transformations&amp;amp;f=false&quot; rel=&quot;nofollow noreferrer&quot;&gt;page 218&lt;/a&gt;, equation 5.47, the author gives the potential energy of a piezoelectric material (static case, no kinetic energy so &lt;span class=&quot;math-container&quot;&gt;$L=-U$&lt;/span&gt;):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  U = \int(\frac{1}{2}S_{i}c_{ij}^D S_{j} - S_ih_{ij}D_j + \frac{1}{2}D_{i}\beta_{ij}^SD_{j}) dV$&lt;/span&gt;&#xA;which is actually derivable from &lt;em&gt;my Lagrangian approach&lt;/em&gt; rather than the one given in the two previous reference examples when you substitute the constitutive relations into the Lagrangian:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ T_i = c_{ij}^DS_j - h_{ij}D_j $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ E_i = - h_{ji}S_j + \beta_{ij}^S D_j $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do we reconcile the contradictory references?&lt;/p&gt;&#xA;" OwnerUserId="26519" LastActivityDate="2020-04-11T12:53:52.423" Title="Lagrangian of a piezoelectric material" Tags="|mechanical-engineering|structural-engineering|materials|modeling|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35117" PostTypeId="2" ParentId="35105" CreationDate="2020-04-11T14:19:05.453" Score="1" Body="&lt;p&gt;Is a cell phone and a pocket calculator the same thing? It's not, but a cell phone is often used as one (using the app). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think it depends on the context. In many multi body simulation solvers, holonomic constraints are the usual way to model joints. Think of it as using &quot;loop closures&quot; in all places, as a convenient way to not set up equations recursively. &#xA;I am not sure about that, but there may also be other ways to close the (algebraic - this what you mean?)  loop. &lt;/p&gt;&#xA;" OwnerUserId="23772" LastActivityDate="2020-04-11T14:19:05.453" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35119" PostTypeId="2" ParentId="35111" CreationDate="2020-04-11T21:32:00.027" Score="0" Body="&lt;p&gt;This has a direct analytical solution that in probably most softwares can be solved in one line...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example mathematica:&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;DSolve[{m x''[t] + d x'[t] + c x[t] == f, x'[T] == a, x[T] == b}, x[t], t]&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/zPMOk.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zPMOk.png&quot; alt=&quot;sol&quot;&gt;&lt;/a&gt;&#xA;Or, if one wanted...by hand using pretty traditional ODE methods.&lt;/p&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2020-04-11T21:32:00.027" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35120" PostTypeId="2" ParentId="35111" CreationDate="2020-04-11T21:52:12.507" Score="0" Body="&lt;p&gt;Your initial second order ODE can also be written as the following system of first order ODE&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{bmatrix}&#xA;\dot{x} \\ \ddot{x}&#xA;\end{bmatrix} = &#xA;\begin{bmatrix}&#xA;0 &amp;amp; I \\ -M^{-1} K &amp;amp; -M^{-1} C&#xA;\end{bmatrix}&#xA;\begin{bmatrix}&#xA;x \\ \dot{x}&#xA;\end{bmatrix} + &#xA;\begin{bmatrix}&#xA;0 \\ M^{-1} f&#xA;\end{bmatrix}, \tag{1}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which can be solved by many known solvers. The terminal condition can also be used as an initial condition when also negating the right-hand-side of &lt;span class=&quot;math-container&quot;&gt;$(1)$&lt;/span&gt;. Namely, given that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{dx}{dt} = f(t,x), \quad x(T)=x_T. \tag{2}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now by using that &lt;span class=&quot;math-container&quot;&gt;$\tau=T-t$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$z(\tau)=x(T-\tau)$&lt;/span&gt; then &lt;span class=&quot;math-container&quot;&gt;$(2)$&lt;/span&gt; would be equivalent to&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{dz}{d\tau} = \frac{dx}{dt} \frac{dt}{d\tau} = -f(T-\tau,z), \quad z(0)=x_T. \tag{3}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, it might be possible that the negated dynamics is unstable, which potentially causes the solution backwards in time to diverge. This would mean that also the analytical solution might also diverge.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2020-04-11T21:52:12.507" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35121" PostTypeId="2" ParentId="35107" CreationDate="2020-04-11T22:29:29.193" Score="2" Body="&lt;p&gt;The idea is mostly sound. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have a load F that is too high for my linear rails....so instead of beefing up my rails, I increase the load sharing across 3 or more rails, thus dividing it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In reality, I have actually seen such things....but only on poorly built machines as a half solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In general one &lt;em&gt;can&lt;/em&gt; do anything they want....but that doesn't mean it's the correct way to do it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Linear rails are specifically designed for uses of 2 rails for a specific type of load and/or torque, in various orientations and environments within specific tolerances and drift or alignment tolerances. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Beyond this, trying to align 3, or four rails along a linear axis accurately, without them binding or causing premature wear on the rails or carriages themselves would be tedious and difficult.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But it is possible...you just wouldn't see anyone do it. &lt;/p&gt;&#xA;" OwnerUserId="14318" LastActivityDate="2020-04-11T22:29:29.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35122" PostTypeId="1" CreationDate="2020-04-11T22:44:46.170" Score="-1" ViewCount="52" Body="&lt;p&gt;What happens to air flow and propeller efficiency if it is moving with Wind perpendicular to rotation axis. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The air is blowing in a direction lying on the Planet of the tips.&lt;/p&gt;&#xA;" OwnerUserId="23424" LastActivityDate="2021-01-12T12:03:05.280" Title="Efficiency of propeller with Wind perpendicular to rotation axis" Tags="|aerodynamics|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="35123" PostTypeId="2" ParentId="35111" CreationDate="2020-04-11T23:38:48.960" Score="0" Body="&lt;p&gt;The problem with stability is caused by the equation itself, not by your attempts at solving it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you integrate backwards, the &quot;damping terms&quot; are &lt;em&gt;adding&lt;/em&gt; energy to the system as the integration proceeds, not dissipating it. Any numerical rounding errors will therefore grow with time. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Even if the numerical errors are negligible, the initial conditions are very poorly determined by the final state of the system.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-04-11T23:38:48.960" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35124" PostTypeId="2" ParentId="35122" CreationDate="2020-04-12T01:56:22.467" Score="0" Body="&lt;p&gt;If the wind is near perpendicular to the propellers axis it will make it turn and creat lift. It is called a gyrocopter. Usually, another fan on the back provides the thrust and the big near-horizontal disk creates the lift.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But if you position a propeller in the exact perpendicular wind it loses the lift and will stall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Unless one designs a custom profile blade specifically designed for maintaining lift under wind coming from a tangent plane. But even then building directional control, authority, efficiency is a different ballpark. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-12T01:56:22.467" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35125" PostTypeId="1" CreationDate="2020-04-12T02:40:30.337" Score="1" ViewCount="292" Body="&lt;p&gt;I am currently doing some lab work involving wire drawing and trying to calculate the % volume change of the wire after wire drawing. I am using the geometrical data (length and diameter) of the wires before and after the drawing process for these calculations. From various sources online I have learned that the volume doesn't change for the wire drawing process. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, my calculations don't seem to show this feature. I have calculated the volume of the wire (before and after the wire drawing process) using the formula: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Volume = &lt;span class=&quot;math-container&quot;&gt;$\small\sf\pi {(radius^2)(length)}$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In my calculations the volume of the wire after each draw gradually decreased.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can some tell me if this is correct or if there is another method to conduct this calculation. Any form of help would be greatly appreciated.&lt;/p&gt;&#xA;" OwnerUserId="26524" LastEditorUserId="90" LastEditDate="2020-04-12T06:00:21.007" LastActivityDate="2020-04-12T14:31:07.660" Title="How to calculate % volume change due to wire drawing" Tags="|mechanical-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35126" PostTypeId="2" ParentId="35125" CreationDate="2020-04-12T03:16:34.157" Score="0" Body="&lt;p&gt;Assuming your wire is steel and the Poisson ratio 0.3, if you pull the wire such that you have 1% elongation then your radius shrinks by 0.033% so your ratio of the volume change is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(0.997)&lt;sup&gt;2&lt;/sup&gt;*1.01 = 1.0039&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means the volume has increased by a factor of 0.0039.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Approximately 4 thousands. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="90" LastEditDate="2020-04-12T06:05:42.393" LastActivityDate="2020-04-12T06:05:42.393" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35128" PostTypeId="1" CreationDate="2020-04-12T11:10:26.083" Score="4" ViewCount="279" Body="&lt;p&gt;In Eurocode, when use buckling curves to calculate the maximum load that a compressive member can be subjected to before it risks buckling. Buckling curves are based on the Perry-Robertson formula. I'm reading a derivation of the formula &lt;a href=&quot;http://www.colincaprani.com/files/notes/Column%20Notes.pdf&quot; rel=&quot;noreferrer&quot;&gt;here.&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The formula is derived basically like this: we assume some initial eccentricity of the column in shape of a sinusoid, and then arrive at this expression for the deflection of the column at the mid-point:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y(\frac{L}{2})=\frac{\sigma_E}{\sigma_E-\sigma}a$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt; is the stress compressing the column, &lt;span class=&quot;math-container&quot;&gt;$\sigma_E$&lt;/span&gt; the Euler critical load of the column and &lt;span class=&quot;math-container&quot;&gt;$a$&lt;/span&gt; the assumed eccentricity of the column at mid-point.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then we use an expression for the maximum stress inside the column, summing the direct compressive stress as well as the stress that occurs from the bending moment:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_{max}=\sigma+\frac{Mc}{Ar^2}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\sigma_{max}$&lt;/span&gt; is the stress on the extreme fiber on the mid-point of the column, &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt; the moment of the column due to load, &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; the distance from the neutral axis to the extreme fiber, &lt;span class=&quot;math-container&quot;&gt;$A$&lt;/span&gt; the cross sectional area and &lt;span class=&quot;math-container&quot;&gt;$r$&lt;/span&gt; the radius of gyration.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Knowing that &lt;span class=&quot;math-container&quot;&gt;$M = Py(\frac{L}{2})$&lt;/span&gt;, we get:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\sigma_{max}=\sigma+\sigma\frac{c}{r^2}\frac{\sigma_E}{\sigma_E-\sigma}a$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It is then said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;At failure the maximum stress is the yield stress, &lt;span class=&quot;math-container&quot;&gt;$\sigma_y$&lt;/span&gt;&#xA;  . &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and we proceed to set &lt;span class=&quot;math-container&quot;&gt;$\sigma_{max}=\sigma_y$&lt;/span&gt; and solve for the compressive stress &lt;span class=&quot;math-container&quot;&gt;$\sigma$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This last step puzzles me.&lt;/strong&gt; Why do we set the stress to yield stress, if we are interested specifically in designing against buckling? If we design a column in this way, then yes we can limit the compressive load so that the extreme fiber (and thus the entire column) never reaches the yield limit anywhere, but how does this prevent a failure by buckling? We know buckling is a failure occuring from sudden sideways deflection of a member. But buckling can very well occur entirely elastically. This formula seems to only ensure that the member does not &lt;em&gt;yield&lt;/em&gt;, but does not seem to say anything about how much the member deflects. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;To me, it would make more sense to take the very first equation, the deflection of the midpoint, and set some limits on that, based on the slenderness of the column. &lt;strong&gt;So don't we really care about how much the member bends sideways as long as it does not yield, or have I misunderstood something?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2020-04-13T21:55:58.457" Title="Why is local yielding considered the limiting case in context of buckling in Eurocode?" Tags="|structural-engineering|buckling|yield-point|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35130" PostTypeId="2" ParentId="35125" CreationDate="2020-04-12T14:31:07.660" Score="1" Body="&lt;p&gt;There are two causes of the change in shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is plastic deformation, which takes place as constant volume.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is the elastic deformation caused by the stress that is &quot;locked in&quot; to the structure when you stretch the material. They usually do cause a change of volume, related to the value of Poisson's Ratio of the material.&#xA;Try measuring the volume of your wire, then annealing it, then measure the volume again. If the wire is a &quot;soft&quot; material like brass or copper, you can anneal it by simply heating it up to red hot for a few seconds and letting it cool naturally in air. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note that if you only measure the diameter at a few places along the length, the calculation for the volume will probably not be accurate because after drawing, the wire probably won't have a constant diameter. It might be more accurate to do something like make a coil of the wire, immerse it in a liquid, and measure the volume directly.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-04-12T14:31:07.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35131" PostTypeId="2" ParentId="35128" CreationDate="2020-04-12T16:14:48.557" Score="2" Body="&lt;p&gt;One of the most fundamental principles in the eurocodes is the distinction between the serviceability limit state (SLS) and the ultimate limit state (ULS). An element that has large, visible deformations but still is able to support the load is ugly and potentially scary, but not dangerous. Therefore, initial buckling is considered an SLS failure, but not necessarily a ULS failure. The column will not fail completely before the deformation becomes great enough for the maximum stress to reach the yield stress, and that means you need different equations for the SLS check and for the ULS check. In the ULS check, we generally don't care how large the deformations are as long as nothing fails.&lt;/p&gt;&#xA;" OwnerUserId="15064" LastActivityDate="2020-04-12T16:14:48.557" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="35132" PostTypeId="1" AcceptedAnswerId="35137" CreationDate="2020-04-12T19:15:31.987" Score="0" ViewCount="221" Body="&lt;p&gt;I was recently reading some of Leonardo da Vinci's &lt;a href=&quot;https://en.wikisource.org/wiki/The_Notebooks_of_Leonardo_Da_Vinci/XIII&quot; rel=&quot;nofollow noreferrer&quot;&gt;notes on architecture&lt;/a&gt;. One passage on arches states:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;An Experiment to show that a weight placed on an arch does not&#xA;  discharge itself entirely on its columns; on the contrary the greater&#xA;  the weight placed on the arches, the less the arch transmits the&#xA;  weight to the columns. The experiment is the following. Let a man be&#xA;  placed on a steel yard in the middle of the shaft of a well, then let&#xA;  him spread out his hands and feet between the walls of the well, and&#xA;  you will see him weigh much less on the steel yard; give him a weight&#xA;  on the shoulders, you will see by experiment, that the greater the&#xA;  weight you give him the greater effort he will make in spreading his&#xA;  arms and legs, and in pressing against the wall and the less weight&#xA;  will be thrown on the steel yard.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;There are two parts to my question:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What are the physical principals behind Leonardo's statements on the distribution of an arch's load to its columns?&lt;/li&gt;&#xA;&lt;li&gt;How does his experiment demonstrate these principals?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;On 1, I interpret his scenario as in the below diagram. There is some arch which is supported at is ends by two columns &lt;span class=&quot;math-container&quot;&gt;$C_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_2$&lt;/span&gt;. There are additionally two abutments &lt;span class=&quot;math-container&quot;&gt;$A_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$A_2$&lt;/span&gt; to the sides of the arch to counter the arch's horizontal thrust. A downward force &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; is applied to the apex, which is directed through internal compression forces to the arch's ends, where we see &lt;span class=&quot;math-container&quot;&gt;$F_1$&lt;/span&gt; being applied to &lt;span class=&quot;math-container&quot;&gt;$A_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$C_1$&lt;/span&gt; (and analogously for the other side). My understanding is that no matter how large &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; is, the vertical components of &lt;span class=&quot;math-container&quot;&gt;$F_1$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$F_2$&lt;/span&gt; must both be &lt;span class=&quot;math-container&quot;&gt;$F/2$&lt;/span&gt; in order to satisfy the static equilibrium constraints.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, how do I reconcile this analysis with Leonardo's claim &quot;the greater the weight placed on the arches, the less the arch transmits the weight to the columns&quot;? Is Leonardo assuming a different setup than I have? Or is one of us wrong?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/g2F44.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/g2F44.jpg&quot; alt=&quot;Arch Diagram&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26535" LastEditorUserId="26535" LastEditDate="2020-04-12T19:28:58.700" LastActivityDate="2020-04-13T04:50:48.347" Title="What does Leonardo da Vinci mean in his statement on arches?" Tags="|statics|building-physics|architecture|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="35134" PostTypeId="2" ParentId="17110" CreationDate="2020-04-12T21:01:54.210" Score="0" Body="&lt;p&gt;It seems like you could use a link and have it follow a path something like a ring cam to control the blade pitch however you like.&lt;/p&gt;&#xA;" OwnerUserId="26538" LastActivityDate="2020-04-12T21:01:54.210" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35137" PostTypeId="2" ParentId="35132" CreationDate="2020-04-13T04:50:48.347" Score="0" Body="&lt;p&gt;I think he may have been thinking two things,&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Friction.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Large deformations&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;If we follow him in his path of organizing the gradual transfer of vertical reactions to tangential friction forces from the wall taking most the vertical load, he must have been thinking of a gradual change in the mechanism of the arch from a regular arch to an arch with a plastic hinge formed on the apex.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That would make it possible for the vertical forces to be supported entirely by a triangular force mechanism with the external force on the top and a cracked or flexed apex and to friction forces on the two lateral bottom sides as the tangential components of the friction force.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Let's assume at the ultimate case when the apex is fully cracked or turned into a plastic hinge, a triangular truss substituting the arch with the two base angles at 45 degrees and the load P and the friction factor of 0.8 just to illustrate his idea.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The horizontal reaction at the walls&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_h= P/2 \quad \text{and the vertical component of friction, Fr} \quad  F_r= 0.8P$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So for an angle of 45 degrees, the wall takes 80% of the load.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-13T04:50:48.347" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35138" PostTypeId="1" CreationDate="2020-04-13T06:13:24.073" Score="1" ViewCount="1061" Body="&lt;p&gt;Every time I have seen a spider at home, in the bathroom, whether it was in a house or in apartments, whether on the bottom or upper floor, I always immediately assumed that it had come &quot;through the pipes&quot;; that is, crawled up through the drain pipe leading into the bathtub, or the drain hole in the floor which takes care of any water that spills out on the floor.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or even that it had swimmed all the way up through the water toilet.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Luckily, it hasn't happened for quite some time now (knock on wood...), but this thought was especially present in my mind when the spider was sighted in the actual bathtub.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My question is: do spiders actually crawl up such pipes, into homes? Or is this a false assumption? Don't spiders tend to come in from opened doors and windows? Not through pipes?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I mean, it's not like the drain pipes lead directly down underground with dirt and creatures living there. It's still connected to the bigger sewage pipes, no? I cannot imagine any creature surviving not only the oxygen-less, long trip through these large sewage/water pipes, but to also have the strength to stop and actively push themselves up and keep crawling all the way up to my apartment/house's bathroom.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But maybe I underestimate those vile creatures of pure evil.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;On the other hand, it seems unlikely that a spider would come from the outside, though a window or door, only to find itself crawling &lt;em&gt;into&lt;/em&gt; the bathttub and be sighted there, so that talks against my theory that it's impossible for insects to enter through these drain pipes...&lt;/p&gt;&#xA;" OwnerUserId="26540" LastActivityDate="2023-08-24T02:21:01.680" Title="Do spiders *actually* get into a house through the sewer/water pipes?" Tags="|drain-waste-vent|" AnswerCount="4" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="35139" PostTypeId="2" ParentId="35138" CreationDate="2020-04-13T06:35:40.070" Score="2" Body="&lt;p&gt;No, they come in through all the gaps and cracks in the building frame. They also come when you have a window open for ventilation or through ventilation grills.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are not swimming against the sewage and coming up through the plug hole.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-13T06:35:40.070" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35141" PostTypeId="1" CreationDate="2020-04-13T11:58:21.953" Score="2" ViewCount="114" Body="&lt;p&gt;I'm reading &lt;a href=&quot;https://www.comsol.se/blogs/buckling-structures-suddenly-collapse/&quot; rel=&quot;nofollow noreferrer&quot;&gt;this article&lt;/a&gt; on buckling.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It first introduces us to linear buckling analysis, and then mentions that linear analysis can be unrealistic as it doesn't take imperfections into account. Then it is said:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;One good strategy is to first perform a linearized buckling analysis and then use the computed mode shape as imperfection. &lt;strong&gt;The idea is that the structure will be most sensitive to this shape.&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;So if I understand correcty, the article suggests that to get a more realistic analysis we first compute the linear buckling modes, use this as initial imperfections and then do a non-linear analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I don't exactly understand why the structure is most sensitive to the buckling modes computed from linear analysis. What does this statement mean exactly? &lt;/p&gt;&#xA;" OwnerUserId="13372" LastActivityDate="2020-04-13T22:39:27.107" Title="Using modes from linear buckling analysis as imperfections" Tags="|structural-engineering|buckling|" AnswerCount="2" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="35142" PostTypeId="1" CreationDate="2020-04-13T13:50:28.667" Score="3" ViewCount="129" Body="&lt;p&gt;I'm trying to understand a function in a SIMULINK example of a path following UAV simulation. The gist is that, the UAV Waypoint Follower block provides a required heading from a list of waypoints and this heading is converted, by the heading control block, into a Target Roll that is used by the autopilot.&lt;a href=&quot;https://i.stack.imgur.com/yC8yy.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yC8yy.png&quot; alt=&quot;fixedWingPathFollowing from the Robotics Blockset&quot;&gt;&lt;/a&gt;&#xA;The heading control block is the following function:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/vTWCv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/vTWCv.png&quot; alt=&quot;Heading Control&quot;&gt;&lt;/a&gt;&#xA;This function in the absence of wind evaluates to:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\phi = \text{atan2}(p(\psi_d - \psi)V,g)&#xA;$$&lt;/span&gt;&#xA;meaning,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ubJbr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ubJbr.png&quot; alt=&quot;atan2&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where, &lt;span class=&quot;math-container&quot;&gt;$\phi$&lt;/span&gt; is the target roll, &lt;span class=&quot;math-container&quot;&gt;$V$&lt;/span&gt; is the current speed, &lt;span class=&quot;math-container&quot;&gt;$\psi$&lt;/span&gt; is current heading, &lt;span class=&quot;math-container&quot;&gt;$\psi_d$&lt;/span&gt; is the desired heading and &lt;span class=&quot;math-container&quot;&gt;$p$&lt;/span&gt; is the proportional gain &quot;PHeadingAngle&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm struggling to understand how that equation works mathematically, and how it transforms a heading to a roll angle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If possible, I'd also like an explanation for how the function deals with wind. I know that it transforms the heading angle and the speed using the wind speeds but am unsure how that transformation works.&lt;/p&gt;&#xA;" OwnerUserId="26544" LastEditorUserId="12754" LastEditDate="2020-04-14T00:44:43.300" LastActivityDate="2020-04-14T00:44:43.300" Title="Mathematical transformation of a heading angle to a roll angle" Tags="|control-engineering|control-theory|mathematics|aerodynamics|matlab|" AnswerCount="0" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35144" PostTypeId="2" ParentId="28211" CreationDate="2020-04-13T18:25:39.350" Score="1" Body="&lt;p&gt;I had this same question but was able to solve it. I've attached a picture of my circuit. I know this was asked but I hope this helps someone. The Supply rail is attached to the rippled power supply. Whether it's stable or rippled doesn't affect the functionality noticeably (as long as it's above the minimum Supply) &lt;a href=&quot;https://i.stack.imgur.com/zfN7b.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/zfN7b.jpg&quot; alt=&quot;op amp based negative voltage regulator&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26551" LastActivityDate="2020-04-13T18:25:39.350" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35146" PostTypeId="2" ParentId="35128" CreationDate="2020-04-13T21:55:58.457" Score="2" Body="&lt;p&gt;Euler buckling is a perfectly elastic behavior: if a beam buckles, it takes a sinusoidal shape of arbitrary (possibly infinite) amplitude. If you then remove the applied force, the beam will return to its original, perfectly-straight shape.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Real-world beams, however, don't behave this way: if you buckle a beam, it gets destroyed. After all, Euler buckling's infinite amplitude implies infinite strain (and therefore stress) throughout the column. So a real-world column buckles until it reaches its yield strain/stress, at which point it collapses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That's why the Perry-Robertson considers the yield stress: while buckling is an elastic behavior, the column's eventual destruction happens at its yield stress (on the extreme fiber).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-04-13T21:55:58.457" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35148" PostTypeId="2" ParentId="35141" CreationDate="2020-04-13T22:09:13.360" Score="2" Body="&lt;p&gt;I briefly glanced this article. Seems interesting but I haven't read it yet - hopefully I'll still be able to help.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First, lets say a few words regarding the non-linear buckling analysis. It is important to realize that contrary to linear analysis, it takes into account the following:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Geometrical non linearity - the model shape changes according to the&#xA;applied loads (deformation) so the load distribution through the&#xA;model is updated accordingly. Step by step&lt;/li&gt;&#xA;&lt;li&gt;Material non linearity - As the stresses reached the yield point,&#xA;the YOUNG modulus changes and so is the model stiffness matrix.&#xA;Yielded zones will not carry excessive loads anymore and the load&#xA;distribution will be updated&lt;/li&gt;&#xA;&lt;li&gt;Imperfections - small deflections that make a difference between a&#xA;&quot;clean&quot; FEA model to a real life case&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;In the sense of it, buckling modes are the shapes a body will tend to wear while being subjected to the a specific load pattern. There are shapes for bending loads, torsional loads and so on. For each mode there is critical load the model could not further withstand and that will eventually cause it to buckle. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;While running a buckling analysis, we apply loads on a model and examine its behavior. We would be able to obtain the deflection it gets and to check whether buckling occurs under these loads. &#xA;If we run a non-linear analysis, it is a good advice to add a small imperfection, in the shape of the relevant mode (i.e, a shape the will contribute to the deflection under the specific case). Do not forget that physically and mathematically, the model will first begin to deform when linearity governs (small deflections, constant stiffness) - so it will try to deform into one of the modes that can be acquired with linear buckling analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hope this helps...&lt;/p&gt;&#xA;" OwnerUserId="16860" LastActivityDate="2020-04-13T22:09:13.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35149" PostTypeId="2" ParentId="35141" CreationDate="2020-04-13T22:39:27.107" Score="1" Body="&lt;p&gt;It's worth remembering that buckling is caused by the bending moment generated by axial loads due to (possibly infinitesimal) imperfections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Given that, it becomes simple to understand what the text is saying. The linear buckling analysis will show you the shape the column or structure will take upon buckling. The particular shape this gives you is, by definition, the shape which does the worst job resisting the bending moment generated in buckling; if another shape were worse, you'd get that shape instead.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if you set your imperfections according to the linear buckling analysis, you'll basically be doing buckling's job for it.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-04-13T22:39:27.107" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35152" PostTypeId="2" ParentId="14091" CreationDate="2020-04-14T01:57:45.580" Score="0" Body="&lt;p&gt;.95mm would be what is used in the slot car industry for this&lt;/p&gt;&#xA;" OwnerUserId="26556" LastActivityDate="2020-04-14T01:57:45.580" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35153" PostTypeId="1" AcceptedAnswerId="49090" CreationDate="2020-04-14T03:04:51.080" Score="1" ViewCount="232" Body="&lt;p&gt;I have a worm gear set with no drawings. I need to model them. The only things I know are&#xA;1) The Gear ratio is 30, and the worm gear only has one thread. It is a single start as opposed to a multi start worm.&#xA;2) The diameter of the worm is 0.7 Inches, and the length of the worm is 0.852 Inch&#xA;3) The diameter of the gear is 1.007 Inches, and it has 30 teeth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know the pressure angle, diametrical pitch or gear module. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;What other piece of missing information would I need to know? How do I go about figuring out the diametrical pitch? How do I figure out the module.&lt;/p&gt;&#xA;" OwnerUserId="24389" LastActivityDate="2022-01-04T15:04:37.817" Title="How to I determine the dimentions for a worm gear?" Tags="|gears|" AnswerCount="2" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="35154" PostTypeId="2" ParentId="35153" CreationDate="2020-04-14T05:01:43.117" Score="1" Body="&lt;p&gt;Lets say to make it simple you have the disk with 30 teeth. So the distance between each is 1.07 *pi /30= P&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now your worm circumfrance is 0.7pi.&#xA;We know the slope of the thread on the worm should be such that 30P = 0.7pi.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dividing by 30 we get P = 0.7pi/30.&#xA;This is the slope of the thread on the worm.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;30 turns on the worm will turn the disk 30 teeth or one turn. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-14T05:01:43.117" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35155" PostTypeId="2" ParentId="33829" CreationDate="2020-04-14T08:15:08.677" Score="0" Body="&lt;p&gt;Since your system is a drone, it is very hard to control the whole attitude at once. Instead, what is standard is controlling each of the three Euler angles, corresponding to rotation in each direction (i.e &lt;span class=&quot;math-container&quot;&gt;$\phi,\theta,\psi$&lt;/span&gt;). Additionally, the dynamics are nonlinear, although you can linearize them which is also standard procedure. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what you should do if you actually want to implement this: look up the dynamics for your system (what are your state variables, what are your control inputs). Then try to do PID controller for pitch and see how that works. You can do the same for roll and yaw, but sometimes what people do is since roll and yaw are highly coupled, they implement a coupled controller for both instead of two separate ones. In any case, you have many choices here but you need a solid model first for analysis. &lt;/p&gt;&#xA;" OwnerUserId="15049" LastActivityDate="2020-04-14T08:15:08.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35156" PostTypeId="2" ParentId="33781" CreationDate="2020-04-14T08:17:57.720" Score="0" Body="&lt;p&gt;One possibility: since you are saying that you know what this load is a priori, you can just add a time dependent function into your dynamics, something like &lt;span class=&quot;math-container&quot;&gt;$P(t)$&lt;/span&gt; where &lt;span class=&quot;math-container&quot;&gt;$P$&lt;/span&gt; represents the load. You will then have a time dependent system, which you can still control. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the load is uncertain, you can add noise to the system dynamics and try experimenting with Kalman Filters and LQG controllers. &lt;/p&gt;&#xA;" OwnerUserId="15049" LastActivityDate="2020-04-14T08:17:57.720" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35157" PostTypeId="2" ParentId="33814" CreationDate="2020-04-14T08:25:49.993" Score="0" Body="&lt;p&gt;If you have a transfer function already, there are a bunch of easy methods you can use to characterize it (as well as for a frequency sweep). Some of the common ones are: root locus plot, Nyquist plot, and Bode plot. These will all tell you in some way characteristics about the stability of your system instead of having to look at settling time or rise time directly. &lt;/p&gt;&#xA;" OwnerUserId="15049" LastActivityDate="2020-04-14T08:25:49.993" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35158" PostTypeId="1" AcceptedAnswerId="35165" CreationDate="2020-04-14T08:32:35.230" Score="2" ViewCount="278" Body="&lt;p&gt;I am reading a report titled ``&lt;em&gt;Definition of safe separation criteria for external store and pilot escape capsules&lt;/em&gt;'' in which the author uses following expressions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Newton's second law:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$ z \left( t\right) = \frac{1}{m} \int \int\limits_{0}^{t}F \left( \tau \right)d\tau dt$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;From Taylor Series,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ F \left(\tau \right) = F \left( 0 \right) + \sum\limits_{n=1}^{\infty} \frac{d^{n}F}{d\tau^n} \Bigg\rvert_{\tau = 0} \frac{\tau^{n}}{n!}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Substituting the Taylor series expansion in first equation, we get&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ z \left( t\right) = \frac{1}{m} \left[ \frac{F \left( 0 \right) t^{2}}{2!} + \sum\limits_{n=1}^{\infty} \frac{d^{n}F}{d\tau^n} \Bigg\rvert_{\tau = 0} \frac{\tau^{n+2}}{\left(n + 2 \right)!}\right] + w \left( 0 \right) t + z \left( 0 \right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Questions:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;I am familiar with differntial equation form of Newton's Second law,&#xA;i.e., &lt;span class=&quot;math-container&quot;&gt;$F = m \frac{d^{2}z}{dt^{2}}$&lt;/span&gt; but I don't know how the&#xA;author obtained integral form, especially the limits of integration.&lt;/li&gt;&#xA;&lt;li&gt;Since the integral with respect to &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; is a definite integral, why the author has added constant of integration when integrating with respect to &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt;.&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="17712" LastActivityDate="2020-04-16T14:51:30.467" Title="Converting Newton's Second Law Differential Form to Integral Form and Integration Limits" Tags="|applied-mechanics|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35159" PostTypeId="2" ParentId="32656" CreationDate="2020-04-14T08:37:44.930" Score="1" Body="&lt;p&gt;In general, the problem of finding a controller that minimizes some objective function falls under an extremely deep field of control theory called “optimal control”. Optimal control theory deals with minimizing functions of the state and controls (in your case it’s the squared error, which would fall under state) subject to some dynamics (in your case the transfer function), and constraints on the system. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are many methods to solve these kinds of problems, such as Hamilton-Jacobi-Bellman (HJB) equations, dynamic programming, Pontryagin’s Maximum Principle (PMP) and many others. &lt;/p&gt;&#xA;" OwnerUserId="15049" LastActivityDate="2020-04-14T08:37:44.930" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35160" PostTypeId="2" ParentId="35158" CreationDate="2020-04-14T09:18:13.420" Score="2" Body="&lt;p&gt;I have never seen such an expression of the Newton second law. &#xA;From the first equation you provide, it seems that there are two different time variables, &lt;span class=&quot;math-container&quot;&gt;$t$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt;, which is obviously not the case. From what I get this notation is just a translation of a double integration in time.&#xA;I agree with you that the limits of the integration are quite confusing. To my mind both integral limits should be 0 and t, and I would rewrite the whole equation this way:&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;z \left( t\right) = \frac{1}{m} \int\limits_{0}^{t} (\int\limits_{0}^{t}F \left( \tau \right)d\tau) d\tau&#xA;$$&lt;/span&gt;&#xA;As for your second question, the constants of integration just reflect the physical initial conditions. &lt;span class=&quot;math-container&quot;&gt;$w(0)$&lt;/span&gt; is the velocity and &lt;span class=&quot;math-container&quot;&gt;$z(0)$&lt;/span&gt; the position at time &lt;span class=&quot;math-container&quot;&gt;$t=0$&lt;/span&gt;, I do not think it has something to do with the integral being definite.&lt;/p&gt;&#xA;" OwnerUserId="26427" LastActivityDate="2020-04-14T09:18:13.420" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35161" PostTypeId="1" CreationDate="2020-04-14T09:47:19.520" Score="2" ViewCount="104" Body="&lt;p&gt;I'm trying to use the Solidworks simulation package to solve for the stresses inside a body as it accelerates (very rapidly). Because I expect some small features to deform quite a bit, I need to use the large displacement option, but because the object translates through space, I need to solve it using the inertia relief option. When I start the simulation with both option checked, I get the error message:  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;em&gt;Inertia Relief with Iterative solver is not supported by Large Displacement. Please use the Direct Sparse Solver.&lt;/em&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the conceptual/math level, why can't I use both the Large Displacement option and the Inertia Relief option with the default FFEPlus solver?&lt;/p&gt;&#xA;" OwnerUserId="26561" LastActivityDate="2020-04-14T09:47:19.520" Title="Solidworks Simulation- Large displacement with Inertia Relief" Tags="|finite-element-method|simulation|solidworks|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35162" PostTypeId="1" AcceptedAnswerId="35163" CreationDate="2020-04-14T10:31:12.003" Score="2" ViewCount="37" Body="&lt;p&gt;According to Wikipedia &lt;a href=&quot;http://en.wikipedia.org/wiki/Heat_exchanger&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://en.wikipedia.org/wiki/Heat_exchanger&lt;/a&gt;, plate exchangers typically serves lower pressure fluids than shell and tube heat exchanger, why?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tubes in plate exchangers are simpler and have larger diameter, seems to be more robust to hold higher pressure fluids.&lt;/p&gt;&#xA;" OwnerUserId="26563" LastEditorUserId="90" LastEditDate="2020-04-14T13:21:51.670" LastActivityDate="2020-04-14T13:21:51.670" Title="Why do plate exchangers typically serve lower pressure fluids than shell and tube heat exchanger?" Tags="|heat-transfer|heat-exchanger|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35163" PostTypeId="2" ParentId="35162" CreationDate="2020-04-14T11:06:59.663" Score="2" Body="&lt;p&gt;Because the plates are usually very thin for their dimensions and tubes tend to have more strength for the pressure loading.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I would check better references such as Simonson: Heat transfer for more thorough detail instead of wiki...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-14T11:06:59.663" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35164" PostTypeId="1" CreationDate="2020-04-14T12:43:41.107" Score="0" ViewCount="339" Body="&lt;p&gt;I have two sets of forces and moments, measured by two 6 DOF force/torque sensors at position A and B on a robot arm (B at end effector). My goal is to use this measurements to relate the forces/moments at position A to those at position B, i.e. find a transformation of  &lt;span class=&quot;math-container&quot;&gt;$F_A, M_A$&lt;/span&gt; to &lt;span class=&quot;math-container&quot;&gt;$F_B, M_B$&lt;/span&gt;. The transformation is static and no other external forces apply.&lt;br&gt;&#xA;The related formulas for that are: &#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;F_B \approx RF_A \\&#xA;M_B \approx RM_A+t\times F_B &#xA;$$&lt;/span&gt;&#xA;Is that correct (it's addaped from &lt;a href=&quot;https://flexiblelearning.auckland.ac.nz/sportsci303/13_3/forceplate_manual.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;this&lt;/a&gt;)?&lt;br&gt;&#xA;My goal is now to find an optimal (in a least squares way) matrix R and vector t. How would I do this? Is the &lt;a href=&quot;https://en.wikipedia.org/wiki/Kabsch_algorithm&quot; rel=&quot;nofollow noreferrer&quot;&gt;Kabsch algorithm&lt;/a&gt; suitable for this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Update:&lt;br&gt;&#xA;Below a small sketch of the setup. It is some sort of handle on the robot arm. A force is applied (and measured) at Sensor A, the goal is to use the measured force at sensor B to find the transformation.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/9HBZL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9HBZL.png&quot; alt=&quot;Sketch of the setup&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26564" LastEditorUserId="26564" LastEditDate="2020-04-15T09:24:22.837" LastActivityDate="2020-04-15T09:24:22.837" Title="Finding transformation of forces and moments" Tags="|mechanical-engineering|mathematics|robotics|moments|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="35165" PostTypeId="2" ParentId="35158" CreationDate="2020-04-14T12:48:44.120" Score="4" Body="&lt;h3&gt;Review of Terms&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;w(t) -&amp;gt; v(t) -&amp;gt; velocity at time t&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;z(t) -&amp;gt; x(t) -&amp;gt; position at time t&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;code&gt;w(0) -&amp;gt; initial velocity&lt;/code&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Derivation&lt;/h3&gt;&#xA;&#xA;&lt;p&gt;We can compute &lt;span class=&quot;math-container&quot;&gt;$w(t)$&lt;/span&gt; from force, since &lt;span class=&quot;math-container&quot;&gt;$\frac{dw}{dt} = F(t) / m$&lt;/span&gt;, so &lt;span class=&quot;math-container&quot;&gt;$$w(t) - w(0) = \int_{0}^{t}\left(\frac{1}{m}F(\tau)d\tau\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Which means &lt;span class=&quot;math-container&quot;&gt;$$w(t) = w(0) + \int_{0}^{t}\left(\frac{1}{m}F(\tau)d\tau\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We can compute &lt;span class=&quot;math-container&quot;&gt;$z(t)$&lt;/span&gt; from velocity: &lt;span class=&quot;math-container&quot;&gt;$$z(t) - z(0)  = \int_{0}^{t} \left( w(\tau) d\tau \right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Where we substitute the above integral expression for &lt;span class=&quot;math-container&quot;&gt;$w(t)$&lt;/span&gt;: &lt;span class=&quot;math-container&quot;&gt;$$z(t) - z(0) = \int_{0}^{t}\left( w(0) + \int_{0}^{t}\frac{1}{m}F(\tau) d\tau\right)d\tau$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since &lt;span class=&quot;math-container&quot;&gt;$w(0)$&lt;/span&gt; is constant, we can compute its integral in closed form: &lt;span class=&quot;math-container&quot;&gt;$w(0) * t$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$z(t) = z(0) + w(0)t + \int_{0}^{t}\left(\int_{0}^{t}\frac{1}{m}F(\tau) d\tau\right)d\tau$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And you can use that Taylor expansion to write an expression for this integral.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;br/&gt;&lt;/p&gt;&#xA;&#xA;&lt;h3&gt;Conclusion&lt;/h3&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;The integral form is the double-integral of the differential equation. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;The integration constants you are seeing come from the fundamental theorem of calculus — they’re the initial conditions. Recall the classic kinematic equation: &lt;span class=&quot;math-container&quot;&gt;$z(t) = z(0) + w_0t + \frac{a_0t^2}{2}$&lt;/span&gt; for the same expression, where acceleration is known to be constant. &lt;span class=&quot;math-container&quot;&gt;$w_0$&lt;/span&gt; is &lt;span class=&quot;math-container&quot;&gt;$w(0)$&lt;/span&gt;. The body’s initial velocity applies a linear offset to its motion&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="3374" LastEditorUserId="3374" LastEditDate="2020-04-16T14:51:30.467" LastActivityDate="2020-04-16T14:51:30.467" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35166" PostTypeId="1" CreationDate="2020-04-14T12:52:05.430" Score="1" ViewCount="55" Body="&lt;p&gt;I recently was riding my e-bike (with a bldc hub motor) and decided to cool the motor (it -the stator- was approx 100-150 celcius I think) in the river. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The controller is on the top of the bike and definitely stayed completely dry.&#xA;The motor has cooling holes drilled in its housing through which the water can flow in. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;cooling holes in motor housing:&#xA;&lt;a href=&quot;https://i.stack.imgur.com/2ZZVU.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/2ZZVU.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This method was very effective for cooling but the drive train (motor/throttle/controller/battery started acting up after riding it again after cooling...)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note: after the motor had dried/warmed up again the issues were gone (not sure what caused the issues so not sure if the water or the sudden change in temperature or something else caused the issues).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Symptoms: &#xA;-weird wining noises from motor when accelerating (much higher pitched and louder than normal)&#xA;-when the motor reaches a certain rpm (above this rpm) it appears as if the coil groups are misfiring (motor vibrates/shakes and appears to be braking rather than accelerating). The severity of this braking effect appears to be random, sometimes it's worse but not sure why..&#xA;The issue appeared to mostly start occurring when  a certain rpm was exceeded but sometimes (less often) the issue also occurred at much lower rpm.&#xA;-the issue was gone after approx 30 minutes of riding when the motor was approx 60 celcius (motor housing)(stator was probably approx 80 Celcius).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no temperature sensor from the motor connected to the bldc motor controller (the controller does not have this feature).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What might cause these issues? &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Can the water entering the motor mess up the signals coming from the hall sensors? If so what causes this? &lt;/li&gt;&#xA;&lt;li&gt;could the sudden temperature change cause issues similar to this? If so please explain how&lt;/li&gt;&#xA;&lt;li&gt;Is there another explanation which I did not think of? &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;video:&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=l2dHn89Dvso&amp;amp;feature=youtu.be&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=l2dHn89Dvso&amp;amp;feature=youtu.be&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you!&lt;/p&gt;&#xA;" OwnerUserId="10328" LastActivityDate="2020-04-14T12:52:05.430" Title="BLDC Motor underwater ; weird behaviour" Tags="|motors|cooling|electromagnetism|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="35167" PostTypeId="1" AcceptedAnswerId="35183" CreationDate="2020-04-14T13:51:39.843" Score="-1" ViewCount="57" Body="&lt;p&gt;I'm trying to spec a stepper motor that would allow me to position a tube, as well as spin the tube continuously in either direction. Ideally at a maximum of 60rpm, but variable speed which be ideal.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tube would be held in the middle. It is 1035mm long and has a 42mm diameter and weighs 1kg.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What are the calculations I would need to be able to work this out please?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Regards&lt;/p&gt;&#xA;" OwnerUserId="26566" LastActivityDate="2020-04-15T04:32:10.360" Title="Torque Required for spinning tube" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35169" PostTypeId="1" CreationDate="2020-04-14T18:24:02.220" Score="2" ViewCount="52" Body="&lt;p&gt;I have some research papers on aircraft controller design that i want to simulate, but no system data is given in the research paper itself. I am also unable to find the aerodynamic data for any commercial/fighter aircraft anywhere.&#xA;So i want to know where do the academic researchers get their data sets from? And is there some way for me to get trimmed or untrimmed aerodynamic data for my simulation?&lt;/p&gt;&#xA;" OwnerUserId="19673" LastActivityDate="2021-07-01T20:04:40.517" Title="Where can i get the aerodynamic data of aircrafts for controller design and testing?" Tags="|control-engineering|aerospace-engineering|aircraft-design|aviation|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35171" PostTypeId="1" AcceptedAnswerId="35175" CreationDate="2020-04-14T19:06:58.613" Score="2" ViewCount="43" Body="&lt;p&gt;Cabins have just been installed on an asphalt car park. The portable cabins weigh 4000 kg with 4 no. 300 mm square spreaders under each foot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The car park is constructed from:&#xA;1&quot; thick rolled asphalt&#xA;2&quot; thick bitumen macadam&#xA;6&quot; thick granular subbase&#xA;well compacted ground&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The effective pressure on the asphalt is:&#xA;4000 kg = 40 kN / (4 × 0.3²) = 111 kN/m²&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The question is whether this load is likely to cause permanent deformation of the asphalt leading to problems with ponding of water in the future.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I know this is a complicated question depending on the semi-solid nature of the asphalt and its rheological characteristics. The temperature for the duration of the portable cabins is not expected to exceed 30°C. The cabins are expected to be in place for approximately a month.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any guidance, rules of thumb or a point in the right direction would be appreciated.&lt;/p&gt;&#xA;" OwnerUserId="10270" LastActivityDate="2020-04-15T01:58:38.060" Title="Permanent uneven deformation of asphalt car park under portable cabins" Tags="|civil-engineering|highway-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35173" PostTypeId="1" AcceptedAnswerId="35176" CreationDate="2020-04-14T20:04:35.317" Score="1" ViewCount="169" Body="&lt;ul&gt;&#xA;&lt;li&gt;This question is specifically about stationary or vehicle mounted crossbows, not man-portable ones&lt;/li&gt;&#xA;&lt;li&gt;The projectile size would measure about 1.8 - 2 meters in length, so the crossbow would have corresponding proportions&lt;/li&gt;&#xA;&lt;li&gt;The materials would be wood and metal/metal compounds, no plastics&lt;/li&gt;&#xA;&lt;li&gt;The goal is to have a siege engine capable of cocking itslef, like a recoil operated firearm would&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;The easiest concept I have in my head for this would be as follows:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;There's a carriage riding on two rails that carries the projectile&lt;/li&gt;&#xA;&lt;li&gt;The carriage has two holes on the sides&lt;/li&gt;&#xA;&lt;li&gt;A wire is threaded through these holes&lt;/li&gt;&#xA;&lt;li&gt;At the front of the crossbow two springs are placed around the wire&lt;/li&gt;&#xA;&lt;li&gt;When the carriage gets released and propells the projectile it hits the springs and gets pushed back into it's resting position&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/PqYPq.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/PqYPq.jpg&quot; alt=&quot;Recoil operated crossbow&quot;&gt;&lt;/a&gt;&#xA;&lt;em&gt;(simple sketch of it)&lt;/em&gt;&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Could the cocking springs provide enough energy to push against the limbs and get the carriage back into it's resting position?&lt;/li&gt;&#xA;&lt;li&gt;Are there any pitfalls with a system like this?&lt;/li&gt;&#xA;&lt;li&gt;Is there a more efficient way to build a recoil operated crossbow?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="26573" LastActivityDate="2020-04-14T20:35:06.763" Title="Are recoil operated crossbows/ballistae feasible?" Tags="|mechanical-engineering|springs|coil-spring|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35174" PostTypeId="1" CreationDate="2020-04-14T20:32:15.843" Score="-1" ViewCount="212" Body="&lt;p&gt;For a copper rod heated by a heat gun, why might the temperature of the rod continue to increase for a short time after the heat gun has been removed as the heat source? With reference to transient heat transfer &lt;/p&gt;&#xA;" OwnerUserId="26356" LastEditorUserId="26356" LastEditDate="2020-04-14T20:41:43.013" LastActivityDate="2020-04-15T01:45:04.703" Title="Heat transfer for a copper rod" Tags="|heat-transfer|temperature|thermal-conduction|convection|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35175" PostTypeId="2" ParentId="35171" CreationDate="2020-04-14T20:33:35.117" Score="4" Body="&lt;p&gt;I wouldn't expect this to be a big problem.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Would you be worried about leaving a car parked in one place on this pavement for a month? Typical car tire pressures are at least 2 bar or 200 kN/m^2, about twice as high as your cabin.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Note, the weight of the car is almost entirely supported by the air pressure in the tires, so the pressure on the pavement equals the tire pressure - and the contact area of a tire would be smaller than your 300mm square spreader, so it is more likely to make a dent.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-04-14T20:33:35.117" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35176" PostTypeId="2" ParentId="35173" CreationDate="2020-04-14T20:35:06.763" Score="2" Body="&lt;p&gt;To make my comment into an actual answer, this design is not feasible because the energy balance doesn't add up. All the energy that you have available is stored in the limbs being pulled back. Now you need to do three things with it:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Accelerate the projectile&lt;/li&gt;&#xA;&lt;li&gt;Reload the crossbow&lt;/li&gt;&#xA;&lt;li&gt;Overcome friction in the system&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;The problem is, that the energy required for reloading is exactly the same as you started out with, but you still want to accelerate your projectile, so both together doesn't work. Not to mention friction losses of the sliding action and in the springs, so even without projectile it would probably not work.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The reason automatic firearms can do this, is that they tap the high pressure gas in the barrel, basically using some of the chemical energy in the gunpowder. &lt;/p&gt;&#xA;" OwnerUserId="14629" LastActivityDate="2020-04-14T20:35:06.763" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="35179" PostTypeId="2" ParentId="35174" CreationDate="2020-04-15T01:45:04.703" Score="2" Body="&lt;p&gt;If your temperature sensor is far away from the heat source, the detected temperature may increase even after the heat gun is turned off due to a preexisting temperature gradient evening itself out.&lt;/p&gt;&#xA;" OwnerUserId="14040" LastActivityDate="2020-04-15T01:45:04.703" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35180" PostTypeId="2" ParentId="35171" CreationDate="2020-04-15T01:58:38.060" Score="1" Body="&lt;p&gt;It depends on the tar and bitumen creeping properties.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Some asphalt perform well under transient loads but slowly creep under steady load. Hence the concrete is material of choice for runways.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Although the max 30 sounds like a mild temperature, some asphalt which has been exposed previously to high temps tend to let the aggregate sink and tar float. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Bottom line,  in the absence of lab tests, one has to try to check for any signs of pavement deterioration most visible around the posts penetration or intersection of the curbside and street asphalt and the potential depression under street lane painting. &lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-15T01:58:38.060" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35181" PostTypeId="1" AcceptedAnswerId="35889" CreationDate="2020-04-15T02:47:36.213" Score="2" ViewCount="58" Body="&lt;p&gt;I am trying to fabricate a very small elliptical nozzle; if possible, 10um x 20um. I am almost certain that this cannot be achieved with regular machining. Ideally, the material would be corrosion resistant (e.g. stainless, brass) although that is not absolutely necessary. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The depth of the hole would depend on the material. I need to have small deflection with a pressure differential of 80kPa. Therefore, a stronger material like SS would maybe have 1-2mm of depth. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The tolerance is less important than the smoothness of the edges. A hole that is a couple of microns off spec is better than one that has edges with roughness of 1-2um. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Could it be done with laser cutting? Are there any other methods that could achieve such feature size? Thanks.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Edit: added material and tolerance requirements.&lt;/p&gt;&#xA;" OwnerUserId="26579" LastEditorUserId="26579" LastEditDate="2020-04-15T13:42:59.847" LastActivityDate="2020-05-21T21:29:58.853" Title="Smallest possible elliptical hole" Tags="|machining|lasers|cutting|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35183" PostTypeId="2" ParentId="35167" CreationDate="2020-04-15T04:18:57.863" Score="0" Body="&lt;p&gt;The rotation at 60rpm creates air drag. It is kind of turbulent and not linear, due to rotation, but we just consider the simple version.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Fd = 1/2C_d  ρ v^2 A \quad \text{for a bar moving into wind perpendicular to long axis}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;because of rotation the speed hence the drag changes from 0 to &lt;span class=&quot;math-container&quot;&gt;$(1.035/2 *2 \pi)/s$&lt;/span&gt; and an average of Va= 1.035pi/s.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Assuming a &lt;span class=&quot;math-container&quot;&gt;$Cd=0.6 and \rho= 1.2kg/m^3 $&lt;/span&gt; and assuming average drag = 1/2 max Fd&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$F_d= 1/2*0.6*1.2*1.01*0.042 \pi^2=0.6/4*1.2*0.042*1.01 \pi^2 *0.50015=0.0376kg \quad \text{at 60RPM}$&lt;/span&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This force works at 2/3*L/2, So torque is,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  \tau= 2/3* 0.5015*0.0375= 0.0126kgm=0.123 Nm$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This torque should be enough to position the bar and rotate it at a maximum of 60 RPM. you can add more if you need a more agile system.&#xA;check my numbers, please.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-04-15T04:32:10.360" LastActivityDate="2020-04-15T04:32:10.360" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35184" PostTypeId="1" CreationDate="2020-04-15T05:01:05.937" Score="2" ViewCount="685" Body="&lt;p&gt;I am working on a pipeline design scenario&lt;a href=&quot;https://i.stack.imgur.com/TTQGG.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TTQGG.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; and I am trying to determine how many pumps I need. In addition to this diagram, I'm given a flow rate range and an acceptable velocity to also determine the diameter of the pipeline, which I have (using a Turbulent Flow III approach). The part that I am stuck on is determining how many pumps I need. Any help will be very appreciated&lt;/p&gt;&#xA;" OwnerUserId="26584" LastActivityDate="2020-04-15T16:20:34.567" Title="How do you determine how many pumps you need in a pipeline?" Tags="|fluid-mechanics|chemical-engineering|pipelines|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35185" PostTypeId="1" CreationDate="2020-04-15T05:24:33.700" Score="1" ViewCount="5105" Body="&lt;p&gt;In many textbooks it's given that gain margin (in db) and phase margin of the system needs to be positive for the system to be stable. I could not find the reason behind this. Could someone explain this?&lt;/p&gt;&#xA;" OwnerUserId="26586" LastEditorUserId="9" LastEditDate="2020-04-15T20:02:42.260" LastActivityDate="2023-07-10T00:08:26.633" Title="Why phase margin and gain margin should be positive?" Tags="|control-theory|stability|frequency-response|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35186" PostTypeId="2" ParentId="35184" CreationDate="2020-04-15T05:51:42.567" Score="2" Body="&lt;p&gt;Find the total pressure drop or losses and that tells you how many pumps you need. Consider where the pumps should be located...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-15T05:51:42.567" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35187" PostTypeId="1" AcceptedAnswerId="35189" CreationDate="2020-04-15T12:25:48.707" Score="1" ViewCount="120" Body="&lt;p&gt;I'm working on 3d printing a planetary gear for LEGO. I've got a constrained set of planetary + sun gears and was looking into designing a custom ring gear.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've read a little bit into the design of gears but am currently confused. This is what I've found.&lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Sun gear: 20 teeth, OD=21.3mm&#xA;Planet gear: 12 teeth, OD=13.38mm&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;p&gt;In both cases, module (m) = 1 (for the gears to mesh the module has to be the same?)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Pressure angle = 20 degrees.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I was just wondering how I would go about designing such a ring gear in CAD, i.e. figuring out the number of teeth, tooth size, height etc. Thanks in Advance.&lt;/p&gt;&#xA;" OwnerUserId="26591" LastActivityDate="2020-04-15T15:50:15.427" Title="Calculating and Designing a Ring Gear for LEGO" Tags="|mechanical-engineering|gears|cad|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35188" PostTypeId="2" ParentId="35181" CreationDate="2020-04-15T13:01:24.547" Score="3" Body="&lt;p&gt;It's unlikely you'll be able to accomplish your goal. Lasers and high pressure water jets are not the tools for the smallest possible features.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Electrical Discharge Machining (EDM) is capable of fairly impressive performance, but even then, your objective is smaller than the &quot;world's record:&quot;&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Engineers and scientists at Cardiff University are appearing in this&#xA;  year's edition of the Guinness World Records book for the smallest&#xA;  hole ever drilled -- at least with human-made tools. By using a&#xA;  process named 'electro-discharge machining' (EDM), they've drilled&#xA;  holes just 22 microns wide.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;The &lt;a href=&quot;https://www.zdnet.com/article/the-worlds-smallest-drilled-hole/&quot; rel=&quot;nofollow noreferrer&quot;&gt;above information&lt;/a&gt; is from 2006 and may be out of date.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another option that has to be rejected would be &lt;a href=&quot;https://www.uweinc.com/photo-etching.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;photo chemical etching&lt;/a&gt;. From the linked site:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;Chemical etching, or photo chemical machining, offers many advantages&#xA;  over other milling techniques may not be as accurate or economical.&#xA;  It:&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Doesn’t introduce any sources of mechanical stress: Each part maintains the mechanical properties of the customer’s selected metal (or metal alloy).&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Dissolves the unwanted metal via chemical reaction: burring, shearing deformation, and ablative deformation typically associated&#xA;  with other machining processes are avoidable.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Improves precision: Chemical etching can be used with materials as thin as .0003 inches. Holes can be as small as .004 inches, while&#xA;  edge dimensions can be within ±10% of a part’s thickness&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;I think one aspect of the above advantages to approach carefully is the plus or minus ten percent. A piece of stainless steel one to two millimeters thick could have a hole with 100 micron variation over its depth. A hole &quot;as small as 0.004 inches&quot; is 100 microns, far from your desired figure.&lt;/p&gt;&#xA;" OwnerUserId="13297" LastEditorUserId="13297" LastEditDate="2020-04-15T15:32:58.853" LastActivityDate="2020-04-15T15:32:58.853" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35189" PostTypeId="2" ParentId="35187" CreationDate="2020-04-15T13:56:05.093" Score="1" Body="&lt;p&gt;The module (or pitch in American units) and pressure angle needs to be constant across the gear mesh for all the gears to mesh properly. 20 degrees is the standard pressure angle for most applications.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Your ring gear number of teeth = # teeth sun gear + twice the # teeth of planet gear. Rest of the geometry is pretty standard derivation from gear formulas. This is a useful link for that: &lt;a href=&quot;https://khkgears.net/new/gear_knowledge/gear_technical_reference/calculation_gear_dimensions.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://khkgears.net/new/gear_knowledge/gear_technical_reference/calculation_gear_dimensions.html&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As for the CAD, there are plenty of online tutorials (YouTube) based on whatever package you're using. It can get a little hairy producing a true involute curve (profile of gear tooth) but there are ways to do it.&lt;/p&gt;&#xA;" OwnerUserId="23871" LastActivityDate="2020-04-15T13:56:05.093" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35190" PostTypeId="2" ParentId="35184" CreationDate="2020-04-15T13:59:37.430" Score="1" Body="&lt;p&gt;Find the flowrate and the variation in flow, check what kind of redundancy you will need. From this you can decide if you want one or more pumps in parallel per pumping station.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then, as per solar mikes answer, find the total head loss and decide if you need an additional pumping station (or several) along the way.&lt;/p&gt;&#xA;" OwnerUserId="61" LastActivityDate="2020-04-15T13:59:37.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35191" PostTypeId="1" AcceptedAnswerId="35194" CreationDate="2020-04-15T14:33:09.817" Score="-1" ViewCount="1456" Body="&lt;p&gt;This is in continuation/related to the following question asked in the below link&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://engineering.stackexchange.com/questions/23741/eulerian-and-lagrangian-descriptions-of-velocity&quot;&gt;Eulerian and Lagrangian descriptions of velocity&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;According to my understanding Lagrangian approach is an approach where you track the individual object of which you want to take measurement of for eg in a fluid flow you track a particular fluid particle and measure the physical quantity of that particular fluid particle.(leading to pathlines)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But in the Eulerian Approach, you basically try to measure the physical properties around a particular area/surface.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In application to continuum mechanics in structure, it is mentioned that the initial/reference position of a body is called as the Lagrangian Configuration and the current position(Mapped from Initial Configuration) is called as Eulerian Approach. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q1) I am not able to understand why is that so? Is it because of the fact the in Lagrangian Configuration/Initial Condition the body is supposed to be rigid and it becomes easier to understand rigid body motion and when there is no rigid body motion ie when deformation starts to occur we shift to Eulerian Approach to study deformation?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Q2) If the answer to the above question is yes, why is the Eulerian approach used to study deformation and if not why do we use two approaches??&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 1:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for your reply. I have been using this particular reference. &lt;a href=&quot;http://web.mit.edu/abeyaratne/Volumes/RCA_Vol_II.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://web.mit.edu/abeyaratne/Volumes/RCA_Vol_II.pdf&lt;/a&gt; in which Eulerian and Lagrangian configurations are explained.(Page 6,  1.3).&#xA;Here it describes Eulerian configuration as  &lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;the representation (1.6) which deals with the positions of the particles in the deformed configuration, (the configuration in which the physical quantity is being characterized,) is called the Eulerian or spatial description.&quot; &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;and for lagrangian&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;If a reference configuration has been introduced we can label a particle by its position&#xA;  x = χref(p) in that configuration, and this, in turn, allows us to describe physical quantities in Lagrangian form.&quot;&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;This is why I am getting confused with Eulerian and Lagrangian.&#xA;Is there any specific reason so as to why its named like that?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;EDIT 2:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding on more reference for the same with the same query(Why are these configurations called Eulerian or Lagrangian ? )&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/52Fdl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/52Fdl.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26592" LastEditorUserId="26592" LastEditDate="2020-04-16T17:09:50.730" LastActivityDate="2020-04-16T17:09:50.730" Title="Eulerian and Lagrangian --configuration-- (continuum mechanics)" Tags="|mechanical-engineering|mechanical|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35192" PostTypeId="2" ParentId="35184" CreationDate="2020-04-15T15:04:46.307" Score="1" Body="&lt;p&gt;Depends on mostly on length, elevation changes, viscosity. But when the input says &quot;tailings&quot; , I think you have bigger problems like settling of solids in the low spots. First you need to know what velocity you need to prevent settling . Also add cleaning pig entry and exits. Actually the first question is &quot;do you know what you are pumping?&quot; . And the first question,&quot;Why do you think more pumping is needed?&quot;.  Frequently the best engineering answer to to use one larger pump ( minimum infrastructure). What type of pump is planned ? Maybe the type of pump should be optimized. &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2020-04-15T16:20:34.567" LastActivityDate="2020-04-15T16:20:34.567" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35193" PostTypeId="2" ParentId="35187" CreationDate="2020-04-15T15:50:15.427" Score="1" Body="&lt;p&gt;As jko suggests, one would search for tutorials based on your selection of CAD software. Fusion 360 has a gear generator as to others. I purchased a &lt;a href=&quot;https://woodgears.ca/gear/index.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;gear generator program&lt;/a&gt; to create SVG files for laser cutting. It also allows one to adjust the parameters to meet requirements.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I used your figures for outside diameter in the program, which resulted in a variety of parameters:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;12 tooth gear:&#xA;11.886 pitch diameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;20 tooth gear:&#xA;19.810 pitch diameter&lt;/p&gt;&#xA;&#xA;&lt;p&gt;shaft spacing 15.848&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The 20 degree pressure angle is taken as a constant.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/S4ATT.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/S4ATT.png&quot; alt=&quot;sun and planet gear&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-04-15T15:50:15.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35194" PostTypeId="2" ParentId="35191" CreationDate="2020-04-15T16:03:30.027" Score="2" Body="&lt;p&gt;I think it is quite not clear for you what is an Eulerian and Lagrangian description.&lt;br&gt;&#xA;As you briefly said at the beginning of your post, in the Eulerian approach, you consider a small volume that is fixed is time and space, and study what happens to this volume: what enters or goes out (convection), if there are some forces applied to it, if it is heated or not etc.&lt;br&gt;&#xA;In the Lagrangian approach, you consider a material particle an follows its evolution. That is why for example there is no convection term in a Lagrangian description of the motion, since there is no material motion to consider within a material particle.&lt;br&gt;&#xA;Lagrangian and Eulerian approaches are thus two different way to describe the evolution of physical quantities, and the equations that result are different.&lt;br&gt;&#xA;In solids mechanics, the Lagrangian description of the motion is very widely used,&#xA;while the Eulerian approach is usually preferred in fluid mechanics.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;br&gt;&#xA;I went through the chapter you mention in the reference you sent. It is a very good book, but quite involved if you start studying continuum mechanics. Section 1.3 is very clear about what is the difference between the Eulerian and Lagrangian &lt;strong&gt;approaches&lt;/strong&gt; (or description, as written in the book): They are two different &quot;mindsets&quot; to describe the evolutions of physical quantities.&lt;br&gt;&#xA;Either you decide that you decide that you sit in a specific location, and look what happens around you. You observe that particles are moving, the temperature is changing etc. You observe the evolution of the physical quantities from that specific location point of view. This is an &lt;strong&gt;Eulerian approach&lt;/strong&gt;.&lt;br&gt;&#xA;Otherwise you decide to follow a specific particle and move with it. You position is changing with time, and you observe the evolution of the physical quantities from that specific particle point of view. This is a &lt;strong&gt;Lagrangian approach&lt;/strong&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I do not know the reference you are using, but having quite extensively studied continuum mechanics, these are the most common appellations for the various configurations:  &lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;Initial configuration = reference configuration = undeformed configuration  &lt;/li&gt;&#xA;&lt;li&gt;Current configuration = deformed configuration   &lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;EDIT:&lt;br&gt;&#xA;I was quite surprise you wrote about &quot;Eulerian configuration&quot; and &quot;Lagrangian configuration&quot; since I never encountered this terminology before. Actually, they are not in the reference you provided neither. The terms that I introduced (reference/current/(un)deformed) are the one that are used as well.&lt;br&gt;&#xA;You have however Eulerian and Lagrangian &lt;strong&gt;coordinates&lt;/strong&gt;, as introduced in the snippet you added in your post. The Lagrangian coordinates &lt;span class=&quot;math-container&quot;&gt;$\mathbf{X}$&lt;/span&gt; are measured in the undeformed configuration, and the Eulerian coordinates &lt;span class=&quot;math-container&quot;&gt;$\mathbf{x}$&lt;/span&gt; are measured in the deformed configuration. You thus have the displacement &lt;span class=&quot;math-container&quot;&gt;$\mathbf{u}$&lt;/span&gt; of a material point, which is the difference between its current position and its original position, that is given by &lt;span class=&quot;math-container&quot;&gt;$\mathbf{u} = \mathbf{x}-\mathbf{X} $&lt;/span&gt;.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Be careful : in your post you say that the current configuration is the Eulerian approach. &quot;Approach&quot; here means a way to describe the motion. &quot;Configuration&quot; relates to the state of the body at a certain time.&lt;br&gt;&#xA;When you decide to write your equations that describe the motion of the domain you want to study, you chose one or the other approach. You cannot switch from one to the other. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think that the questions you asked result from a misunderstanding of these notions, which, I agree, are quite complex.&lt;/p&gt;&#xA;" OwnerUserId="26427" LastEditorUserId="26427" LastEditDate="2020-04-16T10:20:55.667" LastActivityDate="2020-04-16T10:20:55.667" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="35197" PostTypeId="1" CreationDate="2020-04-15T18:11:21.850" Score="1" ViewCount="102" Body="&lt;p&gt;The 3M Aura 9332+ mask is labeled FFP3 which means that it filters at least 99% of airborne particles.&#xA;This &lt;a href=&quot;http://news.mit.edu/2020/gamma-radiation-found-ineffective-in-sterilizing-n95-masks-0410&quot; rel=&quot;nofollow noreferrer&quot;&gt;article&lt;/a&gt; mentions that N95 masks (which filtrate 95% of airborne particles) filtrate the airborne particles with some materials (wool, glass particles, and plastics...)with 1-2 percent copper and/or zinc.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I looked at the  3M Aura 9332+ &lt;a href=&quot;http://multimedia.3m.com/mws/media/1194322O/aura-disposable-respirator-tds.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;datasheet&lt;/a&gt; and I couldn't find any information about their (possible) content in copper/zinc. They only mention that the filter is made of Polypropylene (cf. p2). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my question is: if the 3M Aura 9332+ doesn't contain copper/zinc, how does it filtrate the airborne particles? Do they work like &lt;a href=&quot;https://en.wikipedia.org/wiki/HEPA#Mechanism&quot; rel=&quot;nofollow noreferrer&quot;&gt;HEPA&lt;/a&gt; with a combination of diffusion, interception, impaction, and electrostatic attraction? If yes, how is it possible because the masks are much thinner than HEPA filters?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Side question (which depend on the answer ― special pandemic): &#xA;Can this mechanism be destroyed by humidity (breathing) and/or sweating (salty water)? &lt;/p&gt;&#xA;" OwnerUserId="2708" LastActivityDate="2020-04-15T18:11:21.850" Title="How the 3M Aura 9332+ masks filtrate airborne particles/virus?" Tags="|fluid-filtration|diffusion|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35198" PostTypeId="2" ParentId="35063" CreationDate="2020-04-15T19:04:18.847" Score="0" Body="&lt;p&gt;In your problem, you define the supply rate (I will name it &lt;span class=&quot;math-container&quot;&gt;$s$&lt;/span&gt;) &lt;span class=&quot;math-container&quot;&gt;$s(y,u)=u^{\top}y$&lt;/span&gt;. And it is (strict) passive if &lt;span class=&quot;math-container&quot;&gt;$\dot{V} \leq s(y,u)$&lt;/span&gt;. I also consider that you are working taking the case where this consideration is valid &lt;span class=&quot;math-container&quot;&gt;$u^{\top}y \equiv uy$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Saying that, I believe you can re-arrange your equation as follows:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align*}&#xA;\dot{V}(x) &amp;amp;= uy+\frac{1}{k}y^2-xy \\&#xA;            &amp;amp;=y(u+\frac{1}{k}y) -xy&#xA;\end{align*}&lt;/span&gt;&#xA;then define as the new input &lt;span class=&quot;math-container&quot;&gt;$\nu := u+\frac{1}{k}y$&lt;/span&gt;, in that sense you will have the following equation:&#xA;&lt;span class=&quot;math-container&quot;&gt;\begin{align*}&#xA;\dot{V}(x) &amp;amp;= y\nu-xy\leq \nu y&#xA;\end{align*}&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The system is (strict) passive if &lt;span class=&quot;math-container&quot;&gt;$xh(x)&amp;gt; 0\quad\forall x\in\mathbb{R}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="25296" LastEditorUserId="25296" LastEditDate="2020-04-16T06:59:39.603" LastActivityDate="2020-04-16T06:59:39.603" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35199" PostTypeId="2" ParentId="33781" CreationDate="2020-04-15T20:05:51.170" Score="0" Body="&lt;p&gt;You need to think about how the load is going to affect your system. Then you need to add it to your model, as you said you have already the mathematical model of it. And probably you will notice that the load is going to act as a disturbance of the system. Thus, if you consider the Torque (not the load) as a state you will easily find the state-space equation. An example of a state-space mathematical model of the motor is stated here&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://ctms.engin.umich.edu/CTMS/index.php?example=MotorSpeed&amp;amp;section=SystemModeling&quot; rel=&quot;nofollow noreferrer&quot;&gt;http://ctms.engin.umich.edu/CTMS/index.php?example=MotorSpeed&amp;amp;section=SystemModeling&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, I would recommend you to read &lt;a href=&quot;https://robotics.stackexchange.com/a/10054&quot;&gt;https://robotics.stackexchange.com/a/10054&lt;/a&gt;&#xA;since I think that you are a little confused about how the motor works. Good luck!&lt;/p&gt;&#xA;" OwnerUserId="25296" LastActivityDate="2020-04-15T20:05:51.170" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35201" PostTypeId="2" ParentId="35138" CreationDate="2020-04-15T22:36:28.933" Score="1" Body="&lt;p&gt;The concept of a spider navigating a gravity driven sewer system is not realistic.  The water flows downhill and the spider that could swim against the water current would be all legs (and our overloard).  Water surface tension would mean any spider would have difficulty getting out of the water.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All plumbing systems have a vent stack through the roof so they could get in to the plumbing system, but they cannot get past water traps in the plumbing system meant to keep sewer gas out of homes.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dryer vents or bathroom exhaust fans would be more realistic entry points.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Odds are most spiders are coming in on a person's clothes or something brought into the home.  Laundry drying on clothes lines is a common entry point (for me - catch and release).  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;They end up in tubs because they cannot get out of them due to slippery vertical slopes.  A healthy spider would not fall into a tub, but without a source of food...&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2020-04-15T22:36:28.933" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35202" PostTypeId="1" CreationDate="2020-04-16T07:09:28.993" Score="0" ViewCount="87" Body="&lt;p&gt;In structural analysis, does inelastic analyis mean EXACTLY the same thing as non linear analysis?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If not what is the difference? &lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2020-05-15T02:59:01.440" Title="In structural analysis, does inelastic analyis mean the same thing as non linear analysis?" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35203" PostTypeId="2" ParentId="35202" CreationDate="2020-04-16T09:41:17.570" Score="3" Body="&lt;p&gt;No.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a linear analysis, the strain is always proportional to the stress.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In an elastic analysis, the strain follows the same function of stress for loading and unloading. More importantly, strains are reversible meaning that upon unloading, the material returns to its original state without residual stress/strains. For example, the stress-strain curve of a hyperelastic material in the elastic range is nonlinear. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Therefore, a linear analysis will also be an elastic analysis but the opposite is not true. An elastic analysis may very well be nonlinear.  &lt;/p&gt;&#xA;" OwnerUserId="15064" LastEditorUserId="19642" LastEditDate="2020-05-15T02:59:01.440" LastActivityDate="2020-05-15T02:59:01.440" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35205" PostTypeId="2" ParentId="35138" CreationDate="2020-04-16T14:55:35.537" Score="0" Body="&lt;p&gt;Unlikely but possible, they easily enter under doors and other places. However , I had a scorpion enter once through the sewer. A state Parks and Wildlife guy told me the drop down the stacks ( sewer vents) on the roof. No swimming needed, air all the way to the bathtub and sink vents.I put screens over all my stacks.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-04-16T14:55:35.537" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35206" PostTypeId="1" CreationDate="2020-04-16T16:21:12.150" Score="1" ViewCount="51" Body="&lt;p&gt;For simpler buildings we can use response spectrum analysis. How is this used? The maximum acceleration we get from this , is simply multiplied by the mass of the building to get F=ma , total earthquake force on the building? Simple as that?&lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2020-04-16T18:25:23.043" Title="How is response spectrum result used for earthquake analysis" Tags="|structural-engineering|civil-engineering|structural-analysis|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35207" PostTypeId="2" ParentId="35206" CreationDate="2020-04-16T17:43:35.457" Score="1" Body="&lt;p&gt;Yes, Right. After you verify that your building structure geometry and utility make it qualify for using design spectra charts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;it gives you the acceleration Sa and speed of ground motion, Sv and the amount of motion, Sd which are all related by factors of &lt;span class=&quot;math-container&quot;&gt;$ \omega$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Then you can apply F=ma.&#xA;Ther are inherent errors in this method that a designer must be aware of and use good judgment. Search for -  (Nigam and Jennings,&#xA;1969), to see the article on the errors.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;here is the chart.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/wqmuI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/wqmuI.jpg&quot; alt=&quot;design specra&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-16T17:43:35.457" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="35211" PostTypeId="1" CreationDate="2020-04-16T18:56:19.780" Score="3" ViewCount="75" Body="&lt;p&gt;See the image below for the detail of 4 no. cable ducts that run from a trench in a raft foundation (on the left) through an approximately 1 m square penetration in the raft foundation, through soil (if it had been backfilled) and through an approx 1 m square penetration in the retaining wall for the basement of an industrial structure. At present there is a continuous flow of water through the base of the penetration in the retaining wall.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My questions are:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Given that the structure is approximately 50 years old what are the likely original construction details? My theory is that the cable ducts were fed through, the penetration at either end was sealed with mastic but I am unclear if it would have been possible to backfill around the cable ducts or whether the cable ducts are likely in a void filled with ground water?&lt;/li&gt;&#xA;&lt;li&gt;What approach could be used to prevent water ingress in through the retaining wall penetration. Is it possible that the existing mastic sealant has degraded with age? Would raking it out and reinstating with a modern product be effective? Is there a system that can be applied on the internal face of the retaining wall that would be effective? Is the best approach leak management to a sump to be pumped out?&lt;/li&gt;&#xA;&lt;li&gt;If you were to construct this detail today what approach would you take? My instinct would be to use a &lt;a href=&quot;https://www.flexseal.co.uk/products/pipe-seals/puddle-flanges/&quot; rel=&quot;nofollow noreferrer&quot;&gt;puddle flange&lt;/a&gt; for each cable duct  that was cast into the concrete rather than leaving 1 m square voids to be sealed with mastic. I believe this would have prevented water ingress? What is current best practice for construction these cable duct penetration details?&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;I apologise for the quantity of questions. I could break it into a series of independent questions if required but I felt it better to keep everything together. &lt;a href=&quot;https://i.stack.imgur.com/L2OIv.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/L2OIv.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10270" LastActivityDate="2020-04-16T18:56:19.780" Title="water leak through subterranean cable penetrations in concrete" Tags="|civil-engineering|concrete|seals|waterproofing|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35212" PostTypeId="1" CreationDate="2020-04-16T19:11:42.787" Score="1" ViewCount="41" Body="&lt;p&gt;I am currently working on a project that involves CFD simulation on ANSYS and i need some help as it has got me stuck. I am designing a faucet atomizer and it's functioning is similar to a nozzle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have designed it on CATIA and am trying to analyse it on ANSYS Fluent.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am using the following conditions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Pressure based solver, absolute velocity, steady flow&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Flow characteristics: Laminar at inlet, turbulent at outlet&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Boundary conditions: inlet velocity: 0.041 m/s, inlet pressure: 300000 Pa,outlet pressure: 0 Pa&lt;/p&gt;&#xA;&#xA;&lt;p&gt;-Fluid: water(liquid)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the following doubts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1.)I read that turbulent models give accurate results even if flow is laminar but not vice versa so i chose turbulent but which turbulent model should i apply?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2.)My geometry has swirl, complex curves and rotational flow and i read that k-epsilon, k-omega models do not handle such flows and curves well so which model should i adopt(my geometry is very small in dimensions so swirl may not matter?).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3.)How should i do the meshing of my model? I did it multiple times according to my understanding but due to some reason the solution didn't converge and diverged a couple of times as well. My system also does not have a lot of computational power hence less elements and nodes in mesh.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am attaching some images of my geometry with sectional views, extracted volume and also the meshing done by me. Kindly take a look.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be able to share the geometry file with you if you require.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I will be very grateful if you can help me out.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ON7Pl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ON7Pl.png&quot; alt=&quot;Part&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/nrLTb.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/nrLTb.png&quot; alt=&quot;part&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/H0ZpR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/H0ZpR.png&quot; alt=&quot;Section view&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1FK5Y.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1FK5Y.png&quot; alt=&quot;Extracted volume&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/1dl0J.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1dl0J.png&quot; alt=&quot;Mesh&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/Gwios.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Gwios.png&quot; alt=&quot;mesh&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/4zPCN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/4zPCN.png&quot; alt=&quot;Mesh&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26611" LastActivityDate="2020-04-16T19:11:42.787" Title="Ansys Fluent - Turbulent model and mesh" Tags="|fluid-mechanics|finite-element-method|simulation|ansys|fluent|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35214" PostTypeId="1" CreationDate="2020-04-16T23:40:35.640" Score="1" ViewCount="3756" Body="&lt;p&gt;Looking through texts such as Shigley's &lt;em&gt;Mechanical Engineering Design&lt;/em&gt;, you come across tables of the stress concentration factor (defined as &lt;span class=&quot;math-container&quot;&gt;$K=\sigma_{max}/\sigma_{avg}$&lt;/span&gt;) such as the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/b0GVl.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/b0GVl.png&quot; alt=&quot;Image from Beer, Johnson, and DeWolf&amp;#39;s *Mechanics of Materials*&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This plot is sourced from Pilkey's &lt;em&gt;Peterson's Stress Concentration Factors&lt;/em&gt;, which, unfortunately, I do not have access to (and even less so now, since interlibrary loans are on hold), so I do not know what underlies this plot.  However, the line seems to be related to Kirsch's elastic stress distribution since the value of the stress concentration factor approaches 3 when the ratio r/d approaches zero. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;That exact item, though, is what gets me.  In the limit of radius going to zero, the bar no longer has a hole in it, and according to Saint-Venant's principle, the stress a a point in the middle of a long bar should be equal to the average stress on that cross section.  Hence:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Why does the stress concentration factor go to 3 instead of 1 as &lt;em&gt;r&lt;/em&gt; approaches 0?&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="26612" LastActivityDate="2020-04-17T03:01:11.723" Title="Stress Concentration Factor" Tags="|mechanical-engineering|stresses|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35215" PostTypeId="2" ParentId="35214" CreationDate="2020-04-17T02:50:06.877" Score="1" Body="&lt;p&gt;Let's imagine a strap with a width A and let's assume the hole diameter is =A/2.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At initial configuration, the relative stiffness of the two parts of the strap are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$  (A-A/2) = (A/2)=1/2A \ \text{for two blades}=1/4 A\ \text{per blade}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As the hole diameter shrinks the relative stiffness of the two sides increases, and the strain contour line becomes flatter. meaning at the interface of the side of the hole and the two blades there is a sharper stress concentration. one could say because the strain has a lot of material as back up to maintain it. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Ultimately at a hole size hypothetically infinitely small but still existing, the width of each blade is &lt;span class=&quot;math-container&quot;&gt;$= 1/2 A$&lt;/span&gt; and they are very stiff.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;It may help to think of the way glazing people cut glass. they score it by a thin shallow line, stress concentration is so high there that by a small shake the glass breaks on the scoreline.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-17T02:50:06.877" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35216" PostTypeId="2" ParentId="35214" CreationDate="2020-04-17T03:01:11.723" Score="3" Body="&lt;p&gt;There is a difference between a very small hole (even of radius zero, theoretically) and a solid bar.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The hole causes the axial stress along the line of its center to be zero, because it is normal to the free surface of the hole. That reduction in the local load-carrying capacity has to cause an increase somewhere else, and that is what causes the stress concentration factor to be greater than 1.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The limiting value of 3 is for any size of circular hole in an infinitely wide plate.  For a finite width, the edges of the plate also affect the stress distribution and the stress concentration factor.&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-04-17T03:01:11.723" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35217" PostTypeId="2" ParentId="35090" CreationDate="2020-04-17T04:14:20.767" Score="0" Body="&lt;p&gt;I found my old Machine Design (Machine Design by Norton) textbook, and this is how it puts it:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;If the diameter of the base circle of a gear is increased without limit, the base circle will become a straight line. If the &quot;string&quot; wrapped around this base circle to generate the involute were still in place after the base circle's enlargement to an infinite radius, the string would be pivoted at infinity and would generate an involute that is a straight line.  &lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;" OwnerUserId="26490" LastActivityDate="2020-04-17T04:14:20.767" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35218" PostTypeId="1" CreationDate="2020-04-17T04:23:23.633" Score="1" ViewCount="507" Body="&lt;p&gt;I am currently trying to process some data from a wire drawing (cold drawing) test. I would like to know how I can calculate the external work of plastic deformation due to wire drawing, which is basically the work done by the wire drawing machine. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The data that I have are the length and diameter of the brass wire before and after each wire drawing operation (brass was drawn 3 times) as well as the time and force data for the wire drawing operations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Any form of help would be appreciated. Thanks in advance.&lt;/p&gt;&#xA;" OwnerUserId="26524" LastActivityDate="2020-04-17T05:05:45.310" Title="How to calculate the external work of plastic deformation due to wire drawing (the work done by the wire drawing machine)?" Tags="|mechanical-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35219" PostTypeId="2" ParentId="35218" CreationDate="2020-04-17T04:57:45.137" Score="1" Body="&lt;p&gt;You need to break your elongation in two parts and calculate the work done separately, while measuring the diameter repeatedly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The elastic part work is the force your machine is pulling multiplied by the elongation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The plastic part it is the sum of incremental area measurements multiplied by the yield stress Fy times the particular section elongation or if your sample does not have a pronounced flat area again what the machine gouge reads.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-04-17T05:05:45.310" LastActivityDate="2020-04-17T05:05:45.310" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35220" PostTypeId="1" CreationDate="2020-04-17T06:06:36.343" Score="1" ViewCount="203" Body="&lt;p&gt;Basically can engineers design a machine where , while we are freezing and compressing the air for energy storage, we can also purify it of pollutants?&lt;/p&gt;&#xA;" OwnerUserId="26620" LastActivityDate="2020-04-17T07:48:13.387" Title="Is it possible to create a compressed/freezed air energy storage solution that also purifies the air of pollutants?" Tags="|air|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35221" PostTypeId="1" CreationDate="2020-04-17T07:04:06.023" Score="0" ViewCount="19" Body="&lt;p&gt;Is there a software that exists that takes a macro object for example a MacBook casing and its composition (Aluminum) and allow you to zoom between the macro scale (a fully rendered MacBook) and the atomic scale (a view of the atomic structure).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've played around with VMD, PyMol, ASE, CAD. The first 3 are focused on the molecular level, but CAD is for designing macro scale objects. Can't seem to find a combination of both&lt;/p&gt;&#xA;" OwnerUserId="26622" LastEditorUserId="26622" LastEditDate="2020-04-17T16:26:30.947" LastActivityDate="2020-04-17T16:26:30.947" Title="Macro to atomic scale software" Tags="|chemical-engineering|computer-engineering|software|biomechanics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35222" PostTypeId="2" ParentId="35220" CreationDate="2020-04-17T07:06:04.677" Score="1" Body="&lt;p&gt;So two possibilities:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One, pre filter the air prior to cooling/freezing&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Two, will the impurities freeze and soldify first? If so remove solids before final freezing.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Are you freezing the air to a solid / liquid as this may make a difference.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-17T07:06:04.677" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35223" PostTypeId="2" ParentId="35220" CreationDate="2020-04-17T07:08:48.693" Score="3" Body="&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;Compressing air (for energy storage or other purposes) &lt;a href=&quot;https://en.wikipedia.org/wiki/Adiabatic_process#Adiabatic_heating_and_cooling&quot; rel=&quot;nofollow noreferrer&quot;&gt;heats&lt;/a&gt; it up, expansion cools. For energy storage the heat has to be stored, a detailed description can be found &lt;a href=&quot;https://ease-storage.eu/wp-content/uploads/2016/03/EASE_TD_ACAES.pdf&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;. Cooling air further would require additional energy input - twice, for cooling and then again for reheating upon expansion. Again, see the link for details.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Compression concentrates dust etc (1g dust per 1m³ at 1bar becomes 6g/m³ at 6bar approx). It's certainly viable to have a coarse filter to protect the compressors, then a fine filter for the high pressure air. If this is what you mean by pollutants. This is farily common in compressed air systems for pneumatics: Fine filter and dewatering are installed downstream of the compressors&lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="61" LastEditorUserId="61" LastEditDate="2020-04-17T07:48:13.387" LastActivityDate="2020-04-17T07:48:13.387" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35224" PostTypeId="1" CreationDate="2020-04-17T07:52:01.227" Score="0" ViewCount="14" Body="&lt;p&gt;I am an undergraduate student who is trying to classify motor imagery from EEG data!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no experience working with EEG or any neuroscience background, I only have a very basic knowledge of how EEG, ERD/ERS and frequency domain work from the papers I've tried to read but don't understand and the basic idea of what they are doing. I also have a reasonably basic knowledge of supervised Machine Learning.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I plan to use ERD/ERS to classify Left foot, Right hand and rest to classify motor imagery in real time!&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;What is the easiest way to do this?&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So far I have bandpassed the data between 7-30 Hz for Mu and Beta wavelengths as they are the ones related to the motor imagery as well as removing the baseline and just plotted the Power Spectral Density. I passed this data through a basic SVM; linear and gaussian, optimised parameters with poor results (60% accuracy on average)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/9z21Z.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/9z21Z.png&quot; alt=&quot;Power Spectral Density&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have no idea how to do feature extraction, where to start, how to properly classify EEG data. and how to do this in real time.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;If someone could guide me and provide the steps on how to proceed I would extremely grateful!&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I want to work in the field of Brain Computer Interfaces as I find it fascinating, this is my first step :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Here is the link to the dataset and its details &lt;a href=&quot;http://www.bbci.de/competition/iii/desc_IVc.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;BCI competition: IVc&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26625" LastActivityDate="2020-04-17T07:52:01.227" Title="Classifying motor imagery using EEG data" Tags="|biomedical-engineering|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35226" PostTypeId="1" CreationDate="2020-04-17T09:45:25.447" Score="1" ViewCount="127" Body="&lt;p&gt;Heat pump cost 2-4 times lesser of energy to heat than non-heat pump heater, then why are non-heat pump heater still very common ?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(1) At home, boiler and water radiator is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;(2) For district heating, central boiler is used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why aren't most these replace by heat pump?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For (2) the reason may be there is not enough heat source to pump from to support megawatts of power output.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Possible reasons for home: heat pump is driven by electricity and boiler cost lesser then electricity, then the cost difference is not that much.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I also heard in cold winter air conditioner won't work well as heat pump, is this true and is this a possible reason?&lt;/p&gt;&#xA;" OwnerUserId="26563" LastActivityDate="2023-04-24T23:12:09.980" Title="Why isn't heat pump used for heating for most case" Tags="|heating-systems|" AnswerCount="3" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35227" PostTypeId="1" CreationDate="2020-04-16T15:14:24.640" Score="1" ViewCount="863" Body="&lt;p&gt;It has bugged me for a while. If the earth is a constant 50-60-odd degrees several tens of feet beneath ground, and houses (I assume; ours at least does) don't drop down to lower than those temperatures often, how come a geothermal unit isn't just using supplemental electrical resistance heating (and not the ground loop heat source?) all the time?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is it because the above ground equipment is actually a heat pump (extracting more heat out of the circulated liquid than the temperature it comes out of the ground at)? &lt;/p&gt;&#xA;" OwnerUserId="21393" OwnerDisplayName="user66001" LastEditorUserId="21393" LastEditDate="2020-04-17T13:53:39.170" LastActivityDate="2020-04-17T13:53:39.170" Title="How can Geothermal heat a home using the ~55 degree heat source of underground to a more usual 72 degree inside temperature?" Tags="|heating-systems|cooling|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35228" PostTypeId="2" ParentId="35227" CreationDate="2020-04-16T18:14:46.740" Score="3" Body="&lt;p&gt;Barring a convenient &quot;hot spring&quot; location, &quot;geothermal heat&quot; is, indeed, shorthand for &quot;geothermal heat-pump heat&quot; and typically takes in water at common ground temperatures in the 50F/10C range, drops that water a few degrees to extract heat and sends it on its way, whether back through a loop, out to a pond, or down a different well than it was pumped from.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is much less difficult than the job of cold-climate air-source heat pumps, which have to work over a much wider input temperature range, and need the most heat output just when it's hardest to extract heat from very cold air.&lt;/p&gt;&#xA;" OwnerDisplayName="Ecnerwal" LastActivityDate="2020-04-16T18:14:46.740" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35229" PostTypeId="2" ParentId="35226" CreationDate="2020-04-17T10:19:21.077" Score="1" Body="&lt;p&gt;In most cases, heat pumps are used for heating, at least where I am. They take low grade heat from the atmosphere or underground and use that to heat dwellings. Some are reversible to cool buildings in the summer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Why aren't they more common? well that depends on many factors: what is there already, what the owner is prepared to pay, how much time is available for replacement if the old unit has failed etc etc&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I installed a « puit canadienne » for the incoming ventilation air and as the ground temperature is relatively constant at a 2 metre depth (approx a 3 or 4 degree C change yearly, compared to a 30 degree change for the surface) this help moderate the incoming air both summmer and winter.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="10902" LastEditDate="2020-04-17T10:31:11.193" LastActivityDate="2020-04-17T10:31:11.193" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35230" PostTypeId="2" ParentId="35226" CreationDate="2020-04-17T10:19:58.663" Score="2" Body="&lt;p&gt;The last one is the reason furnaces are still nearly universal where temps drop down to below freezing. People will run an air source heat pump usually down to an outside temp of 40 F, at which point they become very inefficient and require a defrost cycle. Below that, the furnace has the needed capacity and operates cheaply. Water or ground source heat pumps can run in colder weather, but there is about a three month time lag on ground temperature vs air temperature, so even these need a short winter to be viable on there own, else they run terribly in the spring.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastEditorUserId="12754" LastEditDate="2020-04-17T19:33:15.447" LastActivityDate="2020-04-17T19:33:15.447" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35231" PostTypeId="1" CreationDate="2020-04-17T13:21:58.960" Score="0" ViewCount="1228" Body="&lt;p&gt;What is the sign convention for deflection and slope of a beam as shown in the diagram below?&#xA;&lt;a href=&quot;https://i.stack.imgur.com/sqLb9.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/sqLb9.jpg&quot; alt=&quot;Deflected beam&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When I derive the formula using Euler-Bernoulli for deflection and slope myself, I end up with &lt;strong&gt;-&lt;/strong&gt; deflection at y, &lt;strong&gt;-&lt;/strong&gt; slope at A and &lt;strong&gt;+&lt;/strong&gt; slope at B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, using formula I found, for the same scenario:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$y=\frac{Px}{12EI}\ \left(\frac{3l^2}{4}-x^2\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$θ1=θ2=\frac{Pl^2}{16EI}\ $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Gives &lt;strong&gt;+&lt;/strong&gt; deflection at y, &lt;strong&gt;+&lt;/strong&gt; slope at A and &lt;strong&gt;+&lt;/strong&gt; slope at B.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the convention here?&lt;/p&gt;&#xA;" OwnerUserId="23791" LastActivityDate="2020-04-17T13:21:58.960" Title="Sign convention for deflection and slope of beam?" Tags="|beam|deflection|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35234" PostTypeId="1" CreationDate="2020-04-17T19:20:16.643" Score="1" ViewCount="26" Body="&lt;p&gt;I have seen some designs being made for linear generators for WEC(wave energy) applications. A buoy driving a shaft. But what if the shaft inside was a rack and pinion? If you converted the bidirectional input coming through the pinion gear into a single direction drive output using some small pulleys and clutch bearings, you could then add some gear reductions and maintain some rotational velocity with a very small travel on a large-ish displacement. Using this mechanical conversion, how would an output compare to that of the variable output of a linear generator? Isn't the size of your displacement much more easy to change around with a mechanical conversion?&lt;/p&gt;&#xA;" OwnerUserId="26595" LastEditorUserId="26595" LastEditDate="2020-04-26T02:45:47.017" LastActivityDate="2020-04-26T02:45:47.017" Title="WEC linear inputs" Tags="|mechanical-engineering|electrical-engineering|design|torque|stresses|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="35235" PostTypeId="1" CreationDate="2020-04-17T20:33:15.190" Score="1" ViewCount="853" Body="&lt;p&gt;For a simply supported beam with a UDL if the deflection is assumed to be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V=a+bx+cx^2+dx^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;What is the formula for calculating the slope?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I thought it would be:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$θ=dV/dx$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, working through examples where I calculate the values for the constants this does not appear to be the case. At the midpoint of the beam when I should be getting zero deflection, the value I get using dV/dx is completely wrong.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;How do I obtain θ?&lt;/p&gt;&#xA;" OwnerUserId="23791" LastActivityDate="2020-04-17T22:24:42.500" Title="Rayleigh-Ritz method, calculating slope of beam?" Tags="|finite-element-method|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35236" PostTypeId="2" ParentId="35235" CreationDate="2020-04-17T22:24:42.500" Score="3" Body="&lt;p&gt;The Euler-Bernoulli beam equation is as follows:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$q = \dfrac{\partial^2}{\partial x^2}\left(EI\dfrac{\partial^2 \delta}{\partial x^2}\right)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; is the distributed load along the beam, &lt;span class=&quot;math-container&quot;&gt;$\delta$&lt;/span&gt; is the deflection, and &lt;span class=&quot;math-container&quot;&gt;$EI$&lt;/span&gt; is the beam's stiffness (assumed constant). So, the deflection is the fourth integral of the applied load. From this we can derive that, for the case of a uniform load along the entire beam of uniform stiffness:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\begin{align}&#xA;Q &amp;amp;= \int q\text{d}x \\&#xA;&amp;amp;= qx + Q_0 \\&#xA;M &amp;amp;= \int Q\text{d}x \\&#xA;&amp;amp;= \dfrac{1}{2}qx^2 + Q_0x + M_0 \\&#xA;\theta &amp;amp;= \dfrac{1}{EI}\int M\text{d}x \\&#xA;&amp;amp;= \dfrac{1}{6}qx^3 + \dfrac{1}{2}Q_0x^2 + M_0x + \theta_0 \\&#xA;\delta &amp;amp;= \int \theta\text{d}x \\&#xA;&amp;amp;= \dfrac{1}{24}qx^4 + \dfrac{1}{6}Q_0x^3 + \dfrac{1}{2}M_0x^2 + \theta_0x + \delta_0 \\&#xA;\end{align}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$q$&lt;/span&gt; is the constant uniform load and &lt;span class=&quot;math-container&quot;&gt;$Q_0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$M_0$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$\theta_0$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\delta_0$&lt;/span&gt; are the shear force, bending moment, rotation and deflection at the beam's origin, respectively.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, the problem you're having is that your initial equation is incorrect: a beam under a UDL's deflection is a quartic function. And for small deflections and rotations, you are correct: we can approximate the rotations (in radians) as equal to the tangent of the deflection, i.e. its derivative.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-04-17T22:24:42.500" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35240" PostTypeId="1" CreationDate="2020-04-18T02:09:43.577" Score="1" ViewCount="43" Body="&lt;p&gt;I’m designing a hydrogen and corrosion sensor. A thin film changes resistance based on hydrogen concentration. Then the capacitance between two electrodes will change depending on substrate thickness (corrosion). AC analysis can be used to see a phase change in the impedance, giving capacitance changes. I have designed a basic sensor, here it is with a basic model. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;But now I think an interdigitated electrodes might be better to detect the change in corrosion from the steel substrate. I’ve thrown together a basic design. I’m not sure though, has anyone got any thoughts? And what would be the best material for this electrode, I was thinking a copper? Also what would be the best way to go about modelling interdigitated electrodes, assuming lots of parallel capacitors?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Many thanks&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/TPHVK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TPHVK.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/NudNf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/NudNf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26635" LastActivityDate="2020-04-18T16:02:36.777" Title="Design of a hydrogen and corrosion sensor query" Tags="|electrical-engineering|sensors|electrical|ac|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35241" PostTypeId="1" CreationDate="2020-04-18T04:33:24.647" Score="1" ViewCount="26" Body="&lt;p&gt;I was asked to design a simple press mold for a 0.5mm sheet metal part. It has some small but rather deep sinks (5-7mm) which has made me wondering if it can be done in a single stage or it needs more stages. I guess there should be a table to help me through without the hassle of FEM with Ansys or other softwares. I read some books on metal-forming but I couldn't find my answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/5vGvM.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/5vGvM.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="21427" LastActivityDate="2020-04-18T14:43:22.073" Title="How can we tell if a press mold need more stages?" Tags="|metals|molding|metal-folding|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35243" PostTypeId="1" AcceptedAnswerId="35248" CreationDate="2020-04-18T10:05:16.423" Score="0" ViewCount="62" Body="&lt;p&gt;Tunnel boring machines are a common tool in deep tunnel construction. I am curious if they have ever been mass produced in specific types. What are the possible problems in (a possible) mass production of such machines?&lt;/p&gt;&#xA;" OwnerUserId="26638" LastEditorUserId="26638" LastEditDate="2020-04-18T12:35:08.383" LastActivityDate="2020-04-18T20:13:51.687" Title="Have tunnel boring machines been mass produced? What are the difficulties in mass producing a tunnel boring machine?" Tags="|structural-engineering|tunnels|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35245" PostTypeId="2" ParentId="35243" CreationDate="2020-04-18T10:56:32.187" Score="1" Body="&lt;p&gt;Mass produced? Depends what you mean by “mass”&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For the channel tunnel they made 5, 5 of such a specific machine is perhaps “mass”...&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-18T10:56:32.187" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35247" PostTypeId="2" ParentId="35065" CreationDate="2020-04-18T13:10:42.370" Score="0" Body="&lt;p&gt;Easiest way would be creating &lt;em&gt;&lt;code&gt;n&lt;/code&gt;&lt;/em&gt; no of input files with orientation corresponding to &lt;em&gt;&lt;code&gt;orientation_i&lt;/code&gt;&lt;/em&gt; with the help of some programming language and run all of them as a batch either with a simple python script or directly in the command prompt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example the python script would be,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;myJob1 = mdb.JobFromInputFile(&amp;lt;jobName&amp;gt;, &amp;lt;inputFile_i&amp;gt;)&lt;br&gt;&#xA;  myJob1.submit()&lt;br&gt;&#xA;  myJob1.waitForCompletion()&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Repeating this &lt;em&gt;&lt;code&gt;n&lt;/code&gt;&lt;/em&gt; of times as required save as a python script and run it inside Abaqus CAE.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or in command prompt,&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;abaqus job=&amp;lt;inputFile_i&amp;gt; interactive&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;over &lt;em&gt;&lt;code&gt;n&lt;/code&gt;&lt;/em&gt; times&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Another way would be using &lt;code&gt;*PARAMETER&lt;/code&gt; keyword available in Abaqus. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;For more information, look at PARAMETER keyword and JobFromInputFile in abaqus documentation.&lt;/p&gt;&#xA;" OwnerUserId="26534" LastActivityDate="2020-04-18T13:10:42.370" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35248" PostTypeId="2" ParentId="35243" CreationDate="2020-04-18T13:48:03.860" Score="5" Body="&lt;p&gt;One of the main problem with tunnel boring machines is they are required for a specific tunneling job: a certain sewer tunnel, a one off underground rail tunnel, etc. Once the job has been completed the tunnel boring machine may no longer be required. In a particular location there may not a need to continually excavate tunnels. Tunnel boring machines aren't easily mobile. Logistics are required to move them from site to site.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tunnel boring machines are made to construct a tunnel of a specific diameter. They may also be constructed to deal with specific types of ground conditions: clay soils, sand, highly stressed rock, etc. A machine built to perform for a specific set of conditions may not be applicable to other ground conditions.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For a given set of tunneling conditions, tunnel boring machines may not be the most appropriate device to use. In near surface tunnels in sandy soils, a cut and cover approach may be more appropriate technically and economically.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tunnel boring machine manufacturers don't want to have a yard of unsold machines waiting to be purchased. It's better for them financially to make machines as required and to the specifications of the client.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2020-04-18T20:13:51.687" LastActivityDate="2020-04-18T20:13:51.687" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35249" PostTypeId="2" ParentId="35241" CreationDate="2020-04-18T14:43:22.073" Score="2" Body="&lt;p&gt;Depends on the ductility of the &quot;sheet metal&quot;.Commonly steel or aluminum, but that is just part of the answer ; the particular alloy and condition/temper must be known . Also die life- number of parts to be made from a set of dies.And as commented, the force available in the press.&lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-04-18T14:43:22.073" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35250" PostTypeId="2" ParentId="35240" CreationDate="2020-04-18T15:05:56.597" Score="1" Body="&lt;p&gt;Are you making a corrosion sensor or a hydrogen sensor ? Corrosion sensors ( corrosion probes) ,based on metal loss have been around for over 50 years. And hydrogen is a special problem , strongly dependent on the diffusion of the hydrogen. Steel is relatively &quot;transparent&quot; to hydrogen above 400 F. And the solubility below 400 F depends on the steel condition and alloy. Because the surface interaction chemistry has the primary control of hydrogen entry , copper would be relatively useless for determining the hydrogen activity ( fugasity - don't get to used that word much) on a steel surface. Forgot there are hydrogen activity tools; they may be a section of steel with a concentric outer pipe ( sealed) and have just a simple pressure gage and vent valve. Periodically the pressure of hydrogen is noted and the vent opened to release the hydrogen. I may be biased as I pretty much made a career of hydrogen damage to metals.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastEditorUserId="11628" LastEditDate="2020-04-18T16:02:36.777" LastActivityDate="2020-04-18T16:02:36.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35253" PostTypeId="1" AcceptedAnswerId="35293" CreationDate="2020-04-18T15:37:02.557" Score="3" ViewCount="360" Body="&lt;p&gt;Reading about shearing stresses in beams, it is commonly stated that it is necessary to load a section through its shear center, so no twisting is applied to the beam. This is specially important in thin-walled open section, that have very weak torsion strength.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But how do you desing a beam in such a way? For example: if I'm designing a floor beam with a C-shape section, how do I enforce that the shearing forces act through the shear center?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks!&lt;/p&gt;&#xA;" OwnerUserId="202" LastActivityDate="2020-04-21T01:50:45.183" Title="How do you ensure that a beam is loaded through the shear center?" Tags="|mechanical-engineering|structural-engineering|shear|cross-section|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35255" PostTypeId="5" CreationDate="2020-04-18T18:09:57.463" Score="0" Body="&lt;p&gt;For engineering questions involving chemistry. For pure chemistry problems, please visit our sister site &lt;a href=&quot;https://chemistry.stackexchange.com/&quot;&gt;Chemistry Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6767" LastEditorUserId="6767" LastEditDate="2020-04-18T19:22:24.160" LastActivityDate="2020-04-18T19:22:24.160" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35256" PostTypeId="4" CreationDate="2020-04-18T18:09:57.463" Score="0" Body="For engineering questions involving chemistry. For pure chemistry problems, please visit our sister site Chemistry Stack Exchange." OwnerUserId="6767" LastEditorUserId="6767" LastEditDate="2020-04-18T19:22:17.617" LastActivityDate="2020-04-18T19:22:17.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35257" PostTypeId="5" CreationDate="2020-04-18T18:10:45.423" Score="0" Body="&lt;p&gt;For engineering questions involving mathematics. For pure mathematics problems, please visit our sister site &lt;a href=&quot;https://math.stackexchange.com/&quot;&gt;Mathematics Stack Exchange&lt;/a&gt;.&lt;/p&gt;&#xA;" OwnerUserId="6767" LastEditorUserId="6767" LastEditDate="2020-04-18T19:22:11.860" LastActivityDate="2020-04-18T19:22:11.860" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35258" PostTypeId="4" CreationDate="2020-04-18T18:10:45.423" Score="0" Body="For engineering questions involving mathematics. For pure mathematics problems, please visit our sister site Mathematics Stack Exchange." OwnerUserId="6767" LastEditorUserId="6767" LastEditDate="2020-04-18T19:22:21.363" LastActivityDate="2020-04-18T19:22:21.363" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35259" PostTypeId="1" CreationDate="2020-04-18T19:02:24.663" Score="1" ViewCount="108" Body="&lt;p&gt;For a simply supported beam with a UDL, with boundary conditions of X=0, V=0 &amp;amp; X=L, V=0. Using a cubic trial function.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$V=a+bx+cx^2+dx^3$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The following curve is obtained (Red Line). It is shown in comparison to the simple beam theory curve (Blue line).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KD6Gw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KD6Gw.png&quot; alt=&quot;RR Graph&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Is this the level of accuracy to be expected from a Rayleigh-Ritz method for this scenario using these conditions? I don't have an intuitive sense of how accurate the approximation should be so am unsure if I have made a mistake in my calculations or if this is an expected result. &lt;/p&gt;&#xA;" OwnerUserId="23791" LastActivityDate="2020-04-18T19:02:24.663" Title="How accurate should Rayleigh-Ritz approximation be?" Tags="|finite-element-method|" AnswerCount="0" CommentCount="7" ContentLicense="CC BY-SA 4.0" />
  <row Id="35265" PostTypeId="1" CreationDate="2020-04-19T14:43:38.877" Score="-1" ViewCount="485" Body="&lt;p&gt;I am not an engineer, but I thought maybe this could be a very old and classic problem in structural/mechanical engineering.&#xA;I need to build a support structure to hold a uniformly-distributed load of 75 kg (which has the shape of a square box). I can only use pipe and welding (or bolts+nuts). The only force it needs to hold is the weight of the box, there are not lateral forces expected (it will be placed indoors on a hard, level surface and will not be pulled/pushed from any of the 4 sides).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wonder if there is a relatively well known and simple answer to these&#xA;1) what is the strongest structure possible?&#xA;2) what is the optimum case in terms of strongest structure possible using a minimum amount of pipe?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I drew a few options (front and side views) but I guess there could be many more and a large amount of combinations. Is this a well-known case? or maybe it doesn't matter much given the relation between the strength of the steel pipe and the &quot;small&quot; load?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/j0n9n.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/j0n9n.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The distance between the floor and the bottom of the box is 70 cm. &#xA;The box is 65x65x22 cm thus the pipe structure view from above is a 65x65 cm square (or that's how I would initially imagine anyway). &lt;/p&gt;&#xA;" OwnerUserId="14759" LastEditorUserId="14759" LastEditDate="2020-04-19T15:38:24.507" LastActivityDate="2020-04-20T01:02:26.590" Title="What is the strongest structure?" Tags="|structural-engineering|structures|" AnswerCount="1" CommentCount="9" ContentLicense="CC BY-SA 4.0" />
  <row Id="35266" PostTypeId="1" AcceptedAnswerId="35321" CreationDate="2020-04-19T19:46:11.170" Score="1" ViewCount="641" Body="&lt;p&gt;I have a few questions regarding the physical interpretation of the information presented on a Bode plot. I believe it is one of those cases where I &quot;don't know what I don't know&quot;, so please bear with me if I meander.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;1) The magnitude plot of a Bode diagram represents the ratio of a system's output and input amplitudes, in dB. Given any system, is there a threshold dB which represents &quot;too much&quot;, or is it a system-by-system assessment which must be made by the designer? (Note: I believe I know the answer to be the latter, but want to be sure). Could you provide a contrasting example, such as a filter design vs. something with vastly different technical requirements?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;2) What is the physical interpretation of a magnitude plot having infinite gain at &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt; rad/s (for example, the bode plot of a pure integrator)? When an undamped second order system hits resonance, you get a theoretically infinite-dB magnitude response, which is obviously bad. So why does this logic not apply to very low frequencies? Is the infinite gain of a pure integrator simply a mathematical construct?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;3) In controller design, what is the physical interpretation of phase margin and gain margin? I understand the underlying mathematics which brings about those particular properties, but not what they represent. For example, the gain margin is the amount of gain which can be added to the system prior to it going unstable, but why would you want more gain? Is it simply that more controller gain allows it to better track the input signal? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4) How does the gain of the system as represented by the magnitude bode plot relate to the gains of the controller? I feel as though much of my confusion stems from this distinction, where large bode plot magnitudes are not necessarily desirable but large controller gains always seem to be so.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you very much in advance.&lt;/p&gt;&#xA;" OwnerUserId="26657" LastEditorUserId="18238" LastEditDate="2020-04-23T02:32:18.693" LastActivityDate="2020-04-23T02:32:18.693" Title="Physical interpretation of a Bode plot" Tags="|control-engineering|control-theory|stability|frequency-response|feedback-loop|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35268" PostTypeId="1" AcceptedAnswerId="35271" CreationDate="2020-04-19T20:34:29.227" Score="4" ViewCount="1677" Body="&lt;p&gt;What are the degrees of freedom of an internal hinge in a plane beam?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I understand that the bending moment=0, does this mean there is no slope either?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or does it have both degrees of freedom of deflection (V) and slope (θ)?&lt;/p&gt;&#xA;" OwnerUserId="23791" LastActivityDate="2020-04-19T23:52:03.017" Title="Degrees of freedom of an internal hinge in plane beam?" Tags="|beam|stiffness|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35269" PostTypeId="2" ParentId="35268" CreationDate="2020-04-19T22:52:01.343" Score="3" Body="&lt;p&gt;You seem to be mixing up the conditions on the &lt;em&gt;forces and moments&lt;/em&gt; acting across the hinge, and the &lt;em&gt;displacements and slopes&lt;/em&gt; there.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is no bending moment transmitted across the hinge, but there can be equal and opposite shear forces on each side, and the two slopes can be different non-zero values.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The fact that the slopes can be non-zero should be obvious if you think about how a real-life hinge (e.g a door hinge) works!&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-04-19T22:52:01.343" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35271" PostTypeId="2" ParentId="35268" CreationDate="2020-04-19T23:52:03.017" Score="5" Body="&lt;p&gt;Let's start talking not about hinges, but supports. Specifically, why do supports generate the forces (including bending moments, if applicable) they do (or don't)?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Think of a simply supported beam under a downward force. Its supports generate upward reaction forces. Why? Because if they didn't, the system would be unbalanced, and we wouldn't have a structure but a mechanism accelerating according to &lt;span class=&quot;math-container&quot;&gt;$F = ma$&lt;/span&gt;. More specifically, though, the supports create their reactions because the specific nodes they are attached to would move otherwise. For example, you can create structures where the deflection at a given unsupported point is zero. See the example below, where the deflection at midspan is zero and vertical reactions at each support.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/7OXre.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/7OXre.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If you then put a support at the midspan, that support's reaction will be zero; after all, it's not having to resist that point's movement.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/MjDw2.png&quot; rel=&quot;noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/MjDw2.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This is a trivial example of a fundamental relationship between reactions and deflections (including rotations): boundary conditions create reactions when they are inhibiting a deflection. If a given boundary condition doesn't create a reaction, it doesn't resist that deflection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, as you've stated, hinges have zero bending moment. What does that tell us? That hinges don't resist rotations. They also don't have force reactions (though axial and shear forces aren't necessarily zero at hinges; they transferred through the hinges, not absorbed), so we also know hinges don't resist deflections.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, to sum up: hinges can have non-zero deflections and non-zero rotations (specifically, the slope on one side of the hinge can be different from the other side).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-04-19T23:52:03.017" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35272" PostTypeId="2" ParentId="35265" CreationDate="2020-04-20T01:02:26.590" Score="0" Body="&lt;p&gt;Your load is too small to create high stresses.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But as a rule of thumb, your version 2 is the most unstable du to leaving one or more corners wobbly.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first version is the most reasonable and it seems it has the optimal economy in the material used.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If your load was significantly more, then you'd think about locating your 45-degree braces in such a way as to reduce the free extension of the legs against buckling and at the same time watching for the length of the brace itself not to buckle itself.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The third version could redeem its excessive use of material only if the loads are sensitive to vibration and require stabilized steady support.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-20T01:02:26.590" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35273" PostTypeId="1" AcceptedAnswerId="35278" CreationDate="2020-04-20T01:25:12.033" Score="4" ViewCount="281" Body="&lt;p&gt;Diesel exhaust is hot, and there's a lot of it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Diesel radiators also get pretty hot.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In a typical diesel engine, which one is dissipating more heat?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: In response to a request for specificity below, I am specifically interested in heat losses on five different gensets I operate. The engines are a 0.7L Kubota D722 (no turbo), two Lister Petter LPW4 engines (also no turbos), a 2.216L Perkins 404D-22 (also no turbo), and a 4.5L John Deere 4045TF151 (turbocharged). These engines are run at 1800 RPM (except for the kubota, which is 3600 RPM), under roughly half of rated continuous load.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;That is my specific situation, but I am looking for a general understanding of the matter, not just situation-specific advice.&lt;/p&gt;&#xA;" OwnerUserId="25141" LastEditorUserId="25141" LastEditDate="2020-04-20T15:18:25.577" LastActivityDate="2020-04-20T15:18:25.577" Title="In a diesel engine, does more heat leave the engine through the tailpipe or the radiator?" Tags="|heat-transfer|energy-efficiency|heat-exchanger|engines|diesel|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="35274" PostTypeId="2" ParentId="35018" CreationDate="2020-04-20T03:13:43.233" Score="1" Body="&lt;p&gt;When you are Smart Dimensioning, if you enter =&quot;R&quot;, SW will create a global variable &quot;R&quot;, which can be used in equations to redimension as required as long as you use &quot;R&quot; in the new dimensions as formulas.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case, I am making a circle with dimension &quot;R&quot;.  I do not care what R actually is.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/lAZQc.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/lAZQc.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;SW puts sigma before the dimension to indicate it is a equation:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DWuSS.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DWuSS.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;and adds the Equation box to the part tree.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gkw4x.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gkw4x.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So if I add a second circle with Smart Dimension =&quot;R&quot;*2, I get a second circle sized based upon the dimension &quot;R&quot;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/dArho.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/dArho.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/XWAJf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/XWAJf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Right mouse click on Equations and you can manage Equations.  Notice I have changed R to 40 and both are scaled.  This is how I handle scaling.  Usually I adjust a fey key variables and I get good results.  Start simple.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t4c6b.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t4c6b.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="1592" LastActivityDate="2020-04-20T03:13:43.233" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35275" PostTypeId="1" CreationDate="2020-04-20T04:19:45.777" Score="2" ViewCount="297" Body="&lt;p&gt;I did a Civil Engineering course some years ago and this question from my textbook, I have found what seems impossible to get the correct answer of 46.2 m for. By putting this question up I am hoping that I will be able to come to a conclusion on how this should be solved.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;A ship displaces 10,000 metric tonnes and area of its plane of flotation is 1,480 m^2. The centre of mass is 49 m and centre of area of plane of flotation is 55 m from the stern.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The metacentric height for pitching motion about transverse principal axis is 91.5 m. The ship is loaded in sea water with 300 metric tonnes of cargo. Find minimum allowable distance of mass centre of this extra load from the stern if, when ship passes from sea water into a fresh water canal, the stern draft must not increase by more than 0.3 m ?&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Assume metacentric height and area of plane of flotation are not altered by the change in draft, density of sea water is 1,025 kg/m^3 and density of fresh water is 1,000 kg/m^3.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Just looking for some help on this from someone who understands ship stability well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;Here is my attempt at finding a Solution to this Question&lt;/strong&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The first step:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will put the 300 Tonne Cargo Load at the ships centre of gravity of 49 m, so we don't have to correct ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly we will treat this as if the cargo has a weight shift from centre of flotation to centre of gravity to calculate the aft added draft from the cargo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to run the Change of Trim Formula&lt;/p&gt;&#xA;&#xA;&lt;p&gt;by using COT = (100 * LBP * w * d) / (W * GML) with d = 6 m because you want to have initial LCG AT 49m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;COT = (100*LBP*300*6) / (10,300*91.5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aft Added Draft = (LCF / LBP) * COT = (55 / LBP) * [(100*LBP*300*6) / (10,300*91.5)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and LBP cancels out,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so Aft Added Draft = (55*100*300*6) / (10,300*91.5) = 10.5 cm at original 49 m CG&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Second Step:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to take into account sinkage change due to density, for 10,000 Tonne Ship + 300 Tonne Cargo all at 49m CG&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tonnes per centimetre (TPC) = Density of Sea Water * Area of Waterplane * 1/100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TPC = 1.025 * 1,480 * 1/100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TPC = 15.17 Tonnes per cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parallel Sinkage in Fresh Water = 10,300 / 15.17 = 678.97 cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And because the Density Changes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at 49 m CG the Sinkage Change = [(1.025 - 1) / (1)] * (10,300 / 15.17) * (49/55) = 15.12 cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Third and Final Step:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cargo will shift from the ships centre of gravity of 49m, to where it should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this part we will firstly have to calculate the remainder draft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remainder draft = 30 - 10.5 - 15.12 = 4.38 cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like before we need to use Aft added draft formula, but this time we need to find d (distance moved)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now find d in following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change in Draft Aft (from shifted load) = LCF [(100 * w * d) / (W * GML)] = 4.38 cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now solve for d&lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.38 cm = 55 [(100 * 300 * d) / (10,300 * 91.5)] = 55 [30,000 d / 942450]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so d = 2.5m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this gives my answer so far of 46.5 m from stern for Longtitudinal centre of gravity of cargo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is 0.3 m off the answer,&#xA;so this is what I mean by it seems impossible to get the correct answer of 46.2 m. By putting my answer up along with the question, I am hoping that I will get feedback from someone who knows how this works and can point me in the right direction.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8013" LastEditorUserId="8013" LastEditDate="2020-04-20T23:27:26.297" LastActivityDate="2024-01-31T17:03:40.520" Title="Longtitudinal Centre of Gravity for Ship Cargo" Tags="|stability|ships|naval-engineering|" AnswerCount="2" CommentCount="9" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35276" PostTypeId="2" ParentId="35275" CreationDate="2020-04-20T04:19:45.777" Score="0" Body="&lt;p&gt;&lt;strong&gt;The first step:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;We will put the 300 Tonne Cargo Load at the ships centre of gravity of 49 m, so we don't have to correct ourselves.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Firstly we will treat this as if the cargo has a weight shift from centre of flotation to centre of gravity to calculate the aft added draft from the cargo.  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;We need to run the Change of Trim Formula &lt;/p&gt;&#xA;&#xA;&lt;p&gt;by using COT = (100 * LBP * w * d) / (W * GML) with d = 6 m because you want to have initial LCG AT 49m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;COT = (100*LBP*300*6) / (10,300*91.5)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Aft Added Draft = (LCF / LBP) * COT = (55 / LBP) * [(100*LBP*300*6) / (10,300*91.5)]&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and LBP cancels out,&lt;/p&gt;&#xA;&#xA;&lt;p&gt;so Aft Added Draft = (55*100*300*6) / (10,300*91.5) = 10.5 cm at original 49 m CG&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Second Step:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now we need to take into account sinkage change due to density, for 10,000 Tonne Ship + 300 Tonne Cargo all at 49m CG&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Tonnes per centimetre (TPC) = Density of Sea Water * Area of Waterplane * 1/100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TPC = 1.025 * 1,480 * 1/100&lt;/p&gt;&#xA;&#xA;&lt;p&gt;TPC = 15.17 Tonnes per cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Parallel Sinkage in Fresh Water = 10,300 / 15.17 = 678.97 cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;And because the Density Changes&lt;/p&gt;&#xA;&#xA;&lt;p&gt;at 49 m CG the Sinkage Change = [(1.025 - 1) / (1)] * (10,300 / 15.17) * (49/55) = 15.12 cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The Third and Final Step:&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The cargo will shift from the ships centre of gravity of 49m, to where it should be.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For this part we will firstly have to calculate the remainder draft.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The remainder draft = 30 - 10.5 - 15.12 = 4.38 cm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Like before we need to use Aft added draft formula, but this time we need to find d (distance moved) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;So now find d in following formula:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Change in Draft Aft (from shifted load) = LCF [(100 * w * d) / (W * GML)] = 4.38 cm &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now solve for d &lt;/p&gt;&#xA;&#xA;&lt;p&gt;4.38 cm = 55 [(100 * 300 * d) / (10,300 * 91.5)] = 55 [30,000 d / 942450] &lt;/p&gt;&#xA;&#xA;&lt;p&gt;so d = 2.5m&lt;/p&gt;&#xA;&#xA;&lt;p&gt;and this gives my answer so far of &lt;strong&gt;46.5 m&lt;/strong&gt; from stern for Longtitudinal centre of gravity of cargo.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;This is 0.3 m off the answer,&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;so this is what I mean by it seems impossible to get the correct answer of 46.2 m. By putting my answer up along with the question, I am hoping that I will get feedback from someone who knows how this works and can point me in the right direction.&lt;/strong&gt;&lt;/p&gt;&#xA;" OwnerUserId="8013" LastActivityDate="2020-04-20T04:19:45.777" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35277" PostTypeId="1" CreationDate="2020-04-20T11:31:41.623" Score="0" ViewCount="64" Body="&lt;p&gt;I have seen a sort disformation on wall of my room I have attached snapshot of both inside view(from inside room) and out side view(that is out of room)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;inside view is below&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/IkeI7.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/IkeI7.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;out side view is below&lt;/strong&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/TpCph.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/TpCph.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="22801" LastEditorUserId="22801" LastEditDate="2020-04-21T17:07:26.923" LastActivityDate="2020-04-21T17:07:26.923" Title="what is issue with my room brick wall? How can i solve it?" Tags="|civil-engineering|deformation|retaining-wall|home-improvement|maintenance|" AnswerCount="1" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35278" PostTypeId="2" ParentId="35273" CreationDate="2020-04-20T11:41:35.547" Score="7" Body="&lt;p&gt;They are very nearly equal for typical four-stroke non-turbo diesels under load. A turbo diesel under load should have slightly more radiator loss than exhaust loss.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the bottom is a link to the technical spec sheet for a Cat 3412 powered genset. It's a probably a bit bigger than what you had in mind. It is a turbo with aftercooler (A/C in the doc below). Since it is setup for stationary power, the turbo is big, heavy, slow and efficient, and the aftercooler has a substantial heat rejection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;All industrial engines will have sheets like this. They can be customized by the mfg for unusual operating conditions such as high altitude or arctic conditions. This lets the dealer's installers get everything sized and arranged right on the second try (and they just love to fly to northern Alaska to fix stuff in the winter).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For over-the-road diesels, the heat rejection for internal jacket water and exhaust would be closer together. Radiators are draggy and heavy and you design vehicle engines to minimize them. You can reject more heat from surfaces in a vehicle.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Cat has a nice engineering primer on engine cooling - &lt;a href=&quot;http://s7d2.scene7.com/is/content/Caterpillar/CM20160713-53120-13199&quot; rel=&quot;noreferrer&quot;&gt;http://s7d2.scene7.com/is/content/Caterpillar/CM20160713-53120-13199&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Vehicles often use air-to-air aftercooling and oil cooling, so the aftercooler and oil cooler heat doesn't end up in the radiator. Most larger engines use jacket water for aftercooling and oil cooling, so you need to add those values to the internal jacket water heat to size the radiator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;http://s7d2.scene7.com/is/content/Caterpillar/CM20160713-53120-13199&quot; rel=&quot;noreferrer&quot;&gt;Cat 3412 genset technical data sheet&lt;/a&gt;   (see section titled heat rejection on page 2)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'll try to dig up a similar sheet for mobile equipment engines later. Or someone else can ;) &lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2020-04-20T11:41:35.547" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35279" PostTypeId="1" AcceptedAnswerId="35409" CreationDate="2020-04-20T12:28:44.880" Score="5" ViewCount="5270" Body="&lt;p&gt;I understand that for fully developed flow inside a tube (with constant properties), Nusselt number doesn’t change in axial direction but it is surprising that it is also independent of Reynolds number which means that the rate of heat transfer (for constant pipe diameter and fluid properties) doesn’t change with mass flow rate (and velocity).&#xA;So could someone please  give an explanation for this.&lt;/p&gt;&#xA;" OwnerUserId="23043" LastEditorUserId="90" LastEditDate="2023-12-21T00:38:42.940" LastActivityDate="2023-12-23T18:53:38.547" Title="Why Nusselt number for laminar flow in a pipe is independent of Reynolds number" Tags="|fluid-mechanics|heat-transfer|heat-exchanger|" AnswerCount="4" CommentCount="0" FavoriteCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35280" PostTypeId="2" ParentId="35279" CreationDate="2020-04-20T12:42:22.447" Score="2" Body="&lt;p&gt;The Nusselt number is about the ratio of conductive and convective heat transfer across a boundary.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Reynolds number is about how a fluid is moving - usually laminar (below 1700), turbulent (above 2000) or in-between aka critical. However these numbers are not absolute as laminar has been seen above 2000 in &lt;strong&gt;very&lt;/strong&gt; carefully controlled conditions i.e. it is not stable.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, while heat transfer can be affected by the type of fluid flow, those changes may not change in proportion to any change in the Reynolds number.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastEditorUserId="90" LastEditDate="2020-04-20T14:39:05.080" LastActivityDate="2020-04-20T14:39:05.080" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35281" PostTypeId="2" ParentId="35277" CreationDate="2020-04-20T13:35:46.773" Score="5" Body="&lt;p&gt;The wall is experiencing &lt;a href=&quot;https://www.diydoctor.org.uk/projects/rising-damp.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;rising damp&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Basically, moisture in the soil is being absorbed by the wall and traveling up the wall. This can become more prevalent during wet periods and subside during dry periods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The photograph of the outside of the wall appears to show a of line salt crystallizing towards the top of the affected zone. This is not unusual as the rising ground water brings any salt that is within the ground into the wall. As the moisture in the wall evaporates, salt crystallizes on the wall. Over time this can damage the wall. If the wall is made of bricks, it can cause the bricks to crumble.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;During construction of the wall, a damp course should have been incorporated into the wall. This is a waterproof layer which prevents ground moisture from rising into the wall above the damp course.&lt;/p&gt;&#xA;" OwnerUserId="90" LastEditorUserId="90" LastEditDate="2020-04-20T19:23:19.750" LastActivityDate="2020-04-20T19:23:19.750" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35284" PostTypeId="1" CreationDate="2020-04-20T15:37:02.600" Score="0" ViewCount="1544" Body="&lt;p&gt;I am looking for the formula allowing to calculate the deflection of a rectangular plate with a &quot;transverse loading in its center&quot; (maybe the kind of loading will be more clear with this picture : &lt;a href=&quot;https://i.stack.imgur.com/F5yHf.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/F5yHf.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I already found this formula for the case of a rectangular plate with a concentrated load at center (and I know the one I'm looking for is very close).&#xA;&lt;a href=&quot;https://www.engineersedge.com/material_science/rectangular_plate_concentrated_load_13645.htm&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.engineersedge.com/material_science/rectangular_plate_concentrated_load_13645.htm&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But I can't find the coefficients for my precise case with the load uniformly distributed across the width (as in the first picture)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thank you&lt;/p&gt;&#xA;" OwnerUserId="26675" LastActivityDate="2022-05-11T06:07:15.690" Title="Deflection of a rectangular plate simply supported (transverse load)" Tags="|deflection|" AnswerCount="2" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35285" PostTypeId="2" ParentId="35284" CreationDate="2020-04-20T16:05:30.550" Score="1" Body="&lt;p&gt;The behavior of a plate supported on only two opposing sides with loads that are uniform along the width of the plate is identical to that of a beam, so the standard beam deflection equations can be used. The deflection of a beam with a concentrated load at its midspan is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\delta = \dfrac{F\ell^3}{48EI}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; is the total concentrated load at midspan, &lt;span class=&quot;math-container&quot;&gt;$\ell$&lt;/span&gt; is the beam's span length, &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is the material's elastic modulus, and &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; is its moment of inertia. Considering the plate's cross-section of &lt;span class=&quot;math-container&quot;&gt;$w \times t$&lt;/span&gt;, we have&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I = \dfrac{wt^3}{12}$$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-04-20T16:05:30.550" CommentCount="6" ContentLicense="CC BY-SA 4.0" />
  <row Id="35286" PostTypeId="1" CreationDate="2020-04-20T16:45:07.413" Score="1" ViewCount="71" Body="&lt;p&gt;I generated different 2D microstructures with different fibers volumetric fraction (basically a square representing the matrix and a lot of small rectangles inside, with different measures and random orientations, representing the fibers). I'm attaching a figure to be clearer.&#xA;&lt;a href=&quot;https://i.stack.imgur.com/eFyqJ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eFyqJ.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;To run the simulations using Abaqus I need to define the properties of the fibers (the rectangles), but how do I set the properties according to the main axis of each fibers considering that I have random orientations? I cannot manually set the axis orientations for each fiber, it would be too timeconsuming considering that I have configurations with 40% of fibers (more than 400). I need to find a way to set the properties according to the main axis of the fibers (rectangles).&#xA;To be more specific, I created an RSA alghoritm in Matlab to create the random distribution of fibers. Then, I create the main square (representing the matrix of the composite) in Abaqus and I create the fibers by partion, importing the generated fibers distribution as a sketch.&#xA;Thank you all for your attention.&lt;/p&gt;&#xA;" OwnerUserId="26676" LastActivityDate="2020-04-20T16:45:07.413" Title="How do I define properties orientation in a 2D short random fibers composite? ABAQUS" Tags="|abaqus|composite|" AnswerCount="0" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35287" PostTypeId="2" ParentId="2852" CreationDate="2020-04-20T19:33:09.273" Score="0" Body="&lt;p&gt;RE: for your well:&lt;/p&gt;&#xA;&lt;p&gt;In geotech, piezometers measure water depth in monitoring boreholes.&lt;/p&gt;&#xA;&lt;p&gt;Here's a professional sensor:&#xA;&lt;a href=&quot;https://www.geokon.com/4500-Series&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.geokon.com/4500-Series&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/3FCMK.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/3FCMK.png&quot; alt=&quot;piezometer&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Here are possible DIY discussions:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://www.envirodiy.org/topic/monitoring-well-or-piezometer-water-level-sensor/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.envirodiy.org/topic/monitoring-well-or-piezometer-water-level-sensor/&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/iPaal.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/iPaal.png&quot; alt=&quot;DIY&quot; /&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://www.envirodiy.org/construction-of-water-level-monitoring-sensor-station/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.envirodiy.org/construction-of-water-level-monitoring-sensor-station/&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/tDuHw.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/tDuHw.png&quot; alt=&quot;DIY&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26396" LastEditorUserId="26396" LastEditDate="2021-01-09T01:41:49.800" LastActivityDate="2021-01-09T01:41:49.800" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35289" PostTypeId="2" ParentId="35284" CreationDate="2020-04-20T23:11:38.100" Score="0" Body="&lt;p&gt;Roark’s Formulas for Stress and Strain by WARREN C. YOUNG RICHARD G. BUDYNAS&#xA;On pp 435 has a reference to this question.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;A very wide beam of rectangular cross section can be treated as a beam&#xA;  if E is replaced by &lt;span class=&quot;math-container&quot;&gt;$\frac{E}{1-\nu^2}$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I$&lt;/span&gt; by &lt;span class=&quot;math-container&quot;&gt;$t^3/12$&lt;/span&gt; (see Sec. 8.11). It can also&#xA;  be treated as a plate with two opposite edges free as shown in Figs.&#xA;  8.16 and 11.2. For details see Ref. 88.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;If you don't have access to it I suppose I can take picture of the page and give it to you. Because another conditions apply too.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-04-20T23:22:28.840" LastActivityDate="2020-04-20T23:22:28.840" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35290" PostTypeId="1" CreationDate="2020-04-20T23:19:12.303" Score="1" ViewCount="1400" Body="&lt;p&gt;I am designing a flange to be use to maintain a pressure of 10&lt;sup&gt;-2&lt;/sup&gt; Torr for at least 3 months. I am using an o-ring seal from my (device) to the product weldment. &lt;strong&gt;I am looking for feedback on using NPT fittings in a vacuum system.&lt;/strong&gt; I have heard it is unwise. However what I find readily available online is only NPT fitted gauges. I was thinking of using Teflon tape or Loctite. I should add there is a fitting I am using to adapt from NPT to conflat which will be welded. &lt;/p&gt;&#xA;" OwnerUserId="6389" LastEditorUserId="90" LastEditDate="2020-04-21T07:10:10.867" LastActivityDate="2020-04-21T07:10:10.867" Title="Can I use a NPT fastened device to vacuum or 10^-2 Torr?" Tags="|mechanical-engineering|welding|vacuum|fasteners|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35291" PostTypeId="2" ParentId="35290" CreationDate="2020-04-21T00:17:58.427" Score="2" Body="&lt;p&gt;Are you going to continue pumping or have intermittent pumping ?  As a engineering professor of sealing once told me &quot;everything leaks, it is a matter of how much&quot;. Without continued pumping you will not hold that vacuum. For serious leak tight construction I have mostly seen Swageloc connections. NPT has no metal to metal seal , I think metal/metal will be necessary depending on pump.For a start use as many welds as possible to minimize the threaded connections. American Petroleum Institute (API) pipe dope has metal powders in it to seal threads like NPT for moderate pressures so it does not depend on grease-like stuff ( regular pipe dope) to stop leaks; it may not be easy for you to find.  &lt;/p&gt;&#xA;" OwnerUserId="11628" LastActivityDate="2020-04-21T00:17:58.427" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35292" PostTypeId="2" ParentId="35290" CreationDate="2020-04-21T00:18:29.470" Score="1" Body="&lt;p&gt;TLDR: NPT and vacuum - no problem - use correct type. NPT and o-ring - possible (likely) problem landing o-ring if you use a tapered thread style.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Please post a sketch so we know the size of the thing and how the o-ring relates to your flange.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://en.wikipedia.org/wiki/National_pipe_thread&quot; rel=&quot;nofollow noreferrer&quot;&gt;NPT&lt;/a&gt; can be one of a dozen or so thread types. Some make sense and some don't.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The most common NPT type is a tapered thread that has an interference fit on the crest/valley of the mating threads, but has a spiral leak path around the threads. It is also not produced accurate enough to land an o-ring due to differing thread depths cut into different parts. If you are making both mating pieces, this isn't really a problem - you can jig it up so you are consistent. But if you only have control over one of the pieces in the joint, no, go with a non-tapered thread style. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use any of several &quot;dry seal&quot; NPT styles intended for fuels which don't have the spiral leak path.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can use any straight thread type plus your o-ring to accomplish the seal.&lt;/p&gt;&#xA;" OwnerUserId="12754" LastActivityDate="2020-04-21T00:18:29.470" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35293" PostTypeId="2" ParentId="35253" CreationDate="2020-04-21T01:50:45.183" Score="1" Body="&lt;p&gt;For a C channel with the following dimensions:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;b = width of the flange measured to the center of the web, inches or mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;c = Distance to the shear center, inches or mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;h = Height of section measured to the center of flanges, inches or mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I = The second moment of area, in^4 or mm^4&lt;/p&gt;&#xA;&#xA;&lt;p&gt;t = flange thickness, inches or mm&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ c=\frac{b^2h^2t}{4I_x}$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we apply load on this offset falling outside of the channel opposing the flanges there will be no twisting torque.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;But usually in the construction C channels are rarely used as beams, the favorite section fo steel is I beams, or similar shapes because of their symmetry, a large relative I and the fact that they have been designed as beam and already include many requirements to be used as a beam by codes.&#xA;However, C channels in most cases (if ever) are used in situations where they are laterally braced by the floor slabs or roof sheathing so the torque created by loading, not through the shear center is resisted by those members.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-21T01:50:45.183" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35294" PostTypeId="1" AcceptedAnswerId="35296" CreationDate="2020-04-21T05:23:24.093" Score="4" ViewCount="1131" Body="&lt;p&gt;In FEM is symmetry technically a boundary condition?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For example using the FEM on a symmetrically loaded and supported beam. The FEM can be simplified by using symmetry.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;However, is this symmetry just implemented as a boundary condition or is it technically one? Because it's not actually constraining the movement given that the result is the same with or without it.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So my own impression is it's not a boundary condition?&lt;/p&gt;&#xA;" OwnerUserId="23791" LastActivityDate="2020-04-21T07:29:31.127" Title="FEM - is symmetry a boundary condition?" Tags="|finite-element-method|stiffness|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35295" PostTypeId="2" ParentId="35294" CreationDate="2020-04-21T05:52:00.600" Score="3" Body="&lt;p&gt;Symmetry is used to reduce the size of the object and therefore the mesh or allow more mesh to represent that reduced object.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So the “cut plane” of symmetry is not a boundary where the material changes in the real object.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-21T05:52:00.600" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35296" PostTypeId="2" ParentId="35294" CreationDate="2020-04-21T07:29:31.127" Score="5" Body="&lt;p&gt;The symmetry itself is not a boundary condition. It is a property of your system which means that both the geometry and the load are symmetric with respect to an axis or a plane. It allows to reduce the computation to a downsized domain, which leads to considerable computational time saving.&lt;br&gt;&#xA;I guess you are using a FEA software and manually reduced the mesh. You then have to tell the software that some of the boundaries are not the physical boundaries of the domain you study, but are part of the plane of symmetry. Under the hood, the software applies some boundary conditions that are consequences of symmetry.&lt;br&gt;&#xA;For example, if you have a planar symmetry, no displacement on the axis perpendicular to this plane is allowed, whereas only the rotation around this normal axis is permitted. So to answer your question, in &quot;FEA software&quot; language, symmetry is a boundary condition for the reasons I just explained, but this is not very rigorous.&lt;br&gt;&#xA;The fact that you get the same results with and without the symmetry BC may be due to various reasons, one of which may be chance, if your load does not &quot;put into use&quot; the degrees of freedom blocked by the symmetry.&lt;/p&gt;&#xA;" OwnerUserId="26427" LastActivityDate="2020-04-21T07:29:31.127" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35297" PostTypeId="1" CreationDate="2020-04-21T07:58:10.657" Score="3" ViewCount="2249" Body="&lt;p&gt;I have an assembly in Solidworks with gears that I want to make a sketch from it.&#xA;I want to hide the gears teeth and represent the each gear teeth only as center line, like in the picture below on the left:&lt;a href=&quot;https://i.stack.imgur.com/i78oh.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/i78oh.png&quot; alt=&quot;Taken as an example from https://www.cad-notes.com/automate-standard-additional-notes-in-the-drawing/amp/&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have tried every button in Solidworks, but didn't find how to hide the unnecessary tooth at the end.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks for any help.&lt;/p&gt;&#xA;" OwnerUserId="24318" LastActivityDate="2020-04-21T07:58:10.657" Title="Solidworks gear representation in drawing" Tags="|solidworks|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35298" PostTypeId="1" CreationDate="2020-04-21T11:50:37.960" Score="0" ViewCount="24" Body="&lt;p&gt;I am doing an experiment regarding elbow flexion's effect on peak sprinting velocity and we want to measure trunk rotation and shoulder rotation during our trials. We don't have an overhead camera so we have to make do with cameras in the sagittal and frontal planes. Is there a way of estimating the trunk rotation and shoulder rotation this way? Perhaps using angular area and estimating based on the known width of the participants and distance from the camera (although my geometry skills aren't good enough to find a closed form for any such equation)? I can't think of any other way to do it. Any suggestions would be much appreciated.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Thanks&lt;/p&gt;&#xA;" OwnerUserId="26689" LastActivityDate="2020-04-21T11:50:37.960" Title="Estimating trunk rotation using frontal and sagittal views" Tags="|measurements|biomechanics|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35299" PostTypeId="1" CreationDate="2020-04-21T20:24:12.170" Score="0" ViewCount="104" Body="&lt;p&gt;Time history analysis of earthquakes for structures, is a NON-LINEAR dynamic method correct?&#xA;Then why is there also something called &quot;LINEAR response history analysis&quot;? What is the difference of this from non linear time history analysis?&lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2020-04-22T21:14:02.723" Title="Earthquake time history analysis" Tags="|structural-engineering|structural-analysis|structures|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35300" PostTypeId="1" CreationDate="2020-04-21T20:53:09.677" Score="0" ViewCount="44" Body="&lt;p&gt;I'm planning to build a mini backpack gravity kart. Uphill by foot bringing kart in back, downhill driving offroad that kart.&#xA;So, it must be small, light, &quot;flat&quot; (similar to a skateboard).&#xA;I'm thinking about steer. My two choiches are:&#xA;- normal front wheel steering, driven by lateral lever instead of steering wheel&#xA;- fixed wheel, steering made by braking left or right brake&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I prefer the second solution because of simplicity, the kart could be very minimal.&#xA;but I never seen kart like that, what are the cons of that steering system? &lt;/p&gt;&#xA;" OwnerUserId="19708" LastActivityDate="2020-04-21T20:53:09.677" Title="Steering for gravity kart" Tags="|mechanical-engineering|" AnswerCount="0" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35301" PostTypeId="1" CreationDate="2020-04-21T21:35:01.733" Score="1" ViewCount="246" Body="&lt;p&gt;Is &quot;Modal Response Spectrum Analysis&quot; the same as response spectrum analysis? How about modal superposition analysis? This is the same or different? Or all are same?&lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2024-01-02T14:07:50.883" Title="Modal Response Spectrum analysis" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35303" PostTypeId="1" AcceptedAnswerId="35323" CreationDate="2020-04-22T02:35:42.087" Score="1" ViewCount="5385" Body="&lt;p&gt;I'm studying Root Locus method and I still confused. The question below exemplifiques my doubt.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Determine the &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; gain so that the dominant roots have a damping factor equal to &lt;span class=&quot;math-container&quot;&gt;$0.5$&lt;/span&gt;, where &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$G(s) = \frac{K(s^2+5s+9)}{s^2(s+3)}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have successfully draw the root locus from this &lt;span class=&quot;math-container&quot;&gt;$G(s)$&lt;/span&gt;, but I can't figure out how I could calculate the gain &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="26696" LastEditorUserId="18238" LastEditDate="2020-04-22T15:01:40.427" LastActivityDate="2020-04-22T22:20:59.007" Title="Determine the gain K from a root locus" Tags="|control-engineering|control-theory|" AnswerCount="3" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35304" PostTypeId="2" ParentId="35299" CreationDate="2020-04-22T02:50:06.430" Score="2" Body="&lt;p&gt;A linear analysis is ... linear. If you increase the loads by a factor of two, the results increase by a factor of two. If you increase the loads by a factor of 1 million, the results increase by a factor of 1 million, even though the results are likely to be wrong.  The analysis does not understand that other effects occur when the results are so large. (Those other effects are the stiffness of the structure changes, and the direction of the load changes, and so on.)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;A nonlinear analysis does not have a linear relationship between the load and result. Doubling the load may only give results 1.5 times as much (the structure stiffens) or a result 3 times more  (the material yields or gets weaker for example).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Time history indicates it is a dynamic analysis. The loafs and results can change over time. &lt;/p&gt;&#xA;" OwnerUserId="12984" LastActivityDate="2020-04-22T02:50:06.430" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35305" PostTypeId="2" ParentId="35301" CreationDate="2020-04-22T04:29:32.693" Score="0" Body="&lt;p&gt;A building with several degrees of freedom under a certain dynamic vibration, such as earthquake ground motion, will vibrate in several mode shapes. Each with a different frequency.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Modal response spectrum design involves imposing the predetermined response spectra of that particular frequency to that mode and after calculating the mass participation factor of individual modes add them together with as many modes as it takes to have at least 90% or either what the code requires mass participation.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There are several ways to add these modes together.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is the square root of the sum of the squares SRSS method, there are other methods as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;These are linear dynamic methods, ie, structure assumed never go beyond elastic range.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we want to allow the structure to deform beyond yield stress into the plastic range we have to use complicated nonlinear methods.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://ui.adsabs.harvard.edu/abs/1983STIN...8336282S/abstract&quot; rel=&quot;nofollow noreferrer&quot;&gt;Harvard University link&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-04-22T04:48:07.493" LastActivityDate="2020-04-22T04:48:07.493" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35308" PostTypeId="1" AcceptedAnswerId="35375" CreationDate="2020-04-22T11:26:42.417" Score="-1" ViewCount="63" Body="&lt;p&gt;I am reading a paper titled &lt;em&gt;Conditions for safe separation of external stores&lt;/em&gt; by EE Covert&lt;a href=&quot;https://arc.aiaa.org/doi/abs/10.2514/3.44728?journalCode=ja&quot; rel=&quot;nofollow noreferrer&quot;&gt; [Link]&lt;/a&gt;. In Section 1 &lt;em&gt;Introduction&lt;/em&gt; of the paper, the author has given an expression for &lt;span class=&quot;math-container&quot;&gt;$d$&lt;/span&gt;, the relative distance between some point &lt;span class=&quot;math-container&quot;&gt;$r_{p}$&lt;/span&gt; on the store and &lt;span class=&quot;math-container&quot;&gt;$r_{q}$&lt;/span&gt; on the aircraft as:&lt;/p&gt;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$d = d \left( 0 \right) + \int\limits_{0}^{t} \left\{ \left[ R \right] \cdot \left( U + \omega \times r_{p}\right) - \left( V_{A/C} + \Omega \times r_{q} \right)\right\} \mathrm{d}t $$&lt;/span&gt;&lt;/p&gt;&#xA;&lt;p&gt;where,&lt;/p&gt;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ V_{A/C} $&lt;/span&gt;: aircraft translational velocity in aircraft body coordinates.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \Omega $&lt;/span&gt;: aircraft rotational velocity in aircraft body coordinates.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ r_{q} $&lt;/span&gt;: position vector between aircraft center of gravity, and point of interest in aircraft.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ r_{p} $&lt;/span&gt;: position vector between origin and point on store.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \left[ R \right] $&lt;/span&gt;: rotation matrix to transfer store velocities to aircraft coordinate frame.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ U $&lt;/span&gt;: store linear velocity in store coordinates.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \omega $&lt;/span&gt;: store angular velocity in store coordinates.&lt;/p&gt;&#xA;&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&lt;p&gt;I want to derive this expression, but I don't know where to start.&lt;/p&gt;&#xA;" OwnerUserId="17712" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2020-04-25T20:16:22.767" Title="Equation governing relative motion between aircraft and store" Tags="|applied-mechanics|dynamics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35309" PostTypeId="1" AcceptedAnswerId="35319" CreationDate="2020-04-22T13:03:16.960" Score="2" ViewCount="4565" Body="&lt;p&gt;Consider the borehole record below (in particular the &lt;em&gt;Water Depth&lt;/em&gt; column).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a white inverted triangle at a depth of 3.20 m. What is the meaning of this symbol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The white inverted triangle indicates the depth at which water was struck.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There is a black inverted triangle at a depth of 2.50 m. What is the meaning of this symbol?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;The black inverted triangle indicates the depth to which the water rose after 20 minutes.&lt;/strong&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;My new question is why does the water rise above where it was originally struck? What is this implication of this? Is the location of the black inverted triangle the &lt;em&gt;true&lt;/em&gt; location of the water table within the soil? Why was water struck lower than this?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/69cLL.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/69cLL.png&quot; alt=&quot;borehole record&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="10270" LastEditorUserId="10270" LastEditDate="2020-04-22T16:23:37.330" LastActivityDate="2020-04-22T18:43:34.210" Title="What is the meaning of these two inverted triangle water table symbols in a borehole record?" Tags="|civil-engineering|geotechnical-engineering|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35310" PostTypeId="1" CreationDate="2020-04-22T13:53:50.293" Score="2" ViewCount="49" Body="&lt;p&gt;Hi there I found my self with a broken pair of headphones for which I am designing a spare part in Inventor right now. (Using Inventor Professional 2020). &lt;/p&gt;&#xA;&#xA;&lt;p&gt;I have the problem that I have a set of extrusions to move since they are interfering with other parts of the model and because the model itself is not in the right dimensions yet (I went with modelling the part to coarsely resemble the proportions and shape and bother with the dimensions last which is usually a straightforward enough thing to do). Now how can I move the extrusions shown in the below picture? I have tried moving the axis on which I have placed the extrusions but that leaves me with the end of the extrusion hanging in the air which is obviously not to be desired. Moving the layer on which the 2D Shape to be extruded is drawn after moving the axis is not possible since this results in an infinite loop of dependencies (why actually?) &#xA;&lt;a href=&quot;https://i.stack.imgur.com/Dcs2G.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/Dcs2G.jpg&quot; alt=&quot;extrusions to move&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/eVH2R.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/eVH2R.png&quot; alt=&quot;air gap when moved&quot;&gt;&lt;/a&gt;&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1tAxA.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1tAxA.jpg&quot; alt=&quot;error message when redefining the 2d draft layer&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Since I just realized that other than in some other SE(-ish) sites it is not posible to post attachments you find the 3D model of my part here: &lt;a href=&quot;https://drive.google.com/open?id=1Kg2T7q0JwWIVn6-k4mO3M5ZtNvWAFk4N&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://drive.google.com/open?id=1Kg2T7q0JwWIVn6-k4mO3M5ZtNvWAFk4N&lt;/a&gt;&#xA;so you can tinker around with it as needed. If someone sports an older version of Inventor I can provide a version saved in an older format on demand.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Obviously answers like &quot;remodel it completely&quot; are not the type of advice I want to hear. If it is really not possible without going beyond reasonable labour ok then be it that way else not.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;EDIT: As pointed out by Optical Resonator it is not clear that I tried to go with an extrusion that is always orthogonal on the bent surface but this is what I tried to archive...&lt;/p&gt;&#xA;" OwnerUserId="26706" LastEditorUserId="26706" LastEditDate="2020-04-22T14:38:34.820" LastActivityDate="2020-04-22T14:38:34.820" Title="Moving extrusions on bent surface" Tags="|cad|autodesk-inventor|" AnswerCount="0" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="35312" PostTypeId="2" ParentId="35303" CreationDate="2020-04-22T14:21:37.273" Score="1" Body="&lt;p&gt;The closed loop system should have three poles, whose characteristic polynomial can be factored into&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;(s - a)(s^2 + 2\,\zeta\,\omega\,s + \omega^2),&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt; refers to the damping factor. The poles are the roots of that polynomial, which for the second factor can be shown to be &lt;span class=&quot;math-container&quot;&gt;$s=\omega(-\zeta \pm i \sqrt{1 - \zeta^2})$&lt;/span&gt;. It can be noted that for &lt;span class=&quot;math-container&quot;&gt;$|\zeta|\leq 1$&lt;/span&gt; those poles are each others complex conjugate with their magnitude only a function of &lt;span class=&quot;math-container&quot;&gt;$\omega$&lt;/span&gt; and their phase only a function of &lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The given damping factor therefore defines what the phase of complex conjugate poles should be. The corresponding value for &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; can thus be found by looking when the root locus crosses the line with the appropriate angle.&lt;/p&gt;&#xA;" OwnerUserId="1765" LastActivityDate="2020-04-22T14:21:37.273" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35313" PostTypeId="2" ParentId="35303" CreationDate="2020-04-22T15:51:33.453" Score="1" Body="&lt;p&gt;Adding some graphical information to what &lt;code&gt;fibonatic&lt;/code&gt; included in his own answer. The root locus of the trasfer function &lt;span class=&quot;math-container&quot;&gt;$G$&lt;/span&gt; with some random value for &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt;, let's say &lt;span class=&quot;math-container&quot;&gt;$K=1$&lt;/span&gt; (it is almost always good to start with the value &lt;span class=&quot;math-container&quot;&gt;$1$&lt;/span&gt; as starting point since it is like adding no controller to the closed loop and be able to see the behaviour of your system just by closing the loop), is the following where the poles are noted with an &lt;code&gt;x&lt;/code&gt; and the zeros with an &lt;code&gt;o&lt;/code&gt;:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/gQxsI.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/gQxsI.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Next step is to add the lines which correspond to the design requirement of the damping ratio &lt;span class=&quot;math-container&quot;&gt;$ζ = 0.5$&lt;/span&gt; and the following root locus graph is produced:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/t0UBf.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/t0UBf.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, your goal is to pick the gain &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; of your compensator which moves the poles of your closed loop system inside the white area which is defined under the two diagonal lines. This area generally defines a regio where the damping ratio of the system belongs the interval &lt;span class=&quot;math-container&quot;&gt;$ζ&amp;lt;0.5$&lt;/span&gt;. So if you want your damping ratio to be exactly &lt;span class=&quot;math-container&quot;&gt;$ζ=0.5$&lt;/span&gt; you need to place your poles exactly on the diagonal lines where the lines cross the root locus graph (the poles of the closed loop system are noted with a pink &lt;code&gt;o&lt;/code&gt;). Perfoming these modifications lead to the following root locus graph of the closed loop (see that the poles are now placed exactly on the two lines):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/DNbr1.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/DNbr1.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;After the design procedure, it is always useful to obtain the step response of your closed loop tuned system to see if it achieves the required behaviour. So, the step response for this system gives an overshoot of &lt;span class=&quot;math-container&quot;&gt;$$\%OS = 25.5\%$$&lt;/span&gt; which is not what the damping ratio of &lt;span class=&quot;math-container&quot;&gt;$0.5$&lt;/span&gt; explicitly dictates by working out the math of the equation: &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\%PO = 100\cdot e^{({\frac{-ζ\pi}{\sqrt{1-ζ^2}}})}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ζ=0.5 \rightarrow \%PO = 16.3\%$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This happens because you have a third pole which is not that far away of the dominant poles of the system and as a consequence this third pole influences the behaviour of the system (the further away the third pole is compared to the dominant poles the least it influences the behaviour of the system):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/r3WeK.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/r3WeK.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="18238" LastEditorUserId="18238" LastEditDate="2020-04-22T15:57:19.807" LastActivityDate="2020-04-22T15:57:19.807" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35314" PostTypeId="1" CreationDate="2020-04-22T16:16:40.160" Score="1" ViewCount="680" Body="&lt;p&gt;My house has several transfer grilles like this:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fJ7WV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fJ7WV.png&quot; alt=&quot;Transfer Grille&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;Except they have no baffles inside them, so light and sound come through into our bedrooms quite easily. I'd like to put in a sound maze like this to block light and sound:&lt;/p&gt;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/698lN.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/698lN.png&quot; alt=&quot;Sound maze&quot; /&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&lt;p&gt;My concern is that I will be restricting air flow enough to impact the HVAC system's effect on these rooms when the door is closed. Is that a valid concern, and if so, what considerations should one make to maximize air flow through a sound maze like this?&lt;/p&gt;&#xA;" OwnerUserId="26709" LastEditorUserId="-1" LastEditDate="2020-06-17T09:44:36.033" LastActivityDate="2023-08-10T10:52:52.927" Title="How can you maximize air flow through a sound maze in a HVAC transfer grille?" Tags="|airflow|hvac|sound-isolation|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35315" PostTypeId="2" ParentId="35309" CreationDate="2020-04-22T17:17:07.707" Score="0" Body="&lt;p&gt;The Report should have some language explaining that.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;There may have been rains elevating the aquifer level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If there are water wells near the site the change in pumping can affect the water level.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Or it could be a simple as a utility water line has broken near the site.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-22T17:17:07.707" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35316" PostTypeId="1" CreationDate="2020-04-22T17:55:24.150" Score="-1" ViewCount="20" Body="&lt;p&gt;I have a go kart project and this is what I'm trying to do. The kart also has 28-8 tires in the back.I have a 6.5hp motor that can run at max torque at 2500rpm. I need it to move 400 to 500 pounds. With the gearing of a 30 teeth sprocket on the shaft and a 28 teeth sprocket in the rear, &lt;/p&gt;&#xA;" OwnerUserId="26712" LastActivityDate="2020-04-22T20:46:11.907" Title="What is the maximum speed per hours or miles per hour i will be able to hit on a straight? Read description, please help" Tags="|automotive-engineering|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35317" PostTypeId="2" ParentId="35314" CreationDate="2020-04-22T18:10:38.640" Score="1" Body="&lt;p&gt;Yes, it is definitely going to impede the circulation of air. The fan of the HVAC blows the air very gently, so much so that for a large room one register maybe not enough.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For blocking the light you could try hanging a small drape 12 inches in front of the grill.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If the noise is really a nuisance the only effective solution is to connect your grill directly vis a flex duct to the HVAC system.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You can hide it in a cosmetic soffit.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-22T18:10:38.640" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35319" PostTypeId="2" ParentId="35309" CreationDate="2020-04-22T18:43:34.210" Score="3" Body="&lt;p&gt;The water table in some locations may vary during the day or over time. For this reason, the borehole drilling process includes waiting periods to see if there are any changes to the water table.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The depth at which the borehole initially finds the water table may be incorrect, an artifact of the drilling process itself. Therefore, if any changes to the water table level are observed, boring must be halted for a short while (10+ minutes) to let the table stabilize (and this must be included in the report).&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I've found a &lt;a href=&quot;https://www.rbkc.gov.uk/idoxWAM/doc/Other-1448588.pdf?extension=.pdf&amp;amp;id=1448588&amp;amp;location=volume2&amp;amp;contentType=application/pdf&amp;amp;pageCount=1&quot; rel=&quot;nofollow noreferrer&quot;&gt;nice report&lt;/a&gt; (pdf) stored by the UK government which just so happens to explain this on page 7. It also has nice examples of borehole records (i.e. page 8) which indicate differing water levels over time.&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;h2&gt;GROUND WATER&lt;/h2&gt;&#xA;  &#xA;  &lt;p&gt;The depth at which ground water was struck is entered on the Borehole Records. However, this observation may not indicate the true water level at that period. Due to the speed of boring and the relatively small diameter of the borehole, natural ground water may be present at a depth slightly higher than the water strike. Moreover, ground water levels are subject to variations caused by changes in the local drainage conditions and by seasonal effects. When a moderate inflow of water does take place, boring is suspended for at least 10 minutes to enable a more accurate short term water level to be achieved. An estimate of the rate of inflow is also given. This is a relative term and serves only as a guide to the probable flow of water into an excavation.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Further observations of the water level made during the progress of the borehole are shown including end of shift and overnight readings and the depth at which water was sealed off by the borehole casing, if applicable.&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;Whilst drilling through granular soils, it is usually necessary to introduce water into the borehole to permit their extraction. When additional water has been used a remark is made on the Borehole Record and the implications are discussed in the text.&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;What I'm going to say next is based on what I've found in Brazilian codes; I'm not 100% sure if this also applies elsewhere, but I believe it does.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Once the borehole has reached the requested depth, Brazilian codes (NBR 6484:2001 6.5.5-6) require the removal of water from the borehole in an attempt to lower the measured water table as much as possible. After waiting at least 12 hours, a new water table depth measurement is made and registered in the report.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I'm unsure if this was the exact procedure used in your report, but it seems clear that the two marks indicate the water table observed at two different moments. The report really should come with a legend with this information (especially since water table measurements should come with timestamps).&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-04-22T18:43:34.210" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35320" PostTypeId="2" ParentId="35316" CreationDate="2020-04-22T18:53:57.400" Score="0" Body="&lt;p&gt;I have seen similar carts professionally manufactured with similar engines go up to 40-45 mph.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the final analysis, your engine has to deal with two factors both friction type.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;One is the friction loss through your drivetrain.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The other is drag forces induced by the stream of air you go through.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the second one is easier to analyze.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;the well-known equation is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$Fd = 1/2C_d  ρ v^2 A $&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;ul&gt;&#xA;&lt;li&gt;&lt;p&gt;Fd is the drag force in Newtons.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;Cd is the drag factor which for open carts with the rider is roughly 1.2-1.4.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ \rho$&lt;/span&gt; is the air density which is approximate &lt;span class=&quot;math-container&quot;&gt;$1.2kgm^3$&lt;/span&gt;.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$ v$&lt;/span&gt; is the speed of the car m/s.&lt;/p&gt;&lt;/li&gt;&#xA;&lt;li&gt;&lt;p&gt;A is the total surface area of the rider on the cart&lt;span class=&quot;math-container&quot;&gt;$ m^2$&lt;/span&gt;. &lt;/p&gt;&lt;/li&gt;&#xA;&lt;/ul&gt;&#xA;&#xA;&lt;p&gt;you could estimate it by taking a photo from 40 feet away of the cart set in front of a scaled black and white checkered cardboard background, similar to the formula one racing flag.&#xA;The first part is all up to your design and quality of the parts you use, your experience in design.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastEditorUserId="4816" LastEditDate="2020-04-22T20:46:11.907" LastActivityDate="2020-04-22T20:46:11.907" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35321" PostTypeId="2" ParentId="35266" CreationDate="2020-04-22T19:05:13.280" Score="1" Body="&lt;p&gt;I apologize in advance for the length of the text but as you can understand these concepts can't be easily defined in few words. I will do some research about the second question and come back to update the answer.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;1.&lt;/strong&gt; The magnitude bode plot of a system indeed represents the ratio you mentio. However, there is not something like a general &quot;good&quot; requirement for the gain ratio of any given system. Each system represents a different procedure which includes different characteristics. The gain ratio may come from a design specification. Although, the output of any system is very often needed to track some reference either time dependent (like a trajectory or a simple sinusoidal) or constant (a step input of any amplitude). This means you want your gain to be equal to &lt;span class=&quot;math-container&quot;&gt;$1 \ (0dB)$&lt;/span&gt; ideally in order to have zero steady state error. Consider the following first order lowpass filters with transfer functions (they are randomly chosen for demonstration purposes):&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$L(s) = \frac{1}{s+15}$$&lt;/span&gt;&#xA;&lt;span class=&quot;math-container&quot;&gt;$$F(s) = \frac{15}{s+15}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The first one has DC gain &lt;span class=&quot;math-container&quot;&gt;$K_1 = 0.0667$&lt;/span&gt; which means there is much loss of the input. The second one has DC gain &lt;span class=&quot;math-container&quot;&gt;$K_2 = 1$&lt;/span&gt; which is the ideal in general and also what you want by a filter (at least very often). The bode plots of these two filters are the following:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/ftuNy.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/ftuNy.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Notice that while the pase plots are identical the magnitude plots have significant differences. The plot for &lt;span class=&quot;math-container&quot;&gt;$F$&lt;/span&gt; is the ideal case where the gain at low frequencies is &lt;span class=&quot;math-container&quot;&gt;$0dB$&lt;/span&gt; while the gain of &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; at low frequencies is very much undesirable. For a more complex example I place here the bode plot of a closed loop system which corresponds to my diploma thesis and I have designed. The closed loop system consists of the following parts:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$T(s) = \frac{b_0}{s^2+a_1s+a_2} \rightarrow Plant \ Dynamics$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$C(s) = K_p + K_d\frac{15s}{s+15} \rightarrow PD-Controller$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$F(s) = \frac{15}{s+15} \rightarrow Lowpass \ Filter$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/u4VFr.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/u4VFr.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;See that at low frequencies the gain is about (&lt;span class=&quot;math-container&quot;&gt;$0dB$&lt;/span&gt;) with too litle variation and approaches the ideal case. Generally the engineer is responsible for tuning the system in order to achieve certain behaviour, which in many cases dictates that the DC gain should be &lt;span class=&quot;math-container&quot;&gt;$0$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;3.&lt;/strong&gt; The purpose of control systems is to be able to control real mechanical systems and in general procedures which take place in the real world. In order to be able to do so, it is almost always great to have a mathematical model which represents the procedure you are trying to control and force certain behaviours. However, since the real world is not perfect you will never (and believe me never) be able to obtain a mathematical model which represents perfectly the real world procedure. First of all, it is almost always prefered to have a linear model (because we have a bunch of design tools for linear models and they are much simpler than non-linear) although the real world is full of non-linearities which means you only approach the real procedure. You will also need sensors to measure data that correspond to the system's behaviour and measurements have noise. Most importantly though is that there are always disturbances and the so called unmodeled dynamics (dynamics that can't be modeled but we know they exist) that influence your system's behaviour and have not been taken into account when designing your controllers and perfoming simulations since the simulations perfect. All these things (linearization, disturbances, noise, unmodeled dynamics) influence your system and add phase and gain to your system which you are not able to take, as said before, into account when designing your controllers. This means that the more gain and phase margin the more robust your systems to these variations is. For example, if you have a gain margin of &lt;span class=&quot;math-container&quot;&gt;$5dB$&lt;/span&gt; and there are some variations that may add up to &lt;span class=&quot;math-container&quot;&gt;$4.5dB$&lt;/span&gt; gain, your system will remain stable regardless of these variations but if these variations add up to &lt;span class=&quot;math-container&quot;&gt;$6dB$&lt;/span&gt; gain your system most probably will go unstable. The same goes for phase margin as well. That does not mean you make your system &quot;infinite&quot; robust because you pay a cost for more robustness but I will leave it to you to earch for robust control (very interesting subject). I suggest you also to look for the so-called disk margin which can tell you many great things about your system including phase and gain margin. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;strong&gt;4.&lt;/strong&gt; The issue regarding large controller gains is very much debatable. Generally, the larger the controller gains are the more aggresive and fast your system is. For example, if you have a closed loop system with third poles, you want your third pole to be like &lt;span class=&quot;math-container&quot;&gt;$10$&lt;/span&gt; times larger than the two dominant poles in order not to affect the behaviour which is dictated by the dominant poles. This means that the gains of the controller will surely be very large in order to achieve this. However, when controlling a real system you need actuators (dc motors, servo motors, stepper motors etc) which feed the control signal produced by your controller to the real system. These actuators always come with certain limitations. They can produce a certain amount of power, force or torque which are generated by the controller. The larger the controller gains the more torque, for example, is required by the actuator yet the actuator can only provide a certain amount of torque. You need to be very careful because if you do not take into consideration these limitations you can damage the actuator or even the system itself. You always need to saturate the control signal and the more saturation the control signal undergoes the worst your produced results regarding the system's behaviour may be because there are phenomena like integral windup that may come up. &lt;/p&gt;&#xA;" OwnerUserId="18238" LastActivityDate="2020-04-22T19:05:13.280" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35322" PostTypeId="2" ParentId="35299" CreationDate="2020-04-22T21:14:02.723" Score="1" Body="&lt;p&gt;The linear time history analysis of the response of the structure is calculated at small time intervals sequenced with appropriate seismic load, maybe even after the ground motion stops until the structure stops from vibrating. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The FEA's programs either use Duhamel integral or numerical methods to calculate stress and deformations. During the loading and unloading, the structure's stresses don't go beyond elastic linear levers.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The nonlinear time history analysis or Push-over method allows during the short sequence of time intervals again, for stresses and deformations of the structure to go beyond the elastic limit into the plastic limit with high angry absorbing deformations.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This analysis is mostly used to verify the viability of the structure in a very large earthquake to while deforming beyond a repairable state, still, give the occupants time to safely exit the structure.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-22T21:14:02.723" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35323" PostTypeId="2" ParentId="35303" CreationDate="2020-04-22T22:20:59.007" Score="1" Body="&lt;p&gt;An analytical solution.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The closed-loop system is &lt;span class=&quot;math-container&quot;&gt;$G(s)/(1+G(s))$&lt;/span&gt; and its poles are those of &lt;span class=&quot;math-container&quot;&gt;$1+G(s)=0$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In this case that is &lt;span class=&quot;math-container&quot;&gt;$$k \left(s^2+5 s+9\right)+(s+3) s^2=0 \ \ \ \  (1)$$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For general third-order system with a pair of complex dominant poles, the poles are the roots of &lt;span class=&quot;math-container&quot;&gt;$(\alpha +s) \left(s^2 + 2 \zeta  s \omega _n+\omega _n^2\right)=0$&lt;/span&gt;. Here &lt;span class=&quot;math-container&quot;&gt;$\alpha$&lt;/span&gt; is the real pole, &lt;span class=&quot;math-container&quot;&gt;$\zeta$&lt;/span&gt; is the damping factor, and &lt;span class=&quot;math-container&quot;&gt;$\omega _n$&lt;/span&gt; is the natural frequency. In this case &lt;span class=&quot;math-container&quot;&gt;$\zeta=0.5$&lt;/span&gt; and hence the equation becomes &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$(\alpha +s) \left(s^2 +  s \omega _n+\omega _n^2\right)=0  \ \ \ \  (2)$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;If we equate the coefficients of (1) and (2) we will have 3 equations and 3 unknowns of which &lt;span class=&quot;math-container&quot;&gt;$k$&lt;/span&gt; is one, and hence can be computed. &lt;/p&gt;&#xA;&#xA;&lt;pre&gt;&lt;code&gt;Expand[s^2*(s + 3) + k*(s^2 + 5*s + 9) - (s^2 + 2*(1/2)*wn *s + wn^2)*(s + a)]&#xA;eqns = Thread[CoefficientList[%, s] == 0]&#xA;sols = Solve[eqns, {a, k, wn}, Reals]&#xA;k /. sols&#xA;&lt;/code&gt;&lt;/pre&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;9 k + 5 k s + 3 s^2 - a s^2 + k s^2 - a s wn - s^2 wn - a wn^2 - s wn^2&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;{9 k - a wn^2 == 0, 5 k - a wn - wn^2 == 0, 3 - a + k - wn == 0}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;{{a -&gt; 3, k -&gt; 0, wn -&gt; 0}, {a -&gt; 9/2, k -&gt; 9/2, wn -&gt; 3}}&lt;/p&gt;&#xA;  &#xA;  &lt;p&gt;{0, 9/2}&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$k=0$&lt;/span&gt; is an artifact of the calculations. The answer turns out to be &lt;span class=&quot;math-container&quot;&gt;$k=\frac{9}{2}$&lt;/span&gt;.&lt;/p&gt;&#xA;" OwnerUserId="1627" LastActivityDate="2020-04-22T22:20:59.007" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35324" PostTypeId="1" CreationDate="2020-04-23T02:58:29.747" Score="1" ViewCount="56" Body="&lt;p&gt;It many be due to costs or that the brake discs would have to be made out of aluminium/copper for the electromagnetic braking to be effective or there may not be a significant reduction in braking distance. I'm just not sure if these are valid reasons or not.&lt;/p&gt;&#xA;" OwnerUserId="19007" LastActivityDate="2020-04-23T04:11:29.617" Title="Why don't car manufacturer's use a combination of disc brakes and electeomagnetic brakes to reduce braking distances?" Tags="|materials|automotive-engineering|electromagnetism|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35325" PostTypeId="2" ParentId="35324" CreationDate="2020-04-23T04:11:29.617" Score="1" Body="&lt;p&gt;The existing brake systems consisting of hydraulic pumps and calipers and disks have proved to be, reliable, efficient, low maintenance, and fail-safe.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;They are powerful enough that from the moment of deployment they impart maximum practical deceleration right up to the level of the limit stopping force short of skidding. There is even a need for a governor control system ( ABS) to rain in on their performance not to allow the potential lock-up of the wheels.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The possible bottleneck is in the tire's friction and heat resistance capabilities.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Adding any other system will multiply the complexity, increase the chance of failure, and add to the cost of the car and its maintenance.&lt;/p&gt;&#xA;" OwnerUserId="4816" LastActivityDate="2020-04-23T04:11:29.617" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35326" PostTypeId="2" ParentId="23077" CreationDate="2020-04-23T06:09:57.237" Score="2" Body="&lt;p&gt;Between disc 1 &amp;amp; 2, there is a reaction force on disc 1 which is not going to act on the edge rather at the distance of 2nd disc edges. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;This means you need to have the model of 1st disc bending developed for 2 cases,&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Pressure drop&lt;/li&gt;&#xA;&lt;li&gt;load acting at the radius of 2nd disc&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For the 2nd disc, you have 2 cases&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;reaction load from the 1st disc at the edge of second disc&lt;/li&gt;&#xA;&lt;li&gt;reaction load from the 3rd disc acting at the radius of 3rd disc&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;For each disc, the effective deflection is calculating by summing up both cases... Keep in mind the the reaction from disc on top produces negative deflection...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep building the equations and there you will see you end up with system of equations which can be solved by any mathematical methods, gauss jordan or something...&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Finally you get the deflection @ disc 1 on the edge which is the requirement for your analysis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I hope this helps.&lt;/p&gt;&#xA;" OwnerUserId="26717" LastActivityDate="2020-04-23T06:09:57.237" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35327" PostTypeId="1" CreationDate="2020-04-23T07:35:26.123" Score="2" ViewCount="86" Body="&lt;p&gt;I was wondering if there is an intuitive interpretation for the Jakes spectrum? If I look at it in the time domain, which is basically the Bessel function zeroth order, it gives me a reading, how long the channel impulse response stays coherent i.e. phase and amplitude don't vary significantly. Or how long it takes, till the channel is decorrelated however if I look at the Jakes spectrum, the only thing which I can 'see' is how large the Doppler spread is but I can't get my head around, how should I interpret the amplitude (the y-axis). Does it mean, that the frequency components which are closer to the maximum shift are more prominent? And if so, is it just because of the Fourier transformation and there is no way to intuitively understand it?&lt;/p&gt;&#xA;" OwnerUserId="26719" LastEditorUserId="26719" LastEditDate="2020-05-06T08:03:49.317" LastActivityDate="2022-08-08T16:49:17.777" Title="Intuitive interpretation of the U-shaped Jakes Spectrum" Tags="|wireless-communication|" AnswerCount="0" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35328" PostTypeId="1" CreationDate="2020-04-23T10:51:41.303" Score="-1" ViewCount="47" Body="&lt;p&gt;For structure earthquake analysis methods, do these all below mean the same thing, just ways of calling it differently?&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Linear Response History Analysis&lt;/li&gt;&#xA;&lt;li&gt;Linear Elastic Time History Analysis&lt;/li&gt;&#xA;&lt;li&gt;Linear Time History Analysis&lt;/li&gt;&#xA;&lt;li&gt;Response Spectrum Analysis&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;&#xA;&lt;p&gt;Or if there are differences, which one(S) is different?&lt;/p&gt;&#xA;" OwnerUserId="24283" LastEditorUserId="1832" LastEditDate="2020-04-24T17:16:32.233" LastActivityDate="2021-01-21T08:06:20.553" Title="names of some earthquake analysis methods" Tags="|structural-engineering|structural-analysis|" AnswerCount="1" CommentCount="4" ContentLicense="CC BY-SA 4.0" />
  <row Id="35329" PostTypeId="1" CreationDate="2020-04-23T10:53:01.353" Score="-1" ViewCount="353" Body="&lt;p&gt;Are there electric motors that can produce a lot of power at low RPM? Why couldn't a motor produce as much power as needed if you give it enough voltage? Why would an electric motor be able to produce more power at a higher RPM?&lt;/p&gt;&#xA;" OwnerUserId="26722" LastActivityDate="2020-04-26T03:02:34.230" Title="What kinds of electric motors can produce a lot of power at low RPM?" Tags="|motors|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35330" PostTypeId="1" CreationDate="2020-04-23T13:19:27.393" Score="0" ViewCount="4198" Body="&lt;p&gt;From the first law of thermodynamics:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\Delta U = Q_{12} - W_{12}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;where &lt;span class=&quot;math-container&quot;&gt;$\Delta U = m(u_2-u_1)$&lt;/span&gt;, &lt;span class=&quot;math-container&quot;&gt;$W = p(V_2 - V_1)$&lt;/span&gt; with constant pressure, and &lt;span class=&quot;math-container&quot;&gt;$Q = m(h_2-h_1)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;In the energy rate form of the first law simplification for a turbine (or compressor): &lt;span class=&quot;math-container&quot;&gt;$\left(\dfrac{\partial E}{\partial t}=0=-W'+m'[h_1-h_2]\right)$&lt;/span&gt;  &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$W' = m'(h_1-h_2)$&lt;/span&gt; while assuming that there's no heat transfer &lt;span class=&quot;math-container&quot;&gt;$\left(Q'\right)$&lt;/span&gt; and change in potential and kinetic energy is zero.  (&lt;span class=&quot;math-container&quot;&gt;$W'$&lt;/span&gt; is power, or &lt;span class=&quot;math-container&quot;&gt;$\dfrac{\partial W}{\partial t}$&lt;/span&gt;)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I think I must be missing something obvious here, but it seems inconsistent that &lt;span class=&quot;math-container&quot;&gt;$Q' = 0 = m'(h_1-h_2)$&lt;/span&gt; while using &lt;span class=&quot;math-container&quot;&gt;$(h_2-h_1)$&lt;/span&gt; in the equation.  Shouldn't &lt;span class=&quot;math-container&quot;&gt;$h_2 = h_1$&lt;/span&gt; in both equations?&lt;/p&gt;&#xA;" OwnerUserId="16122" LastEditorUserId="16122" LastEditDate="2020-04-23T16:46:09.117" LastActivityDate="2020-04-23T16:46:09.117" Title="When does Q = m(h2 - h1) hold true?" Tags="|mechanical-engineering|thermodynamics|heat-transfer|" AnswerCount="0" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35331" PostTypeId="2" ParentId="14352" CreationDate="2020-04-23T14:00:17.737" Score="0" Body="&lt;p&gt;When you said moment in your question, I got confused. You need to say how to distribute shear. You are not distributing any moment. You are distributing shear forces to shear walls, which would cause &quot;torsional moment&quot; or &quot;torsion&quot; with respect to the geometric center of the floor. Even if you were talking about moment, you are talking about torsional moment here, which you aren't. You are talking about distributing shear forces. &#xA;Other than that yes the shear walls will resist the shear with respect to their stiffnesses. Remember, stiffness attracts force. &lt;/p&gt;&#xA;" OwnerUserId="24283" LastActivityDate="2020-04-23T14:00:17.737" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35332" PostTypeId="1" CreationDate="2020-04-23T14:12:27.197" Score="-1" ViewCount="458" Body="&lt;p&gt;I am trying to make DIY Bubble CPAP that can put a constant pressure on patient.&#xA;I am using an Aquarium Oxygen pump as air source. &#xA;The motor says 38L/Min in the spec. And by googling I found that an average adult can takes up to 7L/min air while resting.&#xA;But when I used it, I felt like I am getting very low air flow. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Can anyone help, what am I missing?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Incase you there is any suggestion please drop it too. &#xA;Thank you.&lt;/p&gt;&#xA;" OwnerUserId="26724" LastActivityDate="2020-04-23T15:16:24.540" Title="What is the amount airflow needed from Aquarium Air pump per minute to make a DIY CPAP?" Tags="|pumps|airflow|biomedical-engineering|medical-devices|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="35334" PostTypeId="2" ParentId="35332" CreationDate="2020-04-23T15:16:24.540" Score="1" Body="&lt;p&gt;A medical device known as an &lt;a href=&quot;https://www.drugs.com/cg/how-to-use-an-incentive-spirometer.html&quot; rel=&quot;nofollow noreferrer&quot;&gt;incentive inspirometer&lt;/a&gt; measures volume of air inhaled. Twenty years ago, I could pull five liters (was serious cyclist) but now I'm down to four liters per breath. Sixteen breaths per minute means sixty-four liters per minute.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;At the five liter value, that's eighty liters per minute, in resting state.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/HpNIr.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/HpNIr.png&quot; alt=&quot;incentive inspirometer image&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="13297" LastActivityDate="2020-04-23T15:16:24.540" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35342" PostTypeId="2" ParentId="33792" CreationDate="2020-04-23T23:51:20.873" Score="0" Body="&lt;p&gt;I seriously doubt anyone else has thought like that.&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;What do the symbols mean? ^ means 'I want to go up', and v means 'I want to go down'. &lt;/li&gt;&#xA;&lt;li&gt;What if the lift car is below me and I want to go down, would I press the ^? No, you don't always get indications of where the lift car is currently located, so if you don't know where it is, how could you input a request for it to move towards you based on where it is?&lt;/li&gt;&#xA;&lt;li&gt;Why do both buttons stay lit? Because there could be 2 (or more) people waiting for a lift car, and each/some want to go in different directions...&lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="26735" LastEditorUserId="1832" LastEditDate="2020-04-24T01:12:15.520" LastActivityDate="2020-04-24T01:12:15.520" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35343" PostTypeId="1" CreationDate="2020-04-24T01:52:01.720" Score="1" ViewCount="89" Body="&lt;p&gt;I am trying using pump to promote circulation of latex solution in a latex dipping tank. Is there any pump that suits this purpose? Also, is there any better ideas to promote circulation of the latex in the latex tank without handling the solution vigorously.&#xA;The explanation of main process can be found in this link.&#xA;&lt;a href=&quot;https://www.youtube.com/watch?v=oTIADlYOkNc&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.youtube.com/watch?v=oTIADlYOkNc&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26738" LastActivityDate="2020-04-24T10:18:41.037" Title="What pump to use to pump latex?" Tags="|pumps|" AnswerCount="1" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35345" PostTypeId="2" ParentId="35329" CreationDate="2020-04-24T05:31:15.660" Score="0" Body="&lt;p&gt;Motors that produce lots of power at low RPM are big and heavy. The optimum solution is to instead use a small, high-speed motor connected to a reduction gearset- which is called a &lt;em&gt;gearmotor&lt;/em&gt;. &lt;/p&gt;&#xA;" OwnerUserId="13363" LastActivityDate="2020-04-24T05:31:15.660" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35347" PostTypeId="2" ParentId="35343" CreationDate="2020-04-24T10:18:41.037" Score="1" Body="&lt;p&gt;You could consider using a flat plate stirrer or a helix depending on what you wish to achieve.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Controlling the speed of rotation will be useful as well.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-24T10:18:41.037" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35348" PostTypeId="2" ParentId="32771" CreationDate="2020-04-24T11:16:06.033" Score="4" Body="&lt;p&gt;I'm going to answer this for a LWR, since I guess that's what you're thinking of.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;First of all, the premise of the question is not quite correct. The power distribution of an LWR is generally sinusoidal, with a peak towards the middle of the fuel. This is true with or without control rods inserted. The most common power reactor design in operation--a conventional PWRs operating in full-power baseload operation--does not have control rods deeply inserted into the core for any significant amount of time. This shape is driven by a few factors: first, fast neutrons tend to 'leak' out of the core without thermalizing more towards the extremities. Additionally, because the top of the fuel has a higher moderator temperature (or maybe even some voids!), this also leads to less thermalization there. So, because of the shape of the power distribution, the tops and bottoms of the fuel assembly have far less power (and therefore less relative depletion) than the 'middle' of the fuel.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;When control rods are more heavily use for power distribution control (e.g., in a BWR, or some PWR designs), this creates a secondary 'rod shadowing' effect that further shifts the power distribution away from one axial extremity of the core. Also, there are different groups of control rods, and not every fuel assembly location has a control rod (if you google, you should be able to find some patterns online for conventional reactor types.) Some of these groups are used only for shutdown, and some are meant for insertion into the core when it is operating for power distribution control. These control rods used for power distribution control are not the same composition as rods used for shutdown--power distribution control typically uses control rods that are much weaker neutron absorbers than shutdown rods. So while they cause local reductions in the neutron flux when inserted, there is still plenty of fission going on in those areas... just less of it than there would be if the control rods were withdrawn. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;(Side note: BWR control rods come from the bottom of the core, and PWR control rods from the top.) &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, to answer your actual question: there are a few ways the fuel is built to economize around having less fission at the top and the bottom of the core, but all are basically built around reducing the amount of reactivity in those regions. You can do some or all of these things just for fuel in the top and bottom of a fuel assembly:&lt;/p&gt;&#xA;&#xA;&lt;ol&gt;&#xA;&lt;li&gt;Lower the enrichment (right down to using natural U)&lt;/li&gt;&#xA;&lt;li&gt;Use annular fuel pellets &lt;/li&gt;&#xA;&lt;li&gt;Not using any neutron poison in the fuel (gadolina etc.) &lt;/li&gt;&#xA;&lt;/ol&gt;&#xA;" OwnerUserId="26744" LastActivityDate="2020-04-24T11:16:06.033" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35350" PostTypeId="1" CreationDate="2020-04-24T14:16:54.583" Score="-1" ViewCount="508" Body="&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/KxziR.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/KxziR.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Hi there, how do you find the moment of area about the x and y axes (at centroid)? &lt;/p&gt;&#xA;&#xA;&lt;p&gt;Dimensions are given as in the picture and thickness &quot;c&quot; is constant throughout.&lt;/p&gt;&#xA;" OwnerUserId="26746" LastActivityDate="2020-04-24T15:36:26.627" Title="Thin Wall Moment of Area" Tags="|solid-mechanics|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35351" PostTypeId="1" CreationDate="2020-04-24T14:33:59.010" Score="2" ViewCount="147" Body="&lt;p&gt;In preparation for a Fluid Mechanics experiment, I am looking for a simple commercial ready made&#xA;apparatus to generate a linear velocity trend to prove a new theory:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://aip.scitation.org/doi/abs/10.1063/1.4964761&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://aip.scitation.org/doi/abs/10.1063/1.4964761&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The load is around 10kg starting from rest to reach linearly say a 0.5m/sec within a distance of say 0.5m..&#xA;Any hints ? :)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;72nd Annual Meeting of the APS Division of Fluid Dynamics&#xA;Volume 64, Number 13&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Leading Edge theory: a new insight into the laminar Boundary Layer by Mohammad Gabr&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Abstract :&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The flow properties at the leading edge of a flat plate represent a singularity to the Blasius laminar boundary layer equations; by applying the diffusion equation where the velocity of a moving flat plate in a stationary fluid is diffused to the far field, the leading edge velocity profiles are studied. Experimental observations as well as the theoretical analysis show an exact Gaussian distribution curve as the original starting profile of the laminar flow.To conclude; the main key results are as follows: [1] The velocity profiles at the leading edge of a flat plate are Gaussian Curves that grow in space and time; whereas the Blasius velocity profile is a part of the general Gaussian curves solution. [2]A new method to calculate the friction drag is successfully tested, based on the displacement area of the leading edge velocity profile. [3] In order to obtain the final physical proof of the new theory it is recommended to carry out experiments using an ultra-thin flat plate moving in a stationary fluid and measuring the velocity profiles at the leading edge with different measurement techniques.&lt;/p&gt;&#xA;" OwnerUserId="26747" LastEditorUserId="26747" LastEditDate="2020-05-02T01:11:52.050" LastActivityDate="2024-02-14T07:00:45.077" Title="Generating Linear Velocity Movement?" Tags="|mechanical-engineering|fluid-mechanics|control-engineering|applied-mechanics|" AnswerCount="2" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="35353" PostTypeId="2" ParentId="35350" CreationDate="2020-04-24T15:36:26.627" Score="1" Body="&lt;p&gt;I'm here going to assume that by moment of area you mean the second moment of area (aka moment of inertia). That being said, the argument I'll be making is valid for the first moment as well.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The planar second moment of area around an axis (say, the x-axis) is calculated by:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$I_x = \iint y^2 \text{d}x\text{d}y$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You'll notice this means the moment of inertia around the x-axis is a measure of the square of the distance from the origin along the y-axis. So, what's the impact of the shape being angled? Well, none. The bottom point is to the left of the centroid, but the moment of inertia around the x-axis doesn't care.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, when calculating the moment of inertia around the x-axis, you can pretend you're dealing with a rectangular beam, where the classic equation &lt;span class=&quot;math-container&quot;&gt;$I_x = bh^3/12$&lt;/span&gt; applies. The same applies when calculating the moment of inertia around the y-axis.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Now, there is a nuance: the beam's thickness &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; is also inclined. So the thickness to be used when calculating &lt;span class=&quot;math-container&quot;&gt;$I_x$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$I_y$&lt;/span&gt; is not &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt;, but the horizontal and vertical projections of &lt;span class=&quot;math-container&quot;&gt;$c$&lt;/span&gt; according to the beam's inclination. (This projection fails if the beam were horizontal, but is otherwise accurate)&lt;/p&gt;&#xA;&#xA;&lt;p&gt;By calculating the beam as rectangular, you are effectively cutting off tiny triangles from the ends of the beam, but those are certainly negligible.&lt;/p&gt;&#xA;" OwnerUserId="1832" LastActivityDate="2020-04-24T15:36:26.627" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35357" PostTypeId="1" CreationDate="2020-04-24T19:26:17.500" Score="1" ViewCount="102" Body="&lt;p&gt;Which kinds of linear actuators offer the lowest levels of noise and vibration?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am looking for an actuator with these properties that can push a load vertically on the cm scale. Velocity, accuracy and max force are secondary. I'd be very glad for a hint to the specific category of actuators that might meet these requirements.&lt;/p&gt;&#xA;" OwnerUserId="23115" LastActivityDate="2020-04-26T01:49:07.883" Title="Linear actuator with low vibration and low noise" Tags="|vibration|actuator|linear-motion|linear-motors|noise|" AnswerCount="1" CommentCount="2" ContentLicense="CC BY-SA 4.0" />
  <row Id="35359" PostTypeId="1" CreationDate="2020-04-24T21:18:52.397" Score="1" ViewCount="75" Body="&lt;p&gt;A similar question was answered &lt;a href=&quot;https://physics.stackexchange.com/questions/130638/air-flow-with-line-fans?newreg=9dc5865a5f94497b8f3265cde80cd7d1&quot;&gt;here&lt;/a&gt;. And while that did help me understand airflow, I have an even more specific question with a little different context.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So what I'm dealing with here is pulling fresh air in a vacuum duct through the front passenger door on a 1984 Volkswagen Vanagon. I have 2 sets of 3 vertically stacked 40mm computer fans, making 6 fans in total. You can read more about (and critique(please)) my ventilation system idea &lt;a href=&quot;https://www.thesamba.com/vw/forum/viewtopic.php?p=9334028#9334028&quot; rel=&quot;nofollow noreferrer&quot;&gt;here&lt;/a&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/yk5YQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/yk5YQ.png&quot; alt=&quot;start of the duct&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/d5ppQ.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/d5ppQ.png&quot; alt=&quot;set of stacked fans&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/fZJEP.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/fZJEP.png&quot; alt=&quot;airflow dilemma&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;As you can see, when the door is closed, air travels through two 90 degree angles. So the question is.... Where is the optimal placement of my fans. Should I put them both just before the first angle to give it the 'OOMPH' it needs? Or should I space them out, one at the beginning of the duct and another just before the angle? Or does it make any difference at all?&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I tested it with just a single set of my stacked fans and I was able to feel a little bit of flow coming out of the vents. Can I expect to double that with my other set added?&lt;/p&gt;&#xA;" OwnerUserId="26753" LastActivityDate="2020-04-25T18:04:23.040" Title="Best Fan Positions for Airflow in a Duct w/ Two 90° Angles" Tags="|pressure|airflow|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35360" PostTypeId="1" CreationDate="2020-04-24T22:12:27.590" Score="1" ViewCount="105" Body="&lt;p&gt;I've seen a number of opinions on conductive gels to ensure the best conductivity between battery terminals and battery cables. I'm interested in knowing an expert's opinion on the matter. I'm particularly interested in products that work well in harsh marine environments.&lt;/p&gt;&#xA;" OwnerUserId="26755" LastActivityDate="2020-04-24T23:10:05.257" Title="What is the best practice for ensuring the best possible connection between a battery cable and a battery terminal" Tags="|electrical|battery|marine-engineering|" AnswerCount="1" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35361" PostTypeId="1" CreationDate="2020-04-24T22:56:59.633" Score="0" ViewCount="246" Body="&lt;p&gt;I have a pressure/temperature sensor for a vehicle, which I decided to tear apart to see exactly how the sensor works.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;This sensor as designed by the manufacturer it is meant to be immersed in oil to detect fluid temperature. The pressure sensor side of it seems pretty straight forward in that there is a hole in the sensor physical body and a PCB.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am stuck on how the temperature sensor works in this specific sensor, if you look at the picture I attached the entire PCB is covered in a glue like substance I am guessing to insulate it from the oil. The temperature sensor is covered in some sort of cement like substance and is bonded only to a metal rod.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am a little loss as to how the temperature sensor would be able to detect change in the fluid temperature if there is all the insulation glue on the board isolating the sensor from any fluid. My only guess is the metal rod acts to conduct heat towards the temperature sensor on the PCB. So to prove this I have tried measuring the temperature signal via a pin on the sensor and oscilloscope. I've immersed the sensor in oil, raised the heat of the oil and applied the proper 5 V to the sensor. For some reason the temperature signal will stay only at 1.3 V regardless of oil temperature.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I am beginning to think maybe the sensor can only read the temperature with a pressurized fluid, and not just sitting in fluid. Any input is appreciated thanks&#xA;&lt;a href=&quot;https://i.stack.imgur.com/1Wwn4.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/1Wwn4.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/xwgUR.jpg&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/xwgUR.jpg&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="26756" LastEditorUserId="26756" LastEditDate="2020-04-24T23:36:42.867" LastActivityDate="2024-02-26T19:04:30.523" Title="Need help with reverse engineering a sensor" Tags="|mechanical-engineering|electrical-engineering|pressure|sensors|temperature|" AnswerCount="2" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35362" PostTypeId="2" ParentId="35360" CreationDate="2020-04-24T23:10:05.257" Score="4" Body="&lt;p&gt;We really need more info on the batteries and type of connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will the connection be permanent? In that case soldering or welding are the best choices. If you're connecting to wire, you'll need good strain relief. Any repetitive motion at the solder joint / weld will cause failure eventually. In a vehicle the battery may have a tendency to jostle around. Make sure that this will not cause flexture of the wire to battery connection.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For permanent connections you can coat the resulting joint (among other things) with conformal coating, which will help inhibit corrosion.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;For removable connections, corrosion of the contacts will be your biggest enemy. Use gold contacts if possible. There are solutions such as &quot;deoxit&quot; which will inhibit corrosion of the contacts, but that won't work as well or last as long as gold plating.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Keep the battery compartment as dry as possible obviously. Consider using a water proof box with a desiccant packet, or a heat source.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Chemicals like bleach will destroy contacts faster than anything in my experience, so be sure to keep them away from your connectors.&lt;/p&gt;&#xA;" OwnerUserId="20490" LastActivityDate="2020-04-24T23:10:05.257" CommentCount="1" ContentLicense="CC BY-SA 4.0" />
  <row Id="35363" PostTypeId="2" ParentId="35361" CreationDate="2020-04-25T00:17:30.033" Score="0" Body="&lt;p&gt;The sensing element of the transmission pressure sensor consists of a membrane that is deformed by the applied pressure.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The deformation is converted into a voltage proportional to the pressure. An electronic evaluation circuit then amplifies and digitizes the voltage.&lt;/p&gt;&#xA;" OwnerUserId="26595" LastEditorUserId="26595" LastEditDate="2020-04-25T14:07:53.433" LastActivityDate="2020-04-25T14:07:53.433" CommentCount="3" ContentLicense="CC BY-SA 4.0" />
  <row Id="35367" PostTypeId="1" CreationDate="2020-04-25T14:35:38.037" Score="1" ViewCount="61" Body="&lt;p&gt;My house was built around 1920, and has field stone foundation. The joists seem to be all 2&quot;x8&quot;. I want to put a 120 gallon(as measured in the United States) fish tank in my home office, which will be roughly 1000lbs, plus a stand, which would be maybe another 100lbs, and the footprint of this  tank is 48&quot;x24&quot;x24&quot;. I  want to reinforce the floor under the tank  in the basement. I plan to place the tank along the exterior wall, about 12&quot; away from the wall, over the first joist. To strengthen the floor, I want to use bridging to connect the joists. I will do this at every third of the  joist, for about  3-4 joists below the tank. I also plan to use two  4&quot;x4&quot;s, that are beneath the joists laying perpendicular to them, and are supported each by a jack post which will sit on the basement slab. I plan to use one jackpost at each third of the tank. Should this be a safe way to support the tank?&lt;/p&gt;&#xA;" OwnerUserId="26766" LastEditorUserId="26766" LastEditDate="2020-04-25T15:46:16.200" LastActivityDate="2020-04-25T15:46:16.200" Title="How to strengthen the floor to support a 120gallon fish tank" Tags="|structural-engineering|" AnswerCount="0" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="35370" PostTypeId="1" AcceptedAnswerId="35373" CreationDate="2020-04-25T17:08:50.427" Score="1" ViewCount="26" Body="&lt;p&gt;I'm reading company training materials and I don't understand a few terms.  I'm hoping that some of you might be able to decode.  The company receives product and delivers it to work sites.  I believe these are forklifts or ?order pickers?.  An excerpt from the slides:&lt;/p&gt;&#xA;&#xA;&lt;blockquote&gt;&#xA;  &lt;p&gt;&quot;if your branch has a personal man up device like the Little Joey or JLF&quot;...&lt;/p&gt;&#xA;&lt;/blockquote&gt;&#xA;&#xA;&lt;p&gt;Can you help me to understand the terms used in that excerpt?  I assume Little Joey is referring to &lt;a href=&quot;https://bigjoeforklifts.com/pages/products&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://bigjoeforklifts.com/pages/products&lt;/a&gt;  but I couldn't find &quot;JLF&quot;; perhaps a trade name.&lt;/p&gt;&#xA;" OwnerUserId="26769" LastActivityDate="2020-04-26T16:55:03.473" Title="Are these &quot;parts pickers&quot;? &quot;JLF&quot; and &quot;Little Joey&quot;" Tags="|manufacturing-engineering|" AnswerCount="2" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35371" PostTypeId="1" CreationDate="2020-04-25T17:51:39.587" Score="2" ViewCount="344" Body="&lt;p&gt;I have this 1-DOF helicopter simulator.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;a href=&quot;https://i.stack.imgur.com/CX0LV.png&quot; rel=&quot;nofollow noreferrer&quot;&gt;&lt;img src=&quot;https://i.stack.imgur.com/CX0LV.png&quot; alt=&quot;enter image description here&quot;&gt;&lt;/a&gt; &lt;/p&gt;&#xA;&#xA;&lt;p&gt;The motor's armature resistance is &lt;span class=&quot;math-container&quot;&gt;$R$&lt;/span&gt;, inductance &lt;span class=&quot;math-container&quot;&gt;$L$&lt;/span&gt; and motor constant &lt;span class=&quot;math-container&quot;&gt;$K$&lt;/span&gt; and reaction torque on the motor shaft is &lt;span class=&quot;math-container&quot;&gt;$τ$&lt;/span&gt;. &lt;span class=&quot;math-container&quot;&gt;$T$&lt;/span&gt; is equal to &lt;span class=&quot;math-container&quot;&gt;$k_1w_2^2$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\tau$&lt;/span&gt; is equal to &lt;span class=&quot;math-container&quot;&gt;$k_2w_2^2$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$w_1 = \dot{\varphi}_1$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$w_2=\dot{\theta}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Τhe differential equation for the current is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial I}{\partial t}=\frac{E - RI - K_1w_2}{L}$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which &lt;span class=&quot;math-container&quot;&gt;$E$&lt;/span&gt; is the input armature voltage and the equation for the torque is&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$\frac{\partial w_2}{\partial t}=\frac{KI - K_2w_2^2}{J} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;in which &lt;span class=&quot;math-container&quot;&gt;$k_2w_2^2$&lt;/span&gt; is the load torque and finally for the last equation I used lagrange method to derive equation of motion of mass &lt;span class=&quot;math-container&quot;&gt;$m$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$M$&lt;/span&gt;. &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ T = \frac{1}{2}mb^2\dot{\varphi}^2 + \frac{1}{2}M\alpha^2\dot{\varphi}^2 + \frac{1}{2}Jw_2^2 $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ U = mgb\sin\varphi - Μg\alpha\sin\varphi $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;which finally gives &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ mb^2\dot{w} + M\alpha^2\dot{w}_1 + bmg\cos\varphi - Mg\alpha\cos\varphi = -Tb = K_1w_2^2b $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{\partial w_1}{\partial t} = \frac{-k_1w_1^2 + Mg\alpha\cos\varphi - mgb\cos\varphi}{M\alpha^2 + mb^2} $$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;I wanna know whether everything here is right and how to linearize it around &lt;span class=&quot;math-container&quot;&gt;$\varphi = 0$&lt;/span&gt; cause when I try to do that &lt;span class=&quot;math-container&quot;&gt;$\dfrac{\partial w_1}{\partial t}$&lt;/span&gt; becomes a constant which is:&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$ \frac{\partial w_1}{\partial t} = \frac{Mg\alpha - mgb}{M\alpha^2 + mb^2} $$&lt;/span&gt;&lt;/p&gt;&#xA;" OwnerUserId="23679" LastEditorUserId="6790" LastEditDate="2020-05-01T14:09:28.000" LastActivityDate="2020-05-01T14:09:28.000" Title="linearizing a non linear state space equation" Tags="|mechanical-engineering|electrical-engineering|aerospace-engineering|modeling|mathematics|" AnswerCount="1" CommentCount="5" ContentLicense="CC BY-SA 4.0" />
  <row Id="35372" PostTypeId="2" ParentId="35359" CreationDate="2020-04-25T18:04:23.040" Score="1" Body="&lt;p&gt;One set of fans creates a pressure difference, so two will create double (assuming identical sets of fans), well close enough to double.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Will it be better to put one set close to the other? probably, but noise might then be an issue.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;So, I may put one at the start, as you show, then the second halfway - depends on spacing etc but testing will give the best answer for the situation.&lt;/p&gt;&#xA;" OwnerUserId="10902" LastActivityDate="2020-04-25T18:04:23.040" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35373" PostTypeId="2" ParentId="35370" CreationDate="2020-04-25T18:34:42.340" Score="2" Body="&lt;p&gt;The company name is &quot;Big Joe&quot;, and &quot;Little Joey&quot; is the brand name of their smallest range of products.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;JLF is the brand name of a range of products similar to this: &lt;a href=&quot;https://www.jung-hebetechnik.de/en/product-details/jlf-5/&quot; rel=&quot;nofollow noreferrer&quot;&gt;https://www.jung-hebetechnik.de/en/product-details/jlf-5/&lt;/a&gt;&lt;/p&gt;&#xA;" OwnerUserId="683" LastActivityDate="2020-04-25T18:34:42.340" CommentCount="0" ContentLicense="CC BY-SA 4.0" />
  <row Id="35374" PostTypeId="2" ParentId="35371" CreationDate="2020-04-25T20:00:12.060" Score="3" Body="&lt;p&gt;The Lagrangian for your 1D pitch system is &lt;span class=&quot;math-container&quot;&gt;$\mathcal{L} := T - U$&lt;/span&gt;, which is &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\mathcal{L} = \frac{1}{2}(mb^2 + Ma^2)\dot{\phi}^2 + \frac{1}{2}J\dot{\theta}^2 + (Ma - mb)g\sin\phi&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;The Euler-Lagrange Equations are&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\frac{d}{dt}\bigg(\frac{\partial\mathcal{L}}{\partial \dot{q}_k}\bigg) + \frac{\partial\mathcal{L}}{\partial q_k} = Q_k&#xA;$$&lt;/span&gt;&#xA;This gives the EOM&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;(mb^2 + Ma^2)\ddot{\phi} + (Ma - mb)g\cos\phi = -Tb = -k_1\dot{\theta}^2b&#xA;$$&lt;/span&gt;&#xA;Under a small angle approximation, &lt;span class=&quot;math-container&quot;&gt;$\cos\phi \approx 1$&lt;/span&gt;, so this leaves &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;(mb^2 + Ma^2)\ddot{\phi} + (Ma - mb)g = -k_1b\dot{\theta}^2,&#xA;$$&lt;/span&gt;&#xA;or in another form,&#xA;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\ddot{\phi} = \frac{mb-Ma}{mb^2 + Ma^2}g - \frac{k_1b}{mb^2 + Ma^2}\dot{\theta}^2 = \beta_1 - \beta_2\dot{\theta}^2.&#xA;$$&lt;/span&gt;&#xA;Now, I don't know if these are the correct equations for your system, but based on what you outlined in your question, this is what the EOM are for &lt;span class=&quot;math-container&quot;&gt;$\phi(t)$&lt;/span&gt;.&lt;/p&gt;&#xA;&#xA;&lt;p&gt;You also have a coupled system of ODE's for &lt;span class=&quot;math-container&quot;&gt;$I(t)$&lt;/span&gt; and &lt;span class=&quot;math-container&quot;&gt;$\theta(t)$&lt;/span&gt; since from &lt;span class=&quot;math-container&quot;&gt;$\dot{\theta} = \omega_2$&lt;/span&gt; it follows that&lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;\ddot{\theta} &amp;amp;= \frac{K}{J}I - \frac{k_2}{J}\dot{\theta}^2\\&#xA;\dot{I} &amp;amp;= \frac{E}{L} - \frac{R}{L}I - \frac{k_1}{L}\dot{\theta}&#xA;\end{align}&#xA;$$&lt;/span&gt;&lt;/p&gt;&#xA;&#xA;&lt;p&gt;Again, this is all based off your model that you wrote in the problem statement. So, in total, you have three ODE's for the three variables &lt;span class=&quot;math-container&quot;&gt;$\theta,\phi,I$&lt;/span&gt;. Given that this system of 3 ODE's is nonlinear, there is no analytical solution. What you can do to simulate this is solve the system in MATLAB for example using ODE45 or just any differential equaton solver.&lt;/p&gt;&#xA;&#xA;&lt;hr&gt;&#xA;&#xA;&lt;p&gt;If you want to linearize this system, you can do the following. First, put the system in state space form, i.e. let &lt;span class=&quot;math-container&quot;&gt;$x = [\phi,\dot{\phi},\theta,\dot{\theta},I]^\intercal$&lt;/span&gt; so that the &lt;em&gt;full&lt;/em&gt; dynamics are &lt;/p&gt;&#xA;&#xA;&lt;p&gt;&lt;span class=&quot;math-container&quot;&gt;$$&#xA;\begin{align}&#xA;\dot{x}_1 &amp;amp;= x_2\\&#xA;\dot{x}_2 &amp;amp;= \beta_1\cos x_1 - \beta_2 x_4^2\\&#xA;\dot{x}_3 &amp;amp;= x_4\\&#xA;\dot{x}_4 &amp;amp;= \frac{K}{J}x_5 - \frac{k_2}{J}x_4^2\\&#xA;\dot{x}_6 &amp;amp;= \frac{E}{L} - \